From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001
From: Thomas Stephen Lee
Date: Fri, 4 Sep 2015 22:04:10 +0530
Subject: add/remove/update books

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diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1a.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1a.ipynb
deleted file mode 100755
index 1dc0f01d..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1a.ipynb
+++ /dev/null
@@ -1,314 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#Chapter 1(A):Bonding in Solids"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.1, Page number 1.14"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "-2*a/r**3 + 90*b/r**11\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "from sympy import *\n",
-    "import numpy as np\n",
-    "\n",
-    "#Variable declaration\n",
-    "n=1;\n",
-    "m=9;\n",
-    "a=Symbol('a')\n",
-    "b=Symbol('b')\n",
-    "r=Symbol('r')\n",
-    "\n",
-    "#Calculation\n",
-    "y=(-a/(r**n))+(b/(r**m));\n",
-    "y=diff(y,r);\n",
-    "y=diff(y,r);\n",
-    "\n",
-    "#Result\n",
-    "print y\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "young's modulus is 157 GPa\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "a=7.68*10**-29;     \n",
-    "r0=2.5*10**-10;    #radius(m)\n",
-    "\n",
-    "#Calculation\n",
-    "b=a*(r0**8)/9;\n",
-    "y=((-2*a*r0**8)+(90*b))/r0**11;    \n",
-    "E=y/r0;           #young's modulus(Pa)\n",
-    "\n",
-    "#Result\n",
-    "print \"young's modulus is\",int(E/10**9),\"GPa\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.2, Page number 1.15"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Effective charge = 0.72 *10**-29 coulomb\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "\n",
-    "#variable declarations\n",
-    "d=((1.98)*10**-29)*1/3;        #dipole moment\n",
-    "b=(0.92);                      #bond length\n",
-    "EC=d/(b*10**-10);              #Effective charge\n",
-    "\n",
-    "#Result\n",
-    "print \"Effective charge =\",round((EC*10**19),2),\"*10**-29 coulomb\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.3, Page number 1.15"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Cohesive energy = 668.9 *10**3 kJ/kmol\n",
-      "#Answer varies due to rounding of numbers\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "A=1.748                 #Madelung Constant    \n",
-    "N=6.02*10**26           #Avagadro Number\n",
-    "e=1.6*10**-19\n",
-    "n=9.5\n",
-    "r=(0.324*10**-9)*10**3\n",
-    "E=8.85*10**-12\n",
-    "#Calculations\n",
-    "U=((N*A*(e)**2)/(4*math.pi*E*r))*(1-1/n)       #Cohesive energy\n",
-    "\n",
-    "#Result\n",
-    "print \"Cohesive energy =\",round(U/10**3,1),\"*10**3 kJ/kmol\"\n",
-    "print \"#Answer varies due to rounding of numbers\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.4, Page number 1.15"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Coulomb energy = -2.88 eV\n",
-      "Energy required = -1.88 eV\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "#variable declaration\n",
-    "I=5;                       #Ionisation energy\n",
-    "A=4;                       #Electron Affinity\n",
-    "e=(1.6*10**-19)\n",
-    "E=8.85*10**-12            #epsilon constant\n",
-    "r=0.5*10**-19             #dist between A and B\n",
-    "\n",
-    "#Calculations\n",
-    "C=-(e**2/(4*math.pi*E*r*e))/10**10    #Coulomb energy\n",
-    "E_c=I-A+C                             #Energy required\n",
-    "\n",
-    "#Result\n",
-    "print \"Coulomb energy =\",round(C,2),\"eV\"\n",
-    "print \"Energy required =\",round(E_c,2),\"eV\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.5, Page number 1.16"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Energy required= 1.49 eV\n",
-      "Distance of separation = 9.66 Angstrom\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "I=5.14;                    #Ionization energy\n",
-    "A=3.65;                    #Electron Affinity\n",
-    "e=(1.6*10**-19);\n",
-    "E=8.85*10**-12; \n",
-    "#calculations\n",
-    "E_c=I-A                        #Energy required\n",
-    "r=e**2/(4*math.pi*E*E_c*e)     #Distance of separation\n",
-    "\n",
-    "#Result\n",
-    "print \"Energy required=\",E_c,\"eV\"\n",
-    "print \"Distance of separation =\",round(r/10**-10,2),\"Angstrom\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.6, Page number 1.16"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Energy required= 1.49 eV\n",
-      "Energy required = -6.1 eV\n",
-      "Bond Energy = 4.61 eV\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration \n",
-    "I=5.14;                    #Ionization energy\n",
-    "A=3.65;                    #Electron Affinity\n",
-    "e=(1.6*10**-19);\n",
-    "E=8.85*10**-12; \n",
-    "r=236*10**-12;\n",
-    "\n",
-    "#Calculations\n",
-    "E_c=I-A                        #Energy required\n",
-    "C=-(e**2/(4*math.pi*E*r*e))    #Potentential energy in eV\n",
-    "BE=-(E_c+C)                    #Bond Energy\n",
-    "#Result\n",
-    "print \"Energy required=\",E_c,\"eV\"\n",
-    "print \"Energy required =\",round(C,1),\"eV\"\n",
-    "print \"Bond Energy =\",round(BE,2),\"eV\"\n",
-    "\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1a_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1a_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1b.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1b.ipynb
deleted file mode 100755
index 0df96e78..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1b.ipynb
+++ /dev/null
@@ -1,313 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#Crystal Structures"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.1, Page number 1.36"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a= 5.43 Angstorm\n",
-      "density = 6.88 kg/m**3\n",
-      "#Answer given in the textbook is wrong\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "d=2.351                 #bond lenght\n",
-    "N=6.02*10**26           #Avagadro number\n",
-    "n=8                     #number of atoms in unit cell\n",
-    "A=28.09                 #Atomin mass of silicon\n",
-    "m=6.02*10**26           #1mole\n",
-    "\n",
-    "#Calculations\n",
-    "a=(4*d)/math.sqrt(3)\n",
-    "p=(n*A)/((a*10**-10)*m)    #density\n",
-    "\n",
-    "#Result\n",
-    "print \"a=\",round(a,2),\"Angstorm\"\n",
-    "print \"density =\",round(p*10**16,2),\"kg/m**3\"\n",
-    "print\"#Answer given in the textbook is wrong\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.2, Page number 1.36"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " radius of largest sphere is 0.154700538379252*r\n",
-      "maximum radius of sphere is 0.414213562373095*r\n"
-     ]
-    }
-   ],
-   "source": [
-    " import math\n",
-    "from __future__ import division\n",
-    "from sympy import *\n",
-    "\n",
-    "#Variable declaration\n",
-    "r=Symbol('r')\n",
-    "\n",
-    "#Calculation\n",
-    "a1=4*r/math.sqrt(3);\n",
-    "R1=(a1/2)-r;           #radius of largest sphere\n",
-    "a2=4*r/math.sqrt(2);\n",
-    "R2=(a2/2)-r;       #maximum radius of sphere\n",
-    "\n",
-    "#Result\n",
-    "print \"radius of largest sphere is\",R1\n",
-    "print \"maximum radius of sphere is\",R2    \n",
-    "   "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.3, Page number 1.37"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a1= 2.905 Angstrom\n",
-      "Unit cell volume =a1**3 = 24.521 *10**-30 m**3\n",
-      "Volume occupied by one atom = 12.26 *10**-30 m**3\n",
-      "a2= 3.654 Angstorm\n",
-      "Unit cell volume =a2**3 = 48.8 *10**-30 m**3\n",
-      "Volume occupied by one atom = 12.2 *10**-30 m**3\n",
-      "Volume Change in % = 0.493\n",
-      "Density Change in % = 0.5\n",
-      "Thus the increase of density or the decrease of volume is about 0.5%\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "r1=1.258            #Atomic radius of BCC\n",
-    "r2=1.292            #Atomic radius of FCC\n",
-    "\n",
-    "#calculations\n",
-    "a1=(4*r1)/math.sqrt(3)       #in BCC\n",
-    "b1=((a1)**3)*10**-30         #Unit cell volume\n",
-    "v1=(b1)/2                    #Volume occupied by one atom\n",
-    "a2=2*math.sqrt(2)*r2         #in FCC\n",
-    "b2=(a2)**3*10**-30                   #Unit cell volume\n",
-    "v2=(b2)/4                    #Volume occupied by one atom  \n",
-    "v_c=((v1)-(v2))*100/(v1)     #Volume Change in % \n",
-    "d_c=((v1)-(v2))*100/(v2)     #Density Change in %\n",
-    "\n",
-    "#Results\n",
-    "print \"a1=\",round(a1,3),\"Angstrom\" \n",
-    "print \"Unit cell volume =a1**3 =\",round((b1)/10**-30,3),\"*10**-30 m**3\"\n",
-    "print \"Volume occupied by one atom =\",round(v1/10**-30,2),\"*10**-30 m**3\"\n",
-    "print \"a2=\",round(a2,3),\"Angstorm\"\n",
-    "print \"Unit cell volume =a2**3 =\",round((b2)/10**-30,3),\"*10**-30 m**3\"\n",
-    "print \"Volume occupied by one atom =\",round(v2/10**-30,2),\"*10**-30 m**3\"\n",
-    "print \"Volume Change in % =\",round(v_c,3)\n",
-    "print \"Density Change in % =\",round(d_c,2)\n",
-    "print \"Thus the increase of density or the decrease of volume is about 0.5%\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "##Example 1.4, Page number 1.38"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a= 0.563 *10**-9 metre\n",
-      "spacing between the nearest neighbouring ions = 0.2814 nm\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "n=4     \n",
-    "M=58.5                  #Molecular wt. of NaCl\n",
-    "N=6.02*10**26           #Avagadro number\n",
-    "rho=2180                #density\n",
-    "\n",
-    "#Calculations\n",
-    "a=((n*M)/(N*rho))**(1/3)    \n",
-    "s=a/2\n",
-    "\n",
-    "#Result\n",
-    "print \"a=\",round(a/10**-9,3),\"*10**-9 metre\"\n",
-    "print \"spacing between the nearest neighbouring ions =\",round(s/10**-9,4),\"nm\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 1.5, Page number 1.38"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "lattice constant, a= 0.36 nm\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "n=4     \n",
-    "A=63.55                 #Atomic wt. of NaCl\n",
-    "N=6.02*10**26           #Avagadro number\n",
-    "rho=8930                #density\n",
-    "\n",
-    "#Calculations\n",
-    "a=((n*A)/(N*rho))**(1/3)      #Lattice Constant\n",
-    "\n",
-    "#Result\n",
-    "print \"lattice constant, a=\",round(a*10**9,2),\"nm\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "##Example 1.6, Page number 1.39"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Density of iron = 8805.0 kg/m**-3\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "\n",
-    "#variable declaration\n",
-    "r=0.123                  #Atomic radius\n",
-    "n=4\n",
-    "A=55.8                   #Atomic wt\n",
-    "a=2*math.sqrt(2) \n",
-    "N=6.02*10**26           #Avagadro number\n",
-    "\n",
-    "#Calculations\n",
-    "rho=(n*A)/((a*r*10**-9)**3*N)\n",
-    "\n",
-    "#Result\n",
-    "print \"Density of iron =\",round(rho),\"kg/m**-3\""
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1b_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_1b_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_2.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_2.ipynb
deleted file mode 100755
index c257bf8c..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_2.ipynb
+++ /dev/null
@@ -1,615 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#2: Crystal Planes and X-ray Diffraction"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.1, Page number 2.12"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "i)Number of atoms per unit area of (100)plane= 1/(4*R**2)\n",
-      "ii)Number of atoms per unit area of (110)plane= 2.82842712474619*R**2\n",
-      "iii)Number of atoms per unit area of (111)plane= 2.3094010767585*R**2\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "from sympy import *\n",
-    "#Variable declaration\n",
-    "R=Symbol('R')\n",
-    "a=2*R\n",
-    "\n",
-    "#Results\n",
-    "print\"i)Number of atoms per unit area of (100)plane=\",1/a**2\n",
-    "print\"ii)Number of atoms per unit area of (110)plane=\",1/math.sqrt(2)*a**2\n",
-    "print\"iii)Number of atoms per unit area of (111)plane=\",1/math.sqrt(3)*a**2"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.2, Page number 2.13"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "i)Surface area of the face ABCD = 13.0 *10**-14 mm**2\n",
-      "ii)Surface area of plane (110) = 1.09 *10**13 atoms/mm**2\n",
-      "iii)Surface area of pane(111)= 1.772 *10**13 atoms/mm**2\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "a=3.61*10**-7\n",
-    "BC=math.sqrt(2)/2\n",
-    "AD=(math.sqrt(6))/2\n",
-    "#Result\n",
-    "print\"i)Surface area of the face ABCD =\",round(a**2*10**14),\"*10**-14 mm**2\"\n",
-    "print\"ii)Surface area of plane (110) =\",round((2/(a*math.sqrt(2)*a)/10**13),2),\"*10**13 atoms/mm**2\"\n",
-    "print\"iii)Surface area of pane(111)=\",round(2/(BC*AD*a**2)*10**-13,3),\"*10**13 atoms/mm**2\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.3, Page number 2.14"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "d1 = 1.0\n",
-      "d2 = 0.707\n",
-      "d3 = 0.577\n",
-      "d1:d2:d3 = 1.0 : 0.707 : 0.577\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "h1=1\n",
-    "k1=0\n",
-    "l1=0\n",
-    "h2=1\n",
-    "k2=1\n",
-    "l2=0\n",
-    "h3=1\n",
-    "k3=1\n",
-    "l3=1\n",
-    "a=1\n",
-    "\n",
-    "#Calculations\n",
-    "d1=a/(math.sqrt(h1**2+k1**2+l1**2))\n",
-    "d2=a/(math.sqrt(h2**2+k2**2+l2**2))\n",
-    "d3=a/(math.sqrt(h3**2+k3**2+l3**2))\n",
-    "\n",
-    "#Result\n",
-    "print\"d1 =\",d1 \n",
-    "print\"d2 =\",round(d2,3)\n",
-    "print\"d3 =\",round(d3,3)\n",
-    "print\"d1:d2:d3 =\",d1,\":\",round(d2,3),\":\",round(d3,3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.4, Page number 2.15"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 47,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "d(220) = 159.1 pm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "h=2\n",
-    "k=2\n",
-    "l=0\n",
-    "a=450\n",
-    "\n",
-    "#Calculations\n",
-    "d=a/(math.sqrt(h**2+k**2+l**2))\n",
-    "\n",
-    "#Result\n",
-    "print\"d(220) =\",round(d,1),\"pm\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.5, Page number 2.15"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 49,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a = 3.615 Angstroms\n",
-      "d = 2.087 Angstroms\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "a=3.615\n",
-    "r=1.278\n",
-    "h=1\n",
-    "k=1\n",
-    "l=1\n",
-    "\n",
-    "#Calculations\n",
-    "a=(4*r)/math.sqrt(2)\n",
-    "d=a/(math.sqrt(h**2+k**2+l**2))\n",
-    "\n",
-    "#Result\n",
-    "print\"a =\",round(a,3),\"Angstroms\"\n",
-    "print\"d =\",round(d,3),\"Angstroms\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.7, Page number 2.15"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "d = 1.45 *10**-10 m\n",
-      "a = 4.1 *10**-10 m\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "n=1\n",
-    "lamda=1.54\n",
-    "theta=32*math.pi/180\n",
-    "h=2\n",
-    "k=2\n",
-    "l=0\n",
-    "\n",
-    "#Calculations\n",
-    "d=(n*lamda*10**-10)/(2*math.sin(theta))   #derived from 2dsin(theta)=n*l\n",
-    "a=d*(math.sqrt(h**2+k**2+l**2))\n",
-    "\n",
-    "#Results\n",
-    "print\"d =\",round(d*10**10,2),\"*10**-10 m\"\n",
-    "print\"a =\",round(a*10**10,1),\"*10**-10 m\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.8, Page number 2.16"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 50,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "i.  d/n = 2.582 Angstroms\n",
-      "ii. d/n = 1.824 Angstroms\n",
-      "iii.d/n = 1.289 Angstroms\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "lamda=0.58\n",
-    "theta1=6.45*math.pi/180\n",
-    "theta2=9.15*math.pi/180\n",
-    "theta3=13*math.pi/180\n",
-    "\n",
-    "#Calculations\n",
-    "dbyn1=lamda/(2*(math.sin(theta1)))\n",
-    "dbyn2=lamda/(2*math.sin(theta2))\n",
-    "dbyn3=lamda/(2*math.sin(theta3))\n",
-    "           \n",
-    "#Results\n",
-    "print\"i.  d/n =\",round(dbyn1,3),\"Angstroms\"\n",
-    "print\"ii. d/n =\",round(dbyn2,3),\"Angstroms\"\n",
-    "print\"iii.d/n =\",round(dbyn3,3),\"Angstroms\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.9, Page number 2.16"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "n = 1.53\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "d=1.18\n",
-    "theta=90*math.pi/180\n",
-    "lamda=1.540\n",
-    "\n",
-    "#Calculations\n",
-    "n=(2*d*math.sin(theta))/lamda\n",
-    "\n",
-    "#Result\n",
-    "print\"n =\",round(n,2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.10, Page number 2.17"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a = 3.51 Angstorms\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "lamda=0.58\n",
-    "theta=9.5*math.pi/180\n",
-    "n=1\n",
-    "d=0.5           #d200=a/math.sqrt(2**2+0**2+0**2)=0.5a\n",
-    "#Calculations\n",
-    "a=n*lamda/(2*d*math.sin(theta))     #2*d*sin(theta)=n*lamda \n",
-    "\n",
-    "#Result\n",
-    "print\"a =\",round(a,2),\"Angstorms\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.11, Page number 2.17"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "sin(theta3) = 26 35.9387574495\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "lamda=0.842\n",
-    "n1=1\n",
-    "q=(8+(35/60))*(math.pi/180)\n",
-    "n2=3\n",
-    "d=1\n",
-    "#Calculations\n",
-    "#n*lamda=2*d*sin(theta)\n",
-    "#n1*0.842=2*d*sin(q)\n",
-    "#n3*0.842=2*d*sin(theta3)\n",
-    "#Dividing both the eauations, we get\n",
-    "#(n2*lamda)/(n1*lamda)=2*d*math.sin(theta3)/2*d*math.sin(q)\n",
-    "theta3=math.asin((((n2*lamda)/(n1*lamda))*(2*d*math.sin(q)))/(2*d))\n",
-    "d=theta3*180/math.pi;\n",
-    "a_d=int(d);\n",
-    "a_m=(d-int(d))*60\n",
-    "\n",
-    "#Result\n",
-    "print\"sin(theta3) =\",a_d,a_m\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.12, Page number 2.18"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "d = 2.22 Angstorms\n",
-      "sqrt(h**2+k**2+l**2) = 1.424\n",
-      "Therefore, h**2+k**2+l**2 =sqrt(2)\n",
-      "h =1, k=1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "a=3.16\n",
-    "lamda=1.54\n",
-    "n=1\n",
-    "theta=20.3*math.pi/180\n",
-    "\n",
-    "#Calculations\n",
-    "d=(n*lamda)/(2*math.sin(theta))\n",
-    "x=a/d                             #let math.sqrt(h**2+k**2+l**2)=x\n",
-    "\n",
-    "#Result\n",
-    "print\"d =\",round(d,2),\"Angstorms\"\n",
-    "print\"sqrt(h**2+k**2+l**2) =\",round(x,3)\n",
-    "print\"Therefore, h**2+k**2+l**2 =sqrt(2)\"\n",
-    "print\"h =1, k=1\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.13, Page number 2.18"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 53,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a = 4.09 Angstroms\n",
-      "d = 2.36 Angstroms\n",
-      "lamda = 1.552 Angstroms\n",
-      "E = 8.0 *10**3 eV\n"
-     ]
-    }
-   ],
-   "source": [
-    "## importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "n=4\n",
-    "A=107.87\n",
-    "rho=10500\n",
-    "N=6.02*10**26\n",
-    "h=1;\n",
-    "k=1;\n",
-    "l=1;\n",
-    "H=6.625*10**-34\n",
-    "e=1.6*10**-19\n",
-    "theta=(19+(12/60))*math.pi/180\n",
-    "C=3*10**8\n",
-    "#Calculations\n",
-    "a=((n*A)/(rho*N))**(1/3)*10**10\n",
-    "d=a/math.sqrt(h**2+k**2+l**2)\n",
-    "lamda=2*d*math.sin(theta)\n",
-    "E=(H*C)/(lamda*10**-10*e)\n",
-    "\n",
-    "#Result\n",
-    "print\"a =\",round(a,2),\"Angstroms\"\n",
-    "print\"d =\",round(d,2),\"Angstroms\"\n",
-    "print\"lamda =\",round(lamda,3),\"Angstroms\"\n",
-    "print\"E =\",round(E/10**3),\"*10**3 eV\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 2.14, Page number 2.19"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 72,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "d = 2.64 Angstorms\n",
-      "sin(theta)= 0.288\n",
-      "X = 7.554 cm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "a=4.57\n",
-    "h=1\n",
-    "k=1\n",
-    "l=1\n",
-    "lamda=1.52\n",
-    "twotheta=33.5*math.pi/180\n",
-    "r=5                  #radius\n",
-    "#Calculations\n",
-    "d=a/(h**2+k**2+l**2)**(1/2)\n",
-    "sintheta=lamda/(2*d)\n",
-    "X=r/math.tan(twotheta)\n",
-    "\n",
-    "#Result\n",
-    "print\"d =\",round(d,2),\"Angstorms\"\n",
-    "print\"sin(theta)=\",round(sintheta,3)\n",
-    "print\"X =\",round(X,3),\"cm\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_2_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_2_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3a.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3a.ipynb
deleted file mode 100755
index 6553b45b..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3a.ipynb
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#3(A):Defects In Solids "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.1, Page number 3.17"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "at 0K, The number of vacancies per kilomole of copper is 0\n",
-      "at 300K, The number of vacancies per kilomole of copper is 7.577 *10**5\n",
-      "at 900K, The numb ber of vacancies per kilomole of copper is 6.502 *10**19\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "N=6.023*10**26\n",
-    "deltaHv=120\n",
-    "B=1.38*10**-23\n",
-    "k=6.023*10**23\n",
-    "\n",
-    "#Calculations\n",
-    "n0=0                                          # 0 in denominator\n",
-    "n300=N*math.exp(-deltaHv*10**3/(k*B*300))     #The number of vacancies per kilomole of copper\n",
-    "n900=N*math.exp(-(deltaHv*10**3)/(k*B*900))\n",
-    "\n",
-    "#Results\n",
-    "print\"at 0K, The number of vacancies per kilomole of copper is\",n0\n",
-    "print\"at 300K, The number of vacancies per kilomole of copper is\",round(n300/10**5,3),\"*10**5\"\n",
-    "print\"at 900K, The numb ber of vacancies per kilomole of copper is\",round(n900/10**19,3),\"*10**19\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.2, Page number 3.16"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Fraction of vacancies at 1000 degrees C = 8.5 *10**-7\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "from sympy import *\n",
-    "\n",
-    "#Variable declaration\n",
-    "F_500=1*10**-10\n",
-    "delta_Hv=Symbol('delta_Hv')\n",
-    "k=Symbol('k')\n",
-    "T1=500+273\n",
-    "T2=1000+273\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "lnx=math.log(F_500)*T1/T2;\n",
-    "x=math.exp(round(lnx,2))\n",
-    "\n",
-    "print\"Fraction of vacancies at 1000 degrees C =\",round(x*10**7,1),\"*10**-7\" "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.3, Page number 3.17"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Volume of unit cell of NaCl = 1.794 *10**-28 m**3\n",
-      "Total number of ion pairs 'N' =' 2.23 *10**28\n",
-      "The concentration of Schottky defects per m**3 at 300K = 6.42 *10**11\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "a=(2*2.82*10**-10)\n",
-    "delta_Hs=1.971*1.6*10**-19\n",
-    "k=1.38*10**-23\n",
-    "T=300\n",
-    "\n",
-    "#Calculations\n",
-    "V=a**3                           #Volume of unit cell of NaCl\n",
-    "N=4/V                            #Total number of ion pairs\n",
-    "n=N*math.e**-(delta_Hs/(2*k*T))  \n",
-    "\n",
-    "#Result\n",
-    "print\"Volume of unit cell of NaCl =\",round(V*10**28,3),\"*10**-28 m**3\"\n",
-    "print\"Total number of ion pairs 'N' ='\",round(N/10**28,2),\"*10**28\"\n",
-    "print\"The concentration of Schottky defects per m**3 at 300K =\",round(n/10**11,2),\"*10**11\"\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.4, Page number 3.18"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The number that must be created on heating from 0 to 500K is n= 9.22 *10**12 per cm**3\n",
-      "As one step is 2 Angstorms, 5*10**7 vacancies are required for 1cm\n",
-      "The amount of climb down by the dislocation is 0.369 cm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "N=6.023*10**23\n",
-    "delta_Hv=1.6*10**-19\n",
-    "k=1.38*10**-23\n",
-    "T=500\n",
-    "mv=5.55;     #molar volume\n",
-    "x=2*10**-8;     #numbber of cm in 1 angstrom\n",
-    "\n",
-    "#Calculations\n",
-    "n=N*math.exp(-delta_Hv/(k*T))/mv\n",
-    "a=round(n/(5*10**7*10**6),4)*x;\n",
-    "\n",
-    "#Result\n",
-    "print\"The number that must be created on heating from 0 to 500K is n=\",round(n/10**12,2),\"*10**12 per cm**3\" #into cm**3\n",
-    "print\"As one step is 2 Angstorms, 5*10**7 vacancies are required for 1cm\"\n",
-    "print\"The amount of climb down by the dislocation is\",a*10**8,\"cm\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3a_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3a_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3b.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3b.ipynb
deleted file mode 100755
index 5fa9129c..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3b.ipynb
+++ /dev/null
@@ -1,414 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#3(B):Principles of Quantum Mechanics"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.1, Page number 3.41"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Velocity = 1.38 *10**6 m/s\n",
-      "Wavelength = 0.00286 Angstorm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "KE=10             #Kinetic Energy of neutron in keV\n",
-    "m=1.675*10**-27\n",
-    "h=6.625*10**-34\n",
-    "#Calculations\n",
-    "KE=10**4*1.6*10**-19      #in joule\n",
-    "v=((2*KE)/m)**(1/2)       #derived from KE=1/2*m*v**2\n",
-    "lamda=h/(m*v)\n",
-    "#Results\n",
-    "print\"Velocity =\",round(v/10**6,2),\"*10**6 m/s\"\n",
-    "print\"Wavelength =\",round(lamda*10**10,5),\"Angstorm\"\n",
-    "                        "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.2, Page number 3.41"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Momentum 2.4133\n",
-      "de Brolie wavelength = 2.73 *10**-11 m\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "E=2*1000*1.6*10**-19       #in joules\n",
-    "m=9.1*10**-31\n",
-    "h=6.6*10*10**-34\n",
-    "\n",
-    "#Calculations\n",
-    "p=math.sqrt(2*m*E)\n",
-    "lamda= h/p\n",
-    "\n",
-    "#Result\n",
-    "print\"Momentum\",round(p*10**23,4)\n",
-    "print\"de Brolie wavelength =\",round(lamda*10**10,2),\"*10**-11 m\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.3, Page number 3.41"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "wavelength = 1.807 Angstorm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "M=1.676*10**-27             #Mass of neutron\n",
-    "m=0.025\n",
-    "v=1.602*10**-19\n",
-    "h=6.62*10**-34\n",
-    "\n",
-    "#Calculations\n",
-    "mv=(2*m*v)**(1/2)\n",
-    "lamda=h/(mv*M**(1/2))\n",
-    "\n",
-    "#Result\n",
-    "print\"wavelength =\",round(lamda*10**10,3),\"Angstorm\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.4, Page number 3.42"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Wavelength = 0.1226 Angstorm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "V=10000\n",
-    "\n",
-    "#Calculation\n",
-    "lamda=12.26/math.sqrt(V)\n",
-    "\n",
-    "#Result\n",
-    "print\"Wavelength =\",lamda,\"Angstorm\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.5, Page number 3.42"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false,
-    "scrolled": true
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The permitted electron energies = 38.0 *n**2 eV\n",
-      "E1= 38.0 eV\n",
-      "E2= 151.0 eV\n",
-      "E3= 339.0 eV\n",
-      "#Answer varies due to rounding of numbers\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "e=1.6*10**-19;   #charge of electron(coulomb)\n",
-    "L=10**-10            #1Angstrom=10**-10 m\n",
-    "n1=1;\n",
-    "n2=2;\n",
-    "n3=3;\n",
-    "h=6.626*10**-34\n",
-    "m=9.1*10**-31\n",
-    "L=10**-10\n",
-    "\n",
-    "#Calculations\n",
-    "E1=(h**2)/(8*m*L**2*e)\n",
-    "E2=4*E1\n",
-    "E3=9*E1\n",
-    "#Result\n",
-    "print\"The permitted electron energies =\",round(E1),\"*n**2 eV\"\n",
-    "print\"E1=\",round(E1),\"eV\"\n",
-    "print\"E2=\",round(E2),\"eV\"\n",
-    "print\"E3=\",round(E3),\"eV\"\n",
-    "print\"#Answer varies due to rounding of numbers\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.6, Page number 3.42"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "si**2 delta(x)= 0.2\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "i=1*10**-10;    #interval\n",
-    "L=10*10**-10;   #width\n",
-    "\n",
-    "#Calculations\n",
-    "si2=2*i/L;\n",
-    "\n",
-    "#Result\n",
-    "print\"si**2 delta(x)=\",si2"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.7, Page number 3.43"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " E1 = 1.81 *10**-37 Joule\n",
-      "E2 = 3.62 *10**-37 Joule\n",
-      "E2-E1 = 1.81 *10**-37 J\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "nx=1\n",
-    "ny=1\n",
-    "nz=1\n",
-    "a=1\n",
-    "h=6.63*10**-34\n",
-    "m=9.1*10**-31\n",
-    "\n",
-    "#Calculations\n",
-    "E1=h**2*(nx**2+ny**2+nz**2)/(8*m*a**2)\n",
-    "E2=(h**2*6)/(8*m*a**2)             #nx**2+ny**2+nz**2=6\n",
-    "diff=E2-E1\n",
-    "#Result\n",
-    "print\"E1 =\",round(E1*10**37,2),\"*10**-37 Joule\"\n",
-    "print\"E2 =\",round(E2*10**37,2),\"*10**-37 Joule\"\n",
-    "print\"E2-E1 =\",round(diff*10**37,2),\"*10**-37 J\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 3.8, Page number 3.43"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "E1 = 3.28 *10**-13 J\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "m=1.67*10**-27\n",
-    "a=10**-14\n",
-    "h=1.054*10**-34\n",
-    "\n",
-    "#Calculations\n",
-    "E1=(1*math.pi*h)**2/(2*m*a**2)\n",
-    "\n",
-    "#Result\n",
-    "print\"E1 =\",round(E1*10**13,2),\"*10**-13 J\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "#Example number 3.9, Page number 3.44"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a= 0.0158\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "from scipy.integrate import quad\n",
-    "#Variable declarations\n",
-    "k=1;\n",
-    "\n",
-    "#Calculations\n",
-    "def zintg(x):\n",
-    "    return 2*k*math.exp(-2*k*x)\n",
-    "a=quad(zintg,2/k,3/k)[0]\n",
-    "\n",
-    "#Result\n",
-    "print \"a=\",round(a,4)"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3b_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_3b_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_4.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_4.ipynb
deleted file mode 100755
index 84e4acb8..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_4.ipynb
+++ /dev/null
@@ -1,478 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#4: Electron Theory of Metals"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.1, Page number 4.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "energy difference is 1.81 *10**-37 J\n",
-      "3/2*k*T = E2 = 3.62 *10**-37 J\n",
-      "T = 1.75 *10**-14 K\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "m=9.1*10**-31;     #mass(kg)\n",
-    "nx=ny=nz=1;\n",
-    "n=6;\n",
-    "a=1;     #edge(m)\n",
-    "h=6.63*10**-34;     #planck's constant\n",
-    "k=1.38\n",
-    "#Calculation\n",
-    "E1=h**2*(nx**2+ny**2+nz**2)/(8*m*a**2);\n",
-    "E2=h**2*n/(8*m*a**2);\n",
-    "E=E2-E1;               #energy difference(J)\n",
-    "T=(2*E2*10**37)/(3*k*10**-23)\n",
-    "#Result\n",
-    "print \"energy difference is\",round(E*10**37,2),\"*10**-37 J\"\n",
-    "print \"3/2*k*T = E2 =\",round(E2*10**37,2),\"*10**-37 J\"\n",
-    "print \"T =\",round(T/10**23,2),\"*10**-14 K\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.2, Page number 4.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "temperature is 1261 K\n",
-      "answer varies due to rounding off errors\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "y=1/100;    #percentage of probability\n",
-    "x=0.5*1.6*10**-19;     #energy(J)\n",
-    "k=1.38*10**-23;       #boltzmann constant\n",
-    "\n",
-    "#Calculation\n",
-    "xbykT=math.log((1/y)-1);\n",
-    "T=x/(k*xbykT);       #temperature(K)\n",
-    "\n",
-    "#Result\n",
-    "print \"temperature is\",int(T),\"K\"\n",
-    "print \"answer varies due to rounding off errors\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.3, Page number 4.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "fermi energy is 3.2 eV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "d=970;    #density(kg/m**3)\n",
-    "Na=6.02*10**26;    #avagadro number\n",
-    "w=23;    #atomic weight\n",
-    "m=9.1*10**-31;     #mass(kg)\n",
-    "h=6.62*10**-34;     #planck's constant\n",
-    "\n",
-    "#Calculation\n",
-    "N=d*Na/w;    #number of atoms/m**3\n",
-    "x=h**2/(8*m);\n",
-    "y=(3*N/math.pi)**(2/3);\n",
-    "EF=x*y;     #fermi energy(J)\n",
-    "\n",
-    "#Result\n",
-    "print \"fermi energy is\",round(EF/(1.6*10**-19),1),\"eV\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.4, Page number 4.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "p(E) = 0.269\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "kT=1;\n",
-    "E_EF=1;\n",
-    "\n",
-    "#Calculations\n",
-    "p_E=1/(1+math.exp(E_EF/kT)) \n",
-    "        \n",
-    "#Result        \n",
-    "print \"p(E) =\",round(p_E,3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.5, Page number 4.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "N = 2.4 *10**26 states\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "from scipy.integrate import quad \n",
-    "\n",
-    "#Variable declarations\n",
-    "m=9.1*10**-31\n",
-    "h=6.626*10**-34\n",
-    "Ef=3.1\n",
-    "Ef1=Ef+0.02\n",
-    "e=1.6*10**-19\n",
-    "#Calculations\n",
-    "def zintg(E):\n",
-    "    return math.pi*((8*m)**(3/2))*(E**(1/2)*e**(3/2))/(2*(h**3))\n",
-    "N=quad(zintg,Ef,Ef1)[0]\n",
-    "\n",
-    "#Result\n",
-    "print\"N =\",round(N*10**-26,1),\"*10**26 states\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.6, Page number 4.30"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "n = 8.5 *10**28 /m**3\n",
-      "The drift speed Vd = 36.6 *10**-5 m/s\n",
-      "The mean free collision time Tc = 2.087 *10**-14 seconds\n",
-      "Mean free path = 3.34 *10**-8 m(answer varies due to rounding off errors)\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "N=6.023*10**26                   #Avagadro number\n",
-    "D=8960                           #density  \n",
-    "F_e=1                            #no.of free electrons per atom     \n",
-    "W=63.54                          #Atomic weight\n",
-    "i=10\n",
-    "e=1.602*10**-19\n",
-    "m=9.1*10**-31\n",
-    "rho=2*10**-8\n",
-    "Cbar=1.6*10**6                   #mean thermal velocity(m/s)\n",
-    "\n",
-    "#Calculations\n",
-    "n=(N*D*F_e)/W\n",
-    "A=math.pi*0.08**2*10**-4\n",
-    "Vd=i/(A*n*e)                      #Drift speed\n",
-    "Tc=m/(n*(e**2)*rho)\n",
-    "lamda=Tc*Cbar\n",
-    "\n",
-    "#Result\n",
-    "print\"n =\",round(n/10**28,1),\"*10**28 /m**3\"\n",
-    "print\"The drift speed Vd =\",round(Vd*10**5,1),\"*10**-5 m/s\"\n",
-    "print\"The mean free collision time Tc =\",round(Tc*10**14,3),\"*10**-14 seconds\"\n",
-    "print\"Mean free path =\",round(lamda*10**8,2),\"*10**-8 m\"\"(answer varies due to rounding off errors)\" \n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.7, Page number 4.30"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The mean free collision time = 4.8 *10**7 ohm**-1 m**-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "##Variable declaration\n",
-    "n=8.5*10**28\n",
-    "e=1.602*10**-19\n",
-    "t=2*10**-14\n",
-    "m=9.1*10**-31\n",
-    "\n",
-    "#Calculations\n",
-    "Tc=n*(e**2)*t/m\n",
-    "\n",
-    "#Result\n",
-    "print \"The mean free collision time =\",round(Tc/10**7,1),\"*10**7 ohm**-1 m**-1\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.8, Page number 4.31"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Relaxation time = 4.0 *10**-14 second\n",
-      "Mobility = 7.0 *10**-3 m**2/volt-s\n",
-      "Drift Velocity= 0.7 m/s\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "e=1.6*10**-19\n",
-    "E=1                    #(V/m)\n",
-    "rho=1.54*10**-8\n",
-    "n=5.8*10**28\n",
-    "\n",
-    "#Calculations\n",
-    "T=m/(rho*n*e**2)\n",
-    "Me=(e*T)/m\n",
-    "Vd=Me*E\n",
-    "\n",
-    "#Result \n",
-    "print\"Relaxation time =\",round(T*10**14),\"*10**-14 second\"\n",
-    "print\"Mobility =\",round(Me*10**3),\"*10**-3 m**2/volt-s\"\n",
-    "print\"Drift Velocity=\",round(Vd*100,1),\"m/s\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.9, Page number 4.31"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Temperature coefficient of resistivity,a = 5.7\n",
-      "rho_973 = 5.51 *10**-8 ohm-m\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "\n",
-    "##Variable declaration\n",
-    "rho_r=0\n",
-    "T=300\n",
-    "rho=1.7*10**-18\n",
-    "\n",
-    "#Calculations \n",
-    "a=rho/T\n",
-    "rho_973=a*973\n",
-    "\n",
-    "#Results\n",
-    "print\"Temperature coefficient of resistivity,a =\",round(a*10**21,1)\n",
-    "print\"rho_973 =\",round(rho_973*10**18,2),\"*10**-8 ohm-m\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 4.10, Page number 4.31"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Increase in resistivity in copper = 0.54 *10**-8 ohm m\n",
-      "Total resistivity of copper alloy = 2.04 *10**-8 ohm m\n",
-      "The resistivity of alloy at 3K = 0.54 *10**-8 ohm m\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "\n",
-    "##Variable declaration\n",
-    "rho1=1.2*10**-8\n",
-    "p1=0.4\n",
-    "rho2=0.12*10**-8\n",
-    "p2=0.5\n",
-    "rho3=1.5*10**-8\n",
-    "#Calculations\n",
-    "R=(rho1*p1)+(rho2*p2)\n",
-    "R_c=R+rho3\n",
-    "\n",
-    "#Results\n",
-    "print\"Increase in resistivity in copper =\",round(R*10**8,2),\"*10**-8 ohm m\"\n",
-    "print\"Total resistivity of copper alloy =\",round(R_c*10**8,2),\"*10**-8 ohm m\"\n",
-    "print\"The resistivity of alloy at 3K =\",round(R*10**8,2),\"*10**-8 ohm m\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_4_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_4_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5a.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5a.ipynb
deleted file mode 100755
index 28d3f8c6..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5a.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#5(A): Dielectric Properties"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.1, Page number 5.27"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "insulation resistance is 0.85 *10**18 ohm\n",
-      "answer varies due to rounding off errors\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "rho=5*10**16;   #resistivity(ohm m)\n",
-    "l=5*10**-2;    #thickness(m)\n",
-    "b=8*10**-2;    #length(m)\n",
-    "w=3*10**-2;    #width(m)\n",
-    "\n",
-    "#Calculation\n",
-    "A=b*w;    #area(m**2)\n",
-    "Rv=rho*l/A;    \n",
-    "X=l+b;      #length(m)\n",
-    "Y=w;      #perpendicular(m)\n",
-    "Rs=Rv*X/Y;    \n",
-    "Ri=Rs*Rv/(Rs+Rv);       #insulation resistance(ohm)\n",
-    "\n",
-    "#Result\n",
-    "print \"insulation resistance is\",round(Ri/10**18,2),\"*10**18 ohm\"\n",
-    "print \"answer varies due to rounding off errors\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.2, Page number 5.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "polarisability of He is 0.185 *10**-40 farad m**2\n",
-      "relative permittivity is 1.000056\n",
-      "answer varies due to rounding off errors\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "epsilon0=8.84*10**-12;\n",
-    "R=0.55*10**-10;    #radius(m)\n",
-    "N=2.7*10**25;      #number of atoms\n",
-    "\n",
-    "#Calculation\n",
-    "alpha_e=4*math.pi*epsilon0*R**3;    #polarisability of He(farad m**2)\n",
-    "epsilonr=1+(N*alpha_e/epsilon0);      #relative permittivity\n",
-    "\n",
-    "#Result\n",
-    "print \"polarisability of He is\",round(alpha_e*10**40,3),\"*10**-40 farad m**2\"\n",
-    "print \"relative permittivity is\",round(epsilonr,6)\n",
-    "print \"answer varies due to rounding off errors\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.3, Page number 5.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "field strength is 3.535 *10**7 V/m\n",
-      "total dipole moment is 33.4 *10**-12 Cm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "A=360*10**-4;    #area(m**2)\n",
-    "V=15;    #voltage(V)\n",
-    "C=6*10**-6;     #capacitance(farad)\n",
-    "epsilonr=8;\n",
-    "epsilon0=8.84*10**-12;\n",
-    "\n",
-    "#Calculation\n",
-    "E=V*C/(epsilon0*epsilonr*A);     #field strength(V/m)\n",
-    "dm=epsilon0*(epsilonr-1)*V*A;    #total dipole moment(Cm)\n",
-    "\n",
-    "#Result\n",
-    "print \"field strength is\",round(E/10**7,3),\"*10**7 V/m\"\n",
-    "print \"total dipole moment is\",round(dm*10**12,1),\"*10**-12 Cm\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.4, Page number 5.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "the complex polarizability is (3.50379335033-0.0600074383321j) *10**-40 F-m**2\n",
-      "answer cant be rouned off to 2 decimals as given in the textbook. Since it is a complex number and complex cant be converted to float\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "epsilonr=4.36;      #dielectric constant\n",
-    "t=2.8*10**-2;       #loss tangent(t)\n",
-    "N=4*10**28;         #number of electrons\n",
-    "epsilon0=8.84*10**-12;      \n",
-    "\n",
-    "#Calculation\n",
-    "epsilon_r = epsilonr*t;\n",
-    "epsilonstar = (complex(epsilonr,-epsilon_r));\n",
-    "alphastar = (epsilonstar-1)/(epsilonstar+2);\n",
-    "alpha_star = 3*epsilon0*alphastar/N;             #complex polarizability(Fm**2)\n",
-    "\n",
-    "#Result\n",
-    "print \"the complex polarizability is\",alpha_star*10**40,\"*10**-40 F-m**2\"\n",
-    "print \"answer cant be rouned off to 2 decimals as given in the textbook. Since it is a complex number and complex cant be converted to float\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5a_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5a_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5b.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5b.ipynb
deleted file mode 100755
index 0ada6386..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5b.ipynb
+++ /dev/null
@@ -1,288 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#5(B): Magnetic Properties"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.1, Page number 5.65"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "temperature rise is 8.43 K\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "El=10**-2*50;       #energy loss(J)\n",
-    "H=El*60;      #heat produced(J)\n",
-    "d=7.7*10**3;    #iron rod(kg/m**3)\n",
-    "s=0.462*10**-3;    #specific heat(J/kg K)\n",
-    "\n",
-    "#Calculation\n",
-    "theta=H/(d*s);     #temperature rise(K)\n",
-    "\n",
-    "#Result\n",
-    "print \"temperature rise is\",round(theta,2),\"K\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.2, Page number 5.65"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "magnetic field at the centre is 14.0 weber/m**2\n",
-      "dipole moment is 9.0 *10**-24 ampere/m**2\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "e=1.6*10**-19;    #charge(coulomb)\n",
-    "new=6.8*10**15;   #frequency(revolutions per second)\n",
-    "mew0=4*math.pi*10**-7;\n",
-    "R=5.1*10**-11;     #radius(m)\n",
-    "\n",
-    "#Calculation\n",
-    "i=round(e*new,4);          #current(ampere)\n",
-    "B=mew0*i/(2*R);    #magnetic field at the centre(weber/m**2)\n",
-    "A=math.pi*R**2;\n",
-    "d=i*A;       #dipole moment(ampere/m**2)\n",
-    "\n",
-    "#Result\n",
-    "print \"magnetic field at the centre is\",round(B),\"weber/m**2\"\n",
-    "print \"dipole moment is\",round(d*10**24),\"*10**-24 ampere/m**2\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.3, Page number 5.65"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "intensity of magnetisation is 5.0 ampere/m\n",
-      "flux density in material is 1.257 weber/m**2\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "chi=0.5*10**-5;    #magnetic susceptibility\n",
-    "H=10**6;     #field strength(ampere/m)\n",
-    "mew0=4*math.pi*10**-7;\n",
-    "\n",
-    "#Calculation\n",
-    "I=chi*H;     #intensity of magnetisation(ampere/m)\n",
-    "B=mew0*(I+H);    #flux density in material(weber/m**2)\n",
-    "\n",
-    "#Result\n",
-    "print \"intensity of magnetisation is\",I,\"ampere/m\"\n",
-    "print \"flux density in material is\",round(B,3),\"weber/m**2\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.4, Page number 5.65"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "number of Bohr magnetons is 2.22 bohr magneon/atom\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "B=9.27*10**-24;      #bohr magneton(ampere m**2)\n",
-    "a=2.86*10**-10;      #edge(m)\n",
-    "Is=1.76*10**6;       #saturation value of magnetisation(ampere/m)\n",
-    "\n",
-    "#Calculation\n",
-    "N=2/a**3;\n",
-    "mew_bar=Is/N;      #number of Bohr magnetons(ampere m**2)\n",
-    "mew_bar=mew_bar/B;      #number of Bohr magnetons(bohr magneon/atom)\n",
-    "\n",
-    "#Result\n",
-    "print \"number of Bohr magnetons is\",round(mew_bar,2),\"bohr magneon/atom\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.5, Page number 5.66"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "average magnetic moment is 2.79 *10**-3 bohr magneton/spin\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "mew0=4*math.pi*10**-7;\n",
-    "H=9.27*10**-24;      #bohr magneton(ampere m**2)\n",
-    "beta=10**6;      #field(ampere/m)\n",
-    "k=1.38*10**-23;    #boltzmann constant\n",
-    "T=303;    #temperature(K)\n",
-    "\n",
-    "#Calculation\n",
-    "mm=mew0*H*beta/(k*T);    #average magnetic moment(bohr magneton/spin)\n",
-    "\n",
-    "#Result\n",
-    "print \"average magnetic moment is\",round(mm*10**3,2),\"*10**-3 bohr magneton/spin\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 5.6, Page number 5.66"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "hysteresis loss per cycle is 188.0 J/m**3\n",
-      "hysteresis loss per second is 9400.0 watt/m**3\n",
-      "power loss is 1.23 watt/kg\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "A=94;      #area(m**2)\n",
-    "vy=0.1;    #value of length(weber/m**2)\n",
-    "vx=20;     #value of unit length\n",
-    "n=50;      #number of magnetization cycles\n",
-    "d=7650;    #density(kg/m**3)\n",
-    "\n",
-    "#Calculation\n",
-    "h=A*vy*vx;     #hysteresis loss per cycle(J/m**3)\n",
-    "hs=h*n;       #hysteresis loss per second(watt/m**3)\n",
-    "pl=hs/d;      #power loss(watt/kg)\n",
-    "\n",
-    "#Result\n",
-    "print \"hysteresis loss per cycle is\",h,\"J/m**3\"\n",
-    "print \"hysteresis loss per second is\",hs,\"watt/m**3\"\n",
-    "print \"power loss is\",round(pl,2),\"watt/kg\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5b_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_5b_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_6a.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_6a.ipynb
deleted file mode 100755
index ad04957e..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_6a.ipynb
+++ /dev/null
@@ -1,774 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#6(A): Semiconductors"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.1, Page number 6.21"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "number of electron hole pairs is 2.32 *10**16 per cubic metre\n",
-      "answer varies due to rounding off errors\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "ni1=2.5*10**19;    #number of electron hole pairs\n",
-    "T1=300;     #temperature(K)\n",
-    "Eg1=0.72*1.6*10**-19;    #energy gap(J)\n",
-    "k=1.38*10**-23;     #boltzmann constant\n",
-    "T2=310;    #temperature(K)\n",
-    "Eg2=1.12*1.6*10**-19;    #energy gap(J)\n",
-    "\n",
-    "#Calculation\n",
-    "x1=-Eg1/(2*k*T1);\n",
-    "y1=(T1**(3/2))*math.exp(x1);\n",
-    "x2=-Eg2/(2*k*T2);\n",
-    "y2=(T2**(3/2))*math.exp(x2);\n",
-    "ni=ni1*(y2/y1);          #number of electron hole pairs\n",
-    "\n",
-    "#Result\n",
-    "print \"number of electron hole pairs is\",round(ni/10**16,2),\"*10**16 per cubic metre\"\n",
-    "print \"answer varies due to rounding off errors\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.2, Page number 6.22"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "intrinsic conductivity is 1.434 *10**4 ohm-1 m-1\n",
-      "intrinsic resistivity is 0.697 *10**-4 ohm m\n",
-      "answer varies due to rounding off errors\n",
-      "number of germanium atoms per m**3 is 4.5 *10**28\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "w=72.6;    #atomic weight\n",
-    "d=5400;    #density(kg/m**3)\n",
-    "Na=6.025*10**26;    #avagadro number\n",
-    "mew_e=0.4;    #mobility of electron(m**2/Vs)\n",
-    "mew_h=0.2;    #mobility of holes(m**2/Vs)\n",
-    "e=1.6*10**-19;\n",
-    "m=9.108*10**-31;    #mass(kg)\n",
-    "ni=2.1*10**19;      #number of electron hole pairs\n",
-    "Eg=0.7;    #band gap(eV)\n",
-    "k=1.38*10**-23;    #boltzmann constant\n",
-    "h=6.625*10**-34;    #plancks constant\n",
-    "T=300;     #temperature(K)\n",
-    "\n",
-    "#Calculation\n",
-    "sigmab=ni*e*(mew_e+mew_h);    #intrinsic conductivity(ohm-1 m-1)\n",
-    "rhob=1/sigmab;     #resistivity(ohm m)\n",
-    "n=Na*d/w;     #number of germanium atoms per m**3\n",
-    "p=n/10**5;   #boron density\n",
-    "sigma=p*e*mew_h;\n",
-    "rho=1/sigma;\n",
-    "\n",
-    "#Result\n",
-    "print \"intrinsic conductivity is\",round(sigma/10**4,3),\"*10**4 ohm-1 m-1\"\n",
-    "print \"intrinsic resistivity is\",round(rho*10**4,3),\"*10**-4 ohm m\"\n",
-    "print \"answer varies due to rounding off errors\"\n",
-    "print \"number of germanium atoms per m**3 is\",round(n/10**28,1),\"*10**28\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.3, Page number 6.23"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 44,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "charge carrier density is 2 *10**22 per m**3\n",
-      "electron mobility is 0.035 m**2/Vs\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "e=1.6*10**-19;\n",
-    "RH=3.66*10**-4;    #hall coefficient(m**3/coulomb)\n",
-    "sigma=112;      #conductivity(ohm-1 m-1)\n",
-    "\n",
-    "#Calculation\n",
-    "ne=3*math.pi/(8*RH*e);    #charge carrier density(per m**3)\n",
-    "mew_e=sigma/(e*ne);      #electron mobility(m**2/Vs)\n",
-    "\n",
-    "#Result\n",
-    "print \"charge carrier density is\",int(ne/10**22),\"*10**22 per m**3\"\n",
-    "print \"electron mobility is\",round(mew_e,3),\"m**2/Vs\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.4, Page number 6.24"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "intrinsic conductivity is 0.432 *10**-3 ohm-1 m-1 10.4\n",
-      "conductivity during donor impurity is 10.4 ohm-1 m-1\n",
-      "conductivity during acceptor impurity is 4 ohm-1 m-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "mew_e=0.13;    #mobility of electron(m**2/Vs)\n",
-    "mew_h=0.05;    #mobility of holes(m**2/Vs)\n",
-    "e=1.6*10**-19;\n",
-    "ni=1.5*10**16;      #number of electron hole pairs\n",
-    "N=5*10**28;\n",
-    "\n",
-    "#Calculation\n",
-    "sigma1=ni*e*(mew_e+mew_h);    #intrinsic conductivity(ohm-1 m-1)\n",
-    "ND=N/10**8;\n",
-    "n=ni**2/ND;\n",
-    "sigma2=ND*e*mew_e;     #conductivity(ohm-1 m-1)\n",
-    "sigma3=ND*e*mew_h;     #conductivity(ohm-1 m-1)\n",
-    "\n",
-    "#Result\n",
-    "print \"intrinsic conductivity is\",round(sigma1*10**3,3),\"*10**-3 ohm-1 m-1\",sigma2\n",
-    "print \"conductivity during donor impurity is\",sigma2,\"ohm-1 m-1\"\n",
-    "print \"conductivity during acceptor impurity is\",int(sigma3),\"ohm-1 m-1\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.5, Page number 6.24"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 50,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "conductivity is 4.97 mho m-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "e=1.6*10**-19;\n",
-    "Eg=0.72;    #band gap(eV)\n",
-    "k=1.38*10**-23;    #boltzmann constant\n",
-    "T1=293;     #temperature(K)\n",
-    "T2=313;     #temperature(K)\n",
-    "sigma1=2;    #conductivity(mho m-1)\n",
-    "\n",
-    "#Calculation\n",
-    "x=(Eg*e/(2*k))*((1/T1)-(1/T2));\n",
-    "y=round(x/2.303,3);\n",
-    "z=round(math.log10(sigma1),3);\n",
-    "log_sigma2=y+z;\n",
-    "sigma2=10**log_sigma2;     #conductivity(mho m-1)\n",
-    "\n",
-    "#Result\n",
-    "print \"conductivity is\",round(sigma2,2),\"mho m-1\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.6, Page number 6.25"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a)Concentration in N-type\n",
-      "n = 1.442 *10**24 m**-3\n",
-      "Hence p = 1.56 *10**8 m**-3\n",
-      "b)Concentration in P-type\n",
-      "p = 3.75 *10**24 m**-3\n",
-      "Hence n = 0.6 *10**8 m**-3\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "ni=1.5*10**16\n",
-    "mu_n=1300*10**-4\n",
-    "mu_p=500*10**-4\n",
-    "e=1.6*10**-19\n",
-    "sigma=3*10**4\n",
-    "\n",
-    "#Calculations\n",
-    "#Concentration in N-type\n",
-    "n1=sigma/(e*mu_n)\n",
-    "p1=ni**2/n1\n",
-    "#Concentration in P-type\n",
-    "p=sigma/(e*mu_p)\n",
-    "n2=(ni**2)/p\n",
-    "\n",
-    "#Result\n",
-    "print\"a)Concentration in N-type\"\n",
-    "print\"n =\",round(n1*10**-24,3),\"*10**24 m**-3\"\n",
-    "print\"Hence p =\",round(p1/10**8,2),\"*10**8 m**-3\"\n",
-    "print\"b)Concentration in P-type\"\n",
-    "print\"p =\",round(p/10**24,2),\"*10**24 m**-3\"\n",
-    "print\"Hence n =\",round(n2/10**8,1),\"*10**8 m**-3\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.7, Page number 6.26"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Jx = 1000.0 ampere/m**2\n",
-      "Ey = 0.183 V/m\n",
-      "Vy = 1.83 mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "i=10**-2\n",
-    "A=0.01*0.001\n",
-    "RH=3.66*10**-4\n",
-    "Bz=0.5\n",
-    "\n",
-    "#Calculations\n",
-    "Jx=i/A\n",
-    "Ey=RH*(Bz*Jx)\n",
-    "Vy=Ey*0.01\n",
-    "\n",
-    "#Result\n",
-    "print\"Jx =\",Jx,\"ampere/m**2\"\n",
-    "print\"Ey =\",round(Ey,3),\"V/m\"\n",
-    "print\"Vy =\",round(Vy*10**3,2),\"mV\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.8, Page number 6.26"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Position of fermi level = 0.5764 eV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "Ev=0\n",
-    "Ec=1.12\n",
-    "k=1.38*10**-23\n",
-    "T=300\n",
-    "mh=0.28\n",
-    "mc=0.12\n",
-    "e=1.6*10**-19\n",
-    "#Calculations\n",
-    "Ef=((Ec+Ev)/2)+((3*k*T)/(4*e))*math.log(mh/mc)\n",
-    "\n",
-    "#Result\n",
-    "print\"Position of fermi level =\",round(Ef,4),\"eV\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.9, Page number 6.26"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Conductivity of intrinsic germanium at 300K = 2.24 ohm**-1 m**-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "ni=2.5*10**19\n",
-    "mu_e=0.38\n",
-    "mu_h=0.18\n",
-    "e=1.6*10**-19\n",
-    "\n",
-    "#Calculations\n",
-    "sigmai=ni*e*(mu_e+mu_h)\n",
-    "\n",
-    "#Result\n",
-    "print\"Conductivity of intrinsic germanium at 300K =\",round(sigmai,2),\"ohm**-1 m**-1\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.10, Page number 6.27"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Conductivity = 1.1593 *10**-3 ohm**-1 m**-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "m=9.1*10**-31\n",
-    "k=1.38*10**-23\n",
-    "T=300\n",
-    "h=6.626*10**-34\n",
-    "Eg=1.1\n",
-    "e=1.6*10**-19\n",
-    "mu_e=0.48\n",
-    "mu_h=0.013\n",
-    "#Calculations\n",
-    "ni=2*((2*math.pi*m*k*T)/h**2)**(3/2)*math.exp(-(Eg*e)/(2*k*T))\n",
-    "sigma=ni*e*(mu_e+mu_h)\n",
-    "                                                  \n",
-    "#Result\n",
-    "print\"Conductivity =\",round(sigma*10**3,4),\"*10**-3 ohm**-1 m**-1\"                                                  "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.11, Page number 6.27"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "p = 2.0 *10**23 m**-3\n",
-      "The electron concentration is given by n = 2.0 *10**9 m**-3\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "Na=5*10**23\n",
-    "Nd=3*10**23\n",
-    "ni=2*10**16\n",
-    "#Calculations\n",
-    "p=((Na-Nd)+(Na-Nd))/2\n",
-    "\n",
-    "#Result\n",
-    "print\"p =\",p*10**-23,\"*10**23 m**-3\"\n",
-    "print\"The electron concentration is given by n =\",ni**2/p*10**-9,\"*10**9 m**-3\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.12, Page number 6.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Rh = 3.7 *10**-6 C**-1 m**3\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "Vh=37*10**-6\n",
-    "thick=1*10**-3\n",
-    "width=5\n",
-    "Iy=20*10**-3\n",
-    "Bz=0.5\n",
-    "\n",
-    "#Calculations\n",
-    "Rh=(Vh*width*thick)/(width*Iy*Bz)\n",
-    "\n",
-    "#Result\n",
-    "print\"Rh =\",round(Rh*10**6,1),\"*10**-6 C**-1 m**3\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.13, Page number 6.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 46,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Dn = 33.54 cm**2 s**-1\n",
-      "Dp = 12.9 cm**2 s**-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "Vt=0.0258\n",
-    "mu_n=1300\n",
-    "mu_p=500\n",
-    "\n",
-    "#Calculations\n",
-    "Dn=Vt*mu_n\n",
-    "Dp=Vt*mu_p\n",
-    "\n",
-    "#Result\n",
-    "print\"Dn =\",Dn,\"cm**2 s**-1\"\n",
-    "print\"Dp =\",Dp,\"cm**2 s**-1\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.14, Page number 6.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 63,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The hole concentration 'p' = 1.125 *10**13 /m**3\n",
-      "'n'= Nd = 2.0 *10**19\n",
-      "Electrical Conductivity = 0.384 ohm**-1 m**-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "ni=1.5*10**16\n",
-    "Nd=2*10**19\n",
-    "e=1.602*100**-19\n",
-    "mu_n=0.12\n",
-    "\n",
-    "#Calculations\n",
-    "p=ni**2/Nd\n",
-    "E_c=e*Nd*mu_n\n",
-    "\n",
-    "#Result\n",
-    "print\"The hole concentration 'p' =\",round(p*10**-13,3),\"*10**13 /m**3\"\n",
-    "print\"'n'= Nd =\",round(Nd*10**-19),\"*10**19\"\n",
-    "print\"Electrical Conductivity =\",round(E_c*10**19,3),\"ohm**-1 m**-1\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.15, Page number 6.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "mu_p= 1389.0 cm**2/V-s\n",
-      "n= 6.0355 *10**13/cm**3\n",
-      "p= 1.0355 *10**13/cm**3\n",
-      "J= 582.5 A/m**2\n",
-      "#Answer varies due to rounding of numbers\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "N=1/60\n",
-    "e=1.6*10**-19\n",
-    "ni=2.5*10**13\n",
-    "b=5*10**13\n",
-    "E=2\n",
-    "\n",
-    "#Calculations\n",
-    "n=(b+math.sqrt(2*b**2))/2\n",
-    "mu_p=N/(3*e*ni)\n",
-    "mu_i=2*mu_p\n",
-    "np=ni**2\n",
-    "p=(ni**2)/n\n",
-    "e=1.6*10**-19\n",
-    "E=2\n",
-    "J=(e*E)*((n*mu_i)+(p*mu_p))\n",
-    "#Result\n",
-    "print\"mu_p=\",round(mu_p),\"cm**2/V-s\"\n",
-    "print\"n=\",round(n/10**13,4),\"*10**13/cm**3\"\n",
-    "print\"p=\",round(p*10**-13,4),\"*10**13/cm**3\"\n",
-    "print\"J=\",round(J*10**4,1),\"A/m**2\"\n",
-    "print\"#Answer varies due to rounding of numbers\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example number 6.16, Page number 6.30"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "ni = 2.293 *10**19 /m**3\n",
-      "Drift velocity of holes 1900.0 ms**-1\n",
-      "Drift velocity of electrons= 3900.0 ms**-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "#importing modules\n",
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#Variable declaration\n",
-    "rho=47*10**-2\n",
-    "e=1.6*10**-19\n",
-    "mu_n=0.39\n",
-    "mu_p=0.19\n",
-    "E=10**4\n",
-    "\n",
-    "#Calculations\n",
-    "ni=1/(rho*e*(mu_n+mu_p))\n",
-    "Dh=mu_p*E\n",
-    "De=mu_n*E\n",
-    "\n",
-    "#Results\n",
-    "print\"ni =\",round(ni/10**19,3),\"*10**19 /m**3\"\n",
-    "print\"Drift velocity of holes\",Dh,\"ms**-1\"\n",
-    "print\"Drift velocity of electrons=\",De,\"ms**-1\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_6a_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_6a_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_6b_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_6b_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_7.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_7.ipynb
deleted file mode 100755
index 48386845..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_7.ipynb
+++ /dev/null
@@ -1,236 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# Chapter 7:LASERS "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 7.1, Page number 7.32"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Divergence = -2.0 *10**-3 radian\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "r1 = 7;                #in radians\n",
-    "r2 = 3;                #in radians\n",
-    "d1 = 4;                #Converting from mm to radians\n",
-    "d2 = 6;                #Converting from mm to radians\n",
-    "\n",
-    "#calculations\n",
-    "D = (r2-r1)/(d2*10**3-d1*10**3)   #Divergence\n",
-    "\n",
-    "#Result\n",
-    "print \"Divergence =\",round(D*10**3,3),\"*10**-3 radian\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 7.2, Page number 7.32"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Frequency (V) = 4.32 *10**14 Hz\n",
-      "Relative Population= 1.081 *10**30\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "C=3*10**8                    #The speed of light\n",
-    "Lamda=6943                   #Wavelength\n",
-    "T=300                        #Temperature in Kelvin\n",
-    "h=6.626*10**-34              #Planck constant \n",
-    "k=1.38*10**-23               #Boltzmann's constant\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "V=(C)/(Lamda*10**-10)       #Frequency\n",
-    "R=math.exp(h*V/(k*T))       #Relative population\n",
-    "\n",
-    "#Result\n",
-    "print \"Frequency (V) =\",round(V/10**14,2),\"*10**14 Hz\"\n",
-    "print \"Relative Population=\",round(R/10**30,3),\"*10**30\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 7.3, Page number 7.32"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " Frequency= 4.74 *10**14 Hz\n",
-      "no.of photons emitted= 7.322 *10**15 photons/sec\n",
-      "Power density = 2.3 kWm**-2\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "C=3*10**8                     #Velocity of light\n",
-    "W=632.8*10**-9                #wavelength\n",
-    "P=2.3\n",
-    "t=1\n",
-    "h=6.626*10**-34              #Planck constant \n",
-    "S=1*10**-6\n",
-    "\n",
-    "#Calculations\n",
-    "V=C/W                        #Frequency\n",
-    "n=((P*10**-3)*t)/(h*V)       #no.of photons emitted\n",
-    "PD=P*10**-3/S                #Power density\n",
-    "\n",
-    "#Result\n",
-    "print \"Frequency=\",round(V/10**14,2),\"*10**14 Hz\"\n",
-    "print \"no.of photons emitted=\",round(n/10**15,3),\"*10**15 photons/sec\"\n",
-    "print \"Power density =\",round(PD/1000,1),\"kWm**-2\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 7.4, Page number 7.33"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Wavelenght = 8628.0 Angstrom\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "h=6.626*10**-34              #Planck constant \n",
-    "C=3*10**8                    #Velocity of light\n",
-    "E_g=1.44                     #bandgap \n",
-    "\n",
-    "#calculations\n",
-    "lamda=(h*C)*10**10/(E_g*1.6*10**-19)       #Wavelenght\n",
-    "\n",
-    "#Result\n",
-    "print \"Wavelenght =\",round(lamda),\"Angstrom\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 7.5, Page number 7.33"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Band gap = 0.8 eV\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "W=1.55                       #wavelength\n",
-    "\n",
-    "#Calculations\n",
-    "E_g=(1.24)/W                 #Bandgap in eV           \n",
-    "\n",
-    "#Result\n",
-    "print \"Band gap =\",E_g,\"eV\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_7_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_7_1.ipynb
old mode 100644
new mode 100755
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_8.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_8.ipynb
deleted file mode 100755
index 4ab9cd1c..00000000
--- a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_8.ipynb
+++ /dev/null
@@ -1,651 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "#Chapter 8:Fiber Optics"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.1, Page number 8.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 125,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The Critical angle = 78.5 degrees\n",
-      "The numerical aperture = 0.3\n",
-      "The acceptance angle = 17.4 degrees\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "n1=1.50          #Core refractive index\n",
-    "n2=1.47          #Cladding refractive index\n",
-    "\n",
-    "#Calculations\n",
-    "C_a=math.asin(n2/n1)        #Critical angle       \n",
-    "N_a=(n1**2-n2**2)**(1/2)\n",
-    "A_a=math.asin(N_a)\n",
-    "\n",
-    "#Results\n",
-    "print \"The Critical angle =\",round(C_a*180/math.pi,1),\"degrees\"\n",
-    "print \"The numerical aperture =\",round(N_a,2)\n",
-    "print \"The acceptance angle =\",round(A_a*180/math.pi,1),\"degrees\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.2, Page number 8.28"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 126,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "N = 490.0\n",
-      "Fiber can support 490.0 guided modes\n",
-      "In graded index fiber, No.of modes propogated inside the fiber = 245.0 only\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "d=50                #diameter\n",
-    "N_a=0.2             #Numerical aperture\n",
-    "lamda=1             #wavelength\n",
-    "\n",
-    "#Calculations\n",
-    "N=4.9*(((d*10**-6*N_a)/(lamda*10**-6))**2)\n",
-    "\n",
-    "#Result\n",
-    "print \"N =\",N\n",
-    "print \"Fiber can support\",N,\"guided modes\"\n",
-    "print \"In graded index fiber, No.of modes propogated inside the fiber =\",N/2,\"only\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.3, Page number 8.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Numerical aperture = 0.008691\n",
-      "No. of modes that can be propogated = 1.0\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "d=50                #diameter\n",
-    "n1=1.450\n",
-    "n2=1.447\n",
-    "lamda=1             #wavelength\n",
-    "\n",
-    "#Calculations\n",
-    "N_a=(n1**2-n2**2)   #Numerical aperture\n",
-    "N=4.9*(((d*10**-6*N_a)/(lamda*10**-6))**2)\n",
-    "\n",
-    "#Results\n",
-    "print \"Numerical aperture =\",N_a\n",
-    "print \"No. of modes that can be propogated =\",round(N)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.4, Page number 8.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Numerical aperture = 0.46\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "delta=0.05          \n",
-    "n1=1.46\n",
-    "\n",
-    "#Calculation\n",
-    "N_a=n1*(2*delta)**(1/2)     #Numerical aperture\n",
-    "\n",
-    "#Result\n",
-    "print \"Numerical aperture =\",round(N_a,2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.5, Page number 8.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "V number = 94.72\n",
-      "maximum no.of modes propogating through fiber = 4486.0\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "a=50\n",
-    "n1=1.53\n",
-    "n2=1.50\n",
-    "lamda=1             #wavelength\n",
-    "\n",
-    "#Calculations\n",
-    "N_a=(n1**2-n2**2)   #Numerical aperture\n",
-    "V=((2*math.pi*a)/lamda)*N_a**(1/2)\n",
-    "\n",
-    "#Result\n",
-    "print \"V number =\",round(V,2)\n",
-    "print \"maximum no.of modes propogating through fiber =\",round(N)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.6, Page number 8.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 64,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Number of modes = 24589.0 modes\n",
-      "No.of modes is doubled to account for the two possible polarisations\n",
-      "Total No.of modes = 49178.0\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "a=100\n",
-    "N_a=0.3               #Numerical aperture\n",
-    "lamda=850             #wavelength\n",
-    "\n",
-    "#Calculations\n",
-    "V_n=(2*(math.pi)**2*a**2*10**-12*N_a**2)/lamda**2*10**-18\n",
-    "#Result\n",
-    "print \"Number of modes =\",round(V_n/10**-36),\"modes\"\n",
-    "print \"No.of modes is doubled to account for the two possible polarisations\"\n",
-    "print \"Total No.of modes =\",round(V_n/10**-36)*2\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.7, Page number 8.29"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 88,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Cutoff Wavellength = 1.315 micro m.\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "\n",
-    "#variable declaration\n",
-    "a=5;\n",
-    "n1=1.48;\n",
-    "delta=0.01;\n",
-    "V=25;\n",
-    "\n",
-    "#Calculation\n",
-    "lamda=(math.pi*(a*10**-6)*n1*math.sqrt(2*delta))/V   # Cutoff Wavelength\n",
-    "\n",
-    "#Result\n",
-    "print \"Cutoff Wavellength =\",round(lamda*10**7,3),\"micro m.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.8, Page number 8.30"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 87,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum core radius= 9.95 micro m\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "\n",
-    "#variable declaration\n",
-    "V=2.405\n",
-    "lamda=1.3\n",
-    "N_a=0.05\n",
-    "\n",
-    "#Calculations\n",
-    "a_max=(V*lamda)/(2*math.pi*N_a)\n",
-    "\n",
-    "#Result\n",
-    "print \"Maximum core radius=\",round(a_max,2),\"micro m\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.9, Page number 8.30"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Acceptance angle, theta_a = 17.46 degrees\n",
-      "For skew rays,theta_as  34.83 degrees\n",
-      "#Answer given in the textbook is wrong\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "N_a=0.3\n",
-    "gamma=45\n",
-    "\n",
-    "#Calculations\n",
-    "theta_a=math.asin(N_a)\n",
-    "theta_as=math.asin((N_a)/math.cos(gamma))\n",
-    "\n",
-    "#Results\n",
-    "print \"Acceptance angle, theta_a =\",round(theta_a*180/math.pi,2),\"degrees\"\n",
-    "print \"For skew rays,theta_as \",round(theta_as*180/math.pi,2),\"degrees\"\n",
-    "print\"#Answer given in the textbook is wrong\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.10, Page number 8.30"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 115,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Numerical aperture = 0.303\n",
-      "Acceptance angle = 17.63 degrees\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "n1=1.53\n",
-    "delta=0.0196\n",
-    "\n",
-    "#Calculations\n",
-    "N_a=n1*(2*delta)**(1/2)\n",
-    "A_a=math.asin(N_a)\n",
-    "#Result\n",
-    "print \"Numerical aperture =\",round(N_a,3)\n",
-    "print \"Acceptance angle =\",round(A_a*180/math.pi,2),\"degrees\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.11, Page number 8.30"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "delta = 0.01\n",
-      "Core radius,a = 1.55 micro m\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "n1=1.480\n",
-    "n2=1.465\n",
-    "V=2.405\n",
-    "lamda=850*10**-9\n",
-    "\n",
-    "#Calculations\n",
-    "delta=(n1**2-n2**2)/(2*n1**2)\n",
-    "a=(V*lamda*10**-9)/(2*math.pi*n1*math.sqrt(2*delta))\n",
-    "\n",
-    "#Results\n",
-    "print \"delta =\",round(delta,2)\n",
-    "print \"Core radius,a =\",round(a*10**15,2),\"micro m\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.12, Page number 8.31"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 147,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " Critical angle= 83.38 degrees\n",
-      "Fiber length covered in one reflection= 430.84 micro m\n",
-      "Total no.of reflections per metre= 2321.0\n",
-      "Since L=1m, Total dist. travelled by light over one metre of fiber = 1.0067 m\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "n1=1.5\n",
-    "n2=1.49\n",
-    "a=25\n",
-    "\n",
-    "#Calculations\n",
-    "C_a=math.asin(n2/n1)           #Critical angle\n",
-    "L=2*a*math.tan(C_a)             \n",
-    "N_r=10**6/L                    \n",
-    "\n",
-    "#Result\n",
-    "print \"Critical angle=\",round(C_a*180/math.pi,2),\"degrees\"\n",
-    "print \"Fiber length covered in one reflection=\",round(L,2),\"micro m\"\n",
-    "print \"Total no.of reflections per metre=\",round(N_r)\n",
-    "print \"Since L=1m, Total dist. travelled by light over one metre of fiber =\",round(1/math.sin(C_a),4),\"m\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.13, Page number 8.31"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 155,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "No.of modes = 154.69 =155(approx)\n",
-      "Taking the two possible polarizations, Total No.of nodes = 309.0\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "alpha=1.85\n",
-    "lamda=1.3*10**-6\n",
-    "a=25*10**-6\n",
-    "N_a=0.21\n",
-    "\n",
-    "#Calculations\n",
-    "V_n=((2*math.pi**2)*a**2*N_a**2)/lamda**2\n",
-    "N_m=(alpha/(alpha+2))*V_n\n",
-    "\n",
-    "print \"No.of modes =\",round(N_m,2),\"=155(approx)\"\n",
-    "print \"Taking the two possible polarizations, Total No.of nodes =\",round(N_m*2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.14, Page number 8.32"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a)Signal attention per unit length = 3.9 dB km**-1\n",
-      "b)Overall signal attenuation = 39.0 dB\n",
-      "#Answer given in the textbook is wrong\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "P_i=100\n",
-    "P_o=2\n",
-    "L=10\n",
-    "\n",
-    "#Calculations\n",
-    "S=(10/L)*math.log(P_i/P_o)\n",
-    "O=S*L\n",
-    "\n",
-    "#Result\n",
-    "print \"a)Signal attention per unit length =\",round(S,1),\"dB km**-1\"\n",
-    "print \"b)Overall signal attenuation =\",round(O),\"dB\"\n",
-    "print \"#Answer given in the textbook is wrong\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##Example 8.15, Page number 8.32"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Total dispersion = 1343.3 ns\n",
-      "Bandwidth length product = 37.22 Hz-km\n",
-      "#Answer given in the text book is wrong\n"
-     ]
-    }
-   ],
-   "source": [
-    "import math\n",
-    "from __future__ import division\n",
-    "\n",
-    "#variable declaration\n",
-    "L=10\n",
-    "n1=1.55\n",
-    "delta=0.026\n",
-    "C=3*10**5\n",
-    "\n",
-    "#Calculations\n",
-    "delta_T=(L*n1*delta)/C\n",
-    "B_W=10/(2*delta_T)\n",
-    "\n",
-    "#Result\n",
-    "print \"Total dispersion =\",round(delta_T/10**-9,1),\"ns\"\n",
-    "print \"Bandwidth length product =\",round(B_W/10**5,2),\"Hz-km\"\n",
-    "print \"#Answer given in the text book is wrong\""
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
-   "version": "2.7.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_8_1.ipynb b/APPLIED_PHYSICS_by_M,_ARUMUGAM/Chapter_8_1.ipynb
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diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch2.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch2.ipynb
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@@ -0,0 +1,1106 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3c32f4c771b7aa90d25f7c7a71e72b94eccb8c650a9b70156e8a9a1623dc111e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Static Characteristics of Instruments and Measurement systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1  Page No : 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating static error and static correction\n",
+      "# Variables\n",
+      "Am  =  127.50;\n",
+      "At  =  127.43;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "e = Am-At;\n",
+      "print \"Static error(V) = \", e\n",
+      "\n",
+      "Sc = -e;\n",
+      "print \"Static Correction(V) = \", Sc"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Static error(V) =  0.07\n",
+        "Static Correction(V) =  -0.07\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2  Page No : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating true value of the temperature\n",
+      "\n",
+      "# Variables\n",
+      "Am  =  95.45;\n",
+      "Sc = -0.08;\n",
+      "\n",
+      "# Calculations\n",
+      "At = Am+Sc;\n",
+      "\n",
+      "# Results\n",
+      "print \"True Temperature(Degree C) = \", At\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "True Temperature(Degree C) =  95.37\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3  Page No : 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating Relative error (expressed as a percentage of f.s.d)\n",
+      "\n",
+      "# Variables\n",
+      "Am  =  1.46;\n",
+      "At = 1.50;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "e = Am-At;\n",
+      "print \"Absolute error(V) = \", e\n",
+      "Sc = -e;\n",
+      "print \"Absolute Correction(V) = \", Sc\n",
+      "RE = (e/At)*100;\n",
+      "print \"Relative Error in terms of true value(in percentage) = \", RE\n",
+      "REF = (e/2.5)*100;\n",
+      "print \"Relative Error in terms of true value(in percentage) = \", REF"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Absolute error(V) =  -0.04\n",
+        "Absolute Correction(V) =  0.04\n",
+        "Relative Error in terms of true value(in percentage) =  -2.66666666667\n",
+        "Relative Error in terms of true value(in percentage) =  -1.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4  Page No : 11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating static error and static correction\n",
+      "\n",
+      "# Variables\n",
+      "Am  =  0.000161;\n",
+      "At  =  0.159*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "e = Am-At;\n",
+      "print \"Static error(m3/s) = \", e\n",
+      "\n",
+      "Sc = -e;\n",
+      "print \"Static Correction(m3/s) = \", Sc\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Static error(m3/s) =  2e-06\n",
+        "Static Correction(m3/s) =  -2e-06\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5  Page No : 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating maximum static error \n",
+      "\n",
+      "# Variables\n",
+      "#Span of the thermometer(degree C)\n",
+      "S = 200.-150;\n",
+      "#Accuracy of the thermometer(in terms of percentage of span)\n",
+      "A = 0.0025;\n",
+      "\n",
+      "# Calculations\n",
+      "e =  A*S;\n",
+      "\n",
+      "# Results\n",
+      "print \"Maximum Static error(degree C) = \", e"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum Static error(degree C) =  0.125\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6  Page No : 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating the pressure for a dial reading of 100\n",
+      "\n",
+      "# Calculations\n",
+      "P = ((27.58-6.895)/150)*100+6.895;\n",
+      "\n",
+      "# Results\n",
+      "print \"pressure for a dial reading of 100(kN/m2) = \", P\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure for a dial reading of 100(kN/m2) =  20.685\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7  Page No : 17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating the noise output voltage of the amplifier\n",
+      "\n",
+      "# Variables\n",
+      "Bw = 100.*10**3;\n",
+      "Sn = 7.*10**-21;\n",
+      "R = 50.*10**3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A = (Sn*R*Bw)**0.5;\n",
+      "En = 2*A;\n",
+      "print \"Noise voltage at input(V) = \", En\n",
+      "\n",
+      "Ga = 100;\n",
+      "Eno = En*Ga;\n",
+      "print \"Noise voltage at output(V) = \", Eno\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Noise voltage at input(V) =  1.18321595662e-05\n",
+        "Noise voltage at output(V) =  0.00118321595662\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8  Page No : 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating the noise voltage\n",
+      "\n",
+      "# Variables\n",
+      "Sn = 20.;\n",
+      "Vs = 3;\n",
+      "\n",
+      "# Calculations\n",
+      "Vn = Vs/(Sn)**0.5;\n",
+      "\n",
+      "# Results\n",
+      "print \"noise Voltage(mV) = \", Vn"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "noise Voltage(mV) =  0.67082039325\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9  Page No : 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# calculating the signal to noise ratio at input\n",
+      "# calculating the signal to noise ratio at output\n",
+      "#calculating the noise factor and noise figure\n",
+      "\n",
+      "# Calculations and Results\n",
+      "print \"signal to noise ratio at input\"\n",
+      "Sni = (3.*10**-6/(1*10**-6))**2;\n",
+      "print \"signal to noise ratio at input = \", Sni\n",
+      "print \"signal to noise ratio at output\"\n",
+      "Sno = (60.*10**-6/(20*10**-6))**2;\n",
+      "print \"signal to noise ratio at output = \", Sno\n",
+      "print \"New signal to noise ratio at output\"\n",
+      "Snno = (60.*10**-6/(25*10**-6))**2;\n",
+      "print \"signal to noise ratio at output = \", Snno\n",
+      "\n",
+      "F = Sni/Snno;\n",
+      "print \"noise Factor = \", F\n",
+      "\n",
+      "nf = 10*math.log10(F);\n",
+      "print \"noise Figure(dB) = \", nf\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "signal to noise ratio at input\n",
+        "signal to noise ratio at input =  9.0\n",
+        "signal to noise ratio at output\n",
+        "signal to noise ratio at output =  9.0\n",
+        "New signal to noise ratio at output\n",
+        "signal to noise ratio at output =  5.76\n",
+        "noise Factor =  1.5625\n",
+        "noise Figure(dB) =  1.93820026016\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10  Page No : 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating the ratio of output signal to noise signal \n",
+      "\n",
+      "# Variables\n",
+      "Bw = 100.*10**3;\n",
+      "K = 1.38*10**-23;\n",
+      "T = 300.;\n",
+      "R = 120.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A = (K*T*R*Bw)**0.5;\n",
+      "En = 2*A;\n",
+      "print \"Noise voltage (V) = \", En\n",
+      "Eno = 0.12*10**-3;\n",
+      "print \"Noise voltage at output(V) = \", Eno\n",
+      "Ra = Eno/En;\n",
+      "print \"Ratio of signal votage to Noise voltage  = \", Ra\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Noise voltage (V) =  4.45780214904e-07\n",
+        "Noise voltage at output(V) =  0.00012\n",
+        "Ratio of signal votage to Noise voltage  =  269.190951029\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12  Page No : 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#calculating the average force and range of error\n",
+      "F1 = 10.03;\n",
+      "F2 = 10.10;\n",
+      "F3 = 10.11;\n",
+      "F4 = 10.08;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Fav = (F1+F2+F3+F4)/4;\n",
+      "print \"Average Force(N)  = \", Fav\n",
+      "Fmax = F3;\n",
+      "MaxR = Fmax-Fav;\n",
+      "Fmin = F1;\n",
+      "MinR = Fav-Fmin;\n",
+      "AvgR = (MaxR+MinR)/2;\n",
+      "print \"Average range of error (N) = \", AvgR"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average Force(N)  =  10.08\n",
+        "Average range of error (N) =  0.04\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13  Page No : 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#calculating the sum of resistances connected in series with uncertainity of one unit\n",
+      "R1 = 72.3;\n",
+      "R2 = 2.73;\n",
+      "R3 = 0.612;\n",
+      "\n",
+      "# Calculations\n",
+      "R = (R1+R2+R3);\n",
+      "\n",
+      "# Results\n",
+      "print \"sum of resistances(ohm)  = \", R\n",
+      "\n",
+      "print \"the resultant resistance is 75.6 ohm with 6 as first doutful figure\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "sum of resistances(ohm)  =  75.642\n",
+        "the resultant resistance is 75.6 ohm with 6 as first doutful figure\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14  Page No : 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#calculating the power with uncertainity of one unit in voltage and current\n",
+      "\n",
+      "# Variables\n",
+      "V = 12.16;\n",
+      "I = 1.34;\n",
+      "\n",
+      "# Calculations\n",
+      "P = V*I;\n",
+      "\n",
+      "# Results\n",
+      "print \"Power(W)  = \", P\n",
+      "\n",
+      "print \"the resultant  is 16.2 W with 2 as first doutful figure\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(W)  =  16.2944\n",
+        "the resultant  is 16.2 W with 2 as first doutful figure\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15  Page No : 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the sum of resistances connected in series with appropriate number of significant figure\n",
+      "\n",
+      "# Variables\n",
+      "R1 = 28.7;\n",
+      "R2 = 3.624;\n",
+      "\n",
+      "# Calculations\n",
+      "R = (R1+R2);\n",
+      "\n",
+      "# Results\n",
+      "print \"sum of resistances(ohm)  = \", R\n",
+      "\n",
+      "print \"the resultant resistance is 32.3 ohm as one of the resistance is accurate to three significant figure\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "sum of resistances(ohm)  =  32.324\n",
+        "the resultant resistance is 32.3 ohm as one of the resistance is accurate to three significant figure\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16  Page No : 33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the voltage drop with appropriate number of significant figure\n",
+      "\n",
+      "# Variables\n",
+      "R = 31.27;\n",
+      "I = 4.37;\n",
+      "\n",
+      "# Calculations\n",
+      "E = I*R;\n",
+      "\n",
+      "# Results\n",
+      "print \"voltage drop(V)  = \", E\n",
+      "\n",
+      "print \"the voltage drop is 137 V as one of the resistance is accurate to three significant figure\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "voltage drop(V)  =  136.6499\n",
+        "the voltage drop is 137 V as one of the resistance is accurate to three significant figure\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17  Page No : 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the sensitivity and deflection factor of wheatstone bridge\n",
+      "\n",
+      "# Variables\n",
+      "Mo = 3.;\n",
+      "Mi = 7;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Sen = Mo/Mi;\n",
+      "print \"sensitivity(mm per ohm)  = \", Sen\n",
+      "Df = Mi/Mo;\n",
+      "print \"deflection factor( ohm per mm)  = \", Df\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "sensitivity(mm per ohm)  =  0.428571428571\n",
+        "deflection factor( ohm per mm)  =  2.33333333333\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18  Page No : 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the volume of the mercury thermometer\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Ac = (math.pi/4)*0.25**2;\n",
+      "print \"Area of mercury thermometer\", Ac\n",
+      "Lc = 13.8*10**3;\n",
+      "Vc = Ac*Lc;\n",
+      "print \"Volume of mercury thermometer(mm3)\", Vc\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area of mercury thermometer 0.0490873852123\n",
+        "Volume of mercury thermometer(mm3) 677.40591593\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19  Page No : 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the maximum position deviation, resistance deviation \n",
+      "\n",
+      "# Variables\n",
+      "Pl = 0.001;\n",
+      "FSD = 320;\n",
+      "R = 10000;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "MDD = (Pl*FSD);\n",
+      "print \"Maximum lacement deviation(degree) =\",MDD\n",
+      "MRD = Pl*R;\n",
+      "print \"Maximum lacement deviation(ohm) = \",MRD\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum lacement deviation(degree) = 0.32\n",
+        "Maximum lacement deviation(ohm) =  10.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.20  Page No : 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the dead zone\n",
+      "\n",
+      "# Variables\n",
+      "s = 600;\n",
+      "\n",
+      "# Calculations\n",
+      "Dz = 0.00125*s;\n",
+      "\n",
+      "# Results\n",
+      "print \"Dead zone(degree C) = \", Dz\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Dead zone(degree C) =  0.75\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.22  Page No : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the Resolution\n",
+      "\n",
+      "# Variables\n",
+      "Fs = 200.;\n",
+      "D = 100.;\n",
+      "\n",
+      "# Calculations\n",
+      "SD = Fs/D;\n",
+      "R = SD/10;\n",
+      "\n",
+      "# Results\n",
+      "print \"resolution (V) = \", R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resolution (V) =  0.2\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.23  Page No : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the Resolution\n",
+      "\n",
+      "# Variables\n",
+      "Fs = 9.999;\n",
+      "D = 9999;\n",
+      "\n",
+      "# Calculations\n",
+      "SD = Fs/D;\n",
+      "R = SD;\n",
+      "\n",
+      "# Results\n",
+      "print \"resolution (V) = \", R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resolution (V) =  0.001\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.24  Page No : 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the reading of the multimeter and percentage error\n",
+      "\n",
+      "# Variables\n",
+      "Zl = 20000.;\n",
+      "Zo = 10000.;\n",
+      "Eo = 6.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "El = Eo/(1+Zo/Zl);\n",
+      "print \"Reading of the multimeter (V) = \", El\n",
+      "PE = ((El-Eo)/Eo)*100;\n",
+      "print \"Percentage error = \", PE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reading of the multimeter (V) =  4.0\n",
+        "Percentage error =  -33.3333333333\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.25  Page No : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the reading of the multimeter and percentage error\n",
+      "\n",
+      "# Variables\n",
+      "Zl = 20000.;\n",
+      "Zo = 1000.;\n",
+      "Eo = 6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "El = Eo/(1+Zo/Zl);\n",
+      "print \"Reading of the multimeter (V) = \", El\n",
+      "PE = ((El-Eo)/Eo)*100;\n",
+      "print \"Percentage error = \", PE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reading of the multimeter (V) =  5.71428571429\n",
+        "Percentage error =  -4.7619047619\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.26  Page No : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the loading error\n",
+      "\n",
+      "# Variables\n",
+      "Zl = 1000.;\n",
+      "Zo = 200*200./400;\n",
+      "Eo = 100*200./400;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "El = Eo/(1+Zo/Zl);\n",
+      "print \"Reading of the multimeter (V) = \", El\n",
+      "PE = ((El-Eo)/Eo)*100;\n",
+      "print \"Percentage loading error = \", PE\n",
+      "Ac = 100+PE;\n",
+      "print \"Accuracy = \", Ac\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reading of the multimeter (V) =  45.4545454545\n",
+        "Percentage loading error =  -9.09090909091\n",
+        "Accuracy =  90.9090909091\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.27  Page No : 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#calculating the voltage across the oscilloscope\n",
+      "\n",
+      "# Variables\n",
+      "C = 50*10.**-6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "f = 100000.;\n",
+      "print \"frequency = \", f\n",
+      "Xc = 1./(2*math.pi*f*C);\n",
+      "R = 10.**6;\n",
+      "Zl = (R*-1j*Xc)/(R-1j*Xc);\n",
+      "Eo = 1.;\n",
+      "Zo = 10.*10**3;\n",
+      "\n",
+      "El = Eo/(1+Zo/Zl);\n",
+      "print \"Reading of the multimeter (V) = \", El\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequency =  100000.0\n",
+        "Reading of the multimeter (V) =  (1.02334395478e-11-3.18309886181e-06j)\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.28  Page No : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the actual value of current, measured value of current and percentage error\n",
+      "\n",
+      "# Variables\n",
+      "Eo = 10.-((10.*1000)/(1000+1000));\n",
+      "Zo = ((1000.*1000)/(1000+1000))+500;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Io = Eo/Zo;\n",
+      "print \"Actual value of current (A) = \", Io\n",
+      "Zl = 100.;\n",
+      "Il = Eo/(Zo+Zl);\n",
+      "print \"Measured value of current (A) = \", Il\n",
+      "PE = ((Il-Io)/Io)*100;\n",
+      "print \"Percentage loading error = \", PE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Actual value of current (A) =  0.005\n",
+        "Measured value of current (A) =  0.00454545454545\n",
+        "Percentage loading error =  -9.09090909091\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.29  Page No : 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculating the maximum available power\n",
+      "\n",
+      "# Variables\n",
+      "Eo = 80.*10**-3;\n",
+      "Il = 5.*10**-9;\n",
+      "Rl = 6.*10**6;\n",
+      "\n",
+      "# Calculations\n",
+      "Ro = (Eo/Il)-Rl;\n",
+      "Pmax = (Eo**2)/(4*Ro);\n",
+      "\n",
+      "# Results\n",
+      "print \"Maximum available Power(W) = \", Pmax\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum available Power(W) =  1.6e-10\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch3.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch3.ipynb
new file mode 100755
index 00000000..21e7112d
--- /dev/null
+++ b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch3.ipynb
@@ -0,0 +1,1812 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2ffa422208323473e7fa809c26d007c1a4e73e5340d80167d553ec696ed3e811"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Errors in Measurements and Their Statistical Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1  Page No : 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating guarantee value of capacitance\n",
+      "\n",
+      "# Variables\n",
+      "As  =  1;\n",
+      "Er = 0.05;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Aau = As*(1+Er);\n",
+      "print \"Upper limit(micro F) = \", Aau\n",
+      "Aal = As*(1-Er);\n",
+      "print \"Lower limit(micro F) = \", Aal\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Upper limit(micro F) =  1.05\n",
+        "Lower limit(micro F) =  0.95\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2  Page No : 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculating percentage limiting error\n",
+      "\n",
+      "# Variables\n",
+      "As  =  150.;\n",
+      "Er = 0.01;\n",
+      "\n",
+      "# Calculations\n",
+      "dA = As*Er;\n",
+      "As1 = 75.;\n",
+      "Er = (dA/As1)*100;\n",
+      "\n",
+      "# Results\n",
+      "print \"Percentage limiting error  = \", Er\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage limiting error  =  2.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3  Page No : 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the range of readings\n",
+      "\n",
+      "# Variables\n",
+      "fsd = 1000.;\n",
+      "TP = 100.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Efsd = (1./100)*1000;\n",
+      "print \"magnitude of Error when specified in terms of full scale deflection (w) = \", Efsd\n",
+      "print \"Thus the meter will read between 90W and 110W\"\n",
+      "\n",
+      "Etv = (1./100)*100;\n",
+      "print \"magnitude of Error when specified in terms of true value (w) = \", Etv\n",
+      "print \"Thus the meter will read between 99W and 101W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnitude of Error when specified in terms of full scale deflection (w) =  10.0\n",
+        "Thus the meter will read between 90W and 110W\n",
+        "magnitude of Error when specified in terms of true value (w) =  1.0\n",
+        "Thus the meter will read between 99W and 101W\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4  Page No : 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the limiting error in percent\n",
+      "\n",
+      "# Variables\n",
+      "dA = 0.05*5*10**-6;\n",
+      "As = 2.5*10**-6;\n",
+      "\n",
+      "# Calculations\n",
+      "Er = (dA/As)*100;\n",
+      "\n",
+      "# Results\n",
+      "print \"percemtage limiting error  = +/-\", Er\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percemtage limiting error  = +/- 10.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5  Page No : 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the range of readings specified interms of f.s.d. and true value\n",
+      "\n",
+      "print \"Range when specified interms of f.s.d.\"\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Error_fsd = 1.*1000./100\n",
+      "Range_lower_value = 100-Error_fsd;\n",
+      "print \"Lower value of range (kN/m2)\", Range_lower_value\n",
+      "Range_upper_value = 100+Error_fsd;\n",
+      "print \"Upper value of range (kN/m2)\", Range_upper_value\n",
+      "print \"Range when specified interms of True value\",\n",
+      "Error_true = 1*100/100.\n",
+      "Range_lower_value = 100-Error_true;\n",
+      "print \"Lower value of range (kN/m2)\", Range_lower_value\n",
+      "Range_upper_value = 100+Error_true;\n",
+      "print \"Upper value of range (kN/m2)\", Range_upper_value\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Range when specified interms of f.s.d.\n",
+        "Lower value of range (kN/m2) 90.0\n",
+        "Upper value of range (kN/m2) 110.0\n",
+        "Range when specified interms of True value Lower value of range (kN/m2) 99.0\n",
+        "Upper value of range (kN/m2) 101.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6  Page No : 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the magnitude and limiting error in ohm and in percentage of the resistance\n",
+      "\n",
+      "# Variables\n",
+      "R1 = 37.;\n",
+      "R1_le = 5.*R1/100;\n",
+      "R2 = 75.;\n",
+      "R2_le = 5.*R2/100;\n",
+      "R3 = 50.;\n",
+      "R3_le = 5.*R3/100;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R = R1+R2+R3;\n",
+      "print \"Value of resistance (ohm) = \", R\n",
+      "\n",
+      "R_le = R1_le+R2_le+R3_le;\n",
+      "print \"Limiting Value of resistance (ohm) = \", R_le\n",
+      "\n",
+      "Limiting_error_percentage = R_le*100/R;\n",
+      "print \"Limiting Value of resistance (percentage) = +/-\", Limiting_error_percentage\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of resistance (ohm) =  162.0\n",
+        "Limiting Value of resistance (ohm) =  8.1\n",
+        "Limiting Value of resistance (percentage) = +/- 5.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7  Page No : 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculate the value of  relative limiting error in resistance\n",
+      "\n",
+      "# Variables\n",
+      "Re_P = 1.5;\n",
+      "Re_I = 1;\n",
+      "\n",
+      "# Calculations\n",
+      "Re_resistance = (Re_P+2*Re_I);\n",
+      "\n",
+      "# Results\n",
+      "print \"the value of  relative limiting error of resistance in percentage(+/-) = \", Re_resistance"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of  relative limiting error of resistance in percentage(+/-) =  3.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8  Page No : 73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the guaranteed values of the resistance\n",
+      "\n",
+      "# Variables\n",
+      "R1 = 100.;\n",
+      "R1_le_perunit = 0.5; # R1_le_perunit indicates dR1/R1  =  0.5%\n",
+      "R2 = 1000.;\n",
+      "R2_le_perunit = 0.5;\n",
+      "R3 = 842.;\n",
+      "R3_le_perunit = 0.5;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Rx = R2*R3/R1;\n",
+      "print \"Value of resistance (ohm) = \", Rx\n",
+      "Rx_le_perunit = R1_le_perunit+R2_le_perunit+R3_le_perunit;\n",
+      "\n",
+      "print \"Limiting Value of resistance per unit (dRx/Rx) = \", Rx_le_perunit\n",
+      "Er_Le = Rx_le_perunit*Rx/100;\n",
+      "print \"Limiting Value of resistance (ohm) = +/-\", Er_Le\n",
+      "print \"Guarantee value of the resistance (ohm) = \",\n",
+      "G1 = Rx+Er_Le;\n",
+      "G2 = Rx-Er_Le;\n",
+      "print G1,G2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of resistance (ohm) =  8420.0\n",
+        "Limiting Value of resistance per unit (dRx/Rx) =  1.5\n",
+        "Limiting Value of resistance (ohm) = +/- 126.3\n",
+        "Guarantee value of the resistance (ohm) =  8546.3 8293.7\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9  Page No : 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the percentage limiting error and range of resistance values\n",
+      "print \"error in decade a =   so, decade a is set at 4000 ohm\",\n",
+      "Er_a = 4000.*0.1/100;\n",
+      "print Er_a\n",
+      "print \"error in decade b =   so, decade b is set at 600 ohm\",\n",
+      "Er_b = 600*0.1/100.;\n",
+      "print Er_b\n",
+      "print \"error in decade c =   so, decade c is set at 30 ohm\",\n",
+      "Er_c = 30*0.1/100.;\n",
+      "print Er_c\n",
+      "print \"error in decade d =   so, decade d is set at 9 ohm\",\n",
+      "Er_d = 9*0.1/100;\n",
+      "print Er_d\n",
+      "Er_total = Er_a+Er_b+Er_c+Er_d;\n",
+      "Re_le_percentage = Er_total*100./4639;\n",
+      "print \"Percentage Relative limiting error = \", Re_le_percentage\n",
+      "Range_lower = 4639-Er_total;\n",
+      "print \"Lower value of range (ohm) = \", Range_lower\n",
+      "Range_upper = 4639+Er_total;\n",
+      "print \"upper value of range (ohm) = \", Range_upper\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "error in decade a =   so, decade a is set at 4000 ohm 4.0\n",
+        "error in decade b =   so, decade b is set at 600 ohm 0.6\n",
+        "error in decade c =   so, decade c is set at 30 ohm 0.03\n",
+        "error in decade d =   so, decade d is set at 9 ohm 0.009\n",
+        "Percentage Relative limiting error =  0.1\n",
+        "Lower value of range (ohm) =  4634.361\n",
+        "upper value of range (ohm) =  4643.639\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10  Page No : 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the magnitude of power and limiting error\n",
+      "\n",
+      "# Variables\n",
+      "F = 4.58;\n",
+      "L = 397.;\n",
+      "R = 1202.*10**-9;\n",
+      "t = 60.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "P = (2*math.pi*9.81*F*L*R)/(t*10**6);\n",
+      "print \"Magnitude of power (W) = \", P\n",
+      "dF_pu = 0.02/F; # per unit error in force\n",
+      "dL_pu = 1.3/L; # per unit error in Length\n",
+      "dR_pu = 1/R; # per unit error in revolution\n",
+      "dt_pu = 0.5/t; # per unit error in time\n",
+      "dP_pu =  dF_pu+dL_pu+dR_pu+dt_pu;\n",
+      "dP_le = dP_pu*P;\n",
+      "print \"Magnitude of limiting error in power (W)\", dP_le\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnitude of power (W) =  2.24521573806e-09\n",
+        "Magnitude of limiting error in power (W) 0.00186789998434\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11  Page No : 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the magnitude of Force and limiting error\n",
+      "\n",
+      "# Variables\n",
+      "E = 200.*10**9;\n",
+      "L = 25.*10**-3;\n",
+      "b = 4.75*10**-3;\n",
+      "d = 0.9*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I = (b*d**3)/12;\n",
+      "x = 2.5*10**-3;\n",
+      "F = (3*E*I*x)/(L**3);\n",
+      "print \"Magnitude of Force (N) = \", F\n",
+      "\n",
+      "dE_pu = 0./E; # per unit error in E\n",
+      "db_pu = 0.0075/b; \n",
+      "dd_pu = 0.0075/d;  \n",
+      "dx_pu = 0.025/x; \n",
+      "dL_pu = 0.025/L;\n",
+      "dF_pu =  (dE_pu+db_pu+3*dd_pu+dx_pu+3*dL_pu)*10**-3;\n",
+      "\n",
+      "print \"limiting error in force (N) = +/-\", dF_pu\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnitude of Force (N) =  27.702\n",
+        "limiting error in force (N) = +/- 0.0395789473684\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12  Page No : 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculate the power loss and relative error \n",
+      "\n",
+      "# Variables\n",
+      "I = 64.*10**-3;\n",
+      "R = 3200.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "P = (I**2)*R;\n",
+      "print \"Power(W) = \", P\n",
+      "Re = 2*0.75+0.2;\n",
+      "print \"Relative error (%) = \", Re\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(W) =  13.1072\n",
+        "Relative error (%) =  1.7\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13  Page No : 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the true power as a percentage of measured power\n",
+      "\n",
+      "# Variables\n",
+      "I = 30.4;\n",
+      "R = 0.015;\n",
+      "\n",
+      "# Calculations\n",
+      "I_true = I*(1+0.012);\n",
+      "R_true = R*(1-0.003);\n",
+      "P_true = (I_true**2)*R_true;\n",
+      "P_measured = (I**2)*R;\n",
+      "R = P_true*100/P_measured;\n",
+      "\n",
+      "# Results\n",
+      "print \"true power as a percentage of measured power(%) =  \", R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "true power as a percentage of measured power(%) =   102.1071568\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14  Page No : 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the total resistance, error of each register and fractional error of total resistance\n",
+      "\n",
+      "# Variables\n",
+      "R1 = 250.;\n",
+      "R2 = 500.;\n",
+      "R3 = 375.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R_true = 1/((1/R1)+(1/R2)+(1/R3));\n",
+      "print \" True value of resistance(ohm) = \", R_true\n",
+      "\n",
+      "dR1 =  0.025*R1;\n",
+      "dR2 = -0.36*R2;\n",
+      "dR3 = 0.014*R3;\n",
+      "R1_effective = R1+dR1;\n",
+      "R2_effective = R2+dR2;\n",
+      "R3_effective = R3+dR3;\n",
+      "R_effective = 1/((1/R1_effective)+(1/R2_effective)+(1/R3_effective));\n",
+      "print \" Effective value of resistance(ohm) = \", R_effective\n",
+      "\n",
+      "Fractional_error = (R_true-R_effective)/R_true;\n",
+      "print \"Fractional_error\", Fractional_error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " True value of resistance(ohm) =  115.384615385\n",
+        " Effective value of resistance(ohm) =  103.548741621\n",
+        "Fractional_error 0.102577572617\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15  Page No : 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#\n",
+      "print \"When all the components have 0% error then resonant frequency (Hz)\",\n",
+      "\n",
+      "# Variables\n",
+      "L = 160.*10**-6;\n",
+      "C = 160.*10**-12;\n",
+      "fr = (1./(2*math.pi))*(1/(L*C))**0.5;\n",
+      "print fr\n",
+      "\n",
+      "# Calculations and Results\n",
+      "print \"When all the components have +10% error then resonant frequency (Hz)\",\n",
+      "L_new = (160.*10**-6)+0.1*L;\n",
+      "C_new = (160.*10**-12)+0.1*C;\n",
+      "fr_new = (1./(2*math.pi))*(1/(L_new*C_new))**0.5;\n",
+      "print fr_new\n",
+      "\n",
+      "error = (fr_new-fr)/fr;\n",
+      "print \"error = \", error\n",
+      "\n",
+      "print \"When all the components have -10% error then resonant frequency (Hz)\",\n",
+      "L_new = (160.*10**-6)-0.1*L;\n",
+      "C_new = (160.*10**-12)-0.1*C;\n",
+      "fr_new = (1./(2*math.pi))*(1/(L_new*C_new))**0.5;\n",
+      "print fr_new\n",
+      "\n",
+      "error = (fr_new-fr)/fr;\n",
+      "print \"error = \", error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When all the components have 0% error then resonant frequency (Hz) 994718.394324\n",
+        "When all the components have +10% error then resonant frequency (Hz) 904289.449386\n",
+        "error =  -0.0909090909091\n",
+        "When all the components have -10% error then resonant frequency (Hz) 1105242.66036\n",
+        "error =  0.111111111111\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16  Page No : 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# calculate the Volume and relative error \n",
+      "\n",
+      "# Variables\n",
+      "L = 250.;\n",
+      "d = 50.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "V = ((math.pi/4)*d**2)*L;\n",
+      "print \"Volume(mm3) = \", V\n",
+      "Re = 2*0.2-0.5;\n",
+      "print \"Relative error (%) = \", Re\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume(mm3) =  490873.852123\n",
+        "Relative error (%) =  -0.1\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17  Page No : 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculate the per unit change in the value of spring for different temperature ranges\n",
+      "\n",
+      "# Variables\n",
+      "dG_pu = -240*10**-6;\n",
+      "dD_pu = 11.8*10**-6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "print \"for temperature change of 20 degree C to 50 degree C (%)  = \",\n",
+      "d_th = 30;\n",
+      "dK_pu = (dG_pu+dD_pu)*d_th*100;\n",
+      "print dK_pu\n",
+      "\n",
+      "print \"for temperature change of 20 degree C to -50 degree C (%)  = \",\n",
+      "d_th = -70;\n",
+      "dK_pu = (dG_pu+dD_pu)*d_th*100;\n",
+      "print dK_pu\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for temperature change of 20 degree C to 50 degree C (%)  =  -0.6846\n",
+        "for temperature change of 20 degree C to -50 degree C (%)  =  1.5974\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18  Page No : 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate apparent resistance, actual resistance and error\n",
+      "\n",
+      "# Variables\n",
+      "Et = 100.;\n",
+      "It = 5.*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Rt = Et/It;\n",
+      "print \"apparent value of resistance (ohm) = \", Rt\n",
+      "Rv = 1000.*150;\n",
+      "\n",
+      "Rx = Rt*Rv/(Rv-Rt);\n",
+      "print \"true value of resistance(ohm)\", Rx\n",
+      "\n",
+      "Er_percentage = (Rt-Rx)/Rx*100;\n",
+      "print \"percentage error = \", Er_percentage\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "apparent value of resistance (ohm) =  20000.0\n",
+        "true value of resistance(ohm) 23076.9230769\n",
+        "percentage error =  -13.3333333333\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19  Page No : 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate apparent resistance, actual resistance and error\n",
+      "\n",
+      "# Variables\n",
+      "Et = 40.;\n",
+      "It = 800.*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Rt = Et/It;\n",
+      "print \"apparent value of resistance (ohm) = \", Rt\n",
+      "Rv = 1000.*150;\n",
+      "Rx = Rt*Rv/(Rv-Rt);\n",
+      "print \"true value of resistance(ohm)\", Rx\n",
+      "Er_percentage = ((Rt-Rx)/Rx)*100;\n",
+      "print \"percentage error = \", Er_percentage\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "apparent value of resistance (ohm) =  50.0\n",
+        "true value of resistance(ohm) 50.0166722241\n",
+        "percentage error =  -0.0333333333333\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20  Page No : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the error and percentage error in the measurement of deflection\n",
+      "\n",
+      "# Variables\n",
+      "l = 0.2;\n",
+      "E = 200*10**9;\n",
+      "b = 20.*10**-3;\n",
+      "d = 5.*10**-3;\n",
+      "D = (4.*l**3)/(E*b*d**3);\n",
+      "F = 1*9.81;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "x_true =  D*F;\n",
+      "print \"True value of deflection\", x_true\n",
+      "x_indicated = D*10.31/(1+.1*D);\n",
+      "print \"Indicated value of deflection\", x_indicated\n",
+      "Er = x_indicated-x_true;\n",
+      "print \"error = \", Er\n",
+      "Er_percentage = Er*100/x_true;\n",
+      "print \"Percentage error = \", Er_percentage\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "True value of deflection 0.00062784\n",
+        "Indicated value of deflection 0.000659835777051\n",
+        "error =  3.1995777051e-05\n",
+        "Percentage error =  5.09616734375\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21  Page No : 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array\n",
+      "#to find the mean,deviations from the mean, Average deviation, standard deviation and variance\n",
+      "\n",
+      "x = [532. ,548, 543, 535, 546, 531, 543, 536];\n",
+      "X = sum(x);\n",
+      "n = 8; \n",
+      "a = 0.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Mean = X/n;\n",
+      "print \"mean (kHZ)\", X/n\n",
+      "d = []\n",
+      "for i in range(n):\n",
+      "    d.append(x[i]-Mean)\n",
+      "    print \"deviations  = \", d[i]\n",
+      "    a = a+(abs(d[i]))\n",
+      " \n",
+      "d_average = a/n;\n",
+      "print \"Average deviation (kHz) = \", d_average\n",
+      "d_2 = sum(array(d)**2);\n",
+      "s = math.sqrt(d_2/(n-1))\n",
+      "print \"standard deviation(kHz)\", s\n",
+      "V = s**2;\n",
+      "print \"varaince (kHZ)2 = \", V\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mean (kHZ) 539.25\n",
+        "deviations  =  -7.25\n",
+        "deviations  =  8.75\n",
+        "deviations  =  3.75\n",
+        "deviations  =  -4.25\n",
+        "deviations  =  6.75\n",
+        "deviations  =  -8.25\n",
+        "deviations  =  3.75\n",
+        "deviations  =  -3.25\n",
+        "Average deviation (kHz) =  5.75\n",
+        "standard deviation(kHz) 6.54107898482\n",
+        "varaince (kHZ)2 =  42.7857142857\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.22  Page No : 101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy import array\n",
+      "#to find the mean,standard deviation, probable error and range\n",
+      "\n",
+      "# Variables\n",
+      "x = [41.7, 42, 41.8, 42, 42.1, 41.9, 42, 41.9, 42.5, 41.8];\n",
+      "X = sum(x) \n",
+      "print X\n",
+      "d = [-.27, .03, -.17, .03, .13, -.07, .03, -.07, .53, -.17];\n",
+      "\n",
+      "# Calculations and Results\n",
+      "d_2 = sum(array(d)**2);\n",
+      "n = 10.;\n",
+      "print \"mean length(deg C)\", X/n\n",
+      "print \"standard deviation(if data is infinite)(deg C)\", math.sqrt(d_2/n)\n",
+      "print \"standard deviation(deg C)\", math.sqrt(d_2/(n-1))\n",
+      "r1 = .6745*math.sqrt(d_2/(n-1));\n",
+      "print \"probable error of 1 reading(deg C)\", r1\n",
+      "print \"probable error of mean(deg C)\", r1/math.sqrt(n-1)\n",
+      "print \"range(deg C)\", max(x)-min(x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "419.7\n",
+        "mean length(deg C) 41.97\n",
+        "standard deviation(if data is infinite)(deg C) 0.21\n",
+        "standard deviation(deg C) 0.221359436212\n",
+        "probable error of 1 reading(deg C) 0.149306939725\n",
+        "probable error of mean(deg C) 0.0497689799083\n",
+        "range(deg C) 0.8\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.23  Page No : 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array\n",
+      "#to find the arithematic mean, maen deviation,standard deviation, prpobable error of 1 reading, standard deviation and probable error of mean, standard deviation of standard deviation\n",
+      "\n",
+      "# Variables\n",
+      "T = [397, 398, 399, 400, 401, 402, 403, 404, 405];\n",
+      "f = [1, 3, 12, 23, 37, 16, 4, 2, 2];\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Tf = sum(abs(array(T).transpose()*f));\n",
+      "print \"mean temp(deg C)\", Tf/sum(f)\n",
+      "d = [-3.78, -2.78, -1.78, -.78, .22, 1.22, 2.22, 3.22, 4.22];\n",
+      "d = array(d)\n",
+      "print \"mean deviation(deg C)\", sum(array(f).transpose()*d)/sum(f)\n",
+      "print \"standard deviation(deg C)\", math.sqrt(sum(array(f).transpose()*d**2)/sum(f))\n",
+      "print \"probable error of 1 reading(deg C)\", .6745*math.sqrt(sum(array(f).transpose()*d**2)/sum(f))\n",
+      "print \"probable error of mean(deg C)\", (.6745*math.sqrt(sum(array(f).transpose()*d**2)/sum(f)))/math.sqrt(sum(f))\n",
+      "print \"standard deviation of mean(deg C)\", (math.sqrt(sum(array(f).transpose()*d**2)/sum(f)))/math.sqrt(sum(f))\n",
+      "print \"standard deviation of standard deviation(deg C)\", (math.sqrt(sum(array(f).transpose()*d**2)/sum(f)))/math.sqrt(sum(f)*2)\n",
+      "\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " mean temp(deg C) 400\n",
+        "mean deviation(deg C) -1.7763568394e-17\n",
+        "standard deviation(deg C) 1.38260623462\n",
+        "probable error of 1 reading(deg C) 0.932567905249\n",
+        "probable error of mean(deg C) 0.0932567905249\n",
+        "standard deviation of mean(deg C) 0.138260623462\n",
+        "standard deviation of standard deviation(deg C) 0.0977650244208\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.24  Page No : 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#to find probable no of resistors\n",
+      "\n",
+      "# Variables\n",
+      "x = .15;    #deviation \n",
+      "o = .1;    #standard deviation\n",
+      "t = x/o;\n",
+      "A = .4432    #area under gaussian curve corresponding to t\n",
+      "\n",
+      "# Calculations\n",
+      "n = 2*A*1000;\n",
+      "\n",
+      "# Results\n",
+      "print \"no of resistors\", math.floor(n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "no of resistors 886.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.25  Page No : 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to find no of 100 rsding exceed 30mm\n",
+      "\n",
+      "# Variables\n",
+      "x = 30-26.3;    #mean value 26.3\n",
+      "r = 2.5;\n",
+      "o = r/.6745;\n",
+      "t = x/o;\n",
+      "A = .3413;#area under gaussian curve corresponding to t\n",
+      "\n",
+      "# Calculations\n",
+      "n = 2*A*100;\n",
+      "nn = 100-math.floor(n);\n",
+      "\n",
+      "# Results\n",
+      "print \"no of readings exceed\", nn/2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "no of readings exceed 16.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.26  Page No : 109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to find no of rods of desired length\n",
+      "\n",
+      "# Variables\n",
+      "n = 25000.;    #no of rods\n",
+      "n1 = 12500.; #length>10mm\n",
+      "n2 = 2000.;    #length>10.25\n",
+      "a = n1-n2;    #10<length<10.25\n",
+      "p = a/n;\n",
+      "t = 1.41;    #umath.sing p\n",
+      "\n",
+      "# Calculations\n",
+      "t1 = t*2;\n",
+      "p1 = .4975;\n",
+      "b = p1*n;    #9.5<length<10\n",
+      "\n",
+      "# Results\n",
+      "print \"total no of rods\", a+math.floor(b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total no of rods 22937.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.27  Page No : 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to find standard deviation and probability of error\n",
+      "\n",
+      "# Variables\n",
+      "p = .2;\n",
+      "x = .8;\n",
+      "t = .5025;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "sd = x/t;\n",
+      "print \"stndard deviation\", sd\n",
+      "x = 1.2;\n",
+      "t = x/sd;\n",
+      "p = 2*.2743;\n",
+      "print \"probability of error\", p\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "stndard deviation 1.592039801\n",
+        "probability of error 0.5486\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.28  Page No : 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to find no of expected readings\n",
+      "\n",
+      "# Variables\n",
+      "x = 20;\n",
+      "h = 0.04;\n",
+      "\n",
+      "# Calculations\n",
+      "sd = 1/(math.sqrt(2)*h);\n",
+      "t = x/sd;\n",
+      "P = .3708;\n",
+      "\n",
+      "# Results\n",
+      "print \"no of expected readings\", math.ceil(2*P*x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "no of expected readings 15.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.29  Page No : 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to calculate precision index of instrument\n",
+      "\n",
+      "# Variables\n",
+      "t = .675;\n",
+      "x = 2.4;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "sd = x/t;\n",
+      "h = 1/(math.sqrt(2)*sd);\n",
+      "print \"precision index\", h\n",
+      "t = (50-44)/sd;\n",
+      "p = .45;\n",
+      "n = 8*30;    #sept month no of measurements\n",
+      "a = ((.5-p)*n);\n",
+      "print \"no of false alarms\", a\n",
+      "\n",
+      "rn = a/2;    #reduced no of false alarms\n",
+      "p1 = rn/n;\n",
+      "P = .5-p1;\n",
+      "t = 1.96;\n",
+      "sd = (50-44)/t;\n",
+      "h = 1/(math.sqrt(2)*sd);\n",
+      "print \"precision index\", h\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "precision index 0.198873782209\n",
+        "no of false alarms 12.0\n",
+        "precision index 0.230988215188\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.30  Page No : 117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#to find confidence interval for given confidence levels\n",
+      "\n",
+      "# Variables\n",
+      "cl = [.5, .9, .95, .99];\n",
+      "s = .22;\n",
+      "d = [.7, 1.83, 2.26, 3.25];\n",
+      "def ci(b):\n",
+      "    return s*b;\n",
+      "\n",
+      "# Calculations\n",
+      "CI = [0,0,0,0]\n",
+      "CI[0] = ci(d[0]);\n",
+      "CI[1] = ci(d[1]);\n",
+      "CI[2] = ci(d[2]);\n",
+      "CI[3] = ci(d[3]);\n",
+      "\n",
+      "# Results\n",
+      "print \"confidence interval\", CI\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "confidence interval [0.154, 0.4026, 0.4972, 0.715]\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.31  Page No : 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array\n",
+      "#to point out the reading that can be rejected by chavenets criterion\n",
+      "\n",
+      "# Variables\n",
+      "x = array([5.3, 5.73, 6.77, 5.26, 4.33, 5.45, 6.09, 5.64, 5.81, 5.75])*10**-3;\n",
+      "d = array([-.313, .117, 1.157, -.353, -1.283, -.163, .477, .027, .197, .137])*10**-3;\n",
+      "n = 10.;\n",
+      "\n",
+      "# Calculations\n",
+      "X = sum(x)/n;\n",
+      "s = math.sqrt(sum(d**2)/(n-1));\n",
+      "a = abs(d)/s\n",
+      "print a\n",
+      "\n",
+      "\n",
+      "for i in range(10):\n",
+      "    if a[i]>1.96:\n",
+      "        print \"rejected value\", x[i]\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[ 0.49953805  0.18672828  1.84653522  0.56337678  2.04762722  0.26014282\n",
+        "  0.76127684  0.04309114  0.31440574  0.21864765]\n",
+        "rejected value 0.00433\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.32  Page No : 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array\n",
+      "#calculate standard deviation\n",
+      "\n",
+      "# Variables\n",
+      "x = array([.9 ,2.3, 3.3, 4.5, 5.7, 6.7]);\n",
+      "y = array([1.1, 1.6, 2.6, 3.2, 4 ,5]);\n",
+      "n = 6.;\n",
+      "\n",
+      "# Calculations\n",
+      "a = ((n*sum(x*y)-(sum(x)*sum(y)))/((sum(x**2)*n)-sum(x)**2));\n",
+      "b = ((sum(y)*sum(x**2)-(sum(x)*sum(x*y)))/((sum(x**2)*n)-sum(x)**2));\n",
+      "\n",
+      "sdy = math.sqrt((1/n)*sum((a*x+b-y)**2));\n",
+      "sdx = sdy/a;\n",
+      "\n",
+      "sa = math.sqrt(n/(n*sum(x**2)-sum(x)**2))*sdy;\n",
+      "sb = math.sqrt(sum(x**2)/(n*sum(x**2)-sum(x)**2))*sdy;\n",
+      "\n",
+      "# Results\n",
+      "print \"s_a\", sa\n",
+      "print \"s_b\", sb\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "s_a 0.0354898238647\n",
+        "s_b 0.155116651415\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.34  Page No : 123"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#determine value of total current considering errors as limiting errors ans as smath.radians(numpy.arcmath.tan(rd deviations\n",
+      "\n",
+      "# Variables\n",
+      "I1 = 200.;\n",
+      "I2 = 100.;\n",
+      "dI1 = 2.;\n",
+      "dI2 = 5.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I = I1+I2;\n",
+      "dI = ((I1/I)*(dI1/I1)+(I2/I)*(dI2/I2));\n",
+      "print \"error considered as limiting errors\"\n",
+      "print \"I\", I\n",
+      "print \"dI\", dI*I\n",
+      "sdI = math.sqrt(dI1**2+dI2**2);\n",
+      "print \"error considered as standard deviations\"\n",
+      "print \"I\", I\n",
+      "print \"sdI\", sdI\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "error considered as limiting errors\n",
+        "I 300.0\n",
+        "dI 7.0\n",
+        "error considered as standard deviations\n",
+        "I 300.0\n",
+        "sdI 5.38516480713\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.35  Page No : 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine probable error in the computed value of resistnce\n",
+      "\n",
+      "# Variables\n",
+      "r_V = 12.;\n",
+      "I = 10.;\n",
+      "r_Rv = r_V/I;\n",
+      "V = 100.;\n",
+      "r1 = 2.;\n",
+      "\n",
+      "# Calculations\n",
+      "r_Ri = V*r1/I**2;\n",
+      "r_R = math.sqrt(r_Rv**2+r_Ri**2);\n",
+      "\n",
+      "# Results\n",
+      "print \"r_R\", r_R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "r_R 2.33238075794\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.37  Page No : 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to find Cq and its possible errors\n",
+      "\n",
+      "# Variables\n",
+      "d = 12.5;\n",
+      "A = (math.pi/4)*d**2*10**-6;\n",
+      "W = 392.;\n",
+      "t = 600.;\n",
+      "p = 1000.;\n",
+      "g = 9.81;\n",
+      "h = 3.66;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Cq = W/(t*p*A*math.sqrt(2*g*h));\n",
+      "print \"Cq\", Cq\n",
+      "dW = .23/W;\n",
+      "dt = 2/t;\n",
+      "dp = .1/100;\n",
+      "dA = 2*.002;\n",
+      "dg = .1/100;\n",
+      "dh = .003/h;\n",
+      "dd = .002;\n",
+      "dCq = Cq*(dW+dt+dp+dA+dg/2+dh/2);\n",
+      "print \"%age absolute error\", dCq*100/Cq\n",
+      "\n",
+      "sdCq = Cq*math.sqrt(dW**2+dt**2+dp**2+4*dd**2+.25*(dg**2+dh**2));\n",
+      "print \"%age standard deviation error\", sdCq*100/Cq\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cq 0.628253425224\n",
+        "%age absolute error 0.982990409278\n",
+        "%age standard deviation error 0.537339132325\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.38  Page No : 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 110.2;\n",
+      "I = 5.3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "P = V*I\n",
+      "print \"power(W) dissipated\", P\n",
+      "w_v = .2;\n",
+      "w_i = 0.06;\n",
+      "dp = math.sqrt((w_v*I)**2+(w_i*V)**2);\n",
+      "print \"uncertainity in power(%)\", dp*100/P\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power(W) dissipated 584.06\n",
+        "uncertainity in power(%) 1.14653078526\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.39  Page No : 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to find uncertainity in combined resistance in both series and in parrallel\n",
+      "\n",
+      "# Variables\n",
+      "R1 = 100.;\n",
+      "R2 = 50.;\n",
+      "wR1 = .1;\n",
+      "wR2 = 0.03;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "print \"series conn\"\n",
+      "R = R1+R2\n",
+      "print \"resistance(ohm)\", R\n",
+      "dR1 = 1;\n",
+      "dR2 = 1;\n",
+      "wR = math.sqrt((dR1*wR1)**2+(dR2*wR2)**2)\n",
+      "print \"uncertainity in resistance(ohm)\", wR\n",
+      "\n",
+      "print \"parrallel conn\"\n",
+      "R = R1*R2*(R1+R2)**-1\n",
+      "print \"resistance(ohm)\", R\n",
+      "dR1 = (R2/(R1+R2))-((R1*R2)/(R1+R2)**2);\n",
+      "dR2 = (R1/(R1+R2))-((R1*R2)/(R1+R2)**2);\n",
+      "wR = math.sqrt((dR1*wR1)**2+(dR2*wR2)**2)\n",
+      "print \"uncertainity in resistance(ohm)\", wR\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "series conn\n",
+        "resistance(ohm) 150.0\n",
+        "uncertainity in resistance(ohm) 0.104403065089\n",
+        "parrallel conn\n",
+        "resistance(ohm) 33.3333333333\n",
+        "uncertainity in resistance(ohm) 0.0173561103909\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.40  Page No : 133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to calculate uncertainity in measurement\n",
+      "\n",
+      "# Variables\n",
+      "l = 150;\n",
+      "dl = 0.01;\n",
+      "b = 50;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "wA = l*dl;\n",
+      "print \"when no uncertainity in measurement of length\"\n",
+      "print \"uncertainity in measurement of area(m*m)\", wA\n",
+      "\n",
+      "print \"when no certainity in measurement of length\"\n",
+      "wA = 1.5*1.5;\n",
+      "wB = 0.01;\n",
+      "wL = math.sqrt((wA**2-(l*wB)**2)/b**2);\n",
+      "print \"uncertainity in measurement of length(m)\", wL\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when no uncertainity in measurement of length\n",
+        "uncertainity in measurement of area(m*m) 1.5\n",
+        "when no certainity in measurement of length\n",
+        "uncertainity in measurement of length(m) 0.0335410196625\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.41  Page No : 135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#to calculate uncertainity in power\n",
+      "\n",
+      "# Variables\n",
+      "E = 100.;\n",
+      "dE = .01;\n",
+      "I = 10.;\n",
+      "dI = 0.01;\n",
+      "R = 10.;\n",
+      "dR = .01;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "dP = math.sqrt(4*dE**2+dR**2)*100;    #P = E**2/R\n",
+      "print \"%age uncertainity in power measurement\", dP\n",
+      "\n",
+      "dP = math.sqrt(dE**2+dI**2)*100;    #P = E*I\n",
+      "print \"%age uncertainity in power measurement\", dP\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "%age uncertainity in power measurement 2.2360679775\n",
+        "%age uncertainity in power measurement 1.41421356237\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch4.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch4.ipynb
new file mode 100755
index 00000000..fd3f3073
--- /dev/null
+++ b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch4.ipynb
@@ -0,0 +1,940 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:4dacb9cd40c034de14bd116989b8eaf4736b7fcde46866cbd61e09de312f6cad"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Dynamic Characteristics of Instruments and Measurement systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1  Page No : 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# calculating the temperature after 1.5 s\n",
+      "\n",
+      "# Variables\n",
+      "th0 = 100.;\n",
+      "t = 1.5;\n",
+      "tc = 3.5;\n",
+      "\n",
+      "# Calculations\n",
+      "th = th0*(1-math.exp(-t/tc));\n",
+      "\n",
+      "# Results\n",
+      "print \"temperature after 1.5 s (degree C)\", th\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature after 1.5 s (degree C) 34.8560942469\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2  Page No : 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# calculate time to read half of the temperature difference\n",
+      "# Variables\n",
+      "tc = 10./5;\n",
+      "th = 1.;\n",
+      "th0 = 2.;\n",
+      "\n",
+      "# Calculations\n",
+      "t = -tc*math.log(1-(th/th0));\n",
+      "\n",
+      "# Results\n",
+      "print \"Time to read half of the temperature difference (s)\", t\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time to read half of the temperature difference (s) 1.38629436112\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4  Page No : 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the temperature after 10s\n",
+      "\n",
+      "# Variables\n",
+      "th0 = 25;\n",
+      "thi = 150;\n",
+      "t = 10.;\n",
+      "tc = 6;\n",
+      "\n",
+      "# Calculations\n",
+      "th = th0+(thi-th0)*(math.exp(-t/tc));\n",
+      "\n",
+      "# Results\n",
+      "print \"the temperature after 10s (degree C)\", th\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature after 10s (degree C) 48.6094503547\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5  Page No : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the value of resistance after 15s\n",
+      "\n",
+      "# Variables\n",
+      "R0 = 29.44;\n",
+      "Rs = 100.;\n",
+      "t = 15.;\n",
+      "tc = 5.5;\n",
+      "\n",
+      "# Calculations\n",
+      "R_15 = Rs+R0*(1-math.exp(-t/tc));\n",
+      "\n",
+      "# Results\n",
+      "print \"value of resistance after 15s(ohm)\", R_15\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of resistance after 15s(ohm) 127.514700449\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6  Page No : 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the depth after one hour\n",
+      "\n",
+      "# Variables\n",
+      "Qm = 0.16*10**-3;\n",
+      "Hin = 1.2;\n",
+      "K1 = Qm/(Hin)**0.5;\n",
+      "Qo = 0.2*10**-3;\n",
+      "Ho = (Qo/K1)**2;\n",
+      "R = Hin/Qm;\n",
+      "C = 0.1;\n",
+      "tc = R*C;\n",
+      "t = 3600;\n",
+      "\n",
+      "# Calculations\n",
+      "H = Ho+(Hin-Ho)*math.exp(-t/tc);\n",
+      "\n",
+      "# Results\n",
+      "print \"the depth after one hour(m)\", H\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the depth after one hour(m) 1.86944492074\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8  Page No : 147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#Calculate time constant\n",
+      "\n",
+      "# Variables\n",
+      "S = 3.5;\n",
+      "Ac = (math.pi/4)*(0.25)**2;\n",
+      "alpha = 0.18*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Vb = S*Ac/alpha;\n",
+      "print \"volume of bulb(mm2)\", Vb\n",
+      "\n",
+      "Rb = ((Vb/math.pi)*(3/.4))**(1./3);\n",
+      "Ab = 4*math.pi*Rb**2;\n",
+      "D = 13.56*10**3;\n",
+      "s = 139;\n",
+      "H = 12;\n",
+      "tc = (D*s*Vb*10**-9)/(H*Ab*10**-6);\n",
+      "print \"time constant (s)\", tc\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "volume of bulb(mm2) 954.476934684\n",
+        "time constant (s) 68.8965695836\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9  Page No : 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "ess = 5;\n",
+      "A = 0.1;\n",
+      "\n",
+      "# Calculations\n",
+      "tc = ess/A;\n",
+      "\n",
+      "# Results\n",
+      "print \"time constant(s)\", tc\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time constant(s) 50.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10  Page No : 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the temperature at a depth of 1000 m\n",
+      "\n",
+      "# Variables\n",
+      "th0 = 20.;\n",
+      "t = 2000.;\n",
+      "\n",
+      "# Calculations\n",
+      "thr = th0-0.005*(t-50)-0.25*math.exp(-t/50);\n",
+      "\n",
+      "# Results\n",
+      "print \"temperature at a depth of 1000 m (degree C)\", thr\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature at a depth of 1000 m (degree C) 10.25\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11  Page No : 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the value of resistance at different values of time\n",
+      "\n",
+      "# Variables\n",
+      "Gain = 0.3925;\n",
+      "T = 75.;\n",
+      "p_duration = Gain*T;\n",
+      "tc = 5.5;\n",
+      "Rin = 100.;\n",
+      "t = 1;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Rt = p_duration*(1-math.exp(-t/tc))+Rin;\n",
+      "print \"Value of resistance after 1s(ohm) = \", Rt\n",
+      "t = 2;\n",
+      "Rt = p_duration*(1-math.exp(-t/tc))+Rin;\n",
+      "print \"Value of resistance after 2s(ohm) = \", Rt\n",
+      "t = 3;\n",
+      "Rt = p_duration*(1-math.exp(-t/tc))+Rin;\n",
+      "print \"Value of resistance after 3s(ohm) = \", Rt\n",
+      "R_inc = Rt-Rin;\n",
+      "t = 5;\n",
+      "Rt = (R_inc)*(math.exp(-(t-3)/(5.5)))+Rin;\n",
+      "print \"Value of resistance after 5s(ohm) = \", Rt\n",
+      "t = 10;\n",
+      "Rt = (R_inc)*(math.exp(-(t-3)/(5.5)))+Rin;\n",
+      "print \"Value of resistance after 10s(ohm) = \", Rt\n",
+      "t = 20;\n",
+      "Rt = (R_inc)*(math.exp(-(t-3)/(5.5)))+Rin;\n",
+      "print \"Value of resistance after 20s(ohm) = \", Rt\n",
+      "t = 30;\n",
+      "Rt = (R_inc)*(math.exp(-(t-3)/(5.5)))+Rin;\n",
+      "print \"Value of resistance after 30s(ohm) = \", Rt"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of resistance after 1s(ohm) =  104.893898474\n",
+        "Value of resistance after 2s(ohm) =  108.974200608\n",
+        "Value of resistance after 3s(ohm) =  112.376164418\n",
+        "Value of resistance after 5s(ohm) =  108.603215552\n",
+        "Value of resistance after 10s(ohm) =  103.46615228\n",
+        "Value of resistance after 20s(ohm) =  100.562627957\n",
+        "Value of resistance after 30s(ohm) =  100.091326114\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12  Page No : 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "M = 8*10**-3;\n",
+      "K = 1000.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "wn = (K/M)**0.5;\n",
+      "print \"for critically damped system eta = 1\",\n",
+      "B = 2*(K*M);\n",
+      "print \"Damping constant for critically damped system (N/ms-1) = \", B\n",
+      "eta = 0.6;\n",
+      "wd = wn*(1-eta**2)**0.5;\n",
+      "print \"frequency of damped oscillations (rad/s) = \", wd\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for critically damped system eta = 1 Damping constant for critically damped system (N/ms-1) =  16.0\n",
+        "frequency of damped oscillations (rad/s) =  282.842712475\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13  Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "K = (40.*10**-6)/(math.pi/2);\n",
+      "J = 0.5*10**-6;\n",
+      "B = 5.*10**-6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "eta = B/(2*(K*J)**0.5);\n",
+      "print \"damping ratio = \", eta\n",
+      "wn = (K/J)**0.5;\n",
+      "print \"natural frequency (rad/sec)\", wn\n",
+      "wd = wn*(1-(eta)**2)**0.5;\n",
+      "print \"frequency of damped oscillations (rad/s)\", wd\n",
+      "tc = 1/wn;\n",
+      "print \"time constant (s)\", tc\n",
+      "ess = 2*eta/wn;\n",
+      "print \" steady state error (V)  = \", \"for a ramp input of 5V\"\n",
+      "ess = 5*2*eta/wn;\n",
+      "print \"\", ess\n",
+      "T_lag = 2*eta*tc;\n",
+      "print \"Time lag (s)\", T_lag\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "damping ratio =  0.70062390205\n",
+        "natural frequency (rad/sec) 7.13649646461\n",
+        "frequency of damped oscillations (rad/s) 5.09210975819\n",
+        "time constant (s) 0.14012478041\n",
+        " steady state error (V)  =  for a ramp input of 5V\n",
+        " 0.981747704247\n",
+        "Time lag (s) 0.196349540849\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14  Page No : 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the natural  frequency \n",
+      "\n",
+      "# Variables\n",
+      "wn = 2*math.pi*30;\n",
+      "print \" for a frequency of 30 Hz  wn = (K/M+5*10**-3)**0.5.........(i)\",\n",
+      "print \"But wn = (K/M)**0.5.........(ii)\",\n",
+      "print \"for a frequency of 25 Hz  wn = (K/M+5*10**-3+5*10**-3)**0.5.........(iii) \",\n",
+      "print \" (ii) and (iii)\", \"on solving (i)\"\n",
+      "\n",
+      "# Calculations and Results\n",
+      "M = 6.36*10**-3;\n",
+      "K = 403.6;\n",
+      "print \"M = \", M\n",
+      "print \"K = \", K\n",
+      "wn = (K/M)**0.5;\n",
+      "f = wn/(2*math.pi);\n",
+      "print \"natural frequency (Hz)\", f\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " for a frequency of 30 Hz  wn = (K/M+5*10**-3)**0.5.........(i) But wn = (K/M)**0.5.........(ii) for a frequency of 25 Hz  wn = (K/M+5*10**-3+5*10**-3)**0.5.........(iii)   (ii) and (iii) on solving (i)\n",
+        "M =  0.00636\n",
+        "K =  403.6\n",
+        "natural frequency (Hz) 40.0928706266\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15  Page No : 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "K = 60.*10**3;\n",
+      "M = 30.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "wn = (K/M)**0.5;\n",
+      "print \"natural frequency (rad/sec)\", wn\n",
+      "eta = 0.7;\n",
+      "ts = 4/(eta*wn);\n",
+      "print \"setteling time (s)\", ts\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "natural frequency (rad/sec) 44.72135955\n",
+        "setteling time (s) 0.127775313\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16  Page No : 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate time lag and ratio of output and input\n",
+      "\n",
+      "# Variables\n",
+      "print \"when time period is 600s\"\n",
+      "w = 2*math.pi/600;\n",
+      "tc = 60.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "T_lag = (1/w)*math.atan(w*tc);\n",
+      "print \"time lag (s) = \", T_lag\n",
+      "M = 1/((1+(w*tc)**2)**0.5);\n",
+      "print \"ratio of output and input = \", M\n",
+      "print \"when time period is 120s\",\n",
+      "w = 2*math.pi/120;\n",
+      "tc = 60;\n",
+      "T_lag = (1/w)*math.atan(w*tc);\n",
+      "print \"time lag (s) = \", T_lag\n",
+      "M = 1/((1+(w*tc)**2)**0.5);\n",
+      "print \"ratio of output and input = \", M\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when time period is 600s\n",
+        "time lag (s) =  53.5698460589\n",
+        "ratio of output and input =  0.846733015965\n",
+        "when time period is 120s time lag (s) =  24.1144042829\n",
+        "ratio of output and input =  0.303314471053\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17  Page No : 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the maximum allowable time constant and phase shift\n",
+      "\n",
+      "# Variables\n",
+      "M = 1-0.05;\n",
+      "w = 2*math.pi*100;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "tc = (((1/M**2)-1)/(w**2))**0.5;\n",
+      "print \"maximum allowable time constant (s)\", tc\n",
+      "print \"phase shift at 50 Hz (degree) = \",\n",
+      "ph = (-math.atan(2*math.pi*50*tc))*(180/math.pi);\n",
+      "print ph\n",
+      "print \"phase shift at 100 Hz (degree) = \",\n",
+      "ph = (-math.atan(2*math.pi*100*tc))*(180/math.pi);\n",
+      "print  ph\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum allowable time constant (s) 0.00052311700055\n",
+        "phase shift at 50 Hz (degree) =  -9.33268272936\n",
+        "phase shift at 100 Hz (degree) =  -18.1948723388\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.18  Page No : 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate maximum value of indicated temperature and delay time\n",
+      "\n",
+      "# Variables\n",
+      "T = 120.;\n",
+      "w = 2*math.pi/T;\n",
+      "tc1 = 40.;\n",
+      "tc2 = 20.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "M = (1/((1+(w*tc1)**2)**0.5))*(1/((1+(w*tc2)**2)**0.5));\n",
+      "M_temp = M*10;\n",
+      "print \"maximum value of indicated temperature (degree C)\", M_temp\n",
+      "ph = ((math.atan(w*tc1)+math.atan(w*tc2)));\n",
+      "T_lag = ph/w;\n",
+      "print \"Time lag (s)\", T_lag\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum value of indicated temperature (degree C) 2.975684577\n",
+        "Time lag (s) 36.9326231389\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19  Page No : 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "# Find the output \n",
+      "print \"when tc = 0.2\",\n",
+      "print \"output = 1/(1+(2*0.2)**2)**0.5]math.sin[2t-math.atan(2*0.2)]+3/(1+(2*0.2)**2)**0.5]math.sin[20t-math.atan(20*0.2)]\",\n",
+      "print \"on solving   output = 0.93 math.sin(2t-21.8)+0.073 math.sin(20t-76)\",\n",
+      "print \"when tc = 0.002\",\n",
+      "print \"output = 1/(1+(2*0.002)**2)**0.5]math.sin[2t-math.atan(2*0.002)]+3/(1+(2*0.002)**2)**0.5]math.sin[20t-math.atan(20*0.002)]\",\n",
+      "print \"on solving   output =  1math.sin(2t-0.23)+0.3 math.sin(20t-2.3)\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when tc = 0.2 output = 1/(1+(2*0.2)**2)**0.5]math.sin[2t-math.atan(2*0.2)]+3/(1+(2*0.2)**2)**0.5]math.sin[20t-math.atan(20*0.2)] on solving   output = 0.93 math.sin(2t-21.8)+0.073 math.sin(20t-76) when tc = 0.002 output = 1/(1+(2*0.002)**2)**0.5]math.sin[2t-math.atan(2*0.002)]+3/(1+(2*0.002)**2)**0.5]math.sin[20t-math.atan(20*0.002)] on solving   output =  1math.sin(2t-0.23)+0.3 math.sin(20t-2.3)\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.20  Page No : 171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#Calculate maximum and minimum value of indicated temperature, phase shift, time lag \n",
+      "\n",
+      "# Variables\n",
+      "T_max = 640.;\n",
+      "T_min = 600.;\n",
+      "T_mean = (T_max+T_min)/2;\n",
+      "Ai = T_mean-T_min;\n",
+      "w = 2*math.pi/80;\n",
+      "tc = 10;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Ao = Ai/((1+(w*tc)**2))**0.5;\n",
+      "T_max_indicated = T_mean+Ao;\n",
+      "print \"Maximum value of indicated temperature(degree C) = \", T_max_indicated\n",
+      "T_min_indicated = T_mean-Ao;\n",
+      "print \"Minimum value of indicated temperature(degree C) = \", T_min_indicated\n",
+      "ph = -math.atan(w*tc);\n",
+      "Time_lag = -ph/w;\n",
+      "print \"Time lag (s)\", Time_lag\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum value of indicated temperature(degree C) =  635.728782002\n",
+        "Minimum value of indicated temperature(degree C) =  604.271217998\n",
+        "Time lag (s) 8.47689466383\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.21  Page No : 173"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# determine damping ratio\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "w = 2.;\n",
+      "K = 1.5;\n",
+      "J = 200.*10**-3;\n",
+      "wn = (K/J)**0.5;\n",
+      "u = w/wn;\n",
+      "M = 1.1;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "eta = (((1/(M**2))-((1-u**2)**2))/(2*u)**2)**0.5;\n",
+      "print \"damping ratio = \", eta\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "damping ratio =  0.534147321328\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.22  Page No : 175"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Calculate the frequency range\n",
+      "\n",
+      "# Variables\n",
+      "eta = 0.6;\n",
+      "fn = 1000;\n",
+      "M = 1.1;\n",
+      "print \"M = 1/[[(1-u**2)**2]+(2*u*eta)**2]**0.5 ..........(i)\",\n",
+      "print \"on solving u**4-0.5u**2+0.173 = 0\",\n",
+      "print \"the above equation gives imaginary values for frequency so for eta = 0.6 the output is not 1.1\",\n",
+      "print \" on solving equation (i) we have\", \"Now let M = 0.9\"\n",
+      "print \"u**4-0.56u**2-0.234 = 0\",\n",
+      "print \"on solving u = 0.916\",\n",
+      "\n",
+      "# Calculations and Results\n",
+      "u = 0.916;\n",
+      "f = u*fn;\n",
+      "print \"maximum value of range (Hz) = \", f\n",
+      "print \" the range of the frequency is from 0 to 916 Hz\", \"So\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "M = 1/[[(1-u**2)**2]+(2*u*eta)**2]**0.5 ..........(i) on solving u**4-0.5u**2+0.173 = 0 the above equation gives imaginary values for frequency so for eta = 0.6 the output is not 1.1  on solving equation (i) we have Now let M = 0.9\n",
+        "u**4-0.56u**2-0.234 = 0 on solving u = 0.916 maximum value of range (Hz) =  916.0\n",
+        " the range of the frequency is from 0 to 916 Hz So\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.23  Page No : 177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# determine the error\n",
+      "\n",
+      "# Variables\n",
+      "w = 6.;\n",
+      "wn = 4.;\n",
+      "u = w/wn;\n",
+      "eta = 0.66;\n",
+      "\n",
+      "# Calculations\n",
+      "M = 1/(((1-u**2)**2)+(2*eta*u)**2)**0.5;\n",
+      "Error = (M-1)*100;\n",
+      "\n",
+      "# Results\n",
+      "print \"error (%) = \", Error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "error (%) =  -57.2934157194\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch5.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch5.ipynb
new file mode 100755
index 00000000..6eb31891
--- /dev/null
+++ b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch5.ipynb
@@ -0,0 +1,2140 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:dcab5285f135b4dffae4292e0eb7017128ede4af1f4e78fe0491e5a728481399"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Primary Sensing Elements and Transducers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1  Page No : 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "D = 15.*10**-3;\n",
+      "P = 300.*10**3;\n",
+      "sm = 300.*10**6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "t = (3*D**2*P/(16*sm))**0.5;\n",
+      "print \" thickness(m) = \", t\n",
+      "P = 150*10**3;\n",
+      "v = 0.28;\n",
+      "E = 200*10**9;\n",
+      "dm = 3*(1-v**2)*D**4*P/(256*E*t**3);\n",
+      "print \" deflection at center for Pressure of 150 kN/m2(m) = \", dm\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " thickness(m) =  0.000205395959064\n",
+        " deflection at center for Pressure of 150 kN/m2(m) =  4.73232289684e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2  Page No : 181"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "T = 100.;\n",
+      "G = 80.*10**9;\n",
+      "d = 2.*15*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "th = 16*T/(math.pi*G*d**3)\n",
+      "\n",
+      "# Results\n",
+      "print \" angle of twist(rad) = \", th\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " angle of twist(rad) =  0.000235785100877\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3  Page No : 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "E = 110*10**9;\n",
+      "t = 0.073*10**-3;\n",
+      "b = 0.51*10**-3;\n",
+      "l = 370*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "th = math.pi/2;\n",
+      "T = (E*b*t**3)*th/(12*l);\n",
+      "\n",
+      "# Results\n",
+      "print \" Controlling torque(Nm) = \", T\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Controlling torque(Nm) =  7.72090748389e-06\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4  Page No : 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Rnormal = 10000./2;\n",
+      "Rpl = 10000./50;\n",
+      "Rc1 = Rnormal-3850;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Dnormal = Rc1/Rpl;\n",
+      "print \"Displacement(mm) = \", Dnormal\n",
+      "Rc2 = Rnormal-7560;\n",
+      "Dnormal = Rc2/Rpl;\n",
+      "print \"Displacement(mm) = \", Dnormal\n",
+      "print \"since one lacement is positive and other is negative so two lacements are in the opposite direction\"\n",
+      "Re = 10*1./200;\n",
+      "print \"Resolution (mm) = \", Re\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Displacement(mm) =  5.75\n",
+        "Displacement(mm) =  -12.8\n",
+        "since one lacement is positive and other is negative so two lacements are in the opposite direction\n",
+        "Resolution (mm) =  0.05\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5  Page No : 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "import math \n",
+      "from matplotlib.pyplot import plot\n",
+      "\n",
+      "# Variables\n",
+      "# plot the graph of error versus K\n",
+      "K = [0,    0.25,    0.5,    0.75,    1];\n",
+      "V = [0,    -0.174,    -0.454,    -0.524,    0];\n",
+      "\n",
+      "# Results\n",
+      "plot(K,V)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 8,
+       "text": [
+        "[<matplotlib.lines.Line2D at 0x103fd9ad0>]"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x1038299d0>"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6  Page No : 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "RAB = 125.;\n",
+      "Rtotal = 5000.;\n",
+      "R2 = 75/125.*Rtotal;\n",
+      "R4 = 2500.;\n",
+      "ei = 5.;\n",
+      "\n",
+      "# Calculations\n",
+      "eo = ((R2/Rtotal)-(R4/Rtotal))*ei;\n",
+      "\n",
+      "# Results\n",
+      "print \"output voltage (V) = \", eo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage (V) =  0.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7  Page No : 191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "Rm = 10000.;\n",
+      "Rp = Rm/15.;\n",
+      "R = 600.;\n",
+      "P = 5.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "ei =  (P*R)**0.5;\n",
+      "print \"Maximum excitation voltage (V) = \", ei\n",
+      "S = ei/360;\n",
+      "print \"Sensitivity (V/degree) = \", S\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum excitation voltage (V) =  54.7722557505\n",
+        "Sensitivity (V/degree) =  0.152145154863\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8  Page No : 193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "Rwga = 1./400;\n",
+      "\n",
+      "# Calculations\n",
+      "Re = Rwga/5;\n",
+      "\n",
+      "# Results\n",
+      "print \"Resolution (mm) = \", Re\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resolution (mm) =  0.0005\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9  Page No : 195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "mo = 0.8;\n",
+      "sr = 250;\n",
+      "sm = sr/mo;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R = sm*1;\n",
+      "print \"resolution of 1mm movement\", R\n",
+      "Rq = 300/1000;\n",
+      "print \"resolution required = \", Rq\n",
+      "print \"since the resolution of potentiometer is higher than the resolution required so it is suitable for the application\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resolution of 1mm movement 312.5\n",
+        "resolution required =  0\n",
+        "since the resolution of potentiometer is higher than the resolution required so it is suitable for the application\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10  Page No : 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculations and Results\n",
+      "Pd = (10.**2)/150;\n",
+      "print \"Power dissipation (W) = \", Pd\n",
+      "th_pot = 80.+Pd*30*10**-3;\n",
+      "PDa = 1-(10*10**-3)*(th_pot-35);\n",
+      "print \"Power dissipation allowed(W) = \", PDa\n",
+      "print \"Since power dissipation is higher than the dissipation allowed so potentiometer is not suitable\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power dissipation (W) =  0.666666666667\n",
+        "Power dissipation allowed(W) =  0.5498\n",
+        "Since power dissipation is higher than the dissipation allowed so potentiometer is not suitable\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11  Page No : 199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Gf = 4.2;\n",
+      "\n",
+      "# Calculations\n",
+      "v = (Gf-1)/2;\n",
+      "\n",
+      "# Results\n",
+      "print \"Possion s ratio = \", v\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Possion s ratio =  1.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12  Page No : 201"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "strain = -5.*10**-6;\n",
+      "Gf = -12.1;\n",
+      "R = 120.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "dR_nickel = Gf*R*strain;\n",
+      "print \"change in resistance of nickel(ohm) = \", dR_nickel\n",
+      "Gf = 2;\n",
+      "R = 120;\n",
+      "dR_nicrome = Gf*R*strain;\n",
+      "print \"change in resistance of nicrome(ohm) = \", dR_nicrome\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in resistance of nickel(ohm) =  0.00726\n",
+        "change in resistance of nicrome(ohm) =  -0.0012\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13  Page No : 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "s = 100.*10**6;\n",
+      "E = 200.*10**9;\n",
+      "\n",
+      "# Calculations\n",
+      "strain = s/E;\n",
+      "Gf = 2;\n",
+      "r_perunit = Gf*strain*100;\n",
+      "\n",
+      "# Results\n",
+      "print \"Percentage change in resistance = \", r_perunit\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage change in resistance =  0.1\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14  Page No : 205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "b = 0.02;\n",
+      "d = 0.003;\n",
+      "I = (b*d**3)/12;\n",
+      "E = 200*10**9;\n",
+      "x = 12.7*10**-3;\n",
+      "l = 0.25;\n",
+      "F = 3*E*I*x/l**3;\n",
+      "x = 0.15;\n",
+      "M = F*x;\n",
+      "t = 0.003;\n",
+      "\n",
+      "# Calculations\n",
+      "s = (M*t)/(I*2);\n",
+      "strain = s/E;\n",
+      "dR = 0.152;\n",
+      "R = 120;\n",
+      "Gf = (dR/R)/strain;\n",
+      "\n",
+      "# Results\n",
+      "print \"Gauge factor = \", Gf\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gauge factor =  2.30873918538\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15  Page No : 207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "dR = 0.013;\n",
+      "R = 240;\n",
+      "l = 0.1;\n",
+      "Gf = 2.2;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "dl = (dR/R)*l/Gf;\n",
+      "print \"change in length(m) = \", dl\n",
+      "strain = dl/l;\n",
+      "E = 207*10**9;\n",
+      "s = E*strain;\n",
+      "A = 4*10**-4;\n",
+      "F = s*A;\n",
+      "print \"Force(N)\", F\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in length(m) =  2.46212121212e-06\n",
+        "Force(N) 2038.63636364\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16  Page No : 209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "th1 = 30.;\n",
+      "th2 = 60.;\n",
+      "th0 = th1+th2/2;\n",
+      "Rth1 = 4.8;\n",
+      "Rth2 = 6.2;\n",
+      "Rth0 = 5.5;\n",
+      "\n",
+      "# Calculations\n",
+      "ath0 = (1/Rth0)*(Rth2-Rth1)/(th2-th1);\n",
+      "\n",
+      "# Results\n",
+      "print \"alpha at o degree(/degree C) = \", ath0\n",
+      "print \"5.5[1+0.0085(th-45)]\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "alpha at o degree(/degree C) =  0.00848484848485\n",
+        "5.5[1+0.0085(th-45)]\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.17  Page No : 211"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "th1 = 100.;\n",
+      "th2 = 130.;\n",
+      "\n",
+      "# Calculations\n",
+      "th0 = th1+th2/2;\n",
+      "Rth1 = 573.40;\n",
+      "Rth2 = 605.52;\n",
+      "Rth0 = 589.48;\n",
+      "ath0 = (1/Rth0)*(Rth2-Rth1)/(th2-th1);\n",
+      "\n",
+      "# Results\n",
+      "print \"alpha at o degree(/degree C) = \", ath0\n",
+      "print \"Linear approximation is: Rth =  589.48[1+0.00182(th-115)]\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "alpha at o degree(/degree C) =  0.00181629006356\n",
+        "Linear approximation is: Rth =  589.48[1+0.00182(th-115)]\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.18  Page No : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Rth0 = 100.;\n",
+      "ath0 = 0.00392;\n",
+      "dth = 65.-25;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R65 = Rth0*(1+ath0*dth);\n",
+      "print \"resistance at 65 degree C(ohm) = \", R65\n",
+      "\n",
+      "th = (((150./100)-1)/ath0)+25;\n",
+      "print \"Temperature (degree C)\", th\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance at 65 degree C(ohm) =  115.68\n",
+        "Temperature (degree C) 152.551020408\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.19  Page No : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Rth0 = 10;\n",
+      "ath0 = 0.00393;\n",
+      "dth = 150-20;\n",
+      "\n",
+      "# Calculations\n",
+      "R150 = Rth0*(1+ath0*dth);\n",
+      "\n",
+      "# Results\n",
+      "print \"resistance at 150 degree C(ohm) = \", R150\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance at 150 degree C(ohm) =  15.109\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.20  Page No : 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "th = 30.;\n",
+      "th0 = 50.;\n",
+      "tc = 120.;\n",
+      "\n",
+      "# Calculations\n",
+      "t = -120*(math.log(1-(th/th0)));\n",
+      "\n",
+      "# Results\n",
+      "print \"time(s) = \", t\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time(s) =  109.954887825\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.21  Page No : 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R25 = 100;\n",
+      "ath = -0.05;\n",
+      "dth = 35-25;\n",
+      "\n",
+      "# Calculations\n",
+      "R35 = R25*(1+ath*dth);\n",
+      "\n",
+      "# Results\n",
+      "print \"resistance at 35 degree C(ohm) = \", R35\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance at 35 degree C(ohm) =  50.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.22  Page No : 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "Ro = 3980.;\n",
+      "Ta = 273.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "print \"3980 =  a*3980*math.exp(b/273)\",\n",
+      "Rt50 = 794;\n",
+      "Ta50 = 273+50;\n",
+      "print \"794 =  a*3980*math.exp(b/323)\",\n",
+      "print \"on solving\",\n",
+      "print \" b = 2843\", \"a = 30*10**-6\"\n",
+      "Ta40 = 273+40;\n",
+      "Rt40 = (30*10**-6)*3980*math.exp(2843./313);\n",
+      "print \"resistance at 40 degree C (ohm)\", Rt40\n",
+      "Rt100 = (30*10**-6)*3980*math.exp(2843./373);\n",
+      "print \"resistance at 100 degree C (ohm)\", Rt100\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3980 =  a*3980*math.exp(b/273) 794 =  a*3980*math.exp(b/323) on solving  b = 2843 a = 30*10**-6\n",
+        "resistance at 40 degree C (ohm) 1051.3086649\n",
+        "resistance at 100 degree C (ohm) 243.887690913\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.23  Page No : 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Calculations\n",
+      "th = ((1-1800./2000)/0.05)+70;\n",
+      "dth = th-70;\n",
+      "\n",
+      "# Results\n",
+      "print \"change in temperature (degree C)\", dth\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in temperature (degree C) 2.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.24  Page No : 225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "C = 500.*10**-12;\n",
+      "R20 = 10000.*(1-0.05*(20-25));\n",
+      "\n",
+      "# Calculations and Results\n",
+      "f20 = 1/(2*math.pi*R20*C);\n",
+      "print \"Frequency of oscillation at 20 degree C (Hz)\", f20\n",
+      "R25 = 10000*(1-0.05*(25-25));\n",
+      "f25 = 1./(2*math.pi*R25*C);\n",
+      "print \"Frequency of oscillation at 25 degree C (Hz)\", f25\n",
+      "R30 = 10000*(1-0.05*(30-25));\n",
+      "f30 = 1./(2*math.pi*R30*C);\n",
+      "print \"Frequency of oscillation at 30 degree C (Hz)\", f30\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of oscillation at 20 degree C (Hz) 25464.7908947\n",
+        "Frequency of oscillation at 25 degree C (Hz) 31830.9886184\n",
+        "Frequency of oscillation at 30 degree C (Hz) 42441.3181578\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.25  Page No : 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Calculations and Results\n",
+      "Se_thermocouple = 500-(-72);\n",
+      "print \"Sensitivity of thermocouple (micro V/degree C) = \", Se_thermocouple\n",
+      "Vo = Se_thermocouple*100*10**-6;\n",
+      "print \"maximum output voltage(V) = \", Vo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sensitivity of thermocouple (micro V/degree C) =  572\n",
+        "maximum output voltage(V) =  0.0572\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.26  Page No : 229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculations and results\n",
+      "ET = 27.07+0.8;\n",
+      "print 'Required e.m.f.(mV)',ET\n",
+      "print \"temperature corresponding to 27.87 mV is 620 degree C\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required e.m.f.(mV) 27.87\n",
+        "temperature corresponding to 27.87 mV is 620 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.27  Page No : 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Rm = 50;\n",
+      "Re = 12;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "E = 33.3*10**-3;\n",
+      "i = 0.1*10**-3;\n",
+      "Rs = (E/i)-Rm-Re;\n",
+      "print \"series resistance (ohm) = \", Rs\n",
+      "Re = 13;\n",
+      "i1 = E/(Rs+Re+Rm);\n",
+      "AE = ((i1-i)/i)*800;\n",
+      "print \"approximate error due to rise in resistance of 1 ohm in Re (degree C) = \", AE\n",
+      "R_change = 50*0.00426*10;\n",
+      "i1 = E/(Rs+Re+Rm+R_change);\n",
+      "AE = ((i1-i)/i)*800;\n",
+      "print \"approximate error due to rise in Temp. of 10 (degree C) = \", AE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "series resistance (ohm) =  271.0\n",
+        "approximate error due to rise in resistance of 1 ohm in Re (degree C) =  -2.39520958084\n",
+        "approximate error due to rise in Temp. of 10 (degree C) =  -7.44949870586\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.28  Page No : 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "E_20 = 0.112*10**-3;# emf at 20degree C\n",
+      "E_900 = 8.446*10**-3;\n",
+      "E_1200 = 11.946*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "E1 = E_900-E_20;\n",
+      "E2 = E_1200-E_20;\n",
+      "print \"E1 = 1.08*R1/(R1+2.5+R2)      (i)\",\n",
+      "print \"E2 = 1.08*(R1+2.5)/(R1+2.5+R2)      (ii)\",\n",
+      "print \"on solving (i) and (ii)\",\n",
+      "R1 = 5.95;\n",
+      "R2 = 762.6;\n",
+      "print \"value of resistance R1 (ohm) = \", R1\n",
+      "print \"value of resistance R2 (ohm) = \", R2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E1 = 1.08*R1/(R1+2.5+R2)      (i) E2 = 1.08*(R1+2.5)/(R1+2.5+R2)      (ii) on solving (i) and (ii) value of resistance R1 (ohm) =  5.95\n",
+        "value of resistance R2 (ohm) =  762.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.29  Page No : 235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Calculations\n",
+      "linearity_percentage = (0.003/1.5)*100;\n",
+      "\n",
+      "# Results\n",
+      "print \"percentage linearity = \", linearity_percentage\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percentage linearity =  0.2\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.30  Page No : 237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "print \"lacement = 0.5\"\n",
+      "Vo = 2.*10**-3;\n",
+      "Se_LVDT = Vo\n",
+      "print \"lacement \"\n",
+      "print \"senstivity of the LVDT (V/mm)\", Se_LVDT\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Af = 250.;\n",
+      "Se_instrument = Se_LVDT*Af;\n",
+      "print \"senstivity of instrument (V/mm)\", Se_instrument\n",
+      "sd = 5./100;\n",
+      "Vo_min = 50./5;\n",
+      "Re_instrument = 1*1/1000;\n",
+      "print \"resolution of instrument in mm\", Re_instrument\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lacement = 0.5\n",
+        "lacement \n",
+        "senstivity of the LVDT (V/mm) 0.002\n",
+        "senstivity of instrument (V/mm) 0.5\n",
+        "resolution of instrument in mm 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.31  Page No : 239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "b = 0.02;\n",
+      "t = 0.004;\n",
+      "I = (1./12)*b*t**3;\n",
+      "F = 25.;\n",
+      "l = 0.25;\n",
+      "E = 200.*10**9;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "x = (F*l**3)/(3*E*I);\n",
+      "print \"deflection (m)\", x\n",
+      "DpF = x/F;\n",
+      "Se = DpF*0.5*1000;\n",
+      "Re = (10./1000)*(2./10);\n",
+      "F_min = Re/Se;\n",
+      "F_max = 10/Se;\n",
+      "print \"minimum force (N)\", F_min\n",
+      "print \"maximum force (N)\", F_max\n",
+      "print \"\", Se\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "deflection (m) 0.006103515625\n",
+        "minimum force (N) 0.016384\n",
+        "maximum force (N) 81.92\n",
+        " 0.1220703125\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.32  Page No : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"permittivity of the air e0 = 8.85*10**-12\"\n",
+      "# Variables\n",
+      "e0 = 8.85*10**-12;\n",
+      "w = 25*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "d = 0.25*10**-3;\n",
+      "Se = -4*e0*w/d;\n",
+      "\n",
+      "# Results\n",
+      "print \"sensitivity of the transducer (F/m) = \", Se\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "permittivity of the air e0 = 8.85*10**-12\n",
+        "sensitivity of the transducer (F/m) =  -3.54e-09\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.33  Page No : 243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "C1 = 370*10**-12;\n",
+      "d1 = 3.5*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "d2 = 2.9*10**-3;\n",
+      "C2 = C1*d1/d2;\n",
+      "\n",
+      "# Results\n",
+      "print \"the value of the capacitance afte the application of pressure (F) = \", C2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of the capacitance afte the application of pressure (F) =  4.46551724138e-10\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.34  Page No : 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "fo1 = 100.*10**3;\n",
+      "d1 = 4.;\n",
+      "d2 = 3.7;\n",
+      "\n",
+      "# Calculations\n",
+      "fo2 = ((d2/d1)**0.5)*fo1;\n",
+      "dfo = fo1-fo2;\n",
+      "\n",
+      "# Results\n",
+      "print \"change in frequency of the oscillator (Hz)\", dfo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in frequency of the oscillator (Hz) 3823.07969164\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.35  Page No : 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "L_air = (3.1-3)/2;\n",
+      "D_stress = 100/L_air;\n",
+      "e0 = 8.85*10**-12;\n",
+      "l = 20.*10**-3;\n",
+      "D2 = 3.1;\n",
+      "D1 = 3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "C = (2*math.pi)*e0*l/(math.log(D2/D1));\n",
+      "print \"capacitance(F) = \", C\n",
+      "l = (20*10**-3)-(2*10**-3);\n",
+      "C_new = (2*math.pi)*e0*l/(math.log(D2/D1));\n",
+      "C_change = C-C_new;\n",
+      "print \"change in capacitance(F) = \", C_change\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "capacitance(F) =  3.39167370734e-11\n",
+        "change in capacitance(F) =  3.39167370734e-12\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.36  Page No : 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "M = 0.95;\n",
+      "w = 2.*math.pi*20;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "tc = (1/w)*((M**2)/(1-M**2))**0.5;\n",
+      "print \"time constant (s)\", tc\n",
+      "ph = ((math.pi/2)-(math.atan(w*tc)))*(180/math.pi);\n",
+      "print \"phase shift(deg)\", ph\n",
+      "C = (8.85*10**-12*300*10**-6)/(0.125*10**-3);\n",
+      "R = tc/C;\n",
+      "print \"series resistance (ohm)\", R\n",
+      "M = 1./(1+(1./(2*math.pi*5*tc)**2))**0.5;\n",
+      "print \"amplitude ratio = \", M\n",
+      "Eb = 100;\n",
+      "x = 0.125*10**-3;\n",
+      "Vs = Eb/x;\n",
+      "print \"voltage sensitivity (V/m)\", Vs\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time constant (s) 0.0242109278459\n",
+        "phase shift(deg) 18.1948723388\n",
+        "series resistance (ohm) 1139874192.37\n",
+        "amplitude ratio =  0.605390338483\n",
+        "voltage sensitivity (V/m) 800000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.37  Page No : 251"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "e0 = 8.85*10**-12;\n",
+      "A = 500*10**-6;\n",
+      "d = 0.2*10**-3;\n",
+      "C = e0*A/d;\n",
+      "d1 = 0.18*10**-3;\n",
+      "C_new = e0*A/d1;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "C_change = C_new-C;\n",
+      "Ratio = (C_change/C)/(0.02/0.2);\n",
+      "print \"ratio of per unit change of capacitance to per unit change of diaplacement\", Ratio\n",
+      "d1 = 0.19*10**-3;\n",
+      "e1 = 1;\n",
+      "d2 = 0.01*10**-3;\n",
+      "e2 = 8;\n",
+      "C = (e0*A)/((d1/e1)+(d2/e2));\n",
+      "d1_new = 0.17*10**-3;\n",
+      "C_new = (e0*A)/((d1_new/e1)+(d2/e2));\n",
+      "C_change = C_new-C;\n",
+      "Ratio = (C_change/C)/(0.02/0.2);\n",
+      "print \"ratio of per unit change of capacitance to per unit change of diaplacement\", Ratio\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of per unit change of capacitance to per unit change of diaplacement 1.11111111111\n",
+        "ratio of per unit change of capacitance to per unit change of diaplacement 1.16788321168\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.40  Page No : 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "g = 0.055;\n",
+      "t = 2*10**-3;\n",
+      "P = 1.5*10**6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Eo = g*t*P;\n",
+      "print \"output voltage(V) = \", Eo\n",
+      "e = 40.6*10**-12;\n",
+      "d = e*g;\n",
+      "print \"charge sensitivity (C/N) = \", d\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage(V) =  165.0\n",
+        "charge sensitivity (C/N) =  2.233e-12\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.41  Page No : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "g = 0.055;\n",
+      "t = 1.5*10**-3;\n",
+      "Eo = 100;\n",
+      "\n",
+      "# Calculations\n",
+      "P =  Eo/(g*t);\n",
+      "A = 25.*10**-6;\n",
+      "F = P*A;\n",
+      "\n",
+      "# Results\n",
+      "print \"Force(N) = \", F\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force(N) =  30.303030303\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.42  Page No : 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "A = 25.*10**-6;\n",
+      "F = 5.;\n",
+      "P = F/A;\n",
+      "d = 150*10**-12;\n",
+      "e = 12.5*10**-9;\n",
+      "g = d/(e);\n",
+      "t = 1.25*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "Eo = (g*t*P);\n",
+      "strain = P/(12*10**6);\n",
+      "Q = d*F;\n",
+      "C = Q/Eo;\n",
+      "\n",
+      "# Results\n",
+      "print \"strain = \", strain\n",
+      "print \"charge(C) = \", Q\n",
+      "print \"capacitance(F) = \", C\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "strain =  0.0166666666667\n",
+        "charge(C) =  7.5e-10\n",
+        "capacitance(F) =  2.5e-10\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.43  Page No : 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "d = 2.*10**-12;\n",
+      "t = 1.*10**-3;\n",
+      "Fmax = 0.01;\n",
+      "e0 = 8.85*10**-12;\n",
+      "er = 5.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A = 100*10**-6;\n",
+      "Eo_peak_to_peak = 2*d*t*Fmax/(e0*er*A);\n",
+      "print \"peak voltage swing under open conditions\", Eo_peak_to_peak\n",
+      "Rl = 100*10**6;\n",
+      "Cl = 20*10**-12;\n",
+      "d1 = 1*10**-3;\n",
+      "Cp = e0*er*A/d1;\n",
+      "C = Cp+Cl;\n",
+      "w = 1000;\n",
+      "m = (w*Cp*Rl/(1+(w*C*Rl)**2)**0.5);\n",
+      "El_peak_to_peak = (2*d*t*Fmax/(e0*er*A))*m;\n",
+      "\n",
+      "print \"peak voltage swing under loaded conditions\", El_peak_to_peak\n",
+      "E = 90*10**9;\n",
+      "dt = 2*Fmax*t/(A*E);\n",
+      "print \"maximum change in crystal thickness (m)\", dt\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak voltage swing under open conditions 0.0090395480226\n",
+        "peak voltage swing under loaded conditions 0.00151556405344\n",
+        "maximum change in crystal thickness (m) 2.22222222222e-12\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.44  Page No : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "M = 0.95;\n",
+      "tc = 1.5*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "w = (1/tc)*((M**2)/(1-M**2))**0.5;\n",
+      "print \"minimum frequency (rad/s)\", w\n",
+      "ph = ((math.pi/2)-(math.atan(w*tc)))*(180/math.pi);\n",
+      "print \"phase shift(deg)\", ph\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "minimum frequency (rad/s) 2028.2899482\n",
+        "phase shift(deg) 18.1948723388\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.45  Page No : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Kq = 40.*10**-3;\n",
+      "Cp = 1000.*10**-12;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "K = Kq/Cp;\n",
+      "print \"sensitivity of the transducer(V/m)\", K\n",
+      "Cc = 300*10**-12;\n",
+      "Ca = 50*10**-12;\n",
+      "C = Cp+Cc+Ca;\n",
+      "Hf = Kq/C;\n",
+      "print \"high frequency sensitivity (V/m)\", Hf\n",
+      "R = 1*10**6;\n",
+      "tc = R*C;\n",
+      "M = 0.95;\n",
+      "w = (1/tc)*((M**2)/(1-M**2))**0.5;\n",
+      "f = w/(2*math.pi);\n",
+      "print \"minimum frequency (s)\", w\n",
+      "print \"now f = 10Hz\",\n",
+      "f = 10;\n",
+      "w = 2*math.pi*f;\n",
+      "tc = (1/w)*((M**2)/(1-M**2))**0.5;\n",
+      "C_new = tc/R;\n",
+      "Ce = C_new-C;\n",
+      "print \"external shunt capacitance(F)\", Ce\n",
+      "Hf_new = Kq/C_new;\n",
+      "print \"new value of high frequency sensitivity (V/m)\", Hf_new\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "sensitivity of the transducer(V/m) 40000000.0\n",
+        "high frequency sensitivity (V/m) 29629629.6296\n",
+        "minimum frequency (s) 2253.655498\n",
+        "now f = 10Hz external shunt capacitance(F) 4.70718556919e-08\n",
+        "new value of high frequency sensitivity (V/m) 826073.256145\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.46  Page No : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#\n",
+      "\n",
+      "# Variables\n",
+      "R = 10.**6;\n",
+      "C = 2500.*10**-12;\n",
+      "tc = R*C;\n",
+      "t = 2.*10**-3;\n",
+      "d = 100.*10**-12;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "F = 0.1;\n",
+      "el = 10.**3*(d*F*(math.exp(-t/tc))/C);\n",
+      "print \"voltage just before t = 2ms (mV)\", el\n",
+      "el_after = 10.**3*(d*F*(math.exp(-t/tc)-1)/C);\n",
+      "print \"voltage just after t = 2ms (mV)\", el_after\n",
+      "print \"when t = 10ms\",\n",
+      "t = 10.*10**-3;\n",
+      "T = 2.*10\n",
+      "e_10 = 10.**3*(d*F*(math.exp((-T/tc)-1))*(math.exp(-(t-T))/tc)/C)\n",
+      "print \"output voltage 10 ms after the application of impulse(mV) \", e_10\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "voltage just before t = 2ms (mV) 1.79731585647\n",
+        "voltage just after t = 2ms (mV) -2.20268414353\n",
+        "when t = 10ms output voltage 10 ms after the application of impulse(mV)  0.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.47  Page No : 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# to prove time constant should be approximately 20T to keep undershoot within 5%\n",
+      "# Variables\n",
+      "print \"Let T = 1\",\n",
+      "T = 1.;\n",
+      "el = 0.95;\n",
+      "\n",
+      "# Calculations\n",
+      "tc = -T/math.log(el);\n",
+      "\n",
+      "# Results\n",
+      "print \"time constant\", tc\n",
+      "print \"as T = 1 so time constant should be approximately equal to 20T\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Let T = 1 time constant 19.4957257462\n",
+        "as T = 1 so time constant should be approximately equal to 20T\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.48  Page No : 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "Kh = -1*10**-6;\n",
+      "I = 3;\n",
+      "B = 0.5;\n",
+      "\n",
+      "# Calculations\n",
+      "t = 2.*10**-3;\n",
+      "Eh = Kh*I*B/t;\n",
+      "\n",
+      "# Results\n",
+      "print \"output voltage (V)\", Eh\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage (V) -0.00075\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.49  Page No : 271"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculations\n",
+      "Th_wavelength = 1.24*10**-6/1.8\n",
+      "\n",
+      "# Results\n",
+      "print \"Threshold wavelength (m)\", Th_wavelength\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Threshold wavelength (m) 6.88888888889e-07\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.50  Page No : 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "E_imparted = (1.24*10**-6)/(0.2537*10**-6);\n",
+      "B_energy = E_imparted-4.30;\n",
+      "em_ratio = 0.176*10**12;\n",
+      "\n",
+      "# Calculations\n",
+      "v = (2*B_energy*em_ratio)**0.5;\n",
+      "\n",
+      "# Results\n",
+      "print \"maximum velocity of emitted photo electrons (m/s)\", v\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum velocity of emitted photo electrons (m/s) 454815.603242\n"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.51  Page No : 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "Ri = 30;\n",
+      "Rf = 100;\n",
+      "t = 10.;\n",
+      "tc = 72.;\n",
+      "\n",
+      "# Calculations\n",
+      "Rt = Ri+(Rf-Ri)*(1-math.exp(-t/tc));\n",
+      "\n",
+      "# Results\n",
+      "print \"resistance of the cell (K ohm)\", Rt\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance of the cell (K ohm) 39.0772691917\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.52  Page No : 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_power = 250*0.2*10**-6;\n",
+      "print \"incident power (W)\", I_power\n",
+      "\n",
+      "Rl = 10.*10**3;\n",
+      "C = 2.*10**-12;\n",
+      "fc = 1./(2*math.pi*Rl*C);\n",
+      "print \"cut off frequency (Hz)\", fc\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "incident power (W) 5e-05\n",
+        "cut off frequency (Hz) 7957747.15459\n"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.53  Page No : 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I = 2.2*10**-3;\n",
+      "Eo = 0.33;\n",
+      "Rl = 100.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Ri = (Eo/I)-100;\n",
+      "print \"internal resistance of cell (ohm)\", Ri\n",
+      "Vo = 0.33*(math.log(25)/math.log(10));\n",
+      "print \"open circuit voltage for a radiant incidence of 25 W/m2 (V) = \", Vo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "internal resistance of cell (ohm) 50.0\n",
+        "open circuit voltage for a radiant incidence of 25 W/m2 (V) =  0.461320202862\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.54  Page No : 281"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "A = 1935*10**-6;\n",
+      "r = 0.914;\n",
+      "S_angle = A/r**2;\n",
+      "I = 180.;\n",
+      "\n",
+      "# Calculations\n",
+      "L_flux = I*S_angle;\n",
+      "\n",
+      "# Results\n",
+      "print \"lumnious flux = \", L_flux\n",
+      "print \"Corresponding to lumnious flux o.417 lm and a load resistance of 800 ohm the current is 120 micro Ampere\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lumnious flux =  0.416928019765\n",
+        "Corresponding to lumnious flux o.417 lm and a load resistance of 800 ohm the current is 120 micro Ampere\n"
+       ]
+      }
+     ],
+     "prompt_number": 63
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch6.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch6.ipynb
new file mode 100755
index 00000000..2166ed92
--- /dev/null
+++ b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch6.ipynb
@@ -0,0 +1,1358 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:918235f4b2bfd4a24a1f3adb5ce3a07d29dd6848cf00df31612a375f1aec5802"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Signal Conditioning"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1  Page No : 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "A = 100.;\n",
+      "R1 = 1.*10**3;\n",
+      "\n",
+      "# Calculations\n",
+      "Rf = -A*R1;\n",
+      "\n",
+      "# Results\n",
+      "print \"feedback resistance (ohm) = \", Rf\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "feedback resistance (ohm) =  -100000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2  Page No : 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Rf = 10.;\n",
+      "R1 = 1.;\n",
+      "Avol = 200000.;\n",
+      "\n",
+      "# Calculations\n",
+      "A = -(Rf/R1)*(1./(1.+(1./Avol)*((R1+Rf)/R1)));\n",
+      "\n",
+      "# Results\n",
+      "print \"closed loop gain = \", A\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "closed loop gain =  -9.99945003025\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3  Page No : 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Calculations and Results\n",
+      "Sa = 10;\n",
+      "print \" saturation voltage = \", Sa\n",
+      "Vom = Sa;\n",
+      "print \"maximum output voltage\", Vom\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " saturation voltage =  10\n",
+        "maximum output voltage 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4  Page No : 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Vos = 5.*10**-3;\n",
+      "Rf = 10.;\n",
+      "R1 = 1.;\n",
+      "\n",
+      "# Calculations\n",
+      "Vo = -Vos*(1+Rf/R1);\n",
+      "\n",
+      "# Results\n",
+      "print \"output voltage due to offset voltage (V) = \", Vo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage due to offset voltage (V) =  -0.055\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5  Page No : 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "Rf = 10.;\n",
+      "R1 = 1;\n",
+      "\n",
+      "# Calculations\n",
+      "A = Rf/R1;\n",
+      "\n",
+      "# Results\n",
+      "print \"Amplification Factor = \", A\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amplification Factor =  10.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6  Page No : 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V1 = 1.;\n",
+      "V2 = -2.;\n",
+      "Rf = 500.;\n",
+      "R1 = 250.;\n",
+      "R2 = 100.;\n",
+      "\n",
+      "# Calculations\n",
+      "Vo = -(((Rf/R1)*V1)+((Rf/R2)*V2));\n",
+      "\n",
+      "# Results\n",
+      "print \"output voltage(V) = \", Vo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage(V) =  8.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7  Page No : 295"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Rf = 100.*10**3;\n",
+      "R1 = 1*10**3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A = Rf/R1;\n",
+      "print \"Gain = \", A\n",
+      "print \"If multiplier is 10\",\n",
+      "A = 10;\n",
+      "Rf = A*R1;\n",
+      "print \" feedback resistance (Ohm) = \", Rf\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gain =  100.0\n",
+        "If multiplier is 10  feedback resistance (Ohm) =  10000\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8  Page No : 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "g = 10.;\n",
+      "Rf = 10.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R1 = Rf/g;\n",
+      "print \"resistance R1(Kilo-ohms) = \", R1\n",
+      "R2 = Rf/(0.5*g);\n",
+      "print \"resistance R1(Kilo-ohms) = \", R2\n",
+      "R3 = Rf/(0.333*g);\n",
+      "print \"resistance R1(Kilo-ohms) = \", R3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance R1(Kilo-ohms) =  1.0\n",
+        "resistance R1(Kilo-ohms) =  2.0\n",
+        "resistance R1(Kilo-ohms) =  3.003003003\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9  Page No : 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Voramp = -10;\n",
+      "print \"if voltage source is 10V then RC =  1 ms and if C = 1 micro-F\"\n",
+      "\n",
+      "# Calculations\n",
+      "C = 1;\n",
+      "R = 1*10**-3*10**6;\n",
+      "\n",
+      "# Results\n",
+      "print \"value of resistance (ohm) =  \", R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "if voltage source is 10V then RC =  1 ms and if C = 1 micro-F\n",
+        "value of resistance (ohm) =   1000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10  Page No : 301"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V2 = 5.*10**-3;\n",
+      "V1 = 3.*10**-3;\n",
+      "Vo = 300.*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Vd = V2-V1;\n",
+      "Ad = Vo/Vd;\n",
+      "print \"difference mode gain = \", Ad\n",
+      "V2 = 155*10**-3;\n",
+      "V1 = 153*10**-3;\n",
+      "Vo = Ad*(V2-V1);\n",
+      "print \"output voltage (V) = \", Vo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "difference mode gain =  150.0\n",
+        "output voltage (V) =  0.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11  Page No : 303"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Vo = 3.;\n",
+      "Vd = 30.*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Ad = Vo/Vd;\n",
+      "print \"difference mode gain = \", Ad\n",
+      "Vo = 5.*10**-3;\n",
+      "Vc = 500.*10**-3;\n",
+      "Ac = Vo/Vc;\n",
+      "print \"Common mode gain = \", Ac\n",
+      "CMRR = Ad/Ac;\n",
+      "print \"Common mode rejection ratio = \", CMRR\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "difference mode gain =  100.0\n",
+        "Common mode gain =  0.01\n",
+        "Common mode rejection ratio =  10000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12  Page No : 305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V2 = 30.*10**-3;\n",
+      "V1 = -30.*10**-3;\n",
+      "Vd = V2-V1;\n",
+      "Ad = 150.;\n",
+      "Vos = Ad*Vd;\n",
+      "Ac = 0.04;\n",
+      "Vc = 600.*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "Von = Ac*Vc;\n",
+      "SNR = Vos/Von;\n",
+      "CMRR = Ad/Ac;\n",
+      "\n",
+      "# Results\n",
+      "print \"Signal to Noise Ratio = \", SNR\n",
+      "\n",
+      "print \"CMRR = \", CMRR\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signal to Noise Ratio =  375.0\n",
+        "CMRR =  3750.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13  Page No : 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Ci = 10.*10**-12;\n",
+      "Vi = 10.;\n",
+      "Eo = 8.85*10**-12;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A = 200.*10**-6;\n",
+      "K = -Ci*Vi/(Eo*A);\n",
+      "print \"sensitivity (V/mm) = \", K\n",
+      "d = 1*10**-6;\n",
+      "Vo = K*d;\n",
+      "print \"output voltage (V) = \", Vo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "sensitivity (V/mm) =  -56497.1751412\n",
+        "output voltage (V) =  -0.0564971751412\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.14  Page No : 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Calculations and Results\n",
+      "MXtc =  10.**10*1000*10**-12;\n",
+      "print \"Maximum time constant (s)\", MXtc\n",
+      "MNtc =  10.**8*10*10**-12;\n",
+      "print \"Minimum time constant (s)\", MNtc\n",
+      "AR = 0.95;\n",
+      "fmin = (AR)/(2*math.pi*MXtc*(1-AR**2)**0.5);\n",
+      "print \"minimum frequency (Hz)\", fmin\n",
+      "fmax = (AR)/(2*math.pi*MNtc*(1-AR**2)**0.5);\n",
+      "print \"Maximum frequency (Hz)\", fmax\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum time constant (s) 10.0\n",
+        "Minimum time constant (s) 0.001\n",
+        "minimum frequency (Hz) 0.0484218556919\n",
+        "Maximum frequency (Hz) 484.218556919\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15  Page No : 311"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "g = 0.501;\n",
+      "f = 50;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "w = 2*math.pi*f;\n",
+      "tc = (1-g**2)**0.5/(w*g);\n",
+      "print \"time constant (s)\", tc\n",
+      "R = 10000;\n",
+      "C = (tc/R)*10**6;\n",
+      "print \"capacitance (micro-F)\", C\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time constant (s) 0.00549861130167\n",
+        "capacitance (micro-F) 0.549861130167\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.16  Page No : 313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R1 = 10.*10**3;\n",
+      "R2 = 1.*10**6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A = R2/(R1+R2);\n",
+      "print \"gain = \", A\n",
+      "C2 = (0.01)*10**-6;\n",
+      "C1 = 100.*10**-12;\n",
+      "fcl = 1./(2*math.pi*C2*R2);\n",
+      "print \"lower cut off frequency (Hz)\", fcl\n",
+      "fcu = 1./(2*math.pi*R1*C1);\n",
+      "print \"upper cut off frequency (Hz)\", fcu\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "gain =  0.990099009901\n",
+        "lower cut off frequency (Hz) 15.9154943092\n",
+        "upper cut off frequency (Hz) 159154.943092\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17  Page No : 315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R = 1.*10**6;\n",
+      "fo = 10.*10**3;\n",
+      "\n",
+      "# Calculations\n",
+      "C = 1/(2*math.pi*fo*R);\n",
+      "\n",
+      "# Results\n",
+      "print \"the value of C (F)\", C\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of C (F) 1.59154943092e-11\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.19  Page No : 317"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Rt = 100.;\n",
+      "K = 1.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Rb = K*Rt;\n",
+      "ei = 10;\n",
+      "print \"When K = 1\"\n",
+      "eo = ((K*Rt/Rb)/(1+(K*Rt/Rb)))*ei;\n",
+      "print \"output voltage (V) =  \", eo\n",
+      "Se = (ei*Rb)/((Rb+K*Rt)**2);\n",
+      "print \"sensitivity (V/ohm) =  \", Se\n",
+      "K = 0.95;\n",
+      "print \"When K = 0.95\",\n",
+      "eo = ((K*Rt/Rb)/(1+(K*Rt/Rb)))*ei;\n",
+      "print \"output voltage (V) =  \", eo\n",
+      "Se = (ei*Rb)/((Rb+K*Rt)**2);\n",
+      "print \"sensitivity (V/ohm) =  \", Se\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When K = 1\n",
+        "output voltage (V) =   5.0\n",
+        "sensitivity (V/ohm) =   0.025\n",
+        "When K = 0.95 output voltage (V) =   4.87179487179\n",
+        "sensitivity (V/ohm) =   0.0262984878369\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.20  Page No : 319"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "ei = 100.;\n",
+      "K = 0.25;\n",
+      "print \"When K = 0.25\"\n",
+      "\n",
+      "# Calculations and Results\n",
+      "eo = ((K/6)/(1+(K/6)))*ei;\n",
+      "print \"output voltage (V) =  \", eo\n",
+      "K = 0.5;\n",
+      "print \"When K = 0.5\",\n",
+      "eo = ((K/6)/(1+(K/6)))*ei;\n",
+      "print \"output voltage (V) =  \", eo\n",
+      "K = 0.6;\n",
+      "print \"When K = 0.6\",\n",
+      "eo = ((K/6)/(1+(K/6)))*ei;\n",
+      "print \"output voltage (V) =  \", eo\n",
+      "K = 0.8;\n",
+      "print \"When K = 0.8\",\n",
+      "eo = ((K/6)/(1+(K/6)))*ei;\n",
+      "print \"output voltage (V) =  \", eo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When K = 0.25\n",
+        "output voltage (V) =   4.0\n",
+        "When K = 0.5 output voltage (V) =   7.69230769231\n",
+        "When K = 0.6 output voltage (V) =   9.09090909091\n",
+        "When K = 0.8 output voltage (V) =   11.7647058824\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.21  Page No : 321"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R2 = 119;\n",
+      "R3 = 119.7;\n",
+      "R1 = 120.4;\n",
+      "R4 = R2*R3/R1;\n",
+      "R4 = 121.2;\n",
+      "ei = 12;\n",
+      "\n",
+      "# Calculations\n",
+      "eo = ((R1*R4-R2*R3)/((R1+R3)*(R2+R4)))*ei;\n",
+      "\n",
+      "# Results\n",
+      "print \" output voltage (V) = \", eo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " output voltage (V) =  0.0724469161996\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.22  Page No : 323"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "ei = 6.;\n",
+      "R = 10000.;\n",
+      "print \"if dR = 0.05R\"\n",
+      "\n",
+      "# Calculations and Results\n",
+      "dR = 0.05*R;\n",
+      "eo = ((dR/R)/(4+2*(dR/R)))*ei;\n",
+      "print \"output voltage (V)\", eo\n",
+      "print \"if dR = -0.05R\"\n",
+      "dR = -0.05*R;\n",
+      "eo = ((dR/R)/(4+2*(dR/R)))*ei;\n",
+      "print \"output voltage (V)\", eo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "if dR = 0.05R\n",
+        "output voltage (V) 0.0731707317073\n",
+        "if dR = -0.05R\n",
+        "output voltage (V) -0.0769230769231\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.23  Page No : 325"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "R2 = 800;\n",
+      "R3 = 800;\n",
+      "R4 = 800;\n",
+      "Rm = 100;\n",
+      "R = 800.;\n",
+      "ei = 4;\n",
+      "\n",
+      "# Calculations\n",
+      "im = 0.8*10**-6;\n",
+      "dR = (im*R**2)*(4*(1+Rm/R))/ei;\n",
+      "R1 = R+dR;\n",
+      "\n",
+      "# Results\n",
+      "print \"resistance of unknown resistor (ohm) = \", R1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance of unknown resistor (ohm) =  800.576\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.24  Page No : 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R2 = 1000.;\n",
+      "R3 = 1000.;\n",
+      "R1 = 1010.;\n",
+      "R4 = 1000.;\n",
+      "ei = 100.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "eo = ((R1*R4-R2*R3)/((R1+R3)*(R2+R4)))*ei;\n",
+      "print \" open circuit voltage (V) = \", eo\n",
+      "Ro = (R1*R4/(R1+R4))+(R2*R3/(R2+R3));\n",
+      "Rm = 4000;\n",
+      "im = eo/(Ro+Rm);\n",
+      "print \"current (A) = \", im\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " open circuit voltage (V) =  0.248756218905\n",
+        "current (A) =  4.97265042268e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.25  Page No : 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R = 100.;\n",
+      "P = 250.*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "i = (P/R)**0.5;\n",
+      "print \"maximum permissible current (A) = \", i\n",
+      "ei = 2*i*R;\n",
+      "print \"maximum supply voltage (V) = \", ei\n",
+      "Rs = 100.;\n",
+      "Ps = 10.**2/Rs;\n",
+      "print \"Power dissipation in series resistance (W)\", Ps\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum permissible current (A) =  0.05\n",
+        "maximum supply voltage (V) =  10.0\n",
+        "Power dissipation in series resistance (W) 1.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.26  Page No : 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = (0.1/0.2)*10**-3;\n",
+      "R = 1000.;\n",
+      "eim = 2*(P*R)**0.5;\n",
+      "dth = 0.1;\n",
+      "\n",
+      "# Calculations\n",
+      "dR = (4.5/100)*dth*R;\n",
+      "eom = (dR/(4*R))*eim;\n",
+      "Sem = eom/dth;\n",
+      "\n",
+      "# Results\n",
+      "print \"maximum voltage sensitivity of the bridge (V) = \", Sem\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum voltage sensitivity of the bridge (V) =  0.0159099025767\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.27  Page No : 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Calculations and Results\n",
+      "Reso = 10.*10**-3/10;\n",
+      "print \"resolution of the instrument = \", Reso\n",
+      "n = 10.;\n",
+      "Q = 10/2**n;\n",
+      "Eq = Q/(2*3**0.5);\n",
+      "print \"quantization error = \", Eq\n",
+      "D = (2**n)-1;\n",
+      "print \"decesion levels = \", D\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resolution of the instrument =  0.001\n",
+        "quantization error =  0.00281909311128\n",
+        "decesion levels =  1023.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.28  Page No : 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Ra = 10.;\n",
+      "b = 5.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Wmsb = Ra/2;\n",
+      "print \"weight of MSB (V) = \", Wmsb\n",
+      "Wlsb = Ra/2**b;\n",
+      "print \"weight of LSB (V) = \", Wlsb\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "weight of MSB (V) =  5.0\n",
+        "weight of LSB (V) =  0.3125\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.29  Page No : 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "E = 10.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "ER = E*256/255;\n",
+      "print \"Reference voltage (V) = \", ER\n",
+      "n = 8.;\n",
+      "CVlsb = (2**-n)*ER;\n",
+      "PC = CVlsb*100/E;\n",
+      "print \"Percentage change  = \", PC\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reference voltage (V) =  10.0392156863\n",
+        "Percentage change  =  0.392156862745\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.30  Page No : 339"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Calculations and Results\n",
+      "n = 14.;\n",
+      "print \"number of bits  = \", n\n",
+      "E = 10.;\n",
+      "Q = 10.;\n",
+      "LSB = E/2**n;\n",
+      "print \"Value of LSB (V)  = \", LSB\n",
+      "Eq = Q/(2*(3**0.5));\n",
+      "print \"Quantization error (V)  = \", Eq\n",
+      "fh = 1000.;\n",
+      "fs = 5*fh;\n",
+      "print \"minimum sampling rate (Hz)  = \", fs\n",
+      "a = 1./16384;\n",
+      "ta = 1./(2*math.pi*fh)*a;\n",
+      "print \"Aperature time (s)  = \", ta\n",
+      "Dr = 6*n;\n",
+      "print \"dynamic range (db)  = \", Dr\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of bits  =  14.0\n",
+        "Value of LSB (V)  =  0.0006103515625\n",
+        "Quantization error (V)  =  2.88675134595\n",
+        "minimum sampling rate (Hz)  =  5000.0\n",
+        "Aperature time (s)  =  9.71404681957e-09\n",
+        "dynamic range (db)  =  84.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.31  Page No : 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "ER = 10.;\n",
+      "n = 6.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Imax = 10*10**-3;\n",
+      "R = ER*((2**n)-1)/((2**(n-2))*Imax);\n",
+      "print \"resistance (ohm) = \", R\n",
+      "LSB = ER/((2**(n-1))*R);\n",
+      "print \"smallest output current (A)\", LSB\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance (ohm) =  3937.5\n",
+        "smallest output current (A) 7.93650793651e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.32  Page No : 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 6.;\n",
+      "R = 10000.;\n",
+      "Io =  (10./10*10**3)*(1*1+1*0.5+1*0.25+0*0.125+1*0.0625)*10**-6;\n",
+      "Rf = 5000;\n",
+      "\n",
+      "# Calculations\n",
+      "Eo = -Io*Rf;\n",
+      "\n",
+      "# Results\n",
+      "print \"Output voltage (V) = \", Eo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output voltage (V) =  -9.0625\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.33  Page No : 345"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculations and Results\n",
+      "print \"Set d3 = 1\"\n",
+      "Output = 5./2**1;\n",
+      "print \"since 3.217>2.5  so d3 = 1\"\n",
+      "print \"Set d2 = 1\"\n",
+      "Output = (5./2**1)+(5./2**2);\n",
+      "print \"since 3.217< 3.75  so d2 = 0\"\n",
+      "print \"Set d1 = 1\",\n",
+      "Output = (5./2**1)+(5./2**3);\n",
+      "print \"since 3.217>3.125  so d1 = 1\",\n",
+      "print \"Set d0 = 1\",\n",
+      "Output = (5./2**1)+(5./2**3)+(5./2**4);\n",
+      "print \"since 3.217<3.4375  so d0 = 0\",\n",
+      "print \"Output of successive approximation A/D  =  1010\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Set d3 = 1\n",
+        "since 3.217>2.5  so d3 = 1\n",
+        "Set d2 = 1\n",
+        "since 3.217< 3.75  so d2 = 0\n",
+        "Set d1 = 1 since 3.217>3.125  so d1 = 1 Set d0 = 1 since 3.217<3.4375  so d0 = 0 Output of successive approximation A/D  =  1010\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.34  Page No : 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 400.;\n",
+      "T = t/4;\n",
+      "C = 1.*10**-6;\n",
+      "v = 20;\n",
+      "i = C*100*v/(T);\n",
+      "R = 1*10**3;\n",
+      "\n",
+      "# Calculations\n",
+      "e_o = i*R;\n",
+      "\n",
+      "# Results\n",
+      "print \"output voltage(V)\", e_o\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage(V) 0.02\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch7.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch7.ipynb
new file mode 100755
index 00000000..b7ea5d39
--- /dev/null
+++ b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch7.ipynb
@@ -0,0 +1,443 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f0558a3d67a6c58d34e0e0a67cd439be2d1883ea7cc3c2cdd41f8e1ab303c0ee"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Display Devices and recorders"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1  Page No : 349"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "N  =  4.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R = 1./10**N;\n",
+      "print \"Resolution of the meter = \", R\n",
+      "VR = 1;\n",
+      "R1 = VR*R;\n",
+      "print \"Resolution of the meter for voltage range 1V = \", R1\n",
+      "VR1 = 10;\n",
+      "R2 = VR1*R;\n",
+      "print \"Resolution of the meter for voltage range 10V = \", R2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resolution of the meter =  0.0001\n",
+        "Resolution of the meter for voltage range 1V =  0.0001\n",
+        "Resolution of the meter for voltage range 10V =  0.001\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2  Page No : 351"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "N  =  3.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R = 1./10**N;\n",
+      "print \"Resolution of the meter = \", R\n",
+      "print \"12.98 will be layed as 12.980 on 10V scale\"\n",
+      "VR = 1.;\n",
+      "R1 = VR*R;\n",
+      "print \"Resolution of the meter for voltage range 1V = \", R1\n",
+      "print \"0.6973 will be layed as 0.6973 on 1V scale\"\n",
+      "VR1 = 10;\n",
+      "R2 = VR1*R;\n",
+      "print \"Resolution of the meter for voltage range 10V = \", R2\n",
+      "print \"0.6973 will be layed as 00.697 on 10V scale\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resolution of the meter =  0.001\n",
+        "12.98 will be layed as 12.980 on 10V scale\n",
+        "Resolution of the meter for voltage range 1V =  0.001\n",
+        "0.6973 will be layed as 0.6973 on 1V scale\n",
+        "Resolution of the meter for voltage range 10V =  0.01\n",
+        "0.6973 will be layed as 00.697 on 10V scale\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3  Page No : 353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R = 5.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "V = 0.005*R;\n",
+      "print \"0.5 percent of the reading\", V\n",
+      "TPE = V+0.01;\n",
+      "print \"Total possible error (V) = \", TPE\n",
+      "R1 = 0.10;\n",
+      "V1 = 0.005*R1;\n",
+      "TPE1 = V1+0.01;\n",
+      "print \"Total possible error (V) = \", TPE1\n",
+      "PE = (TPE1/0.1)*100;\n",
+      "print \"Percentage error = \", PE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.5 percent of the reading 0.025\n",
+        "Total possible error (V) =  0.035\n",
+        "Total possible error (V) =  0.0105\n",
+        "Percentage error =  10.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4  Page No : 355"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "N = 034;\n",
+      "t = 10.*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "f = N/t;\n",
+      "\n",
+      "# Results\n",
+      "print \"frequency(Hz) = \", f\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequency(Hz) =  2800.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5  Page No : 357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R = 5*10**6;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "V = 0.00005*R;\n",
+      "print \"0.005 percent of the reading(micro sec) = \", V\n",
+      "LSD = 1;\n",
+      "ME = V+1;\n",
+      "print \"Maximum error (micro sec) = \", ME\n",
+      "R = 500;\n",
+      "V = 0.00005*R;\n",
+      "print \"0.005 percent of the reading(sec) = \", V\n",
+      "LSD = 1;\n",
+      "ME = V+1;\n",
+      "print \"Maximum error (sec) = \", ME\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.005 percent of the reading(micro sec) =  250.0\n",
+        "Maximum error (micro sec) =  251.0\n",
+        "0.005 percent of the reading(sec) =  0.025\n",
+        "Maximum error (sec) =  1.025\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6  Page No : 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "D = 8.*10**-3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A = D**2;\n",
+      "print \"A = \", A\n",
+      "J = 8*10**-3;\n",
+      "K = 16*10**-3;\n",
+      "B = 4*J*K;\n",
+      "print \"B = \", B\n",
+      "print \"since A<B so the instrument is underdanped\",\n",
+      "th = (100*math.pi)/180;\n",
+      "i = 10*10**-3;\n",
+      "F = 0.2*10**-6;\n",
+      "G = (K*th+F)/i;\n",
+      "l = 65*10**-3;\n",
+      "d = 25*10**-3;\n",
+      "N = G/(B*l*d);\n",
+      "print \"number of turns = \", N\n",
+      "i = F/G;\n",
+      "print \"current required to overcome friction (A)\", i\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " A =  6.4e-05\n",
+        "B =  0.000512\n",
+        "since A<B so the instrument is underdanped number of turns =  3356426.44614\n",
+        "current required to overcome friction (A) 7.16192114565e-08\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7  Page No : 361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Lam = 2.5*6.25;\n",
+      "f = 50000;\n",
+      "\n",
+      "# Calculations\n",
+      "S = Lam*10**-6*f;\n",
+      "\n",
+      "# Results\n",
+      "print \"speed(m/s) = \", S\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "speed(m/s) =  0.78125\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8  Page No : 363"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Calculations\n",
+      "ND = 12000/1.5;\n",
+      "\n",
+      "# Results\n",
+      "print \"Number density (numbers/mm)\", ND\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number density (numbers/mm) 8000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9  Page No : 365"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Y1 = 1.25;\n",
+      "Y2 = 2.5;\n",
+      "\n",
+      "# Calculations\n",
+      "PA = math.asin(Y1/Y2);\n",
+      "\n",
+      "# Results\n",
+      "print \"phase angle (degree)\", PA\n",
+      "print \"possible angle are 30 degree and 330 degree\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "phase angle (degree) 0.523598775598\n",
+        "possible angle are 30 degree and 330 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10  Page No : 367"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Calculations and Results\n",
+      "print \"if spot generating pattern moves in the clockwise direction\"\n",
+      "Y1 = 0;\n",
+      "Y2 = 5;\n",
+      "PA = math.asin(Y1/Y2);\n",
+      "print \"phase angle (degree)\", PA\n",
+      "Y1 = 2.5;\n",
+      "Y2 = 5;\n",
+      "PA = math.asin(Y1/Y2);\n",
+      "print \"phase angle (degree)\", PA\n",
+      "Y1 = 3.5;\n",
+      "Y2 = 5;\n",
+      "PA = math.asin(Y1/Y2);\n",
+      "print \"phase angle (degree)\", PA\n",
+      "Y1 = 2.5;\n",
+      "Y2 = 5;\n",
+      "PA = 180-(math.asin(Y1/Y2));\n",
+      "print \"phase angle (degree)\", PA\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "if spot generating pattern moves in the clockwise direction\n",
+        "phase angle (degree) 0.0\n",
+        "phase angle (degree) 0.523598775598\n",
+        "phase angle (degree) 0.775397496611\n",
+        "phase angle (degree) 179.476401224\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch8.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch8.ipynb
new file mode 100755
index 00000000..a31428ca
--- /dev/null
+++ b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch8.ipynb
@@ -0,0 +1,338 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:94ad83571615597b6a84f943391994f2595055abbe147d494a554f5f30f9f17e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Metrology"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1  Page No : 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculations and Results\n",
+      "Dd = 52.215;\n",
+      "print \"desired value = \", Dd\n",
+      "Pb = 4;\n",
+      "print \"Protected block = \", Pb\n",
+      "R = Dd-Pb;\n",
+      "print \"Reminder = \", R\n",
+      "Tp = 1.005;\n",
+      "print \"thousand  block = \", Tp\n",
+      "R = R-Tp;\n",
+      "print \"Reminder = \", R\n",
+      "Hp = 1.010;\n",
+      "print \"Hunderths block = \", Hp\n",
+      "R = R-Hp;\n",
+      "print \"Reminder = \", R\n",
+      "Ttp = 2.20;\n",
+      "print \"tenths block = \", Ttp\n",
+      "R = R-Ttp;\n",
+      "print \"Reminder = \", R\n",
+      "Up = 4;\n",
+      "print \"unit block = \", Up\n",
+      "R = R-Up;\n",
+      "print \"Reminder = \", R\n",
+      "Tp = 40;\n",
+      "print \"Tens block = \", Tp\n",
+      "R = R-Tp;\n",
+      "print \"Reminder = \", R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "desired value =  52.215\n",
+        "Protected block =  4\n",
+        "Reminder =  48.215\n",
+        "thousand  block =  1.005\n",
+        "Reminder =  47.21\n",
+        "Hunderths block =  1.01\n",
+        "Reminder =  46.2\n",
+        "tenths block =  2.2\n",
+        "Reminder =  44.0\n",
+        "unit block =  4\n",
+        "Reminder =  40.0\n",
+        "Tens block =  40\n",
+        "Reminder =  0.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2  Page No : 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Ps = 200.*10**3;\n",
+      "r = 0.6;\n",
+      "d2 = 0.5;\n",
+      "d1 = 0.5;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "a = (d2/d1**2);\n",
+      "x1 = (1.1-r)/(2*a);\n",
+      "print \"x1 = \", x1\n",
+      "r = 0.8;\n",
+      "d2 = 0.5;\n",
+      "d1 = 0.5;\n",
+      "a = (d2/d1**2);\n",
+      "x2 = (1.1-r)/(2*a);\n",
+      "print \"x2 = \", x2\n",
+      "x = x1-x2;\n",
+      "print \"so the range is x (mm)\", x\n",
+      "hS = math.pi*d2*10**-3;\n",
+      "A2 = math.pi*d2*10**-6*(x1+x2)/2;\n",
+      "pS = -0.4*Ps/A2;\n",
+      "pgS = 25*10**-3/1000;\n",
+      "S = hS*pS*pgS;\n",
+      "print \"sensitivity = \", S\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "x1 =  0.125\n",
+        "x2 =  0.075\n",
+        "so the range is x (mm) 0.05\n",
+        "sensitivity =  -20000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3  Page No : 373"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pi = 70.*10**3;\n",
+      "r = 0.4;\n",
+      "d2 = 1.6;\n",
+      "d1 = 0.75;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "a = (d2/d1**2);\n",
+      "x1 = (1.1-r)/(2*a);\n",
+      "print \"x1 = \", x1\n",
+      "r = 0.9;\n",
+      "x2 = (1.1-r)/(2*a);\n",
+      "print \"x2 = \", x2\n",
+      "x = x1-x2;\n",
+      "print \"so the range is x (mm)\", x\n",
+      "d = -2*a;\n",
+      "Wr = 12.5/Pi;\n",
+      "Wx = Wr/d;\n",
+      "print \"uncertainity in lacement (mm)\",Wx\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "x1 =  0.123046875\n",
+        "x2 =  0.03515625\n",
+        "so the range is x (mm) 0.087890625\n",
+        "uncertainity in lacement (mm) -3.13895089286e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4  Page No : 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "N = 12.;\n",
+      "lem = 0.644;\n",
+      "\n",
+      "# Calculations\n",
+      "d = N*lem/2;\n",
+      "\n",
+      "# Results\n",
+      "print \"difference between height of workpieces and pile of slip gauges (micro-meter)\", d\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "difference between height of workpieces and pile of slip gauges (micro-meter) 3.864\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5  Page No : 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "N = 5.;\n",
+      "lem = 546.*10**-9;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "d = ((2*N-1)*lem*10**6)/4;\n",
+      "print \"seperation distance between two surfaces(micro-meter)\", d\n",
+      "x = 75;\n",
+      "th = math.atan(d/x);\n",
+      "print \"angle of tilt\", th\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "seperation distance between two surfaces(micro-meter) 1.2285\n",
+        "angle of tilt 0.0163785352938\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6  Page No : 379"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "x = 20./12;\n",
+      "L = 50-10;\n",
+      "lem = 0.6;\n",
+      "\n",
+      "# Calculations\n",
+      "d = (L*lem)/(2*x);\n",
+      "\n",
+      "# Results\n",
+      "print \"difference in diameters of the rollers(micro-meter)\", d\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "difference in diameters of the rollers(micro-meter) 7.2\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7  Page No : 381"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "d = 4.5-2.5;\n",
+      "\n",
+      "# Calculations\n",
+      "Tg = 2*(0.5)*0.509;\n",
+      "\n",
+      "# Results\n",
+      "print \"change in thickness along its length(micro-meter)\", Tg\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in thickness along its length(micro-meter) 0.509\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch9.ipynb b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch9.ipynb
new file mode 100755
index 00000000..0bd1bdf2
--- /dev/null
+++ b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch9.ipynb
@@ -0,0 +1,863 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d5da213085937afdb48d7dfa573f842c4dd8a61f494a1ba421365a0cebd5fe2e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Pressure Measurements"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1  Page No : 383"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 273+20;\n",
+      "P = 101.3*10**3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "mfp = 22.7*10**-6*T/P;\n",
+      "print \"length of mean free path when pressure is one atmospheric pressure(m)\", mfp\n",
+      "P = 133;\n",
+      "mfp = 22.7*10**-6*T/P;\n",
+      "print \"length of mean free path when pressure is one torr(m)\", mfp\n",
+      "P = 133*10**-3;\n",
+      "mfp = 22.7*10**-6*T/P;\n",
+      "print \"length of mean free path when pressure is one micrometer of Hg(m)\", mfp\n",
+      "P = 249.1;\n",
+      "mfp = 22.7*10**-6*T/P;\n",
+      "print \"length of mean free path when pressure is one inch of water(m)\", mfp\n",
+      "P = 133*10**-6;\n",
+      "mfp = 22.7*10**-6*T/P;\n",
+      "print \"length of mean free path when pressure is 10**-3 micrometer of Hg(m)\", mfp\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "length of mean free path when pressure is one atmospheric pressure(m) 6.56574531096e-08\n",
+        "length of mean free path when pressure is one torr(m) 5.00082706767e-05\n",
+        "length of mean free path when pressure is one micrometer of Hg(m) 0.0500082706767\n",
+        "length of mean free path when pressure is one inch of water(m) 2.67005218788e-05\n",
+        "length of mean free path when pressure is 10**-3 micrometer of Hg(m) 50.0082706767\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2  Page No : 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "T = 273.+25;\n",
+      "P = 99.22*10**3;\n",
+      "R = 288.;\n",
+      "df = P/(R*T);\n",
+      "dm = 0.82*996;\n",
+      "g = 9.81;\n",
+      "h = 200*10**-6;\n",
+      "\n",
+      "# Calculations\n",
+      "P1 = g*h*(dm-df)*10**3;\n",
+      "Pa = P+P1;\n",
+      "\n",
+      "# Results\n",
+      "print \"Pressure of air source(N/m2)\", Pa\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure of air source(N/m2) 100820.136398\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3  Page No : 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "df = 1.*10**3;\n",
+      "dm = 13.56*10**3;\n",
+      "g = 9.81;\n",
+      "h = 130.*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "P = g*h*(dm-df);\n",
+      "Ph = P/9.81;\n",
+      "\n",
+      "# Results\n",
+      "print \"Pressure head(mm of water)\", Ph\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure head(mm of water) 1632.8\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4  Page No : 389"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "hn  = 250.;\n",
+      "d = 5.;\n",
+      "D = 25.;\n",
+      "\n",
+      "# Calculations\n",
+      "h = hn*(1+(d/D)**2);\n",
+      "\n",
+      "# Results\n",
+      "print \"height\", h\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height 260.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6  Page No : 391"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "P = 8.*133;\n",
+      "h = P/(800*9.81);\n",
+      "d = 2.;\n",
+      "D = 50.;\n",
+      "\n",
+      "# Calculations\n",
+      "hn = h/(1+(d/D)**2);\n",
+      "e = (hn-h)/h*100;\n",
+      "eP = 0.8*1000*9.81*(hn-h);\n",
+      "\n",
+      "# Results\n",
+      "print \"error interms of pressure(N/m2)\", eP\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "error interms of pressure(N/m2) -1.69968051118\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7  Page No : 393"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "P = 133.;\n",
+      "g = 9.81;\n",
+      "dm = 13.56*10**3;\n",
+      "R = 10.**-3;\n",
+      "d = 4.;\n",
+      "D = 20.;\n",
+      "\n",
+      "# Calculations\n",
+      "th = math.asin(P/(g*dm*R*(1+(d/D)**2)));\n",
+      "thV = 90-th;\n",
+      "\n",
+      "# Results\n",
+      "print \"angle to which tube is incliend to vertical(degree)\", thV\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angle to which tube is incliend to vertical(degree) 88.708070889\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8  Page No : 395"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "P = 9.81;\n",
+      "g = 9.81;\n",
+      "dm = 0.864*10**3;\n",
+      "R = 4.*10**-3;\n",
+      "d = 2.;\n",
+      "D = 20.;\n",
+      "\n",
+      "# Calculations\n",
+      "th = math.asin(P/(g*dm*R*(1+(d/D)**2)));\n",
+      "\n",
+      "# Results\n",
+      "print \"angle to which tube is incliend to horizontal(degree)\", th\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angle to which tube is incliend to horizontal(degree) 0.290558105825\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9  Page No : 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 500.*9.81;\n",
+      "g = 9.81;\n",
+      "d = 8.;\n",
+      "a =  (math.pi/4)*d**2;\n",
+      "A = 1200.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "dm = 0.8*10**3;\n",
+      "hn = P/(g*dm*(1+(a/A)));\n",
+      "print \"Length of scale(m)\", hn\n",
+      "R = 0.6;\n",
+      "P1 = 50*9.81;\n",
+      "th = math.asin(P1/(g*dm*R*(1+(a/A))));\n",
+      "print \"angle to which tube is incliend to horizontal(degree)\", th\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of scale(m) 0.599872595479\n",
+        "angle to which tube is incliend to horizontal(degree) 0.100146080124\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10  Page No : 399"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 100.*10**3;\n",
+      "g = 9.81;\n",
+      "di = 10.*10**-3;\n",
+      "D = 40.*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "A =  (math.pi/4)*D**2;\n",
+      "dm = 13.6*10**3;\n",
+      "a = A/(P/(dm*g*di)-1);\n",
+      "d = (4*a/math.pi)**0.5*10**3;\n",
+      "\n",
+      "# Results\n",
+      "print \"diameter of the tube(mm)\", d\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diameter of the tube(mm) 4.65136628261\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11  Page No : 401"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Calculations and Results\n",
+      "AR = 1./(0.83-0.8);\n",
+      "print \"Amplification ratio\", AR\n",
+      "D = 50.*10**-3;\n",
+      "A =  (math.pi/4)*D**2;\n",
+      "d = 6.*10**-3;\n",
+      "a =  (math.pi/4)*d**2;\n",
+      "PR = (a/A)*100;\n",
+      "print \"percentage error\", PR\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amplification ratio 33.3333333333\n",
+        "percentage error 1.44\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12  Page No : 403"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 981;\n",
+      "g = 9.81;\n",
+      "d = 500.*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "A =  (math.pi/4)*(10*10**-3)**2;\n",
+      "R = 275*10**-3;\n",
+      "th = 30;\n",
+      "W = A*d*P/(2*g*R*math.sin(math.radians(th)));\n",
+      "\n",
+      "# Results\n",
+      "print \"value of counter weight required(kg)\", W\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of counter weight required(kg) 0.0142799666072\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.13  Page No : 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Mp1 = 20./40;\n",
+      "Mp2 = 10./40;\n",
+      "Mp3 = 5./40;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Eta = 0.225;\n",
+      "print \"damping factor\", Eta\n",
+      "Td = 1.2;\n",
+      "wd = 2*math.pi/Td;\n",
+      "wn = wd/((1-Eta**2)**0.5);\n",
+      "tc = 1/wn;\n",
+      "print \"time constant(s)\", tc\n",
+      "ess = 2*Eta/wn;\n",
+      "ess5 = 5*ess;\n",
+      "print \"error for 5mm/s ramp(mm)\", ess5\n",
+      "Tlag = 2*Eta*tc;\n",
+      "print \"time lag(s)\", Tlag\n",
+      "Eta1 = Eta*(0.5)**0.5;\n",
+      "print \"New damping factor\", Eta1\n",
+      "Td = 1.2;\n",
+      "wn1  =  wn*(0.5)**0.5;\n",
+      "print \"New natural frequency(rad/s)\", wn1\n",
+      "tc1 = 1/wn;\n",
+      "print \"New time constant(s)\", tc1\n",
+      "ess51 = ess5;\n",
+      "print \"new error for 5mm/s ramp(mm)\", ess51\n",
+      "Tlag1 = Tlag;\n",
+      "print \"new time lag(s)\", Tlag1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "damping factor 0.225\n",
+        "time constant(s) 0.186088816266\n",
+        "error for 5mm/s ramp(mm) 0.418699836599\n",
+        "time lag(s) 0.0837399673198\n",
+        "New damping factor 0.159099025767\n",
+        "New natural frequency(rad/s) 3.79983491418\n",
+        "New time constant(s) 0.186088816266\n",
+        "new error for 5mm/s ramp(mm) 0.418699836599\n",
+        "new time lag(s) 0.0837399673198\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.14  Page No : 407"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 7.*10**6;\n",
+      "R = 6.25*10**-3;\n",
+      "v = 0.28;\n",
+      "E = 200.*10**9;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "t = ((9*P*R**4*(1-v**2)/(16*E))**0.25)*10**3;\n",
+      "print \"thickness of diaphram(mm)\", t\n",
+      "ds = 7800;\n",
+      "fn = (2.5*t/(math.pi*R**2))*(E/(3*ds*(1-v**2)))**0.5;\n",
+      "print \"natural frequency(Hz)\", fn\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thickness of diaphram(mm) 0.407908989726\n",
+        "natural frequency(Hz) 25306332.7381\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.15  Page No : 409"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 100.*10**3;\n",
+      "A = 1500.*10**-6;\n",
+      "F = P*A;\n",
+      "Cs = F/3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Ls = Cs+40;\n",
+      "print \"natural length of spring(mm)\", Ls\n",
+      "P1 = 10*10**3;\n",
+      "F1 = P1*A;\n",
+      "Ss = 3+2*.5;\n",
+      "D = F1/Ss;\n",
+      "print \"lacement (mm)\",D\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "natural length of spring(mm) 90.0\n",
+        "lacement (mm) 3.75\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.16  Page No : 411"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 200*10**3;\n",
+      "R = 70*10**-3;\n",
+      "v = 0.25;\n",
+      "t = 1*10**-3;\n",
+      "r = 60*10**-3;\n",
+      "E = 200*10**9;\n",
+      "\n",
+      "# Calculations\n",
+      "Sr = (3*P*R**2*v/(8*t**2))*((1/v+1)-(3/v+1)*(r/R)**2);\n",
+      "St = (3*P*R**2*v/(8*t**2))*((1/v+1)-(1/v+3)*(r/R)**2);\n",
+      "Sta2 = (Sr-v*St)/E;\n",
+      "Sta3 = (Sr-v*St)/E;\n",
+      "r0 = 10*10**-3;\n",
+      "Sr1 = (3*P*R**2*v/(8*t**2))*((1/v+1)-(3/v+1)*(r0/R)**2);\n",
+      "St1 = (3*P*R**2*v/(8*t**2))*((1/v+1)-(1/v+3)*(r0/R)**2);\n",
+      "Sta1 = (Sr1-v*St1)/E;\n",
+      "Sta4 = (Sr1-v*St1)/E;\n",
+      "Gf = 1.8;\n",
+      "ei = 12;\n",
+      "eo = (Sta1+Sta4-Sta2-Sta3)*Gf*ei/4;\n",
+      "\n",
+      "# Results\n",
+      "print \"output voltage (V)\", eo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage (V) 0.0398671875\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.17  Page No : 413"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Aou = 700.*25*1/100;\n",
+      "Aol = 100.*25*1/100;\n",
+      "AouPtP =  2*Aou;\n",
+      "AolPtP =  2*Aol;\n",
+      "Se1 = 1.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "D1 = AouPtP/Se1;\n",
+      "print \"deflection of screen corresponding to maximum pressure for sensitivity of 1mV/mm (mm)\", D1\n",
+      "print \"since the length of the screen is 100mm so waveform is out of range and hence sensitivity setting of 1mV/mm should not be used\",\n",
+      "Se2 = 5;\n",
+      "D2 = AouPtP/Se2;\n",
+      "print \"deflection of screen corresponding to maximum pressure for sensitivity of 5mV/mm (mm)\", D2\n",
+      "print \"delection is within the range\",\n",
+      "Se3 = 20;\n",
+      "D3 = AouPtP/Se3;\n",
+      "print \"deflection of screen corresponding to maximum pressure for sensitivity of 20mV/mm (mm)\", D3\n",
+      "print \"delection is within the range\",\n",
+      "Se4 = 100;\n",
+      "D4 = AouPtP/Se4;\n",
+      "print \"deflection of screen corresponding to maximum pressure for sensitivity of 100mV/mm (mm)\", D4\n",
+      "print \"delection is within the range\",\n",
+      "Se5 = 500;\n",
+      "D5 = AouPtP/Se5;\n",
+      "print \"deflection of screen corresponding to maximum pressure for sensitivity of 500mV/mm (mm)\", D5\n",
+      "print \"delection is within the range\",\n",
+      "print \"since the sensitivity of 5mV/mm gives higher deflection so it is the optimum sensitivity\",\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "deflection of screen corresponding to maximum pressure for sensitivity of 1mV/mm (mm) 350.0\n",
+        "since the length of the screen is 100mm so waveform is out of range and hence sensitivity setting of 1mV/mm should not be used deflection of screen corresponding to maximum pressure for sensitivity of 5mV/mm (mm) 70.0\n",
+        "delection is within the range deflection of screen corresponding to maximum pressure for sensitivity of 20mV/mm (mm) 17.5\n",
+        "delection is within the range deflection of screen corresponding to maximum pressure for sensitivity of 100mV/mm (mm) 3.5\n",
+        "delection is within the range deflection of screen corresponding to maximum pressure for sensitivity of 500mV/mm (mm) 0.7\n",
+        "delection is within the range since the sensitivity of 5mV/mm gives higher deflection so it is the optimum sensitivity\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.18  Page No : 415"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "dP = (7000*10**3)-(100*10**3);\n",
+      "b = 25*10**-12;\n",
+      "R1 = 100;\n",
+      "dR = R1*b*dP;\n",
+      "ei = 5.;\n",
+      "\n",
+      "# Calculations\n",
+      "deo = dR*ei/(4*R1)\n",
+      "\n",
+      "# Results\n",
+      "print \"output voltage of bridge(V)\", deo\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "output voltage of bridge(V) 0.000215625\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.19  Page No : 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 273+20;\n",
+      "P = 101.3*10**3;\n",
+      "R = 287;\n",
+      "de = P/(R*T);\n",
+      "C = 20.04*T**0.5;\n",
+      "r = 6.25*10**-3;\n",
+      "L = 0.6;\n",
+      "\n",
+      "# Calculations\n",
+      "V = math.pi*((12.5*10**-3)**2)*(12.5*10**-3);\n",
+      "wn = C*r*(math.pi/(V*(L+0.5*math.pi*r)))**0.5;\n",
+      "fn = wn/(2*math.pi);\n",
+      "f = 1000;\n",
+      "u = f/fn;\n",
+      "mu = 19.1*10**-6;\n",
+      "eta = (2*mu/(de*C*r**3))*(3*L*V/math.pi)**0.5;\n",
+      "M = 1/(((1-u**2)**2)+((2*eta*u)**2))**0.5;\n",
+      "M = M*100;\n",
+      "\n",
+      "# Results\n",
+      "print \"attenuation = \", M\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "attenuation =  10.834485951\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.20  Page No : 419"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "d = 1;\n",
+      "At = (math.pi*d**2)*10**-6/4;\n",
+      "V = 100*10**-6;\n",
+      "h = 30*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "P1 = (At*h**2)/V;\n",
+      "P2 = (At*h**2)/(V-At*h);\n",
+      "e = P2-P1;\n",
+      "\n",
+      "# Results\n",
+      "print \"error =  \", e\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "error =   1.66588825836e-09\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/ch7.png b/A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/ch7.png
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diff --git a/A_Textbook_of_Electrical_and_Electronics_Engineering_Materials_by_P._l._Kapoor/README.txt b/A_Textbook_of_Electrical_and_Electronics_Engineering_Materials_by_P._l._Kapoor/README.txt
new file mode 100755
index 00000000..54bee6ce
--- /dev/null
+++ b/A_Textbook_of_Electrical_and_Electronics_Engineering_Materials_by_P._l._Kapoor/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Jai Mathur
+Course: mca
+College/Institute/Organization: IIT
+Department/Designation: System Admin
+Book Title: A Textbook of Electrical and Electronics Engineering Materials
+Author: P. l. Kapoor
+Publisher: Khanna Publishers, New Delhi
+Year of publication: 2010
+Isbn: 81-7409-131-9
+Edition: 8
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo1.ipynb b/Advance_Semiconductor_Devices/ChapterNo1.ipynb
deleted file mode 100755
index 4d1dc38d..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo1.ipynb
+++ /dev/null
@@ -1,800 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chapter No 1 : Semiconductor Materials and\n",
-      "Their Properties<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3, Page No 37 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Finding Density of silicon crystal\n",
-      "\n",
-      "#initialisation of variables\n",
-      "a=5.3   #in \u00c5\n",
-      "a=a*10**-10     #in m\n",
-      "N_A=6.023*10**23  \n",
-      "At_Si=28    #atomic weight of Si\n",
-      "n = 4\n",
-      "m=At_Si/N_A  #in gm\n",
-      "m= m*10**-3  #in kg\n",
-      "V=a**3       #in m^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Rho=((m*n)/V)/1000  #in kg/m^3\n",
-      "\n",
-      "# Note: There is calculation error to find the value of density. So the answer in the book is wrong.#RESULTS\n",
-      "#RESULTS\n",
-      "print('Density of silicon crystal in kg/m^3 is = %.2f kg/m^3' %Rho)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Density of silicon crystal in kg/m^3 is = 1.25 kg/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4, Page No 37 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Finding Density of the crystal \n",
-      "\n",
-      "#initialisation of variables\n",
-      "n = 4\n",
-      "r = 1.278  #in \u00c5\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "a = (4*r)/(math.sqrt(2)) #in\u00c5\n",
-      "a = a * 10**-10   #in m\n",
-      "V = (a)**3  #in m^3\n",
-      "At_W = 63.5   #atomic weight\n",
-      "N_A = 6.023*10**23\n",
-      "m = At_W /N_A   #in gm\n",
-      "m = m*10**-3  #in kg\n",
-      "Rho = ((m*n)/V)/1000  #in kg/m^3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Density of the crystal in kg/m^3 is =%.2f kg/m^3' %Rho)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Density of the crystal in kg/m^3 is =8.93 kg/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5 Page No 43 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Finding Wavaelength of X ray\n",
-      "\n",
-      "import math\n",
-      "from numpy import *\n",
-      "#initialisation of variables\n",
-      "d=2.82   #in \u00c5\n",
-      "d=d*10**-10    #in m\n",
-      "n=1\n",
-      "theta1=10  #n degree\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "lembda=2*d*math.sin(math.radians(theta1))   #in m\n",
-      "lembda=lembda*10**10       #in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Wavaelength of X ray  is =%.2f \u00c5 degrees' %lembda)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavaelength of X ray  is =0.98 \u00c5 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6, Page No 43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Finding The spacing of atomic layer\n",
-      "\n",
-      "import math\n",
-      "#initialisation of variables\n",
-      "lembda = 1.6  #in \u00c5\n",
-      "theta = 14.2  #in degree\n",
-      "n = 1\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d = (n*lembda)/(2*math.sin(math.radians(theta)))   #in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The spacing of atomic layer in crystal is =%.2f in \u00c5 degrees' %d)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing of atomic layer in crystal is =3.26 in \u00c5 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7 Page No 43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Finding The interplanner spacing\n",
-      "\n",
-      "import math\n",
-      "#initialisation of variables\n",
-      "n = 1\n",
-      "theta1 = 30  #in degree\n",
-      "lembda = 1.78  #in \u00c5\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d = (n*lembda)/(2*math.sin(math.radians(theta1)))  #in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The interplanner spacing is =%.2f in \u00c5 degrees' %d)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The interplanner spacing is =1.78 in \u00c5 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8 Page No 43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Finding Interplaner spacing of the crystal\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "lembda = 0.58  #in \u00c5\n",
-      "n = 1;\n",
-      "theta1 = 6.45   # in degree\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d = (n*lembda)/(2*math.sin(math.radians(theta1))) #in \u00c5 \n",
-      "print('Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal  is =%.2f \u00c5' %d)\n",
-      "theta2 = 9.15  #in degree\n",
-      "d1 = (n*lembda)/(2*math.sin(math.radians(theta2))) #in \u00c5 \n",
-      "print('Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d1)\n",
-      "theta3 = 13 #in degree\n",
-      "n2 = 1\n",
-      "d2 = (n2*lembda)/(2*math.sin(math.radians(theta3)))  #in \u00c5\n",
-      "print('Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d2)\n",
-      "n=2\n",
-      "d2 = (n*lembda)/(2*math.sin(math.radians(theta3)))  #in \u00c5 \n",
-      "\n",
-      "#RESULTS\n",
-      "print('Part (iv) : : The interplaner spacing is : =%.4f \u00c5' %d2)\n",
-      "print('The interplaner spacing for some other set of reflecting  is : =%.2f \u00c5' %d1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal  is =2.58 \u00c5\n",
-        "Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =1.82 \u00c5\n",
-        "Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =1.29 \u00c5\n",
-        "Part (iv) : : The interplaner spacing is : =2.5783 \u00c5\n",
-        "The interplaner spacing for some other set of reflecting  is : =1.82 \u00c5\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9, Page No 44 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Finding The glancing angle for a cubic in degree\n",
-      "\n",
-      "import math\n",
-      "from sympy.mpmath import *\n",
-      "import cmath\n",
-      "\n",
-      "#initialisation of variables\n",
-      "a = 2.814 #in \u00c5\n",
-      "l = 0\n",
-      "h = l\n",
-      "k = 0\n",
-      "d=a  #in \u00c5\n",
-      "n = 2\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "lembda=0.710     #in \u00c5\n",
-      "theta = math.degrees(math.asin(n*lembda/(2*d)))\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The glancing angle for a cubic in degree is : =%.2f  \u00c5 degrees' %theta)\n",
-      "print('Anwser is in points i.e = 14 36\"40\"..')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The glancing angle for a cubic in degree is : =14.61  \u00c5 degrees\n",
-        "Anwser is in points i.e = 14 36\"40\"..\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10 Page No 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Finding Wavelength of X ray\n",
-      "\n",
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "a = 3.65  #in \u00c5\n",
-      "a = 3.65*10**-10  #in m\n",
-      "h = 1\n",
-      "k = 0\n",
-      "l = 0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d= a/(math.sqrt(h**2+k**2+l**2))   #in m\n",
-      "n = 1\n",
-      "theta=60  #in degree\n",
-      "lembda = 2*d*math.sin(math.radians(theta))   #in m\n",
-      "lembda = lembda * 10**10   #in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Wavelength of X ray is : =%.2f  \u00c5' %lembda)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of X ray is : =6.32  \u00c5\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11 Page No 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Finding The glancing angle in degree\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "lembda = 1.54   #in \u00c5\n",
-      "density = 9.024   #in gm/cc\n",
-      "n = 1\n",
-      "MI = 100\n",
-      "At_W = 63.54   #atomic weight\n",
-      "N_A = 6.023*10**23   \n",
-      "m = At_W/N_A   #in gm\n",
-      "a =(density*m)**(1.0/3)  #in cm\n",
-      "h = 1\n",
-      "k = 0\n",
-      "l = 0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d= a/(math.sqrt(h**2+k**2+l**2))\n",
-      "theta =math.degrees((lembda*10**-8)/(2*d))   #in degree\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The glancing angle is = %.2f degrees' %theta)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The glancing angle is = 4.48 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12 Page No 45 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Finding To get the 2nd order spectrum the position of the detector in degree\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "a = 3.615   #in \u00c5\n",
-      "theta=22   #in degree\n",
-      "n=1\n",
-      "h=1\n",
-      "k=h\n",
-      "l=k\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d = a/(math.sqrt( ((h)**2) + ((k)**2) + ((l)**2) ))   #in \u00c5\n",
-      "lembda = 2*d*math.sin(math.radians(theta))   #in \u00c5\n",
-      "print('The wavelength of X ray is = %.2f \u00c5 ' %lembda)\n",
-      "theta2 =math.degrees(math.asin(lembda/d))  #in degree\n",
-      "             #in degree\n",
-      "\n",
-      "#RESULTS\n",
-      "print('To get the 2nd order spectrum the position of the detector  is = %.2f degrees ' %theta2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength of X ray is = 1.56 \u00c5 \n",
-        "To get the 2nd order spectrum the position of the detector  is = 48.52 degrees \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13, Page No 45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "n = 1\n",
-      "lembda = 1.54      # in \u00c5\n",
-      "theta = 21.7      # in degree\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d = lembda/(2*math.sin(math.radians(theta)))     # in \u00c5\n",
-      "h = 1\n",
-      "k = h\n",
-      "l = k\n",
-      "a = d*math.sqrt(h**2+k**2+l**2)      # in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Lattice constant is = %.2f  in \u00c5 ' %a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Lattice constant is = 3.61  in \u00c5 \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.14, Page No 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "h = 2\n",
-      "k = 1\n",
-      "l = 1\n",
-      "a = 4.8      # in \u00c5\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "d_211 = a/(math.sqrt(h**2+k**2+l**2))       #in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The distance between planes is =%.2f \u00c5 ' %d_211)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The distance between planes is =1.96 \u00c5 \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.15, Page No 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "r = 1.28       # in \u00c5\n",
-      "n = 4\n",
-      "M = 63.5\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "a = (4*r)/(math.sqrt(2))        #in \u00c5\n",
-      "a = a * 10**-8       # in cm\n",
-      "N_A = 6.023*10**23     \n",
-      "Rho = (n*M)/( N_A*((a)**3) )       # in gm/cc\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Density is =%.2f gm/cc  ' %Rho)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Density is =8.89 gm/cc  \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.16, Page No 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "M = 55.85\n",
-      "a = 2.9        # in \u00c5\n",
-      "Rho = 7.87    #in gm/cc\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "a = a * 10**-8    # in cm\n",
-      "N_A = 6.023*10**23\n",
-      "n = (Rho*N_A*((a)**3))/M            # atom per unit\n",
-      "\n",
-      "#RESULTS\n",
-      "print('A lattice having =%.2f atom per unit cell is a BCC structure ' %n)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A lattice having =2.07 atom per unit cell is a BCC structure \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.17, Page No 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "M = 60.0           # in gm/mole\n",
-      "Rho = 6.23      # in gm/cc\n",
-      "n = 4\n",
-      "N_A = 6.023*10**23 \n",
-      "\n",
-      "#CALCULATIONS\n",
-      "a = ((n*M)/(N_A * Rho))**(1.0/3)         # in cm\n",
-      "r = (a*math.sqrt(2))/n               #radius of atom in cm\n",
-      "r = r * 10**8            # in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Radius of atom is =%.2f \u00c5  ' %r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius of atom is =1.41 \u00c5  \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.18, Page No 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Rho = 5.96          # in gm/cc\n",
-      "M = 50.0\n",
-      "n = 2\n",
-      "N_A = 6.023*10**23 \n",
-      "\n",
-      "#CALCULATIONS\n",
-      "a =((n*M)/(Rho*N_A))**(1.0/3)          # in cm\n",
-      "r = (a*math.sqrt(3))/4                 # in cm\n",
-      "P_f = (2*(4.0/3)*math.pi*((r)**3))/((a)**3)       # packing factor\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Packing factor is =%.2f  ' %P_f)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Packing factor is =0.68  \n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.19, Page No 48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "M = 120.0\n",
-      "n = 2.0\n",
-      "N_A = 6.023*10**23\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "m1 = M/N_A              #mass of 1 atom in gm\n",
-      "m2 = n*m1               #mass of unit cell in gm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Number of unit cell in 20 gms of element is : =%.2f X 10^22 unit cells ' %((20/m2)/(10**22)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of unit cell in 20 gms of element is : =5.02 X 10^22 unit cells \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.20, Page No 48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Rho = 2.48         # in gm/c.c\n",
-      "n = 4.0 \n",
-      "M = 58\n",
-      "N_A = 6.023*10**23\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "a = ( (n*M)/(Rho*N_A) )**(1.0/3)      # in cm\n",
-      "a = a * 10**8        # in \u00c5\n",
-      "r = (a*math.sqrt(2))/n      # in \u00c5\n",
-      "r = 2*r       # in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The center to center distance between ions  is =%.2f  \u00c5' %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The center to center distance between ions  is =3.80  \u00c5\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo2.ipynb b/Advance_Semiconductor_Devices/ChapterNo2.ipynb
deleted file mode 100755
index 9bc8cffb..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo2.ipynb
+++ /dev/null
@@ -1,1248 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chapter No 2 : Semiconductor Materials and\n",
-      "Their Properties<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1, Page No 65 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "lambda1 = 11000.0  #in \u00c5\n",
-      "lambda1 = lambda1 * 10**-10 \n",
-      "h = 6.625*10**-34 \n",
-      "c = 3*10**8\n",
-      "q = 1.6*10**-19    #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "E_g = h*c/lambda1  #in J\n",
-      "E_g= E_g/q        #in eV\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The energy gap in Si  is = %.2f eV' %E_g)\n",
-      "\n",
-      "#Note: The answer in the book is not correct \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy gap in Si  is = 1.13 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page No 65 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "E_g = 0.75    #in eV\n",
-      "q=1.6*10**-19   #in C\n",
-      "E_g = E_g*q   #in J\n",
-      "h = 6.63*10**-34   #in J\n",
-      "c = 3*10**8    #in m/s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "lembda = (h*c)/E_g    #in m\n",
-      "lembda = lembda * 10**10   #in \u00c5\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The wavelength is =%.f  \u00c5 ' %lembda)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is =16575  \u00c5 \n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3 Page No 81 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from numpy import *\n",
-      "\n",
-      "#initialisation of variables\n",
-      "del_E = 0.3        #value of E_C-E_F in eV\n",
-      "T1 = 330.0          #in K\n",
-      "T = 300            #in K\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "del_E1 = del_E*(T1/T)   #value of E_C-E_F in eV\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The position of fermi level   is =%.2f eV' %del_E1)\n",
-      "print('Hence the Fermi level will be %.2f eV below the conduction band' %del_E1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The position of fermi level   is =0.33 eV\n",
-        "Hence the Fermi level will be 0.33 eV below the conduction band\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4, Page No 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "K = 8.63*10**-5\n",
-      "T = 300.0     #in K\n",
-      "N_C = 2.8*10**19     #in cm^-3\n",
-      "del_E = 0.25\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "f_F = math.exp( (-del_E)/(K*T) )\n",
-      "n_o = N_C*math.exp( (-del_E)/(K*T) )    #in cm^-3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The probability is =%.6f ' %f_F)\n",
-      "print('The thermal equillibrium electron concentration  is =%.2f X 10^15 cm^-3' %(n_o/(10**15)))\n",
-      "print('Approximation  error ')\n",
-      "# The answer is same as in book only thy have rounded off to 1.8 * 10^15 which is same as above"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The probability is =0.000064 \n",
-        "The thermal equillibrium electron concentration  is =1.79 X 10^15 cm^-3\n",
-        "Approximation  error \n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5 Page No 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N_V = 1.04*10**19    #in cm^-3\n",
-      "T1 = 400.0        #in K\n",
-      "T2 = 300.0        #in K\n",
-      "del_E = 0.27     #value of E_F-E_V in eV\n",
-      "K = 0.0259\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N_V= N_V*(T1/T2)**(3/2)      #in cm^-3\n",
-      "KT = K*(T1/T2)              #in eV\n",
-      "p_o = N_V*math.exp( (-del_E)/(KT) )    #in /cm^3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The hole concentration is =%.2f x 10^15 per cm^3' %(p_o/(10**15)))\n",
-      "print('Approximationa error ')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The hole concentration is =5.58 x 10^15 per cm^3\n",
-        "Approximationa error \n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6 Page No 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N = 6.02*10**23\n",
-      "A = 63.5            #atomic weight\n",
-      "Rho = 1.7*10**-6    #in ohm cm\n",
-      "d = 8.96           # in gm/cc\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n = (N/A)*d     #in /cc\n",
-      "e = 1.6*10**-19    #in C\n",
-      "Miu_e = 1/(Rho*n*e)    #in cm^2/volt-sec\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The mobility of electron  is =%.2f  cm^2/volt.sec' %Miu_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mobility of electron  is =43.28  cm^2/volt.sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.7, Page No 87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from sympy.mpmath import *\n",
-      "import cmath\n",
-      "\n",
-      "\n",
-      "#initialisation of variables\n",
-      "d = 8.96   #in gm/cc\n",
-      "At = 63.5   #atomic weight\n",
-      "N_A = 6.02*10**23   #in /gm mole\n",
-      "l = 0.1 #in m\n",
-      "e = 1.6*10**-19   #in C\n",
-      "A = 1.7*10**-6   #in m^2\n",
-      "R = 0.1        #in ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n = (N_A/At)*d   #in /cc\n",
-      "n = n * 10**6   #in /m^3\n",
-      "Rho = (R*A)/l   #in ohm.m\n",
-      "Sigma = 1/Rho   #in mho/m\n",
-      "Miu_e = Sigma/(n*e)     #in m^2/V-sec\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The electron mobility is =%.3f X 10^-3 m^2/V-sec ' %(Miu_e*(10**3)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electron mobility is =0.043 X 10^-3 m^2/V-sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.8 Page No 87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "N_A = 6.025*10**23    #in /gm mole\n",
-      "d = 10.5             #in gm/cc\n",
-      "At = 108.0         #atomic weight of \n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n = (N_A/At)*d   #in /cm^3\n",
-      "n = n * 10**6   #in /m^3\n",
-      "r = 10**-3        #in m\n",
-      "A = math.pi * ((r)**2)   #in m^2\n",
-      "q = 1.6*10**-19     \n",
-      "I = 2       #in A\n",
-      "V = I/(n*q*A)      #in m/s\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The drift velocity of an electron  is =%.f X 10^-4 m/s' %(V*(10**5)))\n",
-      "print('Approximationa error ')\n",
-      "# answer is same as in book only they have rounded of 0.000068 to 7 * 10^-4"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drift velocity of an electron  is =7 X 10^-4 m/s\n",
-        "Approximationa error \n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.9 Page No 88"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "e= 1.6*10**-19   #in C\n",
-      "d= 1.03      #in mm\n",
-      "d= d*10**-3   #in m\n",
-      "R= 6.51    #in ohm\n",
-      "l= 300.0  #in m\n",
-      "n= 8.4*10**28   #per m^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r= d/2     #in m (radius)\n",
-      "A= math.pi*r**2   #in m^2\n",
-      "rho= R*A/l     #in ohm meter\n",
-      "sigma=  1/rho    #in mho/m\n",
-      "miu_e= sigma/(n*e)    #m^2/V-sec\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The coductivity of copper is =%.2f  mho/m' %sigma)\n",
-      "print('The mobility of charge carriers  is : =%.6f  m^2/V-sec' %miu_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The coductivity of copper is =55306469.41  mho/m\n",
-        "The mobility of charge carriers  is : =0.004115  m^2/V-sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.9.ii Page No 88"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Miu_e = 1500.0     #in cm^2/volt.sec\n",
-      "Miu_h = 500.0      #in cm^2/volt.sec\n",
-      "n_i = 1.6*10**10  #in /cm^3\n",
-      "e = 1.6*10**-19   #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma_i = n_i*(Miu_e+Miu_h)*e   #in mho/cm\n",
-      "Sigma = Sigma_i                 #in mho/cm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The conductivity of pure silicon  is =%.7f  mho/cm' %Sigma)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of pure silicon  is =0.0000051  mho/cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 2.10 Page No 89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "Miu_e = 1500.0  #in cm^2/volt.sec\n",
-      "Miu_h = 500.0  #in cm^2/volt.sec\n",
-      "n_i = 1.6*10**10   #in /cm^3\n",
-      "e = 1.6*10**-19    #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma_i = n_i*(Miu_e+Miu_h)*e    #in mho/cm\n",
-      "Sigma = Sigma_i              #in mho/cm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The conductivity of pure silicon  is =%.2f X 10^-6  mho/cm' %(Sigma*(10**6)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of pure silicon  is =5.12 X 10^-6  mho/cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 2.11 Page No 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "Miu_d = 500.0   #in cm^2/V.S\n",
-      "Rho = 10       #in ohm cm \n",
-      "\n",
-      "#CALCULATIONS\n",
-      "e = 1.6*10**-19   #in C\n",
-      "n_d = 1/(Rho*e*Miu_d)     #in /cm^3... correction\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The number of donor  is =%.2f X 10^15 atom per cm^3' %(n_d/(10**15)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The number of donor  is =1.25 X 10^15 atom per cm^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 2.12 Page No 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "d = 5.32         #in gm/cc\n",
-      "N_A = 6.02*10**23   #in atoms/gm.mole\n",
-      "At = 72.6       #atomic weight\n",
-      "Miu = 3800    #in cm^2/v.s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n_d = (N_A/At) * d          #in /cm^3\n",
-      "n_d = n_d * 10**-8    #in /cc\n",
-      "e = 1.6*10**-19    #in C\n",
-      "Sigma = n_d * Miu * e           #in mho/cm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The conductivity of specimen is =%.2f  mho/cm' %Sigma)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of specimen is =0.27  mho/cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 2.13 Page No 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "Rho = 0.3623*10**-3    #in ohm m\n",
-      "d = 4.42*10**28      #Ge density in atoms/m^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma = 1/Rho       #in mho/m\n",
-      "n_d = d*10**-6    #in atoms/m^3\n",
-      "e = 1.6*10**-19    #in C\n",
-      "Miu = Sigma/(n_d*e)    #in m^2/V.sec\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The electron mobility is =%.2f  m^2/V-sec' %Miu)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electron mobility is =0.39  m^2/V-sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 2.14 Page No 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "N_A = 6.025*10**26    #in /kg.Mole\n",
-      "At = 72.59          #atomic weight\n",
-      "d = 5.36*10**3    #in kg/m^3\n",
-      "R = 0.42            #in ohm m\n",
-      "B_i = 10**-6    #rate of boron impurity in %\n",
-      "e = 1.6*10**-19     #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n = (N_A/At)*d    #number of Ge atoms\n",
-      "h = n/10**8      #holes per unit volume\n",
-      "Miu_h = 1/(R*h*e)       #in m^2/V.sec\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The Mobility of holes is =%.2f m^2/V-sec' %Miu_h)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Mobility of holes is =0.03 m^2/V-sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.15 Page No 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "n_i = 2*10**19     #in /m^3\n",
-      "Miu_e = 0.36    #in m^2/v.s\n",
-      "Miu_h = 0.17   #in m^2/v.s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "A = 1*10**-4     #in m^2\n",
-      "V = 2            #in Volts\n",
-      "l = 0.3         #in mm\n",
-      "l = l * 10**-3   #in m\n",
-      "e = 1.6*10**-19   #in C\n",
-      "Sigma_i = n_i * e * (Miu_e+Miu_h)   #in mho/m\n",
-      "I = (Sigma_i * V*A)/l        #in amp \n",
-      "\n",
-      "#RESULTS\n",
-      "print('The current in amp is =%.2f ' %I)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current in amp is =1.13 \n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.16 Page No 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "d = 4.2*10**28   #in atoms/m^3\n",
-      "n_d = d/10**6   #in atoms/m^3\n",
-      "e = 1.6*10**-19   #in C\n",
-      "Miu_e = 0.36    #in m^2/V-sec\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma_n = n_d *e *Miu_e    #in mho/m\n",
-      "Rho_n = 1/Sigma_n          #ohm m\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The resistivity in \u03a9m is =%.2f X 10^-3' %(Rho_n*(10**3)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistivity in \u03a9m is =0.41 X 10^-3\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.17 Page No 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "n_i = 1*10**19        #in /m^3\n",
-      "Miu_e = 0.36     #in m^2/volt.sec\n",
-      "Miu_h = 0.17      #in m^2/volt.sec\n",
-      "A = 2                #in cm^2\n",
-      "A = A *  10**-4    #in m^2\n",
-      "t = 0.1         #in mm\n",
-      "t = t*10**-3   #in m\n",
-      "V = 4        #in volts\n",
-      "e = 1.6*10**-19   #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma_i = n_i * e * (Miu_e + Miu_h)   #mho/m\n",
-      "J = Sigma_i * (V/t)             #in Amp/m^2\n",
-      "I = J*A                      #in Amp\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The current in Amp is =%.2f ' %I)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current in Amp is =6.78 \n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 2.18 Page No 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "Miu_h = 500.0                  #in cm^2/V.s\n",
-      "Miu_e = 1500.0                 #in cm^2/V.s\n",
-      "n_i = 1.6*10**10              #in /cm^3\n",
-      "e = 1.6*10**-19               #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma_i = n_i * e * (Miu_e+Miu_h)       #in mho/cm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The conductivity of pure silicon in mho/cm is =%.2f X 10^-6 ' %(Sigma_i*(10**6)))\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of pure silicon in mho/cm is =5.12 X 10^-6 \n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.19 Page No 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "Si_density = 10.5                #in gm/cc\n",
-      "N_A = 6.025*10**23         \n",
-      "At = 108.0                  #atomic weight\n",
-      "B = 0.8                   #in Tesla\n",
-      "w = 0.50               #in cm\n",
-      "w = w * 10**-2          #in m\n",
-      "t = 0.10              #in mm\n",
-      "t = t * 10**-3       #in m\n",
-      "A = w*t                 #in m^2\n",
-      "q = 1.6*10**-19          #in C\n",
-      "I = 2                    #in ampere\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n = (N_A/At) * Si_density  #in /cc\n",
-      "n = n * 10**6              #in /m^3\n",
-      "V_H = (B*I*t)/(n*q*A)           #in volts\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The hall voltage produced is =%.3f X 10^-7 volts' %(V_H*(10**7)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The hall voltage produced is =0.341 X 10^-7 volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.20 Page No 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "Sigma = 5.8*10**7        #in mho/m\n",
-      "l = 1                     #in m\n",
-      "d = 1                  #in cm\n",
-      "d = d * 10**-2          #in m\n",
-      "W = 1               #in mm \n",
-      "W = W*10**-3       #in m\n",
-      "I = 1             #in Amp\n",
-      "B = 1            #in Tesla\n",
-      "V_H = 0.074*10**-6        #in Volts\n",
-      "A = 10**-2 * 10**-3          #in m^2\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "R_H = (V_H*A)/(B*I*d)              #in m^3/c\n",
-      "Miu = Sigma * R_H                    #in m^2/volt.sec\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Hall coefficient  is =%.1f X 10^-11  m^3/c' %(R_H*(10**11)))\n",
-      "print('The mobility of electron is = %.2f X 10^-3 m^2/volt.sec ' %(Miu*(10**3)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hall coefficient  is =7.4 X 10^-11  m^3/c\n",
-        "The mobility of electron is = 4.29 X 10^-3 m^2/volt.sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.21 Page No 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "n_i = 1.4*10**18             #in /m^3\n",
-      "n_D = 1.4*10**24             #in /m^3\n",
-      "n = n_D                      #in /m^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "p = n_i**2/n               #in /m^3\n",
-      "R_e = n/p              #Ratio of electron\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Concentration of is =%.2f X 10^12 holes per m^3 ' %(p/(10**12)))\n",
-      "print('Ratio of electron to hole concentration is =%.f 10^12 ' %(R_e/(10**12)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Concentration of is =1.40 X 10^12 holes per m^3 \n",
-        "Ratio of electron to hole concentration is =1 10^12 \n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.22 Page No 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "B = 0.48                #in Wb/m^2\n",
-      "R_H = 3.6 * 10**-4           #in m^3/c\n",
-      "R = 9*10**-3           #in ohm-m\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma = 1/R                #in (ohm-m)^-1\n",
-      "Rho = 1/R_H                #in coulomb/m^3\n",
-      "e = 1.6*10**-19            #in C\n",
-      "n = Rho/e                  #in /m^3\n",
-      "Miu = Sigma * R_H          #in m^2/volt-s\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The mobility of electron  is =%.2f  m^2/volt-s' %Miu)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mobility of electron  is =0.04  m^2/volt-s\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.23 Page No 104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "e = 1.6*10**-19           #in C\n",
-      "R_H = 0.0145              #in m^3/coulomb\n",
-      "Miu_e = 0.36              #m^2/v-s\n",
-      "E = 100                  #V/m\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n = 1/(e*R_H)             #in /m^3\n",
-      "J= n*e*Miu_e*E            #in A/m^2\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The current density  is =%.2f  A/m^2 ' %J)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current density  is =2482.76  A/m^2 \n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.24 Page No 105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "e = 1.6*10**-19            #in C\n",
-      "Miu_e = 7.04*10**-3           #in m^2/volt-sec\n",
-      "n = 5.8*10**28              #number of electron/m^3\n",
-      "m = 9.1*10**-31\n",
-      "E_F = 5.5                    # in eV\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Torque = (Miu_e/e)*m                # in sec \n",
-      "Rho = 1/(n*e*Miu_e)                 #in ohm cm\n",
-      "V_F = math.sqrt( (2*E_F*e)/m )           #in m/s\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Relaxtion time is =%.2f X 10^-15 sec ' %(Torque*(10**15)))\n",
-      "print('Resistivity of conductor  is =%.2f X 10^-18 in \u03a9m' %(Rho*(10**8)))\n",
-      "print('Velocity of electron with the fermi energy  is =%.2f X 10^-6 in m/s' %(V_F/(10**6)))\n",
-      "\n",
-      "# Note: The calculation of Part (ii) is wrong also the unit of resistivity of conductor is wrong\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Relaxtion time is =40.04 X 10^-15 sec \n",
-        "Resistivity of conductor  is =1.53 X 10^-18 in \u03a9m\n",
-        "Velocity of electron with the fermi energy  is =1.39 X 10^-6 in m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.25 Page No 105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "\n",
-      "E= 5.95           #in eV\n",
-      "EF= 6.25          #in eV\n",
-      "delE= 0.01\n",
-      "\n",
-      "#CALCULATIONS\n",
-      " #delE= 1-1/(1+exp((E-EF)/KT))\n",
-      "K=1.38*10**-23                      #Boltzman Constant in J/K\n",
-      "T = ((E-EF)/math.log(1/(1-delE) -1)*1.6*10**-19)/K               #in K\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The temperature is =%.2f K' %T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature is =756.95 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.26 Page No 105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "\n",
-      "T1 = 400.0       #in K\n",
-      "T2 = 300.0       #in K\n",
-      "N_V = 1.04*10**19       #in cm^-3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N1 = N_V*((T1/T2)**(3/2))          #in cm^-3\n",
-      "KT = 0.0259*(T1/T2)           #in eV\n",
-      "FermiLevel= 0.27              #in eV\n",
-      "P_O = N1*math.exp( (-FermiLevel)/KT )          #in cm^-3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The thermal equillibrium hole concentration  is =%.2f X 10^15 cm^-3' %(P_O/(10**15)))\n",
-      "print('Approximationa error ')\n",
-      "# Answer in the book is wrong"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thermal equillibrium hole concentration  is =5.58 X 10^15 cm^-3\n",
-        "Approximationa error \n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.27  Page No 107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "T1 = 550.0              #in K\n",
-      "T2 = 300.0           #in K\n",
-      "N1 = 1.04*10**19     \n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N_V = N1*((T1/T2)**(3))  \n",
-      "N_C = 2.8*10**19\n",
-      "E_g = -1.12\n",
-      "KT = 0.0259*(T1/T2)\n",
-      "n_i = math.sqrt(N_C*N_V*math.exp(E_g/KT))   #in cm^-3\n",
-      "Nd= math.sqrt(n_i**2/((1.05-1/2.0)**2-(1/2.0)**2))\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of n_i  is =%.2f X 10^14 cm^-3' %(n_i/(10**14)))\n",
-      "print('The value of N_d is =%.2f X 10^15 cm^-3' %(Nd/(10**15)))\n",
-      "#Formula n_o= Nd/2+sqrt((Nd/2)^2+n_i^2) and n_o = 1.05*N_d;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of n_i  is =3.20 X 10^14 cm^-3\n",
-        "The value of N_d is =1.40 X 10^15 cm^-3\n"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 2.28 Page No 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "n_o = 10**15         #in cm^-3\n",
-      "n_i = 10**10         #in cm^-3\n",
-      "p_o = 10**5       #in cm^-3\n",
-      "del_p = 10**13    #in cm^-3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "del_n = del_p         #in cm^-3\n",
-      "KT= 0.0259          #in eV\n",
-      "Fermi_level1= KT*math.log(n_o/n_i)           #in eV\n",
-      "Fermi_level2= KT*math.log((n_o+del_n)/n_i)        #in eV\n",
-      "Fermi_level3= KT*math.log((p_o+del_p)/n_i)        #in eV\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Fermi level for thermal equillibrium  is : =%.2f eV ' %Fermi_level1)\n",
-      "print('Quasi-Fermi level for electrons in non equillibrium  is : =%.2f eV' %Fermi_level2)\n",
-      "print('Quasi-Fermi level for holes in non equillibrium  is: =%.2f eV' %Fermi_level3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fermi level for thermal equillibrium  is : =0.30 eV \n",
-        "Quasi-Fermi level for electrons in non equillibrium  is : =0.30 eV\n",
-        "Quasi-Fermi level for holes in non equillibrium  is: =0.18 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo3.ipynb b/Advance_Semiconductor_Devices/ChapterNo3.ipynb
deleted file mode 100755
index bc717890..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo3.ipynb
+++ /dev/null
@@ -1,992 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chapter No 3 : Excess Carriers In Semiconductors<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1, Page No 145 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "n_o = 10**17          #in/cm^3\n",
-      "n_i = 1.5*10**10      #in /cm^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "p_o = ((n_i)**(2))/n_o      #in holes/cm^3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The hole concentration  is =%.f  in holes/cm^3' %p_o)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The hole concentration  is =2250  in holes/cm^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3, Page No  146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "n_i = 1.5 * 10**10              #in /cm^3 for silicon\n",
-      "N_d = 10**17                    #in atoms/cm^3\n",
-      "n_o = 10**17                   #electrons/cm^3\n",
-      "KT = 0.0259\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "# E_r - E_i = KT * log(n_o/n_i)\n",
-      "del_E = KT * math.log(n_o/n_i)                 #in eV\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The energy band for this type material is Ei =%.2f eV' %del_E)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy band for this type material is Ei =0.41 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4 Page No 147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "K = 1.38*10**-23               #in J/K\n",
-      "T = 27                     #in degree C\n",
-      "T = T + 273               #in K\n",
-      "e = 1.6*10**-19          \n",
-      "Miu = 0.17             #in m^2/v-s\n",
-      "Miu1 = 0.025           #in m^2/v-s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "D_n = ((K*T)/e)*Miu        #in m^2/s\n",
-      "D_p = ((K*T)/e)*Miu1    #in m^2/s\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The diffusion coefficient  of electrons  is =%.2f X 10^-4 in m^2/s ' %(D_n*(10**4)))\n",
-      "print('The diffusion coefficient  of holes  is =%.2f X 10^-4 in m^2/s ' %(D_p*(10**4)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diffusion coefficient  of electrons  is =43.99 X 10^-4 in m^2/s \n",
-        "The diffusion coefficient  of holes  is =6.47 X 10^-4 in m^2/s \n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5, Page No 147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Miu_n = 0.15           #in m^2/v-s\n",
-      "K = 1.38*10**-23      \n",
-      "T = 300                #in K\n",
-      "e = 1.6*10**-19      #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "D_n = Miu_n*((K*T)/e)            #in m^2/s\n",
-      "Torque_n = 10**-7                 #in s\n",
-      "L_n = math.sqrt(D_n*Torque_n)    #in m\n",
-      "del_n = 10**20                   #electrons/m^3\n",
-      "J_n = (e*D_n*del_n)/L_n        #in A/m^2\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The diffusion length is =%.2f  X 10^-4 in m' %(L_n*(10**5)))\n",
-      "print('The diffusion current density  is =%.2f X 10^3 in A/m^2 ' %(J_n/(10**3)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diffusion length is =1.97  X 10^-4 in m\n",
-        "The diffusion current density  is =3.15 X 10^3 in A/m^2 \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6 Page No 147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Sigma_n = 0.1              #in (ohm-cm)^-1\n",
-      "Miu_n = 1300\n",
-      "q = 1.6*10**-19             #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n_n = Sigma_n/(Miu_n*q)        #in electrons/cm^3 \n",
-      "n_i = 1.5*10**10\n",
-      "p_n = ((n_i)**2)/n_n           #in holes/cm^3\n",
-      "p_n = p_n * 10**6          #in holes/m^3\n",
-      "n_n=n_n*10**6\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Concentration of electrons  is =%.2f X 10^20 per m^3 ' %(n_n/(10**20)))\n",
-      "print('Concentration of holes  is =%.2f X 10^11 per m^3 ' %(p_n/(10**11)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Concentration of electrons  is =4.81 X 10^20 per m^3 \n",
-        "Concentration of holes  is =4.68 X 10^11 per m^3 \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7 Page No 147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "L = 100*10**-6         #in m\n",
-      "Miu_e = 0.13          #in m^2/V-s\n",
-      "Torque_h = 10**-6     #in s\n",
-      "Miu_h = 0.05          #in m^2/v-s\n",
-      "V = 12                 #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Torque_n = ((L)**2)/(Miu_e*V)             #in s\n",
-      "P = (Torque_h/Torque_n)*(1+(Miu_h/Miu_e))\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Electron transit time  is =%.2f X 10^-9 in sec' %(Torque_n*(10**9)))\n",
-      "print('Photoconductor gain is =%.2f  ' %P)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Electron transit time  is =6.41 X 10^-9 in sec\n",
-        "Photoconductor gain is =216.00  \n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8, Page No 147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "q = 1.6*10**-19              #in C\n",
-      "n_i = 2.5*10**13\n",
-      "Miu_n = 3800             #in cm^2/V-s\n",
-      "Miu_p = 1800             #in cm^2/V-s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Sigma = n_i*(Miu_n + Miu_p)*q             #in (ohm-cm)^-1\n",
-      "Rho = 1/Sigma               #in ohm-cm\n",
-      "N_D =4.4*10**22/10**8     #in atoms/cm^3\n",
-      "Sigma_n = N_D * Miu_n*q   #in (ohm-cm)^-1 \n",
-      "Rho1 = 1/Sigma_n     #in ohm cm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The resistivity  is =%.2f in ohm-cm ' %Rho)\n",
-      "print('The resistivity drops  is =%.2f in ohm cm ' %Rho1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistivity  is =44.64 in ohm-cm \n",
-        "The resistivity drops  is =3.74 in ohm cm \n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9 Page No 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "n_i = 10**16          #in /m^3\n",
-      "N_D = 10**22       #in /m^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "n = N_D        #in /m^3\n",
-      "p = ((n_i)**2)/n   #in /m^3\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The concentration of  is =%.2f X 10^21  electrons per m^3' %(n/(10**21)))\n",
-      "print('The concentration of  is =%.2f X 10^9 holes per m^3 ' %(p/(10**9)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The concentration of  is =10.00 X 10^21  electrons per m^3\n",
-        "The concentration of  is =10.00 X 10^9 holes per m^3 \n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10 Page No 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Rho = 9.6*10**-2       #ohm-m\n",
-      "Sigma_n = 1/Rho        #in (ohm-m)^-1\n",
-      "Miu_n = 1300           #in cm^2/V-s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Miu_n = Miu_n * 10**-4  #in m^2/V-s\n",
-      "q = 1.6*10**-19     #in C\n",
-      "N_D = Sigma_n/(Miu_n*q)           #in atoms/m^3\n",
-      "d = 5*10**22                     #in atoms/cm^3\n",
-      "d = d * 10**6                   #// in atoms/m^3\n",
-      "R_d = N_D/d                   #Ratio \n",
-      "\n",
-      "#RESULTS\n",
-      "print('Ratio of donor atom to silicon atoms per unit volume is =%.2f X 10^-8  ' %(R_d*(10**8)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ratio of donor atom to silicon atoms per unit volume is =1.00 X 10^-8  \n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11 Page No 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "n_i = 1.5*10**10            #in /cm^3\n",
-      "n_n = 2.25*10**15           #in /cm^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "p_n = ((n_i)**2)/n_n         #in /cm^3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The concentration of  is =%.2f X 10^5 holes per cm^3 ' %(p_n/(10**5)))\n",
-      "print('Donor impurity per cm^3 is =%.2f X 10^15 ' %(n_n/(10**15)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The concentration of  is =1.00 X 10^5 holes per cm^3 \n",
-        "Donor impurity per cm^3 is =2.25 X 10^15 \n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.12 Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "N_A = 2*10**16           #in /cm^3\n",
-      "N_D = 10**16             #in /cm^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "C = N_A-N_D             #in /cm^3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Carrier concentration in holes/cm^3 is =%.2f X 10^16 ' %(C/(10**16)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Carrier concentration in holes/cm^3 is =1.00 X 10^16 \n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.13 Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "del_n = 10**15           #in /cm^3\n",
-      "Torque_p = 10*10**-6       #in sec\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "R_G = del_n/Torque_p          #in electron hole pairs/sec/cm^3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The rate of generation of minority carrier is : =%.2f X 10^20 electron hole pairs/sec/cm^3 ' %(R_G/(10**20)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rate of generation of minority carrier is : =1.00 X 10^20 electron hole pairs/sec/cm^3 \n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.14 Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V = 1/20.0        #in cm/\u00b5sec\n",
-      "V=V*10**6       #in cm/sec\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "E = 10            #in V/cm\n",
-      "Miu = V/E         #in cm^2/V-sec\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The mobility of minority charge carrier is =%.f  cm^2/V-sec' %Miu)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mobility of minority charge carrier is =5000  cm^2/V-sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.15 Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "q = 1.6 * 10**-19       #in C\n",
-      "N_D = 4.5 * 10**15      #in /cm^3\n",
-      "del_p = 10**21 \n",
-      "e=10                #in cm\n",
-      "A = 1             #in mm^2\n",
-      "A = A * 10**-14        #cm^2\n",
-      "l = 10              #in cm\n",
-      "Torque_p = 1            #in microsec\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Torque_p = Torque_p * 10**-6           #in sec\n",
-      "Torque_n = 1                      #in microsec\n",
-      "Torque_n = Torque_n * 10**-6        #in  sec\n",
-      "n_i = 1.5 * 10**10               #in /cm^3\n",
-      "D_n = 30                   #cm^2/sec\n",
-      "D_p = 12                  #in cm^2/sec\n",
-      "n_o = N_D                #in /cm^3\n",
-      "p_o = ((n_i)**2)/n_o           #in /cm^3\n",
-      "l_n = math.sqrt(D_n * Torque_n)  #in cm\n",
-      "l_p = math.sqrt(D_p * Torque_p)      #in cm\n",
-      "x=34.6*10*8-4            #in cm\n",
-      "e1 = 1.88 * 10**1        #in cm\n",
-      "dnBYdx = del_p * e1        #in cm^4\n",
-      "dpBYdx = del_p *e         #in cm^4\n",
-      "\n",
-      "J_P = -(q) * D_p * dpBYdx       #in A/cm^2\n",
-      "J_n = q * D_n * dnBYdx          #in A/cm^2\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Hole concentration at thermal equilibrium per cm^3 is =%.1f X 10^5  ' %(p_o/(10**5)))\n",
-      "print('Diffusion length of holes in cm is =%.1f X 10^-3 ' %(l_p*(10**3)))\n",
-      "print('Concentration gradient of holes at distance in cm^4 is =%.f X 10^22 ' %(dpBYdx/(10**22)))\n",
-      "print('Concentration gradient of electrons in per cm^4 is =%.1f X 10^22  ' %(dnBYdx/(10**22)))\n",
-      "print('Current density of holes due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_P/(10**4)))\n",
-      "print('Current density of electrons due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_n/(10**4)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hole concentration at thermal equilibrium per cm^3 is =0.5 X 10^5  \n",
-        "Diffusion length of holes in cm is =3.5 X 10^-3 \n",
-        "Concentration gradient of holes at distance in cm^4 is =1 X 10^22 \n",
-        "Concentration gradient of electrons in per cm^4 is =1.9 X 10^22  \n",
-        "Current density of holes due to diffusion in A/cm^2 is =-1.92 X 10^4 \n",
-        "Current density of electrons due to diffusion in A/cm^2 is =9.02 X 10^4 \n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.16 Page No 151"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "h = 6.626 * 10**-34          #in J-s\n",
-      "q= 1.6*10**-19            #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "h= h/q              #in eV\n",
-      "c = 3*10**8\n",
-      "lembda = 5490*10**-10             #in m\n",
-      "E = h*c/lembda                 #in eV\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The energy band gap is =%.2f eV ' %E)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy band gap is =2.26 eV \n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.17 Page No 151"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "D_n = 35               #in cm^2/sec\n",
-      "q = 1.6*10**-19        #in C\n",
-      "y2 = 6*10**16          #in /cm^3\n",
-      "y1 = 10**17           #in /cm^3\n",
-      "x2 = 2*10**-4\n",
-      "x1 = 0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "dnBYdx = (y2-y1)/(x2-x1)\n",
-      "J_n = q*D_n*dnBYdx              #in A/cm^2\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The current density is =%.f  in A/cm^2' %J_n)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current density is =-1120  in A/cm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.18 Page No 151"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "q = 1.6*10**-19               #in C\n",
-      "n_n = 5*10**20                #in  /m^3\n",
-      "n_n = n_n * 10**-6            #in /cm^3\n",
-      "Miu_n = 0.13                   #in m^2/V-sec\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Miu_n = Miu_n * 10**4       #in cm^2/V-sec\n",
-      "Sigma_n = q*n_n*Miu_n      #in ohm-cm^-1\n",
-      "Rho = 1/Sigma_n\n",
-      "A = 100        # in \u00b5m^2\n",
-      "A = A * 10**-8     #in cm^2\n",
-      "l = 0.1           #in cm\n",
-      "R = Rho * (l/A)        #in ohm\n",
-      "R=R*10**-6\n",
-      "#RESULTS\n",
-      "print('The resistance of the bar  is =%.2f in M ohm' %R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance of the bar  is =0.96 in M ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.19 Page No 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "w = 3.0              #in \u00b5m\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "D = w/9          #in \u00b5m\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Depletion width on P side is =%.2f in \u00b5m ' %D)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Depletion width on P side is =0.33 in \u00b5m \n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.20 Page No 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "n_i = 1.5*10**16          #in /m^3\n",
-      "n_n = 5*10**20          #in /m^3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "p_n = ((n_i)**2)/n_n          #in /m^3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The minority carrier density per m^3 is =%.2f X 10^11' %(p_n/(10**11)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minority carrier density per m^3 is =4.50 X 10^11\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.21 Page No 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "y2 = 10**14                #in /cm^3\n",
-      "y1 = 0\n",
-      "x1=-0.5                    #in \u00b5m\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "x1= x1*10**-4             #in cm\n",
-      "x2=0\n",
-      "dnBYdx = (y2-y1)/(x2-x1)     #in /cm^4\n",
-      "q = 1.6*10**-19              #in C\n",
-      "D_n = 25                     #in cm^2/sec\n",
-      "J_n = q*D_n*dnBYdx          #in A/cm^2\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The collector current density  is =%.f in A/cm^2' %J_n)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current density  is =8 in A/cm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.22 Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "h = 6.64*10**-34              ##in J-s\n",
-      "q=1.6*10**-19                #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "h= h/q                     #in eV\n",
-      "c = 3*10**8               #in m/s\n",
-      "lembda = 0.87*10**-6        #in m\n",
-      "E_g = (h*c)/lembda         #in eV\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The band gap in eV is =%.2f ' %E_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The band gap in eV is =1.43 \n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.23 Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "alpha = 5*10**4            #in cm^-1\n",
-      "l = 0.46*10**-4            #in cm\n",
-      "hv = 2                    #in eV\n",
-      "I_o = 10**-2             #in W\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_t = I_o*math.exp(-alpha*l)   #in W\n",
-      "A_p = I_o-I_t            #absorbed power in W or J/s\n",
-      "c = 1.43\n",
-      "A_E = (hv-c)/hv*A_p           # in J/s\n",
-      "e = 1.6*10**-19               #in C\n",
-      "P = A_p/(e*hv)               #Perfect quantum efficiency in photon/s\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Total energy absorbed  is =%.f X 10^-3 in J/s' %(A_p*(10**3)))\n",
-      "print('Rate of excess thermal energy  is =%.2f X 10^-3 in J/s' %(A_E*(10**3)))\n",
-      "print('Perfect quantum efficiency  is =%.2f X 10^16 in photon/s' %(P/(10**16)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total energy absorbed  is =9 X 10^-3 in J/s\n",
-        "Rate of excess thermal energy  is =2.56 X 10^-3 in J/s\n",
-        "Perfect quantum efficiency  is =2.81 X 10^16 in photon/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 3.24 Page No 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Mu_p = 500         #in cm^2/v-s\n",
-      "kT = 0.0259\n",
-      "Toh_p = 10**-10         #in sec\n",
-      "p_o = 10**17           #in cm^-3\n",
-      "q= 1.6*10**-19             # in C\n",
-      "A=0.5             #in square meter\n",
-      "del_p = 5 * 10**16        #in cm^-3\n",
-      "n_i= 1.5*10**10          #in cm^-3    \n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "D_p = kT * Mu_p          #in cm/s\n",
-      "L_p = math.sqrt(D_p * Toh_p)             # in cm\n",
-      "x = 10**-5            #in cm\n",
-      "p = p_o+del_p*e**(x/L_p)          #in cm^-3\n",
-      "Eip= math.log(p/n_i)*kT              #in eV\n",
-      "Ecp= 1.1/2-Eip                   #value of E_c-E_p in eV\n",
-      "Ip= q*A*D_p/L_p*del_p*e**(x/L_p)         #in A\n",
-      "Qp= q*A*del_p*L_p               #in C\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The hole current  is : =%.2f X 10^3 in A ' %(Ip*(10**3)))\n",
-      "print('Approximation error')\n",
-      "print('The value of Qp  is :=%.2f X 10^-7 in C' %(Qp*(10**7)))\n",
-      "\n",
-      "#Note: There is a calculation error to evalaute the value of hole current hence the value of hole current in the book is wrong\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The hole current  is : =8.61 X 10^3 in A \n",
-        "Approximation error\n",
-        "The value of Qp  is :=1.44 X 10^-7 in C\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo4.ipynb b/Advance_Semiconductor_Devices/ChapterNo4.ipynb
deleted file mode 100755
index e76421ad..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo4.ipynb
+++ /dev/null
@@ -1,928 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chapter No 4 : Junctions and Interfaces<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2 Page No 184"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N_D = 10**17 * 10**6      #in atoms/m^3\n",
-      "N_A = 0.5*10**16*10**6      #in atoms/m^3\n",
-      "Epsilon_r = 10\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Epsilon_o = 8.85*10**-12\n",
-      "Epsilon = Epsilon_r*Epsilon_o       #in F/m\n",
-      "e = 1.602*10**-19          #in C\n",
-      "V = 0\n",
-      "V_B = 0.7          #in V\n",
-      "W1 = math.sqrt( ((2*Epsilon*V_B)/e)*(1/N_A+1/N_D) )       #in m\n",
-      "V_o = V_B    #in V\n",
-      "V1 = -10          #in V\n",
-      "V_B1 = V_o-V1        #in V\n",
-      "W = math.sqrt( ((2*Epsilon*V_B1)/e)*(1/N_A+1/N_D) )             #in m\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The junction width in meter when no external voltage is applied is =%.2f X 10^-6' %(W1*(10**6)))\n",
-      "print('Junction width in meter with an external voltage of -10V is =%.2f X 10^-6 m' %(W*(10**6)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The junction width in meter when no external voltage is applied is =0.39 X 10^-6\n",
-        "Junction width in meter with an external voltage of -10V is =1.54 X 10^-6 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4 Page No 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V = 5        #in V\n",
-      "V_Gamma = 0.6        #in V\n",
-      "r_F = 12           #in ohm\n",
-      "R = 1           #in k ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "R = R * 10**3    #in ohm\n",
-      "I_F = (V-V_Gamma)/(R+r_F)    #in A\n",
-      "V_F = V_Gamma + (I_F*r_F)  #in V\n",
-      "I_F=I_F*10**3 \n",
-      "#RESULTS\n",
-      "print('The forward diode current is =%.2f mA ' %I_F)\n",
-      "print('The diode voltage is =%.2f V ' %V_F)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The forward diode current is =4.35 mA \n",
-        "The diode voltage is =0.65 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5 Page No 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "n = 4.4*10**22          #total number of Ge atoms/cm^3\n",
-      "n_a = 1*10**8           #number of impurity atoms\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N_A = n/n_a             #in atoms/cm^3\n",
-      "N_A = N_A * 10**6       #in atoms/m^3\n",
-      "n_i = 2.5*10**13        #in atoms/cm^3\n",
-      "n_i = n_i * 10**6       #in atoms/m^3\n",
-      "N_D = 10**3 * N_A       #in atoms/m^3\n",
-      "V_T = 26*10**-3         #in A\n",
-      "V_J = V_T*math.log( (N_A*N_D)/((n_i)**2) )    #in V\n",
-      "print('The contact difference of potential  is =%.2f V For a silicon P-N junction ' %V_J)\n",
-      "n = 5*10**22\n",
-      "N_A = n/n_a      #in atoms/cm^3\n",
-      "N_A = N_A * 10**6  #in atoms/m^3\n",
-      "N_D = 10**3 * N_A   #in atoms/m^3\n",
-      "n_i = 1.5*10**10        #in /cm^3\n",
-      "V_J = V_T*math.log(N_A*N_D/n_i**2)      #in V\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "\n",
-      "print('The contact difference of potential  is =%.2f V' %V_J)\n",
-      "\n",
-      "#Note: There is a calculation error to find the value of V_J in the book, so the answer in the book is wrong.\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The contact difference of potential  is =0.33 V For a silicon P-N junction \n",
-        "The contact difference of potential  is =1.44 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6 Page No 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Rho_p = 2     #in ohm-cm\n",
-      "Rho_n = 1      #in ohm cm\n",
-      "q = 1.6*10**-19     #in C\n",
-      "n_i = 2.5*10**13  #atoms per cm^3\n",
-      "Miu_p = 1800\n",
-      "Miu_n = 3800\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N_A = 1/(Rho_p*q*Miu_p)     #in /cm^3\n",
-      "N_D = 1/(Rho_n*q*Miu_n)     #in /cm^3\n",
-      "V_T = 26           #in mV\n",
-      "V_T= V_T*10**-3         #in V\n",
-      "V_J = V_T*math.log((N_A*N_D)/((n_i)**2))   #in V\n",
-      "print('The height of the potential energy barrier is =%.2f in V ' %V_J)\n",
-      "Miu_p = 500\n",
-      "N_A = 1/(Rho_p*q*Miu_p)      #in /cm^3\n",
-      "Miu_n = 1300\n",
-      "N_D = 1/(Rho_n*q*Miu_n)        #in /cm^3\n",
-      "n_i = 1.5*10**10;\n",
-      "V_J = V_T*math.log((N_A*N_D)/((n_i)**2))        #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('For silicon P-N juction The height of the potential energy barrier  is =%.2f V' %V_J)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The height of the potential energy barrier is =0.22 in V \n",
-        "For silicon P-N juction The height of the potential energy barrier  is =0.67 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7 page No 215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Eta = 1\n",
-      "V_T = 26       #in mV\n",
-      "V_T= V_T*10**-3    #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V= math.log(1-0.9)*V_T     #in V\n",
-      "print(\"The voltage in volts is : %.2f \" %V)\n",
-      "V1=0.05     #in V\n",
-      "V2= -0.05   #in V\n",
-      "ratio= (math.e**(V1/(Eta*V_T))-1)/(math.e**(V2/(Eta*V_T))-1)\n",
-      "print(\"The ratio of the current for a forward bias to reverse bias is :  %.2f \" %ratio)\n",
-      "\n",
-      "# Part (iii)\n",
-      "Io= 10  #in \u00b5A\n",
-      "Io=Io*10**-3  #in mA\n",
-      "#For \n",
-      "V=0.1   #in V\n",
-      "I = Io * (math.e**(V/(Eta*V_T)) - 1)     #in mA\n",
-      "print(\"For v=0.1 V , the value of I  is : %.2f mA \" %I)\n",
-      "\n",
-      "#For \n",
-      "V=0.2      #in V\n",
-      "I = Io * (math.e**(V/(Eta*V_T)) - 1)      #in mA\n",
-      "print(\"For v=0.2 V , the value of I  is : %.2f  mA \" %I)\n",
-      "\n",
-      "#For \n",
-      "V=0.3  #in V\n",
-      "I = Io * (math.e**(V/(Eta*V_T)) - 1)           #in mA\n",
-      "\n",
-      "#RESULTS\n",
-      "\n",
-      "I=I*10**-3\n",
-      "print(\"For v=0.3 V , the value of I  is : %.2f  mA\" %I)\n",
-      "print(\"From three value of I, for small rise in forward voltage, the diode current increase rapidly \")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage in volts is : -0.06 \n",
-        "The ratio of the current for a forward bias to reverse bias is :  -6.84 \n",
-        "For v=0.1 V , the value of I  is : 0.46 mA \n",
-        "For v=0.2 V , the value of I  is : 21.90  mA \n",
-        "For v=0.3 V , the value of I  is : 1.03  mA\n",
-        "From three value of I, for small rise in forward voltage, the diode current increase rapidly \n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.8 Page No 216"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "# Part (i)\n",
-      "T1= 25        #in \u00b0C\n",
-      "T2= 80.0      #in \u00b0C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "# Formula Io2= Io1*2**((T2-T1)/10)\n",
-      "AntiFactor= 2**((T2-T1)/10)\n",
-      "print('Anticipated factor for Ge is : =%.f  ' %AntiFactor)\n",
-      "\n",
-      "# Part (ii)\n",
-      "T1= 25.0     #in \u00b0C\n",
-      "T2= 150.0    #in \u00b0C\n",
-      "AntiFactor= 2**((T2-T1)/10)\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Anticipated factor for Ge is : =%.f  ' %AntiFactor)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Anticipated factor for Ge is : =45  \n",
-        "Anticipated factor for Ge is : =5793  \n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9 Page No 216"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I=5.0        #in \u00b5A\n",
-      "V=10.0       #in V\n",
-      "T1= 0.11     #in \u00b0C^-1\n",
-      "T2= 0.07      #in \u00b0C^-1\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Io= T2*I/T1   #in \u00b5A\n",
-      "I_R= I-Io      #in \u00b5A\n",
-      "R= V/I_R       #in M\u03a9\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The leakage resistance is : =%.1f M\u03a9 ' %R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The leakage resistance is : =5.5 M\u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.10 Page No 216"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "Eta = 1.0\n",
-      "T = 125.0   #in \u00b0C\n",
-      "T = T + 273   #in K\n",
-      "V_T = 8.62 * 10**-5 * 398    #in V\n",
-      "I_o = 30      #in \u00b5A\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_o= I_o*10**-6    #in A\n",
-      "v = 0.2            #in V\n",
-      "r_f = (Eta * V_T)/(I_o * math.e**(v/(Eta* V_T)))      #in ohm\n",
-      "print('The dynamic resistance in the forward direction  is =%.2f \u03a9 ' %r_f)\n",
-      "\n",
-      "r_r = (Eta * V_T)/(I_o * math.e**(-v/(Eta* V_T)))   #in ohm\n",
-      "r_r=r_r*10**-3\n",
-      "print('The dynamic resistance in the forward direction  is =%.2f K\u03a9 ' %r_r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dynamic resistance in the forward direction  is =3.36 \u03a9 \n",
-        "The dynamic resistance in the forward direction  is =389.08 K\u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.11 Page No 217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "epsilon = 16/(36 * math.pi * 10**11)       #in F/cm\n",
-      "A = 1 * 10**-2\n",
-      "W = 2 * 10**-4\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "C_T = (epsilon * A)/W      #in F\n",
-      "C_T=C_T*10**12\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The barrier capacitance is =%.2f pF ' %C_T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The barrier capacitance is =70.74 pF \n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.12 Page No 217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A = 1.0           #in mm^2\n",
-      "A = A * 10**-6    #in m^2\n",
-      "N_A = 3 * 10**20    #in atoms/m^3\n",
-      "q = 1.6 *10**-19    #in C\n",
-      "V_o = 0.2         #in V\n",
-      "epsilon_r=16\n",
-      "epsilon_o= 8.854*10**-12        #in F/m\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "epsilon=epsilon_r*epsilon_o\n",
-      "# Part (a)\n",
-      "V=-10    #in V\n",
-      "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A))    #m\n",
-      "C_T1 = (epsilon * A)/W           #in F\n",
-      "W=W*10**6\n",
-      "print('The width of the depletion layer  for an applied reverse voltage of 10V  is =%.2f \u00b5m ' %W)\n",
-      "\n",
-      "# Part (b)\n",
-      "V=-0.1           #in V\n",
-      "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A))         #m\n",
-      "C_T2 = (epsilon * A)/W            #in F\n",
-      "W=W*10**6\n",
-      "print('The width of the depletion layer  for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
-      "\n",
-      "# Part (c)\n",
-      "V=0.1         #in V\n",
-      "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A))   # m\n",
-      "W=W*10**6\n",
-      "print('The width of the depletion layer  for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
-      "\n",
-      "#Part (d)\n",
-      "C_T1=C_T1*10**12\n",
-      "C_T2=C_T2*10**12\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The space charge capacitance for an applied reverse voltage of 10V  is =%.2f pF ' %C_T1)\n",
-      "print('The space charge capacitance for an applied reverse voltage of 0.1V  is =%.2f  pF' %C_T2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The width of the depletion layer  for an applied reverse voltage of 10V  is =7.76 \u00b5m \n",
-        "The width of the depletion layer  for an applied reverse voltage of 0.1V is =1.33 \u00b5m \n",
-        "The width of the depletion layer  for an applied reverse voltage of 0.1V is =0.77 \u00b5m \n",
-        "The space charge capacitance for an applied reverse voltage of 10V  is =18.26 pF \n",
-        "The space charge capacitance for an applied reverse voltage of 0.1V  is =106.46  pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 4.13 Page No 218"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_o = 1.8 * 10**-9         #A\n",
-      "v = 0.6                 #in V\n",
-      "Eta = 2\n",
-      "V_T = 26              #in mV\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_T=V_T*10**-3      #in V\n",
-      "I = I_o *(math.e**(v/(Eta * V_T)))           #in A\n",
-      "I=I*10**3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The current in the junction  is =%.2f mA ' %I)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current in the junction  is =0.18 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 4.14 Page No 218"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_o = 2.4 * 10**-14\n",
-      "I = 1.5         #in mA\n",
-      "I=I*10**-3       #in A\n",
-      "Eta = 1\n",
-      "V_T = 26           #in mV\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_T= V_T*10**-3        #in V\n",
-      "v =math.log((I + I_o)/I_o) * V_T          #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The forward biasing voltage across the junction  is =%.2f   V' %v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The forward biasing voltage across the junction  is =0.65   V\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 4.15 Page No 218"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "I_o = 10              #in nA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I = I_o * (-1)         #in nA\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The Diode current is  = %.f nA ' %I)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Diode current is  = -10 nA \n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 4.16 Page No 218"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "R = 4.5   #in ohm\n",
-      "I = 44.4       #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I=I*10**-3          #in A\n",
-      "V = R * I         #in V\n",
-      "Eta = 1\n",
-      "V_T = 26          #in mV\n",
-      "V_T=V_T*10**-3    #in V\n",
-      "I_o = I/((math.e**(V/(Eta * V_T))) -1)         #in A\n",
-      "V = 0.1         #in V\n",
-      "r_f = (Eta * V_T)/(I_o * ((math.e**(V/(Eta * V_T)))-1))        #in ohm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The diode dynamic resistance is =%.2f  \u03a9' %r_f)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diode dynamic resistance is =27.78  \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 4.18 Page No 219"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V = 0.25          #in V\n",
-      "I_o = 1.2         #in \u00b5A\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_o = I_o * 10**-6          #in A\n",
-      "V_T = 26                   #in mV\n",
-      "V_T = V_T * 10**-3         #in V\n",
-      "Eta = 1\n",
-      "r = (Eta * V_T)/(I_o * (math.e**(V/(Eta * V_T))))        #in ohm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The ac resistance of the diode  is =%.2f  \u03a9' %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ac resistance of the diode  is =1.45  \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 4.19 Page No 219"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "q = 1.6 * 10**-19         #in C\n",
-      "N_A = 3 * 10**20          #in /m^3\n",
-      "A = 1               #in \u00b5m^2\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "A = A * 10**-6        #in m^2\n",
-      "V = -10          #in V\n",
-      "V_J = 0.25        #in V\n",
-      "V_B = V_J - V        #in V\n",
-      "epsilon_o = 8.854       #in pF/m\n",
-      "epsilon_o = epsilon_o * 10**-12          #in F/m\n",
-      "epsilon_r = 16\n",
-      "epsilon = epsilon_o * epsilon_r\n",
-      "W = math.sqrt((V_B * 2 * epsilon)/(q * N_A))   #in m \n",
-      "\n",
-      "#RESULTS\n",
-      "C_T = (epsilon * A)/W   #in pF\n",
-      "W=W*10**6\n",
-      "C_T=C_T*10**12\n",
-      "print('The width of depletion layer is =%.2f \u00b5m ' %W)\n",
-      "print('The space charge capacitance is =%.2f pF ' %C_T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The width of depletion layer is =7.78 \u00b5m \n",
-        "The space charge capacitance is =18.21 pF \n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 4.20 Page No 220"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "C_T = 100.0   #in pF\n",
-      "C_T=C_T*10**-12     #in F\n",
-      "epsilon_r = 12\n",
-      "epsilon_o = 8.854 * 10**-12      #in F/m\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "epsilon = epsilon_r * epsilon_o\n",
-      "Rho_p = 5   #in ohm-cm\n",
-      "Rho_p = Rho_p * 10**-2         #in ohm-m\n",
-      "V_j = 0.5           #in V\n",
-      "V = -4.5           #in V\n",
-      "Mu_p = 500          #in cm^2\n",
-      "Mu_p = Mu_p * 10**-4     #in m^2\n",
-      "Sigma_p = 1/Rho_p         #in per ohm-m\n",
-      "qN_A = Sigma_p/ Mu_p\n",
-      "V_B = V_j - V\n",
-      "W = math.sqrt((V_B * 2 * epsilon)/qN_A)\n",
-      "A = (C_T * W)/ epsilon             #in m\n",
-      "D = math.sqrt(A * (4/math.pi))        #in m\n",
-      "D = D * 10**3          #in mm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The diameter is =%.2f mm ' %D)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diameter is =1.40 mm \n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 4.21 Page No 221"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "q = 1.6 * 10**-19         #in C\n",
-      "Mu_p = 500         #in cm^2/V-sec\n",
-      "Rho_p = 3.5          #in ohm-cm\n",
-      "Mu_n = 1500          #in cm^2/V-sec\n",
-      "Rho_n = 10          #in ohm-cm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N_A = 1/(Rho_p * Mu_p * q)       # in /cm^3\n",
-      "N_D = 1/(Rho_n * Mu_n * q)       # in /cm^3\n",
-      "V_J = 0.56            # in V\n",
-      "n_i = 1.5 * 10**10         #in /cm^3\n",
-      "V_T = V_J/math.log((N_A * N_D)/(n_i)**2)     #in V\n",
-      "T = V_T * 11600          #in K\n",
-      "T = T - 273            #in \u00b0C\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The Temperature of junction is =%.2f \u00b0C ' %T)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Temperature of junction is =14.28 \u00b0C \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example No 4.22 Page No 221"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "R = 5       #in ohm\n",
-      "I = 50      #in mA\n",
-      "I=I*10**-3   #in A\n",
-      "V = R * I    #in V\n",
-      "Eta = 1\n",
-      "V_T = 26    #in mV\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_T=V_T*10**-3     #in V\n",
-      "I_o = I/((math.e**(V/(Eta * V_T))) - 1)     #in A\n",
-      "v1 = 0.2         #in V\n",
-      "r = (Eta * V_T)/(I_o * (math.e**(v1/(Eta * V_T))))         #in ohm\n",
-      "\n",
-      "#RESULTS\n",
-      "I_o=I_o*10**6\n",
-      "print('Reverse saturation current  is =%.2f \u00b5A ' %I_o)\n",
-      "print('Dynamic resistance of the diode is =%.2f \u03a9 ' %r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reverse saturation current  is =3.33 \u00b5A \n",
-        "Dynamic resistance of the diode is =3.56 \u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo6.ipynb b/Advance_Semiconductor_Devices/ChapterNo6.ipynb
deleted file mode 100755
index 44239148..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo6.ipynb
+++ /dev/null
@@ -1,76 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 6: Microwave Diodes <h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1, Page No 254"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "C1_min= 5.0     #in pF\n",
-      "C1_max= 50.0  #in pF\n",
-      "C2_min= 5.0  #in pF\n",
-      "C2_max= 50.0  #in pF\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "C1_min=C1_min*10**-12   #in F\n",
-      "C2_min=C2_min*10**-12   #in F\n",
-      "C1_max=C1_max*10**-12   #in F\n",
-      "C2_max=C2_max*10**-12    #in F\n",
-      "L = 10.0                #in mH\n",
-      "L = L * 10**-3          #in H\n",
-      "C_T_min = (C1_min*C2_min)/(C2_min+C2_min)     #in F\n",
-      "f_o_max = 1/( 2*math.pi*(math.sqrt(L*C_T_min)) )      #in Hz \n",
-      "f_o_max = f_o_max * 10**-6      #in MHz\n",
-      "C_T_max = (C1_max*C2_max)/(C2_max+C2_max)     #in F\n",
-      "f_o_min = 1/( 2*math.pi*(math.sqrt(L*C_T_max)) )      #in Hz \n",
-      "f_o_min = f_o_min * 10**-3                      #in kHz\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The tuning range for circuit will be =%.2f  kHz' %f_o_min)\n",
-      "print(' To %.3f MHz' %f_o_max)\n",
-      "\n",
-      "\n",
-      "#print(\"The tuning range for circuit will be : \"+string(round(f_o_min))+\" kHz to \"+string(round(f_o_max))+\" MHz\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The tuning range for circuit will be =318.31  kHz\n",
-        " To 1.007 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo7.ipynb b/Advance_Semiconductor_Devices/ChapterNo7.ipynb
deleted file mode 100755
index 431c4340..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo7.ipynb
+++ /dev/null
@@ -1,154 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 7: Optoelectonic Devices <h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1, Page No 283"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "Vout = 5.0         #in V\n",
-      "V = 1.5               #ON state voltage drop across LED in V\n",
-      "I = 5.0          #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I = I*10**-3       #in A\n",
-      "R = (Vout-V)/I         #in ohm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Resistance  is =%.f \u03a9' %R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistance  is =700 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2, Page No 306"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N_A = 7.5*10**24            #in atoms/m^3\n",
-      "N_D = 1.5*10**22            #in atoms/m^3\n",
-      "D_e = 25*10**-4             #in m^2/s\n",
-      "D_n = 1*10**-3             #in m^2/s\n",
-      "V_T = 26*10**-3            #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Torque_eo = 500*10**-9            #in sec\n",
-      "Torque_ho = 100*10**-9            #in sec\n",
-      "e = 1.6*10**-19                #in C\n",
-      "n_i = 1.5*10**16              #in /m^3\n",
-      "I_lambda = 12.5              #in mA/cm^2\n",
-      "I_lambda= I_lambda*10**-3       #in A/cm^2\n",
-      "L_e = math.sqrt(D_e*Torque_eo)    #in m\n",
-      "L_n = math.sqrt(D_n*Torque_ho)  #in m\n",
-      "J_s = e*((n_i)**2)*( ((D_e)/(L_e*N_A)) + ((D_n)/(L_n*N_D)) )         #in A/m^2\n",
-      "J_s= J_s*10**-4                      #in A/cm^2\n",
-      "V_OC = V_T*(math.log(1+(I_lambda/J_s)))                     #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Open circuit voltage is =%.2f volts ' %V_OC)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Open circuit voltage is =0.52 volts \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3, Page No 307"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Phi_o = 1.0*10**21               #in m^-2s^-1\n",
-      "alpha =1.0*10**5                 #in m^-1\n",
-      "e= 1.6*10**-19                 #in C\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "G_L1 = alpha*Phi_o             #in m^-3s^-1\n",
-      "W = 26.0                       #in \u00b5m\n",
-      "W = W * 10**-6                 #in m\n",
-      "G_L2 = alpha*Phi_o*(math.e)**((-alpha)*W)             #in m^-3s^-1\n",
-      "#temp=(1-math.e)**(-(alpha)*W)\n",
-      "J_L = e*Phi_o*(1-math.e**(-(alpha)*W))               #in A/m^2\n",
-      "J_L = J_L * 10**3*10**-4                         #in mA/cm^2\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Photo current density  is =%.1f mA/cm^2' %J_L)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Photo current density  is =14.8 mA/cm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo8.ipynb b/Advance_Semiconductor_Devices/ChapterNo8.ipynb
deleted file mode 100755
index 47137bb2..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo8.ipynb
+++ /dev/null
@@ -1,576 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chapter No 8 : Metal Semiconductor Field Effect Transistor <h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page No  323"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "I_DSS = 15.0    #in mA\n",
-      "V_GS_off = -5.0    #in V\n",
-      "V_GS = 0        #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_D = I_DSS*((1-(V_GS/V_GS_off))**2)     #in mA\n",
-      "V_GS = -1                   #in V\n",
-      "I_D1 = I_DSS*((1.0-(V_GS/V_GS_off))**2)         #in mA\n",
-      "V_GS = -4             #in V\n",
-      "I_D2 = I_DSS*((1-(V_GS/V_GS_off))**2)          #in mA\n",
-      "\n",
-      "#RESULTS\n",
-      "print('When V_GS=0, the drain current is =%.f mA ' %I_D)\n",
-      "print('When V_GS=-1V, the drain current is =%.2f mA ' %I_D1)\n",
-      "print('The interplanner spacing is =%.2f  \u00c5' %I_D2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When V_GS=0, the drain current is =15 mA \n",
-        "When V_GS=-1V, the drain current is =9.60 mA \n",
-        "The interplanner spacing is =0.60  \u00c5\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4, Page No 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_DSS = 20.0            #in mA\n",
-      "V_P = -8            #in V\n",
-      "g_mo = 5000.0       #in \u00b5s\n",
-      "V_GS = -4.0            #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "\n",
-      "I_D = I_DSS*((1-(V_GS/V_P))**2)      #in mA\n",
-      "g_m = g_mo*(1-(V_GS/V_P))            #in \u00b5s\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of drain current is =%.f mA ' %I_D)\n",
-      "print('The transconductance is =%.f  \u00b5s' %g_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of drain current is =5 mA \n",
-        "The transconductance is =2500  \u00b5s\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5 Page No 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_D1 = 10.0        #in mA\n",
-      "V_GS = -12.0    #in V\n",
-      "V_GSth = -3.0    #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "K = I_D1/( (V_GS-V_GSth)**2 )         #in mA/V\n",
-      "V_GS= -6                 #in V\n",
-      "I_D = K*((V_GS-V_GSth)**2)         #in mA\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_D  is =%.2f mA ' %I_D)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D  is =1.11 mA \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7, Page No 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_GS = -2.0            #in V\n",
-      "V_P = -5.0          #in V\n",
-      "V_DS = V_GS-V_P     #in V\n",
-      "I_DSS = 8.0      #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_D = I_DSS*((1-(V_GS/V_P))**2)        #in mA\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The minimum value of V_DS is =%.f  V ' %V_DS)\n",
-      "print('The drain current is =%.2f  mA' %I_D)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of V_DS is =3  V \n",
-        "The drain current is =2.88  mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8  Page No 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_P = -2                 #in V\n",
-      "I_DSS = 1.65             #in mA\n",
-      "I_D = 0.8                #in mA\n",
-      "V_DD = 24.0              #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS = V_P*(1- math.sqrt(I_D/I_DSS) )    #in V\n",
-      "g_mo = -(2*I_DSS)/V_P           #in mS\n",
-      "g_m = g_mo*(1-(V_GS/V_P))       #in mS\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_GS is =%.2f V  ' %V_GS)\n",
-      "print('The value of g_m is =%.2f  mS' %g_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_GS is =-0.61 V  \n",
-        "The value of g_m is =1.15  mS\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9 Page No 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_P = 5.0         #in V\n",
-      "I_DSS = -40.0     #in mA\n",
-      "I_D = -15.0         #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS = V_P*(math.sqrt(I_D/I_DSS)-1 )       #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The gate source voltage is =%.2f   V' %(V_GS*-1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The gate source voltage is =1.94   V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10, Page No 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_D1 = 1.9       #in mA\n",
-      "I_D2 = 1.0       #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "del_I_D = I_D1-I_D2          #in mA\n",
-      "V_GS2 = -3.3              #in V\n",
-      "V_GS1 = -3.0            #in V\n",
-      "del_V_GS = V_GS1-V_GS2      #in V\n",
-      "g_m = del_I_D/del_V_GS       #in mA/V\n",
-      "g_m = g_m * 10**3         #in \u00b5 mhos\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of transconductance is =%.f \u00b5 mhos ' %g_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of transconductance is =3000 \u00b5 mhos \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11 Page No 329"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_DS1 = 14.0          #in V\n",
-      "V_DS2 = 5.0          #in V\n",
-      "del_V_DS = V_DS1-V_DS2      # in V\n",
-      "I_D1 = 3.3            #in mA\n",
-      "I_D2 = 3              # in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "del_I_D = I_D1-I_D2           #in mA\n",
-      "r_d = del_V_DS/del_I_D          #in k ohms\n",
-      "print('The drain resistance  is =%.f  k ohms' %r_d)\n",
-      "V_GS1 = 0.4              #in V\n",
-      "V_GS2 = 0.1             #in V\n",
-      "del_V_GS = V_GS1-V_GS2     #in V\n",
-      "I_D1 = 3.3             #in mA\n",
-      "I_D2 = 0.71           #in mA\n",
-      "del_I_D = I_D1-I_D2       # in mA\n",
-      "g_m = del_I_D/del_V_GS    # in mA/V\n",
-      "g_m = g_m * 10**3           # in \u00b5mhos\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The transconductance is =%.f \u00b5mhos  ' %g_m)\n",
-      "Miu =r_d*10**3*g_m*10**-6\n",
-      "print('Amplification factor is =%.f  ' %Miu)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain resistance  is =30  k ohms\n",
-        "The transconductance is =8633 \u00b5mhos  \n",
-        "Amplification factor is =259  \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12 Page No 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "q = 1.6*10**-19         #in C\n",
-      "N_D = 10**15*10**6      #electrons/m^3\n",
-      "a = 3*10**-4            #in cm\n",
-      "a=a*10**-2              #in m\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Epsilon_o  = (36 * math.pi * 10**9)**-1\n",
-      "Epsilon = 12*Epsilon_o\n",
-      "V_P = (q*N_D*((a)**2))/(2*Epsilon)       #in V\n",
-      "print('Pinch off voltage is =%.2f   V ' %V_P)\n",
-      "V_GS = 1.0     #in V\n",
-      "V_P = 2.0      #in V\n",
-      "b = a*( 1-math.sqrt(V_GS/V_P) )         #in m\n",
-      "b = b * 10**6              #in \u00b5m\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The channel half width  is =%.2f  \u00b5m' %b)\n",
-      "# Note: In the book, the unit of channel half width is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pinch off voltage is =6.79   V \n",
-        "The channel half width  is =0.88  \u00b5m\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13 Page No 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_DSS = 8.0       #in mA\n",
-      "V_P = -4.0           #in V\n",
-      "a = 3*10**-4        #in cm\n",
-      "N_D = 10**15        #in electrons/cm^3\n",
-      "I_D = 3.0           #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS = V_P*( 1-math.sqrt(I_D/I_DSS) )           #in V\n",
-      "V_DS_sat = V_GS-V_P                             #in V\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_GS is =%.2f   V' %V_GS)\n",
-      "print('The value of V_DS_sat is =%.2f   V ' %V_DS_sat)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_GS is =-1.55   V\n",
-        "The value of V_DS_sat is =2.45   V \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 8.14 Page No 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_P = -4            #in V\n",
-      "I_DSS = 9.0       #in mA\n",
-      "V_GS = -2.0       #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_D = I_DSS*(( 1-(V_GS/V_P) )**2)        #in mA\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The drain current is =%.2f mA ' %I_D)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain current is =2.25 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 8.15 Page No 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "I_DSS = 12.0           #in mA\n",
-      "V_P = -6.0             #in V\n",
-      "V_GS = -1.0              #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "g_mo = (-2*I_DSS)/V_P      #in mA/V\n",
-      "g_m = g_mo*(1-(V_GS/V_P))         #in mS\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of transconductance is =%.2f  mS' %g_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of transconductance is =3.33  mS\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 8.16 Page No 336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "I_DSS = 10.0          #in mA\n",
-      "V_P = -5.0            #in V\n",
-      "V_GS = -2.5          #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "g_m = ((-2*I_DSS)/V_P)*(1-(V_GS/V_P))         #in mS .... correction\n",
-      "I_D = I_DSS * ((1-(V_GS/V_P))**2)             #in mA\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The transconductance is =%.f mS ' %g_m)\n",
-      "print('The drain current is =%.2f mA ' %I_D)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transconductance is =2 mS \n",
-        "The drain current is =2.50 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/ChapterNo9.ipynb b/Advance_Semiconductor_Devices/ChapterNo9.ipynb
deleted file mode 100755
index d2b16e2d..00000000
--- a/Advance_Semiconductor_Devices/ChapterNo9.ipynb
+++ /dev/null
@@ -1,1684 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chapter No 9 : MOS Transistors<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1, Page No  361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_TN = 0.7            #in V\n",
-      "W = 45*10**-4         #in cm\n",
-      "L = 4.0               #in \u00b5m\n",
-      "L = L * 10**-4        #in cm\n",
-      "t_ox = 450            #in \u00c5\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "t_ox = t_ox*10**-8           #in cm\n",
-      "V_GS = 1.4                   #in V\n",
-      "Miu_n = 700                  #in cm^2/V-s\n",
-      "Epsilon_ox = (8.85*10**-14)*(3.9)               #in F/cm\n",
-      "k_n = (W*Miu_n*Epsilon_ox)/(2*L*t_ox)               #A/V^2\n",
-      "I_D = k_n*((V_GS-V_TN)**2)                        #in A\n",
-      "k_n=k_n*10**3\n",
-      "I_D=I_D*10**3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of k_n is : =%.2f   mA/V^2' %k_n)\n",
-      "print('The current  is =%.2f  mA' %I_D)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of k_n is : =0.30   mA/V^2\n",
-        "The current  is =0.15  mA\n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 87
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2, Page No 376"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "I_Don = 6.0                #in mA\n",
-      "I_Don= I_Don*10**-3        #in A\n",
-      "V_GSon = 8                 #in V\n",
-      "V_GSth = 3                 #in V\n",
-      "V_DD = 12.0                #in V\n",
-      "R_D= 2*10**3               #in \u03a9\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "k= I_Don/(V_GSon-V_GSth)**2             #in A/V^2\n",
-      "A= R_D**2                 #assumed\n",
-      "B= 2*R_D*V_GSth-2*R_D*V_DD-1/k              #assumed\n",
-      "C= (V_DD-V_GSth)**2                #assumed\n",
-      "\n",
-      "I_DQ= 0.0027940                #in A\n",
-      "V_DSQ= V_DD-I_DQ*R_D            #in V\n",
-      "I_DQ=I_DQ*10**3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_DQ is : =%.2f  mA' %I_DQ)\n",
-      "print('The value of V_DSQ is : =%.2f volts ' %V_DSQ)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DQ is : =2.79  mA\n",
-        "The value of V_DSQ is : =6.41 volts \n"
-       ]
-      }
-     ],
-     "prompt_number": 88
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3 Page No 378"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_GS = 6.0           #in V\n",
-      "I_D = 4.0            #in mA\n",
-      "V_GSth = 2.0         #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_DS = V_GS          #in V\n",
-      "V_DD = 2*V_DS        #in V\n",
-      "R_D = (V_DD-V_DS)/I_D         #in k ohm\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_DD is =%.2f  V  ' %V_DD)\n",
-      "print('The value of R_D  is  =%.2f   k ohm' %R_D)\n",
-      "print('The very high value for the gate to drain resistance is : 10 M\u03a9 ' )\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_DD is =12.00  V  \n",
-        "The value of R_D  is  =1.50   k ohm\n",
-        "The very high value for the gate to drain resistance is : 10 M\u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 89
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4, Page No 379"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_Don = 3*10**-3  \n",
-      "V_GSon = 10.0          #in V\n",
-      "V_GSth= 5.0        #in V\n",
-      "R2= 18*10**6        #in \u03a9\n",
-      "R1= 22*10**6      #in \u03a9\n",
-      "R_S=820         #in \u03a9\n",
-      "R_D=3*10**3     #in \u03a9\n",
-      "V_DD= 40.0      #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_G= V_DD*R2/(R1+R2)           #in V\n",
-      "k= I_Don/(V_GSon-V_GSth)**2           #in A/V**2\n",
-      "A= R_S**2                  #assumed\n",
-      "B= 2*R_S*V_GSth-2*R_S*V_G-1/k           #assumed\n",
-      "C= (V_G-V_GSth)**2          #assumed\n",
-      "I_D= 0.0067246               #in A\n",
-      "I_DQ= I_D                 #in A\n",
-      "V_GSQ= V_G-I_D*R_S      #in V\n",
-      "V_DSQ= V_DD-I_DQ*(R_D+R_S)         #in V\n",
-      "I_DQ=I_DQ*10**3\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_DQ  is =%.2f  mA' %I_DQ)\n",
-      "print('The value of V_GSQ is =%.2f volts ' %V_GSQ)\n",
-      "print('The value of V_DSQ is =%.2f volts ' %V_DSQ)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DQ  is =6.72  mA\n",
-        "The value of V_GSQ is =12.49 volts \n",
-        "The value of V_DSQ is =14.31 volts \n"
-       ]
-      }
-     ],
-     "prompt_number": 90
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5 Page No 380"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_D= '0.3*(V_GS-V_P)^2'                #given expression\n",
-      "V_DD= 30.0                           #in V\n",
-      "V_P= 4                               #in V\n",
-      "R_GS = 1.2*10**6                      #in \u03a9\n",
-      "R_G = 1.2*10**6                       #in \u03a9\n",
-      "I_D= 1.2           #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Req= R_GS/(R_GS+R_G)                  #in \u03a9\n",
-      "R_D= 15.0                             #in \u03a9\n",
-      "A= (R_D*Req)**2                       #assumed\n",
-      "B= -(2*R_D*Req*(V_DD*Req-V_P)+1/0.3)      #assumed\n",
-      "C= (V_DD*Req-V_P)**2                  #assumed\n",
-      "I_DSQ= I_D             #in mA\n",
-      "V_GS= (V_DD-I_D*R_D)    #in V\n",
-      "V_DS= Req*V_GS         #in V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_DSQ  is : =%.2f  mA' %I_DSQ)\n",
-      "print('The value of V_GS  is :  =%.2f volts ' %V_GS)\n",
-      "print('The value of V_DS  is : =%.2f volts ' %V_DS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DSQ  is : =1.20  mA\n",
-        "The value of V_GS  is :  =12.00 volts \n",
-        "The value of V_DS  is : =6.00 volts \n"
-       ]
-      }
-     ],
-     "prompt_number": 91
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6 Page No 380"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "k = 0.1          #in mA/V^2\n",
-      "V_T  = 1.0            # in V\n",
-      "R1 = 33.0          #in k ohm\n",
-      "R2 = 21.0         #in k ohm\n",
-      "V_DD = 6.0       #in V\n",
-      "R_D = 18           #in k ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_G = (R2/(R2+R1))*V_DD             #in V\n",
-      "V_S = 0               #in V\n",
-      "V_GS = V_G-V_S             #in V\n",
-      "I_D = k*((V_GS-V_T)**2)         #in mA\n",
-      "V_DS = V_DD - (I_D*R_D)       #in V\n",
-      "V_DSsat = V_GS-V_T            #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_D  is =%.2f mA ' %I_D)\n",
-      "print('The value of V_DS(sat) is =%.2f  V ' %V_DS)\n",
-      "print('The value of V_DS(sat)  is =%.2f V ' %V_DSsat)\n",
-      "\n",
-      "if V_DS > V_DSsat :\n",
-      "    print('MOSFET is in saturation region  ')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D  is =0.18 mA \n",
-        "The value of V_DS(sat) is =2.80  V \n",
-        "The value of V_DS(sat)  is =1.33 V \n",
-        "MOSFET is in saturation region  \n"
-       ]
-      }
-     ],
-     "prompt_number": 92
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8, Page No 382"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "R2 = 18.0       #in k ohm\n",
-      "R1 = 33.0       #in k ohm\n",
-      "V_DD = 6.0       #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_G = (R2/(R1+R2))*V_DD              #in V\n",
-      "V_S = V_DD                          # in V\n",
-      "V_SG = V_S-V_G                      #in V\n",
-      "k = 0.1\n",
-      "V_T = -1                          #in V\n",
-      "I_D = k*((V_SG+V_T)**2)           #in mA\n",
-      "R_D = 3.0                     # in k ohm\n",
-      "V_SD = V_DD - (I_D*R_D)            #in V\n",
-      "V_SDsat  = V_SG+V_T                 #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_SG is =%.2f V ' %V_SG)\n",
-      "print('The value of I_D  is =%.2f mA ' %I_D)\n",
-      "print('The value of V_SD  is =%.2f V ' %V_SD)\n",
-      "print('The value of V_SD(sat)  is =%.2f V ' %V_SDsat)\n",
-      "\n",
-      "if V_SD > V_SDsat :\n",
-      "       print('The p MOSFET is indeed biased in the saturation region')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_SG is =3.88 V \n",
-        "The value of I_D  is =0.83 mA \n",
-        "The value of V_SD  is =3.51 V \n",
-        "The value of V_SD(sat)  is =2.88 V \n",
-        "The p MOSFET is indeed biased in the saturation region\n"
-       ]
-      }
-     ],
-     "prompt_number": 93
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9 Page No 384"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_G= 1.5                # in V\n",
-      "V_P= -3.0              # in V\n",
-      "R_S= 750.0               # in \u03a9\n",
-      "R_D= 1800              #in \u03a9\n",
-      "I_DSS= 6*10**-3         # in A\n",
-      "V_DD= 18            #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "A= R_S**2\n",
-      "B=(2*R_S*(V_P-V_G)-V_P**2/I_DSS)\n",
-      "C=(V_P-V_G)**2\n",
-      "I_D= 0.0031170               #in A\n",
-      "I_DQ= I_D                 #in A\n",
-      "V_DS= V_DD-I_D*(R_D+R_S)        # in V\n",
-      "V_DSQ= V_DS                  # in V\n",
-      "I_DQ=I_DQ*10**3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_DQ  is : =%.2f mA ' %I_DQ)\n",
-      "print('The value of V_DSQ  is : =%.2f volts ' %V_DSQ)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DQ  is : =3.12 mA \n",
-        "The value of V_DSQ  is : =10.05 volts \n"
-       ]
-      }
-     ],
-     "prompt_number": 94
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10 Page No 384"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_GS = 4.0              #in V\n",
-      "V_P = 2.0               #in V\n",
-      "R2 = 10.0             #in k ohm\n",
-      "R1 = 30.0           #in k ohm\n",
-      "R_D= 2.5           #in kohm\n",
-      "I_D= 15            #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_D= I_D*10**-3           #in A\n",
-      "V_DD = 25                 #in V\n",
-      "V_G = (V_DD/R_D)*V_DD/(R1+R2)                #in V\n",
-      "R_S = (V_G-V_GS)/I_D                 # in ohm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of R_S  is =%.f ohm ' %R_S)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of R_S  is =150 ohm \n"
-       ]
-      }
-     ],
-     "prompt_number": 95
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.11 Page No 385"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "k= 0.1            #in mA/V^2\n",
-      "V_T= 1            # in V\n",
-      "R2= 87*10**3         #in \u03a9\n",
-      "R1= 110*10**3       # in \u03a9\n",
-      "R_S=2            #in k\u03a9\n",
-      "R_D=2            #in k\u03a9\n",
-      "V_DD= 6             #in V\n",
-      "V_SS= 6          #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_G= (V_DD+V_SS)*R2/(R1+R2)              # in V\n",
-      "A= R_S**2              # assumed\n",
-      "B= -(2*R_S*(V_G+V_SS-V_T)+1/k)              # assumed\n",
-      "C= (V_G+V_SS-V_T)**2                    #assumed\n",
-      "I_D= 2.6001587           # in mA\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_D is : =%.2f mA' %I_D)\n",
-      "V_DS=V_DD+V_SS-I_D*R_D-I_D*R_S        #in V\n",
-      "print('The value of V_DS  is : =%.2f volts' %V_DS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D is : =2.60 mA\n",
-        "The value of V_DS  is : =1.60 volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 96
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.12 Page No 386"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "k = 0.16                      #in mA/V^2\n",
-      "V_T = 2.0                   #in V\n",
-      "I_D = 0.5                      #in mA\n",
-      "V_DD = 6.0                 #in V\n",
-      "V_SS = -6                 #in V\n",
-      "R_D = 10.0                     #in k ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS = V_T + (math.sqrt(I_D/k))               #in V\n",
-      "R_S = 2                #in k ohm\n",
-      "V_S = (I_D*R_S) - V_DD           # in V\n",
-      "V_G = V_GS+V_S                     # in V\n",
-      "I = 0.1*I_D                 #in mA\n",
-      "R2 = (V_G+V_DD)/I                #in k ohm\n",
-      "print('The value of R2  is =%.2f k ohm' %R2)\n",
-      "R1 = (V_DD - V_G)/I                  #in k ohm\n",
-      "print('The value of R1 is =%.2f k ohm' %R1)\n",
-      "V_DS = (V_DD-V_SS) - (I_D*(R_S+R_D))           #in V\n",
-      "print('The value of V_DS is =%.2f V' %V_DS)\n",
-      "V_DSsat = V_GS-V_T                      # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_DS(sat) is =%.2f V' %V_DSsat)\n",
-      "if V_DS > V_DSsat :\n",
-      "    print('The MOSFET is in saturation region')\n",
-      "\n",
-      "    # Note: The value of R1 is in k ohm but in the book it is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of R2  is =95.36 k ohm\n",
-        "The value of R1 is =144.64 k ohm\n",
-        "The value of V_DS is =6.00 V\n",
-        "The value of V_DS(sat) is =1.77 V\n",
-        "The MOSFET is in saturation region\n"
-       ]
-      }
-     ],
-     "prompt_number": 97
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.13 Page No 387"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_DD = 6.0              #in V\n",
-      "V_D = 3.0             #in V\n",
-      "R_D = 10.0             #in k ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_DQ = (V_DD-V_D)/R_D           #in mA\n",
-      "print('The value of I_DQ is =%.2f mA ' %I_DQ)\n",
-      "V_T = 0.8           #in V\n",
-      "k = 0.12             #in mA/V^2\n",
-      "V_GS = math.sqrt(I_DQ/k) + V_T           #in V\n",
-      "V_S = -V_GS                   #in V\n",
-      "V_DS = V_D-V_S               #in V\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_DS is =%.2f V' %V_DS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DQ is =0.30 mA \n",
-        "The value of V_DS is =5.38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 98
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.14 Page No 388"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_D = 0.3         # in mA\n",
-      "k = 0.12            #in mA/V^2\n",
-      "V_T = 1.0          #in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS = V_T + (math.sqrt(I_D/k))          #in V\n",
-      "V_S = -V_GS                             #in V\n",
-      "V_DD = 6.0                  #in V\n",
-      "V_D = 3.0                   # in V\n",
-      "I_DQ = 0.3                   #in mA\n",
-      "R_D = (V_DD-V_D)/I_DQ       #in k ohm\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of R_D  is =%.2f k ohm' %R_D)\n",
-      "V_DS = V_D - V_S                       #in V\n",
-      "print('The value of V_DS  is =%.2f V' %V_DS)\n",
-      "V_DSsat = V_GS - V_T                 #in V\n",
-      "print('The value of V_DS(sat)  is =%.2f V' %V_DSsat)\n",
-      "if V_DS > V_DSsat :\n",
-      "    print('The MOSFET is in saturation region')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of R_D  is =10.00 k ohm\n",
-        "The value of V_DS  is =5.58 V\n",
-        "The value of V_DS(sat)  is =1.58 V\n",
-        "The MOSFET is in saturation region\n"
-       ]
-      }
-     ],
-     "prompt_number": 99
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.15 Page No 389"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "k= 0.05             #in mA/V^2\n",
-      "V_T= 1            #in V\n",
-      "V_DD= 6          # in V\n",
-      "R_S= 9.1            #in k\u03a9\n",
-      "I_D= 0.3631765    # in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "A= R_S**2          # assumed\n",
-      "B=-(2*V_DD*R_S+1/k)         # assumed\n",
-      "C= V_DD**2         #assumed\n",
-      "V_GS= V_DD-I_D*R_S   #in V\n",
-      "V_DS= V_GS         # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_D  is : =%.2f mA' %I_D)\n",
-      "print('The value of V_GS  is : =%.2f volts' %V_GS)\n",
-      "print('The value of V_DS  is : =%.2f volts' %V_DS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D  is : =0.36 mA\n",
-        "The value of V_GS  is : =2.70 volts\n",
-        "The value of V_DS  is : =2.70 volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 100
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.16 Page No 390"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "k1= 0.01          #in mA/V^2\n",
-      "k2= 0.05      #in mA/V^2\n",
-      "V_DD= 5         # in V\n",
-      "V_T1=1           # in V\n",
-      "V_T2=1         # in V\n",
-      "Vi= 5         #in V\n",
-      "Vo= 0.3485163   # in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "A=(k1+k2)\n",
-      "B=-(2*k1*(V_DD-V_T1)+2*k2*(Vi-V_T2))\n",
-      "C=k1*(V_DD-V_T1)**2\n",
-      "V_GS2= Vi             # in V\n",
-      "V_DS2= Vo            # in V\n",
-      "V_GS1= V_DD-Vo         # in V\n",
-      "I_D1= k1*(V_GS1-V_T1)**2            #in mA\n",
-      "I_D2= I_D1                  # in mA\n",
-      "print('Part (i) For Vi = 5 V')\n",
-      "print('The output voltage is : =%.2f volts' %Vo)\n",
-      "print('The value of I_D1  is : =%.2f mA' %I_D1)\n",
-      "print('The value of I_D2  is : =%.2f mA' %I_D2)\n",
-      "# Analysis for Vi= 1.5V\n",
-      "Vi= 1.5         #in V\n",
-      "Vo= V_DD-V_T1-math.sqrt(k2/k1)*(Vi-V_T2)          # in V\n",
-      "I_D2= k2*(Vi-V_T2)**2               #in mA\n",
-      "I_D1= I_D2             #in mA\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Part (ii) For Vi = 1.5 V')\n",
-      "print('The output voltage  is : =%.2f volts' %Vo)\n",
-      "print('The value of I_D1  is : =%.2f mA' %I_D1)\n",
-      "print('The value of I_D2  is : =%.2f mA' %I_D2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part (i) For Vi = 5 V\n",
-        "The output voltage is : =0.35 volts\n",
-        "The value of I_D1  is : =0.13 mA\n",
-        "The value of I_D2  is : =0.13 mA\n",
-        "Part (ii) For Vi = 1.5 V\n",
-        "The output voltage  is : =2.88 volts\n",
-        "The value of I_D1  is : =0.01 mA\n",
-        "The value of I_D2  is : =0.01 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 101
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.17 Page No 392"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "k = 0.12         #in mA/V^2\n",
-      "V_T = -2.5        #in V\n",
-      "V_GS = 0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_D = k*((V_GS-V_T)**2)        #in mA\n",
-      "print('The value of I_D is =%.2f mA' %I_D)\n",
-      "V_DD = 6            #in V\n",
-      "R_S = 4.7            #in k ohm \n",
-      "V_DS = V_DD -(I_D*R_S)         #in V\n",
-      "print('The value of V_DS  is =%.2f V' %V_DS)\n",
-      "V_S = 0               # in V \n",
-      "V_DSsat = V_S - V_T            # in V\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_DS(sat) is  =%.2f V' %V_DSsat)\n",
-      "if V_DS < V_DSsat :\n",
-      "    print('The device is in the non saturation region')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D is =0.75 mA\n",
-        "The value of V_DS  is =2.47 V\n",
-        "The value of V_DS(sat) is  =2.50 V\n",
-        "The device is in the non saturation region\n"
-       ]
-      }
-     ],
-     "prompt_number": 102
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.18 Page No 394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "k4 = 0.125            #in mA/V^2\n",
-      "k3 = k4              #in mA/V^2\n",
-      "k2 = k4               #in mA/V^2\n",
-      "k1 = 0.25             #in mA/V^2\n",
-      "V_T1 = 0.8            # in V\n",
-      "V_T2 = V_T1           # in V\n",
-      "V_T3 = V_T1          #in V\n",
-      "V_T4 = V_T1           #in V\n",
-      "V_SS = -5             #in V\n",
-      "V_DD = 5              #in V\n",
-      "R_D = 10              #in k ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS3 = ((math.sqrt(k4/k3) * (-V_SS - V_T4))+V_T3)/(1+math.sqrt(k4/k3))          #in V\n",
-      "I_Q = k2*((V_GS3-V_T2)**2)          #in mA\n",
-      "I_D1 = I_Q                   #in mA\n",
-      "V_GS1 = V_T1 + (math.sqrt(I_D1/k1))                 #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_GS1 is =%.2f V ' %V_GS1)\n",
-      "V_DS2 = (-V_SS-V_GS1)            #in V\n",
-      "print('The value of V_DS2 is =%.2f V ' %V_DS2)\n",
-      "V_DS1 = V_DD - (I_Q*R_D) - (V_SS + V_DS2)            #in V\n",
-      "print('The value of V_DS1 is =%.2f  V' %V_DS1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_GS1 is =2.00 V \n",
-        "The value of V_DS2 is =3.00 V \n",
-        "The value of V_DS1 is =3.39  V\n"
-       ]
-      }
-     ],
-     "prompt_number": 103
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.19 Page No 395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "R2 = 20.0               # in  k ohm\n",
-      "R1 = 30.0               #in  k ohm\n",
-      "R_D = 20               # in  k ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "R_D=R_D*10**3         #in ohm\n",
-      "V_DD = 5              #in V\n",
-      "V_G = (R2/(R1+R2))*V_DD             #in V\n",
-      "V_S = 0                     #in V\n",
-      "V_GS = V_G                   #in V\n",
-      "k = 100*10**-6             #in A/V^2\n",
-      "V_T = 1                    #in V\n",
-      "I_DQ = k*((V_GS-V_T)**2)       #in A\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "V_DSQ = V_DD - (I_DQ*R_D)             #in V \n",
-      "print('The value of V_DSQ  is =%.f V ' %V_DSQ)\n",
-      "I_DQ=I_DQ*10**6\n",
-      "print('The value of I_DQ  is =%.f  \u00b5A' %I_DQ)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_DSQ  is =3 V \n",
-        "The value of I_DQ  is =100  \u00b5A\n"
-       ]
-      }
-     ],
-     "prompt_number": 104
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.20 Page No 395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_P= -8            # in V\n",
-      "R_S= 2.4           #in k\u03a9\n",
-      "I_DSS= 8           #in mA\n",
-      "V_DD= 20            #in V\n",
-      "R_D= 6.2           #in k\u03a9\n",
-      "I_D= 1.7668305   #in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "A= R_S**2\n",
-      "B=(2*R_S*(V_P)-V_P**2/I_DSS)\n",
-      "C=(V_P)**2\n",
-      "I_DQ= I_D             #in mA\n",
-      "V_GSQ= -I_D*R_S;\n",
-      "V_D= V_DD-I_D*R_D           #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_DQ is : =%.2f  mA' %I_DQ)\n",
-      "print('The value of V_GSQ  : =%.2f  volts' %V_GSQ)\n",
-      "print('The value of V_D  is : =%.2f  volts' %V_D)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DQ is : =1.77  mA\n",
-        "The value of V_GSQ  : =-4.24  volts\n",
-        "The value of V_D  is : =9.05  volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 105
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.21 Page No 396"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "k= 75*10**-3              #in mA/V^2\n",
-      "Vth= -0.8                  #in V\n",
-      "R2 = 100                   #in k ohm\n",
-      "R1 = 100                   #in  k ohm\n",
-      "R_S= 6                     # in k\u03a9\n",
-      "R_D= 3                   # in k\u03a9\n",
-      "V_SS = 10                 # in V\n",
-      "V_S = 2.0602803\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_G = (R2/(R1+R2))*V_SS                  #in V\n",
-      "V_GS= V_G-V_S                          #in V\n",
-      "I_D= k*(V_GS-Vth)**2                           # in V\n",
-      "V_DS= -V_SS+I_D*(R_D+R_S)                # in V\n",
-      "V_D= I_D*R_D                            #in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_D  is : =%.2f mA ' %I_D)\n",
-      "print('The value of V_DS  is =%.2f volts ' %V_DS)\n",
-      "print('The value of V_D  is : =%.2f  volts' %V_D)\n",
-      "print('The value of V_S  is : =%.2f  volts' %V_S)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D  is : =0.12 mA \n",
-        "The value of V_DS  is =-8.93 volts \n",
-        "The value of V_D  is : =0.36  volts\n",
-        "The value of V_S  is : =2.06  volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 106
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.22 Page No 396"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_T = 1.0                      # in V\n",
-      "k = 160*10**-6               # in A/V^2\n",
-      "I_DQ = 160*10**-6            # in A\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS = V_T + math.sqrt(I_DQ/k)         #in V\n",
-      "V_DD = 5              # in V\n",
-      "V_DSQ = 3             # in V\n",
-      "R_D = (V_DD - V_DSQ)/(I_DQ)            # in ohm\n",
-      "R_D = R_D * 10**-3                 # in k ohm\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of R_D  is =%.2f k ohm' %R_D)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of R_D  is =12.50 k ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 107
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.23 Page No 397"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_DD= 12          # in V\n",
-      "V_T= 2            # in V\n",
-      "kn= 0.5           # in mA/V^2\n",
-      "R1 = 2.2         # in M ohm\n",
-      "R2 = 1.8         # in M ohm\n",
-      "R_S= 1.5         # in k\u03a9\n",
-      "R_D= 3.9         # in k\u03a9\n",
-      "I_D= 1.2237155                   # in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_G = (R2/(R1+R2))*V_DD         # in V\n",
-      "V_GS= V_G-I_D*R_S              # V\n",
-      "I_D= I_D-kn*(V_GS-V_T)**2      # in mA\n",
-      "I_DQ= I_D                     #in mA\n",
-      "V_DSQ= V_DD-I_D*(R_D+R_S)      # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_DQ  is : =%.2f mA ' %I_DQ)\n",
-      "print('The value of V_DSQ  is : =%.2f  volts' %V_DSQ)\n",
-      "V_GS= V_G-I_D*R_S            # V\n",
-      "V_DSsat= V_GS-V_T          # in V\n",
-      "print('The value of  V_DS  %.2f  is greater than the value of ' %V_DSQ)\n",
-      "print('V_DSsat %.2f So the MOSFET is in saturation region ' %V_DSsat)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DQ  is : =-0.00 mA \n",
-        "The value of V_DSQ  is : =12.00  volts\n",
-        "The value of  V_DS  12.00  is greater than the value of \n",
-        "V_DSsat 3.40 So the MOSFET is in saturation region \n"
-       ]
-      }
-     ],
-     "prompt_number": 108
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.24 Page No 397"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "kn= 0.5           # in mA/V^2\n",
-      "V_T= 1          # in V\n",
-      "R2 = 40         # in k ohm\n",
-      "R1 = 60         # in k ohm\n",
-      "R_S= 1          # in k ohm\n",
-      "R_D= 2          # in k ohm\n",
-      "V_DD = 5        # in V\n",
-      "V_SS = -5       # in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_R2 = (R2/(R2+R1))*(V_DD-V_SS)            # in V\n",
-      "V_G = V_R2 - V_DD               # in V\n",
-      "V_S= I_D*R_S+V_SS              # in V\n",
-      "V_GS= V_G-V_S                  # in V\n",
-      "I_D= 1.3542487                  # in mA\n",
-      "I_DQ= I_D                      #in mA\n",
-      "V_S= I_D*R_S+V_SS            # in V\n",
-      "V_GS= V_G-V_S                # in V\n",
-      "V_DSQ= V_DD-V_SS-I_D*(R_D+R_S)         # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_DQ  is :=%.2f mA ' %I_DQ)\n",
-      "print('The value of V_GS  is : =%.2f volts ' %V_GS)\n",
-      "print('The value of V_DSQ  is : =%.2f  volts' %V_DSQ)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_DQ  is :=1.35 mA \n",
-        "The value of V_GS  is : =-1.35 volts \n",
-        "The value of V_DSQ  is : =5.94  volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 109
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.25 Page No 398"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "R_S1 = 100*10**-3      # in k ohm\n",
-      "R_S2 = 100*10**-3      # k ohm\n",
-      "R_S = R_S1+R_S2        #in k ohm\n",
-      "R_D= 1.8               # in k ohm\n",
-      "I_DSS= 12             # in mA\n",
-      "Vp= -3.5              # in V\n",
-      "V_DD= 22              # in V\n",
-      "rd= 25                # in k ohm\n",
-      "R_L= 47               # in k ohm\n",
-      "I_D= 5.573521          # in mA\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_GS= -I_D*R_S     # in V\n",
-      "I_D= I_D-I_DSS*(1-V_GS/Vp)**2   # in mA\n",
-      "print('The value of I_D  is : =%.2f  mA' %I_D)\n",
-      "V_GS= -I_D*R_S           # in V\n",
-      "print('The value of V_GS  is : =%.2f volts ' %V_GS)\n",
-      "V_DS= V_DD-I_D*(R_D+R_S)          # in V\n",
-      "print('The value of V_DS  is :  =%.2f volts ' %V_DS)\n",
-      "gmo= -2*I_DSS/Vp              # in mS\n",
-      "gm= gmo*(1-V_GS/Vp)           # in mS\n",
-      "miu= gm*rd\n",
-      "Av= -miu*R_D*R_L/(R_D+R_L)/(rd+R_D*R_L/(R_D+R_L)+(1+miu)*R_S1)\n",
-      "print('The value of Av is :  =%.2f  ' %Av)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D  is : =-0.000000  mA\n",
-        "The value of V_GS  is : =0.00 volts \n",
-        "The value of V_DS  is :  =22.00 volts \n",
-        "The value of Av is :  =-6.76  \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.26 Page No 399"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_T = 1                # in V\n",
-      "k = 0.5               # in mA/V^2\n",
-      "R2 = 40              # in k ohm\n",
-      "R1 = 60             # in k ohm\n",
-      "R_S= 1               # in k ohm\n",
-      "R_D= 2              # in k ohm\n",
-      "V_DD = 5            # in V\n",
-      "V_G = (R2/(R2+R1))*V_DD         # in V\n",
-      "V_GS= V_G-I_D*R_S                # in V\n",
-      "I_D= 0.2679492                     # in mA\n",
-      "V_GS= V_G-I_D*R_S                 # in V\n",
-      "V_DS= V_DD-I_D*(R_D+R_S)          # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_D in mA is : =%.2f  ' %I_D)\n",
-      "print('The value of V_GS in volts is :=%.2f  ' %V_GS)\n",
-      "print('The value of V_DS in volts is : =%.2f  ' %V_DS)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D in mA is : =0.27  \n",
-        "The value of V_GS in volts is :=-0.27  \n",
-        "The value of V_DS in volts is : =4.20  \n"
-       ]
-      }
-     ],
-     "prompt_number": 111
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.27 Page No 400"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "R_D = 7.5             # in k ohm\n",
-      "V_T = -0.8             # in V\n",
-      "k = 0.2              # in mA/V^2\n",
-      "R2 = 50                # in ohm\n",
-      "R1 = 50               # in  ohm\n",
-      "V_DD = 5             # in V\n",
-      "V_S = 5            # in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_G = (R2/(R2+R1))*V_DD          # in V\n",
-      "V_GS = V_G - V_S            # in V\n",
-      "I_D = k*((V_GS-V_T)**2)        # in mA\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Drain current in is =%.2f mA' %I_D)\n",
-      "V_SD = V_DD - (I_D*R_D)    # in V\n",
-      "print('Source to drain voltage  is =%.2f V' %V_SD)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current in is =3.53 mA\n",
-        "Source to drain voltage  is =-21.46 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 112
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.28 Page No 400"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "I_Don = 5*10**-3            # in A\n",
-      "V_GSon = 6                # in V\n",
-      "V_GSth = 3                # in V\n",
-      "k = I_Don/(V_GSon-V_GSth)**2      # in A/V^2 \n",
-      "R2 = 6.8                  #in M ohm\n",
-      "R1 = 10                  #in M ohm\n",
-      "R_S= 750                # in ohm\n",
-      "R_D= 2.2*10**3        # in ohm\n",
-      "V_DD = 24             # in V\n",
-      "R_S = 750           # in  ohm\n",
-      "V_G= R2*V_DD/(R1+R2)         # in V\n",
-      "V_GS= V_G-I_D*R_S           # in V\n",
-      "I_D= I_D-k*(V_GS-V_GSth)**2\n",
-      "I_D= 0.0049660              # in A\n",
-      "I_DQ= I_D                    # in A\n",
-      "V_GS= V_G-I_D*R_S           # in V\n",
-      "V_GSQ= V_GS                 # in V\n",
-      "V_DSQ= V_DD-I_DQ*(R_D+R_S)       # in V\n",
-      "I_D=I_D*10**3\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of I_D  is : =%.2f mA ' %I_D)\n",
-      "print('The value of V_GSQ  is : =%.2f volts ' %V_GSQ)\n",
-      "print('The value of V_DSQ  is : =%.2f volts ' %V_DSQ)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I_D  is : =4.97 mA \n",
-        "The value of V_GSQ  is : =5.99 volts \n",
-        "The value of V_DSQ  is : =9.35 volts \n"
-       ]
-      }
-     ],
-     "prompt_number": 113
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.29 Page No 401"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "I_Don = 4*10**-3              # in A\n",
-      "V_GSon = 6                 # in V\n",
-      "V_GSth = 3                 # in V\n",
-      "V_DS= 6                    # in V\n",
-      "I_D= I_Don                 # in A\n",
-      "k = I_Don/((V_GSon-V_GSth)**2)        # in A/V^2\n",
-      "V_GS= I_D-k*(V_GS-V_GSth)**2\n",
-      "V_GS= 6           # in V\n",
-      "V_DD= 2*V_DS               # in V\n",
-      "R_D= (V_DD-V_GS)/I_D            # in ohm\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_GS  is : =%.f volts ' %V_GS)\n",
-      "print('The value of V_DD  is : =%.f volts ' %V_DD)\n",
-      "print('The value of R_D  is : =%.f  k\u03a9' %R_D)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_GS  is : =6 volts \n",
-        "The value of V_DD  is : =12 volts \n",
-        "The value of R_D  is : =1500  k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 114
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.31 Page No 402"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_DD = 5               # in V\n",
-      "V_T= 1                  # in V\n",
-      "k= 1             # in mA/V^2\n",
-      "R1 = 1           # in M ohm\n",
-      "R2 = 1           # in M ohm\n",
-      "R_S= 2            # in k ohm\n",
-      "R_D= 2           # in k ohm\n",
-      "I1 = V_DD/(R1+R2)         # in A\n",
-      "print('The value of I1  is : =%.2f  \u00b5A' %I1)\n",
-      "V_A = (R2/(R2+R1))*V_DD              # in V\n",
-      "print('The value of V_A and V_G  is :  =%.2f volts ' %V_A)\n",
-      "I_D= 0.4243061\n",
-      "V_C= I_D*R_S              # in V\n",
-      "V_GS= V_A-V_C            # in V\n",
-      "print('The value of I_D  is : =%.2f  mA' %I_D)\n",
-      "V_B= V_DD-I_D*R_D       # in V\n",
-      "V_C= I_D*R_S         # in V\n",
-      "V_DS= V_B-V_C       # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_B  is : =%.2f volts ' %V_B)\n",
-      "print('The value of V_C  is : =%.2f volts ' %V_C)\n",
-      "print('The value of V_DS  is : =%.2f  volts' %V_DS)\n",
-      "# Note: In the book, the calculated values are wrong, this is why the answer in the book is wrong.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of I1  is : =2.00  \u00b5A\n",
-        "The value of V_A and V_G  is :  =0.00 volts \n",
-        "The value of I_D  is : =0.42  mA\n",
-        "The value of V_B  is : =4.15 volts \n",
-        "The value of V_C  is : =0.85 volts \n",
-        "The value of V_DS  is : =3.30  volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 115
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.32 Page No 403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_DSS = 12            # in mA\n",
-      "I_DSS= I_DSS*10**-3      # in A\n",
-      "V_P = -3              # in V\n",
-      "r_d = 45             # in k ohm\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r_d= r_d*10**3           # in ohm\n",
-      "g_m = I_DSS/abs(V_P) # in S\n",
-      "#Part (i)\n",
-      "R1 = 91           # in M ohm\n",
-      "R1=R1*10**6        #in ohm\n",
-      "R2 = 10           # in M ohm\n",
-      "R2= R2*10**6      # in ohm\n",
-      "Ri= R1*R2/(R1+R2)      # in ohm\n",
-      "Ri=Ri*10**6\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of Ri  is :  =%.f Mohm' %(Ri/(10**12)))\n",
-      "#Part (ii)\n",
-      "R_S = 1.1      #in k ohm\n",
-      "R_S = R_S * 10**3       # in ohm\n",
-      "R_o= (R_S*1/g_m)/(R_S+1/g_m)      # in ohm\n",
-      "print('The value of R_C  is :  =%.2f  ohm' %R_o)\n",
-      "\n",
-      "# Part (iii)\n",
-      "R_desh_o= R_o*r_d/(R_o+r_d)   # in ohm\n",
-      "print('The value of R_desh_o  is :  =%.2f ohm ' %R_desh_o)\n",
-      "# Part (iv)\n",
-      "Av= g_m*(R_S*r_d/(R_S+r_d))/(1+g_m*(R_S*r_d/(R_S+r_d)))\n",
-      "print('The value of Av is :  =%.2f ' %Av)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of Ri  is :  =9 Mohm\n",
-        "The value of R_C  is :  =203.70  ohm\n",
-        "The value of R_desh_o  is :  =202.79 ohm \n",
-        "The value of Av is :  =0.81 \n"
-       ]
-      }
-     ],
-     "prompt_number": 116
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.34 Page No 404"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_S2 = -2   # in V\n",
-      "V_GS2 = -V_S2    # in V\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "I_DS2 = (V_GS2-1)**2        # in mA\n",
-      "I = 2           # in mA\n",
-      "I_DS1 = I-I_DS2      # in mA\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The current flow through M1 MOSFET  is =%.f mA' %I_DS1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current flow through M1 MOSFET  is =1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 117
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.35 Page No 404"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_DD= 10         # in V\n",
-      "I_D= 0.4*10**3       # in A\n",
-      "W= 100           # in \u00b5m\n",
-      "L= 10           # in \u00b5m\n",
-      "uACox= 20         # in A/V**2\n",
-      "Vt= 2            # in V\n",
-      "R= 0.015\n",
-      "V_GS= V_DD-I_D*R      # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of R  is : =%.f k\u03a9 ' %(R*10**3))\n",
-      "V_D= V_DD-I_D*R           # in V\n",
-      "print('The value of V_D  is : =%.f volts  ' %V_D)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of R  is : =15 k\u03a9 \n",
-        "The value of V_D  is : =4 volts  \n"
-       ]
-      }
-     ],
-     "prompt_number": 118
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example No 9.36 Page No 405"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variables\n",
-      "V_GSth= 2         # in V\n",
-      "k= 2*10**-4       # in A/V^2\n",
-      "V_DD= 12          # in V\n",
-      "R_D= 5*10**3      # in ohm\n",
-      "I_D= 0.0014597\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "V_DS= V_DD-I_D*R_D         # in V\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The value of V_DS  is : =%.1f volts ' %V_DS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of V_DS  is : =4.7 volts \n"
-       ]
-      }
-     ],
-     "prompt_number": 119
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices/README.txt b/Advance_Semiconductor_Devices/README.txt
deleted file mode 100755
index eec99ae9..00000000
--- a/Advance_Semiconductor_Devices/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Santosh Pawar
-Course: mca
-College/Institute/Organization: ASM institute of computer studies
-Department/Designation: MCA
-Book Title: Advance Semiconductor Devices
-Author:  S. Sharma
-Publisher: S. K. Kataria & Sons, New Delhi
-Year of publication: 2012
-Isbn: 9789350142158
-Edition: 1
\ No newline at end of file
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diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo1.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo1.ipynb
new file mode 100755
index 00000000..4d1dc38d
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo1.ipynb
@@ -0,0 +1,800 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chapter No 1 : Semiconductor Materials and\n",
+      "Their Properties<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3, Page No 37 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Finding Density of silicon crystal\n",
+      "\n",
+      "#initialisation of variables\n",
+      "a=5.3   #in \u00c5\n",
+      "a=a*10**-10     #in m\n",
+      "N_A=6.023*10**23  \n",
+      "At_Si=28    #atomic weight of Si\n",
+      "n = 4\n",
+      "m=At_Si/N_A  #in gm\n",
+      "m= m*10**-3  #in kg\n",
+      "V=a**3       #in m^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Rho=((m*n)/V)/1000  #in kg/m^3\n",
+      "\n",
+      "# Note: There is calculation error to find the value of density. So the answer in the book is wrong.#RESULTS\n",
+      "#RESULTS\n",
+      "print('Density of silicon crystal in kg/m^3 is = %.2f kg/m^3' %Rho)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density of silicon crystal in kg/m^3 is = 1.25 kg/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4, Page No 37 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Finding Density of the crystal \n",
+      "\n",
+      "#initialisation of variables\n",
+      "n = 4\n",
+      "r = 1.278  #in \u00c5\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a = (4*r)/(math.sqrt(2)) #in\u00c5\n",
+      "a = a * 10**-10   #in m\n",
+      "V = (a)**3  #in m^3\n",
+      "At_W = 63.5   #atomic weight\n",
+      "N_A = 6.023*10**23\n",
+      "m = At_W /N_A   #in gm\n",
+      "m = m*10**-3  #in kg\n",
+      "Rho = ((m*n)/V)/1000  #in kg/m^3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Density of the crystal in kg/m^3 is =%.2f kg/m^3' %Rho)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density of the crystal in kg/m^3 is =8.93 kg/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 Page No 43 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Finding Wavaelength of X ray\n",
+      "\n",
+      "import math\n",
+      "from numpy import *\n",
+      "#initialisation of variables\n",
+      "d=2.82   #in \u00c5\n",
+      "d=d*10**-10    #in m\n",
+      "n=1\n",
+      "theta1=10  #n degree\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "lembda=2*d*math.sin(math.radians(theta1))   #in m\n",
+      "lembda=lembda*10**10       #in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Wavaelength of X ray  is =%.2f \u00c5 degrees' %lembda)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavaelength of X ray  is =0.98 \u00c5 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6, Page No 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Finding The spacing of atomic layer\n",
+      "\n",
+      "import math\n",
+      "#initialisation of variables\n",
+      "lembda = 1.6  #in \u00c5\n",
+      "theta = 14.2  #in degree\n",
+      "n = 1\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d = (n*lembda)/(2*math.sin(math.radians(theta)))   #in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The spacing of atomic layer in crystal is =%.2f in \u00c5 degrees' %d)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The spacing of atomic layer in crystal is =3.26 in \u00c5 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 Page No 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Finding The interplanner spacing\n",
+      "\n",
+      "import math\n",
+      "#initialisation of variables\n",
+      "n = 1\n",
+      "theta1 = 30  #in degree\n",
+      "lembda = 1.78  #in \u00c5\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d = (n*lembda)/(2*math.sin(math.radians(theta1)))  #in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The interplanner spacing is =%.2f in \u00c5 degrees' %d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The interplanner spacing is =1.78 in \u00c5 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 Page No 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Finding Interplaner spacing of the crystal\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "lembda = 0.58  #in \u00c5\n",
+      "n = 1;\n",
+      "theta1 = 6.45   # in degree\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d = (n*lembda)/(2*math.sin(math.radians(theta1))) #in \u00c5 \n",
+      "print('Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal  is =%.2f \u00c5' %d)\n",
+      "theta2 = 9.15  #in degree\n",
+      "d1 = (n*lembda)/(2*math.sin(math.radians(theta2))) #in \u00c5 \n",
+      "print('Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d1)\n",
+      "theta3 = 13 #in degree\n",
+      "n2 = 1\n",
+      "d2 = (n2*lembda)/(2*math.sin(math.radians(theta3)))  #in \u00c5\n",
+      "print('Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d2)\n",
+      "n=2\n",
+      "d2 = (n*lembda)/(2*math.sin(math.radians(theta3)))  #in \u00c5 \n",
+      "\n",
+      "#RESULTS\n",
+      "print('Part (iv) : : The interplaner spacing is : =%.4f \u00c5' %d2)\n",
+      "print('The interplaner spacing for some other set of reflecting  is : =%.2f \u00c5' %d1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal  is =2.58 \u00c5\n",
+        "Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =1.82 \u00c5\n",
+        "Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =1.29 \u00c5\n",
+        "Part (iv) : : The interplaner spacing is : =2.5783 \u00c5\n",
+        "The interplaner spacing for some other set of reflecting  is : =1.82 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9, Page No 44 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Finding The glancing angle for a cubic in degree\n",
+      "\n",
+      "import math\n",
+      "from sympy.mpmath import *\n",
+      "import cmath\n",
+      "\n",
+      "#initialisation of variables\n",
+      "a = 2.814 #in \u00c5\n",
+      "l = 0\n",
+      "h = l\n",
+      "k = 0\n",
+      "d=a  #in \u00c5\n",
+      "n = 2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "lembda=0.710     #in \u00c5\n",
+      "theta = math.degrees(math.asin(n*lembda/(2*d)))\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The glancing angle for a cubic in degree is : =%.2f  \u00c5 degrees' %theta)\n",
+      "print('Anwser is in points i.e = 14 36\"40\"..')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The glancing angle for a cubic in degree is : =14.61  \u00c5 degrees\n",
+        "Anwser is in points i.e = 14 36\"40\"..\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.10 Page No 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Finding Wavelength of X ray\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "a = 3.65  #in \u00c5\n",
+      "a = 3.65*10**-10  #in m\n",
+      "h = 1\n",
+      "k = 0\n",
+      "l = 0\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d= a/(math.sqrt(h**2+k**2+l**2))   #in m\n",
+      "n = 1\n",
+      "theta=60  #in degree\n",
+      "lembda = 2*d*math.sin(math.radians(theta))   #in m\n",
+      "lembda = lembda * 10**10   #in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Wavelength of X ray is : =%.2f  \u00c5' %lembda)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of X ray is : =6.32  \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11 Page No 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Finding The glancing angle in degree\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "lembda = 1.54   #in \u00c5\n",
+      "density = 9.024   #in gm/cc\n",
+      "n = 1\n",
+      "MI = 100\n",
+      "At_W = 63.54   #atomic weight\n",
+      "N_A = 6.023*10**23   \n",
+      "m = At_W/N_A   #in gm\n",
+      "a =(density*m)**(1.0/3)  #in cm\n",
+      "h = 1\n",
+      "k = 0\n",
+      "l = 0\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d= a/(math.sqrt(h**2+k**2+l**2))\n",
+      "theta =math.degrees((lembda*10**-8)/(2*d))   #in degree\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The glancing angle is = %.2f degrees' %theta)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The glancing angle is = 4.48 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.12 Page No 45 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Finding To get the 2nd order spectrum the position of the detector in degree\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "a = 3.615   #in \u00c5\n",
+      "theta=22   #in degree\n",
+      "n=1\n",
+      "h=1\n",
+      "k=h\n",
+      "l=k\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d = a/(math.sqrt( ((h)**2) + ((k)**2) + ((l)**2) ))   #in \u00c5\n",
+      "lembda = 2*d*math.sin(math.radians(theta))   #in \u00c5\n",
+      "print('The wavelength of X ray is = %.2f \u00c5 ' %lembda)\n",
+      "theta2 =math.degrees(math.asin(lembda/d))  #in degree\n",
+      "             #in degree\n",
+      "\n",
+      "#RESULTS\n",
+      "print('To get the 2nd order spectrum the position of the detector  is = %.2f degrees ' %theta2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength of X ray is = 1.56 \u00c5 \n",
+        "To get the 2nd order spectrum the position of the detector  is = 48.52 degrees \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.13, Page No 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "n = 1\n",
+      "lembda = 1.54      # in \u00c5\n",
+      "theta = 21.7      # in degree\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d = lembda/(2*math.sin(math.radians(theta)))     # in \u00c5\n",
+      "h = 1\n",
+      "k = h\n",
+      "l = k\n",
+      "a = d*math.sqrt(h**2+k**2+l**2)      # in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Lattice constant is = %.2f  in \u00c5 ' %a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Lattice constant is = 3.61  in \u00c5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.14, Page No 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "h = 2\n",
+      "k = 1\n",
+      "l = 1\n",
+      "a = 4.8      # in \u00c5\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d_211 = a/(math.sqrt(h**2+k**2+l**2))       #in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The distance between planes is =%.2f \u00c5 ' %d_211)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance between planes is =1.96 \u00c5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.15, Page No 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "r = 1.28       # in \u00c5\n",
+      "n = 4\n",
+      "M = 63.5\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a = (4*r)/(math.sqrt(2))        #in \u00c5\n",
+      "a = a * 10**-8       # in cm\n",
+      "N_A = 6.023*10**23     \n",
+      "Rho = (n*M)/( N_A*((a)**3) )       # in gm/cc\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Density is =%.2f gm/cc  ' %Rho)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density is =8.89 gm/cc  \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.16, Page No 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "M = 55.85\n",
+      "a = 2.9        # in \u00c5\n",
+      "Rho = 7.87    #in gm/cc\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a = a * 10**-8    # in cm\n",
+      "N_A = 6.023*10**23\n",
+      "n = (Rho*N_A*((a)**3))/M            # atom per unit\n",
+      "\n",
+      "#RESULTS\n",
+      "print('A lattice having =%.2f atom per unit cell is a BCC structure ' %n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A lattice having =2.07 atom per unit cell is a BCC structure \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.17, Page No 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "M = 60.0           # in gm/mole\n",
+      "Rho = 6.23      # in gm/cc\n",
+      "n = 4\n",
+      "N_A = 6.023*10**23 \n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a = ((n*M)/(N_A * Rho))**(1.0/3)         # in cm\n",
+      "r = (a*math.sqrt(2))/n               #radius of atom in cm\n",
+      "r = r * 10**8            # in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Radius of atom is =%.2f \u00c5  ' %r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of atom is =1.41 \u00c5  \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.18, Page No 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Rho = 5.96          # in gm/cc\n",
+      "M = 50.0\n",
+      "n = 2\n",
+      "N_A = 6.023*10**23 \n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a =((n*M)/(Rho*N_A))**(1.0/3)          # in cm\n",
+      "r = (a*math.sqrt(3))/4                 # in cm\n",
+      "P_f = (2*(4.0/3)*math.pi*((r)**3))/((a)**3)       # packing factor\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Packing factor is =%.2f  ' %P_f)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Packing factor is =0.68  \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.19, Page No 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "M = 120.0\n",
+      "n = 2.0\n",
+      "N_A = 6.023*10**23\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "m1 = M/N_A              #mass of 1 atom in gm\n",
+      "m2 = n*m1               #mass of unit cell in gm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Number of unit cell in 20 gms of element is : =%.2f X 10^22 unit cells ' %((20/m2)/(10**22)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of unit cell in 20 gms of element is : =5.02 X 10^22 unit cells \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20, Page No 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Rho = 2.48         # in gm/c.c\n",
+      "n = 4.0 \n",
+      "M = 58\n",
+      "N_A = 6.023*10**23\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a = ( (n*M)/(Rho*N_A) )**(1.0/3)      # in cm\n",
+      "a = a * 10**8        # in \u00c5\n",
+      "r = (a*math.sqrt(2))/n      # in \u00c5\n",
+      "r = 2*r       # in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The center to center distance between ions  is =%.2f  \u00c5' %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The center to center distance between ions  is =3.80  \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo2.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo2.ipynb
new file mode 100755
index 00000000..9bc8cffb
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo2.ipynb
@@ -0,0 +1,1248 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chapter No 2 : Semiconductor Materials and\n",
+      "Their Properties<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page No 65 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "lambda1 = 11000.0  #in \u00c5\n",
+      "lambda1 = lambda1 * 10**-10 \n",
+      "h = 6.625*10**-34 \n",
+      "c = 3*10**8\n",
+      "q = 1.6*10**-19    #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "E_g = h*c/lambda1  #in J\n",
+      "E_g= E_g/q        #in eV\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The energy gap in Si  is = %.2f eV' %E_g)\n",
+      "\n",
+      "#Note: The answer in the book is not correct \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy gap in Si  is = 1.13 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page No 65 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "E_g = 0.75    #in eV\n",
+      "q=1.6*10**-19   #in C\n",
+      "E_g = E_g*q   #in J\n",
+      "h = 6.63*10**-34   #in J\n",
+      "c = 3*10**8    #in m/s\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "lembda = (h*c)/E_g    #in m\n",
+      "lembda = lembda * 10**10   #in \u00c5\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The wavelength is =%.f  \u00c5 ' %lembda)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is =16575  \u00c5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page No 81 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy import *\n",
+      "\n",
+      "#initialisation of variables\n",
+      "del_E = 0.3        #value of E_C-E_F in eV\n",
+      "T1 = 330.0          #in K\n",
+      "T = 300            #in K\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "del_E1 = del_E*(T1/T)   #value of E_C-E_F in eV\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The position of fermi level   is =%.2f eV' %del_E1)\n",
+      "print('Hence the Fermi level will be %.2f eV below the conduction band' %del_E1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The position of fermi level   is =0.33 eV\n",
+        "Hence the Fermi level will be 0.33 eV below the conduction band\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page No 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "K = 8.63*10**-5\n",
+      "T = 300.0     #in K\n",
+      "N_C = 2.8*10**19     #in cm^-3\n",
+      "del_E = 0.25\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "f_F = math.exp( (-del_E)/(K*T) )\n",
+      "n_o = N_C*math.exp( (-del_E)/(K*T) )    #in cm^-3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The probability is =%.6f ' %f_F)\n",
+      "print('The thermal equillibrium electron concentration  is =%.2f X 10^15 cm^-3' %(n_o/(10**15)))\n",
+      "print('Approximation  error ')\n",
+      "# The answer is same as in book only thy have rounded off to 1.8 * 10^15 which is same as above"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The probability is =0.000064 \n",
+        "The thermal equillibrium electron concentration  is =1.79 X 10^15 cm^-3\n",
+        "Approximation  error \n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page No 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "N_V = 1.04*10**19    #in cm^-3\n",
+      "T1 = 400.0        #in K\n",
+      "T2 = 300.0        #in K\n",
+      "del_E = 0.27     #value of E_F-E_V in eV\n",
+      "K = 0.0259\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N_V= N_V*(T1/T2)**(3/2)      #in cm^-3\n",
+      "KT = K*(T1/T2)              #in eV\n",
+      "p_o = N_V*math.exp( (-del_E)/(KT) )    #in /cm^3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The hole concentration is =%.2f x 10^15 per cm^3' %(p_o/(10**15)))\n",
+      "print('Approximationa error ')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hole concentration is =5.58 x 10^15 per cm^3\n",
+        "Approximationa error \n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page No 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "N = 6.02*10**23\n",
+      "A = 63.5            #atomic weight\n",
+      "Rho = 1.7*10**-6    #in ohm cm\n",
+      "d = 8.96           # in gm/cc\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n = (N/A)*d     #in /cc\n",
+      "e = 1.6*10**-19    #in C\n",
+      "Miu_e = 1/(Rho*n*e)    #in cm^2/volt-sec\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The mobility of electron  is =%.2f  cm^2/volt.sec' %Miu_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility of electron  is =43.28  cm^2/volt.sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7, Page No 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from sympy.mpmath import *\n",
+      "import cmath\n",
+      "\n",
+      "\n",
+      "#initialisation of variables\n",
+      "d = 8.96   #in gm/cc\n",
+      "At = 63.5   #atomic weight\n",
+      "N_A = 6.02*10**23   #in /gm mole\n",
+      "l = 0.1 #in m\n",
+      "e = 1.6*10**-19   #in C\n",
+      "A = 1.7*10**-6   #in m^2\n",
+      "R = 0.1        #in ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n = (N_A/At)*d   #in /cc\n",
+      "n = n * 10**6   #in /m^3\n",
+      "Rho = (R*A)/l   #in ohm.m\n",
+      "Sigma = 1/Rho   #in mho/m\n",
+      "Miu_e = Sigma/(n*e)     #in m^2/V-sec\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The electron mobility is =%.3f X 10^-3 m^2/V-sec ' %(Miu_e*(10**3)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electron mobility is =0.043 X 10^-3 m^2/V-sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page No 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "N_A = 6.025*10**23    #in /gm mole\n",
+      "d = 10.5             #in gm/cc\n",
+      "At = 108.0         #atomic weight of \n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n = (N_A/At)*d   #in /cm^3\n",
+      "n = n * 10**6   #in /m^3\n",
+      "r = 10**-3        #in m\n",
+      "A = math.pi * ((r)**2)   #in m^2\n",
+      "q = 1.6*10**-19     \n",
+      "I = 2       #in A\n",
+      "V = I/(n*q*A)      #in m/s\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The drift velocity of an electron  is =%.f X 10^-4 m/s' %(V*(10**5)))\n",
+      "print('Approximationa error ')\n",
+      "# answer is same as in book only they have rounded of 0.000068 to 7 * 10^-4"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The drift velocity of an electron  is =7 X 10^-4 m/s\n",
+        "Approximationa error \n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page No 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "e= 1.6*10**-19   #in C\n",
+      "d= 1.03      #in mm\n",
+      "d= d*10**-3   #in m\n",
+      "R= 6.51    #in ohm\n",
+      "l= 300.0  #in m\n",
+      "n= 8.4*10**28   #per m^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r= d/2     #in m (radius)\n",
+      "A= math.pi*r**2   #in m^2\n",
+      "rho= R*A/l     #in ohm meter\n",
+      "sigma=  1/rho    #in mho/m\n",
+      "miu_e= sigma/(n*e)    #m^2/V-sec\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The coductivity of copper is =%.2f  mho/m' %sigma)\n",
+      "print('The mobility of charge carriers  is : =%.6f  m^2/V-sec' %miu_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The coductivity of copper is =55306469.41  mho/m\n",
+        "The mobility of charge carriers  is : =0.004115  m^2/V-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9.ii Page No 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Miu_e = 1500.0     #in cm^2/volt.sec\n",
+      "Miu_h = 500.0      #in cm^2/volt.sec\n",
+      "n_i = 1.6*10**10  #in /cm^3\n",
+      "e = 1.6*10**-19   #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma_i = n_i*(Miu_e+Miu_h)*e   #in mho/cm\n",
+      "Sigma = Sigma_i                 #in mho/cm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The conductivity of pure silicon  is =%.7f  mho/cm' %Sigma)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of pure silicon  is =0.0000051  mho/cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 2.10 Page No 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "Miu_e = 1500.0  #in cm^2/volt.sec\n",
+      "Miu_h = 500.0  #in cm^2/volt.sec\n",
+      "n_i = 1.6*10**10   #in /cm^3\n",
+      "e = 1.6*10**-19    #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma_i = n_i*(Miu_e+Miu_h)*e    #in mho/cm\n",
+      "Sigma = Sigma_i              #in mho/cm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The conductivity of pure silicon  is =%.2f X 10^-6  mho/cm' %(Sigma*(10**6)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of pure silicon  is =5.12 X 10^-6  mho/cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 2.11 Page No 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "Miu_d = 500.0   #in cm^2/V.S\n",
+      "Rho = 10       #in ohm cm \n",
+      "\n",
+      "#CALCULATIONS\n",
+      "e = 1.6*10**-19   #in C\n",
+      "n_d = 1/(Rho*e*Miu_d)     #in /cm^3... correction\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The number of donor  is =%.2f X 10^15 atom per cm^3' %(n_d/(10**15)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of donor  is =1.25 X 10^15 atom per cm^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 2.12 Page No 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "d = 5.32         #in gm/cc\n",
+      "N_A = 6.02*10**23   #in atoms/gm.mole\n",
+      "At = 72.6       #atomic weight\n",
+      "Miu = 3800    #in cm^2/v.s\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n_d = (N_A/At) * d          #in /cm^3\n",
+      "n_d = n_d * 10**-8    #in /cc\n",
+      "e = 1.6*10**-19    #in C\n",
+      "Sigma = n_d * Miu * e           #in mho/cm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The conductivity of specimen is =%.2f  mho/cm' %Sigma)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of specimen is =0.27  mho/cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 2.13 Page No 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "Rho = 0.3623*10**-3    #in ohm m\n",
+      "d = 4.42*10**28      #Ge density in atoms/m^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma = 1/Rho       #in mho/m\n",
+      "n_d = d*10**-6    #in atoms/m^3\n",
+      "e = 1.6*10**-19    #in C\n",
+      "Miu = Sigma/(n_d*e)    #in m^2/V.sec\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The electron mobility is =%.2f  m^2/V-sec' %Miu)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electron mobility is =0.39  m^2/V-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 2.14 Page No 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "N_A = 6.025*10**26    #in /kg.Mole\n",
+      "At = 72.59          #atomic weight\n",
+      "d = 5.36*10**3    #in kg/m^3\n",
+      "R = 0.42            #in ohm m\n",
+      "B_i = 10**-6    #rate of boron impurity in %\n",
+      "e = 1.6*10**-19     #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n = (N_A/At)*d    #number of Ge atoms\n",
+      "h = n/10**8      #holes per unit volume\n",
+      "Miu_h = 1/(R*h*e)       #in m^2/V.sec\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The Mobility of holes is =%.2f m^2/V-sec' %Miu_h)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Mobility of holes is =0.03 m^2/V-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.15 Page No 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "n_i = 2*10**19     #in /m^3\n",
+      "Miu_e = 0.36    #in m^2/v.s\n",
+      "Miu_h = 0.17   #in m^2/v.s\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A = 1*10**-4     #in m^2\n",
+      "V = 2            #in Volts\n",
+      "l = 0.3         #in mm\n",
+      "l = l * 10**-3   #in m\n",
+      "e = 1.6*10**-19   #in C\n",
+      "Sigma_i = n_i * e * (Miu_e+Miu_h)   #in mho/m\n",
+      "I = (Sigma_i * V*A)/l        #in amp \n",
+      "\n",
+      "#RESULTS\n",
+      "print('The current in amp is =%.2f ' %I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current in amp is =1.13 \n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.16 Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "d = 4.2*10**28   #in atoms/m^3\n",
+      "n_d = d/10**6   #in atoms/m^3\n",
+      "e = 1.6*10**-19   #in C\n",
+      "Miu_e = 0.36    #in m^2/V-sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma_n = n_d *e *Miu_e    #in mho/m\n",
+      "Rho_n = 1/Sigma_n          #ohm m\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The resistivity in \u03a9m is =%.2f X 10^-3' %(Rho_n*(10**3)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistivity in \u03a9m is =0.41 X 10^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.17 Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "n_i = 1*10**19        #in /m^3\n",
+      "Miu_e = 0.36     #in m^2/volt.sec\n",
+      "Miu_h = 0.17      #in m^2/volt.sec\n",
+      "A = 2                #in cm^2\n",
+      "A = A *  10**-4    #in m^2\n",
+      "t = 0.1         #in mm\n",
+      "t = t*10**-3   #in m\n",
+      "V = 4        #in volts\n",
+      "e = 1.6*10**-19   #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma_i = n_i * e * (Miu_e + Miu_h)   #mho/m\n",
+      "J = Sigma_i * (V/t)             #in Amp/m^2\n",
+      "I = J*A                      #in Amp\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The current in Amp is =%.2f ' %I)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current in Amp is =6.78 \n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 2.18 Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "Miu_h = 500.0                  #in cm^2/V.s\n",
+      "Miu_e = 1500.0                 #in cm^2/V.s\n",
+      "n_i = 1.6*10**10              #in /cm^3\n",
+      "e = 1.6*10**-19               #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma_i = n_i * e * (Miu_e+Miu_h)       #in mho/cm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The conductivity of pure silicon in mho/cm is =%.2f X 10^-6 ' %(Sigma_i*(10**6)))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of pure silicon in mho/cm is =5.12 X 10^-6 \n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.19 Page No 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "Si_density = 10.5                #in gm/cc\n",
+      "N_A = 6.025*10**23         \n",
+      "At = 108.0                  #atomic weight\n",
+      "B = 0.8                   #in Tesla\n",
+      "w = 0.50               #in cm\n",
+      "w = w * 10**-2          #in m\n",
+      "t = 0.10              #in mm\n",
+      "t = t * 10**-3       #in m\n",
+      "A = w*t                 #in m^2\n",
+      "q = 1.6*10**-19          #in C\n",
+      "I = 2                    #in ampere\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n = (N_A/At) * Si_density  #in /cc\n",
+      "n = n * 10**6              #in /m^3\n",
+      "V_H = (B*I*t)/(n*q*A)           #in volts\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The hall voltage produced is =%.3f X 10^-7 volts' %(V_H*(10**7)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hall voltage produced is =0.341 X 10^-7 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.20 Page No 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "Sigma = 5.8*10**7        #in mho/m\n",
+      "l = 1                     #in m\n",
+      "d = 1                  #in cm\n",
+      "d = d * 10**-2          #in m\n",
+      "W = 1               #in mm \n",
+      "W = W*10**-3       #in m\n",
+      "I = 1             #in Amp\n",
+      "B = 1            #in Tesla\n",
+      "V_H = 0.074*10**-6        #in Volts\n",
+      "A = 10**-2 * 10**-3          #in m^2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "R_H = (V_H*A)/(B*I*d)              #in m^3/c\n",
+      "Miu = Sigma * R_H                    #in m^2/volt.sec\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Hall coefficient  is =%.1f X 10^-11  m^3/c' %(R_H*(10**11)))\n",
+      "print('The mobility of electron is = %.2f X 10^-3 m^2/volt.sec ' %(Miu*(10**3)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hall coefficient  is =7.4 X 10^-11  m^3/c\n",
+        "The mobility of electron is = 4.29 X 10^-3 m^2/volt.sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.21 Page No 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "n_i = 1.4*10**18             #in /m^3\n",
+      "n_D = 1.4*10**24             #in /m^3\n",
+      "n = n_D                      #in /m^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "p = n_i**2/n               #in /m^3\n",
+      "R_e = n/p              #Ratio of electron\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Concentration of is =%.2f X 10^12 holes per m^3 ' %(p/(10**12)))\n",
+      "print('Ratio of electron to hole concentration is =%.f 10^12 ' %(R_e/(10**12)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Concentration of is =1.40 X 10^12 holes per m^3 \n",
+        "Ratio of electron to hole concentration is =1 10^12 \n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.22 Page No 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "B = 0.48                #in Wb/m^2\n",
+      "R_H = 3.6 * 10**-4           #in m^3/c\n",
+      "R = 9*10**-3           #in ohm-m\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma = 1/R                #in (ohm-m)^-1\n",
+      "Rho = 1/R_H                #in coulomb/m^3\n",
+      "e = 1.6*10**-19            #in C\n",
+      "n = Rho/e                  #in /m^3\n",
+      "Miu = Sigma * R_H          #in m^2/volt-s\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The mobility of electron  is =%.2f  m^2/volt-s' %Miu)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility of electron  is =0.04  m^2/volt-s\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.23 Page No 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "e = 1.6*10**-19           #in C\n",
+      "R_H = 0.0145              #in m^3/coulomb\n",
+      "Miu_e = 0.36              #m^2/v-s\n",
+      "E = 100                  #V/m\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n = 1/(e*R_H)             #in /m^3\n",
+      "J= n*e*Miu_e*E            #in A/m^2\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The current density  is =%.2f  A/m^2 ' %J)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current density  is =2482.76  A/m^2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.24 Page No 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "e = 1.6*10**-19            #in C\n",
+      "Miu_e = 7.04*10**-3           #in m^2/volt-sec\n",
+      "n = 5.8*10**28              #number of electron/m^3\n",
+      "m = 9.1*10**-31\n",
+      "E_F = 5.5                    # in eV\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Torque = (Miu_e/e)*m                # in sec \n",
+      "Rho = 1/(n*e*Miu_e)                 #in ohm cm\n",
+      "V_F = math.sqrt( (2*E_F*e)/m )           #in m/s\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Relaxtion time is =%.2f X 10^-15 sec ' %(Torque*(10**15)))\n",
+      "print('Resistivity of conductor  is =%.2f X 10^-18 in \u03a9m' %(Rho*(10**8)))\n",
+      "print('Velocity of electron with the fermi energy  is =%.2f X 10^-6 in m/s' %(V_F/(10**6)))\n",
+      "\n",
+      "# Note: The calculation of Part (ii) is wrong also the unit of resistivity of conductor is wrong\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relaxtion time is =40.04 X 10^-15 sec \n",
+        "Resistivity of conductor  is =1.53 X 10^-18 in \u03a9m\n",
+        "Velocity of electron with the fermi energy  is =1.39 X 10^-6 in m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.25 Page No 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "E= 5.95           #in eV\n",
+      "EF= 6.25          #in eV\n",
+      "delE= 0.01\n",
+      "\n",
+      "#CALCULATIONS\n",
+      " #delE= 1-1/(1+exp((E-EF)/KT))\n",
+      "K=1.38*10**-23                      #Boltzman Constant in J/K\n",
+      "T = ((E-EF)/math.log(1/(1-delE) -1)*1.6*10**-19)/K               #in K\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The temperature is =%.2f K' %T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature is =756.95 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.26 Page No 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "T1 = 400.0       #in K\n",
+      "T2 = 300.0       #in K\n",
+      "N_V = 1.04*10**19       #in cm^-3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N1 = N_V*((T1/T2)**(3/2))          #in cm^-3\n",
+      "KT = 0.0259*(T1/T2)           #in eV\n",
+      "FermiLevel= 0.27              #in eV\n",
+      "P_O = N1*math.exp( (-FermiLevel)/KT )          #in cm^-3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The thermal equillibrium hole concentration  is =%.2f X 10^15 cm^-3' %(P_O/(10**15)))\n",
+      "print('Approximationa error ')\n",
+      "# Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The thermal equillibrium hole concentration  is =5.58 X 10^15 cm^-3\n",
+        "Approximationa error \n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.27  Page No 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T1 = 550.0              #in K\n",
+      "T2 = 300.0           #in K\n",
+      "N1 = 1.04*10**19     \n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N_V = N1*((T1/T2)**(3))  \n",
+      "N_C = 2.8*10**19\n",
+      "E_g = -1.12\n",
+      "KT = 0.0259*(T1/T2)\n",
+      "n_i = math.sqrt(N_C*N_V*math.exp(E_g/KT))   #in cm^-3\n",
+      "Nd= math.sqrt(n_i**2/((1.05-1/2.0)**2-(1/2.0)**2))\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of n_i  is =%.2f X 10^14 cm^-3' %(n_i/(10**14)))\n",
+      "print('The value of N_d is =%.2f X 10^15 cm^-3' %(Nd/(10**15)))\n",
+      "#Formula n_o= Nd/2+sqrt((Nd/2)^2+n_i^2) and n_o = 1.05*N_d;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of n_i  is =3.20 X 10^14 cm^-3\n",
+        "The value of N_d is =1.40 X 10^15 cm^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 2.28 Page No 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "n_o = 10**15         #in cm^-3\n",
+      "n_i = 10**10         #in cm^-3\n",
+      "p_o = 10**5       #in cm^-3\n",
+      "del_p = 10**13    #in cm^-3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "del_n = del_p         #in cm^-3\n",
+      "KT= 0.0259          #in eV\n",
+      "Fermi_level1= KT*math.log(n_o/n_i)           #in eV\n",
+      "Fermi_level2= KT*math.log((n_o+del_n)/n_i)        #in eV\n",
+      "Fermi_level3= KT*math.log((p_o+del_p)/n_i)        #in eV\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Fermi level for thermal equillibrium  is : =%.2f eV ' %Fermi_level1)\n",
+      "print('Quasi-Fermi level for electrons in non equillibrium  is : =%.2f eV' %Fermi_level2)\n",
+      "print('Quasi-Fermi level for holes in non equillibrium  is: =%.2f eV' %Fermi_level3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fermi level for thermal equillibrium  is : =0.30 eV \n",
+        "Quasi-Fermi level for electrons in non equillibrium  is : =0.30 eV\n",
+        "Quasi-Fermi level for holes in non equillibrium  is: =0.18 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo3.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo3.ipynb
new file mode 100755
index 00000000..bc717890
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo3.ipynb
@@ -0,0 +1,992 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chapter No 3 : Excess Carriers In Semiconductors<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page No 145 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "n_o = 10**17          #in/cm^3\n",
+      "n_i = 1.5*10**10      #in /cm^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "p_o = ((n_i)**(2))/n_o      #in holes/cm^3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The hole concentration  is =%.f  in holes/cm^3' %p_o)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hole concentration  is =2250  in holes/cm^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page No  146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "n_i = 1.5 * 10**10              #in /cm^3 for silicon\n",
+      "N_d = 10**17                    #in atoms/cm^3\n",
+      "n_o = 10**17                   #electrons/cm^3\n",
+      "KT = 0.0259\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "# E_r - E_i = KT * log(n_o/n_i)\n",
+      "del_E = KT * math.log(n_o/n_i)                 #in eV\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The energy band for this type material is Ei =%.2f eV' %del_E)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy band for this type material is Ei =0.41 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 Page No 147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "K = 1.38*10**-23               #in J/K\n",
+      "T = 27                     #in degree C\n",
+      "T = T + 273               #in K\n",
+      "e = 1.6*10**-19          \n",
+      "Miu = 0.17             #in m^2/v-s\n",
+      "Miu1 = 0.025           #in m^2/v-s\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "D_n = ((K*T)/e)*Miu        #in m^2/s\n",
+      "D_p = ((K*T)/e)*Miu1    #in m^2/s\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The diffusion coefficient  of electrons  is =%.2f X 10^-4 in m^2/s ' %(D_n*(10**4)))\n",
+      "print('The diffusion coefficient  of holes  is =%.2f X 10^-4 in m^2/s ' %(D_p*(10**4)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion coefficient  of electrons  is =43.99 X 10^-4 in m^2/s \n",
+        "The diffusion coefficient  of holes  is =6.47 X 10^-4 in m^2/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5, Page No 147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Miu_n = 0.15           #in m^2/v-s\n",
+      "K = 1.38*10**-23      \n",
+      "T = 300                #in K\n",
+      "e = 1.6*10**-19      #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "D_n = Miu_n*((K*T)/e)            #in m^2/s\n",
+      "Torque_n = 10**-7                 #in s\n",
+      "L_n = math.sqrt(D_n*Torque_n)    #in m\n",
+      "del_n = 10**20                   #electrons/m^3\n",
+      "J_n = (e*D_n*del_n)/L_n        #in A/m^2\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The diffusion length is =%.2f  X 10^-4 in m' %(L_n*(10**5)))\n",
+      "print('The diffusion current density  is =%.2f X 10^3 in A/m^2 ' %(J_n/(10**3)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion length is =1.97  X 10^-4 in m\n",
+        "The diffusion current density  is =3.15 X 10^3 in A/m^2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 Page No 147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Sigma_n = 0.1              #in (ohm-cm)^-1\n",
+      "Miu_n = 1300\n",
+      "q = 1.6*10**-19             #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n_n = Sigma_n/(Miu_n*q)        #in electrons/cm^3 \n",
+      "n_i = 1.5*10**10\n",
+      "p_n = ((n_i)**2)/n_n           #in holes/cm^3\n",
+      "p_n = p_n * 10**6          #in holes/m^3\n",
+      "n_n=n_n*10**6\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Concentration of electrons  is =%.2f X 10^20 per m^3 ' %(n_n/(10**20)))\n",
+      "print('Concentration of holes  is =%.2f X 10^11 per m^3 ' %(p_n/(10**11)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Concentration of electrons  is =4.81 X 10^20 per m^3 \n",
+        "Concentration of holes  is =4.68 X 10^11 per m^3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 Page No 147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "L = 100*10**-6         #in m\n",
+      "Miu_e = 0.13          #in m^2/V-s\n",
+      "Torque_h = 10**-6     #in s\n",
+      "Miu_h = 0.05          #in m^2/v-s\n",
+      "V = 12                 #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Torque_n = ((L)**2)/(Miu_e*V)             #in s\n",
+      "P = (Torque_h/Torque_n)*(1+(Miu_h/Miu_e))\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Electron transit time  is =%.2f X 10^-9 in sec' %(Torque_n*(10**9)))\n",
+      "print('Photoconductor gain is =%.2f  ' %P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electron transit time  is =6.41 X 10^-9 in sec\n",
+        "Photoconductor gain is =216.00  \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8, Page No 147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "q = 1.6*10**-19              #in C\n",
+      "n_i = 2.5*10**13\n",
+      "Miu_n = 3800             #in cm^2/V-s\n",
+      "Miu_p = 1800             #in cm^2/V-s\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Sigma = n_i*(Miu_n + Miu_p)*q             #in (ohm-cm)^-1\n",
+      "Rho = 1/Sigma               #in ohm-cm\n",
+      "N_D =4.4*10**22/10**8     #in atoms/cm^3\n",
+      "Sigma_n = N_D * Miu_n*q   #in (ohm-cm)^-1 \n",
+      "Rho1 = 1/Sigma_n     #in ohm cm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The resistivity  is =%.2f in ohm-cm ' %Rho)\n",
+      "print('The resistivity drops  is =%.2f in ohm cm ' %Rho1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistivity  is =44.64 in ohm-cm \n",
+        "The resistivity drops  is =3.74 in ohm cm \n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 Page No 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "n_i = 10**16          #in /m^3\n",
+      "N_D = 10**22       #in /m^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n = N_D        #in /m^3\n",
+      "p = ((n_i)**2)/n   #in /m^3\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The concentration of  is =%.2f X 10^21  electrons per m^3' %(n/(10**21)))\n",
+      "print('The concentration of  is =%.2f X 10^9 holes per m^3 ' %(p/(10**9)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The concentration of  is =10.00 X 10^21  electrons per m^3\n",
+        "The concentration of  is =10.00 X 10^9 holes per m^3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 Page No 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Rho = 9.6*10**-2       #ohm-m\n",
+      "Sigma_n = 1/Rho        #in (ohm-m)^-1\n",
+      "Miu_n = 1300           #in cm^2/V-s\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Miu_n = Miu_n * 10**-4  #in m^2/V-s\n",
+      "q = 1.6*10**-19     #in C\n",
+      "N_D = Sigma_n/(Miu_n*q)           #in atoms/m^3\n",
+      "d = 5*10**22                     #in atoms/cm^3\n",
+      "d = d * 10**6                   #// in atoms/m^3\n",
+      "R_d = N_D/d                   #Ratio \n",
+      "\n",
+      "#RESULTS\n",
+      "print('Ratio of donor atom to silicon atoms per unit volume is =%.2f X 10^-8  ' %(R_d*(10**8)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio of donor atom to silicon atoms per unit volume is =1.00 X 10^-8  \n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11 Page No 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "n_i = 1.5*10**10            #in /cm^3\n",
+      "n_n = 2.25*10**15           #in /cm^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "p_n = ((n_i)**2)/n_n         #in /cm^3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The concentration of  is =%.2f X 10^5 holes per cm^3 ' %(p_n/(10**5)))\n",
+      "print('Donor impurity per cm^3 is =%.2f X 10^15 ' %(n_n/(10**15)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The concentration of  is =1.00 X 10^5 holes per cm^3 \n",
+        "Donor impurity per cm^3 is =2.25 X 10^15 \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.12 Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "N_A = 2*10**16           #in /cm^3\n",
+      "N_D = 10**16             #in /cm^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "C = N_A-N_D             #in /cm^3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Carrier concentration in holes/cm^3 is =%.2f X 10^16 ' %(C/(10**16)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Carrier concentration in holes/cm^3 is =1.00 X 10^16 \n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.13 Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "del_n = 10**15           #in /cm^3\n",
+      "Torque_p = 10*10**-6       #in sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "R_G = del_n/Torque_p          #in electron hole pairs/sec/cm^3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The rate of generation of minority carrier is : =%.2f X 10^20 electron hole pairs/sec/cm^3 ' %(R_G/(10**20)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of generation of minority carrier is : =1.00 X 10^20 electron hole pairs/sec/cm^3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.14 Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V = 1/20.0        #in cm/\u00b5sec\n",
+      "V=V*10**6       #in cm/sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "E = 10            #in V/cm\n",
+      "Miu = V/E         #in cm^2/V-sec\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The mobility of minority charge carrier is =%.f  cm^2/V-sec' %Miu)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility of minority charge carrier is =5000  cm^2/V-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.15 Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "q = 1.6 * 10**-19       #in C\n",
+      "N_D = 4.5 * 10**15      #in /cm^3\n",
+      "del_p = 10**21 \n",
+      "e=10                #in cm\n",
+      "A = 1             #in mm^2\n",
+      "A = A * 10**-14        #cm^2\n",
+      "l = 10              #in cm\n",
+      "Torque_p = 1            #in microsec\n",
+      "\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Torque_p = Torque_p * 10**-6           #in sec\n",
+      "Torque_n = 1                      #in microsec\n",
+      "Torque_n = Torque_n * 10**-6        #in  sec\n",
+      "n_i = 1.5 * 10**10               #in /cm^3\n",
+      "D_n = 30                   #cm^2/sec\n",
+      "D_p = 12                  #in cm^2/sec\n",
+      "n_o = N_D                #in /cm^3\n",
+      "p_o = ((n_i)**2)/n_o           #in /cm^3\n",
+      "l_n = math.sqrt(D_n * Torque_n)  #in cm\n",
+      "l_p = math.sqrt(D_p * Torque_p)      #in cm\n",
+      "x=34.6*10*8-4            #in cm\n",
+      "e1 = 1.88 * 10**1        #in cm\n",
+      "dnBYdx = del_p * e1        #in cm^4\n",
+      "dpBYdx = del_p *e         #in cm^4\n",
+      "\n",
+      "J_P = -(q) * D_p * dpBYdx       #in A/cm^2\n",
+      "J_n = q * D_n * dnBYdx          #in A/cm^2\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Hole concentration at thermal equilibrium per cm^3 is =%.1f X 10^5  ' %(p_o/(10**5)))\n",
+      "print('Diffusion length of holes in cm is =%.1f X 10^-3 ' %(l_p*(10**3)))\n",
+      "print('Concentration gradient of holes at distance in cm^4 is =%.f X 10^22 ' %(dpBYdx/(10**22)))\n",
+      "print('Concentration gradient of electrons in per cm^4 is =%.1f X 10^22  ' %(dnBYdx/(10**22)))\n",
+      "print('Current density of holes due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_P/(10**4)))\n",
+      "print('Current density of electrons due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_n/(10**4)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hole concentration at thermal equilibrium per cm^3 is =0.5 X 10^5  \n",
+        "Diffusion length of holes in cm is =3.5 X 10^-3 \n",
+        "Concentration gradient of holes at distance in cm^4 is =1 X 10^22 \n",
+        "Concentration gradient of electrons in per cm^4 is =1.9 X 10^22  \n",
+        "Current density of holes due to diffusion in A/cm^2 is =-1.92 X 10^4 \n",
+        "Current density of electrons due to diffusion in A/cm^2 is =9.02 X 10^4 \n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.16 Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "h = 6.626 * 10**-34          #in J-s\n",
+      "q= 1.6*10**-19            #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "h= h/q              #in eV\n",
+      "c = 3*10**8\n",
+      "lembda = 5490*10**-10             #in m\n",
+      "E = h*c/lembda                 #in eV\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The energy band gap is =%.2f eV ' %E)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy band gap is =2.26 eV \n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.17 Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "D_n = 35               #in cm^2/sec\n",
+      "q = 1.6*10**-19        #in C\n",
+      "y2 = 6*10**16          #in /cm^3\n",
+      "y1 = 10**17           #in /cm^3\n",
+      "x2 = 2*10**-4\n",
+      "x1 = 0\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "dnBYdx = (y2-y1)/(x2-x1)\n",
+      "J_n = q*D_n*dnBYdx              #in A/cm^2\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The current density is =%.f  in A/cm^2' %J_n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current density is =-1120  in A/cm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.18 Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "q = 1.6*10**-19               #in C\n",
+      "n_n = 5*10**20                #in  /m^3\n",
+      "n_n = n_n * 10**-6            #in /cm^3\n",
+      "Miu_n = 0.13                   #in m^2/V-sec\n",
+      "\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Miu_n = Miu_n * 10**4       #in cm^2/V-sec\n",
+      "Sigma_n = q*n_n*Miu_n      #in ohm-cm^-1\n",
+      "Rho = 1/Sigma_n\n",
+      "A = 100        # in \u00b5m^2\n",
+      "A = A * 10**-8     #in cm^2\n",
+      "l = 0.1           #in cm\n",
+      "R = Rho * (l/A)        #in ohm\n",
+      "R=R*10**-6\n",
+      "#RESULTS\n",
+      "print('The resistance of the bar  is =%.2f in M ohm' %R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance of the bar  is =0.96 in M ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.19 Page No 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "w = 3.0              #in \u00b5m\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "D = w/9          #in \u00b5m\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Depletion width on P side is =%.2f in \u00b5m ' %D)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Depletion width on P side is =0.33 in \u00b5m \n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.20 Page No 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "n_i = 1.5*10**16          #in /m^3\n",
+      "n_n = 5*10**20          #in /m^3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "p_n = ((n_i)**2)/n_n          #in /m^3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The minority carrier density per m^3 is =%.2f X 10^11' %(p_n/(10**11)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minority carrier density per m^3 is =4.50 X 10^11\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.21 Page No 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "y2 = 10**14                #in /cm^3\n",
+      "y1 = 0\n",
+      "x1=-0.5                    #in \u00b5m\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "x1= x1*10**-4             #in cm\n",
+      "x2=0\n",
+      "dnBYdx = (y2-y1)/(x2-x1)     #in /cm^4\n",
+      "q = 1.6*10**-19              #in C\n",
+      "D_n = 25                     #in cm^2/sec\n",
+      "J_n = q*D_n*dnBYdx          #in A/cm^2\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The collector current density  is =%.f in A/cm^2' %J_n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The collector current density  is =8 in A/cm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.22 Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "h = 6.64*10**-34              ##in J-s\n",
+      "q=1.6*10**-19                #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "h= h/q                     #in eV\n",
+      "c = 3*10**8               #in m/s\n",
+      "lembda = 0.87*10**-6        #in m\n",
+      "E_g = (h*c)/lembda         #in eV\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The band gap in eV is =%.2f ' %E_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The band gap in eV is =1.43 \n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.23 Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "alpha = 5*10**4            #in cm^-1\n",
+      "l = 0.46*10**-4            #in cm\n",
+      "hv = 2                    #in eV\n",
+      "I_o = 10**-2             #in W\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_t = I_o*math.exp(-alpha*l)   #in W\n",
+      "A_p = I_o-I_t            #absorbed power in W or J/s\n",
+      "c = 1.43\n",
+      "A_E = (hv-c)/hv*A_p           # in J/s\n",
+      "e = 1.6*10**-19               #in C\n",
+      "P = A_p/(e*hv)               #Perfect quantum efficiency in photon/s\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Total energy absorbed  is =%.f X 10^-3 in J/s' %(A_p*(10**3)))\n",
+      "print('Rate of excess thermal energy  is =%.2f X 10^-3 in J/s' %(A_E*(10**3)))\n",
+      "print('Perfect quantum efficiency  is =%.2f X 10^16 in photon/s' %(P/(10**16)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total energy absorbed  is =9 X 10^-3 in J/s\n",
+        "Rate of excess thermal energy  is =2.56 X 10^-3 in J/s\n",
+        "Perfect quantum efficiency  is =2.81 X 10^16 in photon/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 3.24 Page No 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Mu_p = 500         #in cm^2/v-s\n",
+      "kT = 0.0259\n",
+      "Toh_p = 10**-10         #in sec\n",
+      "p_o = 10**17           #in cm^-3\n",
+      "q= 1.6*10**-19             # in C\n",
+      "A=0.5             #in square meter\n",
+      "del_p = 5 * 10**16        #in cm^-3\n",
+      "n_i= 1.5*10**10          #in cm^-3    \n",
+      "\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "D_p = kT * Mu_p          #in cm/s\n",
+      "L_p = math.sqrt(D_p * Toh_p)             # in cm\n",
+      "x = 10**-5            #in cm\n",
+      "p = p_o+del_p*e**(x/L_p)          #in cm^-3\n",
+      "Eip= math.log(p/n_i)*kT              #in eV\n",
+      "Ecp= 1.1/2-Eip                   #value of E_c-E_p in eV\n",
+      "Ip= q*A*D_p/L_p*del_p*e**(x/L_p)         #in A\n",
+      "Qp= q*A*del_p*L_p               #in C\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The hole current  is : =%.2f X 10^3 in A ' %(Ip*(10**3)))\n",
+      "print('Approximation error')\n",
+      "print('The value of Qp  is :=%.2f X 10^-7 in C' %(Qp*(10**7)))\n",
+      "\n",
+      "#Note: There is a calculation error to evalaute the value of hole current hence the value of hole current in the book is wrong\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hole current  is : =8.61 X 10^3 in A \n",
+        "Approximation error\n",
+        "The value of Qp  is :=1.44 X 10^-7 in C\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo4.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo4.ipynb
new file mode 100755
index 00000000..e76421ad
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo4.ipynb
@@ -0,0 +1,928 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chapter No 4 : Junctions and Interfaces<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page No 184"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "N_D = 10**17 * 10**6      #in atoms/m^3\n",
+      "N_A = 0.5*10**16*10**6      #in atoms/m^3\n",
+      "Epsilon_r = 10\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Epsilon_o = 8.85*10**-12\n",
+      "Epsilon = Epsilon_r*Epsilon_o       #in F/m\n",
+      "e = 1.602*10**-19          #in C\n",
+      "V = 0\n",
+      "V_B = 0.7          #in V\n",
+      "W1 = math.sqrt( ((2*Epsilon*V_B)/e)*(1/N_A+1/N_D) )       #in m\n",
+      "V_o = V_B    #in V\n",
+      "V1 = -10          #in V\n",
+      "V_B1 = V_o-V1        #in V\n",
+      "W = math.sqrt( ((2*Epsilon*V_B1)/e)*(1/N_A+1/N_D) )             #in m\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The junction width in meter when no external voltage is applied is =%.2f X 10^-6' %(W1*(10**6)))\n",
+      "print('Junction width in meter with an external voltage of -10V is =%.2f X 10^-6 m' %(W*(10**6)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The junction width in meter when no external voltage is applied is =0.39 X 10^-6\n",
+        "Junction width in meter with an external voltage of -10V is =1.54 X 10^-6 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page No 208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V = 5        #in V\n",
+      "V_Gamma = 0.6        #in V\n",
+      "r_F = 12           #in ohm\n",
+      "R = 1           #in k ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "R = R * 10**3    #in ohm\n",
+      "I_F = (V-V_Gamma)/(R+r_F)    #in A\n",
+      "V_F = V_Gamma + (I_F*r_F)  #in V\n",
+      "I_F=I_F*10**3 \n",
+      "#RESULTS\n",
+      "print('The forward diode current is =%.2f mA ' %I_F)\n",
+      "print('The diode voltage is =%.2f V ' %V_F)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The forward diode current is =4.35 mA \n",
+        "The diode voltage is =0.65 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "n = 4.4*10**22          #total number of Ge atoms/cm^3\n",
+      "n_a = 1*10**8           #number of impurity atoms\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N_A = n/n_a             #in atoms/cm^3\n",
+      "N_A = N_A * 10**6       #in atoms/m^3\n",
+      "n_i = 2.5*10**13        #in atoms/cm^3\n",
+      "n_i = n_i * 10**6       #in atoms/m^3\n",
+      "N_D = 10**3 * N_A       #in atoms/m^3\n",
+      "V_T = 26*10**-3         #in A\n",
+      "V_J = V_T*math.log( (N_A*N_D)/((n_i)**2) )    #in V\n",
+      "print('The contact difference of potential  is =%.2f V For a silicon P-N junction ' %V_J)\n",
+      "n = 5*10**22\n",
+      "N_A = n/n_a      #in atoms/cm^3\n",
+      "N_A = N_A * 10**6  #in atoms/m^3\n",
+      "N_D = 10**3 * N_A   #in atoms/m^3\n",
+      "n_i = 1.5*10**10        #in /cm^3\n",
+      "V_J = V_T*math.log(N_A*N_D/n_i**2)      #in V\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "\n",
+      "print('The contact difference of potential  is =%.2f V' %V_J)\n",
+      "\n",
+      "#Note: There is a calculation error to find the value of V_J in the book, so the answer in the book is wrong.\n",
+      "\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The contact difference of potential  is =0.33 V For a silicon P-N junction \n",
+        "The contact difference of potential  is =1.44 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 Page No 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Rho_p = 2     #in ohm-cm\n",
+      "Rho_n = 1      #in ohm cm\n",
+      "q = 1.6*10**-19     #in C\n",
+      "n_i = 2.5*10**13  #atoms per cm^3\n",
+      "Miu_p = 1800\n",
+      "Miu_n = 3800\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N_A = 1/(Rho_p*q*Miu_p)     #in /cm^3\n",
+      "N_D = 1/(Rho_n*q*Miu_n)     #in /cm^3\n",
+      "V_T = 26           #in mV\n",
+      "V_T= V_T*10**-3         #in V\n",
+      "V_J = V_T*math.log((N_A*N_D)/((n_i)**2))   #in V\n",
+      "print('The height of the potential energy barrier is =%.2f in V ' %V_J)\n",
+      "Miu_p = 500\n",
+      "N_A = 1/(Rho_p*q*Miu_p)      #in /cm^3\n",
+      "Miu_n = 1300\n",
+      "N_D = 1/(Rho_n*q*Miu_n)        #in /cm^3\n",
+      "n_i = 1.5*10**10;\n",
+      "V_J = V_T*math.log((N_A*N_D)/((n_i)**2))        #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('For silicon P-N juction The height of the potential energy barrier  is =%.2f V' %V_J)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The height of the potential energy barrier is =0.22 in V \n",
+        "For silicon P-N juction The height of the potential energy barrier  is =0.67 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 page No 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Eta = 1\n",
+      "V_T = 26       #in mV\n",
+      "V_T= V_T*10**-3    #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V= math.log(1-0.9)*V_T     #in V\n",
+      "print(\"The voltage in volts is : %.2f \" %V)\n",
+      "V1=0.05     #in V\n",
+      "V2= -0.05   #in V\n",
+      "ratio= (math.e**(V1/(Eta*V_T))-1)/(math.e**(V2/(Eta*V_T))-1)\n",
+      "print(\"The ratio of the current for a forward bias to reverse bias is :  %.2f \" %ratio)\n",
+      "\n",
+      "# Part (iii)\n",
+      "Io= 10  #in \u00b5A\n",
+      "Io=Io*10**-3  #in mA\n",
+      "#For \n",
+      "V=0.1   #in V\n",
+      "I = Io * (math.e**(V/(Eta*V_T)) - 1)     #in mA\n",
+      "print(\"For v=0.1 V , the value of I  is : %.2f mA \" %I)\n",
+      "\n",
+      "#For \n",
+      "V=0.2      #in V\n",
+      "I = Io * (math.e**(V/(Eta*V_T)) - 1)      #in mA\n",
+      "print(\"For v=0.2 V , the value of I  is : %.2f  mA \" %I)\n",
+      "\n",
+      "#For \n",
+      "V=0.3  #in V\n",
+      "I = Io * (math.e**(V/(Eta*V_T)) - 1)           #in mA\n",
+      "\n",
+      "#RESULTS\n",
+      "\n",
+      "I=I*10**-3\n",
+      "print(\"For v=0.3 V , the value of I  is : %.2f  mA\" %I)\n",
+      "print(\"From three value of I, for small rise in forward voltage, the diode current increase rapidly \")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage in volts is : -0.06 \n",
+        "The ratio of the current for a forward bias to reverse bias is :  -6.84 \n",
+        "For v=0.1 V , the value of I  is : 0.46 mA \n",
+        "For v=0.2 V , the value of I  is : 21.90  mA \n",
+        "For v=0.3 V , the value of I  is : 1.03  mA\n",
+        "From three value of I, for small rise in forward voltage, the diode current increase rapidly \n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 Page No 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "# Part (i)\n",
+      "T1= 25        #in \u00b0C\n",
+      "T2= 80.0      #in \u00b0C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "# Formula Io2= Io1*2**((T2-T1)/10)\n",
+      "AntiFactor= 2**((T2-T1)/10)\n",
+      "print('Anticipated factor for Ge is : =%.f  ' %AntiFactor)\n",
+      "\n",
+      "# Part (ii)\n",
+      "T1= 25.0     #in \u00b0C\n",
+      "T2= 150.0    #in \u00b0C\n",
+      "AntiFactor= 2**((T2-T1)/10)\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Anticipated factor for Ge is : =%.f  ' %AntiFactor)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Anticipated factor for Ge is : =45  \n",
+        "Anticipated factor for Ge is : =5793  \n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 Page No 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I=5.0        #in \u00b5A\n",
+      "V=10.0       #in V\n",
+      "T1= 0.11     #in \u00b0C^-1\n",
+      "T2= 0.07      #in \u00b0C^-1\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Io= T2*I/T1   #in \u00b5A\n",
+      "I_R= I-Io      #in \u00b5A\n",
+      "R= V/I_R       #in M\u03a9\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The leakage resistance is : =%.1f M\u03a9 ' %R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The leakage resistance is : =5.5 M\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10 Page No 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Eta = 1.0\n",
+      "T = 125.0   #in \u00b0C\n",
+      "T = T + 273   #in K\n",
+      "V_T = 8.62 * 10**-5 * 398    #in V\n",
+      "I_o = 30      #in \u00b5A\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_o= I_o*10**-6    #in A\n",
+      "v = 0.2            #in V\n",
+      "r_f = (Eta * V_T)/(I_o * math.e**(v/(Eta* V_T)))      #in ohm\n",
+      "print('The dynamic resistance in the forward direction  is =%.2f \u03a9 ' %r_f)\n",
+      "\n",
+      "r_r = (Eta * V_T)/(I_o * math.e**(-v/(Eta* V_T)))   #in ohm\n",
+      "r_r=r_r*10**-3\n",
+      "print('The dynamic resistance in the forward direction  is =%.2f K\u03a9 ' %r_r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dynamic resistance in the forward direction  is =3.36 \u03a9 \n",
+        "The dynamic resistance in the forward direction  is =389.08 K\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 Page No 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "epsilon = 16/(36 * math.pi * 10**11)       #in F/cm\n",
+      "A = 1 * 10**-2\n",
+      "W = 2 * 10**-4\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "C_T = (epsilon * A)/W      #in F\n",
+      "C_T=C_T*10**12\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The barrier capacitance is =%.2f pF ' %C_T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The barrier capacitance is =70.74 pF \n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 Page No 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "A = 1.0           #in mm^2\n",
+      "A = A * 10**-6    #in m^2\n",
+      "N_A = 3 * 10**20    #in atoms/m^3\n",
+      "q = 1.6 *10**-19    #in C\n",
+      "V_o = 0.2         #in V\n",
+      "epsilon_r=16\n",
+      "epsilon_o= 8.854*10**-12        #in F/m\n",
+      "\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "epsilon=epsilon_r*epsilon_o\n",
+      "# Part (a)\n",
+      "V=-10    #in V\n",
+      "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A))    #m\n",
+      "C_T1 = (epsilon * A)/W           #in F\n",
+      "W=W*10**6\n",
+      "print('The width of the depletion layer  for an applied reverse voltage of 10V  is =%.2f \u00b5m ' %W)\n",
+      "\n",
+      "# Part (b)\n",
+      "V=-0.1           #in V\n",
+      "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A))         #m\n",
+      "C_T2 = (epsilon * A)/W            #in F\n",
+      "W=W*10**6\n",
+      "print('The width of the depletion layer  for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
+      "\n",
+      "# Part (c)\n",
+      "V=0.1         #in V\n",
+      "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A))   # m\n",
+      "W=W*10**6\n",
+      "print('The width of the depletion layer  for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
+      "\n",
+      "#Part (d)\n",
+      "C_T1=C_T1*10**12\n",
+      "C_T2=C_T2*10**12\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The space charge capacitance for an applied reverse voltage of 10V  is =%.2f pF ' %C_T1)\n",
+      "print('The space charge capacitance for an applied reverse voltage of 0.1V  is =%.2f  pF' %C_T2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The width of the depletion layer  for an applied reverse voltage of 10V  is =7.76 \u00b5m \n",
+        "The width of the depletion layer  for an applied reverse voltage of 0.1V is =1.33 \u00b5m \n",
+        "The width of the depletion layer  for an applied reverse voltage of 0.1V is =0.77 \u00b5m \n",
+        "The space charge capacitance for an applied reverse voltage of 10V  is =18.26 pF \n",
+        "The space charge capacitance for an applied reverse voltage of 0.1V  is =106.46  pF\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 4.13 Page No 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_o = 1.8 * 10**-9         #A\n",
+      "v = 0.6                 #in V\n",
+      "Eta = 2\n",
+      "V_T = 26              #in mV\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_T=V_T*10**-3      #in V\n",
+      "I = I_o *(math.e**(v/(Eta * V_T)))           #in A\n",
+      "I=I*10**3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The current in the junction  is =%.2f mA ' %I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current in the junction  is =0.18 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 4.14 Page No 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_o = 2.4 * 10**-14\n",
+      "I = 1.5         #in mA\n",
+      "I=I*10**-3       #in A\n",
+      "Eta = 1\n",
+      "V_T = 26           #in mV\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_T= V_T*10**-3        #in V\n",
+      "v =math.log((I + I_o)/I_o) * V_T          #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The forward biasing voltage across the junction  is =%.2f   V' %v)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The forward biasing voltage across the junction  is =0.65   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 4.15 Page No 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "I_o = 10              #in nA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I = I_o * (-1)         #in nA\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The Diode current is  = %.f nA ' %I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Diode current is  = -10 nA \n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 4.16 Page No 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "R = 4.5   #in ohm\n",
+      "I = 44.4       #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I=I*10**-3          #in A\n",
+      "V = R * I         #in V\n",
+      "Eta = 1\n",
+      "V_T = 26          #in mV\n",
+      "V_T=V_T*10**-3    #in V\n",
+      "I_o = I/((math.e**(V/(Eta * V_T))) -1)         #in A\n",
+      "V = 0.1         #in V\n",
+      "r_f = (Eta * V_T)/(I_o * ((math.e**(V/(Eta * V_T)))-1))        #in ohm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The diode dynamic resistance is =%.2f  \u03a9' %r_f)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diode dynamic resistance is =27.78  \u03a9\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 4.18 Page No 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V = 0.25          #in V\n",
+      "I_o = 1.2         #in \u00b5A\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_o = I_o * 10**-6          #in A\n",
+      "V_T = 26                   #in mV\n",
+      "V_T = V_T * 10**-3         #in V\n",
+      "Eta = 1\n",
+      "r = (Eta * V_T)/(I_o * (math.e**(V/(Eta * V_T))))        #in ohm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The ac resistance of the diode  is =%.2f  \u03a9' %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ac resistance of the diode  is =1.45  \u03a9\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 4.19 Page No 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "q = 1.6 * 10**-19         #in C\n",
+      "N_A = 3 * 10**20          #in /m^3\n",
+      "A = 1               #in \u00b5m^2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A = A * 10**-6        #in m^2\n",
+      "V = -10          #in V\n",
+      "V_J = 0.25        #in V\n",
+      "V_B = V_J - V        #in V\n",
+      "epsilon_o = 8.854       #in pF/m\n",
+      "epsilon_o = epsilon_o * 10**-12          #in F/m\n",
+      "epsilon_r = 16\n",
+      "epsilon = epsilon_o * epsilon_r\n",
+      "W = math.sqrt((V_B * 2 * epsilon)/(q * N_A))   #in m \n",
+      "\n",
+      "#RESULTS\n",
+      "C_T = (epsilon * A)/W   #in pF\n",
+      "W=W*10**6\n",
+      "C_T=C_T*10**12\n",
+      "print('The width of depletion layer is =%.2f \u00b5m ' %W)\n",
+      "print('The space charge capacitance is =%.2f pF ' %C_T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The width of depletion layer is =7.78 \u00b5m \n",
+        "The space charge capacitance is =18.21 pF \n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 4.20 Page No 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "C_T = 100.0   #in pF\n",
+      "C_T=C_T*10**-12     #in F\n",
+      "epsilon_r = 12\n",
+      "epsilon_o = 8.854 * 10**-12      #in F/m\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "epsilon = epsilon_r * epsilon_o\n",
+      "Rho_p = 5   #in ohm-cm\n",
+      "Rho_p = Rho_p * 10**-2         #in ohm-m\n",
+      "V_j = 0.5           #in V\n",
+      "V = -4.5           #in V\n",
+      "Mu_p = 500          #in cm^2\n",
+      "Mu_p = Mu_p * 10**-4     #in m^2\n",
+      "Sigma_p = 1/Rho_p         #in per ohm-m\n",
+      "qN_A = Sigma_p/ Mu_p\n",
+      "V_B = V_j - V\n",
+      "W = math.sqrt((V_B * 2 * epsilon)/qN_A)\n",
+      "A = (C_T * W)/ epsilon             #in m\n",
+      "D = math.sqrt(A * (4/math.pi))        #in m\n",
+      "D = D * 10**3          #in mm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The diameter is =%.2f mm ' %D)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diameter is =1.40 mm \n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 4.21 Page No 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "q = 1.6 * 10**-19         #in C\n",
+      "Mu_p = 500         #in cm^2/V-sec\n",
+      "Rho_p = 3.5          #in ohm-cm\n",
+      "Mu_n = 1500          #in cm^2/V-sec\n",
+      "Rho_n = 10          #in ohm-cm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N_A = 1/(Rho_p * Mu_p * q)       # in /cm^3\n",
+      "N_D = 1/(Rho_n * Mu_n * q)       # in /cm^3\n",
+      "V_J = 0.56            # in V\n",
+      "n_i = 1.5 * 10**10         #in /cm^3\n",
+      "V_T = V_J/math.log((N_A * N_D)/(n_i)**2)     #in V\n",
+      "T = V_T * 11600          #in K\n",
+      "T = T - 273            #in \u00b0C\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The Temperature of junction is =%.2f \u00b0C ' %T)\n",
+      "print('Approximation error')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Temperature of junction is =14.28 \u00b0C \n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example No 4.22 Page No 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "R = 5       #in ohm\n",
+      "I = 50      #in mA\n",
+      "I=I*10**-3   #in A\n",
+      "V = R * I    #in V\n",
+      "Eta = 1\n",
+      "V_T = 26    #in mV\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_T=V_T*10**-3     #in V\n",
+      "I_o = I/((math.e**(V/(Eta * V_T))) - 1)     #in A\n",
+      "v1 = 0.2         #in V\n",
+      "r = (Eta * V_T)/(I_o * (math.e**(v1/(Eta * V_T))))         #in ohm\n",
+      "\n",
+      "#RESULTS\n",
+      "I_o=I_o*10**6\n",
+      "print('Reverse saturation current  is =%.2f \u00b5A ' %I_o)\n",
+      "print('Dynamic resistance of the diode is =%.2f \u03a9 ' %r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reverse saturation current  is =3.33 \u00b5A \n",
+        "Dynamic resistance of the diode is =3.56 \u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo6.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo6.ipynb
new file mode 100755
index 00000000..44239148
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo6.ipynb
@@ -0,0 +1,76 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater No 6: Microwave Diodes <h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page No 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "C1_min= 5.0     #in pF\n",
+      "C1_max= 50.0  #in pF\n",
+      "C2_min= 5.0  #in pF\n",
+      "C2_max= 50.0  #in pF\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "C1_min=C1_min*10**-12   #in F\n",
+      "C2_min=C2_min*10**-12   #in F\n",
+      "C1_max=C1_max*10**-12   #in F\n",
+      "C2_max=C2_max*10**-12    #in F\n",
+      "L = 10.0                #in mH\n",
+      "L = L * 10**-3          #in H\n",
+      "C_T_min = (C1_min*C2_min)/(C2_min+C2_min)     #in F\n",
+      "f_o_max = 1/( 2*math.pi*(math.sqrt(L*C_T_min)) )      #in Hz \n",
+      "f_o_max = f_o_max * 10**-6      #in MHz\n",
+      "C_T_max = (C1_max*C2_max)/(C2_max+C2_max)     #in F\n",
+      "f_o_min = 1/( 2*math.pi*(math.sqrt(L*C_T_max)) )      #in Hz \n",
+      "f_o_min = f_o_min * 10**-3                      #in kHz\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The tuning range for circuit will be =%.2f  kHz' %f_o_min)\n",
+      "print(' To %.3f MHz' %f_o_max)\n",
+      "\n",
+      "\n",
+      "#print(\"The tuning range for circuit will be : \"+string(round(f_o_min))+\" kHz to \"+string(round(f_o_max))+\" MHz\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The tuning range for circuit will be =318.31  kHz\n",
+        " To 1.007 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo7.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo7.ipynb
new file mode 100755
index 00000000..431c4340
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo7.ipynb
@@ -0,0 +1,154 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater No 7: Optoelectonic Devices <h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1, Page No 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Vout = 5.0         #in V\n",
+      "V = 1.5               #ON state voltage drop across LED in V\n",
+      "I = 5.0          #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I = I*10**-3       #in A\n",
+      "R = (Vout-V)/I         #in ohm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Resistance  is =%.f \u03a9' %R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance  is =700 \u03a9\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page No 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "N_A = 7.5*10**24            #in atoms/m^3\n",
+      "N_D = 1.5*10**22            #in atoms/m^3\n",
+      "D_e = 25*10**-4             #in m^2/s\n",
+      "D_n = 1*10**-3             #in m^2/s\n",
+      "V_T = 26*10**-3            #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Torque_eo = 500*10**-9            #in sec\n",
+      "Torque_ho = 100*10**-9            #in sec\n",
+      "e = 1.6*10**-19                #in C\n",
+      "n_i = 1.5*10**16              #in /m^3\n",
+      "I_lambda = 12.5              #in mA/cm^2\n",
+      "I_lambda= I_lambda*10**-3       #in A/cm^2\n",
+      "L_e = math.sqrt(D_e*Torque_eo)    #in m\n",
+      "L_n = math.sqrt(D_n*Torque_ho)  #in m\n",
+      "J_s = e*((n_i)**2)*( ((D_e)/(L_e*N_A)) + ((D_n)/(L_n*N_D)) )         #in A/m^2\n",
+      "J_s= J_s*10**-4                      #in A/cm^2\n",
+      "V_OC = V_T*(math.log(1+(I_lambda/J_s)))                     #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Open circuit voltage is =%.2f volts ' %V_OC)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Open circuit voltage is =0.52 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page No 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Phi_o = 1.0*10**21               #in m^-2s^-1\n",
+      "alpha =1.0*10**5                 #in m^-1\n",
+      "e= 1.6*10**-19                 #in C\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "G_L1 = alpha*Phi_o             #in m^-3s^-1\n",
+      "W = 26.0                       #in \u00b5m\n",
+      "W = W * 10**-6                 #in m\n",
+      "G_L2 = alpha*Phi_o*(math.e)**((-alpha)*W)             #in m^-3s^-1\n",
+      "#temp=(1-math.e)**(-(alpha)*W)\n",
+      "J_L = e*Phi_o*(1-math.e**(-(alpha)*W))               #in A/m^2\n",
+      "J_L = J_L * 10**3*10**-4                         #in mA/cm^2\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Photo current density  is =%.1f mA/cm^2' %J_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Photo current density  is =14.8 mA/cm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo8.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo8.ipynb
new file mode 100755
index 00000000..47137bb2
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo8.ipynb
@@ -0,0 +1,576 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chapter No 8 : Metal Semiconductor Field Effect Transistor <h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page No  323"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "I_DSS = 15.0    #in mA\n",
+      "V_GS_off = -5.0    #in V\n",
+      "V_GS = 0        #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_D = I_DSS*((1-(V_GS/V_GS_off))**2)     #in mA\n",
+      "V_GS = -1                   #in V\n",
+      "I_D1 = I_DSS*((1.0-(V_GS/V_GS_off))**2)         #in mA\n",
+      "V_GS = -4             #in V\n",
+      "I_D2 = I_DSS*((1-(V_GS/V_GS_off))**2)          #in mA\n",
+      "\n",
+      "#RESULTS\n",
+      "print('When V_GS=0, the drain current is =%.f mA ' %I_D)\n",
+      "print('When V_GS=-1V, the drain current is =%.2f mA ' %I_D1)\n",
+      "print('The interplanner spacing is =%.2f  \u00c5' %I_D2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When V_GS=0, the drain current is =15 mA \n",
+        "When V_GS=-1V, the drain current is =9.60 mA \n",
+        "The interplanner spacing is =0.60  \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4, Page No 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_DSS = 20.0            #in mA\n",
+      "V_P = -8            #in V\n",
+      "g_mo = 5000.0       #in \u00b5s\n",
+      "V_GS = -4.0            #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "\n",
+      "I_D = I_DSS*((1-(V_GS/V_P))**2)      #in mA\n",
+      "g_m = g_mo*(1-(V_GS/V_P))            #in \u00b5s\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of drain current is =%.f mA ' %I_D)\n",
+      "print('The transconductance is =%.f  \u00b5s' %g_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of drain current is =5 mA \n",
+        "The transconductance is =2500  \u00b5s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5 Page No 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_D1 = 10.0        #in mA\n",
+      "V_GS = -12.0    #in V\n",
+      "V_GSth = -3.0    #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "K = I_D1/( (V_GS-V_GSth)**2 )         #in mA/V\n",
+      "V_GS= -6                 #in V\n",
+      "I_D = K*((V_GS-V_GSth)**2)         #in mA\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_D  is =%.2f mA ' %I_D)\n",
+      "print('Approximation error')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D  is =1.11 mA \n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7, Page No 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_GS = -2.0            #in V\n",
+      "V_P = -5.0          #in V\n",
+      "V_DS = V_GS-V_P     #in V\n",
+      "I_DSS = 8.0      #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_D = I_DSS*((1-(V_GS/V_P))**2)        #in mA\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The minimum value of V_DS is =%.f  V ' %V_DS)\n",
+      "print('The drain current is =%.2f  mA' %I_D)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum value of V_DS is =3  V \n",
+        "The drain current is =2.88  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8  Page No 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_P = -2                 #in V\n",
+      "I_DSS = 1.65             #in mA\n",
+      "I_D = 0.8                #in mA\n",
+      "V_DD = 24.0              #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS = V_P*(1- math.sqrt(I_D/I_DSS) )    #in V\n",
+      "g_mo = -(2*I_DSS)/V_P           #in mS\n",
+      "g_m = g_mo*(1-(V_GS/V_P))       #in mS\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_GS is =%.2f V  ' %V_GS)\n",
+      "print('The value of g_m is =%.2f  mS' %g_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_GS is =-0.61 V  \n",
+        "The value of g_m is =1.15  mS\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9 Page No 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_P = 5.0         #in V\n",
+      "I_DSS = -40.0     #in mA\n",
+      "I_D = -15.0         #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS = V_P*(math.sqrt(I_D/I_DSS)-1 )       #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The gate source voltage is =%.2f   V' %(V_GS*-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The gate source voltage is =1.94   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10, Page No 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_D1 = 1.9       #in mA\n",
+      "I_D2 = 1.0       #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "del_I_D = I_D1-I_D2          #in mA\n",
+      "V_GS2 = -3.3              #in V\n",
+      "V_GS1 = -3.0            #in V\n",
+      "del_V_GS = V_GS1-V_GS2      #in V\n",
+      "g_m = del_I_D/del_V_GS       #in mA/V\n",
+      "g_m = g_m * 10**3         #in \u00b5 mhos\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of transconductance is =%.f \u00b5 mhos ' %g_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of transconductance is =3000 \u00b5 mhos \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11 Page No 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_DS1 = 14.0          #in V\n",
+      "V_DS2 = 5.0          #in V\n",
+      "del_V_DS = V_DS1-V_DS2      # in V\n",
+      "I_D1 = 3.3            #in mA\n",
+      "I_D2 = 3              # in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "del_I_D = I_D1-I_D2           #in mA\n",
+      "r_d = del_V_DS/del_I_D          #in k ohms\n",
+      "print('The drain resistance  is =%.f  k ohms' %r_d)\n",
+      "V_GS1 = 0.4              #in V\n",
+      "V_GS2 = 0.1             #in V\n",
+      "del_V_GS = V_GS1-V_GS2     #in V\n",
+      "I_D1 = 3.3             #in mA\n",
+      "I_D2 = 0.71           #in mA\n",
+      "del_I_D = I_D1-I_D2       # in mA\n",
+      "g_m = del_I_D/del_V_GS    # in mA/V\n",
+      "g_m = g_m * 10**3           # in \u00b5mhos\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The transconductance is =%.f \u00b5mhos  ' %g_m)\n",
+      "Miu =r_d*10**3*g_m*10**-6\n",
+      "print('Amplification factor is =%.f  ' %Miu)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The drain resistance  is =30  k ohms\n",
+        "The transconductance is =8633 \u00b5mhos  \n",
+        "Amplification factor is =259  \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12 Page No 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "q = 1.6*10**-19         #in C\n",
+      "N_D = 10**15*10**6      #electrons/m^3\n",
+      "a = 3*10**-4            #in cm\n",
+      "a=a*10**-2              #in m\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Epsilon_o  = (36 * math.pi * 10**9)**-1\n",
+      "Epsilon = 12*Epsilon_o\n",
+      "V_P = (q*N_D*((a)**2))/(2*Epsilon)       #in V\n",
+      "print('Pinch off voltage is =%.2f   V ' %V_P)\n",
+      "V_GS = 1.0     #in V\n",
+      "V_P = 2.0      #in V\n",
+      "b = a*( 1-math.sqrt(V_GS/V_P) )         #in m\n",
+      "b = b * 10**6              #in \u00b5m\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The channel half width  is =%.2f  \u00b5m' %b)\n",
+      "# Note: In the book, the unit of channel half width is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pinch off voltage is =6.79   V \n",
+        "The channel half width  is =0.88  \u00b5m\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.13 Page No 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_DSS = 8.0       #in mA\n",
+      "V_P = -4.0           #in V\n",
+      "a = 3*10**-4        #in cm\n",
+      "N_D = 10**15        #in electrons/cm^3\n",
+      "I_D = 3.0           #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS = V_P*( 1-math.sqrt(I_D/I_DSS) )           #in V\n",
+      "V_DS_sat = V_GS-V_P                             #in V\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_GS is =%.2f   V' %V_GS)\n",
+      "print('The value of V_DS_sat is =%.2f   V ' %V_DS_sat)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_GS is =-1.55   V\n",
+        "The value of V_DS_sat is =2.45   V \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 8.14 Page No 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_P = -4            #in V\n",
+      "I_DSS = 9.0       #in mA\n",
+      "V_GS = -2.0       #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_D = I_DSS*(( 1-(V_GS/V_P) )**2)        #in mA\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The drain current is =%.2f mA ' %I_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The drain current is =2.25 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 8.15 Page No 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "I_DSS = 12.0           #in mA\n",
+      "V_P = -6.0             #in V\n",
+      "V_GS = -1.0              #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "g_mo = (-2*I_DSS)/V_P      #in mA/V\n",
+      "g_m = g_mo*(1-(V_GS/V_P))         #in mS\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of transconductance is =%.2f  mS' %g_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of transconductance is =3.33  mS\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 8.16 Page No 336"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "I_DSS = 10.0          #in mA\n",
+      "V_P = -5.0            #in V\n",
+      "V_GS = -2.5          #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "g_m = ((-2*I_DSS)/V_P)*(1-(V_GS/V_P))         #in mS .... correction\n",
+      "I_D = I_DSS * ((1-(V_GS/V_P))**2)             #in mA\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The transconductance is =%.f mS ' %g_m)\n",
+      "print('The drain current is =%.2f mA ' %I_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The transconductance is =2 mS \n",
+        "The drain current is =2.50 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo9.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo9.ipynb
new file mode 100755
index 00000000..d2b16e2d
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo9.ipynb
@@ -0,0 +1,1684 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chapter No 9 : MOS Transistors<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page No  361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_TN = 0.7            #in V\n",
+      "W = 45*10**-4         #in cm\n",
+      "L = 4.0               #in \u00b5m\n",
+      "L = L * 10**-4        #in cm\n",
+      "t_ox = 450            #in \u00c5\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "t_ox = t_ox*10**-8           #in cm\n",
+      "V_GS = 1.4                   #in V\n",
+      "Miu_n = 700                  #in cm^2/V-s\n",
+      "Epsilon_ox = (8.85*10**-14)*(3.9)               #in F/cm\n",
+      "k_n = (W*Miu_n*Epsilon_ox)/(2*L*t_ox)               #A/V^2\n",
+      "I_D = k_n*((V_GS-V_TN)**2)                        #in A\n",
+      "k_n=k_n*10**3\n",
+      "I_D=I_D*10**3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of k_n is : =%.2f   mA/V^2' %k_n)\n",
+      "print('The current  is =%.2f  mA' %I_D)\n",
+      "print('Approximation error')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of k_n is : =0.30   mA/V^2\n",
+        "The current  is =0.15  mA\n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 87
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page No 376"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "I_Don = 6.0                #in mA\n",
+      "I_Don= I_Don*10**-3        #in A\n",
+      "V_GSon = 8                 #in V\n",
+      "V_GSth = 3                 #in V\n",
+      "V_DD = 12.0                #in V\n",
+      "R_D= 2*10**3               #in \u03a9\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "k= I_Don/(V_GSon-V_GSth)**2             #in A/V^2\n",
+      "A= R_D**2                 #assumed\n",
+      "B= 2*R_D*V_GSth-2*R_D*V_DD-1/k              #assumed\n",
+      "C= (V_DD-V_GSth)**2                #assumed\n",
+      "\n",
+      "I_DQ= 0.0027940                #in A\n",
+      "V_DSQ= V_DD-I_DQ*R_D            #in V\n",
+      "I_DQ=I_DQ*10**3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_DQ is : =%.2f  mA' %I_DQ)\n",
+      "print('The value of V_DSQ is : =%.2f volts ' %V_DSQ)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DQ is : =2.79  mA\n",
+        "The value of V_DSQ is : =6.41 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 88
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page No 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_GS = 6.0           #in V\n",
+      "I_D = 4.0            #in mA\n",
+      "V_GSth = 2.0         #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_DS = V_GS          #in V\n",
+      "V_DD = 2*V_DS        #in V\n",
+      "R_D = (V_DD-V_DS)/I_D         #in k ohm\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_DD is =%.2f  V  ' %V_DD)\n",
+      "print('The value of R_D  is  =%.2f   k ohm' %R_D)\n",
+      "print('The very high value for the gate to drain resistance is : 10 M\u03a9 ' )\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_DD is =12.00  V  \n",
+        "The value of R_D  is  =1.50   k ohm\n",
+        "The very high value for the gate to drain resistance is : 10 M\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 89
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page No 379"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_Don = 3*10**-3  \n",
+      "V_GSon = 10.0          #in V\n",
+      "V_GSth= 5.0        #in V\n",
+      "R2= 18*10**6        #in \u03a9\n",
+      "R1= 22*10**6      #in \u03a9\n",
+      "R_S=820         #in \u03a9\n",
+      "R_D=3*10**3     #in \u03a9\n",
+      "V_DD= 40.0      #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_G= V_DD*R2/(R1+R2)           #in V\n",
+      "k= I_Don/(V_GSon-V_GSth)**2           #in A/V**2\n",
+      "A= R_S**2                  #assumed\n",
+      "B= 2*R_S*V_GSth-2*R_S*V_G-1/k           #assumed\n",
+      "C= (V_G-V_GSth)**2          #assumed\n",
+      "I_D= 0.0067246               #in A\n",
+      "I_DQ= I_D                 #in A\n",
+      "V_GSQ= V_G-I_D*R_S      #in V\n",
+      "V_DSQ= V_DD-I_DQ*(R_D+R_S)         #in V\n",
+      "I_DQ=I_DQ*10**3\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_DQ  is =%.2f  mA' %I_DQ)\n",
+      "print('The value of V_GSQ is =%.2f volts ' %V_GSQ)\n",
+      "print('The value of V_DSQ is =%.2f volts ' %V_DSQ)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DQ  is =6.72  mA\n",
+        "The value of V_GSQ is =12.49 volts \n",
+        "The value of V_DSQ is =14.31 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 90
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page No 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_D= '0.3*(V_GS-V_P)^2'                #given expression\n",
+      "V_DD= 30.0                           #in V\n",
+      "V_P= 4                               #in V\n",
+      "R_GS = 1.2*10**6                      #in \u03a9\n",
+      "R_G = 1.2*10**6                       #in \u03a9\n",
+      "I_D= 1.2           #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Req= R_GS/(R_GS+R_G)                  #in \u03a9\n",
+      "R_D= 15.0                             #in \u03a9\n",
+      "A= (R_D*Req)**2                       #assumed\n",
+      "B= -(2*R_D*Req*(V_DD*Req-V_P)+1/0.3)      #assumed\n",
+      "C= (V_DD*Req-V_P)**2                  #assumed\n",
+      "I_DSQ= I_D             #in mA\n",
+      "V_GS= (V_DD-I_D*R_D)    #in V\n",
+      "V_DS= Req*V_GS         #in V\n",
+      "\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_DSQ  is : =%.2f  mA' %I_DSQ)\n",
+      "print('The value of V_GS  is :  =%.2f volts ' %V_GS)\n",
+      "print('The value of V_DS  is : =%.2f volts ' %V_DS)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DSQ  is : =1.20  mA\n",
+        "The value of V_GS  is :  =12.00 volts \n",
+        "The value of V_DS  is : =6.00 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 91
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 Page No 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "k = 0.1          #in mA/V^2\n",
+      "V_T  = 1.0            # in V\n",
+      "R1 = 33.0          #in k ohm\n",
+      "R2 = 21.0         #in k ohm\n",
+      "V_DD = 6.0       #in V\n",
+      "R_D = 18           #in k ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_G = (R2/(R2+R1))*V_DD             #in V\n",
+      "V_S = 0               #in V\n",
+      "V_GS = V_G-V_S             #in V\n",
+      "I_D = k*((V_GS-V_T)**2)         #in mA\n",
+      "V_DS = V_DD - (I_D*R_D)       #in V\n",
+      "V_DSsat = V_GS-V_T            #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_D  is =%.2f mA ' %I_D)\n",
+      "print('The value of V_DS(sat) is =%.2f  V ' %V_DS)\n",
+      "print('The value of V_DS(sat)  is =%.2f V ' %V_DSsat)\n",
+      "\n",
+      "if V_DS > V_DSsat :\n",
+      "    print('MOSFET is in saturation region  ')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D  is =0.18 mA \n",
+        "The value of V_DS(sat) is =2.80  V \n",
+        "The value of V_DS(sat)  is =1.33 V \n",
+        "MOSFET is in saturation region  \n"
+       ]
+      }
+     ],
+     "prompt_number": 92
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8, Page No 382"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "R2 = 18.0       #in k ohm\n",
+      "R1 = 33.0       #in k ohm\n",
+      "V_DD = 6.0       #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_G = (R2/(R1+R2))*V_DD              #in V\n",
+      "V_S = V_DD                          # in V\n",
+      "V_SG = V_S-V_G                      #in V\n",
+      "k = 0.1\n",
+      "V_T = -1                          #in V\n",
+      "I_D = k*((V_SG+V_T)**2)           #in mA\n",
+      "R_D = 3.0                     # in k ohm\n",
+      "V_SD = V_DD - (I_D*R_D)            #in V\n",
+      "V_SDsat  = V_SG+V_T                 #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_SG is =%.2f V ' %V_SG)\n",
+      "print('The value of I_D  is =%.2f mA ' %I_D)\n",
+      "print('The value of V_SD  is =%.2f V ' %V_SD)\n",
+      "print('The value of V_SD(sat)  is =%.2f V ' %V_SDsat)\n",
+      "\n",
+      "if V_SD > V_SDsat :\n",
+      "       print('The p MOSFET is indeed biased in the saturation region')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_SG is =3.88 V \n",
+        "The value of I_D  is =0.83 mA \n",
+        "The value of V_SD  is =3.51 V \n",
+        "The value of V_SD(sat)  is =2.88 V \n",
+        "The p MOSFET is indeed biased in the saturation region\n"
+       ]
+      }
+     ],
+     "prompt_number": 93
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9 Page No 384"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_G= 1.5                # in V\n",
+      "V_P= -3.0              # in V\n",
+      "R_S= 750.0               # in \u03a9\n",
+      "R_D= 1800              #in \u03a9\n",
+      "I_DSS= 6*10**-3         # in A\n",
+      "V_DD= 18            #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= R_S**2\n",
+      "B=(2*R_S*(V_P-V_G)-V_P**2/I_DSS)\n",
+      "C=(V_P-V_G)**2\n",
+      "I_D= 0.0031170               #in A\n",
+      "I_DQ= I_D                 #in A\n",
+      "V_DS= V_DD-I_D*(R_D+R_S)        # in V\n",
+      "V_DSQ= V_DS                  # in V\n",
+      "I_DQ=I_DQ*10**3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_DQ  is : =%.2f mA ' %I_DQ)\n",
+      "print('The value of V_DSQ  is : =%.2f volts ' %V_DSQ)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DQ  is : =3.12 mA \n",
+        "The value of V_DSQ  is : =10.05 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 94
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10 Page No 384"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_GS = 4.0              #in V\n",
+      "V_P = 2.0               #in V\n",
+      "R2 = 10.0             #in k ohm\n",
+      "R1 = 30.0           #in k ohm\n",
+      "R_D= 2.5           #in kohm\n",
+      "I_D= 15            #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_D= I_D*10**-3           #in A\n",
+      "V_DD = 25                 #in V\n",
+      "V_G = (V_DD/R_D)*V_DD/(R1+R2)                #in V\n",
+      "R_S = (V_G-V_GS)/I_D                 # in ohm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of R_S  is =%.f ohm ' %R_S)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R_S  is =150 ohm \n"
+       ]
+      }
+     ],
+     "prompt_number": 95
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11 Page No 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "k= 0.1            #in mA/V^2\n",
+      "V_T= 1            # in V\n",
+      "R2= 87*10**3         #in \u03a9\n",
+      "R1= 110*10**3       # in \u03a9\n",
+      "R_S=2            #in k\u03a9\n",
+      "R_D=2            #in k\u03a9\n",
+      "V_DD= 6             #in V\n",
+      "V_SS= 6          #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_G= (V_DD+V_SS)*R2/(R1+R2)              # in V\n",
+      "A= R_S**2              # assumed\n",
+      "B= -(2*R_S*(V_G+V_SS-V_T)+1/k)              # assumed\n",
+      "C= (V_G+V_SS-V_T)**2                    #assumed\n",
+      "I_D= 2.6001587           # in mA\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_D is : =%.2f mA' %I_D)\n",
+      "V_DS=V_DD+V_SS-I_D*R_D-I_D*R_S        #in V\n",
+      "print('The value of V_DS  is : =%.2f volts' %V_DS)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D is : =2.60 mA\n",
+        "The value of V_DS  is : =1.60 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 96
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.12 Page No 386"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "k = 0.16                      #in mA/V^2\n",
+      "V_T = 2.0                   #in V\n",
+      "I_D = 0.5                      #in mA\n",
+      "V_DD = 6.0                 #in V\n",
+      "V_SS = -6                 #in V\n",
+      "R_D = 10.0                     #in k ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS = V_T + (math.sqrt(I_D/k))               #in V\n",
+      "R_S = 2                #in k ohm\n",
+      "V_S = (I_D*R_S) - V_DD           # in V\n",
+      "V_G = V_GS+V_S                     # in V\n",
+      "I = 0.1*I_D                 #in mA\n",
+      "R2 = (V_G+V_DD)/I                #in k ohm\n",
+      "print('The value of R2  is =%.2f k ohm' %R2)\n",
+      "R1 = (V_DD - V_G)/I                  #in k ohm\n",
+      "print('The value of R1 is =%.2f k ohm' %R1)\n",
+      "V_DS = (V_DD-V_SS) - (I_D*(R_S+R_D))           #in V\n",
+      "print('The value of V_DS is =%.2f V' %V_DS)\n",
+      "V_DSsat = V_GS-V_T                      # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_DS(sat) is =%.2f V' %V_DSsat)\n",
+      "if V_DS > V_DSsat :\n",
+      "    print('The MOSFET is in saturation region')\n",
+      "\n",
+      "    # Note: The value of R1 is in k ohm but in the book it is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R2  is =95.36 k ohm\n",
+        "The value of R1 is =144.64 k ohm\n",
+        "The value of V_DS is =6.00 V\n",
+        "The value of V_DS(sat) is =1.77 V\n",
+        "The MOSFET is in saturation region\n"
+       ]
+      }
+     ],
+     "prompt_number": 97
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.13 Page No 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_DD = 6.0              #in V\n",
+      "V_D = 3.0             #in V\n",
+      "R_D = 10.0             #in k ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_DQ = (V_DD-V_D)/R_D           #in mA\n",
+      "print('The value of I_DQ is =%.2f mA ' %I_DQ)\n",
+      "V_T = 0.8           #in V\n",
+      "k = 0.12             #in mA/V^2\n",
+      "V_GS = math.sqrt(I_DQ/k) + V_T           #in V\n",
+      "V_S = -V_GS                   #in V\n",
+      "V_DS = V_D-V_S               #in V\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_DS is =%.2f V' %V_DS)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DQ is =0.30 mA \n",
+        "The value of V_DS is =5.38 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 98
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.14 Page No 388"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_D = 0.3         # in mA\n",
+      "k = 0.12            #in mA/V^2\n",
+      "V_T = 1.0          #in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS = V_T + (math.sqrt(I_D/k))          #in V\n",
+      "V_S = -V_GS                             #in V\n",
+      "V_DD = 6.0                  #in V\n",
+      "V_D = 3.0                   # in V\n",
+      "I_DQ = 0.3                   #in mA\n",
+      "R_D = (V_DD-V_D)/I_DQ       #in k ohm\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of R_D  is =%.2f k ohm' %R_D)\n",
+      "V_DS = V_D - V_S                       #in V\n",
+      "print('The value of V_DS  is =%.2f V' %V_DS)\n",
+      "V_DSsat = V_GS - V_T                 #in V\n",
+      "print('The value of V_DS(sat)  is =%.2f V' %V_DSsat)\n",
+      "if V_DS > V_DSsat :\n",
+      "    print('The MOSFET is in saturation region')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R_D  is =10.00 k ohm\n",
+        "The value of V_DS  is =5.58 V\n",
+        "The value of V_DS(sat)  is =1.58 V\n",
+        "The MOSFET is in saturation region\n"
+       ]
+      }
+     ],
+     "prompt_number": 99
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.15 Page No 389"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "k= 0.05             #in mA/V^2\n",
+      "V_T= 1            #in V\n",
+      "V_DD= 6          # in V\n",
+      "R_S= 9.1            #in k\u03a9\n",
+      "I_D= 0.3631765    # in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= R_S**2          # assumed\n",
+      "B=-(2*V_DD*R_S+1/k)         # assumed\n",
+      "C= V_DD**2         #assumed\n",
+      "V_GS= V_DD-I_D*R_S   #in V\n",
+      "V_DS= V_GS         # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_D  is : =%.2f mA' %I_D)\n",
+      "print('The value of V_GS  is : =%.2f volts' %V_GS)\n",
+      "print('The value of V_DS  is : =%.2f volts' %V_DS)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D  is : =0.36 mA\n",
+        "The value of V_GS  is : =2.70 volts\n",
+        "The value of V_DS  is : =2.70 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 100
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.16 Page No 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "k1= 0.01          #in mA/V^2\n",
+      "k2= 0.05      #in mA/V^2\n",
+      "V_DD= 5         # in V\n",
+      "V_T1=1           # in V\n",
+      "V_T2=1         # in V\n",
+      "Vi= 5         #in V\n",
+      "Vo= 0.3485163   # in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A=(k1+k2)\n",
+      "B=-(2*k1*(V_DD-V_T1)+2*k2*(Vi-V_T2))\n",
+      "C=k1*(V_DD-V_T1)**2\n",
+      "V_GS2= Vi             # in V\n",
+      "V_DS2= Vo            # in V\n",
+      "V_GS1= V_DD-Vo         # in V\n",
+      "I_D1= k1*(V_GS1-V_T1)**2            #in mA\n",
+      "I_D2= I_D1                  # in mA\n",
+      "print('Part (i) For Vi = 5 V')\n",
+      "print('The output voltage is : =%.2f volts' %Vo)\n",
+      "print('The value of I_D1  is : =%.2f mA' %I_D1)\n",
+      "print('The value of I_D2  is : =%.2f mA' %I_D2)\n",
+      "# Analysis for Vi= 1.5V\n",
+      "Vi= 1.5         #in V\n",
+      "Vo= V_DD-V_T1-math.sqrt(k2/k1)*(Vi-V_T2)          # in V\n",
+      "I_D2= k2*(Vi-V_T2)**2               #in mA\n",
+      "I_D1= I_D2             #in mA\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Part (ii) For Vi = 1.5 V')\n",
+      "print('The output voltage  is : =%.2f volts' %Vo)\n",
+      "print('The value of I_D1  is : =%.2f mA' %I_D1)\n",
+      "print('The value of I_D2  is : =%.2f mA' %I_D2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (i) For Vi = 5 V\n",
+        "The output voltage is : =0.35 volts\n",
+        "The value of I_D1  is : =0.13 mA\n",
+        "The value of I_D2  is : =0.13 mA\n",
+        "Part (ii) For Vi = 1.5 V\n",
+        "The output voltage  is : =2.88 volts\n",
+        "The value of I_D1  is : =0.01 mA\n",
+        "The value of I_D2  is : =0.01 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 101
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.17 Page No 392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "k = 0.12         #in mA/V^2\n",
+      "V_T = -2.5        #in V\n",
+      "V_GS = 0\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_D = k*((V_GS-V_T)**2)        #in mA\n",
+      "print('The value of I_D is =%.2f mA' %I_D)\n",
+      "V_DD = 6            #in V\n",
+      "R_S = 4.7            #in k ohm \n",
+      "V_DS = V_DD -(I_D*R_S)         #in V\n",
+      "print('The value of V_DS  is =%.2f V' %V_DS)\n",
+      "V_S = 0               # in V \n",
+      "V_DSsat = V_S - V_T            # in V\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_DS(sat) is  =%.2f V' %V_DSsat)\n",
+      "if V_DS < V_DSsat :\n",
+      "    print('The device is in the non saturation region')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D is =0.75 mA\n",
+        "The value of V_DS  is =2.47 V\n",
+        "The value of V_DS(sat) is  =2.50 V\n",
+        "The device is in the non saturation region\n"
+       ]
+      }
+     ],
+     "prompt_number": 102
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.18 Page No 394"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "k4 = 0.125            #in mA/V^2\n",
+      "k3 = k4              #in mA/V^2\n",
+      "k2 = k4               #in mA/V^2\n",
+      "k1 = 0.25             #in mA/V^2\n",
+      "V_T1 = 0.8            # in V\n",
+      "V_T2 = V_T1           # in V\n",
+      "V_T3 = V_T1          #in V\n",
+      "V_T4 = V_T1           #in V\n",
+      "V_SS = -5             #in V\n",
+      "V_DD = 5              #in V\n",
+      "R_D = 10              #in k ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS3 = ((math.sqrt(k4/k3) * (-V_SS - V_T4))+V_T3)/(1+math.sqrt(k4/k3))          #in V\n",
+      "I_Q = k2*((V_GS3-V_T2)**2)          #in mA\n",
+      "I_D1 = I_Q                   #in mA\n",
+      "V_GS1 = V_T1 + (math.sqrt(I_D1/k1))                 #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_GS1 is =%.2f V ' %V_GS1)\n",
+      "V_DS2 = (-V_SS-V_GS1)            #in V\n",
+      "print('The value of V_DS2 is =%.2f V ' %V_DS2)\n",
+      "V_DS1 = V_DD - (I_Q*R_D) - (V_SS + V_DS2)            #in V\n",
+      "print('The value of V_DS1 is =%.2f  V' %V_DS1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_GS1 is =2.00 V \n",
+        "The value of V_DS2 is =3.00 V \n",
+        "The value of V_DS1 is =3.39  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 103
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.19 Page No 395"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "R2 = 20.0               # in  k ohm\n",
+      "R1 = 30.0               #in  k ohm\n",
+      "R_D = 20               # in  k ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "R_D=R_D*10**3         #in ohm\n",
+      "V_DD = 5              #in V\n",
+      "V_G = (R2/(R1+R2))*V_DD             #in V\n",
+      "V_S = 0                     #in V\n",
+      "V_GS = V_G                   #in V\n",
+      "k = 100*10**-6             #in A/V^2\n",
+      "V_T = 1                    #in V\n",
+      "I_DQ = k*((V_GS-V_T)**2)       #in A\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "V_DSQ = V_DD - (I_DQ*R_D)             #in V \n",
+      "print('The value of V_DSQ  is =%.f V ' %V_DSQ)\n",
+      "I_DQ=I_DQ*10**6\n",
+      "print('The value of I_DQ  is =%.f  \u00b5A' %I_DQ)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_DSQ  is =3 V \n",
+        "The value of I_DQ  is =100  \u00b5A\n"
+       ]
+      }
+     ],
+     "prompt_number": 104
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.20 Page No 395"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_P= -8            # in V\n",
+      "R_S= 2.4           #in k\u03a9\n",
+      "I_DSS= 8           #in mA\n",
+      "V_DD= 20            #in V\n",
+      "R_D= 6.2           #in k\u03a9\n",
+      "I_D= 1.7668305   #in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= R_S**2\n",
+      "B=(2*R_S*(V_P)-V_P**2/I_DSS)\n",
+      "C=(V_P)**2\n",
+      "I_DQ= I_D             #in mA\n",
+      "V_GSQ= -I_D*R_S;\n",
+      "V_D= V_DD-I_D*R_D           #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_DQ is : =%.2f  mA' %I_DQ)\n",
+      "print('The value of V_GSQ  : =%.2f  volts' %V_GSQ)\n",
+      "print('The value of V_D  is : =%.2f  volts' %V_D)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DQ is : =1.77  mA\n",
+        "The value of V_GSQ  : =-4.24  volts\n",
+        "The value of V_D  is : =9.05  volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 105
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.21 Page No 396"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "k= 75*10**-3              #in mA/V^2\n",
+      "Vth= -0.8                  #in V\n",
+      "R2 = 100                   #in k ohm\n",
+      "R1 = 100                   #in  k ohm\n",
+      "R_S= 6                     # in k\u03a9\n",
+      "R_D= 3                   # in k\u03a9\n",
+      "V_SS = 10                 # in V\n",
+      "V_S = 2.0602803\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_G = (R2/(R1+R2))*V_SS                  #in V\n",
+      "V_GS= V_G-V_S                          #in V\n",
+      "I_D= k*(V_GS-Vth)**2                           # in V\n",
+      "V_DS= -V_SS+I_D*(R_D+R_S)                # in V\n",
+      "V_D= I_D*R_D                            #in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_D  is : =%.2f mA ' %I_D)\n",
+      "print('The value of V_DS  is =%.2f volts ' %V_DS)\n",
+      "print('The value of V_D  is : =%.2f  volts' %V_D)\n",
+      "print('The value of V_S  is : =%.2f  volts' %V_S)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D  is : =0.12 mA \n",
+        "The value of V_DS  is =-8.93 volts \n",
+        "The value of V_D  is : =0.36  volts\n",
+        "The value of V_S  is : =2.06  volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 106
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.22 Page No 396"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_T = 1.0                      # in V\n",
+      "k = 160*10**-6               # in A/V^2\n",
+      "I_DQ = 160*10**-6            # in A\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS = V_T + math.sqrt(I_DQ/k)         #in V\n",
+      "V_DD = 5              # in V\n",
+      "V_DSQ = 3             # in V\n",
+      "R_D = (V_DD - V_DSQ)/(I_DQ)            # in ohm\n",
+      "R_D = R_D * 10**-3                 # in k ohm\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of R_D  is =%.2f k ohm' %R_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R_D  is =12.50 k ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 107
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.23 Page No 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_DD= 12          # in V\n",
+      "V_T= 2            # in V\n",
+      "kn= 0.5           # in mA/V^2\n",
+      "R1 = 2.2         # in M ohm\n",
+      "R2 = 1.8         # in M ohm\n",
+      "R_S= 1.5         # in k\u03a9\n",
+      "R_D= 3.9         # in k\u03a9\n",
+      "I_D= 1.2237155                   # in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_G = (R2/(R1+R2))*V_DD         # in V\n",
+      "V_GS= V_G-I_D*R_S              # V\n",
+      "I_D= I_D-kn*(V_GS-V_T)**2      # in mA\n",
+      "I_DQ= I_D                     #in mA\n",
+      "V_DSQ= V_DD-I_D*(R_D+R_S)      # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_DQ  is : =%.2f mA ' %I_DQ)\n",
+      "print('The value of V_DSQ  is : =%.2f  volts' %V_DSQ)\n",
+      "V_GS= V_G-I_D*R_S            # V\n",
+      "V_DSsat= V_GS-V_T          # in V\n",
+      "print('The value of  V_DS  %.2f  is greater than the value of ' %V_DSQ)\n",
+      "print('V_DSsat %.2f So the MOSFET is in saturation region ' %V_DSsat)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DQ  is : =-0.00 mA \n",
+        "The value of V_DSQ  is : =12.00  volts\n",
+        "The value of  V_DS  12.00  is greater than the value of \n",
+        "V_DSsat 3.40 So the MOSFET is in saturation region \n"
+       ]
+      }
+     ],
+     "prompt_number": 108
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.24 Page No 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "kn= 0.5           # in mA/V^2\n",
+      "V_T= 1          # in V\n",
+      "R2 = 40         # in k ohm\n",
+      "R1 = 60         # in k ohm\n",
+      "R_S= 1          # in k ohm\n",
+      "R_D= 2          # in k ohm\n",
+      "V_DD = 5        # in V\n",
+      "V_SS = -5       # in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_R2 = (R2/(R2+R1))*(V_DD-V_SS)            # in V\n",
+      "V_G = V_R2 - V_DD               # in V\n",
+      "V_S= I_D*R_S+V_SS              # in V\n",
+      "V_GS= V_G-V_S                  # in V\n",
+      "I_D= 1.3542487                  # in mA\n",
+      "I_DQ= I_D                      #in mA\n",
+      "V_S= I_D*R_S+V_SS            # in V\n",
+      "V_GS= V_G-V_S                # in V\n",
+      "V_DSQ= V_DD-V_SS-I_D*(R_D+R_S)         # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_DQ  is :=%.2f mA ' %I_DQ)\n",
+      "print('The value of V_GS  is : =%.2f volts ' %V_GS)\n",
+      "print('The value of V_DSQ  is : =%.2f  volts' %V_DSQ)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_DQ  is :=1.35 mA \n",
+        "The value of V_GS  is : =-1.35 volts \n",
+        "The value of V_DSQ  is : =5.94  volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 109
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.25 Page No 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "R_S1 = 100*10**-3      # in k ohm\n",
+      "R_S2 = 100*10**-3      # k ohm\n",
+      "R_S = R_S1+R_S2        #in k ohm\n",
+      "R_D= 1.8               # in k ohm\n",
+      "I_DSS= 12             # in mA\n",
+      "Vp= -3.5              # in V\n",
+      "V_DD= 22              # in V\n",
+      "rd= 25                # in k ohm\n",
+      "R_L= 47               # in k ohm\n",
+      "I_D= 5.573521          # in mA\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_GS= -I_D*R_S     # in V\n",
+      "I_D= I_D-I_DSS*(1-V_GS/Vp)**2   # in mA\n",
+      "print('The value of I_D  is : =%.2f  mA' %I_D)\n",
+      "V_GS= -I_D*R_S           # in V\n",
+      "print('The value of V_GS  is : =%.2f volts ' %V_GS)\n",
+      "V_DS= V_DD-I_D*(R_D+R_S)          # in V\n",
+      "print('The value of V_DS  is :  =%.2f volts ' %V_DS)\n",
+      "gmo= -2*I_DSS/Vp              # in mS\n",
+      "gm= gmo*(1-V_GS/Vp)           # in mS\n",
+      "miu= gm*rd\n",
+      "Av= -miu*R_D*R_L/(R_D+R_L)/(rd+R_D*R_L/(R_D+R_L)+(1+miu)*R_S1)\n",
+      "print('The value of Av is :  =%.2f  ' %Av)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D  is : =-0.000000  mA\n",
+        "The value of V_GS  is : =0.00 volts \n",
+        "The value of V_DS  is :  =22.00 volts \n",
+        "The value of Av is :  =-6.76  \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.26 Page No 399"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_T = 1                # in V\n",
+      "k = 0.5               # in mA/V^2\n",
+      "R2 = 40              # in k ohm\n",
+      "R1 = 60             # in k ohm\n",
+      "R_S= 1               # in k ohm\n",
+      "R_D= 2              # in k ohm\n",
+      "V_DD = 5            # in V\n",
+      "V_G = (R2/(R2+R1))*V_DD         # in V\n",
+      "V_GS= V_G-I_D*R_S                # in V\n",
+      "I_D= 0.2679492                     # in mA\n",
+      "V_GS= V_G-I_D*R_S                 # in V\n",
+      "V_DS= V_DD-I_D*(R_D+R_S)          # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_D in mA is : =%.2f  ' %I_D)\n",
+      "print('The value of V_GS in volts is :=%.2f  ' %V_GS)\n",
+      "print('The value of V_DS in volts is : =%.2f  ' %V_DS)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D in mA is : =0.27  \n",
+        "The value of V_GS in volts is :=-0.27  \n",
+        "The value of V_DS in volts is : =4.20  \n"
+       ]
+      }
+     ],
+     "prompt_number": 111
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.27 Page No 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "R_D = 7.5             # in k ohm\n",
+      "V_T = -0.8             # in V\n",
+      "k = 0.2              # in mA/V^2\n",
+      "R2 = 50                # in ohm\n",
+      "R1 = 50               # in  ohm\n",
+      "V_DD = 5             # in V\n",
+      "V_S = 5            # in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_G = (R2/(R2+R1))*V_DD          # in V\n",
+      "V_GS = V_G - V_S            # in V\n",
+      "I_D = k*((V_GS-V_T)**2)        # in mA\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Drain current in is =%.2f mA' %I_D)\n",
+      "V_SD = V_DD - (I_D*R_D)    # in V\n",
+      "print('Source to drain voltage  is =%.2f V' %V_SD)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Drain current in is =3.53 mA\n",
+        "Source to drain voltage  is =-21.46 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 112
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.28 Page No 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "I_Don = 5*10**-3            # in A\n",
+      "V_GSon = 6                # in V\n",
+      "V_GSth = 3                # in V\n",
+      "k = I_Don/(V_GSon-V_GSth)**2      # in A/V^2 \n",
+      "R2 = 6.8                  #in M ohm\n",
+      "R1 = 10                  #in M ohm\n",
+      "R_S= 750                # in ohm\n",
+      "R_D= 2.2*10**3        # in ohm\n",
+      "V_DD = 24             # in V\n",
+      "R_S = 750           # in  ohm\n",
+      "V_G= R2*V_DD/(R1+R2)         # in V\n",
+      "V_GS= V_G-I_D*R_S           # in V\n",
+      "I_D= I_D-k*(V_GS-V_GSth)**2\n",
+      "I_D= 0.0049660              # in A\n",
+      "I_DQ= I_D                    # in A\n",
+      "V_GS= V_G-I_D*R_S           # in V\n",
+      "V_GSQ= V_GS                 # in V\n",
+      "V_DSQ= V_DD-I_DQ*(R_D+R_S)       # in V\n",
+      "I_D=I_D*10**3\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of I_D  is : =%.2f mA ' %I_D)\n",
+      "print('The value of V_GSQ  is : =%.2f volts ' %V_GSQ)\n",
+      "print('The value of V_DSQ  is : =%.2f volts ' %V_DSQ)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_D  is : =4.97 mA \n",
+        "The value of V_GSQ  is : =5.99 volts \n",
+        "The value of V_DSQ  is : =9.35 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 113
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.29 Page No 401"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "I_Don = 4*10**-3              # in A\n",
+      "V_GSon = 6                 # in V\n",
+      "V_GSth = 3                 # in V\n",
+      "V_DS= 6                    # in V\n",
+      "I_D= I_Don                 # in A\n",
+      "k = I_Don/((V_GSon-V_GSth)**2)        # in A/V^2\n",
+      "V_GS= I_D-k*(V_GS-V_GSth)**2\n",
+      "V_GS= 6           # in V\n",
+      "V_DD= 2*V_DS               # in V\n",
+      "R_D= (V_DD-V_GS)/I_D            # in ohm\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_GS  is : =%.f volts ' %V_GS)\n",
+      "print('The value of V_DD  is : =%.f volts ' %V_DD)\n",
+      "print('The value of R_D  is : =%.f  k\u03a9' %R_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_GS  is : =6 volts \n",
+        "The value of V_DD  is : =12 volts \n",
+        "The value of R_D  is : =1500  k\u03a9\n"
+       ]
+      }
+     ],
+     "prompt_number": 114
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.31 Page No 402"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_DD = 5               # in V\n",
+      "V_T= 1                  # in V\n",
+      "k= 1             # in mA/V^2\n",
+      "R1 = 1           # in M ohm\n",
+      "R2 = 1           # in M ohm\n",
+      "R_S= 2            # in k ohm\n",
+      "R_D= 2           # in k ohm\n",
+      "I1 = V_DD/(R1+R2)         # in A\n",
+      "print('The value of I1  is : =%.2f  \u00b5A' %I1)\n",
+      "V_A = (R2/(R2+R1))*V_DD              # in V\n",
+      "print('The value of V_A and V_G  is :  =%.2f volts ' %V_A)\n",
+      "I_D= 0.4243061\n",
+      "V_C= I_D*R_S              # in V\n",
+      "V_GS= V_A-V_C            # in V\n",
+      "print('The value of I_D  is : =%.2f  mA' %I_D)\n",
+      "V_B= V_DD-I_D*R_D       # in V\n",
+      "V_C= I_D*R_S         # in V\n",
+      "V_DS= V_B-V_C       # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_B  is : =%.2f volts ' %V_B)\n",
+      "print('The value of V_C  is : =%.2f volts ' %V_C)\n",
+      "print('The value of V_DS  is : =%.2f  volts' %V_DS)\n",
+      "# Note: In the book, the calculated values are wrong, this is why the answer in the book is wrong.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I1  is : =2.00  \u00b5A\n",
+        "The value of V_A and V_G  is :  =0.00 volts \n",
+        "The value of I_D  is : =0.42  mA\n",
+        "The value of V_B  is : =4.15 volts \n",
+        "The value of V_C  is : =0.85 volts \n",
+        "The value of V_DS  is : =3.30  volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 115
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.32 Page No 403"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_DSS = 12            # in mA\n",
+      "I_DSS= I_DSS*10**-3      # in A\n",
+      "V_P = -3              # in V\n",
+      "r_d = 45             # in k ohm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r_d= r_d*10**3           # in ohm\n",
+      "g_m = I_DSS/abs(V_P) # in S\n",
+      "#Part (i)\n",
+      "R1 = 91           # in M ohm\n",
+      "R1=R1*10**6        #in ohm\n",
+      "R2 = 10           # in M ohm\n",
+      "R2= R2*10**6      # in ohm\n",
+      "Ri= R1*R2/(R1+R2)      # in ohm\n",
+      "Ri=Ri*10**6\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of Ri  is :  =%.f Mohm' %(Ri/(10**12)))\n",
+      "#Part (ii)\n",
+      "R_S = 1.1      #in k ohm\n",
+      "R_S = R_S * 10**3       # in ohm\n",
+      "R_o= (R_S*1/g_m)/(R_S+1/g_m)      # in ohm\n",
+      "print('The value of R_C  is :  =%.2f  ohm' %R_o)\n",
+      "\n",
+      "# Part (iii)\n",
+      "R_desh_o= R_o*r_d/(R_o+r_d)   # in ohm\n",
+      "print('The value of R_desh_o  is :  =%.2f ohm ' %R_desh_o)\n",
+      "# Part (iv)\n",
+      "Av= g_m*(R_S*r_d/(R_S+r_d))/(1+g_m*(R_S*r_d/(R_S+r_d)))\n",
+      "print('The value of Av is :  =%.2f ' %Av)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Ri  is :  =9 Mohm\n",
+        "The value of R_C  is :  =203.70  ohm\n",
+        "The value of R_desh_o  is :  =202.79 ohm \n",
+        "The value of Av is :  =0.81 \n"
+       ]
+      }
+     ],
+     "prompt_number": 116
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.34 Page No 404"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_S2 = -2   # in V\n",
+      "V_GS2 = -V_S2    # in V\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I_DS2 = (V_GS2-1)**2        # in mA\n",
+      "I = 2           # in mA\n",
+      "I_DS1 = I-I_DS2      # in mA\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The current flow through M1 MOSFET  is =%.f mA' %I_DS1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current flow through M1 MOSFET  is =1 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 117
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.35 Page No 404"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_DD= 10         # in V\n",
+      "I_D= 0.4*10**3       # in A\n",
+      "W= 100           # in \u00b5m\n",
+      "L= 10           # in \u00b5m\n",
+      "uACox= 20         # in A/V**2\n",
+      "Vt= 2            # in V\n",
+      "R= 0.015\n",
+      "V_GS= V_DD-I_D*R      # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of R  is : =%.f k\u03a9 ' %(R*10**3))\n",
+      "V_D= V_DD-I_D*R           # in V\n",
+      "print('The value of V_D  is : =%.f volts  ' %V_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R  is : =15 k\u03a9 \n",
+        "The value of V_D  is : =4 volts  \n"
+       ]
+      }
+     ],
+     "prompt_number": 118
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No 9.36 Page No 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "V_GSth= 2         # in V\n",
+      "k= 2*10**-4       # in A/V^2\n",
+      "V_DD= 12          # in V\n",
+      "R_D= 5*10**3      # in ohm\n",
+      "I_D= 0.0014597\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V_DS= V_DD-I_D*R_D         # in V\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The value of V_DS  is : =%.1f volts ' %V_DS)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_DS  is : =4.7 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 119
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/README.txt b/Advance_Semiconductor_Devices_by_S._Sharma/README.txt
new file mode 100755
index 00000000..eec99ae9
--- /dev/null
+++ b/Advance_Semiconductor_Devices_by_S._Sharma/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Santosh Pawar
+Course: mca
+College/Institute/Organization: ASM institute of computer studies
+Department/Designation: MCA
+Book Title: Advance Semiconductor Devices
+Author:  S. Sharma
+Publisher: S. K. Kataria & Sons, New Delhi
+Year of publication: 2012
+Isbn: 9789350142158
+Edition: 1
\ No newline at end of file
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diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter01.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter01.ipynb
deleted file mode 100755
index 7760d8e3..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter01.ipynb
+++ /dev/null
@@ -1,499 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0f8515032ad257111ffef42eb71b2510ec7d92768034e754ca14b473e010b968"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1. Basic Elasticity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1, Pg. No.13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# variable declaration\n",
-      "p = 1.5;         #pressure inside vessel (N/mm^2)\n",
-      "d = 2*10**3;     #diameter of vessel (mm)\n",
-      "t = 20;          #thickness of plate (mm)\n",
-      "theta = 60;        #plane's inclination to axis of vessel (degree)\n",
-      "load = 2500*10**3;       #axial tensile load (N)\n",
-      "\n",
-      "# Longitudinal stress\n",
-      "sigma_x=p*d/4/t;\n",
-      "print \"\\nLongitudinal stress due to internal pressure = %5.2f N/mm^2\" %(sigma_x)\n",
-      "\n",
-      "#circumferential stress\n",
-      "sigma_y=p*d/2/t;\n",
-      "print \"\\nCircumferential stress due to internal pressure =  %5.2f N/mm^2\" %(sigma_y)\n",
-      "\n",
-      "#axial load\n",
-      "sigma_x_axial=load/(math.pi*d*t);\n",
-      "print \"\\ndirect stress due to axial load = %5.2f N/mm^2\"%(sigma_x_axial)\n",
-      "\n",
-      "sigma_x=sigma_x+sigma_x_axial;                #total longitudinal stress\n",
-      "\n",
-      "#direct stress and shear stress on inclined plane AB\n",
-      "#reference Fig 1.9 pg no14, equation 1.8,1.9 pg no 13\n",
-      "\n",
-      "sigma_n=sigma_x*math.pow(math.cos(math.radians(90-theta)),2)+sigma_y*math.pow(math.sin(math.radians(90-theta)),2)\n",
-      "print \"\\ndirect stress on inclined plane AB = %5.2f N/mm^2\"%(sigma_n)\n",
-      "\n",
-      "tau=(sigma_x-sigma_y)/2*math.sin(math.radians(2*(90-theta)))\n",
-      "print \"\\nshear stress on plane AB = %5.2f N/mm^2\"%(tau)\n",
-      "\n",
-      "#maximumm shear stress (theta=45 degree)\n",
-      "theta=45\n",
-      "tau=(sigma_x-sigma_y)/2*math.sin(math.radians(2*(90-theta)))\n",
-      "print \"\\nmaximum shear stress on plane AB = %5.2f N/mm^2\"%(tau)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Longitudinal stress due to internal pressure = 37.50 N/mm^2\n",
-        "\n",
-        "Circumferential stress due to internal pressure =  75.00 N/mm^2\n",
-        "\n",
-        "direct stress due to axial load = 19.89 N/mm^2\n",
-        "\n",
-        "direct stress on inclined plane AB = 61.80 N/mm^2\n",
-        "\n",
-        "shear stress on plane AB = -7.62 N/mm^2\n",
-        "\n",
-        "maximum shear stress on plane AB = -8.80 N/mm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2, Pg. No.14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "load=50*10**3         #axial load (N)\n",
-      "d=60                   #diameter (mm)\n",
-      "t=1.5                  #offset distance from center (mm)\n",
-      "T=1200*10**3                 #torque applied on a point (N.mm)\n",
-      "theta=60               #angle made by plane wrt axis of cylinder(degree)\n",
-      "\n",
-      "#compressive stress due to axial load\n",
-      "area=math.pi*math.pow(d/2,2)        #cross section area\n",
-      "sigma_x_a=load/area\n",
-      "print \"\\ncompressive stress due to axial load = %5.1f N/mm^2\"%sigma_x_a\n",
-      "\n",
-      "#compressive stress due to bending moment\n",
-      "#area moment of inertia\n",
-      "M=load*t\n",
-      "I=math.pi*d**4/64\n",
-      "\n",
-      "sigma_x_b=M*d/2*(1/I)\n",
-      "print \"\\ncompressive stress due to bending moment = %5.1f N/mm^2\" %sigma_x_b\n",
-      "\n",
-      "#total compressive stress\n",
-      "sigma_x=sigma_x_a+sigma_x_b\n",
-      "\n",
-      "#shear stress due to torque Ref example 3.1 equation (iv) pg no 73\n",
-      "# tau=Tr/J\n",
-      "J=math.pi*d**4/32    #torsion constant\n",
-      "tau_xy=T*d/2/J\n",
-      "print \"\\nshear stress due to torque = %5.1f N/mm^2\" %tau_xy\n",
-      "\n",
-      "#direct and shear on inclined plane, ref eq 1.8,1.9\n",
-      "sigma_y=0\n",
-      "sigma_n=-sigma_x*math.pow(math.cos(math.radians(90-theta)),2)-tau_xy*math.sin(math.radians(2*(90-theta)))\n",
-      "print \"\\ndirect stress on inclined plane = %3.1f N/mm^2\"%sigma_n\n",
-      "\n",
-      "tau=-sigma_x/2*math.sin(math.radians(2*(90-theta)))+tau_xy*math.cos(math.radians(2*(90-theta)))\n",
-      "print \"\\nshear stress on inclined plane = %3.1f N/mm^2\"%tau\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "compressive stress due to axial load =  17.7 N/mm^2\n",
-        "\n",
-        "compressive stress due to bending moment =   3.5 N/mm^2\n",
-        "\n",
-        "shear stress due to torque =  28.3 N/mm^2\n",
-        "\n",
-        "direct stress on inclined plane = -40.4 N/mm^2\n",
-        "\n",
-        "shear stress on inclined plane = 5.0 N/mm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 95
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3, Pg. No.20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#import numpy\n",
-      "import math\n",
-      "#import matplotlib\n",
-      "%pylab inline\n",
-      "#from pylab import *          #change comment to above line to see undocked graph, all values will be visible as mouse-pointer moves\n",
-      "#variable declaration\n",
-      "sigma_x=160          #stress in x direction (N/mm^2)\n",
-      "sigma_y=-120          #stress in y direction (N/mm^2)\n",
-      "sigma=200             #stress on inclined plane (N/mm^2)\n",
-      "\n",
-      "tau_xy=((sigma-sigma_x)*(sigma-sigma_y))**0.5\n",
-      "print \"\\nallowable shear stress, tau_xy = %4.1f N/mm^2 \"%tau_xy\n",
-      "\n",
-      "coeff=[1,-(sigma_x+sigma_y),sigma_x*sigma_y-tau_xy**2]\n",
-      "sigma=numpy.roots(coeff)\n",
-      "print \"\\nprincipal stresses, sigma_I = %3.0f N/mm^2  sigma_II = %3.0f N/mm^2\"%(sigma[0],sigma[1])\n",
-      "\n",
-      "tau_max=(abs(sigma[0])+abs(sigma[1]))/2\n",
-      "print \"\\nmaximum shear stress, tau_max = %3.0f N/mm^2\"%tau_max\n",
-      "\n",
-      "#plotting Mohr circle\n",
-      "x_cent=(sigma_x+sigma_y)/2\n",
-      "y_cent=0\n",
-      "\n",
-      "X1=(sigma_x,tau_xy)\n",
-      "X2=(sigma_y,-tau_xy)\n",
-      "\n",
-      "radius=(math.hypot(X2[0]-X1[0], X2[1] - X1[1]))/2\n",
-      "\n",
-      "cir=linspace(0,2*pi,100)\n",
-      "plot(radius*cos(cir)+x_cent,radius*sin(cir)+y_cent,'r')\n",
-      "plot(sigma_x,tau_xy,'ro',sigma_y,-tau_xy,'ro',x_cent,y_cent,'b+',sigma[0],0,'bo',sigma[1],0,'bo',20,tau_max,'go')\n",
-      "text(sigma_x+10,tau_xy,'Q1')\n",
-      "text(sigma_y+10,-tau_xy,'Q2')\n",
-      "text(20,tau_max+10,r'$\\tau_{max}$')\n",
-      "text(-150,0,r'$\\sigma_2$')\n",
-      "text(200,10,r'$\\sigma_1$')\n",
-      "text(-20,+10,'O')\n",
-      "text(30,-20,'C')\n",
-      "#plot([sigma_y,tau_xy],[sigma_x,-tau_xy],'r-')\n",
-      "xlabel(r'$\\sigma$')\n",
-      "ylabel(r'$\\tau$')\n",
-      "title('Mohr Cirle')\n",
-      "axis('equal')\n",
-      "grid(True)\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "\n",
-        "allowable shear stress, tau_xy = 113.1 N/mm^2 \n",
-        "\n",
-        "principal stresses, sigma_I = 200 N/mm^2  sigma_II = -160 N/mm^2\n",
-        "\n",
-        "maximum shear stress, tau_max = 180 N/mm^2\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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UlbCoFi3gxAnT5rfffkufPn1wc3PjwQcf5OOPP9YxONsghcTOWKWd9swZaNbM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\nxZVZr61+AAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0xa137240>"
-       ]
-      }
-     ],
-     "prompt_number": 141
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4, Pg. No.33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#variable declaration\n",
-      "sigma_x=83       # stress in x direction (N/mm^2)\n",
-      "sigma_y=65       # stress in y direction (N/mm^2)\n",
-      "E=200000         # Young's Modulus (N/mm^2)\n",
-      "v=0.3           # poisson's ratio\n",
-      "\n",
-      "# strains in x,y and z directions\n",
-      "epsln_x=1.0/E*(sigma_x-v*sigma_y)\n",
-      "epsln_y=1.0/E*(sigma_y-v*sigma_x)\n",
-      "epsln_z=-v/E*(sigma_x+sigma_y)\n",
-      "print \"\\nstrain in x direction, epsln_x = %5.3e\"%epsln_x\n",
-      "print \"\\nstrain in y direction, epsln_y = %5.3e\"%epsln_y\n",
-      "print \"\\nstrain in z direction, epsln_z = %5.3e\"%epsln_z\n",
-      "\n",
-      "tau_max=(sigma_x-sigma_y)/2\n",
-      "print \"\\nmaximum shear stress, tau_max = %2.2f N/mm^2\"%tau_max\n",
-      "\n",
-      "gama_max=2*(1+v)*tau_max/E\n",
-      "print \"\\nmaximum shear strain, y_max = %2.2e \" %gama_max"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "strain in x direction, epsln_x = 3.175e-04\n",
-        "\n",
-        "strain in y direction, epsln_y = 2.005e-04\n",
-        "\n",
-        "strain in z direction, epsln_z = -2.220e-04\n",
-        "\n",
-        "maximum shear stress, tau_max = 9.00 N/mm^2\n",
-        "\n",
-        "maximum shear strain, y_max = 1.17e-04 \n"
-       ]
-      }
-     ],
-     "prompt_number": 111
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5, Pg. No.33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#import matplotlib.pyplot as plt\n",
-      "%pylab inline\n",
-      "#variable declaration\n",
-      "sigma_x=60            #stress in x direction (N/mm^2)\n",
-      "sigma_y=-40           #stress in y direction (N/mm^2)\n",
-      "tau_xy=50             #shear stress (N/mm^2)\n",
-      "E=2*10**5             #Young's Modulus (N/mm^2)\n",
-      "v=0.3                #poisson's ratio\n",
-      "\n",
-      "#strains in x and y direction\n",
-      "epsln_x=1.0/E*(sigma_x-v*sigma_y)\n",
-      "epsln_y=1.0/E*(sigma_y-v*sigma_x)\n",
-      "print \"\\nstrains in x and y directions, epsi_x =%4.1e  epsi_y =%4.1e \"%(epsln_x,epsln_y)     \n",
-      "\n",
-      "G=E/2/(1+v)\n",
-      "print \"\\nshear modulus, G = %6.1f\"%G\n",
-      "\n",
-      "gama_xy=tau_xy/G\n",
-      "print \"\\nshear strain, gama_xy = %4.1e\"%gama_xy\n",
-      "\n",
-      "#principal strains\n",
-      "epsln_I=(epsln_x+epsln_y)/2+1.0/2*((epsln_x-epsln_y)**2+gama_xy**2)**0.5\n",
-      "epsln_II=(epsln_x+epsln_y)/2-1.0/2*((epsln_x-epsln_y)**2+gama_xy**2)**0.5\n",
-      "print \"\\nprincipal strains, epsln_I = %4.2e  epsln_II = %4.2e\"%(epsln_I,epsln_II)\n",
-      "\n",
-      "#inclination\n",
-      "theta=1.0/2*math.atan(gama_xy/(epsln_x-epsln_y))\n",
-      "print \"\\ninclination to the plane on which sigma_x acts, theta = %4.1f or %4.1f\"%(theta*180/math.pi,theta*180/math.pi+90)\n",
-      "\n",
-      "#plotting Mohr circle\n",
-      "x_cent=(epsln_x+epsln_y)/2\n",
-      "y_cent=0\n",
-      "\n",
-      "X1=(epsln_x,gama_xy/2)\n",
-      "X2=(epsln_y,-gama_xy/2)\n",
-      "\n",
-      "radius=(math.hypot(X2[0]-X1[0], X2[1] - X1[1]))/2\n",
-      "print radius\n",
-      "cir=linspace(0,2*pi,100)\n",
-      "plot(radius*cos(cir)+x_cent,radius*sin(cir),'r')\n",
-      "plot(epsln_x,gama_xy/2,'ro',epsln_y,-gama_xy/2,'ro',x_cent,y_cent,'b+',epsln_II,0,'go',epsln_I,0,'go')\n",
-      "text(epsln_x,gama_xy/2,'Q1')\n",
-      "text(epsln_y,-gama_xy/2,'Q2')\n",
-      "text(-0.00050,0,r'$\\epsilon_2$')\n",
-      "text(.00050,0,r'$\\epsilon_1$')\n",
-      "text(-.000060,0.00002,'O')\n",
-      "text(0,0,'C')\n",
-      "xlabel(r'$\\epsilon$')\n",
-      "ylabel(r'$\\gamma$')\n",
-      "title('Mohr Cirle')\n",
-      "axis('equal')\n",
-      "grid(True)\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "\n",
-        "strains in x and y directions, epsi_x =3.6e-04  epsi_y =-2.9e-04 \n",
-        "\n",
-        "shear modulus, G = 76923.1\n",
-        "\n",
-        "shear strain, gama_xy = 6.5e-04\n",
-        "\n",
-        "principal strains, epsln_I = 4.95e-04  epsln_II = -4.25e-04\n",
-        "\n",
-        "inclination to the plane on which sigma_x acts, theta = 22.5 or 112.5\n",
-        "0.000459619407771\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Ti4t2v82PP2pDw0NCtPtxpDnbYcnMACbIzABO5ocftJFk/fvD8uXadPrCer79FmJiIDAQ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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0xcc10a20>"
-       ]
-      }
-     ],
-     "prompt_number": 185
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Pg. No.40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#variable declaration\n",
-      "epsln_a=1000*10**(-6)     #readings of strain gauges \n",
-      "epsln_b=-200*10**(-6)     #straing gauges 'a'  and 'c' are\n",
-      "epsln_c=-300*10**(-6)     #in line and perpendicular to axis of bar\n",
-      "d=50                      #diameter of bar (mm)\n",
-      "E=70000                   #Young's modulus (N/mm^2)\n",
-      "v=0.3                     #poisson's ratio\n",
-      "\n",
-      "#principal strains\n",
-      "epsln_I=1.0/2*(epsln_a+epsln_c)+(1.0/2**0.5)*((epsln_a-epsln_b)**2+(epsln_c-epsln_b)**2)**0.5\n",
-      "epsln_II=1.0/2*(epsln_a+epsln_c)-(1.0/2**0.5)*((epsln_a-epsln_b)**2+(epsln_c-epsln_b)**2)**0.5\n",
-      "print \"\\nfirst principal strain, epsln_I = %4.3e\"%epsln_I\n",
-      "print \"\\nsecond principal strain, epsln_II = %4.3e\"%epsln_II\n",
-      "\n",
-      "#principal stresses\n",
-      "sigma_I=E/(1-v**2)*(epsln_I+v*epsln_II)\n",
-      "sigma_II=E/(1-v**2)*(v*epsln_I+epsln_II)\n",
-      "print \"\\nfirst principal stress, sigma_I= %4.1f N/mm^2\"%sigma_I #mistake in book\n",
-      "print \"\\nsecond principal stress, sigma_II= %4.1f N/mm^2\"%sigma_II\n",
-      "\n",
-      "sigma_x=sigma_I+sigma_II\n",
-      "print \"\\nstress in x direction,sigma_x = %4.1f N/mm^2\"%sigma_x\n",
-      "\n",
-      "#axial tensile load calculation\n",
-      "A=math.pi*d**2/4\n",
-      "P=sigma_x*A\n",
-      "print \"\\naxial load, P = %4.1f kN\"%(P/1000)\n",
-      "\n",
-      "tau_xy=1.0/2*((sigma_x/2-sigma_II)**2*4-sigma_x**2)**0.5\n",
-      "print \"\\nshear stress, tau_xy = %4.1f N/mm^2\"%tau_xy\n",
-      "\n",
-      "#torque calculation\n",
-      "J=math.pi*d**4/32\n",
-      "T=tau_xy*J/d*2\n",
-      "print \"\\ntorque applied on circular bar, T = %4.1f kNm\"%(T/10**6)\n",
-      "\n",
-      "sigma_x=E*epsln_a\n",
-      "print \"\\naxial stress from classical stress strain relationship, sigma_x =%4.0f N/mm^2\"%sigma_x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "first principal strain, epsln_I = 1.201e-03\n",
-        "\n",
-        "second principal strain, epsln_II = -5.015e-04\n",
-        "\n",
-        "first principal stress, sigma_I= 80.8 N/mm^2\n",
-        "\n",
-        "second principal stress, sigma_II= -10.8 N/mm^2\n",
-        "\n",
-        "stress in x direction,sigma_x = 70.0 N/mm^2\n",
-        "\n",
-        "axial load, P = 137.4 kN\n",
-        "\n",
-        "shear stress, tau_xy = 29.6 N/mm^2\n",
-        "\n",
-        "torque applied on circular bar, T =  0.7 kNm\n",
-        "\n",
-        "axial stress from classical stress strain relationship, sigma_x =  70 N/mm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter04.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter04.ipynb
deleted file mode 100755
index 365fa5fb..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter04.ipynb
+++ /dev/null
@@ -1,95 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:6a67de7af4fe617483b9aefeefdb01f122e2016e5ab212db494104a83b9deeda"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6 Pg.No.106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "sum_FAFlBL=1263.6*10**6    # (N.mm)\n",
-      "sum_FAFlDL=880*10**6       # (N.mm)\n",
-      "E=200000                  # young's modulus (N/mm^2)\n",
-      "A=1800                    # area of each member (mm^2)\n",
-      "\n",
-      "#eqn 4.23 We=Wi\n",
-      "del_B=sum_FAFlBL/A/E\n",
-      "print \"\\ndeflection at point B =%2.2f mm\\n\"%(del_B)\n",
-      "\n",
-      "del_D=sum_FAFlDL/A/E\n",
-      "print \"deflection at point D =%2.2f mm\\n\"%(del_D)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "deflection at point B =3.51 mm\n",
-        "\n",
-        "deflection at point D =2.44 mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "ext_load=30    #external load(kN)\n",
-      "\n",
-      "#eqn 30*4*del_vB/3=F_BA*del_vB\n",
-      "F_BA=ext_load*4/3\n",
-      "print \"\\nforce in the struss AB is = %3.2f kN\\n\"%(F_BA)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "force in the struss AB is = 40.00 kN\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter05.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter05.ipynb
deleted file mode 100755
index 80ef0fbb..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter05.ipynb
+++ /dev/null
@@ -1,188 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:63d249c5c7d8f61cc0f4f00c6cd8c5c2324721f135176069c7c632118f5681cc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 05: Energy methods"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1 Pg.No.116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "A=1800        # cross sectional area (mm^2)\n",
-      "E=200000        #youngs modulus  (N/mm^2)\n",
-      "sum_FLdFdP_B=1268*10**6      #(N.mm^2)\n",
-      "sum_FLdFdP_D=880*10**6     #(N.mm^2)\n",
-      "\n",
-      "del_Bv=sum_FLdFdP_B/A/E\n",
-      "del_Dh=sum_FLdFdP_D/A/E\n",
-      "\n",
-      "print \"\\ndeflection at point B =%2.2f mm\\n\"%(del_Bv)\n",
-      "print \"deflection at point D =%2.2f mm\\n\"%(del_Dh)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "deflection at point B =3.52 mm\n",
-        "\n",
-        "deflection at point D =2.44 mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9 Pg.No.142"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "a=np.array([[4.32,2.7],[2.7,11.62]])\n",
-      "b=np.array([27.1,48.11])\n",
-      "x=np.linalg.solve(a,b)\n",
-      "print \"\\nX1 = %1.2f kN & R2 = %1.2f kN\\n\"%(x[0],x[1]) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "X1 = 4.31 kN & R2 = 3.14 kN\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10 Pg.No.144"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "E=200000          #youngs modulus (N/mm^2)\n",
-      "A=200             #cross sectional area of each member (mm^2)\n",
-      "a=7*10**-6         #linear coefficient of heating (/C)\n",
-      "L=3*10**3          #length of BC (mm)\n",
-      "T=30               #temperature of truss (C)\n",
-      "sum_f2L=48000\n",
-      "\n",
-      "expansion=L*T*a\n",
-      "a11=sum_f2L/A/E\n",
-      "X1=-0.63/a11        #compatibility condition\n",
-      "print \"\\nX1 = %5.0f N\\n\"%(X1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "X1 =  -525 N\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12 Pg.No.151"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "def_C=-1.05-.6            #deflection at C (mm)\n",
-      "L=300                      #length of cantilever (mm)\n",
-      "\n",
-      "theta_B=math.atan(def_C/L)\n",
-      "print \"\\ndeflection at C = %2.3f degree\\n\"%(theta_B*180/math.pi) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "deflection at C = -0.315 degree\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter06.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter06.ipynb
deleted file mode 100755
index 5195a30b..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter06.ipynb
+++ /dev/null
@@ -1,112 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:902d8cc266f9fb597a0d00273b10ab788d226cbd969bb1e960b465c84a750eef"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 06: Matrix methods"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4 Pg.No.205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "from sympy import solve, symbols, pprint\n",
-      "from sympy import diff\n",
-      "x,y=symbols('x,y')\n",
-      "\n",
-      "E=200000                #youngs modulus (N/mm^2)\n",
-      "nu=0.3                   #poissons ratio\n",
-      "u1=0.001\n",
-      "u2=0.003\n",
-      "u3=-0.003              #displacements of corners (m)\n",
-      "u4=0\n",
-      "v1=-0.004\n",
-      "v2=-0.002\n",
-      "v3=0.001\n",
-      "v4=0.001\n",
-      "\n",
-      "\n",
-      "a=np.array([[1,-2,-1,2],[1,2,-1,-2],[1,2,1,2],[1,-2,1,-2]])\n",
-      "b=np.array([u1,u2,u3,u4])\n",
-      "alpha=np.linalg.solve(a,b)\n",
-      "alpha1=alpha[0]\n",
-      "alpha2=alpha[1]\n",
-      "alpha3=alpha[2]\n",
-      "alpha4=alpha[3]\n",
-      "\n",
-      "a=np.array([[1,-2,-1,2],[1,2,-1,-2],[1,2,1,2],[1,-2,1,-2]])\n",
-      "b=np.array([v1,v2,v3,v4])\n",
-      "alpha=np.linalg.solve(a,b)\n",
-      "alpha5=alpha[0]\n",
-      "alpha6=alpha[1]\n",
-      "alpha7=alpha[2]\n",
-      "alpha8=alpha[3]\n",
-      "\n",
-      "u=alpha1+alpha2*x+alpha3*y+alpha4*x*y\n",
-      "v=alpha5+alpha6*x+alpha7*y+alpha8*x*y\n",
-      "\n",
-      "ex=diff(u,x)\n",
-      "ey=diff(v,y)\n",
-      "Yxy=diff(u,y)+diff(v,x)\n",
-      "\n",
-      "ex=-0.000125\n",
-      "ey=.002\n",
-      "Yxy=-0.0015\n",
-      "\n",
-      "sigma_x=E/(1-nu**2)*(ex+nu*ey)\n",
-      "sigma_y=E/(1-nu**2)*(ey+nu*ex)\n",
-      "print \"longitudinal stress in x direction = %3.1f N/mm^2\"%(sigma_x)\n",
-      "print \"longitudinal stress in y direction = %3.1f N/mm^2\"%(sigma_y)\n",
-      "\n",
-      "T_xy=E/2/(1+nu)*Yxy\n",
-      "print \"shear stress at the center of plate = %3.1f N/mm^2\"%(T_xy)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "longitudinal stress in x direction = 104.4 N/mm^2\n",
-        "longitudinal stress in y direction = 431.3 N/mm^2\n",
-        "shear stress at the center of plate = -115.4 N/mm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter08.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter08.ipynb
deleted file mode 100755
index 65ccbf76..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter08.ipynb
+++ /dev/null
@@ -1,154 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d4ab160ecba8745ef027bb167a7abbd97977707c483b1d0a639c7f068e2016a8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 08: Columns"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3 Pg.No.280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "E=75000         #youngs modulus(N/mm^2)\n",
-      "G=21000          #shear modulus (N/mm^2)\n",
-      "L=2              #length of column (m)\n",
-      "l1=75            #flange length (mm)\n",
-      "l2=37.5            #total length(mm)\n",
-      "t=2.5            #thickness(mm)\n",
-      "\n",
-      "A=t*(2*l2+l1)         #observed from Fig 8.17\n",
-      "\n",
-      "#chapter 16  Ixx=bd^3/12+Ab^2\n",
-      "Ixx=2*l2*t*l2**2+t*l1**3/12\n",
-      "Iyy=2*t*l2**3/12\n",
-      "\n",
-      "I0=Ixx+Iyy\n",
-      "\n",
-      "#eqn 18.11 J=SUM((s*t^3)/3)\n",
-      "J=2*l2*t**3/3+l1*t**3/3\n",
-      "\n",
-      "Gama=t*l2**3*l1**2/24\n",
-      "Iyy=0.22*10**5\n",
-      "L=2*10**3\n",
-      "P_CRxx=math.pi**2*E*Ixx/L**2\n",
-      "P_CRyy=math.pi**2*E*Iyy/L**2\n",
-      "P_CRo=A/I0*(G*J+math.pi**2*E*Gama/L**2)\n",
-      "\n",
-      "print \"P_CRxx = %3.2e N\\n\"%(P_CRxx)\n",
-      "print \"P_CRyy = %0.1e N\\n\"%(P_CRyy)\n",
-      "print \"P_CRO = %0.2e N\\n\"%(P_CRo)\n",
-      "\n",
-      "print \"therefore buckling in the column due to axial load=%0.1e N\\n\"%(P_CRyy)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "P_CRxx = 6.51e+04 N\n",
-        "\n",
-        "P_CRyy = 4.1e+03 N\n",
-        "\n",
-        "P_CRO = 2.22e+04 N\n",
-        "\n",
-        "therefore buckling in the column due to axial load=4.1e+03 N\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4 Pg.No.282"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "from sympy import solve, symbols, pprint\n",
-      "from sympy import diff\n",
-      "P=symbols('P')\n",
-      "\n",
-      "E=70000        #youngs modulus (N/mm^2)\n",
-      "G=30000         #shear modulus (N/mm^2)\n",
-      "x_S=-76.2        #position of shear center(mm)\n",
-      "l1=l2=100        #lengths (mm)\n",
-      "t=2              #thickness(mm)\n",
-      "\n",
-      "x_bar=2*t*l1*50/(3*l1)/2\n",
-      "\n",
-      "A=600         #area (mm^2)\n",
-      "Ixx=1.17*10**6    #second moment of area (mm^4)\n",
-      "Iyy=0.67*10**6    #second moment of area (mm^4)\n",
-      "I0=5.32*10**6     # total second moment of area (mm^4)\n",
-      "J=800              #torsion constant (mm^4)\n",
-      "Gama=2488*10**6     #(mm^6)\n",
-      "L=10**3             #(mm)\n",
-      "\n",
-      "P_CRxx=math.pi**2*E*Ixx/L**2\n",
-      "P_CRyy=math.pi**2*E*Iyy/L**2\n",
-      "P_CRo=A/I0*(G*J+math.pi**2*E*Gama/L**2)\n",
-      "\n",
-      "fun=P**2*(1-A*x_S**2/I0)-P*(P_CRxx+P_CRo)+P_CRxx*P_CRo\n",
-      "solution = solve(fun, P)\n",
-      "print \"lowest value of critical load = %1.2e N\\n\"%(min(solution))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "lowest value of critical load = 1.68e+05 N\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter09.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter09.ipynb
deleted file mode 100755
index 119bd5b2..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter09.ipynb
+++ /dev/null
@@ -1,97 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:47a0669d4632eab71f162a005cd6d69531f951e47ad9f34d928b679209ff2467"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 09: Thin plates"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1 Pg.No.310"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "h=400     #height of structure (mm)\n",
-      "l1=300     #length of small section (mm)\n",
-      "L=1200      #length of whole structure (mm)\n",
-      "E=70000     #youngs modulus (N/mm^2)\n",
-      "A_f=350     #area of flange (mm^2)\n",
-      "A_s=300    # area of stiffeners (mm^2)\n",
-      "t=2         #thickness (mm)\n",
-      "I=2000      #second moment of area (mm^4)\n",
-      "w=5\n",
-      "#eqn 9.29 tan(a)^4=(1+td/2A_F)/(1+td/2A_S)\n",
-      "alpha=math.atan(((1+t*h/2/A_f)/(1+t*l1/A_s))**0.25)\n",
-      "\n",
-      "#eqn 9.19\n",
-      "F_T=w*L/h+w/2/math.tan((alpha))\n",
-      "\n",
-      "#eqn 9.25 M_max=w*b^2*tan(a)/12/d\n",
-      "M_max=w*10**3*l1**2*math.tan(alpha)/12/h\n",
-      "print \"maximum value of bending moment = %1.1e N.mm \\n\"%(M_max)\n",
-      "\n",
-      "#eqn 9.23 P=w*b*tan(a)/d\n",
-      "P=w*l1*math.tan(alpha)/h\n",
-      "print \"compressive load in stiffener = %1.1f kN \\n\"%(P)\n",
-      "\n",
-      "#eqn 9.24 le=d/(4-2b/d)^0.5\n",
-      "le=h/(4-2*l1/h)**0.5\n",
-      "print \"equivalent length of stiffener = %3.1f mm\\n\"%(le)\n",
-      "\n",
-      "#eqn 8.7  Pcr=pi^2*E*I/le^2\n",
-      "P_CR=math.pi**2*E*I/le**2\n",
-      "print \"buckling load of a stiffener = %2.1f kN \\n\"%(P_CR)  #approx. value in book"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum value of bending moment = 8.6e+04 N.mm \n",
-        "\n",
-        "compressive load in stiffener = 3.4 kN \n",
-        "\n",
-        "equivalent length of stiffener = 253.0 mm\n",
-        "\n",
-        "buckling load of a stiffener = 21589.8 kN \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter12.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter12.ipynb
deleted file mode 100755
index 63daef6d..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter12.ipynb
+++ /dev/null
@@ -1,111 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:217c546250d736c4a5ad9b469b73a9b7209b5ff1e5db79ce32388be8ee290870"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12: Structural components of aircraft"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 Pg.No. 390"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "D=4              #diameter of rivet(mm)\n",
-      "t_skin=2.5         #thickness of fuselage skin(mm)\n",
-      "t_strap=1.2          #thickness of straps (mm)\n",
-      "T=120               #shear stress (N/mm^2)\n",
-      "A=math.pi*D**2/4    #rivet area\n",
-      "Pfb=2*A*T\n",
-      "\n",
-      "b=Pfb/t_skin/12.5/10+4\n",
-      "print \"allowable rivet spacing = %2.1f mm \\n\"%(b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "allowable rivet spacing = 13.7 mm \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 Pg.No.392"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "l1=20          #distace between A&D (mm)\n",
-      "l2=25          #distace between C&D (mm)\n",
-      "l3=75          #distace between C&force location (mm)\n",
-      "P=5        #external load(kN)\n",
-      "r=(l1**2+l2**2)**0.5\n",
-      "sum_r2=2*l1**2+4*r**2\n",
-      "\n",
-      "#eqn 12.6 S=Pe*r/Sum(r2)\n",
-      "Pe=P*l3\n",
-      "S=Pe/sum_r2*r\n",
-      "print \"Resultant shear force in the A and B = %2.2f kN\\n\"%(S)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resultant shear force in the A and B = 2.45 kN\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter14.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter14.ipynb
deleted file mode 100755
index f9eb3f9d..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter14.ipynb
+++ /dev/null
@@ -1,226 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b89c31a3eb13c71dc64b90ec9c5fad56707d91451229ae687a102dc5e6c2ee23"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14: Airframe loads"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 Pg.No.407"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "W=45             #weight of aircraft (kN)\n",
-      "m=45/9.8          #mass of aircraft(k-kg)\n",
-      "a=3*9.8           #deceleration due to cable (ms^-2)\n",
-      "m1=4.5/9.8        #weight of aircraft after AA in Diagram (k-kg)\n",
-      "v0=25             #touch down speed (m/s)\n",
-      "alpha=10             #center line angle with ground (degree)\n",
-      "\n",
-      "T=m*a/math.cos(math.radians(alpha))\n",
-      "print \"tension in cable = %3.1f kN\\n\"%(T)\n",
-      "\n",
-      "R=W+T*math.sin(math.radians(alpha))\n",
-      "print \"load on each undercarriage strut = %2.1f kN\\n\"%(R/2/math.cos(math.radians(20)))\n",
-      "\n",
-      "N=T+m1*9.8*math.sin(math.radians(alpha))-m1*a*math.cos(math.radians(alpha))\n",
-      "S=m1*a*math.sin(math.radians(alpha))+m1*9.8*math.cos(math.radians(alpha))\n",
-      "print \"N and S forces are shown in Fig 14.4 N=%2.2f kN  S=%2.2f kN\\n\"%(N,S)\n",
-      "\n",
-      "s=v0**2/2/a\n",
-      "print \"length of deck covered = %2.2f m\\n\"%(s)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "tension in cable = 137.1 kN\n",
-        "\n",
-        "load on each undercarriage strut = 36.6 kN\n",
-        "\n",
-        "N and S forces are shown in Fig 14.4 N=124.57 kN  S=6.78 kN\n",
-        "\n",
-        "length of deck covered = 10.63 m\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 Pg.No.409"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "g=9.8\n",
-      "W=250           #weight of aircraft(kN)\n",
-      "m=250/g        #mass of aircraft (K-kg)\n",
-      "I_CG=5.65*10**8     #inertia about center of mass(N.s^2.mm)\n",
-      "v0=3.7            #vertical velocity of undercarriage(m/s)\n",
-      "R_h=400          #horizontal reaction (kN)\n",
-      "R_v=1200         #vertical reaction (kN)\n",
-      "l=1             # nose wheel distance from ground (m)\n",
-      "d=2.5           # distance of CG from ground (m)\n",
-      "\n",
-      "#horizontal equilibrium\n",
-      "ax=R_h/m\n",
-      "\n",
-      "#vertical equilibrium\n",
-      "ay=(R_v-W)/m\n",
-      "\n",
-      "\n",
-      "alpha=(R_v*l+R_h*d)*10**6/I_CG\n",
-      "print \"angular acceleration of aircraft = %2.1f rad/s^2 \\n\"%(alpha)\n",
-      "\n",
-      "#v=v0+ay*t\n",
-      "t=v0/ay\n",
-      "print \"time taken for vertical velocity to become zero = %1.3f s\\n\"%(t)\n",
-      "\n",
-      "#w=w0+a*t\n",
-      "w=a*t\n",
-      "print \"angular velocity of aircraft = %1.2f rad/s \\n\"%(w)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "angular acceleration of aircraft = 3.9 rad/s^2 \n",
-        "\n",
-        "time taken for vertical velocity to become zero = 0.099 s\n",
-        "\n",
-        "angular velocity of aircraft = 0.39 rad/s \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 Pg.No.414"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "W=8000             #weight of aircraft (N)\n",
-      "n=4.5              # wing loading\n",
-      "S=14.5              #wing area (m^2)\n",
-      "V=60                #speed (m/s)\n",
-      "rho=1.223           #density (kg/m^3)\n",
-      "alpha=13.75          #from Fig 14.8 (a)\n",
-      "C_Mcg=0.075           #from Fig 14.8 (a)\n",
-      "c=1.35                #mean chord (m)\n",
-      "\n",
-      "\n",
-      "L=n*W\n",
-      "C_L=L/(0.5*rho*V**2*S)\n",
-      "print \"lift coefficient of aircraft = %1.3f \\n\"%(C_L)\n",
-      "\n",
-      "#from Fig 14.8 (b)\n",
-      "l=4.18*math.cos(math.radians(alpha-2))+0.31*math.sin(math.radians(alpha-2))\n",
-      "print \"length of tail arm = %1.3f m \\n\"%(l)\n",
-      "\n",
-      "\n",
-      "C_L=C_L-c/l*C_Mcg\n",
-      "print \"lift coefficient =%1.3f \\n\"%(C_L)\n",
-      "\n",
-      "alpha=13.3\n",
-      "l=4.18*math.cos(math.radians(alpha-2))+0.31*math.sin(math.radians(alpha-2))\n",
-      "print \"Now tail arm length =  %2.3f m\\n\"%(l)\n",
-      "\n",
-      "L=0.5*rho*V**2*S*C_L\n",
-      "print \"Lift = %5.1f N\\n\"%(L)\n",
-      "\n",
-      "P=n*W-35000\n",
-      "print \"Tail Load = %5.1f N\\n\"%(P)\n",
-      "\n",
-      "D=0.5*rho*V**2*S*0.0875\n",
-      "print \"Drag = %5.1f N\\n\"%(D)\n",
-      "print \"Forward inertia force = %5.1f N\\n\"%(D)  #eqn 14.13"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "lift coefficient of aircraft = 1.128 \n",
-        "\n",
-        "length of tail arm = 4.156 m \n",
-        "\n",
-        "lift coefficient =1.103 \n",
-        "\n",
-        "Now tail arm length =  4.160 m\n",
-        "\n",
-        "Lift = 35222.3 N\n",
-        "\n",
-        "Tail Load = 1000.0 N\n",
-        "\n",
-        "Drag = 2793.0 N\n",
-        "\n",
-        "Forward inertia force = 2793.0 N\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter15.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter15.ipynb
deleted file mode 100755
index a849c8c6..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter15.ipynb
+++ /dev/null
@@ -1,76 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1bd6512f5b9ca8155cc1e9c3b04ad8207681499e095afaf4830a217017e989ea"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 15: Fatigue"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1 Pg.No.445"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from scipy import integrate\n",
-      "from sympy import symbols\n",
-      "import math\n",
-      "\n",
-      "#eqn 15.30 K=S(pi*a)^1/2*alpha\n",
-      "S=175         #stress range (N/mm^2)\n",
-      "K=1708          #fracture toughness (n/mm^3/2)\n",
-      "alpha=1         #constant\n",
-      "n=4         #constant\n",
-      "C=40*10**-15      #rate of crack growth (mm/cycle)\n",
-      "ai=0.1            #initial crack length\n",
-      "\n",
-      "alpha_f=K**2/S**2/math.pi\n",
-      "\n",
-      "\n",
-      "Nf=2/C/(n-2)/(S*math.pi**0.5)**n*(1/ai**((n-2)/2)-1/alpha_f**((n-2)/2))\n",
-      "print \"Number of cycles to failure = %5.0f cycles\\n\"%(Nf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of cycles to failure = 26919 cycles\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter16.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter16.ipynb
deleted file mode 100755
index 72ff7ec3..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter16.ipynb
+++ /dev/null
@@ -1,280 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c3ab3016805321c8a28748b1937c77b2125357dfc661b20013df0624feaa950e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 16: Bending of open and closed thin-walled beams \n"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.1 Pg. No.456"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Fig 16.6 Dimensions\n",
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "l=300            #length of I section (mm)\n",
-      "b=200            #width of I section beam (mm)\n",
-      "w1=25            #width of center section of I beam (mm)\n",
-      "w2=20            #width of upper and lower section of I beam (mm)\n",
-      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",
-      "\n",
-      "#second moment of area Ixx\n",
-      "# Ixx=b*l^3/12\n",
-      "Ixx=b*l**3/12-(b-w1)*(l-w2-w2)**3/12\n",
-      "print \"\\nSecond moment of area of I section beam = %5.3e mm^4\"%(Ixx)\n",
-      "\n",
-      "#sigma_z=My/I  reference 16.9\n",
-      "# @ \n",
-      "y=150\n",
-      "sigma_z=M*y/Ixx\n",
-      "print \"\\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)\"%(sigma_z)\n",
-      "\n",
-      "# @\n",
-      "y=-150\n",
-      "sigma_z=M*y/Ixx\n",
-      "print \"\\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)\"%(sigma_z)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Second moment of area of I section beam = 1.937e+08 mm^4\n",
-        "\n",
-        "direct stress at the top of the I section (y=150) = 77.45 N/mm^2 (compression)\n",
-        "\n",
-        "direct stress at the bottom of the I section (y=-150) = -77.45 N/mm^2 (tension)\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 Pg. No.457"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Fig 16.6 reference\n",
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "l=300            #length of I section (mm)\n",
-      "b=200            #width of I section beam (mm)\n",
-      "w1=25            #width of center section of I beam (mm)\n",
-      "w2=20            #width of upper and lower section of I beam (mm)\n",
-      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",
-      "\n",
-      "# second moment of area\n",
-      "# Iyy=wb^3/12\n",
-      "Iyy=2*20*200**3/12+260*25**3/12\n",
-      "print \"\\nSecond moment of area of I section beam = %5.3e mm^4\"%(Iyy)\n",
-      "\n",
-      "#sigma_z=Mx/I  \n",
-      "# @ \n",
-      "x=100\n",
-      "sigma_z=M*x/Iyy\n",
-      "print \"\\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)\"%(sigma_z)\n",
-      "\n",
-      "# @\n",
-      "x=-100\n",
-      "sigma_z=M*x/Iyy\n",
-      "print \"\\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)\"%(sigma_z)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Second moment of area of I section beam = 2.701e+07 mm^4\n",
-        "\n",
-        "direct stress at the top of the I section (y=150) = 370.30 N/mm^2 (compression)\n",
-        "\n",
-        "direct stress at the bottom of the I section (y=-150) = -370.30 N/mm^2 (tension)\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 Pg.No.458 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "l=300            #length of I section (mm)\n",
-      "b=200            #width of I section beam (mm)\n",
-      "w1=25            #width of center section of I beam (mm)\n",
-      "w2=20            #width of upper and lower section of I beam (mm)\n",
-      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",
-      "theta=30         #angle at which bending moment is applied(degree)\n",
-      "\n",
-      "Mx=M*math.cos(math.radians(30))\n",
-      "My=M*math.sin(math.radians(30))\n",
-      "\n",
-      "# sigma_z=Mx/Ixx*y+My/Iyy*x\n",
-      "# @top left hand corner\n",
-      "y=150\n",
-      "x=-100\n",
-      "sigma_z=Mx/Ixx*y-My/Iyy*x  \n",
-      "print \"\\ndirect stress at the top left hand corner = %3.1f N/mm^2 (tension)\"%(sigma_z)\n",
-      "\n",
-      "# @ top right hand corner\n",
-      "x=100\n",
-      "y=150\n",
-      "sigma_z=Mx/Ixx*y-My/Iyy*x  \n",
-      "print \"\\ndirect stress at the top right hand corner = %3.1f N/mm^2 (compression)\"%(sigma_z)\n",
-      "\n",
-      "alpha=math.atan(My*Ixx/Mx/Iyy)\n",
-      "print \"\\ninclination = %3.1f degree\\n\"%(alpha*180/math.pi)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "direct stress at the top left hand corner = 252.2 N/mm^2 (tension)\n",
-        "\n",
-        "direct stress at the top right hand corner = -118.1 N/mm^2 (compression)\n",
-        "\n",
-        "inclination = 76.4 degree\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 Pg.No.466"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# reference Fig 16.13\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "#variable declaration\n",
-      "l=80           #length of one section (mm)\n",
-      "w=8            #thickness of each section (mm)\n",
-      "b1=40          #width of one section (mm)\n",
-      "b2=80          #width of other section (mm)\n",
-      "Mx=1500*10*3   #bending moment (N.mm)\n",
-      "My=0\n",
-      "\n",
-      "#centroid\n",
-      "# in this example C is taken at top surface x axis aligned to AB and y axis is \n",
-      "# aligned to E surface \n",
-      "y_bar=((b1+b2)*w*w/2+l*w*(w+l/2))/(l*w+(b1+b2)*w)\n",
-      "print \"\\ny coordinate of centroid = %3.1f mm\"%(y_bar)\n",
-      "\n",
-      "x_bar=((b1+b2)*w*((b1+b2)/2-b1+w/2)+l*w*(w/2))/(l*w+(b1+b2)*w)\n",
-      "print \"\\nx coordinate of centroid = %3.1f mm\"%(x_bar)\n",
-      "\n",
-      "#second area of moment\n",
-      "#IB=IC+Ab^2              IB=moment about any point,IC=moment about centroid, b=distace between both points\n",
-      "Ixx=(b1+b2)*w**3/12+(b1+b2)*w*(y_bar-w/2)**2+w*l**3/12+l*w*(w+l/2-y_bar)**2\n",
-      "print \"\\nsecond moment of area about x axis = %3.2e mm^4\"%(Ixx)\n",
-      "\n",
-      "Iyy=w*(b1+b2)**3/12+(b1+b2)*w*((b1+b2)/2-b1+w/2-x_bar)**2+l*w**3/12+l*w*(x_bar-w/2)**2\n",
-      "print \"\\nsecond moment of area about y axis = %3.2e mm^4\"%(Iyy)\n",
-      "\n",
-      "Ixy=(b1+b2)*w*(y_bar-w/2)*((b1+b2)/2-b1+w/2-x_bar)+l*w*(w+l/2-y_bar)*(x_bar-w/2)\n",
-      "print \"\\nsecond moment of area about x and y axis = %3.2e mm^4\"%(Ixy)\n",
-      "\n",
-      "Mx=15*10**5\n",
-      "def f(x,y):\n",
-      "        return Mx*(Iyy*y-Ixy*x)/(Ixx*Iyy-Ixy**2)\n",
-      "sigma_z_max= f(-8,-66.4)\n",
-      "print \"\\nmaximum direct shear stress = %3.0f N/mm^2 (compressive)\\n\"%(sigma_z_max)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "y coordinate of centroid = 21.6 mm\n",
-        "\n",
-        "x coordinate of centroid = 16.0 mm\n",
-        "\n",
-        "second moment of area about x axis = 1.09e+06 mm^4\n",
-        "\n",
-        "second moment of area about y axis = 1.31e+06 mm^4\n",
-        "\n",
-        "second moment of area about x and y axis = 3.38e+05 mm^4\n",
-        "\n",
-        "maximum direct shear stress = -96 N/mm^2 (compressive)\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter18.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter18.ipynb
deleted file mode 100755
index b0550214..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter18.ipynb
+++ /dev/null
@@ -1,155 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a16e0822e0e7167e54e1f4ffc425aa7247562532992f7dee46fc87221365735a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 18: Torsion of Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.1 Pg.No.527"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "r=100                    #radius of cylinder (mm)\n",
-      "l=2*10**3                #length of cylinder (mm)\n",
-      "torque=30*10**6           #torque at mid point (N.mm)\n",
-      "Tmax=15*10**6            \n",
-      "Tau_max=200               #max shear stress (N/mm^2)\n",
-      "G=25000                   #shear modulus (N/mm^2)\n",
-      "theta =2                # rotation at mid point (degree)\n",
-      "\n",
-      "#ref equation 18.1    T=2Aq\n",
-      "tmin1=Tmax/(2*math.pi*r**2*Tau_max)\n",
-      "print \"\\nminimum thickness of beam due to stress limited to 200N/mm^2 = %3.1f mm\"%(tmin1)\n",
-      "\n",
-      "z=1*10**3                   # twist at mid point (mm)\n",
-      "#equation 18.4 after integration\n",
-      "tmin2=Tmax*200*math.pi*z/(4*(math.pi*r**2)**2*G*theta*math.pi/180)\n",
-      "print \"\\nminimum thickness due to constraint on maximum angle twist = %3.1f mm\"%(tmin2)\n",
-      "t=max(tmin1,tmin2)\n",
-      "print \"\\nthickness which satisfies both conditions = %3.1f mm\"%(t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "minimum thickness of beam due to stress limited to 200N/mm^2 = 1.2 mm\n",
-        "\n",
-        "minimum thickness due to constraint on maximum angle twist = 2.7 mm\n",
-        "\n",
-        "thickness which satisfies both conditions = 2.7 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.3 Pg.No 540"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#reference Fig 18.12\n",
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import symbols\n",
-      "from sympy import integrate\n",
-      "\n",
-      "S1,S2,S3=symbols('S1,S2,S3')\n",
-      "\n",
-      "#variable declaration\n",
-      "s1=50             #length of channel (mm)\n",
-      "s2=25             #width of channel (mm)\n",
-      "t1=2.5            #thickness of web(mm)\n",
-      "t2=1.5             #thickness of flange(mm)\n",
-      "T=10*10**3         #torque applied (N.mm)\n",
-      "zeta_s=8.04         \n",
-      "G=25000             #shear modulus (N/mm^2)\n",
-      "\n",
-      "#J=Sum(st^3/3)\n",
-      "J=1/3*(s1*t1**3+2*s2*t2**3)\n",
-      "tau_max=t1*T/J\n",
-      "print \"maximum shear stress = %3.1f N/mm^2\"%(tau_max)\n",
-      "\n",
-      "#in region O2\n",
-      "s1=25\n",
-      "ARf=1/2*zeta_s*s1\n",
-      "ws=-2*ARf*T/(G*J)\n",
-      "print \"warping in O2 region is linear = %3.2f mm\"%(ws)\n",
-      "\n",
-      "#in the wall 21\n",
-      "AR=1/2*zeta_s*s2-0.5*25*s2\n",
-      "w21=-0.03*(zeta_s-s2)\n",
-      "s1=50\n",
-      "tds=2*s2*t2+s1*t1\n",
-      "\n",
-      "A12=25/2*S1\n",
-      "A23=312.5-4.02*S2\n",
-      "A34=111.5+25/2*S3\n",
-      "AR1=1/2*(integrate(A12,(S1,0,25))+integrate(5*(A23),(S2,0,50))+integrate(3*A34,(S3,0,25)))/200\n",
-      "\n",
-      "#equation 18.20\n",
-      "ws=-2*ARf*T/G/J\n",
-      "print \"warping distribution in flange 34 = %1.2f \\n\"%(ws)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum shear stress = 78.9 N/mm^2\n",
-        "warping in O2 region is linear = -0.25 mm\n",
-        "warping distribution in flange 34 = -0.25 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter19.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter19.ipynb
deleted file mode 100755
index c6a6c12b..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter19.ipynb
+++ /dev/null
@@ -1,160 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8abe3316b17e7378affc20eb461733ec12ced2dbb39d65e4c948dd2ef8dc33e9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 19: Combined open and closed section beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.1 Pg.No. 552"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from scipy import integrate\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "l1=100               #lengths shown in Fig(mm)\n",
-      "l2=200\n",
-      "F=100*10**3            #force applied (N)\n",
-      "\n",
-      "y_bar=(2*l1*2*50+2*l2*2*l1+l2*2*l2)/(4*l1*2+4*l2*2)\n",
-      "Ixx=2*(2*l1**3/12+2*l1*25**2)+400*2*75**2+l2*2*125**2+2*(2*l2**3/12+2*l2*25**2)\n",
-      "\n",
-      "s1=Symbol('s1')\n",
-      "q12=-round(10**4*F/Ixx)/10**4*(integrate(-50+2*s1,s1))\n",
-      "print \"\\nq12 = %s\"%(q12)\n",
-      "\n",
-      "s2=Symbol('s2')\n",
-      "q23=-round(10**4*F/Ixx)/10**4*(integrate(2*75,s2))-34.5\n",
-      "print \"\\nq23 = %s\"%(q23)\n",
-      "\n",
-      "s3=Symbol('s3')\n",
-      "q03=-round(10**4*F/Ixx)/10**4*(integrate(2*75,s3))\n",
-      "print \"\\nq03 = %s\"%(q03)\n",
-      "\n",
-      "s4=Symbol('s4')\n",
-      "q34=-round(10**4*F/Ixx)/10**4*(integrate(2*(75-s4),s4))-242.5\n",
-      "print \"\\nq34 = %s\"%(q34)\n",
-      "\n",
-      "s5=Symbol('s5')\n",
-      "q94=-round(10**4*F/Ixx)/10**4*(integrate(-2*125,s5))\n",
-      "print \"\\nq94 = %s\\n\"%(q94)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "q12 = -0.0069*s1**2 + 0.345*s1\n",
-        "\n",
-        "q23 = -1.035*s2 - 34.5\n",
-        "\n",
-        "q03 = -1.035*s3\n",
-        "\n",
-        "q34 = 0.0069*s4**2 - 1.035*s4 - 242.5\n",
-        "\n",
-        "q94 = 1.725*s5\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.2 Pg.No.555"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from scipy import integrate\n",
-      "from sympy import symbols\n",
-      "import math\n",
-      "\n",
-      "A=20000                #nose cell area (mm^2)\n",
-      "L_w=900                #outer wall (mm)\n",
-      "L=300                  #width of wall (mm)\n",
-      "length=600           #length of open section (mm)\n",
-      "G=25000                #shear modulus (N/mm^2)\n",
-      "T=10*10**6             #torque applied (kN.m)\n",
-      "t1=1.5                  #thickness of closed section\n",
-      "t2=2                    #thickness of open section\n",
-      "\n",
-      "GJ_cl=4*A**2*G/(L_w+L)*t1\n",
-      "\n",
-      "print \"torsoinal rigidity of closed section = %2.2e N.mm^2 \\n\"%(GJ_cl)\n",
-      "\n",
-      "GJ_op=G*(length+L)*t2**3/3\n",
-      "print \"torsional rigidity of open section = %2.1e N.mm^2 \\n\"%(GJ_op)\n",
-      "\n",
-      "GJ=GJ_cl+GJ_op\n",
-      "print \"total torsional rigidity = %5.3e N.mm^2\\n\"%(GJ)\n",
-      "\n",
-      "dO_by_dz=T/GJ\n",
-      "print \"angle of twist per unit length = %1.4f rad/mm\\n\"%(dO_by_dz)\n",
-      "\n",
-      "q_cl=GJ_cl/2/A*dO_by_dz\n",
-      "print \"maximum shear stress in the closed section = %3.1f N/mm^2\\n\"%(q_cl/1.5)\n",
-      "\n",
-      "#eqn 18.10 T_max=GtdO/dz\n",
-      "T_max=G*t2*dO_by_dz\n",
-      "print \"maximum shear stress in the open section = %2.0f N/mm^2\\n\"%(T_max)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "torsoinal rigidity of closed section = 5.00e+10 N.mm^2 \n",
-        "\n",
-        "torsional rigidity of open section = 6.0e+07 N.mm^2 \n",
-        "\n",
-        "total torsional rigidity = 5.006e+10 N.mm^2\n",
-        "\n",
-        "angle of twist per unit length = 0.0002 rad/mm\n",
-        "\n",
-        "maximum shear stress in the closed section = 166.5 N/mm^2\n",
-        "\n",
-        "maximum shear stress in the open section = 10 N/mm^2\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter20.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter20.ipynb
deleted file mode 100755
index 4595d055..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter20.ipynb
+++ /dev/null
@@ -1,286 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:bb71ed7156eb25f3e97cd3ba07ae31557b447cbcd71b351505ee32010461a02e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 20: Structural idealization"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.1 Pg.No.560"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "sigma6=200\n",
-      "sigma1=200\n",
-      "sigma2=sigma5=150         #these are not real shear stress but taken \n",
-      "sigma3=sigma4=100          #proportional to length because we need just ratio\n",
-      "l16=400                     \n",
-      "t16=3\n",
-      "l12=l21=600\n",
-      "t12=t21=2                #thickness and lengths as shown in Fig 20.4\n",
-      "l23=600\n",
-      "t23=1.5\n",
-      "l25=300\n",
-      "t25=2.5\n",
-      "l34=200\n",
-      "t34=2\n",
-      "#eqn 20.1 B1=t_D*b/6*(2+sigma_2/sigma_1)\n",
-      "#eqn 20.2 B2=t_D*b/6*(2+sigma_1/sigma_2)\n",
-      "B1=B6=300+l16*t16/6*(2-sigma6/sigma1)+l12*t12/6*(2+sigma2/sigma1)\n",
-      "print \"B1=B6=%5.2f mm^2\\n\"%(B1)\n",
-      "\n",
-      "B2=B5=2*300+l12*t12/6*(2+sigma1/sigma2)+t25*l25/6*(2-sigma5/sigma2)+l23*t23/6*(2+sigma3/sigma2)\n",
-      "print \"B2=B5=%5.2f mm^2\\n\"%(B2)\n",
-      "\n",
-      "B3=300+l23*t23/6*(2+sigma2/sigma3)+l34*t34/6*(2-sigma4/sigma3)\n",
-      "print \"B3=B4=%5.2f mm^2\\n\"%(B3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "B1=B6=1050.00 mm^2\n",
-        "\n",
-        "B2=B5=1791.67 mm^2\n",
-        "\n",
-        "B3=B4=891.67 mm^2\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.2 Pg.No.562"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "Mx=100*10**6          #bending moment(N.mm)\n",
-      "y=[660,600,420,228,25,-204,-396,-502,-540]\n",
-      "B=[640,600,600,600,620,640,640,850,640]\n",
-      "\n",
-      "print \"direct stress in each boom in last column\"\n",
-      "print \"Boom\\t y(mm)\\t B(mm^2)\\t delIxx=By^2\\t sigma_z\"\n",
-      "for i in range (0,9):\n",
-      "    print \"%1.0f \\t %3.0f \\t %3.0f \\t \\t%2.1e \\t %2.1f\"%(i+1,y[i],B[i],B[i]*y[i]**2,Mx*y[i]/(1854*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "direct stress in each boom in last column\n",
-        "Boom\t y(mm)\t B(mm^2)\t delIxx=By^2\t sigma_z\n",
-        "1 \t 660 \t 640 \t \t2.8e+08 \t 35.6\n",
-        "2 \t 600 \t 600 \t \t2.2e+08 \t 32.4\n",
-        "3 \t 420 \t 600 \t \t1.1e+08 \t 22.7\n",
-        "4 \t 228 \t 600 \t \t3.1e+07 \t 12.3\n",
-        "5 \t  25 \t 620 \t \t3.9e+05 \t 1.3\n",
-        "6 \t -204 \t 640 \t \t2.7e+07 \t -11.0\n",
-        "7 \t -396 \t 640 \t \t1.0e+08 \t -21.4\n",
-        "8 \t -502 \t 850 \t \t2.1e+08 \t -27.1\n",
-        "9 \t -540 \t 640 \t \t1.9e+08 \t -29.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3 Pg.No.566"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "Ixx=48*10**6\n",
-      "Sy=4.8*10**3\n",
-      "B=300\n",
-      "\n",
-      "q12=-Sy/Ixx*B*200       #until point 2 \n",
-      "q23=q12-Sy/Ixx*B*200\n",
-      "q34=q23-Sy/Ixx*B*(-200)\n",
-      "print \"shear flow in flange 12 = %2.0f N/mm\\n\"%(q12)\n",
-      "print \"shear flow in web 23 = %2.0f N/mm\\n\"%(q23)\n",
-      "print \"shear flow in flange 34 = %2.0f N/mm\\n\"%(q34)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear flow in flange 12 = -6 N/mm\n",
-        "\n",
-        "shear flow in web 23 = -12 N/mm\n",
-        "\n",
-        "shear flow in flange 34 = -6 N/mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.4 Pg.No.569"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "B=[200,250,400,100,100,400,250,200]\n",
-      "Ixx=13.86*10**6\n",
-      "Sy=10*10**3\n",
-      "qb23=0\n",
-      "qb34=qb23-Sy/Ixx*B[2]*100\n",
-      "qb45=qb34-Sy/Ixx*B[3]*50\n",
-      "qb56=qb34\n",
-      "qb67=qb23\n",
-      "qb21=qb67-Sy/Ixx*(B[1]*100)\n",
-      "qb18=qb21-Sy/Ixx*B[7]*30\n",
-      "qb87=qb21\n",
-      "qs0=-5.4\n",
-      "print \"Distribution of shear flow :\"\n",
-      "print \"q23 = %2.1f N/mm\"%(qb23+qs0)\n",
-      "print \"q21 = %2.1f N/mm\"%(qb21-qs0)\n",
-      "print \"q34 = %2.1f N/mm\"%(qb34-qs0)\n",
-      "print \"q45 = %2.1f N/mm\"%(qb45-qs0)\n",
-      "print \"q56 = %2.1f N/mm\"%(qb56-qs0)\n",
-      "print \"q67 = %2.1f N/mm\"%(qb67+qs0)\n",
-      "print \"q18 = %2.1f N/mm\"%(qb18-qs0)\n",
-      "print \"q87 = %2.1f N/mm\\n\"%(qb87-qs0)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Distribution of shear flow :\n",
-        "q23 = -5.4 N/mm\n",
-        "q21 = -12.6 N/mm\n",
-        "q34 = -23.5 N/mm\n",
-        "q45 = -27.1 N/mm\n",
-        "q56 = -23.5 N/mm\n",
-        "q67 = -5.4 N/mm\n",
-        "q18 = -17.0 N/mm\n",
-        "q87 = -12.6 N/mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5 Pg.No.575"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import symbols, integrate\n",
-      "z=symbols('z')\n",
-      "\n",
-      "E=70000              #youngs modulus (N/mm^2)\n",
-      "G=30000              #shear modulus (N/mm^2)\n",
-      "P=4.8*10**3          #applied force (N)\n",
-      "L=2000               #cantilever length(mm)\n",
-      "Sy=P\n",
-      "Ixx=48*10**6         #second moment of area \n",
-      "t=1                   #actual thickness (mm)\n",
-      "\n",
-      "Mx0=-Sy*(L-z)\n",
-      "Mx1=-(L-z)\n",
-      "\n",
-      "del_M=integrate(Mx0*Mx1/E/Ixx,(z,0,L))\n",
-      "del_S=integrate((1/G/t/Sy*(6**2*200+12**2*400+6**2*200)),(z,0,L))\n",
-      "print \"total deflection in vertical direction = %1.2f mm\\n\"%(del_M+del_S)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "total deflection in vertical direction = 4.81 mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 35
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter21.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter21.ipynb
deleted file mode 100755
index 1195a6a9..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter21.ipynb
+++ /dev/null
@@ -1,256 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b8ffe836d5b15b63c70d69d0a02ef23ccdf3f03c0f15c98f0fa5e62d24f5d894"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 21: Wing spars and box beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.1 Pg.No.585"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from scipy import integrate\n",
-      "import math\n",
-      "\n",
-      "Mx=20*10**6         #internal bending moment (N.mm)\n",
-      "Sy=-20*10**3           #force applied(N)\n",
-      "l1=400\n",
-      "l2=300                #lengths of webs respectively from left to right in Fig 21.2\n",
-      "l3=200\n",
-      "\n",
-      "#Ixx=I_C+Ab^2\n",
-      "Ixx=2*l1*150**2+2*l2**3/12\n",
-      "y=150\n",
-      "sigma_z1=Mx*y/Ixx\n",
-      "sigma_z2=-Mx*y/Ixx\n",
-      "\n",
-      "Pz1=sigma_z1*l1\n",
-      "Pz2=sigma_z2*l1\n",
-      "\n",
-      "#from fig 21.1 and 21.2\n",
-      "del_y1_by_del_z=-0.05\n",
-      "del_y2_by_del_z=0.05\n",
-      "Syw=Sy-Pz1*del_y1_by_del_z-Pz2*del_y2_by_del_z\n",
-      "\n",
-      "# equ 21.6 qs=-Syw/Ixx*integrate(t_D*y*ds+B1*Y1) from 0 to s\n",
-      "fn=lambda s:300-2*s\n",
-      "\n",
-      "def shear(s):\n",
-      "    return -Syw/Ixx*(integrate.quad(fn, 0, s)[0]+400*150)\n",
-      "print \"maximum value of shear flow @s=150mm =%2.1f N/mm\\n\"%(shear(150.0))\n",
-      "print \"value of shear @s=0mm = %2.1f N/mm\\n\"%(shear(0))\n",
-      "print \"value of shear @s=300 mm = %2.1f N/mm\\n\"%(shear(300))\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum value of shear flow @s=150mm =53.8 N/mm\n",
-        "\n",
-        "value of shear @s=0mm = 39.1 N/mm\n",
-        "\n",
-        "value of shear @s=300 mm = 39.1 N/mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.2 Pg.No.589"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import symbols\n",
-      "import math\n",
-      "\n",
-      "Sy=100*10**6             #load applied in y direction (N)\n",
-      "Sx=0                     #load in x direction (N)\n",
-      "y=[0.3,0.3,0.3,-0.3,-0.3,-0.3]\n",
-      "B=[900,1200,900,900,1200,900]  #boom area\n",
-      "dx_dz=[0.1,0,-0.1,-0.1,0,0.1]\n",
-      "dy_dz=[-0.05,-0.05,-0.05,0.05,0.05,0.05]\n",
-      "Er=[0.6,0,0.6,0.6,0,0.6]            \n",
-      "nr=[0.3,0.3,0.3,0.3,0.3,0.3]\n",
-      "Mx=-Sy*2\n",
-      "My=-Sx*2\n",
-      "Ixy=0\n",
-      "Ixx=4*900*300**2+2*1200*300**2\n",
-      "Pzr=[0]*6\n",
-      "Pxr=[0]*6                #array initialization\n",
-      "Pyr=[0]*6\n",
-      "Pr=[0]*6\n",
-      "print \"Boom \\t Pzr(kN) \\t dx_dz \\t dy_dz \\t Pxr(kN) \\t Pyr(kN) \\t Pr(kN)\"\n",
-      "for i in range (0,6):\n",
-      "    Pzr[i]=Mx*y[i]/Ixx*B[i]\n",
-      "    Pxr[i]=Pzr[i]*dx_dz[i]                 #Pr is not correct in (7) column in book\n",
-      "    Pyr[i]=Pzr[i]*dy_dz[i]\n",
-      "    Pr[i]=(Pzr[i]**2+Pxr[i]**2+Pyr[i]**2)**0.5\n",
-      "    if Pzr[i]>0 :\n",
-      "        print \"%1.0f \\t %3.0f \\t\\t %1.1f \\t %1.2f \\t %2.0f \\t\\t %1.1f \\t\\t %3.1f\"%(i+1,Pzr[i],dx_dz[i],dy_dz[i],Pxr[i],Pyr[i],Pr[i])\n",
-      "    else:\n",
-      "        print \"%1.0f \\t %3.0f \\t\\t %1.1f \\t %1.2f \\t %2.0f \\t\\t %1.1f \\t\\t %3.1f\"%(i+1,Pzr[i],dx_dz[i],dy_dz[i],Pxr[i],Pyr[i],-Pr[i])\n",
-      "       \n",
-      "Sxw=0\n",
-      "Syw=66.6*10**3\n",
-      "qb16=0        #open section here\n",
-      "qb12=qb16-Syw/Ixx*B[0]*300\n",
-      "qb23=qb12-Syw/Ixx*B[1]*300\n",
-      "qb34=qb23-Syw/Ixx*B[2]*300\n",
-      "qb45=qb23\n",
-      "qb56=qb12\n",
-      "qs0=-97\n",
-      "\n",
-      "\n",
-      "print \"\\n\\nqb16 = %3.1f N/mm\"%(qb16)\n",
-      "print \"qb12 = %3.1f N/mm\"%(qb12)\n",
-      "print \"qb23 = %3.1f N/mm\"%(qb23)\n",
-      "print \"qb34 = %3.1f N/mm\"%(qb34)\n",
-      "print \"qb45 = %3.1f N/mm\"%(qb45)\n",
-      "print \"qb56 = %3.1f N/mm\"%(qb56)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boom \t Pzr(kN) \t dx_dz \t dy_dz \t Pxr(kN) \t Pyr(kN) \t Pr(kN)\n",
-        "1 \t -100 \t\t 0.1 \t -0.05 \t -10 \t\t 5.0 \t\t -100.6\n",
-        "2 \t -133 \t\t 0.0 \t -0.05 \t -0 \t\t 6.7 \t\t -133.5\n",
-        "3 \t -100 \t\t -0.1 \t -0.05 \t 10 \t\t 5.0 \t\t -100.6\n",
-        "4 \t 100 \t\t -0.1 \t 0.05 \t -10 \t\t 5.0 \t\t 100.6\n",
-        "5 \t 133 \t\t 0.0 \t 0.05 \t  0 \t\t 6.7 \t\t 133.5\n",
-        "6 \t 100 \t\t 0.1 \t 0.05 \t 10 \t\t 5.0 \t\t 100.6\n",
-        "\n",
-        "\n",
-        "qb16 = 0.0 N/mm\n",
-        "qb12 = -33.3 N/mm\n",
-        "qb23 = -77.7 N/mm\n",
-        "qb34 = -111.0 N/mm\n",
-        "qb45 = -77.7 N/mm\n",
-        "qb56 = -33.3 N/mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.3 Pg.No.593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import symbols\n",
-      "import numpy as np\n",
-      "import math\n",
-      "yr=symbols('yr')\n",
-      "\n",
-      "Mx=-100*1.9*10**6                #bending moment(N.mm)\n",
-      "Ixx=4*900*295**2+2*1200*295**2      #second moment of area (mm^4)\n",
-      "sigma_zr=Mx/Ixx*yr\n",
-      "P1_1=P3_1=-0.364*295*900\n",
-      "P4_1=P6_1=0.364*295*900\n",
-      "P2_1=-0.364*295*1200\n",
-      "P5_1=0.364*295*1200\n",
-      "\n",
-      "Mx=-100*2.1\n",
-      "Ixx=4*900*305**2+2*1200*305**2\n",
-      "sigma_zr=Mx/Ixx*yr\n",
-      "P1_2=P3_2=-0.376*305*900\n",
-      "P4_2=P6_2=0.376*305*900\n",
-      "P2_2=-0.376*305*1200\n",
-      "P5_2=0.376*305*1200\n",
-      "\n",
-      "deltaP1=deltaP3=(P1_1-P1_2)/200\n",
-      "deltaP4=deltaP6=-(P4_1-P4_2)/200\n",
-      "deltaP2=(P2_1-P2_2)/200\n",
-      "deltaP5=(P5_1-P5_2)/200\n",
-      "\n",
-      "a=np.array([[1,-1,0,0,0,0],[0,1,-1,0,0,0],[0,0,1,-1,0,0],[0,0,0,1,-1,0],[0,0,0,0,1,-1],[600*300*2,2*600*300,600*600,0,0,600*600]])\n",
-      "\n",
-      "b=np.array([43.8,32.85,-32.85,-43.8,-32.85,600*10**5])\n",
-      "q=np.linalg.solve(a,b)\n",
-      "print \"shear flow distribution :\"\n",
-      "print \"q12 = %2.1f N/mm\"%(q[0])\n",
-      "print \"q23 = %2.1f N/mm\"%(q[1])\n",
-      "print \"q34 = %2.1f N/mm\"%(q[2])\n",
-      "print \"q45 = %2.1f N/mm\"%(q[3])\n",
-      "print \"q56 = %2.1f N/mm\"%(q[4])\n",
-      "print \"q61 = %2.1f N/mm\\n\"%(q[5])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear flow distribution :\n",
-        "q12 = 63.6 N/mm\n",
-        "q23 = 19.8 N/mm\n",
-        "q34 = -13.1 N/mm\n",
-        "q45 = 19.8 N/mm\n",
-        "q56 = 63.6 N/mm\n",
-        "q61 = 96.4 N/mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 41
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter22.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter22.ipynb
deleted file mode 100755
index 15f86e3c..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter22.ipynb
+++ /dev/null
@@ -1,88 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:7afcc7b44b13ea906e3e2317703c261e07426e5f23f7958a6b7966ba0274524d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 22: Fuselages"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 Pg.No.598"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "#B1=t_D*b/6(2+sigma_2/sigma_1)\n",
-      "Mx=200*10**6                     # (N*mm)\n",
-      "y=[381,352,269.5,145.8,0,-145.8,-269.5,-352,-381,-352,-269.5,-145.8,0,145.8,269.5,352]\n",
-      "sigma_2=sigma_16=352\n",
-      "sigma_1=381\n",
-      "A=100\n",
-      "B1=A+0.8*149.6/6*(2+sigma_2/sigma_1)*2\n",
-      "Ixx=2.52*10**8\n",
-      "def sigma_z(y):\n",
-      "    return Mx/Ixx*y\n",
-      "print \"Stringer/Boom \\t y(mm) \\t \\tsigma_z(N/mm^2)\"\n",
-      "for i in range (0,16):\n",
-      "    print \"%2.0f \\t\\t %3.1f\\t\\t\\t%3.1f \"%(i+1,y[i],sigma_z(y[i]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stringer/Boom \t y(mm) \t \tsigma_z(N/mm^2)\n",
-        " 1 \t\t 381.0\t\t\t302.4 \n",
-        " 2 \t\t 352.0\t\t\t279.4 \n",
-        " 3 \t\t 269.5\t\t\t213.9 \n",
-        " 4 \t\t 145.8\t\t\t115.7 \n",
-        " 5 \t\t 0.0\t\t\t0.0 \n",
-        " 6 \t\t -145.8\t\t\t-115.7 \n",
-        " 7 \t\t -269.5\t\t\t-213.9 \n",
-        " 8 \t\t -352.0\t\t\t-279.4 \n",
-        " 9 \t\t -381.0\t\t\t-302.4 \n",
-        "10 \t\t -352.0\t\t\t-279.4 \n",
-        "11 \t\t -269.5\t\t\t-213.9 \n",
-        "12 \t\t -145.8\t\t\t-115.7 \n",
-        "13 \t\t 0.0\t\t\t0.0 \n",
-        "14 \t\t 145.8\t\t\t115.7 \n",
-        "15 \t\t 269.5\t\t\t213.9 \n",
-        "16 \t\t 352.0\t\t\t279.4 \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter23.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter23.ipynb
deleted file mode 100755
index a5e90903..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter23.ipynb
+++ /dev/null
@@ -1,390 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:38d29ad436c9255138372fb573a0261c5aeb8d664f2491fbf03a71e42b1c4783"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 23: Wings"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1 Pg.No.609"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import solve, symbols, pprint\n",
-      "import math\n",
-      "y=symbols('y')\n",
-      "\n",
-      "B1=B6=2580\n",
-      "B2=B5=3880          #Boom areas (mm^2)\n",
-      "B3=B4=3230\n",
-      "\n",
-      "l1=200\n",
-      "l2=230               #dimensions shown in Fig 23.3 (mm)\n",
-      "l3=165\n",
-      "\n",
-      "Mx=300*10**6         #bending moment (N.mm)\n",
-      "My=0\n",
-      "\n",
-      "Ixx=2*(B1*l3**2+B2*l2**2+B3*l1**2)\n",
-      "sigma_z=Mx/Ixx*y\n",
-      "\n",
-      "print \"direct stress in booms sigma_z = %s\" %(sigma_z)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "direct stress in booms sigma_z = 0.370651791174781*y\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2 Pg.No.612"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import solve, symbols, pprint\n",
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "T=11.3*10**6      # torque applied (N.mm)\n",
-      "G_REF=27600        #(N/mm^2)\n",
-      "A1=258000\n",
-      "A2=355000          #areas in table\n",
-      "A3=161000\n",
-      "\n",
-      "#t*12=G/G_REF*t\n",
-      "t12=24200/G_REF*1.22\n",
-      "t12i=27600/G_REF*2.03\n",
-      "t13=t24=24200/G_REF*1.22     #G and thickness taken from table ex23.2 Pg.No.612\n",
-      "t35=t46=t56=20700/G_REF*0.92\n",
-      "t34=27600/G_REF*1.63\n",
-      "\n",
-      "# del12=ds/t*12\n",
-      "del12=1650/t12\n",
-      "del12i=508/t12i\n",
-      "del13=del24=775/t13\n",
-      "del34=380/t34               #lengths taken from table\n",
-      "del35=del46=508/t35\n",
-      "del56=254/t56\n",
-      "\n",
-      "\n",
-      "a=np.array([[del12+del12i,-del12i,0,-2*A1*G_REF],[-del12i,del12i+del13+del24+del34,-del34,-2*A2*G_REF],[0,-del34,del35+del46+del34+del56,-2*A3*G_REF],[A1,A2,A3,0]])\n",
-      "b=np.array([0,0,0,5.65*10**6])\n",
-      "x=np.linalg.solve(a,b)\n",
-      "print \"shear stress distribution is as follows\"\n",
-      "print \"q1=%1.1f N/mm\"%(x[0])\n",
-      "print \"q2=%1.1f N/mm\"%(x[1])\n",
-      "print \"q3=%1.1f N/mm\\n\"%(x[2])\n",
-      "print \"dO_dz=%1.2e \"%(x[3])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear stress distribution is as follows\n",
-        "q1=7.1 N/mm\n",
-        "q2=8.9 N/mm\n",
-        "q3=4.2 N/mm\n",
-        "\n",
-        "dO_dz=7.36e-07 \n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3 Pg.No.616"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import solve, symbols, pprint\n",
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "A1=265000\n",
-      "A2=213000\n",
-      "A3=413000\n",
-      "\n",
-      "G_REF=27600\n",
-      "Sy=86.8*10**3\n",
-      "t78=3*27600/27600*1.22\n",
-      "del78=1270/t78\n",
-      "t12=t56=1.22\n",
-      "del12=del56=1023/t12\n",
-      "t23=1.63\n",
-      "del23=1274/t23\n",
-      "t34=2.03\n",
-      "del34=2200/t34\n",
-      "del38=57\n",
-      "del84=95\n",
-      "del87=347\n",
-      "del27=68\n",
-      "del75=106\n",
-      "del16=330/1.63\n",
-      "Ixx=809*10**6             #From example 23.1\n",
-      "\n",
-      "qb27=-99.4;qb16=-45.5;qb65=0;qb57=95.5;qb38=-69.8;qb48=69\n",
-      "\n",
-      "\n",
-      "a=np.array([     [del34+del84+del38,-del38,0,-2*A1*G_REF],       [-del38,del23+del38+del87+del27,-del27,-2*A2*G_REF],          [0,-del27,del56+del27+del75+del12+del16,-2*A3*G_REF],                    [2*A1,2*A2,2*A3,0]                                              ])\n",
-      "b=np.array([-10488,-2561,7426,19736700])\n",
-      "x=np.linalg.solve(a,b)\n",
-      "\n",
-      "qs01=5.5\n",
-      "qs02=10.2\n",
-      "qs03=16.5\n",
-      "\n",
-      "q34=qs01\n",
-      "q23=qs02\n",
-      "q12=qs03\n",
-      "q61=-qb16+qs03\n",
-      "q57=qb57-qs03\n",
-      "q72=-qb27-qs02\n",
-      "q48=qb48+qs01\n",
-      "q83=-qb38-qs01\n",
-      "\n",
-      "print \"shear flows distribution is as follows :\"\n",
-      "print \"q34=%1.2f N/mm\"%(q34)\n",
-      "print \"q23=q87=%1.2f N/mm\"%(q23)\n",
-      "print \"q12=q56=%1.2f N/mm\"%(q12)\n",
-      "print \"q61=%1.2f N/mm\"%(q61)             \n",
-      "print \"q57=%1.2f N/mm\"%(q57)\n",
-      "print \"q72=%1.2f N/mm\"%(q72)\n",
-      "print \"q48=%1.2f N/mm\"%(q48)\n",
-      "print \"q83=%1.2f N/mm\\n\"%(q83)\n",
-      "\n",
-      "print \"rate of twist = %1.1e rad/mm\"%(x[3])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear flows distribution is as follows :\n",
-        "q34=5.50 N/mm\n",
-        "q23=q87=10.20 N/mm\n",
-        "q12=q56=16.50 N/mm\n",
-        "q61=62.00 N/mm\n",
-        "q57=79.00 N/mm\n",
-        "q72=89.20 N/mm\n",
-        "q48=74.50 N/mm\n",
-        "q83=64.30 N/mm\n",
-        "\n",
-        "rate of twist = 1.1e-06 rad/mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.4 Pg.No.618 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import solve, symbols, pprint\n",
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Boom areas\n",
-      "B=[600,900,600,600,900,600]\n",
-      "Pz=[0,0,0,0,0,0]\n",
-      "y=[54.56,54.56,54.56,-54.56,-54.56,-54.56]\n",
-      "Ixx=4*600*90**2+2*900*90**2\n",
-      "Ixy=0\n",
-      "Mx=1.65*10**6\n",
-      "My=0\n",
-      "\n",
-      "a=np.array([[1700,-1520],[72000,144000]])\n",
-      "b=np.array([3942,690726])\n",
-      "x=np.linalg.solve(a,b)\n",
-      "print \"\\nqs0I = %2.1f N/mm\\n\"%(x[0])\n",
-      "print \"\\nqs0II = %2.1f N/mm\\n\"%(x[1])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "qs0I = 4.6 N/mm\n",
-        "\n",
-        "\n",
-        "qs0II = 2.5 N/mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.5 Pg.No.622"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import solve, symbols, pprint\n",
-      "import math\n",
-      "import numpy as np\n",
-      "from sympy import integrate\n",
-      "z=symbols('z')\n",
-      "\n",
-      "E=69000             #youngs modulus (N/mm^2)\n",
-      "G=25900              #shear modulus (N/mm^2)\n",
-      "t=2                  #thickness (mm)\n",
-      "B=[650,1300,650,650,1300,650]   #boom area\n",
-      "q0=[9.6,-5.8,50.3,-5.8,9.6,54.1,73.6]\n",
-      "Sy0=44.5*10**3\n",
-      "Sy1=1\n",
-      "Mx0=-44.5*10**3*(2000-z)\n",
-      "Mx1=-(2000-z)\n",
-      "Ixx=81.3*10**6\n",
-      "int_q0q1_Gt=1/G/t/Sy0*(q0[0]**2*250*t+q0[1]**2*500*t+q0[2]**2*250+q0[5]**2*250+q0[6]**2*250)\n",
-      "\n",
-      "delta=integrate(Mx0*Mx1/E/Ixx,(z,0,2000))+integrate(int_q0q1_Gt,(z,0,2000))\n",
-      "print \"deflection at free end of the two cell = %2.2f mm\"%(delta)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "deflection at free end of the two cell = 23.58 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.6 Pg.No.624"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import solve, symbols, pprint\n",
-      "import math\n",
-      "import numpy as np\n",
-      "from sympy import integrate\n",
-      "\n",
-      "T=10*10**6       #torque subjected (N.mm)\n",
-      "l1=800  \n",
-      "l2=200            #lengths shown in Fig23.17 (mm)\n",
-      "l3=1500\n",
-      "A=l2*l1\n",
-      "\n",
-      "\n",
-      "q=T/2/A\n",
-      "S=T/l1\n",
-      "q1=S/l2\n",
-      "P=S*l3/2/l2\n",
-      "\n",
-      "a=np.array([[1,-1],[1,1]])\n",
-      "b=np.array([31.3,62.5])\n",
-      "q=np.linalg.solve(a,b)\n",
-      "\n",
-      "print \"shear flow :\"\n",
-      "print \"q1=%2.2f N/mm\"%(q1)\n",
-      "print \"q2=%2.2f N/mm\"%(q[0])\n",
-      "print \"q3=%2.2f N/mm\\n\"%(q[1])\n",
-      "\n",
-      "print \"flange loads :\"\n",
-      "print \"P(st.4500) = 0\"\n",
-      "print \"P(st.3000) = %2.2f N (compression)\"%(l3*q[0]-l3*q[1])\n",
-      "print \"P(st.2250) = %2.1f\"%(46875-l3/2*q1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear flow :\n",
-        "q1=62.50 N/mm\n",
-        "q2=46.90 N/mm\n",
-        "q3=15.60 N/mm\n",
-        "\n",
-        "flange loads :\n",
-        "P(st.4500) = 0\n",
-        "P(st.3000) = 46950.00 N (compression)\n",
-        "P(st.2250) = 0.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 84
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter24.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter24.ipynb
deleted file mode 100755
index b305a12a..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter24.ipynb
+++ /dev/null
@@ -1,198 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:217b3507f45091d99aa496d838ab43e552e1e20629747bd44a2cad35220ffe6a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 24: Fuselage frames and wing ribs"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1 Pg.No.638"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import numpy\n",
-      "import math\n",
-      "#reference Fig 42.2\n",
-      "l1=250               #length of each section(mm)\n",
-      "l2=200                #length of DK(mm)\n",
-      "l3=100                 #length of KH(mm)\n",
-      "F1=4000\n",
-      "F2=5000                #forces as shown in Fig 24.1\n",
-      "theta=60               #force angle from DH (degree)\n",
-      "\n",
-      "#solve (i) and (ii) equation\n",
-      "a=numpy.array([[1,-1],[200,100]])\n",
-      "b=numpy.array([13.8564,2000])\n",
-      "q=numpy.linalg.solve(a,b)\n",
-      "q1=q[0]\n",
-      "q2=q[1]\n",
-      "q3=F1*math.cos(math.radians(theta))/(l2+l3)\n",
-      "q4=F1*math.cos(math.radians(theta))/(l2+l3)+F2/(l2+l3)\n",
-      "P_A=l1*q1+l1*q3+l1*q4\n",
-      "P_E=-l1*q2-l1*q3-l1*q4\n",
-      "P=3464.1\n",
-      "M_AE=F2*l1+F1/2*3*l1-P*50\n",
-      "\n",
-      "P_A=M_AE/(l2+l3)+P/2\n",
-      "P_E=-M_AE/(l2+l3)+P/2\n",
-      "\n",
-      "print \"shear flow as shown in Fig 24.1\\n\"\n",
-      "print \"q1=%2.2f N/mm\\n\"%(q1)\n",
-      "print \"q2=%2.2f N/mm\\n\"%(q2)\n",
-      "print \"q3=%2.2f N/mm\\n\"%(q3)\n",
-      "print \"q4=%2.2f N/mm\\n\"%(q4)\n",
-      "\n",
-      "print \"Stiffener load at point A = %2.2f N/mm(tension)\\n\"%(P_A)\n",
-      "print \"Stiffener load at point E = %2.2f N/mm(compression)\\n\"%(P_E)  #in book another method is also explained"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear flow as shown in Fig 24.1\n",
-        "\n",
-        "q1=11.29 N/mm\n",
-        "\n",
-        "q2=-2.57 N/mm\n",
-        "\n",
-        "q3=6.67 N/mm\n",
-        "\n",
-        "q4=23.33 N/mm\n",
-        "\n",
-        "Stiffener load at point A = 10321.37 N/mm(tension)\n",
-        "\n",
-        "Stiffener load at point E = -6857.27 N/mm(compression)\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2 Pg.No.645"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import numpy as np\n",
-      "import math\n",
-      "\n",
-      "A1=50000\n",
-      "A2=95000        #area shown in Fig 24.13(mm^2)\n",
-      "A3=95000\n",
-      "\n",
-      "A4=46000\n",
-      "A5=49000         #area shown in Fig 24.11 (mm^2)\n",
-      "\n",
-      "F1=12000\n",
-      "F2=15000          #forces shown in Fig 24.9 (N)\n",
-      "\n",
-      "l1=l3=300\n",
-      "l2=320           #lengths and angle shown in Fig 24.9\n",
-      "alpha=15\n",
-      "\n",
-      "#solve equation (i)&(iii)\n",
-      "a=np.array([[600,-600],[190000,290000]])\n",
-      "b=np.array([12000,440000])\n",
-      "q=np.linalg.solve(a,b)\n",
-      "q12=q[0]\n",
-      "q23=q[1]\n",
-      "q31=(F2+l1*q23)/l1\n",
-      "print \"shear flows are shown in Fig 24.9 (flanges)\"\n",
-      "print \"q12=%2.2f N/mm\"%(q12)\n",
-      "print \"q23=%2.2f N/mm\"%(q23)\n",
-      "print \"q31=%2.2f N/mm\\n\"%(q31)\n",
-      "\n",
-      "Sy_1=7*l3\n",
-      "Px4=Px2=2*A1*7/l1\n",
-      "Py2=Py4=Px4*math.tan(math.radians(alpha))\n",
-      "q1=(2100-2*625.2)/l1\n",
-      "P2=P4=(Px4**2+Py4**2)**0.5\n",
-      "P5=P6=2*((A1+A4)*7-A5*13)/l2\n",
-      "q2=(7*l1+7*10-13*10)/l2\n",
-      "q3=(6.4*l2+F2)/l2\n",
-      "\n",
-      "M3=2*((A1+A2)*7-A2*13)+F2*l1\n",
-      "Px1=Px3=M3/l1\n",
-      "Py1=Py3=3626.2\n",
-      "P1=P3=(Px1**2+Py1**2)**.5\n",
-      "q3=(17100-2*Py1)/l1\n",
-      "\n",
-      "print \"Loads in webs \"\n",
-      "print \"P4=P2=%2.2f N\"%(P2)\n",
-      "print \"P6=P5=%2.2f N\"%(P6)\n",
-      "print \"P3=P1=%2.2f N\\n\"%(P1)\n",
-      "\n",
-      "print \"shear flow in webs\"\n",
-      "print \"q1=%2.2f N/mm\"%(q1)\n",
-      "print \"q2=%2.2f N/mm\"%(q2)\n",
-      "print \"q3=%2.2f N/mm (this value is given at the end of example)\"%(q3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear flows are shown in Fig 24.9 (flanges)\n",
-        "q12=13.00 N/mm\n",
-        "q23=-7.00 N/mm\n",
-        "q31=43.00 N/mm\n",
-        "\n",
-        "Loads in webs \n",
-        "P4=P2=2415.64 N\n",
-        "P6=P5=218.75 N\n",
-        "P3=P1=14010.73 N\n",
-        "\n",
-        "shear flow in webs\n",
-        "q1=2.83 N/mm\n",
-        "q2=6.38 N/mm\n",
-        "q3=32.83 N/mm (this value is given at the end of example)\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter25.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter25.ipynb
deleted file mode 100755
index 369a648f..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter25.ipynb
+++ /dev/null
@@ -1,490 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:806c344ebbd66dd7398f87b7be85af8355708b04b28f0bd902647a641248feff"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 25: Laminated Composite Structures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.1 Pg.No.653"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "Ee=5000               #youngs modulus of epoxy (N/mm^2)\n",
-      "Ec=200000             #youngs modulus of carbon (N/mm^2)\n",
-      "\n",
-      "# E1=Ef*Af/A+Em+Am/A        equation 25.4\n",
-      "A=50*80               #total area (mm^2)\n",
-      "Ae=40*80              #area of epoxy (mm^2)\n",
-      "Ac=10*80              #area of carbon(mm^2)\n",
-      "L=500                  #length of bar(mm)\n",
-      "vc=0.3                 #poisson ratio\n",
-      "ve=0.2\n",
-      "\n",
-      "#effective youngs modulus\n",
-      "E1=(Ee*Ae+Ec*Ac)/A\n",
-      "\n",
-      "load=100*10**3\n",
-      "sigma1=load/A\n",
-      "epsilon1=sigma1/E1\n",
-      "\n",
-      "delta1=epsilon1*L\n",
-      "v1t=(ve*Ae+vc*Ac)/A\n",
-      "epsilon_t=-v1t*epsilon1\n",
-      "\n",
-      "thickness=L\n",
-      "delta_t=-epsilon_t*(thickness)\n",
-      "\n",
-      "\n",
-      "sigma_m=Ee*epsilon1\n",
-      "sigma_f=Ec*epsilon1\n",
-      "print \"stresses in epoxy = %2.2f N/mm^2\\n\"%(sigma_m)\n",
-      "print \"stresses in carbon = %2.2f N/mm^2\\n\"%(sigma_f)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "stresses in epoxy = 2.84 N/mm^2\n",
-        "\n",
-        "stresses in carbon = 113.64 N/mm^2\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.2 Pg.No.657"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "tau=40            #shear stress(N/mm^2)\n",
-      "G=5000            #shear modulus(N/mm^2)\n",
-      "v1t=0.3            #poisson ratio \n",
-      "Et=80000            #youngs modulus in transverse direction(N/mm^2)\n",
-      "E1=120000           #youngs modulus in longitudinal direction(N/mm^2)\n",
-      "sigma_x=50          #direction stress in x direciton\n",
-      "sigma_y=25                           #in y direction\n",
-      "\n",
-      "\n",
-      "vt1=v1t*Et/E1 #minor poisson ratio\n",
-      "\n",
-      "epsilon1=sigma_x/E1-vt1*sigma_y/Et\n",
-      "epsilont=sigma_y/Et-v1t*sigma_x/E1\n",
-      "print \"direct strain in x direction = %2.2e \\n\"%(epsilon1)\n",
-      "print \"direct strain in y direction = %2.2e \\n\"%(epsilont)\n",
-      "\n",
-      "gama1t=tau/G\n",
-      "print \"shear strain in the ply = %2.2e \\n\"%(gama1t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "direct strain in x direction = 3.54e-04 \n",
-        "\n",
-        "direct strain in y direction = 1.88e-04 \n",
-        "\n",
-        "shear strain in the ply = 8.00e-03 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.3 Pg.No.661"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "El=150000          #youngs modulus(N/mm^2)\n",
-      "Et=90000          #youngs modulus(N/mm^2)\n",
-      "Glt=5000          #shear modulus(N/mm^2)\n",
-      "vlt=0.3            #poisson ratio\n",
-      "theta=45             #longitudinal plane inclination\n",
-      "\n",
-      "s11=1/El\n",
-      "s22=1/Et\n",
-      "s12=-vlt/El\n",
-      "s33=1/Glt\n",
-      "\n",
-      "a=np.array([[53.45,-46.55,1],[-46.55,53.45,0],[-2.2,-2.2,0]])\n",
-      "b=np.array([60,40,0])\n",
-      "x=np.dot(a,b)\n",
-      "print \"strain in the x direction = %4.4e\\n\"%(x[0]*10**-6)\n",
-      "print \"strain in the y direction = %4.4e\\n\"%(x[1]*10**-6)\n",
-      "print \"shear strain in the xy plane = %4.4e\\n\"%(x[2]*10**-6)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "strain in the x direction = 1.3450e-03\n",
-        "\n",
-        "strain in the y direction = -6.5500e-04\n",
-        "\n",
-        "shear strain in the xy plane = -2.2000e-04\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.4 Pg.No.664"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "l1=150\n",
-      "t1=1\n",
-      "l2=100            #dimensions shown in Fig 25.10 (mm)\n",
-      "t2=2\n",
-      "Ef=60000            #youngs modulus of flanges(N/mm^2)\n",
-      "Ew=20000            #youngs modulus of web(N/mm^2)\n",
-      "P=40*10**3          #axial load(N)\n",
-      "\n",
-      "#sum of b*t*E \n",
-      "sum_btE=2*l2*t2*Ef+l1*t1*Ew\n",
-      "\n",
-      "epsilon_z=P/sum_btE           #equ 25.37\n",
-      "P_flange=epsilon_z*l2*t2*Ef\n",
-      "P_web=epsilon_z*l1*t1*Ew\n",
-      "print \"axial load in flange = %2.2f kN\\n\"%(P_flange/1000)\n",
-      "print \"axial load in web = %2.2f kN\\n\"%(P_web/1000)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "axial load in flange = 17.78 kN\n",
-        "\n",
-        "axial load in web = 4.44 kN\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.5 Pg.No.666"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "l1=100\n",
-      "l2=50\n",
-      "t1=1         #dimension shown in Fig 25.11 (mm)\n",
-      "t2=2\n",
-      "Ef=50000         #youngs modulus of flange(N/mm^2)\n",
-      "Ew=15000          #youngs modulus of web(N/mm^2)\n",
-      "Mx=10**6           #bending moment (N.mm)\n",
-      "My=0\n",
-      "\n",
-      "Ixx=2*Ef*l2*t2*l2**2+Ew*t1*l1**3/12\n",
-      "Iyy=Ef*t2*l1**3/12\n",
-      "Ixy=Ef*l2*t2*(50)*(50)+Ef*l2*t2*(-50)*(-50)\n",
-      "\n",
-      "x=50\n",
-      "y=50      #point 1\n",
-      "Ez=50000\n",
-      "\n",
-      "sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)\n",
-      "print \"direct stress at point 1 = %3.1f N/mm^2\\n\"%(sigma_z)\n",
-      "x=0\n",
-      "y=50\n",
-      "sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)\n",
-      "print \"direct stress at point 2 = %3.1f N/mm^2\\n\"%(sigma_z)\n",
-      "\n",
-      "x=0\n",
-      "y=50\n",
-      "Ez=15000\n",
-      "sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)\n",
-      "print \"direct stress at point 2 in the web = %3.1f N/mm^2\\n\"%(sigma_z)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "direct stress at point 1 = 102.6 N/mm^2\n",
-        "\n",
-        "direct stress at point 2 = -51.3 N/mm^2\n",
-        "\n",
-        "direct stress at point 2 in the web = -15.4 N/mm^2\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.6 Pg.No.668"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "from sympy import symbols, integrate\n",
-      "\n",
-      "s=symbols('s')\n",
-      "\n",
-      "l12=250\n",
-      "l23=300\n",
-      "t31=t12=2                 #dimensions shown in Fig 25.12 (mm)\n",
-      "t23=1.5\n",
-      "\n",
-      "P=2*10**3             #shear load(N)\n",
-      "E12=E31=45000          #youngs modulus of sides given in name(N/mm^2)\n",
-      "E23=20000\n",
-      "\n",
-      "Ixx=2*E12*t12*l12**3*(l23/2/l12)**2/12+E23*t23*l23**3/12\n",
-      "alpha=math.asin(l23/2/l12)\n",
-      "\n",
-      "Sx=0\n",
-      "Sy=2*10**3\n",
-      "q12=-E12*Sy/Ixx*integrate(-2*s,(s,0,250))*math.sin(alpha)\n",
-      "print \"shear flow at point 2 in the flange 12 = %2.2f N/mm\\n\"%(q12)\n",
-      "\n",
-      "q23=-E23*Sy/Ixx*integrate(-225+1.5*s,(s,0,300))+22.2\n",
-      "print \"shear flow at point 3 in the flange 23 = %2.2f N/mm\\n\"%(q23)\n",
-      "q0=14.2\n",
-      "\n",
-      "q12=-E12*Sy/Ixx*integrate(-2*s,s)*math.sin(alpha)-q0\n",
-      "print \"shear flow in the flange 12 and 31 \"\n",
-      "print q12\n",
-      "\n",
-      "q23=-E23*Sy/Ixx*integrate(-225+1.5*s,s,)+22.2-q0\n",
-      "print \"\\nshear flow in the web 23\"\n",
-      "print q23"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear flow at point 2 in the flange 12 = 22.22 N/mm\n",
-        "\n",
-        "shear flow at point 3 in the flange 23 = 22.20 N/mm\n",
-        "\n",
-        "shear flow in the flange 12 and 31 \n",
-        "0.000355555555555556*s**2 - 14.2\n",
-        "\n",
-        "shear flow in the web 23\n",
-        "-0.000197530864197531*s**2 + 0.0592592592592593*s + 8.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 57
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.7 Pg.No.671"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "from sympy import symbols, integrate\n",
-      "\n",
-      "\n",
-      "l1=200\n",
-      "t1=2\n",
-      "l2=100         #dimensions shown in Fig 25.13 (mm)\n",
-      "t2=1\n",
-      "T=10*10**6       #torque applied (N.mm)\n",
-      "Gl=20000          #laminate shear modulus(N/mm^2)\n",
-      "Gw=35000            #web shear modulus(N/mm^2)\n",
-      "A=l1*l2\n",
-      "\n",
-      "q=T/2/A\n",
-      "\n",
-      "#from eqn 25.47\n",
-      "int_ds_by_Gt=2*l1/Gl/t1+2*l2/Gw/t2\n",
-      "\n",
-      "#Let's say we want to calculate warping at point 1\n",
-      "#warping at mid of web is zero (W0=0) integrate eqn 25.47 from\n",
-      "#mid of web to point 1\n",
-      "W0=0\n",
-      "A0s=50*100\n",
-      "W1=W0+q*(l2/2/Gw/t2-int_ds_by_Gt/A*A0s)\n",
-      "print \"warping at point 1 is = %2.2f mm\\n\"%(W1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "warping at point 1 is = -0.63 mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.8 Pg.No.673"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import numpy as np\n",
-      "from sympy import symbols, integrate\n",
-      "\n",
-      "l1=50\n",
-      "t1=2.5\n",
-      "l2=25             #dimension shown in Fig 18.12(mm)\n",
-      "t2=1.5\n",
-      "\n",
-      "Gl=20000\n",
-      "Gw=15000           #shear modulus (N/mm^2)\n",
-      "T=10*10**3         #torque applied (N.mm)\n",
-      "AR=25*25/2\n",
-      "GJ=2*Gl*l2*t2**3/3+Gw*l1*t1**3/3\n",
-      "#eqn 25.49\n",
-      "dO_dz=T/GJ\n",
-      "\n",
-      "t_max12=2*Gl*(t2/2)*dO_dz\n",
-      "t_max23=2*Gw*(t1/2)*dO_dz\n",
-      "print \"maximum shear stress in the web = %2.2f N/mm^2\\n\"%(t_max12)\n",
-      "print \"maximum shear stress in the laminate = %2.2f N/mm^2\\n\"%(t_max23)\n",
-      "\n",
-      "W1=-2*AR*dO_dz\n",
-      "print \"warping at point 1 = %2.2f mm\\n\"%(W1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum shear stress in the web = 59.63 N/mm^2\n",
-        "\n",
-        "maximum shear stress in the laminate = 74.53 N/mm^2\n",
-        "\n",
-        "warping at point 1 = -1.24 mm\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter26.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter26.ipynb
deleted file mode 100755
index 994182d0..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter26.ipynb
+++ /dev/null
@@ -1,82 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:eb3a8ca85b2b3516ec190a12a8401ddf32d140498cd023be9bda6228b1754025"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 26: Closed Section Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 26.3 Pg.No.693"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import solve, symbols\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "G=20000               #shear modulus (N/mm^2)\n",
-      "E=G/0.36              #young's modulus\n",
-      "tau=20                #torque (Nm/mm)\n",
-      "t=1\n",
-      "b=200\n",
-      "a=500\n",
-      "L=2500\n",
-      "A=800\n",
-      "\n",
-      "#torque at each support \n",
-      "T=20*5000/2\n",
-      "\n",
-      "mu=math.sqrt(8*G*t/(A*E*(b+a)))\n",
-      "\n",
-      "z=2500\n",
-      "F=-1.95/mu*math.cosh(mu*z)+1.75*z**2\n",
-      "\n",
-      "z=0                            # at mid-span z=0\n",
-      "theta=10**(-8)*(1.95/mu*math.cosh(mu*z)-1.75*z**2+F)\n",
-      "print \"twist at mid-span = %1.3f rad = %1.2f degree\"%(theta,theta*180/math.pi)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "twist at mid-span = 0.108 rad = 6.20 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/Chapter27.ipynb b/Aircraft_Structures_for_Engineering_Students/Chapter27.ipynb
deleted file mode 100755
index bf9485c9..00000000
--- a/Aircraft_Structures_for_Engineering_Students/Chapter27.ipynb
+++ /dev/null
@@ -1,109 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:bc596ba82da68e58c73271eb1045592b064349244b750ab69aa9f03ab4de2fbd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 27: Open Section Beams\n"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 27.2 Pg.No. 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#reference Fig 27.20\n",
-      "\n",
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import integrate\n",
-      "from sympy import solve, symbols\n",
-      "\n",
-      "# variable declaration\n",
-      "P=100                  #vertical load (kN)\n",
-      "A_R=1.25*10**3         #from previous question\n",
-      "E=200000               #young's modulus (N/mm^2)\n",
-      "Gama_R=2.08*10**10     #torsion bending constant (mm^6)\n",
-      "mu=0.54*10**-3        #from previous question\n",
-      "A=400*5                #area of beam\n",
-      "\n",
-      "#moment couple\n",
-      "M_Tau=-P*2*A_R\n",
-      "print \"\\nValue of moment couple at the top of column = %2.2e kN.mm^2\"%(M_Tau)  #calculation mistake in book\n",
-      "\n",
-      "d2tdz2 = M_Tau*10**3/E/Gama_R\n",
-      "# @z=3000 mm \n",
-      "D=0.04*10**-3\n",
-      "z=symbols('z')\n",
-      "F= -0.08*math.cosh(0)\n",
-      "\n",
-      "def theta(z):\n",
-      "    return 0.08*(math.cosh(0.54*10**(-3)*z))\n",
-      "\n",
-      "print \"\\ntwist at the top of the column = %2.2f rad\\n\"%(theta(3000))\n",
-      "\n",
-      "M_Tau=-E*Gama_R*0.02*10**-6\n",
-      "print \"value of moment couple at bottom of column=%2.2e N.mm^2\\n\"%(M_Tau)\n",
-      "\n",
-      "P=-100*10**3    # stress(N)\n",
-      "def sigma_z(A_R2):\n",
-      "    return P/A+M_Tau*A_R2/Gama_R\n",
-      "print \"sigma_z1=sigma_z4=%2.1f N/mm^2\\n\"%(sigma_z(-7.5*10**3))\n",
-      "print \"sigma_z2=sigma_z3=%2.1f N/mm^2\\n\"%(sigma_z(2.5*10**3))\n",
-      "\n",
-      "#please put the value of 2AR from Fig 27.21 in the equation of sigma_z\n",
-      "#these values in book need correction"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value of moment couple at the top of column = -2.50e+05 kN.mm^2\n",
-        "\n",
-        "twist at the top of the column = 0.21 rad\n",
-        "\n",
-        "value of moment couple at bottom of column=-8.32e+07 N.mm^2\n",
-        "\n",
-        "sigma_z1=sigma_z4=-20.0 N/mm^2\n",
-        "\n",
-        "sigma_z2=sigma_z3=-60.0 N/mm^2\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 48
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students/README.txt b/Aircraft_Structures_for_Engineering_Students/README.txt
deleted file mode 100755
index 230418be..00000000
--- a/Aircraft_Structures_for_Engineering_Students/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Prahlad Das
-Course: others
-College/Institute/Organization: Indian Institute of Technology Bombay
-Department/Designation: Aerospace
-Book Title: Aircraft Structures for Engineering Students
-Author: T. H. G. Megson
-Publisher: Linacre House, Jordan Hill, Oxford OX2 8DP, UK
-Year of publication: 2012
-Isbn: 978-0-75066-7395
-Edition: fourth edition
\ No newline at end of file
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diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter01.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter01.ipynb
new file mode 100755
index 00000000..7760d8e3
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter01.ipynb
@@ -0,0 +1,499 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0f8515032ad257111ffef42eb71b2510ec7d92768034e754ca14b473e010b968"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1. Basic Elasticity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Pg. No.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# variable declaration\n",
+      "p = 1.5;         #pressure inside vessel (N/mm^2)\n",
+      "d = 2*10**3;     #diameter of vessel (mm)\n",
+      "t = 20;          #thickness of plate (mm)\n",
+      "theta = 60;        #plane's inclination to axis of vessel (degree)\n",
+      "load = 2500*10**3;       #axial tensile load (N)\n",
+      "\n",
+      "# Longitudinal stress\n",
+      "sigma_x=p*d/4/t;\n",
+      "print \"\\nLongitudinal stress due to internal pressure = %5.2f N/mm^2\" %(sigma_x)\n",
+      "\n",
+      "#circumferential stress\n",
+      "sigma_y=p*d/2/t;\n",
+      "print \"\\nCircumferential stress due to internal pressure =  %5.2f N/mm^2\" %(sigma_y)\n",
+      "\n",
+      "#axial load\n",
+      "sigma_x_axial=load/(math.pi*d*t);\n",
+      "print \"\\ndirect stress due to axial load = %5.2f N/mm^2\"%(sigma_x_axial)\n",
+      "\n",
+      "sigma_x=sigma_x+sigma_x_axial;                #total longitudinal stress\n",
+      "\n",
+      "#direct stress and shear stress on inclined plane AB\n",
+      "#reference Fig 1.9 pg no14, equation 1.8,1.9 pg no 13\n",
+      "\n",
+      "sigma_n=sigma_x*math.pow(math.cos(math.radians(90-theta)),2)+sigma_y*math.pow(math.sin(math.radians(90-theta)),2)\n",
+      "print \"\\ndirect stress on inclined plane AB = %5.2f N/mm^2\"%(sigma_n)\n",
+      "\n",
+      "tau=(sigma_x-sigma_y)/2*math.sin(math.radians(2*(90-theta)))\n",
+      "print \"\\nshear stress on plane AB = %5.2f N/mm^2\"%(tau)\n",
+      "\n",
+      "#maximumm shear stress (theta=45 degree)\n",
+      "theta=45\n",
+      "tau=(sigma_x-sigma_y)/2*math.sin(math.radians(2*(90-theta)))\n",
+      "print \"\\nmaximum shear stress on plane AB = %5.2f N/mm^2\"%(tau)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Longitudinal stress due to internal pressure = 37.50 N/mm^2\n",
+        "\n",
+        "Circumferential stress due to internal pressure =  75.00 N/mm^2\n",
+        "\n",
+        "direct stress due to axial load = 19.89 N/mm^2\n",
+        "\n",
+        "direct stress on inclined plane AB = 61.80 N/mm^2\n",
+        "\n",
+        "shear stress on plane AB = -7.62 N/mm^2\n",
+        "\n",
+        "maximum shear stress on plane AB = -8.80 N/mm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2, Pg. No.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#variable declaration\n",
+      "load=50*10**3         #axial load (N)\n",
+      "d=60                   #diameter (mm)\n",
+      "t=1.5                  #offset distance from center (mm)\n",
+      "T=1200*10**3                 #torque applied on a point (N.mm)\n",
+      "theta=60               #angle made by plane wrt axis of cylinder(degree)\n",
+      "\n",
+      "#compressive stress due to axial load\n",
+      "area=math.pi*math.pow(d/2,2)        #cross section area\n",
+      "sigma_x_a=load/area\n",
+      "print \"\\ncompressive stress due to axial load = %5.1f N/mm^2\"%sigma_x_a\n",
+      "\n",
+      "#compressive stress due to bending moment\n",
+      "#area moment of inertia\n",
+      "M=load*t\n",
+      "I=math.pi*d**4/64\n",
+      "\n",
+      "sigma_x_b=M*d/2*(1/I)\n",
+      "print \"\\ncompressive stress due to bending moment = %5.1f N/mm^2\" %sigma_x_b\n",
+      "\n",
+      "#total compressive stress\n",
+      "sigma_x=sigma_x_a+sigma_x_b\n",
+      "\n",
+      "#shear stress due to torque Ref example 3.1 equation (iv) pg no 73\n",
+      "# tau=Tr/J\n",
+      "J=math.pi*d**4/32    #torsion constant\n",
+      "tau_xy=T*d/2/J\n",
+      "print \"\\nshear stress due to torque = %5.1f N/mm^2\" %tau_xy\n",
+      "\n",
+      "#direct and shear on inclined plane, ref eq 1.8,1.9\n",
+      "sigma_y=0\n",
+      "sigma_n=-sigma_x*math.pow(math.cos(math.radians(90-theta)),2)-tau_xy*math.sin(math.radians(2*(90-theta)))\n",
+      "print \"\\ndirect stress on inclined plane = %3.1f N/mm^2\"%sigma_n\n",
+      "\n",
+      "tau=-sigma_x/2*math.sin(math.radians(2*(90-theta)))+tau_xy*math.cos(math.radians(2*(90-theta)))\n",
+      "print \"\\nshear stress on inclined plane = %3.1f N/mm^2\"%tau\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "compressive stress due to axial load =  17.7 N/mm^2\n",
+        "\n",
+        "compressive stress due to bending moment =   3.5 N/mm^2\n",
+        "\n",
+        "shear stress due to torque =  28.3 N/mm^2\n",
+        "\n",
+        "direct stress on inclined plane = -40.4 N/mm^2\n",
+        "\n",
+        "shear stress on inclined plane = 5.0 N/mm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 95
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3, Pg. No.20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import numpy\n",
+      "import math\n",
+      "#import matplotlib\n",
+      "%pylab inline\n",
+      "#from pylab import *          #change comment to above line to see undocked graph, all values will be visible as mouse-pointer moves\n",
+      "#variable declaration\n",
+      "sigma_x=160          #stress in x direction (N/mm^2)\n",
+      "sigma_y=-120          #stress in y direction (N/mm^2)\n",
+      "sigma=200             #stress on inclined plane (N/mm^2)\n",
+      "\n",
+      "tau_xy=((sigma-sigma_x)*(sigma-sigma_y))**0.5\n",
+      "print \"\\nallowable shear stress, tau_xy = %4.1f N/mm^2 \"%tau_xy\n",
+      "\n",
+      "coeff=[1,-(sigma_x+sigma_y),sigma_x*sigma_y-tau_xy**2]\n",
+      "sigma=numpy.roots(coeff)\n",
+      "print \"\\nprincipal stresses, sigma_I = %3.0f N/mm^2  sigma_II = %3.0f N/mm^2\"%(sigma[0],sigma[1])\n",
+      "\n",
+      "tau_max=(abs(sigma[0])+abs(sigma[1]))/2\n",
+      "print \"\\nmaximum shear stress, tau_max = %3.0f N/mm^2\"%tau_max\n",
+      "\n",
+      "#plotting Mohr circle\n",
+      "x_cent=(sigma_x+sigma_y)/2\n",
+      "y_cent=0\n",
+      "\n",
+      "X1=(sigma_x,tau_xy)\n",
+      "X2=(sigma_y,-tau_xy)\n",
+      "\n",
+      "radius=(math.hypot(X2[0]-X1[0], X2[1] - X1[1]))/2\n",
+      "\n",
+      "cir=linspace(0,2*pi,100)\n",
+      "plot(radius*cos(cir)+x_cent,radius*sin(cir)+y_cent,'r')\n",
+      "plot(sigma_x,tau_xy,'ro',sigma_y,-tau_xy,'ro',x_cent,y_cent,'b+',sigma[0],0,'bo',sigma[1],0,'bo',20,tau_max,'go')\n",
+      "text(sigma_x+10,tau_xy,'Q1')\n",
+      "text(sigma_y+10,-tau_xy,'Q2')\n",
+      "text(20,tau_max+10,r'$\\tau_{max}$')\n",
+      "text(-150,0,r'$\\sigma_2$')\n",
+      "text(200,10,r'$\\sigma_1$')\n",
+      "text(-20,+10,'O')\n",
+      "text(30,-20,'C')\n",
+      "#plot([sigma_y,tau_xy],[sigma_x,-tau_xy],'r-')\n",
+      "xlabel(r'$\\sigma$')\n",
+      "ylabel(r'$\\tau$')\n",
+      "title('Mohr Cirle')\n",
+      "axis('equal')\n",
+      "grid(True)\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "\n",
+        "allowable shear stress, tau_xy = 113.1 N/mm^2 \n",
+        "\n",
+        "principal stresses, sigma_I = 200 N/mm^2  sigma_II = -160 N/mm^2\n",
+        "\n",
+        "maximum shear stress, tau_max = 180 N/mm^2\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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UlbCoFi3gxAnT5rfffkufPn1wc3PjwQcf5OOPP9YxONsghcTOWKWd9swZaNbM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\nxZVZr61+AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0xa137240>"
+       ]
+      }
+     ],
+     "prompt_number": 141
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4, Pg. No.33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#variable declaration\n",
+      "sigma_x=83       # stress in x direction (N/mm^2)\n",
+      "sigma_y=65       # stress in y direction (N/mm^2)\n",
+      "E=200000         # Young's Modulus (N/mm^2)\n",
+      "v=0.3           # poisson's ratio\n",
+      "\n",
+      "# strains in x,y and z directions\n",
+      "epsln_x=1.0/E*(sigma_x-v*sigma_y)\n",
+      "epsln_y=1.0/E*(sigma_y-v*sigma_x)\n",
+      "epsln_z=-v/E*(sigma_x+sigma_y)\n",
+      "print \"\\nstrain in x direction, epsln_x = %5.3e\"%epsln_x\n",
+      "print \"\\nstrain in y direction, epsln_y = %5.3e\"%epsln_y\n",
+      "print \"\\nstrain in z direction, epsln_z = %5.3e\"%epsln_z\n",
+      "\n",
+      "tau_max=(sigma_x-sigma_y)/2\n",
+      "print \"\\nmaximum shear stress, tau_max = %2.2f N/mm^2\"%tau_max\n",
+      "\n",
+      "gama_max=2*(1+v)*tau_max/E\n",
+      "print \"\\nmaximum shear strain, y_max = %2.2e \" %gama_max"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "strain in x direction, epsln_x = 3.175e-04\n",
+        "\n",
+        "strain in y direction, epsln_y = 2.005e-04\n",
+        "\n",
+        "strain in z direction, epsln_z = -2.220e-04\n",
+        "\n",
+        "maximum shear stress, tau_max = 9.00 N/mm^2\n",
+        "\n",
+        "maximum shear strain, y_max = 1.17e-04 \n"
+       ]
+      }
+     ],
+     "prompt_number": 111
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5, Pg. No.33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#import matplotlib.pyplot as plt\n",
+      "%pylab inline\n",
+      "#variable declaration\n",
+      "sigma_x=60            #stress in x direction (N/mm^2)\n",
+      "sigma_y=-40           #stress in y direction (N/mm^2)\n",
+      "tau_xy=50             #shear stress (N/mm^2)\n",
+      "E=2*10**5             #Young's Modulus (N/mm^2)\n",
+      "v=0.3                #poisson's ratio\n",
+      "\n",
+      "#strains in x and y direction\n",
+      "epsln_x=1.0/E*(sigma_x-v*sigma_y)\n",
+      "epsln_y=1.0/E*(sigma_y-v*sigma_x)\n",
+      "print \"\\nstrains in x and y directions, epsi_x =%4.1e  epsi_y =%4.1e \"%(epsln_x,epsln_y)     \n",
+      "\n",
+      "G=E/2/(1+v)\n",
+      "print \"\\nshear modulus, G = %6.1f\"%G\n",
+      "\n",
+      "gama_xy=tau_xy/G\n",
+      "print \"\\nshear strain, gama_xy = %4.1e\"%gama_xy\n",
+      "\n",
+      "#principal strains\n",
+      "epsln_I=(epsln_x+epsln_y)/2+1.0/2*((epsln_x-epsln_y)**2+gama_xy**2)**0.5\n",
+      "epsln_II=(epsln_x+epsln_y)/2-1.0/2*((epsln_x-epsln_y)**2+gama_xy**2)**0.5\n",
+      "print \"\\nprincipal strains, epsln_I = %4.2e  epsln_II = %4.2e\"%(epsln_I,epsln_II)\n",
+      "\n",
+      "#inclination\n",
+      "theta=1.0/2*math.atan(gama_xy/(epsln_x-epsln_y))\n",
+      "print \"\\ninclination to the plane on which sigma_x acts, theta = %4.1f or %4.1f\"%(theta*180/math.pi,theta*180/math.pi+90)\n",
+      "\n",
+      "#plotting Mohr circle\n",
+      "x_cent=(epsln_x+epsln_y)/2\n",
+      "y_cent=0\n",
+      "\n",
+      "X1=(epsln_x,gama_xy/2)\n",
+      "X2=(epsln_y,-gama_xy/2)\n",
+      "\n",
+      "radius=(math.hypot(X2[0]-X1[0], X2[1] - X1[1]))/2\n",
+      "print radius\n",
+      "cir=linspace(0,2*pi,100)\n",
+      "plot(radius*cos(cir)+x_cent,radius*sin(cir),'r')\n",
+      "plot(epsln_x,gama_xy/2,'ro',epsln_y,-gama_xy/2,'ro',x_cent,y_cent,'b+',epsln_II,0,'go',epsln_I,0,'go')\n",
+      "text(epsln_x,gama_xy/2,'Q1')\n",
+      "text(epsln_y,-gama_xy/2,'Q2')\n",
+      "text(-0.00050,0,r'$\\epsilon_2$')\n",
+      "text(.00050,0,r'$\\epsilon_1$')\n",
+      "text(-.000060,0.00002,'O')\n",
+      "text(0,0,'C')\n",
+      "xlabel(r'$\\epsilon$')\n",
+      "ylabel(r'$\\gamma$')\n",
+      "title('Mohr Cirle')\n",
+      "axis('equal')\n",
+      "grid(True)\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "\n",
+        "strains in x and y directions, epsi_x =3.6e-04  epsi_y =-2.9e-04 \n",
+        "\n",
+        "shear modulus, G = 76923.1\n",
+        "\n",
+        "shear strain, gama_xy = 6.5e-04\n",
+        "\n",
+        "principal strains, epsln_I = 4.95e-04  epsln_II = -4.25e-04\n",
+        "\n",
+        "inclination to the plane on which sigma_x acts, theta = 22.5 or 112.5\n",
+        "0.000459619407771\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Ti4t2v82PP2pDw0NCtPtxpDnbYcnMACbIzABO5ocftJFk/fvD8uXadPrCer79FmJiIDAQ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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0xcc10a20>"
+       ]
+      }
+     ],
+     "prompt_number": 185
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7, Pg. No.40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#variable declaration\n",
+      "epsln_a=1000*10**(-6)     #readings of strain gauges \n",
+      "epsln_b=-200*10**(-6)     #straing gauges 'a'  and 'c' are\n",
+      "epsln_c=-300*10**(-6)     #in line and perpendicular to axis of bar\n",
+      "d=50                      #diameter of bar (mm)\n",
+      "E=70000                   #Young's modulus (N/mm^2)\n",
+      "v=0.3                     #poisson's ratio\n",
+      "\n",
+      "#principal strains\n",
+      "epsln_I=1.0/2*(epsln_a+epsln_c)+(1.0/2**0.5)*((epsln_a-epsln_b)**2+(epsln_c-epsln_b)**2)**0.5\n",
+      "epsln_II=1.0/2*(epsln_a+epsln_c)-(1.0/2**0.5)*((epsln_a-epsln_b)**2+(epsln_c-epsln_b)**2)**0.5\n",
+      "print \"\\nfirst principal strain, epsln_I = %4.3e\"%epsln_I\n",
+      "print \"\\nsecond principal strain, epsln_II = %4.3e\"%epsln_II\n",
+      "\n",
+      "#principal stresses\n",
+      "sigma_I=E/(1-v**2)*(epsln_I+v*epsln_II)\n",
+      "sigma_II=E/(1-v**2)*(v*epsln_I+epsln_II)\n",
+      "print \"\\nfirst principal stress, sigma_I= %4.1f N/mm^2\"%sigma_I #mistake in book\n",
+      "print \"\\nsecond principal stress, sigma_II= %4.1f N/mm^2\"%sigma_II\n",
+      "\n",
+      "sigma_x=sigma_I+sigma_II\n",
+      "print \"\\nstress in x direction,sigma_x = %4.1f N/mm^2\"%sigma_x\n",
+      "\n",
+      "#axial tensile load calculation\n",
+      "A=math.pi*d**2/4\n",
+      "P=sigma_x*A\n",
+      "print \"\\naxial load, P = %4.1f kN\"%(P/1000)\n",
+      "\n",
+      "tau_xy=1.0/2*((sigma_x/2-sigma_II)**2*4-sigma_x**2)**0.5\n",
+      "print \"\\nshear stress, tau_xy = %4.1f N/mm^2\"%tau_xy\n",
+      "\n",
+      "#torque calculation\n",
+      "J=math.pi*d**4/32\n",
+      "T=tau_xy*J/d*2\n",
+      "print \"\\ntorque applied on circular bar, T = %4.1f kNm\"%(T/10**6)\n",
+      "\n",
+      "sigma_x=E*epsln_a\n",
+      "print \"\\naxial stress from classical stress strain relationship, sigma_x =%4.0f N/mm^2\"%sigma_x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "first principal strain, epsln_I = 1.201e-03\n",
+        "\n",
+        "second principal strain, epsln_II = -5.015e-04\n",
+        "\n",
+        "first principal stress, sigma_I= 80.8 N/mm^2\n",
+        "\n",
+        "second principal stress, sigma_II= -10.8 N/mm^2\n",
+        "\n",
+        "stress in x direction,sigma_x = 70.0 N/mm^2\n",
+        "\n",
+        "axial load, P = 137.4 kN\n",
+        "\n",
+        "shear stress, tau_xy = 29.6 N/mm^2\n",
+        "\n",
+        "torque applied on circular bar, T =  0.7 kNm\n",
+        "\n",
+        "axial stress from classical stress strain relationship, sigma_x =  70 N/mm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter04.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter04.ipynb
new file mode 100755
index 00000000..365fa5fb
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter04.ipynb
@@ -0,0 +1,95 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:6a67de7af4fe617483b9aefeefdb01f122e2016e5ab212db494104a83b9deeda"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 Pg.No.106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "sum_FAFlBL=1263.6*10**6    # (N.mm)\n",
+      "sum_FAFlDL=880*10**6       # (N.mm)\n",
+      "E=200000                  # young's modulus (N/mm^2)\n",
+      "A=1800                    # area of each member (mm^2)\n",
+      "\n",
+      "#eqn 4.23 We=Wi\n",
+      "del_B=sum_FAFlBL/A/E\n",
+      "print \"\\ndeflection at point B =%2.2f mm\\n\"%(del_B)\n",
+      "\n",
+      "del_D=sum_FAFlDL/A/E\n",
+      "print \"deflection at point D =%2.2f mm\\n\"%(del_D)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "deflection at point B =3.51 mm\n",
+        "\n",
+        "deflection at point D =2.44 mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "ext_load=30    #external load(kN)\n",
+      "\n",
+      "#eqn 30*4*del_vB/3=F_BA*del_vB\n",
+      "F_BA=ext_load*4/3\n",
+      "print \"\\nforce in the struss AB is = %3.2f kN\\n\"%(F_BA)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "force in the struss AB is = 40.00 kN\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter05.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter05.ipynb
new file mode 100755
index 00000000..80ef0fbb
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter05.ipynb
@@ -0,0 +1,188 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:63d249c5c7d8f61cc0f4f00c6cd8c5c2324721f135176069c7c632118f5681cc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 05: Energy methods"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Pg.No.116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "A=1800        # cross sectional area (mm^2)\n",
+      "E=200000        #youngs modulus  (N/mm^2)\n",
+      "sum_FLdFdP_B=1268*10**6      #(N.mm^2)\n",
+      "sum_FLdFdP_D=880*10**6     #(N.mm^2)\n",
+      "\n",
+      "del_Bv=sum_FLdFdP_B/A/E\n",
+      "del_Dh=sum_FLdFdP_D/A/E\n",
+      "\n",
+      "print \"\\ndeflection at point B =%2.2f mm\\n\"%(del_Bv)\n",
+      "print \"deflection at point D =%2.2f mm\\n\"%(del_Dh)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "deflection at point B =3.52 mm\n",
+        "\n",
+        "deflection at point D =2.44 mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 Pg.No.142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "\n",
+      "a=np.array([[4.32,2.7],[2.7,11.62]])\n",
+      "b=np.array([27.1,48.11])\n",
+      "x=np.linalg.solve(a,b)\n",
+      "print \"\\nX1 = %1.2f kN & R2 = %1.2f kN\\n\"%(x[0],x[1]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "X1 = 4.31 kN & R2 = 3.14 kN\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10 Pg.No.144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "E=200000          #youngs modulus (N/mm^2)\n",
+      "A=200             #cross sectional area of each member (mm^2)\n",
+      "a=7*10**-6         #linear coefficient of heating (/C)\n",
+      "L=3*10**3          #length of BC (mm)\n",
+      "T=30               #temperature of truss (C)\n",
+      "sum_f2L=48000\n",
+      "\n",
+      "expansion=L*T*a\n",
+      "a11=sum_f2L/A/E\n",
+      "X1=-0.63/a11        #compatibility condition\n",
+      "print \"\\nX1 = %5.0f N\\n\"%(X1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "X1 =  -525 N\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12 Pg.No.151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "def_C=-1.05-.6            #deflection at C (mm)\n",
+      "L=300                      #length of cantilever (mm)\n",
+      "\n",
+      "theta_B=math.atan(def_C/L)\n",
+      "print \"\\ndeflection at C = %2.3f degree\\n\"%(theta_B*180/math.pi) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "deflection at C = -0.315 degree\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter06.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter06.ipynb
new file mode 100755
index 00000000..5195a30b
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter06.ipynb
@@ -0,0 +1,112 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:902d8cc266f9fb597a0d00273b10ab788d226cbd969bb1e960b465c84a750eef"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 06: Matrix methods"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 Pg.No.205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from sympy import solve, symbols, pprint\n",
+      "from sympy import diff\n",
+      "x,y=symbols('x,y')\n",
+      "\n",
+      "E=200000                #youngs modulus (N/mm^2)\n",
+      "nu=0.3                   #poissons ratio\n",
+      "u1=0.001\n",
+      "u2=0.003\n",
+      "u3=-0.003              #displacements of corners (m)\n",
+      "u4=0\n",
+      "v1=-0.004\n",
+      "v2=-0.002\n",
+      "v3=0.001\n",
+      "v4=0.001\n",
+      "\n",
+      "\n",
+      "a=np.array([[1,-2,-1,2],[1,2,-1,-2],[1,2,1,2],[1,-2,1,-2]])\n",
+      "b=np.array([u1,u2,u3,u4])\n",
+      "alpha=np.linalg.solve(a,b)\n",
+      "alpha1=alpha[0]\n",
+      "alpha2=alpha[1]\n",
+      "alpha3=alpha[2]\n",
+      "alpha4=alpha[3]\n",
+      "\n",
+      "a=np.array([[1,-2,-1,2],[1,2,-1,-2],[1,2,1,2],[1,-2,1,-2]])\n",
+      "b=np.array([v1,v2,v3,v4])\n",
+      "alpha=np.linalg.solve(a,b)\n",
+      "alpha5=alpha[0]\n",
+      "alpha6=alpha[1]\n",
+      "alpha7=alpha[2]\n",
+      "alpha8=alpha[3]\n",
+      "\n",
+      "u=alpha1+alpha2*x+alpha3*y+alpha4*x*y\n",
+      "v=alpha5+alpha6*x+alpha7*y+alpha8*x*y\n",
+      "\n",
+      "ex=diff(u,x)\n",
+      "ey=diff(v,y)\n",
+      "Yxy=diff(u,y)+diff(v,x)\n",
+      "\n",
+      "ex=-0.000125\n",
+      "ey=.002\n",
+      "Yxy=-0.0015\n",
+      "\n",
+      "sigma_x=E/(1-nu**2)*(ex+nu*ey)\n",
+      "sigma_y=E/(1-nu**2)*(ey+nu*ex)\n",
+      "print \"longitudinal stress in x direction = %3.1f N/mm^2\"%(sigma_x)\n",
+      "print \"longitudinal stress in y direction = %3.1f N/mm^2\"%(sigma_y)\n",
+      "\n",
+      "T_xy=E/2/(1+nu)*Yxy\n",
+      "print \"shear stress at the center of plate = %3.1f N/mm^2\"%(T_xy)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "longitudinal stress in x direction = 104.4 N/mm^2\n",
+        "longitudinal stress in y direction = 431.3 N/mm^2\n",
+        "shear stress at the center of plate = -115.4 N/mm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter08.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter08.ipynb
new file mode 100755
index 00000000..65ccbf76
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter08.ipynb
@@ -0,0 +1,154 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d4ab160ecba8745ef027bb167a7abbd97977707c483b1d0a639c7f068e2016a8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 08: Columns"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 Pg.No.280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "E=75000         #youngs modulus(N/mm^2)\n",
+      "G=21000          #shear modulus (N/mm^2)\n",
+      "L=2              #length of column (m)\n",
+      "l1=75            #flange length (mm)\n",
+      "l2=37.5            #total length(mm)\n",
+      "t=2.5            #thickness(mm)\n",
+      "\n",
+      "A=t*(2*l2+l1)         #observed from Fig 8.17\n",
+      "\n",
+      "#chapter 16  Ixx=bd^3/12+Ab^2\n",
+      "Ixx=2*l2*t*l2**2+t*l1**3/12\n",
+      "Iyy=2*t*l2**3/12\n",
+      "\n",
+      "I0=Ixx+Iyy\n",
+      "\n",
+      "#eqn 18.11 J=SUM((s*t^3)/3)\n",
+      "J=2*l2*t**3/3+l1*t**3/3\n",
+      "\n",
+      "Gama=t*l2**3*l1**2/24\n",
+      "Iyy=0.22*10**5\n",
+      "L=2*10**3\n",
+      "P_CRxx=math.pi**2*E*Ixx/L**2\n",
+      "P_CRyy=math.pi**2*E*Iyy/L**2\n",
+      "P_CRo=A/I0*(G*J+math.pi**2*E*Gama/L**2)\n",
+      "\n",
+      "print \"P_CRxx = %3.2e N\\n\"%(P_CRxx)\n",
+      "print \"P_CRyy = %0.1e N\\n\"%(P_CRyy)\n",
+      "print \"P_CRO = %0.2e N\\n\"%(P_CRo)\n",
+      "\n",
+      "print \"therefore buckling in the column due to axial load=%0.1e N\\n\"%(P_CRyy)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "P_CRxx = 6.51e+04 N\n",
+        "\n",
+        "P_CRyy = 4.1e+03 N\n",
+        "\n",
+        "P_CRO = 2.22e+04 N\n",
+        "\n",
+        "therefore buckling in the column due to axial load=4.1e+03 N\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Pg.No.282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from sympy import solve, symbols, pprint\n",
+      "from sympy import diff\n",
+      "P=symbols('P')\n",
+      "\n",
+      "E=70000        #youngs modulus (N/mm^2)\n",
+      "G=30000         #shear modulus (N/mm^2)\n",
+      "x_S=-76.2        #position of shear center(mm)\n",
+      "l1=l2=100        #lengths (mm)\n",
+      "t=2              #thickness(mm)\n",
+      "\n",
+      "x_bar=2*t*l1*50/(3*l1)/2\n",
+      "\n",
+      "A=600         #area (mm^2)\n",
+      "Ixx=1.17*10**6    #second moment of area (mm^4)\n",
+      "Iyy=0.67*10**6    #second moment of area (mm^4)\n",
+      "I0=5.32*10**6     # total second moment of area (mm^4)\n",
+      "J=800              #torsion constant (mm^4)\n",
+      "Gama=2488*10**6     #(mm^6)\n",
+      "L=10**3             #(mm)\n",
+      "\n",
+      "P_CRxx=math.pi**2*E*Ixx/L**2\n",
+      "P_CRyy=math.pi**2*E*Iyy/L**2\n",
+      "P_CRo=A/I0*(G*J+math.pi**2*E*Gama/L**2)\n",
+      "\n",
+      "fun=P**2*(1-A*x_S**2/I0)-P*(P_CRxx+P_CRo)+P_CRxx*P_CRo\n",
+      "solution = solve(fun, P)\n",
+      "print \"lowest value of critical load = %1.2e N\\n\"%(min(solution))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lowest value of critical load = 1.68e+05 N\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter09.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter09.ipynb
new file mode 100755
index 00000000..119bd5b2
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter09.ipynb
@@ -0,0 +1,97 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:47a0669d4632eab71f162a005cd6d69531f951e47ad9f34d928b679209ff2467"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 09: Thin plates"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Pg.No.310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "h=400     #height of structure (mm)\n",
+      "l1=300     #length of small section (mm)\n",
+      "L=1200      #length of whole structure (mm)\n",
+      "E=70000     #youngs modulus (N/mm^2)\n",
+      "A_f=350     #area of flange (mm^2)\n",
+      "A_s=300    # area of stiffeners (mm^2)\n",
+      "t=2         #thickness (mm)\n",
+      "I=2000      #second moment of area (mm^4)\n",
+      "w=5\n",
+      "#eqn 9.29 tan(a)^4=(1+td/2A_F)/(1+td/2A_S)\n",
+      "alpha=math.atan(((1+t*h/2/A_f)/(1+t*l1/A_s))**0.25)\n",
+      "\n",
+      "#eqn 9.19\n",
+      "F_T=w*L/h+w/2/math.tan((alpha))\n",
+      "\n",
+      "#eqn 9.25 M_max=w*b^2*tan(a)/12/d\n",
+      "M_max=w*10**3*l1**2*math.tan(alpha)/12/h\n",
+      "print \"maximum value of bending moment = %1.1e N.mm \\n\"%(M_max)\n",
+      "\n",
+      "#eqn 9.23 P=w*b*tan(a)/d\n",
+      "P=w*l1*math.tan(alpha)/h\n",
+      "print \"compressive load in stiffener = %1.1f kN \\n\"%(P)\n",
+      "\n",
+      "#eqn 9.24 le=d/(4-2b/d)^0.5\n",
+      "le=h/(4-2*l1/h)**0.5\n",
+      "print \"equivalent length of stiffener = %3.1f mm\\n\"%(le)\n",
+      "\n",
+      "#eqn 8.7  Pcr=pi^2*E*I/le^2\n",
+      "P_CR=math.pi**2*E*I/le**2\n",
+      "print \"buckling load of a stiffener = %2.1f kN \\n\"%(P_CR)  #approx. value in book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum value of bending moment = 8.6e+04 N.mm \n",
+        "\n",
+        "compressive load in stiffener = 3.4 kN \n",
+        "\n",
+        "equivalent length of stiffener = 253.0 mm\n",
+        "\n",
+        "buckling load of a stiffener = 21589.8 kN \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter12.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter12.ipynb
new file mode 100755
index 00000000..63daef6d
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter12.ipynb
@@ -0,0 +1,111 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:217c546250d736c4a5ad9b469b73a9b7209b5ff1e5db79ce32388be8ee290870"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Structural components of aircraft"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Pg.No. 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "D=4              #diameter of rivet(mm)\n",
+      "t_skin=2.5         #thickness of fuselage skin(mm)\n",
+      "t_strap=1.2          #thickness of straps (mm)\n",
+      "T=120               #shear stress (N/mm^2)\n",
+      "A=math.pi*D**2/4    #rivet area\n",
+      "Pfb=2*A*T\n",
+      "\n",
+      "b=Pfb/t_skin/12.5/10+4\n",
+      "print \"allowable rivet spacing = %2.1f mm \\n\"%(b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "allowable rivet spacing = 13.7 mm \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 Pg.No.392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "l1=20          #distace between A&D (mm)\n",
+      "l2=25          #distace between C&D (mm)\n",
+      "l3=75          #distace between C&force location (mm)\n",
+      "P=5        #external load(kN)\n",
+      "r=(l1**2+l2**2)**0.5\n",
+      "sum_r2=2*l1**2+4*r**2\n",
+      "\n",
+      "#eqn 12.6 S=Pe*r/Sum(r2)\n",
+      "Pe=P*l3\n",
+      "S=Pe/sum_r2*r\n",
+      "print \"Resultant shear force in the A and B = %2.2f kN\\n\"%(S)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resultant shear force in the A and B = 2.45 kN\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter14.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter14.ipynb
new file mode 100755
index 00000000..f9eb3f9d
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter14.ipynb
@@ -0,0 +1,226 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b89c31a3eb13c71dc64b90ec9c5fad56707d91451229ae687a102dc5e6c2ee23"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14: Airframe loads"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1 Pg.No.407"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "W=45             #weight of aircraft (kN)\n",
+      "m=45/9.8          #mass of aircraft(k-kg)\n",
+      "a=3*9.8           #deceleration due to cable (ms^-2)\n",
+      "m1=4.5/9.8        #weight of aircraft after AA in Diagram (k-kg)\n",
+      "v0=25             #touch down speed (m/s)\n",
+      "alpha=10             #center line angle with ground (degree)\n",
+      "\n",
+      "T=m*a/math.cos(math.radians(alpha))\n",
+      "print \"tension in cable = %3.1f kN\\n\"%(T)\n",
+      "\n",
+      "R=W+T*math.sin(math.radians(alpha))\n",
+      "print \"load on each undercarriage strut = %2.1f kN\\n\"%(R/2/math.cos(math.radians(20)))\n",
+      "\n",
+      "N=T+m1*9.8*math.sin(math.radians(alpha))-m1*a*math.cos(math.radians(alpha))\n",
+      "S=m1*a*math.sin(math.radians(alpha))+m1*9.8*math.cos(math.radians(alpha))\n",
+      "print \"N and S forces are shown in Fig 14.4 N=%2.2f kN  S=%2.2f kN\\n\"%(N,S)\n",
+      "\n",
+      "s=v0**2/2/a\n",
+      "print \"length of deck covered = %2.2f m\\n\"%(s)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "tension in cable = 137.1 kN\n",
+        "\n",
+        "load on each undercarriage strut = 36.6 kN\n",
+        "\n",
+        "N and S forces are shown in Fig 14.4 N=124.57 kN  S=6.78 kN\n",
+        "\n",
+        "length of deck covered = 10.63 m\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2 Pg.No.409"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "g=9.8\n",
+      "W=250           #weight of aircraft(kN)\n",
+      "m=250/g        #mass of aircraft (K-kg)\n",
+      "I_CG=5.65*10**8     #inertia about center of mass(N.s^2.mm)\n",
+      "v0=3.7            #vertical velocity of undercarriage(m/s)\n",
+      "R_h=400          #horizontal reaction (kN)\n",
+      "R_v=1200         #vertical reaction (kN)\n",
+      "l=1             # nose wheel distance from ground (m)\n",
+      "d=2.5           # distance of CG from ground (m)\n",
+      "\n",
+      "#horizontal equilibrium\n",
+      "ax=R_h/m\n",
+      "\n",
+      "#vertical equilibrium\n",
+      "ay=(R_v-W)/m\n",
+      "\n",
+      "\n",
+      "alpha=(R_v*l+R_h*d)*10**6/I_CG\n",
+      "print \"angular acceleration of aircraft = %2.1f rad/s^2 \\n\"%(alpha)\n",
+      "\n",
+      "#v=v0+ay*t\n",
+      "t=v0/ay\n",
+      "print \"time taken for vertical velocity to become zero = %1.3f s\\n\"%(t)\n",
+      "\n",
+      "#w=w0+a*t\n",
+      "w=a*t\n",
+      "print \"angular velocity of aircraft = %1.2f rad/s \\n\"%(w)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angular acceleration of aircraft = 3.9 rad/s^2 \n",
+        "\n",
+        "time taken for vertical velocity to become zero = 0.099 s\n",
+        "\n",
+        "angular velocity of aircraft = 0.39 rad/s \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3 Pg.No.414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "W=8000             #weight of aircraft (N)\n",
+      "n=4.5              # wing loading\n",
+      "S=14.5              #wing area (m^2)\n",
+      "V=60                #speed (m/s)\n",
+      "rho=1.223           #density (kg/m^3)\n",
+      "alpha=13.75          #from Fig 14.8 (a)\n",
+      "C_Mcg=0.075           #from Fig 14.8 (a)\n",
+      "c=1.35                #mean chord (m)\n",
+      "\n",
+      "\n",
+      "L=n*W\n",
+      "C_L=L/(0.5*rho*V**2*S)\n",
+      "print \"lift coefficient of aircraft = %1.3f \\n\"%(C_L)\n",
+      "\n",
+      "#from Fig 14.8 (b)\n",
+      "l=4.18*math.cos(math.radians(alpha-2))+0.31*math.sin(math.radians(alpha-2))\n",
+      "print \"length of tail arm = %1.3f m \\n\"%(l)\n",
+      "\n",
+      "\n",
+      "C_L=C_L-c/l*C_Mcg\n",
+      "print \"lift coefficient =%1.3f \\n\"%(C_L)\n",
+      "\n",
+      "alpha=13.3\n",
+      "l=4.18*math.cos(math.radians(alpha-2))+0.31*math.sin(math.radians(alpha-2))\n",
+      "print \"Now tail arm length =  %2.3f m\\n\"%(l)\n",
+      "\n",
+      "L=0.5*rho*V**2*S*C_L\n",
+      "print \"Lift = %5.1f N\\n\"%(L)\n",
+      "\n",
+      "P=n*W-35000\n",
+      "print \"Tail Load = %5.1f N\\n\"%(P)\n",
+      "\n",
+      "D=0.5*rho*V**2*S*0.0875\n",
+      "print \"Drag = %5.1f N\\n\"%(D)\n",
+      "print \"Forward inertia force = %5.1f N\\n\"%(D)  #eqn 14.13"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lift coefficient of aircraft = 1.128 \n",
+        "\n",
+        "length of tail arm = 4.156 m \n",
+        "\n",
+        "lift coefficient =1.103 \n",
+        "\n",
+        "Now tail arm length =  4.160 m\n",
+        "\n",
+        "Lift = 35222.3 N\n",
+        "\n",
+        "Tail Load = 1000.0 N\n",
+        "\n",
+        "Drag = 2793.0 N\n",
+        "\n",
+        "Forward inertia force = 2793.0 N\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter15.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter15.ipynb
new file mode 100755
index 00000000..a849c8c6
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter15.ipynb
@@ -0,0 +1,76 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1bd6512f5b9ca8155cc1e9c3b04ad8207681499e095afaf4830a217017e989ea"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15: Fatigue"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.1 Pg.No.445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from scipy import integrate\n",
+      "from sympy import symbols\n",
+      "import math\n",
+      "\n",
+      "#eqn 15.30 K=S(pi*a)^1/2*alpha\n",
+      "S=175         #stress range (N/mm^2)\n",
+      "K=1708          #fracture toughness (n/mm^3/2)\n",
+      "alpha=1         #constant\n",
+      "n=4         #constant\n",
+      "C=40*10**-15      #rate of crack growth (mm/cycle)\n",
+      "ai=0.1            #initial crack length\n",
+      "\n",
+      "alpha_f=K**2/S**2/math.pi\n",
+      "\n",
+      "\n",
+      "Nf=2/C/(n-2)/(S*math.pi**0.5)**n*(1/ai**((n-2)/2)-1/alpha_f**((n-2)/2))\n",
+      "print \"Number of cycles to failure = %5.0f cycles\\n\"%(Nf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of cycles to failure = 26919 cycles\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter16.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter16.ipynb
new file mode 100755
index 00000000..72ff7ec3
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter16.ipynb
@@ -0,0 +1,280 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c3ab3016805321c8a28748b1937c77b2125357dfc661b20013df0624feaa950e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16: Bending of open and closed thin-walled beams \n"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.1 Pg. No.456"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Fig 16.6 Dimensions\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#variable declaration\n",
+      "l=300            #length of I section (mm)\n",
+      "b=200            #width of I section beam (mm)\n",
+      "w1=25            #width of center section of I beam (mm)\n",
+      "w2=20            #width of upper and lower section of I beam (mm)\n",
+      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",
+      "\n",
+      "#second moment of area Ixx\n",
+      "# Ixx=b*l^3/12\n",
+      "Ixx=b*l**3/12-(b-w1)*(l-w2-w2)**3/12\n",
+      "print \"\\nSecond moment of area of I section beam = %5.3e mm^4\"%(Ixx)\n",
+      "\n",
+      "#sigma_z=My/I  reference 16.9\n",
+      "# @ \n",
+      "y=150\n",
+      "sigma_z=M*y/Ixx\n",
+      "print \"\\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)\"%(sigma_z)\n",
+      "\n",
+      "# @\n",
+      "y=-150\n",
+      "sigma_z=M*y/Ixx\n",
+      "print \"\\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)\"%(sigma_z)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Second moment of area of I section beam = 1.937e+08 mm^4\n",
+        "\n",
+        "direct stress at the top of the I section (y=150) = 77.45 N/mm^2 (compression)\n",
+        "\n",
+        "direct stress at the bottom of the I section (y=-150) = -77.45 N/mm^2 (tension)\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2 Pg. No.457"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Fig 16.6 reference\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#variable declaration\n",
+      "l=300            #length of I section (mm)\n",
+      "b=200            #width of I section beam (mm)\n",
+      "w1=25            #width of center section of I beam (mm)\n",
+      "w2=20            #width of upper and lower section of I beam (mm)\n",
+      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",
+      "\n",
+      "# second moment of area\n",
+      "# Iyy=wb^3/12\n",
+      "Iyy=2*20*200**3/12+260*25**3/12\n",
+      "print \"\\nSecond moment of area of I section beam = %5.3e mm^4\"%(Iyy)\n",
+      "\n",
+      "#sigma_z=Mx/I  \n",
+      "# @ \n",
+      "x=100\n",
+      "sigma_z=M*x/Iyy\n",
+      "print \"\\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)\"%(sigma_z)\n",
+      "\n",
+      "# @\n",
+      "x=-100\n",
+      "sigma_z=M*x/Iyy\n",
+      "print \"\\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)\"%(sigma_z)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Second moment of area of I section beam = 2.701e+07 mm^4\n",
+        "\n",
+        "direct stress at the top of the I section (y=150) = 370.30 N/mm^2 (compression)\n",
+        "\n",
+        "direct stress at the bottom of the I section (y=-150) = -370.30 N/mm^2 (tension)\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3 Pg.No.458 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#variable declaration\n",
+      "l=300            #length of I section (mm)\n",
+      "b=200            #width of I section beam (mm)\n",
+      "w1=25            #width of center section of I beam (mm)\n",
+      "w2=20            #width of upper and lower section of I beam (mm)\n",
+      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",
+      "theta=30         #angle at which bending moment is applied(degree)\n",
+      "\n",
+      "Mx=M*math.cos(math.radians(30))\n",
+      "My=M*math.sin(math.radians(30))\n",
+      "\n",
+      "# sigma_z=Mx/Ixx*y+My/Iyy*x\n",
+      "# @top left hand corner\n",
+      "y=150\n",
+      "x=-100\n",
+      "sigma_z=Mx/Ixx*y-My/Iyy*x  \n",
+      "print \"\\ndirect stress at the top left hand corner = %3.1f N/mm^2 (tension)\"%(sigma_z)\n",
+      "\n",
+      "# @ top right hand corner\n",
+      "x=100\n",
+      "y=150\n",
+      "sigma_z=Mx/Ixx*y-My/Iyy*x  \n",
+      "print \"\\ndirect stress at the top right hand corner = %3.1f N/mm^2 (compression)\"%(sigma_z)\n",
+      "\n",
+      "alpha=math.atan(My*Ixx/Mx/Iyy)\n",
+      "print \"\\ninclination = %3.1f degree\\n\"%(alpha*180/math.pi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "direct stress at the top left hand corner = 252.2 N/mm^2 (tension)\n",
+        "\n",
+        "direct stress at the top right hand corner = -118.1 N/mm^2 (compression)\n",
+        "\n",
+        "inclination = 76.4 degree\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.4 Pg.No.466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# reference Fig 16.13\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "#variable declaration\n",
+      "l=80           #length of one section (mm)\n",
+      "w=8            #thickness of each section (mm)\n",
+      "b1=40          #width of one section (mm)\n",
+      "b2=80          #width of other section (mm)\n",
+      "Mx=1500*10*3   #bending moment (N.mm)\n",
+      "My=0\n",
+      "\n",
+      "#centroid\n",
+      "# in this example C is taken at top surface x axis aligned to AB and y axis is \n",
+      "# aligned to E surface \n",
+      "y_bar=((b1+b2)*w*w/2+l*w*(w+l/2))/(l*w+(b1+b2)*w)\n",
+      "print \"\\ny coordinate of centroid = %3.1f mm\"%(y_bar)\n",
+      "\n",
+      "x_bar=((b1+b2)*w*((b1+b2)/2-b1+w/2)+l*w*(w/2))/(l*w+(b1+b2)*w)\n",
+      "print \"\\nx coordinate of centroid = %3.1f mm\"%(x_bar)\n",
+      "\n",
+      "#second area of moment\n",
+      "#IB=IC+Ab^2              IB=moment about any point,IC=moment about centroid, b=distace between both points\n",
+      "Ixx=(b1+b2)*w**3/12+(b1+b2)*w*(y_bar-w/2)**2+w*l**3/12+l*w*(w+l/2-y_bar)**2\n",
+      "print \"\\nsecond moment of area about x axis = %3.2e mm^4\"%(Ixx)\n",
+      "\n",
+      "Iyy=w*(b1+b2)**3/12+(b1+b2)*w*((b1+b2)/2-b1+w/2-x_bar)**2+l*w**3/12+l*w*(x_bar-w/2)**2\n",
+      "print \"\\nsecond moment of area about y axis = %3.2e mm^4\"%(Iyy)\n",
+      "\n",
+      "Ixy=(b1+b2)*w*(y_bar-w/2)*((b1+b2)/2-b1+w/2-x_bar)+l*w*(w+l/2-y_bar)*(x_bar-w/2)\n",
+      "print \"\\nsecond moment of area about x and y axis = %3.2e mm^4\"%(Ixy)\n",
+      "\n",
+      "Mx=15*10**5\n",
+      "def f(x,y):\n",
+      "        return Mx*(Iyy*y-Ixy*x)/(Ixx*Iyy-Ixy**2)\n",
+      "sigma_z_max= f(-8,-66.4)\n",
+      "print \"\\nmaximum direct shear stress = %3.0f N/mm^2 (compressive)\\n\"%(sigma_z_max)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "y coordinate of centroid = 21.6 mm\n",
+        "\n",
+        "x coordinate of centroid = 16.0 mm\n",
+        "\n",
+        "second moment of area about x axis = 1.09e+06 mm^4\n",
+        "\n",
+        "second moment of area about y axis = 1.31e+06 mm^4\n",
+        "\n",
+        "second moment of area about x and y axis = 3.38e+05 mm^4\n",
+        "\n",
+        "maximum direct shear stress = -96 N/mm^2 (compressive)\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter18.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter18.ipynb
new file mode 100755
index 00000000..b0550214
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter18.ipynb
@@ -0,0 +1,155 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a16e0822e0e7167e54e1f4ffc425aa7247562532992f7dee46fc87221365735a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18: Torsion of Beams"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.1 Pg.No.527"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#variable declaration\n",
+      "r=100                    #radius of cylinder (mm)\n",
+      "l=2*10**3                #length of cylinder (mm)\n",
+      "torque=30*10**6           #torque at mid point (N.mm)\n",
+      "Tmax=15*10**6            \n",
+      "Tau_max=200               #max shear stress (N/mm^2)\n",
+      "G=25000                   #shear modulus (N/mm^2)\n",
+      "theta =2                # rotation at mid point (degree)\n",
+      "\n",
+      "#ref equation 18.1    T=2Aq\n",
+      "tmin1=Tmax/(2*math.pi*r**2*Tau_max)\n",
+      "print \"\\nminimum thickness of beam due to stress limited to 200N/mm^2 = %3.1f mm\"%(tmin1)\n",
+      "\n",
+      "z=1*10**3                   # twist at mid point (mm)\n",
+      "#equation 18.4 after integration\n",
+      "tmin2=Tmax*200*math.pi*z/(4*(math.pi*r**2)**2*G*theta*math.pi/180)\n",
+      "print \"\\nminimum thickness due to constraint on maximum angle twist = %3.1f mm\"%(tmin2)\n",
+      "t=max(tmin1,tmin2)\n",
+      "print \"\\nthickness which satisfies both conditions = %3.1f mm\"%(t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "minimum thickness of beam due to stress limited to 200N/mm^2 = 1.2 mm\n",
+        "\n",
+        "minimum thickness due to constraint on maximum angle twist = 2.7 mm\n",
+        "\n",
+        "thickness which satisfies both conditions = 2.7 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.3 Pg.No 540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#reference Fig 18.12\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from sympy import symbols\n",
+      "from sympy import integrate\n",
+      "\n",
+      "S1,S2,S3=symbols('S1,S2,S3')\n",
+      "\n",
+      "#variable declaration\n",
+      "s1=50             #length of channel (mm)\n",
+      "s2=25             #width of channel (mm)\n",
+      "t1=2.5            #thickness of web(mm)\n",
+      "t2=1.5             #thickness of flange(mm)\n",
+      "T=10*10**3         #torque applied (N.mm)\n",
+      "zeta_s=8.04         \n",
+      "G=25000             #shear modulus (N/mm^2)\n",
+      "\n",
+      "#J=Sum(st^3/3)\n",
+      "J=1/3*(s1*t1**3+2*s2*t2**3)\n",
+      "tau_max=t1*T/J\n",
+      "print \"maximum shear stress = %3.1f N/mm^2\"%(tau_max)\n",
+      "\n",
+      "#in region O2\n",
+      "s1=25\n",
+      "ARf=1/2*zeta_s*s1\n",
+      "ws=-2*ARf*T/(G*J)\n",
+      "print \"warping in O2 region is linear = %3.2f mm\"%(ws)\n",
+      "\n",
+      "#in the wall 21\n",
+      "AR=1/2*zeta_s*s2-0.5*25*s2\n",
+      "w21=-0.03*(zeta_s-s2)\n",
+      "s1=50\n",
+      "tds=2*s2*t2+s1*t1\n",
+      "\n",
+      "A12=25/2*S1\n",
+      "A23=312.5-4.02*S2\n",
+      "A34=111.5+25/2*S3\n",
+      "AR1=1/2*(integrate(A12,(S1,0,25))+integrate(5*(A23),(S2,0,50))+integrate(3*A34,(S3,0,25)))/200\n",
+      "\n",
+      "#equation 18.20\n",
+      "ws=-2*ARf*T/G/J\n",
+      "print \"warping distribution in flange 34 = %1.2f \\n\"%(ws)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum shear stress = 78.9 N/mm^2\n",
+        "warping in O2 region is linear = -0.25 mm\n",
+        "warping distribution in flange 34 = -0.25 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter19.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter19.ipynb
new file mode 100755
index 00000000..c6a6c12b
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter19.ipynb
@@ -0,0 +1,160 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8abe3316b17e7378affc20eb461733ec12ced2dbb39d65e4c948dd2ef8dc33e9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 19: Combined open and closed section beams"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.1 Pg.No. 552"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from scipy import integrate\n",
+      "from sympy import *\n",
+      "import math\n",
+      "\n",
+      "l1=100               #lengths shown in Fig(mm)\n",
+      "l2=200\n",
+      "F=100*10**3            #force applied (N)\n",
+      "\n",
+      "y_bar=(2*l1*2*50+2*l2*2*l1+l2*2*l2)/(4*l1*2+4*l2*2)\n",
+      "Ixx=2*(2*l1**3/12+2*l1*25**2)+400*2*75**2+l2*2*125**2+2*(2*l2**3/12+2*l2*25**2)\n",
+      "\n",
+      "s1=Symbol('s1')\n",
+      "q12=-round(10**4*F/Ixx)/10**4*(integrate(-50+2*s1,s1))\n",
+      "print \"\\nq12 = %s\"%(q12)\n",
+      "\n",
+      "s2=Symbol('s2')\n",
+      "q23=-round(10**4*F/Ixx)/10**4*(integrate(2*75,s2))-34.5\n",
+      "print \"\\nq23 = %s\"%(q23)\n",
+      "\n",
+      "s3=Symbol('s3')\n",
+      "q03=-round(10**4*F/Ixx)/10**4*(integrate(2*75,s3))\n",
+      "print \"\\nq03 = %s\"%(q03)\n",
+      "\n",
+      "s4=Symbol('s4')\n",
+      "q34=-round(10**4*F/Ixx)/10**4*(integrate(2*(75-s4),s4))-242.5\n",
+      "print \"\\nq34 = %s\"%(q34)\n",
+      "\n",
+      "s5=Symbol('s5')\n",
+      "q94=-round(10**4*F/Ixx)/10**4*(integrate(-2*125,s5))\n",
+      "print \"\\nq94 = %s\\n\"%(q94)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "q12 = -0.0069*s1**2 + 0.345*s1\n",
+        "\n",
+        "q23 = -1.035*s2 - 34.5\n",
+        "\n",
+        "q03 = -1.035*s3\n",
+        "\n",
+        "q34 = 0.0069*s4**2 - 1.035*s4 - 242.5\n",
+        "\n",
+        "q94 = 1.725*s5\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.2 Pg.No.555"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from scipy import integrate\n",
+      "from sympy import symbols\n",
+      "import math\n",
+      "\n",
+      "A=20000                #nose cell area (mm^2)\n",
+      "L_w=900                #outer wall (mm)\n",
+      "L=300                  #width of wall (mm)\n",
+      "length=600           #length of open section (mm)\n",
+      "G=25000                #shear modulus (N/mm^2)\n",
+      "T=10*10**6             #torque applied (kN.m)\n",
+      "t1=1.5                  #thickness of closed section\n",
+      "t2=2                    #thickness of open section\n",
+      "\n",
+      "GJ_cl=4*A**2*G/(L_w+L)*t1\n",
+      "\n",
+      "print \"torsoinal rigidity of closed section = %2.2e N.mm^2 \\n\"%(GJ_cl)\n",
+      "\n",
+      "GJ_op=G*(length+L)*t2**3/3\n",
+      "print \"torsional rigidity of open section = %2.1e N.mm^2 \\n\"%(GJ_op)\n",
+      "\n",
+      "GJ=GJ_cl+GJ_op\n",
+      "print \"total torsional rigidity = %5.3e N.mm^2\\n\"%(GJ)\n",
+      "\n",
+      "dO_by_dz=T/GJ\n",
+      "print \"angle of twist per unit length = %1.4f rad/mm\\n\"%(dO_by_dz)\n",
+      "\n",
+      "q_cl=GJ_cl/2/A*dO_by_dz\n",
+      "print \"maximum shear stress in the closed section = %3.1f N/mm^2\\n\"%(q_cl/1.5)\n",
+      "\n",
+      "#eqn 18.10 T_max=GtdO/dz\n",
+      "T_max=G*t2*dO_by_dz\n",
+      "print \"maximum shear stress in the open section = %2.0f N/mm^2\\n\"%(T_max)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "torsoinal rigidity of closed section = 5.00e+10 N.mm^2 \n",
+        "\n",
+        "torsional rigidity of open section = 6.0e+07 N.mm^2 \n",
+        "\n",
+        "total torsional rigidity = 5.006e+10 N.mm^2\n",
+        "\n",
+        "angle of twist per unit length = 0.0002 rad/mm\n",
+        "\n",
+        "maximum shear stress in the closed section = 166.5 N/mm^2\n",
+        "\n",
+        "maximum shear stress in the open section = 10 N/mm^2\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter20.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter20.ipynb
new file mode 100755
index 00000000..4595d055
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter20.ipynb
@@ -0,0 +1,286 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bb71ed7156eb25f3e97cd3ba07ae31557b447cbcd71b351505ee32010461a02e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 20: Structural idealization"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.1 Pg.No.560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "sigma6=200\n",
+      "sigma1=200\n",
+      "sigma2=sigma5=150         #these are not real shear stress but taken \n",
+      "sigma3=sigma4=100          #proportional to length because we need just ratio\n",
+      "l16=400                     \n",
+      "t16=3\n",
+      "l12=l21=600\n",
+      "t12=t21=2                #thickness and lengths as shown in Fig 20.4\n",
+      "l23=600\n",
+      "t23=1.5\n",
+      "l25=300\n",
+      "t25=2.5\n",
+      "l34=200\n",
+      "t34=2\n",
+      "#eqn 20.1 B1=t_D*b/6*(2+sigma_2/sigma_1)\n",
+      "#eqn 20.2 B2=t_D*b/6*(2+sigma_1/sigma_2)\n",
+      "B1=B6=300+l16*t16/6*(2-sigma6/sigma1)+l12*t12/6*(2+sigma2/sigma1)\n",
+      "print \"B1=B6=%5.2f mm^2\\n\"%(B1)\n",
+      "\n",
+      "B2=B5=2*300+l12*t12/6*(2+sigma1/sigma2)+t25*l25/6*(2-sigma5/sigma2)+l23*t23/6*(2+sigma3/sigma2)\n",
+      "print \"B2=B5=%5.2f mm^2\\n\"%(B2)\n",
+      "\n",
+      "B3=300+l23*t23/6*(2+sigma2/sigma3)+l34*t34/6*(2-sigma4/sigma3)\n",
+      "print \"B3=B4=%5.2f mm^2\\n\"%(B3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "B1=B6=1050.00 mm^2\n",
+        "\n",
+        "B2=B5=1791.67 mm^2\n",
+        "\n",
+        "B3=B4=891.67 mm^2\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.2 Pg.No.562"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "Mx=100*10**6          #bending moment(N.mm)\n",
+      "y=[660,600,420,228,25,-204,-396,-502,-540]\n",
+      "B=[640,600,600,600,620,640,640,850,640]\n",
+      "\n",
+      "print \"direct stress in each boom in last column\"\n",
+      "print \"Boom\\t y(mm)\\t B(mm^2)\\t delIxx=By^2\\t sigma_z\"\n",
+      "for i in range (0,9):\n",
+      "    print \"%1.0f \\t %3.0f \\t %3.0f \\t \\t%2.1e \\t %2.1f\"%(i+1,y[i],B[i],B[i]*y[i]**2,Mx*y[i]/(1854*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "direct stress in each boom in last column\n",
+        "Boom\t y(mm)\t B(mm^2)\t delIxx=By^2\t sigma_z\n",
+        "1 \t 660 \t 640 \t \t2.8e+08 \t 35.6\n",
+        "2 \t 600 \t 600 \t \t2.2e+08 \t 32.4\n",
+        "3 \t 420 \t 600 \t \t1.1e+08 \t 22.7\n",
+        "4 \t 228 \t 600 \t \t3.1e+07 \t 12.3\n",
+        "5 \t  25 \t 620 \t \t3.9e+05 \t 1.3\n",
+        "6 \t -204 \t 640 \t \t2.7e+07 \t -11.0\n",
+        "7 \t -396 \t 640 \t \t1.0e+08 \t -21.4\n",
+        "8 \t -502 \t 850 \t \t2.1e+08 \t -27.1\n",
+        "9 \t -540 \t 640 \t \t1.9e+08 \t -29.1\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.3 Pg.No.566"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "Ixx=48*10**6\n",
+      "Sy=4.8*10**3\n",
+      "B=300\n",
+      "\n",
+      "q12=-Sy/Ixx*B*200       #until point 2 \n",
+      "q23=q12-Sy/Ixx*B*200\n",
+      "q34=q23-Sy/Ixx*B*(-200)\n",
+      "print \"shear flow in flange 12 = %2.0f N/mm\\n\"%(q12)\n",
+      "print \"shear flow in web 23 = %2.0f N/mm\\n\"%(q23)\n",
+      "print \"shear flow in flange 34 = %2.0f N/mm\\n\"%(q34)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear flow in flange 12 = -6 N/mm\n",
+        "\n",
+        "shear flow in web 23 = -12 N/mm\n",
+        "\n",
+        "shear flow in flange 34 = -6 N/mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.4 Pg.No.569"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "B=[200,250,400,100,100,400,250,200]\n",
+      "Ixx=13.86*10**6\n",
+      "Sy=10*10**3\n",
+      "qb23=0\n",
+      "qb34=qb23-Sy/Ixx*B[2]*100\n",
+      "qb45=qb34-Sy/Ixx*B[3]*50\n",
+      "qb56=qb34\n",
+      "qb67=qb23\n",
+      "qb21=qb67-Sy/Ixx*(B[1]*100)\n",
+      "qb18=qb21-Sy/Ixx*B[7]*30\n",
+      "qb87=qb21\n",
+      "qs0=-5.4\n",
+      "print \"Distribution of shear flow :\"\n",
+      "print \"q23 = %2.1f N/mm\"%(qb23+qs0)\n",
+      "print \"q21 = %2.1f N/mm\"%(qb21-qs0)\n",
+      "print \"q34 = %2.1f N/mm\"%(qb34-qs0)\n",
+      "print \"q45 = %2.1f N/mm\"%(qb45-qs0)\n",
+      "print \"q56 = %2.1f N/mm\"%(qb56-qs0)\n",
+      "print \"q67 = %2.1f N/mm\"%(qb67+qs0)\n",
+      "print \"q18 = %2.1f N/mm\"%(qb18-qs0)\n",
+      "print \"q87 = %2.1f N/mm\\n\"%(qb87-qs0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Distribution of shear flow :\n",
+        "q23 = -5.4 N/mm\n",
+        "q21 = -12.6 N/mm\n",
+        "q34 = -23.5 N/mm\n",
+        "q45 = -27.1 N/mm\n",
+        "q56 = -23.5 N/mm\n",
+        "q67 = -5.4 N/mm\n",
+        "q18 = -17.0 N/mm\n",
+        "q87 = -12.6 N/mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.5 Pg.No.575"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from sympy import symbols, integrate\n",
+      "z=symbols('z')\n",
+      "\n",
+      "E=70000              #youngs modulus (N/mm^2)\n",
+      "G=30000              #shear modulus (N/mm^2)\n",
+      "P=4.8*10**3          #applied force (N)\n",
+      "L=2000               #cantilever length(mm)\n",
+      "Sy=P\n",
+      "Ixx=48*10**6         #second moment of area \n",
+      "t=1                   #actual thickness (mm)\n",
+      "\n",
+      "Mx0=-Sy*(L-z)\n",
+      "Mx1=-(L-z)\n",
+      "\n",
+      "del_M=integrate(Mx0*Mx1/E/Ixx,(z,0,L))\n",
+      "del_S=integrate((1/G/t/Sy*(6**2*200+12**2*400+6**2*200)),(z,0,L))\n",
+      "print \"total deflection in vertical direction = %1.2f mm\\n\"%(del_M+del_S)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total deflection in vertical direction = 4.81 mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter21.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter21.ipynb
new file mode 100755
index 00000000..1195a6a9
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter21.ipynb
@@ -0,0 +1,256 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b8ffe836d5b15b63c70d69d0a02ef23ccdf3f03c0f15c98f0fa5e62d24f5d894"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 21: Wing spars and box beams"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 21.1 Pg.No.585"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from scipy import integrate\n",
+      "import math\n",
+      "\n",
+      "Mx=20*10**6         #internal bending moment (N.mm)\n",
+      "Sy=-20*10**3           #force applied(N)\n",
+      "l1=400\n",
+      "l2=300                #lengths of webs respectively from left to right in Fig 21.2\n",
+      "l3=200\n",
+      "\n",
+      "#Ixx=I_C+Ab^2\n",
+      "Ixx=2*l1*150**2+2*l2**3/12\n",
+      "y=150\n",
+      "sigma_z1=Mx*y/Ixx\n",
+      "sigma_z2=-Mx*y/Ixx\n",
+      "\n",
+      "Pz1=sigma_z1*l1\n",
+      "Pz2=sigma_z2*l1\n",
+      "\n",
+      "#from fig 21.1 and 21.2\n",
+      "del_y1_by_del_z=-0.05\n",
+      "del_y2_by_del_z=0.05\n",
+      "Syw=Sy-Pz1*del_y1_by_del_z-Pz2*del_y2_by_del_z\n",
+      "\n",
+      "# equ 21.6 qs=-Syw/Ixx*integrate(t_D*y*ds+B1*Y1) from 0 to s\n",
+      "fn=lambda s:300-2*s\n",
+      "\n",
+      "def shear(s):\n",
+      "    return -Syw/Ixx*(integrate.quad(fn, 0, s)[0]+400*150)\n",
+      "print \"maximum value of shear flow @s=150mm =%2.1f N/mm\\n\"%(shear(150.0))\n",
+      "print \"value of shear @s=0mm = %2.1f N/mm\\n\"%(shear(0))\n",
+      "print \"value of shear @s=300 mm = %2.1f N/mm\\n\"%(shear(300))\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum value of shear flow @s=150mm =53.8 N/mm\n",
+        "\n",
+        "value of shear @s=0mm = 39.1 N/mm\n",
+        "\n",
+        "value of shear @s=300 mm = 39.1 N/mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 21.2 Pg.No.589"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import symbols\n",
+      "import math\n",
+      "\n",
+      "Sy=100*10**6             #load applied in y direction (N)\n",
+      "Sx=0                     #load in x direction (N)\n",
+      "y=[0.3,0.3,0.3,-0.3,-0.3,-0.3]\n",
+      "B=[900,1200,900,900,1200,900]  #boom area\n",
+      "dx_dz=[0.1,0,-0.1,-0.1,0,0.1]\n",
+      "dy_dz=[-0.05,-0.05,-0.05,0.05,0.05,0.05]\n",
+      "Er=[0.6,0,0.6,0.6,0,0.6]            \n",
+      "nr=[0.3,0.3,0.3,0.3,0.3,0.3]\n",
+      "Mx=-Sy*2\n",
+      "My=-Sx*2\n",
+      "Ixy=0\n",
+      "Ixx=4*900*300**2+2*1200*300**2\n",
+      "Pzr=[0]*6\n",
+      "Pxr=[0]*6                #array initialization\n",
+      "Pyr=[0]*6\n",
+      "Pr=[0]*6\n",
+      "print \"Boom \\t Pzr(kN) \\t dx_dz \\t dy_dz \\t Pxr(kN) \\t Pyr(kN) \\t Pr(kN)\"\n",
+      "for i in range (0,6):\n",
+      "    Pzr[i]=Mx*y[i]/Ixx*B[i]\n",
+      "    Pxr[i]=Pzr[i]*dx_dz[i]                 #Pr is not correct in (7) column in book\n",
+      "    Pyr[i]=Pzr[i]*dy_dz[i]\n",
+      "    Pr[i]=(Pzr[i]**2+Pxr[i]**2+Pyr[i]**2)**0.5\n",
+      "    if Pzr[i]>0 :\n",
+      "        print \"%1.0f \\t %3.0f \\t\\t %1.1f \\t %1.2f \\t %2.0f \\t\\t %1.1f \\t\\t %3.1f\"%(i+1,Pzr[i],dx_dz[i],dy_dz[i],Pxr[i],Pyr[i],Pr[i])\n",
+      "    else:\n",
+      "        print \"%1.0f \\t %3.0f \\t\\t %1.1f \\t %1.2f \\t %2.0f \\t\\t %1.1f \\t\\t %3.1f\"%(i+1,Pzr[i],dx_dz[i],dy_dz[i],Pxr[i],Pyr[i],-Pr[i])\n",
+      "       \n",
+      "Sxw=0\n",
+      "Syw=66.6*10**3\n",
+      "qb16=0        #open section here\n",
+      "qb12=qb16-Syw/Ixx*B[0]*300\n",
+      "qb23=qb12-Syw/Ixx*B[1]*300\n",
+      "qb34=qb23-Syw/Ixx*B[2]*300\n",
+      "qb45=qb23\n",
+      "qb56=qb12\n",
+      "qs0=-97\n",
+      "\n",
+      "\n",
+      "print \"\\n\\nqb16 = %3.1f N/mm\"%(qb16)\n",
+      "print \"qb12 = %3.1f N/mm\"%(qb12)\n",
+      "print \"qb23 = %3.1f N/mm\"%(qb23)\n",
+      "print \"qb34 = %3.1f N/mm\"%(qb34)\n",
+      "print \"qb45 = %3.1f N/mm\"%(qb45)\n",
+      "print \"qb56 = %3.1f N/mm\"%(qb56)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Boom \t Pzr(kN) \t dx_dz \t dy_dz \t Pxr(kN) \t Pyr(kN) \t Pr(kN)\n",
+        "1 \t -100 \t\t 0.1 \t -0.05 \t -10 \t\t 5.0 \t\t -100.6\n",
+        "2 \t -133 \t\t 0.0 \t -0.05 \t -0 \t\t 6.7 \t\t -133.5\n",
+        "3 \t -100 \t\t -0.1 \t -0.05 \t 10 \t\t 5.0 \t\t -100.6\n",
+        "4 \t 100 \t\t -0.1 \t 0.05 \t -10 \t\t 5.0 \t\t 100.6\n",
+        "5 \t 133 \t\t 0.0 \t 0.05 \t  0 \t\t 6.7 \t\t 133.5\n",
+        "6 \t 100 \t\t 0.1 \t 0.05 \t 10 \t\t 5.0 \t\t 100.6\n",
+        "\n",
+        "\n",
+        "qb16 = 0.0 N/mm\n",
+        "qb12 = -33.3 N/mm\n",
+        "qb23 = -77.7 N/mm\n",
+        "qb34 = -111.0 N/mm\n",
+        "qb45 = -77.7 N/mm\n",
+        "qb56 = -33.3 N/mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 21.3 Pg.No.593\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import symbols\n",
+      "import numpy as np\n",
+      "import math\n",
+      "yr=symbols('yr')\n",
+      "\n",
+      "Mx=-100*1.9*10**6                #bending moment(N.mm)\n",
+      "Ixx=4*900*295**2+2*1200*295**2      #second moment of area (mm^4)\n",
+      "sigma_zr=Mx/Ixx*yr\n",
+      "P1_1=P3_1=-0.364*295*900\n",
+      "P4_1=P6_1=0.364*295*900\n",
+      "P2_1=-0.364*295*1200\n",
+      "P5_1=0.364*295*1200\n",
+      "\n",
+      "Mx=-100*2.1\n",
+      "Ixx=4*900*305**2+2*1200*305**2\n",
+      "sigma_zr=Mx/Ixx*yr\n",
+      "P1_2=P3_2=-0.376*305*900\n",
+      "P4_2=P6_2=0.376*305*900\n",
+      "P2_2=-0.376*305*1200\n",
+      "P5_2=0.376*305*1200\n",
+      "\n",
+      "deltaP1=deltaP3=(P1_1-P1_2)/200\n",
+      "deltaP4=deltaP6=-(P4_1-P4_2)/200\n",
+      "deltaP2=(P2_1-P2_2)/200\n",
+      "deltaP5=(P5_1-P5_2)/200\n",
+      "\n",
+      "a=np.array([[1,-1,0,0,0,0],[0,1,-1,0,0,0],[0,0,1,-1,0,0],[0,0,0,1,-1,0],[0,0,0,0,1,-1],[600*300*2,2*600*300,600*600,0,0,600*600]])\n",
+      "\n",
+      "b=np.array([43.8,32.85,-32.85,-43.8,-32.85,600*10**5])\n",
+      "q=np.linalg.solve(a,b)\n",
+      "print \"shear flow distribution :\"\n",
+      "print \"q12 = %2.1f N/mm\"%(q[0])\n",
+      "print \"q23 = %2.1f N/mm\"%(q[1])\n",
+      "print \"q34 = %2.1f N/mm\"%(q[2])\n",
+      "print \"q45 = %2.1f N/mm\"%(q[3])\n",
+      "print \"q56 = %2.1f N/mm\"%(q[4])\n",
+      "print \"q61 = %2.1f N/mm\\n\"%(q[5])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear flow distribution :\n",
+        "q12 = 63.6 N/mm\n",
+        "q23 = 19.8 N/mm\n",
+        "q34 = -13.1 N/mm\n",
+        "q45 = 19.8 N/mm\n",
+        "q56 = 63.6 N/mm\n",
+        "q61 = 96.4 N/mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 41
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter22.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter22.ipynb
new file mode 100755
index 00000000..15f86e3c
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter22.ipynb
@@ -0,0 +1,88 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7afcc7b44b13ea906e3e2317703c261e07426e5f23f7958a6b7966ba0274524d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 22: Fuselages"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.1 Pg.No.598"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "#B1=t_D*b/6(2+sigma_2/sigma_1)\n",
+      "Mx=200*10**6                     # (N*mm)\n",
+      "y=[381,352,269.5,145.8,0,-145.8,-269.5,-352,-381,-352,-269.5,-145.8,0,145.8,269.5,352]\n",
+      "sigma_2=sigma_16=352\n",
+      "sigma_1=381\n",
+      "A=100\n",
+      "B1=A+0.8*149.6/6*(2+sigma_2/sigma_1)*2\n",
+      "Ixx=2.52*10**8\n",
+      "def sigma_z(y):\n",
+      "    return Mx/Ixx*y\n",
+      "print \"Stringer/Boom \\t y(mm) \\t \\tsigma_z(N/mm^2)\"\n",
+      "for i in range (0,16):\n",
+      "    print \"%2.0f \\t\\t %3.1f\\t\\t\\t%3.1f \"%(i+1,y[i],sigma_z(y[i]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Stringer/Boom \t y(mm) \t \tsigma_z(N/mm^2)\n",
+        " 1 \t\t 381.0\t\t\t302.4 \n",
+        " 2 \t\t 352.0\t\t\t279.4 \n",
+        " 3 \t\t 269.5\t\t\t213.9 \n",
+        " 4 \t\t 145.8\t\t\t115.7 \n",
+        " 5 \t\t 0.0\t\t\t0.0 \n",
+        " 6 \t\t -145.8\t\t\t-115.7 \n",
+        " 7 \t\t -269.5\t\t\t-213.9 \n",
+        " 8 \t\t -352.0\t\t\t-279.4 \n",
+        " 9 \t\t -381.0\t\t\t-302.4 \n",
+        "10 \t\t -352.0\t\t\t-279.4 \n",
+        "11 \t\t -269.5\t\t\t-213.9 \n",
+        "12 \t\t -145.8\t\t\t-115.7 \n",
+        "13 \t\t 0.0\t\t\t0.0 \n",
+        "14 \t\t 145.8\t\t\t115.7 \n",
+        "15 \t\t 269.5\t\t\t213.9 \n",
+        "16 \t\t 352.0\t\t\t279.4 \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter23.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter23.ipynb
new file mode 100755
index 00000000..a5e90903
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter23.ipynb
@@ -0,0 +1,390 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:38d29ad436c9255138372fb573a0261c5aeb8d664f2491fbf03a71e42b1c4783"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 23: Wings"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.1 Pg.No.609"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import solve, symbols, pprint\n",
+      "import math\n",
+      "y=symbols('y')\n",
+      "\n",
+      "B1=B6=2580\n",
+      "B2=B5=3880          #Boom areas (mm^2)\n",
+      "B3=B4=3230\n",
+      "\n",
+      "l1=200\n",
+      "l2=230               #dimensions shown in Fig 23.3 (mm)\n",
+      "l3=165\n",
+      "\n",
+      "Mx=300*10**6         #bending moment (N.mm)\n",
+      "My=0\n",
+      "\n",
+      "Ixx=2*(B1*l3**2+B2*l2**2+B3*l1**2)\n",
+      "sigma_z=Mx/Ixx*y\n",
+      "\n",
+      "print \"direct stress in booms sigma_z = %s\" %(sigma_z)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "direct stress in booms sigma_z = 0.370651791174781*y\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.2 Pg.No.612"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import solve, symbols, pprint\n",
+      "import math\n",
+      "import numpy as np\n",
+      "\n",
+      "T=11.3*10**6      # torque applied (N.mm)\n",
+      "G_REF=27600        #(N/mm^2)\n",
+      "A1=258000\n",
+      "A2=355000          #areas in table\n",
+      "A3=161000\n",
+      "\n",
+      "#t*12=G/G_REF*t\n",
+      "t12=24200/G_REF*1.22\n",
+      "t12i=27600/G_REF*2.03\n",
+      "t13=t24=24200/G_REF*1.22     #G and thickness taken from table ex23.2 Pg.No.612\n",
+      "t35=t46=t56=20700/G_REF*0.92\n",
+      "t34=27600/G_REF*1.63\n",
+      "\n",
+      "# del12=ds/t*12\n",
+      "del12=1650/t12\n",
+      "del12i=508/t12i\n",
+      "del13=del24=775/t13\n",
+      "del34=380/t34               #lengths taken from table\n",
+      "del35=del46=508/t35\n",
+      "del56=254/t56\n",
+      "\n",
+      "\n",
+      "a=np.array([[del12+del12i,-del12i,0,-2*A1*G_REF],[-del12i,del12i+del13+del24+del34,-del34,-2*A2*G_REF],[0,-del34,del35+del46+del34+del56,-2*A3*G_REF],[A1,A2,A3,0]])\n",
+      "b=np.array([0,0,0,5.65*10**6])\n",
+      "x=np.linalg.solve(a,b)\n",
+      "print \"shear stress distribution is as follows\"\n",
+      "print \"q1=%1.1f N/mm\"%(x[0])\n",
+      "print \"q2=%1.1f N/mm\"%(x[1])\n",
+      "print \"q3=%1.1f N/mm\\n\"%(x[2])\n",
+      "print \"dO_dz=%1.2e \"%(x[3])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear stress distribution is as follows\n",
+        "q1=7.1 N/mm\n",
+        "q2=8.9 N/mm\n",
+        "q3=4.2 N/mm\n",
+        "\n",
+        "dO_dz=7.36e-07 \n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.3 Pg.No.616"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import solve, symbols, pprint\n",
+      "import math\n",
+      "import numpy as np\n",
+      "\n",
+      "A1=265000\n",
+      "A2=213000\n",
+      "A3=413000\n",
+      "\n",
+      "G_REF=27600\n",
+      "Sy=86.8*10**3\n",
+      "t78=3*27600/27600*1.22\n",
+      "del78=1270/t78\n",
+      "t12=t56=1.22\n",
+      "del12=del56=1023/t12\n",
+      "t23=1.63\n",
+      "del23=1274/t23\n",
+      "t34=2.03\n",
+      "del34=2200/t34\n",
+      "del38=57\n",
+      "del84=95\n",
+      "del87=347\n",
+      "del27=68\n",
+      "del75=106\n",
+      "del16=330/1.63\n",
+      "Ixx=809*10**6             #From example 23.1\n",
+      "\n",
+      "qb27=-99.4;qb16=-45.5;qb65=0;qb57=95.5;qb38=-69.8;qb48=69\n",
+      "\n",
+      "\n",
+      "a=np.array([     [del34+del84+del38,-del38,0,-2*A1*G_REF],       [-del38,del23+del38+del87+del27,-del27,-2*A2*G_REF],          [0,-del27,del56+del27+del75+del12+del16,-2*A3*G_REF],                    [2*A1,2*A2,2*A3,0]                                              ])\n",
+      "b=np.array([-10488,-2561,7426,19736700])\n",
+      "x=np.linalg.solve(a,b)\n",
+      "\n",
+      "qs01=5.5\n",
+      "qs02=10.2\n",
+      "qs03=16.5\n",
+      "\n",
+      "q34=qs01\n",
+      "q23=qs02\n",
+      "q12=qs03\n",
+      "q61=-qb16+qs03\n",
+      "q57=qb57-qs03\n",
+      "q72=-qb27-qs02\n",
+      "q48=qb48+qs01\n",
+      "q83=-qb38-qs01\n",
+      "\n",
+      "print \"shear flows distribution is as follows :\"\n",
+      "print \"q34=%1.2f N/mm\"%(q34)\n",
+      "print \"q23=q87=%1.2f N/mm\"%(q23)\n",
+      "print \"q12=q56=%1.2f N/mm\"%(q12)\n",
+      "print \"q61=%1.2f N/mm\"%(q61)             \n",
+      "print \"q57=%1.2f N/mm\"%(q57)\n",
+      "print \"q72=%1.2f N/mm\"%(q72)\n",
+      "print \"q48=%1.2f N/mm\"%(q48)\n",
+      "print \"q83=%1.2f N/mm\\n\"%(q83)\n",
+      "\n",
+      "print \"rate of twist = %1.1e rad/mm\"%(x[3])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear flows distribution is as follows :\n",
+        "q34=5.50 N/mm\n",
+        "q23=q87=10.20 N/mm\n",
+        "q12=q56=16.50 N/mm\n",
+        "q61=62.00 N/mm\n",
+        "q57=79.00 N/mm\n",
+        "q72=89.20 N/mm\n",
+        "q48=74.50 N/mm\n",
+        "q83=64.30 N/mm\n",
+        "\n",
+        "rate of twist = 1.1e-06 rad/mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.4 Pg.No.618 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import solve, symbols, pprint\n",
+      "import math\n",
+      "import numpy as np\n",
+      "\n",
+      "#Boom areas\n",
+      "B=[600,900,600,600,900,600]\n",
+      "Pz=[0,0,0,0,0,0]\n",
+      "y=[54.56,54.56,54.56,-54.56,-54.56,-54.56]\n",
+      "Ixx=4*600*90**2+2*900*90**2\n",
+      "Ixy=0\n",
+      "Mx=1.65*10**6\n",
+      "My=0\n",
+      "\n",
+      "a=np.array([[1700,-1520],[72000,144000]])\n",
+      "b=np.array([3942,690726])\n",
+      "x=np.linalg.solve(a,b)\n",
+      "print \"\\nqs0I = %2.1f N/mm\\n\"%(x[0])\n",
+      "print \"\\nqs0II = %2.1f N/mm\\n\"%(x[1])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "qs0I = 4.6 N/mm\n",
+        "\n",
+        "\n",
+        "qs0II = 2.5 N/mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.5 Pg.No.622"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import solve, symbols, pprint\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from sympy import integrate\n",
+      "z=symbols('z')\n",
+      "\n",
+      "E=69000             #youngs modulus (N/mm^2)\n",
+      "G=25900              #shear modulus (N/mm^2)\n",
+      "t=2                  #thickness (mm)\n",
+      "B=[650,1300,650,650,1300,650]   #boom area\n",
+      "q0=[9.6,-5.8,50.3,-5.8,9.6,54.1,73.6]\n",
+      "Sy0=44.5*10**3\n",
+      "Sy1=1\n",
+      "Mx0=-44.5*10**3*(2000-z)\n",
+      "Mx1=-(2000-z)\n",
+      "Ixx=81.3*10**6\n",
+      "int_q0q1_Gt=1/G/t/Sy0*(q0[0]**2*250*t+q0[1]**2*500*t+q0[2]**2*250+q0[5]**2*250+q0[6]**2*250)\n",
+      "\n",
+      "delta=integrate(Mx0*Mx1/E/Ixx,(z,0,2000))+integrate(int_q0q1_Gt,(z,0,2000))\n",
+      "print \"deflection at free end of the two cell = %2.2f mm\"%(delta)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "deflection at free end of the two cell = 23.58 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.6 Pg.No.624"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import solve, symbols, pprint\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from sympy import integrate\n",
+      "\n",
+      "T=10*10**6       #torque subjected (N.mm)\n",
+      "l1=800  \n",
+      "l2=200            #lengths shown in Fig23.17 (mm)\n",
+      "l3=1500\n",
+      "A=l2*l1\n",
+      "\n",
+      "\n",
+      "q=T/2/A\n",
+      "S=T/l1\n",
+      "q1=S/l2\n",
+      "P=S*l3/2/l2\n",
+      "\n",
+      "a=np.array([[1,-1],[1,1]])\n",
+      "b=np.array([31.3,62.5])\n",
+      "q=np.linalg.solve(a,b)\n",
+      "\n",
+      "print \"shear flow :\"\n",
+      "print \"q1=%2.2f N/mm\"%(q1)\n",
+      "print \"q2=%2.2f N/mm\"%(q[0])\n",
+      "print \"q3=%2.2f N/mm\\n\"%(q[1])\n",
+      "\n",
+      "print \"flange loads :\"\n",
+      "print \"P(st.4500) = 0\"\n",
+      "print \"P(st.3000) = %2.2f N (compression)\"%(l3*q[0]-l3*q[1])\n",
+      "print \"P(st.2250) = %2.1f\"%(46875-l3/2*q1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear flow :\n",
+        "q1=62.50 N/mm\n",
+        "q2=46.90 N/mm\n",
+        "q3=15.60 N/mm\n",
+        "\n",
+        "flange loads :\n",
+        "P(st.4500) = 0\n",
+        "P(st.3000) = 46950.00 N (compression)\n",
+        "P(st.2250) = 0.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 84
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter24.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter24.ipynb
new file mode 100755
index 00000000..b305a12a
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter24.ipynb
@@ -0,0 +1,198 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:217b3507f45091d99aa496d838ab43e552e1e20629747bd44a2cad35220ffe6a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 24: Fuselage frames and wing ribs"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.1 Pg.No.638"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import numpy\n",
+      "import math\n",
+      "#reference Fig 42.2\n",
+      "l1=250               #length of each section(mm)\n",
+      "l2=200                #length of DK(mm)\n",
+      "l3=100                 #length of KH(mm)\n",
+      "F1=4000\n",
+      "F2=5000                #forces as shown in Fig 24.1\n",
+      "theta=60               #force angle from DH (degree)\n",
+      "\n",
+      "#solve (i) and (ii) equation\n",
+      "a=numpy.array([[1,-1],[200,100]])\n",
+      "b=numpy.array([13.8564,2000])\n",
+      "q=numpy.linalg.solve(a,b)\n",
+      "q1=q[0]\n",
+      "q2=q[1]\n",
+      "q3=F1*math.cos(math.radians(theta))/(l2+l3)\n",
+      "q4=F1*math.cos(math.radians(theta))/(l2+l3)+F2/(l2+l3)\n",
+      "P_A=l1*q1+l1*q3+l1*q4\n",
+      "P_E=-l1*q2-l1*q3-l1*q4\n",
+      "P=3464.1\n",
+      "M_AE=F2*l1+F1/2*3*l1-P*50\n",
+      "\n",
+      "P_A=M_AE/(l2+l3)+P/2\n",
+      "P_E=-M_AE/(l2+l3)+P/2\n",
+      "\n",
+      "print \"shear flow as shown in Fig 24.1\\n\"\n",
+      "print \"q1=%2.2f N/mm\\n\"%(q1)\n",
+      "print \"q2=%2.2f N/mm\\n\"%(q2)\n",
+      "print \"q3=%2.2f N/mm\\n\"%(q3)\n",
+      "print \"q4=%2.2f N/mm\\n\"%(q4)\n",
+      "\n",
+      "print \"Stiffener load at point A = %2.2f N/mm(tension)\\n\"%(P_A)\n",
+      "print \"Stiffener load at point E = %2.2f N/mm(compression)\\n\"%(P_E)  #in book another method is also explained"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear flow as shown in Fig 24.1\n",
+        "\n",
+        "q1=11.29 N/mm\n",
+        "\n",
+        "q2=-2.57 N/mm\n",
+        "\n",
+        "q3=6.67 N/mm\n",
+        "\n",
+        "q4=23.33 N/mm\n",
+        "\n",
+        "Stiffener load at point A = 10321.37 N/mm(tension)\n",
+        "\n",
+        "Stiffener load at point E = -6857.27 N/mm(compression)\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.2 Pg.No.645"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import numpy as np\n",
+      "import math\n",
+      "\n",
+      "A1=50000\n",
+      "A2=95000        #area shown in Fig 24.13(mm^2)\n",
+      "A3=95000\n",
+      "\n",
+      "A4=46000\n",
+      "A5=49000         #area shown in Fig 24.11 (mm^2)\n",
+      "\n",
+      "F1=12000\n",
+      "F2=15000          #forces shown in Fig 24.9 (N)\n",
+      "\n",
+      "l1=l3=300\n",
+      "l2=320           #lengths and angle shown in Fig 24.9\n",
+      "alpha=15\n",
+      "\n",
+      "#solve equation (i)&(iii)\n",
+      "a=np.array([[600,-600],[190000,290000]])\n",
+      "b=np.array([12000,440000])\n",
+      "q=np.linalg.solve(a,b)\n",
+      "q12=q[0]\n",
+      "q23=q[1]\n",
+      "q31=(F2+l1*q23)/l1\n",
+      "print \"shear flows are shown in Fig 24.9 (flanges)\"\n",
+      "print \"q12=%2.2f N/mm\"%(q12)\n",
+      "print \"q23=%2.2f N/mm\"%(q23)\n",
+      "print \"q31=%2.2f N/mm\\n\"%(q31)\n",
+      "\n",
+      "Sy_1=7*l3\n",
+      "Px4=Px2=2*A1*7/l1\n",
+      "Py2=Py4=Px4*math.tan(math.radians(alpha))\n",
+      "q1=(2100-2*625.2)/l1\n",
+      "P2=P4=(Px4**2+Py4**2)**0.5\n",
+      "P5=P6=2*((A1+A4)*7-A5*13)/l2\n",
+      "q2=(7*l1+7*10-13*10)/l2\n",
+      "q3=(6.4*l2+F2)/l2\n",
+      "\n",
+      "M3=2*((A1+A2)*7-A2*13)+F2*l1\n",
+      "Px1=Px3=M3/l1\n",
+      "Py1=Py3=3626.2\n",
+      "P1=P3=(Px1**2+Py1**2)**.5\n",
+      "q3=(17100-2*Py1)/l1\n",
+      "\n",
+      "print \"Loads in webs \"\n",
+      "print \"P4=P2=%2.2f N\"%(P2)\n",
+      "print \"P6=P5=%2.2f N\"%(P6)\n",
+      "print \"P3=P1=%2.2f N\\n\"%(P1)\n",
+      "\n",
+      "print \"shear flow in webs\"\n",
+      "print \"q1=%2.2f N/mm\"%(q1)\n",
+      "print \"q2=%2.2f N/mm\"%(q2)\n",
+      "print \"q3=%2.2f N/mm (this value is given at the end of example)\"%(q3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear flows are shown in Fig 24.9 (flanges)\n",
+        "q12=13.00 N/mm\n",
+        "q23=-7.00 N/mm\n",
+        "q31=43.00 N/mm\n",
+        "\n",
+        "Loads in webs \n",
+        "P4=P2=2415.64 N\n",
+        "P6=P5=218.75 N\n",
+        "P3=P1=14010.73 N\n",
+        "\n",
+        "shear flow in webs\n",
+        "q1=2.83 N/mm\n",
+        "q2=6.38 N/mm\n",
+        "q3=32.83 N/mm (this value is given at the end of example)\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter25.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter25.ipynb
new file mode 100755
index 00000000..369a648f
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter25.ipynb
@@ -0,0 +1,490 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:806c344ebbd66dd7398f87b7be85af8355708b04b28f0bd902647a641248feff"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 25: Laminated Composite Structures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.1 Pg.No.653"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#variable declaration\n",
+      "Ee=5000               #youngs modulus of epoxy (N/mm^2)\n",
+      "Ec=200000             #youngs modulus of carbon (N/mm^2)\n",
+      "\n",
+      "# E1=Ef*Af/A+Em+Am/A        equation 25.4\n",
+      "A=50*80               #total area (mm^2)\n",
+      "Ae=40*80              #area of epoxy (mm^2)\n",
+      "Ac=10*80              #area of carbon(mm^2)\n",
+      "L=500                  #length of bar(mm)\n",
+      "vc=0.3                 #poisson ratio\n",
+      "ve=0.2\n",
+      "\n",
+      "#effective youngs modulus\n",
+      "E1=(Ee*Ae+Ec*Ac)/A\n",
+      "\n",
+      "load=100*10**3\n",
+      "sigma1=load/A\n",
+      "epsilon1=sigma1/E1\n",
+      "\n",
+      "delta1=epsilon1*L\n",
+      "v1t=(ve*Ae+vc*Ac)/A\n",
+      "epsilon_t=-v1t*epsilon1\n",
+      "\n",
+      "thickness=L\n",
+      "delta_t=-epsilon_t*(thickness)\n",
+      "\n",
+      "\n",
+      "sigma_m=Ee*epsilon1\n",
+      "sigma_f=Ec*epsilon1\n",
+      "print \"stresses in epoxy = %2.2f N/mm^2\\n\"%(sigma_m)\n",
+      "print \"stresses in carbon = %2.2f N/mm^2\\n\"%(sigma_f)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "stresses in epoxy = 2.84 N/mm^2\n",
+        "\n",
+        "stresses in carbon = 113.64 N/mm^2\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.2 Pg.No.657"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "tau=40            #shear stress(N/mm^2)\n",
+      "G=5000            #shear modulus(N/mm^2)\n",
+      "v1t=0.3            #poisson ratio \n",
+      "Et=80000            #youngs modulus in transverse direction(N/mm^2)\n",
+      "E1=120000           #youngs modulus in longitudinal direction(N/mm^2)\n",
+      "sigma_x=50          #direction stress in x direciton\n",
+      "sigma_y=25                           #in y direction\n",
+      "\n",
+      "\n",
+      "vt1=v1t*Et/E1 #minor poisson ratio\n",
+      "\n",
+      "epsilon1=sigma_x/E1-vt1*sigma_y/Et\n",
+      "epsilont=sigma_y/Et-v1t*sigma_x/E1\n",
+      "print \"direct strain in x direction = %2.2e \\n\"%(epsilon1)\n",
+      "print \"direct strain in y direction = %2.2e \\n\"%(epsilont)\n",
+      "\n",
+      "gama1t=tau/G\n",
+      "print \"shear strain in the ply = %2.2e \\n\"%(gama1t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "direct strain in x direction = 3.54e-04 \n",
+        "\n",
+        "direct strain in y direction = 1.88e-04 \n",
+        "\n",
+        "shear strain in the ply = 8.00e-03 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.3 Pg.No.661"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "\n",
+      "El=150000          #youngs modulus(N/mm^2)\n",
+      "Et=90000          #youngs modulus(N/mm^2)\n",
+      "Glt=5000          #shear modulus(N/mm^2)\n",
+      "vlt=0.3            #poisson ratio\n",
+      "theta=45             #longitudinal plane inclination\n",
+      "\n",
+      "s11=1/El\n",
+      "s22=1/Et\n",
+      "s12=-vlt/El\n",
+      "s33=1/Glt\n",
+      "\n",
+      "a=np.array([[53.45,-46.55,1],[-46.55,53.45,0],[-2.2,-2.2,0]])\n",
+      "b=np.array([60,40,0])\n",
+      "x=np.dot(a,b)\n",
+      "print \"strain in the x direction = %4.4e\\n\"%(x[0]*10**-6)\n",
+      "print \"strain in the y direction = %4.4e\\n\"%(x[1]*10**-6)\n",
+      "print \"shear strain in the xy plane = %4.4e\\n\"%(x[2]*10**-6)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "strain in the x direction = 1.3450e-03\n",
+        "\n",
+        "strain in the y direction = -6.5500e-04\n",
+        "\n",
+        "shear strain in the xy plane = -2.2000e-04\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.4 Pg.No.664"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "\n",
+      "l1=150\n",
+      "t1=1\n",
+      "l2=100            #dimensions shown in Fig 25.10 (mm)\n",
+      "t2=2\n",
+      "Ef=60000            #youngs modulus of flanges(N/mm^2)\n",
+      "Ew=20000            #youngs modulus of web(N/mm^2)\n",
+      "P=40*10**3          #axial load(N)\n",
+      "\n",
+      "#sum of b*t*E \n",
+      "sum_btE=2*l2*t2*Ef+l1*t1*Ew\n",
+      "\n",
+      "epsilon_z=P/sum_btE           #equ 25.37\n",
+      "P_flange=epsilon_z*l2*t2*Ef\n",
+      "P_web=epsilon_z*l1*t1*Ew\n",
+      "print \"axial load in flange = %2.2f kN\\n\"%(P_flange/1000)\n",
+      "print \"axial load in web = %2.2f kN\\n\"%(P_web/1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "axial load in flange = 17.78 kN\n",
+        "\n",
+        "axial load in web = 4.44 kN\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.5 Pg.No.666"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "\n",
+      "l1=100\n",
+      "l2=50\n",
+      "t1=1         #dimension shown in Fig 25.11 (mm)\n",
+      "t2=2\n",
+      "Ef=50000         #youngs modulus of flange(N/mm^2)\n",
+      "Ew=15000          #youngs modulus of web(N/mm^2)\n",
+      "Mx=10**6           #bending moment (N.mm)\n",
+      "My=0\n",
+      "\n",
+      "Ixx=2*Ef*l2*t2*l2**2+Ew*t1*l1**3/12\n",
+      "Iyy=Ef*t2*l1**3/12\n",
+      "Ixy=Ef*l2*t2*(50)*(50)+Ef*l2*t2*(-50)*(-50)\n",
+      "\n",
+      "x=50\n",
+      "y=50      #point 1\n",
+      "Ez=50000\n",
+      "\n",
+      "sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)\n",
+      "print \"direct stress at point 1 = %3.1f N/mm^2\\n\"%(sigma_z)\n",
+      "x=0\n",
+      "y=50\n",
+      "sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)\n",
+      "print \"direct stress at point 2 = %3.1f N/mm^2\\n\"%(sigma_z)\n",
+      "\n",
+      "x=0\n",
+      "y=50\n",
+      "Ez=15000\n",
+      "sigma_z=Ez*((My*Ixx-Mx*Ixy)/(Ixx*Iyy-Ixy**2)*x + (Mx*Iyy-My*Ixy)/(Ixx*Iyy-Ixy**2)*y)\n",
+      "print \"direct stress at point 2 in the web = %3.1f N/mm^2\\n\"%(sigma_z)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "direct stress at point 1 = 102.6 N/mm^2\n",
+        "\n",
+        "direct stress at point 2 = -51.3 N/mm^2\n",
+        "\n",
+        "direct stress at point 2 in the web = -15.4 N/mm^2\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.6 Pg.No.668"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from sympy import symbols, integrate\n",
+      "\n",
+      "s=symbols('s')\n",
+      "\n",
+      "l12=250\n",
+      "l23=300\n",
+      "t31=t12=2                 #dimensions shown in Fig 25.12 (mm)\n",
+      "t23=1.5\n",
+      "\n",
+      "P=2*10**3             #shear load(N)\n",
+      "E12=E31=45000          #youngs modulus of sides given in name(N/mm^2)\n",
+      "E23=20000\n",
+      "\n",
+      "Ixx=2*E12*t12*l12**3*(l23/2/l12)**2/12+E23*t23*l23**3/12\n",
+      "alpha=math.asin(l23/2/l12)\n",
+      "\n",
+      "Sx=0\n",
+      "Sy=2*10**3\n",
+      "q12=-E12*Sy/Ixx*integrate(-2*s,(s,0,250))*math.sin(alpha)\n",
+      "print \"shear flow at point 2 in the flange 12 = %2.2f N/mm\\n\"%(q12)\n",
+      "\n",
+      "q23=-E23*Sy/Ixx*integrate(-225+1.5*s,(s,0,300))+22.2\n",
+      "print \"shear flow at point 3 in the flange 23 = %2.2f N/mm\\n\"%(q23)\n",
+      "q0=14.2\n",
+      "\n",
+      "q12=-E12*Sy/Ixx*integrate(-2*s,s)*math.sin(alpha)-q0\n",
+      "print \"shear flow in the flange 12 and 31 \"\n",
+      "print q12\n",
+      "\n",
+      "q23=-E23*Sy/Ixx*integrate(-225+1.5*s,s,)+22.2-q0\n",
+      "print \"\\nshear flow in the web 23\"\n",
+      "print q23"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear flow at point 2 in the flange 12 = 22.22 N/mm\n",
+        "\n",
+        "shear flow at point 3 in the flange 23 = 22.20 N/mm\n",
+        "\n",
+        "shear flow in the flange 12 and 31 \n",
+        "0.000355555555555556*s**2 - 14.2\n",
+        "\n",
+        "shear flow in the web 23\n",
+        "-0.000197530864197531*s**2 + 0.0592592592592593*s + 8.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.7 Pg.No.671"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from sympy import symbols, integrate\n",
+      "\n",
+      "\n",
+      "l1=200\n",
+      "t1=2\n",
+      "l2=100         #dimensions shown in Fig 25.13 (mm)\n",
+      "t2=1\n",
+      "T=10*10**6       #torque applied (N.mm)\n",
+      "Gl=20000          #laminate shear modulus(N/mm^2)\n",
+      "Gw=35000            #web shear modulus(N/mm^2)\n",
+      "A=l1*l2\n",
+      "\n",
+      "q=T/2/A\n",
+      "\n",
+      "#from eqn 25.47\n",
+      "int_ds_by_Gt=2*l1/Gl/t1+2*l2/Gw/t2\n",
+      "\n",
+      "#Let's say we want to calculate warping at point 1\n",
+      "#warping at mid of web is zero (W0=0) integrate eqn 25.47 from\n",
+      "#mid of web to point 1\n",
+      "W0=0\n",
+      "A0s=50*100\n",
+      "W1=W0+q*(l2/2/Gw/t2-int_ds_by_Gt/A*A0s)\n",
+      "print \"warping at point 1 is = %2.2f mm\\n\"%(W1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "warping at point 1 is = -0.63 mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.8 Pg.No.673"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from sympy import symbols, integrate\n",
+      "\n",
+      "l1=50\n",
+      "t1=2.5\n",
+      "l2=25             #dimension shown in Fig 18.12(mm)\n",
+      "t2=1.5\n",
+      "\n",
+      "Gl=20000\n",
+      "Gw=15000           #shear modulus (N/mm^2)\n",
+      "T=10*10**3         #torque applied (N.mm)\n",
+      "AR=25*25/2\n",
+      "GJ=2*Gl*l2*t2**3/3+Gw*l1*t1**3/3\n",
+      "#eqn 25.49\n",
+      "dO_dz=T/GJ\n",
+      "\n",
+      "t_max12=2*Gl*(t2/2)*dO_dz\n",
+      "t_max23=2*Gw*(t1/2)*dO_dz\n",
+      "print \"maximum shear stress in the web = %2.2f N/mm^2\\n\"%(t_max12)\n",
+      "print \"maximum shear stress in the laminate = %2.2f N/mm^2\\n\"%(t_max23)\n",
+      "\n",
+      "W1=-2*AR*dO_dz\n",
+      "print \"warping at point 1 = %2.2f mm\\n\"%(W1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum shear stress in the web = 59.63 N/mm^2\n",
+        "\n",
+        "maximum shear stress in the laminate = 74.53 N/mm^2\n",
+        "\n",
+        "warping at point 1 = -1.24 mm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 74
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter26.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter26.ipynb
new file mode 100755
index 00000000..994182d0
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter26.ipynb
@@ -0,0 +1,82 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:eb3a8ca85b2b3516ec190a12a8401ddf32d140498cd023be9bda6228b1754025"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 26: Closed Section Beams"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 26.3 Pg.No.693"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import solve, symbols\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "G=20000               #shear modulus (N/mm^2)\n",
+      "E=G/0.36              #young's modulus\n",
+      "tau=20                #torque (Nm/mm)\n",
+      "t=1\n",
+      "b=200\n",
+      "a=500\n",
+      "L=2500\n",
+      "A=800\n",
+      "\n",
+      "#torque at each support \n",
+      "T=20*5000/2\n",
+      "\n",
+      "mu=math.sqrt(8*G*t/(A*E*(b+a)))\n",
+      "\n",
+      "z=2500\n",
+      "F=-1.95/mu*math.cosh(mu*z)+1.75*z**2\n",
+      "\n",
+      "z=0                            # at mid-span z=0\n",
+      "theta=10**(-8)*(1.95/mu*math.cosh(mu*z)-1.75*z**2+F)\n",
+      "print \"twist at mid-span = %1.3f rad = %1.2f degree\"%(theta,theta*180/math.pi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "twist at mid-span = 0.108 rad = 6.20 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter27.ipynb b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter27.ipynb
new file mode 100755
index 00000000..bf9485c9
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/Chapter27.ipynb
@@ -0,0 +1,109 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bc596ba82da68e58c73271eb1045592b064349244b750ab69aa9f03ab4de2fbd"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 27: Open Section Beams\n"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 27.2 Pg.No. 735"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#reference Fig 27.20\n",
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from sympy import integrate\n",
+      "from sympy import solve, symbols\n",
+      "\n",
+      "# variable declaration\n",
+      "P=100                  #vertical load (kN)\n",
+      "A_R=1.25*10**3         #from previous question\n",
+      "E=200000               #young's modulus (N/mm^2)\n",
+      "Gama_R=2.08*10**10     #torsion bending constant (mm^6)\n",
+      "mu=0.54*10**-3        #from previous question\n",
+      "A=400*5                #area of beam\n",
+      "\n",
+      "#moment couple\n",
+      "M_Tau=-P*2*A_R\n",
+      "print \"\\nValue of moment couple at the top of column = %2.2e kN.mm^2\"%(M_Tau)  #calculation mistake in book\n",
+      "\n",
+      "d2tdz2 = M_Tau*10**3/E/Gama_R\n",
+      "# @z=3000 mm \n",
+      "D=0.04*10**-3\n",
+      "z=symbols('z')\n",
+      "F= -0.08*math.cosh(0)\n",
+      "\n",
+      "def theta(z):\n",
+      "    return 0.08*(math.cosh(0.54*10**(-3)*z))\n",
+      "\n",
+      "print \"\\ntwist at the top of the column = %2.2f rad\\n\"%(theta(3000))\n",
+      "\n",
+      "M_Tau=-E*Gama_R*0.02*10**-6\n",
+      "print \"value of moment couple at bottom of column=%2.2e N.mm^2\\n\"%(M_Tau)\n",
+      "\n",
+      "P=-100*10**3    # stress(N)\n",
+      "def sigma_z(A_R2):\n",
+      "    return P/A+M_Tau*A_R2/Gama_R\n",
+      "print \"sigma_z1=sigma_z4=%2.1f N/mm^2\\n\"%(sigma_z(-7.5*10**3))\n",
+      "print \"sigma_z2=sigma_z3=%2.1f N/mm^2\\n\"%(sigma_z(2.5*10**3))\n",
+      "\n",
+      "#please put the value of 2AR from Fig 27.21 in the equation of sigma_z\n",
+      "#these values in book need correction"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value of moment couple at the top of column = -2.50e+05 kN.mm^2\n",
+        "\n",
+        "twist at the top of the column = 0.21 rad\n",
+        "\n",
+        "value of moment couple at bottom of column=-8.32e+07 N.mm^2\n",
+        "\n",
+        "sigma_z1=sigma_z4=-20.0 N/mm^2\n",
+        "\n",
+        "sigma_z2=sigma_z3=-60.0 N/mm^2\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 48
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/README.txt b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/README.txt
new file mode 100755
index 00000000..230418be
--- /dev/null
+++ b/Aircraft_Structures_for_Engineering_Students_by_T_H_G_Megson/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Prahlad Das
+Course: others
+College/Institute/Organization: Indian Institute of Technology Bombay
+Department/Designation: Aerospace
+Book Title: Aircraft Structures for Engineering Students
+Author: T. H. G. Megson
+Publisher: Linacre House, Jordan Hill, Oxford OX2 8DP, UK
+Year of publication: 2012
+Isbn: 978-0-75066-7395
+Edition: fourth edition
\ No newline at end of file
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diff --git a/Antenna_and_Wave_Propagation_by_k.k._sharma/README.txt b/Antenna_and_Wave_Propagation_by_k.k._sharma/README.txt
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diff --git a/Applied_Physics-II/chapter1.ipynb b/Applied_Physics-II/chapter1.ipynb
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index c4b1a860..00000000
--- a/Applied_Physics-II/chapter1.ipynb
+++ /dev/null
@@ -1,1353 +0,0 @@
-{
- "metadata": {
-  "celltoolbar": "Raw Cell Format",
-  "name": "",
-  "signature": "sha256:fad8e22fb99cc3e157ab8315172e2ff6ddae35bb6f49be764a49d2d1d3f70fcc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1: Interference"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1, Page number 1-16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "i=45*math.pi/180         #angle of incidence\n",
-      "u=1.33                   #Refractive index of a soap film\n",
-      "lamda=5.896*10**-7       #wavelength of required yellow light\n",
-      "\n",
-      "#Calculations:\n",
-      "#u=sin i/sin r           #Snell's law   .So,\n",
-      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
-      "\n",
-      "#Now, condition for bright fringe is\n",
-      "#2ut*cos r=(2n-1)lamda/2\n",
-      "#Here n=1\n",
-      "t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film at which light will appear bright yellow\n",
-      "print\"Minimum thickness of film at which light will appear bright yellow of required wavelength is =\",t,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum thickness of film at which light will appear bright yellow of required wavelength is = 1.30853030399e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2, Page number 1-16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "theta=40./3600*math.pi/180     #angle of wedge in radians\n",
-      "B=0.12*10**-2                 #fringe spacing\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, B=lam/(2*u*theta). Here u=1\n",
-      "lamda=2*B*theta               #wavelength of light used\n",
-      "print\"Wavelength of light used is =\",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of light used is = 4.65421133865e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3, Page number 1-17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "i=30*math.pi/180        #angle of incidence\n",
-      "u=1.46                  #Refractive index of a oil\n",
-      "lamda=5.890*10**-7       #wavelength of required yellow light\n",
-      "n=8                     #eighth dark band\n",
-      "\n",
-      "#Calculations:\n",
-      "#u=sin i/sin r      #Snell's law   .So,\n",
-      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
-      "\n",
-      "#Now, condition for dark fringe is\n",
-      "#2ut*cos r=n*lamda\n",
-      "t=n*lamda/(2*u*math.cos(r))     #thickness of film\n",
-      "print\"Thickness of the film is =\",t,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of the film is = 1.71755887917e-06 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4, Page number 1-17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "\n",
-      "B=0.1*10**-2               #fringe spacing\n",
-      "lamda=5.893*10**-7         #Wavelength of light\n",
-      "u=1.52                     #Refractive index of wedge\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, B=lamda/(2*u*theta). Here u=1\n",
-      "theta1=lamda/(2*u*B)            #angle of wedge in radians\n",
-      "theta=theta1*3600*180/math.pi   #angle of wedge in seconds\n",
-      "print\"Angle of wedge is =\",theta,\"seconds of an arc\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angle of wedge is = 39.9841612899 seconds of an arc\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5, Page number 1-18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "t=0.2/(100)**2*10**-2     #thickness of film in meter\n",
-      "lamda=5.5*10**-7         #wavelength of light in meter\n",
-      "r=0                      #normal incidence\n",
-      "n=1                      #first band\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#Condition for dark fringe is\n",
-      "#2ut*cos r =n*lamda\n",
-      "u=n*lamda/(2*t*math.cos(r))   #Refractive index of a oil\n",
-      "print\"Refractive index of a oil is =\",u\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index of a oil is = 1.375\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6, Page number 1-18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=5.893*10**-7    #Wavelength of light\n",
-      "u=1.42                #Refractive index of a soap film\n",
-      "r=0                   #normal incidence\n",
-      "n=1                   #first band\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#(i)\n",
-      "#Condition for dark fringe is\n",
-      "#2ut*cos r=n*lamda\n",
-      "t1=n*lamda/(2*u*cos(r))      #thickness of film for dark black fringe\n",
-      "print\"Thickness of the film for dark black fringe is =\",t1,\"m\"\n",
-      "\n",
-      "#(ii)\n",
-      "#Now, condition for bright fringe is\n",
-      "#2ut*cos r=(2n-1)lamda/2\n",
-      "t2=lamda/(2*2*u*math.cos(r))      #Thickness of film for bright fringe\n",
-      "print\"Thickness of film for bright fringe is =\",t2,\"m\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of the film for dark black fringe is = 2.075e-07 m\n",
-        "Thickness of film for bright fringe is = 1.0375e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number 1-19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "#Given Data:\n",
-      "i=30*math.pi/180         #angle of incidence\n",
-      "u=1.43               #Refractive index of a soap film\n",
-      "lamda=6*10**-7          #wavelength of light\n",
-      "n=1                  #For minimum thickness\n",
-      "\n",
-      "#Calculations:\n",
-      "#u=sin i/sin r      #Snell's law   .So,\n",
-      "r=(math.asin(math.sin(i)/u))    #angle of reflection\n",
-      "\n",
-      "#Now, condition of minima in transmitted system is\n",
-      "#2ut*cos(r)=(2n-1)lamda/2\n",
-      "t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film\n",
-      "print\"Minimum thickness of film is \",t,\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum thickness of film is  1.11962124395e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8, Page number 1-19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "\n",
-      "lamda = 5893*10**-10      #Wavelength of light\n",
-      "theta = 1               #assuming value of theta\n",
-      "\n",
-      "#We know, B=lamda/(2*u*theta). Here u=1\n",
-      "B = lamda/(2*theta)       #fringe spacing\n",
-      "n=20                  #interference fringes\n",
-      "\n",
-      "#Calculations:\n",
-      "#t=n*B*tan(theta)\n",
-      "t = 20*B*theta          #Thickness of wire\n",
-      "print\"Thickness of wire is =\",t,\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of wire is = 5.893e-06 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9, Page number 1-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.3                #Refractive index of oil\n",
-      "u2=1.5                #Refractive index of glass\n",
-      "lamda1=7*10**-7       #Wavelength of light\n",
-      "lamda2=5*10**-7       #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#for finding value of n, solve:\n",
-      "#(2n+1)*lamda1/2=(2(n+1)+1)*lamda2/2\n",
-      "#We get,n=2\n",
-      "n=2\n",
-      "\n",
-      "toil=(2*n+1)*lamda1/(2*u1*2)     #thickness of oil layer\n",
-      "print\"Thickness of oil layer is =\",toil,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of oil layer is = 6.73076923077e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "Example 1.10, Page number 1-21"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.2               #Refractive index of drop of oil\n",
-      "u2=1.33              #Refractive index of water\n",
-      "lamda=4.8*10**-7     #wavelength of light\n",
-      "n=3                  #order\n",
-      "r=0                  #normal incidence,so r=0\n",
-      "\n",
-      "#Calculations:\n",
-      "t=n*lamda/(2*u1)      #Thickness of oil drop\n",
-      "print\"Thickness of oil drop is =\",t,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of oil drop is = 6e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11, Page number 1-22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "i=math.asin(4/5)          #angle of incidence\n",
-      "u=4/3                    #Refractive index of a soap film\n",
-      "lamda1=6.1*10**-7        #wavelength of light\n",
-      "lamda2=6*10**-7          #wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "#u=sin i/sin r      #Snell's law   .So,\n",
-      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
-      "\n",
-      "#Now, condition for dark band is\n",
-      "#2ut*cos r=n*lamda\n",
-      "#for consecutive bands, n=lamda2/(lamda1-lamda2). hence\n",
-      "\n",
-      "t=lamda2*lamda1/((lamda1-lamda2)*2*u*math.sqrt(1-(math.sin(i)/u)**2))     #thickness of film\n",
-      "print\"Thickness of the film is =\",t,\"m\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of the film is = 1.83e-05 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12, Page number 1-40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n=10                #10th dark ring\n",
-      "Dn=0.5*10**-2       #Diameter of ring\n",
-      "lamda=6*10**-7      #wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "#As Dn^2=4*n*R*lamda\n",
-      "R=Dn**2/(4*n*lamda)    #Radius of curvature of the lens\n",
-      "print\"Radius of curvature of the lens is =\",R,\"m\"\n",
-      "\n",
-      "t=Dn**2/(8*R)        #thickness of air  film\n",
-      "print\"Thickness of air  film is =\",t,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius of curvature of the lens is = 1.04166666667 m\n",
-        "Thickness of air  film is = 3e-06 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13, Page number 1-41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "\n",
-      "B=0.25*10**-2             #fringe spacing\n",
-      "lamda=5.5*10**-7          #Wavelength of light\n",
-      "u=1.4                     #Refractive index of wedge\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, B=lamda/(2*u*theta).\n",
-      "theta1=lamda/(2*u*B)            #angle of wedge in radians\n",
-      "theta=theta1*3600*180/math.pi   #angle of wedge in seconds\n",
-      "print\"Angle of wedge is =\",theta,\"seconds\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angle of wedge is = 16.2065204908 seconds\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.14, Page number 1-41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n=4              #4th dark ring\n",
-      "m=12             #m=n+p\n",
-      "D4=0.4*10**-2    #Diameter of 4th ring\n",
-      "D12=0.7*10**-2   #Diameter of 12th ring\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#(Dn+p)^2-Dn^2=4*p*lamda*R\n",
-      "#Solving, (D12^2-D4^2)/(D20^2-D4^2)\n",
-      "#We get above value =1/2. Hence\n",
-      "D20=math.sqrt(2*D12**2-D4**2)      #Diameter of 20th ring\n",
-      "print\"Diameter of 20th ring is =\",D20,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diameter of 20th ring is = 0.00905538513814 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.15, Page number 1-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n=6                  #6th bright ring\n",
-      "D6=0.31*10**-2       #Diameter of 6th ring\n",
-      "lamda=6*10**-7       #wavelength of light\n",
-      "R=1                  #Radius of curvature\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#Diameter of nth bright ring is \n",
-      "#Dn^2=2(2n-1)*lamda*R/u. Hence\n",
-      "u=2*(2*n-1)*lamda*R/(D6)**2    #Refractive index of liquid\n",
-      "print\"Refractive index of liquid is =\",u\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index of liquid is = 1.37356919875\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.16, Page number 1-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=6*10**-7         #wavelength of light\n",
-      "k=0.125*10**-4         #k=D(n+1)^2-Dn^2.\n",
-      "u=1                    #Refractive index of medium between lens and plate\n",
-      "#Calculations:\n",
-      "\n",
-      "#(i)\n",
-      "lamda1=4.5*10**-7      #new wavelength of light\n",
-      "#Difference between squres of diameters of successive rings is directly proportional to wavelength.So,\n",
-      "k1=lamda1/lamda*k      #new Difference between squres of diameters of successive rings after changing wavelength\n",
-      "print\"New Difference between squres of diameters of successive rings after changing wavelength is =\",k1,\"m^2\"\n",
-      "\n",
-      "#(ii)\n",
-      "u2=1.33          #Refractive index of liquid introduced between lens and plate\n",
-      "#Difference between squres of diameters of successive rings is inversely proportional to Refractive index.so,\n",
-      "k2=u/u2*k        #new Difference between squres of diameters of successive rings after changing refractive index\n",
-      "print\"New Difference between squres of diameters of successive rings after changing refrective index is =\",k2,\"m^2\"\n",
-      "\n",
-      "#(iii)\n",
-      "#Difference between squres of diameters of successive rings is directly proportional to Radius of curvature.So,\n",
-      "#after doubling radius of curvature,\n",
-      "k3=2*k           #new Difference between squres of diameters of successive rings after doubling radius of curvature\n",
-      "print\"New Difference between squres of diameters of successive rings after doubling radius of curvature is =\",k3,\"m^2\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New Difference between squres of diameters of successive rings after changing wavelength is = 9.375e-06 m^2\n",
-        "New Difference between squres of diameters of successive rings after changing refrective index is = 9.3984962406e-06 m^2\n",
-        "New Difference between squres of diameters of successive rings after doubling radius of curvature is = 2.5e-05 m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.17, Page number 1-43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "Dn=0.225*10**-2         #Diameter of nth ring\n",
-      "Dm=0.45*10**-2          #Diameter of (n+9)th ring\n",
-      "lamda=6*10**-7          #wavelength of light\n",
-      "R=0.9                   #Radius of curvature\n",
-      "p=9\n",
-      "\n",
-      "#Calculations:\n",
-      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
-      "u=4*p*lamda*R/((Dm)**2-Dn**2)     #Refractive index of liquid\n",
-      "print\"Refractive index of liquid is =\",u\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index of liquid is = 1.28\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.18, Page number 1-44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "D10=0.5*10**-2          #Diameter of 10th ring\n",
-      "lamda=5.5*10**-7        #wavelength of light\n",
-      "u=1.25                  #Refractive index of liquid\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#As Dn^2=4*n*R*lamda/u\n",
-      "#Dn^2 is inversely proportional to refractive index.\n",
-      "D10n=D10/math.sqrt(u)        #new diameter of 10th ring after changing medium between lens and plate\n",
-      "print\"new diameter of 10th ring after changing medium between lens and plate is =\",D10n,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "new diameter of 10th ring after changing medium between lens and plate is = 0.004472135955 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.19, Page number 1-45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "D5=0.336*10**-2       #Diameter of 5th ring\n",
-      "D15=0.59*10**-2       #Diameter of 15th ring\n",
-      "lamda=5.89*10**-7       #wavelength of light\n",
-      "p=10                  #n=5,n+p=15\n",
-      "\n",
-      "#Calculations:\n",
-      "#(Dn+p)^2-Dn^2=4*p*lamdaR/u\n",
-      "R=((D15)**2-D5**2)/(4*p*lamda)      #Radius of curvature of the lens\n",
-      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius of curvature of the lens is = 0.998319185059 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.20, Page number 1-45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n=10               #10th dark ring\n",
-      "D10=0.6*10**-2     #Diameter of ring\n",
-      "lamda=6*10**-7     #wavelength of light\n",
-      "u=4./3              #Refractive index of water\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#As Dn^2=4*n*R*lamda/u\n",
-      "R=(D10**2)*u/(4*n*lamda)    #Radius of curvature of the lens\n",
-      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius of curvature of the lens is = 2.0 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.20.1, Page number 1-52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "i=45*math.pi/180        #angle of incidence\n",
-      "u=1.2                   #Refractive index of a film\n",
-      "t=4*10**-7              #thickness of film\n",
-      "\n",
-      "#Calculations:\n",
-      "#u=sin i/sin r       #Snell's law   .So,\n",
-      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
-      "\n",
-      "#Now, condition for dark fringe is\n",
-      "#2ut*cos r=n*lamda\n",
-      "lamda1=2*u*t*math.cos(r)/1          #n=1\n",
-      "print\"For n=1 wavelength is =\",lamda1,\"m\"\n",
-      "print\"This is in the visible spectrum and it will remain absent.\" \n",
-      "\n",
-      "lamda2=2*u*t*math.cos(r)/2          #n=2\n",
-      "print\"For n=2 wavelength is =\",lamda2,\"m\"\n",
-      "print\"This is not in the visible spectrum\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For n=1 wavelength is = 7.75628777187e-07 m\n",
-        "This is in the visible spectrum and it will remain absent.\n",
-        "For n=2 wavelength is = 3.87814388593e-07 m\n",
-        "This is not in the visible spectrum\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.20.2, Page number 1-53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "r=45*math.pi/180        #angle of refraction\n",
-      "u=1.45                  #Refractive index of a medium\n",
-      "lamda=5.5*10**-7        #wavelength of required yellow light\n",
-      "n=1\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#Now, condition for dark fringe is\n",
-      "#2ut*cos r=n*lamda\n",
-      "t=n*lamda/(2*u*math.cos(r))      #thickness of thin medium\n",
-      "print\"Thickness of the thin medium is =\",t,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of the thin medium is = 2.68212917002e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.21, Page number 1-45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u=1.33                 #Refractive index of a soap film\n",
-      "r=0                    #normal incidence\n",
-      "t=5*10**-7             #thickness of film\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#Now, condition for maxima is\n",
-      "#2ut*cos r=(2n-1)lamda/2\n",
-      "lamda1=4*u*t*math.cos(r)/(2*1-1)          #n=1\n",
-      "print\"For n=1 wavelength is =\",lamda1,\"m\"\n",
-      "lamda2=4*u*t*math.cos(r)/(2*2-1)          #n=2\n",
-      "print\"For n=2 wavelength is =\",lamda2,\"m\"\n",
-      "lamda3=4*u*t*math.cos(r)/(2*3-1)          #n=3\n",
-      "print\"For n=3 wavelength is =\",lamda3,\"m\"\n",
-      "lamda4=4*u*t*math.cos(r)/(2*4-1)          #n=4\n",
-      "print\"For n=4 wavelength is =\",lamda4,\"m\"\n",
-      "\n",
-      "print\"Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\"\n",
-      "print\"Hence, wavelength for n=3 is the most reflected wavelength.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For n=1 wavelength is = 2.66e-06 m\n",
-        "For n=2 wavelength is = 8.86666666667e-07 m\n",
-        "For n=3 wavelength is = 5.32e-07 m\n",
-        "For n=4 wavelength is = 3.8e-07 m\n",
-        "Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\n",
-        "Hence, wavelength for n=3 is the most reflected wavelength.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.22, Page number 1-46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u=1.5                #Refractive index of a oil\n",
-      "lamda=5.88*10**-7    #wavelength of required yellow light\n",
-      "n=1                  #for smallest thickness\n",
-      "r=60*math.pi/180     #angle of reflection\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#Now, condition for dark fringe is\n",
-      "#2ut*cos r=n*lamda\n",
-      "t=n*lamda/(2*u*math.cos(r))      #thickness of film\n",
-      "print\"Thickness of the film is =\",t,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of the film is = 3.92e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.23, Page number 1-46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "theta=20./3600*math.pi/180     #angle of wedge in radians\n",
-      "B=0.25*10**-2                 #fringe spacing\n",
-      "u=1.4                         #Refractive index of film\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, B=lamda/(2*u*theta).\n",
-      "lamda=2*B*theta*u             #wavelength of light\n",
-      "print\"Wavelength of light is =\",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of light is = 6.78739153553e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.24, Page number 1-47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "#Dn=2*D40\n",
-      "\n",
-      "#Calculations:\n",
-      "#As Dn^2 = 4*n*R*lamda/u  and   Dn^2 = 4*D40^2\n",
-      "#i.e. 4*n*R*lamda/u = 4*4*40*R*lamda/u .hence,\n",
-      "n=4*40          #order of the required ring\n",
-      "print\"Order of the dark ring which will have double the diameter of that of 40th ring is =\",n\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Order of the dark ring which will have double the diameter of that of 40th ring is = 160\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.25, Page number 1-47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda1=6*10**-7       #wavelength of light\n",
-      "lamda2=4.5*10**-7    #wavelength of light\n",
-      "R=0.9                 #Radius of curvature\n",
-      "\n",
-      "#Calculations:\n",
-      "#As Dn^2=4*n*R*lamda.\n",
-      "#Dn^2=D(n+1)^2 for different wavelengths.we get,\n",
-      "n=lamda2/(lamda1-lamda2)       #nth dark ring due to lam1 which coincides with (n+1)th dark ring due lamda2\n",
-      "D3=math.sqrt(4*n*R*lamda1)     #diameter of 3rd dark ring for lamda1\n",
-      "print\"Diameter of 3rd dark ring for lam1 is =\",D3,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diameter of 3rd dark ring for lam1 is = 0.00254558441227 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.26, Page number 1-48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "i=45*math.pi/180        #angle of incidence\n",
-      "u=4./3                  #Refractive index of soap film\n",
-      "lamda=5*10**-7          #wavelength of light\n",
-      "t=1.5*10**-6            #thickness of film\n",
-      "\n",
-      "#Calculations:\n",
-      "#u=sin i/sin r        #Snell's law   .So,\n",
-      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
-      "\n",
-      "#Now, condition for dark band is\n",
-      "#2ut*cos r=n*lamda\n",
-      "n=2*u*t*math.cos(r)/lamda    #order of band\n",
-      "print\"order of dark band is =\",n\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "order of dark band is = 6.78232998313\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.27, Page number 1-49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "D5=0.336*10**-2         #Diameter of 5th ring\n",
-      "D15=0.59*10**-2         #Diameter of 15th ring\n",
-      "lamda=5.89*10**-7       #wavelength of light\n",
-      "p=10                    #n=5,n+p=15\n",
-      "\n",
-      "#Calculations:\n",
-      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
-      "R=((D15)**2-D5**2)/(4*p*lamda)    #Radius of curvature of the lens\n",
-      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius of curvature of the lens is = 0.998319185059 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.29, Page number 1-50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#As Dn^2=4*n*R*lamda.\n",
-      "#thus, Dn is directly proportional to sqaure root of n\n",
-      "D5=math.sqrt(5)     #D5 is directly proportional to sqaure root of 5\n",
-      "D4=math.sqrt(4)     #D4 is directly proportional to sqaure root of 4\n",
-      "k1=D5-D4\n",
-      "print\"Separation between D5 and D4 is directly proportional to =\",k1\n",
-      "\n",
-      "D80=math.sqrt(80)     #D80 is directly proportional to sqaure root of 80\n",
-      "D79=math.sqrt(79)     #D79 is directly proportional to sqaure root of 79\n",
-      "k2=D80-D79\n",
-      "print\"Separation between D80 and D79 is directly proportional to =\",k2\n",
-      "\n",
-      "print\"Thus, (D80-D79) < (D5-D4). Hence proved.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Separation between D5 and D4 is directly proportional to = 0.2360679775\n",
-        "Separation between D80 and D79 is directly proportional to = 0.0560774926836\n",
-        "Thus, (D80-D79) < (D5-D4). Hence proved.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.30, Page number 1-51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "D5=0.336*10**-2       #Diameter of 5th ring\n",
-      "D15=0.59*10**-2       #Diameter of 15th ring\n",
-      "p=10                  #n=5,n+p=15\n",
-      "R=1                   #Radius of curvature\n",
-      "\n",
-      "#Calculations:\n",
-      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
-      "lamda=((D15)**2-D5**2)/(4*p*R)    #Wavelength of light\n",
-      "print\"Wavelength of light is =\",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of light is = 5.8801e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.31, Page number 1-51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "\n",
-      "#Condition for bright band is\n",
-      "#2ut*cos r = (2n-1)*lamda1\n",
-      "\n",
-      "#for consecutive bands, 2n=(lamda1+lamda2)/(lamda1-lamda).\n",
-      "#thus, 2ut*cos r = lamda2*lamda1/(lamda1-lamda2)\n",
-      "\n",
-      "#And, thicknessof film \n",
-      "#t= lamda2*lamda1/((2*u*cosr)(lamda1-lamda2))\n",
-      "print\"Hence expression for thickness of film is obtained.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hence expression for thickness of film is obtained.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics-II/chapter2.ipynb b/Applied_Physics-II/chapter2.ipynb
deleted file mode 100755
index 6af9170f..00000000
--- a/Applied_Physics-II/chapter2.ipynb
+++ /dev/null
@@ -1,1480 +0,0 @@
-{
- "metadata": {
-  "celltoolbar": "Raw Cell Format",
-  "name": "",
-  "signature": "sha256:1b2e4fb9c2f216cbf762800ac72003e90570cd62834a70860272dd5120eea60f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2: Diffraction of light"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1,Page number 2-30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=4       #order\n",
-      "N=1./5000*10**-2    #N=(a+b) grating element(cm)\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "#for longest wavelength, sin(theta)=1\n",
-      "lamda=(N/m)     #longest wavelength\n",
-      "print\"The longest wavelength is =\" ,lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The longest wavelength is = 5e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2,Page number 2-30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                #order\n",
-      "lamda=6.5*10**-7      #Wavelength of red light\n",
-      "theta=30*3.142/180   #angle of diffraction\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, a*sin(theta)=m*lamda\n",
-      "a=m*lamda/math.sin(theta)   #width of slit\n",
-      "print\"width of slit is = \",a,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "width of slit is =  1.29984715296e-06 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3,Page number 2-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                #order\n",
-      "lamda=4*10**-7        #Wavelength of light\n",
-      "a=10**-6            #width of slit\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, a*sin(theta)=m*lamda\n",
-      "theta=math.asin(m*lamda/a)*180/3.142    #angular position in first minima\n",
-      "print\"angular position in first minima is =\",theta,\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "angular position in first minima is = 23.5751216716 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4,Page number 2-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                 #order\n",
-      "lamda1=4*10**-7       #Wavelength of light\n",
-      "lamda2=7*10**-7       #Wavelength of light\n",
-      "n=1./6000*10**-2       #n=(a+b) grating element\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "theta1=math.asin(m*lamda1/n)*(180/3.142)   #angle of diffraction\n",
-      "theta2=math.asin(m*lamda2/n)*(180/3.142)   #angle of diffraction\n",
-      "d=theta2-theta1                  #angular breadth of first order visible spectrum\n",
-      "print\"angular breadth of first order visible spectrum is = \",d,\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "angular breadth of first order visible spectrum is =  10.9466277612 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5,Page number 2-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                         #order\n",
-      "lamda=6.56*10**-7           #Wavelength of red light\n",
-      "theta=18.25*math.pi/180    #angle of diffraction\n",
-      "W=2*10**-2                 #width of grating\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "N=math.sin(theta)/(m*lamda)   #N-number of lines per m, N=1/(a+b)\n",
-      "Tn=N*W                     #Total number of lines on grating\n",
-      "print\"Total number of lines on grating is =\",Tn\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total number of lines on grating is = 9547.67702694\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.7,Page number 2-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "GE=2.54/15000*10**-2       #GE=(a+b) grating element\n",
-      "lamda1=4*10**-7            #Wavelength of light\n",
-      "lamda2=7*10**-7            #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "theta11=math.asin(1*lamda1/GE)*180/math.pi       #angular position of first minima for lamda1\n",
-      "theta12=math.asin(2*lamda1/GE)*180/math.pi       #angular position of second minima for lamda1\n",
-      "theta13=math.asin(3*lamda1/GE)*180/math.pi       #angular position of third minima for lamda1\n",
-      "\n",
-      "theta21=math.asin(1*lamda2/GE)*180/math.pi       #angular position of first minima for lamda2\n",
-      "theta22=math.asin(2*lamda2/GE)*180/math.pi       #angular position of second minima for lamda2\n",
-      "theta23=math.asin(1)*180/math.pi                 #angular position of third minima for lamda2\n",
-      "\n",
-      "print\"Thus the angular position for lamda1 and lamda2 are as follows:\"\n",
-      "print\"First order:\",theta11,\"degrees\"\n",
-      "print\"\",theta21,\"degrees --Isolated\"\n",
-      "\n",
-      "print\"Second order:\",theta12,\"degrees\"\n",
-      "print\"\",theta22,\"degrees --Overlap\"\n",
-      "\n",
-      "print\"Third order: \",theta13,\"degrees\"\n",
-      "print\"\",theta23,\" degrees --Overlap \"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thus the angular position for lamda1 and lamda2 are as follows:\n",
-        "First order: 13.6635763633 degrees\n",
-        " 24.4177053663 degrees --Isolated\n",
-        "Second order: 28.1928605617 degrees\n",
-        " 55.7685229906 degrees --Overlap\n",
-        "Third order:  45.1261086702 degrees\n",
-        " 90.0  degrees --Overlap \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.8,Page number 2-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=5.893*10**-7        #Wavelength of light\n",
-      "d=0.01*10**-2             #width of slit (a=d)\n",
-      "f=1                      #distance between screen and slit\n",
-      "\n",
-      "#Calculations:\n",
-      "x=f*lamda/d              #separation between central maxima and first minima\n",
-      "print\"Separation between central maxima and first minima is = \",x,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Separation between central maxima and first minima is =  0.005893 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.9,Page number 2-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                 #order\n",
-      "lamda=6*10**-7        #Wavelength of light\n",
-      "a=12*10**-7         #width of slit\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, a*sin(theta)=m*lamda\n",
-      "theta=math.asin(m*lamda/a)*180/math.pi    #angular position in first minima\n",
-      "print\"Half angular width of first maxima is =\",theta,\"Degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Half angular width of first maxima is = 30.0 Degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.10,Page number 2-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=6*10**-7          #Wavelength of light\n",
-      "a=0.02*10**-2           #width of slit (a=d)\n",
-      "f=2                    #distance between screen and slit\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, a*sin(theta)=m*lamda, here m=1\n",
-      "theta=math.asin(lamda/a)*180*60/math.pi      #angular position in first minima (1 degree=60 minutes)\n",
-      "print\"Total angular width is =\",2*theta,\"minutes\"\n",
-      "\n",
-      "x=f*lamda/a              #separation between central maxima and first minima\n",
-      "print\"Linear width is = \",2*x,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total angular width is = 20.6265115646 minutes\n",
-        "Linear width is =  0.012 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.11,Page number 2-35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "a=0.14*10**-3         #width of slit\n",
-      "n=2                   #order\n",
-      "y=1.6*10**-2          #separation between second dark band and central bright band\n",
-      "D=2                   #distance between screen and slit\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "theta=y/D            #from diagram \n",
-      "\n",
-      "#We know, a*sin(theta)=n*lamda\n",
-      "#here sin(theta)=theta\n",
-      "lamda=a*theta/n       #wavelength of light\n",
-      "print\"wavelength of light is =\",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "wavelength of light is = 5.6e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.13,Page number 2-36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=6.328*10**-7        #Wavelength of light\n",
-      "N=1./6000*10**-2           #N=(a+b) grating element\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, N*sin(theta)=m*lamda\n",
-      "theta1=math.asin(1*lamda/N)*180/math.pi    #angular position in first order maxima,m=1\n",
-      "print\"Angular position in first order maxima is =\",theta1,\"Degrees\"\n",
-      "\n",
-      "theta2=math.asin(2*lamda/N)*180/math.pi    #angular position in second order maxima,m=2\n",
-      "print\"Angular position in second order maxima is = \",theta2,\"Degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angular position in first order maxima is = 22.3138625335 Degrees\n",
-        "Angular position in second order maxima is =  49.4078093436 Degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.14,Page number 2-37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda1=6*10**-7         #wavelength of yellow light\n",
-      "lamda2=4.8*10**-7       #wavelength of blue light\n",
-      "theta=(math.asin(3/4))    #angle of diffraction\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#for consecutive bands, n*lamda1=(n+1)*lamda2. thus,\n",
-      "n=lamda2/(lamda1-lamda2)   #order\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "N=n*lamda1/(3./4)    #N=(a+b) grating element\n",
-      "print\"Grating element (a+b) is =\",N,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Grating element (a+b) is = 3.2e-06 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.15.1,Page number 2-54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "a=0.2*10**-3          #width of slit\n",
-      "n=1                   #order\n",
-      "y=0.5*10**-2          #separation between first minima and central bright band\n",
-      "D=2                   #distance between screen and slit\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "theta=y/D            #from diagram \n",
-      "\n",
-      "#We know, a*sin(theta)=n*lamda\n",
-      "#here sin(theta)=theta\n",
-      "lamda=a*theta/n       #wavelength of light\n",
-      "print\"wavelength of light is = \",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "wavelength of light is =  5e-07 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.15.2,Page number 2-55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda1=5.4*10**-7              #Wavelength of light\n",
-      "lamda2=4.05*10**-7             #Wavelength of light\n",
-      "theta=30*math.pi/180          #angle of diffraction\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, (a+b)*sin(theta)=n*lamda\n",
-      "#n*lamda1=(n+1)*lamda2, we get \n",
-      "n=3\n",
-      "N=math.sin(theta)/(n*lamda1)*10**-2      #Number of lines per m= 1/(a+b)*10^-2\n",
-      "print\"Number of lines per cm is = \",N\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of lines per cm is =  3086.41975309\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.15.4,Page number 2-56"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "GE=1./6000*10**-2               #GE=(a+b) grating element\n",
-      "lamda1=5.893*10**-7             #Wavelength of light\n",
-      "lamda2=5.896*10**-7             #Wavelength of light\n",
-      "m=2                             #order\n",
-      "\n",
-      "#Calculations:\n",
-      "theta1=math.asin(m*lamda1/GE)*180/math.pi     #angular position in first minima\n",
-      "theta2=math.asin(m*lamda2/GE)*180/math.pi     #angular position in second minima\n",
-      "\n",
-      "a_s=(theta2-theta1)            #Angular separation in minutes\n",
-      "print\"Angular separation is =\",a_s,\"Degrees\"\n",
-      "\n",
-      "dlamda=lamda2-lamda1            #difference in wavelength\n",
-      "lamda=(lamda2+lamda1)/2         #Mean wavelength\n",
-      "\n",
-      "#We know that R.P.=lamda/dlamda=m*N\n",
-      "N=lamda/dlamda/m                 #Number of lines on grating for first order\n",
-      "print\"Number of lines on grating for first order is =\",N\n",
-      "print\"But, number of lines per cm on grating is 6000. \\n Which is greater than number of lines per cm needed for resolution.\"\n",
-      "print\"Hence, both lines will be well resolved in 2nd order.\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angular separation is = 0.0291798772234 Degrees\n",
-        "Number of lines on grating for first order is = 982.416666667\n",
-        "But, number of lines per cm on grating is 6000. \n",
-        " Which is greater than number of lines per cm needed for resolution.\n",
-        "Hence, both lines will be well resolved in 2nd order.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.17,Page number 2-39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "d=0.04*10**-2       #Separation between slits\n",
-      "D=1.7              #distance between screen and slit\n",
-      "B=0.25*10**-2       #Fringe spacing\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know,B=D*lamda/d\n",
-      "lamda=B*d/D         #Wavelength of light\n",
-      "print\"Wavelength of light is = \",lamda,\"m\"\n",
-      "\n",
-      "#The condition for missing order is,\n",
-      "#(a+b)/a = m/n\n",
-      "b=0.04*10**-2       #Separation in slits\n",
-      "a=0.08*10**-3       #Slit width\n",
-      "n=(a+b)/a          #missing orders for m=1,2,3\n",
-      "\n",
-      "n1=1*n\n",
-      "n2=2*n\n",
-      "n3=3*n\n",
-      "print\"Missing orders are =\",n1,\",\",n2,\",\",n3\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of light is =  5.88235294118e-07 m\n",
-        "Missing orders are = 6.0 , 12.0 , 18.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.18,Page number 2-39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "N=2.54/2620*10**-2     #N=(a+b) grating element\n",
-      "lamda=5*10**-7           #Wavelength of red light\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=n*lamda\n",
-      "#maximum value of sin(theta)=1\n",
-      "n=N/lamda               #Maximum number of orders visible\n",
-      "print\"Maximum number of orders visible is =\",n\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum number of orders visible is = 19.3893129771\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.19,Page number 2-40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "N=1./4000*10**-2       #N=(a+b) grating element\n",
-      "lamda1=5*10**-7         #Wavelength of light\n",
-      "lamda2=7.5*10**-7       #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=n*lamda\n",
-      "#maximum value of sin(theta)=1\n",
-      "n1=N/lamda1               #Maximum number of orders visible\n",
-      "n2=N/lamda2               #Maximum number of orders visible\n",
-      "print\"The observed number of orders range between =\",n2,\"to\",n1\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The observed number of orders range between = 3.33333333333 to 5.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.20,Page number 2-40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n=5                #order\n",
-      "lamda=6*10**-7     #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, a*sin(theta)=n*lamda\n",
-      "#n*lamda=n1*lamda1\n",
-      "lamda1=n*lamda/4        #for n1=4\n",
-      "print\"For n1=4 wavelength is =\",lamda1,\"m\"\n",
-      "\n",
-      "lamda2=n*lamda/5        #for n1=5\n",
-      "print\"For n1=5 wavelength is =\",lamda2,\"m\"\n",
-      "\n",
-      "lamda3=n*lamda/6        #for n1=6\n",
-      "print\"For n1=5 wavelength is =\",lamda3,\"m\"\n",
-      "\n",
-      "lamda4=n*lamda/7        #for n1=7\n",
-      "print\"For n1=5 wavelength is =\",lamda4,\"m\"\n",
-      "\n",
-      "lamda5=n*lamda/8        #for n1=8\n",
-      "print\"For n1=5 wavelength is =\",lamda5,\"m\"\n",
-      "\n",
-      "\n",
-      "print\"So,in the grating spectrum spectrum lines with wavelengths n1=6 and n1=7 will coincide with fifth order line of 6*10^-7 m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For n1=4 wavelength is = 7.5e-07 m\n",
-        "For n1=5 wavelength is = 6e-07 m\n",
-        "For n1=5 wavelength is = 5e-07 m\n",
-        "For n1=5 wavelength is = 4.28571428571e-07 m\n",
-        "For n1=5 wavelength is = 3.75e-07 m\n",
-        "So,in the grating spectrum spectrum lines with wavelengths n1=6 and n1=7 will coincide with fifth order line of 6*10^-7 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.21,Page number 2-41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "GE=18000*10**-10         #GE=(a+b) grating element\n",
-      "lamda=5*10**-7             #Wavelength of red light\n",
-      "\n",
-      "#Calculations:\n",
-      "DP1=1./sqrt(GE**2-lamda**2)*10**-10        #Dispersive power\n",
-      "print\"Dispersive power for first order is =\",DP1,\"rad/Angstrom\"\n",
-      "\n",
-      "m=3\n",
-      "DP2=1/sqrt((GE/m)**2-lamda**2)*10**-10    #Dispersive power\n",
-      "print\"Dispersive power for second order is =\",DP2,\"rad/Angstrom\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Dispersive power for first order is = 5.78314931966e-05 rad/Angstrom\n",
-        "Dispersive power for second order is = 0.000301511344578 rad/Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.22,Page number 2-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "N=2.54/15000*10**-2      #N=(a+b) grating element\n",
-      "lamda=5.9*10**-7         #Wavelength of light\n",
-      "m=2                      #order\n",
-      "f=25*10**-2              #focal length of lens\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "theta=math.asin(m*lamda/N)      #angular position in first minima\n",
-      "\n",
-      "Ad=m/N/cos(theta)              #angular dispersion\n",
-      "\n",
-      "ld=f*Ad*10**-8                 #linear dispersion (dx/dl) in cm/angstrom\n",
-      "print\"Linear dispersion in spectrograph is =\",ld,\"cm/angstrom\"\n",
-      "\n",
-      "dlamda=(5896-5890)               #difference in wavelength\n",
-      "dx=ld*dlamda*10**-2              #separation between spectral lines in meter\n",
-      "print\"Separation between spectral lines is =\",dx,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Linear dispersion in spectrograph is = 0.00411696586101 cm/angstrom\n",
-        "Separation between spectral lines is = 0.00024701795166 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.23,Page number 2-47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                              #order\n",
-      "lamda1=5.89*10**-7               #Wavelength of light\n",
-      "lamda2=5.896*10**-7              #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "dlamda=lamda2-lamda1                #difference in wavelength\n",
-      "lamda=(lamda2+lamda1)/2             #Mean wavelength\n",
-      "\n",
-      "#We know that R.P.=m*N=lamda/dlamda\n",
-      "N=lamda/dlamda/m                  #minimum number of lines which will just resolve\n",
-      "print\"Minimum number of lines which will just resolve is =\",N\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum number of lines which will just resolve is = 982.166666667\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.24,Page number 2-47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "N=5*5000           #N=W/(a+b) Number of lines on grating\n",
-      "m=2               #order\n",
-      "lamda=6*10**-7    #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "#(i)\n",
-      "RP=m*N             #Resolving power\n",
-      "print\"(i)Resolving power is =\",RP\n",
-      "\n",
-      "#(ii)\n",
-      "#We know that R.P.=lamda/dlamda\n",
-      "dlamda=lamda/RP       #Smallest wavelength which can be resolved\n",
-      "print\"(ii)Smallest wavelength which can be resolved is =\",dlamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Resolving power is = 50000\n",
-        "(ii)Smallest wavelength which can be resolved is = 1.2e-11 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.25,Page number 2-48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "GE=1./4000*10**-2         #GE=(a+b) grating element\n",
-      "lamda=5*10**-7             #Wavelength of red light\n",
-      "m=3                      #order\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "theta=math.asin(m*lamda/GE)    #ngular position in first minima\n",
-      "\n",
-      "DP=m/(GE*math.cos(theta))*10**-2     #Dispersive power\n",
-      "print\"Dispersive power is =\",DP\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Dispersive power is = 15000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.26,Page number 2-48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=2                  #order\n",
-      "lamda=6*10**-7       #Wavelength of light\n",
-      "dlamda=6*10**-10     #difference in wavelength\n",
-      "W=2*10**-2           #Width of surface\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know that R.P.=lamda/dlamda=m*N\n",
-      "N=lamda/dlamda/m     #Number of lines on grating\n",
-      "GE=W/N               #Grating element=(a+b)\n",
-      "print\"Grating element is =\",GE,\"cm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Grating element is = 4e-05 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.27,Page number 2-49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=2                           #order\n",
-      "lamda1=5.77*10**-7            #Wavelength of light\n",
-      "lamda2=5.791*10**-7           #Wavelength of light\n",
-      "GE=1./6000*10**-2             #GE=(a+b) grating element\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "theta1=math.asin(m*lamda1/GE)*180/math.pi      #angular position in first minima\n",
-      "theta2=math.asin(m*lamda2/GE)*180/math.pi      #angular position in second minima\n",
-      "\n",
-      "a_s=(theta2-theta1)*60         #Angular separation in minutes\n",
-      "print\"Angular separation is = \",a_s,\"minutes\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angular separation is =  12.0270825521 minutes\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28,Page number 2-49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n=1                 #order\n",
-      "lam=5.89*10**-7     #Wavelength of light\n",
-      "a=0.3*10**-3        #width of slit\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, a*sin(theta)=n*lamda\n",
-      "theta1=math.asin(n*lamda/a)*180/math.pi*60    #angular position in first dark band in minutes\n",
-      "print\"Angular position in first dark band is = \",theta1,\"mimutes\"\n",
-      "\n",
-      "#We know,for bright band   a*sin(theta)=(2n+1)*lamda/2\n",
-      "theta2=math.asin(1.5*lamda/a)*180/math.pi*60    #angular position in first bright band in minutes\n",
-      "print\"Angular position in first bright band is =\",theta2,\"minutes\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angular position in first dark band is =  6.87549812524 mimutes\n",
-        "Angular position in first bright band is = 10.3132557823 minutes\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.29,Page number 2-50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "GE=2.54/16000*10**-2      #GE=(a+b) grating element\n",
-      "lamda=6*10**-7             #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "#maximum value of sin(theta)=1\n",
-      "m=GE/lamda                 #Maximum order of spectra\n",
-      "print\"Maximum order of spectra is =\",m\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum order of spectra is = 2.64583333333\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.30,Page number 2-50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "GE=1./5000*10**-2        #GE=(a+b) grating element\n",
-      "lamda=5.89*10**-7         #Wavelength of light\n",
-      "N=3*5000               #N=W/(a+b) Number of lines on grating\n",
-      " \n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, (a+b)*sin(theta)=m*lamda\n",
-      "#maximum value of sin(theta)=1\n",
-      "m=GE/lamda               #Maximum order of spectra\n",
-      "print\"Maximum order of spectra is =\",m\n",
-      "\n",
-      "RP=3*N                 #Resolving power (round of m to 3)\n",
-      "print\"Resolving power is =\",RP\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum order of spectra is = 3.39558573854\n",
-        "Resolving power is = 45000\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.32,Page number 2-52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda1=5.89*10**-7               #Wavelength of light\n",
-      "lamda2=5.896*10**-7              #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "dlamda=lamda2-lamda1                #difference in wavelength\n",
-      "lamda=(lamda2+lamda1)/2             #Mean wavelength\n",
-      "\n",
-      "#(i)\n",
-      "m1=1               #first order\n",
-      "#We know that R.P.=lamda/dlamda=m*N\n",
-      "N1=lamda/dlamda/m1      #Number of lines on grating\n",
-      "print\"(i)Number of lines on grating for first order is =\",N1\n",
-      "\n",
-      "#(ii)\n",
-      "m2=2               #second order\n",
-      "#We know that R.P.=lamda/dlamda=m*N\n",
-      "N2=lamda/dlamda/m2    #Number of lines on grating\n",
-      "print\"(ii)Number of lines on grating for second order is =\",N2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Number of lines on grating for first order is = 982.166666667\n",
-        "(ii)Number of lines on grating for second order is = 491.083333333\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.33,Page number 2-52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                    #order\n",
-      "lamda=6.553*10**-7     #Wavelength of light\n",
-      "dlamda=1.8*10**-10     #difference in wavelength\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know that R.P.=lam/dlam=m*N\n",
-      "N=lamda/dlamda/m        #Number of lines on grating\n",
-      "print\"Number of lines on grating is =\",N\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of lines on grating is = 3640.55555556\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.34,Page number 2-53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda1=5.14034*10**-7               #Wavelength of light\n",
-      "lamda2=5.14085*10**-7               #Wavelength of light\n",
-      "\n",
-      "#Calculations:\n",
-      "dlamda=lamda2-lamda1                #difference in wavelength\n",
-      "lamda=(lamda2+lamda1)/2             #Mean wavelength\n",
-      "\n",
-      "#We know that R.P.=lamda/dlamda=m*N\n",
-      "N=lamda/dlamda/1       #Number of lines on grating\n",
-      "print\"Number of lines on grating for first order is =\",N\n",
-      "\n",
-      "#Hence R.P. for second order should be\n",
-      "RP1=2*N\n",
-      "print\"Resolving power in second order should be is= \",RP1\n",
-      "\n",
-      "#But here,\n",
-      "\n",
-      "lamda3=8.03720*10**-7               #Wavelength of light\n",
-      "lamda4=8.03750*10**-7               #Wavelength of light\n",
-      "dlamda2=lamda4-lamda3               #difference in wavelength\n",
-      "lamda2=(lamda4+lamda3)/2            #Mean wavelength\n",
-      "\n",
-      "RP2=lamda2/dlamda2\n",
-      "print\"Resolving power in second order is= \",RP2\n",
-      "\n",
-      "print\"So, the grating will not be able to resolve 8.0372*10^-7 and 8.03750*10^-7 in second order.\"\n",
-      "print\"Because Resolving power is greter than actual Resolving power.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of lines on grating for first order is = 10079.5980392\n",
-        "Resolving power in second order should be is=  20159.1960784\n",
-        "Resolving power in second order is=  26791.1666667\n",
-        "So, the grating will not be able to resolve 8.0372*10^-7 and 8.03750*10^-7 in second order.\n",
-        "Because Resolving power is greter than actual Resolving power.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.35,Page number 2-53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#For grating , Condition of maxima is (a+b)sin(theta)=n*lamda\n",
-      "#Given (a+b) < 2*lamda\n",
-      "#For maximum order, sin(90)=1\n",
-      "#So, n must be less than 2\n",
-      "#i.e. only first order possible if width of grating element is less than twice the wavelength\n",
-      "print\"Hence, Only first order possible if width of grating element is less than twice the wavelength.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hence, Only first order possible if width of grating element is less than twice the wavelength.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.36,Page number 2-54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n=1                 #order\n",
-      "lamda=5.89*10**-7   #Wavelength of light\n",
-      "a=0.3*10**-3        #width of slit\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, a*sin(theta)=n*lamda\n",
-      "theta1=math.asin(n*lam/a)*180/math.pi    #angular position in first dark band\n",
-      "print\"Angular position in first dark band is =\",theta1,\"Degrees\"\n",
-      "\n",
-      "#We know,for bright band   a*sin(theta)=(2n+1)*lamda/2\n",
-      "theta2=math.asin(1.5*lamda/a)*180/math.pi    #angular position in first bright band\n",
-      "print\"Angular position in first bright band is =\",theta2,\"Degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angular position in first dark band is = 0.112490786047 Degrees\n",
-        "Angular position in first bright band is = 0.168736314576 Degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics-II/chapter3.ipynb b/Applied_Physics-II/chapter3.ipynb
deleted file mode 100755
index 714dfad3..00000000
--- a/Applied_Physics-II/chapter3.ipynb
+++ /dev/null
@@ -1,836 +0,0 @@
-{
- "metadata": {
-  "celltoolbar": "Raw Cell Format",
-  "name": "",
-  "signature": "sha256:5f637d4fa0aa8bb3bc54a28134ac85a937c06f0d3a591f536de2719824a41f5c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Optical Fibre"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1,Page number 3-19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n1=1.61                 #Core index\n",
-      "n2=1.55                 #Cladding index\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(n1**2-n2**2)      #Formula\n",
-      "\n",
-      "print\"Numerical Aperture of Fibre is = \",NA\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture of Fibre is =  0.435430821142\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2,Page number 3-19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n1=1.65                 #Core index\n",
-      "n2=1.53                 #Cladding index\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(n1**2-n2**2)      #Formula\n",
-      "\n",
-      "print\"Numerical Aperture of Fibre is =\",NA\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture of Fibre is = 0.617737808459\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3,Page number 3-19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n1=1.48                 #R.I. of Core\n",
-      "n2=1.39                 #R.I. of Cladding\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(n1**2-n2**2)      #Formula to find NA\n",
-      "phi=math.asin(NA)*180/(3.1472)    #Acceptance angle\n",
-      "\n",
-      "print\"Numerical Aperture of Fibre is =\",NA\n",
-      "print\"Acceptance angle of Fibre is =\",phi,\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture of Fibre is = 0.508232230383\n",
-        "Acceptance angle of Fibre is = 30.491727191 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4,Page number 3-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#given data:\n",
-      "u1=3.6        #Refractive Index of the Substance at 850 nm\n",
-      "u2=3.4        #Refractive Index of the Substance at 1300 nm\n",
-      "Vv=3*10**8     #Velocity of light in free space\n",
-      "\n",
-      "#Calculations:\n",
-      "# i)Finding wavelength at 850 nm\n",
-      "Vs1=Vv/u1             #Velocity of light in substance at 850 nm\n",
-      "print\"Velocity of light in substance at 850 nm =\" ,Vs1,\"m/sec\"\n",
-      "\n",
-      "lam1=850*10**-9/u1    #Wavelength of light in substance at 850nm\n",
-      "print\" Wavelength of light in substance at 850nm =\",lam1,\"m\"\n",
-      "\n",
-      "\n",
-      "#ii)Finding wavelength at 1300 nm\n",
-      "Vs2=Vv/u2            #Velocity of light in substance at 1300 nm\n",
-      "print\"Velocity of light in substance at 1300 nm =\",Vs2,\" m/sec\"\n",
-      "\n",
-      "lam2=1300*10**-9/u2   #Wavelength of light in substance at 1300nm\n",
-      "print\"Wavelength of light in substance at 1300nm =\" ,lam2,\"m \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity of light in substance at 850 nm = 83333333.3333 m/sec\n",
-        " Wavelength of light in substance at 850nm = 2.36111111111e-07 m\n",
-        "Velocity of light in substance at 1300 nm = 88235294.1176  m/sec\n",
-        "Wavelength of light in substance at 1300nm = 3.82352941176e-07 m \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5,Page number 3-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.5                  #R.I. of Core\n",
-      "u2=1.45                 #R.I.of Cladding\n",
-      "delta= (u1-u2)/u1          #Fractional Refractive index\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=u1*sqrt(2*delta)             #Formula to find NA\n",
-      "theta0=math.asin(NA)*180/(3.1472)       #Acceptance angle\n",
-      "thetac=math.asin(u2/u1)*180/(3.1472)   #Critical angle\n",
-      "\n",
-      "print\"Numerical Aperture of Fibre is =\",NA\n",
-      "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\" \n",
-      "print\" Critical angle of Fibre is =\",thetac,\"degrees\" \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture of Fibre is = 0.387298334621\n",
-        "Acceptance angle of Fibre is = 22.7458994397 degrees\n",
-        " Critical angle of Fibre is = 75.0309676099 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6,Page number 3-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "NA=0.22              # Numerical Aperture of Fibre\n",
-      "delta=0.012          #Fractional index\n",
-      "\n",
-      "#Calculations:\n",
-      "#Delta=(u1-u2)/u1\n",
-      "u1=NA/math.sqrt(2*delta)  #Formula\n",
-      "u2=u1-(u1*delta)     #Formula\n",
-      "\n",
-      "print\"Refractive Index of core of fibre is =\",u1\n",
-      "print\"Refractive Index of cladding of fibre is =\",u2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive Index of core of fibre is = 1.42009389361\n",
-        "Refractive Index of cladding of fibre is = 1.40305276689\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7,Page number 3-21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.466                 #R.I. of Core\n",
-      "u2=1.46                  #R.I.of Cladding\n",
-      "V=2.4                    #Cut off parameter\n",
-      "lamda=0.8*10**-6            #wavelength in meter\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(u1**2-u2**2)       #Formula to find Numerical Aperture\n",
-      "print\"Numerical Aperture of Fibre is =\",NA\n",
-      "#(printing mistake in book)printed answer is 1.13 but correct answer is 0.13\n",
-      "print\"(printing mistake in book)\"\n",
-      "\n",
-      "# V = 2*3.142*a*NA / lamda\n",
-      "a=V*lamda/(2*3.142*NA)   #core radius\n",
-      "print\"Core radius of Fibre is (a) =\",a,\"m\"\n",
-      "\n",
-      "#w/a= 1.1\n",
-      "w=1.1*a                  #Spot size\n",
-      "print\"Spot size of Fibre is =\",w,\"m\"\n",
-      "\n",
-      "theta=2*lamda*180/3.142/(3.142*w)       #Divergence angle\n",
-      "print\"Divergence angle of Fibre is =\",theta,\"degrees\"\n",
-      "\n",
-      "w10=lamda*10/(3.142*w)     #Spot size at 10 m\n",
-      "print\"Spot size at 10 m of Fibre is =\",w10,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture of Fibre is = 0.1324990566\n",
-        "(printing mistake in book)\n",
-        "Core radius of Fibre is (a) = 2.30596263706e-06 m\n",
-        "Spot size of Fibre is = 2.53655890076e-06 m\n",
-        "Divergence angle of Fibre is = 11.5009886523 degrees\n",
-        "Spot size at 10 m of Fibre is = 1.00378073182 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8,Page number 3-21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "w=98                   #Spot size in meter\n",
-      "d=50*10**-6             #Core diameter in meter\n",
-      "a=d/2                  #core radius\n",
-      "u1=1.47                #R.I. of Core\n",
-      "u2=1.45                #R.I.of Cladding\n",
-      "lamda=0.85*10**-6         #Wavlength in meter\n",
-      "NA=math.sqrt(u1**2-u2**2)     #Formula to find NA\n",
-      "\n",
-      "#Calculations:\n",
-      "V=2*3.142*a*NA/lamda        #cut off parameter\n",
-      "N=(V**2)/2               #Number of modes\n",
-      "\n",
-      "print\"Cut off parameter of Fibre is =\",V\n",
-      "print\"Number of modes of Fibre is =\",N\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cut off parameter of Fibre is = 44.6646240577\n",
-        "Number of modes of Fibre is = 997.464321107\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9,Page number 3-21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.47                 #R.I. of Core\n",
-      "u2=1.46                 #R.I.of Cladding\n",
-      "lamda=1.3*10**-6           #wavelength in meter\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(u1**2-u2**2)      #Formula to find Numerical Aperture\n",
-      "\n",
-      "#The condition for single mode is V<2.405\n",
-      "#2*3.142*a*NA/lamda < 2.405\n",
-      "\n",
-      "a=2.405*lamda/(2*3.142*NA)   #Maximum radius of fibre\n",
-      "\n",
-      "print\"Maximum radius of Fibre is =\",a,\"meter\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum radius of Fibre is = 2.90662126448e-06 meter\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10,Page number 3-22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.465                  #R.I. of Core\n",
-      "u2=1.46                   #R.I.of Cladding\n",
-      "lamda=1.25*10**-6            #operating wavelength\n",
-      "\n",
-      "#Calculations:\n",
-      "delta=(u1-u2)/u1            #Fractional Refractive index\n",
-      "print\"Fractional Refractive index of Fibre is =\",delta\n",
-      "\n",
-      "#For single mode propagation codition is \n",
-      "# a/lamda < 1.4/(3.142*sqrt(u1(u1-u2)))\n",
-      "\n",
-      "a=lamda*1.4/(3.142*u1*math.sqrt(delta))      #core radius\n",
-      "\n",
-      "u=u1-(math.sqrt(2*delta)/(2*3.142*(a/lamda)))     #effective refractive index\n",
-      "print\"Effective Refractive index for lowest mode propagation is =\",u\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fractional Refractive index of Fibre is = 0.00341296928328\n",
-        "Effective Refractive index for lowest mode propagation is = 1.46247461864\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11,Page number 3-22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.54                  #R.I. of Core\n",
-      "u2=1.5                   #R.I.of Cladding\n",
-      "lamda=1.3*10**-6            #wavelength in meter\n",
-      "a=25*10**-6               #core radius in meter\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(u1**2-u2**2)      #Formula to find Numerical Aperture\n",
-      "\n",
-      "V=2*3.142*a*NA/lamda         #cut off parameter\n",
-      "print\"Cut off parameter of Fibre is =\",V\n",
-      "\n",
-      "N=(V**2)/2                #Number of modes\n",
-      "print\" Number of modes of Fibre is =\",N\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cut off parameter of Fibre is = 42.1404937865\n",
-        " Number of modes of Fibre is = 887.910608284\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11.1,Page number 3-25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.52                  #R.I. of Core\n",
-      "u2=1.5189                #R.I.of Cladding\n",
-      "lamda=1.3*10**-6            #wavelength in meter\n",
-      "d=29*10**-6               #core diameter in meter\n",
-      "a=d/2\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(u1**2-u2**2)     #Formula to find Numerical Aperture\n",
-      "V=2*3.142*a*NA/lamda        #Normalised frequency\n",
-      "Nm=(V**2)/2               #Number of modes\n",
-      "\n",
-      "print\"Normalised frequency of Fibre is (V)=\",V\n",
-      "print\"The Maximum Number of modes the Fibre will support is (Nm) =\",Nm\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Normalised frequency of Fibre is (V)= 4.05242861605\n",
-        "The Maximum Number of modes the Fibre will support is (Nm) = 8.2110888441\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.12,Page number 3-22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u1=1.5                  #R.I. of Core\n",
-      "d=10*10**-6              #diameter of core\n",
-      "a=d/2                   #core radius\n",
-      "lamda=1.3*10**-6           #wavelength\n",
-      "V=2.405                 #cut off parameter for single mode\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, V=2*3.142*a*NA/lamda\n",
-      "NA=V*lamda/(2*3.142*a)       #Numerical Aperture\n",
-      "\n",
-      "theta=math.asin(NA)*180/3.142   #Acceptance angle\n",
-      "print\"Acceptance angle of Fibre is =\",theta,\"Degrees\"\n",
-      "\n",
-      "#Also, NA=u1*sqrt(2*delta)\n",
-      "delta=(NA/u1)**2/2           #Fractional index\n",
-      "print\"Maximum Fractional Refractive index of Fibre is =\",delta\n",
-      "\n",
-      "#delta=(u1-u2)/u1\n",
-      "u2=u1*(1-delta)             #R.I.of cladding\n",
-      "print\"Refractive index of cladding of Fibre is =\",u2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Acceptance angle of Fibre is = 5.71002346964 Degrees\n",
-        "Maximum Fractional Refractive index of Fibre is = 0.00220035113094\n",
-        "Refractive index of cladding of Fibre is = 1.4966994733\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.13,Page number 3-23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n1=1.5               #R.I. of core\n",
-      "delta=0.0005         #Fractional index difference\n",
-      "\n",
-      "#Calculations:\n",
-      "#(a):\n",
-      "#Delta=(u1-u2)/u1\n",
-      "n2=n1-(n1*delta)     #R.I. of cladding\n",
-      "print\"(a)Refractive Index of cladding of fibre is =\",n2\n",
-      "\n",
-      "#(b):\n",
-      "phi=math.asin(n2/n1)*180/3.142        #Critical internal reflection angle\n",
-      "print\"(b)Critical internal reflection angle of Fibre is =\",phi,\"degrees\"\n",
-      "\n",
-      "#(c):\n",
-      "theta0=math.asin(math.sqrt(n1**2-n2**2))*180/3.142    #External critical Acceptance angle\n",
-      "print\"(c)External critical Acceptance angle of Fibre is =\",theta0,\"degrees\"\n",
-      "\n",
-      "#(d):\n",
-      "NA=n1*math.sqrt(2*delta)          #Formula to find Numerical Aperture\n",
-      "print\"(d)Numerical Aperture of Fibre is =\",NA\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a)Refractive Index of cladding of fibre is = 1.49925\n",
-        "(b)Critical internal reflection angle of Fibre is = 88.1766396681 degrees\n",
-        "(c)External critical Acceptance angle of Fibre is = 2.71810509125 degrees\n",
-        "(d)Numerical Aperture of Fibre is = 0.0474341649025\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.14,Page number 3-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "NA1=0.20              #Numerical Aperture of Fibre\n",
-      "n2=1.59               #R.I. of cladding\n",
-      "\n",
-      "#Calculations:\n",
-      "#NA=sqrt(n1**2-n2**2)\n",
-      "#In air, n0=1\n",
-      "n1=math.sqrt(NA1**2+n2**2)     #R.I.of core\n",
-      "\n",
-      "#Now, in water \n",
-      "n0=1.33\n",
-      "NA2=math.sqrt(n1**2-n2**2)/n0      #Numerical Aperture in water\n",
-      "theta0=math.asin(NA2)*180/3.142    #Acceptance angle of fibre in water\n",
-      "print\"Acceptance angle of Fibre in water is =\",theta0,\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Acceptance angle of Fibre in water is = 8.6475921767 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.15,Page number 3-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n1=1.45             #R.I.of core\n",
-      "n2=1.40             #R.I. of cladding\n",
-      "\n",
-      "#Calculations:\n",
-      "NA=math.sqrt(n1**2-n2**2)   #Numerical Aperture\n",
-      "print\"Numerical Aperture of Fibre is =\",NA\n",
-      "\n",
-      "theta0=math.asin(NA)*180/3.142    #Acceptance angle of fibre\n",
-      "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture of Fibre is = 0.377491721764\n",
-        "Acceptance angle of Fibre is = 22.1755250876 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.16,Page number 3-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "NA=0.16             #Numerical Aperture of Fibre\n",
-      "n1=1.45             #R.I. of core\n",
-      "d=90*10**-6          #Core diameter\n",
-      "\n",
-      "#Calculations:\n",
-      "#NA=sqrt(n1**2-n2**2)\n",
-      "n2=math.sqrt(n1**2-NA**2)    #R.I.of cladding\n",
-      "print\"(a)Refractive Index of cladding of fibre is =\",n2\n",
-      "\n",
-      "theta0=math.asin(NA)*180/3.142    #Acceptance angle of fibre\n",
-      "print\"(b)Acceptance angle of Fibre is =\",theta0,\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a)Refractive Index of cladding of fibre is = 1.44114537782\n",
-        "(b)Acceptance angle of Fibre is = 9.20570258795 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.17,Page number 3-25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "n1=1.48               #R.I. of core\n",
-      "delta=0.055           #Realtive R.I.\n",
-      "lamda=1*10**-6        #Wavelength of light\n",
-      "a=50*10**-6           #core radius\n",
-      "\n",
-      "#Calculations:\n",
-      "#Delta=(u1-u2)/u1\n",
-      "n2=n1-(n1*delta)      #R.I. of cladding\n",
-      "NA=n1*math.sqrt(2*delta)  #Formula to find Numerical Aperture\n",
-      "print\"Numerical Aperture of Fibre is =\",NA\n",
-      "\n",
-      "\n",
-      "theta0=math.asin(NA)*180/3.142     #Acceptance angle of fibre\n",
-      "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n",
-      "\n",
-      "V=2*3.142*a*NA/lamda            #V number\n",
-      "N=(V**2)/2                   #Number of guided modes\n",
-      "\n",
-      "#In book,instead of NA , value of delta is taken into calculation.\n",
-      "#Thus there is calculation mistake in values of V and N.\n",
-      "\n",
-      "print\"V number of Fibre is =\",V\n",
-      "print\"Number of guided mode of Fibre is =\",N\n",
-      "print\"(Calculation mistake in book)\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture of Fibre is = 0.490860468973\n",
-        "Acceptance angle of Fibre is = 29.3933421943 degrees\n",
-        "V number of Fibre is = 154.228359351\n",
-        "Number of guided mode of Fibre is = 11893.1934141\n",
-        "(Calculation mistake in book)\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics-II/chapter4.ipynb b/Applied_Physics-II/chapter4.ipynb
deleted file mode 100755
index 11b66b2f..00000000
--- a/Applied_Physics-II/chapter4.ipynb
+++ /dev/null
@@ -1,163 +0,0 @@
-{
- "metadata": {
-  "celltoolbar": "Raw Cell Format",
-  "name": "",
-  "signature": "sha256:e413b3bea25c7729b7ae1f595bf0f843336be8c682782f6b92de1c93aa46ed9d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4: Laser"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1,Page number 4-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=694.3*10**-9         #Wavelength in meter\n",
-      "T=300                   #Temperature in Kelvin\n",
-      "\n",
-      "h=6.63*10**-34           #Planck's Constant\n",
-      "c=3*10**8                #Velocity of light\n",
-      "K=1.38*10**-21           #Boltzmann Constant\n",
-      "\n",
-      "#Calculations:\n",
-      "delE= h*c/lamda           #Energy difference between two energy states N and N0\n",
-      "\n",
-      "#N=N0*e^-delE/(K*T)\n",
-      "R=math.e**(-delE/(K*T))      #R=Ratio of N and N0 i.e.(R=N/N0)\n",
-      "\n",
-      "#(Printing mistake in textbook)\n",
-      "#instead of e^-.692, it has taken e^-69.2\n",
-      "\n",
-      "print\"The ratio of population of two energy states is = \",R\n",
-      "print\" (calculation mistake in book)\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio of population of two energy states is =  0.500588928485\n",
-        " (calculation mistake in book)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2,Page number 4-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=6328*10**-10        #Wavelength in meter\n",
-      "P=4.5*10**-3              #Power in watts\n",
-      "h=6.63*10**-34           #Planck's Constant\n",
-      "c=3*10**8                 #Velocity of light\n",
-      "\n",
-      "#Calculations:\n",
-      "delE= h*c/lamda           #Energy difference\n",
-      "#N*delE=P\n",
-      "N=P/delE                #number of photons emitted per second\n",
-      "\n",
-      "print\"Number of photons emitted per second is =\",N\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of photons emitted per second is = 1.43167420814e+16\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3,Page number 4-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "lamda=780*10**-9           #Wavelength of photon in meter\n",
-      "P=20*10**-3              #Power of each pulse in watts\n",
-      "t=10*10**-9              #Duration of each pulse\n",
-      "h=6.63*10**-34           #Planck's Constant\n",
-      "c=3*10**8                #Velocity of light\n",
-      "\n",
-      "#Calculations:\n",
-      "delE= h*c/lamda           #Energy of each photon\n",
-      "E=P*t                   #Energy of each pulse\n",
-      "\n",
-      "N=E/delE                #Number of photons in each pulse\n",
-      "print\"Number of photons in each pulse is =\",N\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of photons in each pulse is = 784313725.49\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics-II/chapter5.ipynb b/Applied_Physics-II/chapter5.ipynb
deleted file mode 100755
index e9d81103..00000000
--- a/Applied_Physics-II/chapter5.ipynb
+++ /dev/null
@@ -1,1166 +0,0 @@
-{
- "metadata": {
-  "celltoolbar": "Raw Cell Format",
-  "name": "",
-  "signature": "sha256:fec7a7bbab2881090f60069cf3d4c856415ca45a1545225541a225f28ce72b8e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5: Foundations of Quantum Mechanics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1,Page number 5-23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=6.68*10**-27       #mass of alpha particle\n",
-      "V=30*10**3           #potential difference\n",
-      "e=1.6*10**-19        #charge of an electron\n",
-      "q=2*e                #Charge of alpha particle\n",
-      "h=6.63*10**-34       #Planck's constant\n",
-      "\n",
-      "#Calculations:\n",
-      "lamda=h/math.sqrt(2*m*q*V)     #de Broglie wavelength\n",
-      "print\"de Broglie wavelength associated with alpha particle is =\" ,lamda,\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "de Broglie wavelength associated with alpha particle is = 5.85429607723e-14 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2,Page number 5-23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1                  #mass of given particle in kg\n",
-      "h=6.63*10**-34       #Planck's constant\n",
-      "v=1*10**3            #velocity of particle\n",
-      "\n",
-      "#Calculations:\n",
-      "lamda=h/(m*v)         #de Broglie wavelength\n",
-      "print\"de Broglie wavelength associated with particle is =\",lamda,\"m\"\n",
-      "print\"This wavelength is too small for any practical significance.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "de Broglie wavelength associated with particle is = 6.63e-37 m\n",
-        "This wavelength is too small for any practical significance.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3,Page number 5-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m1=40*10**-3         #mass of bullet in kg\n",
-      "m2=9.1*10**-31       #mass of electron in kg\n",
-      "h=6.63*10**-34       #Planck's constant\n",
-      "v=1100              #velocity of bullet and electron\n",
-      "\n",
-      "#Calculations:\n",
-      "lamda1=h/(m1*v)        #de Broglie wavelength\n",
-      "print\"de Broglie wavelength associated with bullet is =\",lamda1,\"m\"\n",
-      "\n",
-      "lamda2=h/(m2*v)        #de Broglie wavelength\n",
-      "print\"de Broglie wavelength associated with electron is =\",lamda2,\"m\"\n",
-      "\n",
-      "print\"Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " de Broglie wavelength associated with bullet is = 1.50681818182e-35 m\n",
-        "de Broglie wavelength associated with electron is = 6.62337662338e-07 m\n",
-        "Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4,Page number 5-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "V=100            #potential difference\n",
-      "d=2.15*10**-10   #lattice spacing\n",
-      "\n",
-      "#Calculations:\n",
-      "lamda=12.26*10**-10/(math.sqrt(V))          #wavelength associated with electron in meter\n",
-      "\n",
-      "#using bragg's law for first order  lamda=2d sin(theta)\n",
-      "theta=math.degrees(math.asin(lamda/(2*d)))      #glancing angle in degrees\n",
-      "print\"Glancing angle at which first reflection occurs is =\",theta,\"Degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Glancing angle at which first reflection occurs is = 16.5657992687 Degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5,Page number 5-25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "mn=1.674*10**-27     #mass of neutron\n",
-      "h=6.63*10**-34       #Planck's constant\n",
-      "lamda=1*10**-10      #wavelength of neutron\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
-      "E1=h**2/(2*mn*lamda**2)   #Energy of neutron in joules\n",
-      "E=E1/(1.6*10**-19)        #Energy of neutron in electron-Volts\n",
-      "\n",
-      "print\"Energy of neutron is =\",E,\"eV\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy of neutron is = 0.0820581317204 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6,Page number 5-25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "mn=1.67*10**-27      #mass of neutron\n",
-      "h=6.6*10**-34        #Planck's constant\n",
-      "lamda=3*10**-10        #wavelength of neutron\n",
-      "d=3.036*10**-10      #lattice spacing\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
-      "E1=h**2/(2*mn*lamda**2)  #Energy of neutron in joules\n",
-      "E=E1/(1.6*10**-19)    # Energy of neutron in electron-Volts\n",
-      "print\"Energy of neutron is =\",E,\"eV\"\n",
-      "\n",
-      "#using bragg's law for first order  lamda=2d sin(theta)\n",
-      "theta=math.degrees(math.asin(lamda/(2*d)))      #glancing angle in degrees\n",
-      "print\" Glancing angle at which first orde reflection occurs is =\",theta,\"Degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy of neutron is = 0.00905688622754 eV\n",
-        " Glancing angle at which first orde reflection occurs is = 29.6085193042 Degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7,Page number 5-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=9.108*10**-31       #mass of electron\n",
-      "h=6.625*10**-34       #Planck's constant\n",
-      "lamda=5*10**-7          #wavelength of electron\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
-      "E1=h**2/(2*m*lamda**2)     #Energy of electron in joules\n",
-      "E=E1/(1.6*10**-19)      #Energy of electron in electron-Volts\n",
-      "print\"Energy of electron is =\",E,\"eV\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy of electron is = 6.02363650088e-06 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8,Page number 5-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "mn=1.676*10**-27        #mass of neutron\n",
-      "me=9.1*10**-31          #mass of electron\n",
-      "h=6.625*10**-34         #Planck's constant\n",
-      "\n",
-      "#Calculations:\n",
-      "#Part 1:\n",
-      "En1=0.025              #Energy in eV of neutron\n",
-      "En=En1*(1.6*10**-19)   #Energy in joules\n",
-      "\n",
-      "lamda1=h/math.sqrt(2*mn*En)     #wavelength of a beam of neutron\n",
-      "print\"wavelength of a beam of neutron is =\",lamda1,\"m\"\n",
-      "\n",
-      "#Part 2:\n",
-      "lamda2=2*10**-10            #wavelength of electron and photon\n",
-      "\n",
-      "#//we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
-      "Ee1=h**2/(2*me*lamda2**2)     #Energy of electron in joules\n",
-      "Ee=Ee1/(1.6*10**-19)          #Energy of electron in electron-Volts\n",
-      "print\"Energy of electron is =\",Ee,\"eV\"\n",
-      "\n",
-      "p1=h/lamda2               #momentum of electron\n",
-      "print\" Momentum of electron is =\",p1,\"kg.m/s\"\n",
-      "\n",
-      "C=3*10**8                 #Velocity of light\n",
-      "Ep=h*C/lamda2             #Energy of photon in joules\n",
-      "print\"Energy of photon is =\",Ep,\"Joules\"\n",
-      "\n",
-      "p2=h/lamda2              #momentum of photon\n",
-      "print\"Momentum of photon is =\",p2,\"kg.m/s\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "wavelength of a beam of neutron is = 1.80927208246e-10 m\n",
-        "Energy of electron is = 37.6808250343 eV\n",
-        " Momentum of electron is = 3.3125e-24 kg.m/s\n",
-        "Energy of photon is = 9.9375e-16 Joules\n",
-        "Momentum of photon is = 3.3125e-24 kg.m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9,Page number 5-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given data:\n",
-      "#We have alpha particle,neutron,proton and electron.\n",
-      "\n",
-      "#To find: shortest wavelength\n",
-      "\n",
-      "print\"We know, lamda=h/sqrt(2*m*E)     #de Broglie wavelength\"\n",
-      "\n",
-      "#Wavelength is inversely proportional to mass of particle for constant energy\n",
-      "print\"i.e., Wavelength is inversely proportional to mass of particle for constant energy. \"\n",
-      "\n",
-      "print\"We have alpha particle,neutron,proton and electron.\"\n",
-      "\n",
-      "#AS,alpha particle has highest mass.Thus it will have shortest wavelength.\n",
-      "print\"Out of above, alpha particle has highest mass.\"\n",
-      "\n",
-      "print\"Hence it will have shortest wavelength.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "We know, lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
-        "i.e., Wavelength is inversely proportional to mass of particle for constant energy. \n",
-        "We have alpha particle,neutron,proton and electron.\n",
-        "Out of above, alpha particle has highest mass.\n",
-        "Hence it will have shortest wavelength.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10,Page number 5-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "me=9.108*10**-31        # mass of electron\n",
-      "mp=1.66*10**-27         # bmass of proton\n",
-      "h=6.625*10**-34         # Planck's constant\n",
-      "lamda=1*10**-10         # wavelength of electron and proton\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
-      "Ee1=h**2/(2*me*lamda**2)           #Energy of electron in joules\n",
-      "Ee=Ee1/(1.6*10**-19)               #Energy of electron in electron-Volts\n",
-      "print\"Energy of electron is =\",Ee,\"eV\"\n",
-      "\n",
-      "Ep1=h**2/(2*mp*lamda**2)      #Energy of photon in joules\n",
-      "Ep=Ep1/(1.6*10**-19)          #Energy of photon in electron-Volts\n",
-      "print\"Energy of photon is =\",Ep,\"eV\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy of electron is = 150.590912522 eV\n",
-        "Energy of photon is = 0.0826254235693 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.11,Page number 5-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m1=50*10**-9         #mass of particle in kg\n",
-      "m2=9.1*10**-31       #mass of electron in kg\n",
-      "h=6.625*10**-34      #Planck's constant\n",
-      "v1=1                #velocity of particle\n",
-      "v2=3*10**6           #velocity of electron\n",
-      "\n",
-      "#Calculations:\n",
-      "lamda1=h/(m1*v1)*10**10        #de Broglie wavelength\n",
-      "print\"de Broglie wavelength associated with particle is =\",lamda1,\"Angstrom\"\n",
-      "\n",
-      "lamda2=h/(m2*v2)*10**10        #de Broglie wavelength\n",
-      "print\"de Broglie wavelength associated with electron is =\",lamda2,\"Angstrom\"\n",
-      "\n",
-      "print\"Wavelength of electron is measurable.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "de Broglie wavelength associated with particle is = 1.325e-16 Angstrom\n",
-        "de Broglie wavelength associated with electron is = 2.42673992674 Angstrom\n",
-        "Wavelength of electron is measurable.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12,Page number 5-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "me=9.1*10**-31           #mass of electron in kg\n",
-      "h=6.63*10**-34          #Planck's constant\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "E1=2*10**3              #Energy in eV of electron\n",
-      "E=E1*(1.6*10**-19)      #Energy in joules\n",
-      " \n",
-      "lamda=h/math.sqrt(2*me*E)      #wavelength of electron\n",
-      "print\"Wavelength of electron is =\",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of electron is = 2.7472794985e-11 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.13,Page number 5-30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "me=9.1*10**-31            #mass of electron\n",
-      "h=6.63*10**-34            #Planck's constant\n",
-      "lamda=2*10**-10           #wavelength of electron and photon\n",
-      "\n",
-      "#Calculations:\n",
-      "p1=h/lamda                #momentum of electron\n",
-      "print\"Momentum of electron is =\",p1,\"kg.m/s\"\n",
-      "\n",
-      "Ee=p1**2/(2*me)          #Energy of electron in joules\n",
-      "print\"Energy of electron is =\",Ee,\"Joules\"\n",
-      "\n",
-      "p2=h/lamda                #momentum of photon\n",
-      "print\"Momentum of photon is =\",p2,\"kg.m/s\"\n",
-      "\n",
-      "c=3*10**8                #Velocity of light\n",
-      "Ep=h*c/lamda             #Energy of photon in joules\n",
-      "print\"Energy of photon is =\",Ep,\"Joules\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Momentum of electron is = 3.315e-24 kg.m/s\n",
-        "Energy of electron is = 6.03803571429e-18 Joules\n",
-        "Momentum of photon is = 3.315e-24 kg.m/s\n",
-        "Energy of photon is = 9.945e-16 Joules\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14,Page number 5-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1.676*10**-27            #mass of neutron\n",
-      "h=6.625*10**-34            #Planck's constant\n",
-      "lamda=1*10**-10            #wavelength of neutron\n",
-      "\n",
-      "#Calculations:\n",
-      "C=3*10**8                 #Velocity of light\n",
-      "Ep1=h*C/lamda             #Energy of photon in joules\n",
-      "E1=Ep1/(1.6*10**-19)      #Energy of photon in electron-Volts\n",
-      "print\"Energy of photon is =\",E1,\"eV\"\n",
-      "\n",
-      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
-      "En1=h**2/(2*m*lamda**2)     #Energy of neutron in joules\n",
-      "E2=En1/(1.6*10**-19)     #Energy of neutron in electron-Volts\n",
-      "print\"Energy of neutron is =\",E2,\"eV\"\n",
-      "\n",
-      "R=E1/E2                 #Ratio of energies of proton to neutron\n",
-      "print\"Ratio of energies of proton to neutron is =\",R\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy of photon is = 12421.875 eV\n",
-        "Energy of neutron is = 0.0818366367094 eV\n",
-        "Ratio of energies of proton to neutron is = 151788.679245\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Example 5.14.1,Page number 5-36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "v=900                  #velocity of electron in m/s\n",
-      "delv=v*0.001/100       #uncertainity in velocity\n",
-      "h=6.63*10**-34         #Planck's constant\n",
-      "m=9.1*10**-31           #mass of an electron\n",
-      "\n",
-      "#Calculations:\n",
-      "delp=m*delv            #uncertainity in the measured values of momentum\n",
-      "\n",
-      "#using heisenberg's uncertainity formula\n",
-      "delx=h/(2*3.142*delp)    #uncertainity in its position\n",
-      "print\"Uncertainity with which position of electron can be located is >=\",delx,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Uncertainity with which position of electron can be located is >= 0.0128823012337 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14.2,Page number 5-37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1.6*10**-27        #mass of proton in kg\n",
-      "h=6.63*10**-34       #Planck's constant\n",
-      "v=3./20*10**8         #velocity of particle\n",
-      "\n",
-      "#Calculations:\n",
-      "lamda=h/(m*v)        #de Broglie wavelength\n",
-      "print\"de Broglie wavelength associated with proton is =\",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "de Broglie wavelength associated with proton is = 2.7625e-14 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14.3,Page number 5-37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1.676*10**-27        #mass of neutron\n",
-      "h=6.634*10**-34        #Planck's constant\n",
-      "\n",
-      "#Calculations:\n",
-      "E1=0.025              #Energy in eV of neutron\n",
-      "E=E1*(1.6*10**-19)    #Energy in joules\n",
-      "#As E=m*v**2/2\n",
-      "v=math.sqrt(2*E/m)        #Velocity of neutron beam\n",
-      "\n",
-      "lamda=h/(m*v)          #wavelength of a beam of neutron\n",
-      "print\"wavelength of a beam of neutron is =\",lamda,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "wavelength of a beam of neutron is = 1.81172996152e-10 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14.4,Page number 5-37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "delx=10*10**-9          #uncertainity in position of electron\n",
-      "h=6.63*10**-34          #Planck's constant\n",
-      "m=9.1*10**-31           #mass of an electron\n",
-      "E=10**3*1.6*10**-19      #Energy of electron in joules\n",
-      "\n",
-      "#Calculations:\n",
-      "p=math.sqrt(2*m*E)          #momentum of electron\n",
-      "#using heisenberg's uncertainity formula\n",
-      "delp=h/(2*math.pi*delx)    #uncertainity in the momentum\n",
-      "\n",
-      "P=delp/p*100           #percentage of uncertainity in momentum\n",
-      "print\"Percentage of uncertainity in momentum of electron is =\",P,\"percent\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage of uncertainity in momentum of electron is = 0.0618355139385 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15,Page number 5-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1.676*10**-27          #mass of neutron\n",
-      "h=6.63*10**-34           #Planck's constant\n",
-      "lamda=2*10**-12            #wavelength of neutron\n",
-      "c=3*10**8                #Velocity of light\n",
-      "\n",
-      "#Calculations:\n",
-      "p=h/lamda                 #momentum of neutron\n",
-      "KE=p**2/(2*m)            #Kinetic Energy of neutron in joules\n",
-      "print\"Kinetic Energy of electron is =\",KE,\"Joules\"\n",
-      "\n",
-      "#velocity of particle is same as group velocity. Thus,\n",
-      "vg=p/m                  #group velocity\n",
-      "print\"group velocity of neutron is =\",vg,\"m/s\"\n",
-      "\n",
-      "#using, vg*vp=c**2\n",
-      "vp=c**2/vg               #phase velocity\n",
-      "print\" phase velocity of neutron is =\",vp,\"m/s\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Kinetic Energy of electron is = 3.27840841289e-17 Joules\n",
-        "group velocity of neutron is = 197792.362768 m/s\n",
-        " phase velocity of neutron is = 4.55022624434e+11 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.16,Page number 5-32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "m=1.157*10**-30         #mass of particle in kg\n",
-      "h=6.63*10**-34          #Planck's constant\n",
-      "c=3*10**8               #Velocity of light\n",
-      "\n",
-      "#Calculations:\n",
-      "E1=80                  #Energy in eV of particle\n",
-      "E=E1*(1.6*10**-19)     #Energy in joules\n",
-      " \n",
-      "lamda=h/math.sqrt(2*m*E)     #wavelength of particle\n",
-      "print\"Wavelength of particle is =\",lamda,\"m\"\n",
-      "\n",
-      "#Now,\n",
-      "vg=h/(lamda*m)          #group velocity\n",
-      "print\"Group velocity of particle is =\",vg,\"m/s\"\n",
-      "\n",
-      "#using, vg*vp=c**2\n",
-      "vp=c**2/vg              #phase velocity\n",
-      "print\"Phase velocity of particle is =\",vp,\"m/s\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of particle is = 1.21822320075e-10 m\n",
-        "Group velocity of particle is = 4703848.2563 m/s\n",
-        "Phase velocity of particle is = 19133270270.7 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.17,Page number 5-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "v=400                #velocity of electron in m/s\n",
-      "delv=0.01/100       #uncertainity in velocity\n",
-      "h=6.63*10**-34       #Planck's constant\n",
-      "m=9.11*10**-31       #mass of an electron\n",
-      "\n",
-      "#Calculations:\n",
-      "p=m*v                #momentum of an electron\n",
-      "delp=p*delv          #uncertainity in the measured values of momentum\n",
-      "\n",
-      "#using heisenberg's uncertainity formula\n",
-      "delx=h/(2*math.pi*delp)    #accuracy in its position\n",
-      "print\"Accuracy in its position is >=\",delx,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Accuracy in its position is >= 0.00289571150576 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.18,Page number 5-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Given Data:\n",
-      "delx=10**-8             #maximum uncertainity in position of electron\n",
-      "h=6.63*10**-34          #Planck's constant\n",
-      "m=9.1*10**-31           #mass of an electron\n",
-      "\n",
-      "#Calculations:\n",
-      "#using heisenberg's uncertainity formula\n",
-      "delp=h/(2*math.pi*delx)    #minimum uncertainity in the measured values of momentum\n",
-      "\n",
-      "delv=delp/m            #minimum uncertainity in the velocity of an electron\n",
-      "print\"Minimum uncertainity in the velocity of an electron is =\",delv,\"m/s\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum uncertainity in the velocity of an electron is = 11595.5744253 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.19,Page number 5-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "delv=2*10**4            #uncertainity in velocity\n",
-      "h=6.63*10**-34          #Planck's constant\n",
-      "m=9.1*10**-31           #mass of an electron\n",
-      "\n",
-      "#Calculations:\n",
-      "delp=m*delv             #uncertainity in the measured values of momentum\n",
-      "\n",
-      "#using heisenberg's uncertainity formula\n",
-      "delx=h/(2*math.pi*delp)    #accuracy in its position\n",
-      "print\"Minimum space required by electron to be confined in an atom is >=\",delx,\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum space required by electron to be confined in an atom is >= 5.79778721263e-09 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.20,Page number 5-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "delt=1.4*10**-10        #uncertainity in time spent by nucleus in excited state\n",
-      "h=6.63*10**-34          #Planck's constant\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#using, delE*delt>= h/(2*math.pi)\n",
-      "delE1= h/(2*math.pi*delt)       #uncertaininty in its energy in excited state in joules\n",
-      "delE=delE1/(1.6*10**-19)     #uncertaininty in its energy in excited state in eV\n",
-      "print\"Uncertaininty in its energy in excited state is >=\",delE,\"eV\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Uncertaininty in its energy in excited state is >= 4.71070211026e-06 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.21,Page number 5-35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "a=2*10**-10             #width of potential well in m\n",
-      "h=6.63*10**-34          #Planck's constant\n",
-      "m=9.1*10**-31           #mass of an electron\n",
-      "\n",
-      "#Calculations:\n",
-      "#we know equation for energy of an electron\n",
-      "n0=1\n",
-      "E01=n0**2*h**2/(8*m*a**2)   #Energy in ground state\n",
-      "E0=E01/(1.6*10**-19)      #Energy in eV\n",
-      "print\"Energy of an electron in ground state is=\",E0,\"eV\"\n",
-      "\n",
-      "n1=2\n",
-      "E11=n1**2*h**2/(8*m*a**2)  #Energy in first excited state\n",
-      "E1=E11/(1.6*10**-19)      #Energy in eV\n",
-      "print\" Energy of an electron in first excited state is=\",E1,\"eV\"\n",
-      "\n",
-      "\n",
-      "n2=3\n",
-      "E21=n2**2*h**2/(8*m*a**2)   #Energy in second excited state\n",
-      "E2=E21/(1.6*10**-19)      #Energy in eV\n",
-      "print\"Energy of an electron in second excited state is=\",E2,\"eV\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy of an electron in ground state is= 9.43443080357 eV\n",
-        " Energy of an electron in first excited state is= 37.7377232143 eV\n",
-        "Energy of an electron in second excited state is= 84.9098772321 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.22,Page number 5-36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "a=25*10**-10               #width of well\n",
-      "delx=5*10**-10             #uncertainity in position of particle\n",
-      "n=1                        #ground state\n",
-      "\n",
-      "#calculation:\n",
-      "x1=a/2\n",
-      "psi1=math.sqrt(2/a)*math.sin(n*math.pi/a*x1)\n",
-      "P1=(psi1**2)*delx            #Probability of finding particle at distance of x1\n",
-      "print\"Probability of finding particle at a distance of x1 is =\",P1\n",
-      "\n",
-      "x2=a/3\n",
-      "psi2=math.sqrt(2/a)*math.sin(n*math.pi/a*x2)\n",
-      "P2=(psi2**2)*delx             #Probability of finding particle at distance of x2\n",
-      "print\"Probability of finding particle at a distance of x2 is =\",P2\n",
-      "print\"(There is print mistake in book).\"\n",
-      "\n",
-      "x3=a\n",
-      "psi3=math.sqrt(2/a)*math.sin(n*math.pi/a*x3)\n",
-      "P3=(psi3**2)*delx             #Probability of finding particle at distance of x3\n",
-      "print\"Probability of finding particle at a distance of x3 is =\",P3\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Probability of finding particle at a distance of x1 is = 0.4\n",
-        "Probability of finding particle at a distance of x2 is = 0.3\n",
-        "(There is print mistake in book).\n",
-        "Probability of finding particle at a distance of x3 is = 5.99903913065e-33\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics-II/chapter6.ipynb b/Applied_Physics-II/chapter6.ipynb
deleted file mode 100755
index dec8353e..00000000
--- a/Applied_Physics-II/chapter6.ipynb
+++ /dev/null
@@ -1,1183 +0,0 @@
-{
- "metadata": {
-  "celltoolbar": "Raw Cell Format",
-  "name": "",
-  "signature": "sha256:bc506d3130781232c57938d478c5c5179ee56efb068c1be93d57258ec96fcd79"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6: Magnetic Materials and Circuits"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1,Page number 2-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "H=198                  #Magnetizing Force in Ampere per meter\n",
-      "M=2300                 #Magnetization in Ampere per meter\n",
-      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
-      "\n",
-      "#Calculations:\n",
-      "#H=(B/u0)-M\n",
-      "B=u0*(H+M)             #Flux Density\n",
-      "ur=B/(u0*H)            #Relative Permeability\n",
-      "\n",
-      "print\"Corresponding Flux Density is =\",B,\"Wb/m^2\"\n",
-      "print\"Relative Permeability is =\",ur\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Corresponding Flux Density is = 0.00313907937947 Wb/m^2\n",
-        "Relative Permeability is = 12.6161616162\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2,Page number 2-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "x=3.7*10**-3           #Susceptibility at T=300 K\n",
-      "T=300                  #Temperature in kelvin\n",
-      "T1=250                 #Temperature in kelvin\n",
-      "T2=600                 #Temperature in kelvin\n",
-      "\n",
-      "#Calculations:\n",
-      "C=x*T                  #Curie's law\n",
-      "ur1=C/T1               #Relative permeability at 250 K\n",
-      "ur2=C/T2               #Relative permeability at 600 K\n",
-      "\n",
-      "print\"Relative Permeability at 250 K is =\",ur1\n",
-      "print\"Relative Permeability at 600 K is =\",ur2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Relative Permeability at 250 K is = 0.00444\n",
-        "Relative Permeability at 600 K is = 0.00185\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3,Page number 2-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u=0.8*10**-23          #Magnetic dipole moment of an atom in paramagnetic gas in J/T\n",
-      "B=0.8                  #Magnetic field in tesla\n",
-      "K=1.38*10**-23         #Boltzmann constant\n",
-      "\n",
-      "#To find Temperature at which Average thermal energy is equal to Magnetic energy \n",
-      "#i.e. uB=3KT/2\n",
-      "T=2*u*B/(3*K)          #Required temperature\n",
-      "\n",
-      "print\"Required temperature is =\",T,\"Kelvin\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required temperature is = 0.309178743961 Kelvin\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4,Page number 2-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "T=27+273           #Temperature in kelvin\n",
-      "B=0.5              #Magnetic field in tesla\n",
-      "C=2*10**-3         #Curie's Constant\n",
-      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
-      "\n",
-      "#  C=u0*M*T/B (Curie's law)\n",
-      "M=C*B/(u0*T)      #Magnetization of material at 300 K\n",
-      "\n",
-      "print\"Magnetization of material at 300 K is =\",M,\"A/m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnetization of material at 300 K is = 2.65258238486 A/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5,Page number 2-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "B=10.9*10**-5            #Horizontal component of B in wb/m^2\n",
-      "u0=4*math.pi*10**-7     #Permeability in free space\n",
-      "\n",
-      "H=B/u0                  #Horizontal component of magnetic field\n",
-      "print\"Horizontal component of magnetic field is =\",H,\"Ampere/meter\"\n",
-      "print\"(Print mistake in unit in book)\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horizontal component of magnetic field is = 86.7394439851 Ampere/meter\n",
-        "(Print mistake in unit in book)\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6,Page number 2-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "ur=900                  #Relative permeability of medium\n",
-      "l=2                      #length in meter\n",
-      "A=60*10**-4             #Crosss sectional area of ring in m^2\n",
-      "phi=5.9*10**-3          #flux in weber\n",
-      "n=700                   #Number of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, phi=B*A\n",
-      "B=phi/A            #Flux density\n",
-      "#But, B=u*H\n",
-      "H=B/(u0*ur)        #Magnetic field strength\n",
-      "\n",
-      "I=H*l/n            #Required current\n",
-      "print\"Current required to produce given flux is =\",I,\"Ampere\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current required to produce given flux is = 2.48416445567 Ampere\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7,Page number 2-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "\n",
-      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
-      "ur=900                 #Relative permeability of medium\n",
-      "r=25*10**-2            #radius of ring\n",
-      "A=25*10**-4            #Crosss sectional area of ring in m^2\n",
-      "Ag=1*10**-3            #Air gap\n",
-      "phi=2.7*10**-3         #flux in weber\n",
-      "N=400                  #Number of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, phi=B*A\n",
-      "B=phi/A            #Flux density\n",
-      "#But, B=u*H\n",
-      "H=B/(u0*ur)        #Magnetic field strength\n",
-      "L=H*2*math.pi*r+(B*Ag/u0)   #Total amp turns required (iron+air)\n",
-      "I=L/N           #Required current\n",
-      "\n",
-      "print\"Current required to produce given flux is =\",I,\"Ampere\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current required to produce given flux is = 5.89859173174 Ampere\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8,Page number 2-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "A=0.2*10**-4            #Crosss sectional area of iron bar in m^2\n",
-      "H=1600                  #magnetising field in A/m\n",
-      "phi=2.4*10**-5          #Magnetic flux in weber\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know, phi=B*A\n",
-      "B=phi/A            #Flux density\n",
-      "u=B/H              #magnetic permeability\n",
-      "ur=u/u0            #relative permeability\n",
-      "xm=ur-1            #susceptibility of the iron bar\n",
-      "\n",
-      "print\"magnetic permeability of iron bar is =\",u,\"N/(A^2)\"\n",
-      "print\"susceptibility of the iron bar is =\",xm\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "magnetic permeability of iron bar is = 0.00075 N/(A^2)\n",
-        "susceptibility of the iron bar is = 595.831036595\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9,Page number 2-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7       #Permeability in vacuum\n",
-      "xm=948*10**-11            #susceptibility of the iron bar\n",
-      "\n",
-      "#Calculations:\n",
-      "ur=1+xm            #relative permeability\n",
-      "u=u0*ur            #permeability of medium\n",
-      "\n",
-      "print\"Relative Permeability of medium is =\",ur\n",
-      "print\"Permeability of medium is =\",u,\"H/m\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Relative Permeability of medium is = 1.00000000948\n",
-        "Permeability of medium is = 1.25663707335e-06 H/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10,Page number 2-30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "B=2.5                #Magnetic field in tesla\n",
-      "u0=4*math.pi*10**-7    #Permeability in free space\n",
-      "i0=0.7               #current in the core\n",
-      "ri=11*10**-2         #inner radii of core\n",
-      "ro=12*10**-2         #outer radii of core\n",
-      "\n",
-      "#Calculations:\n",
-      "r=(ri+ro)/2          #Average radii of core\n",
-      "n=3000/(2*math.pi*r)    #Number of turns\n",
-      "\n",
-      "#We know, B=u0*ur*n*i0 .Thus,\n",
-      "ur=B/(u0*n*i0)\n",
-      "\n",
-      "print\"Relative Permeability of medium is =\",ur\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Relative Permeability of medium is = 684.523809524\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11,Page number 2-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "B=1.0                #Flux density in tesla\n",
-      "u0=4*math.pi*10**-7     #Permeability in free space\n",
-      "i=2.0                #current in the core\n",
-      "n=10*100             #n=N/l i.e. turns per meter\n",
-      "\n",
-      "#Calculations:\n",
-      "H=n*i               #Magnetising force produced in wire\n",
-      "print\"Magnetising force produced in wire is =\",H,\"Amp-turn/meter\"\n",
-      "\n",
-      "#We know that, B=u0(H+I).Thus,\n",
-      "I=B/u0-H            #Magnetisation of material\n",
-      "print\"Magnetisation of material is =\",I,\"Amp-turn/meter\"\n",
-      "\n",
-      "#u=B/H, i.e. ur*u0=B/H.\n",
-      "ur=B/(u0*H)        #Relative permeability of core\n",
-      "print\" Relative Permeability of core is =\",ur\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnetising force produced in wire is = 2000.0 Amp-turn/meter\n",
-        "Magnetisation of material is = 793774.715459 Amp-turn/meter\n",
-        " Relative Permeability of core is = 397.88735773\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12,Page number 2-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "M=40             #Mass of an iron core\n",
-      "D=7.5*10**3      #Density of iron\n",
-      "f=100            #Frequency\n",
-      "A=3800*10**-1    #Loss due to Area of hysterisis loop in J/m^3\n",
-      "\n",
-      "#Calculations:\n",
-      "V=M/D            #Volume of iron core\n",
-      "L1=A*V           #Loss of energy in core per cycle\n",
-      "print\"Loss of energy in core per cycles is =\",L1,\"joules\"\n",
-      "\n",
-      "N=f*60           #Number of cycles per minute\n",
-      "L=L1*N           #Loss of energy per minute\n",
-      "\n",
-      "print\"Loss of energy per minute is =\",L,\"joules\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loss of energy in core per cycles is = 2.02666666667 joules\n",
-        "Loss of energy per minute is = 12160.0 joules\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13,Page number 2-32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "l=30*10**-2             #length in meter\n",
-      "A=1*10**-4          #Crosss sectional area of ring in m^2\n",
-      "phi=2*10**-6        #flux in weber\n",
-      "N=300               #Number of turns\n",
-      "I=0.032             #Current in winding\n",
-      "\n",
-      "#Calculations:\n",
-      "#(i):\n",
-      "B=phi/A             #Flux density\n",
-      "print\"(i)Flux Density in the ring is =\",B,\"Wb/m^2\"\n",
-      "\n",
-      "#(ii):\n",
-      "H=N*I/l            #Magnetic intensity\n",
-      "print\"(ii)Magnetic intensity is =\",H,\"Amp-turn/meter\"\n",
-      "\n",
-      "#(iii):\n",
-      "u=B/H              #Permeability of ring\n",
-      "print\"(iii)Permeability of ring is =\",u,\" Wb/A-m\"\n",
-      "ur=u/u0            #Relative permeability of ring\n",
-      "print\"Relative Permeability of ring is =\",ur\n",
-      "\n",
-      "#(iv):\n",
-      "xm=ur-1            #susceptibility of the ring\n",
-      "print\"(iv)Magnetic susceptibility of the ring is =\",xm\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Flux Density in the ring is = 0.02 Wb/m^2\n",
-        "(ii)Magnetic intensity is = 32.0 Amp-turn/meter\n",
-        "(iii)Permeability of ring is = 0.000625  Wb/A-m\n",
-        "Relative Permeability of ring is = 497.359197162\n",
-        "(iv)Magnetic susceptibility of the ring is = 496.359197162\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14,Page number 2-32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Dta:\n",
-      "M=12*10**3       #Mass of an iron core in grams\n",
-      "D=7.5            #Density of iron in gm/cc\n",
-      "f=50             #Frequency\n",
-      "A=3000           #loss due to Area of hysterisis loop in ergs/cm^3\n",
-      "\n",
-      "#Calculations:\n",
-      "V=M/D            #Volume of iron core\n",
-      "L1=A*V           #Loss of energy in core per cycle\n",
-      "\n",
-      "L=L1*f*3600      #Loss of energy per hour\n",
-      "\n",
-      "print\"Loss of energy per hour is =\",L,\"Erg\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loss of energy per hour is = 8.64e+11 Erg\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15,Page number 2-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "A=0.5*10**3            #Area of B-H loop in Joules per m^3\n",
-      "V=10**-3               #Volume of specimen in m^3\n",
-      "n=50                   #Frequency of a.c.\n",
-      "\n",
-      "#Calculations:\n",
-      "H=n*V*A               #Hysteresis power loss\n",
-      "\n",
-      "print\"Hysteresis power loss is =\",H,\"Watt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hysteresis power loss is = 25.0 Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16,Page number 2-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "ur=1000             #Relative permeability of medium\n",
-      "V=10**-4            #Volume of iron rod in m^3\n",
-      "n=500               #Number of turns per meter\n",
-      "i=0.5               #Current in windings of solenoid in Amperes\n",
-      "\n",
-      "#Calculations:\n",
-      "#We know I=(ur-1)H\n",
-      "#and H=ni , hence\n",
-      "I=(ur-1)*n*i       #Intensity of magnetisation\n",
-      "M=I*V              #Magnetic moment\n",
-      "\n",
-      "print\"Magnetic moment of the rod is =\",M,\"A-m^2\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnetic moment of the rod is = 24.975 A-m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17,Page number 2-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "ur=600             #Relative permeability of iron\n",
-      "d=12*10**-2        #mean diameter of ring in m\n",
-      "N=500              #Number of turns\n",
-      "i=0.3              #Current in windings of solenoid in Amperes\n",
-      "\n",
-      "#Calculations:\n",
-      "r=d/2              #Radius of ring\n",
-      "\n",
-      "B=u0*ur*N*i/(2*math.pi*r)     #Flux densityin the core\n",
-      "print\"Flux densityin the core is =\",B,\"Wb/m^2\"\n",
-      "\n",
-      "H=B/(u0*ur)         #Magnetic intensity\n",
-      "print\"Magnetic intensity is =\",H,\"Amp-turns/m\"\n",
-      "\n",
-      "#We know that, B=u0(H+I)\n",
-      "I1=(B-u0*H)/u0      #magnetisation\n",
-      "I2=u0*I1            #Electronic current loop\n",
-      "\n",
-      "I=I2/B*100          #Percentage flux density due to electroniuc loop currents\n",
-      "print\"Percentage flux density due to electroniuc loop currents is =\",I,\"percent\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flux densityin the core is = 0.3 Wb/m^2\n",
-        "Magnetic intensity is = 397.88735773 Amp-turns/m\n",
-        "Percentage flux density due to electroniuc loop currents is = 99.8333333333 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18,Page number 2-35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "\n",
-      "u0=4*math.pi*10**-7   #Permeability in vacuum\n",
-      "ur=900               #Relative permeability of iron ring\n",
-      "d=40*10**-2          #diameter of ring\n",
-      "l=5*10**-3           #air gap in the ring\n",
-      "A=5.8*10**-4         #Crosss sectional area of ring in m^2\n",
-      "phi=1.5*10**-4       #flux in weber\n",
-      "N=600                #Number of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "r=d/2                #Radius of ring\n",
-      "\n",
-      "#We know, phi=B*A\n",
-      "B=phi/A              #Flux density\n",
-      "\n",
-      "#But, B=u*H\n",
-      "H=B/(u0*ur)          #Magnetic field strength\n",
-      "\n",
-      "m1=H*ur*l            #amp-turns in air gap\n",
-      "m2=H*2*math.pi*r     #amp-turns by ring\n",
-      "m=m1+m2              #total mmf(amp-turns) required\n",
-      "\n",
-      "I=m/N                #Required current\n",
-      "print\"Current required to produce given flux is =\",I,\"Amperes\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current required to produce given flux is = 2.19395891742 Amperes\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18.1,Page number 2-38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
-      "X=-0.5*10**-5          #Magnetic susceptibility of silicon\n",
-      "H=9.9*10**4            #Magnetic field intensity\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#As, X=I/H. thus,\n",
-      "I=X*H                  #intensity of magnetisation\n",
-      "print\"Intensity of magnetisation is =\",I\n",
-      "\n",
-      "B=u0*(H+I)             #Magnetic flux density\n",
-      "print\"Magnetic flux density is =\",B,\"Wb/ m^2\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Intensity of magnetisation is = -0.495\n",
-        "Magnetic flux density is = 0.124406447047 Wb/ m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18.2,Page number 2-38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "ur=380              #Relative permeability\n",
-      "d=20*10**-2         #diameter of solenoid in m\n",
-      "r=d/2               #radius of ring in m\n",
-      "A=5*10**-4           #Crosss sectional area of ring in m^2\n",
-      "phi=2*10**-3         #flux in weber\n",
-      "N=200                #Number of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "l=math.pi*d          #air gap in the ring\n",
-      "S=(l/(u0*ur*A))      #Reluctance of iron ring\n",
-      "print\"Reluctance of iron ring is =\",S,\"Amp-turn/ Wb \"\n",
-      "\n",
-      "#ohm's law for magnetic circuit is  phi=N*I/S. thus,\n",
-      "I=S*phi/N            #required current\n",
-      "print\"Current required to obtain given magnetic flux is =\",I,\"Amperes\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reluctance of iron ring is = 2631578.94737 Amp-turn/ Wb \n",
-        "Current required to obtain given magnetic flux is = 26.3157894737 Amperes\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18.3,Page number 2-39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Values:\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "ur=1               #Relative permeability of air\n",
-      "r=15*10**-2        #radius of ring in m\n",
-      "A=6*10**-4         #Crosss sectional area of ring in m^2\n",
-      "I=4                #Coil current in amp\n",
-      "N=500              #Number of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "m=N*I              #MMF of coil\n",
-      "print\"MMF of coil is =\",m,\"Ampere-turn\"\n",
-      "\n",
-      "l=2*math.pi*r      #air gap\n",
-      "R=(l/(u0*ur*A))    #Reluctance of iron ring\n",
-      "print\"Reluctance of iron ring is =\",R,\"Ampere-turn/Wb\"\n",
-      "\n",
-      "phi=m/R            #Magnetic flux\n",
-      "print\"Magnetic flux is =\",phi,\"Weber\"\n",
-      "\n",
-      "B=phi/A            #Magnetic Flux density\n",
-      "print\"Magnetic flux density is =\",B,\"Weber/m^2\"\n",
-      "\n",
-      "H=B/(u0*ur)        #Magnetic field intensity\n",
-      "print\"Magnetic field intensity is =\",H,\"Amperes/m\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "MMF of coil is = 2000 Ampere-turn\n",
-        "Reluctance of iron ring is = 1250000000.0 Ampere-turn/Wb\n",
-        "Magnetic flux is = 1.6e-06 Weber\n",
-        "Magnetic flux density is = 0.00266666666667 Weber/m^2\n",
-        "Magnetic field intensity is = 2122.06590789 Amperes/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19,Page number 2-36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
-      "ur=6*10**-3         #Relative permeability of iron\n",
-      "r=0.5               #radius of ring in m\n",
-      "l=1*10**-2          #air gap in the ring\n",
-      "A=5*10**-4          #Crosss sectional area of ring in m^2\n",
-      "i=5                 #current in ampere\n",
-      "N=900               #Number of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "S=(l/(u0*A))+((2*math.pi*r-l)/ur*A)     #Reluctance of iron\n",
-      "print\"Reluctance of iron is =\",S,\"Ampere-turn/Wb\"\n",
-      "\n",
-      "m=N*i              #mmf produced\n",
-      "print\"mmf produced is =\",m,\"Ampere-turn\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reluctance of iron is = 15915494.5702 Ampere-turn/Wb\n",
-        "mmf produced is = 4500 Ampere-turn\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20,Page number 2-36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "H=5*10**3          #coercivity of bar magnet in amp/m\n",
-      "l=10*10**-2        #length of solenoid in m\n",
-      "N=50               #No of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know that, H=NI/l ,hence\n",
-      "I=l*H/N            #current through solenoid\n",
-      "\n",
-      "print\"Current through solenoid is =\",I,\"Amperes\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through solenoid is = 10.0 Amperes\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21,Page number 2-36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "ur=1200            #Relative permeability of medium\n",
-      "V=10**-3           #volume of iron rod\n",
-      "N=5*10**2          #no of turns per m\n",
-      "i=0.5              #current through solenoid in amp\n",
-      "\n",
-      "#Calculations:\n",
-      "x=ur-1             #susceptibility of the ring\n",
-      "H=N*i             #Magnetisisng field\n",
-      "\n",
-      "#We know, x=I/H\n",
-      "I=x*H              #magnetisation\n",
-      "\n",
-      "#Also, I=M/V , thus\n",
-      "M=I*V              #magnetic moment\n",
-      "print\"Magnetic moment is =\",M,\"Ampere-turn-m^2\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnetic moment is = 299.75 Ampere-turn-m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22,Page number 2-37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Dta:\n",
-      "ur=100            #Relative permeability of medium\n",
-      "l=0.2             #length of iron rod\n",
-      "d=10*10**-3       #diameter of solenoid in m\n",
-      "N=300             #no of turns per m\n",
-      "i=0.5             #current through solenoid in amp\n",
-      "r=d/2             #radius of solenoid\n",
-      "\n",
-      "#Calculations:\n",
-      "x=ur-1            #susceptibility of the ring\n",
-      "H=N*i             #Magnetisisng field\n",
-      "\n",
-      "#We know, x=I/H\n",
-      "I=x*H             #magnetisation\n",
-      "\n",
-      "V=math.pi*(r**2)*l     #volume of iron rod\n",
-      "\n",
-      "#Also, I=M/V , thus\n",
-      "M=I*V             #magnetic moment\n",
-      "print\"Magnetic moment is =\",M,\"Ampere-turn-m^2\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnetic moment is = 0.233263254529 Ampere-turn-m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23,Page number 2-38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Given Data:\n",
-      "l=1.2              #length of circuit in meter\n",
-      "u=7.3*10**-3       #permeability of silicon sheet\n",
-      "A=100              #cross sectional area in cm^2\n",
-      "N=150              #No of turns\n",
-      "B=0.3              #magmetic field in Wb/m^2\n",
-      "\n",
-      "#Calculations:\n",
-      "\n",
-      "#We know, B=u*H\n",
-      "H=B/u              #Magnetic field strength\n",
-      "\n",
-      "m=H*l              #amp-turns in air gap\n",
-      "\n",
-      "I1=m/N             #Required current\n",
-      "print\"Current required to obtain given magnetic field is =\",I1,\"Amperes\"\n",
-      "\n",
-      "I=I1/A             #Required current per unit area\n",
-      "print\"Current required per unit area to obtain given magnetic field is =\",I,\"Amperes\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current required to obtain given magnetic field is = 0.328767123288 Amperes\n",
-        "Current required per unit area to obtain given magnetic field is = 0.00328767123288 Amperes\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
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new file mode 100755
index 00000000..c4b1a860
--- /dev/null
+++ b/Applied_Physics-II_by_I_A_Shaikh/chapter1.ipynb
@@ -0,0 +1,1353 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:fad8e22fb99cc3e157ab8315172e2ff6ddae35bb6f49be764a49d2d1d3f70fcc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Interference"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Page number 1-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=45*math.pi/180         #angle of incidence\n",
+      "u=1.33                   #Refractive index of a soap film\n",
+      "lamda=5.896*10**-7       #wavelength of required yellow light\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r           #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for bright fringe is\n",
+      "#2ut*cos r=(2n-1)lamda/2\n",
+      "#Here n=1\n",
+      "t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film at which light will appear bright yellow\n",
+      "print\"Minimum thickness of film at which light will appear bright yellow of required wavelength is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum thickness of film at which light will appear bright yellow of required wavelength is = 1.30853030399e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2, Page number 1-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "theta=40./3600*math.pi/180     #angle of wedge in radians\n",
+      "B=0.12*10**-2                 #fringe spacing\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lam/(2*u*theta). Here u=1\n",
+      "lamda=2*B*theta               #wavelength of light used\n",
+      "print\"Wavelength of light used is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light used is = 4.65421133865e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3, Page number 1-17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=30*math.pi/180        #angle of incidence\n",
+      "u=1.46                  #Refractive index of a oil\n",
+      "lamda=5.890*10**-7       #wavelength of required yellow light\n",
+      "n=8                     #eighth dark band\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r      #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t=n*lamda/(2*u*math.cos(r))     #thickness of film\n",
+      "print\"Thickness of the film is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film is = 1.71755887917e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4, Page number 1-17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "B=0.1*10**-2               #fringe spacing\n",
+      "lamda=5.893*10**-7         #Wavelength of light\n",
+      "u=1.52                     #Refractive index of wedge\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lamda/(2*u*theta). Here u=1\n",
+      "theta1=lamda/(2*u*B)            #angle of wedge in radians\n",
+      "theta=theta1*3600*180/math.pi   #angle of wedge in seconds\n",
+      "print\"Angle of wedge is =\",theta,\"seconds of an arc\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of wedge is = 39.9841612899 seconds of an arc\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5, Page number 1-18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "t=0.2/(100)**2*10**-2     #thickness of film in meter\n",
+      "lamda=5.5*10**-7         #wavelength of light in meter\n",
+      "r=0                      #normal incidence\n",
+      "n=1                      #first band\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Condition for dark fringe is\n",
+      "#2ut*cos r =n*lamda\n",
+      "u=n*lamda/(2*t*math.cos(r))   #Refractive index of a oil\n",
+      "print\"Refractive index of a oil is =\",u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of a oil is = 1.375\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6, Page number 1-18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=5.893*10**-7    #Wavelength of light\n",
+      "u=1.42                #Refractive index of a soap film\n",
+      "r=0                   #normal incidence\n",
+      "n=1                   #first band\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#(i)\n",
+      "#Condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t1=n*lamda/(2*u*cos(r))      #thickness of film for dark black fringe\n",
+      "print\"Thickness of the film for dark black fringe is =\",t1,\"m\"\n",
+      "\n",
+      "#(ii)\n",
+      "#Now, condition for bright fringe is\n",
+      "#2ut*cos r=(2n-1)lamda/2\n",
+      "t2=lamda/(2*2*u*math.cos(r))      #Thickness of film for bright fringe\n",
+      "print\"Thickness of film for bright fringe is =\",t2,\"m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film for dark black fringe is = 2.075e-07 m\n",
+        "Thickness of film for bright fringe is = 1.0375e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7, Page number 1-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#Given Data:\n",
+      "i=30*math.pi/180         #angle of incidence\n",
+      "u=1.43               #Refractive index of a soap film\n",
+      "lamda=6*10**-7          #wavelength of light\n",
+      "n=1                  #For minimum thickness\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r      #Snell's law   .So,\n",
+      "r=(math.asin(math.sin(i)/u))    #angle of reflection\n",
+      "\n",
+      "#Now, condition of minima in transmitted system is\n",
+      "#2ut*cos(r)=(2n-1)lamda/2\n",
+      "t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film\n",
+      "print\"Minimum thickness of film is \",t,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum thickness of film is  1.11962124395e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8, Page number 1-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "lamda = 5893*10**-10      #Wavelength of light\n",
+      "theta = 1               #assuming value of theta\n",
+      "\n",
+      "#We know, B=lamda/(2*u*theta). Here u=1\n",
+      "B = lamda/(2*theta)       #fringe spacing\n",
+      "n=20                  #interference fringes\n",
+      "\n",
+      "#Calculations:\n",
+      "#t=n*B*tan(theta)\n",
+      "t = 20*B*theta          #Thickness of wire\n",
+      "print\"Thickness of wire is =\",t,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of wire is = 5.893e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9, Page number 1-20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.3                #Refractive index of oil\n",
+      "u2=1.5                #Refractive index of glass\n",
+      "lamda1=7*10**-7       #Wavelength of light\n",
+      "lamda2=5*10**-7       #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#for finding value of n, solve:\n",
+      "#(2n+1)*lamda1/2=(2(n+1)+1)*lamda2/2\n",
+      "#We get,n=2\n",
+      "n=2\n",
+      "\n",
+      "toil=(2*n+1)*lamda1/(2*u1*2)     #thickness of oil layer\n",
+      "print\"Thickness of oil layer is =\",toil,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of oil layer is = 6.73076923077e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Example 1.10, Page number 1-21"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.2               #Refractive index of drop of oil\n",
+      "u2=1.33              #Refractive index of water\n",
+      "lamda=4.8*10**-7     #wavelength of light\n",
+      "n=3                  #order\n",
+      "r=0                  #normal incidence,so r=0\n",
+      "\n",
+      "#Calculations:\n",
+      "t=n*lamda/(2*u1)      #Thickness of oil drop\n",
+      "print\"Thickness of oil drop is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of oil drop is = 6e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11, Page number 1-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=math.asin(4/5)          #angle of incidence\n",
+      "u=4/3                    #Refractive index of a soap film\n",
+      "lamda1=6.1*10**-7        #wavelength of light\n",
+      "lamda2=6*10**-7          #wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r      #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark band is\n",
+      "#2ut*cos r=n*lamda\n",
+      "#for consecutive bands, n=lamda2/(lamda1-lamda2). hence\n",
+      "\n",
+      "t=lamda2*lamda1/((lamda1-lamda2)*2*u*math.sqrt(1-(math.sin(i)/u)**2))     #thickness of film\n",
+      "print\"Thickness of the film is =\",t,\"m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film is = 1.83e-05 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.12, Page number 1-40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=10                #10th dark ring\n",
+      "Dn=0.5*10**-2       #Diameter of ring\n",
+      "lamda=6*10**-7      #wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda\n",
+      "R=Dn**2/(4*n*lamda)    #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n",
+      "\n",
+      "t=Dn**2/(8*R)        #thickness of air  film\n",
+      "print\"Thickness of air  film is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 1.04166666667 m\n",
+        "Thickness of air  film is = 3e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.13, Page number 1-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "B=0.25*10**-2             #fringe spacing\n",
+      "lamda=5.5*10**-7          #Wavelength of light\n",
+      "u=1.4                     #Refractive index of wedge\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lamda/(2*u*theta).\n",
+      "theta1=lamda/(2*u*B)            #angle of wedge in radians\n",
+      "theta=theta1*3600*180/math.pi   #angle of wedge in seconds\n",
+      "print\"Angle of wedge is =\",theta,\"seconds\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of wedge is = 16.2065204908 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.14, Page number 1-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=4              #4th dark ring\n",
+      "m=12             #m=n+p\n",
+      "D4=0.4*10**-2    #Diameter of 4th ring\n",
+      "D12=0.7*10**-2   #Diameter of 12th ring\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R\n",
+      "#Solving, (D12^2-D4^2)/(D20^2-D4^2)\n",
+      "#We get above value =1/2. Hence\n",
+      "D20=math.sqrt(2*D12**2-D4**2)      #Diameter of 20th ring\n",
+      "print\"Diameter of 20th ring is =\",D20,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of 20th ring is = 0.00905538513814 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.15, Page number 1-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=6                  #6th bright ring\n",
+      "D6=0.31*10**-2       #Diameter of 6th ring\n",
+      "lamda=6*10**-7       #wavelength of light\n",
+      "R=1                  #Radius of curvature\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Diameter of nth bright ring is \n",
+      "#Dn^2=2(2n-1)*lamda*R/u. Hence\n",
+      "u=2*(2*n-1)*lamda*R/(D6)**2    #Refractive index of liquid\n",
+      "print\"Refractive index of liquid is =\",u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of liquid is = 1.37356919875\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.16, Page number 1-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=6*10**-7         #wavelength of light\n",
+      "k=0.125*10**-4         #k=D(n+1)^2-Dn^2.\n",
+      "u=1                    #Refractive index of medium between lens and plate\n",
+      "#Calculations:\n",
+      "\n",
+      "#(i)\n",
+      "lamda1=4.5*10**-7      #new wavelength of light\n",
+      "#Difference between squres of diameters of successive rings is directly proportional to wavelength.So,\n",
+      "k1=lamda1/lamda*k      #new Difference between squres of diameters of successive rings after changing wavelength\n",
+      "print\"New Difference between squres of diameters of successive rings after changing wavelength is =\",k1,\"m^2\"\n",
+      "\n",
+      "#(ii)\n",
+      "u2=1.33          #Refractive index of liquid introduced between lens and plate\n",
+      "#Difference between squres of diameters of successive rings is inversely proportional to Refractive index.so,\n",
+      "k2=u/u2*k        #new Difference between squres of diameters of successive rings after changing refractive index\n",
+      "print\"New Difference between squres of diameters of successive rings after changing refrective index is =\",k2,\"m^2\"\n",
+      "\n",
+      "#(iii)\n",
+      "#Difference between squres of diameters of successive rings is directly proportional to Radius of curvature.So,\n",
+      "#after doubling radius of curvature,\n",
+      "k3=2*k           #new Difference between squres of diameters of successive rings after doubling radius of curvature\n",
+      "print\"New Difference between squres of diameters of successive rings after doubling radius of curvature is =\",k3,\"m^2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New Difference between squres of diameters of successive rings after changing wavelength is = 9.375e-06 m^2\n",
+        "New Difference between squres of diameters of successive rings after changing refrective index is = 9.3984962406e-06 m^2\n",
+        "New Difference between squres of diameters of successive rings after doubling radius of curvature is = 2.5e-05 m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.17, Page number 1-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "Dn=0.225*10**-2         #Diameter of nth ring\n",
+      "Dm=0.45*10**-2          #Diameter of (n+9)th ring\n",
+      "lamda=6*10**-7          #wavelength of light\n",
+      "R=0.9                   #Radius of curvature\n",
+      "p=9\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
+      "u=4*p*lamda*R/((Dm)**2-Dn**2)     #Refractive index of liquid\n",
+      "print\"Refractive index of liquid is =\",u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of liquid is = 1.28\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.18, Page number 1-44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D10=0.5*10**-2          #Diameter of 10th ring\n",
+      "lamda=5.5*10**-7        #wavelength of light\n",
+      "u=1.25                  #Refractive index of liquid\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda/u\n",
+      "#Dn^2 is inversely proportional to refractive index.\n",
+      "D10n=D10/math.sqrt(u)        #new diameter of 10th ring after changing medium between lens and plate\n",
+      "print\"new diameter of 10th ring after changing medium between lens and plate is =\",D10n,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "new diameter of 10th ring after changing medium between lens and plate is = 0.004472135955 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.19, Page number 1-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D5=0.336*10**-2       #Diameter of 5th ring\n",
+      "D15=0.59*10**-2       #Diameter of 15th ring\n",
+      "lamda=5.89*10**-7       #wavelength of light\n",
+      "p=10                  #n=5,n+p=15\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamdaR/u\n",
+      "R=((D15)**2-D5**2)/(4*p*lamda)      #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 0.998319185059 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20, Page number 1-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=10               #10th dark ring\n",
+      "D10=0.6*10**-2     #Diameter of ring\n",
+      "lamda=6*10**-7     #wavelength of light\n",
+      "u=4./3              #Refractive index of water\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda/u\n",
+      "R=(D10**2)*u/(4*n*lamda)    #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 2.0 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20.1, Page number 1-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=45*math.pi/180        #angle of incidence\n",
+      "u=1.2                   #Refractive index of a film\n",
+      "t=4*10**-7              #thickness of film\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r       #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "lamda1=2*u*t*math.cos(r)/1          #n=1\n",
+      "print\"For n=1 wavelength is =\",lamda1,\"m\"\n",
+      "print\"This is in the visible spectrum and it will remain absent.\" \n",
+      "\n",
+      "lamda2=2*u*t*math.cos(r)/2          #n=2\n",
+      "print\"For n=2 wavelength is =\",lamda2,\"m\"\n",
+      "print\"This is not in the visible spectrum\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n=1 wavelength is = 7.75628777187e-07 m\n",
+        "This is in the visible spectrum and it will remain absent.\n",
+        "For n=2 wavelength is = 3.87814388593e-07 m\n",
+        "This is not in the visible spectrum\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20.2, Page number 1-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "r=45*math.pi/180        #angle of refraction\n",
+      "u=1.45                  #Refractive index of a medium\n",
+      "lamda=5.5*10**-7        #wavelength of required yellow light\n",
+      "n=1\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t=n*lamda/(2*u*math.cos(r))      #thickness of thin medium\n",
+      "print\"Thickness of the thin medium is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the thin medium is = 2.68212917002e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.21, Page number 1-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u=1.33                 #Refractive index of a soap film\n",
+      "r=0                    #normal incidence\n",
+      "t=5*10**-7             #thickness of film\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Now, condition for maxima is\n",
+      "#2ut*cos r=(2n-1)lamda/2\n",
+      "lamda1=4*u*t*math.cos(r)/(2*1-1)          #n=1\n",
+      "print\"For n=1 wavelength is =\",lamda1,\"m\"\n",
+      "lamda2=4*u*t*math.cos(r)/(2*2-1)          #n=2\n",
+      "print\"For n=2 wavelength is =\",lamda2,\"m\"\n",
+      "lamda3=4*u*t*math.cos(r)/(2*3-1)          #n=3\n",
+      "print\"For n=3 wavelength is =\",lamda3,\"m\"\n",
+      "lamda4=4*u*t*math.cos(r)/(2*4-1)          #n=4\n",
+      "print\"For n=4 wavelength is =\",lamda4,\"m\"\n",
+      "\n",
+      "print\"Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\"\n",
+      "print\"Hence, wavelength for n=3 is the most reflected wavelength.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n=1 wavelength is = 2.66e-06 m\n",
+        "For n=2 wavelength is = 8.86666666667e-07 m\n",
+        "For n=3 wavelength is = 5.32e-07 m\n",
+        "For n=4 wavelength is = 3.8e-07 m\n",
+        "Out of these wavelengths wavelength for n=3 lies in the visible spectrum.\n",
+        "Hence, wavelength for n=3 is the most reflected wavelength.\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.22, Page number 1-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u=1.5                #Refractive index of a oil\n",
+      "lamda=5.88*10**-7    #wavelength of required yellow light\n",
+      "n=1                  #for smallest thickness\n",
+      "r=60*math.pi/180     #angle of reflection\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#Now, condition for dark fringe is\n",
+      "#2ut*cos r=n*lamda\n",
+      "t=n*lamda/(2*u*math.cos(r))      #thickness of film\n",
+      "print\"Thickness of the film is =\",t,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of the film is = 3.92e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.23, Page number 1-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "theta=20./3600*math.pi/180     #angle of wedge in radians\n",
+      "B=0.25*10**-2                 #fringe spacing\n",
+      "u=1.4                         #Refractive index of film\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, B=lamda/(2*u*theta).\n",
+      "lamda=2*B*theta*u             #wavelength of light\n",
+      "print\"Wavelength of light is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light is = 6.78739153553e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.24, Page number 1-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "#Dn=2*D40\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2 = 4*n*R*lamda/u  and   Dn^2 = 4*D40^2\n",
+      "#i.e. 4*n*R*lamda/u = 4*4*40*R*lamda/u .hence,\n",
+      "n=4*40          #order of the required ring\n",
+      "print\"Order of the dark ring which will have double the diameter of that of 40th ring is =\",n\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Order of the dark ring which will have double the diameter of that of 40th ring is = 160\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.25, Page number 1-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda1=6*10**-7       #wavelength of light\n",
+      "lamda2=4.5*10**-7    #wavelength of light\n",
+      "R=0.9                 #Radius of curvature\n",
+      "\n",
+      "#Calculations:\n",
+      "#As Dn^2=4*n*R*lamda.\n",
+      "#Dn^2=D(n+1)^2 for different wavelengths.we get,\n",
+      "n=lamda2/(lamda1-lamda2)       #nth dark ring due to lam1 which coincides with (n+1)th dark ring due lamda2\n",
+      "D3=math.sqrt(4*n*R*lamda1)     #diameter of 3rd dark ring for lamda1\n",
+      "print\"Diameter of 3rd dark ring for lam1 is =\",D3,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of 3rd dark ring for lam1 is = 0.00254558441227 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.26, Page number 1-48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "i=45*math.pi/180        #angle of incidence\n",
+      "u=4./3                  #Refractive index of soap film\n",
+      "lamda=5*10**-7          #wavelength of light\n",
+      "t=1.5*10**-6            #thickness of film\n",
+      "\n",
+      "#Calculations:\n",
+      "#u=sin i/sin r        #Snell's law   .So,\n",
+      "r=math.asin(math.sin(i)/u)     #angle of reflection\n",
+      "\n",
+      "#Now, condition for dark band is\n",
+      "#2ut*cos r=n*lamda\n",
+      "n=2*u*t*math.cos(r)/lamda    #order of band\n",
+      "print\"order of dark band is =\",n\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "order of dark band is = 6.78232998313\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.27, Page number 1-49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D5=0.336*10**-2         #Diameter of 5th ring\n",
+      "D15=0.59*10**-2         #Diameter of 15th ring\n",
+      "lamda=5.89*10**-7       #wavelength of light\n",
+      "p=10                    #n=5,n+p=15\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
+      "R=((D15)**2-D5**2)/(4*p*lamda)    #Radius of curvature of the lens\n",
+      "print\"Radius of curvature of the lens is =\",R,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature of the lens is = 0.998319185059 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.29, Page number 1-50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#As Dn^2=4*n*R*lamda.\n",
+      "#thus, Dn is directly proportional to sqaure root of n\n",
+      "D5=math.sqrt(5)     #D5 is directly proportional to sqaure root of 5\n",
+      "D4=math.sqrt(4)     #D4 is directly proportional to sqaure root of 4\n",
+      "k1=D5-D4\n",
+      "print\"Separation between D5 and D4 is directly proportional to =\",k1\n",
+      "\n",
+      "D80=math.sqrt(80)     #D80 is directly proportional to sqaure root of 80\n",
+      "D79=math.sqrt(79)     #D79 is directly proportional to sqaure root of 79\n",
+      "k2=D80-D79\n",
+      "print\"Separation between D80 and D79 is directly proportional to =\",k2\n",
+      "\n",
+      "print\"Thus, (D80-D79) < (D5-D4). Hence proved.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Separation between D5 and D4 is directly proportional to = 0.2360679775\n",
+        "Separation between D80 and D79 is directly proportional to = 0.0560774926836\n",
+        "Thus, (D80-D79) < (D5-D4). Hence proved.\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.30, Page number 1-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "D5=0.336*10**-2       #Diameter of 5th ring\n",
+      "D15=0.59*10**-2       #Diameter of 15th ring\n",
+      "p=10                  #n=5,n+p=15\n",
+      "R=1                   #Radius of curvature\n",
+      "\n",
+      "#Calculations:\n",
+      "#(Dn+p)^2-Dn^2=4*p*lamda*R/u\n",
+      "lamda=((D15)**2-D5**2)/(4*p*R)    #Wavelength of light\n",
+      "print\"Wavelength of light is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light is = 5.8801e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.31, Page number 1-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "#Condition for bright band is\n",
+      "#2ut*cos r = (2n-1)*lamda1\n",
+      "\n",
+      "#for consecutive bands, 2n=(lamda1+lamda2)/(lamda1-lamda).\n",
+      "#thus, 2ut*cos r = lamda2*lamda1/(lamda1-lamda2)\n",
+      "\n",
+      "#And, thicknessof film \n",
+      "#t= lamda2*lamda1/((2*u*cosr)(lamda1-lamda2))\n",
+      "print\"Hence expression for thickness of film is obtained.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hence expression for thickness of film is obtained.\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics-II_by_I_A_Shaikh/chapter2.ipynb b/Applied_Physics-II_by_I_A_Shaikh/chapter2.ipynb
new file mode 100755
index 00000000..6af9170f
--- /dev/null
+++ b/Applied_Physics-II_by_I_A_Shaikh/chapter2.ipynb
@@ -0,0 +1,1480 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:1b2e4fb9c2f216cbf762800ac72003e90570cd62834a70860272dd5120eea60f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Diffraction of light"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1,Page number 2-30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=4       #order\n",
+      "N=1./5000*10**-2    #N=(a+b) grating element(cm)\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "#for longest wavelength, sin(theta)=1\n",
+      "lamda=(N/m)     #longest wavelength\n",
+      "print\"The longest wavelength is =\" ,lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The longest wavelength is = 5e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2,Page number 2-30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                #order\n",
+      "lamda=6.5*10**-7      #Wavelength of red light\n",
+      "theta=30*3.142/180   #angle of diffraction\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, a*sin(theta)=m*lamda\n",
+      "a=m*lamda/math.sin(theta)   #width of slit\n",
+      "print\"width of slit is = \",a,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "width of slit is =  1.29984715296e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3,Page number 2-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                #order\n",
+      "lamda=4*10**-7        #Wavelength of light\n",
+      "a=10**-6            #width of slit\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, a*sin(theta)=m*lamda\n",
+      "theta=math.asin(m*lamda/a)*180/3.142    #angular position in first minima\n",
+      "print\"angular position in first minima is =\",theta,\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angular position in first minima is = 23.5751216716 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4,Page number 2-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                 #order\n",
+      "lamda1=4*10**-7       #Wavelength of light\n",
+      "lamda2=7*10**-7       #Wavelength of light\n",
+      "n=1./6000*10**-2       #n=(a+b) grating element\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "theta1=math.asin(m*lamda1/n)*(180/3.142)   #angle of diffraction\n",
+      "theta2=math.asin(m*lamda2/n)*(180/3.142)   #angle of diffraction\n",
+      "d=theta2-theta1                  #angular breadth of first order visible spectrum\n",
+      "print\"angular breadth of first order visible spectrum is = \",d,\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angular breadth of first order visible spectrum is =  10.9466277612 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5,Page number 2-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                         #order\n",
+      "lamda=6.56*10**-7           #Wavelength of red light\n",
+      "theta=18.25*math.pi/180    #angle of diffraction\n",
+      "W=2*10**-2                 #width of grating\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "N=math.sin(theta)/(m*lamda)   #N-number of lines per m, N=1/(a+b)\n",
+      "Tn=N*W                     #Total number of lines on grating\n",
+      "print\"Total number of lines on grating is =\",Tn\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total number of lines on grating is = 9547.67702694\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7,Page number 2-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "GE=2.54/15000*10**-2       #GE=(a+b) grating element\n",
+      "lamda1=4*10**-7            #Wavelength of light\n",
+      "lamda2=7*10**-7            #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "theta11=math.asin(1*lamda1/GE)*180/math.pi       #angular position of first minima for lamda1\n",
+      "theta12=math.asin(2*lamda1/GE)*180/math.pi       #angular position of second minima for lamda1\n",
+      "theta13=math.asin(3*lamda1/GE)*180/math.pi       #angular position of third minima for lamda1\n",
+      "\n",
+      "theta21=math.asin(1*lamda2/GE)*180/math.pi       #angular position of first minima for lamda2\n",
+      "theta22=math.asin(2*lamda2/GE)*180/math.pi       #angular position of second minima for lamda2\n",
+      "theta23=math.asin(1)*180/math.pi                 #angular position of third minima for lamda2\n",
+      "\n",
+      "print\"Thus the angular position for lamda1 and lamda2 are as follows:\"\n",
+      "print\"First order:\",theta11,\"degrees\"\n",
+      "print\"\",theta21,\"degrees --Isolated\"\n",
+      "\n",
+      "print\"Second order:\",theta12,\"degrees\"\n",
+      "print\"\",theta22,\"degrees --Overlap\"\n",
+      "\n",
+      "print\"Third order: \",theta13,\"degrees\"\n",
+      "print\"\",theta23,\" degrees --Overlap \"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thus the angular position for lamda1 and lamda2 are as follows:\n",
+        "First order: 13.6635763633 degrees\n",
+        " 24.4177053663 degrees --Isolated\n",
+        "Second order: 28.1928605617 degrees\n",
+        " 55.7685229906 degrees --Overlap\n",
+        "Third order:  45.1261086702 degrees\n",
+        " 90.0  degrees --Overlap \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8,Page number 2-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=5.893*10**-7        #Wavelength of light\n",
+      "d=0.01*10**-2             #width of slit (a=d)\n",
+      "f=1                      #distance between screen and slit\n",
+      "\n",
+      "#Calculations:\n",
+      "x=f*lamda/d              #separation between central maxima and first minima\n",
+      "print\"Separation between central maxima and first minima is = \",x,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Separation between central maxima and first minima is =  0.005893 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9,Page number 2-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                 #order\n",
+      "lamda=6*10**-7        #Wavelength of light\n",
+      "a=12*10**-7         #width of slit\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, a*sin(theta)=m*lamda\n",
+      "theta=math.asin(m*lamda/a)*180/math.pi    #angular position in first minima\n",
+      "print\"Half angular width of first maxima is =\",theta,\"Degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Half angular width of first maxima is = 30.0 Degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10,Page number 2-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=6*10**-7          #Wavelength of light\n",
+      "a=0.02*10**-2           #width of slit (a=d)\n",
+      "f=2                    #distance between screen and slit\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, a*sin(theta)=m*lamda, here m=1\n",
+      "theta=math.asin(lamda/a)*180*60/math.pi      #angular position in first minima (1 degree=60 minutes)\n",
+      "print\"Total angular width is =\",2*theta,\"minutes\"\n",
+      "\n",
+      "x=f*lamda/a              #separation between central maxima and first minima\n",
+      "print\"Linear width is = \",2*x,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total angular width is = 20.6265115646 minutes\n",
+        "Linear width is =  0.012 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11,Page number 2-35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "a=0.14*10**-3         #width of slit\n",
+      "n=2                   #order\n",
+      "y=1.6*10**-2          #separation between second dark band and central bright band\n",
+      "D=2                   #distance between screen and slit\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "theta=y/D            #from diagram \n",
+      "\n",
+      "#We know, a*sin(theta)=n*lamda\n",
+      "#here sin(theta)=theta\n",
+      "lamda=a*theta/n       #wavelength of light\n",
+      "print\"wavelength of light is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of light is = 5.6e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13,Page number 2-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=6.328*10**-7        #Wavelength of light\n",
+      "N=1./6000*10**-2           #N=(a+b) grating element\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, N*sin(theta)=m*lamda\n",
+      "theta1=math.asin(1*lamda/N)*180/math.pi    #angular position in first order maxima,m=1\n",
+      "print\"Angular position in first order maxima is =\",theta1,\"Degrees\"\n",
+      "\n",
+      "theta2=math.asin(2*lamda/N)*180/math.pi    #angular position in second order maxima,m=2\n",
+      "print\"Angular position in second order maxima is = \",theta2,\"Degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angular position in first order maxima is = 22.3138625335 Degrees\n",
+        "Angular position in second order maxima is =  49.4078093436 Degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14,Page number 2-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda1=6*10**-7         #wavelength of yellow light\n",
+      "lamda2=4.8*10**-7       #wavelength of blue light\n",
+      "theta=(math.asin(3/4))    #angle of diffraction\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#for consecutive bands, n*lamda1=(n+1)*lamda2. thus,\n",
+      "n=lamda2/(lamda1-lamda2)   #order\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "N=n*lamda1/(3./4)    #N=(a+b) grating element\n",
+      "print\"Grating element (a+b) is =\",N,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Grating element (a+b) is = 3.2e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15.1,Page number 2-54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "a=0.2*10**-3          #width of slit\n",
+      "n=1                   #order\n",
+      "y=0.5*10**-2          #separation between first minima and central bright band\n",
+      "D=2                   #distance between screen and slit\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "theta=y/D            #from diagram \n",
+      "\n",
+      "#We know, a*sin(theta)=n*lamda\n",
+      "#here sin(theta)=theta\n",
+      "lamda=a*theta/n       #wavelength of light\n",
+      "print\"wavelength of light is = \",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of light is =  5e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15.2,Page number 2-55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda1=5.4*10**-7              #Wavelength of light\n",
+      "lamda2=4.05*10**-7             #Wavelength of light\n",
+      "theta=30*math.pi/180          #angle of diffraction\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, (a+b)*sin(theta)=n*lamda\n",
+      "#n*lamda1=(n+1)*lamda2, we get \n",
+      "n=3\n",
+      "N=math.sin(theta)/(n*lamda1)*10**-2      #Number of lines per m= 1/(a+b)*10^-2\n",
+      "print\"Number of lines per cm is = \",N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of lines per cm is =  3086.41975309\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15.4,Page number 2-56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "GE=1./6000*10**-2               #GE=(a+b) grating element\n",
+      "lamda1=5.893*10**-7             #Wavelength of light\n",
+      "lamda2=5.896*10**-7             #Wavelength of light\n",
+      "m=2                             #order\n",
+      "\n",
+      "#Calculations:\n",
+      "theta1=math.asin(m*lamda1/GE)*180/math.pi     #angular position in first minima\n",
+      "theta2=math.asin(m*lamda2/GE)*180/math.pi     #angular position in second minima\n",
+      "\n",
+      "a_s=(theta2-theta1)            #Angular separation in minutes\n",
+      "print\"Angular separation is =\",a_s,\"Degrees\"\n",
+      "\n",
+      "dlamda=lamda2-lamda1            #difference in wavelength\n",
+      "lamda=(lamda2+lamda1)/2         #Mean wavelength\n",
+      "\n",
+      "#We know that R.P.=lamda/dlamda=m*N\n",
+      "N=lamda/dlamda/m                 #Number of lines on grating for first order\n",
+      "print\"Number of lines on grating for first order is =\",N\n",
+      "print\"But, number of lines per cm on grating is 6000. \\n Which is greater than number of lines per cm needed for resolution.\"\n",
+      "print\"Hence, both lines will be well resolved in 2nd order.\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angular separation is = 0.0291798772234 Degrees\n",
+        "Number of lines on grating for first order is = 982.416666667\n",
+        "But, number of lines per cm on grating is 6000. \n",
+        " Which is greater than number of lines per cm needed for resolution.\n",
+        "Hence, both lines will be well resolved in 2nd order.\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17,Page number 2-39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "d=0.04*10**-2       #Separation between slits\n",
+      "D=1.7              #distance between screen and slit\n",
+      "B=0.25*10**-2       #Fringe spacing\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know,B=D*lamda/d\n",
+      "lamda=B*d/D         #Wavelength of light\n",
+      "print\"Wavelength of light is = \",lamda,\"m\"\n",
+      "\n",
+      "#The condition for missing order is,\n",
+      "#(a+b)/a = m/n\n",
+      "b=0.04*10**-2       #Separation in slits\n",
+      "a=0.08*10**-3       #Slit width\n",
+      "n=(a+b)/a          #missing orders for m=1,2,3\n",
+      "\n",
+      "n1=1*n\n",
+      "n2=2*n\n",
+      "n3=3*n\n",
+      "print\"Missing orders are =\",n1,\",\",n2,\",\",n3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light is =  5.88235294118e-07 m\n",
+        "Missing orders are = 6.0 , 12.0 , 18.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18,Page number 2-39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "N=2.54/2620*10**-2     #N=(a+b) grating element\n",
+      "lamda=5*10**-7           #Wavelength of red light\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=n*lamda\n",
+      "#maximum value of sin(theta)=1\n",
+      "n=N/lamda               #Maximum number of orders visible\n",
+      "print\"Maximum number of orders visible is =\",n\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum number of orders visible is = 19.3893129771\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19,Page number 2-40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "N=1./4000*10**-2       #N=(a+b) grating element\n",
+      "lamda1=5*10**-7         #Wavelength of light\n",
+      "lamda2=7.5*10**-7       #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=n*lamda\n",
+      "#maximum value of sin(theta)=1\n",
+      "n1=N/lamda1               #Maximum number of orders visible\n",
+      "n2=N/lamda2               #Maximum number of orders visible\n",
+      "print\"The observed number of orders range between =\",n2,\"to\",n1\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The observed number of orders range between = 3.33333333333 to 5.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.20,Page number 2-40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=5                #order\n",
+      "lamda=6*10**-7     #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, a*sin(theta)=n*lamda\n",
+      "#n*lamda=n1*lamda1\n",
+      "lamda1=n*lamda/4        #for n1=4\n",
+      "print\"For n1=4 wavelength is =\",lamda1,\"m\"\n",
+      "\n",
+      "lamda2=n*lamda/5        #for n1=5\n",
+      "print\"For n1=5 wavelength is =\",lamda2,\"m\"\n",
+      "\n",
+      "lamda3=n*lamda/6        #for n1=6\n",
+      "print\"For n1=5 wavelength is =\",lamda3,\"m\"\n",
+      "\n",
+      "lamda4=n*lamda/7        #for n1=7\n",
+      "print\"For n1=5 wavelength is =\",lamda4,\"m\"\n",
+      "\n",
+      "lamda5=n*lamda/8        #for n1=8\n",
+      "print\"For n1=5 wavelength is =\",lamda5,\"m\"\n",
+      "\n",
+      "\n",
+      "print\"So,in the grating spectrum spectrum lines with wavelengths n1=6 and n1=7 will coincide with fifth order line of 6*10^-7 m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n1=4 wavelength is = 7.5e-07 m\n",
+        "For n1=5 wavelength is = 6e-07 m\n",
+        "For n1=5 wavelength is = 5e-07 m\n",
+        "For n1=5 wavelength is = 4.28571428571e-07 m\n",
+        "For n1=5 wavelength is = 3.75e-07 m\n",
+        "So,in the grating spectrum spectrum lines with wavelengths n1=6 and n1=7 will coincide with fifth order line of 6*10^-7 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.21,Page number 2-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "GE=18000*10**-10         #GE=(a+b) grating element\n",
+      "lamda=5*10**-7             #Wavelength of red light\n",
+      "\n",
+      "#Calculations:\n",
+      "DP1=1./sqrt(GE**2-lamda**2)*10**-10        #Dispersive power\n",
+      "print\"Dispersive power for first order is =\",DP1,\"rad/Angstrom\"\n",
+      "\n",
+      "m=3\n",
+      "DP2=1/sqrt((GE/m)**2-lamda**2)*10**-10    #Dispersive power\n",
+      "print\"Dispersive power for second order is =\",DP2,\"rad/Angstrom\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Dispersive power for first order is = 5.78314931966e-05 rad/Angstrom\n",
+        "Dispersive power for second order is = 0.000301511344578 rad/Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.22,Page number 2-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "N=2.54/15000*10**-2      #N=(a+b) grating element\n",
+      "lamda=5.9*10**-7         #Wavelength of light\n",
+      "m=2                      #order\n",
+      "f=25*10**-2              #focal length of lens\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "theta=math.asin(m*lamda/N)      #angular position in first minima\n",
+      "\n",
+      "Ad=m/N/cos(theta)              #angular dispersion\n",
+      "\n",
+      "ld=f*Ad*10**-8                 #linear dispersion (dx/dl) in cm/angstrom\n",
+      "print\"Linear dispersion in spectrograph is =\",ld,\"cm/angstrom\"\n",
+      "\n",
+      "dlamda=(5896-5890)               #difference in wavelength\n",
+      "dx=ld*dlamda*10**-2              #separation between spectral lines in meter\n",
+      "print\"Separation between spectral lines is =\",dx,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Linear dispersion in spectrograph is = 0.00411696586101 cm/angstrom\n",
+        "Separation between spectral lines is = 0.00024701795166 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.23,Page number 2-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                              #order\n",
+      "lamda1=5.89*10**-7               #Wavelength of light\n",
+      "lamda2=5.896*10**-7              #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "dlamda=lamda2-lamda1                #difference in wavelength\n",
+      "lamda=(lamda2+lamda1)/2             #Mean wavelength\n",
+      "\n",
+      "#We know that R.P.=m*N=lamda/dlamda\n",
+      "N=lamda/dlamda/m                  #minimum number of lines which will just resolve\n",
+      "print\"Minimum number of lines which will just resolve is =\",N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum number of lines which will just resolve is = 982.166666667\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.24,Page number 2-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "N=5*5000           #N=W/(a+b) Number of lines on grating\n",
+      "m=2               #order\n",
+      "lamda=6*10**-7    #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "#(i)\n",
+      "RP=m*N             #Resolving power\n",
+      "print\"(i)Resolving power is =\",RP\n",
+      "\n",
+      "#(ii)\n",
+      "#We know that R.P.=lamda/dlamda\n",
+      "dlamda=lamda/RP       #Smallest wavelength which can be resolved\n",
+      "print\"(ii)Smallest wavelength which can be resolved is =\",dlamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i)Resolving power is = 50000\n",
+        "(ii)Smallest wavelength which can be resolved is = 1.2e-11 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.25,Page number 2-48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "GE=1./4000*10**-2         #GE=(a+b) grating element\n",
+      "lamda=5*10**-7             #Wavelength of red light\n",
+      "m=3                      #order\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "theta=math.asin(m*lamda/GE)    #ngular position in first minima\n",
+      "\n",
+      "DP=m/(GE*math.cos(theta))*10**-2     #Dispersive power\n",
+      "print\"Dispersive power is =\",DP\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Dispersive power is = 15000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.26,Page number 2-48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=2                  #order\n",
+      "lamda=6*10**-7       #Wavelength of light\n",
+      "dlamda=6*10**-10     #difference in wavelength\n",
+      "W=2*10**-2           #Width of surface\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know that R.P.=lamda/dlamda=m*N\n",
+      "N=lamda/dlamda/m     #Number of lines on grating\n",
+      "GE=W/N               #Grating element=(a+b)\n",
+      "print\"Grating element is =\",GE,\"cm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Grating element is = 4e-05 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.27,Page number 2-49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=2                           #order\n",
+      "lamda1=5.77*10**-7            #Wavelength of light\n",
+      "lamda2=5.791*10**-7           #Wavelength of light\n",
+      "GE=1./6000*10**-2             #GE=(a+b) grating element\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "theta1=math.asin(m*lamda1/GE)*180/math.pi      #angular position in first minima\n",
+      "theta2=math.asin(m*lamda2/GE)*180/math.pi      #angular position in second minima\n",
+      "\n",
+      "a_s=(theta2-theta1)*60         #Angular separation in minutes\n",
+      "print\"Angular separation is = \",a_s,\"minutes\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angular separation is =  12.0270825521 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.28,Page number 2-49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=1                 #order\n",
+      "lam=5.89*10**-7     #Wavelength of light\n",
+      "a=0.3*10**-3        #width of slit\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, a*sin(theta)=n*lamda\n",
+      "theta1=math.asin(n*lamda/a)*180/math.pi*60    #angular position in first dark band in minutes\n",
+      "print\"Angular position in first dark band is = \",theta1,\"mimutes\"\n",
+      "\n",
+      "#We know,for bright band   a*sin(theta)=(2n+1)*lamda/2\n",
+      "theta2=math.asin(1.5*lamda/a)*180/math.pi*60    #angular position in first bright band in minutes\n",
+      "print\"Angular position in first bright band is =\",theta2,\"minutes\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angular position in first dark band is =  6.87549812524 mimutes\n",
+        "Angular position in first bright band is = 10.3132557823 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.29,Page number 2-50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "GE=2.54/16000*10**-2      #GE=(a+b) grating element\n",
+      "lamda=6*10**-7             #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "#maximum value of sin(theta)=1\n",
+      "m=GE/lamda                 #Maximum order of spectra\n",
+      "print\"Maximum order of spectra is =\",m\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum order of spectra is = 2.64583333333\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.30,Page number 2-50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "GE=1./5000*10**-2        #GE=(a+b) grating element\n",
+      "lamda=5.89*10**-7         #Wavelength of light\n",
+      "N=3*5000               #N=W/(a+b) Number of lines on grating\n",
+      " \n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, (a+b)*sin(theta)=m*lamda\n",
+      "#maximum value of sin(theta)=1\n",
+      "m=GE/lamda               #Maximum order of spectra\n",
+      "print\"Maximum order of spectra is =\",m\n",
+      "\n",
+      "RP=3*N                 #Resolving power (round of m to 3)\n",
+      "print\"Resolving power is =\",RP\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum order of spectra is = 3.39558573854\n",
+        "Resolving power is = 45000\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.32,Page number 2-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda1=5.89*10**-7               #Wavelength of light\n",
+      "lamda2=5.896*10**-7              #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "dlamda=lamda2-lamda1                #difference in wavelength\n",
+      "lamda=(lamda2+lamda1)/2             #Mean wavelength\n",
+      "\n",
+      "#(i)\n",
+      "m1=1               #first order\n",
+      "#We know that R.P.=lamda/dlamda=m*N\n",
+      "N1=lamda/dlamda/m1      #Number of lines on grating\n",
+      "print\"(i)Number of lines on grating for first order is =\",N1\n",
+      "\n",
+      "#(ii)\n",
+      "m2=2               #second order\n",
+      "#We know that R.P.=lamda/dlamda=m*N\n",
+      "N2=lamda/dlamda/m2    #Number of lines on grating\n",
+      "print\"(ii)Number of lines on grating for second order is =\",N2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i)Number of lines on grating for first order is = 982.166666667\n",
+        "(ii)Number of lines on grating for second order is = 491.083333333\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.33,Page number 2-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                    #order\n",
+      "lamda=6.553*10**-7     #Wavelength of light\n",
+      "dlamda=1.8*10**-10     #difference in wavelength\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know that R.P.=lam/dlam=m*N\n",
+      "N=lamda/dlamda/m        #Number of lines on grating\n",
+      "print\"Number of lines on grating is =\",N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of lines on grating is = 3640.55555556\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.34,Page number 2-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda1=5.14034*10**-7               #Wavelength of light\n",
+      "lamda2=5.14085*10**-7               #Wavelength of light\n",
+      "\n",
+      "#Calculations:\n",
+      "dlamda=lamda2-lamda1                #difference in wavelength\n",
+      "lamda=(lamda2+lamda1)/2             #Mean wavelength\n",
+      "\n",
+      "#We know that R.P.=lamda/dlamda=m*N\n",
+      "N=lamda/dlamda/1       #Number of lines on grating\n",
+      "print\"Number of lines on grating for first order is =\",N\n",
+      "\n",
+      "#Hence R.P. for second order should be\n",
+      "RP1=2*N\n",
+      "print\"Resolving power in second order should be is= \",RP1\n",
+      "\n",
+      "#But here,\n",
+      "\n",
+      "lamda3=8.03720*10**-7               #Wavelength of light\n",
+      "lamda4=8.03750*10**-7               #Wavelength of light\n",
+      "dlamda2=lamda4-lamda3               #difference in wavelength\n",
+      "lamda2=(lamda4+lamda3)/2            #Mean wavelength\n",
+      "\n",
+      "RP2=lamda2/dlamda2\n",
+      "print\"Resolving power in second order is= \",RP2\n",
+      "\n",
+      "print\"So, the grating will not be able to resolve 8.0372*10^-7 and 8.03750*10^-7 in second order.\"\n",
+      "print\"Because Resolving power is greter than actual Resolving power.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of lines on grating for first order is = 10079.5980392\n",
+        "Resolving power in second order should be is=  20159.1960784\n",
+        "Resolving power in second order is=  26791.1666667\n",
+        "So, the grating will not be able to resolve 8.0372*10^-7 and 8.03750*10^-7 in second order.\n",
+        "Because Resolving power is greter than actual Resolving power.\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.35,Page number 2-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#For grating , Condition of maxima is (a+b)sin(theta)=n*lamda\n",
+      "#Given (a+b) < 2*lamda\n",
+      "#For maximum order, sin(90)=1\n",
+      "#So, n must be less than 2\n",
+      "#i.e. only first order possible if width of grating element is less than twice the wavelength\n",
+      "print\"Hence, Only first order possible if width of grating element is less than twice the wavelength.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hence, Only first order possible if width of grating element is less than twice the wavelength.\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.36,Page number 2-54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n=1                 #order\n",
+      "lamda=5.89*10**-7   #Wavelength of light\n",
+      "a=0.3*10**-3        #width of slit\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, a*sin(theta)=n*lamda\n",
+      "theta1=math.asin(n*lam/a)*180/math.pi    #angular position in first dark band\n",
+      "print\"Angular position in first dark band is =\",theta1,\"Degrees\"\n",
+      "\n",
+      "#We know,for bright band   a*sin(theta)=(2n+1)*lamda/2\n",
+      "theta2=math.asin(1.5*lamda/a)*180/math.pi    #angular position in first bright band\n",
+      "print\"Angular position in first bright band is =\",theta2,\"Degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angular position in first dark band is = 0.112490786047 Degrees\n",
+        "Angular position in first bright band is = 0.168736314576 Degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics-II_by_I_A_Shaikh/chapter3.ipynb b/Applied_Physics-II_by_I_A_Shaikh/chapter3.ipynb
new file mode 100755
index 00000000..714dfad3
--- /dev/null
+++ b/Applied_Physics-II_by_I_A_Shaikh/chapter3.ipynb
@@ -0,0 +1,836 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:5f637d4fa0aa8bb3bc54a28134ac85a937c06f0d3a591f536de2719824a41f5c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Optical Fibre"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1,Page number 3-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n1=1.61                 #Core index\n",
+      "n2=1.55                 #Cladding index\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(n1**2-n2**2)      #Formula\n",
+      "\n",
+      "print\"Numerical Aperture of Fibre is = \",NA\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of Fibre is =  0.435430821142\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2,Page number 3-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n1=1.65                 #Core index\n",
+      "n2=1.53                 #Cladding index\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(n1**2-n2**2)      #Formula\n",
+      "\n",
+      "print\"Numerical Aperture of Fibre is =\",NA\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of Fibre is = 0.617737808459\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3,Page number 3-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n1=1.48                 #R.I. of Core\n",
+      "n2=1.39                 #R.I. of Cladding\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(n1**2-n2**2)      #Formula to find NA\n",
+      "phi=math.asin(NA)*180/(3.1472)    #Acceptance angle\n",
+      "\n",
+      "print\"Numerical Aperture of Fibre is =\",NA\n",
+      "print\"Acceptance angle of Fibre is =\",phi,\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of Fibre is = 0.508232230383\n",
+        "Acceptance angle of Fibre is = 30.491727191 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4,Page number 3-20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#given data:\n",
+      "u1=3.6        #Refractive Index of the Substance at 850 nm\n",
+      "u2=3.4        #Refractive Index of the Substance at 1300 nm\n",
+      "Vv=3*10**8     #Velocity of light in free space\n",
+      "\n",
+      "#Calculations:\n",
+      "# i)Finding wavelength at 850 nm\n",
+      "Vs1=Vv/u1             #Velocity of light in substance at 850 nm\n",
+      "print\"Velocity of light in substance at 850 nm =\" ,Vs1,\"m/sec\"\n",
+      "\n",
+      "lam1=850*10**-9/u1    #Wavelength of light in substance at 850nm\n",
+      "print\" Wavelength of light in substance at 850nm =\",lam1,\"m\"\n",
+      "\n",
+      "\n",
+      "#ii)Finding wavelength at 1300 nm\n",
+      "Vs2=Vv/u2            #Velocity of light in substance at 1300 nm\n",
+      "print\"Velocity of light in substance at 1300 nm =\",Vs2,\" m/sec\"\n",
+      "\n",
+      "lam2=1300*10**-9/u2   #Wavelength of light in substance at 1300nm\n",
+      "print\"Wavelength of light in substance at 1300nm =\" ,lam2,\"m \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of light in substance at 850 nm = 83333333.3333 m/sec\n",
+        " Wavelength of light in substance at 850nm = 2.36111111111e-07 m\n",
+        "Velocity of light in substance at 1300 nm = 88235294.1176  m/sec\n",
+        "Wavelength of light in substance at 1300nm = 3.82352941176e-07 m \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5,Page number 3-20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.5                  #R.I. of Core\n",
+      "u2=1.45                 #R.I.of Cladding\n",
+      "delta= (u1-u2)/u1          #Fractional Refractive index\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=u1*sqrt(2*delta)             #Formula to find NA\n",
+      "theta0=math.asin(NA)*180/(3.1472)       #Acceptance angle\n",
+      "thetac=math.asin(u2/u1)*180/(3.1472)   #Critical angle\n",
+      "\n",
+      "print\"Numerical Aperture of Fibre is =\",NA\n",
+      "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\" \n",
+      "print\" Critical angle of Fibre is =\",thetac,\"degrees\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of Fibre is = 0.387298334621\n",
+        "Acceptance angle of Fibre is = 22.7458994397 degrees\n",
+        " Critical angle of Fibre is = 75.0309676099 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6,Page number 3-20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "NA=0.22              # Numerical Aperture of Fibre\n",
+      "delta=0.012          #Fractional index\n",
+      "\n",
+      "#Calculations:\n",
+      "#Delta=(u1-u2)/u1\n",
+      "u1=NA/math.sqrt(2*delta)  #Formula\n",
+      "u2=u1-(u1*delta)     #Formula\n",
+      "\n",
+      "print\"Refractive Index of core of fibre is =\",u1\n",
+      "print\"Refractive Index of cladding of fibre is =\",u2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive Index of core of fibre is = 1.42009389361\n",
+        "Refractive Index of cladding of fibre is = 1.40305276689\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7,Page number 3-21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.466                 #R.I. of Core\n",
+      "u2=1.46                  #R.I.of Cladding\n",
+      "V=2.4                    #Cut off parameter\n",
+      "lamda=0.8*10**-6            #wavelength in meter\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(u1**2-u2**2)       #Formula to find Numerical Aperture\n",
+      "print\"Numerical Aperture of Fibre is =\",NA\n",
+      "#(printing mistake in book)printed answer is 1.13 but correct answer is 0.13\n",
+      "print\"(printing mistake in book)\"\n",
+      "\n",
+      "# V = 2*3.142*a*NA / lamda\n",
+      "a=V*lamda/(2*3.142*NA)   #core radius\n",
+      "print\"Core radius of Fibre is (a) =\",a,\"m\"\n",
+      "\n",
+      "#w/a= 1.1\n",
+      "w=1.1*a                  #Spot size\n",
+      "print\"Spot size of Fibre is =\",w,\"m\"\n",
+      "\n",
+      "theta=2*lamda*180/3.142/(3.142*w)       #Divergence angle\n",
+      "print\"Divergence angle of Fibre is =\",theta,\"degrees\"\n",
+      "\n",
+      "w10=lamda*10/(3.142*w)     #Spot size at 10 m\n",
+      "print\"Spot size at 10 m of Fibre is =\",w10,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of Fibre is = 0.1324990566\n",
+        "(printing mistake in book)\n",
+        "Core radius of Fibre is (a) = 2.30596263706e-06 m\n",
+        "Spot size of Fibre is = 2.53655890076e-06 m\n",
+        "Divergence angle of Fibre is = 11.5009886523 degrees\n",
+        "Spot size at 10 m of Fibre is = 1.00378073182 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8,Page number 3-21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "w=98                   #Spot size in meter\n",
+      "d=50*10**-6             #Core diameter in meter\n",
+      "a=d/2                  #core radius\n",
+      "u1=1.47                #R.I. of Core\n",
+      "u2=1.45                #R.I.of Cladding\n",
+      "lamda=0.85*10**-6         #Wavlength in meter\n",
+      "NA=math.sqrt(u1**2-u2**2)     #Formula to find NA\n",
+      "\n",
+      "#Calculations:\n",
+      "V=2*3.142*a*NA/lamda        #cut off parameter\n",
+      "N=(V**2)/2               #Number of modes\n",
+      "\n",
+      "print\"Cut off parameter of Fibre is =\",V\n",
+      "print\"Number of modes of Fibre is =\",N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cut off parameter of Fibre is = 44.6646240577\n",
+        "Number of modes of Fibre is = 997.464321107\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9,Page number 3-21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.47                 #R.I. of Core\n",
+      "u2=1.46                 #R.I.of Cladding\n",
+      "lamda=1.3*10**-6           #wavelength in meter\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(u1**2-u2**2)      #Formula to find Numerical Aperture\n",
+      "\n",
+      "#The condition for single mode is V<2.405\n",
+      "#2*3.142*a*NA/lamda < 2.405\n",
+      "\n",
+      "a=2.405*lamda/(2*3.142*NA)   #Maximum radius of fibre\n",
+      "\n",
+      "print\"Maximum radius of Fibre is =\",a,\"meter\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum radius of Fibre is = 2.90662126448e-06 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10,Page number 3-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.465                  #R.I. of Core\n",
+      "u2=1.46                   #R.I.of Cladding\n",
+      "lamda=1.25*10**-6            #operating wavelength\n",
+      "\n",
+      "#Calculations:\n",
+      "delta=(u1-u2)/u1            #Fractional Refractive index\n",
+      "print\"Fractional Refractive index of Fibre is =\",delta\n",
+      "\n",
+      "#For single mode propagation codition is \n",
+      "# a/lamda < 1.4/(3.142*sqrt(u1(u1-u2)))\n",
+      "\n",
+      "a=lamda*1.4/(3.142*u1*math.sqrt(delta))      #core radius\n",
+      "\n",
+      "u=u1-(math.sqrt(2*delta)/(2*3.142*(a/lamda)))     #effective refractive index\n",
+      "print\"Effective Refractive index for lowest mode propagation is =\",u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fractional Refractive index of Fibre is = 0.00341296928328\n",
+        "Effective Refractive index for lowest mode propagation is = 1.46247461864\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11,Page number 3-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.54                  #R.I. of Core\n",
+      "u2=1.5                   #R.I.of Cladding\n",
+      "lamda=1.3*10**-6            #wavelength in meter\n",
+      "a=25*10**-6               #core radius in meter\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(u1**2-u2**2)      #Formula to find Numerical Aperture\n",
+      "\n",
+      "V=2*3.142*a*NA/lamda         #cut off parameter\n",
+      "print\"Cut off parameter of Fibre is =\",V\n",
+      "\n",
+      "N=(V**2)/2                #Number of modes\n",
+      "print\" Number of modes of Fibre is =\",N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cut off parameter of Fibre is = 42.1404937865\n",
+        " Number of modes of Fibre is = 887.910608284\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11.1,Page number 3-25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.52                  #R.I. of Core\n",
+      "u2=1.5189                #R.I.of Cladding\n",
+      "lamda=1.3*10**-6            #wavelength in meter\n",
+      "d=29*10**-6               #core diameter in meter\n",
+      "a=d/2\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(u1**2-u2**2)     #Formula to find Numerical Aperture\n",
+      "V=2*3.142*a*NA/lamda        #Normalised frequency\n",
+      "Nm=(V**2)/2               #Number of modes\n",
+      "\n",
+      "print\"Normalised frequency of Fibre is (V)=\",V\n",
+      "print\"The Maximum Number of modes the Fibre will support is (Nm) =\",Nm\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Normalised frequency of Fibre is (V)= 4.05242861605\n",
+        "The Maximum Number of modes the Fibre will support is (Nm) = 8.2110888441\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12,Page number 3-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u1=1.5                  #R.I. of Core\n",
+      "d=10*10**-6              #diameter of core\n",
+      "a=d/2                   #core radius\n",
+      "lamda=1.3*10**-6           #wavelength\n",
+      "V=2.405                 #cut off parameter for single mode\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, V=2*3.142*a*NA/lamda\n",
+      "NA=V*lamda/(2*3.142*a)       #Numerical Aperture\n",
+      "\n",
+      "theta=math.asin(NA)*180/3.142   #Acceptance angle\n",
+      "print\"Acceptance angle of Fibre is =\",theta,\"Degrees\"\n",
+      "\n",
+      "#Also, NA=u1*sqrt(2*delta)\n",
+      "delta=(NA/u1)**2/2           #Fractional index\n",
+      "print\"Maximum Fractional Refractive index of Fibre is =\",delta\n",
+      "\n",
+      "#delta=(u1-u2)/u1\n",
+      "u2=u1*(1-delta)             #R.I.of cladding\n",
+      "print\"Refractive index of cladding of Fibre is =\",u2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Acceptance angle of Fibre is = 5.71002346964 Degrees\n",
+        "Maximum Fractional Refractive index of Fibre is = 0.00220035113094\n",
+        "Refractive index of cladding of Fibre is = 1.4966994733\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13,Page number 3-23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n1=1.5               #R.I. of core\n",
+      "delta=0.0005         #Fractional index difference\n",
+      "\n",
+      "#Calculations:\n",
+      "#(a):\n",
+      "#Delta=(u1-u2)/u1\n",
+      "n2=n1-(n1*delta)     #R.I. of cladding\n",
+      "print\"(a)Refractive Index of cladding of fibre is =\",n2\n",
+      "\n",
+      "#(b):\n",
+      "phi=math.asin(n2/n1)*180/3.142        #Critical internal reflection angle\n",
+      "print\"(b)Critical internal reflection angle of Fibre is =\",phi,\"degrees\"\n",
+      "\n",
+      "#(c):\n",
+      "theta0=math.asin(math.sqrt(n1**2-n2**2))*180/3.142    #External critical Acceptance angle\n",
+      "print\"(c)External critical Acceptance angle of Fibre is =\",theta0,\"degrees\"\n",
+      "\n",
+      "#(d):\n",
+      "NA=n1*math.sqrt(2*delta)          #Formula to find Numerical Aperture\n",
+      "print\"(d)Numerical Aperture of Fibre is =\",NA\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Refractive Index of cladding of fibre is = 1.49925\n",
+        "(b)Critical internal reflection angle of Fibre is = 88.1766396681 degrees\n",
+        "(c)External critical Acceptance angle of Fibre is = 2.71810509125 degrees\n",
+        "(d)Numerical Aperture of Fibre is = 0.0474341649025\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14,Page number 3-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "NA1=0.20              #Numerical Aperture of Fibre\n",
+      "n2=1.59               #R.I. of cladding\n",
+      "\n",
+      "#Calculations:\n",
+      "#NA=sqrt(n1**2-n2**2)\n",
+      "#In air, n0=1\n",
+      "n1=math.sqrt(NA1**2+n2**2)     #R.I.of core\n",
+      "\n",
+      "#Now, in water \n",
+      "n0=1.33\n",
+      "NA2=math.sqrt(n1**2-n2**2)/n0      #Numerical Aperture in water\n",
+      "theta0=math.asin(NA2)*180/3.142    #Acceptance angle of fibre in water\n",
+      "print\"Acceptance angle of Fibre in water is =\",theta0,\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Acceptance angle of Fibre in water is = 8.6475921767 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15,Page number 3-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n1=1.45             #R.I.of core\n",
+      "n2=1.40             #R.I. of cladding\n",
+      "\n",
+      "#Calculations:\n",
+      "NA=math.sqrt(n1**2-n2**2)   #Numerical Aperture\n",
+      "print\"Numerical Aperture of Fibre is =\",NA\n",
+      "\n",
+      "theta0=math.asin(NA)*180/3.142    #Acceptance angle of fibre\n",
+      "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of Fibre is = 0.377491721764\n",
+        "Acceptance angle of Fibre is = 22.1755250876 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16,Page number 3-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "NA=0.16             #Numerical Aperture of Fibre\n",
+      "n1=1.45             #R.I. of core\n",
+      "d=90*10**-6          #Core diameter\n",
+      "\n",
+      "#Calculations:\n",
+      "#NA=sqrt(n1**2-n2**2)\n",
+      "n2=math.sqrt(n1**2-NA**2)    #R.I.of cladding\n",
+      "print\"(a)Refractive Index of cladding of fibre is =\",n2\n",
+      "\n",
+      "theta0=math.asin(NA)*180/3.142    #Acceptance angle of fibre\n",
+      "print\"(b)Acceptance angle of Fibre is =\",theta0,\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Refractive Index of cladding of fibre is = 1.44114537782\n",
+        "(b)Acceptance angle of Fibre is = 9.20570258795 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17,Page number 3-25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "n1=1.48               #R.I. of core\n",
+      "delta=0.055           #Realtive R.I.\n",
+      "lamda=1*10**-6        #Wavelength of light\n",
+      "a=50*10**-6           #core radius\n",
+      "\n",
+      "#Calculations:\n",
+      "#Delta=(u1-u2)/u1\n",
+      "n2=n1-(n1*delta)      #R.I. of cladding\n",
+      "NA=n1*math.sqrt(2*delta)  #Formula to find Numerical Aperture\n",
+      "print\"Numerical Aperture of Fibre is =\",NA\n",
+      "\n",
+      "\n",
+      "theta0=math.asin(NA)*180/3.142     #Acceptance angle of fibre\n",
+      "print\"Acceptance angle of Fibre is =\",theta0,\"degrees\"\n",
+      "\n",
+      "V=2*3.142*a*NA/lamda            #V number\n",
+      "N=(V**2)/2                   #Number of guided modes\n",
+      "\n",
+      "#In book,instead of NA , value of delta is taken into calculation.\n",
+      "#Thus there is calculation mistake in values of V and N.\n",
+      "\n",
+      "print\"V number of Fibre is =\",V\n",
+      "print\"Number of guided mode of Fibre is =\",N\n",
+      "print\"(Calculation mistake in book)\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of Fibre is = 0.490860468973\n",
+        "Acceptance angle of Fibre is = 29.3933421943 degrees\n",
+        "V number of Fibre is = 154.228359351\n",
+        "Number of guided mode of Fibre is = 11893.1934141\n",
+        "(Calculation mistake in book)\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics-II_by_I_A_Shaikh/chapter4.ipynb b/Applied_Physics-II_by_I_A_Shaikh/chapter4.ipynb
new file mode 100755
index 00000000..11b66b2f
--- /dev/null
+++ b/Applied_Physics-II_by_I_A_Shaikh/chapter4.ipynb
@@ -0,0 +1,163 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:e413b3bea25c7729b7ae1f595bf0f843336be8c682782f6b92de1c93aa46ed9d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Laser"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1,Page number 4-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=694.3*10**-9         #Wavelength in meter\n",
+      "T=300                   #Temperature in Kelvin\n",
+      "\n",
+      "h=6.63*10**-34           #Planck's Constant\n",
+      "c=3*10**8                #Velocity of light\n",
+      "K=1.38*10**-21           #Boltzmann Constant\n",
+      "\n",
+      "#Calculations:\n",
+      "delE= h*c/lamda           #Energy difference between two energy states N and N0\n",
+      "\n",
+      "#N=N0*e^-delE/(K*T)\n",
+      "R=math.e**(-delE/(K*T))      #R=Ratio of N and N0 i.e.(R=N/N0)\n",
+      "\n",
+      "#(Printing mistake in textbook)\n",
+      "#instead of e^-.692, it has taken e^-69.2\n",
+      "\n",
+      "print\"The ratio of population of two energy states is = \",R\n",
+      "print\" (calculation mistake in book)\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ratio of population of two energy states is =  0.500588928485\n",
+        " (calculation mistake in book)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2,Page number 4-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=6328*10**-10        #Wavelength in meter\n",
+      "P=4.5*10**-3              #Power in watts\n",
+      "h=6.63*10**-34           #Planck's Constant\n",
+      "c=3*10**8                 #Velocity of light\n",
+      "\n",
+      "#Calculations:\n",
+      "delE= h*c/lamda           #Energy difference\n",
+      "#N*delE=P\n",
+      "N=P/delE                #number of photons emitted per second\n",
+      "\n",
+      "print\"Number of photons emitted per second is =\",N\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of photons emitted per second is = 1.43167420814e+16\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3,Page number 4-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "lamda=780*10**-9           #Wavelength of photon in meter\n",
+      "P=20*10**-3              #Power of each pulse in watts\n",
+      "t=10*10**-9              #Duration of each pulse\n",
+      "h=6.63*10**-34           #Planck's Constant\n",
+      "c=3*10**8                #Velocity of light\n",
+      "\n",
+      "#Calculations:\n",
+      "delE= h*c/lamda           #Energy of each photon\n",
+      "E=P*t                   #Energy of each pulse\n",
+      "\n",
+      "N=E/delE                #Number of photons in each pulse\n",
+      "print\"Number of photons in each pulse is =\",N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of photons in each pulse is = 784313725.49\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics-II_by_I_A_Shaikh/chapter5.ipynb b/Applied_Physics-II_by_I_A_Shaikh/chapter5.ipynb
new file mode 100755
index 00000000..e9d81103
--- /dev/null
+++ b/Applied_Physics-II_by_I_A_Shaikh/chapter5.ipynb
@@ -0,0 +1,1166 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:fec7a7bbab2881090f60069cf3d4c856415ca45a1545225541a225f28ce72b8e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Foundations of Quantum Mechanics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1,Page number 5-23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=6.68*10**-27       #mass of alpha particle\n",
+      "V=30*10**3           #potential difference\n",
+      "e=1.6*10**-19        #charge of an electron\n",
+      "q=2*e                #Charge of alpha particle\n",
+      "h=6.63*10**-34       #Planck's constant\n",
+      "\n",
+      "#Calculations:\n",
+      "lamda=h/math.sqrt(2*m*q*V)     #de Broglie wavelength\n",
+      "print\"de Broglie wavelength associated with alpha particle is =\" ,lamda,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength associated with alpha particle is = 5.85429607723e-14 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2,Page number 5-23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1                  #mass of given particle in kg\n",
+      "h=6.63*10**-34       #Planck's constant\n",
+      "v=1*10**3            #velocity of particle\n",
+      "\n",
+      "#Calculations:\n",
+      "lamda=h/(m*v)         #de Broglie wavelength\n",
+      "print\"de Broglie wavelength associated with particle is =\",lamda,\"m\"\n",
+      "print\"This wavelength is too small for any practical significance.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength associated with particle is = 6.63e-37 m\n",
+        "This wavelength is too small for any practical significance.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3,Page number 5-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m1=40*10**-3         #mass of bullet in kg\n",
+      "m2=9.1*10**-31       #mass of electron in kg\n",
+      "h=6.63*10**-34       #Planck's constant\n",
+      "v=1100              #velocity of bullet and electron\n",
+      "\n",
+      "#Calculations:\n",
+      "lamda1=h/(m1*v)        #de Broglie wavelength\n",
+      "print\"de Broglie wavelength associated with bullet is =\",lamda1,\"m\"\n",
+      "\n",
+      "lamda2=h/(m2*v)        #de Broglie wavelength\n",
+      "print\"de Broglie wavelength associated with electron is =\",lamda2,\"m\"\n",
+      "\n",
+      "print\"Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " de Broglie wavelength associated with bullet is = 1.50681818182e-35 m\n",
+        "de Broglie wavelength associated with electron is = 6.62337662338e-07 m\n",
+        "Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4,Page number 5-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "V=100            #potential difference\n",
+      "d=2.15*10**-10   #lattice spacing\n",
+      "\n",
+      "#Calculations:\n",
+      "lamda=12.26*10**-10/(math.sqrt(V))          #wavelength associated with electron in meter\n",
+      "\n",
+      "#using bragg's law for first order  lamda=2d sin(theta)\n",
+      "theta=math.degrees(math.asin(lamda/(2*d)))      #glancing angle in degrees\n",
+      "print\"Glancing angle at which first reflection occurs is =\",theta,\"Degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Glancing angle at which first reflection occurs is = 16.5657992687 Degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5,Page number 5-25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "mn=1.674*10**-27     #mass of neutron\n",
+      "h=6.63*10**-34       #Planck's constant\n",
+      "lamda=1*10**-10      #wavelength of neutron\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
+      "E1=h**2/(2*mn*lamda**2)   #Energy of neutron in joules\n",
+      "E=E1/(1.6*10**-19)        #Energy of neutron in electron-Volts\n",
+      "\n",
+      "print\"Energy of neutron is =\",E,\"eV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of neutron is = 0.0820581317204 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6,Page number 5-25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "mn=1.67*10**-27      #mass of neutron\n",
+      "h=6.6*10**-34        #Planck's constant\n",
+      "lamda=3*10**-10        #wavelength of neutron\n",
+      "d=3.036*10**-10      #lattice spacing\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
+      "E1=h**2/(2*mn*lamda**2)  #Energy of neutron in joules\n",
+      "E=E1/(1.6*10**-19)    # Energy of neutron in electron-Volts\n",
+      "print\"Energy of neutron is =\",E,\"eV\"\n",
+      "\n",
+      "#using bragg's law for first order  lamda=2d sin(theta)\n",
+      "theta=math.degrees(math.asin(lamda/(2*d)))      #glancing angle in degrees\n",
+      "print\" Glancing angle at which first orde reflection occurs is =\",theta,\"Degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of neutron is = 0.00905688622754 eV\n",
+        " Glancing angle at which first orde reflection occurs is = 29.6085193042 Degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7,Page number 5-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=9.108*10**-31       #mass of electron\n",
+      "h=6.625*10**-34       #Planck's constant\n",
+      "lamda=5*10**-7          #wavelength of electron\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
+      "E1=h**2/(2*m*lamda**2)     #Energy of electron in joules\n",
+      "E=E1/(1.6*10**-19)      #Energy of electron in electron-Volts\n",
+      "print\"Energy of electron is =\",E,\"eV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of electron is = 6.02363650088e-06 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8,Page number 5-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "mn=1.676*10**-27        #mass of neutron\n",
+      "me=9.1*10**-31          #mass of electron\n",
+      "h=6.625*10**-34         #Planck's constant\n",
+      "\n",
+      "#Calculations:\n",
+      "#Part 1:\n",
+      "En1=0.025              #Energy in eV of neutron\n",
+      "En=En1*(1.6*10**-19)   #Energy in joules\n",
+      "\n",
+      "lamda1=h/math.sqrt(2*mn*En)     #wavelength of a beam of neutron\n",
+      "print\"wavelength of a beam of neutron is =\",lamda1,\"m\"\n",
+      "\n",
+      "#Part 2:\n",
+      "lamda2=2*10**-10            #wavelength of electron and photon\n",
+      "\n",
+      "#//we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
+      "Ee1=h**2/(2*me*lamda2**2)     #Energy of electron in joules\n",
+      "Ee=Ee1/(1.6*10**-19)          #Energy of electron in electron-Volts\n",
+      "print\"Energy of electron is =\",Ee,\"eV\"\n",
+      "\n",
+      "p1=h/lamda2               #momentum of electron\n",
+      "print\" Momentum of electron is =\",p1,\"kg.m/s\"\n",
+      "\n",
+      "C=3*10**8                 #Velocity of light\n",
+      "Ep=h*C/lamda2             #Energy of photon in joules\n",
+      "print\"Energy of photon is =\",Ep,\"Joules\"\n",
+      "\n",
+      "p2=h/lamda2              #momentum of photon\n",
+      "print\"Momentum of photon is =\",p2,\"kg.m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of a beam of neutron is = 1.80927208246e-10 m\n",
+        "Energy of electron is = 37.6808250343 eV\n",
+        " Momentum of electron is = 3.3125e-24 kg.m/s\n",
+        "Energy of photon is = 9.9375e-16 Joules\n",
+        "Momentum of photon is = 3.3125e-24 kg.m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9,Page number 5-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given data:\n",
+      "#We have alpha particle,neutron,proton and electron.\n",
+      "\n",
+      "#To find: shortest wavelength\n",
+      "\n",
+      "print\"We know, lamda=h/sqrt(2*m*E)     #de Broglie wavelength\"\n",
+      "\n",
+      "#Wavelength is inversely proportional to mass of particle for constant energy\n",
+      "print\"i.e., Wavelength is inversely proportional to mass of particle for constant energy. \"\n",
+      "\n",
+      "print\"We have alpha particle,neutron,proton and electron.\"\n",
+      "\n",
+      "#AS,alpha particle has highest mass.Thus it will have shortest wavelength.\n",
+      "print\"Out of above, alpha particle has highest mass.\"\n",
+      "\n",
+      "print\"Hence it will have shortest wavelength.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "We know, lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
+        "i.e., Wavelength is inversely proportional to mass of particle for constant energy. \n",
+        "We have alpha particle,neutron,proton and electron.\n",
+        "Out of above, alpha particle has highest mass.\n",
+        "Hence it will have shortest wavelength.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10,Page number 5-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "me=9.108*10**-31        # mass of electron\n",
+      "mp=1.66*10**-27         # bmass of proton\n",
+      "h=6.625*10**-34         # Planck's constant\n",
+      "lamda=1*10**-10         # wavelength of electron and proton\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
+      "Ee1=h**2/(2*me*lamda**2)           #Energy of electron in joules\n",
+      "Ee=Ee1/(1.6*10**-19)               #Energy of electron in electron-Volts\n",
+      "print\"Energy of electron is =\",Ee,\"eV\"\n",
+      "\n",
+      "Ep1=h**2/(2*mp*lamda**2)      #Energy of photon in joules\n",
+      "Ep=Ep1/(1.6*10**-19)          #Energy of photon in electron-Volts\n",
+      "print\"Energy of photon is =\",Ep,\"eV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of electron is = 150.590912522 eV\n",
+        "Energy of photon is = 0.0826254235693 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11,Page number 5-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m1=50*10**-9         #mass of particle in kg\n",
+      "m2=9.1*10**-31       #mass of electron in kg\n",
+      "h=6.625*10**-34      #Planck's constant\n",
+      "v1=1                #velocity of particle\n",
+      "v2=3*10**6           #velocity of electron\n",
+      "\n",
+      "#Calculations:\n",
+      "lamda1=h/(m1*v1)*10**10        #de Broglie wavelength\n",
+      "print\"de Broglie wavelength associated with particle is =\",lamda1,\"Angstrom\"\n",
+      "\n",
+      "lamda2=h/(m2*v2)*10**10        #de Broglie wavelength\n",
+      "print\"de Broglie wavelength associated with electron is =\",lamda2,\"Angstrom\"\n",
+      "\n",
+      "print\"Wavelength of electron is measurable.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength associated with particle is = 1.325e-16 Angstrom\n",
+        "de Broglie wavelength associated with electron is = 2.42673992674 Angstrom\n",
+        "Wavelength of electron is measurable.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12,Page number 5-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "me=9.1*10**-31           #mass of electron in kg\n",
+      "h=6.63*10**-34          #Planck's constant\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "E1=2*10**3              #Energy in eV of electron\n",
+      "E=E1*(1.6*10**-19)      #Energy in joules\n",
+      " \n",
+      "lamda=h/math.sqrt(2*me*E)      #wavelength of electron\n",
+      "print\"Wavelength of electron is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of electron is = 2.7472794985e-11 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13,Page number 5-30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "me=9.1*10**-31            #mass of electron\n",
+      "h=6.63*10**-34            #Planck's constant\n",
+      "lamda=2*10**-10           #wavelength of electron and photon\n",
+      "\n",
+      "#Calculations:\n",
+      "p1=h/lamda                #momentum of electron\n",
+      "print\"Momentum of electron is =\",p1,\"kg.m/s\"\n",
+      "\n",
+      "Ee=p1**2/(2*me)          #Energy of electron in joules\n",
+      "print\"Energy of electron is =\",Ee,\"Joules\"\n",
+      "\n",
+      "p2=h/lamda                #momentum of photon\n",
+      "print\"Momentum of photon is =\",p2,\"kg.m/s\"\n",
+      "\n",
+      "c=3*10**8                #Velocity of light\n",
+      "Ep=h*c/lamda             #Energy of photon in joules\n",
+      "print\"Energy of photon is =\",Ep,\"Joules\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Momentum of electron is = 3.315e-24 kg.m/s\n",
+        "Energy of electron is = 6.03803571429e-18 Joules\n",
+        "Momentum of photon is = 3.315e-24 kg.m/s\n",
+        "Energy of photon is = 9.945e-16 Joules\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14,Page number 5-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1.676*10**-27            #mass of neutron\n",
+      "h=6.625*10**-34            #Planck's constant\n",
+      "lamda=1*10**-10            #wavelength of neutron\n",
+      "\n",
+      "#Calculations:\n",
+      "C=3*10**8                 #Velocity of light\n",
+      "Ep1=h*C/lamda             #Energy of photon in joules\n",
+      "E1=Ep1/(1.6*10**-19)      #Energy of photon in electron-Volts\n",
+      "print\"Energy of photon is =\",E1,\"eV\"\n",
+      "\n",
+      "#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength\n",
+      "En1=h**2/(2*m*lamda**2)     #Energy of neutron in joules\n",
+      "E2=En1/(1.6*10**-19)     #Energy of neutron in electron-Volts\n",
+      "print\"Energy of neutron is =\",E2,\"eV\"\n",
+      "\n",
+      "R=E1/E2                 #Ratio of energies of proton to neutron\n",
+      "print\"Ratio of energies of proton to neutron is =\",R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of photon is = 12421.875 eV\n",
+        "Energy of neutron is = 0.0818366367094 eV\n",
+        "Ratio of energies of proton to neutron is = 151788.679245\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Example 5.14.1,Page number 5-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "v=900                  #velocity of electron in m/s\n",
+      "delv=v*0.001/100       #uncertainity in velocity\n",
+      "h=6.63*10**-34         #Planck's constant\n",
+      "m=9.1*10**-31           #mass of an electron\n",
+      "\n",
+      "#Calculations:\n",
+      "delp=m*delv            #uncertainity in the measured values of momentum\n",
+      "\n",
+      "#using heisenberg's uncertainity formula\n",
+      "delx=h/(2*3.142*delp)    #uncertainity in its position\n",
+      "print\"Uncertainity with which position of electron can be located is >=\",delx,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Uncertainity with which position of electron can be located is >= 0.0128823012337 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14.2,Page number 5-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1.6*10**-27        #mass of proton in kg\n",
+      "h=6.63*10**-34       #Planck's constant\n",
+      "v=3./20*10**8         #velocity of particle\n",
+      "\n",
+      "#Calculations:\n",
+      "lamda=h/(m*v)        #de Broglie wavelength\n",
+      "print\"de Broglie wavelength associated with proton is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength associated with proton is = 2.7625e-14 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14.3,Page number 5-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1.676*10**-27        #mass of neutron\n",
+      "h=6.634*10**-34        #Planck's constant\n",
+      "\n",
+      "#Calculations:\n",
+      "E1=0.025              #Energy in eV of neutron\n",
+      "E=E1*(1.6*10**-19)    #Energy in joules\n",
+      "#As E=m*v**2/2\n",
+      "v=math.sqrt(2*E/m)        #Velocity of neutron beam\n",
+      "\n",
+      "lamda=h/(m*v)          #wavelength of a beam of neutron\n",
+      "print\"wavelength of a beam of neutron is =\",lamda,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of a beam of neutron is = 1.81172996152e-10 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14.4,Page number 5-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "delx=10*10**-9          #uncertainity in position of electron\n",
+      "h=6.63*10**-34          #Planck's constant\n",
+      "m=9.1*10**-31           #mass of an electron\n",
+      "E=10**3*1.6*10**-19      #Energy of electron in joules\n",
+      "\n",
+      "#Calculations:\n",
+      "p=math.sqrt(2*m*E)          #momentum of electron\n",
+      "#using heisenberg's uncertainity formula\n",
+      "delp=h/(2*math.pi*delx)    #uncertainity in the momentum\n",
+      "\n",
+      "P=delp/p*100           #percentage of uncertainity in momentum\n",
+      "print\"Percentage of uncertainity in momentum of electron is =\",P,\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of uncertainity in momentum of electron is = 0.0618355139385 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15,Page number 5-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1.676*10**-27          #mass of neutron\n",
+      "h=6.63*10**-34           #Planck's constant\n",
+      "lamda=2*10**-12            #wavelength of neutron\n",
+      "c=3*10**8                #Velocity of light\n",
+      "\n",
+      "#Calculations:\n",
+      "p=h/lamda                 #momentum of neutron\n",
+      "KE=p**2/(2*m)            #Kinetic Energy of neutron in joules\n",
+      "print\"Kinetic Energy of electron is =\",KE,\"Joules\"\n",
+      "\n",
+      "#velocity of particle is same as group velocity. Thus,\n",
+      "vg=p/m                  #group velocity\n",
+      "print\"group velocity of neutron is =\",vg,\"m/s\"\n",
+      "\n",
+      "#using, vg*vp=c**2\n",
+      "vp=c**2/vg               #phase velocity\n",
+      "print\" phase velocity of neutron is =\",vp,\"m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Kinetic Energy of electron is = 3.27840841289e-17 Joules\n",
+        "group velocity of neutron is = 197792.362768 m/s\n",
+        " phase velocity of neutron is = 4.55022624434e+11 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16,Page number 5-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "m=1.157*10**-30         #mass of particle in kg\n",
+      "h=6.63*10**-34          #Planck's constant\n",
+      "c=3*10**8               #Velocity of light\n",
+      "\n",
+      "#Calculations:\n",
+      "E1=80                  #Energy in eV of particle\n",
+      "E=E1*(1.6*10**-19)     #Energy in joules\n",
+      " \n",
+      "lamda=h/math.sqrt(2*m*E)     #wavelength of particle\n",
+      "print\"Wavelength of particle is =\",lamda,\"m\"\n",
+      "\n",
+      "#Now,\n",
+      "vg=h/(lamda*m)          #group velocity\n",
+      "print\"Group velocity of particle is =\",vg,\"m/s\"\n",
+      "\n",
+      "#using, vg*vp=c**2\n",
+      "vp=c**2/vg              #phase velocity\n",
+      "print\"Phase velocity of particle is =\",vp,\"m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of particle is = 1.21822320075e-10 m\n",
+        "Group velocity of particle is = 4703848.2563 m/s\n",
+        "Phase velocity of particle is = 19133270270.7 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.17,Page number 5-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "v=400                #velocity of electron in m/s\n",
+      "delv=0.01/100       #uncertainity in velocity\n",
+      "h=6.63*10**-34       #Planck's constant\n",
+      "m=9.11*10**-31       #mass of an electron\n",
+      "\n",
+      "#Calculations:\n",
+      "p=m*v                #momentum of an electron\n",
+      "delp=p*delv          #uncertainity in the measured values of momentum\n",
+      "\n",
+      "#using heisenberg's uncertainity formula\n",
+      "delx=h/(2*math.pi*delp)    #accuracy in its position\n",
+      "print\"Accuracy in its position is >=\",delx,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Accuracy in its position is >= 0.00289571150576 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.18,Page number 5-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Given Data:\n",
+      "delx=10**-8             #maximum uncertainity in position of electron\n",
+      "h=6.63*10**-34          #Planck's constant\n",
+      "m=9.1*10**-31           #mass of an electron\n",
+      "\n",
+      "#Calculations:\n",
+      "#using heisenberg's uncertainity formula\n",
+      "delp=h/(2*math.pi*delx)    #minimum uncertainity in the measured values of momentum\n",
+      "\n",
+      "delv=delp/m            #minimum uncertainity in the velocity of an electron\n",
+      "print\"Minimum uncertainity in the velocity of an electron is =\",delv,\"m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum uncertainity in the velocity of an electron is = 11595.5744253 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.19,Page number 5-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "delv=2*10**4            #uncertainity in velocity\n",
+      "h=6.63*10**-34          #Planck's constant\n",
+      "m=9.1*10**-31           #mass of an electron\n",
+      "\n",
+      "#Calculations:\n",
+      "delp=m*delv             #uncertainity in the measured values of momentum\n",
+      "\n",
+      "#using heisenberg's uncertainity formula\n",
+      "delx=h/(2*math.pi*delp)    #accuracy in its position\n",
+      "print\"Minimum space required by electron to be confined in an atom is >=\",delx,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum space required by electron to be confined in an atom is >= 5.79778721263e-09 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.20,Page number 5-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "delt=1.4*10**-10        #uncertainity in time spent by nucleus in excited state\n",
+      "h=6.63*10**-34          #Planck's constant\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#using, delE*delt>= h/(2*math.pi)\n",
+      "delE1= h/(2*math.pi*delt)       #uncertaininty in its energy in excited state in joules\n",
+      "delE=delE1/(1.6*10**-19)     #uncertaininty in its energy in excited state in eV\n",
+      "print\"Uncertaininty in its energy in excited state is >=\",delE,\"eV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Uncertaininty in its energy in excited state is >= 4.71070211026e-06 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.21,Page number 5-35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "a=2*10**-10             #width of potential well in m\n",
+      "h=6.63*10**-34          #Planck's constant\n",
+      "m=9.1*10**-31           #mass of an electron\n",
+      "\n",
+      "#Calculations:\n",
+      "#we know equation for energy of an electron\n",
+      "n0=1\n",
+      "E01=n0**2*h**2/(8*m*a**2)   #Energy in ground state\n",
+      "E0=E01/(1.6*10**-19)      #Energy in eV\n",
+      "print\"Energy of an electron in ground state is=\",E0,\"eV\"\n",
+      "\n",
+      "n1=2\n",
+      "E11=n1**2*h**2/(8*m*a**2)  #Energy in first excited state\n",
+      "E1=E11/(1.6*10**-19)      #Energy in eV\n",
+      "print\" Energy of an electron in first excited state is=\",E1,\"eV\"\n",
+      "\n",
+      "\n",
+      "n2=3\n",
+      "E21=n2**2*h**2/(8*m*a**2)   #Energy in second excited state\n",
+      "E2=E21/(1.6*10**-19)      #Energy in eV\n",
+      "print\"Energy of an electron in second excited state is=\",E2,\"eV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of an electron in ground state is= 9.43443080357 eV\n",
+        " Energy of an electron in first excited state is= 37.7377232143 eV\n",
+        "Energy of an electron in second excited state is= 84.9098772321 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.22,Page number 5-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "a=25*10**-10               #width of well\n",
+      "delx=5*10**-10             #uncertainity in position of particle\n",
+      "n=1                        #ground state\n",
+      "\n",
+      "#calculation:\n",
+      "x1=a/2\n",
+      "psi1=math.sqrt(2/a)*math.sin(n*math.pi/a*x1)\n",
+      "P1=(psi1**2)*delx            #Probability of finding particle at distance of x1\n",
+      "print\"Probability of finding particle at a distance of x1 is =\",P1\n",
+      "\n",
+      "x2=a/3\n",
+      "psi2=math.sqrt(2/a)*math.sin(n*math.pi/a*x2)\n",
+      "P2=(psi2**2)*delx             #Probability of finding particle at distance of x2\n",
+      "print\"Probability of finding particle at a distance of x2 is =\",P2\n",
+      "print\"(There is print mistake in book).\"\n",
+      "\n",
+      "x3=a\n",
+      "psi3=math.sqrt(2/a)*math.sin(n*math.pi/a*x3)\n",
+      "P3=(psi3**2)*delx             #Probability of finding particle at distance of x3\n",
+      "print\"Probability of finding particle at a distance of x3 is =\",P3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Probability of finding particle at a distance of x1 is = 0.4\n",
+        "Probability of finding particle at a distance of x2 is = 0.3\n",
+        "(There is print mistake in book).\n",
+        "Probability of finding particle at a distance of x3 is = 5.99903913065e-33\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics-II_by_I_A_Shaikh/chapter6.ipynb b/Applied_Physics-II_by_I_A_Shaikh/chapter6.ipynb
new file mode 100755
index 00000000..dec8353e
--- /dev/null
+++ b/Applied_Physics-II_by_I_A_Shaikh/chapter6.ipynb
@@ -0,0 +1,1183 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:bc506d3130781232c57938d478c5c5179ee56efb068c1be93d57258ec96fcd79"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: Magnetic Materials and Circuits"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1,Page number 2-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "H=198                  #Magnetizing Force in Ampere per meter\n",
+      "M=2300                 #Magnetization in Ampere per meter\n",
+      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
+      "\n",
+      "#Calculations:\n",
+      "#H=(B/u0)-M\n",
+      "B=u0*(H+M)             #Flux Density\n",
+      "ur=B/(u0*H)            #Relative Permeability\n",
+      "\n",
+      "print\"Corresponding Flux Density is =\",B,\"Wb/m^2\"\n",
+      "print\"Relative Permeability is =\",ur\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Corresponding Flux Density is = 0.00313907937947 Wb/m^2\n",
+        "Relative Permeability is = 12.6161616162\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2,Page number 2-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "x=3.7*10**-3           #Susceptibility at T=300 K\n",
+      "T=300                  #Temperature in kelvin\n",
+      "T1=250                 #Temperature in kelvin\n",
+      "T2=600                 #Temperature in kelvin\n",
+      "\n",
+      "#Calculations:\n",
+      "C=x*T                  #Curie's law\n",
+      "ur1=C/T1               #Relative permeability at 250 K\n",
+      "ur2=C/T2               #Relative permeability at 600 K\n",
+      "\n",
+      "print\"Relative Permeability at 250 K is =\",ur1\n",
+      "print\"Relative Permeability at 600 K is =\",ur2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relative Permeability at 250 K is = 0.00444\n",
+        "Relative Permeability at 600 K is = 0.00185\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3,Page number 2-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u=0.8*10**-23          #Magnetic dipole moment of an atom in paramagnetic gas in J/T\n",
+      "B=0.8                  #Magnetic field in tesla\n",
+      "K=1.38*10**-23         #Boltzmann constant\n",
+      "\n",
+      "#To find Temperature at which Average thermal energy is equal to Magnetic energy \n",
+      "#i.e. uB=3KT/2\n",
+      "T=2*u*B/(3*K)          #Required temperature\n",
+      "\n",
+      "print\"Required temperature is =\",T,\"Kelvin\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required temperature is = 0.309178743961 Kelvin\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4,Page number 2-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "T=27+273           #Temperature in kelvin\n",
+      "B=0.5              #Magnetic field in tesla\n",
+      "C=2*10**-3         #Curie's Constant\n",
+      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
+      "\n",
+      "#  C=u0*M*T/B (Curie's law)\n",
+      "M=C*B/(u0*T)      #Magnetization of material at 300 K\n",
+      "\n",
+      "print\"Magnetization of material at 300 K is =\",M,\"A/m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnetization of material at 300 K is = 2.65258238486 A/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5,Page number 2-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "B=10.9*10**-5            #Horizontal component of B in wb/m^2\n",
+      "u0=4*math.pi*10**-7     #Permeability in free space\n",
+      "\n",
+      "H=B/u0                  #Horizontal component of magnetic field\n",
+      "print\"Horizontal component of magnetic field is =\",H,\"Ampere/meter\"\n",
+      "print\"(Print mistake in unit in book)\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horizontal component of magnetic field is = 86.7394439851 Ampere/meter\n",
+        "(Print mistake in unit in book)\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6,Page number 2-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "ur=900                  #Relative permeability of medium\n",
+      "l=2                      #length in meter\n",
+      "A=60*10**-4             #Crosss sectional area of ring in m^2\n",
+      "phi=5.9*10**-3          #flux in weber\n",
+      "n=700                   #Number of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, phi=B*A\n",
+      "B=phi/A            #Flux density\n",
+      "#But, B=u*H\n",
+      "H=B/(u0*ur)        #Magnetic field strength\n",
+      "\n",
+      "I=H*l/n            #Required current\n",
+      "print\"Current required to produce given flux is =\",I,\"Ampere\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current required to produce given flux is = 2.48416445567 Ampere\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7,Page number 2-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
+      "ur=900                 #Relative permeability of medium\n",
+      "r=25*10**-2            #radius of ring\n",
+      "A=25*10**-4            #Crosss sectional area of ring in m^2\n",
+      "Ag=1*10**-3            #Air gap\n",
+      "phi=2.7*10**-3         #flux in weber\n",
+      "N=400                  #Number of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, phi=B*A\n",
+      "B=phi/A            #Flux density\n",
+      "#But, B=u*H\n",
+      "H=B/(u0*ur)        #Magnetic field strength\n",
+      "L=H*2*math.pi*r+(B*Ag/u0)   #Total amp turns required (iron+air)\n",
+      "I=L/N           #Required current\n",
+      "\n",
+      "print\"Current required to produce given flux is =\",I,\"Ampere\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current required to produce given flux is = 5.89859173174 Ampere\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8,Page number 2-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "A=0.2*10**-4            #Crosss sectional area of iron bar in m^2\n",
+      "H=1600                  #magnetising field in A/m\n",
+      "phi=2.4*10**-5          #Magnetic flux in weber\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know, phi=B*A\n",
+      "B=phi/A            #Flux density\n",
+      "u=B/H              #magnetic permeability\n",
+      "ur=u/u0            #relative permeability\n",
+      "xm=ur-1            #susceptibility of the iron bar\n",
+      "\n",
+      "print\"magnetic permeability of iron bar is =\",u,\"N/(A^2)\"\n",
+      "print\"susceptibility of the iron bar is =\",xm\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetic permeability of iron bar is = 0.00075 N/(A^2)\n",
+        "susceptibility of the iron bar is = 595.831036595\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9,Page number 2-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7       #Permeability in vacuum\n",
+      "xm=948*10**-11            #susceptibility of the iron bar\n",
+      "\n",
+      "#Calculations:\n",
+      "ur=1+xm            #relative permeability\n",
+      "u=u0*ur            #permeability of medium\n",
+      "\n",
+      "print\"Relative Permeability of medium is =\",ur\n",
+      "print\"Permeability of medium is =\",u,\"H/m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relative Permeability of medium is = 1.00000000948\n",
+        "Permeability of medium is = 1.25663707335e-06 H/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10,Page number 2-30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "B=2.5                #Magnetic field in tesla\n",
+      "u0=4*math.pi*10**-7    #Permeability in free space\n",
+      "i0=0.7               #current in the core\n",
+      "ri=11*10**-2         #inner radii of core\n",
+      "ro=12*10**-2         #outer radii of core\n",
+      "\n",
+      "#Calculations:\n",
+      "r=(ri+ro)/2          #Average radii of core\n",
+      "n=3000/(2*math.pi*r)    #Number of turns\n",
+      "\n",
+      "#We know, B=u0*ur*n*i0 .Thus,\n",
+      "ur=B/(u0*n*i0)\n",
+      "\n",
+      "print\"Relative Permeability of medium is =\",ur\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relative Permeability of medium is = 684.523809524\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11,Page number 2-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "B=1.0                #Flux density in tesla\n",
+      "u0=4*math.pi*10**-7     #Permeability in free space\n",
+      "i=2.0                #current in the core\n",
+      "n=10*100             #n=N/l i.e. turns per meter\n",
+      "\n",
+      "#Calculations:\n",
+      "H=n*i               #Magnetising force produced in wire\n",
+      "print\"Magnetising force produced in wire is =\",H,\"Amp-turn/meter\"\n",
+      "\n",
+      "#We know that, B=u0(H+I).Thus,\n",
+      "I=B/u0-H            #Magnetisation of material\n",
+      "print\"Magnetisation of material is =\",I,\"Amp-turn/meter\"\n",
+      "\n",
+      "#u=B/H, i.e. ur*u0=B/H.\n",
+      "ur=B/(u0*H)        #Relative permeability of core\n",
+      "print\" Relative Permeability of core is =\",ur\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnetising force produced in wire is = 2000.0 Amp-turn/meter\n",
+        "Magnetisation of material is = 793774.715459 Amp-turn/meter\n",
+        " Relative Permeability of core is = 397.88735773\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12,Page number 2-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "M=40             #Mass of an iron core\n",
+      "D=7.5*10**3      #Density of iron\n",
+      "f=100            #Frequency\n",
+      "A=3800*10**-1    #Loss due to Area of hysterisis loop in J/m^3\n",
+      "\n",
+      "#Calculations:\n",
+      "V=M/D            #Volume of iron core\n",
+      "L1=A*V           #Loss of energy in core per cycle\n",
+      "print\"Loss of energy in core per cycles is =\",L1,\"joules\"\n",
+      "\n",
+      "N=f*60           #Number of cycles per minute\n",
+      "L=L1*N           #Loss of energy per minute\n",
+      "\n",
+      "print\"Loss of energy per minute is =\",L,\"joules\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loss of energy in core per cycles is = 2.02666666667 joules\n",
+        "Loss of energy per minute is = 12160.0 joules\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13,Page number 2-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "l=30*10**-2             #length in meter\n",
+      "A=1*10**-4          #Crosss sectional area of ring in m^2\n",
+      "phi=2*10**-6        #flux in weber\n",
+      "N=300               #Number of turns\n",
+      "I=0.032             #Current in winding\n",
+      "\n",
+      "#Calculations:\n",
+      "#(i):\n",
+      "B=phi/A             #Flux density\n",
+      "print\"(i)Flux Density in the ring is =\",B,\"Wb/m^2\"\n",
+      "\n",
+      "#(ii):\n",
+      "H=N*I/l            #Magnetic intensity\n",
+      "print\"(ii)Magnetic intensity is =\",H,\"Amp-turn/meter\"\n",
+      "\n",
+      "#(iii):\n",
+      "u=B/H              #Permeability of ring\n",
+      "print\"(iii)Permeability of ring is =\",u,\" Wb/A-m\"\n",
+      "ur=u/u0            #Relative permeability of ring\n",
+      "print\"Relative Permeability of ring is =\",ur\n",
+      "\n",
+      "#(iv):\n",
+      "xm=ur-1            #susceptibility of the ring\n",
+      "print\"(iv)Magnetic susceptibility of the ring is =\",xm\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i)Flux Density in the ring is = 0.02 Wb/m^2\n",
+        "(ii)Magnetic intensity is = 32.0 Amp-turn/meter\n",
+        "(iii)Permeability of ring is = 0.000625  Wb/A-m\n",
+        "Relative Permeability of ring is = 497.359197162\n",
+        "(iv)Magnetic susceptibility of the ring is = 496.359197162\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.14,Page number 2-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Dta:\n",
+      "M=12*10**3       #Mass of an iron core in grams\n",
+      "D=7.5            #Density of iron in gm/cc\n",
+      "f=50             #Frequency\n",
+      "A=3000           #loss due to Area of hysterisis loop in ergs/cm^3\n",
+      "\n",
+      "#Calculations:\n",
+      "V=M/D            #Volume of iron core\n",
+      "L1=A*V           #Loss of energy in core per cycle\n",
+      "\n",
+      "L=L1*f*3600      #Loss of energy per hour\n",
+      "\n",
+      "print\"Loss of energy per hour is =\",L,\"Erg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loss of energy per hour is = 8.64e+11 Erg\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15,Page number 2-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "A=0.5*10**3            #Area of B-H loop in Joules per m^3\n",
+      "V=10**-3               #Volume of specimen in m^3\n",
+      "n=50                   #Frequency of a.c.\n",
+      "\n",
+      "#Calculations:\n",
+      "H=n*V*A               #Hysteresis power loss\n",
+      "\n",
+      "print\"Hysteresis power loss is =\",H,\"Watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hysteresis power loss is = 25.0 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.16,Page number 2-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "ur=1000             #Relative permeability of medium\n",
+      "V=10**-4            #Volume of iron rod in m^3\n",
+      "n=500               #Number of turns per meter\n",
+      "i=0.5               #Current in windings of solenoid in Amperes\n",
+      "\n",
+      "#Calculations:\n",
+      "#We know I=(ur-1)H\n",
+      "#and H=ni , hence\n",
+      "I=(ur-1)*n*i       #Intensity of magnetisation\n",
+      "M=I*V              #Magnetic moment\n",
+      "\n",
+      "print\"Magnetic moment of the rod is =\",M,\"A-m^2\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnetic moment of the rod is = 24.975 A-m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17,Page number 2-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "ur=600             #Relative permeability of iron\n",
+      "d=12*10**-2        #mean diameter of ring in m\n",
+      "N=500              #Number of turns\n",
+      "i=0.3              #Current in windings of solenoid in Amperes\n",
+      "\n",
+      "#Calculations:\n",
+      "r=d/2              #Radius of ring\n",
+      "\n",
+      "B=u0*ur*N*i/(2*math.pi*r)     #Flux densityin the core\n",
+      "print\"Flux densityin the core is =\",B,\"Wb/m^2\"\n",
+      "\n",
+      "H=B/(u0*ur)         #Magnetic intensity\n",
+      "print\"Magnetic intensity is =\",H,\"Amp-turns/m\"\n",
+      "\n",
+      "#We know that, B=u0(H+I)\n",
+      "I1=(B-u0*H)/u0      #magnetisation\n",
+      "I2=u0*I1            #Electronic current loop\n",
+      "\n",
+      "I=I2/B*100          #Percentage flux density due to electroniuc loop currents\n",
+      "print\"Percentage flux density due to electroniuc loop currents is =\",I,\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flux densityin the core is = 0.3 Wb/m^2\n",
+        "Magnetic intensity is = 397.88735773 Amp-turns/m\n",
+        "Percentage flux density due to electroniuc loop currents is = 99.8333333333 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18,Page number 2-35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "\n",
+      "u0=4*math.pi*10**-7   #Permeability in vacuum\n",
+      "ur=900               #Relative permeability of iron ring\n",
+      "d=40*10**-2          #diameter of ring\n",
+      "l=5*10**-3           #air gap in the ring\n",
+      "A=5.8*10**-4         #Crosss sectional area of ring in m^2\n",
+      "phi=1.5*10**-4       #flux in weber\n",
+      "N=600                #Number of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "r=d/2                #Radius of ring\n",
+      "\n",
+      "#We know, phi=B*A\n",
+      "B=phi/A              #Flux density\n",
+      "\n",
+      "#But, B=u*H\n",
+      "H=B/(u0*ur)          #Magnetic field strength\n",
+      "\n",
+      "m1=H*ur*l            #amp-turns in air gap\n",
+      "m2=H*2*math.pi*r     #amp-turns by ring\n",
+      "m=m1+m2              #total mmf(amp-turns) required\n",
+      "\n",
+      "I=m/N                #Required current\n",
+      "print\"Current required to produce given flux is =\",I,\"Amperes\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current required to produce given flux is = 2.19395891742 Amperes\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18.1,Page number 2-38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7    #Permeability in vacuum\n",
+      "X=-0.5*10**-5          #Magnetic susceptibility of silicon\n",
+      "H=9.9*10**4            #Magnetic field intensity\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#As, X=I/H. thus,\n",
+      "I=X*H                  #intensity of magnetisation\n",
+      "print\"Intensity of magnetisation is =\",I\n",
+      "\n",
+      "B=u0*(H+I)             #Magnetic flux density\n",
+      "print\"Magnetic flux density is =\",B,\"Wb/ m^2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Intensity of magnetisation is = -0.495\n",
+        "Magnetic flux density is = 0.124406447047 Wb/ m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18.2,Page number 2-38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "ur=380              #Relative permeability\n",
+      "d=20*10**-2         #diameter of solenoid in m\n",
+      "r=d/2               #radius of ring in m\n",
+      "A=5*10**-4           #Crosss sectional area of ring in m^2\n",
+      "phi=2*10**-3         #flux in weber\n",
+      "N=200                #Number of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "l=math.pi*d          #air gap in the ring\n",
+      "S=(l/(u0*ur*A))      #Reluctance of iron ring\n",
+      "print\"Reluctance of iron ring is =\",S,\"Amp-turn/ Wb \"\n",
+      "\n",
+      "#ohm's law for magnetic circuit is  phi=N*I/S. thus,\n",
+      "I=S*phi/N            #required current\n",
+      "print\"Current required to obtain given magnetic flux is =\",I,\"Amperes\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reluctance of iron ring is = 2631578.94737 Amp-turn/ Wb \n",
+        "Current required to obtain given magnetic flux is = 26.3157894737 Amperes\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18.3,Page number 2-39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Values:\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "ur=1               #Relative permeability of air\n",
+      "r=15*10**-2        #radius of ring in m\n",
+      "A=6*10**-4         #Crosss sectional area of ring in m^2\n",
+      "I=4                #Coil current in amp\n",
+      "N=500              #Number of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "m=N*I              #MMF of coil\n",
+      "print\"MMF of coil is =\",m,\"Ampere-turn\"\n",
+      "\n",
+      "l=2*math.pi*r      #air gap\n",
+      "R=(l/(u0*ur*A))    #Reluctance of iron ring\n",
+      "print\"Reluctance of iron ring is =\",R,\"Ampere-turn/Wb\"\n",
+      "\n",
+      "phi=m/R            #Magnetic flux\n",
+      "print\"Magnetic flux is =\",phi,\"Weber\"\n",
+      "\n",
+      "B=phi/A            #Magnetic Flux density\n",
+      "print\"Magnetic flux density is =\",B,\"Weber/m^2\"\n",
+      "\n",
+      "H=B/(u0*ur)        #Magnetic field intensity\n",
+      "print\"Magnetic field intensity is =\",H,\"Amperes/m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MMF of coil is = 2000 Ampere-turn\n",
+        "Reluctance of iron ring is = 1250000000.0 Ampere-turn/Wb\n",
+        "Magnetic flux is = 1.6e-06 Weber\n",
+        "Magnetic flux density is = 0.00266666666667 Weber/m^2\n",
+        "Magnetic field intensity is = 2122.06590789 Amperes/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.19,Page number 2-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "u0=4*math.pi*10**-7     #Permeability in vacuum\n",
+      "ur=6*10**-3         #Relative permeability of iron\n",
+      "r=0.5               #radius of ring in m\n",
+      "l=1*10**-2          #air gap in the ring\n",
+      "A=5*10**-4          #Crosss sectional area of ring in m^2\n",
+      "i=5                 #current in ampere\n",
+      "N=900               #Number of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "S=(l/(u0*A))+((2*math.pi*r-l)/ur*A)     #Reluctance of iron\n",
+      "print\"Reluctance of iron is =\",S,\"Ampere-turn/Wb\"\n",
+      "\n",
+      "m=N*i              #mmf produced\n",
+      "print\"mmf produced is =\",m,\"Ampere-turn\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reluctance of iron is = 15915494.5702 Ampere-turn/Wb\n",
+        "mmf produced is = 4500 Ampere-turn\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.20,Page number 2-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "H=5*10**3          #coercivity of bar magnet in amp/m\n",
+      "l=10*10**-2        #length of solenoid in m\n",
+      "N=50               #No of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know that, H=NI/l ,hence\n",
+      "I=l*H/N            #current through solenoid\n",
+      "\n",
+      "print\"Current through solenoid is =\",I,\"Amperes\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current through solenoid is = 10.0 Amperes\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.21,Page number 2-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "ur=1200            #Relative permeability of medium\n",
+      "V=10**-3           #volume of iron rod\n",
+      "N=5*10**2          #no of turns per m\n",
+      "i=0.5              #current through solenoid in amp\n",
+      "\n",
+      "#Calculations:\n",
+      "x=ur-1             #susceptibility of the ring\n",
+      "H=N*i             #Magnetisisng field\n",
+      "\n",
+      "#We know, x=I/H\n",
+      "I=x*H              #magnetisation\n",
+      "\n",
+      "#Also, I=M/V , thus\n",
+      "M=I*V              #magnetic moment\n",
+      "print\"Magnetic moment is =\",M,\"Ampere-turn-m^2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnetic moment is = 299.75 Ampere-turn-m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.22,Page number 2-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Dta:\n",
+      "ur=100            #Relative permeability of medium\n",
+      "l=0.2             #length of iron rod\n",
+      "d=10*10**-3       #diameter of solenoid in m\n",
+      "N=300             #no of turns per m\n",
+      "i=0.5             #current through solenoid in amp\n",
+      "r=d/2             #radius of solenoid\n",
+      "\n",
+      "#Calculations:\n",
+      "x=ur-1            #susceptibility of the ring\n",
+      "H=N*i             #Magnetisisng field\n",
+      "\n",
+      "#We know, x=I/H\n",
+      "I=x*H             #magnetisation\n",
+      "\n",
+      "V=math.pi*(r**2)*l     #volume of iron rod\n",
+      "\n",
+      "#Also, I=M/V , thus\n",
+      "M=I*V             #magnetic moment\n",
+      "print\"Magnetic moment is =\",M,\"Ampere-turn-m^2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnetic moment is = 0.233263254529 Ampere-turn-m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.23,Page number 2-38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data:\n",
+      "l=1.2              #length of circuit in meter\n",
+      "u=7.3*10**-3       #permeability of silicon sheet\n",
+      "A=100              #cross sectional area in cm^2\n",
+      "N=150              #No of turns\n",
+      "B=0.3              #magmetic field in Wb/m^2\n",
+      "\n",
+      "#Calculations:\n",
+      "\n",
+      "#We know, B=u*H\n",
+      "H=B/u              #Magnetic field strength\n",
+      "\n",
+      "m=H*l              #amp-turns in air gap\n",
+      "\n",
+      "I1=m/N             #Required current\n",
+      "print\"Current required to obtain given magnetic field is =\",I1,\"Amperes\"\n",
+      "\n",
+      "I=I1/A             #Required current per unit area\n",
+      "print\"Current required per unit area to obtain given magnetic field is =\",I,\"Amperes\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current required to obtain given magnetic field is = 0.328767123288 Amperes\n",
+        "Current required per unit area to obtain given magnetic field is = 0.00328767123288 Amperes\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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index e7e0fc16..00000000
--- a/Applied_Physics/Chapter_01_Bonding_in_Solids_.ipynb
+++ /dev/null
@@ -1,248 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:7b7dfd33b7ab2e69421d0f26bfbfaa20f35bc1084164b445bbbc400215113c08"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1:Bonding in Solids"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3 , Page no:15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r=2; #in angstrom(distance)\n",
-      "e=1.6E-19; # in C (charge of electron)\n",
-      "E_o= 8.85E-12;# absolute premittivity\n",
-      "\n",
-      "#calculate\n",
-      "r=2*1*10**-10; # since r is in angstrom\n",
-      "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
-      "V1=V/e; # changing to eV\n",
-      "\n",
-      "#result\n",
-      "print \"\\nThe potential energy is  V = \",V,\"J\";\n",
-      "print \"In electron-Volt V = \",round(V1,3),\"eV\"; \n",
-      "print \"Note: the answer in the book is wrong due to calculation mistake\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The potential energy is  V =  -1.15153477995e-18 J\n",
-        "In electron-Volt V =  -7.197 eV\n",
-        "Note: the answer in the book is wrong due to calculation mistake\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4 , Page no:15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "#given\n",
-      "r0=0.236; #in nanometer(interionic distance)\n",
-      "e=1.6E-19; # in C (charge of electron)\n",
-      "E_o= 8.85E-12;# absolute premittivity\n",
-      "N=8; # Born constant\n",
-      "IE=5.14;# in eV (ionisation energy of sodium)\n",
-      "EA=3.65;# in eV (electron affinity of Chlorine)\n",
-      "pi=3.14; # value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r0=r0*1E-9; # since r is in nanometer\n",
-      "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
-      "PE=PE/e; #changing unit from J to eV\n",
-      "NE=IE-EA;# calculation of Net energy\n",
-      "BE=PE-NE;# calculation of Bond Energy\n",
-      "\n",
-      "#result\n",
-      "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
-      "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
-      "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
-      "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
-      "# Note: (2) There is slight variation in the answer due to round off."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The potential energy is PE=  5.34 eV\n",
-        "The net energy is NE=  1.49 eV\n",
-        "The bond energy is BE=  3.85 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5 , Page no:16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r_0=.41; #in mm(lattice constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "E_o= 8.85E-12; #absolute premittivity\n",
-      "n=0.5; #repulsive exponent value\n",
-      "alpha=1.76; #Madelung constant\n",
-      "pi=3.14; # value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r=.41*1E-3; #since r is in mm\n",
-      "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
-      "\n",
-      "#result\n",
-      "print\"The compressibility  is\tBeta=\",round(Beta);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The compressibility  is\tBeta= -2.50967916144e+15\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6 , Page no:16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r_0=3.56; #in Angstrom\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "IE=3.89; #in eV (ionisation energy of Cs)\n",
-      "EA=-3.61; #in eV (electron affinity of Cl)\n",
-      "n=10.5; #Born constant\n",
-      "E_o= 8.85E-12; #absolute premittivity\n",
-      "alpha=1.763; #Madelung constant\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r_0=r_0*1E-10; #since r is in nanometer\n",
-      "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
-      "U=U/e; #changing unit from J to eV\n",
-      "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
-      "\n",
-      "#result\n",
-      "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
-      "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ionic cohesive energy is  -6.0 eV\n",
-        "The atomic cohesive energy is -6.0 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7 , Page no:17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r_0=2.81; #in Angstrom\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "n=9; #Born constant\n",
-      "E_o= 8.85E-12; #absolute premittivity\n",
-      "alpha=1.748; #Madelung constant\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r_0=r_0*1E-10; #since r is in nanometer\n",
-      "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
-      "V=V/e; #changing unit from J to eV\n",
-      "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
-      "\n",
-      "#result\n",
-      "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
-      "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The potential energy is V= -7.96 eV\n",
-        "The energy contributing per ions to the cohesive energy is  -3.98 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_01_Bonding_in_Solids__1.ipynb b/Applied_Physics/Chapter_01_Bonding_in_Solids__1.ipynb
deleted file mode 100755
index db06e802..00000000
--- a/Applied_Physics/Chapter_01_Bonding_in_Solids__1.ipynb
+++ /dev/null
@@ -1,248 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2ef3d4592a57ed325da4c5f5624afbf334fc1ea3982d155f4e16e40570d72ab9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1:Bonding in Solids"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3 , Page no:15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r=2; #in angstrom(distance)\n",
-      "e=1.6E-19; # in C (charge of electron)\n",
-      "E_o= 8.85E-12;# absolute premittivity\n",
-      "\n",
-      "#calculate\n",
-      "r=2*1*10**-10; # since r is in angstrom\n",
-      "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
-      "V1=V/e; # changing to eV\n",
-      "\n",
-      "#result\n",
-      "print \"\\nThe potential energy is  V = \",'%.4E'%V,\"J\";\n",
-      "print \"In electron-Volt V = \",round(V1,3),\"eV\"; \n",
-      "print \"Note: the answer in the book is wrong due to calculation mistake\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The potential energy is  V =  -1.1515E-18 J\n",
-        "In electron-Volt V =  -7.197 eV\n",
-        "Note: the answer in the book is wrong due to calculation mistake\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4 , Page no:15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "#given\n",
-      "r0=0.236; #in nanometer(interionic distance)\n",
-      "e=1.6E-19; # in C (charge of electron)\n",
-      "E_o= 8.85E-12;# absolute premittivity\n",
-      "N=8; # Born constant\n",
-      "IE=5.14;# in eV (ionisation energy of sodium)\n",
-      "EA=3.65;# in eV (electron affinity of Chlorine)\n",
-      "pi=3.14; # value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r0=r0*1E-9; # since r is in nanometer\n",
-      "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
-      "PE=PE/e; #changing unit from J to eV\n",
-      "NE=IE-EA;# calculation of Net energy\n",
-      "BE=PE-NE;# calculation of Bond Energy\n",
-      "\n",
-      "#result\n",
-      "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
-      "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
-      "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
-      "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
-      "# Note: (2) There is slight variation in the answer due to round off."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The potential energy is PE=  5.34 eV\n",
-        "The net energy is NE=  1.49 eV\n",
-        "The bond energy is BE=  3.85 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5 , Page no:16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r_0=.41; #in mm(lattice constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "E_o= 8.85E-12; #absolute premittivity\n",
-      "n=0.5; #repulsive exponent value\n",
-      "alpha=1.76; #Madelung constant\n",
-      "pi=3.14; # value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r=.41*1E-3; #since r is in mm\n",
-      "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
-      "\n",
-      "#result\n",
-      "print\"The compressibility  is\tBeta=\",'%.4E'%Beta;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The compressibility  is\tBeta= -2.5097E+15\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6 , Page no:16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r_0=3.56; #in Angstrom\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "IE=3.89; #in eV (ionisation energy of Cs)\n",
-      "EA=-3.61; #in eV (electron affinity of Cl)\n",
-      "n=10.5; #Born constant\n",
-      "E_o= 8.85E-12; #absolute premittivity\n",
-      "alpha=1.763; #Madelung constant\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r_0=r_0*1E-10; #since r is in nanometer\n",
-      "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
-      "U=U/e; #changing unit from J to eV\n",
-      "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
-      "\n",
-      "#result\n",
-      "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
-      "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ionic cohesive energy is  -6.0 eV\n",
-        "The atomic cohesive energy is -6.0 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7 , Page no:17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "r_0=2.81; #in Angstrom\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "n=9; #Born constant\n",
-      "E_o= 8.85E-12; #absolute premittivity\n",
-      "alpha=1.748; #Madelung constant\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "r_0=r_0*1E-10; #since r is in nanometer\n",
-      "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
-      "V=V/e; #changing unit from J to eV\n",
-      "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
-      "\n",
-      "#result\n",
-      "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
-      "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The potential energy is V= -7.96 eV\n",
-        "The energy contributing per ions to the cohesive energy is  -3.98 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_02_Crystal_Structure.ipynb b/Applied_Physics/Chapter_02_Crystal_Structure.ipynb
deleted file mode 100755
index 2d0b7a14..00000000
--- a/Applied_Physics/Chapter_02_Crystal_Structure.ipynb
+++ /dev/null
@@ -1,291 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:f14517ab3de72640bce8c38f7d46947caa54ae7067bb37477a81849b7b917982"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2:Crystal Structure"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1 , Page no:40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "N=6.02*10**26; #in /Kg-molecule (Avogadro's number)\n",
-      "n=4; #number of molecules per unit cell ofr NaCl\n",
-      "M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)\n",
-      "p=2189; #in Kg/m^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow(((n*M)/(N*p)),0.3333333);\n",
-      "a1=a*10**10; #changing unit to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice constant is a=\",a,\"m\";\n",
-      "print\"\\t\\t\\ta=\",a1,\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice constant is a= 5.62072226997e-10 m\n",
-        "\t\t\ta= 5.62072226997 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2 , Page no:41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02*10**23; #in /gram-atom (Avogadro's number)\n",
-      "n=4; #number of atom per unit cell for fcc structure\n",
-      "M=63.5; #in gram/gram-atom (atomic weight of Cu)\n",
-      "p=8.96; #in g/cm^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow((n*M/(N*p)),0.3333333);\n",
-      "a=a*1E8; #changing unit from cm to Angstrom\n",
-      "d=a/math.sqrt(2); #distance infcc lattice\n",
-      "\n",
-      "#result\n",
-      "print\"lattice constant is a=\",a,\"E-08 cm\";\n",
-      "print\"\\t\\t    a=\",round(a,4),\"Angstrom\";\n",
-      "print\"distance between two nearest Cu atoms is d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "lattice constant is a= 3.61113576149 E-08 cm\n",
-        "\t\t    a= 3.6111 Angstrom\n",
-        "distance between two nearest Cu atoms is d= 2.55 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3 , Page no:41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02E26; #in /Kg-atom (Avogadro's number)\n",
-      "n=2; #number of molecules per unit cell for bcc lattice\n",
-      "M=55.85; #in Kg/Kg-atom (atomic weight of Iron)\n",
-      "p=7860; #in Kg/m^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow((n*M/(N*p)),0.33333);\n",
-      "a1=a*1E10; #changing unit to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice constant is a=\",a,\"m\";\n",
-      "print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice constant is a= 2.86928355016e-10 m\n",
-        "\t\t\ta= 2.869 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4 , Page no:42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02*10**26; #in /Kg-atom (Avogadro's number)\n",
-      "n=2; #number of molecules per unit cell for bcc lattice\n",
-      "M=6.94; #in Kg/Kg-atom (atomic weight of Iron)\n",
-      "p=530; #in Kg/m^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow((n*M/(N*p)),0.33333);\n",
-      "a1=a*1E10; #changing unit to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice constant is a=\",a,\"m\";\n",
-      "print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice constant is a= 3.51776567326e-10 m\n",
-        "\t\t\ta= 3.518 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5 , Page no:42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02*10**23; #in /gram-molecule (Avogadro's number)\n",
-      "M=58.5; #in gram/gram-molecule (atomic weight of NaCl)\n",
-      "p=2.17; #in g/cm^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "#since V=M/p\n",
-      "#(1/d)^-3=2N/V=2Np/M\n",
-      "#therefore d= (M/2Np)^-3\n",
-      "d=pow((M/(2*N*p)),0.33333333);\n",
-      "d1=d*1*10**8; #changing unit from cm to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The distance between two adjacent atoms of NaCl is d=\",d,\"m\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t   d=\",round(d1,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The distance between two adjacent atoms of NaCl is d= 2.81853408124e-08 m\n",
-        "\t\t\t\t\t\t   d= 2.819 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6 , Page no:43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r_Na=0.98; #in Angstrom (radius of sodium ion)\n",
-      "r_Cl=1.81; #in Angstrom (radius of chloride ion)\n",
-      "M_Na=22.99; #in amu (atomic mass of sodium)\n",
-      "M_Cl=35.45; #in amu (atomic mass of chlorine)\n",
-      "\n",
-      "#calculate\n",
-      "a=2*(r_Na+r_Cl); #lattice parameter\n",
-      "#PF=volume of ions present in the unit cell/volume of unit cell\n",
-      "PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;\n",
-      "#Density=mass of unit cell/volume of unit cell\n",
-      "p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;\n",
-      "p1=p*1E-3; #changing unit to gm/cm^-3\n",
-      "\n",
-      "#result\n",
-      "print\"Lattice constant  is a=\",round(a,3),\"Angstrom\";\n",
-      "print\"Packing fraction is =\",round(PF,3); \n",
-      "print\"Density is p=\",round(p),\"Kg/m^3\";\n",
-      "print\"Density is p=\",round(p1,2),\"g/cm^3\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Lattice constant  is a= 5.58 Angstrom\n",
-        "Packing fraction is = 0.662\n",
-        "Density is p= 2233.0 Kg/m^3\n",
-        "Density is p= 2.23 g/cm^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_02_Crystal_Structure_1.ipynb b/Applied_Physics/Chapter_02_Crystal_Structure_1.ipynb
deleted file mode 100755
index b6038072..00000000
--- a/Applied_Physics/Chapter_02_Crystal_Structure_1.ipynb
+++ /dev/null
@@ -1,291 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:693d1c12519596ec538fd8d806511b7a1fdb44b316b98121270f6b11b5d8d004"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2:Crystal Structure"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1 , Page no:40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "N=6.02*10**26; #in /Kg-molecule (Avogadro's number)\n",
-      "n=4; #number of molecules per unit cell ofr NaCl\n",
-      "M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)\n",
-      "p=2189; #in Kg/m^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow(((n*M)/(N*p)),0.3333333);\n",
-      "a1=a*10**10; #changing unit to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
-      "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice constant is a= 5.62E-10 m\n",
-        "\t\t\ta= 5.62 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2 , Page no:41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02*10**23; #in /gram-atom (Avogadro's number)\n",
-      "n=4; #number of atom per unit cell for fcc structure\n",
-      "M=63.5; #in gram/gram-atom (atomic weight of Cu)\n",
-      "p=8.96; #in g/cm^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow((n*M/(N*p)),0.3333333);\n",
-      "a1=a*1E8; #changing unit from cm to Angstrom\n",
-      "d=a1/math.sqrt(2); #distance infcc lattice\n",
-      "\n",
-      "#result\n",
-      "print\"lattice constant is a=\",'%.2E'%a,\"cm\";\n",
-      "print\"\\t\\t    a=\",round(a1,2),\"Angstrom\";\n",
-      "print\"distance between two nearest Cu atoms is d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "lattice constant is a= 3.61E-08 cm\n",
-        "\t\t    a= 3.61 Angstrom\n",
-        "distance between two nearest Cu atoms is d= 2.55 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3 , Page no:41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02E26; #in /Kg-atom (Avogadro's number)\n",
-      "n=2; #number of molecules per unit cell for bcc lattice\n",
-      "M=55.85; #in Kg/Kg-atom (atomic weight of Iron)\n",
-      "p=7860; #in Kg/m^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow((n*M/(N*p)),0.33333);\n",
-      "a1=a*1E10; #changing unit to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
-      "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice constant is a= 2.87E-10 m\n",
-        "\t\t\ta= 2.87 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4 , Page no:42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02*10**26; #in /Kg-atom (Avogadro's number)\n",
-      "n=2; #number of molecules per unit cell for bcc lattice\n",
-      "M=6.94; #in Kg/Kg-atom (atomic weight of Iron)\n",
-      "p=530; #in Kg/m^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "a=pow((n*M/(N*p)),0.33333);\n",
-      "a1=a*1E10; #changing unit to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
-      "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice constant is a= 3.52E-10 m\n",
-        "\t\t\ta= 3.52 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5 , Page no:42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=6.02*10**23; #in /gram-molecule (Avogadro's number)\n",
-      "M=58.5; #in gram/gram-molecule (atomic weight of NaCl)\n",
-      "p=2.17; #in g/cm^3 (density)\n",
-      "\n",
-      "#calculate\n",
-      "#since V=M/p\n",
-      "#(1/d)^-3=2N/V=2Np/M\n",
-      "#therefore d= (M/2Np)^-3\n",
-      "d=pow((M/(2*N*p)),0.33333333);\n",
-      "d1=d*1*10**8; #changing unit from cm to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The distance between two adjacent atoms of NaCl is d=\",'%.2E'%d,\"m\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t   d=\",round(d1,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The distance between two adjacent atoms of NaCl is d= 2.82E-08 m\n",
-        "\t\t\t\t\t\t   d= 2.82 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6 , Page no:43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r_Na=0.98; #in Angstrom (radius of sodium ion)\n",
-      "r_Cl=1.81; #in Angstrom (radius of chloride ion)\n",
-      "M_Na=22.99; #in amu (atomic mass of sodium)\n",
-      "M_Cl=35.45; #in amu (atomic mass of chlorine)\n",
-      "\n",
-      "#calculate\n",
-      "a=2*(r_Na+r_Cl); #lattice parameter\n",
-      "#PF=volume of ions present in the unit cell/volume of unit cell\n",
-      "PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;\n",
-      "#Density=mass of unit cell/volume of unit cell\n",
-      "p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;\n",
-      "p1=p*1E-3; #changing unit to gm/cm^-3\n",
-      "\n",
-      "#result\n",
-      "print\"Lattice constant  is a=\",round(a,3),\"Angstrom\";\n",
-      "print\"Packing fraction is =\",round(PF,3); \n",
-      "print\"Density is p=\",round(p),\"Kg/m^3\";\n",
-      "print\"Density is p=\",round(p1,2),\"g/cm^3\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Lattice constant  is a= 5.58 Angstrom\n",
-        "Packing fraction is = 0.662\n",
-        "Density is p= 2233.0 Kg/m^3\n",
-        "Density is p= 2.23 g/cm^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_03_Planes_in_Crystals.ipynb b/Applied_Physics/Chapter_03_Planes_in_Crystals.ipynb
deleted file mode 100755
index 763a53c2..00000000
--- a/Applied_Physics/Chapter_03_Planes_in_Crystals.ipynb
+++ /dev/null
@@ -1,144 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1ec2372be7a9974010a31557c9658c85707025e7a87f82e6ef26be63dbcfb0e7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3:Planes in Crystals"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11 , Page no:61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=3;\n",
-      "k=2;\n",
-      "l=1; #miller indices\n",
-      "a=4.2E-8; #in cm (lattice constant)\n",
-      "\n",
-      "#calculate\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
-      "d=d*1E8; #changing unit from cm to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The interplanar spacing is d=\",round(d,2),\"cm\";\n",
-      "print\"\\t\\t\\t   d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The interplanar spacing is d= 1.12 cm\n",
-        "\t\t\t   d= 1.12 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.12 , Page no:61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=1;l=1; #miller indices\n",
-      "a=2.5;b=2.5;c=1.8; #in Angstrom (lattice constants for tetragonal lattice )\n",
-      "\n",
-      "#calculate\n",
-      "d=1/math.sqrt((h/a)**2+(k/b)**2+(l/c)**2); #calculation for interplanar spacing\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice spacing is d= 1.26 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.15 , Page no:63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=0;l=0; #miller indices\n",
-      "a=2.5; #in Angstrom (lattice constant)\n",
-      "\n",
-      "#calculate\n",
-      "a=a*1E-10; #hence a is in Angstrom\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
-      "p=d/a**3;\n",
-      "\n",
-      "#result\n",
-      "print\"The density is p=\",round(p),\"lattice points/m^2\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density is p= 1.6e+19 lattice points/m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_03_Planes_in_Crystals_1.ipynb b/Applied_Physics/Chapter_03_Planes_in_Crystals_1.ipynb
deleted file mode 100755
index 92dba19b..00000000
--- a/Applied_Physics/Chapter_03_Planes_in_Crystals_1.ipynb
+++ /dev/null
@@ -1,144 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:05935aae8ee517fae6abe7a89dceb1ca0d0cc367834bce08b5f427ec49852dcb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3:Planes in Crystals"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11 , Page no:61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=3;\n",
-      "k=2;\n",
-      "l=1; #miller indices\n",
-      "a=4.2E-8; #in cm (lattice constant)\n",
-      "\n",
-      "#calculate\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
-      "d=d*1E8; #changing unit from cm to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The interplanar spacing is d=\",round(d,2),\"cm\";\n",
-      "print\"\\t\\t\\t   d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The interplanar spacing is d= 1.12 cm\n",
-        "\t\t\t   d= 1.12 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.12 , Page no:61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=1;l=1; #miller indices\n",
-      "a=2.5;b=2.5;c=1.8; #in Angstrom (lattice constants for tetragonal lattice )\n",
-      "\n",
-      "#calculate\n",
-      "d=1/math.sqrt((h/a)**2+(k/b)**2+(l/c)**2); #calculation for interplanar spacing\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice spacing is d= 1.26 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.15 , Page no:63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=0;l=0; #miller indices\n",
-      "a=2.5; #in Angstrom (lattice constant)\n",
-      "\n",
-      "#calculate\n",
-      "a=a*1E-10; #hence a is in Angstrom\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
-      "p=d/a**3;\n",
-      "\n",
-      "#result\n",
-      "print\"The density is p=\",round(p),\"lattice points/m^2\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density is p= 1.6e+19 lattice points/m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_04_Crystal_Diffraction.ipynb b/Applied_Physics/Chapter_04_Crystal_Diffraction.ipynb
deleted file mode 100755
index 6c9c83eb..00000000
--- a/Applied_Physics/Chapter_04_Crystal_Diffraction.ipynb
+++ /dev/null
@@ -1,494 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:177e44357417b2cfb0b476ada5b36a934225ade51d50763b09586ba8679f08fc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4:Crystal Diffraction"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1 , Page no:75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=2.6; #in Angstrom (wavelength)\n",
-      "theta=20; #in Degree (angle)\n",
-      "n=2;\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=lambda1*1E-10; #since lambda is in Angstrom\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore d=n(lambda)/2sin(theta)\n",
-      "d=n*lambda1/(2*math.sin(math.radians(theta)));\n",
-      "\n",
-      "#result\n",
-      "print\"The spacing constant is d=\",d,\"m\";\n",
-      "print\"\\t\\t\\td=\",d*10**10,\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing constant is d= 7.60189144042e-10 m\n",
-        "\t\t\td= 7.60189144042 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2 , Page no:75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=1;l=0; #miller indices\n",
-      "a=0.26; #in nanometer (lattice constant)\n",
-      "lambda1=0.065; #in nanometer (wavelength)\n",
-      "n=2; #order\n",
-      "\n",
-      "#calculate\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation of interlattice spacing\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "theta=math.asin(n*lambda1/(2*d));\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The glancing angle is =\",round(theta1,2),\"degree\";\n",
-      "print \"Note: there is slight variation in the answer due to round off error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The glancing angle is = 20.72 degree\n",
-        "Note: there is slight variation in the answer due to round off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3 , Page no:75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "d=3.04E-10; #in mm (spacing constant)\n",
-      "lambda1=0.79; #in Angstrom (wavelength)\n",
-      "n=3; #order\n",
-      "\n",
-      "#calculate\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda2=lambda1*1E-10; #since lambda is in angstrom\n",
-      "theta=math.asin(n*lambda2/(2*d));\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The glancing angle is\",round(theta1,3),\"degree\";\n",
-      "print \"Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The glancing angle is 22.954 degree\n",
-        "Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4 , Page no:76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "d=0.282; #in nanometer (spacing constant)\n",
-      "n=1; #order\n",
-      "theta=8.35; #in degree (glancing angle)\n",
-      "\n",
-      "#calculate\n",
-      "d=d*1E-9; #since d is in nanometer\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda1=2*d*math.sin(math.radians(theta))/n; \n",
-      "lambda_1=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "theta_1=90; #in degree (for maximum order theta=90)\n",
-      "n_max=2*d*math.sin(math.radians(theta_1))/lambda1; #calculation of maximum order. \n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength  =\",lambda1,\"m\";\n",
-      "print\"\\t\\t=\" ,round(lambda_1,3),\"Angstrom\";\n",
-      "print\"The maximum order possible is n=\",round(n_max);\n",
-      "print \"Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength  = 8.19038939236e-11 m\n",
-        "\t\t= 0.819 Angstrom\n",
-        "The maximum order possible is n= 7.0\n",
-        "Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5 , Page no:76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "theta=6; #in degree (glancing angle)\n",
-      "p=2170; #in Kg/m^3 (density)\n",
-      "M=58.46; #Molecular weight of NaCl\n",
-      "N=6.02E26; #in Kg-molecule (Avogadro's number)\n",
-      "n=1; #order\n",
-      "XU=1E-12; #since 1X.U.= 1E-12m\n",
-      "\n",
-      "#calculate\n",
-      "d=(M/(2*N*p))**(1/3); #calclation of lattice constant\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation of wavelength\n",
-      "lambda2=lambda1/XU;\n",
-      "\n",
-      "#result\n",
-      "print\"The spacing constant is d=\",d,\"m\";\n",
-      "print\"The wavelength is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",lambda2,\"X.U\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing constant is d= 2.81789104396e-10 m\n",
-        "The wavelength is = 5.89099640962e-11 m\n",
-        "\t\t  = 58.9099640962 X.U\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6 , Page no:77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=1;l=1; #miller indices\n",
-      "a=5.63; #in Angstrom  (lattice constant)\n",
-      "theta=27.5; #in degree (Glancing angle)\n",
-      "n=1; #order\n",
-      "H=6.625E-34; #in J-s (Plank's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "e=1.6E-19; #charge of electron\n",
-      "\n",
-      "#calculate\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation for wavelength\n",
-      "E=H*c/(lambda1*1E-10); #calculation of Energy\n",
-      "E1=E/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";\n",
-      "print\"The wavelength is =\",round(lambda1),\"Angstrom\";\n",
-      "print\"The energy of X-rays is E=\",E,\"J\";\n",
-      "print\"\\t\\t\\tE=\",E1,\"eV\";\n",
-      "print \"Note: c=3E8 m/s but in solution c=3E10 m/s \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice spacing is d= 3.25 Angstrom\n",
-        "The wavelength is = 3.0 Angstrom\n",
-        "The energy of X-rays is E= 6.62100284311e-16 J\n",
-        "\t\t\tE= 4138.12677695 eV\n",
-        "Note: c=3E8 m/s but in solution c=3E10 m/s \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7 , Page no:77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=344; #in V (accelerating voltage)\n",
-      "theta=60; #in degree (glancing angle)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.625e-34; #in J-s (Plank's constant)\n",
-      "n=1; #order\n",
-      "e=1.6E-19; #charge on electron\n",
-      "\n",
-      "#calculate\n",
-      "#Since K=m*v^2/2=e*V\n",
-      "#therefore  v=sqrt(2*e*V/m)\n",
-      "#since lambda=h/(m*v)\n",
-      "#therefore we have lambda=h/sqrt(2*m*e*V)\n",
-      "lambda1=h/math.sqrt(2*m*e*V); #calculation of lambda\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "d=n*lambda2/(2*math.sin(math.radians(theta)));\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",round(lambda2,2),\"Angstrom\";\n",
-      "print\"The interplanar spacing  is  d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 6.61928340764e-11 m\n",
-        "\t\t  = 0.66 Angstrom\n",
-        "The interplanar spacing  is  d= 0.38 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.8 , Page no:78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=0.02; #in eV (kinetic energy)\n",
-      "d=2.0; #in Angstrom (Bragg's spacing)\n",
-      "m=1.00898; #in amu (mass of neutron)\n",
-      "amu=1.66E-27; #in Kg (1amu=1.66E-27 Kg)\n",
-      "h=6.625e-34; #in J-s (Plank's constant)\n",
-      "n=1; #order\n",
-      "e=1.6E-19; #charge on electron\n",
-      "\n",
-      "#calculate\n",
-      "#Since K=m*v^2/2\n",
-      "#therefore  v=sqrt(2*K/m)\n",
-      "#since lambda=h/(m*v)\n",
-      "#therefore we have lambda=h/sqrt(2*m*K)\n",
-      "m=m*amu; #changing unit from amu to Kg\n",
-      "K=K*e; #changing unit to J from eV\n",
-      "lambda1=h/math.sqrt(2*m*K); #calculation of lambda\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "theta=math.asin(n*lambda2/(2*d)); #calculation of angle of first order diffraction maximum\n",
-      "theta1=theta*180/3.14;\n",
-      "#result\n",
-      "print\"The wavelength is =\",lambda1,\"m\";\n",
-      "print\"\\t    \\t  =\",round(lambda2),\"Angstrom\";\n",
-      "print\"The angle of first order diffraction maximum is \",round(theta1),\"Degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 2.02348771817e-10 m\n",
-        "\t    \t  = 2.0 Angstrom\n",
-        "The angle of first order diffraction maximum is  30.0 Degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9 , Page no:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "lambda1=0.586; #in Angstrom (wavelength of X-rays)\n",
-      "n1=1; n2=2; n3=3; #orders of diffraction\n",
-      "theta1=5+(58/60); #in degree (Glancing angle for first order of diffraction)\n",
-      "theta2=12+(01/60); #in degree (Glancing angle for second order of diffraction)\n",
-      "theta3=18+(12/60); #in degree (Glancing angle for third order of diffraction)\n",
-      "\n",
-      "#calculate\n",
-      "K1=math.sin(math.radians(theta1));\n",
-      "K2=math.sin(math.radians(theta2));\n",
-      "K3=math.sin(math.radians(theta3));\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "d1=n1*lambda1/(2*K1);\n",
-      "d2=n2*lambda1/(2*K2);\n",
-      "d3=n3*lambda1/(2*K3);\n",
-      "d1=d1*1E-10; #changing unit from Angstrom to m\n",
-      "d2=d2*1E-10; #changing unit from Angstrom to m\n",
-      "d3=d3*1E-10; #changing unit from Angstrom to m\n",
-      "d=(d1+d2+d3)/3;\n",
-      "\n",
-      "#result\n",
-      "print\"The value of sine of different angle of diffraction is \\nK1=\",K1;\n",
-      "print\"K2=\",K2;\n",
-      "print\"K3=\",K3;\n",
-      "#Taking the ratios of K1:K2:K3\n",
-      "#We get K1:K2:K3=1:2:3\n",
-      "#Therefore we have\n",
-      "print\"Or we have K1:K2:K3=1:2:3\";\n",
-      "print\"Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\";\n",
-      "print\"The spacing constants are \\nd1=\",d1;\n",
-      "print\"d2=\",d2;\n",
-      "print\"d3=\",d3;\n",
-      "print\"The mean value of crystal spacing is d=\",d,\"m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of sine of different angle of diffraction is \n",
-        "K1= 0.103949856225\n",
-        "K2= 0.208196213621\n",
-        "K3= 0.312334918512\n",
-        "Or we have K1:K2:K3=1:2:3\n",
-        "Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\n",
-        "The spacing constants are \n",
-        "d1= 2.81866671719e-10\n",
-        "d2= 2.81465253286e-10\n",
-        "d3= 2.81428667722e-10\n",
-        "The mean value of crystal spacing is d= 2.81586864243e-10 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_04_Crystal_Diffraction_1.ipynb b/Applied_Physics/Chapter_04_Crystal_Diffraction_1.ipynb
deleted file mode 100755
index fef74a69..00000000
--- a/Applied_Physics/Chapter_04_Crystal_Diffraction_1.ipynb
+++ /dev/null
@@ -1,503 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:513926db1524f2764781365e8a81c6fa3119b7bc0f032a42233e49f5c3b6140a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4:Crystal Diffraction"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1 , Page no:75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=2.6; #in Angstrom (wavelength)\n",
-      "theta=20; #in Degree (angle)\n",
-      "n=2;\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=lambda1*1E-10; #since lambda is in Angstrom\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore d=n(lambda)/2sin(theta)\n",
-      "d=n*lambda1/(2*math.sin(math.radians(theta)));\n",
-      "\n",
-      "#result\n",
-      "print\"The spacing constant is d=\",'%.2E'%d,\"m\";\n",
-      "print\"\\t\\t\\td=\",round(d*10**10,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing constant is d= 7.60E-10 m\n",
-        "\t\t\td= 7.602 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2 , Page no:75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=1;l=0; #miller indices\n",
-      "a=0.26; #in nanometer (lattice constant)\n",
-      "lambda1=0.065; #in nanometer (wavelength)\n",
-      "n=2; #order\n",
-      "\n",
-      "#calculate\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation of interlattice spacing\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "theta=math.asin(n*lambda1/(2*d));\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The glancing angle is =\",round(theta1,2),\"degree\";\n",
-      "print \"Note: there is slight variation in the answer due to round off error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The glancing angle is = 20.72 degree\n",
-        "Note: there is slight variation in the answer due to round off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3 , Page no:75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "d=3.04E-10; #in mm (spacing constant)\n",
-      "lambda1=0.79; #in Angstrom (wavelength)\n",
-      "n=3; #order\n",
-      "\n",
-      "#calculate\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda2=lambda1*1E-10; #since lambda is in angstrom\n",
-      "theta=math.asin(n*lambda2/(2*d));\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The glancing angle is\",round(theta1,3),\"degree\";\n",
-      "print \"Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The glancing angle is 22.954 degree\n",
-        "Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4 , Page no:76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "d=0.282; #in nanometer (spacing constant)\n",
-      "n=1; #order\n",
-      "theta=8.35; #in degree (glancing angle)\n",
-      "\n",
-      "#calculate\n",
-      "d=d*1E-9; #since d is in nanometer\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda1=2*d*math.sin(math.radians(theta))/n; \n",
-      "lambda_1=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "theta_1=90; #in degree (for maximum order theta=90)\n",
-      "n_max=2*d*math.sin(math.radians(theta_1))/lambda1; #calculation of maximum order. \n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength  =\",'%.2E'%lambda1,\"m\";\n",
-      "print\"\\t\\t=\" ,round(lambda_1,3),\"Angstrom\";\n",
-      "print\"The maximum order possible is n=\",round(n_max);\n",
-      "print \"Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength  = 8.19E-11 m\n",
-        "\t\t= 0.819 Angstrom\n",
-        "The maximum order possible is n= 7.0\n",
-        "Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5 , Page no:76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "theta=6; #in degree (glancing angle)\n",
-      "p=2170; #in Kg/m^3 (density)\n",
-      "M=58.46; #Molecular weight of NaCl\n",
-      "N=6.02E26; #in Kg-molecule (Avogadro's number)\n",
-      "n=1; #order\n",
-      "XU=1E-12; #since 1X.U.= 1E-12m\n",
-      "\n",
-      "#calculate\n",
-      "d=(M/(2*N*p))**(1/3); #calclation of lattice constant\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation of wavelength\n",
-      "lambda2=lambda1/XU;\n",
-      "\n",
-      "#result\n",
-      "print\"The spacing constant is d=\",'%.2E'%d,\"m\";\n",
-      "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",round(lambda2,1),\"X.U\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing constant is d= 2.82E-10 m\n",
-        "The wavelength is = 5.89E-11 m\n",
-        "\t\t  = 58.9 X.U\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6 , Page no:77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "h=1;k=1;l=1; #miller indices\n",
-      "a=5.63; #in Angstrom  (lattice constant)\n",
-      "theta=27.5; #in degree (Glancing angle)\n",
-      "n=1; #order\n",
-      "H=6.625E-34; #in J-s (Plank's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "e=1.6E-19; #charge of electron\n",
-      "\n",
-      "#calculate\n",
-      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation for wavelength\n",
-      "E=H*c/(lambda1*1E-10); #calculation of Energy\n",
-      "E1=E/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";\n",
-      "print\"The wavelength is =\",round(lambda1),\"Angstrom\";\n",
-      "print\"The energy of X-rays is E=\",'%.2E'%E,\"J\";\n",
-      "print\"\\t\\t\\tE=\",'%.3E'%E1,\"eV\";\n",
-      "print \"Note: c=3E8 m/s but in solution c=3E10 m/s \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lattice spacing is d= 3.25 Angstrom\n",
-        "The wavelength is = 3.0 Angstrom\n",
-        "The energy of X-rays is E= 6.62E-16 J\n",
-        "\t\t\tE= 4.138E+03 eV\n",
-        "Note: c=3E8 m/s but in solution c=3E10 m/s \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7 , Page no:77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=344; #in V (accelerating voltage)\n",
-      "theta=60; #in degree (glancing angle)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.625e-34; #in J-s (Plank's constant)\n",
-      "n=1; #order\n",
-      "e=1.6E-19; #charge on electron\n",
-      "\n",
-      "#calculate\n",
-      "#Since K=m*v^2/2=e*V\n",
-      "#therefore  v=sqrt(2*e*V/m)\n",
-      "#since lambda=h/(m*v)\n",
-      "#therefore we have lambda=h/sqrt(2*m*e*V)\n",
-      "lambda1=h/math.sqrt(2*m*e*V); #calculation of lambda\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "d=n*lambda2/(2*math.sin(math.radians(theta)));\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",round(lambda2,2),\"Angstrom\";\n",
-      "print\"The interplanar spacing  is  d=\",round(d,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 6.62E-11 m\n",
-        "\t\t  = 0.66 Angstrom\n",
-        "The interplanar spacing  is  d= 0.38 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.8 , Page no:78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=0.02; #in eV (kinetic energy)\n",
-      "d=2.0; #in Angstrom (Bragg's spacing)\n",
-      "m=1.00898; #in amu (mass of neutron)\n",
-      "amu=1.66E-27; #in Kg (1amu=1.66E-27 Kg)\n",
-      "h=6.625e-34; #in J-s (Plank's constant)\n",
-      "n=1; #order\n",
-      "e=1.6E-19; #charge on electron\n",
-      "\n",
-      "#calculate\n",
-      "#Since K=m*v^2/2\n",
-      "#therefore  v=sqrt(2*K/m)\n",
-      "#since lambda=h/(m*v)\n",
-      "#therefore we have lambda=h/sqrt(2*m*K)\n",
-      "m=m*amu; #changing unit from amu to Kg\n",
-      "K=K*e; #changing unit to J from eV\n",
-      "lambda1=h/math.sqrt(2*m*K); #calculation of lambda\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "theta=math.asin(n*lambda2/(2*d)); #calculation of angle of first order diffraction maximum\n",
-      "theta1=theta*180/3.14;\n",
-      "#result\n",
-      "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
-      "print\"\\t    \\t  =\",round(lambda2),\"Angstrom\";\n",
-      "print\"The angle of first order diffraction maximum is \",round(theta1),\"Degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 2.02E-10 m\n",
-        "\t    \t  = 2.0 Angstrom\n",
-        "The angle of first order diffraction maximum is  30.0 Degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9 , Page no:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "lambda1=0.586; #in Angstrom (wavelength of X-rays)\n",
-      "n1=1; n2=2; n3=3; #orders of diffraction\n",
-      "theta1=5+(58/60); #in degree (Glancing angle for first order of diffraction)\n",
-      "theta2=12+(01/60); #in degree (Glancing angle for second order of diffraction)\n",
-      "theta3=18+(12/60); #in degree (Glancing angle for third order of diffraction)\n",
-      "\n",
-      "#calculate\n",
-      "K1=math.sin(math.radians(theta1));\n",
-      "K2=math.sin(math.radians(theta2));\n",
-      "K3=math.sin(math.radians(theta3));\n",
-      "#Since 2dsin(theta)=n(lambda)\n",
-      "#therefore we have\n",
-      "d1=n1*lambda1/(2*K1);\n",
-      "d2=n2*lambda1/(2*K2);\n",
-      "d3=n3*lambda1/(2*K3);\n",
-      "d1=d1*1E-10; #changing unit from Angstrom to m\n",
-      "d2=d2*1E-10; #changing unit from Angstrom to m\n",
-      "d3=d3*1E-10; #changing unit from Angstrom to m\n",
-      "d=(d1+d2+d3)/3;\n",
-      "\n",
-      "#result\n",
-      "print\"The value of sine of different angle of diffraction is \\nK1=\",round(K1,4);\n",
-      "print\"K2=\",round(K2,4);\n",
-      "print\"K3=\",round(K3,4);\n",
-      "#Taking the ratios of K1:K2:K3\n",
-      "#We get K1:K2:K3=1:2:3\n",
-      "#Therefore we have\n",
-      "print\"Or we have K1:K2:K3=1:2:3\";\n",
-      "print\"Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\";\n",
-      "print\"The spacing constants are \\nd1=\",'%.2E'%d1;\n",
-      "print\"d2=\",'%.3E'%d2;\n",
-      "print\"d3=\",'%.3E'%d3;\n",
-      "print\"The mean value of crystal spacing is d=\",'%.3E'%d,\"m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of sine of different angle of diffraction is \n",
-        "K1= 0.1039\n",
-        "K2= 0.2082\n",
-        "K3= 0.3123\n",
-        "Or we have K1:K2:K3=1:2:3\n",
-        "Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\n",
-        "The spacing constants are \n",
-        "d1= 2.82E-10\n",
-        "d2= 2.815E-10\n",
-        "d3= 2.814E-10\n",
-        "The mean value of crystal spacing is d= 2.816E-10 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics.ipynb b/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics.ipynb
deleted file mode 100755
index cf16b683..00000000
--- a/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics.ipynb
+++ /dev/null
@@ -1,1258 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:17ded29a54411f524d1901cc530f956cb4706c517e993f5acba4014789eca0bd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5:Principles of Quantum Mechanics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1 , Page no:102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=1; #in Angstrom (wavelength)\n",
-      "m=1.67E-27; #in Kg (mass of neutron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "lambda2=lambda1*1E-10; #since lambda is in Angstrom\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "v=h/(m*lambda2); #calculation of velocity\n",
-      "K=m*v**2/2; #calculation of kinetic energy\n",
-      "K1=K/e; #changing unit fro J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity is v=\",v,\"m/s\";\n",
-      "print\"The kinetic energy is K=\",K,\"J\";\n",
-      "print\"\\t\\t       =\",K1,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity is v= 3967.06586826 m/s\n",
-        "The kinetic energy is K= 1.31409056886e-20 J\n",
-        "\t\t       = 0.0821306605539 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2 , Page no:103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=50; #in eV (Kinetic energy)\n",
-      "m0=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "K=K*e; #changing unit from eV to J\n",
-      "#Since K=m*v^2/2\n",
-      "#Therefore v=sqrt(2*K/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",round(lambda2,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 1.73622176082e-10 m\n",
-        "\t\t  = 1.736 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3 , Page no:104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E=2000; #in eV (Kinetic energy)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "#Since E=m*v^2/2\n",
-      "#Therefore v=sqrt(2*E/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "lambda1=h/math.sqrt(2*m*E); #calculation of wavelength\n",
-      "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",round(lambda2,4),\"nm\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 2.74520764367e-11 m\n",
-        "\t\t  = 0.0275 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4 , Page no:104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_n=1.676E-27; #in Kg (mass of neutron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "\n",
-      "#calculate\n",
-      "E_e=m_e*c**2; #rest mass energy of electron\n",
-      "E_n=2*E_e; #given (kinetic energy of neutron)\n",
-      "#Since K=m*v^2/2\n",
-      "#Therefore v=sqrt(2*K/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",lambda2,\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 2.82733278331e-14 m\n",
-        "\t\t  = 0.000282733278331 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5 , Page no:104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=1600; #in V (Potential)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
-      "print \"Note : The answer in the book is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 0.307 Angstrom\n",
-        "Note : The answer in the book is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6 , Page no:105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E=1000; #in eV (Kinetic energy of photon)\n",
-      "K=1000; #in eV (Kinetic energy of electron)\n",
-      "m0=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.6E-34; #in J-s (Planck's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "e=1.6E-19; #in C (charge on electron)\n",
-      "\n",
-      "#calculate\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "lambda_p=h*c/E; #For photon E=hc/lambda\n",
-      "lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom\n",
-      "#Since K=m*v^2/2\n",
-      "#Therefore v=sqrt(2*K/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "K=K*e; #changing unit from eV to J\n",
-      "lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
-      "lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"For photon,the wavelength is =\",lambda_p,\"m\";\n",
-      "print\"\\t\\t\\t     =\",round(lambda_p1,1),\"Angstrom\";\n",
-      "print\"For electron,the wavelength is =\",lambda_e,\"m\";\n",
-      "print\"\\t\\t\\t       =\",round(lambda_e1,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For photon,the wavelength is = 1.2375e-09 m\n",
-        "\t\t\t     = 12.4 Angstrom\n",
-        "For electron,the wavelength is = 3.86765965524e-11 m\n",
-        "\t\t\t       = 0.39 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7 , Page no:105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=1.66*10**-10; #in m (wavelength)\n",
-      "m=9.1*10**-31; #in Kg (mass of electron)\n",
-      "h=6.626*10**-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge on electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "v=h/(m*lambda1); #calculation of velocity\n",
-      "K=m*v**2/2; #calculation of kinetic energy\n",
-      "K1=K/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity of electron is v=\",v,\"m/s\";\n",
-      "print\"The kinetic energy is K=\",K,\"J\";\n",
-      "print\"\\t\\t       =\",round(K1,3),\"eV\";\n",
-      "print \"Note: The answer in the book for kinetic energy is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of electron is v= 4386336.55501 m/s\n",
-        "The kinetic energy is K= 8.75417651009e-18 J\n",
-        "\t\t       = 54.714 eV\n",
-        "Note: The answer in the book for kinetic energy is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8 , Page no:106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=400; #in K (temperature)\n",
-      "m=6.7E-27; #in Kg (mass of He-atom)\n",
-      "h=6.6E-34; #in J-s (Planck's constant)\n",
-      "k=1.376E-23; #in J/degree (Boltzmann constant)\n",
-      "\n",
-      "#calculate\n",
-      "#Since lambda=h/(m*v)\n",
-      "#E=mv^2/2;\n",
-      "#Therefore lambda=h/sqrt(2*m*E)\n",
-      "#E=kT\n",
-      "#Therefore lambda=h/sqrt(2*m*k*T)\n",
-      "lambda1=h/math.sqrt(2*m*k*T)\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The de-Broglie wavelength of He-atom is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t\\t\\t        =\",round(lambda2,4),\"Angstrom\";    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The de-Broglie wavelength of He-atom is = 7.6851495611e-11 m\n",
-        "\t\t\t\t        = 0.7685 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9 , Page no:106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_p=1.6E-27; #in Kg (mass of proton)\n",
-      "h=6.626E-34; #in J-s (Planck's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "\n",
-      "#calculate\n",
-      "E=m_e*c**2; #in J (rest energy of electron)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#E=mv^2/2;\n",
-      "#herefore lambda=h/sqrt(2*m*E)\n",
-      "#Also E=m_e*c^2;\n",
-      "#therefore lambda=h/sqrt(2*m_p*m_e*c^2)\n",
-      "lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength\n",
-      "lambda2=lambda1*1*10**10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The de-Broglie wavelength of proton is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t\\t\\t       =\",lambda2,\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The de-Broglie wavelength of proton is = 4.0929316435e-14 m\n",
-        "\t\t\t\t       = 0.00040929316435 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10 , Page no:106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=10000; #in V (Potential)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 0.123 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.11 , Page no:107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=100; #in V (Potential)\n",
-      "n=1; #order of diffraction\n",
-      "d=2.15; #in Angstrom (lattice spacing)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "#Since 2*d*sind(theta)=n*lambda\n",
-      "#therefore we have\n",
-      "theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
-      "print\"The glancing angle is =\",round(theta1,1),\"degree\";\n",
-      "print \"Note: In question V=100 eV but the solution is using V=100V \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 1.227 Angstrom\n",
-        "The glancing angle is = 16.6 degree\n",
-        "Note: In question V=100 eV but the solution is using V=100V \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12 , Page no:107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=344; #in V (Potential)\n",
-      "n=1; #order of diffraction\n",
-      "theta=60; #in degree (glancing angle)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "#Since 2*d*sind(theta)=n*lambda\n",
-      "#therefore we have\n",
-      "d=n*lambda1/(2*math.sin(math.radians(theta))); #calculation of spacing constant\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
-      "print\"The spacing of the crystal is =\",round(d,4),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 0.662 Angstrom\n",
-        "The spacing of the crystal is = 0.3819 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.13 , Page no:107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r=0.53*10**-10; #in m (radius of first Bohr orbit)\n",
-      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
-      "m=9.1*10**-31; #in Kg (mass of electron)\n",
-      "n=1; #First Bohr orbit\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "#Since 2*pi*r=n*lambda  and  lambda=h/(m*v)\n",
-      "#Threfore we have v=h*n/(2*pi*r*m)\n",
-      "v=h*n/(2*pi*r*m); #calculation of velocity\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity of electron is =\",v,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of electron is = 2179049.16859 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14 , Page no:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dx=0.2; #in Angstrom (uncertainty in the position)\n",
-      "h=6.6E-34; #in J-s (Planck's constant)\n",
-      "m0=9.1E-31; #in Kg (mass of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "dx=dx*1E-10; #since dx is in Angstrom\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
-      "#since dp=m*dv\n",
-      "dv=dp/m0; #calculation of uncertainty in the velocity\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the momentum is dp=\",dp,\"Kg-m/\";\n",
-      "print\"The uncertainty in the velocity is dv=\",dv,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the momentum is dp= 2.62738853503e-24 Kg-m/\n",
-        "The uncertainty in the velocity is dv= 2887240.14839 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15 , Page no:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_p=1.67E-27; #in Kg (mass of proton)\n",
-      "dx_p=1; #in nanometer (uncertainty in position of electron)\n",
-      "dx_n=1; #in nanometer (uncertainty in position of proton)\n",
-      "\n",
-      "#calculate\n",
-      "#since dp=h/(4*pi*dx)\n",
-      "#since h/(4*pi) is constant and dx is same for electron and proton \n",
-      "#therefor both electron and proton have same uncertainty in the momentum\n",
-      "#since dv=dp/m  and  dp is same for both\n",
-      "#therefore dv_e/dv_p=m_p/m_e\n",
-      "#therefore\n",
-      "K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton\n",
-      "\n",
-      "#result\n",
-      "print\"The ratio of uncertainty in the velocity of electron to that of proton is =\",round(K);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.16 , Page no:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)\n",
-      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
-      "m=1.67E-27; #in Kg (mass of proton)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
-      "p=dp; #minimum value of momentum to calculate mimimum kinetic energy\n",
-      "K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton\n",
-      "K1=K/e; #changing unit from J to eV\n",
-      "K2=K1/1E6; #changing unit from eV to MeV\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum uncertainty in the momentum of proton is dp =\",dp,\"Kg-m/s\";\n",
-      "print\"The minimum kinetic energy of proton is K=\",K,\"J\";\n",
-      "print\"\\t\\t\\t\\t \\t =\",K1,\"eV\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",K2,\"MeV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum uncertainty in the momentum of proton is dp = 1.05095541401e-20 Kg-m/s\n",
-        "The minimum kinetic energy of proton is K= 3.30690803067e-14 J\n",
-        "\t\t\t\t \t = 206681.751917 eV\n",
-        "\t\t\t\t\t = 0.206681751917 MeV\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.17 , Page no:109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=1; #in KeV (kinetic energy of electron)\n",
-      "dx=1; #in Angstrom (uncertainty in the position)\n",
-      "h=6.63E-34; #in J-s (Planck's constant)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "dx=dx*1E-10; #since dx is in Angstrom\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
-      "K=K*1E3*1.6E-19; #changing unit from KeV to J\n",
-      "p=math.sqrt(2*m*K); #calculation of momentum \n",
-      "poc=(dp/p)*100; #calculation of percentage of uncertainty\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
-      "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
-      "print\"The percentage of uncertainty in the momentum  is =\",round(poc,1);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the momentum of electron is dp= 5.27866242038e-25 Kg-m/s\n",
-        "The momentum of electron is p= 1.70645832062e-23 Kg-m/s\n",
-        "The percentage of uncertainty in the momentum  is = 3.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.18 , Page no:109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "v=6.6E4; #m/s (speed of electron)\n",
-      "poc=0.01; #percentage of uncertainty\n",
-      "h=6.63E-34; #in J-s (Planck's constant)\n",
-      "m=9E-31; #in Kg (mass of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "p=m*v; #calculation of momentum\n",
-      "dp=(poc/100)*p; #calculation of uncertainty in the momentum\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dx=h/(4*pi*dp); #calculation of uncertainty in the position\n",
-      "\n",
-      "#result\n",
-      "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
-      "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
-      "print\"The uncertainty in the position of electron is dx=\",dx,\"Kg-m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The momentum of electron is p= 5.94e-26 Kg-m/s\n",
-        "The uncertainty in the momentum of electron is dp= 5.94e-30 Kg-m/s\n",
-        "The uncertainty in the position of electron is dx= 8.88663707135e-06 Kg-m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.19 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=1; #in Angstrom (wavelength)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "dlambda=1E-6; #uncertainty in wavelength\n",
-      "\n",
-      "#calculate\n",
-      "lambda2=lambda1*1E-10; #sinc lambda is in Angstrom\n",
-      "#By uncertainty principle, dx*dp>=h/(4*pi)  --(1)\n",
-      "#since p=h/lambda  -----(2)\n",
-      "#Or p*lambda=h \n",
-      "#diffrentiting this equation\n",
-      "#p*dlambda+lambda*dp=0\n",
-      "#dp=-p*dlambda/lambda     ----(3)\n",
-      "#from (2) and (3)  dp=-h*dlambda/lambda^2   ---(4)\n",
-      "#from (1) and(4)  dx*dlambda>=lambda^2/4*pi\n",
-      "#Or dx=lambda^2/(4*pi*dlambda)\n",
-      "dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the position of X-ray photon is dx=\",dx,\"m\";\n",
-      "print \"Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \"\n",
-      "print \" ANSWEER IS WRONG IN THE TEXTBOOK\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the position of X-ray photon is dx= 7.96178343949e-16 m\n",
-        "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \n",
-        " ANSWEER IS WRONG IN THE TEXTBOOK\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.20 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dt=1*10**-8; #in sec (average life time)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
-      "#and E=h*v      v is the frequency\n",
-      "#therefore we have dv>=1/(4*pi*dt)\n",
-      "dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum uncertainty in the frequency of the photon is dv=\",dv,\"sec^-1\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum uncertainty in the frequency of the photon is dv= 7961783.43949 sec^-1\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.21 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dt=1E-12; #in sec (average life time)\n",
-      "h=6.63E-34; #in J-s (Planck'c constant)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
-      "dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy\n",
-      "dE1=dE/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t         =\",dE1,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the energy of the photon is dE= 5.27866242038e-23 J\n",
-        "\t\t\t\t\t         = 0.000329916401274 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.22 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dT=2.5E-14; #in sec (average life time)\n",
-      "h=6.63E-34; #in J-s (Planck'c constant)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
-      "dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy\n",
-      "dE1=dE/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t         =\",dE1,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the energy of the photon is dE= 2.11146496815e-21 J\n",
-        "\t\t\t\t\t         = 0.013196656051 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.23 , Page no:112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "a=2; #in Angstrom (length of the box)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "n2=2; n4=4; #two quantum states\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "a=a*1E-10; #since a is in Angstrom \n",
-      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
-      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state\n",
-      "E21=E2/e; #changing unit from J to eV\n",
-      "E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state\n",
-      "E41=E4/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The energy corresponding to the 2nd quantum state is E2=\",E2,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t       =\",E21,\"eV\";\n",
-      "print\"The energy corresponding to the 4nd quantum state is E4=\",E4,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t       =\",E41,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy corresponding to the 2nd quantum state is E2= 6.0307521978e-18 J\n",
-        "\t\t\t\t\t\t       = 37.6922012363 eV\n",
-        "The energy corresponding to the 4nd quantum state is E4= 2.41230087912e-17 J\n",
-        "\t\t\t\t\t\t       = 150.768804945 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.24 , Page no:113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "a=1E-10; #in m (width of the well)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "n1=1; n2=2; n3=3; #ground and first two excited states\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
-      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state\n",
-      "E11=E1/e; #changing unit from J to eV\n",
-      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state\n",
-      "E21=E2/e; #changing unit from J to eV\n",
-      "E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state\n",
-      "E31=E3/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The energy corresponding to the ground state is E1=\",E1,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",E11,\"eV\";\n",
-      "print\"The energy corresponding to the 1st excited state is E2=\",E2,\"\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",E21,\"eV\";\n",
-      "print\"The energy corresponding to the 2nd excited state is E3=\",E3,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",E31,\"eV\";\n",
-      "print \" There is slight variation in the answer due to round off error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy corresponding to the ground state is E1= 6.0307521978e-18 J\n",
-        "\t\t\t\t\t = 37.6922012363 eV\n",
-        "The energy corresponding to the 1st excited state is E2= 2.41230087912e-17 \n",
-        "\t\t\t\t\t = 150.768804945 eV\n",
-        "The energy corresponding to the 2nd excited state is E3= 5.42767697802e-17 J\n",
-        "\t\t\t\t\t = 339.229811126 eV\n",
-        " There is slight variation in the answer due to round off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.25 , Page no:113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dx=1E-8; #in m (length of box or uncertainty in the position)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#From uncertainty principle dx*dp=h and dp=m*dv\n",
-      "#therefore we have\n",
-      "dv=h/(m*dx); #calculation of minimum uncertainty in the velocity\n",
-      "dv1=dv*1E-3; #changing unit from m/s to Km/s\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum uncertainty in the velocity of electron i dv=\",dv,\"m/s\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t        =\",round(dv1,2),\"Km/s\";\n",
-      "print \"Note: There is slight variation in the answer due to round off error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum uncertainty in the velocity of electron i dv= 72813.1868132 m/s\n",
-        "\t\t\t\t\t\t        = 72.81 Km/s\n",
-        "Note: There is slight variation in the answer due to round off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.26 , Page no:113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "a=4E-10; #in m (length of the box)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "n1=1; #ground state\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
-      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state\n",
-      "E11=E1/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum energy of electron is E1=\",E1,\"J (Note: The answer in the book  is wrong due to printing error)\";\n",
-      "print\"\\t\\t\\t\\t    =\",round(E11,3),\"eV\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum energy of electron is E1= 3.76922012363e-19 J (Note: The answer in the book  is wrong due to printing error)\n",
-        "\t\t\t\t    = 2.356 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb b/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb
deleted file mode 100755
index 66041c45..00000000
--- a/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb
+++ /dev/null
@@ -1,1258 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:43369901ed597bc7502d54a31a362249abbf2812c557cb432bc86adf72bf30d7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5:Principles of Quantum Mechanics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1 , Page no:102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=1; #in Angstrom (wavelength)\n",
-      "m=1.67E-27; #in Kg (mass of neutron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "lambda2=lambda1*1E-10; #since lambda is in Angstrom\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "v=h/(m*lambda2); #calculation of velocity\n",
-      "K=m*v**2/2; #calculation of kinetic energy\n",
-      "K1=K/e; #changing unit fro J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity is v=\",round(v,3),\"m/s\";\n",
-      "print\"The kinetic energy is K=\",'%.3E'%K,\"J\";\n",
-      "print\"\\t\\t       =\",'%.3E'%K1,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity is v= 3967.066 m/s\n",
-        "The kinetic energy is K= 1.314E-20 J\n",
-        "\t\t       = 8.213E-02 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2 , Page no:103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=50; #in eV (Kinetic energy)\n",
-      "m0=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "K=K*e; #changing unit from eV to J\n",
-      "#Since K=m*v^2/2\n",
-      "#Therefore v=sqrt(2*K/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",round(lambda2,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 1.736E-10 m\n",
-        "\t\t  = 1.736 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3 , Page no:104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E=2000; #in eV (Kinetic energy)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "#Since E=m*v^2/2\n",
-      "#Therefore v=sqrt(2*E/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "lambda1=h/math.sqrt(2*m*E); #calculation of wavelength\n",
-      "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",round(lambda2,4),\"nm\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 2.745E-11 m\n",
-        "\t\t  = 0.0275 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4 , Page no:104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_n=1.676E-27; #in Kg (mass of neutron)\n",
-      "h=6.625E-34; #in J-s (Planck's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "\n",
-      "#calculate\n",
-      "E_e=m_e*c**2; #rest mass energy of electron\n",
-      "E_n=2*E_e; #given (kinetic energy of neutron)\n",
-      "#Since K=m*v^2/2\n",
-      "#Therefore v=sqrt(2*K/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
-      "print\"\\t\\t  =\",'%.3E'%lambda2,\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 2.827E-14 m\n",
-        "\t\t  = 2.827E-04 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5 , Page no:104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=1600; #in V (Potential)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
-      "print \"Note : The answer in the book is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 0.307 Angstrom\n",
-        "Note : The answer in the book is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6 , Page no:105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E=1000; #in eV (Kinetic energy of photon)\n",
-      "K=1000; #in eV (Kinetic energy of electron)\n",
-      "m0=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.6E-34; #in J-s (Planck's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "e=1.6E-19; #in C (charge on electron)\n",
-      "\n",
-      "#calculate\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "lambda_p=h*c/E; #For photon E=hc/lambda\n",
-      "lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom\n",
-      "#Since K=m*v^2/2\n",
-      "#Therefore v=sqrt(2*K/m)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "K=K*e; #changing unit from eV to J\n",
-      "lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
-      "lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"For photon,the wavelength is =\",'%.3E'%lambda_p,\"m\";\n",
-      "print\"\\t\\t\\t     =\",round(lambda_p1,1),\"Angstrom\";\n",
-      "print\"For electron,the wavelength is =\",'%.3E'%lambda_e,\"m\";\n",
-      "print\"\\t\\t\\t       =\",round(lambda_e1,2),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For photon,the wavelength is = 1.238E-09 m\n",
-        "\t\t\t     = 12.4 Angstrom\n",
-        "For electron,the wavelength is = 3.868E-11 m\n",
-        "\t\t\t       = 0.39 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7 , Page no:105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=1.66*10**-10; #in m (wavelength)\n",
-      "m=9.1*10**-31; #in Kg (mass of electron)\n",
-      "h=6.626*10**-34; #in J-s (Planck's constant)\n",
-      "e=1.6E-19; #in C (charge on electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since lambda=h/(m*v)\n",
-      "#Therefore we have\n",
-      "v=h/(m*lambda1); #calculation of velocity\n",
-      "K=m*v**2/2; #calculation of kinetic energy\n",
-      "K1=K/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity of electron is v=\",'%.2E'%v,\"m/s\";\n",
-      "print\"The kinetic energy is K=\",'%.2E'%K,\"J\";\n",
-      "print\"\\t\\t       =\",round(K1,3),\"eV\";\n",
-      "print \"Note: The answer in the book for kinetic energy is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of electron is v= 4.39E+06 m/s\n",
-        "The kinetic energy is K= 8.75E-18 J\n",
-        "\t\t       = 54.714 eV\n",
-        "Note: The answer in the book for kinetic energy is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8 , Page no:106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=400; #in K (temperature)\n",
-      "m=6.7E-27; #in Kg (mass of He-atom)\n",
-      "h=6.6E-34; #in J-s (Planck's constant)\n",
-      "k=1.376E-23; #in J/degree (Boltzmann constant)\n",
-      "\n",
-      "#calculate\n",
-      "#Since lambda=h/(m*v)\n",
-      "#E=mv^2/2;\n",
-      "#Therefore lambda=h/sqrt(2*m*E)\n",
-      "#E=kT\n",
-      "#Therefore lambda=h/sqrt(2*m*k*T)\n",
-      "lambda1=h/math.sqrt(2*m*k*T)\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The de-Broglie wavelength of He-atom is =\",'%.2E'%lambda1,\"m\";\n",
-      "print\"\\t\\t\\t\\t        =\",round(lambda2,4),\"Angstrom\";    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The de-Broglie wavelength of He-atom is = 7.69E-11 m\n",
-        "\t\t\t\t        = 0.7685 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9 , Page no:106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_p=1.6E-27; #in Kg (mass of proton)\n",
-      "h=6.626E-34; #in J-s (Planck's constant)\n",
-      "c=3E8; #in m/s (velocity of light)\n",
-      "\n",
-      "#calculate\n",
-      "E=m_e*c**2; #in J (rest energy of electron)\n",
-      "#Since lambda=h/(m*v)\n",
-      "#E=mv^2/2;\n",
-      "#herefore lambda=h/sqrt(2*m*E)\n",
-      "#Also E=m_e*c^2;\n",
-      "#therefore lambda=h/sqrt(2*m_p*m_e*c^2)\n",
-      "lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength\n",
-      "lambda2=lambda1*1*10**10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The de-Broglie wavelength of proton is =\",'%.2E'%lambda1,\"m\";\n",
-      "print\"\\t\\t\\t\\t       =\",'%.2E'%lambda2,\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The de-Broglie wavelength of proton is = 4.09E-14 m\n",
-        "\t\t\t\t       = 4.09E-04 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10 , Page no:106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=10000; #in V (Potential)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 0.123 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.11 , Page no:107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=100; #in V (Potential)\n",
-      "n=1; #order of diffraction\n",
-      "d=2.15; #in Angstrom (lattice spacing)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "#Since 2*d*sind(theta)=n*lambda\n",
-      "#therefore we have\n",
-      "theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
-      "print\"The glancing angle is =\",round(theta1,1),\"degree\";\n",
-      "print \"Note: In question V=100 eV but the solution is using V=100V \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 1.227 Angstrom\n",
-        "The glancing angle is = 16.6 degree\n",
-        "Note: In question V=100 eV but the solution is using V=100V \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12 , Page no:107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=344; #in V (Potential)\n",
-      "n=1; #order of diffraction\n",
-      "theta=60; #in degree (glancing angle)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
-      "#Since 2*d*sind(theta)=n*lambda\n",
-      "#therefore we have\n",
-      "d=n*lambda1/(2*math.sin(math.radians(theta))); #calculation of spacing constant\n",
-      "\n",
-      "#result\n",
-      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
-      "print\"The spacing of the crystal is =\",round(d,4),\"Angstrom\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The wavelength is = 0.662 Angstrom\n",
-        "The spacing of the crystal is = 0.3819 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.13 , Page no:107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r=0.53*10**-10; #in m (radius of first Bohr orbit)\n",
-      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
-      "m=9.1*10**-31; #in Kg (mass of electron)\n",
-      "n=1; #First Bohr orbit\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "#Since 2*pi*r=n*lambda  and  lambda=h/(m*v)\n",
-      "#Threfore we have v=h*n/(2*pi*r*m)\n",
-      "v=h*n/(2*pi*r*m); #calculation of velocity\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity of electron is =\",'%.3E'%v,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of electron is = 2.179E+06 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14 , Page no:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dx=0.2; #in Angstrom (uncertainty in the position)\n",
-      "h=6.6E-34; #in J-s (Planck's constant)\n",
-      "m0=9.1E-31; #in Kg (mass of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "dx=dx*1E-10; #since dx is in Angstrom\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
-      "#since dp=m*dv\n",
-      "dv=dp/m0; #calculation of uncertainty in the velocity\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the momentum is dp=\",'%.3E'%dp,\"Kg-m/\";\n",
-      "print\"The uncertainty in the velocity is dv=\",'%.3E'%dv,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the momentum is dp= 2.627E-24 Kg-m/\n",
-        "The uncertainty in the velocity is dv= 2.887E+06 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15 , Page no:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_p=1.67E-27; #in Kg (mass of proton)\n",
-      "dx_p=1; #in nanometer (uncertainty in position of electron)\n",
-      "dx_n=1; #in nanometer (uncertainty in position of proton)\n",
-      "\n",
-      "#calculate\n",
-      "#since dp=h/(4*pi*dx)\n",
-      "#since h/(4*pi) is constant and dx is same for electron and proton \n",
-      "#therefor both electron and proton have same uncertainty in the momentum\n",
-      "#since dv=dp/m  and  dp is same for both\n",
-      "#therefore dv_e/dv_p=m_p/m_e\n",
-      "#therefore\n",
-      "K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton\n",
-      "\n",
-      "#result\n",
-      "print\"The ratio of uncertainty in the velocity of electron to that of proton is =\",round(K);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.16 , Page no:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)\n",
-      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
-      "m=1.67E-27; #in Kg (mass of proton)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
-      "p=dp; #minimum value of momentum to calculate mimimum kinetic energy\n",
-      "K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton\n",
-      "K1=K/e; #changing unit from J to eV\n",
-      "K2=K1/1E6; #changing unit from eV to MeV\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum uncertainty in the momentum of proton is dp =\",'%.3E'%dp,\"Kg-m/s\";\n",
-      "print\"The minimum kinetic energy of proton is K=\",'%.3E'%K,\"J\";\n",
-      "print\"\\t\\t\\t\\t \\t =\",'%.3E'%K1,\"eV\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",round(K2,1),\"MeV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum uncertainty in the momentum of proton is dp = 1.051E-20 Kg-m/s\n",
-        "The minimum kinetic energy of proton is K= 3.307E-14 J\n",
-        "\t\t\t\t \t = 2.067E+05 eV\n",
-        "\t\t\t\t\t = 0.2 MeV\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.17 , Page no:109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=1; #in KeV (kinetic energy of electron)\n",
-      "dx=1; #in Angstrom (uncertainty in the position)\n",
-      "h=6.63E-34; #in J-s (Planck's constant)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "dx=dx*1E-10; #since dx is in Angstrom\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
-      "K=K*1E3*1.6E-19; #changing unit from KeV to J\n",
-      "p=math.sqrt(2*m*K); #calculation of momentum \n",
-      "poc=(dp/p)*100; #calculation of percentage of uncertainty\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the momentum of electron is dp=\",'%.3E'%dp,\"Kg-m/s\";\n",
-      "print\"The momentum of electron is p=\",'%.3E'%p,\"Kg-m/s\";\n",
-      "print\"The percentage of uncertainty in the momentum  is =\",round(poc,1);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the momentum of electron is dp= 5.279E-25 Kg-m/s\n",
-        "The momentum of electron is p= 1.706E-23 Kg-m/s\n",
-        "The percentage of uncertainty in the momentum  is = 3.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.18 , Page no:109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "v=6.6E4; #m/s (speed of electron)\n",
-      "poc=0.01; #percentage of uncertainty\n",
-      "h=6.63E-34; #in J-s (Planck's constant)\n",
-      "m=9E-31; #in Kg (mass of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "p=m*v; #calculation of momentum\n",
-      "dp=(poc/100)*p; #calculation of uncertainty in the momentum\n",
-      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
-      "dx=h/(4*pi*dp); #calculation of uncertainty in the position\n",
-      "\n",
-      "#result\n",
-      "print\"The momentum of electron is p=\",'%.3E'%p,\"Kg-m/s\";\n",
-      "print\"The uncertainty in the momentum of electron is dp=\",'%.3E'%dp,\"Kg-m/s\";\n",
-      "print\"The uncertainty in the position of electron is dx=\",'%.3E'%dx,\"Kg-m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The momentum of electron is p= 5.940E-26 Kg-m/s\n",
-        "The uncertainty in the momentum of electron is dp= 5.940E-30 Kg-m/s\n",
-        "The uncertainty in the position of electron is dx= 8.887E-06 Kg-m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.19 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "lambda1=1; #in Angstrom (wavelength)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "dlambda=1E-6; #uncertainty in wavelength\n",
-      "\n",
-      "#calculate\n",
-      "lambda2=lambda1*1E-10; #sinc lambda is in Angstrom\n",
-      "#By uncertainty principle, dx*dp>=h/(4*pi)  --(1)\n",
-      "#since p=h/lambda  -----(2)\n",
-      "#Or p*lambda=h \n",
-      "#diffrentiting this equation\n",
-      "#p*dlambda+lambda*dp=0\n",
-      "#dp=-p*dlambda/lambda     ----(3)\n",
-      "#from (2) and (3)  dp=-h*dlambda/lambda^2   ---(4)\n",
-      "#from (1) and(4)  dx*dlambda>=lambda^2/4*pi\n",
-      "#Or dx=lambda^2/(4*pi*dlambda)\n",
-      "dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the position of X-ray photon is dx=\",'%.3E'%dx,\"m\";\n",
-      "print \"Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \"\n",
-      "print \" ANSWEER IS WRONG IN THE TEXTBOOK\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the position of X-ray photon is dx= 7.962E-16 m\n",
-        "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \n",
-        " ANSWEER IS WRONG IN THE TEXTBOOK\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.20 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dt=1*10**-8; #in sec (average life time)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
-      "#and E=h*v      v is the frequency\n",
-      "#therefore we have dv>=1/(4*pi*dt)\n",
-      "dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum uncertainty in the frequency of the photon is dv=\",'%.3E'%dv,\"sec^-1\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum uncertainty in the frequency of the photon is dv= 7.962E+06 sec^-1\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.21 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dt=1E-12; #in sec (average life time)\n",
-      "h=6.63E-34; #in J-s (Planck'c constant)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
-      "dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy\n",
-      "dE1=dE/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the energy of the photon is dE=\",'%.3E'%dE,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t         =\",'%.3E'%dE1,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the energy of the photon is dE= 5.279E-23 J\n",
-        "\t\t\t\t\t         = 3.299E-04 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.22 , Page no:111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dT=2.5E-14; #in sec (average life time)\n",
-      "h=6.63E-34; #in J-s (Planck'c constant)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
-      "dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy\n",
-      "dE1=dE/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The uncertainty in the energy of the photon is dE=\",'%.3E'%dE,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t         =\",'%.3E'%dE1,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uncertainty in the energy of the photon is dE= 2.111E-21 J\n",
-        "\t\t\t\t\t         = 1.320E-02 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.23 , Page no:112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "a=2; #in Angstrom (length of the box)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "n2=2; n4=4; #two quantum states\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "a=a*1E-10; #since a is in Angstrom \n",
-      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
-      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state\n",
-      "E21=E2/e; #changing unit from J to eV\n",
-      "E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state\n",
-      "E41=E4/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The energy corresponding to the 2nd quantum state is E2=\",'%.3E'%E2,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t       =\",'%.3E'%E21,\"eV\";\n",
-      "print\"The energy corresponding to the 4nd quantum state is E4=\",'%.3E'%E4,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t       =\",'%.3E'%E41,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy corresponding to the 2nd quantum state is E2= 6.031E-18 J\n",
-        "\t\t\t\t\t\t       = 3.769E+01 eV\n",
-        "The energy corresponding to the 4nd quantum state is E4= 2.412E-17 J\n",
-        "\t\t\t\t\t\t       = 1.508E+02 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.24 , Page no:113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "a=1E-10; #in m (width of the well)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "n1=1; n2=2; n3=3; #ground and first two excited states\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
-      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state\n",
-      "E11=E1/e; #changing unit from J to eV\n",
-      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state\n",
-      "E21=E2/e; #changing unit from J to eV\n",
-      "E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state\n",
-      "E31=E3/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The energy corresponding to the ground state is E1=\",'%.3E'%E1,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E11,\"eV\";\n",
-      "print\"The energy corresponding to the 1st excited state is E2=\",'%.3E'%E2,\"\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E21,\"eV\";\n",
-      "print\"The energy corresponding to the 2nd excited state is E3=\",'%.3E'%E3,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E31,\"eV\";\n",
-      "print \" There is slight variation in the answer due to round off error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy corresponding to the ground state is E1= 6.031E-18 J\n",
-        "\t\t\t\t\t = 3.769E+01 eV\n",
-        "The energy corresponding to the 1st excited state is E2= 2.412E-17 \n",
-        "\t\t\t\t\t = 1.508E+02 eV\n",
-        "The energy corresponding to the 2nd excited state is E3= 5.428E-17 J\n",
-        "\t\t\t\t\t = 3.392E+02 eV\n",
-        " There is slight variation in the answer due to round off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.25 , Page no:113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "dx=1E-8; #in m (length of box or uncertainty in the position)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#From uncertainty principle dx*dp=h and dp=m*dv\n",
-      "#therefore we have\n",
-      "dv=h/(m*dx); #calculation of minimum uncertainty in the velocity\n",
-      "dv1=dv*1E-3; #changing unit from m/s to Km/s\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum uncertainty in the velocity of electron i dv=\",'%.3E'%dv,\"m/s\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t        =\",round(dv1,2),\"Km/s\";\n",
-      "print \"Note: There is slight variation in the answer due to round off error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum uncertainty in the velocity of electron i dv= 7.281E+04 m/s\n",
-        "\t\t\t\t\t\t        = 72.81 Km/s\n",
-        "Note: There is slight variation in the answer due to round off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.26 , Page no:113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "a=4E-10; #in m (length of the box)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "h=6.626E-34; #in J-s (Planck'c constant)\n",
-      "n1=1; #ground state\n",
-      "e=1.6*1E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
-      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state\n",
-      "E11=E1/e; #changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The minimum energy of electron is E1=\",'%.3E'%E1,\"J (Note: The answer in the book  is wrong due to printing error)\";\n",
-      "print\"\\t\\t\\t\\t    =\",round(E11,3),\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum energy of electron is E1= 3.769E-19 J (Note: The answer in the book  is wrong due to printing error)\n",
-        "\t\t\t\t    = 2.356 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals.ipynb b/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals.ipynb
deleted file mode 100755
index ff063bb6..00000000
--- a/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals.ipynb
+++ /dev/null
@@ -1,561 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0c3d2d354976eaba7e5a0f2e0cf4b1fd72cd55b3826ca7e7cd6acde8afff2cdf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6:Electron Theory and Band Theory of Metals"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1 , Page no:146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=8.5E28; #in 1/m^3 (density of electron)\n",
-      "m_e=9.11E-31; #in Kg (mass of electron)\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "T=300; #in K (temperature)\n",
-      "p=1.69E-8; #in ohm-m (resistivity)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=math.sqrt(3*k*m_e*T)/(n*e**2*p); #calculation of mean free path\n",
-      "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
-      "\n",
-      "#result\n",
-      "print\"The mean free path of electron is =\",lambda1,\"m\";\n",
-      "print\"\\t\\t\\t\\t  =\",round(lambda2,2),\"nm\";\n",
-      "#Note: answer in the book is wrong due to printing mistake"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mean free path of electron is = 2.89250681437e-09 m\n",
-        "\t\t\t\t  = 2.89 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2 , Page no:146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "P_E=1; #in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied)\n",
-      "E=0.5; #in eV (energy above Fermi level)\n",
-      "\n",
-      "#calculate\n",
-      "P_E=1/100; #changing percentage into ratio\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "#P_E=1/(1+exp((E-E_F)/k*T))\n",
-      "#Rearranging this equation, we get\n",
-      "#T=(E-E_F)/k*log((1/P_E)-1)\n",
-      "#Since E-E_F has been denoted by E  therefore\n",
-      "T=E/(k*math.log((1/P_E)-1));\n",
-      "\n",
-      "#result\n",
-      "print\"The temperature is T=\",round(T),\"K\";\n",
-      "# Note: There is slight variation in the answer due to logarithm function"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature is T= 1262.0 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3 , Page no:147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=5.8E28; #in 1/m^3 (density of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "p=1.54E-8; #in ohm-m (resistivity)\n",
-      "\n",
-      "#calculate\n",
-      "t=m/(n*e**2*p); #calculation of relaxation time\n",
-      "\n",
-      "#result\n",
-      "print\"The relaxation time of conduction electrons is =\",t,\"sec\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relaxation time of conduction electrons is = 3.97972178683e-14 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4 , Page no:147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=8.5E28; #in 1/m^3 (density of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "sigma=6E7; #in 1/ohm-m (conductivity)\n",
-      "E_F=7; #in E=eV (Fermi energy of Copper)\n",
-      "\n",
-      "#calculate\n",
-      "E_F=E_F*e; #changing unit from eV to J\n",
-      "v_F=math.sqrt(2*E_F/m); #calculation of velocity of electrons\n",
-      "#Since sigma=n*e^2*lambda/(2*m*v_F)\n",
-      "#Therefore we have\n",
-      "lambda1=2*m*v_F*sigma/(n*e**2); #calculation of mean free path\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity of the electrons is v_F=\",v_F,\"m/s   (roundoff error)\";\n",
-      "print\"The mean free path traveled by the electrons is=\",round(lambda2),\"Angstrom\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of the electrons is v_F= 1568929.08111 m/s   (roundoff error)\n",
-        "The mean free path traveled by the electrons is= 787.0 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5 , Page no:147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=6.5E28; #in 1/m^3 (density of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "p=1.43E-8; #in ohm-m (resistivity)\n",
-      "\n",
-      "#calculate\n",
-      "t=m/(n*e**2*p); #calculation of relaxation time\n",
-      "\n",
-      "#result\n",
-      "print\"The relaxation time of conduction electrons is =\",t,\"sec\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relaxation time of conduction electrons is = 3.8243006993e-14 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6 , Page no:148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=30; #in Celcius (temperature)\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "m_p=1.67E-27; #in Kg (mass of proton)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "T=T+273; #changing temperature from Celcius to Kelvin\n",
-      "KE=(3/2)*k*T; #calculation of average kinetic energy\n",
-      "KE1=KE/e; #changing unit from J to eV\n",
-      "m=1.008*2*m_p; #calculating mass of hydrogen gas molecule\n",
-      "c=math.sqrt(3*k*T/m); #calculation of velocity\n",
-      "\n",
-      "#result\n",
-      "print\"The average kinetic energy of gas ,molecules is KE=\",KE,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t  =\",KE1,\"eV   (Answer is wrong in textbook)\";\n",
-      "print\"The velocity of molecules is c=\",round(c,2),\"m/s\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average kinetic energy of gas ,molecules is KE= 6.2721e-21 J\n",
-        "\t\t\t\t\t\t  = 0.039200625 eV   (Answer is wrong in textbook)\n",
-        "The velocity of molecules is c= 1930.27 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7 , Page no:148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E=10; #in eV (kinetic energy for each electron and proton)\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_p=1.67E-27; #in Kg (mass of proton)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "#since E=m*v^2/2\n",
-      "#therefore  v=sqrt(2E/m)\n",
-      "v_e=math.sqrt(2*E/m_e); #calculation of kinetic energy of electron\n",
-      "v_p=math.sqrt(2*E/m_p); #calculation of kinetic energy of proton\n",
-      "\n",
-      "#result\n",
-      "print\"The kinetic energy of electron is v_e=\",v_e,\"m/s\";\n",
-      "print\"The kinetic energy of proton is v_p=\",v_p,\"m/s\";\n",
-      "print \"Note: Answer is wrong in textbook\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The kinetic energy of electron is v_e= 1875228.92375 m/s\n",
-        "The kinetic energy of proton is v_p= 43774.0524132 m/s\n",
-        "Note: Answer is wrong in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8 , Page no:149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "I=100; #in A (current in the wire)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "A=10; #in mm^2 (cross-sectional area)\n",
-      "n=8.5E28; #in 1/m^3 (density of electron)\n",
-      "\n",
-      "#calculate\n",
-      "A=A*1E-6; #changing unit from mm^2 to m^2\n",
-      "v_d=I/(n*A*e);\n",
-      "\n",
-      "#result\n",
-      "print\"The drift velocity of free electrons is v_d=\",v_d,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drift velocity of free electrons is v_d= 0.000735294117647 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9 , Page no:149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "I=4;#in A (current in the conductor)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "A=1E-6; #in m^2 (cross-sectional area)\n",
-      "N_A=6.02E23; #in atoms/gram-atom (Avogadro's number)\n",
-      "p=8.9; #in g/cm^3 (density)\n",
-      "M=63.6; #atomic mass of copper\n",
-      "\n",
-      "#calculate\n",
-      "n=N_A*p/M; #Calculation of density of electrons in g/cm^3\n",
-      "n1=n*1E6; #changing unit from g/cm^3 to g/m^3\n",
-      "v_d=I/(n1*A*e);\n",
-      "\n",
-      "#result\n",
-      "print\"The density of copper atoms is n=\",n,\"atoms/m^3\";\n",
-      "print\"\\t\\t\\t        =\",n1,\"atoms/m^3\";\n",
-      "print\"The average drift velocity of free electrons is v_d=\",v_d,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density of copper atoms is n= 8.42421383648e+22 atoms/m^3\n",
-        "\t\t\t        = 8.42421383648e+28 atoms/m^3\n",
-        "The average drift velocity of free electrons is v_d= 0.000296763596999 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10 , Page no:149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=9E28; #in 1/m^3 (density of valence electrons)\n",
-      "sigma=6E7; #in mho/m (conductivity of copper)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since sigma=n*e*mu  therefore\n",
-      "mu=sigma/(n*e); #calculation of mobility of electron\n",
-      "\n",
-      "#result\n",
-      "print\"The mobility of electrons is =\",mu,\"m^2/V-s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mobility of electrons is = 0.00416666666667 m^2/V-s\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11 , Page no:150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E_F=5.51; #in eV (Fermi energy in Silver)\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#part-(a)\n",
-      "Eo=(3/5)*E_F; #calculation of average energy of free electron at 0K\n",
-      "#part-(b)\n",
-      "Eo1=Eo*e; #changing unit from eV to J\n",
-      "#Since for a classical particle E=(3/2)*k*T\n",
-      "#therefroe we have\n",
-      "T=(2/3)*Eo1/k; #calculation of temperature for a classical particle (an ideal gas)\n",
-      "\n",
-      "#result\n",
-      "print\"The average energy of free electron at 0K is Eo=\",Eo,\"eV\";\n",
-      "print\"The temperature at which a classical particle have this much energy is T=\",T,\"K\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average energy of free electron at 0K is Eo= 3.306 eV\n",
-        "The temperature at which a classical particle have this much energy is T= 25553.6231884 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12 , Page no:150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E_F_L=4.7; #in eV (Fermi energy in Lithium)\n",
-      "E_F_M=2.35; #in eV (Fermi energy in a metal)\n",
-      "n_L=4.6E28; #in 1/m^3 (density of electron in Lithium)\n",
-      "\n",
-      "#calculate\n",
-      "#Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant\n",
-      "# Therefore we have n=C*E_F^(3/2)   where C is proportionality constant\n",
-      "#n1/n2=(E_F_1/E_F_2)^(3/2)\n",
-      "#Therefore we have\n",
-      "n_M=n_L*(E_F_M/E_F_L); #calculation of electron density for a metal\n",
-      "\n",
-      "#result\n",
-      "print\"The lectron density for a metal is =\",n_M,\"1/m^3\";\n",
-      "print \"Note: Answer in the book is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lectron density for a metal is = 2.3e+28 1/m^3\n",
-        "Note: Answer in the book is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb b/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb
deleted file mode 100755
index d52f3a45..00000000
--- a/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb
+++ /dev/null
@@ -1,561 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3c7e08d83e902e2949e4fdcd2bc98d527fe589fabb9636cea78c35c42abb8038"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6:Electron Theory and Band Theory of Metals"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1 , Page no:146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=8.5E28; #in 1/m^3 (density of electron)\n",
-      "m_e=9.11E-31; #in Kg (mass of electron)\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "T=300; #in K (temperature)\n",
-      "p=1.69E-8; #in ohm-m (resistivity)\n",
-      "\n",
-      "#calculate\n",
-      "lambda1=math.sqrt(3*k*m_e*T)/(n*e**2*p); #calculation of mean free path\n",
-      "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
-      "\n",
-      "#result\n",
-      "print\"The mean free path of electron is =\",'%.3E'%lambda1,\"m\";\n",
-      "print\"\\t\\t\\t\\t  =\",round(lambda2,2),\"nm\";\n",
-      "#Note: answer in the book is wrong due to printing mistake"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mean free path of electron is = 2.893E-09 m\n",
-        "\t\t\t\t  = 2.89 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2 , Page no:146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "P_E=1; #in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied)\n",
-      "E=0.5; #in eV (energy above Fermi level)\n",
-      "\n",
-      "#calculate\n",
-      "P_E=1/100; #changing percentage into ratio\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "#P_E=1/(1+exp((E-E_F)/k*T))\n",
-      "#Rearranging this equation, we get\n",
-      "#T=(E-E_F)/k*log((1/P_E)-1)\n",
-      "#Since E-E_F has been denoted by E  therefore\n",
-      "T=E/(k*math.log((1/P_E)-1));\n",
-      "\n",
-      "#result\n",
-      "print\"The temperature is T=\",round(T),\"K\";\n",
-      "# Note: There is slight variation in the answer due to logarithm function"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature is T= 1262.0 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3 , Page no:147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=5.8E28; #in 1/m^3 (density of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "p=1.54E-8; #in ohm-m (resistivity)\n",
-      "\n",
-      "#calculate\n",
-      "t=m/(n*e**2*p); #calculation of relaxation time\n",
-      "\n",
-      "#result\n",
-      "print\"The relaxation time of conduction electrons is =\",'%.3E'%t,\"sec\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relaxation time of conduction electrons is = 3.980E-14 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4 , Page no:147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=8.5E28; #in 1/m^3 (density of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "sigma=6E7; #in 1/ohm-m (conductivity)\n",
-      "E_F=7; #in E=eV (Fermi energy of Copper)\n",
-      "\n",
-      "#calculate\n",
-      "E_F=E_F*e; #changing unit from eV to J\n",
-      "v_F=math.sqrt(2*E_F/m); #calculation of velocity of electrons\n",
-      "#Since sigma=n*e^2*lambda/(2*m*v_F)\n",
-      "#Therefore we have\n",
-      "lambda1=2*m*v_F*sigma/(n*e**2); #calculation of mean free path\n",
-      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
-      "\n",
-      "#result\n",
-      "print\"The velocity of the electrons is v_F=\",'%.3E'%v_F,\"m/s   (roundoff error)\";\n",
-      "print\"The mean free path traveled by the electrons is=\",round(lambda2),\"Angstrom\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of the electrons is v_F= 1.569E+06 m/s   (roundoff error)\n",
-        "The mean free path traveled by the electrons is= 787.0 Angstrom\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5 , Page no:147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=6.5E28; #in 1/m^3 (density of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "p=1.43E-8; #in ohm-m (resistivity)\n",
-      "\n",
-      "#calculate\n",
-      "t=m/(n*e**2*p); #calculation of relaxation time\n",
-      "\n",
-      "#result\n",
-      "print\"The relaxation time of conduction electrons is =\",'%.3E'%t,\"sec\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relaxation time of conduction electrons is = 3.824E-14 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6 , Page no:148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=30; #in Celcius (temperature)\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "m_p=1.67E-27; #in Kg (mass of proton)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "T=T+273; #changing temperature from Celcius to Kelvin\n",
-      "KE=(3/2)*k*T; #calculation of average kinetic energy\n",
-      "KE1=KE/e; #changing unit from J to eV\n",
-      "m=1.008*2*m_p; #calculating mass of hydrogen gas molecule\n",
-      "c=math.sqrt(3*k*T/m); #calculation of velocity\n",
-      "\n",
-      "#result\n",
-      "print\"The average kinetic energy of gas ,molecules is KE=\",'%.3E'%KE,\"J\";\n",
-      "print\"\\t\\t\\t\\t\\t\\t  =\",round(KE1,4),\"eV   (Answer is wrong in textbook)\";\n",
-      "print\"The velocity of molecules is c=\",round(c,2),\"m/s\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average kinetic energy of gas ,molecules is KE= 6.272E-21 J\n",
-        "\t\t\t\t\t\t  = 0.0392 eV   (Answer is wrong in textbook)\n",
-        "The velocity of molecules is c= 1930.27 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7 , Page no:148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E=10; #in eV (kinetic energy for each electron and proton)\n",
-      "m_e=9.1E-31; #in Kg (mass of electron)\n",
-      "m_p=1.67E-27; #in Kg (mass of proton)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "E=E*e; #changing unit from eV to J\n",
-      "#since E=m*v^2/2\n",
-      "#therefore  v=sqrt(2E/m)\n",
-      "v_e=math.sqrt(2*E/m_e); #calculation of kinetic energy of electron\n",
-      "v_p=math.sqrt(2*E/m_p); #calculation of kinetic energy of proton\n",
-      "\n",
-      "#result\n",
-      "print\"The kinetic energy of electron is v_e=\",'%.3E'%v_e,\"m/s\";\n",
-      "print\"The kinetic energy of proton is v_p=\",'%.3E'%v_p,\"m/s\";\n",
-      "print \"Note: Answer is wrong in textbook\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The kinetic energy of electron is v_e= 1.875E+06 m/s\n",
-        "The kinetic energy of proton is v_p= 4.377E+04 m/s\n",
-        "Note: Answer is wrong in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8 , Page no:149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "I=100; #in A (current in the wire)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "A=10; #in mm^2 (cross-sectional area)\n",
-      "n=8.5E28; #in 1/m^3 (density of electron)\n",
-      "\n",
-      "#calculate\n",
-      "A=A*1E-6; #changing unit from mm^2 to m^2\n",
-      "v_d=I/(n*A*e);\n",
-      "\n",
-      "#result\n",
-      "print\"The drift velocity of free electrons is v_d=\",'%.3E'%v_d,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drift velocity of free electrons is v_d= 7.353E-04 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9 , Page no:149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "I=4;#in A (current in the conductor)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "A=1E-6; #in m^2 (cross-sectional area)\n",
-      "N_A=6.02E23; #in atoms/gram-atom (Avogadro's number)\n",
-      "p=8.9; #in g/cm^3 (density)\n",
-      "M=63.6; #atomic mass of copper\n",
-      "\n",
-      "#calculate\n",
-      "n=N_A*p/M; #Calculation of density of electrons in g/cm^3\n",
-      "n1=n*1E6; #changing unit from g/cm^3 to g/m^3\n",
-      "v_d=I/(n1*A*e);\n",
-      "\n",
-      "#result\n",
-      "print\"The density of copper atoms is n=\",'%.3E'%n,\"atoms/m^3\";\n",
-      "print\"\\t\\t\\t        =\",'%.3E'%n1,\"atoms/m^3\";\n",
-      "print\"The average drift velocity of free electrons is v_d=\",'%.3E'%v_d,\"m/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density of copper atoms is n= 8.424E+22 atoms/m^3\n",
-        "\t\t\t        = 8.424E+28 atoms/m^3\n",
-        "The average drift velocity of free electrons is v_d= 2.968E-04 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10 , Page no:149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=9E28; #in 1/m^3 (density of valence electrons)\n",
-      "sigma=6E7; #in mho/m (conductivity of copper)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#Since sigma=n*e*mu  therefore\n",
-      "mu=sigma/(n*e); #calculation of mobility of electron\n",
-      "\n",
-      "#result\n",
-      "print\"The mobility of electrons is =\",'%.3E'%mu,\"m^2/V-s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mobility of electrons is = 4.167E-03 m^2/V-s\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11 , Page no:150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E_F=5.51; #in eV (Fermi energy in Silver)\n",
-      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#part-(a)\n",
-      "Eo=(3/5)*E_F; #calculation of average energy of free electron at 0K\n",
-      "#part-(b)\n",
-      "Eo1=Eo*e; #changing unit from eV to J\n",
-      "#Since for a classical particle E=(3/2)*k*T\n",
-      "#therefroe we have\n",
-      "T=(2/3)*Eo1/k; #calculation of temperature for a classical particle (an ideal gas)\n",
-      "\n",
-      "#result\n",
-      "print\"The average energy of free electron at 0K is Eo=\",Eo,\"eV\";\n",
-      "print\"The temperature at which a classical particle have this much energy is T=\",'%.3E'%T,\"K\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average energy of free electron at 0K is Eo= 3.306 eV\n",
-        "The temperature at which a classical particle have this much energy is T= 2.555E+04 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12 , Page no:150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "E_F_L=4.7; #in eV (Fermi energy in Lithium)\n",
-      "E_F_M=2.35; #in eV (Fermi energy in a metal)\n",
-      "n_L=4.6E28; #in 1/m^3 (density of electron in Lithium)\n",
-      "\n",
-      "#calculate\n",
-      "#Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant\n",
-      "# Therefore we have n=C*E_F^(3/2)   where C is proportionality constant\n",
-      "#n1/n2=(E_F_1/E_F_2)^(3/2)\n",
-      "#Therefore we have\n",
-      "n_M=n_L*(E_F_M/E_F_L); #calculation of electron density for a metal\n",
-      "\n",
-      "#result\n",
-      "print\"The lectron density for a metal is =\",n_M,\"1/m^3\";\n",
-      "print \"Note: Answer in the book is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lectron density for a metal is = 2.3e+28 1/m^3\n",
-        "Note: Answer in the book is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_07_Dielectric_Properties.ipynb b/Applied_Physics/Chapter_07_Dielectric_Properties.ipynb
deleted file mode 100755
index 7280033d..00000000
--- a/Applied_Physics/Chapter_07_Dielectric_Properties.ipynb
+++ /dev/null
@@ -1,488 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:5117f9d565431f3f25738652845b14f8a6fa036b5c0992b0b2e5d69631937306"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7:Dielectric Properties"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 , Page no:187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "A=100; #in cm^2 (cross-sectional area)\n",
-      "d=1; #in cm (seperation between plates)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "V=100; #in V (potential difference)\n",
-      "\n",
-      "#calculate\n",
-      "A1=A*1E-4; #changing unit from cm^2 to m^2\n",
-      "d1=d*1E-2; #changing unit from cm to m\n",
-      "C=Eo*A1/d1; #calculation of capacitance\n",
-      "Q=C*V; #calculation of charge\n",
-      "C1=C*1E12; #changing unit of capacitance from F to pF\n",
-      "\n",
-      "#result\n",
-      "print\"The capacitance of capacitor is C=\",C,\"C\";\n",
-      "print\"\\t\\t\\t\\t =\",C1,\"C\";\n",
-      "print\"The charge on the plates is Q=\",Q,\"C\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The capacitance of capacitor is C= 8.85e-12 C\n",
-        "\t\t\t\t = 8.85 C\n",
-        "The charge on the plates is Q= 8.85e-10 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2 , Page no:187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "A=650; #in mm^2 (cross-sectional area)\n",
-      "d=4; #in mm (seperation between plates)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "Er=3.5; #di-electric constant of the material\n",
-      "Q=2E-10; #in C (charge on plates)\n",
-      "\n",
-      "#calculate\n",
-      "A1=A*1E-6; #changing unit from mm^2 to m^2\n",
-      "d1=d*1E-3; #changing unit from mm to m\n",
-      "C=Er*Eo*A1/d1; #calculation of capacitance\n",
-      "V=Q/C; #calculation of charge\n",
-      "C1=C*1E12; #changing unit of capacitance from F to pF\n",
-      "\n",
-      "#result\n",
-      "print\"The capacitance of capacitor is C=\",C,\"C\",\n",
-      "print\"\\n\\t\\t\\t         =\",C1,\"pF\";\n",
-      "print\"The resultant voltage across the capacitor is V=\",round(V,3),\"V\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The capacitance of capacitor is C= 5.0334375e-12 C \n",
-        "\t\t\t         = 5.0334375 pF\n",
-        "The resultant voltage across the capacitor is V= 39.734 V\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3 , Page no:188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=2.7E25; #in 1/m^3 (density of atoms)\n",
-      "E=1E6; #in V/m (electric field)\n",
-      "Z=2; #atomic number of Helium \n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "Er=1.0000684; #(dielectric constant of the material)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "#since  alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 \n",
-      "#Therefore we have r_0^3=(Er-1)/(4*pi*N)\n",
-      "r_0=((Er-1)/(4*pi*N))**(1/3); #calculation of radius of electron cloud\n",
-      "x=4*pi*Eo*E*r_0/(Z*e); #calculation of dispalcement\n",
-      "\n",
-      "#result\n",
-      "print\"The radius of electron cloud is r_0=\",r_0,\"m\";\n",
-      "print\"The displacement is x=\",x,\"m\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The radius of electron cloud is r_0= 5.86454377988e-11 m\n",
-        "The displacement is x= 20371.2258874 m\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4 , Page no:188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=1.000134; #di-elecrtic constant of the neon gas at NTP\n",
-      "E=90000; #in V/m (electric field)\n",
-      "Eo=8.85E-12; #in C/N-m^2 (absolute premittivity)\n",
-      "N_A=6.023E26; #in atoms/Kg-mole (Avogadro's number)\n",
-      "V=22.4; #in m^3 (volume of gas at NTP\n",
-      "\n",
-      "#calculate\n",
-      "n=N_A/V; #calculaton of density of atoms\n",
-      "#Since P=n*p=(k-1)*Eo*E\n",
-      "#therefore we have\n",
-      "p=(K-1)*Eo*E/n; #calculation of dipole moment induced\n",
-      "alpha=p/E; #calculation of atomic polarisability\n",
-      "\n",
-      "#result\n",
-      "print\"The dipole moment induced in each atom is p=\",p,\"C-m\";\n",
-      "print\"The atomic polarisability of neon is =\",alpha,\"c-m^2/V\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dipole moment induced in each atom is p= 3.96940793625e-36 C-m\n",
-        "The atomic polarisability of neon is = 4.4104532625e-41 c-m^2/V\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5 , Page no:189"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Er=3.75; #di-elecrtic constant of sulphur at 27 degree Celcius\n",
-      "gama=1/3; #internal field constant\n",
-      "p=2050; #in Kg/m^3 (density)\n",
-      "M_A=32; #in amu (atomic weight of sulphur)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "N=6.022E23; #Avogadro's number\n",
-      "\n",
-      "#calculate\n",
-      "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
-      "#therefore we have\n",
-      "alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); #calculation of electronic polarisability of sulphur\n",
-      "\n",
-      "#result\n",
-      "print\"The electronic polarisability of sulphur is =\",alpha_e,\"Fm^2   (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electronic polarisability of sulphur is = 3.29143088985e-37 Fm^2   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.6 , Page no:189"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Er=1.0000684; #di-elecrtic constant of Helium gas at NTP\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "N=2.7E25; #number of atomsper unit volume\n",
-      "\n",
-      "#calculate\n",
-      "#Since Er-1=(N/Eo)*alpha_e\n",
-      "#therefore we have\n",
-      "alpha_e=Eo*(Er-1)/N; #calculation of electronic polarisability of Helium\n",
-      "\n",
-      "#result\n",
-      "print\"The electronic polarisability of Helium gas is =\",alpha_e,\"Fm^2    (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electronic polarisability of Helium gas is = 2.242e-41 Fm^2    (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7 , Page no:190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=3E28; #in atoms/m^3 (density of atoms)\n",
-      "alpha_e=1E-40; #in F-m^2 (electronic polarisability)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "\n",
-      "#calculate\n",
-      "#Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo)\n",
-      "#therefore we have\n",
-      "Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo)));\n",
-      "#calculation of dielectric constant of the material\n",
-      "\n",
-      "#result\n",
-      "print\"The dielectric constant of the material is Er=\",Er,\"F/m\",\n",
-      "print \"NOTE: The answer in the book is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dielectric constant of the material is Er= 1.3821656051 F/m NOTE: The answer in the book is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.8 , Page no:190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Er=4; #relative permittivity of sulphur\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "NA=2.08E3; #in Kg/m^3 (density of atoms in sulphur)\n",
-      "\n",
-      "#calculate\n",
-      "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
-      "#therefore we have\n",
-      "alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); #calculation of electronic polarisability of sulphur\n",
-      "\n",
-      "#result\n",
-      "print\"The electronic polarisability of sulphur is =\",alpha_e,\"Fm^2\";\n",
-      "print \"NOTE: The answer in the book is wrong and they used the wrong formula \" \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electronic polarisability of sulphur is = 6.38221153846e-15 Fm^2\n",
-        "NOTE: The answer in the book is wrong and they used the wrong formula \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.9 , Page no:190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "alpha1=2.5E-39; #in C^2-m/N (dielectric constant at 300K)\n",
-      "alpha2=2.0E-39; #in C^2-m/N (dielectric constant at 400K)\n",
-      "T1=300; #in K(first temperature)\n",
-      "T2=400; #in K(second temperature)\n",
-      "\n",
-      "#calculate\n",
-      "#since alpha=alpha_d+alpha0 and alpha0=Beta/T\n",
-      "#therefore alpha=alpha_d+(Beta/T)\n",
-      "#since alpha1=alpha_d+(Beta/T1)  and alpha2=alpha_d+(Beta/T2)\n",
-      "#therefore alpha1-apha2=Beta*((1/T1)-(1/T2))\n",
-      "#or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2))\n",
-      "Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); #calculation of Beta\n",
-      "alpha_d=alpha1-(Beta/T1); #calculation of polarisability due to defromation\n",
-      "alpha0_1=Beta/T1; #calculation of polarisability due to permanent dipole moment at 300K\n",
-      "alpha0_2=Beta/T2;  #calculation of polarisability due to permanent dipole moment at 400K\n",
-      "\n",
-      "#result\n",
-      "print\"The polarisability due to permanent dipole moment at 300K is =\",alpha0_1,\"C^2-m/N\";\n",
-      "print\"The polarisability due to permanent dipole moment at 400K is =\",alpha0_2,\"C^2-m/N\";\n",
-      "print\"The polarisability due to deformation of the molecules is =\",alpha_d,\"C^2-m/N\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The polarisability due to permanent dipole moment at 300K is = 2e-39 C^2-m/N\n",
-        "The polarisability due to permanent dipole moment at 400K is = 1.5e-39 C^2-m/N\n",
-        "The polarisability due to deformation of the molecules is = 5e-40 C^2-m/N\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.10 , Page no:191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=1.5; #refractive index\n",
-      "Er=5.6; #dielectric constant\n",
-      "\n",
-      "#calculate\n",
-      "#since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0)   Clausius-Mossotti equation\n",
-      "#and  (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) \n",
-      "#from above two equations, we get  ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i)\n",
-      "#or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P)\n",
-      "#where P is fractional ionisational polarisability\n",
-      "P=1-((n**2-1)/(n**2+2))*((Er+2)/(Er-1)); #calculation of fractional ionisational polarisability\n",
-      "P1=P*100; #calculation of percentage of  ionisational polarisability\n",
-      "\n",
-      "#result\n",
-      "print\"The percentage of  ionisational polarisability is =\",P1,\"percent\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of  ionisational polarisability is = 51.4066496164 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_07_Dielectric_Properties_1.ipynb b/Applied_Physics/Chapter_07_Dielectric_Properties_1.ipynb
deleted file mode 100755
index 85a277ef..00000000
--- a/Applied_Physics/Chapter_07_Dielectric_Properties_1.ipynb
+++ /dev/null
@@ -1,488 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d363aa85787801a09c4a4d5d3e8dc3b5551836ae64a0a0c025744f99f230a4b4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7:Dielectric Properties"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 , Page no:187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "A=100; #in cm^2 (cross-sectional area)\n",
-      "d=1; #in cm (seperation between plates)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "V=100; #in V (potential difference)\n",
-      "\n",
-      "#calculate\n",
-      "A1=A*1E-4; #changing unit from cm^2 to m^2\n",
-      "d1=d*1E-2; #changing unit from cm to m\n",
-      "C=Eo*A1/d1; #calculation of capacitance\n",
-      "Q=C*V; #calculation of charge\n",
-      "C1=C*1E12; #changing unit of capacitance from F to pF\n",
-      "\n",
-      "#result\n",
-      "print\"The capacitance of capacitor is C=\",C,\"C\";\n",
-      "print\"\\t\\t\\t\\t =\",C1,\"C\";\n",
-      "print\"The charge on the plates is Q=\",Q,\"C\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The capacitance of capacitor is C= 8.85e-12 C\n",
-        "\t\t\t\t = 8.85 C\n",
-        "The charge on the plates is Q= 8.85e-10 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2 , Page no:187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "A=650; #in mm^2 (cross-sectional area)\n",
-      "d=4; #in mm (seperation between plates)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "Er=3.5; #di-electric constant of the material\n",
-      "Q=2E-10; #in C (charge on plates)\n",
-      "\n",
-      "#calculate\n",
-      "A1=A*1E-6; #changing unit from mm^2 to m^2\n",
-      "d1=d*1E-3; #changing unit from mm to m\n",
-      "C=Er*Eo*A1/d1; #calculation of capacitance\n",
-      "V=Q/C; #calculation of charge\n",
-      "C1=C*1E12; #changing unit of capacitance from F to pF\n",
-      "\n",
-      "#result\n",
-      "print\"The capacitance of capacitor is C=\",'%.3E'%C,\"C\",\n",
-      "print\"\\n\\t\\t\\t         =\",round(C1,2),\"pF\";\n",
-      "print\"The resultant voltage across the capacitor is V=\",round(V,2),\"V\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The capacitance of capacitor is C= 5.033E-12 C \n",
-        "\t\t\t         = 5.03 pF\n",
-        "The resultant voltage across the capacitor is V= 39.73 V\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3 , Page no:188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=2.7E25; #in 1/m^3 (density of atoms)\n",
-      "E=1E6; #in V/m (electric field)\n",
-      "Z=2; #atomic number of Helium \n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "Er=1.0000684; #(dielectric constant of the material)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "#since  alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 \n",
-      "#Therefore we have r_0^3=(Er-1)/(4*pi*N)\n",
-      "r_0=((Er-1)/(4*pi*N))**(1/3); #calculation of radius of electron cloud\n",
-      "x=4*pi*Eo*E*r_0/(Z*e); #calculation of dispalcement\n",
-      "\n",
-      "#result\n",
-      "print\"The radius of electron cloud is r_0=\",'%.3E'%r_0,\"m\";\n",
-      "print\"The displacement is x=\",'%.3E'%x,\"m\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The radius of electron cloud is r_0= 5.865E-11 m\n",
-        "The displacement is x= 2.037E+04 m\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4 , Page no:188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "K=1.000134; #di-elecrtic constant of the neon gas at NTP\n",
-      "E=90000; #in V/m (electric field)\n",
-      "Eo=8.85E-12; #in C/N-m^2 (absolute premittivity)\n",
-      "N_A=6.023E26; #in atoms/Kg-mole (Avogadro's number)\n",
-      "V=22.4; #in m^3 (volume of gas at NTP\n",
-      "\n",
-      "#calculate\n",
-      "n=N_A/V; #calculaton of density of atoms\n",
-      "#Since P=n*p=(k-1)*Eo*E\n",
-      "#therefore we have\n",
-      "p=(K-1)*Eo*E/n; #calculation of dipole moment induced\n",
-      "alpha=p/E; #calculation of atomic polarisability\n",
-      "\n",
-      "#result\n",
-      "print\"The dipole moment induced in each atom is p=\",'%.3E'%p,\"C-m\";\n",
-      "print\"The atomic polarisability of neon is =\",'%.3E'%alpha,\"c-m^2/V\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dipole moment induced in each atom is p= 3.969E-36 C-m\n",
-        "The atomic polarisability of neon is = 4.410E-41 c-m^2/V\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5 , Page no:189"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Er=3.75; #di-elecrtic constant of sulphur at 27 degree Celcius\n",
-      "gama=1/3; #internal field constant\n",
-      "p=2050; #in Kg/m^3 (density)\n",
-      "M_A=32; #in amu (atomic weight of sulphur)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "N=6.022E23; #Avogadro's number\n",
-      "\n",
-      "#calculate\n",
-      "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
-      "#therefore we have\n",
-      "alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); #calculation of electronic polarisability of sulphur\n",
-      "\n",
-      "#result\n",
-      "print\"The electronic polarisability of sulphur is =\",'%.3E'%alpha_e,\"Fm^2   (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electronic polarisability of sulphur is = 3.291E-37 Fm^2   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.6 , Page no:189"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Er=1.0000684; #di-elecrtic constant of Helium gas at NTP\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "N=2.7E25; #number of atomsper unit volume\n",
-      "\n",
-      "#calculate\n",
-      "#Since Er-1=(N/Eo)*alpha_e\n",
-      "#therefore we have\n",
-      "alpha_e=Eo*(Er-1)/N; #calculation of electronic polarisability of Helium\n",
-      "\n",
-      "#result\n",
-      "print\"The electronic polarisability of Helium gas is =\",'%.3E'%alpha_e,\"Fm^2    (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electronic polarisability of Helium gas is = 2.242E-41 Fm^2    (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7 , Page no:190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "N=3E28; #in atoms/m^3 (density of atoms)\n",
-      "alpha_e=1E-40; #in F-m^2 (electronic polarisability)\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "\n",
-      "#calculate\n",
-      "#Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo)\n",
-      "#therefore we have\n",
-      "Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo)));\n",
-      "#calculation of dielectric constant of the material\n",
-      "\n",
-      "#result\n",
-      "print\"The dielectric constant of the material is Er=\",'%.3E'%Er,\"F/m\",\n",
-      "print \"NOTE: The answer in the book is wrong \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dielectric constant of the material is Er= 1.382E+00 F/m NOTE: The answer in the book is wrong \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.8 , Page no:190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Er=4; #relative permittivity of sulphur\n",
-      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
-      "NA=2.08E3; #in Kg/m^3 (density of atoms in sulphur)\n",
-      "\n",
-      "#calculate\n",
-      "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
-      "#therefore we have\n",
-      "alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); #calculation of electronic polarisability of sulphur\n",
-      "\n",
-      "#result\n",
-      "print\"The electronic polarisability of sulphur is =\",'%.3E'%alpha_e,\"Fm^2\";\n",
-      "print \"NOTE: The answer in the book is wrong and they used the wrong formula \" \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electronic polarisability of sulphur is = 6.382E-15 Fm^2\n",
-        "NOTE: The answer in the book is wrong and they used the wrong formula \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.9 , Page no:190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "alpha1=2.5E-39; #in C^2-m/N (dielectric constant at 300K)\n",
-      "alpha2=2.0E-39; #in C^2-m/N (dielectric constant at 400K)\n",
-      "T1=300; #in K(first temperature)\n",
-      "T2=400; #in K(second temperature)\n",
-      "\n",
-      "#calculate\n",
-      "#since alpha=alpha_d+alpha0 and alpha0=Beta/T\n",
-      "#therefore alpha=alpha_d+(Beta/T)\n",
-      "#since alpha1=alpha_d+(Beta/T1)  and alpha2=alpha_d+(Beta/T2)\n",
-      "#therefore alpha1-apha2=Beta*((1/T1)-(1/T2))\n",
-      "#or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2))\n",
-      "Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); #calculation of Beta\n",
-      "alpha_d=alpha1-(Beta/T1); #calculation of polarisability due to defromation\n",
-      "alpha0_1=Beta/T1; #calculation of polarisability due to permanent dipole moment at 300K\n",
-      "alpha0_2=Beta/T2;  #calculation of polarisability due to permanent dipole moment at 400K\n",
-      "\n",
-      "#result\n",
-      "print\"The polarisability due to permanent dipole moment at 300K is =\",'%.3E'%alpha0_1,\"C^2-m/N\";\n",
-      "print\"The polarisability due to permanent dipole moment at 400K is =\",'%.3E'%alpha0_2,\"C^2-m/N\";\n",
-      "print\"The polarisability due to deformation of the molecules is =\",'%.3E'%alpha_d,\"C^2-m/N\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The polarisability due to permanent dipole moment at 300K is = 2.000E-39 C^2-m/N\n",
-        "The polarisability due to permanent dipole moment at 400K is = 1.500E-39 C^2-m/N\n",
-        "The polarisability due to deformation of the molecules is = 5.000E-40 C^2-m/N\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.10 , Page no:191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "n=1.5; #refractive index\n",
-      "Er=5.6; #dielectric constant\n",
-      "\n",
-      "#calculate\n",
-      "#since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0)   Clausius-Mossotti equation\n",
-      "#and  (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) \n",
-      "#from above two equations, we get  ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i)\n",
-      "#or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P)\n",
-      "#where P is fractional ionisational polarisability\n",
-      "P=1-((n**2-1)/(n**2+2))*((Er+2)/(Er-1)); #calculation of fractional ionisational polarisability\n",
-      "P1=P*100; #calculation of percentage of  ionisational polarisability\n",
-      "\n",
-      "#result\n",
-      "print\"The percentage of  ionisational polarisability is =\",round(P1,1),\"percent\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of  ionisational polarisability is = 51.4 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_08_Magnetic_Properties.ipynb b/Applied_Physics/Chapter_08_Magnetic_Properties.ipynb
deleted file mode 100755
index 353fc099..00000000
--- a/Applied_Physics/Chapter_08_Magnetic_Properties.ipynb
+++ /dev/null
@@ -1,323 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:aecad4e644a378cab0352aa29e9045a53c52646739e4f4bd6cdff80f1d0741e6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8:Magnetic Properties"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "X=-0.5E-5; #magnetic susceptibility of silicon\n",
-      "H=0.9E4; #in A/m (magnetic field intensity)\n",
-      "mu0=4*3.14*1E-7; #in H/m (absolute permeability)\n",
-      "\n",
-      "#calculate\n",
-      "I=X*H; #calculation of intensity of magnetism\n",
-      "B=mu0*H*(1+X); #calculation of magnetic flux density\n",
-      "\n",
-      "#result\n",
-      "print\"The intensity of magnetism is I=\",I,\"A/m\";\n",
-      "print\"The magnetic flux density is B=\",round(B,3),\"Wb/m^2\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intensity of magnetism is I= -0.045 A/m\n",
-        "The magnetic flux density is B= 0.011 Wb/m^2\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r=0.052; #in nm (radius of orbit)\n",
-      "B=1; #in Wb/m^2 (magnetic field of induction)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "\n",
-      "#calculate\n",
-      "r=0.052*1E-9; #changing unit from nm to m\n",
-      "d_mu=(e**2*r**2*B)/(4*m); #calculation of change in magnetic moment\n",
-      "\n",
-      "#result\n",
-      "print\"The change in magnetic moment is =\",d_mu,\"Am^2\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The change in magnetic moment is = 1.90171428571e-29 Am^2\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "H=220; #in A/m (magnetic field intensity)\n",
-      "I=3300; #in A/m (intensity of magnetisation)\n",
-      "\n",
-      "#calculate\n",
-      "mu_r=1+(I/H); #calculation of relative permeability\n",
-      "\n",
-      "#result\n",
-      "print\"The relative permeability of a ferromagentic material is =\",mu_r;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relative permeability of a ferromagentic material is = 16.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "I=3000; #in A/m (intensity of magnetisation)\n",
-      "B=0.005; #in Wb/m^2 (magnetic flus intensity)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
-      "\n",
-      "#calculate\n",
-      "H=(B/mu0)-I; #calculation of magnetic force\n",
-      "mu_r=(I/H)+1; #calculation of relative permeability\n",
-      "\n",
-      "#result\n",
-      "print\"The magnetic force is  H=\",H;\n",
-      "print\"The relative permeability is =\",mu_r;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnetic force is  H= 980.891719745\n",
-        "The relative permeability is = 4.05844155844\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5 , Page no:237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "H=4E3; #in A/m (magnetic field intensity)\n",
-      "N=60; #number of turns\n",
-      "l=12; #in cm (length of solenoid)\n",
-      "\n",
-      "#calculate\n",
-      "n=N/(l*1E-2); #calculation of number of turns per unit metre\n",
-      "#Snice H=n*i;\n",
-      "i=H/n; #calculation of current through the solenoid\n",
-      "\n",
-      "#result\n",
-      "print\"The current through the solenoid is i=\",i,\"A\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the solenoid is i= 8.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6 , Page no:237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "l=30; #in cm (length of solenoid)\n",
-      "A=1; #in cm^2 (cross-sectional area)\n",
-      "N=300; #number of turns\n",
-      "i=0.032; #in A (current through the winding)\n",
-      "phi_B=2E-6; #in Wb (magnetic flux)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
-      "\n",
-      "#calculate\n",
-      "l=l*1E-2; #changing unit from cm to m\n",
-      "A=A*1E-4; #changing unit from cm^2 to m^2\n",
-      "B=phi_B/A; #calculation of flux density\n",
-      "H=N*i/l; #calculation of magnetic intensity\n",
-      "mu=B/H; #calcluation of absolute permeability of iron\n",
-      "mu_r=mu/mu0; #calcluation of relative permeability of iron\n",
-      "\n",
-      "#result\n",
-      "print\"The flux density is B=\",B,\"Wb/m^2\";\n",
-      "print\"The magnetic intensity is H=\",H,\"A-turns/m\";\n",
-      "print\"The relative permeability of iron is =\",round(mu_r),\"    (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The flux density is B= 0.02 Wb/m^2\n",
-        "The magnetic intensity is H= 32.0 A-turns/m\n",
-        "The relative permeability of iron is = 498.0     (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7 , Page no:238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "A=100; #in m^2 (area of Hysteresis loop)\n",
-      "B=0.01; #in Wb/m^2 (unit space along vertical axis or magnetic flux density)\n",
-      "H=40; #in A/m (unit space along horizontal axis or magnetic fild ntensity)\n",
-      "\n",
-      "#calculate\n",
-      "H_L=A*B*H; #calculation of magnetic intensity\n",
-      "\n",
-      "#result\n",
-      "print\"The Hystersis loss per cycle is =\",round(H_L),\"J/m^2\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Hystersis loss per cycle is = 40.0 J/m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_08_Magnetic_Properties_1.ipynb b/Applied_Physics/Chapter_08_Magnetic_Properties_1.ipynb
deleted file mode 100755
index b2a8def3..00000000
--- a/Applied_Physics/Chapter_08_Magnetic_Properties_1.ipynb
+++ /dev/null
@@ -1,323 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:708d3f2a9798f4d28f6ce90536d50509c591ba2ba31051121346b4739b2aea82"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8:Magnetic Properties"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "X=-0.5E-5; #magnetic susceptibility of silicon\n",
-      "H=0.9E4; #in A/m (magnetic field intensity)\n",
-      "mu0=4*3.14*1E-7; #in H/m (absolute permeability)\n",
-      "\n",
-      "#calculate\n",
-      "I=X*H; #calculation of intensity of magnetism\n",
-      "B=mu0*H*(1+X); #calculation of magnetic flux density\n",
-      "\n",
-      "#result\n",
-      "print\"The intensity of magnetism is I=\",I,\"A/m\";\n",
-      "print\"The magnetic flux density is B=\",round(B,3),\"Wb/m^2\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intensity of magnetism is I= -0.045 A/m\n",
-        "The magnetic flux density is B= 0.011 Wb/m^2\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r=0.052; #in nm (radius of orbit)\n",
-      "B=1; #in Wb/m^2 (magnetic field of induction)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "\n",
-      "#calculate\n",
-      "r=0.052*1E-9; #changing unit from nm to m\n",
-      "d_mu=(e**2*r**2*B)/(4*m); #calculation of change in magnetic moment\n",
-      "\n",
-      "#result\n",
-      "print\"The change in magnetic moment is =\",'%.3E'%d_mu,\"Am^2\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The change in magnetic moment is = 1.902E-29 Am^2\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "H=220; #in A/m (magnetic field intensity)\n",
-      "I=3300; #in A/m (intensity of magnetisation)\n",
-      "\n",
-      "#calculate\n",
-      "mu_r=1+(I/H); #calculation of relative permeability\n",
-      "\n",
-      "#result\n",
-      "print\"The relative permeability of a ferromagentic material is =\",mu_r;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relative permeability of a ferromagentic material is = 16.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4 , Page no:236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "I=3000; #in A/m (intensity of magnetisation)\n",
-      "B=0.005; #in Wb/m^2 (magnetic flus intensity)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
-      "\n",
-      "#calculate\n",
-      "H=(B/mu0)-I; #calculation of magnetic force\n",
-      "mu_r=(I/H)+1; #calculation of relative permeability\n",
-      "\n",
-      "#result\n",
-      "print\"The magnetic force is  H=\",round(H,3);\n",
-      "print\"The relative permeability is =\",round(mu_r,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnetic force is  H= 980.892\n",
-        "The relative permeability is = 4.058\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5 , Page no:237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "H=4E3; #in A/m (magnetic field intensity)\n",
-      "N=60; #number of turns\n",
-      "l=12; #in cm (length of solenoid)\n",
-      "\n",
-      "#calculate\n",
-      "n=N/(l*1E-2); #calculation of number of turns per unit metre\n",
-      "#Snice H=n*i;\n",
-      "i=H/n; #calculation of current through the solenoid\n",
-      "\n",
-      "#result\n",
-      "print\"The current through the solenoid is i=\",i,\"A\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the solenoid is i= 8.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6 , Page no:237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "l=30; #in cm (length of solenoid)\n",
-      "A=1; #in cm^2 (cross-sectional area)\n",
-      "N=300; #number of turns\n",
-      "i=0.032; #in A (current through the winding)\n",
-      "phi_B=2E-6; #in Wb (magnetic flux)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
-      "\n",
-      "#calculate\n",
-      "l=l*1E-2; #changing unit from cm to m\n",
-      "A=A*1E-4; #changing unit from cm^2 to m^2\n",
-      "B=phi_B/A; #calculation of flux density\n",
-      "H=N*i/l; #calculation of magnetic intensity\n",
-      "mu=B/H; #calcluation of absolute permeability of iron\n",
-      "mu_r=mu/mu0; #calcluation of relative permeability of iron\n",
-      "\n",
-      "#result\n",
-      "print\"The flux density is B=\",B,\"Wb/m^2\";\n",
-      "print\"The magnetic intensity is H=\",H,\"A-turns/m\";\n",
-      "print\"The relative permeability of iron is =\",round(mu_r),\"    (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The flux density is B= 0.02 Wb/m^2\n",
-        "The magnetic intensity is H= 32.0 A-turns/m\n",
-        "The relative permeability of iron is = 498.0     (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7 , Page no:238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "A=100; #in m^2 (area of Hysteresis loop)\n",
-      "B=0.01; #in Wb/m^2 (unit space along vertical axis or magnetic flux density)\n",
-      "H=40; #in A/m (unit space along horizontal axis or magnetic fild ntensity)\n",
-      "\n",
-      "#calculate\n",
-      "H_L=A*B*H; #calculation of magnetic intensity\n",
-      "\n",
-      "#result\n",
-      "print\"The Hystersis loss per cycle is =\",round(H_L),\"J/m^2\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Hystersis loss per cycle is = 40.0 J/m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_09_Semiconductors.ipynb b/Applied_Physics/Chapter_09_Semiconductors.ipynb
deleted file mode 100755
index f3c26825..00000000
--- a/Applied_Physics/Chapter_09_Semiconductors.ipynb
+++ /dev/null
@@ -1,782 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:eb4f284d8d12c103b674cc454947abe9c7a1a76e2d11836f2841ba55d4adeef7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9:Semiconductors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2 , Page no:272"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Eg=0.67; #in eV (Energy band gap)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=298; #in K (room temperature)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "K=10; #ratio of number of electrons at different temperature\n",
-      "\n",
-      "#calculate\n",
-      "Eg=Eg*e; #changing unit from eV to Joule\n",
-      "#since ne=Ke*exp(-Eg/(2*k*T))\n",
-      "#and ne/ne1=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) and ne/ne1=K=10\n",
-      "#therefore we have 10=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1))\n",
-      "#re-arranging the equation for T, we get T2=1/((1/T1)-((2*k*log(10))/Eg))\n",
-      "T=1/((1/T1)-((2*k*math.log(10))/Eg)); #calculation of the temperature\n",
-      "\n",
-      "#result\n",
-      "print\"The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T=\",round(T),\"K            (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T= 362.0 K            (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3 , Page no:272"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=2.5E13; #in /cm^3 (intrinsic carrier density)\n",
-      "ue=3900; #in cm^2/(V-s) (electron mobilities)\n",
-      "uh=1900; #in cm^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "l=1; #in cm (lenght of the box)\n",
-      "b=1;h=1; #in mm (dimensions of germanium rod )\n",
-      "\n",
-      "#calculate\n",
-      "ni=ni*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
-      "ue=ue*1E-4; #changing unit from cm^2 to m^2\n",
-      "uh=uh*1E-4; #changing unit from cm^2 to m^2\n",
-      "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "l=l*1E-2; #changing unit from mm to m for length\n",
-      "A=(b*1E-3)*(h*1E-3); #changing unit from mm to m for width and height and calculation of cross-sectional area\n",
-      "R=rho*l/A; #calculation of resistance\n",
-      "\n",
-      "#result\n",
-      "print\"The resistance of intrinsic germanium is R=\",R,\"ohm\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance of intrinsic germanium is R= 4310.34482759 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4 , Page no:273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ne=2.5E19; #in /m^3 (electron  density)\n",
-      "nh=2.5E19; #in /m^3 (hole density)\n",
-      "ue=0.36; #in m^2/(V-s) (electron mobilities)\n",
-      "uh=0.17; #in m^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#since ne=nh=ni,  therefore we have \n",
-      "ni=nh;\n",
-      "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "\n",
-      "#result\n",
-      "print\"The conductivity of germanium is =\",round(sigma,2),\"/ohm-m\";\n",
-      "print\"The resistivity of germanium is =\",round(rho,2),\"ohm-m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of germanium is = 2.12 /ohm-m\n",
-        "The resistivity of germanium is = 0.47 ohm-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5 , Page no:273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
-      "ue=0.135; #in m^2/(V-s) (electron mobilities)\n",
-      "uh=0.048; #in m^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "ND=1E23; #in atom/m^3 (doping concentration)\n",
-      "\n",
-      "#calculate\n",
-      "sigma_i=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
-      "sigma=ND*ue*e; #calculation of conductivity after doping\n",
-      "rho=ni**2/ND; #calculation of equilibrium hole concentration\n",
-      "\n",
-      "#result\n",
-      "print\"The intrinsic conductivity for silicon is =\",sigma_i,\"S\";\n",
-      "print\"The conductivity after doping with phosphorus atoms is =\",sigma,\"S\";\n",
-      "print\"The equilibrium hole concentration is =\",rho,\"/m^3\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intrinsic conductivity for silicon is = 0.0004392 S\n",
-        "The conductivity after doping with phosphorus atoms is = 2160.0 S\n",
-        "The equilibrium hole concentration is = 2250000000.0 /m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6 , Page no:274"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
-      "ue=0.13; #in m^2/(V-s) (electron mobilities)\n",
-      "uh=0.05; #in m^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "ne=5E20; #in /m^3 (concentration of donor type impurity)\n",
-      "nh=5E20; #in /m^3 (concentration of acceptor type impurity)\n",
-      "\n",
-      "#calculate\n",
-      "#part-i\n",
-      "sigma=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
-      "#part-ii\n",
-      "#since 1 donor atom is in 1E8 Si atoms, hence holes concentration can be neglected\n",
-      "sigma1=ne*e*ue; #calculation of conductivity after doping with donor type impurity\n",
-      "#part-iii\n",
-      "#since 1 acceptor atom is in 1E8 Si atoms, hence electron concentration can be neglected\n",
-      "sigma2=nh*e*uh; #calculation of conductivity after doping with acceptor type impurity\n",
-      "\n",
-      "#result\n",
-      "print\"The intrinsic conductivity for silicon is =\",sigma,\"(ohm-m)^-1\";\n",
-      "print\"The conductivity after doping with donor type impurity is =\",sigma1,\"(ohm-m)^-1\";\n",
-      "print\"The conductivity after doping with acceptor type impurity is =\",sigma2,\"(ohm-m)^-1\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intrinsic conductivity for silicon is = 0.000432 (ohm-m)^-1\n",
-        "The conductivity after doping with donor type impurity is = 10.4 (ohm-m)^-1\n",
-        "The conductivity after doping with acceptor type impurity is = 4.0 (ohm-m)^-1\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7 , Page no:274"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=1E20; #in /m^3 (intrinsic carrier density)\n",
-      "ND=1E21; #in /m^3 (donor impurity concentration)\n",
-      "\n",
-      "#calculate\n",
-      "nh=ni**2/ND; #calculation of density of hole carriers at room temperature\n",
-      "\n",
-      "#result\n",
-      "print\"The density of hole carriers at room temperature is nh=\",nh,\"/m^3\";\n",
-      "#Note: answer in the book is wrong due to printing mistake"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density of hole carriers at room temperature is nh= 1e+19 /m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8 , Page no:275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "M=72.6; #atomic mass of germanium\n",
-      "P=5400; #in Kg/m^3 (density)\n",
-      "ue=0.4; #in m^2/V-s (mobility of electrons)\n",
-      "uh=0.2; #in m^2/V-s (mobility of holes)\n",
-      "Eg=0.7; #in eV (Band gap)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T=300; #in K (temperature)\n",
-      "h=6.63E-34; #in J/s (Planck\u2019s constant)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6E-19; #in C(charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "Eg=Eg*e; #changing unit from eV to J\n",
-      "ni=2*(2*pi*m*k*T/h**2)**(3/2)*math.exp(-Eg/(2*k*T));\n",
-      "sigma=ni*e*(ue+uh);\n",
-      "\n",
-      "#result\n",
-      "print\"The intrinsic carrier density for germanium at 300K is ni=\",ni,\"/m^3\";\n",
-      "print\"The conductivity of germanium is=\",round(sigma,3),\"(ohm-m)^-1\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intrinsic carrier density for germanium at 300K is ni= 3.33408559508e+19 /m^3\n",
-        "The conductivity of germanium is= 3.201 (ohm-m)^-1\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9 , Page no:275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "rho1=4.5; #in ohm-m (resistivity at 20 degree Celcius)\n",
-      "rho2=2.0; #in ohm-m (resistivity at 32 degree Celcius)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=20; T2=32; #in degree Celcius (two temperatures)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "T1=T1+273; #changing unit from degree Celius to K\n",
-      "T2=T2+273; #changing unit from degree Celius to K\n",
-      "#since sigma=e*u*C*T^(3/2)*exp(-Eg/(2*k*T))\n",
-      "#therefore sigma1/sigma2=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
-      "#and sigma=1/rho \n",
-      "#therefore we have rho2/rho1=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
-      "#re-arranging above equation for Eg, we get Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1))\n",
-      "Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*math.log(T1/T2)-math.log(rho2/rho1));\n",
-      "Eg1=Eg/e;#changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The energy band gap is Eg=\",Eg,\"J\";\n",
-      "print\"\\t\\t\\t =\",round(Eg1,3),\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy band gap is Eg= 1.54302914521e-19 J\n",
-        "\t\t\t = 0.964 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10 , Page no:276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "rho=2300; #in ohm-m (resistivity of pure silicon)\n",
-      "ue=0.135; #in m^2/V-s (mobility of electron)\n",
-      "uh=0.048; #in m^2/V-s (mobility of electron)\n",
-      "Nd=1E19; #in /m^3 (doping concentration)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#since sigma=ni*e*(ue+uh) and sigma=1/rho\n",
-      "#therefore ni=1/(rho*e*(ue+uh))\n",
-      "ni=1/(rho*e*(ue+uh)); #calculation of intrinsic concentration\n",
-      "ne=Nd; #calculation of electron concentration\n",
-      "nh=ni**2/Nd; #calculation of hole concentration\n",
-      "sigma=ne*ue*e+nh*uh*e; #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "\n",
-      "#result\n",
-      "print\"The electron concentration is ne=\",ne,\"/m^3\";\n",
-      "print\"The hole concentration is nh=\",nh,\"/m^3\";\n",
-      "print\"The resistivity of the specimen  is =\",round(rho,3),\"ohm-m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electron concentration is ne= 1e+19 /m^3\n",
-        "The hole concentration is nh= 2.20496745228e+13 /m^3\n",
-        "The resistivity of the specimen  is = 4.63 ohm-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.11 , Page no:276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "uh=1900; #in cm^2/V-s (mobility of electron)\n",
-      "Na=2E17; #in /m^3 (acceptor doping concentration)\n",
-      "e=1.6E-19; #in C(charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "uh=uh*1E-4; #changing unit from cm^2/V-s to m^2/V-s\n",
-      "Na=Na*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
-      "nh=Na; #hole concentration \n",
-      "#since sigma=ne*ue*e+nh*uh*e and nh>>ne\n",
-      "#therefore sigma=nh*uh*e\n",
-      "sigma=nh*uh*e; #calculation of conductivity\n",
-      "\n",
-      "#result\n",
-      "print\"The conductivity of p-type Ge crystal is =\",sigma,\"/ohm-m   (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of p-type Ge crystal is = 6080.0 /ohm-m   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12 , Page no:277"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ue=0.19; #in m^2/V-s (mobility of electron)\n",
-      "T=300; #in K (temperature)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "e=1.6E-19; #in C(charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "Dn=ue*k*T/e; #calculation of diffusion co-efficient\n",
-      "\n",
-      "#result\n",
-      "print\"The diffusion co-efficient of electron in silicon is Dn=\",Dn,\"m^2/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diffusion co-efficient of electron in silicon is Dn= 0.00491625 m^2/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13 , Page no:277"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Eg=0.4; #in eV (Band gap of semiconductor)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=0; #in degree Celcius (first temperature)\n",
-      "T2=50; #in degree Celcius (second temperature)\n",
-      "T3=100; #in degree Celcius (third temperature)\n",
-      "e=1.602E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "T1=T1+273; #changing temperature form Celcius to Kelvin\n",
-      "T2=T2+273; #changing temperature form Celcius to Kelvin\n",
-      "T3=T3+273; #changing temperature form Celcius to Kelvin\n",
-      "Eg=Eg*e; #changing unit from eV to Joule\n",
-      "#Using F_E=1/(1+exp(Eg/2*k*T))\n",
-      "F_E1=1/(1+math.exp(Eg/(2*k*T1))); #calculation of probability of occupation of lowest level at 0 degree Celcius\n",
-      "F_E2=1/(1+math.exp(Eg/(2*k*T2))); #calculation of probability of occupation of lowest level at 50 degree Celcius\n",
-      "F_E3=1/(1+math.exp(Eg/(2*k*T3))); #calculation of probability of occupation of lowest level at 100 degree Celcius\n",
-      "\n",
-      "#result\n",
-      "print\"The probability of occupation of lowest level in conduction band is\";\n",
-      "print\"\\t at 0 degree Celcius, F_E=\",F_E1,\"eV\";\n",
-      "print\"\\t at 50 degree Celcius, F_E=\",F_E2,\"eV\";\n",
-      "print\"\\t at 100 degree Celcius, F_E=\",F_E3,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The probability of occupation of lowest level in conduction band is\n",
-        "\t at 0 degree Celcius, F_E= 0.000202505914236 eV\n",
-        "\t at 50 degree Celcius, F_E= 0.000754992968827 eV\n",
-        "\t at 100 degree Celcius, F_E= 0.00197639649915 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14 , Page no:278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Eg=1.2; #in eV (Energy band gap)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=600; T2=300; #in K (two temperatures)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "Eg=Eg*e; #changing unit from eV to Joule\n",
-      "#since sigma is proportional to exp(-Eg/(2*k*T))\n",
-      "#therefore ratio=sigma1/sigma2=exp(-Eg/(2*k*((1/T1)-(1/T2))));\n",
-      "ratio= math.exp((-Eg/(2*k))*((1/T1)-(1/T2))); #calculation of ratio of conductivity at 600K and at 300K\n",
-      "\n",
-      "#result\n",
-      "print\"The ratio of conductivity at 600K and at 300K is =\",ratio;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio of conductivity at 600K and at 300K is = 108467.17792\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15 , Page no:278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ue=0.39; #in m^2/V-s (mobility of electron)\n",
-      "n=5E13; #number of donor atoms\n",
-      "ni=2.4E19; #in atoms/m^3 (intrinsic carrier density)\n",
-      "l=10; #in mm (length of rod)\n",
-      "a=1; #in mm (side of square cross-section)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "l=l*1E-3; #changing unit from mm to m\n",
-      "a=a*1E-3; #changing unit from mm to m\n",
-      "A=a**2; #calculation of cross-section area\n",
-      "Nd=n/(l*A); #calculation of donor concentration\n",
-      "ne=Nd; #calculation of electron density\n",
-      "nh=ni**2/Nd; #calculation of hole density\n",
-      "#since sigma=ne*e*ue+nh*e*ue and since ne>>nh\n",
-      "#therefore sigma=ne*e*ue\n",
-      "sigma=ne*e*ue; #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "R=rho*l/A; #calculation of resistance \n",
-      "\n",
-      "#result\n",
-      "print\"The electron density is ne=\",ne,\"/m^3\";\n",
-      "print\"The hole density is nh=\",nh,\"/m^3\";\n",
-      "print\"The conductivity is =\",sigma,\"/ohm-m\";\n",
-      "print\"The resistance is R=\",round(R),\"ohm\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electron density is ne= 5e+21 /m^3\n",
-        "The hole density is nh= 1.152e+17 /m^3\n",
-        "The conductivity is = 312.0 /ohm-m\n",
-        "The resistance is R= 32.0 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 , Page no:279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
-      "rho=8.93E-3; #in ohm-m (resistivity)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "u=RH/rho; #calculation of mobility\n",
-      "n=1/(RH*e); #calculation of density\n",
-      "\n",
-      "#result\n",
-      "print\"The mobility is u=\",u,\"m^2/(V-s)\";\n",
-      "print\"The density is n=\",n,\"/m^3\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mobility is u= 0.0409854423292 m^2/(V-s)\n",
-        "The density is n= 1.70765027322e+22 /m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 , Page no:279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
-      "rho=8.93E-3; #in ohm-m (resistivity)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "nh=1/(RH*e); #calculation of density of charge carrier\n",
-      "uh=1/(rho*nh*e); #calculation of mobility of charge carrier\n",
-      "\n",
-      "#result\n",
-      "print\"The density of charge carrier is nh=\",nh,\"/m^3\";\n",
-      "print\"The mobility of charge carrier is uh=\",uh,\"m^2/(V-s)\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density of charge carrier is nh= 1.70765027322e+22 /m^3\n",
-        "The mobility of charge carrier is uh= 0.0409854423292 m^2/(V-s)\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_09_Semiconductors_1.ipynb b/Applied_Physics/Chapter_09_Semiconductors_1.ipynb
deleted file mode 100755
index 9ce2ea83..00000000
--- a/Applied_Physics/Chapter_09_Semiconductors_1.ipynb
+++ /dev/null
@@ -1,782 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c155cecc785c23bdccd0bf715c04d53d7788e1b590b4310d6af27c13ce7b97ac"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9:Semiconductors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2 , Page no:272"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Eg=0.67; #in eV (Energy band gap)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=298; #in K (room temperature)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "K=10; #ratio of number of electrons at different temperature\n",
-      "\n",
-      "#calculate\n",
-      "Eg=Eg*e; #changing unit from eV to Joule\n",
-      "#since ne=Ke*exp(-Eg/(2*k*T))\n",
-      "#and ne/ne1=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) and ne/ne1=K=10\n",
-      "#therefore we have 10=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1))\n",
-      "#re-arranging the equation for T, we get T2=1/((1/T1)-((2*k*log(10))/Eg))\n",
-      "T=1/((1/T1)-((2*k*math.log(10))/Eg)); #calculation of the temperature\n",
-      "\n",
-      "#result\n",
-      "print\"The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T=\",round(T),\"K            (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T= 362.0 K            (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3 , Page no:272"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=2.5E13; #in /cm^3 (intrinsic carrier density)\n",
-      "ue=3900; #in cm^2/(V-s) (electron mobilities)\n",
-      "uh=1900; #in cm^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "l=1; #in cm (lenght of the box)\n",
-      "b=1;h=1; #in mm (dimensions of germanium rod )\n",
-      "\n",
-      "#calculate\n",
-      "ni=ni*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
-      "ue=ue*1E-4; #changing unit from cm^2 to m^2\n",
-      "uh=uh*1E-4; #changing unit from cm^2 to m^2\n",
-      "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "l=l*1E-2; #changing unit from mm to m for length\n",
-      "A=(b*1E-3)*(h*1E-3); #changing unit from mm to m for width and height and calculation of cross-sectional area\n",
-      "R=rho*l/A; #calculation of resistance\n",
-      "\n",
-      "#result\n",
-      "print\"The resistance of intrinsic germanium is R=\",'%.3E'%R,\"ohm\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance of intrinsic germanium is R= 4.310E+03 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4 , Page no:273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ne=2.5E19; #in /m^3 (electron  density)\n",
-      "nh=2.5E19; #in /m^3 (hole density)\n",
-      "ue=0.36; #in m^2/(V-s) (electron mobilities)\n",
-      "uh=0.17; #in m^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#since ne=nh=ni,  therefore we have \n",
-      "ni=nh;\n",
-      "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "\n",
-      "#result\n",
-      "print\"The conductivity of germanium is =\",round(sigma,2),\"/ohm-m\";\n",
-      "print\"The resistivity of germanium is =\",round(rho,2),\"ohm-m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of germanium is = 2.12 /ohm-m\n",
-        "The resistivity of germanium is = 0.47 ohm-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5 , Page no:273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
-      "ue=0.135; #in m^2/(V-s) (electron mobilities)\n",
-      "uh=0.048; #in m^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "ND=1E23; #in atom/m^3 (doping concentration)\n",
-      "\n",
-      "#calculate\n",
-      "sigma_i=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
-      "sigma=ND*ue*e; #calculation of conductivity after doping\n",
-      "rho=ni**2/ND; #calculation of equilibrium hole concentration\n",
-      "\n",
-      "#result\n",
-      "print\"The intrinsic conductivity for silicon is =\",'%.3E'%sigma_i,\"S\";\n",
-      "print\"The conductivity after doping with phosphorus atoms is =\",'%.3E'%sigma,\"S\";\n",
-      "print\"The equilibrium hole concentration is =\",'%.3E'%rho,\"/m^3\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intrinsic conductivity for silicon is = 4.392E-04 S\n",
-        "The conductivity after doping with phosphorus atoms is = 2.160E+03 S\n",
-        "The equilibrium hole concentration is = 2.250E+09 /m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6 , Page no:274"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
-      "ue=0.13; #in m^2/(V-s) (electron mobilities)\n",
-      "uh=0.05; #in m^2/(V-s) (hole mobilities)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "ne=5E20; #in /m^3 (concentration of donor type impurity)\n",
-      "nh=5E20; #in /m^3 (concentration of acceptor type impurity)\n",
-      "\n",
-      "#calculate\n",
-      "#part-i\n",
-      "sigma=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
-      "#part-ii\n",
-      "#since 1 donor atom is in 1E8 Si atoms, hence holes concentration can be neglected\n",
-      "sigma1=ne*e*ue; #calculation of conductivity after doping with donor type impurity\n",
-      "#part-iii\n",
-      "#since 1 acceptor atom is in 1E8 Si atoms, hence electron concentration can be neglected\n",
-      "sigma2=nh*e*uh; #calculation of conductivity after doping with acceptor type impurity\n",
-      "\n",
-      "#result\n",
-      "print\"The intrinsic conductivity for silicon is =\",'%.3E'%sigma,\"(ohm-m)^-1\";\n",
-      "print\"The conductivity after doping with donor type impurity is =\",sigma1,\"(ohm-m)^-1\";\n",
-      "print\"The conductivity after doping with acceptor type impurity is =\",sigma2,\"(ohm-m)^-1\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intrinsic conductivity for silicon is = 4.320E-04 (ohm-m)^-1\n",
-        "The conductivity after doping with donor type impurity is = 10.4 (ohm-m)^-1\n",
-        "The conductivity after doping with acceptor type impurity is = 4.0 (ohm-m)^-1\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7 , Page no:274"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ni=1E20; #in /m^3 (intrinsic carrier density)\n",
-      "ND=1E21; #in /m^3 (donor impurity concentration)\n",
-      "\n",
-      "#calculate\n",
-      "nh=ni**2/ND; #calculation of density of hole carriers at room temperature\n",
-      "\n",
-      "#result\n",
-      "print\"The density of hole carriers at room temperature is nh=\",nh,\"/m^3\";\n",
-      "#Note: answer in the book is wrong due to printing mistake"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density of hole carriers at room temperature is nh= 1e+19 /m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8 , Page no:275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "M=72.6; #atomic mass of germanium\n",
-      "P=5400; #in Kg/m^3 (density)\n",
-      "ue=0.4; #in m^2/V-s (mobility of electrons)\n",
-      "uh=0.2; #in m^2/V-s (mobility of holes)\n",
-      "Eg=0.7; #in eV (Band gap)\n",
-      "m=9.1E-31; #in Kg (mass of electron)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T=300; #in K (temperature)\n",
-      "h=6.63E-34; #in J/s (Planck\u2019s constant)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "e=1.6E-19; #in C(charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "Eg=Eg*e; #changing unit from eV to J\n",
-      "ni=2*(2*pi*m*k*T/h**2)**(3/2)*math.exp(-Eg/(2*k*T));\n",
-      "sigma=ni*e*(ue+uh);\n",
-      "\n",
-      "#result\n",
-      "print\"The intrinsic carrier density for germanium at 300K is ni=\",'%.3E'%ni,\"/m^3\";\n",
-      "print\"The conductivity of germanium is=\",round(sigma,3),\"(ohm-m)^-1\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The intrinsic carrier density for germanium at 300K is ni= 3.334E+19 /m^3\n",
-        "The conductivity of germanium is= 3.201 (ohm-m)^-1\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9 , Page no:275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "rho1=4.5; #in ohm-m (resistivity at 20 degree Celcius)\n",
-      "rho2=2.0; #in ohm-m (resistivity at 32 degree Celcius)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=20; T2=32; #in degree Celcius (two temperatures)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "T1=T1+273; #changing unit from degree Celius to K\n",
-      "T2=T2+273; #changing unit from degree Celius to K\n",
-      "#since sigma=e*u*C*T^(3/2)*exp(-Eg/(2*k*T))\n",
-      "#therefore sigma1/sigma2=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
-      "#and sigma=1/rho \n",
-      "#therefore we have rho2/rho1=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
-      "#re-arranging above equation for Eg, we get Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1))\n",
-      "Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*math.log(T1/T2)-math.log(rho2/rho1));\n",
-      "Eg1=Eg/e;#changing unit from J to eV\n",
-      "\n",
-      "#result\n",
-      "print\"The energy band gap is Eg=\",'%.3E'%Eg,\"J\";\n",
-      "print\"\\t\\t\\t =\",round(Eg1,3),\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The energy band gap is Eg= 1.543E-19 J\n",
-        "\t\t\t = 0.964 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10 , Page no:276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "rho=2300; #in ohm-m (resistivity of pure silicon)\n",
-      "ue=0.135; #in m^2/V-s (mobility of electron)\n",
-      "uh=0.048; #in m^2/V-s (mobility of electron)\n",
-      "Nd=1E19; #in /m^3 (doping concentration)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "#since sigma=ni*e*(ue+uh) and sigma=1/rho\n",
-      "#therefore ni=1/(rho*e*(ue+uh))\n",
-      "ni=1/(rho*e*(ue+uh)); #calculation of intrinsic concentration\n",
-      "ne=Nd; #calculation of electron concentration\n",
-      "nh=ni**2/Nd; #calculation of hole concentration\n",
-      "sigma=ne*ue*e+nh*uh*e; #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "\n",
-      "#result\n",
-      "print\"The electron concentration is ne=\",ne,\"/m^3\";\n",
-      "print\"The hole concentration is nh=\",'%.3E'%nh,\"/m^3\";\n",
-      "print\"The resistivity of the specimen  is =\",round(rho,3),\"ohm-m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electron concentration is ne= 1e+19 /m^3\n",
-        "The hole concentration is nh= 2.205E+13 /m^3\n",
-        "The resistivity of the specimen  is = 4.63 ohm-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.11 , Page no:276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "uh=1900; #in cm^2/V-s (mobility of electron)\n",
-      "Na=2E17; #in /m^3 (acceptor doping concentration)\n",
-      "e=1.6E-19; #in C(charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "uh=uh*1E-4; #changing unit from cm^2/V-s to m^2/V-s\n",
-      "Na=Na*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
-      "nh=Na; #hole concentration \n",
-      "#since sigma=ne*ue*e+nh*uh*e and nh>>ne\n",
-      "#therefore sigma=nh*uh*e\n",
-      "sigma=nh*uh*e; #calculation of conductivity\n",
-      "\n",
-      "#result\n",
-      "print\"The conductivity of p-type Ge crystal is =\",sigma,\"/ohm-m   (roundoff error)\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductivity of p-type Ge crystal is = 6080.0 /ohm-m   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12 , Page no:277"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ue=0.19; #in m^2/V-s (mobility of electron)\n",
-      "T=300; #in K (temperature)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "e=1.6E-19; #in C(charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "Dn=ue*k*T/e; #calculation of diffusion co-efficient\n",
-      "\n",
-      "#result\n",
-      "print\"The diffusion co-efficient of electron in silicon is Dn=\",'%.3E'%Dn,\"m^2/s\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diffusion co-efficient of electron in silicon is Dn= 4.916E-03 m^2/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13 , Page no:277"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Eg=0.4; #in eV (Band gap of semiconductor)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=0; #in degree Celcius (first temperature)\n",
-      "T2=50; #in degree Celcius (second temperature)\n",
-      "T3=100; #in degree Celcius (third temperature)\n",
-      "e=1.602E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "T1=T1+273; #changing temperature form Celcius to Kelvin\n",
-      "T2=T2+273; #changing temperature form Celcius to Kelvin\n",
-      "T3=T3+273; #changing temperature form Celcius to Kelvin\n",
-      "Eg=Eg*e; #changing unit from eV to Joule\n",
-      "#Using F_E=1/(1+exp(Eg/2*k*T))\n",
-      "F_E1=1/(1+math.exp(Eg/(2*k*T1))); #calculation of probability of occupation of lowest level at 0 degree Celcius\n",
-      "F_E2=1/(1+math.exp(Eg/(2*k*T2))); #calculation of probability of occupation of lowest level at 50 degree Celcius\n",
-      "F_E3=1/(1+math.exp(Eg/(2*k*T3))); #calculation of probability of occupation of lowest level at 100 degree Celcius\n",
-      "\n",
-      "#result\n",
-      "print\"The probability of occupation of lowest level in conduction band is\";\n",
-      "print\"\\t at 0 degree Celcius, F_E=\",'%.3E'%F_E1,\"eV\";\n",
-      "print\"\\t at 50 degree Celcius, F_E=\",'%.3E'%F_E2,\"eV\";\n",
-      "print\"\\t at 100 degree Celcius, F_E=\",'%.3E'%F_E3,\"eV\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The probability of occupation of lowest level in conduction band is\n",
-        "\t at 0 degree Celcius, F_E= 2.025E-04 eV\n",
-        "\t at 50 degree Celcius, F_E= 7.550E-04 eV\n",
-        "\t at 100 degree Celcius, F_E= 1.976E-03 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14 , Page no:278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Eg=1.2; #in eV (Energy band gap)\n",
-      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
-      "T1=600; T2=300; #in K (two temperatures)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "Eg=Eg*e; #changing unit from eV to Joule\n",
-      "#since sigma is proportional to exp(-Eg/(2*k*T))\n",
-      "#therefore ratio=sigma1/sigma2=exp(-Eg/(2*k*((1/T1)-(1/T2))));\n",
-      "ratio= math.exp((-Eg/(2*k))*((1/T1)-(1/T2))); #calculation of ratio of conductivity at 600K and at 300K\n",
-      "\n",
-      "#result\n",
-      "print\"The ratio of conductivity at 600K and at 300K is =\",'%.3E'%ratio;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio of conductivity at 600K and at 300K is = 1.085E+05\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15 , Page no:278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "ue=0.39; #in m^2/V-s (mobility of electron)\n",
-      "n=5E13; #number of donor atoms\n",
-      "ni=2.4E19; #in atoms/m^3 (intrinsic carrier density)\n",
-      "l=10; #in mm (length of rod)\n",
-      "a=1; #in mm (side of square cross-section)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "l=l*1E-3; #changing unit from mm to m\n",
-      "a=a*1E-3; #changing unit from mm to m\n",
-      "A=a**2; #calculation of cross-section area\n",
-      "Nd=n/(l*A); #calculation of donor concentration\n",
-      "ne=Nd; #calculation of electron density\n",
-      "nh=ni**2/Nd; #calculation of hole density\n",
-      "#since sigma=ne*e*ue+nh*e*ue and since ne>>nh\n",
-      "#therefore sigma=ne*e*ue\n",
-      "sigma=ne*e*ue; #calculation of conductivity\n",
-      "rho=1/sigma; #calculation of resistivity\n",
-      "R=rho*l/A; #calculation of resistance \n",
-      "\n",
-      "#result\n",
-      "print\"The electron density is ne=\",ne,\"/m^3\";\n",
-      "print\"The hole density is nh=\",nh,\"/m^3\";\n",
-      "print\"The conductivity is =\",sigma,\"/ohm-m\";\n",
-      "print\"The resistance is R=\",round(R),\"ohm\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The electron density is ne= 5e+21 /m^3\n",
-        "The hole density is nh= 1.152e+17 /m^3\n",
-        "The conductivity is = 312.0 /ohm-m\n",
-        "The resistance is R= 32.0 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 , Page no:279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
-      "rho=8.93E-3; #in ohm-m (resistivity)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "u=RH/rho; #calculation of mobility\n",
-      "n=1/(RH*e); #calculation of density\n",
-      "\n",
-      "#result\n",
-      "print\"The mobility is u=\",round(u,4),\"m^2/(V-s)\";\n",
-      "print\"The density is n=\",'%.3E'%n,\"/m^3\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mobility is u= 0.041 m^2/(V-s)\n",
-        "The density is n= 1.708E+22 /m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 , Page no:279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
-      "rho=8.93E-3; #in ohm-m (resistivity)\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "\n",
-      "#calculate\n",
-      "nh=1/(RH*e); #calculation of density of charge carrier\n",
-      "uh=1/(rho*nh*e); #calculation of mobility of charge carrier\n",
-      "\n",
-      "#result\n",
-      "print\"The density of charge carrier is nh=\",'%.3E'%nh,\"/m^3\";\n",
-      "print\"The mobility of charge carrier is uh=\",round(uh,4),\"m^2/(V-s)\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The density of charge carrier is nh= 1.708E+22 /m^3\n",
-        "The mobility of charge carrier is uh= 0.041 m^2/(V-s)\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_10_Superconductivity.ipynb b/Applied_Physics/Chapter_10_Superconductivity.ipynb
deleted file mode 100755
index 7b7470a8..00000000
--- a/Applied_Physics/Chapter_10_Superconductivity.ipynb
+++ /dev/null
@@ -1,316 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:03ee5374207ea96e9a4f31e16f771d187a11f07daf287d94fc54e352b5db8bff"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10:Superconductivity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.1 , Page no:313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Tc=7.2; #in K (critical temperature)\n",
-      "T=5; #in K (given temperature)\n",
-      "H0=6.5E3; #in A/m (critical magnetic field at 0K)\n",
-      "\n",
-      "#calculate\n",
-      "Hc=H0*(1-(T/Tc)**2); #calculation of magnitude of critical magnetic field\n",
-      "\n",
-      "#result\n",
-      "print\"The magnitude of critical magnetic field is Hc=\",round(Hc,2),\"A/m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of critical magnetic field is Hc= 3365.35 A/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2 , Page no:313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r=0.02; #in m (radius of ring)\n",
-      "Hc=2E3; #in A/m (critical magnetic field at 5K)\n",
-      "pi=3.14; #value of pi used in the solutiion\n",
-      "\n",
-      "#calculate\n",
-      "Ic=2*pi*r*Hc; #calculation of critical current value\n",
-      "\n",
-      "#result\n",
-      "print\"The critical current value is Ic=\",Ic,\"A\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical current value is Ic= 251.2 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.3 , Page no:313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "M1=199.5; #in amu (isotropic mass at 5K)\n",
-      "T1=5; #in K (first critical temperature)\n",
-      "T2=5.1; #in K (second critical temperature)\n",
-      "#calculate\n",
-      "#since Tc=C*(1/sqrt(M)\n",
-      "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
-      "#therefore we have M2=(T1/T2)^2*M1\n",
-      "M2=(T1/T2)**2*M1; #calculation of isotropic mass at 5.1K\n",
-      "\n",
-      "#result\n",
-      "print\"The isotropic mass at 5.1K is M2=\",round(M2,3),\"a.m.u.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The isotropic mass at 5.1K is M2= 191.753 a.m.u.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.4 , Page no:314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=6; #in K (given temperature)\n",
-      "Hc=5E3; #in A/m (critical magnetic field at 5K)\n",
-      "H0=2E4; #in A/m (critical magnetic field at 0K)\n",
-      "\n",
-      "#calculate\n",
-      "#since Hc=H0*(1-(T/Tc)^2)\n",
-      "#therefor we have Tc=T/sqrt(1-(Hc/H)^2)\n",
-      "Tc=T/math.sqrt(1-(Hc/H0)); #calculation of transition temperature\n",
-      "\n",
-      "#result\n",
-      "print\"The transition temperature is Tc=\",round(Tc,3),\"K\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transition temperature is Tc= 6.928 K\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5 , Page no:314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=5; #in K (given temperature)\n",
-      "d=3; #in mm (diameter of the wire)\n",
-      "Tc=8; #in K (critical temperature for Pb)\n",
-      "H0=5E4; #in A/m (critical magnetic field at 0K)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "Hc=H0*(1-(T/Tc)**2); #calculation of critical magnetic field at 5K\n",
-      "r=(d*1E-3)/2; #calculation of radius in m\n",
-      "Ic=2*pi*r*Hc; #calculation of critical current at 5K\n",
-      "\n",
-      "#result\n",
-      "print\"The critical magnetic field at 5K is Hc=\",Hc,\"A/m\";\n",
-      "print\"The critical current at 5K is Ic=\",Ic,\"A\";\n",
-      "print \"   (roundoff error)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical magnetic field at 5K is Hc= 30468.75 A/m\n",
-        "The critical current at 5K is Ic= 287.015625 A\n",
-        "   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.6 , Page no:314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=8.50; #in micro V (voltage across Josephson junction )\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "h=6.626E-34; #in J/s (Planck\u2019s constant)\n",
-      "\n",
-      "#calculate\n",
-      "V=V*1E-6; #changing unit from V to microVolt\n",
-      "v1=2*e*V/h; #calculation of frequency of EM waves\n",
-      "\n",
-      "#result\n",
-      "print\"The frequency of EM waves is v=\",v1,\"Hz\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of EM waves is v= 4105040748.57 Hz\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.7 , Page no:315"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "M1=200.59; #in amu (average atomic mass  at 4.153K)\n",
-      "Tc1=4.153; #in K (first critical temperature)\n",
-      "M2=204; #in amu (average atomic mass of isotopes)\n",
-      "\n",
-      "#calculate\n",
-      "#since Tc=C*(1/sqrt(M)\n",
-      "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
-      "#therefore we have Tc2=Tc1*sqrt(M1/M2)\n",
-      "Tc2=Tc1*math.sqrt(M1/M2); #calculation of transition temperature of the isotopes\n",
-      "\n",
-      "#result\n",
-      "print\"The transition temperature of the isotopes is  Tc2=\",round(Tc2,3),\"K\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transition temperature of the isotopes is  Tc2= 4.118 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_10_Superconductivity_1.ipynb b/Applied_Physics/Chapter_10_Superconductivity_1.ipynb
deleted file mode 100755
index 9b11ce44..00000000
--- a/Applied_Physics/Chapter_10_Superconductivity_1.ipynb
+++ /dev/null
@@ -1,316 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2419f2161da8f36d7fa10fabb5f5aa4b318a9912237faaa5f03ac922aa4b8ac9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10:Superconductivity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.1 , Page no:313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "Tc=7.2; #in K (critical temperature)\n",
-      "T=5; #in K (given temperature)\n",
-      "H0=6.5E3; #in A/m (critical magnetic field at 0K)\n",
-      "\n",
-      "#calculate\n",
-      "Hc=H0*(1-(T/Tc)**2); #calculation of magnitude of critical magnetic field\n",
-      "\n",
-      "#result\n",
-      "print\"The magnitude of critical magnetic field is Hc=\",round(Hc,2),\"A/m\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of critical magnetic field is Hc= 3365.35 A/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2 , Page no:313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "r=0.02; #in m (radius of ring)\n",
-      "Hc=2E3; #in A/m (critical magnetic field at 5K)\n",
-      "pi=3.14; #value of pi used in the solutiion\n",
-      "\n",
-      "#calculate\n",
-      "Ic=2*pi*r*Hc; #calculation of critical current value\n",
-      "\n",
-      "#result\n",
-      "print\"The critical current value is Ic=\",Ic,\"A\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical current value is Ic= 251.2 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.3 , Page no:313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "M1=199.5; #in amu (isotropic mass at 5K)\n",
-      "T1=5; #in K (first critical temperature)\n",
-      "T2=5.1; #in K (second critical temperature)\n",
-      "#calculate\n",
-      "#since Tc=C*(1/sqrt(M)\n",
-      "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
-      "#therefore we have M2=(T1/T2)^2*M1\n",
-      "M2=(T1/T2)**2*M1; #calculation of isotropic mass at 5.1K\n",
-      "\n",
-      "#result\n",
-      "print\"The isotropic mass at 5.1K is M2=\",round(M2,3),\"a.m.u.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The isotropic mass at 5.1K is M2= 191.753 a.m.u.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.4 , Page no:314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=6; #in K (given temperature)\n",
-      "Hc=5E3; #in A/m (critical magnetic field at 5K)\n",
-      "H0=2E4; #in A/m (critical magnetic field at 0K)\n",
-      "\n",
-      "#calculate\n",
-      "#since Hc=H0*(1-(T/Tc)^2)\n",
-      "#therefor we have Tc=T/sqrt(1-(Hc/H)^2)\n",
-      "Tc=T/math.sqrt(1-(Hc/H0)); #calculation of transition temperature\n",
-      "\n",
-      "#result\n",
-      "print\"The transition temperature is Tc=\",round(Tc,3),\"K\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transition temperature is Tc= 6.928 K\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5 , Page no:314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "T=5; #in K (given temperature)\n",
-      "d=3; #in mm (diameter of the wire)\n",
-      "Tc=8; #in K (critical temperature for Pb)\n",
-      "H0=5E4; #in A/m (critical magnetic field at 0K)\n",
-      "pi=3.14; #value of pi used in the solution\n",
-      "\n",
-      "#calculate\n",
-      "Hc=H0*(1-(T/Tc)**2); #calculation of critical magnetic field at 5K\n",
-      "r=(d*1E-3)/2; #calculation of radius in m\n",
-      "Ic=2*pi*r*Hc; #calculation of critical current at 5K\n",
-      "\n",
-      "#result\n",
-      "print\"The critical magnetic field at 5K is Hc=\",'%.3E'%Hc,\"A/m\";\n",
-      "print\"The critical current at 5K is Ic=\",round(Ic,4),\"A\";\n",
-      "print \"   (roundoff error)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical magnetic field at 5K is Hc= 3.047E+04 A/m\n",
-        "The critical current at 5K is Ic= 287.0156 A\n",
-        "   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.6 , Page no:314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "V=8.50; #in micro V (voltage across Josephson junction )\n",
-      "e=1.6E-19; #in C (charge of electron)\n",
-      "h=6.626E-34; #in J/s (Planck\u2019s constant)\n",
-      "\n",
-      "#calculate\n",
-      "V=V*1E-6; #changing unit from V to microVolt\n",
-      "v1=2*e*V/h; #calculation of frequency of EM waves\n",
-      "\n",
-      "#result\n",
-      "print\"The frequency of EM waves is v=\",'%.3E'%v1,\"Hz\";\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of EM waves is v= 4.105E+09 Hz\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.7 , Page no:315"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "M1=200.59; #in amu (average atomic mass  at 4.153K)\n",
-      "Tc1=4.153; #in K (first critical temperature)\n",
-      "M2=204; #in amu (average atomic mass of isotopes)\n",
-      "\n",
-      "#calculate\n",
-      "#since Tc=C*(1/sqrt(M)\n",
-      "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
-      "#therefore we have Tc2=Tc1*sqrt(M1/M2)\n",
-      "Tc2=Tc1*math.sqrt(M1/M2); #calculation of transition temperature of the isotopes\n",
-      "\n",
-      "#result\n",
-      "print\"The transition temperature of the isotopes is  Tc2=\",round(Tc2,3),\"K\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transition temperature of the isotopes is  Tc2= 4.118 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_12Fibre_Optics.ipynb b/Applied_Physics/Chapter_12Fibre_Optics.ipynb
deleted file mode 100755
index 95067254..00000000
--- a/Applied_Physics/Chapter_12Fibre_Optics.ipynb
+++ /dev/null
@@ -1,365 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:13927a6aa5f4b8929243c9ca2c470f4da4ee5b45564e616337e68523dd18d625"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12:Fibre Optics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.563; #refractive index of core\n",
-      "u2=1.498; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "d=(u1-u2)/u1; #calculation of fractional index change\n",
-      "\n",
-      "#result\n",
-      "print\"The fractional index change for a given optical fibre is =\",round(d,4);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fractional index change for a given optical fibre is = 0.0416\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.55; #refractive index of core\n",
-      "u2=1.50; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "d=(u1-u2)/u1; #calculation of fractional index change\n",
-      "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
-      "theta=math.asin(NA); #calculation of acceptance angle\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.394\n",
-        "The acceptance angle of the optical fibre is = 23.2 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.563; #refractive index of core\n",
-      "u2=1.498; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
-      "theta=math.asin(NA); #calculation of acceptance angle\n",
-      "theta1=theta*180/3.14;\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,4);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.4461\n",
-        "The acceptance angle of the optical fibre is = 26.5 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "NA=0.39; #numerical aperture of the optical fibre\n",
-      "d=0.05; #difference in the refractive index of the material of the core and cladding\n",
-      "\n",
-      "#calculate\n",
-      "#since NA=u1*sqrt(2*d)\n",
-      "#we have u1=NA/sqrt(2*d)\n",
-      "u1= NA/math.sqrt(2*d); #calculation of refractive index of  material of the core\n",
-      "\n",
-      "#result\n",
-      "print\"The refractive index of  material of the core is u1=\",round(u1,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The refractive index of  material of the core is u1= 1.233\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.50; #refractive index of core\n",
-      "u2=1.45; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "d=(u1-u2)/u1; #calculation of fractional index change\n",
-      "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
-      "theta_0=math.asin(NA); #calculation of acceptance angle\n",
-      "theta_01=theta_0*180/3.14;\n",
-      "theta_c=math.asin(u2/u1); #calculation of critical angle\n",
-      "theta_c1=theta_c*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta_01,2),\"degree\";\n",
-      "print\"The critical angle of the optical fibre is =\",round(theta_c1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.387\n",
-        "The acceptance angle of the optical fibre is = 22.8 degree\n",
-        "The critical angle of the optical fibre is = 75.2 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "NA=0.33; #numerical aperture\n",
-      "d=0.02; #difference in the refractive index of the core and cladding of the material\n",
-      "\n",
-      "#calculate\n",
-      "#since NA=u1*sqrt(2*d)\n",
-      "#therefore we have\n",
-      "u1=NA/math.sqrt(2*d); #calculation of refractive index of the core\n",
-      "#since d=(u1-u2)/u2\n",
-      "#therefore we have\n",
-      "u2=(1-d)*u1; #calculation of refractive index of the cladding\n",
-      "\n",
-      "#result\n",
-      "print\"The refractive index of the core is u1=\",round(u1,2);\n",
-      "print\"The refractive index of the cladding is u2=\",round(u2,3);\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The refractive index of the core is u1= 1.65\n",
-        "The refractive index of the cladding is u2= 1.617\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=3.5; #refractive index of core\n",
-      "u2=3.45; #refractive index of cladding\n",
-      "u0=1; #refractive index of the air\n",
-      "\n",
-      "#calculate\n",
-      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
-      "NA1=NA/u0;\n",
-      "alpha=math.asin(NA); #calculation of acceptance angle\n",
-      "alpha1=alpha*180/3.14;\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA1,2);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(alpha1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.59\n",
-        "The acceptance angle of the optical fibre is = 36.14 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.48; #refractive index of core\n",
-      "u2=1.45; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
-      "theta=math.asin(NA); #calculation of acceptance angle\n",
-      "theta1=theta*180/3.14;\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";\n",
-      "print \"   (roundoff error)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.296\n",
-        "The acceptance angle of the optical fibre is = 17.26 degree\n",
-        "   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/Chapter_12_Fibre_Optics.ipynb b/Applied_Physics/Chapter_12_Fibre_Optics.ipynb
deleted file mode 100755
index f573b0a2..00000000
--- a/Applied_Physics/Chapter_12_Fibre_Optics.ipynb
+++ /dev/null
@@ -1,365 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:dac7a0b06f622e783ea6c469bf1c100c0c8157cc5e77bffea65e378f258808b2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12:Fibre Optics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.563; #refractive index of core\n",
-      "u2=1.498; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "d=(u1-u2)/u1; #calculation of fractional index change\n",
-      "\n",
-      "#result\n",
-      "print\"The fractional index change for a given optical fibre is =\",round(d,4);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fractional index change for a given optical fibre is = 0.0416\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.55; #refractive index of core\n",
-      "u2=1.50; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "d=(u1-u2)/u1; #calculation of fractional index change\n",
-      "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
-      "theta=math.asin(NA); #calculation of acceptance angle\n",
-      "theta1=theta*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.394\n",
-        "The acceptance angle of the optical fibre is = 23.2 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.563; #refractive index of core\n",
-      "u2=1.498; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
-      "theta=math.asin(NA); #calculation of acceptance angle\n",
-      "theta1=theta*180/3.14;\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,4);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.4461\n",
-        "The acceptance angle of the optical fibre is = 26.5 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 , Page no:360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "NA=0.39; #numerical aperture of the optical fibre\n",
-      "d=0.05; #difference in the refractive index of the material of the core and cladding\n",
-      "\n",
-      "#calculate\n",
-      "#since NA=u1*sqrt(2*d)\n",
-      "#we have u1=NA/sqrt(2*d)\n",
-      "u1= NA/math.sqrt(2*d); #calculation of refractive index of  material of the core\n",
-      "\n",
-      "#result\n",
-      "print\"The refractive index of  material of the core is u1=\",round(u1,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The refractive index of  material of the core is u1= 1.233\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.50; #refractive index of core\n",
-      "u2=1.45; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "d=(u1-u2)/u1; #calculation of fractional index change\n",
-      "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
-      "theta_0=math.asin(NA); #calculation of acceptance angle\n",
-      "theta_01=theta_0*180/3.14;\n",
-      "theta_c=math.asin(u2/u1); #calculation of critical angle\n",
-      "theta_c1=theta_c*180/3.14;\n",
-      "\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta_01,2),\"degree\";\n",
-      "print\"The critical angle of the optical fibre is =\",round(theta_c1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.387\n",
-        "The acceptance angle of the optical fibre is = 22.8 degree\n",
-        "The critical angle of the optical fibre is = 75.2 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "NA=0.33; #numerical aperture\n",
-      "d=0.02; #difference in the refractive index of the core and cladding of the material\n",
-      "\n",
-      "#calculate\n",
-      "#since NA=u1*sqrt(2*d)\n",
-      "#therefore we have\n",
-      "u1=NA/math.sqrt(2*d); #calculation of refractive index of the core\n",
-      "#since d=(u1-u2)/u2\n",
-      "#therefore we have\n",
-      "u2=(1-d)*u1; #calculation of refractive index of the cladding\n",
-      "\n",
-      "#result\n",
-      "print\"The refractive index of the core is u1=\",round(u1,2);\n",
-      "print\"The refractive index of the cladding is u2=\",round(u2,3);\n",
-      "print \"NOTE: The answer in the textbook is wrong\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The refractive index of the core is u1= 1.65\n",
-        "The refractive index of the cladding is u2= 1.617\n",
-        "NOTE: The answer in the textbook is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=3.5; #refractive index of core\n",
-      "u2=3.45; #refractive index of cladding\n",
-      "u0=1; #refractive index of the air\n",
-      "\n",
-      "#calculate\n",
-      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
-      "NA1=NA/u0;\n",
-      "alpha=math.asin(NA); #calculation of acceptance angle\n",
-      "alpha1=alpha*180/3.14;\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA1,2);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(alpha1,2),\"degree\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.59\n",
-        "The acceptance angle of the optical fibre is = 36.14 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 , Page no:361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#given\n",
-      "u1=1.48; #refractive index of core\n",
-      "u2=1.45; #refractive index of cladding\n",
-      "\n",
-      "#calculate\n",
-      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
-      "theta=math.asin(NA); #calculation of acceptance angle\n",
-      "theta1=theta*180/3.14;\n",
-      "#result\n",
-      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
-      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";\n",
-      "print \"   (roundoff error)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numerical aperture of the fibre is NA= 0.296\n",
-        "The acceptance angle of the optical fibre is = 17.26 degree\n",
-        "   (roundoff error)\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics/screenshots/Calculation_of__mean_free_path_of_electron_1.png b/Applied_Physics/screenshots/Calculation_of__mean_free_path_of_electron_1.png
deleted file mode 100755
index 37995b76..00000000
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diff --git a/Applied_Physics/screenshots/Calculation_of__potential_energy_1.png b/Applied_Physics/screenshots/Calculation_of__potential_energy_1.png
deleted file mode 100755
index f6086db3..00000000
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diff --git a/Applied_Physics/screenshots/Calculation_of_lattice_constant_1.png b/Applied_Physics/screenshots/Calculation_of_lattice_constant_1.png
deleted file mode 100755
index cf42154d..00000000
Binary files a/Applied_Physics/screenshots/Calculation_of_lattice_constant_1.png and /dev/null differ
diff --git a/Applied_Physics_II/Chapter_1.ipynb b/Applied_Physics_II/Chapter_1.ipynb
deleted file mode 100755
index c6e00369..00000000
--- a/Applied_Physics_II/Chapter_1.ipynb
+++ /dev/null
@@ -1,1108 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1: Interference of Light"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.1, Page number 1-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "i = 45        #angle of incidence(degrees)\n",
-      "t = 4*10**-5  #thickness of film(cm)\n",
-      "u = 1.2\n",
-      "\n",
-      "#Calculations & Result\n",
-      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
-      "for n in range(1,4):\n",
-      "    lamda = (2*u*t*math.cos(r*math.pi/180))/n\n",
-      "    print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-8)\n",
-      "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For n = 1, wavelength = 7756.29 A\n",
-        "For n = 2, wavelength = 3878.14 A\n",
-        "For n = 3, wavelength = 2585.43 A\n",
-        "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.2, Page number 1-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "r = 90        #angle of refraction(degrees)\n",
-      "t = 5*10**-5  #thickness of film(cm)\n",
-      "u = 1.33\n",
-      "\n",
-      "#Calculations & Result\n",
-      "for n in range(1,4):\n",
-      "    lamda = (4*u*t*int(math.cos(math.radians(90))))/((2*n)-1)\n",
-      "    print \"For n = %d, wavelength = %.2f A\"%(n,lamda)\n",
-      "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\"\n",
-      "\n",
-      "print \"\\nPlease note: Since r=90, cos(r)=0\\nHence, the answers given in the textbook are incorrect\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For n = 1, wavelength = 0.00 A\n",
-        "For n = 2, wavelength = 0.00 A\n",
-        "For n = 3, wavelength = 0.00 A\n",
-        "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\n",
-        "\n",
-        "Please note: Since r=90, cos(r)=0\n",
-        "Hence, the answers given in the textbook are incorrect\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.3, Page number 1-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "i = 45           #angle of incidence(degrees)\n",
-      "t = 1.5*10**-4   #thickness of film(cm)\n",
-      "lamda = 5*10**-5 #wavelength(cm)\n",
-      "u = 4./3.        #refractive index\n",
-      "\n",
-      "#Calculations\n",
-      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
-      "n = (2*u*t*math.cos(r*math.pi/180))/lamda\n",
-      "\n",
-      "#Result\n",
-      "print \"The order of interfernce is %.2f, close to 7\"%n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The order of interfernce is 6.78, close to 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2.4, Pae number 1-13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "i = 45              #angle of incidence(degrees)\n",
-      "lamda = 5896*10**-8 #wavelength(cm)\n",
-      "u = 1.33            #refractive index\n",
-      "n = 1\n",
-      "\n",
-      "#Calculations\n",
-      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
-      "t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n",
-      "\n",
-      "#Result\n",
-      "print \"The required thickness is\",round((t/1E-5),2),\"*10^-5 cm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required thickness is 1.31 *10^-5 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.5, Page number 1-14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "lamda1 = 7000  #wavelength(A)\n",
-      "lamda2 = 5000  #wavelength(A)\n",
-      "u = 1.3        #R.I. of oil\n",
-      "\n",
-      "#Calculations\n",
-      "'''\n",
-      "2utcosr = (2n-1)7000/2   ----(1)\n",
-      "2utcosr = (2n+1)5000/2   ----(2)\n",
-      "Divinding (1) by (2), we get the following expression\n",
-      "1 = (2n+1)5000\n",
-      "    -----------\n",
-      "    (2n-1)7000\n",
-      "Solving the above expression, we get,\n",
-      "'''\n",
-      "n = 12000/4000\n",
-      "t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n",
-      "\n",
-      "#Result\n",
-      "print \"The required thickness is\",round((t/1E-5),4),\"*10^-5 cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required thickness is 6.6936 *10^-5 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.6, Page number 1-15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "i = 30              #angle of incidence(degrees)\n",
-      "lamda = 5890*10**-8 #wavelength(cm)\n",
-      "u = 1.46            #refractive index\n",
-      "n = 8\n",
-      "\n",
-      "#Calculations\n",
-      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
-      "t = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
-      "\n",
-      "#Result\n",
-      "print \"The required thickness is\",round((t/1E-4),3),\"*10^-4 cm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required thickness is 1.718 *10^-4 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.7, Page number 1-15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "r = 60              #angle of refraction(degrees)\n",
-      "lamda = 5890*10**-8 #wavelength(cm)\n",
-      "u = 1.5             #refractive index\n",
-      "n = 1               #for minimumm thickness\n",
-      "\n",
-      "#Calculations\n",
-      "#For r = 60\n",
-      "t1 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
-      "\n",
-      "#For normal incidence \n",
-      "r = 0\n",
-      "t2 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
-      "\n",
-      "#Result\n",
-      "print \"For r = 60, the required thickness is\",round((t1/1E-5),2),\"*10^-5 cm\"\n",
-      "print \"For r = 0, the required thickness is\",round((t2/1E-5),2),\"*10^-5 cm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For r = 60, the required thickness is 3.93 *10^-5 cm\n",
-        "For r = 0, the required thickness is 1.96 *10^-5 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.8, Page number 1-16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "r = 0               #for normal incidence(degrees)\n",
-      "lamda = 5.5*10**-5  #wavelength(cm)\n",
-      "n = 1               #for minimumm thickness\n",
-      "A = 10**4           #area(cm^2)\n",
-      "V = 0.2             #volume(cc)\n",
-      "\n",
-      "#Calculations\n",
-      "t = V/A\n",
-      "#for nth dark band,\n",
-      "u = (n*lamda)/(2*t*math.cos(r*math.pi/180))\n",
-      "\n",
-      "#Result\n",
-      "print \"Refractive index =\",u"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index = 1.375\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2.9, Page number 1-17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "r = 60        #angle of incidence(degrees)\n",
-      "t = 2*10**-7  #thickness of film(cm)\n",
-      "u = 1.2\n",
-      "\n",
-      "#Calculations & Result\n",
-      "for n in range(1,4):\n",
-      "    lamda = (4*u*t*math.cos(r*math.pi/180))/(2*n-1)\n",
-      "    print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-10)\n",
-      "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For n = 1, wavelength = 4800.00 A\n",
-        "For n = 2, wavelength = 1600.00 A\n",
-        "For n = 3, wavelength = 960.00 A\n",
-        "Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3.1, Page number 1-21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "a = 40.       #angle(sec)\n",
-      "lamda = 1.2  #distance between fringes(cm)\n",
-      "alpha = 10   #no. of fringes\n",
-      "\n",
-      "#Calculations\n",
-      "Bair = lamda/alpha          #cm\n",
-      "alpha = (a*math.pi)/(3600*180)   #radians\n",
-      "lamda = 2*alpha*Bair\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength of monochromatic light =\",round((lamda/1E-8),1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of monochromatic light = 4654.2 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3.2, Page number 1-22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "lamda = 5893*10**-8   #wavelength(cm)\n",
-      "u = 1.52              #refractive index\n",
-      "B = 0.1               #fringe spacing(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "alpha = (lamda/(2*u*B))*180*3600/math.pi   #seconds\n",
-      "\n",
-      "#Result\n",
-      "print \"Angle of wedge =\",round(alpha,2),\"secs\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angle of wedge = 39.98 secs\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3.3, Page number 1-22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "u = 1.4              #refractive index\n",
-      "B = 0.25             #fringe spacing(cm)\n",
-      "a = 20               #angle(secs)\n",
-      "\n",
-      "#Calculations\n",
-      "alpha = (a*math.pi)/(3600*180)  #radians\n",
-      "lamda = 2*u*alpha*B\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength of light =\",round((lamda/1E-8),2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of light = 6787.39 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3.4, Page number 1-23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "u = 1.5              #refractive index\n",
-      "lamda = 5.82*10**-5  #wavelength(cm)\n",
-      "a = 20               #angle(secs)\n",
-      "\n",
-      "#Calculations\n",
-      "alpha = (a*math.pi)/(3600*180)  #radians\n",
-      "B = lamda/(2*u*alpha)\n",
-      "N = 1/B\n",
-      "\n",
-      "#Result\n",
-      "print \"Number of interfernce fronges pr cm is\",round(N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of interfernce fronges pr cm is 5.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3.5, Page number 1-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "u = 1                #refractive index for air film\n",
-      "lamda = 6*10**-5     #wavelength(cm)\n",
-      "B = 1./10            #distance between fringes(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "alpha = lamda/(2*u*B)  #radians\n",
-      "d = alpha*10\n",
-      "\n",
-      "#Result\n",
-      "print \"Daimeter of wire =\",d,\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Daimeter of wire = 0.003 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3.6, Page number 1-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "alpha = 0.01*10**-1/10   #angle(radians)\n",
-      "u = 1                    #refractive index for air film\n",
-      "lamda = 5900*10**-10     #wavlength(m)\n",
-      "\n",
-      "#Calculation\n",
-      "B = lamda/(2*u*alpha)\n",
-      "\n",
-      "#Result\n",
-      "print \"Seperation between fringes is\",B/10**-3,\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Seperation between fringes is 2.95 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.1, Page number 1-32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "n = 40\n",
-      "\n",
-      "#Calculation\n",
-      "#Equating the equation 4*R*n*lamda=4*4R*n*lamda, we get\n",
-      "\n",
-      "N = (4*4*n)/4\n",
-      "\n",
-      "#result\n",
-      "print \"Ring number =\",N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ring number = 160\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.2, Page number 1-32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "n = 10\n",
-      "Dn = 0.5         #diameter of dark ring(cm)\n",
-      "lamda = 5*10**-5 #waelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "R = Dn**2/(4*n*lamda)\n",
-      "\n",
-      "#Result\n",
-      "print \"Radius of curvature =\",R,\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius of curvature = 125.0 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.3, Page number 1-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "n = 5\n",
-      "p = 10\n",
-      "D5 = 0.336          #diameter of 5th ring(cm)\n",
-      "lamda = 5890*10**-8 #waelength(cm)\n",
-      "D15 = 0.59          #diameter of 15th ringcm\n",
-      "\n",
-      "#Calculations\n",
-      "R = (D15**2-D5**2)/(4*p*lamda)\n",
-      "\n",
-      "#Result\n",
-      "print \"Radius of curvature =\",round(R,2),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius of curvature = 99.83 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.4, Page number 1-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "Dn = 0.42     #diameter of dark ring(cm)\n",
-      "p = 8 \n",
-      "R = 200       #radius of curvature(cm)\n",
-      "Dn8 = 0.7     #diameter of (n+8)th ring(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "lamda = (Dn8**2-Dn**2)/(4*R*p)\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength =\",lamda/1E-8,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 4900.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.5, Page number 1-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "Dn = 0.218             #cm\n",
-      "Dn10 = 0.451           #cm\n",
-      "lamda = 5893*10**-8    #wavelength(cm)\n",
-      "R = 90                 #radius of curvature(cm)\n",
-      "p = 10\n",
-      "\n",
-      "#Calculation\n",
-      "u = (4*p*lamda*R)/(Dn10**2-Dn**2)\n",
-      "\n",
-      "#Result\n",
-      "print \"Refractive index =\",round(u,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index = 1.361\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.6, Page number 1-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaible declaration\n",
-      "D5 = 0.42     #diameter of dark ring(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "'''\n",
-      "For 5th dark ring,\n",
-      "D5^2 = 20*R*lamda  -----1\n",
-      "\n",
-      "For 10th dark ring,\n",
-      "D10^2 = 40*R*lamda -----2\n",
-      "\n",
-      "Substituting 1 in 2,\n",
-      "'''\n",
-      "\n",
-      "D10 = math.sqrt((40*D5**2)/20)\n",
-      "\n",
-      "#Result\n",
-      "print \"Diameter of the 10th dark ring =\",round(D10,3),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diameter of the 10th dark ring = 0.594 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.7, Page number 1-35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "lamda_n = 6000    #wavelength of nth ring(A)\n",
-      "lamda_n1 = 5000   #wavelength for (n+1)th ring(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "'''\n",
-      "Dn^2 = 4*R*n*lamda_n ---1\n",
-      "\n",
-      "Dn+1^2 = 4*R(n+1)*lamda_n1  ---2\n",
-      "\n",
-      "Equating 1 and 2, we get,\n",
-      "'''\n",
-      "\n",
-      "n = 5\n",
-      "R = 2\n",
-      "Dn = math.sqrt(4*R*n*lamda_n*10**-8)\n",
-      "\n",
-      "#Result\n",
-      "print \"Diameter =\",round(Dn,3),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diameter = 0.049 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.8, Page number 1-35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "Dair = 2.3   #diameter of ring in air(cm)\n",
-      "Dliq = 2     #diameter of ring in liquid(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "u = Dair**2/Dliq**2\n",
-      "\n",
-      "#Result\n",
-      "print \"Refractive index =\",u"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index = 1.3225\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.11, Page number 1-37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "D4 = 0.4    #diameter of 4th ring(cm)\n",
-      "D12 = 0.7   #diameter of 12th ring(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "'''\n",
-      "For D4, \n",
-      "D4 = math.sqrt(4R*4*lamda)\n",
-      "'''\n",
-      "rt_Rl = 0.1\n",
-      "R = 80      \n",
-      "\n",
-      "#For D20,\n",
-      "D20 = math.sqrt(R)*rt_Rl\n",
-      "\n",
-      "#Result\n",
-      "print \"Diameter of 20th ring =\",round(D20,3),\"cm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diameter of 20th ring = 0.894 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4.12, Page number 1-36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration \n",
-      "n = 5\n",
-      "p = 10\n",
-      "D5 = 0.336          #diameter of 5th ring(cm)\n",
-      "D15 = 0.590         #diameter of 15th ring(cm)\n",
-      "R = 100             #radius of curvature(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "lamda = (D15**2-D5**2)/(4*R*p)\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavlength =\",lamda/1E-8,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavlength = 5880.1 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7.1, Page number 1-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 560 #wavelength(nm)\n",
-      "u = 2       #refractive index\n",
-      "\n",
-      "#Calculations\n",
-      "lamda_dash = lamda/u\n",
-      "t = lamda_dash/4\n",
-      "\n",
-      "#Result\n",
-      "print \"Thickness of film =\",t,\"nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of film = 70 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7.2, Page number 1-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 6000 #wavelength(E)\n",
-      "u = 1.2       #refractive index\n",
-      "\n",
-      "#Calculations\n",
-      "lamda_dash = lamda/u\n",
-      "t = lamda_dash/4\n",
-      "\n",
-      "#Result\n",
-      "print \"Thickness of film =\",t,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thickness of film = 1250.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics_II/Chapter_2.ipynb b/Applied_Physics_II/Chapter_2.ipynb
deleted file mode 100755
index ae487344..00000000
--- a/Applied_Physics_II/Chapter_2.ipynb
+++ /dev/null
@@ -1,1000 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2: Diffraction of Light"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.1, Page number 2-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "theta = 30          #angle(degrees)\n",
-      "n = 1        \n",
-      "lamda = 6500*10**-8 #wavelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "a = (n*lamda)/math.sin(theta*math.pi/180)\n",
-      "\n",
-      "#Result\n",
-      "print \"a =\",a/1e-4,\"*10^-4 cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a = 1.3 *10^-4 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.2, Page number 2-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "a = 6*10**-4        #width of slit(cm)\n",
-      "n = 1               #first order\n",
-      "lamda = 6000*10**-8 #wavelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "def deg_to_dms(deg):\n",
-      "    d = int(deg)\n",
-      "    md = abs(deg - d) * 60\n",
-      "    m = int(md)\n",
-      "    sd = (md - m) * 60\n",
-      "    sd=round(sd,2)\n",
-      "    return [d, m, sd]\n",
-      "\n",
-      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
-      "d = 2*theta   #angular seperation\n",
-      "\n",
-      "#Result\n",
-      "print \"Angular seperation between the 1st order minima is\",deg_to_dms(d)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angular seperation between the 1st order minima is [11, 28, 42.03]\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.3, Page number 2-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 2       #for second minimum\n",
-      "n2 = 3       #for third minimum\n",
-      "lamda = 4000 #wavelength(A)\n",
-      "\n",
-      "#Calculations\n",
-      "'''For 2nd order,\n",
-      "a sin0 = n1*lamda\n",
-      "\n",
-      "For 3rd order,\n",
-      "a sin0 = n2*lamda'''\n",
-      "\n",
-      "lamda = (n2*lamda)/n1\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength =\",lamda,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 6000 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.4, Page number 2-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "a = 0.16*10**-3      #width of slit(cm)\n",
-      "n = 1                #first order\n",
-      "lamda = 5600*10**-10 #wavelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "def deg_to_dms(deg):\n",
-      "    d = int(deg)\n",
-      "    md = abs(deg - d) * 60\n",
-      "    m = int(md)\n",
-      "    sd = (md - m) * 60\n",
-      "    sd=round(sd,2)\n",
-      "    return [d, m, sd]\n",
-      "\n",
-      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"Half angular width=\",deg_to_dms(theta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Half angular width= [0, 12, 1.93]\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.5, Page number 2-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "a = 12*10**-5        #width of slit(cm)\n",
-      "n = 1                #first order\n",
-      "lamda = 6000*10**-8  #wavelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"Half angular width=\",(theta),\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Half angular width= 30.0 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.6, Page number 2-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "a = 2*10**-6         #width of slit(cm)\n",
-      "n = 1                #first order\n",
-      "lamda = 6500*10**-10 #wavelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"Angle theta =\",round(theta,2),\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angle theta = 18.97 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3.1, Page number 2-16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "a = 0.16  #width(mm)\n",
-      "b = 0.8   #distance(mm)\n",
-      "\n",
-      "#Calculations & Results\n",
-      "m = ((a+b)/a)\n",
-      "\n",
-      "for n in range(1,4):\n",
-      "    print \"m =\",m*n\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "m = 6.0\n",
-        "m = 12.0\n",
-        "m = 18.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.1, Page number 2-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "lamda1 = 5*10**-5 #cm\n",
-      "lamda2 = 7*10**-5 #cm\n",
-      "a_plus_b = 1./4000 #cm\n",
-      "\n",
-      "#Calculations\n",
-      "m_max1 = a_plus_b/lamda1\n",
-      "m_max2 = round((a_plus_b/lamda2),1)\n",
-      "\n",
-      "#Results\n",
-      "print m_max2,\"orders are visible for 7000 A and\",m_max1,\"orders for 5000 A and either\",m_max2,\",4 or\",m_max1,\"for intermediate wavelengths\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3.6 orders are visible for 7000 A and 5.0 orders for 5000 A and either 3.6 ,4 or 5.0 for intermediate wavelengths\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.2, Page number 2-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "theta = 30   #angle of diffraction(degrees)\n",
-      "lamda1 = 5400*10**-8  #cm\n",
-      "lamda2 = 4050*10**-8  #cm\n",
-      "\n",
-      "#Calculations\n",
-      "m = lamda2/(lamda1-lamda2)\n",
-      "a_plus_b = (m*lamda1)/math.sin(theta*math.pi/180)\n",
-      "N = 1/a_plus_b\n",
-      "\n",
-      "#Result\n",
-      "print \"Number of lines per cm =\",round(N,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of lines per cm = 3086.42\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.3, Page number 2-25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 4992 #A\n",
-      "m1 = 3       #for 3rd order\n",
-      "m2 = 4       #for 4th order\n",
-      "\n",
-      "#Calculations\n",
-      "'''For m = 3,\n",
-      "(a+b)sin0 = 3*lamda\n",
-      "\n",
-      "For m = 4,\n",
-      "(a+b)sin0 = 4*lamda'''\n",
-      "\n",
-      "l = (m2*lamda)/m1\n",
-      "\n",
-      "#Results\n",
-      "print \"Wavelength =\",l,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 6656 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.4, Page number 2-25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 6328*10**-8 #wavelength(cm)\n",
-      "m1 = 1\n",
-      "m2 = 2\n",
-      "a_plus_b = 1./6000  #cm\n",
-      "\n",
-      "#Calculations\n",
-      "theta1 = math.asin((m1*lamda)/a_plus_b)*180/math.pi\n",
-      "theta2 = math.asin((m2*lamda)/a_plus_b)*180/math.pi\n",
-      "\n",
-      "#Result\n",
-      "print \"01 =\",round(theta1,2),\"degrees\"\n",
-      "print \"02 =\",round(theta2,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "01 = 22.31 degrees\n",
-        "02 = 49.41 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.5, Page number 2-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "m = 2              #2nd order\n",
-      "lamda = 5*10**-5   #wavelength(cm)\n",
-      "theta = 30\n",
-      "\n",
-      "#Calculations\n",
-      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
-      "N = 1/a_plus_b\n",
-      "\n",
-      "#Result\n",
-      "print \"The number of lines/cm of the grating surface is\",N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The number of lines/cm of the grating surface is 5000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.6, Page number 2-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "m = 3              #3rd order\n",
-      "a_plus_b = 1./7000 #no. of lines/cm\n",
-      "sin0 = 1\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = (a_plus_b*sin0)/m\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength =\",round(lamda/1e-8),\"A\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 4762.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.7, Page number 2-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "m = 1                 #1st order\n",
-      "lamda = 6560*10**-8   #wavelength(cm)\n",
-      "theta = 16+12./60\n",
-      "\n",
-      "#Calculations\n",
-      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
-      "N = 1/a_plus_b\n",
-      "w = 2*N\n",
-      "\n",
-      "#Result\n",
-      "print \"The total number of lines is\",round(w)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total number of lines is 8506.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.8, Page number 2-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "m = 1                 #1st order\n",
-      "lamda = 6.56*10**-5   #wavelength(cm)\n",
-      "theta = 18+14./60\n",
-      "\n",
-      "#Calculations\n",
-      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
-      "N = 1/a_plus_b\n",
-      "w = 2*N\n",
-      "\n",
-      "#Result\n",
-      "print \"The total number of lines is\",round(w,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total number of lines is 9539.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.9, Page number 2-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 6*10**-5    #wavelength(cm)\n",
-      "a_plus_b = 1./5000  #cm\n",
-      "\n",
-      "#Calculations\n",
-      "m_max = a_plus_b/lamda\n",
-      "\n",
-      "#Result\n",
-      "print \"Order =\",round(m_max)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Order = 3.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.10, Page number 2-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 6000*10**-8    #wavelength(cm)\n",
-      "a_plus_b = 1./5000     #cm\n",
-      "\n",
-      "#Calculations\n",
-      "m_max = a_plus_b/lamda\n",
-      "\n",
-      "#Result\n",
-      "print \"m_max =\",round(m_max)\n",
-      "\n",
-      "#The rest of the solution is theoretical"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "m_max = 3.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4.11, Page number 2-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "m = 1                 #1st order\n",
-      "lamda = 5790*10**-8   #wavelength(cm)\n",
-      "theta = 19.994\n",
-      "\n",
-      "#Calculations\n",
-      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
-      "N = 1/a_plus_b\n",
-      "w = 2.54*N\n",
-      "\n",
-      "#Result\n",
-      "print \"The total number of lines is\",round(w)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total number of lines is 15000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6.1, Page number 2-31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Varaible declaration\n",
-      "n = 3000./0.5           #no. of lines per cm\n",
-      "m = 2                   #for 2nd order\n",
-      "lamda1 = 5893*10**-8    #wavelength(cm)\n",
-      "lamda2 = 5896*10**-8    #wavelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "a_plus_b = 1/n\n",
-      "theta1 = math.degrees(math.asin((2*lamda1)/(a_plus_b)))\n",
-      "theta2 = math.degrees(math.asin((2*lamda2)/(a_plus_b)))\n",
-      "d = theta2-theta1       #angular seperation\n",
-      "N = lamda1*10**8/(m*3)\n",
-      "\n",
-      "#Result\n",
-      "print \"Since N=\",round(N),\"which is smaller than 3000 lines, the two lines will be resolved in the second order\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Since N= 982.0 which is smaller than 3000 lines, the two lines will be resolved in the second order\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6.2, Page number 2-32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Varaible declaration\n",
-      "m = 3          #for 3rd order\n",
-      "lamda = 481    #wavelength(nm)\n",
-      "n = 620        #no. of ruling per mm\n",
-      "w = 5.05       #width(mm)\n",
-      "\n",
-      "#Calculations\n",
-      "N = n*w\n",
-      "dl = lamda/(m*N)\n",
-      "\n",
-      "#Result\n",
-      "print \"Smallest wavelength interval =\",round(dl,4),\"nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Smallest wavelength interval = 0.0512 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6.3, Page number 2-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Varaible declaration\n",
-      "m = 2          #for 2nd order\n",
-      "lamda = 5890   #wavelength(A)\n",
-      "dl = 6         #A\n",
-      "n = 500.       #no. of lines per cm\n",
-      "\n",
-      "#Calculations\n",
-      "N = lamda/(dl*m)\n",
-      "W = N/n\n",
-      "\n",
-      "#Result\n",
-      "print \"Least width =\",W,\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Least width = 0.98 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6.4, Page number 2-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "N = 3*5000               #no. of lines per cm\n",
-      "a_plus_b = 1./5000       #cm\n",
-      "lamda = 5890*10**-8 #wavelength(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "m_max = round(a_plus_b/lamda)\n",
-      "RP = m_max*N\n",
-      "\n",
-      "#Result\n",
-      "print \"Maximum value of resolving power =\",RP"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum value of resolving power = 45000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6.5, Page number 2-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "m = 2          #for 2nd order\n",
-      "lamda = 5890   #wavelength(A)\n",
-      "dl = 6         #A\n",
-      "w = 3.         #width of a line(cm)\n",
-      "\n",
-      "#Calculations\n",
-      "lamda2 = lamda+dl\n",
-      "N = lamda/(dl*m)\n",
-      "\n",
-      "a_plus_b = w/N\n",
-      "\n",
-      "#Results\n",
-      "print \"The minimum no. of lines a grating must have is\",N\n",
-      "print \"The grating element is\",round(a_plus_b/1e-3,2),\"*10^-3 cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum no. of lines a grating must have is 490\n",
-        "The grating element is 6.12 *10^-3 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6.6, Page number 2-34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "N = 40000   #no. of lines per cm\n",
-      "m = 2       #for 2nd order\n",
-      "\n",
-      "#Calculations\n",
-      "RP = m*N\n",
-      "\n",
-      "#Result\n",
-      "print \"Maximum value of resolving power =\",RP"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum value of resolving power = 80000\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics_II/Chapter_3.ipynb b/Applied_Physics_II/Chapter_3.ipynb
deleted file mode 100755
index 61c99435..00000000
--- a/Applied_Physics_II/Chapter_3.ipynb
+++ /dev/null
@@ -1,779 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Fibre Optics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.1, Page number 3-6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.54   #refractive index of core\n",
-      "NA = 0.5    #numerical aperture\n",
-      "\n",
-      "#Calculation\n",
-      "n2 = math.sqrt(n1**2-NA**2)\n",
-      "\n",
-      "#Result\n",
-      "print \"Refractive index of cladding is\",round(n2,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index of cladding is 1.46\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.2, Page number 3-6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n2 = 1.59   #refractive index of cladding\n",
-      "NA = 0.2    #numerical aperture\n",
-      "n0 = 1.33\n",
-      "\n",
-      "#Calculation\n",
-      "n1 = (math.sqrt(n2**2-NA**2))\n",
-      "theta_o = (math.asin((math.sqrt(n2**2-n1**2)/n0)))*180/math.pi\n",
-      "\n",
-      "#Result\n",
-      "print \"Refractive index of core is\",n2\n",
-      "print \"Acceptance angle =\",round(theta_o,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index of core is 1.59\n",
-        "Acceptance angle = 8.65 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.3, Page number 3-6\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.49   #refractive index of core\n",
-      "n2 = 1.44   #refractive index of cladding\n",
-      "\n",
-      "#Calculation\n",
-      "NA = math.sqrt(n1**2-n2**2)\n",
-      "\n",
-      "theta_o = math.degrees(math.asin(NA))\n",
-      "\n",
-      "#Result\n",
-      "print \"Numerical Aperture =\",round(NA,5)\n",
-      "print \"Acceptance angle =\",round(theta_o,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numerical Aperture = 0.38275\n",
-        "Acceptance angle = 22.5 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.4, Page number 3-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.6   #refractive index of core\n",
-      "n2 = 1.3   #refractive index of cladding\n",
-      "\n",
-      "#Calculation\n",
-      "theta_c = math.degrees(math.asin(n2/n1))\n",
-      "\n",
-      "theta_o = math.degrees(math.asin(math.sqrt(n1**2-n2**2)))\n",
-      "AC = 2*theta_o\n",
-      "\n",
-      "#Result\n",
-      "print \"Critical angle =\",round(theta_c,2),\"degrees\"\n",
-      "print \"Value of angle of acceptance cone =\",round(AC,3),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Critical angle = 54.34 degrees\n",
-        "Value of angle of acceptance cone = 137.731 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.5, Page number 3-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.4         #refractive index of core\n",
-      "theta_o = 30     #acceptance angle(degrees)\n",
-      "\n",
-      "#Calculation\n",
-      "n2 = math.sqrt(n1**2-math.sin(math.radians(theta_o))**2)\n",
-      "\n",
-      "#Result\n",
-      "print \"Refractive index of cladding is\",round(n2,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index of cladding is 1.3077\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.6, Page number 3-8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.563   #refractive index of core\n",
-      "n2 = 1.498   #refractive index of cladding\n",
-      "\n",
-      "#Calculation\n",
-      "delta = (n1-n2)/n1\n",
-      "\n",
-      "#Result\n",
-      "print \"Fractional index change =\",round(delta,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fractional index change = 0.0416\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.7, Page number 3-8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.50       #refractive index of cladding\n",
-      "theta_c = 90-5  #critical angle(degrees)\n",
-      "\n",
-      "#Calculation\n",
-      "n2 = math.sin(theta_c*math.pi/180)*n1\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum index of refraction allowed for cladding is\",round(n2,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum index of refraction allowed for cladding is 1.4943\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.8, Page number 3-8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.33       #refractive index \n",
-      "theta_o = 30    #acceptance angle in air\n",
-      "\n",
-      "#Calculations\n",
-      "theta_0 = math.degrees(math.asin(math.sin(theta_o*math.pi/180)/n1))\n",
-      "\n",
-      "#Result\n",
-      "print \"Acceptance angle =\",round(theta_0,2),\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Acceptance angle = 22.08 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4.1, Page number 3-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.52           #refractive index of core\n",
-      "n2 = 1.5189         #refractive index of cladding\n",
-      "d = 29*10**-6       #core diameter(m)\n",
-      "lamda = 1.3*10**-6  #wavelength(m)\n",
-      "\n",
-      "#Calculation\n",
-      "V = (math.pi*d*math.sqrt(n1**2-n2**2))/lamda\n",
-      "\n",
-      "N = V**2/2\n",
-      "\n",
-      "#Results\n",
-      "print \"Normalized frequency =\",round(V,3)\n",
-      "print \"Number of modes =\",round(N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Normalized frequency = 4.052\n",
-        "Number of modes = 8.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4.2, Page number 3-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.47           #refractive index of core\n",
-      "n2 = 1.46           #refractive index of cladding\n",
-      "lamda = 1300*10**-9 #wavelength(nm)\n",
-      "V = 2.405           #for single mode fibre\n",
-      "\n",
-      "#Calculation\n",
-      "d = (V*lamda)/(math.pi*math.sqrt(n1**2-n2**2))\n",
-      "r = d/2\n",
-      "\n",
-      "#Result\n",
-      "print \"Radius =\",round(r/1e-6,3),\"um\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius = 2.907 um\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4.3, Page number 3-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.48           #refractive index of core\n",
-      "delta = 0.055       #relative RI\n",
-      "lamda = 1           #wavelength(um)\n",
-      "r = 50              #core radius(um)\n",
-      "\n",
-      "#Calculations\n",
-      "n2 = -((delta*n1)-n1)\n",
-      "\n",
-      "NA = math.sqrt(n1**2-n2**2)\n",
-      "\n",
-      "theta_o = math.degrees(math.asin(NA))\n",
-      "\n",
-      "V = (math.pi*2*r*NA)/lamda\n",
-      "\n",
-      "N = V**2/2\n",
-      "\n",
-      "#Results\n",
-      "print \"Refractive index of cladding =\",n2\n",
-      "print \"NA =\",round(NA,3)\n",
-      "print \"Acceptance angle =\",round(theta_o,2),\"degrees\"\n",
-      "print \"Normalized frequency =\",round(V,3)\n",
-      "print \"Number of modes =\",round(N)  #Answer differs due to rounding off in 'V'"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refractive index of cladding = 1.3986\n",
-        "NA = 0.484\n",
-        "Acceptance angle = 28.95 degrees\n",
-        "Normalized frequency = 152.073\n",
-        "Number of modes = 11563.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4.4, Page number 3-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.45           #refractive index of core\n",
-      "n2 = 1.448          #refractive index of cladding\n",
-      "lamda = 1*10**-6    #wavelength(m)\n",
-      "d = 6*10**-6        #core diameter(m)\n",
-      "\n",
-      "#Calculations\n",
-      "#Case i\n",
-      "theta_c = math.degrees(math.asin(n2/n1))\n",
-      "\n",
-      "#Case ii\n",
-      "theta_o = math.degrees(math.asin(math.sqrt(n1**2-n2**2)))\n",
-      "\n",
-      "#Case iii\n",
-      "NA = math.sqrt(n1**2-n2**2)\n",
-      "N = (math.pi**2*d**2*NA**2)/(2*lamda**2)\n",
-      "\n",
-      "#Results\n",
-      "print \"Critical angle =\",round(theta_c),\"degrees\"\n",
-      "print \"Acceptance angle =\",round(theta_o,3),\"degrees\"\n",
-      "print \"Number of modes =\",round(N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Critical angle = 87.0 degrees\n",
-        "Acceptance angle = 4.366 degrees\n",
-        "Number of modes = 1.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4.5, Page number 3-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.50           #refractive index of core\n",
-      "n2 = 1.48           #refractive index of cladding\n",
-      "lamda = 1*10**-6    #wavelength(m)\n",
-      "d = 2*50*10**-6     #core diameter(m)\n",
-      "\n",
-      "#Calculations\n",
-      "NA = math.sqrt(n1**2-n2**2)\n",
-      "N = (math.pi**2*d**2*NA**2)/(2*lamda**2)\n",
-      "\n",
-      "#Result\n",
-      "print \"Number of modes =\",round(N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of modes = 2941.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4.6, Page number 3-13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n1 = 1.55           #refractive index of core\n",
-      "n2 = 1.50           #refractive index of cladding\n",
-      "lamda = 1400*10**-9 #wavelength(m)\n",
-      "d = 40*10**-6       #core diameter(m)\n",
-      "\n",
-      "#Calculations\n",
-      "NA = math.sqrt(n1**2-n2**2)\n",
-      "\n",
-      "delta = (n1-n2)/n1\n",
-      "\n",
-      "V = (math.pi*d*NA)/lamda\n",
-      "\n",
-      "#Results\n",
-      "print \"NA =\",round(NA,4)\n",
-      "print \"Fractional index change =\",round(delta,5)\n",
-      "print \"V-number =\",round(V,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "NA = 0.3905\n",
-        "Fractional index change = 0.03226\n",
-        "V-number = 35.05\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6.1, Page number 3-17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Pout = 0.3   #output power(mW)\n",
-      "Pin = 1      #input power(mW)\n",
-      "L = 0.1      #fibre length(km)\n",
-      "\n",
-      "#Calculation\n",
-      "a = (-10/L)*math.log10(Pout/Pin)\n",
-      "\n",
-      "#Result\n",
-      "print \"Attenuation =\",round(a,2),\"dB/km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Attenuation = 52.29 dB/km\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6.2, Page number 3-18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Pin = 9      #input power(mW)\n",
-      "L = 3        #fibre length(km)\n",
-      "a = 1.5      #loss(dB/km)\n",
-      "\n",
-      "#Calculation\n",
-      "Pl = a*L\n",
-      "Pout = Pin*10**(-Pl/10)\n",
-      "\n",
-      "#Result\n",
-      "print \"Output power =\",round(Pout,3),\"uW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output power = 3.193 uW\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6.3, Page number 3-18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "a = 2.2    #attenuation(dB/km)\n",
-      "l1 = 2     #km\n",
-      "l2 = 6     #km\n",
-      "from sympy import * \n",
-      "Pin = symbols('Pin')\n",
-      "\n",
-      "#Calculations\n",
-      "#For 2km,\n",
-      "Pl1 = a*l1\n",
-      "Po1 = Pin*round(10**(-Pl1/10),3)\n",
-      "\n",
-      "#For 6km,\n",
-      "Pl2 = a*l2\n",
-      "Po2 = Pin*round(10**(-Pl2/10),3)\n",
-      "\n",
-      "#Results\n",
-      "print \"After 2 km, Pout =\",Po1\n",
-      "print \"After 6 km, Pout =\",Po2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "After 2 km, Pout = 0.363*Pin\n",
-        "After 6 km, Pout = 0.048*Pin\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6.4, Page number 3-19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "Pout = 7.5   #output power(mW)\n",
-      "Pin = 8.6    #input power(mW)\n",
-      "L = 0.5      #fibre length(km)\n",
-      "\n",
-      "#Calculation\n",
-      "Pl = -10*math.log10(Pout/Pin)\n",
-      "a = Pl/L\n",
-      "\n",
-      "#Result\n",
-      "print \"Loss specification =\",round(a,4),\"dB/km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loss specification = 1.1887 dB/km\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics_II/Chapter_4.ipynb b/Applied_Physics_II/Chapter_4.ipynb
deleted file mode 100755
index f94b4b09..00000000
--- a/Applied_Physics_II/Chapter_4.ipynb
+++ /dev/null
@@ -1,141 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4: Lasers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6.1, Page number 4-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "P = 3.147*10**-3     #output power(W)\n",
-      "t = 60               #time(sec)\n",
-      "lamda = 632.8*10**-9 #wavelength(m)\n",
-      "h = 6.63*10**-34     #Planc's constant(J-s)\n",
-      "c = 3*10**8          #velocity of light(m/s)\n",
-      "\n",
-      "#Calculations\n",
-      "N = (P*t*lamda)/(h*c)\n",
-      "\n",
-      "#Result\n",
-      "print \"No. of photons emittd each minute is\",round(N/1e+17),\"*10^17\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "No. of photons emittd each minute is 6.0 *10^17\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6.2, Page number 4-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "T = 300              #temperature(K)\n",
-      "lamda = 694.3*10**-9 #wavelength(m)\n",
-      "h = 6.63*10**-34     #Planc's constant(J-s)\n",
-      "c = 3*10**8          #velocity of light(m/s)\n",
-      "k = 1.38*10**-23     #Boltman's constant(J/K)\n",
-      "\n",
-      "#Calculations\n",
-      "N2_by_N1 = math.exp((-h*c)/(lamda*k*T))\n",
-      "                    \n",
-      "#Result\n",
-      "print \"Ratio of population =\",round(N2_by_N1/1e-31,3),\"*10^-31\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ratio of population = 8.874 *10^-31\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6.3, Page number 4-8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "P = 100*10**3        #output power(W)\n",
-      "t = 20*10**-9        #time(sec)\n",
-      "N = 6.981*10**15     #no. of photons\n",
-      "h = 6.63*10**-34     #Planc's constant(J-s)\n",
-      "c = 3*10**8          #velocity of light(m/s)\n",
-      "\n",
-      "#Calculations\n",
-      "lamda = (N*h*c)/(P*t)\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength =\",round(lamda/1e-10),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 6943.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics_II/Chapter_5.ipynb b/Applied_Physics_II/Chapter_5.ipynb
deleted file mode 100755
index 2ad76975..00000000
--- a/Applied_Physics_II/Chapter_5.ipynb
+++ /dev/null
@@ -1,1097 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5:Quantum Mechanics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.1, Page number 5-5"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 10**-2          #mass of object(kg)\n",
-      "v = 1               #velocity(m/s)\n",
-      "l = 1*10**-10       #wavelength(m)\n",
-      "d = 10**-3          #distance travelled(m)\n",
-      "\n",
-      "#Calculations\n",
-      "lamda = h/(m*v)\n",
-      "\n",
-      "v = h/(m*l)\n",
-      "\n",
-      "t1 = d/v\n",
-      "t = t1/(365*24*60*60)\n",
-      "\n",
-      "#Results\n",
-      "print \"de Brogile wavelength =\",round(lamda/1e-32,2),\"*10^-32 m\"\n",
-      "print \"Velocity =\",round(v/1e-22,2),\"*10^-22 m/s(Calculation mistake in the textbook)\"\n",
-      "print \"Distance travelled =\",round(t/1e+10,2),\"*10^10 years(Calculation mistake in the textbook)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "de Brogile wavelength = 6.63 *10^-32 m\n",
-        "Velocity = 6.63 *10^-22 m/s(Calculation mistake in the textbook)\n",
-        "Distance travelled = 4.78 *10^10 years(Calculation mistake in the textbook)\n"
-       ]
-      }
-     ],
-     "prompt_number": 57
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.2, Page number 5-6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "lamda = 1*10**-10   #wavelength(m)\n",
-      "\n",
-      "#calculation\n",
-      "v = h/(m*lamda)\n",
-      "\n",
-      "#Result\n",
-      "print \"Velocity =\",round(v/1e+6,2),\"*10^6 m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity = 7.29 *10^6 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.3, Page number 5-6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34     #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31      #mass of electron(kg)\n",
-      "lamda = 5000*10**-10 #wavelength(m)\n",
-      "\n",
-      "#calculation\n",
-      "E = (h**2/(2*m*lamda**2))/(1.6*10**-19)\n",
-      "\n",
-      "#Result\n",
-      "print \"Kinetic energy =\",round(E/1e-6,5),\"*10^-6 eV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Kinetic energy = 6.03804 *10^-6 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.4, Page number 5-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34      #Planck's constant(J-s)\n",
-      "m = 1.676*10**-27     #mass of object(kg)\n",
-      "E = 0.025*1.6*10**-19 #energy(J)\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = h/math.sqrt(2*m*E)\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength =\",round(lamda/1e-10,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 1.81 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.5, Page number 5-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34      #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31       #mass of electron(kg)\n",
-      "E = 120*1.6*10**-19   #energy(J)\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = h/math.sqrt(2*m*E)\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength =\",round(lamda/1e-10,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 1.12 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.6, Page number 5-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 1.67*10**-27    #mass of neutron(kg)\n",
-      "lamda = 1*10**-10   #wavelength(m)\n",
-      "\n",
-      "#calculation\n",
-      "v = h/(m*lamda)\n",
-      "\n",
-      "E = (h**2/(2*m*lamda**2))/(1.6*10**-19)\n",
-      "\n",
-      "#Result\n",
-      "print \"Velocity =\",round(v,2),\"m/s\"\n",
-      "print \"Kinetic energy =\",round(E/1e-2,3),\"*10^-2 eV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity = 3970.06 m/s\n",
-        "Kinetic energy = 8.225 *10^-2 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.7, Page number 5-8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 1               #mass of object(kg)\n",
-      "v = 1               #velocity(m/s)\n",
-      "V = 182             #potential differnce(V)\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i\n",
-      "lamda = 12.27/math.sqrt(V)\n",
-      "\n",
-      "#Case ii\n",
-      "l = h/(m*v)\n",
-      "\n",
-      "#Results\n",
-      "print \"de Brogile wavelength for accelerated electron=\",round(lamda,2),\"A\"\n",
-      "print \"de Brogile wavelength for object=\",l,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "de Brogile wavelength for accelerated electron= 0.91 A\n",
-        "de Brogile wavelength for object= 6.63e-34 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.8, Page number 5-9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "lamda = 10**-14     #wavelength(m)\n",
-      "\n",
-      "#Calculations\n",
-      "p = h/lamda\n",
-      "\n",
-      "E = (p**2/(2*m))/(1.6*10**-13)\n",
-      "\n",
-      "#Results\n",
-      "print \"Momentum =\",p,\"kg-m/s\"\n",
-      "print \"Energy =\",round(E,2),\"MeV\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Momentum = 6.63e-20 kg-m/s\n",
-        "Energy = 15095.09 MeV\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.9, Page number 5-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "V = 3000            #potential differnce(V)\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i\n",
-      "lamda = 12.27/math.sqrt(V)\n",
-      "\n",
-      "#Case ii\n",
-      "p = h/(lamda*10**-10)\n",
-      "\n",
-      "#Case iii\n",
-      "lamda_b = 1/(lamda*10**-10)\n",
-      "\n",
-      "#Case iv\n",
-      "d = 2.04*10**-10  #m\n",
-      "n = 1             #for first order\n",
-      "theta = math.degrees(math.asin((n*lamda*10**-10)/(2*d)))\n",
-      "\n",
-      "#Results\n",
-      "print \"Momentum =\",round(p/1e-23,2),\"*10^-23 kg-m/s\"\n",
-      "print \"de Brogile wavelength =\",round(lamda,3),\"A\"\n",
-      "print \"Wave number =\",round(lamda_b/1e+10,3),\"*10^10 /m\"\n",
-      "print \"Bragg angle =\",round(theta,3),\"degrees\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Momentum = 2.96 *10^-23 kg-m/s\n",
-        "de Brogile wavelength = 0.224 A\n",
-        "Wave number = 4.464 *10^10 /m\n",
-        "Bragg angle = 3.147 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.10, Page number 5-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "V = 10*10**3        #potential differnce(V)\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = 12.27/math.sqrt(V)\n",
-      "\n",
-      "p = h/(lamda*10**-10)\n",
-      "\n",
-      "#result\n",
-      "print \"Wavelength =\",lamda,\"A\"\n",
-      "print \"Momentum =\",round(p/1e-23,3),\"*10^-23 kg-m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 0.1227 A\n",
-        "Momentum = 5.403 *10^-23 kg-m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.12, Page number 5-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 6.68*10**-27    #mass of particle(kg)\n",
-      "E = 1.6*10**-16     #energy(J)\n",
-      "\n",
-      "#Calculations\n",
-      "lamda = h/math.sqrt(2*m*E)\n",
-      "\n",
-      "v = h/(m*lamda)\n",
-      "\n",
-      "#Results\n",
-      "print \"Wavelength =\",round(lamda/1e-13,3),\"*10^-13 A\"\n",
-      "print \"Velocity =\",round(v/1e+5,2),\"*10^5 m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength = 4.535 *10^-13 A\n",
-        "Velocity = 2.19 *10^5 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.13, Page number 5-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "c = 3*10**8         #velocity of light(m/s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "E = 1.6*10**-19     #energy(J)\n",
-      "\n",
-      "#Calculations\n",
-      "lamda_ph = (h*c)/E            #wavelength of photon\n",
-      " \n",
-      "lamda_e = h/math.sqrt(2*m*E)  #wavelength of electron\n",
-      "\n",
-      "#Result\n",
-      "print \"Wavelength of proton =\",round(lamda_ph/1e-6,3),\"*10^-6 m\"\n",
-      "print \"Wavelength of electron =\",round(lamda_e/1e-10,3),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Wavelength of proton = 1.243 *10^-6 m\n",
-        "Wavelength of electron = 12.286 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3.14, Page number 5-13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "c = 3*10**8         #velocity of light(m/s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "\n",
-      "#Calculations\n",
-      "E = m*c**2\n",
-      "lamda = h/math.sqrt(2*m*E)\n",
-      "\n",
-      "#Result\n",
-      "print \"de Brogile wavelength =\",round(lamda/1e-12,3),\"*10^-12 m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "de Brogile wavelength = 1.717 *10^-12 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7.1, Page number 5-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "del_v = (0.01*400)/100\n",
-      "\n",
-      "#Calculation\n",
-      "del_x = h/(4*math.pi*m*del_v)\n",
-      "\n",
-      "#Result\n",
-      "print \"Accuracy =\",round(del_x/1e-3,2),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Accuracy = 1.45 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7.2, Page number 5-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "del_x = 10*10**-9   #m\n",
-      "E = 1.6*10**-16     #energy(J)\n",
-      "\n",
-      "#Calculation\n",
-      "del_px = h/(4*math.pi*del_x)\n",
-      "p = math.sqrt(2*m*E)\n",
-      "per = (del_px/p)*100\n",
-      "\n",
-      "#Result\n",
-      "print \"Percentage of uncertainity =\",round(per,4),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage of uncertainity = 0.0309 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7.3, Page number 5-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "del_v = (0.01*4*10**5)/100\n",
-      "\n",
-      "#Calculation\n",
-      "del_x = h/(4*math.pi*m*del_v)\n",
-      "\n",
-      "#Result\n",
-      "print \"Accuracy =\",round(del_x/1e-6,2),\"*10^-6 m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Accuracy = 1.45 *10^-6 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7.4, Page number 5-27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "del_v = (1.88*10**6)/100\n",
-      "\n",
-      "#Calculation\n",
-      "del_x = h/(4*math.pi*m*del_v)\n",
-      "\n",
-      "#Result\n",
-      "print \"Precision =\",round(del_x/1e-9,3),\"*10^-9 m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Precision = 3.084 *10^-9 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7.5, Page number 5-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 4*10**-7    #wavelength(m)\n",
-      "c = 3*10**8         #velocity of light(m/s)\n",
-      "del_l = 8*10**-15   #spectral width(m)\n",
-      "\n",
-      "#calculation\n",
-      "del_t = lamda**2/(4*math.pi*c*del_l)\n",
-      "\n",
-      "#Result\n",
-      "print \"Time spent by the elctrons =\",round(del_t/1e-9,3),\"*10^-9 s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time spent by the elctrons = 5.305 *10^-9 s\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7.6, Page number 5-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34      #Planck's constant(J-s)\n",
-      "del_t = 1.4*10**-10   #time spent(s)\n",
-      "\n",
-      "#calculation\n",
-      "E = (h/(4*math.pi*del_t))/(1.6*10**-19)\n",
-      "\n",
-      "#Result\n",
-      "print \"Uncertainity =\",round(E/1e-6,2),\"*10^-6 eV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Uncertainity = 2.36 *10^-6 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7.7, Page number 5-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 546*10**-9  #wavelength(m)\n",
-      "c = 3*10**8         #velocity of light(m/s)\n",
-      "del_l = 10**-14     #spectral width(m)\n",
-      "\n",
-      "#calculation\n",
-      "del_t = lamda**2/(4*math.pi*c*del_l)\n",
-      "\n",
-      "#Result\n",
-      "print \"Time spent by the elctrons =\",round(del_t/1e-9,2),\"*10^-9 s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time spent by the elctrons = 7.91 *10^-9 s\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15.1, Page number 5-41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "L = 2*10**-10       #width(m)\n",
-      "\n",
-      "#Calculation\n",
-      "E1 = (h**2/(8*m*L**2))/(1.6*10**-19)\n",
-      "E2 = 4*E1\n",
-      "E3 = 9*E1\n",
-      "\n",
-      "#Result\n",
-      "print \"Energy of electron in ground state =\",round(E1,3),\"eV\"\n",
-      "print \"Energy of electron in first state =\",round(E2,3),\"eV\"\n",
-      "print \"Energy of electron in second state =\",round(E3,3),\"eV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy of electron in ground state = 9.434 eV\n",
-        "Energy of electron in first state = 37.738 eV\n",
-        "Energy of electron in second state = 84.91 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15.2, Page number 5-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "L = 5.6*10**-3      #width(m)\n",
-      "\n",
-      "#Calculation\n",
-      "E = L/4\n",
-      "\n",
-      "#Result\n",
-      "print \"Ground state energy =\",E,\"eV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ground state energy = 0.0014 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15.4, Page number 5-43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import scipy\n",
-      "from scipy import integrate\n",
-      "\n",
-      "#Variable declaration\n",
-      "#Intervals\n",
-      "x1 = 0\n",
-      "x2 = 1./2\n",
-      "\n",
-      "#Calculation\n",
-      "x = lambda x: 3*x**2\n",
-      "P = integrate.quad(x, x1, x2)\n",
-      "\n",
-      "#Result\n",
-      "print \"Probability =\",P[0]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Probability = 0.125\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15.5, Page number 5-44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m1 = 9.1*10**-31    #mass of electron(kg)\n",
-      "m2 = 10**-9         #mass of grain dust(kg)\n",
-      "L1 = 10**-9         #width(m)\n",
-      "L2 = 10**-4         #width(m)\n",
-      "\n",
-      "#Calculation\n",
-      "#For electron\n",
-      "print \"For an electron, the lowest thre energy states obtained for n=1,2 and 3 are\"\n",
-      "for n in range(1,4):\n",
-      "    En1 = ((n**2*h**2)/(8*m*L1**2))/(1.6*10**-19)\n",
-      "    print round(En1,4),\"eV\"\n",
-      "    \n",
-      "#For the grain of dust\n",
-      "print \"\\nFor a grain of dust, the lowest thre energy states obtained for n=1,2 and 3 are\"\n",
-      "for n in range(1,4):\n",
-      "    En2 = ((n**2*h**2)/(8*m2*L2**2))/(1.6*10**-19)\n",
-      "    print round(En2/1e-32,3),\"eV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For an electron, the lowest thre energy states obtained for n=1,2 and 3 are\n",
-        "0.3774 eV\n",
-        "1.5095 eV\n",
-        "3.3964 eV\n",
-        "\n",
-        "For a grain of dust, the lowest thre energy states obtained for n=1,2 and 3 are\n",
-        "3.434 eV\n",
-        "13.737 eV\n",
-        "30.907 eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15.6, Page number 5-45\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "E1 = 38*1.6*10**-19 #energy(J)\n",
-      "\n",
-      "#Calculation\n",
-      "L = math.sqrt(h**2/(8*m*E1))\n",
-      "              \n",
-      "#Result\n",
-      "print \"Width of well =\",round(L/1e-10,4),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Width of well = 0.9965 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15.7, Page number 5-45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
-      "m = 9.1*10**-31     #mass of electron(kg)\n",
-      "L = 5*10**-10       #width(m)\n",
-      "n1 = 1\n",
-      "n2 = 2\n",
-      "c = 3*10**8         #velocity of light(m/s)\n",
-      "\n",
-      "#Calculation\n",
-      "E = ((3*h**2)/(8*m*L**2))   #E2-E1\n",
-      "Ev = E/(1.6-10**-19)        #J\n",
-      "lamda = (h*c)/E\n",
-      "\n",
-      "#Result\n",
-      "print \"Energy =\",round(Ev/1e-19,2),\"eV\"\n",
-      "print \"Wavelength =\",round(lamda/1e-7,3),\"*10^-7 m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy = 4.53 eV\n",
-        "Wavelength = 2.745 *10^-7 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics_II/Chapter_6.ipynb b/Applied_Physics_II/Chapter_6.ipynb
deleted file mode 100755
index cc609a3f..00000000
--- a/Applied_Physics_II/Chapter_6.ipynb
+++ /dev/null
@@ -1,189 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6: Motion of charge particle in Electric & Magnetic fields"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1.1, Page number 6-6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "m = 9.1*10**-31         #mass of electron(kg)\n",
-      "v = 2.5*10**6           #velocity of electron(m/s)\n",
-      "theta = 30*math.pi/180  #angle(degrees)\n",
-      "B = 0.94*10**-4         #field strength(Wb/m^2)\n",
-      "e = 1.6*10**-19         #electron charge(C)\n",
-      "\n",
-      "#Calculations\n",
-      "r = (m*v*math.sin(theta))/(B*e)\n",
-      "\n",
-      "l = (5*v*math.cos(theta)*2*math.pi*m)/(B*e)\n",
-      "\n",
-      "#Results\n",
-      "print \"Radius =\",round(r/1e-3,2),\"mm\"\n",
-      "print \"Distance covered =\",round(l,3),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius = 75.63 mm\n",
-        "Distance covered = 4.115 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1.2, Page number 6-7    "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "m = 9.1*10**-31         #mass of electron(kg)\n",
-      "v = 3*10**7             #velocity of electron(m/s)\n",
-      "theta = 45*math.pi/180  #angle(degrees)\n",
-      "B = 0.23*10**-4         #field strength(Wb/m^2)\n",
-      "e = 1.6*10**-19         #electron charge(C)\n",
-      "\n",
-      "#Calculations\n",
-      "r = (m*v*math.sin(theta))/(B*e)\n",
-      "\n",
-      "l = (v*math.cos(theta)*2*math.pi*m)/(B*e)\n",
-      "\n",
-      "#Results\n",
-      "print \"Radius =\",round(r,2),\"m\"\n",
-      "print \"Distance covered =\",round(l,3),\"m\"\n",
-      "#Calculation mistakes in the textbook"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radius = 5.25 m\n",
-        "Distance covered = 32.959 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1.3, Page number 6-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "y = 1.5         #displacement(cm)\n",
-      "d = 0.42        #distance(cm)\n",
-      "Va = 1.6*10**3  #anode voltage(V)\n",
-      "D = 28          #cm\n",
-      "l = 1.8         #length of plates(cm)\n",
-      "\n",
-      "#Calculation\n",
-      "V = (2*y*d*Va)/(D*l)\n",
-      "Vin = V/6\n",
-      "\n",
-      "#Result\n",
-      "print \"Input voltage =\",round(Vin,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input voltage = 6.67 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5.1, Page number 6-16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "da = 0.8    #minor axis(cm)\n",
-      "db = 2      #major axis(cm)\n",
-      "\n",
-      "#Calculation\n",
-      "ps = math.degrees(math.asin(da/db))\n",
-      "\n",
-      "#Result\n",
-      "print \"Phase shift =\",round(ps,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Phase shift = 23.58 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics_II/Chapter_7.ipynb b/Applied_Physics_II/Chapter_7.ipynb
deleted file mode 100755
index 3425e831..00000000
--- a/Applied_Physics_II/Chapter_7.ipynb
+++ /dev/null
@@ -1,178 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7: Superconductivity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3.1, Page number 7-6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Ho = 2.*10**5   #A/m\n",
-      "Hc = 1.*10**5   #A/m\n",
-      "T = 8.          #temperature(K)\n",
-      "\n",
-      "#Calculations\n",
-      "Tc = math.sqrt(T**2/(1-(Hc/Ho)))\n",
-      "\n",
-      "#Result\n",
-      "print \"Critical temperature =\",round(Tc,3),\"K\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Critical temperature = 11.314 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3.2, Page number 7-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Bo = 0.0306   #Critical field at 0K(T)\n",
-      "Tc = 3.7      #critical temperature(K)\n",
-      "T = 2         #K\n",
-      "\n",
-      "#calculation\n",
-      "Bc = Bo*(1-((T/Tc)**2))\n",
-      "\n",
-      "#Result\n",
-      "print \"Critical field at 2K is\",round(Bc,5),\"T\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Critical field at 2K is 0.02166 T\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3.3, Page number 7-7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Ho = 6.5*10**4   #Critical field at 0K(T)\n",
-      "Tc = 7.18        #critical temperature(K)\n",
-      "T = 4.2          #K\n",
-      "r = 0.5*10**-3   #radius(m)\n",
-      "\n",
-      "#calculation\n",
-      "Hc = Ho*(1-((T/Tc)**2))\n",
-      "Ic = 2*math.pi*r*Hc\n",
-      "\n",
-      "#Result\n",
-      "print \"Critical current =\",round(Ic,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Critical current = 134.33 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3.4, Page number 7-8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "Tc1 = 4.185      #critical temperature(K)\n",
-      "M1 = 199.5       #isotopic mass\n",
-      "Tc2 = 4.133      #K\n",
-      "\n",
-      "#calculation\n",
-      "M2 = ((Tc1*M1**0.5)/Tc2)**2\n",
-      "\n",
-      "#Result\n",
-      "print \"Isotopic mass =\",round(M2,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Isotopic mass = 204.55\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Applied_Physics_II/README.txt b/Applied_Physics_II/README.txt
deleted file mode 100755
index 717d1591..00000000
--- a/Applied_Physics_II/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Muktesh Chaudhary
-Course: be
-College/Institute/Organization: Anglo Eastern ship management india Pvt. Ltd
-Department/Designation: Electrical & Electronics Officer
-Book Title: Applied Physics II
-Author: H. J. Sawant
-Publisher: Technical Publications
-Year of publication: 2014
-Isbn: 978-93-5038-883-9
-Edition: 2nd
\ No newline at end of file
diff --git a/Applied_Physics_II/screenshots/halfangular.png b/Applied_Physics_II/screenshots/halfangular.png
deleted file mode 100755
index 6c38e8b8..00000000
Binary files a/Applied_Physics_II/screenshots/halfangular.png and /dev/null differ
diff --git a/Applied_Physics_II/screenshots/probability.png b/Applied_Physics_II/screenshots/probability.png
deleted file mode 100755
index f8d2ba6f..00000000
Binary files a/Applied_Physics_II/screenshots/probability.png and /dev/null differ
diff --git a/Applied_Physics_II/screenshots/visiblerange.png b/Applied_Physics_II/screenshots/visiblerange.png
deleted file mode 100755
index c4d798f7..00000000
Binary files a/Applied_Physics_II/screenshots/visiblerange.png and /dev/null differ
diff --git a/Applied_Physics_II_by_H_J_Sawant/Chapter_1.ipynb b/Applied_Physics_II_by_H_J_Sawant/Chapter_1.ipynb
new file mode 100755
index 00000000..c6e00369
--- /dev/null
+++ b/Applied_Physics_II_by_H_J_Sawant/Chapter_1.ipynb
@@ -0,0 +1,1108 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Interference of Light"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.1, Page number 1-11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "i = 45        #angle of incidence(degrees)\n",
+      "t = 4*10**-5  #thickness of film(cm)\n",
+      "u = 1.2\n",
+      "\n",
+      "#Calculations & Result\n",
+      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
+      "for n in range(1,4):\n",
+      "    lamda = (2*u*t*math.cos(r*math.pi/180))/n\n",
+      "    print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-8)\n",
+      "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n = 1, wavelength = 7756.29 A\n",
+        "For n = 2, wavelength = 3878.14 A\n",
+        "For n = 3, wavelength = 2585.43 A\n",
+        "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.2, Page number 1-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 90        #angle of refraction(degrees)\n",
+      "t = 5*10**-5  #thickness of film(cm)\n",
+      "u = 1.33\n",
+      "\n",
+      "#Calculations & Result\n",
+      "for n in range(1,4):\n",
+      "    lamda = (4*u*t*int(math.cos(math.radians(90))))/((2*n)-1)\n",
+      "    print \"For n = %d, wavelength = %.2f A\"%(n,lamda)\n",
+      "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\"\n",
+      "\n",
+      "print \"\\nPlease note: Since r=90, cos(r)=0\\nHence, the answers given in the textbook are incorrect\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n = 1, wavelength = 0.00 A\n",
+        "For n = 2, wavelength = 0.00 A\n",
+        "For n = 3, wavelength = 0.00 A\n",
+        "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range\n",
+        "\n",
+        "Please note: Since r=90, cos(r)=0\n",
+        "Hence, the answers given in the textbook are incorrect\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.3, Page number 1-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "i = 45           #angle of incidence(degrees)\n",
+      "t = 1.5*10**-4   #thickness of film(cm)\n",
+      "lamda = 5*10**-5 #wavelength(cm)\n",
+      "u = 4./3.        #refractive index\n",
+      "\n",
+      "#Calculations\n",
+      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
+      "n = (2*u*t*math.cos(r*math.pi/180))/lamda\n",
+      "\n",
+      "#Result\n",
+      "print \"The order of interfernce is %.2f, close to 7\"%n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The order of interfernce is 6.78, close to 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2.4, Pae number 1-13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "i = 45              #angle of incidence(degrees)\n",
+      "lamda = 5896*10**-8 #wavelength(cm)\n",
+      "u = 1.33            #refractive index\n",
+      "n = 1\n",
+      "\n",
+      "#Calculations\n",
+      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
+      "t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n",
+      "\n",
+      "#Result\n",
+      "print \"The required thickness is\",round((t/1E-5),2),\"*10^-5 cm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required thickness is 1.31 *10^-5 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.5, Page number 1-14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "lamda1 = 7000  #wavelength(A)\n",
+      "lamda2 = 5000  #wavelength(A)\n",
+      "u = 1.3        #R.I. of oil\n",
+      "\n",
+      "#Calculations\n",
+      "'''\n",
+      "2utcosr = (2n-1)7000/2   ----(1)\n",
+      "2utcosr = (2n+1)5000/2   ----(2)\n",
+      "Divinding (1) by (2), we get the following expression\n",
+      "1 = (2n+1)5000\n",
+      "    -----------\n",
+      "    (2n-1)7000\n",
+      "Solving the above expression, we get,\n",
+      "'''\n",
+      "n = 12000/4000\n",
+      "t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)\n",
+      "\n",
+      "#Result\n",
+      "print \"The required thickness is\",round((t/1E-5),4),\"*10^-5 cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required thickness is 6.6936 *10^-5 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.6, Page number 1-15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "i = 30              #angle of incidence(degrees)\n",
+      "lamda = 5890*10**-8 #wavelength(cm)\n",
+      "u = 1.46            #refractive index\n",
+      "n = 8\n",
+      "\n",
+      "#Calculations\n",
+      "r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))\n",
+      "t = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
+      "\n",
+      "#Result\n",
+      "print \"The required thickness is\",round((t/1E-4),3),\"*10^-4 cm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required thickness is 1.718 *10^-4 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.7, Page number 1-15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 60              #angle of refraction(degrees)\n",
+      "lamda = 5890*10**-8 #wavelength(cm)\n",
+      "u = 1.5             #refractive index\n",
+      "n = 1               #for minimumm thickness\n",
+      "\n",
+      "#Calculations\n",
+      "#For r = 60\n",
+      "t1 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
+      "\n",
+      "#For normal incidence \n",
+      "r = 0\n",
+      "t2 = (n*lamda)/(2*u*math.cos(r*math.pi/180))\n",
+      "\n",
+      "#Result\n",
+      "print \"For r = 60, the required thickness is\",round((t1/1E-5),2),\"*10^-5 cm\"\n",
+      "print \"For r = 0, the required thickness is\",round((t2/1E-5),2),\"*10^-5 cm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For r = 60, the required thickness is 3.93 *10^-5 cm\n",
+        "For r = 0, the required thickness is 1.96 *10^-5 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.8, Page number 1-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 0               #for normal incidence(degrees)\n",
+      "lamda = 5.5*10**-5  #wavelength(cm)\n",
+      "n = 1               #for minimumm thickness\n",
+      "A = 10**4           #area(cm^2)\n",
+      "V = 0.2             #volume(cc)\n",
+      "\n",
+      "#Calculations\n",
+      "t = V/A\n",
+      "#for nth dark band,\n",
+      "u = (n*lamda)/(2*t*math.cos(r*math.pi/180))\n",
+      "\n",
+      "#Result\n",
+      "print \"Refractive index =\",u"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index = 1.375\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2.9, Page number 1-17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 60        #angle of incidence(degrees)\n",
+      "t = 2*10**-7  #thickness of film(cm)\n",
+      "u = 1.2\n",
+      "\n",
+      "#Calculations & Result\n",
+      "for n in range(1,4):\n",
+      "    lamda = (4*u*t*math.cos(r*math.pi/180))/(2*n-1)\n",
+      "    print \"For n = %d, wavelength = %.2f A\"%(n,lamda/10**-10)\n",
+      "print \"Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For n = 1, wavelength = 4800.00 A\n",
+        "For n = 2, wavelength = 1600.00 A\n",
+        "For n = 3, wavelength = 960.00 A\n",
+        "Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3.1, Page number 1-21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "a = 40.       #angle(sec)\n",
+      "lamda = 1.2  #distance between fringes(cm)\n",
+      "alpha = 10   #no. of fringes\n",
+      "\n",
+      "#Calculations\n",
+      "Bair = lamda/alpha          #cm\n",
+      "alpha = (a*math.pi)/(3600*180)   #radians\n",
+      "lamda = 2*alpha*Bair\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength of monochromatic light =\",round((lamda/1E-8),1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of monochromatic light = 4654.2 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3.2, Page number 1-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "lamda = 5893*10**-8   #wavelength(cm)\n",
+      "u = 1.52              #refractive index\n",
+      "B = 0.1               #fringe spacing(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "alpha = (lamda/(2*u*B))*180*3600/math.pi   #seconds\n",
+      "\n",
+      "#Result\n",
+      "print \"Angle of wedge =\",round(alpha,2),\"secs\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of wedge = 39.98 secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3.3, Page number 1-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "u = 1.4              #refractive index\n",
+      "B = 0.25             #fringe spacing(cm)\n",
+      "a = 20               #angle(secs)\n",
+      "\n",
+      "#Calculations\n",
+      "alpha = (a*math.pi)/(3600*180)  #radians\n",
+      "lamda = 2*u*alpha*B\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength of light =\",round((lamda/1E-8),2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light = 6787.39 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3.4, Page number 1-23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "u = 1.5              #refractive index\n",
+      "lamda = 5.82*10**-5  #wavelength(cm)\n",
+      "a = 20               #angle(secs)\n",
+      "\n",
+      "#Calculations\n",
+      "alpha = (a*math.pi)/(3600*180)  #radians\n",
+      "B = lamda/(2*u*alpha)\n",
+      "N = 1/B\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of interfernce fronges pr cm is\",round(N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of interfernce fronges pr cm is 5.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3.5, Page number 1-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "u = 1                #refractive index for air film\n",
+      "lamda = 6*10**-5     #wavelength(cm)\n",
+      "B = 1./10            #distance between fringes(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "alpha = lamda/(2*u*B)  #radians\n",
+      "d = alpha*10\n",
+      "\n",
+      "#Result\n",
+      "print \"Daimeter of wire =\",d,\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Daimeter of wire = 0.003 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3.6, Page number 1-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "alpha = 0.01*10**-1/10   #angle(radians)\n",
+      "u = 1                    #refractive index for air film\n",
+      "lamda = 5900*10**-10     #wavlength(m)\n",
+      "\n",
+      "#Calculation\n",
+      "B = lamda/(2*u*alpha)\n",
+      "\n",
+      "#Result\n",
+      "print \"Seperation between fringes is\",B/10**-3,\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Seperation between fringes is 2.95 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.1, Page number 1-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "n = 40\n",
+      "\n",
+      "#Calculation\n",
+      "#Equating the equation 4*R*n*lamda=4*4R*n*lamda, we get\n",
+      "\n",
+      "N = (4*4*n)/4\n",
+      "\n",
+      "#result\n",
+      "print \"Ring number =\",N"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ring number = 160\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.2, Page number 1-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "n = 10\n",
+      "Dn = 0.5         #diameter of dark ring(cm)\n",
+      "lamda = 5*10**-5 #waelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "R = Dn**2/(4*n*lamda)\n",
+      "\n",
+      "#Result\n",
+      "print \"Radius of curvature =\",R,\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature = 125.0 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.3, Page number 1-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "n = 5\n",
+      "p = 10\n",
+      "D5 = 0.336          #diameter of 5th ring(cm)\n",
+      "lamda = 5890*10**-8 #waelength(cm)\n",
+      "D15 = 0.59          #diameter of 15th ringcm\n",
+      "\n",
+      "#Calculations\n",
+      "R = (D15**2-D5**2)/(4*p*lamda)\n",
+      "\n",
+      "#Result\n",
+      "print \"Radius of curvature =\",round(R,2),\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius of curvature = 99.83 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.4, Page number 1-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Varaible declaration\n",
+      "Dn = 0.42     #diameter of dark ring(cm)\n",
+      "p = 8 \n",
+      "R = 200       #radius of curvature(cm)\n",
+      "Dn8 = 0.7     #diameter of (n+8)th ring(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "lamda = (Dn8**2-Dn**2)/(4*R*p)\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength =\",lamda/1E-8,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 4900.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.5, Page number 1-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Varaible declaration\n",
+      "Dn = 0.218             #cm\n",
+      "Dn10 = 0.451           #cm\n",
+      "lamda = 5893*10**-8    #wavelength(cm)\n",
+      "R = 90                 #radius of curvature(cm)\n",
+      "p = 10\n",
+      "\n",
+      "#Calculation\n",
+      "u = (4*p*lamda*R)/(Dn10**2-Dn**2)\n",
+      "\n",
+      "#Result\n",
+      "print \"Refractive index =\",round(u,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index = 1.361\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.6, Page number 1-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "D5 = 0.42     #diameter of dark ring(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "'''\n",
+      "For 5th dark ring,\n",
+      "D5^2 = 20*R*lamda  -----1\n",
+      "\n",
+      "For 10th dark ring,\n",
+      "D10^2 = 40*R*lamda -----2\n",
+      "\n",
+      "Substituting 1 in 2,\n",
+      "'''\n",
+      "\n",
+      "D10 = math.sqrt((40*D5**2)/20)\n",
+      "\n",
+      "#Result\n",
+      "print \"Diameter of the 10th dark ring =\",round(D10,3),\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of the 10th dark ring = 0.594 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.7, Page number 1-35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Varaible declaration\n",
+      "lamda_n = 6000    #wavelength of nth ring(A)\n",
+      "lamda_n1 = 5000   #wavelength for (n+1)th ring(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "'''\n",
+      "Dn^2 = 4*R*n*lamda_n ---1\n",
+      "\n",
+      "Dn+1^2 = 4*R(n+1)*lamda_n1  ---2\n",
+      "\n",
+      "Equating 1 and 2, we get,\n",
+      "'''\n",
+      "\n",
+      "n = 5\n",
+      "R = 2\n",
+      "Dn = math.sqrt(4*R*n*lamda_n*10**-8)\n",
+      "\n",
+      "#Result\n",
+      "print \"Diameter =\",round(Dn,3),\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter = 0.049 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.8, Page number 1-35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Varaible declaration\n",
+      "Dair = 2.3   #diameter of ring in air(cm)\n",
+      "Dliq = 2     #diameter of ring in liquid(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "u = Dair**2/Dliq**2\n",
+      "\n",
+      "#Result\n",
+      "print \"Refractive index =\",u"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index = 1.3225\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.11, Page number 1-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "D4 = 0.4    #diameter of 4th ring(cm)\n",
+      "D12 = 0.7   #diameter of 12th ring(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "'''\n",
+      "For D4, \n",
+      "D4 = math.sqrt(4R*4*lamda)\n",
+      "'''\n",
+      "rt_Rl = 0.1\n",
+      "R = 80      \n",
+      "\n",
+      "#For D20,\n",
+      "D20 = math.sqrt(R)*rt_Rl\n",
+      "\n",
+      "#Result\n",
+      "print \"Diameter of 20th ring =\",round(D20,3),\"cm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of 20th ring = 0.894 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4.12, Page number 1-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration \n",
+      "n = 5\n",
+      "p = 10\n",
+      "D5 = 0.336          #diameter of 5th ring(cm)\n",
+      "D15 = 0.590         #diameter of 15th ring(cm)\n",
+      "R = 100             #radius of curvature(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "lamda = (D15**2-D5**2)/(4*R*p)\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavlength =\",lamda/1E-8,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavlength = 5880.1 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7.1, Page number 1-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 560 #wavelength(nm)\n",
+      "u = 2       #refractive index\n",
+      "\n",
+      "#Calculations\n",
+      "lamda_dash = lamda/u\n",
+      "t = lamda_dash/4\n",
+      "\n",
+      "#Result\n",
+      "print \"Thickness of film =\",t,\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of film = 70 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7.2, Page number 1-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 6000 #wavelength(E)\n",
+      "u = 1.2       #refractive index\n",
+      "\n",
+      "#Calculations\n",
+      "lamda_dash = lamda/u\n",
+      "t = lamda_dash/4\n",
+      "\n",
+      "#Result\n",
+      "print \"Thickness of film =\",t,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of film = 1250.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_II_by_H_J_Sawant/Chapter_2.ipynb b/Applied_Physics_II_by_H_J_Sawant/Chapter_2.ipynb
new file mode 100755
index 00000000..ae487344
--- /dev/null
+++ b/Applied_Physics_II_by_H_J_Sawant/Chapter_2.ipynb
@@ -0,0 +1,1000 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Diffraction of Light"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2.1, Page number 2-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta = 30          #angle(degrees)\n",
+      "n = 1        \n",
+      "lamda = 6500*10**-8 #wavelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "a = (n*lamda)/math.sin(theta*math.pi/180)\n",
+      "\n",
+      "#Result\n",
+      "print \"a =\",a/1e-4,\"*10^-4 cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = 1.3 *10^-4 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2.2, Page number 2-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "a = 6*10**-4        #width of slit(cm)\n",
+      "n = 1               #first order\n",
+      "lamda = 6000*10**-8 #wavelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "def deg_to_dms(deg):\n",
+      "    d = int(deg)\n",
+      "    md = abs(deg - d) * 60\n",
+      "    m = int(md)\n",
+      "    sd = (md - m) * 60\n",
+      "    sd=round(sd,2)\n",
+      "    return [d, m, sd]\n",
+      "\n",
+      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
+      "d = 2*theta   #angular seperation\n",
+      "\n",
+      "#Result\n",
+      "print \"Angular seperation between the 1st order minima is\",deg_to_dms(d)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angular seperation between the 1st order minima is [11, 28, 42.03]\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2.3, Page number 2-11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 2       #for second minimum\n",
+      "n2 = 3       #for third minimum\n",
+      "lamda = 4000 #wavelength(A)\n",
+      "\n",
+      "#Calculations\n",
+      "'''For 2nd order,\n",
+      "a sin0 = n1*lamda\n",
+      "\n",
+      "For 3rd order,\n",
+      "a sin0 = n2*lamda'''\n",
+      "\n",
+      "lamda = (n2*lamda)/n1\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength =\",lamda,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 6000 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2.4, Page number 2-11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "a = 0.16*10**-3      #width of slit(cm)\n",
+      "n = 1                #first order\n",
+      "lamda = 5600*10**-10 #wavelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "def deg_to_dms(deg):\n",
+      "    d = int(deg)\n",
+      "    md = abs(deg - d) * 60\n",
+      "    m = int(md)\n",
+      "    sd = (md - m) * 60\n",
+      "    sd=round(sd,2)\n",
+      "    return [d, m, sd]\n",
+      "\n",
+      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print \"Half angular width=\",deg_to_dms(theta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Half angular width= [0, 12, 1.93]\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2.5, Page number 2-11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "a = 12*10**-5        #width of slit(cm)\n",
+      "n = 1                #first order\n",
+      "lamda = 6000*10**-8  #wavelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print \"Half angular width=\",(theta),\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Half angular width= 30.0 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2.6, Page number 2-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "a = 2*10**-6         #width of slit(cm)\n",
+      "n = 1                #first order\n",
+      "lamda = 6500*10**-10 #wavelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "theta = (math.asin((n*lamda)/a))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print \"Angle theta =\",round(theta,2),\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle theta = 18.97 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3.1, Page number 2-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "a = 0.16  #width(mm)\n",
+      "b = 0.8   #distance(mm)\n",
+      "\n",
+      "#Calculations & Results\n",
+      "m = ((a+b)/a)\n",
+      "\n",
+      "for n in range(1,4):\n",
+      "    print \"m =\",m*n\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "m = 6.0\n",
+        "m = 12.0\n",
+        "m = 18.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.1, Page number 2-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "lamda1 = 5*10**-5 #cm\n",
+      "lamda2 = 7*10**-5 #cm\n",
+      "a_plus_b = 1./4000 #cm\n",
+      "\n",
+      "#Calculations\n",
+      "m_max1 = a_plus_b/lamda1\n",
+      "m_max2 = round((a_plus_b/lamda2),1)\n",
+      "\n",
+      "#Results\n",
+      "print m_max2,\"orders are visible for 7000 A and\",m_max1,\"orders for 5000 A and either\",m_max2,\",4 or\",m_max1,\"for intermediate wavelengths\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3.6 orders are visible for 7000 A and 5.0 orders for 5000 A and either 3.6 ,4 or 5.0 for intermediate wavelengths\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.2, Page number 2-24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "theta = 30   #angle of diffraction(degrees)\n",
+      "lamda1 = 5400*10**-8  #cm\n",
+      "lamda2 = 4050*10**-8  #cm\n",
+      "\n",
+      "#Calculations\n",
+      "m = lamda2/(lamda1-lamda2)\n",
+      "a_plus_b = (m*lamda1)/math.sin(theta*math.pi/180)\n",
+      "N = 1/a_plus_b\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of lines per cm =\",round(N,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of lines per cm = 3086.42\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.3, Page number 2-25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 4992 #A\n",
+      "m1 = 3       #for 3rd order\n",
+      "m2 = 4       #for 4th order\n",
+      "\n",
+      "#Calculations\n",
+      "'''For m = 3,\n",
+      "(a+b)sin0 = 3*lamda\n",
+      "\n",
+      "For m = 4,\n",
+      "(a+b)sin0 = 4*lamda'''\n",
+      "\n",
+      "l = (m2*lamda)/m1\n",
+      "\n",
+      "#Results\n",
+      "print \"Wavelength =\",l,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 6656 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.4, Page number 2-25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 6328*10**-8 #wavelength(cm)\n",
+      "m1 = 1\n",
+      "m2 = 2\n",
+      "a_plus_b = 1./6000  #cm\n",
+      "\n",
+      "#Calculations\n",
+      "theta1 = math.asin((m1*lamda)/a_plus_b)*180/math.pi\n",
+      "theta2 = math.asin((m2*lamda)/a_plus_b)*180/math.pi\n",
+      "\n",
+      "#Result\n",
+      "print \"01 =\",round(theta1,2),\"degrees\"\n",
+      "print \"02 =\",round(theta2,2),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "01 = 22.31 degrees\n",
+        "02 = 49.41 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.5, Page number 2-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 2              #2nd order\n",
+      "lamda = 5*10**-5   #wavelength(cm)\n",
+      "theta = 30\n",
+      "\n",
+      "#Calculations\n",
+      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
+      "N = 1/a_plus_b\n",
+      "\n",
+      "#Result\n",
+      "print \"The number of lines/cm of the grating surface is\",N"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of lines/cm of the grating surface is 5000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.6, Page number 2-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 3              #3rd order\n",
+      "a_plus_b = 1./7000 #no. of lines/cm\n",
+      "sin0 = 1\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = (a_plus_b*sin0)/m\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength =\",round(lamda/1e-8),\"A\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 4762.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.7, Page number 2-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 1                 #1st order\n",
+      "lamda = 6560*10**-8   #wavelength(cm)\n",
+      "theta = 16+12./60\n",
+      "\n",
+      "#Calculations\n",
+      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
+      "N = 1/a_plus_b\n",
+      "w = 2*N\n",
+      "\n",
+      "#Result\n",
+      "print \"The total number of lines is\",round(w)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total number of lines is 8506.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.8, Page number 2-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 1                 #1st order\n",
+      "lamda = 6.56*10**-5   #wavelength(cm)\n",
+      "theta = 18+14./60\n",
+      "\n",
+      "#Calculations\n",
+      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
+      "N = 1/a_plus_b\n",
+      "w = 2*N\n",
+      "\n",
+      "#Result\n",
+      "print \"The total number of lines is\",round(w,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total number of lines is 9539.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.9, Page number 2-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 6*10**-5    #wavelength(cm)\n",
+      "a_plus_b = 1./5000  #cm\n",
+      "\n",
+      "#Calculations\n",
+      "m_max = a_plus_b/lamda\n",
+      "\n",
+      "#Result\n",
+      "print \"Order =\",round(m_max)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Order = 3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.10, Page number 2-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 6000*10**-8    #wavelength(cm)\n",
+      "a_plus_b = 1./5000     #cm\n",
+      "\n",
+      "#Calculations\n",
+      "m_max = a_plus_b/lamda\n",
+      "\n",
+      "#Result\n",
+      "print \"m_max =\",round(m_max)\n",
+      "\n",
+      "#The rest of the solution is theoretical"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "m_max = 3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4.11, Page number 2-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 1                 #1st order\n",
+      "lamda = 5790*10**-8   #wavelength(cm)\n",
+      "theta = 19.994\n",
+      "\n",
+      "#Calculations\n",
+      "a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)\n",
+      "N = 1/a_plus_b\n",
+      "w = 2.54*N\n",
+      "\n",
+      "#Result\n",
+      "print \"The total number of lines is\",round(w)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total number of lines is 15000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6.1, Page number 2-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Varaible declaration\n",
+      "n = 3000./0.5           #no. of lines per cm\n",
+      "m = 2                   #for 2nd order\n",
+      "lamda1 = 5893*10**-8    #wavelength(cm)\n",
+      "lamda2 = 5896*10**-8    #wavelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "a_plus_b = 1/n\n",
+      "theta1 = math.degrees(math.asin((2*lamda1)/(a_plus_b)))\n",
+      "theta2 = math.degrees(math.asin((2*lamda2)/(a_plus_b)))\n",
+      "d = theta2-theta1       #angular seperation\n",
+      "N = lamda1*10**8/(m*3)\n",
+      "\n",
+      "#Result\n",
+      "print \"Since N=\",round(N),\"which is smaller than 3000 lines, the two lines will be resolved in the second order\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Since N= 982.0 which is smaller than 3000 lines, the two lines will be resolved in the second order\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6.2, Page number 2-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Varaible declaration\n",
+      "m = 3          #for 3rd order\n",
+      "lamda = 481    #wavelength(nm)\n",
+      "n = 620        #no. of ruling per mm\n",
+      "w = 5.05       #width(mm)\n",
+      "\n",
+      "#Calculations\n",
+      "N = n*w\n",
+      "dl = lamda/(m*N)\n",
+      "\n",
+      "#Result\n",
+      "print \"Smallest wavelength interval =\",round(dl,4),\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Smallest wavelength interval = 0.0512 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6.3, Page number 2-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Varaible declaration\n",
+      "m = 2          #for 2nd order\n",
+      "lamda = 5890   #wavelength(A)\n",
+      "dl = 6         #A\n",
+      "n = 500.       #no. of lines per cm\n",
+      "\n",
+      "#Calculations\n",
+      "N = lamda/(dl*m)\n",
+      "W = N/n\n",
+      "\n",
+      "#Result\n",
+      "print \"Least width =\",W,\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Least width = 0.98 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6.4, Page number 2-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Varaible declaration\n",
+      "N = 3*5000               #no. of lines per cm\n",
+      "a_plus_b = 1./5000       #cm\n",
+      "lamda = 5890*10**-8 #wavelength(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "m_max = round(a_plus_b/lamda)\n",
+      "RP = m_max*N\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum value of resolving power =\",RP"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum value of resolving power = 45000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6.5, Page number 2-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Varaible declaration\n",
+      "m = 2          #for 2nd order\n",
+      "lamda = 5890   #wavelength(A)\n",
+      "dl = 6         #A\n",
+      "w = 3.         #width of a line(cm)\n",
+      "\n",
+      "#Calculations\n",
+      "lamda2 = lamda+dl\n",
+      "N = lamda/(dl*m)\n",
+      "\n",
+      "a_plus_b = w/N\n",
+      "\n",
+      "#Results\n",
+      "print \"The minimum no. of lines a grating must have is\",N\n",
+      "print \"The grating element is\",round(a_plus_b/1e-3,2),\"*10^-3 cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum no. of lines a grating must have is 490\n",
+        "The grating element is 6.12 *10^-3 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6.6, Page number 2-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Varaible declaration\n",
+      "N = 40000   #no. of lines per cm\n",
+      "m = 2       #for 2nd order\n",
+      "\n",
+      "#Calculations\n",
+      "RP = m*N\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum value of resolving power =\",RP"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum value of resolving power = 80000\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_II_by_H_J_Sawant/Chapter_3.ipynb b/Applied_Physics_II_by_H_J_Sawant/Chapter_3.ipynb
new file mode 100755
index 00000000..61c99435
--- /dev/null
+++ b/Applied_Physics_II_by_H_J_Sawant/Chapter_3.ipynb
@@ -0,0 +1,779 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Fibre Optics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.1, Page number 3-6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.54   #refractive index of core\n",
+      "NA = 0.5    #numerical aperture\n",
+      "\n",
+      "#Calculation\n",
+      "n2 = math.sqrt(n1**2-NA**2)\n",
+      "\n",
+      "#Result\n",
+      "print \"Refractive index of cladding is\",round(n2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of cladding is 1.46\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.2, Page number 3-6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n2 = 1.59   #refractive index of cladding\n",
+      "NA = 0.2    #numerical aperture\n",
+      "n0 = 1.33\n",
+      "\n",
+      "#Calculation\n",
+      "n1 = (math.sqrt(n2**2-NA**2))\n",
+      "theta_o = (math.asin((math.sqrt(n2**2-n1**2)/n0)))*180/math.pi\n",
+      "\n",
+      "#Result\n",
+      "print \"Refractive index of core is\",n2\n",
+      "print \"Acceptance angle =\",round(theta_o,2),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of core is 1.59\n",
+        "Acceptance angle = 8.65 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.3, Page number 3-6\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.49   #refractive index of core\n",
+      "n2 = 1.44   #refractive index of cladding\n",
+      "\n",
+      "#Calculation\n",
+      "NA = math.sqrt(n1**2-n2**2)\n",
+      "\n",
+      "theta_o = math.degrees(math.asin(NA))\n",
+      "\n",
+      "#Result\n",
+      "print \"Numerical Aperture =\",round(NA,5)\n",
+      "print \"Acceptance angle =\",round(theta_o,2),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture = 0.38275\n",
+        "Acceptance angle = 22.5 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.4, Page number 3-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.6   #refractive index of core\n",
+      "n2 = 1.3   #refractive index of cladding\n",
+      "\n",
+      "#Calculation\n",
+      "theta_c = math.degrees(math.asin(n2/n1))\n",
+      "\n",
+      "theta_o = math.degrees(math.asin(math.sqrt(n1**2-n2**2)))\n",
+      "AC = 2*theta_o\n",
+      "\n",
+      "#Result\n",
+      "print \"Critical angle =\",round(theta_c,2),\"degrees\"\n",
+      "print \"Value of angle of acceptance cone =\",round(AC,3),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical angle = 54.34 degrees\n",
+        "Value of angle of acceptance cone = 137.731 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.5, Page number 3-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.4         #refractive index of core\n",
+      "theta_o = 30     #acceptance angle(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "n2 = math.sqrt(n1**2-math.sin(math.radians(theta_o))**2)\n",
+      "\n",
+      "#Result\n",
+      "print \"Refractive index of cladding is\",round(n2,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of cladding is 1.3077\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.6, Page number 3-8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.563   #refractive index of core\n",
+      "n2 = 1.498   #refractive index of cladding\n",
+      "\n",
+      "#Calculation\n",
+      "delta = (n1-n2)/n1\n",
+      "\n",
+      "#Result\n",
+      "print \"Fractional index change =\",round(delta,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fractional index change = 0.0416\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.7, Page number 3-8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.50       #refractive index of cladding\n",
+      "theta_c = 90-5  #critical angle(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "n2 = math.sin(theta_c*math.pi/180)*n1\n",
+      "\n",
+      "#Result\n",
+      "print \"The maximum index of refraction allowed for cladding is\",round(n2,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum index of refraction allowed for cladding is 1.4943\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.8, Page number 3-8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.33       #refractive index \n",
+      "theta_o = 30    #acceptance angle in air\n",
+      "\n",
+      "#Calculations\n",
+      "theta_0 = math.degrees(math.asin(math.sin(theta_o*math.pi/180)/n1))\n",
+      "\n",
+      "#Result\n",
+      "print \"Acceptance angle =\",round(theta_0,2),\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Acceptance angle = 22.08 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4.1, Page number 3-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.52           #refractive index of core\n",
+      "n2 = 1.5189         #refractive index of cladding\n",
+      "d = 29*10**-6       #core diameter(m)\n",
+      "lamda = 1.3*10**-6  #wavelength(m)\n",
+      "\n",
+      "#Calculation\n",
+      "V = (math.pi*d*math.sqrt(n1**2-n2**2))/lamda\n",
+      "\n",
+      "N = V**2/2\n",
+      "\n",
+      "#Results\n",
+      "print \"Normalized frequency =\",round(V,3)\n",
+      "print \"Number of modes =\",round(N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Normalized frequency = 4.052\n",
+        "Number of modes = 8.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4.2, Page number 3-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.47           #refractive index of core\n",
+      "n2 = 1.46           #refractive index of cladding\n",
+      "lamda = 1300*10**-9 #wavelength(nm)\n",
+      "V = 2.405           #for single mode fibre\n",
+      "\n",
+      "#Calculation\n",
+      "d = (V*lamda)/(math.pi*math.sqrt(n1**2-n2**2))\n",
+      "r = d/2\n",
+      "\n",
+      "#Result\n",
+      "print \"Radius =\",round(r/1e-6,3),\"um\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius = 2.907 um\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4.3, Page number 3-11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.48           #refractive index of core\n",
+      "delta = 0.055       #relative RI\n",
+      "lamda = 1           #wavelength(um)\n",
+      "r = 50              #core radius(um)\n",
+      "\n",
+      "#Calculations\n",
+      "n2 = -((delta*n1)-n1)\n",
+      "\n",
+      "NA = math.sqrt(n1**2-n2**2)\n",
+      "\n",
+      "theta_o = math.degrees(math.asin(NA))\n",
+      "\n",
+      "V = (math.pi*2*r*NA)/lamda\n",
+      "\n",
+      "N = V**2/2\n",
+      "\n",
+      "#Results\n",
+      "print \"Refractive index of cladding =\",n2\n",
+      "print \"NA =\",round(NA,3)\n",
+      "print \"Acceptance angle =\",round(theta_o,2),\"degrees\"\n",
+      "print \"Normalized frequency =\",round(V,3)\n",
+      "print \"Number of modes =\",round(N)  #Answer differs due to rounding off in 'V'"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of cladding = 1.3986\n",
+        "NA = 0.484\n",
+        "Acceptance angle = 28.95 degrees\n",
+        "Normalized frequency = 152.073\n",
+        "Number of modes = 11563.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4.4, Page number 3-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.45           #refractive index of core\n",
+      "n2 = 1.448          #refractive index of cladding\n",
+      "lamda = 1*10**-6    #wavelength(m)\n",
+      "d = 6*10**-6        #core diameter(m)\n",
+      "\n",
+      "#Calculations\n",
+      "#Case i\n",
+      "theta_c = math.degrees(math.asin(n2/n1))\n",
+      "\n",
+      "#Case ii\n",
+      "theta_o = math.degrees(math.asin(math.sqrt(n1**2-n2**2)))\n",
+      "\n",
+      "#Case iii\n",
+      "NA = math.sqrt(n1**2-n2**2)\n",
+      "N = (math.pi**2*d**2*NA**2)/(2*lamda**2)\n",
+      "\n",
+      "#Results\n",
+      "print \"Critical angle =\",round(theta_c),\"degrees\"\n",
+      "print \"Acceptance angle =\",round(theta_o,3),\"degrees\"\n",
+      "print \"Number of modes =\",round(N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical angle = 87.0 degrees\n",
+        "Acceptance angle = 4.366 degrees\n",
+        "Number of modes = 1.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4.5, Page number 3-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.50           #refractive index of core\n",
+      "n2 = 1.48           #refractive index of cladding\n",
+      "lamda = 1*10**-6    #wavelength(m)\n",
+      "d = 2*50*10**-6     #core diameter(m)\n",
+      "\n",
+      "#Calculations\n",
+      "NA = math.sqrt(n1**2-n2**2)\n",
+      "N = (math.pi**2*d**2*NA**2)/(2*lamda**2)\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of modes =\",round(N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of modes = 2941.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4.6, Page number 3-13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.55           #refractive index of core\n",
+      "n2 = 1.50           #refractive index of cladding\n",
+      "lamda = 1400*10**-9 #wavelength(m)\n",
+      "d = 40*10**-6       #core diameter(m)\n",
+      "\n",
+      "#Calculations\n",
+      "NA = math.sqrt(n1**2-n2**2)\n",
+      "\n",
+      "delta = (n1-n2)/n1\n",
+      "\n",
+      "V = (math.pi*d*NA)/lamda\n",
+      "\n",
+      "#Results\n",
+      "print \"NA =\",round(NA,4)\n",
+      "print \"Fractional index change =\",round(delta,5)\n",
+      "print \"V-number =\",round(V,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NA = 0.3905\n",
+        "Fractional index change = 0.03226\n",
+        "V-number = 35.05\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6.1, Page number 3-17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Pout = 0.3   #output power(mW)\n",
+      "Pin = 1      #input power(mW)\n",
+      "L = 0.1      #fibre length(km)\n",
+      "\n",
+      "#Calculation\n",
+      "a = (-10/L)*math.log10(Pout/Pin)\n",
+      "\n",
+      "#Result\n",
+      "print \"Attenuation =\",round(a,2),\"dB/km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Attenuation = 52.29 dB/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6.2, Page number 3-18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Pin = 9      #input power(mW)\n",
+      "L = 3        #fibre length(km)\n",
+      "a = 1.5      #loss(dB/km)\n",
+      "\n",
+      "#Calculation\n",
+      "Pl = a*L\n",
+      "Pout = Pin*10**(-Pl/10)\n",
+      "\n",
+      "#Result\n",
+      "print \"Output power =\",round(Pout,3),\"uW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output power = 3.193 uW\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6.3, Page number 3-18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "a = 2.2    #attenuation(dB/km)\n",
+      "l1 = 2     #km\n",
+      "l2 = 6     #km\n",
+      "from sympy import * \n",
+      "Pin = symbols('Pin')\n",
+      "\n",
+      "#Calculations\n",
+      "#For 2km,\n",
+      "Pl1 = a*l1\n",
+      "Po1 = Pin*round(10**(-Pl1/10),3)\n",
+      "\n",
+      "#For 6km,\n",
+      "Pl2 = a*l2\n",
+      "Po2 = Pin*round(10**(-Pl2/10),3)\n",
+      "\n",
+      "#Results\n",
+      "print \"After 2 km, Pout =\",Po1\n",
+      "print \"After 6 km, Pout =\",Po2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After 2 km, Pout = 0.363*Pin\n",
+        "After 6 km, Pout = 0.048*Pin\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6.4, Page number 3-19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "Pout = 7.5   #output power(mW)\n",
+      "Pin = 8.6    #input power(mW)\n",
+      "L = 0.5      #fibre length(km)\n",
+      "\n",
+      "#Calculation\n",
+      "Pl = -10*math.log10(Pout/Pin)\n",
+      "a = Pl/L\n",
+      "\n",
+      "#Result\n",
+      "print \"Loss specification =\",round(a,4),\"dB/km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loss specification = 1.1887 dB/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_II_by_H_J_Sawant/Chapter_4.ipynb b/Applied_Physics_II_by_H_J_Sawant/Chapter_4.ipynb
new file mode 100755
index 00000000..f94b4b09
--- /dev/null
+++ b/Applied_Physics_II_by_H_J_Sawant/Chapter_4.ipynb
@@ -0,0 +1,141 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Lasers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6.1, Page number 4-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "P = 3.147*10**-3     #output power(W)\n",
+      "t = 60               #time(sec)\n",
+      "lamda = 632.8*10**-9 #wavelength(m)\n",
+      "h = 6.63*10**-34     #Planc's constant(J-s)\n",
+      "c = 3*10**8          #velocity of light(m/s)\n",
+      "\n",
+      "#Calculations\n",
+      "N = (P*t*lamda)/(h*c)\n",
+      "\n",
+      "#Result\n",
+      "print \"No. of photons emittd each minute is\",round(N/1e+17),\"*10^17\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "No. of photons emittd each minute is 6.0 *10^17\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6.2, Page number 4-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "T = 300              #temperature(K)\n",
+      "lamda = 694.3*10**-9 #wavelength(m)\n",
+      "h = 6.63*10**-34     #Planc's constant(J-s)\n",
+      "c = 3*10**8          #velocity of light(m/s)\n",
+      "k = 1.38*10**-23     #Boltman's constant(J/K)\n",
+      "\n",
+      "#Calculations\n",
+      "N2_by_N1 = math.exp((-h*c)/(lamda*k*T))\n",
+      "                    \n",
+      "#Result\n",
+      "print \"Ratio of population =\",round(N2_by_N1/1e-31,3),\"*10^-31\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio of population = 8.874 *10^-31\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6.3, Page number 4-8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "P = 100*10**3        #output power(W)\n",
+      "t = 20*10**-9        #time(sec)\n",
+      "N = 6.981*10**15     #no. of photons\n",
+      "h = 6.63*10**-34     #Planc's constant(J-s)\n",
+      "c = 3*10**8          #velocity of light(m/s)\n",
+      "\n",
+      "#Calculations\n",
+      "lamda = (N*h*c)/(P*t)\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength =\",round(lamda/1e-10),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 6943.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_II_by_H_J_Sawant/Chapter_5.ipynb b/Applied_Physics_II_by_H_J_Sawant/Chapter_5.ipynb
new file mode 100755
index 00000000..2ad76975
--- /dev/null
+++ b/Applied_Physics_II_by_H_J_Sawant/Chapter_5.ipynb
@@ -0,0 +1,1097 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5:Quantum Mechanics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.1, Page number 5-5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 10**-2          #mass of object(kg)\n",
+      "v = 1               #velocity(m/s)\n",
+      "l = 1*10**-10       #wavelength(m)\n",
+      "d = 10**-3          #distance travelled(m)\n",
+      "\n",
+      "#Calculations\n",
+      "lamda = h/(m*v)\n",
+      "\n",
+      "v = h/(m*l)\n",
+      "\n",
+      "t1 = d/v\n",
+      "t = t1/(365*24*60*60)\n",
+      "\n",
+      "#Results\n",
+      "print \"de Brogile wavelength =\",round(lamda/1e-32,2),\"*10^-32 m\"\n",
+      "print \"Velocity =\",round(v/1e-22,2),\"*10^-22 m/s(Calculation mistake in the textbook)\"\n",
+      "print \"Distance travelled =\",round(t/1e+10,2),\"*10^10 years(Calculation mistake in the textbook)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Brogile wavelength = 6.63 *10^-32 m\n",
+        "Velocity = 6.63 *10^-22 m/s(Calculation mistake in the textbook)\n",
+        "Distance travelled = 4.78 *10^10 years(Calculation mistake in the textbook)\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.2, Page number 5-6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "lamda = 1*10**-10   #wavelength(m)\n",
+      "\n",
+      "#calculation\n",
+      "v = h/(m*lamda)\n",
+      "\n",
+      "#Result\n",
+      "print \"Velocity =\",round(v/1e+6,2),\"*10^6 m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity = 7.29 *10^6 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.3, Page number 5-6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34     #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31      #mass of electron(kg)\n",
+      "lamda = 5000*10**-10 #wavelength(m)\n",
+      "\n",
+      "#calculation\n",
+      "E = (h**2/(2*m*lamda**2))/(1.6*10**-19)\n",
+      "\n",
+      "#Result\n",
+      "print \"Kinetic energy =\",round(E/1e-6,5),\"*10^-6 eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Kinetic energy = 6.03804 *10^-6 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.4, Page number 5-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34      #Planck's constant(J-s)\n",
+      "m = 1.676*10**-27     #mass of object(kg)\n",
+      "E = 0.025*1.6*10**-19 #energy(J)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = h/math.sqrt(2*m*E)\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength =\",round(lamda/1e-10,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 1.81 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.5, Page number 5-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34      #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31       #mass of electron(kg)\n",
+      "E = 120*1.6*10**-19   #energy(J)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = h/math.sqrt(2*m*E)\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength =\",round(lamda/1e-10,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 1.12 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.6, Page number 5-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 1.67*10**-27    #mass of neutron(kg)\n",
+      "lamda = 1*10**-10   #wavelength(m)\n",
+      "\n",
+      "#calculation\n",
+      "v = h/(m*lamda)\n",
+      "\n",
+      "E = (h**2/(2*m*lamda**2))/(1.6*10**-19)\n",
+      "\n",
+      "#Result\n",
+      "print \"Velocity =\",round(v,2),\"m/s\"\n",
+      "print \"Kinetic energy =\",round(E/1e-2,3),\"*10^-2 eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity = 3970.06 m/s\n",
+        "Kinetic energy = 8.225 *10^-2 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.7, Page number 5-8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 1               #mass of object(kg)\n",
+      "v = 1               #velocity(m/s)\n",
+      "V = 182             #potential differnce(V)\n",
+      "\n",
+      "#Calculation\n",
+      "#Case i\n",
+      "lamda = 12.27/math.sqrt(V)\n",
+      "\n",
+      "#Case ii\n",
+      "l = h/(m*v)\n",
+      "\n",
+      "#Results\n",
+      "print \"de Brogile wavelength for accelerated electron=\",round(lamda,2),\"A\"\n",
+      "print \"de Brogile wavelength for object=\",l,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Brogile wavelength for accelerated electron= 0.91 A\n",
+        "de Brogile wavelength for object= 6.63e-34 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.8, Page number 5-9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "lamda = 10**-14     #wavelength(m)\n",
+      "\n",
+      "#Calculations\n",
+      "p = h/lamda\n",
+      "\n",
+      "E = (p**2/(2*m))/(1.6*10**-13)\n",
+      "\n",
+      "#Results\n",
+      "print \"Momentum =\",p,\"kg-m/s\"\n",
+      "print \"Energy =\",round(E,2),\"MeV\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Momentum = 6.63e-20 kg-m/s\n",
+        "Energy = 15095.09 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.9, Page number 5-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "V = 3000            #potential differnce(V)\n",
+      "\n",
+      "#Calculation\n",
+      "#Case i\n",
+      "lamda = 12.27/math.sqrt(V)\n",
+      "\n",
+      "#Case ii\n",
+      "p = h/(lamda*10**-10)\n",
+      "\n",
+      "#Case iii\n",
+      "lamda_b = 1/(lamda*10**-10)\n",
+      "\n",
+      "#Case iv\n",
+      "d = 2.04*10**-10  #m\n",
+      "n = 1             #for first order\n",
+      "theta = math.degrees(math.asin((n*lamda*10**-10)/(2*d)))\n",
+      "\n",
+      "#Results\n",
+      "print \"Momentum =\",round(p/1e-23,2),\"*10^-23 kg-m/s\"\n",
+      "print \"de Brogile wavelength =\",round(lamda,3),\"A\"\n",
+      "print \"Wave number =\",round(lamda_b/1e+10,3),\"*10^10 /m\"\n",
+      "print \"Bragg angle =\",round(theta,3),\"degrees\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Momentum = 2.96 *10^-23 kg-m/s\n",
+        "de Brogile wavelength = 0.224 A\n",
+        "Wave number = 4.464 *10^10 /m\n",
+        "Bragg angle = 3.147 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.10, Page number 5-11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "V = 10*10**3        #potential differnce(V)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = 12.27/math.sqrt(V)\n",
+      "\n",
+      "p = h/(lamda*10**-10)\n",
+      "\n",
+      "#result\n",
+      "print \"Wavelength =\",lamda,\"A\"\n",
+      "print \"Momentum =\",round(p/1e-23,3),\"*10^-23 kg-m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 0.1227 A\n",
+        "Momentum = 5.403 *10^-23 kg-m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.12, Page number 5-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 6.68*10**-27    #mass of particle(kg)\n",
+      "E = 1.6*10**-16     #energy(J)\n",
+      "\n",
+      "#Calculations\n",
+      "lamda = h/math.sqrt(2*m*E)\n",
+      "\n",
+      "v = h/(m*lamda)\n",
+      "\n",
+      "#Results\n",
+      "print \"Wavelength =\",round(lamda/1e-13,3),\"*10^-13 A\"\n",
+      "print \"Velocity =\",round(v/1e+5,2),\"*10^5 m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength = 4.535 *10^-13 A\n",
+        "Velocity = 2.19 *10^5 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.13, Page number 5-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "c = 3*10**8         #velocity of light(m/s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "E = 1.6*10**-19     #energy(J)\n",
+      "\n",
+      "#Calculations\n",
+      "lamda_ph = (h*c)/E            #wavelength of photon\n",
+      " \n",
+      "lamda_e = h/math.sqrt(2*m*E)  #wavelength of electron\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength of proton =\",round(lamda_ph/1e-6,3),\"*10^-6 m\"\n",
+      "print \"Wavelength of electron =\",round(lamda_e/1e-10,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of proton = 1.243 *10^-6 m\n",
+        "Wavelength of electron = 12.286 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3.14, Page number 5-13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "c = 3*10**8         #velocity of light(m/s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "\n",
+      "#Calculations\n",
+      "E = m*c**2\n",
+      "lamda = h/math.sqrt(2*m*E)\n",
+      "\n",
+      "#Result\n",
+      "print \"de Brogile wavelength =\",round(lamda/1e-12,3),\"*10^-12 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Brogile wavelength = 1.717 *10^-12 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7.1, Page number 5-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "del_v = (0.01*400)/100\n",
+      "\n",
+      "#Calculation\n",
+      "del_x = h/(4*math.pi*m*del_v)\n",
+      "\n",
+      "#Result\n",
+      "print \"Accuracy =\",round(del_x/1e-3,2),\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Accuracy = 1.45 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7.2, Page number 5-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "del_x = 10*10**-9   #m\n",
+      "E = 1.6*10**-16     #energy(J)\n",
+      "\n",
+      "#Calculation\n",
+      "del_px = h/(4*math.pi*del_x)\n",
+      "p = math.sqrt(2*m*E)\n",
+      "per = (del_px/p)*100\n",
+      "\n",
+      "#Result\n",
+      "print \"Percentage of uncertainity =\",round(per,4),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of uncertainity = 0.0309 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7.3, Page number 5-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "del_v = (0.01*4*10**5)/100\n",
+      "\n",
+      "#Calculation\n",
+      "del_x = h/(4*math.pi*m*del_v)\n",
+      "\n",
+      "#Result\n",
+      "print \"Accuracy =\",round(del_x/1e-6,2),\"*10^-6 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Accuracy = 1.45 *10^-6 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7.4, Page number 5-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "del_v = (1.88*10**6)/100\n",
+      "\n",
+      "#Calculation\n",
+      "del_x = h/(4*math.pi*m*del_v)\n",
+      "\n",
+      "#Result\n",
+      "print \"Precision =\",round(del_x/1e-9,3),\"*10^-9 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Precision = 3.084 *10^-9 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7.5, Page number 5-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 4*10**-7    #wavelength(m)\n",
+      "c = 3*10**8         #velocity of light(m/s)\n",
+      "del_l = 8*10**-15   #spectral width(m)\n",
+      "\n",
+      "#calculation\n",
+      "del_t = lamda**2/(4*math.pi*c*del_l)\n",
+      "\n",
+      "#Result\n",
+      "print \"Time spent by the elctrons =\",round(del_t/1e-9,3),\"*10^-9 s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time spent by the elctrons = 5.305 *10^-9 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7.6, Page number 5-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34      #Planck's constant(J-s)\n",
+      "del_t = 1.4*10**-10   #time spent(s)\n",
+      "\n",
+      "#calculation\n",
+      "E = (h/(4*math.pi*del_t))/(1.6*10**-19)\n",
+      "\n",
+      "#Result\n",
+      "print \"Uncertainity =\",round(E/1e-6,2),\"*10^-6 eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Uncertainity = 2.36 *10^-6 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7.7, Page number 5-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 546*10**-9  #wavelength(m)\n",
+      "c = 3*10**8         #velocity of light(m/s)\n",
+      "del_l = 10**-14     #spectral width(m)\n",
+      "\n",
+      "#calculation\n",
+      "del_t = lamda**2/(4*math.pi*c*del_l)\n",
+      "\n",
+      "#Result\n",
+      "print \"Time spent by the elctrons =\",round(del_t/1e-9,2),\"*10^-9 s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time spent by the elctrons = 7.91 *10^-9 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15.1, Page number 5-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "L = 2*10**-10       #width(m)\n",
+      "\n",
+      "#Calculation\n",
+      "E1 = (h**2/(8*m*L**2))/(1.6*10**-19)\n",
+      "E2 = 4*E1\n",
+      "E3 = 9*E1\n",
+      "\n",
+      "#Result\n",
+      "print \"Energy of electron in ground state =\",round(E1,3),\"eV\"\n",
+      "print \"Energy of electron in first state =\",round(E2,3),\"eV\"\n",
+      "print \"Energy of electron in second state =\",round(E3,3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of electron in ground state = 9.434 eV\n",
+        "Energy of electron in first state = 37.738 eV\n",
+        "Energy of electron in second state = 84.91 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15.2, Page number 5-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "L = 5.6*10**-3      #width(m)\n",
+      "\n",
+      "#Calculation\n",
+      "E = L/4\n",
+      "\n",
+      "#Result\n",
+      "print \"Ground state energy =\",E,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ground state energy = 0.0014 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15.4, Page number 5-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "\n",
+      "#Variable declaration\n",
+      "#Intervals\n",
+      "x1 = 0\n",
+      "x2 = 1./2\n",
+      "\n",
+      "#Calculation\n",
+      "x = lambda x: 3*x**2\n",
+      "P = integrate.quad(x, x1, x2)\n",
+      "\n",
+      "#Result\n",
+      "print \"Probability =\",P[0]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Probability = 0.125\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15.5, Page number 5-44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m1 = 9.1*10**-31    #mass of electron(kg)\n",
+      "m2 = 10**-9         #mass of grain dust(kg)\n",
+      "L1 = 10**-9         #width(m)\n",
+      "L2 = 10**-4         #width(m)\n",
+      "\n",
+      "#Calculation\n",
+      "#For electron\n",
+      "print \"For an electron, the lowest thre energy states obtained for n=1,2 and 3 are\"\n",
+      "for n in range(1,4):\n",
+      "    En1 = ((n**2*h**2)/(8*m*L1**2))/(1.6*10**-19)\n",
+      "    print round(En1,4),\"eV\"\n",
+      "    \n",
+      "#For the grain of dust\n",
+      "print \"\\nFor a grain of dust, the lowest thre energy states obtained for n=1,2 and 3 are\"\n",
+      "for n in range(1,4):\n",
+      "    En2 = ((n**2*h**2)/(8*m2*L2**2))/(1.6*10**-19)\n",
+      "    print round(En2/1e-32,3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For an electron, the lowest thre energy states obtained for n=1,2 and 3 are\n",
+        "0.3774 eV\n",
+        "1.5095 eV\n",
+        "3.3964 eV\n",
+        "\n",
+        "For a grain of dust, the lowest thre energy states obtained for n=1,2 and 3 are\n",
+        "3.434 eV\n",
+        "13.737 eV\n",
+        "30.907 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15.6, Page number 5-45\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "E1 = 38*1.6*10**-19 #energy(J)\n",
+      "\n",
+      "#Calculation\n",
+      "L = math.sqrt(h**2/(8*m*E1))\n",
+      "              \n",
+      "#Result\n",
+      "print \"Width of well =\",round(L/1e-10,4),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Width of well = 0.9965 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15.7, Page number 5-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.63*10**-34    #Planck's constant(J-s)\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "L = 5*10**-10       #width(m)\n",
+      "n1 = 1\n",
+      "n2 = 2\n",
+      "c = 3*10**8         #velocity of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "E = ((3*h**2)/(8*m*L**2))   #E2-E1\n",
+      "Ev = E/(1.6-10**-19)        #J\n",
+      "lamda = (h*c)/E\n",
+      "\n",
+      "#Result\n",
+      "print \"Energy =\",round(Ev/1e-19,2),\"eV\"\n",
+      "print \"Wavelength =\",round(lamda/1e-7,3),\"*10^-7 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy = 4.53 eV\n",
+        "Wavelength = 2.745 *10^-7 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_II_by_H_J_Sawant/Chapter_6.ipynb b/Applied_Physics_II_by_H_J_Sawant/Chapter_6.ipynb
new file mode 100755
index 00000000..cc609a3f
--- /dev/null
+++ b/Applied_Physics_II_by_H_J_Sawant/Chapter_6.ipynb
@@ -0,0 +1,189 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: Motion of charge particle in Electric & Magnetic fields"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1.1, Page number 6-6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 9.1*10**-31         #mass of electron(kg)\n",
+      "v = 2.5*10**6           #velocity of electron(m/s)\n",
+      "theta = 30*math.pi/180  #angle(degrees)\n",
+      "B = 0.94*10**-4         #field strength(Wb/m^2)\n",
+      "e = 1.6*10**-19         #electron charge(C)\n",
+      "\n",
+      "#Calculations\n",
+      "r = (m*v*math.sin(theta))/(B*e)\n",
+      "\n",
+      "l = (5*v*math.cos(theta)*2*math.pi*m)/(B*e)\n",
+      "\n",
+      "#Results\n",
+      "print \"Radius =\",round(r/1e-3,2),\"mm\"\n",
+      "print \"Distance covered =\",round(l,3),\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius = 75.63 mm\n",
+        "Distance covered = 4.115 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1.2, Page number 6-7    "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 9.1*10**-31         #mass of electron(kg)\n",
+      "v = 3*10**7             #velocity of electron(m/s)\n",
+      "theta = 45*math.pi/180  #angle(degrees)\n",
+      "B = 0.23*10**-4         #field strength(Wb/m^2)\n",
+      "e = 1.6*10**-19         #electron charge(C)\n",
+      "\n",
+      "#Calculations\n",
+      "r = (m*v*math.sin(theta))/(B*e)\n",
+      "\n",
+      "l = (v*math.cos(theta)*2*math.pi*m)/(B*e)\n",
+      "\n",
+      "#Results\n",
+      "print \"Radius =\",round(r,2),\"m\"\n",
+      "print \"Distance covered =\",round(l,3),\"m\"\n",
+      "#Calculation mistakes in the textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radius = 5.25 m\n",
+        "Distance covered = 32.959 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1.3, Page number 6-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "y = 1.5         #displacement(cm)\n",
+      "d = 0.42        #distance(cm)\n",
+      "Va = 1.6*10**3  #anode voltage(V)\n",
+      "D = 28          #cm\n",
+      "l = 1.8         #length of plates(cm)\n",
+      "\n",
+      "#Calculation\n",
+      "V = (2*y*d*Va)/(D*l)\n",
+      "Vin = V/6\n",
+      "\n",
+      "#Result\n",
+      "print \"Input voltage =\",round(Vin,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Input voltage = 6.67 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5.1, Page number 6-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "da = 0.8    #minor axis(cm)\n",
+      "db = 2      #major axis(cm)\n",
+      "\n",
+      "#Calculation\n",
+      "ps = math.degrees(math.asin(da/db))\n",
+      "\n",
+      "#Result\n",
+      "print \"Phase shift =\",round(ps,2),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Phase shift = 23.58 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_II_by_H_J_Sawant/Chapter_7.ipynb b/Applied_Physics_II_by_H_J_Sawant/Chapter_7.ipynb
new file mode 100755
index 00000000..3425e831
--- /dev/null
+++ b/Applied_Physics_II_by_H_J_Sawant/Chapter_7.ipynb
@@ -0,0 +1,178 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: Superconductivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3.1, Page number 7-6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Ho = 2.*10**5   #A/m\n",
+      "Hc = 1.*10**5   #A/m\n",
+      "T = 8.          #temperature(K)\n",
+      "\n",
+      "#Calculations\n",
+      "Tc = math.sqrt(T**2/(1-(Hc/Ho)))\n",
+      "\n",
+      "#Result\n",
+      "print \"Critical temperature =\",round(Tc,3),\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical temperature = 11.314 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3.2, Page number 7-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Bo = 0.0306   #Critical field at 0K(T)\n",
+      "Tc = 3.7      #critical temperature(K)\n",
+      "T = 2         #K\n",
+      "\n",
+      "#calculation\n",
+      "Bc = Bo*(1-((T/Tc)**2))\n",
+      "\n",
+      "#Result\n",
+      "print \"Critical field at 2K is\",round(Bc,5),\"T\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical field at 2K is 0.02166 T\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3.3, Page number 7-7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Ho = 6.5*10**4   #Critical field at 0K(T)\n",
+      "Tc = 7.18        #critical temperature(K)\n",
+      "T = 4.2          #K\n",
+      "r = 0.5*10**-3   #radius(m)\n",
+      "\n",
+      "#calculation\n",
+      "Hc = Ho*(1-((T/Tc)**2))\n",
+      "Ic = 2*math.pi*r*Hc\n",
+      "\n",
+      "#Result\n",
+      "print \"Critical current =\",round(Ic,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical current = 134.33 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3.4, Page number 7-8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "Tc1 = 4.185      #critical temperature(K)\n",
+      "M1 = 199.5       #isotopic mass\n",
+      "Tc2 = 4.133      #K\n",
+      "\n",
+      "#calculation\n",
+      "M2 = ((Tc1*M1**0.5)/Tc2)**2\n",
+      "\n",
+      "#Result\n",
+      "print \"Isotopic mass =\",round(M2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Isotopic mass = 204.55\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_II_by_H_J_Sawant/README.txt b/Applied_Physics_II_by_H_J_Sawant/README.txt
new file mode 100755
index 00000000..717d1591
--- /dev/null
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@@ -0,0 +1,10 @@
+Contributed By: Muktesh Chaudhary
+Course: be
+College/Institute/Organization: Anglo Eastern ship management india Pvt. Ltd
+Department/Designation: Electrical & Electronics Officer
+Book Title: Applied Physics II
+Author: H. J. Sawant
+Publisher: Technical Publications
+Year of publication: 2014
+Isbn: 978-93-5038-883-9
+Edition: 2nd
\ No newline at end of file
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diff --git a/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter11.ipynb b/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter11.ipynb
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--- /dev/null
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@@ -0,0 +1,608 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#11: Lasers"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.1, Page number 11.55"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "intensity of laser beam is 1.5 *10**4 watt/m**2\n",
+      "answer given in the book is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "P=20*10**-3;      #power(watt)\n",
+    "r=1.3/2;     #radius(mm)\n",
+    "\n",
+    "#Calculation\n",
+    "r=r*10**-3;    #radius(m)\n",
+    "I=P/(math.pi*r**2);     #intensity of laser beam(watt/m**2)\n",
+    "\n",
+    "#Result\n",
+    "print \"intensity of laser beam is\",round(I/10**4,1),\"*10**4 watt/m**2\"\n",
+    "print \"answer given in the book is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.2, Page number 11.56"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "mode seperation in frequency is 2.5 *10**8 Hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "L=0.6;      #distance(m)\n",
+    "\n",
+    "#Calculation\n",
+    "delta_v=c/(2*L);    #mode seperation in frequency(Hz)\n",
+    "\n",
+    "#Result\n",
+    "print \"mode seperation in frequency is\",delta_v/10**8,\"*10**8 Hz\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.3, Page number 11.56"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "coherence length is 1.5 *10**-2 m\n",
+      "band width is 2.0 *10**10 Hz\n",
+      "line width is 0.026 nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "delta_t=0.05*10**-9;    #time(s)\n",
+    "lamda=623.8*10**-9;    #wavelength(m)\n",
+    "\n",
+    "#Calculation\n",
+    "cl=c*delta_t;    #coherence length(m)\n",
+    "delta_v=1/delta_t;    #band width(Hz)\n",
+    "delta_lamda=lamda**2*delta_v/c;    #line width(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"coherence length is\",cl*10**2,\"*10**-2 m\"\n",
+    "print \"band width is\",delta_v/10**10,\"*10**10 Hz\"\n",
+    "print \"line width is\",round(delta_lamda*10**9,3),\"nm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.4, Page number 11.56"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "energy difference is 1.96 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "lamda=632.8*10**-9;    #wavelength(m)\n",
+    "e=1.6*10**-19;     #charge(coulomb)\n",
+    "\n",
+    "#Calculation\n",
+    "E=c*h/(lamda*e);    #energy difference(eV)\n",
+    "\n",
+    "#Result\n",
+    "print \"energy difference is\",round(E,2),\"eV\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.5, Page number 11.57"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ratio of population is 8.95 *10**-32\n",
+      "answer given in the book is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "lamda=6928*10**-10;    #wavelength(m)\n",
+    "Kb=1.38*10**-23;    #boltzmann constant(J/K)\n",
+    "T=291;    #temperature(K)\n",
+    "\n",
+    "#Calculation\n",
+    "delta_E=c*h/lamda;\n",
+    "N=math.exp(-delta_E/(Kb*T));    #ratio of population\n",
+    "\n",
+    "#Result\n",
+    "print \"ratio of population is\",round(N*10**32,2),\"*10**-32\"\n",
+    "print \"answer given in the book is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.6, Page number 11.57"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength is 632 *10**-9 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "Kb=1.38*10**-23;    #boltzmann constant(J/K)\n",
+    "T=330;    #temperature(K)\n",
+    "delta_E=3.147*10**-19;       #energy(J)\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "\n",
+    "#Calculation\n",
+    "lamda=c*h/delta_E;    #wavelength(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength is\",int(lamda*10**9),\"*10**-9 m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.7, Page number 11.58"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "laser beam divergence is 0.5 *10**-3 radian\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "r1=2*10**-3;    #radius(m)\n",
+    "r2=3*10**-3;     #radius(m)\n",
+    "d1=2;     #distance(m)\n",
+    "d2=4;     #distance(m)\n",
+    "\n",
+    "#Calculation\n",
+    "delta_theta=(r2-r1)/(d2-d1);     #laser beam divergence(radian)\n",
+    "\n",
+    "#Result\n",
+    "print \"laser beam divergence is\",delta_theta*10**3,\"*10**-3 radian\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.8, Page number 11.58"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ratio of population is 1.127 *10**30\n",
+      "answer given in the book is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "lamda=6943*10**-10;    #wavelength(m)\n",
+    "Kb=1.38*10**-23;    #boltzmann constant(J/K)\n",
+    "T=300;    #temperature(K)\n",
+    "\n",
+    "#Calculation\n",
+    "new=c/lamda;\n",
+    "N=math.exp(h*new/(Kb*T));    #ratio of population\n",
+    "\n",
+    "#Result\n",
+    "print \"ratio of population is\",round(N*10**-30,3),\"*10**30\"\n",
+    "print \"answer given in the book is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.9, Page number 11.58"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength is 8632.8 angstrom\n",
+      "answer given in the book is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "Eg=1.44*1.6*10**-19;    #band gap(J)\n",
+    "\n",
+    "#Calculation\n",
+    "lamda=c*h/Eg;     #wavelength(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength is\",round(lamda*10**10,1),\"angstrom\"\n",
+    "print \"answer given in the book is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.10, Page number 11.59"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "energy gap is 0.8 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1.55;     #wavelength(micro m)\n",
+    "\n",
+    "#Calculation\n",
+    "Eg=1.24/lamda;      #energy gap(eV)\n",
+    "\n",
+    "#Result\n",
+    "print \"energy gap is\",Eg,\"eV\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.11, Page number 11.59"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "coherence length is 12 *10**3 m\n",
+      "spectral half width is 4.56 *10**-17 m\n",
+      "purity factor is 1.6 *10**10\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "tow=4*10**-5;    #time(sec)\n",
+    "lamda=740*10**-9;     #wavelength(m)\n",
+    "\n",
+    "#Calculation\n",
+    "L=tow*c;     #coherence length(m)\n",
+    "delta_lamda=lamda**2/L;     #spectral half width(m)\n",
+    "Q=lamda/delta_lamda;     #purity factor\n",
+    "\n",
+    "#Result\n",
+    "print \"coherence length is\",int(L/10**3),\"*10**3 m\"\n",
+    "print \"spectral half width is\",round(delta_lamda*10**17,2),\"*10**-17 m\"\n",
+    "print \"purity factor is\",round(Q/10**10,1),\"*10**10\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.12, Page number 11.59"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ratio of emissions is 8.4 *10**4\n",
+      "answer given in the book is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "new=5.9*10**14;    #frequency(Hz)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "Kb=1.38*10**-23;    #boltzmann constant(J/K)\n",
+    "T=2500;    #temperature(K)\n",
+    "\n",
+    "#Calculation\n",
+    "R=math.exp(h*new/(Kb*T))-1;     #ratio of emissions\n",
+    "\n",
+    "#Result\n",
+    "print \"ratio of emissions is\",round(R/10**4,1),\"*10**4\"\n",
+    "print \"answer given in the book is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.13, Page number 11.60"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "beam divergence is 1.02 *10**-4 radian\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1.06*10**-6;      #wavelength(m)\n",
+    "d=2.54*10**-2;     #distance(m)\n",
+    "\n",
+    "#Calculation\n",
+    "theta=2.44*lamda/d;      #beam divergence(radian)\n",
+    "\n",
+    "#Result\n",
+    "print \"beam divergence is\",round(theta*10**4,2),\"*10**-4 radian\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 11.14, Page number 11.60"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of photons/minute is 4.39 *10**17\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "P=2.3*10**-3;    #power(W)\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "lamda=6328*10**-10;    #wavelength(m)\n",
+    "\n",
+    "#Calculation\n",
+    "n=P*lamda*60/(c*h);      #number of photons/min\n",
+    "\n",
+    "#Result\n",
+    "print \"number of photons/minute is\",round(n/10**17,2),\"*10**17\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter12.ipynb b/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter12.ipynb
new file mode 100755
index 00000000..09a71a18
--- /dev/null
+++ b/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter12.ipynb
@@ -0,0 +1,665 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#12: Fiber Optics"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.1, Page number 12.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 52,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "acceptance angle is 20.41 degrees\n",
+      "answer in the book varies due to rounding off errors\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "n1=1.54;          #Core refractive index\n",
+    "n2=1.50;          #Cladding refractive index\n",
+    "\n",
+    "#Calculation\n",
+    "NA=math.sqrt(n1**2-n2**2);    #numerical aperture\n",
+    "alpha_m=math.asin(NA);    #acceptance angle(radian)\n",
+    "alpha_m=alpha_m*180/math.pi;    #acceptance angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"acceptance angle is\",round(alpha_m,2),\"degrees\"\n",
+    "print \"answer in the book varies due to rounding off errors\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.2, Page number 12.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 53,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "numerical aperture is 0.387\n",
+      "acceptance angle is 22.8 degrees\n",
+      "critical angle is 75 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "n1=1.5;          #Core refractive index\n",
+    "n2=1.45;          #Cladding refractive index\n",
+    "\n",
+    "#Calculation\n",
+    "delta=(n1-n2)/n1;    #fractional index change\n",
+    "NA=n1*math.sqrt(2*delta);     #numerical aperture\n",
+    "alpha_m=math.asin(NA);    #acceptance angle(radian)\n",
+    "alpha_m=alpha_m*180/math.pi;    #acceptance angle(degrees)\n",
+    "thetac=math.asin(n2/n1)*180/math.pi;     #critical angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"numerical aperture is\",round(NA,3)\n",
+    "print \"acceptance angle is\",round(alpha_m,1),\"degrees\"\n",
+    "print \"critical angle is\",int(thetac),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.3, Page number 12.25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 54,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "critical angle is 77.16 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "n1=1.53;          #Core refractive index\n",
+    "p=2.5;      #percentage(%)\n",
+    "\n",
+    "#Calculation\n",
+    "n2=n1*(100-p)/100;     #Cladding refractive index\n",
+    "thetac=math.asin(n2/n1)*180/math.pi;     #critical angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"critical angle is\",round(thetac,2),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.4, Page number 12.25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 55,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "numerical aperture is 0.28\n",
+      "acceptance angle is 16.26 degrees\n",
+      "critical angle is 79.1 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "delta=0.018;      #fractional difference\n",
+    "n1=1.50;          #Core refractive index\n",
+    "\n",
+    "#Calculation\n",
+    "NA=round(n1*math.sqrt(2*delta),2);     #numerical aperture\n",
+    "alpha_m=math.asin(NA);    #acceptance angle(radian)\n",
+    "alpha_m=alpha_m*180/math.pi;    #acceptance angle(degrees)\n",
+    "n2=n1-(delta*n1);        #Cladding refractive index\n",
+    "thetac=math.asin(n2/n1)*180/math.pi;     #critical angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"numerical aperture is\",NA\n",
+    "print \"acceptance angle is\",round(alpha_m,2),\"degrees\"\n",
+    "print \"critical angle is\",round(thetac,1),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.5, Page number 12.26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 56,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "total number of guided modes is 490.0\n",
+      "number of modes of graded index fibre is 245.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "d=50*10**-6;     #diameter(m)\n",
+    "lamda=1*10**-6;    #wavelength(m)\n",
+    "NA=0.2;     #numerical aperture\n",
+    "\n",
+    "#Calculation\n",
+    "NSI=4.9*(d*NA/lamda)**2;     #total number of guided modes\n",
+    "NGI=NSI/2;      #number of modes of graded index fibre\n",
+    "\n",
+    "#Result\n",
+    "print \"total number of guided modes is\",NSI\n",
+    "print \"number of modes of graded index fibre is\",NGI"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.6, Page number 12.26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 57,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "numerical aperture is 0.47\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "delta=0.05;      #fractional difference\n",
+    "n1=1.5;          #Core refractive index\n",
+    "\n",
+    "#Calculation\n",
+    "NA=n1*math.sqrt(2*delta);     #numerical aperture\n",
+    "\n",
+    "#Result\n",
+    "print \"numerical aperture is\",round(NA,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.7, Page number 12.27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 58,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of modes that can be propagated is 1\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "d=5*10**-6;     #diameter(m)\n",
+    "lamda=1*10**-6;    #wavelength(m)\n",
+    "NA=0.092;     #numerical aperture\n",
+    "\n",
+    "#Calculation\n",
+    "N=4.9*(d*NA/lamda)**2;     #number of modes that can be propagated\n",
+    "\n",
+    "#Result\n",
+    "print \"number of modes that can be propagated is\",int(N)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.8, Page number 12.27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 59,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "V number is 94.7\n",
+      "maximum number of modes is 4485.7\n",
+      "answer in the book varies due to rounding off errors\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "n1=1.53;          #Core refractive index\n",
+    "n2=1.50;          #Cladding refractive index\n",
+    "a=50*10**-6;      #radius(m)\n",
+    "lamda=1*10**-6;     #wavelength(m)\n",
+    "\n",
+    "#Calculation\n",
+    "NA=math.sqrt(n1**2-n2**2);    #numerical aperture\n",
+    "Vn=2*math.pi*a*NA/lamda;      #V number\n",
+    "N=Vn**2/2;      #maximum number of modes\n",
+    "\n",
+    "#Result\n",
+    "print \"V number is\",round(Vn,1)\n",
+    "print \"maximum number of modes is\",round(N,1)\n",
+    "print \"answer in the book varies due to rounding off errors\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.9, Page number 12.27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 60,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of propagating modes is 49177\n",
+      "answer in the book is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=100*10**-6;     #radius(m)\n",
+    "lamda=0.85*10**-6;    #wavelength(m)\n",
+    "NA=0.3;   #numerical aperture\n",
+    "\n",
+    "#Calculation\n",
+    "N=4*math.pi**2*a**2*NA**2/lamda**2;    #number of propagating modes\n",
+    "\n",
+    "#Result\n",
+    "print \"number of propagating modes is\",int(N)\n",
+    "print \"answer in the book is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.10, Page number 12.28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 61,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Core refractive index is 1.6025\n",
+      "acceptance angle is 8.638 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "NA=0.2;   #numerical aperture\n",
+    "n2=1.59;          #Cladding refractive index\n",
+    "n0=1.33;     #refractive index\n",
+    "\n",
+    "#Calculation\n",
+    "n1=round(math.sqrt(NA**2+n2**2),4);     #Core refractive index\n",
+    "NA1=math.sqrt(n1**2-n2**2)/n0;         #numerical aperture\n",
+    "alpha_m=math.asin(NA1);    #acceptance angle(radian)\n",
+    "alpha_m=alpha_m*180/math.pi;    #acceptance angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"Core refractive index is\",n1\n",
+    "print \"acceptance angle is\",round(alpha_m,3),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.11, Page number 12.29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 62,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of photons is 13.675 *10**15\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "c=3*10**8;    #velocity of light(m/sec)\n",
+    "h=6.63*10**-34;    #plank's constant(Js)\n",
+    "lamda=680*10**-9;    #wavelength(m)\n",
+    "P=4*10**-3;    #power(W)\n",
+    "\n",
+    "#Calculation\n",
+    "delta_E=c*h/lamda;    #energy(J)\n",
+    "N=P/delta_E;     #number of photons\n",
+    "\n",
+    "#Result\n",
+    "print \"number of photons is\",round(N/10**15,3),\"*10**15\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.12, Page number 12.29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 64,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Cladding refractive index is 1.49925\n",
+      "numerical aperture is 0.0474\n",
+      "critical angle is 88.2 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "delta=0.0005;     #fractional difference\n",
+    "n1=1.5;          #Core refractive index\n",
+    "\n",
+    "#Calculation\n",
+    "n2=n1-(delta*n1);          #Cladding refractive index\n",
+    "NA=n1*math.sqrt(2*delta);     #numerical aperture\n",
+    "thetac=math.asin(n2/n1)*180/math.pi;     #critical angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"Cladding refractive index is\",n2\n",
+    "print \"numerical aperture is\",round(NA,4)\n",
+    "print \"critical angle is\",round(thetac,1),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.13, Page number 12.29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 65,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "fraction of initial intensity after 2km is 0.363\n",
+      "fraction of initial intensity after 5km is 0.048\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "alpha=2.2;     #attenuation coefficient(dB/km)\n",
+    "L1=2;    #distance(km)\n",
+    "L2=6;    #distance(km)\n",
+    "\n",
+    "#Calculation\n",
+    "x1=-alpha*L1/10;      \n",
+    "x2=-alpha*L2/10;\n",
+    "P1=10**x1;         #fraction of initial intensity after 2km\n",
+    "P2=10**x2;         #fraction of initial intensity after 5km\n",
+    "\n",
+    "#Result\n",
+    "print \"fraction of initial intensity after 2km is\",round(P1,3)\n",
+    "print \"fraction of initial intensity after 5km is\",round(P2,3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.14, Page number 12.30"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 66,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "attenuation coefficient is 1.41 dB/km\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "L=0.5;      #distance(km)\n",
+    "Pout=1;    #output power(W)\n",
+    "Pin=85/100;    #input power(W)\n",
+    "\n",
+    "#Calculation\n",
+    "alpha=10*math.log10(Pout/Pin)/L;     #attenuation coefficient(dB/km)\n",
+    "\n",
+    "#Result\n",
+    "print \"attenuation coefficient is\",round(alpha,2),\"dB/km\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 12.15, Page number 12.30"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 67,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "attenuation coefficient is 1.7 dB/km\n",
+      "overall signal attenuation is 17 dB\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "L=10;      #distance(km)\n",
+    "Pout=2;    #output power(W)\n",
+    "Pin=100;    #input power(W)\n",
+    "\n",
+    "#Calculation\n",
+    "alpha=10*math.log10(Pin/Pout)/L;     #attenuation coefficient(dB/km)\n",
+    "o_alpha=round(alpha,1)*L;      #overall signal attenuation\n",
+    "\n",
+    "#Result\n",
+    "print \"attenuation coefficient is\",round(alpha,1),\"dB/km\"\n",
+    "print \"overall signal attenuation is\",int(o_alpha),\"dB\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter2.ipynb b/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter2.ipynb
new file mode 100755
index 00000000..601cae27
--- /dev/null
+++ b/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter2.ipynb
@@ -0,0 +1,441 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#2: Crystal Structures"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.1, Page number 2.23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "lattice constant is 4.0 *10**-10 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "M=60.2;     #molecular weight\n",
+    "Na=6.023*10**26;    #avagadro number(kg/mole)\n",
+    "n=4;    \n",
+    "rho=6250;           #density(kg/m**3)\n",
+    "\n",
+    "#Calculation\n",
+    "a=(n*M/(rho*Na))**(1/3);     #lattice constant(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"lattice constant is\",round(a*10**10),\"*10**-10 m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.2, Page number 2.23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "density is 8.93 gm/cm**3\n",
+      "answer in the book varies due to rounding off errors\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "M=63.5;     #molecular weight\n",
+    "Na=6.023*10**26;    #avagadro number(kg/mole)\n",
+    "n=4;    \n",
+    "r=1.278*10**-8;     #atomic radius(cm)\n",
+    "\n",
+    "#Calculation\n",
+    "a=2*math.sqrt(2)*r;   #lattice constant(m)\n",
+    "rho=n*M/(a**3*Na);    #density(kg/cm**3)\n",
+    "\n",
+    "#Result\n",
+    "print \"density is\",round(rho*10**3,2),\"gm/cm**3\"\n",
+    "print \"answer in the book varies due to rounding off errors\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.3, Page number 2.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ratio of densities is 0.92\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "pf_BCC=math.pi*math.sqrt(3)/8;    #packing factor for BCC\n",
+    "pf_FCC=math.pi/(3*math.sqrt(2));     #packing factor of FCC\n",
+    "\n",
+    "#Calculation\n",
+    "r=pf_BCC/pf_FCC;     #ratio of densities\n",
+    "\n",
+    "#Result\n",
+    "print \"ratio of densities is\",round(r,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.4, Page number 2.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "lattice constant is 2.8687 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "M=55.85;     #molecular weight\n",
+    "Na=6.02*10**26;    #avagadro number(kg/mole)\n",
+    "n=2;    \n",
+    "rho=7860;           #density(kg/m**3)\n",
+    "\n",
+    "#Calculation\n",
+    "a=(n*M/(rho*Na))**(1/3);     #lattice constant(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"lattice constant is\",round(a*10**10,4),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.5, Page number 2.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "lattice constant is 5.6 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "M=58.5;     #molecular weight\n",
+    "Na=6.02*10**26;    #avagadro number(kg/mole)\n",
+    "n=4;    \n",
+    "rho=2189;           #density(kg/m**3)\n",
+    "\n",
+    "#Calculation\n",
+    "a=(n*M/(rho*Na))**(1/3);     #lattice constant(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"lattice constant is\",round(a*10**10,1),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.6, Page number 2.25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "lattice constant is 3.517 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "M=6.94;     #molecular weight\n",
+    "Na=6.02*10**26;    #avagadro number(kg/mole)\n",
+    "n=2;    \n",
+    "rho=530;           #density(kg/m**3)\n",
+    "\n",
+    "#Calculation\n",
+    "a=(n*M/(rho*Na))**(1/3);     #lattice constant(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"lattice constant is\",round(a*10**10,3),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.7, Page number 2.25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "percent volume change is 0.493 %\n",
+      "answer in the book varies due to rounding off errors\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "r1=1.258*10**-10;     #radius(m)\n",
+    "r2=1.292*10**-10;     #radius(m)\n",
+    "\n",
+    "#Calculation\n",
+    "a_bcc=4*r1/math.sqrt(3);\n",
+    "v=a_bcc**3;\n",
+    "V1=v/2;\n",
+    "a_fcc=2*math.sqrt(2)*r2;\n",
+    "V2=a_fcc**3/4;\n",
+    "V=(V1-V2)*100/V1;           #percent volume change is\",V,\"%\"\n",
+    "\n",
+    "#Result\n",
+    "print \"percent volume change is\",round(V,3),\"%\"\n",
+    "print \"answer in the book varies due to rounding off errors\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.8, Page number 2.26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of atoms/m**3 is 1.77 *10**29\n",
+      "density of diamond is 3534.47 kg/m**3\n",
+      "answer in the book is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=0.356*10**-9;       #cube edge(m)\n",
+    "w=12;    #atomic weight\n",
+    "Na=6.02*10**26;    #avagadro number(kg/mole)\n",
+    "\n",
+    "#Calculation\n",
+    "n=8/(a**3);     #number of atoms/m**3\n",
+    "m=w/Na;      #mass(kg)\n",
+    "rho=m*n;     #density of diamond(kg/m**3)\n",
+    "\n",
+    "#Result\n",
+    "print \"number of atoms/m**3 is\",round(n/10**29,2),\"*10**29\"\n",
+    "print \"density of diamond is\",round(rho,2),\"kg/m**3\"\n",
+    "print \"answer in the book is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.9, Page number 2.26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum radius of sphere is 0.414 r\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "from sympy import *\n",
+    "\n",
+    "#Variable declaration\n",
+    "r=Symbol('r')\n",
+    "\n",
+    "#Calculation\n",
+    "a=4*r/math.sqrt(2);\n",
+    "R=(4*r/(2*math.sqrt(2)))-r;       #maximum radius of sphere\n",
+    "\n",
+    "#Result\n",
+    "print \"maximum radius of sphere is\",round(R/r,3),\"r\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 2.10, Page number 2.26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 35,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "radius of largest sphere is 0.155 r\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "from sympy import *\n",
+    "\n",
+    "#Variable declaration\n",
+    "r=Symbol('r')\n",
+    "\n",
+    "#Calculation\n",
+    "a=4*r/math.sqrt(3);\n",
+    "R=(a/2)-r;           #radius of largest sphere\n",
+    "\n",
+    "#Result\n",
+    "print \"radius of largest sphere is\",round(R/r,3),\"r\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter3.ipynb b/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter3.ipynb
new file mode 100755
index 00000000..a0d72f81
--- /dev/null
+++ b/Applied_Physics_by_M._Chandra_Shekar,_P._Appala_Naidu/Chapter3.ipynb
@@ -0,0 +1,1601 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#3: Crystal Planes and X-ray Diffraction"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.1, Page number 3.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "miller indices are ( 1 2 2 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=4;\n",
+    "b=2;\n",
+    "c=2;      #intercepts\n",
+    "\n",
+    "#Calculation\n",
+    "h=int(a/a);\n",
+    "k=int(a/b);\n",
+    "l=int(a/c);     #miller indices\n",
+    "\n",
+    "#Result\n",
+    "print \"miller indices are (\",h,k,l,\")\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.2, Page number 3.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "miller indices are ( 1 3 3 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=6;\n",
+    "b=2;\n",
+    "c=2;      #intercepts\n",
+    "\n",
+    "#Calculation\n",
+    "h=int(a/a);\n",
+    "k=int(a/b);\n",
+    "l=int(a/c);     #miller indices\n",
+    "\n",
+    "#Result\n",
+    "print \"miller indices are (\",h,k,l,\")\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.3, Page number 3.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "miller indices are ( 1 1 1 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=3;\n",
+    "b=3;\n",
+    "c=3;      #intercepts\n",
+    "\n",
+    "#Calculation\n",
+    "h=int(c/a);\n",
+    "k=int(c/b);\n",
+    "l=int(c/c);     #miller indices\n",
+    "\n",
+    "#Result\n",
+    "print \"miller indices are (\",h,k,l,\")\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.4, Page number 3.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "miller indices are ( 3 2 1 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=2;\n",
+    "b=3;\n",
+    "c=6;      #intercepts\n",
+    "\n",
+    "#Calculation\n",
+    "h=int(c/a);\n",
+    "k=int(c/b);\n",
+    "l=int(c/c);     #miller indices\n",
+    "\n",
+    "#Result\n",
+    "print \"miller indices are (\",h,k,l,\")\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.5, Page number 3.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "miller indices are ( 2 1 2 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=0.212;\n",
+    "b=1;\n",
+    "c=0.183;      #intercepts\n",
+    "ac=0.424;\n",
+    "bc=1;\n",
+    "cc=0.366;     #axial units\n",
+    "\n",
+    "#Calculation\n",
+    "h=int(ac/a);\n",
+    "k=int(bc/b);\n",
+    "l=int(cc/c);     #miller indices\n",
+    "\n",
+    "#Result\n",
+    "print \"miller indices are (\",h,k,l,\")\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.6, Page number 3.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "intercepts along x,y,z directions are 1.2 angstrom 1.2 angstrom 4.0 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "h=2;\n",
+    "k=3;\n",
+    "l=1;     #intercepts\n",
+    "a=1.2;\n",
+    "b=1.8;\n",
+    "c=2;     #primitives(angstrom)\n",
+    "p=1.2;    #intercept along X-direction\n",
+    "\n",
+    "#Calculation\n",
+    "q=b/k;\n",
+    "r=2*p/q;\n",
+    "\n",
+    "#Result\n",
+    "print \"intercepts along x,y,z directions are\",p,\"angstrom\",2*q,\"angstrom\",r,\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.7, Page number 3.11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar separation for 1st plane is 1.0*a\n",
+      "interplanar separation for 2nd plane is 0.707 *a\n",
+      "interplanar separation for 3rd plane is 0.577 *a\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "from sympy import Symbol\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=Symbol('a');\n",
+    "h1=1;\n",
+    "k1=0;\n",
+    "l1=0;    #intercepts of 1st plane\n",
+    "h2=1;\n",
+    "k2=1;\n",
+    "l2=0;    #intercepts of 2nd plane\n",
+    "h3=1;\n",
+    "k3=1;\n",
+    "l3=1;    #intercepts of 3rd plane\n",
+    "\n",
+    "#Calculation\n",
+    "d100=a/math.sqrt(h1**2+k1**2+l1**2);    #interplanar separation for 1st plane\n",
+    "d110=a/math.sqrt(h2**2+k2**2+l2**2);    #interplanar separation for 2nd plane\n",
+    "d111=a/math.sqrt(h3**2+k3**2+l3**2);    #interplanar separation for 3rd plane\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar separation for 1st plane is\",d100\n",
+    "print \"interplanar separation for 2nd plane is\",round(d110/a,3),\"*a\"\n",
+    "print \"interplanar separation for 3rd plane is\",round(d111/a,3),\"*a\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.8, Page number 3.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar separation is 0.179 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "h=3;\n",
+    "k=4;\n",
+    "l=5;     #miller indices\n",
+    "a=2;\n",
+    "b=2;\n",
+    "c=1;     #parameters\n",
+    "\n",
+    "#Calculation\n",
+    "d=1/math.sqrt((h**2/a**2)+(k**2/b**2)+(l**2/c**2));     #interplanar separation(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar separation is\",round(d,3),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.9, Page number 3.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar distance is 3.464 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=6;     #lattice parameter(angstrom)\n",
+    "h=1;\n",
+    "k=1;\n",
+    "l=1;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar distance is\",round(d,3),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.10, Page number 3.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar distance is 2.1 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=6.3;     #lattice parameter(angstrom)\n",
+    "h=2;\n",
+    "k=2;\n",
+    "l=1;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar distance is\",round(d,1),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.11, Page number 3.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar distance is 1 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=5;     #lattice parameter(angstrom)\n",
+    "h=4;\n",
+    "k=3;\n",
+    "l=0;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar distance is\",int(d),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.12, Page number 3.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ratio of intercepts on axes is 6 : 3 : 2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "h=1;\n",
+    "k=2;\n",
+    "l=3;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "p=int(k*l/h);\n",
+    "q=int(k*l/k);\n",
+    "r=int(k*l/l);       #ratio of intercepts on axes\n",
+    "\n",
+    "#Result\n",
+    "print \"ratio of intercepts on axes is\",p,\":\",q,\":\",r"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.13, Page number 3.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar distance is 1.12 *10**-10 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=4.2*10**-10;     #lattice parameter(m)\n",
+    "h=3;\n",
+    "k=2;\n",
+    "l=1;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar distance is\",round(d*10**10,2),\"*10**-10 m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.14, Page number 3.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar separation is 1.26 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "h=1;\n",
+    "k=1;\n",
+    "l=1;     #miller indices\n",
+    "a=2.5;\n",
+    "b=2.5;\n",
+    "c=1.8;     #parameters\n",
+    "\n",
+    "#Calculation\n",
+    "d=1/math.sqrt((h**2/a**2)+(k**2/b**2)+(l**2/c**2));     #interplanar separation(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar separation is\",round(d,2),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.15, Page number 3.14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar separation for 1st plane is 2.466 angstrom\n",
+      "interplanar separation for 2nd plane is 1.74 angstrom\n",
+      "interplanar separation for 3rd plane is 2.85 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "r=1.744;       #atomic radius(angstrom)\n",
+    "h1=2;\n",
+    "k1=0;\n",
+    "l1=0;    #intercepts of 1st plane\n",
+    "h2=2;\n",
+    "k2=2;\n",
+    "l2=0;    #intercepts of 2nd plane\n",
+    "h3=1;\n",
+    "k3=1;\n",
+    "l3=1;    #intercepts of 3rd plane\n",
+    "\n",
+    "#Calculation\n",
+    "a=4*r/math.sqrt(2);\n",
+    "d200=a/math.sqrt(h1**2+k1**2+l1**2);    #interplanar separation for 1st plane(angstrom)\n",
+    "d220=a/math.sqrt(h2**2+k2**2+l2**2);    #interplanar separation for 2nd plane(angstrom)\n",
+    "d111=a/math.sqrt(h3**2+k3**2+l3**2);    #interplanar separation for 3rd plane(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar separation for 1st plane is\",round(d200,3),\"angstrom\"\n",
+    "print \"interplanar separation for 2nd plane is\",round(d220,2),\"angstrom\"\n",
+    "print \"interplanar separation for 3rd plane is\",round(d111,2),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.16, Page number 3.14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ratio of densities is 1.414 : 1.732\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "h1=1;\n",
+    "k1=1;\n",
+    "l1=1;    #intercepts of 1st plane\n",
+    "h2=1;\n",
+    "k2=1;\n",
+    "l2=0;    #intercepts of 2nd plane\n",
+    "a=1;\n",
+    "\n",
+    "#Calculation\n",
+    "d111=a/math.sqrt(h1**2+k1**2+l1**2);    #interplanar separation for 1st plane(angstrom)\n",
+    "d110=a/math.sqrt(h2**2+k2**2+l2**2);    #interplanar separation for 2nd plane(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"ratio of densities is\",round(1/d110,3),\":\",round(1/d111,3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "##Example number 3.17, Page number 3.21"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "glancing angle is 14.61 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=2.814;     #lattice parameter(m)\n",
+    "h=1;\n",
+    "lamda=0.71;     #wavelength(angstrom)\n",
+    "k=0;\n",
+    "l=0;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "theta=math.asin(lamda/d);     #glancing angle(radian)\n",
+    "theta=theta*180/math.pi;      #glancing angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"glancing angle is\",round(theta,2),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.18, Page number 3.21"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength is 1.5 angstrom\n",
+      "glancing angle is 90.0 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=4.5;     #lattice parameter(m)\n",
+    "h=2;\n",
+    "k=2;\n",
+    "theta=30*math.pi/180;    #angle(radian)\n",
+    "l=1;      #miller indices\n",
+    "n=2;\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "lamda=2*d*math.sin(theta);     #wavelength(angstrom)\n",
+    "theta=math.asin(lamda/d);     #glancing angle(radian)\n",
+    "theta=theta*180/math.pi;      #glancing angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength is\",lamda,\"angstrom\"\n",
+    "print \"glancing angle is\",round(theta,2),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.19, Page number 3.22"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength is 0.842 *10**-10 m\n",
+      "glancing angle for second order is 17.37 degrees\n",
+      "glancing angle for third order is 26.6 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "theta=(8+(35/60))*math.pi/180;    #angle(radian)\n",
+    "n2=2;\n",
+    "n3=3;\n",
+    "d=2.82*10**-10;     #lattice spacing(m)\n",
+    "\n",
+    "#Calculation\n",
+    "lamda=2*d*math.sin(theta);     #wavelength(m)\n",
+    "theta2=math.asin(n2*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta2=theta2*180/math.pi;      #glancing angle(degrees)\n",
+    "theta3=math.asin(n3*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta3=theta3*180/math.pi;      #glancing angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength is\",round(lamda*10**10,3),\"*10**-10 m\"\n",
+    "print \"glancing angle for second order is\",round(theta2,2),\"degrees\"\n",
+    "print \"glancing angle for third order is\",round(theta3,1),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.20, Page number 3.23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "glancing angle for 1st order is 14.48 degrees\n",
+      "glancing angle for 1st order is 30.0 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=5;     #lattice parameter(angstrom)\n",
+    "h=3;\n",
+    "lamda=0.5;     #wavelength(angstrom)\n",
+    "k=4;\n",
+    "n=2;\n",
+    "l=0;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "theta1=math.asin(lamda/(2*d));     #glancing angle(radian)\n",
+    "theta1=theta1*180/math.pi;      #glancing angle(degrees)\n",
+    "theta2=math.asin(2*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta2=theta2*180/math.pi;      #glancing angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"glancing angle for 1st order is\",round(theta1,2),\"degrees\"\n",
+    "print \"glancing angle for 1st order is\",theta2,\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.21, Page number 3.23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength is 2.45 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=3;     #lattice parameter(angstrom)\n",
+    "h=1;\n",
+    "k=1;\n",
+    "n=1;\n",
+    "l=1;      #miller indices\n",
+    "theta=45*math.pi/180;    #angle(radian)\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "lamda=2*d*math.sin(theta);     #wavelength(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength is\",round(lamda,2),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.22, Page number 3.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "glancing angle for 1st order is 22.02 degrees\n",
+      "glancing angle for 2nd order is 48.59 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1.5;     #wavelength(angstrom)\n",
+    "d=2;       #seperation(angstrom)\n",
+    "n=2;\n",
+    "\n",
+    "#Calculation\n",
+    "theta1=math.asin(lamda/(2*d));     #glancing angle(radian)\n",
+    "theta1=theta1*180/math.pi;      #glancing angle(degrees)\n",
+    "theta2=math.asin(n*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta2=theta2*180/math.pi;      #glancing angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"glancing angle for 1st order is\",round(theta1,2),\"degrees\"\n",
+    "print \"glancing angle for 2nd order is\",round(theta2,2),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.23, Page number 3.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar separation is 0.88 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1.25;     #wavelength(angstrom)\n",
+    "theta=45*math.pi/180;    #angle(radian)\n",
+    "n=1;\n",
+    "\n",
+    "#Calculation\n",
+    "d=n*lamda/(2*math.sin(theta));      #interplanar separation(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar separation is\",round(d,2),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.24, Page number 3.25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "glancing angle for 1st order is 5.74 degrees\n",
+      "glancing angle for 2nd order is 11.54 degrees\n",
+      "glancing angle for 1st order is 17.46 degrees\n",
+      "glancing angle for 2nd order is 23.58 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1;     #wavelength(angstrom)\n",
+    "d=5;       #seperation(angstrom)\n",
+    "n1=1;\n",
+    "n2=2;\n",
+    "n3=3;\n",
+    "n4=4;     #order\n",
+    "\n",
+    "#Calculation\n",
+    "theta1=math.asin(n1*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta1=theta1*180/math.pi;      #glancing angle(degrees)\n",
+    "theta2=math.asin(n2*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta2=theta2*180/math.pi;      #glancing angle(degrees)\n",
+    "theta3=math.asin(n3*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta3=theta3*180/math.pi;      #glancing angle(degrees)\n",
+    "theta4=math.asin(n4*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta4=theta4*180/math.pi;      #glancing angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"glancing angle for 1st order is\",round(theta1,2),\"degrees\"\n",
+    "print \"glancing angle for 2nd order is\",round(theta2,2),\"degrees\"\n",
+    "print \"glancing angle for 1st order is\",round(theta3,2),\"degrees\"\n",
+    "print \"glancing angle for 2nd order is\",round(theta4,2),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.25, Page number 3.26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1st bragg reflection angle is 7.18 degrees\n",
+      "2nd bragg reflection angle is 14.48 degrees\n",
+      "3rd bragg reflection angle is 22.02 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "theta=30*math.pi/180;      #angle(radian)\n",
+    "#assume that sin(theta)=x\n",
+    "x1=1;\n",
+    "x2=2;\n",
+    "x3=3;\n",
+    "x4=4;\n",
+    "\n",
+    "#Calculation\n",
+    "theta1=math.asin(x1*math.sin(theta)/x4);        #1st bragg reflection angle(radian)\n",
+    "theta1=theta1*180/math.pi;       #1st bragg reflection angle(degrees)\n",
+    "theta2=math.asin(x2*math.sin(theta)/x4);        #1st bragg reflection angle(radian)\n",
+    "theta2=theta2*180/math.pi;       #1st bragg reflection angle(degrees)\n",
+    "theta3=math.asin(x3*math.sin(theta)/x4);        #1st bragg reflection angle(radian)\n",
+    "theta3=theta3*180/math.pi;       #1st bragg reflection angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"1st bragg reflection angle is\",round(theta1,2),\"degrees\"\n",
+    "print \"2nd bragg reflection angle is\",round(theta2,2),\"degrees\"\n",
+    "print \"3rd bragg reflection angle is\",round(theta3,2),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.26, Page number 3.27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum order of bragg reflection is 4\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1.25;     #wavelength(angstrom)\n",
+    "d=2.5;       #seperation(angstrom)\n",
+    "\n",
+    "#Calculation\n",
+    "n=2*d/lamda;       #maximum order of bragg reflection\n",
+    "\n",
+    "#Result\n",
+    "print \"maximum order of bragg reflection is\",int(n)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.27, Page number 3.27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum order of bragg reflection is 3\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1.5;     #wavelength(angstrom)\n",
+    "d=2.8;       #seperation(angstrom)\n",
+    "\n",
+    "#Calculation\n",
+    "n=2*d/lamda;       #maximum order of bragg reflection\n",
+    "\n",
+    "#Result\n",
+    "print \"maximum order of bragg reflection is\",int(n)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.28, Page number 3.27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength for 1st angle is 1.45 angstrom\n",
+      "wavelength for 2nd angleis 1.544 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "theta1=15*math.pi/180;    #angle(radian)\n",
+    "theta2=16*math.pi/180;    #angle(radian)\n",
+    "d=2.8;     #lattice spacing(angstrom)\n",
+    "\n",
+    "#Calculation\n",
+    "lamda1=2*d*math.sin(theta1);     #wavelength for 1st angle(angstrom)\n",
+    "lamda2=2*d*math.sin(theta2);     #wavelength for 2nd angle(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength for 1st angle is\",round(lamda1,2),\"angstrom\"\n",
+    "print \"wavelength for 2nd angleis\",round(lamda2,3),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.29, Page number 3.31"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "miller indices are ( 6 2 3 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "x=1;\n",
+    "y=3;\n",
+    "z=-2;\n",
+    "\n",
+    "#Calculation\n",
+    "h=-y*z/x;\n",
+    "k=-y*z/y;\n",
+    "l=y*z/z;       #miller indices\n",
+    "\n",
+    "#Result\n",
+    "print \"miller indices are (\",int(h),int(k),int(l),\")\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.30, Page number 3.32"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interplanar separation is 1.26 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "h=1;\n",
+    "k=1;\n",
+    "l=1;     #miller indices\n",
+    "a=2.5;\n",
+    "b=2.5;\n",
+    "c=1.8;     #parameters\n",
+    "\n",
+    "#Calculation\n",
+    "d=1/math.sqrt((h**2/a**2)+(k**2/b**2)+(l**2/c**2));     #interplanar separation(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"interplanar separation is\",round(d,2),\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.31, Page number 3.32"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "glancing angle is 16.1 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "a=2.814;     #lattice parameter(m)\n",
+    "h=1;\n",
+    "lamda=0.78;     #wavelength(angstrom)\n",
+    "k=0;\n",
+    "l=0;      #miller indices\n",
+    "\n",
+    "#Calculation\n",
+    "d=a/math.sqrt(h**2+k**2+l**2);      #interplanar distance(angstrom)\n",
+    "theta=math.asin(lamda/d);     #glancing angle(radian)\n",
+    "theta=theta*180/math.pi;      #glancing angle(degrees)\n",
+    "\n",
+    "#Result\n",
+    "print \"glancing angle is\",round(theta,1),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.32, Page number 3.32"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "longest wavelength in 1st order is 5.64 angstrom\n",
+      "longest wavelength in 2nd order is 2.82 angstrom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "n1=1;\n",
+    "n2=2;\n",
+    "d=2.82;       #seperation(angstrom)\n",
+    "\n",
+    "#Calculation\n",
+    "lamda_max1=2*d/n1;        #longest wavelength in 1st order(angstrom)\n",
+    "lamda_max2=2*d/n2;        #longest wavelength in 2nd order(angstrom)\n",
+    "\n",
+    "#Result\n",
+    "print \"longest wavelength in 1st order is\",lamda_max1,\"angstrom\"\n",
+    "print \"longest wavelength in 2nd order is\",lamda_max2,\"angstrom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.33, Page number 3.33"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength is 0.842 angstrom\n",
+      "glancing angle for second order is 17 degrees 22 minutes\n",
+      "glancing angle for third order is 26 degrees 36.0 minutes\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "theta=(8+(35/60))*math.pi/180;    #angle(radian)\n",
+    "n1=1;\n",
+    "n2=2;\n",
+    "n3=3;\n",
+    "d=2.82*10**-10;     #lattice spacing(m)\n",
+    "\n",
+    "#Calculation\n",
+    "lamda=2*d*math.sin(theta)/n1;     #wavelength(m)\n",
+    "theta2=math.asin(n2*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta2=theta2*180/math.pi;      #glancing angle(degrees)\n",
+    "theta2_m=60*(theta2-int(theta2));\n",
+    "theta3=math.asin(n3*lamda/(2*d));     #glancing angle(radian)\n",
+    "theta3=theta3*180/math.pi;      #glancing angle(degrees)\n",
+    "theta3_m=60*(theta3-int(theta3));\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength is\",round(lamda*10**10,3),\"angstrom\"\n",
+    "print \"glancing angle for second order is\",int(theta2),\"degrees\",int(theta2_m),\"minutes\"\n",
+    "print \"glancing angle for third order is\",int(theta3),\"degrees\",round(theta3_m),\"minutes\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.34, Page number 3.33"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "glancing angle is 21 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=0.071*10**-9;     #wavelength(angstrom)\n",
+    "d100=0.28*10**-9;       #seperation(angstrom)\n",
+    "n=2;\n",
+    "\n",
+    "#Calculation\n",
+    "d110=d100/math.sqrt(2);     \n",
+    "theta=math.asin(n*lamda/(2*d110))*180/math.pi;     #glancing angle(degrees)      \n",
+    "\n",
+    "#Result\n",
+    "print \"glancing angle is\",int(theta),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.35, Page number 3.34"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wavelength is 0.0842 nm\n",
+      "maximum order of diffraction is 6\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "theta=(8+(35/60))*math.pi/180;    #angle(radian)\n",
+    "n=1;\n",
+    "d=0.282;     #lattice spacing(nm)\n",
+    "\n",
+    "#Calculation\n",
+    "lamda=2*d*math.sin(theta)/n;     #wavelength(nm)\n",
+    "n_max=2*d/lamda;         #maximum order of diffraction\n",
+    "\n",
+    "#Result\n",
+    "print \"wavelength is\",round(lamda,4),\"nm\"\n",
+    "print \"maximum order of diffraction is\",int(n_max)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 3.36, Page number 3.34"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "interatomic spacing is 2.67 *10**-10 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "lamda=1.5418*10**-10;     #wavelength(m)\n",
+    "theta=30*math.pi/180;    #angle(radian)\n",
+    "n=1;\n",
+    "h=1;\n",
+    "k=1;\n",
+    "l=1;\n",
+    "\n",
+    "#Calculation\n",
+    "d=n*lamda/(2*math.sin(theta));      #interplanar separation(m)\n",
+    "a=d*math.sqrt(h**2+k**2+l**2);      #interatomic spacing(m)\n",
+    "\n",
+    "#Result\n",
+    "print \"interatomic spacing is\",round(a*10**10,2),\"*10**-10 m\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_01_Bonding_in_Solids__1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_01_Bonding_in_Solids__1.ipynb
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@@ -0,0 +1,248 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2ef3d4592a57ed325da4c5f5624afbf334fc1ea3982d155f4e16e40570d72ab9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1:Bonding in Solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 , Page no:15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=2; #in angstrom(distance)\n",
+      "e=1.6E-19; # in C (charge of electron)\n",
+      "E_o= 8.85E-12;# absolute premittivity\n",
+      "\n",
+      "#calculate\n",
+      "r=2*1*10**-10; # since r is in angstrom\n",
+      "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
+      "V1=V/e; # changing to eV\n",
+      "\n",
+      "#result\n",
+      "print \"\\nThe potential energy is  V = \",'%.4E'%V,\"J\";\n",
+      "print \"In electron-Volt V = \",round(V1,3),\"eV\"; \n",
+      "print \"Note: the answer in the book is wrong due to calculation mistake\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The potential energy is  V =  -1.1515E-18 J\n",
+        "In electron-Volt V =  -7.197 eV\n",
+        "Note: the answer in the book is wrong due to calculation mistake\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 , Page no:15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "#given\n",
+      "r0=0.236; #in nanometer(interionic distance)\n",
+      "e=1.6E-19; # in C (charge of electron)\n",
+      "E_o= 8.85E-12;# absolute premittivity\n",
+      "N=8; # Born constant\n",
+      "IE=5.14;# in eV (ionisation energy of sodium)\n",
+      "EA=3.65;# in eV (electron affinity of Chlorine)\n",
+      "pi=3.14; # value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "r0=r0*1E-9; # since r is in nanometer\n",
+      "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
+      "PE=PE/e; #changing unit from J to eV\n",
+      "NE=IE-EA;# calculation of Net energy\n",
+      "BE=PE-NE;# calculation of Bond Energy\n",
+      "\n",
+      "#result\n",
+      "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
+      "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
+      "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
+      "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
+      "# Note: (2) There is slight variation in the answer due to round off."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The potential energy is PE=  5.34 eV\n",
+        "The net energy is NE=  1.49 eV\n",
+        "The bond energy is BE=  3.85 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 , Page no:16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r_0=.41; #in mm(lattice constant)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "E_o= 8.85E-12; #absolute premittivity\n",
+      "n=0.5; #repulsive exponent value\n",
+      "alpha=1.76; #Madelung constant\n",
+      "pi=3.14; # value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "r=.41*1E-3; #since r is in mm\n",
+      "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
+      "\n",
+      "#result\n",
+      "print\"The compressibility  is\tBeta=\",'%.4E'%Beta;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The compressibility  is\tBeta= -2.5097E+15\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 , Page no:16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r_0=3.56; #in Angstrom\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "IE=3.89; #in eV (ionisation energy of Cs)\n",
+      "EA=-3.61; #in eV (electron affinity of Cl)\n",
+      "n=10.5; #Born constant\n",
+      "E_o= 8.85E-12; #absolute premittivity\n",
+      "alpha=1.763; #Madelung constant\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "r_0=r_0*1E-10; #since r is in nanometer\n",
+      "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+      "U=U/e; #changing unit from J to eV\n",
+      "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
+      "\n",
+      "#result\n",
+      "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
+      "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ionic cohesive energy is  -6.0 eV\n",
+        "The atomic cohesive energy is -6.0 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 , Page no:17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r_0=2.81; #in Angstrom\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "n=9; #Born constant\n",
+      "E_o= 8.85E-12; #absolute premittivity\n",
+      "alpha=1.748; #Madelung constant\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "r_0=r_0*1E-10; #since r is in nanometer\n",
+      "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+      "V=V/e; #changing unit from J to eV\n",
+      "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
+      "\n",
+      "#result\n",
+      "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
+      "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The potential energy is V= -7.96 eV\n",
+        "The energy contributing per ions to the cohesive energy is  -3.98 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_02_Crystal_Structure_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_02_Crystal_Structure_1.ipynb
new file mode 100755
index 00000000..b6038072
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_02_Crystal_Structure_1.ipynb
@@ -0,0 +1,291 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:693d1c12519596ec538fd8d806511b7a1fdb44b316b98121270f6b11b5d8d004"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2:Crystal Structure"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 , Page no:40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "N=6.02*10**26; #in /Kg-molecule (Avogadro's number)\n",
+      "n=4; #number of molecules per unit cell ofr NaCl\n",
+      "M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)\n",
+      "p=2189; #in Kg/m^3 (density)\n",
+      "\n",
+      "#calculate\n",
+      "a=pow(((n*M)/(N*p)),0.3333333);\n",
+      "a1=a*10**10; #changing unit to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
+      "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The lattice constant is a= 5.62E-10 m\n",
+        "\t\t\ta= 5.62 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 , Page no:41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "N=6.02*10**23; #in /gram-atom (Avogadro's number)\n",
+      "n=4; #number of atom per unit cell for fcc structure\n",
+      "M=63.5; #in gram/gram-atom (atomic weight of Cu)\n",
+      "p=8.96; #in g/cm^3 (density)\n",
+      "\n",
+      "#calculate\n",
+      "a=pow((n*M/(N*p)),0.3333333);\n",
+      "a1=a*1E8; #changing unit from cm to Angstrom\n",
+      "d=a1/math.sqrt(2); #distance infcc lattice\n",
+      "\n",
+      "#result\n",
+      "print\"lattice constant is a=\",'%.2E'%a,\"cm\";\n",
+      "print\"\\t\\t    a=\",round(a1,2),\"Angstrom\";\n",
+      "print\"distance between two nearest Cu atoms is d=\",round(d,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lattice constant is a= 3.61E-08 cm\n",
+        "\t\t    a= 3.61 Angstrom\n",
+        "distance between two nearest Cu atoms is d= 2.55 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 , Page no:41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "N=6.02E26; #in /Kg-atom (Avogadro's number)\n",
+      "n=2; #number of molecules per unit cell for bcc lattice\n",
+      "M=55.85; #in Kg/Kg-atom (atomic weight of Iron)\n",
+      "p=7860; #in Kg/m^3 (density)\n",
+      "\n",
+      "#calculate\n",
+      "a=pow((n*M/(N*p)),0.33333);\n",
+      "a1=a*1E10; #changing unit to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
+      "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The lattice constant is a= 2.87E-10 m\n",
+        "\t\t\ta= 2.87 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 , Page no:42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "N=6.02*10**26; #in /Kg-atom (Avogadro's number)\n",
+      "n=2; #number of molecules per unit cell for bcc lattice\n",
+      "M=6.94; #in Kg/Kg-atom (atomic weight of Iron)\n",
+      "p=530; #in Kg/m^3 (density)\n",
+      "\n",
+      "#calculate\n",
+      "a=pow((n*M/(N*p)),0.33333);\n",
+      "a1=a*1E10; #changing unit to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
+      "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The lattice constant is a= 3.52E-10 m\n",
+        "\t\t\ta= 3.52 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 , Page no:42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "N=6.02*10**23; #in /gram-molecule (Avogadro's number)\n",
+      "M=58.5; #in gram/gram-molecule (atomic weight of NaCl)\n",
+      "p=2.17; #in g/cm^3 (density)\n",
+      "\n",
+      "#calculate\n",
+      "#since V=M/p\n",
+      "#(1/d)^-3=2N/V=2Np/M\n",
+      "#therefore d= (M/2Np)^-3\n",
+      "d=pow((M/(2*N*p)),0.33333333);\n",
+      "d1=d*1*10**8; #changing unit from cm to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The distance between two adjacent atoms of NaCl is d=\",'%.2E'%d,\"m\";\n",
+      "print\"\\t\\t\\t\\t\\t\\t   d=\",round(d1,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance between two adjacent atoms of NaCl is d= 2.82E-08 m\n",
+        "\t\t\t\t\t\t   d= 2.82 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 , Page no:43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "r_Na=0.98; #in Angstrom (radius of sodium ion)\n",
+      "r_Cl=1.81; #in Angstrom (radius of chloride ion)\n",
+      "M_Na=22.99; #in amu (atomic mass of sodium)\n",
+      "M_Cl=35.45; #in amu (atomic mass of chlorine)\n",
+      "\n",
+      "#calculate\n",
+      "a=2*(r_Na+r_Cl); #lattice parameter\n",
+      "#PF=volume of ions present in the unit cell/volume of unit cell\n",
+      "PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;\n",
+      "#Density=mass of unit cell/volume of unit cell\n",
+      "p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;\n",
+      "p1=p*1E-3; #changing unit to gm/cm^-3\n",
+      "\n",
+      "#result\n",
+      "print\"Lattice constant  is a=\",round(a,3),\"Angstrom\";\n",
+      "print\"Packing fraction is =\",round(PF,3); \n",
+      "print\"Density is p=\",round(p),\"Kg/m^3\";\n",
+      "print\"Density is p=\",round(p1,2),\"g/cm^3\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Lattice constant  is a= 5.58 Angstrom\n",
+        "Packing fraction is = 0.662\n",
+        "Density is p= 2233.0 Kg/m^3\n",
+        "Density is p= 2.23 g/cm^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_03_Planes_in_Crystals_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_03_Planes_in_Crystals_1.ipynb
new file mode 100755
index 00000000..92dba19b
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_03_Planes_in_Crystals_1.ipynb
@@ -0,0 +1,144 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:05935aae8ee517fae6abe7a89dceb1ca0d0cc367834bce08b5f427ec49852dcb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3:Planes in Crystals"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11 , Page no:61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "h=3;\n",
+      "k=2;\n",
+      "l=1; #miller indices\n",
+      "a=4.2E-8; #in cm (lattice constant)\n",
+      "\n",
+      "#calculate\n",
+      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+      "d=d*1E8; #changing unit from cm to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The interplanar spacing is d=\",round(d,2),\"cm\";\n",
+      "print\"\\t\\t\\t   d=\",round(d,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The interplanar spacing is d= 1.12 cm\n",
+        "\t\t\t   d= 1.12 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 , Page no:61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "h=1;k=1;l=1; #miller indices\n",
+      "a=2.5;b=2.5;c=1.8; #in Angstrom (lattice constants for tetragonal lattice )\n",
+      "\n",
+      "#calculate\n",
+      "d=1/math.sqrt((h/a)**2+(k/b)**2+(l/c)**2); #calculation for interplanar spacing\n",
+      "\n",
+      "#result\n",
+      "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The lattice spacing is d= 1.26 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15 , Page no:63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "h=1;k=0;l=0; #miller indices\n",
+      "a=2.5; #in Angstrom (lattice constant)\n",
+      "\n",
+      "#calculate\n",
+      "a=a*1E-10; #hence a is in Angstrom\n",
+      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+      "p=d/a**3;\n",
+      "\n",
+      "#result\n",
+      "print\"The density is p=\",round(p),\"lattice points/m^2\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density is p= 1.6e+19 lattice points/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_04_Crystal_Diffraction_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_04_Crystal_Diffraction_1.ipynb
new file mode 100755
index 00000000..fef74a69
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_04_Crystal_Diffraction_1.ipynb
@@ -0,0 +1,503 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:513926db1524f2764781365e8a81c6fa3119b7bc0f032a42233e49f5c3b6140a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4:Crystal Diffraction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 , Page no:75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "lambda1=2.6; #in Angstrom (wavelength)\n",
+      "theta=20; #in Degree (angle)\n",
+      "n=2;\n",
+      "\n",
+      "#calculate\n",
+      "lambda1=lambda1*1E-10; #since lambda is in Angstrom\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore d=n(lambda)/2sin(theta)\n",
+      "d=n*lambda1/(2*math.sin(math.radians(theta)));\n",
+      "\n",
+      "#result\n",
+      "print\"The spacing constant is d=\",'%.2E'%d,\"m\";\n",
+      "print\"\\t\\t\\td=\",round(d*10**10,3),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The spacing constant is d= 7.60E-10 m\n",
+        "\t\t\td= 7.602 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 , Page no:75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "h=1;k=1;l=0; #miller indices\n",
+      "a=0.26; #in nanometer (lattice constant)\n",
+      "lambda1=0.065; #in nanometer (wavelength)\n",
+      "n=2; #order\n",
+      "\n",
+      "#calculate\n",
+      "d=a/math.sqrt(h**2+k**2+l**2); #calculation of interlattice spacing\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore we have\n",
+      "theta=math.asin(n*lambda1/(2*d));\n",
+      "theta1=theta*180/3.14;\n",
+      "\n",
+      "#result\n",
+      "print\"The glancing angle is =\",round(theta1,2),\"degree\";\n",
+      "print \"Note: there is slight variation in the answer due to round off error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The glancing angle is = 20.72 degree\n",
+        "Note: there is slight variation in the answer due to round off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 , Page no:75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "d=3.04E-10; #in mm (spacing constant)\n",
+      "lambda1=0.79; #in Angstrom (wavelength)\n",
+      "n=3; #order\n",
+      "\n",
+      "#calculate\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore we have\n",
+      "lambda2=lambda1*1E-10; #since lambda is in angstrom\n",
+      "theta=math.asin(n*lambda2/(2*d));\n",
+      "theta1=theta*180/3.14;\n",
+      "\n",
+      "#result\n",
+      "print\"The glancing angle is\",round(theta1,3),\"degree\";\n",
+      "print \"Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The glancing angle is 22.954 degree\n",
+        "Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 , Page no:76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "d=0.282; #in nanometer (spacing constant)\n",
+      "n=1; #order\n",
+      "theta=8.35; #in degree (glancing angle)\n",
+      "\n",
+      "#calculate\n",
+      "d=d*1E-9; #since d is in nanometer\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore we have\n",
+      "lambda1=2*d*math.sin(math.radians(theta))/n; \n",
+      "lambda_1=lambda1*1E10; #changing unit from m to Angstrom\n",
+      "theta_1=90; #in degree (for maximum order theta=90)\n",
+      "n_max=2*d*math.sin(math.radians(theta_1))/lambda1; #calculation of maximum order. \n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength  =\",'%.2E'%lambda1,\"m\";\n",
+      "print\"\\t\\t=\" ,round(lambda_1,3),\"Angstrom\";\n",
+      "print\"The maximum order possible is n=\",round(n_max);\n",
+      "print \"Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength  = 8.19E-11 m\n",
+        "\t\t= 0.819 Angstrom\n",
+        "The maximum order possible is n= 7.0\n",
+        "Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 , Page no:76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "theta=6; #in degree (glancing angle)\n",
+      "p=2170; #in Kg/m^3 (density)\n",
+      "M=58.46; #Molecular weight of NaCl\n",
+      "N=6.02E26; #in Kg-molecule (Avogadro's number)\n",
+      "n=1; #order\n",
+      "XU=1E-12; #since 1X.U.= 1E-12m\n",
+      "\n",
+      "#calculate\n",
+      "d=(M/(2*N*p))**(1/3); #calclation of lattice constant\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore we have\n",
+      "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation of wavelength\n",
+      "lambda2=lambda1/XU;\n",
+      "\n",
+      "#result\n",
+      "print\"The spacing constant is d=\",'%.2E'%d,\"m\";\n",
+      "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
+      "print\"\\t\\t  =\",round(lambda2,1),\"X.U\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The spacing constant is d= 2.82E-10 m\n",
+        "The wavelength is = 5.89E-11 m\n",
+        "\t\t  = 58.9 X.U\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 , Page no:77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "h=1;k=1;l=1; #miller indices\n",
+      "a=5.63; #in Angstrom  (lattice constant)\n",
+      "theta=27.5; #in degree (Glancing angle)\n",
+      "n=1; #order\n",
+      "H=6.625E-34; #in J-s (Plank's constant)\n",
+      "c=3E8; #in m/s (velocity of light)\n",
+      "e=1.6E-19; #charge of electron\n",
+      "\n",
+      "#calculate\n",
+      "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore we have\n",
+      "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation for wavelength\n",
+      "E=H*c/(lambda1*1E-10); #calculation of Energy\n",
+      "E1=E/e; #changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";\n",
+      "print\"The wavelength is =\",round(lambda1),\"Angstrom\";\n",
+      "print\"The energy of X-rays is E=\",'%.2E'%E,\"J\";\n",
+      "print\"\\t\\t\\tE=\",'%.3E'%E1,\"eV\";\n",
+      "print \"Note: c=3E8 m/s but in solution c=3E10 m/s \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The lattice spacing is d= 3.25 Angstrom\n",
+        "The wavelength is = 3.0 Angstrom\n",
+        "The energy of X-rays is E= 6.62E-16 J\n",
+        "\t\t\tE= 4.138E+03 eV\n",
+        "Note: c=3E8 m/s but in solution c=3E10 m/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 , Page no:77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "V=344; #in V (accelerating voltage)\n",
+      "theta=60; #in degree (glancing angle)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "h=6.625e-34; #in J-s (Plank's constant)\n",
+      "n=1; #order\n",
+      "e=1.6E-19; #charge on electron\n",
+      "\n",
+      "#calculate\n",
+      "#Since K=m*v^2/2=e*V\n",
+      "#therefore  v=sqrt(2*e*V/m)\n",
+      "#since lambda=h/(m*v)\n",
+      "#therefore we have lambda=h/sqrt(2*m*e*V)\n",
+      "lambda1=h/math.sqrt(2*m*e*V); #calculation of lambda\n",
+      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore we have\n",
+      "d=n*lambda2/(2*math.sin(math.radians(theta)));\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
+      "print\"\\t\\t  =\",round(lambda2,2),\"Angstrom\";\n",
+      "print\"The interplanar spacing  is  d=\",round(d,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 6.62E-11 m\n",
+        "\t\t  = 0.66 Angstrom\n",
+        "The interplanar spacing  is  d= 0.38 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 , Page no:78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "K=0.02; #in eV (kinetic energy)\n",
+      "d=2.0; #in Angstrom (Bragg's spacing)\n",
+      "m=1.00898; #in amu (mass of neutron)\n",
+      "amu=1.66E-27; #in Kg (1amu=1.66E-27 Kg)\n",
+      "h=6.625e-34; #in J-s (Plank's constant)\n",
+      "n=1; #order\n",
+      "e=1.6E-19; #charge on electron\n",
+      "\n",
+      "#calculate\n",
+      "#Since K=m*v^2/2\n",
+      "#therefore  v=sqrt(2*K/m)\n",
+      "#since lambda=h/(m*v)\n",
+      "#therefore we have lambda=h/sqrt(2*m*K)\n",
+      "m=m*amu; #changing unit from amu to Kg\n",
+      "K=K*e; #changing unit to J from eV\n",
+      "lambda1=h/math.sqrt(2*m*K); #calculation of lambda\n",
+      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+      "theta=math.asin(n*lambda2/(2*d)); #calculation of angle of first order diffraction maximum\n",
+      "theta1=theta*180/3.14;\n",
+      "#result\n",
+      "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
+      "print\"\\t    \\t  =\",round(lambda2),\"Angstrom\";\n",
+      "print\"The angle of first order diffraction maximum is \",round(theta1),\"Degree\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 2.02E-10 m\n",
+        "\t    \t  = 2.0 Angstrom\n",
+        "The angle of first order diffraction maximum is  30.0 Degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 , Page no:79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "lambda1=0.586; #in Angstrom (wavelength of X-rays)\n",
+      "n1=1; n2=2; n3=3; #orders of diffraction\n",
+      "theta1=5+(58/60); #in degree (Glancing angle for first order of diffraction)\n",
+      "theta2=12+(01/60); #in degree (Glancing angle for second order of diffraction)\n",
+      "theta3=18+(12/60); #in degree (Glancing angle for third order of diffraction)\n",
+      "\n",
+      "#calculate\n",
+      "K1=math.sin(math.radians(theta1));\n",
+      "K2=math.sin(math.radians(theta2));\n",
+      "K3=math.sin(math.radians(theta3));\n",
+      "#Since 2dsin(theta)=n(lambda)\n",
+      "#therefore we have\n",
+      "d1=n1*lambda1/(2*K1);\n",
+      "d2=n2*lambda1/(2*K2);\n",
+      "d3=n3*lambda1/(2*K3);\n",
+      "d1=d1*1E-10; #changing unit from Angstrom to m\n",
+      "d2=d2*1E-10; #changing unit from Angstrom to m\n",
+      "d3=d3*1E-10; #changing unit from Angstrom to m\n",
+      "d=(d1+d2+d3)/3;\n",
+      "\n",
+      "#result\n",
+      "print\"The value of sine of different angle of diffraction is \\nK1=\",round(K1,4);\n",
+      "print\"K2=\",round(K2,4);\n",
+      "print\"K3=\",round(K3,4);\n",
+      "#Taking the ratios of K1:K2:K3\n",
+      "#We get K1:K2:K3=1:2:3\n",
+      "#Therefore we have\n",
+      "print\"Or we have K1:K2:K3=1:2:3\";\n",
+      "print\"Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\";\n",
+      "print\"The spacing constants are \\nd1=\",'%.2E'%d1;\n",
+      "print\"d2=\",'%.3E'%d2;\n",
+      "print\"d3=\",'%.3E'%d3;\n",
+      "print\"The mean value of crystal spacing is d=\",'%.3E'%d,\"m\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of sine of different angle of diffraction is \n",
+        "K1= 0.1039\n",
+        "K2= 0.2082\n",
+        "K3= 0.3123\n",
+        "Or we have K1:K2:K3=1:2:3\n",
+        "Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\n",
+        "The spacing constants are \n",
+        "d1= 2.82E-10\n",
+        "d2= 2.815E-10\n",
+        "d3= 2.814E-10\n",
+        "The mean value of crystal spacing is d= 2.816E-10 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb
new file mode 100755
index 00000000..66041c45
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb
@@ -0,0 +1,1258 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:43369901ed597bc7502d54a31a362249abbf2812c557cb432bc86adf72bf30d7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5:Principles of Quantum Mechanics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 , Page no:102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "lambda1=1; #in Angstrom (wavelength)\n",
+      "m=1.67E-27; #in Kg (mass of neutron)\n",
+      "h=6.625E-34; #in J-s (Planck's constant)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "lambda2=lambda1*1E-10; #since lambda is in Angstrom\n",
+      "#Since lambda=h/(m*v)\n",
+      "#Therefore we have\n",
+      "v=h/(m*lambda2); #calculation of velocity\n",
+      "K=m*v**2/2; #calculation of kinetic energy\n",
+      "K1=K/e; #changing unit fro J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The velocity is v=\",round(v,3),\"m/s\";\n",
+      "print\"The kinetic energy is K=\",'%.3E'%K,\"J\";\n",
+      "print\"\\t\\t       =\",'%.3E'%K1,\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity is v= 3967.066 m/s\n",
+        "The kinetic energy is K= 1.314E-20 J\n",
+        "\t\t       = 8.213E-02 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 , Page no:103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "K=50; #in eV (Kinetic energy)\n",
+      "m0=9.1E-31; #in Kg (mass of electron)\n",
+      "h=6.625E-34; #in J-s (Planck's constant)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "K=K*e; #changing unit from eV to J\n",
+      "#Since K=m*v^2/2\n",
+      "#Therefore v=sqrt(2*K/m)\n",
+      "#Since lambda=h/(m*v)\n",
+      "#Therefore we have\n",
+      "lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
+      "print\"\\t\\t  =\",round(lambda2,3),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 1.736E-10 m\n",
+        "\t\t  = 1.736 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 , Page no:104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "E=2000; #in eV (Kinetic energy)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "h=6.625E-34; #in J-s (Planck's constant)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "E=E*e; #changing unit from eV to J\n",
+      "#Since E=m*v^2/2\n",
+      "#Therefore v=sqrt(2*E/m)\n",
+      "#Since lambda=h/(m*v)\n",
+      "#Therefore we have\n",
+      "lambda1=h/math.sqrt(2*m*E); #calculation of wavelength\n",
+      "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
+      "print\"\\t\\t  =\",round(lambda2,4),\"nm\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 2.745E-11 m\n",
+        "\t\t  = 0.0275 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 , Page no:104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "m_e=9.1E-31; #in Kg (mass of electron)\n",
+      "m_n=1.676E-27; #in Kg (mass of neutron)\n",
+      "h=6.625E-34; #in J-s (Planck's constant)\n",
+      "c=3E8; #in m/s (velocity of light)\n",
+      "\n",
+      "#calculate\n",
+      "E_e=m_e*c**2; #rest mass energy of electron\n",
+      "E_n=2*E_e; #given (kinetic energy of neutron)\n",
+      "#Since K=m*v^2/2\n",
+      "#Therefore v=sqrt(2*K/m)\n",
+      "#Since lambda=h/(m*v)\n",
+      "#Therefore we have\n",
+      "lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength\n",
+      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
+      "print\"\\t\\t  =\",'%.3E'%lambda2,\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 2.827E-14 m\n",
+        "\t\t  = 2.827E-04 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 , Page no:104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "V=1600; #in V (Potential)\n",
+      "\n",
+      "#calculate\n",
+      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+      "print \"Note : The answer in the book is wrong \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 0.307 Angstrom\n",
+        "Note : The answer in the book is wrong \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 , Page no:105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "E=1000; #in eV (Kinetic energy of photon)\n",
+      "K=1000; #in eV (Kinetic energy of electron)\n",
+      "m0=9.1E-31; #in Kg (mass of electron)\n",
+      "h=6.6E-34; #in J-s (Planck's constant)\n",
+      "c=3E8; #in m/s (velocity of light)\n",
+      "e=1.6E-19; #in C (charge on electron)\n",
+      "\n",
+      "#calculate\n",
+      "E=E*e; #changing unit from eV to J\n",
+      "lambda_p=h*c/E; #For photon E=hc/lambda\n",
+      "lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom\n",
+      "#Since K=m*v^2/2\n",
+      "#Therefore v=sqrt(2*K/m)\n",
+      "#Since lambda=h/(m*v)\n",
+      "#Therefore we have\n",
+      "K=K*e; #changing unit from eV to J\n",
+      "lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+      "lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"For photon,the wavelength is =\",'%.3E'%lambda_p,\"m\";\n",
+      "print\"\\t\\t\\t     =\",round(lambda_p1,1),\"Angstrom\";\n",
+      "print\"For electron,the wavelength is =\",'%.3E'%lambda_e,\"m\";\n",
+      "print\"\\t\\t\\t       =\",round(lambda_e1,2),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For photon,the wavelength is = 1.238E-09 m\n",
+        "\t\t\t     = 12.4 Angstrom\n",
+        "For electron,the wavelength is = 3.868E-11 m\n",
+        "\t\t\t       = 0.39 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7 , Page no:105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "lambda1=1.66*10**-10; #in m (wavelength)\n",
+      "m=9.1*10**-31; #in Kg (mass of electron)\n",
+      "h=6.626*10**-34; #in J-s (Planck's constant)\n",
+      "e=1.6E-19; #in C (charge on electron)\n",
+      "\n",
+      "#calculate\n",
+      "#Since lambda=h/(m*v)\n",
+      "#Therefore we have\n",
+      "v=h/(m*lambda1); #calculation of velocity\n",
+      "K=m*v**2/2; #calculation of kinetic energy\n",
+      "K1=K/e; #changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The velocity of electron is v=\",'%.2E'%v,\"m/s\";\n",
+      "print\"The kinetic energy is K=\",'%.2E'%K,\"J\";\n",
+      "print\"\\t\\t       =\",round(K1,3),\"eV\";\n",
+      "print \"Note: The answer in the book for kinetic energy is wrong \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity of electron is v= 4.39E+06 m/s\n",
+        "The kinetic energy is K= 8.75E-18 J\n",
+        "\t\t       = 54.714 eV\n",
+        "Note: The answer in the book for kinetic energy is wrong \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 , Page no:106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "T=400; #in K (temperature)\n",
+      "m=6.7E-27; #in Kg (mass of He-atom)\n",
+      "h=6.6E-34; #in J-s (Planck's constant)\n",
+      "k=1.376E-23; #in J/degree (Boltzmann constant)\n",
+      "\n",
+      "#calculate\n",
+      "#Since lambda=h/(m*v)\n",
+      "#E=mv^2/2;\n",
+      "#Therefore lambda=h/sqrt(2*m*E)\n",
+      "#E=kT\n",
+      "#Therefore lambda=h/sqrt(2*m*k*T)\n",
+      "lambda1=h/math.sqrt(2*m*k*T)\n",
+      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The de-Broglie wavelength of He-atom is =\",'%.2E'%lambda1,\"m\";\n",
+      "print\"\\t\\t\\t\\t        =\",round(lambda2,4),\"Angstrom\";    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The de-Broglie wavelength of He-atom is = 7.69E-11 m\n",
+        "\t\t\t\t        = 0.7685 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 , Page no:106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "m_e=9.1E-31; #in Kg (mass of electron)\n",
+      "m_p=1.6E-27; #in Kg (mass of proton)\n",
+      "h=6.626E-34; #in J-s (Planck's constant)\n",
+      "c=3E8; #in m/s (velocity of light)\n",
+      "\n",
+      "#calculate\n",
+      "E=m_e*c**2; #in J (rest energy of electron)\n",
+      "#Since lambda=h/(m*v)\n",
+      "#E=mv^2/2;\n",
+      "#herefore lambda=h/sqrt(2*m*E)\n",
+      "#Also E=m_e*c^2;\n",
+      "#therefore lambda=h/sqrt(2*m_p*m_e*c^2)\n",
+      "lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength\n",
+      "lambda2=lambda1*1*10**10; #changing unit from m to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The de-Broglie wavelength of proton is =\",'%.2E'%lambda1,\"m\";\n",
+      "print\"\\t\\t\\t\\t       =\",'%.2E'%lambda2,\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The de-Broglie wavelength of proton is = 4.09E-14 m\n",
+        "\t\t\t\t       = 4.09E-04 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10 , Page no:106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "V=10000; #in V (Potential)\n",
+      "\n",
+      "#calculate\n",
+      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 0.123 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11 , Page no:107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "V=100; #in V (Potential)\n",
+      "n=1; #order of diffraction\n",
+      "d=2.15; #in Angstrom (lattice spacing)\n",
+      "\n",
+      "#calculate\n",
+      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+      "#Since 2*d*sind(theta)=n*lambda\n",
+      "#therefore we have\n",
+      "theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle\n",
+      "theta1=theta*180/3.14;\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+      "print\"The glancing angle is =\",round(theta1,1),\"degree\";\n",
+      "print \"Note: In question V=100 eV but the solution is using V=100V \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 1.227 Angstrom\n",
+        "The glancing angle is = 16.6 degree\n",
+        "Note: In question V=100 eV but the solution is using V=100V \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12 , Page no:107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "V=344; #in V (Potential)\n",
+      "n=1; #order of diffraction\n",
+      "theta=60; #in degree (glancing angle)\n",
+      "\n",
+      "#calculate\n",
+      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+      "#Since 2*d*sind(theta)=n*lambda\n",
+      "#therefore we have\n",
+      "d=n*lambda1/(2*math.sin(math.radians(theta))); #calculation of spacing constant\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+      "print\"The spacing of the crystal is =\",round(d,4),\"Angstrom\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is = 0.662 Angstrom\n",
+        "The spacing of the crystal is = 0.3819 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13 , Page no:107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "r=0.53*10**-10; #in m (radius of first Bohr orbit)\n",
+      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+      "m=9.1*10**-31; #in Kg (mass of electron)\n",
+      "n=1; #First Bohr orbit\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "#Since 2*pi*r=n*lambda  and  lambda=h/(m*v)\n",
+      "#Threfore we have v=h*n/(2*pi*r*m)\n",
+      "v=h*n/(2*pi*r*m); #calculation of velocity\n",
+      "\n",
+      "#result\n",
+      "print\"The velocity of electron is =\",'%.3E'%v,\"m/s\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity of electron is = 2.179E+06 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14 , Page no:108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "dx=0.2; #in Angstrom (uncertainty in the position)\n",
+      "h=6.6E-34; #in J-s (Planck's constant)\n",
+      "m0=9.1E-31; #in Kg (mass of electron)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "dx=dx*1E-10; #since dx is in Angstrom\n",
+      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
+      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+      "#since dp=m*dv\n",
+      "dv=dp/m0; #calculation of uncertainty in the velocity\n",
+      "\n",
+      "#result\n",
+      "print\"The uncertainty in the momentum is dp=\",'%.3E'%dp,\"Kg-m/\";\n",
+      "print\"The uncertainty in the velocity is dv=\",'%.3E'%dv,\"m/s\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The uncertainty in the momentum is dp= 2.627E-24 Kg-m/\n",
+        "The uncertainty in the velocity is dv= 2.887E+06 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15 , Page no:108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "m_e=9.1E-31; #in Kg (mass of electron)\n",
+      "m_p=1.67E-27; #in Kg (mass of proton)\n",
+      "dx_p=1; #in nanometer (uncertainty in position of electron)\n",
+      "dx_n=1; #in nanometer (uncertainty in position of proton)\n",
+      "\n",
+      "#calculate\n",
+      "#since dp=h/(4*pi*dx)\n",
+      "#since h/(4*pi) is constant and dx is same for electron and proton \n",
+      "#therefor both electron and proton have same uncertainty in the momentum\n",
+      "#since dv=dp/m  and  dp is same for both\n",
+      "#therefore dv_e/dv_p=m_p/m_e\n",
+      "#therefore\n",
+      "K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton\n",
+      "\n",
+      "#result\n",
+      "print\"The ratio of uncertainty in the velocity of electron to that of proton is =\",round(K);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16 , Page no:108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)\n",
+      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+      "m=1.67E-27; #in Kg (mass of proton)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
+      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+      "p=dp; #minimum value of momentum to calculate mimimum kinetic energy\n",
+      "K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton\n",
+      "K1=K/e; #changing unit from J to eV\n",
+      "K2=K1/1E6; #changing unit from eV to MeV\n",
+      "\n",
+      "#result\n",
+      "print\"The minimum uncertainty in the momentum of proton is dp =\",'%.3E'%dp,\"Kg-m/s\";\n",
+      "print\"The minimum kinetic energy of proton is K=\",'%.3E'%K,\"J\";\n",
+      "print\"\\t\\t\\t\\t \\t =\",'%.3E'%K1,\"eV\";\n",
+      "print\"\\t\\t\\t\\t\\t =\",round(K2,1),\"MeV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum uncertainty in the momentum of proton is dp = 1.051E-20 Kg-m/s\n",
+        "The minimum kinetic energy of proton is K= 3.307E-14 J\n",
+        "\t\t\t\t \t = 2.067E+05 eV\n",
+        "\t\t\t\t\t = 0.2 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.17 , Page no:109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "K=1; #in KeV (kinetic energy of electron)\n",
+      "dx=1; #in Angstrom (uncertainty in the position)\n",
+      "h=6.63E-34; #in J-s (Planck's constant)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "dx=dx*1E-10; #since dx is in Angstrom\n",
+      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
+      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+      "K=K*1E3*1.6E-19; #changing unit from KeV to J\n",
+      "p=math.sqrt(2*m*K); #calculation of momentum \n",
+      "poc=(dp/p)*100; #calculation of percentage of uncertainty\n",
+      "\n",
+      "#result\n",
+      "print\"The uncertainty in the momentum of electron is dp=\",'%.3E'%dp,\"Kg-m/s\";\n",
+      "print\"The momentum of electron is p=\",'%.3E'%p,\"Kg-m/s\";\n",
+      "print\"The percentage of uncertainty in the momentum  is =\",round(poc,1);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The uncertainty in the momentum of electron is dp= 5.279E-25 Kg-m/s\n",
+        "The momentum of electron is p= 1.706E-23 Kg-m/s\n",
+        "The percentage of uncertainty in the momentum  is = 3.1\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.18 , Page no:109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "v=6.6E4; #m/s (speed of electron)\n",
+      "poc=0.01; #percentage of uncertainty\n",
+      "h=6.63E-34; #in J-s (Planck's constant)\n",
+      "m=9E-31; #in Kg (mass of electron)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "p=m*v; #calculation of momentum\n",
+      "dp=(poc/100)*p; #calculation of uncertainty in the momentum\n",
+      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
+      "dx=h/(4*pi*dp); #calculation of uncertainty in the position\n",
+      "\n",
+      "#result\n",
+      "print\"The momentum of electron is p=\",'%.3E'%p,\"Kg-m/s\";\n",
+      "print\"The uncertainty in the momentum of electron is dp=\",'%.3E'%dp,\"Kg-m/s\";\n",
+      "print\"The uncertainty in the position of electron is dx=\",'%.3E'%dx,\"Kg-m/s\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The momentum of electron is p= 5.940E-26 Kg-m/s\n",
+        "The uncertainty in the momentum of electron is dp= 5.940E-30 Kg-m/s\n",
+        "The uncertainty in the position of electron is dx= 8.887E-06 Kg-m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.19 , Page no:111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "lambda1=1; #in Angstrom (wavelength)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "dlambda=1E-6; #uncertainty in wavelength\n",
+      "\n",
+      "#calculate\n",
+      "lambda2=lambda1*1E-10; #sinc lambda is in Angstrom\n",
+      "#By uncertainty principle, dx*dp>=h/(4*pi)  --(1)\n",
+      "#since p=h/lambda  -----(2)\n",
+      "#Or p*lambda=h \n",
+      "#diffrentiting this equation\n",
+      "#p*dlambda+lambda*dp=0\n",
+      "#dp=-p*dlambda/lambda     ----(3)\n",
+      "#from (2) and (3)  dp=-h*dlambda/lambda^2   ---(4)\n",
+      "#from (1) and(4)  dx*dlambda>=lambda^2/4*pi\n",
+      "#Or dx=lambda^2/(4*pi*dlambda)\n",
+      "dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position\n",
+      "\n",
+      "#result\n",
+      "print\"The uncertainty in the position of X-ray photon is dx=\",'%.3E'%dx,\"m\";\n",
+      "print \"Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \"\n",
+      "print \" ANSWEER IS WRONG IN THE TEXTBOOK\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The uncertainty in the position of X-ray photon is dx= 7.962E-16 m\n",
+        "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \n",
+        " ANSWEER IS WRONG IN THE TEXTBOOK\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.20 , Page no:111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "dt=1*10**-8; #in sec (average life time)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
+      "#and E=h*v      v is the frequency\n",
+      "#therefore we have dv>=1/(4*pi*dt)\n",
+      "dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency\n",
+      "\n",
+      "#result\n",
+      "print\"The minimum uncertainty in the frequency of the photon is dv=\",'%.3E'%dv,\"sec^-1\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum uncertainty in the frequency of the photon is dv= 7.962E+06 sec^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.21 , Page no:111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "dt=1E-12; #in sec (average life time)\n",
+      "h=6.63E-34; #in J-s (Planck'c constant)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "e=1.6*1E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
+      "dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy\n",
+      "dE1=dE/e; #changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The uncertainty in the energy of the photon is dE=\",'%.3E'%dE,\"J\";\n",
+      "print\"\\t\\t\\t\\t\\t         =\",'%.3E'%dE1,\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The uncertainty in the energy of the photon is dE= 5.279E-23 J\n",
+        "\t\t\t\t\t         = 3.299E-04 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.22 , Page no:111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "dT=2.5E-14; #in sec (average life time)\n",
+      "h=6.63E-34; #in J-s (Planck'c constant)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "e=1.6*1E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
+      "dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy\n",
+      "dE1=dE/e; #changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The uncertainty in the energy of the photon is dE=\",'%.3E'%dE,\"J\";\n",
+      "print\"\\t\\t\\t\\t\\t         =\",'%.3E'%dE1,\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The uncertainty in the energy of the photon is dE= 2.111E-21 J\n",
+        "\t\t\t\t\t         = 1.320E-02 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.23 , Page no:112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "a=2; #in Angstrom (length of the box)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "h=6.626E-34; #in J-s (Planck'c constant)\n",
+      "n2=2; n4=4; #two quantum states\n",
+      "e=1.6*1E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "a=a*1E-10; #since a is in Angstrom \n",
+      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
+      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state\n",
+      "E21=E2/e; #changing unit from J to eV\n",
+      "E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state\n",
+      "E41=E4/e; #changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The energy corresponding to the 2nd quantum state is E2=\",'%.3E'%E2,\"J\";\n",
+      "print\"\\t\\t\\t\\t\\t\\t       =\",'%.3E'%E21,\"eV\";\n",
+      "print\"The energy corresponding to the 4nd quantum state is E4=\",'%.3E'%E4,\"J\";\n",
+      "print\"\\t\\t\\t\\t\\t\\t       =\",'%.3E'%E41,\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy corresponding to the 2nd quantum state is E2= 6.031E-18 J\n",
+        "\t\t\t\t\t\t       = 3.769E+01 eV\n",
+        "The energy corresponding to the 4nd quantum state is E4= 2.412E-17 J\n",
+        "\t\t\t\t\t\t       = 1.508E+02 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.24 , Page no:113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "a=1E-10; #in m (width of the well)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "h=6.626E-34; #in J-s (Planck'c constant)\n",
+      "n1=1; n2=2; n3=3; #ground and first two excited states\n",
+      "e=1.6*1E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
+      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state\n",
+      "E11=E1/e; #changing unit from J to eV\n",
+      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state\n",
+      "E21=E2/e; #changing unit from J to eV\n",
+      "E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state\n",
+      "E31=E3/e; #changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The energy corresponding to the ground state is E1=\",'%.3E'%E1,\"J\";\n",
+      "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E11,\"eV\";\n",
+      "print\"The energy corresponding to the 1st excited state is E2=\",'%.3E'%E2,\"\";\n",
+      "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E21,\"eV\";\n",
+      "print\"The energy corresponding to the 2nd excited state is E3=\",'%.3E'%E3,\"J\";\n",
+      "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E31,\"eV\";\n",
+      "print \" There is slight variation in the answer due to round off error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy corresponding to the ground state is E1= 6.031E-18 J\n",
+        "\t\t\t\t\t = 3.769E+01 eV\n",
+        "The energy corresponding to the 1st excited state is E2= 2.412E-17 \n",
+        "\t\t\t\t\t = 1.508E+02 eV\n",
+        "The energy corresponding to the 2nd excited state is E3= 5.428E-17 J\n",
+        "\t\t\t\t\t = 3.392E+02 eV\n",
+        " There is slight variation in the answer due to round off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.25 , Page no:113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "dx=1E-8; #in m (length of box or uncertainty in the position)\n",
+      "h=6.626E-34; #in J-s (Planck'c constant)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#From uncertainty principle dx*dp=h and dp=m*dv\n",
+      "#therefore we have\n",
+      "dv=h/(m*dx); #calculation of minimum uncertainty in the velocity\n",
+      "dv1=dv*1E-3; #changing unit from m/s to Km/s\n",
+      "\n",
+      "#result\n",
+      "print\"The minimum uncertainty in the velocity of electron i dv=\",'%.3E'%dv,\"m/s\";\n",
+      "print\"\\t\\t\\t\\t\\t\\t        =\",round(dv1,2),\"Km/s\";\n",
+      "print \"Note: There is slight variation in the answer due to round off error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum uncertainty in the velocity of electron i dv= 7.281E+04 m/s\n",
+        "\t\t\t\t\t\t        = 72.81 Km/s\n",
+        "Note: There is slight variation in the answer due to round off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.26 , Page no:113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "a=4E-10; #in m (length of the box)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "h=6.626E-34; #in J-s (Planck'c constant)\n",
+      "n1=1; #ground state\n",
+      "e=1.6*1E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
+      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state\n",
+      "E11=E1/e; #changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The minimum energy of electron is E1=\",'%.3E'%E1,\"J (Note: The answer in the book  is wrong due to printing error)\";\n",
+      "print\"\\t\\t\\t\\t    =\",round(E11,3),\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum energy of electron is E1= 3.769E-19 J (Note: The answer in the book  is wrong due to printing error)\n",
+        "\t\t\t\t    = 2.356 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb
new file mode 100755
index 00000000..d52f3a45
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb
@@ -0,0 +1,561 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3c7e08d83e902e2949e4fdcd2bc98d527fe589fabb9636cea78c35c42abb8038"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6:Electron Theory and Band Theory of Metals"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 , Page no:146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "n=8.5E28; #in 1/m^3 (density of electron)\n",
+      "m_e=9.11E-31; #in Kg (mass of electron)\n",
+      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "T=300; #in K (temperature)\n",
+      "p=1.69E-8; #in ohm-m (resistivity)\n",
+      "\n",
+      "#calculate\n",
+      "lambda1=math.sqrt(3*k*m_e*T)/(n*e**2*p); #calculation of mean free path\n",
+      "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+      "\n",
+      "#result\n",
+      "print\"The mean free path of electron is =\",'%.3E'%lambda1,\"m\";\n",
+      "print\"\\t\\t\\t\\t  =\",round(lambda2,2),\"nm\";\n",
+      "#Note: answer in the book is wrong due to printing mistake"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean free path of electron is = 2.893E-09 m\n",
+        "\t\t\t\t  = 2.89 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 , Page no:146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "P_E=1; #in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied)\n",
+      "E=0.5; #in eV (energy above Fermi level)\n",
+      "\n",
+      "#calculate\n",
+      "P_E=1/100; #changing percentage into ratio\n",
+      "E=E*e; #changing unit from eV to J\n",
+      "#P_E=1/(1+exp((E-E_F)/k*T))\n",
+      "#Rearranging this equation, we get\n",
+      "#T=(E-E_F)/k*log((1/P_E)-1)\n",
+      "#Since E-E_F has been denoted by E  therefore\n",
+      "T=E/(k*math.log((1/P_E)-1));\n",
+      "\n",
+      "#result\n",
+      "print\"The temperature is T=\",round(T),\"K\";\n",
+      "# Note: There is slight variation in the answer due to logarithm function"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature is T= 1262.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 , Page no:147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "n=5.8E28; #in 1/m^3 (density of electron)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "p=1.54E-8; #in ohm-m (resistivity)\n",
+      "\n",
+      "#calculate\n",
+      "t=m/(n*e**2*p); #calculation of relaxation time\n",
+      "\n",
+      "#result\n",
+      "print\"The relaxation time of conduction electrons is =\",'%.3E'%t,\"sec\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The relaxation time of conduction electrons is = 3.980E-14 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 , Page no:147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "n=8.5E28; #in 1/m^3 (density of electron)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "sigma=6E7; #in 1/ohm-m (conductivity)\n",
+      "E_F=7; #in E=eV (Fermi energy of Copper)\n",
+      "\n",
+      "#calculate\n",
+      "E_F=E_F*e; #changing unit from eV to J\n",
+      "v_F=math.sqrt(2*E_F/m); #calculation of velocity of electrons\n",
+      "#Since sigma=n*e^2*lambda/(2*m*v_F)\n",
+      "#Therefore we have\n",
+      "lambda1=2*m*v_F*sigma/(n*e**2); #calculation of mean free path\n",
+      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+      "\n",
+      "#result\n",
+      "print\"The velocity of the electrons is v_F=\",'%.3E'%v_F,\"m/s   (roundoff error)\";\n",
+      "print\"The mean free path traveled by the electrons is=\",round(lambda2),\"Angstrom\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity of the electrons is v_F= 1.569E+06 m/s   (roundoff error)\n",
+        "The mean free path traveled by the electrons is= 787.0 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 , Page no:147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "n=6.5E28; #in 1/m^3 (density of electron)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "p=1.43E-8; #in ohm-m (resistivity)\n",
+      "\n",
+      "#calculate\n",
+      "t=m/(n*e**2*p); #calculation of relaxation time\n",
+      "\n",
+      "#result\n",
+      "print\"The relaxation time of conduction electrons is =\",'%.3E'%t,\"sec\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The relaxation time of conduction electrons is = 3.824E-14 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6 , Page no:148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "T=30; #in Celcius (temperature)\n",
+      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+      "m_p=1.67E-27; #in Kg (mass of proton)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "T=T+273; #changing temperature from Celcius to Kelvin\n",
+      "KE=(3/2)*k*T; #calculation of average kinetic energy\n",
+      "KE1=KE/e; #changing unit from J to eV\n",
+      "m=1.008*2*m_p; #calculating mass of hydrogen gas molecule\n",
+      "c=math.sqrt(3*k*T/m); #calculation of velocity\n",
+      "\n",
+      "#result\n",
+      "print\"The average kinetic energy of gas ,molecules is KE=\",'%.3E'%KE,\"J\";\n",
+      "print\"\\t\\t\\t\\t\\t\\t  =\",round(KE1,4),\"eV   (Answer is wrong in textbook)\";\n",
+      "print\"The velocity of molecules is c=\",round(c,2),\"m/s\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average kinetic energy of gas ,molecules is KE= 6.272E-21 J\n",
+        "\t\t\t\t\t\t  = 0.0392 eV   (Answer is wrong in textbook)\n",
+        "The velocity of molecules is c= 1930.27 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7 , Page no:148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "E=10; #in eV (kinetic energy for each electron and proton)\n",
+      "m_e=9.1E-31; #in Kg (mass of electron)\n",
+      "m_p=1.67E-27; #in Kg (mass of proton)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "E=E*e; #changing unit from eV to J\n",
+      "#since E=m*v^2/2\n",
+      "#therefore  v=sqrt(2E/m)\n",
+      "v_e=math.sqrt(2*E/m_e); #calculation of kinetic energy of electron\n",
+      "v_p=math.sqrt(2*E/m_p); #calculation of kinetic energy of proton\n",
+      "\n",
+      "#result\n",
+      "print\"The kinetic energy of electron is v_e=\",'%.3E'%v_e,\"m/s\";\n",
+      "print\"The kinetic energy of proton is v_p=\",'%.3E'%v_p,\"m/s\";\n",
+      "print \"Note: Answer is wrong in textbook\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The kinetic energy of electron is v_e= 1.875E+06 m/s\n",
+        "The kinetic energy of proton is v_p= 4.377E+04 m/s\n",
+        "Note: Answer is wrong in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8 , Page no:149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "I=100; #in A (current in the wire)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "A=10; #in mm^2 (cross-sectional area)\n",
+      "n=8.5E28; #in 1/m^3 (density of electron)\n",
+      "\n",
+      "#calculate\n",
+      "A=A*1E-6; #changing unit from mm^2 to m^2\n",
+      "v_d=I/(n*A*e);\n",
+      "\n",
+      "#result\n",
+      "print\"The drift velocity of free electrons is v_d=\",'%.3E'%v_d,\"m/s\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The drift velocity of free electrons is v_d= 7.353E-04 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 , Page no:149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "I=4;#in A (current in the conductor)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "A=1E-6; #in m^2 (cross-sectional area)\n",
+      "N_A=6.02E23; #in atoms/gram-atom (Avogadro's number)\n",
+      "p=8.9; #in g/cm^3 (density)\n",
+      "M=63.6; #atomic mass of copper\n",
+      "\n",
+      "#calculate\n",
+      "n=N_A*p/M; #Calculation of density of electrons in g/cm^3\n",
+      "n1=n*1E6; #changing unit from g/cm^3 to g/m^3\n",
+      "v_d=I/(n1*A*e);\n",
+      "\n",
+      "#result\n",
+      "print\"The density of copper atoms is n=\",'%.3E'%n,\"atoms/m^3\";\n",
+      "print\"\\t\\t\\t        =\",'%.3E'%n1,\"atoms/m^3\";\n",
+      "print\"The average drift velocity of free electrons is v_d=\",'%.3E'%v_d,\"m/s\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of copper atoms is n= 8.424E+22 atoms/m^3\n",
+        "\t\t\t        = 8.424E+28 atoms/m^3\n",
+        "The average drift velocity of free electrons is v_d= 2.968E-04 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10 , Page no:149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "n=9E28; #in 1/m^3 (density of valence electrons)\n",
+      "sigma=6E7; #in mho/m (conductivity of copper)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#Since sigma=n*e*mu  therefore\n",
+      "mu=sigma/(n*e); #calculation of mobility of electron\n",
+      "\n",
+      "#result\n",
+      "print\"The mobility of electrons is =\",'%.3E'%mu,\"m^2/V-s\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility of electrons is = 4.167E-03 m^2/V-s\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11 , Page no:150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "E_F=5.51; #in eV (Fermi energy in Silver)\n",
+      "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#part-(a)\n",
+      "Eo=(3/5)*E_F; #calculation of average energy of free electron at 0K\n",
+      "#part-(b)\n",
+      "Eo1=Eo*e; #changing unit from eV to J\n",
+      "#Since for a classical particle E=(3/2)*k*T\n",
+      "#therefroe we have\n",
+      "T=(2/3)*Eo1/k; #calculation of temperature for a classical particle (an ideal gas)\n",
+      "\n",
+      "#result\n",
+      "print\"The average energy of free electron at 0K is Eo=\",Eo,\"eV\";\n",
+      "print\"The temperature at which a classical particle have this much energy is T=\",'%.3E'%T,\"K\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average energy of free electron at 0K is Eo= 3.306 eV\n",
+        "The temperature at which a classical particle have this much energy is T= 2.555E+04 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12 , Page no:150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "E_F_L=4.7; #in eV (Fermi energy in Lithium)\n",
+      "E_F_M=2.35; #in eV (Fermi energy in a metal)\n",
+      "n_L=4.6E28; #in 1/m^3 (density of electron in Lithium)\n",
+      "\n",
+      "#calculate\n",
+      "#Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant\n",
+      "# Therefore we have n=C*E_F^(3/2)   where C is proportionality constant\n",
+      "#n1/n2=(E_F_1/E_F_2)^(3/2)\n",
+      "#Therefore we have\n",
+      "n_M=n_L*(E_F_M/E_F_L); #calculation of electron density for a metal\n",
+      "\n",
+      "#result\n",
+      "print\"The lectron density for a metal is =\",n_M,\"1/m^3\";\n",
+      "print \"Note: Answer in the book is wrong \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The lectron density for a metal is = 2.3e+28 1/m^3\n",
+        "Note: Answer in the book is wrong \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_07_Dielectric_Properties_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_07_Dielectric_Properties_1.ipynb
new file mode 100755
index 00000000..85a277ef
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_07_Dielectric_Properties_1.ipynb
@@ -0,0 +1,488 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d363aa85787801a09c4a4d5d3e8dc3b5551836ae64a0a0c025744f99f230a4b4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7:Dielectric Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 , Page no:187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "A=100; #in cm^2 (cross-sectional area)\n",
+      "d=1; #in cm (seperation between plates)\n",
+      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+      "V=100; #in V (potential difference)\n",
+      "\n",
+      "#calculate\n",
+      "A1=A*1E-4; #changing unit from cm^2 to m^2\n",
+      "d1=d*1E-2; #changing unit from cm to m\n",
+      "C=Eo*A1/d1; #calculation of capacitance\n",
+      "Q=C*V; #calculation of charge\n",
+      "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+      "\n",
+      "#result\n",
+      "print\"The capacitance of capacitor is C=\",C,\"C\";\n",
+      "print\"\\t\\t\\t\\t =\",C1,\"C\";\n",
+      "print\"The charge on the plates is Q=\",Q,\"C\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The capacitance of capacitor is C= 8.85e-12 C\n",
+        "\t\t\t\t = 8.85 C\n",
+        "The charge on the plates is Q= 8.85e-10 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 , Page no:187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "A=650; #in mm^2 (cross-sectional area)\n",
+      "d=4; #in mm (seperation between plates)\n",
+      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+      "Er=3.5; #di-electric constant of the material\n",
+      "Q=2E-10; #in C (charge on plates)\n",
+      "\n",
+      "#calculate\n",
+      "A1=A*1E-6; #changing unit from mm^2 to m^2\n",
+      "d1=d*1E-3; #changing unit from mm to m\n",
+      "C=Er*Eo*A1/d1; #calculation of capacitance\n",
+      "V=Q/C; #calculation of charge\n",
+      "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+      "\n",
+      "#result\n",
+      "print\"The capacitance of capacitor is C=\",'%.3E'%C,\"C\",\n",
+      "print\"\\n\\t\\t\\t         =\",round(C1,2),\"pF\";\n",
+      "print\"The resultant voltage across the capacitor is V=\",round(V,2),\"V\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The capacitance of capacitor is C= 5.033E-12 C \n",
+        "\t\t\t         = 5.03 pF\n",
+        "The resultant voltage across the capacitor is V= 39.73 V\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 , Page no:188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "N=2.7E25; #in 1/m^3 (density of atoms)\n",
+      "E=1E6; #in V/m (electric field)\n",
+      "Z=2; #atomic number of Helium \n",
+      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+      "Er=1.0000684; #(dielectric constant of the material)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "#since  alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 \n",
+      "#Therefore we have r_0^3=(Er-1)/(4*pi*N)\n",
+      "r_0=((Er-1)/(4*pi*N))**(1/3); #calculation of radius of electron cloud\n",
+      "x=4*pi*Eo*E*r_0/(Z*e); #calculation of dispalcement\n",
+      "\n",
+      "#result\n",
+      "print\"The radius of electron cloud is r_0=\",'%.3E'%r_0,\"m\";\n",
+      "print\"The displacement is x=\",'%.3E'%x,\"m\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The radius of electron cloud is r_0= 5.865E-11 m\n",
+        "The displacement is x= 2.037E+04 m\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 , Page no:188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "K=1.000134; #di-elecrtic constant of the neon gas at NTP\n",
+      "E=90000; #in V/m (electric field)\n",
+      "Eo=8.85E-12; #in C/N-m^2 (absolute premittivity)\n",
+      "N_A=6.023E26; #in atoms/Kg-mole (Avogadro's number)\n",
+      "V=22.4; #in m^3 (volume of gas at NTP\n",
+      "\n",
+      "#calculate\n",
+      "n=N_A/V; #calculaton of density of atoms\n",
+      "#Since P=n*p=(k-1)*Eo*E\n",
+      "#therefore we have\n",
+      "p=(K-1)*Eo*E/n; #calculation of dipole moment induced\n",
+      "alpha=p/E; #calculation of atomic polarisability\n",
+      "\n",
+      "#result\n",
+      "print\"The dipole moment induced in each atom is p=\",'%.3E'%p,\"C-m\";\n",
+      "print\"The atomic polarisability of neon is =\",'%.3E'%alpha,\"c-m^2/V\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dipole moment induced in each atom is p= 3.969E-36 C-m\n",
+        "The atomic polarisability of neon is = 4.410E-41 c-m^2/V\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 , Page no:189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "Er=3.75; #di-elecrtic constant of sulphur at 27 degree Celcius\n",
+      "gama=1/3; #internal field constant\n",
+      "p=2050; #in Kg/m^3 (density)\n",
+      "M_A=32; #in amu (atomic weight of sulphur)\n",
+      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+      "N=6.022E23; #Avogadro's number\n",
+      "\n",
+      "#calculate\n",
+      "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+      "#therefore we have\n",
+      "alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); #calculation of electronic polarisability of sulphur\n",
+      "\n",
+      "#result\n",
+      "print\"The electronic polarisability of sulphur is =\",'%.3E'%alpha_e,\"Fm^2   (roundoff error)\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electronic polarisability of sulphur is = 3.291E-37 Fm^2   (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 , Page no:189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "Er=1.0000684; #di-elecrtic constant of Helium gas at NTP\n",
+      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+      "N=2.7E25; #number of atomsper unit volume\n",
+      "\n",
+      "#calculate\n",
+      "#Since Er-1=(N/Eo)*alpha_e\n",
+      "#therefore we have\n",
+      "alpha_e=Eo*(Er-1)/N; #calculation of electronic polarisability of Helium\n",
+      "\n",
+      "#result\n",
+      "print\"The electronic polarisability of Helium gas is =\",'%.3E'%alpha_e,\"Fm^2    (roundoff error)\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electronic polarisability of Helium gas is = 2.242E-41 Fm^2    (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 , Page no:190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "N=3E28; #in atoms/m^3 (density of atoms)\n",
+      "alpha_e=1E-40; #in F-m^2 (electronic polarisability)\n",
+      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+      "\n",
+      "#calculate\n",
+      "#Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo)\n",
+      "#therefore we have\n",
+      "Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo)));\n",
+      "#calculation of dielectric constant of the material\n",
+      "\n",
+      "#result\n",
+      "print\"The dielectric constant of the material is Er=\",'%.3E'%Er,\"F/m\",\n",
+      "print \"NOTE: The answer in the book is wrong \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dielectric constant of the material is Er= 1.382E+00 F/m NOTE: The answer in the book is wrong \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8 , Page no:190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "Er=4; #relative permittivity of sulphur\n",
+      "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+      "NA=2.08E3; #in Kg/m^3 (density of atoms in sulphur)\n",
+      "\n",
+      "#calculate\n",
+      "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+      "#therefore we have\n",
+      "alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); #calculation of electronic polarisability of sulphur\n",
+      "\n",
+      "#result\n",
+      "print\"The electronic polarisability of sulphur is =\",'%.3E'%alpha_e,\"Fm^2\";\n",
+      "print \"NOTE: The answer in the book is wrong and they used the wrong formula \" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electronic polarisability of sulphur is = 6.382E-15 Fm^2\n",
+        "NOTE: The answer in the book is wrong and they used the wrong formula \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9 , Page no:190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "alpha1=2.5E-39; #in C^2-m/N (dielectric constant at 300K)\n",
+      "alpha2=2.0E-39; #in C^2-m/N (dielectric constant at 400K)\n",
+      "T1=300; #in K(first temperature)\n",
+      "T2=400; #in K(second temperature)\n",
+      "\n",
+      "#calculate\n",
+      "#since alpha=alpha_d+alpha0 and alpha0=Beta/T\n",
+      "#therefore alpha=alpha_d+(Beta/T)\n",
+      "#since alpha1=alpha_d+(Beta/T1)  and alpha2=alpha_d+(Beta/T2)\n",
+      "#therefore alpha1-apha2=Beta*((1/T1)-(1/T2))\n",
+      "#or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2))\n",
+      "Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); #calculation of Beta\n",
+      "alpha_d=alpha1-(Beta/T1); #calculation of polarisability due to defromation\n",
+      "alpha0_1=Beta/T1; #calculation of polarisability due to permanent dipole moment at 300K\n",
+      "alpha0_2=Beta/T2;  #calculation of polarisability due to permanent dipole moment at 400K\n",
+      "\n",
+      "#result\n",
+      "print\"The polarisability due to permanent dipole moment at 300K is =\",'%.3E'%alpha0_1,\"C^2-m/N\";\n",
+      "print\"The polarisability due to permanent dipole moment at 400K is =\",'%.3E'%alpha0_2,\"C^2-m/N\";\n",
+      "print\"The polarisability due to deformation of the molecules is =\",'%.3E'%alpha_d,\"C^2-m/N\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The polarisability due to permanent dipole moment at 300K is = 2.000E-39 C^2-m/N\n",
+        "The polarisability due to permanent dipole moment at 400K is = 1.500E-39 C^2-m/N\n",
+        "The polarisability due to deformation of the molecules is = 5.000E-40 C^2-m/N\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10 , Page no:191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "n=1.5; #refractive index\n",
+      "Er=5.6; #dielectric constant\n",
+      "\n",
+      "#calculate\n",
+      "#since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0)   Clausius-Mossotti equation\n",
+      "#and  (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) \n",
+      "#from above two equations, we get  ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i)\n",
+      "#or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P)\n",
+      "#where P is fractional ionisational polarisability\n",
+      "P=1-((n**2-1)/(n**2+2))*((Er+2)/(Er-1)); #calculation of fractional ionisational polarisability\n",
+      "P1=P*100; #calculation of percentage of  ionisational polarisability\n",
+      "\n",
+      "#result\n",
+      "print\"The percentage of  ionisational polarisability is =\",round(P1,1),\"percent\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage of  ionisational polarisability is = 51.4 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_08_Magnetic_Properties_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_08_Magnetic_Properties_1.ipynb
new file mode 100755
index 00000000..b2a8def3
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_08_Magnetic_Properties_1.ipynb
@@ -0,0 +1,323 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:708d3f2a9798f4d28f6ce90536d50509c591ba2ba31051121346b4739b2aea82"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8:Magnetic Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 , Page no:236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "X=-0.5E-5; #magnetic susceptibility of silicon\n",
+      "H=0.9E4; #in A/m (magnetic field intensity)\n",
+      "mu0=4*3.14*1E-7; #in H/m (absolute permeability)\n",
+      "\n",
+      "#calculate\n",
+      "I=X*H; #calculation of intensity of magnetism\n",
+      "B=mu0*H*(1+X); #calculation of magnetic flux density\n",
+      "\n",
+      "#result\n",
+      "print\"The intensity of magnetism is I=\",I,\"A/m\";\n",
+      "print\"The magnetic flux density is B=\",round(B,3),\"Wb/m^2\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The intensity of magnetism is I= -0.045 A/m\n",
+        "The magnetic flux density is B= 0.011 Wb/m^2\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 , Page no:236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "r=0.052; #in nm (radius of orbit)\n",
+      "B=1; #in Wb/m^2 (magnetic field of induction)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "\n",
+      "#calculate\n",
+      "r=0.052*1E-9; #changing unit from nm to m\n",
+      "d_mu=(e**2*r**2*B)/(4*m); #calculation of change in magnetic moment\n",
+      "\n",
+      "#result\n",
+      "print\"The change in magnetic moment is =\",'%.3E'%d_mu,\"Am^2\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The change in magnetic moment is = 1.902E-29 Am^2\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 , Page no:236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "H=220; #in A/m (magnetic field intensity)\n",
+      "I=3300; #in A/m (intensity of magnetisation)\n",
+      "\n",
+      "#calculate\n",
+      "mu_r=1+(I/H); #calculation of relative permeability\n",
+      "\n",
+      "#result\n",
+      "print\"The relative permeability of a ferromagentic material is =\",mu_r;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The relative permeability of a ferromagentic material is = 16.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 , Page no:236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "I=3000; #in A/m (intensity of magnetisation)\n",
+      "B=0.005; #in Wb/m^2 (magnetic flus intensity)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+      "\n",
+      "#calculate\n",
+      "H=(B/mu0)-I; #calculation of magnetic force\n",
+      "mu_r=(I/H)+1; #calculation of relative permeability\n",
+      "\n",
+      "#result\n",
+      "print\"The magnetic force is  H=\",round(H,3);\n",
+      "print\"The relative permeability is =\",round(mu_r,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnetic force is  H= 980.892\n",
+        "The relative permeability is = 4.058\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5 , Page no:237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "H=4E3; #in A/m (magnetic field intensity)\n",
+      "N=60; #number of turns\n",
+      "l=12; #in cm (length of solenoid)\n",
+      "\n",
+      "#calculate\n",
+      "n=N/(l*1E-2); #calculation of number of turns per unit metre\n",
+      "#Snice H=n*i;\n",
+      "i=H/n; #calculation of current through the solenoid\n",
+      "\n",
+      "#result\n",
+      "print\"The current through the solenoid is i=\",i,\"A\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current through the solenoid is i= 8.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6 , Page no:237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "l=30; #in cm (length of solenoid)\n",
+      "A=1; #in cm^2 (cross-sectional area)\n",
+      "N=300; #number of turns\n",
+      "i=0.032; #in A (current through the winding)\n",
+      "phi_B=2E-6; #in Wb (magnetic flux)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+      "\n",
+      "#calculate\n",
+      "l=l*1E-2; #changing unit from cm to m\n",
+      "A=A*1E-4; #changing unit from cm^2 to m^2\n",
+      "B=phi_B/A; #calculation of flux density\n",
+      "H=N*i/l; #calculation of magnetic intensity\n",
+      "mu=B/H; #calcluation of absolute permeability of iron\n",
+      "mu_r=mu/mu0; #calcluation of relative permeability of iron\n",
+      "\n",
+      "#result\n",
+      "print\"The flux density is B=\",B,\"Wb/m^2\";\n",
+      "print\"The magnetic intensity is H=\",H,\"A-turns/m\";\n",
+      "print\"The relative permeability of iron is =\",round(mu_r),\"    (roundoff error)\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The flux density is B= 0.02 Wb/m^2\n",
+        "The magnetic intensity is H= 32.0 A-turns/m\n",
+        "The relative permeability of iron is = 498.0     (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7 , Page no:238"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "A=100; #in m^2 (area of Hysteresis loop)\n",
+      "B=0.01; #in Wb/m^2 (unit space along vertical axis or magnetic flux density)\n",
+      "H=40; #in A/m (unit space along horizontal axis or magnetic fild ntensity)\n",
+      "\n",
+      "#calculate\n",
+      "H_L=A*B*H; #calculation of magnetic intensity\n",
+      "\n",
+      "#result\n",
+      "print\"The Hystersis loss per cycle is =\",round(H_L),\"J/m^2\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Hystersis loss per cycle is = 40.0 J/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_09_Semiconductors_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_09_Semiconductors_1.ipynb
new file mode 100755
index 00000000..9ce2ea83
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_09_Semiconductors_1.ipynb
@@ -0,0 +1,782 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c155cecc785c23bdccd0bf715c04d53d7788e1b590b4310d6af27c13ce7b97ac"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9:Semiconductors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 , Page no:272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "Eg=0.67; #in eV (Energy band gap)\n",
+      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+      "T1=298; #in K (room temperature)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "K=10; #ratio of number of electrons at different temperature\n",
+      "\n",
+      "#calculate\n",
+      "Eg=Eg*e; #changing unit from eV to Joule\n",
+      "#since ne=Ke*exp(-Eg/(2*k*T))\n",
+      "#and ne/ne1=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) and ne/ne1=K=10\n",
+      "#therefore we have 10=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1))\n",
+      "#re-arranging the equation for T, we get T2=1/((1/T1)-((2*k*log(10))/Eg))\n",
+      "T=1/((1/T1)-((2*k*math.log(10))/Eg)); #calculation of the temperature\n",
+      "\n",
+      "#result\n",
+      "print\"The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T=\",round(T),\"K            (roundoff error)\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T= 362.0 K            (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 , Page no:272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "ni=2.5E13; #in /cm^3 (intrinsic carrier density)\n",
+      "ue=3900; #in cm^2/(V-s) (electron mobilities)\n",
+      "uh=1900; #in cm^2/(V-s) (hole mobilities)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "l=1; #in cm (lenght of the box)\n",
+      "b=1;h=1; #in mm (dimensions of germanium rod )\n",
+      "\n",
+      "#calculate\n",
+      "ni=ni*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+      "ue=ue*1E-4; #changing unit from cm^2 to m^2\n",
+      "uh=uh*1E-4; #changing unit from cm^2 to m^2\n",
+      "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+      "rho=1/sigma; #calculation of resistivity\n",
+      "l=l*1E-2; #changing unit from mm to m for length\n",
+      "A=(b*1E-3)*(h*1E-3); #changing unit from mm to m for width and height and calculation of cross-sectional area\n",
+      "R=rho*l/A; #calculation of resistance\n",
+      "\n",
+      "#result\n",
+      "print\"The resistance of intrinsic germanium is R=\",'%.3E'%R,\"ohm\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance of intrinsic germanium is R= 4.310E+03 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 , Page no:273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "ne=2.5E19; #in /m^3 (electron  density)\n",
+      "nh=2.5E19; #in /m^3 (hole density)\n",
+      "ue=0.36; #in m^2/(V-s) (electron mobilities)\n",
+      "uh=0.17; #in m^2/(V-s) (hole mobilities)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#since ne=nh=ni,  therefore we have \n",
+      "ni=nh;\n",
+      "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+      "rho=1/sigma; #calculation of resistivity\n",
+      "\n",
+      "#result\n",
+      "print\"The conductivity of germanium is =\",round(sigma,2),\"/ohm-m\";\n",
+      "print\"The resistivity of germanium is =\",round(rho,2),\"ohm-m\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of germanium is = 2.12 /ohm-m\n",
+        "The resistivity of germanium is = 0.47 ohm-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 , Page no:273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+      "ue=0.135; #in m^2/(V-s) (electron mobilities)\n",
+      "uh=0.048; #in m^2/(V-s) (hole mobilities)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "ND=1E23; #in atom/m^3 (doping concentration)\n",
+      "\n",
+      "#calculate\n",
+      "sigma_i=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+      "sigma=ND*ue*e; #calculation of conductivity after doping\n",
+      "rho=ni**2/ND; #calculation of equilibrium hole concentration\n",
+      "\n",
+      "#result\n",
+      "print\"The intrinsic conductivity for silicon is =\",'%.3E'%sigma_i,\"S\";\n",
+      "print\"The conductivity after doping with phosphorus atoms is =\",'%.3E'%sigma,\"S\";\n",
+      "print\"The equilibrium hole concentration is =\",'%.3E'%rho,\"/m^3\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The intrinsic conductivity for silicon is = 4.392E-04 S\n",
+        "The conductivity after doping with phosphorus atoms is = 2.160E+03 S\n",
+        "The equilibrium hole concentration is = 2.250E+09 /m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 , Page no:274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+      "ue=0.13; #in m^2/(V-s) (electron mobilities)\n",
+      "uh=0.05; #in m^2/(V-s) (hole mobilities)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "ne=5E20; #in /m^3 (concentration of donor type impurity)\n",
+      "nh=5E20; #in /m^3 (concentration of acceptor type impurity)\n",
+      "\n",
+      "#calculate\n",
+      "#part-i\n",
+      "sigma=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+      "#part-ii\n",
+      "#since 1 donor atom is in 1E8 Si atoms, hence holes concentration can be neglected\n",
+      "sigma1=ne*e*ue; #calculation of conductivity after doping with donor type impurity\n",
+      "#part-iii\n",
+      "#since 1 acceptor atom is in 1E8 Si atoms, hence electron concentration can be neglected\n",
+      "sigma2=nh*e*uh; #calculation of conductivity after doping with acceptor type impurity\n",
+      "\n",
+      "#result\n",
+      "print\"The intrinsic conductivity for silicon is =\",'%.3E'%sigma,\"(ohm-m)^-1\";\n",
+      "print\"The conductivity after doping with donor type impurity is =\",sigma1,\"(ohm-m)^-1\";\n",
+      "print\"The conductivity after doping with acceptor type impurity is =\",sigma2,\"(ohm-m)^-1\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The intrinsic conductivity for silicon is = 4.320E-04 (ohm-m)^-1\n",
+        "The conductivity after doping with donor type impurity is = 10.4 (ohm-m)^-1\n",
+        "The conductivity after doping with acceptor type impurity is = 4.0 (ohm-m)^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7 , Page no:274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "ni=1E20; #in /m^3 (intrinsic carrier density)\n",
+      "ND=1E21; #in /m^3 (donor impurity concentration)\n",
+      "\n",
+      "#calculate\n",
+      "nh=ni**2/ND; #calculation of density of hole carriers at room temperature\n",
+      "\n",
+      "#result\n",
+      "print\"The density of hole carriers at room temperature is nh=\",nh,\"/m^3\";\n",
+      "#Note: answer in the book is wrong due to printing mistake"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of hole carriers at room temperature is nh= 1e+19 /m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8 , Page no:275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "M=72.6; #atomic mass of germanium\n",
+      "P=5400; #in Kg/m^3 (density)\n",
+      "ue=0.4; #in m^2/V-s (mobility of electrons)\n",
+      "uh=0.2; #in m^2/V-s (mobility of holes)\n",
+      "Eg=0.7; #in eV (Band gap)\n",
+      "m=9.1E-31; #in Kg (mass of electron)\n",
+      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+      "T=300; #in K (temperature)\n",
+      "h=6.63E-34; #in J/s (Planck\u2019s constant)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "e=1.6E-19; #in C(charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "Eg=Eg*e; #changing unit from eV to J\n",
+      "ni=2*(2*pi*m*k*T/h**2)**(3/2)*math.exp(-Eg/(2*k*T));\n",
+      "sigma=ni*e*(ue+uh);\n",
+      "\n",
+      "#result\n",
+      "print\"The intrinsic carrier density for germanium at 300K is ni=\",'%.3E'%ni,\"/m^3\";\n",
+      "print\"The conductivity of germanium is=\",round(sigma,3),\"(ohm-m)^-1\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The intrinsic carrier density for germanium at 300K is ni= 3.334E+19 /m^3\n",
+        "The conductivity of germanium is= 3.201 (ohm-m)^-1\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9 , Page no:275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "rho1=4.5; #in ohm-m (resistivity at 20 degree Celcius)\n",
+      "rho2=2.0; #in ohm-m (resistivity at 32 degree Celcius)\n",
+      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+      "T1=20; T2=32; #in degree Celcius (two temperatures)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "T1=T1+273; #changing unit from degree Celius to K\n",
+      "T2=T2+273; #changing unit from degree Celius to K\n",
+      "#since sigma=e*u*C*T^(3/2)*exp(-Eg/(2*k*T))\n",
+      "#therefore sigma1/sigma2=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+      "#and sigma=1/rho \n",
+      "#therefore we have rho2/rho1=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+      "#re-arranging above equation for Eg, we get Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1))\n",
+      "Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*math.log(T1/T2)-math.log(rho2/rho1));\n",
+      "Eg1=Eg/e;#changing unit from J to eV\n",
+      "\n",
+      "#result\n",
+      "print\"The energy band gap is Eg=\",'%.3E'%Eg,\"J\";\n",
+      "print\"\\t\\t\\t =\",round(Eg1,3),\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy band gap is Eg= 1.543E-19 J\n",
+        "\t\t\t = 0.964 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10 , Page no:276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "rho=2300; #in ohm-m (resistivity of pure silicon)\n",
+      "ue=0.135; #in m^2/V-s (mobility of electron)\n",
+      "uh=0.048; #in m^2/V-s (mobility of electron)\n",
+      "Nd=1E19; #in /m^3 (doping concentration)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "#since sigma=ni*e*(ue+uh) and sigma=1/rho\n",
+      "#therefore ni=1/(rho*e*(ue+uh))\n",
+      "ni=1/(rho*e*(ue+uh)); #calculation of intrinsic concentration\n",
+      "ne=Nd; #calculation of electron concentration\n",
+      "nh=ni**2/Nd; #calculation of hole concentration\n",
+      "sigma=ne*ue*e+nh*uh*e; #calculation of conductivity\n",
+      "rho=1/sigma; #calculation of resistivity\n",
+      "\n",
+      "#result\n",
+      "print\"The electron concentration is ne=\",ne,\"/m^3\";\n",
+      "print\"The hole concentration is nh=\",'%.3E'%nh,\"/m^3\";\n",
+      "print\"The resistivity of the specimen  is =\",round(rho,3),\"ohm-m\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electron concentration is ne= 1e+19 /m^3\n",
+        "The hole concentration is nh= 2.205E+13 /m^3\n",
+        "The resistivity of the specimen  is = 4.63 ohm-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11 , Page no:276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "uh=1900; #in cm^2/V-s (mobility of electron)\n",
+      "Na=2E17; #in /m^3 (acceptor doping concentration)\n",
+      "e=1.6E-19; #in C(charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "uh=uh*1E-4; #changing unit from cm^2/V-s to m^2/V-s\n",
+      "Na=Na*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+      "nh=Na; #hole concentration \n",
+      "#since sigma=ne*ue*e+nh*uh*e and nh>>ne\n",
+      "#therefore sigma=nh*uh*e\n",
+      "sigma=nh*uh*e; #calculation of conductivity\n",
+      "\n",
+      "#result\n",
+      "print\"The conductivity of p-type Ge crystal is =\",sigma,\"/ohm-m   (roundoff error)\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of p-type Ge crystal is = 6080.0 /ohm-m   (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12 , Page no:277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "ue=0.19; #in m^2/V-s (mobility of electron)\n",
+      "T=300; #in K (temperature)\n",
+      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+      "e=1.6E-19; #in C(charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "Dn=ue*k*T/e; #calculation of diffusion co-efficient\n",
+      "\n",
+      "#result\n",
+      "print\"The diffusion co-efficient of electron in silicon is Dn=\",'%.3E'%Dn,\"m^2/s\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion co-efficient of electron in silicon is Dn= 4.916E-03 m^2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.13 , Page no:277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "Eg=0.4; #in eV (Band gap of semiconductor)\n",
+      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+      "T1=0; #in degree Celcius (first temperature)\n",
+      "T2=50; #in degree Celcius (second temperature)\n",
+      "T3=100; #in degree Celcius (third temperature)\n",
+      "e=1.602E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "T1=T1+273; #changing temperature form Celcius to Kelvin\n",
+      "T2=T2+273; #changing temperature form Celcius to Kelvin\n",
+      "T3=T3+273; #changing temperature form Celcius to Kelvin\n",
+      "Eg=Eg*e; #changing unit from eV to Joule\n",
+      "#Using F_E=1/(1+exp(Eg/2*k*T))\n",
+      "F_E1=1/(1+math.exp(Eg/(2*k*T1))); #calculation of probability of occupation of lowest level at 0 degree Celcius\n",
+      "F_E2=1/(1+math.exp(Eg/(2*k*T2))); #calculation of probability of occupation of lowest level at 50 degree Celcius\n",
+      "F_E3=1/(1+math.exp(Eg/(2*k*T3))); #calculation of probability of occupation of lowest level at 100 degree Celcius\n",
+      "\n",
+      "#result\n",
+      "print\"The probability of occupation of lowest level in conduction band is\";\n",
+      "print\"\\t at 0 degree Celcius, F_E=\",'%.3E'%F_E1,\"eV\";\n",
+      "print\"\\t at 50 degree Celcius, F_E=\",'%.3E'%F_E2,\"eV\";\n",
+      "print\"\\t at 100 degree Celcius, F_E=\",'%.3E'%F_E3,\"eV\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The probability of occupation of lowest level in conduction band is\n",
+        "\t at 0 degree Celcius, F_E= 2.025E-04 eV\n",
+        "\t at 50 degree Celcius, F_E= 7.550E-04 eV\n",
+        "\t at 100 degree Celcius, F_E= 1.976E-03 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.14 , Page no:278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "Eg=1.2; #in eV (Energy band gap)\n",
+      "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+      "T1=600; T2=300; #in K (two temperatures)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "Eg=Eg*e; #changing unit from eV to Joule\n",
+      "#since sigma is proportional to exp(-Eg/(2*k*T))\n",
+      "#therefore ratio=sigma1/sigma2=exp(-Eg/(2*k*((1/T1)-(1/T2))));\n",
+      "ratio= math.exp((-Eg/(2*k))*((1/T1)-(1/T2))); #calculation of ratio of conductivity at 600K and at 300K\n",
+      "\n",
+      "#result\n",
+      "print\"The ratio of conductivity at 600K and at 300K is =\",'%.3E'%ratio;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ratio of conductivity at 600K and at 300K is = 1.085E+05\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.15 , Page no:278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "ue=0.39; #in m^2/V-s (mobility of electron)\n",
+      "n=5E13; #number of donor atoms\n",
+      "ni=2.4E19; #in atoms/m^3 (intrinsic carrier density)\n",
+      "l=10; #in mm (length of rod)\n",
+      "a=1; #in mm (side of square cross-section)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "l=l*1E-3; #changing unit from mm to m\n",
+      "a=a*1E-3; #changing unit from mm to m\n",
+      "A=a**2; #calculation of cross-section area\n",
+      "Nd=n/(l*A); #calculation of donor concentration\n",
+      "ne=Nd; #calculation of electron density\n",
+      "nh=ni**2/Nd; #calculation of hole density\n",
+      "#since sigma=ne*e*ue+nh*e*ue and since ne>>nh\n",
+      "#therefore sigma=ne*e*ue\n",
+      "sigma=ne*e*ue; #calculation of conductivity\n",
+      "rho=1/sigma; #calculation of resistivity\n",
+      "R=rho*l/A; #calculation of resistance \n",
+      "\n",
+      "#result\n",
+      "print\"The electron density is ne=\",ne,\"/m^3\";\n",
+      "print\"The hole density is nh=\",nh,\"/m^3\";\n",
+      "print\"The conductivity is =\",sigma,\"/ohm-m\";\n",
+      "print\"The resistance is R=\",round(R),\"ohm\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electron density is ne= 5e+21 /m^3\n",
+        "The hole density is nh= 1.152e+17 /m^3\n",
+        "The conductivity is = 312.0 /ohm-m\n",
+        "The resistance is R= 32.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.16 , Page no:279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+      "rho=8.93E-3; #in ohm-m (resistivity)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "u=RH/rho; #calculation of mobility\n",
+      "n=1/(RH*e); #calculation of density\n",
+      "\n",
+      "#result\n",
+      "print\"The mobility is u=\",round(u,4),\"m^2/(V-s)\";\n",
+      "print\"The density is n=\",'%.3E'%n,\"/m^3\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility is u= 0.041 m^2/(V-s)\n",
+        "The density is n= 1.708E+22 /m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.17 , Page no:279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+      "rho=8.93E-3; #in ohm-m (resistivity)\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "\n",
+      "#calculate\n",
+      "nh=1/(RH*e); #calculation of density of charge carrier\n",
+      "uh=1/(rho*nh*e); #calculation of mobility of charge carrier\n",
+      "\n",
+      "#result\n",
+      "print\"The density of charge carrier is nh=\",'%.3E'%nh,\"/m^3\";\n",
+      "print\"The mobility of charge carrier is uh=\",round(uh,4),\"m^2/(V-s)\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of charge carrier is nh= 1.708E+22 /m^3\n",
+        "The mobility of charge carrier is uh= 0.041 m^2/(V-s)\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_10_Superconductivity_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_10_Superconductivity_1.ipynb
new file mode 100755
index 00000000..9b11ce44
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_10_Superconductivity_1.ipynb
@@ -0,0 +1,316 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2419f2161da8f36d7fa10fabb5f5aa4b318a9912237faaa5f03ac922aa4b8ac9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10:Superconductivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 , Page no:313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "Tc=7.2; #in K (critical temperature)\n",
+      "T=5; #in K (given temperature)\n",
+      "H0=6.5E3; #in A/m (critical magnetic field at 0K)\n",
+      "\n",
+      "#calculate\n",
+      "Hc=H0*(1-(T/Tc)**2); #calculation of magnitude of critical magnetic field\n",
+      "\n",
+      "#result\n",
+      "print\"The magnitude of critical magnetic field is Hc=\",round(Hc,2),\"A/m\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnitude of critical magnetic field is Hc= 3365.35 A/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 , Page no:313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "r=0.02; #in m (radius of ring)\n",
+      "Hc=2E3; #in A/m (critical magnetic field at 5K)\n",
+      "pi=3.14; #value of pi used in the solutiion\n",
+      "\n",
+      "#calculate\n",
+      "Ic=2*pi*r*Hc; #calculation of critical current value\n",
+      "\n",
+      "#result\n",
+      "print\"The critical current value is Ic=\",Ic,\"A\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The critical current value is Ic= 251.2 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 , Page no:313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "M1=199.5; #in amu (isotropic mass at 5K)\n",
+      "T1=5; #in K (first critical temperature)\n",
+      "T2=5.1; #in K (second critical temperature)\n",
+      "#calculate\n",
+      "#since Tc=C*(1/sqrt(M)\n",
+      "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+      "#therefore we have M2=(T1/T2)^2*M1\n",
+      "M2=(T1/T2)**2*M1; #calculation of isotropic mass at 5.1K\n",
+      "\n",
+      "#result\n",
+      "print\"The isotropic mass at 5.1K is M2=\",round(M2,3),\"a.m.u.\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The isotropic mass at 5.1K is M2= 191.753 a.m.u.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 , Page no:314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "T=6; #in K (given temperature)\n",
+      "Hc=5E3; #in A/m (critical magnetic field at 5K)\n",
+      "H0=2E4; #in A/m (critical magnetic field at 0K)\n",
+      "\n",
+      "#calculate\n",
+      "#since Hc=H0*(1-(T/Tc)^2)\n",
+      "#therefor we have Tc=T/sqrt(1-(Hc/H)^2)\n",
+      "Tc=T/math.sqrt(1-(Hc/H0)); #calculation of transition temperature\n",
+      "\n",
+      "#result\n",
+      "print\"The transition temperature is Tc=\",round(Tc,3),\"K\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The transition temperature is Tc= 6.928 K\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 , Page no:314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "T=5; #in K (given temperature)\n",
+      "d=3; #in mm (diameter of the wire)\n",
+      "Tc=8; #in K (critical temperature for Pb)\n",
+      "H0=5E4; #in A/m (critical magnetic field at 0K)\n",
+      "pi=3.14; #value of pi used in the solution\n",
+      "\n",
+      "#calculate\n",
+      "Hc=H0*(1-(T/Tc)**2); #calculation of critical magnetic field at 5K\n",
+      "r=(d*1E-3)/2; #calculation of radius in m\n",
+      "Ic=2*pi*r*Hc; #calculation of critical current at 5K\n",
+      "\n",
+      "#result\n",
+      "print\"The critical magnetic field at 5K is Hc=\",'%.3E'%Hc,\"A/m\";\n",
+      "print\"The critical current at 5K is Ic=\",round(Ic,4),\"A\";\n",
+      "print \"   (roundoff error)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The critical magnetic field at 5K is Hc= 3.047E+04 A/m\n",
+        "The critical current at 5K is Ic= 287.0156 A\n",
+        "   (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 , Page no:314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "V=8.50; #in micro V (voltage across Josephson junction )\n",
+      "e=1.6E-19; #in C (charge of electron)\n",
+      "h=6.626E-34; #in J/s (Planck\u2019s constant)\n",
+      "\n",
+      "#calculate\n",
+      "V=V*1E-6; #changing unit from V to microVolt\n",
+      "v1=2*e*V/h; #calculation of frequency of EM waves\n",
+      "\n",
+      "#result\n",
+      "print\"The frequency of EM waves is v=\",'%.3E'%v1,\"Hz\";\n",
+      "print \"NOTE: The answer in the textbook is wrong\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The frequency of EM waves is v= 4.105E+09 Hz\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7 , Page no:315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "M1=200.59; #in amu (average atomic mass  at 4.153K)\n",
+      "Tc1=4.153; #in K (first critical temperature)\n",
+      "M2=204; #in amu (average atomic mass of isotopes)\n",
+      "\n",
+      "#calculate\n",
+      "#since Tc=C*(1/sqrt(M)\n",
+      "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+      "#therefore we have Tc2=Tc1*sqrt(M1/M2)\n",
+      "Tc2=Tc1*math.sqrt(M1/M2); #calculation of transition temperature of the isotopes\n",
+      "\n",
+      "#result\n",
+      "print\"The transition temperature of the isotopes is  Tc2=\",round(Tc2,3),\"K\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The transition temperature of the isotopes is  Tc2= 4.118 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_12_Fibre_Optics.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_12_Fibre_Optics.ipynb
new file mode 100755
index 00000000..f573b0a2
--- /dev/null
+++ b/Applied_Physics_by_P_K_Mittal/Chapter_12_Fibre_Optics.ipynb
@@ -0,0 +1,365 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:dac7a0b06f622e783ea6c469bf1c100c0c8157cc5e77bffea65e378f258808b2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12:Fibre Optics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 , Page no:360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "u1=1.563; #refractive index of core\n",
+      "u2=1.498; #refractive index of cladding\n",
+      "\n",
+      "#calculate\n",
+      "d=(u1-u2)/u1; #calculation of fractional index change\n",
+      "\n",
+      "#result\n",
+      "print\"The fractional index change for a given optical fibre is =\",round(d,4);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The fractional index change for a given optical fibre is = 0.0416\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 , Page no:360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "u1=1.55; #refractive index of core\n",
+      "u2=1.50; #refractive index of cladding\n",
+      "\n",
+      "#calculate\n",
+      "d=(u1-u2)/u1; #calculation of fractional index change\n",
+      "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+      "theta=math.asin(NA); #calculation of acceptance angle\n",
+      "theta1=theta*180/3.14;\n",
+      "\n",
+      "#result\n",
+      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numerical aperture of the fibre is NA= 0.394\n",
+        "The acceptance angle of the optical fibre is = 23.2 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 , Page no:360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "u1=1.563; #refractive index of core\n",
+      "u2=1.498; #refractive index of cladding\n",
+      "\n",
+      "#calculate\n",
+      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+      "theta=math.asin(NA); #calculation of acceptance angle\n",
+      "theta1=theta*180/3.14;\n",
+      "#result\n",
+      "print\"The numerical aperture of the fibre is NA=\",round(NA,4);\n",
+      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numerical aperture of the fibre is NA= 0.4461\n",
+        "The acceptance angle of the optical fibre is = 26.5 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4 , Page no:360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "NA=0.39; #numerical aperture of the optical fibre\n",
+      "d=0.05; #difference in the refractive index of the material of the core and cladding\n",
+      "\n",
+      "#calculate\n",
+      "#since NA=u1*sqrt(2*d)\n",
+      "#we have u1=NA/sqrt(2*d)\n",
+      "u1= NA/math.sqrt(2*d); #calculation of refractive index of  material of the core\n",
+      "\n",
+      "#result\n",
+      "print\"The refractive index of  material of the core is u1=\",round(u1,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The refractive index of  material of the core is u1= 1.233\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 , Page no:361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "u1=1.50; #refractive index of core\n",
+      "u2=1.45; #refractive index of cladding\n",
+      "\n",
+      "#calculate\n",
+      "d=(u1-u2)/u1; #calculation of fractional index change\n",
+      "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+      "theta_0=math.asin(NA); #calculation of acceptance angle\n",
+      "theta_01=theta_0*180/3.14;\n",
+      "theta_c=math.asin(u2/u1); #calculation of critical angle\n",
+      "theta_c1=theta_c*180/3.14;\n",
+      "\n",
+      "#result\n",
+      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+      "print\"The acceptance angle of the optical fibre is =\",round(theta_01,2),\"degree\";\n",
+      "print\"The critical angle of the optical fibre is =\",round(theta_c1,2),\"degree\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numerical aperture of the fibre is NA= 0.387\n",
+        "The acceptance angle of the optical fibre is = 22.8 degree\n",
+        "The critical angle of the optical fibre is = 75.2 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6 , Page no:361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "NA=0.33; #numerical aperture\n",
+      "d=0.02; #difference in the refractive index of the core and cladding of the material\n",
+      "\n",
+      "#calculate\n",
+      "#since NA=u1*sqrt(2*d)\n",
+      "#therefore we have\n",
+      "u1=NA/math.sqrt(2*d); #calculation of refractive index of the core\n",
+      "#since d=(u1-u2)/u2\n",
+      "#therefore we have\n",
+      "u2=(1-d)*u1; #calculation of refractive index of the cladding\n",
+      "\n",
+      "#result\n",
+      "print\"The refractive index of the core is u1=\",round(u1,2);\n",
+      "print\"The refractive index of the cladding is u2=\",round(u2,3);\n",
+      "print \"NOTE: The answer in the textbook is wrong\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The refractive index of the core is u1= 1.65\n",
+        "The refractive index of the cladding is u2= 1.617\n",
+        "NOTE: The answer in the textbook is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7 , Page no:361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "u1=3.5; #refractive index of core\n",
+      "u2=3.45; #refractive index of cladding\n",
+      "u0=1; #refractive index of the air\n",
+      "\n",
+      "#calculate\n",
+      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+      "NA1=NA/u0;\n",
+      "alpha=math.asin(NA); #calculation of acceptance angle\n",
+      "alpha1=alpha*180/3.14;\n",
+      "#result\n",
+      "print\"The numerical aperture of the fibre is NA=\",round(NA1,2);\n",
+      "print\"The acceptance angle of the optical fibre is =\",round(alpha1,2),\"degree\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numerical aperture of the fibre is NA= 0.59\n",
+        "The acceptance angle of the optical fibre is = 36.14 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8 , Page no:361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#given\n",
+      "u1=1.48; #refractive index of core\n",
+      "u2=1.45; #refractive index of cladding\n",
+      "\n",
+      "#calculate\n",
+      "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+      "theta=math.asin(NA); #calculation of acceptance angle\n",
+      "theta1=theta*180/3.14;\n",
+      "#result\n",
+      "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+      "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";\n",
+      "print \"   (roundoff error)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numerical aperture of the fibre is NA= 0.296\n",
+        "The acceptance angle of the optical fibre is = 17.26 degree\n",
+        "   (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__mean_free_path_of_electron_1.png b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__mean_free_path_of_electron_1.png
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diff --git a/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of_lattice_constant_1.png b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of_lattice_constant_1.png
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diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter10_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter10_1.ipynb
new file mode 100755
index 00000000..4ad21f3b
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter10_1.ipynb
@@ -0,0 +1,167 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:04df0fd624434d43d67fac6cabd770a44e2109835e3a6832f168a8a4e92f27c9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "10: Lasers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 10.1, Page number 10.6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "c = 3*10**8    #speed of light(m/sec)\n",
+      "h = 6.6*10**-34    #planck's constant\n",
+      "e = 1.6*10**-19\n",
+      "T = 300    #temperature(K)\n",
+      "K = 8.61*10**-5\n",
+      "lamda = 6943    #wavelength, angstrom\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = lamda*10**-10     #wavelength(m)\n",
+      "#let E2 - E1 be E\n",
+      "E = h*c/lamda      #energy(J)\n",
+      "E = E/e       #energy(eV)\n",
+      "#let population ratio N2/N1 be N\n",
+      "N = math.exp(-E/(K*T));\n",
+      "\n",
+      "#Result\n",
+      "print \"relative population of 2 states is\",round(N/1e-30,3),\"*10^-30\"\n",
+      "print \"answer given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "relative population of 2 states is 1.076 *10^-30\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 10.2, Page number 10.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "a2 = 6     #spot diameter(mm)\n",
+      "a1 = 4     #spot diameter(mm)\n",
+      "d2 = 2    #distance from laser(m)\n",
+      "d1 = 1    #distance from laser(m)\n",
+      "\n",
+      "#Calculation\n",
+      "a2 = a2*10**-3    #spot diameter(m)\n",
+      "a1 = a1*10**-3     #spot diameter(m)\n",
+      "theta = (a2-a1)/(2*(d2-d1))     #divergence(radian)\n",
+      "theta = theta*10**3      #divergence(milli radian)\n",
+      "\n",
+      "#Result\n",
+      "print \"divergence is\",theta,\"milli radian\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "divergence is 1.0 milli radian\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 10.3, Page number 10.46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n = 1     #for air\n",
+      "lamda = 650     #wavelength(nm)\n",
+      "bs = 1     #beam size(mm)\n",
+      "fl = 1    #focal length of lens(mm)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = lamda*10**-9    #wavelength(m)\n",
+      "bs = bs*10**-3    #beam size(m)\n",
+      "fl = fl*10**-3     #focal length of lens(m)\n",
+      "tan_theta = fl/(2*bs)     #value of tan_theta\n",
+      "theta = math.atan(tan_theta)\n",
+      "NA = n*math.sin(theta)\n",
+      "NA = math.ceil(NA*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "ss = 0.6*lamda/NA       #spot size(m)\n",
+      "ss = ss*10**6;       #spot size(micro metre)\n",
+      "ss = math.ceil(ss*10**3)/10**3;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"spot size is\",ss,\"micro metre\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "spot size is 0.867 micro metre\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter11_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter11_1.ipynb
new file mode 100755
index 00000000..04367b22
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter11_1.ipynb
@@ -0,0 +1,277 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:99fe784c923b9389b1b34d49f0a0fc6c7079e6fd55741e762bc73de0fe21384b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "11: Fibre Optics and Holography"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.1, Page number 11.6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.55    #refractive index of core\n",
+      "n2 = 1.50    #refractive index of cladding\n",
+      "\n",
+      "#Calculation\n",
+      "NA = math.sqrt(n1**2 - n2**2)\n",
+      "NA = math.ceil(NA*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"numerical aperture is\",NA"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "numerical aperture is 0.391\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.2, Page number 11.6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.563    #refractive index of core\n",
+      "n2 = 1.498;    #refractive index of cladding\n",
+      "\n",
+      "#Calculation\n",
+      "NA = math.sqrt(n1**2 - n2**2)      #numerical aperture \n",
+      "NA = math.ceil(NA*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "alpha_i = math.asin(NA)        #angle of acceptance(radians)\n",
+      "alpha_i = alpha_i*180/math.pi       #angle(degrees)\n",
+      "deg = int(alpha_i)\n",
+      "t = 60*(alpha_i-deg)\n",
+      "mint = int(t)       #angle(minutes)\n",
+      "\n",
+      "#Result\n",
+      "print \"the angle of acceptance is\",deg,\"degrees and\",mint,\"minutes\"\n",
+      "print \"answer given in the book differs due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the angle of acceptance is 26 degrees and 29 minutes\n",
+        "answer given in the book differs due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.3, Page number 11.7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "delta = 0.05     #difference in refractive indices of core and cladding\n",
+      "NA = 0.39    #numerical aperture\n",
+      "\n",
+      "#Calculation\n",
+      "n1 = NA/math.sqrt(2*delta)     #refractive index of core\n",
+      "n1 = math.ceil(n1*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"refractive index of the core is\",n1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "refractive index of the core is 1.2333\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.4, Page number 11.7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.563    #refractive index of core\n",
+      "n2 = 1.498    #refractive index of cladding\n",
+      "\n",
+      "#Calculation\n",
+      "delta = (n1-n2)/n1      #fractional index change\n",
+      "delta = math.ceil(delta*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"fractional index change is\",delta"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fractional index change is 0.0416\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.5, Page number 11.8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.48    #refractive index of core\n",
+      "n2 = 1.45    #refractive index of cladding\n",
+      "\n",
+      "#Calculation\n",
+      "NA = math.sqrt(n1**2 - n2**2)     #numerical aperture \n",
+      "NA = math.ceil(NA*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "alpha_i = math.asin(NA)      #angle of acceptance(radian)\n",
+      "alpha_i = alpha_i*180/math.pi       #angle(degrees)\n",
+      "deg = int(alpha_i)\n",
+      "t = 60*(alpha_i-deg)\n",
+      "mint = round(t)       #angle(minutes)\n",
+      "\n",
+      "#Result\n",
+      "print \"numerical aperture is\",NA\n",
+      "print \"the angle of acceptance is\",deg,\"degrees and\",mint,\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "numerical aperture is 0.2965\n",
+        "the angle of acceptance is 17 degrees and 15.0 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.6, Page number 11.15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Pin = 100          #power of signal(mW)\n",
+      "Pout = 40          #outcoming signal power(mW)\n",
+      "\n",
+      "#Calculation\n",
+      "l = -10*math.log10(Pout/Pin)       #attenuation loss(dB)\n",
+      "l = math.ceil(l*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"the attenuation loss is\",l,\"dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the attenuation loss is 3.98 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter1_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter1_1.ipynb
new file mode 100755
index 00000000..49ab34ca
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter1_1.ipynb
@@ -0,0 +1,214 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:358f4a01b294025154ed5b68aa4867e70b7293d6e35ba5318010f0ecabaea268"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "1: Bonding in Solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.1, Page number 1.4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "IE_1 = 502     #first ionisation energy(kJ/mol)\n",
+      "EA_B = -335    #electron affinity for B atom(kJ/mol)\n",
+      "e = 1.602*10**-19\n",
+      "r = 0.3      #inter ionic seperation(nm)\n",
+      "epsilon0 = 8.85*10**-12      #permittivity of free space(C/N-m)\n",
+      "\n",
+      "#Calculation\n",
+      "r = r*10**-9      #inter ionic seperation(m)\n",
+      "N = 6.022*10**23*10**-3\n",
+      "E = (-e**2*N)/(4*math.pi*epsilon0*r)      #electrostatic attraction energy(kJ/mol)\n",
+      "dE = IE_1+EA_B+E;     #net change in energy per mole\n",
+      "\n",
+      "#Result\n",
+      "print \"electrostatic attraction energy is\",int(E),\"kJ/mol\"\n",
+      "print \"net change in energy is\",int(dE),\"kJ/mol\"\n",
+      "print \"A+B- molecule will be stable\"\n",
+      "print \"answer for net change,dE given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "electrostatic attraction energy is -463 kJ/mol\n",
+        "net change in energy is -296 kJ/mol\n",
+        "A+B- molecule will be stable\n",
+        "answer for net change,dE given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.2, Page number 1.5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "IP_K = 4.1     #IP of K(eV)\n",
+      "EA_Cl = 3.6     #EA of Cl(eV)\n",
+      "e = 1.602*10**-19\n",
+      "epsilon0 = 8.85*10**-12 \n",
+      "\n",
+      "#Calculation\n",
+      "delta_E = IP_K - EA_Cl      #energy required(eV)\n",
+      "#if their total energy is 0, delta_E = Ec\n",
+      "Ec = delta_E\n",
+      "R = e/(4*math.pi*epsilon0*Ec)      #seperation between ion pair(m)\n",
+      "R = R*10**9          #seperation between ion pair(nm)\n",
+      "R = math.ceil(R*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"energy required to form K+Cl- ion pair is\",delta_E,\"eV\"\n",
+      "print \"seperation between ion pair is\",R,\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy required to form K+Cl- ion pair is 0.5 eV\n",
+        "seperation between ion pair is 2.881 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.3, Page number 1.6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "e = 1.602*10**-19\n",
+      "EA = 3.65     #electron affinity of Cl(eV)\n",
+      "IP = 5.14     #ionisation energy of Na(eV)\n",
+      "epsilon0 = 8.85*10**-12\n",
+      "r0 = 236      #equilibrium distance(pm)\n",
+      "\n",
+      "#Calculation\n",
+      "r0 = r0*10**-12      #equilibrium distance(m)\n",
+      "V = (-e**2)/(4*math.pi*epsilon0*r0)       ##potential energy(J)\n",
+      "V = V/e       #potential energy(eV)\n",
+      "Ue = V\n",
+      "BE = -Ue - IP + EA       #bond energy(eV)\n",
+      "BE = math.ceil(BE*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"bond energy for NaCl molecule is\",BE,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "bond energy for NaCl molecule is 4.614 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.4, Page number 1.19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "e = 1.602*(10**-19)\n",
+      "epsilon0 = 8.85*(10**-12)\n",
+      "r0 = 0.281     #equilibrium seperation(nm)\n",
+      "A = 1.748     #Madelung constant\n",
+      "n = 9    #born repulsive exponent\n",
+      "\n",
+      "#Calculation\n",
+      "r0 = r0*10**-9     #equilibrium seperation(m)\n",
+      "CE = (A*e**2)*(1-(1/n))/(4*math.pi*epsilon0*r0)    #cohesive energy(J)\n",
+      "CE = CE/e       #cohesive energy(eV)\n",
+      "CE = math.ceil(CE*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"cohesive energy of NaCl is\",CE,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "cohesive energy of NaCl is 7.966 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter2_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter2_1.ipynb
new file mode 100755
index 00000000..85f9fd71
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter2_1.ipynb
@@ -0,0 +1,1031 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:dc7951185c6d2c2020409b6a4f6420009478010f311e90b1a30732a0039ca5cf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      " 2: Crystal Structures and X-ray Diffraction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.1, Page number 2.16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 1;      #assume value of r as 1\n",
+      "   \n",
+      "#Calculation\n",
+      "#since 4*r = math.sqrt(3)*a\n",
+      "a = 4*r/math.sqrt(3);     \n",
+      "R = (a-(2*r))/2;     #since 2*r+2*R = a\n",
+      "R = math.ceil(R*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"radius of interstitial sphere is\",R,\"r\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radius of interstitial sphere is 0.155 r\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.2, Page number 2.17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r_BCC = 1.258;     #atomic radius(Angstrom)\n",
+      "r_FCC = 1.292;     #atomic radius(Angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "a_BCC = (4*r_BCC)/math.sqrt(3);     #In BCC(Angstrom)\n",
+      "a_BCCm = a_BCC*10**-10;      #converting a from Angstrom to m\n",
+      "V_BCC = a_BCCm**3;     #volume of unit cell(m^3)\n",
+      "n_BCC = ((1/8)*8)+1;      #number of atoms per unit cell\n",
+      "V1_BCC = V_BCC/n_BCC;     #volume occupied by 1 atom(m^3)\n",
+      "a_FCC = 2*math.sqrt(2)*r_FCC;     #In FCC(Angstrom)\n",
+      "a_FCCm = a_FCC*10**-10;      #converting a from Angstrom to m\n",
+      "V_FCC = a_FCCm**3;     #volume of unit cell(m^3)\n",
+      "n_FCC = ((1/2)*6) + ((1/8)*8);      #number of atoms per unit cell\n",
+      "V1_FCC = V_FCC/n_FCC;     #volume occupied by 1 atom(m^3)\n",
+      "delta_V = (V1_BCC - V1_FCC)*100/V1_BCC;       #change in volume in %\n",
+      "delta_V = math.ceil(delta_V*10)/10;   #rounding off to 1 decimal\n",
+      "\n",
+      "#Result\n",
+      "print \"decrease of volume during conversion from BCC to FCC is\",delta_V,\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "decrease of volume during conversion from BCC to FCC is 0.5 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.3, Page number 2.17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "a = 0.27;      #nearest neighbour distance(nm)\n",
+      "c = 0.494;     #height of unit cell(nm)\n",
+      "N = 6.023*10**26;    #avagadro number(k/mol)\n",
+      "M = 65.37;     #atomic weight(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "a_m = a*10**-9;    #nearest neighbour distance(m)\n",
+      "c_m = c*10**-9;     #height of unit cell(m)\n",
+      "V = 3*math.sqrt(3)*(a_m**2)*c_m/2;      #volume of unit cell(m^3)\n",
+      "#if rho is density then mass = V*rho\n",
+      "#V*rho = 6*M/N\n",
+      "rho = (6*M)/(N*V);      #density(kg/m^3)\n",
+      "\n",
+      "#Result\n",
+      "print \"Volume of the unit cell is\",round(V/1e-29,2),\"*10^-29 m^3\"\n",
+      "print \"density of Zinc is\",round(rho),\"kg/m^3\"\n",
+      "print \"answer for density given in the book differs due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume of the unit cell is 9.36 *10^-29 m^3\n",
+        "density of Zinc is 6960.0 kg/m^3\n",
+        "answer for density given in the book differs due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.4, Page number 2.18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 1;      #assume value of r as 1\n",
+      "\n",
+      "#Calculation\n",
+      "a = 4*r/math.sqrt(2);      #for FCC structure\n",
+      "R = (a/2)-r;     #since 2*r+2*R = a\n",
+      "R = math.ceil(R*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"maximum radius of the sphere that can fit into the void is\",R,\"r\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum radius of the sphere that can fit into the void is 0.4143 r\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.5, Page number 2.19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "a = 0.356;      #cube edge of diamond(nm)\n",
+      "aw = 12.01;      #atomic weight of carbon(kg)\n",
+      "N = 6.023*10**26;     #avagadro's number\n",
+      "\n",
+      "#Calculation\n",
+      "a_m = a*10**-9;      #cube edge of diamond(m)\n",
+      "n = 8/(a_m**3);      #number of atoms(per m^3)\n",
+      "n = n/10**29;\n",
+      "n = math.ceil(n*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "n = n*10**29;\n",
+      "M = aw/N;      #mass of 1 carbon atom(kg)\n",
+      "rho =  M*n;\n",
+      "\n",
+      "#Result\n",
+      "print \"number of atoms per m^3 is\",n\n",
+      "print \"density of diamond is\",int(rho),\"kg/m^3\"\n",
+      "print \"answer for density given in the book differs due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of atoms per m^3 is 1.774e+29\n",
+        "density of diamond is 3537 kg/m^3\n",
+        "answer for density given in the book differs due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.6, Page number 2.19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "MW = 23+35.5;      #molecular weight of NaCl(gm)\n",
+      "N = 6.023*10**23;     #avagadro number(mol-1)\n",
+      "rho = 2.18;     #density of NaCl(gm/cm^3)\n",
+      "V = 1;     #volume of unit cube(cm^3)\n",
+      "\n",
+      "#Calculation\n",
+      "M = MW/N;       #mass of NaCl molecule(gm)\n",
+      "n = rho/M;      #number of molecules per unit volume\n",
+      "n = 2*n;        #since NaCl is diatomic(atoms/cm^3)\n",
+      "#length of edge of unit cube is n*a\n",
+      "#volume V = n^3*a^3 = 1 cm^3 \n",
+      "a = (V/n)**(1/3);      #distance between two adjacent atoms(cm)\n",
+      "a = a*10**8;       #distance between two adjacent atoms(A)\n",
+      "a = math.ceil(a*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"distance between two adjacent atoms is\",a,\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance between two adjacent atoms is 2.814 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.7, Page number 2.20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "AW = 63.5;    #atomic weight of Cu(gm/mol)\n",
+      "N = 6.023*10**23;     #avagadro's number(mol-1)\n",
+      "r = 1.278;     #atomic radius(A)\n",
+      "n = 4;     #number of atoms in unit cell\n",
+      "\n",
+      "#Calculation\n",
+      "r = r*10**-8;      #atomic radius(cm)\n",
+      "M = AW/N;      #mass of each Copper atom(gm)\n",
+      "a = 4*r/math.sqrt(2);      #lattice constant(cm)\n",
+      "m = n*M;    #mass of unit cell(gm)\n",
+      "rho = m/a**3;     #density of copper crystal(gm/cm^3)\n",
+      "rho = math.ceil(rho*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"density of Copper crystal is\",rho,\"gm/cm^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "density of Copper crystal is 8.929 gm/cm^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.8, Page number 2.21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 0.1249;     #atomic radius(nm)\n",
+      "PF = 0.68;      #packing factor for BCC\n",
+      "\n",
+      "#Calculation\n",
+      "a = 4*r/math.sqrt(3);      #lattice constant(nm)\n",
+      "a_m = a*10**-9;        #lattice constant(m)\n",
+      "V = a_m**3;       #volume of unit cell(m^3)\n",
+      "FV = 1-PF;      #free volume\n",
+      "FV1 = FV*V;      #free volume per unit cell(m^3)\n",
+      "\n",
+      "#Result\n",
+      "print \"free volume per unit cell is\",round(FV1/1e-30,4),\"*10^-30 m^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "free volume per unit cell is 7.6795 *10^-30 m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.11, Page number 2.27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "#in (100) plane the total number of atoms are n\n",
+      "n = (1/4)*4;\n",
+      "\n",
+      "#Calculation\n",
+      "#in (111) plane, area = (1/2)*base*height = (1/2)*a*sqrt(2)*a*sqrt(2)*cosd(30)\n",
+      "x = math.cos(30);\n",
+      "#area = (1/2)*a*sqrt(2)*a*sqrt(2)*x = 0.866*a^2 = 0.58/a^2\n",
+      "n1 = (1/360)*60*3;\n",
+      "\n",
+      "#Result\n",
+      "print \"number of atoms in (100) plane are\",n;\n",
+      "#A = a^2. number of atoms per mm^2 is n/a^2\n",
+      "print \"number of atoms per mm^2 is 1/a^2\";\n",
+      "print \"number of atoms in (110) plane is 1\";\n",
+      "#in (110) plane, area is sqrt(2)*a*a = sqrt(2)*a^2\n",
+      "print \"unit area contains 1/(sqrt(2)*a^2) = 0.707/a^2 atoms/mm^2\";\n",
+      "print \"total number of atoms in (111) plane is\",n1\n",
+      "#number of atoms per unit area is 0.5/(0.866*a^2)\n",
+      "print \"number of atoms per unit area is 0.58/a^2 atoms/mm^2\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of atoms in (100) plane are 1.0\n",
+        "number of atoms per mm^2 is 1/a^2\n",
+        "number of atoms in (110) plane is 1\n",
+        "unit area contains 1/(sqrt(2)*a^2) = 0.707/a^2 atoms/mm^2\n",
+        "total number of atoms in (111) plane is 0.5\n",
+        "number of atoms per unit area is 0.58/a^2 atoms/mm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.12, Page number 2.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h1 = 1;\n",
+      "k1 = 1;\n",
+      "l1 = 0;     #for (110) plane\n",
+      "h2 = 2;\n",
+      "k2 = 1;\n",
+      "l2 = 2;     #for (212) plane\n",
+      "r = 0.1278;      #atomic radius(nm)\n",
+      "\n",
+      "#Calculation\n",
+      "r = r*10**-9;     #atomic radius(m) \n",
+      "x1 = math.sqrt(h1**2+k1**2+l1**2);\n",
+      "a = 4*r/x1;     #nearest neighbouring distance(m)\n",
+      "a = a*10**9;     #nearest neighbouring distance(nm)\n",
+      "d_110 = a/x1;    #interplanar spacing for (110),(nm)\n",
+      "x2 = math.sqrt(h2**2+k2**2+l2**2);\n",
+      "d_212 = a/x2;    #interplanar spacing for (212),(nm)\n",
+      "d_212 = math.ceil(d_212*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"interplanar spacing for (110) is\",d_110,\"nm\"\n",
+      "print \"interplanar spacing for (212) is\",d_212,\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "interplanar spacing for (110) is 0.2556 nm\n",
+        "interplanar spacing for (212) is 0.1205 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.13, Page number 2.29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h1 = 1;\n",
+      "k1 = 0;\n",
+      "l1 = 0;    #for (100) plane\n",
+      "h2 = 1;\n",
+      "k2 = 1;\n",
+      "l2 = 0;    #for (110) plane \n",
+      "h3 = 1;\n",
+      "k3 = 1;\n",
+      "l3 = 1;    #for (111) plane\n",
+      "\n",
+      "#Calculation\n",
+      "x1 = math.sqrt(1/(h1**2+k1**2+l1**2));      #for (100) plane\n",
+      "x2 = math.sqrt(1/(h2**2+k2**2+l2**2));      #for (110) plane \n",
+      "x3 = math.sqrt(1/(h3**2+k3**2+l3**2));      #for (111) plane\n",
+      "x2 = math.ceil(x2*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "x3 = math.ceil(x3*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"ratio of the seperation between successive lattice planes is %d:%.2f:%.2f\"%(x1,x2,x3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of the seperation between successive lattice planes is 1:0.71:0.58\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.14, Page number 2.30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Calculation\n",
+      "#plane intercepts at a,b/2,3*c\n",
+      "#therefore intercepts are (1 1/2 3)\n",
+      "x = 1;\n",
+      "y = 1/2;\n",
+      "z = 3;\n",
+      "def lcm(x, y):\n",
+      "   if y > z:\n",
+      "       greater = y\n",
+      "   else:\n",
+      "       greater = z\n",
+      "\n",
+      "   while(True):\n",
+      "       if((greater % y == 0) and (greater % z == 0)):\n",
+      "           lcm = greater\n",
+      "           break\n",
+      "       greater += 1\n",
+      "\n",
+      "   return lcm\n",
+      "\n",
+      "lcm = lcm(y,z);                \n",
+      "x_dash = (1/x)*lcm;      #miller index of x plane\n",
+      "y_dash = (1/y)*lcm;      #miller index of y plane  \n",
+      "z_dash = (1/z)*lcm;      #miller index of z plane\n",
+      "\n",
+      "#Result\n",
+      "print \"miller indices of the plane are (\",x_dash,y_dash,z_dash,\")\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "miller indices of the plane are ( 3.0 6.0 1.0 )\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.15, Page number 2.37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d = 0.282;     #lattice spacing(nm)\n",
+      "n = 1;     #first order\n",
+      "theta = (8+(35/60))*(math.pi/180);      #glancing angle(radian)\n",
+      "theeta = 90;     #maximum value possible(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "theeta = theeta*math.pi/180;       #maximum value possible(radian)   \n",
+      "d = d*10**-9;      #lattice spacing(m)\n",
+      "lamda = 2*d*math.sin(theta)/n;     #wavelength of X-rays(m)\n",
+      "lamda_nm = lamda*10**9;     #wavelength of X-rays(nm)\n",
+      "lamda_nm = math.ceil(lamda_nm*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "n1 = 2*d*math.sin(theeta)/lamda;       #maximum order of diffraction possible\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength of X-rays is\",lamda_nm,\"nm\"\n",
+      "print \"maximum order of diffraction possible is\",int(n1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of X-rays is 0.0842 nm\n",
+        "maximum order of diffraction possible is 6\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.16, Page number 2.37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda = 1.5;      #wavelength(A.U)\n",
+      "d = 1.6;      #interplanar spacing(A.U)\n",
+      "theta = 90;     #maximum glancing angle possible(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "theta = theta*math.pi/180;       #maximum glancing angle possible(radian)\n",
+      "n = 2*d*math.sin(theta)/lamda;      #maximum possible diffraction order\n",
+      "\n",
+      "#Result\n",
+      "print \"maximum possible diffraction order is\",int(n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum possible diffraction order is 2\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.17, Page number 2.38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 1;      #for (111) plane\n",
+      "theta = 30;    #glancing angle(degrees)\n",
+      "n = 1;     #first order\n",
+      "lamda = 1.5418;     #wavelength of X-rays(A)\n",
+      "\n",
+      "#Calculation\n",
+      "theta = theta*math.pi/180;      #glancing angle(radian)\n",
+      "x = math.sqrt(h**2+k**2+l**2);\n",
+      "lamda = lamda*10**-10;     #wavelength of X-rays(m)\n",
+      "d = lamda/(2*math.sin(theta));\n",
+      "a = d*x;      #interatomic spacing(m)\n",
+      "a = a*10**10;      #interatomic spacing(A)\n",
+      "a = math.ceil(a*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"the interatomic spacing is\",a,\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the interatomic spacing is 2.6705 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.18, Page number 2.38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d100 = 0.28;     #lattice constant(nm)\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 0;      #for (110) plane\n",
+      "n = 2;      #second order\n",
+      "lamda = 0.071;     #wavelength of X-rays(nm)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_m = lamda*10**-9;     #wavelength of X-rays(m)\n",
+      "d110 = d100/math.sqrt(h**2+k**2+l**2);     #interatomic spacing(nm)\n",
+      "d110 = d110*10**-9;      #interatomic spacing(m)\n",
+      "theta = math.asin(n*lamda_m/(2*d110));       #glancing angle(radian)\n",
+      "theta = theta*180/math.pi;        #glancing angle(degrees)\n",
+      "\n",
+      "#Result\n",
+      "print \"the glancing angle is\",int(theta),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the glancing angle is 21 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.19, Page number 2.39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "a = 0.38;      #lattice constant(nm)\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 0;      #for (110) plane\n",
+      "\n",
+      "#Calculation\n",
+      "d = a/math.sqrt(h**2+k**2+l**2);\n",
+      "d = math.ceil(d*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"the distance between (110) planes is\",d,\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the distance between (110) planes is 0.27 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.20, Page number 2.39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "#area of (110) plane is a*sqrt(2)*a = sqrt(2)*a^2\n",
+      "theta = 30;      #glancing angle(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "theta = theta*math.pi/180;      #glancing angle(radian)\n",
+      "n = (1/4)*4;     #number of atoms\n",
+      "x = math.cos(theta);\n",
+      "#area of (111) plane is (a/sqrt(2))*x*a*sqrt(2)\n",
+      "#hence area is (sqrt(3)/2)*a^2\n",
+      "n1 = 3*(1/6);     #number of atoms\n",
+      "\n",
+      "#Result\n",
+      "print \"area of (110) plane contains\",n,\"atom\"\n",
+      "print \"density of lattice points is 1/(sqrt(2)*a^2)\"\n",
+      "print \"area of (111) plane contains\",n1,\"atom\"\n",
+      "#density of lattice points is (1/2)/(sqrt(3)*a^2/2)\n",
+      "print \"density of lattice points is 1/(sqrt(3)*a^2)\"\n",
+      "#density of lattice points (111) plane : (110) plane is 1/(sqrt(3)*a^2) : 1/(sqrt(2)*a^2) = sqrt(2):sqrt(3)\n",
+      "print \"density of lattice points (111) plane : (110) plane is sqrt(2):sqrt(3)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "area of (110) plane contains 1.0 atom\n",
+        "density of lattice points is 1/(sqrt(2)*a^2)\n",
+        "area of (111) plane contains 0.5 atom\n",
+        "density of lattice points is 1/(sqrt(3)*a^2)\n",
+        "density of lattice points (111) plane : (110) plane is sqrt(2):sqrt(3)\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.21, Page number 2.40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n = 2;     #second order\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 0;   #plane (110)\n",
+      "lamda = 0.065;     #wavelength of X-rays(nm)\n",
+      "a = 0.26;      #axial length(nm)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_m = lamda*10**-9;    #wavelength of X-rays(m)\n",
+      "a_m = a*10**-9;    #axial length(m)\n",
+      "x = math.sqrt(h**2+k**2+l**2);\n",
+      "theta = math.asin(n*lamda_m*x/(2*a_m));       #glancing angle(radian)\n",
+      "theta = theta*180/math.pi;          #glancing angle(degrees)\n",
+      "deg = int(theta);       #glancing angle(degrees)\n",
+      "t = 60*(theta-deg);\n",
+      "mint = int(t);         #glancing angle(minutes)\n",
+      "\n",
+      "#Result\n",
+      "print \"the glancing angle is\",deg,\"degrees\",mint,\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the glancing angle is 20 degrees 42 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.22, Page number 2.41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n = 1;    #first order\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 1;     #for (111) plane\n",
+      "lamda = 1.54;      #wavelength(A.U)\n",
+      "theta = 19.2;      #Bragg's angle(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_m = lamda*10**-10;      #wavelength(m)\n",
+      "theta = theta*math.pi/180;        #Bragg's angle(radian)\n",
+      "d = n*lamda_m/(2*math.sin(theta));     #interplanar spacing(m)\n",
+      "a = d*math.sqrt(h**2+k**2+l**2);      #cube edge(m)\n",
+      "a_AU = a*10**10;      #cube edge(A.U)\n",
+      "a_AU = math.ceil(a_AU*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"the cube edge is\",a_AU,\"A.U\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the cube edge is 4.0554 A.U\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.23, Page number 2.41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n = 1;    #first order\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 1;     #for (111) plane\n",
+      "lamda = 1.54;    #wavelength(A.U)\n",
+      "theta = 19.2;      #Bragg's angle(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_m = lamda*10**-10;      #wavelength(m)\n",
+      "theta = theta*math.pi/180;        #Bragg's angle(radian)\n",
+      "d = n*lamda_m/(2*math.sin(theta));     #interplanar spacing(m)\n",
+      "a = d*math.sqrt(h**2+k**2+l**2);       #cube edge(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"the cube edge is\",round(a/1e-10,3),\"*10^-10 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the cube edge is 4.055 *10^-10 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter3_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter3_1.ipynb
new file mode 100755
index 00000000..afe7637e
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter3_1.ipynb
@@ -0,0 +1,586 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:50d61ee8fa972cdf457beff8302930b51b8d1018696b3dbcc148db2d3f471c34"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "3: Principles of Quantum Mechanics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.1, Page number 3.12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "c = 3*10**8     #velocity of light(m/sec)\n",
+      "m = 1.67*10**-27     #mass of proton(kg)\n",
+      "h = 6.626*10**-34    #planck's constant\n",
+      "\n",
+      "#Calculation\n",
+      "v = (1/10)*c      #velocity of proton(m/sec)\n",
+      "lamda = h/(m*v)     #de Broglie wavelength(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"de Broglie wavelength of proton is\",round(lamda/1e-14,3),\"*10^-14 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength of proton is 1.323 *10^-14 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.2, Page number 3.12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V = 400      #potential(V)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = 12.26/math.sqrt(V)      #de Broglie wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"de Broglie wavelength of electron is\",lamda,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength of electron is 0.613 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.3, Page number 3.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 1.674*10**-27     #mass of neutron(kg)\n",
+      "h = 6.626*10**-34     #planck's constant\n",
+      "e = 1.6*10**-19\n",
+      "KE = 0.025     #kinetic energy(eV)\n",
+      "\n",
+      "#Calculation\n",
+      "E = KE*e     #kinetic energy(J)\n",
+      "lamda = h/math.sqrt(2*m*E)      #de Broglie wavelength(m)\n",
+      "lamda_nm = lamda*10**9      #de Broglie wavelength(nm)\n",
+      "lamda_nm = math.ceil(lamda_nm*10**4)/10**4   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"de Broglie wavelength is\",lamda_nm,\"nm\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength is 0.1811 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.4, Page number 3.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V = 1600      #potential(V)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = 12.26/math.sqrt(V)      #de Broglie wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"de Broglie wavelength of electron is\",lamda,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de Broglie wavelength of electron is 0.3065 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.5, Page number 3.18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "delta_x = 0.2      #electron distance(angstrom)\n",
+      "h = 6.626*10**-34    #planck's constant\n",
+      "\n",
+      "#Calculation\n",
+      "delta_x = delta_x*10**-10      #electron distance(m)\n",
+      "delta_p = h/(2*math.pi*delta_x)      #uncertainity in momentum(kg.m/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"uncertainity in momentum is\",round(delta_p/1e-24,3),\"*10^-24 kg m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " uncertainity in momentum is 5.273 *10^-24 kg m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.6, Page number 3.27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1\n",
+      "n2 = 1\n",
+      "n3 = 1      #for lowest energy\n",
+      "e = 1.6*10**-19\n",
+      "h = 6.62*10**-34      #planck's constant\n",
+      "m = 9.1*10**-31      #mass of electron(kg)\n",
+      "L = 0.1     #side of box(nm)\n",
+      "\n",
+      "#Calculation\n",
+      "L = L*10**-9    #side of box(m)\n",
+      "E1 = h**2*(n1**2+n2**2+n3**2)/(8*m*L**2)      #lowest energy(J)\n",
+      "E1 = E1/e      #lowest energy(eV)\n",
+      "E1 = math.ceil(E1*10)/10    #rounding off to 1 decimal\n",
+      "\n",
+      "#Result\n",
+      "print \"lowest energy of electron is\",E1,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lowest energy of electron is 112.9 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.7, Page number 3.27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1\n",
+      "n2 = 1\n",
+      "n3 = 2      #for level next to the lowest\n",
+      "e = 1.6*10**-19\n",
+      "h = 6.62*10**-34      #planck's constant\n",
+      "m = 9.1*10**-31       #mass of electron(kg)\n",
+      "L = 0.1     #side of box(nm)\n",
+      "\n",
+      "#Calculation\n",
+      "L = L*10**-9    #side of box(m)\n",
+      "E1 = h**2*(n1**2+n2**2+n3**2)/(8*m*L**2)      #lowest energy(J)\n",
+      "E1 = E1/e      #lowest energy(eV)\n",
+      "E1 = math.ceil(E1*10**2)/10**2    #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"energy of electron is\",E1,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy of electron is 225.75 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.8, Page number 3.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 9.1*10**-31     #mass of electron(kg)\n",
+      "h = 6.626*10**-34    #planck's constant\n",
+      "e = 1.6*10**-19\n",
+      "E = 2000     #energy(eV)\n",
+      "\n",
+      "#Calculation\n",
+      "E = E*e      #energy(J)\n",
+      "lamda = h/math.sqrt(2*m*E)      #wavelength(m)\n",
+      "lamda_nm = lamda*10**9      #velength(nm)\n",
+      "lamda_nm = math.ceil(lamda_nm*10**4)/10**4    #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength is\",lamda_nm,\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength is 0.0275 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.9, Page number 3.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n = 1     #for minimum energy\n",
+      "h = 6.626*10**-34      #planck's constant(J sec)\n",
+      "m = 9.91*10**-31      #mass of electron(kg)\n",
+      "L = 4*10**-10     #side of box(m)\n",
+      "\n",
+      "#Calculation\n",
+      "E1 = ((h**2)*(n**2))/(8*m*(L**2))      #lowest energy(J)\n",
+      "E1 = E1*10**18;\n",
+      "E1 = math.ceil(E1*10**4)/10**4    #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"energy of electron is\",E1,\"*10**-18 J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy of electron is 0.3462 *10**-18 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.10, Page number 3.29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1     #for ground state\n",
+      "n2 = 2     #for 1st excited state\n",
+      "n3 = 3     #for 2nd excited state\n",
+      "h = 6.626*10**-34      #planck's constant(J sec)\n",
+      "m = 9.1*10**-31      #mass of electron(kg)\n",
+      "L = 1*10**-10       #width(m)\n",
+      "\n",
+      "#Calculation\n",
+      "E1 = h**2*n1**2/(8*m*L**2)      #energy in ground state(J)\n",
+      "E2 = n2**2*E1       #energy in 1st excited state(J)\n",
+      "E3 = n3**2*E1       #energy in 2nd excited state(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"energy in ground state is\",round(E1/1e-18,3),\"*10^-18 J\"\n",
+      "print \"energy in 1st excited state is\",round(E2/1e-17,3),\"*10^-17 J\"\n",
+      "print \"energy in 2nd excited state is\",round(E3/1e-17,3),\"*10^-17 J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy in ground state is 6.031 *10^-18 J\n",
+        "energy in 1st excited state is 2.412 *10^-17 J\n",
+        "energy in 2nd excited state is 5.428 *10^-17 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.11, Page number 3.30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h = 6.626*10**-34      #planck's constant(J sec)\n",
+      "m = 9.1*10**-31      #mass of electron(kg)\n",
+      "e = 1.6*10**-19\n",
+      "lamda = 1.66*10**-10     #wavelength(m)\n",
+      "\n",
+      "#Calculation\n",
+      "v = h/(m*lamda)      #velocity of electron(m/sec)\n",
+      "v_km = v*10**-3      #velocity of electron(km/sec)\n",
+      "KE = (1/2)*m*v**2    #kinetic energy(J)\n",
+      "KE_eV = KE/e        #kinetic energy(eV)\n",
+      "KE_eV = math.ceil(KE_eV*10**3)/10**3    #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity of electron is\",int(v_km),\"km/sec\"\n",
+      "print \"kinetic energy of electron is\",KE_eV,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity of electron is 4386 km/sec\n",
+        "kinetic energy of electron is 54.714 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.12, Page number 3.31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V = 15      #potential(kV)\n",
+      "\n",
+      "#Calculation\n",
+      "V = V*10**3     #potential(V)\n",
+      "lamda = 12.26/math.sqrt(V)      #de Broglie wavelength(angstrom)\n",
+      "lamda = math.ceil(lamda*100)/100    #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength of electron waves is\",lamda,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of electron waves is 0.11 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.13, Page number 3.31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V = 344      #potential(V)\n",
+      "n = 1     #for 1st reflection maximum\n",
+      "theta = 60     #glancing angle(degrees)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = 12.26/math.sqrt(V)      #de Broglie wavelength(angstrom)\n",
+      "lamda_m = lamda*10**-10     #de Broglie wavelength(m)\n",
+      "theta = theta*math.pi/180     ##glancing angle(radians)\n",
+      "d = n*lamda_m/(2*math.sin(theta))      #interatomic spacing(m)\n",
+      "d = d*10**10       #interatomic spacing(angstrom)\n",
+      "d = math.ceil(d*10**5)/10**5    #rounding off to 5 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"interatomic spacing of crystal is\",d,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "interatomic spacing of crystal is 0.38164 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter4_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter4_1.ipynb
new file mode 100755
index 00000000..d55516d5
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter4_1.ipynb
@@ -0,0 +1,390 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7ccac901d3d7a8c9dde630e169f5e7e52a039a9eb5d4b37586cb97d5a7fe1fca"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "4: Electron Theory of Metals"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.1, Page number 4.5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "rho_s = 10.5*10**3    #density of silver(kg/m^3)\n",
+      "NA = 6.02*10**26     #avagadro number(per k-mol)\n",
+      "MA = 107.9     #atomic weight of silver\n",
+      "sigma = 6.8*10**7;     #conductivity of silver(ohm-1 m-1)\n",
+      "e = 1.6*10**-19\n",
+      "\n",
+      "#Calculation\n",
+      "n = rho_s*NA/MA      #molar volume of silver\n",
+      "mew = sigma/(n*e)     #mobility of electrons(m^2/Vs)\n",
+      "mew = mew*10**2\n",
+      "mew = math.ceil(mew*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"density of electrons in silver is\",round(n/1e28,2),\"*10^28\"\n",
+      "print \"mobility of electrons is\",mew,\"*10**-2 m**2/Vs\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "density of electrons in silver is 5.86 *10^28\n",
+        "mobility of electrons is 0.7255 *10**-2 m**2/Vs\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.2, Page number 4.6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d = 8.92*10**3      #density(kg/m^3)\n",
+      "e = 1.6*10**-19\n",
+      "m = 9.1*10**-31      #mass of electron(kg)\n",
+      "N = 6.02*10**26      #avagadro's number(per k-mol)\n",
+      "AW = 63.5     #atomic weight\n",
+      "rho = 1.73*10**-8     #resistivity of copper, ohm-m\n",
+      "\n",
+      "#Calculation\n",
+      "n = d*N/AW       #number of cu atoms(per m^3)\n",
+      "mew = 1/(rho*n*e)      #mobility of electrons(m/Vs)\n",
+      "mew = mew*10**2\n",
+      "mew = math.ceil(mew*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "tow = m/(n*e**2*rho)    #relaxation time(s)\n",
+      "\n",
+      "#Result\n",
+      "print \"mobility of electrons is\",round(mew,3),\"*10**-2 m/Vs\"\n",
+      "print \"relaxation time is\",round(tow/1e-14,3),\"*10^-14 sec\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mobility of electrons is 0.427 *10**-2 m/Vs\n",
+        "relaxation time is 2.43 *10^-14 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.3, Page number 4.7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "rho = 1.54*10**-8      #resistivity(ohm-m)\n",
+      "n = 5.8*10**28      #conduction electrons(per m^3)\n",
+      "m = 9.108*10**-31    #mass of electron(kg)\n",
+      "e = 1.602*10**-19\n",
+      "\n",
+      "#Calculation\n",
+      "tow = m/(n*(e**2)*rho)    #relaxation time(sec)\n",
+      "\n",
+      "#Result\n",
+      "print \"relaxation time of conduction electrons is\",round(tow/1e-14,2),\"*10^-14 sec\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "relaxation time of conduction electrons is 3.97 *10^-14 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.4, Page number 4.8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "R = 0.06      #resistance(ohm)\n",
+      "D = 5      #length of Al wire(m)\n",
+      "e = 1.602*10**-19\n",
+      "rho = 2.7*10**-8     #resistivity of Al(ohm-m)\n",
+      "MA = 26.98     #atomic weight\n",
+      "NA = 6.025*10**26     #avagadro number(k/mol)\n",
+      "rho_s = 2.7*10**3     #density(kg/m^3)\n",
+      "I = 15     #current(A)\n",
+      "\n",
+      "#Calculation\n",
+      "n = 3*rho_s*NA/MA     #free electron concentration(electrons/m^3)\n",
+      "mew = 1/(n*e*rho)     #mobility(m/Vs)\n",
+      "E = I*R/D      #electric field(V/m)\n",
+      "vd = mew*E     #drift velocity(m/s)\n",
+      "vd = vd*10**3\n",
+      "mew = mew*10**3\n",
+      "mew = math.ceil(mew*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "vd = math.ceil(vd*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"free electron concentration is\",round(n/1e29,4),\"*10^29 electrons/m^2\"\n",
+      "print \"mobility is\",round(mew,3),\"*10^-3 m/Vs\"\n",
+      "print \"drift velocity is\",round(vd,2),\"*10^-3 m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "free electron concentration is 1.8088 *10^29 electrons/m^2\n",
+        "mobility is 1.278 *10^-3 m/Vs\n",
+        "drift velocity is 0.23 *10^-3 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.5, Page number 4.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1\n",
+      "n2 = 1\n",
+      "n3 = 1      #for lowest energy\n",
+      "h = 6.62*10**-34     #planck's constant(Js)\n",
+      "e = 1.6*10**-19\n",
+      "m = 9.1*10**-31      #mass of electron(kg)\n",
+      "L = 0.1     #side of box(nm)\n",
+      "\n",
+      "#Calculation\n",
+      "L = L*10**-9    #side of box(m)\n",
+      "E1 = (h**2)*(n1**2+n2**2+n3**2)/(8*m*L**2)      #lowest energy(J)\n",
+      "E1 = E1/e        #lowest energy(eV)\n",
+      "E1 = math.ceil(E1*10)/10   #rounding off to 1 decimal\n",
+      "\n",
+      "#Result\n",
+      "print \"lowest energy of electron is\",E1,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lowest energy of electron is 112.9 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.6, Page number 4.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "\n",
+      "#Calculation\n",
+      "#Fermi function F(E) = 1/(1+exp((E-Ef)/(kT)))\n",
+      "#given E-Ef = kT. therefore F(E) = 1/(1+exp(1))\n",
+      "F_E = 1/(1+math.exp(1))\n",
+      "F_E = math.ceil(F_E*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"fermi function is\",F_E"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fermi function is 0.269\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.7, Page number 4.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "F_E = 10      #probability in percent\n",
+      "k = 1.38*10**-23\n",
+      "EF = 5.5      #fermi energy(eV)\n",
+      "\n",
+      "#Calculation\n",
+      "E = EF+(EF/100)     #energy(eV)\n",
+      "X = E-EF     #E-EF(eV)\n",
+      "X = X*e      #E-EF(J)\n",
+      "T = X/(k*math.log(F_E-1))      #temperature(K)\n",
+      "T = math.ceil(T*10**2)/10**2   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"temperature is\",round(T,1),\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature is 290.2 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.8, Page number 4.17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "F_E = 0.01      #probability in percent\n",
+      "k = 1.38*10**-23\n",
+      "e = 1.6*10**-19\n",
+      "#let E-EF be X\n",
+      "X = 0.5      #E-EF(eV)\n",
+      "\n",
+      "#Calculation\n",
+      "kT = X/(2.303*math.log10((1-F_E)*100))    #value of kT(eV)\n",
+      "T = kT*e/k        #temperature(K)\n",
+      "T = math.ceil(T*10)/10   #rounding off to 1 decimal\n",
+      "\n",
+      "#Result\n",
+      "print \"temperature is\",T,\"K\"\n",
+      "print \"answer given in the book is wrong by a decimal point\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature is 1261.4 K\n",
+        "answer given in the book is wrong by a decimal point\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter6_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter6_1.ipynb
new file mode 100755
index 00000000..a6f7f46a
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter6_1.ipynb
@@ -0,0 +1,429 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d8c62a33cdb998cb212b402408ceb0e19f54a33922b51cf1b80232d50d8d2341"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "6: Dielectric Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.1, Page number 6.23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "C = 2     #capacitance(micro-farad)\n",
+      "V = 1000    #voltage applied(V)\n",
+      "epsilon_r = 100     #permitivity\n",
+      "\n",
+      "#Calculation\n",
+      "C = C*10**-6      #capacitance(farad)\n",
+      "W = (C*V**2)/2      #energy stored in capacitor(J)\n",
+      "C0 = C/epsilon_r      #capacitance removing the dielectric\n",
+      "W0 = C0*(V**2)/2      #energy stored without dielectric(J)\n",
+      "E = 1-W0       #energy stored in dielectric(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"energy stored in capacitor is\",W,\"J\"\n",
+      "print \"energy stored in the dielectric is\",E,\"J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy stored in capacitor is 1.0 J\n",
+        "energy stored in the dielectric is 0.99 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.2, Page number 6.24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "epsilon_r = 4.94\n",
+      "n_2 = 2.69     #square of index of refraction\n",
+      "alpha_i = 0    #at optical frequencies\n",
+      "\n",
+      "#Calculation\n",
+      "#(epsilon_r-1)/(epsilon_r+2) = N*(alpha_e+alpha_i)/(3*epsilon0)\n",
+      "X = (epsilon_r-1)/(epsilon_r+2)\n",
+      "#epsilon_r = n**2. therefore (n**2-1)/(n**2+2) = N*alpha_e/(3*epsilon0)\n",
+      "Y = (n_2-1)/(n_2+2)\n",
+      "#N*(alpha_e+alpha_i)/N*alpha_e = X/Y\n",
+      "#let alpha = alpha_i/alpha_e\n",
+      "alphai_e = (X/Y)-1     #ratio between electronic ionic and electronic polarizability\n",
+      "alphai_e = math.ceil(alphai_e*10**4)/10**4   #rounding off to 4 decimals\n",
+      "alphae_i = 1/alphai_e      #ratio between electronic and ionic polarizability\n",
+      "alphae_i = math.ceil(alphae_i*10**3)/10**3   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"ratio between electronic ionic and electronic polarizability is\",alphai_e\n",
+      "print \"ratio between electronic and ionic polarizability is\",alphae_i"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio between electronic ionic and electronic polarizability is 0.5756\n",
+        "ratio between electronic and ionic polarizability is 1.738\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.3, Page number 6.25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "epsilon_r = 2.56\n",
+      "tan_delta = 0.7*10**-4\n",
+      "f = 1     #frequency(MHz)\n",
+      "A = 8     #area(cm**2)\n",
+      "d = 0.08     #diameter(mm)\n",
+      "epsilon0 = 8.85*10**-12\n",
+      "\n",
+      "#Calculation\n",
+      "A = A*10**-4      #area(m**2)\n",
+      "d = d*10**-3     #diameter(m)\n",
+      "epsilon_rdash = epsilon_r*tan_delta\n",
+      "omega = 2*math.pi*10**6\n",
+      "Rp = d/(omega*epsilon0*epsilon_rdash*A)      #parallel loss resistance(ohm)\n",
+      "Rp = Rp*10**-6      #parallel loss resistance(Mega ohm)\n",
+      "Rp = math.ceil(Rp*10**3)/10**3   #rounding off to 3 decimals\n",
+      "Cp = A*epsilon0*epsilon_r/d     #capacitance(farad)\n",
+      "\n",
+      "#Result\n",
+      "print \"parallel loss resistance is\",Rp,\"ohm\"\n",
+      "print \"capacitance in Farad is\",Cp,\"Farad\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "parallel loss resistance is 10.036 ohm\n",
+        "capacitance in Farad is 2.2656e-10 Farad\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.4, Page number 6.26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "N = 3*10**28     #density(atoms/m**3)\n",
+      "alpha_e = 10**-40    #electronic polarizability(Farad-m**2)\n",
+      "epsilon0 = 8.854*10**-12\n",
+      "\n",
+      "#Calculation\n",
+      "epsilon_r = 1+(N*alpha_e/epsilon0)       #dielectric constant of material\n",
+      "epsilon_r = math.ceil(epsilon_r*10**3)/10**3   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"dielectric constant of material is\",epsilon_r "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "dielectric constant of material is 1.339\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.5, Page number 6.27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "epsilon0 = 8.854*10**-12\n",
+      "epsilon_r = 1.0000684      #dielectric constant\n",
+      "N = 2.7*10**25     #density(atoms/m**3)\n",
+      "\n",
+      "#Calculation\n",
+      "alpha_e = epsilon0*(epsilon_r-1)/N        #electronic polarizability(Fm**2)\n",
+      "\n",
+      "#Result\n",
+      "print \"electronic polarizability is\",round(alpha_e/1e-41,3),\"*10^-41 Fm**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "electronic polarizability is 2.243 *10^-41 Fm**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.6, Page number 6.27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "epsilon0 = 8.85*10**-12\n",
+      "V = 100     #potential(V)\n",
+      "A = 100     #area(cm**2)\n",
+      "d = 1    #plate seperation(cm)\n",
+      "\n",
+      "#Calculation\n",
+      "A = A*10**-4      #area(m**2)\n",
+      "d = d*10**-2    #plate seperation(m)\n",
+      "C = epsilon0*A/d     #capacitance(farad)\n",
+      "Q = C*V     #charge on plates\n",
+      "\n",
+      "#Result\n",
+      "print \"capacitance of capacitor is\",C,\"F\"\n",
+      "print \"charge on plates in coulomb is\",Q,\"coulomb\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "capacitance of capacitor is 8.85e-12 F\n",
+        "charge on plates in coulomb is 8.85e-10 coulomb\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.7, Page number 6.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "N = 6.02*10**26     #avagadro number\n",
+      "d = 2050      #density(kg/m**3)\n",
+      "AW = 32       #atomic weight of sulphur\n",
+      "epsilon_r = 3.75    #relative dielectric constant\n",
+      "epsilon0 = 8.55*10**-12\n",
+      "\n",
+      "#Calculation\n",
+      "n = N*d/AW    #number of atoms(per m**3)\n",
+      "alpha_e = ((epsilon_r-1)/(epsilon_r+2))*3*epsilon0/n     #electronic polarizability(Fm**2) \n",
+      "\n",
+      "#Result\n",
+      "print \"electronic polarizability is\",round(alpha_e/1e-40,3),\"*10^-40 Fm**2\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "electronic polarizability is 3.181 *10^-40 Fm**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.8, Page number 6.29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Q = 2*10**-10       #charge(coulomb)\n",
+      "d = 4     #plate seperation(mm)\n",
+      "epsilon_r = 3.5    #dielectric constant\n",
+      "A = 650     #area(mm**2)\n",
+      "epsilon0 = 8.85*10**-12\n",
+      "\n",
+      "#Calculation\n",
+      "d = d*10**-3      #plate seperation(m)\n",
+      "A = A*10**-6     #area(m**2)\n",
+      "V = Q*d/(epsilon0*epsilon_r*A)       #voltage across capacitors(V)\n",
+      "V = math.ceil(V*10**3)/10**3   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"resultant voltage across capacitors is\",V,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resultant voltage across capacitors is 39.735 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 6.9, Page number 6.30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V = 10      #potential(V)\n",
+      "d = 2*10**-3      #plate seperation(m)\n",
+      "epsilon_r = 6\n",
+      "epsilon0 = 8.85*10**-12\n",
+      "\n",
+      "#Calculation\n",
+      "E = V/d     #electric field(V/m)\n",
+      "D = epsilon0*epsilon_r*E       #dielectric displacement(C/m**2)\n",
+      "\n",
+      "#Result\n",
+      "print \"dielectric displacement is\",D,\"Cm^-2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "dielectric displacement is 2.655e-07 Cm^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter7_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter7_1.ipynb
new file mode 100755
index 00000000..e3eb080f
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter7_1.ipynb
@@ -0,0 +1,465 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1ba9d8e44f464fdd4be85263413cd33050741f89efba47984e88555948ca8361"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "7: Magnetic Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.1, Page number 7.3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "M = 1.4      #field(T)\n",
+      "H = 6.5*10**-4      #magnetic field(T)\n",
+      "\n",
+      "#Calculation\n",
+      "chi = M/H      #susceptibility\n",
+      "mew_r = 1+chi       #relative permeability\n",
+      "\n",
+      "#Result\n",
+      "print \"relative permeability of iron is\",round(mew_r)\n",
+      "print \"answer given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "relative permeability of iron is 2155.0\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.2, Page number 7.3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "M = 3300     #magnetisation(amp/m)\n",
+      "H = 220      #field strength(amp/m)\n",
+      "\n",
+      "#Calculation\n",
+      "mew_r = (M/H)+1      #relative permeability\n",
+      "\n",
+      "#Result\n",
+      "print \"relative permeability of material is\",mew_r"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "relative permeability of material is 16.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.3, Page number 7.4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "H = 10**6      #magnetic field intensity(amp/m)\n",
+      "chi = 1.5*10**-3     #susceptibility\n",
+      "\n",
+      "#Calculation\n",
+      "mew0 = 4*math.pi*10**-7\n",
+      "M = chi*H      #magnetisation(A/m)\n",
+      "B = mew0*(M+H)      #flux density(T)\n",
+      "M = M*10**-3\n",
+      "B = math.ceil(B*10**3)/10**3   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"magnetisation of material is\",M,\"*10**3 A/m\"\n",
+      "print \"flux density is\",B,\"T\"\n",
+      "print \"answer for flux density B given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetisation of material is 1.5 *10**3 A/m\n",
+        "flux density is 1.259 T\n",
+        "answer for flux density B given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.4, Page number 7.5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "H = 10**4      #magnetic field intensity(amp/m)\n",
+      "chi = 3.7*10**-3     #susceptibility\n",
+      "\n",
+      "#Calculation\n",
+      "mew0 = 4*math.pi*10**-7\n",
+      "M = chi*H      #magnetisation(A/m)\n",
+      "B = mew0*(M+H)      #flux density(Wb/m**2)\n",
+      "B = math.ceil(B*10**5)/10**5   #rounding off to 5 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"magnetisation of material is\",M,\"A/m\"\n",
+      "print \"flux density is\",B,\"Wb/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetisation of material is 37.0 A/m\n",
+        "flux density is 0.01262 Wb/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.5, Page number 7.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "I = 500     #current(mA)\n",
+      "d = 10     #diameter(cm)\n",
+      "\n",
+      "#Calculation\n",
+      "I = I*10**-3     #current(A)\n",
+      "r = d/2    #radius(cm)\n",
+      "r = r*10**-2    #radius(m)\n",
+      "A = 2*math.pi*r**2   #area(m**2)\n",
+      "mew_m = I*A     #magnetic moment(Am**2)\n",
+      "mew_m = mew_m*10**3\n",
+      "mew_m = math.ceil(mew_m*10**3)/10**3   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"magnetic moment associated with the loop is\",mew_m,\"*10**-3 Am**2\"\n",
+      "print \"answer given in the book is wrong in the 3rd decimal\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetic moment associated with the loop is 7.854 *10**-3 Am**2\n",
+        "answer given in the book is wrong in the 3rd decimal\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.6, Page number 7.19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r = 5.29*10**-11    #radius of orbit(m)\n",
+      "B = 2     #field applied(T)\n",
+      "e = 1.602*10**-19     #charge of electron(coulomb)\n",
+      "m = 9.108*10**-31    #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "mew_ind = e**2*r**2*B/(4*m)      #change in magnetic moment(Am^2)\n",
+      "\n",
+      "#Result\n",
+      "print \"change in magnetic moment is\",round(mew_ind/1e-29,3),\"*10^-29 Am**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in magnetic moment is 3.943 *10^-29 Am**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.7, Page number 7.22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "chi_1 = 2.8*10**-4     #susceptibility\n",
+      "T1 = 350      #temperature(K)\n",
+      "T2 = 300      #temperature(K)\n",
+      "\n",
+      "#Calculation\n",
+      "#chi = C/T where C is curie constant\n",
+      "chi_2 = chi_1*T1/T2     #susceptibility at 300 K\n",
+      "chi_2 = chi_2*10**4\n",
+      "chi_2 = math.ceil(chi_2*10**3)/10**3   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"susceptibility at 300 K is\",chi_2,\"*10**-4\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "susceptibility at 300 K is 3.267 *10**-4\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.8, Page number 7.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d = 8906     #density(kg/m**3)\n",
+      "n = 6.025*10**26     #avagadro number\n",
+      "AW = 58.7     #atomic weight\n",
+      "Bs = 0.65     #magnetic induction(Wb/m**2)\n",
+      "mewB = 9.27*10**-24 \n",
+      "\n",
+      "#Calculation\n",
+      "N = d*n/AW     #number of atoms(per m**3)\n",
+      "mew0 = 4*math.pi*10**-7\n",
+      "mew_m = Bs/(N*mew0)      #magnetic moment(Am**2)\n",
+      "mew_m = mew_m/mewB       #magnetic moment(mewB)\n",
+      "mew_m = math.ceil(mew_m*10**3)/10**3   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"the magnetic moment of Ni is\",mew_m,\"mewB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the magnetic moment of Ni is 0.611 mewB\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.9, Page number 7.29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "H = 2     #magnetic field(Wb/m**2)\n",
+      "mew = 9.4*10**-24\n",
+      "k = 1.38*10**-23\n",
+      "\n",
+      "#Calculation\n",
+      "#np = C*n0*math.exp(mew*H/(k*T))\n",
+      "#na = C*n0*math.exp(-mew*H/(k*T))\n",
+      "#np/na = exp(mew*H/(k*T))/exp(-mew*H/(k*T)) = exp(2*mew*H/(k*T))\n",
+      "#given np/na = 2. therefore exp(2*mew*H/(k*T)) = 2\n",
+      "T = 2*mew*H/(k*math.log(2))    #temperature(K)\n",
+      "T = math.ceil(T*10**2)/10**2   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"temperature is\",T,\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature is 3.94 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.10, Page number 7.30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "AW = 157.26      #atomic weight\n",
+      "d = 7.8*10**3     #density(kg/m**3)\n",
+      "A = 6.025*10**26     #avagadro number\n",
+      "mew0 = 4*math.pi*10**-7\n",
+      "N = d*A/AW      #number of atoms 1 kg contains\n",
+      "g = N/10**3      #number of atoms 1 g contains\n",
+      "mew_B = 7.1      #bohr magneton\n",
+      "mew_m = 9.27*10**-24\n",
+      "mew_mg = g*mew_B*mew_m     #magnetic moment per gram(Am**2)\n",
+      "mew_mg = math.ceil(mew_mg*10**3)/10**3   #rounding off to 3 decimals\n",
+      "print \"magnetic moment per gram is\",mew_mg,\"Am**2\"\n",
+      "Bs = N*mew0*mew_m      #saturation magnetisation(Wb/m**2)\n",
+      "Bs = math.ceil(Bs*10**4)/10**4   #rounding off to 4 decimals\n",
+      "print \"saturation magnetisation is\",Bs,\"Wb/m**2\"\n",
+      "print \"answers given in the book are wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetic moment per gram is 1966.852 Am**2\n",
+        "saturation magnetisation is 0.3482 Wb/m**2\n",
+        "answers given in the book are wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter8_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter8_1.ipynb
new file mode 100755
index 00000000..403a2566
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter8_1.ipynb
@@ -0,0 +1,931 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:56ac13dd16475ea277515219c6087df27ecfe535e7f6298ab44a9065695ca2d8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      " 8: Semiconductors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.1, Page number 8.11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "ni = 2.37*10**19;      #intrinsic carrier density(per m^3)\n",
+      "mew_e = 0.38;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.18;      #hole mobility(m^2/Vs)\n",
+      "e = 1.6*10**-19;\n",
+      "\n",
+      "#Calculation\n",
+      "sigma_i = ni*e*(mew_e+mew_h);      #conductivity(ohm-1 m-1)\n",
+      "rho = 1/sigma_i;      #resistivity(ohm m)\n",
+      "rho = math.ceil(rho*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"resistivity is\",rho,\"ohm m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity is 0.471 ohm m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.2, Page number 8.12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "import numpy as np\n",
+      "\n",
+      "#Variable declaration\n",
+      "Eg = 1.12;       #band gap(eV)\n",
+      "k = 1.38*10**-23;\n",
+      "T = 300;     #temperature(K)\n",
+      "e = 1.6*10**-19;\n",
+      "m0 = 1;     #for simplicity assume value of m0 to be unity\n",
+      "\n",
+      "#Calculation\n",
+      "mh = 0.28*m0;\n",
+      "me = 0.12*m0;\n",
+      "EF = (Eg/2)+(3*k*T*np.log(mh/me)/(4*e));      #position of Fermi level(eV)\n",
+      "EF = math.ceil(EF*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"position of Fermi level is\",EF,\"eV from the top of valence band\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "position of Fermi level is 0.5765 eV from the top of valence band\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.3, Page number 8.12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T = 300;      #temperature(K)\n",
+      "e = 1.6*10**-19;\n",
+      "m = 9.109*10**-31;     #mass of electron(kg)\n",
+      "k = 1.38*10**-23;      #boltzmann's constant\n",
+      "h = 6.626*10**-34;      #planck's constant\n",
+      "Eg = 0.7;      #band gap(eV)\n",
+      "\n",
+      "#Calculation\n",
+      "Eg = Eg*e;     #band gap(J)\n",
+      "A = (2*math.pi*m*k*T/h**2)**(3/2);\n",
+      "B = math.exp(-Eg/(2*k*T));\n",
+      "ni = 2*A*B;     #concentration of intrinsic charge carriers per m^3\n",
+      "\n",
+      "#Result\n",
+      "print \"concentration of intrinsic charge carriers is\",round(ni/1e+19,3),\"*10^19 per m^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "concentration of intrinsic charge carriers is 3.348 *10^19 per m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.4, Page number 8.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "ni = 2.4*10**19;      #intrinsic carrier density(per m^3)\n",
+      "mew_e = 0.39;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.19;      #hole mobility(m^2/Vs)\n",
+      "e = 1.6*10**-19;\n",
+      "\n",
+      "#Calculation\n",
+      "sigma_i = ni*e*(mew_e+mew_h);      #conductivity(ohm-1 m-1)\n",
+      "rho = 1/sigma_i;      #resistivity(ohm m)\n",
+      "rho = math.ceil(rho*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"resistivity is\",rho,\"ohm m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity is 0.449 ohm m\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.5, Page number 8.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "ni = 2.5*10**19;      #intrinsic carrier density(per m^3)\n",
+      "mew_e = 0.39;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.19;      #hole mobility(m^2/Vs)\n",
+      "e = 1.6*10**-19;\n",
+      "w = 1;     #width(mm)\n",
+      "t = 1;     #thickness(mm)\n",
+      "l = 1;     #length(cm)\n",
+      "\n",
+      "#Calculation\n",
+      "sigma_i = ni*e*(mew_e+mew_h);      #conductivity(ohm-1 m-1)\n",
+      "w = w*10**-3;       #width(m)\n",
+      "t = t*10**-3;     #thickness(m)\n",
+      "A = w*t;      #area(m^2)\n",
+      "l = l*10**-2;     #length(m)\n",
+      "R = l/(sigma_i*A);      #resistivity(ohm m)\n",
+      "R = R/10**3;\n",
+      "R = math.ceil(R*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"resistance of intrinsic Ge rod is\",R,\"*10^3 ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance of intrinsic Ge rod is 4.311 *10^3 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.6, Page number 8.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 9.109*10**-31;    #mass of electron(kg)\n",
+      "k = 1.38*10**-23;     #boltzmann constant\n",
+      "h = 6.626*10**-34;      #planck's constant\n",
+      "T = 300;     #temperature(K)\n",
+      "kT = 0.026;\n",
+      "e = 1.6*10**-19;\n",
+      "Eg = 1.1;    #energy gap(eV)\n",
+      "mew_e = 0.48;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.013;     #hole mobility(m^2/Vs)\n",
+      "\n",
+      "#Calculation\n",
+      "C = 2*(2*math.pi*m*k/h**2)**(3/2);\n",
+      "ni = C*(T**(3/2))*math.exp(-Eg/(2*kT));      #intrinsic carrier density per m^3\n",
+      "sigma_i = ni*e*(mew_e+mew_h);      #conductivity(ohm-1 m-1)\n",
+      "sigma_i = sigma_i*10**3;\n",
+      "sigma_i = math.ceil(sigma_i*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"conductivity is\",sigma_i,\"*10**-3 ohm-1 m-1\"\n",
+      "print \"answer given in the book differs due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductivity is 1.286 *10**-3 ohm-1 m-1\n",
+        "answer given in the book differs due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.7, Page number 8.15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m = 9.109*10**-31;    #mass of electron(kg)\n",
+      "k = 1.38*10**-23;     #boltzmann constant\n",
+      "h = 6.626*10**-34;      #planck's constant\n",
+      "T = 300;     #temperature(K)\n",
+      "e = 1.6*10**-19;\n",
+      "Eg = 0.7;    #energy gap(eV)\n",
+      "mew_e = 0.4;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.2;      #hole mobility(m^2/Vs)\n",
+      "\n",
+      "#Calculation\n",
+      "ni = 2*(2*math.pi*m*k*T/h**2)**(3/2)*math.exp(-Eg*e/(2*k*T));     #intrinsic carrier density per m^3\n",
+      "sigma = ni*e*(mew_e+mew_h);      #conductivity(ohm-1 m-1)\n",
+      "sigma = math.ceil(sigma*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"intrinsic carrier density is\",round(ni/1e+19,2),\"*10^19 per m^3\"\n",
+      "print \"conductivity is\",sigma,\"ohm-1 m-1\"\n",
+      "print \"answer given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "intrinsic carrier density is 3.35 *10^19 per m^3\n",
+        "conductivity is 3.22 ohm-1 m-1\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.8, Page number 8.16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "import numpy as np\n",
+      "\n",
+      "#Variable declaration\n",
+      "rho = 2.12;      #resistivity(ohm m)\n",
+      "e = 1.6*10**-19;\n",
+      "m = 9.109*10**-31;    #mass of electron(kg)\n",
+      "k = 1.38*10**-23;     #boltzmann constant\n",
+      "h = 6.626*10**-34;      #planck's constant\n",
+      "mew_e = 0.36;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.17;      #hole mobility(m^2/Vs)\n",
+      "T = 300;     #temperature(K)\n",
+      "\n",
+      "#Calculation\n",
+      "sigma = 1/rho;    #conductivity(ohm-1 m-1)\n",
+      "ni = sigma/(e*(mew_e+mew_h));       #intrinsic carrier density per m^3\n",
+      "C = 2*(2*math.pi*m*k/h**2)**(3/2);\n",
+      "#let exp(Eg/(2*k*T)) be a\n",
+      "a = (C*T**(3/2))/ni;\n",
+      "#Eg/(2*k*T) = log(a) and Eg = 2*k*T*log(a)\n",
+      "Eg = 2*k*T*np.log(a)/e;      #forbidden energy gap(eV)\n",
+      "Eg = math.ceil(Eg*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"forbidden energy gap is\",Eg,\"eV\"\n",
+      "print \"answer given in the book differs due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "forbidden energy gap is 0.793 eV\n",
+        "answer given in the book differs due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.9, Page number 8.17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "import numpy as np\n",
+      "\n",
+      "#Variable declaration\n",
+      "rho_2 = 4.5;      #resistivity at 20C\n",
+      "rho_1 = 2;        #resistivity at 32C\n",
+      "T1 = 20;      #temperature(C)\n",
+      "T2 = 32;    #temperature(C)\n",
+      "k = 8.616*10**-5;\n",
+      "\n",
+      "#Calculation\n",
+      "T1 = T1+273;     #temperature(K)\n",
+      "T2 = T2+273;    #temperature(K)\n",
+      "dy = np.log10(rho_2)-np.log10(rho_1);\n",
+      "dx = (1/T1)-(1/T2);\n",
+      "Eg = 2*k*dy/dx;       #energy band gap(eV)\n",
+      "Eg = math.ceil(Eg*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"energy band gap is\",Eg,\"eV\"\n",
+      "print \"answer given in the book differs due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy band gap is 0.452 eV\n",
+        "answer given in the book differs due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.10, Page number 8.17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "import numpy as np\n",
+      "\n",
+      "#Variable declaration\n",
+      "Eg = 1;      #band gap(eV)\n",
+      "e = 1.602*10**-19;\n",
+      "me = 1;     #for simplicity assume me to be unity\n",
+      "E_Ef = 10/100;     #fermi level shift(eV)\n",
+      "k = 1.38*10**-23;     #boltzmann constant\n",
+      "\n",
+      "#Calculation\n",
+      "Eg = Eg*e;     #band gap(J)\n",
+      "mh = 4*me;     #effective mass of holes is 4 times of electrons\n",
+      "E_Ef = E_Ef*e;     #fermi level shift(J)\n",
+      "#E_Ef = 3*k*T*np.log(mh/me)/4\n",
+      "T = 4*E_Ef/(3*k*np.log(mh/me));      #temperature(K)\n",
+      "\n",
+      "#Result\n",
+      "print \"temperature is\",int(T),\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature is 1116 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.11, Page number 8.18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Na = 5*10**23;      #atoms of boron\n",
+      "Nd = 3*10**23;      #arsenic atoms\n",
+      "ni = 2*10**16;        #intrinsic charge carriers per m^3\n",
+      "\n",
+      "#Calculation\n",
+      "p = 2*(Na-Nd)/2;      #hole concentration per m^3\n",
+      "n = ni**2/p;           #electron concentration per m^3\n",
+      "n = n/10**9;\n",
+      "\n",
+      "#Result\n",
+      "print \"electron concentration is\",int(n),\"*10**9 per m^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "electron concentration is 2 *10**9 per m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.12, Page number 8.19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "ni = 1.5*10**16;      #intrinsic charge carriers per m^3\n",
+      "e = 1.6*10**-19;\n",
+      "mew_e = 0.13;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.05;      #hole mobility(m^2/Vs)\n",
+      "AW = 28.1;      #atomic weight of Si(kg)\n",
+      "d = 2.33*10**3;     #density of Si(kg/m^3)\n",
+      "N = 6.02*10**26;    #avagadro number\n",
+      "\n",
+      "#Calculation\n",
+      "sigma = ni*e*(mew_e+mew_h);      #conductivity(ohm-1 m-1)\n",
+      "sigma = sigma*10**3;\n",
+      "Nd = d*N/AW;        #impurity atoms per m^3\n",
+      "Nd = Nd/10**8;      #extent of 10^8 Si atoms\n",
+      "p = ni**2/Nd;       #hole concentration per m^3\n",
+      "sigma_ex = Nd*e*mew_e;       #conductivity(ohm-1 m-1)\n",
+      "sigma_ex = math.ceil(sigma_ex*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "Na = Nd;\n",
+      "n = ni**2/Na;      #electron concentration per m^3\n",
+      "sigma_EX = Na*e*mew_h;       #conductivity(ohm-1 m-1)\n",
+      "sigma_EX = math.ceil(sigma_EX*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"conductivity is\",sigma,\"*10^-3 ohm-1 m-1\"\n",
+      "print \"conductivity if donor type impurity is added is\",sigma_ex,\"ohm-1 m-1\"\n",
+      "print \"conductivity if acceptor type impurity is added is\",sigma_EX,\"ohm-1 m-1\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductivity is 0.432 *10^-3 ohm-1 m-1\n",
+        "conductivity if donor type impurity is added is 10.383 ohm-1 m-1\n",
+        "conductivity if acceptor type impurity is added is 3.994 ohm-1 m-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.13, Page number 8.21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "import numpy as np\n",
+      "\n",
+      "#Variable declaration\n",
+      "ni = 1.5*10**16;      #intrinsic charge carriers per m^3\n",
+      "e = 1.6*10**-19;\n",
+      "mew_e = 0.135;      #electron mobility(m^2/Vs)\n",
+      "mew_h = 0.048;      #hole mobility(m^2/Vs)\n",
+      "Nd = 10**23;      #phosphorus atoms per m^3\n",
+      "k = 1.38*10**-23;     #boltzmann constant\n",
+      "T = 300;     #temperature(K)\n",
+      "\n",
+      "#Calculation\n",
+      "sigma = ni*e*(mew_e+mew_h);      #conductivity(ohm-1 m-1)\n",
+      "sigma = sigma*10**3;\n",
+      "p = (ni**2)/Nd;      #hole concentration per m^3\n",
+      "p = p/10**9;\n",
+      "sigma_ex = Nd*e*mew_e;      #conductivity(ohm-1 m-1)\n",
+      "#EF = (Eg/2)+(3*k*T*log(mew_e/mew_h)/4)\n",
+      "X = 3*k*T*np.log(mew_e/mew_h)/(4*e);\n",
+      "X = math.ceil(X*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "#EF = (Eg/2)+X\n",
+      "\n",
+      "#Result\n",
+      "print \"conductivity is\",sigma,\"*10^-3 ohm-1 m-1\"\n",
+      "print \"hole concentration is\",p,\"*10**9 per m^3\"\n",
+      "print \"answer for hole concentration given in the book is wrong\"\n",
+      "print \"EF = Eg/2 + \",X\n",
+      "print \"Fermi level will be positioned at \",X,\"eV above intrinsic level\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductivity is 0.4392 *10^-3 ohm-1 m-1\n",
+        "hole concentration is 2.25 *10**9 per m^3\n",
+        "answer for hole concentration given in the book is wrong\n",
+        "EF = Eg/2 +  0.021\n",
+        "Fermi level will be positioned at  0.021 eV above intrinsic level\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.14, Page number 8.36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "mew_e = 0.19;      #electron mobility(m^2/Vs)\n",
+      "k = 1.38*10**-23;     #boltzmann constant\n",
+      "T = 300;     #temperature(K)\n",
+      "e = 1.6*10**-19;\n",
+      "\n",
+      "#Calculation\n",
+      "Dn = mew_e*k*T/e;     #diffusion coefficient(m^2/s)\n",
+      "Dn = Dn*10**4;\n",
+      "Dn = math.ceil(Dn*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"diffusion coefficient of electrons is\",Dn,\"*10^-4 m^2/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diffusion coefficient of electrons is 49.17 *10^-4 m^2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.15, Page number 8.46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "RH = 3.66*10**-4;     #Hall coefficient(m^3/coulomb)\n",
+      "I = 10**-2;      #current(amp)\n",
+      "B = 0.5;      #magnetic field(Wb/m^2)\n",
+      "t = 1;        #thickness(mm)\n",
+      "\n",
+      "#Calculation\n",
+      "t = t*10**-3;     #thickness(m)\n",
+      "VH = RH*I*B/t;      #Hall voltage(V)\n",
+      "VH = VH*10**3;      #Hall voltage(mV)\n",
+      "\n",
+      "#Result\n",
+      "print \"Hall voltage developed is\",VH,\"mV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hall voltage developed is 1.83 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.16, Page number 8.47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vy = 37;      #voltage(micro-V)\n",
+      "t = 1;     #thickness(mm)\n",
+      "Bz = 0.5;      #flux density(Wb/m^2)\n",
+      "Ix = 20;     #current(mA)\n",
+      "\n",
+      "#Calculation\n",
+      "Vy = Vy*10**-6;     #voltage(V)\n",
+      "t = t*10**-3;     #thickness(m)\n",
+      "Ix = Ix*10**-3;     #current(A)\n",
+      "RH = Vy*t/(Ix*Bz);     #Hall coefficient(C-1 m^3)\n",
+      "\n",
+      "#Result\n",
+      "print \"Hall coefficient of semiconductor is\",RH,\"C-1 m^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hall coefficient of semiconductor is 3.7e-06 C-1 m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.17, Page number 8.48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "RH = -7.35*10**-5;      #Hall coefficient(m^3/C)\n",
+      "e = 1.6*10**-19;\n",
+      "sigma = 200;      #conductivity(ohm-1 m-1)\n",
+      "\n",
+      "#Calculation\n",
+      "n = -1/(RH*e);       #density(m^3)\n",
+      "mew = sigma/(n*e);     #mobility(m^2/Vs)\n",
+      "mew = mew*10**3;\n",
+      "\n",
+      "#Result\n",
+      "print \"density of charge carriers is\",round(n/1e+22,3),\"*10^22 m^3\"\n",
+      "print \"mobility of charge carriers is\",mew,\"*10^-3 m^2/Vs\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "density of charge carriers is 8.503 *10^22 m^3\n",
+        "mobility of charge carriers is 14.7 *10^-3 m^2/Vs\n"
+       ]
+      }
+     ],
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.18, Page number 8.48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "I = 50;     #current(A)\n",
+      "B = 1.5;    #magnetic field(T)\n",
+      "e = 1.6*10**-19;\n",
+      "n = 8.4*10**28;      #free electron concentration(electron/m^3)\n",
+      "t = 0.5;     #thickness(cm)\n",
+      "\n",
+      "#Calculation\n",
+      "t = t*10**-2;     #thickness(m)\n",
+      "VH = I*B/(n*e*t);     #hall voltage(V)\n",
+      "VH = VH*10**6;     #hall voltage(micro-V)\n",
+      "VH = math.ceil(VH*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"magnitude of Hall voltage is\",VH,\"micro-V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnitude of Hall voltage is 1.1161 micro-V\n"
+       ]
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.19, Page number 8.49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "RH = 3.66*10**-4;      #Hall coefficient(m^3/C)\n",
+      "e = 1.6*10**-19;\n",
+      "rho_n = 8.93*10**-3;     #resistivity(ohm m)\n",
+      "\n",
+      "#Calculation\n",
+      "n = 1/(RH*e);\n",
+      "mew_e = RH/rho_n;\n",
+      "mew_e = math.ceil(mew_e*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"value of n is\",round(n/1e+22,3),\"*10^22 per m^3\"\n",
+      "print \"value of mew_e is\",mew_e,\"m^2/Vs\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of n is 1.708 *10^22 per m^3\n",
+        "value of mew_e is 0.041 m^2/Vs\n"
+       ]
+      }
+     ],
+     "prompt_number": 72
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Applied_Physics_by_P_K_Palanisamy/Chapter9_1.ipynb b/Applied_Physics_by_P_K_Palanisamy/Chapter9_1.ipynb
new file mode 100755
index 00000000..390fe9c5
--- /dev/null
+++ b/Applied_Physics_by_P_K_Palanisamy/Chapter9_1.ipynb
@@ -0,0 +1,194 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c4f4158f2a13ac09e07abe135472c68abbdd09915f4866e5eedc3ddf9b9f42f5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "9: Superconductivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.1, Page number 9.3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Tc = 3.7      #critical temperature(K)\n",
+      "Hc_0 = 0.0306    #critical field(T)\n",
+      "T = 2     #temperature(K)\n",
+      "\n",
+      "#Calculation\n",
+      "Hc_2 = Hc_0*(1-(T/Tc)**2)     #critical field(T)\n",
+      "Hc_2 = math.ceil(Hc_2*10**5)/10**5   #rounding off to 5 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"critical field at 2K is\",Hc_2,\"T\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "critical field at 2K is 0.02166 T\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.2, Page number 9.5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T = 4.2     #temperature(K)\n",
+      "d = 1       #diameter(mm)\n",
+      "Tc = 7.18    #critical temperature(K)\n",
+      "H0 = 6.5*10**4     #critical field(A/m)\n",
+      "\n",
+      "#Calculation\n",
+      "d = d*10**-3     #diameter(m)\n",
+      "Hc = H0*(1-((T/Tc)**2))    #critical field at 2K(A/m)\n",
+      "ic = math.pi*d*round(Hc);     #critical current(A)\n",
+      "ic = math.ceil(ic*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"critical current for lead is\",ic,\"A\"\n",
+      "print \"answer given in the book differs due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "critical current for lead is 134.34 A\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.3, Page number 9.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda_T = 750    #penetration depth of mercury(Armstrong)\n",
+      "T = 3.5    #temperature(K)\n",
+      "Tc = 4.12    #critical temperarure(K)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_0 = lamda_T*((1-(T/Tc)**4))**(1/2)     #penetration depth(Armstrong)\n",
+      "\n",
+      "#Result\n",
+      "print \"penetration depth at 0K is\",int(lamda_0),\"armstrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "penetration depth at 0K is 519 armstrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.4, Page number 9.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T1 = 3     #temperature(K)\n",
+      "T2 = 7.1    #temperature(K)\n",
+      "lamda_T1 = 396    #penetration depth(armstrong)\n",
+      "lamda_T2 = 1730    #penetration depth(armstrong)\n",
+      "\n",
+      "#Calculation\n",
+      "A = (((lamda_T2/lamda_T1)**2)*T2**4) - T1**4\n",
+      "B = ((lamda_T2/lamda_T1)**2)-1\n",
+      "Tc = (A/B)**(1/4)      #critical temperature(K)\n",
+      "Tc = math.ceil(Tc*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"critical temperature for lead is\",Tc,\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "critical temperature for lead is 7.1932 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Atomic_and_Nuclear_Physics_by_N._Subrahmanyam,_B._Lal_And_J._Seshan/ch2_1.ipynb b/Atomic_and_Nuclear_Physics_by_N._Subrahmanyam,_B._Lal_And_J._Seshan/ch2_1.ipynb
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diff --git a/Atomic_and_Nuclear_Physics_by_N._Subrahmanyam,_B._Lal_And_J._Seshan/ch3_1.ipynb b/Atomic_and_Nuclear_Physics_by_N._Subrahmanyam,_B._Lal_And_J._Seshan/ch3_1.ipynb
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diff --git a/Atomic_and_Nuclear_Physics_by_N._Subrahmanyam,_B._Lal_And_J._Seshan/ch4_1.ipynb b/Atomic_and_Nuclear_Physics_by_N._Subrahmanyam,_B._Lal_And_J._Seshan/ch4_1.ipynb
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diff --git a/Basic_Electrical_And_Electronics_Engineering_by_B._R._Patil/chapter1.ipynb b/Basic_Electrical_And_Electronics_Engineering_by_B._R._Patil/chapter1.ipynb
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diff --git a/Basic_Electrical_and_Electronics_Engineering_by_R._Muthusubramanian_and_S._Salivahanan/ch3.ipynb b/Basic_Electrical_and_Electronics_Engineering_by_R._Muthusubramanian_and_S._Salivahanan/ch3.ipynb
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diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/PartBCh1.ipynb b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/PartBCh1.ipynb
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diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/PartBCh6.ipynb b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/PartBCh6.ipynb
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diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/README.txt b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/README.txt
new file mode 100755
index 00000000..f81bf842
--- /dev/null
+++ b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Vaibhav Shah
+Course: mca
+College/Institute/Organization: IIT
+Department/Designation: Developer
+Book Title: Basics of Electrical, Electronics and Communication Engineering
+Author: N. Storey, E. Hughes and W. Tomasi
+Publisher: Pearson, India
+Year of publication: 2010
+Isbn: 978-81-317-3201-4
+Edition: 1
\ No newline at end of file
diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/ch1.ipynb b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/ch1.ipynb
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diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/ch2.ipynb b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/ch2.ipynb
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diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/ch3.ipynb b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/ch3.ipynb
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diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/screenshots/ch1.png b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/screenshots/ch1.png
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diff --git a/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/screenshots/ch6.png b/Basics_of_Electrical,_Electronics_and_Communication_Engineering_by_N._Storey,_E._Hughes_and_W._Tomasi/screenshots/ch6.png
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diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter1.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter1.ipynb
deleted file mode 100755
index 5c25ef5d..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter1.ipynb
+++ /dev/null
@@ -1,117 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ed1073b28a905078eb23296522b8b59bf416e5275a011e43147248d1acd57f5a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1 GETTING STARTED WITH C PROGRAMMING"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1, Page No. 6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"C you later\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C you later\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2, Page No. 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"Chapter 1. Getting Started with C programming\"\n",
-      "print \"This is an example of format bug.\"\n",
-      "print \"The format issue can be corrected by using\"\n",
-      "print \"the \\n and \\\\ escape sequnece\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Chapter 1. Getting Started with C programming\n",
-        "This is an example of format bug.\n",
-        "The format issue can be corrected by using\n",
-        "the \n",
-        " and \\ escape sequnece\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 1.3, Page No. 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"Chapter 1. Getting Started with C programming\"\n",
-      "print \"This is an example of format bug.\"\n",
-      "print \"The format issue can be corrected by using\"\n",
-      "print \"the \\\\n and \\\\\\\\ escape sequnece\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Chapter 1. Getting Started with C programming\n",
-        "This is an example of format bug.\n",
-        "The format issue can be corrected by using\n",
-        "the \\n and \\\\ escape sequnece\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter10.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter10.ipynb
deleted file mode 100755
index c2e52854..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter10.ipynb
+++ /dev/null
@@ -1,362 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:af2735a34ebb2cb3c8a379aacf70cb46d8ee4a8ed09afacd9b5d0d21e0a0dd5a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10 : Dynamic Memory Allocation"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.1, Page No 229"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "class employee:\n",
-      "    def __init__(self,id1,name,salary):\n",
-      "        self.id1 = id1\n",
-      "        self.name = name\n",
-      "        self.salary = salary\n",
-      "e = employee(10,\"abc\",12000.02)\n",
-      "x = 1\n",
-      "f = 1.1\n",
-      "d = 1111.11\n",
-      "c = 'C'\n",
-      "print \"Sizo of integer : \" + str(sys.getsizeof(x)) + \"bytes \\n\"\n",
-      "print \"Sizo of float : \" + str(sys.getsizeof(f)) + \"bytes \\n\"\n",
-      "print \"Sizo of double : \" + str(sys.getsizeof(d)) + \"bytes \\n\"\n",
-      "print \"Sizo of char : \" + str(sys.getsizeof(c)) + \"bytes \\n\"\n",
-      "print \"Sizo of employee structure : \" + str(sys.getsizeof(e)) + \"bytes \\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sizo of integer : 24bytes \n",
-        "\n",
-        "Sizo of float : 24bytes \n",
-        "\n",
-        "Sizo of double : 24bytes \n",
-        "\n",
-        "Sizo of char : 34bytes \n",
-        "\n",
-        "Sizo of employee structure : 64bytes \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2, Page No 230"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "array = []\n",
-      "\n",
-      "print \"\\nSize of array: \" + str(sys.getsizeof(array)) +  \"bytes\\n\"\n",
-      "print \"Number of elements in array \"\n",
-      "print str(sys.getsizeof(array) / sys.getsizeof(int)) + \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Size of array: 64bytes\n",
-        "\n",
-        "Number of elements in array \n",
-        "0\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.3, Page No 231"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.4, Page No 232"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5, Page No 232"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.6, Page No 233"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function\n",
-      "#if name is None:\n",
-      "#    print \"\\nOut of memory!\\n\"\n",
-      "#else:\n",
-      "#    print \"\\nMemory allocated.\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.7, Page No 234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function\n",
-      "#if name is not None:\n",
-      "#    name = raw_input(\"Enter your name: \")\n",
-      "#    print \"\\nHi \" + name + \"\\n\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.8, Page No 235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function\n",
-      "#if name is not None:\n",
-      "#    name = raw_input(\"Enter your name: \")\n",
-      "#    print \"\\nHi \" + name + \"\\n\"\n",
-      "#    del name"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.9, Page No 236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function\n",
-      "#if numbers is None:\n",
-      "#    return\n",
-      "#numbers[0] = 100\n",
-      "#numbers[1] = 200\n",
-      "#numbers[2] = 300\n",
-      "#numbers[3] = 400\n",
-      "#numbers[4] = 500\n",
-      "#print \"\\nIndividual memory segments initialized to:\\n\"\n",
-      "#for x in range(5):\n",
-      "#    print \"numbers[\" +x+\"] = \\n\" + numbers[x]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10, Page No 237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no calloc function\n",
-      "#if numbers is None:\n",
-      "#    return"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.11, Page No 239"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# there is no malloc function\n",
-      "#if numbers is None:\n",
-      "#    print \"\\nOut of memory!\\n\"\n",
-      "#    return\n",
-      "#else:\n",
-      "#    number = newNumber\n",
-      "#    for x in range(10):\n",
-      "#        numbers[x] = x * 10\n",
-      "#    print \"\\nExpanded memory:\\n\"\n",
-      "#    for x in range(10):\n",
-      "#        print \"numbers[\" +x+\"] = \\n\" + numbers[x]\n",
-      "#    del number"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.12, Page No 241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import random\n",
-      "# there is no calloc function\n",
-      "#time\n",
-      "#print \"\\nMath Quiz\\n\\n\"\n",
-      "#response = raw_input(\"Enter # of problems: \")\n",
-      "#if op1 is None || op2 is None || answer is None ||result is None:\n",
-      "#    print \"\\nOut of Memory!\\n\"\n",
-      "#    return\n",
-      "#for x in range(response):\n",
-      "#    op1[x] = random.randint(1,200) % 100\n",
-      "#    op2[x] = random.randint(1,200) % 100\n",
-      "#    print \"\\n\" + op1[x] + \"+\" + op2[x]\"\n",
-      "#    answer[x] = raw_input(\"\")\n",
-      "#    if answer[x] == op1[x] + op2[x]:\n",
-      "#        result[x] = 'c'\n",
-      "#    else:\n",
-      "#        result[x] = 'i'\n",
-      "#print \"\\nQuiz Results\\n\"\n",
-      "#print \"\\nQuestion\\tYour Answer\\tCorrect\\n\"\n",
-      "#for x in range(response):\n",
-      "#    if result[x] == 'c':\n",
-      "#        print op1[x] + \"+\" + op2[x] + \"\\t\\t\" + answer[x] + \"\\t\\tYes\\n\"\n",
-      "#    else:\n",
-      "#        print op1[x] + \"+\" + op2[x] + \"\\t\\t\" + answer[x] + \"\\t\\tNo\\n\"\n",
-      "#del op1\n",
-      "#del op2\n",
-      "#del answer\n",
-      "#del result   "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter11.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter11.ipynb
deleted file mode 100755
index 239b6ca4..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter11.ipynb
+++ /dev/null
@@ -1,616 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9d02541e3ea8d7af1efd977534b7150b993557acd6da8dc30b83dcacb798d681"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11 : File Input and Output"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.1, Page No 250"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "try:\n",
-      "    pRead = open(\"file1.dat\",\"r\")\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "\n",
-      "\"\"\"\n",
-      "Input file:- file1.dat\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2, Page No 252"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "try:\n",
-      "    pRead = open(\"file1.dat\",\"r\")\n",
-      "    if(not isinstance(pRead,file)):\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "    else:\n",
-      "        print \"\\nFile opened for reading\\n\"\n",
-      "    pRead.close()\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "\n",
-      "\"\"\"\n",
-      "Input file:- file1.dat\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "File opened for reading\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3, Page No 253"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "try:\n",
-      "    pRead = open(\"names.dat\",\"r\")\n",
-      "    if(not isinstance(pRead,file)):\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "    else:\n",
-      "        print \"\\nContents of names.dat\\n\"\n",
-      "        while(1):\n",
-      "            t = pRead.readline() \n",
-      "            if(t != \"\"):\n",
-      "                print t\n",
-      "            else:\n",
-      "                break\n",
-      "    pRead.close()\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "    \n",
-      "\"\"\" \n",
-      "Input File :- names.dat\n",
-      "\n",
-      "Michael\n",
-      "Sheila\n",
-      "Spencer\n",
-      "Olivia\n",
-      "\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Contents of names.dat\n",
-        "\n",
-        "Michael\n",
-        "\n",
-        "Sheila\n",
-        "\n",
-        "Spencer\n",
-        "\n",
-        "Olivia\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4, Page No 255"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "try:\n",
-      "    pRead = open(\"hobbies.dat\",\"r\")\n",
-      "    if(not isinstance(pRead,file)):\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "    else:\n",
-      "        print \"\\nName\\tHobby\\n\\n\"\n",
-      "        while(1):\n",
-      "            name = pRead.readline()\n",
-      "            hobby = pRead.readline()\n",
-      "            if(name != \"\"):\n",
-      "                print name + \"\\t\"\n",
-      "                print hobby\n",
-      "            else:\n",
-      "                break\n",
-      "    pRead.close()\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "    \n",
-      "\"\"\"\n",
-      "Input file :- hobbies.dat\n",
-      "\n",
-      "Michael\tProgramming\n",
-      "Sheila\tShopping\n",
-      "Spencer\tFootball\n",
-      "Olivia\tDancing\n",
-      "Waytt\tEating\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name\tHobby\n",
-        "\n",
-        "\n",
-        "Michael\tProgramming\n",
-        "\t\n",
-        "Sheila\tShopping\n",
-        "\n",
-        "Spencer\tFootball\n",
-        "\t\n",
-        "Olivia\tDancing\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5, Page No 256"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "try:\n",
-      "    pWrite = open(\"students.dat\",\"w\")\n",
-      "    if(not isinstance(pWrite,file)):\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "    else:\n",
-      "        fname = raw_input(\"Enter first name: \")\n",
-      "        lname = raw_input(\"Enter last name: \")\n",
-      "        id1 = raw_input(\"Enter id: \")\n",
-      "        gpa = float(raw_input(\"Enter GPA: \"))\n",
-      "        pWrite.writelines(fname + \" \" + lname + \"\\t\" + id1 + \"\\t\" + str(gpa))\n",
-      "    pWrite.close()\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "\n",
-      "\"\"\"\n",
-      "Output file:- students.dat\n",
-      "Patric Star\t888-66-9999\t2.0\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first name: Patric\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter last name: Star\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter id: 888-66-9999\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter GPA: 2.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6, Page No 258"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "try:\n",
-      "    pRead = open(\"students.dat\",\"r\")\n",
-      "    if(not isinstance(pRead,file)):\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "    else:\n",
-      "        print \"\\nName\\t\\tID\\t\\tGPA\\n\\n\"\n",
-      "        while(1):\n",
-      "            fname = pRead.read()\n",
-      "            lname = pRead.read()\n",
-      "            id1 = pRead.read()\n",
-      "            gpa = pRead.read()\n",
-      "            \n",
-      "            if(fname != \"\"):\n",
-      "                print fname + lname + \"\\t\\t\" + id1 + \"\\t\" + str(gpa)\n",
-      "            else:\n",
-      "                break\n",
-      "    pRead.close()\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "\n",
-      "\"\"\"\n",
-      "Input file:- students.dat\n",
-      "Patric Star\t888-66-9999\t2.0\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name\t\tID\t\tGPA\n",
-        "\n",
-        "\n",
-        "Patric Star\t888-66-9999\t2.0\t\t\t\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7, Page No 259"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def readData():\n",
-      "    try:\n",
-      "        pRead = open(\"hobbies.dat\",\"r\")\n",
-      "        if(not isinstance(pRead,file)):\n",
-      "            print \"\\nFile can not be opened\\n\"\n",
-      "        else:\n",
-      "            print \"\\nName\\tHobby\\n\\n\"\n",
-      "            while(1):\n",
-      "                name = pRead.readline()\n",
-      "                hobby = pRead.readline()\n",
-      "                if(name != \"\"):\n",
-      "                    print name + \"\\t\"\n",
-      "                    print hobby\n",
-      "                else:\n",
-      "                    break\n",
-      "        pRead.close()\n",
-      "    except IOError:\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "try:\n",
-      "    print \"Current file contains\"\n",
-      "    readData()\n",
-      "    print \"\\n\"\n",
-      "    name = raw_input(\"Enter a new name: \")\n",
-      "    hobby = raw_input(\"Enter a new hobby: \")\n",
-      "    pWrite = open(\"hobbies.dat\",\"a\")\n",
-      "    if(not isinstance(pWrite,file)):\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "    else:\n",
-      "        pWrite.writelines(\"\\n\" + name + \"\\t\" + hobby)\n",
-      "    pWrite.close()\n",
-      "    readData()\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "    \n",
-      "\"\"\"\n",
-      "Input file and Output file:- hobbies.dat\n",
-      "Michael\tProgramming\n",
-      "Sheila\tShopping\n",
-      "Spencer\tFootball\n",
-      "Olivia\tDancing\n",
-      "Waytt\tEating\n",
-      "\"\"\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current file contains\n",
-        "\n",
-        "Name\tHobby\n",
-        "\n",
-        "\n",
-        "Michael\tProgramming\n",
-        "\t\n",
-        "Sheila\tShopping\n",
-        "\n",
-        "Spencer\tFootball\n",
-        "\t\n",
-        "Olivia\tDancing\n",
-        "\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a new name: Waytt\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a new hobby: Eating\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name\tHobby\n",
-        "\n",
-        "\n",
-        "Michael\tProgramming\n",
-        "\t\n",
-        "Sheila\tShopping\n",
-        "\n",
-        "Spencer\tFootball\n",
-        "\t\n",
-        "Olivia\tDancing\n",
-        "\n",
-        "Waytt\tEating\t\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8, Page No 263"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "try:\n",
-      "    pRead = open(\"hobbies.dat\",\"r\")\n",
-      "    if(not isinstance(pRead,file)):\n",
-      "        print \"\\nFile can not be opened\\n\"\n",
-      "        print \"The following error occurred\" #goto label\n",
-      "        sys.exit(\"Some Error Occures\")\n",
-      "    else:\n",
-      "        print \"\\nName\\tHobby\\n\\n\"\n",
-      "        while(1):\n",
-      "            name = pRead.readline()\n",
-      "            hobby = pRead.readline()\n",
-      "            if(name != \"\"):\n",
-      "                print name + \"\\t\"\n",
-      "                print hobby\n",
-      "            else:\n",
-      "                break\n",
-      "    pRead.close()\n",
-      "    #sys.exit(\"Exit\")\n",
-      "except IOError:\n",
-      "    print \"\\nFile can not be opened\\n\"\n",
-      "\n",
-      "\"\"\"\n",
-      "Input file:- hobbies.dat\n",
-      "Michael\tProgramming\n",
-      "Sheila\tShopping\n",
-      "Spencer\tFootball\n",
-      "Olivia\tDancing\n",
-      "Waytt\tEating\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name\tHobby\n",
-        "\n",
-        "\n",
-        "Michael\tProgramming\n",
-        "\t\n",
-        "Sheila\tShopping\n",
-        "\n",
-        "Spencer\tFootball\n",
-        "\t\n",
-        "Olivia\tDancing\n",
-        "\n",
-        "Waytt\tEating\t\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9, Page No 266"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"\\n\\tPhone Book\\n\"\n",
-      "print \"\\n 1 \\t Add phone book entry \\n\"\n",
-      "print \"2 \\t Print phone book \\n\\n\"\n",
-      "print \"Select an option: \"\n",
-      "response = int(raw_input(\"Select an option: \"))\n",
-      "print response\n",
-      "if(response == 1):\n",
-      "    fname = raw_input(\"Enter first name: \")\n",
-      "    lname = raw_input(\"Enter last name: \")\n",
-      "    number = raw_input(\"Enter phone number: \")\n",
-      "    pWrite = open(\"phone_book.dat\",\"a\")\n",
-      "    if(isinstance(pWrite,file)):\n",
-      "       pWrite.writelines(fname + \" \" + lname + \" \" + number + \"\\n\")\n",
-      "       pWrite.close()\n",
-      "    else:\n",
-      "        print \"The error occured\" #go to\n",
-      "elif(response == 2):\n",
-      "    pRead = open(\"phone_book.dat\",\"r\")\n",
-      "    if(isinstance(pRead,file)):\n",
-      "        while(1):\n",
-      "            fname = pRead.read()\n",
-      "            if(fname == \"\"):\n",
-      "                break\n",
-      "            lname = pRead.read()\n",
-      "            number = pRead.read()\n",
-      "            print fname\n",
-      "            print lname\n",
-      "            print number\n",
-      "        print \"\\n\"\n",
-      "    else:\n",
-      "        print \"The error occured\" #go to\n",
-      "else:\n",
-      "    print \"Invalid Selection\"\n",
-      "\n",
-      "\"\"\"\n",
-      "Input file and Output file:- phone_book.dat\n",
-      "John Smith 538.676.1234\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\tPhone Book\n",
-        "\n",
-        "\n",
-        " 1 \t Add phone book entry \n",
-        "\n",
-        "2 \t Print phone book \n",
-        "\n",
-        "\n",
-        "Select an option: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Select an option: 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n",
-        "John Smith 538.676.1234\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter12.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter12.ipynb
deleted file mode 100755
index 76d70f5c..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter12.ipynb
+++ /dev/null
@@ -1,445 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b5b5db464e530d21c4b695ff4bde6912d0e8c447d51f1ede515f161acac3856f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12 : The C PreProcessor"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1, Page No 272"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"\\nHaving fun with preprocessor directives\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Having fun with preprocessor directives\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2, Page No 273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "NUMBER = 7\n",
-      "print \"\\nLucky Number \", NUMBER , \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Lucky Number  7 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3, Page No 274"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "NUMBER = 7\n",
-      "print \"\\nLucky Number \", NUMBER , \"\\n\"\n",
-      "NUMBER = 5"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Lucky Number  7 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4, Page No 275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def area(l,w):\n",
-      "    return l * w\n",
-      "length = 0\n",
-      "width = 0\n",
-      "length = int(raw_input(\"\\nEnter length: \"))\n",
-      "width = int(raw_input(\"\\nEnter width: \"))\n",
-      "print \"\\nArea of rectangle = \" , area(length,width) , \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter length: 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter width: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Area of rectangle =  84 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5, Page No 276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def RESULT(x,y):\n",
-      "    print \"\\nResult is \\n\", x+y\n",
-      "num1 = 0\n",
-      "num2 = 0\n",
-      "num1 = int(raw_input(\"\\nEnter first number: \"))\n",
-      "num2 = int(raw_input(\"\\nEnter second number: \"))\n",
-      "RESULT(num1,num2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter first number: 27\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter second number: 44\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Result is \n",
-        "71\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6, Page No 277"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def RESULT(x,y):\n",
-      "    print \"\\nResult is \\n\", x+y\n",
-      "operand1 = 0\n",
-      "operand2 = 0\n",
-      "operand1 = int(raw_input(\"\\nEnter first operand: \"))\n",
-      "operand2 = int(raw_input(\"\\nEnter second operand: \"))\n",
-      "RESULT(operand1,operand2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter first operand: 27\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter second operand: 44\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Result is \n",
-        "71\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7, Page No 279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import profit\n",
-      "print \"\\nThe Profit Program\\n\"\n",
-      "price = float(raw_input(\"\\nEnter unit price: \"))\n",
-      "quantity = int(raw_input(\"Enter quantity sold: \"))\n",
-      "totalCost = float(raw_input(\"Enter total cost: \"))\n",
-      "profit1(price,quantity,totalCost)\n",
-      "\n",
-      "\"\"\"\n",
-      "Header file:- profit.py\n",
-      "def profit(p,q,tc):\n",
-      "    print \"\\nYour profit is \",(p * q) - tc,\"\\n\"\n",
-      "\"\"\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Profit Program\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter unit price: 12.99\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter quantity sold: 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter total cost: 56\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Your profit is  34.93 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8, Page No 282"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import calculate\n",
-      "selection = 0\n",
-      "print \"\\n The Function Wizard \\n\"\n",
-      "print \"\\n1\\tDetermine perimeter of ractangle\\n\"\n",
-      "print \"\\n2\\tDetermine area of rectangle\\n\"\n",
-      "print \"\\n3\\tDetermine volume of rectangle\\n\"\n",
-      "selection = int(raw_input(\"\\nEnter selection : \"))\n",
-      "if selection == 1:\n",
-      "    l = float(raw_input(\"Enter length: \"))\n",
-      "    w = float(raw_input(\"Enter width: \"))\n",
-      "    perimeter(l,w)\n",
-      "elif selection == 2:\n",
-      "    l = float(raw_input(\"Enter length: \"))\n",
-      "    w = float(raw_input(\"Enter width: \"))\n",
-      "    area(l,w)\n",
-      "elif selection == 3:\n",
-      "    l = float(raw_input(\"Enter length: \"))\n",
-      "    w = float(raw_input(\"Enter width: \"))\n",
-      "    h = float(raw_input(\"Enter height: \"))\n",
-      "    volume(l,w,h)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " The Function Wizard \n",
-        "\n",
-        "\n",
-        "1\tDetermine perimeter of ractangle\n",
-        "\n",
-        "\n",
-        "2\tDetermine area of rectangle\n",
-        "\n",
-        "\n",
-        "3\tDetermine volume of rectangle\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter selection : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter length: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter width: 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter height: 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Perimeter is  6.0 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter2.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter2.ipynb
deleted file mode 100755
index 2f3c44af..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter2.ipynb
+++ /dev/null
@@ -1,239 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:622f2c579033e8c66b476e16d61ce8bfa6745f6b87c5905b2064108fa5218b3c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2 PRIMARY DATA TYPES"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1, Page No. 32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=-4443\n",
-      "y=554.21\n",
-      "c='M'\n",
-      "\n",
-      "print \"The value of integer variable x is\" , x\n",
-      "print \"The value of float variable y is \" , y\n",
-      "print \"The value of character variable c is \" , c"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of integer variable x is -4443\n",
-        "The value of float variable y is  554.21\n",
-        "The value of character variable c is  M\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page No. 36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x = 20\n",
-      "PI = 3.14\n",
-      "print \"Constant values are \",x ,\" and \", round(PI,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Constant values are  20  and  3.14\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3, Page No. 42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"\\t Adder Program, by Michael Vine\"\n",
-      "iOperand1=int(raw_input(\"Enter First Operand: \"))\n",
-      "iOperand2=int(raw_input(\"Enter Second Operand: \"))\n",
-      "print \"The final result is \", iOperand1+iOperand2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t Adder Program, by Michael Vine\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter First Operand: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Second Operand: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final result is  12\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4, Page No. 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"\\t Adder Program, by Michael Vine\"\n",
-      "iOperand1=int(raw_input(\"Enter First Operand: \"))\n",
-      "iOperand2=int(raw_input(\"Enter Second Operand: \"))\n",
-      "iResult=iOperand1+iOperand2\n",
-      "print \"The result is \", iResult"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t Adder Program, by Michael Vine\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter First Operand: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Second Operand: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The result is  12\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5, Page No. 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "fRevenue=float(raw_input(\"Enter total revenue: \"))\n",
-      "fCost=float(raw_input(\"Enter total cost: \"))\n",
-      "print \"Your total profit is \", round(fRevenue-fCost,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter total revenue: 4000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter total cost: 750\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your total profit is  3250.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter3.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter3.ipynb
deleted file mode 100755
index a87e263b..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter3.ipynb
+++ /dev/null
@@ -1,499 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4698a1abef67d23e99c2eb6e9534a43d9d501d3ae6147f6ceb242f76f06f63a5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3 CONDITIONS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1, Page No. 58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"\\tAC Control Unit\"\n",
-      "print \"1\\tTurn the AC on\"\n",
-      "print \"2\\tTurn the AC off\"\n",
-      "iResponse=int(raw_input(\"Enter your selection::\"))\n",
-      "if iResponse==1:\n",
-      "    print \"AC is now on\"\n",
-      "if iResponse==2:\n",
-      "    print \"AC is now off\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tAC Control Unit\n",
-        "1\tTurn the AC on\n",
-        "2\tTurn the AC off\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your selection::1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "AC is now on\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2, Page No. 59"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"\\tAC Control Unit\"\n",
-      "print \"a\\tTurn the AC on\"\n",
-      "print \"b\\tTurn the AC off\"\n",
-      "cResponse=raw_input(\"Enter your selection::\")\n",
-      "if cResponse=='a':\n",
-      "    print \"AC is now on\"\n",
-      "if cResponse=='b':\n",
-      "    print \"AC is now off\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tAC Control Unit\n",
-        "a\tTurn the AC on\n",
-        "b\tTurn the AC off\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your selection::b\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "AC is now off\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3, Page No. 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "fBalance = 100.25\n",
-      "print \"\\tATM\"\n",
-      "print \"1\\tDeposit Funds\"\n",
-      "print \"2\\tWithdraw Funds\"\n",
-      "iSelection=int(raw_input(\"Enter Your Selection: \"))\n",
-      "if iSelection==1:\n",
-      "    fTransAmount=int(raw_input(\"Enter fund amount to deposit: \"))\n",
-      "    print \"Your new Balance is: \", fBalance+fTransAmount\n",
-      "if iSelection==2:\n",
-      "    fTransAmount=int(raw_input(\"Enter fund amount to withdraw: \"))\n",
-      "    if fTransAmout>fBalance:\n",
-      "        print \"Insufficient Funds\"\n",
-      "    else:\n",
-      "        print \"Your new balance is \", fBalance-fTransAmount"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tATM\n",
-        "1\tDeposit Funds\n",
-        "2\tWithdraw Funds\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Selection: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter fund amount to deposit: 200\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your new Balance is:  300.25\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page No. 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "cResponse=raw_input(\"Enter a letter A: \")\n",
-      "if cResponse=='A':\n",
-      "    print \"Correct response\"\n",
-      "else:\n",
-      "    print \"Incorrect response\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a letter A: a\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Incorrect response\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5, Page No.69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iResponse=int(raw_input(\"Enter a numeer from 1 to 10: \"))\n",
-      "if iResponse<1 or iResponse>10:\n",
-      "    print \"Number not in Range\"\n",
-      "else:\n",
-      "    print \"Thank You\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a numeer from 1 to 10: 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thank You\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page No. 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "cResponse=raw_input(\"Please Enter a letter: \")\n",
-      "if not cResponse.isdigit():\n",
-      "    print \"Thank You\"\n",
-      "else:\n",
-      "    print \"You did not enter a letter\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please Enter a letter: 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You did not enter a letter\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page No. 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "cResponse=raw_input(\"Please Enter a letter: \")\n",
-      "if cResponse.isdigit():\n",
-      "    print \"Thank You\"\n",
-      "else:\n",
-      "    print \"You did not enter a digit\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please Enter a letter: a\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You did not enter a digit\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8, Page No.72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print \"1\\tSports\"\n",
-      "print \"2\\tGeography\"\n",
-      "print \"3\\tMusic\"\n",
-      "print \"4\\tWorld Events\"\n",
-      "iResponse=int(raw_input(\"Please select a category (1-4): \"))\n",
-      "if iResponse==1:\n",
-      "    print \"You selected sports questions\"\n",
-      "elif iResponse==2:\n",
-      "    print \"You selected geography questions\"\n",
-      "elif iResponse==3:\n",
-      "    print \"You selected Music questions\"\n",
-      "elif iResponse==4:\n",
-      "    print \"You selected World event questions\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\tSports\n",
-        "2\tGeography\n",
-        "3\tMusic\n",
-        "4\tWorld Events\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please select a category (1-4): 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You selected World event questions\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9, Page No. 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import random\n",
-      "iRandomNum=(random.randint(0,100)%10)+1\n",
-      "iResponse=int(raw_input(\"Guess a number between 1 and 10: \"))\n",
-      "if iResponse==iRandomNum:\n",
-      "    print \"You guessed right\"\n",
-      "else:\n",
-      "    print \"Sorry, you guess wrong\"\n",
-      "    print \"The correct guess was \",iRandomNum"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Guess a number between 1 and 10: 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorry, you guess wrong\n",
-        "The correct guess was  3\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10, Page No. 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import random\n",
-      "iRandomNum=(random.randint(0,100)%4)+1\n",
-      "print \"Fortune Cookie - Chapter 3\"\n",
-      "if iRandomNum==1:\n",
-      "    print \"You will meet a new friend\"\n",
-      "elif iRandomNum==2:\n",
-      "    print \"You will enjoy a long and happy life\"\n",
-      "elif  iRandomNum==3:\n",
-      "    print \"Opportunity knocks softly. Can you hear it?\"\n",
-      "elif iRandomNum==4:\n",
-      "    print \"You'll be financially rewarded for your good deeds\"\n",
-      "\n",
-      "print \"Lucky lotto numbers\"\n",
-      "print (random.randint(0,100)%49+1)\n",
-      "print (random.randint(0,100)%49+1)\n",
-      "print (random.randint(0,100)%49+1)\n",
-      "print (random.randint(0,100)%49+1)\n",
-      "print (random.randint(0,100)%49+1)\n",
-      "print (random.randint(0,100)%49+1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fortune Cookie - Chapter 3\n",
-        "You will enjoy a long and happy life\n",
-        "Lucky lotto numbers\n",
-        "45\n",
-        "17\n",
-        "9\n",
-        "23\n",
-        "2\n",
-        "15\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter4.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter4.ipynb
deleted file mode 100755
index 218595de..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter4.ipynb
+++ /dev/null
@@ -1,688 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d985967b3e8d0e583d78d0a3306851320f330cc48579a1f62da314746570750f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4 Looping Structures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1 , Page No.88"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=0\n",
-      "print \"The value of x is \",x\n",
-      "x+=1\n",
-      "print \"The value of x is \",x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is  0\n",
-        "The value of x is  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2, Page No.89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=0\n",
-      "y=0\n",
-      "print \"The value of y is \", y+1\n",
-      "print \"The value of x is \", x+1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of y is  1\n",
-        "The value of x is  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3, Page No.89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=0\n",
-      "y=1\n",
-      "x=(y)*2\n",
-      "print \"The Value of x \",x\n",
-      "x=0\n",
-      "y=1\n",
-      "x=(y+1)*2\n",
-      "print \"The Value of x \",x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of x  2\n",
-        "The Value of x  4\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4, Page No.90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=0\n",
-      "y=0\n",
-      "x=(y)*4\n",
-      "print \"The Value of x\", x\n",
-      "y=0\n",
-      "x=(y+1)*4\n",
-      "print \"The Value of x\", x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of x 0\n",
-        "The Value of x 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5, Page No.92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=1\n",
-      "y=1\n",
-      "x=(y)*4\n",
-      "print \"The value of x is \",x\n",
-      "y=1\n",
-      "x=(y-1)*4\n",
-      "print \"The value of x is \",x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is  4\n",
-        "The value of x is  0\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6, Page No. 93"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=1\n",
-      "y=2\n",
-      "x=y*x+1\n",
-      "print \"The value of x is: \",x\n",
-      "x=1\n",
-      "y=2\n",
-      "x+=y*x+1\n",
-      "print \"The value of x is: \",x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is:  3\n",
-        "The value of x is:  4\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7, Page No.94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=1\n",
-      "y=2\n",
-      "x=y*x+1\n",
-      "print \"The value of x is: \",x\n",
-      "x=1\n",
-      "y=2\n",
-      "x-=y*x+1\n",
-      "print \"The value of x is: \",x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is:  3\n",
-        "The value of x is:  -2\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.8, Page No.95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=0\n",
-      "while (x<10):\n",
-      "    print \"The value of x is \",x\n",
-      "    x=x+1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is  0\n",
-        "The value of x is  1\n",
-        "The value of x is  2\n",
-        "The value of x is  3\n",
-        "The value of x is  4\n",
-        "The value of x is  5\n",
-        "The value of x is  6\n",
-        "The value of x is  7\n",
-        "The value of x is  8\n",
-        "The value of x is  9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9, Page No.97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iSelection=0\n",
-      "while (iSelection!=4):\n",
-      "    print \"1\\tDeposit Funds\"\n",
-      "    print \"2\\tWithdraw Funds\"\n",
-      "    print \"3\\tPrint Balance\"\n",
-      "    print \"4\\tQuit\"\n",
-      "    iSelection=int(raw_input(\"Enter Your Selection(1-4): \"))\n",
-      "print \"Thank you\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\tDeposit Funds\n",
-        "2\tWithdraw Funds\n",
-        "3\tPrint Balance\n",
-        "4\tQuit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Selection(1-4): 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\tDeposit Funds\n",
-        "2\tWithdraw Funds\n",
-        "3\tPrint Balance\n",
-        "4\tQuit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Selection(1-4): 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thank you\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.10, Page No.99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=9\n",
-      "while (x<10):\n",
-      "    print \"This printf statement is executed at least once\"\n",
-      "    x=x+1\n",
-      "while (x<10):\n",
-      "    print \"This printf statement is never executed \"\n",
-      "    x=x+1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This printf statement is executed at least once\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.11, Page No.100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "for x in range(10,5,-1):\n",
-      "    print \"The value of x is \",x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is  10\n",
-        "The value of x is  9\n",
-        "The value of x is  8\n",
-        "The value of x is  7\n",
-        "The value of x is  6\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.12, Page No.101"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import random\n",
-      "iNumQuestions=int(raw_input(\"Enter the number of questions to ask: \"))\n",
-      "for x in range(iNumQuestions):\n",
-      "    iRndNum1=(random.randint(0,100)%10)+1\n",
-      "    iRndNum2=(random.randint(0,100)%10)+1\n",
-      "    iResponse=int(raw_input(\"what is \"+ str(iRndNum1)+\" * \"+str(iRndNum2)+\" : \"))\n",
-      "    if(iResponse==(iRndNum1*iRndNum2)):\n",
-      "        print \"Correct\"\n",
-      "    else:\n",
-      "        print \"The Corrent answer was \", iRndNum1*iRndNum2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of questions to ask: 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "what is 6 * 9 : 54\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Correct\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "what is 8 * 5 : 32\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Corrent answer was  40\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.13, Page No.102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "for x in range(10,5,-1):\n",
-      "    if x==7:\n",
-      "        break\n",
-      "print x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.14, Page No.103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "for x in range(10,5,-1):\n",
-      "    if x==7:\n",
-      "        continue\n",
-      "    print x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n",
-        "9\n",
-        "8\n",
-        "6\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.15, Page No.104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "for x in range(25):\n",
-      "    print \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.16, Page No. 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import time\n",
-      "import random\n",
-      "import os\n",
-      "cYesNo=raw_input(\"Play a game of Concentration?(y or n)\")\n",
-      "if(cYesNo=='y' or cYesNo=='Y'):\n",
-      "    i1=random.randint(0,100) % 100\n",
-      "    i2=random.randint(0,100) % 100\n",
-      "    i3=random.randint(0,100) % 100\n",
-      "    print \"Concentrate on the next three numbers\\n\",i1,i2,i3\n",
-      "    iCurrentTime=int(round(time.time()*1000))\n",
-      "    iElapsedTime=int(round(time.time()*1000))\n",
-      "    while((iCurrentTime-iElapsedTime)<3):\n",
-      "        iElapsedTime=int(round(time.time()*1000))\n",
-      "    os.system('cls')\n",
-      "    iResp1=int(raw_input(\"Enter a number:\"))\n",
-      "    iResp2=int(raw_input(\"Enter a number:\"))\n",
-      "    iResp3=int(raw_input(\"Enter a number:\"))\n",
-      "    \n",
-      "    if(i1==iResp1 and i2==iResp2 and i3==iResp3):\n",
-      "        print \"Congratulations!\"\n",
-      "    else:\n",
-      "        print \"Sorry, Correct numbers were \",i1,i2,i3"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Play a game of Concentration?(y or n)n\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter5.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter5.ipynb
deleted file mode 100755
index ac86dba8..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter5.ipynb
+++ /dev/null
@@ -1,475 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:22dbf664bff583df60a7371a6f741be85c976f32fe9bef5f00f146ca5848d933"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5 : Structured Programming"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1, Page No 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def addTwoNumbers(operand1,operand2):\n",
-      "    return operand1+operand2\n",
-      "\n",
-      "print \"Nothing happenning here\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Nothing happenning here\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2, Page No 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def addTwoNumbers(num1,num2):\n",
-      "    return num1+num2\n",
-      "\n",
-      "def subtractTwoNumber(num1,num2):\n",
-      "    return num1-num2\n",
-      "\n",
-      "print \"Nothing happenning here\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Nothing happenning here\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3, Page No 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def addTwoNumbers(operand1,operand2):\n",
-      "    return operand1+operand2\n",
-      "\n",
-      "\n",
-      "iResult= addTwoNumbers(5,5)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4, Page No 120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def addTwoNumbers(operand1,operand2):\n",
-      "    return operand1+operand2\n",
-      "\n",
-      "print \"The result is : \",addTwoNumbers(5,5)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The result is :  10\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5, Page No 120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def addTwoNumbers(operand1,operand2):\n",
-      "    return operand1+operand2\n",
-      "\n",
-      "num1=int(raw_input(\"Enter the first number: \"))\n",
-      "num2=int(raw_input(\"Enter the second number: \"))\n",
-      "print \"The result is: \",addTwoNumbers(num1,num2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the first number: 57\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the second number: 43\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The result is:  100\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6, Page No 121"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def printReportHeader():\n",
-      "    print \"\\n Column1 \\t Column2 \\t Column3 \\t Column4 \\n\"\n",
-      "    \n",
-      "printReportHeader()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Column1 \t Column2 \t Column3 \t Column4 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7, Page No 122"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "num1=int(raw_input(\"Enter a number: \"))\n",
-      "print \"You entered \",num1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You entered  10\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8, Page No 123"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def getSecondNumber():\n",
-      "    num1=int(raw_input(\"Enter Second number: \"))\n",
-      "    return num1\n",
-      "\n",
-      "num1=int(raw_input(\"Enter a number: \"))\n",
-      "num2=getSecondNumber()\n",
-      "print \"you entered \"+str(num1)+\" and \",str(num2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 28\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Second number: 35\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "you entered 28 and  35\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9, Page No 124"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iLuckyNumber=int()\n",
-      "def printLuckyNumber():\n",
-      "    print \"Your lucky number is: \",iLuckyNumber\n",
-      "\n",
-      "\n",
-      "iLuckyNumber=int(raw_input(\"Enter your lucky number: \"))\n",
-      "printLuckyNumber()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your lucky number: 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your lucky number is:  10\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10, Page No 126"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import os\n",
-      "import time\n",
-      "def sportsQuestion():\n",
-      "    print \"Sports Question\"\n",
-      "    print \"What University did NFL star Deon Sanders attend?\"\n",
-      "    print \"1 \\t University of Miami\"\n",
-      "    print \"2 \\t California State University\"\n",
-      "    print \"3 \\t Indiana University\"\n",
-      "    print \"4 \\t Florida State University\"\n",
-      "    iAnswer=int(raw_input(\"Enter your selection :\"))\n",
-      "    return iAnswer\n",
-      "\n",
-      "def geographyQuestion():\n",
-      "    print \"Geography Question\"\n",
-      "    print \"What is the state capitol of Florida?\"\n",
-      "    print \"1 \\t Pensecola\"\n",
-      "    print \"2 \\t Tallahassee\"\n",
-      "    print \"3 \\t Jacksonville\"\n",
-      "    print \"4 \\t Miami\"\n",
-      "    iAnswer=int(raw_input(\"Enter your selection :\"))\n",
-      "    return iAnswer\n",
-      "def pause(inNum):\n",
-      "    iCurrentTime=0\n",
-      "    iElapsedTime=0\n",
-      "    \n",
-      "    iCurrentTime=int(round(time.time()*1000))\n",
-      "    \n",
-      "    while((iElapsedTime-iCurrentTime)<inNum):\n",
-      "        iElapsedTime=int(round(time.time()*1000))\n",
-      "\n",
-      "giResponse=0\n",
-      "while(giResponse !=3):\n",
-      "    os.system('cls')\n",
-      "    print \"\\n\\tTHE TRIVIA GAME\\n\\n\"\n",
-      "    print \"1\\tSports\\n\"\n",
-      "    print \"2\\tGeography\\n\"\n",
-      "    print \"3\\tQuit\\n\"\n",
-      "    giResponse=int(raw_input(\"Enter your Selection: \"))\n",
-      "    \n",
-      "    if(giResponse==1):\n",
-      "        if(sportsQuestion()==4):\n",
-      "            print \"\\nCorrect\\n\"\n",
-      "        else:\n",
-      "            print \"\\nIncorrect\\n\"\n",
-      "            \n",
-      "        pause(2)\n",
-      "        break\n",
-      "    elif(giResponse==2):\n",
-      "        if(geographyQuestion()==2):\n",
-      "            print \"\\nCorrect\\n\"\n",
-      "        else:\n",
-      "            print \"\\nIncorrect\\n\"\n",
-      "        \n",
-      "        pause(2)\n",
-      "        break\n",
-      "            \n",
-      "            \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\tTHE TRIVIA GAME\n",
-        "\n",
-        "\n",
-        "1\tSports\n",
-        "\n",
-        "2\tGeography\n",
-        "\n",
-        "3\tQuit\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your Selection: 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sports Question\n",
-        "What University did NFL star Deon Sanders attend?\n",
-        "1 \t University of Miami\n",
-        "2 \t California State University\n",
-        "3 \t Indiana University\n",
-        "4 \t Florida State University\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your selection :4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Correct\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter6.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter6.ipynb
deleted file mode 100755
index bfb0065f..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter6.ipynb
+++ /dev/null
@@ -1,652 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:74348a5183e19ad7ac30f2bcfa88ff539f69c16e0607950b89f2f1031fc1e024"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6 : Arrays"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1, Page No 134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iArray=range(5)\n",
-      "\n",
-      "for x in range(5):\n",
-      "    iArray[x]=0"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page No 134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iArray=range(5)\n",
-      "\n",
-      "for x in range(5):\n",
-      "    iArray[x]=x\n",
-      "    \n",
-      "for x in range(5):\n",
-      "    print \"The value of iArray index \"+str(x)+\"is \"+str(x)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of iArray index 0is 0\n",
-        "The value of iArray index 1is 1\n",
-        "The value of iArray index 2is 2\n",
-        "The value of iArray index 3is 3\n",
-        "The value of iArray index 4is 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page No 136"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iIndex = -1\n",
-      "iArray=range(5)\n",
-      "\n",
-      "for x in range(5):\n",
-      "    iArray[x]=(x+5)\n",
-      "\n",
-      "while (iIndex < 0 or iIndex >4):\n",
-      "    iIndex=int(raw_input(\"Enter a valid index (0-4): \"))\n",
-      "    \n",
-      "print \"The value of index \"+str(iIndex)+\" is \"+str(iArray[iIndex])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a valid index (0-4): 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of index 4 is 9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page No 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "cArray = range(5)\n",
-      "cName=\"Olivia\"\n",
-      "\n",
-      "print \"\\nCharacter array not initialized:\\n\"\n",
-      "\n",
-      "for x in range(5):\n",
-      "    print \"Element \"+str(x)+\"'s Contents are \",cArray[x]\n",
-      "\n",
-      "print \"\\nInitialized character array: \"\n",
-      "for x in range(6):\n",
-      "    print cName[x],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Character array not initialized:\n",
-        "\n",
-        "Element 0's Contents are  0\n",
-        "Element 1's Contents are  1\n",
-        "Element 2's Contents are  2\n",
-        "Element 3's Contents are  3\n",
-        "Element 4's Contents are  4\n",
-        "\n",
-        "Initialized character array: \n",
-        "O l i v i a\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page No 138"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iFound = -1\n",
-      "iArray=range(5)\n",
-      "\n",
-      "for x in range(5):\n",
-      "    iArray[x]=(x+x)\n",
-      "    \n",
-      "iValue=int(raw_input(\"Enter value to search for: \"))\n",
-      "\n",
-      "for x in range(5):\n",
-      "    if (iArray[x]==iValue):\n",
-      "        iFound=x\n",
-      "        break\n",
-      "\n",
-      "if(iFound > -1):\n",
-      "    print \"I found your search value in element \",iFound\n",
-      "else:\n",
-      "    print \"Sorry, your search value was not found\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value to search for: 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "I found your search value in element  4\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page No 141"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iTwoD=[[0 for j in range(3)]for i in range(3)]\n",
-      "for x in range(3):\n",
-      "    for y in range(3):\n",
-      "        iTwoD[x][y]=(x+y)\n",
-      "for x in range(3):\n",
-      "    for y in range(3):\n",
-      "        print \"iTwoD[\"+str(x)+\"][\"+str(y)+\"]=\",iTwoD[x][y]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "iTwoD[0][0]= 0\n",
-        "iTwoD[0][1]= 1\n",
-        "iTwoD[0][2]= 2\n",
-        "iTwoD[1][0]= 1\n",
-        "iTwoD[1][1]= 2\n",
-        "iTwoD[1][2]= 3\n",
-        "iTwoD[2][0]= 2\n",
-        "iTwoD[2][1]= 3\n",
-        "iTwoD[2][2]= 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page No 143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iTwoD=[[1,2,3],[4,5,6],[7,8,9]]\n",
-      "iFoundAt=[0,0]\n",
-      "\n",
-      "iValue=0\n",
-      "iFound=0\n",
-      "\n",
-      "iValue=int(raw_input(\"Enter your search value: \"))\n",
-      "\n",
-      "for x in range(3):\n",
-      "    for y in range(3):\n",
-      "        if(iTwoD[x][y]==iValue):\n",
-      "            iFound=1\n",
-      "            iFoundAt[0]=x\n",
-      "            iFoundAt[1]=y\n",
-      "            break\n",
-      "            \n",
-      "if (iFound==1):\n",
-      "    print \"Found value in iTwoD [\"+str(iFoundAt[0])+\"][\"+str(iFoundAt[1])+\"]\"\n",
-      "else:\n",
-      "    print \"Value not found\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your search value: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Found value in iTwoD [2][0]\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page No 145"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "board=range(9)\n",
-      "cWhoWon=' '\n",
-      "iCurrentPlayer=0\n",
-      "\n",
-      "def displayBoard():\n",
-      "    print \"\\n\\t|\\t|\\n\"\n",
-      "    print \"\\t|\\t|\\n\"\n",
-      "    print \"\"+str(board[0])+\"\\t|\"+str(board[1])+\"\\t|\"+str(board[2])+\"\\n\"\n",
-      "    print \"--------|-------|--------\\n\"\n",
-      "    print \"\"+str(board[3])+\"\\t|\"+str(board[4])+\"\\t|\"+str(board[5])+\"\\n\"\n",
-      "    print \"--------|-------|--------\\n\"\n",
-      "    print \"\\t|\\t|\\n\"\n",
-      "    print \"\"+str(board[6])+\"\\t|\"+str(board[7])+\"\\t|\"+str(board[8])+\"\\n\"\n",
-      "    print \"\\t|\\t|\\n\"\n",
-      "\n",
-      "def verifySelection(iSquare,iPlayer):\n",
-      "    if(board[iSquare-1]==' ' and (iPlayer==1 or iPlayer==0)):\n",
-      "        board[iSquare-1]='X'\n",
-      "        return 0\n",
-      "    elif(board[iSquare-1]==' ' and iPlayer==2):\n",
-      "        board[iSquare-1]='0'\n",
-      "        return 0\n",
-      "    else:\n",
-      "        return 1\n",
-      "    \n",
-      "def checkForWin():\n",
-      "    global cWhoWon\n",
-      "    if(board[0]=='X' and board[1]=='X' and board[2]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[3]=='X' and board[4]=='X' and board[5]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[6]=='X' and board[7]=='X' and board[8]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[0]=='X' and board[3]=='X' and board[6]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[1]=='X' and board[4]=='X' and board[7]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[2]=='X' and board[5]=='X' and board[8]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[0]=='X' and board[5]=='X' and board[8]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[2]=='X' and board[5]=='X' and board[6]=='X'):\n",
-      "        cWhoWon='X'\n",
-      "    elif(board[0]=='0' and board[1]=='0' and board[2]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    elif(board[3]=='0' and board[4]=='0' and board[5]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    elif(board[6]=='0' and board[7]=='0' and board[8]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    elif(board[0]=='0' and board[3]=='0' and board[6]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    elif(board[1]=='0' and board[4]=='0' and board[7]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    elif(board[2]=='0' and board[5]=='0' and board[8]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    elif(board[0]=='0' and board[5]=='0' and board[8]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    elif(board[2]=='0' and board[5]=='0' and board[6]=='0'):\n",
-      "        cWhoWon='0'\n",
-      "    \n",
-      "    if(cWhoWon!=' '):\n",
-      "        if(cWhoWon=='X'):\n",
-      "            print \"\\nX Wins!\\n\"\n",
-      "            return\n",
-      "        \n",
-      "        if(cWhoWon=='0'):\n",
-      "            print \"\\n0 Wins!\\n\"\n",
-      "            return\n",
-      "    else:\n",
-      "            return\n",
-      "    \n",
-      "    for x in range(9):\n",
-      "        if(board[x]!=' '):\n",
-      "            catTotal=catTotal+1\n",
-      "            \n",
-      "    if(catTotal==9):\n",
-      "        cWhoWon='C'\n",
-      "        print \"\\nCAT Game!\\n\"\n",
-      "        return\n",
-      "iSquareNum=0\n",
-      "\n",
-      "for x in range(8):\n",
-      "    board[x]=' '\n",
-      "    \n",
-      "displayBoard()\n",
-      "\n",
-      "while(cWhoWon==' '):\n",
-      "    print \"\\n\"+cWhoWon+\"\\n\"\n",
-      "    \n",
-      "    if(iCurrentPlayer==1 or iCurrentPlayer==0):\n",
-      "        print \"\\nPLAYER X\\n\"\n",
-      "        iSquareNum=int(raw_input(\"Enter an available square number (1-9): \"))\n",
-      "        \n",
-      "        if(verifySelection(iSquareNum,iCurrentPlayer)==1):\n",
-      "            iCurrentPlayer=1\n",
-      "        else:\n",
-      "            iCurrentPlayer=2\n",
-      "    else:\n",
-      "        print \"\\nPLAYER 0\\n\"\n",
-      "        iSquareNum=int(raw_input(\"Enter an available square number (1-9): \"))\n",
-      "        \n",
-      "        if(verifySelection(iSquareNum,iCurrentPlayer)==1):\n",
-      "            iCurrentPlayer=2\n",
-      "        else:\n",
-      "            iCurrentPlayer=1\n",
-      "        displayBoard()\n",
-      "        checkForWin()\n",
-      "        \n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        " \t| \t| \n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        " \t| \t| \n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        " \t| \t|8\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\n",
-        " \n",
-        "\n",
-        "\n",
-        "PLAYER X\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an available square number (1-9): 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " \n",
-        "\n",
-        "\n",
-        "PLAYER 0\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an available square number (1-9): 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "X\t|0\t| \n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        " \t| \t| \n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        " \t| \t|8\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\n",
-        " \n",
-        "\n",
-        "\n",
-        "PLAYER X\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an available square number (1-9): 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " \n",
-        "\n",
-        "\n",
-        "PLAYER 0\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an available square number (1-9): 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "X\t|0\t|0\n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        "X\t| \t| \n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        " \t| \t|8\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\n",
-        " \n",
-        "\n",
-        "\n",
-        "PLAYER X\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an available square number (1-9): 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " \n",
-        "\n",
-        "\n",
-        "PLAYER 0\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an available square number (1-9): 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "X\t|0\t|0\n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        "X\t|0\t| \n",
-        "\n",
-        "--------|-------|--------\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "X\t| \t|8\n",
-        "\n",
-        "\t|\t|\n",
-        "\n",
-        "\n",
-        "X Wins!\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter7.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter7.ipynb
deleted file mode 100755
index c1085a33..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter7.ipynb
+++ /dev/null
@@ -1,582 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:be75d77e91d7005f064cd2c560347aec768e900f2e81ef47804d4b096bc083a6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7 : Pointers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1, Page No 157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=5\n",
-      "iPtr=id(x)\n",
-      "iPtr=7"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2, Page No 157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=1\n",
-      "iPtr=id(x)\n",
-      "\n",
-      "iPtr=5\n",
-      "\n",
-      "print \"\\n*iPtr = \"+str(iPtr)+\"\\n &x = \"+str(id(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "*iPtr = 5\n",
-        " &x = 20176608\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3, Page No 158"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x=5\n",
-      "y=10\n",
-      "iPtr=None\n",
-      "\n",
-      "print \"Pointer points to: \",iPtr\n",
-      "iPtr=id(y)\n",
-      "print \"Ptr now points to: \",iPtr\n",
-      "x=iPtr\n",
-      "print \"The value of x is now\",x\n",
-      "iPtr=15\n",
-      "print \"The value of y is now\",iPtr"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pointer points to:  None\n",
-        "Ptr now points to:  20373108\n",
-        "The value of x is now 20373108\n",
-        "The value of y is now 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4, Page No 159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def addTwoNumbers(x,y):\n",
-      "    return x+y\n",
-      "\n",
-      "\n",
-      "x=0\n",
-      "y=0\n",
-      "\n",
-      "x=int(raw_input(\"Enter first number:\"))\n",
-      "y=int(raw_input(\"Enter second number:\"))\n",
-      "\n",
-      "print \"Result is: \",addTwoNumbers(x,y)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number:10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number:15\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result is:  25\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5, Page No 161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def demoPassByValue(x):\n",
-      "    x+=5\n",
-      "    print \"The value of x is:\",x\n",
-      "\n",
-      "x=0\n",
-      "x=int(raw_input(\"Enter a number:\"))\n",
-      "demoPassByValue(x)\n",
-      "print \"The original value of x did not change:\",x"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number:5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is: 10\n",
-        "The original value of x did not change: 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.6, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def demoPassByReference(x):\n",
-      "    x+=5\n",
-      "    print \"The value of x is now\",x\n",
-      "\n",
-      "x=0\n",
-      "x=int(raw_input(\"Enter a number:\"))\n",
-      "demoPassByReference(x)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number:10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of x is now 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7, Page No 164"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "iArray=[1,2,3,4,5]\n",
-      "iPtr=id(iArray)\n",
-      "print \"Address of pointer: \",iPtr\n",
-      "print \"First address of array: \",id(iArray[0])\n",
-      "print \"Pointer points to: \",iArray[0]\n",
-      "print \"First element of array contains\",iArray[0]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Address of pointer:  28621960\n",
-        "First address of array:  19652320\n",
-        "Pointer points to:  1\n",
-        "First element of array contains 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.8, Page No 165"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def nameLength(name):\n",
-      "    x=0\n",
-      "    while(name[x] != '\\0'):\n",
-      "        x+=1;\n",
-      "    return x\n",
-      "aName='\\0'\n",
-      "aName=raw_input(\"Enter your first name: \")\n",
-      "aName=aName+'\\0'\n",
-      "print \"Your first name contains\"\n",
-      "print str(nameLength(aName))+\" characters\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your first name: mehta\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your first name contains\n",
-        "5 characters\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.9, Page No 167"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def squareNumber(num):\n",
-      "    for x in range(3):\n",
-      "        num[x]=num[x]*num[x]\n",
-      "\n",
-      "iNumbers=[2,4,6]\n",
-      "print \"The current array value are:\"\n",
-      "\n",
-      "for x in range(3):\n",
-      "    print iNumbers[x]\n",
-      "\n",
-      "print \"\\n\"\n",
-      "\n",
-      "squareNumber(iNumbers)\n",
-      "\n",
-      "print \"The modified array values are :\"\n",
-      "\n",
-      "for x in range(3):\n",
-      "    print iNumbers[x]\n",
-      "\n",
-      "print \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current array value are:\n",
-        "2\n",
-        "4\n",
-        "6\n",
-        "\n",
-        "\n",
-        "The modified array values are :\n",
-        "4\n",
-        "16\n",
-        "36\n",
-        "\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.10, Page No 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def printArgument(num):\n",
-      "    print \"Read only argument is: \",num\n",
-      "    \n",
-      "iNumber=5\n",
-      "printArgument(iNumber)\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read only argument is:  5\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.11, Page No 169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def printArray(num):\n",
-      "    for x in range(3):\n",
-      "        print num[x]\n",
-      "\n",
-      "iNumbers=[2,4,6]\n",
-      "printArray(iNumbers)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n",
-        "4\n",
-        "6\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.12, Page No 170"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def modifyArray(num):\n",
-      "    for x in range(3):\n",
-      "        num[x]=num[x]*num[x]\n",
-      "        \n",
-      "iNumbers=[2,4,6]\n",
-      "modifyArray(iNumbers)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.13, Page No 173"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import random\n",
-      "def encrypt(sMessage,random):\n",
-      "    x=0\n",
-      "    while(sMessage[x]):\n",
-      "        sMessage+=str(random)\n",
-      "        x+=1\n",
-      "    x=0\n",
-      "    print \"Encrypted message is: \"\n",
-      "    \n",
-      "    while(sMessage[x]):\n",
-      "        print sMessage[x]\n",
-      "        x+=1\n",
-      "\n",
-      "def decrypt(sMessage,random):\n",
-      "    x=0\n",
-      "    while(sMessage[x]):\n",
-      "        sMessage[x]=sMessage[x]-str(random)\n",
-      "        x+=1\n",
-      "    x=0\n",
-      "    while(sMessage[x]):\n",
-      "        print sMessage[x]\n",
-      "        x+=1\n",
-      "        \n",
-      "myString=range(21)\n",
-      "iSelection=0\n",
-      "iRand= (random.randint(1,4)%4)+1\n",
-      "\n",
-      "while(iSelection != 4):\n",
-      "    print \"\\n\\n1\\t Encrypt Clear Text\\n\"\n",
-      "    print \"2\\tDecrypt Cipher Text\\n\"\n",
-      "    print \"3\\tGenerate New Key\\n\"\n",
-      "    print \"4\\tQuit\\n\"\n",
-      "    iSelection=int(raw_input(\"\\nSelect a Cryptography Option: \"))\n",
-      "    \n",
-      "    if(iSelection ==1):\n",
-      "        myString=raw_input(\"\\nEnter one word as clear text to encrypt: \")\n",
-      "        encrypt(myString,iRand)\n",
-      "        break\n",
-      "    elif(iSelection==2):\n",
-      "        myString=raw_input(\"\\nEnter cipher text to decrypt: \")\n",
-      "        decrypt(myString,iRand)\n",
-      "        break\n",
-      "    elif(iSelection==3):\n",
-      "        iRand= (random.randrange(1,4,1)%4)+1\n",
-      "        print \"\\nNew Key Generated\\n\"\n",
-      "        break\n",
-      "    else:\n",
-      "            exit(0)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "1\t Encrypt Clear Text\n",
-        "\n",
-        "2\tDecrypt Cipher Text\n",
-        "\n",
-        "3\tGenerate New Key\n",
-        "\n",
-        "4\tQuit\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Select a Cryptography Option: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter one word as clear text to encrypt: bhavin\n"
-       ]
-      }
-     ]
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter8.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter8.ipynb
deleted file mode 100755
index bca914f3..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter8.ipynb
+++ /dev/null
@@ -1,703 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2192643aa60481aa37dedf877c77d457bede1bb6978dcdba9008684a80b17ad1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8 Strings"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page No. 181"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "myString=\"MIKE\"\n",
-      "print \"The pointer variable's value is: \", id(myString)\n",
-      "print \"The pointer variable points to: \",myString\n",
-      "print \"The memory locations for each character are::\"\n",
-      "for x in range(0,4):\n",
-      "    print id(myString[x]),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The pointer variable's value is:  44562688\n",
-        "The pointer variable points to:  MIKE\n",
-        "The memory locations for each character are::\n",
-        "19548160 20083536 20086752 20285008\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2, Page No.183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "color=raw_input(\"Enter your favorite color: \")\n",
-      "print \"You entered: \",color"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your favorite color: Blue\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You entered:  Blue\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3, Page No.183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "color=raw_input(\"Enter your favorite color: \")\n",
-      "print \"You entered: \",color"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your favorite color: Black\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You entered:  Black\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4, Page No. 185"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "names = []\n",
-      "names.append(\"Michael\")\n",
-      "names.append(\"Sheila\")\n",
-      "names.append(\"Spencer\")\n",
-      "names.append(\"Hunter\")\n",
-      "names.append(\"Kenya\")\n",
-      "print \"\\nNames in pointer array of type char:\\n\"\n",
-      "for i in range(5):\n",
-      "    print names[i]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Names in pointer array of type char:\n",
-        "\n",
-        "Michael\n",
-        "Sheila\n",
-        "Spencer\n",
-        "Hunter\n",
-        "Kenya\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5, Page No.186"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "colors = []\n",
-      "print \"Enter 3 colors: \"\n",
-      "for i in range(3):\n",
-      "    colors.append(raw_input())\n",
-      "print \"Your entered: \"\n",
-      "for i in range(3):\n",
-      "    print colors[i]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 3 colors: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "blue\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "black\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "white\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your entered: \n",
-        "blue\n",
-        "black\n",
-        "white\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6, Page No.187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str1=\"123.79\"\n",
-      "str2=\"55\"\n",
-      "\n",
-      "print \"String1 is:\\\"\",str1,\"\\\"\"\n",
-      "print \"String2 is:\\\"\",str2,\"\\\"\"\n",
-      "\n",
-      "x=float(str1)\n",
-      "y=int(str2)\n",
-      "\n",
-      "print \"String 1 converted to a float is \",round(x,2)\n",
-      "print \"String 2 converted to an integer is \",y"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String1 is:\" 123.79 \"\n",
-        "String2 is:\" 55 \"\n",
-        "String 1 converted to a float is  123.79\n",
-        "String 2 converted to an integer is  55\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7, Page No.188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str1=\"37\"\n",
-      "str2=\"20\"\n",
-      "\n",
-      "print \"String1 + String2 is \", str1+str2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String1 + String2 is  3720\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8, Page No. 189"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str1=\"37\"\n",
-      "str2=\"20\"\n",
-      "iResult=int(str1)+int(str2)\n",
-      "print \"String1 + String2 is \", iResult"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String1 + String2 is  57\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9, Page No. 190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str1=\"Michael\"\n",
-      "str2=\"Vine\"\n",
-      "print \"The length of string1 is \",len(str1)\n",
-      "print \"The length of string2 is \",len(str2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The length of string1 is  7\n",
-        "The length of string2 is  4\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10, Page No.191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def convertL(char):\n",
-      "    char=char.lower()\n",
-      "    print \"The first name converted to lower case is: \",char\n",
-      "def convertU(char):\n",
-      "    char=char.upper()\n",
-      "    print \"The first name converted to Upper case is: \",char\n",
-      "name1=\"Michael\"\n",
-      "name2=\"Vine\"\n",
-      "\n",
-      "convertL(name1)\n",
-      "convertU(name2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first name converted to lower case is:  michael\n",
-        "The first name converted to Upper case is:  VINE\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11, Page No.193"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str2=\"C Language\"\n",
-      "str1=str2\n",
-      "print \"String 1 now contains \", str1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 1 now contains  C Language\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12, Page No.194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str1=\"Computer Science \"\n",
-      "str2=\"is applied mathematics\"\n",
-      "print str1+str2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Computer Science is applied mathematics\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 8.13, Page No.195"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str1=\"A\"\n",
-      "str2=\"A\"\n",
-      "str3=\"!\"\n",
-      "print str1\n",
-      "print str2\n",
-      "print str3\n",
-      "if str1==str2:\n",
-      "    print \"Letter A is equal to letter A\"\n",
-      "if str1>str3:\n",
-      "    print \"Letter A is greater than character !\"\n",
-      "if str3<str1:\n",
-      "    print \"Character ! is greater than letter A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A\n",
-        "A\n",
-        "!\n",
-        "Letter A is equal to letter A\n",
-        "Letter A is greater than character !\n",
-        "Character ! is greater than letter A\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 8.14, Page No.197"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "str1=\"Analyzing strings with with the strstr() function\"\n",
-      "str2=\"ing\"\n",
-      "str3=\"xyz\"\n",
-      "\n",
-      "print str1\n",
-      "print str2\n",
-      "print str3\n",
-      "\n",
-      "if((str1.find(str2))!=-1):\n",
-      "    print \"Str2 was found in str1\"\n",
-      "else: \n",
-      "    print \"Str2 was not found in str1\"\n",
-      "\n",
-      "if((str1.find(str3))!=-1):\n",
-      "     print \"Str3 was found in str1\"\n",
-      "else: \n",
-      "    print \"Str3 was not found in str1\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Analyzing strings with with the strstr() function\n",
-        "ing\n",
-        "xyz\n",
-        "Str2 was found in str1\n",
-        "Str3 was not found in str1\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 8.15, Page No.198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import time\n",
-      "import os\n",
-      "from array import array\n",
-      "\n",
-      "\n",
-      "def checkAnswer(string1,string2):\n",
-      "    for x in range(len(string2)):\n",
-      "        string2=string2.upper()\n",
-      "    if(set(string2).intersection(string1.split())):\n",
-      "        print \"Great job\"\n",
-      "    else :\n",
-      "        print \"Sorry no word found...\"\n",
-      "   \n",
-      "starGame=[]\n",
-      "starGame.append(\"ADELANGUAGEFERVZOPIBMOU\")\n",
-      "starGame.append(\"ZBPOINTERSKLMLOOPMNOCOT\")\n",
-      "starGame.append(\"PODSTRINGGDIWHIEEICERLS\")\n",
-      "starGame.append(\"YVCPROGRAMMERWQKNULTHMD\")\n",
-      "starGame.append(\"UKUNIXFIMWXIZEQZINPUTEX\")\n",
-      "\n",
-      "displayed=0\n",
-      "startTime=0\n",
-      "os.system('cls')\n",
-      "print \"word find\"\n",
-      "startTime=int(round(time.time()*1000))\n",
-      "for x in range(0,5):\n",
-      "    while ((startTime+3)>int(round(time.time()*1000))):\n",
-      "        if displayed==0:\n",
-      "            print \"find a word in:\"\n",
-      "            print \"\",starGame[x]\n",
-      "            displayed=1\n",
-      "    os.system('cls')\n",
-      "    answer=raw_input(\"Enter word found:\")\n",
-      "    checkAnswer(starGame[x],answer)\n",
-      "    displayed=0\n",
-      "    startTime=int(round(time.time()*1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "word find\n",
-        "find a word in:\n",
-        " ADELANGUAGEFERVZOPIBMOU\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter word found:U\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorry no word found...\n",
-        "find a word in:\n",
-        " ZBPOINTERSKLMLOOPMNOCOT\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter word found:T\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorry no word found...\n",
-        "find a word in:\n",
-        " PODSTRINGGDIWHIEEICERLS\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter word found:S\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorry no word found...\n",
-        "find a word in:\n",
-        " YVCPROGRAMMERWQKNULTHMD\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter word found:YV\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorry no word found...\n",
-        "find a word in:\n",
-        " UKUNIXFIMWXIZEQZINPUTEX\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter word found:UKU\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorry no word found...\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/Chapter9.ipynb b/C_Programming_for_the_Absolute_Beginner/Chapter9.ipynb
deleted file mode 100755
index 583e1e8f..00000000
--- a/C_Programming_for_the_Absolute_Beginner/Chapter9.ipynb
+++ /dev/null
@@ -1,549 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c7a7a5fed3d237b23773c146200cb1c66c8b51b03326b1d931d288bf060c18a1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9 : Introduction to Data Structures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1, Page No 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class employee:\n",
-      "    def __init__(self,fname,lname,id1,salary):\n",
-      "        self.fname = fname\n",
-      "        self.lname = lname\n",
-      "        self.id1 = id1\n",
-      "        self.salary = salary\n",
-      "if __name__ == '__main__':\n",
-      "    emp1 = employee(\"Michael\",\"Vine\",123,50000.00)\n",
-      "    print \"\\n First name: \" + emp1.fname + \"\\n\"\n",
-      "    print \"Last name: \" + emp1.lname + \"\\n\"\n",
-      "    print \"Employee ID: \" + str(emp1.id1) + \"\\n\"\n",
-      "    print \"Salary: \" + str(emp1.salary) + \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " First name: Michael\n",
-        "\n",
-        "Last name: Vine\n",
-        "\n",
-        "Employee ID: 123\n",
-        "\n",
-        "Salary: 50000.0\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2, Page No 206"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class employee:\n",
-      "    def __init__(self,fname,lname,id1,salary):\n",
-      "        self.fname = fname\n",
-      "        self.lname = lname\n",
-      "        self.id1 = id1\n",
-      "        self.salary = salary\n",
-      "if __name__ == '__main__':\n",
-      "    emp1 = employee(\"Michael\",\"Vine\",123,50000.00)\n",
-      "    print \"\\n First name: \" + emp1.fname + \"\\n\"\n",
-      "    print \"Last name: \" + emp1.lname + \"\\n\"\n",
-      "    print \"Employee ID: \" + str(emp1.id1) + \"\\n\"\n",
-      "    print \"Salary: \" + str(emp1.salary) + \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " First name: Michael\n",
-        "\n",
-        "Last name: Vine\n",
-        "\n",
-        "Employee ID: 123\n",
-        "\n",
-        "Salary: 50000.0\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3, Page No 209"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class scores:\n",
-      "    def __init__(self,name,score):\n",
-      "        self.name = name\n",
-      "        self.score = score\n",
-      "if __name__ == '__main__':\n",
-      "    highScores = []\n",
-      "    highScores.append(scores(\"Hunter\",40768))\n",
-      "    highScores.append(scores(\"Kenya\",38568))\n",
-      "    highScores.append(scores(\"Apollo\",35985))\n",
-      "    for x in range(3):\n",
-      "        print \"\\n\" + highScores[x].name + \"\\t\" + str(highScores[x].score) + \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Hunter\t40768\n",
-        "\n",
-        "\n",
-        "Kenya\t38568\n",
-        "\n",
-        "\n",
-        "Apollo\t35985\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4, Page No 211"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class employee:\n",
-      "    def __init__(self,id1,name,salary):\n",
-      "        self.id1 = id1\n",
-      "        self.name = name\n",
-      "        self.salary = salary\n",
-      "def processEmp(emp1):\n",
-      "    emp1.id1 = 123\n",
-      "    emp1.name = \"Sheila\"\n",
-      "    emp1.salary = 65000.00\n",
-      "if __name__ == '__main__':\n",
-      "    emp1 = employee(0,0,0)\n",
-      "    processEmp(emp1)\n",
-      "    print \"\\n Id : \" + str(emp1.id1) + \"\\n\"\n",
-      "    print \"Name : \" + emp1.name + \"\\n\"\n",
-      "    print \"Salary : \" + str(emp1.salary) + \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Id : 123\n",
-        "\n",
-        "Name : Sheila\n",
-        "\n",
-        "Salary : 65000.0\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5, Page No 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class player:\n",
-      "    def __init__(self,name,score):\n",
-      "        self.name = name\n",
-      "        self.score = score\n",
-      "if __name__ == '__main__':\n",
-      "    aPlayer = player(0,0)\n",
-      "    ptrPlayer = id(aPlayer)\n",
-      "    aPlayer.name = \"Pinball Wizard\"\n",
-      "    aPlayer.score = 1000000.00\n",
-      "    print \"\\nPlayer : \" + aPlayer.name + \"\\n\"\n",
-      "    print \"Score : \" + str(aPlayer.score)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Player : Pinball Wizard\n",
-        "\n",
-        "Score : 1000000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6, Page No 213"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class employee:\n",
-      "    def __init__(self,id1,name,salary):\n",
-      "        self.id1 = id1\n",
-      "        self.name = name\n",
-      "        self.salary = salary\n",
-      "def processEmp(emp1):\n",
-      "    emp1.id1 = 123\n",
-      "    emp1.name = \"Sheila\"\n",
-      "    emp1.salary = 65000.00\n",
-      "if __name__ == '__main__':\n",
-      "    emp1 = employee(0,0,0)\n",
-      "    ptrEmp = id(emp1)\n",
-      "    processEmp(emp1)\n",
-      "    print \"\\nId : \" + str(emp1.id1) + \"\\n\"\n",
-      "    print \"Name : \" + emp1.name + \"\\n\"\n",
-      "    print \"Salary : \" + str(emp1.salary) + \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Id : 123\n",
-        "\n",
-        "Name : Sheila\n",
-        "\n",
-        "Salary : 65000.0\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7, Page No 215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class employee:\n",
-      "    def __init__(self,id1,name,salary):\n",
-      "        self.id1 = id1\n",
-      "        self.name = name\n",
-      "        self.salary = salary\n",
-      "def processEmp(emp1 = []):\n",
-      "    emp1.append(employee(123,\"Sheila\",65000.00))\n",
-      "    emp1.append(employee(234,\"Hunter\",28000.00))\n",
-      "    emp1.append(employee(456,\"Kenya\",48000.00))\n",
-      "    return emp1\n",
-      "if __name__ == '__main__':\n",
-      "    emp1 = []\n",
-      "    emp1 = employee(0,0,0)\n",
-      "    emp1 = processEmp(emp1 = [])\n",
-      "    for x in range(3):\n",
-      "        print \"\\nId : \" + str(emp1[x].id1) + \"\\n\" \n",
-      "        print \"Name : \" + emp1[x].name + \"\\n\"\n",
-      "        print \"Salary : \" + str(emp1[x].salary)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Id : 123\n",
-        "\n",
-        "Name : Sheila\n",
-        "\n",
-        "Salary : 65000.0\n",
-        "\n",
-        "Id : 234\n",
-        "\n",
-        "Name : Hunter\n",
-        "\n",
-        "Salary : 28000.0\n",
-        "\n",
-        "Id : 456\n",
-        "\n",
-        "Name : Kenya\n",
-        "\n",
-        "Salary : 48000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8, Page No 217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "class phonebook:\n",
-      "    def __init__(self,name,number,address):\n",
-      "        self.name = name\n",
-      "        self.number = number\n",
-      "        self.address = address\n",
-      "class magazine:\n",
-      "    def __init__(self,name,author,isbn):\n",
-      "        self.name = name\n",
-      "        self.author = author\n",
-      "        self.isbn = isbn\n",
-      "if __name__ == '__main__':\n",
-      "    aBook = phonebook(\"John\",123,\"John Smith\")\n",
-      "    aMagazine = magazine(\"C Programming\",\"Michael Vine\",1-59863-480-1)\n",
-      "    print \"\\nUnion Details\\n\"\n",
-      "    print \"Address for aBook.name: \" +  str(id(aBook.name)) + \"\\n\"\n",
-      "    print \"Address for aBook.number: \", str(id(aBook.number)) + \"\\n\"\n",
-      "    print \"Address for aBook.address: \", str(id(aBook.address)) + \"\\n\"\n",
-      "    print \"\\nStructure Details\\n\"\n",
-      "    print \"Address for aMagazine.name: \", str(id(aMagazine.name)) + \"\\n\"\n",
-      "    print \"Address for aMagazine.author: \", str(id(aMagazine.author)) + \"\\n\"\n",
-      "    print \"Address for aMagazine.isbn: \", str(id(aMagazine.isbn)) + \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        "Union Details\n",
-        "\n",
-        "Address for aBook.name: 66178424\n",
-        "\n",
-        "Address for aBook.number:  30796656\n",
-        "\n",
-        "Address for aBook.address:  66151792\n",
-        "\n",
-        "\n",
-        "Structure Details\n",
-        "\n",
-        "Address for aMagazine.name:  66151744\n",
-        "\n",
-        "Address for aMagazine.author:  66150976\n",
-        "\n",
-        "Address for aMagazine.isbn:  64689272\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9, Page No 219"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "x = 12\n",
-      "y = 5\n",
-      "print \"\\nWithout Type-Casting\\n\"\n",
-      "print \"12 \\\\ 5 = \" + str(x/y) + \"\\n\"\n",
-      "print \"\\nWith Type-Casting\\n\"\n",
-      "print \"12 \\\\ 5 = \" + str(float(x)/float(y)) + \"\\n\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Without Type-Casting\n",
-        "\n",
-        "12 \\ 5 = 2\n",
-        "\n",
-        "\n",
-        "With Type-Casting\n",
-        "\n",
-        "12 \\ 5 = 2.4\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10, Page No 220"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "number = 86\n",
-      "letter = 'M'\n",
-      "print \"\\n86 type-casted to char is: \" + chr(number) + \"\\n\"\n",
-      "print \"\\n'M' type-casted to int is: \" + str(ord(letter)) + \"\\n \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "86 type-casted to char is: V\n",
-        "\n",
-        "\n",
-        "'M' type-casted to int is: 77\n",
-        " \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.11, Page No 222"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import random\n",
-      "class dec:\n",
-      "    def __init__(self,type1,used,value):\n",
-      "        self.type1 = type1\n",
-      "        self.used = used\n",
-      "        self.value = value\n",
-      "def shuffel(thisDeck):\n",
-      "    found = 0\n",
-      "    print \"\\nYour five cards are: \\n\\n\"\n",
-      "    while found < 5:\n",
-      "        #iRnd = random.randint(1, 100) % 51 when we use this it gives error out of range\n",
-      "        if thisDeck[iRnd].used == 'n':\n",
-      "            if thisDeck[iRnd].value == 12:\n",
-      "                print \"Ace of : \" + str(thisDeck[iRnd].type1) + \"\\n\"\n",
-      "            elif thisDeck[iRnd].value == 11:\n",
-      "                print \"King of : \" + str(thisDeck[iRnd].type1) + \"\\n\"\n",
-      "            elif thisDeck[iRnd].value == 10:\n",
-      "                print \"Queen of : \" + str(thisDeck[iRnd].type1) + \"\\n\"\n",
-      "            elif thisDeck[iRnd].value == 9:\n",
-      "                print \"Jack of : \" + str(thisDeck[iRnd].type1) + \"\\n\"\n",
-      "            else:\n",
-      "                print str(thisDeck[iRnd].value + 2) + \"of\"\n",
-      "                print str(thisDeck[iRnd].type1) + \"\\n\"     \n",
-      "if __name__ == '__main__':\n",
-      "    myDeck = []\n",
-      "    for x in range(3):\n",
-      "        for y  in range(13):\n",
-      "            if x==0:\n",
-      "                myDeck.append(dec(\"dimonds\",\"y\",\"n\"))\n",
-      "            elif x==1:\n",
-      "                myDeck.append(dec(\"clubs\",\"y\",\"n\"))\n",
-      "            elif x==2:\n",
-      "                myDeck.append(dec(\"hearts\",\"y\",\"n\"))\n",
-      "            elif x==3:\n",
-      "                myDeck.append(dec(\"spades\",\"y\",\"n\"))\n",
-      "    shuffel(myDeck)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner/README.txt b/C_Programming_for_the_Absolute_Beginner/README.txt
deleted file mode 100755
index 9bf2614f..00000000
--- a/C_Programming_for_the_Absolute_Beginner/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Bhavin Mehta
-Course: mca
-College/Institute/Organization: Shri Brahmanand Institute of Computer Science - Chaparda
-Department/Designation: Assistant Professor
-Book Title: C Programming for the Absolute Beginner
-Author: Michael Vine
-Publisher: Thomson Course Technology United States of America
-Year of publication: 2008
-Isbn: 1-59863-480-1
-Edition: 2nd edition
\ No newline at end of file
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diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter10_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter10_1.ipynb
new file mode 100755
index 00000000..c2e52854
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter10_1.ipynb
@@ -0,0 +1,362 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:af2735a34ebb2cb3c8a379aacf70cb46d8ee4a8ed09afacd9b5d0d21e0a0dd5a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Dynamic Memory Allocation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1, Page No 229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "class employee:\n",
+      "    def __init__(self,id1,name,salary):\n",
+      "        self.id1 = id1\n",
+      "        self.name = name\n",
+      "        self.salary = salary\n",
+      "e = employee(10,\"abc\",12000.02)\n",
+      "x = 1\n",
+      "f = 1.1\n",
+      "d = 1111.11\n",
+      "c = 'C'\n",
+      "print \"Sizo of integer : \" + str(sys.getsizeof(x)) + \"bytes \\n\"\n",
+      "print \"Sizo of float : \" + str(sys.getsizeof(f)) + \"bytes \\n\"\n",
+      "print \"Sizo of double : \" + str(sys.getsizeof(d)) + \"bytes \\n\"\n",
+      "print \"Sizo of char : \" + str(sys.getsizeof(c)) + \"bytes \\n\"\n",
+      "print \"Sizo of employee structure : \" + str(sys.getsizeof(e)) + \"bytes \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sizo of integer : 24bytes \n",
+        "\n",
+        "Sizo of float : 24bytes \n",
+        "\n",
+        "Sizo of double : 24bytes \n",
+        "\n",
+        "Sizo of char : 34bytes \n",
+        "\n",
+        "Sizo of employee structure : 64bytes \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2, Page No 230"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "array = []\n",
+      "\n",
+      "print \"\\nSize of array: \" + str(sys.getsizeof(array)) +  \"bytes\\n\"\n",
+      "print \"Number of elements in array \"\n",
+      "print str(sys.getsizeof(array) / sys.getsizeof(int)) + \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Size of array: 64bytes\n",
+        "\n",
+        "Number of elements in array \n",
+        "0\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3, Page No 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4, Page No 232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5, Page No 232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6, Page No 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function\n",
+      "#if name is None:\n",
+      "#    print \"\\nOut of memory!\\n\"\n",
+      "#else:\n",
+      "#    print \"\\nMemory allocated.\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7, Page No 234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function\n",
+      "#if name is not None:\n",
+      "#    name = raw_input(\"Enter your name: \")\n",
+      "#    print \"\\nHi \" + name + \"\\n\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.8, Page No 235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function\n",
+      "#if name is not None:\n",
+      "#    name = raw_input(\"Enter your name: \")\n",
+      "#    print \"\\nHi \" + name + \"\\n\"\n",
+      "#    del name"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.9, Page No 236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function\n",
+      "#if numbers is None:\n",
+      "#    return\n",
+      "#numbers[0] = 100\n",
+      "#numbers[1] = 200\n",
+      "#numbers[2] = 300\n",
+      "#numbers[3] = 400\n",
+      "#numbers[4] = 500\n",
+      "#print \"\\nIndividual memory segments initialized to:\\n\"\n",
+      "#for x in range(5):\n",
+      "#    print \"numbers[\" +x+\"] = \\n\" + numbers[x]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.10, Page No 237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no calloc function\n",
+      "#if numbers is None:\n",
+      "#    return"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.11, Page No 239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# there is no malloc function\n",
+      "#if numbers is None:\n",
+      "#    print \"\\nOut of memory!\\n\"\n",
+      "#    return\n",
+      "#else:\n",
+      "#    number = newNumber\n",
+      "#    for x in range(10):\n",
+      "#        numbers[x] = x * 10\n",
+      "#    print \"\\nExpanded memory:\\n\"\n",
+      "#    for x in range(10):\n",
+      "#        print \"numbers[\" +x+\"] = \\n\" + numbers[x]\n",
+      "#    del number"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.12, Page No 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "# there is no calloc function\n",
+      "#time\n",
+      "#print \"\\nMath Quiz\\n\\n\"\n",
+      "#response = raw_input(\"Enter # of problems: \")\n",
+      "#if op1 is None || op2 is None || answer is None ||result is None:\n",
+      "#    print \"\\nOut of Memory!\\n\"\n",
+      "#    return\n",
+      "#for x in range(response):\n",
+      "#    op1[x] = random.randint(1,200) % 100\n",
+      "#    op2[x] = random.randint(1,200) % 100\n",
+      "#    print \"\\n\" + op1[x] + \"+\" + op2[x]\"\n",
+      "#    answer[x] = raw_input(\"\")\n",
+      "#    if answer[x] == op1[x] + op2[x]:\n",
+      "#        result[x] = 'c'\n",
+      "#    else:\n",
+      "#        result[x] = 'i'\n",
+      "#print \"\\nQuiz Results\\n\"\n",
+      "#print \"\\nQuestion\\tYour Answer\\tCorrect\\n\"\n",
+      "#for x in range(response):\n",
+      "#    if result[x] == 'c':\n",
+      "#        print op1[x] + \"+\" + op2[x] + \"\\t\\t\" + answer[x] + \"\\t\\tYes\\n\"\n",
+      "#    else:\n",
+      "#        print op1[x] + \"+\" + op2[x] + \"\\t\\t\" + answer[x] + \"\\t\\tNo\\n\"\n",
+      "#del op1\n",
+      "#del op2\n",
+      "#del answer\n",
+      "#del result   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter11_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter11_1.ipynb
new file mode 100755
index 00000000..239b6ca4
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter11_1.ipynb
@@ -0,0 +1,616 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:9d02541e3ea8d7af1efd977534b7150b993557acd6da8dc30b83dcacb798d681"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : File Input and Output"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1, Page No 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "try:\n",
+      "    pRead = open(\"file1.dat\",\"r\")\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "\n",
+      "\"\"\"\n",
+      "Input file:- file1.dat\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page No 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "try:\n",
+      "    pRead = open(\"file1.dat\",\"r\")\n",
+      "    if(not isinstance(pRead,file)):\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "    else:\n",
+      "        print \"\\nFile opened for reading\\n\"\n",
+      "    pRead.close()\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "\n",
+      "\"\"\"\n",
+      "Input file:- file1.dat\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "File opened for reading\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3, Page No 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "try:\n",
+      "    pRead = open(\"names.dat\",\"r\")\n",
+      "    if(not isinstance(pRead,file)):\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "    else:\n",
+      "        print \"\\nContents of names.dat\\n\"\n",
+      "        while(1):\n",
+      "            t = pRead.readline() \n",
+      "            if(t != \"\"):\n",
+      "                print t\n",
+      "            else:\n",
+      "                break\n",
+      "    pRead.close()\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "    \n",
+      "\"\"\" \n",
+      "Input File :- names.dat\n",
+      "\n",
+      "Michael\n",
+      "Sheila\n",
+      "Spencer\n",
+      "Olivia\n",
+      "\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Contents of names.dat\n",
+        "\n",
+        "Michael\n",
+        "\n",
+        "Sheila\n",
+        "\n",
+        "Spencer\n",
+        "\n",
+        "Olivia\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4, Page No 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "try:\n",
+      "    pRead = open(\"hobbies.dat\",\"r\")\n",
+      "    if(not isinstance(pRead,file)):\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "    else:\n",
+      "        print \"\\nName\\tHobby\\n\\n\"\n",
+      "        while(1):\n",
+      "            name = pRead.readline()\n",
+      "            hobby = pRead.readline()\n",
+      "            if(name != \"\"):\n",
+      "                print name + \"\\t\"\n",
+      "                print hobby\n",
+      "            else:\n",
+      "                break\n",
+      "    pRead.close()\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "    \n",
+      "\"\"\"\n",
+      "Input file :- hobbies.dat\n",
+      "\n",
+      "Michael\tProgramming\n",
+      "Sheila\tShopping\n",
+      "Spencer\tFootball\n",
+      "Olivia\tDancing\n",
+      "Waytt\tEating\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name\tHobby\n",
+        "\n",
+        "\n",
+        "Michael\tProgramming\n",
+        "\t\n",
+        "Sheila\tShopping\n",
+        "\n",
+        "Spencer\tFootball\n",
+        "\t\n",
+        "Olivia\tDancing\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5, Page No 256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "try:\n",
+      "    pWrite = open(\"students.dat\",\"w\")\n",
+      "    if(not isinstance(pWrite,file)):\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "    else:\n",
+      "        fname = raw_input(\"Enter first name: \")\n",
+      "        lname = raw_input(\"Enter last name: \")\n",
+      "        id1 = raw_input(\"Enter id: \")\n",
+      "        gpa = float(raw_input(\"Enter GPA: \"))\n",
+      "        pWrite.writelines(fname + \" \" + lname + \"\\t\" + id1 + \"\\t\" + str(gpa))\n",
+      "    pWrite.close()\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "\n",
+      "\"\"\"\n",
+      "Output file:- students.dat\n",
+      "Patric Star\t888-66-9999\t2.0\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first name: Patric\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter last name: Star\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter id: 888-66-9999\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter GPA: 2.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6, Page No 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "try:\n",
+      "    pRead = open(\"students.dat\",\"r\")\n",
+      "    if(not isinstance(pRead,file)):\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "    else:\n",
+      "        print \"\\nName\\t\\tID\\t\\tGPA\\n\\n\"\n",
+      "        while(1):\n",
+      "            fname = pRead.read()\n",
+      "            lname = pRead.read()\n",
+      "            id1 = pRead.read()\n",
+      "            gpa = pRead.read()\n",
+      "            \n",
+      "            if(fname != \"\"):\n",
+      "                print fname + lname + \"\\t\\t\" + id1 + \"\\t\" + str(gpa)\n",
+      "            else:\n",
+      "                break\n",
+      "    pRead.close()\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "\n",
+      "\"\"\"\n",
+      "Input file:- students.dat\n",
+      "Patric Star\t888-66-9999\t2.0\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name\t\tID\t\tGPA\n",
+        "\n",
+        "\n",
+        "Patric Star\t888-66-9999\t2.0\t\t\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7, Page No 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def readData():\n",
+      "    try:\n",
+      "        pRead = open(\"hobbies.dat\",\"r\")\n",
+      "        if(not isinstance(pRead,file)):\n",
+      "            print \"\\nFile can not be opened\\n\"\n",
+      "        else:\n",
+      "            print \"\\nName\\tHobby\\n\\n\"\n",
+      "            while(1):\n",
+      "                name = pRead.readline()\n",
+      "                hobby = pRead.readline()\n",
+      "                if(name != \"\"):\n",
+      "                    print name + \"\\t\"\n",
+      "                    print hobby\n",
+      "                else:\n",
+      "                    break\n",
+      "        pRead.close()\n",
+      "    except IOError:\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "try:\n",
+      "    print \"Current file contains\"\n",
+      "    readData()\n",
+      "    print \"\\n\"\n",
+      "    name = raw_input(\"Enter a new name: \")\n",
+      "    hobby = raw_input(\"Enter a new hobby: \")\n",
+      "    pWrite = open(\"hobbies.dat\",\"a\")\n",
+      "    if(not isinstance(pWrite,file)):\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "    else:\n",
+      "        pWrite.writelines(\"\\n\" + name + \"\\t\" + hobby)\n",
+      "    pWrite.close()\n",
+      "    readData()\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "    \n",
+      "\"\"\"\n",
+      "Input file and Output file:- hobbies.dat\n",
+      "Michael\tProgramming\n",
+      "Sheila\tShopping\n",
+      "Spencer\tFootball\n",
+      "Olivia\tDancing\n",
+      "Waytt\tEating\n",
+      "\"\"\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current file contains\n",
+        "\n",
+        "Name\tHobby\n",
+        "\n",
+        "\n",
+        "Michael\tProgramming\n",
+        "\t\n",
+        "Sheila\tShopping\n",
+        "\n",
+        "Spencer\tFootball\n",
+        "\t\n",
+        "Olivia\tDancing\n",
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a new name: Waytt\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a new hobby: Eating\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name\tHobby\n",
+        "\n",
+        "\n",
+        "Michael\tProgramming\n",
+        "\t\n",
+        "Sheila\tShopping\n",
+        "\n",
+        "Spencer\tFootball\n",
+        "\t\n",
+        "Olivia\tDancing\n",
+        "\n",
+        "Waytt\tEating\t\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8, Page No 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "try:\n",
+      "    pRead = open(\"hobbies.dat\",\"r\")\n",
+      "    if(not isinstance(pRead,file)):\n",
+      "        print \"\\nFile can not be opened\\n\"\n",
+      "        print \"The following error occurred\" #goto label\n",
+      "        sys.exit(\"Some Error Occures\")\n",
+      "    else:\n",
+      "        print \"\\nName\\tHobby\\n\\n\"\n",
+      "        while(1):\n",
+      "            name = pRead.readline()\n",
+      "            hobby = pRead.readline()\n",
+      "            if(name != \"\"):\n",
+      "                print name + \"\\t\"\n",
+      "                print hobby\n",
+      "            else:\n",
+      "                break\n",
+      "    pRead.close()\n",
+      "    #sys.exit(\"Exit\")\n",
+      "except IOError:\n",
+      "    print \"\\nFile can not be opened\\n\"\n",
+      "\n",
+      "\"\"\"\n",
+      "Input file:- hobbies.dat\n",
+      "Michael\tProgramming\n",
+      "Sheila\tShopping\n",
+      "Spencer\tFootball\n",
+      "Olivia\tDancing\n",
+      "Waytt\tEating\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name\tHobby\n",
+        "\n",
+        "\n",
+        "Michael\tProgramming\n",
+        "\t\n",
+        "Sheila\tShopping\n",
+        "\n",
+        "Spencer\tFootball\n",
+        "\t\n",
+        "Olivia\tDancing\n",
+        "\n",
+        "Waytt\tEating\t\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9, Page No 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"\\n\\tPhone Book\\n\"\n",
+      "print \"\\n 1 \\t Add phone book entry \\n\"\n",
+      "print \"2 \\t Print phone book \\n\\n\"\n",
+      "print \"Select an option: \"\n",
+      "response = int(raw_input(\"Select an option: \"))\n",
+      "print response\n",
+      "if(response == 1):\n",
+      "    fname = raw_input(\"Enter first name: \")\n",
+      "    lname = raw_input(\"Enter last name: \")\n",
+      "    number = raw_input(\"Enter phone number: \")\n",
+      "    pWrite = open(\"phone_book.dat\",\"a\")\n",
+      "    if(isinstance(pWrite,file)):\n",
+      "       pWrite.writelines(fname + \" \" + lname + \" \" + number + \"\\n\")\n",
+      "       pWrite.close()\n",
+      "    else:\n",
+      "        print \"The error occured\" #go to\n",
+      "elif(response == 2):\n",
+      "    pRead = open(\"phone_book.dat\",\"r\")\n",
+      "    if(isinstance(pRead,file)):\n",
+      "        while(1):\n",
+      "            fname = pRead.read()\n",
+      "            if(fname == \"\"):\n",
+      "                break\n",
+      "            lname = pRead.read()\n",
+      "            number = pRead.read()\n",
+      "            print fname\n",
+      "            print lname\n",
+      "            print number\n",
+      "        print \"\\n\"\n",
+      "    else:\n",
+      "        print \"The error occured\" #go to\n",
+      "else:\n",
+      "    print \"Invalid Selection\"\n",
+      "\n",
+      "\"\"\"\n",
+      "Input file and Output file:- phone_book.dat\n",
+      "John Smith 538.676.1234\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\tPhone Book\n",
+        "\n",
+        "\n",
+        " 1 \t Add phone book entry \n",
+        "\n",
+        "2 \t Print phone book \n",
+        "\n",
+        "\n",
+        "Select an option: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Select an option: 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n",
+        "John Smith 538.676.1234\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter12_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter12_1.ipynb
new file mode 100755
index 00000000..76d70f5c
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter12_1.ipynb
@@ -0,0 +1,445 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b5b5db464e530d21c4b695ff4bde6912d0e8c447d51f1ede515f161acac3856f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : The C PreProcessor"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1, Page No 272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"\\nHaving fun with preprocessor directives\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Having fun with preprocessor directives\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2, Page No 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "NUMBER = 7\n",
+      "print \"\\nLucky Number \", NUMBER , \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Lucky Number  7 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3, Page No 274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "NUMBER = 7\n",
+      "print \"\\nLucky Number \", NUMBER , \"\\n\"\n",
+      "NUMBER = 5"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Lucky Number  7 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4, Page No 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def area(l,w):\n",
+      "    return l * w\n",
+      "length = 0\n",
+      "width = 0\n",
+      "length = int(raw_input(\"\\nEnter length: \"))\n",
+      "width = int(raw_input(\"\\nEnter width: \"))\n",
+      "print \"\\nArea of rectangle = \" , area(length,width) , \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter length: 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter width: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Area of rectangle =  84 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5, Page No 276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def RESULT(x,y):\n",
+      "    print \"\\nResult is \\n\", x+y\n",
+      "num1 = 0\n",
+      "num2 = 0\n",
+      "num1 = int(raw_input(\"\\nEnter first number: \"))\n",
+      "num2 = int(raw_input(\"\\nEnter second number: \"))\n",
+      "RESULT(num1,num2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter first number: 27\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter second number: 44\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Result is \n",
+        "71\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6, Page No 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def RESULT(x,y):\n",
+      "    print \"\\nResult is \\n\", x+y\n",
+      "operand1 = 0\n",
+      "operand2 = 0\n",
+      "operand1 = int(raw_input(\"\\nEnter first operand: \"))\n",
+      "operand2 = int(raw_input(\"\\nEnter second operand: \"))\n",
+      "RESULT(operand1,operand2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter first operand: 27\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter second operand: 44\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Result is \n",
+        "71\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7, Page No 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import profit\n",
+      "print \"\\nThe Profit Program\\n\"\n",
+      "price = float(raw_input(\"\\nEnter unit price: \"))\n",
+      "quantity = int(raw_input(\"Enter quantity sold: \"))\n",
+      "totalCost = float(raw_input(\"Enter total cost: \"))\n",
+      "profit1(price,quantity,totalCost)\n",
+      "\n",
+      "\"\"\"\n",
+      "Header file:- profit.py\n",
+      "def profit(p,q,tc):\n",
+      "    print \"\\nYour profit is \",(p * q) - tc,\"\\n\"\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Profit Program\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter unit price: 12.99\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter quantity sold: 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter total cost: 56\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Your profit is  34.93 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8, Page No 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import calculate\n",
+      "selection = 0\n",
+      "print \"\\n The Function Wizard \\n\"\n",
+      "print \"\\n1\\tDetermine perimeter of ractangle\\n\"\n",
+      "print \"\\n2\\tDetermine area of rectangle\\n\"\n",
+      "print \"\\n3\\tDetermine volume of rectangle\\n\"\n",
+      "selection = int(raw_input(\"\\nEnter selection : \"))\n",
+      "if selection == 1:\n",
+      "    l = float(raw_input(\"Enter length: \"))\n",
+      "    w = float(raw_input(\"Enter width: \"))\n",
+      "    perimeter(l,w)\n",
+      "elif selection == 2:\n",
+      "    l = float(raw_input(\"Enter length: \"))\n",
+      "    w = float(raw_input(\"Enter width: \"))\n",
+      "    area(l,w)\n",
+      "elif selection == 3:\n",
+      "    l = float(raw_input(\"Enter length: \"))\n",
+      "    w = float(raw_input(\"Enter width: \"))\n",
+      "    h = float(raw_input(\"Enter height: \"))\n",
+      "    volume(l,w,h)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The Function Wizard \n",
+        "\n",
+        "\n",
+        "1\tDetermine perimeter of ractangle\n",
+        "\n",
+        "\n",
+        "2\tDetermine area of rectangle\n",
+        "\n",
+        "\n",
+        "3\tDetermine volume of rectangle\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter selection : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter length: 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter width: 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter height: 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Perimeter is  6.0 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter5_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter5_1.ipynb
new file mode 100755
index 00000000..ac86dba8
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter5_1.ipynb
@@ -0,0 +1,475 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:22dbf664bff583df60a7371a6f741be85c976f32fe9bef5f00f146ca5848d933"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Structured Programming"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page No 116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def addTwoNumbers(operand1,operand2):\n",
+      "    return operand1+operand2\n",
+      "\n",
+      "print \"Nothing happenning here\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Nothing happenning here\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page No 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def addTwoNumbers(num1,num2):\n",
+      "    return num1+num2\n",
+      "\n",
+      "def subtractTwoNumber(num1,num2):\n",
+      "    return num1-num2\n",
+      "\n",
+      "print \"Nothing happenning here\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Nothing happenning here\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3, Page No 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def addTwoNumbers(operand1,operand2):\n",
+      "    return operand1+operand2\n",
+      "\n",
+      "\n",
+      "iResult= addTwoNumbers(5,5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page No 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def addTwoNumbers(operand1,operand2):\n",
+      "    return operand1+operand2\n",
+      "\n",
+      "print \"The result is : \",addTwoNumbers(5,5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The result is :  10\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5, Page No 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def addTwoNumbers(operand1,operand2):\n",
+      "    return operand1+operand2\n",
+      "\n",
+      "num1=int(raw_input(\"Enter the first number: \"))\n",
+      "num2=int(raw_input(\"Enter the second number: \"))\n",
+      "print \"The result is: \",addTwoNumbers(num1,num2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the first number: 57\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the second number: 43\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The result is:  100\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6, Page No 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def printReportHeader():\n",
+      "    print \"\\n Column1 \\t Column2 \\t Column3 \\t Column4 \\n\"\n",
+      "    \n",
+      "printReportHeader()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Column1 \t Column2 \t Column3 \t Column4 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page No 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "num1=int(raw_input(\"Enter a number: \"))\n",
+      "print \"You entered \",num1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You entered  10\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8, Page No 123"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def getSecondNumber():\n",
+      "    num1=int(raw_input(\"Enter Second number: \"))\n",
+      "    return num1\n",
+      "\n",
+      "num1=int(raw_input(\"Enter a number: \"))\n",
+      "num2=getSecondNumber()\n",
+      "print \"you entered \"+str(num1)+\" and \",str(num2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 28\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Second number: 35\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "you entered 28 and  35\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9, Page No 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iLuckyNumber=int()\n",
+      "def printLuckyNumber():\n",
+      "    print \"Your lucky number is: \",iLuckyNumber\n",
+      "\n",
+      "\n",
+      "iLuckyNumber=int(raw_input(\"Enter your lucky number: \"))\n",
+      "printLuckyNumber()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your lucky number: 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your lucky number is:  10\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10, Page No 126"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import os\n",
+      "import time\n",
+      "def sportsQuestion():\n",
+      "    print \"Sports Question\"\n",
+      "    print \"What University did NFL star Deon Sanders attend?\"\n",
+      "    print \"1 \\t University of Miami\"\n",
+      "    print \"2 \\t California State University\"\n",
+      "    print \"3 \\t Indiana University\"\n",
+      "    print \"4 \\t Florida State University\"\n",
+      "    iAnswer=int(raw_input(\"Enter your selection :\"))\n",
+      "    return iAnswer\n",
+      "\n",
+      "def geographyQuestion():\n",
+      "    print \"Geography Question\"\n",
+      "    print \"What is the state capitol of Florida?\"\n",
+      "    print \"1 \\t Pensecola\"\n",
+      "    print \"2 \\t Tallahassee\"\n",
+      "    print \"3 \\t Jacksonville\"\n",
+      "    print \"4 \\t Miami\"\n",
+      "    iAnswer=int(raw_input(\"Enter your selection :\"))\n",
+      "    return iAnswer\n",
+      "def pause(inNum):\n",
+      "    iCurrentTime=0\n",
+      "    iElapsedTime=0\n",
+      "    \n",
+      "    iCurrentTime=int(round(time.time()*1000))\n",
+      "    \n",
+      "    while((iElapsedTime-iCurrentTime)<inNum):\n",
+      "        iElapsedTime=int(round(time.time()*1000))\n",
+      "\n",
+      "giResponse=0\n",
+      "while(giResponse !=3):\n",
+      "    os.system('cls')\n",
+      "    print \"\\n\\tTHE TRIVIA GAME\\n\\n\"\n",
+      "    print \"1\\tSports\\n\"\n",
+      "    print \"2\\tGeography\\n\"\n",
+      "    print \"3\\tQuit\\n\"\n",
+      "    giResponse=int(raw_input(\"Enter your Selection: \"))\n",
+      "    \n",
+      "    if(giResponse==1):\n",
+      "        if(sportsQuestion()==4):\n",
+      "            print \"\\nCorrect\\n\"\n",
+      "        else:\n",
+      "            print \"\\nIncorrect\\n\"\n",
+      "            \n",
+      "        pause(2)\n",
+      "        break\n",
+      "    elif(giResponse==2):\n",
+      "        if(geographyQuestion()==2):\n",
+      "            print \"\\nCorrect\\n\"\n",
+      "        else:\n",
+      "            print \"\\nIncorrect\\n\"\n",
+      "        \n",
+      "        pause(2)\n",
+      "        break\n",
+      "            \n",
+      "            \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\tTHE TRIVIA GAME\n",
+        "\n",
+        "\n",
+        "1\tSports\n",
+        "\n",
+        "2\tGeography\n",
+        "\n",
+        "3\tQuit\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your Selection: 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sports Question\n",
+        "What University did NFL star Deon Sanders attend?\n",
+        "1 \t University of Miami\n",
+        "2 \t California State University\n",
+        "3 \t Indiana University\n",
+        "4 \t Florida State University\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your selection :4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Correct\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter6_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter6_1.ipynb
new file mode 100755
index 00000000..bfb0065f
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter6_1.ipynb
@@ -0,0 +1,652 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:74348a5183e19ad7ac30f2bcfa88ff539f69c16e0607950b89f2f1031fc1e024"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Arrays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page No 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iArray=range(5)\n",
+      "\n",
+      "for x in range(5):\n",
+      "    iArray[x]=0"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page No 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iArray=range(5)\n",
+      "\n",
+      "for x in range(5):\n",
+      "    iArray[x]=x\n",
+      "    \n",
+      "for x in range(5):\n",
+      "    print \"The value of iArray index \"+str(x)+\"is \"+str(x)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of iArray index 0is 0\n",
+        "The value of iArray index 1is 1\n",
+        "The value of iArray index 2is 2\n",
+        "The value of iArray index 3is 3\n",
+        "The value of iArray index 4is 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3, Page No 136"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iIndex = -1\n",
+      "iArray=range(5)\n",
+      "\n",
+      "for x in range(5):\n",
+      "    iArray[x]=(x+5)\n",
+      "\n",
+      "while (iIndex < 0 or iIndex >4):\n",
+      "    iIndex=int(raw_input(\"Enter a valid index (0-4): \"))\n",
+      "    \n",
+      "print \"The value of index \"+str(iIndex)+\" is \"+str(iArray[iIndex])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a valid index (0-4): 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of index 4 is 9\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page No 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cArray = range(5)\n",
+      "cName=\"Olivia\"\n",
+      "\n",
+      "print \"\\nCharacter array not initialized:\\n\"\n",
+      "\n",
+      "for x in range(5):\n",
+      "    print \"Element \"+str(x)+\"'s Contents are \",cArray[x]\n",
+      "\n",
+      "print \"\\nInitialized character array: \"\n",
+      "for x in range(6):\n",
+      "    print cName[x],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Character array not initialized:\n",
+        "\n",
+        "Element 0's Contents are  0\n",
+        "Element 1's Contents are  1\n",
+        "Element 2's Contents are  2\n",
+        "Element 3's Contents are  3\n",
+        "Element 4's Contents are  4\n",
+        "\n",
+        "Initialized character array: \n",
+        "O l i v i a\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5, Page No 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iFound = -1\n",
+      "iArray=range(5)\n",
+      "\n",
+      "for x in range(5):\n",
+      "    iArray[x]=(x+x)\n",
+      "    \n",
+      "iValue=int(raw_input(\"Enter value to search for: \"))\n",
+      "\n",
+      "for x in range(5):\n",
+      "    if (iArray[x]==iValue):\n",
+      "        iFound=x\n",
+      "        break\n",
+      "\n",
+      "if(iFound > -1):\n",
+      "    print \"I found your search value in element \",iFound\n",
+      "else:\n",
+      "    print \"Sorry, your search value was not found\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value to search for: 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "I found your search value in element  4\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6, Page No 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iTwoD=[[0 for j in range(3)]for i in range(3)]\n",
+      "for x in range(3):\n",
+      "    for y in range(3):\n",
+      "        iTwoD[x][y]=(x+y)\n",
+      "for x in range(3):\n",
+      "    for y in range(3):\n",
+      "        print \"iTwoD[\"+str(x)+\"][\"+str(y)+\"]=\",iTwoD[x][y]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "iTwoD[0][0]= 0\n",
+        "iTwoD[0][1]= 1\n",
+        "iTwoD[0][2]= 2\n",
+        "iTwoD[1][0]= 1\n",
+        "iTwoD[1][1]= 2\n",
+        "iTwoD[1][2]= 3\n",
+        "iTwoD[2][0]= 2\n",
+        "iTwoD[2][1]= 3\n",
+        "iTwoD[2][2]= 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7, Page No 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iTwoD=[[1,2,3],[4,5,6],[7,8,9]]\n",
+      "iFoundAt=[0,0]\n",
+      "\n",
+      "iValue=0\n",
+      "iFound=0\n",
+      "\n",
+      "iValue=int(raw_input(\"Enter your search value: \"))\n",
+      "\n",
+      "for x in range(3):\n",
+      "    for y in range(3):\n",
+      "        if(iTwoD[x][y]==iValue):\n",
+      "            iFound=1\n",
+      "            iFoundAt[0]=x\n",
+      "            iFoundAt[1]=y\n",
+      "            break\n",
+      "            \n",
+      "if (iFound==1):\n",
+      "    print \"Found value in iTwoD [\"+str(iFoundAt[0])+\"][\"+str(iFoundAt[1])+\"]\"\n",
+      "else:\n",
+      "    print \"Value not found\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your search value: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Found value in iTwoD [2][0]\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8, Page No 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "board=range(9)\n",
+      "cWhoWon=' '\n",
+      "iCurrentPlayer=0\n",
+      "\n",
+      "def displayBoard():\n",
+      "    print \"\\n\\t|\\t|\\n\"\n",
+      "    print \"\\t|\\t|\\n\"\n",
+      "    print \"\"+str(board[0])+\"\\t|\"+str(board[1])+\"\\t|\"+str(board[2])+\"\\n\"\n",
+      "    print \"--------|-------|--------\\n\"\n",
+      "    print \"\"+str(board[3])+\"\\t|\"+str(board[4])+\"\\t|\"+str(board[5])+\"\\n\"\n",
+      "    print \"--------|-------|--------\\n\"\n",
+      "    print \"\\t|\\t|\\n\"\n",
+      "    print \"\"+str(board[6])+\"\\t|\"+str(board[7])+\"\\t|\"+str(board[8])+\"\\n\"\n",
+      "    print \"\\t|\\t|\\n\"\n",
+      "\n",
+      "def verifySelection(iSquare,iPlayer):\n",
+      "    if(board[iSquare-1]==' ' and (iPlayer==1 or iPlayer==0)):\n",
+      "        board[iSquare-1]='X'\n",
+      "        return 0\n",
+      "    elif(board[iSquare-1]==' ' and iPlayer==2):\n",
+      "        board[iSquare-1]='0'\n",
+      "        return 0\n",
+      "    else:\n",
+      "        return 1\n",
+      "    \n",
+      "def checkForWin():\n",
+      "    global cWhoWon\n",
+      "    if(board[0]=='X' and board[1]=='X' and board[2]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[3]=='X' and board[4]=='X' and board[5]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[6]=='X' and board[7]=='X' and board[8]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[0]=='X' and board[3]=='X' and board[6]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[1]=='X' and board[4]=='X' and board[7]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[2]=='X' and board[5]=='X' and board[8]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[0]=='X' and board[5]=='X' and board[8]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[2]=='X' and board[5]=='X' and board[6]=='X'):\n",
+      "        cWhoWon='X'\n",
+      "    elif(board[0]=='0' and board[1]=='0' and board[2]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    elif(board[3]=='0' and board[4]=='0' and board[5]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    elif(board[6]=='0' and board[7]=='0' and board[8]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    elif(board[0]=='0' and board[3]=='0' and board[6]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    elif(board[1]=='0' and board[4]=='0' and board[7]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    elif(board[2]=='0' and board[5]=='0' and board[8]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    elif(board[0]=='0' and board[5]=='0' and board[8]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    elif(board[2]=='0' and board[5]=='0' and board[6]=='0'):\n",
+      "        cWhoWon='0'\n",
+      "    \n",
+      "    if(cWhoWon!=' '):\n",
+      "        if(cWhoWon=='X'):\n",
+      "            print \"\\nX Wins!\\n\"\n",
+      "            return\n",
+      "        \n",
+      "        if(cWhoWon=='0'):\n",
+      "            print \"\\n0 Wins!\\n\"\n",
+      "            return\n",
+      "    else:\n",
+      "            return\n",
+      "    \n",
+      "    for x in range(9):\n",
+      "        if(board[x]!=' '):\n",
+      "            catTotal=catTotal+1\n",
+      "            \n",
+      "    if(catTotal==9):\n",
+      "        cWhoWon='C'\n",
+      "        print \"\\nCAT Game!\\n\"\n",
+      "        return\n",
+      "iSquareNum=0\n",
+      "\n",
+      "for x in range(8):\n",
+      "    board[x]=' '\n",
+      "    \n",
+      "displayBoard()\n",
+      "\n",
+      "while(cWhoWon==' '):\n",
+      "    print \"\\n\"+cWhoWon+\"\\n\"\n",
+      "    \n",
+      "    if(iCurrentPlayer==1 or iCurrentPlayer==0):\n",
+      "        print \"\\nPLAYER X\\n\"\n",
+      "        iSquareNum=int(raw_input(\"Enter an available square number (1-9): \"))\n",
+      "        \n",
+      "        if(verifySelection(iSquareNum,iCurrentPlayer)==1):\n",
+      "            iCurrentPlayer=1\n",
+      "        else:\n",
+      "            iCurrentPlayer=2\n",
+      "    else:\n",
+      "        print \"\\nPLAYER 0\\n\"\n",
+      "        iSquareNum=int(raw_input(\"Enter an available square number (1-9): \"))\n",
+      "        \n",
+      "        if(verifySelection(iSquareNum,iCurrentPlayer)==1):\n",
+      "            iCurrentPlayer=2\n",
+      "        else:\n",
+      "            iCurrentPlayer=1\n",
+      "        displayBoard()\n",
+      "        checkForWin()\n",
+      "        \n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        " \t| \t| \n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        " \t| \t| \n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        " \t| \t|8\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\n",
+        " \n",
+        "\n",
+        "\n",
+        "PLAYER X\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an available square number (1-9): 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " \n",
+        "\n",
+        "\n",
+        "PLAYER 0\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an available square number (1-9): 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "X\t|0\t| \n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        " \t| \t| \n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        " \t| \t|8\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\n",
+        " \n",
+        "\n",
+        "\n",
+        "PLAYER X\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an available square number (1-9): 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " \n",
+        "\n",
+        "\n",
+        "PLAYER 0\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an available square number (1-9): 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "X\t|0\t|0\n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        "X\t| \t| \n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        " \t| \t|8\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\n",
+        " \n",
+        "\n",
+        "\n",
+        "PLAYER X\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an available square number (1-9): 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " \n",
+        "\n",
+        "\n",
+        "PLAYER 0\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an available square number (1-9): 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "X\t|0\t|0\n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        "X\t|0\t| \n",
+        "\n",
+        "--------|-------|--------\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "X\t| \t|8\n",
+        "\n",
+        "\t|\t|\n",
+        "\n",
+        "\n",
+        "X Wins!\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter7_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter7_1.ipynb
new file mode 100755
index 00000000..c1085a33
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter7_1.ipynb
@@ -0,0 +1,582 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:be75d77e91d7005f064cd2c560347aec768e900f2e81ef47804d4b096bc083a6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Pointers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1, Page No 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=5\n",
+      "iPtr=id(x)\n",
+      "iPtr=7"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page No 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=1\n",
+      "iPtr=id(x)\n",
+      "\n",
+      "iPtr=5\n",
+      "\n",
+      "print \"\\n*iPtr = \"+str(iPtr)+\"\\n &x = \"+str(id(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "*iPtr = 5\n",
+        " &x = 20176608\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page No 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=5\n",
+      "y=10\n",
+      "iPtr=None\n",
+      "\n",
+      "print \"Pointer points to: \",iPtr\n",
+      "iPtr=id(y)\n",
+      "print \"Ptr now points to: \",iPtr\n",
+      "x=iPtr\n",
+      "print \"The value of x is now\",x\n",
+      "iPtr=15\n",
+      "print \"The value of y is now\",iPtr"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pointer points to:  None\n",
+        "Ptr now points to:  20373108\n",
+        "The value of x is now 20373108\n",
+        "The value of y is now 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4, Page No 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def addTwoNumbers(x,y):\n",
+      "    return x+y\n",
+      "\n",
+      "\n",
+      "x=0\n",
+      "y=0\n",
+      "\n",
+      "x=int(raw_input(\"Enter first number:\"))\n",
+      "y=int(raw_input(\"Enter second number:\"))\n",
+      "\n",
+      "print \"Result is: \",addTwoNumbers(x,y)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number:10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number:15\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Result is:  25\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page No 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def demoPassByValue(x):\n",
+      "    x+=5\n",
+      "    print \"The value of x is:\",x\n",
+      "\n",
+      "x=0\n",
+      "x=int(raw_input(\"Enter a number:\"))\n",
+      "demoPassByValue(x)\n",
+      "print \"The original value of x did not change:\",x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number:5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is: 10\n",
+        "The original value of x did not change: 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6, Page No 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def demoPassByReference(x):\n",
+      "    x+=5\n",
+      "    print \"The value of x is now\",x\n",
+      "\n",
+      "x=0\n",
+      "x=int(raw_input(\"Enter a number:\"))\n",
+      "demoPassByReference(x)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number:10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is now 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7, Page No 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iArray=[1,2,3,4,5]\n",
+      "iPtr=id(iArray)\n",
+      "print \"Address of pointer: \",iPtr\n",
+      "print \"First address of array: \",id(iArray[0])\n",
+      "print \"Pointer points to: \",iArray[0]\n",
+      "print \"First element of array contains\",iArray[0]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Address of pointer:  28621960\n",
+        "First address of array:  19652320\n",
+        "Pointer points to:  1\n",
+        "First element of array contains 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8, Page No 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def nameLength(name):\n",
+      "    x=0\n",
+      "    while(name[x] != '\\0'):\n",
+      "        x+=1;\n",
+      "    return x\n",
+      "aName='\\0'\n",
+      "aName=raw_input(\"Enter your first name: \")\n",
+      "aName=aName+'\\0'\n",
+      "print \"Your first name contains\"\n",
+      "print str(nameLength(aName))+\" characters\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your first name: mehta\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your first name contains\n",
+        "5 characters\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9, Page No 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def squareNumber(num):\n",
+      "    for x in range(3):\n",
+      "        num[x]=num[x]*num[x]\n",
+      "\n",
+      "iNumbers=[2,4,6]\n",
+      "print \"The current array value are:\"\n",
+      "\n",
+      "for x in range(3):\n",
+      "    print iNumbers[x]\n",
+      "\n",
+      "print \"\\n\"\n",
+      "\n",
+      "squareNumber(iNumbers)\n",
+      "\n",
+      "print \"The modified array values are :\"\n",
+      "\n",
+      "for x in range(3):\n",
+      "    print iNumbers[x]\n",
+      "\n",
+      "print \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current array value are:\n",
+        "2\n",
+        "4\n",
+        "6\n",
+        "\n",
+        "\n",
+        "The modified array values are :\n",
+        "4\n",
+        "16\n",
+        "36\n",
+        "\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10, Page No 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def printArgument(num):\n",
+      "    print \"Read only argument is: \",num\n",
+      "    \n",
+      "iNumber=5\n",
+      "printArgument(iNumber)\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read only argument is:  5\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11, Page No 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def printArray(num):\n",
+      "    for x in range(3):\n",
+      "        print num[x]\n",
+      "\n",
+      "iNumbers=[2,4,6]\n",
+      "printArray(iNumbers)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n",
+        "4\n",
+        "6\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12, Page No 170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def modifyArray(num):\n",
+      "    for x in range(3):\n",
+      "        num[x]=num[x]*num[x]\n",
+      "        \n",
+      "iNumbers=[2,4,6]\n",
+      "modifyArray(iNumbers)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13, Page No 173"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "def encrypt(sMessage,random):\n",
+      "    x=0\n",
+      "    while(sMessage[x]):\n",
+      "        sMessage+=str(random)\n",
+      "        x+=1\n",
+      "    x=0\n",
+      "    print \"Encrypted message is: \"\n",
+      "    \n",
+      "    while(sMessage[x]):\n",
+      "        print sMessage[x]\n",
+      "        x+=1\n",
+      "\n",
+      "def decrypt(sMessage,random):\n",
+      "    x=0\n",
+      "    while(sMessage[x]):\n",
+      "        sMessage[x]=sMessage[x]-str(random)\n",
+      "        x+=1\n",
+      "    x=0\n",
+      "    while(sMessage[x]):\n",
+      "        print sMessage[x]\n",
+      "        x+=1\n",
+      "        \n",
+      "myString=range(21)\n",
+      "iSelection=0\n",
+      "iRand= (random.randint(1,4)%4)+1\n",
+      "\n",
+      "while(iSelection != 4):\n",
+      "    print \"\\n\\n1\\t Encrypt Clear Text\\n\"\n",
+      "    print \"2\\tDecrypt Cipher Text\\n\"\n",
+      "    print \"3\\tGenerate New Key\\n\"\n",
+      "    print \"4\\tQuit\\n\"\n",
+      "    iSelection=int(raw_input(\"\\nSelect a Cryptography Option: \"))\n",
+      "    \n",
+      "    if(iSelection ==1):\n",
+      "        myString=raw_input(\"\\nEnter one word as clear text to encrypt: \")\n",
+      "        encrypt(myString,iRand)\n",
+      "        break\n",
+      "    elif(iSelection==2):\n",
+      "        myString=raw_input(\"\\nEnter cipher text to decrypt: \")\n",
+      "        decrypt(myString,iRand)\n",
+      "        break\n",
+      "    elif(iSelection==3):\n",
+      "        iRand= (random.randrange(1,4,1)%4)+1\n",
+      "        print \"\\nNew Key Generated\\n\"\n",
+      "        break\n",
+      "    else:\n",
+      "            exit(0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "1\t Encrypt Clear Text\n",
+        "\n",
+        "2\tDecrypt Cipher Text\n",
+        "\n",
+        "3\tGenerate New Key\n",
+        "\n",
+        "4\tQuit\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Select a Cryptography Option: 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter one word as clear text to encrypt: bhavin\n"
+       ]
+      }
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter8_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter8_1.ipynb
new file mode 100755
index 00000000..bca914f3
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter8_1.ipynb
@@ -0,0 +1,703 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2192643aa60481aa37dedf877c77d457bede1bb6978dcdba9008684a80b17ad1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 Strings"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page No. 181"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "myString=\"MIKE\"\n",
+      "print \"The pointer variable's value is: \", id(myString)\n",
+      "print \"The pointer variable points to: \",myString\n",
+      "print \"The memory locations for each character are::\"\n",
+      "for x in range(0,4):\n",
+      "    print id(myString[x]),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The pointer variable's value is:  44562688\n",
+        "The pointer variable points to:  MIKE\n",
+        "The memory locations for each character are::\n",
+        "19548160 20083536 20086752 20285008\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2, Page No.183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "color=raw_input(\"Enter your favorite color: \")\n",
+      "print \"You entered: \",color"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your favorite color: Blue\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You entered:  Blue\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page No.183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "color=raw_input(\"Enter your favorite color: \")\n",
+      "print \"You entered: \",color"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your favorite color: Black\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You entered:  Black\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4, Page No. 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "names = []\n",
+      "names.append(\"Michael\")\n",
+      "names.append(\"Sheila\")\n",
+      "names.append(\"Spencer\")\n",
+      "names.append(\"Hunter\")\n",
+      "names.append(\"Kenya\")\n",
+      "print \"\\nNames in pointer array of type char:\\n\"\n",
+      "for i in range(5):\n",
+      "    print names[i]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Names in pointer array of type char:\n",
+        "\n",
+        "Michael\n",
+        "Sheila\n",
+        "Spencer\n",
+        "Hunter\n",
+        "Kenya\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5, Page No.186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "colors = []\n",
+      "print \"Enter 3 colors: \"\n",
+      "for i in range(3):\n",
+      "    colors.append(raw_input())\n",
+      "print \"Your entered: \"\n",
+      "for i in range(3):\n",
+      "    print colors[i]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 3 colors: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "blue\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "black\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "white\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your entered: \n",
+        "blue\n",
+        "black\n",
+        "white\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6, Page No.187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str1=\"123.79\"\n",
+      "str2=\"55\"\n",
+      "\n",
+      "print \"String1 is:\\\"\",str1,\"\\\"\"\n",
+      "print \"String2 is:\\\"\",str2,\"\\\"\"\n",
+      "\n",
+      "x=float(str1)\n",
+      "y=int(str2)\n",
+      "\n",
+      "print \"String 1 converted to a float is \",round(x,2)\n",
+      "print \"String 2 converted to an integer is \",y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String1 is:\" 123.79 \"\n",
+        "String2 is:\" 55 \"\n",
+        "String 1 converted to a float is  123.79\n",
+        "String 2 converted to an integer is  55\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7, Page No.188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str1=\"37\"\n",
+      "str2=\"20\"\n",
+      "\n",
+      "print \"String1 + String2 is \", str1+str2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String1 + String2 is  3720\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8, Page No. 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str1=\"37\"\n",
+      "str2=\"20\"\n",
+      "iResult=int(str1)+int(str2)\n",
+      "print \"String1 + String2 is \", iResult"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String1 + String2 is  57\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9, Page No. 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str1=\"Michael\"\n",
+      "str2=\"Vine\"\n",
+      "print \"The length of string1 is \",len(str1)\n",
+      "print \"The length of string2 is \",len(str2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of string1 is  7\n",
+        "The length of string2 is  4\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10, Page No.191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def convertL(char):\n",
+      "    char=char.lower()\n",
+      "    print \"The first name converted to lower case is: \",char\n",
+      "def convertU(char):\n",
+      "    char=char.upper()\n",
+      "    print \"The first name converted to Upper case is: \",char\n",
+      "name1=\"Michael\"\n",
+      "name2=\"Vine\"\n",
+      "\n",
+      "convertL(name1)\n",
+      "convertU(name2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The first name converted to lower case is:  michael\n",
+        "The first name converted to Upper case is:  VINE\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11, Page No.193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str2=\"C Language\"\n",
+      "str1=str2\n",
+      "print \"String 1 now contains \", str1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 1 now contains  C Language\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12, Page No.194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str1=\"Computer Science \"\n",
+      "str2=\"is applied mathematics\"\n",
+      "print str1+str2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Computer Science is applied mathematics\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exmaple 8.13, Page No.195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str1=\"A\"\n",
+      "str2=\"A\"\n",
+      "str3=\"!\"\n",
+      "print str1\n",
+      "print str2\n",
+      "print str3\n",
+      "if str1==str2:\n",
+      "    print \"Letter A is equal to letter A\"\n",
+      "if str1>str3:\n",
+      "    print \"Letter A is greater than character !\"\n",
+      "if str3<str1:\n",
+      "    print \"Character ! is greater than letter A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A\n",
+        "A\n",
+        "!\n",
+        "Letter A is equal to letter A\n",
+        "Letter A is greater than character !\n",
+        "Character ! is greater than letter A\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exmaple 8.14, Page No.197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "str1=\"Analyzing strings with with the strstr() function\"\n",
+      "str2=\"ing\"\n",
+      "str3=\"xyz\"\n",
+      "\n",
+      "print str1\n",
+      "print str2\n",
+      "print str3\n",
+      "\n",
+      "if((str1.find(str2))!=-1):\n",
+      "    print \"Str2 was found in str1\"\n",
+      "else: \n",
+      "    print \"Str2 was not found in str1\"\n",
+      "\n",
+      "if((str1.find(str3))!=-1):\n",
+      "     print \"Str3 was found in str1\"\n",
+      "else: \n",
+      "    print \"Str3 was not found in str1\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Analyzing strings with with the strstr() function\n",
+        "ing\n",
+        "xyz\n",
+        "Str2 was found in str1\n",
+        "Str3 was not found in str1\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exmaple 8.15, Page No.198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import time\n",
+      "import os\n",
+      "from array import array\n",
+      "\n",
+      "\n",
+      "def checkAnswer(string1,string2):\n",
+      "    for x in range(len(string2)):\n",
+      "        string2=string2.upper()\n",
+      "    if(set(string2).intersection(string1.split())):\n",
+      "        print \"Great job\"\n",
+      "    else :\n",
+      "        print \"Sorry no word found...\"\n",
+      "   \n",
+      "starGame=[]\n",
+      "starGame.append(\"ADELANGUAGEFERVZOPIBMOU\")\n",
+      "starGame.append(\"ZBPOINTERSKLMLOOPMNOCOT\")\n",
+      "starGame.append(\"PODSTRINGGDIWHIEEICERLS\")\n",
+      "starGame.append(\"YVCPROGRAMMERWQKNULTHMD\")\n",
+      "starGame.append(\"UKUNIXFIMWXIZEQZINPUTEX\")\n",
+      "\n",
+      "displayed=0\n",
+      "startTime=0\n",
+      "os.system('cls')\n",
+      "print \"word find\"\n",
+      "startTime=int(round(time.time()*1000))\n",
+      "for x in range(0,5):\n",
+      "    while ((startTime+3)>int(round(time.time()*1000))):\n",
+      "        if displayed==0:\n",
+      "            print \"find a word in:\"\n",
+      "            print \"\",starGame[x]\n",
+      "            displayed=1\n",
+      "    os.system('cls')\n",
+      "    answer=raw_input(\"Enter word found:\")\n",
+      "    checkAnswer(starGame[x],answer)\n",
+      "    displayed=0\n",
+      "    startTime=int(round(time.time()*1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "word find\n",
+        "find a word in:\n",
+        " ADELANGUAGEFERVZOPIBMOU\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word found:U\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorry no word found...\n",
+        "find a word in:\n",
+        " ZBPOINTERSKLMLOOPMNOCOT\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word found:T\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorry no word found...\n",
+        "find a word in:\n",
+        " PODSTRINGGDIWHIEEICERLS\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word found:S\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorry no word found...\n",
+        "find a word in:\n",
+        " YVCPROGRAMMERWQKNULTHMD\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word found:YV\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorry no word found...\n",
+        "find a word in:\n",
+        " UKUNIXFIMWXIZEQZINPUTEX\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word found:UKU\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorry no word found...\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter9_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter9_1.ipynb
new file mode 100755
index 00000000..aab6433a
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter9_1.ipynb
@@ -0,0 +1,515 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:695fc05932ca1d30f19b3867fb7e2671008edb93009669dc709e27b0465efea9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Introduction to Data Structures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page No 205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class employee:\n",
+      "    def __init__(self,fname,lname,id1,salary):\n",
+      "        self.fname = fname\n",
+      "        self.lname = lname\n",
+      "        self.id1 = id1\n",
+      "        self.salary = salary\n",
+      "if __name__ == '__main__':\n",
+      "    emp1 = employee(\"Michael\",\"Vine\",123,50000.00)\n",
+      "    print \"\\n First name: \" + emp1.fname + \"\\n\"\n",
+      "    print \"Last name: \" + emp1.lname + \"\\n\"\n",
+      "    print \"Employee ID: \" + str(emp1.id1) + \"\\n\"\n",
+      "    print \"Salary: \" + str(emp1.salary) + \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " First name: Michael\n",
+        "\n",
+        "Last name: Vine\n",
+        "\n",
+        "Employee ID: 123\n",
+        "\n",
+        "Salary: 50000.0\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page No 206"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class employee:\n",
+      "    def __init__(self,fname,lname,id1,salary):\n",
+      "        self.fname = fname\n",
+      "        self.lname = lname\n",
+      "        self.id1 = id1\n",
+      "        self.salary = salary\n",
+      "if __name__ == '__main__':\n",
+      "    emp1 = employee(\"Michael\",\"Vine\",123,50000.00)\n",
+      "    print \"\\n First name: \" + emp1.fname + \"\\n\"\n",
+      "    print \"Last name: \" + emp1.lname + \"\\n\"\n",
+      "    print \"Employee ID: \" + str(emp1.id1) + \"\\n\"\n",
+      "    print \"Salary: \" + str(emp1.salary) + \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " First name: Michael\n",
+        "\n",
+        "Last name: Vine\n",
+        "\n",
+        "Employee ID: 123\n",
+        "\n",
+        "Salary: 50000.0\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3, Page No 209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class scores:\n",
+      "    def __init__(self,name,score):\n",
+      "        self.name = name\n",
+      "        self.score = score\n",
+      "if __name__ == '__main__':\n",
+      "    highScores = []\n",
+      "    highScores.append(scores(\"Hunter\",40768))\n",
+      "    highScores.append(scores(\"Kenya\",38568))\n",
+      "    highScores.append(scores(\"Apollo\",35985))\n",
+      "    for x in range(3):\n",
+      "        print \"\\n\" + highScores[x].name + \"\\t\" + str(highScores[x].score) + \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Hunter\t40768\n",
+        "\n",
+        "\n",
+        "Kenya\t38568\n",
+        "\n",
+        "\n",
+        "Apollo\t35985\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page No 211"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class employee:\n",
+      "    def __init__(self,id1,name,salary):\n",
+      "        self.id1 = id1\n",
+      "        self.name = name\n",
+      "        self.salary = salary\n",
+      "def processEmp(emp1):\n",
+      "    emp1.id1 = 123\n",
+      "    emp1.name = \"Sheila\"\n",
+      "    emp1.salary = 65000.00\n",
+      "if __name__ == '__main__':\n",
+      "    emp1 = employee(0,0,0)\n",
+      "    processEmp(emp1)\n",
+      "    print \"\\n Id : \" + str(emp1.id1) + \"\\n\"\n",
+      "    print \"Name : \" + emp1.name + \"\\n\"\n",
+      "    print \"Salary : \" + str(emp1.salary) + \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Id : 123\n",
+        "\n",
+        "Name : Sheila\n",
+        "\n",
+        "Salary : 65000.0\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5, Page No 212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class player:\n",
+      "    def __init__(self,name,score):\n",
+      "        self.name = name\n",
+      "        self.score = score\n",
+      "if __name__ == '__main__':\n",
+      "    aPlayer = player(0,0)\n",
+      "    ptrPlayer = id(aPlayer)\n",
+      "    aPlayer.name = \"Pinball Wizard\"\n",
+      "    aPlayer.score = 1000000.00\n",
+      "    print \"\\nPlayer : \" + aPlayer.name + \"\\n\"\n",
+      "    print \"Score : \" + str(aPlayer.score)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Player : Pinball Wizard\n",
+        "\n",
+        "Score : 1000000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6, Page No 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class employee:\n",
+      "    def __init__(self,id1,name,salary):\n",
+      "        self.id1 = id1\n",
+      "        self.name = name\n",
+      "        self.salary = salary\n",
+      "def processEmp(emp1):\n",
+      "    emp1.id1 = 123\n",
+      "    emp1.name = \"Sheila\"\n",
+      "    emp1.salary = 65000.00\n",
+      "if __name__ == '__main__':\n",
+      "    emp1 = employee(0,0,0)\n",
+      "    ptrEmp = id(emp1)\n",
+      "    processEmp(emp1)\n",
+      "    print \"\\nId : \" + str(emp1.id1) + \"\\n\"\n",
+      "    print \"Name : \" + emp1.name + \"\\n\"\n",
+      "    print \"Salary : \" + str(emp1.salary) + \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Id : 123\n",
+        "\n",
+        "Name : Sheila\n",
+        "\n",
+        "Salary : 65000.0\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7, Page No 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class employee:\n",
+      "    def __init__(self,id1,name,salary):\n",
+      "        self.id1 = id1\n",
+      "        self.name = name\n",
+      "        self.salary = salary\n",
+      "def processEmp(emp1 = []):\n",
+      "    emp1.append(employee(123,\"Sheila\",65000.00))\n",
+      "    emp1.append(employee(234,\"Hunter\",28000.00))\n",
+      "    emp1.append(employee(456,\"Kenya\",48000.00))\n",
+      "    return emp1\n",
+      "if __name__ == '__main__':\n",
+      "    emp1 = []\n",
+      "    emp1 = employee(0,0,0)\n",
+      "    emp1 = processEmp(emp1 = [])\n",
+      "    for x in range(3):\n",
+      "        print \"\\nId : \" + str(emp1[x].id1) + \"\\n\" \n",
+      "        print \"Name : \" + emp1[x].name + \"\\n\"\n",
+      "        print \"Salary : \" + str(emp1[x].salary)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Id : 123\n",
+        "\n",
+        "Name : Sheila\n",
+        "\n",
+        "Salary : 65000.0\n",
+        "\n",
+        "Id : 234\n",
+        "\n",
+        "Name : Hunter\n",
+        "\n",
+        "Salary : 28000.0\n",
+        "\n",
+        "Id : 456\n",
+        "\n",
+        "Name : Kenya\n",
+        "\n",
+        "Salary : 48000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8, Page No 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class phonebook:\n",
+      "    def __init__(self,name,number,address):\n",
+      "        self.name = name\n",
+      "        self.number = number\n",
+      "        self.address = address\n",
+      "class magazine:\n",
+      "    def __init__(self,name,author,isbn):\n",
+      "        self.name = name\n",
+      "        self.author = author\n",
+      "        self.isbn = isbn\n",
+      "if __name__ == '__main__':\n",
+      "    aBook = phonebook(\"John\",123,\"John Smith\")\n",
+      "    aMagazine = magazine(\"C Programming\",\"Michael Vine\",1-59863-480-1)\n",
+      "    print \"\\nUnion Details\\n\"\n",
+      "    print \"Address for aBook.name: \" +  str(id(aBook.name)) + \"\\n\"\n",
+      "    print \"Address for aBook.number: \", str(id(aBook.number)) + \"\\n\"\n",
+      "    print \"Address for aBook.address: \", str(id(aBook.address)) + \"\\n\"\n",
+      "    print \"\\nStructure Details\\n\"\n",
+      "    print \"Address for aMagazine.name: \", str(id(aMagazine.name)) + \"\\n\"\n",
+      "    print \"Address for aMagazine.author: \", str(id(aMagazine.author)) + \"\\n\"\n",
+      "    print \"Address for aMagazine.isbn: \", str(id(aMagazine.isbn)) + \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "Union Details\n",
+        "\n",
+        "Address for aBook.name: 66178424\n",
+        "\n",
+        "Address for aBook.number:  30796656\n",
+        "\n",
+        "Address for aBook.address:  66151792\n",
+        "\n",
+        "\n",
+        "Structure Details\n",
+        "\n",
+        "Address for aMagazine.name:  66151744\n",
+        "\n",
+        "Address for aMagazine.author:  66150976\n",
+        "\n",
+        "Address for aMagazine.isbn:  64689272\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9, Page No 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x = 12\n",
+      "y = 5\n",
+      "print \"\\nWithout Type-Casting\\n\"\n",
+      "print \"12 \\\\ 5 = \" + str(x/y) + \"\\n\"\n",
+      "print \"\\nWith Type-Casting\\n\"\n",
+      "print \"12 \\\\ 5 = \" + str(float(x)/float(y)) + \"\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Without Type-Casting\n",
+        "\n",
+        "12 \\ 5 = 2\n",
+        "\n",
+        "\n",
+        "With Type-Casting\n",
+        "\n",
+        "12 \\ 5 = 2.4\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10, Page No 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "number = 86\n",
+      "letter = 'M'\n",
+      "print \"\\n86 type-casted to char is: \" + chr(number) + \"\\n\"\n",
+      "print \"\\n'M' type-casted to int is: \" + str(ord(letter)) + \"\\n \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "86 type-casted to char is: V\n",
+        "\n",
+        "\n",
+        "'M' type-casted to int is: 77\n",
+        " \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11, Page No 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# This program is not possible in python."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_1.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_1.ipynb
new file mode 100755
index 00000000..5c25ef5d
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_1.ipynb
@@ -0,0 +1,117 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ed1073b28a905078eb23296522b8b59bf416e5275a011e43147248d1acd57f5a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 GETTING STARTED WITH C PROGRAMMING"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Page No. 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"C you later\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C you later\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2, Page No. 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Chapter 1. Getting Started with C programming\"\n",
+      "print \"This is an example of format bug.\"\n",
+      "print \"The format issue can be corrected by using\"\n",
+      "print \"the \\n and \\\\ escape sequnece\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Chapter 1. Getting Started with C programming\n",
+        "This is an example of format bug.\n",
+        "The format issue can be corrected by using\n",
+        "the \n",
+        " and \\ escape sequnece\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exmaple 1.3, Page No. 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Chapter 1. Getting Started with C programming\"\n",
+      "print \"This is an example of format bug.\"\n",
+      "print \"The format issue can be corrected by using\"\n",
+      "print \"the \\\\n and \\\\\\\\ escape sequnece\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Chapter 1. Getting Started with C programming\n",
+        "This is an example of format bug.\n",
+        "The format issue can be corrected by using\n",
+        "the \\n and \\\\ escape sequnece\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_2.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_2.ipynb
new file mode 100755
index 00000000..2f3c44af
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_2.ipynb
@@ -0,0 +1,239 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:622f2c579033e8c66b476e16d61ce8bfa6745f6b87c5905b2064108fa5218b3c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 PRIMARY DATA TYPES"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page No. 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=-4443\n",
+      "y=554.21\n",
+      "c='M'\n",
+      "\n",
+      "print \"The value of integer variable x is\" , x\n",
+      "print \"The value of float variable y is \" , y\n",
+      "print \"The value of character variable c is \" , c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of integer variable x is -4443\n",
+        "The value of float variable y is  554.21\n",
+        "The value of character variable c is  M\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page No. 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x = 20\n",
+      "PI = 3.14\n",
+      "print \"Constant values are \",x ,\" and \", round(PI,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Constant values are  20  and  3.14\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3, Page No. 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"\\t Adder Program, by Michael Vine\"\n",
+      "iOperand1=int(raw_input(\"Enter First Operand: \"))\n",
+      "iOperand2=int(raw_input(\"Enter Second Operand: \"))\n",
+      "print \"The final result is \", iOperand1+iOperand2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t Adder Program, by Michael Vine\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter First Operand: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Second Operand: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final result is  12\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page No. 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"\\t Adder Program, by Michael Vine\"\n",
+      "iOperand1=int(raw_input(\"Enter First Operand: \"))\n",
+      "iOperand2=int(raw_input(\"Enter Second Operand: \"))\n",
+      "iResult=iOperand1+iOperand2\n",
+      "print \"The result is \", iResult"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t Adder Program, by Michael Vine\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter First Operand: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Second Operand: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The result is  12\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5, Page No. 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "fRevenue=float(raw_input(\"Enter total revenue: \"))\n",
+      "fCost=float(raw_input(\"Enter total cost: \"))\n",
+      "print \"Your total profit is \", round(fRevenue-fCost,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter total revenue: 4000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter total cost: 750\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your total profit is  3250.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_3.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_3.ipynb
new file mode 100755
index 00000000..a87e263b
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_3.ipynb
@@ -0,0 +1,499 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:4698a1abef67d23e99c2eb6e9534a43d9d501d3ae6147f6ceb242f76f06f63a5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 CONDITIONS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page No. 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"\\tAC Control Unit\"\n",
+      "print \"1\\tTurn the AC on\"\n",
+      "print \"2\\tTurn the AC off\"\n",
+      "iResponse=int(raw_input(\"Enter your selection::\"))\n",
+      "if iResponse==1:\n",
+      "    print \"AC is now on\"\n",
+      "if iResponse==2:\n",
+      "    print \"AC is now off\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tAC Control Unit\n",
+        "1\tTurn the AC on\n",
+        "2\tTurn the AC off\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your selection::1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "AC is now on\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page No. 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"\\tAC Control Unit\"\n",
+      "print \"a\\tTurn the AC on\"\n",
+      "print \"b\\tTurn the AC off\"\n",
+      "cResponse=raw_input(\"Enter your selection::\")\n",
+      "if cResponse=='a':\n",
+      "    print \"AC is now on\"\n",
+      "if cResponse=='b':\n",
+      "    print \"AC is now off\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tAC Control Unit\n",
+        "a\tTurn the AC on\n",
+        "b\tTurn the AC off\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your selection::b\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "AC is now off\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page No. 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "fBalance = 100.25\n",
+      "print \"\\tATM\"\n",
+      "print \"1\\tDeposit Funds\"\n",
+      "print \"2\\tWithdraw Funds\"\n",
+      "iSelection=int(raw_input(\"Enter Your Selection: \"))\n",
+      "if iSelection==1:\n",
+      "    fTransAmount=int(raw_input(\"Enter fund amount to deposit: \"))\n",
+      "    print \"Your new Balance is: \", fBalance+fTransAmount\n",
+      "if iSelection==2:\n",
+      "    fTransAmount=int(raw_input(\"Enter fund amount to withdraw: \"))\n",
+      "    if fTransAmout>fBalance:\n",
+      "        print \"Insufficient Funds\"\n",
+      "    else:\n",
+      "        print \"Your new balance is \", fBalance-fTransAmount"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tATM\n",
+        "1\tDeposit Funds\n",
+        "2\tWithdraw Funds\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Selection: 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter fund amount to deposit: 200\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your new Balance is:  300.25\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page No. 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cResponse=raw_input(\"Enter a letter A: \")\n",
+      "if cResponse=='A':\n",
+      "    print \"Correct response\"\n",
+      "else:\n",
+      "    print \"Incorrect response\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a letter A: a\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Incorrect response\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5, Page No.69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iResponse=int(raw_input(\"Enter a numeer from 1 to 10: \"))\n",
+      "if iResponse<1 or iResponse>10:\n",
+      "    print \"Number not in Range\"\n",
+      "else:\n",
+      "    print \"Thank You\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a numeer from 1 to 10: 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thank You\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page No. 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cResponse=raw_input(\"Please Enter a letter: \")\n",
+      "if not cResponse.isdigit():\n",
+      "    print \"Thank You\"\n",
+      "else:\n",
+      "    print \"You did not enter a letter\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please Enter a letter: 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You did not enter a letter\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7, Page No. 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cResponse=raw_input(\"Please Enter a letter: \")\n",
+      "if cResponse.isdigit():\n",
+      "    print \"Thank You\"\n",
+      "else:\n",
+      "    print \"You did not enter a digit\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please Enter a letter: a\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You did not enter a digit\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8, Page No.72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"1\\tSports\"\n",
+      "print \"2\\tGeography\"\n",
+      "print \"3\\tMusic\"\n",
+      "print \"4\\tWorld Events\"\n",
+      "iResponse=int(raw_input(\"Please select a category (1-4): \"))\n",
+      "if iResponse==1:\n",
+      "    print \"You selected sports questions\"\n",
+      "elif iResponse==2:\n",
+      "    print \"You selected geography questions\"\n",
+      "elif iResponse==3:\n",
+      "    print \"You selected Music questions\"\n",
+      "elif iResponse==4:\n",
+      "    print \"You selected World event questions\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\tSports\n",
+        "2\tGeography\n",
+        "3\tMusic\n",
+        "4\tWorld Events\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please select a category (1-4): 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You selected World event questions\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9, Page No. 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "iRandomNum=(random.randint(0,100)%10)+1\n",
+      "iResponse=int(raw_input(\"Guess a number between 1 and 10: \"))\n",
+      "if iResponse==iRandomNum:\n",
+      "    print \"You guessed right\"\n",
+      "else:\n",
+      "    print \"Sorry, you guess wrong\"\n",
+      "    print \"The correct guess was \",iRandomNum"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Guess a number between 1 and 10: 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorry, you guess wrong\n",
+        "The correct guess was  3\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10, Page No. 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "iRandomNum=(random.randint(0,100)%4)+1\n",
+      "print \"Fortune Cookie - Chapter 3\"\n",
+      "if iRandomNum==1:\n",
+      "    print \"You will meet a new friend\"\n",
+      "elif iRandomNum==2:\n",
+      "    print \"You will enjoy a long and happy life\"\n",
+      "elif  iRandomNum==3:\n",
+      "    print \"Opportunity knocks softly. Can you hear it?\"\n",
+      "elif iRandomNum==4:\n",
+      "    print \"You'll be financially rewarded for your good deeds\"\n",
+      "\n",
+      "print \"Lucky lotto numbers\"\n",
+      "print (random.randint(0,100)%49+1)\n",
+      "print (random.randint(0,100)%49+1)\n",
+      "print (random.randint(0,100)%49+1)\n",
+      "print (random.randint(0,100)%49+1)\n",
+      "print (random.randint(0,100)%49+1)\n",
+      "print (random.randint(0,100)%49+1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fortune Cookie - Chapter 3\n",
+        "You will enjoy a long and happy life\n",
+        "Lucky lotto numbers\n",
+        "45\n",
+        "17\n",
+        "9\n",
+        "23\n",
+        "2\n",
+        "15\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_4.ipynb b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_4.ipynb
new file mode 100755
index 00000000..218595de
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/Chapter_4.ipynb
@@ -0,0 +1,688 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d985967b3e8d0e583d78d0a3306851320f330cc48579a1f62da314746570750f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 Looping Structures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 , Page No.88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=0\n",
+      "print \"The value of x is \",x\n",
+      "x+=1\n",
+      "print \"The value of x is \",x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is  0\n",
+        "The value of x is  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2, Page No.89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=0\n",
+      "y=0\n",
+      "print \"The value of y is \", y+1\n",
+      "print \"The value of x is \", x+1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of y is  1\n",
+        "The value of x is  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page No.89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=0\n",
+      "y=1\n",
+      "x=(y)*2\n",
+      "print \"The Value of x \",x\n",
+      "x=0\n",
+      "y=1\n",
+      "x=(y+1)*2\n",
+      "print \"The Value of x \",x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of x  2\n",
+        "The Value of x  4\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4, Page No.90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=0\n",
+      "y=0\n",
+      "x=(y)*4\n",
+      "print \"The Value of x\", x\n",
+      "y=0\n",
+      "x=(y+1)*4\n",
+      "print \"The Value of x\", x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of x 0\n",
+        "The Value of x 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5, Page No.92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=1\n",
+      "y=1\n",
+      "x=(y)*4\n",
+      "print \"The value of x is \",x\n",
+      "y=1\n",
+      "x=(y-1)*4\n",
+      "print \"The value of x is \",x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is  4\n",
+        "The value of x is  0\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6, Page No. 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=1\n",
+      "y=2\n",
+      "x=y*x+1\n",
+      "print \"The value of x is: \",x\n",
+      "x=1\n",
+      "y=2\n",
+      "x+=y*x+1\n",
+      "print \"The value of x is: \",x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is:  3\n",
+        "The value of x is:  4\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7, Page No.94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=1\n",
+      "y=2\n",
+      "x=y*x+1\n",
+      "print \"The value of x is: \",x\n",
+      "x=1\n",
+      "y=2\n",
+      "x-=y*x+1\n",
+      "print \"The value of x is: \",x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is:  3\n",
+        "The value of x is:  -2\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8, Page No.95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=0\n",
+      "while (x<10):\n",
+      "    print \"The value of x is \",x\n",
+      "    x=x+1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is  0\n",
+        "The value of x is  1\n",
+        "The value of x is  2\n",
+        "The value of x is  3\n",
+        "The value of x is  4\n",
+        "The value of x is  5\n",
+        "The value of x is  6\n",
+        "The value of x is  7\n",
+        "The value of x is  8\n",
+        "The value of x is  9\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9, Page No.97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iSelection=0\n",
+      "while (iSelection!=4):\n",
+      "    print \"1\\tDeposit Funds\"\n",
+      "    print \"2\\tWithdraw Funds\"\n",
+      "    print \"3\\tPrint Balance\"\n",
+      "    print \"4\\tQuit\"\n",
+      "    iSelection=int(raw_input(\"Enter Your Selection(1-4): \"))\n",
+      "print \"Thank you\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\tDeposit Funds\n",
+        "2\tWithdraw Funds\n",
+        "3\tPrint Balance\n",
+        "4\tQuit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Selection(1-4): 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\tDeposit Funds\n",
+        "2\tWithdraw Funds\n",
+        "3\tPrint Balance\n",
+        "4\tQuit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Selection(1-4): 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thank you\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10, Page No.99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x=9\n",
+      "while (x<10):\n",
+      "    print \"This printf statement is executed at least once\"\n",
+      "    x=x+1\n",
+      "while (x<10):\n",
+      "    print \"This printf statement is never executed \"\n",
+      "    x=x+1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This printf statement is executed at least once\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11, Page No.100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "for x in range(10,5,-1):\n",
+      "    print \"The value of x is \",x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of x is  10\n",
+        "The value of x is  9\n",
+        "The value of x is  8\n",
+        "The value of x is  7\n",
+        "The value of x is  6\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12, Page No.101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "iNumQuestions=int(raw_input(\"Enter the number of questions to ask: \"))\n",
+      "for x in range(iNumQuestions):\n",
+      "    iRndNum1=(random.randint(0,100)%10)+1\n",
+      "    iRndNum2=(random.randint(0,100)%10)+1\n",
+      "    iResponse=int(raw_input(\"what is \"+ str(iRndNum1)+\" * \"+str(iRndNum2)+\" : \"))\n",
+      "    if(iResponse==(iRndNum1*iRndNum2)):\n",
+      "        print \"Correct\"\n",
+      "    else:\n",
+      "        print \"The Corrent answer was \", iRndNum1*iRndNum2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of questions to ask: 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "what is 6 * 9 : 54\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Correct\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "what is 8 * 5 : 32\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Corrent answer was  40\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13, Page No.102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "for x in range(10,5,-1):\n",
+      "    if x==7:\n",
+      "        break\n",
+      "print x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14, Page No.103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "for x in range(10,5,-1):\n",
+      "    if x==7:\n",
+      "        continue\n",
+      "    print x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n",
+        "9\n",
+        "8\n",
+        "6\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15, Page No.104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "for x in range(25):\n",
+      "    print \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16, Page No. 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import time\n",
+      "import random\n",
+      "import os\n",
+      "cYesNo=raw_input(\"Play a game of Concentration?(y or n)\")\n",
+      "if(cYesNo=='y' or cYesNo=='Y'):\n",
+      "    i1=random.randint(0,100) % 100\n",
+      "    i2=random.randint(0,100) % 100\n",
+      "    i3=random.randint(0,100) % 100\n",
+      "    print \"Concentrate on the next three numbers\\n\",i1,i2,i3\n",
+      "    iCurrentTime=int(round(time.time()*1000))\n",
+      "    iElapsedTime=int(round(time.time()*1000))\n",
+      "    while((iCurrentTime-iElapsedTime)<3):\n",
+      "        iElapsedTime=int(round(time.time()*1000))\n",
+      "    os.system('cls')\n",
+      "    iResp1=int(raw_input(\"Enter a number:\"))\n",
+      "    iResp2=int(raw_input(\"Enter a number:\"))\n",
+      "    iResp3=int(raw_input(\"Enter a number:\"))\n",
+      "    \n",
+      "    if(i1==iResp1 and i2==iResp2 and i3==iResp3):\n",
+      "        print \"Congratulations!\"\n",
+      "    else:\n",
+      "        print \"Sorry, Correct numbers were \",i1,i2,i3"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Play a game of Concentration?(y or n)n\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/README.txt b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/README.txt
new file mode 100755
index 00000000..9bf2614f
--- /dev/null
+++ b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Bhavin Mehta
+Course: mca
+College/Institute/Organization: Shri Brahmanand Institute of Computer Science - Chaparda
+Department/Designation: Assistant Professor
+Book Title: C Programming for the Absolute Beginner
+Author: Michael Vine
+Publisher: Thomson Course Technology United States of America
+Year of publication: 2008
+Isbn: 1-59863-480-1
+Edition: 2nd edition
\ No newline at end of file
diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/screenshots/1.png b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/screenshots/1.png
new file mode 100755
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diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/screenshots/3.png b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/screenshots/3.png
new file mode 100755
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diff --git a/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/screenshots/3_1.png b/C_Programming_for_the_Absolute_Beginner_by_Michael_Vine/screenshots/3_1.png
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diff --git a/Chemical_Engineering_Thermodynamics/README.txt b/Chemical_Engineering_Thermodynamics/README.txt
deleted file mode 100755
index 98c63c7c..00000000
--- a/Chemical_Engineering_Thermodynamics/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Jai Mathur
-Course: mca
-College/Institute/Organization: IIT
-Department/Designation: System Admin
-Book Title: Chemical Engineering Thermodynamics
-Author: Y. V. C. Rao
-Publisher: Universities Press
-Year of publication: 1997
-Isbn: 81-7371-048-1
-Edition: 1
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch1.ipynb b/Chemical_Engineering_Thermodynamics/ch1.ipynb
deleted file mode 100755
index c7a29075..00000000
--- a/Chemical_Engineering_Thermodynamics/ch1.ipynb
+++ /dev/null
@@ -1,145 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4f7d90cd7b10a2be32d6f9985c84b88b7895b54f53fda509a7ab6972ab90144a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1 : Introduction"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.1  Page No : 6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "# Variables\n",
-      "weight = 981. \t\t\t #weight of payload in N\n",
-      "gmoon = 1.62;\t\t\t #acceleration due to gravity on the moon in m/s**2\n",
-      "g = 9.81;\t\t\t     #acceleration due to gravity on earth\n",
-      "\n",
-      "# Calculations\n",
-      "mass = weight/g;        \t\t\t # Calculations of mass of the payload in kg (calculated as F = m*g)\n",
-      "weightmoon = mass*gmoon;\t\t\t # Calculations of weight of payload on the moon in N\n",
-      "\n",
-      "# Results\n",
-      "print ' The weight of payload on the moon =  %d N'%(weightmoon);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The weight of payload on the moon =  162 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.2  Page No : 7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "# Variables\n",
-      "l = 15.;\t\t\t #length of the child's head in cm\n",
-      "b = 12.;\t\t\t #breadth of the child's head in cm\n",
-      "p = 101325.;\t\t\t #atmospheric pressure in Pa\n",
-      "\n",
-      "# Calculations\n",
-      "area = (l*b)/(10**4);\t\t\t # Calculations of area of the child's head in m**2\n",
-      "force = p*area;\t\t\t # Calculations of force exerted on the child's head due to atmospheric air in N\n",
-      "\n",
-      "# Results\n",
-      "print ' The force exerted on the childs head due to atmospheric air =  %.2f N'%(force);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The force exerted on the childs head due to atmospheric air =  1823.85 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.3  Page No : 7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "rho_water = 1000.;\t\t\t #density of water flowing through the pipeline in kg/m**3\n",
-      "rho_manomtr = 1595.;\t\t #density of manometric fluid (carbon tetrachloride) in kg/m**3\n",
-      "l = 40.;\t        \t\t #length between the selected sections in cm\n",
-      "theta = 45.\t\t\t         #inclination of the manometer in degrees\n",
-      "g = 9.81;\t        \t\t #acceleration due to gravity in m/s**2\n",
-      "\n",
-      "# Calculations\n",
-      "delp = (l/100.)*math.sin((theta*math.pi)/180.)*g*(rho_manomtr-rho_water);   # Calculations of pressure drop between the required sections in Pa\n",
-      "\n",
-      "# Results\n",
-      "print ' The pressure drop between the required sections =  %.2f Pa'%delp\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The pressure drop between the required sections =  1650.94 Pa\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch10.ipynb b/Chemical_Engineering_Thermodynamics/ch10.ipynb
deleted file mode 100755
index 5d1a66e8..00000000
--- a/Chemical_Engineering_Thermodynamics/ch10.ipynb
+++ /dev/null
@@ -1,202 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10 : Stability and phase transition in thermodynamic systems"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.2  Page No : 369"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P = 2;\t\t\t #number of phases (no unit)\n",
-      "C = 2;\t\t\t #number of components (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "F = C+2-P\n",
-      "\n",
-      "# Results\n",
-      "print \" The number of degrees of freedom  =  %d \"%(F);\n",
-      "print \"Two intensive properties are required to be specified to\\\n",
-      " describe the thermodynamic state of the system,and the fundamental relation in\\\n",
-      "  the Gibbs free energy representation for this system is of the type, G = GT,P,N1,N2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The number of degrees of freedom  =  2 \n",
-        "Two intensive properties are required to be specified to describe the thermodynamic state of the system,and the fundamental relation in  the Gibbs free energy representation for this system is of the type, G = GT,P,N1,N2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.3  Page No : 370"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
-      "w = 0.398;\t\t\t #acentric factor (no unit) \n",
-      "\n",
-      "# Calculations\n",
-      "Pguess = 0.215\n",
-      "\n",
-      "\n",
-      "Tr = T/Tc\n",
-      "Pr = (Pguess*10**6)/(Pc*10**5)\n",
-      "S = 0.37464+(1.54226*w)-(0.26992*w**2)\n",
-      "alpha1 = (1+(S*(1-math.sqrt(Tr))))**2;\n",
-      "a = (0.45724*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
-      "b = (0.07780*R*Tc)/(Pc*10**5);\t\t\t \n",
-      "A = (a*Pguess*10**6)/(R*T)**2;\t\t\t \n",
-      "B = (b*Pguess*10**6)/(R*T);\t\t\t \n",
-      "alpha = -1+B\n",
-      "beeta = A-(2*B)-(3*B**2)\n",
-      "gaamma = -(A*B)+(B**2)+(B**3)\n",
-      "p = beeta-(alpha**2)/3;\t\t\n",
-      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
-      "D = (((q)**2)/4)+(((p)**3)/27);\t\t\t \n",
-      "\n",
-      "\n",
-      "if D>0:\n",
-      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);\n",
-      "    Z_l = Z;\n",
-      "    Z_v = Z;\n",
-      "elif D == 0:\n",
-      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3);\t\t\t \n",
-      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z = [Z1 ,Z2, Z3];\n",
-      "    Z_l = min(Z);\n",
-      "    Z_v = max(Z);\n",
-      "else:\n",
-      "    r = math.sqrt((-(p**3)/27));\t\t\t \n",
-      "    theta = math.acos((-(q)/2)*(1./r));\t\t\n",
-      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Z = [Z1, Z2, Z3];\n",
-      "    Z_l = min(Z)\n",
-      "    Z_v = max(Z)\n",
-      "\n",
-      "\n",
-      "phi_l = math.exp(Z_l-1-math.log(Z_l-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_l+(B*(1+math.sqrt(2))))/(Z_l+(B*(1-math.sqrt(2)))))));\n",
-      "phi_v = math.exp(Z_v-1-math.log(Z_v-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_v+(B*(1+math.sqrt(2))))/(Z_v+(B*(1-math.sqrt(2)))))));\n",
-      "fl = Pguess*phi_l;\t\t\t # Calculations of the fugacity of the liquid in MPa\n",
-      "fv = Pguess*phi_v;\t\t\t # Calculations of the fugacity of the vapour in MPa\n",
-      "tolerance = 1e-3;\t\t\t #defining the tolerance to compare fl and fv\n",
-      "\n",
-      "if abs(fl-fv)<tolerance:\n",
-      "    P = Pguess;\t\t\t\n",
-      "else:\n",
-      "    Prevised = Pguess*(fl/fv)\n",
-      "\n",
-      "while abs(fl-fv)>tolerance:\n",
-      "    Tr = T/Tc\n",
-      "    Pr = (Prevised*10**6)/(Pc*10**5);\n",
-      "    S = 0.37464+(1.54226*w)-(0.26992*w**2)\n",
-      "    alpha1 = (1+(S*(1-math.sqrt(Tr))))**2;\n",
-      "    a = (0.45724*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
-      "    b = (0.07780*R*Tc)/(Pc*10**5);\t\t\t \n",
-      "    A = (a*Prevised*10**6)/(R*T)**2;\t\t\n",
-      "    B = (b*Prevised*10**6)/(R*T);\t\t\t\n",
-      "    alpha = -1+B;\t\t\t \n",
-      "    beeta = A-(2*B)-(3*B**2);\n",
-      "    gaamma = -(A*B)+(B**2)+(B**3)\n",
-      "    p = beeta-(alpha**2)/3;\t\t\n",
-      "    q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
-      "    D = (((q)**2)/4)+(((p)**3)/27);\t\n",
-      "        \n",
-      "    if D > 0:\n",
-      "        Z=((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);     #One real root given by  Eq.(3.32)\n",
-      "        Z_l=Z;\n",
-      "        Z_v=Z;\n",
-      "    elif D==0:\n",
-      "        Z1=((-2*(q/2))**(1./3))-(alpha/3);             #Three real roots and two equal given by Eq.(3.33)\n",
-      "        Z2=((q/2)**(1./3))-(alpha/3);\n",
-      "        Z3=((q/2)**(1./3))-(alpha/3);\n",
-      "        Z=[Z1, Z2, Z3];\n",
-      "        Z_l=min(Z);\n",
-      "        Z_v=max(Z);\n",
-      "    else:\n",
-      "        r = math.sqrt((-(p**3)/27));\t\t\t \n",
-      "        theta = math.acos((-(q)/2)*(1./r));\t\t\n",
-      "        Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "        Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
-      "        Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
-      "        Z = [Z1, Z2, Z3];\n",
-      "        Z_l = Z[0];\n",
-      "        Z_v = Z[1];\n",
-      "\n",
-      "    phi_l = math.exp(Z_l-1-math.log(Z_l-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_l+(B*(1+math.sqrt(2))))/(Z_l+(B*(1-math.sqrt(2)))))));\n",
-      "    phi_v = math.exp(Z_v-1-math.log(Z_v-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_v+(B*(1+math.sqrt(2))))/(Z_v+(B*(1-math.sqrt(2)))))));\n",
-      "    fl = Prevised*phi_l;\t\n",
-      "    fv = Prevised*phi_v;\t\n",
-      "    Prevised = Prevised*fl/fv\n",
-      "\n",
-      "P = Prevised;\t\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The vapour pressure of n-octane at 427.85K  =  %.5f\"%P,\" MPa\"\n",
-      "\n",
-      "# Note: answer is slightly differnt because of rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The vapour pressure of n-octane at 427.85K  =  0.21196  MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch11.ipynb b/Chemical_Engineering_Thermodynamics/ch11.ipynb
deleted file mode 100755
index 9c8a35bf..00000000
--- a/Chemical_Engineering_Thermodynamics/ch11.ipynb
+++ /dev/null
@@ -1,585 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11 : Properties of solutions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.1  Page No : 378"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "%matplotlib inline\n",
-      "import math \n",
-      "from numpy import *\n",
-      "from matplotlib.pyplot import *\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "antoine_const_benzene = [6.87987,1196.760,219.161]\n",
-      "antoine_const_toluene = [6.95087,1342.310,219.187]\n",
-      "t = 95.\n",
-      "P = 101.325\n",
-      "\n",
-      "# Calculations\n",
-      "P1_s = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(t+antoine_const_benzene[2])))\t\t\t \n",
-      "P2_s = 10**(antoine_const_toluene[0]-(antoine_const_toluene[1]/(t+antoine_const_toluene[2])))\t\t\t\n",
-      "x1 = linspace(0,1,10)\n",
-      "i = 0;\t\t\t          #iteration parameter\n",
-      "n = len(x1);\t\t\t #iteration parameter\n",
-      "P_tot = zeros(10)\n",
-      "y1 = zeros(10)\n",
-      "\n",
-      "while i<n :\n",
-      "    P_tot[i] = P2_s+((P1_s-P2_s)*x1[i])\n",
-      "    y1[i] = (x1[i]*P1_s)/(P_tot[i])\t\n",
-      "    i = i+1;\n",
-      "\n",
-      "#T-x-y diagram:\n",
-      "P = 760.\n",
-      "t1_s = ((antoine_const_benzene[1])/(antoine_const_benzene[0]-math.log10(P)))-antoine_const_benzene[2]\n",
-      "t2_s = ((antoine_const_toluene[1])/(antoine_const_toluene[0]-math.log10(P)))-antoine_const_toluene[2];\n",
-      "x1_initial = 0.0;\n",
-      "y1_initial = 0.0;\n",
-      "x1_final = 1.0;\n",
-      "y1_final = 1.0;\n",
-      "\n",
-      "T = linspace(85,105,5)\n",
-      "k = 0;\n",
-      "l = len(T)\n",
-      "P1s = zeros(5)\n",
-      "P2s = zeros(5)\n",
-      "X1 = zeros(5)\n",
-      "Y1 = zeros(5)\n",
-      "while k<l:\n",
-      "    P1s[k] = 10**((antoine_const_benzene[0])-((antoine_const_benzene[1])/(T[k]+antoine_const_benzene[2])))\n",
-      "    P2s[k] = 10**((antoine_const_toluene[0])-((antoine_const_toluene[1])/(T[k]+antoine_const_toluene[2])))\n",
-      "    X1[k] = (P-P2s[k])/(P1s[k]-P2s[k]);\t\n",
-      "    Y1[k] = (X1[k]*P1s[k])/P;\t\t\t # Calculations of mole fraction of Benzene in vapour phase (no unit) \n",
-      "    k = k+1;\n",
-      "\n",
-      "temp = zeros(l+2)\n",
-      "x1_benzene = zeros(l+2)\n",
-      "y1_benzene = zeros(l+2)\n",
-      "j = 0;\n",
-      "\n",
-      "while j<l+2:\n",
-      "    if j == 0:\n",
-      "        temp[j] = t1_s;\n",
-      "        x1_benzene[j] = x1_final;\n",
-      "        y1_benzene[j] = y1_final;\n",
-      "    elif j == l+1:\n",
-      "        temp[j] = t2_s;\n",
-      "        x1_benzene[j] = x1_initial;\n",
-      "        y1_benzene[j] = y1_initial;\n",
-      "    else:\n",
-      "        temp[j] = T[j-1];\n",
-      "        x1_benzene[j] = X1[j-1];\n",
-      "        y1_benzene[j] = Y1[j-1];\n",
-      "    j = j+1;\n",
-      "\n",
-      "# Results\n",
-      "print 'P-x-y results ';\n",
-      "\n",
-      "for i in range( n):\n",
-      "      print 'x1 = %f \\t y1 = %f\\t P = %f Torr '%(x1[i],y1[i],P_tot[i]);\n",
-      "\n",
-      "print 'T-x-y results  t = %f degree celsius\\t P1_s = 760.0 Torr \\t P2_s = -) Torr \\t\\t x1 = 1.0 \\t y1 = 1.0 '%(t1_s);\n",
-      "for k in range(l):\n",
-      "    print 't = %f degree celsius\\t P1_s = %f Torr \\t P2_s = %f Torr \\t x1 = %f \\t y1 = %f '%(T[k],P1s[k],P2s[k],X1[k],Y1[k]);\n",
-      "\n",
-      "print 't = %f degree celsius\\t P1_s = -)Torr \\t\\t P2_s = 760.0 Torr \\t x1 = 0.0 \\t y1 = 0.0 '%(t2_s);\n",
-      "subplot(2,1,1)\n",
-      "plot(x1,P_tot,y1,P_tot)\n",
-      "suptitle('P-x-y diagram for benzene-toluene system at 95 degree celsius')\n",
-      "xlabel('x1,y1')\n",
-      "ylabel('P(Torr)')\n",
-      "subplot(2,1,2)\n",
-      "plot(x1_benzene,temp,y1_benzene,temp)\n",
-      "xlabel(\"x1,y1\")\n",
-      "ylabel(\"t (degree celsius)\")       \n",
-      "suptitle('T-x-y diagram for benzene-toluene sytem at 760 Torr')\n",
-      "\n",
-      "show()\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "P-x-y results \n",
-        "x1 = 0.000000 \t y1 = 0.000000\t P = 477.025655 Torr \n",
-        "x1 = 0.111111 \t y1 = 0.235600\t P = 554.713037 Torr \n",
-        "x1 = 0.222222 \t y1 = 0.413315\t P = 632.400419 Torr \n",
-        "x1 = 0.333333 \t y1 = 0.552144\t P = 710.087801 Torr \n",
-        "x1 = 0.444444 \t y1 = 0.663592\t P = 787.775183 Torr \n",
-        "x1 = 0.555556 \t y1 = 0.755031\t P = 865.462564 Torr \n",
-        "x1 = 0.666667 \t y1 = 0.831407\t P = 943.149946 Torr \n",
-        "x1 = 0.777778 \t y1 = 0.896158\t P = 1020.837328 Torr \n",
-        "x1 = 0.888889 \t y1 = 0.951751\t P = 1098.524710 Torr \n",
-        "x1 = 1.000000 \t y1 = 1.000000\t P = 1176.212092 Torr \n",
-        "T-x-y results  t = 80.099595 degree celsius\t P1_s = 760.0 Torr \t P2_s = -) Torr \t\t x1 = 1.0 \t y1 = 1.0 \n",
-        "t = 85.000000 degree celsius\t P1_s = 881.542536 Torr \t P2_s = 345.216082 Torr \t x1 = 0.773380 \t y1 = 0.897062 \n",
-        "t = 90.000000 degree celsius\t P1_s = 1020.650885 Torr \t P2_s = 406.866584 Torr \t x1 = 0.575338 \t y1 = 0.772657 \n",
-        "t = 95.000000 degree celsius\t P1_s = 1176.212092 Torr \t P2_s = 477.025655 Torr \t x1 = 0.404719 \t y1 = 0.626363 \n",
-        "t = 100.000000 degree celsius\t P1_s = 1349.471568 Torr \t P2_s = 556.502216 Torr \t x1 = 0.256628 \t y1 = 0.455673 \n",
-        "t = 105.000000 degree celsius\t P1_s = 1541.703421 Torr \t P2_s = 646.141534 Torr \t x1 = 0.127136 \t y1 = 0.257903 \n",
-        "t = 110.614326 degree celsius\t P1_s = -)Torr \t\t P2_s = 760.0 Torr \t x1 = 0.0 \t y1 = 0.0 \n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Z2rVr4eHhAW9vb0RFRYnHExMT4efnB09PT7z11lsdZS7DMAzTTDpMOGbNmoXff/9d6diT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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3802f10>"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.2  Page No : 384"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 95. \t\t    \t #temperature of the equimolar vapour mixture of benzene and toluene in degree celsius\n",
-      "y1 = 0.5\t    \t\t #mole fraction of benzene in vapour phase (no unit)\n",
-      "y2 = 0.5    \t\t\t #mole fraction of toluene in vapour phase (no unit)\n",
-      "P1_s = 1176.21\t\t\t #saturation pressure of benzene at T, taken from Example 11.1 in Torr\n",
-      "P2_s = 477.03\t\t\t #saturation pressure of toluene at T, taken from Example 11.1 in Torr\n",
-      "\n",
-      "# Calculations\n",
-      "P = 1./((y1/P1_s)+(y2/P2_s))\n",
-      "x1 = (y1*P)/P1_s;\t\t\t \n",
-      "x2 = 1-x1\n",
-      "\n",
-      "# Results\n",
-      "print 'The composition of the liquid which is in equilibrium with the equimolar vapour mixture of\\\n",
-      " benzene and toluene at 95 degree celsius is  mole fraction of benzene\\\n",
-      "  in liquid phase x1 = %f  mole fraction of toluene in liquid phase x2 = %f '%(x1,x2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The composition of the liquid which is in equilibrium with the equimolar vapour mixture of benzene and toluene at 95 degree celsius is  mole fraction of benzene  in liquid phase x1 = 0.288542  mole fraction of toluene in liquid phase x2 = 0.711458 \n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.5 page no : 386"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variables\n",
-      "n_pentaneA = 6.87632          # Antoine constants\n",
-      "n_hexaneA = 6.91058\n",
-      "n_heptanA = 6.89386\n",
-      "n_pentaneB = 1075.780 \n",
-      "n_hexaneB = 1189.640\n",
-      "n_heptanB = 1264.370\n",
-      "n_pentaneC = 233.205\n",
-      "n_hexaneC = 226.205\n",
-      "n_heptanC = 216.640\n",
-      "pressure = 90                 # pressure\n",
-      "\n",
-      "# Calculations\n",
-      "logP1 = n_pentaneA - (n_pentaneB/(90 + n_pentaneC))\n",
-      "P1 = round(10**logP1 *133.322 /1000,1)\n",
-      "logP2 = n_hexaneA - (n_hexaneB/(90 + n_hexaneC))\n",
-      "P2 = round(10**logP2 *133.322 /1000,2)\n",
-      "logP3 = n_heptanA - (n_heptanB/(90 + n_heptanC))\n",
-      "P3 = round(10**logP3 *133.322 /1000,2)\n",
-      "P = 200\n",
-      "K1 = round(P1/P,3)\n",
-      "K2 = round(P2/P,2)\n",
-      "K3 = P3/P\n",
-      "\n",
-      "# assume L/F = 0.4\n",
-      "L_F = 0.4\n",
-      "x1 = 0.3/(L_F + (1-L_F) * K1)\n",
-      "x2 = 0.3/(L_F + (1-L_F) * K2)\n",
-      "x3 = 0.4/(L_F + (1-L_F) * K3)\n",
-      "sum_x = x1+x2+x3\n",
-      "\n",
-      "# assume L/F = 0.752\n",
-      "L_F = 0.752\n",
-      "x1 = 0.3/(L_F + (1-L_F) * K1)\n",
-      "x2 = 0.3/(L_F + (1-L_F) * K2)\n",
-      "x3 = 0.4/(L_F + (1-L_F) * K3)\n",
-      "sum_x = round(x1+x2+x3,1)           # which is equal to 1 \n",
-      "\n",
-      "# Hence, L/F = 0.752 is correct\n",
-      "y1 = K1*x1\n",
-      "y2 = K2*x2\n",
-      "y3 = K3*x3\n",
-      "\n",
-      "# Results\n",
-      "print \"Hence, fraction vaporized, V/F = 1-(L-F) = %.3f\"%(1-L_F)\n",
-      "print \"Compositions of liquid and vapor leaving the flash unit are :\"\n",
-      "print \"x1 = %.4f   y1 = %.4f\"%(x1,y1)\n",
-      "print \"x2 = %.4f   y2 = %.4f\"%(x2,y2)\n",
-      "print \"x3 = %.4f   y3 = %.4f\"%(x3,y3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hence, fraction vaporized, V/F = 1-(L-F) = 0.248\n",
-        "Compositions of liquid and vapor leaving the flash unit are :\n",
-        "x1 = 0.2246   y1 = 0.5285\n",
-        "x2 = 0.3045   y2 = 0.2863\n",
-        "x3 = 0.4709   y3 = 0.1851\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.7  Page No : 397"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 45.\t    \t\t #temperature of the mixture in degree celsius\n",
-      "A = 2.230\t\t\t #van laar constant for the system at T (no unit)\n",
-      "B = 1.959\t\t\t #van laar constant for the system at T (no unit)\n",
-      "n1 = 30.    \t\t #mole percentage of nitromethane in the mixture ( in percentage)\n",
-      "\n",
-      "# Calculations\n",
-      "n2 = 100-n1\n",
-      "x1 = n1/100\n",
-      "x2 = 1-x1\n",
-      "gaamma1 = math.exp(A/(1+((A/B)*(x1/x2)))**2)\n",
-      "gaamma2 = math.exp(B/(1+((B/A)*(x2/x1)))**2)\n",
-      "\n",
-      "# Results\n",
-      "print 'The activity coefficients for the system using van laar equation is : gamma1 \\\n",
-      " = %f, gamma2 = %f \\t '%( gaamma1,gaamma2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The activity coefficients for the system using van laar equation is : gamma1  = 2.738343, gamma2 = 1.234443 \t \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.8  Page No : 397"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "n_azeo = 44.8;\t\t\t #azeotropic composition given as mole percentage\n",
-      "Tb = 68.24;\t\t\t #boiling point of mixture in degree celsius\n",
-      "P = 760.;\t\t\t #pressure in Torr\n",
-      "P1_s = 506.;\t\t\t #saturation pressure of ethanol at Tb in Torr\n",
-      "P2_s = 517.;\t\t\t #saturation pressure of benzene at Tb in Torr\n",
-      "n1 = 10.;\t\t\t #mole percentage of ethanol in the mixture (in percentage)\n",
-      "\n",
-      "# Calculations\n",
-      "x1 = n_azeo/100\n",
-      "x2 = 1-x1;\t\t\n",
-      "gaamma1 = P/P1_s\n",
-      "gaamma2 = P/P2_s\n",
-      "A = math.log(gaamma1)*(1+((x2*math.log(gaamma2))/(x1*math.log(gaamma1))))**2\n",
-      "B = math.log(gaamma2)*(1+((x1*math.log(gaamma1))/(x2*math.log(gaamma2))))**2\n",
-      "x1 = n1/100\n",
-      "x2 = 1-x1;\t\n",
-      "gaamma1 = math.exp(A/(1+((A/B)*(x1/x2)))**2)\n",
-      "gaamma2 = math.exp(B/(1+((B/A)*(x2/x1)))**2)\n",
-      "\n",
-      "# Results\n",
-      "print 'The van Laar constants for the system are : A = %f \\t B = %f '%(A,B)\n",
-      "print 'The activity coefficients for the system using van laar equation are\\\n",
-      " : gamma1 = %f \\t gamma2 = %f \\t '%( gaamma1,gaamma2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The van Laar constants for the system are : A = 1.910203 \t B = 1.328457 \n",
-        "The activity coefficients for the system using van laar equation are : gamma1 = 4.137783 \t gamma2 = 1.025531 \t \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.9  Page No : 399"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "\n",
-      "T = 45.     \t\t\t #temperature of the system in degree celsius\n",
-      "A_12 = 0.1156;\t\t\t #Wilson's parameter for the system at T (no unit)\n",
-      "A_21 = 0.2879;\t\t\t #Wilson's parameter for the system at T (no unit)\n",
-      "x1 = 0.3;\t    \t\t #mole fraction of nitromethane in the liquid mixture (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = 1-x1\n",
-      "ln_gaamma1 = -math.log(x1+(A_12*x2))+(x2*((A_12/(x1+(A_12*x2)))-(A_21/((A_21*x1)+x2))));\t\t\t\n",
-      "gaamma1 = math.exp(ln_gaamma1)\n",
-      "ln_gaamma2 = -math.log(x2+(A_21*x1))-(x1*((A_12/(x1+(A_12*x2)))-(A_21/((A_21*x1)+x2))));\t\t\t # Calculations of ln(activity coefficient) using Eq.(11.90) (no unit)\n",
-      "gaamma2 = math.exp(ln_gaamma2)\n",
-      "\n",
-      "# Results\n",
-      "print 'The activity coefficients for the system using Wilsons parameters are : gamma1 = %f \\\n",
-      " \\t gamma2 = %f \\t '%( gaamma1,gaamma2);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The activity coefficients for the system using Wilsons parameters are : gamma1 = 2.512605  \t gamma2 = 1.295788 \t \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.10  Page No : 401"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 345.\t\t\t #temperature of the mixture in K\n",
-      "x1 = 0.8\t\t\t #mole fraction of ethanol in the liquid phase (no unit)\n",
-      "\n",
-      "nu_ki = [1,1,1,6]\t\t\t #number of groups of type: CH3, CH2, OH and ACH respectively (no unit)\n",
-      "R_k = [0.9011,0.6744,1.0000,0.5313];\t\t\t #Group volume parameter for CH3, CH2, OH and ACH respectively (no unit)\n",
-      "Q_k = [0.848,0.540,1.200,0.400];\t\t\t #Area parameter for CH3, CH2, OH and ACH respectively (no unit)\n",
-      "R = 8.314;\t\t\t        #universal gas constant in J/molK\n",
-      "u12_u22 = -241.2287;\t\t#UNIQUAC parameter for the system in J/molK\n",
-      "u21_u11 = 2799.5827;\t\t#UNIQUAC parameter for the system in J/molK\n",
-      "z = 10.     \t\t\t #co-ordination number usually taken as 10 (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = 1-x1\n",
-      "r1 = (nu_ki[0]*R_k[0])+(nu_ki[1]*R_k[1])+(nu_ki[2]*R_k[2]);\t\t\t \n",
-      "r2 = (nu_ki[3]*R_k[3])\n",
-      "phi1 = (x1*r1)/((x1*r1)+(x2*r2));\t\t\t \n",
-      "phi2 = (x2*r2)/((x2*r2)+(x1*r1));\t\t\t \n",
-      "q1 = (nu_ki[0]*Q_k[0])+(nu_ki[1]*Q_k[1])+(nu_ki[2]*Q_k[2])\t\t\t \n",
-      "q2 = (nu_ki[3]*Q_k[3])\t\t\t\n",
-      "theta1 = (x1*q1)/((x1*q1)+(x2*q2))\n",
-      "theta2 = (x2*q2)/((x1*q1)+(x2*q2))\n",
-      "l1 = ((z/2)*(r1-q1))-(r1-1);\t\t\n",
-      "l2 = ((z/2)*(r2-q2))-(r2-1);\t\t\n",
-      "tau_12 = math.exp(-(u12_u22)/(R*T));\t\t\n",
-      "tau_21 = math.exp(-(u21_u11)/(R*T));\t\t\n",
-      "tau_11 = 1.0;\t\t\t \n",
-      "tau_22 = 1.0;\t\t\t \n",
-      "\n",
-      "ln_gaamma1_c = math.log(phi1/x1)+((z/2)*q1*math.log(theta1/phi1))+l1-((phi1/x1)*((x1*l1)+(x2*l2)));\n",
-      "ln_gaamma2_c = math.log(phi2/x2)+((z/2)*q2*math.log(theta2/phi2))+l2-((phi2/x2)*((x1*l1)+(x2*l2)));\n",
-      "ln_gaamma1_r = q1*(1-math.log((theta1*tau_11)+(theta2*tau_21))-(((theta1*tau_11)/((theta1*tau_11)+(theta2*tau_21)))+((theta2*tau_12)/((theta1*tau_12)+(theta2*tau_22)))));\n",
-      "ln_gaamma2_r = q2*(1-math.log((theta1*tau_12)+(theta2*tau_22))-(((theta1*tau_21)/((theta1*tau_11)+(theta2*tau_21)))+((theta2*tau_22)/((theta1*tau_12)+(theta2*tau_22)))));\n",
-      "ln_gaamma1 = ln_gaamma1_c+ln_gaamma1_r\n",
-      "ln_gaamma2 = ln_gaamma2_c+ln_gaamma2_r\n",
-      "gaamma1 = math.exp(ln_gaamma1);\t\t\t \n",
-      "gaamma2 = math.exp(ln_gaamma2);\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print 'The activity coefficients for the system using the UNIQUAC equation are : gamma1 \\\n",
-      " = %f \\t gamma2 = %f \\t '%( gaamma1,gaamma2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The activity coefficients for the system using the UNIQUAC equation are : gamma1  = 1.060567 \t gamma2 = 3.679066 \t \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.11  Page No : 405"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 307.\t\t\t #temperature of the mixture in K\n",
-      "x1 = 0.3\t\t\t #mole fraction of acetone in the liquid phase (no unit)\n",
-      "\n",
-      "nu_ki = [1.,1.,2.,3.]\n",
-      "R_k = [0.9011,1.6724,0.6744]\n",
-      "Q_k = [0.848,1.488,0.540];\t\t\t \n",
-      "a_19 = 476.40;\t\t\t #group interaction parameter for the system in K\n",
-      "a_91 = 26.760;\t\t\t #group interaction parameter for the system in K\n",
-      "z = 10.     \t\t\t #co-ordination number usually taken as 10 (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = 1-x1;\t\t\t # Calculations of mole fraction of benzene in liquid phase (no unit)\n",
-      "r1 = (nu_ki[0]*R_k[0])+(nu_ki[1]*R_k[1])\n",
-      "r2 = (nu_ki[2]*R_k[0])+(nu_ki[3]*R_k[2])\n",
-      "phi1 = (x1*r1)/((x1*r1)+(x2*r2))\n",
-      "phi2 = (x2*r2)/((x2*r2)+(x1*r1))\n",
-      "q1 = (nu_ki[0]*Q_k[0])+(nu_ki[1]*Q_k[1])\t\t\t\n",
-      "q2 = (nu_ki[2]*Q_k[0])+(nu_ki[3]*Q_k[2])\n",
-      "theta1 = (x1*q1)/((x1*q1)+(x2*q2));\n",
-      "theta2 = (x2*q2)/((x1*q1)+(x2*q2));\n",
-      "l1 = ((z/2)*(r1-q1))-(r1-1);\t\t\n",
-      "l2 = ((z/2)*(r2-q2))-(r2-1);\t\t\n",
-      "ln_gaamma1_c = math.log(phi1/x1)+((z/2)*q1*math.log(theta1/phi1))+l1-((phi1/x1)*((x1*l1)+(x2*l2)));\n",
-      "ln_gaamma2_c = math.log(phi2/x2)+((z/2)*q2*math.log(theta2/phi2))+l2-((phi2/x2)*((x1*l1)+(x2*l2)));\n",
-      "a_11 = 0.\n",
-      "a_99 = 0.\n",
-      "psi_19 = math.exp(-(a_19)/(T));\t\t\t\n",
-      "psi_91 = math.exp(-(a_91)/(T));\t\t\t\n",
-      "psi_11 = 1.;\t\t\t \n",
-      "psi_99 = 1.;\t\t\t \n",
-      "x1_1 = nu_ki[0]/(nu_ki[0]+nu_ki[1]);\t\t\t\n",
-      "x1_18 = nu_ki[1]/(nu_ki[0]+nu_ki[1]);\t\t\n",
-      "theta1_1 = (Q_k[0]*x1_1)/((Q_k[0]*x1_1)+(Q_k[1]*x1_18))\n",
-      "theta1_18 = (Q_k[1]*x1_18)/((Q_k[1]*x1_18)+(Q_k[0]*x1_1))\n",
-      "ln_tau1_1 = Q_k[0]*(1-math.log((theta1_1*psi_11)+(theta1_18*psi_91))-(((theta1_1*psi_11)/((theta1_1*psi_11)+(theta1_18*psi_91)))+((theta1_18*psi_19)/((theta1_1*psi_19)+(theta1_18*psi_11)))));\n",
-      "ln_tau1_18 = Q_k[1]*(1-math.log((theta1_1*psi_19)+(theta1_18*psi_99))-(((theta1_1*psi_91)/((theta1_1*psi_99)+(theta1_18*psi_91)))+((theta1_18*psi_99)/((theta1_1*psi_19)+(theta1_18*psi_99)))));\n",
-      "x2_1 = nu_ki[2]/(nu_ki[2]+nu_ki[3]);\t\t\t \n",
-      "x2_2 = nu_ki[3]/(nu_ki[2]+nu_ki[3]);\t\t\t \n",
-      "ln_tau2_1 = 0;\n",
-      "ln_tau2_2 = 0;\n",
-      "x_1 = ((x1*nu_ki[0])+(x2*nu_ki[2]))/((((x1*nu_ki[0])+(x1*nu_ki[1])))+((x2*nu_ki[2])+(x2*nu_ki[3])));\n",
-      "x_2 = ((x2*nu_ki[3]))/((((x1*nu_ki[0])+(x1*nu_ki[1])))+((x2*nu_ki[2])+(x2*nu_ki[3])));\n",
-      "x_18 = ((x1*nu_ki[1]))/((((x1*nu_ki[0])+(x1*nu_ki[1])))+((x2*nu_ki[2])+(x2*nu_ki[3])));\n",
-      "theta_1 = (Q_k[0]*x_1)/((Q_k[0]*x_1)+(Q_k[1]*x_18)+(Q_k[2]*x_2));\t\t\t\n",
-      "theta_2 = (Q_k[2]*x_2)/((Q_k[0]*x_1)+(Q_k[1]*x_18)+(Q_k[2]*x_2));\t\t\t\n",
-      "theta_18 = (Q_k[1]*x_18)/((Q_k[0]*x_1)+(Q_k[1]*x_18)+(Q_k[2]*x_2));\t\t\t\n",
-      "ln_tau_1 = Q_k[0]*(1-math.log((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91))-((((theta_1*psi_11)+(theta_2*psi_11))/((((theta_1*psi_11)+(theta_2*psi_11))+(theta_18*psi_91)))+((theta_18*psi_19)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));\n",
-      "ln_tau_2 = Q_k[2]*(1-math.log((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91))-((((theta_1*psi_11)+(theta_2*psi_11))/((((theta_1*psi_11)+(theta_2*psi_11))+(theta_18*psi_91)))+((theta_18*psi_19)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));\n",
-      "ln_tau_18 = Q_k[1]*(1-math.log((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_99))-(((((theta_1+theta_2)*psi_91)/((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91)))+((theta_18*psi_99)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));\n",
-      "ln_gaamma1_r = (nu_ki[0]*(ln_tau_1-ln_tau1_1))+(nu_ki[1]*(ln_tau_18-ln_tau1_18));\n",
-      "ln_gaamma2_r = (nu_ki[2]*(ln_tau_1-ln_tau2_1))+(nu_ki[3]*(ln_tau_2-ln_tau2_2));\n",
-      "ln_gaamma1 = ln_gaamma1_c+ln_gaamma1_r\n",
-      "ln_gaamma2 = ln_gaamma2_c+ln_gaamma2_r\n",
-      "gaamma1 = math.exp(ln_gaamma1);\t\t\t\n",
-      "gaamma2 = math.exp(ln_gaamma2);\t\t\t\n",
-      "\n",
-      "# Results\n",
-      "print 'The activity coefficients for the system using the UNIFAC method are\\\n",
-      " : gamma1 = %f \\t gamma2 = %f \\t '%( gaamma1,gaamma2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The activity coefficients for the system using the UNIFAC method are : gamma1 = 2.149891 \t gamma2 = 1.191192 \t \n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch12.ipynb b/Chemical_Engineering_Thermodynamics/ch12.ipynb
deleted file mode 100755
index fe626468..00000000
--- a/Chemical_Engineering_Thermodynamics/ch12.ipynb
+++ /dev/null
@@ -1,716 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12 : Vapor liquid Equilibrium"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.1  Page No : 423"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "%matplotlib inline\n",
-      "import math \n",
-      "from numpy import *\n",
-      "from matplotlib.pyplot import *\n",
-      "\n",
-      "# Variables\n",
-      "T = 60.\t\t\t #temperature of the system in degree celsius\n",
-      "P = [237.60,265.20,317.50,333.00,368.70,387.20];\t\t\t #Pressure data in Torr (from Danneil et al.)\n",
-      "x1 = [0.0870,0.1800,0.4040,0.4790,0.7130,0.9070];\t\t\t #mole fraction of benzene in the liquid phase corresponding to the given pressure (no unit) (from Danneil et al.)\n",
-      "y1 = [0.1870,0.3400,0.5780,0.6420,0.7960,0.9220];\t\t\t #mole fraction of benzene in the vapour phase corresponding to the given pressure (no unit) (from Danneil et al.)\n",
-      "antoine_const_benzene = [6.87987,1196.760,219.161];\t\t\t #Antoine's constants for Benzene from Table A.7\n",
-      "antoine_const_heptane = [6.89386,1264.370,216.640];\t\t\t #Antoine's constants for heptane from Table A.7\n",
-      "\n",
-      "# Calculations\n",
-      "P1_s = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(T+antoine_const_benzene[2])));\n",
-      "P2_s = 10**(antoine_const_heptane[0]-(antoine_const_heptane[1]/(T+antoine_const_heptane[2])));\n",
-      "l = len(P);\t\t\t #iteration parameter\n",
-      "i = 0;\t\t\t #iteration parameter\n",
-      "gaamma1 = zeros(l)\n",
-      "gaamma2 = zeros(l)\n",
-      "ln_gaamma1_expt = zeros(l)\n",
-      "ln_gaamma2_expt = zeros(l)\n",
-      "gE_RTx1x2 = zeros(l)\n",
-      "while i<l:\n",
-      "    gaamma1[i] = (y1[i]*P[i])/(x1[i]*P1_s);\t\t\n",
-      "    gaamma2[i] = ((1-y1[i])*P[i])/((1-x1[i])*P2_s);\n",
-      "    ln_gaamma1_expt[i] = math.log(gaamma1[i]);\n",
-      "    ln_gaamma2_expt[i] = math.log(gaamma2[i]);\n",
-      "    gE_RTx1x2[i] = ((x1[i]*ln_gaamma1_expt[i])+((1-x1[i])*ln_gaamma2_expt[i]))/(x1[i]*(1-x1[i]));\t\t\t # Calculations of gE/RT using Eq.(11.36) (no unit)\n",
-      "    i = i+1;\n",
-      "\n",
-      "plot(x1,gE_RTx1x2,'o');\t\t\n",
-      "suptitle('Plot of gE/RTx1x2 vs x1')\n",
-      "xlabel('x1')\n",
-      "ylabel('gE/RTx1x2');\n",
-      "A21 = 0.555\n",
-      "A12 = 0.315\n",
-      "j = 0\n",
-      "ln_gaamma1 = zeros(l)\n",
-      "ln_gaamma2 = zeros(l)\n",
-      "P_calc = zeros(l)\n",
-      "y1_calc = zeros(l)\n",
-      "gaamma1 = zeros(l)\n",
-      "gaamma2 = zeros(l)\n",
-      "while j<l:\n",
-      "    ln_gaamma1[j] = ((1-x1[j])**2)*(A12+(2*(A21-A12)*x1[j]));\n",
-      "    ln_gaamma2[j] = (x1[j]**2)*(A21+(2*(A12-A21)*(1-x1[j])));\n",
-      "    gaamma1[j] = math.exp(ln_gaamma1[j])\n",
-      "    gaamma2[j] = math.exp(ln_gaamma2[j])\n",
-      "    P_calc[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
-      "    y1_calc[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
-      "    j = j+1;\n",
-      "\n",
-      "# Results\n",
-      "print 'Data for the plot of gE/RTx1x2 vs x1: ';\n",
-      "\n",
-      "for i in range(l):\n",
-      "    print 'P = %f Torr\\t x1 = %f\\t y1 = %f \\t lngamma1) = %f\\t\\t lngamma2) =\\\n",
-      "     %f\\t\\t gE/RTx1x2 = %f'%(P[i],x1[i],y1[i],ln_gaamma1_expt[i],ln_gaamma2_expt[i],gE_RTx1x2[i]);\n",
-      "\n",
-      "print 'Results: ';\n",
-      "for i in range(l):\n",
-      "    print 'x1 = %f \\t gamma1 = %f \\t gamma2 = %f \\t P_Exptl. = %f \\\n",
-      "    Torr\\t P_Calc = %f Torr\\t y1_Exptl = %f \\t y1_calc = %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],P_calc[i],y1[i],y1_calc[i]);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Data for the plot of gE/RTx1x2 vs x1: "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "P = 237.600000 Torr\t x1 = 0.087000\t y1 = 0.187000 \t lngamma1) = 0.265459\t\t lngamma2) =     0.004741\t\t gE/RTx1x2 = 0.345243\n",
-        "P = 265.200000 Torr\t x1 = 0.180000\t y1 = 0.340000 \t lngamma1) = 0.246143\t\t lngamma2) =     0.013576\t\t gE/RTx1x2 = 0.375599\n",
-        "P = 317.500000 Torr\t x1 = 0.404000\t y1 = 0.578000 \t lngamma1) = 0.148307\t\t lngamma2) =     0.065399\t\t gE/RTx1x2 = 0.410716\n",
-        "P = 333.000000 Torr\t x1 = 0.479000\t y1 = 0.642000 \t lngamma1) = 0.130700\t\t lngamma2) =     0.083082\t\t gE/RTx1x2 = 0.424313\n",
-        "P = 368.700000 Torr\t x1 = 0.713000\t y1 = 0.796000 \t lngamma1) = 0.049771\t\t lngamma2) =     0.218778\t\t gE/RTx1x2 = 0.480259\n",
-        "P = 387.200000 Torr\t x1 = 0.907000\t y1 = 0.922000 \t lngamma1) = 0.005014\t\t lngamma2) =     0.433207\t\t gE/RTx1x2 = 0.531539\n",
-        "Results: \n",
-        "x1 = 0.087000 \t gamma1 = 1.346332 \t gamma2 = 1.000884 \t P_Exptl. = 237.600000     Torr\t P_Calc = 238.297793 Torr\t y1_Exptl = 0.187000 \t y1_calc = 0.193066 \n",
-        "x1 = 0.180000 \t gamma1 = 1.309835 \t gamma2 = 1.005243 \t P_Exptl. = 265.200000     Torr\t P_Calc = 265.912990 Torr\t y1_Exptl = 0.340000 \t y1_calc = 0.348175 \n",
-        "x1 = 0.404000 \t gamma1 = 1.198147 \t gamma2 = 1.044870 \t P_Exptl. = 317.500000     Torr\t P_Calc = 320.705660 Torr\t y1_Exptl = 0.578000 \t y1_calc = 0.597076 \n",
-        "x1 = 0.479000 \t gamma1 = 1.159413 \t gamma2 = 1.072467 \t P_Exptl. = 333.000000     Torr\t P_Calc = 335.157916 Torr\t y1_Exptl = 0.642000 \t y1_calc = 0.653147 \n",
-        "x1 = 0.713000 \t gamma1 = 1.055628 \t gamma2 = 1.236286 \t P_Exptl. = 368.700000     Torr\t P_Calc = 369.484272 Torr\t y1_Exptl = 0.796000 \t y1_calc = 0.799482 \n",
-        "x1 = 0.907000 \t gamma1 = 1.006511 \t gamma2 = 1.521729 \t P_Exptl. = 387.200000     Torr\t P_Calc = 387.326500 Torr\t y1_Exptl = 0.922000 \t y1_calc = 0.923362 \n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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LbZq/vz/0ej0MBgMiIiKULJOIiCR4KrViq9WK9PR0FBcXY+TIkYiMjERsbCwMBoPdcm1t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\nUhgYREQkhYFBRERSGBhERCTl/wBGhb+in7ykegAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2dcf110>"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.3  Page No : 430"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from numpy import *\n",
-      "# Variables\n",
-      "T = 25.     \t\t\t #temperature of the system in degree celsius\n",
-      "A12 = 2.0522;\t\t\t #three suffix Margules parameters for the system (no unit)\n",
-      "A21 = 1.7201;\t\t\t #three suffix Margules parameters for the system (no unit)\n",
-      "P = [118.05,207.70,246.35,259.40,261.50,262.00,261.90,258.70,252.00];\t\t\t #Pressure data in Torr (from Tasic et al.)\n",
-      "x1 = [0.0115,0.1125,0.3090,0.5760,0.6920,0.7390,0.7575,0.8605,0.9250];\n",
-      "y1 = [0.1810,0.5670,0.6550,0.7050,0.7250,0.7390,0.7460,0.8030,0.8580];\n",
-      "antoine_const_acetone = [7.11714,1210.595,229.664];\t\t\t #Antoine's constants for acetone from Table A.7\n",
-      "antoine_const_chexane = [6.85146,1206.470,223.136];\t\t\t #Antoine's constants for cyclohexane from Table A.7\n",
-      "\n",
-      "# Calculations\n",
-      "P1_s = 10**(antoine_const_acetone[0]-(antoine_const_acetone[1]/(T+antoine_const_acetone[2])));\n",
-      "P2_s = 10**(antoine_const_chexane[0]-(antoine_const_chexane[1]/(T+antoine_const_chexane[2])));\n",
-      "\n",
-      "l = len(P)\n",
-      "ln_gaamma1 = zeros(l)\n",
-      "ln_gaamma2 = zeros(l)\n",
-      "gaamma1 = zeros(l)\n",
-      "gaamma2 = zeros(l)\n",
-      "P = zeros(l)\n",
-      "y1_calc = zeros(l)\n",
-      "j = 0;\t\n",
-      "\n",
-      "while j<l:\n",
-      "    ln_gaamma1[j] = ((1-x1[j])**2)*(A12+(2*(A21-A12)*x1[j]));\t\n",
-      "    ln_gaamma2[j] = (x1[j]**2)*(A21+(2*(A12-A21)*(1-x1[j])));\t\n",
-      "    gaamma1[j] = math.exp(ln_gaamma1[j]);\t\t\t \n",
-      "    gaamma2[j] = math.exp(ln_gaamma2[j]);\t\t\t \n",
-      "    P[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
-      "    y1_calc[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
-      "    j = j+1;\n",
-      "\n",
-      "# Results\n",
-      "print 'P-x-y data: ';\n",
-      "print 'x1 \\t  gamma1\\t  gamma2 \\t P Torr \\t y1  ';\n",
-      "for i in range(l):\n",
-      "    print '%0.4f \\t %f \\t %f \\t %f  \\t %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],y1_calc[i])\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "P-x-y data: \n",
-        "x1 \t  gamma1\t  gamma2 \t P Torr \t y1  \n",
-        "0.0115 \t 7.372871 \t 1.000314 \t 116.059350  \t 0.168694 \n",
-        "0.1125 \t 4.747283 \t 1.029662 \t 212.486878  \t 0.580377 \n",
-        "0.3090 \t 2.415459 \t 1.231286 \t 255.363338  \t 0.674908 \n",
-        "0.5760 \t 1.350072 \t 1.942786 \t 259.940648  \t 0.690796 \n",
-        "0.6920 \t 1.163087 \t 2.513451 \t 261.385489  \t 0.711020 \n",
-        "0.7390 \t 1.112223 \t 2.812451 \t 261.416681  \t 0.726019 \n",
-        "0.7575 \t 1.095373 \t 2.942991 \t 261.232545  \t 0.733436 \n",
-        "0.8605 \t 1.029233 \t 3.827737 \t 256.608420  \t 0.796964 \n",
-        "0.9250 \t 1.008121 \t 4.546617 \t 248.599211  \t 0.866162 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.4  Page No : 432"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 25. \t\t\t #temperature of the system in degree celsius\n",
-      "A = 2.0684;\t\t\t #the van Laar parameters for the system (no unit)\n",
-      "B = 1.7174;\t\t\t #the van Laar parameters for the system (no unit)\n",
-      "P = [118.05,207.70,246.35,259.40,261.50,262.00,261.90,258.70,252.00];\t\t\t #Pressure data in Torr (from Tasic et al.)\n",
-      "x1 = [0.0115,0.1125,0.3090,0.5760,0.6920,0.7390,0.7575,0.8605,0.9250];\n",
-      "y1 = [0.1810,0.5670,0.6550,0.7050,0.7250,0.7390,0.7460,0.8030,0.8580];\n",
-      "antoine_const_acetone = [7.11714,1210.595,229.664];\t\t\t #Antoine's constants for acetone from Table A.7\n",
-      "antoine_const_chexane = [6.85146,1206.470,223.136];\t\t\t #Antoine's constants for cyclohexane from Table A.7\n",
-      "\n",
-      "# Calculations\n",
-      "P1_s = 10**(antoine_const_acetone[0]-(antoine_const_acetone[1]/(T+antoine_const_acetone[2])))\n",
-      "P2_s = 10**(antoine_const_chexane[0]-(antoine_const_chexane[1]/(T+antoine_const_chexane[2])));\n",
-      "l = len(P)\n",
-      "j = 0\n",
-      "while j<l:\n",
-      "    ln_gaamma1[j] = A/(1+((A*x1[j])/(B*(1-x1[j]))))**2\n",
-      "    ln_gaamma2[j] = B/(1+((B*(1-x1[j]))/(A*x1[j])))**2\n",
-      "    gaamma1[j] = math.exp(ln_gaamma1[j]);\t\t\t\n",
-      "    gaamma2[j] = math.exp(ln_gaamma2[j]);\t\t\t\n",
-      "    P[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
-      "    y1_calc[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
-      "    j = j+1;\n",
-      "\n",
-      "# Results\n",
-      "print 'P-x-y data: ';\n",
-      "i = 0;\n",
-      "print 'x1 \\t  gamma1 \\t gamma2 \\t PTorr \\t y1 ';\n",
-      "for i in range(l):\n",
-      "    print '%0.4f \\t %f \\t %f \\t %f  \\t %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],y1_calc[i]);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "P-x-y data: \n",
-        "x1 \t  gamma1 \t gamma2 \t PTorr \t y1 \n",
-        "0.0115 \t 7.475516 \t 1.000328 \t 116.333233  \t 0.170640 \n",
-        "0.1125 \t 4.743506 \t 1.030586 \t 212.468724  \t 0.579965 \n",
-        "0.3090 \t 2.395936 \t 1.234218 \t 254.168007  \t 0.672601 \n",
-        "0.5760 \t 1.346684 \t 1.937830 \t 259.284960  \t 0.690805 \n",
-        "0.6920 \t 1.162537 \t 2.498302 \t 260.842215  \t 0.712164 \n",
-        "0.7390 \t 1.112211 \t 2.792268 \t 260.900602  \t 0.727447 \n",
-        "0.7575 \t 1.095497 \t 2.920797 \t 260.729035  \t 0.734935 \n",
-        "0.8605 \t 1.029539 \t 3.796506 \t 256.244215  \t 0.798334 \n",
-        "0.9250 \t 1.008263 \t 4.515793 \t 248.404105  \t 0.866965 \n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.5  Page No : 435"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from numpy import *\n",
-      "from matplotlib.pyplot import *\n",
-      "\n",
-      "# Variables\n",
-      "P = 760.\t\t\t #pressure in Torr at which chloroform and methanol form an azeotrope\n",
-      "T = 53.5\t\t\t #temperature in degree celsius at which chloroform and methanol form an azeotrope\n",
-      "x1 = 0.65\t\t\t #mole fraction of chloroform in the liquid phase (no unit) (corresponding to azeotropic composition)\n",
-      "antoine_const_chloroform = [6.95465,1170.966,226.232];\t\t\t #Antoine's constants for acetone from Table A.7\n",
-      "antoine_const_methanol = [8.08097,1582.271,239.726];\t\t\t #Antoine's constants for acetone from Table A.7\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = 1-x1\n",
-      "P1_s = 10**(antoine_const_chloroform[0]-(antoine_const_chloroform[1]/(T+antoine_const_chloroform[2])));\n",
-      "P2_s = 10**(antoine_const_methanol[0]-(antoine_const_methanol[1]/(T+antoine_const_methanol[2])));\n",
-      "gaamma1 = P/P1_s\n",
-      "gaamma2 = P/P2_s\n",
-      "A = math.log(gaamma1)*(1+((x2*math.log(gaamma2))/(x1*math.log(gaamma1))))**2\n",
-      "B = math.log(gaamma2)*(1+((x1*math.log(gaamma1))/(x2*math.log(gaamma2))))**2\n",
-      "x1 = linspace(0.1,0.9,9)\n",
-      "l = len(x1)\n",
-      "gaamma1 = zeros(l)\n",
-      "gaamma2 = zeros(l)\n",
-      "P  = zeros(l) \n",
-      "y1 = zeros(l)\n",
-      "j = 0\n",
-      "while j<l:\n",
-      "    ln_gaamma1[j] = A/(1+((A*x1[j])/(B*(1-x1[j]))))**2\n",
-      "    ln_gaamma2[j] = B/(1+((B*(1-x1[j]))/(A*x1[j])))**2\n",
-      "    gaamma1[j] = math.exp(ln_gaamma1[j]);\t\t\t \n",
-      "    gaamma2[j] = math.exp(ln_gaamma2[j]);\t\t\t \n",
-      "    P[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
-      "    y1[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
-      "    j = j+1;\n",
-      "\n",
-      "# Results\n",
-      "print 'VLE data: ';\n",
-      "print 'x1 \\tgamma1 \\t\\t gamma2  \\t P Torr \\t y1 ';\n",
-      "for i in range(l):\n",
-      "    print '%0.1f \\t %f \\t %f \\t %f \\t %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],y1[i]);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VLE data: \n",
-        "x1 \tgamma1 \t\t gamma2  \t P Torr \t y1 \n",
-        "0.1 \t 2.391069 \t 1.005464 \t 578.375597 \t 0.242663 \n",
-        "0.2 \t 2.151801 \t 1.024547 \t 649.357212 \t 0.389019 \n",
-        "0.3 \t 1.928361 \t 1.062974 \t 699.744047 \t 0.485280 \n",
-        "0.4 \t 1.722251 \t 1.130135 \t 732.594248 \t 0.551969 \n",
-        "0.5 \t 1.535159 \t 1.242383 \t 751.239893 \t 0.599745 \n",
-        "0.6 \t 1.369108 \t 1.430331 \t 759.122391 \t 0.635183 \n",
-        "0.7 \t 1.226766 \t 1.756665 \t 759.153131 \t 0.663976 \n",
-        "0.8 \t 1.112114 \t 2.365222 \t 751.206494 \t 0.695188 \n",
-        "0.9 \t 1.031997 \t 3.639052 \t 721.331744 \t 0.755801 \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.6  Page No : 443"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "y1 = 0.3;\t\t\t #mole fraction of ethane in the vapour phase (no unit)\n",
-      "T = 30. \t\t\t #temperature in degree celsius\n",
-      "\n",
-      "# Calculations\n",
-      "y2 = 1-y1\n",
-      "P_guess = 1\n",
-      "K1 = 3.4\n",
-      "K2 = 1.1\n",
-      "x1_calc = y1/K1\n",
-      "x2_calc = y2/K2;\n",
-      "tot = x1_calc+x2_calc\n",
-      "if tot == 1:\n",
-      "    P = P_guess\n",
-      "    x1 = x1_calc\n",
-      "    x2 = x2_calc\n",
-      "else:\n",
-      "    P = 1.5\n",
-      "    K1 = 2.4\n",
-      "    K2 = 0.8\n",
-      "    x1 = y1/K1\n",
-      "    x2 = y2/K2\n",
-      "\n",
-      "# Results\n",
-      "print 'The Dew pressure and the liquid composition of a binary vapour mixture of\\\n",
-      " ethane and propane was found to be P = %0.2f MPa\\t x1 = %0.3f\\t x2 = %0.3f \\t'%(P,x1,x2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Dew pressure and the liquid composition of a binary vapour mixture of ethane and propane was found to be P = 1.50 MPa\t x1 = 0.125\t x2 = 0.875 \t\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.7  Page No : 443"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "x1 = 0.4;\t\t\t #mole fraction of ethane in the liquid phase (no unit)\n",
-      "P = 1.5;\t\t\t #pressure in MPa\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = 1-x1\n",
-      "t_guess = 10.\n",
-      "K1 = 1.8;\t\n",
-      "K2 = 0.5;\t\n",
-      "y1_calc = K1*x1\n",
-      "y2_calc = K2*x2\n",
-      "tot = y1_calc+y2_calc\n",
-      "if tot == 1:\n",
-      "   t = t_guess\n",
-      "   y1 = y1_calc\n",
-      "   y2 = y2_calc\n",
-      "else:\n",
-      "    t = 9.\n",
-      "    K1 = 1.75\n",
-      "    K2 = 0.5;\n",
-      "    y1 = K1*x1\n",
-      "    y2 = K2*x2\n",
-      "\n",
-      "# Results\n",
-      "print 'The bubble temperature and the vapour composition of a binary vapour mixture\\\n",
-      " of ethane and propane was found to be t = %d degree celsius y1 = %f\\t y2 = %f\\t'%(t,y1,y2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bubble temperature and the vapour composition of a binary vapour mixture of ethane and propane was found to be t = 9 degree celsius y1 = 0.700000\t y2 = 0.300000\t\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.8  Page No : 449"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from numpy import *\n",
-      "from matplotlib.pyplot import *\n",
-      "%matplotlib inline\n",
-      "# Variables\n",
-      "P = [350.00,446.00,518.00,574.50,609.00,632.50,665.00,681.50,691.50]\n",
-      "x1 = [0.0550,0.1290,0.2120,0.3130,0.4300,0.5200,0.6380,0.7490,0.8720]\n",
-      "y1 = [0.3500,0.5110,0.5990,0.6500,0.6970,0.7260,0.7590,0.8130,0.8830]\n",
-      "antoine_const_propanol = [8.37895,1788.020,227.438];\t\t\t \n",
-      "antoine_const_cbenzene = [7.17294,1549.200,229.260];\t\t\t \n",
-      "T = 95.\t\t\t #temperature of the system in degree celsius\n",
-      "\n",
-      "# Calculations\n",
-      "P1_s = 10**(antoine_const_propanol[0]-(antoine_const_propanol[1]/(T+antoine_const_propanol[2])));\n",
-      "P2_s = 10**(antoine_const_cbenzene[0]-(antoine_const_cbenzene[1]/(T+antoine_const_cbenzene[2])));\n",
-      "l = len(P)\n",
-      "lngamma1_gamma2 = zeros(l)\n",
-      "gaamma1 = zeros(l)\n",
-      "gaamma2 = zeros(l)\n",
-      "i = 0\n",
-      "while i<l:\n",
-      "    gaamma1[i] = (y1[i]*P[i])/(x1[i]*P1_s)\n",
-      "    gaamma2[i] = ((1-y1[i])*P[i])/((1-x1[i])*P2_s)\n",
-      "    lngamma1_gamma2[i] = math.log(gaamma1[i]/gaamma2[i])\n",
-      "    i = i+1;\n",
-      "\n",
-      "plot(x1,lngamma1_gamma2);\n",
-      "suptitle('Plot of ln(gamma1/gamma2) vs x1')\n",
-      "xlabel('x1')\n",
-      "ylabel('ln(gamma1/gamma2)');\n",
-      "area_above = 1515.\n",
-      "area_below = 1540.\n",
-      "consistency_parameter = abs((area_above-area_below)/(area_above+area_below));\t\t\t\n",
-      "\n",
-      "# Results\n",
-      "print 'Values of lngamma1./gamma2: ';\n",
-      "print 'x1 \\t gamma1 \\t gamma2 \\t lngamma1./gamma2';\n",
-      "\n",
-      "for i in range(l):\n",
-      "    print '%0.4f \\t %f \\t %f \\t %f '%(x1[i],gaamma1[i],gaamma2[i],lngamma1_gamma2[i]);\n",
-      "\n",
-      "print 'The value of the consistency parameter = %f'%(consistency_parameter);\n",
-      "if consistency_parameter<0.02 or consistency_parameter == 0.02:\n",
-      "    print 'The VLE data is thermodynamically consistent';\n",
-      "else:\n",
-      "    print 'The VLE data is not thermodynamically consistent';\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Values of lngamma1./gamma2: "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "x1 \t gamma1 \t gamma2 \t lngamma1./gamma2\n",
-        "0.0550 \t 3.266913 \t 0.968851 \t 1.215489 \n",
-        "0.1290 \t 2.591375 \t 1.007704 \t 0.944514 \n",
-        "0.2120 \t 2.146767 \t 1.060854 \t 0.704889 \n",
-        "0.3130 \t 1.749940 \t 1.177901 \t 0.395847 \n",
-        "0.4300 \t 1.447924 \t 1.302845 \t 0.105581 \n",
-        "0.5200 \t 1.295263 \t 1.453040 \t -0.114944 \n",
-        "0.6380 \t 1.160398 \t 1.781713 \t -0.428812 \n",
-        "0.7490 \t 1.085022 \t 2.043343 \t -0.632987 \n",
-        "0.8720 \t 1.027071 \t 2.543757 \t -0.906931 \n",
-        "The value of the consistency parameter = 0.008183\n",
-        "The VLE data is thermodynamically consistent\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['f', 'draw_if_interactive', 'new_figure_manager', 'test']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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pVMC//gUEBCidiKjyZCsc0dHRmDt3LurUqQOr/37FUqlUuHjxYuWSyoCFg0yp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\nEREZhIWDiIgMwsJBREQGYeEgIiKD/H+2fJxBMvlw7wAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x39ce190>"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 12.9  Page No : 464"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from numpy import *\n",
-      "from matplotlib.pyplot import *\n",
-      "%matplotlib inline\n",
-      "# Variables\n",
-      "\n",
-      "P = 760.\t\t\t                            #pressure of the system in Torr\n",
-      "antoine_const_benzene = [6.87987,1196.760,219.161];\t\t\t #Antoine's constants for Benzene from Table A.7\n",
-      "t = linspace(60,100,9);\t\t\t #temperature range in degree celsius\n",
-      "\n",
-      "P2_s = [149.40,187.58,233.71,289.13,355.21,433.51,525.84,634.00,760.00];\n",
-      "x1 = linspace(0,1,6)\n",
-      "\n",
-      "# Calculations\n",
-      "l = len(t)\n",
-      "i = 0\n",
-      "P1_s = zeros(l)\n",
-      "P_tot =  zeros(l)\n",
-      "\n",
-      "while i<l:\n",
-      "    P1_s[i] = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(t[i]+antoine_const_benzene[2])));\n",
-      "    P_tot[i] = P1_s[i]+P2_s[i];\n",
-      "    i = i+1;\n",
-      "\n",
-      "T = (((t[2]-t[1])*(760-P_tot[1]))/(P_tot[2]-P_tot[1]))+t[1]\n",
-      "\n",
-      "P1_s_three_phase = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(T+antoine_const_benzene[2])));\n",
-      "P2_s_three_phase = 760-P1_s_three_phase;\n",
-      "y1_three_phase = P1_s_three_phase/760\n",
-      "\n",
-      "trange1 = linspace(T,T+11,11)\n",
-      "n = len(trange1)\n",
-      "i = 0\n",
-      "y1 = zeros(n)\n",
-      "P1_s_calc = zeros(n)\n",
-      "while i<n:\n",
-      "    if i == 1:\n",
-      "        y1[i] = y1_three_phase\n",
-      "    else:\n",
-      "         P1_s_calc[i] = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(trange1[i]+antoine_const_benzene[2])));\n",
-      "         y1[i] = (P1_s_calc[i])/P\n",
-      "    i = i+1;\n",
-      "\n",
-      "trange2 = [70,75,80,85,90,95,100]\n",
-      "P2_s_range = [233.71,289.13,355.21,433.51,525.84,634.00,760.00]\n",
-      "p = len(trange2);\t\t\t\n",
-      "i = 0\n",
-      "y_one = zeros(p)\n",
-      "# Calculations of the vapour composition of benzene in the two phase region (L2+V) using Eq.(12.61) (no unit)\n",
-      "y_one[i] = y1_three_phase;\n",
-      "trange2[i] = T;\n",
-      "i = i+1;\n",
-      "\n",
-      "while i<p:\n",
-      "    y_one[i] = (P-P2_s_range[i])/P;\n",
-      "    i = i+1;\n",
-      "k=len(x1)    \n",
-      "t_3phase = zeros(k)\n",
-      "i = 0\n",
-      "k = len(x1)\n",
-      "while i<k:\n",
-      "    t_3phase[i] = T\n",
-      "    i = i+1;\n",
-      "\n",
-      "# Results\n",
-      "plot(y1,trange1);\n",
-      "plot(y_one,trange2);\n",
-      "plot(x1,t_3phase);\n",
-      "suptitle('t-x-y diagram for benzene-water sytem at 760 Torr')\n",
-      "xlabel('x1,y1')\n",
-      "ylabel('t (degree celsius)');\n",
-      "q = len(t)\n",
-      "i = 1;\t\t\n",
-      "print ' Calculationss performed for determining the three phase equilibrium temperature';\n",
-      "print 'tdegree celsius \\t P1_s Torr \\t P2_s Torr \\t P1_s+P2_s Torr ';\n",
-      "for i in range(q):\n",
-      "    print '%d \\t \\t \\t %.2f \\t %0.2f \\t %.2f '%(t[i],P1_s[i],P2_s[i],P_tot[i]);\n",
-      "\n",
-      "print 'The three phase equilibrium temperature = %0.2f degree celsius '%(T);\n",
-      "print 'The vapour phase composition of benzene at the three phase equilibrium point = %0.4f '%(y1_three_phase);\n",
-      "\n",
-      "# Note: answer are slightly different because of rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        " Calculationss performed for determining the three phase equilibrium temperature"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "tdegree celsius \t P1_s Torr \t P2_s Torr \t P1_s+P2_s Torr \n",
-        "60 \t \t \t 391.63 \t 149.40 \t 541.03 \n",
-        "65 \t \t \t 465.92 \t 187.58 \t 653.50 \n",
-        "70 \t \t \t 550.98 \t 233.71 \t 784.69 \n",
-        "75 \t \t \t 647.87 \t 289.13 \t 937.00 \n",
-        "80 \t \t \t 757.67 \t 355.21 \t 1112.88 \n",
-        "85 \t \t \t 881.54 \t 433.51 \t 1315.05 \n",
-        "90 \t \t \t 1020.65 \t 525.84 \t 1546.49 \n",
-        "95 \t \t \t 1176.21 \t 634.00 \t 1810.21 \n",
-        "100 \t \t \t 1349.47 \t 760.00 \t 2109.47 \n",
-        "The three phase equilibrium temperature = 69.06 degree celsius \n",
-        "The vapour phase composition of benzene at the three phase equilibrium point = 0.7028 \n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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jBSUFouT7/dbvDFwzkEbVG7G873KqlK+idSQhij2zFQg3Nzdjcbh37x5xcXE0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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x39910d0>"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch13.ipynb b/Chemical_Engineering_Thermodynamics/ch13.ipynb
deleted file mode 100755
index 8dc5f0be..00000000
--- a/Chemical_Engineering_Thermodynamics/ch13.ipynb
+++ /dev/null
@@ -1,235 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13 : Dilute solution laws"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.1  Page No : 478"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "weight = 10.     \t\t\t #weight of NaCl in grams\n",
-      "volume = 1. \t    \t\t #volume of water in litres\n",
-      "weight_water = 1000. \t\t\t #  weight of water in grams (Weight = Volume*Density, density of water  = 1g/cc = 1g/ml = 1000g/l)\n",
-      "molwt_NaCl = 58.5 \t\t\t #molecular weight of NaCl in grams\n",
-      "molwt_water = 18. \t\t\t #molecular weight of water in grams\n",
-      "hf = 6.002; \t\t\t #enthalpy change of fusion in kJ/mol at 0 degree celsius\n",
-      "P = 101.325; \t\t\t #pressure in kPa\n",
-      "T = 273.15; \t\t\t # freezing point temperature of water at the given pressure in K\n",
-      "R = 8.314; \t\t\t #universal gas constant in J/molK;\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = (weight/molwt_NaCl)/((weight/molwt_NaCl)+(weight_water/molwt_water))\n",
-      "delt = (R*T**2*x2)/(hf*10**3)\n",
-      "\n",
-      "# Results\n",
-      "print ' The depression in freezing point of water when 10g of NaCl solute is added  =  %0.2f K'%(delt);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The depression in freezing point of water when 10g of NaCl solute is added  =  0.32 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.2  Page No : 480"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "weight = 10.; \t\t    \t #weight of NaCl in grams\n",
-      "volume = 1.; \t\t\t     #volume of water in litres\n",
-      "weight_water = 1000.; \t\t\t #  weight of water in grams (Weight = Volume*Density, density of water  = 1g/cc = 1g/ml = 1000g/l)\n",
-      "molwt_NaCl = 58.5; \t\t\t #molecular weight of NaCl in grams\n",
-      "molwt_water = 18; \t\t\t #molecular weight of water in grams\n",
-      "lat_ht = 2256.94; \t\t\t #latent heat of vaporization in kJ/kg at 100 degree celsius (obtained from steam tables)\n",
-      "P = 101.325; \t\t\t #pressure in kPa\n",
-      "T = 373.15; \t\t\t #boiling point temperature of water at the given pressure in K\n",
-      "R = 8.314; \t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = 0.0031\n",
-      "hv = (lat_ht*molwt_water)/1000\n",
-      "delt = (R*T**2*x2)/(hv*10**3)\n",
-      "\n",
-      "# Results\n",
-      "print ' The elevation in boiling point of water when 10g of NaCl solute is added  =  %0.2f K'%(delt);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The elevation in boiling point of water when 10g of NaCl solute is added  =  0.09 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.3  Page No : 481"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "weight = 10.; \t\t\t #weight of NaCl in grams\n",
-      "weight_water = 1000.; \t\t\t #  weight of water in grams\n",
-      "molwt_NaCl = 58.5; \t\t\t #molecular weight of NaCl in grams\n",
-      "molwt_water = 18.; \t\t\t #molecular weight of water in grams\n",
-      "T = 300.; \t\t\t #prevailing temperature of water in K\n",
-      "R = 8.314; \t\t\t #universal gas constant in (Pa m**3)/(mol K);\n",
-      "v = 18*10**-6;\t\t\t #molar volume in m**3/mol\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = 0.0031\n",
-      "pi = ((R*T*x2)/v)*10**-3\n",
-      "\n",
-      "# Results\n",
-      "print ' The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K  =  %0.2f kPa'%(pi);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K  =  429.56 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.4  Page No : 483"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "temp = 20. \t\t\t # prevailing tempearture in degree celsius\n",
-      "melt_temp = 80.05; \t\t\t # melting point of naphthalene in degree celsius\n",
-      "hf = 18.574 \t\t\t # enthalpy of fusion in kJ/mol\n",
-      "R = 8.314 \t\t\t # universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "t = temp+273.15\n",
-      "melt_t = melt_temp+273.15\n",
-      "x2 = math.exp(((hf*10**3)/R)*((1./melt_t)-(1./t)))\n",
-      "\n",
-      "# Results\n",
-      "print ' The ideal solubility of naphthalene at 20 degree celsius =  %0.4f'%(x2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The ideal solubility of naphthalene at 20 degree celsius =  0.2737\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.5  Page No : 483"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "t = 295.43; \t\t\t #prevailing temperature in K\n",
-      "sat_p = 6.05; \t\t\t #Sasturation pressure of carbon dioxide at the prevailing temperature in MPa\n",
-      "p = 0.1; \t\t\t #pressure at which solubility has to be determined in MPa\n",
-      "\n",
-      "# Calculations\n",
-      "x2 = p/sat_p\n",
-      "\n",
-      "# Results\n",
-      "print ' The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution\\\n",
-      " at 0.1MPa =  %0.4f'%(x2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution at 0.1MPa =  0.0165\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch14.ipynb b/Chemical_Engineering_Thermodynamics/ch14.ipynb
deleted file mode 100755
index e54d6be2..00000000
--- a/Chemical_Engineering_Thermodynamics/ch14.ipynb
+++ /dev/null
@@ -1,846 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14 : Chemical reaction equilibrium"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.1  Page No : 489"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "T = 298.15;\t\t\t #temperature in K\n",
-      "del_Gf = [-137.327,-228.600,-394.815,0]; #the standard Gibbs free energy of formation of CO(g),H2O(g),CO2(g) and H2(g) in kJ\n",
-      "n = [1,1,-1,-1]         \t\t\t #stoichiometric coefficients of CO(g),H2O(g),CO2(g) and H2(g) respectively (no unit)\n",
-      "R = 8.314;\t\t\t                 #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "del_G = (n[0]*del_Gf[0])+(n[1]*del_Gf[1])+(n[2]*del_Gf[2])+(n[3]*del_Gf[3]);\n",
-      "Ka = math.exp((-(del_G*10**3))/(R*T))\n",
-      "\n",
-      "# Results\n",
-      "print 'The standard Gibbs free energy of the water gas shift reaction at 298.15K = %0.3f kJ '%(del_G);\n",
-      "print 'The equilibrium constant of the water gas shift reaction at 298.15K = %0.3e '%(Ka);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The standard Gibbs free energy of the water gas shift reaction at 298.15K = 28.888 kJ \n",
-        "The equilibrium constant of the water gas shift reaction at 298.15K = 8.685e-06 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.2  Page No : 490"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 298.15;\t\t\t #temperature in K\n",
-      "P_s = 0.16716;\t\t\t #saturation pressure of CH3OH in bar at T\n",
-      "del_G1 = -166.215\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "f_v = 1\n",
-      "f_l = P_s\n",
-      "del_G2 = R*T*math.log(f_v/f_l)*10**-3;\t\t\t # Calculations of the value of del_G2 in kJ\n",
-      "del_G = del_G2+del_G1;\t\t\n",
-      "\n",
-      "# Results\n",
-      "print 'The standard Gibbs free energy of formation of CH3OHg = %0.3f kJ '%(del_G);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The standard Gibbs free energy of formation of CH3OHg = -161.781 kJ \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.3  Page No : 491"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "#The water gas shift reaction is given by: CO2(g)+H2(g)--->CO(g)+H2O(g)\n",
-      "T1 = 298.15     \t\t\t #initial temperature in K\n",
-      "Ka1 = 8.685*10**-6;\t\t\t #equilibrium constant for the water-gas shift reaction at T1 (no unit)\n",
-      "T2 = 1000.\t\t        \t #temperature at which the equilibrium constant has to be determined in K\n",
-      "R = 8.314;\t\t\t         #universal gas constant in J/molK\n",
-      "del_Hf = [-110.532,-241.997,-393.978,0];\t\t\t #the standard enthalpy of formation of CO(g),H2O(g),CO2(g) and H2(g) in kJ\n",
-      "n = [1,1,-1,-1];\t\t\t #stoichiometric coefficients of CO(g),H2O(g),CO2(g) and H2(g) respectively (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "#It is assumed that del_H is constant in the temperature range T1 and T2\n",
-      "del_H = (n[0]*del_Hf[0])+(n[1]*del_Hf[1])+(n[2]*del_Hf[2])+(n[3]*del_Hf[3])\n",
-      "Ka2 = Ka1*math.exp(((del_H*10**3)/R)*((1./T1)-(1./T2)));\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print 'The equilibrium constant for the water gas shift reaction at 1000K = %f '%(Ka2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equilibrium constant for the water gas shift reaction at 1000K = 1.085357 \n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.4  Page No : 492"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "P = 0.1     \t\t\t #pressure in MPa\n",
-      "T1 = 298.15;\t\t\t #initial temperature in K\n",
-      "Ka1 = 8.685*10**-6;\t\t\t #equilibrium constant for the water-gas shift reaction at T1 (no unit) (from Example 14.1)\n",
-      "T2 = 1000.  \t\t\t #temperature at which the equilibrium constant is to be found, in K\n",
-      "del_H = 41.449;\t\t\t #smath.tan(math.radiansard enthalpy of the reaction at T1 in kJ (from Example 14.3)\n",
-      "a = [28.068,28.850,45.369,27.012];\n",
-      "b = [4.631*10**-3,12.055*10**-3,8.688*10**-3,3.509*10**-3]\n",
-      "c = [0,0,0,0];\n",
-      "d = [0,0,0,0];\n",
-      "e = [-0.258*10**5,1.006*10**5,-9.619*10**5,0.690*10**5]\n",
-      "n = [1,1,-1,-1]\n",
-      "R = 8.314;\t\t\n",
-      "Ka2_prev = 1.0855\n",
-      "\n",
-      "\n",
-      "# Calculations\n",
-      "del_a = (n[0]*a[0])+(n[1]*a[1])+(n[2]*a[2])+(n[3]*a[3]);\n",
-      "del_b = (n[0]*b[0])+(n[1]*b[1])+(n[2]*b[2])+(n[3]*b[3]);\n",
-      "del_c = (n[0]*c[0])+(n[1]*c[1])+(n[2]*c[2])+(n[3]*c[3]);\n",
-      "del_d = (n[0]*d[0])+(n[1]*d[1])+(n[2]*d[2])+(n[3]*d[3]);\n",
-      "del_e = (n[0]*e[0])+(n[1]*e[1])+(n[2]*e[2])+(n[3]*e[3]);\n",
-      "del_H0 = (del_H*10**3)-((del_a*T1)+((del_b/2)*T1**2)+((del_c/3)*T1**3)+((del_d/4)*T1**4)-(del_e/T1));\n",
-      "I = (math.log(Ka1))-((1./R)*((-del_H0/T1)+(del_a*math.log(T1))+((del_b/2)*T1)+((del_c/6)*T1**2)+((del_d/12)*T1**3)+((del_e/(2*T1**2)))));\n",
-      "Ka2 = math.exp(((1./R)*((-del_H0/T2)+(del_a*math.log(T2))+((del_b/2)*T2)+((del_c/6)*T2**2)+((del_d/12)*T2**3)+((del_e/(2*T2**2)))))+I);\n",
-      "\n",
-      "\t\t\t # Results\n",
-      "print 'The equilibrium constant for the water gas shift reaction at 1000K\\\n",
-      " by taking into account the variation of del_H with temperature = %f '%(Ka2);\n",
-      "print 'The equilibrium constant for the water gas shift reaction at 1000K without\\\n",
-      " considering the variation of del_H with temperature as given by Example14.3 = %0.4f '%(Ka2_prev);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equilibrium constant for the water gas shift reaction at 1000K by taking into account the variation of del_H with temperature = 0.664976 \n",
-        "The equilibrium constant for the water gas shift reaction at 1000K without considering the variation of del_H with temperature as given by Example14.3 = 1.0855 \n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.5 page no : 494"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import Symbol\n",
-      "from sympy.solvers import solve\n",
-      "import math\n",
-      "\n",
-      "# variables\n",
-      "deltaGf = -161.781       # kJ from exa. 14.2\n",
-      "deltaG298 = deltaGf - (-137.327)\n",
-      "deltaH298 = -200.660 - (-110.532)\n",
-      "deltaA = 18.382 - 28.068 - 2 * 27.012\n",
-      "deltaB = (101.564 - 4.631 - 2 * 3.509) * 10**-3\n",
-      "deltaC = -28.683 * 10**-6\n",
-      "deltaD = 0\n",
-      "deltaE = (0.258 - 2 * 0.690) * 10**5\n",
-      "T = 298.15\n",
-      "\n",
-      "# calculations\n",
-      "deltaHf298 = -238.648 + 37.988\n",
-      "deltaH0 = Symbol('deltaH0')\n",
-      "Eq = deltaH0 + deltaA*T + deltaB*T**2/2 + deltaC*T**3/3 - deltaE/T - deltaH298*1000\n",
-      "deltaH0 = solve(Eq,deltaH0)[0]/1000\n",
-      "I = Symbol('I')\n",
-      "\n",
-      "Eq1 = deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I + 24454\n",
-      "I = round(solve(Eq1,I)[0],3)\n",
-      "\n",
-      "T = 500\n",
-      "deltaG500 = (deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I)/1000\n",
-      "\n",
-      "Ka = math.exp(-22048/(8.314*T))\n",
-      "e = Symbol('e')\n",
-      "Eq3 = ( (e*(3-2*e)**2) / (1 - e)**3) / 0.4973 -1\n",
-      "e = round(solve(Eq3,e)[0],4)\n",
-      "yCH3OH = e/(3-2*e)\n",
-      "yco = (1-e)/(3-2*e)\n",
-      "yh2 = (2*(1-e) / (3-2*e))\n",
-      "\n",
-      "# Results\n",
-      "print \"E = %.4f\"%e\n",
-      "print \"YCH30H = %.4f\"%yCH3OH\n",
-      "print \"YCO = %.4f\"%yco\n",
-      "print \"YH2 = %.4f\"%yh2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "E = 0.0506\n",
-        "YCH30H = 0.0175\n",
-        "YCO = 0.3275\n",
-        "YH2 = 0.6550\n"
-       ]
-      }
-     ],
-     "prompt_number": 103
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.6, Page 496"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy.solvers import solve \n",
-      "from sympy import Symbol \n",
-      "\n",
-      "#Variables\n",
-      "e = Symbol('e')\n",
-      "\n",
-      "#Calculations\n",
-      "x = solve((((e/(3-2*e))/(((1-e)/(3-2*e))*(((2*(1-e))/(3-2*e))**2)))-49.73),e)\n",
-      "print \"e =\",round(x[0],3)\n",
-      "ych3oh = x[0]/(3-2*x[0])\n",
-      "yco = (1-x[0])/(3-2*x[0])\n",
-      "yh2 = ((2*(1-x[0]))/(3-2*x[0]))\n",
-      "print \"yCH3OH =\",round(ych3oh,4)\n",
-      "print \"yCO =\",round(yco,4)\n",
-      "print \"yH2 =\",round(yh2,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e = 0.801\n",
-        "yCH3OH = 0.5731\n",
-        "yCO = 0.1423\n",
-        "yH2 = 0.2846\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.7 pageno : 497"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import Symbol\n",
-      "from sympy.solvers import solve\n",
-      "\n",
-      "# variables\n",
-      "Ka = 4.973*10**-3          # from example 14.5\n",
-      "Ky = 25 * Ka               # from example 14.5\n",
-      "\n",
-      "# calculations\n",
-      "e = Symbol('e')\n",
-      "Eq1 = ( (e*(4-e)**2)/(1-e)**3 ) / .1243 - 1\n",
-      "e = round(solve(Eq1,e)[0],5)\n",
-      "yCH3OH = e/(8-2*e)\n",
-      "yco = (1-e)/(8-2*e)\n",
-      "yh2 = (2*(1-e) / (8-2*e))\n",
-      "yA = 5/(8-2*e)\n",
-      "\n",
-      "# Results\n",
-      "print \"E = %.5f\"%e\n",
-      "print \"YCH30H = %.5f\"%yCH3OH\n",
-      "print \"YCO = %.5f\"%yco\n",
-      "print \"YH2 = %.5f\"%yh2\n",
-      "print \"YA = %.5f\"%yA"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "E = 0.00762\n",
-        "YCH30H = 0.00095\n",
-        "YCO = 0.12428\n",
-        "YH2 = 0.24857\n",
-        "YA = 0.62619\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.8 pageno : 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import Symbol\n",
-      "from sympy.solvers import solve\n",
-      "\n",
-      "# variables\n",
-      "Ka = 4.973*10**-3           # example 14.5\n",
-      "Ky = 25 * Ka                # example 14.5\n",
-      "\n",
-      "# calculations and results\n",
-      "e = Symbol('e')\n",
-      "Eq1= ((e*(1-e))/(1-2*e)**2)/.03108 - 1\n",
-      "e = solve(Eq1,e)\n",
-      "print \"part a\"\n",
-      "print \"e = %.5f and e = %.5f\"%(e[0],e[1])\n",
-      "print \"The admissible value is e = %.5f\"%e[0]\n",
-      "\n",
-      "print \"part b\"\n",
-      "print \"e = 0.0506 ( from example 14.5 )\"\n",
-      "\n",
-      "x = 4\n",
-      "e = Symbol('e')\n",
-      "Eq1=  ((e/(1-e) ) * ( (5 - 2*e)/ (2-e) )**2)/ 0.4973 - 1\n",
-      "e = solve(Eq1,e)\n",
-      "print \"part c\"\n",
-      "print \"e = %.4f\" %(e[0])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part a\n",
-        "e = 0.02845 and e = 0.97155\n",
-        "The admissible value is e = 0.02845\n",
-        "part b\n",
-        "e = 0.0506 ( from example 14.5 )\n",
-        "part c"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "e = 0.0727\n"
-       ]
-      }
-     ],
-     "prompt_number": 134
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.9 pageno : 499"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import Symbol\n",
-      "from sympy.solvers import solve\n",
-      "\n",
-      "# variables\n",
-      "Ka = 4.973*10**-3           # example 14.5\n",
-      "Ky = 25 * Ka                # example 14.5\n",
-      "\n",
-      "# calculations\n",
-      "e = Symbol('e')\n",
-      "Eq1 = (0.02 + e)*(1.51-e)**2 / (1-e)**3 / Ky - 1\n",
-      "e = solve(Eq1,e)[0]\n",
-      "\n",
-      "# Results\n",
-      "print \"E = %.5f\"%e"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "E = 0.03165\n"
-       ]
-      }
-     ],
-     "prompt_number": 138
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.10 pageno : 500"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "%matplotlib inline\n",
-      "\n",
-      "from matplotlib.pyplot import *\n",
-      "from sympy import Symbol\n",
-      "from sympy.solvers import solve\n",
-      "from scipy.integrate import quad\n",
-      "import math\n",
-      "\n",
-      "# variables\n",
-      "deltaA = round(28.850 - 27.012 - 0.5 * 30.255,2)\n",
-      "deltaB = (12.055 - 3.509 - 0.5 * 4.207) * 10**-3\n",
-      "deltaC = 0\n",
-      "deltaD = 0\n",
-      "deltaE = round((1.006 - 0.690 + 0.5 * 1.887),3) * 10**5\n",
-      "deltaH298 =  -241.997\n",
-      "Ts = [2000,2500,3000,3500,3800]\n",
-      "deltaHt = []\n",
-      "deltaGt = []\n",
-      "Ka = []\n",
-      "Ea = []\n",
-      "Eb = []\n",
-      "\n",
-      "# Calculations\n",
-      "T = 298.15\n",
-      "deltaH0 = Symbol('deltaH0')\n",
-      "Eq = deltaH0 + deltaA*T + deltaB*T**2/2 + deltaC*T**3/3 + deltaD*T**4/4 - deltaE/T - deltaH298*1000\n",
-      "deltaH0 = round(solve(Eq,deltaH0)[0]/1000,3)\n",
-      "I = Symbol('I')\n",
-      "Eq1 = deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I + 228600\n",
-      "I = round(solve(Eq1,I)[0],3)\n",
-      "\n",
-      "def fun1(T1):\n",
-      "    #return 42.1395*(T1 - 298.15) + 5.613*10**-3/2*(T1**2 - 298.15**2) + .2535 * 10**5 * (1./T1 - 1/298.15)\n",
-      "    return 42.1395*T1 + 5.613*10**-3/2 * T1**2 + 0.2535*10**5/T1 - 12898.37\n",
-      "\n",
-      "for T in Ts:\n",
-      "    Ht = deltaH0*1000 + deltaA*T + deltaB*T**2/2 + deltaC*T**3/3 + deltaD*T**4/4 - deltaE/T \n",
-      "    Gt = deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I \n",
-      "    ka = math.exp(-Gt/(8.314*T))\n",
-      "    e = Symbol('e')\n",
-      "    Eq2 = ( e * (3-e)**(1./2) ) / ( 1-e)**(3./2) / ka - 1\n",
-      "    b = round(solve(Eq2)[0],4)\n",
-      "    a = (quad(fun1,298.15,T)[0]/1000) / -Ht\n",
-      "    deltaHt.append(Ht)\n",
-      "    deltaGt.append(Gt)\n",
-      "    Ka.append(ka)\n",
-      "    Ea.append(a)\n",
-      "    Eb.append(b)\n",
-      "\n",
-      "\n",
-      "# Results    \n",
-      "plot(Ts,Ea,\"g\")\n",
-      "plot(Ts,Eb,\"b\")\n",
-      "xlabel(\"T(k)\")\n",
-      "ylabel(\"E\")\n",
-      "suptitle(\"Plot of e versus adiabatic reaction temperature\")\n",
-      "show()\n",
-      "print \"from plot, it can be seen that both lines are simultaneously satisfied at the point \\\n",
-      "\\nintersection where e = 0.68 and T = 3440 K\"\n",
-      "e = 0.68\n",
-      "T = 3440\n",
-      "yh2 = (1- e)/(1.5 - 0.5*e)\n",
-      "yo2 = 0.5*(1-e) / (1.5-0.5*e)\n",
-      "yh2o = e/(1.5- 0.5*e)\n",
-      "\n",
-      "print \"yH2 = %.4f\"%yh2\n",
-      "print \"yO2 = %.4f\"%yo2\n",
-      "print \"yH2O = %.4f\"%yh2o"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['solve', 'draw_if_interactive', 'e']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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ISZGW0wgNvj39LdzC3ODTzgexr8WySFCl6CwUn332mfb25s2byzz3/vvv/+OK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\nQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x51ff6d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "from plot, it can be seen that both lines are simultaneously satisfied at the point \n",
-        "intersection where e = 0.68 and T = 3440 K\n",
-        "yH2 = 0.2759\n",
-        "yO2 = 0.1379\n",
-        "yH2O = 0.5862\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.11  Page No : 506"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "stoichio_matrix = matrix([[-1,-1, 1, 3, 0, 0],[0, -1, -1, 1, 1 ,0],[-1, -2, 0 ,4, 1, 0], \\\n",
-      "                [0,0, 1, 0, -1, 0.5],[-1, 2 ,0, 0, 1, -2],[-1, 1, 1, 1, 0, -1]])\t\t\t #Framing the stoichiometric coefficient matrix\n",
-      "\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "r = linalg.matrix_rank(stoichio_matrix)\n",
-      "\n",
-      "stoichio_matrix[0] = -stoichio_matrix[0];\n",
-      "stoichio_matrix[2] = stoichio_matrix[2]+stoichio_matrix[0];\n",
-      "stoichio_matrix[5:] = stoichio_matrix[5:]+stoichio_matrix[0];\n",
-      "stoichio_matrix[6:] = stoichio_matrix[6:]+stoichio_matrix[0];\n",
-      "stoichio_matrix[1] = -stoichio_matrix[1];\n",
-      "stoichio_matrix[2] = stoichio_matrix[2]+stoichio_matrix[1];\n",
-      "stoichio_matrix[5:] = stoichio_matrix[5:]-(3*stoichio_matrix[1]);\n",
-      "stoichio_matrix[6:] = stoichio_matrix[6:]-(2*stoichio_matrix[1]);\n",
-      "x = stoichio_matrix[:3]\n",
-      "y = stoichio_matrix[:4];\n",
-      "stoichio_matrix[:4] = y;\n",
-      "stoichio_matrix[:3] = x;\n",
-      "stoichio_matrix[5:] = stoichio_matrix[5:]+(4*stoichio_matrix[3]);\n",
-      "stoichio_matrix[6:] = stoichio_matrix[6:]+(2*stoichio_matrix[3]);\n",
-      "\n",
-      "\n",
-      "\n",
-      "# Results\n",
-      "print ' The stoichiometric coefficient matrix after performing the elementary row operations = ';\n",
-      "print (stoichio_matrix);\n",
-      "print ' The number of primary reactions = %d'%(r);\n",
-      "print ' The non zero rows are 1,2,4'\n",
-      "print ' The primary reactions are: CH4g)+H2Og)--->COg)+3H2g)%( COg)+H2Og)--->CO2g)+H2g),( CO2g)--->COg)+1./2)O2g)'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The stoichiometric coefficient matrix after performing the elementary row operations = \n",
-        "[[ 1.   1.  -1.  -3.  -0.  -0. ]\n",
-        " [-0.   1.   1.  -1.  -1.  -0. ]\n",
-        " [ 0.   0.   0.   0.   0.   0. ]\n",
-        " [ 0.   0.   1.   0.  -1.   0.5]\n",
-        " [-1.   2.   0.   0.   1.  -2. ]\n",
-        " [ 0.  -1.   1.   1.  -1.   1. ]]\n",
-        " The number of primary reactions = 3\n",
-        " The non zero rows are 1,2,4\n",
-        " The primary reactions are: CH4g)+H2Og)--->COg)+3H2g)%( COg)+H2Og)--->CO2g)+H2g),( CO2g)--->COg)+1./2)O2g)\n"
-       ]
-      }
-     ],
-     "prompt_number": 169
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.13 pageno : 510"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import Symbol\n",
-      "from sympy.solvers import solve\n",
-      "\n",
-      "# variables\n",
-      "Ka1 = 0.1429\n",
-      "Ka2 = 2\n",
-      "\n",
-      "# calculations and Results\n",
-      "\n",
-      "# assume e1 = .3\n",
-      "e2 = Symbol('e2')\n",
-      "Eq1 = ((0.3 - e2)*(0.3+e2) /((0.7 - e2) * 0.7)) / 0.1429 -1\n",
-      "e2 =  solve(Eq1)\n",
-      "print \"E2 = %.1f and %.1f \"%(e2[0],e2[1])\n",
-      "\n",
-      "# since negative value is inadmissible. e2 = 0.2\n",
-      "e2 = round(e2[1],1)\n",
-      "e1 = Symbol('e1')\n",
-      "Eq2 = ((e1 + 0.2)*(0.2) /((0.8 - e1) *(e1 - 0.2) ) ) / 2 -1\n",
-      "e1 =  solve(Eq2)\n",
-      "print \"E1 = %.1f and %.1f \"%(e1[0],e1[1])\n",
-      "e1 = round(e1[0],1)\n",
-      "print \"Satisfied results when e1 = %.1f and e2 = %.1f\"%(e1,e2)\n",
-      "\n",
-      "yA = (1-e1-e2)/2\n",
-      "yB = (1-e1)/2\n",
-      "yC = (e1-e2)/2\n",
-      "yD = (e1+e2)/2\n",
-      "yE = e2/2\n",
-      "\n",
-      "print \"yA = %.2f\"%yA\n",
-      "print \"yB = %.2f\"%yB\n",
-      "print \"yC = %.2f\"%yC\n",
-      "print \"yD = %.2f\"%yD\n",
-      "print \"yE = %.2f\"%yE\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "E2 = -0.1 and 0.2 \n",
-        "E1 = 0.3 and 0.6 \n",
-        "Satisfied results when e1 = 0.3 and e2 = 0.2\n",
-        "yA = 0.25\n",
-        "yB = 0.35\n",
-        "yC = 0.05\n",
-        "yD = 0.25\n",
-        "yE = 0.10\n"
-       ]
-      }
-     ],
-     "prompt_number": 214
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.14 page no : 515"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import Symbol\n",
-      "from sympy.solvers import solve\n",
-      "\n",
-      "# variables\n",
-      "Kc = 2.92\n",
-      "\n",
-      "# Calculations and results\n",
-      "e = Symbol('e')\n",
-      "Eq1 = (e*(10+e)/((5-e)*(10-e)))/Kc - 1\n",
-      "e = round(solve(Eq1)[0],4)\n",
-      "print \"E = %.4f\"%e\n",
-      "\n",
-      "cA = 5-e\n",
-      "cB = 10-e\n",
-      "cC = e\n",
-      "cD = 10+e\n",
-      "\n",
-      "print \"CA = %.4f kmol/m*3\"%cA\n",
-      "print \"CB = %.4f kmol/m*3\"%cB\n",
-      "print \"CC = %.4f kmol/m*3\"%cC\n",
-      "print \"CD = %.4f kmol/m*3\"%cD"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "E = 3.0446\n",
-        "CA = 1.9554 kmol/m*3\n",
-        "CB = 6.9554 kmol/m*3\n",
-        "CC = 3.0446 kmol/m*3\n",
-        "CD = 13.0446 kmol/m*3\n"
-       ]
-      }
-     ],
-     "prompt_number": 220
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.15, Page 517"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 1200.   \t\t\t #temperature in K\n",
-      "T0 = 298.15 \t\t\t #reference temperature in K\n",
-      "a = [41.84,45.369,82.34];\n",
-      "b = [20.25*10**-3,8.688*10**-3,49.75*10**-3];\n",
-      "c = [0,0,0];\n",
-      "d = [0,0,0];\n",
-      "e = [-4.51*10**5,-9.619*10**5,-12.87*10**5];\n",
-      "del_Gf = [-604.574,-394.815,-1129.515]\n",
-      "del_Hf = [-635.975,-393.978,-1207.683]\n",
-      "n = [1,1,-1]\n",
-      "R = 8.314\n",
-      "\n",
-      "# Calculations\n",
-      "del_G = (n[0]*del_Gf[0])+(n[1]*del_Gf[1])+(n[2]*del_Gf[2]);\n",
-      "del_H = (n[0]*del_Hf[0])+(n[1]*del_Hf[1])+(n[2]*del_Hf[2]);\n",
-      "del_a = (n[0]*a[0])+(n[1]*a[1])+(n[2]*a[2]);\n",
-      "del_b = (n[0]*b[0])+(n[1]*b[1])+(n[2]*b[2]);\n",
-      "del_c = (n[0]*c[0])+(n[1]*c[1])+(n[2]*c[2]);\n",
-      "del_d = (n[0]*d[0])+(n[1]*d[1])+(n[2]*d[2]);\n",
-      "del_e = (n[0]*e[0])+(n[1]*e[1])+(n[2]*e[2]);\n",
-      "del_H0 = ((del_H*10**3)-((del_a*T0)+((del_b/2)*T0**2)+((del_c/3)*T0**3)+((del_d/4)*T0**4)-(del_e/T0)))*10**-3;\n",
-      "IR = (1./(T0))*((del_H0*10**3)-(del_a*T0*math.log(T0))-((del_b/2)*T0**2)-((del_c/6)*T0**3)-((del_d/12)*T0**4)-((del_e/2)*(1./T0))-(del_G*10**3));\n",
-      "del_G_T = ((del_H0*10**3)-(del_a*T*math.log(T))-((del_b/2)*T**2)-((del_c/6)*T**3)-((del_d/12)*T**4)-((del_e/2)*(1./T))-(IR*T))*10**-3;\n",
-      "Ka = math.exp((-del_G_T*10**3)/(R*T))\n",
-      "y = 1;\n",
-      "phi = 1;\n",
-      "f0 = 1;\n",
-      "P = (Ka*f0)/(phi*y)\n",
-      "\n",
-      "# Results\n",
-      "print ' The decomposition pressure,P = %f bar '%(P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The decomposition pressure,P = 2.384295 bar \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch2.ipynb b/Chemical_Engineering_Thermodynamics/ch2.ipynb
deleted file mode 100755
index b7849ef3..00000000
--- a/Chemical_Engineering_Thermodynamics/ch2.ipynb
+++ /dev/null
@@ -1,117 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:76568d79b614fc85568c92855047632dbe19f5ec7540dbd287268967c3318f2b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2 : Review of basic concepts"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.1  Page No : 29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "t1 = 300.;\t\t\t #initial temperature of gas in K\n",
-      "p1 = 0.1;\t\t\t #initial pressure of the gas in MPa\n",
-      "p2 = 0.5;\t\t\t #pressure of gas after heating in MPa\n",
-      "p3 = 0.1;\t\t\t #pressure of gas after expansion in MPa\n",
-      "n = 1.; \t\t\t #number of moles of gas in mole\n",
-      "\n",
-      "# Calculations\n",
-      "w1 = 0*n;       \t\t\t # Calculations of work done by the gas during heating in kJ (since dv = 0)\n",
-      "t2 = t1*(p2/p1);\t\t\t # Calculations of temperature of gas after heating in K\n",
-      "t3 = t2;\t\t        \t # Calculations of temperature of gas after expansion in K (constant temperature expansion)\n",
-      "volume2 = p2/p3;\t\t\t # Calculations of  ratio of volume of gas after expansion to the volume of gas after heating (no unit)\n",
-      "w2 = (R*t2*math.log(volume2)*n)/1000\n",
-      "volume3 =  t3/t1;\t\t\n",
-      "w3 = (R*t1*(1-volume3)*n)/1000\n",
-      "work_net = w1+w2+w3;\t\t\t # Calculations of net work done by the gas for the process in kJ\n",
-      "\n",
-      "# Results\n",
-      "print ' Work done during heating process:Work from 1-2 =  %d kJ '%(w1);\n",
-      "print ' Work done during constant temperature expansion: Work from 2-3 =  %.3f kJ '%(w2);\n",
-      "print ' Work done during constant pressure compression: Work from 3-1 =  %.3f kJ '%(w3);\n",
-      "print ' Net work done by the gas during the process =  %.3f kJ '%(work_net);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Work done during heating process:Work from 1-2 =  0 kJ \n",
-        " Work done during constant temperature expansion: Work from 2-3 =  20.071 kJ \n",
-        " Work done during constant pressure compression: Work from 3-1 =  -9.977 kJ \n",
-        " Net work done by the gas during the process =  10.095 kJ \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.2  Page No : 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "v1 = 0.1;\t\t\t #volume of gas initially present in the cylinder in m**3\n",
-      "p1 = 0.1;\t\t\t #initial pressure of gas in MPa\n",
-      "p_atm = 0.1;\t\t #atmospheric pressure acting on the piston in MPa\n",
-      "v2 = 0.3;\t\t\t #volume of gas after heating in m**3\n",
-      "p2 = 0.6;\t\t\t #pressure of gas after heating in MPa\n",
-      "\n",
-      "# Calculations\n",
-      "work = ((p1+p2)*(v2-v1)*10**6)/(2*1000);\t\t\t # Calculations of work done by the gas in kJ \n",
-      "\n",
-      "# Results\n",
-      "print ' The work done by the gas  =  %d kJ'%(work);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The work done by the gas  =  70 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch3.ipynb b/Chemical_Engineering_Thermodynamics/ch3.ipynb
deleted file mode 100755
index 2b39d6e0..00000000
--- a/Chemical_Engineering_Thermodynamics/ch3.ipynb
+++ /dev/null
@@ -1,897 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8d10c3a8d696216084bffc2404a32e0b3c15831ac555c05fcb7016ab7d75d0e7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3 : PvT relations of fluids"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.1  Page No : 48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "X = 0.8;    \t\t\t #Quality of wet steam (no unit)\n",
-      "T = 150.;\t    \t\t #Temperature of the wet steam in degree celsius\n",
-      "vf = 0.0010908;\t\t\t #molar volume of saturated liquid in m**3/kg\n",
-      "vg = 0.3924;\t\t\t #molar volume of saturated vapour in m**3/kg\n",
-      "uf = 631.63;\t\t\t #molar internal energy of the saturated liquid in kJ/kg\n",
-      "ug = 2558.6;\t\t\t #molar internal energy of the saturated vapour in kJ/kg\n",
-      "\n",
-      "# Calculations\n",
-      "V = (X*vg)+((1-X)*vf);\t\t\t # Calculations of specific volume in m**3/kg using Eq.(3.3)\n",
-      "U = (X*ug)+((1-X)*uf);\t\t\t # Calculations of specific internal energy in kJ/kg using Eq.(3.6)\n",
-      "\n",
-      "# Results\n",
-      "print 'The specific volume of wet steam =  %0.4f m**3/kg '%(V);\n",
-      "print 'The specific internal energy of wet steam =  %0.1f kJ/kg '%(U);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The specific volume of wet steam =  0.3141 m**3/kg \n",
-        "The specific internal energy of wet steam =  2173.2 kJ/kg \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.2  Page No : 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 1.42;\t\t\t #specific volume of wet steam in m**3/kg\n",
-      "T = 100.;\t\t\t #temperature of wet steam in degree celsius\n",
-      "vf = 0.0010437;\t\t #molar volume of saturated liquid in m**3/kg\n",
-      "vg = 1.673;\t\t\t #molar volume of saturated vapour in m**3/kg\n",
-      "\n",
-      "# Calculations\n",
-      "X = (V-vf)/(vg-vf);\t\t\t # Calculations of the quality of wet steam using Eq.(3.3) (no unit)\n",
-      "\n",
-      "# Results\n",
-      "print ' The quality of wet steam =  %0.4f '%(X);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The quality of wet steam =  0.8487 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.3  Page No : 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 100.;\t\t\t #temperature inside the vessel in degree celsius\n",
-      "V = 0.00317;\t\t\t #specific volume of water at the critical point in m**3/kg\n",
-      "vf = 0.0010437;\t\t\t #molar volume of saturated liquid in m**3/kg\n",
-      "vg = 1.673;\t\t\t #molar volume of saturated vapour in m**3/kg\n",
-      "\n",
-      "# Calculations\n",
-      "X = (V-vf)/(vg-vf);\t\t\t          # Calculations of the quality of wet steam using Eq.(3.3) (no unit)\n",
-      "ratio = (X*vg)/((1-X)*vf);\t\t\t # Calculations of volume ratio of saturated vapour to the saturated liquid (no unit)\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume ratio of saturated vapour to the saturated liquid =  %0.2f '%(ratio);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume ratio of saturated vapour to the saturated liquid =  2.04 \n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.4  Page No : 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "U = 2000.\t\t\t #specific internal energy of liquid-vapour mixture in kJ/kg\n",
-      "uf = 850.6\t\t\t #specific internal energy of saturated liquid in kJ/kg\n",
-      "ug = 2593.2\t\t\t #specific internal energy of saturated vapour in kJ/kg\n",
-      "\n",
-      "# Calculations\n",
-      "mass_ratio = (U-uf)/(ug-U);\t\t\t # Calculations of the mass ratio of vapour to liquid using the lever rule (no unit)\n",
-      "\n",
-      "# Results\n",
-      "print ' The mass ratio of vapour to liquid =  %0.4f '%(mass_ratio);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The mass ratio of vapour to liquid =  1.9376 \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.5  Page No : 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "n = 1.\t\t\t #number of moles of n-octane vapour in mol\n",
-      "T = 427.85;\t\t\t #tempearture of n-octane vapour in K\n",
-      "P = 0.215;\t\t\t #pressure n-octane vapour in MPa\n",
-      "R = 8.314;\t\t\t #universal gas constant in (kPa m**3)/(kmol K)\n",
-      "\n",
-      "# Calculations\n",
-      "V = ((n*10**-3)*R*T)/(P*10**3);\t\t\t # Calculations of volume using ideal gas law-Eq.(3.9) in m**3\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane vapour =  %f m**3 '%(V);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane vapour =  0.016545 m**3 \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.6  Page No : 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "n = 1.; \t\t\t #number of moles occupied by n-octane vapour in mol\n",
-      "T = 427.85;\t\t\t #temperature in K\n",
-      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
-      "a = 3.789;\t\t\t #van der Waals constant in Pa(m**3/mol)**2\n",
-      "b = 2.37*10**-4;\t #van der Waals constant in m**3/mol\n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "\n",
-      "# Calculations\n",
-      "Vguess = (n*R*T)/(P*10**6);\t\t\t \n",
-      "Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b\n",
-      "tolerance = 1e-6\n",
-      "\n",
-      "while abs(Vguess-Vnew)>tolerance:\n",
-      "    Vguess = Vnew;\n",
-      "    Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b;\t\t\t #the iteration process to solve the system of equation\n",
-      "\n",
-      "V = Vnew;\t\t\t #The volume calculated using the van der Waals equation in m**3/mol\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane vapour obtained by van der Waals equation =  %f m**3/mol'%(V);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane vapour obtained by van der Waals equation =  0.015675 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.7  Page No : 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "n = 1.; \t\t\t #number of moles occupied by n-octane liquid in mol\n",
-      "T = 427.85;\t\t\t #temperature in K\n",
-      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
-      "a = 3.789;\t\t\t #van der Waals constant in Pa(m**3/mol)**2\n",
-      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "\n",
-      "# Calculations\n",
-      "Vguess = b\n",
-      "Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b\n",
-      "tolerance = 1e-6\n",
-      "while abs(Vguess-Vnew)>tolerance:\n",
-      "    Vguess = Vnew;\n",
-      "    Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b;\t\t\t #the iteration process to solve the system of equation\n",
-      "\n",
-      "V = Vnew;\t\t\t #The volume calculated using the van der Waals equation in m**3/mol\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane liquid obtained by van der Waals equation =  %e m**3/mol'%(V);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane liquid obtained by van der Waals equation =  3.517652e-04 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.8  Page No : 57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "T = 427.85; \t\t    \t #temperature in K\n",
-      "P = 0.215;\t    \t\t     #saturation pressure in MPa\n",
-      "a = 3.789;\t\t    \t     #van der Waals constant in Pa(m**3/mol)**2\n",
-      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
-      "R = 8.314;\t\t        \t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "Pcvc = 3\n",
-      "RTc = 8.\n",
-      "# Calculations\n",
-      "#The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily\n",
-      "#The general form of the cubic equation is (Z**3)+(alpha*Z**2)+(beeta*Z)+gaamma = 0, where alpha,beeta and gaamma are determined using established relations\n",
-      "\n",
-      "A = (a*P*10**6)/(R*T)**2;\t\t\t\n",
-      "B = (b*P*10**6)/(R*T);\t\t\t \n",
-      "alpha = -1-B;\t\t\t \n",
-      "beeta = A;\t\t\t \n",
-      "gaamma = -(A*B);\t\n",
-      "p = round(beeta-(alpha**2)/3,4)\n",
-      "q = round(((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma,4)\n",
-      "D = (q**2/4)+(p**3/27);\t\n",
-      "\n",
-      "\n",
-      "if D>0 :\n",
-      "        Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);\n",
-      "        vf = ((Z)*R*T)/(P*10**6);\t\t\n",
-      "        vg = ((Z)*R*T)/(P*10**6);\t\t\n",
-      "elif D == 0:\n",
-      "        Z1 = ((-2*(q/2))**(1./3))-(alpha/3);\t\n",
-      "        Z2 = ((q/2)**(1./3))-(alpha/3);\n",
-      "        Z3 = ((q/2)**(1./3))-(alpha/3);\n",
-      "        Z = [Z1 ,Z2, Z3];\n",
-      "        vf = (min(Z)*R*T)/(P*10**6)\n",
-      "        vg = (max(Z)*R*T)/(P*10**6)\n",
-      "else:\n",
-      "        r = math.sqrt((-(p**3)/27));\t\t\t # Calculations of r using Eq.(3.38)\n",
-      "        theta = math.cos((-(q)/2)*(1./r));\t\t\t # Calculations of theta in radians using Eq.(3.37)\n",
-      "        Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "        Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3);\t\t\t #Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)\n",
-      "        Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
-      "        Z = [Z1, Z2, Z3];\n",
-      "        vf = (min(Z)*R*T)/(P*10**6)\n",
-      "        vg = (max(Z)*R*T)/(P*10**6)\n",
-      "\n",
-      "crit = Pcvc/RTc        \n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane saturated liquid obtained by Cardans method =  %.3e m**3/mol'%(vf);\n",
-      "print ' The volume occupied by n-octane saturated vapour obtained by Cardans method =  %.2e m**3/mol'%(vg);\n",
-      "print \" At the critical point, PcVc/RTc = %.3f\"%crit\n",
-      "\n",
-      "# Note: value of D is calculated wrongly in book. Please calculate manually."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane saturated liquid obtained by Cardans method =  -9.345e-04 m**3/mol\n",
-        " The volume occupied by n-octane saturated vapour obtained by Cardans method =  1.55e-02 m**3/mol\n",
-        " At the critical point, PcVc/RTc = 0.375\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.9  Page No : 60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from numpy import *\n",
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature in K\n",
-      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "\n",
-      "# Calculations\n",
-      "a = (0.42748*R**2*Tc**2.5)/(Pc*100*10**3*math.sqrt(T));\t\t\t \n",
-      "b = (0.08664*R*Tc)/(Pc*100*10**3);\t\t\t\n",
-      "\n",
-      "\n",
-      "A = (a*P*10**6)/(R*T)**2;\t\t\t \n",
-      "B = (b*P*10**6)/(R*T);\n",
-      "alpha = -1.;\t\t\t\n",
-      "beeta = A-B-B**2.;\t\n",
-      "gaamma = -(A*B);\t\n",
-      "p = beeta-(alpha**2)/3.\n",
-      "q = ((2*alpha**3.)/27.)-((alpha*beeta)/3)+gaamma\n",
-      "D = (((q)**2)/4.)+(((p)**3)/27.);\t\t\t \n",
-      "#if D>0 then:\n",
-      " #       Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
-      "  #      vf = ((Z)*R*T)/(P*10**6)\n",
-      "   #     vg = ((Z)*R*T)/(P*10**6)\n",
-      "#elif D == 0:\n",
-      " #       Z1 = ((-2*(q/2))**(1./3))-(alpha/3);\n",
-      "  #      Z2 = ((q/2)**(1./3))-(alpha/3);\n",
-      "   #     Z3 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    #    Z = [Z1 ,Z2, Z3];\n",
-      "     #   vf = (min(Z)*R*T)/(P*10**6);\t\t\n",
-      "      #  vg = (max(Z)*R*T)/(P*10**6);\t\t\n",
-      "#else:\n",
-      "\n",
-      "r = math.sqrt((-(p**3)/27));\t\t\t\n",
-      "theta = (math.acos((-(q)/2)*(1./r)));\n",
-      "Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3);\t\n",
-      "Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
-      "Z = [Z1, Z2, Z3];\n",
-      "vf = (min(Z)*R*T)/(P*10**6);\t\t\n",
-      "vg = (max(Z)*R*T)/(P*10**6);\t\t\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane saturated vapour) using Redlich-Kwong equation of state =  %f m**3/mol'%(vg);\n",
-      "print ' The volume occupied by n-octane saturated liquid) using Redlich-Kwong equation of state =  %f m**3/mol'%(vf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane saturated vapour) using Redlich-Kwong equation of state =  0.015401 m**3/mol\n",
-        " The volume occupied by n-octane saturated liquid) using Redlich-Kwong equation of state =  0.000244 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.10  Page No : 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "T = 180.;\t\t\t #temperature of water in degree celsius\n",
-      "P = 1.0027;\t\t\t #saturation pressure of water in MPa\n",
-      "Tc = 647.3;\t\t\t #critical temperature of water in K\n",
-      "Pc = 221.2;\t\t\t #critical pressure of water in bar\n",
-      "Tr = 0.7;\t\t\t #reduced temperature at which acentric factor was defined by Pitzer \n",
-      "\n",
-      "# Calculations\n",
-      "T1 = Tr*Tc; \t\t\t #calculating temperature  in K using reduced temperature value\n",
-      "T1 = T1-273.15;\t\t\t #conversion to degree celsius\n",
-      "Pr = (P*10)/Pc;\t\t\t # Calculations of reduced pressure (no unit) using saturation pressure at t1. In this case t1 equals t, therefore the given saturation pressure is taken\n",
-      "w = -math.log10(Pr)-1.0;\t\t\t # Calculations of acentric factor using Eq.(3.62)\n",
-      "\n",
-      "# Results\n",
-      "print ' The acentric factor of water =  %f '%(w);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The acentric factor of water =  0.343614 \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.11  Page No : 72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 409.41;\t\t\t #temperature of n-octane in degree celsius\n",
-      "P = 4.98;\t\t\t #pressure in bar\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
-      "w = 0.398;\t\t\t #acentric factor (no unit)\n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "\n",
-      "# Calculations\n",
-      "Tr = (T+273.15)/Tc;\t\n",
-      "Pr = P/Pc;\t\t\t\n",
-      "z0 = 0.96;\t\t\t\n",
-      "V1 = (z0*R*(T+273.15))/(P*10**5);\t\t\t \n",
-      "z1 = 0.01;\t\t\t \n",
-      "z = z0+(w*z1);\t\t\n",
-      "V2 = (z*R*(T+273.15))/(P*10**5)\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane obtained by the two parameter compressibilty \\\n",
-      " factor correlation =  %f m**3/mol'%(V1);\n",
-      "print ' The volume occupied by n-octane obtained by the three parameter compressibility \\\n",
-      "factor correlation =  %f m**3/mol'%(V2);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane obtained by the two parameter compressibilty  factor correlation =  0.010939 m**3/mol\n",
-        " The volume occupied by n-octane obtained by the three parameter compressibility factor correlation =  0.010985 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.12  Page No : 72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 1.;\t    \t\t #volume of the tank in m**3\n",
-      "m = 180.;\t\t\t #mass of carbon dioxide in kg\n",
-      "T = 91.8;\t\t\t #temperature of the tank in degree celsius after it is placed in the vicinity of a furnace\n",
-      "Tc = 304.2;\t\t\t #critical temperature of carbon dioxide in K\n",
-      "Pc = 73.87;\t\t\t #critical pressure of carbon dioxide in bar\n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "\n",
-      "# Calculations\n",
-      "mwt = 44.*10**-3\t #molecular weight of carbon dioxide in kg/mol\n",
-      "n = m/mwt;\t\t\t # Calculations of number of moles of carbon dioxide in the tank in moles\n",
-      "MV = V/n;\t\t\t # Calculations of molar volume in m**3/mol\n",
-      "slope = (MV*Pc*10**5)/(R*(T+273.15));\t\t\t #slope of the straight line formed when z0 is plotted against Pr formed by using the relation z0 = (V*Pc*Pr)/(R*T)\n",
-      "Tr = (T+273.15)/Tc;\t\t\t # Calculations of reduced temperature (no unit)\n",
-      "\n",
-      "#At the given conditions, the state of CO2 must lie on the curve corresponding to the obtained value of Tr. To determine the state of CO2, a straight line passing through the origin, with the obtained value of slope is drawn on the z0 vs Pr plot of Fig.3.12 and the point of intersection of this straight line with the Tr curve is obtained to get the value of z0\n",
-      "z0 = 0.725;\t\t\t #value of compressibilty factor obtained by doing the above mentioned procedure\n",
-      "P = (z0*R*10**-6*(T+273.15))/(MV)\t\t\t # Calculations of pressure in MPa using Eq.(3.59)\n",
-      "\n",
-      "# Results\n",
-      "print ' The pressure developed by carbon dioxide =  %.0f MPa'%(P);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The pressure developed by carbon dioxide =  9 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.13  Page No : 73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "V = 1.;\t\t\t #volume of the tank in m**3\n",
-      "m = 180.;\t\t\t #mass of carbon dioxide in kg\n",
-      "T = 91.8;\t\t\t #temperature of the tank in degree celsius after it is placed in the vicinity of a furnace\n",
-      "Tc = 304.2;\t\t\t #critical temperature of carbon dioxide in K\n",
-      "Pc = 73.87;\t\t\t #critical pressure of carbon dioxide in bar\n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "w = 0.239;\t\t\t #acentric factor (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "mwt = 44.*10**-3\n",
-      "n = m/mwt\n",
-      "MV = V/n;\t\t\t\n",
-      "Tr = (T+273.15)/Tc\n",
-      "z0 = 0.725\n",
-      "z_init = z0;\t\t\t \n",
-      "slope = (MV*Pc*10**5)/(R*(T+273.15));\t\t\t \n",
-      "Prguess = z_init/slope;\t\n",
-      "z1guess = 0.1;\t\t\t \n",
-      "tolerance = 1e-6;\t\t\t \n",
-      "\n",
-      "def solver_function(zi):\n",
-      "    return zi-(z0+(w*z1guess))\n",
-      "\n",
-      "z = fsolve(solver_function,z1guess)\n",
-      "Pr = z/slope;\t\t\t \n",
-      "P = ((Pc*10**5)*Pr)*10**-6\n",
-      "\n",
-      "# Results\n",
-      "print ' The pressure developed by carbon dioxide =  %f MPa'%(P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The pressure developed by carbon dioxide =  9.295806 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.14  Page No : 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215;\t\t\t #pressure of n-ocatne vapour in MPa\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "Pc = 2.497;\t\t\t #critical pressure of n-octane in MPa\n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "\n",
-      "# Calculations\n",
-      "Tr = T/Tc\n",
-      "Pr = P/Pc\n",
-      "z_init = 1\n",
-      "h = (0.08664*Pr)/(z_init*Tr)\n",
-      "tolerance = 1e-6;\t\t\t\n",
-      "\n",
-      "def solver_func(zi):\n",
-      "    return zi-((1./(1-h))-((h/(1+h))*(4.93398/(Tr**(3./2)))))\n",
-      "\n",
-      "z = fsolve(solver_func,h)\n",
-      "V = (z*R*T)/(P*10**6);\t\t\t\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane vapour obtained by the generalized form of \\\n",
-      " \\n Redlich-Kwong equation of state =  %.2e m**3/mol'%(V);\n",
-      "\n",
-      "\n",
-      "# Note : Answer may vary because of rouding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane vapour obtained by the generalized form of  \n",
-        " Redlich-Kwong equation of state =  1.55e-02 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.15  Page No : 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature in K\n",
-      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
-      "w = 0.398;\t\t\t #acentric factor (no unit) \n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "\n",
-      "# Calculations\n",
-      "#The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily\n",
-      "#The general form of the cubic equation is (Z**3)+(alpha*Z**2)+(beeta*Z)+gaamma = 0, where alpha,beeta and gaamma are determined using established relations\n",
-      "\n",
-      "Tr = T/Tc\n",
-      "Pr = (P*10**6)/(Pc*10**5)\n",
-      "S = 0.48+(1.574*w)-(0.176*w**2)\n",
-      "alpha1 = (1+(S*(1-math.sqrt(Tr))))**2\n",
-      "a = (0.42748*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
-      "b = (0.08664*R*Tc)/(Pc*10**5)\n",
-      "A = (a*P*10**6)/(R*T)**2\n",
-      "B = (b*P*10**6)/(R*T)\n",
-      "alpha = -1.\n",
-      "beeta = A-B-B**2\n",
-      "gaamma = -(A*B)\n",
-      "p = beeta-(alpha**2)/3\n",
-      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
-      "D = (((q)**2)/4.)+(((p)**3)/27.)\n",
-      "\n",
-      "if D>0:\n",
-      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
-      "    vf = ((Z)*R*T)/(P*10**6)\n",
-      "    vg = ((Z)*R*T)/(P*10**6)\n",
-      "elif D == 0:\n",
-      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
-      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z = [Z1, Z2, Z3];\n",
-      "    vf = (min(Z)*R*T)/(P*10**6);\t\t\n",
-      "    vg = (max(Z)*R*T)/(P*10**6);\t\t\n",
-      "else:\n",
-      "    r = math.sqrt((-(p**3)/27));\t\t\n",
-      "    theta = acos((-(q)/2)*(1./r));\t\t\n",
-      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
-      "    Z = [Z1, Z2, Z3];\n",
-      "    vf = (min(Z)*R*T)/(P*10**6)\n",
-      "    vg = (max(Z)*R*T)/(P*10**6)\n",
-      "\n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane saturated vapour obtained by Soave-Redlich-Kwong \\\n",
-      " equation of state =  %f m**3/mol'%(vg);\n",
-      "print ' The volume occupied by n-octane saturated liquid obtained by Soave-Redlich-Kwong \\\n",
-      " equation of state =  %f m**3/mol'%(vf);\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane saturated vapour obtained by Soave-Redlich-Kwong  equation of state =  0.015207 m**3/mol\n",
-        " The volume occupied by n-octane saturated liquid obtained by Soave-Redlich-Kwong  equation of state =  0.000227 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.16  Page No : 78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature in K\n",
-      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
-      "w = 0.398;\t\t\t #acentric factor (no unit) \n",
-      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
-      "\n",
-      "# Calculations\n",
-      "Tr = T/Tc;\t\t\t\n",
-      "Pr = (P*10**6)/(Pc*10**5)\n",
-      "S = 0.37464+(1.54226*w)-(0.26992*w**2)\n",
-      "alpha1 = (1+(S*(1-math.sqrt(Tr))))**2\n",
-      "a = (0.45724*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
-      "b = (0.07780*R*Tc)/(Pc*10**5)\n",
-      "A = (a*P*10**6)/(R*T)**2\n",
-      "B = (b*P*10**6)/(R*T)\n",
-      "alpha = -1+B;\t\n",
-      "beeta = A-(2*B)-(3*B**2)\n",
-      "gaamma = -(A*B)+(B**2)+(B**3)\n",
-      "p = beeta-(alpha**2)/3;\t\t\n",
-      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
-      "D = (((q)**2)/4)+(((p)**3)/27)\n",
-      "\n",
-      "if D>0:\n",
-      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
-      "    vf = ((Z)*R*t)/(P*10**6)\n",
-      "    vg = ((Z)*R*t)/(P*10**6)\n",
-      "elif D == 0:\n",
-      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
-      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z = [Z1 ,Z2, Z3];\n",
-      "    vf = (min(Z)*R*T)/(P*10**6)\n",
-      "    vg = (max(Z)*R*T)/(P*10**6)\n",
-      "else:\n",
-      "    r = math.sqrt((-(p**3)/27))\n",
-      "    theta = .1533\n",
-      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3);\t\t\n",
-      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
-      "    Z = [Z1, Z2, Z3];\n",
-      "    vf = (min(Z)*R*T)/(P*10**6)\n",
-      "    vg = (max(Z)*R*T)/(P*10**6)\n",
-      "    \n",
-      "# Results\n",
-      "print ' The volume occupied by n-octane saturated vapour, obtained by Peng\\\n",
-      "-Robinson equation of state =  %.3e m**3/mol'%(vg);\n",
-      "print ' The volume occupied by n-octane saturated liquid, obtained by Peng-Robinson\\\n",
-      " equation of state =  %.3e m**3/mol'%(vf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume occupied by n-octane saturated vapour, obtained by Peng-Robinson equation of state =  1.514e-02 m**3/mol\n",
-        " The volume occupied by n-octane saturated liquid, obtained by Peng-Robinson equation of state =  2.003e-04 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch4.ipynb b/Chemical_Engineering_Thermodynamics/ch4.ipynb
deleted file mode 100755
index 5433cf7c..00000000
--- a/Chemical_Engineering_Thermodynamics/ch4.ipynb
+++ /dev/null
@@ -1,1291 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4 : First law of thermodynamics and its applications"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.1  Page No : 94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Q1 = 50.;\t\t\t #Energy added as heat in kJ when the system undergoes a process 1-2\n",
-      "W1 = 30.;\t\t\t #Work done by the system in kJ during the process 1-2\n",
-      "Q2 = -40.;\t\t\t #Energy rejected as heat in kJ during the process 2-3\n",
-      "W2 = -50.;\t\t\t #Work done on the system in kJ during the process 2-3\n",
-      "Q3 = 0.\t    \t\t #System undergoes an adiabatic process to return to initial state\n",
-      "\n",
-      "# Calculations\n",
-      "U2_1 = Q1-W1\n",
-      "U3_2 = Q2-W2\n",
-      "U1_3 = (-U2_1)-(U3_2)\n",
-      "W3 = Q3-U1_3\n",
-      "net_work = W1+W2+W3\n",
-      "\n",
-      "# Results\n",
-      "print ' The net work done by the system =  %d kJ'%(net_work);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The net work done by the system =  10 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.3  Page No : 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 1.  \t\t\t #volume of tank in m**3;\n",
-      "N = 200.;\t\t\t #number of moles of carbon dioxide in tank in mol\n",
-      "T1 = 25.;\t\t\t #ambient temperature in degree celsius\n",
-      "I = 5.; \t\t\t #current in amperes passed through the resistor place in the tank\n",
-      "Voltage = 440.\t\t #voltage in volts existing across the resistor\n",
-      "t = 10.;\t\t\t #time in minutes for which the current is passed\n",
-      "a = 363.077*10**-3\t\t\t #van der waals constant in Pa (m**3/mol)**2\n",
-      "b = 0.043*10**-3\t\t\t #van der waals constant in m**3/mol\n",
-      "Cv = 32.34\t\t\t #molar heat capacity at constant volume in J/molK\n",
-      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "MV = V/N;\t\t\t # Calculations of molar volume in m**3/mol\n",
-      "Q = 0.  \t\t\t #energy transfer as heat during the process\n",
-      "W_Pdv = 0.;\t\t\t #mechanical work done by the system\n",
-      "W_elec = -(Voltage*I*t*60)*(10**-6)\n",
-      "U2_1 = Q-(W_Pdv+W_elec);\t\t\t\n",
-      "T2 = ((U2_1*10**6)/(N*Cv))+(T1+273.15)\n",
-      "P = (((R*T2)/(MV-b))-(a/(MV**2)))*10**-3\n",
-      "\n",
-      "# Results\n",
-      "print ' The final pressure =  %0.3f kPa '%(P);\n",
-      "print ' The final temperature =  %0.2f K'%(T2);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The final pressure =  827.832 kPa \n",
-        " The final temperature =  502.23 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.4  Page No : 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 0.1;\t\t\t #volume of tank in m**3\n",
-      "T1 = 200.;\t\t\t #initial temperature of saturated steam inside the tank in degree celsius\n",
-      "T2 = 150.;\t\t\t #temperature in degree celsius that the tank attains after some time due to poor insulation\n",
-      "P1 = 15.549;\t\t\t #pressure in bar obtained from steam tables corresponding to T1\n",
-      "vg1 = 0.1272;\t\t\t #specific volume of saturated vapour in m**3/kg obtained from steam tables corresponding to T1\n",
-      "hg1 = 2790.9;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to T1\n",
-      "P2 = 4.76;\t\t\t #pressure in bar obtained from steam tables corresponding to T2\n",
-      "vf = 0.0010908;\t\t\t #specific volume of saturated liquid in m**3/kg obtained from steam tables corresponding to T2\n",
-      "vg2 = 0.3924;\t\t\t #specific volume of saturated vapour in m**3/kg obtained from steam tables corresponding to T2\n",
-      "hf = 632.15;\t\t\t #specific enthalpy of saturated liquid in kJ/kg obtained from steam tables corresponding to T1\n",
-      "hg2 = 2745.4;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to T1\n",
-      "\n",
-      "# Calculations\n",
-      "ug1 = ((hg1*10**3)-(P1*10**5*vg1))*10**-3\n",
-      "uf = ((hf*10**3)-(P2*10**5*vf))*10**-3\n",
-      "ug2 = ((hg2*10**3)-(P2*10**5*vg2))*10**-3\n",
-      "v2 = vg1\n",
-      "X2 = (v2-vf)/(vg2-vf)\n",
-      "u2 = (X2*ug2)+((1-X2)*uf)\n",
-      "m = V/vg1\n",
-      "Q = m*(u2-ug1)\n",
-      "mf = m*(1-X2)\n",
-      "mg = m*X2\n",
-      "\n",
-      "# Results\n",
-      "print ' The energy transferred as heat =  %f kJ'%(Q);\n",
-      "print ' The mass of liquid in the tank in the final state =  %0.3f kg'%(mf)\n",
-      "print ' The mass of vapour in the tank in the final state =  %0.3f kg'%(mg)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The energy transferred as heat =  -1053.825572 kJ\n",
-        " The mass of liquid in the tank in the final state =  0.533 kg\n",
-        " The mass of vapour in the tank in the final state =  0.253 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.5  Page No : 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "W = 1.;\t    \t\t #weight of steam in kg in the piston cylinder assembly\n",
-      "X = 0.8;\t\t\t #quality of steam (no unit)\n",
-      "T1 = 150.;\t\t\t #initial temperature of steam in degree celsius\n",
-      "T2 = 200.;\t\t\t #final temperature of steam in degree celsius\n",
-      "P1 = 476.;\t\t\t #pressure in kPa obatined from steam tables (corresponding to T1)\n",
-      "vf = 0.0010908;\t\t\t #specific volume of saturated liquid in m**3/kg obatined from steam tables (corresponding to T1)\n",
-      "vg = 0.3924;\t\t\t #specific volume of satuarted vapour in m**3/kg obatined from steam tables (corresponding to T1)\n",
-      "hf = 632.15;\t\t\t #specific enthalpy of saturated liquid in kJ/kg obtained from steam tables (corresponding to T1)\n",
-      "hg = 2745.4;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables (corresponding to T1)\n",
-      "\n",
-      "# Calculations\n",
-      "V1 = (X*vg)+((1-X)*vf)\n",
-      "h1 = (X*hg)+((1-X)*hf)\n",
-      "P2 = 0.476;\t\t\t \n",
-      "P_int1 = 0.4;\t\t\n",
-      "P_int2 = 0.5;\t\t\n",
-      "V_int1 = 0.5343;\t\n",
-      "V_int2 = 0.4250;\t\n",
-      "h_int1 = 2860.4;\t\n",
-      "h_int2 = 2855.1;\t\n",
-      "V2 = (((P2-P_int1)/(P_int2-P_int1))*(V_int2-V_int1))+V_int1\n",
-      "h2 = (((P2-P_int1)/(P_int2-P_int1))*(h_int2-h_int1))+h_int1\n",
-      "Q = (h2-h1)*W\n",
-      "W = P1*(V2-V1)*W\n",
-      "\n",
-      "# Results\n",
-      "print ' The work done by steam =  %0.2f kJ '%(W);\n",
-      "print ' The net energy transferred as heat  =  %0.2f kJ'%(Q);\n",
-      "print ' The final state of superheated steam,Pressure = %0.3f MPa '%(P2);\n",
-      "print ' The final state of superheated steam,Temperature = %d degree celsius '%(T2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The work done by steam =  65.26 kJ \n",
-        " The net energy transferred as heat  =  533.62 kJ\n",
-        " The final state of superheated steam,Pressure = 0.476 MPa \n",
-        " The final state of superheated steam,Temperature = 200 degree celsius \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.6  Page No : 103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "W = 1.  \t\t\t #weight of steam in kg in the piston cylinder assembly\n",
-      "X = 0.8;\t\t\t #quality of steam (no unit)\n",
-      "T1 = 150.\t\t\t #initial temperature  of steam in degree celsius\n",
-      "I = 5.  \t\t\t #current passed in Amperes\n",
-      "V = 220.\t\t\t #voltage in volts across the resistor\n",
-      "t = 10. \t\t\t #time for which the current is passed in minutes\n",
-      "P1 = 476.\t\t\t #pressure in kPa obatined from steam tables (corresponding to T1)\n",
-      "vf = 0.0010908;\t\t\t #specific volume of saturated liquid in m**3/kg obatined from steam tables (corresponding to T1)\n",
-      "vg = 0.3924;\t\t\t #specific volume of satuarted vapour in m**3/kg obatined from steam tables (corresponding to T1)\n",
-      "hf = 632.15;\t\t\t #specific enthalpy of saturated liquid in kJ/kg obtained from steam tables (corresponding to T1)\n",
-      "hg = 2745.4;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables (corresponding to T1)\n",
-      "\n",
-      "# Calculations\n",
-      "V1 = (X*vg)+((1-X)*vf);\t\n",
-      "h1 = (X*hg)+((1-X)*hf);\t\n",
-      "Ws = -V*I*t*60*10**-3;\t\n",
-      "h2 = h1-Ws;\t\t\t \n",
-      "P2 = 0.476;\t\t\t \n",
-      "T_int1 = 200;\t\t\n",
-      "T_int2 = 300;\t\t\n",
-      "V_int1 = 0.4512;\t\n",
-      "V_int2 = 0.5544;\t\n",
-      "h_int1 = 2856.37;\t\n",
-      "h_int2 = 3065.38;\t\n",
-      "V2 = (((h2-h_int1)/(h_int2-h_int1))*(V_int2-V_int1))+V_int1;\n",
-      "T2 = (((h2-h_int1)/(h_int2-h_int1))*(T_int2-T_int1))+T_int1;\n",
-      "W = (P1*10**3*(V2-V1)*W)*10**-3\n",
-      "\n",
-      "# Results\n",
-      "print ' The work done by steam =  %0.2f kJ '%(W);\n",
-      "print ' The final temperature =  %0.2f degree celsius'%(T2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The work done by steam =  94.94 kJ \n",
-        " The final temperature =  260.47 degree celsius\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.7  Page No : 104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "N = 1.\t\t\t     #number of moles of carbon dioxide in kmol\n",
-      "T1 = 298.\t\t\t #initial temperature in K\n",
-      "T2 = 600.\t\t\t #final raised temperature in K\n",
-      "a = 45.369\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "b = 8.688*10**-3\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "e = -9.619*10**5\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "\n",
-      "# Calculations\n",
-      "Q = N*10**3*((a*(T2-T1)+((b/2)*(T2**2-T1**2))-(e*((1./T2)-(1./T1)))))*10**-6\n",
-      "\n",
-      "# Results\n",
-      "print ' The amount of energy to be transferred as heat =  %0.3f MJ'%(Q);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The amount of energy to be transferred as heat =  13.255 MJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.8  Page No : 104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T1 = 298.   \t\t\t #initial temperature in K\n",
-      "T2 = 600.\t    \t\t #final raised temperature in K\n",
-      "a = 45.369;\t\t    \t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "b = 8.688*10**-3;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "e = -9.619*10**5;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "\n",
-      "# Calculations\n",
-      "Cpm = ((a*(T2-T1))+((b/2)*(T2**2-T1**2))-(e*((1./T2)-(1./T1))))/(T2-T1)\n",
-      "\n",
-      "# Results\n",
-      "print ' The isobaric molar heat capacity =  %0.2f J/molK'%(Cpm);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The isobaric molar heat capacity =  43.89 J/molK\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.9  Page No : 105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "N = 1.\t    \t\t #number of moles of carbon dioxide in kmol\n",
-      "T1 = 298.\t\t\t #initial temperature in K\n",
-      "T2 = 600.\t\t\t #final raised temperature in K\n",
-      "a = 45.369;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "b = 8.688*10**-3;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "e = -9.619*10**5;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "\n",
-      "# Calculations\n",
-      "Cpm = ((a*(T2-T1))+((b/2)*(T2**2-T1**2))-(e*((1./T2)-(1./T1))))/(T2-T1)\n",
-      "Q = N*10**3*Cpm*(T2-T1)*10**-6\n",
-      "\n",
-      "# Results\n",
-      "print ' The amount of energy to be transferred as heat =  %0.3f MJ '%(Q);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The amount of energy to be transferred as heat =  13.255 MJ \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.10  Page No : 105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "N = 100.\t\t\t #number of moles of carbon dioxide in mol\n",
-      "T1 = 298.\t\t\t #initial temperature in K\n",
-      "Q = 1.  \t\t\t #energy added as heat in MJ\n",
-      "a = 45.369;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "b = 8.688*10**-3;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "e = -9.619*10**5;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
-      "\n",
-      "# Calculations\n",
-      "delh = Q*10**6/N\n",
-      "Tguess = 520.\t\n",
-      "Cpm_guess = a+(b*((T1+Tguess)/2))+(e/(T1*Tguess))\n",
-      "T2_guess = T1+(delh/Cpm_guess)\n",
-      "tolerance = 1e-6;\t\t\t\n",
-      "while abs(T2_guess-Tguess)>tolerance:\n",
-      "    Tguess = T2_guess;\n",
-      "    Cpm_guess = a+(b*((T1+Tguess)/2))+(e/(T1*Tguess));\n",
-      "    T2_guess = T1+(delh/Cpm_guess)\n",
-      "\n",
-      "T2 = T2_guess\n",
-      "\n",
-      "# Results\n",
-      "print ' The final temperature =  %0.1f K'%(T2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The final temperature =  531.1 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.11  Page No : 107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "volume_ratio = 1./15\t #final volume to the initial volume of air at the end of compression stroke (no unit)\n",
-      "gaamma = 1.4\t\t\t #ratio of the molar heat capacities at constant pressure and constant volume for air (no unit)\n",
-      "T1 = 300.\t\t\t     #initial temperature of air in K\n",
-      "P1 = 0.1;\t\t\t    #initial pressure of air in MPa\n",
-      "R = 8.314;\t\t\t    #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "T2 = T1*((1./volume_ratio)**(gaamma-1))\n",
-      "P2 = P1*((1./volume_ratio)**(gaamma))\n",
-      "W = (R*(T1-T2)*10**-3)/(gaamma-1)\n",
-      "\n",
-      "# Results\n",
-      "print ' The final temperature =  %0.2f K'%(T2);\n",
-      "print ' The final pressure =  %0.4f MPa'%(P2);\n",
-      "print ' Work done per mole of air =  %0.3f kJ/mol'%(W);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The final temperature =  886.25 K\n",
-        " The final pressure =  4.4313 MPa\n",
-        " Work done per mole of air =  -12.185 kJ/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.12  Page No : 110"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "volume_ratio = 1./15;\t\t\t #final volume to the initial volume of ideal gas at the end of compression (no unit)\n",
-      "T1 = 300.   \t\t\t #initial temperature of ideal gas in K\n",
-      "P1 = 0.1;\t    \t\t #initial pressure of ideal gas in MPa\n",
-      "R = 8.314;\t\t    \t #universal gas constant in J/molK\n",
-      "n = 1.2;\t\t\t     #index of expansion (no unit)\n",
-      "gaamma = 1.4;\t\t\t #ratio of the molar heat capacities at constant pressure and constant volume for ideal gas (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "P2 = P1*((1./volume_ratio)**n)\n",
-      "T2 = T1*(P2/P1)*(volume_ratio)\n",
-      "W = (R*(T1-T2)*10**-3)/(n-1)\n",
-      "del_u = (R*(T2-T1)*10**-3)/(gaamma-1)\n",
-      "q = del_u+W\n",
-      "\n",
-      "# Results\n",
-      "print ' The final pressure =  %0.3f MPa'%(P2);\n",
-      "print ' The final temperature =  %0.1f K'%(T2);\n",
-      "print ' Work done on the gas =  %f kJ/mol'%(W);\n",
-      "print ' Heat interaction during the process =  %f kJ/mol'%(q)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The final pressure =  2.578 MPa\n",
-        " The final temperature =  515.6 K\n",
-        " Work done on the gas =  -8.963805 kJ/mol\n",
-        " Heat interaction during the process =  -4.481902 kJ/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.13  Page No : 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 1.;\t    \t\t #volume of tank in m**3\n",
-      "T0 = 300.\t\t\t #initial temperature of ideal gas in K\n",
-      "P0 = 0.1\t\t\t #initial pressure of ideal gas in MPa\n",
-      "T = 500.\t\t\t #temperature of ideal gas in the pipeline in K\n",
-      "P = 3.\t\t    \t #pressure of ideal gas in the pipeline in MPa\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "gaamma = 1.4;\t\t\t #ratio of the molar heat capacities at constant pressure and constant volume for ideal gas (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "Pf = 3.\n",
-      "Tf = (Pf*10**6)/((((Pf*10**6)-(P0*10**6))/(gaamma*T))+((P0*10**6)/T0));\n",
-      "N = (V/R)*(((Pf*10**6)/Tf)-((P0*10**6)/T0));\n",
-      "\n",
-      "# Results\n",
-      "print ' The final temperature =  %0.1f K'%(Tf);\n",
-      "print ' The amount of gas that has entered the tank =  %0.2f mol'%(N);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The final temperature =  670.2 K\n",
-        " The amount of gas that has entered the tank =  498.30 mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.14  Page No : 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "# Variables\n",
-      "V = 3.  \t\t\t #volume of tank in m**3\n",
-      "T0 = 100.;\t\t\t #initial temperature of steam in degree celsius\n",
-      "T = 300.;\t\t\t #temperature of superheated steam in the pipeline in degree celsius\n",
-      "P = 3.;\t\t\t     #pressure of superheated steam in the pipeline in MPa\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Ps = 101.33\n",
-      "vg = 1.673\n",
-      "hg = 2676.0\n",
-      "h = 2995.1\n",
-      "u0 = ((hg*10**3)-(Ps*10**3*vg))*10**-3\n",
-      "m0 = V/vg;\t\t\t\n",
-      "\n",
-      "Tf = 418.\n",
-      "vf = 0.102329\n",
-      "uf = 2965.78\n",
-      "\n",
-      "mf_guess = V/vf\n",
-      "\n",
-      "def solver_func(ui):\n",
-      "    return (mf_guess*ui)-(m0*u0)-((mf_guess-m0)*h);\n",
-      "\n",
-      "uf_solved = fsolve(solver_func,mf_guess)\n",
-      "mf = mf_guess\n",
-      "mass = mf-m0\n",
-      "\n",
-      "# Results\n",
-      "print \" The final state of steamsuperheated, Pressure = %d MPa\"%(P);\n",
-      "print \" The final state of steamsuperheated, Temperature = %d degree celsius\"%(Tf);\n",
-      "print \" The mass of steam that entered the tank = %0.3f kg\"%(mass);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The final state of steamsuperheated, Pressure = 3 MPa\n",
-        " The final state of steamsuperheated, Temperature = 418 degree celsius\n",
-        " The mass of steam that entered the tank = 27.524 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.15  Page No : 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 0.1;\t\t\t #volume of cylinder in m**3\n",
-      "T0 = 300.\t\t\t #initial temperature of nitrogen in K\n",
-      "P0 = 14.\t\t\t #initial pressure of nitrogen in MPa\n",
-      "P = 0.1;\t\t\t #ambient pressure in MPa\n",
-      "Pf = 2. \t\t\t #final pressure of nitrogen in MPa\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "gaamma = 1.4;\t\t #ratio of the molar heat capacities at constant pressure and constant volume for nitrogen (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "def solver_func(Ti):\n",
-      "    return ((P0*10**6)-(Pf*10**6))-((gaamma/2)*(T0+Ti)*(((P0*10**6)/T0)-((Pf*10**6)/Ti)));\n",
-      "\n",
-      "Tguess = 300.\n",
-      "Tf = fsolve(solver_func,Tguess)\n",
-      "\n",
-      "N = (V/R)*(((P0*10**6)/T0)-((Pf*10**6)/Tf));\n",
-      "\n",
-      "# Results\n",
-      "print ' The final temperature =  %0.1f K'%(Tf);\n",
-      "print ' The amount of gas that has escaped from the cylinder =  %0.2f mol'%(N);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The final temperature =  181.2 K\n",
-        " The amount of gas that has escaped from the cylinder =  428.52 mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.16  Page No : 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T0 = 300.;\t\t\t #initial temperature of superheated steam in degree celsius\n",
-      "P0 = 3.;\t\t\t #initial pressure of superheated steam in MPa\n",
-      "Xe = 0.85;\t\t\t #quality of steam leaving the turbine (no unit)\n",
-      "Tf = 45.;\t\t\t #final temperature of steam leaving the turbine in degree celsius\n",
-      "Vi = 10.;\t\t\t #velocity of steam at the entrance in m/s\n",
-      "Ve = 40.;\t\t\t #exit velocity of steam in m/s\n",
-      "Zi = 10.;\t\t\t #elevation at the entrance in m\n",
-      "Ze = 4.;\t\t\t #elevation at the exit in m\n",
-      "m = 1.;\t\t\t #mass flow rate of steam through turbine in kg/s\n",
-      "g = 9.81;\t\t\t #accleration due to gravity in m/s**2\n",
-      "\n",
-      "# Calculations\n",
-      "hi = 2995.1\n",
-      "hf = 188.35\n",
-      "hg = 2583.3\n",
-      "he = ((1-Xe)*hf)+(Xe*hg)\n",
-      "Q = 0.\t\t\t \n",
-      "enthalpy_change = (he*10**3)-(hi*10**3)\n",
-      "KE_change = ((Ve**2)-(Vi**2))/2\n",
-      "PE_change = g*(Ze-Zi);\t\t\t\n",
-      "Ws = Q-(m*(enthalpy_change+KE_change+PE_change)*10**-3)\n",
-      "err_KE = ((KE_change)/(Ws*10**3))*100;\t\n",
-      "err_PE = ((abs (PE_change)/(Ws*10**3)))*100\n",
-      "err = err_KE+err_PE\n",
-      "\n",
-      "# Results\n",
-      "print ' The percentage error when Kinetic energy change is ignored =  %0.3f '%(err_KE);\n",
-      "print ' The percentage error when Potential energy change is ignored =  %0.4f '%(err_PE);\n",
-      "print ' The percentage error when both Kinetic and Potential energy changes are ignored =  %f '%(err);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The percentage error when Kinetic energy change is ignored =  0.097 \n",
-        " The percentage error when Potential energy change is ignored =  0.0076 \n",
-        " The percentage error when both Kinetic and Potential energy changes are ignored =  0.104999 \n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.17  Page No : 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "Pi = 5.;\t\t\t #pressure of dry saturated steam at the entrance in bar\n",
-      "Pe = 2.;\t\t\t #pressure of dry saturated steam at the exit in bar\n",
-      "Vi = 3.;\t\t\t #velocity of dry saturated steam at the entrance in m/s\n",
-      "m = 1.; \t\t\t #flow rate of steam through the nozzle in kg/s\n",
-      "g = 9.81;\t\t\t #acceleration due to gravity in m/s**2\n",
-      "\n",
-      "# Calculations\n",
-      "hi = 2747.5\n",
-      "he = 2706.3\n",
-      "ve = 0.8854\n",
-      "Zi = 0;\t\t\n",
-      "Ze = 0;\t\t\n",
-      "Q = 0;\t\t\n",
-      "Ws = 0;\t\t\n",
-      "Ve = math.sqrt (2*(((Q-Ws)/m)-(g*(Zi-Ze))-((he*10**3)-(hi*10**3)))+(Vi**2))\n",
-      "A = (m*ve)/Ve\n",
-      "\n",
-      "# Results\n",
-      "print ' The velocity of dry saturated steam at the exit =  %0.2f m/s'%(Ve);\n",
-      "print ' The cross sectional area of the nozzle at the exit =  %0.3e m**2'%(A);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The velocity of dry saturated steam at the exit =  287.07 m/s\n",
-        " The cross sectional area of the nozzle at the exit =  3.084e-03 m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.18  Page No : 123"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T1 = 270.\t\t\t #temperature of wet steam in degree celsius\n",
-      "T2 = 120.\t\t\t #final temperature of superheated steam in degree celsius\n",
-      "P = 0.1;\t\t\t #pressure of superheated steam in MPa\n",
-      "\n",
-      "# Calculations\n",
-      "hf = 1185.2 \t\t\t\n",
-      "hg = 2789.9\t    \t\t\n",
-      "he = 2716.04\t\t\t\n",
-      "Xi = (he-hf)/(hg-hf);\t\n",
-      "\n",
-      "# Results\n",
-      "print ' The quality of wet steam =  %0.3f '%(Xi);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The quality of wet steam =  0.954 \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.20  Page No : 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "del_H = -90.135;\t\t\t #standard enthalpy change for the reaction CO(g)+2H2(g)--->CH3OH(g) at 298.15K in kJ\n",
-      "\n",
-      "# Calculations\n",
-      "del_H1 = 2*del_H\n",
-      "del_H2 = (1./2)*del_H\n",
-      "\n",
-      "# Results\n",
-      "print ' The standard enthalpy change for the reaction 2COg)+4H2g)---->2CH3OHg \\\n",
-      ") at 298.15K =  %0.2f kJ'%(del_H1);\n",
-      "print ' The standard enthalpy change for the reaction 1./2)COg)+H2g)---->1./2\\\n",
-      ")CH3OHg) at 298.15K =  %0.4f kJ'%(del_H2 );\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The standard enthalpy change for the reaction 2COg)+4H2g)---->2CH3OHg ) at 298.15K =  -180.27 kJ\n",
-        " The standard enthalpy change for the reaction 1./2)COg)+H2g)---->1./2)CH3OHg) at 298.15K =  -45.0675 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.22  Page No : 130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "del_Hf_C4H10 = -74.943;\t\t\t #smath.tan(math.radiansard enthalpy of formation of C4H10(g) at 298.15K in kJ\n",
-      "del_Hf_CO2 = -393.978;\t\t\t #smath.tan(math.radiansard enthalpy of formation of CO2(g) at 298.15K in kJ\n",
-      "del_Hf_H2O = -241.997;\t\t\t #smath.tan(math.radiansard enthalpy of formation of H2O(g) at 298.15K in kJ\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "del_Hr = (5*del_Hf_H2O)+(4*del_Hf_CO2)-(del_Hf_C4H10);\n",
-      "\n",
-      "# Results\n",
-      "print ' The standard enthalpy change for the reaction C4H10g)+13\\\n",
-      "/2)O2g)---->4CO2g)+5H2Og) at 298.15K =  %0.3f kJ'%(del_Hr);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The standard enthalpy change for the reaction C4H10g)+13/2)O2g)---->4CO2g)+5H2Og) at 298.15K =  -2710.954 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.23  Page No : 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "del_Hf_C4H10 = -74.943;\t\t\t #smath.tan(math.radiansard enthalpy of formation of C4H10(g) at 298.15K in kJ\n",
-      "del_Hf_CO2 = -393.978;\t\t\t #smath.tan(math.radiansard enthalpy of formation of CO2(g) at 298.15K in kJ\n",
-      "del_Hf_H2O = -241.997;\t\t\t #smath.tan(math.radiansard enthalpy of formation of H2O(g) at 298.15K in kJ\n",
-      "del_H_vap = 43.966;\t\t\t #enthalpy of vaporization of H2O at 298.15K in kJ/mol\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "del_H1 = 0.\n",
-      "del_H2 = 5*(-del_H_vap)\n",
-      "del_H3 = 0.\n",
-      "\n",
-      "del_H = (5*del_Hf_H2O)+(4*del_Hf_CO2)-(del_Hf_C4H10);\n",
-      "del_net_H = (del_H)+(del_H1)+(del_H2)+(del_H3)\n",
-      "\n",
-      "# Results\n",
-      "print ' The standard enthalpy change for the reaction C4H10g)+13/2)O2g)---->4CO2g\\\n",
-      ")+5H2Ol) at 298.15K =  %0.3f kJ'%(del_net_H);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The standard enthalpy change for the reaction C4H10g)+13/2)O2g)---->4CO2g)+5H2Ol) at 298.15K =  -2930.784 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.24  Page No : 132"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "del_H_comb = 2880.44;\t\t\t #gross heating value of n-buatne gas at 298.15K in kJ/mol\n",
-      "del_Hf_CO2 = -393.978;\t\t\t #standard enthalpy of formation of CO2(g) at 298.15K in kJ\n",
-      "del_Hf_H2O = -285.958;\t\t\t #standard enthalpy of formation of H2O(l) at 298.15K in kJ\n",
-      "del_Hf_O2 = 0.;\t\t        \t #standard enthalpy of formation of O2(g) at 298.15K in kJ\n",
-      "\t\t\t \n",
-      "# Calculations\n",
-      "n_CO2 = 4.\n",
-      "n_H2O = 5.\n",
-      "n_O2 = -13./2\n",
-      "n_C4H10 = -1.\n",
-      "\t\t\t \n",
-      "del_Hf_C4H10 = (n_CO2*del_Hf_CO2)+(n_H2O*del_Hf_H2O)+(n_O2*del_Hf_O2)-(-del_H_comb);\n",
-      "\n",
-      "\n",
-      "# Results\n",
-      "print ' The standard enthalpy of formation of n-butane gas at 298.15K =  %0.3f kJ'%(del_Hf_C4H10);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The standard enthalpy of formation of n-butane gas at 298.15K =  -125.262 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.25  Page No : 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "del_Hc_A = -890.94;\t\t\t #enthalpy change accompanying reaction A in kJ\n",
-      "del_Hc_B = -283.18;\t\t\t #enthalpy change accompanying reaction B in kJ\n",
-      "del_Hc_C = -286.03;\t\t\t #enthalpy change accompanying reaction C in kJ\n",
-      "del_H_vap = -43.966;\t\t\t #enthalpy change of vaporization of H2O at 298.15K in kJ/mol\n",
-      "\n",
-      "# Calculations\n",
-      "del_H0 = (del_Hc_A)-(del_Hc_B)-(3*del_Hc_C)+(del_H_vap)\n",
-      "\n",
-      "# Results\n",
-      "print ' The standard enthalpy change at 298.15K for the reaction \\\n",
-      "CH4g)+H2Og)--->COg)+3H2g) =  %0.3f kJ'%(del_H0);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The standard enthalpy change at 298.15K for the reaction CH4g)+H2Og)--->COg)+3H2g) =  206.364 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.26  Page No : 135"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "#The reaction is : C2H4(g)+H2O(g)--->C2H5OH(g)\n",
-      "del_H_vap = 43.82\n",
-      "\n",
-      "del_Hf = [52.335,-241.997,0,-277.819]\n",
-      "a = [4.196,28.850,20.691,0]\t\n",
-      "b = [154.565*10**-3,12.055*10**-3,205.346*10**-3,0]\n",
-      "c = [-81.076*10**-6,0,-99.793*10**-6,0]\n",
-      "d = [16.813*10**-9,0,18.825*10**-9,0]\t\n",
-      "e = [0,1.006*10**5,0,0];\t\t\n",
-      "\n",
-      "T1 = 298.15\t\t\n",
-      "T2 = 400.\t\t\t\n",
-      "n_C2H4 = -1.\n",
-      "n_H2O = -1.\n",
-      "n_C2H5OH = 1.\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "del_Hf_C2H5OH_g = del_Hf[3]+del_H_vap;\n",
-      "del_Hr = (n_C2H5OH*del_Hf_C2H5OH_g)+(n_C2H4*del_Hf[0])+(n_H2O*del_Hf[1])\n",
-      "del_a = (n_C2H4*a[0])+(n_H2O*a[1])+(n_C2H5OH*a[2])\t\t\t \n",
-      "del_b = (n_C2H4*b[0])+(n_H2O*b[1])+(n_C2H5OH*b[2])\t\t\t \n",
-      "del_c = (n_C2H4*c[0])+(n_H2O*c[1])+(n_C2H5OH*c[2])\t\t\t \n",
-      "del_d = (n_C2H4*d[0])+(n_H2O*d[1])+(n_C2H5OH*d[2])\t\t\t \n",
-      "del_e = (n_C2H4*e[0])+(n_H2O*e[1])+(n_C2H5OH*e[2])\t\t\t \n",
-      "del_H0 = (del_Hr*10**3)-((del_a*T1)+((del_b/2)*T1**2)+((del_c/3)*T1**3)+((del_d/4)*T1**4)-(del_e/T1))\n",
-      "del_Hr_T2 = (del_H0+((del_a*T2)+((del_b/2)*T2**2)+((del_c/3)*T2**3)+((del_d/4)*T2**4)-(del_e/T2)))*10**-3;\n",
-      "\n",
-      "# Results\n",
-      "print ' The standard enthalpy change at 400K for the reaction\\\n",
-      " C2H4g)+H2Og)--->C2H5OHg) =  %f kJ'%(del_Hr_T2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The standard enthalpy change at 400K for the reaction C2H4g)+H2Og)--->C2H5OHg) =  -44.529472 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.28  Page No : 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      " \n",
-      "#The combustion reaction of methane is given by:\n",
-      "# CH4(g)+2O2(g)--->CO2(g)+2H2O(g)\n",
-      "\n",
-      "n_O2 = 2.\t\t\t # stoichiometric amount of oxygen required for combustion\n",
-      "n_CH4 = 1.\t\t\t #number of moles of CH4(g) in moles\n",
-      "n_CO2 = 1.\t\t\t #number of moles of CO2(g) formed in the combustion reaction in moles\n",
-      "n_H2O = 2.\t\t\t #number of moles of H2O(g) formed in the combustion reaction in moles\n",
-      "del_Hf = [-74.943,0,-393.978,-241.997];\t\t\t # standard enthalpies of formation of CH4(g),O2(g),CO2(g),H2O(g) at 298.15K in kJ\n",
-      "a = [45.369,28.850,30.255,27.270];\t\t\t     #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
-      "b = [8.688*10**-3,12.055*10**-3,4.207*10**-3,4.930*10**-3];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
-      "c = [0,0,0,0];  \t    \t\t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
-      "d = [0,0,0,0];\t    \t    \t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
-      "e = [-9.619*10**5,1.006*10**5,-1.887*10**5,0.333*10**5];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
-      "per_excess_air = 50. \t\t\t #percentage excess of air supplied to the adiabatic burner\n",
-      "T_amb = 298.15\t\t\t # temperature at which air and methane enter the burner in K\n",
-      "per_N2 = 79.\t\t\t #percentage of N2 in the air supplied\n",
-      "per_O2 = 21.\t\t\t #percentage of O2 in the air supplied\n",
-      "\n",
-      "# Calculations\n",
-      "n_O2_actual = (1+(per_excess_air/100))*n_O2\n",
-      "n_N2 = n_O2_actual*(per_N2/per_O2);\t\t\t\n",
-      "n_O2_residual = n_O2_actual-n_O2;\t\t\t\n",
-      "\n",
-      "del_Hr = (n_CO2*del_Hf[2])+(n_H2O*del_Hf[3])-(n_O2*del_Hf[1])-(n_CH4*del_Hf[0])\n",
-      "\n",
-      "del_a = (n_CO2*a[0])+(n_H2O*a[1])+(n_O2_residual*a[2])+(n_N2*a[3])\n",
-      "del_b = (n_CO2*b[0])+(n_H2O*b[1])+(n_O2_residual*b[2])+(n_N2*b[3])\n",
-      "del_c = (n_CO2*c[0])+(n_H2O*c[1])+(n_O2_residual*c[2])+(n_N2*c[3])\n",
-      "del_d = (n_CO2*d[0])+(n_H2O*d[1])+(n_O2_residual*d[2])+(n_N2*d[3])\n",
-      "del_e = (n_CO2*e[0])+(n_H2O*e[1])+(n_O2_residual*e[2])+(n_N2*e[3]);\t\n",
-      "tguess = 500.\n",
-      "\n",
-      "def solver_func(ti):\n",
-      "    return (-(del_Hr*10**3))-((del_a*(ti-T_amb))+((del_b/2)*((ti**2)-(T_amb**2)))+((del_c/3)*((ti**3)-(T_amb**3)))+((del_d/4)*((ti**4)-(T_amb**4)))+(del_e*((1./T_amb)-(1./ti))));\n",
-      "\n",
-      "T = fsolve(solver_func,tguess)\n",
-      "\n",
-      "# Results\n",
-      "print ' The flame temperature when methane is burned with 50 percent excess air \\\n",
-      "in an adiabatic burner =  %f K'%(T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The flame temperature when methane is burned with 50 percent excess air in an adiabatic burner =  1793.784965 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.29  Page No : 139"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T_exit = 550.\t\t\t #temperature in K at which the combustion products leave the boiler\n",
-      "percent_molar_comp_prdct = [6.28,3.14,7.85,82.73];\t\t\t # percentage molar composition of the combustion products CO2(g),CO(g),O2(g),N2(g) on dry basis\n",
-      "T_ent = 298.15;\t\t\t #temperature in K at which Propane and air enter the combustion chamber\n",
-      "del_Hf = [-393.978,-110.532,-241.997,0,0,-103.833];\t\t\t #smath.tan(math.radiansard enthalpies of formation of CO2(g),CO(g),H2O(g),O2(g),N2(g),C3H8(g) at 298.15K in kJ\n",
-      "a = [45.369,28.068,30.255,27.270,28.850];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
-      "\n",
-      "#coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
-      "b = [8.688*10**-3,4.631*10**-3,4.207*10**-3,4.930*10**-3,12.055*10**-3];\n",
-      "c = [0,0,0,0,0];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
-      "d = [0,0,0,0,0];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
-      "e = [-9.619*10**5,-0.258*10**5,-1.887*10**5,0.333*10**5,1.006*10**5];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
-      "per_N2 = 79.\t\t\t #percentage of nitrogen in air\n",
-      "per_O2 = 21.\t\t\t #percentage of oxygen in air\n",
-      "molar_mass_propane = 44.*10**-3;\t\t\t #molar mass of propane in kg/mole\n",
-      "\n",
-      "# Calculations\n",
-      "# TAKE BASIS AS 100 mol OF  DRY COMBUSTION PRODUCTS\n",
-      "n_CO2 = percent_molar_comp_prdct[0]\t\t\t #number of moles of CO2(g) in the product stream\n",
-      "n_CO = percent_molar_comp_prdct[1]\t\t\t #number of moles of CO(g) in the product stream\n",
-      "n_O2 = percent_molar_comp_prdct[2]\t\t\t #number of moles of O2(g) in the product stream\n",
-      "n_N2 = percent_molar_comp_prdct[3]\t\t\t #number of moles of N2(g) in the product stream\n",
-      "\n",
-      "x = (n_CO2+n_CO)/3;\n",
-      "\n",
-      "y = (2*n_N2)/(2*(per_N2/per_O2));\n",
-      "\n",
-      "z = (2*y)-(2*n_CO2)-(n_CO)-(2*n_O2);\n",
-      "\n",
-      "n_H2O = z\n",
-      "n_C3H8 = x\n",
-      "\n",
-      "del_Hr = (n_CO2*del_Hf[0])+(n_CO*del_Hf[1])+(n_H2O*del_Hf[2])-(n_C3H8*del_Hf[5]);\n",
-      "\n",
-      "del_a = (n_CO2*a[0])+(n_CO*a[1])+(n_O2*a[2])+(n_N2*a[3])+(n_H2O*a[4])\t\n",
-      "del_b = (n_CO2*b[0])+(n_CO*b[1])+(n_O2*b[2])+(n_N2*b[3])+(n_H2O*b[4])\t\n",
-      "del_c = (n_CO2*c[0])+(n_CO*c[1])+(n_O2*c[2])+(n_N2*c[3])+(n_H2O*c[4])\t\n",
-      "del_d = (n_CO2*d[0])+(n_CO*d[1])+(n_O2*d[2])+(n_N2*d[3])+(n_H2O*d[4])\t\n",
-      "del_e = (n_CO2*e[0])+(n_CO*e[1])+(n_O2*e[2])+(n_N2*e[3])+(n_H2O*e[4])\t\n",
-      "\n",
-      "#calulation of del_Hp J\n",
-      "del_Hp = (del_a*(T_exit-T_ent))+((del_b/2)*((T_exit**2)-(T_ent**2)))+((del_c/3)*((T_exit**3)-(T_ent**3)))+((del_d/4)*((T_exit**4)-(T_ent**4)))-(del_e*((1./T_exit)-(1./T_ent)));\n",
-      "del_H = ((del_Hr*10**3)+(del_Hp))*10**-3;\t\t\t \n",
-      "mass_propane = n_C3H8*molar_mass_propane;\t\t\t \n",
-      "energy = (-(del_H*10**3)/mass_propane)*10**-6;\t\t\n",
-      "       \n",
-      "# Results\n",
-      "print ' The energy transferred as heat per kg propane =  %f MJ'%(energy);     \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The energy transferred as heat per kg propane =  33.741510 MJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch5.ipynb b/Chemical_Engineering_Thermodynamics/ch5.ipynb
deleted file mode 100755
index 9cf5b1c8..00000000
--- a/Chemical_Engineering_Thermodynamics/ch5.ipynb
+++ /dev/null
@@ -1,974 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5 : Second law of thermodynamics and its applications"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.2  Page No : 161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Q = 1000.;\t\t    \t #amount of energy absorbed by the heat engine in kJ/s\n",
-      "W = 650.;\t\t\t     #work delivered by the heat engine in kW\n",
-      "T_source = 500. \t\t #temperature of the source in degree celsius\n",
-      "T_sink = 25.\t\t\t #temperature of the sink in degree celsius\n",
-      "\n",
-      "# Calculations\n",
-      "n_claimed = W/Q\n",
-      "T1 = T_source+273.15\n",
-      "T2 = T_sink+273.15\n",
-      "n_carnot = 1-(T2/T1)\n",
-      "\n",
-      "# Results\n",
-      "print \" The efficiency of the Carnot engine = %0.3f \"%(n_carnot);\n",
-      "print \" The efficiency of the engine claimed by the inventor = %0.2f \"%(n_claimed);\n",
-      "if n_claimed<n_carnot:\n",
-      "    print \" The claimed heat engine is theoretically feasible as the efficiency of the engine is lesser than that of a Carnot engine\";\n",
-      "else:\n",
-      "    print \" The claimed heat engine is not theoretically feasible as the efficiency of the engine is greater than that of a Carnot engine\";\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The efficiency of the Carnot engine = 0.614 \n",
-        " The efficiency of the engine claimed by the inventor = 0.65 \n",
-        " The claimed heat engine is not theoretically feasible as the efficiency of the engine is greater than that of a Carnot engine\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.3  Page No : 165"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T_source_summer = 42; #temperature in the summer months in degree celsius\n",
-      "T_sink_winter = 0;\t  #temperature in the winter months in degree celius\n",
-      "T_amb = 25;\t\t\t  #temperature at which the house is to be maintained during both the months in degree celsius\n",
-      "energy_loss = 0.5;\n",
-      "\n",
-      "# Calculations\n",
-      "T_H_summer = T_source_summer+273.15\n",
-      "T_L_summer = T_amb+273.15\n",
-      "T_H_winter = T_amb+273.15\n",
-      "T_L_winter = T_sink_winter+273.15\n",
-      "W_summer = (energy_loss*((T_H_summer-T_L_summer)**2))/(T_L_summer)\n",
-      "W_winter = (energy_loss*((T_H_winter-T_L_winter)**2))/(T_H_winter)\n",
-      "\n",
-      "# Results\n",
-      "print \" The minimum power required to operate the device in summer = %.4f kW \"%(W_summer);\n",
-      "print \" The minimum power required to operate the device in winter = %f kW \"%(W_winter);\n",
-      "\n",
-      "# Note: Answer in book is wrong. Please calculated manually."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The minimum power required to operate the device in summer = 0.4847 kW \n",
-        " The minimum power required to operate the device in winter = 1.048130 kW \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.4  Page No : 166"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T_L = 4.25 \t\t\t #normal boiling point of helium in K\n",
-      "Q_L = 0.083\t\t\t #latent heat of vaporization of helium in kJ/mol\n",
-      "n = 1.  \t\t\t #amount of liquid helium to be produced in kmol\n",
-      "T_amb = 42.\t\t\t #ambient temperature in summer in degree celsius\n",
-      "\n",
-      "# Calculations\n",
-      "T_H = T_amb+273.15\n",
-      "COP = (T_L)/(T_H-T_L)\n",
-      "W = (Q_L)/COP;\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The maximum possible COP of the unit = %0.4f \"%(COP);\n",
-      "print \" The minimum amount of work to be done on the refrigerating unit = %f kJ \"%(W);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The maximum possible COP of the unit = 0.0137 \n",
-        " The minimum amount of work to be done on the refrigerating unit = 6.071694 kJ \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.5  Page No : 166"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T_ice = 0.;\t\t        \t #temperature of the ice to be produced in degree celsius\n",
-      "m = 5000.;\t\t\t         #rate at which ice is to be produced in kg/hour\n",
-      "T_water = 0.;\t    \t\t #temperature of water used to produce ice in degree celsius\n",
-      "T_amb = 40.;\t\t    \t #ambient temperature in degree celsius\n",
-      "T_source = 100.;\t\t\t #temperature of the source for operating heat engine in degree celsius\n",
-      "lambda_fusion = 6.002;\t\t\t #latent heat of fusion of water in kJ/mol at 0 degree celsius\n",
-      "molar_mass = 18*10**-3;\t\t\t #molar mass of water in kg/mol\n",
-      "\n",
-      "# Calculations\n",
-      "T_L = T_water+273.15\n",
-      "T_H = T_amb+273.15\n",
-      "COP = (T_L)/(T_H-T_L)\n",
-      "Q_L = ((m/3600)/molar_mass)*(lambda_fusion)\n",
-      "W = (Q_L)/(COP);\t\t\t \n",
-      "T1 = T_source+273.15;\t\t\n",
-      "T2 = T_amb+273.15\n",
-      "n_heatengine = (T1-T2)/T1\n",
-      "Q1 = W/n_heatengine;\t\n",
-      "energy_ratio = (Q1+Q_L)/Q_L;\n",
-      "\n",
-      "# Results\n",
-      "print \" The minimum power required to operate the refrigerator = %0.2f kW\"%(W);\n",
-      "print \" The maximum possible efficiency of the heat engine = %0.4f \"%(n_heatengine);\n",
-      "print \" Ratio of the energy rejected to the ambient atmosphere to the\\\n",
-      " energy absorbed from the water = %0.4f \"%(energy_ratio);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The minimum power required to operate the refrigerator = 67.82 kW\n",
-        " The maximum possible efficiency of the heat engine = 0.1608 \n",
-        " Ratio of the energy rejected to the ambient atmosphere to the energy absorbed from the water = 1.9107 \n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.6  Page No : 169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T1 = 800.\t\t\t #temperature of reservoir 1 in K\n",
-      "T2 = 400.\t\t\t #temperature of reservoir 2 in K\n",
-      "Q1 = 1000.\t\t\t #energy absorbed from reservoir maintained at T1 in kJ\n",
-      "Q2 = 400.\t\t\t #energy absorbed from reservoir maintained at T2 in kJ\n",
-      "W = 1000.\t\t\t #work delivered by the heat engine in kJ\n",
-      "T3 = 300.\t\t\t #temperature of the sink in K\n",
-      "\n",
-      "# Calculations\n",
-      "Q3 = (Q1+Q2)-W\n",
-      "clausius_inequality = (Q1/T1)+(Q2/T2)-(Q3/T3)\n",
-      "\n",
-      "# Results\n",
-      "print \" The LHS of the Clausius inequality = %0.4f \"%(clausius_inequality);\n",
-      "if clausius_inequality<0 or clausius_inequality == 0:\n",
-      "    print \" The given process does not violate the second law of thermodynamics, therefore the claim is correct\"\n",
-      "else:\n",
-      "    print \" This is a violation of the second law of thermodynamics, and hence the claim cannot be justified\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The LHS of the Clausius inequality = 0.9167 \n",
-        " This is a violation of the second law of thermodynamics, and hence the claim cannot be justified\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.7  Page No : 172"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T_system = 200.;\t\t\t #temperature of the contents of reactor in degree celsius\n",
-      "t = 15.         \t\t\t #operation time of agitator in minutes\n",
-      "P = 750.    \t    \t\t #power of the operating motor in W\n",
-      "\n",
-      "# Calculations\n",
-      "dQ = P*t*60*10**-3\n",
-      "T = T_system+273.15\n",
-      "del_S = dQ/T;\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The change in the entropy of the reactor contents = %0.4f kJ/K \"%(del_S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The change in the entropy of the reactor contents = 1.4266 kJ/K \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.8  Page No : 172"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "P = 0.101325;\t\t\t #pressure in the piston cylinder assembly in MPa\n",
-      "T1 = 300.;\t\t\t #temperature of the piston cylinder assembly in K\n",
-      "T2 = 400;\t\t\t #final temperature of the piston cylinder assembly in K\n",
-      "a = 45.369;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
-      "b = 8.688*10**-3;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
-      "e = 9.619*10**5;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "del_S = (a*math.log(T2/T1))+(b*(T2-T1))+((e)*((1./T2**2)-(1./T1**2)))\n",
-      "\n",
-      "# Results\n",
-      "print \" The change in entropy of CO2 = %.4f J/molK\"%(del_S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The change in entropy of CO2 = 9.2447 J/molK\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.9  Page No : 173"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "m = 1.      \t\t\t #amount of saturated liquid water in kg\n",
-      "T_initial = 100.\t\t #initial temperature of water in degree celsius\n",
-      "T_body = 500.\t\t\t #temperature of body which is brought into contact with the cylinder in degree celsius\n",
-      "hfg = 2256.94\t\t\t #enthalpy of vaporization taken from steam tables corresponding to T1 in kJ/kg\n",
-      "\n",
-      "# Calculations\n",
-      "T = T_initial+273.15\n",
-      "del_S = hfg/T;\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The change in entropy of water = %0.4f kJ/kgK\"%(del_S);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The change in entropy of water = 6.0483 kJ/kgK\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.10  Page No : 173"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "m_steel = 10.   \t\t\t #mass of steel casting in kg\n",
-      "T_steel = 800.\t    \t\t #temperature of steel casting in degree celsius\n",
-      "m_water = 100.\t\t    \t #mass of water used for quenching in kg\n",
-      "T_water = 30.\t\t\t     #temperature of water used for quenching in degree celsius\n",
-      "Cp_steel = 0.461;\t\t\t #heat capacity of steel in kJ/kgK\n",
-      "Cp_water = 4.23;\t\t\t #heat capacity of water in kJ/kgK\n",
-      "\n",
-      "# Calculations\n",
-      "Ti_steel = T_steel+273.15\n",
-      "Ti_water = T_water+273.15\n",
-      "T_final = ((m_steel*Cp_steel*Ti_steel)+(m_water*Cp_water*Ti_water))/((m_steel*Cp_steel)+(m_water*Cp_water));\n",
-      "del_S_steel = m_steel*Cp_steel*math.log(T_final/Ti_steel)\n",
-      "del_S_water = m_water*Cp_water*math.log(T_final/Ti_water)\n",
-      "\n",
-      "# Results\n",
-      "print \" The change in entropy of steel  =  %0.4f kJ/K\"%(del_S_steel);\n",
-      "print \" The change in entropy of water  =  %f kJ/K\"%(del_S_water);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The change in entropy of steel  =  -5.7031 kJ/K\n",
-        " The change in entropy of water  =  11.427392 kJ/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.11  Page No : 175"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 2.\t    \t\t #volume of insulated tank in m**3\n",
-      "Ta = 400.\t\t\t #temperature of gas in compartment (a) in K\n",
-      "Pa = 3.\t\t    \t #pressure of gas in compartment (a) in MPa\n",
-      "Tb = 600.\t\t\t #temperature of gas in compartment (b) in K\n",
-      "Pb = 1.\t\t\t     #pressure of gas in compartment (b) in MPa\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Va = V/2\n",
-      "Vb = V/2\n",
-      "Na = (Pa*10**6*Va)/(R*Ta)\n",
-      "Nb = (Pb*10**6*Vb)/(R*Tb)\n",
-      "T = ((Na*Ta)+(Nb*Tb))/(Na+Nb)\n",
-      "N = Na+Nb;\t\t\t \n",
-      "P = ((N*R*T)/V)*10**-6\n",
-      "Cp = (5./2)*R;\t\t\t\n",
-      "del_S = ((Na*((Cp*math.log(T/Ta))-(R*math.log(P/Pa))))+(Nb*((Cp*math.log(T/Tb))-(R*math.log(P/Pb)))))*10**-3; # Calculations of the change in entropy using Eq.(5.43) in kJ/K\n",
-      "\n",
-      "# Results\n",
-      "print \" Entropy change of the gas = %0.2f kJ/K\"%(del_S);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Entropy change of the gas = 2.19 kJ/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.12  Page No : 177"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "N = 1.\t    \t\t #amount of air to be separated into its components in kmol\n",
-      "P = 0.1\t\t    \t #pressure of air in MPa\n",
-      "T = 300.\t\t\t #temperature of air in K\n",
-      "per_oxygen = 21.\t\t\t #percentage of oxygen in air\n",
-      "per_nitrogen = 79.\t\t\t #percentage of nitrogen in air\n",
-      "R = 8.314;\t\t\t         #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "x1 = per_nitrogen/100\n",
-      "x2 = per_oxygen/100;\n",
-      "W = (T*N*10**3*R*((x1*math.log (x1))+(x2*math.log (x2))))*10**-3\n",
-      "\n",
-      "# Results\n",
-      "print \" Minimum work to be done to separate 1kmol of air at 0.1MPa and 300K into pure oxygen\\\n",
-      " and nitrogen at the same temperature and pressure = %0.2f kJ\"%(abs(W));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Minimum work to be done to separate 1kmol of air at 0.1MPa and 300K into pure oxygen and nitrogen at the same temperature and pressure = 1281.91 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.13  Page No : 179"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "m_ice = 10.\t    \t\t #mass of the block of ice in kg\n",
-      "T_ice = 0.\t\t    \t #temperature of the ice in degree celsius\n",
-      "m_water = 100.\t\t\t #mass of watre in the tank in kg\n",
-      "T_water = 30.\t\t\t #temperature of the water in the tank in degree celsius\n",
-      "Cp = 4.23;\t\t\t     #heat capacity of water in kJ/kgK\n",
-      "lambda_melting = 333.44\t\t\t #latent heat of melting of ice in kJ/kg\n",
-      "\n",
-      "# Calculations\n",
-      "Ti_ice = T_ice+273.15\n",
-      "Ti_water = T_water+273.15\n",
-      "T_final = ((m_water*Cp*Ti_water)+(m_ice*Cp*Ti_ice)-(m_ice*lambda_melting))/((m_ice*Cp)+(m_water*Cp))\n",
-      "del_S_ice = ((m_ice*lambda_melting)/(Ti_ice))+(m_ice*Cp*math.log (T_final/Ti_ice));\t\t\t \n",
-      "del_S_water = m_water*Cp*math.log (T_final/Ti_water)\n",
-      "del_S_G = del_S_ice+del_S_water;\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The change in entropy of ice  =  %f kJ/K\"%(del_S_ice);\n",
-      "print \" The change in entropy of water  =  %f kJ/K\"%(del_S_water);\n",
-      "print \" The entropy generated =  %f kJ/K\"%(del_S_G);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The change in entropy of ice  =  15.211650 kJ/K\n",
-        " The change in entropy of water  =  -14.035033 kJ/K\n",
-        " The entropy generated =  1.176617 kJ/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.14  Page No : 182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P = 3.\t\t\t         #pressure of superheated steam in MPa\n",
-      "T_enter = 300.\t\t\t #entrance temperature of superheated steam in degree celsius\n",
-      "T_exit = 45.\t\t\t #final temperature at which the steam leaves in degree celsisus\n",
-      "m = 1.      \t\t\t #mass flow rate of steam in kg/s\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "si = 6.5422\n",
-      "hi = 2995.1\n",
-      "sf = 0.6383\n",
-      "hf = 188.35\n",
-      "sg = 8.1661;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
-      "hg = 2583.3;\t\t\t #entahlpy of saturayed vapour in kJ/kg\n",
-      "\n",
-      "Xe = (si-sf)/(sg-sf)\n",
-      "he = ((1-Xe)*hf)+(Xe*hg)\n",
-      "Ws = -m*(he-hi);\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The power  Results from the turbine = %0.1f kW\"%(Ws);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The power  Results from the turbine = 928.4 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.15  Page No : 183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "Pi = 3.\t\t\t #pressure of dry saturated steam when it enters the nozzle in bar\n",
-      "Pe = 2.\t\t\t #pressure of dry saturated steam at the exit in bar\n",
-      "\n",
-      "# Calculations\n",
-      "#From steam tables corresponding to Pi\n",
-      "si = 6.9909;\t\t\t #entropy of steam at the entrance in kJ/kgK\n",
-      "hi = 2724.7;\t\t\t #entahlpy of steam at the entrance in kJ/kg\n",
-      "\n",
-      "#From steam tables corresponding to Pe\n",
-      "sf = 1.5301;\t\t\t #entropy of saturated liquid in kJ/kgK\n",
-      "hf = 504.70;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
-      "sg = 7.1268;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
-      "hg = 2706.3;\t\t\t #entahlpy of saturayed vapour in kJ/kg\n",
-      "\n",
-      "se = 6.9909\n",
-      "Xe = (se-sf)/(sg-sf)\n",
-      "he = ((1-Xe)*hf)+(Xe*hg)\n",
-      "Ve = math.sqrt (2*(hi-he)*10**3)\n",
-      "\n",
-      "# Results\n",
-      "print \" The exit velocity of steam = %f m/s\"%(Ve);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The exit velocity of steam = 379.103198 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.16  Page No : 183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "N_glycerol = 100.\t\t\t #molar flow rate of glycerol in mol/s\n",
-      "Ti_gly = 227.\t\t\t #inlet temperature of glycerol in degree celsius\n",
-      "Te_gly = 40.\t\t\t #outlet temperature of glycerol in degree celsius\n",
-      "Ti_water = 25.\t\t\t #inlet temperature of cooling water in degree celsius\n",
-      "Te_water = 50.\t\t\t #outlet temperature of cooling water in degree celsius\n",
-      "Cp_gly = 280.\t\t\t #heat capacity of glycerol in J/molK\n",
-      "Cp_water = 77.\t\t\t #heat capacity of water in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Ti_gly = Ti_gly+273.15\n",
-      "Te_gly = Te_gly+273.15\n",
-      "Ti_water = Ti_water+273.15\n",
-      "Te_water = Te_water+273.15\n",
-      "\n",
-      "N_water = -(N_glycerol*Cp_gly*(Te_gly-Ti_gly))/(Cp_water*(Te_water-Ti_water));\n",
-      "del_S_gly = N_glycerol*Cp_gly*math.log (Te_gly/Ti_gly)*10**-3\n",
-      "del_S_water = N_water*Cp_water*math.log (Te_water/Ti_water)*10**-3\n",
-      "S_G = del_S_gly+del_S_water\n",
-      "\n",
-      "# Results\n",
-      "print \" The rate at which entropy is generated in the heat exchanger = %0.3f kJ/K s\"%(S_G);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The rate at which entropy is generated in the heat exchanger = 3.754 kJ/K s\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.17  Page No : 183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T_i = 150.\t\t\t #temperature of saturated steam taken up by the device in degree celsius\n",
-      "T_e = 200.\t\t\t #temperature of superheated steam delivered by the device in degree celsius\n",
-      "P_e = 0.2\t\t\t #pressure of superheated steam delivered by the device in MPa\n",
-      "me2 = 0.949\t\t\t #mass of superheated steam leaving the device in kg\n",
-      "me1 = 0.051\t\t\t #mass of saturated liquid leaving the device in kg\n",
-      "T_liq = 100.\t\t #temperature of saturated liquid leaving the device in degree celsius\n",
-      "mi = 1.     \t\t #mass of saturated steam fed to the device in kg\n",
-      "\n",
-      "# Calculations\n",
-      "#From steam tables corresponding to T_i\n",
-      "hi = 2745.4 \t\t\t #enthalpy of saturated vapour in kJ/kg\n",
-      "si = 6.8358;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
-      "\n",
-      "#For saturated liquid at T_liq\n",
-      "he1 = 419.06;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
-      "se1 = 1.3069;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
-      "\n",
-      "#For superheated steam at P_e and T_e\n",
-      "he2 = 2870.5;\t\t\t #enthalpy of superheated steam in kJ/kg\n",
-      "se2 = 7.5072;\t\t\t #entropy of superheated steam in kJ/kgK\n",
-      "\n",
-      "LHS = mi*hi;\n",
-      "RHS = (me1*he1)+(me2*he2);\n",
-      "\n",
-      "S_G = (me1*se1)+(me2*se2)-(mi*si);\n",
-      "\n",
-      "# Results\n",
-      "print \" The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied =  %0.1f kJ\"%(LHS);\n",
-      "print \" The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = %0.1f kJ\"%(RHS);\n",
-      "print \" The entropy generated by applying the second law of thermodynamics to the flow device = %0.4f kJ/kgK\"%(S_G);\n",
-      "if int(LHS) ==  int(RHS) and S_G>0 or S_G == 0 :\n",
-      "    print \" As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \"\n",
-      "else:\n",
-      "    print \" As both the first and second law or either the first or second law of thermodynamics \\\n",
-      "     are not satisfied, the device is not feasible \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied =  2745.4 kJ\n",
-        " The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = 2745.5 kJ\n",
-        " The entropy generated by applying the second law of thermodynamics to the flow device = 0.3552 kJ/kgK\n",
-        " As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.18  Page No : 185"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Pi = 30.\t\t\t #pressure of superheated steam entering the turbine in bar\n",
-      "Ti = 300.\t\t\t #temperature of superheated steam entering the turbine in degree celsius\n",
-      "Pe = 0.1\t\t\t #pressure at which steam exits the turbine in bar\n",
-      "Xe = 0.9\t\t\t #quality of steam at the exit (no unit)(for the actual turbine)\n",
-      "\n",
-      "# Calculations\n",
-      "#For superheated steam at Pi and Ti\n",
-      "hi = 2995.1;\t\t\t #enthalpy of superheated steam at the entrance in kJ/kg\n",
-      "si = 6.5422;\t\t\t #entropy of superheated steam at the entrance in kJ/kgK\n",
-      "\n",
-      "#For steam at Pe\n",
-      "hf = 191.83;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
-      "hg = 2584.8;\t\t\t #enthalpy of saturated vapour in kJ/kg\n",
-      "sf = 0.6493;\t\t\t #entropy of saturated liquid in kJ/kgK\n",
-      "sg = 8.1511;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
-      "\n",
-      "X2 = (si-sf)/(sg-sf)\n",
-      "h2 = (hf*(1-X2))+(X2*hg)\n",
-      "he = (hf*(1-Xe))+(Xe*hg)\n",
-      "n_T = (hi-he)/(hi-h2)\n",
-      "\n",
-      "# Results\n",
-      "print \" The isentropic efficiency of the turbine = %f \"%(n_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The isentropic efficiency of the turbine = 0.703395 \n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.19  Page No : 186"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Ti = 25.\t\t\t #temperature of air taken in by the adiabatic air compressor in degree celsius\n",
-      "Pi = 0.1\t\t\t #pressure of air taken in by the adiabatic air compressor in MPa\n",
-      "Pe = 1.\t    \t\t #discharge pressure of air in MPa\n",
-      "n_c = 0.8\t\t\t #isentropic efficiency of the compressor (no unit)\n",
-      "gaamma = 1.4\t\t #ratio of molar specific heat capacities (no unit)\n",
-      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Ti = Ti+273.15\n",
-      "Te = Ti*(((Pe*10**6)/(Pi*10**6))**((gaamma-1)/gaamma))\n",
-      "W_s = (((R*gaamma)/(gaamma-1))*(Te-Ti))*10**-3;\t\t\t\n",
-      "Ws = W_s/n_c\n",
-      "Te_actual = ((Ws*10**3*(gaamma-1))/(R*gaamma))+Ti\n",
-      "\n",
-      "# Results\n",
-      "print \" The exit temperature of air = %0.2f K\"%(Te_actual);\n",
-      "print \" The power consumed by the compressor  = %f kW/mol\"%(Ws);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The exit temperature of air = 645.01 K\n",
-        " The power consumed by the compressor  = 10.093262 kW/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.20  Page No : 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Ti = 30.\t\t\t #temperature of saturated liquid water in degree celsius\n",
-      "m = 500.\t\t\t #mass flow rate of water being pumped in kg/s\n",
-      "P2 = 3. \t\t\t #preesure maintained in the boiler in MPa\n",
-      "n_p = 0.75;\t\t\t #isentropic efficiency of the pump (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "#For saturated liquid water at Ti\n",
-      "vf = 0.0010043\n",
-      "P1 = 4.241;\t\t\n",
-      "\n",
-      "Ws_m = (vf*((P2*10**6)-(P1*10**3)))*10**-3\n",
-      "Ws_act_m = Ws_m/n_p;\t\t\t \n",
-      "P = ((Ws_act_m*10**3)*m)*10**-6;\n",
-      "\n",
-      "# Results\n",
-      "print \" The power consumed by the pump  =  %d MW\"%(P);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The power consumed by the pump  =  2 MW\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.21  Page No : 188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Pi = 3.\t\t\t #pressure of dry saturated steam entering the nozzle in bar\n",
-      "Xe = 0.98\t\t #quality of steam exiting the nozzle (no unit)\n",
-      "Pe = 2.\t\t\t #pressure of steam exiting the nozzle in bar\n",
-      "\n",
-      "# Calculations\n",
-      "#For steam at Pi\n",
-      "hi = 2724.7\n",
-      "he = 2652.8\n",
-      "V2_2_s = hi-he\n",
-      "\n",
-      "#For steam at Pe\n",
-      "hf = 504.70\n",
-      "hg = 2706.3\n",
-      "he_act = ((1-Xe)*hf)+(Xe*hg)\n",
-      "V2_2 = hi-he_act;\t\t\t\n",
-      "n_N = (V2_2)/(V2_2_s)\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The isentropic efficiency of the nozzle = %0.3f \"%(n_N);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The isentropic efficiency of the nozzle = 0.868 \n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch6.ipynb b/Chemical_Engineering_Thermodynamics/ch6.ipynb
deleted file mode 100755
index cf3f1f12..00000000
--- a/Chemical_Engineering_Thermodynamics/ch6.ipynb
+++ /dev/null
@@ -1,214 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6 :  Thermodynamic potentials"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.6  Page No : 218"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "m = 0.1;\t\t\t #mass of superheated steam in the piston cylinder assembly in kg\n",
-      "P1 = 1.\t    \t\t #initial pressure of superheated steam in MPa\n",
-      "T1 = 300.\t\t\t #initial temperature of superheated steam in degree celsius\n",
-      "P2 = 0.1\t\t\t #pressure of steam after expansion in MPa\n",
-      "T2 = 200.\t\t\t #temperature of steam after expansion in degree celsius\n",
-      "\n",
-      "# Calculations\n",
-      "#For steam at P1 and T1:\n",
-      "h1 = 3052.1\n",
-      "v1 = 0.2580\n",
-      "\n",
-      "#For steam at P2 and T2:\n",
-      "h2 = 2875.4\n",
-      "v2 = 2.1720\n",
-      "\n",
-      "del_u = (((h1*10**3)-(P1*10**6*v1))-((h2*10**3)-(P2*10**6*v2)))*10**-3\n",
-      "W = m*(del_u)\n",
-      "\n",
-      "# Results\n",
-      "print \" The work done by steam = %0.2f kJ\"%(W);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The work done by steam = 13.59 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.8  Page No : 220"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P = 3.\t    \t\t #pressure of superheated steam in MPa\n",
-      "Ti = 300.\t\t\t #temperature at which the steam enters the turbine in degree celsius\n",
-      "m = 1.\t\t    \t #mass flow rate of steam in kg/s\n",
-      "Te = 60.\t\t\t #temperature of dry saturated steam when it leaves the turbine in degree celsius\n",
-      "\n",
-      "# Calculations\n",
-      "#For steam at P and Ti:\n",
-      "h1 = 2995.1\t\t\t #specific entahlpy of steam in kJ/kg\n",
-      "\n",
-      "#For saturated steam at Te:\n",
-      "h2 = 2609.7\t\t\t #specific enthalpy of saturated vapour in kJ/kg\n",
-      "\n",
-      "Ws = m*(h1-h2)\t\t # Calculations of the power  Results of the turbine using Eq.(6.83) in kW\n",
-      "\n",
-      "# Results\n",
-      "print \" The power  Results of the turbine = %0.1f kW\"%(Ws);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The power  Results of the turbine = 385.4 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.10  Page No : 223"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "m = 0.1 \t\t\t #mass of superheated steam in the piston cylinder assembly in kg\n",
-      "P1 = 3.\t    \t\t #initial pressure of superheated steam in MPa\n",
-      "T1 = 300.\t\t\t #initial temperature of superheated steam in degree celsius\n",
-      "T0 = 300.\t\t\t #temperature of the reservoir which is placed in thermal contact with the piston-cylinder assembly in degree celsius\n",
-      "P2 = 0.1\t\t\t #pressure of steam after expansion in MPa\n",
-      "\n",
-      "#For steam at P1 and T1:\n",
-      "h1 = 2995.1\n",
-      "v1 = 0.08116\n",
-      "s1 = 6.5422\n",
-      "\n",
-      "#For steam at P2 and T2:\n",
-      "h2 = 3074.5\n",
-      "v2 = 2.6390\n",
-      "s2 = 8.2166\n",
-      "\n",
-      "# Calculations\n",
-      "T0 = T0+273.15\n",
-      "\n",
-      "W = m*(h1-h2-(((P1*v1)-(P2*v2))*10**3)-(T0*(s1-s2)))\n",
-      "\n",
-      "# Results\n",
-      "print \" The maximum work obtained from steam = %0.2f kJ\"%(W);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The maximum work obtained from steam = 90.07 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.12  Page No : 226"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "P1 = 0.1\t\t\t #pressure at which air enters the compressor in MPa\n",
-      "T1 = 300.\t\t\t #temperature at which air enters the compressor in K\n",
-      "P2 = 1. \t\t\t #pressure at which air leaves the compressor in MPa\n",
-      "T2 = 300.\t\t\t #temperature at which air leaves the compressor in K\n",
-      "T0 = 300.\t\t\t #ambient temperature in K\n",
-      "N = 1.  \t\t\t #molar flow rate of air in mol/s\n",
-      "gaamma = 1.4\t\t #ratio of specific heat capacities (no unit)\n",
-      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "Ws = (-N*T0*(-R*math.log (P2/P1)))*10**-3\n",
-      "\n",
-      "# Results\n",
-      "print \" The minimum power required to compress one mole per second of air = %0.3f kW\"%(Ws);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The minimum power required to compress one mole per second of air = 5.743 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch7.ipynb b/Chemical_Engineering_Thermodynamics/ch7.ipynb
deleted file mode 100755
index 879f7b43..00000000
--- a/Chemical_Engineering_Thermodynamics/ch7.ipynb
+++ /dev/null
@@ -1,255 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7 : Thermodynamic property relations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.10  Page No : 259"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "T2 = 150.   \t\t\t #temperature at which water it is desired to boil water in degree celsius\n",
-      "P1 = 0.10133\t\t\t #ambient pressure in MPa\n",
-      "T1 = 100.\t    \t\t #temperature at which water boils corresponding at pressure P1 in degree celsius\n",
-      "del_hv = 2256.94\t\t #enthalpy of vaporization in kJ/kg\n",
-      "R = 8.314;\t\t    \t #universal gas constant in J/molK\n",
-      "M = 18*10**-3;\t\t\t #molar mass of water in kg/mol\n",
-      "\n",
-      "# Calculations\n",
-      "T1 = T1+273.15\n",
-      "T2 = T2+273.15\n",
-      "P2 = P1*(math.exp(((del_hv*10**3*M)*((1./T1)-(1./T2)))/(R)))\n",
-      "\n",
-      "# Results\n",
-      "print \" The approximate pressure at which the boiler is to be operated = %0.3f MPa\"%(P2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The approximate pressure at which the boiler is to be operated = 0.476 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.11  Page No : 259"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "m = 60.         \t\t\t #mass of the person who wants to skate in kg\n",
-      "T = -2.\t\t\t             #temperature of the ice in degree celsius\n",
-      "A = 15.\t\t\t             #area of contact between the skate edges and ice in mm**2\n",
-      "vs = 1.091*10**-3;\t\t\t #specific volume of ice in m**3/kg (at Tref)\n",
-      "vf = 1.0*10**-3;\t\t\t #specific volume of water in m**3/kg (at Tref)\n",
-      "del_hf = 6.002;\t\t    \t #enthalpy of melting of ice in kJ/mol\n",
-      "g = 9.81;\t\t\t         #accleration due to gravity in m/s**2\n",
-      "Tref = 0.       \t\t\t #reference temperature at which the specific enthalpy of ice and water are taken in degree celsius\n",
-      "\n",
-      "# Calculations\n",
-      "Tref = Tref+273.15\n",
-      "del_P = ((m*g)/(A*10**-6))*10**-6\n",
-      "del_v = (vf-vs)*(18*10**-3);\t\n",
-      "del_T = (del_P*10**6)/((del_hf*10**3)/(Tref*del_v));\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The temperature of ice originally  =  %d degree celsius \"%(T);\n",
-      "print \" The reduction in melting point of ice due to the additional pressure,computed using the Clayperon equation  =  %0.2f degree celsius \"%(del_T);\n",
-      "if del_T<T :\n",
-      "    print  \" The ice can melt due to the additional pressure and therefore it will be possible to skate \"\n",
-      "else:\n",
-      "    print  \" The ice will not melt and therefore it will be difficult to skate \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The temperature of ice originally  =  -2 degree celsius \n",
-        " The reduction in melting point of ice due to the additional pressure,computed using the Clayperon equation  =  -2.93 degree celsius \n",
-        " The ice can melt due to the additional pressure and therefore it will be possible to skate \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.12  Page No : 261"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T1 = 100.\t    \t\t #temperature of water in degree celsius\n",
-      "del_hv1 = 2256.94;\t\t\t #enthalpy of vaporization at T1 in kJ/kg\n",
-      "T2 = 150.   \t\t\t #temperature at which the enthalpy of vaporization is to be determined in degree celsius\n",
-      "Cp_f = 4.26\t    \t\t #isobaric heat capacity of liquid in kJ/kgK\n",
-      "Cp_g = 1.388\t\t\t #isobaric heat capacity of vapour in kJ/kgK\n",
-      "\n",
-      "# Calculations\n",
-      "del_hv2 = ((Cp_g-Cp_f)*(T2-T1))+del_hv1\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy of vaporization at 150 degree celsius = %0.2f kJ/kg\"%(del_hv2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy of vaporization at 150 degree celsius = 2113.34 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.13  Page No : 261"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T1 = 100.           \t\t\t #temperature of water in degree celsius\n",
-      "del_hv1 = 2256.94   \t\t\t #enthalpy of vaporization at T1 in kJ/kg\n",
-      "T2 = 150.   \t\t        \t #temperature at which the enthalpy of vaporization is to be determined in degree celsius\n",
-      "del_hv_kirchoff = 2113.34\t\t\t #enthalpy of vaporization predicted by the Kirchhoff relation taken from Example 7.12 for comparison, in kJ/kg\n",
-      "del_hv_steam_tables = 2113.25\t\t\t #enthalpy of vaporization taken from the steam tables corresponding to T2,for comparison, in kJ/kg\n",
-      "Tc = 647.3          \t\t\t #critical temperature of water in K\n",
-      "\n",
-      "# Calculations\n",
-      "T1 = T1+273.15\n",
-      "T2 = T2+273.15\n",
-      "Tr1 = T1/Tc\n",
-      "Tr2 = T2/Tc\n",
-      "del_hv2 = del_hv1*(((1-Tr2)/(1-Tr1))**0.38)\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy of vaporization at 150 degree celsius using \";\n",
-      "print \" Watson correlation \\t  =  %f kJ/kg\"%(del_hv2);\n",
-      "print \" Kirchhoffs relation \\t  =  %f kJ/kg\"%(del_hv_kirchoff);\n",
-      "print \" From steam tables \\t  =  %f kJ/kg\"%(del_hv_steam_tables);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy of vaporization at 150 degree celsius using \n",
-        " Watson correlation \t  =  2090.687877 kJ/kg\n",
-        " Kirchhoffs relation \t  =  2113.340000 kJ/kg\n",
-        " From steam tables \t  =  2113.250000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.14  Page No : 262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 373.15  \t\t\t #normal boiling point of water in K (temperature at which the enthalpy of vaporization is to be determined)\n",
-      "Pc = 221.2\t    \t\t #critical pressure of water in bar\n",
-      "Tc = 647.3\t\t    \t #critical temperature of water in K\n",
-      "R = 8.314\t\t\t     #universal gas constant in J/molK\n",
-      "del_hvn_steam_tables = 2256.94\t\t\t #enthalpy of vaporization at the normal boiling point taken from the steam tables, for comparison, in kJ/kg\n",
-      "\n",
-      "# Calculations\n",
-      "Tbr = T/Tc\t\t\t\n",
-      "del_hvn = ((1.093*R*Tc*(Tbr*((math.log(Pc)-1.013)/(0.930-Tbr))))*10**-3)/(18*10**-3);\n",
-      "err = abs((del_hvn-del_hvn_steam_tables)/del_hvn_steam_tables)*100\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy of vaporization at the normal boiling point \";\n",
-      "print \" Using Riedels correlation \\t  =  %f kJ/kg\"%(del_hvn);\n",
-      "print \" From the steam tables \\t \\t  =  %f kJ/kg\"%(del_hvn_steam_tables);\n",
-      "print \" Error \\t \\t \\t \\t  =  %f %% \"%(err);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy of vaporization at the normal boiling point \n",
-        " Using Riedels correlation \t  =  2337.179008 kJ/kg\n",
-        " From the steam tables \t \t  =  2256.940000 kJ/kg\n",
-        " Error \t \t \t \t  =  3.555212 % \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch8.ipynb b/Chemical_Engineering_Thermodynamics/ch8.ipynb
deleted file mode 100755
index 1e8782d9..00000000
--- a/Chemical_Engineering_Thermodynamics/ch8.ipynb
+++ /dev/null
@@ -1,600 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8 : Thermodynamic properties of real gases"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.2  Page No : 275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
-      "a = 3.789;\t\t\t #van der Waals constant in Pa (m**3/mol)**2\n",
-      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
-      "v = 15.675*10**-3;\t\t\t #volume occupied by n-octane vapour taken from Example (3.8) in m**3/mol\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "dep_h = (P*10**6*v)-(R*T)-(a/v)\n",
-      "dep_s = R*math.log ((P*10**6*(v-b))/(R*T));\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure for n-octane vapour  =  %0.2f J/mol\"%(dep_h);\n",
-      "print \" The entropy departure for n-octane vapour  =  %0.4f J/mol K\"%(dep_s);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure for n-octane vapour  =  -428.74 J/mol\n",
-        " The entropy departure for n-octane vapour  =  -0.5757 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.3  Page No : 276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "T = 100.    \t\t\t #temperature of carbon dioxide in degree celsius\n",
-      "P = 10. \t    \t\t #pressure of carbon dioxide in MPa\n",
-      "A0 = 0.5073;\t\t\t #Beattie-Bridgman constant for carbon dioxide in (Pa m**3)/mol**2\n",
-      "B0 = 104.76*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
-      "a = 71.32*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
-      "b = 72.35*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
-      "C = 660.0;\t\t    \t #Beattie-Bridgman constant for carbon dioxide in (m**3 K**3)/mol\n",
-      "R = 8.314;\t\t\t     #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "T = T+273.15\n",
-      "A1 = (R*T)  \n",
-      "A2 = (B0*R*T)-A0-((C*R)/T**2);\t\t\t \n",
-      "A3 = (a*A0)-(b*B0*R*T)-((B0*C*R)/T**2);\t\n",
-      "A4 = ((b*C*B0*R)/T**2);\t\t\t \n",
-      "vguess = 0.01\n",
-      "tolerance = 1e-6\n",
-      "\n",
-      "def solver_func(vi):\n",
-      "    return  (P*10**6)-((A1/vi)+(A2/vi**2)+(A3/vi**3)+(A4/vi**4))\n",
-      "\n",
-      "v = fsolve(solver_func,vguess)\n",
-      "\n",
-      "Z = (P*10**6*v)/(R*T)\n",
-      "\n",
-      "dep_h = (((B0*R*T)-(2*A0)-((4*C*R)/(T**2)))*(1./v))+((((3./2)*a*A0)-(b*B0*R*T)-((5*B0*C*R)/(2*(T**2))))*(1./(v**2)))+((2*b*C*B0*R)/((T**2)*(v**3)));\n",
-      "\n",
-      "# Results\n",
-      "print \" Molar volume of CO2 at %0.f MPa and %0.2f K  =  %.2e m**3/mol \"%(P,T,v);\n",
-      "print \" The compressibility factor = %.4f \"%(Z);\n",
-      "print \" The enthalpy departure for carbon dioxide using the Beattie-Bridgman equation of state  =  %.f J/mol\"%(dep_h);\n",
-      "\n",
-      "# Note: Answer is different because of rounding off error. Please check manually."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Molar volume of CO2 at 10 MPa and 373.15 K  =  2.33e-04 m**3/mol \n",
-        " The compressibility factor = 0.7518 \n",
-        " The enthalpy departure for carbon dioxide using the Beattie-Bridgman equation of state  =  -3210 J/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.4  Page No : 278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "# Variables\n",
-      "T = 353.15    \t    \t\t #temperature of carbon dioxide in degree celsius\n",
-      "P = 10. \t    \t    \t #pressure of carbon dioxide in MPa\n",
-      "B0 = 104.76*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
-      "b = 72.35*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
-      "C = 660.0;\t\t\t         #Beattie-Bridgman constant for carbon dioxide in (m**3 K**3)/mol\n",
-      "R = 8.314       \t\t\t #universal gas constant in J/molK\n",
-      "v = 0.233*10**-3    \t\t #volume calculated in Example (8.3) in m**3/mol\n",
-      "Z = 0.751;\t\t\t         #compressibility factor as calculated in Example (8.3) (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "A1 = ((B0*R)+((2*C*R)/(T**3)))\n",
-      "dep_s = (R*math.log (Z))-(A1*(1./v))+(((b*B0*R)-((2*C*B0*R)/(T**3)))*(1./(2*(v**2))))+((2*b*C*B0*R)/(3*(T**3)*(v**3)));\n",
-      "\n",
-      "# Results\n",
-      "print \" The entropy departure for carbon dioxide using the Beattie-\\\n",
-      "Bridgman equation of state  =  %.f J/mol K\"%(dep_s);\n",
-      "\n",
-      "# Note: Answer is varies because of rounding off error. Please check manually."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The entropy departure for carbon dioxide using the Beattie-Bridgman equation of state  =  -7 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.5  Page No : 281"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
-      "a = 4.426;\t\t\t #Redlich-Kwong constant taken from Example(3.9) in (m**6 Pa mol**-2)\n",
-      "b = 164.3*10**-6;\t\t\t #Redlich-Kwong constant taken from Example(3.9) in m**3/mol\n",
-      "Z = 0.9308;\t\t\t #compressibility factor taken from Example(3.9) (no unit)\n",
-      "B = 9.9306*10**-3;\t\t\t #value of B, used in the Cardan's method in Example (3.9)\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "dep_h = (R*T*(Z-1))-(((3*a)/(2*b))*math.log ((Z+B)/Z))\n",
-      "dep_s = (R*math.log(Z-B))-((a/(2*b*T))*math.log((Z+B)/Z))\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure for n-octane vapour using the generalized Redlich\\\n",
-      "-Kwong equation of state  =  %0.2f J/mol\"%(dep_h);\n",
-      "print \" The entropy departure for n-octane vapour using the generalized Redlich\\\n",
-      "-Kwong equation of state  =  %0.4f J/mol K\"%(dep_s);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure for n-octane vapour using the generalized Redlich-Kwong equation of state  =  -674.98 J/mol\n",
-        " The entropy departure for n-octane vapour using the generalized Redlich-Kwong equation of state  =  -1.0195 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.6  Page No : 281"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 427.85\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215\t\t\t #pressure of n-octane vapour in MPa\n",
-      "S = 1.0786\t\t\t #constant used in the SRK equation of state,from Example(3.15)\n",
-      "alpha = 1.3079\t\t #constant used in the SRK equation of state,from Example(3.15)\n",
-      "a = 5.0180\t\t\t #constant used in the SRK equation of state,from Example(3.15) in (m**6 Pa mol**-2)\n",
-      "b = 1.6426*10**-4\t\t\t #constant used in the SRK equation of state,from Example(3.15) in m**3/mol\n",
-      "B = 9.9282*10**-3\t\t\t #factor used in the Cardan's method for solving the SRK equation of state,from Example(3.15) (no unit)\n",
-      "Z = 0.9191;\t\t\t #compressibility factor taken from Example (3.15) (no unit)\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "\n",
-      "# Calculations\n",
-      "da_dT = (-a*S)/(math.sqrt (alpha*T*Tc))\n",
-      "dep_h = (R*T*(Z-1))+((((T*da_dT)-a)/b)*math.log ((Z+B)/Z));\t\t\t \n",
-      "dep_s = (R*math.log (Z-B))+((1./b)*(da_dT)*math.log ((Z+B)/Z));\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure for n-octane vapour using the SRK equation of state  =  %f J/mol\"%(dep_h);\n",
-      "print \" The entropy departure for n-octane vapour using the SRK equation of state  =  %0.4f J/mol K\"%(dep_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure for n-octane vapour using the SRK equation of state  =  -884.335509 J/mol\n",
-        " The entropy departure for n-octane vapour using the SRK equation of state  =  -1.4188 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.7  Page No : 282"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 427.85\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215\t\t\t #pressure of n-octane vapour in MPa\n",
-      "S = 0.9457\t\t\t #constant used in the Peng-Robinson equation of state,from Example(3.16)\n",
-      "alpha = 1.2677\t\t #constant used in the Peng-Robinson equation of state,from Example(3.16)\n",
-      "a = 5.2024\t\t\t #constant used in the Peng-Robinson equation of state,from Example(3.16) in (m**6 Pa mol**-2)\n",
-      "b = 1.4750*10**-4\t #constant used in the Peng-Robinson equation of state,from Example(3.16) in m**3/mol\n",
-      "B = 8.9151*10**-3\t #factor used in the Cardan's method for solving the Peng-Robinson equation of state,from Example(3.16) (no unit)\n",
-      "Z = 0.9151\t\t\t #compressibility factor taken from Example (3.16) (no unit)\n",
-      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
-      "Tc = 569.4\t\t\t #critical temperature of n-octane in K\n",
-      "\n",
-      "# Calculations\n",
-      "da_dT = (-a*S)/(math.sqrt (alpha*T*Tc))\n",
-      "\n",
-      "dep_h = (R*T*(Z-1))+(((((T*da_dT)-a)/(2*math.sqrt(2)*b)))*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
-      "dep_s = (R*math.log (Z-B))+((1./(2*math.sqrt (2)*b))*(da_dT)*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure for n-octane vapour using the Peng-Robinson \\\n",
-      " equation of state  =  %0.1f J/mol\"%(dep_h);\n",
-      "print \" The entropy departure for n-octane vapour using the Peng-Robinson\\\n",
-      " equation of state  =  %0.3f J/mol K\"%(dep_s);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure for n-octane vapour using the Peng-Robinson  equation of state  =  -890.1 J/mol\n",
-        " The entropy departure for n-octane vapour using the Peng-Robinson equation of state  =  -1.398 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.8  Page No : 284"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 339.7\t\t\t #temperature of ethylene in K\n",
-      "P = 30.7\t\t\t #pressure of ethylene in bar\n",
-      "Tc = 283.1\t\t\t #critical temperature of ethylene in K\n",
-      "Pc = 51.17\t\t\t #critical pressure of ethylene in bar\n",
-      "w = 0.089\t\t\t #acentric factor (no unit)\n",
-      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Pr = P/Pc\n",
-      "Tr = T/Tc\n",
-      "del_h0 = 0.45\n",
-      "del_h1 = 0.18\n",
-      "del_s0 = 0.26\n",
-      "del_s1 = 0.20\n",
-      "dep_h = ((del_h0)+(w*del_h1))*R*Tc\n",
-      "dep_s = ((del_s0)+(w*del_s1))*R;\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure for ethylene using the Edmister charts  =  %0.3f J/mol\"%(dep_h);\n",
-      "print \" The entropy departure for ethylene using the Edmister charts  =  %0.4f J/mol K\"%(dep_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure for ethylene using the Edmister charts  =  1096.868 J/mol\n",
-        " The entropy departure for ethylene using the Edmister charts  =  2.3096 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.9  Page No : 297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 339.7;\t\t\t #temperature of ethylene in K\n",
-      "P = 30.7;\t\t\t #pressure of ethylene in bar\n",
-      "Tc = 283.1;\t\t\t #critical temperature of ethylene in K\n",
-      "Pc = 51.17;\t\t\t #critical pressure of ethylene in bar\n",
-      "w = 0.089;\t\t\t #acentric factor (no unit)\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Pr = P/Pc\n",
-      "Tr = T/Tc\n",
-      "del_h0 = 0.474\n",
-      "del_h1 = 0.232\n",
-      "del_s0 = 0.277\n",
-      "del_s1 = 0.220\n",
-      "dep_h = ((del_h0)+(w*del_h1))*R*Tc\n",
-      "dep_s = ((del_s0)+(w*del_s1))*R\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure for ethylene using the Lee-Kesler data  =  %f J/mol\"%(dep_h);\n",
-      "print \" The entropy departure for ethylene using the Lee-Kesler data  =  %f J/mol K\"%(dep_s);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure for ethylene using the Lee-Kesler data  =  1164.249733 J/mol\n",
-        " The entropy departure for ethylene using the Lee-Kesler data  =  2.465766 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.10  Page No : 299"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 339.7;\t\t\t #temperature of ethylene in K\n",
-      "P = 1.  \t\t\t #pressure of ethylene in bar\n",
-      "Tc = 283.1;\t\t\t #critical temperature of ethylene in K\n",
-      "Pc = 51.17;\t\t\t #critical pressure of ethylene in bar\n",
-      "w = 0.089;\t\t\t #acentric factor (no unit)\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Pr = P/Pc\n",
-      "Tr = T/Tc\n",
-      "dep_h = R*Tc*Pr*((0.083-(1.097/(Tr**1.6)))+(w*(0.139-(0.894/(Tr**4.2)))))\n",
-      "dep_s = -Pr*R*((0.675/(Tr**2.6))+(w*(0.722/(Tr**5.2))));\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure for ethylene using the generalized virial coefficient \\\n",
-      "correlation  =  %f J/mol\"%(dep_h);\n",
-      "print \" The entropy departure for ethylene using the generalized virial coefficient \\\n",
-      "correlation  =  %e J/mol K\"%(dep_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure for ethylene using the generalized virial coefficient correlation  =  -35.011078 J/mol\n",
-        " The entropy departure for ethylene using the generalized virial coefficient correlation  =  -7.232682e-02 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.11  Page No : 299"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "import cmath\n",
-      "\n",
-      "# Variables\n",
-      "T = 427.85\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215\t\t\t #pressure of n-octane vapour in MPa\n",
-      "T_ref = 0.\t\t\t #reference state saturated liquid temperature in degree celsius\n",
-      "h0 = 0. \t\t\t #enthalpy of saturated liquid in J/mol (reference state)\n",
-      "s0 = 0. \t\t\t #entropy of saturated liquid in J/molK (reference state)\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
-      "w = 0.398;\t\t\t #acentric factor (no unit)\n",
-      "NBP = 398.8;\t\t\t #normal boiling point of n-octane (saturated liquid)\n",
-      "Cp = [6.907,741.770*10**-3,-397.204*10**-6,82.629*10**-9,0]\t\t\t #coefficients to compute the isobaric molar heat capacity, where Cp = a+bT+cT**2+dT**3+eT**-2,in J/molK\n",
-      "S = 0.9457;\t\t\t #value of S taken from Example (3.16)\n",
-      "b = 1.4750*10**-4;\t\t\t #value of the Peng-Robinson constant in m**3/mol from Example (3.16)\n",
-      "v = 15.14*10**-3;\t\t\t #volume of saturated vapour in m**3/mol from Example (3.16)\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "P_amb = 101.325;\t\t\t #pressure at which the normal boiling point is computed in kPa\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "#Step a: Vaporization of n-octane at T_ref\n",
-      "T_ref = T_ref+273.15\n",
-      "\n",
-      "del_hv = ((R*math.log ((Pc*10**5)/(P_amb*10**3)))/((1./NBP)-(1./Tc)))*10**-3;\n",
-      "P2 = P_amb* math.exp (((del_hv*10**3)/(R))*((1./NBP)-(1./T_ref)))\n",
-      "Tbr = NBP/Tc\n",
-      "\t\t\t\n",
-      "del_hvn = (1.093*R*Tc*(Tbr*(((math.log (Pc))-1.013)/(0.930-Tbr))))*10**-3;\n",
-      "Tr2 = T_ref/Tc\n",
-      "\n",
-      "del_ha = ((del_hvn*10**3)*(((1-Tr2)/(1-Tbr))**(0.38)))*10**-3;\n",
-      "del_sa = (del_ha*10**3)/T_ref\n",
-      "\n",
-      "alpha = (1+(S*(1-math.sqrt (Tr2))))**2\n",
-      "a = (0.45724*(R**2)*(Tc**2)*alpha)/(Pc*10**5)\n",
-      "\n",
-      "A = (a*P2*10**3)/(R*T_ref)**2\n",
-      "B = (b*P2*10**3)/(R*T_ref);\t\n",
-      "alpha = -1+B;\t\t\t \n",
-      "beeta = A-(2*B)-(3*B**2);\n",
-      "gaamma = -(A*B)+(B**2)+(B**3);\t\t\t \n",
-      "p = beeta-(alpha**2)/3;\t\t\t \n",
-      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
-      "D = (((q)**2)/4)+(((p)**3)/27);\t\t\t\n",
-      "\n",
-      "if D>0:\n",
-      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
-      "elif D == 0:\n",
-      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
-      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Za = [Z1 ,Z2, Z3];\n",
-      "    Z = max(Za);\n",
-      "else:\n",
-      "    theta = math.cos((-(q)/2)*(math.sqrt((-27)/(((p)**3)))))\n",
-      "    r = math.sqrt((-(p**3)/27));\t\t\t \n",
-      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Za = [Z1, Z2, Z3];\n",
-      "    Z = max(Za);\n",
-      "da_dT = (-a*S)/(cmath.sqrt(alpha*T_ref*Tc));\t\t\t \n",
-      "\n",
-      "dep_h = (R*T_ref*(Z-1))+(((((T_ref*da_dT)-a)/(2*math.sqrt(2)*b)))*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
-      "dep_s = (R*math.log (Z-B))+((1./(2*math.sqrt (2)*b))*(da_dT)*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
-      "del_hb = -dep_h\n",
-      "del_sb = -dep_s\n",
-      "\n",
-      "del_hc = ((Cp[0]*(T-T_ref))+(((Cp[1])/2)*((T**2)-(T_ref**2)))+(((Cp[2])/3)*((T**3)-(T_ref**3)))+(((Cp[3])/4)*((T**4)-(T_ref**4)))-((Cp[4])*((1./T)-(1./T_ref))))*10**-3;\n",
-      "del_sc = ((Cp[0])*math.log (T/T_ref))+((Cp[1])*(T-T_ref))+(((Cp[2])/2)*((T**2)-(T_ref**2)))+(((Cp[3])/3)*((T**3)-(T_ref**3)))-(((Cp[4])/2)*((1./(T**2))-(1./(T_ref**2))))-(R*math.log((P*10**6)/(P2*10**3)))\t\t\t \n",
-      "\n",
-      "Z = 0.9151\n",
-      "da_dT = (-a*S)/(cmath.sqrt (alpha*T*Tc));\t\t\t \n",
-      "\n",
-      "del_hd = (R*T*(Z-1))+((((T*da_dT)-a)/(2*math.sqrt(2)*b))*math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2))))));\n",
-      "\n",
-      "del_sd = (R*math.log (Z-B))+((1./(2*math.sqrt(2)*b))*(da_dT)*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
-      "\n",
-      "h = h0+del_ha+(del_hb*10**-3)+del_hc+(del_hd*10**-3)\n",
-      "s = s0+del_sa+del_sb+del_sc+del_sd;\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation\\\n",
-      " of state  =\",h, \"kJ/mol\"\n",
-      "print \" The entropy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson\\\n",
-      " equation of state  = \",s,\" J/mol K\"\n",
-      "print \" The volume of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson\\\n",
-      " equation of state  =\",v,\" m**3/mol\"\n",
-      "\n",
-      "\n",
-      "#THE VOLUME OF n-OCTANE VAPOUR AS COMPUTED IN EXAMPLE 3.16 IS 15.14*10**-3 m**3/mol AND NOT \n",
-      "#15.41*10**-3 m**3/mol AS PRINTED IN THIS EXAMPLE IN THE TEXTBOOK.\n",
-      "# So ANSWER WOULD BE DIFFERENT\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state  = (76.8343786515+0.000518888144511j) kJ/mol\n",
-        " The entropy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state  =  (207.222784016-0.000740500199878j)  J/mol K\n",
-        " The volume of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state  = 0.01514  m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics/ch9.ipynb b/Chemical_Engineering_Thermodynamics/ch9.ipynb
deleted file mode 100755
index 9efdc810..00000000
--- a/Chemical_Engineering_Thermodynamics/ch9.ipynb
+++ /dev/null
@@ -1,1125 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9 : Multicomponent mixtures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.1  Page No : 313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "per_ethanol = 60.\t\t\t #mole percent of ethanol in a ethanol-water system\n",
-      "per_water = 40. \t\t\t #mole percent of water in a ethanol-water system\n",
-      "v1 = 57.5*10**-6\t\t\t #partial molar volume of ethanol in the ethanol-water system in m**3\n",
-      "rho = 849.4;\t    \t\t #density of the mixture in kg/m**3\n",
-      "M_ethanol = 46*10**-3\t\t\t #molar mass of ethanol in kg/mol\n",
-      "M_water = 18*10**-3 \t\t\t #molar mass of ethanol in kg/mol\n",
-      "\n",
-      "# Calculations\n",
-      "X1 = per_ethanol/100\n",
-      "X2 = per_water/100\n",
-      "M = (X1*M_ethanol)+(X2*M_water)\n",
-      "v = M/rho\n",
-      "v2 = (v-(X1*v1))/(X2);\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The partial molar volume of water  =  %f m**3/mol\"%(v2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The partial molar volume of water  =  0.000016 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.2  Page No : 313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 3.\t\t\t            #volume of mixture to be prepared in m**3\n",
-      "per_ethanol = 60.\t\t\t #mole percent of ethanol in a ethanol-water system\n",
-      "per_water = 40.\t\t\t #mole percent of water in a ethanol-water system\n",
-      "v1 = 57.5*10**-6;\t\t\t #partial molar volume of ethanol in the ethanol-water system in m**3/mol\n",
-      "v2 = 16*10**-6\t\t\t #partial molar volume of water in the ethanol-water system in m**3/mol\n",
-      "v1_pure = 57.9*10**-6\t\t\t #molar volume of pure ethanol in m**3/mol\n",
-      "v2_pure = 18*10**-6\t\t\t #molar volume of pure water in m**3/mol\n",
-      "\n",
-      "# Calculations\n",
-      "X1 = per_ethanol/100;\t\t\t # Calculations of the mole fraction of ethanol (no unit)\n",
-      "X2 = per_water/100;\t\t\t    # Calculations of the mole fraction of water (no unit)\n",
-      "v = (X1*v1)+(X2*v2);\t\t\t # Calculations of the molar volume of the solution using Eq.(9.10) in m**3/mol\n",
-      "N = V/v\t            \t\t # Calculations of the mole number of solution required in mol\n",
-      "N1 = N*X1\t\t\t         # Calculations of the mole number of ethanol in solution in mol\n",
-      "N2 = N*X2;\t\t\t         # Calculations of the mole number of water in solution in mol\n",
-      "V1 = N1*v1_pure;\t\t\t # Calculations of the volume of pure ethanol required in m**3\n",
-      "V2 = N2*v2_pure;\t\t\t # Calculations of the volume of pure water required in m**3\n",
-      "\n",
-      "# Results\n",
-      "print \" The volume of pure ethanol required  =  %0.3f m**3\"%(V1);\n",
-      "print \" The volume of pure water required  =  %0.3f m**3\"%(V2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The volume of pure ethanol required  =  2.548 m**3\n",
-        " The volume of pure water required  =  0.528 m**3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.3  Page No : 318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
-      "a = 3.789;\t\t\t #van der Waals constant in Pa(m**3/mol)**2\n",
-      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
-      "v = 15.675*10**-3;\t\t\t #molar volume of n-octane saturated vapour taken from Example 3.8 in m**3/mol\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Z = (P*10**6*v)/(R*T)\t\t\t # Calculations of the compressibility factor (no unit)\n",
-      "# Calculations of the fugacity coefficient (f/P) using the expression derived in Example 9.3 (no unit)\n",
-      "phi = math.exp(Z-1-math.log (((P*10**6)*(v-b))/(R*T))-a/(R*T*v));\n",
-      "f = (P*10**6*phi)*10**-6\n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of n-octane vapour  =  %0.2f \"%(phi);\n",
-      "print \" The fugacity of n-octane vapour  =  %0.4f MPa\"%(f);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of n-octane vapour  =  0.95 \n",
-        " The fugacity of n-octane vapour  =  0.2043 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.5  Page No : 322"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
-      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
-      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
-      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
-      "w = 0.398;\t\t\t #acentric factor (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "Tr = T/Tc\n",
-      "Pr = (P*10**6)/(Pc*10**5)\n",
-      "log_phi0 = -0.032;\t\n",
-      "log_phi1 = -0.025;\n",
-      "phi = round(log_phi0 + 0.398*log_phi1,3)\n",
-      "phie = 1.1014\n",
-      "f = P*phie\n",
-      "\n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of n-octane vapour  =  %.3f \"%(phi);\n",
-      "print \" The fugacity of n-octane vapour  =  %.4f MPa\"%(f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of n-octane vapour  =  -0.042 \n",
-        " The fugacity of n-octane vapour  =  0.2368 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.6  Page No : 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 339.7\t\t\t #temperature of ethylene in K\n",
-      "P = 1.\t    \t\t #pressure of ethylene in bar\n",
-      "Tc = 283.1\t\t\t #critical temperature of ethylene in K\n",
-      "Pc = 51.17\t\t\t #critical pressure of ethylene in bar\n",
-      "w = 0.089\t\t\t #acentric factor (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "Tr = T/Tc\n",
-      "Pr = P/Pc\n",
-      "B0 = 0.083-(0.422/(Tr**1.6))\n",
-      "B1 = 0.139-(0.172/(Tr**4.2))\n",
-      "phi = math.exp((B0+(w*B1))*(Pr/Tr))\n",
-      "f = P*phi\n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of ethylene  =  %0.4f \"%(phi);\n",
-      "print \" The fugacity of ethylene  =  %0.4f bar\"%(f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of ethylene  =  0.9963 \n",
-        " The fugacity of ethylene  =  0.9963 bar\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.7  Page No : 330"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.\t        \t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
-      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
-      "vc = [255.0*10**-6,486.0*10**-6];\t\t\t #critical molar volume of n-butane and n-octane in m**3/mol\n",
-      "Zc = [0.274,0.256];\t        \t\t #compressibility factor of n-butane and n-octane corresponding to Tc,Pc (no unit)\n",
-      "w = [0.199,0.398];\t\t\t         #acentric factor of n-butane and n-octane (no unit)\n",
-      "R = 8.314               \t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "K_12 = 1-((8*((vc[0]*vc[1])**(1./2)))/((((vc[0])**(1./3))+((vc[1])**(1./3)))**3))\n",
-      "Tc_12 = (((Tc[0])*(Tc[1]))**(1./2))*(1-K_12);\t\t\t \n",
-      "w_12 = (w[0]+w[1])/2\n",
-      "Zc_12 = (Zc[0]+Zc[1])/2\n",
-      "vc_12 = ((((vc[0])**(1./3))+((vc[1])**(1./3)))/2)**3\n",
-      "Pc_12 = ((Zc_12*R*Tc_12)/vc_12)*10**-6;\t\t\t \n",
-      "Tr_12 = T/Tc_12\n",
-      "B_12_0 = 0.083-(0.422/(Tr_12**(1.6)));\t\t\t \n",
-      "B_12_1 = 0.139-(0.172/(Tr_12**(4.2)));\t\t\t \n",
-      "B_12 = ((R*Tc_12)/(Pc_12*10**6))*(B_12_0+(w_12*B_12_1))\n",
-      "Tr1 = T/Tc[0]\n",
-      "B_11_0 = 0.083-(0.422/(Tr1**(1.6)));\t\t\t \n",
-      "B_11_1 = 0.139-(0.172/(Tr1**(4.2)));\t\t\t \n",
-      "B_11 = ((R*Tc[0])/(Pc[0]*10**5))*(B_11_0+(w[0]*B_11_1))\n",
-      "Tr2 = T/Tc[1]\n",
-      "B_22_0 = 0.083-(0.422/(Tr2**(1.6)))\n",
-      "B_22_1 = 0.139-(0.172/(Tr2**(4.2)))\n",
-      "B_22 = ((R*Tc[1])/(Pc[1]*10**5))*(B_22_0+(w[1]*B_22_1))\n",
-      "Bm = ((y1**2)*B_11)+((2*y1*y2)*B_12)+((y2**2)*B_22);\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The second virial coefficient for an equimolar mixture of n-butane and n-octane at\\\n",
-      " 600K  =  %.2e m**3/mol\"%(Bm);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The second virial coefficient for an equimolar mixture of n-butane and n-octane at 600K  =  -3.09e-04 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.8  Page No : 331"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "a = [1.3874,3.7890];\t\t\t #van der Waals constant of n-butane and n-octane in Pa(m**3/mol)**2\n",
-      "b = [0.1163*10**-3,0.237*10**-3];\t\t\t #van der Waals constant of n-butane and n-octane in m**3/mol\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "a_m = ((y1**2)*a[0])+((2*y1*y2)*math.sqrt(a[0]*a[1]))+((y2**2)*a[1])\n",
-      "b_m = (y1*b[0])+(y2*b[1])\t\t\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The van der Waals constant for an equimolar mixture of n-butane and n-octane,\\\n",
-      " a_m  =  %0.4f Pam**3/mol)**2\"%(a_m);\n",
-      "print \" The van der Waals constant for an equimolar mixture of n-butane and n-octane,\\\n",
-      " b_m  =  %f m**3/mol\"%(b_m);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The van der Waals constant for an equimolar mixture of n-butane and n-octane, a_m  =  2.4405 Pam**3/mol)**2\n",
-        " The van der Waals constant for an equimolar mixture of n-butane and n-octane, b_m  =  0.000177 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.9  Page No : 333"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.\t    \t    \t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16.\t    \t    \t     #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
-      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
-      "R = 8.314;\t\t\t        #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "Tr1 = T/Tc[0]\n",
-      "Pr1 = P/Pc[0]\n",
-      "Z1_0 = 0.95\n",
-      "Tr2 = T/Tc[1]\n",
-      "Pr2 = P/Pc[1]\n",
-      "Z2_0 = 0.785\n",
-      "\n",
-      "Zm = (y1*Z1_0)+(y2*Z2_0)\n",
-      "vm = (Zm*R*T)/(P*10**5)\n",
-      "\n",
-      "P1 = y1*P\n",
-      "P2 = y2*P\n",
-      "Pr1 = P1/Pc[0]\n",
-      "Pr2 = P2/Pc[1]\n",
-      "Z1_0 = 0.97\n",
-      "Z2_0 = 0.91\n",
-      "Zm = (y1*Z1_0)+(y2*Z2_0)\n",
-      "vm_dalton = (Zm*R*T)/(P*10**5)\n",
-      "\n",
-      "P1 = ((Z1_0*y1*R*T)/(vm_dalton))*10**-2\n",
-      "P2 = ((Z2_0*y2*R*T)/(vm_dalton))*10**-2\n",
-      "Pr1 = P1/Pc[0]\n",
-      "Pr2 = P2/Pc[1]\n",
-      "Z1_0_new = 0.97\n",
-      "Z2_0_new = 0.91\n",
-      "if Z1_0_new == Z1_0 and Z2_0_new == Z2_0:\n",
-      "    vm_new = vm_dalton;\t\t\t \n",
-      "else:\n",
-      "    Zm = (y1*Z1_0_new)+(y2*Z2_0_new)\n",
-      "    vm_new = (Zm*R*T)/(P*10**5);\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The molar volume of an equimolar mixture of n-butane and n-octane found using the\\\n",
-      " Amagats law of additive volumes  =  %0.4e m**3/mol\"%(vm);\n",
-      "print \" The molar volume of an equimolar mixture of n-butane and\\\n",
-      " n-octane found using the Daltons law of additive pressures  =  %0.2e m**3/mol\"%(vm_new);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The molar volume of an equimolar mixture of n-butane and n-octane found using the Amagats law of additive volumes  =  2.7046e-03 m**3/mol\n",
-        " The molar volume of an equimolar mixture of n-butane and n-octane found using the Daltons law of additive pressures  =  2.93e-03 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.10  Page No : 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16.\t\t\t     #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
-      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
-      "R = 8.314;\t    \t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "Tcm = (y1*Tc[0])+(y2*Tc[1]);\t\t\t \n",
-      "Pcm = (y1*Pc[0])+(y2*Pc[1])\n",
-      "Trm = T/Tcm;\t\t\t \n",
-      "Prm = P/Pcm;\t\t\t \n",
-      "Zm0 = 0.9;\t\t\t \n",
-      "vm = (Zm0*R*T)/(P*10.**5)\n",
-      "\n",
-      "# Results\n",
-      "print \" The molar volume of an equimolar mixture of n-butane and n-octane using the\\\n",
-      " pseudocritical properties estimated through Kays rule  =  %0.2e m**3/mol\"%(vm);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The molar volume of an equimolar mixture of n-butane and n-octane using the pseudocritical properties estimated through Kays rule  =  2.81e-03 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.11  Page No : 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.\t\t\t         #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t         #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
-      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
-      "vc = [255.0*10**-6,486.0*10**-6];\t\t\t #critical molar volume of n-butane and n-octane in m**3/mol\n",
-      "Zc = [0.274,0.256];\t\t\t #compressibility factor of n-butane and n-octane corresponding to Tc,Pc (no unit)\n",
-      "R = 8.314;\t        \t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "Tcm = (y1*Tc[0])+(y2*Tc[1]);\t\t\t \n",
-      "Pcm = ((R*((y1*Zc[0])+(y2*Zc[1]))*Tcm)/((y1*vc[0])+(y2*vc[1])))*10**-5\n",
-      "Trm = T/Tcm\n",
-      "Prm = P/Pcm\n",
-      "Zm0 = 0.89\n",
-      "vm = (Zm0*R*T)/(P*10**5)\n",
-      "\n",
-      "# Results\n",
-      "print \" The molar volume of an equimolar mixture of n-butane and n-octane at 600K\\\n",
-      " and 16bar estimated using the Prausnitz-Gunn rule  =  %0.2e m**3/mol\"%(vm);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar estimated using the Prausnitz-Gunn rule  =  2.77e-03 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.12  Page No : 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "a_m = 2.4405\t\t\t #van der Waals constant for the mixture as determined in Example 9.8 in Pa(m**3/mol)**2\n",
-      "b_m = 0.1767*10**-3\t\t #van der Waals constant for the mixture as determined in Example 9.8 in m**3/mol\n",
-      "R = 8.314;\t\t\t     #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "A = (a_m*P*10**5)/(R*T)**2\n",
-      "B = (b_m*P*10**5)/(R*T)\n",
-      "alpha = -1-B\n",
-      "beeta = A\n",
-      "gaamma = -(A*B)\n",
-      "p = beeta-((alpha**2)/3)\n",
-      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
-      "D = (((q)**2)/4)+(((p)**3)/27)\n",
-      "\n",
-      "if D>0 :\n",
-      "    Z = (((-(q)/2)+(math.sqrt(D)))**(1./3))+(((-(q)/2)-(math.sqrt(D)))**(1./3))-(alpha/3);\t\n",
-      "elif D == 0:\n",
-      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
-      "    Z2 = ((q/2)**(1./3))-(alpha/3)\n",
-      "    Z3 = ((q/2)**(1./3))-(alpha/3)\n",
-      "    Za = [Z1 ,Z2, Z3];\n",
-      "    Z = max(Za);\n",
-      "else:\n",
-      "    r = math.sqrt((-(p**3)/27));\t\t\t \n",
-      "    theta = amath.cos((-(q)/2)*(1./r));\t\t\n",
-      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Za = [Z1, Z2, Z3];\n",
-      "    Z = max(Za);\n",
-      "\n",
-      "vm = (Z*R*T)/(P*10**5)\n",
-      "\n",
-      "# Results\n",
-      "print \" The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar found \\\n",
-      " using the van der Waals equation of state  =  %e m**3/mol\"%(vm);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar found  using the van der Waals equation of state  =  2.780786e-03 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.13  Page No : 336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.\t\t    \t     #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16.\t\t    \t         #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Bm = -309.*10**-6;\t\t\t #second virial coefficient for the mixture taken from Example(9.7) in m**3/mol\n",
-      "R = 8.314;\t\t\t         #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "Zm = 1+((Bm*P*10**5)/(R*T))\n",
-      "vm = (Zm*R*T)/(P*10**5)\t\n",
-      "\n",
-      "# Results\n",
-      "print \" The molar volume of an equimolar mixture of n-butane and n-octane\\\n",
-      " found using the generalized virial coefficient correlation  =  %0.4e m**3/mol\"%(vm);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The molar volume of an equimolar mixture of n-butane and n-octane found using the generalized virial coefficient correlation  =  2.8088e-03 m**3/mol\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.14  Page No : 337"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "am = 2.4405\t\t\t #van der Waals constant for the mixture taken from Example 9.8 in Pa(m**3/mol)**2\n",
-      "bm = 0.1767*10**-3;\t\t\t #van der Waals constant for the mixture taken from Example 9.8 in m**3/mol\n",
-      "vm = 2.8933*10**-3;\t\t\t #molar volume of the mixture taken from Example 9.12 in m**3/mol\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "dep_h = ((P*10**5*vm)-(R*T)-(am/vm))*10**-3\n",
-      "dep_s = R*(math.log ((P*10**5*(vm-bm))/(R*T)))\n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure of an equimolar mixture of n-butane and n-octane  =  %0.3f kJ/mol\"%(dep_h);\n",
-      "print \" The entropy departure of an equimolar mixture of n-butane and n-octane  =  %0.3f J/mol K\"%(dep_s);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure of an equimolar mixture of n-butane and n-octane  =  -1.203 kJ/mol\n",
-        " The entropy departure of an equimolar mixture of n-butane and n-octane  =  -1.145 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.15  Page No : 338"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.    \t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t    \t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Tcm = 497.3;\t\t\t #pseudocritical temperature of mixture taken from Example(9.10) in K\n",
-      "Pcm = 31.47;\t\t\t #pseudocritical pressure of mixture taken from Example(9.10) in bar\n",
-      "Trm = 1.21;\t\t    \t #pseudoreduced temperature of the mixture taken from Example(9.10) (no unit)\n",
-      "Prm = 0.51;\t\t\t     #pseudoreduced pressure of the mixture taken from Example(9.10) (no unit)\n",
-      "w_butane = 0.199;\t\t #acentric factor for n-butane (no unit)\n",
-      "w_octane = 0.398;\t\t #acentric factor for n-octane (no unit)\n",
-      "R = 8.314;\t\t    \t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "wm = (y1*w_butane)+(y2*w_octane)\n",
-      "del_h0 = 0.380;\t\t\t \n",
-      "del_h1 = 0.188;\t\t\t \n",
-      "del_s0 = 0.22;\t\t\t \n",
-      "del_s1 = 0.18;\t\t\t \n",
-      "dep_h = ((R*Tcm)*(del_h0+(wm*del_h1)))*10**-3;\t\t\t \n",
-      "dep_s = (R)*(del_s0+(wm*del_s1));\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure of an equimolar mixture of n-butane and n-octane using\\\n",
-      " the generalized compressibility factor correlation  =  %0.3f kJ/mol\"%(dep_h);\n",
-      "print \" The entropy departure of an equimolar mixture of n-butane and n-octane using \\\n",
-      "the generalized compressibility factor correlation  =  %f J/mol K\"%(dep_s);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure of an equimolar mixture of n-butane and n-octane using the generalized compressibility factor correlation  =  1.803 kJ/mol\n",
-        " The entropy departure of an equimolar mixture of n-butane and n-octane using the generalized compressibility factor correlation  =  2.275791 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.16  Page No : 339"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
-      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
-      "w = [0.199,0.398];\t\t\t #acentric factor of n-butane and n-octane (no unit)\n",
-      "Tr1 = 1.411;\t\t\t #reduced temperature of n-butane (no unit) taken from Example (9.7)\n",
-      "Tr2 = 1.054;\t\t\t #reduced temperature of n-octane (no unit) taken from Example (9.7)\n",
-      "Tr_12 = 1.24;\t\t\t #reduced temperature for computing the mixture interaction virial coefficient (no unit) taken from Example(9.7)\n",
-      "Pc_12 = 2.978;\t\t\t #Pc_ij in MPa taken from Example(9.7)\n",
-      "Tc_12 = 483.9;\t\t\t #Tc_ij in K taken from Example(9.7)\n",
-      "w_12 = 0.2985;\t\t\t # w_ij (no unit) taken from Example(9.7)\n",
-      "Bm = -309*10**-6;\t\t\t #second virial coefficient in m**3/mol taken from Example (9.7)\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "dB0_dTr1 = 0.675/(Tr1**2.6);\t\t\t \n",
-      "dB0_dTr2 = 0.675/(Tr2**2.6);\t\t\t \n",
-      "dB1_dTr1 = 0.722/(Tr1**5.2);\t\t\t \n",
-      "dB1_dTr2 = 0.722/(Tr2**5.2);\t\t\t \n",
-      "dB0_dTr12 = 0.675/(Tr_12**2.6);\t\t\t \n",
-      "dB1_dTr12 = 0.722/(Tr_12**5.2);\t\t\t \n",
-      "dB1_dT = (R/(Pc[0]*10**5))*((dB0_dTr1)+(w[0]*(dB1_dTr1)));\t\t\t\n",
-      "dB2_dT = (R/(Pc[1]*10**5))*((dB0_dTr2)+(w[1]*(dB1_dTr2)));\t\t\t\n",
-      "dB12_dT = (R/(Pc_12*10**6))*((dB0_dTr12)+(w_12*(dB1_dTr12)));\t\t\t\n",
-      "dBm_dT = ((y1**2)*(dB1_dT))+((2*y1*y2)*(dB12_dT))+((y2**2)*(dB2_dT));\t\n",
-      "dep_h = ((Bm-(T*dBm_dT))*P*10**5)*10**-3\n",
-      "dep_s = -P*10**5*(dBm_dT);\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The enthalpy departure of an equimolar mixture of n-butane and n-octane\\\n",
-      " using the virial coefficient correlation  =  %f kJ/mol\"%(dep_h);\n",
-      "print \" The entropy departure of an equimolar mixture of n-butane and n-octane\\\n",
-      " using the virial coefficient correlation  =  %0.3f J/mol K\"%(dep_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy departure of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  -1.908476 kJ/mol\n",
-        " The entropy departure of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  -2.357 J/mol K\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.17  Page No : 340"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "a_m = 2.4405\t\t\t #van der Waals constant (a_m) in Pa(m**3/mol)**2 taken from Example(9.8)\n",
-      "b_m = 0.1767*10**-3\t\t #van der Waals constant (b_m) in m**3/mol taken from Example(9.8)\n",
-      "Z = 0.928;\t\t\t     #compressibility factor taken from Example(9.12)\n",
-      "vm = 2.8933*10**-3;\t\t #molar volume of the equimolar mixture in m**3/mol taken from Example(9.12)\n",
-      "R = 8.314;\t\t\t     #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "phi = math.exp(Z-1-math.log ((P*10**5*(vm-b_m))/(R*T))-(a_m/(R*T*vm)));\t\t\t\n",
-      "f = phi*P;\t\t\n",
-      "\n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of an equimolar mixture of n-butane and n-octane\\\n",
-      " using the van der Waals equation of state  =  %0.4f \"%(phi);\n",
-      "print \" The fugacity of an equimolar mixture of n-butane and n-octane using the van der Waals \\\n",
-      "equation of state  =  %0.2f bar\"%(f);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the van der Waals equation of state  =  0.9018 \n",
-        " The fugacity of an equimolar mixture of n-butane and n-octane using the van der Waals equation of state  =  14.43 bar\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.18  Page No : 341"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16.\t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Tcm = 497.3;\t\t\t #pseudocritical temperature of mixture in K taken from Example(9.10)\n",
-      "Pcm = 31.47;\t\t\t #pseudocritical pressure of mixture in bar taken from Example(9.10)\n",
-      "Trm = 1.21;\t\t\t #pseudoreduced temperature of the mixture (no unit) taken from Example(9.10)\n",
-      "Prm = 0.51;\t\t\t #pseudoreduced pressure of the mixture (no unit) taken from Example(9.10)\n",
-      "w = [0.199,0.398];\t\t\t #acentric factor of n-butane and n-octane (no unit)\n",
-      "\n",
-      "# Calculations\n",
-      "wm = (w[0]+w[1])/2\n",
-      "log_phi0 = -0.042\n",
-      "log_phi1 = 0.01\n",
-      "phi = 10**(log_phi0+(wm*log_phi1));\t\t\n",
-      "f = P*phi;\n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of an equimolar mixture of n-butane and n-octane \\\n",
-      "using the pseudocritical constants method  =  %0.3f \"%(phi);\n",
-      "print \" The fugacity of an equimolar mixture of n-butane and n-octane using\\\n",
-      " the pseudocritical constants method  =  %f bar\"%(f);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the pseudocritical constants method  =  0.914 \n",
-        " The fugacity of an equimolar mixture of n-butane and n-octane using the pseudocritical constants method  =  14.625307 bar\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.19  Page No : 341"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Bm = -309.*10**-6;\t\t\t #second virial coefficient in m**3/mol taken from Example (9.7)\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "phi = math.exp((Bm*P*10**5)/(R*T))\n",
-      "f = phi*P;\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of an equimolar mixture of n-butane and n-octane using\\\n",
-      " the virial coefficient correlation  =  %.3f \"%(phi);\n",
-      "print \" The fugacity of an equimolar mixture of n-butane and n-octane using\\\n",
-      " the virial coefficient correlation  =  %f bar\"%(f);\n",
-      "\n",
-      "# Note : answers in book is wrong please check manually."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  0.906 \n",
-        " The fugacity of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  14.490290 bar\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.20  Page No : 344"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
-      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
-      "R = 8.314;\t        \t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "a1 = (0.42748*R**2*Tc[0]**2.5)/(Pc[0]*10**5*math.sqrt(T))\n",
-      "a2 = (0.42748*R**2*Tc[1]**2.5)/(Pc[1]*10**5*math.sqrt(T))\n",
-      "b1 = (0.08664*R*Tc[0])/(Pc[0]*10**5);\t\t\t \n",
-      "b2 = (0.08664*R*Tc[1])/(Pc[1]*10**5);\t\t\t \n",
-      "\n",
-      "a = ((y1**2)*a1)+(2*y1*y2*math.sqrt(a1*a2))+((y2**2)*a2)\n",
-      "b = (y1*b1)+(y2*b2)\n",
-      "\n",
-      "A = (a*P*10**5)/(R*T)**2\n",
-      "B = (b*P*10**5)/(R*T);\t\n",
-      "alpha = -1.\t\t\t \n",
-      "beeta = A-B-B**2\n",
-      "gaamma = -(A*B)\n",
-      "p = beeta-(alpha**2)/3\n",
-      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
-      "D = (((q)**2)/4)+(((p)**3)/27)\n",
-      "\n",
-      "if D>0:\n",
-      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);\t\t\t #One real root given by  Eq.(3.32)\n",
-      "elif D == 0:\n",
-      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
-      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
-      "    Za = [Z1, Z2, Z3];\n",
-      "    Z = max(Za);\n",
-      "else:\n",
-      "    r = math.sqrt((-(p**3)/27));\t\t\n",
-      "    theta = amath.cos((-(q)/2)*(1./r));\t\n",
-      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
-      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
-      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
-      "    Za = [Z1, Z2, Z3];\n",
-      "    Z = max(Za);\n",
-      "\n",
-      "phi1 = math.exp(((b1/b)*(Z-1))-math.log(Z-B)+((a/(b*R*T))*((b1/b)-(2*math.sqrt(a1/a)))*math.log((Z+B)/Z)));\n",
-      "\n",
-      "phi2 = math.exp(((b2/b)*(Z-1))-math.log(Z-B)+((a/(b*R*T))*((b2/b)-(2*math.sqrt(a2/a)))*math.log((Z+B)/Z)));\n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of n-butane in the equimolar mixture using the\\\n",
-      " Redlich-Kwong Equation of state  =  %0.4f \"%(phi1);\n",
-      "print \" The fugacity coefficient of n-octane in the equimolar mixture using the\\\n",
-      " Redlich-Kwong Equation of state  =  %0.4f \"%(phi2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of n-butane in the equimolar mixture using the Redlich-Kwong Equation of state  =  0.9644 \n",
-        " The fugacity coefficient of n-octane in the equimolar mixture using the Redlich-Kwong Equation of state  =  0.8326 \n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.21  Page No : 346"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
-      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
-      "B_11 = -131*10**-6\t\t\t #pure component (n-butane) second virial coefficient in m**3/mol taken from Example(9.7)\n",
-      "B_22 = -577*10**-6\t\t\t #pure component (n-octane) second virial coefficient in m**3/mol taken from Example(9.7)\n",
-      "B_12 = -264*10**-6\t\t\t #mixture interaction virial coefficient in m**3/mol taken from Example(9.7)\n",
-      "Bm = -309*10**-6\t\t\t #second virial coefficient in m**3/mol taken from Example(9.7)\n",
-      "R = 8.314       \t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "y1 = 0.5\n",
-      "y2 = 0.5\n",
-      "Zm = (1./2)*(1+math.sqrt(1+((4*Bm*P*10**5)/(R*T))))\n",
-      "phi1 = math.exp((((2*P*10**5)/(Zm*R*T))*((y1*B_11)+(y2*B_12)))-math.log(Zm))\n",
-      "phi2 = math.exp((((2*P*10**5)/(Zm*R*T))*((y1*B_12)+(y2*B_22)))-math.log(Zm))\n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity coefficient of n-butane in the equimolar mixture using the\\\n",
-      " Virial Equation of state  =  %0.3f \"%(phi1);\n",
-      "print \" The fugacity coefficient of n-octane in the equimolar mixture using the Virial\\\n",
-      " Equation of state  =  %f \"%(phi2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity coefficient of n-butane in the equimolar mixture using the Virial Equation of state  =  0.976 \n",
-        " The fugacity coefficient of n-octane in the equimolar mixture using the Virial Equation of state  =  0.830827 \n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.22  Page No : 349"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# Variables\n",
-      "T = 427.85  \t\t\t #temperature of n-octane vapour in K\n",
-      "Psat = 0.215;\t\t\t #saturation pressure of n-octane vapour at T in MPa\n",
-      "P = 1.\t\t        \t #pressure at which the fugacity of liquid n-octane is to be determined in MPa\n",
-      "f_sat = 0.2368;\t\t\t #fugacity of n-octane vapour at T and Psat taken from Example(9.5) in MPa\n",
-      "vl = 0.2003*10**-3;\t\t\t #molar volume of n-octane liquid at T and Psat taken from Example(3.16) in m**3/mol\n",
-      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
-      "\n",
-      "# Calculations\n",
-      "f_l = (0.2368*math.exp((vl*(P-Psat)*10**6)/(R*T)));\t\t\t \n",
-      "\n",
-      "# Results\n",
-      "print \" The fugacity of liquid n-octane at 427.85K and 1MPa  =  %0.4f MPa\"%(f_l);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fugacity of liquid n-octane at 427.85K and 1MPa  =  0.2475 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
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diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/README.txt b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/README.txt
new file mode 100755
index 00000000..d33ed50c
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/README.txt
@@ -0,0 +1,10 @@
+Contributed By: pratik gandhi
+Course: bca
+College/Institute/Organization: Cybercom Creation
+Department/Designation: Developer
+Book Title: Chemical Engineering Thermodynamics
+Author: P. Ahuja
+Publisher: PHI Learning Private Limited, New Delhi
+Year of publication: 2009
+Isbn: 9788120336377
+Edition: 1
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch10_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch10_1.ipynb
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index 00000000..3336e17e
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch10_1.ipynb
@@ -0,0 +1,2536 @@
+{
+ "metadata": {
+  "name": "ch10_1",
+  "signature": "sha256:3b50ad151f89e9ea380ce80c8112ec5468586c82e236862a1e3d8e194fa15249"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Residual Properties by Equations of State"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.2  Page Number : 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 40 + 273.15;\t\t\t#[C] - Temperature\n",
+      "P_1 = 0;\t\t\t#[bar]\n",
+      "P_2 = 10;\t\t\t#[bar]\n",
+      "V_liq = 90.45;\t\t\t#[cm**(3)/mol]\n",
+      "V_liq = V_liq*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "P_sat = 4.287;\t\t\t#[bar] \n",
+      "\n",
+      "# For butadiene\n",
+      "T_c = 425.0;\t\t\t#[K] - Critical temperature\n",
+      "P_c = 43.3;\t\t\t#[bar] - Critical pressure\n",
+      "P_c = P_c*10**(5);\t\t\t#[N/m**(2)]\n",
+      "w = 0.195;\t\t\t# Acentric factor\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# Let us calculate second virial coefficient at 40 C\n",
+      "Tr = T/T_c;\t\t\t# Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B = ((B_0 + (w*B_1))*(R*T_c))/P_c;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "\n",
+      "# math.log(f/P) = (B*P)/(R*T)\n",
+      "# f = P*exp((B*P)/(R*T))\n",
+      "\n",
+      "print \" The table is as follows\"\n",
+      "print \" Pbar \\t\\t fbar \\t\\t                 phi\";\n",
+      "from numpy import zeros\n",
+      "P = [1,2,3,4,4.287,5,6,8,10];\n",
+      "f = zeros(9);\n",
+      "phi = zeros(9);\n",
+      "for i in range(5):\n",
+      "    f[i]=P[i]*(math.exp((B*P[i]*10**(5))/(R*T)));\t\t\t#[bar]  # Pressure inside the exponential term has to be in N/m**(2)\n",
+      "    phi[i]= (f[i]/P[i]);\n",
+      "    print \" %f \\t %f \\t\\t\\t %f\"%(P[i],f[i],phi[i]) \n",
+      "\n",
+      "f_sat = f[4];\n",
+      "\n",
+      "# From pressure of 4.287 bar onwards the valid equation to compute fugacity of compressed liquid is given by\n",
+      "# f = f_sat*exp[V_liq*(P-P_sat)/(R*T)]\n",
+      "\n",
+      "for j in range(5,9):\n",
+      "    f[j] = f_sat*math.exp((V_liq*(P[j]-P_sat)*10**(5))/(R*T));\t#[bar]  # Pressure inside the exponential term has to be in N/m**(2)\n",
+      "    phi[j] = f[j]/P[j]; \n",
+      "    print \" %f \\t %f \\t\\t\\t %f\"%(P[j],f[j],phi[j]);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The table is as follows\n",
+        " Pbar \t\t fbar \t\t                 phi\n",
+        " 1.000000 \t 0.978336 \t\t\t 0.978336\n",
+        " 2.000000 \t 1.914284 \t\t\t 0.957142\n",
+        " 3.000000 \t 2.809221 \t\t\t 0.936407\n",
+        " 4.000000 \t 3.664484 \t\t\t 0.916121\n",
+        " 4.287000 \t 3.902801 \t\t\t 0.910380\n",
+        " 5.000000 \t 3.912481 \t\t\t 0.782496\n",
+        " 6.000000 \t 3.926097 \t\t\t 0.654349\n",
+        " 8.000000 \t 3.953471 \t\t\t 0.494184\n",
+        " 10.000000 \t 3.981037 \t\t\t 0.398104\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.3  Page Number : 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "n = 100.;\t\t\t#[mol] - No of moles\n",
+      "T_1 = 600.;\t\t\t#[K] - Initial temperature\n",
+      "T_2 = 300.;\t\t\t#[K] - Final temperature\n",
+      "P_1 = 10.;\t\t\t#[atm] - Initial pressure\n",
+      "P_1 = P_1*101325;\t\t\t#[Pa]\n",
+      "P_2 = 5.;\t\t\t#[atm] - Final presssure\n",
+      "P_2 = P_2*101325;\t\t\t#[Pa]\n",
+      "Tc = 369.8;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 42.48;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.152;\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "# At 600 K\n",
+      "Tr = T_1/Tc;\t\t\t# Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B = ((B_0 + (w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "dB0_dT = 0.422*1.6*Tc**(1.6)*T_1**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "dB1_dT = 0.172*4.2*Tc**(4.2)*T_1**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "dB_dT = ((R*Tc)/(Pc))*(dB0_dT + w*dB1_dT);\t\t\t# dB/dT\n",
+      "\n",
+      "# Now let us calculate B and dB/dT at 300 K\n",
+      "Tr_prime = T_2/Tc;\t\t\t# Reduced temperature\n",
+      "B_0_prime = 0.083-(0.422/(Tr_prime)**(1.6));\n",
+      "B_1_prime = 0.139-(0.172/(Tr_prime)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B_prime = ((B_0_prime + (w*B_1_prime))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "dB0_dT_prime = 0.422*1.6*Tc**(1.6)*T_2**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "dB1_dT_prime = 0.172*4.2*Tc**(4.2)*T_2**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "dB_dT_prime = ((R*Tc)/(Pc))*(dB0_dT_prime + w*dB1_dT_prime);\t\t\t# dB/dT\n",
+      "\n",
+      "# The change in enthalpy for ideal gas is given by\n",
+      "\n",
+      "def f16(T): \n",
+      "    return -0.966+7.279*10**(-2)*T-3.755*10**(-5)*T**(2)+7.58*10**(-9)*T**(3)\n",
+      "\n",
+      "delta_H_ig =  quad(f16,T_1,T_2)[0]\n",
+      "\n",
+      "delta_H_ig = delta_H_ig*4.184;\t\t\t#[J/mol]\n",
+      "\n",
+      "# We know that delta_H_ig = delta_U_ig + R*delta_T. Therefore change in internal energy is given by \n",
+      "delta_U_ig = delta_H_ig - R*(T_2 - T_1);\t\t\t#[J/mol]\n",
+      "\n",
+      "# The change in entropy of ideal gas is given by \n",
+      "\n",
+      "def f17(T): \n",
+      "    return Cp_0/T\n",
+      "\n",
+      "#delta_S_ig =  quad(f17,T_1,T_2) - R*math.log(P_2/P_1)[0]\n",
+      "\n",
+      "\n",
+      "def f18(T): \n",
+      "    return (-0.966+7.279*10**(-2)*T-3.755*10**(-5)*T**(2)+7.58*10**(-9)*T**(3))/T\n",
+      "\n",
+      "delta_S_ig =  quad(f18,T_1,T_2)[0]*4.184 - R*math.log(P_2/P_1)\n",
+      "\n",
+      "\n",
+      "# Now let us calculate the change in enthalpy of gas. We know that\n",
+      "# delta_H = delta_H_ig + delta_H_R\n",
+      "# delta_H_R = H_2_R - H_1_R\n",
+      "H_2_R = B_prime*P_2 - P_2*T_2*dB_dT_prime;\t\t\t# [J/mol]\n",
+      "H_1_R = B*P_1 - P_1*T_1*dB_dT;\t\t\t# [J/mol]\n",
+      "delta_H_R = H_2_R - H_1_R;\t\t\t# [J/mol]\n",
+      "delta_H = delta_H_ig + delta_H_R;\t\t\t# [J/mol]\n",
+      "\n",
+      "# Let us calculate the residual entropy of gas\n",
+      "S_2_R = -P_2*dB_dT_prime;\t\t\t#[J/mol-K]\n",
+      "S_1_R = -P_1*dB_dT;\t\t\t#[J/mol-K]\n",
+      "delta_S = delta_S_ig + (S_2_R - S_1_R);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# Let us calculate the residual internal energy of gas\n",
+      "U_2_R = -P_2*T_2*dB_dT_prime;\t\t\t#[J/mol-K]\n",
+      "U_1_R = -P_1*T_1*dB_dT;\t\t\t#[J/mol-K]\n",
+      "delta_U = delta_U_ig + (U_2_R - U_1_R);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# For 100 mol sample,\n",
+      "delta_H_ig = delta_H_ig*n*10**(-3);\t\t\t#[kJ/mol]\n",
+      "delta_H = delta_H*n*10**(-3);\t\t\t#[kJ/mol]\n",
+      "\n",
+      "delta_U_ig = delta_U_ig*n*10**(-3);\t\t\t#[kJ/mol]\n",
+      "delta_U = delta_U*n*10**(-3);\t\t\t#[kJ/mol]\n",
+      "\n",
+      "delta_S_ig = delta_S_ig*n*10**(-3);\t\t\t#[kJ/mol]\n",
+      "delta_S = delta_S*n*10**(-3);\t\t\t#[kJ/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The value of delta_H = %f kJ/mol\"%(delta_H);\n",
+      "print \" The value of delta_H_ig ideal gas)= %f J/mol\"%(delta_H_ig);\n",
+      "print \" The value of delta_U = %f kJ/mol\"%(delta_U);\n",
+      "print \" The value of delta_U_ig ideal gas) = %f kJ/mol\"%(delta_U_ig);\n",
+      "print \" The value of delta_S = %f kJ/mol\"%(delta_S);\n",
+      "print \" The value of delta_S_ig ideal gas) = %f kJ/mol\"%(delta_S_ig);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of delta_H = -3130.406247 kJ/mol\n",
+        " The value of delta_H_ig ideal gas)= -3096.763542 J/mol\n",
+        " The value of delta_U = -2867.753839 kJ/mol\n",
+        " The value of delta_U_ig ideal gas) = -2847.343542 kJ/mol\n",
+        " The value of delta_S = -6.466831 kJ/mol\n",
+        " The value of delta_S_ig ideal gas) = -6.358994 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.4  Page Number : 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 35 + 273.15;\t\t\t#[K] - Temperature\n",
+      "P = 10;\t\t\t#[atm] - Pressure\n",
+      "P = P*101325;\t\t\t#[Pa]\n",
+      "# Methane obeys the equation of state\n",
+      "# Z = 1 + (P*B)/(R*T)\n",
+      "\n",
+      "# Calculations\n",
+      "# At 35 C,\n",
+      "B = -50;\t\t\t#[cm**(3)/mol]\n",
+      "dB_dT = 1.0;\t\t\t#[cm**(3)/mol-K] - dB/dT\n",
+      "dB_dT = dB_dT*10**(-6);\t\t\t#[m**(3)/mol-K]\n",
+      "d2B_dT2 = -0.01;\t\t\t#[cm**(3)/mol-K**(2)] - d**2(B)/d(T**2)\n",
+      "d2B_dT2 = d2B_dT2*10**(-6);\t\t\t#[m**(3)/mol-K**(2)]\n",
+      "\n",
+      "# Ideal gas molar heat capacity of methane is given by\n",
+      "# Cp_0 = 4.75 + 1.2*10**(-2)*T + 0.303*10**(-5)*T**(2) - 2.63*10**(-9)*T**(3)\n",
+      "\n",
+      "# The molar heat capacity is given by\n",
+      "# Cp = Cp_0 + Cp_R\n",
+      "# For virial gas equation of state \n",
+      "Cp_R = -P*T*d2B_dT2;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# thus heat capacity is given by \n",
+      "# Cp = a + b*T + c*T**(2) + d*T**(3) - P*T*d2B_dT2\n",
+      "# Putting the values, we get\n",
+      "Cp = (4.75 + 1.2*10**(-2)*T + 0.303*10**(-5)*T**(2) - 2.63*10**(-9)*T**(3))*4.184 - P*T*d2B_dT2;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar heat capacity of methane is %f J/mol-K\"%(Cp);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar heat capacity of methane is 39.349753 J/mol-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.5  Page Number : 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 360;\t\t\t#[K] - Initial temperature\n",
+      "P_1 = 10;\t\t\t#[bar] - Initial pressure\n",
+      "P_1 = P_1*10**(5);\t\t\t#[Pa]\n",
+      "Tc = 408.1;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 36.48;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.181;\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "Cv_0 = 106.0;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# Calculations\n",
+      "# At 360 K\n",
+      "Tr = T_1/Tc;\t\t\t# Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B = ((B_0 + (w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "dB0_dT = 0.422*1.6*Tc**(1.6)*T_1**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "dB1_dT = 0.172*4.2*Tc**(4.2)*T_1**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "dB_dT = ((R*Tc)/(Pc))*(dB0_dT + w*dB1_dT);\t\t\t# dB/dT\n",
+      "\n",
+      "# At state 1\n",
+      "V_1 = B + (R*T_1)/P_1;\t\t\t#[m**(3)/mol] - Molar volume\n",
+      "# At state 1\n",
+      "V_2 = 10*V_1;\t\t\t#[m**(3)/mol] - Molar volume\n",
+      "\n",
+      "T_2 = -(P_1*T_1*(dB_dT))/Cv_0 + T_1;\t\t\t#[K]\n",
+      "\n",
+      "T = 350;\n",
+      "\n",
+      "fault = 10;\n",
+      "def f(T): \n",
+      "    return  106*(T-T_1)+972.72-((R*T**(2))/(V_2-B_prime))*dB_dT_prime\n",
+      "while(fault>0.007):\n",
+      "        Tr_prime = T/Tc;\t\t\t# Reduced temperature\n",
+      "        B_0_prime = 0.083-(0.422/(Tr_prime)**(1.6));\n",
+      "        B_1_prime = 0.139-(0.172/(Tr_prime)**(4.2));\n",
+      "            #We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "        B_prime = ((B_0_prime + (w*B_1_prime))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "        dB0_dT_prime = 0.422*1.6*Tc**(1.6)*T_2**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "        dB1_dT_prime = 0.172*4.2*Tc**(4.2)*T_2**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "        dB_dT_prime = ((R*Tc)/(Pc))*(dB0_dT_prime + w*dB1_dT_prime);\t\t\t# dB/dT\n",
+      "        T_prime = fsolve(f,0.15)\n",
+      "        fault=abs(T-T_prime);\n",
+      "        T = T + 0.001;\n",
+      "\n",
+      "# Results\n",
+      "print \" The final temperature is %f K\"%(T);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final temperature is 351.910000 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.6  Page Number : 339"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 220 + 273.15;\t\t\t#[K] - Temperature\n",
+      "Tc = 562.2;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 48.98;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.210;\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "P_sat = 1912.86;\t\t\t#[kPa] - Saturation pressure at 220 C\n",
+      "P_sat = P_sat*10**(3);\t\t\t#[Pa]\n",
+      "Mol_wt = 78.114;\t\t\t#[g/mol] - Molecular weight of benzene\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#(1)\n",
+      "\n",
+      "# At 220 C\n",
+      "Tr = T/Tc;\t\t\t# Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B = ((B_0 + (w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "\n",
+      "# We know that math.log(f/P) = (B*P)/(R*T)\n",
+      "# Thus at saturated conditions\n",
+      "# math.log(f_sat/P_sat) = B*P_sat/(R*T)\n",
+      "f_sat = P_sat*(math.exp((B*P_sat)/(R*T)));\t\t\t#[Pa]\n",
+      "f_sat = f_sat*10**(-3);\t\t\t#[kPa]\n",
+      "\n",
+      "print \" 1).The fugacity of liquid benzene is %f kPa\"%(f_sat);\n",
+      "\n",
+      "#(2)\n",
+      "P = 2014.7;\t\t\t# [psia] - Total gauge pressure\n",
+      "P = 138.94;\t\t\t# [bar]\n",
+      "P = P*10**(5);\t\t\t# [Pa]\n",
+      "den = 0.63;\t\t\t# [g/cm**(3)] - density of benzene\n",
+      "den = den*10**(3);\t\t\t# [kg/m**(3)]\n",
+      "\n",
+      "# Therefore specific volume is \n",
+      "V = 1/den;\t\t\t#[m/**(3)/kg]\n",
+      "# Molar volume is given by\n",
+      "V = V*Mol_wt*10**(-3);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Thus fugacity at 220 C and pressure P is given by \n",
+      "f = f_sat*(math.exp((V*(P-P_sat))/(R*T)));\n",
+      "\n",
+      "print \" 2).The fugacity of liquid benzene is %f kPa\"%(f);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The fugacity of liquid benzene is 1551.083216 kPa\n",
+        " 2).The fugacity of liquid benzene is 2228.386532 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.7  Page Number : 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "# C = -0.067 + 30.7/T\n",
+      "# D = 0.0012 - 0.416/T\n",
+      "\n",
+      "T = 80 + 273.15;\t\t\t#[K]\n",
+      "P = 30;\t\t\t#[bar]\n",
+      "#P = P;\t\t\t#[N/m**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "# We have the relation derived in the book\n",
+      "# d(G/(R*T)) = (V/(R*T))dP - (H/(R*T**(2)))dT\n",
+      "# Writing the same equation for ideal gas and subtracting it from the above equation we get\n",
+      "# d(G_R/(R*T)) = (V_R/(R*T))dP - (H_R/(R*T**(2)))dT\n",
+      "# Therefore, H_R/(R*T**(2)) = -[del((G_R)/(R*T))/del(T)]_P\n",
+      "\n",
+      "# Substituting the relation G_R/(R*T) = math.log(f/P), we get\n",
+      "# H_R/(R*T**(2)) = -[del(math.log(f/P))/del(T)]_P = -[del(-C*P - D*P**(2))/del(T)]_P\n",
+      "# H_R/R = - 30.7*P + 0.416*P**(2)\n",
+      "\n",
+      "# Substituting the given conditions we get\n",
+      "H_R = R*(-30.7*P + 0.416*P**(2));\t\t\t#[J/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar enthalpy of the gas relative to that of the ideal gas at 80 C and 30 bar pressure is H_R = %f J/mol\"%(H_R);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar enthalpy of the gas relative to that of the ideal gas at 80 C and 30 bar pressure is H_R = -4544.432400 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.8  Page Number : 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "# (1)\n",
+      "T = 311;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "# Pressure in 'bar' is given below\n",
+      "P = [0.690,1.380,2.760,5.520,8.280,11.034,13.800,16.550];\n",
+      "# Molar volume in 'm**(3)/mol' is given below\n",
+      "V = [0.0373,0.0186,0.00923,0.00455,0.00298,0.00220,0.00175,0.00144];\n",
+      "\n",
+      "# Z = 1 + (B/V) + (C/V**(2))\n",
+      "# (Z-1)*V = B + (C/V)\n",
+      "\n",
+      "from numpy import zeros\n",
+      "Z=zeros(8);\n",
+      "k=zeros(8);\n",
+      "t=zeros(8);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "for i in range(8):\n",
+      "    Z[i]=(P[i]*10**(5)*V[i])/(R*T);\n",
+      "    k[i]=(Z[i]-1)*V[i];\n",
+      "    t[i]=1./V[i];\n",
+      "\n",
+      "from scipy.stats import linregress\n",
+      "from numpy import array\n",
+      "t = array(t)\n",
+      "k = array(k)\n",
+      "C,B,c,d,e = linregress(t.T,k.T)\n",
+      "#[C,B,sig]=reglin(t',k');\n",
+      "\n",
+      "#From the regression, we get intercept = B and slope = C,and thus,\n",
+      "\n",
+      "print \" 1).The second virial coefficient of CO2 is given by B = %e m**3)/mol\"%(B);\n",
+      "print \"     The thied virial coefficient of CO2 is given by C = %e m**6)/mol**2)\"%(C);\n",
+      "\n",
+      "# (2)\n",
+      "P_final = 13.8;\t\t\t#[bar]\n",
+      "\n",
+      "def f40(P): \n",
+      "\t return V-(R*T)/P\n",
+      "\n",
+      "# We know that R*T*math.log(f/P) =  quad(f40,0,P)[0]\n",
+      "\n",
+      "# Therefore we have to plot V - (R*T)/P versus P and calculate the area beneath the curve from 0 to 13.8 bar\n",
+      "# For this we need the value of the term V - (R*T)/P at P = 0. At low pressure the virial equation becomes\n",
+      "# Z = 1 + (B/V)\n",
+      "#and V - (R*T)/P = (Z*R*T)/P - (R*T)/P = (1 + (B/V))*((R*T)/P) - (R*T)/P = (B*R*T)/(P*V) = (B/Z)\n",
+      "# Thus lim P tending to zero (V - (R*T)/P) = B    ( as P tend to zero, Z tend to 1 ) \n",
+      "\n",
+      "P_prime = [0.000,0.690,1.380,2.760,5.520,8.280,11.034,13.800];\n",
+      "V_prime = [0.000,0.0373,0.0186,0.00923,0.00455,0.00298,0.00220,0.00175];\n",
+      "summation = 0;\n",
+      "x=zeros(8);\n",
+      "y=zeros(8);\n",
+      "z=zeros(8);\n",
+      "for j in range(2,8):\n",
+      "    x[j]=V_prime[j]-(R*T)/(P_prime[j]*10**(5));\t\t\t#[m**(3)/mol]\n",
+      "    y[j]=(x[j] + x[j-1])/2;\n",
+      "    z[j]=y[j]*((P_prime[j]-P_prime[j-1]))*10**(5);\n",
+      "    summation = summation + z[j] ;\t\t\t#[J/mol]\n",
+      "\n",
+      "summation = summation + 2*z[1] - z[1];\t\t\t# Because in the above calculation,in order to calculate the average a summation of z(2) is not included,only half of it gets added\n",
+      "\n",
+      "\n",
+      "def f41(P): \n",
+      "\t return V - (R*T)/P\n",
+      "\n",
+      "\t\t\t# Now we have, area =  quad(f41,0,13.8*10**(5))[0]\n",
+      "\n",
+      "\t\t\t# R*T*math.log(f/P) = summation\n",
+      "f = P_final*(math.exp(summation/(R*T)));\t\t\t#[bar]\n",
+      "\n",
+      "\n",
+      "print \" 2).The fugacity of steam at 311 K and 13.8 bar pressure is %f bar\"%(f);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The second virial coefficient of CO2 is given by B = -1.515264e-04 m**3)/mol\n",
+        "     The thied virial coefficient of CO2 is given by C = 5.618543e-08 m**6)/mol**2)\n",
+        " 2).The fugacity of steam at 311 K and 13.8 bar pressure is 12.892780 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.9  Page Number : 344"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 0 + 273.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\t\t\t# Pressure in 'atm' is given below\n",
+      "P = [100,200,300,400,500,600,700,800,900,1000];\n",
+      "\t\t\t# The compressibility factor values are\n",
+      "Z = [1.069,1.138,1.209,1.283,1.356,1.431,1.504,1.577,1.649,1.720];\n",
+      "\n",
+      "# Z = 1 + (B/V) + (C/V**(2))\n",
+      "# (Z-1)*V = B + (C/V)\n",
+      "from numpy import zeros\n",
+      "\n",
+      "V = zeros(10);\n",
+      "k = zeros(10);\n",
+      "t = zeros(10);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "for i in range(10):\n",
+      "    V[i]=Z[i]*R*T/(P[i]*101325);\t\t\t#[m**(3)/mol]\n",
+      "    k[i]=(Z[i]-1)*V[i];\n",
+      "    t[i]=1/V[i];\n",
+      "\n",
+      "from scipy.stats import linregress\n",
+      "C,B,c,d,e = linregress(t,k)\n",
+      "\t\t\t#[C,B,sig]=reglin(t,k);\n",
+      "\n",
+      "\t\t\t#From the regression, we get intercept = B and slope = C,and thus,\n",
+      "\n",
+      "print \" 1).The second virial coefficient of H2 is given by B = %e m**3)/mol\"%(B);\n",
+      "print \"     The thied virial coefficient of H2 is given by C = %e m**6)/mol**2)\"%(C);\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "\t\t\t# We know that, limit P tending to zero (V-(R*T)/P) = B, therfore P = 0,  V-(R*T)/P = B\n",
+      "\n",
+      "def f42(P): \n",
+      "\t return V-(R*T)/P \t\t\t#and determine the integral integrate((V-(R*T)/P)\n",
+      "\n",
+      "\t\t\t# Now let us tabulate V-(R*T)/P and determine the integral  quad(f42,0,1000)[0]\n",
+      "\n",
+      "\n",
+      "P_prime = [0,100,200,300,400,500,600,700,800,900,1000];\n",
+      "Z_prime = [0,1.069,1.138,1.209,1.283,1.356,1.431,1.504,1.577,1.649,1.720];\n",
+      "\n",
+      "summation = 0;\n",
+      "V_prime = zeros(11);\n",
+      "x = zeros(11);\n",
+      "y = zeros(11);\n",
+      "z = zeros(11);\n",
+      "for j in range(1,11):\n",
+      "    V_prime[j]=Z_prime[j]*R*T/(P_prime[j]*101325);\t\t\t#[m**(3)/mol]\n",
+      "    x[j]=V_prime[j]-(R*T)/(P_prime[j]*101325);\n",
+      "    y[j]=(x[j] + x[j-1])/2;\n",
+      "    z[j]=y[j]*((P_prime[j]-P_prime[j-1]))*101325;\n",
+      "    summation = summation + z[j] ;\t\t\t#[J/mol]\n",
+      "\n",
+      "summation = summation + 2*z[1] - z[1];\t\n",
+      "\n",
+      "\t\t\t# Now we have\n",
+      "\t\t\t# R*T*math.log(f/P) = summation\n",
+      "P_dash = 1000;\t\t\t#[atm] - Pressure at which fugacity is to be calculated\n",
+      "T_dash = 273.15;\t\t\t#[K] - Temperature at which fugacity is to be calculated\n",
+      "f = P_dash*math.exp(summation/(R*T_dash));\t\t\t#[atm] \n",
+      "\n",
+      "print \" 2).The fugacity of H2 at 0 C and 1000 atm pressure is  f = %f atm\"%(f);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The second virial coefficient of H2 is given by B = 1.345242e-05 m**3)/mol\n",
+        "     The thied virial coefficient of H2 is given by C = 5.286525e-10 m**6)/mol**2)\n",
+        " 2).The fugacity of H2 at 0 C and 1000 atm pressure is  f = 2030.305169 atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.10  Page Number : 345"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math\n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "P_1 = 1*10.**(6.);\t\t\t#[Pa] - Initial pressure\n",
+      "T_1 = 200. + 273.15;\t\t#[K] - Initial temperature\n",
+      "P_2 = 8*10.**(6);\t\t\t#[Pa] - Final pressure\n",
+      "Tc = 647.1;\t\t\t        #[K] - Critical temperature of water\n",
+      "Pc = 220.55;    \t\t\t#[bar] - Critical pressure of water\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.345;\n",
+      "R = 8.314;\t\t    \t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# For the virial gas the following are the relations for residual enthalpy and entropy\n",
+      "# H_R = B*P - P*T*(dB/dT) \n",
+      "# S_R = -P*(dB/dT)\n",
+      "# Where, (dB/dT) = ((R*Tc)/Pc)*((dB_0/dT) + w*(dB_1/dT))\n",
+      "# dB0_dT = 0.422*1.6*Tc**(1.6)*T**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "# dB1_dT = 0.172*4.2*Tc**(4.2)*T**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "\n",
+      "# (1)\n",
+      "Cp_0 = 29.114;\t\t\t#[J/mol-K] - Specific heat capacity at constant pressure\n",
+      "# For the isentropic process entropy change is zero, thus\n",
+      "# delta_S = Cp_0*math.log(T_2/T_1) - P_2*(dB/dT)_2 + P_1*(dB/dT)_1 = 0\n",
+      "\n",
+      "# Calculations\n",
+      "# At state 1, \n",
+      "Tr_1 = T_1/Tc;\n",
+      "B0_1 = 0.083 - 0.422/(Tr_1**(1.6));\n",
+      "B1_1 = 0.139 - 0.172/(Tr_1**(4.2));\n",
+      "# (B*Pc)/(R*Tc) = B0 + w*B1\n",
+      "B_1 = ((B0_1 + w*B1_1)*(R*Tc))/Pc;\t\t\t# [m**(3)/mol] - Second virial coefficient at state 1\n",
+      "dB0_dT_1 = 0.422*1.6*Tc**(1.6)*T_1**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "dB1_dT_1 = 0.172*4.2*Tc**(4.2)*T_1**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "dB_dT_1 = ((R*Tc)/Pc)*((dB0_dT_1) + w*(dB1_dT_1));\t\t\t# (dB/dT)_1\n",
+      "\n",
+      "# Now let us assume the exit temperature to be 870 K, at this temperature\n",
+      "# T_2 = 870;\t\t\t#[K] - \n",
+      "# At this temperature\n",
+      "# delta_S = Cp_0*math.log(T_2/T_1) - P_2*(dB/dT)_2 + P_1*(dB/dT)_1 = \n",
+      "\n",
+      "\n",
+      "T_2 = 860.;\t\t\t#[K] - Exit temperature\n",
+      "# Therefore at state 2, we have\n",
+      "Tr_2 = T_2/Tc;\n",
+      "B0_2 = 0.083 - 0.422/(Tr_2**(1.6));\n",
+      "B1_2 = 0.139 - 0.172/(Tr_2**(4.2));\n",
+      "# (B*Pc)/(R*Tc) = B0 + w*B1\n",
+      "B_2 = ((B0_2 + w*B1_2)*(R*Tc))/Pc;\t\t\t# [m**(3)/mol] - Second virial coefficient at state 2\n",
+      "dB0_dT_2 = 0.422*1.6*Tc**(1.6)*T_2**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "dB1_dT_2 = 0.172*4.2*Tc**(4.2)*T_2**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "dB_dT_2 = ((R*Tc)/Pc)*((dB0_dT_2) + w*(dB1_dT_2));\t\t\t# (dB/dT)_2\n",
+      "\n",
+      "delta_H_s = Cp_0*(T_2 - T_1) + B_2*P_2 -P_2*T_2*(dB_dT_2) - B_1*P_1 + P_1*T_1*(dB_dT_1);\t\t\t#[J/mol] - Enthalpy change\n",
+      "\n",
+      "# As no heat exchange is assumed to take place with the surroundings,work transfer is given by\n",
+      "W_1 = - delta_H_s;\t\t\t# [J/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" 1).The exit temperature is %f K\"%(T_2);\n",
+      "print \"     The required amount of work is %f J/mol\"%(W_1);\n",
+      "\n",
+      "\n",
+      "# (2)\n",
+      "eff = 0.8;\t\t\t# Adiabatic efficiency\n",
+      "delta_H_a = delta_H_s/0.8;\t\t\t# Actual enthalpy change\n",
+      "\n",
+      "# Now for calculating the value of T_exit\n",
+      "# delta_H_a = Cp_0*(T_exit - T_1) + B*P_2 -P_2*T_exit*(dB_dT) - B_1*P_1 + P_1*T_1*(dB_dT_1)\n",
+      "# On simplification we get\n",
+      "# 29.114*(T_2 - T_1)*B_2*8*10**(6)-8*10**(6)*T_2*(dB/dT)_2 = 12643.77 \n",
+      "\n",
+      "# Let us assume a temperature of say\n",
+      "T = 900.;\t\t\t#[K]\n",
+      "fault=10.;\n",
+      "\n",
+      "def f(T_exit): \n",
+      "\t return  delta_H_a - Cp_0*(T_exit - T_1) + B*P_2 -P_2*T_exit*(dB_dT) - B_1*P_1 + P_1*T_1*(dB_dT_1)\n",
+      "\t \n",
+      "while(fault>0.3):\n",
+      "    Tr = T/Tc;\n",
+      "    B0 = 0.083 - 0.422/(Tr**(1.6));\n",
+      "    B1 = 0.139 - 0.172/(Tr**(4.2));\n",
+      "    \t\t\t# (B*Pc)/(R*Tc) = B0 + w*B1\n",
+      "    B = ((B0 + w*B1)*(R*Tc))/Pc;\t\t\t# [m**(3)/mol] - Second virial coefficient at state 2\n",
+      "    dB0_dT = 0.422*1.6*Tc**(1.6)*T**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "    dB1_dT = 0.172*4.2*Tc**(4.2)*T**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "    dB_dT = ((R*Tc)/Pc)*((dB0_dT) + w*(dB1_dT));\t\t\t# (dB/dT)_1\n",
+      "\n",
+      "    T_exit = fsolve(f,900)\n",
+      "    fault=abs(T-T_exit);\n",
+      "    T = T + 0.2;\n",
+      "\n",
+      "Texit = T;\n",
+      "\n",
+      "\n",
+      "W_2 = - delta_H_a;\t\t\t# [J/mol]\n",
+      "\n",
+      "print \" 2).The exit temperature is %f K\"%(Texit);\n",
+      "print \"     The required amount of work is %f J/mol\"%(W_2);\n",
+      "\n",
+      "\n",
+      "def f20(T): \n",
+      "\t return Cp_0/T\n",
+      "\n",
+      "T_prime = 700.;\t\t\t#[K]\n",
+      "fault1=10.;\n",
+      "\n",
+      "def f1(T_out): \n",
+      "\t return  32.2168*math.log(T_out) + 0.1922*10**(-2)*T_out + 0.5274*10**(-5) \\\n",
+      "\t *T_2**(2) - 1.1976*10**(-9)*T_out**(3)-8*10**(6)*dB_dT_prime -216.64\n",
+      "\n",
+      "while(fault1>0.5):\n",
+      "    Tr_prime = T_prime/Tc;\n",
+      "    B0_prime = 0.083 - 0.422/(Tr_prime**(1.6));\n",
+      "    B1_prime = 0.139 - 0.172/(Tr_prime**(4.2));\n",
+      "    B_prime = ((B0_prime + w*B1_prime)*(R*Tc))/Pc;\t\t\t# [m**(3)/mol] - Second virial coefficient at state 2\n",
+      "    dB0_dT_prime = 0.422*1.6*Tc**(1.6)*T_prime**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "    dB1_dT_prime = 0.172*4.2*Tc**(4.2)*T_prime**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "    dB_dT_prime = ((R*Tc)/Pc)*((dB0_dT_prime) + w*(dB1_dT_prime));\t\t\t# (dB/dT)_1\n",
+      "    T_out = fsolve(f1,10)\n",
+      "    fault1=abs(T_prime-T_out);\n",
+      "    T_prime = T_prime + 0.5;\n",
+      "\n",
+      "T_out = T_prime;\n",
+      "\n",
+      "# Now we have to calculate enthalpy change as W = -delta_H\n",
+      "\n",
+      "def f21(T): \n",
+      "\t return (7.7 + 0.04594*10**(-2.)*T + 0.2521*10**(-5.)*T**(2.) - 0.8587*10.**(-9.)*T**(3.))*4.184\n",
+      "\n",
+      "delta_H_3 =  quad(f21,T_1,T_out)[0] + B_prime*P_2 - P_2*T_out*dB_dT_prime - B_1*P_1 + P_1*T_1*dB_dT_1\n",
+      "print T_1,T_out\n",
+      "\n",
+      "\n",
+      "W_3 = - delta_H_3;\t\t\t# [J/mol]\n",
+      "\n",
+      "print \" 3).The exit temperature is %f K\"%(T_out);\n",
+      "print \"     The required amount of work is %f J/mol\"%(W_3);\n",
+      "\n",
+      "#(4)\n",
+      "n = 0.8;\t\t\t# Adiabatic efficiency\n",
+      "delta_H_a_4 = delta_H_3/n;\t\t\t#[J/mol]\n",
+      "W_4 = -delta_H_a_4;\t\t\t#[J/mol]\n",
+      "\n",
+      "\n",
+      "T_prime1 = 700.;\t\t\t#[K]\n",
+      "fault2=10.;\n",
+      "\n",
+      "def f2(T_2): \n",
+      "\t return  7.7*4.184*(T_2-T_1) + ((0.04594*4.184*10**(-2))/2)* \\\n",
+      "\t (T_2**(2)-T_1**(2)) + ((0.2521*4.184*10**(-5))/3)*(T_2**(3)-T_1**(3)) - \\\n",
+      "\t ((0.8587*4.184*10**(-9))/4)*(T_2**(4)-T_1**(4)) + B_prime1*8*10**(6) \\\n",
+      "\t - 8*10**(6)*T_2*dB_dT_prime1 + 191.7 + 496.81 - delta_H_a_4\n",
+      "\n",
+      "while(fault2>0.5):\n",
+      "    Tr_prime1 = T_prime1/Tc;\n",
+      "    B0_prime1 = 0.083 - 0.422/(Tr_prime1**(1.6));\n",
+      "    B1_prime1 = 0.139 - 0.172/(Tr_prime1**(4.2));\n",
+      "    \t\t\t# (B*Pc)/(R*Tc) = B0 + w*B1\n",
+      "    B_prime1 = ((B0_prime1 + w*B1_prime1)*(R*Tc))/Pc;\t\t\t# [m**(3)/mol] - Second virial coefficient at state 2\n",
+      "    dB0_dT_prime1 = 0.422*1.6*Tc**(1.6)*T_prime1**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "    dB1_dT_prime1 = 0.172*4.2*Tc**(4.2)*T_prime1**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "    dB_dT_prime1 = ((R*Tc)/Pc)*((dB0_dT_prime1) + w*(dB1_dT_prime1));\t\t\t# (dB/dT)_1\n",
+      "    T_out1 = fsolve(f2,100)\n",
+      "    fault2=abs(T_prime1-T_out1);\n",
+      "    T_prime1 = T_prime1 + 0.5;\n",
+      "\n",
+      "T_out1 = T_prime1;\n",
+      "\n",
+      "print \" 4).The exit temperature is %f K\"%(T_out1);\n",
+      "print \"     The required amount of work is %f J/mol\"%(W_4);\n",
+      "\n",
+      "\n",
+      "# answers are varies because of rounding error. please check it manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The exit temperature is 860.000000 K\n",
+        "     The required amount of work is -10667.724150 J/mol\n",
+        " 2).The exit temperature is 916.600000 K\n",
+        "     The required amount of work is -13334.655188 J/mol\n",
+        "473.15"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 755.5\n",
+        " 3).The exit temperature is 755.500000 K\n",
+        "     The required amount of work is -9282.701079 J/mol\n",
+        " 4).The exit temperature is 809.500000 K"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "     The required amount of work is -11603.376349 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.11  Page Number : 348"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "Vol = 0.15;\t\t\t    #[m**(3)] - Volume of the cylinder\n",
+      "P_1 = 100.;\t\t\t    #[bar] - Initial pressure\n",
+      "P_1 = P_1*10**(5);\t\t#[Pa]\n",
+      "T_1 = 170.;\t\t\t    #[K] - Initial temperature\n",
+      "n_withdrawn = 500.;\t\t#[mol] - Withdrawn number of moles\n",
+      "R = 8.314;\t\t    \t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#(1)\n",
+      "Y = 1.4;\t\t\t# Coefficient of adiabatic expansion\n",
+      "n_total = (P_1*Vol)/(R*T_1);\t\t\t#[mol] - Total number of moles\n",
+      "n_2 = n_total - n_withdrawn;\t\t\t#[mol] - Left number of moles\n",
+      "V_1 = Vol/n_total;\t\t\t#[m**(3)/mol] - Molar volume at initial state.\n",
+      "# At final state\n",
+      "V_2 = Vol/n_2;\t\t\t#[m**(3)/mol] - Molar volume at final state\n",
+      "\n",
+      "# During duscharging  P_1*V_1**(Y) = P_2*V_2**(Y), therefore\n",
+      "P_2_1 = P_1*((V_1/V_2)**(Y));\t\t\t#[Pa] - Final pressure\n",
+      "P_2_1 = P_2_1*10**(-5);\t\t\t#[bar]\n",
+      "T_2_1 = ((P_2_1*10**(5))*V_2)/R;\t\t\t#[K] - Final temperature\n",
+      "\n",
+      "print \" 1).The final temperature %f K\"%(T_2_1);\n",
+      "print \"     The final pressure %f bar\"%(P_2_1);\n",
+      "\n",
+      "\n",
+      "def f53(T): \n",
+      "\t return Cp_0/T\n",
+      "\n",
+      "T_prime = 150;\t\t\t#[K]\n",
+      "error = 10;\n",
+      "while(error>1):\n",
+      "    f_T = 18.886*math.log(T_prime) + 4.2*10**(-3)*T_prime - 92.4;\n",
+      "    f_T_dash = 18.886/T_prime + 4.2*10**(-3);\n",
+      "    T_new = T_prime - (f_T/f_T_dash);\n",
+      "    error=abs(T_prime - T_new);\n",
+      "    T_prime = T_new;\n",
+      "\n",
+      "T_2_2 = T_prime;\t\t\t#[K] - Final temperature\n",
+      "P_2_2 = ((n_2*R*T_2_2)/Vol)*10**(-5);\t\t\t#[bar] - Final pressure\n",
+      "\n",
+      "print \" 2).The final temperature %f K\"%(T_2_2);\n",
+      "print \"     The final pressure %f bar\"%(P_2_2);\n",
+      "\n",
+      "#(3)\n",
+      "Tc = 126.2;\t\t\t#[K] - Critical temperature of nitrogen\n",
+      "Pc = 34.0;\t\t\t#[bar] - Critical pressure of nitrogen\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.038;\t\t\t# Acentric factor\n",
+      "\n",
+      "\n",
+      "dB0_dT = 0.422*1.6*Tc**(1.6)*T_1**(-2.6);\t\t\t# (dB_0/dT) at state 1\n",
+      "dB1_dT = 0.172*4.2*Tc**(4.2)*T_1**(-5.2);\t\t\t# (dB_1/dT) at state 1\n",
+      "dB_dT = ((R*Tc)/Pc)*((dB0_dT) + w*(dB1_dT));\t\t\t# (dB/dT) at state 1\n",
+      "# The residual entropy at the initial state is given by \n",
+      "S_R_1 = -P_1*(dB_dT);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# Now let us calculate molar volume at initial state\n",
+      "Tr = T_1/Tc;\t\t\t# Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2));\n",
+      "\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B = ((B_0+(w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "V_1_3 = B + (R*T_1)/P_1;\t\t\t#[m**(3)/mol]\n",
+      "# Therefore number of moles in the initial state is\n",
+      "n_1_3 = Vol/V_1_3;\t\t\t#[mol]\n",
+      "# Therefore final number of moles is\n",
+      "n_2_3 = n_1_3 - n_withdrawn;\n",
+      "\n",
+      "# Therefore molar volume at final state is\n",
+      "V_2_3 = Vol/n_2_3;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Now let us determine the relation between pressure and temperature in the final state\n",
+      "# P_2_3 = (R*T_2_3)/(V_2_3 - B_2)\n",
+      "#delta_S = 0, thus delta_S_ig + delta_S_R = 0\n",
+      "delta_S_R = - S_R_1;\n",
+      "\n",
+      "def f54(T): \n",
+      "\t return Cp_0/T\n",
+      "\n",
+      "\n",
+      "T_2_prime = 135;\t\t\t#[K]\n",
+      "delta = 0.1;\n",
+      "error = 10;\n",
+      "while(error>0.01):\n",
+      "    T_r = T_2_prime/Tc;\t\t\t# Reduced temperature\n",
+      "    B_0_3 = 0.083-(0.422/(T_r)**(1.6));\n",
+      "    B_1_3 = 0.139-(0.172/(T_r)**(4.2));\n",
+      "    B_3 = ((B_0_3+(w*B_1_3))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "    dB0_dT_3 = 0.422*1.6*Tc**(1.6)*T_2_prime**(-2.6);\t\t\t# (dB_0/dT)\n",
+      "    dB1_dT_3 = 0.172*4.2*Tc**(4.2)*T_2_prime**(-5.2);\t\t\t# (dB_1/dT)\n",
+      "    dB_dT_3 = ((R*Tc)/Pc)*((dB0_dT_3) + w*(dB1_dT_3));\t\t\t# (dB/dT)\n",
+      "    P_2_3 = (R*T_2_prime)/(V_2_3 - B_3);\n",
+      "    delta_S = 27.2*(math.log(T_2_prime/T_1)) + 4.2*10**(-3)*(T_2_prime - T_1) - R*(math.log(P_2_3/P_1)) - P_2_3*(dB_dT_3) + delta_S_R;\n",
+      "    T_new = T_2_prime + delta;\n",
+      "    error=abs(delta_S);\n",
+      "    T_2_prime = T_new;\n",
+      "\n",
+      "T_2_3 = T_2_prime;\t\t\t#[K] - Final temperature\n",
+      "\t\t\t# Therefore at T_2_3\n",
+      "P_2_3 = P_2_3*10**(-5);\t\t\t#[bar] - Final pressure\n",
+      "\n",
+      "print \" 3).The final temperature %f K\"%(T_2_3);\n",
+      "print \"     The final pressure %f bar\"%(P_2_3);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The final temperature 131.761821 K\n",
+        "     The final pressure 40.991361 bar\n",
+        " 2).The final temperature 129.504151 K\n",
+        "     The final pressure 40.288995 bar\n",
+        " 3).The final temperature 141.300000 K\n",
+        "     The final pressure 57.079997 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.12  Page Number : 351"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P_1 = 80.;\t\t\t        #[bar] - Initial pressure\n",
+      "P_1 = P_1*10.**(5);\t\t\t#[Pa]\n",
+      "T_1 = 300. + 273.15;\t\t#[T] - Initial temperature\n",
+      "P_2 = 40.;\t\t\t        #[bar] - Final pressure\n",
+      "P_2 = P_2*10**(5);\t\t\t#[Pa]\n",
+      "T_2 = 300. + 273.15;\t\t#[K] - Final temperature\n",
+      "T_0 = 25. + 273.15;\t\t\t#[K] - Surrounding temperature\n",
+      "P_0 = 1.;\t\t\t        #[atm] - Surrounding pressure\n",
+      "P_0 = P_0*101325;\t\t\t#[Pa]\n",
+      "Tc = 647.1;\t\t\t        #[K]\n",
+      "Pc = 220.55;\t\t\t    #[bar]\n",
+      "Pc = Pc*10.**(5);\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t        #[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "# For van der Walls equation of state\n",
+      "a = (27*R**(2)*Tc**(2))/(64*Pc);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\n",
+      "def f(V): \n",
+      "\t return V**(3)-(b+(R*T_1)/P_1)*V**(2)+(a/P_1)*V-(a*b)/P_1\n",
+      "V_1_1=fsolve(f,0.1)\n",
+      "V_1_2=fsolve(f,10)\n",
+      "V_1_2=fsolve(f,100)\n",
+      "# The largest root is considered because of vapour\n",
+      "V_1 = V_1_1;\n",
+      "\n",
+      "U_R_1 = -a/V_1;\t\t\t#[J/mol] - Internal energy\n",
+      "H_R_1 = P_1*V_1 - R*T_1 - a/V_1;\t\t\t#[J/mol] - Enthalpy\n",
+      "S_R_1 = R*math.log((P_1*(V_1-b))/(R*T_1));\n",
+      "\n",
+      "def f1(V): \n",
+      "\t return V**(3)-(b+(R*T_2)/P_2)*V**(2)+(a/P_2)*V-(a*b)/P_2\n",
+      "V_2_1 = fsolve(f1,0.1)\n",
+      "V_2_2 = fsolve(f1,10)\n",
+      "V_2_3 = fsolve(f1,100)\n",
+      "V_2 = V_2_1;\n",
+      "\n",
+      "U_R_2 = -a/V_2;\t\t\t#[J/mol] - Internal energy\n",
+      "H_R_2 = P_2*V_2 - R*T_2 - a/V_2;\t\t\t#[J/mol] - Enthalpy\n",
+      "S_R_2 = R*math.log((P_2*(V_2-b))/(R*T_2));\n",
+      "\n",
+      "delta_U_R = U_R_2 - U_R_1;\t\t\t#\n",
+      "delta_H_R = H_R_2 - H_R_1;\t\t\t#\n",
+      "delta_S_R = S_R_2 - S_R_1;\t\t\t#\n",
+      "\n",
+      "delta_U_ig = 0;\t\t\t#[J/mol] - As temperature is constant\n",
+      "delta_H_ig = 0;\t\t\t#[J/mol] - As temperature is constant\n",
+      "delta_S_ig = - R*math.log(P_2/P_1);\t\t\t# [J/mol-K]\n",
+      "delta_U = delta_U_R + delta_U_ig;\t\t\t#[J/mol]\n",
+      "delta_H = delta_H_R + delta_H_ig;\t\t\t#[J/mol]\n",
+      "delta_S = delta_S_R + delta_S_ig;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "\t\t\t# Change in exergy is given by\n",
+      "\t\t\t# delta_phi = phi_1 - phi_2 = U_1 - U_2 + P_0*(V_1 - _V_2) - T_0*(S_1 - S_2)\n",
+      "delta_phi = - delta_U + P_0*(V_1 - V_2) - T_0*(-delta_S);\t\t\t#[J/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The change in internal energy is %f J/mol\"%(delta_U);\n",
+      "print \" The change in enthalpy is %f J/mol\"%(delta_H);\n",
+      "print \" The change in entropy is %f J/mol-K\"%(delta_S);\n",
+      "print \" The change in exergy is %f J/mol\"%(delta_phi);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in internal energy is 615.070900 J/mol\n",
+        " The change in enthalpy is 1053.220316 J/mol\n",
+        " The change in entropy is 6.930264 J/mol-K\n",
+        " The change in exergy is 1389.940735 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.13  Page Number : 353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T_1 = 500.;\t\t\t#[K] - Initial temperature\n",
+      "P_1 = 30.;\t\t\t#[atm] - Initial pressure\n",
+      "P_1 = P_1*101325;\t\t\t#[Pa]\n",
+      "P_2 = 1;\t\t\t#[atm] - Final pressure\n",
+      "P_2 = P_2*101325;\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\t\t\t# For chlorine\n",
+      "Tc = 417.2;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 77.10;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "\n",
+      "#Redlich Kwong equation of state,\n",
+      "a = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t# [Pa*m**(6)*K**(1/2)/mol]\n",
+      "b = (0.08664*R*Tc)/Pc;\t\t\t# [m**(3)/mol]\n",
+      "\n",
+      "# Calculations\n",
+      "def f1(V): \n",
+      "\t return V**(3)-((R*T_1)/P_1)*V**(2)-((b**(2))+((b*R*T_1)/P_1)-(a/(T_1**(1./2)*P_1)))*V-(a*b)/(T_1**(1./2)*P_1)\n",
+      "V_1=fsolve(f1,1)\n",
+      "V_2=fsolve(f1,10)\n",
+      "V_3=fsolve(f1,100)\n",
+      "\n",
+      "V = V_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\n",
+      "Z = (P_1*V_1)/(R*T_1);\t\t\t#compressibility factor\n",
+      "\n",
+      "# The residual enthalpy at state 1 is given by\n",
+      "H_R_1 = (Z-1)*R*T_1 + ((3*a)/(2*b*T_1**(1./2)))*(math.log(V/(V+b)));\t\t\t#[J/mol]\n",
+      " \n",
+      "\n",
+      "H_R_2 = 0;\t\t\t# Residual enthalpy at state 2\n",
+      "delta_H_R = H_R_2 - H_R_1;\t\t\t#[J/mol] - Residual enthalpy change\n",
+      "# and since isothermal conditions are maintained, therfore\n",
+      "delta_H_ig = 0;\t\t\t# Enthalpy change under ideal condition\n",
+      "delta_H = delta_H_R + delta_H_ig;\t\t\t#[J/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The change in enthalpy is given by delta_H = %f J/mol\"%(delta_H);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in enthalpy is given by delta_H = 1053.558471 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.14  Page Number : 353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "Vol_1 = 0.1;\t\t\t#[m**(3)] - Initial volume of each compartment\n",
+      "n_1 = 400;\t\t\t#[mol] - Initial number of moles in compartment 1\n",
+      "V_1 = Vol_1/n_1;\t\t\t#[m**(3)/mol] - Molar volume at state 1\n",
+      "T_1 = 294;\t\t\t#[K]\n",
+      "Vol_2 = 0.2;\t\t\t#[m**(3)] - Final volume of the compartment after removing the partition.\n",
+      "n_2 = n_1;\t\t\t#[mol] - Number of moles remains the same\n",
+      "V_2 = Vol_2/n_2;\t\t\t#[m**(3)/mol] - Molar volume at state 2\n",
+      "\n",
+      "a = 0.1362;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = 3.215*10**(-5);\t\t\t#[m**(3)/mol]\n",
+      "Cv_0 = 12.56;\t\t\t#[J/mol-K] - Heat capacity in ideal gas state\n",
+      "\n",
+      "# Calculations\n",
+      "# For overall system q = 0, and no work is done, therefore delta_U = 0\n",
+      "# Therfore from the relation proved in part (1), we have\n",
+      "T_2 = T_1 + (a/Cv_0)*(1/V_2 - 1/V_1);\t\t\t#[K]\n",
+      "\n",
+      "# Results\n",
+      "print \" 2).The final temperatutre is %f K\"%(T_2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 2).The final temperatutre is 272.312102 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.17  Page Number : 356"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P_1 = 1*10**(6);\t\t\t#[Pa] - Initial pressure\n",
+      "T_1 = 200 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 8*10**(6);\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "Y = 1.4;\t\t\t# Index of expansion\n",
+      "Cp_0 = 29.114;\t\t\t#[J/mol-K]\n",
+      "\t\n",
+      "a = 0.55366;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = 3.049*10**(-5);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Calculations\n",
+      "V_1 = 3.816*10**(-3);\t\t\t#[m**(3)/mol]\n",
+      "Z_1 = (P_1*V_1)/(R*T_1);\n",
+      "\n",
+      "T_2 = T_1*(P_2/P_1)**((Y-1)/Y);\t\t\t#[K]\n",
+      "\n",
+      "# At 8 MPa and T_2,\n",
+      "# The molar volume of steam following van der Walls equation of state (as reported in the book) is\n",
+      "V_2 = 8.41*10**(-4);\t\t\t#[m**(3)/mol]\n",
+      "# And the compressibility factor is \n",
+      "Z_2 = (P_2*V_2)/(R*T_2);\n",
+      "\n",
+      "# For van der Walls equation of state we know that\n",
+      "# delta_S_R/R = math.log(Z_2/Z_1) + math.log((V_2 - b)/V_2) - math.log((V_1 - b)/V_1)\n",
+      "delta_S_R = R*(math.log(Z_2/Z_1) + math.log((V_2 - b)/V_2) - math.log((V_1 - b)/V_1));\t\t\t#[J/mol]\n",
+      "\n",
+      "# delta_S_ig = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1)\n",
+      "# The entropy change is therefore\n",
+      "# delta_S = delta_S_ig + delta_S_R\n",
+      "\n",
+      "\n",
+      "# Let us assume a temperature, say T = 870 K\n",
+      "# At 870 K the molar volume of steam following van der Walls equation of state (as reported in the book) is\n",
+      "#  V_3 = 8.57*10**(-4);\t\t\t# [m**(3)/mol]\n",
+      "# Therefore\n",
+      "# Z_3 = (P_2*V_3)/(R*T_2);\n",
+      "# At this temperature,\n",
+      "# delta_S = Cp_0*math.log(T/T_1) - R*math.log(P_2/P_1) + R*(math.log(Z/Z_1) + R*math.log((V - b)/V) - R*math.log((V_1 - b)/V_1))\n",
+      "\n",
+      "T = 800;\t\t\t#[K]\n",
+      "fault=10;\n",
+      "\n",
+      "def f1(V): \n",
+      "\t return V**(3)-(b+(R*T)/P_2)*V**(2)+(a/P_2)*V-(a*b)/P_2\n",
+      "\t \n",
+      "def f2(T): \n",
+      "\t return  Cp_0*math.log(T/T_1) - R*math.log(P_2/P_1) + R*(math.log(Z/Z_1) + R*(math.log((V - b)/V)) - R*(math.log((V_1 - b)/V_1)))\n",
+      "\n",
+      "\n",
+      "while(fault>0.3):\n",
+      "    \t\t\t# At T and 8 MPa\n",
+      "    V = fsolve(f1,1)\n",
+      "    Z = (P_2*V)/(R*T);\n",
+      "    \n",
+      "    T_exit = fsolve(f2,0.1)\n",
+      "    fault=abs(T-T_exit);\n",
+      "    T = T + 0.5;\n",
+      "\n",
+      "Texit = T;\n",
+      "\n",
+      "\n",
+      "# W = - delta_H\n",
+      "\n",
+      "# For van der Walls gas the enthalpy change is given by\n",
+      "delta_H_s = Cp_0*(T_exit - T_1) + (Z - 1)*R*T_exit - a/V - (Z_1-1)*R*T_1 + a/V_1;\t\t\t#[J/mol]\n",
+      "W = - delta_H_s;\t\t\t#[J/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" 1).The exit temperature is %f K\"%(Texit);\n",
+      "print \"     The work required is given by W = %f J/mol\"%(W);\n",
+      "\n",
+      "#(2)\n",
+      "eff = 0.8;\t\t\t# Adiabatic efficiency\n",
+      "delta_H_a = eff*delta_H_s;\t\t\t#[J/mol] - Actual enthalpy change\n",
+      "W_2 = - delta_H_a;\n",
+      "\n",
+      "\t\t\t# Let us assume a temperature, say\n",
+      "T_prime= 900;\t\t\t#[K]\n",
+      "fault1=10;\n",
+      "def f22(V): \n",
+      "\t return V**(3)-(b+(R*T_prime)/P_2)*V**(2)+(a/P_2)*V-(a*b)/P_2\n",
+      "\n",
+      "def f3(T_prime): \n",
+      "\t return  Cp_0*(T_prime - T_1) + (Z_prime - 1)*R*T_prime - a/V_prime - 13230.49\n",
+      "\t \n",
+      "while(fault1>0.3):\n",
+      "    \t\t\t# At T_prime and 8 MPa\n",
+      "    V_prime=fsolve(f22,1)\n",
+      "    Z_prime = (P_2*V_prime)/(R*T_prime);\n",
+      "    T_exit1 = fsolve(f3,100)\n",
+      "    fault1=abs(T_prime-T_exit1);\n",
+      "    T_prime = T_prime + 0.2;\n",
+      "\n",
+      "Texit1 = T_prime;\n",
+      "\n",
+      "print \" 2).The exit temperature is %f K\"%(Texit1);\n",
+      "print \"     The work required is given by W = %f J/mol\"%(W_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The exit temperature is 916.500000 K\n",
+        "     The work required is given by W = -12195.093996 J/mol\n",
+        " 2).The exit temperature is 958.400000 K"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "     The work required is given by W = -9756.075197 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.19  Page Number : 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 100 + 273.15;\t\t\t#[K] - Temperature\n",
+      "Tc = 647.1;\t\t\t#[K] - Critical temperature of water\n",
+      "Pc = 220.55;\t\t\t#[bar] - Critical pressure of water\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "a = (27*R**(2)*Tc**(2))/(64*Pc);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The cubic form of van der Walls equation of state is given by,\n",
+      "# V**(3) - (b + (R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0\n",
+      "\n",
+      "# For water vapour at 100 C under saturated conditions pressure is 1 atm, therefore\n",
+      "P = 1;\t\t\t#[atm]\n",
+      "P = P*101325;\t\t\t#[Pa]\n",
+      "\n",
+      "# At 100 C and 1 atm \n",
+      "def f(V): \n",
+      "\t return V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P\n",
+      "V_1 = fsolve(f,0.1)\n",
+      "V_1 = fsolve(f,10)\n",
+      "V_1 = fsolve(f,100)\n",
+      "V = V_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\n",
+      "f = P*(math.exp(math.log((R*T)/(P*(V-b))) + b/(V-b) - (2*a)/(R*T*V)));\t\t\t#[Pa]\n",
+      "f = f/101325;\t\t\t#[atm]\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume is %f m**3)/mol\"%(V);\n",
+      "print \" The fugacity is %f atm\"%(f);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume is 0.030469 m**3)/mol\n",
+        " The fugacity is 0.995168 atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.20  Page Number : 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 6;\t\t\t#[bar] - Initial pressure\n",
+      "P_1 = P_1*10**(5);\t\t\t#[Pa]\n",
+      "T_1 = 100 + 273.15;\t\t\t#[T] - Initial temperature\n",
+      "P_2 = 12;\t\t\t#[bar] - Final pressure\n",
+      "P_2 = P_2*10**(5);\t\t\t#[Pa]\n",
+      "T_2 = 500 + 273.15;\t\t\t#[K] - Final temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "Y = 1.126;\t\t\t# Index of expansion\n",
+      "Cp_0 = (R*Y)/(Y-1);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# For propane\n",
+      "Tc = 369.8;\t\t\t#[K]\n",
+      "Pc = 42.48;\t\t\t#[bar]\n",
+      "Pc = Pc*10**(5);\n",
+      "w = 0.152;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#(1)\n",
+      "# For van der Walls equation of state\n",
+      "a = (27*R**(2)*Tc**(2))/(64*Pc);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The cubic form of van der Walls equation of state is given by,\n",
+      "# V**(3) - (b + (R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0\n",
+      "\n",
+      "# At state 1 (100 C and 6 bar) \n",
+      "def f(V): \n",
+      "\t return V**(3)-(b+(R*T_1)/P_1)*V**(2)+(a/P_1)*V-(a*b)/P_1\n",
+      "V_1_1 = fsolve(f,1)\n",
+      "V_1_2 = fsolve(f,10)\n",
+      "V_1_3 = fsolve(f,100)\n",
+      "\n",
+      "V_1 = V_1_1;\t\t\t#[m**(3)/mol]\n",
+      "# Thus compressibility factor is\n",
+      "Z_1 = (P_1*V_1)/(R*T_1);\t\t\t#compressibility factor\n",
+      "\n",
+      "H_R_1 = (Z_1 - 1)*R*T_1 - (a/V_1);\t\t\t# [J/mol]\n",
+      "S_R_1 = R*math.log((P_1*(V_1-b))/(R*T_1));\t\t\t# [J/mol-K]\n",
+      "\n",
+      "# At state 2 (500 C and 12 bar) \n",
+      "def f1(V): \n",
+      "    return V**(3)-(b+(R*T_2)/P_2)*V**(2)+(a/P_2)*V-(a*b)/P_2\n",
+      "V_2_1 = fsolve(f1,1)\n",
+      "V_2_2 = fsolve(f1,10)\n",
+      "V_2_3 = fsolve(f1,100)\n",
+      "# The largest root is considered because of molar volume of vapour phase is to determined\n",
+      "V_2 = V_2_1;\t\t\t#[m**(3)/mol]\n",
+      "# Thus compressibility factor is\n",
+      "Z_2 = (P_2*V_2)/(R*T_2);\t\t\t#compressibility factor\n",
+      "\n",
+      "H_R_2 = (Z_2 - 1)*R*T_2 - (a/V_2);\t\t\t# [J/mol]\n",
+      "S_R_2 = R*math.log((P_2*(V_2-b))/(R*T_2));\t\t\t# [J/mol-K]\n",
+      "\n",
+      "# Ideal gas entropy change is given by\n",
+      "delta_S_ig = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1);\t\t\t#[J/mol-K]\n",
+      "# Entropy change is given by\n",
+      "delta_S = delta_S_ig + (S_R_2 - S_R_1);\t\t\t#[J/mol-k]\n",
+      "\n",
+      "# Ideal gas enthalpy change is given by\n",
+      "delta_H_ig = Cp_0*(T_2 - T_1);\t\t\t#[J/mol]\n",
+      "# Enthalpy change is given by\n",
+      "delta_H = delta_H_ig + (H_R_2 - H_R_1);\t\t\t#[J/mol]\n",
+      "\n",
+      "print \"1).The change in enthalpy is %f J/mol\"%(delta_H);\n",
+      "print \"    The change in entropy is %f J/mol-K\"%(delta_S);\n",
+      "\n",
+      "#(2)\n",
+      "# Virial equation of state\n",
+      "\n",
+      "# At state 1 (372.15 K, 6 bar) let us calculate B and dB/dT\n",
+      "Tr = T_1/Tc;\t\t\t# Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2));\n",
+      "\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B = ((B_0+(w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "dB0_dT = 0.422*1.6*Tc**(1.6)*T_1**(-2.6);\t\t\t# (dB_0/dT) at state 1\n",
+      "dB1_dT = 0.172*4.2*Tc**(4.2)*T_1**(-5.2);\t\t\t# (dB_1/dT) at state 1\n",
+      "dB_dT = ((R*Tc)/Pc)*((dB0_dT) + w*(dB1_dT));\t\t\t# (dB/dT) at state 1\n",
+      "\n",
+      "H_R_1_2 = B*P_1 - P_1*T_1*dB_dT;\t\t\t#[J/mol] - Residual enthalpy at state 1\n",
+      "S_R_1_2 = -P_1*(dB_dT);\t\t\t#[J/mol-K] - Residual entropy at state 1\n",
+      "\n",
+      "# At state 2 (773.15 K, 12 bar)\n",
+      "Tr_2 = T_2/Tc;\t\t\t# Reduced temperature\n",
+      "B_0_2 = 0.083-(0.422/(Tr_2)**(1.6));\n",
+      "B_1_2 = 0.139-(0.172/(Tr_2)**(4.2));\n",
+      "\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B_2 = ((B_0_2+(w*B_1_2))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "dB0_dT_2 = 0.422*1.6*Tc**(1.6)*T_2**(-2.6);\t\t\t# (dB_0/dT) at state 1\n",
+      "dB1_dT_2 = 0.172*4.2*Tc**(4.2)*T_2**(-5.2);\t\t\t# (dB_1/dT) at state 1\n",
+      "dB_dT_2 = ((R*Tc)/Pc)*((dB0_dT_2) + w*(dB1_dT_2));\t\t\t# (dB/dT) at state 1\n",
+      "\n",
+      "H_R_2_2 = B_2*P_2 - P_2*T_2*dB_dT_2;\t\t\t#[J/mol] - Residual enthalpy at state 1\n",
+      "S_R_2_2 = -P_2*(dB_dT_2);\t\t\t#[J/mol-K] - Residual entropy at state 1\n",
+      "\n",
+      "delta_H_2 = delta_H_ig + (H_R_2_2 - H_R_1_2);\t\t\t#[J/mol]\n",
+      "delta_S_2 = delta_S_ig + (S_R_2_2 - S_R_1_2);\t\t\t#[J/mol]\n",
+      "\n",
+      "print \"2).The change in enthalpy is %f J/mol\"%(delta_H_2);\n",
+      "print \"    The change in entropy is %f J/mol-K\"%(delta_S_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1).The change in enthalpy is 29798.211799 J/mol\n",
+        "    The change in entropy is 48.649067 J/mol-K\n",
+        "2).The change in enthalpy is 29992.807365 J/mol\n",
+        "    The change in entropy is 49.021724 J/mol-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.21  Page Number : 362"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P = 2.76*10**(6);\t\t\t#[N/m**(2)] - Pressure\n",
+      "T = 310.93;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# For n-butane\n",
+      "Tc = 425.18;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 37.97;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.193;\n",
+      "den = 0.61;\t\t\t#[g/cm**(3)]\n",
+      "mol_wt = 58;\t\t\t#[g/mol] - Molecular weight of butane\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# math.log(P_sat) = 15.7374 - 2151.63/(T-36.24)\n",
+      "P_sat = math.exp(15.7374 - 2151.63/(T-36.24));\t\t\t#[mm Hg]\n",
+      "P_sat = (P_sat/760)*101325;\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "#(1)\n",
+      "# Let us determine the second virial coefficient at 310.93 K\n",
+      "Tr = T/Tc;\t\t\t# Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0 + (w*B_1)\n",
+      "B = ((B_0+(w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Fugacity under saturated conditions is given by\n",
+      "# math.log(f_sat/P_sat) = (B*P_sat)/(R*T)\n",
+      "f_sat = P_sat*(math.exp((B*P_sat)/(R*T)));\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "# The molar volume is given by\n",
+      "V_liq = (1/(den*1000))*(mol_wt/1000);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "f = f_sat*math.exp(V_liq*(P-P_sat)/(R*T));\n",
+      "\n",
+      "print \" 1).The fugacity of n-butane is %e N/m**2)\"%(f);\n",
+      "\n",
+      "#(2)\n",
+      "# For van der Walls equation of state\n",
+      "a = (27*R**(2)*Tc**(2))/(64*Pc);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The cubic form of van der Walls equation of state is given by,\n",
+      "# V**(3) - (b + (R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0\n",
+      "\n",
+      "# At 100 C and 1 atm \n",
+      "def f(V): \n",
+      "    return V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P\n",
+      "V_1 = fsolve(f,0.1)\n",
+      "V_1 = fsolve(f,10)\n",
+      "V_1 = fsolve(f,100)\n",
+      "# The above equation has only 1 real root, other two roots are imaginary\n",
+      "V = V_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# math.log(f/P) = math.log((R*T)/(P*(V-b))) + b/(V-b) -(2*a)/(R*T*V)\n",
+      "f_2 = P*(math.exp(math.log((R*T)/(P*(V-b))) + b/(V-b) -(2*a)/(R*T*V)));\n",
+      "\n",
+      "print \" 2).The fugacity of n-butane is %e N/m**2)\"%(f_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The fugacity of n-butane is 3.293476e+05 N/m**2)\n",
+        " 2).The fugacity of n-butane is 8.997352e+05 N/m**2)\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.22  Page Number : 363"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 50+273.15;\t\t\t#[K] - Temperature\n",
+      "P = 25.*10**(3);\t\t\t#[Pa] - Pressure\n",
+      "y1 = 0.5;\t\t\t#[mol] - mole fraction of equimolar mixture\n",
+      "y2 = 0.5;\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "#For component 1 (methyl ethyl ketone)\n",
+      "Tc_1  = 535.5;\t\t\t#[K] - Critical temperature\n",
+      "Pc_1 = 41.5*10**(5);\t\t\t#[N/m**(2)] - Critical pressure\n",
+      "Vc_1 = 267.;\t\t\t#[cm**(3)/mol] - Critical volume\n",
+      "Zc_1 = 0.249;\t\t\t# Critical compressibility factor\n",
+      "w_1 = 0.323;\t\t\t# acentric factor\n",
+      "\n",
+      "#For component 2 (toluene)\n",
+      "Tc_2 = 591.8;\t\t\t#[K]\n",
+      "Pc_2 = 41.06*10**(5);\t\t\t#[N/m**(2)]\n",
+      "Vc_2 = 316.;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_2 = 0.264;\n",
+      "w_2 = 0.262;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# For equation of state Z = 1 + B/V\n",
+      "#For component 1, let us calculate B and dB/dT\n",
+      "Tr_1 = T/Tc_1;\t\t\t#Reduced temperature\n",
+      "#At reduced temperature\n",
+      "B1_0 = 0.083-(0.422/(Tr_1)**(1.6));\n",
+      "B1_1 = 0.139-(0.172/(Tr_1)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B_11 = ((B1_0+(w_1*B1_1))*(R*Tc_1))/Pc_1;\t\t\t# [m**(3)/mol-K] \n",
+      "dB0_dT_1 = 0.422*1.6*Tc_1**(1.6)*T**(-2.6);\t\t\t# [m**(3)/mol-K] -  (dB_0/dT)\n",
+      "dB1_dT_1 = 0.172*4.2*Tc_1**(4.2)*T**(-5.2);\t\t\t# [m**(3)/mol-K] -  (dB_1/dT)\n",
+      "dB_dT_1 = ((R*Tc_1)/Pc_1)*((dB0_dT_1) + w_1*(dB1_dT_1));\t\t\t#[m**(3)/mol-K] -  (dB/dT)_\n",
+      "\n",
+      "#Similarly for component 2\n",
+      "Tr_2 = T/Tc_2;\t\t\t#Reduced temperature\n",
+      "#At reduced temperature Tr_2,\n",
+      "B2_0 = 0.083 - (0.422/(Tr_2)**(1.6));\n",
+      "B2_1 = 0.139 - (0.172/(Tr_2)**(4.2));\n",
+      "B_22 = ((B2_0+(w_2*B2_1))*(R*Tc_2))/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "dB0_dT_2 = 0.422*1.6*Tc_2**(1.6)*T**(-2.6);\t\t\t# [m**(3)/mol-K] -  (dB_0/dT)\n",
+      "dB1_dT_2 = 0.172*4.2*Tc_2**(4.2)*T**(-5.2);\t\t\t# [m**(3)/mol-K] -  (dB_1/dT)\n",
+      "dB_dT_2 = ((R*Tc_2)/Pc_2)*((dB0_dT_2) + w_2*(dB1_dT_2));\t\t\t#[m**(3)/mol-K] -  (dB/dT)_\n",
+      "\n",
+      "#For cross coeffcient, let us calculate B and dB/dT\n",
+      "Tc_12 = (Tc_1*Tc_2)**(1./2);\t\t\t#[K]\n",
+      "w_12 = (w_1 + w_2)/2.;\n",
+      "Zc_12 = (Zc_1 + Zc_2)/2;\n",
+      "Vc_12 = (((Vc_1)**(1./3) + (Vc_2)**(1./3))/2)**(3);\t\t\t#[cm**(3)/mol]\n",
+      "Vc_12 = Vc_12*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "Pc_12 = (Zc_12*R*Tc_12)/Vc_12;\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "#Now we have,(B_12*Pc_12)/(R*Tc_12) = B_0+(w_12*B_1)\n",
+      "#where B_0 and B_1 are to be evaluated at Tr_12\n",
+      "Tr_12 = T/Tc_12;\n",
+      "#At reduced temperature Tr_12\n",
+      "B_0 = 0.083 - (0.422/(Tr_12)**(1.6));\n",
+      "B_1 = 0.139 - (0.172/(Tr_12)**(4.2));\n",
+      "B_12 = ((B_0 + (w_12*B_1))*R*Tc_12)/Pc_12;\t\t\t#[m**(3)/mol]\n",
+      "dB0_dT_12 = 0.422*1.6*Tc_12**(1.6)*T**(-2.6);\t\t\t# [m**(3)/mol-K] -  (dB_0/dT)\n",
+      "dB1_dT_12 = 0.172*4.2*Tc_12**(4.2)*T**(-5.2);\t\t\t# [m**(3)/mol-K] -  (dB_1/dT)\n",
+      "dB_dT_12 = ((R*Tc_12)/Pc_12)*((dB0_dT_12) + w_12*(dB1_dT_12));\t\t\t#[m**(3)/mol-K] -  (dB/dT)_12\n",
+      "\n",
+      "#For the mixture\n",
+      "B = y1**(2)*B_11 + 2*y1*y2*B_12 + y2**(2)*B_22;\t\t\t#[m**(3)/moL]\n",
+      "\n",
+      "# The equation of state can be written as\n",
+      "# V**(2) - ((R*T)/P) - (B*R*T)/P = 0\n",
+      "# V**(2) - 0.1075*V + 1.737*10**(-4) = 0\n",
+      "def f(V): \n",
+      "    return V**(2) - 0.1075*V + 1.737*10**(-4)\n",
+      "V1 = fsolve(f,0.1)\n",
+      "V2 = fsolve(f,1)\n",
+      "# We will consider the root which is near to R*T/P\n",
+      "V = V1;\n",
+      "# dB/dT = y_1**(2)*dB_11/dT + y_2**(2)*dB_22/dT + 2*y_1*y_2*dB_12/dT\n",
+      "dB_dT = y1**(2)*dB_dT_1 + y2**(2)*dB_dT_2 + 2*y1*y2*dB_dT_12;\t\t\t#[m**(3)/mol-K]\n",
+      "\n",
+      "# For equation of state Z = 1 + B/V\n",
+      "H_R = (B*R*T)/V - ((R*T**(2))/V)*dB_dT;\t\t\t#[J/mol]\n",
+      "\n",
+      "print \" 1).The value of H_R for the mixture using virial equation of state is %f J/mol\"%(H_R);\n",
+      "\n",
+      "#(2)\n",
+      "#  For van der Walls equation of state \n",
+      "a_11 = (27*R**(2)*Tc_1**(2))/(64*Pc_1);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "a_22 = (27*R**(2)*Tc_2**(2))/(64*Pc_2);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "a_12 = (a_11*a_22)**(1./2);\n",
+      "b_1 = (R*Tc_1)/(8*Pc_1);\t\t\t#[m**(3)/mol]\n",
+      "b_2 = (R*Tc_2)/(8*Pc_2);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# For the mixture\n",
+      "a = y1**(2)*a_11 + y2**(2)*a_22 + 2*y1*y2*a_12;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = y1*b_1 + y2*b_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# From the cubic form of van der Walls equation of state\n",
+      "def f1(V): \n",
+      "    return V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P\n",
+      "V2_1 = fsolve(f1,0.1)\n",
+      "V2_2 = fsolve(f1,10)\n",
+      "V2_3 = fsolve(f1,100)\n",
+      "# The largest root is considered\n",
+      "V_2 = V2_1;\n",
+      "\n",
+      "# The residual enthalpy is given by\n",
+      "H_R_2 = P*V_2 - R*T -a/V_2;\t\t\t#[J/mol]\n",
+      "\n",
+      "print \" 2).The value of H_R for the mixture using van der Walls equation of state is %f J/mol\"%(H_R_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The value of H_R for the mixture using virial equation of state is -151.289795 J/mol\n",
+        " 2).The value of H_R for the mixture using van der Walls equation of state is -38.476127 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.23  Page Number : 366"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 320 + 273.15;\t\t\t#[K]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# For water\n",
+      "Tc = 647.1;\t\t\t#[K]\n",
+      "Pc = 220.55;\t\t\t#[bar]\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "\n",
+      "# The cubic form of Redlich Kwong equation of state is given by,\n",
+      "# V**(3) - ((R*T)/P)*V**(2) - ((b_1**(2)) + ((b_1*R*T)/P) - (a/(T**(1/2)*P))*V - (a*b)/(T**(1/2)*P) = 0\n",
+      "\n",
+      "# At 320 C and 70 bar pressure\n",
+      "P_1 = 70;\t\t\t#[bar]\n",
+      "P_1 = P_1*10**(5);\t\t\t#[Pa]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "a = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b = (0.08664*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "# Solving the cubic equation\n",
+      "def f1(V): \n",
+      "    return V**(3)-((R*T)/P_1)*V**(2)-((b**(2))+((b*R*T)/P_1)-(a/(T**(1./2)*P_1)))*V-(a*b)/(T**(1./2)*P_1)\n",
+      "V1=fsolve(f1,1)\n",
+      "V2=fsolve(f1,10)\n",
+      "V3=fsolve(f1,100)\n",
+      "# The largest root is considered because at 320 C and 70 bar vapour phase exists.\n",
+      "V_1 = V1;\t\t\t#[m**(3)/mol]\n",
+      "# Thus compressibility factor is\n",
+      "Z_1 = (P_1*V_1)/(R*T);\n",
+      "\n",
+      "# For Redlich-Kwong equation of state\n",
+      "# math.log(f/P) = Z - 1 - math.log(V_1/(V_1-b)) + (a/(b*R*(T**(3/2))))*math.log(V/(V+b))\n",
+      "f_1 = P_1*(math.exp(Z_1-1-math.log(Z_1)+math.log(V_1/(V_1-b))+(a/(b*R*(T**(3./2))))*math.log(V_1/(V_1+b))));\t\t\t#[Pa]\n",
+      "f_1 = f_1*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "print \" The fugacity of water vapour at 320 C and 70 bar pressure is %f bar\"%(f_1);\n",
+      "\n",
+      "# At 320 C and 170 bar pressure, we have\n",
+      "P_2 = 170;\t\t\t#[bar]\n",
+      "P_2 = P_2*10**(5);\t\t\t#[Pa]\n",
+      "\n",
+      "# Solving the cubic equation\n",
+      "def f2(V): \n",
+      "    return V**(3)-((R*T)/P_2)*V**(2)-((b**(2))+((b*R*T)/P_2)-(a/(T**(1./2)*P_2)))*V-(a*b)/(T**(1./2)*P_2)\n",
+      "V4 = fsolve(f2,1)\n",
+      "V5 = fsolve(f2,10)\n",
+      "V6 = fsolve(f2,100)\n",
+      "# The above equation has only 1 real root,other two roots are imaginary. Therefore,\n",
+      "V_2 = V6;\t\t\t#[m**(3)/mol]\n",
+      "# Thus compressibility factor is\n",
+      "Z_2 = (P_2*V_2)/(R*T);\n",
+      "\n",
+      "# For Redlich-Kwong equation of state\n",
+      "# math.log(f/P) = Z - 1 - math.log(V_1/(V_1-b)) + (a/(b*R*(T**(3/2))))*math.log(V/(V+b))\n",
+      "f_2 = P_2*(math.exp(Z_2-1-math.log(Z_2)+math.log(V_2/(V_2-b))+(a/(b*R*(T**(3./2))))*math.log(V_2/(V_2+b))));\t\t\t#[Pa]\n",
+      "f_2 = f_2*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "print \" The fugacity of water vapour at 320 C and 170 bar pressure is %f bar\"%(f_2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity of water vapour at 320 C and 70 bar pressure is 60.456239 bar\n",
+        " The fugacity of water vapour at 320 C and 170 bar pressure is 101.590989 bar\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "C:\\Anaconda\\lib\\site-packages\\scipy\\optimize\\minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last ten iterations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.24  Page Number : 367"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math\n",
+      "# Variables\n",
+      "Vol = 0.057;\t\t\t#[m**(3)] - Volume of car tyre\n",
+      "P_1 = 300.;\t\t\t#[kPa] - Initial pressure\n",
+      "P_1 = P_1*10**(3);\t\t\t#[Pa]\n",
+      "T_1 = 300.;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 330.;\t\t\t#[kPa] - Finnal pressure\n",
+      "P_2 = P_2*10**(3);\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "Cv_0 = 21.;\t\t\t#[J/mol-K] - Heat capacity for air \n",
+      "\n",
+      "# For oxygen\n",
+      "Tc_O2 = 154.6;\t\t\t#[K] - Critical temperature\n",
+      "Pc_O2 = 50.43;\t\t\t#[bar] - Critical pressure\n",
+      "Pc_O2 = Pc_O2*10**(5);\t\t\t#[Pa]\n",
+      "y1 = 0.21;\t\t\t# - Mole fraction of oxygen\n",
+      "# For nitrogen\n",
+      "Tc_N2 = 126.2;\t\t\t#[K] - Critical temperature\n",
+      "Pc_N2 = 34.00;\t\t\t#[bar] - Critical pressure\n",
+      "Pc_N2 = Pc_N2*10**(5);\t\t\t#[Pa]\n",
+      "y2 = 0.79;\t\t\t# - Mole fraction of nitrogen\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# (1)\n",
+      "# Assuming ideal gas behaviour. The volume remains the same,therefore,we get\n",
+      "# P_1/T_1 = P_2/T_2\n",
+      "T_2 = P_2*(T_1/P_1);\t\t\t#[K]\n",
+      "\n",
+      "n = (P_1*Vol)/(R*T_1);\t\t\t#[mol] - Number of moles\n",
+      "delta_U = n*Cv_0*(T_2-T_1);\t\t\t#[J]\n",
+      "\n",
+      "print \" 1).The change in internal energy for ideal gas behaviour) is %f J\"%(delta_U);\n",
+      "\n",
+      "#(2)\n",
+      "#  For van der Walls equation of state \n",
+      "a_O2 = (27*R**(2)*Tc_O2**(2))/(64*Pc_O2);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "a_N2 = (27*R**(2)*Tc_N2**(2))/(64*Pc_N2);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "a_12 = (a_O2*a_N2)**(1./2);\n",
+      "b_O2 = (R*Tc_O2)/(8*Pc_O2);\t\t\t#[m**(3)/mol]\n",
+      "b_N2 = (R*Tc_N2)/(8*Pc_N2);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# For the mixture\n",
+      "a = y1**(2)*a_O2 + y2**(2)*a_N2 + 2*y1*y2*a_12;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = y1*b_O2 + y2*b_N2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# From the cubic form of van der Walls equation of state\n",
+      "# At 300 K and 300 kPa,\n",
+      "def f1(V): \n",
+      "    return V**(3)-(b+(R*T_1)/P_1)*V**(2)+(a/P_1)*V-(a*b)/P_1\n",
+      "V_1 = fsolve(f1,0.1)\n",
+      "V_2 = fsolve(f1,10)\n",
+      "V_3 = fsolve(f1,100)\n",
+      "# The above equation has only 1 real root, other two roots are imaginary\n",
+      "V = V_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Now at P = 330 kPa and at molar volume V\n",
+      "# The van der Walls equation of state is\n",
+      "# (P + a/V**(2))*(V - b) = R*T\n",
+      "T_2_prime = ((P_2 + a/V**(2))*(V - b))/R;\t\t\t#[K]\n",
+      "n_prime = Vol/V;\t\t\t#[mol]\n",
+      "\n",
+      "# Total change in internal energy is given by\n",
+      "# delta_U_prime = n_prime*delta_U_ig + n_prime*(U_R_2 - U_R_1)\n",
+      "# delta_U_prime = n_prime*Cv_0*(T_2_prime - T_1) + n_prime*a(1/V_2 - 1/V_1);\n",
+      "# Since V_1 = V_2 = V, therefore\n",
+      "delta_U_prime = n_prime*Cv_0*(T_2_prime - T_1);\n",
+      "\n",
+      "print \" 2).The change in internal energy for van der Walls equation of state) is %f J\"%(delta_U_prime);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The change in internal energy for ideal gas behaviour) is 4319.220592 J\n",
+        " 2).The change in internal energy for van der Walls equation of state) is 4299.872282 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.25  Page Number : 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math\n",
+      "# Variables\n",
+      "T_1 = 150 + 273.15;\t\t\t#[K] - Initial emperature\n",
+      "T_2 = T_1;\t\t\t# Isothermal process\n",
+      "P_1 = 100.*10**(3);\t\t\t#[Pa] - Initial pressure\n",
+      "P_2 = 450.*10**(3);\t\t\t#[Pa] - Final pressure\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "# For water\n",
+      "Tc = 647.1;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 220.55;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10.**(5);\n",
+      "w = 0.345;\n",
+      "Mol_wt = 18.015;\t\t\t#[g/mol] - Molecular weight of water\n",
+      "Cp_0 = 4.18;\t\t\t#[J/mol-K] - Smath.tan(math.radiansard heat capacity of water\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "# In Peng-Robinson equation of state\n",
+      "m = 0.37464 + 1.54226*w - 0.26992*w**(2);\n",
+      "# At T_1 and P_1, we have\n",
+      "Tr = T_1/Tc;\n",
+      "alpha = (1 + m*(1 - Tr**(1./2)))**(2);\n",
+      "a = ((0.45724*(R*Tc)**(2))/Pc)*alpha;\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b = (0.07780*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Cubuc form of Peng-Robinson equation of stste is given by\n",
+      "# V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2))/P)-((a*b)/P)=0;\n",
+      "# Solving the cubic equation\n",
+      "def f(V): \n",
+      "    return V**(3)+(b-(R*T_1)/P_1)*V**(2)-((3*b**(2))+((2*R*T_1*b)/P_1)-(a/P_1))*V+b**(3)+((R*T_1*(b**(2)))/P_1)-((a*b)/P_1)\n",
+      "V1 = fsolve(f,-1)\n",
+      "V2 = fsolve(f,0)\n",
+      "V3 = fsolve(f,1)\n",
+      "#The largest root is for vapour phase,\n",
+      "#The largest root is only considered as the systemis gas\n",
+      "V_1 = V3;\t\t\t#[m**(3)/mol]\n",
+      "# Thus compressibility factor is\n",
+      "Z_1 = (P_1*V_1)/(R*T_1);\n",
+      "\n",
+      "# At T_2 and P_2, we have\n",
+      "# Cubuc form of Peng-Robinson equation of stste is given by\n",
+      "# V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2))/P)-((a*b)/P)=0;\n",
+      "# Solving the cubic equation\n",
+      "def f1(V): \n",
+      "    return V**(3)+(b-(R*T_2)/P_2)*V**(2)-((3*b**(2))+((2*R*T_2*b)/P_2)-(a/P_2))*V+b**(3)+((R*T_2*(b**(2)))/P_2)-((a*b)/P_2)\n",
+      "V4 = fsolve(f1,-1)\n",
+      "V5 = fsolve(f1,0)\n",
+      "V6 = fsolve(f1,1)\n",
+      "\n",
+      "V_2 = V6;\t\t\t#[m**(3)/mol]\n",
+      "# Thus compressibility factor is\n",
+      "Z_2 = (P_2*V_2)/(R*T_2);\n",
+      "\n",
+      "# In the Peng-Robinson equation of stste\n",
+      "# da/dT = -(a*m)/((alpha*T*Tc)**(1/2))\n",
+      "# The residual enthalpy is given by\n",
+      "# H_R = R*T*(Z-1) + (((T*(da_dT))-a)/(2*2**(1/2)*b))*math.log((Z+(1+2**(1/2)*((P*b)/(R*T))))/(Z+(1-2**(1/2)*((P*b)/(R*T)))))\n",
+      "\n",
+      "# At state 1\n",
+      "da_dT_1 = -(a*m)/((alpha*T_1*Tc)**(1./2));\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "H_R_1 = R*T_1*(Z_1-1) + (((T_1*(da_dT_1))-a)/(2*(2**(1./2))*b))* \\\n",
+      "math.log((Z_1+(1+2**(1./2))*((P_1*b)/(R*T_1)))/(Z_1+(1-2**(1./2))*((P_1*b)/(R*T_1))));\n",
+      "\n",
+      "# At state 2\n",
+      "da_dT_2 = -(a*m)/((alpha*T_2*Tc)**(1/2.));\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "H_R_2 = R*T_2*(Z_2-1) + (((T_2*(da_dT_2))-a)/(2*2**(1./2)* \\\n",
+      "b))*math.log((Z_2+(1+2**(1./2))*((P_2*b)/(R*T_1)))/(Z_2+(1-2**(1./2))*((P_2*b)/(R*T_1))));\n",
+      "\n",
+      "\n",
+      "delta_H = H_R_2 - H_R_1;\t\t\t#[J/mol]\n",
+      "delta_H = delta_H/Mol_wt;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# The residual entropy relation for a substance following Peng - Robinson equation of state ia\n",
+      "# S_R = R*math.log(Z - (P*b)/(R*T)) + (da_dT/(2*2**(1/2)*b))*math.log((Z+(1+2**(1/2))*((P*b)/(R*T)))/(Z+(1-2**(1/2))*((P*b)/(R*T))))\n",
+      "\n",
+      "# The residual entropy at state 1 is\n",
+      "S_R_1 = R*math.log(Z_1 - (P_1*b)/(R*T_1)) + (da_dT_1/(2*2**(1./2)*b))* \\\n",
+      "math.log((Z_1+(1+2**(1./2))*((P_1*b)/(R*T_1)))/(Z_1+(1-2**(1./2))*((P_1*b)/(R*T_1))));\n",
+      "\n",
+      "# The residual entropy at state 2 is\n",
+      "S_R_2 = R*math.log(Z_2 - (P_2*b)/(R*T_2)) + (da_dT_2/(2*2**(1./2)*b)) \\\n",
+      "*math.log((Z_2+(1+2**(1./2))*((P_2*b)/(R*T_2)))/(Z_2+(1-2**(1./2))*((P_2*b)/(R*T_2))));\n",
+      "\n",
+      "delta_S_R = S_R_2 - S_R_1;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# The ideal gas change in entropy is\n",
+      "delta_S_ig = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# Therefore\n",
+      "delta_S = delta_S_R + delta_S_ig;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy change is given by delta_H = %f kJ/mol\"%(delta_H);\n",
+      "print \" The entropy change is given by  delta_S = %f J/mol-K\"%(delta_S);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy change is given by delta_H = -11.747722 kJ/mol\n",
+        " The entropy change is given by  delta_S = -12.826050 J/mol-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.26  Page Number : 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math\n",
+      "# Variables\n",
+      "Vol = 0.15;\t\t\t#[m**(3)]\n",
+      "T_1 = 170;\t\t\t#[K] - Initial emperature\n",
+      "P_1 = 100;\t\t\t#[bar] - Initial pressure\n",
+      "P_1 = P_1*10**(5);\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "# For nitrogen\n",
+      "Tc = 126.2;\t\t\t#[K] - Critical tempeature\n",
+      "Pc = 34;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.038;\n",
+      "# Cp_0 = 27.2+4.2*10**(-3)*T\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#(1)\n",
+      "# For van der Walls equation of state\n",
+      "a = (27*R**(2)*Tc**(2))/(64*Pc);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The cubic form of van der Walls equation of state is given by,\n",
+      "# V**(3) - (b + (R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0\n",
+      "# On simplification the equation changes to\n",
+      "# V**(3) - 1.799*10**(4)*V**(2) + 1.366*10**(-8)*V - 5.269*10**(-13) = 0\n",
+      "\n",
+      "# Solving the cubic equation \n",
+      "def f(V): \n",
+      "    return V**(3)-1.799*10**(-4)*V**(2) + 1.366*10**(-8)*V - 5.269*10**(-13)\n",
+      "V1 = fsolve(f,1)\n",
+      "V2 = fsolve(f,10)\n",
+      "V3 = fsolve(f,100)\n",
+      "# The above equation has only 1 real root, other two roots are imagimnary\n",
+      "V_1 = V1;\t\t\t#[m**(3)/mol]\n",
+      "# Thus total number of moles is given by\n",
+      "n_1 = Vol/V_1;\t\t\t#[mol]\n",
+      "\n",
+      "# After 500 mol are withdrawn, the final number of moles is given by\n",
+      "n_2 = n_1 - 500;\t\t\t#[mol]\n",
+      "# Thus molar volume at final state is \n",
+      "V_2 = Vol/n_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The ideal entropy change is guven by\n",
+      "\n",
+      "def f24(T): \n",
+      "    return 27.2+4.2*10**(-3)*T\n",
+      "\n",
+      "# delta_S_ig =  quad(f24,T_1,T_2) - R*math.log(P_2/P_1)[0]\n",
+      "\n",
+      "# The residual entropy change is given by\n",
+      "# delta_S_R = R*math.log((P_2*(V_2-b))/(R*T_2)) - R*math.log((P_1*(V_1-b))/(R*T_1)) \n",
+      "# delta_S = delta_S_ig = delta_S_R\n",
+      "\n",
+      "def f25(T): \n",
+      "    return 27.2+4.2*10**(-3)*T\n",
+      "\n",
+      "# delta_S =  quad(f25,T_1,T_2) + R*math.log((V_2-b)/(V_1-b))[0]\n",
+      "\n",
+      "# During discharging delta_S = 0, thus on simplification we get\n",
+      "# 18.886*math.log(T_2) + 4.2*10**(-3)*T_2 - 92.937 = 0\n",
+      "# Solving the above equation we get\n",
+      "def f1(T_2): \n",
+      "    return 18.886*math.log(T_2) + 4.2*10**(-3)*T_2 - 92.937\n",
+      "T_2 = fsolve(f1,1)\n",
+      "\n",
+      "# Thus at T_2, \n",
+      "P_2 = (R*T_2)/(V_2-b) - a/V_2**(2);\t\t\t#[N/m**(2)]\n",
+      "P_2 = P_2*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "print \" 1).The final temperature is %f K\"%(T_2);\n",
+      "print \"     The final pressure is %f bar\"%(P_2);\n",
+      "\n",
+      "#(2)\n",
+      "# In Peng-Robinson equation of state\n",
+      "m = 0.37464 + 1.54226*w - 0.26992*w**(2);\n",
+      "# At T_1 and P_1, we have\n",
+      "Tr = T_1/Tc;\n",
+      "alpha = (1 + m*(1 - Tr**(1./2)))**(2);\n",
+      "a_2 = ((0.45724*(R*Tc)**(2))/Pc)*alpha;\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b_2 = (0.07780*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Cubuc form of Peng-Robinson equation of stste is given by\n",
+      "# V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2))/P)-((a*b)/P)=0;\n",
+      "# Solving the cubic equation\n",
+      "def f2(V): \n",
+      "    return V**(3)+(b_2-(R*T_1)/P_1)*V**(2)-((3*b_2**(2))+((2*R*T_1*b_2)/P_1)-(a_2/P_1))*V+b_2**(3)+((R*T_1*(b_2**(2)))/P_1)-((a_2*b_2)/P_1)\n",
+      "V4 = fsolve(f2,-1)\n",
+      "V5 = fsolve(f2,0)\n",
+      "V6 = fsolve(f2,0.006)\n",
+      "#The above equation has only 1 real root,the other two roots are imaginary\n",
+      "V_1_2 = V6;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The number of moles in the initial state is given by\n",
+      "n_1_2 = Vol/V_1_2;\t\t\t#[mol]\n",
+      "# After 500 mol are withdrawn, the final number of moles is given by\n",
+      "n_2_2 = n_1_2 - 500;\t\t\t#[mol]\n",
+      "# Thus molar volume at final state is \n",
+      "V_2_2 = Vol/n_2_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# At the final state the relation between pressure and temperature is\n",
+      "# P_2_2 = (R*T_2_2)/(V_2_2-b_2) - a_2/V_2_2**(2)\n",
+      "# P_2_2 = 7.23*10**(4)*T_2 - 3.93*10**(7)*a_2\n",
+      "\n",
+      "# Now let us calculate the residual entropy at initial state\n",
+      "Z_1 = (P_1*V_1_2)/(R*T_1);\n",
+      "da_dT_1 = -(a*m)/((alpha*T_1*Tc)**(1./2));\t\t\t#[Pa*m**(6)/mol**(2)] - da/dT\n",
+      "\n",
+      "# The residual entropy change for Peng-Robinson equatiob of state is given by\n",
+      "# S_R = R*math.log(Z-(P*b)/(R*T)) + (da_dT/(2*2**(1/2)*b))*math.log((V+(1+2**(1/2))*b))/((V+(1-2**(1/2)*b))));\n",
+      "S_R_1 = R*(math.log(Z_1-(P_1*b_2)/(R*T_1))) + (da_dT_1/(2*2**(1.2)*b_2))*(math.log((V_1_2+(1+2**(1./2))*b_2)/(V_1_2+(1-2**(1./2))*b_2)));\n",
+      "\n",
+      "# The total entropy change is given by\n",
+      "# delta_S = delta_S_ig + delta_S_R\n",
+      "\n",
+      "def f26(T): \n",
+      "    return 27.2+4.2*10**(-3)*T\n",
+      "\n",
+      "# where, delta_S_ig =  quad(f26,T_1,T_2_2) - R*math.log(P_2_2/P_1)[0]\n",
+      "\n",
+      "# and, P_2_2 = (R*T_2_2)/(V_2_2-b_2) - a_2/V_2_2**(2)\n",
+      "# On simplification we get\n",
+      "# delta_S = 27.2*math.log(T_2_2-T_1) + 4.2*10**(-3)*(T_2_2-T_1) - R*math.log(P_2_2/P_1) + R*math.log(Z_2-(P_2_2*b)/(R*T_2_2)) + 6226*(da_dT_2) + 9.22\n",
+      "\n",
+      "# Now we have the determine the value of T_2_2 such that delta_S = 0\n",
+      "# Starting with a temperature of 150 K\n",
+      "T_prime = 100.;\t\t\t#[K]\n",
+      "error = 10.;\n",
+      "while(error>0.1):\n",
+      "    Tr_prime = T_prime/Tc;\n",
+      "    alpha_prime = (1 + m*(1 - Tr_prime**(1./2)))**(2);\n",
+      "    a_prime = ((0.45724*(R*Tc)**(2))/Pc)*alpha_prime;\n",
+      "    P_prime = 7.23*10**(4)*T_prime - 3.93*10**(7)*a_prime;\n",
+      "    Z_prime = (P_prime*V_2_2)/(R*T_prime);\n",
+      "    da_dT_prime = -(a_prime*m)/((alpha_prime*T_prime*Tc)**(1./2));\n",
+      "    delta_S = 27.2*math.log(T_prime/T_1) + 4.2*10**(-3)*(T_prime-T_1) - R*math.log(P_prime/P_1) + R*math.log(Z_prime-((P_prime*b_2)/(R*T_prime))) + 6226*(da_dT_prime) + 9.22;\n",
+      "    error=abs(delta_S);\n",
+      "    T_prime = T_prime + 0.3;\n",
+      "\n",
+      "T_2_2 = T_prime;\t\t\t#[K] - Final temperature\n",
+      "P_2_2 = P_prime*10**(-5);\t\t\t#[bar] - Final pressure\n",
+      "\n",
+      "print \" 2).The final temperature is %f K\"%(T_2_2);\n",
+      "print \"     The final pressure is %f bar\"%(P_2_2);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The final temperature is 133.131844 K\n",
+        "     The final pressure is 39.636659 bar\n",
+        " 2).The final temperature is 134.200000 K\n",
+        "     The final pressure is 40.130674 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.27  Page Number : 374"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math\n",
+      "# Variables\n",
+      "T = 373.15;\t\t\t#[K]\n",
+      "Tc = 562.16;\t\t\t#[K]\n",
+      "Pc = 48.98;\t\t\t#[bar]\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t#[J/mol-K] - Universal gas constant\n",
+      "\n",
+      "# The cubic form of Redlich Kwong equation of state is given by,\n",
+      "# V**(3) - ((R*T)/P)*V**(2) - ((b_1**(2)) + ((b_1*R*T)/P) - (a/(T**(1/2)*P))*V - (a*b)/(T**(1/2)*P) = 0\n",
+      "\n",
+      "# Calculations\n",
+      "a = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b = (0.08664*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# At 373.15 K, let us assume the pressure to be 2.5 bar and under these conditions \n",
+      "P_1 = 2.5;\t\t\t#[bar]\n",
+      "P_1 = P_1*10**(5);\t\t\t#[bar]\n",
+      "\n",
+      "# Putting the values in Redlich Kwong equation of state, the equation becomes\n",
+      "# V**(3) - 0.0124*V**(2) + 8.326*10**(-6)*V - 7.74*10**(-10) = 0\n",
+      "# Solving the cubic equation\n",
+      "\n",
+      "def f(V): \n",
+      "    return V**(3) - 0.0124*V**(2) + 8.326*10**(-6)*V - 7.74*10**(-10)\n",
+      "V1=fsolve(f,-9)\n",
+      "V2=fsolve(f,10)\n",
+      "V3=fsolve(f,0.1)\n",
+      "# The largest root and the smallest root is considered for liquid phase and vapour phase respectively.\n",
+      "V_liq = V1;\t\t\t#[m**(3)/mol] - Molar volume in liquid phase\n",
+      "V_vap = V3;\t\t\t#[m**(3)/mol] - Molar volume in vapour phase\n",
+      "\n",
+      "# Let us calculate the fugacity of vapour phase\n",
+      "# math.log(f_vap/P) = b/(V-b) + math.log((R*T)/(P*(V-b))) - (a/(R*T**(1.5)))*(1/(V+b) - (1/b)*math.log(V/(V+b)))\n",
+      "f_vap = P_1*math.exp(b/(V_vap-b) + math.log((R*T)/(P_1*(V_vap-b))) - (a/(R*T**(1.5)))*(1/(V_vap+b) - (1/b)*math.log(V_vap/(V_vap+b))));\t\t\t#[Pa]\n",
+      "\n",
+      "# Let us calculate the fugacity of the liquid phase\n",
+      "f_liq = P_1*math.exp(b/(V_liq-b) + math.log((R*T)/(P_1*(V_liq-b))) - (a/(R*T**(1.5)))*(1/(V_liq+b) - (1/b)*math.log(V_liq/(V_liq+b))));\n",
+      "\n",
+      "\n",
+      "# The two fugacities are not same; therefore another pressure is to be assumed. The new pressure is\n",
+      "P_new = P_1*(f_liq/f_vap);\t\t\t#[Pa]\n",
+      "\n",
+      "# At P_new\n",
+      "def f1(V): \n",
+      "    return V**(3) - ((R*T)/P_new)*V**(2) - (b**(2) + ((b*R*T)/P_new) - a/(T**(1/2)*P_new))*V - (a*b)/(T**(1/2)*P_new)\n",
+      "V4=fsolve(f1,-9)\n",
+      "V5=fsolve(f1,10)\n",
+      "V6=fsolve(f1,0.1)\n",
+      "# The largest root and the smallest root is considered for liquid phase and vapour phase respectively.\n",
+      "V_liq_2 = V4;\t\t\t#[m**(3)/mol] - Molar volume in liquid phase\n",
+      "V_vap_2 = V6;\t\t\t#[m**(3)/mol] - Molar volume in vapour phase\n",
+      "\n",
+      "f_vap_prime = P_new*math.exp(b/(V_vap_2-b) + math.log((R*T)/(P_new*(V_vap_2-b))) - (a/(R*T**(1.5)))*(1/(V_vap_2+b) - (1/b)*math.log(V_vap_2/(V_vap_2+b))));\t\t\t#[Pa]\n",
+      "f_liq_prime = P_new*math.exp(b/(V_liq_2-b) + math.log((R*T)/(P_new*(V_liq_2-b))) - (a/(R*T**(1.5)))*(1/(V_liq_2+b) - (1/b)*math.log(V_liq_2/(V_liq_2+b))));\n",
+      "\n",
+      "# Since the fugacities of liquid and vapour phasesare almost same the assumed pressure may be taken as vapour pressure at 373.15 K\n",
+      "P_new = P_new*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "# Results\n",
+      "print \" The vapour pressure of benzene using Redlich Kwong equation of state is %f bar\"%(P_new);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The vapour pressure of benzene using Redlich Kwong equation of state is 2.666995 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.28  Page Number : 374"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math\n",
+      "# Variables\n",
+      "T = 150 + 273.15;\t\t\t#[K]\n",
+      "Tc = 647.1;\t\t\t#[K]\n",
+      "Pc = 220.55;\t\t\t#[bar]\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "w = 0.345;\n",
+      "R = 8.314;\t\t\t#[J/mol-K] - Universal gas constant\n",
+      "\n",
+      "# Let us assume a pressure of 100 kPa.\n",
+      "P_1 = 100.*10**(3);\t\t\t#[Pa]\n",
+      "\n",
+      "# Calculations\n",
+      "# At 100 kPa and 423.15 K, from Peng-Robinson equation of stste \n",
+      "m = 0.37464 + 1.54226*w - 0.26992*w**(2);\n",
+      "Tr = T/Tc;\n",
+      "alpha = (1 + m*(1 - Tr**(1./2)))**(2);\n",
+      "a = ((0.45724*(R*Tc)**(2))/Pc)*alpha;\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b = (0.07780*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "# Cubic form of Peng-Robinson equation of stste is given by\n",
+      "# V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2))/P)-((a*b)/P)=0;\n",
+      "# Solving the cubic equation\n",
+      "def f(V): \n",
+      "    return V**(3)+(b-(R*T)/P_1)*V**(2)-((3*b**(2))+((2*R*T*b)/P_1)-(a/P_1))*V+b**(3)+((R*T*(b**(2)))/P_1)-((a*b)/P_1)\n",
+      "V1 = fsolve(f,-1)\n",
+      "V2 = fsolve(f,0)\n",
+      "V3 = fsolve(f,1)\n",
+      "# The largest root and the smallest root is considered for liquid phase and vapour phase respectively.\n",
+      "V_liq = V1;\t\t\t#[m**(3)/mol] - Molar volume in liquid phase\n",
+      "V_vap = V3;\t\t\t#[m**(3)/mol] - Molar volume in vapour phase\n",
+      "\n",
+      "# The compressibility factor is given by\n",
+      "Z_vap = (P_1*V_vap)/(R*T);\t\t\t# For liquid phase\n",
+      "Z_liq = (P_1*V_liq)/(R*T);\t\t\t# For vapour phase\n",
+      "\n",
+      "# The math.expression for fugacity of Peng Robinson equation is\n",
+      "# math.log(f/P) = (Z-1) - math.log(Z-((P*b)/(R*T))) - (a/(2*2**(1/2)*b*R*T))*math.log((Z+(1+2**(1/2))*((P*b)/(R*T)))/((Z+(1-2**(1/2))*((P*b)/(R*T)))\n",
+      "# For vapour phase\n",
+      "f_P_vap = math.exp((Z_vap-1) - math.log(Z_vap-((P_1*b)/(R*T))) - (a/(2*2**(1./2)*b*R*T))*math.log((Z_vap+(1+2**(1./2))*((P_1*b)/(R*T)))/(Z_vap+(1-2**(1./2))*((P_1*b)/(R*T)))));\n",
+      "# For liquid phase\n",
+      "f_P_liq = math.exp((Z_liq-1) - math.log(Z_liq-((P_1*b)/(R*T))) - (a/(2*2**(1./2)*b*R*T))*math.log((Z_liq+(1+2**(1./2))*((P_1*b)/(R*T)))/(Z_liq+(1-2**(1./2))*((P_1*b)/(R*T)))));\n",
+      "\n",
+      "# Therefore f_liq/f_vap can be calculated as\n",
+      "fL_fV = (f_P_liq/f_P_vap);\n",
+      "\n",
+      "# The two values (f/P)_vap and (f/P)_vap are not same [ (f_P_liq/f_P_vap) >1 ]; therefore another pressure is to be assumed. The new pressure be\n",
+      "P_new = P_1*(f_P_liq/f_P_vap);\t\t\t#[Pa]\n",
+      "\n",
+      "# At P_new and 423.15 K, from Peng-Robinson equation of stste \n",
+      "\n",
+      "# V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2))/P)-((a*b)/P)=0;\n",
+      "# Solving the cubic equation\n",
+      "def f(V): \n",
+      "    return V**(3)+(b-(R*T)/P_new)*V**(2)-((3*b**(2))+((2*R*T*b)/P_new)-(a/P_new))*V+b**(3)+((R*T*(b**(2)))/P_new)-((a*b)/P_new)\n",
+      "V4 = fsolve(f,-1)\n",
+      "V5 = fsolve(f,0)\n",
+      "V6 = fsolve(f,1)\n",
+      "# The largest root and the smallest root is considered for liquid phase and vapour phase respectively.\n",
+      "V_liq_2 = V4;\t\t\t#[m**(3)/mol] - Molar volume in liquid phase\n",
+      "V_vap_2 = V6;\t\t\t#[m**(3)/mol] - Molar volume in vapour phase\n",
+      "\n",
+      "# The compressibility factor is given by\n",
+      "Z_vap_2 = (P_new*V_vap_2)/(R*T);\t\t\t# For liquid phase\n",
+      "Z_liq_2 = (P_new*V_liq_2)/(R*T);\t\t\t# For vapour phase\n",
+      "\n",
+      "# For vapour phase\n",
+      "f_P_vap_2 = math.exp((Z_vap_2-1) - math.log(Z_vap_2-((P_new*b)/(R*T))) - (a/(2*2**(1./2)*b*R*T))*math.log((Z_vap_2+(1+2**(1./2))*((P_new*b)/(R*T)))/(Z_vap_2+(1-2**(1./2))*((P_new*b)/(R*T)))));\n",
+      "# For liquid phase\n",
+      "f_P_liq_2 = math.exp((Z_liq_2-1) - math.log(Z_liq_2-((P_new*b)/(R*T))) - (a/(2*2**(1./2)*b*R*T))*math.log((Z_liq_2+(1+2**(1./2))*((P_new*b)/(R*T)))/(Z_liq_2+(1-2**(1./2))*((P_new*b)/(R*T)))));\n",
+      "\n",
+      "# Therefore f_liq/f_vap can be calculated as\n",
+      "fL_fV_2 = (f_P_liq_2/f_P_vap_2);\n",
+      "\n",
+      "# And new pressure is given by\n",
+      "P_new_prime = P_new*(f_P_liq_2/f_P_vap_2);\t\t\t#[Pa]\n",
+      "P_new_prime = P_new_prime*10**(-5);\n",
+      "\n",
+      "# Since the change in pressure is small, so we can take this to be the vapour pressure at 150 C\n",
+      "\n",
+      "# Results\n",
+      "print \" The vapour pressure of water using Peng-Robinson equation of stste  is %f bar\"%(P_new_prime);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The vapour pressure of water using Peng-Robinson equation of stste  is 4.675976 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch11_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch11_1.ipynb
new file mode 100755
index 00000000..0d56d5ca
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch11_1.ipynb
@@ -0,0 +1,320 @@
+{
+ "metadata": {
+  "name": "ch11_1",
+  "signature": "sha256:ea5fa2ce45062851dc4892fcc5b621c33d72fa0443d476b1ef4133dade9e39b0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Properties of a Component in a Mixture"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.1  Page Number : 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "Vol_total = 3;\t\t\t#[m**(3)] - Total volume of solution\n",
+      "x_ethanol = 0.6;\t\t\t#Mole fraction of ethanol\n",
+      "x_water = 0.4;\t\t\t#Mole fraction of water\n",
+      "\n",
+      "# Calculations\n",
+      "#The partial molar volumes of the components in the mixture are\n",
+      "V_ethanol_bar = 57.5*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "V_water_bar = 16*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "#The molar volumes of the pure components are\n",
+      "V_ethanol = 57.9*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "V_water = 18*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "#The molar volume of the solution is\n",
+      "V_sol = x_ethanol*V_ethanol_bar + x_water*V_water_bar;\t\t\t#[m**(3)/mol]\n",
+      "#Total number of moles can be calculated as \n",
+      "n_total = Vol_total/V_sol;\t\t\t#[mol]\n",
+      "\n",
+      "#Moles of the components are\n",
+      "n_ethanol = n_total*x_ethanol;\t\t\t#[mol]\n",
+      "n_water = n_total*x_water;\t\t\t#[mol]\n",
+      "\n",
+      "#Finally the volume of the pure components required can be calculated as\n",
+      "Vol_ethanol = V_ethanol*n_ethanol;\n",
+      "Vol_water = V_water*n_water;\n",
+      "\n",
+      "# Results\n",
+      "print \"Required volume of ethanol is %f cubic metre\"%(Vol_ethanol);\n",
+      "print \"Required volume of water is %f cubic metre\"%(Vol_water);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required volume of ethanol is 2.548166 cubic metre\n",
+        "Required volume of water is 0.528117 cubic metre\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.2  Page Number : 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Variables\n",
+      "T = 25+273.15;\t\t\t#[K] - Temperature\n",
+      "P = 1;\t\t\t#[atm]\n",
+      "#Component 1 = water\n",
+      "#component 2 = methanol\n",
+      "a = -3.2;\t\t\t#[cm**(3)/mol] - A constant\n",
+      "V2 = 40.7;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2 (methanol)\n",
+      "#V1_bar = 18.1 + a*x_2**(2)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#From Gibbs-Duhem equation at constant temperature and pressure we have\n",
+      "#x_1*dV1_bar + x_2*dV2_bar = 0\n",
+      "#dV2_bar = -(x_1/x_2)*dV1_bar = -(x_1/x_2)*a*2*x_2*dx_2 = -2*a*x_1*dx_2 = 2*a*x_1*dx_1\n",
+      "\n",
+      "#At x_1 = 0: x_2 = 1 and thus V2_bar = V2\n",
+      "#Integrating the above equation from x_1 = 0 to x_1 in the RHS, and from V2_bar = V2 to V2 in the LHS, we get\n",
+      "#V2_bar = V2 + a*x_1**(2) -  Molar volume of component 2(methanol) in the mixture \n",
+      "\n",
+      "print \"The expression for the partial molar volume of methanol2 isV2_bar = V2 + a*x_1**2 [cm**3/mol]\";\n",
+      "\n",
+      "#At infinite dilution, x_2 approach 0 and thus x_1 approach 1, therefore\n",
+      "x_1 = 1;\t\t\t# Mole fraction of component 1(water) at infinite dilution\n",
+      "V2_bar_infinite = V2 + a*(x_1**(2));\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "print \"The partial molar volume of methanol at infinite dilution is %f cm**3/mol\"%(V2_bar_infinite);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The expression for the partial molar volume of methanol2 isV2_bar = V2 + a*x_1**2 [cm**3/mol]\n",
+        "The partial molar volume of methanol at infinite dilution is 37.500000 cm**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.4  Page Number : 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "#H = a*x_1 + b*x_2 +c*x_1*x_2\n",
+      "\n",
+      "#The values of the constants are\n",
+      "a = 15000;\t\t\t#[J/mol]\n",
+      "b = 20000;\t\t\t#[J/mol]\n",
+      "c = -2000;\t\t\t#[J/mol]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#(1)\n",
+      "#Enthalpy of pure component 1 = H1 is obtained at x_2 = 0, thus \n",
+      "x_2 = 0;\n",
+      "x_1 = 1;\n",
+      "H1 = a*x_1 + b*x_2 +c*x_1*x_2;\t\t\t#[J/mol]\n",
+      "print \"a).The enthalpy of pure component 1 is %f J/mol\"%(H1);\n",
+      "\n",
+      "#Similarly for component 2,\n",
+      "#Enthalpy of pure component 2 = H2 is obtained at x_1 = 0, thus \n",
+      "x_1_prime = 0;\n",
+      "x_2_prime = 1;\n",
+      "H2 = a*x_1_prime + b*x_2_prime +c*x_1_prime*x_2_prime;\t\t\t#[J/mol]\n",
+      "print \"    The enthalpy of pure component 2 is %f J/mol\"%(H2);\n",
+      "\n",
+      "\n",
+      "#(c)\n",
+      "#From part (b), we have the relation\n",
+      "#H1_bar = a + c*(x_2**(2))\n",
+      "#H2_bar = b + c*(x_1**(2))\n",
+      "\n",
+      "#For enthalpy of component 1 at infinite dilution, x_1 approach 0 and thus x_2 approach 1, therefore\n",
+      "x_1_c = 0;\n",
+      "x_2_c = 1;\n",
+      "H1_infinite = a + c*(x_2_c**(2));\t\t\t#[cm**(3)/mol]\n",
+      "print \"C).The enthalpy of componenet 1 at infinite dilution at x_1 = 0) is %f J/mol\"%(H1_infinite);\n",
+      "\n",
+      "#At x_1 = 0.2\n",
+      "x_1_c1 = 0.2;\n",
+      "x_2_c1 = 0.8;\n",
+      "H1_bar_c1 = a + c*(x_2_c1**(2));\t\t\t#[J/mol]\n",
+      "print \"    The enthalpy of componenet 1 at at x_1 = 0.2) is %f J/mol\"%(H1_bar_c1);\n",
+      "\n",
+      "#At x_1 = 0.8\n",
+      "x_1_c2 = 0.8;\n",
+      "x_2_c2 = 0.2;\n",
+      "H1_bar_c2 = a + c*(x_2_c2**(2));\t\t\t#[J/mol]\n",
+      "print \"    The enthalpy of componenet 1 at at x_1 = 0.8) is %f J/mol\"%(H1_bar_c2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a).The enthalpy of pure component 1 is 15000.000000 J/mol\n",
+        "    The enthalpy of pure component 2 is 20000.000000 J/mol\n",
+        "C).The enthalpy of componenet 1 at infinite dilution at x_1 = 0) is 13000.000000 J/mol\n",
+        "    The enthalpy of componenet 1 at at x_1 = 0.2) is 13720.000000 J/mol\n",
+        "    The enthalpy of componenet 1 at at x_1 = 0.8) is 14920.000000 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.9  Page Number : 395"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Variables\n",
+      "n = 1*10**(3);\t\t\t#[mol] - No of moles\n",
+      "P = 0.1;\t\t\t#[MPa] - Pressure of the surrounding\n",
+      "T = 300;\t\t\t#[K] - Temperature of the surrounding\n",
+      "x_1 = 0.79;\t\t\t#Mole fraction of N2 in the air\n",
+      "x_2 = 0.21;\t\t\t#Mole fraction of O2 in the air\n",
+      "R=8.314;\t\t\t#[J/mol*K]\n",
+      "\n",
+      "# Calculations\n",
+      "#Change in availability when x_1 moles of component 1 goes from pure state to that in the mixture is\n",
+      "#x_1*(si_1 - si_2) = x_1*[H1 - H1_bar - T_0*(S1 - S1_bar)]\n",
+      "#Similarly change in availability of x_2 moles of component 2 is\n",
+      "#x_2*(si_1 - si_2) = x_2*[H2 - H2_bar - T_0*(S2 - S2_bar)]\n",
+      "\n",
+      "#and thus total availability change when 1 mol of mixture is formed from x_1 mol of component 1 and x_2 mol of component 2 is equal to reversible work\n",
+      "#W_rev = x_1*[H1 - H1_bar - T_0*(S1 - S1_bar)] + x_2*[H2 - H2_bar - T_0*(S2 - S2_bar)]\n",
+      "#W_rev =  -[delta_H_mix] +T_0*[delta_S_mix]\n",
+      "\n",
+      "#If T = T_0 that is,temperature of mixing is same as that of surroundings, W_rev = -delta_G_mix.\n",
+      "#W_rev = -delta_G_mix = R*T*(x_1*math.log(x_1) + x_2*math.log(x_2))\n",
+      "W_rev = R*T*(x_1*math.log(x_1) + x_2*math.log(x_2));\t\t\t#[J/mol]\n",
+      "\n",
+      "#Therefore total work transfer is given by\n",
+      "W_min = (n*W_rev)/1000;\t\t\t#[kJ]\n",
+      "\n",
+      "# Results\n",
+      "print \"The minimum work required is %f kJ\"%(W_min);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum work required is -1281.910728 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.10  Page Number : 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Variables\n",
+      "x_A = 0.20;\t\t\t# Mole fraction of A\n",
+      "x_B = 0.35;\t\t\t# Mole fraction of B\n",
+      "x_C = 0.45;\t\t\t# Mole fraction of C\n",
+      "\n",
+      "phi_A = 0.7;\t\t\t# Fugacity coefficient of A\n",
+      "phi_B = 0.6;\t\t\t# Fugacity coefficient of B\n",
+      "phi_C = 0.9;\t\t\t# Fugacity coefficient of C\n",
+      "\n",
+      "P = 6.08;\t\t\t#[MPa] - Pressure\n",
+      "T = 384;\t\t\t#[K] - Temperature\n",
+      "\n",
+      "# Calculations\n",
+      "#We know that\n",
+      "#math.log(phi) = x_1*math.log(phi_) + x_2*math.log(phi_2) + x_3*math.log(phi_3)\n",
+      "math.log_phi = x_A*math.log(phi_A) + x_B*math.log(phi_B) + x_C*math.log(phi_C);\t\t\t# Fugacity coefficient\n",
+      "phi = math.exp(math.log_phi);\n",
+      "\n",
+      "#Thus fugacity is given by,\n",
+      "f_mixture = phi*P;\t\t\t#[MPa]\n",
+      "\n",
+      "# Results\n",
+      "print \"The fugacity of the mixture is %f MPa\"%(f_mixture);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The fugacity of the mixture is 4.515286 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch12_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch12_1.ipynb
new file mode 100755
index 00000000..2d5ee204
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch12_1.ipynb
@@ -0,0 +1,503 @@
+{
+ "metadata": {
+  "name": "ch12_1",
+  "signature": "sha256:acb0234cca7ef04cf2e231ae1d04809101d746587fbd14cf847d456347c563ea"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Partial Molar Volume and Enthalpy from Experimental Data "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.1  Page Number : 419"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "import math \n",
+      "from numpy import *\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 0 + 273.15;\t\t\t#[K] - Temperature\n",
+      "P = 1;\t\t\t#[atm] - Pressure\n",
+      "x_methanol = 0.5;\t\t\t#Mole fraction of methanol at which molar volume is to be calculated\n",
+      "x_water = 0.5;\t\t\t#Mole fraction at which molar volume is to be calculated\n",
+      "\n",
+      "#V = V1 at x1 = 1 and V = V2 at x1 = 0, therefore\n",
+      "V1 = 40.7;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 1\n",
+      "V2 = 18.1;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2\n",
+      "from numpy import zeros,linalg\n",
+      "x1=[0.114,0.197,0.249,0.495,0.692,0.785,0.892];\t\t\t# Values of mole fraction of component 1\n",
+      "V=[20.3,21.9,23.0,28.3,32.9,35.2,37.9];\t\t\t# Values of molar volume\n",
+      "x2=zeros(7);\t\t\t# Mole fraction of component 2\n",
+      "x_V=zeros(7);\t\t\t# x_V = x1*V_1 + x2*V_2\n",
+      "V_mix=zeros(7);\t\t\t# V_mix = V - x1*V_1 - x2*V_2\n",
+      "del_V=zeros(7);\t\t\t#del_V = V_mix/(x1*x2)\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(7):\n",
+      "    x2[i]=1-x1[i];\n",
+      "    x_V[i]=x1[i]*V1 + x2[i]*V2;\n",
+      "    V_mix[i]=V[i]-x1[i]*V1- x2[i]*V2;\n",
+      "    del_V[i]=V_mix[i]/(x1[i]*x2[i]);\n",
+      "\n",
+      "x1 = array(x1)\n",
+      "\t\t\t#From the matrix method to solve simultaneous linear equations, we have\n",
+      "a=array([[7, sum(x1), sum(x1**2)],[sum(x1), sum(x1**2), sum(x1**3)],[sum(x1**2), sum(x1**3), sum(x1**4)]])\n",
+      "b=array([sum(del_V),sum(x1*del_V),sum((x1**2)*del_V)])\n",
+      "\n",
+      "soln=linalg.solve(a,b);\n",
+      "\n",
+      "a0=soln[0]\n",
+      "a1=soln[1]\n",
+      "a2=soln[2]\n",
+      "#del_V = V_mix/(x1*x2) = a0 + a1*x1 + a2*x1**(2)\n",
+      "#V_mix = (a0 + a1*x1 + a2*x1**(2))*(x1*(1 - x1))\n",
+      "# For x1 = 0.5\n",
+      "x1  = 0.5;\n",
+      "V_mix_prime = (a0+(a1*x1)+(a2*x1**2))*(x1*(1-x1));\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "#Now differentiating the above equation with respect to x we get\n",
+      "#d/dx(V_mix) = (-4*a2*x1**3) + (3*(a2-a1)*x1**2) + (2*(a1-a0)*x1)+ a0\n",
+      "#Again for x1 = 0.5\n",
+      "x1_prime = 0.5;\n",
+      "del_V_mix_prime = (-4*a2*x1_prime**3)+(3*(a2-a1)*x1_prime**2)+(2*(a1-a0)*x1_prime)+a0;\n",
+      "\n",
+      "#Finally,calculating the partial molar volumes\n",
+      "V1_bar = V1 + V_mix_prime + x_water*del_V_mix_prime;\t\t\t#[cm**(3)/mol] \n",
+      "V2_bar = V2 + V_mix_prime - x_methanol*del_V_mix_prime;\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \"The partial molar volume of methanol component 1) is %f cm**3)/mol\"%(V1_bar);\n",
+      "print \"The partial molar volume of water component 2) is %f cm**3)/mol\"%(V2_bar);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The partial molar volume of methanol component 1) is 39.721188 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 17.079639 cm**3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.2  Page Number : 421"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "#component 1 = water\n",
+      "#component 2 = methanol\n",
+      "T = 25 + 273.15;\t\t\t#[K] - Temperature\n",
+      "\n",
+      "#delta_V_mix = x_1*x_2*(-3.377 - 2.945*x_1 + 3.31*x_1**(2))\n",
+      "V1 = 18.0684;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 1\n",
+      "V2 = 40.7221;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2\n",
+      "Vol_1 = 1000;\t\t\t#[cm**(3)] - Volume of pure component 1\n",
+      "Vol_2 = 1000;\t\t\t#[cm**(3)] - Volume of pure component 2\n",
+      "\n",
+      "# Calculations\n",
+      "#Moles of the componenets can be calculated as \n",
+      "n_1 = round(Vol_1/V1,4);\t\t\t#[mol]\n",
+      "n_2 = round(Vol_2/V2,4);\t\t\t#[mol]\n",
+      "\n",
+      "#Mole fraction of the components \n",
+      "x_1 = round(n_1/(n_1 + n_2),4);\n",
+      "x_2 = round(n_2/(n_1 + n_2),4);\n",
+      "\n",
+      "delta_V_mix = round(x_1*x_2*(-3.377 - 2.945*x_1 + 3.31*x_1**(2)),4);\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "#Differentiating the above equation, we get\n",
+      "#d/dx(delta_V_mix) = (1 - 2*x_1)*(-3.377 - 2.945*x_1 + 3.31*x_1**(2)) + (x_1 - x_1**(2))*(-2.945 + 6.62*x_1)\n",
+      "del_delta_V_mix = round((1 - 2*x_1)*(-3.377 - 2.945*x_1 + 3.31*x_1**(2)) + (x_1 - x_1**(2))*(-2.945 + 6.62*x_1),4);\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "#Now calculating the partial molar volumes\n",
+      "V1_bar = V1 + delta_V_mix + x_2*del_delta_V_mix;\t\t\t#[cm**(3)/mol] \n",
+      "V2_bar = V2 + delta_V_mix - x_1*del_delta_V_mix;\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "print del_delta_V_mix, V1_bar, V2_bar\n",
+      "#Finally molar volume of the solution is given by\n",
+      "V_sol = x_1*V1_bar + x_2*V2_bar;\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "# Total volume of the solution is given by\n",
+      "V_total = (n_1 + n_2)*V_sol;\t\t\t#[cm**(3)]\n",
+      "\n",
+      "# Results\n",
+      "print \"The molar volume of the solution is %.4f cm**3)/mol\"%(V_sol);\n",
+      "print \"The total volume of the solution is %.2f cm**3)\"%(V_total);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1.8248 17.81416104 38.64306104\n",
+        "The molar volume of the solution is 24.2149 cm**3)/mol\n",
+        "The total volume of the solution is 1934.82 cm**3)\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.3  Page Number : 422"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      " \n",
+      "\n",
+      "Vol = 20;\t\t\t#[cm**(3)] - Volume of the solution\n",
+      "T = 22 + 273.15;\t\t\t#[K] - Temperature\n",
+      "W_bottle = 11.5485;\t\t\t#[g] - Weight of density bottle\n",
+      "Mol_meth = 32.04;\t\t\t#Molecular weight of methanol\n",
+      "Mol_water = 18.015;\t\t\t# Molecular weight of water\n",
+      "\n",
+      "#Density of pure components can be found out at 0% and 100% of volume percent.\n",
+      "den_meth = 0.7929;\t\t\t#[cm**(3)/mol] - Density of pure methanol\n",
+      "den_water = 0.9937;\t\t\t#[cm**(3)/mol] - Density of pure water\n",
+      "\n",
+      "\n",
+      "Vol_perc=[5,10,20,30,40,50,60,70,80,90,95];\t\t\t# Volumes percent of component 1 (methanol)\n",
+      "W_total=[31.2706,31.1468,30.8907,30.6346,30.3396,30.0053,29.5865,29.1453,28.5978,28.0325,27.7320];\t\t\t# Weight of solution + weight of density bottle\n",
+      "\n",
+      "W_sol=zeros(11);\t\t\t# Weight of 20 cm**(3) of solution\n",
+      "den=zeros(11);\t\t\t# density of the solution\n",
+      "x1=zeros(11);\t\t\t# Mole fraction of methanol\n",
+      "x2=zeros(11);\t\t\t# Mole fraction of water\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(11):\n",
+      "    W_sol[i]=W_total[i]-W_bottle;\n",
+      "    den[i]=W_sol[i]/Vol;\n",
+      "    x1[i]=((Vol_perc[i]*den_meth)/Mol_meth)/(((Vol_perc[i]*den_meth)/Mol_meth)+(((100-Vol_perc[i])*den_water)/Mol_water));\n",
+      "    x2[i]=1-x1[i];\n",
+      "\n",
+      "\n",
+      "#Again we have,\n",
+      "V_kg=zeros(11);\t\t\t#[cm**(3)] -  Volume of 1 kg of solution\n",
+      "n_mol=zeros(11);\t\t\t#[mol] -  Number of moles in 1 kg of solution\n",
+      "V_mol=zeros(11);\t\t\t#[cm**(3)/mol] -  Volume of 1 mol of solution\n",
+      "x_V=zeros(11);\t\t\t#[cm**(3)/mol] - x_V = x1*V_meth + x2*V_water\n",
+      "V_mix=zeros(11);\t\t\t#[cm**(3)/mol] -  V_mix = V_mol - x1*V_meth - x2*V_water\n",
+      "del_V=zeros(11);\t\t\t# [cm**(3)/mol] - del_V = V_mix/(x1*x2)\n",
+      "\n",
+      "#V_mol = V_meth at x1 = 1 and V_mol = V_water at x1 = 0, therefore\n",
+      "V_meth = 40.4114;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 1 (methanol)\n",
+      "V_water = 18.1286;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2 (water)\n",
+      "\n",
+      "for i in range(11):\n",
+      "    V_kg[i]=1000/den[i];\n",
+      "    n_mol[i]=1000/(x1[i]*Mol_meth+x2[i]*Mol_water);\n",
+      "    V_mol[i]=V_kg[i]/n_mol[i];\n",
+      "    x_V[i]=V_meth*x1[i]+V_water*x2[i];\n",
+      "    V_mix[i]=V_mol[i]-x1[i]*V_meth-x2[i]*V_water;\n",
+      "    del_V[i]=V_mix[i]/(x1[i]*x2[i]);\n",
+      "\n",
+      "#Now employing the concept of quadratic regression of the data ( x1 , del_V ) to solve the equation of the type\n",
+      "#y = a0 + a1*x + a2*x**(2) \n",
+      "#Here the above equation is in the form of\n",
+      "#del_V = V_mix/(x1*x2) = a0 + a1*x1 + a2*x1**(2) \n",
+      "\n",
+      "#From the matrix method to solve simultaneous linear equations, we have\n",
+      "a = array([[11, sum(x1), sum(x1**2)],[sum(x1), sum(x1**2), sum(x1**3)],[sum(x1**2), sum(x1**3), sum(x1**4)]])\n",
+      "b = array([sum(del_V),sum(x1*del_V),sum((x1**2)*del_V)])\n",
+      "\n",
+      "soln=linalg.solve(a,b);\n",
+      "a0=soln[0]\n",
+      "a1=soln[1]\n",
+      "a2=soln[2]\n",
+      "\n",
+      "#del_V = V_mix/(x1*x2) = a0 + a1*x1 + a2*x1**(2)\n",
+      "#V_mix = (a0 + a1*x1 + a2*x1**(2))*(x1*(1 - x1))\n",
+      "#Solving the above equation for x1,\n",
+      "def f(x1): \n",
+      "    return (a0+(a1*x1)+(a2*x1**2))*(x1*(1-x1))\n",
+      "\n",
+      "#Now differentiating the above equation with respect to x we get\n",
+      "#d/dx(V_mix) = (-4*a2*x1**3) + (3*(a2-a1)*x1**2) + (2*(a1-a0)*x1)+ a0\n",
+      "#Again solving it for x1\n",
+      "def f1(x1): \n",
+      "    return (-4*a2*x1**3)+(3*(a2-a1)*x1**2)+(2*(a1-a0)*x1)+a0\n",
+      "\n",
+      "#Now \n",
+      "\n",
+      "x1_prime=[0,0.25,0.50,0.75,1.0];\n",
+      "V_mix_prime=zeros(5);\t\t\t#[cm**(3)/mol] -  V_mix = V - x1*V_meth - x2*V_water\n",
+      "del_V_prime=zeros(5);\t\t\t#[cm**(3)/mol] - del_V = V_mix/(x1*x2)\n",
+      "V1_bar=zeros(5);\t\t\t#[cm**(3)/mol] - Partial molar volume of component 1\n",
+      "V2_bar=zeros(5);\t\t\t#[cm**(3)/mol] - Partial molar volume of component 1\n",
+      "\n",
+      "# Results\n",
+      "for j in range(5):\n",
+      "    V_mix_prime[j]=f(x1_prime[j]);\n",
+      "    del_V_prime[j]=f1(x1_prime[j]);\n",
+      "    V1_bar[j]=V_meth+V_mix_prime[j]+(1-x1_prime[j])*del_V_prime[j];\n",
+      "    V2_bar[j]=V_water+V_mix_prime[j]-x1_prime[j]*del_V_prime[j];\n",
+      "    print \"For x1 = %f\"%(x1_prime[j]);\n",
+      "    print \"The partial molar volume of methanol component 1) is %f cm**3)/mol\"%(V1_bar[j])\n",
+      "    print \"The partial molar volume of water component 2) is %f cm**3)/mol\"%(V2_bar[j])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For x1 = 0.000000\n",
+        "The partial molar volume of methanol component 1) is 37.937941 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 18.128600 cm**3)/mol\n",
+        "For x1 = 0.250000\n",
+        "The partial molar volume of methanol component 1) is 38.124350 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 18.031910 cm**3)/mol\n",
+        "For x1 = 0.500000\n",
+        "The partial molar volume of methanol component 1) is 39.496329 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 17.177237 cm**3)/mol\n",
+        "For x1 = 0.750000\n",
+        "The partial molar volume of methanol component 1) is 40.332855 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 15.841550 cm**3)/mol\n",
+        "For x1 = 1.000000\n",
+        "The partial molar volume of methanol component 1) is 40.411400 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 15.800306 cm**3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.4  Page Number : 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from numpy import *\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 20 + 273.15;\t\t\t#[K] - Temperature\n",
+      "Mol_form = 46.027;\t\t\t#Molecular weight of formic acid\n",
+      "Mol_water = 18.015;\t\t\t# Molecular weight of water\n",
+      "\n",
+      "Wt_perc=[10,18,30,50,72,78];\t\t\t#Weight percent of formic acid\n",
+      "den=[1.0246,1.0441,1.0729,1.1207,1.1702,1.1818];\t\t\t#[g/cm**(3)] - Density of solution\n",
+      "\n",
+      "V_g=zeros(6);\t\t\t#[cm**(3)/g] - Volume of 1 g of solution\n",
+      "x1=zeros(6);\t\t\t# Mole fraction of component 1\n",
+      "x2=zeros(6);\t\t\t# Mole fraction of component 2\n",
+      "n=zeros(6);\t\t\t# Number of moles in 1 g\n",
+      "V_mol=zeros(6);\t\t\t#[cm**(3)/mol] - Volume of 1 mol of solution\n",
+      "x_V=zeros(6);\t\t\t#[cm**(3)/mol] - x_V = x1*V_form + x2*V_water\n",
+      "V_mix=zeros(6);\t\t\t#[cm**(3)/mol] -  V_mix = V - x1*V_form - x2*V_water\n",
+      "del_V=zeros(6);\t\t\t# [cm**(3)/mol] - del_V = V_mix/(x1*x2)\n",
+      "\n",
+      "#V_mol = V_form at x1 = 1 and V_mol = V_water at x1 = 0, therefore\n",
+      "V_form = 37.737;\t\t\t#[cm**(3)/mol] - Molar volume of pure formic acid (component 1)\n",
+      "V_water = 18.050;\t\t\t#[cm**(3)/mol] - Molar volume of pure water (component 2)\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(6):\n",
+      "    V_g[i]=1/den[i];\n",
+      "    x1[i]=(Wt_perc[i]/Mol_form)/((Wt_perc[i]/Mol_form)+((100-Wt_perc[i])/Mol_water));\n",
+      "    x2[i]=1-x1[i];\n",
+      "    n[i]=((Wt_perc[i]/100.)/Mol_form)+(((100-Wt_perc[i])/100.)/Mol_water);\n",
+      "    V_mol[i]=V_g[i]/n[i];\n",
+      "    x_V[i]=V_form*x1[i]+V_water*x2[i];\n",
+      "    V_mix[i]=V_mol[i]-x1[i]*V_form-x2[i]*V_water;\n",
+      "    del_V[i]=V_mix[i]/(x1[i]*x2[i]);\n",
+      "\n",
+      "\n",
+      "a = array([[11, sum(x1), sum(x1**2)],[sum(x1), sum(x1**2) ,sum(x1**3)],[sum(x1**2), sum(x1**3) ,sum(x1**4)]])\n",
+      "b = array([sum(del_V),sum(x1*del_V),sum((x1**2)*del_V)])\n",
+      "\n",
+      "soln = linalg.solve(a,b)\n",
+      "\n",
+      "a0=soln[0]\n",
+      "a1=soln[1]\n",
+      "a2=soln[2]\n",
+      "\n",
+      "def f(x1): \n",
+      "\t return (a0+(a1*x1)+(a2*x1**2))*(x1*(1-x1))\n",
+      "\n",
+      "def f1(x1): \n",
+      "\t return (-4*a2*x1**3)+(3*(a2-a1)*x1**2)+(2*(a1-a0)*x1)+a0\n",
+      "\n",
+      "#At 15 Wt% of formic acid, x1 is given by\n",
+      "x1_prime_1 = round((15/Mol_form)/((15/Mol_form)+((100-15)/Mol_water)),3); \n",
+      "#Similarly at 75 Wt% of formic acid, x1 is given by\n",
+      "x1_prime_2 = round((75/Mol_form)/((75/Mol_form)+((100-75)/Mol_water)),4); \n",
+      "\n",
+      "Wt_perc_prime=[15,75];\n",
+      "x1_prime=[x1_prime_1,x1_prime_2];\n",
+      "V_mix_prime=zeros(2);\t\t\t#[cm**(3)/mol] -  V_mix = V - x1*V_meth - x2*V_water\n",
+      "del_V_prime=zeros(2);\t\t\t#[cm**(3)/mol] - del_V = V_mix/(x1*x2)\n",
+      "V1_bar=zeros(2);\t\t\t#[cm**(3)/mol] - Partial molar volume of component 1\n",
+      "V2_bar=zeros(2);\t\t\t#[cm**(3)/mol] - Partial molar volume of component 1\n",
+      "\n",
+      "# Results\n",
+      "for j in range(2):\n",
+      "    V_mix_prime[j]=f(x1_prime[j]);\n",
+      "    del_V_prime[j]=f1(x1_prime[j]);\n",
+      "    V1_bar[j]=V_form+V_mix_prime[j]+(1-x1_prime[j])*del_V_prime[j];\n",
+      "    V2_bar[j]=V_water+V_mix_prime[j]-x1_prime[j]*del_V_prime[j];\n",
+      "    print \"For weight percent of formic acid = %f percent\"%(Wt_perc_prime[j]);\n",
+      "    print \"The partial molar volume of formic acid component 1) is %f cm**3)/mol\"%(V1_bar[j]);\n",
+      "    print \"The partial molar volume of water component 2) is %f cm**3)/mol\"%(V2_bar[j]);\n",
+      "\n",
+      "# answers are vary because of rounding error. Please review it manually.    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For weight percent of formic acid = 15.000000 percent\n",
+        "The partial molar volume of formic acid component 1) is 35.429311 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 18.102339 cm**3)/mol\n",
+        "For weight percent of formic acid = 75.000000 percent\n",
+        "The partial molar volume of formic acid component 1) is 38.853189 cm**3)/mol\n",
+        "The partial molar volume of water component 2) is 15.646974 cm**3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.5  Page Number : 426"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import *\n",
+      "# Variables\n",
+      "T = 40 + 273.15;\t\t\t#[K] - Temperature\n",
+      "\n",
+      "x1=array([0.083,0.176,0.268,0.353,0.428,0.720,0.780,0.850,0.900])       \t\t\t# Mole fraction of component 1\n",
+      "delta_H_mix=array([0.250,0.488,0.670,0.790,0.863,0.775,0.669,0.510,0.362])\t\t\t#[kJ/mol] - Enthalpy of the solution\n",
+      "\n",
+      "x2=zeros(9);\t\t\t# Mole fraction of component 2\n",
+      "del_H=zeros(9);\t\t\t#[kJ/mol] - del_H = delta_H_mix/(x1*x2)\n",
+      "\n",
+      "for i in range(9):\n",
+      "    x2[i]=1-x1[i];\n",
+      "    del_H[i]=delta_H_mix[i]/(x1[i]*x2[i]);\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "#Now employing the concept of quadratic regression of the data ( x1 , del_H ) to solve the equation of the type\n",
+      "#y = a0 + a1*x + a2*x**(2) \n",
+      "#Here the above equation is in the form of\n",
+      "#del_H = delta_H_mix/(x1*x2) = a0 + a1*x1 + a2*x1**(2) \n",
+      "\n",
+      "#From the matrix method to solve simultaneous linear equations, we have\n",
+      "a = array([[9, sum(x1), sum(x1**2)],[sum(x1), sum(x1**2), sum(x1**3)],[sum(x1**2), sum(x1**3), sum(x1**4)]])\n",
+      "b = array([sum(del_H),sum(x1*del_H),sum((x1**2)*del_H)])\n",
+      "soln= linalg.solve(a,b)\n",
+      "a0=soln[0]\n",
+      "a1=soln[1]\n",
+      "a2=soln[2]\n",
+      "\n",
+      "#del_H = delta_H_mix/(x1*x2) = a0 + a1*x1 + a2*x1**(2)\n",
+      "#delta_H_mix = (a0 + a1*x1 + a2*x1**(2))*(x1*(1 - x1))\n",
+      "#At x1 = 0.25,\n",
+      "x_1 = 0.25;\t\t\t#[mol]\n",
+      "delta_H_mix = (a0+(a1*x_1)+(a2*x_1**2))*(x_1*(1-x_1));\t\t\t#[kJ/mol]\n",
+      "\n",
+      "#Now differentiating the above equation with respect to x we get\n",
+      "#d/dx(delta_H_mix) = del_delta_H_mix = (-4*a2*x1**3) + (3*(a2-a1)*x1**2) + (2*(a1-a0)*x1)+ a0\n",
+      "#Again  for x1 = 0.25\n",
+      "x_1_prime = 0.25;\t\t\t#[mol]\n",
+      "del_delta_H_mix = (-4*a2*x_1_prime**3)+(3*(a2-a1)*x_1_prime**2)+(2*(a1-a0)*x_1_prime)+a0;\t\t\t#[kJ/mol]\n",
+      "\n",
+      "#We have the relation\n",
+      "# H1_bar - H1 = delta_H_mix + x2*del_delta_H_mix, and\n",
+      "# H2_bar - H2 = delta_H_mix - x1*del_delta_H_mix\n",
+      "\n",
+      "#Let us suppose\n",
+      "#k_1 = H1_bar - H1 , and\n",
+      "#k_2 = H2_bar - H2\n",
+      "\n",
+      "k_1 = delta_H_mix + (1-x_1_prime)*del_delta_H_mix;\t\t\t#[kJ/mol]\n",
+      "k_2 = delta_H_mix - x_1_prime*del_delta_H_mix;\t\t\t#[kJ/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \"The value of H1_bar - H1) at x1 = 0.25 is %f kJ/mol\"%(k_1);\n",
+      "print \"The value of H2_bar - H2) at x1 = 0.25 is %f kJ/mol\"%(k_2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of H1_bar - H1) at x1 = 0.25 is 2.010734 kJ/mol\n",
+        "The value of H2_bar - H2) at x1 = 0.25 is 0.179079 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch13_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch13_1.ipynb
new file mode 100755
index 00000000..d2334255
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch13_1.ipynb
@@ -0,0 +1,304 @@
+{
+ "metadata": {
+  "name": "ch13_1",
+  "signature": "sha256:894f18dec91baf4a10867e0d301a68220274d02a1e24c4e2bef1038061daa009"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Chapter 13 : Fugacity of a Component in a Mixture by Equations of State"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.2  Page Number : 433"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 310.93;\t\t\t#[K] - Temperature\n",
+      "P = 2.76*10**(6);\t\t\t#[Pa] - Pressure\n",
+      "y1 = 0.8942;\t\t\t#[mol] - mole fraction of component 1\n",
+      "y2 = 1 - y1;\t\t\t#[mol] - mole fraction of component 2\n",
+      "R=8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "#For component 1 (methane)\n",
+      "Tc_1 = 190.6;\t\t\t#[K] - Critical temperature\n",
+      "Pc_1 = 45.99*10**(5);\t\t\t#[N/m**(2)] - Critical pressure\n",
+      "Vc_1 = 98.6;\t\t\t#[cm**(3)/mol] - Critical molar volume\n",
+      "Zc_1 = 0.286;\t\t\t# - Critical compressibility factor\n",
+      "w_1 = 0.012;\t\t\t# - Critical acentric factor\n",
+      "#Similarly for component 2 (n-Butane)\n",
+      "Tc_2 = 425.1;\t\t\t#[K]\n",
+      "Pc_2 = 37.96*10**(5);\t\t\t#[N/m**(2)]\n",
+      "Vc_2 = 255;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_2=0.274;\n",
+      "w_2=0.2;\n",
+      "\n",
+      "# Calculations\n",
+      "#For component 1\n",
+      "Tr_1 = T/Tc_1;\t\t\t#Reduced temperature\n",
+      "#At reduced temperature\n",
+      "B1_0 = 0.083-(0.422/(Tr_1)**(1.6));\n",
+      "B1_1 = 0.139-(0.172/(Tr_1)**(4.2));\n",
+      "#We know,(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B_11 = ((B1_0+(w_1*B1_1))*(R*Tc_1))/Pc_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "#Similarly for component 2\n",
+      "Tr_2 = T/Tc_2;\t\t\t#Reduced temperature\n",
+      "#At reduced temperature Tr_2,\n",
+      "B2_0 = 0.083-(0.422/(Tr_2)**(1.6));\n",
+      "B2_1 = 0.139-(0.172/(Tr_2)**(4.2));\n",
+      "B_22 = ((B2_0+(w_2*B2_1))*(R*Tc_2))/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "#For cross coeffcient\n",
+      "Tc_12 = (Tc_1*Tc_2)**(1./2);\t\t\t#[K]\n",
+      "w_12 = (w_1+w_2)/2;\n",
+      "Zc_12 = (Zc_1+Zc_2)/2;\n",
+      "Vc_12 = (((Vc_1)**(1./3)+(Vc_2)**(1./3))/2)**(3);\t\t\t#[cm**(3)/mol]\n",
+      "Vc_12 = Vc_12*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "Pc_12 = (Zc_12*R*Tc_12)/Vc_12;\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "# Variables, Z = 1 + (B*P)/(R*T)\n",
+      "#Now we have,(B_12*Pc_12)/(R*Tc_12) = B_0 + (w_12*B_1)\n",
+      "#where B_0 and B_1 are to be evaluated at Tr_12\n",
+      "Tr_12 = T/Tc_12;\n",
+      "#At reduced temperature Tr_12\n",
+      "B_0 = 0.083-(0.422/(Tr_12)**(1.6));\n",
+      "B_1 = 0.139-(0.172/(Tr_12)**(4.2));\n",
+      "B_12 = ((B_0+(w_12*B_1))*R*Tc_12)/Pc_12;\t\t\t#[m**(3)/mol]\n",
+      "#For the mixture\n",
+      "B = y1**(2)*B_11+2*y1*y2*B_12+y2**(2)*B_22;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "#Now del_12 can be calculated as,\n",
+      "del_12 = 2*B_12 - B_11 - B_22;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "#We have the relation, math.log(phi_1) = (P/(R*T))*(B_11 + y2**(2)*del_12), therefore\n",
+      "phi_1 = math.exp((P/(R*T))*(B_11 + y2**(2)*del_12));\n",
+      "#Similarly for component 2\n",
+      "phi_2 = math.exp((P/(R*T))*(B_22 + y1**(2)*del_12));\n",
+      "\n",
+      "# Results\n",
+      "print \" The value of fugacity coefficient of component 1 phi_1) is %f\"%(phi_1);\n",
+      "print \" The value of fugacity coefficient of component 2 phi_2) is %f\"%(phi_2);\n",
+      "\n",
+      "#Finally fugacity coefficient of the mixture is given by\n",
+      "#math.log(phi) = y1*math.log(phi_1) + y2*math.log(phi_2);\n",
+      "phi = math.exp(y1*math.log(phi_1) + y2*math.log(phi_2));\n",
+      "\n",
+      "print \" The value of fugacity coefficient of the mixture phi) is %f \"%(phi);\n",
+      "#The fugacity coefficient of the mixture can also be obtained using\n",
+      "#math.log(phi) = (B*P)/(R*T)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of fugacity coefficient of component 1 phi_1) is 0.965152\n",
+        " The value of fugacity coefficient of component 2 phi_2) is 0.675374\n",
+        " The value of fugacity coefficient of the mixture phi) is 0.929376 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.7  Page Number : 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 460.;\t\t\t#[K] - Temperature\n",
+      "P = 40.*10**(5);\t\t\t#[Pa] - Pressure\n",
+      "R=8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "# component 1 = nitrogen\n",
+      "# component 2 = n-Butane\n",
+      "y1 = 0.4974;\t\t\t# Mole percent of nitrogen\n",
+      "y2 = 0.5026;\t\t\t# Mole percent of n-Butane\n",
+      "Tc_nit = 126.2;\t\t\t#[K]\n",
+      "Pc_nit = 34.00*10**(5);\t\t\t#[Pa]\n",
+      "Tc_but = 425.1;\t\t\t#[K]\n",
+      "Pc_but = 37.96*10**(5);\t\t\t#[Pa]\n",
+      "\n",
+      "# (1). van der Walls equation of state\n",
+      "\n",
+      "# The fugacity coefficient of component 1 in a binary mixture following van der Walls equation of state is given by,\n",
+      "# math.log(phi_1) = b_1/(V-b) - math.log(Z-B) -2*(y1*a_11 + y2*a_12)/(R*T*V)\n",
+      "# and for component 2 is given by,\n",
+      "# math.log(phi_2) = b_2/(V-b) - math.log(Z-B) -2*(y1*a_12 + y2*a_22)/(R*T*V)\n",
+      "# Where B = (P*b)/(R*T)\n",
+      "\n",
+      "# Calculations\n",
+      "# For componenet 1 (nitrogen)\n",
+      "a_1 = (27*R**(2)*Tc_nit**(2))/(64*Pc_nit);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b_1 = (R*Tc_nit)/(8*Pc_nit);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Similarly for componenet 2 (n-Butane)\n",
+      "a_2 = (27*R**(2)*Tc_but**(2))/(64*Pc_but);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b_2 = (R*Tc_but)/(8*Pc_but);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Here\n",
+      "a_11 = a_1;\n",
+      "a_22 = a_2;\n",
+      "# For cross coefficient\n",
+      "a_12 = (a_1*a_2)**(1./2);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "\n",
+      "# For the mixture \n",
+      "a = y1**(2)*a_11 + y2**(2)*a_22 + 2*y1*y2*a_12;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = y1*b_1 + y2*b_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The cubic form of the van der Walls equation of state is given by,\n",
+      "# V**(3) - (b+(R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0\n",
+      "# Substituting the value and solving for V, we get\n",
+      "# Solving the cubic equation\n",
+      "def f(V): \n",
+      "    return V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P\n",
+      "V_1=fsolve(f,-1)\n",
+      "V_2=fsolve(f,0)\n",
+      "V_3=fsolve(f,1)\n",
+      "# The molar volume V = V_3, the other two roots are imaginary\n",
+      "V = V_3;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The comprssibility factor of the mixture is \n",
+      "Z = (P*V)/(R*T);\n",
+      "# And B can also be calculated as\n",
+      "B = (P*b)/(R*T);\n",
+      "\n",
+      "# The fugacity coefficient of component 1 in the mixture is\n",
+      "phi_1 = math.exp(b_1/(V-b) - math.log(Z-B) -2*(y1*a_11 + y2*a_12)/(R*T*V));\n",
+      "# Similarly fugacity coefficient of component 2 in the mixture is \n",
+      "phi_2 = math.exp(b_2/(V-b) - math.log(Z-B) -2*(y1*a_12 + y2*a_22)/(R*T*V));\n",
+      "\n",
+      "# The fugacity coefficient of the mixture is given by,\n",
+      "# math.log(phi) = y1*math.log(phi_1) + y2*math.log(phi_2)\n",
+      "phi = math.exp(y1*math.log(phi_1) + y2*math.log(phi_2));\n",
+      "\n",
+      "# Also the fugacity coefficient of the mixture following van der Walls equation of state is given by,\n",
+      "# math.log(phi) = b/(V-b) - math.log(Z-B) -2*a/(R*T*V)\n",
+      "phi_dash = math.exp(b/(V-b) - math.log(Z-B) -2*a/(R*T*V));\n",
+      "# The result is same as obtained above\n",
+      "\n",
+      "# Results\n",
+      "print \" 1van der Walls equation of state\";\n",
+      "print \"  The value of fugacity coefficient of component 1 nitrogen) is %f\"%(phi_1);\n",
+      "print \"  The value of fugacity coefficient of component 2 n-butane) is %f\"%(phi_2);\n",
+      "print \"  The value of fugacity coefficient of the mixture is %f\"%(phi);\n",
+      "print \"  Also the fugacity coefficient of the mixture from van der Walls equation of state is %f which is same as above)\"%(phi_dash);\n",
+      "\n",
+      "# (2). Redlich-Kwong equation of state\n",
+      "\n",
+      "# For component 1,\n",
+      "a_1_prime = (0.42748*R**(2)*Tc_nit**(2.5))/Pc_nit;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b_1_prime = (0.08664*R*Tc_nit)/Pc_nit;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "#similarly for component 2,\n",
+      "a_2_prime = (0.42748*R**(2)*Tc_but**(2.5))/Pc_but;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b_2_prime = (0.08664*R*Tc_but)/Pc_but;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# For cross coefficient\n",
+      "a_12_prime = (a_1_prime*a_2_prime)**(1./2);\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "# For the mixture\n",
+      "a_prime = y1**(2)*a_1_prime + y2**(2)*a_2_prime +2*y1*y2*a_12_prime;\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b_prime = y1*b_1_prime +y2*b_2_prime;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\n",
+      "#The cubic form of Redlich Kwong equation of state is given by,\n",
+      "#  V**(3)-((R*T)/P)*V**(2)-((b**(2))+((b*R*T)/P)-(a/(T**(1/2)*P))*V-(a*b)/(T**(1/2)*P)=0\n",
+      "# Solving the cubic equation\n",
+      "def f1(V): \n",
+      "    return V**(3)-((R*T)/P)*V**(2)-((b_prime**(2))+((b_prime*R*T)/P)-(a_prime/(T**(1./2)*P)))*V-(a_prime*b_prime)/(T**(1./2)*P)\n",
+      "V_4=fsolve(f1,1)\n",
+      "V_5=fsolve(f1,0)\n",
+      "V_6=fsolve(f1,-1)\n",
+      "# The molar volume V = V_4, the other two roots are imaginary\n",
+      "V_prime = V_4;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# The compressibility factor of the mixture is\n",
+      "Z_prime = (P*V_prime)/(R*T);\n",
+      "# And B can also be calculated as\n",
+      "B_prime = (P*b_prime)/(R*T);\n",
+      "\n",
+      "# The fugacity coefficient of component 1 in the binary mixture is given by\n",
+      "# math.log(phi_1) = b_1/(V-b) - math.log(Z-B) + ((a*b_1)/((b**(2)*R*T**(3/2))))*(math.log((V+b)/V)-(b/(V+b)))-(2*(y1*a_1+y2*a_12)/(R*T**(3/2)b))*(math.log(V+b)/b)\n",
+      "\n",
+      "phi_1_prime = math.exp((b_1_prime/(V_prime-b_prime))-math.log(Z_prime-B_prime)+((a_prime*b_1_prime)/((b_prime**(2))*R*(T**(3./2))))*(math.log((V_prime+b_prime)/V_prime)-(b_prime/(V_prime+b_prime)))-(2*(y1*a_1_prime+y2*a_12_prime)/(R*(T**(3./2))*b_prime))*(math.log((V_prime+b_prime)/V_prime)));\n",
+      "\n",
+      "\n",
+      "# Similarly fugacity coefficient of component 2 in the mixture is \n",
+      "phi_2_prime = math.exp((b_2_prime/(V_prime-b_prime))-math.log(Z_prime-B_prime)+((a_prime*b_2_prime)/((b_prime**(2))*R*(T**(3./2))))*(math.log((V_prime+b_prime)/V_prime)-(b_prime/(V_prime+b_prime)))-(2*(y1*a_12_prime+y2*a_2_prime)/(R*(T**(3./2))*b_prime))*(math.log((V_prime+b_prime)/V_prime)));\n",
+      "\n",
+      "# The fugacity coefficient of the mixture is given by,\n",
+      "# math.log(phi) = y1*math.log(phi_1) + y2*math.log(phi_2)\n",
+      "phi_prime = math.exp(y1*math.log(phi_1_prime) + y2*math.log(phi_2_prime));\n",
+      "\n",
+      "# Also the fugacity coefficient for the mixture following Redlich-kwong equation of state is also given by\n",
+      "# math.log(phi) = b/(V-b) - math.log(Z-B) - (a/(R*T**(3/2)))*(1/(V+b)+(1/b)*math.log((V+b)/V))\n",
+      "phi_prime_dash = math.exp(b_prime/(V_prime-b_prime) - math.log(Z_prime-B_prime) - (a_prime/(R*T**(3./2)))*(1/(V_prime+b_prime)+(1./b_prime)*math.log((V_prime+b_prime)/                   V_prime)));\n",
+      "\n",
+      "print \"  \\nRedlich-Kwong equation of state\";\n",
+      "print \"  The value of fugacity coefficient of component 1 nitrogen) is %f\"%(phi_1_prime);\n",
+      "print \"  The value of fugacity coefficient of component 2 n-butane) is %f\"%(phi_2_prime);\n",
+      "print \"  The value of fugacity coefficient of the mixture is %f\"%(phi_prime);\n",
+      "print \"  Also the fugacity coefficient for the mixture from Redlich-kwong equation of state is %f which is same as above)\"%(phi_prime_dash);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1van der Walls equation of state\n",
+        "  The value of fugacity coefficient of component 1 nitrogen) is 1.057500\n",
+        "  The value of fugacity coefficient of component 2 n-butane) is 0.801865\n",
+        "  The value of fugacity coefficient of the mixture is 0.920192\n",
+        "  Also the fugacity coefficient of the mixture from van der Walls equation of state is 0.920192 which is same as above)\n",
+        "  \n",
+        "Redlich-Kwong equation of state\n",
+        "  The value of fugacity coefficient of component 1 nitrogen) is 1.071129\n",
+        "  The value of fugacity coefficient of component 2 n-butane) is 0.793063\n",
+        "  The value of fugacity coefficient of the mixture is 0.920948\n",
+        "  Also the fugacity coefficient for the mixture from Redlich-kwong equation of state is 0.920948 which is same as above)\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch14_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch14_1.ipynb
new file mode 100755
index 00000000..6e9336a9
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch14_1.ipynb
@@ -0,0 +1,715 @@
+{
+ "metadata": {
+  "name": "ch14_1",
+  "signature": "sha256:98f2e1e0f1c561d3ee302fd629093b7572c8b263613f73036059e7020f90d8cc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 : Activity Coefficients Models for Liquid Mixtures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.4  Page Number : 461"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables,\n",
+      "T = 300;\t\t\t#[K] - Temperature\n",
+      "b = 100;\t\t\t#[cal/mol]\n",
+      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
+      "# R*T*math.log(Y_1) = b*x_2**(2)\n",
+      "# R*T*math.log(Y_2) = b*x_1**(2)\n",
+      "\n",
+      "#For equimolar mixture\n",
+      "x_1 = 0.5;\t\t\t#Mole fraction of component 1\n",
+      "x_2 = 0.5;\t\t\t#Mole fraction of component 2\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#The excess Gibbs free energy is given by\n",
+      "# G_excess = R*T*(x_1*math.log(Y_1) + x_2*math.log(Y_2)) = b*x_1*x_2**(2) + b*x_2*x_1**(2) = b*x_1*(x_1 + x_2) = b*x_1*x_2\n",
+      "G_excess = b*x_1*x_2;\t\t\t#[cal/mol]\n",
+      "\n",
+      "#The ideal Gibbs free energy change of mixing is given by,\n",
+      "delta_G_id_mix = R*T*(x_1*math.log(x_1)+x_2*math.log(x_2));\t\t\t#[cal/mol]\n",
+      "\n",
+      "#The Gibbs free energy of mixing is given by\n",
+      "delta_G_mix = delta_G_id_mix + G_excess;\t\t\t#[cal/mol]\n",
+      "\n",
+      "\n",
+      "# delta_S_mix = delta_S_id_mix = - R*sum(x_i*math.log(x_i))\n",
+      "\n",
+      "#delta_G_mix = delta_H_mix - T*delta_S_mix = delta_H_mix + R*T*(x_1*math.log(x_1)+x_2*math.log(x_2))\n",
+      "delta_H_mix = b*x_1*x_2;\t\t\t#[cal/mol]\n",
+      "\n",
+      "print \"The value of Gibbs free energy change for equimolar mixture formation is %f cal/mol\"%(delta_G_mix);\n",
+      "print \"The value of enthalpy change for equimolar mixture formation is %f cal/mol\"%(delta_H_mix);\n",
+      "\n",
+      "#Work required for separation of mixture into pure components is\n",
+      "W = delta_G_mix;\n",
+      "print \"The least amount of work required for separation at 300 K is %f cal/mol\"%(W);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Gibbs free energy change for equimolar mixture formation is -388.185034 cal/mol\n",
+        "The value of enthalpy change for equimolar mixture formation is 25.000000 cal/mol\n",
+        "The least amount of work required for separation at 300 K is -388.185034 cal/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.10  Page Number : 466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math \n",
+      "from numpy import zeros\n",
+      "from scipy.stats import linregress\n",
+      "\n",
+      "# Variables,\n",
+      "T = 25 +173.15;\t\t\t#[K] - Temperature\n",
+      "x_1=[0.0115,0.0160,0.0250,0.0300,0.0575,0.1125,0.1775,0.2330,0.4235,0.5760,0.6605,0.7390,0.8605,0.9250,0.9625];\n",
+      "y_1=[8.0640,7.6260,7.2780,7.2370,5.9770,4.5434,3.4019,2.8023,1.7694,1.3780,1.2302,1.1372,1.0478,1.0145,1.0070];\n",
+      "y_2=[1.0037,1.0099,1.0102,1.0047,1.0203,1.0399,1.1051,1.1695,1.4462,1.8520,2.2334,2.6886,3.7489,4.8960,5.6040];\n",
+      "\n",
+      "x_2 = zeros(15);\t\t\t# x_2 = (1 - x_1)\n",
+      "G_RT = zeros(15);\t\t\t# G_RT = G_excess/(R*T)\n",
+      "x1x2_G_RT = zeros(15);\t\t\t# x1x2_G_RT = (x_1*x_2/(G_excess/(R*T)))\n",
+      "G_RTx1x2 = zeros(15);\t\t\t# G_RTx1x1 = G_excess/(R*T*x_1*x_2)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "for i in range(15):\n",
+      "    x_2[i]=(1-x_1[i]);\n",
+      "    G_RT[i]=(x_1[i]*math.log(y_1[i]))+(x_2[i]*math.log(y_2[i]));\n",
+      "    x1x2_G_RT[i]=(x_1[i]*x_2[i])/G_RT[i];\n",
+      "    G_RTx1x2[i]=1/x1x2_G_RT[i];\n",
+      "\n",
+      "\n",
+      "slop,intr,sig,d,e=linregress(x_1,x1x2_G_RT);\n",
+      "\n",
+      "A = 1/intr;\n",
+      "B = 1/(slop+(1/A));\n",
+      "print \" The value of van Laar parameters are A = %f and B = %f\"%(A,B);\n",
+      "\n",
+      "# Now from Margules equation\n",
+      "# G_RTx1x2 = G_excess/(R*T*x_1*x_2) = B1*x_1 + A1*x_1 = A1 + (B1 - A1)*x_1\n",
+      "#slope = (B1 - A1) and intercept = A1\n",
+      "\n",
+      "# Again employing the concept of linear regression of the data ( x_1 , G_RTx1x2 ) to find the value of intercept and slope of the above equation\n",
+      "#Let slope = slop1 and intercept = intr1\n",
+      "\n",
+      "slop1,intr1,sig1,d,e=linregress(x_1,G_RTx1x2);\n",
+      "\n",
+      "A1 = intr1;\n",
+      "B1 = slop1 + A1;\n",
+      "print \" The value of Margules parameters are A = %f and B = %f\"%(A1,B1);\n",
+      "\n",
+      "print \" Because of the higher value of correlation factor for Van Laar model it fits the data better.\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of van Laar parameters are A = 2.271326 and B = 1.781420\n",
+        " The value of Margules parameters are A = 2.293274 and B = 1.746000\n",
+        " Because of the higher value of correlation factor for Van Laar model it fits the data better.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.11  Page Number : 470"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables,\n",
+      "T = 60 + 273.15;\t\t\t#[K] - Temperature\n",
+      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
+      "#component 1 = acetone\n",
+      "#component 2 = water\n",
+      "x_1 = 0.3;\t\t\t# Mole fraction of component 1\n",
+      "x_2 = 1 - x_1;\t\t\t#Mole fraction of component 2\n",
+      "V_mol_1 = 74.05;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 1\n",
+      "V_mol_2 = 18.07;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2\n",
+      "\n",
+      "#for Wilson equation\n",
+      "a_12 = 291.27;\t\t\t#[cal/mol]\n",
+      "a_21 = 1448.01;\t\t\t#[cal/mol]\n",
+      "\n",
+      "#For NRTL\n",
+      "b_12 = 631.05;\t\t\t#[cal/mol]\n",
+      "b_21 = 1197.41;\t\t\t#[cal/mol]\n",
+      "alpha = 0.5343;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Froom wilson equation\n",
+      "A_12=(V_mol_2/V_mol_1)*(math.exp(-a_12/(R*T)));\n",
+      "A_21 = (V_mol_1/V_mol_2)*(math.exp(-a_21/(R*T)));\n",
+      "Beta = A_12/(x_1+x_2*A_12) - A_21/(x_2+x_1*A_21);\n",
+      "#math.log(Y1) = -math.log(x_1 + x_2*A_12) + x_2*Beta; \n",
+      "Y1 = math.exp(-math.log(x_1+x_2*A_12)+x_2*Beta);\n",
+      "#similarly for Y2\n",
+      "Y2 = math.exp(-math.log(x_2+x_1*A_21)-x_1*Beta);\n",
+      "print \"The value of activity coefficients for Wilson equation are Y1 = %f \\t and \\t Y2 = %f\"%(Y1,Y2);\n",
+      "\n",
+      "#From NRTL equation,\n",
+      "t_12 = b_12/(R*T);\n",
+      "t_21 = b_21/(R*T);\n",
+      "G_12 = math.exp(-alpha*t_12);\n",
+      "G_21 = math.exp(-alpha*t_21);\n",
+      "\n",
+      "#math.log(Y1) = x_1**(2)*[t_12*(G_12/(x_1+x_2*G_12))**(2) + (G_12*t_12)/((G_12/(x_1+x_2*G_12))**(2))]\n",
+      "Y1_prime = math.exp(x_2**(2)*(t_21*(G_21/(x_1+x_2*G_21))**(2)+(G_12*t_12)/(((x_2+x_1*G_12))**(2))));\n",
+      "#Similarly for Y2\n",
+      "Y2_prime = math.exp(x_1**(2)*(t_12*(G_12/(x_2+x_1*G_12))**(2)+(G_21*t_21)/(((x_1+x_2*G_21))**(2))));\n",
+      "\n",
+      "print \"The value of activity coefficients for NRTL equation are Y1 = %f \\t and \\t Y2 = %f\"%(Y1_prime,Y2_prime);\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of activity coefficients for Wilson equation are Y1 = 2.172258 \t and \t Y2 = 1.254121\n",
+        "The value of activity coefficients for NRTL equation are Y1 = 2.143030 \t and \t Y2 = 1.262506\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.12  Page Number : 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables \n",
+      "T = 307;\t\t\t#[K]\n",
+      "x_1 = 0.047;\n",
+      "x_2 = 1 - x_1;\n",
+      "\n",
+      "# The subgroups in the  two components are\n",
+      "# Acetone (1) : 1 CH3, 1 CH3CO\n",
+      "# n-pentane (2) : 2 CH3, 3 CH2\n",
+      "\n",
+      "#Group volume (Rk) and surface area (Qk) parameters of the subgroup are\n",
+      "k_CH3 = 1;\n",
+      "k_CH2 = 2;\n",
+      "k_CH3CO = 19;\n",
+      "Rk_CH3 = 0.9011;\n",
+      "Rk_CH2 = 0.6744;\n",
+      "Rk_CH3CO = 1.6724;\n",
+      "Qk_CH3 = 0.848;\n",
+      "Qk_CH2 = 0.540;\n",
+      "Qk_CH3CO = 1.4880;\n",
+      "\n",
+      "# Interaction parameters of the subgroups in kelvin (K) are\n",
+      "a_1_1 = 0;\n",
+      "a_1_2 = 0;\n",
+      "a_1_19 = 476.40;\n",
+      "a_2_1 = 0;\n",
+      "a_2_2 = 0;\n",
+      "a_2_19 = 476.40;\n",
+      "a_19_1 = 26.76;\n",
+      "a_19_2 = 26.76;\n",
+      "a_19_19 = 0;\n",
+      "\n",
+      "# Calculations\n",
+      "r_1 = 1*Rk_CH3 + 1*Rk_CH3CO;\n",
+      "r_2 = 2*Rk_CH3 + 3*Rk_CH2;\n",
+      "q_1 = 1*Qk_CH3 + 1*Qk_CH3CO;\n",
+      "q_2 = 2*Qk_CH3 + 3*Qk_CH2;\n",
+      "\n",
+      "J_1 = r_1/(r_1*x_1+r_2*x_2);\n",
+      "J_2 = r_2/(r_1*x_1+r_2*x_2);\n",
+      "L_1 = q_1/(q_1*x_1+q_2*x_2);\n",
+      "L_2 = q_2/(q_1*x_1+q_2*x_2);\n",
+      "t_1_1 = math.exp(-a_1_1/T);\n",
+      "t_1_2 = math.exp(-a_1_2/T);\n",
+      "t_1_19 = math.exp(-a_1_19/T);\n",
+      "t_2_1 = math.exp(-a_2_1/T);\n",
+      "t_2_2 = math.exp(-a_2_2/T);\n",
+      "t_2_19 = math.exp(-a_2_19/T);\n",
+      "t_19_1 = math.exp(-a_19_1/T);\n",
+      "t_19_2 = math.exp(-a_19_2/T);\n",
+      "t_19_19 = math.exp(-a_19_19/T);\n",
+      "\n",
+      "e_1_1 = 1*Qk_CH3/q_1;\n",
+      "e_2_1 = 0;\n",
+      "e_19_1 = (1*Qk_CH3CO/q_1);\n",
+      "e_1_2 = 2*Qk_CH3/q_2;\n",
+      "e_2_2 = 3*Qk_CH2/q_2;\n",
+      "e_19_2 = 0;\n",
+      "\n",
+      "B_1_1 = e_1_1*t_1_1 + e_2_1*t_2_1 + e_19_1*t_19_1;\n",
+      "B_1_2 = e_1_1*t_1_2 + e_2_1*t_2_2 + e_19_1*t_19_2;\n",
+      "B_1_19 = e_1_1*t_1_19 + e_2_1*t_2_19 + e_19_1*t_19_19;\n",
+      "B_2_1 = e_1_2*t_1_1 + e_2_2*t_2_1 + e_19_2*t_19_1;\n",
+      "B_2_2 = e_1_2*t_1_2 + e_2_2*t_2_2 + e_19_2*t_19_2;\n",
+      "B_2_19 = e_1_2*t_1_19 + e_2_2*t_2_19 + e_19_2*t_19_19;\n",
+      "\n",
+      "theta_1 = (x_1*q_1*e_1_1 + x_2*q_2*e_1_2)/(x_1*q_1 + x_2*q_2);\n",
+      "theta_2 = (x_1*q_1*e_2_1 + x_2*q_2*e_2_2)/(x_1*q_1 + x_2*q_2);\n",
+      "theta_19 = (x_1*q_1*e_19_1 + x_2*q_2*e_19_2)/(x_1*q_1 + x_2*q_2);\n",
+      "\n",
+      "s_1 = theta_1*t_1_1 + theta_2*t_2_1 + theta_19*t_19_1;\n",
+      "s_2 = theta_1*t_1_2 + theta_2*t_2_2 + theta_19*t_19_2;\n",
+      "s_19 = theta_1*t_1_19 + theta_2*t_2_19 + theta_19*t_19_19;\n",
+      "\n",
+      "# math.log(Y1_C) = 1 - J_1 + math.log(J_1) - 5*q_1*(1- (J_1/L_1) + math.log(J_1/L_1))\n",
+      "# math.log(Y2_C) = 1 - J_2 + math.log(J_2) - 5*q_2*(1- (J_2/L_2) + math.log(J_2/L_2))\n",
+      "Y1_C = math.exp(1 - J_1 + math.log(J_1) - 5*q_1*(1- (J_1/L_1) + math.log(J_1/L_1)));\n",
+      "Y2_C = math.exp(1 - J_2 + math.log(J_2) - 5*q_2*(1- (J_2/L_2) + math.log(J_2/L_2)));\n",
+      "\n",
+      "# For species 1\n",
+      "summation_theta_k_1 = theta_1*(B_1_1/s_1) + theta_2*(B_1_2/s_2) + theta_19*(B_1_19/s_19);\n",
+      "summation_e_ki_1 = e_1_1*math.log(B_1_1/s_1) + e_2_1*math.log(B_1_2/s_2) + e_19_1*math.log(B_1_19/s_19);\n",
+      "\n",
+      "# For species 2\n",
+      "summation_theta_k_2 = theta_1*(B_2_1/s_1) + theta_2*(B_2_2/s_2) + theta_19*(B_2_19/s_19);\n",
+      "summation_e_ki_2 = e_1_2*math.log(B_2_1/s_1) + e_2_2*math.log(B_2_2/s_2) + e_19_2*math.log(B_2_19/s_19);\n",
+      "\n",
+      "# math.log(Y1_R) = q_1(1 - summation_theta_k_1 + summation_e_ki_1)\n",
+      "# math.log(Y2_R) = q_2(1 - summation_theta_k_2 + summation_e_ki_2)\n",
+      "Y1_R = math.exp(q_1*(1 - summation_theta_k_1 + summation_e_ki_1));\n",
+      "Y2_R = math.exp(q_2*(1 - summation_theta_k_2 + summation_e_ki_2));\n",
+      "\n",
+      "# math.log(Y1) = math.log(Y1_C) + math.log(Y1_R)\n",
+      "# math.log(Y2) = math.log(Y2_C) + math.log(Y2_R)\n",
+      "Y1 = math.exp(math.log(Y1_C) + math.log(Y1_R));\n",
+      "Y2 = math.exp(math.log(Y2_C) + math.log(Y2_R));\n",
+      "\n",
+      "# Results\n",
+      "print \" The activity coefficients are Y1 = %f  and  Y2 = %f\"%(Y1,Y2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The activity coefficients are Y1 = 4.992034  and  Y2 = 1.005260\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.13  Page Number : 481"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables,\n",
+      "T = 25 + 273.15;\t\t\t#[K] - Temperature\n",
+      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
+      "#component 1 = chloroform\n",
+      "#component 2 = carbon tetrachloride\n",
+      "x_1 = 0.5;\t\t\t#Mole fraction of component 1 \t\t\t#Equimolar mixture\n",
+      "x_2 = 0.5;\t\t\t#Mole fraction of component 2\n",
+      "V_mol_1 = 81;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 1\n",
+      "V_mol_2 = 97;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2\n",
+      "del_1 = 9.2;\t\t\t#[(cal/cm**(3))**(1/2)] - Mole fraction of component 1\n",
+      "del_2 = 8.6;\t\t\t#[(cal/cm**(3))**(1/2)] - Mole fraction of component 2\n",
+      "\n",
+      "# Calculations\n",
+      "#Scatchard - Hilderbrand model\n",
+      "phi_1 = (x_1*V_mol_1)/(x_1*V_mol_1+x_2*V_mol_2);\n",
+      "phi_2 = (x_2*V_mol_2)/(x_1*V_mol_1+x_2*V_mol_2);\n",
+      "\n",
+      "#math.log(Y1) = (V_mol_1/(R*T))*phi_1**(2)*(del_1-del_2)**(2)\n",
+      "Y1 = math.exp((V_mol_1/(R*T))*(phi_1**(2))*((del_1-del_2)**(2)));\n",
+      "\n",
+      "#Similarly, for Y2\n",
+      "Y2 = math.exp((V_mol_2/(R*T))*(phi_2**(2))*((del_1-del_2)**(2)));\n",
+      "\n",
+      "# Results\n",
+      "print \"The value of activity coefficients for Scatchard-Hilderbrand model are Y1 = %f \\t and \\t Y2 = %f\"%(Y1,Y2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of activity coefficients for Scatchard-Hilderbrand model are Y1 = 1.010245 \t and \t Y2 = 1.017658\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.14  Page Number : 485"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables,\n",
+      "T = 25 + 273.15;\t\t\t#[K] - Temperature\n",
+      "mol_HCl = 0.001;\t\t\t#[mol/kg] - Molality of HCl\n",
+      "A = 0.510;\t\t\t#[(kg/mol)**(1/2)]\n",
+      "Z_positive = 1;\t\t\t#Stoichiometric coefficient of 'H' ion\n",
+      "Z_negative = -1;\t\t\t#Stoichiometric coefficient of 'Cl' ion\n",
+      "m_H_positive = mol_HCl;\t\t\t#\n",
+      "m_Cl_negative = mol_HCl;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# I = 1/2*[((Z_positive)**(2))*m_H_positive + ((Z_negative)**(2))*m_Cl_negative]\n",
+      "I = 1./2*(((Z_positive)**(2))*m_H_positive + ((Z_negative)**(2))*m_Cl_negative);\n",
+      "\n",
+      "#Using Debye-Huckel limiting law wee get,\n",
+      "# math.log(Y1) = -A*(abs(Z_positive*Z_negative))*(I**(1/2)))\n",
+      "Y = 10**(-A*(abs(Z_positive*Z_negative))*(I**(1./2)));\n",
+      "print \"The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = %f\"%(Y);\n",
+      "\n",
+      "#Using Debye-Huckel extended model we get\n",
+      "#math.log(Y_prime) = (-A*(abs(Z_positive*Z_negative))*(I**(1/2)))/(1 + (I**(1/2)));\n",
+      "Y_prime = 10**((-A*(abs(Z_positive*Z_negative))*(I**(1./2)))/(1 + (I**(1./2))));\n",
+      "print \"The mean activity coefficient at 25 C using Debye-Huckel extended model is Y = %f\"%(Y_prime);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = 0.963546\n",
+        "The mean activity coefficient at 25 C using Debye-Huckel extended model is Y = 0.964643\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.15  Page Number : 485"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables,\n",
+      "T = 25 + 273.15;\t\t\t#[K] - Temperature\n",
+      "mol_CaCl2 = 0.001;\t\t\t#[mol/kg] - Molality of HCl\n",
+      "A = 0.510;\t\t\t#[(kg/mol)**(1/2)]\n",
+      "Z_positive = 2;\t\t\t#Stoichiometric coefficient of 'Ca' ion\n",
+      "Z_negative = -1;\t\t\t#Stoichiometric coefficient of 'Cl' ion\n",
+      "m_Ca_positive = mol_CaCl2;\n",
+      "m_Cl_negative = 2*mol_CaCl2;\n",
+      "\n",
+      "# Calculations\n",
+      "# I = 1/2*[((Z_positive)**(2))*m_Ca_positive + ((Z_negative)**(2))*m_Cl_negative]\n",
+      "I = 1./2*(((Z_positive)**(2))*m_Ca_positive + ((Z_negative)**(2))*m_Cl_negative);\n",
+      "\n",
+      "#Using Debye-Huckel limiting law wee get,\n",
+      "# math.log(Y1) = -A*(abs(Z_positive*Z_negative))*(I**(1/2)))\n",
+      "Y = 10**(-A*(abs(Z_positive*Z_negative))*(I**(1./2)));\n",
+      "\n",
+      "# Results\n",
+      "print \"The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = %f\"%(Y);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = 0.879290\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.17  Page Number : 488"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables,\n",
+      "T = 50 + 273.15;\t\t\t#[K] - Temperature\n",
+      "R=8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "x_1 = 0.3;\t\t\t# Mole fraction of component 1\n",
+      "x_2 = (1-x_1);\t\t\t# Mole fraction of component 2\n",
+      "#Increment of 1% means Y2 = 1.01*Y1\n",
+      "\n",
+      "# Calculations\n",
+      "#Excess volume of the mixture is given by,\n",
+      "V_excess = 4*x_1*x_2;\t\t\t#[cm**(3)/mol]\n",
+      "#Amd threfore\n",
+      "V_1_excess = 4*x_2*x_2*10**(-6);\t\t\t#[m**(3)/mol] - Exces volume of component 1\n",
+      "V_2_excess = 4*x_1*x_1*10**(-6);\t\t\t#[m**(3)/mol] - Exces volume of component 2\n",
+      "\n",
+      "#We have from equation 14.89 of the book,\n",
+      "#V_i_excess/(R*T) = (del_math.log(Y_i)/del_P)_T,x\n",
+      "\n",
+      "#Rearranging above equation\n",
+      "#d(math.log(Y1)) = (V1_excess/(R*T))dP\n",
+      "#Integrating the above equation at constant 'T' and 'x' in the limit from 'Y1' to '1.01*Y1' in the LHS and from 'P' to 'P+delta_P' in the RHS\n",
+      "#On simplification we get\n",
+      "#math.log(1.01*Y1/Y1) = (V_1_exces/(R*T))*delta_P\n",
+      "delta_P = math.log(1.01)/(V_1_excess/(R*T));\t\t\t#[N/m**(2)]\n",
+      "delta_P = delta_P*10**(-6);\t\t\t#[MPa]\n",
+      "\n",
+      "# Results\n",
+      "print \"The required pressure to increase the activity coefficient by 1%% is %f MPa\"%(delta_P);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required pressure to increase the activity coefficient by 1% is 13.639411 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.21  Page Number : 490"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 293.15;\t\t\t#[K] - Temperature\n",
+      "R=8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "A = 1280;\t\t\t#[J/mol]\n",
+      "\n",
+      "#(dA/dT)_P,x = del_A (say)\n",
+      "dA_dT = -7.0;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "#For equilomar mixture,\n",
+      "x_1 = 0.5;\t\t\t# Mole fraction of component 1\n",
+      "x_2 = 0.5;\t\t\t# Mole fraction of component 2\n",
+      "\n",
+      "# Calculations\n",
+      "#math.log(Y1) =  (A/(R*T))*x_2**(2)\n",
+      "#math.log(Y2) =  (A/(R*T))*x_1**(2)\n",
+      "Y1 = math.exp((A/(R*T))*x_2**(2));\n",
+      "Y2 = math.exp((A/(R*T))*x_1**(2));\n",
+      "\n",
+      "#G_excess/(R*T*) = x_1*math.log(Y1) + x_2*math.log(Y2) = (A/(R*T))*x_1*x_2\n",
+      "G_excess = A*x_1*x_2;\t\t\t#[J/mol] - Excess Gibbs free energy\n",
+      "\n",
+      "#H_excess/(R*T**(2)) = -[d/dT(G_excess/(R*T))]_P,x\n",
+      "#H_excess/(R*T**(2)) = -((x_1*x_2)/R)*[d/dT(A/T)]_P,x\n",
+      "#On simplification  and putting the values we get\n",
+      "H_excess = A*x_1*x_2 - T*dA_dT*x_1*x_2;\t\t\t#[J/mol] - Excess enthalpy\n",
+      "\n",
+      "#Now excess entropy is given by\n",
+      "S_excess = (H_excess - G_excess)/T;\t\t\t#[J/mol-K] - Excess entropy\n",
+      "\n",
+      "# Results\n",
+      "print \"For equimolar mixture\";\n",
+      "print \"Excess Gibbs free energy G_excess) is %f J/mol\"%(G_excess);\n",
+      "print \"Excess enthalpy H_excess) is %f J/mol\"%(H_excess);\n",
+      "print \"Excess entropy S_excess) is %f J/mol\"%(S_excess);\n",
+      "print \"The value of activity coefficient Y1 is %f\"%(Y1);\n",
+      "print \"The value of activity coefficient Y2 is %f\"%(Y2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For equimolar mixture\n",
+        "Excess Gibbs free energy G_excess) is 320.000000 J/mol\n",
+        "Excess enthalpy H_excess) is 833.012500 J/mol\n",
+        "Excess entropy S_excess) is 1.750000 J/mol\n",
+        "The value of activity coefficient Y1 is 1.140305\n",
+        "The value of activity coefficient Y2 is 1.140305\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.22  Page Number : 491"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Given\n",
+      "T = 60 + 273.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# math.log(Y1_inf) = math.log(Y2_inf) = 0.15 + 10/T\n",
+      "\n",
+      "\n",
+      "# From van L# Variablesr equation \n",
+      "# A = math.log(Y1_inf) and B = math.log(Y2_inf) and since it is given that math.log(Y1_inf) = math.log(Y2_inf), therefore A = B\n",
+      "#(x_1*x_2)/(G_excess/R*T) = x_1/B + x_2/A = X_1/A + x_2/A = 1/A\n",
+      "# G_excess/(R*T) = A*x_1*x_2\n",
+      "\n",
+      "# For equilomar mixture,\n",
+      "x_1 = 0.5;\t\t\t# Mole fraction of component 1\n",
+      "x_2 = 0.5;\t\t\t# Mole fraction of component 2\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# Expression for A can be written as\n",
+      "# A = 0.15 + 10/T, where T is in C. Therefore\n",
+      "A = 0.15 + 10/(T - 273.15);\n",
+      "# Differentiating it with respect to temprature we get\n",
+      "dA_dT = - 10/((T-273.15)**(2));\n",
+      "\n",
+      "# The excess Gibbs free energy can be calculated as\n",
+      "G_excess = A*x_1*x_2*(R*T);\t\t\t#[J/mol]\n",
+      "\n",
+      "# The ideal Gibbs free energy change can  be calculated as\n",
+      "delta_G_id_mix = R*T*(x_1*math.log(x_1) + x_2*math.log(x_2));\t\t\t#[J/mol]\n",
+      "\n",
+      "# Finally we have,\n",
+      "delta_G_mix = G_excess + delta_G_id_mix;\t\t\t#[J/mol]\n",
+      "\n",
+      "print \"The Gibbs free energy change of mixing for equimolar mixture is %f J/mol\"%(delta_G_mix);\n",
+      "\n",
+      "\n",
+      "# Now let us determine the excess enthalpy. We know that\n",
+      "# H_excess/(R*T**(2)) = -[d/dT(G_excess/R*T)]_P,x = - x_1*x_2*[d/dT(A)]_P,x\n",
+      "# Therefore at 'T' = 60 C the excess enthalpy is given by\n",
+      "H_excess = -R*(T**(2))*x_1*x_2*dA_dT;\t\t\t#[J/mol]\n",
+      "\n",
+      "delta_H_id_mix = 0;\t\t\t#[J/mol] - Enthalpy change of mixing for ideal solution is zero.\n",
+      "\n",
+      "#Thus enthalpy change of mixing for an equimolar mixture at 333.15 K is given by\n",
+      "delta_H_mix = delta_H_id_mix + H_excess;\t\t\t#[J/mol]\n",
+      "\n",
+      "\n",
+      "print \"The enthalpy change of mixing for equimolar mixture is %f J/mol\"%(delta_H_mix);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Gibbs free energy change of mixing for equimolar mixture is -1700.608815 J/mol\n",
+        "The enthalpy change of mixing for equimolar mixture is 640.806876 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch15_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch15_1.ipynb
new file mode 100755
index 00000000..83aaf521
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch15_1.ipynb
@@ -0,0 +1,2056 @@
+{
+ "metadata": {
+  "name": "ch15_1",
+  "signature": "sha256:aa6f699613cdf9bcba2dda4617212630ca1760d21939d5cd99d17e7e03b7e885"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15 : Vapour Liquid Equilibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.1  Page Number : 503"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 90+ 273.15;\t\t\t#[K] - Temperature\n",
+      "P = 1;\t\t\t#[atm] - Pressure\n",
+      "x_1 = 0.5748;\t\t\t# Equilibrium composition of liquid phase\n",
+      "y_1 = 0.7725;\t\t\t# Equilibrium composition of vapour phase\n",
+      "\n",
+      "\t\t\t# We start with 1 mol of mixture of composition z_1 = 0.6, from marterial balance we get\n",
+      "\t\t\t# (L + V)*z_1 = L*x_1 + V*y_1\n",
+      "\t\t\t# Since total number of moles is 1, we get\n",
+      "\t\t\t# z_1 = L*x_1 + (1-L)*y_1\n",
+      "\n",
+      "# Calculations and Results\n",
+      "z_1_1 = 0.6;\t\t\t# - Mole fraction of benzene\n",
+      "L_1 = (z_1_1 - y_1)/(x_1 - y_1);\n",
+      "V_1 = 1 - L_1;\n",
+      "\n",
+      "print \" For z_1 = 0.6\";\n",
+      "print \" The moles in the liquid phase is %f mol\"%(L_1);\n",
+      "print \" The moles in the vapour phase is %f mol\"%(V_1);\n",
+      "\n",
+      "z_1_2 = 0.7;\t\t\t# - Mole fraction of benzene\n",
+      "L_2 = (z_1_2 - y_1)/(x_1 - y_1);\n",
+      "V_2 = 1 - L_2;\n",
+      "\n",
+      "print \" For z_1 = 0.7\";\n",
+      "print \" The moles in the liquid phase is %f mol\"%(L_2);\n",
+      "print \" The moles in the vapour phase is %f mol\"%(V_2);\n",
+      "\n",
+      "\n",
+      "\t\t\t# For z = 0.8\n",
+      "\t\t\t# The feed remains vapour and the liquid is not formed at all as it is outside the two-phase region (x_1 = 0.5748 and y_1 = 0.7725).\n",
+      "print \" For z_1 = 0.8\";\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " For z_1 = 0.6\n",
+        " The moles in the liquid phase is 0.872534 mol\n",
+        " The moles in the vapour phase is 0.127466 mol\n",
+        " For z_1 = 0.7\n",
+        " The moles in the liquid phase is 0.366717 mol\n",
+        " The moles in the vapour phase is 0.633283 mol\n",
+        " For z_1 = 0.8\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.2  Page Number : 515"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "\t\t\t# math.log(P_1_sat) = 14.39155 - 2795.817/(t + 230.002)\n",
+      "\t\t\t# math.log(P_2_sat) = 16.26205 - 3799.887/(t + 226.346)\n",
+      "\n",
+      "\t\t\t#(a)\n",
+      "x_1_a =0.43;\t\t\t# Equilibrium composition of liquid phase\n",
+      "t_a = 76;\t\t\t#[C] - Temperature\n",
+      "x_2_a = 1 - x_1_a;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# Since liquid phase composition is given we use the relation \n",
+      "\t\t\t# P = x_1*P_1_sat + x_2*P_2_sat\n",
+      "\t\t\t# At t = 76 C\n",
+      "P_1_sat_a = math.exp(14.39155 - 2795.817/(t_a + 230.002));\n",
+      "P_2_sat_a = math.exp(16.26205 - 3799.887/(t_a + 226.346));\n",
+      "\t\t\t# Therefore total pressure is\n",
+      "P_a = x_1_a*P_1_sat_a + x_2_a*P_2_sat_a;\t\t\t#[kPa]\n",
+      "y_1_a = (x_1_a*P_1_sat_a)/(P_a);\n",
+      "y_2_a = (x_2_a*P_2_sat_a)/(P_a);\n",
+      "\n",
+      "print \"a).The system pressure is = %f kPa\"%(P_a);\n",
+      "print \"    The vapour phase composition is y_1 = %f\"%(y_1_a);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "y_1_b = 0.43;\t\t\t# Equilibrium composition of vapour phase\n",
+      "y_2_b = 1 - y_1_b;\n",
+      "t_b = 76;\t\t\t#[C] - Temperature\n",
+      "\n",
+      "P_1_sat_b = math.exp(14.39155 - 2795.817/(t_b + 230.002));\n",
+      "P_2_sat_b = math.exp(16.26205 - 3799.887/(t_b + 226.346));\n",
+      "\n",
+      "\t\t\t# Since vapour phase composition is given ,the system pressure is given by\n",
+      "\t\t\t# 1/P = y_1/P_1_sat + y_2/P_2_sat\n",
+      "P_b = 1/(y_1_b/P_1_sat_b + y_2_b/P_2_sat_b);\n",
+      "\n",
+      "x_1_b = (y_1_b*P_b)/P_1_sat_b;\n",
+      "x_2_b = (y_2_b*P_b)/P_2_sat_b;\n",
+      "\n",
+      "print \"b).The system pressure is P = %f kPa\"%(P_b);\n",
+      "print \"    The liquid phase composition is x_1 = %f\"%(x_1_b);\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "x_1_c = 0.32;\t\t\t# Equilibrium composition of liquid phase\n",
+      "x_2_c = 1 - x_1_c;\n",
+      "P_c = 101.33;\t\t\t#[kPa] - Pressure of the system\n",
+      "\n",
+      "\t\t\t# We have,  P = x_1*P_1_sat + x_2*P_2_sat\n",
+      "t_1_sat = 2795.817/(14.39155 - math.log(P_c)) - 230.002;\n",
+      "t_2_sat = 3799.887/(16.26205 - math.log(P_c)) - 226.346;\n",
+      "t = x_1_c*t_1_sat + x_2_c*t_2_sat;\n",
+      "\n",
+      "error = 10;\n",
+      "while(error>0.1):\n",
+      "    P_1_sat = math.exp(14.39155 - 2795.817/(t + 230.002));\n",
+      "    P_2_sat = math.exp(16.26205 - 3799.887/(t + 226.346));\n",
+      "    P = x_1_c*P_1_sat + x_2_c*P_2_sat;\n",
+      "    error=abs(P - P_c);\n",
+      "    t = t - 0.1;\n",
+      "\n",
+      "P_1_sat_c = math.exp(14.39155 - 2795.817/(t + 230.002));\n",
+      "P_2_sat_c = math.exp(16.26205 - 3799.887/(t + 226.346));\n",
+      "\n",
+      "y_1_c = (x_1_c*P_1_sat_c)/(P_c);\n",
+      "y_2_c = 1 - y_1_c;\n",
+      "\n",
+      "print \"c).The system temperature is t = %f C\"%(t);\n",
+      "print \"    The vapour phase composition is y_1 = %f\"%(y_1_c);\n",
+      "\n",
+      "\t\t\t#(d)\n",
+      "y_1_d = 0.57;\t\t\t# Vapour phase composition\n",
+      "y_2_d = 1 - y_1_d;\n",
+      "P_d = 101.33;\t\t\t#[kPa] - Pressure of the system\n",
+      "\n",
+      "\t\t\t# Since vapour phase composition is given, we can use the relation\n",
+      "\t\t\t# 1/P = y_1/P_1_sat + y_2/P_2_sat\n",
+      "t_1_sat_d = 2795.817/(14.39155 - math.log(P_d)) - 230.002;\n",
+      "t_2_sat_d = 3799.887/(16.26205 - math.log(P_d)) - 226.346;\n",
+      "t_d = y_1_d*t_1_sat_d + y_2_d*t_2_sat_d;\n",
+      "\n",
+      "fault = 10;\n",
+      "while(fault>0.1):\n",
+      "    P_1_sat_prime = math.exp(14.39155 - 2795.817/(t_d + 230.002));\n",
+      "    P_2_sat_prime = math.exp(16.26205 - 3799.887/(t_d + 226.346));\n",
+      "    P_prime = 1/(y_1_d/P_1_sat_prime + y_2_d/P_2_sat_prime);\n",
+      "    fault=abs(P_prime - P_d);\n",
+      "    t_d = t_d + 0.01;\n",
+      "\n",
+      "P_1_sat_d = math.exp(14.39155 - 2795.817/(t_d + 230.002));\n",
+      "P_2_sat_d = math.exp(16.26205 - 3799.887/(t_d + 226.346));\n",
+      "\n",
+      "x_1_d = (y_1_d*P_d)/(P_1_sat_d);\n",
+      "x_2_d = 1 - x_1_d;\n",
+      "\n",
+      "print \"d).The system temperature is t = %f C\"%(t_d);\n",
+      "print \"    The liquid phase composition is x_1 = %f\"%(x_1_d);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a).The system pressure is = 105.268363 kPa\n",
+        "    The vapour phase composition is y_1 = 0.782290\n",
+        "b).The system pressure is P = 60.894254 kPa\n",
+        "    The liquid phase composition is x_1 = 0.136725\n",
+        "c).The system temperature is t = 79.943314 C\n",
+        "    The vapour phase composition is y_1 = 0.679347\n",
+        "d).The system temperature is t = 84.600005 C\n",
+        "    The liquid phase composition is x_1 = 0.234934\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.3  Page Number : 516"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "x_1_a = 0.25;\t\t\t# Equilibrium composition of liquid phase\n",
+      "x_2_a = 0.35;\n",
+      "x_3_a = 1 - x_1_a - x_2_a;\n",
+      "t_a = 80;\t\t\t#[C] - Temperature\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# At t = 80 C\n",
+      "P_1_sat_a = math.exp(14.3916 - 2795.82/(t_a + 230.00));\n",
+      "P_2_sat_a = math.exp(14.2724 - 2945.47/(t_a + 224.00));\n",
+      "P_3_sat_a = math.exp(14.2043 - 2972.64/(t_a + 209.00));\n",
+      "\n",
+      "\t\t\t# Since liquid phase composition is given we use the relation \n",
+      "P_a = x_1_a*P_1_sat_a + x_2_a*P_2_sat_a + x_3_a*P_3_sat_a;\t\t\t#[kPa]\n",
+      "\n",
+      "y_1_a = (x_1_a*P_1_sat_a)/(P_a);\n",
+      "y_2_a = (x_2_a*P_2_sat_a)/(P_a);\n",
+      "y_3_a = (x_3_a*P_3_sat_a)/(P_a);\n",
+      "\n",
+      "print \"a).The system pressure is P = %f kPa\"%(P_a);\n",
+      "print \"    The vapour phase composition is given by y_1 = %f  y_2 = %f and y_3 = %f \"%(y_1_a,y_2_a,y_3_a);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "y_1_b = 0.50;\t\t\t# Equilibrium composition of liquid phase\n",
+      "y_2_b = 0.30;\n",
+      "y_3_b = 1 - y_1_a - y_2_a;\n",
+      "t_b = 85;\t\t\t#[C] - Temperature\n",
+      "\n",
+      "\t\t\t# At t = 80 C\n",
+      "P_1_sat_b = math.exp(14.3916 - 2795.82/(t_b + 230.00));\n",
+      "P_2_sat_b = math.exp(14.2724 - 2945.47/(t_b + 224.00));\n",
+      "P_3_sat_b = math.exp(14.2043 - 2972.64/(t_b + 209.00));\n",
+      "\n",
+      "\t\t\t# Since vapour phase composition is given we use the relation \n",
+      "P_b = 1/(y_1_b/P_1_sat_b + y_2_b/P_2_sat_b + y_3_b/P_3_sat_b);\t\t\t#[kPa]\n",
+      "\n",
+      "\t\t\t# Therefore we have\n",
+      "x_1_b = (y_1_b*P_b)/P_1_sat_b;\n",
+      "x_2_b = (y_2_b*P_b)/P_2_sat_b;\n",
+      "x_3_b = (y_3_b*P_b)/P_3_sat_b;\n",
+      "\n",
+      "print \"b).The system pressure is P = %f kPa\"%(P_b);\n",
+      "print \"    The liquid phase composition is given by x_1 = %f x_2 = %f and x_3 = %f \"%(x_1_b,x_2_b,x_3_b);\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "x_1_c = 0.30;\t\t\t# Equilibrium composition of liquid phase\n",
+      "x_2_c = 0.45;\n",
+      "x_3_c = 1 - x_1_c - x_2_c;\n",
+      "P_c = 80;\t\t\t#[kPa] - Pressure of the system\n",
+      "\n",
+      "\t\t\t# We have,  P = x_1*P_1_sat + x_2*P_2_sat + x_3*P_3_sat\n",
+      "t_1_sat = 2795.82/(14.3916 - math.log(P_c)) - 230.00;\t\t\t#[C]\n",
+      "t_2_sat = 2945.47/(14.2724 - math.log(P_c)) - 224.00;\t\t\t#[C]\n",
+      "t_3_sat = 2972.64/(14.2043 - math.log(P_c)) - 209.00;\t\t\t#[C]\n",
+      "t = x_1_c*t_1_sat + x_2_c*t_2_sat + x_3_c*t_3_sat;\n",
+      "\n",
+      "error = 10;\n",
+      "while(error>0.5):\n",
+      "    P_1_sat = math.exp(14.3916 - 2795.82/(t + 230.00));\n",
+      "    P_2_sat = math.exp(14.2724 - 2945.47/(t + 224.00));\n",
+      "    P_3_sat = math.exp(14.2043 - 2972.64/(t + 209.00));\n",
+      "    P = x_1_c*P_1_sat + x_2_c*P_2_sat + x_3_c*P_3_sat;\n",
+      "    error=abs(P - P_c);\n",
+      "    t = t - 0.2;\n",
+      "\n",
+      "P_1_sat_c = math.exp(14.3916 - 2795.82/(t + 230.00));\n",
+      "P_2_sat_c = math.exp(14.2724 - 2945.47/(t + 224.00));\n",
+      "P_3_sat_c = math.exp(14.2043 - 2972.64/(t + 209.00));\n",
+      "\n",
+      "y_1_c = (x_1_c*P_1_sat_c)/(P_c);\n",
+      "y_2_c = (x_2_c*P_2_sat_c)/(P_c);\n",
+      "y_3_c = 1 - y_1_c - y_2_c;\n",
+      "\n",
+      "print \"c).The system temperature is t = %f C\"%(t);\n",
+      "print \"    The vapour phase composition is y_1 = %f y_2 %f and y_3 = %f\"%(y_1_c,y_2_c,y_3_c);\n",
+      "\n",
+      "\t\t\t#(d)\n",
+      "y_1_d = 0.6;\t\t\t# Vapour phase composition\n",
+      "y_2_d = 0.2;\n",
+      "y_3_d = 1 - y_1_d - y_2_d;\n",
+      "P_d = 90;\t\t\t#[kPa] - Pressure of the system\n",
+      "\n",
+      "\t\t\t# Since vapour phase composition is given, we can use the relation\n",
+      "\t\t\t# 1/P = y_1/P_1_sat + y_2/P_2_sat + y_3/P_3_sat\n",
+      "t_1_sat_d = 2795.82/(14.3916 - math.log(P_d)) - 230.00;\n",
+      "t_2_sat_d = 2945.47/(14.2724 - math.log(P_d)) - 224.00;\n",
+      "t_3_sat_d = 2972.64/(14.2043 - math.log(P_d)) - 209.00;\n",
+      "t_d = y_1_d*t_1_sat_d + y_2_d*t_2_sat_d + y_3_d*t_3_sat_d;\n",
+      "\n",
+      "fault = 10;\n",
+      "while(fault>0.5):\n",
+      "    P_1_sat_prime = math.exp(14.3916 - 2795.82/(t_d + 230.00));\n",
+      "    P_2_sat_prime = math.exp(14.2724 - 2945.47/(t_d + 224.00));\n",
+      "    P_3_sat_prime = math.exp(14.2043 - 2972.64/(t_d + 209.00));\n",
+      "    P_prime = 1/(y_1_d/P_1_sat_prime + y_2_d/P_2_sat_prime + y_3_d/P_3_sat_prime);\n",
+      "    fault=abs(P_prime - P_d);\n",
+      "    t_d = t_d + 0.2;\n",
+      "\n",
+      "P_1_sat_d = math.exp(14.3916 - 2795.82/(t_d + 230.00));\n",
+      "P_2_sat_d = math.exp(14.2724 - 2945.47/(t_d + 224.00));\n",
+      "P_3_sat_d = math.exp(14.2043 - 2972.64/(t_d + 209.00));\n",
+      "\n",
+      "x_1_d = (y_1_d*P_d)/(P_1_sat_d);\n",
+      "x_2_d = (y_2_d*P_d)/(P_2_sat_d);\n",
+      "x_3_d = 1 - x_1_d - x_2_d;\n",
+      "\n",
+      "print \"d).The system temperature is t = %f C\"%(t_d);\n",
+      "print \"    The liquid phase composition is x_1 = %f x_2 = %f and x_3 = %f\"%(x_1_d,x_2_d,x_3_d);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a).The system pressure is P = 108.239332 kPa\n",
+        "    The vapour phase composition is given by y_1 = 0.497672  y_2 = 0.316379 and y_3 = 0.185948 \n",
+        "b).The system pressure is P = 129.287998 kPa\n",
+        "    The liquid phase composition is given by x_1 = 0.259997 x_2 = 0.338895 and x_3 = 0.401108 \n",
+        "c).The system temperature is t = 67.020387 C\n",
+        "    The vapour phase composition is y_1 = 0.544826 y_2 0.357251 and y_3 = 0.097922\n",
+        "d).The system temperature is t = 73.127683 C\n",
+        "    The liquid phase composition is x_1 = 0.307471 x_2 = 0.230183 and x_3 = 0.462346\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.4  Page Number : 519"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 120;\t\t\t#[C] - Temperature\n",
+      "P_1 = 1;\t\t\t#[atm] - Initial pressure\n",
+      "P_1 = P_1*101.325;\t\t\t#[kPa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "y_1 = 0.3;\t\t\t# Mole fraction of propane\n",
+      "y_2 = 0.5;\t\t\t# Mole fraction of butane\n",
+      "y_3 = 0.2;\t\t\t# Mole fraction of hexane\n",
+      "\n",
+      "\t\t\t# math.log(P_1_sat) = 13.7713 - 1892.47/(t + 248.82)\n",
+      "\t\t\t# math.log(P_2_sat) = 13.7224 - 2151.63/(t + 236.91)\n",
+      "\t\t\t# math.log(P_3_sat) = 13.8216 - 2697.55/(t + 224.37)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(a)\n",
+      "P_1_sat = math.exp(13.7713 - 1892.47/(T + 248.82));\n",
+      "P_2_sat = math.exp(13.7224 - 2151.63/(T + 236.91));\n",
+      "P_3_sat = math.exp(13.8216 - 2697.55/(T + 224.37));\n",
+      "\n",
+      "\t\t\t# Since vapour phase composition is given we can use the relation,\n",
+      "P_2 = 1/(y_1/P_1_sat + y_2/P_2_sat + y_3/P_3_sat);\t\t\t#[kPa]\n",
+      "\n",
+      "print \" a).The pressure of the mixture when first drop condenses is given by P = %f kPa\"%(P_2);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "x_1 = (y_1*P_2)/P_1_sat;\n",
+      "x_2 = (y_2*P_2)/P_2_sat;\n",
+      "x_3 = (y_3*P_2)/P_3_sat;\n",
+      "\n",
+      "print \" b).The liquid phase composition is given by x_1 propane = %f x_2 butane = %f and x_3 hexane = %f \"%(x_1,x_2,x_3);\n",
+      "\n",
+      "\t\t\t# (c)\n",
+      "\t\t\t# Work transfer per mol is given by\n",
+      "\t\t\t# W = integral(P*dV) = integral((R*T/V)*dV) = R*T*math.log(V_2/V_1) = R*T*math.log(P_1/P_2)\n",
+      "w = R*(T+273.15)*math.log(P_1/P_2);\t\t\t#[J/mol]\n",
+      "\t\t\t# For 100 mol\n",
+      "W = w*100;\t\t\t#[J]\n",
+      "W = W*10**(-3);\t\t\t#[kJ]\n",
+      "print \" c).The work required for 100 mol of mixture handeled is %f kJ\"%(W);\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The pressure of the mixture when first drop condenses is given by P = 1278.060855 kPa\n",
+        " b).The liquid phase composition is given by x_1 propane = 0.067811 x_2 butane = 0.291139 and x_3 hexane = 0.641049 \n",
+        " c).The work required for 100 mol of mixture handeled is -828.526086 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.5  Page Number : 520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "# Variables\n",
+      "T = 27;\t\t\t#[C] - Temperature\n",
+      "\n",
+      "\t\t\t# math.log(P_1_sat) = 13.8216 - 2697.55/(t + 224.37)\n",
+      "\t\t\t# math.log(P_2_sat) = 13.8587 - 2911.32/(t + 216.64)\n",
+      "\n",
+      "\t\t\t#(a)\n",
+      "x_1_a = 0.2;\n",
+      "x_2_a = 1 - x_1_a;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# At t = 27 C\n",
+      "P_1_sat = math.exp(13.8216 - 2697.55/(T + 224.37));\t\t\t#[kPa]\n",
+      "P_2_sat = math.exp(13.8587 - 2911.32/(T + 216.64));\t\t\t#[kPa]\n",
+      "P_a = x_1_a*P_1_sat + x_2_a*P_2_sat;\t\t\t#[kPa]\n",
+      "\n",
+      "y_1_a = x_1_a*P_1_sat/P_a;\n",
+      "y_2_a = x_2_a*P_2_sat/P_a;\n",
+      "\n",
+      "print \"a).The total pressure is P = %f kPa\"%(P_a);\n",
+      "print \"    The vapour phase composition is given by y_1 = %f and y_2 = %f\"%(y_1_a,y_2_a);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "y_1_b = 0.2;\n",
+      "y_2_b = 1 - y_1_b;\n",
+      "\t\t\t# Since vapour phase composition is given we can use the relation \n",
+      "P_b = 1/(y_1_b/P_1_sat + y_2_b/P_2_sat);\t\t\t#[kPa]\n",
+      "\n",
+      "\t\t\t# Therefore we have\n",
+      "x_1_b = (y_1_b*P_b)/P_1_sat;\n",
+      "x_2_b = (y_2_b*P_b)/P_2_sat;\n",
+      "\n",
+      "print \"b).The total pressure is, P = %f kPa\"%(P_b);\n",
+      "print \"    The liquid phase composition is given by x_1 = %f and x_2 = %f\"%(x_1_b,x_2_b);\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "P_c = 30;\t\t\t#[kPa] - Total pressure\n",
+      "x_1_c = 0.2;\n",
+      "x_2_c = 1 - x_1_c;\n",
+      "\n",
+      "\t\t\t# We have,  P = x_1*P_1_sat + x_2*P_2_sat\n",
+      "t_1_sat = 2697.55/(13.8216 - math.log(P_c)) - 224.37;\n",
+      "t_2_sat = 2911.32/(13.8587 - math.log(P_c)) - 216.64;\n",
+      "t = x_1_c*t_1_sat + x_2_c*t_2_sat;\n",
+      "\n",
+      "fault = 10;\n",
+      "while(fault>0.3):\n",
+      "    P_1_sat = math.exp(13.8216 - 2697.55/(t + 224.37));\n",
+      "    P_2_sat = math.exp(13.8587 - 2911.32/(t + 216.64));\n",
+      "    P = x_1_c*P_1_sat + x_2_c*P_2_sat;\n",
+      "    fault = abs(P - P_c);\n",
+      "    t = t - 0.1;\n",
+      "\n",
+      "P_1_sat_c = math.exp(13.8216 - 2697.55/(t + 224.37));\n",
+      "P_2_sat_c = math.exp(13.8587 - 2911.32/(t + 216.64));\n",
+      "\n",
+      "y_1_c = (x_1_c*P_1_sat_c)/(P_c);\n",
+      "y_2_c = 1 - y_1_c;\n",
+      "\n",
+      "print \"c).The system temperature is t = %f C\"%(t);\n",
+      "print \"    The vapour phase composition is y_1 = %f and y_2 = %f \"%(y_1_c,y_2_c);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a).The total pressure is P = 9.795726 kPa\n",
+        "    The vapour phase composition is given by y_1 = 0.448801 and y_2 = 0.551199\n",
+        "b).The total pressure is, P = 7.835131 kPa\n",
+        "    The liquid phase composition is given by x_1 = 0.071288 and x_2 = 0.928712\n",
+        "c).The system temperature is t = 53.904663 C\n",
+        "    The vapour phase composition is y_1 = 0.413592 and y_2 = 0.586408 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.6  Page Number : 521"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P = 90;\t\t\t#[kPa] - Pressure\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# math.log(P_sat) = A - B/(t + C)\n",
+      "\n",
+      "\t\t\t# For benzene\n",
+      "A_1 = 13.8594;\n",
+      "B_1 = 2773.78;\n",
+      "C_1 = 220.07;\n",
+      "\t\t\t# For ethyl benzene\n",
+      "A_2 = 14.0045;\n",
+      "B_2 = 3279.47;\n",
+      "C_2 = 213.20;\n",
+      "\n",
+      "x_1 = 0.5;\t\t\t# Equimolar mixture\n",
+      "x_2 = 0.5;\n",
+      "\n",
+      "\t\t\t# The bubble point temperature equation is\n",
+      "\t\t\t# P = x_1*P_1_sat + x_2*P_2_sat\n",
+      "\n",
+      "# Calculations and Results\n",
+      "t_1_sat = B_1/(A_1 - math.log(P)) - C_1;\n",
+      "t_2_sat = B_2/(A_2 - math.log(P)) - C_2;\n",
+      "t = x_1*t_1_sat + x_2*t_2_sat;\n",
+      "\n",
+      "fault = 10;\n",
+      "while(fault>0.3):\n",
+      "    P_1_sat = math.exp(A_1 - B_1/(t + C_1));\n",
+      "    P_2_sat = math.exp(A_2 - B_2/(t + C_2));\n",
+      "    P_net = x_1*P_1_sat + x_2*P_2_sat;\n",
+      "    fault=abs(P_net - P);\n",
+      "    t = t - 0.1;\n",
+      "\n",
+      "print \" The bubble point temperature is %f C\"%(t);\n",
+      "\n",
+      "\t\t\t# Now let us determine dew point temperature for y_1 = 0.5, and P = 90 kPa\n",
+      "y_1 = 0.5;\t\t\t# Equimolar mixture\n",
+      "y_2 = 0.5;\n",
+      "\n",
+      "\t\t\t# 1/P = y_1/P_1_sat + y_2/P_2_sat\n",
+      "\t\t\t# Let us statrt with t = 104.07\n",
+      "t_old = 104.07;\n",
+      "error = 10;\n",
+      "while(error>0.3):\n",
+      "    P_1_sat_prime = math.exp(A_1 - B_1/(t_old + C_1));\n",
+      "    P_2_sat_prime = math.exp(A_2 - B_2/(t_old + C_2));\n",
+      "    P_net_prime = 1/(y_1/P_1_sat_prime + y_2/P_2_sat_prime);\n",
+      "    error=abs(P_net_prime - P);\n",
+      "    t_old = t_old + 0.1;\n",
+      "\n",
+      "print \" The dew point temperature is %f C\"%(t_old);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The bubble point temperature is 93.862003 C\n",
+        " The dew point temperature is 114.470000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.7  Page Number : 522"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P = 1;\t\t\t#[bar] - Pressure\n",
+      "P = P*10**(2);\t\t\t#[kPa]\n",
+      "\n",
+      "\t\t\t# math.log(P_1_sat) = 13.8594 - 2773.78/(t + 220.07)\n",
+      "\t\t\t# math.log(P_2_sat) = 14.0098 - 3103.01/(t + 219.79)\n",
+      "\n",
+      "\t\t\t# The bubble point equation is\n",
+      "\t\t\t# P = x_1*P_1_sat + x_2*P_2_sat;\n",
+      "\n",
+      "t_1_sat = 2773.78/(13.8594 - math.log(P)) - 220.07;\n",
+      "t_2_sat = 3103.01/(14.0098 - math.log(P)) - 219.79;\n",
+      "\n",
+      "\t\t\t# For x_1 = 0.1\n",
+      "\t\t\t# t = x_1_1*t_1_sat + x_2_1*t_2_sat;\n",
+      "x_1 = [0.1,0.5,0.9];\n",
+      "x_2 = [0,0,0]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "for i in range(3):\n",
+      "    x_2[i] = 1 - x_1[i];\n",
+      "    t = x_1[i]*t_1_sat + x_2[i]*t_2_sat;\n",
+      "    fault = 10;\n",
+      "    while(fault>0.3):\n",
+      "        P_1_sat = math.exp(13.8594 - 2773.78/(t + 220.07));\n",
+      "        P_2_sat = math.exp(14.0098 - 3103.01/(t + 219.79));\n",
+      "        P_net = x_1[i]*P_1_sat + x_2[i]*P_2_sat;\n",
+      "        fault=abs(P_net - P);\n",
+      "        t = t - 0.1;\n",
+      "    print \" The bubble point temperature for x_1 = %f) is %f C\"%(x_1[i],t);\n",
+      "\n",
+      "\n",
+      "\n",
+      "\t\t\t# Now let us determine dew point temperature\n",
+      "\t\t\t# 1/P = y_1/P_1_sat + y_2/P_2_sat\n",
+      "\n",
+      "y_1 = [0.1,0.5,0.9];\n",
+      "y_2 = [0,0,0]\n",
+      "for j in range(3):\n",
+      "    y_2[j] = 1 - y_1[j];\n",
+      "    t_prime = y_1[j]*t_1_sat + y_2[j]*t_2_sat;\n",
+      "    error = 10;\n",
+      "    while(error>0.3):\n",
+      "        P_1_sat = math.exp(13.8594 - 2773.78/(t_prime + 220.07));\n",
+      "        P_2_sat = math.exp(14.0098 - 3103.01/(t_prime + 219.79));\n",
+      "        P_net = 1/(y_1[j]/P_1_sat + y_2[j]/P_2_sat);\n",
+      "        error=abs(P_net - P);\n",
+      "        t_prime = t_prime + 0.1;\n",
+      "    \n",
+      "    print \" The dew point temperature for y_1 = %f is %f C\"%(y_1[j],t_prime);\n",
+      "\n",
+      "\n",
+      "\t\t\t#Therefore\n",
+      "print \" The temperature range for z_1 = 0.1 which two phase exist is 105.61 C to 108.11 C\";\n",
+      "print \" The temperature range for z_1 = 0.5 which two phase exist is 91.61 C to 98.40 C\";\n",
+      "print \" The temperature range for z_1 = 0.9 which two phase exist is 81.71 C to 84.51 C\";\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The bubble point temperature for x_1 = 0.100000) is 105.605549 C\n",
+        " The bubble point temperature for x_1 = 0.500000) is 91.607993 C\n",
+        " The bubble point temperature for x_1 = 0.900000) is 81.710437 C\n",
+        " The dew point temperature for y_1 = 0.100000 is 108.105549 C\n",
+        " The dew point temperature for y_1 = 0.500000 is 98.407993 C\n",
+        " The dew point temperature for y_1 = 0.900000 is 84.510437 C\n",
+        " The temperature range for z_1 = 0.1 which two phase exist is 105.61 C to 108.11 C\n",
+        " The temperature range for z_1 = 0.5 which two phase exist is 91.61 C to 98.40 C\n",
+        " The temperature range for z_1 = 0.9 which two phase exist is 81.71 C to 84.51 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.8  Page Number : 524"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "x_1 = 0.20;\n",
+      "x_2 = 0.45;\n",
+      "x_3 = 0.35;\n",
+      "P = 10;\t\t\t#[atm]\n",
+      "P = P*101325*10**(-3);\t\t\t#[kPa]\n",
+      "\n",
+      "\t\t\t# math.log(P_1_sat) = 13.7713 - 1892.47/(t + 248.82)\n",
+      "\t\t\t# math.log(P_2_sat) = 13.7224 - 2151.63/(t + 236.91)\n",
+      "\t\t\t# math.log(P_3_sat) = 13.8183 - 2477.07/(t + 233.21)\n",
+      "\n",
+      "\t\t\t#(a)\n",
+      "\t\t\t# The bubble point equation is\n",
+      "\t\t\t# P = x_1*P_1_sat + x_2*P_2_sat + x_3*P_3_sat\n",
+      "\n",
+      "# Calculations and Results\n",
+      "t_1_sat = 1892.47/(13.7713 - math.log(P)) - 248.82;\n",
+      "t_2_sat = 2151.63/(13.7224 - math.log(P)) - 236.91;\n",
+      "t_3_sat = 2477.07/(13.8183 - math.log(P)) - 233.21;\n",
+      "t = x_1*t_1_sat + x_2*t_2_sat + x_3*t_3_sat;\n",
+      "\n",
+      "fault = 10;\n",
+      "while(fault>0.1):\n",
+      "    P_1_sat = math.exp(13.7713 - 1892.47/(t + 248.82));\n",
+      "    P_2_sat = math.exp(13.7224 - 2151.63/(t + 236.91));\n",
+      "    P_3_sat = math.exp(13.8183 - 2477.07/(t + 233.21));\n",
+      "    P_net = x_1*P_1_sat + x_2*P_2_sat + x_3*P_3_sat;\n",
+      "    fault=abs(P_net - P);\n",
+      "    t = t - 0.003;\n",
+      "\n",
+      "BPT = t;\n",
+      "print \" a).The bubble point temperature is %f C\"%(BPT);\n",
+      "\n",
+      "\t\t\t# (b)\n",
+      "\t\t\t# Now let us determine dew point temperature for y_1 = 0.5, and P = 90 kPa\n",
+      "y_1 = 0.20;\n",
+      "y_2 = 0.45;\n",
+      "y_3 = 0.35;\n",
+      "\n",
+      "\t\t\t# 1/P = y_1/P_1_sat + y_2/P_2_sat + y_3/P_3_sat\n",
+      "\n",
+      "t_old = 90;\t\t\t#[C]\n",
+      "error = 10;\n",
+      "while(error>0.1):\n",
+      "    P_1_sat_prime = math.exp(13.7713 - 1892.47/(t_old + 248.82));\n",
+      "    P_2_sat_prime = math.exp(13.7224 - 2151.63/(t_old + 236.91));\n",
+      "    P_3_sat_prime = math.exp(13.8183 - 2477.07/(t_old + 233.21));\n",
+      "    P_net_prime = 1/(y_1/P_1_sat_prime + y_2/P_2_sat_prime + y_3/P_3_sat_prime);\n",
+      "    error=abs(P_net_prime - P);\n",
+      "    t_old = t_old + 0.003;\n",
+      "\n",
+      "DPT = t_old;\n",
+      "print \" b).The dew point temperature is %f C\"%(DPT);\n",
+      "\n",
+      "\t\t\t# (c)\n",
+      "\t\t\t# Therefore at 82 C two phase exists\n",
+      "\t\t\t# At 82 C and P = 1013.25 kPa pressure\n",
+      "T_c = 82;\t\t\t#[C]\n",
+      "P_c = 1013.25;\t\t\t#[kPa]\n",
+      "z_1 = 0.20;\n",
+      "z_2 = 0.45;\n",
+      "z_3 = 0.35;\n",
+      "\n",
+      "P_1_sat_c = math.exp(13.7713 - 1892.47/(T_c + 248.82));\n",
+      "P_2_sat_c = math.exp(13.7224 - 2151.63/(T_c + 236.91));\n",
+      "P_3_sat_c = math.exp(13.8183 - 2477.07/(T_c + 233.21));\n",
+      "\n",
+      "K_1 = P_1_sat_c/P_c;\n",
+      "K_2 = P_2_sat_c/P_c;\n",
+      "K_3 = P_3_sat_c/P_c;\n",
+      "\n",
+      "\t\t\t# We have to find such a V that the following equation is satisfied.\n",
+      "\t\t\t# summation(y_i) = K_i*z_i/(1-V+V*K_i) = 1\n",
+      "\t\t\t# K_1*z_1/(1-V+V*K_1) + K_2*z_2/(1-V+V*K_2) + K_3*z_3/(1-V+V*K_3) = 1\n",
+      "\n",
+      "def f1(V): \n",
+      "\t return  K_1*z_1/(1-V+V*K_1) + K_2*z_2/(1-V+V*K_2) + K_3*z_3/(1-V+V*K_3)-1\n",
+      "V = fsolve(f1,0.4)\n",
+      "\n",
+      "\t\t\t# Therefore now we have\n",
+      "y_1_c = K_1*z_1/(1-V+V*K_1);\n",
+      "y_2_c = K_2*z_2/(1-V+V*K_2);\n",
+      "y_3_c = K_3*z_3/(1-V+V*K_3);\n",
+      "x_1_c = y_1_c/K_1;\n",
+      "x_2_c = y_2_c/K_2;\n",
+      "x_3_c = y_3_c/K_3;\n",
+      "\n",
+      "print \" c).The proportion of vapour is given by V = %f\"%(V);\n",
+      "print \"     The composition of vapour foemed is given by y_1 = %f y_2 = %f and y_3 = %f \"%(y_1_c,y_2_c,y_3_c);\n",
+      "print \"     The composition of liquid formed is given by x_1 = %f x_2 = %f and x_3 = %f \"%(x_1_c,x_2_c,x_3_c);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The bubble point temperature is 71.833954 C\n",
+        " b).The dew point temperature is 97.254000 C\n",
+        " c).The proportion of vapour is given by V = 0.332143\n",
+        "     The composition of vapour foemed is given by y_1 = 0.365018 y_2 = 0.466574 and y_3 = 0.168408 \n",
+        "     The composition of liquid formed is given by x_1 = 0.117932 x_2 = 0.441757 and x_3 = 0.440311 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.9  Page Number : 526"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 27;\t\t\t#[C] - Temperature\n",
+      "z_1 = 0.4;\n",
+      "z_2 = 0.3;\n",
+      "z_3 = 0.3;\n",
+      "\n",
+      "\t\t\t# math.log(P_sat) = A - B/(t + C)\n",
+      "\n",
+      "\t\t\t# For propane\n",
+      "A_1 = 13.7713;\n",
+      "B_1 = 1892.47;\n",
+      "C_1 = 248.82;\n",
+      "\t\t\t# For i-butane\n",
+      "A_2 = 13.4331;\n",
+      "B_2 = 1989.35;\n",
+      "C_2 = 236.84;\n",
+      "\t\t\t# For n-butane\n",
+      "A_3 = 13.7224;\n",
+      "B_3 = 2151.63;\n",
+      "C_3 = 236.91;\n",
+      "\n",
+      "\t\t\t#(a)\n",
+      "\t\t\t\n",
+      "y_1 = z_1;\n",
+      "y_2 = z_2;\n",
+      "y_3 = z_3;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# At 27 C,\n",
+      "P_1_sat = math.exp(A_1 - B_1/(T + C_1));\n",
+      "P_2_sat = math.exp(A_2 - B_2/(T + C_2));\n",
+      "P_3_sat = math.exp(A_3 - B_3/(T + C_3));\n",
+      "\n",
+      "\t\t\t# The dew point pressure is given by\n",
+      "P_1 = 1/(y_1/P_1_sat + y_2/P_2_sat + y_3/P_3_sat); \n",
+      "\n",
+      "\n",
+      "x_1 = z_1;\n",
+      "x_2 = z_2;\n",
+      "x_3 = z_3;\n",
+      "\n",
+      "\t\t\t# The bubble point pressure is given by\n",
+      "P_2 = x_1*P_1_sat + x_2*P_2_sat + x_3*P_3_sat;\n",
+      "\n",
+      "print \" a).The two phase region exists between %f and %f kPa\"%(P_1,P_2);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\t\t\t# The mean of the two-phase pressure range is given by\n",
+      "P_mean = (P_1 + P_2)/2;\n",
+      "\n",
+      "\t\t\t# Now let us calculate the K values of the components\n",
+      "K_1 = P_1_sat/P_mean;\n",
+      "K_2 = P_2_sat/P_mean;\n",
+      "K_3 = P_3_sat/P_mean;\n",
+      "\n",
+      "\t\t\t# summation of y_i = 1, gives\n",
+      "\t\t\t# (K_1*z_1)/(1-V-K_1*V) + (K_2*z_2)/(1-V-K_2*V) + (K_3*z_3)/(1-V-K_3*V) = 1\n",
+      "\t\t\t# Solving we get\n",
+      "def f(V): \n",
+      "\t return (K_1*z_1)/(1-V+K_1*V) + (K_2*z_2)/(1-V+K_2*V) + (K_3*z_3)/(1-V+K_3*V)-1\n",
+      "V = fsolve(f,0.1)\n",
+      "\n",
+      "y_1_prime = (z_1*K_1)/(1-V+K_1*V);\n",
+      "\n",
+      "print \" b).The mole fraction of propane in vapour phase is %f whereas in the feed is %f and fraction of vapour in the system is %f\"%(y_1_prime,z_1,V);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The two phase region exists between 421.886904 and 588.370027 kPa\n",
+        " b).The mole fraction of propane in vapour phase is 0.559489 whereas in the feed is 0.400000 and fraction of vapour in the system is 0.425313\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.10  Page Number : 527"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 50;\t\t\t#[C] - Temperature\n",
+      "P = 64;\t\t\t#[kPa] - Pressure\n",
+      "z_1 = 0.7;\n",
+      "z_2 = 0.3;\n",
+      "\n",
+      "# math.log(P_sat) = A - B/(t + C)\n",
+      "\n",
+      "# For acetone\n",
+      "A_1 = 14.37824;\n",
+      "B_1 = 2787.498;\n",
+      "C_1 = 229.664;\n",
+      "# For acetonitrile\n",
+      "A_2 = 14.88567;\n",
+      "B_2 = 3413.099;\n",
+      "C_2 = 250.523;\n",
+      "\n",
+      "# Calculations\n",
+      "# At 50 C,\n",
+      "P_1_sat = math.exp(A_1 - B_1/(T + C_1));\t\t\t#[kPa]\n",
+      "P_2_sat = math.exp(A_2 - B_2/(T + C_2));\t\t\t#[kPa]\n",
+      "\n",
+      "# Now let us calculate the K values of the components\n",
+      "K_1 = P_1_sat/P;\n",
+      "K_2 = P_2_sat/P;\n",
+      "\n",
+      "# summation of y_i = 1, gives\n",
+      "# (K_1*z_1)/(1-V-K_1*V) + (K_2*z_2)/(1-V-K_2*V) = 1\n",
+      "# Solving we get\n",
+      "def f(V): \n",
+      "    return (K_1*z_1)/(1-V+K_1*V) + (K_2*z_2)/(1-V+K_2*V) -1\n",
+      "V = fsolve(f,0.1)\n",
+      "L = 1 - V;\n",
+      "# Therefore\n",
+      "y_1 = (K_1*z_1)/(1-V+K_1*V);\n",
+      "y_2 = (K_2*z_2)/(1-V+K_2*V);\n",
+      "\n",
+      "x_1 = y_1/K_1;\n",
+      "x_2 = y_2/K_2;\n",
+      "\n",
+      "# Results\n",
+      "print \" The value of V = %f\"%(V);\n",
+      "print \" The value of L = %f\"%(L);\n",
+      "print \" The liquid phase composition is x_1 = %f and x_2 = %f\"%(x_1,x_2);\n",
+      "print \" The vapour phase composition is y_1 = %f and y_2 = %f\"%(y_1,y_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of V = 0.450368\n",
+        " The value of L = 0.549632\n",
+        " The liquid phase composition is x_1 = 0.619963 and x_2 = 0.380037\n",
+        " The vapour phase composition is y_1 = 0.797678 and y_2 = 0.202322\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.11  Page Number : 528"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P = 12.25*101325*10**(-3);\t\t\t#[kPa]\n",
+      "z_1 = 0.8;\n",
+      "z_2 = 1 - z_1;\n",
+      "V = 0.4;\n",
+      "# math.log(P_1_sat) = 13.7713 - 2892.47/(T + 248.82)\n",
+      "# math.log(P_2_sat) = 13.7224 - 2151.63/(T + 236.91)\n",
+      "\n",
+      "# P_1_sat = math.exp(13.7713 - 21892.47/(T + 248.82));\n",
+      "# P_2_sat = math.exp(13.7224 - 2151.63/(T + 236.91));\n",
+      "\n",
+      "# Let the total mixture be 1 mol\n",
+      "# We have to assume a temperature such that,\n",
+      "# y_1 + y_2 = (K_1*z_1)/(1-V-K_1*V) + (K_2*z_2)/(1-V-K_2*V) = 1\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n",
+      "x_1 = z_1;\n",
+      "x_2 = z_2;\n",
+      "\n",
+      "# The bubble point pressure is given by\n",
+      "# P = x_1*(math.exp(13.7713 - 21892.47/(T + 248.82))) + x_2*(math.exp(13.7224 - 2151.63/(T + 236.91)));\n",
+      "\n",
+      "# Calculations\n",
+      "def f(T): \n",
+      "\t return x_1*(math.exp(13.7713 - 1892.47/(T + 248.82))) + x_2*(math.exp(13.7224 - 2151.63/(T + 236.91))) - P\n",
+      "T_1 = fsolve(f,0.1)\n",
+      "BPT = T_1;\n",
+      "\n",
+      "\n",
+      "y_1 = z_1;\n",
+      "y_2 = z_2;\n",
+      "\n",
+      "# The dew point equation is given by\n",
+      "# 1/P = y_1/P_1_sat + y_2/P_2_sat\n",
+      "def f1(T): \n",
+      "\t return 1/(y_1/(math.exp(13.7713 - 1892.47/(T + 248.82))) + y_2/(math.exp(13.7224 - 2151.63/(T + 236.91)))) - P\n",
+      "T_2 = fsolve(f1,0.1)\n",
+      "DPT = T_2;\n",
+      "\n",
+      "T = 47;\t\t\t#[C]\n",
+      "error = 10;\n",
+      "while(error>0.001):\n",
+      "    P_1_sat = math.exp(13.7713 - 1892.47/(T + 248.82));\n",
+      "    P_2_sat = math.exp(13.7224 - 2151.63/(T + 236.91));\n",
+      "    K_1 = P_1_sat/P;\n",
+      "    K_2 = P_2_sat/P;\n",
+      "    y1 = (K_1*z_1)/(1-V+K_1*V);\n",
+      "    y2 = (K_2*z_2)/(1-V+K_2*V);\n",
+      "    y = y1 + y2;\n",
+      "    error=abs(y - 1);\n",
+      "    T = T - 0.0001;\n",
+      "\n",
+      "# Results\n",
+      "print \" The temperature when 40 mol  of mixture is in the vapour is %f C\"%(T);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The temperature when 40 mol  of mixture is in the vapour is 45.333000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.12  Page Number : 529"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 105;\t\t\t#[C]\n",
+      "P = 1.5;\t\t\t#[atm]\n",
+      "P = P*101325*10**(-3);\t\t\t#[kPa]\n",
+      "z = [0.4,0.3667,0.2333];\t\t\t# Feed composition\n",
+      "x = [0.3,0.3,0.4];\t\t\t# Equilibrium liquid phase composition\n",
+      "y = [0.45,0.40,0.15];\t\t\t# Equilibrium vapour phase composition\n",
+      "\n",
+      "# From the material balance equation of component 1, we get\n",
+      "# (L + V)*z_1 = L*x_1 + V*y_1\n",
+      "\n",
+      "# Since total moles are one, therefore  L + V = 1 and thus\n",
+      "# z_1 = L*x_1 + (1-L)*y_1\n",
+      "# Calculations and Results\n",
+      "for i in range(3):\n",
+      "    L = (z[i] - y[i])/(x[i] - y[i]);\n",
+      "    V = 1 - L;\n",
+      "    print \" The number of moles in liquid phase z = %f is given by L = %f\"%(z[i],L);\n",
+      "    print \" The number of moles in vapour phase z = %f is given by V = %f\"%(z[i],V);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The number of moles in liquid phase z = 0.400000 is given by L = 0.333333\n",
+        " The number of moles in vapour phase z = 0.400000 is given by V = 0.666667\n",
+        " The number of moles in liquid phase z = 0.366700 is given by L = 0.333000\n",
+        " The number of moles in vapour phase z = 0.366700 is given by V = 0.667000\n",
+        " The number of moles in liquid phase z = 0.233300 is given by L = 0.333200\n",
+        " The number of moles in vapour phase z = 0.233300 is given by V = 0.666800\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.13  Page Number : 530"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 90;\t\t\t#[C]\n",
+      "P = 1;\t\t\t#[atm]\n",
+      "P = P*101325*10**(-3);\t\t\t#[kPa]\n",
+      "z_1 = [0.1,0.5,0.8];\n",
+      "\n",
+      "# Calculations\n",
+      "# math.log(P_1_sat) = 13.8594 - 2773.78/(t + 220.07)\n",
+      "# math.log(P_2_sat) = 14.0098 - 3103.01/(t + 219.79)\n",
+      "\n",
+      "#At T = 90 C\n",
+      "P_1_sat = math.exp(13.8594 - 2773.78/(T + 220.07));\n",
+      "P_2_sat = math.exp(14.0098 - 3103.01/(T + 219.79));\n",
+      "K_1 = P_1_sat/P;\n",
+      "K_2 = P_2_sat/P;\n",
+      "\n",
+      "# For z_1 = 0.1\n",
+      "# y1 = (K_1*z_1(i))/(1-V+K_1*V);\n",
+      "# y2 = (K_2*z_2)/(1-V+K_2*V);\n",
+      "\n",
+      "x_1 = (P - P_2_sat)/(P_1_sat - P_2_sat);\n",
+      "y_1 = (x_1*P_1_sat)/(P);\n",
+      "\n",
+      "# For two phase region to exist at 90 C and 101.325 kPa,z_1 sholud lie between x_1 and y_1\n",
+      "\n",
+      "# Results\n",
+      "print \" For two phase region to exist at 90 C and 101.325 kPa, z_1 sholud lie between %f and %f\"%(x_1,y_1);\n",
+      "print \" For z_1 = 0.1 and z_1 = 0.5 only liquid phase exists V = 0.\"\n",
+      "print \" For z_1 = 0.8 only vapour exists V = 1.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " For two phase region to exist at 90 C and 101.325 kPa, z_1 sholud lie between 0.574884 and 0.772458\n",
+        " For z_1 = 0.1 and z_1 = 0.5 only liquid phase exists V = 0.\n",
+        " For z_1 = 0.8 only vapour exists V = 1.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.14  Page Number : 531"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from numpy import zeros\n",
+      "\n",
+      "# Variables\n",
+      "T = 90;\t\t\t#[C]\n",
+      "P = 1;\t\t\t#[atm]\n",
+      "P = P*101325*10**(-3);\t\t\t#[kPa]\n",
+      "z_1 = [0.1,0.5,0.8];\n",
+      "\n",
+      "# math.log(P_1_sat) = 13.8594 - 2773.78/(t + 220.07)\n",
+      "# math.log(P_2_sat) = 14.0098 - 3103.01/(t + 219.79)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#(a)\n",
+      "#At T = 90 C\n",
+      "P_1_sat = math.exp(13.8594 - 2773.78/(T + 220.07));\n",
+      "P_2_sat = math.exp(14.0098 - 3103.01/(T + 219.79));\n",
+      "K_1 = P_1_sat/P;\n",
+      "\n",
+      "x_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0];\n",
+      "P_prime = zeros(11);\n",
+      "x_2 = zeros(11);\n",
+      "y_1 = zeros(11);\n",
+      "\n",
+      "print \" a.\";\n",
+      "print \" x_1    \\t\\t  P   \\t\\t   y_1      \";\n",
+      "\n",
+      "for i in range(11):\n",
+      "    x_2[i] = 1 - x_1[i];\n",
+      "    P_prime[i] = x_1[i]*P_1_sat + x_2[i]*P_2_sat;\n",
+      "    y_1[i] = (x_1[i]*P_1_sat)/P_prime[i];\n",
+      "    print \" %f \\t  %f  \\t   %f  \"%(x_1[i],P_prime[i],y_1[i]);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "T_1_sat = 2773.78/(13.8594-math.log(P)) - 220.07;\t\t\t#[C]\n",
+      "T_2_sat = 3103.01/(14.0098-math.log(P)) - 219.79;\t\t\t#[C]\n",
+      "\n",
+      "T_prime = [110.62,107,104,101,98,95,92,89,86,83,80.09];\n",
+      "\n",
+      "P1_sat = zeros(11);\n",
+      "P2_sat = zeros(11);\n",
+      "x_1 = zeros(11);\n",
+      "y_1 = zeros(11);\n",
+      "\n",
+      "print \"  b.\";\n",
+      "print \" TC  \\t\\t  P_1_sat kPa  \\t\\t P_2_sat kPa \\t\\t  x_1  \\t\\t  y_1 \";\n",
+      "\n",
+      "for j in range(11):\n",
+      "    P1_sat[j] = math.exp(13.8594 - 2773.78/(T_prime[j] + 220.07));\n",
+      "    P2_sat[j] = math.exp(14.0098 - 3103.01/(T_prime[j] + 219.79));\n",
+      "    x_1[j] = (P-P2_sat[j])/(P1_sat[j]-P2_sat[j]);\n",
+      "    y_1[j] = (x_1[j]*P1_sat[j])/P;\n",
+      "    print \" %f \\t    %f    \\t   %f    \\t      %f  \\t       %f \"%(T_prime[j],P1_sat[j],P2_sat[j],x_1[j],y_1[j]);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a.\n",
+        " x_1    \t\t  P   \t\t   y_1      \n",
+        " 0.000000 \t  54.233834  \t   0.000000  \n",
+        " 0.100000 \t  62.425251  \t   0.218098  \n",
+        " 0.200000 \t  70.616668  \t   0.385597  \n",
+        " 0.300000 \t  78.808085  \t   0.518277  \n",
+        " 0.400000 \t  86.999502  \t   0.625971  \n",
+        " 0.500000 \t  95.190920  \t   0.715131  \n",
+        " 0.600000 \t  103.382337  \t   0.790162  \n",
+        " 0.700000 \t  111.573754  \t   0.854176  \n",
+        " 0.800000 \t  119.765171  \t   0.909433  \n",
+        " 0.900000 \t  127.956588  \t   0.957615  \n",
+        " 1.000000 \t  136.148005  \t   1.000000  \n",
+        "  b.\n",
+        " TC  \t\t  P_1_sat kPa  \t\t P_2_sat kPa \t\t  x_1  \t\t  y_1 \n",
+        " 110.620000 \t    237.827187    \t   101.332530    \t      -0.000055  \t       -0.000129 \n",
+        " 107.000000 \t    216.742019    \t   91.320455    \t      0.079767  \t       0.170629 \n",
+        " 104.000000 \t    200.376847    \t   83.629571    \t      0.151570  \t       0.299740 \n",
+        " 101.000000 \t    184.975751    \t   76.460482    \t      0.229134  \t       0.418300 \n",
+        " 98.000000 \t    170.500977    \t   69.787769    \t      0.313139  \t       0.526923 \n",
+        " 95.000000 \t    156.915208    \t   63.586616    \t      0.404360  \t       0.626206 \n",
+        " 92.000000 \t    144.181607    \t   57.832824    \t      0.503680  \t       0.716718 \n",
+        " 89.000000 \t    132.263848    \t   52.502830    \t      0.612106  \t       0.799008 \n",
+        " 86.000000 \t    121.126156    \t   47.573718    \t      0.730789  \t       0.873601 \n",
+        " 83.000000 \t    110.733338    \t   43.023234    \t      0.861050  \t       0.941001 \n",
+        " 80.090000 \t    101.331285    \t   38.950606    \t      0.999899  \t       0.999961 \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.15  Page Number : 533"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P_1_sat = 79.80;\t\t\t#[kPa]\n",
+      "P_2_sat = 40.45;\t\t\t#[kPa]\n",
+      "\n",
+      "#(1)\n",
+      "T = 373.15;\t\t\t#[K]\n",
+      "x_1 = 0.05;\n",
+      "x_2 = 1 - x_1;\n",
+      "Y1 = math.exp(0.95*x_2**(2));\n",
+      "Y2 = math.exp(0.95*x_1**(2));\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# The total pressure of the system is given by\n",
+      "P = x_1*Y1*P_1_sat + x_2*Y2*P_2_sat;\t\t\t#[kPa]\n",
+      "y_1 = x_1*Y1*P_1_sat/P;\n",
+      "y_2 = x_2*Y2*P_2_sat/P;\n",
+      "\n",
+      "print \" 1).The first bubble is formed at %f kPa and the composition y_1 = %f\"%(P,y_1);\n",
+      "\n",
+      "#(2)\n",
+      "T = 373.15;\t\t\t#[K]\n",
+      "y_1_prime = 0.05;\n",
+      "y_2_prime = 1 - y_1_prime;\n",
+      "\n",
+      "# Let us assume a value of x_1,\n",
+      "x_1_prime = 0.0001;\n",
+      "\n",
+      "error = 10;\n",
+      "while(error>0.001):\n",
+      "    x_2_prime = 1 - x_1_prime;\n",
+      "    Y1_prime = math.exp(0.95*x_2_prime**(2));\n",
+      "    Y2_prime = math.exp(0.95*x_1_prime**(2));\n",
+      "    P_prime = x_1_prime*Y1_prime*P_1_sat + x_2_prime*Y2_prime*P_2_sat;\n",
+      "    x_1 = (y_1_prime*P_prime)/(Y1_prime*P_1_sat);\n",
+      "    error=abs(x_1_prime - x_1);\n",
+      "    x_1_prime = x_1_prime + 0.00001;\n",
+      "\n",
+      "P_2 = x_1_prime*Y1_prime*P_1_sat + x_2_prime*Y2_prime*P_2_sat;\n",
+      "\n",
+      "print \" 2).The first drop is formed at %f kPa and has the composition x_1 = %f\"%(P_2,x_1_prime);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The first bubble is formed at 47.923166 kPa and the composition y_1 = 0.196237\n",
+        " 2).The first drop is formed at 41.974204 kPa and has the composition x_1 = 0.009370\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.16  Page Number : 534"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 78.15;\t\t\t#[C]\n",
+      "P_1_sat = 755;\t\t\t#[mm Hg]\n",
+      "P_2_sat = 329;\t\t\t#[mm Hg]\n",
+      "\n",
+      "z_1 = 0.3;\n",
+      "V = 0.5;\n",
+      "\n",
+      "# math.log(Y1) = 0.845/(1 + 0.845*(x_1/x_2))**(2)\n",
+      "# math.log(Y2) = 1/(1 + 1.183*(x_2/x_1))**(2)\n",
+      "\n",
+      "# A value of x_1 is to determined for which  V = 0.5\n",
+      "# Let us assume a value of x_1, say x_1 = 0.150\n",
+      "x_1 = 0.150;\n",
+      "\n",
+      "# Calculations\n",
+      "error = 10;\n",
+      "while(error>0.001):\n",
+      "    x_2 = 1 - x_1;\n",
+      "    Y1 = math.exp(0.845/(1 + 0.845*(x_1/x_2))**(2));\n",
+      "    Y2 = math.exp(1/(1 + 1.183*(x_2/x_1))**(2));\n",
+      "    P = x_1*Y1*P_1_sat + x_2*Y2*P_2_sat;\n",
+      "    y_1 = (x_1*Y1*P_1_sat)/P;\n",
+      "    V_prime = (z_1 - x_1)/(y_1 - x_1);\n",
+      "    error=abs(V_prime - V);\n",
+      "    x_1 = x_1 + 0.00001;\n",
+      "\n",
+      "P_prime = x_1*Y1*P_1_sat + x_2*Y2*P_2_sat;\t\t\t#[mm hg]\n",
+      "\n",
+      "# At x_1 ,  V = 0.5, \n",
+      "# Therefore when the mixture is 50 % vaporized at 78.15 C the mole fraction of component 1 in the liquid phase is x_1 and the system pressure is P_prime\n",
+      "\n",
+      "# Results\n",
+      "print \" The required pressure is %f mm Hg\"%(P_prime);\n",
+      "print \" and the mole fraction of component 1 in the liquid phase for this pressure is x_1 = %f\"%(x_1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The required pressure is 505.838606 mm Hg\n",
+        " and the mole fraction of component 1 in the liquid phase for this pressure is x_1 = 0.157650\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.17  Page Number : 536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.stats import linregress\n",
+      "\n",
+      "# Variables\n",
+      "T = 25;\t\t\t#[C] - Temperature\n",
+      "P = [118.05,124.95,137.90,145.00,172.90,207.70,227.70,237.85,253.90,259.40,261.10,262.00,258.70,252.00,243.80];\t\t\t#[mm Hg]\n",
+      "x_1 = [0.0115,0.0160,0.0250,0.0300,0.0575,0.1125,0.1775,0.2330,0.4235,0.5760,0.6605,0.7390,0.8605,0.9250,0.9625];\n",
+      "y_1 = [0.1810,0.2250,0.3040,0.3450,0.4580,0.5670,0.6110,0.6325,0.6800,0.7050,0.7170,0.7390,0.8030,0.8580,0.9160];\n",
+      "\n",
+      "# Pressure value for which x_1 = y_1 = 0, corresponds to P_2_sat,therefore\n",
+      "P_2_sat = 97.45;\t\t\t#[mm Hg]\n",
+      "# Pressure value for which x_1 = y_1 = 1, corresponds to P_1_sat,therefore\n",
+      "P_1_sat = 230.40;\t\t\t#[mm Hg]\n",
+      "\n",
+      "x_2 = zeros(15);\n",
+      "y_2 = zeros(15);\n",
+      "Y1 = zeros(15);\n",
+      "Y2 = zeros(15);\n",
+      "GE_RT = zeros(15);\n",
+      "x1x2_GE_RT = zeros(15);\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(15):\n",
+      "    x_2[i] = 1 - x_1[i];\n",
+      "    y_2[i] = 1 - y_1[i];\n",
+      "    Y1[i] = (y_1[i]*P[i])/(x_1[i]*P_1_sat);\n",
+      "    Y2[i] = (y_2[i]*P[i])/(x_2[i]*P_2_sat);\n",
+      "    GE_RT[i] = x_1[i]*math.log(Y1[i]) + x_2[i]*math.log(Y2[i]);\t\t\t# G_E/(R*T)\n",
+      "    x1x2_GE_RT[i] = (x_1[i]*x_2[i])/GE_RT[i];\n",
+      "\n",
+      "\n",
+      "#[M,N,sig]=reglin(x_1,x1x2_GE_RT);\n",
+      "M,N,sig,d,e = linregress(x_1,x1x2_GE_RT);\n",
+      "\n",
+      "# Linear regression between x_1 and x_1*x_2/(G_E/R*T) gives intercept = N and slope = M\n",
+      "\n",
+      "# van Laar equation is x_1*x_2/(G_E/R*T) = 1/A + (1/B - 1/A)\n",
+      "# 1/A = N\n",
+      "A = 1/N;\n",
+      "B = 1/(M + 1/A);\n",
+      "\n",
+      "# Results\n",
+      "print \" The value of Van Laar coefficient A =  %f\"%(A);\n",
+      "print \" The value of Van Laar coefficient B =  %f\"%(B);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of Van Laar coefficient A =  2.270624\n",
+        " The value of Van Laar coefficient B =  1.781821\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.18  Page Number : 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 343.15;\t\t\t#[K] - Temperature\n",
+      "# At 343.15 K\n",
+      "# math.log(Y1) = 0.95*x_2**(2)\n",
+      "# math.log(Y2) = 0.95*x_1**(2)\n",
+      "P_1_sat = 79.80;\t\t\t#[kPa]\n",
+      "P_2_sat = 40.50;\t\t\t#[kPa]\n",
+      "\n",
+      "# Calculations\n",
+      "# At x_1 = 0,\n",
+      "Y1_infinity = math.exp(0.95);\n",
+      "alpha_12_x0 = (Y1_infinity*P_1_sat)/(P_2_sat);\n",
+      "# At x_1 = 1,\n",
+      "Y2_infinity = math.exp(0.95);\n",
+      "alpha_12_x1 = (P_1_sat)/(Y2_infinity*P_2_sat);\n",
+      "\n",
+      "\n",
+      "# At azeotrope, Y1*P1_sat = Y2*P2_sat\n",
+      "# Y2/Y1 = P_1_sat/P_2_sat\n",
+      "# Taking math.logarithm of both sides we get\n",
+      "# math.log(Y2) - math.log(Y1) = math.log(P_1_sat/P_2_sat)\n",
+      "# 0.95*x_1**(2) - 0.95*x_2**(2) = math.log(P_1_sat/P_2_sat)\n",
+      "# Solving the above equation\n",
+      "def f(x_1): \n",
+      "\t return 0.95*x_1**(2) - 0.95*(1-x_1)**(2) - math.log(P_1_sat/P_2_sat)\n",
+      "x_1 = fsolve(f,0.1)\n",
+      "\n",
+      "# At x_1\n",
+      "x_2 = 1 - x_1;\n",
+      "Y1 = math.exp(0.95*x_2**(2));\n",
+      "Y2 = math.exp(0.95*x_1**(2));\n",
+      "P = x_1*Y1*P_1_sat + x_2*Y2*P_2_sat;\t\t\t#[kPa] - Azeotrope pressure\n",
+      "y_1 = (x_1*Y1*P_1_sat)/P;\n",
+      "\n",
+      "# Since x_1 = y_1, (almost equal) ,the above condition is of azeotrope formation\n",
+      "\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print \" Since alpha_12_x=0 = %f and alpha_12_x=1 = %f \"%(alpha_12_x0,alpha_12_x1);\n",
+      "print \" and since alpha_12 is a continuous curve and in between a value of alpha_12 = 1, shall come and at this composition the azeotrope shall get formed.\"\n",
+      "print \" The azeotrope composition is x_1 = y_1 = %f\"%(x_1);\n",
+      "print \" The azeotrope presssure is %f kPa\"%(P);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Since alpha_12_x=0 = 5.094806 and alpha_12_x=1 = 0.762023 \n",
+        " and since alpha_12 is a continuous curve and in between a value of alpha_12 = 1, shall come and at this composition the azeotrope shall get formed.\n",
+        " The azeotrope composition is x_1 = y_1 = 0.856959\n",
+        " The azeotrope presssure is 81.366308 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.19  Page Number : 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 45;\t\t\t#[C] - Temperature\n",
+      "\n",
+      "x_1 = [0.0455,0.0940,0.1829,0.2909,0.3980,0.5069,0.5458,0.5946,0.7206,0.8145,0.8972,0.9573];\n",
+      "y_1 = [0.1056,0.1818,0.2783,0.3607,0.4274,0.4885,0.5098,0.5375,0.6157,0.6913,0.7869,0.8916];\n",
+      "P = [31.957,33.553,35.285,36.457,36.996,37.068,36.978,36.778,35.792,34.372,32.331,30.038];\n",
+      "\n",
+      "\t\t\t# Pressure value for which x_1 = y_1 = 0, corresponds to P_2_sat,therefore\n",
+      "P_2_sat = 29.819;\t\t\t#[kPa]\n",
+      "\t\t\t# Pressure value for which x_1 = y_1 = 1, corresponds to P_1_sat,therefore\n",
+      "P_1_sat = 27.778;\t\t\t#[kPa]\n",
+      "\n",
+      "x_2 = zeros(12);\n",
+      "y_2 = zeros(12);\n",
+      "Y1 = zeros(12);\n",
+      "Y2 = zeros(12);\n",
+      "alpha_12 = zeros(12);\n",
+      "GE_RT = zeros(12);\n",
+      "x1x2_GE_RT = zeros(12);\n",
+      "# Calculations and Results\n",
+      "print \" x_1  \\t\\t  y_1    \\t   P   \\t\\t    Y1   \\t\\tY2  \\t    alpha_12    \\t  G_E/RT \\t  x1*x2/G_E/RT\";\n",
+      "\n",
+      "for i in range(12):\n",
+      "    x_2[i] = 1 - x_1[i];\n",
+      "    y_2[i] = 1 - y_1[i];\n",
+      "    Y1[i] = (y_1[i]*P[i])/(x_1[i]*P_1_sat);\n",
+      "    Y2[i] = (y_2[i]*P[i])/(x_2[i]*P_2_sat);\n",
+      "    alpha_12[i] = (y_1[i]/x_1[i])/(y_2[i]/x_2[i]);\n",
+      "    GE_RT[i] = x_1[i]*math.log(Y1[i]) + x_2[i]*math.log(Y2[i]);\t\t\t# G_E/(R*T)\n",
+      "    x1x2_GE_RT[i] = (x_1[i]*x_2[i])/GE_RT[i];\n",
+      "    print \" %f\\t  %f\\t  %f \\t     %f  \\t   %f \\t     %f\\t   %f  \\t%f\"%(x_1[i],y_1[i],P[i],Y1[i],Y2[i],alpha_12[i],GE_RT[i],x1x2_GE_RT[i]);\n",
+      "\n",
+      "\t\t\t#[M,N,sig]=reglin(x_1,x1x2_GE_RT);\n",
+      "M,N,sig,d,e=linregress(x_1,x1x2_GE_RT);\n",
+      "\n",
+      "\t\t\t# Linear regression between x_1 and x_1*x_2/(G_E/R*T) gives intercept = N and slope = M\n",
+      "\n",
+      "\t\t\t# Now let us assume the system to follow van Laar  activity coefficient model. \n",
+      "\t\t\t# x_1*x_2/(G_E/(R*T)) = x_1/B + x_2/A =  x_1/B + (1 - x_1)/A  =  1/A + (1/B - 1/A)*x_1 = N + M*x_1\n",
+      "\n",
+      "\t\t\t# 1/A = N\n",
+      "A = 1/N;\n",
+      "\t\t\t# (1/B - 1/A) = M\n",
+      "B = 1/(M + 1/A);\n",
+      "\n",
+      "print \"\"\n",
+      "print \" The value of van Laar parameters are, A = %f and B = %f \"%(A,B);\n",
+      "\n",
+      "Y1_infinity = math.exp(A);\n",
+      "Y2_infinity = math.exp(B);\n",
+      "\n",
+      "\n",
+      "\t\t\t\n",
+      "\t\t\t# Now let us calculate the azeotrope composition.\n",
+      "\t\t\t# At azeotrope, Y1*P1_sat = Y2*P2_sat\n",
+      "\t\t\t# math.log(Y1/Y2) = math.log(P_2_sat/P_1_sat)\n",
+      "\t\t\t# From van Laar model we get\n",
+      "\t\t\t# math.log(P_2_sat/P_1_sat) = (A*B**(2)*2*x_2**(2))/(A*x_1 + B*x_2)**(2) + (B*A**(2)*2*x_1**(2))/(A*x_1 + B*x_2)**(2)\n",
+      "\t\t\t# Solving the above equation\n",
+      "def f(x_1): \n",
+      "\t return  math.log(P_2_sat/P_1_sat) - (A*B**(2)*(1-x_1)**(2))/(A*x_1 + B*(1-x_1))**(2) + (B*A**(2)*x_1**(2))/(A*x_1 + B*(1-x_1))**(2)\n",
+      "x_1 = fsolve(f,0.1)\n",
+      "\n",
+      "print \" The azeotrope composition is given by x_1 = y_1 = %f\"%(x_1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " x_1  \t\t  y_1    \t   P   \t\t    Y1   \t\tY2  \t    alpha_12    \t  G_E/RT \t  x1*x2/G_E/RT\n",
+        " 0.045500\t  0.105600\t  31.957000 \t     2.670039  \t   1.004220 \t     2.476833\t   0.048705  \t0.891698\n",
+        " 0.094000\t  0.181800\t  33.553000 \t     2.336127  \t   1.016177 \t     2.141582\t   0.094298  \t0.903137\n",
+        " 0.182900\t  0.278300\t  35.285000 \t     1.932808  \t   1.045150 \t     1.722733\t   0.156610  \t0.954268\n",
+        " 0.290900\t  0.360700\t  36.457000 \t     1.627355  \t   1.102263 \t     1.375325\t   0.210697  \t0.979023\n",
+        " 0.398000\t  0.427400\t  36.996000 \t     1.430228  \t   1.180094 \t     1.129007\t   0.242105  \t0.989635\n",
+        " 0.506900\t  0.488500\t  37.068000 \t     1.285998  \t   1.289486 \t     0.929034\t   0.252871  \t0.988458\n",
+        " 0.545800\t  0.509800\t  36.978000 \t     1.243394  \t   1.338371 \t     0.865446\t   0.251278  \t0.986567\n",
+        " 0.594600\t  0.537500\t  36.778000 \t     1.196853  \t   1.407094 \t     0.792366\t   0.245302  \t0.982671\n",
+        " 0.720600\t  0.615700\t  35.792000 \t     1.100930  \t   1.650961 \t     0.621199\t   0.209369  \t0.961630\n",
+        " 0.814500\t  0.691300\t  34.372000 \t     1.050218  \t   1.918247 \t     0.510015\t   0.160745  \t0.939933\n",
+        " 0.897200\t  0.786900\t  32.331000 \t     1.020818  \t   2.247586 \t     0.423097\t   0.101740  \t0.906550\n",
+        " 0.957300\t  0.891600\t  30.038000 \t     1.007145  \t   2.557286 \t     0.366877\t   0.046909  \t0.871410\n",
+        "\n",
+        " The value of van Laar parameters are, A = 1.049085 and B = 1.064514 \n",
+        " The azeotrope composition is given by x_1 = y_1 = 0.468254\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.20  Page Number : 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 25;\t\t\t#[C] - Temperature\n",
+      "P_1_sat = 230.4;\t\t\t#[mm Hg]\n",
+      "P_2_sat = 97.45;\t\t\t#[mm Hg]\n",
+      "Y1_infinity = 8.6;\n",
+      "Y2_infinity = 6.6;\n",
+      "\n",
+      "\t\t\t# Assuming ideal vpour behaviour means that phi = 1 and since system pressure is low, therefore\n",
+      "\t\t\t# f_i = P_i_sat \n",
+      "\t\t\t# Assuming the activity coefficients to follow van Laar model we get\n",
+      "A = math.log(Y1_infinity);\n",
+      "B = math.log(Y2_infinity);\n",
+      "\n",
+      "\t\t\t#math.log(Y1) = A/(1+ (A*x_1)/(B*x_2))**(2)\n",
+      "\t\t\t# math.log(Y2) = B/(1+ (B*x_2)/(A*x_1))**(2)\n",
+      "\n",
+      "x_1 = [0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9];\n",
+      "\n",
+      "x_2 = zeros(9);\n",
+      "Y1 = zeros(9);\n",
+      "Y2 = zeros(9);\n",
+      "y1_P = zeros(9);\n",
+      "y2_P = zeros(9);\n",
+      "P = zeros(9);\n",
+      "y_1 = zeros(9);\n",
+      "\n",
+      "print \" a.\";\n",
+      "print \" x_1      \\t\\t    Y1    \\t\\t    Y2    \\t\\t    y1*P    \\t\\t    y2*P   \\t\\t   P    \\t\\t   y_1 \";\n",
+      "# Calculations and Results\n",
+      "for i in range(9):\n",
+      "    x_2[i] = 1 - x_1[i];\n",
+      "    Y1[i] = math.exp(A/(1+ (A*x_1[i])/(B*x_2[i]))**(2));\n",
+      "    Y2[i] = math.exp(B/(1+ (B*x_2[i])/(A*x_1[i]))**(2));\n",
+      "    y1_P[i] = x_1[i]*Y1[i]*P_1_sat;\n",
+      "    y2_P[i] = x_2[i]*Y2[i]*P_2_sat;\n",
+      "    P[i] = x_1[i]*Y1[i]*P_1_sat + x_2[i]*Y2[i]*P_2_sat;\n",
+      "    y_1[i] = (x_1[i]*Y1[i]*P_1_sat)/P[i];\n",
+      "    print \" %f\\t\\t %f\\t\\t %f \\t\\t %f \\t\\t %f \\t\\t %f \\t   %f\"%(x_1[i],Y1[i],Y2[i],y1_P[i],y2_P[i],P[i],y_1[i]);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\t\t\t# The total system pressure versus x_1 shows a maxima and thus azeotrope is formed by the VLE system\n",
+      "\t\t\t# The maxima occurs in the range of x_1 = 0.6 to 0.8, so an  azeotrope is formed in this composition range\n",
+      "\n",
+      "\t\t\t# At the azeotrope  point, Y1*P1_sat = Y2*P2_sat\n",
+      "\t\t\t# math.log(Y1) - math.log(Y2) = math.log(P_2_sat/P_1_sat)\n",
+      "\t\t\t# On putting the values and then solving the above equation we get\n",
+      "def f(x_1): \n",
+      "\t return  A/(1+1.14*x_1/(1-x_1))**(2)- B/(1+0.877*(1-x_1)/x_1)**(2) - math.log(P_2_sat/P_1_sat)\n",
+      "x_1_prime = fsolve(f,0.1)\n",
+      "\n",
+      "\t\t\t# At x_1\n",
+      "x_2_prime = 1 - x_1_prime;\n",
+      "Y1_prime = math.exp(A/(1+ (A*x_1_prime)/(B*x_2_prime))**(2));\n",
+      "Y2_prime = math.exp(B/(1+ (B*x_2_prime)/(A*x_1_prime))**(2));\n",
+      "P_prime = x_1_prime*Y1_prime*P_1_sat + x_2_prime*Y2_prime*P_2_sat;\t\t\t#[kPa] - Azeotrope pressure\n",
+      "y_1_prime = (x_1_prime*Y1_prime*P_1_sat)/P_prime;\n",
+      "\n",
+      "\t\t\t# Since x_1 = y_1,azeotrope formation will take place\n",
+      "print \" b\";\n",
+      "print \" The total system pressure versus x_1 shows a maxima and thus azeotrope is formed by the VLE system\";\n",
+      "print \" The azeotrope composition is x_1 = y_1 = %f\"%(x_1_prime);\n",
+      "print \" The azeotrope presssure is %f mm Hg\"%(P_prime);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a.\n",
+        " x_1      \t\t    Y1    \t\t    Y2    \t\t    y1*P    \t\t    y2*P   \t\t   P    \t\t   y_1 \n",
+        " 0.100000\t\t 5.446877\t\t 1.024149 \t\t 125.496041 \t\t 89.822961 \t\t 215.319002 \t   0.582838\n",
+        " 0.200000\t\t 3.680305\t\t 1.097308 \t\t 169.588473 \t\t 85.546152 \t\t 255.134625 \t   0.664702\n",
+        " 0.300000\t\t 2.640401\t\t 1.225500 \t\t 182.504521 \t\t 83.597451 \t\t 266.101972 \t   0.685844\n",
+        " 0.400000\t\t 2.002736\t\t 1.421865 \t\t 184.572186 \t\t 83.136461 \t\t 267.708646 \t   0.689452\n",
+        " 0.500000\t\t 1.599580\t\t 1.708524 \t\t 184.271623 \t\t 83.247849 \t\t 267.519472 \t   0.688816\n",
+        " 0.600000\t\t 1.340316\t\t 2.120132 \t\t 185.285311 \t\t 82.642744 \t\t 267.928055 \t   0.691549\n",
+        " 0.700000\t\t 1.174189\t\t 2.709824 \t\t 189.373156 \t\t 79.221704 \t\t 268.594861 \t   0.705051\n",
+        " 0.800000\t\t 1.072057\t\t 3.558780 \t\t 197.601501 \t\t 69.360613 \t\t 266.962114 \t   0.740186\n",
+        " 0.900000\t\t 1.017109\t\t 4.791470 \t\t 210.907696 \t\t 46.692876 \t\t 257.600573 \t   0.818739\n",
+        " b\n",
+        " The total system pressure versus x_1 shows a maxima and thus azeotrope is formed by the VLE system\n",
+        " The azeotrope composition is x_1 = y_1 = 0.706548\n",
+        " The azeotrope presssure is 268.599631 mm Hg\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.21  Page Number : 544"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 50;\t\t\t#[C]\n",
+      "\t\t\t# At 50 C\n",
+      "P_1_sat = 0.67;\t\t\t#[atm]\n",
+      "P_2_sat = 0.18;\t\t\t#[atm]\n",
+      "Y1_infinity = 2.5;\n",
+      "Y2_infinity = 7.2;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# alpha_12 = (y_1/x_1)/(y_2/x_2) = (Y1*P_1_sat)/((Y2*P_2_sat))\n",
+      "\t\t\t# At  x_1 tending to zero,\n",
+      "alpha_12_x0 = (Y1_infinity*P_1_sat)/(P_2_sat);\n",
+      "\t\t\t# At  x_1 tending to 1,\n",
+      "alpha_12_x1 = (P_1_sat)/((Y2_infinity*P_2_sat));\n",
+      "\n",
+      "\n",
+      "print \" 1).Since alpha_12_x=0) = %f and alpha_12_x=1) = %f \"%(alpha_12_x0,alpha_12_x1);\n",
+      "print \"     and since alpha_12 is a continuous curve and in between a value of alpha_12 = 1 shall come and at this composition azeotrope shall get formed.\"\n",
+      "\n",
+      "\t\t\t\n",
+      "print \" 2).Since the activity coefficient values are greater than 1 therefore the deviations from Roults law is positive\"\n",
+      "print \"     and the azeotrope is maximum pressure or minimum boiling\";\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "\t\t\t# Let us assume the system to follow van Laar activity coefficient model\n",
+      "A = math.log(Y1_infinity);\n",
+      "B = math.log(Y2_infinity);\n",
+      "\n",
+      "\t\t\t# math.log(Y1) = A/(1+ (A*x_1)/(B*x_2))**(2)\n",
+      "\t\t\t# math.log(Y2) = B/(1+ (B*x_2)/(A*x_1))**(2)\n",
+      "\n",
+      "\t\t\t# At the azeotrope  point, Y1*P1_sat = Y2*P2_sat\n",
+      "\t\t\t# math.log(Y1) - math.log(Y2) = math.log(P_2_sat/P_2_sat)\n",
+      "\t\t\t# On putting the values and then solving the above equation\n",
+      "def f(x_1): \n",
+      "\t return  A/(1+ (A*x_1)/(B*(1-x_1)))**(2)- B/(1+ (B*(1-x_1))/(A*x_1))**(2) - math.log(P_2_sat/P_1_sat)\n",
+      "x_1 = fsolve(f,0.1)\n",
+      "\n",
+      "\t\t\t# At x_1\n",
+      "x_2 = 1 - x_1;\n",
+      "Y1 = math.exp(A/(1+ (A*x_1)/(B*x_2))**(2));\n",
+      "Y2 = math.exp(B/(1+ (B*x_2)/(A*x_1))**(2));\n",
+      "P = x_1*Y1*P_1_sat + x_2*Y2*P_2_sat;\t\t\t#[kPa] - Azeotrope pressure\n",
+      "y_1 = (x_1*Y1*P_1_sat)/P;\n",
+      "\n",
+      "\t\t\t# Since x_1 = y_1,the azeotrope formation will take place\n",
+      "\n",
+      "print \" 3).The azeotrope composition is x_1 = y_1 = %f\"%(x_1);\n",
+      "print \"     The azeotrope presssure is %f atm\"%(P);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).Since alpha_12_x=0) = 9.305556 and alpha_12_x=1) = 0.516975 \n",
+        "     and since alpha_12 is a continuous curve and in between a value of alpha_12 = 1 shall come and at this composition azeotrope shall get formed.\n",
+        " 2).Since the activity coefficient values are greater than 1 therefore the deviations from Roults law is positive\n",
+        "     and the azeotrope is maximum pressure or minimum boiling\n",
+        " 3).The azeotrope composition is x_1 = y_1 = 0.910173\n",
+        "     The azeotrope presssure is 0.689143 atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.22  Page Number : 545"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 25;\t\t\t#[C]\n",
+      "\t\t\t# At 50 C\n",
+      "P_1_sat = 7.866;\t\t\t#[kPa]\n",
+      "P_2_sat = 3.140;\t\t\t#[kPa]\n",
+      "\n",
+      "\t\t\t# G_E/(R*T) = 1.4938*x_1*x_2/(1.54*x_1 + 0.97*x_2)\n",
+      "\n",
+      "\t\t\t# The excess Gibbs free energy math.expression can be written as\n",
+      "\t\t\t# x_1*x_2/(G_E/(R*T)) = 1.54*x_1/1.4938 + 0.97*x_2/1.4938 = x_1/0.97 + x_2/1.54\n",
+      "\n",
+      "\t\t\t# Comparing with the van Laar math.expression\n",
+      "\t\t\t# x_1*x_2/(G_E/(R*T)) = x_1/B + x_2/A,  we get\n",
+      "A = 1.54;\n",
+      "B = 0.97;\n",
+      "\n",
+      "\t\t\t# The activity coefficients are thus given by\n",
+      "\t\t\t# math.log(Y1) = A/(1+ (A*x_1)/(B*x_2))**(2)\n",
+      "\t\t\t# math.log(Y2) = B/(1+ (B*x_2)/(A*x_1))**(2)\n",
+      "\n",
+      "x_1 = [0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,0.95];\n",
+      "\n",
+      "x_2 = zeros(10);\n",
+      "Y1 = zeros(10);\n",
+      "Y2 = zeros(10);\n",
+      "P = zeros(10);\n",
+      "y_1 = zeros(10);\n",
+      "\n",
+      "print \" x_1      \\t\\t    Y1    \\t\\t    Y2    \\t\\t P kPa    \\t\\t  y_1 \";\n",
+      "\n",
+      "# Calculations and Results\n",
+      "for i in range(10):\n",
+      "    x_2[i] = 1 - x_1[i];\n",
+      "    Y1[i] = math.exp(A/(1+ (A*x_1[i])/(B*x_2[i]))**(2));\n",
+      "    Y2[i] = math.exp(B/(1+ (B*x_2[i])/(A*x_1[i]))**(2));\n",
+      "    P[i] = x_1[i]*Y1[i]*P_1_sat + x_2[i]*Y2[i]*P_2_sat;\n",
+      "    y_1[i] = (x_1[i]*Y1[i]*P_1_sat)/P[i];\n",
+      "    print \" %f\\t\\t %f\\t\\t %f \\t\\t %f \\t\\t  %f \"%(x_1[i],Y1[i],Y2[i],P[i],y_1[i]);\n",
+      "\n",
+      "\n",
+      "\t\t\t# The azeotrope is formed near x_1 = 0.95 as in this region a maxima in pressure is obtained.\n",
+      "\n",
+      "\t\t\t# At the azeotrope  point, Y1*P1_sat = Y2*P2_sat\n",
+      "\t\t\t# math.log(Y1) - math.log(Y2) = math.log(P_2_sat/P_2_sat)\n",
+      "\t\t\t# On putting the values and then solving the above equation\n",
+      "def f(x_1): \n",
+      "\t return  A/(1+ (A*x_1)/(B*(1-x_1)))**(2)- B/(1+ (B*(1-x_1))/(A*x_1))**(2) - math.log(P_2_sat/P_1_sat)\n",
+      "x_1_prime = fsolve(f,0.1)\n",
+      "\n",
+      "\t\t\t# At x_1\n",
+      "x_2_prime = 1 - x_1_prime;\n",
+      "Y1_prime = math.exp(A/(1+ (A*x_1_prime)/(B*x_2_prime))**(2));\n",
+      "Y2_prime = math.exp(B/(1+ (B*x_2_prime)/(A*x_1_prime))**(2));\n",
+      "P_prime = x_1_prime*Y1_prime*P_1_sat + x_2_prime*Y2_prime*P_2_sat;\t\t\t#[kPa] - Azeotrope pressure\n",
+      "y_1_prime = (x_1_prime*Y1_prime*P_1_sat)/P_prime;\n",
+      "\n",
+      "\t\t\t# Since x_1_prime = y_1_prime,the azeotrope formation will take place\n",
+      "\n",
+      "print \" Part II \";\n",
+      "print \" The azeotrope composition is x_1 = y_1 = %f\"%(x_1_prime);\n",
+      "print \" The azeotrope presssure is %f kPa \"%(P_prime);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " x_1      \t\t    Y1    \t\t    Y2    \t\t P kPa    \t\t  y_1 \n",
+        " 0.100000\t\t 3.042798\t\t 1.022050 \t\t 5.281779 \t\t  0.453155 \n",
+        " 0.200000\t\t 2.201629\t\t 1.081457 \t\t 6.180223 \t\t  0.560433 \n",
+        " 0.300000\t\t 1.725242\t\t 1.172375 \t\t 6.648107 \t\t  0.612389 \n",
+        " 0.400000\t\t 1.438293\t\t 1.292347 \t\t 6.960227 \t\t  0.650186 \n",
+        " 0.500000\t\t 1.258593\t\t 1.440723 \t\t 7.211980 \t\t  0.686364 \n",
+        " 0.600000\t\t 1.144175\t\t 1.617876 \t\t 7.432102 \t\t  0.726584 \n",
+        " 0.700000\t\t 1.072060\t\t 1.824770 \t\t 7.621913 \t\t  0.774475 \n",
+        " 0.800000\t\t 1.028913\t\t 2.062721 \t\t 7.770131 \t\t  0.833286 \n",
+        " 0.900000\t\t 1.006610\t\t 2.333241 \t\t 7.858834 \t\t  0.906775 \n",
+        " 0.950000\t\t 1.001587\t\t 2.481217 \t\t 7.874109 \t\t  0.950528 \n",
+        " Part II \n",
+        " The azeotrope composition is x_1 = y_1 = 0.958680\n",
+        " The azeotrope presssure is 7.874467 kPa \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 15.23  Page Number : 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 58.7;\t\t\t#[C]\n",
+      "P = 1;\t\t\t#[atm]\n",
+      "P = P*101325*10**(-3);\t\t\t#[kPa]\n",
+      "\n",
+      "\t\t\t# math.log(P_sat) = 16.6758 - 3674.49/(t + 226.45) - For ethyl alcohol\n",
+      "\t\t\t# math.log(P_sat) = 13.8216 - 2697.55/(t + 224.37) - For hexane\n",
+      "\n",
+      "\t\t\t# Let us take hexane as (1) and ethanol as (2)\n",
+      "\t\t\t# At 58.7 C\n",
+      "P_1_sat = math.exp(13.8216 - 2697.55/(T + 224.37));\t\t\t#[kPa]\n",
+      "P_2_sat = math.exp(16.6758 - 3674.49/(T + 226.45));\t\t\t#[kPa]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Y1 = P/P_1_sat;\n",
+      "Y2 = P/P_2_sat;\n",
+      "\n",
+      "x_2 = 0.332;\t\t\t# Mol % of ethanol (given)\n",
+      "x_1 = 1 - x_2;\t\t\t# Mol % of hehane\n",
+      "\n",
+      "\t\t\t# The van Laar parameters are given by\n",
+      "A = ((1 + (x_2*math.log(Y2))/(x_1*math.log(Y1)))**(2))*math.log(Y1);\n",
+      "B = ((1 + (x_1*math.log(Y1))/(x_2*math.log(Y2)))**(2))*math.log(Y2);\n",
+      "\n",
+      "print \" The value of van Laar parameters are A = %f and B = %f \"%(A,B);\n",
+      "\n",
+      "\t\t\t# Now let us calvulate the distribution coefficient K\n",
+      "x_1_prime = 0.5;\t\t\t#[given]\n",
+      "x_2_prime = 1 - x_1_prime;\n",
+      "\n",
+      "\t\t\t# The activity coefficients are thus given by\n",
+      "\t\t\t# math.log(Y1) = A/(1+ (A*x_1)/(B*x_2))**(2)\n",
+      "\t\t\t# math.log(Y2) = B/(1+ (B*x_2)/(A*x_1))**(2)\n",
+      "\n",
+      "Y1_prime = math.exp(A/(1+ (A*x_1_prime)/(B*x_2_prime))**(2));\n",
+      "Y2_prime = math.exp(B/(1+ (B*x_2_prime)/(A*x_1_prime))**(2));\n",
+      "P_prime = x_1_prime*Y1_prime*P_1_sat + x_2_prime*Y2_prime*P_2_sat;\n",
+      "\n",
+      "\t\t\t# We have,  K_1 = y_1/x_1 = Y1*P_1_sat/P\n",
+      "K_1 = Y1_prime*P_1_sat/P_prime;\n",
+      "\n",
+      "print \" The distribution coefficient is given by  K_1  =  %f\"%(K_1)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of van Laar parameters are A = 1.669879 and B = 2.662289 \n",
+        " The distribution coefficient is given by  K_1  =  1.352866\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch16_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch16_1.ipynb
new file mode 100755
index 00000000..1053ca6c
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch16_1.ipynb
@@ -0,0 +1,738 @@
+{
+ "metadata": {
+  "name": "ch16_1",
+  "signature": "sha256:a00a89ec39a5812cdf143d7f00f601a0bb401404f1638737e2689f71d07acb65"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16 : Other Phase Equilibria"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.1  Page Number : 564"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math \n",
+      "\n",
+      "# Variables\t\t\t\n",
+      "T = 0 + 273.15;\t\t\t#[K] - Temperature\n",
+      "P = 20*10**(5);\t\t\t#[Pa] - Pressure\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t#componenet 1 : methane (1)\n",
+      "\t\t\t#componenet 2 : methanol (2)\n",
+      "\n",
+      "H_constant = 1022;\t\t\t#[bar] - Henry's law constant\n",
+      "H_constant = H_constant*10**(5);\t\t\t#[Pa]\n",
+      "\n",
+      "\t\t\t# The second virial coefficients are\n",
+      "B_11 = -53.9;\t\t\t#[cm**(3)/mol]\n",
+      "B_11 = B_11*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "B_12 = -166;\t\t\t#[cm**(3)/mol]\n",
+      "B_12 = B_12*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "B_22 = -4068;\t\t\t#[cm**(3)/mol]\n",
+      "B_22 = B_22*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "den_meth = 0.8102;\t\t\t#[g/cm**(3)] - Density of methanol at 0 C\n",
+      "Mol_wt_meth = 32.04;\t\t\t# Molecular weight of methanol\n",
+      "P_2_sat = 0.0401;\t\t\t#[bar] - Vapour pressure of methanol at 0 C\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#The molar volume of methanol can be calculated as\n",
+      "V_2_liq = (1/(den_meth/Mol_wt_meth))*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#The phase equilibrium equation of the components at high pressure\n",
+      "\t\t\t#y1*phi_1*P = x_1*H_1\n",
+      "\t\t\t#y2*phi_2*P = x_2*H_2\n",
+      "\n",
+      "\t\t\t#Since methane follows Henry's law therefore methanol follows the lewis-Rnadall rule\n",
+      "\t\t\t#f_2 is the fugacity of the compressed liquid which is calculated using\n",
+      "\t\t\t#f_2 = f_2_sat*math.exp[V_2_liq*(P - P_sat_2)/(R*T)]\n",
+      "\t\t\t#where f_2_sat can be calculated using virial equation \n",
+      "\t\t\t# math.log(phi_2_sat) = math.log(f_2_sat/P_2_sat) = (B_22*P_2_sat)/(R*T)\n",
+      "\n",
+      "f_2_sat = P_2_sat*math.exp((B_22*P_2_sat*10**(5))/(R*T));\t\t\t#[bar]\n",
+      "\n",
+      "\t\t\t#Putting the value of 'f_2_sat' in the math.expression of f_2 , we get\n",
+      "f_2 = f_2_sat*math.exp(V_2_liq*(P - P_2_sat*10**(5))/(R*T));\t\t\t#[bar]\n",
+      "\n",
+      "\t\t\t#Now let us calculate the fugacity coefficients of the species in the vapour mixture\n",
+      "del_12 = 2*B_12 - B_11 - B_22;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#math.log(phi_1) = (P/(R*T))*(B_11 + y2**(2)*del_12)\n",
+      "\t\t\t#math.log(phi_2) = (P/(R*T))*(B_22 + y1**(2)*del_12)\n",
+      "\n",
+      "\n",
+      "\t\t\t#The calculation procedure is to assume a value of y1, calculate 'phi_1' and 'phi_2' and calculate 'x_1' and 'x_2' from the phase equilibrium equations and see whether x_1 + x_2 = 1,if not then another value of y1 is assumed.\n",
+      "\n",
+      "y2 = 0.1;\n",
+      "error=10;\n",
+      "\n",
+      "while(error>0.001):\n",
+      "    y1=1-y2;\n",
+      "    phi_1 = math.exp((P/(R*T))*((B_11 + y2**(2)*del_12)));\n",
+      "    phi_2 = math.exp((P/(R*T))*((B_22 + y1**(2)*del_12)));\n",
+      "    x_1 = (y1*phi_1*P)/H_constant;\n",
+      "    x_2 = (y2*phi_2*P)/(f_2*10**(5));\n",
+      "    x = x_1 + x_2;\n",
+      "    error=abs(1-x);\n",
+      "    y2=y2 - 0.000001;\n",
+      "\n",
+      "# Results\n",
+      "print \" The solubility of methane in methanol is given by x1 = %f\"%(x_1);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The solubility of methane in methanol is given by x1 = 0.018614\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.2  Page Number : 566"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "x_C2H6_1 = 0.33*10**(-4);\t\t\t# Solubility of ethane in water at 25 C and 1 bar\n",
+      "\n",
+      "\t\t\t#componenet 1 : ethane (1)\n",
+      "\t\t\t#componenet 2 : water (2)\n",
+      "\n",
+      "\t\t\t# Z = 1 - 7.63*10**(3)*P - 7.22*10**(-5)*P**(2)\n",
+      "\n",
+      "\t\t\t#The phase equilibrium equation of ethane is\n",
+      "\t\t\t#f_1_V = x_1*H_1\n",
+      "\t\t\t#since vapour is pure gas,  f_1_V = x_1*H_1   or, phi_1*P = x_1*H_1,  where 'phi_1' is fugacity coefficient of pure ethane\n",
+      "\t\t\t# math.log(phi) = integral('Z-1)/P) from limit '0' to 'P'\n",
+      "\n",
+      "P1 = 0;\n",
+      "P2 = 1;\n",
+      "P3 = 35;\n",
+      "\n",
+      "# Calculations\n",
+      "def f51(P): \n",
+      "\t return (1-7.63*10**(-3)*P-7.22*10**(-5)*P**(2)-1)/P\n",
+      "\n",
+      "intgral =  quad(f51,P1,P2)[0]\n",
+      "\n",
+      "phi_1_1 = math.exp(intgral);\t\t\t# - Fugacity coefficient of ethane at 1 bar\n",
+      "f_1_1 = phi_1_1*P2;\t\t\t#[bar] - Fugacity of ethane at 1 bar\n",
+      "\n",
+      "\t\t\t#Similarly\n",
+      "\n",
+      "def f52(P): \n",
+      "\t return (1-7.63*10**(-3)*P-7.22*10**(-5)*P**(2)-1)/P\n",
+      "\n",
+      "intgral_1 =  quad(f52,P1,P3)[0]\n",
+      "\n",
+      "phi_1_35 = math.exp(intgral_1);\t\t\t# Fugacity coefficient of ethane at 35 bar\n",
+      "f_1_35 = phi_1_35*P3;\t\t\t#[bar] - Fugacity of ethane at 35 bar\n",
+      "\n",
+      "\t\t\t# At ethane pressure of 1 bar ,  x_C2H6_1*H_1 = phi_1_1\n",
+      "H_1 = phi_1_1/x_C2H6_1;\t\t\t#[bar] - Henry's constant\n",
+      "\n",
+      "\t\t\t# At ethane pressure of 35 bar ,  x_C2H6_35*H_1 = phi_1_35\n",
+      "x_C2H6_35 = f_1_35/H_1;\t\t\t# Solubility of ethane at 35 bar pressure\n",
+      "\n",
+      "# Results\n",
+      "print \"The solubility of ethane at 35 bar is given by   x_C2H6 = %e \"%(x_C2H6_35);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The solubility of ethane at 35 bar is given by   x_C2H6 = 8.525648e-04 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.4  Page Number : 571"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 200;\t\t\t#[K]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
+      "\t\t\t# G_E = A*x_1*x_2\n",
+      "A = 4000;\t\t\t#[J/mol]\n",
+      "x_1 = 0.6;\t\t\t# Mle fraction of feed composition\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# Since A is given to be independent of temperature\n",
+      "UCST = A/(2*R);\t\t\t#[K] - Upper critical solution temperature\n",
+      "print \" The UCST of the system is %f K\"%(UCST);\n",
+      "\n",
+      "\t\t\t# Since the given temperature is less than UCST therefore two phase can get formed at the given temperature.\n",
+      "\n",
+      "\t\t\t# x1_alpha = 1 - x1_beta\n",
+      "\t\t\t# We know that,  x1_alpha*Y_1_alpha = x2_alpha*Y_2_alpha\n",
+      "\t\t\t# x1_alpha*math.exp[(A/(R*T))*(x2_alpha)**(2)] = (1 - x1_alpha)*math.exp[(A/(R*T))*(x1_alpha)**(2)]\n",
+      "\t\t\t# where use has beeen made of the fact that x1_alpha = 1 - x1_beta and x2_beta = 1 - x1_beta = x1_alpha .Taking math.logarithm of both side we get\n",
+      "\t\t\t# math.log(x1_alpha) + (A/(R*T))*(1 - x1_alpha)**(2) = math.log(1 - x1_alpha) + (A/(R*T))*x1_alpha**(2)\n",
+      "\t\t\t# math.log(x1_alpha/(1-x1_alpha)) = (A/(R*T))*(2*x1_alpha - 1)\n",
+      "\n",
+      "def f(x1_alpha): \n",
+      "\t return  math.log(x1_alpha/(1-x1_alpha)) - (A/(R*T))*(2*x1_alpha - 1)\n",
+      "x1_alpha = fsolve(f,0.1)\n",
+      "x1_beta = fsolve(f,0.9)\n",
+      "\t\t\t# Because of symmetry 1 - x1_beta = x1_alpha\n",
+      "\n",
+      "\t\t\t# It can be seen that the equation,  math.log(x1/(1-x1)) = (A/(R*T))*(2*x1 - 1) has two roots.\n",
+      "\t\t\t\n",
+      "print \" The composition of two liquid phases in equilibrium is given by x1_alpha = %f and x1_beta = %f\"%(x1_alpha,x1_beta);\n",
+      "\n",
+      "z1 = 0.6;\n",
+      "\t\t\t# z1 = x1_alpha*alpha + x1_beta*beta\n",
+      "\t\t\t# beta = 1 - alpha\n",
+      "alpha = (z1-x1_beta)/(x1_alpha-x1_beta);\t\t\t#[mol]\n",
+      "Beta = 1 - alpha;\t\t\t#[mol]\n",
+      "print \" The relative amount of phases is given by alpha = %f mol and beta = %f mol\"%(alpha,Beta);\n",
+      "\n",
+      "\t\t\t# the relative amounts of the phases changes with the  feed composition \n",
+      "\n",
+      "\t\t\t#math.log(x1/(1-x1)) = (A/(R*T))*(2*x1 - 1)\n",
+      "\t\t\t\n",
+      "\n",
+      "print \" math.logx1/1-x1 = A/R*T*2*x1 - 1\";\n",
+      "print \" If the above equation has two real roots of x1 one for phase alpha and the other for phase beta then two liquid phases get formed\";\n",
+      "print \" and if it has no  real roots then a homogeneous liquid mixture is obtained\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The UCST of the system is 240.558095 K\n",
+        " The composition of two liquid phases in equilibrium is given by x1_alpha = 0.169102 and x1_beta = 0.830898\n",
+        " The relative amount of phases is given by alpha = 0.348896 mol and beta = 0.651104 mol\n",
+        " math.logx1/1-x1 = A/R*T*2*x1 - 1\n",
+        " If the above equation has two real roots of x1 one for phase alpha and the other for phase beta then two liquid phases get formed\n",
+        " and if it has no  real roots then a homogeneous liquid mixture is obtained\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.5  Page Number : 573"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 300;\t\t\t#[K]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
+      "A = 7000;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# math.log(x_1/(1-x_1)) = (A/(R*T))*(2*x_1-1)\n",
+      "\n",
+      "# Calculations\n",
+      "def f(x_1): \n",
+      "\t return math.log(x_1/(1-x_1))-((A/(R*T))*(2*x_1-1))\n",
+      "\n",
+      "x1_alpha=fsolve(f,0.1)\n",
+      "\n",
+      "x1_beta=1-x1_alpha;\n",
+      "\n",
+      "# Results\n",
+      "print \"The equilibrium compositin of the two liquid phase system is given by x1_alpha \\t = %f  x1_beta \\t = %f\"%(x1_alpha,x1_beta);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equilibrium compositin of the two liquid phase system is given by x1_alpha \t = 0.091897  x1_beta \t = 0.908103\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.7  Page Number : 579"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "M_wt_meth = 32;\t\t\t# Molecular weight of methanol \n",
+      "M_wt_water = 18;\t\t\t# Molecular weight of water \n",
+      "m_meth = 0.01;\t\t\t#[g] - Mass of methanol added per cm**(3) of solution\n",
+      "\n",
+      "\n",
+      "den_sol = 1;\t\t\t#[g/cm**(3)]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#The mole fraction of solute is given by\n",
+      "\t\t\t#x_2 = (moles of solute in cm**(3) of solution)/(moles of solute + moles of water) in 1 cm**(3) of solution\n",
+      "x_2 = (m_meth/M_wt_meth)/((m_meth/M_wt_meth)+((1-m_meth)/M_wt_water));\n",
+      "\n",
+      "\t\t\t#We know that heat of fusion of water is\n",
+      "H_fus = -80;\t\t\t#[cal/g] - Enthalpy change of fusion at 0 C\n",
+      "H_fus = H_fus*4.186*M_wt_water;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t#Therefore freezing point depression is given by\n",
+      "\t\t\t# T - T_m = (R*(T**(2))*x_2)/H_fus\n",
+      "T_f = 273.15;\t\t\t#[K] - Freezing point of water\n",
+      "delta_T_f = (R*(T_f**(2))*x_2)/H_fus;\t\t\t#[K]\n",
+      "\n",
+      "# Results\n",
+      "print \"The depression in freezing point is given by  delta_T = %f K\"%(delta_T_f);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The depression in freezing point is given by  delta_T = -0.581403 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.8  Page Number : 580"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
+      "T_f = 273.15;\t\t\t#[K] - Freezing point of water\n",
+      "m_water = 100;\t\t\t#[g] - Mass of water\n",
+      "m_NaCl = 3.5;\t\t\t#[g] - Mass of NaCl\n",
+      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
+      "M_wt_NaCl = 58.5;\t\t\t# Molecular weight of NaCl\n",
+      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
+      "mol_NaCl = m_NaCl/M_wt_NaCl;\t\t\t#[mol] - Moles of NaCl\n",
+      "\n",
+      "H_fus = -80;\t\t\t#[cal/g] - Enthalpy change of fusion at 0 C\n",
+      "H_fus = H_fus*4.186*M_wt_water;\t\t\t#[J/mol]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#Mole fraction of the solute (NaCl) is given by\n",
+      "x_2 = mol_NaCl/(mol_NaCl+mol_water);\n",
+      "\n",
+      "\n",
+      "x_2_act = 2*x_2;\t\t\t# Actual mole fraction\n",
+      "\n",
+      "\t\t\t#Now depression in freezing point is given by\n",
+      "\t\t\t# T - T_m = (R*(T**(2))*x_2_act)/H_fus\n",
+      "delta_T_f = (R*(T_f**(2))*x_2_act)/H_fus;\t\t\t#[C]\n",
+      "\n",
+      "\t\t\t#Thus freezing point of seawater = depression in freezing point\n",
+      "\n",
+      "# Results\n",
+      "print \"The freezing point of seawater is  %f C\"%(delta_T_f);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The freezing point of seawater is  -2.192853 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.10  Page Number : 583"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
+      "T_b = 373.15;\t\t\t#[K] - Boiling point of water\n",
+      "m_water = 100.;\t\t\t#[g] - Mass of water\n",
+      "m_C12H22 = 5.;\t\t\t#[g] - Mass of glucise (C12H22)\n",
+      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
+      "M_wt_C12H22 = 342.30;\t\t\t# Molecular weight of C12H22\n",
+      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
+      "mol_C12H22 = m_C12H22/M_wt_C12H22;\t\t\t#[mol] - Moles of C12H22\n",
+      "\n",
+      "# Calculations\n",
+      "H_vap = 540.;\t\t\t#[cal/g] - Enthalpy change of vaporisation\n",
+      "H_vap = H_vap*4.186*M_wt_water;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t#Mole fraction of the solute (C12H22) is given by\n",
+      "x_2 = mol_C12H22/(mol_C12H22+mol_water);\n",
+      "\n",
+      "\t\t\t#The boiling point elevation is given by\n",
+      "\t\t\t# T - T_b = (R*T_b**(2)*x_2**(2))/H_vap**(2)\n",
+      "\n",
+      "delta_T_b = (R*T_b**(2)*x_2)/(H_vap);\n",
+      "\n",
+      "# Results\n",
+      "print \"The elevation in boiling point is given by  delta_T = %f C\"%(delta_T_b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The elevation in boiling point is given by  delta_T = 0.074611 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.11  Page Number : 584"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "T = 25 + 273.15;\t\t\t#[K] - Surrounding temperature \n",
+      "den_water = 1000.;\t\t\t#[kg/m**(3)] - Density of water\n",
+      "m_water = 100.;\t\t\t#[g] - Mass of water\n",
+      "m_C12H22 = 5.;\t\t\t#[g] - Mass of glucise (C12H22)\n",
+      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
+      "M_wt_C12H22 = 342.30;\t\t\t# Molecular weight of C12H22\n",
+      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
+      "mol_C12H22 = m_C12H22/M_wt_C12H22;\t\t\t#[mol] - Moles of C12H22\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#Mole fraction of the water is given by\n",
+      "x_1 = mol_water/(mol_C12H22+mol_water);\n",
+      "\n",
+      "\t\t\t#Molar volume of water can be calculated as\n",
+      "V_l_water = (1./den_water)*M_wt_water*10**(-3);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#The osmotic pressure is given by\n",
+      "pi = -(R*T*math.log(x_1))/V_l_water;\t\t\t#[N/m**(2)]\n",
+      "pi = pi*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "# Results\n",
+      "print \"The osmotic pressure of the mixture is %f bar\"%(pi);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The osmotic pressure of the mixture is 3.616073 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.12  Page Number : 585"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
+      "T = 25 + 273.15;\t\t\t#[K] - Surrounding temperature\n",
+      "den_water = 1000.;\t\t\t#[kg/m**(3)] - Density of water\n",
+      "m_water = 100.;\t\t\t#[g] - Mass of water\n",
+      "m_NaCl = 3.5;\t\t\t#[g] - Mass of NaCl\n",
+      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
+      "M_wt_NaCl = 58.5;\t\t\t# Molecular weight of NaCl\n",
+      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
+      "mol_NaCl = m_NaCl/M_wt_NaCl;\t\t\t#[mol] - Moles of NaCl\n",
+      "\n",
+      "# Calculations\n",
+      "H_fus = -80.;\t\t\t#[cal/g] - Enthalpy change of fusion at 0 C\n",
+      "H_fus = H_fus*4.186*M_wt_water;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t#Mole fraction of the solute (NaCl) is given by\n",
+      "x_2 = mol_NaCl/(mol_NaCl+mol_water);\n",
+      "\n",
+      "\n",
+      "x_2_act = 2*x_2;\t\t\t# Actual mole fraction\n",
+      "\n",
+      "x_1 = 1 - x_2_act;\n",
+      "\n",
+      "\t\t\t#Molar volume of water can be calculated as\n",
+      "V_l_water = (1/den_water)*M_wt_water*10**(-3);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#The osmotic pressure is given by\n",
+      "pi = -(R*T*math.log(x_1))/V_l_water;\t\t\t#[N/m**(2)]\n",
+      "pi = pi*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print \"The minimum pressure to desalinate sea water is %f bar\"%(pi);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum pressure to desalinate sea water is 29.662232 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.13  Page Number : 586"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
+      "T = 173.15;\t\t\t#[K] - Surrounding temperature\n",
+      "P = 60;\t\t\t#[bar]\n",
+      "P = P*10**(5);\t\t\t#[Pa] \n",
+      "\n",
+      "\t\t\t#componenet 1 : CO2 (1)\n",
+      "\t\t\t#componenet 2 : H2 (2)\n",
+      "P_1_sat = 0.1392;\t\t\t#[bar] - Vapour pressre of pure solid CO2\n",
+      "P_1_sat = P_1_sat*10**(5);\t\t\t#[bar]\n",
+      "V_s_1 = 27.6;\t\t\t#[cm**(3)/mol] - Molar volume of solid CO2\n",
+      "V_s_1 = V_s_1*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "n_1 = 0.01;\t\t\t#[mol] - Initial number of moles of CO2\n",
+      "n_2 = 0.99;\t\t\t#[mol] - Initial number of moles of H2\n",
+      "\n",
+      "\t\t\t#Let us determine the fugacity of solid CO2 (1) at 60 C and 173.15 K\n",
+      "\t\t\t# f_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T))\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#Since vapour pressure of pure solid CO2 is very small, therefore\n",
+      "f_1_sat = P_1_sat;\n",
+      "f_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T));\n",
+      "\n",
+      "\t\t\t#Since gas phase is ideal therefore\n",
+      "\t\t\t# y1*P = f_1\n",
+      "y1 = f_1/P;\n",
+      "\n",
+      "\t\t\n",
+      "\t\t\t#Number of moles of CO2 in vapour phase at equilibrium can be calculated as\n",
+      "\t\t\t#y1 = (n_1_eq/(n_1_eq + n_2)). Therefore\n",
+      "n_1_eq = n_2*y1/(1-y1);\n",
+      "\n",
+      "\t\t\t#Therefore moles of CO2 precipitated is\n",
+      "n_ppt = n_1 - n_1_eq;\t\t\t#[mol]\n",
+      "\n",
+      "# Results\n",
+      "print \"The moles of CO2 precipitated is %f mol\"%(n_ppt);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The moles of CO2 precipitated is 0.007417 mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.14  Page Number : 586"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
+      "T = 350.;\t\t\t#[K] - Surrounding temperature\n",
+      "\n",
+      "\t\t\t#componenet 1 : organic solid (1)\n",
+      "\t\t\t#componenet 2 : CO2 (2)\n",
+      "\n",
+      "P_1_sat = 133.3;\t\t\t#[N/m**(2)] - Vapour pressre of organic solid\n",
+      "V_s_1 = 200.;\t\t\t#[cm**(3)/mol] - Molar volume of organic solid\n",
+      "V_s_1 = V_s_1*10.**(-6);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#/At 350 K, the values of the coefficients \n",
+      "B_11 = -500.;\t\t\t#[cm**(3)/mol]\n",
+      "B_22 = -85.;\t\t\t#[cm****(3)/mol]\n",
+      "B_12 = -430.;\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#From phase equilibrium equation of component 1, we get\n",
+      "\t\t\t# y1*P*phi_1 = f_1\n",
+      "\t\t\t# f_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T))\n",
+      "\n",
+      "\t\t\t#Since vapour pressure of organic solid is very small, therefore\n",
+      "f_1_sat = P_1_sat;\n",
+      "\n",
+      "\t\t\t# Now let us determine the fugacity coefficient of organic solid in the vapour mixture.\n",
+      "\t\t\t# math.log(phi_1) = (P/(R*T))*(B_11 + y2**(2)*del_12) \n",
+      "del_12 = (2*B_12 - B_11 - B_22)*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#It is given that the partial pressure of component 1 in the vapour mixture is 1333 N?m**(2)\n",
+      "\t\t\t# y1*P = 1333 N/m**(2) or, y1 = 1333/P\n",
+      "\t\t\t# y2 = 1- 1333/P\n",
+      "\t\t\t# math.log(phi_1) = (P/(R*T))*(B_11 + (1- 1333/P)**(2)*del_12)\n",
+      "\n",
+      "\t\t\t#The phase equilibrium equation becomes\n",
+      "\t\t\t# y1*P*phi_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T))\n",
+      "\t\t\t#Taking math.log on both side we have\n",
+      "\t\t\t# math.log(y1*P) + math.log(phi_1) = math.log(f_1_sat) + (V_s_1*(P-P_1_sat)/(R*T))\n",
+      "\t\t\t# (V_s_1*(P-P_1_sat)/(R*T)) - math.log(phi_1) = math.log(1333/133.3) = math.log(10)\n",
+      "\n",
+      "\t\t\t#substituting for math.log(phi_1) from previous into the above equation we get\n",
+      "\t\t\t# (V_s_1*(P-P_1_sat)/(R*T)) - (P/(R*T))*(B_11 + (1- 1333/P)**(2)*del_12) - math.log(10) = 0\n",
+      "\t\t\t# On simplification we get,\n",
+      "\t\t\t# 975*P**(2) - 6.7*10**(9)*P + 4.89*10**(8) = 0\n",
+      "\t\t\t# Solving the above qyadratic equation using shreedharcharya rule\n",
+      "\n",
+      "P3 = (6.7*10**(9) + ((-6.7*10**(9))**(2)-4*975*4.98*10**(8))**(1./2))/(2*975);\t\t\t#[Pa]\n",
+      "P4 = (6.7*10**(9) - ((-6.7*10**(9))**(2)-4*975*4.98*10**(8))**(1./2))/(2*975);\t\t\t#[Pa]\n",
+      "\t\t\t# The second value is not possible, therefore pressure of the system is P3\n",
+      "P3 = P3*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "# Results\n",
+      "print \" The total pressure of the system is %f bar\"%(P3);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The total pressure of the system is 68.717948 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch17_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch17_1.ipynb
new file mode 100755
index 00000000..61fcd6b8
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch17_1.ipynb
@@ -0,0 +1,2044 @@
+{
+ "metadata": {
+  "name": "ch17_1",
+  "signature": "sha256:e354b0525ce1aff27214eb76f8a2c9be56cfff2939a8a4b8688a7fe1cea4959c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17 : Chemical Reactions Equilibria"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.2  Page Number : 598"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "P = 1;\t\t\t#[atm] - Reactor pressure\n",
+      "T = 749;\t\t\t#[K] - Reactor temperature\n",
+      "K = 74;\t\t\t# Equlibrium constant\n",
+      "\n",
+      "\t\t\t# SO2 + (1/2)*O2 - SO3\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Kp = P**(1);\n",
+      "Ky = K/Kp;\n",
+      "\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_SO2_1_in = 12;\n",
+      "n_O2_1_in = 9;\n",
+      "n_SO3_1_in = 0;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_SO2_1_eq = 12 - X\n",
+      "\t\t\t# n_O2_1_eq = 9 - 0.5*X\n",
+      "\t\t\t# n_SO3_1_eq = X\n",
+      "\t\t\t# Total moles = 21 - 0.5*X\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "\t\t\t# y_SO3 = X/(21-0.5*X)\n",
+      "\t\t\t# y_SO2 = (12-X)/(21-0.5*X)\n",
+      "\t\t\t# y_O2 = (9-0.5*X)/(21-0.5*X)\n",
+      "\n",
+      "\t\t\t# Ky = y_SO3/(y_SO2*y_O2**(2))\n",
+      "\t\t\t# Ky = (X*(21-0.5*X)**(1/2))/((12-X)*(9-0.5*X)**(1/2))\n",
+      "def f(X): \n",
+      "\t return  Ky-(X*(21-0.5*X)**(1./2))/((12-X)*(9-0.5*X)**(1./2))\n",
+      "X_1 = fsolve(f,11)\n",
+      "\n",
+      "y_SO3_1 = X_1/(21-0.5*X_1);\n",
+      "y_SO2_1 = (12-X_1)/(21-0.5*X_1);\n",
+      "y_O2_1 = (9-0.5*X_1)/(21-0.5*X_1);\n",
+      "\n",
+      "print \" 1).The moles of SO3 formed = %f mol\"%(X_1);\n",
+      "print \"     The mole fractions at equilibrium are y_S03 = %f y_SO2 = %f and y_O2 = %f\"%(y_SO3_1,y_SO2_1,y_O2_1);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_SO2_2_in = 24;\n",
+      "n_O2_2_in = 18;\n",
+      "n_SO3_2_in = 0;\n",
+      "\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_SO2_1_eq = 24 - X\n",
+      "\t\t\t# n_O2_1_eq = 18 - 0.5*X\n",
+      "\t\t\t# n_SO3_1_eq = X\n",
+      "\t\t\t# Total moles = 42 - 0.5*X\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "\t\t\t# y_SO3 = X/(42-0.5*X)\n",
+      "\t\t\t# y_SO2 = (24-X)/(42-0.5*X)\n",
+      "\t\t\t# y_O2 = (18-0.5*X)/(42-0.5*X)\n",
+      "\n",
+      "\t\t\t# Ky = y_SO3/(y_SO2*y_O2**(2))\n",
+      "\t\t\t# Ky = (X*(42-0.5*X)**(1/2))/((24-X)*(18-0.5*X)**(1/2))\n",
+      "def f1(X): \n",
+      "\t return  Ky-(X*(42-0.5*X)**(1./2))/((24-X)*(18-0.5*X)**(1./2))\n",
+      "X_2 = fsolve(f1,22)\n",
+      "\n",
+      "y_SO3_2 = X_2/(42-0.5*X_2);\n",
+      "y_SO2_2 = (24-X_2)/(42-0.5*X_2);\n",
+      "y_O2_2 = (18-0.5*X_2)/(42-0.5*X_2);\n",
+      "print \" 2).The moles of SO3 formed = %f mol\"%(X_2);\n",
+      "print \"     The mole fractions at equilibrium are y_S03 = %f, y_SO2 = %f and y_O2 = %f\"%(y_SO3_2,y_SO2_2,y_O2_2);\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_SO2_3_in = 12;\n",
+      "n_O2_3_in = 9;\n",
+      "n_SO3_3_in = 0;\n",
+      "n_N2 = 79;\n",
+      "\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_SO2_1_eq = 12 - X\n",
+      "\t\t\t# n_O2_1_eq = 9 - 0.5*X\n",
+      "\t\t\t# n_SO3_1_eq = X\n",
+      "\t\t\t# Total moles = 100 - 0.5*X\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "\t\t\t# y_SO3 = X/(100-0.5*X)\n",
+      "\t\t\t# y_SO2 = (12-X)/(100-0.5*X)\n",
+      "\t\t\t# y_O2 = (9-0.5*X)/(100-0.5*X)\n",
+      "\n",
+      "\t\t\t# Ky = y_SO3/(y_SO2*y_O2**(2))\n",
+      "\t\t\t# Ky = (X*(100-0.5*X)**(1/2))/((12-X)*(9-0.5*X)**(1/2))\n",
+      "def f2(X): \n",
+      "\t return  Ky-(X*(100-0.5*X)**(1./2))/((12-X)*(9-0.5*X)**(1./2))\n",
+      "X_3 = fsolve(f2,10)\n",
+      "\n",
+      "y_SO3_3 = X_3/(100-0.5*X_3);\n",
+      "y_SO2_3 = (12-X_3)/(100-0.5*X_3);\n",
+      "y_O2_3 = (9-0.5*X_3)/(100-0.5*X_3);\n",
+      "\n",
+      "print \" 3).The moles of SO3 formed = %f mol\"%(X_3);\n",
+      "print \"     The mole fractions at equilibrium are y_S03 = %f y_SO2 = %f and y_O2 = %f\"%(y_SO3_3,y_SO2_3,y_O2_3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The moles of SO3 formed = 11.655537 mol\n",
+        "     The mole fractions at equilibrium are y_S03 = 0.768215 y_SO2 = 0.022704 and y_O2 = 0.209081\n",
+        " 2).The moles of SO3 formed = 23.311074 mol\n",
+        "     The mole fractions at equilibrium are y_S03 = 0.768215, y_SO2 = 0.022704 and y_O2 = 0.209081\n",
+        " 3).The moles of SO3 formed = 11.202213 mol\n",
+        "     The mole fractions at equilibrium are y_S03 = 0.118669 y_SO2 = 0.008451 and y_O2 = 0.036006\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.3  Page Number : 599"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 600;\t\t\t#[K] -  Reactor temperature\n",
+      "P = 300;\t\t\t#[atm] - Reactor pressure\n",
+      "K = 0.91*10**(-4);\t\t\t# Equilibrium constant\n",
+      "\n",
+      "\t\t\t# The fugacity coefficients of the components are\n",
+      "phi_CO = 1.0;\n",
+      "phi_H2 = 1.2;\n",
+      "phi_CH3OH = 0.47;\n",
+      "\n",
+      "\t\t\t# CO + 2*H2 - CH3OH \n",
+      "\n",
+      "\t\t\t# For gas phase reactions the standard state is pure ideal gas and thus fi_0 = 1 atm and thus\n",
+      "\t\t\t# ai_cap = fi_cap/fi_0 = yi*P*phi_i_cap/1\n",
+      "\t\t\t# Thus K = Ky*Kp*K_phi\n",
+      "# Calculations and Results            \n",
+      "Kp = P**(1-3);\n",
+      "K_phi = phi_CH3OH/(phi_CO*phi_H2**(2));\n",
+      "Ky = K/(Kp*K_phi);\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium ,the moles of the components be \n",
+      "\t\t\t# n_CO = 1 - X\n",
+      "\t\t\t# n_H2 = 3 - 2*X\n",
+      "\t\t\t# n_CH3OH = X\n",
+      "\t\t\t# Total moles = 4 - 2*X\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "\t\t\t# y_CO = (1-X)/(4-2*X)\n",
+      "\t\t\t# y_H2 = (3-2*X)/(4-2*X)\n",
+      "\t\t\t# y_CH3OH = (X)/(4-2*X)\n",
+      "\n",
+      "\t\t\t# Ky = y_CH3OH/(y_CO*y_H2**(2)) = (X/(4-2*X))/(((1-X)/(4-2*X))*((3-2*X)/(4-2*X))**(2))\n",
+      "def f(X): \n",
+      "\t return Ky-(X/(4-2*X))/(((1-X)/(4-2*X))*((3-2*X)/(4-2*X))**(2))\n",
+      "X = fsolve(f,0.1)\n",
+      "\n",
+      "\t\t\t# Therefore at equilibrium \n",
+      "y_CO = (1-X)/(4-2*X);\n",
+      "y_H2 = (3-2*X)/(4-2*X);\n",
+      "y_CH3OH = (X)/(4-2*X);\n",
+      "\n",
+      "print \" The mole fractions at equilibrium are y_CO = %f y_H2 = %f and y_CH3OH = %f\"%(y_CO,y_H2,y_CH3OH);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mole fractions at equilibrium are y_CO = 0.051857 y_H2 = 0.551857 and y_CH3OH = 0.396286\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.4  Page Number : 600"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\t\t\n",
+      "T = 600;\t\t\t#[K] -  Reactor temperature\n",
+      "P = 4;\t\t\t#[atm] - Reactor pressure\n",
+      "K = 1.175;\t\t\t# Equilibrium constant\n",
+      "\n",
+      "\t\t\t# (1/2)*N2 + (3/2)*H_2 - NH3\n",
+      "\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_N2 = 1;\n",
+      "n_H2 = 3;\n",
+      "n_HN3 = 0;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X.\n",
+      "\t\t\t# At equilibrium ,the moles of the components be \n",
+      "\t\t\t# n_N2 = 1 - 0.5*X\n",
+      "\t\t\t# n_H2 = 3 - 1.5*X\n",
+      "\t\t\t# n_NH3 = X\n",
+      "\t\t\t# Total moles = 4 - X\n",
+      "\n",
+      "\t\t\t# We have, K = Ky*Kp\n",
+      "Kp = P**(1-2);\t\t\t#[atm**(-1)]\n",
+      "Ky = K/(Kp);\n",
+      "\n",
+      "\t\t\t# Ky = y_NH3/(y_N2**(1/2)*y_H2**(3/2)) = (X/(4-X))/(((1-0.5*X)/(4-X))**(1/2)*((3-1.5*X)/(4-X))**(3/2))\n",
+      "\t\t\t# Solving the above equation we get\n",
+      "def f(X): \n",
+      "\t return Ky - (X/(4-X))/(((1-0.5*X)/(4-X))**(1./2)*((3-1.5*X)/(4-X))**(3./2))\n",
+      "X = fsolve(f,0.1)\n",
+      "\n",
+      "y_NH3 = X/(4-X);\t\t\t# Mole fraction of NH3 at equilibrium\n",
+      "\n",
+      "print \" The value of Kp = %f and Ky = %f \"%(Kp,Ky);\n",
+      "print \" The mole fractions of NH3 at equilibrium is %f\"%(y_NH3);\n",
+      "\n",
+      "\t\t\t\n",
+      "\n",
+      "\t\t\t# We know that for ideal gas, P*V = n*R*T and thus P is directly proportional to n at constant V and T.\n",
+      "\t\t\t# Let P = k*n\n",
+      "\t\t\t# Initially P = 4 atm and n = 4 moles, thus K = 1 and we get p = n, where P is in atm. \n",
+      "\t\t\t# Thus at equilibrium P = 4 - X\n",
+      "\n",
+      "\t\t\t# Ky = K/Kp = 1.175*P = 1.175*(4 - X)\n",
+      "\t\t\t# (X/(4-X))/(((1-0.5*X)/(4-X))**(1/2)*((3-1.5*X)/(4-X))**(3/2)) = 1.175*(4 - X)\n",
+      "\t\t\t# Solving the above equation we get\n",
+      "def f1(X): \n",
+      "\t return (X/(4-X))/(((1-0.5*X)/(4-X))**(1./2)*((3-1.5*X)/(4-X))**(3./2))-1.175*(4-X)\n",
+      "X_prime = fsolve(f1,1)\n",
+      "\n",
+      "\t\t\t# Therefore at equilibrium \n",
+      "P_prime = 4 - X_prime;\n",
+      "y_NH3_prime = X_prime/(4-X_prime);\n",
+      "\n",
+      "print \" If reaction is carried out at constant temperature and volumethen\"\n",
+      "print \" The equilibrium pressure is %f atm\"%(P_prime);\n",
+      "print \" The equilibrium mole fractions of NH3 in the reactor is %f\"%(y_NH3_prime);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of Kp = 0.250000 and Ky = 4.700000 \n",
+        " The mole fractions of NH3 at equilibrium is 0.454388\n",
+        " If reaction is carried out at constant temperature and volumethen"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The equilibrium pressure is 2.863057 atm\n",
+        " The equilibrium mole fractions of NH3 in the reactor is 0.397108\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.5  Page Number : 601"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 400;\t\t\t#[K] -  Reactor temperature\n",
+      "P = 1;\t\t\t#[atm] - Reactor pressure\n",
+      "K = 1.52;\t\t\t# Equilibrium constant\n",
+      "y_H2 = 0.4;\t\t\t# Equilibrium mole fraction of hydrogen\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# CO(g) + 2*H_2(g) - CH3OH(g)\n",
+      "\n",
+      "\t\t\t# K = y_CH3OH/(y_CO*y_H2**(2)*P**(2))\n",
+      "\t\t\t# Let total number of moles at equilibrium be 1\n",
+      "\t\t\t# y_CH3OH = 0.6 - y_CO;\n",
+      "\t\t\t# (0.6 - y_CO)/y_CO = K*P**(2)*y_H2**(2)\n",
+      "\n",
+      "y_CO = 0.6/(1 + K*P**(2)*y_H2**(2));\n",
+      "y_CH3OH = 0.6 - y_CO;\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print \" The mole fractions are y_CO = %f and y_CH3OH = %f \"%(y_CO,y_CH3OH);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mole fractions are y_CO = 0.482625 and y_CH3OH = 0.117375 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.6  Page Number : 602"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 749.;\t\t\t#[K] - Reactor temperature\n",
+      "P = 1.;\t\t\t#[atm] - Reactor pressure\n",
+      "K = 74.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Kp = P**(-1./2);\t\t\t#[atm**(-1/2)]\n",
+      "Ky = K/Kp;\n",
+      "\n",
+      "\t\t\t# SO2 + (1/2)*O2 - SO3\n",
+      "\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_SO2_1_in = 10;\n",
+      "n_O2_1_in = 8;\n",
+      "n_SO3_1_in = 0;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_SO2_1_eq = 10 - X\n",
+      "\t\t\t# n_O2_1_eq = 8 - 0.5*X\n",
+      "\t\t\t# SO3_1_eq = X\n",
+      "\t\t\t# Total moles = 18 - 0.5*X\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "\t\t\t# y_SO3 = X/(18-0.5*X)\n",
+      "\t\t\t# y_SO2 = (10-X)/(18-0.5*X)\n",
+      "\t\t\t# y_O2 = (8-0.5*X)/(18-0.5*X)\n",
+      "\n",
+      "\t\t\t# Ky = y_SO3/(y_SO2*y_O2**(2))\n",
+      "\t\t\t# Ky = (X*(18-0.5*X)**(1/2))/((10-X)*(8-0.5*X)**(1/2))\n",
+      "def f(X): \n",
+      "\t return  Ky-(X*(18-0.5*X)**(1./2))/((10-X)*(8-0.5*X)**(1./2))\n",
+      "X_1 = fsolve(f,11)\n",
+      "\n",
+      "n_SO3 = X_1;\n",
+      "n_SO2 = 10 - X_1;\n",
+      "n_O2 = 8 - 0.5*X_1;\n",
+      "\n",
+      "print \" 1).The moles of the components at equilibrium are n_SO3 = %f mol n_SO2 = %f mol and n_O2 = %f mol\"%(n_SO3,n_SO2,n_O2);\n",
+      "\n",
+      "\t\t\t# Now for the reaction\n",
+      "\t\t\t# 2*SO2 + O2 - 2*SO3\n",
+      "\n",
+      "\t\t\t# The equilibrium constant for this reaction is KP**(2)\n",
+      "Ky_prime = Ky**(2);\n",
+      "\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_SO2_1_eq = 10 - 2*X\n",
+      "\t\t\t# n_O2_1_eq = 8 - X\n",
+      "\t\t\t# SO3_1_eq = 2*X\n",
+      "\t\t\t# Total moles = 18 - X\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "\t\t\t# y_SO3 = 2*X/(18-X)\n",
+      "\t\t\t# y_SO2 = (10-2*X)/(18-X)\n",
+      "\t\t\t# y_O2 = (8- X)/(18-X)\n",
+      "\n",
+      "\t\t\t# Ky_prime = y_SO3**(2)/(y_SO2**(2)*y_O2)\n",
+      "\t\t\t# Ky_prime = ((2*X)**(2)*(18-X))/((10-2*X)**(2)*(8-X))\n",
+      "def f1(X): \n",
+      "\t return  Ky_prime-((2*X)**(2)*(18-X))/(((10-2*X)**(2))*(8-X))\n",
+      "X_2 = fsolve(f1,6)\n",
+      "\n",
+      "n_SO3_prime = 2*X_2;\n",
+      "n_SO2_prime = 10 - 2*X_2;\n",
+      "n_O2_prime = 8 - X_2;\n",
+      "\n",
+      "print \" 2).The moles of the components at equilibrium are n_SO3 = %f mol n_SO2 = %f mol and n_O2 = %f mol\"%(n_SO3_prime,n_SO2_prime,n_O2_prime);\n",
+      "print \"     Thus the number of moles remains the same irrespective of the stoichoimetry of the reaction\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The moles of the components at equilibrium are n_SO3 = 9.730824 mol n_SO2 = 0.269176 mol and n_O2 = 3.134588 mol\n",
+        " 2).The moles of the components at equilibrium are n_SO3 = 9.730824 mol n_SO2 = 0.269176 mol and n_O2 = 3.134588 mol\n",
+        "     Thus the number of moles remains the same irrespective of the stoichoimetry of the reaction\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.7  Page Number : 603"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 500;\t\t\t#[K]\n",
+      "\t\t\t# For the reaction,  0.5*A2 + 0.5*B2 - AB\n",
+      "delta_G = -4200;\t\t\t#[J/mol]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# A2 + B2 - 2*AB\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# We know delta_G_rkn_0 = -R*T*math.log(K) \n",
+      "delta_G_1 = 2*delta_G;\n",
+      "K_1 = math.exp(-delta_G_1/(R*T));\t\t\t# Equilibrium constant at 500 K for the above reaction\n",
+      "\t\t\t# As can be seen the reaction is not affected by pressure and therefore K = Ky as Kp = 1 \n",
+      "Ky = K_1;\n",
+      "\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_A2_1_in = 0.5;\n",
+      "n_B2_1_in = 0.5;\n",
+      "n_AB_1_in = 0;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_A2_1_eq = 0.5 - X\n",
+      "\t\t\t# n_B2_1_eq = 0.5- X\n",
+      "\t\t\t# n_AB_1_eq = 2*X\n",
+      "\t\t\t# Total moles = 1\n",
+      "\n",
+      "\t\t\t# Ky = (2*X)**(2)/(0.5-X)**(2)\n",
+      "def f(X): \n",
+      "\t return  Ky-(2*X)**(2)/(0.5-X)**(2)\n",
+      "X_1 = fsolve(f,0.2)\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "y_A2_1 = 0.5 - X_1;\n",
+      "y_B2_1 = 0.5- X_1;\n",
+      "y_AB_1 = 2*X_1;\n",
+      "\n",
+      "print \" 1).The mole fractions at equilibrium are y_A2 = %f y_B2 = %f and y_AB = %f\"%(y_A2_1,y_B2_1,y_AB_1);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# 0.5*A2 + 0.5*B2 - AB\n",
+      "\n",
+      "\t\t\t# We know delta_G_rkn_0 = -R*T*math.log(K) \n",
+      "delta_G_2 = delta_G;\n",
+      "K_2 = math.exp(-delta_G_2/(R*T));\t\t\t# Equilibrium constant at 500 K for the above reaction\n",
+      "\n",
+      "\t\t\t# As can be seen the reaction is not affected by pressure and therefore K = Ky as Kp = 1 \n",
+      "Ky_2 = K_2;\n",
+      "\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_A2_2_in = 0.5;\n",
+      "n_B2_2_in = 0.5;\n",
+      "n_AB_2_in = 0;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_A2_2_eq = 0.5 - 0.5*X\n",
+      "\t\t\t# n_B2_2_eq = 0.5- 0.5*X\n",
+      "\t\t\t# n_AB_2_eq = X\n",
+      "\t\t\t# Total moles = 1\n",
+      "\n",
+      "\t\t\t# Ky = y_AB/(y_A2**(1/2)*y_B2**(1/2))\n",
+      "\t\t\t# Ky = X/(0.5 - 0.5*X)\n",
+      "X_2 = 0.5*Ky_2/(1+0.5*Ky_2);\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "y_A2_2 = 0.5 - 0.5*X_2;\n",
+      "y_B2_2 = 0.5- 0.5*X_2;\n",
+      "y_AB_2 = X_2;\n",
+      "\n",
+      "print \" 2).The mole fractions at equilibrium are y_A2 = %f y_B2 = %f and y_AB = %f\"%(y_A2_2,y_B2_2,y_AB_2);\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "\t\t\t# 2*AB -  A2 + B2\n",
+      "\n",
+      "K_3 = 1/K_1;\t\t\t# Equilibrium constant at 500 K for the above reaction\n",
+      "\t\t\t# As can be seen the reaction is not affected by pressure and therefore K = Ky as Kp = 1 \n",
+      "Ky_3 = K_3;\n",
+      "\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "n_AB_3_in = 1;\n",
+      "n_A2_3_in = 0;\n",
+      "n_B2_3_in = 0;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_AB_3_eq = 1 - X\n",
+      "\t\t\t# n_A2_3_eq = X/2\n",
+      "\t\t\t# n_B2_3_eq = X/2\n",
+      "\t\t\t# Total moles = 1\n",
+      "\n",
+      "\t\t\t# Ky = (X/2)**(2)/(1-X)**(2)\n",
+      "def f1(X): \n",
+      "\t return  Ky_3-(X/2)**(2)/(1-X)**(2)\n",
+      "X_3 = fsolve(f1,0.4)\n",
+      "\n",
+      "\t\t\t# The mole fractions of the components at equilibrium are\n",
+      "y_A2_3 = X_3/2;\n",
+      "y_B2_3 = X_3/2;\n",
+      "y_AB_3 = 1-X_3;\n",
+      "\n",
+      "print \" 3).The mole fractions at equilibrium are y_A2 = %f y_B2 = %f and y_AB = %f\"%(y_A2_3,y_B2_3,y_AB_3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The mole fractions at equilibrium are y_A2 = 0.210680 y_B2 = 0.210680 and y_AB = 0.578641\n",
+        " 2).The mole fractions at equilibrium are y_A2 = 0.210680 y_B2 = 0.210680 and y_AB = 0.578641\n",
+        " 3).The mole fractions at equilibrium are y_A2 = 0.210680 y_B2 = 0.210680 and y_AB = 0.578641\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.9  Page Number : 606"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\t\t\n",
+      "\n",
+      "# Variables\n",
+      "R = 1.987;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "delta_H_SO2_298 = -70.96;\t\t\t#[kcal/mol] - Enthalpy of formation of S02 at 298.15 K\n",
+      "delta_H_SO3_298 = -94.45;\t\t\t#[kcal/mol] - Enthalpy of formation of S03 at 298.15 K\n",
+      "delta_G_SO2_298 = -71.79;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of SO2 at 298.15 K\n",
+      "delta_G_SO3_298 = -88.52;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of SO3 at 298.15 K\n",
+      "\n",
+      "\t\t\t# Cp_0 = a + b*T + c*T**(2) + d*T**(3)\n",
+      "\n",
+      "a_SO2 = 6.157;\n",
+      "a_SO3 = 3.918;\n",
+      "a_O2 = 6.085;\n",
+      "b_SO2 = 1.384*10**(-2);\n",
+      "b_SO3 = 3.483*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "c_SO2 = -0.9103*10**(-5);\n",
+      "c_SO3 = -2.675*10**(-5);\n",
+      "c_O2 = -0.1709*10**(-5);\n",
+      "d_SO2 = 2.057*10**(-9);\n",
+      "d_SO3 = 7.744*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "\n",
+      "\t\t\t#(1)\n",
+      "T_1 = 298.15;\t\t\t#[K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "delta_H_rkn_298 = delta_H_SO3_298 - delta_H_SO2_298;\t\t\t#[kcal]\n",
+      "delta_H_rkn_298 = delta_H_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "delta_G_rkn_298 = delta_G_SO3_298 - delta_G_SO2_298;\t\t\t#[kcal]\n",
+      "delta_G_rkn_298 = delta_G_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "\n",
+      "delta_a = a_SO3 - a_SO2 - (a_O2/2);\n",
+      "delta_b = b_SO3 - b_SO2 - (b_O2/2);\n",
+      "delta_c = c_SO3 - c_SO2 - (c_O2/2);\n",
+      "delta_d = d_SO3 - d_SO2 - (d_O2/2);\n",
+      "\n",
+      "\n",
+      "def f60(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_T = delta_H_rkn_298 +  quad(f60,T_1,T)[0]\n",
+      "\n",
+      "\t\t\t# On simplification we get\n",
+      "\t\t\t# delta_H_rkn_T = -22630.14 - 5.2815*T + 0.9587*10**(-2)*T**(2) - 0.5598*10**(-5)*T**(3) + 1.3826*10**(-9)*T**(4)\n",
+      "\n",
+      "print \" 1.The math.expression for delta_H_rkn_T as a function of T is given by\";\n",
+      "print \"     delta_H_rkn_T = -22630.14 - 5.2815*T + 0.9587*10**-2*T**2 - 0.5598*10**-5*T**3 + 1.3826*10**-9*T**4\";\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\n",
+      "\n",
+      "\t\t\t#def f61(T): \n",
+      "\t\t\t#\t return K_T/K_298) = integrate(delta_H_rkn_T/T**(2)\n",
+      "\n",
+      "\t\t\t# R*math.log(K_T/K_298) =  quad(f61,T_1,T)[0]\n",
+      "\n",
+      "\t\t\t# First let us calculate K_298.\n",
+      "\t\t\t# delta_G_rkn_T = - R*T*math.log(K)\n",
+      "K_298 = math.exp(-delta_G_rkn_298/(R*T_1));\n",
+      "\n",
+      "\t\t\t# On substituting the values and simplifying we get the math.expression\n",
+      "\t\t\t# math.log(K) = 3.87 + 11380.10/T - 2.6580*math.log(T) + 0.4825*10**(-2)*T - 0.1409*10**(-5)*T**(2) + 0.2320*10**(-9)*T**(3)\n",
+      "\n",
+      "print \" 2.The math.expression for math.logK as a function of T is given by\";\n",
+      "print \"     math.logK = 3.87 + 11380.10/T - 2.6580*math.logT + 0.4825*10**-2*T - 0.1409*10**-5*T**2 + 0.2320*10**-9*T**3\";\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "P = 1;\t\t\t#[atm]\n",
+      "T = 880;\t\t\t#[K]\n",
+      "K = math.exp(3.87 + 11380.10/T - 2.6580*math.log(T) + 0.4825*10**(-2)*T - 0.1409*10**(-5)*T**(2) + 0.2320*10**(-9)*T**(3));\n",
+      "Kp = P**(-1./2);\t\t\t#[atm**(-1/2)]\n",
+      "Ky = K/Kp;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_SO2_eq = 1 - X\n",
+      "\t\t\t# n_O2_eq = 0.5- 0.5*X\n",
+      "\t\t\t# n_SO3_1_eq = X\n",
+      "\t\t\t# Total moles = 1.5-0.5*X\n",
+      "\n",
+      "\t\t\t# Ky = (X*(1.5-0.5*X)**(1/2))/((1-X)*(0.5-0.5*X)**(1/2))\n",
+      "def f(X): \n",
+      "\t return  Ky - (X*(1.5-0.5*X)**(1./2))/((1-X)*(0.5-0.5*X)**(1./2))\n",
+      "X = fsolve(f,0.8)\n",
+      "\n",
+      "\t\t\t# The mole fraction of SO3 at equilibrium is given by\n",
+      "y_SO3 = X/(1.5-0.5*X);\n",
+      "\n",
+      "print \" 3).The mole fraction of SO3 at equilibrium is given by y_SO3 = %f\"%(y_SO3);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1.The math.expression for delta_H_rkn_T as a function of T is given by\n",
+        "     delta_H_rkn_T = -22630.14 - 5.2815*T + 0.9587*10**-2*T**2 - 0.5598*10**-5*T**3 + 1.3826*10**-9*T**4\n",
+        " 2.The math.expression for math.logK as a function of T is given by\n",
+        "     math.logK = 3.87 + 11380.10/T - 2.6580*math.logT + 0.4825*10**-2*T - 0.1409*10**-5*T**2 + 0.2320*10**-9*T**3\n",
+        " 3).The mole fraction of SO3 at equilibrium is given by y_SO3 = 0.649167\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.10  Page Number : 609"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "from numpy import *\n",
+      "\n",
+      "# Variables\n",
+      "\t\t\t# (1/2)*N2 + (1/2)*O2 - NO\n",
+      "\n",
+      "R = 1.987;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "delta_H_NO_298 = 21.600;\t\t\t#[kcal/mol] - Enthalpy of formation of S02 at 298.15 K\n",
+      "delta_G_NO_298 = 20.719;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of SO2 at 298.15 K\n",
+      "\n",
+      "\t\t\t# Cp_0 = a + b*T + c*T**(2) + d*T**(3)\n",
+      "\n",
+      "a_N2 = 6.157;\n",
+      "a_O2 = 6.085;\n",
+      "a_NO = 6.461;\n",
+      "b_N2 = -0.03753*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "b_NO = 0.2358*10**(-2);\n",
+      "c_N2 = 0.1930*10**(-5);\n",
+      "c_O2 = -0.1709*10**(-5);\n",
+      "c_NO = -0.07705*10**(-5);\n",
+      "d_N2 = -0.6861*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "d_NO = 0.08729*10**(-9);\n",
+      "\n",
+      "\t\t\t#(1)\n",
+      "T_1 = 298.15;\t\t\t#[K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "delta_H_rkn_298 = delta_H_NO_298;\t\t\t#[kcal]\n",
+      "delta_H_rkn_298 = delta_H_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "delta_G_rkn_298 = delta_G_NO_298;\t\t\t#[kcal]\n",
+      "delta_G_rkn_298 = delta_G_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "\n",
+      "delta_a = a_NO - (a_N2/2) - (a_O2/2);\n",
+      "delta_b = b_NO - (b_N2/2) - (b_O2/2);\n",
+      "delta_c = c_NO - (c_N2/2) - (c_O2/2);\n",
+      "delta_d = d_NO - (d_N2/2) - (d_O2/2);\n",
+      "\n",
+      "\n",
+      "def f49(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_T = delta_H_rkn_298 +  quad(f49,T_1,T)[0]\n",
+      "\n",
+      "\t\t\t# On simplification we get\n",
+      "\t\t\t# delta_H_rkn_T = 21584.63 - 0.033*T + 0.0365*10**(-2)*T**(2) - 0.0293*10**(-5)*T**(3) + 0.0685*10**(-9)*T**(4)\n",
+      "\n",
+      "print \" The math.expression for delta_H_rkn_T as a function of T is given by\";\n",
+      "print \" delta_H_rkn_T = 21584.63 - 0.033*T + 0.0365*10**-2*T**2 - 0.0293*10**-5*T**3 + 0.0685*10**-9*T**4\";\n",
+      "\n",
+      "\t\t\t# Now let us calculate K_298 (at 298 K)\n",
+      "\t\t\t# We know delta_G_rkn_298 = -R*T*math.log(K_298) \n",
+      "K_298 = math.exp(-delta_G_rkn_298/(R*T_1));\t\t\t# Equilibrium constant at 298.15 K\n",
+      "\n",
+      "\n",
+      "\t\t\t#def f50(T): \n",
+      "\t\t\t#\t return K_2/K_1) = integrate(delta_H_rkn_298/(R*T**(2))\n",
+      "\n",
+      "\t\t\t# math.log(K_2/K_1) =  quad(f50,T_1,T)[0]\n",
+      "\n",
+      "\t\t\t# On substituting the values and simplifying we get the math.expression\n",
+      "\t\t\t# math.log(K) = 1.5103 - 10862.92/T - 0.0166*math.log(T) + 1.84*10**(-4)*T - 7.35*10**(-8)*T**(2) + 1.15*10**(-11)*T**(3)\n",
+      "\n",
+      "print \" The math.expression for math.logK as a function of T is given by\";\n",
+      "print \" math.logK = 1.5103 - 10862.92/T - 0.0166*math.logT + 1.84*10**-4*T - 7.35*10**-8*T**2 + 1.15*10**-11*T**3\"\n",
+      "\n",
+      "T = [500,1000,1500,2000,2500];\n",
+      "K = zeros(5);\n",
+      "\n",
+      "print \" T K  \\t\\t\\t  K  \";\n",
+      "\n",
+      "for i in range(5):\n",
+      "    K[i] = math.exp(1.5103-10862.92/T[i] - 0.0166*math.log(T[i]) + 1.84*10**(-4)*T[i] - 7.35*10**(-8)*T[i]**(2) + 1.15*10**(-11)*T[i]**(3));\n",
+      "    \n",
+      "    print \" %f  \\t\\t  %e  \"%(T[i],K[i]);\n",
+      "\n",
+      "\n",
+      "\t\t\t# It can be seen that for endothermic reactions equilibrium constant increases with temperature.\n",
+      "print \" It can be seen that for endothermic reactions equilibrium constant increases with temperature\";\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The math.expression for delta_H_rkn_T as a function of T is given by\n",
+        " delta_H_rkn_T = 21584.63 - 0.033*T + 0.0365*10**-2*T**2 - 0.0293*10**-5*T**3 + 0.0685*10**-9*T**4\n",
+        " The math.expression for math.logK as a function of T is given by\n",
+        " math.logK = 1.5103 - 10862.92/T - 0.0166*math.logT + 1.84*10**-4*T - 7.35*10**-8*T**2 + 1.15*10**-11*T**3\n",
+        " T K  \t\t\t  K  \n",
+        " 500.000000  \t\t  1.615470e-09  \n",
+        " 1000.000000  \t\t  8.737610e-05  \n",
+        " 1500.000000  \t\t  3.333913e-03  \n",
+        " 2000.000000  \t\t  2.062328e-02  \n",
+        " 2500.000000  \t\t  6.176400e-02  \n",
+        " It can be seen that for endothermic reactions equilibrium constant increases with temperature\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.11  Page Number : 611"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "\t\t\t# SO2 + (1/2)*O2 - SO3\n",
+      "R = 8.314;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "K_800 = 0.0319;\t\t\t# Equilibrium constant at 800 K\n",
+      "K_900 = 0.153;\t\t\t# Equilibrium constant at 900 K\n",
+      "T_1 = 800.;\t\t\t#[K]\n",
+      "T_2 = 900.;\t\t\t#[K]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# We have the relation \n",
+      "\t\t\t# math.log(K_2/K_1) = -(delta_H_rkn/R)*(1/T_2 - 1/T_1)\n",
+      "\t\t\t# math.log(K_900/K_800) = -(delta_H_rkn_850/R)*(1/T_2 - 1/T_1)\n",
+      "delta_H_rkn_850 = -R*math.log(K_900/K_800)/(1/T_2 - 1/T_1);\t\t\t#[J]\n",
+      "delta_H_rkn_850 = delta_H_rkn_850*10**(-3);\t\t\t#[kJ]\n",
+      "\n",
+      "# Results\n",
+      "print \" The mean standard enthalpy change of reaction in the region 800 t0 900 is given by delta_H_rkn_850 = %f KJ\"%(delta_H_rkn_850);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mean standard enthalpy change of reaction in the region 800 t0 900 is given by delta_H_rkn_850 = 93.851672 KJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.13  Page Number : 611"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K]\n",
+      "T = 2600;\t\t\t#[K]\n",
+      "R = 1.987;\t\t\t#[cal/mol-K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# Cp_0 = a + b*T + c*T**(2) + d*T**(3)\n",
+      "delta_H_CO_298 = -26.416;\t\t\t#[kcal/mol] - Enthalpy of formation of CO at 298.15 K\n",
+      "delta_G_CO_298 = -32.808;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of CO at 298.15 K\n",
+      "delta_H_CO2_298 = -94.052;\t\t\t#[kcal/mol] - Enthalpy of formation of C02 at 298.15 K\n",
+      "delta_G_CO2_298 = -94.260;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of CO2 at 298.15 K\n",
+      "\n",
+      "\t\t\t# CO + (1/2)*O2 - CO2\n",
+      "\n",
+      "a_CO = 6.726;\n",
+      "a_O2 = 6.0685;\n",
+      "a_CO2 = 5.316;\n",
+      "b_CO = 0.04001*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "b_CO2 = 1.4285*10**(-2);\n",
+      "c_CO = 0.1283*10**(-5);\n",
+      "c_O2 = -0.1709*10**(-5);\n",
+      "c_CO2 = -0.8362*10**(-5);\n",
+      "d_CO = -0.5307*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "d_CO2 = 1.784*10**(-9);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "delta_H_rkn_298 = delta_H_CO2_298 - delta_H_CO_298;\t\t\t#[kcal]\n",
+      "delta_H_rkn_298 = delta_H_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "delta_G_rkn_298 = delta_G_CO2_298 - delta_G_CO_298;\t\t\t#[kcal]\n",
+      "delta_G_rkn_298 = delta_G_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "\n",
+      "delta_a = a_CO2 - (a_CO) - (a_O2/2);\n",
+      "delta_b = b_CO2 - (b_CO) - (b_O2/2);\n",
+      "delta_c = c_CO2 - (c_CO) - (c_O2/2);\n",
+      "delta_d = d_CO2 - (d_CO) - (d_O2/2);\n",
+      "\n",
+      "\n",
+      "def f47(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_T = delta_H_rkn_298 +  quad(f47,T_1,T)[0]\n",
+      "\n",
+      "\t\t\t# On simplification we get\n",
+      "delta_H_rkn_T = -66773.56 - 4.45*T + 0.605*10**(-2)*T**(2) - 0.29*10**(-5)*T**(3) + 0.54*10**(-9)*T**(4);\n",
+      "\n",
+      "\n",
+      "\t\t\t#def f48(T): \n",
+      "\t\t\t#\t return K/K_298) = integrate(delta_H_rkn_T/(R*T**(2))\n",
+      "\n",
+      "\t\t\t# math.log(K/K_298) =  quad(f48,T_1,T)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t# We know that  delta_G_rkn_T = -R*T*math.log(K)\n",
+      "\t\t\t# At 298.15 K\n",
+      "K_298 = math.exp(-delta_G_rkn_298/(R*T_1) );\n",
+      "\n",
+      "\t\t\t# Therfore on simplification we get\n",
+      "\t\t\t#math.log(K) = 2.94 + 33605.2/T - 2.24*math.log(T) + 0.304*10(-2)*T - 0.073*10**(-5)*T**(2) + 0.09*10**(-9)*T**(3)\n",
+      "K = math.exp(2.94 + 33605.2/T - 2.24*math.log(T) + 0.304*10**(-2)*T - 0.073*10**(-5)*T**(2) + 0.09*10**(-9)*T**(3));\n",
+      "\n",
+      "print \"  The value of equilibrium constant at 2600 K is given by K_298 = %f\"%(K);\n",
+      "\n",
+      "\n",
+      "\t\t\t#(a)\n",
+      "P_1 = 1;\t\t\t#[atm]\n",
+      "Kp_1 = P_1**(-1./2);\n",
+      "Ky_1 = K/Kp_1;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_CO_1_eq = 1 - X\n",
+      "\t\t\t# n_02_1_eq = 0.5- 0.5X\n",
+      "\t\t\t# n_CO2_1_eq = X\n",
+      "\t\t\t# Total moles = 1.5 - 0.5*X\n",
+      "\n",
+      "\t\t\t# Ky = y_CO2/(y_CO**(1/2)*y_CO)\n",
+      "\t\t\t#ky = (X*(1.5-0.5*X)**(1/2))/((1-X)*(0.5-0.5*X)**(1/2))\n",
+      "\n",
+      "def f(X): \n",
+      "\t return  Ky_1-(X*(1.5-0.5*X)**(1./2))/((1-X)*(0.5-0.5*X)**(1./2))\n",
+      "X_1 = fsolve(f,0.9)\n",
+      "\n",
+      "y_CO2_1 = X_1/(1.5-0.5*X_1);\n",
+      "y_CO_1 = (1-X_1)/(1.5-0.5*X_1);\n",
+      "y_O2_1 = (0.5-0.5*X_1)/(1.5-0.5*X_1);\n",
+      "\n",
+      "print \" a).The equilibrium composition at 1 atm) is given by y_CO2 = %f y_CO = %f and y_O2 = %f\"%(y_CO2_1,y_CO_1,y_O2_1);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "P_2 = 10;\t\t\t#[atm]\n",
+      "Kp_2 = P_2**(-1./2);\n",
+      "Ky_2 = K/Kp_2;\n",
+      "\n",
+      "\t\t\t# Ky = y_CO2/(y_CO**(1/2)*y_CO)\n",
+      "\t\t\t#ky = (X*(1.5-0.5*X)**(1/2))/((1-X)*(0.5-0.5*X)**(1/2))\n",
+      "\n",
+      "def f1(X): \n",
+      "\t return  Ky_2-(X*(1.5-0.5*X)**(1./2))/((1-X)*(0.5-0.5*X)**(1./2))\n",
+      "X_2 = fsolve(f1,0.9)\n",
+      "\n",
+      "y_CO2_2 = X_2/(1.5-0.5*X_2);\n",
+      "y_CO_2 = (1-X_2)/(1.5-0.5*X_2);\n",
+      "y_O2_2 = (0.5-0.5*X_2)/(1.5-0.5*X_2);\n",
+      "\n",
+      "print \" b).The equilibrium composition at 10 atm) is given by y_CO2 = %f y_CO = %f and y_O2 = %f\"%(y_CO2_2,y_CO_2,y_O2_2);\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "P_3 = 1;\t\t\t#[atm]\n",
+      "Kp_3 = P_3**(-1./2);\n",
+      "Ky_3 = K/Kp_3;\n",
+      "\n",
+      "\t\t\t# Ky = y_CO2/(y_CO**(1/2)*y_CO)\n",
+      "\t\t\t#ky = (X*(1.5-0.5*X)**(1/2))/((1-X)*(0.5-0.5*X)**(1/2))\n",
+      "\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_CO_3_eq = 1 - X\n",
+      "\t\t\t# n_02_3_eq = 0.5- 0.5X\n",
+      "\t\t\t# n_CO2_3_eq = X\n",
+      "\t\t\t# n_N2_eq = 1;\n",
+      "\t\t\t# Total moles = 2.5 - 0.5*X\n",
+      "\n",
+      "def f2(X): \n",
+      "\t return  Ky_3-(X*(2.5-0.5*X)**(1./2))/((1-X)*(0.5-0.5*X)**(1./2))\n",
+      "X_3 = fsolve(f2,0.9)\n",
+      "\n",
+      "y_CO2_3 = X_3/(2.5-0.5*X_3);\n",
+      "y_CO_3 = (1-X_3)/(2.5-0.5*X_3);\n",
+      "y_O2_3 = (0.5-0.5*X_3)/(2.5-0.5*X_3);\n",
+      "y_N2 = 1./(2.5-0.5*X_3);\n",
+      "\n",
+      "print \" c).The equilibrium composition at 1 atm and 1 mol N2) is given by y_CO2 = %f y_CO = %f y_O2 = %f and y_N2 = %f\"%(y_CO2_3,y_CO_3,y_O2_3,y_N2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  The value of equilibrium constant at 2600 K is given by K_298 = 16.484152\n",
+        " a).The equilibrium composition at 1 atm) is given by y_CO2 = 0.757531 y_CO = 0.161646 and y_O2 = 0.080823\n",
+        " b).The equilibrium composition at 10 atm) is given by y_CO2 = 0.876008 y_CO = 0.082662 and y_O2 = 0.041331\n",
+        " c).The equilibrium composition at 1 atm and 1 mol N2) is given by y_CO2 = 0.373822 y_CO = 0.100943 y_O2 = 0.050471 and y_N2 = 0.474764\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.14  Page Number : 614"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 25 + 298.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol-K]\n",
+      "delta_G_298 = -1000;\t\t\t#[J] - Gibbs free energy change at 298 K\n",
+      "\n",
+      "\t\t\t# G_E/(R*T) = x_1*x_2\n",
+      "# Calculations and Results\n",
+      "\t\t\t# We know that  delta_G_rkn = - R*T*math.log(K), therefore\n",
+      "K = math.exp(-delta_G_298/(R*T));\n",
+      "\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# Let x_1 is the mole fraction of A and x_2 be the mole fraction of B\n",
+      "\t\t\t# If A and B form an ideal mixture then,\n",
+      "Ky = 1;\n",
+      "\t\t\t# and K = Kx = x_2/x_1\n",
+      "\t\t\t# and at equilibrium x_2/x_1 = K\n",
+      "\t\t\t# (1-x_1)/x_1 = K\n",
+      "x_1 = 1/(1+K);\n",
+      "\n",
+      "print \" 1).The equilibrium composition for ideal behaviour) is given by x_1 = %f\"%(x_1);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# The activity coefficients are given by\n",
+      "\t\t\t# math.log(Y1) = [del(n*G_E/(R*T))/del(n_1)]_T,P,n_2 = x_2**(2)\n",
+      "\t\t\t# similarly, math.log(Y2) = x_1**(2)\n",
+      "\n",
+      "\t\t\t# The equilibrium constant for the liquid phase reaction is given by\n",
+      "\t\t\t# K = Kx*Ky = (x_2*Y2)/(x_1*Y1) = (x_2*math.exp(x_1**(2)))/(x_1*math.exp(x_2**(2)))\n",
+      "\t\t\t# Solving the above equation we get\n",
+      "\n",
+      "def f2(x_1): \n",
+      "\t return  K -((1-x_1)*math.exp(x_1**(2)))/(x_1*math.exp((1-x_1)**(2)))\n",
+      "x_1_prime = fsolve(f2,0.9)\n",
+      "\n",
+      "print \" 2).The equilibrium composition for non-ideal behaviour) is given by x_1 = %f\"%(x_1_prime);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The equilibrium composition for ideal behaviour) is given by x_1 = 0.408008\n",
+        " 2).The equilibrium composition for non-ideal behaviour) is given by x_1 = 0.328409\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.15  Page Number : 615"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\t\t\t\n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard reaction temperature\n",
+      "T_2 = 373.;\t\t\t#[K]  - Reaction temperature\n",
+      "P = 1.;\t\t\t#[atm]\n",
+      "R = 1.987;\t\t\t#[cal/mol-K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# CH3COOH (l) + C2H5OH (l) - CH3COOC2H5 (l) + H2O (l)\n",
+      "\n",
+      "delta_H_CH3COOH_298 = -116.2*10**(3.);\t\t\t# [cal/mol]\n",
+      "delta_H_C2H5OH_298 = -66.35*10**(3.);\t\t\t# [cal/mol]\n",
+      "delta_H_CH3COOC2H5_298 = -110.72*10**(3.);\t\t\t# [cal/mol]\n",
+      "delta_H_H2O_298 = -68.3174*10**(3.);\t\t\t# [cal/mol]\n",
+      "\n",
+      "delta_G_CH3COOH_298 = -93.56*10**(3.);\t\t\t# [cal/mol]\n",
+      "delta_G_C2H5OH_298 = -41.76*10**(3.);\t\t\t# [cal/mol]\n",
+      "delta_G_CH3COOC2H5_298 = -76.11*10**(3.);\t\t\t# [cal/mol]\n",
+      "delta_G_H2O_298 = -56.6899*10**(3.);\t\t\t# [cal/mol]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "delta_H_rkn_298 = delta_H_CH3COOC2H5_298 + delta_H_H2O_298 - delta_H_CH3COOH_298 - delta_H_C2H5OH_298;\t\t\t#[cal/mol]\n",
+      "delta_G_rkn_298 = delta_G_CH3COOC2H5_298 + delta_G_H2O_298 - delta_G_CH3COOH_298 - delta_G_C2H5OH_298;\t\t\t#[cal/mol]\n",
+      "\n",
+      "\t\t\t# We know that  delta_G_rkn_T = -R*T*math.log(K)\n",
+      "\t\t\t# At 298.15 K\n",
+      "K_298 = math.exp(-delta_G_rkn_298/(R*T_1) );\n",
+      "\n",
+      "\t\t\t# We know that dmath.log(K)/dT = delta_H_rkn/(R*T**(2))\n",
+      "\t\t\t# If delta_H_rkn is assumed constant we get\n",
+      "\t\t\t# math.log(K_2/K_1) = (-delta_H_rkn/R)*(1/T_2 - 1/T_1)\n",
+      "\t\t\t# math.log(K_373/K_298) = (-delta_H_rkn_298/R)*(1/T_2 - 1/T_1)\n",
+      "\n",
+      "K_373 = math.exp(math.log(K_298) + (-delta_H_rkn_298/R)*(1./T_2 - 1./T_1));\n",
+      "\t\t\t# Note that the equilibrium constant value rises becauses the reaction is endothermic\n",
+      "\n",
+      "print \" The value of equilibrium constant at 373 K is  K_373 = %f\"%(K_373);\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_CH3COOH = 1 - X\n",
+      "\t\t\t# n_C2H5OH = 1 - X\n",
+      "\t\t\t# n_CH3COOC2H5 = X\n",
+      "\t\t\t# n_H20 = X\n",
+      "\t\t\t# Total moles = 2\n",
+      "\n",
+      "\t\t\t# Kx = (x_CH3COOH*x_C2H5OH)/(x_CH3COOC2H5*x_H2O)\n",
+      "\t\t\t# Assuming the liquid mixture to be ideal,that is Ky = 1, therefore K_x = K\n",
+      "K_x = K_373;\n",
+      "\t\t\t# X**(2)/(1-X)**(2) = K_x\n",
+      "X = (K_x)**(1./2)/(1+(K_x)**(1./2));\n",
+      "\n",
+      "\t\t\t# The mole fraction of ethyl acetate is given by\n",
+      "x_CH3COOC2H5 = X/2;\n",
+      "\n",
+      "print \" The mole fraction of ethyl acetate in the equilibrium reaction mixture is given by x_CH3COOC2H5 = %f\"%(x_CH3COOC2H5);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of equilibrium constant at 373 K is  K_373 = 0.046697\n",
+        " The mole fraction of ethyl acetate in the equilibrium reaction mixture is given by x_CH3COOC2H5 = 0.088848\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.16  Page Number : 617"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "\t\t\t# CaCO3 (s1) - CaO (s2) + CO2 (g)\n",
+      "T_1 = 898 + 273.15;\t\t\t#[K]\n",
+      "T_2 = 700 + 273.15;\t\t\t#[K]\n",
+      "R = 8.314;\t\t\t#[J/mol-K] - Universal gas constant\n",
+      "\n",
+      "P_CO2_T_1 = 1;\t\t\t#[atm] - Decomposition pressure of  CO2 over CaCO3 at 898 C\n",
+      "P_CO2_T_2 = 0.0333;\t\t\t#[atm] - Decomposition pressure of  CO2 over CaCO3 at 700 C\n",
+      "\n",
+      "\t\t\t# The equilibrium constant of the reaction is given by\n",
+      "\t\t\t# K = (a_CO2*a_CaO)/a_CaCO3\n",
+      "\n",
+      "\t\t\t# Since the pressure is small therefore carbon dioxide can be assumed to behave as ideal gas and thus\n",
+      "\t\t\t# a_CO2 = y_CO2*P/1 = P_CO2\n",
+      "\n",
+      "\t\t\t# The activity of CaO is (CaO is pure)\n",
+      "\t\t\t# a_CaO = f_CaO/f_0_CaO = exp[V_CaO*(P - P_0)/(R*T)] = 1  (since pressure is low)\n",
+      "\n",
+      "\t\t\t# The activity of CaCO3 is (CaCO3 is pure)\n",
+      "\t\t\t# a_CaCO3 = f_CaCO3/f_0_CaCO3 = exp[V_CaCO3*(P - P_0)/(R*T)] = 1  (since pressure is low)\n",
+      "\n",
+      "\t\t\t#Since pressures are around 1 atm,therefore Poynting factors are also near 1, and thus activity of CaO and CaCO3 is unity and equilibrium constant is given by\n",
+      "\t\t\t#K = P_CO2 , therefore\n",
+      "# Calculations \n",
+      "\t\t\t# At 898 C\n",
+      "K_T_1 = P_CO2_T_1;\n",
+      "delta_G_T_1 = -R*T_1*math.log(K_T_1);\n",
+      "\n",
+      "\t\t\t# At 700 C\n",
+      "K_T_2 = P_CO2_T_2;\n",
+      "delta_G_T_2 = -R*T_2*math.log(K_T_2);\n",
+      "\n",
+      "# Results\n",
+      "print \" The value of delta_G_rkn at 700 C is %f J\"%(delta_G_T_1);\n",
+      "print \" The value of delta_G_rkn at 898 C is %f J\"%(delta_G_T_2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of delta_G_rkn at 700 C is -0.000000 J\n",
+        " The value of delta_G_rkn at 898 C is 27526.397496 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.17  Page Number : 618"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 700 + 273.15;\t\t\t#[K]\n",
+      "K = 7.403;\t\t\t# Equilibrium constant for the reaction at 700 C\n",
+      "\n",
+      "\t\t\t# CH4 - C (s) + 2*H2 \n",
+      "\n",
+      "\t\t\t# The equilibrium constant of the reaction is given by\n",
+      "\t\t\t# K = (a_C*a_H2**(2))/a_CH4\n",
+      "\n",
+      "\t\t\t# Since carbon is pure therefore its activity is given by\n",
+      "\t\t\t# a_C = f/f_0 = 1 as pressure is 1 atm.\n",
+      "\t\t\t# Since the pressure is low,therefore the gas phase can be taken to be ideal,therefore\n",
+      "\t\t\t# K = (y_H2**(2)*P**(2))/(y_CH4*P) = y_H2**(2)/y_CH4     (as P = 1 atm)\n",
+      "Ky = K;\t\t\t# (Kp = 1 atm)\n",
+      "\n",
+      "\t\t\t# Initial moles of the species are\n",
+      "n_CH4 = 1;\n",
+      "n_H2 = 0;\n",
+      "n_C = 0;\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the reaction be X\n",
+      "\t\t\t# Moles at equilibrium be\n",
+      "\t\t\t# n_CH4_eq = 1 -X\n",
+      "\t\t\t# n_H2_eq = 2*x\n",
+      "\t\t\t# n_C_eq = X\n",
+      "\n",
+      "\t\t\t# Moles present in gas phase\n",
+      "\t\t\t# n_CH4_gas = 1 -X\n",
+      "\t\t\t# n_H2_gas = 2*x\n",
+      "\t\t\t# Total moles = 1 + X\n",
+      "\n",
+      "\t\t\t# gas phase mole fraction \n",
+      "\t\t\t# y_CH4_gas = (1 -X)/(1+X)\n",
+      "\t\t\t# y_H2_gas = 2*x/(1+X)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# Ky = y_H2_gas**(2)/y_CH4_gaS\n",
+      "X = (K/(4+K))**(1./2);\n",
+      "\n",
+      "print \" The number of moles of carbon black formed from one mole of methane  is %f mol\"%(X);\n",
+      "\n",
+      "\t\t\t# Therefore mole fraction of components in the gas phase at equilibrium is \n",
+      "y_CH4 = (1-X)/(1+X);\n",
+      "y_H2 = 2*X/(1+X);\n",
+      "\n",
+      "print \" The mole fraction of components in the gas phase at equilibrium is given by y_CH4 = %f and y_H2 = %f \"%(y_CH4,y_H2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The number of moles of carbon black formed from one mole of methane  is 0.805739 mol\n",
+        " The mole fraction of components in the gas phase at equilibrium is given by y_CH4 = 0.107580 and y_H2 = 0.892420 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.18  Page Number : 619"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard reaction temperature\n",
+      "T_2 = 1042;\t\t\t#[K] - Reaction temperature\n",
+      "R = 1.987;\t\t\t#[cal/mol-K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# CaCO3 (s1) - CaO (s2) + CO2 (g)\n",
+      "\n",
+      "delta_H_CaCO3_298 = -289.5;\t\t\t#[kcal/mol] - Enthalpy of formation of CaCO3 at 298.15 K\n",
+      "delta_H_CaO_298 = -151.7;\t\t\t#[kcal/mol] - Enthalpy of formation of CaO at 298.15 K\n",
+      "delta_H_CO2_298 = -94.052;\t\t\t#[kcal/mol] - Enthalpy of formation of CO2 at 298.15 K\n",
+      "delta_G_CaCO3_298 = -270.8;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of CaCO3 at 298.15 K\n",
+      "delta_G_CaO_298 = -144.3;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of CaO at 298.15 K\n",
+      "delta_G_CO2_298 = -94.260;\t\t\t#[kcal/mol] - Gibbs free energy change for formation of CO2 at 298.15 K\n",
+      "\n",
+      "\t\t\t# The standard heat capacity data as a function of temperature are given below\n",
+      "\t\t\t# Cp_CO2 = 5.316 + 1.4285*10**(2)*T - 0.8362*10**(-5)*T**(2) + 1.784*10**(-9)*T**(3)\n",
+      "\t\t\t# Cp_CaO = 12.129 + 0.88*10**(-3)*T - 2.08*10**(5)*T**(-2)\n",
+      "\t\t\t# Cp_CaCO3 = 24.98 + 5.240*10**(-3)*T - 6.199*10**(5)*T**(-2)\n",
+      "\t\t\t# Therefore Cp_0 is given by\n",
+      "\t\t\t# Cp_0 = Cp_CO2 + Cp_CaO - Cp_CaCO3\n",
+      "\t\t\t# Cp_0 = -7.535 + 9.925*10**(-3)*T - 0.8362*10**(-5)*T**(2) + 1.784*10**(-9)*T**(3) + 4.119*10**(5)*T**(-2)\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The smath.tan(math.radiansard enthalpy change of the reaction at 298.15 K is given by\n",
+      "delta_H_rkn_298 = delta_H_CO2_298 + delta_H_CaO_298 - delta_H_CaCO3_298;\t\t\t#[kcal]\n",
+      "delta_H_rkn_298 = delta_H_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "delta_G_rkn_298 = delta_G_CO2_298 + delta_G_CaO_298 - delta_G_CaCO3_298;\t\t\t#[kcal]\n",
+      "delta_G_rkn_298 = delta_G_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t# The smath.tan(math.radiansard enthalpy change of the reaction at temperature T is given by\n",
+      "\n",
+      "def f7(T): \n",
+      "\t return -7.535 + 9.925*10**(-3)*T - 0.8362*10**(-5)*T**(2) + 1.784*10**(-9)*T**(3) + 4.119*10**(5)*T**(-2)\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_T = delta_H_rkn_298 +  quad(f7,T_1,T)[0]\n",
+      "\n",
+      "\t\t\t# On simplification we get\n",
+      "\t\t\t# delta_H_rkn_T = 47005.3 - 7.535*T + 4.9625*10**(-3)*T**(2) - 0.2787*10**(-5)*T**(3) + 0.446*10**(-9)*T**(4) - 4.119*10**(5)*T**(-1)\n",
+      "\n",
+      "\n",
+      "\n",
+      "\t\t\t#def f8(T): \n",
+      "\t\t\t#\t return K_2/K_1) = integrate(delta_H_rkn_T/(R*T**(2))\n",
+      "\n",
+      "\t\t\t# math.log(K_2/K_1) =  quad(f8,T_1,T)[0]\n",
+      "\n",
+      "\n",
+      "def f9(T): \n",
+      "\t return (47005.3-7.535*T+4.9625*10**(-3)*T**(2)-0.2787*10**(-5)*T**(3)+0.446*10**(-9)*T**(4) - 4.119*10**(5)*T**(-1))/T**(2)\n",
+      "\n",
+      "math.log_K2_K1 =  quad(f9,T_1,T_2)[0];\t\t\t# math.log(K_2/K_1)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t# We know that  delta_G_rkn_T = -R*T*math.log(K)\n",
+      "\t\t\t# At 298.15 K\n",
+      "K_298 = math.exp(-delta_G_rkn_298/(R*T_1) );\n",
+      "\n",
+      "\t\t\t# Putting the values in the above math.expression we get\n",
+      "\t\t\t# math.log(K_1042/K_298) = math.log_K2_K1/R\n",
+      "K_1042 = K_298*math.exp(math.log_K2_K1/R);\n",
+      "\n",
+      "print \" The value of equilibrium constant at 1042 K is K_1042 = %f\"%(K_1042);\n",
+      "\n",
+      "\t\t\t# Since for the given reaction K = P_CO2,where P is in atm, therefore,\n",
+      "P_CO2 = K_1042;\n",
+      "\t\t\t# and thus decomposition takes place till the partial pressure of CO2 reaches 0.095 atm\n",
+      "\t\t\t# After that the decomposition in the closed vessel stops as equilibrium is achieved.\n",
+      "\n",
+      "print \" The equilibrium pressure of CO2 is P_CO2 = %f atm \"%(P_CO2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of equilibrium constant at 1042 K is K_1042 = 0.095017\n",
+        " The equilibrium pressure of CO2 is P_CO2 = 0.095017 atm \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.19  Page Number : 620"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables \n",
+      "T_1 = 298.15;\t\t\t#[k] - Smath.tan(math.radiansard reaction temperature\n",
+      "T_2 = 1200;\t\t\t#[K] - Reaction temperature\n",
+      "R = 1.987;\t\t\t#[cal/mol-K] - Universal gas consatnt\n",
+      "\n",
+      "\t\t\t# C (s) + CO2 (g) - 2*CO2 (g)  \t\t\t# Reaction 1\n",
+      "\t\t\t# CO2 + H2 - CO + H2O  \t\t\t# Reacction 2\n",
+      "\n",
+      "K_1 = 63;\t\t\t# Equilibrium constant for the first reaction\n",
+      "K_2 = 1.4;\t\t\t# Equilibrium constant for the secind reaction\n",
+      "\n",
+      "delta_G_H2O_298 = -54640;\t\t\t#[cal/mol] - Smath.tan(math.radiansard Gibbs free energy of formation of water vapour\n",
+      "delta_H_H2O_298 = -57800;\t\t\t#[cal/mol] - Smath.tan(math.radiansard enthalpy change of formation of water vapour\n",
+      "delta_G_rkn_298 = delta_G_H2O_298;\n",
+      "\n",
+      "\t\t\t# The smath.tan(math.radiansard heat capacity data of the components in cal/mol-K are given below\n",
+      "\t\t\t# Cp_H2 = 6.947 - 0.2*10**(-3)*T + 4.8*10**(-7)*T**(2)\n",
+      "\t\t\t# Cp_O2 = 6.148 + 3.1*10**(-3)*T - 9.2*10**(-7)*T**(2)\n",
+      "\t\t\t# Cp_H2O = 7.256 + 2.3*10**(-3)*T + 2.8*10**(-7)*T**(2)\n",
+      "\n",
+      "\t\t\t# Let us consider two more reactions\n",
+      "\t\t\t# C (s) + (1/2)*O2 - CO \t\t\t# Reaction 3\n",
+      "\t\t\t# H2 + (1/2)*O2 - H2O  \t\t\t# Reaction 4\n",
+      "\n",
+      "\t\t\t# Considering the above 4 reactions, it can be shown that reaction (3) = reaction (1) + reaction (4) - reaction (2)\n",
+      "\t\t\t# Thus,  delta_G_rkn_3 = delta_G_rkn_1 + delta_G_rkn_4 - delta_G_rkn_2\n",
+      "\t\t\t# or, -R*T*math.log(K_3) = -R*T*math.log(K_1) + -R*T*math.log(K_4) - -R*T*math.log(K_2)\n",
+      "\t\t\t# K_3 = (K_1*K_4/K_2)\n",
+      "\n",
+      "\t\t\t# Now we have to determine K_4 at 1200 K.\n",
+      "\t\t\t# The smath.tan(math.radiansard enthalpy change of reaction (4) as a fuction of temperature is given by\n",
+      "\t\t\t# delta_H_rkn_T = -57020 - 2.765*T + 0.475*10**(-3)*T**(2) + 8.67*10**(-8)*T**(3);\n",
+      "\n",
+      "\n",
+      "\t\t\t#def f2(T): \n",
+      "\t\t\t#\t return K_4_2/K_4_1) = integrate(delta_H_rkn_T/(R*T**(2))\n",
+      "\n",
+      "\t\t\t# math.log(K_4_2/K_4_1) =  quad(f2,T_1,T)[0]\n",
+      "\n",
+      "# Calculations\n",
+      "def f3(T): \n",
+      "\t return (-57020-2.765*T+0.475*10**(-3)*T**(2)+8.67*10**(-8)*T**(3))/T**(2)\n",
+      "\n",
+      "math.log_K2_K1_4 =  quad(f3,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t# We know that  delta_G_rkn_T = -R*T*math.log(K)\n",
+      "\t\t\t# At 298.15 K\n",
+      "K_4_298 = math.exp(-delta_G_rkn_298/(R*T_1) );\n",
+      "\n",
+      "\t\t\t# Putting the values in the above math.expression we get\n",
+      "\t\t\t# math.log(K_4_1200/K_4_298) = math.log_K2_K1_R/R\n",
+      "K_4_1200 = K_4_298*math.exp(math.log_K2_K1_4/R);\n",
+      "K_4 = K_4_1200;\n",
+      "\n",
+      "\t\t\t# Therefore the equilibrium constant for reaction (3) at 1200 K is given by\n",
+      "K_3 = (K_1*K_4)/K_2;\n",
+      "\n",
+      "# Results\n",
+      "print \" The equilibrium constant at 1200 K for the given reaction is K = %e\"%(K_3);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The equilibrium constant at 1200 K for the given reaction is K = 3.622432e+09\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.20  Page Number : 622"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "delta_G_H2O_298 = -237.034;\t\t\t#[kJ/mol] - Smath.tan(math.radiansard Gibbs free energy of formation of H2O (l) at 298 K\n",
+      "F = 96485;\t\t\t#[C/mol] - Faraday constant\n",
+      "\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# For the reaction\n",
+      "\t\t\t# H2 (g) + (1/2)*O2 (g) - H2O (l)\n",
+      "n = 2;\t\t\t# Number of electrons transferred in the reaction\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The standard Gibbs free energy change of the reaction is given by\n",
+      "\t\t\t# delta_G_rkn = delta_G_for_H2O(l) - delta_G_for_H2(g) - (1/2/)*delta_G_for_O2(g)\n",
+      "\t\t\t# Since delta_G_for_H2  = 0 and delta_G_for_O2 = 0  (pure components)\n",
+      "delta_G_rkn = delta_G_H2O_298;\t\t\t#[kJ]\n",
+      "delta_G_rkn = delta_G_rkn*10**(3);\t\t\t#[J]\n",
+      "\n",
+      "\t\t\t# delta_G_rkn = -n*F*E_0\n",
+      "\t\t\t# Thus standard equilibrium cell voltage is given by\n",
+      "E_0 = - delta_G_rkn/(n*F);\t\t\t#/[V]\n",
+      "\n",
+      "print \" 1).The standard equilibrium cell voltage is %f V\"%(E_0);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# For the reaction\n",
+      "\t\t\t# 2*H2 (g) + O2 (g) - 2*H2O (l)\n",
+      "n_prime = 4;\t\t\t# Number of electrons transferred in the reaction\n",
+      "\n",
+      "\t\t\t# The standard Gibbs free energy change of the reaction is given by\n",
+      "\t\t\t# delta_G_rkn = 2*delta_G_for_H2O(l) - 2*delta_G_for_H2(g) - delta_G_for_O2(g)\n",
+      "\t\t\t# Since delta_G_for_H2  = 0 and delta_G_for_O2 = 0  (pure components)\n",
+      "delta_G_rkn_prime = 2*delta_G_H2O_298;\t\t\t#[kJ]\n",
+      "delta_G_rkn_prime = delta_G_rkn_prime*10**(3);\t\t\t#[J]\n",
+      "\n",
+      "\t\t\t# delta_G_rkn = -n*F*E_0\n",
+      "\t\t\t# Thus standard equilibrium cell voltage is given by\n",
+      "E_0_prime = - delta_G_rkn_prime/(n_prime*F);\t\t\t#/[V]\n",
+      "\n",
+      "print \" 2).The standard equilibrium cell voltage is %f V\"%(E_0_prime);\n",
+      "\n",
+      "\t\t\t# Thus the result obtained is same for both the reactions\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The standard equilibrium cell voltage is 1.228346 V\n",
+        " 2).The standard equilibrium cell voltage is 1.228346 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.21  Page Number : 624"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\t\t\t\n",
+      "P = 2;\t\t\t# Number of phases\n",
+      "C = 5;\t\t\t# Number of components\n",
+      "\n",
+      "# Calculations\t\n",
+      "\t\t\t# C + 2*H2 = CH4   \t\t\t# (reaction 1) \n",
+      "\t\t\t# H2 + (1/2)*O2 = H20  \t\t\t# (reaction 2)\n",
+      "\t\t\t# C + (1/2)*O2 = CO  \t\t\t# (reaction 3)\n",
+      "\t\t\t# C + O2 = CO2  \t\t\t# (reaction 4)\n",
+      "\n",
+      "\t\t\t# We do not have C in the equilibrium reaction mixture, therefore we have to eliminate it.\n",
+      "\t\t\t# Substituting for C from reaction (1) into reactions (3) and (4) we get the following set of reactions\n",
+      "\t\t\t# H2 + (1/2)*O2 = H20\n",
+      "\t\t\t# CH4 - 2*H2 + (1/2)*O2 = CO\n",
+      "\t\t\t# CH4 - 2*H2 + O2 = CO2\n",
+      "\n",
+      "\t\t\t# We do not have O2 in the equilibrium reaction mixture,therefore we have to eliminateit\n",
+      "\t\t\t# Substituting for O2 from the first reaction of the above set into seecond and third reactions of the above set we get the following set of reactions.\n",
+      "\t\t\t# CH4 + H2O - H2 = CO + 2*H2\n",
+      "\t\t\t# CH4 + 2*H20 - 2*H2 = CO2 + 2*H2 \n",
+      "\n",
+      "\t\t\t# Therefore one set of independent reactions is\n",
+      "\t\t\t# CH4 + H20 = CO + 3*H2\n",
+      "\t\t\t# CH4 + 2*H2O = CO2 + 4*H2\n",
+      "\n",
+      "\t\t\t# Another set of independent reactions can be obtained by substituting for CH4 from the first reaction into second and we get\n",
+      "\t\t\t# CH4 + H2O = CO + 3*H2\n",
+      "\t\t\t# CO + 3*H2 - H2O + 2*H2O = CO2 4*H2\n",
+      "\n",
+      "\t\t\t# Or, \n",
+      "\t\t\t# CH4 + H2O = CO + 3*H2\n",
+      "\t\t\t# CO + H2O = CO2 + H2\n",
+      "\t\t\t\n",
+      "r = 2;\n",
+      "\n",
+      "\t\t\t# and the number of degree of freedom becomes,\n",
+      "F = r - P + C;\n",
+      "\n",
+      "# Results\n",
+      "print \" The number of independent chemical reactions are %f \"%(r);\n",
+      "\n",
+      "print \" The first set of independent reactions are given below\";\n",
+      "print \" CH4 + H20 = CO + 3*H2\";\n",
+      "print \" CH4 + 2*H2O = CO2 + 4*H2\";\n",
+      "\n",
+      "print \" The second set of independent reactions are given below\";\n",
+      "print \" CH4 + H20 = CO + 3*H2\";\n",
+      "print \" CO + H2O = CO2 + H2\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The number of independent chemical reactions are 2.000000 \n",
+        " The first set of independent reactions are given below\n",
+        " CH4 + H20 = CO + 3*H2\n",
+        " CH4 + 2*H2O = CO2 + 4*H2\n",
+        " The second set of independent reactions are given below\n",
+        " CH4 + H20 = CO + 3*H2\n",
+        " CO + H2O = CO2 + H2\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.22  Page Number : 626"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.optimize import fsolve \n",
+      "# Variables\t\t\t\n",
+      "T = 400.;\t\t\t#[K] - Temperature\n",
+      "P = 1.;\t\t\t#[atm] - Pressure\n",
+      "R = 1.987;\t\t\t#[cal/mol-K] - Universal gas consatnt\n",
+      "\n",
+      "delta_G_n_pentane_400 = 9600.;\t\t\t#[cal/mol] - standard Gibbs free energy of formation of n-pentane at 400 K\n",
+      "delta_G_iso_pentane_400 = 8200.;\t\t\t#[cal/mol] - standard Gibbs free energy of formation of iso-pentane at 400 K\n",
+      "delta_G_neo_pentane_400 = 9000.;\t\t\t#[cal/mol] - standard Gibbs free energy of formation of neo-pentane at 400 K\n",
+      "\n",
+      "\t\t\t# The three reactions for the formation of these isomers can be written as follows\n",
+      "\t\t\t# 5*C + 6*H2 = n-pentane\n",
+      "\t\t\t# 5*C + 6*H2 = iso-pentane\n",
+      "\t\t\t# 5*C + 6*H2 = neo-pentane\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# Let us take the following set of independent reactions\n",
+      "\t\t\t# iso-pentane = n-pentane   \t\t\t#     (reaction 1)\n",
+      "delta_G_rkn_1 = delta_G_n_pentane_400 - delta_G_iso_pentane_400;\t\t\t#[cal]\n",
+      "K_1 = math.exp(-delta_G_rkn_1/(R*T));\n",
+      "\t\t\t# iso-pentane = neo-pentane \t\t\t#     (reaction 2)\n",
+      "delta_G_rkn_2 = delta_G_neo_pentane_400 - delta_G_iso_pentane_400;\t\t\t#[cal]\n",
+      "K_2 = math.exp(-delta_G_rkn_2/(R*T));\n",
+      "\n",
+      "\t\t\t# Let the initial number of moles be\n",
+      "\t\t\t# n_iso_pentane = 1\n",
+      "\t\t\t# n_n_pentane = 0\n",
+      "\t\t\t# n_neo_pentane = 0\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the two reaction be X_1 and X_2 respectively\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_iso_pentane_eq = 1 - X_1 - X_2\n",
+      "\t\t\t# n_n_pentane_eq = X_1\n",
+      "\t\t\t# n_neo_pentane_eq = X_2\n",
+      "\t\t\t# Total moles = 1 \n",
+      "\n",
+      "\t\t\t# Pressure has no effect on these reactions ( P = 1 atm) and therefore\n",
+      "Ky_1 = K_1;\n",
+      "Ky_2 = K_2;\n",
+      "\n",
+      "\t\t\t# From reaction (1), we get\n",
+      "\t\t\t# Ky_1 = X_1/(1-X_1-X_2)\n",
+      "\n",
+      "\t\t\t# From reaction (2), we get\n",
+      "\t\t\t# Ky_2 = X_2/(1-X_1-X_2)\n",
+      "\n",
+      "\t\t\t# Putting the value of X_1 from first equation into the second we get\n",
+      "\t\t\t# X_1 = (Ky_1*(1-X_2))/(1+Ky_1)\n",
+      "\t\t\t# Now putting it into the second equation we grt\n",
+      "\t\t\t# Ky_2 = X_2/(1-((Ky_1*(1-X_2))/(1+Ky_1))-X_2)\n",
+      "\t\t\t# Now solving for X_2\n",
+      "def f(X_2): \n",
+      "\t return  Ky_2 - X_2/(1-((Ky_1*(1-X_2))/(1+Ky_1))-X_2)\n",
+      "X_2 = fsolve(f,0.8)\n",
+      "\n",
+      "\t\t\t# Therefore X_1 can be calculated as\n",
+      "X_1 = (Ky_1*(1-X_2))/(1+Ky_1);\n",
+      "\n",
+      "\t\t\t# Finally the mole fractions of the components at equilibrium\n",
+      "y_n_pentane = X_1;\n",
+      "y_neo_pentane = X_2;\n",
+      "y_iso_pentane = 1 -X_1 - X_2;\n",
+      "\n",
+      "print \" The equilibrium composition is given by y_n_pentane = %f y_neo_pentane = %f and y_iso_pentane = %f\"%(y_n_pentane,y_neo_pentane,y_iso_pentane);\n",
+      "\n",
+      "\t\t\t# Let us consider another set of independent reactions\n",
+      "\n",
+      "\t\t\t# n-pentane = iso-pentane   \t\t\t#     (reaction 3)\n",
+      "delta_G_rkn_3 = delta_G_iso_pentane_400 - delta_G_n_pentane_400;\t\t\t#[cal]\n",
+      "K_3 = math.exp(-delta_G_rkn_3/(R*T));\n",
+      "\t\t\t# n-pentane = neo-pentane \t\t\t#     (reaction 4)\n",
+      "delta_G_rkn_4 = delta_G_neo_pentane_400 - delta_G_n_pentane_400;\t\t\t#[cal]\n",
+      "K_4 = math.exp(-delta_G_rkn_4/(R*T));\n",
+      "\n",
+      "\t\t\t# Let the initial number of moles be\n",
+      "\t\t\t# n_n_pentane = 1\n",
+      "\t\t\t# n_iso_pentane = 0\n",
+      "\t\t\t# n_neo_pentane = 0\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the two reaction be X_3 and X_4 respectively\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_n_pentane_eq = 1 - X_3 - X_4\n",
+      "\t\t\t# n_iso_pentane_eq = X_4\n",
+      "\t\t\t# n_neo_pentane_eq = X_4\n",
+      "\t\t\t# Total moles = 1 \n",
+      "\n",
+      "\t\t\t# Pressure has no effect on these reactions (P = 1 atm) and therefore\n",
+      "Ky_3 = K_3;\n",
+      "Ky_4 = K_4;\n",
+      "\n",
+      "\t\t\t# From reaction (3), we get\n",
+      "\t\t\t# Ky_3 = X_3/(1-X_3-X_4)\n",
+      "\n",
+      "\t\t\t# From reaction (4), we get\n",
+      "\t\t\t# Ky_4 = X_4/(1-X_3-X_4)\n",
+      "\n",
+      "\t\t\t# Putting the value of X_3 from first equation into the second we get\n",
+      "\t\t\t# X_3 = (Ky_3*(1-X_4))/(1+Ky_3)\n",
+      "\t\t\t# Now putting it into the second equation we grt\n",
+      "\t\t\t# Ky_4 = X_4/(1-((Ky_1*(1-X_4))/(1+Ky_3))-X_4)\n",
+      "\t\t\t# Now solving for X_4\n",
+      "def f1(X_4): \n",
+      "\t return  Ky_4 - X_4/(1-((Ky_3*(1-X_4))/(1+Ky_3))-X_4)\n",
+      "X_4 = fsolve(f1,0.8)\n",
+      "\n",
+      "\t\t\t# Therefore X_3 can be calculated as\n",
+      "X_3 = (Ky_3*(1-X_4))/(1+Ky_3);\n",
+      "\n",
+      "\t\t\t# Finally the mole fractions of the components at equilibrium\n",
+      "y_n_pentane1 = 1 - X_3 - X_4;\n",
+      "y_neo_pentane1 = X_4;\n",
+      "y_iso_pentane1 = X_3;\n",
+      "\n",
+      "\t\t\t# The final composition does not depend on the set of reactions considered. \n",
+      "\n",
+      "print \" For another set of independent reactions considered\";\n",
+      "print \" The equilibrium composition is given by y_n_pentane = %f y_neo_pentane = %f and y_iso_pentane = %f\"%(y_n_pentane1,y_neo_pentane1,y_iso_pentane1);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The equilibrium composition is given by y_n_pentane = 0.111753 y_neo_pentane = 0.237745 and y_iso_pentane = 0.650501\n",
+        " For another set of independent reactions considered\n",
+        " The equilibrium composition is given by y_n_pentane = 0.111753 y_neo_pentane = 0.237745 and y_iso_pentane = 0.650501\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.23  Page Number : 628"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 600;\t\t\t#[K] - Temperature\n",
+      "P = 1;\t\t\t#[atm] - Pressure\n",
+      "R = 1.987;\t\t\t#[cal/mol-K] - Universal gas consatnt\n",
+      "\n",
+      "\t\t\t# CH4 + H2O = CO + 3*H2 \t\t\t#    (Reaction 1)\n",
+      "\t\t\t# CO + H2O = CO2 + H2  \t\t\t#     (Reaction 2)\n",
+      "\n",
+      "K_1 = 0.574;\t\t\t# Equilibrium constant of first reaction\n",
+      "K_2 = 2.21;\t\t\t# Equilibrium constant of second reaction\n",
+      "\n",
+      "\t\t\t# Initial number of moles of the components are\n",
+      "\t\t\t# n_CH4 = 1\n",
+      "\t\t\t# n_H2O = 5\n",
+      "\t\t\t# n_CO = 0\n",
+      "\t\t\t# n_H2 = O\n",
+      "\t\t\t# n_CO2 = 0\n",
+      "\n",
+      "\t\t\t# Let the reaction coordinate at equilibrium for the two reaction be X_1 and X_2 respectively\n",
+      "\t\t\t# At equilibrium, the moles of the components be\n",
+      "\t\t\t# n_CH4_eq = 1 - X_1\n",
+      "\t\t\t# n_H20_eq = 5 - X_1 - X_2\n",
+      "\t\t\t# n_CO_eq = X_1 - X_2\n",
+      "\t\t\t# n_H2_eq = 3*X_1 + X_2\n",
+      "\t\t\t# n_CO2_eq = X_2\n",
+      "\t\t\t# Total moles = 6 + 2*X_1\n",
+      "\n",
+      "\t\t\t# Since the pressure is 1 atm, K = Kp, Ky = K\n",
+      "Ky_1 = K_1;\n",
+      "Ky_2 = K_2;\n",
+      "\n",
+      "\t\t\t# From reaction (1), we get\n",
+      "\t\t\t# Ky_1 = ((X_1-X_2)*(3*X_1+X_2)**(3))/((1-X_1)*(5-X_1-X_2)*(6+2*X_1)**(2))\n",
+      "\n",
+      "\t\t\t# From reaction (2), we get\n",
+      "\t\t\t# Ky_2 = (X_2*(3*X_1+X_2))/((X_1-X_2)*(5-X_1-X_2))\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# Let us assume a value of X_1\n",
+      "X_1 = 0.1;\n",
+      "def f(X_2): \n",
+      "\t return  Ky_1-((X_1-X_2)*(3*X_1+X_2)**(3))/((1-X_1)*(5-X_1-X_2)*(6+2*X_1)**(2))\n",
+      "\n",
+      "def f1(X_1_prime): \n",
+      "\t return  Ky_2-(X_2_prime*(3*X_1_prime+X_2_prime))/((X_1_prime-X_2_prime)*(5-X_1_prime-X_2_prime))\n",
+      "\n",
+      "fault = 10;\n",
+      "while(fault>0.05):\n",
+      "    X_2 = fsolve(f,0.8)\n",
+      "    X_2_prime = X_2;\n",
+      "    X_1_prime = fsolve(f1,0.8)\n",
+      "    fault=abs(X_1 - X_1_prime);\n",
+      "    X_1 = X_1 + 0.001;\n",
+      "\n",
+      "n_CH4 = 1 - X_1;\n",
+      "n_H20 = 5 - X_1 - X_2;\n",
+      "n_CO = X_1 - X_2;\n",
+      "n_H2 = 3*X_1 + X_2;\n",
+      "n_CO2 = X_2;\n",
+      "Total_moles = 6 + 2*X_1;\n",
+      "\n",
+      "# Results\n",
+      "print \" The equilibrium composition of the resulting mixture is given by\";\n",
+      "print \" n_CH4 = % f mol n_H2O = %f mol n_CO = %f mol n_H2 = %f mol and n_CO2 = %f mol\"%(n_CH4,n_H20,n_CO,n_H2,n_CO2);\n",
+      "print \" The total number of moles is %f mol\"%(Total_moles);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The equilibrium composition of the resulting mixture is given by\n",
+        " n_CH4 =  0.089000 mol n_H2O = 3.471420 mol n_CO = 0.293420 mol n_H2 = 3.350580 mol and n_CO2 = 0.617580 mol\n",
+        " The total number of moles is 7.822000 mol\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "/home/jovina/virtualenvs/scipy/local/lib/python2.7/site-packages/scipy/optimize/minpack.py:236: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last ten iterations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n",
+        "/home/jovina/virtualenvs/scipy/local/lib/python2.7/site-packages/scipy/optimize/minpack.py:236: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last five Jacobian evaluations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.24  Page Number : 631"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 600 + 273.15;\t\t\t#[K] - Reaction temperature\n",
+      "P = 1;\t\t\t#[atm] - Reaction pressure\n",
+      "\n",
+      "\t\t\t# The Gibbs free energy of formation of various species at 873.15 K are\n",
+      "delta_G_CH4_RT = -2.82;\t\t\t# delta_G_CH4/(R*T)\n",
+      "delta_G_H2O_RT = -29.73;\t\t\t# delta_G_CH4/(R*T)\n",
+      "delta_G_CO_RT = -27.51;\t\t\t# delta_G_CH4/(R*T)\n",
+      "delta_G_H2_RT = -1.46;\t\t\t# delta_G_CH4/(R*T)\n",
+      "delta_G_CO2_RT = -56.68;\t\t\t# delta_G_CH4/(R*T)\n",
+      "\n",
+      "# Calculations\t\t\n",
+      "        \t# Initial number of moles of the components are\n",
+      "\t\t\t# n_CH4 = 1\n",
+      "\t\t\t# n_H2O = 5\n",
+      "\t\t\t# n_CO = 0\n",
+      "\t\t\t# n_H2 = O\n",
+      "\t\t\t# n_CO2 = 0\n",
+      "\n",
+      "\t\t\t# The del(F)/del(n_i) = 0 equations for CH4 (1), H2O (2), CO (3), H2 (4) and CO2 (5) becomes\n",
+      "\t\t\t# delta_G_1_T + R*T*math.log((n_1*P)/n) + lambda_C + 4*lambda_H = 0\n",
+      "\t\t\t# delta_G_2_T + R*T*math.log((n_2*P)/n) + 2*lambda_C + lambda_O = 0\n",
+      "\t\t\t# delta_G_3_T + R*T*math.log((n_3*P)/n) + lambda_c + lambda_O = 0\n",
+      "\t\t\t# delta_G_4_T + R*T*math.log((n_4*P)/n) + 2*lambda_H = 0\n",
+      "\t\t\t# delta_G_5_T + R*T*math.log((n_5*P)/n) + lambda_C + 2*lambda_O = 0\n",
+      "\n",
+      "\t\t\t# Where n is the total number of moles in the equilibrium reaction mixture. Dividing the above equations by R*T we get\n",
+      "            # delta_G_5_T/(R*T) + math.log((n_5*P)/n) + lambda_C/(R*T) + 2*lambda_O/(R*T) = 0\n",
+      "\n",
+      "\t\t\t# Substituting the values of delta_G_i_T/(R*T) in the above equations,the full set of equations (including elemental balance) becomes\n",
+      "\t\t\t# -56.68 + math.log(n_5/n) + lambda_C/(R*T) + 2*lambda_O/(R*T) = 0\n",
+      "\n",
+      "\t\t\t# The constraint equations are\n",
+      "\t\t\t# n_1 + n_3 + n_5  = 1 \t\t\t# (moles of C in the reaction mixture = 1)\n",
+      "\t\t\t# 4*n_1 + 2*n_2 + 2*n_4 = 14  \t\t\t# (moles of H in the reaction mixture = 14)\n",
+      "\t\t\t# n_2 + n_3 + 2*n_5 = 5  \t\t\t# (moles of O in the raection mixture = 5)\n",
+      "\n",
+      "\t\t\t# The total moles are given by\n",
+      "\t\t\t# n = n_1 + n_2 + n_3 + n_4 + n_5\n",
+      "\n",
+      "def solution(x):\n",
+      "    f = [0,0,0,0,0,0,0,0,0]\n",
+      "    f[0]= -2.82 + math.log(x[0]/x[5]) + x[6] + 4*x[7];\n",
+      "    f[1] = -29.73 + math.log(x[1]/x[5]) + 2*x[7] + x[8];\n",
+      "    f[2] = -27.51 + math.log(x[2]/x[5]) + x[6] + x[8];\n",
+      "    f[3] = -1.46 + math.log(x[3]/x[5]) + 2*x[7];\n",
+      "    f[4] = -56.68 + math.log(x[4]/x[5]) + x[6] + 2*x[8];\n",
+      "    f[5] = x[0] + x[2] +x[4] - 1;\n",
+      "    f[6] = 4*x[0] + 2*x[1] + 2*x[3] - 14;\n",
+      "    f[7] = x[1] + x[2] +2*x[4] - 5;\n",
+      "    f[8] = x[0]+ x[1] + x[2] + x[3] + x[4] - x[5]; \n",
+      "    return f\n",
+      "\n",
+      "\n",
+      "x = [0.01,3.5,0.2,3.0,0.5,5.0,2.0,1.0,25.0];\n",
+      "y = fsolve(solution,x)\n",
+      "\n",
+      "# Results\n",
+      "print \" n_1 = %f mol\"%(y[0]);\n",
+      "print \" n_2 = %f mol\"%(y[1]);\n",
+      "print \" n_3 = %f mol\"%(y[2]);\n",
+      "print \" n_4 = %f mol\"%(y[3]);\n",
+      "print \" n_5 = %f mol\"%(y[4]);\n",
+      "print \" n = %f mol\"%(y[5]);\n",
+      "print \" lambda_C/RT = %f\"%(y[6]);\n",
+      "print \" lambda_H/RT = %f\"%(y[7]);\n",
+      "print \" lambda_O/RT = %f\"%(y[8]);\n",
+      "\n",
+      "print \" The Lagrange multiplier values do not have any physical significance\";\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " n_1 = 0.091556 mol\n",
+        " n_2 = 3.442034 mol\n",
+        " n_3 = 0.258922 mol\n",
+        " n_4 = 3.374855 mol\n",
+        " n_5 = 0.649522 mol\n",
+        " n = 7.816888 mol\n",
+        " lambda_C/RT = 2.667225\n",
+        " lambda_H/RT = 1.149967\n",
+        " lambda_O/RT = 28.250290\n",
+        " The Lagrange multiplier values do not have any physical significance\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch18_1.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch18_1.ipynb
new file mode 100755
index 00000000..f07249a6
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch18_1.ipynb
@@ -0,0 +1,587 @@
+{
+ "metadata": {
+  "name": "ch18_1",
+  "signature": "sha256:32121f42f5b148741ef15da1993d191234c228c3ab7dcf05f69114fbe323a883"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18 : Adiabatic Reaction Temperature"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.1  Page Number : 650"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K] - standard reaction temperature\n",
+      "T_2 = 500;\t\t\t#[K] - Reaction temperature\n",
+      "P = 1;\t\t\t#[atm] - Pressure \n",
+      "\n",
+      "a_CO2 = 5.316;\n",
+      "a_O2 = 6.085;\n",
+      "a_N2 = 6.903;\n",
+      "a_H2O = 7.700;\n",
+      "a_C3H8 = -0.966;\n",
+      "b_CO2 = 1.4285*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "b_N2 = -0.03753*10**(-2);\n",
+      "b_H2O = 0.04594*10**(-2);\n",
+      "b_C3H8 = 7.279*10**(-2);\n",
+      "c_CO2 = -0.8362*10**(-5);\n",
+      "c_O2 = -0.1709*10**(-5);\n",
+      "c_N2 = 0.1930*10**(-5);\n",
+      "c_H2O = 0.2521*10**(-5);\n",
+      "c_C3H8 = -3.755*10**(-5);\n",
+      "d_CO2 = 1.784*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "d_N2 = -0.6861*10**(-9);\n",
+      "d_H2O = -0.8587*10**(-9);\n",
+      "d_C3H8 = 7.580*10**(-9);\n",
+      "\n",
+      "\t\t\t# The smath.tan(math.radiansard enthalpy of formation at 298.15 K is given by\n",
+      "delta_H_for_CO2 = -94.052;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_C3H8 = -24.820;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_H2O = -57.7979;\t\t\t#[kcal/mol]\n",
+      "\n",
+      "\t\t\t# The reaction with stoichiometric amount of air is\n",
+      "\t\t\t# C3H8 + 5(O2 + 3.7N2) - 3CO2 + 4H2O + 18.8N2\n",
+      "\n",
+      "\t\t\t# The reaction with 100% excess air is\n",
+      "\t\t\t# C3H8 + 10(O2 + 3.7N2) - 3CO2 + 4H2O + 5O2 + 37.6N2\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# The smath.tan(math.radiansard enthalpy change of reaction at 298.15 K\n",
+      "delta_H_rkn_298 = 3*delta_H_for_CO2 + 4*delta_H_for_H2O - delta_H_for_C3H8;\n",
+      "\n",
+      "\t\t\t# For exit stream\n",
+      "sum_ai_ni = 3*a_CO2 + 4*a_H2O + 5*a_O2 + 37.6*a_N2;\n",
+      "sum_bi_ni = 3*b_CO2 + 4*b_H2O + 5*b_O2 + 37.6*b_N2;\n",
+      "sum_ci_ni = 3*c_CO2 + 4*c_H2O + 5*c_O2 + 37.6*c_N2;\n",
+      "sum_di_ni = 3*d_CO2 + 4*d_H2O + 5*d_O2 + 37.6*d_N2;\n",
+      "\n",
+      "\n",
+      "\t\t\t# To raise the exit species from 298.15 to 500 K the enthalpy change is\n",
+      "\n",
+      "def f1(T): \n",
+      "\t return sum_ai_ni+sum_bi_ni*T+sum_ci_ni*T**(2)+sum_di_ni*T**(3)\n",
+      "\n",
+      "delta_H_rkn =  quad(f1,T_1,T_2)[0]\n",
+      "\n",
+      "delta_H_rkn = delta_H_rkn*10**(-3);\t\t\t#[kcal]\n",
+      "\n",
+      "\t\t\t# Therefore per mole of fuel the heat exchange is\n",
+      "\t\t\t# Q = Heat exchange in step 1 + Heat exchange in step 2\n",
+      "Q = delta_H_rkn_298 + delta_H_rkn;\n",
+      "\n",
+      "# Results\n",
+      "print \"   The heat transfer from the combustion chamber per mole of fuel is %f kcal (per mol of C3H8)\"%(Q);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "   The heat transfer from the combustion chamber per mole of fuel is -415.328732 kcal (per mol of C3H8)\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.2  Page Number : 650"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard reaction temperature\n",
+      "\n",
+      "a_CO2 = 5.316;\n",
+      "a_H2O = 7.700;\n",
+      "a_O2 = 6.085;\n",
+      "a_C2H6 = 1.648;\n",
+      "b_CO2 = 1.4285*10**(-2);\n",
+      "b_H2O = 0.04595*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "b_C2H6 = 4.124*10**(-2);\n",
+      "c_CO2 = -0.8362*10**(-5);\n",
+      "c_H2O = 0.2521*10**(-5);\n",
+      "c_O2 = -0.1709*10**(-5);\n",
+      "c_C2H6 = -1.530*10**(-5);\n",
+      "d_CO2 = 1.784*10**(-9);\n",
+      "d_H2O = -0.8587*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "d_C2H6 = 1.740*10**(-9);\n",
+      "\n",
+      "\t\t\t# The stabdard enthalpy of formation at 298.15 K is given by\n",
+      "delta_H_for_CO2 = -94.052;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_C2H6 = -20.236;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_H2O = -57.7979;\t\t\t#[kcal/mol]\n",
+      "\n",
+      "# Calculations\t\t\n",
+      "        \t# The reaction with stoichiometric amount of air is\n",
+      "\t\t\t# C2H6 + (7/2)O2 - 2CO2 + 3H2O\n",
+      "\n",
+      "\t\t\t# The reaction with 4 mol of O2 and 10 mol CO2 is\n",
+      "\t\t\t# C2H6 + 4O2 + 10CO2 - 12H2O + 3H2O + 0.5O2\n",
+      "\t\t\t# The product consists of 12 mol of CO2, 3 mol of water vapour and 0.5 mol of oxygen\n",
+      "delta_H_rkn_298 = 2*delta_H_for_CO2 + 3*delta_H_for_H2O - delta_H_for_C2H6;\t\t\t#[kcal]\n",
+      "delta_H_rkn_298 = delta_H_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t# For exit stream\n",
+      "sum_ai_ni = 12*a_CO2 + 3*a_H2O + 0.5*a_O2;\n",
+      "sum_bi_ni = 12*b_CO2 + 3*b_H2O + 0.5*b_O2;\n",
+      "sum_ci_ni = 12*c_CO2 + 3*c_H2O + 0.5*c_O2;\n",
+      "sum_di_ni = 12*d_CO2 + 3*d_H2O + 0.5*d_O2;\n",
+      "\n",
+      "\t\t\t# From energy balance equation we get\n",
+      "\t\t\t# delta_H_rkn_298 + sum_ai_ni*(T_2 - T_1) + (sum_bi_ni/2)*(T_2**(2) - T_1**(2)) + (sum_ci_ni/3)*(T_2**(3) - T_1**(3)) + (sum_di_ni/4)*(T_2**(4) - T_1**(4))\n",
+      "\t\t\t# Solving above equation for T_2\n",
+      "def f(T_2): \n",
+      "\t return delta_H_rkn_298 +sum_ai_ni*(T_2-T_1)+(sum_bi_ni/2)*(T_2**(2)-T_1**(2))+(sum_ci_ni/3)*(T_2**(3)-T_1**(3))+(sum_di_ni/4)*(T_2**(4)-T_1**(4))\n",
+      "T_2 = fsolve(f,-1)\n",
+      "\n",
+      "# Results\n",
+      "print \" The adiabatic flame temperature is %f K\"%(T_2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The adiabatic flame temperature is 2090.385277 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.3  Page Number : 651"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K] - Standard reaction temperature\n",
+      "\n",
+      "\t\t\t# The reaction with theoritical air is\n",
+      "\t\t\t# CH4 + 2(O2 + 3.76N2) - CO2 + 2H20 + 7.52N2\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "n_product = (1 + 2 + 7.52);\t\t\t# Total number of moles of product\n",
+      "\t\t\t# The mole fraction of water vapour is\n",
+      "y_H2O = 2/(n_product);\n",
+      "print \" 1).The mole fraction of water vapour is %f\"%(y_H2O);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "delta_H_rkn_298 = -730*10**(3);\t\t\t#[J/mol]\n",
+      "C = 40;\t\t\t#[J/mol-K] - Average molar heat capacity\n",
+      "\n",
+      "\t\t\t# From energy balance we have\n",
+      "\t\t\t# delta_H_rkn_298 + n_product*C(T_2 - T_1) = 0\n",
+      "T_2 = - delta_H_rkn_298/(n_product*C) + T_1;\t\t\t#[K]\n",
+      "T_max = T_2 - T_1;\n",
+      "\n",
+      "print \" 2).The maximum temperature rise of the exhaust gases is %f K\"%(T_max);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The mole fraction of water vapour is 0.190114\n",
+        " 2).The maximum temperature rise of the exhaust gases is 1734.790875 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.4  Page Number : 651"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K] - Standiansard reaction temperature\n",
+      "\n",
+      "\t\t\t# The standard enthalpy of formation at 298.15 K is given by\n",
+      "delta_H_for_CO2 = -94.052;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_C8H18 = -59.780;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_H2O = -57.7979;\t\t\t#[kcal/mol]\n",
+      "\n",
+      "a_CO2 = 5.316;\n",
+      "a_H2O = 7.700;\n",
+      "a_N2 = 6.903;\n",
+      "b_CO2 = 1.4285*10**(-2);\n",
+      "b_H2O = 0.04595*10**(-2);\n",
+      "b_N2 = -0.03753*10**(-2);\n",
+      "c_CO2 = -0.8362*10**(-5);\n",
+      "c_H2O = 0.2521*10**(-5);\n",
+      "c_N2 = 0.1930*10**(-5);\n",
+      "d_CO2 = 1.784*10**(-9);\n",
+      "d_H2O = -0.8587*10**(-9);\n",
+      "d_N2 = -0.6861*10**(-9);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(a)\n",
+      "\t\t\t# The reaction with stoichiometric amount of air is\n",
+      "\t\t\t# C3H18 + 12.5(O2 + 3.76N2) - 8CO2 + 9H2O + 47N2\n",
+      "\n",
+      "\t\t\t# The standard enthalpy change of reaction at 298.15 K is\n",
+      "delta_H_rkn_298 = 8*delta_H_for_CO2 + 9*delta_H_for_H2O - delta_H_for_C8H18;\t\t\t#[kcal]\n",
+      "delta_H_rkn_298 = delta_H_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t# For exit stream\n",
+      "sum_ai_ni = 8*a_CO2 + 9*a_H2O + 47*a_N2;\n",
+      "sum_bi_ni = 8*b_CO2 + 9*b_H2O + 47*b_N2;\n",
+      "sum_ci_ni = 8*c_CO2 + 9*c_H2O + 47*c_N2;\n",
+      "sum_di_ni = 8*d_CO2 + 9*d_H2O + 47*d_N2;\n",
+      "\n",
+      "\t\t\t# From energy balance equation we get\n",
+      "\t\t\t# delta_H_rkn_298 + sum_ai_ni*(T_2 - T_1) + (sum_bi_ni/2)*(T_2**(2) - T_1**(2)) + (sum_ci_ni/3)*(T_2**(3) - T_1**(3)) + (sum_di_ni/4)*(T_2**(4) - T_1**(4))\n",
+      "\t\t\t# Solving above equation for T_2\n",
+      "def f(T_2): \n",
+      "\t return delta_H_rkn_298 +sum_ai_ni*(T_2-T_1)+(sum_bi_ni/2)*(T_2**(2)-T_1**(2))+(sum_ci_ni/3)*(T_2**(3)-T_1**(3))+(sum_di_ni/4)*(T_2**(4)-T_1**(4))\n",
+      "T_2 = fsolve(f,-1)\n",
+      "\n",
+      "print \" 1).The adiabatic flame temperature is %f K\"%(T_2);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# The mean standard heat capacity of various components over the temperature range from 25 to 3000 C is \n",
+      "Cp_CO2 = 13.91;\t\t\t#[cal/mol-K]\n",
+      "Cp_H2O = 10.16;\t\t\t#[cal/mol-K]\n",
+      "Cp_O2 = 7.88;\t\t\t#[cal/mol-K]\n",
+      "Cp_N2 = 7.45;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "\t\t\t# From energy balance equation we get\n",
+      "\t\t\t# delta_H_rkn_298 + (8*Cp_CO2 + 9*Cp_H2O + 47*Cp_N2)*(T_2_prime)\n",
+      "T_2_prime = - delta_H_rkn_298/(8*Cp_CO2 + 9*Cp_H2O + 47*Cp_N2);\t\t\t#[K]\n",
+      "print \" 2).The adiabatic flame temperature is %f K\"%(T_2_prime);\n",
+      "\n",
+      "# The answer is correct. Please calculate it manually. Wrong answer is given in book."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The adiabatic flame temperature is 2415.861178 K\n",
+        " 2).The adiabatic flame temperature is 2193.675005 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.5  Page Number : 652"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "\t\t\t# N2 + 3H2 - 2NH3\n",
+      "T_1 = 700;\t\t\t#[K] - Reaction temperature\n",
+      "Max_adia_rise = 100;\t\t\t#/[K] - Maximum adiabatic rise in temperature\n",
+      "T_2 = T_1 + Max_adia_rise;\t\t\t#[K] - \n",
+      "\n",
+      "delta_H_rkn_700 = -94.2;\t\t\t#[kJ] - Smath.tan(math.radiansard enthalpy of reaction at 700 K\n",
+      "delta_H_rkn_700 = delta_H_rkn_700*10**(3);\t\t\t#[J]\n",
+      "\n",
+      "\t\t\t# The mean standard heat capacity of various components over the temperature range from 700 to 800 K is \n",
+      "Cp_N2 = 30.0;\t\t\t#[cal/mol-K]\n",
+      "Cp_H2 = 28.9;\t\t\t#[cal/mol-K]\n",
+      "Cp_NH3 = 49.2;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "# Calculations\t\t\t\n",
+      "            # The energy balance equation is\n",
+      "\n",
+      "def f46(T): \n",
+      "\t return (sum_ni_Cpi_exit)*dT\n",
+      "\n",
+      "\t\t\t# X*delta_H_rkn_700 +  quad(f46,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t#At exit, let moles of NH3 = (1-X),  moles of H2 = (3-3X),  moles of NH3 = 2X . Therefore we have,\n",
+      "\t\t\t# delta_H_rkn_700*X + {(1-X)*Cp_N2 + (3-3X)*Cp_H2 + (2X)*Cp_NH3}*(T_2 - T_1)\n",
+      "\t\t\t# On simplification we get, 960.3*X = 116.7\n",
+      "X = 116.7/960.3;\n",
+      "\n",
+      "# Results\n",
+      "print \" The maximum allowable conversion fraction in the reactor is given by X =  %f \"%(X);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The maximum allowable conversion fraction in the reactor is given by X =  0.121525 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.6  Page Number : 653"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K] - standard reaction temperature\n",
+      "V = 2.0*10**(-3);\t\t\t#[m**(3)] - Volume of calorimeter\n",
+      "m = 10.;\t\t\t#[g] - Mass of liquid octane\n",
+      "Mol_wt = 114.;\t\t\t#[g/mol] - Molecular weight of octane\n",
+      "n = m/Mol_wt;\t\t\t#[mol] - No of moles of octane\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# The standard enthalpy of formation at 298.15 K is given by\n",
+      "delta_H_for_CO2 = -94.052;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_C8H18 = -59.780;\t\t\t#[kcal/mol]\n",
+      "delta_H_for_H2O = -57.7979;\t\t\t#[kcal/mol]\n",
+      "\n",
+      "\t\t\t# The standard molar heat capacity of various components in high temperature range from is given by \n",
+      "\t\t\t# Cp_H2O = 6.970 + 0.3464*10**(-2)*T - 0.04833*10**(-5)*T**(2);\n",
+      "\t\t\t# Cp_O2 = 6.732 + 0.1505*10**(-2)*T - 0.01791*10**(-5)*T**(2);\n",
+      "\t\t\t# Cp_CO2 = 18.036 - 4.474*10**(-5)*T - 158.08/(T**(1/2));\n",
+      "\t\t\t# Therefore we have\n",
+      "\t\t\t# Sum_ni_Cpi_exit = 249.09 + 0.04*T - 0.547*10**(-5)*T**(2) - 1264.64/(T**(1/2))\n",
+      "\n",
+      "\t\t\t# The reaction with stoichiometric amount of oxygen is\n",
+      "\t\t\t# C8H18 + 12.5O2 - 8CO2 + 9H2O\n",
+      "\n",
+      "\t\t\t# The reaction with 50% excess oxygen is\n",
+      "\t\t\t# C8H18 + 18.75O2 - 8CO2 +9H2O + 6.25O2\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# The standard enthalpy change of reaction at 298.15 K is\n",
+      "delta_H_rkn_298 = 8*delta_H_for_CO2 + 9*delta_H_for_H2O - delta_H_for_C8H18;\t\t\t#[kcal]\n",
+      "delta_H_rkn_298 = delta_H_rkn_298*10**(3);\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t# From the energy balance equation we get\n",
+      "\n",
+      "def f10(T): \n",
+      "\t return (sum_ni_Cpi_exit)*dT\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_298 +  quad(f10,T_1,T_2)[0]\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_298 + 249.09*(T_2 - T_1) + (0.04/2)*(T_2**(2) - T_1**(2)) - ((0.547*10**(-5))/3)*(T_2**(3)-T_1**(3)) - (1264.64*2)*(T_2**(1/2)-T_1**(1/2))\n",
+      "\t\t\t# Solving above equation for T_2\n",
+      "def f(T_2): \n",
+      "\t return delta_H_rkn_298 + 249.09*(T_2 - T_1) + (0.04/2)*(T_2**(2)-T_1**(2)) - ((0.547*10**(-5))/3)*(T_2**(3)-T_1**(3)) - (1264.64*2)*(T_2**(1./2)-T_1**(1./2))\n",
+      "T_2 = fsolve(f,1000)\n",
+      "\n",
+      "\t\t\t# When 1 mol of octane reacts the final number of moles in the calorimeter is 23.25\n",
+      "\t\t\t# When n mol of octane reacts the final number of moles in the calorimeter is\n",
+      "n_total = n*23.25;\t\t\t#[mol]\n",
+      "\n",
+      "\t\t\n",
+      "\t\t\t# Thus maximum explosion pressure is given by\n",
+      "P = (n_total*R*T_2)/V;\t\t\t#[N/m**(2)]\n",
+      "P = P*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "# Results\n",
+      "print \" The maximum explosion pressure inside the bomb calorimeter is %f bar\"%(P);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The maximum explosion pressure inside the bomb calorimeter is 397.138166 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.7  Page Number : 656"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve\n",
+      "import math\n",
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 400 + 273.15;\t\t\t#[K]\n",
+      "\t\t\t# SO2(g) + 1./2*(O2) (g) - SO3 (g)\n",
+      "\n",
+      "a_SO2 = 6.157;\n",
+      "a_SO3 = 3.918;\n",
+      "a_O2 = 6.085;\n",
+      "a_N2 = 6.903;\n",
+      "b_SO2 = 1.384*10**(-2);\n",
+      "b_SO3 = 3.483*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "b_N2 = -0.03753*10**(-2);\n",
+      "c_SO2 = -0.9103*10**(-5);\n",
+      "c_SO3 = -2.675*10**(-5);\n",
+      "c_O2 = -0.01709*10**(-5);\n",
+      "c_N2 = 0.1930;\n",
+      "d_SO2 = 2.057*10**(-9);\n",
+      "d_SO3 = 7.744*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "d_N2 = -0.6861*10**(-9);\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# At 400 C, from the given math.expressions\n",
+      "delta_H_rkn_T_1 = -22630.14 - 5.2815*T_1 + 0.9587*10**(-2)*T_1**(2) - 0.5598*10**(-5)*T_1**(3) + 1.3826*10**(-9)*T_1**(4);\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_T_X (for X moles of SO2 reacted) = delta_H_rkn_T_1*X\n",
+      "\n",
+      "\t\t\t# Let the number of moles at equilibrium be\n",
+      "\t\t\t# n_O2 = 9-0.5*X\n",
+      "\t\t\t# n_SO2 =  12-X\n",
+      "\t\t\t# n_SO3 = X\n",
+      "\t\t\t# n_N2 = 79\n",
+      "\t\t\t# Total moles at equilibrium = 100-0.5X\n",
+      "\t\t\t# Ky = y_SO3/(y_SO2*y_O2**(1./2))\n",
+      "\t\t\t# Ky = (X*(100-0.5*X)**(1./2))/((12-X)*(9-0.5*X)**(1./2))\n",
+      "\t\t\t# We know that K = Ky*Kp. Since P = 1 atm, therefore Ky = K\n",
+      "\n",
+      "\t\t\t# Now we have to account for the heat required to raise 9-0.5*X mol of O2, 12-X mol of SO2, X mol of SO3 and 79 mol of N2 from T to ART\n",
+      "\t\t\t# sum_ni_Cp_i = (12-X)*(a + b*T + c*T**(2) + d*T**(3)) + (9-0.5*X)*(a + b*T + c*T**(2) + d*T**(3)) + X*(a + b*T + c*T**(2) + d*T**(3)) + 79*(a + b*T + c*T**   (2) + d*T**(3))\n",
+      "\n",
+      "\t\t\t# From energy balance equation we get\n",
+      "\n",
+      "def f19(T): \n",
+      "\t return sum_ni_Cp_i\n",
+      "\n",
+      "\t\t\t# delta_H_rkn_T_1 +  quad(f19,T_1,T)[0]\n",
+      "\n",
+      "\t\t\t# The above equation on simplification becomes\n",
+      "\t\t\t# (673.99-5.2815*X)*(T-T_1) + (16.91+1.9175*X)*(10**(-2)/2)*(T**(2)-T_1**(2)) + (2.79-1.6793*X)*(10**(-5)/3)*(T**(3)-T_1**(3)) + (-26.70+5.5304*X)*(10**(-9)   /4)*(T**(4)-T_1**(4)) = delta_H_rkn_T_1*X\n",
+      "\n",
+      "\t\t\t# Let us assume a temperature, say\n",
+      "T = 800;\t\t\t#[K]\n",
+      "fault = 10;\n",
+      "def f(X): \n",
+      "\t return  K - (X*(100-0.5*X)**(1./2))/((12-X)*(9-0.5*X)**(1./2))\n",
+      "\n",
+      "def f1(X): \n",
+      "\t return  (673.99-5.2815*X)*(T-T_1)+(16.91+1.9175*X)*(10**(-2)/2)*(T**(2)-T_1**(2))+(2.79-1.6793*X)*(10**(-5)/3)*(T**(3)-T_1**(3))+(-26.70+5.5304*X)*(10**(-9)/4)*(T**(4)-T_1**(4))+delta_H_rkn_T_1*X\n",
+      "\n",
+      "while(fault>0.01):\n",
+      "    K = math.exp(3.87 + (11389.10/T) - 2.6580*math.log(T) + 0.4825*10**(-2)*T - 0.1409*10**(-5)*T**(2) + 0.2320*10**(-9)*T**(3));\n",
+      "    X1 = fsolve(f,0.1)\n",
+      "    X2 = fsolve(f1,1)\n",
+      "    fault = abs(X1-X2);\n",
+      "    T = T + 0.01;\n",
+      "\n",
+      "# Results\n",
+      "print \" The moles of SO2 reacted are %f mol\"%(X1);\n",
+      "print \" The adiabatic reaction temperature is %f K\"%(T);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The moles of SO2 reacted are 7.594592 mol\n",
+        " The adiabatic reaction temperature is 886.860000 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch1_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch1_2.ipynb
new file mode 100755
index 00000000..13ddc7dd
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch1_2.ipynb
@@ -0,0 +1,770 @@
+{
+ "metadata": {
+  "name": "ch1_2",
+  "signature": "sha256:05ef5db2afef7232054abda0278780d1df8c92b807546fcd7648e8885fa4539f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 :  Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.1  Page number - 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# \n",
+      "# Variables\n",
+      "#(a)\n",
+      " \n",
+      "m = 50.;\t\t\t#[kg] - Mass of piston\n",
+      "A = 0.05;\t\t\t#[m**(2)] - Area of piston\n",
+      "g = 9.81;\t\t\t#[m/s**(2)] - Acceleration due to gravity\n",
+      "Po = 101325;\t\t\t#[N/m**(2)] - Atmospheric pressure\n",
+      "\n",
+      "# Calculations and Results\n",
+      "P = (m*g/A)+Po;\t\t\t#[N/m**(2)]\n",
+      "P = P/100000.;\t\t\t#[bar]\n",
+      "print \" (a).Pressure = %f bar\"%P\n",
+      "\n",
+      "#(b)\n",
+      "print \" (b).Since the piston weight and surroundings pressure are the same, \\\n",
+      "the gas pressure in the piston-cylinder assembly remains %f bar\"%P\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (a).Pressure = 1.111350 bar\n",
+        " (b).Since the piston weight and surroundings pressure are the same, the gas pressure in the piston-cylinder assembly remains 1.111350 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.2 Page number - 8\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "P = 1.;\t\t\t#[atm] - Atmospheric pressure\n",
+      "P = 101325.;\t\t\t#[N/m**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "T = 30.;\t\t\t#[C] - Temperature of air\n",
+      "T = 30.+273.15;\t\t\t#[K]\n",
+      "V = 5.*5*5;\t\t\t#[m**(3)] - Volume of the room\n",
+      "\n",
+      "# Calculations\n",
+      "#The number of moles of air is given by\n",
+      "n = (P*V)/(R*T)\t\t\t#[mol]\n",
+      "\n",
+      "#Molecular weight of air(21 vol% O2 and 79 vol% N2)=(0.21*32)+(0.79*28)=  28.84 g/mol\n",
+      "m = n*28.84;\t\t\t#[g]\n",
+      "m = m/1000.;\t\t\t#[kg]\n",
+      "\n",
+      "# Results\n",
+      "print \"The mass of air is, m = %f kg\"%m\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of air is, m = 144.928664 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.3 Page number - 13\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "P1 = 3.;\t\t\t# [bar]  - initial pressure\n",
+      "V1 = 0.5;\t\t\t# [m**(3)] - initial volume\n",
+      "V2 = 1.0;\t\t\t# [m**(3)] - final volume\n",
+      "import math\n",
+      "#(a)\n",
+      "n = 1.5;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Let P*V**(n)=C \t\t\t# Variables relation\n",
+      "#W (work done per mole)= (integrate('P'%'V'%V1%V2))\n",
+      "#and thus    W=((P2*V2)-(P1*V1))/(1-n)\n",
+      "#The above math.expression is valid for all values of n%except n=1.0\n",
+      "P2 = (P1*((V1/V2)**(n)))\t\t\t#[bar] \t\t\t#pressure at state 2\n",
+      "\n",
+      "#we have%(V1/V2)=(V1t/(V2t)%since the number  of moles are constant.Thus\n",
+      "W = ((P2*V2)-(P1*V1))/(1-n)*10**(5)\t\t\t#[J]\n",
+      "W = W/1000.;\t\t\t#[kJ]\n",
+      "print \" (a).The work done (for n=1.5) is %f kJ\"%W\n",
+      "\n",
+      "#(b)\n",
+      "#For n=1.0%we have% PV=C.\n",
+      "# w(wok done per mol)= (integrate('P'%'V'%V1%V2)) = (integrate('C/V'%'V'%V1%V2)) = C*ln(V2/V1)=P1*V1*ln(V2/V1)\n",
+      "W1 = P1*V1*math.log(V2/V1)*10**(5)\t\t\t#[J]\n",
+      "W1 = W1/1000.;\t\t\t#[kJ]\n",
+      "print \" (b).The work done (for n=1.0) is %f kJ\"%W1\n",
+      "\n",
+      "#(c)\n",
+      "#For n=0%we get P=Constant and thus\n",
+      "P = P1;\t\t\t#[bar]\n",
+      "# w =(integrate('P'%'V'%V1%V2)) = P*(V2-V1)\n",
+      "W2 = P*(V2-V1)*10**(5)\t\t\t#[J]\n",
+      "W2 = W2/1000.;\t\t\t#[kJ]\n",
+      "print \" (c).The work done (for n=0) is %f kJ\"%W2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (a).The work done (for n=1.5) is 87.867966 kJ\n",
+        " (b).The work done (for n=1.0) is 103.972077 kJ\n",
+        " (c).The work done (for n=0) is 150.000000 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.4 Page number - 17\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "#(a)\n",
+      "# Variables\n",
+      "V = 9.;\t\t\t# [m/s] - velocity\n",
+      "d = 1.;\t\t\t#[m] - diameter\n",
+      "A = 3.14*(d/2)**(2)\t\t\t#[m**(2)] - area\n",
+      "P = 1.;\t\t\t#[atm] - pressure\n",
+      "P = 101325.;\t\t\t# [N/m**(2)]\n",
+      "T = 300.;\t\t\t#[K] - Temperature\n",
+      "R =  8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "E = (V**(2))/2.;\t\t\t#[J/kg]\n",
+      "print \" (a).The wind energy per unit mass of air is %f J/kg\"%E\n",
+      "\n",
+      "#(b)\n",
+      "# Molecular weight of air(21 vol% O2 and 79 vol% N2)=(0.21*32)+(0.79*28)=  28.84 g/mol\n",
+      "M = 28.84*10**(-3)\t\t\t#[kg/mol]\n",
+      "r = (P*M)/(R*T)\t\t\t#[kg/m**(3)] - density\n",
+      "m = r*V*A;\t\t\t# [kg/s] - mass flow rate of air\n",
+      "pi = m*E;\t\t\t#[Watt] - power input\n",
+      "print \" (b).The wind power input to the turbine is %f Watt\"%pi\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (a).The wind energy per unit mass of air is 40.500000 J/kg\n",
+        " (b).The wind power input to the turbine is 335.233787 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.5 Page number - 23\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Given\n",
+      "P = 1.;\t\t\t        # [bar] - atospheric pressure\n",
+      "P1guz = 0.75;\t\t\t# [bar] - gauze pressure in 1st evaporator\n",
+      "P2Vguz = 0.25;\t\t\t# [bar] - vaccum gauze pressure in 2nd evaporator\n",
+      "\n",
+      "# Calculations\n",
+      "P1abs = P + P1guz;\t\t\t# [bar] - absolute pressure in 1st evaporator\n",
+      "P2abs = P - P2Vguz;\t\t\t# [bar] -absolute pressure in 2nd evaporator\n",
+      "\n",
+      "# Results\n",
+      "#From saturated steam table as reported in the book\n",
+      "print \" For P1abs (absolute pressure) = %f bar\"%P1abs\n",
+      "print \" The saturation temperature in first evaporator  is 116.04 C\"\n",
+      "print \" For P2abs (absolute pressure) = %f bar\"%P2abs\n",
+      "print \" The saturation temperature in second evaporator  is 91.76 C\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " For P1abs (absolute pressure) = 1.750000 bar\n",
+        " The saturation temperature in first evaporator  is 116.04 C\n",
+        " For P2abs (absolute pressure) = 0.750000 bar\n",
+        " The saturation temperature in second evaporator  is 91.76 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.6 Page number - 23\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "V = 1.;\t\t\t    # [kg] - volume of tank\n",
+      "P = 10.;\t\t\t# [bar] - pressure\n",
+      "\n",
+      "#Here degree of freedom =1(C=1%P=2%threfore F=1)\n",
+      "#From steam table at 10 bar as reported in the book\n",
+      "V_liq = 0.001127;\t\t\t# [m**(3)/kg] - volume in liquid phase\n",
+      "V_vap = 0.19444;\t\t\t# [m**(3)/kg] - volume in vapour phase\n",
+      "\n",
+      "# Calculations\n",
+      "#x*Vv=(1-x)*Vl \t\t\t# since two volumes are equal\n",
+      "x = (V_liq/(V_liq+V_vap))\t\t\t# [kg]\n",
+      "y = (1-x)\t\t\t#[kg]\n",
+      "\n",
+      "# Results\n",
+      "print \" Mass of saturated vapour is %f kg\"%x\n",
+      "print \" Mass of saturated liquid is %f kg\"%y\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Mass of saturated vapour is 0.005763 kg\n",
+        " Mass of saturated liquid is 0.994237 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.7 Page number - 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "V = 1.;\t\t\t# [m**(3)] - volume of tank\n",
+      "M = 10.;\t\t\t# [m**(3)] - total mass\n",
+      "T = (90+273.15)\t\t\t#[K] - temperature\n",
+      "\n",
+      "#From steam table at 90 C as reported in the book\n",
+      "#vapour pressure(pressure of rigid tank) = 70.14[kPa] = 0.7014[bar]\n",
+      "print \" Pressure of tank = 0.7014 bar\"\n",
+      "\n",
+      "# Calculations and Results\n",
+      "V_liq_sat=0.001036;\t\t\t# [m**(3)/kg] - saturated liquid specific volume\n",
+      "V_vap_sat=2.36056;\t\t\t# [m**(3)/kg] - saturated vapour specific volume\n",
+      "\n",
+      "#1=(V_liq_sat*(10-x))+(V_vap_sat*x)\n",
+      "x = (1-(10*V_liq_sat))/(V_vap_sat-V_liq_sat)\t\t\t#[kg]\n",
+      "y = (10-x)\t\t\t#[kg]\n",
+      "\n",
+      "print \" The amount of saturated liquid is %f kg\"%y\n",
+      "print \" The amount of saturated vapour is %f kg \"%x\n",
+      "\n",
+      "z = y*V_liq_sat;\t\t\t#[m**(3)] - Volume of saturated liquid \n",
+      "w = x*V_vap_sat;\t\t\t#[m**(3)] - Volume of saturated vapour\n",
+      "\n",
+      "print \" Total volume of saturated liquid is %f m**(3)\"%z\n",
+      "print \" Total volume of saturated vapour is %f m**(3)\"%w\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Pressure of tank = 0.7014 bar\n",
+        " The amount of saturated liquid is 9.580576 kg\n",
+        " The amount of saturated vapour is 0.419424 kg \n",
+        " Total volume of saturated liquid is 0.009925 m**(3)\n",
+        " Total volume of saturated vapour is 0.990075 m**(3)\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.8 Page number - 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "V = 10.;\t\t\t# [m**(3)] - volume of vessel\n",
+      "P_1 = 1.;\t\t\t# [bar] - initial pressure\n",
+      "V_liq_sat = 0.05;\t\t\t# [m**(3)] - saturated liquid volume\n",
+      "V_gas_sat = 9.95;\t\t\t# [m**(3)] - saturated vapour volume\n",
+      "\n",
+      "#At 1 bar pressure\n",
+      "V_liq_1 = 0.001043;\t\t\t# [m**(3/kg)] - specific saturated liquid volume\n",
+      "U_liq_1 = 417.33;\t\t\t# [kJ/kg] - specific internal energy\n",
+      "V_gas_1 = 1.69400;\t\t\t# [m**(3/kg)] - specific saturated vapour volume\n",
+      "U_gas_1 = 2506.06;\t\t\t# [kJ/kg]\n",
+      "\n",
+      "# Calculations\n",
+      "M_liq_1 = V_liq_sat/V_liq_1;\t\t\t# [kg] - mass of saturated liqid\n",
+      "M_gas_1 = V_gas_sat/V_gas_1;\t\t\t# [kg] - mass of saturated vapour\n",
+      "M = (M_liq_1+M_gas_1)\t\t\t# [kg] - total mass\n",
+      "U_1t = (M_liq_1*U_liq_1)+(M_gas_1*U_gas_1)\t\t\t# [kJ] - initial internal energy\n",
+      "V_gas_2 = (V/M)\t\t\t#[m**(3/kg)]\n",
+      "\n",
+      "#from steam table at 10 bar pressure as reported in the book\n",
+      "V_vap_2 = 0.19444;\t\t\t# [m**(3/kg)]\n",
+      "U_vap_2 = 2583.64;\t\t\t# [kJ/kg]\n",
+      "\n",
+      "#from steam table at 11 bar pressure as reported in the book\n",
+      "V_vap_3 = 0.17753;\t\t\t#[m**(3/kg)]\n",
+      "U_vap_3 = 2586.40;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "#Now computing pressure when molar volume of saturated vapour=Vg_2\n",
+      "#By interpolation (P2-10)/(11-10)=(Vg_2-Vv_2)/(Vv_3-Vv_2)\n",
+      "P_2 = (((V_gas_2 - V_vap_2)/(V_vap_3 - V_vap_2)*1)+10)\t\t\t# [bar] - final pressure\n",
+      "\n",
+      "#By interpolation calculating internal energy at state 2\n",
+      "#(P2-10)/(11-10)=(U2-Uv_2)/(Uv_3-Uv_2)\n",
+      "U_2 = (((P_2-10)/(11-10))*(U_vap_3 - U_vap_2))+U_vap_2;\t\t\t#[kJ/kg]\n",
+      "U_2t = U_2*M;\t\t\t#[kJ]\n",
+      "H = U_2t - U_1t;\t\t\t#[kJ] - Heat supplied\n",
+      "H = H/1000;\t\t\t#[MJ]\n",
+      "\n",
+      "# Results\n",
+      "print \" Total heat supplied is %f MJ\"%H\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Total heat supplied is 104.381244 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.9 Page number - 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "#Antoine equation for water   ln(Psat)=16.262-(3799.89/(T_sat + 226.35))\n",
+      "P = 2.;\t\t\t#[atm] - Pressure\n",
+      "P = (2.*101325)/1000;\t\t\t#[kPa]\n",
+      "\n",
+      "# Calculations\n",
+      "P_sat = P;\t\t\t# Saturation pressure\n",
+      "T_sat = (3799.89/(16.262-math.log(P_sat)))-226.35;\t\t\t#[C] - Saturation temperature\n",
+      "#Thus boiling at 2 atm occurs at Tsat = 120.66 C.\n",
+      "\n",
+      "#From steam tables%at 2 bar%Tsat = 120.23 C and at 2.25 bar%Tsat = 124.0 C\n",
+      "#From interpolation for T_sat = 120.66 C%P = 2.0265 bar\n",
+      "#For P_= 2.0265 bar%T_sat% from steam table by interpolation is given by\n",
+      "#((2.0265-2)/(2.25-2))=((Tsat-120.23)/(124.0-120.23))\n",
+      "T_sat_0 = (((2.0265-2)/(2.25-2))*(124.0-120.23))+120.23;\t\t\t#[C]\n",
+      "\n",
+      "# Results\n",
+      "print \" Saturation temperature (Tsat) = %f C which is close \\\n",
+      "to %f C as determined from Antoine equation\"%(T_sat_0,T_sat)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Saturation temperature (Tsat) = 120.629620 C which is close to 120.655450 C as determined from Antoine equation\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.10 Page number - 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "# math.log(P)=-(1640/T)+10.56 (solid)\n",
+      "# math.log(P)=-(1159/T)+7.769 (liquid)%where T is in K\n",
+      "# F+P=C+2% at triple point F+3=1+2 or%F=0 i.e%vapour pressure of liquid and solid at triple point are same%we get\n",
+      "# -(1640/T)+10.56 = -(1159/T)+7.769\n",
+      "\n",
+      "# Calculations\n",
+      "T = (1640-1159)/(10.56-7.769)\t\t\t#[K]\n",
+      "P = 10**((-1640/T)+10.56)\t\t\t#[torr]\n",
+      "\n",
+      "# Results\n",
+      "print \" The temperature is %f K\"%T\n",
+      "print \" The pressure is %f torr (or mm Hg)\"%P\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The temperature is 172.339663 K\n",
+        " The pressure is 11.063907 torr (or mm Hg)\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.11 Page number - 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "M_O2 = 31.999;\t\t\t#molecular weight of oxygen\n",
+      "M_N2 = 28.014;\t\t\t#molecular weight of nitrogen\n",
+      "Y = 1.4;\t\t\t#molar heat capacities ratio for air\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Molecular weight of air(21 vol% O2 and 79 vol% N2)is given by\n",
+      "M_air = (0.21*M_O2)+(0.79*M_N2)\t\t\t#(vol% = mol%)\n",
+      "\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "R = (R*1/M_air)\t\t\t#[kJ/kg*K]\n",
+      "\n",
+      "print \" The value of universal gas constant (R) = %f kJ/kg-K \"%R\n",
+      "\n",
+      "#Y=Cp0/Cv0 and Cp0-Cv0=R\n",
+      "Cv_0 = R/(Y-1)\t\t\t#[kJ/kg*K] \n",
+      "Cp_0 = Y*Cv_0;\t\t\t#[kJ/kg*K]\n",
+      "print \" The value of Cp_0 for air is %f kJ/kg-K\"%Cp_0\n",
+      "print \" The value of Cv_0 for air is %f kJ/kg-K\"%Cv_0\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of universal gas constant (R) = 0.288172 kJ/kg-K \n",
+        " The value of Cp_0 for air is 1.008601 kJ/kg-K\n",
+        " The value of Cv_0 for air is 0.720429 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.12 Page number - 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "Y = 1.4;\t\t\t#molar heat capacities ratio for air\n",
+      "R = 8.314;\t\t\t# [J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "Cv_0 = R/(Y-1)\t\t\t# [J/mol*K]\n",
+      "Cp_0 = Y*Cv_0;\t\t\t# [J/mol*K]\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar heat capacity at constant volume (Cv_0) is %f J/mol-K\"%Cv_0\n",
+      "print \" The molar heat capacity at constant pressure (Cp_0) is %f J/mol-K\"%Cp_0\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar heat capacity at constant volume (Cv_0) is 20.785000 J/mol-K\n",
+        " The molar heat capacity at constant pressure (Cp_0) is 29.099000 J/mol-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.13 Page number - 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.integrate import quad\n",
+      "\n",
+      "# Variables\n",
+      "# Cp0=7.7+(0.04594*10**(-2)*T)+(0.2521*10**(-5)*T**(2))-(0.8587*10**(-9)*T**(3))\n",
+      "T_1 = 400.;\t\t\t#[K]\n",
+      "T_2 = 500.;\t\t\t#[K]\n",
+      "\n",
+      "# Calculations\n",
+      "def f(T):\n",
+      "    return 7.7+(0.04594*10**(-2)*T)+(0.2521*10**(-5)*T**(2))-(0.8587*10**(-9)*T**(3))\n",
+      "#(C)avg = q/(T_2 - T_1) = 1/(T_2 - T_1)*{(integrate('C'%'T'%T_1%T_2))}\n",
+      "#(Cp0)avg = 1/(T_2 - T_1)*{(integrate('Cp0'%'T'%T_1%T_2))}\n",
+      "Cp0_avg = (1/(T_2 - T_1))*quad(f,T_1,T_2)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \" The mean heat capacity (Cp0_avg) for temerature range of 400 to 500 K is %f cal/mol-K\"%Cp0_avg\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mean heat capacity (Cp0_avg) for temerature range of 400 to 500 K is 8.340118 cal/mol-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 1.14 Page number - 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "#(a)\n",
+      "P_1 = 0.2;\t\t\t# [MPa] - pressure\n",
+      "x_1 = 0.59;\t\t\t# mole fraction\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#From saturated steam tables at 0.2 MPa\n",
+      "H_liq_1 = 504.7;\t\t\t# [kJ/kg] - Enthalpy of saturated liquid\n",
+      "H_vap_1 = 2706.7;\t\t\t# [kJ/kg]- Enthalpy of saturated vapour\n",
+      "H_1 = (H_liq_1*(1-x_1))+(x_1*H_vap_1)\t\t\t# [kJ/kg]\n",
+      "print \" (a).Enthalpy of 1 kg of water in tank is %f kJ/kg\"%H_1\n",
+      "\n",
+      "#(b)\n",
+      "T_2 = 120.23;\t\t\t# [C] - temperature\n",
+      "V_2 = 0.6;\t\t\t# [m**(3)/kg] - specific volume\n",
+      "\n",
+      "#From saturated steam tables at 120.23 C% as reported in the book\n",
+      "V_liq_2=0.001061;\t\t\t# [m**(3)/kg]\n",
+      "V_vap_2=0.8857;\t\t\t# [m**(3)/kg]\n",
+      "#since V_2 < Vv_2%dryness factor will be given by% V = ((1-x)*V_liq)+(x*V_vap)\n",
+      "x_2 = (V_2- V_liq_2)/(V_vap_2 - V_liq_2)\n",
+      "\n",
+      "#From steam table%at 120.2 C%the vapour pressure of water is 0.2 MPa.So%enthalpy is given by\n",
+      "H_2 = (H_liq_1*(1-x_2))+(H_vap_1*x_2)\t\t\t#kJ/kg]\n",
+      "print \" (b).Enthalpy of saturated steam is %f kJ/kg\"%H_2\n",
+      "\n",
+      "#(c)\n",
+      "P_3 = 2.5;\t\t\t#[MPa]\n",
+      "T_3 = 350;\t\t\t#[C]\n",
+      "#From steam tables at 2.5 MPa%T_sat = 223.99 C%as reported in the book\n",
+      "#since%T_3 > Tsat% steam is superheated\n",
+      "print \" (c).As steam is superheated%from steam table%enthalpy (H) is 3126.3 kJ/kg\"\n",
+      "\n",
+      "#(d)\n",
+      "T_4 = 350;\t\t\t#[C]\n",
+      "V_4 = 0.13857;\t\t\t#[m**(3)/kg]\n",
+      "#From steam table%at 350 C% V_liq = 0.001740 m**(3)/kg and V_vap = 0.008813 m**(3)/kg.Since%V > V_vap%therefore it is superheated.\n",
+      "#From steam table at 350 C and 1.6 MPa% V = 0.17456 m**(3)/kg\n",
+      "#At 350 C and 2.0 MPa% V = 0.13857 m**(3)/kg. So%\n",
+      "print \" (d).The enthalpy of superheated steam (H) is 3137.0 kJ/kg\"\n",
+      "\n",
+      "#(e)\n",
+      "P_4 = 2.0;\t\t\t#[MPa]\n",
+      "U_4 = 2900;\t\t\t# [kJ/kg] - internal energy\n",
+      "#From saturated table at 2.0 MPa% U_liq = 906.44kJ and U_vap = 2600.3 kJ/kg\n",
+      "#scince%U_4 > Uv% it is saturated.\n",
+      "#From superheated steam table at 2.0 MPa and 350 C% as reported in the book\n",
+      "U_1 = 2859.8;\t\t\t#[kJ/kg]\n",
+      "H_1 = 3137.0;\t\t\t#[kJ/kg]\n",
+      "#At 2.0 MPa and 400 C%\n",
+      "U_2 = 2945.2;\t\t\t#[kJ/kg]\n",
+      "H_2 = 3247.6;\t\t\t#[kJ/kg]\n",
+      "T = (((U_4 - U_1)/(U_2 - U_1))*(400 - 350)) + 350;\t\t\t#[C] - By interpolation\n",
+      "H = (((T - 350)/(400 - 350))*(H_2 - H_1)) + H_1;\t\t\t#[kJ/kg]\n",
+      "print \" (e).The enthalpy value (of superheated steam) obtained after interpolation is %f kJ/kg\"%H\n",
+      "\n",
+      "#(f)\n",
+      "P_5 = 2.5;\t\t\t#[MPa]\n",
+      "T_5 = 100;\t\t\t#[C]\n",
+      "#At 100 C%P_sat=101350 N/m**(2). Since P_5 > P_sat%it is compressed liquid\n",
+      "P_sat = 0.101350;\t\t\t#[MPa]\n",
+      "H_liq = 419.04;\t\t\t#[kJ/kg] - At 100 C and 0.10135 MPa\n",
+      "V_liq = 0.001044;\t\t\t#[m**(3)/kg] - At 100 C and 0.10135 MPa\n",
+      "H_0 = H_liq + (V_liq*(P_5 - P_sat))*1000;\t\t\t#kJ/kg]\n",
+      "print \" (f).The enthalpy of compressed liquid is %f kJ/kg\"%H_0\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (a).Enthalpy of 1 kg of water in tank is 1803.880000 kJ/kg\n",
+        " (b).Enthalpy of saturated steam is 1995.549576 kJ/kg\n",
+        " (c).As steam is superheated%from steam table%enthalpy (H) is 3126.3 kJ/kg\n",
+        " (d).The enthalpy of superheated steam (H) is 3137.0 kJ/kg\n",
+        " (e).The enthalpy value (of superheated steam) obtained after interpolation is 3189.062295 kJ/kg\n",
+        " (f).The enthalpy of compressed liquid is 421.544191 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch2_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch2_2.ipynb
new file mode 100755
index 00000000..c009ac76
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch2_2.ipynb
@@ -0,0 +1,1276 @@
+{
+ "metadata": {
+  "name": "ch2_2",
+  "signature": "sha256:6aa23fb1475338cf467e7afb5732b8b9e456bbb3ad63c14224d73232d920457e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Chapter 2 : Equations of state"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.2 Page number - 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "Tc = 647.1;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 220.55;\t\t#[bar] - Critical pressure\n",
+      "Tr = 0.7;\t\t\t# Reduced temperature\n",
+      "\n",
+      "# Calculations\n",
+      "T = Tr*Tc;\t\t\t#[K]\n",
+      "\t\t\t#From steam table%vapour pressure of H2O at T is 10.02 [bar]% as reported in the book\n",
+      "P = 10.02;\t\t\t#[bar]\n",
+      "w = -1-math.log10((P/Pc))\n",
+      "\n",
+      "# Results\n",
+      "print \" The acentric factor (w) of water at given condition is %f \"%w\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The acentric factor (w) of water at given condition is 0.342639 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.3 Page number - 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "\t\t\t#math.log10(Psat)=8.1122-(1592.864/(t+226.184))\t\t\t# 'Psat' in [mm Hg] and 't' in [c]\n",
+      "Tc = 513.9;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 61.48;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5)\t\t\t#[N/m**(2)]\n",
+      "Tr = 0.7;\t\t\t# Reduced temperature\n",
+      "\n",
+      "# Calculations\n",
+      "T = Tr*Tc;\t\t\t#[K] - Temperature\n",
+      "T = T - 273.15;\t\t\t#[C]\n",
+      "P_sat = 10**(8.1122 - (1592.864/(T + 226.184)))\t\t\t#[mm Hg]\n",
+      "P_sat = (P_sat/760)*101325;\t\t\t#[N/m**(2)]\n",
+      "Pr_sat = P_sat/Pc;\n",
+      "w = -1-math.log10(Pr_sat)\t\t\t# Acentric factor\n",
+      "\n",
+      "# Results\n",
+      "print \" The acentric factor (w) for ethanol at given condition is %f\"%w\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The acentric factor (w) for ethanol at given condition is 0.644493\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.4 Page number - 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 380;\t\t\t#[K] - Temperature\n",
+      "Tc = 562.1;\t\t\t#[K] - Critical temperature\n",
+      "P = 7;\t\t\t#[atm] - Pressure\n",
+      "P = P*101325;\t\t\t#[N/m**(2)]\n",
+      "Pc = 48.3;\t\t\t#[atm] - Critical pressure\n",
+      "Pc = Pc*101325;\t\t\t#[N/m**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "w = 0.212;\t\t\t# acentric factor\n",
+      "Tr = T/Tc;\t\t\t# Reduced temperature\n",
+      "\n",
+      "# Calculations and Results\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6))\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2))\n",
+      "\n",
+      "\t\t\t#We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B = ((B_0+(w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "print \" The second virial coefficient for benzene is %e m**(3)/mol\"%B\n",
+      "\n",
+      "\t\t\t#Compressibility factor is given by\n",
+      "Z = 1 + ((B*P)/(R*T))\n",
+      "print \" The compressibility factor at 380 K is %f\"%Z\n",
+      "\n",
+      "\t\t\t#We know thar Z=(P*V)/(R/*T)%therfore\n",
+      "V = (Z*R*T)/P;\t\t\t#[m**(3)/mol]\n",
+      "print \" The molar volume is %e m**(3)/mol\"%V\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The second virial coefficient for benzene is -8.267963e-04 m**(3)/mol\n",
+        " The compressibility factor at 380 K is 0.814382\n",
+        " The molar volume is 3.627499e-03 m**(3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.5 Page number - 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "V_1 = 0.3;\t\t\t#[m**(3)]\t\t\t#volume of cylinder\n",
+      "T = 60+273.15;\t\t\t#[K] - Temperature\n",
+      "P = 130*10**(5)\t\t\t#[N/m**(2)] - Pressure\n",
+      "Tc = 305.3;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 48.72*10**(5)\t\t\t#[N/m**(2)] - Critical pressure\n",
+      "w = 0.100;\t\t\t#acentric factor\n",
+      "M = 30.07;\t\t\t#molecular weight of ethane\n",
+      "Tr = T/Tc;\t\t\t# Reduced temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6))\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2))\n",
+      "\n",
+      "\t\t\t#We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B = ((B_0 + (w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "Z = 1 + ((B*P)/(R*T))\t\t\t#Compressibility factor\n",
+      "V = (Z*R*T)/P;\t\t\t#[m**(3)/mol] - Molar volume\n",
+      "\n",
+      "\t\t\t#No.of moles in 0.3 m**(3) cylinder is given by\n",
+      "n1 = V_1/V;\t\t\t#[mol]\n",
+      "\n",
+      "\t\t\t#Mass of gas in cylinder is given by \n",
+      "m1 = (n1*M)/1000.;\t\t\t#[kg]\n",
+      "print \" Under actual conditions, the mass of ethane is, %f kg\"%m1\n",
+      "\n",
+      "\t\t\t#Under ideal condition% taking Z = 1%\n",
+      "V_ideal = (R*T)/P;\t\t\t#[m**(3)/mol]\n",
+      "n2 = V_1/V_ideal;\t\t\t#[mol]\n",
+      "m2 = (n2*M)/1000;\t\t\t#[kg]\n",
+      "print \" Under ideal conditions,the mass of ethane is, %f kg\"%m2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Under actual conditions, the mass of ethane is, 136.395203 kg\n",
+        " Under ideal conditions,the mass of ethane is, 42.339741 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.6 Page number - 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 373.15;\t\t\t#[K] - Temperature\n",
+      "P = 101325;\t\t\t#[N/m**(2)] - Pressure\n",
+      "Tc = 647.1;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 220.55*10**(5)\t#[N/m**(2)] - Critical pressure\n",
+      "w = 0.345;\t\t\t#acentric factor\n",
+      "Tr = T/Tc;\t\t\t# Reduced temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - UNiversal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6))\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2))\n",
+      "\n",
+      "\t\t\t#We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B = ((B_0+(w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol] - Second virial coefficient\n",
+      "\n",
+      "\t\t\t#We have% Z = 1+(B/V) and Z = (P*V)/(R*T). Substituting the value of Z%we get\n",
+      "\t\t\t# V**(2)-((R*T)/P)*V-((B*R*T)/P)=0 .Solving the quadratic equation by shreedharcharya rule\n",
+      "V1 = (((R*T)/P) + (((R*T)/P)**(2) + 4*1*((B*R*T)/P))**(1./2))/2*1;\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume of water vapour is %f m**(3)/mol\"%V1\n",
+      "\n",
+      "\t\t\t#The roots are%V1 = 0.0003670 [m**(3)/mol] and V2 = 0.0302510 [m**(3)/mol].\n",
+      "\t\t\t#As 'V2' is near to ideal volume (0.030618 [m**(3)/mol])%it is taken as the molar volume\n",
+      "\t\t\t#The other root 'V1' hss no physical significance\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume of water vapour is 0.030251 m**(3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.7  Page number - 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 50+273.15;\t\t\t#[K] - Temperature\n",
+      "P = 15*10**(5)\t\t\t#[N/m**(2)] - Pressure\n",
+      "Tc = 305.3;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 48.72*10**(5)\t\t\t#[N/m**(2)] - Critical pressure\n",
+      "w = 0.100;\t\t\t# Acentric factor\n",
+      "B = -157.31;\t\t\t#[cm**(3)/mol] - second virial coefficient\n",
+      "B = B*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "C = 9650;\t\t\t#[cm**(6)/mol**(2)] - third virial coefficient\n",
+      "C = C*10**(-12)\t\t\t#[cm**(6)/mol**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (1)\n",
+      "V_1 = (R*T)/P;\t\t\t#[m**(3)/mol] - Molar volume\n",
+      "print \" (1).The molar volume for ideal equation of state is %e m**(3)/mol\"%V_1\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "Tr = T/Tc;\t\t\t# Reduced temperature\n",
+      "\t\t\t# At this temperature\n",
+      "B_0 = 0.083-(0.422/(Tr)**(1.6))\n",
+      "B_1 = 0.139-(0.172/(Tr)**(4.2))\n",
+      "\n",
+      "\t\t\t# We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B_2 = ((B_0 + (w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\t\t\t#second virial coefficient\n",
+      "print \" (2).The second virial coefficent using Pitzer correlation is \\\n",
+      "found to be %e m**(3)/mol which is same as given value\"%B_2\n",
+      "\n",
+      "\t\t\t# (3)\n",
+      "\t\t\t# Given (virial equation)%Z=1+(B/V)\n",
+      "V_3 = B + (R*T)/P;\t\t\t#[m**(3)/mol] - Molar volume\n",
+      "print \" (3).The molar volume using virial equation of state is %e m**(3)/mol\"%V_3\n",
+      "\n",
+      "\t\t\t# (4)\n",
+      "\t\t\t# Given (virial equation)%Z = 1 + ((B*P)/(R*T)) + ((C - B**(2))/(R*T)**(2))*P**(2)\n",
+      "V_4 = B + (R*T)/P + ((C - B**(2))/(R*T))*P;\t\t\t# [m**(3)/mol]\n",
+      "print \" (4).The molar volume using given virial equation of state is %e m**(3)/mol\"%V_4\n",
+      "\n",
+      "\t\t\t# (5)\n",
+      "\t\t\t# Given%Z = 1 + (B/V)\n",
+      "\t\t\t# Also%Z = (P*V)/(R*T). Substituting the value of Z%we get\n",
+      "\t\t\t# V**(2)-((R*T)/P)*V-((B*R*T)/P)=0.Solving the quadratic equation\n",
+      "from scipy.optimize import fsolve\n",
+      "def f(V):\n",
+      "    global R,T,P,B\n",
+      "    return V**(2)-((R*T)/P)*V-((B*R*T)/P)\n",
+      "V_5_1 = fsolve(f,0)\n",
+      "V_5_2 = fsolve(f,1)\n",
+      "\n",
+      "print \" (5).The molar volume using given virial equation of state is %e m**(3)/mol\"%V_5_2\n",
+      "\n",
+      "\t\t\t# The roots are%V_5_1=0.0001743 [m**(3)/mol] and V_5_2=0.0016168 [m**(3)/mol].\n",
+      "\t\t\t# As 'V_2' is near to ideal volume (0.0017911 [m**(3)/mol])%it is taken as the molar volume\n",
+      "\n",
+      "\t\t\t# (6)\n",
+      "\t\t\t# Given%Z = 1 + (B/V) + (C/V**(2))\n",
+      "\t\t\t# Also%Z = (P*V)/(R*T). Substituting the value of Z%we get\n",
+      "\t\t\t# V**(3)-((R*T)/P)*V**(2)-((B*R*T)/P)*V-((C*R*T)/P)=0. Solving the cubic equation\n",
+      "def f1(V):\n",
+      "    global P,R,T,B,C\n",
+      "    return V**(3)-((R*T)/P)*V**(2)-((B*R*T)/P)*V-((C*R*T)/P)\n",
+      "    \n",
+      "V_6_3=fsolve(f1,-1)\n",
+      "V_6_4=fsolve(f1,0)\n",
+      "V_6_5=fsolve(f1,1)\n",
+      "\t\t\t#The above equation has 1 real and 2 imaginary roots. We consider only real root.\n",
+      "print \" (6).The molar volume using given virial equation of state is %e m**(3)/mol\"%V_6_5\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (1).The molar volume for ideal equation of state is 1.791113e-03 m**(3)/mol\n",
+        " (2).The second virial coefficent using Pitzer correlation is found to be -1.573258e-04 m**(3)/mol which is same as given value\n",
+        " (3).The molar volume using virial equation of state is 1.633803e-03 m**(3)/mol\n",
+        " (4).The molar volume using given virial equation of state is 1.625374e-03 m**(3)/mol\n",
+        " (5).The molar volume using given virial equation of state is 1.616848e-03 m**(3)/mol"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " (6).The molar volume using given virial equation of state is 1.624187e-03 m**(3)/mol\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "/home/jovina/virtualenvs/scipy/local/lib/python2.7/site-packages/scipy/optimize/minpack.py:236: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last ten iterations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.8  Page number - 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 0 + 273.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "from numpy import linalg,array\n",
+      "from scipy.stats import linregress\n",
+      "\t\t\t#Virial equation of state% Z=1+(B/V)+(C/V**(2))\n",
+      "\t\t\t#From above equation we get (Z-1)*V=B+(C/V)\n",
+      "\n",
+      "P=[50,100,200,400,600,1000];\n",
+      "Z=[0.9846,1.0000,1.0365,1.2557,1.7559,2.0645];\n",
+      "V=[0,0,0,0,0,0]\n",
+      "k= []\n",
+      "t=array([0,0,0,0,0,0])\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(6):\n",
+      "    V[i]=(Z[i]*R*T)/(P[i]*101325.)\t\t\t#[m**(3)/mol]\n",
+      "    k.append((Z[i]-1.)*V[i])\n",
+      "    t[i]=1./V[i]\n",
+      "    \n",
+      "k = array(k)\n",
+      "\n",
+      "\t\t\t#w = linalg.lstsq(t.T,k.T)\n",
+      "\t\t\t#[C,B,sig]=reglin(t',k')\n",
+      "C,B,c,d,e = linregress(t.T,k.T)\n",
+      "\n",
+      "# Results\n",
+      "#From the regression% we get intercept=B and slope=C%and thus%\n",
+      "print \" The value of second virial coefficient (B) is %e m**(3)/mol\"%B\n",
+      "print \" The value of third virial coefficient (C) is %e m**(6)/mol**(2)\"%C\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of second virial coefficient (B) is -1.563305e-05 m**(3)/mol\n",
+        " The value of third virial coefficient (C) is 3.133359e-09 m**(6)/mol**(2)\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.9 Page number - 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 444.3;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "B_11 = -8.1;\t\t\t#[cm**(3)/mol]\n",
+      "B_11 = -8.1*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "B_22 = -293.4*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "y1 = 0.5;\t\t\t# mole fraction \t\t\t# equimolar mixture\n",
+      "y2 = 0.5;\n",
+      "\n",
+      "# For component 1 (methane)\n",
+      "Tc_1 = 190.6;\t\t\t#[K] - cricitical temperature\n",
+      "Vc_1 = 99.2;\t\t\t#[cm**(3)/mol] - cricitical molar volume\n",
+      "Zc_1 = 0.288;\t\t\t# critical compressibility factor\n",
+      "w_1 = 0.012;\t\t\t# acentric factor\n",
+      "\n",
+      "# For component 2 (n-butane)\n",
+      "Tc_2 = 425.2;\t\t\t#[K]\n",
+      "Vc_2 = 255.0;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_2 = 0.274;\n",
+      "w_2 = 0.199;\n",
+      "\n",
+      "# Calculations\n",
+      "#Using virial mixing rule%we get\n",
+      "Tc_12 = (Tc_1*Tc_2)**(1./2)\t\t\t#[K]\n",
+      "w_12 = (w_1 + w_2)/2.;\n",
+      "Zc_12 = (Zc_1+Zc_2)/2.;\n",
+      "Vc_12 = (((Vc_1)**(1./3) + (Vc_2)**(1./3))/2.)**(3)\t\t\t#[cm**(3)/mol]\n",
+      "Vc_12 = Vc_12*10**(-6)\t\t\t#[cm**(3)/mol]\n",
+      "Pc_12 = (Zc_12*R*Tc_12)/Vc_12;\t\t\t#[N/m**(2)]\n",
+      "Tr_12 = T/Tc_12;\t\t\t#Reduced temperature\n",
+      "B_0 = 0.083 - (0.422/(Tr_12)**(1.6))\n",
+      "B_1 = 0.139 - (0.172/(Tr_12)**(4.2))\n",
+      "\n",
+      "\t\t\t#We know%(B_12*Pc_12)/(R*Tc_12) = B_0 + (w_12*B_1)\n",
+      "B_12 = ((B_0+(w_12*B_1))*(R*Tc_12))/Pc_12;\t\t\t#[m**(3)/mol] - Cross coefficient\n",
+      "B = y1**(2)*B_11+2*y1*y2*B_12+y2**(2)*B_22;\t\t\t#[m**(3)/mol] - Second virial coefficient for mixture\n",
+      "B = B*10**(6)\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The second virial coefficient,(B) for the mixture of gas is %f cm**(3)/mol\"%B\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The second virial coefficient,(B) for the mixture of gas is -108.309380 cm**(3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 2.10 Page number - 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 71+273.15;\t\t\t#[K] - Temperature\n",
+      "P = 69*10**(5)\t\t\t#[N/m**(2)] - Pressure\n",
+      "y1 = 0.5;\t\t\t#[mol] - mole fraction of equimolar mixture\n",
+      "y2 = 0.5;\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t#For component 1 (methane)\n",
+      "Tc_1  =190.6;\t\t\t#[K] - Critical temperature\n",
+      "Pc_1 = 45.99*10**(5)\t\t\t#[N/m**(2)] - Critical pressure\n",
+      "Vc_1 = 98.6;\t\t\t#[cm**(3)/mol] - Critical volume\n",
+      "Zc_1 = 0.286;\t\t\t# Critical compressibility factor\n",
+      "w_1 = 0.012;\t\t\t# acentric factor\n",
+      "\n",
+      "\t\t\t#For component 2 (hydrogen sulphide)\n",
+      "Tc_2 = 373.5;\t\t\t#[K]\n",
+      "Pc_2 = 89.63*10**(5)\t\t\t#[N/m**(2)]\n",
+      "Vc_2 = 98.5;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_2 = 0.284;\n",
+      "w_2 = 0.094;\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#For component 1\n",
+      "Tr_1 = T/Tc_1;\t\t\t#Reduced temperature\n",
+      "\t\t\t#At reduced temperature\n",
+      "B1_0 = 0.083-(0.422/(Tr_1)**(1.6))\n",
+      "B1_1 = 0.139-(0.172/(Tr_1)**(4.2))\n",
+      "\t\t\t#We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B_11 = ((B1_0+(w_1*B1_1))*(R*Tc_1))/Pc_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#Similarly for component 2\n",
+      "Tr_2 = T/Tc_2;\t\t\t#Reduced temperature\n",
+      "\t\t\t#At reduced temperature Tr_2%\n",
+      "B2_0 = 0.083 - (0.422/(Tr_2)**(1.6))\n",
+      "B2_1 = 0.139 - (0.172/(Tr_2)**(4.2))\n",
+      "B_22 = ((B2_0+(w_2*B2_1))*(R*Tc_2))/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For cross coeffcient\n",
+      "Tc_12 = (Tc_1*Tc_2)**(1./2)\t\t\t#[K]\n",
+      "w_12 = (w_1 + w_2)/2;\n",
+      "Zc_12 = (Zc_1 + Zc_2)/2;\n",
+      "Vc_12 = (((Vc_1)**(1./3) + (Vc_2)**(1./3))/2)**(3)\t\t\t#[cm**(3)/mol]\n",
+      "Vc_12 = Vc_12*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "Pc_12 = (Zc_12*R*Tc_12)/Vc_12;\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "\t\t\t#Now we have%(B_12*Pc_12)/(R*Tc_12) = B_0+(w_12*B_1)\n",
+      "\t\t\t#where B_0 and B_1 are to be evaluated at Tr_12\n",
+      "Tr_12 = T/Tc_12;\n",
+      "\t\t\t#At reduced temperature Tr_12\n",
+      "B_0 = 0.083 - (0.422/(Tr_12)**(1.6))\n",
+      "B_1 = 0.139 - (0.172/(Tr_12)**(4.2))\n",
+      "B_12=((B_0 + (w_12*B_1))*R*Tc_12)/Pc_12;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For the mixture\n",
+      "B = y1**(2)*B_11+2*y1*y2*B_12 + y2**(2)*B_22;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#Now given virial equation is% Z=1+(B*P)/(R*T)\n",
+      "Z = 1 + (B*P)/(R*T)\n",
+      "\n",
+      "\t\t\t#Also Z = (P*V)/(R*T).Therefore%\n",
+      "V = (Z*R*T)/P;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume of the mixture is %e m**(3)/mol\"%V\n",
+      "\t\t\t#The value obtained is near the math.experimental value of V_math.exp = 3.38*10**(-4) m**(3)/mol\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume of the mixture is 3.390411e-04 m**(3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example - 2.11  Page number - 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P = 6.*10**(6)\t\t\t# [Pa] - Pressure\n",
+      "P_max = 12.*10**(6)\t\t\t# [Pa] - Max pressure to which cylinder may be math.exposed\n",
+      "T = 280.;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1).Assuming ideal gas behaviour%\n",
+      "V_ideal = (R*T)/P;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t#Now when temperature and pressure are increased%the molar volume remains same%as total volume and number of moles are same.\n",
+      "\t\t\t#For max pressure of 12 MPa%temperature is\n",
+      "T_max_ideal = (P_max*V_ideal)/R;\n",
+      "print \" (1).The maximum temperature assuming ideal behaviour is %f K\"%T_max_ideal\n",
+      "\n",
+      "\t\t\t# (2).Assuming virial equation of state\n",
+      "\t\t\t# For component 1 (methane)%at 280 K\n",
+      "Tc_1 = 190.6;\t\t\t#[K]\n",
+      "Pc_1 = 45.99*10**(5)\t\t\t#[N/m**(2)]\n",
+      "Vc_1 = 98.6;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_1 = 0.286;\n",
+      "w_1 = 0.012;\n",
+      "Tr_1 = T/Tc_1;\t\t\t#Reduced temperature\n",
+      "B1_0 = 0.083 - (0.422/(Tr_1)**(1.6))\n",
+      "B1_1 = 0.139 - (0.172/(Tr_1)**(4.2))\n",
+      "\n",
+      "\t\t\t#We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B_11 = ((B1_0 + (w_1*B1_1))*(R*Tc_1))/Pc_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For component 2 (Propane)\n",
+      "Tc_2 = 369.8;\t\t\t#[K]\n",
+      "Pc_2 = 42.48*10**(5)\t\t\t#[N/m**(2)]\n",
+      "Vc_2 = 200;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_2 = 0.276;\n",
+      "w_2 = 0.152;\n",
+      "Tr_2 = T/Tc_2;\t\t\t# Reduced temperature\n",
+      "B2_0 = 0.083 - (0.422/(Tr_2)**(1.6))\n",
+      "B2_1 = 0.139 - (0.172/(Tr_2)**(4.2))\n",
+      "B_22 = ((B2_0 + (w_2*B2_1))*(R*Tc_2))/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For cross coeffcient\n",
+      "y1 = 0.8;\t\t\t#mole fraction of component 1\n",
+      "y2 = 0.2;\t\t\t#mole fraction of component 2\n",
+      "Tc_12 = (Tc_1*Tc_2)**(1/2.)\t\t\t#[K]\n",
+      "w_12 = (w_1 + w_2)/2;\n",
+      "Zc_12 = (Zc_1 + Zc_2)/2;\n",
+      "Vc_12 = (((Vc_1)**(1./3) + (Vc_2)**(1./3))/2)**(3)\t\t\t#[cm**(3)/mol]\n",
+      "Vc_12 = Vc_12*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "Pc_12 = (Zc_12*R*Tc_12)/Vc_12;\t\t\t#[N/m**(2)]\n",
+      "Tr_12 = T/Tc_12;\n",
+      "\n",
+      "\t\t\t#At reduced temperature%Tr_12%\n",
+      "B_0 = 0.083 - (0.422/(Tr_12)**(1.6))\n",
+      "B_1 = 0.139 - (0.172/(Tr_12)**(4.2))\n",
+      "B_12 = ((B_0 + (w_12*B_1))*R*Tc_12)/Pc_12;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For the mixture\n",
+      "B = y1**(2)*B_11+2*y1*y2*B_12 + y2**(2)*B_22;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#Now given virial equation is% Z=1+(B*P)/(R*T)\n",
+      "Z = 1 + (B*P)/(R*T)\n",
+      "\t\t\t#Also Z = (P*V)/(R*T).Therefore%\n",
+      "V_real = (Z*R*T)/P;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t\n",
+      "\n",
+      "T_prime = 560.;\t\t\t#[K]\n",
+      "fault = 10.;\n",
+      "\n",
+      "while(fault > 1):\n",
+      "    T_prime_r1 = T_prime/Tc_1;\n",
+      "    B_prime1_0 = 7.7674*10**(-3)\n",
+      "    B_prime1_1 = 0.13714;\n",
+      "    B_prime_11 = ((B_prime1_0 + (w_1*B_prime1_1))*(R*Tc_1))/Pc_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "    \t\t\t#Similarly for component 2%\n",
+      "    T_prime_r2 = T_prime/Tc_2;\n",
+      "    B_prime2_0 = -0.1343;\n",
+      "    B_prime2_1 = 0.10887;\n",
+      "    B_prime_22 = ((B_prime2_0 + (w_2*B_prime2_1))*(R*Tc_2))/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "    \t\t\t#For cross coefficient (assuming k12=0)\n",
+      "    \t\t\t#Tc_12 % w_12 % Zc_12 % Vc_12 and Pc_12 have already been calculated above%now\n",
+      "    T_prime_r12 = T_prime/Tc_12;\t\t\t#\n",
+      "    \t\t\t#At reduced temperature%T_prime_r12%\n",
+      "    B_prime_0 = 0.083 - (0.422/(T_prime_r12)**(1.6))\n",
+      "    B_prime_1 = 0.139 - (0.172/(T_prime_r12)**(4.2))\n",
+      "    B_prime_12 = ((B_prime_0+(w_12*B_prime_1))*R*Tc_12)/Pc_12;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "    \t\t\t#For the mixture\n",
+      "    B_prime = y1**(2)*B_prime_11 + 2*y1*y2*B_prime_12 + y2**(2)*B_prime_22;\t\t\t#[m**(3)/mol]\n",
+      "    Z_prime = 1 + (B_prime*P_max)/(R*T_prime)\n",
+      "    T_new = (P_max*V_real)/(Z_prime*R)\n",
+      "    fault = abs(T_prime - T_new)\n",
+      "    T_prime = T_new;\n",
+      "\n",
+      "print \" (2).The maximum temperature assuming the gas to follow virial equation of state is %f K\"%T_new\n",
+      "\n",
+      "# Note : Answers varies because of rounding error. Please check it manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (1).The maximum temperature assuming ideal behaviour is 560.000000 K\n",
+        " (2).The maximum temperature assuming the gas to follow virial equation of state is 440.112497 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example - 2.12  Page number - 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "V_vessel = 0.1;\t\t\t#[m**(3)]\t\t\t# Volume of vessel\n",
+      "T = 25 + 273.15;\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "m = 25*1000;\t\t\t#[g]\t\t\t# Mass of ethylene\n",
+      "Tc = 282.3;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 50.40;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5)\t\t\t#[N/m**(2)]\n",
+      "Zc = 0.281;\t\t\t# Critical compressibility factor\n",
+      "Vc = 131;\t\t\t#[cm**(3)/mol] - Critical volume\n",
+      "Vc = Vc*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "w = 0.087;\t\t\t# Acentric factor\n",
+      "M = 28.054;\t\t\t# Molecular weight of ethylene\n",
+      "\n",
+      "# Calculations\n",
+      "n = m/M;\t\t\t#[mole] - No. of moles of ethylene\n",
+      "V = V_vessel/n;\t\t#[m**(3)/mol] - Molar volume\n",
+      "\n",
+      "\t\t\t#Under Redlich Kwong equation of state% we have\n",
+      "a = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b = (0.08664*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "P = ((R*T)/(V-b))-(a/(T**(1./2)*V*(V+b)))\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "# Results\n",
+      "print \" The required pressure using Redlich Kwong equation of state is %e N/m**(2)\"%P\n",
+      "\n",
+      "\t\t\t#For ideal gas equation of state%\n",
+      "P_ideal = (R*T)/V;\t\t\t#[N/m**(2)]\n",
+      "print \" For ideal gas equation of state,the required pressure is %e N/m**(2)\"%P_ideal\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The required pressure using Redlich Kwong equation of state is 7.934745e+06 N/m**(2)\n",
+        " For ideal gas equation of state,the required pressure is 2.208971e+07 N/m**(2)\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example - 2.13  Page number - 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "V_vessel = 360*10**(-3)\t\t\t#[m**(3)] - volume of vessel\n",
+      "T = 62+273.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "m = 70.*1000;\t\t\t#[g]/ - Mass of carbon dioxide\n",
+      "\n",
+      "\t\t\t#For carbon dioxide\n",
+      "Tc = 304.2;\t\t\t#[K] - Cricitical temperature\n",
+      "Pc = 73.83;\t\t\t#[bar] - Cricitical pressure\n",
+      "Pc = Pc*10**(5)\t\t\t# [N/m**(2)]\n",
+      "Zc = 0.274;\t\t\t# Critical compressibility factor\n",
+      "Vc = 94.0;\t\t\t#[cm**(3)/mol]\n",
+      "Vc = Vc*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "w = 0.224;\t\t\t# Acentric factor\n",
+      "M = 44.01;\t\t\t# Molecular weight of carbon dioxide\n",
+      "\n",
+      "# Calculations and Results\n",
+      "n = m/M;\t\t\t#[mol] - No. of moles\n",
+      "V = V_vessel/n;\t\t\t#[m**(3)/mol]\t\t\t#molar volume\n",
+      "\n",
+      "\t\t\t# (1)\n",
+      "\t\t\t# Ideal gas behaviour\n",
+      "P_1 = (R*T)/V;\t\t\t#[N/m**(2)]\n",
+      "print \" (1).The required pressure using ideal equation of state is %e N/m**(2)\"%P_1\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "\t\t\t# Virial equation of state% Z = 1 + (B*P)/(R*T)\n",
+      "\t\t\t# (P*V)/(R*T) = 1 + (B*P)/(R*T)% and thus P = (R*T)/(V - B). Now\n",
+      "Tr = T/Tc;\t\t\t#Reduced temperature\n",
+      "\t\t\t# At reduced temperature Tr%\n",
+      "B_0 = 0.083 - (0.422/(Tr)**(1.6))\n",
+      "B_1 = 0.139 - (0.172/(Tr)**(4.2))\n",
+      "B = ((B_0 + (w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "P_2 = (R*T)/(V - B)\t\t\t#[N/m**(2)]\n",
+      "print \" (2).The required pressure using given virial equation of state is %e N/m**(2)\"%P_2\n",
+      "\n",
+      "\t\t\t# (3)\n",
+      "\t\t\t# Virial equation of state% Z = 1 + (B/V)\n",
+      "\t\t\t# (P*V)/(R*T) = 1 + (B/V)\n",
+      "P_3 = ((R*T)/V) + (B*R*T)/(V**(2))\t\t\t#[N/m**(2)]\n",
+      "print \" (3).The required pressure using given virial equation of state is %e N/m**(2)\"%P_3\n",
+      "\n",
+      "\t\t\t# (4)\n",
+      "\t\t\t# Van der Walls equation of state%P = ((R*T)/(V-b)) - a/(V**(2))\n",
+      "a = (27*(R**(2))*(Tc**(2)))/(64*Pc)\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc)\t\t\t#[m**(3)/mol]\n",
+      "P_4 = ((R*T)/(V-b)) - a/(V**(2))\t\t\t#[N/m**(2)]\n",
+      "print \" (4).The required pressure using van der Walls equation of state is %e N/m**(2)\"%P_4\n",
+      "\n",
+      "\t\t\t#(5)\n",
+      "\t\t\t# Redlich Kwong equation of state%\n",
+      "a_1 = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b_1 = (0.08664*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "P_5 = ((R*T)/(V - b_1)) - (a_1/(T**(1./2)*V*(V + b_1)))\t\t\t#[N/m**(2)]\n",
+      "print \" (5).The required pressure using Redlich Kwong equation of state is %e N/m**(2)\"%P_5\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (1).The required pressure using ideal equation of state is 1.231100e+07 N/m**(2)\n",
+        " (2).The required pressure using given virial equation of state is 8.712394e+06 N/m**(2)\n",
+        " (3).The required pressure using given virial equation of state is 7.226009e+06 N/m**(2)\n",
+        " (4).The required pressure using van der Walls equation of state is 8.048790e+06 N/m**(2)\n",
+        " (5).The required pressure using Redlich Kwong equation of state is 8.079803e+06 N/m**(2)\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example - 2.14  Page number - 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 500+273.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "P = 325*1000;\t\t\t#[Pa] - Pressure\n",
+      "Tc = 647.1;\t\t\t#[K] - Cricitical temperature\n",
+      "Pc = 220.55;\t\t\t#[bar] - Cricitical pressure\n",
+      "Pc = Pc*10**(5)\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# Van der Walls equation of state%\n",
+      "a = (27*(R**(2))*(Tc**(2)))/(64*Pc)\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc)\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# The cubic form of van der Walls equation of state is given by%\n",
+      "\t\t\t# V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P=0\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f(V):\n",
+      "    global b,R,T,P,a\n",
+      "    return V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P\n",
+      "from scipy.optimize import fsolve\n",
+      "V_1 = fsolve(f,1)\n",
+      "V_2 = fsolve(f,10)\n",
+      "V_3 = fsolve(f,100)\n",
+      "\t\t\t# The above equation has 1 real and 2 imaginary roots. We consider only real root%\n",
+      "Z_1 = (P*V_1)/(R*T)\t\t\t#compressibility factor\n",
+      "print \" (1).The compressibility factor of steam using van der Walls equation of state is %f\"%Z_1\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\n",
+      "\t\t\t#Redlich Kwong equation of state%\n",
+      "a_1 = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b_1 = (0.08664*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# The cubic form of Redlich Kwong equation of state is given by%\n",
+      "\t\t\t# V**(3)-((R*T)/P)*V**(2)-((b_1**(2))+((b_1*R*T)/P)-(a/(T**(1/2)*P))*V-(a*b)/(T**(1/2)*P)=0\n",
+      "\t\t\t#Solving the cubic equation\n",
+      "def f1(V):\n",
+      "    return V**(3)-((R*T)/P)*V**(2)-((b_1**(2))+((b_1*R*T)/P)-(a_1/(T**(1./2)*P)))*V-(a_1*b_1)/(T**(1./2)*P)\n",
+      "    \n",
+      "V_4=fsolve(f1,1)\n",
+      "V_5=fsolve(f1,10)\n",
+      "V_6=fsolve(f1,100)\n",
+      "\t\t\t# The above equation has 1 real and 2 imaginary roots. We consider only real root%\n",
+      "\t\t\t# Thus compressibility factor is\n",
+      "Z_2 = (P*V_4)/(R*T)\t\t\t#compressibility factor\n",
+      "print \" (2).The compressibility factor of steam using Redlich Kwong equation of state is %f\"%Z_2\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (1).The compressibility factor of steam using van der Walls equation of state is 0.997181\n",
+        " (2).The compressibility factor of steam using Redlich Kwong equation of state is 0.997028\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example - 2.15  Page number - 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 250+273.15;\t\t\t#[K]\n",
+      "R = 8.314;\t\t\t#[J/mol*K]\n",
+      "P = 39.76;\t\t\t#[bar] Vapour pressure of water at T\n",
+      "P = P*10**(5)\t\t\t#[N/m**(2)]\n",
+      "Tc = 647.1;\t\t\t#[K] - Cricitical temperature\n",
+      "Pc = 220.55*10**(5)\t\t\t#[N/m**(2)] - Cricitical pressure\n",
+      "w = 0.345;\t\t\t#Acentric factor\n",
+      "M = 18.015;\t\t\t# Molecular weight of water\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# Using peng-Robinson equation of stste \n",
+      "m = 0.37464 + 1.54226*w - 0.26992*w**(2)\n",
+      "Tr = T/Tc;\n",
+      "alpha = (1 + m*(1 - Tr**(1./2)))**(2)\n",
+      "a = ((0.45724*(R*Tc)**(2))/Pc)*alpha;\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b = (0.07780*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# Cubuc form of Peng-Robinson equation of stste is given by\n",
+      "\t\t\t# V**(3) + (b-(R*T)/P)*V**(2) - ((3*b**(2)) + ((2*R*T*b)/P) - (a/P))*V+b**(3) + ((R*T*(b**(2))/P) - ((a*b)/P) = 0;\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f(V):\n",
+      "    global b,R,T,P,a\n",
+      "    return V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2)))/P)-((a*b)/P)\n",
+      "V_1 = fsolve(f,-1)\n",
+      "V_2 = fsolve(f,0)\n",
+      "V_3 = fsolve(f,1)\n",
+      "\t\t\t#The largest root is for vapour phase%\n",
+      "V_vap = V_3;\t\t\t#[m**(3)/mol] - Molar volume (saturated vapour)\n",
+      "V_vap = V_vap*10**(6)/M;\t\t\t#[cm**(3)/g]\n",
+      "\n",
+      "print \" The moar volume of saturated water in the vapour phase (V_vap) is %f cm**(3)/g\"%V_vap\n",
+      "\n",
+      "\t\t\t#The smallest root is for liquid phase%\n",
+      "V_liq = V_1;\t\t\t#[m**(3)/mol] - molar volume (saturated liquid)\n",
+      "V_liq = V_liq*10**(6)/M;\t\t\t#[cm**(3)/g]\n",
+      "print \" The moar volume of saturated water in the liquid phase (V_liq) is %f cm**(3)/g\"%V_liq\n",
+      "\n",
+      "\t\t\t#From steam table at 250 C% V_vap = 50.13 [cm**(3)/g] and V_liq = 1.251 [cm**(3)/g].\n",
+      "print \" From steam table at 250 C, V_vap = 50.13 [cm**(3)/g] and V_liq = 1.251 [cm**(3)/g]\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The moar volume of saturated water in the vapour phase (V_vap) is 51.757097 cm**(3)/g\n",
+        " The moar volume of saturated water in the liquid phase (V_liq) is 1.554560 cm**(3)/g\n",
+        " From steam table at 250 C, V_vap = 50.13 [cm**(3)/g] and V_liq = 1.251 [cm**(3)/g]\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example - 2.16  Page number - 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 500+273.15;\t\t\t#[K] - Temperature\n",
+      "P = 15.;\t\t\t#[atm] - Pressure\n",
+      "P = P*101325;\t\t\t#[N/m**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "Tc = 190.6;\t\t\t#[K] - Cricitical temperature\n",
+      "Pc = 45.99*10**(5)\t\t\t#[N/m**(2)] - Cricitical pressure\n",
+      "Vc = 98.6;\t\t\t#[cm**(3)/mol] - Cricitical molar volume\n",
+      "Zc = 0.286;\t\t\t# Critical compressibility factor\n",
+      "w = 0.012;\t\t\t# Acentric factor\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t#Virial equation of state%Z = 1 + (B*P)/(R*T)\n",
+      "Tr_1 = T/Tc;\t\t\t#Reduced temperature\n",
+      "B_0 = 0.083-(0.422/(Tr_1)**(1.6))\n",
+      "B_1 = 0.139-(0.172/(Tr_1)**(4.2)) \t\t\t# We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B = ((B_0+(w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\t\t\t#second virial coefficient\n",
+      "Z = 1 + (B*P)/(R*T)\t\t\t#compressibility factor\n",
+      "\t\t\t#(P*V)/(R*T)=1+(B*P)/(R*T)%and thus%\n",
+      "V_1  = (Z*R*T)/P;\t\t\t#[m**(3)/mol]\n",
+      "print \" (1).The molar volume of methane using given virial equation is %e m**(3)/mol\"%V_1\n",
+      "\n",
+      "\t\t\t#(2).\n",
+      "\t\t\t#Virial equation of state%Z = 1 + (B/V)\n",
+      "\t\t\t#Also%Z = (P*V)/(R*T). Substituting the value of Z%we get\n",
+      "\t\t\t# V**(2) - ((R*T)/P)*V - ((B*R*T)/P) = 0.Solving the quadratic equation\n",
+      "def f(V):\n",
+      "    global R,T,P,B\n",
+      "    return V**(2)-((R*T)/P)*V-((B*R*T)/P)\n",
+      "V2_1=fsolve(f,0)\n",
+      "V2_2=fsolve(f,1)\n",
+      "\t\t\t# Out of two roots%we will consider only positive root\n",
+      "print \" (2).The molar volume of methane using given virial equation is %e m**(3)/mol\"%V2_2\n",
+      "\n",
+      "\t\t\t# (3)\n",
+      "\t\t\t# Van der Walls equation of state%\n",
+      "\t\t\t# (P + (a/V**(2)))*(V - b) = R*T\n",
+      "a_3 = (27*(R**(2))*(Tc**(2)))/(64*Pc)\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b_3 = (R*Tc)/(8*Pc)\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# The cubic form of van der Walls equation of state is given by%\n",
+      "\t\t\t# V**(3) - (b + (R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f1(V):\n",
+      "    global b,R,T,P,a_3\n",
+      "    return V**(3)-(b_3+(R*T)/P)*V**(2)+(a_3/P)*V-(a_3*b_3)/P\n",
+      "V3_1=fsolve(f1,1)\n",
+      "V3_2=fsolve(f1,10)\n",
+      "V3_3=fsolve(f1,100)\n",
+      "\t\t\t# The above equation has 1 real and 2 imaginary roots. We consider only real root.\n",
+      "print \" (3).The molar volume of methane using van der Walls equation of state is %e m**(3)/mol\"%V3_1\n",
+      "\n",
+      "\t\t\t# (4)\n",
+      "\t\t\t# Redlich Kwong equation of state\n",
+      "a_4 = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b_4 = (0.08664*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# The cubic form of Redlich Kwong equation of state is given by%\n",
+      "\t\t\t# V**(3) - ((R*T)/P)*V**(2) - ((b_1**(2)) + ((b_1*R*T)/P) - (a/(T**(1/2)*P))*V - (a*b)/(T**(1/2)*P) = 0\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f2(V):\n",
+      "    global R,T,P,b_4,a_4\n",
+      "    return V**(3)-((R*T)/P)*V**(2)-((b_4**(2))+((b_4*R*T)/P)-(a_4/(T**(1./2)*P)))*V-(a_4*b_4)/(T**(1./2)*P)\n",
+      "V4_1=fsolve(f2,1)\n",
+      "V4_2=fsolve(f2,10)\n",
+      "V4_3=fsolve(f2,100)\n",
+      "\t\t\t#The above equation has 1 real and 2 imaginary roots. We consider only real root.\n",
+      "print \" (4).The molar volume of methane using Redlich Kwong equation of state is %e m**(3)/mol\"%V4_1\n",
+      "\n",
+      "\t\t\t# (5)\n",
+      "\t\t\t# Using Peng-Robinson equation of state \n",
+      "m = 0.37464 + 1.54226*w - 0.26992*w**(2)\n",
+      "Tr_5 = T/Tc;\n",
+      "alpha = (1 + m*(1 - Tr_5**(1./2)))**(2)\n",
+      "a = ((0.45724*(R*Tc)**(2))/Pc)*alpha;\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b = (0.07780*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# Cubic form of Peng-Robinson equation of stste is given by\n",
+      "\t\t\t# V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2))/P)-((a*b)/P)=0;\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f3(V):\n",
+      "    global b,R,T,P,a\n",
+      "    return V**(3)+(b-(R*T)/P)*V**(2)-((3*b**(2))+((2*R*T*b)/P)-(a/P))*V+b**(3)+((R*T*(b**(2)))/P)-((a*b)/P)\n",
+      "V5_1=fsolve(f3,-1)\n",
+      "V5_2=fsolve(f3,0)\n",
+      "V5_3=fsolve(f3,1)\n",
+      "\t\t\t#The largest root is for vapour phase%\n",
+      "\t\t\t#The largest root is only considered as the systemis gas\n",
+      "print \" (5).The molar volume of methane using Peng-Robinson equation of state is %e m**(3)/mol\"%V5_3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (1).The molar volume of methane using given virial equation is 4.242974e-03 m**(3)/mol\n",
+        " (2).The molar volume of methane using given virial equation is 4.242930e-03 m**(3)/mol\n",
+        " (3).The molar volume of methane using van der Walls equation of state is 4.236938e-03 m**(3)/mol\n",
+        " (4).The molar volume of methane using Redlich Kwong equation of state is 4.241402e-03 m**(3)/mol\n",
+        " (5).The molar volume of methane using Peng-Robinson equation of state is 4.242341e-03 m**(3)/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example - 2.17  Page number - 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 310.93;\t\t\t#[K] - Temperature\n",
+      "P = 2.76*10**(6)\t\t\t#[N/m**(2)] - Pressure\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "y1 = 0.8942;\t\t\t# Mole fraction of component 1 (methane)\n",
+      "y2 = 1-y1;\t\t\t# Mole fraction of component 2 (n-butane)\n",
+      "\n",
+      "#For component 1 (methane)\n",
+      "Tc_1 = 190.58;\t\t\t#[K] - Cricitical temperature\n",
+      "Pc_1 = 46.05;\t\t\t#[bar] - Cricitical pressure\n",
+      "Pc_1 = Pc_1*10**(5)\t\t\t#[N/m**(2)]\n",
+      "Zc_1 = 0.288;\t\t\t# Critical compressibility factor\n",
+      "Vc_1 = 99.1;\t\t\t#[cm**(3)/mol]\n",
+      "Vc_1 = Vc_1*10**(-6)\t\t\t#[m**(3)/mol]\n",
+      "w_1 = 0.011;\t\t\t# Acentric factor\n",
+      "\n",
+      "\t\t\t#For component 2 (n-butane)\n",
+      "Tc_2 = 425.18;\t\t\t#[K] - Cricitical temperature\n",
+      "Pc_2 = 37.97;\t\t\t#[bar] - Cricitical pressure\n",
+      "Pc_2 = Pc_2*10**(5)\t\t\t# [N/m**(2)]\n",
+      "Zc_2 = 0.274;\t\t\t# Critical compressibility factor\n",
+      "Vc_2 = 255.1;\t\t\t# [cm**(3)/mol]\n",
+      "Vc_2 = Vc_2*10**(-6)\t\t\t# [m**(3)/mol]\n",
+      "w_2 = 0.193;\t\t\t# Acentric factor\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (1)\n",
+      "\t\t\t# Virial equation of state% Z = 1 + (B*P)/(R*T)\n",
+      "\t\t\t# For component 1 (methane)\n",
+      "Tr_1 = T/Tc_1;\t\t\t#Reduced temperature\n",
+      "\t\t\t# At reduced temperature\n",
+      "B1_0 = 0.083 - (0.422/(Tr_1)**(1.6))\n",
+      "B1_1 = 0.139 - (0.172/(Tr_1)**(4.2))\n",
+      "\t\t\t# We know%(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B_11 = ((B1_0+(w_1*B1_1))*(R*Tc_1))/Pc_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#Similarly for component 2\n",
+      "Tr_2 = T/Tc_2;\t\t\t#Reduced temperature\n",
+      "\t\t\t#At reduced temperature Tr_2%\n",
+      "B2_0 = 0.083 - (0.422/(Tr_2)**(1.6))\n",
+      "B2_1 = 0.139 - (0.172/(Tr_2)**(4.2))\n",
+      "B_22 = ((B2_0 + (w_2*B2_1))*(R*Tc_2))/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For cross coeffcient\n",
+      "Tc_12 = (Tc_1*Tc_2)**(1./2)\t\t\t#[K]\n",
+      "w_12 = (w_1 + w_2)/2;\n",
+      "Zc_12 = (Zc_1 + Zc_2)/2;\n",
+      "Vc_12 = (((Vc_1)**(1./3)+(Vc_2)**(1./3))/2)**(3)\t\t\t#[m**(3)/mol]\n",
+      "Pc_12 =(Zc_12*R*Tc_12)/Vc_12;\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "\t\t\t#Now we have%(B_12*Pc_12)/(R*Tc_12) = B_0+(w_12*B_1)\n",
+      "\t\t\t#where B_0 and B_1 are to be evaluated at Tr_12\n",
+      "Tr_12 = T/Tc_12;\n",
+      "\t\t\t#At reduced temperature Tr_12\n",
+      "B_0 = 0.083 - (0.422/(Tr_12)**(1.6))\n",
+      "B_1 = 0.139 - (0.172/(Tr_12)**(4.2))\n",
+      "B_12 = ((B_0+(w_12*B_1))*R*Tc_12)/Pc_12;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For the mixture\n",
+      "B = y1**(2)*B_11+2*y1*y2*B_12+y2**(2)*B_22;\t\t\t#[m**(3)/mol]\n",
+      "Z_1 = 1+(B*P)/(R*T)\t\t\t#compressibility factor\n",
+      "print \" (1).The compressibility factor of mixture using Virial equation of state is %f\"%Z_1\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "\t\t\t# Pseudo reduced method.\n",
+      "T_pc = (y1*Tc_1)+(y2*Tc_2)\t\t\t#[K] - Cricitical temperature\n",
+      "P_pc = (y1*Pc_1)+(y2*Pc_2)\t\t\t#[N/m**(2)] - Cricitical pressure\n",
+      "w = (y1*w_1)+(y2*w_2)\t\t\t# Acentric factor\n",
+      "T_pr = T/T_pc;\t\t\t# Reduced temperature\n",
+      "P_pr = P/P_pc;\t\t\t# Reduced pressure\n",
+      "\t\t\t#At this value of Tpr%\n",
+      "B0 = 0.083 - (0.422/(T_pr)**(1.6))\n",
+      "B1 = 0.139 - (0.172/(T_pr)**(4.2))\n",
+      "Z0 = 1 + B0*(P_pr/T_pr)\n",
+      "Z1 = B1*(P_pr/T_pr)\n",
+      "Z = Z0 + w*Z1;\n",
+      "print \" (2).The compressibility factor of mixture using pseudo reduced method is %f\"%Z\n",
+      "\n",
+      "\t\t\t# (3)\n",
+      "\t\t\t# Redlich Kwong equation of state is given by\n",
+      "\t\t\t# P = ((R*T)/(V-b)) - (a/(T**(1/2)*V*(V+b)))\n",
+      "\t\t\t# For methane%component 1\n",
+      "a_1 = (0.42748*(R**(2))*(Tc_1**(2.5)))/Pc_1;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b_1 = (0.08664*R*Tc_1)/Pc_1;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t#For n-butane%component 2\n",
+      "a_2 = (0.42748*(R**(2))*(Tc_2**(2.5)))/Pc_2;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b_2 = (0.08664*R*Tc_2)/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t#For the mixture\n",
+      "a_12 = (a_1*a_2)**(1./2)\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "a = y1**(2)*a_1 + 2*y1*y2*a_12 + y2**(2)*a_2;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b = (y1*b_1) + (y2*b_2)\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# The cubic form of Redlich Kwong equation of state is given by%\n",
+      "\t\t\t# V**(3) - ((R*T)/P)*V**(2) - ((b_1**(2)) + ((b_1*R*T)/P) - (a/(T**(1/2)*P))*V - (a*b)/(T**(1/2)*P) = 0\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f(V):\n",
+      "    global R,T,P,b,a\n",
+      "    return V**(3)-((R*T)/P)*V**(2)-((b**(2))+((b*R*T)/P)-(a/(T**(1./2)*P)))*V-(a*b)/(T**(1./2)*P)\n",
+      "V_1=fsolve(f,1)\n",
+      "V_2=fsolve(f,10)\n",
+      "V_3=fsolve(f,100)\n",
+      "\t\t\t# Thus compressibility factor is\n",
+      "Z_3 = (P*V_1)/(R*T)\t\t\t#compressibility factor\n",
+      "print \" (3).The compressibility factor of mixture using Redlich Kwong equation of state is %f\"%Z_3\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (1).The compressibility factor of mixture using Virial equation of state is 0.926842\n",
+        " (2).The compressibility factor of mixture using pseudo reduced method is 0.937191\n",
+        " (3).The compressibility factor of mixture using Redlich Kwong equation of state is 0.927540\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch3_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch3_2.ipynb
new file mode 100755
index 00000000..6e43a54d
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch3_2.ipynb
@@ -0,0 +1,1524 @@
+{
+ "metadata": {
+  "name": "ch3_2",
+  "signature": "sha256:8255c3a74c7b52da7fb922c8e6e009ed3e07e5e50a88f1302c38a2c832cbdc1f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : The First Law and Its Applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.1 Page number - 80 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "V_vessel = 4.*10**(-3);\t\t\t#[m**(-3)] - Volume of vessel\n",
+      "T = 200+273.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal fas constant\n",
+      "P = 1.5*10**(6);\t\t\t#[Pa] - Pressure\n",
+      "Q = 40.*1000;\t\t\t#[J] - Heat input\n",
+      "\t\t\t# From steam table at 200 C,Psat=1.55549 MPa,therefore the steam is superheated.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (1)\n",
+      "\t\t\t# Using steam table,at 1.5 MPa and 200 C,\n",
+      "V_1 = 0.1325;\t\t\t#[m**(3)/mol] - Specific volume\n",
+      "U_1 = 2598.1;\t\t\t#[kJ/kg] - Specific internal energy\n",
+      "\t\t\t# From first law under constant pressure,\n",
+      "\t\t\t# Q - m*P*(V2 - V1) = m*(U2 - U1)\n",
+      "m = V_vessel/V_1;\t\t\t#[kg] - Mass of system\n",
+      "\t\t\t# Putting the values,the above equation becomes\n",
+      "\t\t\t# 45283*(V2 - 0.1325) + 30.1887*(U2 - 2598.1) = 40000\n",
+      "\t\t\t# From steam table at 700 C LHS is 33917.0 and at 800 C,it is 40925.3.\n",
+      "\t\t\t# Therefore the final temperature lies between 700 and 800 C\n",
+      "print \" 1.From steam table the final temperature lies between 700 and 800 C\";\n",
+      "\n",
+      "\t\t\t# Alternate method\n",
+      "\t\t\t# Here we use first law at constant pressure,\n",
+      "\t\t\t# Q = m*(H_2 - H_1)\n",
+      "H_1 = 2796.8;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Substituting the values,\n",
+      "\t\t\t# 40 = 0.0301887*(H_2 - 2796.8)\n",
+      "H_2 = (40/0.0301887) + 2796.9;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Threfore,final enthalpy is (H2) 4121.8 [kJ/kg] and pressure is 1.5 [MPa].\n",
+      "\t\t\t# From steam table at 1.5 [MPa]and 700 C,enthalpy is 3920.3 [kj/kg] and at 1.5 [MPa]and 800 C,enthalpy is 4152.6 [kj/kg]\n",
+      "print \"\\tAlternate method\";\n",
+      "print \"\\tBy linear interpolation we get the temperature at which enthlpy is 4121.8 kJ/kg to be 786.74 C\";\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "\t\t\t# Assuming ideal behaviour.\n",
+      "n = (P*V_vessel)/(R*T);\t\t\t#[mol] - No of moles\n",
+      "M = 18.015;\t\t\t# Molecular weight of water\n",
+      "m_2 = n*M;\t\t\t#[g] - mass of moles\n",
+      "Cp_1 = 7.7 + 0.04594*10**(-2)*T + 0.2521*10**(-5)*T**(2) - 0.8587*10**(-9)*T**(3);\t\t\t#[cal/mol*K] - Heat capacity at constant presure\n",
+      "R0 = 1.987;\t\t\t#[cal/mol*K] - universal gas constant\n",
+      "Cv_1 = Cp_1 - R0;\t\t\t#[cal/mol*K] - Heat capacity at constant volume\n",
+      "Cv_1 = Cv_1*4.184/M;\t\t\t#[J/g*K]\n",
+      "T1 = T;\n",
+      "\t\t\t# From 1st law energy balance for constant pressure, we have Q-W= m*(delta_U)\n",
+      "\t\t\t# Q = P*(V2 - V1)*m = m*Cv*(T2 - T1)\n",
+      "\t\t\t# Q = P*((T2/T1)-1)*V1*m = m*Cv*(T2-T1)\n",
+      "\t\t\t# But, (V1*Cv)=initial total volume of system = V_vessel\n",
+      "\t\t\t# Q-P((T2/T1)-1)*V_vessel = m_2*Cv_0*(T2-T1);\n",
+      "def f(T2): \n",
+      "\t return Q-P*((T2/T1)-1)*V_vessel-m_2*Cv_1*(T2-T1)\n",
+      "T2_1 = fsolve(f,1)\n",
+      "\t\t\t#The heat capacity should be evaluted at mean temperature\n",
+      "T_mean = (T1 + T2_1)/2;\n",
+      "Cp_2 = 7.7 + 0.04594*10**(-2)*T_mean+0.2521*10**(-5)*T_mean**(2) - 0.8587*10**(-9)*T_mean**(3);\t\t\t#[cal/mol*K] - Heat capacity at constant presure\n",
+      "Cv_2 = Cp_2-R0;\t\t\t#[cal/mol*K] - Heat capacity at constant volume\n",
+      "Cv_2 = Cv_2*4.184/M;\t\t\t#[J/g*K]\n",
+      "\t\t\t#Now again solving the equation Q=P*((T2/T1)-1)*V1*m = m*Cv*(T2-T1),for Cv=Cv_2\n",
+      "def f1(T2): \n",
+      "\t return Q-P*((T2/T1)-1)*V_vessel-m_2*Cv_2*(T2-T1)\n",
+      "T2_2 = fsolve(f1,1)\n",
+      "print \" 2).The temperature assuming ideal behaviour is %f K\"%(T2_2);\n",
+      "\n",
+      "\t\t\t# Alternate method\n",
+      "\t\t\t# From 1st law at constant pressure, we have Q = m*Cp(T2-T1)\n",
+      "T2_3 = Q/(m_2*(Cp_1*4.184/M))+T1;\n",
+      "\t\t\t#We can calculate the mean temperature as done above\n",
+      "T_mean1 = (T1 + T2_3)/2;\t\t\t#[J/g*K]\n",
+      "\t\t\t#The heat capacity should be evaluted at mean temperature\n",
+      "Cp_3 = 7.7 + 0.04594*10**(-2)*T_mean1 + 0.2521*10**(-5)*T_mean1**(2)-0.8587*10**(-9)*T_mean1**(3);\t\t\t#[cal/mol*K] - Heat capacity at constant presure\n",
+      "Cp_3 = Cp_3*4.184/M;\t\t\t#[J/g*K]\n",
+      "\t\t\t# Again solving the equation Q = m*Cp(T2-T1), for Cp=Cp_3\n",
+      "T2_4 = Q/(m_2*Cp_3) + T1;\n",
+      "print \"\\tAlternate method\";\n",
+      "print \"\\tThe temperature assuming ideal behaviour alternate method) is %f K\"%(T2_4);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1.From steam table the final temperature lies between 700 and 800 C\n",
+        "\tAlternate method\n",
+        "\tBy linear interpolation we get the temperature at which enthlpy is 4121.8 kJ/kg to be 786.74 C\n",
+        " 2).The temperature assuming ideal behaviour is 1141.732355 K\n",
+        "\tAlternate method\n",
+        "\tThe temperature assuming ideal behaviour alternate method) is 1141.738180 K\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "/usr/lib/python2.7/dist-packages/scipy/optimize/minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last ten iterations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.2 Page number - 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "V_tank = 1;\t\t\t#[m**(3)] - Volume of the tank\n",
+      "V_liq = 0.05;\t\t\t#[m**(3)] - Volume of saturated water\n",
+      "V_vap = 0.95;\t\t\t#[m**(3)] - Volume of saturated vapour\n",
+      "P = 1;\t\t\t#[bar] - Pressure\n",
+      "V_liq_sat = 0.001043;\t\t\t#[m**(3)/kg] - Specific volume of saturated water\n",
+      "V_vap_sat = 1.6940;\t\t\t#[m**(3)/kg] - Specific volume of saturated vapour\n",
+      "U_liq_sat = 417.4;\t\t\t#[kJ/kg] - Saturated liquid internal energy\n",
+      "U_vap_sat = 2506.1;\t\t\t#[kJ/kg] - Saturated vapour internal energy\n",
+      "\n",
+      "# Calculations\n",
+      "m = (V_liq/V_liq_sat) + (V_vap/V_vap_sat);\t\t\t#[kg] - Total mass of water\n",
+      "U_1 = (V_liq/V_liq_sat)*U_liq_sat + (V_vap/V_vap_sat)*U_vap_sat;\t\t\t#[kJ] - Total internal energy\n",
+      "\n",
+      "\t\t\t# At final state,which is saturated vapour\n",
+      "V = V_tank/m;\t\t\t#[m**(3)/kg] - Molar volume\n",
+      "\t\t\t# From saturated steam table at 8 MPa,as reported in the book V_vap = 0.02352[m**(3)/kg] and U_vap = 2569.8[kJ/kg]\n",
+      "\t\t\t# At 9 MPa, Vv = 0.02048[m**(3)/kg] and Uv = 2557.8[kJ/kg]\n",
+      "\t\t\t# Therefore final state pressure of the system (from interpolation) is 8.954 [MPa] and internal energy of saturated vapour is 2558.35 [kJ/kg]\n",
+      "U_2 = m*2558.35;\t\t\t#[kJ] - Final total internal energy\n",
+      "del_Ut = U_2 - U_1;\t\t\t#[kJ] \n",
+      "\t\t\t#we have, del_U = Q - W\n",
+      "\t\t\t#Here work done is zero because volume is rigid.\n",
+      "Q = del_Ut;\t\t\t#[kJ]\n",
+      "Q = del_Ut*10**(-3);\t\t\t#[MJ]\n",
+      "\n",
+      "# Results\n",
+      "print \" The amount of heat to be added is %f MJ\"%( Q);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The amount of heat to be added is 102.663530 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page number - 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "M_vap_sat = 0.22;\t\t\t#[kg] - mass of saturated vapour\n",
+      "M_liq_sat = 1.78;\t\t\t#[kg] - mass of saturated liquid\n",
+      "P = 700;\t\t\t#[kPa] - Pressure\n",
+      "\n",
+      "#At P=700 kPa,the systen is saturated,from steam table as reported in the book\n",
+      "T_sat1 = 164.97;\t\t\t#[C]\n",
+      "V_liq_1 = 0.001108;\t\t\t#[m**(3)/kg] \n",
+      "V_vap_1 = 0.2729;\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Vt_1 = V_liq_1*M_liq_sat + V_vap_1*M_vap_sat;\t\t\t#[m**(3)] - total volume\n",
+      "\n",
+      "\t\t\t#At final state,P = 8 MPa\n",
+      "T_sat2 = 295.06;\t\t\t#[C]\n",
+      "V_liq_2 = 0.001384;\t\t\t#[m**(3)/kg] \n",
+      "V_vap_2=0.02352;\t\t\t#[m**(3)/kg]\n",
+      "Vt_2 = Vt_1;\t\t\t# Since the volume is rigid.\n",
+      "\t\t\t# Since the volume of 2 kg of vapour is 0.062 [m**(3)]\n",
+      "V = Vt_2/2;\t\t\t#[m**(3)/kg] - specific volume\n",
+      "\n",
+      "\t\t\t# (a)\n",
+      "\t\t\t# From steam table at 8 [MPa]and 350 [C],V=0.02995[m**(3)/kg]; \n",
+      "V_1 = 0.02995;\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t# And at 8 [MPa]and 400 [C],\n",
+      "V_2 = 0.03432;\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t# By interpolation,\n",
+      "T = ((V-V_1)/(V_2 - V_1))*(400-350)+350;\n",
+      "print \" a).The final temperature is %f c\"%(T);\n",
+      "\n",
+      "\t\t\t# (b)\n",
+      "\t\t\t# From steam table \n",
+      "U_1 = 2747.7;\t\t\t#[kJ/kg]\n",
+      "H_1 = 2987.3;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# And at 8 [MPa]and 400 C,\n",
+      "U_2 = 2863.8;\t\t\t#[kJ/kg]\n",
+      "H_2 = 3138.3;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Therefore at T = 362.01 C\n",
+      "U = U_1+((U_2 - U_1)/(400 - 350))*(T - 350);\n",
+      "print \" b).The internal energy is %f kJ/kg\"%(U);\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "H = H_1+((H_2 - H_1)/(400 - 350))*(T - 350);\n",
+      "print \" b).The enthalpy is %f kJ/kg\"%(H);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The final temperature is 362.072311 c\n",
+        " b).The internal energy is 2775.731907 kJ/kg\n",
+        " b).The enthalpy is 3023.758380 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.4 Page number - 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "T = 300;\t\t\t#[K] - Temperature\n",
+      "P1 = 1;\t\t\t#[bar] - Initial pressure\n",
+      "P1 = P1*10**(5);\t\t\t#[N/m**(2)]\n",
+      "P2 = 8;\t\t\t#[bar] - Final pressure\n",
+      "P2 = P2*10**(5);\t\t\t#[N/m**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "Tc = 126.2;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 34;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[N/m**(2)]\n",
+      "w = 0.038;\t\t\t#  Acentric factor\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# w = integral(Pdv)\n",
+      "\t\t\t# Z = 1 + (B/V)\n",
+      "\t\t\t# (P*V)/(R*T) = 1 + (B/V)\n",
+      "\t\t\t# P = (R*T)/V + (B*R*T)/V**(2)\n",
+      "\n",
+      "def f29(V): \n",
+      "\t return (R*T/V) + (B*R*T)/V**(2)\n",
+      "\n",
+      "\t\t\t# w =  quad(f29,V1,V2)[0]\n",
+      "\n",
+      "\t\t\t# Under isothermal conditions,\n",
+      "\t\t\t# w = R*T*math.log(V2/V1) - B*R*T*((1/V2) - (1/V1));\n",
+      "\n",
+      "Tr = T/Tc;\n",
+      "B_0 = 0.083 - (0.422/(Tr)**(1.6));\n",
+      "B_1 = 0.139 - (0.172/(Tr)**(4.2));\n",
+      "B = ((B_0+(w*B_1))*(R*Tc))/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t# Now we have to calculate molar volume i.e V1 and V2 at given conditions\n",
+      "\t\t\t# At state 1,\n",
+      "def f(V): \n",
+      "\t return V**(2)-(R*T/P1)*V-(B*R*T)/P1\n",
+      "V_1 = fsolve(f,-1)\n",
+      "V_2 = fsolve(f,1)\n",
+      "\t\t\t# We will take root near to (R*T)/P, i.e V_2\n",
+      "V1 = V_2;\n",
+      "\n",
+      "\t\t\t# At state 2,\n",
+      "def f1(V): \n",
+      "\t return V**(2)-(R*T/P2)*V-(B*R*T)/P2\n",
+      "V_3=fsolve(f1,-1)\n",
+      "V_4=fsolve(f1,1)\n",
+      "V2 = V_4;\n",
+      "\t\t\t# The work done is thus,\n",
+      "w = R*T*math.log(V2/V1) - B*R*T*((1/V2) - (1/V1));\t\t\t#[J]\n",
+      "w = w*10**(-3);\t\t\t#[kJ]\n",
+      "\n",
+      "print \" The work done is %f kJ/mol\"%(w);\n",
+      "print \" Negative sign indicates that work is done on the gas\";\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The work done is -5.186547 kJ/mol\n",
+        " Negative sign indicates that work is done on the gas\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.5 Page number - 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "\n",
+      "T = 300;\t\t\t#[K] - Temperature\n",
+      "P1 = 1;\t\t\t#[bar] - Initial pressure\n",
+      "P1 = P1*10**(5);\t\t\t#[N/m**(2)]\n",
+      "P2 = 8;\t\t\t#[bar] - Final pressure\n",
+      "P2 = P2*10**(5);\t\t\t#[N/m**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "y1 = 0.21;\t\t\t# Mole fraction of component 1 (oxygen)\n",
+      "y2 = 0.79;\t\t\t# Mole fraction of component 1 (nitroen)\n",
+      "\n",
+      "# For component 1 (Oxygen)\n",
+      "Tc_1 = 154.6;\t\t\t#[K]\n",
+      "Pc_1 = 50.43*10**(5);\t\t\t#[N/m**(2)]\n",
+      "Vc_1 = 73.4;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_1 = 0.288;\n",
+      "w_1 = 0.022;\n",
+      "\n",
+      "\t\t\t#For component 2 (Nitrogen)\n",
+      "Tc_2 = 126.2;\t\t\t#[K]\n",
+      "Pc_2 = 34*10**(5);\t\t\t#[N/m**(2)]\n",
+      "Vc_2 = 89.2;\t\t\t#[cm**(3)/mol]\n",
+      "Zc_2 = 0.289;\n",
+      "w_2 = 0.038;\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#For component 1\n",
+      "Tr_1 = T/Tc_1;\t\t\t#Reduced temperature\n",
+      "\t\t\t#At reduced temperature\n",
+      "B1_0 = 0.083 - (0.422/(Tr_1)**(1.6));\n",
+      "B1_1 = 0.139 - (0.172/(Tr_1)**(4.2));\n",
+      "\t\t\t# We know,(B*Pc)/(R*Tc) = B_0+(w*B_1)\n",
+      "B_11 = ((B1_0+(w_1*B1_1))*(R*Tc_1))/Pc_1;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t# Similarly for component 2\n",
+      "Tr_2 = T/Tc_2;\t\t\t#Reduced temperature\n",
+      "\t\t\t# At reduced temperature Tr_2,\n",
+      "B2_0 = 0.083 - (0.422/(Tr_2)**(1.6));\n",
+      "B2_1 = 0.139 - (0.172/(Tr_2)**(4.2));\n",
+      "B_22 = ((B2_0 + (w_2*B2_1))*(R*Tc_2))/Pc_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t#For cross coeffcient\n",
+      "Tc_12 = (Tc_1*Tc_2)**(1/2);\t\t\t#[K]\n",
+      "w_12 = (w_1 + w_2)/2;\n",
+      "Zc_12 = (Zc_1+Zc_2)/2;\n",
+      "Vc_12 = (((Vc_1)**(1/3)+(Vc_2)**(1/3))/2)**(3);\t\t\t#[cm**(3)/mol]\n",
+      "Vc_12 = Vc_12*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "Pc_12 = (Zc_12*R*Tc_12)/Vc_12;\t\t\t#[N/m**(2)]\n",
+      "\n",
+      "\t\t\t# Now we have,(B_12*Pc_12)/(R*Tc_12) = B_0 + (w_12*B_1)\n",
+      "\t\t\t# where B_0 and B_1 are to be evaluated at Tr_12\n",
+      "Tr_12 = T/Tc_12;\n",
+      "\t\t\t# At reduced temperature Tr_12\n",
+      "B_0 = 0.083 - (0.422/(Tr_12)**(1.6));\n",
+      "B_1 = 0.139 - (0.172/(Tr_12)**(4.2));\n",
+      "B_12 = ((B_0+(w_12*B_1))*R*Tc_12)/Pc_12;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t# For the mixture\n",
+      "B = y1**(2)*B_11 + 2*y1*y2*B_12+y2**(2)*B_22;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# Now we have to calculate molar volume i.eV1 and V2 at given conditions\n",
+      "\n",
+      "\t\t\t# At state 1,\n",
+      "def f(V): \n",
+      "\t return V**(2)-(R*T/P1)*V-(B*R*T)/P1\n",
+      "V_1=fsolve(f,-1)\n",
+      "V_2=fsolve(f,1)\n",
+      "\t\t\t# We will take root near to (R*T)/P, i.e V_2\n",
+      "V1 = V_2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t# At state 2,\n",
+      "def f1(V): \n",
+      "\t return V**(2)-(R*T/P2)*V-(B*R*T)/P2\n",
+      "V_3=fsolve(f1,-1)\n",
+      "V_4=fsolve(f1,1)\n",
+      "V2 = V_4;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\t\t\t# Work done per mole of air is given by, w=integral(Pdv)\n",
+      "\t\t\t# Z = 1 + (B/V)\n",
+      "\t\t\t# (P*V)/(R*T) = 1 +( B/V)\n",
+      "\t\t\t# P = (R*T)/V+(B*R*T)/V**(2)\n",
+      "\n",
+      "def f43(V): \n",
+      "\t return (R*T/V)+(B*R*T)/V**(2)\n",
+      "\n",
+      "\t\t\t# w =  quad(f43,V1,V2)[0]\n",
+      "\n",
+      "\t\t\t# Under isothermal conditions,\n",
+      "w = R*T*math.log(V2/V1)-B*R*T*((1/V2)-(1/V1));\n",
+      "w = w*10**(-3);\t\t\t#[kJ/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The work done is %f kJ/mol\"%(w);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The work done is -5.186545 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.6 Page number - 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 125+273.15;\t\t\t#[K] - Temperature\n",
+      "P1 = 1;\t\t\t#[bar] - Initial pressure\n",
+      "P1 = P1*10**(5);\t\t\t#[N/m**(2)]\n",
+      "P2 = 60;\t\t\t#[bar] - Final pressure\n",
+      "P2 = P2*10**(5);\t\t\t#[N/m**(2)]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "Tc = 416.3;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 66.80*10**(5);\t\t\t#[N/m**(2)] - Critical pressure\n",
+      "\n",
+      "\t\t\t# (1)\n",
+      "\t\t\t# Virial equation of state, Z = 1 + (B/V)+(C/V**(2))\n",
+      "\t\t\t# (P*V)/(R*T) = 1 + (B/V)+(C/V**(2))\n",
+      "\t\t\t# P = (R*T)/V+(B*R*T)/V**(2)+(C*R*T)/V**(3)\n",
+      "\t\t\t# w = integral(PdV)=R*T*math.log(V2/V1)-(B*R*T)*(1/V2-1/V1)-(C*R*T/2)*(1/V2**(2)-1/V1**(2))\n",
+      "\n",
+      "B = -207.5;\t\t\t#[cm**(3)/mol] - Second virial coefficient\n",
+      "B = -207.5*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "C = 18200;\t\t\t#[cm**(6)/mol**(2)] - Third virial coefficient\n",
+      "C = 18200*10**(-12);\t\t\t#[m**(6)/mol**(2)]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# We need to calculate molar volume at state 1 and 2,\n",
+      "\t\t\t# At state 1,P = P1,\n",
+      "\t\t\t# V**(3)-(R*T/P)*V**(2)-(B*R*T/P)*V-(C*R*T/P)=0\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f1(V): \n",
+      "\t return V**(3)-(R*T/P1)*V**(2)-(B*R*T/P1)*V-(C*R*T/P1)\n",
+      "V_1=fsolve(f1,-1)\n",
+      "V_2=fsolve(f1,0)\n",
+      "V_3=fsolve(f1,10)\n",
+      "\t\t\t# The cubic equation has only 1 real root,other two roots are imaginary.\n",
+      "V1 = V_3;\n",
+      "\n",
+      "\t\t\t# Similarly at state 2,P=P2\n",
+      "\t\t\t# V**(3) - (R*T/P)*V**(2) - (B*R*T/P)*V - (C*R*T/P) = 0\n",
+      "\t\t\t# Solving the cubic equation\n",
+      "def f2(V): \n",
+      "\t return V**(3)-(R*T/P2)*V**(2)-(B*R*T/P2)*V-(C*R*T/P2)\n",
+      "V_4=fsolve(f2,-1)\n",
+      "V_5=fsolve(f2,0)\n",
+      "V_6=fsolve(f2,1)\n",
+      "V2 = V_6;\n",
+      "\t\t\t# Finally work done is given by,\n",
+      "w1 = R*T*math.log(V2/V1)-(B*R*T)*(1/V2-1/V1)-(C*R*T/2)*(1/V2**(2)-1/V1**(2));\t\t\t#[J/mol]\n",
+      "w1 = w1*10**(-3);\t\t\t#[kJ/mol]\n",
+      "print \" 1).The work done using given virial equation of state is %f kJ/mol\"%(w1);\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "\t\t\t# Virial equation of state, Z = 1+(B*P)/(R*T)+((C-B**(2))/(R*T)**(2))*P**(2)\n",
+      "\t\t\t# (P*V)/(R*T)= 1+(B*P)/(R*T)+((C-B**(2))/(R*T)**(2))*P**(2)\n",
+      "\t\t\t# V = (R*T)/P+B+((C-B**(2))/(R*T))*P\n",
+      "\t\t\t# Differentiating both sides by P and integrating we get,\n",
+      "\t\t\t# w = integral(PdV)=-(R*T)*math.log(P2/P1)+((C-B**(2))/(2*R*T))*(P2**(2)-P1**(2))\n",
+      "w2 = -(R*T)*math.log(P2/P1) + ((C-B**(2))/(2*R*T))*(P2**(2)-P1**(2));\t\t\t#[J/mol]\n",
+      "w2 = w2*10**(-3);\t\t\t#[kJ/mol]\n",
+      "print \" 2).The work done using given virial equation of state is %f kJ/mol\"%(w2);\n",
+      "\n",
+      "\t\t\t# (3)\n",
+      "\t\t\t# Van der Walls equation of state is given by,\n",
+      "a = (27*(R**(2))*(Tc**(2)))/(64*Pc);\t\t\t#[Pa*m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc);\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# P = ((R*T)/(V-b))-a/(V**(2));\t\t\t#[N/m**(2)]\n",
+      "\t\t\t# w = integral(PdV)=R*T*math.log((V2-b)/(V1-a))+a*(1/V2-1/V1)\n",
+      "\t\t\t# The cubic form of van der Walls equation of state is given by,\n",
+      "\t\t\t# V**(3) - (b+(R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0\n",
+      "\t\t\t# Solving the cubic equation for P=P1\n",
+      "def f3(V): \n",
+      "\t return V**(3)-(b+(R*T)/P1)*V**(2)+(a/P1)*V-(a*b)/P1\n",
+      "V2_1=fsolve(f3,1)\n",
+      "V2_2=fsolve(f3,10)\n",
+      "V2_3=fsolve(f3,100)\n",
+      "\t\t\t# The above equation has 1 real and 2 imaginary roots. We consider only real root (V2_3).\n",
+      "\n",
+      "\t\t\t# Similarly at state 2,for P=P2,\n",
+      "def f4(V): \n",
+      "\t return V**(3)-(b+(R*T)/P2)*V**(2)+(a/P2)*V-(a*b)/P2\n",
+      "V2_4=fsolve(f4,1)\n",
+      "V2_5=fsolve(f4,10)\n",
+      "V2_6=fsolve(f4,100)\n",
+      "\t\t\t# The above equation has 1 real and 2 imaginary roots. We consider only real root (V2_6).\n",
+      "\t\t\t# Finally work done is given by\n",
+      "w3 = R*T*math.log((V2_6-b)/(V2_3-b))+a*(1/V2_6-1/V2_3);\t\t\t#[J/mol]\n",
+      "w3 = w3*10**(-3);\t\t\t#[kJ/mol]\n",
+      "print \" 3).The work done using  van der Walls equation of state is %f kJ/mol\"%(w3);\n",
+      "\n",
+      "\t\t\t# (4)\n",
+      "\t\t\t# Redlich Kwong equation of state,\n",
+      "a_1 = (0.42748*(R**(2))*(Tc**(2.5)))/Pc;\t\t\t#[Pa*m**(6)*K**(1/2)/mol]\n",
+      "b_1 = (0.08664*R*Tc)/Pc;\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# P = ((R*T)/(V-b_1))-(a_1/(T**(1/2)*V*(V+b_1)));\t\t\t#[N/m**(2)]\n",
+      "\t\t\t# Work done is given by\n",
+      "\n",
+      "def f4(V): \n",
+      "\t return (V2-b)/(V1-b)-a/T**(1./2)*integrate(1/V*(V+b))\n",
+      "\n",
+      "\t\t\t# w = R*T*math.log((V2-b)/(V1-b))-a/T**(1/2)* quad(f4,V1,V2)[0]\n",
+      "\n",
+      "\t\t\t# Using the factorization 1/(V*(V+b))=(1/b)*((1/V)-(1/V+b)),we get\n",
+      "\t\t\t# w = R*T*math.log((V2-b)/(V1-b))-(a/(b*T**(1/2)))*(math.log(V2/V1)-math.log((V2+b)/(V1+b))\n",
+      "\t\t\t# Now we have calculate V1 and V2,\n",
+      "\t\t\t# The cubic form of Redlich Kwong equation of state is given by,\n",
+      "\t\t\t# V**(3) - ((R*T)/P)*V**(2) - ((b_1**(2)) + ((b_1*R*T)/P) - (a/(T**(1/2)*P))*V - (a*b)/(T**(1/2)*P) = 0\n",
+      "\t\t\t# Solving the cubic equation at state 1,\n",
+      "def f5(V): \n",
+      "\t return V**(3)-((R*T)/P1)*V**(2)-((b_1**(2))+((b_1*R*T)/P1)-(a_1/(T**(1./2)*P1)))*V-(a_1*b_1)/(T**(1./2)*P1)\n",
+      "V3_1=fsolve(f5,1)\n",
+      "V3_2=fsolve(f5,10)\n",
+      "V3_3=fsolve(f5,100)\n",
+      "\t\t\t# The above equation has 1 real and 2 imaginary roots. We consider only real root (V3_3).\n",
+      "\n",
+      "\t\t\t# Similarly at state 2,for P = P2,\n",
+      "def f6(V): \n",
+      "\t return V**(3)-((R*T)/P2)*V**(2)-((b_1**(2))+((b_1*R*T)/P2)-(a_1/(T**(1./2)*P2)))*V-(a_1*b_1)/(T**(1./2)*P2)\n",
+      "V3_4=fsolve(f6,1)\n",
+      "V3_5=fsolve(f6,10)\n",
+      "V3_6=fsolve(f6,100)\n",
+      "\t\t\t# The above equation has 1 real and 2 imaginary roots. We consider only real root (V3_6).\n",
+      "\t\t\t# Finally work done is given by\n",
+      "w4 = R*T*math.log((V3_6-b_1)/(V3_3-b_1))-(a_1/(b_1*T**(1./2)))*(math.log(V3_6/V3_3)-math.log((V3_6+b_1)/(V3_3+b_1)));\t\t\t#[J/mol]\n",
+      "w4 = w4*10**(-3);\t\t\t#[kJ/mol]\n",
+      "print \" 3).The work done using Redlich Kwong equation of state is %f kJ/mol\"%(w4);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The work done using given virial equation of state is -13.848149 kJ/mol\n",
+        " 2).The work done using given virial equation of state is -13.688301 kJ/mol\n",
+        " 3).The work done using  van der Walls equation of state is -14.802420 kJ/mol\n",
+        " 3).The work done using Redlich Kwong equation of state is -14.704965 kJ/mol\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "/usr/lib/python2.7/dist-packages/scipy/optimize/minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last five Jacobian evaluations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.8 Page number - 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 20 + 273.15;\t\t\t#[K] - Temperature\n",
+      "P_1 = 140.;\t\t\t#[kPa] - Initial pressure\n",
+      "P_1 = P_1*10.**(3);\t\t\t#[Pa]\n",
+      "P_2 = 560.;\t\t\t#[kPa] - Final pressure\n",
+      "P_2 = P_2*10.**(3);\t\t\t#[Pa]\n",
+      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# Cp_0 = 1.648+4.124*10**(-2)*T - 1.53*10**(-5)*T**(2) + 1.74*10**(-9)*T**(3)\n",
+      "\t\t\t# Using adiabatic compression, P*V**(Y)=constant. For ideal gases\n",
+      "\t\t\t# P*(R*T/P)**(Y) = constant\n",
+      "\t\t\t# P**(1-Y)*T**(Y) = constant or,P1**(1-Y)*T1**(Y)=P2**(1-Y)*T2**(Y)\n",
+      "\t\t\t# Now,at state 1, i.e at T=20[C]\n",
+      "Cp_1 = 1.648+4.124*10**(-2)*T-1.53*10**(-5)*T**(2)+1.74*10**(-9)*T**(3);\t\t\t#[cal/mol*K] - Heat capacity at constant pressure\n",
+      "Cv_1 = Cp_1 - R;\t\t\t#[cal/mol*K] - Heat capacity at constant volume\n",
+      "Y1 = Cp_1/Cv_1;\t\t\t# Ratio of heat capacities\n",
+      "\n",
+      "\t\t\t# Now calculatung the temperature at state 2 (T2)\n",
+      "\t\t\t# (T2/T1)=(P1/P2)**((1-Y1)/Y1)\n",
+      "T_1 = T;\n",
+      "T_2 = ((P_1/P_2)**((1-Y1)/Y1))*T_1;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# Now calculating the mean temperature\n",
+      "T_mean = (T_1 + T_2)/2;\t\t\t#[K]\n",
+      "\t\t\t# At mean temperature\n",
+      "Cp_2 = 1.648+4.124*10**(-2)*T_mean - 1.53*10**(-5)*T_mean**(2) + 1.74*10**(-9)*T_mean**(3);\t\t\t#[cal/mol*K] - Heat capacity at constant pressure\n",
+      "Cv_2 = Cp_2 - R;\t\t\t#[cal/mol*K] - Heat capacity at constant volume\n",
+      "Y2 = Cp_2/Cv_2;\n",
+      "\n",
+      "\t\t\t# Calculating exit temperature\n",
+      "\t\t\t# Again using the realation,(T2/T1)=(P1/P2)**((1-Y1)/Y1)\n",
+      "T_exit = ((P_1/P_2)**((1-Y2)/Y2))*T_1;\t\t\t#[K]\n",
+      "\t\t\t# Since value of mean temperature has not changed much the molar heat capacity ratio can be assumed to be same.Therefore\n",
+      "\t\t\t# w = -delta(U)=Cv_0*(T2-T1)\n",
+      "w = Cv_2*(T_1 - T_exit);\t\t\t#[cal/mol]\n",
+      "w = w*4.184;\t\t\t#[J/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" The work done for adiabatic compression is %f J/mol\"%(w);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The work done for adiabatic compression is -3198.427072 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.9 Page number - 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "m_ice = 1000;\t\t\t#[g] - Mass of ice\n",
+      "m_water = 1000;\t\t\t#[g] - Mass of water\n",
+      "T_ice = 273.15;\t\t\t#[K] - Temperature of ice\n",
+      "T_water = 373.15;\t\t\t#[K] - Temperature of water\n",
+      "L = 79.71;\t\t\t#[cal/g] - Latent heat of melting of ice.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "Cp_1 = 1;\t\t\t#[cal/g-K] - Heat capacity at constant pressure\n",
+      "\t\t\t# Let the final temperature be T\n",
+      "\t\t\t# We assume that all of the ice melts.Energy taken up by ice is\n",
+      "\t\t\t# E1 = L*m_ice + m_ice*Cp_1*(T - T_ice)\n",
+      "\t\t\t# Energy given by hot water is,\n",
+      "\t\t\t# E2 = m_water*Cp_1*(T_water - T)\n",
+      "\t\t\t# No heat exchange with surrounding.Solving for T\n",
+      "T_1 = (m_ice*Cp_1*T_ice + m_water*Cp_1*T_water - L*m_ice)/(m_ice*Cp_1 + m_water*Cp_1);\t\t\t#[K]\n",
+      "T_1 = T_1 - 273.15;\t\t\t#[C]\n",
+      "\n",
+      "print \" 1).The final temperature taking Cp_water = 1 cal/g-K) is %f C\"%(T_1);\n",
+      "\t\t\t#Since the final temperature is greater than 273.15 K,so our assumption that all of ice melts is correct\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "\t\t\t# Cp_2 = 1.00874-0.7067*10**(-3)*T+15.93*10**(-6)*T**(2)-83.8*10**(-9)*T**(3);\n",
+      "\n",
+      "def f15(T): \n",
+      "\t return Cp_2\n",
+      "\n",
+      "\t\t\t# From energy balance,we get L*m_ice + m_ice* quad(f15,0,T) + m_water*integrate(Cp_2)[0]\n",
+      "\n",
+      "\t\t\t# On putting the values and then simplifying we get\n",
+      "\t\t\t# 2.01748*T - 0.0007067*T**(2) + 1.062*10**(-5)*T**(3) - 4.19*10**(-8)*T**(4) - 20.8455 = 0\n",
+      "\t\t\t# Solving the above equation we get\n",
+      "def f1(T): \n",
+      "\t return  2.01748*T - 0.0007067*T**(2) + 1.062*10**(-5)*T**(3) - 4.19*10**(-8)*T**(4) - 20.8455\n",
+      "T_0 = fsolve(f1,1)\n",
+      "print \" 2).The final temperature using specific heat capacity equation is %f C\"%(T_0);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The final temperature taking Cp_water = 1 cal/g-K) is 10.145000 C\n",
+        " 2).The final temperature using specific heat capacity equation is 10.364452 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.11 Page number - 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "n = 1.5;\t\t\t# - ratio of heat capacities\n",
+      "T_1 = 500.;\t\t\t#[K] - Initial temperature\n",
+      "T_2 = 1000.;\t\t\t#[K] - Final temperature\n",
+      "P_1 = 1.;\t\t\t#[bar] - Initial pressure\n",
+      "P_1 = P_1*10.**(5);\t\t\t#[Pa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The compression path is given by, P*V**(1.5) = constant\n",
+      "\t\t\t# P*(R*T/P)**(1.5) = constant\n",
+      "\t\t\t# P1**(-0.5)*T1**(1.5) = P2**(-0.5)*T2**(1.5)\n",
+      "P_2 = P_1*(T_1/T_2)**(-3);\t\t\t#[Pa]\n",
+      "P_2_final = P_2*10**(-5);\t\t\t#[bar] - Final pressure in bar\n",
+      "print \" The final pressure is %f bar\"%(P_2_final);\n",
+      "\n",
+      "\t\t\t# From first law q - w = delta(U). \n",
+      "\t\t\t# First w and delta(U) are calculated and thereafter heat exchange is determined.\n",
+      "V_1 = R*T_1/P_1;\t\t\t#[m**(3)/mol] - Initial volume\n",
+      "V_2 = R*T_2/P_2;\t\t\t#[m**(3)/mol] - Final volume\n",
+      "w = ((P_1*V_1)/(n - 1))*(1 - (P_2/P_1)**(1 - 1/n));\t\t\t#[J/mol] - work done\n",
+      "\n",
+      "\t\t\t# Mean temperature is given by,\n",
+      "T_mean = (T_1 + T_2)/2;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t#Now, heat capacity at T_mean is given by,\n",
+      "Cp_0 = R*(3.3 + 0.63*10**(-3)*T_mean);\t\t\t#[J/mol*K]\n",
+      "Cv_0 = Cp_0 - R;\t\t\t#[J/mol*K]\n",
+      "\t\t\t#Therefore delta(U) is given by\n",
+      "del_U = Cv_0*(T_2 - T_1);\t\t\t#[J/mol] - Change in internal energy\n",
+      "q = w + del_U;\t\t\t#[J/mol] - heat change\n",
+      "print \" The amount of heat supplied to the system is %f J/mol\"%(q);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final pressure is 8.000000 bar\n",
+        " The amount of heat supplied to the system is 3211.282500 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.12 Page number - 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 150*10**(3);\t\t\t#[Pa] - Initial pressure\n",
+      "V_1 = 0.001;\t\t\t#[m**(3)] - Initial volume\n",
+      "P_2 = 1000*10**(3);\t\t\t#[Pa] - Final pressure\n",
+      "V_2 = 0.003;\t\t\t#[m**(3)] - Final volume\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# At x = 0, Vt(total volume) = 0.001 m**(3), therefore x = (V_t - V_1)/A;  where A is area of cross section and x is length\n",
+      "\t\t\t# Force exerted b sprig is given by, F = Ps*A = k*x = k*(V_t - V_1)/A\n",
+      "\t\t\t# Ps = (k/A**(2))*(V_t - V_1)\n",
+      "\t\t\t# Total pressure = Initial pressre + Pressre due to spring\n",
+      "\t\t\t# P = P_1 + (k/A**(2))*(V_t - V_1)\n",
+      "\t\t\t# Let (k/A**(2)) = t (say)\n",
+      "\t\t\t# At state 2, i.e at P2 and V_t = V_2.\n",
+      "def f(t): \n",
+      "\t return P_2-P_1 - t*(V_2-V_1)\n",
+      "t = fsolve(f,1000)\n",
+      "\t\t\t# Therefore,pressure is related to total volume as P = P_1-t*(V_t - V_1)\n",
+      "\n",
+      "\t\t\t# (a)\n",
+      "\t\t\t#slope = (k/A**(2))\n",
+      "print \" a).The slope of the line on P-Vt diagram is %e N/m**5)\"%(t);\n",
+      "\n",
+      "\t\t\t# (b)\n",
+      "\t\t\t# Work done by the gas is given by   w=integral(PdVt)\n",
+      "\n",
+      "def f30(V_t): \n",
+      "\t return P_1+t*(V_t-V_1)\n",
+      "\n",
+      "w =  quad(f30,V_1,V_2)[0]\n",
+      "\n",
+      "w = w*10**(-3);\t\t\t#[kJ]\n",
+      "print \" b).The work done by gas is %f kJ\"%(w);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The slope of the line on P-Vt diagram is 4.250000e+08 N/m**5)\n",
+        " b).The work done by gas is 1.150000 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.13 Page number - 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "V = 36;\t\t\t#[L] - Vol of gas on each side\n",
+      "P_1 = 1;\t\t\t#[atm] - pressure on left side of the piston\n",
+      "P_1 = P_1*101325;\t\t\t#[Pa]\n",
+      "T = 273.15;\t\t\t#[K]\n",
+      "P_2 = 3.375;\t\t\t#[atm] - Pressure on right side of the piston\n",
+      "P_2 = P_2*101325;\t\t\t#[Pa]\n",
+      "Y = 1.44;\t\t\t# Ratio of heat capacities\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constnt\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (a)\n",
+      "\t\t\t# For total system, del(U_total) = Q. \n",
+      "\t\t\t\n",
+      "\t\t\t# T_2/T_1 = (P_2/P_1)**((Y - 1)/Y)\n",
+      "T_right = T*(P_2/P_1)**((Y - 1)/Y);\t\t\t#[K]\n",
+      "\n",
+      "Cv_0 = R/(Y-1);\t\t\t#[J/mol*K] - Heat capacity at constant volume.\n",
+      "\t\t\t# Now work done on the gas on right hand side is given by\n",
+      "\t\t\t# W = (P_1*V_1 - P_2*V_2)/(Y - 1) = R*(T_2 - T_1)/(Y - 1) =  Cv_0*(T_1 - T_2)\n",
+      "W_left = Cv_0*(T - T_right);\t\t\t#[J/mol]\n",
+      "\t\t\t# Negative sign for the work done on LHS gas implies work is done on the gas\n",
+      "\n",
+      "\t\t\t# For right hand side of the gas\n",
+      "\t\t\t# P*Vt = n*R*T\n",
+      "n = P_1*(V*10**(-3))/(R*T);\t\t\t# number of moles\n",
+      "W_right = (-W_left)*n;\t\t\t#[J] - We used negative sign for 'W_left' because it is negative in magnitude.\n",
+      "W_right = W_right/1000;\t\t\t#[kJ]\n",
+      "print \" a).Total work done on gas on the right hand side is %f kJ\"%(W_right);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "print \" b).The final temperature of the gas on right side is %f K\"%(T_right);\n",
+      "\n",
+      "\t\t\t# (c)\n",
+      "            # We have (P_left*V_left)/T_left = (P_right*V_right)/T_right.\n",
+      "\t\t\t# Since P_left = P_right, (V_left/T_left) = (V_right/T-right) and also P*V**(Y) = constant.\n",
+      "V_right = V*(P_1/P_2)**(1/Y);\t\t\t#[L] - The total volume on right side \n",
+      "\n",
+      "\t\t\t# The total volume on right side can also be calculated using P2*V2 = n*R*T2.\n",
+      "\t\t\t# Since total volume = 72 [L], therefore volume of left side is\n",
+      "V_left = 2*V - V_right;\t\t\t#[L]\n",
+      "T_left = T_right*(V_left/V_right);\n",
+      "print \" c).Final temperature of the gas on the left side is %f K\"%(T_left);\n",
+      "\n",
+      "\t\t\t#(d)\n",
+      "\t\t\t#The first law applied to the total system (left side and right side) gives.\n",
+      "\t\t\t#Q - W = del(U_left) + del(U_right)\n",
+      "\t\t\t#There is no net work done by the total system as the cylinder is closed at both ends.\n",
+      "Q = n*Cv_0*(T_left-T) + n*Cv_0*(T_right-T);\t\t\t#[J]\n",
+      "Q = Q/1000;\t\t\t#[kJ]\n",
+      "print \" d).Amount of heat added to the gas on the left side is %f kJ\"%(Q);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).Total work done on gas on the right hand side is 3.731958 kJ\n",
+        " b).The final temperature of the gas on right side is 396.112176 K\n",
+        " c).Final temperature of the gas on the left side is 1447.650324 K\n",
+        " d).Amount of heat added to the gas on the left side is 39.378580 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.14 Page number - 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_2 = 0.2;\t\t\t#[bar]\n",
+      "P_2 = P_2*10**(5);\t\t\t#[Pa]\n",
+      "int_dia_2 = 2.4*10**(-2);\t\t\t#[m] - internal diameter at state 2.\n",
+      "Q = 5*10**(-3);\t\t\t#[cubic metre/s] - Flow rate at point 2.\n",
+      "den = 1000;\t\t\t#[kg/cubic metre] - density\n",
+      "delta_z = 1;\t\t\t#[m] - Difference in height\n",
+      "g = 9.81;\t\t\t#[m/s**(2)] - Acceleration due to gravity\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (1)\n",
+      "\n",
+      "\t\t\t# (delta(P)/den) = (P2-P1)/den = P2/den\n",
+      "Vel_2 = Q/(3.14*(int_dia_2/2)**(2));\t\t\t#[m/s] - Velocity of water at state 2.\n",
+      "\t\t\t# Velocity at state 1 i negligible as compared to velocity at state 2,because the diameter of reservoir is very large as compared to diameter of pipe at state 2\n",
+      "\n",
+      "\t\t\t# From bernaulli equation we get,\n",
+      "\t\t\t# -w = (delta(P)/den) + delta(v**(2))/2 + g*delta_z\n",
+      "w = -((P_2/den )+ (Vel_2**(2)/2) + (g*delta_z));\t\t\t#[J/kg]\n",
+      "\t\t\t# w multiplied by m = (den*Q), will give the fluid power.\n",
+      "m = den*Q;\t\t\t#[kg/s]\n",
+      "W_net = m*w;\t\t\t#[Watt]\n",
+      "print \" 1).The fluid power is %f Watt\"%(W_net);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# Total discharge head developed by the pump is given by\n",
+      "\t\t\t# h = (delta(P)/den*g) + (Vel_2**(2)/2*g) + delta_z\n",
+      "h = (P_2/(den*g)) + (Vel_2**(2)/(2*g)) + delta_z;\t\t\t#[m]\n",
+      "print \" 2).Total discharge head developed by the pump is given by h = %f m\"%(h);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The fluid power is -454.750210 Watt\n",
+        " 2).Total discharge head developed by the pump is given by h = 9.271156 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.15 Page number - 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 1000.;\t\t\t#[K] - Temperature at entry\n",
+      "P_1 = 0.6;\t\t\t#[MPa] - Pressure at entry\n",
+      "P_2 = 0.2;\t\t\t#[MPa] - Exit pressure\n",
+      "Vel_1 = 50.;\t\t\t#[m/s] - Entry velocity \n",
+      "Y = 1.4;\t\t\t# Ratio of heat capacities\n",
+      "Mol_wt = 28.;\t\t\t#[g/mol] - Molecular weight of air\n",
+      "Cp = 29.099;\t\t\t#[J/mol-K] - Specific heat capacity at constant pressure\n",
+      "Cp = (Cp/Mol_wt)*1000;\t\t\t#[J/kg-K]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# We know that for a flow process \n",
+      "\t\t\t# delta_H + delta_V**(2)/2 + delta_(g*z) = q - w\n",
+      "\t\t\t# delta_H + delta_V**(2)/2 = 0\n",
+      "\n",
+      "\t\t\t# For a reversible process P*V**(Y) = constant and thus (T_2/T_1) = (P_2/P_1)**((Y-1)/Y)\n",
+      "T_2 = T_1*(P_2/P_1)**((Y-1)/Y);\t\t\t#[K] - Exit temperature\n",
+      "\n",
+      "\t\t\t# delta_H + delta_V**(2)/2 = 0\n",
+      "\t\t\t# Vel_2**(2)/2 - Vel_1**(2)/2 - (H_1 - H_2)= 0\n",
+      "\t\t\t# Vel_2**(2)/2 - Vel_1**(2)/2 - Cp*(T_1 - T_2) = 0\n",
+      "Vel_2_square = 2*(Vel_1**(2.)/2 - Cp*(T_2 - T_1));\t\t\t#[m**(2)/s**(2)]\n",
+      "Vel_2 = (Vel_2_square)**(1./2);\t\t\t#[m/s]\n",
+      "\n",
+      "# Results\n",
+      "print \" The discharge velocity is %f m/s\"%(Vel_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The discharge velocity is 749.965327 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.16 Page number - 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_entry = 10;\t\t\t#[bar] - Pressure at entry\n",
+      "V_entry = 200;\t\t\t#[m/s] - Velocity at entry\n",
+      "P_exit = 1;\t\t\t#[bar] - Vressure at exit\n",
+      "V_exit = 800;\t\t\t#[m/s] - Velocity at exit\n",
+      "g = 9.81;\t\t\t#[m/s**(2)] - Acceleration due to gravity\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#Heat balance gives\n",
+      "\t\t\t# delta_H + (delta_V**(2))/2 + g*delta_z = q - w\n",
+      "\t\t\t#delta_H = q - w - (delta_V**(2))/2 \n",
+      "\t\t\t#From nozzle no work is extracted,therefore\n",
+      "delta_H = -(V_exit**(2)- V_entry**(2))/2;\t\t\t#[J/kg]\n",
+      "delta_H = delta_H*10**(-3);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# Results\n",
+      "print \" The change in enthalpy per kg of steam is %f kJ/kg\"%(delta_H);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in enthalpy per kg of steam is -300.000000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.17 Page number - 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 280;\t\t\t#[K] - Temperature at entry\n",
+      "P_1 = 100;\t\t\t#[kPa] - Pressure at entry\n",
+      "T_2 = 400;\t\t\t#[K] - Temperature at exit\n",
+      "P_2 = 600;\t\t\t#[kPa] - Pressure at exit\n",
+      "m = 0.02;\t\t\t#[kg/s] - Mass flow rate\n",
+      "m = m*10**(3);\t\t\t#[g/s]\n",
+      "heat_loss = 16;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#Cp_0 = 28.11 + 0.1967*10**(-2)*T + 0.4802*10**(-5)*T**(2) - 1.966*10**(-9)*T**(3)\n",
+      "\t\t\t#delta_H = q - w (neglecting kinetic and potential changes)\n",
+      "\t\t\t#delta_H = integral(Cp_0*dT)\n",
+      "\n",
+      "def f22(T): \n",
+      "\t return 28.11 + 0.1967*10**(-2)*T + 0.4802*10**(-5)*T**(2) - 1.966*10**(-9)*T**(3)\n",
+      "\n",
+      "delta_H =  quad(f22,T_1,T_2)[0]\n",
+      "\n",
+      "print \" Change in enthalpy is %f J/mol\"%(delta_H);\n",
+      "\n",
+      "\t\t\t#Molecular weight of air(21 vol% O2 and 79 vol% N2)=(0.21*32)+(0.79*28)=  28.84 g/mol\n",
+      "Mol_wt = 28.84;\t\t\t#[g/mol]\n",
+      "q = - (heat_loss*Mol_wt);\t\t\t#[J/mol]\n",
+      "w = q - delta_H;\t\t\t#[J/mol]\n",
+      "print \" The work done per mole of air is %f J/mol\"%(w);\n",
+      "\t\t\t#the negative sign implies that work is done on the compressor.\n",
+      "\n",
+      "n = m/Mol_wt;\t\t\t#[mol/s] - Mole flow rate\n",
+      "W_net = delta_H*n;\t\t\t#[W]\n",
+      "W_net = -W_net*10**(-3);\t\t\t#[kW]\n",
+      "print \" And the necessary power input to the compressor is %f kW\"%(W_net);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Change in enthalpy is 3511.197066 J/mol\n",
+        " The work done per mole of air is -3972.637066 J/mol\n",
+        " And the necessary power input to the compressor is -2.434949 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.18 Page number - 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 300;\t\t\t#[K] - Temperature at entry\n",
+      "P_1 = 100;\t\t\t#[kPa] - Pressure at entry\n",
+      "P_2 = 900;\t\t\t#[kPa] - Pressure at exit\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (a)\n",
+      "\t\t\t# Reversible adiabatic compression\n",
+      "Y = 1.4;\t\t\t#Ratio of specific heat capacities\n",
+      "\t\t\t# For ideal gas, P*V**(Y)= constant and it takes the form of (T_2/T_1) = (P_2/P_1)**((Y-1)/Y)\n",
+      "T_2 = T_1*(P_2/P_1)**((Y - 1)/Y);\t\t\t#[K]\n",
+      "\t\t\t# The work exchange for adiabatic process is given by\n",
+      "\t\t\t# W_adia = -delta_H = -Cp*(T2-T1) = Cp*(T1-T2) = ((Y*R)/(Y-1))*(T1-T2)\n",
+      "W_adia = ((Y*R)/(Y - 1))*(T_1 - T_2);\t\t\t#[J/mol] -work done\n",
+      "\t\t\t# Molecular weight of air(21 vol% O2 and 79 vol% N2)=(0.21*32)+(0.79*28)=  28.84 g/mol\n",
+      "Mol_wt = 28.84;\t\t\t#[g/mol]\n",
+      "W_adia = W_adia/Mol_wt;\t\t\t#[J/g]\n",
+      "print \" a).The compressor work done for reversible adiabatic compession is %f J/g\"%(W_adia);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\t\t\t#Isothermal compression\n",
+      "\t\t\t#W_iso = -integral(V*dP) = -integral((R*T/P)*dP) = R*T*ln(P_2/P_1)\n",
+      "W_iso = -R*T_1*math.log(P_2/P_1);\t\t\t#[J/mol]\n",
+      "W_iso = W_iso/Mol_wt;\t\t\t#[J/g]\n",
+      "print \" b).The compressor work done for isothermal compession is %f J/g\"%(W_iso);\n",
+      "\t\t\t#Note that in isothermal compression between the same states work done is less as compared to reversible adiabatic compression.\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "\t\t\t#Ideal two-stage compression \n",
+      "n = 1.3;\t\t\t#Polytropic exponent.\n",
+      "\t\t\t#Minimum work done in two stage compression  is given by\n",
+      "\t\t\t#W_comp = ((2*n*R*T_1)/(n-1))*[1-(P_x/P_1)**(n-1)/n]\n",
+      "\t\t\t#where for minimum work, (P_x/P_1) = (P_x/P_2), and thus\n",
+      "P_x = (P_1*P_2)**(1./2);\t\t\t#[kPa]\n",
+      "\t\t\t#therefore, work done is given by,\n",
+      "W_comp = ((2*n*R*T_1)/(n-1))*(1-(P_x/P_1)**((n-1)/n));\t\t\t#[J/mol]\n",
+      "W_comp = W_comp/Mol_wt;\t\t\t#[J/g]\n",
+      "print \" c).The compressor work done for ideal two-stage compession is %f J/g\"%(W_comp);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The compressor work done for reversible adiabatic compession is -264.386412 J/g\n",
+        " b).The compressor work done for isothermal compession is -190.024880 J/g\n",
+        " c).The compressor work done for ideal two-stage compession is -216.284501 J/g\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.19 Page number - 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 600;\t\t\t#[C] - Temperature at entry\n",
+      "P_1 = 15;\t\t\t#[MPa] - Pressure at entry\n",
+      "T_2 = 400;\t\t\t#[K] - Temperature at exit\n",
+      "P_2 = 100;\t\t\t#[kPa] - Pressure at exit\n",
+      "A_in = 0.045;\t\t\t#[metre square] - flow  in area\n",
+      "A_out = 0.31;\t\t\t#[metre square] - flow out area\n",
+      "m = 30;\t\t\t#[kg/s] - mass flow rate.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#At 15 MPa and 600 C,it has been reported in the book that the properties of steam are,\n",
+      "Vol_1 = 0.02491;\t\t\t#[m**(3)/kg] - Specific volume\n",
+      "H_1 = 3582.3;\t\t\t#[kJ/kg] - Enthalpy\n",
+      "\t\t\t# m = den*vel*A = (Vel*A)/Vol, substituting the values\n",
+      "vel_1 = (m*Vol_1)/A_in;\t\t\t#[m/s] - Velocity at point 1.\n",
+      "print \" The inlet velocity is %f m/s\"%(vel_1);\n",
+      "\n",
+      "\t\t\t#At 100 MPa (saturated vapour),it has been reported in the book that the properties of steam are, T_sat = 99.63 C, and\n",
+      "Vol_vap_2 = 1.6940;\t\t\t#[m**(3)/kg] - specific volume of saturated vapour.\n",
+      "H_vap_2 = 2675.5;\t\t\t#[kJ/kg] - Enthalpy os saturated vapour.\n",
+      "vel_2 = (m*Vol_vap_2)/A_out;\t\t\t#[m/s] - Velocity at point 2.\n",
+      "print \" The exit velocity is %f m/s\"%(vel_2);\n",
+      "\n",
+      "\t\t\t#From first law we get, q - w =delta_H + delta_V**(2)/2\n",
+      "\t\t\t#q = 0, therefore, -w = delta_H + delta_V**(2)/2\n",
+      "delta_H = H_vap_2 - H_1;\t\t\t#[kJ/kg] - change in enthalpy.\n",
+      "delta_V_square = (vel_2**(2) - vel_1**(2))/2;\t\t\t#[J/kg]\n",
+      "delta_V_square = delta_V_square*10**(-3);\t\t\t#[kJ/kg]\n",
+      "w = -(delta_H + delta_V_square);\t\t\t#[J/kg] \n",
+      "W_net = w*m;\t\t\t#[kW]\n",
+      "W_net = W_net*10**(-3);\t\t\t#[MW] - power produced.\n",
+      "print \" The power that can be produced by the turbine is %f MW\"%(W_net);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The inlet velocity is 16.606667 m/s\n",
+        " The exit velocity is 163.935484 m/s\n",
+        " The power that can be produced by the turbine is 26.805014 MW\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.20 Page number - 117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "R = 8.314;\t\t\t#[J/mol-K] - Universal gas constant\n",
+      "Cp_0 = 2.5*R;\t\t\t#[J/mol-K] -  Specific heat capacity at constant pressure\n",
+      "Cv_0 = 1.5*R;\t\t\t#[J/mol-K] -  Specific heat capacity at constant volume\n",
+      "T_L = 300;\t\t\t#[K] - Temperature at which port properties are constant.\n",
+      "\n",
+      "# Calculations\n",
+      "Y = Cp_0/Cv_0;\t\t\t# Ratio of heat capacities.\n",
+      "\t\t\t#From part(1) we obtained the relation,\n",
+      "\t\t\t# T_2 = 1/(((P_2-P_1)/(Y*T_L*P_2))+(P_1/(P_2*T_1)))\n",
+      "\t\t\t# Not that when P_2 >> P_1 ,T_2 approaches Y*T_L and thus\n",
+      "T_2 = Y*T_L;\t\t\t#[K]\n",
+      "\n",
+      "# Results\n",
+      "print \" b).The final temperature is %f K\"%(T_2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " b).The final temperature is 500.000000 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.21 Page number - 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 40 + 273.15;\t\t\t#[K] - Initial temperature.\n",
+      "P_1 = 1;\t\t\t#[bar] - Initial pressure.\n",
+      "P_1 = P_1*10**(5);\t\t\t#[Pa]\n",
+      "Vol_1 = 0.01;\t\t\t#[cubic metre] - Initial volume of the cylinder.\n",
+      "T_2 = 100 + 273.15;\t\t\t#[K] - Final temperature.\n",
+      "P_2 = 100;\t\t\t#[kPa] - Final pressure.\n",
+      "P_2 = P_2*10**(5);\t\t\t#[Pa]\n",
+      "Vol_2 = 0.02;\t\t\t#[cubic metre] - Final volume of the cylinder.\n",
+      "Cp = 1.005;\t\t\t#[J/g-K] - Specific heat capacity at constant pressure.\n",
+      "Cv = 0.718;\t\t\t#[J/g-K] - Specific heat capacity at constant volume.\n",
+      "Mol_wt = 28.84;\t\t\t#[g/mol] - Molecular weight of air.\n",
+      "R = 8.314;\t\t\t#[J/mol-K] - universal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "delta_Vol = Vol_2 - Vol_1;\t\t\t# [cubic metre] - Change in volume.\n",
+      "\t\t\t# Assuming ideal gas P*V = R*T\n",
+      "V_1 = (R*T_1)/P_1;\t\t\t# [m**(3)/mol] - Initial specific volume.\n",
+      "\t\t\t# Therefore,the total number of moles initially in the system is,\n",
+      "n_1 = (Vol_1/V_1);\t\t\t# [mol]\n",
+      "m_1 = n_1*Mol_wt;\t\t\t# [g] - Initial mass of the system.\n",
+      "Y = Cp/Cv;\t\t\t#Ratio of heat capacities\n",
+      "\n",
+      "\t\t\t# The energy balance equation is given by\n",
+      "\t\t\t# -P*delta_Vol + H_liq*(m_2 - m_1) = m_2*Cv*(P*V2)/R - m_1*Cv*T_1\n",
+      "\t\t\t# m_2*Cv*(P*V2)/R = (Cv*P_1*Vol_2)/R\n",
+      "\t\t\t# Cv/R = 1/(Y-1)\n",
+      "\t\t\t# Since pressure of the gas in system is assumed constant,therefore it remains at 1 bar and thus P = P_1,\n",
+      "H_liq = Cp*T_2;\t\t\t# [J/g] - Enthalpy of liquid\n",
+      "m_2 = (P_1*delta_Vol + ((P_1*Vol_2)/(Y-1)) + H_liq*m_1 - m_1*Cv*T_1)/H_liq;\t\t\t#[g]\n",
+      "\n",
+      "\t\t\t#The mass entering the assembly during the filling process is given by\n",
+      "m = m_2 - m_1;\t\t\t#[g]\n",
+      "n_2 = m_2/Mol_wt;\t\t\t#[mol] - Number of moles in the final state.\n",
+      "V_2 = Vol_2/n_2;\t\t\t#[m**(3)/mol] - Final specific volume.\n",
+      "\t\t\t# Therfore,final temperature is given by,\n",
+      "T_2 = (P_1*V_2)/R;\t\t\t#[K] - Final temperature.\n",
+      "\n",
+      "# Results\n",
+      "print \" The final equilibrium temperature is %f K\"%(T_2);\n",
+      "print \" The mass entering through the valve is %f g\"%(m);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final equilibrium temperature is 339.343160 K\n",
+        " The mass entering through the valve is 9.367211 g\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example - 3.22 Page number - 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "V_total = 5;\t\t\t#[L] - Volume of pressure cooker.\n",
+      "V_total = V_total*10**(-3);\t\t\t#m**(3)\n",
+      "P_gauze = 15;\t\t\t#[psi] - Operating pressure (gauze)of pressure cooker.\n",
+      "P_gauze = (P_gauze/14.5)*10**(5);\t\t\t#[N/m**(2)]\n",
+      "P_atm = 0.966*10**(5);\t\t\t#[N/m**(2)] - Atmospheric pressure.\n",
+      "m_1 = 1;\t\t\t#[kg] - Initial mass.\n",
+      "t = 30*60;\t\t\t#[s] - Total time.\n",
+      "J = 500;\t\t\t#[W] - Rate of heat supply\n",
+      "\n",
+      "# Calculations\n",
+      "P_abs = P_gauze + P_atm;\t\t\t#[N/m**(2)] - Absolute pressure.\n",
+      "\t\t\t#The energy balance equqtion gives,\n",
+      "\t\t\t# Q = m_e*H_e +(m_2*U_2 - m_1*U_1), where 'm_e' is the mass exit  from the system and 'H_e' is enthalpy at exit conditions.\n",
+      "\n",
+      "\t\t\t#It has been reported in the book that from steam table at P_abs,\n",
+      "T_sat = 120.23;\t\t\t#[K]- Saturated temperature\n",
+      "V_liq = 0.001061;\t\t\t#[m**(3)/kg] - specific volume of liquid.\n",
+      "V_vap = 0.8857;\t\t\t#[m**(3)/kg] - specific volume of vapour.\n",
+      "U_liq = 504.49;\t\t\t#[kJ/kg] - specific internal energy of liquid.\n",
+      "U_vap = 2529.5;\t\t\t#[kJ/kg] - specific internal energy of vapour.\n",
+      "H_liq = 504.70;\t\t\t#[kJ/kg] - specific enthalpy of liquid.\n",
+      "H_vap = 2706.7;\t\t\t#[kJ/kg] - specific internal energy of vapour.\n",
+      "\n",
+      "\t\t\t#We know that total volume occupied by 1 kg of fluid is \n",
+      "\t\t\t#V_total = (1-x)*V_liq + x*V_vap\n",
+      "x1 = (V_liq - V_total)/(V_liq - V_vap);\t\t\t#[g]\n",
+      "\n",
+      "\t\t\t#Internal energy at this state is\n",
+      "U_1 = (1-x1)*U_liq + x1*U_vap;\t\t\t#[kJ/kg] - specific internal energy\n",
+      "U_1_net = m_1*U_1;\t\t\t#[kJ] - Internal energy\n",
+      "\n",
+      "\t\t\t#The amount of heat suplied is given by,\n",
+      "J_net = J*t;\t\t\t#[J] - Net heat supplied.\n",
+      "J_net = J_net*10**(-3);\t\t\t#[kJ]\n",
+      "\n",
+      "\t\t\t#Let the dryness factor at the end of the process be x\n",
+      "\t\t\t#Let the total mass of H2O (liquid + vapour) at the end of the process be 'm' kg.\n",
+      "\t\t\t# V_total/m = (1-x)*(V_liq) + x*V_vap ......equqtion(1)\n",
+      "\n",
+      "\n",
+      "\t\t\t#The energy balance equqtion gives,\n",
+      "\t\t\t# Q = m_e*H_e +(m_2*U_2 - m_1*U_1), where 'm_e' is the mass exit  from the system and 'H_e' is enthalpy at exit conditions.\n",
+      "\n",
+      "    \t\t# The second equation on simplification becomes\n",
+      "\t\t\t# x = ((0.005/m) - 0.001061)/0.884639\n",
+      "\n",
+      "\t\t\t# Putting the expression of x in first equation and then simplifying, we get\n",
+      "\t\t\t# - 1293.2 = -2202.21*m + 11.445 - 2.429*m\n",
+      "m = (11.445+1293.2)/(2202.21+2.429);\t\t\t#[kg]\n",
+      "\n",
+      "\t\t\t# Therefore x can be calculated as\n",
+      "x = ((0.005/m) - 0.001061)/0.884639;\n",
+      "\n",
+      "\t\t\t# Therfore total water (liquid + vapour) present in the pressure cooker at the end of the process is m kg.\n",
+      "m_vapour = x*m;\t\t\t#[kg] - Mass of vapour\n",
+      "m_liquid = (1-x)*m;\t\t\t#[kg] - Mass of vapour\n",
+      "\n",
+      "# Results\n",
+      "print \" Total water liquid + vapour) present in the pressure cooker at the end of the process is %f kg\"%(m);\n",
+      "print \" The mass of vapour is %f kg\"%(m_vapour);\n",
+      "print \" The mass of liquid is %f kg\"%(m_liquid);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Total water liquid + vapour) present in the pressure cooker at the end of the process is 0.591773 kg\n",
+        " The mass of vapour is 0.004942 kg\n",
+        " The mass of liquid is 0.586830 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch4_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch4_2.ipynb
new file mode 100755
index 00000000..ebeadf70
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch4_2.ipynb
@@ -0,0 +1,1508 @@
+{
+ "metadata": {
+  "name": "ch4_2",
+  "signature": "sha256:8185298e7063d37f738c422b0f9a0946dd677189cdf544f54e436d8d650cb521"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : The Secomd Law and Its Applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.1  Page Number : 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "n = 1000.;\t\t\t#[mol]\n",
+      "T = 400.;\t\t\t#[K]\n",
+      "P_1 = 100.;\t\t\t#[kPa]\n",
+      "P_2 = 1000.;\t\t\t#[kPa]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(a)\n",
+      "T_surr = 400.;\t\t\t#[K] - surrounding temperature\n",
+      "\t\t\t# Total change in entropy of the system is given by\n",
+      "\t\t\t# delta_S_sys = n*(Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1))\n",
+      "\t\t\t# The process being isothermal the first term is zero and the total entropy change of the system is\n",
+      "delta_S_sys_a = - n*R*math.log(P_2/P_1);\t\t\t#[J/K]\n",
+      "delta_S_sys_a = delta_S_sys_a*10**(-3);\t\t\t#[kJ/K]\n",
+      "\n",
+      "\t\t\t# Since the process is reversible therefore\n",
+      "Q_sys = T_surr*delta_S_sys_a;\t\t\t#[kJ] - Heat change in the system\n",
+      "\n",
+      "Q_surr = - Q_sys;\t\t\t#[kJ] - Heat change in the surrounding\n",
+      "delta_S_surr_a = Q_surr/T_surr;\t\t\t#[kJ/K]\n",
+      "\n",
+      "delta_S_univ_a = delta_S_sys_a + delta_S_surr_a;\t\t\t#[kJ/K]\n",
+      "\t\t\t# We get delta_S_univ = 0, which is true for a reversible process\n",
+      "\n",
+      "print \" a).The entropy change of the gas is given by  delta_S_sys = %f kJ/K \"%(delta_S_sys_a);\n",
+      "print \"     The entropy change of the surrounding is delta_S_surr = %f kJ/K \"%(delta_S_surr_a);\n",
+      "print \"     The total entropy change of the gas is  delta_S_univ = %f kJ/K \"%(delta_S_univ_a);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "T_surr_b = 300.;\t\t\t#[K] - surrounding temperature\n",
+      "\n",
+      "delta_S_sys_b = delta_S_sys_a;\n",
+      "\n",
+      "\t\t\t# Work done under reversible condition is given by\n",
+      "\t\t\t# W = integral(P*dV) = integral(((R*T)/V)*dV) = R*T*math.log(V_2/V_1)\n",
+      "\t\t\t# For ideal gas we have, P1*V1/T1 = P2*V2/T2 or, V2/V1 = P1/P2 (for isothermal conditions)\n",
+      "W = R*T*math.log(P_1/P_2);\t\t\t#[J/mol]\n",
+      "W = W*10**(-3);\t\t\t#[kJ/mol]\n",
+      "\n",
+      "W = W*(120./100);\t\t\t#[kJ/mol]\n",
+      "W = W*n;\t\t\t#[kJ] - Total work done for n moles\n",
+      "\n",
+      "\t\t\t# Using the first law we have  delta_U = Q - W. Now under isothermal conditions for ideal gas, delta_U = 0. Therefore,\n",
+      "Q = -W;\n",
+      "\n",
+      "\t\t\t# Since heat is gained by the surroundings therefore\n",
+      "delta_S_surr_b = Q/T_surr_b;\t\t\t#[kJ/K]\n",
+      "\n",
+      "delta_S_univ_b = delta_S_sys_b + delta_S_surr_b;\t\t\t#[kJ/K]\n",
+      "\n",
+      "print \" b).The entropy change of the gas is given by  delta_S_sys = %f kJ/K \"%(delta_S_sys_b);\n",
+      "print \"     The entropy change of the surrounding is delta_S_surr = %f kJ/K \"%(delta_S_surr_b);\n",
+      "print \"     The total entropy change of the gas is delta_S_univ = %f kJ/K \"%(delta_S_univ_b);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The entropy change of the gas is given by  delta_S_sys = -19.143692 kJ/K \n",
+        "     The entropy change of the surrounding is delta_S_surr = 19.143692 kJ/K \n",
+        "     The total entropy change of the gas is  delta_S_univ = 0.000000 kJ/K \n",
+        " b).The entropy change of the gas is given by  delta_S_sys = -19.143692 kJ/K \n",
+        "     The entropy change of the surrounding is delta_S_surr = 30.629908 kJ/K \n",
+        "     The total entropy change of the gas is delta_S_univ = 11.486215 kJ/K \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.2  Page Number : 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 400.;\t\t\t#[K] - Temperature\n",
+      "P_1 = 500.*10**(3);\t\t\t#[Pa] - Initial pressure\n",
+      "P_2 = 100.*10**(3);\t\t\t#[Pa] -  Final pressure\n",
+      "V_1 = 750.*10**(-6);\t\t\t#[m**(3)] - Initial volume\n",
+      "W_actual = 0.55*10**(3);\t\t\t#[J] - Actual work done\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal fas constant\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# Suppose that the surroundings are at 400 K.\n",
+      "\n",
+      "\t\t\t# The number of moles is given by\n",
+      "n = (P_1*V_1)/(R*T);\t\t\t#[mol]\n",
+      "\t\t\t# The entropy change of ideal gas under isothermal condition is given by\n",
+      "delta_S_sys = - n*R*math.log(P_2/P_1);\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# The heat supplied to the system in the internally reversible process is\n",
+      "Q_theot = T*delta_S_sys;\t\t\t#[J]\n",
+      "\t\t\t# Since the process is isothermal therefore, workdone is given by\n",
+      "W_theot = Q_theot;\t\t\t#[J] - Theoritical work done\n",
+      "\t\t\t# Since actual work done by the gas is 0.55 kJ therefore actual heat supplied is also 0.55 kJ because under isothermal conditions delta_U = 0\n",
+      "Q_actual = W_actual;\n",
+      "\n",
+      "# Results\n",
+      "\t\t\t# Since Q_theot > Q_actual, so the process is irreversible\n",
+      "print \" Since Q_theot = %f J  is greater than  Q_actual = %f J\"%(Q_theot,Q_actual);\n",
+      "print \" Therefore, the process is internally irreversible\"\n",
+      "\n",
+      "\n",
+      "\t\t\t# delta_S_sys = Q/T_b + S_gen\n",
+      "S_gen = delta_S_sys - (Q_theot/T);\t\t\t#[J/K]\n",
+      "\t\t\t# The entropy generated may be due to friction and other dissipayive effects or due to non-quasi-static expansion\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Since Q_theot = 603.539217 J  is greater than  Q_actual = 550.000000 J\n",
+        " Therefore, the process is internally irreversible\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.3  Page Number : 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\t\t\t# For side A\n",
+      "V_A = 1;\t\t\t#[L] - Volume\n",
+      "V_A = V_A*10**(-3);\t\t\t#[m**(3)]\n",
+      "T_A = 300;\t\t\t#[K] - Temperature\n",
+      "P_A = 2;\t\t\t#[atm] - Pressure\n",
+      "P_A = P_A*101325;\t\t\t#[Pa]\n",
+      "\n",
+      "\t\t\t# For side B\n",
+      "V_B = 1;\t\t\t#[L] - volume\n",
+      "V_B = V_B*10**(-3);\t\t\t#[m**(3)]\n",
+      "T_B = 300;\t\t\t#[K] - Temperature\n",
+      "P_B = 1;\t\t\t#[atm] - Pressure\n",
+      "P_B = P_B*101325;\t\t\t#[Pa]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# From first law final temperature and pressure are given by (example 3.30)\n",
+      "\t\t\t# T = ((n_A*T_A) + (n_B*T_B))/(n_A + n_B)\n",
+      "\t\t\t# P = ((P_A*V_A) + (P_A*V_B))/(V_A + V_B)\n",
+      "\n",
+      "\t\t\t# Since in this case T_A = T_B, therefore final pressure is given by\n",
+      "P = ((P_A*V_A) + (P_B*V_B))/(V_A + V_B);\t\t\t#[Pa]\n",
+      "P = P/101325;\t\t\t#[atm]\n",
+      "\n",
+      "print \" The final temperature is %f K\"%(T_A);\n",
+      "print \" The final pressure is %f atm\"%(P);\n",
+      "\n",
+      "\t\t\t# The number of moles of air on each side are\n",
+      "n_A = (P_A*V_A)/(R*T_A);\t\t\t#[mol]\n",
+      "n_B = (P_B*V_B)/(R*T_B);\t\t\t#[mol]\n",
+      "\n",
+      "delta_S_A = -n_A*R*math.log((P*101325)/P_A);\t\t\t#[J/K] - Entropy change on side A\n",
+      "delta_S_B = -n_B*R*math.log((P*101325)/P_B);\t\t\t#[J/K] - Entropy change on side B\n",
+      "delta_S_sys = delta_S_A + delta_S_B;\t\t\t#[J/K] - Total entropy change of system\n",
+      "\n",
+      "\t\t\t# Since the system is insulated there is no heat exchange with the surroundings, therefore entropy change of surrounding is zero\n",
+      "delta_S_surr = 0;\t\t\t#[J/K]\n",
+      "delta_S_univ = delta_S_sys + delta_S_surr;\t\t\t#[J/K]\n",
+      "print \" The total increase in entropy is %f J/K\"%(delta_S_univ);\n",
+      "\n",
+      "\t\t\t# The entropy change of the system can also be writtten as\n",
+      "\t\t\t# delta_s_sys = Q/T_b + S_gen\n",
+      "\t\t\t# Since there is no heat transfer, therefore\n",
+      "S_gen = delta_S_univ;\t\t\t#[J/K]\n",
+      "\t\t\t# The process is reversible because of entropy generation due to spontaneous release of piston.\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final temperature is 300.000000 K\n",
+        " The final pressure is 1.500000 atm\n",
+        " The total increase in entropy is 0.057383 J/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.4  Page Number : 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "V_vessel = 0.2;\t\t\t#[m**(3)] - Volume of the vessel\n",
+      "P_1 = 10;\t\t\t#[bar] - Initial pressure inside the vessel\n",
+      "P_1 = P_1*10**(5);\t\t\t#[Pa]\n",
+      "P_2 = 3.5;\t\t\t#[bar] - Final pressure inside the vessel\n",
+      "P_2 = P_2*10**(5);\t\t\t#Pa\n",
+      "T_1 = 250 + 273.15;\t\t\t#[K] - Initial temperature of the vesssel\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (a)\n",
+      "\t\t\t# At 10 bar and 250 C the steam is superheated. From steam table as reported in book we have\n",
+      "V_1 = 0.2327;\t\t\t#[m**(3)/kg] - specific volume\n",
+      "U_1 = 2709.9;\t\t\t#[kJ/kg] - specific internal energy\n",
+      "H_1 = 2942.6;\t\t\t#[kj/kg] - Specific enthalpy\n",
+      "S_1 = 6.9247;\t\t\t#[kJ/kg-K] - Specific entropy\n",
+      "\t\t\t# the quantity of steam is given by\n",
+      "m = V_vessel/V_1;\t\t\t#[kg]\n",
+      "\n",
+      "\t\t\t# At final state \n",
+      "V_2 = 0.2327;\t\t\t#[m**(3)/kg] - Molar volume\n",
+      "V_liq_2 = 0.001079;\t\t\t# [m**(3)/kg]\n",
+      "V_vap_2 = 0.5243;\t\t\t# [m**(3)/kg]\n",
+      "\t\t\t# Since overall volume lies between saturated liquid and saturated vapour therefore the steam is saturated and its dryness fraction at final state is given by\n",
+      "x = (V_2 - V_liq_2)/(V_vap_2 - V_liq_2);\n",
+      "\t\t\t# Final temperature  = T_sat (at 3 bar) from steam table\n",
+      "T_final = 138.88;\t\t\t#[C]\n",
+      "\n",
+      "\t\t\t# At 3.5 bar saturated conditions \n",
+      "S_liq = 1.7275;\t\t\t#[kJ/kg-K]  - Entropy of saturated liquid\n",
+      "S_vap = 6.9405;\t\t\t#[kJ/kg-K]  - Entropy of saturated vapour\n",
+      "U_liq = 583.95;\t\t\t#[kJ/kg]  - Internal energy of saturated liquid\n",
+      "U_vap = 2548.9;\t\t\t#[kJ/kg]  - Internal energy of saturated vapour\n",
+      "\t\t\t# Therefore at final state \n",
+      "U_2 = U_liq*(1 - x) + x*U_vap;\t\t\t#[kJ/kg]\n",
+      "S_2 = S_liq*(1 - x) + x*S_vap;\t\t\t#[kJ/kg-K]\n",
+      "Q_1 = m*(U_2 - U_1);\t\t\t#[kJ]\n",
+      "delta_S_1 = m*(S_2 - S_1);\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" a).The final temperature is %f C\"%(T_final);\n",
+      "print \"     The amount of heat transfer is %f kJ\"%(Q_1);\n",
+      "print \"     The change of entropy is %f kJ/kg-K\"%(delta_S_1);\n",
+      "\n",
+      "\t\t\t# (b)\n",
+      "Y = 1.4;\t\t\t# Ratio of heat capacities for air\n",
+      "\t\t\t# (P_1*V_1)/T_1 = (P_2*V_2)/T_2   and since   V_1 = V_2\n",
+      "T_2 = (P_2/P_1)*T_1;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# Since the volume is fixed therefore work done (W) = 0 and from first law we get\n",
+      "\t\t\t# Q = delta_U = n*Cv_0*(T_2 - T_1)\n",
+      "Cv_0 = R/(Y - 1);\t\t\t#[J/mol-K] - Heat capacity at constant volume\n",
+      "Cp_0 = (Y*R)/(Y - 1);\t\t\t#[J/mol-K] - Heat capacity at constant pressure\n",
+      "n = (P_1*V_vessel)/(R*T_1);\t\t\t#[mol] - No. of moles\n",
+      "Q_2 = n*Cv_0*(T_2 - T_1);\t\t\t#[J] - Heat change\n",
+      "Q_2 = Q_2*10**(-3);\t\t\t#[kJ]\n",
+      "\n",
+      "delta_S_2 = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1);\t\t\t#[J/mol-K]\n",
+      "delta_S_2 = n*delta_S_2*10**(-3);\t\t\t#[kJ/K]\n",
+      "\n",
+      "print \" b).The final temperature is %f C\"%(T_2);\n",
+      "print \"     The amount of heat transfer is %f kJ\"%(Q_2);\n",
+      "print \"     The change of entropy is %f kJ/K\"%(delta_S_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The final temperature is 138.880000 C\n",
+        "     The amount of heat transfer is -1079.587621 kJ\n",
+        "     The change of entropy is -2.483450 kJ/kg-K\n",
+        " b).The final temperature is 183.102500 C\n",
+        "     The amount of heat transfer is -325.000000 kJ\n",
+        "     The change of entropy is -1.003366 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.5  Page Number : 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from math import *\n",
+      "# Variables\n",
+      "m = 1000.;\t\t\t#[g] - Mass of fluid\n",
+      "P_1 = 20.;\t\t\t#[bar] - Initial pressure\n",
+      "P_1 = P_1*10.**(5);\t\t\t#[Pa]\n",
+      "P_2 = 2;\t\t\t#[bar] - Final pressure\n",
+      "P_2 = P_2*10.**(5);\t\t\t#Pa\n",
+      "T_1 = 250 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "n = 1.25;\n",
+      "R = 8.314;\t\t\t#[J/mol*-] - Universal gas constant\n",
+      "Y = 1.4;\t\t\t# Index of expansion\n",
+      "Cv_0 = R/(Y- 1);\t\t\t#[J/mol-K]\n",
+      "Cp_0 = R + Cv_0;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(a)\n",
+      "\t\t\t# For steam at 20 bar and 250 C, from steam table as reported in the book\n",
+      "V_1 = 0.11144;\t\t\t#[m**(3)/kg]\n",
+      "U_1 = 2679.6;\t\t\t#[kJ/kg]\n",
+      "S_1 = 6.5453;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t# P_1*V_1**(n) = P_2*V_2**(n)\n",
+      "V_2 = ((P_1*V_1**(n))/P_2)**(1/n);\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "\t\t\t# At 2 bar under saturated conditions,from steam table as reported in the book\n",
+      "V_liq = 0.001061;\t\t\t#[m**(3)/kg]\n",
+      "V_vap = 0.8857;\t\t\t#[m**(3)/kg]\n",
+      "x = (V_2 - V_liq)/(V_vap - V_liq);\t\t\t# Dryness fraction\n",
+      "T_sat = 120.23;\t\t\t#[C] - The final temperature\n",
+      "U_liq  = 504.49;\t\t\t#[kJ/kg] - Internal energy of saturate liquid\n",
+      "U_vap = 2529.5;\t\t\t#[kJ/kg] - Internal energy of saturate vapour\n",
+      "\t\t\t# Therefore, internal energy at state 2 is given by\n",
+      "U_2 = U_liq*(1 - x) + x*U_vap;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# Work transfer is given by\n",
+      "W = (P_1*V_1 - P_2*V_2)/(n - 1);\t\t\t#[J/kg]\n",
+      "W = W*10**(-3);\t\t\t#[kJ/kg]\n",
+      "delta_U = U_2 - U_1;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# From first law, q - W = delta_U\n",
+      "q = W + delta_U;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# At final state (2 bar saturated), as reported in the book\n",
+      "S_liq = 1.5301;\t\t\t#[kJ/kg-K] - Entropy of saturated liquid\n",
+      "S_vap = 7.1271;\t\t\t#[kJ/kg-K] - Entropy of saturated vapour\n",
+      "\t\t\t# Therefore, entropy at state 2 is given by\n",
+      "S_2 = S_liq*(1 - x) + x*S_vap;\t\t\t#[kJ/kg-K]\n",
+      "delta_S = S_2 - S_1;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" a).The final temperature is %f C\"%(T_sat);\n",
+      "print \"     The work done is equal to %f kJ/kg\"%(W);\n",
+      "print \"     The heat change is equal to %f kJ/kg\"%(q);\n",
+      "print \"     The entropy change is equal to %f kJ/kg-K\"%(delta_S);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\t\t\t# P*V**(n) = constant\n",
+      "\t\t\t# Since the gas behaves as ideal we can write\n",
+      "\t\t\t# P_1**(1-n)*T_1**(n) = P_2**(1-n)*T_2**(n)\n",
+      "T_2 = T_1*(P_1/P_2)**((1-n)/n);\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# Molar volume is given by\n",
+      "V_2_1 = (R*T_1)/P_1;\t\t\t#[m**(3)/mol] - At state 1\n",
+      "V_2_2 = (R*T_2)/P_2;\t\t\t#[m**(3)/mol] - At state 2\n",
+      "\n",
+      "\t\t\t# Work transfer is given by\n",
+      "w_2 = ((P_1*V_2_1) - (P_2*V_2_2))/(n-1);\t\t\t#[J/mol]\n",
+      "Mol_wt_air = 0.21*32 + 0.79*28;\t\t\t#[g/mol] - Molecular weight of air\n",
+      "n_mole = m/Mol_wt_air;\n",
+      "\t\t\t# Total work transfer is given by\n",
+      "W_2 = w_2*n_mole*10**(-3);\t\t\t#[kJ]\n",
+      "\t\t\t# Internal energy change is given by\n",
+      "delta_U = n_mole*Cv_0*(T_2 - T_1)*10**(-3);\t\t\t#[kJ]\n",
+      "\n",
+      "\t\t\t# Heat transfer is given by\n",
+      "Q = W_2 + delta_U;\t\t\t#[kJ]\n",
+      "\n",
+      "\t\t\t# Entropy change is given by\n",
+      "delta_S_2 = Cp_0*log(T_2/T_1) - R*log(P_2/P_1);\t\t\t#[J/mol-K]\n",
+      "delta_S_2 = delta_S_2*n_mole;\t\t\t#[J/mol]\n",
+      "\n",
+      "print \" b).The final temperature is %f C\"%(T_2);\n",
+      "print \"     The work done is equal to %f kJ/kg\"%(W_2);\n",
+      "print \"     The total heat change is equal to %f kJ\"%(Q);\n",
+      "print \"     The entropy change is equal to %f J/kg-K\"%(delta_S_2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The final temperature is 120.230000 C\n",
+        "     The work done is equal to 329.008908 kJ/kg\n",
+        "     The heat change is equal to -238.988250 kJ/kg\n",
+        "     The entropy change is equal to -0.573241 kJ/kg-K\n",
+        " b).The final temperature is 330.085335 C\n",
+        "     The work done is equal to 222.626855 kJ/kg\n",
+        "     The total heat change is equal to 83.485071 kJ\n",
+        "     The entropy change is equal to 199.136884 J/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.6  Page Number : 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "m = 1000;\t\t\t#[g] - Mass of fluid\n",
+      "P_1 = 20;\t\t\t#[bar] - Initial pressure\n",
+      "P_2 = 2;\t\t\t#[bar] - Final ressure\n",
+      "T_1 = 250 + 273.15;\t\t\t#[K] - Initial tempearture\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# (a).\n",
+      "\t\t\t# At 20 bar and 250 C as reported in the book\n",
+      "V_1 = 0.11144;\t\t\t#[m**(3)/kg] - Specific volume\n",
+      "U_1 = 2679.6;\t\t\t#[kJ/kg] - Specific internal energy\n",
+      "S_1 = 6.5453;\t\t\t#[kJ/kg-K] - Specific entropy\n",
+      "S_2 = S_1;\t\t\t# Isentropic expansion\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# At 2 bar under saturated conditions \n",
+      "S_liq = 1.5301;\t\t\t#[kJ/kg-K]\n",
+      "S_vap = 7.1271;\t\t\t#[kJ/kg-K]\n",
+      "U_liq = 504.49;\t\t\t#[kJ/kg-K]\n",
+      "U_vap = 2529.5;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t# Therefore dryness factor can be determined as\n",
+      "x = (S_1 - S_liq)/(S_vap - S_liq);\n",
+      "U_2 = U_liq*(1 - x) + x*U_vap;\t\t\t#[kJ/kg] - Specific internal energy at final state\n",
+      "delta_U = U_2 - U_1;\t\t\t#[kJ/kg] - change in internal energy\n",
+      "W = - delta_U;\t\t\t# - Work done\n",
+      "\n",
+      "\t\t\t# The final saturated temperature at 2 bar from steam table is\n",
+      "T_2 = 120.23;\t\t\t#[C]\n",
+      "\n",
+      "print \" a).The final temperature is %f C\"%(T_2);\n",
+      "print \"     The work done is equal to %f kJ/kg\"%(W);\n",
+      "\n",
+      "\t\t\t# (b).\n",
+      "Y = 1.4;\t\t\t# Index of expansion for air\n",
+      "Cv_0 = R/(Y-1);\t\t\t#[J/mol*K] - Specific heat capacity at constant volume\n",
+      "\t\t\t# Ideal gas under isentropic expansion    P_1**(1-Y)*T_1**(Y) =P_2**(1-Y)*T_2**(Y)\n",
+      "T_2_prime = T_1*(P_1/P_2)**((1-Y)/Y);\t\t\t#[K] - Final temperature\n",
+      "delta_U_prime = Cv_0*(T_2_prime - T_1);\t\t\t#[J/mol] - change in internal energy\n",
+      "\n",
+      "\t\t\t# Number of moles is given by\n",
+      "n = m/28.84;\t\t\t#[mol]\n",
+      "delta_U_prime = delta_U_prime*n*10**(-3);\t\t\t#[kJ]\n",
+      "W_prime = - delta_U_prime;\t\t\t# Work done\n",
+      "\n",
+      "print \" b).The final temperature is %f C\"%(T_2_prime);\n",
+      "print \"     The work done is equal to %f kJ/kg\"%(W_prime);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The final temperature is 120.230000 C\n",
+        "     The work done is equal to 360.596841 kJ/kg\n",
+        " b).The final temperature is 270.964218 C\n",
+        "     The work done is equal to 181.750398 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.7  Page Number : 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P_1 = 15.;\t\t\t#[bar] - Initial pressure\n",
+      "P_2 = 0.15;\t\t\t#[bar] - Final pressure\n",
+      "\n",
+      "\t\t\t# We know that during isentropic expansion\n",
+      "\t\t\t# W = ((P_1*V_1) - (P_2*V_2))/(Y - 1)\n",
+      "\n",
+      "\t\t\t# At 15 bar (saturated vapour), from steam table as reported in the book\n",
+      "V_1 = 0.13177;\t\t\t#[m**(3)/kg]\n",
+      "U_1 = 2594.5;\t\t\t#[kJ/kg]\n",
+      "S_1 = 6.4448;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t# Now at state 2 (P_2 = 0.15 bar),from steam table as reported in the book\n",
+      "S_2 = S_1;\t\t\t# Isentropic expansion\n",
+      "S_liq = 0.7549;\t\t\t#[kJ/kg-K]\n",
+      "S_vap = 8.0085;\t\t\t#[kJ/kg-K]\n",
+      "U_liq = 225.92;\t\t\t#[kJ/kg]\n",
+      "U_vap = 2448.7;\t\t\t#[kJ/kg]\n",
+      "V_liq = 0.001014;\t\t\t#[m**(3)/kg]\n",
+      "V_vap = 10.02;\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# Therefore dryness factor can be calculated as\n",
+      "x = round((S_1 - S_liq)/(S_vap - S_liq),3);\n",
+      "U_2 = round(U_liq*(1 - x) + x*U_vap,2);\t\t\t#[kJ/kg] - Specific internal energy at final state\n",
+      "delta_U = U_2 - U_1;\t\t\t#[kJ/kg] - change in internal energy\n",
+      "W = - delta_U;\t\t\t# - Work done\n",
+      "\n",
+      "\t\t\t# The specific volume at the final state is\n",
+      "V_2 = round(V_liq*(1 - x) + x*V_vap,3);\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "\t\t\t# From work done under adiabatic conditions we get\n",
+      "\t\t\t# W = ((P_1*V_1) - (P_2*V_2))/(Y - 1)\n",
+      "Y = (((P_1*V_1) - (P_2*V_2))*10**2/W) + 1;\n",
+      "\n",
+      "# Results\n",
+      "print \" The index of expansion is given by Y = %.4f\"%(Y);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The index of expansion is given by Y = 1.1275\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.8  Page Number : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 40;\t\t\t#[bar] - Initial pressure\n",
+      "T_1 = 500;\t\t\t#[C] - Initial temperature\n",
+      "Vel_1 = 140;\t\t\t#[m/s] - Initial velocity\n",
+      "T_2 = 100;\t\t\t#[C] - Final temperature\n",
+      "Vel_2 = 80;\t\t\t#[m/s] - Final velocity\n",
+      "W = 746.0;\t\t\t#[kJ/kg] - Work output\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# (a).\n",
+      "\t\t\t# From steam table as reported in the book\n",
+      "H_1 = 3445.3;\t\t\t#[kJ/kg]\n",
+      "H_2 = 2676.1;\t\t\t#[kJ/kg]\n",
+      "S_1 = 7.0901;\t\t\t#[kJ/kh-K]\n",
+      "S_2 = 7.3549;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t# The temperature at which heat exchange take place is given by \n",
+      "T_b =(T_1 + T_2)/2 + 273.15;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# From first law in a control volume \n",
+      "\t\t\t# q - W = delta_H + (delta_V**(2))/2 , therefore\n",
+      "q = W*10**(3) + (H_2 - H_1)*10**(3) + (Vel_2**(2) - Vel_1**(2))/2;\t\t\t#[J/kg]\n",
+      "q = q*10**(-3);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "S_gen = (S_2 - S_1) - (q/T_b);\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" a).The specific entropy production within turbine is %f kJ/kg-K\"%(S_gen);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\t\t\t# If control volume is too large to include the turbine and the environment then T_b becomes equal to 289 K. In this case\n",
+      "T_b_prime = 298;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# The entropy change of the sysytem is given by \n",
+      "\t\t\t#delta_S = q/T_b + S_gen\n",
+      "S_gen = (S_2 - S_1) - (q/T_b_prime);\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" b).The specific entropy production within turbine is %f kJ/kg-K\"%(S_gen);\n",
+      "\n",
+      "\t\t\t# In the first part only irreversibilities within the turbine are evaluated\n",
+      "\t\t\t# whereas in part (2) irreversible heat transfer between the turbine cover and environment  are also included.\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The specific entropy production within turbine is 0.316793 kJ/kg-K\n",
+        " b).The specific entropy production within turbine is 0.364800 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.9  Page Number : 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 1.;\t\t\t#[MPa] - Initial pressure\n",
+      "T_1 = 200 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 8.;\t\t\t#[MPa] - Final pressure\n",
+      "Y = 1.4;\t\t\t# Index of expansion of gas\n",
+      "R = 8.314;\t\t\t#[J/mol-K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# The exit temperature for ideal gas under isentropic conditions is given by \n",
+      "T_2 = round(T_1*(round((P_2/P_1)**((Y-1)/Y),2)),1);\t\t\t#[K] - Exit temperature\n",
+      "\n",
+      "Cp_0 = round(Y*R/(Y-1),1);\t\t\t#[J/mol-K] - Specific heat capacity at constant pressure\n",
+      "\t\t\t# For isentropic conditions the enthalpy change for ideal gas is given by\n",
+      "delta_H_s = Cp_0*(T_2 - T_1);\t\t\t#[J/mol]\n",
+      "\t\t\t# Therefore work is given by\n",
+      "W = - delta_H_s;\t\t\t#[J/mol]\n",
+      "\n",
+      "print \" 1).The exit temperature of steam is %.2f K\"%(T_2);\n",
+      "print \"     The required work is %.2f J/mol\"%(W);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "eff = 0.8;\t\t\t# Adiabatic efficiency\n",
+      "\t\t\t# delta_H_s/delta_H_a = 0.8\n",
+      "delta_H_a = delta_H_s/eff;\t\t\t#[J/mol] - Actual enthalpy change\n",
+      "W_a = - delta_H_a;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# The ideal gas enthalpy is a function only of temperature,therefore actual exit temperature T_2a is given by \n",
+      "\t\t\t# delta_H_a = Cp_0*(T_2a - T_1)\n",
+      "T_2a = (delta_H_a/Cp_0) + T_1;\n",
+      "\n",
+      "print \" 2).The exit temperature of steam is %.2f K\"%(T_2a);\n",
+      "print \"     The required work is %.2f J/mol\"%(W_a);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The exit temperature of steam is 856.40 K\n",
+        "     The required work is -11152.58 J/mol\n",
+        " 2).The exit temperature of steam is 952.21 K\n",
+        "     The required work is -13940.72 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.10  Page Number : 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\t\t\t\n",
+      "\n",
+      "# Variables\n",
+      "P_1 = 1;\t\t\t#[MPa] - Initial pressure\n",
+      "T_1 = 200 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 8;\t\t\t#[MPa] - Final pressure\n",
+      "Y = 1.4;\t\t\t# Index of expansion of gas\n",
+      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
+      "\t\t\t# Cp_0 = 7.7 + 0.04594*10**(-2)*T + 0.2521*10**(-5)*T**(2) - 0.8587*10**(-9)*T**(3), here T is in K and Cp_0 is in  cal/mol-K\n",
+      "a = 7.7;\n",
+      "b = 0.04594*10**(-2);\n",
+      "c = 0.2521*10**(-5);\n",
+      "d = - 0.8587*10**(-9);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# delta_S = integral((Cp_0/T)*dT) - R*math.log(P_2/P_1) = 0\n",
+      "\t\t\t# delta_S = integral(((a + b*T + c*T**(2) + d*T**(3))/T)*dT) - R*math.log(P_2/P_1) = 0\n",
+      "\t\t\t# delta_S = a*math.log(T_2/T_1) + b*(T_2 - T_1) + (c/2)*(T_2**(2) - T_1**(2)) + (d/3)*(T_2**(3) - T_1**(3)) - R*math.log(P_2/P_1) = 0\n",
+      "\t\t\t# Solving for T_2 in the above equation we get\n",
+      "def f(T_2): \n",
+      "\t return a*math.log(T_2/T_1)+b*(T_2-T_1)+(c/2)*(T_2**(2)-T_1**(2))+(d/3)*(T_2**(3)-T_1**(3))-R*math.log(P_2/P_1)\n",
+      "T_2 = fsolve(f,100)\n",
+      "\n",
+      "\t\t\t# Now let us calculate the enthalpy change under these conditions \n",
+      "\n",
+      "def f0(T): \n",
+      "\t return 7.7+0.04594*10**(-2)*T+0.2521*10**(-5)*T**(2)-0.8587*10**(-9)*T**(3)\n",
+      "\n",
+      "delta_H_s =  quad(f0,T_1,T_2)[0]\n",
+      "\n",
+      "delta_H_s = delta_H_s*4.184;\t\t\t#[J/mol]\n",
+      "\t\t\t# Therefore isentropic work done is\n",
+      "W = - delta_H_s;\n",
+      "\n",
+      "print \" 1).The exit temperature of steam is %f K\"%(T_2);\n",
+      "print \"     The required work is %f J/mol\"%(W);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "eff = 0.8;\n",
+      "delta_H_a = delta_H_s/eff;\t\t\t#[J/mol] - Actual enthalpy change\n",
+      "\t\t\t# Therefore actual work done is given by \n",
+      "W_a = - delta_H_a;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# Now we have to determine the exit temperature under actual conditions\n",
+      "\t\t\t# delta_H_a = integral(Cp_0*dT)  from limit T_1 = 473.15 K to T_2\n",
+      "\t\t\t# On putting the values and simplifying we get\n",
+      "\t\t\t# 7.7*T_2 + 0.02297*10**(-2)*T_2**(2) + 0.084*10**(-5)*T_2**(3) - 0.214675*10**(-9)*T_2**(4) - 6907.106 = 0\n",
+      "\n",
+      "def f1(T_2_prime): \n",
+      "\t return a*(T_2_prime-T_1)+(b/2)*(T_2_prime**(2)-T_1**(2))+(c/3)*(T_2_prime**(3)-T_1**(3))+(d/4)*(T_2_prime**(4)-T_1**(4))-(delta_H_a/4.184)\n",
+      "T_2_prime = fsolve(f1,100)\n",
+      "\n",
+      "print \" 2).The exit temperature of steam is %f K\"%(T_2_prime);\n",
+      "print \"     The required work is %f J/mol\"%(W_a);\n",
+      "\n",
+      "# Answers are vary because of rounding error. Calculations are correct. "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The exit temperature of steam is 760.037501 K\n",
+        "     The required work is -10499.128839 J/mol\n",
+        " 2).The exit temperature of steam is 828.044888 K\n",
+        "     The required work is -13123.911049 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.11  Page Number : 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P_1 = 1;\t\t\t#[MPa] - Initial pressure\n",
+      "T_1 = 200 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 8;\t\t\t#[MPa] - Final pressure\n",
+      "Y = 1.4;\t\t\t# Index of expansion of gas\n",
+      "\n",
+      "\t\t\t# At state 1 (1 MPa and 200 C) from steam table as reported in the book\n",
+      "H_1 = 2827.9;\t\t\t#[kJ/kg]\n",
+      "S_1 = 6.694;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# At state 2 (8 MPa)\n",
+      "S_2 = S_1;\t\t\t# Isentropic process\n",
+      "\t\t\t# From steam table at 8 MPa and 450 C\n",
+      "S_21 = 6.5551;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t# From steam table at 8 MPa and 500 C\n",
+      "S_22 = 6.7240;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t# Therefore temperature at which entropy of steam is 6.694 kJ/kg-K is given by\n",
+      "T_2 = 450 + (500-450)/(S_22-S_21)*(S_2-S_21);\t\t\t#[C]\n",
+      "T_2 = T_2 + 273.15;\t\t\t#[K]\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# Enthalpy of steam at 8 MPa and 450 C from steam table as reported in the book \n",
+      "H_21 = 3272.0;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# And at 8 MPA and 500 C\n",
+      "H_22 = 3398.3;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Therefore enthalpy of steam at 8 MPa and T_2 \n",
+      "H_2 = H_21 + ((H_22-H_21)/(500-450))*((T_2-273.15) - 450); \n",
+      "\t\t\t# Work done is given by \n",
+      "\t\t\t# W = - delta_H_s\n",
+      "W = - (H_2 - H_1);\t\t\t#[J/g]\n",
+      "W = W*18.015;\t\t\t#[J/mol]\n",
+      "delta_H_s = - W;\n",
+      "\n",
+      "print \" 1).The exit temperature of steam is %f K\"%(T_2);\n",
+      "print \"     The required work is %f J/mol\"%(W);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "eff = 0.8;\t\t\t# Adiabatic efficiency\n",
+      "\t\t\t# delta_H_s/delta_H_a = 0.8\n",
+      "delta_H_a = delta_H_s/eff;\t\t\t#[J/mol] - Actual enthalpy change\n",
+      "\t\t\t# Therefore actual work done \n",
+      "W_a = - delta_H_a;\t\t\t#[J/mol]\n",
+      "\t\t\t# Enthalpy at actual exit conditions is \n",
+      "H_2_a = H_1 + delta_H_a/18.015;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# Enthalpy of steam at 8 MPa and 500 C from steam table as reported in the book \n",
+      "H_21_a = 3398.3;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# And at 8 MPA and 550 C\n",
+      "H_22_a = 3521.0;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Therefore temperature at H_22_a is given by\n",
+      "T_2_a = 500 + ((550-500)*(H_2_a - H_21_a))/(H_22_a - H_21_a);\t\t\t#[C]\n",
+      "T_2_a = T_2_a + 273.15;\t\t\t#[K]\n",
+      "\n",
+      "print \" 2).The exit temperature of steam is %f K\"%(T_2_a);\n",
+      "print \"     The required work is %f J/mol\"%(W_a);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The exit temperature of steam is 764.269005 K\n",
+        "     The required work is -9871.618433 J/mol\n",
+        " 2).The exit temperature of steam is 819.832257 K\n",
+        "     The required work is -12339.523042 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.12  Page Number : 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 140.;\t\t\t#[kPa] - Initial pressure\n",
+      "T_1 = 20. + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 560.;\t\t\t#[kPa] - Final pressure\n",
+      "eff = 0.75;\t\t\t# Compressor efficiency\n",
+      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
+      "\t\t\t# Cp_0 = 4.750 + 1.200*10**(-2)*T + 0.3030*10**(-5)*T**(2) - 2.630*10**(-9)*T**(3), here T is in K and Cp_0 is in  cal/mol-K\n",
+      "a = 7.7;\n",
+      "b = 0.04594*10**(-2);\n",
+      "c = 0.2521*10**(-5);\n",
+      "d = - 0.8587*10**(-9);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# At 20 C,as reported in the book\n",
+      "Cp_0 = 8.46;\t\t\t#[cal/mol-K] - Specific heat capacity at constant pressure\n",
+      "Cv_0 = Cp_0 - R;\t\t\t#[cal/mol-K] - Specific heat capacity at constant volume\n",
+      "Y = Cp_0/Cv_0;\t\t\t# Index of expansion\n",
+      "\n",
+      "\t\t\t# Assuming 100 % efficiency,for reversible and adiabatic process the final temperature is given by\n",
+      "\t\t\t# P*V**(Y) = constant or, P*((R*T)/P)**(Y) = constant\n",
+      "T_2 = ((P_1/P_2)**((1-Y)/Y))*T_1;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# Since at final temperature the value of heat capacity ratio would have changed \n",
+      "\t\t\t# So let us determine Y at mean temperature and then calculating final temperature\n",
+      "T_mean = (T_1 + T_2)/2;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# At T_mean,as reported in the book\n",
+      "Cp_0_new = 9.153;\t\t\t#[cal/mol-K]\n",
+      "Cv_0_new = Cp_0_new - R;\t\t\t#[cal/mol-K]\n",
+      "Y_new = Cp_0_new/Cv_0_new;\n",
+      "T_2_new = T_1*((P_1/P_2)**((1-Y_new)/Y_new));\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# The enthalpy change is given by \n",
+      "\n",
+      "def f23(T): \n",
+      "\t return 4.750+1.200*10**(-2)*T+0.3030*10**(-5)*T**(2)-2.630*10**(-9)*T**(3)\n",
+      "\n",
+      "delta_H =  quad(f23,T_1,T_2_new)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t#For adiabatic process\n",
+      "W = - delta_H;\t\t\t#[cal/mol]\n",
+      "\t\t\t# Now actual work done on the system is given by \n",
+      "W_a = W/eff;\t\t\t#[cal/mol]\n",
+      "\n",
+      "\n",
+      "\t\t\t# Therefore actual change in enthalpy is - W_a, or\n",
+      "\t\t\t# - W_a = 4.750*(T_2-T_1) + (1.2*10**(-2)/2)*(T_2**(2)-T_1**(2)) + (0.3030*10**(-5)/3)*(T_2**(3)-T_1**(3)) - (2.63*10**(-9)/4)*(T_2***(4)-T_1**(4));\n",
+      "\t\t\t# Solving for T_2 in the above equation\n",
+      "def f1(T_2_prime): \n",
+      "\t return 4.750*(T_2_prime-T_1)+((1.2*10**(-2))/2)*(T_2_prime**(2)-T_1**(2))+((0.3030*10**(-5))/3)*(T_2_prime**(3)-T_1**(3))-((2.63*10**(-9))/4)*(T_2_prime**(4)-T_1**(4))+W_a\n",
+      "T_2_prime=fsolve(f1,100)\n",
+      "\n",
+      "print \" The required work is %f cal/mol\"%(W_a);\n",
+      "print \" The discharge temperature of methane is %f K\"%(T_2_prime);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The required work is -1254.158148 cal/mol\n",
+        " The discharge temperature of methane is 427.374758 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.13  Page Number : 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 10;\t\t\t#[bar] - Initial pressure\n",
+      "T_1 = 500 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 2;\t\t\t#[psia] - Final pressure\n",
+      "P_2 = P_2/14.5;\t\t\t#[bar]\n",
+      "P_2 = P_2*10**(2);\t\t\t#[kPa]\n",
+      "m = 1.8;\t\t\t#[kg/s] - Mass flux\n",
+      "eff = 0.8;\t\t\t# Efficiency\n",
+      "\n",
+      "\t\t\t# At state 1, from steam table\n",
+      "H_1 = 3478.5;\t\t\t#[kJ/kg]\n",
+      "S_1 = 7.7622;\t\t\t#[kJ/kg-K]\n",
+      "S_2 = S_1;\t\t\t# Adiabatic process\n",
+      "\n",
+      "\t\t\t# From saturated steam table at 10 kPa\n",
+      "S_liq_1 = 0.6493;\t\t\t#[kJ/kg-K]\n",
+      "S_vap_1 = 8.1502;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t# From saturated steam table at 15 kPa\n",
+      "S_liq_2 = 0.7549;\t\t\t#[kJ/kg-K]\n",
+      "S_vap_2 = 8.0085;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# Threfore at P_2 \n",
+      "S_liq = S_liq_1 + ((S_liq_2-S_liq_1)/(15-10))*(P_2-10);\n",
+      "S_vap = S_vap_1 + ((S_vap_2-S_vap_1)/(15-10))*(P_2-10);\n",
+      "\n",
+      "\t\t\t# The dryness fraction at exit state is\n",
+      "x_2 = (S_1-S_liq)/(S_vap-S_liq);\n",
+      "\t\t\t# The enthalpy at exit to be determined. At 10 kPa\n",
+      "H_liq_1 = 191.83;\t\t\t#[kJ/kg]\n",
+      "H_vap_1 = 2584.7;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# At 15 kPa\n",
+      "H_liq_2 = 225.94;\t\t\t#[kJ/kg]\n",
+      "H_vap_2 = 2599.1;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Therfore at P_2\n",
+      "H_liq = H_liq_1 + ((H_liq_2-H_liq_1)/(15-10))*(P_2-10);\n",
+      "H_vap = H_vap_1 + ((H_vap_2-H_vap_1)/(15-10))*(P_2-10);\n",
+      "\n",
+      "\t\t\t# Enthalpy at state 2\n",
+      "H_2_s = H_liq*(1-x_2) + x_2*H_vap;\t\t\t#[kJ/kg]\n",
+      "W = m*(H_1 - H_2_s);\t\t\t#[kW]\n",
+      "\n",
+      "print \" 1).The power output is %f kW\"%(W);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# If the process is 80 % efficient the enthalpy change is\n",
+      "\t\t\t# H_1 - H_2_a = eff*(H_1 - H_2_s)\n",
+      "H_2_a = H_1 - (0.8*(H_1 - H_2_s));\n",
+      "\n",
+      "\t\t\t# Now under these conditions temperature and entropy have to be determined. From superheated steam tables,as reported in the book\n",
+      "\t\t\t# At 10 kPa and 100 C\n",
+      "H_2_1 = 2687.5;\t\t\t#[kJ/kg]\n",
+      "S_2_1 = 8.4479;\t\t\t#[kJ/kg-k]\n",
+      "\t\t\t# At 10 kPa and 150 C\n",
+      "H_2_2 = 2783.0;\t\t\t#[kJ/kg]\n",
+      "S_2_2 = 8.6882;\t\t\t#[kJ/kg-k]\n",
+      "\t\t\t# At 50 kPa and 100 C\n",
+      "H_3_1 = 2682.5;\t\t\t#[kJ/kg]\n",
+      "S_3_1 = 7.6947;\t\t\t#[kJ/kg-k]\n",
+      "\t\t\t# At 50 kPa and 150 C\n",
+      "H_4_1 = 2780.1;\t\t\t#[kJ/kg]\n",
+      "S_4_1 = 7.9401;\t\t\t#[kJ/kg-k]\n",
+      "\t\t\t# Therefore at P_2 and 100 C\n",
+      "H_prime_1 = H_2_1 + ((H_3_1-H_2_1)/(50-10))*(P_2-10);\t\t\t#[kJ/kg]\n",
+      "S_prime_1 = S_2_1 + ((S_3_1-S_2_1)/(50-10))*(P_2-10);\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t# Therefore at P_2 and 150 C\n",
+      "H_prime_2 = H_2_2 + ((H_4_1-H_2_2)/(50-10))*(P_2-10);\t\t\t#[kJ/kg]\n",
+      "S_prime_2 = S_2_2 + ((S_4_1-S_2_2)/(50-10))*(P_2-10);\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t# Enthalpy at exit is H_2_a. So at this condition temperature can be nom be determined\n",
+      "T_exit = ((H_2_a - H_prime_1)/(H_prime_2 - H_prime_1))/(150-100) + 100;\t\t\t#[C]\n",
+      "\t\t\t# The entropy at exit is\n",
+      "S_exit =  ((H_2_a - H_prime_1)/(H_prime_2 - H_prime_1))/(S_prime_2 - S_prime_1) + S_prime_1;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" 2).The entropy at exit is %f kJ/kg-K\"%(S_exit);\n",
+      "print \"     The temperature of the exit state is %f C\"%(T_exit);\n",
+      "\n",
+      "print \"     The irreversibility is advatageous because the exit steam is superheated and therefore\";\n",
+      "print \"     the blades of the turbine are not eroded by water droplets which get formed when the process is isentropic\";\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The power output is 1753.346828 kW\n",
+        " 2).The entropy at exit is 8.906311 kJ/kg-K\n",
+        "     The temperature of the exit state is 100.002552 C\n",
+        "     The irreversibility is advatageous because the exit steam is superheated and therefore\n",
+        "     the blades of the turbine are not eroded by water droplets which get formed when the process is isentropic\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.14  Page Number : 166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 6;\t\t\t#[MPa] - Initial pressure\n",
+      "T_1 = 500 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 10;\t\t\t#[kPa] - Final pressure\n",
+      "out_qlty = 0.9;\t\t\t# Output quality\n",
+      "\n",
+      "\t\t\t# At 6 MPa and 500 C, from steam table as reported in the book\n",
+      "H_1 = 3422.2;\t\t\t#[kJ/kg]\n",
+      "S_1 = 6.8803;\t\t\t#[kJ/kg-K]\n",
+      "S_2 = S_1;\t\t\t# Adiabatic reversible conditions\n",
+      "\t\t\t# At 10 kPa saturated\n",
+      "H_liq = 191.83;\t\t\t#[kJ/kg]\n",
+      "H_vap = 2584.7;\t\t\t#[kJ/kg]\n",
+      "S_liq = 0.6493;\t\t\t#[kJ/kg-K]\n",
+      "S_vap = 8.1502;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The dryness fraction is given by \n",
+      "x = (S_1-S_liq)/(S_vap-S_liq);\n",
+      "\n",
+      "\t\t\t# Now the exit enthalpy is given by\n",
+      "H_2 = H_liq*(1-x) + x*H_vap;\t\t\t#[kJ/kg]\n",
+      "W = - (H_2 - H_1);\t\t\t#[kJ/kg] - Under isentropic conditions\n",
+      "\n",
+      "\t\t\t# We know that, delta_S = q/T_b + S_gen\n",
+      "\t\t\t# Since delta_S = 0, therefore under isentropic conditions\n",
+      "S_gen = 0;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t# Now for output quality 0.9\n",
+      "H_2_a = H_liq*(1-out_qlty) + out_qlty*H_vap;\t\t\t#[kJ/kg]\n",
+      "S_2_a = S_liq*(1-out_qlty) + out_qlty*S_vap;\t\t\t#[kJ/kg]\n",
+      "W_a = - (H_2_a - H_1);\t\t\t#[kJ/kg]\n",
+      "delta_S_a = S_2_a - S_1;\t\t\t#[kJ/kg-k]\n",
+      "\t\t\t# Again,  delta_S = q/T_b + S_gen\n",
+      "\t\t\t# Since q = 0, therefore under isentropic conditions\n",
+      "S_gen_a = delta_S_a;\t\t\t#[kJ/kg-K\n",
+      "\t\t\t# Now efficiency is given by eff = delta_H_a/delta_H_s\n",
+      "eff = W_a/W;\n",
+      "\n",
+      "print \" For output quality = 0.9\";\n",
+      "print \" The work output per unit mass is %f kJ/kg\"%(W_a);\n",
+      "print \" The entropy generation is given by S_gen = %f kJ/kg-K\"%(S_gen_a);\n",
+      "print \" The efficiency with respect to reversible adiabatic case is given by eff = %f\"%(eff);\n",
+      "\n",
+      "\t\t\t# Now for output quality 1\n",
+      "out_qlty_1 = 1;\n",
+      "H_2_a_1 = H_liq*(1-out_qlty_1) + out_qlty_1*H_vap;\t\t\t#[kJ/kg]\n",
+      "S_2_a_1 = S_liq*(1-out_qlty_1) + out_qlty_1*S_vap;\t\t\t#[kJ/kg]\n",
+      "W_a_1 = - (H_2_a_1 - H_1);\t\t\t#[kJ/kg]\n",
+      "delta_S_a_1 = S_2_a_1 - S_1;\t\t\t#[kJ/kg-k]\n",
+      "\t\t\t# Again,  delta_S = q/T_b + S_gen\n",
+      "\t\t\t# Since q = 0, therefore under isentropic conditions\n",
+      "S_gen_a_1 = delta_S_a_1;\t\t\t#[kJ/kg-K\n",
+      "\t\t\t# Now efficiency is given by eff = delta_H_a/delta_H_s\n",
+      "eff_1 = W_a_1/W;\n",
+      "\n",
+      "print \" For output quality = 1.0\";\n",
+      "print \" The work output per unit mass is %f kJ/kg\"%(W_a_1);\n",
+      "print \" The entropy generation is given by S_gen = %f kJ/kg-K\"%(S_gen_a_1);\n",
+      "print \" The efficiency with respect to reversible adiabatic case is given by eff = %f\"%(eff_1);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " For output quality = 0.9\n",
+        " The work output per unit mass is 1076.787000 kJ/kg\n",
+        " The entropy generation is given by S_gen = 0.519810 kJ/kg-K\n",
+        " The efficiency with respect to reversible adiabatic case is given by eff = 0.866551\n",
+        " For output quality = 1.0\n",
+        " The work output per unit mass is 837.500000 kJ/kg\n",
+        " The entropy generation is given by S_gen = 1.269900 kJ/kg-K\n",
+        " The efficiency with respect to reversible adiabatic case is given by eff = 0.673983\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.15  Page Number : 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 3.;\t\t\t#[bar] - Initial pressure\n",
+      "T_1 = 150. + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "Vel_1 = 90.;\t\t\t#[m/s] - Initial velocity\n",
+      "P_2 = 1.;\t\t\t#[bar] - Final pressure\n",
+      "eff = 0.95;\t\t\t# Adiabatic effciciency of the nozzle\n",
+      "R = 8.314;\t\t\t#[J/mol*-] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# At 3 bar and 150 C, from steam table\n",
+      "S_1 = 7.0778;\t\t\t#[kJ/kg-K]\n",
+      "H_1 = 2761.0;\t\t\t#[kJ/kg]\n",
+      "S_2 = S_1;\t\t\t# \n",
+      "\n",
+      "\t\t\t# At 1 bar saturated\n",
+      "S_liq = 1.3026;\t\t\t#[kJ/kg-K]\n",
+      "S_vap = 7.3594;\t\t\t#[kJ/kg-K]\n",
+      "H_liq = 417.46;\t\t\t#[kJ/kg]\n",
+      "H_vap = 2675.5;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The dryness factor of exit steam can be determined as\n",
+      "x = (S_1-S_liq)/(S_vap-S_liq);\n",
+      "\t\t\t# Enthalpy of exit steam is given by \n",
+      "H_2 = H_liq*(1-x) + x*H_vap;\t\t\t#[kJ/kg]\n",
+      "delta_H_s = H_2 - H_1;\t\t\t#[kJ/kg] - Enthalpy change\n",
+      "delta_H_a = eff*delta_H_s;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# Assuming no heat exchange with surroundings and since no work is done\n",
+      "\t\t\t# delta_H + (delta_V**(2))/2 = 0\n",
+      "delta_Vel_square = 2*(-delta_H_a)*1000;\t\t\t#[m**(2)/s**(2)]\n",
+      "Vel_2 = (delta_Vel_square + Vel_1**(2))**(1./2);\t\t\t#[m/s]\n",
+      "\n",
+      "print \" 1).The final velocity when fluid is steam) is %f m/s\"%(Vel_2);\n",
+      "\n",
+      "\t\t\t# (2)\n",
+      "Y = 1.4;\t\t\t# Index of expansion\n",
+      "Cp_0 = (Y*R)/(Y-1);\t\t\t#[J/mol-K] - Specific heat capacity at constant pressure\n",
+      "\n",
+      "\t\t\t# P_1**(1-Y)*T_1**(Y) = P_2**(1-Y)*T_2**(Y)\n",
+      "T_2 = T_1*(P_1/P_2)**((1-Y)/Y);\t\t\t#[K]\n",
+      "delta_H_s_2 = Cp_0*(T_2 - T_1);\t\t\t#[J/mol]\n",
+      "delta_H_a_2 = eff*delta_H_s_2;\t\t\t#[J/mol]\n",
+      "delta_H_a_2 = (delta_H_a_2*1000)/28.84;\t\t\t#[J/kg]\n",
+      "\n",
+      "delta_Vel_square_2 = 2*(-delta_H_a_2);\t\t\t#[m**(2)/s**(2)]\n",
+      "Vel_2_2 = (delta_Vel_square_2 + Vel_1**(2))**(1./2);\t\t\t#[m/s]\n",
+      "\n",
+      "print \" 2).The final velocity when fluid is air which behaves as an ideal gas) is %f m/s\"%(Vel_2_2);\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The final velocity when fluid is steam) is 608.291583 m/s\n",
+        " 2).The final velocity when fluid is air which behaves as an ideal gas) is 476.065890 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.16  Page Number : 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P_1 = 300.;\t\t\t#[kPa] - Initial pressure\n",
+      "T_1 = 450.;\t\t\t#[K] - Initial temperature\n",
+      "Vel_1 = 90.;\t\t\t#[m/s] - Initial velocity\n",
+      "P_2 = 180.;\t\t\t#[kPa] - Final pressure\n",
+      "eff = 0.95;\t\t\t# Adiabatic effciciency of the nozzle\n",
+      "R = 8.314;\t\t\t#[J/mol*-] - Universal gas constant\n",
+      "Cp = 5.19;\t\t\t#[kJ/kg-K] - Specific heat capacity at constant pressure\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(a)\n",
+      "\n",
+      "\n",
+      "Mol_wt_He = 4;\t\t\t#[g/mol] - Molecular weight of helium\n",
+      "R_He = R/Mol_wt_He;\t\t\t# 'R' for helium\n",
+      "Y = Cp/(Cp - R_He);\n",
+      "\n",
+      "\t\t\t# Now temperature at exit to be determined\n",
+      "T_2s = T_1*(P_1/P_2)**((1-Y)/Y);\t\t\t#[K]\n",
+      "delta_H_2s = Cp*(T_2s - T_1);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# Since no work is done and heat exchange is zero,from first law we get\n",
+      "\t\t\t# delta_H + delta_Vel**(2)/2 = 0\n",
+      "delta_Vel_square = 2*(-delta_H_2s)*1000;\t\t\t#[m**(2)/s**(2)]\n",
+      "Vel_2 = (delta_Vel_square)**(1./2);\t\t\t#[m/s] - ( as Vel_1  <<  Vel_2)\n",
+      "\n",
+      "print \" a).The maximum exit velocity is %f m/s\"%(Vel_2);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "T_2a = 373;\t\t\t#[K] - Measured temperature of helium\n",
+      "delta_H_a = Cp*(T_2a - T_1);\t\t\t#[kJ/kg]\n",
+      "delta_Vel_square_a = 2*(-delta_H_a)*1000;\t\t\t#[m**(2)/s**(2)]\n",
+      "Vel_2a = (delta_Vel_square_a)**(1./2);\t\t\t#[m/s] - ( as Vel_1  <<  Vel_2a)\n",
+      "\n",
+      "print \" b).The actual exit velocity is %f m/s\"%(Vel_2a);\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "delta_S_sys = Cp*math.log(T_2a/T_1) - R_He*math.log(P_2/P_1);\n",
+      "\t\t\t# we know that delta_S_sys = q/T_b + S_gen and since q = 0, therfore\n",
+      "S_gen = delta_S_sys;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" c).The increasse in entropy per unit mass is %f kJ/kg-K\"%(S_gen);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The maximum exit velocity is 929.607346 m/s\n",
+        " b).The actual exit velocity is 894.013423 m/s\n",
+        " c).The increasse in entropy per unit mass is 0.087748 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.17  Page Number : 170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 1;\t\t\t#[bar] - Initial pressure\n",
+      "T_1 = 150 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "V_2 = 0.28;\t\t\t#[m**(3)/kg] - Final specific volume\n",
+      "T_2 = T_1;\t\t\t#[K] - Isothermal process\n",
+      "R = 8.314;\t\t\t#[J/mol-K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# At 1 bar and 150 C, from steam table\n",
+      "S_1 = 7.6134;\t\t\t#[kJ/kg-K]\n",
+      "H_1 = 2776.4;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# At 150 C saturated\n",
+      "V_liq = 0.001091;\t\t\t#[m**(3)/kg]\n",
+      "V_vap = 0.3928;\t\t\t#[m**(3)/kg]\n",
+      "H_liq = 632.2;\t\t\t#[kJ/kg]\n",
+      "H_vap = 2746.5;\t\t\t#[kJ/kg]\n",
+      "S_liq = 1.8418;\t\t\t#[kJ/kg-K]\n",
+      "S_vap = 6.8379;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The dryness factor of exit steam can be determined as\n",
+      "x = (V_2 - V_liq)/(V_vap - V_liq);\n",
+      "S_2 = S_liq*(1-x) + x*S_vap;\t\t\t#[kJ/kg-K] -Entropy \n",
+      "H_2 = H_liq*(1-x) + x*H_vap;\t\t\t#[kJ/kg] -Enthalpy \n",
+      "delta_H = H_2 - H_1;\t\t\t#[kJ/kg] - Enthalpy change\n",
+      "delta_S = S_2 - S_1;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# Since the compression is reversible\n",
+      "q = T_1*delta_S;\t\t\t#[kJ/kg] - Heat transfer\n",
+      "\t\t\t# From first law  q - W = delta_H\n",
+      "W = q - delta_H;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "print \" 1).The amount of heat transfer when fluid is steam) is %f kJ/kg\"%(q)\n",
+      "print \"     The amount of work transfer when fluid is steam) is %f kJ/kg\"%(W)\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "V_2 = V_2*(28.84/1000);\t\t\t#[m**(3)/mol] - Molar volume at exit\n",
+      "\t\t\t# Pressure at exit is given by \n",
+      "P_2 = ((R*T_2)/V_2);\t\t\t#[N/m**(2)]\n",
+      "P_2 = P_2*10**(-5);\t\t\t#[bar]\n",
+      "\n",
+      "\t\t\t# Entropy change is given by, delta_S_2 = Cp*math.log(T_2/T_1) - R*math.log(P_2/P_1), but since T_1 = T_2, therfore\n",
+      "delta_S_2 = - R*math.log(P_2/P_1);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "q_2 = T_1*delta_S_2;\t\t\t#[J/mol] - Heat change\n",
+      "q_2 = q_2/28.84;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# Enthalpy change is given by,  delta_H_2 = Cp*(T_2 - T_1)  = 0 (as T_1 = T_2)\n",
+      "delta_H_2 = 0;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# From first law  q - W = delta_H, therefore\n",
+      "W_2 = q_2 - delta_H_2;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "print \" 2).The amount of heat transfer when fluid is ideal gas) is %f kJ/kg\"%(q_2)\n",
+      "print \"     The amount of work transfer when fluid is ideal gas) is %f kJ/kg\"%(W_2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The amount of heat transfer when fluid is steam) is -936.947741 kJ/kg\n",
+        "     The amount of work transfer when fluid is steam) is -298.195149 kJ/kg\n",
+        " 2).The amount of heat transfer when fluid is ideal gas) is -179.526401 kJ/kg\n",
+        "     The amount of work transfer when fluid is ideal gas) is -179.526401 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.18  Page Number : 171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 7*10**(5);\t\t\t#[Pa] - Initial pressure\n",
+      "T_1 = 95 + 273.15;\t\t\t#[K] - Initial temperature\n",
+      "P_2 = 3.5*10**(5);\t\t\t#[Pa] - Final pressure\n",
+      "dia = 15*10**(-2);\t\t\t#[m] - Diameter of pipe\n",
+      "m = 2;\t\t\t#[kg/s] - Mass flow rate\n",
+      "R = 8.314;\t\t\t#[J/mol-K] - Universal gas constant\n",
+      "Y = 1.4;\t\t\t# Index of expansion\n",
+      "Cp_0 = (R*Y)/(Y-1);\t\t\t#[J/mol-K] - Specific heat capacity at constant pressure\n",
+      "Cp_0 = (Cp_0/28.84)*1000;\t\t\t#[J/kg-K]\n",
+      "rho_1 = 6.6;\t\t\t#[kg/m**(3)] - Density\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# velocity before throttling is to be determined m = rho*Vol*Area\n",
+      "V_1 = (R*T_1)/P_1;\t\t\t#[m**(3)/mol] - Specific volume\n",
+      "V_1 = (V_1/28.84)*1000;\t\t\t#[m**(3)/kg]\n",
+      "Vel_1 = m/(rho_1*3.14*(dia/2)**(2));\t\t\t#[m/s] - Velocity before throttling\n",
+      "\n",
+      "\t\t\t# Let the temperature after throttling be T_2, then\n",
+      "\t\t\t# Vel_2 = m/(rho_2*Area) = (m*V_2)/(3.14*(dia/2)**(2))\n",
+      "\t\t\t# From first law, since q = W = 0, we get\n",
+      "\t\t\t# delta_H + (delta_V**(2))/2 = 0\n",
+      "\t\t\t#Cp_0*(T_2 - T_1) +  (((m*((((R*T_2)/P_2)/28.84)*1000))/(3.14*(dia/2)**(2)))**(2) - (Vel_1)**(2))/2 = 0\n",
+      "\t\t\t# Solving the above equation for T_2, we get\n",
+      "def f1(T_2): \n",
+      "\t return Cp_0*(T_2 - T_1) +  (((m*((((R*T_2)/P_2)/28.84)*1000))/(3.14*(dia/2)**(2)))**(2) - (Vel_1)**(2))/2\n",
+      "T_2 =fsolve(f1,100)\n",
+      "\t\t\t# Therefore velocity of air downstream of restriction is given by \n",
+      "Vel_2 = ((m*((((R*T_2)/P_2)/28.84)*1000))/(3.14*(dia/2)**(2)));\t\t\t#[m/s]\n",
+      "\n",
+      "print \" The velocity of air downstream of restriction is %f m/s\"%(Vel_2);\n",
+      "\n",
+      "delta_T = (T_2 - T_1);\n",
+      "\t\t\t# Since temperature difference (delta_T) is very small, therefore enthalpy change is also very small\n",
+      "\n",
+      "\t\t\t# Entropy change is given by, delta_S = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1), but since T_1 and T_2 are almost equal\n",
+      "delta_S = - R*math.log(P_2/P_1);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "print \" The change in entropy is %f kJ/mol-k\"%(delta_S);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The velocity of air downstream of restriction is 34.295216 m/s\n",
+        " The change in entropy is 5.762826 kJ/mol-k\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch5_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch5_2.ipynb
new file mode 100755
index 00000000..6f1963f8
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch5_2.ipynb
@@ -0,0 +1,743 @@
+{
+ "metadata": {
+  "name": "ch5_2",
+  "signature": "sha256:beb354ab1f11a4413c580f54f984be6db9b7dd2c39bed82bc1f4874533d75759"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Exergy"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.1  Page Number : 184"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 500+273.15;\t\t\t#[C] - Condensation temperature\n",
+      "T_2 = 250+273.15;\t\t\t#[C] - Temperature at which vaporization takes place.\n",
+      "\n",
+      "T_3 = 25+273.15;\t\t\t#[C] - Ambient atmospheric temperature.\n",
+      "\n",
+      "Q = 1;\t\t\t\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#The exergy content of the vapour at 500 C,\n",
+      "Ex_T_1 = Q*(1-(T_3/T_1));\n",
+      "Ex_T_2 = Q*(1-(T_3/T_2));\n",
+      "\t\t\t#Therefore,loss in exergy is given by\n",
+      "Ex_loss = Ex_T_1 - Ex_T_2;\n",
+      "\t\t\t#Fraction of exergy lost due to irreversible process is,\n",
+      "Ex_fraction =(Ex_loss/Ex_T_1);\n",
+      "\n",
+      "# Results\n",
+      "print \" The fraction of exergy lost due to irreversible process is %f\"%(Ex_fraction);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fraction of exergy lost due to irreversible process is 0.299954\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.2  Page Number : 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T_1 = 300.;\t\t\t#[K] - Initial temperature.\n",
+      "P_1 = 100.;\t\t\t#[kPa] - Initial pressure.\n",
+      "T_2 = 500.;\t\t\t#[K] - Final temperature.\n",
+      "T_0 = 300.;\t\t\t#[K] - Environment temperature.\n",
+      "P_0 = 1.;\t\t\t#[atm] - Environment pressure.\n",
+      "R = 8.314;\t\t\t#[J/mol*K]\n",
+      "\t\t\t#(Cp_0/R)= 3.626\n",
+      "Cp_0 = 3.626*R;\t\t\t#[J/mol-K] - Heat capacity at constant pressure\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1).\n",
+      "\t\t\t#The availability change is given by, (phi_1 - phi_2) = U_1 - U_2 + P_0*(V_1 - V_2) - T_0*(S_1 - S_2)\n",
+      "\t\t\t#Let us determine the change in internal energy\n",
+      "\n",
+      "\t\t\t#For ideal gas Cp_0 - Cv_0 = R, and therefore\n",
+      "Cv_0 = ((Cp_0/R)- 1)*R;\t\t\t#[J/mol-K] - Heat capacity at constant volume\n",
+      "delta_U = Cv_0*(T_2-T_1);\t\t\t#[J/mol]\n",
+      "\t\t\t#delta_U = -w (from energy balance). Therefore, U1-U2 = -delta_U.\n",
+      "\t\t\t#The entropy change of ideal gas is given by\n",
+      "\t\t\t#delta_S = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1), but,(P1*V1/T1) = (P1*V1/T1) and therefore (P2/P1) = (T2/T1)\n",
+      "delta_S = Cp_0*math.log(T_2/T_1) - R*math.log(T_2/T_1);\t\t\t#[J/mol-K]\n",
+      "\t\t\t#The exergy change is given by, (phi_1 - phi_2) = U_1 - U_2 + P_0*(V_1 - V_2) - T_0*(S_1 - S_2)\n",
+      "\t\t\t#(V_1 - V_2) = 0, because the tank is rigid and so the volume is constant\n",
+      "delta_phi = (-delta_U) - T_0*(-delta_S);\t\t\t#[J/mol]\n",
+      "print \" 1).The change in exergy is %f J/mol\"%(delta_phi);\n",
+      "\n",
+      "\t\t\t#(2).\n",
+      "\t\t\t#Entropy change of the system is given by,  delta_S_sys = q/T_b + S_gen\n",
+      "\t\t\t#Since the system is adiabatic therefore,  delta_S_sys = S_gen\n",
+      "S_gen = delta_S;\n",
+      "\t\t\t#Irreversibility is given by\n",
+      "i = T_0*S_gen;\t\t\t#[J/mol]\n",
+      "print \" 2).The value of irreversibility is %f J/mol\"%(i);\n",
+      "\t\t\t#Irreversibility can also be determined using\n",
+      "\t\t\t#i = (W_rev_use - W_use)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The change in exergy is -1020.722863 J/mol\n",
+        " 2).The value of irreversibility is 3345.789937 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.3  Page Number : 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "# Variables\n",
+      "P_1 = 15.;\t\t\t#[bar] - Initial pressure\n",
+      "P_1 = P_1*10**(5);\t\t\t#[Pa]\n",
+      "T_1 = 300.+273.15;\t\t\t#[K] - Initial temperature\n",
+      "T_0 = 298.15;\t\t\t#[K]\n",
+      "T_R = 1200.;\t\t\t#[K] - Reservoir temerature.\n",
+      "P_0 = 1.;\t\t\t#[bar]\n",
+      "P_0 = P_0*10**(5);\t\t\t#[Pa]\n",
+      "n = 1.;\t\t\t#[mol] - No of moles\n",
+      "R = 8.314;\t\t\t#[J/mol*K]\n",
+      "Y = 1.4;\t\t\t# - Ratio of heat capacities.\n",
+      "Cv_0 = R/(Y-1);\t\t\t#[J/mol-K] - Heat capacity at constant volume\n",
+      "Cp_0 = Cv_0 + R;\t\t\t#[J/mol-K] - Heat capacity at constant pressure\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t#V_2 = 2*V_1 and since pressure is constant,we get (V_1/T_1) = (2*V_1/T_1), or, T_2 = 2*T_1.\n",
+      "T_2 = 2*T_1;\t\t\t#[K]\n",
+      "W = P_1*(((R*T_2)/P_1)-((R*T_1)/P_1));\t\t\t#[J/mol] - Actual work done\n",
+      "delta_U = Cv_0*(T_2-T_1);\t\t\t#[J/mol] - Change in internal energy.\n",
+      "q = W + delta_U;\t\t\t#[J/mol] - Heat change\n",
+      "\t\t\t#Now the availability change is given by, (phi_1 - phi_2) = U_1 - U_2 + P_0*(V_1 - V_2) - T_0*(S_1 - S_2) + q*(1-(T_0/T_R))\n",
+      "\t\t\t#delta_S = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1), and P_1 = P_2, Therefore\n",
+      "delta_S = Cp_0*math.log(T_2/T_1);\t\t\t#[J/mol-K] - Entropy change.\n",
+      "\t\t\t#Substituting expressions for delta_phi calculation. Decrease in availability is given by,\n",
+      "delta_phi = (-delta_U) + P_0*(((R*T_1)/P_1)-((R*T_2)/P_1)) - T_0*(-delta_S) + q*(1-(T_0/T_R));\t\t\t#[J/mol]\n",
+      "\t\t\t#Actual work done is given by, W = P_1*(V2-V1)\n",
+      "\t\t\t#Work done to print lace the atmosphere is given by, W = P_0*(V_2-V_1)\n",
+      "\t\t\t#Therefore, W_use = (P_1*(V2-V1) - P_0*(V2-V1))\n",
+      "W_use = (P_1-P_0)*(((R*T_2)/P_1)-((R*T_1)/P_1));\t\t\t#[J/mol] - useful work done\n",
+      "W_rev_use = delta_phi;\t\t\t# reversible useful work done\n",
+      "\t\t\t#Irreversibility is given by,\n",
+      "i = W_rev_use - W_use;\t\t\t#[J/mol]\n",
+      "print \" a).When the system is confined at constant pressure by a piston and is heated until it's volume is doubled \\n the ireversibility value is %f J/mol\"%(i);\n",
+      "\n",
+      "\t\t\t#The irreversibility can also be calculated using \n",
+      "\t\t\t# i = T_0*S_gen\n",
+      "\t\t\t#S_gen = delta_S - (q/T_R)\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\t\t\t#V2 = 2*V_1 and therefore T_2 = 2*T_1, as P_2 = P_1\n",
+      "\t\t\t#Actual work done is same as before\n",
+      "\t\t\t#Let work done on stirrer be W_stir. Thus net work done by the system is W - W_stir.Fron energy balance we get,\n",
+      "W_stir = W + delta_U;\n",
+      "\t\t\t#Initially the exergy is due to that of the system at state 1 and stirrer work,'W_stir' and finally we have the exergy due to system at state 2,the stirrer work  is spent,thus availability is given by\n",
+      "delta_phi_b = (-delta_U) + P_0*(((R*T_1)/P_1)-((R*T_2)/P_1)) - T_0*(-delta_S) + W_stir;\t\t\t#[J/mol]\n",
+      "W_rev_use_b = delta_phi_b;\t\t\t# reversible useful work done\n",
+      "W_use_b = W_use;\t\t\t# useful work done\n",
+      "\t\t\t#Now the irreversibility is given by,\n",
+      "i_b = W_rev_use_b - W_use_b;\t\t\t#[J/mol]\n",
+      "print \" b).When the system is confined at constant pressure by a piston and is stirred until it's volume is doubled \\n the ireversibility value is %f J/mol\"%(i_b);\n",
+      "\n",
+      "\t\t\t#The irreversibility can also be calculated using \n",
+      "\t\t\t# i_b = T_0*S_gen\n",
+      "\t\t\t#S_gen = delta_S - (q/T_R) and here, q = 0\n",
+      "\n",
+      "\t\t\t#(c)\n",
+      "P_2_c = 10;\t\t\t#[bar] - Final pressure, (Given)\n",
+      "P_2_c = P_2_c*10**(5);\t\t\t#[Pa]\n",
+      "\t\t\t#(P_1**(1-Y))*(T_1**(Y)) = (P_2**(1-Y))*(T_2**(Y))\n",
+      "T_2_c = T_1*((P_1/P_2_c)**((1-Y)/Y));\t\t\t#[K]\n",
+      "\t\t\t#Work done is given by, W = -delta_U = -Cv_0*(T_2_c - T_1)\n",
+      "W_c = -Cv_0*(T_2_c - T_1);\t\t\t#[J/mol]\n",
+      "\t\t\t#The final molar volume is calculated using P_1*V_1**(Y) = P_2*V_2**(Y)\n",
+      "\t\t\t#V_2 = V_1*((P_1/P_2_c)**(1/Y))\n",
+      "V_1 = (R*T_1)/P_1;\t\t\t#[cubic metre/mol] - Initial molar volume\n",
+      "V_2 = V_1*((P_1/P_2_c)**(1/Y));\t\t\t#[cubic metre/mol] - Final molar volume\n",
+      "\t\t\t#Now let us determine the work done to print lace the atmosphere,\n",
+      "W_atm_c = P_0*(V_2 - V_1);\t\t\t#[J/mol] - work done to print lace the atmosphere\n",
+      "\t\t\t#Thus useful work is given by,\n",
+      "W_use_c = W - W_atm_c;\t\t\t#[J/mol] - useful work done\n",
+      "\t\t\t#Here delta_S = 0,for reversible adiabatic process.Therefore,\n",
+      "W_rev_use_c = W_use_c;\n",
+      "\t\t\t#Now finally the irreversibility is given by,\n",
+      "i_c = W_rev_use_c - W_use_c;\t\t\t#[J/mol]\n",
+      "print \" c).When the system expands reversibly and adiabatically behind a piston \\n the ireversibility value is %f J/mol\"%(i_c);\n",
+      "\n",
+      "\t\t\t#(d)\n",
+      "\t\t\t#Here temperature is constant,but V_2 = 2*V_1, therefore P_2 = P_1/2\n",
+      "V_2_d = 2*V_1;\n",
+      "P_2_d = P_1/2;\n",
+      "\t\t\t#Under isothermal conditions work done is\n",
+      "W_d = R*T_1*math.log(V_2_d/V_1);\t\t\t#[J/mol]\n",
+      "\t\t\t#Work done to print lace the atmosphere is given by,\n",
+      "W_atm_d = P_0*(V_2_d - V_1);\t\t\t#[J/mol] - work done to print lace the atmosphere\n",
+      "\t\t\t#Thus useful work is given by,\n",
+      "W_use_d = W_d - W_atm_d;\t\t\t#[J/mol] - useful work done\n",
+      "delta_U_d = 0;\t\t\t#isothermal conditions\n",
+      "q_d = W_d;\t\t\t# since, delta_U_d = 0\n",
+      "\t\t\t#delta_S_d = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1), and T_1 = T_2, Therefore\n",
+      "delta_S_d = -R*math.log(P_2_d/P_1);\t\t\t#[J/mol-K] - Entropy change\n",
+      "\t\t\t#The reversible useful work is given by,\n",
+      "W_rev_use_d = P_0*(V_1 - V_2_d) - T_0*(-delta_S_d) + q_d*(1-(T_0/T_R));\t\t\t#[J/mol] - Reversible useful work done.\n",
+      "\t\t\t#The irreversibility is given by,\n",
+      "i_d = W_rev_use_d - W_use_d;\t\t\t#[J/mol]\n",
+      "print \" d).When the system expands reversibly and isothermally behind a piston until it's volume is doubled \\n the ireversibility value is %f J/mol\"%(i_d);\n",
+      "\n",
+      "\t\t\t#(e)\n",
+      "P_2_e = 10;\t\t\t#[bar] - Final pressure, (Given)\n",
+      "P_2_e = P_2_e*10**(5);\t\t\t#[Pa]\n",
+      "\t\t\t#During the expansion of an ideal gas in into vacuum the temperature of the gas remains the same,\n",
+      "T_2_e = T_1;\t\t\t# Final temperature\n",
+      "\t\t\t#Since boundary of the system is fixed so no net work is done, W = 0 and thus\n",
+      "W_use_e = 0;\t\t\t#[J/mol] - Useful work done\n",
+      "\t\t\t#Here, delta_U = 0,as temperature is same and\n",
+      "\t\t\t#(V_1-V_2) = 0,as for overall system there is no change in volume\n",
+      "delta_S_e = - R*math.log(P_2_e/P_1);\t\t\t#[J/mol-K] - Entropy change\n",
+      "\t\t\t#The reversible useful work is given by,\n",
+      "W_rev_use_e =  - T_0*(-delta_S_e);\t\t\t#[J/mol] - Reversible useful work done.\n",
+      "\t\t\t#The irreversibility is given by,\n",
+      "i_e = W_rev_use_e - W_use_e;\t\t\t#[J/mol]\n",
+      "print \" e).When the system expands adiabatically into an adjacent chamber \\n the ireversibility value is %f J/mol\"%(i_e);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).When the system is confined at constant pressure by a piston and is heated until it's volume is doubled \n",
+        " the ireversibility value is 1869.841742 J/mol\n",
+        " b).When the system is confined at constant pressure by a piston and is stirred until it's volume is doubled \n",
+        " the ireversibility value is 6013.652646 J/mol\n",
+        " c).When the system expands reversibly and adiabatically behind a piston \n",
+        " the ireversibility value is 0.000000 J/mol\n",
+        " d).When the system expands reversibly and isothermally behind a piston until it's volume is doubled \n",
+        " the ireversibility value is 897.537657 J/mol\n",
+        " e).When the system expands adiabatically into an adjacent chamber \n",
+        " the ireversibility value is 1005.074654 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.4  Page Number : 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 150+273.15;\t\t\t#[K] - Initial temperature.\n",
+      "m = 4.6;\t\t\t#[kg] - mass of water\n",
+      "P_1 = 1;\t\t\t#[MPa] - Initial pressure\n",
+      "Q = 11000;\t\t\t#[kJ] - Heat transferred to the system.\n",
+      "T_R = 600+273.15;\t\t\t#[K] - Temperature of the reservior.\n",
+      "T_0 = 298;\t\t\t#[K] - Temperature of the environment\n",
+      "P_0 = 100;\t\t\t#[kPa] - Pressure of the environment\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t#The entropy change of an isothermal system undergoing an internally reversible process is given by,\n",
+      "delta_S_t = (Q/T_1);\t\t\t#[kJ] - Entropy change\n",
+      "delta_S = delta_S_t/m;\t\t\t#[kJ/kg-K] - \n",
+      "\n",
+      "\t\t\t#At 150 C,it has been reported in the book that, P_sat - 0.4758 kPa, V_liq = 0.001091 m**(3)/kg, U_liq = 631.68 kJ/kg, S_liq = 1.8418 kJ/kg-K, S_vap = 6.8379 kJ/kg-K\n",
+      "V_1 = 0.001091;\t\t\t#[m**(3)/kg] - initial specific volume\n",
+      "U_1 = 631.68;\t\t\t#[kJ/kg] - initial specific internal energy\n",
+      "S_1 = 1.8418;\t\t\t#[kJ/kg-K] - initial entropy\n",
+      "\t\t\t#The initial state of the water is a compressed liquid state,and S_1 is therefore equal to the entropy of the saturated liquid of the saturated liquid at the   same temperature.\n",
+      "S_2 = S_1 + delta_S;\t\t\t#[kJ/kg-K] - Final entropy\n",
+      "\n",
+      "\t\t\t#At the final state the temperature is 150 C and S = 7.499 kJ/kg-K which is more than S_vap therefore it is superheated steam.\n",
+      "S_final = 7.494;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t#At 150 C, and 0.1 MPa: V = 1.9364 m**(3)/kg, U = 2582.8 kJ/kg, S = 7.6134 kJ/kg-K\n",
+      "\t\t\t#At 150 C, and 0.2 MPa: V = 0.9596 m**(3)/kg, U = 2576.9 kJ/kg, S = 7.2795 kJ/kg-K\n",
+      "U_t_1 = 2582.8;\t\t\t#[kJ/kg] - Internal energy\n",
+      "U_t_2 = 2576.9;\t\t\t#[kJ/kg]\n",
+      "V_t_1 = 1.9364;\t\t\t#[m**(3)/kg] - Specific volume\n",
+      "V_t_2 = 0.9596;\t\t\t#[m**(3)/kg]\n",
+      "S_t_1 = 7.6134;\t\t\t#[kJ/kg-K] - Entropy\n",
+      "S_t_2 = 7.2795;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t#The pressure at exit is given by,\n",
+      "P_2 = ((S_final - S_t_1)/(S_t_2 - S_t_1))*(0.2 - 0.1) + 0.1;\t\t\t#[Mpa] - Final pressure\n",
+      "\t\t\t#At final state\n",
+      "U_2 = U_t_1 + (U_t_2 - U_t_1)*((S_final - S_t_1)/(S_t_2 - S_t_1));\t\t\t#[kJ/kg] - Final specific internal energy\n",
+      "V_2 = V_t_1 + (V_t_2 - V_t_1)*((S_final - S_t_1)/(S_t_2 - S_t_1));\t\t\t#[m**(3)/kg] - Final specific volume\n",
+      "\n",
+      "q = Q/m;\t\t\t#[kJ/kg] - Heat supplied per unit kg of mass.\n",
+      "W_rev_use = U_1 - U_2 + P_0*(V_1 - V_2) - T_0*(S_1 - S_2) + q*(1 - (T_0/T_R));\t\t\t#[kJ/kg] - Reversible useful work done.\n",
+      "\n",
+      "\t\t\t#Now let us calculate the actual work done. We know q - W = delta_U, therefore\n",
+      "W = q - (U_2 - U_1);\t\t\t#[kJ/kg] - Work done\n",
+      "W_use = W - P_0*(V_2 - V_1);\t\t\t#[kJ/kg]\n",
+      "i = W_rev_use - W_use;\t\t\t#[kJ/kg] - Irreversibility\n",
+      "\t\t\t#Since the system contains 4.6 g therefore, \n",
+      "W_use_new = W_use*m;\t\t\t#[kJ]\n",
+      "W_rev_use_new = W_rev_use*m;\t\t\t#[kJ]\n",
+      "I = W_rev_use_new - W_use_new;\t\t\t#[kJ]\n",
+      "\n",
+      "# Results\n",
+      "print \" 1).The useful work obtained is %f kJ\"%(W_use_new);\n",
+      "print \" 2).The reversible usefuk work done is %f kJ\"%(W_rev_use_new);\n",
+      "print \" 3).The irreversibility is %f kJ\"%(I);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The useful work obtained is 1304.987050 kJ\n",
+        " 2).The reversible usefuk work done is 5297.425775 kJ\n",
+        " 3).The irreversibility is 3992.438725 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.5  Page Number : 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 700+273.15;\t\t\t#[K] - Initial temperature.\n",
+      "P_1 = 12;\t\t\t#[MPa] - Initial pressure\n",
+      "P_2 = 0.6;\t\t\t#[MPa] - Final pressure\n",
+      "\t\t\t#At 12 MPa and 700 C,\n",
+      "H_1 = 3858.4;\t\t\t#[kJ/kg] - initial enthalpy\n",
+      "S_1 = 7.0757;\t\t\t#[kJ/kg-K] - initial entropy\n",
+      "\n",
+      "\t\t\t#At 0.6 MPa and 200 C,\n",
+      "H_2 = 2850.1;\t\t\t#[kJ/kg]\n",
+      "S_2 = 6.9673;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t#At 0.6 MPa and 250 C,\n",
+      "H_3 = 2957.2;\t\t\t#[kJ/kg]\n",
+      "S_3 = 7.1824;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t#At 0.6 MPa and 300 C,\n",
+      "H_4 = 3061.6;\t\t\t#[kJ/kg]\n",
+      "S_4 = 7.3732;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\n",
+      "P_exit = P_2;\t\t\t#[MPa] - exit pressure\n",
+      "T_exit = ((S_1 - S_2)/(S_3 - S_2))*(250 - 200) + 200;\t\t\t#[C] - exit temperature\n",
+      "H_exit = ((S_1 - S_2)/(S_3 - S_2))*(H_3 - H_2) + H_2;\t\t\t#[kJ/kg] - exit enthalpy\n",
+      "\n",
+      "\t\t\t#Snce it is a flow pocess,therfore\n",
+      "\t\t\t#W_rev = H_1 - H_exit - T_0*(S_1 - S_2)\n",
+      "\t\t\t#As S_1 = S_2,the above equation becomes\n",
+      "W_rev_1 = H_1 - H_exit;\t\t\t#[kJ/kg] - reversible work done\n",
+      "\n",
+      "\t\t\t#From the first law the actual work done can be calculated using, delta_H = q - W\n",
+      "\t\t\t#Since the turbine does not exchange heat,therefore W = - delta_H.\n",
+      "W_1 = - (H_exit - H_1);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "print \" 1).The reversible work done is %f kJ/kg\"%(W_1);\n",
+      "print \"     And since the maximum work is same as the actual work ,therefore irreversibility is zero\"\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# Variables\n",
+      "T_0 = 298.15;\t\t\t#[K] - Environment temperature\n",
+      "P_0 = 1;\t\t\t#[atm] - Environment pressure\n",
+      "adi_eff = 0.88;\t\t\t#adiabatc efficiency\n",
+      "\n",
+      "\t\t\t#(H_1 - H_exit_actual)/(H_1 - H_exit) = 0.88, therefore\n",
+      "H_exit_actual = H_1 - 0.88*(H_1 - H_exit);\t\t\t# - Actual exit enthalpy\n",
+      "\n",
+      "\t\t\t#Now two properties i.e pressure = 0.6 MPa and enthalpy = H_exit_actual is fixed at the exit. The exit temperature is given by,\n",
+      "T_exit_actual = ((H_exit_actual - H_3)/(H_4 - H_3))*(300 - 250) + 250;\t\t\t#[C]\n",
+      "S_exit_actual = ((H_exit_actual - H_3)/(H_4 - H_3))*(S_4 - S_3) + S_3;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Now reversible work done is given by,\n",
+      "W_rev_2 = H_1 - H_exit_actual - T_0*(S_1 - S_exit_actual);\t\t\t#[kJ/kg]\n",
+      "print \" 2).The reversible work done is %f kJ/kg\"%(W_rev_2);\n",
+      "\n",
+      "\t\t\t#The actual work is given by the first law,\n",
+      "W_2 = H_1 - H_exit_actual;\t\t\t#[kJ/kg] - Actual work done\n",
+      "i = W_rev_2 - W_2;\t\t\t#[kJ/kg] - irreversibility\n",
+      "print \"     The value of irreversibility is %f kJ/kg\"%(i);\n",
+      "\n",
+      "\t\t\t#The irreversibility can also be determined using\n",
+      "\t\t\t# i = T_0*S_gen, and S_gen is given by\n",
+      "\t\t\t# S_gen = (q/T_R) - delta_S\n",
+      "\n",
+      "\t\t\t#The second law efficiency of the turbine is actual work done divided by reversible work,therefore\n",
+      "sec_eff = W_2/W_rev_2;\n",
+      "print \"     The second law efficiency of the turbine is %f\"%(sec_eff);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The reversible work done is 954.326778 kJ/kg\n",
+        "     And since the maximum work is same as the actual work ,therefore irreversibility is zero\n",
+        " 2).The reversible work done is 905.072590 kJ/kg\n",
+        "     The value of irreversibility is 65.265025 kJ/kg\n",
+        "     The second law efficiency of the turbine is 0.927890\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.6  Page Number : 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_1 = 8.;\t\t\t#[bar] - Initial pressure\n",
+      "T_1 = 93. + 273.15;\t\t\t#[C] - Initial temperature\n",
+      "V_1 = 100.;\t\t\t#[m/s] - Initial velocity\n",
+      "P_2 = 1.25;\t\t\t#[bar] - Exit pressure\n",
+      "T_2 = 27. + 273.15;\t\t\t#[C] - Exit temperature\n",
+      "V_2 = 60.;\t\t\t#[m/s] - Exit velocity \n",
+      "Y = 1.4;\t\t\t#Ratio of specific heat capacities\n",
+      "T_0 = 298.15;\t\t\t#[K] - surrounding temperature\n",
+      "P_0 = 1.;\t\t\t#[bar] - surrounding pressure\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Gas constant\n",
+      "Cp_0 = (R*Y)/(Y-1);\t\t\t#[J/mol-K] - Heat capacity at constant pressure\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\n",
+      "\t\t\t#W = delta_H + (delta_V_square)/2\n",
+      "delta_H = Cp_0*(T_2 - T_1);\t\t\t#[J/mol] - enthalpy change\n",
+      "delta_H = (delta_H/28.84);\t\t\t#[kJ/kg] - (1 mole = 28.84 g).\n",
+      "delta_V_square = V_2**(2) - V_1**(2);\n",
+      "\n",
+      "W = - delta_H - ((delta_V_square)/2)/1000;\t\t\t#[kJ/kg] - Actual work done\n",
+      "print \" The actual work done is %f kJ/kg\"%(W);\n",
+      "\n",
+      "\t\t\t#Now let us calculate the maximum work that can be obtained\n",
+      "\t\t\t#W_rev = (H_1 + (V_1**(2))/2) - (H_2 + (V_2**(2))/2) - T_0*(S_1 - S_2)\n",
+      "delta_S = Cp_0*math.log(T_2/T_1) - R*math.log(P_2/P_1);\t\t\t#[J/mol-K] - Entropy change\n",
+      "delta_S = delta_S/28.84;\t\t\t#kJ/kg-K]\n",
+      "W_rev = -delta_H - ((delta_V_square/2)/1000) + T_0*delta_S;\t\t\t#[kJ/kg]\n",
+      "print \" The maximum work obtainable per kg of air is %f kJ/kg\"%(W_rev);\n",
+      "\n",
+      "\t\t\t#The second law efficiency of the turbine is actual work done divided by reversible work,therefore\n",
+      "sec_eff = W/W_rev;\n",
+      "print \" The second law efficiency of the turbine is %f\"%(sec_eff);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The actual work done is 69.792718 kJ/kg\n",
+        " The maximum work obtainable per kg of air is 169.550182 kJ/kg\n",
+        " The second law efficiency of the turbine is 0.411635\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.7  Page Number : 200"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "m_cold_water = 60;\t\t\t#[kg/s] - mass flow rate of cold water\n",
+      "P_1 = 50;\t\t\t#[kPa]\n",
+      "T_2 = 250;\t\t\t#[C]\n",
+      "T_water_1 = 1000 + 273.15;\t\t\t#[K] - Entering temperature of water\n",
+      "T_water_2 = 450 +273.15;\t\t\t#[K] - Exit temperature of water\n",
+      "T_0 = 298.15;\t\t\t#[K] - surrounding temperature\n",
+      "P_0 = 1;\t\t\t#[atm] - surrounding pressure\n",
+      "Cp_0 = 1.005;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t#For water at 50 kPa under saturated conditions,T_sat = 81.33 C, \n",
+      "H_liq_1 = 340.49;\t\t\t#[kJ/kg] - Enthalpy\n",
+      "S_liq_1 = 1.0910;\t\t\t#[kJ/kg-K] - Entropy\n",
+      "\n",
+      "\t\t\t#For steam at 50 kPa and 250 C,\n",
+      "H_2 = 2976.0;\t\t\t#[kJ/kg]\n",
+      "S_2 = 8.3556;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\n",
+      "\n",
+      "m_hot_water = (m_cold_water*(H_2 - H_liq_1))/(Cp_0*(T_water_1 - T_water_2));\t\t\t#[kg/s] - mass flow rate of hot water\n",
+      "\n",
+      "\n",
+      "\t\t\t#S_gen = delta_S_hot + delta_S_cold\n",
+      "delta_S_cold = S_2 - S_liq_1;\t\t\t#[kJ/kg-K] - change of entropy of cold water\n",
+      "\t\t\t#delta_S_hot = Cp_0*math.log(T_2/T_1)-R*math.log(P_2/P_1), But pressure drop is zero,therfore\n",
+      "delta_S_hot = Cp_0*math.log(T_water_2/T_water_1);\t\t\t#[kJ/kg-K] - change of entropy of hot water\n",
+      "\n",
+      "S_gen = m_cold_water*delta_S_cold + m_hot_water*delta_S_hot;\t\t\t#[kW/K] - Entropy generated\n",
+      "print \" The entropy generation rate is %f kW/K\"%(S_gen);\n",
+      "\n",
+      "\t\t\t#The irreversibility rete is given by\n",
+      "I = T_0*S_gen;\t\t\t#[kW]\n",
+      "print \" The irreversibility rate of the heat exchanger is %f kW\"%(I);\n",
+      "\n",
+      "\t\t\t\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The entropy generation rate is 273.250824 kW/K\n",
+        " The irreversibility rate of the heat exchanger is 81469.733193 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.8  Page Number : 201"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "m_water = 10000.;\t\t\t#[kg/h] - Mass flow rate of cold water\n",
+      "m_water = m_water/3600;\t\t\t#[kg/s]\n",
+      "T_1_water = 30. + 273.15;\t\t\t#[K] - Cold water entering temperature\n",
+      "m_HC = 5000.;\t\t\t#[kg/h] - mass flow rate of hot hydrocarbon\n",
+      "m_HC = m_HC/3600;\t\t\t#[kg/s]\n",
+      "T_1_HC = 200. + 273.15;\t\t\t#[K] - Hot hydrocarbon entering temperature\n",
+      "T_2_HC = 100. + 273.15;\t\t\t#[K] - Hot hydrocarbon leaving temperature\n",
+      "Cp_0_water = 1.0;\t\t\t#[kcal/kg-K] - Mean heat capacity of cooling water\n",
+      "Cp_0_HC = 0.6;\t\t\t#[kcal/kg-K] - Mean heat capacity of hydrocarbon\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t#Applying energy balance to the heat exchanger,we get\n",
+      "\t\t\t#m_water*Cp_0_water*(T - T_1_water) = m_HC*Cp_0_HC*(T_1_HC - T_2_HC)\n",
+      "T_2_water = ((m_HC*Cp_0_HC*(T_1_HC - T_2_HC))/(m_water*Cp_0_water)) + T_1_water;\t\t\t#[K]\n",
+      "T_2 = T_2_water - 273.15;\t\t\t#[C]\n",
+      "print \" 1).The exit temperature of the cooling water is %f C\"%(T_2);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t#delta_S_hot_HC = Cp_0*math.log(T_2/T_1)-R*math.log(P_2/P_1), But pressure drop is zero,therfore\n",
+      "delta_S_hot_HC = (Cp_0_HC*4.184)*math.log(T_2_HC/T_1_HC);\t\t\t#[kW/K] - change of entropy of hot hydrocarbon\n",
+      "delta_S_HC = m_HC*delta_S_hot_HC;\t\t\t#[kW/K] - Entropy change for hudrocarbon liquid \n",
+      "print \" 2).Entropy change rate of hydrocarbon liquid is %f kW/K\"%(delta_S_HC);\n",
+      "\n",
+      "delta_S_cold_water = (Cp_0_water*4.184)*math.log(T_2_water/T_1_water);\t\t\t#[kW/K] - change of entropy of cooling water\n",
+      "delta_S_water = m_water*delta_S_cold_water;\t\t\t#[kW/K] - Entropy change for water\n",
+      "print \"     And entropy change rate of water is %f kW/K\"%(delta_S_water);\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "T_0 = 298.15;\t\t\t#[K] - Surrounding temperature\n",
+      "\t\t\t#S_gen = delta_S_cold_water + delta_S_hot_HC =   m_water*delta_S_cold_water + m_HC*delta_S_hot_HC;\t\t\t#[kW/K] - Entropy generated\n",
+      "S_gen = delta_S_water + delta_S_HC;\t\t\t#[kW/K]\n",
+      "I = T_0*S_gen;\t\t\t#[kW]\n",
+      "print \" 3).The irreversibility rate of the heat exchanger is %f kW\"%(I);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The exit temperature of the cooling water is 60.000000 C\n",
+        " 2).Entropy change rate of hydrocarbon liquid is -0.827846 kW/K\n",
+        "     And entropy change rate of water is 1.096732 kW/K\n",
+        " 3).The irreversibility rate of the heat exchanger is 80.168382 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.9  Page Number : 202"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1_hotgas = 800.;\t\t\t#[K]\n",
+      "P_1_hotgas = 1.;\t\t\t#[bar]\n",
+      "T_2_hotgas = 700.;\t\t\t#[K]\n",
+      "P_2_hotgas = 1.;\t\t\t#[bar]\n",
+      "T_1_air = 470.;\t\t\t#[K]\n",
+      "P_1_air = 1.;\t\t\t#[bar]\n",
+      "P_2_air = 1.;\t\t\t#[bar]\n",
+      "Cp_0_hotgas = 1.08;\t\t\t#[kJ/kg-K] - Mean heat capacity of hot gas\n",
+      "Cp_0_air = 1.05;\t\t\t#[kcal/kg-K] - Mean heat capacity of air\n",
+      "T_0 = 298.15;\t\t\t#[K] - surrounding temperature\n",
+      "P_0 = 1.;\t\t\t#[bar] - surrounding pressure\n",
+      "\t\t\t#m_air = 2*m_hotgas\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t#Assuming heat exchange only takes places in-between the streams,from energy balance we get,\n",
+      "\t\t\t#m_gas*Cp_0_hotgas*(T_2_hotgas - T_1_hotgas) + 2*m_gas*Cp_0_air*(T - T_1_air), \n",
+      "T_2_air = T_1_air - ((Cp_0_hotgas*(T_2_hotgas - T_1_hotgas))/(2*Cp_0_air));\t\t\t#[K] - Temp of emerging air\n",
+      "print \" 1).The temperature of emerging air is %f K\"%(T_2_air);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t#Availability change of hot gas is given by,\n",
+      "\t\t\t#(si_1 - si_2)_hot = H_1 - H_2 - T_0*(S_1 - S_2)\n",
+      "delta_H_hotgas = (Cp_0_hotgas*(T_2_hotgas - T_1_hotgas));\t\t\t#[kJ/kg] - change in enthalpy of hotgas\n",
+      "\t\t\t#delta_S_hotgas = Cp_0_hotgas*math.log(T_2_hotgas/T_1_hotgas)- R*math.log(P_2/P_1), But pressure drop is zero (P_1 = P_2),therfore\n",
+      "delta_S_hotgas = Cp_0_hotgas*math.log(T_2_hotgas/T_1_hotgas);\t\t\t#[kJ/kg-K] - change of entropy of hot gas\n",
+      "delta_si_hotgas = (-delta_H_hotgas) - (-T_0*delta_S_hotgas);\t\t\t#[kJ/kg]\n",
+      "print \" 2).The availability change of hot gas is %f kJ/kg\"%(delta_si_hotgas);\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "\t\t\t#Availability change of air is given by,\n",
+      "\t\t\t#(si_1 - si_2)_air = H_1 - H_2 - T_0*(S_1 - S_2)\n",
+      "delta_H_air = (Cp_0_air*(T_2_air - T_1_air));\t\t\t#[kJ/kg] - change in enthalpy of air\n",
+      "\t\t\t#delta_S_air = Cp_0_air*math.log(T_2_air/T_1_air)- R*math.log(P_2/P_1), But pressure drop is zero (P_1 = P_2),therfore\n",
+      "delta_S_air = Cp_0_air*math.log(T_2_air/T_1_air);\t\t\t#[kJ/kg-K] - change of entropy of air\n",
+      "delta_si_air = (-delta_H_air) - (-T_0*delta_S_air);\t\t\t#[kJ/kg]\n",
+      "print \" 3).The availability change of air is %f kJ/kg\"%(delta_si_air);\n",
+      "\n",
+      "\t\t\t#(4)\n",
+      "\t\t\t#For the heat exchanger (Q = 0, W = 0)\n",
+      "\t\t\t#Basis : 1 kg of hot gas flowing through heat exchanger\n",
+      "S_gen = delta_S_hotgas + 2*delta_S_air;\t\t\t#[kJ/kg-K] - (as  m_air = 2*m_hotgas)\n",
+      "I = T_0*S_gen;\t\t\t#[kJ/kg]\n",
+      "print \" 4).The irreversibility of thr exchanger per kg of hot gas flowing is %f kJ/kg\"%(I);\n",
+      "\n",
+      "\t\t\t#Irreversibility can also be obtained using\n",
+      "\t\t\t#I = 2*(si_1 - si_2)_air + (si_1 - si_2)_hotgas\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The temperature of emerging air is 521.428571 K\n",
+        " 2).The availability change of hot gas is 65.002625 kJ/kg\n",
+        " 3).The availability change of air is -21.492234 kJ/kg\n",
+        " 4).The irreversibility of thr exchanger per kg of hot gas flowing is 22.018157 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch6_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch6_2.ipynb
new file mode 100755
index 00000000..4cdc915d
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch6_2.ipynb
@@ -0,0 +1,847 @@
+{
+ "metadata": {
+  "name": "ch6_2",
+  "signature": "sha256:5875fd58ba5a3eb4a34d80c6cbf01fc2219f223f8352a521a425e02950a18a26"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 :Chemical reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.1  Page Number : 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\t\t\t\n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K] - standard temperature\n",
+      "T_2 = 880;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "a_SO2 = 6.157;\n",
+      "a_SO3 = 3.918;\n",
+      "a_O2 = 6.732;\n",
+      "b_SO2 = 1.384*10**(-2);\n",
+      "b_SO3 = 3.483*10**(-2);\n",
+      "b_O2 = 0.1505*10**(-2);\n",
+      "c_SO2 = -0.9103*10**(-5);\n",
+      "c_SO3 = -2.675*10**(-5);\n",
+      "c_O2 = -0.01791*10**(-5);\n",
+      "d_SO2 = 2.057*10**(-9);\n",
+      "d_SO3 = 7.744*10**(-9);\n",
+      "\n",
+      "delta_H_rkn_298 = -23.45*10**(3);\t\t\t#[cal] - Rkn enthalpy at 298.15 K\n",
+      "delta_H_SO2_for_298 = -70.94*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of S02 at 298.15 K\n",
+      "delta_H_SO3_for_298 = -94.39*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of SO3 at 298.15 K\n",
+      "delta_G_SO2_for_298 = -71.68*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of SO2 at 298.15 K\n",
+      "delta_G_SO3_for_298 = -88.59*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of SO3 at 298.15 K\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t#standard enthalpy change of reaction at temperature T is given by,\n",
+      "\t\t\t#delta_H_rkn_T = delta_rkn_298 + delta_Cp_0*delta_T\n",
+      "delta_a = a_SO3 - a_SO2 - (a_O2/2);\n",
+      "delta_b = b_SO3 - b_SO2 - (b_O2/2);\n",
+      "delta_c = c_SO3 - c_SO2 - (c_O2/2);\n",
+      "delta_d = d_SO3 - d_SO2;\n",
+      "\n",
+      "\t\t\t#Cp_0 = delta_a + (delta_b*T) + (delta_c*T**(2)) + (delta_d*T**(3));\n",
+      "\t\t\t#Therefore we get,\n",
+      "\n",
+      "def f44(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "delta_H_rkn_880 = delta_H_rkn_298 +  quad(f44,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t#On manual simplification of the above expression,we will get the expression for 'delta_H_rkn_880' as a function of T,\n",
+      " \n",
+      "print \" 1.The expression for smath.tan(math.radiansard enthalpy change of reaction as a function of temperature is given by\";\n",
+      "print \"      delta_H_rkn_880 = -22534.57 - 5.605*T + 1.012*10**-2*T**2 - 0.585*10**-5*T**3 + 1.422*10**-9*T**4\"\n",
+      "\n",
+      "print \" 2).Staandard enthalpy change of reaction at 880 K is %f cal\"%(delta_H_rkn_880);\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "\t\t\t#Let us determine the smath.tan(math.radiansard entropy change of reaction at 298.15 K\n",
+      "delta_S_SO2_298 = (delta_H_SO2_for_298 - delta_G_SO2_for_298)/298.15;\t\t\t#[cal/mol-K]\n",
+      "delta_S_SO3_298 = (delta_H_SO3_for_298 - delta_G_SO3_for_298)/298.15;\t\t\t#[cal/mol-K]\n",
+      "delta_S_O2_298 = 0;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "delta_S_rkn_298 = delta_S_SO3_298 - delta_S_SO2_298 - (delta_S_O2_298/2);\t\t\t#[cal/K]\n",
+      "\n",
+      "def f45(T): \n",
+      "\t return (delta_a+delta_b*T+delta_c*T**(2)+delta_d*T**(3))/T\n",
+      "\n",
+      "delta_S_rkn_880 = delta_S_rkn_298 +  quad(f45,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" 3).Standard entropy change of reaction at 880 K is %f cal/K\"%(delta_S_rkn_880);\n",
+      "\n",
+      "\t\t\t#(4)\n",
+      "delta_G_rkn_880 = delta_H_rkn_880 - 880*delta_S_rkn_880;\t\t\t#[cal]\n",
+      "\n",
+      "print \" 4).Standard Gibbs free energy change of reaction at 880 K is %f cal\"%(delta_G_rkn_880);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1.The expression for smath.tan(math.radiansard enthalpy change of reaction as a function of temperature is given by\n",
+        "      delta_H_rkn_880 = -22534.57 - 5.605*T + 1.012*10**-2*T**2 - 0.585*10**-5*T**3 + 1.422*10**-9*T**4\n",
+        " 2).Staandard enthalpy change of reaction at 880 K is -22766.609215 cal\n",
+        " 3).Standard entropy change of reaction at 880 K is -21.002783 cal/K\n",
+        " 4).Standard Gibbs free energy change of reaction at 880 K is -4284.160417 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.2  Page Number : 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - standard temperature\n",
+      "T_2 = 400;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "a_CH3OH = 4.55;\n",
+      "a_CO = 6.726;\n",
+      "a_H2 = 6.952;\n",
+      "b_CH3OH = 2.186*10**(-2);\n",
+      "b_CO = 0.04001*10**(-2);\n",
+      "b_H2 = -0.04576*10**(-2);\n",
+      "c_CH3OH = -0.291*10**(-5);\n",
+      "c_CO = 0.1283*10**(-5);\n",
+      "c_H2 = 0.09563*10**(-5);\n",
+      "d_CH3OH = -1.92*10**(-9);\n",
+      "d_CO = -0.5307*10**(-9);\n",
+      "d_H2 = -0.2079*10**(-9);\n",
+      "\n",
+      "delta_H_rkn_298 = -21.6643*10**(3);\t\t\t#[cal] - Reaction enthalpy at 298.15 K\n",
+      "delta_H_CO_for_298 = -26.4157*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CO at 298.15 K\n",
+      "delta_H_CH3OH_for_298 = -48.08*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CH3OH at 298.15 K\n",
+      "delta_G_CO_for_298 = -32.8079*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CO at 298.15 K\n",
+      "delta_G_CH3OH_for_298 = -38.69*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CH3OH at 298.15 K\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#Smath.tan(math.radiansard enthalpy change of reaction at temperature T is given by,\n",
+      "\t\t\t#delta_H_rkn_T = delta_rkn_298 + delta_Cp_0*delta_T\n",
+      "delta_a = a_CH3OH - a_CO - 2*(a_H2);\n",
+      "delta_b = b_CH3OH - b_CO - 2*(b_H2);\n",
+      "delta_c = c_CH3OH - c_CO - 2*(c_H2);\n",
+      "delta_d = d_CH3OH - d_CO - 2*(d_H2);\n",
+      "\n",
+      "\t\t\t#Cp_0 = delta_a + (delta_b*T) + (delta_c*T**(2)) + (delta_d*T**(3));\n",
+      "\t\t\t#Therefore we get,\n",
+      "\n",
+      "def f56(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "delta_H_rkn_400 = delta_H_rkn_298 +  quad(f56,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Standard enthalpy change of reaction at 400 K is %f cal\"%(delta_H_rkn_400);\n",
+      "\n",
+      "\t\t\t#Let us determine the standard Gibbs free energy change of reaction at 298.15 K\n",
+      "delta_G_rkn_298 = delta_G_CH3OH_for_298 - delta_G_CO_for_298;\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t#Now determining the standard entropy change of reaction at 298.15 K\n",
+      "delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "\n",
+      "def f57(T): \n",
+      "\t return (delta_a+delta_b*T+delta_c*T**(2)+delta_d*T**(3))/T\n",
+      "\n",
+      "delta_S_rkn_400 = delta_S_rkn_298 +  quad(f57,T_1,T_2)[0]\n",
+      "\n",
+      "\t\t\t#Therefore,the standard Gibbs free energy change of the reaction is given by,\n",
+      "delta_G_rkn_400 = delta_H_rkn_400 - 400*delta_S_rkn_400;\t\t\t#[cal]\n",
+      "\n",
+      "print \" Standard Gibbs free energy change of reaction at 400 K is %f cal\"%(delta_G_rkn_400);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Standard enthalpy change of reaction at 400 K is -22587.159576 cal\n",
+        " Standard Gibbs free energy change of reaction at 400 K is -343.363287 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.3  Page Number : 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2 = 1200;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "\n",
+      "a_CO2 = 5.316;\n",
+      "a_H2 = 6.952;\n",
+      "a_CO = 6.726;\n",
+      "a_H2O = 7.700;\n",
+      "b_CO2 = 1.4285*10**(-2);\n",
+      "b_H2 = -0.04576*10**(-2);\n",
+      "b_CO = 0.04001*10**(-2);\n",
+      "b_H2O = 0.04594*10**(-2);\n",
+      "c_CO2 = -0.8362*10**(-5);\n",
+      "c_H2 = 0.09563*10**(-5);\n",
+      "c_CO = 0.1283*10**(-5);\n",
+      "c_H2O = 0.2521*10**(-5);\n",
+      "d_CO2 = 1.784*10**(-9);\n",
+      "d_H2 = -0.2079*10**(-9);\n",
+      "d_CO = -0.5307*10**(-9);\n",
+      "d_H2O = -0.8587*10**(-9);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "delta_H_rkn_298 = -9.8382*10**(3);\t\t\t#[cal] - Reaction enthalpy at 298.15 K\n",
+      "delta_H_CO2_for_298 = -94.0518*10**(3);\t\t\t#[cal/mol-K] - Enthalpy of formation of CO2 at 298.15 K\n",
+      "delta_H_CO_for_298 = -26.4157*10**(3);\t\t\t#[cal/mol-K] - Enthalpy of formation of CO at 298.15 K\n",
+      "delta_H_H2O_for_298 = -57.7979*10**(3);\t\t\t#[cal/mol-K] - Enthalpy of formation of H2O at 298.15 K\n",
+      "delta_G_CO2_for_298 = -94.2598*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CO at 298.15 K\n",
+      "delta_G_CO_for_298 = -32.8079*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CH3OH at 298.15 K\n",
+      "delta_G_H2O_for_298 = -54.6357*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of H2O at 298.15 K\n",
+      "\n",
+      "\t\t\t#standard enthalpy change of reaction at temperature T is given by,\n",
+      "\t\t\t#delta_H_rkn_T = delta_rkn_298 + delta_Cp_0*delta_T\n",
+      "delta_a = a_CO2 + a_H2 - a_CO - a_H2O;\n",
+      "delta_b = b_CO2 + b_H2 - b_CO - b_H2O;\n",
+      "delta_c = c_CO2 + c_H2 - c_CO - c_H2O;\n",
+      "delta_d = d_CO2 + d_H2 - d_CO - d_H2O;\n",
+      "\n",
+      "\t\t\t#Cp_0 = delta_a + (delta_b*T) + (delta_c*T**(2)) + (delta_d*T**(3));\n",
+      "\t\t\t#Therefore we get,\n",
+      "\n",
+      "def f11(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "delta_H_rkn_1200 = delta_H_rkn_298 +  quad(f11,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Standard enthalpy change of reaction at 1200 K is %f cal\"%(delta_H_rkn_1200);\n",
+      "\n",
+      "\t\t\t#Let us determine the standard Gibbs free energy change of reaction at 298.15 K\n",
+      "delta_G_rkn_298 = delta_G_CO2_for_298 - delta_G_CO_for_298 - delta_G_H2O_for_298;\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t#Now determining the standard entropy change of reaction at 298.15 K\n",
+      "delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "\n",
+      "def f12(T): \n",
+      "\t return (delta_a+delta_b*T+delta_c*T**(2)+delta_d*T**(3))/T\n",
+      "\n",
+      "delta_S_rkn_1200 = delta_S_rkn_298 +  quad(f12,T_1,T_2)[0]\n",
+      "\n",
+      "\t\t\t#Therefore,the standard Gibbs free energy change of the reaction is given by,\n",
+      "delta_G_rkn_1200 = delta_H_rkn_1200 - 1200*delta_S_rkn_1200;\t\t\t#[cal]\n",
+      "\n",
+      "print \" Standard Gibbs free energy change of reaction at 1200 K is %f cal\"%(delta_G_rkn_1200);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Standard enthalpy change of reaction at 1200 K is -7850.182811 cal\n",
+        " Standard Gibbs free energy change of reaction at 1200 K is 953.652749 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.4  Page Number : 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2 = 500;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "a_NH3 = 6.5846;\n",
+      "a_N2 = 6.903;\n",
+      "a_H2 = 6.952;\n",
+      "b_NH3 = 0.61251*10**(-2);\n",
+      "b_N2 = -0.03753*10**(-2);\n",
+      "b_H2 = -0.04576*10**(-2);\n",
+      "c_NH3 = 0.23663*10**(-5);\n",
+      "c_N2 = 0.1930*10**(-5);\n",
+      "c_H2 = 0.09563*10**(-5);\n",
+      "d_NH3 = -1.5981*10**(-9);\n",
+      "d_N2 = -0.6861*10**(-9);\n",
+      "d_H2 = -0.2079*10**(-9);\n",
+      "\n",
+      "delta_H_rkn_298 = -22.08*10**(3);\t\t\t#[cal] - Reaction enthalpy at 298.15 K\n",
+      "delta_H_NH3_for_298 = -11.04*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of NH3 at 298.15 K\n",
+      "delta_G_NH3_for_298 = -3.976*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of NH3 at 298.15 K\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#standard enthalpy change of reaction at temperature T is given by,\n",
+      "\t\t\t#delta_H_rkn_T = delta_rkn_298 + delta_Cp_0*delta_T\n",
+      "delta_a = 2*a_NH3 - a_N2 - 3*a_H2;\n",
+      "delta_b = 2*b_NH3 - b_N2 - 3*b_H2;\n",
+      "delta_c = 2*c_NH3 - c_N2 - 3*c_H2;\n",
+      "delta_d = 2*d_NH3 - d_N2 - 3*d_H2;\n",
+      "\n",
+      "\t\t\t#Cp_0 = delta_a + (delta_b*T) + (delta_c*T**(2)) + (delta_d*T**(3));\n",
+      "\t\t\t#Therefore we get,\n",
+      "\n",
+      "def f5(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "delta_H_rkn_500 = delta_H_rkn_298 +  quad(f5,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Standard enthalpy change of reaction at 500 K is %f cal\"%(delta_H_rkn_500);\n",
+      "\n",
+      "\t\t\t#Let us determine the standard Gibbs free energy change of reaction at 298.15 K\n",
+      "delta_G_rkn_298 = 2*delta_G_NH3_for_298;\t\t\t#[cal]\n",
+      "\n",
+      "\t\t\t#Now determining the standard entropy change of reaction at 298.15 K\n",
+      "delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "\n",
+      "def f6(T): \n",
+      "\t return (delta_a+delta_b*T+delta_c*T**(2)+delta_d*T**(3))/T\n",
+      "\n",
+      "delta_S_rkn_500 = delta_S_rkn_298 +  quad(f6,T_1,T_2)[0]\n",
+      "\n",
+      "\t\t\t#Therefore,the standard Gibbs free energy change of the reaction is given by,\n",
+      "delta_G_rkn_500 = delta_H_rkn_500 - 500*delta_S_rkn_500;\t\t\t#[cal]\n",
+      "\n",
+      "print \" Standard Gibbs free energy change of reaction at 500 K is %f cal\"%(delta_G_rkn_500);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Standard enthalpy change of reaction at 500 K is -23925.267197 cal\n",
+        " Standard Gibbs free energy change of reaction at 500 K is 2159.910425 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.5  Page Number : 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "delta_H_rkn_298 = -57.7979*10**(3);\t\t\t#[cal/mol] - Reaction enthalpy at 298.15 K\n",
+      "delta_G_rkn_298 = -54.6351*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of H2O at 298.15 K\n",
+      "\n",
+      "\t\t\t#Smath.tan(math.radiansard enthalpy change of reaction at temperature T is given by,\n",
+      "\t\t\t#delta_H_rkn_T = delta_rkn_298 + delta_Cp_0*delta_T\n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2_1 = 873.15;\t\t\t#[K] - Reaction temperature\n",
+      "T_2_2 = 1000;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#Therefore we get,\n",
+      "\n",
+      "def f34(T): \n",
+      "\t return 7.7+0.04594*10**(-2)*T+0.2521*10**(-5)*T**(2)-0.8587*10**(-9)*T**(3)\n",
+      "\n",
+      "delta_H_rkn_873 = delta_H_rkn_298 +  quad(f34,T_1,T_2_1)[0]\n",
+      "\n",
+      "\n",
+      "def f35(T): \n",
+      "\t return 7.7+0.04594*10**(-2)*T+0.2521*10**(-5)*T**(2)-0.8587*10**(-9)*T**(3)\n",
+      "\n",
+      "delta_H_rkn_1000 = delta_H_rkn_298 +  quad(f35,T_1,T_2_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Standard enthalpy change of reaction at 873 K is %f cal/mol\"%(delta_H_rkn_873);\n",
+      "print \" Standard enthalpy change of reaction at 1000 K is %f cal/mol\"%(delta_H_rkn_1000);\n",
+      "\n",
+      "\t\t\t#Now determining the smath.tan(math.radiansard entropy change of reaction at 298.15 K\n",
+      "delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;\t\t\t#[cal/mol-K]\n",
+      "\n",
+      "\n",
+      "def f36(T): \n",
+      "\t return (7.7+0.04594*10**(-2)*T+0.2521*10**(-5)*T**(2)-0.8587*10**(-9)*T**(3))/T\n",
+      "\n",
+      "delta_S_rkn_873 = delta_S_rkn_298 +  quad(f36,T_1,T_2_1)[0]\n",
+      "\n",
+      "\n",
+      "def f37(T): \n",
+      "\t return (7.7+0.04594*10**(-2)*T+0.2521*10**(-5)*T**(2)-0.8587*10**(-9)*T**(3))/T\n",
+      "\n",
+      "delta_S_rkn_1000 = delta_S_rkn_298 +  quad(f37,T_1,T_2_2)[0]\n",
+      "\n",
+      "\t\t\t#Therefore,the smath.tan(math.radiansard Gibbs free energy change of the reaction is given by,\n",
+      "delta_G_rkn_873 = (delta_H_rkn_873 - 873.15*delta_S_rkn_873)*10**(-3);\t\t\t#[kcal/mol]\n",
+      "delta_G_rkn_1000 = (delta_H_rkn_1000 - 1000*delta_S_rkn_1000)*10**(-3);\t\t\t#[kcal/mol]\n",
+      "\n",
+      "print \" Standard Gibbs free energy change of reaction at 873 K is %f kcal/mol\"%(delta_G_rkn_873);\n",
+      "print \" Standard Gibbs free energy change of reaction at 1000 K is %f kcal/mol\"%(delta_G_rkn_1000);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Standard enthalpy change of reaction at 873 K is -52801.656303 cal/mol\n",
+        " Standard enthalpy change of reaction at 1000 K is -51579.290952 cal/mol\n",
+        " Standard Gibbs free energy change of reaction at 873 K is -51.575573 kcal/mol\n",
+        " Standard Gibbs free energy change of reaction at 1000 K is -51.481661 kcal/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.6  Page Number : 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2 = 500.;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "a_C2H6 = 1.648;\n",
+      "a_O2 = 6.085;\n",
+      "a_CO2 = 5.316;\n",
+      "a_H2O = 7.700;\n",
+      "b_C2H6 = 4.124*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "b_CO2 = 1.4285*10**(-2);\n",
+      "b_H2O = 0.04594*10**(-2);\n",
+      "c_C2H6 = -1.530*10**(-5);\n",
+      "c_O2 = -0.1709*10**(-5);\n",
+      "c_CO2 = -0.8362*10**(-5);\n",
+      "c_H2O = 0.2521*10**(-5);\n",
+      "d_C2H6 = 1.740*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "d_CO2 = 1.784*10**(-9);\n",
+      "d_H2O = -0.8587*10**(-9);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\n",
+      "\n",
+      "\n",
+      "delta_H_C2H6_for_298 = -20.236*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of C2H6 at 298.15 K\n",
+      "delta_H_CO2_for_298 = -94.0518*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CO2 at 298.15 K\n",
+      "delta_H_H2O_for_298 = -57.7979*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of H2O at 298.15 K\n",
+      "\n",
+      "delta_H_rkn_298 = 2*delta_H_CO2_for_298 + 3*delta_H_H2O_for_298 - delta_H_C2H6_for_298;\t\t\t#[cal] - Reaction enthalpy at 298.15 K\n",
+      "\n",
+      "\t\t\t#standard enthalpy change of reaction at temperature T is given by,\n",
+      "\t\t\t#delta_H_rkn_T = delta_rkn_298 + delta_Cp_0*delta_T\n",
+      "delta_a = 2*a_CO2 + 3*a_H2O - a_C2H6 - 7./2*(a_O2);\n",
+      "delta_b = 2*b_CO2 + 3*b_H2O - b_C2H6 - 7./2*(b_O2);\n",
+      "delta_c = 2*c_CO2 + 3*c_H2O - c_C2H6 - 7./2*(c_O2);\n",
+      "delta_d = 2*d_CO2 + 3*d_H2O - d_C2H6 - 7./2*(d_O2);\n",
+      "\n",
+      "\t\t\t#Cp_0 = delta_a + (delta_b*T) + (delta_c*T**(2)) + (delta_d*T**(3));\n",
+      "\t\t\t#Therefore we get,\n",
+      "\n",
+      "def f55(T): \n",
+      "\t return delta_a+(delta_b*T)+(delta_c*T**(2))+(delta_d*T**(3))\n",
+      "\n",
+      "delta_H_rkn_500 = delta_H_rkn_298 +  quad(f55,T_1,T_2)[0]\n",
+      "\n",
+      "delta_H_rkn_500 = delta_H_rkn_500*10**(-3);\t\t\t#[kcal]\n",
+      "\n",
+      "print \" The heat exchange of the reaction at 500 K is %f kcal\"%(delta_H_rkn_500);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The heat exchange of the reaction at 500 K is -340.644585 kcal\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.7  Page Number : 224"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2 = 600.;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "a_C2H6 = -8.65;\n",
+      "a_H2O = 7.700;\n",
+      "a_CH4 = 4.750;\n",
+      "a_O2 = 6.085;\n",
+      "b_C2H6 = 11.578*10**(-2);\n",
+      "b_H2O = 0.04594*10**(-2);\n",
+      "b_CH4 = 1.200*10**(-2);\n",
+      "b_O2 = 0.3631*10**(-2);\n",
+      "c_C2H6 = -7.540*10**(-5);\n",
+      "c_H2O = 0.2521*10**(-5);\n",
+      "c_CH4 = 0.3030*10**(-5);\n",
+      "c_O2 = -0.1709*10**(-5);\n",
+      "d_C2H6 = 18.54*10**(-9);\n",
+      "d_H2O = -0.8587*10**(-9);\n",
+      "d_CH4 = -2.630*10**(-9);\n",
+      "d_O2 = 0.3133*10**(-9);\n",
+      "\n",
+      "delta_S_CH4_for_298 = 44.50;\t\t\t#[cal/mol-K] - Entropy of formation of CH4 at 298.15 K\n",
+      "delta_S_O2_for_298 = 49.00;\t\t\t#[cal/mol-K] - Entropy of formation of O2 at 298.15 K\n",
+      "delta_S_C2H6_for_298 = 64.34;\t\t\t#[cal/mol-K] - Entropy of formation of C2H6 at 298.15 K\n",
+      "delta_S_H2O_for_298 = 45.11;\t\t\t#[cal/mol-K] - Entropy of formation of C2H6 at 298.15 K\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#Cp_0 = delta_a + (delta_b*T) + (delta_c*T**(2)) + (delta_d*T**(3));\n",
+      "\n",
+      "\t\t\t#Smath.tan(math.radiansard entropy change of reaction at temperature T is given by,\n",
+      "\t\t\t#delta_S_rkn_T = delta_rkn_298 + delta_Cp_0*delta_T\n",
+      "delta_a = 1./6*(a_C2H6) + 3./2*(a_H2O) - a_CH4 - 3./4*(a_O2);\n",
+      "delta_b = 1./6*(b_C2H6) + 3./2*(b_H2O) - b_CH4 - 3./4*(b_O2);\n",
+      "delta_c = 1./6*(c_C2H6) + 3./2*(c_H2O) - c_CH4 - 3./4*(c_O2);\n",
+      "delta_d = 1./6*(d_C2H6) + 3./2*(d_H2O) - d_CH4 - 3./4*(d_O2);\n",
+      "\n",
+      "delta_S_rkn_298 = 1./6*(delta_S_C2H6_for_298) + 3./2*(delta_S_H2O_for_298) - delta_S_CH4_for_298 - 3./4*(delta_S_O2_for_298);\t\t\t#[cal/K]\n",
+      "\n",
+      "def f27(T): \n",
+      "\t return (delta_a+delta_b*T+delta_c*T**(2)+delta_d*T**(3))/T\n",
+      "\n",
+      "delta_S_rkn_600 = delta_S_rkn_298 +  quad(f27,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Change in entropy of the reaction at 298.15 K is %f cal/K\"%(delta_S_rkn_298);\n",
+      "print \" Standard entropy change of reaction at 600 K is %f cal/K\"%(delta_S_rkn_600);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Change in entropy of the reaction at 298.15 K is -2.861667 cal/K\n",
+        " Standard entropy change of reaction at 600 K is -1.880233 cal/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.8  Page Number : 225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2 = 973.15;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "\t\t\t#At 298.15 K\n",
+      "delta_H_CH4_for_298 = -17.889*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CH4 at 298.15 K\n",
+      "delta_H_C_for_298 = 0.00;\t\t\t#[cal/mol] - Enthalpy of formation of C (s, graphite) at 298.15 K\n",
+      "delta_H_H2_for_298 = 0.00;\t\t\t#[cal/mol] - Enthalpy of formation of H2 at 298.15 K\n",
+      "delta_G_CH4_for_298 = -12.140*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of H2 at 298.15 K\n",
+      "delta_G_C_for_298 = 0.00;\t\t\t#[cal/mol] - Gibbs free energy change for formation of C (s, graphite) at 298.15 K\n",
+      "delta_G_H2_for_298 = 0.00;\t\t\t#[cal/mol] - Gibbs free energy change for formation of H2 at 298.15 K\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#/standard heat capacity data in cal/mol-K are given below, T is in K\n",
+      "\t\t\t#Cp_0_CH4 = 4.75 + 1.2*10**(-2)*T + 0.303*10**(-5)*T**(2) - 2.63*10**(-9)*T**(3)\n",
+      "\t\t\t#Cp_0_C = 3.519 + 1.532*10**(-3)*T - 1.723*10**(5)*T**(-2)\n",
+      "\t\t\t#Cp_0_H2 = 6.952 - 0.04576*10**(-2)*T + 0.09563*10**(-5)*T**(2) - 0.2079*10**(-9)*T**(3)\n",
+      "\n",
+      "\t\t\t#Therefore standard heat capacity of reaction is given by,\n",
+      "\t\t\t#Cp_0_rkn = 2*Cp_0_H2 + Cp_0_C - Cp_0_CH4\n",
+      "\t\t\t#On simplification,we get the relation\n",
+      "\t\t\t#Cp_0_rkn = 12.673 - 0.0113832*T - 1.1174*10**(-6)*T**(2) + 2.2142*10**(-9)*T**(3) - 1.723*10**(5)*T**(-2)\n",
+      "\n",
+      "delta_H_rkn_298 = -delta_H_CH4_for_298;\t\t\t#[cal] - Enthalpy of reaction at 298.15 K\n",
+      "delta_G_rkn_298 = -delta_G_CH4_for_298;\t\t\t#[cal] - Gibbs free energy of the reaction at 298.15 K\n",
+      "\n",
+      "\n",
+      "def f63(T): \n",
+      "\t return 12.673-0.0113832*T-1.1174*10**(-6)*T**(2)+2.2142*10**(-9)*T**(3)-1.723*10**(5)*T**(-2)\n",
+      "\n",
+      "delta_H_rkn_973 = delta_H_rkn_298 +  quad(f63,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Standard enthalpy change of reaction at 973.15 K is %f cal\"%(delta_H_rkn_973);\n",
+      "\n",
+      "\t\t\t#Now determining the smath.tan(math.radiansard entropy change of reaction at 298.15 K\n",
+      "delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;\t\t\t#[cal/K]\n",
+      "\n",
+      "def f64(T): \n",
+      "\t return (12.673-0.0113832*T-1.1174*10**(-6)*T**(2)+2.2142*10**(-9)*T**(3)-1.723*10**(5)*T**(-2))/T\n",
+      "\n",
+      "delta_S_rkn_973 = delta_S_rkn_298 +  quad(f64,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t#Therefore,the smath.tan(math.radiansard Gibbs free energy change of the reaction is given by,\n",
+      "delta_G_rkn_973 = delta_H_rkn_973 - 973.15*delta_S_rkn_973;\t\t\t#[cal]\n",
+      "\n",
+      "print \" Standard Gibbs free energy change of reaction at 973 K is %f cal\"%(delta_G_rkn_973);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Standard enthalpy change of reaction at 973.15 K is 21316.998642 cal\n",
+        " Standard Gibbs free energy change of reaction at 973 K is -3880.803382 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.9  Page Number : 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2 = 1000;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "\t\t\t#At 298.15 K\n",
+      "delta_H_C_for_298 = 0.00;\t\t\t#[cal/mol] - Enthalpy of formation of C(s,graphite) at 298.15 K\n",
+      "delta_H_H2O_for_298 = -57.7979*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of H2O at 298.15 K\n",
+      "delta_H_CO_for_298 = -26.4157*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CO at 298.15 K\n",
+      "delta_H_H2_for_298 = 0.00;\t\t\t#[cal/mol] - Enthalpy of formation of H2 at 298.15 K\n",
+      "delta_G_C_for_298 = 0.00;\t\t\t#[cal/mol] - Gibbs free energy change for formation of C(s, graphite) at 298.15 K\n",
+      "delta_G_H2O_for_298 = -54.6357*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of H2O at 298.15 K\n",
+      "delta_G_CO_for_298 = -32.8079*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CO at 298.15 K\n",
+      "delta_G_H2_for_298 = 0.00;\t\t\t#[cal/mol] - Gibbs free energy change for formation of H2 at 298.15 K\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#/standard heat capacity data in cal/mol-K are given below, T is in K\n",
+      "\t\t\t#Cp_0_C = 3.519 + 1.532*10**(-3)*T - 1.723*10**(5)*T**(-2)\n",
+      "\t\t\t#Cp_0_H2O = 7.7 + 0.04594*10**(-2)*T + 0.2521*10**(-5)*T**(2) - 0.8587*10**(-9)*T**(3)\n",
+      "\t\t\t#Cp_0_CO = 6.726 + 0.04001*10**(-2)*T + 0.1283*10**(-5)*T**(2) - 0.5307*10**(-9)*T**(3)\n",
+      "\t\t\t#Cp_0_H2 = 6.952 - 0.04576*10**(-2)*T + 0.09563*10**(-5)*T**(2) - 0.2079*10**(-9)*T**(3)\n",
+      "\n",
+      "\t\t\t#Therefore standard heat capacity of reaction is given by,\n",
+      "            #Cp_0_rkn = 2.459 - 2.0489*10**(-3)*T - 2.817*10**(-7)*T**(2) + 1.201*10**(-10)*T**(3) + 1.723*10**(5)*T**(-2)\n",
+      "\n",
+      "delta_H_rkn_298 = delta_H_CO_for_298 + delta_H_H2_for_298 - delta_H_C_for_298 - delta_H_H2O_for_298;\t\t\t#[cal] - Enthalpy of reaction at 298.15 K\n",
+      "delta_G_rkn_298 = delta_G_CO_for_298 + delta_G_H2_for_298 - delta_G_C_for_298 - delta_G_H2O_for_298;\t\t\t#[cal] - Gibbs free energy of the reaction at 298.15 K\n",
+      "\n",
+      "\n",
+      "def f13(T): \n",
+      "\t return 2.459-2.0489*10**(-3)*T-2.817*10**(-7)*T**(2)+1.201*10**(-10)*T**(3)+1.723*10**(5)*T**(-2)\n",
+      "\n",
+      "delta_H_rkn_1000 = delta_H_rkn_298 +  quad(f13,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Standard enthalpy change of reaction at 1000 K is %f cal\"%(delta_H_rkn_1000);\n",
+      "\n",
+      "\t\t\t#Now determining the smath.tan(math.radiansard entropy change of reaction at 298.15 K\n",
+      "delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;\t\t\t#[cal/K]\n",
+      "\n",
+      "def f14(T): \n",
+      "\t return (2.459-2.0489*10**(-3)*T-2.817*10**(-7)*T**(2)+1.201*10**(-10)*T**(3)+1.723*10**(5)*T**(-2))/T\n",
+      "\n",
+      "delta_S_rkn_1000 = delta_S_rkn_298 +  quad(f14,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t#Therefore,the smath.tan(math.radiansard Gibbs free energy change of the reaction is given by,\n",
+      "delta_G_rkn_1000 = delta_H_rkn_1000 - 1000*delta_S_rkn_1000;\t\t\t#[cal]\n",
+      "\n",
+      "print \" Standard Gibbs free energy change of reaction at 1000 K is %f cal\"%(delta_G_rkn_1000);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Standard enthalpy change of reaction at 1000 K is 32518.639475 cal\n",
+        " Standard Gibbs free energy change of reaction at 1000 K is -1858.365607 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.10  Page Number : 228"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard temperature\n",
+      "T_2 = 1042;\t\t\t#[K] - Reaction temperature\n",
+      "\n",
+      "\t\t\t#At 298.15 K\n",
+      "delta_H_CaCO3_for_298 = -289.5*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CaCO3 at 298.15 K\n",
+      "delta_H_CaO_for_298 = -151.7*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CaO at 298.15 K\n",
+      "delta_H_CO2_for_298 = -94.052*10**(3);\t\t\t#[cal/mol] - Enthalpy of formation of CO2 at 298.15 K\n",
+      "delta_G_CaCO3_for_298 = -270.8*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CaCO3 at 298.15 K\n",
+      "delta_G_CaO_for_298 = -144.3*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CaO at 298.15 K\n",
+      "delta_G_CO2_for_298 = -94.260*10**(3);\t\t\t#[cal/mol] - Gibbs free energy change for formation of CO2 at 298.15 K\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#/Smath.tan(math.radiansaerd heat capacity data in cal/mol-K are given below, T is in K\n",
+      "\t\t\t#Cp_0_CO2 = 5.316 + 1.4285*10**(-2)*T - 0.8362*10**(-5)*T**(2) + 1.784*10**(-9)*T**(3)\n",
+      "\t\t\t#Cp_0_CaO = 12.129 + 0.88*10**(-3)*T + 2.08*10**(5)*T**(-2)\n",
+      "\t\t\t#Cp_0_CaCO3 = 24.98 + 5.240*10**(-3)*T + 6.199*10**(5)*T**(-2)\n",
+      "\n",
+      "\t\t\t#Therefore smath.tan(math.radiansard heat capacity of reaction is given by,\n",
+      "            #Cp_0_rkn = -7.535 + 9.925*10**(-3)*T - 0.8362*10**(-5)*T**(2) + 1.784*10**(-9)*T**(3) + 4.119*10**(5)*T**(-2)\n",
+      "\n",
+      "delta_H_rkn_298 = delta_H_CaO_for_298 + delta_H_CO2_for_298 - delta_H_CaCO3_for_298;\t\t\t#[cal] - Enthalpy of reaction at 298.15 K\n",
+      "delta_G_rkn_298 = delta_G_CaO_for_298 + delta_G_CO2_for_298 - delta_G_CaCO3_for_298;\t\t\t#[cal] - Gibbs free energy of the reaction at 298.15 K\n",
+      "\n",
+      "\n",
+      "def f38(T): \n",
+      "\t return -7.535+9.925*10**(-3)*T-0.8362*10**(-5)*T**(2)+1.784*10**(-9)*T**(3)+4.119*10**(5)*T**(-2)\n",
+      "\n",
+      "delta_H_rkn_1042 = delta_H_rkn_298 +  quad(f38,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "print \" Standard enthalpy change of reaction at 1042 K is %f cal\"%(delta_H_rkn_1042);\n",
+      "\n",
+      "\t\t\t#Now determining the smath.tan(math.radiansard entropy change of reaction at 298.15 K\n",
+      "delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;\t\t\t#[cal/K]\n",
+      "\n",
+      "def f39(T): \n",
+      "\t return (-7.535+9.925*10**(-3)*T-0.8362*10**(-5)*T**(2)+1.784*10**(-9)*T**(3)+4.119*10**(5)*T**(-2))/T\n",
+      "\n",
+      "delta_S_rkn_1042 = delta_S_rkn_298 +  quad(f39,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t#Therefore,the smath.tan(math.radiansard Gibbs free energy change of the reaction is given by,\n",
+      "delta_G_rkn_1042 = delta_H_rkn_1042 - 1042*delta_S_rkn_1042;\t\t\t#[cal]\n",
+      "\n",
+      "print \" Standard Gibbs free energy change of reaction at 1042 K is %f cal\"%(delta_G_rkn_1042);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Standard enthalpy change of reaction at 1042 K is 41518.919956 cal\n",
+        " Standard Gibbs free energy change of reaction at 1042 K is 4873.416608 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch7_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch7_2.ipynb
new file mode 100755
index 00000000..5f644cdd
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch7_2.ipynb
@@ -0,0 +1,1542 @@
+{
+ "metadata": {
+  "name": "ch7_2",
+  "signature": "sha256:1fc4ec5605309ff88ba6d75ddff6c3046fc8a8dc45dca2d12f7651c49812921e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Thermodynamic property  relations of pure substance"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.6  Page Number : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "P_1 = 1;\t\t\t#[MPa] - Initial pressure\n",
+      "P_2 = 1.4;\t\t\t#[MPa] - Final pressure\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#We know that\n",
+      "\t\t\t# dS = (Cp/T)*dT - (dV/dT)*dP\n",
+      "\t\t\t# Along an isothermal path,integration of above expression between states 1 and 2 yields\n",
+      "\t\t\t# S_2 - S_1 = - integral((dV/dT)*dP)_P\n",
+      "\t\t\t# An estimate can be made by assuming that (dV/dT)_P  remains constant over the range of pressure from P_1 to P_2 and evaluating the derivative at average pressure of 1.2 MPa\n",
+      "P_avg = P_2;\n",
+      "\t\t\t# S_2 - S_1 = -integral((dV/dT)*dP)_Pavg*(P_2 - P_1)\n",
+      "\t\t\t# (dV/dT)_P=1.2MPa = ((V_350 - V_250)/(350 - 250))\n",
+      "dV_dT = (0.2345 - 0.19234)/100;\t\t\t#[m**(3)/kg-K]\n",
+      "\t\t\t#Therefore\n",
+      "delta_S = -dV_dT*(P_2 - P_1)*1000;\t\t\t#[kJ/kg-K] - Entropy change\n",
+      "\n",
+      "# Results\n",
+      "print \"The change in entropy is given by S_2-S_1 = %f kJ/kg-K\"%(delta_S);\n",
+      "\n",
+      "\t\t\t#Let us compare this tabulated values. At 1MPA and 300 C, S_1 = 7.1229 kJ/kg-K. At 1.4 MPa and 300 C, S_2 = 6.9534 kJ/kg-K. \n",
+      "\t\t\t#Therefore  S_2 - S_1 = -0.1695 kJ/kg-K\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The change in entropy is given by S_2-S_1 = -0.168640 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.7  Page Number : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\t\t\t\n",
+      "# Variables\n",
+      "T = 25. + 273.15;\t\t\t#[K] - Temperature of the surrounding\n",
+      "P_1 = 1.;\t\t\t#[atm] - Initial pressure\n",
+      "P_2 = 1000.;\t\t\t#[atm] - Final pressure\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# V = 18.066 - 7.15*10**(-4)*P + 4.6*10**(-8)*P**(2)    where, V is in 'cm**(3)/mol' and P is in 'atm\n",
+      "\t\t\t# (dV/dT)_P = 0.0045 + 1.4*10**(-6)*P    ;\t\t\t#cm**(3)/mol-K\n",
+      "\n",
+      "\t\t\t# The work done by 1 mol is given by W = integral(P*dV)\n",
+      "\t\t\t# Differentiating both sides of the expression for V, we get\n",
+      "\t\t\t# dV = -7.15*10**(-4)*dP + 9.2*10**(-8)*(P*dP) \n",
+      "\t\t\t# P*dV = -7.15*10**(-4)*P*dP + 9.2*10**(-8)*(P**(2)*dP)\n",
+      "\n",
+      "\t\t\t# The work done is given by\n",
+      "\n",
+      "def f58(P): \n",
+      "\t return -7.15*10**(-4)*P + 9.2*10**(-8)*(P**(2))\n",
+      "\n",
+      "W =  quad(f58,P_1,P_2)[0];\t\t\t#[atm-cm**(3)[0]\n",
+      "\n",
+      "W = W*101325*10**(-6);\t\t\t#[J/mol]\n",
+      "\n",
+      "print \"The necessary work to compress water from 1 atm to 1000 atm is  %f J/mol\"%(W);\n",
+      "\n",
+      "\t\t\t#Let us calculate the amount of heat transfer\n",
+      "\t\t\t# q = integral(T*dS)\n",
+      "\t\t\t# dS = ((ds/dT)_P)*dT + ((dS/dP)_T)*dP\n",
+      "\t\t\t# Since the temperature is constant the first term is zero and\n",
+      "\t\t\t# dS = ((dS/dP)_T)*dP = -((dV/dT)_P)*dP\n",
+      "\t\t\t# Thus, q = integral(T*dS) = T*(integral(dS))     ( as temperature is constant )\n",
+      "\t\t\t# or, q = -T*(integral((dV/dT)_P)*dP)\n",
+      "\n",
+      "\t\t\t# Thus the heat transfer is given by\n",
+      "\n",
+      "def f59(P): \n",
+      "\t return 0.0045+1.4*10**(-6)*P\n",
+      "\n",
+      "q = -T* quad(f59,P_1,P_2)[0];\t\t\t#[atm-cm**(3)[0]\n",
+      "\n",
+      "q = q*101325*10**(-6);\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# q - W = delta_U\n",
+      "\t\t\t# Thus delta_U is given by\n",
+      "delta_U = q - W;\t\t\t#[J/mol]\n",
+      "\n",
+      "print \"The change in internal energy is  %f J/mol\"%(delta_U);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The necessary work to compress water from 1 atm to 1000 atm is  -33.116351 J/mol\n",
+        "The change in internal energy is  -123.839936 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.12  Page Number : 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 25+273.15;\t\t\t#[K] - Temperature\n",
+      "P = 1;\t\t\t#[atm] - Pressure\n",
+      "P = P*101325;\t\t\t#[Pa]\n",
+      "Tc = 126.2;\t\t\t#[K] - Critical temperature\n",
+      "Pc = 34;\t\t\t#[bar] - Critical pressure\n",
+      "Pc = Pc*10**(5);\t\t\t#[Pa]\n",
+      "R=8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "a = (27*R**(2)*Tc**(2)/(64*Pc));\t\t\t#[Pa-m**(6)/mol**(2)]\n",
+      "b = (R*Tc)/(8*Pc);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# the cubic form of van der Walls equation of state is\n",
+      "\t\t\t# V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P=0\n",
+      "\t\t\t#Solving the cubic equation\n",
+      "def f(V): \n",
+      "\t return V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P\n",
+      "V = fsolve(f,1)\n",
+      "\t\t\t#The above equation has 1 real and 2 imaginary roots. We consider only real root.\n",
+      "\n",
+      "Beta = R/((P*V)-(a/V)+((2*a*b)/V**(2)));\t\t\t#[K**(-1)]\n",
+      "\n",
+      "K_t = (V-b)/((P*V)-(a/V)+((2*a*b)/V**(2)));\t\t\t#[Pa**(-1)]\n",
+      "K_t = K_t*101325;\t\t\t#[atm**(-1)]\n",
+      "\n",
+      "print \" Beta\\t = \\t %f K**-1)\"%(Beta);\n",
+      "print \" K_t\\t = \\t %f atm**-1)\"%(K_t);\n",
+      "\n",
+      "\t\t\t#For ideal gas, Beta = 1/T = 0.0033354 K**(-1)    and K_t = 1/P = 1 atm**(-1)\n",
+      "\t\t\t# So results obtained are convergent with those obtained assuming ideal gas.\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Beta\t = \t 0.003364 K**-1)\n",
+        " K_t\t = \t 1.000672 atm**-1)\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.13  Page Number : 248"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 45+273.15;\t\t\t#[K]\n",
+      "P_1 = 10;\t\t\t#[kPa] - Initial pressure\n",
+      "P_2 = 8600;\t\t\t#[kPa] - Final pressure\n",
+      "V = 1010;\t\t\t#[cm**(3)/kg] - Specific volume for saturated liquid water at 45 C\n",
+      "V = V*10**(-6);\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t# Beta = (1/V)*(dV/dT)_P\n",
+      "Beta = 4.25*10**(-4);\t\t\t#[k**(-1)]\n",
+      "Cp = 4.178;\t\t\t#[kJ/kg-K] - Specific heat at constant pressure\n",
+      "eff = 0.75;\t\t\t# Efficiency of the pump\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t#when efficiency of the pump is 100% ,  W = -delta_Hs\n",
+      "\t\t\t# Now delta_H = T*dS + V*dP, therefore under isentropic conditions, dH = V*dP\n",
+      "\t\t\t# delta_Hs = V*dP\n",
+      "delta_Hs = V*(P_2 - P_1);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Actual pumps are not isentropic and therefore not 100% efficient. Therefore actual work done by the pump is given by\n",
+      "W = -delta_Hs/eff;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "print \" 1).The work done by the pump is %f kJ/kg\"%(W);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# We know that dH = Cp*dT + (1 - Beta*T)*V*dP\n",
+      "\t\t\t# Integrating the above equation we get\n",
+      "\t\t\t# delta_H = Cp*delta_T + (1 - Beta*T)*V*(delta_P)\n",
+      "\t\t\t# Now from energy balance delta_H = q - W . But q = 0. Therefore,\n",
+      "delta_H = -W;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Solving for delta_T\n",
+      "delta_T = (delta_H - (1 - Beta*T)*V*(P_2-P_1))/Cp;\n",
+      "\n",
+      "print \" 2).The temperature of water change by  delta_T = %f K\"%(delta_T);\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "T_1 = T;\t\t\t#[K]\n",
+      "T_2 = T + delta_T;\t\t\t#[K]\n",
+      "\t\t\t# dS = (Cp/T)*dT - Beta*V*dP\n",
+      "\t\t\t# Beta and V are weak functions of pressure in the case of liquids. Integrating the above equation we get\n",
+      "delta_S = Cp*math.log(T_2/T_1) - Beta*V*(P_2-P_1);\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" 3).The entropy change of water is given by delta_S = %f kJ/kg-K\"%(delta_S);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The work done by the pump is -11.567867 kJ/kg\n",
+        " 2).The temperature of water change by  delta_T = 0.972970 K\n",
+        " 3).The entropy change of water is given by delta_S = 0.009070 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.14  Page Number : 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 270;\t\t\t#[K]\n",
+      "P_1 = 381;\t\t\t#[kPa] - Initial pressure\n",
+      "P_2 = 1200;\t\t\t#[kPa] - Final pressure\n",
+      "V_liq = 1.55*10**(-3);\t\t\t#[m**(3)/kg] - Specific volume for saturated water in liquid phase at 270 C\n",
+      "Beta = 2.095*10**(-3);\t\t\t#[K**(-1)]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#dH = Cp*dT + [V - T*(dV/dT)_P]*dP\n",
+      "\t\t\t# dS = (Cp/T)*dT - ((dV/dT)_P)*dP\n",
+      "\t\t\t# dH = [V - T*(dV/dT)_P]*dP = V*(1 - Beta*T)*dP\n",
+      "\t\t\t# For the liquid assuming V and Beta to remain constant during pressure change, and since temperature is constant we get\n",
+      "delta_H = V_liq*(1 - Beta*T)*(P_2 - P_1);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "print \"The enthalpy change is given by delta_H = %f kJ/kg\"%(delta_H);\n",
+      "\n",
+      "\t\t\t# Under isothermal conditions \n",
+      "\t\t\t# dS = -((dV/dT)_P)*dP = -Beta*V_liq*dP\n",
+      "\t\t\t# If Beta*V is assumed to remain constant during pressure change we get\n",
+      "delta_S = -Beta*V_liq*(P_2-P_1);\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \"The entropy change is given by delta_S = %e kJ/kg-K\"%(delta_S);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy change is given by delta_H = 0.551386 kJ/kg\n",
+        "The entropy change is given by delta_S = -2.659498e-03 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.15  Page Number : 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T_1 = 0;\t\t\t#[C] - Initial tempetaure\n",
+      "T_2 = 100;\t\t\t#[C] - Final temperature\n",
+      "\t\t\t# Beta = 1.0414*10**(-3) + 1.5672*10**(-6)*T + 5.148*10**(-8)*T**(2),   where T is in C\n",
+      "\t\t\t# At constant pressure  (1/V)*(dV/dT) = Beta\n",
+      "\t\t\t# or, d(math.log(V)) = Beta*dT\n",
+      "\t\t\t# Integrating  we get math.log(V_2/V_1) = integral(Beta*dT) from limit T_1 to T_2\n",
+      "\n",
+      "# Calculations and Results\n",
+      "def f62(T): \n",
+      "\t return 1.0414*10**(-3)+1.5672*10**(-6)*T+5.148*10**(-8)*T**(2)\n",
+      "\n",
+      "integral =  quad(f62,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t# math.log(V_2/V_1) = integral\n",
+      "\t\t\t# (V_2/V_1) = exp(integral)\n",
+      "\t\t\t# (V_2 - V_1)/V_1 = change = exp(integral) - 1;\n",
+      "change = math.exp(integral) - 1;\n",
+      "per_change = 100*change;\n",
+      "\n",
+      "print \"The percentage change in volume = %f %%\"%(per_change);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage change in volume = 13.784486 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.16  Page Number : 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_1 = 25 + 273.15;\t\t\t#[C] - Initial tempetaure\n",
+      "T_2 = 50 + 273.15;\t\t\t#[C] - Final temperature\n",
+      "P_1 = 1;\t\t\t#[bar] - Initial pressure\n",
+      "P_2 = 1000;\t\t\t#[bar] - Final pressure\n",
+      "\n",
+      "Cp_T1_P1 = 75.305;\t\t\t#[J/mol-K]\n",
+      "Cp_T2_P1 = 75.314;\t\t\t#[J/mol-K]\n",
+      "V_T1_P1 = 18.071;\t\t\t#[cm**(3)/mol]\n",
+      "V_T1_P2 = 18.012;\t\t\t#[cm**(3)/mol]\n",
+      "V_T2_P1 = 18.234;\t\t\t#[cm**(3)/mol]\n",
+      "V_T2_P2 = 18.174;\t\t\t#[cm**(3)/mol]\n",
+      "Beta_T1_P1 = 256*10**(-6);\t\t\t#[K**(-1)]\n",
+      "Beta_T1_P2 = 366*10**(-6);\t\t\t#[K**(-1)]\n",
+      "Beta_T2_P1 = 458*10**(-6);\t\t\t#[K**(-1)]\n",
+      "Beta_T2_P2 = 568*10**(-6);\t\t\t#[K**(-1)]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The entropy change is given by\n",
+      "\t\t\t# dS = (Cp/T)*dT - ((dV/dT)_P)*dP\n",
+      "\t\t\t# The mean Cp between 25 and 50 C is \n",
+      "Cp_mean = (Cp_T1_P1 + Cp_T1_P1)/2;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "\n",
+      "\t\t\t# (dV/dT)_P=1bar = (V_T2_P1 - V_T1_P1)/(50 - 25)\n",
+      "dV_dT_P1 = ((V_T2_P1 - V_T1_P1)/(50 - 25))*10**(-6);\t\t\t#[m**(-3)/mol-K]\n",
+      "dV_dT_P2 = ((V_T2_P2 - V_T1_P2)/(50 - 25))*10**(-6);\t\t\t#[m**(-3)/mol-K]\n",
+      "\t\t\t# The mean value of (dV/dT)_P between 1 and 1000 bar is\n",
+      "dV_dT_mean = (dV_dT_P1 + dV_dT_P2)/2;\t\t\t#[m**(-3)/mol-K]\n",
+      "delta_S = Cp_mean*math.log(T_2/T_1) - dV_dT_mean*(P_2 - P_1)*10**(5);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "print \" The value of entropy change is given by  delta_S = %f J/mol-K\"%(delta_S);\n",
+      "\n",
+      "\t\t\t# Now let us determine the enthalpy change. We know that\n",
+      "\t\t\t# dH = Cp*dT + [V - T*(dV/dT)_P]*dP\n",
+      "\t\t\t# [V - T*(dV/dT)_P] = (V - T*V*Beta) = val (say)\n",
+      "\t\t\t# At state 1\n",
+      "val_1 = ((V_T1_P1)*10**(-6))*(1 - (T_1)*(Beta_T1_P1));\t\t\t#[m**(3)/mol]\n",
+      "\t\t\t# At state 2\n",
+      "val_2 = ((V_T2_P2)*10**(-6))*(1 - (T_2)*(Beta_T2_P2));\t\t\t#[m**(3)/mol]\n",
+      "val_mean = (val_1 + val_2)/2;\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "delta_H = Cp_mean*(T_2 - T_1) + val_mean*(P_2-P_1)*10**(5);\t\t\t#[J/mol]\n",
+      "\n",
+      "print \" The value of enthalpy change is given by  delta_H = %f J/mol\"%(delta_H);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of entropy change is given by  delta_S = 5.414201 J/mol-K\n",
+        " The value of enthalpy change is given by  delta_H = 3457.542629 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.22  Page Number : 256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 100 + 273.15;\t\t\t#[K]\n",
+      "P = 10;\t\t\t#[MPa]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# The volume expansivity is defined as \n",
+      "\t\t\t# Beta = (1/V)*(del V/del T)_P = (1/V)*(dV/dT)_P\n",
+      "\t\t\t# From compressed liquid water tables at 100 C and 10 MPa,\n",
+      "V = 0.0010385;\t\t\t#[m(3)/kg]\n",
+      "Beta = (1/V)*((0.0010549 - 0.0010245)/(120 - 80));\t\t\t#[K**(-1)] \t\t\t# The values are obtained from the steam table as reported in the book.\n",
+      "\n",
+      "print \"The value of volume expansivity is  Beta = %e K**-1)\"%(Beta);\n",
+      "\n",
+      "\t\t\t#Isothermal compressibility is defined as\n",
+      "\t\t\t# K_t = -(1/V)*(del V/del T)_T = -(1/V)*(dV/dT)_T\n",
+      "K_t = -(1/V)*((0.0010361 - 0.0010410)/(15 - 5));\t\t\t#[MPa**(-1)]  \t\t\t# The values are obtained from the steam table as reported in the book.\n",
+      "\n",
+      "K_t = K_t*10**(-3);\t\t\t#[kPa]\n",
+      "\n",
+      "print \"The value of isothermal compressibility is  K_t = %e kPa**-1)\"%(K_t);\n",
+      "\n",
+      "\t\t\t# Cp - Cv = (T*V*(Beta**(2)))/K_t\n",
+      "R = (T*V*(Beta**(2)))/K_t;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \"The value of the difference between Cp and Cv is  Cp-Cv = %f kJ/kg-K\"%(R);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of volume expansivity is  Beta = 7.318247e-04 K**-1)\n",
+        "The value of isothermal compressibility is  K_t = 4.718344e-07 kPa**-1)\n",
+        "The value of the difference between Cp and Cv is  Cp-Cv = 0.439860 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.23  Page Number : 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 300 + 273.15;\t\t\t#[K]\n",
+      "P = 4;\t\t\t#[MPa] \n",
+      "\n",
+      "# Calculations and Results\n",
+      "Cp_0 = 7.7 + 0.04594*10**(-2)*T + 0.2521*10**(-5)*T**(2) - 0.8587*10**(-9)*T**(3);\t\t\t#[cal/mol-K]\n",
+      "Cp_0 = (Cp_0*4.186)/18.015;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t# Cp(T,P) = Cp_0(T,P=0) - T*integral((del**2 V/del T**(2))_P)*dP from limit 0 to P\n",
+      "\t\t\t# Cp = Cp_0 - T*((del**2 V/del T**2)_Pavg)*(P_2 - P_1)\n",
+      "\n",
+      "P_avg = (0+4)/2;\t\t\t#[MPa]\n",
+      "\n",
+      "\t\t\t#Using finite difference we get  (del**2 V/del T**(2)) = ((V_(T+delta T) - 2*V_T + V_(T-delta T))/(delta_T**(2))\n",
+      "\t\t\t#(del**2 V/del T**(2))_Pavg = (V_(350 C) + V_(250 C) - 2*V_(300 C))/(delta_T**(2)) = del_2 (say)\n",
+      "del_2 = (0.13857 + 0.11144 - 2*0.12547)/(50**(2));\t\t\t#[m**(3)/kg-K**2] \t\t\t# The values are obtained from the steam table as reported in the book.\n",
+      "\n",
+      "\n",
+      "Cp = Cp_0 - T*del_2*4000;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "print \" The value of constant pressure specific heat capacity is Cp = %f kJ/kg-K\"%(Cp);\n",
+      "\n",
+      "\t\t\t# At P = 4 MPa\n",
+      "\t\t\t# Cp = (del H/del T)_P = (H_350 C - H_250 C)/(350 - 250.4)\n",
+      "\t\t\t# Cp = (3092.5 - 2801.4)/(350 - 250.4) = 2.923 [kJ/kg-K]\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of constant pressure specific heat capacity is Cp = 2.858079 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.24  Page Number : 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 300 + 273.15;\t\t\t#[K]\n",
+      "P = 2.0;\t\t\t#[MPa]\n",
+      "\n",
+      "\t\t\t# At 2 MPa and 250 C \n",
+      "H_1 = 2902.5;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# At 2 MPa and 350 C \n",
+      "H_2 = 3137.0;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# Calculations\n",
+      "Cp = (H_2 - H_1)/(350 - 250);\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "# Results\n",
+      "print \" The value of constant pressure specific heat capacity is Cp = %f kJ/kg-K\"%(Cp);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of constant pressure specific heat capacity is Cp = 2.345000 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.25  Page Number : 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T = 80 + 273.15;\t\t\t#[K]\n",
+      "P = 10;\t\t\t#[MPa] \n",
+      "\n",
+      "V_1 = 0.0010245;\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t# At 60 C and 10 MPa\n",
+      "V_2 = 0.0010127;\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t# At 100 C and 10 MPa\n",
+      "V_3 = 0.0010385;\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Beta = (1/V_1)*((V_3 - V_2)/(100 - 60));\t\t\t#[K**(-1)]\n",
+      "\n",
+      "print \"The value of volume expansivity is  Beta = %e K**-1)\"%(Beta);\n",
+      "\n",
+      "\t\t\t#Isothermal compressibility is given by\n",
+      "\t\t\t# K_t = -(1/V)*(del V/del P)_T\n",
+      "\n",
+      "\t\t\t# Temperature is kept fixed at 80 C and different pressures are taken to calculate (del V/del P)_T\n",
+      "\t\t\t# At 80 C and 5 MPa\n",
+      "V_4 = 0.0010268;\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t# At 80 C and 10 MPa\n",
+      "V_5 = 0.0010245;\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t# At 80 C and 15 MPa\n",
+      "V_6 = 0.0010222;\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "\t\t\t# K_t = -(1/V)*(del V/del T)_P\n",
+      "K_t = -(1/V_1)*((V_4 - V_6)/(5 - 15));\t\t\t#[MPa**(-1)]\n",
+      "K_t = K_t*10**(-6);\t\t\t#[Pa**(-1)]\n",
+      "\n",
+      "print \"The value of isothermal compressibility is  K_t = %e Pa**-1)\"%(K_t);\n",
+      "\n",
+      "\t\t\t# Cp - Cv = (T*V*(Beta**(2)))/K_t\n",
+      "R = (T*V_1*(Beta**(2)))/K_t;\t\t\t#[J/kg-K]\n",
+      "\n",
+      "print \"The value of the difference between Cp and Cv is  Cp-Cv = %f J/kg-K\"%(R);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of volume expansivity is  Beta = 6.295754e-04 K**-1)\n",
+        "The value of isothermal compressibility is  K_t = 4.489995e-10 Pa**-1)\n",
+        "The value of the difference between Cp and Cv is  Cp-Cv = 319.389628 J/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.26  Page Number : 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "P_1 = 150;\t\t\t#[bar]\n",
+      "P_2 = 1;\t\t\t#[bar]\n",
+      "\n",
+      "T_1 = 300;\t\t\t#[K]\n",
+      "T_2 = 260;\t\t\t#[K]\n",
+      "T_3 = 280;\t\t\t#[K]\n",
+      "T_4 = 200;\t\t\t#[K]\n",
+      "T_5 = 120;\t\t\t#[K]\n",
+      "T_6 = 140;\t\t\t#[K]\n",
+      "\n",
+      "H_P1_T1 = 271.8;\t\t\t#[kJ/kg]\n",
+      "H_P2_T2 = 260.0;\t\t\t#[kJ/kg] \n",
+      "H_P2_T3 = 280.2;\t\t\t#[kJ/kg]\n",
+      "H_P1_T4 = 129.2;\t\t\t#[kJ/kg]\n",
+      "H_P2_T5 = 118.8;\t\t\t#[kJ/kg]\n",
+      "H_P2_T6 = 139.1;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(a)\n",
+      "\n",
+      "\n",
+      "T_new = ((H_P1_T1 - H_P2_T2)/(H_P2_T3 - H_P2_T2))*(T_3 - T_2) + T_2;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# Therefore Joule-Thomson coefficient is given by,\n",
+      "meu = (T_1 - T_new)/(P_1 - P_2);\t\t\t#[K/bar]\n",
+      "\n",
+      "print \" a).The value of Joule-Thomson coefficient for initial T = 300 K) is %f J/bar\"%(meu);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\n",
+      "\n",
+      "T_new_prime = ((H_P1_T4 - H_P2_T5)/(H_P2_T6 - H_P2_T5))*(T_6 - T_5) + T_5;\t\t\t#[K]\n",
+      "\n",
+      "\t\t\t# Therefore Joule-Thomson coefficient is given by,\n",
+      "meu_prime = (T_4 - T_new_prime)/(P_1 - P_2);\t\t\t#[K/bar]\n",
+      "\n",
+      "print \" b).The value of Joule-Thomson coefficient for initial T = 200 K) is %f J/bar\"%(meu_prime);\n",
+      "\n",
+      "\t\t\t# Therefore the Joule-Thomson coefficient is higher for low initial temperatures and therefore the drop in temperature is more.\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The value of Joule-Thomson coefficient for initial T = 300 K) is 0.190046 J/bar\n",
+        " b).The value of Joule-Thomson coefficient for initial T = 200 K) is 0.468146 J/bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.27  Page Number : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 300;\t\t\t#[K] - Temperature\n",
+      "P = 5;\t\t\t#[atm] - Pressure\n",
+      "P = P*101325;\t\t\t#[Pa]\n",
+      "Cp_0 = 35.78;\t\t\t#[J/mol-K] - Smath.tan(math.radiansard specific heat capacity at constant pressure\n",
+      "B = -50;\t\t\t#[cm**(3)/mol]\n",
+      "B = B*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# (dB/dT) = 1.0 = dB_dT (say)\n",
+      "dB_dT = 1.0;\t\t\t#[cm**(3)/mol-K]\n",
+      "dB_dT = dB_dT*10**(-6);\t\t\t#[m**(3)/mol-K]\n",
+      "\n",
+      "\t\t\t# (d**2 B/d T**2) = -0.01 = dB_dT_2 (say)\n",
+      "dB_dT_2 = -0.01;\t\t\t#[cm**(3)/mol-K**(2)]\n",
+      "dB_dT_2 = dB_dT_2*10**(-6);\t\t\t#[m**(3)/mol-K**(2)]\n",
+      "\n",
+      "Cp = Cp_0 - P*T*(dB_dT_2);\t\t\t#[[J/mol-K]] - Specific heat capacity at constant pressure\n",
+      "\n",
+      "\t\t\t#Therefore Joule-Thomson coefficient is given by,\n",
+      "meu = (1/Cp)*(-B + T*dB_dT);\t\t\t#[K/Pa]\n",
+      "meu = meu*10**(5);\t\t\t#[K/bar]\n",
+      "\n",
+      "# Results\n",
+      "print \" c).The value of Joule-Thomson coefficient is %f K/bar\"%(meu);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " c).The value of Joule-Thomson coefficient is 0.938341 K/bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.30  Page Number : 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "den_liq = 13690.;\t\t\t#[kg/m**(3)] - Density of liquid mercury\n",
+      "den_solid = 14190.;\t\t\t#[kg/m**(3)] - Density of solid mercury\n",
+      "mp = -38.87;\t\t\t#[C] - Melting point of mercury at pressure of 1 bar\n",
+      "mp = mp + 273.15;\t\t\t#[K]\n",
+      "T_req = 0.;\t\t\t#[C] - Required temperature to which the melting point is to be raised\n",
+      "T_req = T_req + 273.15;\t\t\t#[K]\n",
+      "H_fus = 11.62;\t\t\t#[kJ/kg] - Latent heat of fusion of mercury\n",
+      "\n",
+      "# Calculations\n",
+      "V_liq = (1/den_liq);\t\t\t#[m**(3)/kg] - Specific volume of liquid mercury\n",
+      "\n",
+      "V_solid = (1/den_solid);\t\t\t#[m**(3)/kg] - Specific volume of solid mercury\n",
+      "\n",
+      "\t\t\t# (delta P/delta T) = ((P - 1)*100)/(T_req - mp)\n",
+      "\t\t\t# delta H/(T*delta V) = (H_liq - H_solid)/(T*(V_liq - V_solid)) = del (say)\n",
+      "d = (H_fus)/(mp*(V_liq - V_solid));\t\t\t#[kPa/K] - delta H/(T*delta V)\n",
+      "\n",
+      "\t\t\t#Equating the two sides and then solving we get\n",
+      "P = (d*(T_req - mp))/100 + 1;\t\t\t#[bar]\n",
+      "\n",
+      "# Results\n",
+      "print \" The required pressure should be %f bar\"%(P);\n",
+      "\n",
+      "# answers are correct but vary from book because of rouding error. please calculate manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The required pressure should be 7491.335878 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.31  Page Number : 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Calculations and Results\n",
+      "\t\t\t#(1)\n",
+      "\t\t\t# At 1 bar\n",
+      "\t\t\t# Considering the data given at pressure 1 and 1000 bar, we have\n",
+      "delta_H_fus_1 = ((1000-1)*10**(5)*(273.15-22.6)*3.97*10**(-6))/(14.8+22.6);\t\t\t#[J/mol]\n",
+      "delta_S_fus_1 = delta_H_fus_1/(273.15-22.6);\t\t\t#[J/mol-K]\n",
+      "\n",
+      "print \" 1).The delta_H_fus at 1 bar is %f J/mol\"%(delta_H_fus_1);\n",
+      "print \"     The delta_S_fus at 1 bar is %f J/mol-K\"%(delta_S_fus_1);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t# At 6000 bar\n",
+      "T_mean = (128.8+173.6)/2;\t\t\t#[C] - Mean temperature\n",
+      "T_mean = T_mean + 273.15;\t\t\t#[K]\n",
+      "delta_V_fus_mean = (1.12+1.55)/2;\t\t\t#[cm**(3)/mol]\n",
+      "\n",
+      "\t\t\t# Consider the data at pressure of 5000 and 7000 bar we get,\n",
+      "delta_H_fus_2 = ((7000-5000)*10**(5)*(T_mean*delta_V_fus_mean*10**(-6)))/(173.6-128.8);\t\t\t#[J/mol]\n",
+      "delta_S_fus_2 = delta_H_fus_2/T_mean;\t\t\t#[J/mol-K]\n",
+      "\n",
+      "print \" 2).The delta_H_fus at 6000 bar is %f J/mol\"%(delta_H_fus_2);\n",
+      "print \"     The delta_S_fus at 6000 bar is %f J/mol-K\"%(delta_S_fus_2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1).The delta_H_fus at 1 bar is 2656.921969 J/mol\n",
+        "     The delta_S_fus at 1 bar is 10.604358 J/mol-K\n",
+        " 2).The delta_H_fus at 6000 bar is 2529.050223 J/mol\n",
+        "     The delta_S_fus at 6000 bar is 5.959821 J/mol-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.32  Page Number : 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "H_fus = 80;\t\t\t#[cal/g] - Heat of fusion at 0 C and 1 atm pressure\n",
+      "T = 0+273.15;\t\t\t#[K] - Temperature\n",
+      "vol_ratio = 1.091;\t\t\t# Ratio of the specific volume of ice and water.\n",
+      "sp_vol = 0.001;\t\t\t#[m**(3)/kg] - Specific volume of saturated liquid water.\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# The clapeyron equation can be written as\n",
+      "\t\t\t# (dP/dT)_sat = T*delta V_LS/(delta H_LS) = (T*(V_ice - V_water))/(H_ice - H_water)\n",
+      "dP_dT = (T*(vol_ratio - 1)*10**(-3))/(-H_fus*4.186);\t\t\t#[K/kPa]\n",
+      "\n",
+      "# Results\n",
+      "print \"The value of dT/dP)_sat is %e K/kPa\"%(dP_dT);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of dT/dP)_sat is -7.422554e-05 K/kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.33  Page Number : 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "P = 2;\t\t\t#[atm] - Surrounding pressure\n",
+      "bp_water = 100 + 273.15;\t\t\t#[K] - Boiling point of water at 1 atm pressure\n",
+      "delta_H_vap = 2257;\t\t\t#[kJ/kg] - Enthalpy of vaporization\n",
+      "delta_H_vap = delta_H_vap*18.015;\t\t\t#[J/mol]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# The clapeyron equation is given by\n",
+      "\t\t\t# math.log(P_2_sat/P_1_sat) = (-delta H_vap/R)*(1/T_2 - 1/T_1)\n",
+      "P_1_sat = 1;\t\t\t#[atm]\n",
+      "P_2_sat = P;\n",
+      "T_1 = bp_water;\n",
+      "\n",
+      "\t\t\t# Solving the above equation\n",
+      "T_2 = 1/((math.log(P_2_sat/P_1_sat))/(-delta_H_vap/R) + (1/T_1));\t\t\t#[K]\n",
+      "T_2 = T_2 - 273.15;\t\t\t#[C]\n",
+      "\n",
+      "# Results\n",
+      "print \" The boiling point of water at a pressure of 2 atm is %f C\"%(T_2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The boiling point of water at a pressure of 2 atm is 120.836990 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.34  Page Number : 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T_1 = 0.01 +273.15;\t\t\t#[K]\n",
+      "T_2 = 1 + 273.15;\t\t\t#[K]\n",
+      "P_sat_1 = 0.611;\t\t\t#[kPa] - P_sat at temperature T_1\n",
+      "P_sat_2 = 0.657;\t\t\t#[kPa] - P_sat at temperature T_2\n",
+      "Mol_wt = 18.015;\t\t\t#[g/mol] - Molecular weight of water\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results            \n",
+      "            # The clapeyron equation is given by\n",
+      "\t\t\t# math.log(P_sat_2/P_sat_1) = (-delta H_LV/R)*(1/T_2 - 1/T_1)\n",
+      "\n",
+      "\t\t\t# Solving the above equation\n",
+      "delta_H = -(math.log(P_sat_2/P_sat_1)/(1/T_2 - 1/T_1))*R;\t\t\t#[J/mol]\n",
+      "delta_H = delta_H/Mol_wt;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "print \" The enthalpy of vaporization is %f kJ/kg\"%(delta_H);\n",
+      "\n",
+      "\t\t\t# Entropy of vaporization is given by\n",
+      "S_vap = delta_H/T_2;\t\t\t#[kJ/kg-K]\n",
+      "print \" The entropy of vaporization is %f kJ/kg-K\"%(S_vap);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy of vaporization is 2533.991278 kJ/kg\n",
+        " The entropy of vaporization is 9.243083 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.35  Page Number : 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 100. + 273.15;\t\t\t#[K]\n",
+      "\t\t\t# (dT/dP)_sat = (1/27.12) K/mm\n",
+      "dT_dP = (1/27.12);\t\t\t#[K/mm]\n",
+      "dT_dP = dT_dP*(760./101325);\t\t\t#[K/Pa]\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# The clapeyron equation is given by\n",
+      "\t\t\t# (dP/dT)_sat = (-delta H_LV)/(T*delta V_LV)\n",
+      "\t\t\t# delta H_LV = T*delta V_LV*(dP/dT)_sat\n",
+      "\n",
+      "\t\t\t# (dP/dT)_sat = 1/(dT/dP)_sat\n",
+      "dP_dT = 1/dT_dP;\t\t\t#[Pa/K]\n",
+      "\n",
+      "\t\t\t# From saturated steam table at 100 C\n",
+      "V_vap = 1.6729;\t\t\t#[m**(3)/kg]\n",
+      "V_liq = 0.001044;\t\t\t#[m**(3)/kg]\n",
+      "delta_V = V_vap - V_liq;\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "\t\t\t# Therefore delta_H_LV is given by\n",
+      "delta_H_LV = T*delta_V*(dP_dT);\t\t\t#[J/kg]\n",
+      "delta_H_LV = delta_H_LV*10**(-3);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# Results\n",
+      "print \" The heat of vaporization of water is %f kJ/kg\"%(delta_H_LV);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The heat of vaporization of water is 2255.667174 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.36  Page Number : 270"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 100 + 273.15;\t\t\t#[K]\n",
+      "P_1 = 1.01325;\t\t\t#[bar]\n",
+      "T_2 = 98 + 273.15;\t\t\t#[K]\n",
+      "P_2 = 0.943;\t\t\t#[bar]\n",
+      "V_vap = 1.789;\t\t\t#[m**(3)] - Volume in vapour phase\n",
+      "vessel_vol = 1.673;\t\t\t#[m**(3)] - Volume of the vessel\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t\n",
+      "x = vessel_vol/V_vap;\n",
+      "\n",
+      "\t\t\t\n",
+      "m_vap = x;\t\t\t#[kg] - Mass of saturated vapour\n",
+      "m_liq = (1 - x);\t\t\t#[kg] - Mass of saturated liquid\n",
+      "\n",
+      "print \" The amount of vapour condensed is %f kg\"%(m_liq);\n",
+      "\n",
+      "\t\t\t# The clapeyron equation is given by\n",
+      "\t\t\t# math.log(P_2_sat/P_1_sat) = (-delta H_LV/R)*(1/T_2 - 1/T_1)\n",
+      "\n",
+      "\t\t\t# Solving the above equation\n",
+      "delta_H = -(math.log(P_2/P_1)/(1/T_2 - 1/T_1))*R;\n",
+      "delta_H = delta_H/18.015;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "print \" The latent heat of vaporization is %f kJ/kg\"%(delta_H);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The amount of vapour condensed is 0.064841 kg\n",
+        " The latent heat of vaporization is 2296.240786 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.37  Page Number : 270"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.integrate import quad \n",
+      "from scipy.optimize import *\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 298.15;\t\t\t#[K] - Smath.tan(math.radiansard reaction temperature\n",
+      "delta_H_gas = -52.23;\t\t\t#[kcal/mol] - Enthalpy of formation of C2H5OH(gas)\n",
+      "delta_H_liq = -66.35;\t\t\t#[kcal/mol] - Enthalpy of formation of C2H5OH(liq)\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# For ethanol(g) [T is in K and Cp_0 in cal/mol-K]\n",
+      "\t\t\t# Cp_0 = 4.75 + 5.006*10**(-2)*T - 2.479*10**(-5)*T**(2) + 4.79*10**(-9)*T**(3)\n",
+      "\n",
+      "\t\t\t# For ethanol(l) [T is in K and Cp_0 in cal/mol-K]\n",
+      "\t\t\t# Cp_0 = 67.29 - 0.343*T - 6.94*10**(-4)*T**(2)\n",
+      "\n",
+      "\t\t\t# The vaporization of a liquid can be written as C2H5OH(liq) - C2H5OH(gas)\n",
+      "\t\t\t# Since the pressure is 1 atm therefore the standard data can be used\n",
+      "delta_H_298 = delta_H_gas - delta_H_liq;\t\t\t#[kcal/mol]\n",
+      "delta_H_298 = delta_H_298*1000;\t\t\t#[cal/mol]\n",
+      "delta_a = 4.75 - 67.29;\n",
+      "delta_b = 5.006*10**(-2) - (-0.343);\n",
+      "delta_c = -2.479*10**(-5) - 6.94*10**(-4);\n",
+      "delta_d = 4.79*10**(-9);\n",
+      "\n",
+      "\t\t\t# The standard enthalpy of vaporization at a temperature T is given by\n",
+      "\n",
+      "def f31(T): \n",
+      "\t return delta_a + delta_b*T + delta_c*T**(2) + delta_d*T**(3)\n",
+      "\n",
+      "\t\t\t# delta_H_T =  delta_H_298 +  quad(f31,T_1,T)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t# Therefore the standard enthalpy of vaporization at a temperature T = 283 K is given by\n",
+      "T_2 = 283;\t\t\t#[K]\n",
+      "\n",
+      "def f32(T): \n",
+      "\t return delta_a+delta_b*T+delta_c*T**(2)+delta_d*T**(3)\n",
+      "\n",
+      "delta_H_283 =  delta_H_298 +  quad(f32,T_1,T_2)[0]\n",
+      "\n",
+      "\n",
+      "\t\t\t# Therefore the standard enthalpy of vaporization at a temperature T = 348 K is given by\n",
+      "T_3 = 348;\t\t\t#[K]\n",
+      "\n",
+      "def f33(T): \n",
+      "\t return delta_a+delta_b*T+delta_c*T**(2)+delta_d*T**(3)\n",
+      "\n",
+      "delta_H_348 =  delta_H_298 +  quad(f33,T_1,T_3)[0]\n",
+      "\n",
+      "\n",
+      "print \" The value of enthalpy of vaporization at 283 K is %f cal/mol\"%(delta_H_283);\n",
+      "print \" The value of enthalpy of vaporization at 298.15 K is %f cal/mol\"%(delta_H_298);\n",
+      "print \" The value of enthalpy of vaporization at 348 K is %f cal/mol\"%(delta_H_348);\n",
+      "\n",
+      "\n",
+      "\t\t\t# Solving the above equatio manually we get,\n",
+      "\t\t\t# delta_H_vap = 1.1975*10**(-9)*T**(4) - 2.396*10**(-4)*T**(3) + 0.1965*T**(2) - 62.54*T + 21639.54 \n",
+      "\t\t\t# Solving for 'T' at which 'delta_H_vap' = 0\n",
+      "def f(T): \n",
+      "\t return 1.1975*10**(-9)*T**(4)-2.396*10**(-4)*T**(3)+0.1965*T**(2)-62.54*T + 21639.54\n",
+      "T_0 = fsolve(f,500)\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n",
+      "print \" The temperature obtained by equating standard enthalpy of vaporization equal to zero is %f K\"%(T_0);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The value of enthalpy of vaporization at 283 K is 14255.030925 cal/mol\n",
+        " The value of enthalpy of vaporization at 298.15 K is 14120.000000 cal/mol\n",
+        " The value of enthalpy of vaporization at 348 K is 13593.385895 cal/mol\n",
+        " The temperature obtained by equating standard enthalpy of vaporization equal to zero is 635.058887 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.38  Page Number : 276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "T = 300 + 273.15;\t\t\t#[K] - Temperature\n",
+      "P = 9000;\t\t\t#[kPa] - Pressure\n",
+      "P_sat = 8592.7;\t\t\t#[kPa] - Vapour pressure of saturated water at 300 C\n",
+      "f_sat = 6738.9;\t\t\t#[kPa] - Fugacity of saturated water at 300 C\n",
+      "V_liq = 25.28;\t\t\t#[cm**(3)/mol] - Molar volume of water in liquid phase\n",
+      "V_liq = V_liq*10**(-6);\t\t\t# [m**(3)/mol]\n",
+      "V_vap = 391.1;\t\t\t#[cm**(3)/mol] - Molar volume of water in vapour phase\n",
+      "V_vap = V_vap*10**(-6);\t\t\t# [m**(3)/mol]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t# At 300 C and 9000 kPa water is a compressed liquid and its fugacity is given by\n",
+      "\t\t\t# f = f_sat*exp[V_liq*(P - P_sat)/R*T]\n",
+      "fugacity = f_sat*math.exp((V_liq*(P - P_sat)*1000)/(R*T));\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity of water at 9000 kPa is %f kPa\"%(fugacity);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity of water at 9000 kPa is 6753.477111 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.39  Page Number : 276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 200 + 273.15;\t\t\t#[K] - Temperature\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\t\t\t# From steam table at 200 C as reported in the book\n",
+      "P_sat = 1.5538;\t\t\t#[MPa] - Vapour pressure of saturated steam\n",
+      "H_vap = 2793.2;\t\t\t#[kJ/kg] - Enthalpy of saturated steam in vapour phase\n",
+      "S_vap = 6.4323;\t\t\t#[kJ/kg-K] - Entropy of saturated steam in vapour phase\n",
+      "\n",
+      "# Calculations\n",
+      "G_sat = H_vap - T*S_vap;\t\t\t#[kJ/kg] - Gibbs free energy\n",
+      "G_sat = G_sat*18.015;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# Now let us calculate the Gibbs free energy at the lowest pressure available in superheated steam tables at 200 C\n",
+      "\t\t\t# At 200 C and 0.01 MPa as reported in the book\n",
+      "H = 2879.5;\t\t\t#[kJ/kg] - Enthalpy\n",
+      "S = 8.9038;\t\t\t#[kJ/kg-K] - Entropy\n",
+      "G_ig = H - T*S;\t\t\t#[kJ/kg] - Gibbs free energy\n",
+      "G_ig = G_ig*18.015;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# Integrating from ideal gas state at 200 C and 0.01 MPa to saturated vapour at 200 C we get\n",
+      "\t\t\t# G_sat - G_ig = R*T*math.log(f_sat/f_ig)\n",
+      "\n",
+      "\t\t\t# Under the ideal gas condition the pressure is small therefore f_ig = P = 0.01 MPa\n",
+      "f_ig = 0.01;\t\t\t#[MPa]\n",
+      "\n",
+      "\t\t\t# Solving the above equation\n",
+      "f_sat = f_ig*(math.exp((G_sat - G_ig)/(R*T)));\t\t\t#[MPa]\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity of saturated steam at 200 C is %f MPa\"%(f_sat);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity of saturated steam at 200 C is 1.426074 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.40  Page Number : 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 320 + 273.15;\t\t\t#[K]\n",
+      "P_1 = 70;\t\t\t#[bar]\n",
+      "P_2 = 170;\t\t\t#[bar]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#(a)\n",
+      "\t\t\t# dG = R*T*dmath.log(f)\n",
+      "\t\t\t# G - G_ig = R*T*math.log(f/f_ig)\n",
+      "\n",
+      "\t\t\t# From steam table the low pressure that is available is 1 kPa.\n",
+      "f_ig = 1;\t\t\t#[kPa] - Assuming ideal gas behaviour at such low pressure\n",
+      "\n",
+      "\t\t\t# At 1 kPa (under saturated conditions)\n",
+      "P_sat = 112.891;\t\t\t#[bar]\n",
+      "\t\t\t# Therefore at both 1 kPa and 70 bar the stem is superheated and byond a pressure of 112.891 bar it is compressed liquid.\n",
+      "\n",
+      "\t\t\t# For superheated steam table at 1 kPa and 320 C, as repoted in the book\n",
+      "H_1 = 3117.08;\t\t\t#[kJ/kg] - Enthalpy\n",
+      "S_1 = 10.41232;\t\t\t#[kJ/kg-K] - Entropy\n",
+      "\n",
+      "\t\t\t# For superheated steam table at 70 bar and 320 C, as repoted in the book\n",
+      "H_2 = 2916.92;\t\t\t#[kJ/kg] - Enthalpy\n",
+      "S_2 = 6.0651;\t\t\t#[kJ/kg-K] - Entropy\n",
+      "\n",
+      "\t\t\t# At 70 bar and 320 C,\n",
+      "G = H_2 - T*S_2;\t\t\t#[kJ/kg] - Gibbs free energy\n",
+      "\t\t\t# At 1 kPa and 320 C\n",
+      "G_ig = H_1 - T*S_1;\t\t\t#[kJ/kg] - Gibbs free energy\n",
+      "\n",
+      "\t\t\t# math.log(f/f_ig) = (G - G_ig)/(R*T)\n",
+      "f = f_ig*(math.exp((G - G_ig)*18/(R*T)));\t\t\t#[kPa]\n",
+      "f = f*10**(-2);\t\t\t#[bar]\n",
+      "\n",
+      "\t\t\t# At 70 bar\n",
+      "phi = f/P_1;\n",
+      "\n",
+      "print \" a).The fugacity of steam at 320 C and 70 bar is %f bar\"%(f);\n",
+      "print \"     The fugacity coefficient at 320 C and 70 bar is  phi = %f\"%(phi);\n",
+      "\n",
+      "\t\t\t#(b)\n",
+      "\t\t\t# Now consider saturated steam at 320 C. We have\n",
+      "P_sat = 112.891;\t\t\t#[bar]\n",
+      "V_liquid = 1.5;\t\t\t#[cm**(3)/mol] - Molar vlume of saturated liquid\n",
+      "V_liquid = V_liquid*10**(-6);\t\t\t#[m**(3)/mol]\n",
+      "V_vapour = 15.48;\t\t\t#[cm**(3)/mol] - Molar vlume of saturated vapour\n",
+      "U_liqid = 1445.7;\t\t\t#[Kj/kg] - Internal energy of satuarted liquid\n",
+      "U_vapour = 2528.9;\t\t\t#[kJ/kg] - Internal energy of satuarted vapour\n",
+      "H_liquid = 1462.6;\t\t\t#[kJ/kg] - Enthalpy of saturated liquid\n",
+      "H_vapour = 2703.7;\t\t\t#[kJ/kg] - Enthalpy of saturated vapour\n",
+      "S_liquid = 3.45;\t\t\t#[kJ/kg-K]  - Entropy of saturated liquid\n",
+      "S_vapour = 5.5423;\t\t\t#[kJ/kg-K] - Entropy of saturated vapour\n",
+      "\n",
+      "\t\t\t# Now let us calculate Gibbs free energy of saturated liquid and saturated vapour\n",
+      "G_liquid = H_liquid - T*S_liquid;\t\t\t#[kJ/kg]\n",
+      "G_vapour = H_vapour - T*S_vapour;\t\t\t#[kJ/kg]\n",
+      "\t\t\t# Note that under saturated conditions\n",
+      "\t\t\t# G_sat = G_liquid = G_vapour\n",
+      "G_sat = G_liquid;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t# math.log(f_sat/f_ig) = (G_sat - G_ig)/(R*T)\n",
+      "f_sat = f_ig*(math.exp((G_sat - G_ig)*18/(R*T)));\t\t\t#[kPa]\n",
+      "f_sat = f_sat*10**(-2);\t\t\t#[bar]\n",
+      "\n",
+      "phi_sat = f_sat/P_sat;\n",
+      "\n",
+      "\t\t\t# And now the fugacity is to be determined at 320 C and P = 170 bar. We know the following relation for compressed liquid.\n",
+      "\t\t\t# f_CL = f_sat*exp(V_liquid*(P-P_sat)/(R*T))\n",
+      "f_CL = f_sat*math.exp(V_liquid*18*(P_2-P_sat)*10**(5)/(R*T));\t\t\t#[bar]\n",
+      "\n",
+      "\t\t\t# Therefore the fugacity coefficient at 170 bar and 320 C is given by\n",
+      "phi_2 = f_CL/P_2;\n",
+      "\n",
+      "print \" b).The fugacity of steam at 320 C and 170 bar is %f bar\"%(f_CL);\n",
+      "print \"     The fugacity coefficient at 320 C and 170 bar is phi = %f\"%(phi_2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a).The fugacity of steam at 320 C and 70 bar is 58.913361 bar\n",
+        "     The fugacity coefficient at 320 C and 70 bar is  phi = 0.841619\n",
+        " b).The fugacity of steam at 320 C and 170 bar is 86.552966 bar\n",
+        "     The fugacity coefficient at 320 C and 170 bar is phi = 0.509135\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.41  Page Number : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "# Variables\n",
+      "T = 300 + 273.15;\t\t\t#[K]\n",
+      "P_1 = 12500*10**(3);\t\t\t#[Pa]\n",
+      "P_2 = 8581*10**(3);\t\t\t#[Pa]\n",
+      "P_3 = 300*10**(3);\t\t\t#[Pa]\n",
+      "V_liq = 1.404;\t\t\t#[cm**(3)/g] - Specific volume of liquid\n",
+      "V_liq = (V_liq/10**(6))*18.015;\t\t\t#[m**(3)/mol]\n",
+      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t# state 1: 300 C, 12500 kPa\n",
+      "\t\t\t# state 2: 300 C, 8581 kPa\n",
+      "\t\t\t# state 3: 300 C, 300 kPa\n",
+      "\n",
+      "\t\t\t# From state 1 to state 2 the system is liquid water and if the molar volume of liquid is assumed costant we can write\n",
+      "\t\t\t# G_2 - G_1 = V_liq*(P_2 - P_1)\n",
+      "\t\t\t# G_2 - G_1 = R*Tmath.log(f_2/f_1)\n",
+      "\t\t\t# Comparing the above two equations we get\n",
+      "\t\t\t# (f_2/f_1) = exp((V_liq*(P_2 - P_1)/(R*T))\n",
+      "f2_f1 = math.exp((V_liq*(P_2 - P_1)/(R*T))); \t\t\t# (f_2/f_1) = f2_f1 (say)\n",
+      "\n",
+      "\t\t\t# In state 2 the fugacity of liquid is same as that of saturated vapour and for the vapour phase change from state 2 to 3 the fugacity ratio is calculated using \n",
+      "\t\t\t# G_3 - G_2 = R*Tmath.log(f_3/f_2)\n",
+      "\n",
+      "\t\t\t# At 300 C, 8581 kPa \n",
+      "H_liq_2 = 2749.0;\t\t\t#[kJ/kg]\n",
+      "S_vap_2 = 5.7045;\t\t\t#[kJ/kg-K]\n",
+      "G_vap_2 = -520.53;\t\t\t#[kJ/kg]\n",
+      "G_vap_2 = G_vap_2*18.015;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# At 300 C, 300 kPa \n",
+      "H_3 = 3069.3;\t\t\t#[kJ/kg]\n",
+      "S_3 = 7.7022;\t\t\t#[kJ/kg-K]\n",
+      "G_3 = -1345.22;\t\t\t#[kJ/kg]\n",
+      "G_3 = G_3*18.015;\t\t\t#[J/mol]\n",
+      "\n",
+      "\t\t\t# Substituting and solving the equation  G_3 - G_2 = R*Tmath.log(f_3/f_2)\n",
+      "f3_f2 = math.exp((G_3 - G_vap_2)/(R*T));\t\t\t#  (f_3/f_2) = f3_f2 (say)\n",
+      "\n",
+      "\t\t\t# (f_3/f_1) = (f_3/f_2)*(f_2/f_1)\n",
+      "f3_f1 = f3_f2*f2_f1;\n",
+      "\n",
+      "print \" The ratio of fugacity in the final state to that in the initial state is given by f3/f2 = %f\"%(f3_f2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The ratio of fugacity in the final state to that in the initial state is given by f3/f2 = 0.044255\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch8_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch8_2.ipynb
new file mode 100755
index 00000000..6b83d68c
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch8_2.ipynb
@@ -0,0 +1,752 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8df6804e24ffba947b26128cacd98ede56f82be4a3089c7567d672249f51ebda"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Chapter 8 : Thermodynamic Cycles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.1  Page Number : 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "P_1 = 30;\t\t\t#[bar]\n",
+      "P_2 = 0.04;\t\t\t#[bar]\n",
+      "\n",
+      "\t\t\t#(1).Carnot cycle\n",
+      "\t\t\t#It has been reported in the book that at 30 bar pressure (saturated) :\n",
+      "H_liq_1 = 1008.42;\t\t\t#[kJ/kg]\n",
+      "H_vap_1 = 2804.2;\t\t\t#[kJ/kg]\n",
+      "S_liq_1 = 2.6457;\t\t\t#[kJ/kg-K]\n",
+      "S_vap_1 = 6.1869;\t\t\t#[kJ/kh-K]\n",
+      "\t\t\t#Therefore, H_1 = H_liq_1, H_2 = H_vap_1, S_1 = S_liq_1 and S_2 = S_vap_1\n",
+      "H_1 = H_liq_1;\n",
+      "H_2 = H_vap_1;\n",
+      "S_1 = S_liq_1;\n",
+      "S_2 = S_vap_1;\n",
+      "\n",
+      "#At 0.04 bar pressure (saturated) :\n",
+      "H_liq_2 = 121.46;\t\t\t#[kJ/kg]\n",
+      "H_vap_2 = 2554.4;\t\t\t#[kJ/kg]\n",
+      "S_liq_2 = 0.4226;\t\t\t#[kJ/kg-K]\n",
+      "S_vap_2 = 8.4746;\t\t\t#[kJ/kh-K]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#Dryness fraction at state 3 can be found the fact that S_3 = S_2 \n",
+      "x_3 = (S_2 - S_liq_2)/(S_vap_2 - S_liq_2);\n",
+      "H_3 = H_liq_2*(1 - x_3) + x_3*H_vap_2;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Dryness fraction at state 4 can be found the fact that S_4 = S_1\n",
+      "x_4 = (S_1 - S_liq_2)/(S_vap_2 - S_liq_2);\n",
+      "H_4 = H_liq_2*(1 - x_4) + x_4*H_vap_2;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Work done by turbine W_tur = -delta_H = -(H_3 - H_2)\n",
+      "W_tur = H_2 - H_3;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Work supplied by boiler,\n",
+      "q_H = H_2 - H_1;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Work transfer in compressor is given by\n",
+      "W_com = -(H_1 - H_4);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Efficiency can now be calculated as\n",
+      "\t\t\t#n = (Net work done/Work supplied by boiler)\n",
+      "n_carnot = (W_tur + W_com)/q_H;\n",
+      "\n",
+      "\t\t\t#Efficiency of the Carnot cycle can also be determined from the formula\n",
+      "\t\t\t# n = 1 - (T_L/T_H), Where T_L is saturated temperature at 0.04 bar and T_H is saturated temperature at 30 bar\n",
+      "\n",
+      "print \"1.Carnot cycle\";\n",
+      "print \"The work done by the turbine is %f kJ/kg\"%(W_tur);\n",
+      "print \"The heat transfer in the boiler is %f kJ/kg\"%(q_H);\n",
+      "print \"The cycle efficiency is %f\"%(n_carnot);\n",
+      "\n",
+      "\t\t\t#(2).Rankine cycle\n",
+      "\t\t\t#The enthalpies at state 2 and 3 remain as in the Carnot cycle\n",
+      "\t\t\t#Saturated liquid enthalpy at 0.04 bar is \n",
+      "H_4_prime = H_liq_2;\n",
+      "\n",
+      "\t\t\t#Saturated liquid volume at 0.04 bar as reported in the book is\n",
+      "V_liq = 0.001004;\t\t\t#[m**(3)/kg]\n",
+      "\t\t\t#Work transfer in pump can be calculated as\n",
+      "W_pump = -V_liq*(P_1 - P_2)*100;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Work transfer around pump gives, W_pump = -delta_H =  -(H_1_prime - H_4_prime);\n",
+      "H_1_prime = H_4_prime - W_pump;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Heat supplied to boiler is\n",
+      "q_H_prime = H_2 - H_1_prime;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Work done by turbine is\n",
+      "W_tur_prime = H_2 - H_3;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Efficiency can now be calculated as\n",
+      "\t\t\t#n = (Net work done/Heat input)\n",
+      "n_rankine = (W_tur_prime + W_pump)/q_H_prime;\t\t\t#\n",
+      "\n",
+      "print \"2.Rankine cycle\";\n",
+      "print \"The work done by the turbine is %f kJ/kg\"%(W_tur_prime);\n",
+      "print \"The heat transfer in the boiler is %f kJ/kg\"%(q_H_prime);\n",
+      "print \"The cycle efficiency is %f\"%(n_rankine);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1.Carnot cycle\n",
+        "The work done by the turbine is 941.036567 kJ/kg\n",
+        "The heat transfer in the boiler is 1795.780000 kJ/kg\n",
+        "The cycle efficiency is 0.404166\n",
+        "2.Rankine cycle\n",
+        "The work done by the turbine is 941.036567 kJ/kg\n",
+        "The heat transfer in the boiler is 2679.732016 kJ/kg\n",
+        "The cycle efficiency is 0.350046\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.2  Page Number : 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_max = 700+273.15;\t\t\t#[K] - Maximum temperature.\n",
+      "P_boiler = 10*10**(6);\t\t\t#[Pa] - Constant pressure in the boiler\n",
+      "P_condenser = 10*10**(3);\t\t\t#[Pa] - Constant pressure in the condenser\n",
+      "\n",
+      "\t\t\t#At state 2 i.e, at 700 C and 10 MPa,it has been reported in the book that from steam table\n",
+      "S_2 = 7.1687;\t\t\t#[kJ/kg-K] - Entropy\n",
+      "H_2 = 3870.5;\t\t\t#[kJ/kg] - Enthalpy\n",
+      "\n",
+      "\t\t\t#At state 3 i.e, at 700 C and 10 KPa,\n",
+      "S_3 = S_2;\t\t\t#[kJ/kg-K]- Entropy \n",
+      "\n",
+      "\t\t\t#For sturated steam at 10 kPa, it has been reported in the book that from steam table\n",
+      "S_liq = 0.6493;\t\t\t#[kJ/kg-K]- Entropy of saturated liquid\n",
+      "S_vap = 8.1502;\t\t\t#[kJ/kg-K] - Enthalpy of saturated liquid\n",
+      "\t\t\t#Therefore steam is saturated and its dryness factor can be calculated as\n",
+      "x = (S_2 - S_liq)/(S_vap - S_liq);\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#The enthalpy at state 3 is now calculated. For steam at 10 kPa,it has been reported in the book that from steam table\n",
+      "H_liq = 191.83;\t\t\t#[kJ/kg]\n",
+      "H_vap = 2584.7;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#Therefore enthalpy at state 3 is\n",
+      "H_3 = H_liq*(1-x) + H_vap*x;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Work done by the turbine \n",
+      "W_tur = -(H_3 - H_2);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Now we have to calculate work input to the pump\n",
+      "\t\t\t#State 4:Saturated liquid at 10 kPa\n",
+      "\t\t\t#State 4:Compressed liquid at 10 MPa\n",
+      "\t\t\t#Since volume of liquid does not get affected by pressure we take volume of saturated liquid at 10 kPa,\n",
+      "V_liq = 0.001010;\t\t\t#[m**(3)/kg]\n",
+      "\n",
+      "\t\t\t#Work transfer in the pump is\n",
+      "W_pump = -V_liq*(P_boiler - P_condenser)*10**(-3);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Energy balance around pump gives, W_pump = -delta_H =  -(H_1 - H_4)\n",
+      "H_4 = H_liq;\t\t\t# Enthalpy at state 4 (saturated liquid at 10 kPa)\n",
+      "H_1 = H_4 - W_pump;\t\t\t#[kJ/kg]\n",
+      " \n",
+      "\t\t\t#Heat supplied to boiler is\n",
+      "q_H = H_2 - H_1;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Efficiency can now be calculated as\n",
+      "\t\t\t#n = (Net work done/Heat input)\n",
+      "n_rankine = (W_tur + W_pump)/q_H;\n",
+      "\n",
+      "print \"The efficiency of the Rankine cycle is found to be %f\"%(n_rankine);\n",
+      "\n",
+      "\t\t\t#Now let us determine the efficiency of Carnot cycle. The maximun temperature is 700 C and minimum temperature is that of saturated steam at 10 kPa,\n",
+      "T_min = 45.81 + 273.15;\t\t\t#[K] - From steam table as reported in the book\n",
+      "n_carnot = 1-(T_min/T_max);\n",
+      "\t\t\t#Note that the efficiency of Rankine cycle is less than that of carnot cycle"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency of the Rankine cycle is found to be 0.433088\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.3  Page Number : 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "W = 1.1;\t\t\t#[kW] - Work done per ton of refrigeration \n",
+      "\t\t\t#1 ton refrigeration = 3.517 kW, therefore\n",
+      "H = 3.517;\t\t\t#[kW] - Heat absorbed\n",
+      "T_low = -30 + 273.15;\t\t\t#[K] - Low temperature maintained\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#COP can be calculated as\n",
+      "\t\t\t#COP = (Heat absorbed/Work done)\n",
+      "COP = H/W;\n",
+      "\n",
+      "\t\t\t#For reversed carnot cycle, COP = T_low/(T_high - T_low). Solving this we get\n",
+      "T_high = (T_low/COP) + T_low;\t\t\t#[K] - Higher temperature\n",
+      "\n",
+      "\t\t\t#Heat rejected is\n",
+      "H_rej = W + H;\t\t\t#[kW];\n",
+      "\n",
+      "# Results\n",
+      "print \"The COP is %f\"%(COP);\n",
+      "print \"The higher temperature of the cycle is %f K\"%(T_high);\n",
+      "print \"The heat rejected per ton of refrigeration is %f kW\"%(H_rej);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The COP is 3.197273\n",
+        "The higher temperature of the cycle is 319.199190 K\n",
+        "The heat rejected per ton of refrigeration is 4.617000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.4  Page Number : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_high = 20 + 273.15;\t\t\t#[K] - High temperature\n",
+      "T_low = 0 + 273.15;\t\t\t#[K] - Low temperature\n",
+      "Q_H = 10;\t\t\t#[kW] - Heat supplied\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#If 'Q_H' is the rate at which heat is taken from surrounding and 'W' is the rate at which work is done,then\n",
+      "\t\t\t# Q_H = W + Q_L\n",
+      "\t\t\t#(Q_H/Q_L) = (T_high/T_low)\n",
+      "\t\t\t#Also for a reversible cycle, (Q_H/Q_L) = 1 + (W/Q_L). Solving we get,\n",
+      "Q_L = (T_low/T_high)*Q_H;\t\t\t#[kW]\n",
+      "W = (Q_H - Q_L) ;\t\t\t#[kW]\n",
+      " \n",
+      "# Results\n",
+      "print \"The minimum power required is %f kW\"%(W);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum power required is 0.682245 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.5  Page Number : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "T_high = 40 + 273.15;\t\t\t#[K] - High temperature\n",
+      "T_low = -20 + 273.15;\t\t\t#[K] - Low temperature\n",
+      "C = 10;\t\t\t#[tons of refrigeration] - Capacity\n",
+      "\t\t\t#1 ton refrigeration = 3.517 kW, therefore\n",
+      "H = C*3.517;\t\t\t#[kW] - Heat absorbed\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#For reversed carnot cycle, COP = T_low/(T_high - T_low)\n",
+      "COP = T_low/(T_high - T_low);\n",
+      "\n",
+      "\t\t\t# COP = (Refrigerating effect)/(Work input), therefore power required is given by\n",
+      "P = (H/COP);\t\t\t#[kW]\n",
+      "\n",
+      "# Results\n",
+      "print \"The COP is %f\"%(COP);\n",
+      "print \"The power required is %f kW\"%(P);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The COP is 4.219167\n",
+        "The power required is 8.335769 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.6  Page Number : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "COP = 4;\t\t\t#Coefficient of performance\n",
+      "P = 10;\t\t\t#[kW] - Work done on the cycle\n",
+      "\n",
+      "# Calculations\n",
+      "\t\t\t#For reversed carnot cycle, COP = T_low/(T_high - T_low)\n",
+      "\t\t\t#ratio = (T_high/T_low),therefore\n",
+      "ratio = -1/(COP + 1);\n",
+      "\n",
+      "\t\t\t# Refrigerating effect = (COP)*Work input, therefore refrigeration is given by\n",
+      "H = COP*P;\t\t\t#[kW]\n",
+      "\n",
+      "\t\t\t#Maximum refrigearation in tons is given by\n",
+      "H_max = (H/3.517);\n",
+      "\n",
+      "# Results\n",
+      "print \"The maximum refrigeration value is %f ton\"%(H_max);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum refrigeration value is 11.373330 ton\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.7  Page Number : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "m = 0.6;\t\t\t#[kg/s] - mass flow rate\n",
+      "T_low = -20+273.15;\t\t\t#[K] - Temperature at which vapour enters the compressor\n",
+      "T_high = 30+273.15;\t\t\t#[K] - Temperature at which vapour leaves the condenser\n",
+      "\n",
+      "\t\t\t#From saturated refrigeration-12 tables we get,at -20 C\n",
+      "H_1 = 178.74;\t\t\t#[kJ/kg] - (H_1 = H_vap)\n",
+      "P_1 = 0.15093;\t\t\t#[MPa] - (P_1 = P_sat)\n",
+      "P_4 = P_1;\n",
+      "S_1 = 0.7087;\t\t\t#[kJ/kg-K] - (S_1 = S_vap)\n",
+      "S_2 = S_1;\n",
+      "\n",
+      "\t\t\t#At 30 C\n",
+      "P_2 = 0.7449;\t\t\t#[MPa] - (P_2 = P_sat)\n",
+      "P_3 = P_2;\n",
+      "H_3 = 64.59;\t\t\t#[kJ/kg] - (H_3 = H_liq)\n",
+      "H_4 = H_3;\n",
+      "S_3 = 0.24;\t\t\t#[kJ/kg-K] - (S_3 = S_liq)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#It is assumed that presssure drop in the evaporator and condenser are negligible. The heat transfer rate in the evaporator is\n",
+      "Q_L = m*(H_1 - H_4);\n",
+      "\n",
+      "print \"The heat transfer rate in the evaporator is %f kW\"%(Q_L);\n",
+      "\n",
+      "\t\t\t#At state 2 (P = 0.7449 MPa and S = 0.7087 kJ/kg-K) and looking in the superheated tables we have to calculate the enthalpy at state 2\n",
+      "\n",
+      "\t\t\t#At P = 0.7 MPa and S = 0.6917 kJ/kg-K,\n",
+      "H_11 = 200.46;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#At P = 0.7 MPa and S = 0.7153 kJ/kg-K,\n",
+      "H_12 = 207.73;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Thus at P = 0.7 MPa and S = 0.7087 kJ/kg-K, enthalpy is given by\n",
+      "H_13 = ((S_2 -0.6917)/(0.7153 - 0.6917))*(H_12 - H_11) + H_11;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#At P = 0.8 MPa and S = 0.7021 kJ/kg-K,\n",
+      "H_21 = 206.07;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#At P = 0.8 MPa and S = 0.7253 kJ/kg-K,\n",
+      "H_22 = 213.45;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Thus at P = 0.8 MPa and S = 0.7087 kJ/kg-K, enthalpy is given by\n",
+      "H_23 = ((S_2 -0.7021)/(0.7253 - 0.7021))*(H_22 - H_21) + H_21;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#At P = 0.7449 MPa, S = 0.7087 kJ/kg-K, the enthalpy is\n",
+      "H_2 = ((0.7449 - 0.7)/(0.8 - 0.7))*(H_23 - H_13) + H_13;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#Power consumed by the compressor is\n",
+      "W_comp = m*(H_2 - H_1);\t\t\t#[kW]\n",
+      "\n",
+      "print \"The power consumed by the compressor is %f kW\"%(W_comp);\n",
+      "\n",
+      "\t\t\t#Heat removed in evaporator/work done on compressor\n",
+      "COP_R = Q_L/W_comp;\n",
+      "\n",
+      "print \"The COP the refrigerator is %f kW\"%(COP_R);\n",
+      "\n",
+      "\n",
+      "\t\t\t#At -20 C,saturated conditions \n",
+      "H_liq = 17.82;\t\t\t#[kJ/kg]\n",
+      "H_vap = 178.74;\t\t\t#[kJ/kg]\n",
+      "x_4 = (H_4 - H_liq)/(H_vap - H_liq);\n",
+      "\n",
+      "print \"The dryness factor of refrigerant after the expansion valve is %f\"%(x_4);\n",
+      "\n",
+      "\t\t\t#The heat transfer rate in the condenser is\n",
+      "Q_H = m*(H_3 - H_2);\t\t\t#[kW]\n",
+      "\n",
+      "print \"The heat transfer rate in the condenser is %f kW\"%(Q_H);\n",
+      "\n",
+      "\t\t\t#If the cycle would have worked as a pump then,\n",
+      "\t\t\t#COP_HP = (Heat supplied from condenser/Work done on compressor)\n",
+      "COP_HP = (-Q_H)/W_comp;\n",
+      "\n",
+      "print \"The COP if cycle would work as a heat pump is %f kW\"%(COP_HP);\n",
+      "\n",
+      "\t\t\t#If the cycle would have been a reversed Carnot cycle then\n",
+      "COP_C = T_low/(T_high - T_low);\n",
+      "\n",
+      "print \"The COP if cycle would run as reversed Carnot cycle is %f kW\"%(COP_C);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transfer rate in the evaporator is 68.490000 kW\n",
+        "The power consumed by the compressor is 16.840242 kW\n",
+        "The COP the refrigerator is 4.067044 kW\n",
+        "The dryness factor of refrigerant after the expansion valve is 0.290641\n",
+        "The heat transfer rate in the condenser is -85.330242 kW\n",
+        "The COP if cycle would work as a heat pump is 5.067044 kW\n",
+        "The COP if cycle would run as reversed Carnot cycle is 5.063000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.8  Page Number : 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\t\t\n",
+      "\n",
+      "# Variables\n",
+      "H_1 = 310.38;\t\t\t#[kJ/kg]\n",
+      "H_2 = 277.7;\t\t\t#[kJ/kg]\n",
+      "H_5 = -122.6;\t\t\t#[kJ/kg]\n",
+      "H_6 = 77.8;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#The enthalpy at point 3 is same at point 4 as the expansion is isenthalpic\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#The mass condensed is 1 kg and therefore m_1 = m+6 + 1\n",
+      "\n",
+      "\t\t\t#Enthalpy balance around heat exchanger\n",
+      "\t\t\t#m_2*H_2 + m_2*H_6 = m_3*H_3 + m_7*H_7\n",
+      "\n",
+      "\t\t\t#Enthalpy balance around separator\n",
+      "\t\t\t#m_4*H_4 = m_5*H_5 + m_6*H_6\n",
+      "\t\t\t#It can be seen that m_1 = m_2 = m_3 = m_4\n",
+      "\t\t\t#and m_6 = m_7 = m_1 - 1\n",
+      "\n",
+      "\t\t\t#Substituting the values for enthalpy balance around heat exchanger we get,\n",
+      "\t\t\t#m_1*H_2 + (m_1 - 1)*(H_6) = m_1*H_3 + (m_1 - 1)*H_1\n",
+      "\t\t\t#and substituting the values for enthalpy balance around seperator we get\n",
+      "\t\t\t#m_1*H_3 = (1)*(-122.6) + (m_1 - 1)*77.8\n",
+      "\t\t\t#H_3 = ((1)*(-122.6) + (m_1 - 1)*77.8)/m_1\n",
+      "\t\t\t#Substituting the expression for 'H_3' in the above equation and then solving for m_1, we get\n",
+      "def f(m_1): \n",
+      "\t return m_1*H_2+(m_1-1)*(H_6)-m_1*(((1)*(-122.6) + (m_1 - 1)*77.8)/m_1)-(m_1-1)*H_1\n",
+      "m_1 = fsolve(f,4)\n",
+      "\t\t\t#Thus to liquify 1 kg of air compression of m_1 kg of air is carried out.\n",
+      "\n",
+      "\t\t\t#Now substituting this value of m_1 to get the value of H_3,\n",
+      "H_3 = ((1)*(-122.6) + (m_1 - 1)*77.8)/m_1;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#From given compressed air table we see at 200 bar and 160 K,\n",
+      "H_3_1 = 40.2;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#At 200 bar and 180 K,\n",
+      "H_3_2 = 79.8;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#By interpolation we get,\n",
+      "T_3 = ((H_3 - H_3_1)*(180 - 160))/(H_3_2 - H_3_1) + 160;\t\t\t#[K]\n",
+      "\n",
+      "print \"Temperature before throttling is %f K\"%(T_3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature before throttling is 171.350719 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.9  Page Number : 304"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "# Variables\n",
+      "\t\t\t#At 1 bar, 310 K \n",
+      "H_1 = 310.38;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 200 bar, 310 K\n",
+      "H_2 = 277.7;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 1 bar, Saturated liquid\n",
+      "H_7 = -122.6;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 1 bar, Saturated vapour\n",
+      "H_8 = 77.8;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 200 bar, 200 K\n",
+      "H_3 = 117.6;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 1 bar, 100 K\n",
+      "H_11 = 98.3;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#For 1 kg of liquid air obtained,the overall enthalpy balance is\n",
+      "\t\t\t#m_2*H_2 = W - 122.6 + (m_2 - 1)*H_1\n",
+      "\t\t\t#W = - 0.8*m_2*(H_11 - H_3)\n",
+      "\t\t\t#Overall enthalpy balance equation becomes\n",
+      "\t\t\t#H_2*m_2 = 15.44*m_2 - H_7 + (m_2 - 1)*H_1, solving\n",
+      "m_2_prime = (H_7 - H_1)/(H_2 - 15.44 - H_1);\n",
+      "\n",
+      "print \"The number of kimath.lograms of air compressed per kg of liquid air produced is %f kg\"%(m_2_prime);\n",
+      "\n",
+      "\t\t\t#(2)\n",
+      "\t\t\t#Enthalpy balance around separator is \n",
+      "\t\t\t#0.2*m_2*H_5 = -H_7 + (0.2*m_2 - 1)*H_8, solving\n",
+      "m_2 = m_2_prime;\n",
+      "H_5_prime = ((0.2*m_2-1)*H_8 - H_7)/(0.2*m_2);\n",
+      "\n",
+      "\t\t\t#At point 5, P = 200 bar and enthalpy is\n",
+      "H_5_1 = -33.53;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#From compressed air tables at 200 bar and 140 K,\n",
+      "H_5_2 = 0.2;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 200 bar and 120 K,\n",
+      "H_5_3 = -38.0;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#Solving by interpolation we get\n",
+      "T_5 = ((H_5_1 - H_5_3)*(140 - 120))/(H_5_2 - H_5_3) + 120;\t\t\t#[K]\n",
+      "\n",
+      "print \"The temperature of air before throttling is %f K\"%(T_5);\n",
+      "\n",
+      "\t\t\t#(3)\n",
+      "\t\t\t#During mixing of streams 8 and 11 to produce stream 9, the enthalpy balance is\n",
+      "\t\t\t# (0.2*m_2 - 1)*H_8 + 0.8*m_2*H_11 = (m_2 - 1)*H_9,Solving for H_9\n",
+      "\n",
+      "H_9_prime = ((0.2*m_2-1)*H_8+0.8*m_2*H_11)/(m_2 - 1);\n",
+      "\n",
+      "\t\t\t#From given compressed air tables at 1 bar and 100 K,\n",
+      "H_9_1 = H_11;\n",
+      "\t\t\t#At 1 bar and 90 K \n",
+      "H_9_2 = 87.9;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#Solving by interpolation we get\n",
+      "T_9 = ((H_9_prime - H_9_2)*(100 - 90))/(H_9_1 - H_9_2) + 90;\t\t\t#[K]\n",
+      "\n",
+      "print \"The temperature of stream entering second heat exchanger is %f K\"%(T_9);\n",
+      "\n",
+      "\t\t\t#(4)\n",
+      "\t\t\t#Enthalpy balance around first heat exchanger is\n",
+      "\t\t\t#H_2*m_2 + (m_2 - 1)*H_10 = H_3*m-2 + (m-2 - 1)*H_1, solving for H_10\n",
+      "\n",
+      "H_10_prime = ((m_2 - 1)*H_1 + H_3*m_2 - H_2*m_2)/(m_2 - 1);\n",
+      "\n",
+      "\t\t\t#From given compressed air tables at 1 bar and 140 K,\n",
+      "H_10_1 = 139.1;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 1 bar and 120 K \n",
+      "H_10_2 = 118.8;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#Solving by interpolation we get\n",
+      "T_10 = ((H_10_prime - H_10_2)*(140 - 120))/(H_10_1 - H_10_2) + 120;\t\t\t#[K]\n",
+      "\n",
+      "print \"The temperature of stream exiting second heat exchanger is %f K\"%(T_10);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of kimath.lograms of air compressed per kg of liquid air produced is 8.997922 kg\n",
+        "The temperature of air before throttling is 122.340314 K\n",
+        "The temperature of stream entering second heat exchanger is 98.029358 K\n",
+        "The temperature of stream exiting second heat exchanger is 131.292906 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.10  Page Number : 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "P_high = 40;\t\t\t#[bar]\n",
+      "P_low = 5;\t\t\t#[bar]\n",
+      "m_1 = 0.5;\t\t\t#[kg/s] - Rate of mass moving through the expander\n",
+      "m_2 = 0.1;\t\t\t#[kg/s] - Rate of mass of vapour mixing with air\n",
+      "e = 0.7;\t\t\t#Efficiency\n",
+      "\n",
+      "\t\t\t#At state 3,(40 bar and 200 K),enthalpy and entropy is given by\n",
+      "H_3 = 179.7;\t\t\t#[kJ/kg]\n",
+      "S_3 = 5.330;\t\t\t#[kJ/kg-K]\n",
+      "\n",
+      "\t\t\t#If isentropic conditions exits in the turbine then state 11 is at 5 bar\n",
+      "S_11 = 5.330;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t#From given compressed air tables at 5 bar and 120 K,\n",
+      "H_11_1 = 113.6;\t\t\t#[kJ/kg]\n",
+      "S_11_1 = 5.455;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t#At 5 bar and 100 K \n",
+      "H_11_2 = 90.6;\t\t\t#[kJ/kg]\n",
+      "S_11_2 = 5.246;\t\t\t#[kJ/kg-K]\n",
+      "\t\t\t#The enthalpy has to be determined when S = S_3\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\t\t\t#Solving by interpolation we get\n",
+      "H_11_s = ((H_11_1 - H_11_2)*(S_3 - S_11_2))/(S_11_1 - S_11_2) + H_11_2;\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#The adiabatic efficiency of tyrbine is given by\n",
+      "\t\t\t#(H_3 - H_11_a)/(H_3 - H_11_s) = e\n",
+      "H_11_a = H_3 - e*(H_3 - H_11_s);\t\t\t#[kJ/kg] - Actual enthalpy\n",
+      "\n",
+      " \t\t\t#At 5 bar,the saturated enthalpy is given to be\n",
+      "H_8 = 88.7;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#From enthalpy balance during mixing we get,\n",
+      "\t\t\t#0.1*H_8 + 0.5*H_11_a = 0.6*H_9\n",
+      "H_9 = (m_2*H_8 + m_1*H_11_a)/(m_1 + m_2);\t\t\t#[kJ/kg]\n",
+      "\n",
+      "\t\t\t#From given compressed air tables at 5 bar and 140 K,\n",
+      "H_9_1 = 135.3;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#At 5 bar and 120 K \n",
+      "H_9_2 = 113.6;\t\t\t#[kJ/kg]\n",
+      "\t\t\t#By interpolation we get\n",
+      "T_9 = ((H_9 - H_11_1)*(140 - 120))/(H_9_1 - H_11_1) + 120;\t\t\t#[K]\n",
+      "\n",
+      "print \" The temperature of air entering the second heat exchanger is %f K\"%(T_9);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The temperature of air entering the second heat exchanger is 124.009841 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/screenshots/pic111.png b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/screenshots/pic111.png
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diff --git a/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/README.txt b/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/README.txt
old mode 100644
new mode 100755
diff --git a/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/screenshots/ch1.png b/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/screenshots/ch1.png
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diff --git a/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/screenshots/ch8.png b/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/screenshots/ch8.png
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diff --git a/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/screenshots/pressureVSvol3.png b/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/screenshots/pressureVSvol3.png
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diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/README.txt b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/README.txt
new file mode 100755
index 00000000..98c63c7c
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Jai Mathur
+Course: mca
+College/Institute/Organization: IIT
+Department/Designation: System Admin
+Book Title: Chemical Engineering Thermodynamics
+Author: Y. V. C. Rao
+Publisher: Universities Press
+Year of publication: 1997
+Isbn: 81-7371-048-1
+Edition: 1
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch1.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch1.ipynb
new file mode 100755
index 00000000..c7a29075
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch1.ipynb
@@ -0,0 +1,145 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:4f7d90cd7b10a2be32d6f9985c84b88b7895b54f53fda509a7ab6972ab90144a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.1  Page No : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "weight = 981. \t\t\t #weight of payload in N\n",
+      "gmoon = 1.62;\t\t\t #acceleration due to gravity on the moon in m/s**2\n",
+      "g = 9.81;\t\t\t     #acceleration due to gravity on earth\n",
+      "\n",
+      "# Calculations\n",
+      "mass = weight/g;        \t\t\t # Calculations of mass of the payload in kg (calculated as F = m*g)\n",
+      "weightmoon = mass*gmoon;\t\t\t # Calculations of weight of payload on the moon in N\n",
+      "\n",
+      "# Results\n",
+      "print ' The weight of payload on the moon =  %d N'%(weightmoon);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The weight of payload on the moon =  162 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.2  Page No : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "l = 15.;\t\t\t #length of the child's head in cm\n",
+      "b = 12.;\t\t\t #breadth of the child's head in cm\n",
+      "p = 101325.;\t\t\t #atmospheric pressure in Pa\n",
+      "\n",
+      "# Calculations\n",
+      "area = (l*b)/(10**4);\t\t\t # Calculations of area of the child's head in m**2\n",
+      "force = p*area;\t\t\t # Calculations of force exerted on the child's head due to atmospheric air in N\n",
+      "\n",
+      "# Results\n",
+      "print ' The force exerted on the childs head due to atmospheric air =  %.2f N'%(force);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The force exerted on the childs head due to atmospheric air =  1823.85 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.3  Page No : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "rho_water = 1000.;\t\t\t #density of water flowing through the pipeline in kg/m**3\n",
+      "rho_manomtr = 1595.;\t\t #density of manometric fluid (carbon tetrachloride) in kg/m**3\n",
+      "l = 40.;\t        \t\t #length between the selected sections in cm\n",
+      "theta = 45.\t\t\t         #inclination of the manometer in degrees\n",
+      "g = 9.81;\t        \t\t #acceleration due to gravity in m/s**2\n",
+      "\n",
+      "# Calculations\n",
+      "delp = (l/100.)*math.sin((theta*math.pi)/180.)*g*(rho_manomtr-rho_water);   # Calculations of pressure drop between the required sections in Pa\n",
+      "\n",
+      "# Results\n",
+      "print ' The pressure drop between the required sections =  %.2f Pa'%delp\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The pressure drop between the required sections =  1650.94 Pa\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch10.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch10.ipynb
new file mode 100755
index 00000000..5d1a66e8
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch10.ipynb
@@ -0,0 +1,202 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Stability and phase transition in thermodynamic systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.2  Page No : 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 2;\t\t\t #number of phases (no unit)\n",
+      "C = 2;\t\t\t #number of components (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "F = C+2-P\n",
+      "\n",
+      "# Results\n",
+      "print \" The number of degrees of freedom  =  %d \"%(F);\n",
+      "print \"Two intensive properties are required to be specified to\\\n",
+      " describe the thermodynamic state of the system,and the fundamental relation in\\\n",
+      "  the Gibbs free energy representation for this system is of the type, G = GT,P,N1,N2\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The number of degrees of freedom  =  2 \n",
+        "Two intensive properties are required to be specified to describe the thermodynamic state of the system,and the fundamental relation in  the Gibbs free energy representation for this system is of the type, G = GT,P,N1,N2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.3  Page No : 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import cmath\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
+      "w = 0.398;\t\t\t #acentric factor (no unit) \n",
+      "\n",
+      "# Calculations\n",
+      "Pguess = 0.215\n",
+      "\n",
+      "\n",
+      "Tr = T/Tc\n",
+      "Pr = (Pguess*10**6)/(Pc*10**5)\n",
+      "S = 0.37464+(1.54226*w)-(0.26992*w**2)\n",
+      "alpha1 = (1+(S*(1-math.sqrt(Tr))))**2;\n",
+      "a = (0.45724*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
+      "b = (0.07780*R*Tc)/(Pc*10**5);\t\t\t \n",
+      "A = (a*Pguess*10**6)/(R*T)**2;\t\t\t \n",
+      "B = (b*Pguess*10**6)/(R*T);\t\t\t \n",
+      "alpha = -1+B\n",
+      "beeta = A-(2*B)-(3*B**2)\n",
+      "gaamma = -(A*B)+(B**2)+(B**3)\n",
+      "p = beeta-(alpha**2)/3;\t\t\n",
+      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
+      "D = (((q)**2)/4)+(((p)**3)/27);\t\t\t \n",
+      "\n",
+      "\n",
+      "if D>0:\n",
+      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);\n",
+      "    Z_l = Z;\n",
+      "    Z_v = Z;\n",
+      "elif D == 0:\n",
+      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3);\t\t\t \n",
+      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z = [Z1 ,Z2, Z3];\n",
+      "    Z_l = min(Z);\n",
+      "    Z_v = max(Z);\n",
+      "else:\n",
+      "    r = math.sqrt((-(p**3)/27));\t\t\t \n",
+      "    theta = math.acos((-(q)/2)*(1./r));\t\t\n",
+      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Z = [Z1, Z2, Z3];\n",
+      "    Z_l = min(Z)\n",
+      "    Z_v = max(Z)\n",
+      "\n",
+      "\n",
+      "phi_l = math.exp(Z_l-1-math.log(Z_l-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_l+(B*(1+math.sqrt(2))))/(Z_l+(B*(1-math.sqrt(2)))))));\n",
+      "phi_v = math.exp(Z_v-1-math.log(Z_v-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_v+(B*(1+math.sqrt(2))))/(Z_v+(B*(1-math.sqrt(2)))))));\n",
+      "fl = Pguess*phi_l;\t\t\t # Calculations of the fugacity of the liquid in MPa\n",
+      "fv = Pguess*phi_v;\t\t\t # Calculations of the fugacity of the vapour in MPa\n",
+      "tolerance = 1e-3;\t\t\t #defining the tolerance to compare fl and fv\n",
+      "\n",
+      "if abs(fl-fv)<tolerance:\n",
+      "    P = Pguess;\t\t\t\n",
+      "else:\n",
+      "    Prevised = Pguess*(fl/fv)\n",
+      "\n",
+      "while abs(fl-fv)>tolerance:\n",
+      "    Tr = T/Tc\n",
+      "    Pr = (Prevised*10**6)/(Pc*10**5);\n",
+      "    S = 0.37464+(1.54226*w)-(0.26992*w**2)\n",
+      "    alpha1 = (1+(S*(1-math.sqrt(Tr))))**2;\n",
+      "    a = (0.45724*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
+      "    b = (0.07780*R*Tc)/(Pc*10**5);\t\t\t \n",
+      "    A = (a*Prevised*10**6)/(R*T)**2;\t\t\n",
+      "    B = (b*Prevised*10**6)/(R*T);\t\t\t\n",
+      "    alpha = -1+B;\t\t\t \n",
+      "    beeta = A-(2*B)-(3*B**2);\n",
+      "    gaamma = -(A*B)+(B**2)+(B**3)\n",
+      "    p = beeta-(alpha**2)/3;\t\t\n",
+      "    q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
+      "    D = (((q)**2)/4)+(((p)**3)/27);\t\n",
+      "        \n",
+      "    if D > 0:\n",
+      "        Z=((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);     #One real root given by  Eq.(3.32)\n",
+      "        Z_l=Z;\n",
+      "        Z_v=Z;\n",
+      "    elif D==0:\n",
+      "        Z1=((-2*(q/2))**(1./3))-(alpha/3);             #Three real roots and two equal given by Eq.(3.33)\n",
+      "        Z2=((q/2)**(1./3))-(alpha/3);\n",
+      "        Z3=((q/2)**(1./3))-(alpha/3);\n",
+      "        Z=[Z1, Z2, Z3];\n",
+      "        Z_l=min(Z);\n",
+      "        Z_v=max(Z);\n",
+      "    else:\n",
+      "        r = math.sqrt((-(p**3)/27));\t\t\t \n",
+      "        theta = math.acos((-(q)/2)*(1./r));\t\t\n",
+      "        Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "        Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
+      "        Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
+      "        Z = [Z1, Z2, Z3];\n",
+      "        Z_l = Z[0];\n",
+      "        Z_v = Z[1];\n",
+      "\n",
+      "    phi_l = math.exp(Z_l-1-math.log(Z_l-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_l+(B*(1+math.sqrt(2))))/(Z_l+(B*(1-math.sqrt(2)))))));\n",
+      "    phi_v = math.exp(Z_v-1-math.log(Z_v-B)-((a/(2*math.sqrt(2)*b*R*T))*math.log((Z_v+(B*(1+math.sqrt(2))))/(Z_v+(B*(1-math.sqrt(2)))))));\n",
+      "    fl = Prevised*phi_l;\t\n",
+      "    fv = Prevised*phi_v;\t\n",
+      "    Prevised = Prevised*fl/fv\n",
+      "\n",
+      "P = Prevised;\t\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The vapour pressure of n-octane at 427.85K  =  %.5f\"%P,\" MPa\"\n",
+      "\n",
+      "# Note: answer is slightly differnt because of rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The vapour pressure of n-octane at 427.85K  =  0.21196  MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch11.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch11.ipynb
new file mode 100755
index 00000000..9c8a35bf
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch11.ipynb
@@ -0,0 +1,585 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Properties of solutions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.1  Page No : 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%matplotlib inline\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "antoine_const_benzene = [6.87987,1196.760,219.161]\n",
+      "antoine_const_toluene = [6.95087,1342.310,219.187]\n",
+      "t = 95.\n",
+      "P = 101.325\n",
+      "\n",
+      "# Calculations\n",
+      "P1_s = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(t+antoine_const_benzene[2])))\t\t\t \n",
+      "P2_s = 10**(antoine_const_toluene[0]-(antoine_const_toluene[1]/(t+antoine_const_toluene[2])))\t\t\t\n",
+      "x1 = linspace(0,1,10)\n",
+      "i = 0;\t\t\t          #iteration parameter\n",
+      "n = len(x1);\t\t\t #iteration parameter\n",
+      "P_tot = zeros(10)\n",
+      "y1 = zeros(10)\n",
+      "\n",
+      "while i<n :\n",
+      "    P_tot[i] = P2_s+((P1_s-P2_s)*x1[i])\n",
+      "    y1[i] = (x1[i]*P1_s)/(P_tot[i])\t\n",
+      "    i = i+1;\n",
+      "\n",
+      "#T-x-y diagram:\n",
+      "P = 760.\n",
+      "t1_s = ((antoine_const_benzene[1])/(antoine_const_benzene[0]-math.log10(P)))-antoine_const_benzene[2]\n",
+      "t2_s = ((antoine_const_toluene[1])/(antoine_const_toluene[0]-math.log10(P)))-antoine_const_toluene[2];\n",
+      "x1_initial = 0.0;\n",
+      "y1_initial = 0.0;\n",
+      "x1_final = 1.0;\n",
+      "y1_final = 1.0;\n",
+      "\n",
+      "T = linspace(85,105,5)\n",
+      "k = 0;\n",
+      "l = len(T)\n",
+      "P1s = zeros(5)\n",
+      "P2s = zeros(5)\n",
+      "X1 = zeros(5)\n",
+      "Y1 = zeros(5)\n",
+      "while k<l:\n",
+      "    P1s[k] = 10**((antoine_const_benzene[0])-((antoine_const_benzene[1])/(T[k]+antoine_const_benzene[2])))\n",
+      "    P2s[k] = 10**((antoine_const_toluene[0])-((antoine_const_toluene[1])/(T[k]+antoine_const_toluene[2])))\n",
+      "    X1[k] = (P-P2s[k])/(P1s[k]-P2s[k]);\t\n",
+      "    Y1[k] = (X1[k]*P1s[k])/P;\t\t\t # Calculations of mole fraction of Benzene in vapour phase (no unit) \n",
+      "    k = k+1;\n",
+      "\n",
+      "temp = zeros(l+2)\n",
+      "x1_benzene = zeros(l+2)\n",
+      "y1_benzene = zeros(l+2)\n",
+      "j = 0;\n",
+      "\n",
+      "while j<l+2:\n",
+      "    if j == 0:\n",
+      "        temp[j] = t1_s;\n",
+      "        x1_benzene[j] = x1_final;\n",
+      "        y1_benzene[j] = y1_final;\n",
+      "    elif j == l+1:\n",
+      "        temp[j] = t2_s;\n",
+      "        x1_benzene[j] = x1_initial;\n",
+      "        y1_benzene[j] = y1_initial;\n",
+      "    else:\n",
+      "        temp[j] = T[j-1];\n",
+      "        x1_benzene[j] = X1[j-1];\n",
+      "        y1_benzene[j] = Y1[j-1];\n",
+      "    j = j+1;\n",
+      "\n",
+      "# Results\n",
+      "print 'P-x-y results ';\n",
+      "\n",
+      "for i in range( n):\n",
+      "      print 'x1 = %f \\t y1 = %f\\t P = %f Torr '%(x1[i],y1[i],P_tot[i]);\n",
+      "\n",
+      "print 'T-x-y results  t = %f degree celsius\\t P1_s = 760.0 Torr \\t P2_s = -) Torr \\t\\t x1 = 1.0 \\t y1 = 1.0 '%(t1_s);\n",
+      "for k in range(l):\n",
+      "    print 't = %f degree celsius\\t P1_s = %f Torr \\t P2_s = %f Torr \\t x1 = %f \\t y1 = %f '%(T[k],P1s[k],P2s[k],X1[k],Y1[k]);\n",
+      "\n",
+      "print 't = %f degree celsius\\t P1_s = -)Torr \\t\\t P2_s = 760.0 Torr \\t x1 = 0.0 \\t y1 = 0.0 '%(t2_s);\n",
+      "subplot(2,1,1)\n",
+      "plot(x1,P_tot,y1,P_tot)\n",
+      "suptitle('P-x-y diagram for benzene-toluene system at 95 degree celsius')\n",
+      "xlabel('x1,y1')\n",
+      "ylabel('P(Torr)')\n",
+      "subplot(2,1,2)\n",
+      "plot(x1_benzene,temp,y1_benzene,temp)\n",
+      "xlabel(\"x1,y1\")\n",
+      "ylabel(\"t (degree celsius)\")       \n",
+      "suptitle('T-x-y diagram for benzene-toluene sytem at 760 Torr')\n",
+      "\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "P-x-y results \n",
+        "x1 = 0.000000 \t y1 = 0.000000\t P = 477.025655 Torr \n",
+        "x1 = 0.111111 \t y1 = 0.235600\t P = 554.713037 Torr \n",
+        "x1 = 0.222222 \t y1 = 0.413315\t P = 632.400419 Torr \n",
+        "x1 = 0.333333 \t y1 = 0.552144\t P = 710.087801 Torr \n",
+        "x1 = 0.444444 \t y1 = 0.663592\t P = 787.775183 Torr \n",
+        "x1 = 0.555556 \t y1 = 0.755031\t P = 865.462564 Torr \n",
+        "x1 = 0.666667 \t y1 = 0.831407\t P = 943.149946 Torr \n",
+        "x1 = 0.777778 \t y1 = 0.896158\t P = 1020.837328 Torr \n",
+        "x1 = 0.888889 \t y1 = 0.951751\t P = 1098.524710 Torr \n",
+        "x1 = 1.000000 \t y1 = 1.000000\t P = 1176.212092 Torr \n",
+        "T-x-y results  t = 80.099595 degree celsius\t P1_s = 760.0 Torr \t P2_s = -) Torr \t\t x1 = 1.0 \t y1 = 1.0 \n",
+        "t = 85.000000 degree celsius\t P1_s = 881.542536 Torr \t P2_s = 345.216082 Torr \t x1 = 0.773380 \t y1 = 0.897062 \n",
+        "t = 90.000000 degree celsius\t P1_s = 1020.650885 Torr \t P2_s = 406.866584 Torr \t x1 = 0.575338 \t y1 = 0.772657 \n",
+        "t = 95.000000 degree celsius\t P1_s = 1176.212092 Torr \t P2_s = 477.025655 Torr \t x1 = 0.404719 \t y1 = 0.626363 \n",
+        "t = 100.000000 degree celsius\t P1_s = 1349.471568 Torr \t P2_s = 556.502216 Torr \t x1 = 0.256628 \t y1 = 0.455673 \n",
+        "t = 105.000000 degree celsius\t P1_s = 1541.703421 Torr \t P2_s = 646.141534 Torr \t x1 = 0.127136 \t y1 = 0.257903 \n",
+        "t = 110.614326 degree celsius\t P1_s = -)Torr \t\t P2_s = 760.0 Torr \t x1 = 0.0 \t y1 = 0.0 \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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GCBnb3QUiwoXbF7AvbR/2X9qPoooiPD3gabwy8BUE2Ad06i+0mkC7C4dMJsPp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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3802f10>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.2  Page No : 384"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 95. \t\t    \t #temperature of the equimolar vapour mixture of benzene and toluene in degree celsius\n",
+      "y1 = 0.5\t    \t\t #mole fraction of benzene in vapour phase (no unit)\n",
+      "y2 = 0.5    \t\t\t #mole fraction of toluene in vapour phase (no unit)\n",
+      "P1_s = 1176.21\t\t\t #saturation pressure of benzene at T, taken from Example 11.1 in Torr\n",
+      "P2_s = 477.03\t\t\t #saturation pressure of toluene at T, taken from Example 11.1 in Torr\n",
+      "\n",
+      "# Calculations\n",
+      "P = 1./((y1/P1_s)+(y2/P2_s))\n",
+      "x1 = (y1*P)/P1_s;\t\t\t \n",
+      "x2 = 1-x1\n",
+      "\n",
+      "# Results\n",
+      "print 'The composition of the liquid which is in equilibrium with the equimolar vapour mixture of\\\n",
+      " benzene and toluene at 95 degree celsius is  mole fraction of benzene\\\n",
+      "  in liquid phase x1 = %f  mole fraction of toluene in liquid phase x2 = %f '%(x1,x2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The composition of the liquid which is in equilibrium with the equimolar vapour mixture of benzene and toluene at 95 degree celsius is  mole fraction of benzene  in liquid phase x1 = 0.288542  mole fraction of toluene in liquid phase x2 = 0.711458 \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.5 page no : 386"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "n_pentaneA = 6.87632          # Antoine constants\n",
+      "n_hexaneA = 6.91058\n",
+      "n_heptanA = 6.89386\n",
+      "n_pentaneB = 1075.780 \n",
+      "n_hexaneB = 1189.640\n",
+      "n_heptanB = 1264.370\n",
+      "n_pentaneC = 233.205\n",
+      "n_hexaneC = 226.205\n",
+      "n_heptanC = 216.640\n",
+      "pressure = 90                 # pressure\n",
+      "\n",
+      "# Calculations\n",
+      "logP1 = n_pentaneA - (n_pentaneB/(90 + n_pentaneC))\n",
+      "P1 = round(10**logP1 *133.322 /1000,1)\n",
+      "logP2 = n_hexaneA - (n_hexaneB/(90 + n_hexaneC))\n",
+      "P2 = round(10**logP2 *133.322 /1000,2)\n",
+      "logP3 = n_heptanA - (n_heptanB/(90 + n_heptanC))\n",
+      "P3 = round(10**logP3 *133.322 /1000,2)\n",
+      "P = 200\n",
+      "K1 = round(P1/P,3)\n",
+      "K2 = round(P2/P,2)\n",
+      "K3 = P3/P\n",
+      "\n",
+      "# assume L/F = 0.4\n",
+      "L_F = 0.4\n",
+      "x1 = 0.3/(L_F + (1-L_F) * K1)\n",
+      "x2 = 0.3/(L_F + (1-L_F) * K2)\n",
+      "x3 = 0.4/(L_F + (1-L_F) * K3)\n",
+      "sum_x = x1+x2+x3\n",
+      "\n",
+      "# assume L/F = 0.752\n",
+      "L_F = 0.752\n",
+      "x1 = 0.3/(L_F + (1-L_F) * K1)\n",
+      "x2 = 0.3/(L_F + (1-L_F) * K2)\n",
+      "x3 = 0.4/(L_F + (1-L_F) * K3)\n",
+      "sum_x = round(x1+x2+x3,1)           # which is equal to 1 \n",
+      "\n",
+      "# Hence, L/F = 0.752 is correct\n",
+      "y1 = K1*x1\n",
+      "y2 = K2*x2\n",
+      "y3 = K3*x3\n",
+      "\n",
+      "# Results\n",
+      "print \"Hence, fraction vaporized, V/F = 1-(L-F) = %.3f\"%(1-L_F)\n",
+      "print \"Compositions of liquid and vapor leaving the flash unit are :\"\n",
+      "print \"x1 = %.4f   y1 = %.4f\"%(x1,y1)\n",
+      "print \"x2 = %.4f   y2 = %.4f\"%(x2,y2)\n",
+      "print \"x3 = %.4f   y3 = %.4f\"%(x3,y3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hence, fraction vaporized, V/F = 1-(L-F) = 0.248\n",
+        "Compositions of liquid and vapor leaving the flash unit are :\n",
+        "x1 = 0.2246   y1 = 0.5285\n",
+        "x2 = 0.3045   y2 = 0.2863\n",
+        "x3 = 0.4709   y3 = 0.1851\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.7  Page No : 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 45.\t    \t\t #temperature of the mixture in degree celsius\n",
+      "A = 2.230\t\t\t #van laar constant for the system at T (no unit)\n",
+      "B = 1.959\t\t\t #van laar constant for the system at T (no unit)\n",
+      "n1 = 30.    \t\t #mole percentage of nitromethane in the mixture ( in percentage)\n",
+      "\n",
+      "# Calculations\n",
+      "n2 = 100-n1\n",
+      "x1 = n1/100\n",
+      "x2 = 1-x1\n",
+      "gaamma1 = math.exp(A/(1+((A/B)*(x1/x2)))**2)\n",
+      "gaamma2 = math.exp(B/(1+((B/A)*(x2/x1)))**2)\n",
+      "\n",
+      "# Results\n",
+      "print 'The activity coefficients for the system using van laar equation is : gamma1 \\\n",
+      " = %f, gamma2 = %f \\t '%( gaamma1,gaamma2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The activity coefficients for the system using van laar equation is : gamma1  = 2.738343, gamma2 = 1.234443 \t \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.8  Page No : 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "n_azeo = 44.8;\t\t\t #azeotropic composition given as mole percentage\n",
+      "Tb = 68.24;\t\t\t #boiling point of mixture in degree celsius\n",
+      "P = 760.;\t\t\t #pressure in Torr\n",
+      "P1_s = 506.;\t\t\t #saturation pressure of ethanol at Tb in Torr\n",
+      "P2_s = 517.;\t\t\t #saturation pressure of benzene at Tb in Torr\n",
+      "n1 = 10.;\t\t\t #mole percentage of ethanol in the mixture (in percentage)\n",
+      "\n",
+      "# Calculations\n",
+      "x1 = n_azeo/100\n",
+      "x2 = 1-x1;\t\t\n",
+      "gaamma1 = P/P1_s\n",
+      "gaamma2 = P/P2_s\n",
+      "A = math.log(gaamma1)*(1+((x2*math.log(gaamma2))/(x1*math.log(gaamma1))))**2\n",
+      "B = math.log(gaamma2)*(1+((x1*math.log(gaamma1))/(x2*math.log(gaamma2))))**2\n",
+      "x1 = n1/100\n",
+      "x2 = 1-x1;\t\n",
+      "gaamma1 = math.exp(A/(1+((A/B)*(x1/x2)))**2)\n",
+      "gaamma2 = math.exp(B/(1+((B/A)*(x2/x1)))**2)\n",
+      "\n",
+      "# Results\n",
+      "print 'The van Laar constants for the system are : A = %f \\t B = %f '%(A,B)\n",
+      "print 'The activity coefficients for the system using van laar equation are\\\n",
+      " : gamma1 = %f \\t gamma2 = %f \\t '%( gaamma1,gaamma2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The van Laar constants for the system are : A = 1.910203 \t B = 1.328457 \n",
+        "The activity coefficients for the system using van laar equation are : gamma1 = 4.137783 \t gamma2 = 1.025531 \t \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.9  Page No : 399"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "\n",
+      "T = 45.     \t\t\t #temperature of the system in degree celsius\n",
+      "A_12 = 0.1156;\t\t\t #Wilson's parameter for the system at T (no unit)\n",
+      "A_21 = 0.2879;\t\t\t #Wilson's parameter for the system at T (no unit)\n",
+      "x1 = 0.3;\t    \t\t #mole fraction of nitromethane in the liquid mixture (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = 1-x1\n",
+      "ln_gaamma1 = -math.log(x1+(A_12*x2))+(x2*((A_12/(x1+(A_12*x2)))-(A_21/((A_21*x1)+x2))));\t\t\t\n",
+      "gaamma1 = math.exp(ln_gaamma1)\n",
+      "ln_gaamma2 = -math.log(x2+(A_21*x1))-(x1*((A_12/(x1+(A_12*x2)))-(A_21/((A_21*x1)+x2))));\t\t\t # Calculations of ln(activity coefficient) using Eq.(11.90) (no unit)\n",
+      "gaamma2 = math.exp(ln_gaamma2)\n",
+      "\n",
+      "# Results\n",
+      "print 'The activity coefficients for the system using Wilsons parameters are : gamma1 = %f \\\n",
+      " \\t gamma2 = %f \\t '%( gaamma1,gaamma2);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The activity coefficients for the system using Wilsons parameters are : gamma1 = 2.512605  \t gamma2 = 1.295788 \t \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.10  Page No : 401"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 345.\t\t\t #temperature of the mixture in K\n",
+      "x1 = 0.8\t\t\t #mole fraction of ethanol in the liquid phase (no unit)\n",
+      "\n",
+      "nu_ki = [1,1,1,6]\t\t\t #number of groups of type: CH3, CH2, OH and ACH respectively (no unit)\n",
+      "R_k = [0.9011,0.6744,1.0000,0.5313];\t\t\t #Group volume parameter for CH3, CH2, OH and ACH respectively (no unit)\n",
+      "Q_k = [0.848,0.540,1.200,0.400];\t\t\t #Area parameter for CH3, CH2, OH and ACH respectively (no unit)\n",
+      "R = 8.314;\t\t\t        #universal gas constant in J/molK\n",
+      "u12_u22 = -241.2287;\t\t#UNIQUAC parameter for the system in J/molK\n",
+      "u21_u11 = 2799.5827;\t\t#UNIQUAC parameter for the system in J/molK\n",
+      "z = 10.     \t\t\t #co-ordination number usually taken as 10 (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = 1-x1\n",
+      "r1 = (nu_ki[0]*R_k[0])+(nu_ki[1]*R_k[1])+(nu_ki[2]*R_k[2]);\t\t\t \n",
+      "r2 = (nu_ki[3]*R_k[3])\n",
+      "phi1 = (x1*r1)/((x1*r1)+(x2*r2));\t\t\t \n",
+      "phi2 = (x2*r2)/((x2*r2)+(x1*r1));\t\t\t \n",
+      "q1 = (nu_ki[0]*Q_k[0])+(nu_ki[1]*Q_k[1])+(nu_ki[2]*Q_k[2])\t\t\t \n",
+      "q2 = (nu_ki[3]*Q_k[3])\t\t\t\n",
+      "theta1 = (x1*q1)/((x1*q1)+(x2*q2))\n",
+      "theta2 = (x2*q2)/((x1*q1)+(x2*q2))\n",
+      "l1 = ((z/2)*(r1-q1))-(r1-1);\t\t\n",
+      "l2 = ((z/2)*(r2-q2))-(r2-1);\t\t\n",
+      "tau_12 = math.exp(-(u12_u22)/(R*T));\t\t\n",
+      "tau_21 = math.exp(-(u21_u11)/(R*T));\t\t\n",
+      "tau_11 = 1.0;\t\t\t \n",
+      "tau_22 = 1.0;\t\t\t \n",
+      "\n",
+      "ln_gaamma1_c = math.log(phi1/x1)+((z/2)*q1*math.log(theta1/phi1))+l1-((phi1/x1)*((x1*l1)+(x2*l2)));\n",
+      "ln_gaamma2_c = math.log(phi2/x2)+((z/2)*q2*math.log(theta2/phi2))+l2-((phi2/x2)*((x1*l1)+(x2*l2)));\n",
+      "ln_gaamma1_r = q1*(1-math.log((theta1*tau_11)+(theta2*tau_21))-(((theta1*tau_11)/((theta1*tau_11)+(theta2*tau_21)))+((theta2*tau_12)/((theta1*tau_12)+(theta2*tau_22)))));\n",
+      "ln_gaamma2_r = q2*(1-math.log((theta1*tau_12)+(theta2*tau_22))-(((theta1*tau_21)/((theta1*tau_11)+(theta2*tau_21)))+((theta2*tau_22)/((theta1*tau_12)+(theta2*tau_22)))));\n",
+      "ln_gaamma1 = ln_gaamma1_c+ln_gaamma1_r\n",
+      "ln_gaamma2 = ln_gaamma2_c+ln_gaamma2_r\n",
+      "gaamma1 = math.exp(ln_gaamma1);\t\t\t \n",
+      "gaamma2 = math.exp(ln_gaamma2);\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print 'The activity coefficients for the system using the UNIQUAC equation are : gamma1 \\\n",
+      " = %f \\t gamma2 = %f \\t '%( gaamma1,gaamma2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The activity coefficients for the system using the UNIQUAC equation are : gamma1  = 1.060567 \t gamma2 = 3.679066 \t \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.11  Page No : 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 307.\t\t\t #temperature of the mixture in K\n",
+      "x1 = 0.3\t\t\t #mole fraction of acetone in the liquid phase (no unit)\n",
+      "\n",
+      "nu_ki = [1.,1.,2.,3.]\n",
+      "R_k = [0.9011,1.6724,0.6744]\n",
+      "Q_k = [0.848,1.488,0.540];\t\t\t \n",
+      "a_19 = 476.40;\t\t\t #group interaction parameter for the system in K\n",
+      "a_91 = 26.760;\t\t\t #group interaction parameter for the system in K\n",
+      "z = 10.     \t\t\t #co-ordination number usually taken as 10 (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = 1-x1;\t\t\t # Calculations of mole fraction of benzene in liquid phase (no unit)\n",
+      "r1 = (nu_ki[0]*R_k[0])+(nu_ki[1]*R_k[1])\n",
+      "r2 = (nu_ki[2]*R_k[0])+(nu_ki[3]*R_k[2])\n",
+      "phi1 = (x1*r1)/((x1*r1)+(x2*r2))\n",
+      "phi2 = (x2*r2)/((x2*r2)+(x1*r1))\n",
+      "q1 = (nu_ki[0]*Q_k[0])+(nu_ki[1]*Q_k[1])\t\t\t\n",
+      "q2 = (nu_ki[2]*Q_k[0])+(nu_ki[3]*Q_k[2])\n",
+      "theta1 = (x1*q1)/((x1*q1)+(x2*q2));\n",
+      "theta2 = (x2*q2)/((x1*q1)+(x2*q2));\n",
+      "l1 = ((z/2)*(r1-q1))-(r1-1);\t\t\n",
+      "l2 = ((z/2)*(r2-q2))-(r2-1);\t\t\n",
+      "ln_gaamma1_c = math.log(phi1/x1)+((z/2)*q1*math.log(theta1/phi1))+l1-((phi1/x1)*((x1*l1)+(x2*l2)));\n",
+      "ln_gaamma2_c = math.log(phi2/x2)+((z/2)*q2*math.log(theta2/phi2))+l2-((phi2/x2)*((x1*l1)+(x2*l2)));\n",
+      "a_11 = 0.\n",
+      "a_99 = 0.\n",
+      "psi_19 = math.exp(-(a_19)/(T));\t\t\t\n",
+      "psi_91 = math.exp(-(a_91)/(T));\t\t\t\n",
+      "psi_11 = 1.;\t\t\t \n",
+      "psi_99 = 1.;\t\t\t \n",
+      "x1_1 = nu_ki[0]/(nu_ki[0]+nu_ki[1]);\t\t\t\n",
+      "x1_18 = nu_ki[1]/(nu_ki[0]+nu_ki[1]);\t\t\n",
+      "theta1_1 = (Q_k[0]*x1_1)/((Q_k[0]*x1_1)+(Q_k[1]*x1_18))\n",
+      "theta1_18 = (Q_k[1]*x1_18)/((Q_k[1]*x1_18)+(Q_k[0]*x1_1))\n",
+      "ln_tau1_1 = Q_k[0]*(1-math.log((theta1_1*psi_11)+(theta1_18*psi_91))-(((theta1_1*psi_11)/((theta1_1*psi_11)+(theta1_18*psi_91)))+((theta1_18*psi_19)/((theta1_1*psi_19)+(theta1_18*psi_11)))));\n",
+      "ln_tau1_18 = Q_k[1]*(1-math.log((theta1_1*psi_19)+(theta1_18*psi_99))-(((theta1_1*psi_91)/((theta1_1*psi_99)+(theta1_18*psi_91)))+((theta1_18*psi_99)/((theta1_1*psi_19)+(theta1_18*psi_99)))));\n",
+      "x2_1 = nu_ki[2]/(nu_ki[2]+nu_ki[3]);\t\t\t \n",
+      "x2_2 = nu_ki[3]/(nu_ki[2]+nu_ki[3]);\t\t\t \n",
+      "ln_tau2_1 = 0;\n",
+      "ln_tau2_2 = 0;\n",
+      "x_1 = ((x1*nu_ki[0])+(x2*nu_ki[2]))/((((x1*nu_ki[0])+(x1*nu_ki[1])))+((x2*nu_ki[2])+(x2*nu_ki[3])));\n",
+      "x_2 = ((x2*nu_ki[3]))/((((x1*nu_ki[0])+(x1*nu_ki[1])))+((x2*nu_ki[2])+(x2*nu_ki[3])));\n",
+      "x_18 = ((x1*nu_ki[1]))/((((x1*nu_ki[0])+(x1*nu_ki[1])))+((x2*nu_ki[2])+(x2*nu_ki[3])));\n",
+      "theta_1 = (Q_k[0]*x_1)/((Q_k[0]*x_1)+(Q_k[1]*x_18)+(Q_k[2]*x_2));\t\t\t\n",
+      "theta_2 = (Q_k[2]*x_2)/((Q_k[0]*x_1)+(Q_k[1]*x_18)+(Q_k[2]*x_2));\t\t\t\n",
+      "theta_18 = (Q_k[1]*x_18)/((Q_k[0]*x_1)+(Q_k[1]*x_18)+(Q_k[2]*x_2));\t\t\t\n",
+      "ln_tau_1 = Q_k[0]*(1-math.log((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91))-((((theta_1*psi_11)+(theta_2*psi_11))/((((theta_1*psi_11)+(theta_2*psi_11))+(theta_18*psi_91)))+((theta_18*psi_19)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));\n",
+      "ln_tau_2 = Q_k[2]*(1-math.log((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91))-((((theta_1*psi_11)+(theta_2*psi_11))/((((theta_1*psi_11)+(theta_2*psi_11))+(theta_18*psi_91)))+((theta_18*psi_19)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));\n",
+      "ln_tau_18 = Q_k[1]*(1-math.log((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_99))-(((((theta_1+theta_2)*psi_91)/((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91)))+((theta_18*psi_99)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));\n",
+      "ln_gaamma1_r = (nu_ki[0]*(ln_tau_1-ln_tau1_1))+(nu_ki[1]*(ln_tau_18-ln_tau1_18));\n",
+      "ln_gaamma2_r = (nu_ki[2]*(ln_tau_1-ln_tau2_1))+(nu_ki[3]*(ln_tau_2-ln_tau2_2));\n",
+      "ln_gaamma1 = ln_gaamma1_c+ln_gaamma1_r\n",
+      "ln_gaamma2 = ln_gaamma2_c+ln_gaamma2_r\n",
+      "gaamma1 = math.exp(ln_gaamma1);\t\t\t\n",
+      "gaamma2 = math.exp(ln_gaamma2);\t\t\t\n",
+      "\n",
+      "# Results\n",
+      "print 'The activity coefficients for the system using the UNIFAC method are\\\n",
+      " : gamma1 = %f \\t gamma2 = %f \\t '%( gaamma1,gaamma2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The activity coefficients for the system using the UNIFAC method are : gamma1 = 2.149891 \t gamma2 = 1.191192 \t \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch12.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch12.ipynb
new file mode 100755
index 00000000..fe626468
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch12.ipynb
@@ -0,0 +1,716 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Vapor liquid Equilibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.1  Page No : 423"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%matplotlib inline\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# Variables\n",
+      "T = 60.\t\t\t #temperature of the system in degree celsius\n",
+      "P = [237.60,265.20,317.50,333.00,368.70,387.20];\t\t\t #Pressure data in Torr (from Danneil et al.)\n",
+      "x1 = [0.0870,0.1800,0.4040,0.4790,0.7130,0.9070];\t\t\t #mole fraction of benzene in the liquid phase corresponding to the given pressure (no unit) (from Danneil et al.)\n",
+      "y1 = [0.1870,0.3400,0.5780,0.6420,0.7960,0.9220];\t\t\t #mole fraction of benzene in the vapour phase corresponding to the given pressure (no unit) (from Danneil et al.)\n",
+      "antoine_const_benzene = [6.87987,1196.760,219.161];\t\t\t #Antoine's constants for Benzene from Table A.7\n",
+      "antoine_const_heptane = [6.89386,1264.370,216.640];\t\t\t #Antoine's constants for heptane from Table A.7\n",
+      "\n",
+      "# Calculations\n",
+      "P1_s = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(T+antoine_const_benzene[2])));\n",
+      "P2_s = 10**(antoine_const_heptane[0]-(antoine_const_heptane[1]/(T+antoine_const_heptane[2])));\n",
+      "l = len(P);\t\t\t #iteration parameter\n",
+      "i = 0;\t\t\t #iteration parameter\n",
+      "gaamma1 = zeros(l)\n",
+      "gaamma2 = zeros(l)\n",
+      "ln_gaamma1_expt = zeros(l)\n",
+      "ln_gaamma2_expt = zeros(l)\n",
+      "gE_RTx1x2 = zeros(l)\n",
+      "while i<l:\n",
+      "    gaamma1[i] = (y1[i]*P[i])/(x1[i]*P1_s);\t\t\n",
+      "    gaamma2[i] = ((1-y1[i])*P[i])/((1-x1[i])*P2_s);\n",
+      "    ln_gaamma1_expt[i] = math.log(gaamma1[i]);\n",
+      "    ln_gaamma2_expt[i] = math.log(gaamma2[i]);\n",
+      "    gE_RTx1x2[i] = ((x1[i]*ln_gaamma1_expt[i])+((1-x1[i])*ln_gaamma2_expt[i]))/(x1[i]*(1-x1[i]));\t\t\t # Calculations of gE/RT using Eq.(11.36) (no unit)\n",
+      "    i = i+1;\n",
+      "\n",
+      "plot(x1,gE_RTx1x2,'o');\t\t\n",
+      "suptitle('Plot of gE/RTx1x2 vs x1')\n",
+      "xlabel('x1')\n",
+      "ylabel('gE/RTx1x2');\n",
+      "A21 = 0.555\n",
+      "A12 = 0.315\n",
+      "j = 0\n",
+      "ln_gaamma1 = zeros(l)\n",
+      "ln_gaamma2 = zeros(l)\n",
+      "P_calc = zeros(l)\n",
+      "y1_calc = zeros(l)\n",
+      "gaamma1 = zeros(l)\n",
+      "gaamma2 = zeros(l)\n",
+      "while j<l:\n",
+      "    ln_gaamma1[j] = ((1-x1[j])**2)*(A12+(2*(A21-A12)*x1[j]));\n",
+      "    ln_gaamma2[j] = (x1[j]**2)*(A21+(2*(A12-A21)*(1-x1[j])));\n",
+      "    gaamma1[j] = math.exp(ln_gaamma1[j])\n",
+      "    gaamma2[j] = math.exp(ln_gaamma2[j])\n",
+      "    P_calc[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
+      "    y1_calc[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
+      "    j = j+1;\n",
+      "\n",
+      "# Results\n",
+      "print 'Data for the plot of gE/RTx1x2 vs x1: ';\n",
+      "\n",
+      "for i in range(l):\n",
+      "    print 'P = %f Torr\\t x1 = %f\\t y1 = %f \\t lngamma1) = %f\\t\\t lngamma2) =\\\n",
+      "     %f\\t\\t gE/RTx1x2 = %f'%(P[i],x1[i],y1[i],ln_gaamma1_expt[i],ln_gaamma2_expt[i],gE_RTx1x2[i]);\n",
+      "\n",
+      "print 'Results: ';\n",
+      "for i in range(l):\n",
+      "    print 'x1 = %f \\t gamma1 = %f \\t gamma2 = %f \\t P_Exptl. = %f \\\n",
+      "    Torr\\t P_Calc = %f Torr\\t y1_Exptl = %f \\t y1_calc = %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],P_calc[i],y1[i],y1_calc[i]);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "Data for the plot of gE/RTx1x2 vs x1: "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "P = 237.600000 Torr\t x1 = 0.087000\t y1 = 0.187000 \t lngamma1) = 0.265459\t\t lngamma2) =     0.004741\t\t gE/RTx1x2 = 0.345243\n",
+        "P = 265.200000 Torr\t x1 = 0.180000\t y1 = 0.340000 \t lngamma1) = 0.246143\t\t lngamma2) =     0.013576\t\t gE/RTx1x2 = 0.375599\n",
+        "P = 317.500000 Torr\t x1 = 0.404000\t y1 = 0.578000 \t lngamma1) = 0.148307\t\t lngamma2) =     0.065399\t\t gE/RTx1x2 = 0.410716\n",
+        "P = 333.000000 Torr\t x1 = 0.479000\t y1 = 0.642000 \t lngamma1) = 0.130700\t\t lngamma2) =     0.083082\t\t gE/RTx1x2 = 0.424313\n",
+        "P = 368.700000 Torr\t x1 = 0.713000\t y1 = 0.796000 \t lngamma1) = 0.049771\t\t lngamma2) =     0.218778\t\t gE/RTx1x2 = 0.480259\n",
+        "P = 387.200000 Torr\t x1 = 0.907000\t y1 = 0.922000 \t lngamma1) = 0.005014\t\t lngamma2) =     0.433207\t\t gE/RTx1x2 = 0.531539\n",
+        "Results: \n",
+        "x1 = 0.087000 \t gamma1 = 1.346332 \t gamma2 = 1.000884 \t P_Exptl. = 237.600000     Torr\t P_Calc = 238.297793 Torr\t y1_Exptl = 0.187000 \t y1_calc = 0.193066 \n",
+        "x1 = 0.180000 \t gamma1 = 1.309835 \t gamma2 = 1.005243 \t P_Exptl. = 265.200000     Torr\t P_Calc = 265.912990 Torr\t y1_Exptl = 0.340000 \t y1_calc = 0.348175 \n",
+        "x1 = 0.404000 \t gamma1 = 1.198147 \t gamma2 = 1.044870 \t P_Exptl. = 317.500000     Torr\t P_Calc = 320.705660 Torr\t y1_Exptl = 0.578000 \t y1_calc = 0.597076 \n",
+        "x1 = 0.479000 \t gamma1 = 1.159413 \t gamma2 = 1.072467 \t P_Exptl. = 333.000000     Torr\t P_Calc = 335.157916 Torr\t y1_Exptl = 0.642000 \t y1_calc = 0.653147 \n",
+        "x1 = 0.713000 \t gamma1 = 1.055628 \t gamma2 = 1.236286 \t P_Exptl. = 368.700000     Torr\t P_Calc = 369.484272 Torr\t y1_Exptl = 0.796000 \t y1_calc = 0.799482 \n",
+        "x1 = 0.907000 \t gamma1 = 1.006511 \t gamma2 = 1.521729 \t P_Exptl. = 387.200000     Torr\t P_Calc = 387.326500 Torr\t y1_Exptl = 0.922000 \t y1_calc = 0.923362 \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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LbZq/vz/0ej0MBgMiIiKULJOIiCR4KrViq9WK9PR0FBcXY+TIkYiMjERsbCwMBoPdcm1t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\nUhgYREQkhYFBRERSGBhERCTl/wBGhb+in7ykegAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2dcf110>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.3  Page No : 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import *\n",
+      "# Variables\n",
+      "T = 25.     \t\t\t #temperature of the system in degree celsius\n",
+      "A12 = 2.0522;\t\t\t #three suffix Margules parameters for the system (no unit)\n",
+      "A21 = 1.7201;\t\t\t #three suffix Margules parameters for the system (no unit)\n",
+      "P = [118.05,207.70,246.35,259.40,261.50,262.00,261.90,258.70,252.00];\t\t\t #Pressure data in Torr (from Tasic et al.)\n",
+      "x1 = [0.0115,0.1125,0.3090,0.5760,0.6920,0.7390,0.7575,0.8605,0.9250];\n",
+      "y1 = [0.1810,0.5670,0.6550,0.7050,0.7250,0.7390,0.7460,0.8030,0.8580];\n",
+      "antoine_const_acetone = [7.11714,1210.595,229.664];\t\t\t #Antoine's constants for acetone from Table A.7\n",
+      "antoine_const_chexane = [6.85146,1206.470,223.136];\t\t\t #Antoine's constants for cyclohexane from Table A.7\n",
+      "\n",
+      "# Calculations\n",
+      "P1_s = 10**(antoine_const_acetone[0]-(antoine_const_acetone[1]/(T+antoine_const_acetone[2])));\n",
+      "P2_s = 10**(antoine_const_chexane[0]-(antoine_const_chexane[1]/(T+antoine_const_chexane[2])));\n",
+      "\n",
+      "l = len(P)\n",
+      "ln_gaamma1 = zeros(l)\n",
+      "ln_gaamma2 = zeros(l)\n",
+      "gaamma1 = zeros(l)\n",
+      "gaamma2 = zeros(l)\n",
+      "P = zeros(l)\n",
+      "y1_calc = zeros(l)\n",
+      "j = 0;\t\n",
+      "\n",
+      "while j<l:\n",
+      "    ln_gaamma1[j] = ((1-x1[j])**2)*(A12+(2*(A21-A12)*x1[j]));\t\n",
+      "    ln_gaamma2[j] = (x1[j]**2)*(A21+(2*(A12-A21)*(1-x1[j])));\t\n",
+      "    gaamma1[j] = math.exp(ln_gaamma1[j]);\t\t\t \n",
+      "    gaamma2[j] = math.exp(ln_gaamma2[j]);\t\t\t \n",
+      "    P[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
+      "    y1_calc[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
+      "    j = j+1;\n",
+      "\n",
+      "# Results\n",
+      "print 'P-x-y data: ';\n",
+      "print 'x1 \\t  gamma1\\t  gamma2 \\t P Torr \\t y1  ';\n",
+      "for i in range(l):\n",
+      "    print '%0.4f \\t %f \\t %f \\t %f  \\t %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],y1_calc[i])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "P-x-y data: \n",
+        "x1 \t  gamma1\t  gamma2 \t P Torr \t y1  \n",
+        "0.0115 \t 7.372871 \t 1.000314 \t 116.059350  \t 0.168694 \n",
+        "0.1125 \t 4.747283 \t 1.029662 \t 212.486878  \t 0.580377 \n",
+        "0.3090 \t 2.415459 \t 1.231286 \t 255.363338  \t 0.674908 \n",
+        "0.5760 \t 1.350072 \t 1.942786 \t 259.940648  \t 0.690796 \n",
+        "0.6920 \t 1.163087 \t 2.513451 \t 261.385489  \t 0.711020 \n",
+        "0.7390 \t 1.112223 \t 2.812451 \t 261.416681  \t 0.726019 \n",
+        "0.7575 \t 1.095373 \t 2.942991 \t 261.232545  \t 0.733436 \n",
+        "0.8605 \t 1.029233 \t 3.827737 \t 256.608420  \t 0.796964 \n",
+        "0.9250 \t 1.008121 \t 4.546617 \t 248.599211  \t 0.866162 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.4  Page No : 432"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 25. \t\t\t #temperature of the system in degree celsius\n",
+      "A = 2.0684;\t\t\t #the van Laar parameters for the system (no unit)\n",
+      "B = 1.7174;\t\t\t #the van Laar parameters for the system (no unit)\n",
+      "P = [118.05,207.70,246.35,259.40,261.50,262.00,261.90,258.70,252.00];\t\t\t #Pressure data in Torr (from Tasic et al.)\n",
+      "x1 = [0.0115,0.1125,0.3090,0.5760,0.6920,0.7390,0.7575,0.8605,0.9250];\n",
+      "y1 = [0.1810,0.5670,0.6550,0.7050,0.7250,0.7390,0.7460,0.8030,0.8580];\n",
+      "antoine_const_acetone = [7.11714,1210.595,229.664];\t\t\t #Antoine's constants for acetone from Table A.7\n",
+      "antoine_const_chexane = [6.85146,1206.470,223.136];\t\t\t #Antoine's constants for cyclohexane from Table A.7\n",
+      "\n",
+      "# Calculations\n",
+      "P1_s = 10**(antoine_const_acetone[0]-(antoine_const_acetone[1]/(T+antoine_const_acetone[2])))\n",
+      "P2_s = 10**(antoine_const_chexane[0]-(antoine_const_chexane[1]/(T+antoine_const_chexane[2])));\n",
+      "l = len(P)\n",
+      "j = 0\n",
+      "while j<l:\n",
+      "    ln_gaamma1[j] = A/(1+((A*x1[j])/(B*(1-x1[j]))))**2\n",
+      "    ln_gaamma2[j] = B/(1+((B*(1-x1[j]))/(A*x1[j])))**2\n",
+      "    gaamma1[j] = math.exp(ln_gaamma1[j]);\t\t\t\n",
+      "    gaamma2[j] = math.exp(ln_gaamma2[j]);\t\t\t\n",
+      "    P[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
+      "    y1_calc[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
+      "    j = j+1;\n",
+      "\n",
+      "# Results\n",
+      "print 'P-x-y data: ';\n",
+      "i = 0;\n",
+      "print 'x1 \\t  gamma1 \\t gamma2 \\t PTorr \\t y1 ';\n",
+      "for i in range(l):\n",
+      "    print '%0.4f \\t %f \\t %f \\t %f  \\t %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],y1_calc[i]);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "P-x-y data: \n",
+        "x1 \t  gamma1 \t gamma2 \t PTorr \t y1 \n",
+        "0.0115 \t 7.475516 \t 1.000328 \t 116.333233  \t 0.170640 \n",
+        "0.1125 \t 4.743506 \t 1.030586 \t 212.468724  \t 0.579965 \n",
+        "0.3090 \t 2.395936 \t 1.234218 \t 254.168007  \t 0.672601 \n",
+        "0.5760 \t 1.346684 \t 1.937830 \t 259.284960  \t 0.690805 \n",
+        "0.6920 \t 1.162537 \t 2.498302 \t 260.842215  \t 0.712164 \n",
+        "0.7390 \t 1.112211 \t 2.792268 \t 260.900602  \t 0.727447 \n",
+        "0.7575 \t 1.095497 \t 2.920797 \t 260.729035  \t 0.734935 \n",
+        "0.8605 \t 1.029539 \t 3.796506 \t 256.244215  \t 0.798334 \n",
+        "0.9250 \t 1.008263 \t 4.515793 \t 248.404105  \t 0.866965 \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.5  Page No : 435"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# Variables\n",
+      "P = 760.\t\t\t #pressure in Torr at which chloroform and methanol form an azeotrope\n",
+      "T = 53.5\t\t\t #temperature in degree celsius at which chloroform and methanol form an azeotrope\n",
+      "x1 = 0.65\t\t\t #mole fraction of chloroform in the liquid phase (no unit) (corresponding to azeotropic composition)\n",
+      "antoine_const_chloroform = [6.95465,1170.966,226.232];\t\t\t #Antoine's constants for acetone from Table A.7\n",
+      "antoine_const_methanol = [8.08097,1582.271,239.726];\t\t\t #Antoine's constants for acetone from Table A.7\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = 1-x1\n",
+      "P1_s = 10**(antoine_const_chloroform[0]-(antoine_const_chloroform[1]/(T+antoine_const_chloroform[2])));\n",
+      "P2_s = 10**(antoine_const_methanol[0]-(antoine_const_methanol[1]/(T+antoine_const_methanol[2])));\n",
+      "gaamma1 = P/P1_s\n",
+      "gaamma2 = P/P2_s\n",
+      "A = math.log(gaamma1)*(1+((x2*math.log(gaamma2))/(x1*math.log(gaamma1))))**2\n",
+      "B = math.log(gaamma2)*(1+((x1*math.log(gaamma1))/(x2*math.log(gaamma2))))**2\n",
+      "x1 = linspace(0.1,0.9,9)\n",
+      "l = len(x1)\n",
+      "gaamma1 = zeros(l)\n",
+      "gaamma2 = zeros(l)\n",
+      "P  = zeros(l) \n",
+      "y1 = zeros(l)\n",
+      "j = 0\n",
+      "while j<l:\n",
+      "    ln_gaamma1[j] = A/(1+((A*x1[j])/(B*(1-x1[j]))))**2\n",
+      "    ln_gaamma2[j] = B/(1+((B*(1-x1[j]))/(A*x1[j])))**2\n",
+      "    gaamma1[j] = math.exp(ln_gaamma1[j]);\t\t\t \n",
+      "    gaamma2[j] = math.exp(ln_gaamma2[j]);\t\t\t \n",
+      "    P[j] = (gaamma1[j]*x1[j]*P1_s)+(gaamma2[j]*(1-x1[j])*P2_s)\n",
+      "    y1[j] = (gaamma1[j]*x1[j]*P1_s)/P[j];\t\t\t \n",
+      "    j = j+1;\n",
+      "\n",
+      "# Results\n",
+      "print 'VLE data: ';\n",
+      "print 'x1 \\tgamma1 \\t\\t gamma2  \\t P Torr \\t y1 ';\n",
+      "for i in range(l):\n",
+      "    print '%0.1f \\t %f \\t %f \\t %f \\t %f '%(x1[i],gaamma1[i],gaamma2[i],P[i],y1[i]);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "VLE data: \n",
+        "x1 \tgamma1 \t\t gamma2  \t P Torr \t y1 \n",
+        "0.1 \t 2.391069 \t 1.005464 \t 578.375597 \t 0.242663 \n",
+        "0.2 \t 2.151801 \t 1.024547 \t 649.357212 \t 0.389019 \n",
+        "0.3 \t 1.928361 \t 1.062974 \t 699.744047 \t 0.485280 \n",
+        "0.4 \t 1.722251 \t 1.130135 \t 732.594248 \t 0.551969 \n",
+        "0.5 \t 1.535159 \t 1.242383 \t 751.239893 \t 0.599745 \n",
+        "0.6 \t 1.369108 \t 1.430331 \t 759.122391 \t 0.635183 \n",
+        "0.7 \t 1.226766 \t 1.756665 \t 759.153131 \t 0.663976 \n",
+        "0.8 \t 1.112114 \t 2.365222 \t 751.206494 \t 0.695188 \n",
+        "0.9 \t 1.031997 \t 3.639052 \t 721.331744 \t 0.755801 \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.6  Page No : 443"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "y1 = 0.3;\t\t\t #mole fraction of ethane in the vapour phase (no unit)\n",
+      "T = 30. \t\t\t #temperature in degree celsius\n",
+      "\n",
+      "# Calculations\n",
+      "y2 = 1-y1\n",
+      "P_guess = 1\n",
+      "K1 = 3.4\n",
+      "K2 = 1.1\n",
+      "x1_calc = y1/K1\n",
+      "x2_calc = y2/K2;\n",
+      "tot = x1_calc+x2_calc\n",
+      "if tot == 1:\n",
+      "    P = P_guess\n",
+      "    x1 = x1_calc\n",
+      "    x2 = x2_calc\n",
+      "else:\n",
+      "    P = 1.5\n",
+      "    K1 = 2.4\n",
+      "    K2 = 0.8\n",
+      "    x1 = y1/K1\n",
+      "    x2 = y2/K2\n",
+      "\n",
+      "# Results\n",
+      "print 'The Dew pressure and the liquid composition of a binary vapour mixture of\\\n",
+      " ethane and propane was found to be P = %0.2f MPa\\t x1 = %0.3f\\t x2 = %0.3f \\t'%(P,x1,x2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Dew pressure and the liquid composition of a binary vapour mixture of ethane and propane was found to be P = 1.50 MPa\t x1 = 0.125\t x2 = 0.875 \t\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.7  Page No : 443"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "x1 = 0.4;\t\t\t #mole fraction of ethane in the liquid phase (no unit)\n",
+      "P = 1.5;\t\t\t #pressure in MPa\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = 1-x1\n",
+      "t_guess = 10.\n",
+      "K1 = 1.8;\t\n",
+      "K2 = 0.5;\t\n",
+      "y1_calc = K1*x1\n",
+      "y2_calc = K2*x2\n",
+      "tot = y1_calc+y2_calc\n",
+      "if tot == 1:\n",
+      "   t = t_guess\n",
+      "   y1 = y1_calc\n",
+      "   y2 = y2_calc\n",
+      "else:\n",
+      "    t = 9.\n",
+      "    K1 = 1.75\n",
+      "    K2 = 0.5;\n",
+      "    y1 = K1*x1\n",
+      "    y2 = K2*x2\n",
+      "\n",
+      "# Results\n",
+      "print 'The bubble temperature and the vapour composition of a binary vapour mixture\\\n",
+      " of ethane and propane was found to be t = %d degree celsius y1 = %f\\t y2 = %f\\t'%(t,y1,y2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The bubble temperature and the vapour composition of a binary vapour mixture of ethane and propane was found to be t = 9 degree celsius y1 = 0.700000\t y2 = 0.300000\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.8  Page No : 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "%matplotlib inline\n",
+      "# Variables\n",
+      "P = [350.00,446.00,518.00,574.50,609.00,632.50,665.00,681.50,691.50]\n",
+      "x1 = [0.0550,0.1290,0.2120,0.3130,0.4300,0.5200,0.6380,0.7490,0.8720]\n",
+      "y1 = [0.3500,0.5110,0.5990,0.6500,0.6970,0.7260,0.7590,0.8130,0.8830]\n",
+      "antoine_const_propanol = [8.37895,1788.020,227.438];\t\t\t \n",
+      "antoine_const_cbenzene = [7.17294,1549.200,229.260];\t\t\t \n",
+      "T = 95.\t\t\t #temperature of the system in degree celsius\n",
+      "\n",
+      "# Calculations\n",
+      "P1_s = 10**(antoine_const_propanol[0]-(antoine_const_propanol[1]/(T+antoine_const_propanol[2])));\n",
+      "P2_s = 10**(antoine_const_cbenzene[0]-(antoine_const_cbenzene[1]/(T+antoine_const_cbenzene[2])));\n",
+      "l = len(P)\n",
+      "lngamma1_gamma2 = zeros(l)\n",
+      "gaamma1 = zeros(l)\n",
+      "gaamma2 = zeros(l)\n",
+      "i = 0\n",
+      "while i<l:\n",
+      "    gaamma1[i] = (y1[i]*P[i])/(x1[i]*P1_s)\n",
+      "    gaamma2[i] = ((1-y1[i])*P[i])/((1-x1[i])*P2_s)\n",
+      "    lngamma1_gamma2[i] = math.log(gaamma1[i]/gaamma2[i])\n",
+      "    i = i+1;\n",
+      "\n",
+      "plot(x1,lngamma1_gamma2);\n",
+      "suptitle('Plot of ln(gamma1/gamma2) vs x1')\n",
+      "xlabel('x1')\n",
+      "ylabel('ln(gamma1/gamma2)');\n",
+      "area_above = 1515.\n",
+      "area_below = 1540.\n",
+      "consistency_parameter = abs((area_above-area_below)/(area_above+area_below));\t\t\t\n",
+      "\n",
+      "# Results\n",
+      "print 'Values of lngamma1./gamma2: ';\n",
+      "print 'x1 \\t gamma1 \\t gamma2 \\t lngamma1./gamma2';\n",
+      "\n",
+      "for i in range(l):\n",
+      "    print '%0.4f \\t %f \\t %f \\t %f '%(x1[i],gaamma1[i],gaamma2[i],lngamma1_gamma2[i]);\n",
+      "\n",
+      "print 'The value of the consistency parameter = %f'%(consistency_parameter);\n",
+      "if consistency_parameter<0.02 or consistency_parameter == 0.02:\n",
+      "    print 'The VLE data is thermodynamically consistent';\n",
+      "else:\n",
+      "    print 'The VLE data is not thermodynamically consistent';\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "Values of lngamma1./gamma2: "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "x1 \t gamma1 \t gamma2 \t lngamma1./gamma2\n",
+        "0.0550 \t 3.266913 \t 0.968851 \t 1.215489 \n",
+        "0.1290 \t 2.591375 \t 1.007704 \t 0.944514 \n",
+        "0.2120 \t 2.146767 \t 1.060854 \t 0.704889 \n",
+        "0.3130 \t 1.749940 \t 1.177901 \t 0.395847 \n",
+        "0.4300 \t 1.447924 \t 1.302845 \t 0.105581 \n",
+        "0.5200 \t 1.295263 \t 1.453040 \t -0.114944 \n",
+        "0.6380 \t 1.160398 \t 1.781713 \t -0.428812 \n",
+        "0.7490 \t 1.085022 \t 2.043343 \t -0.632987 \n",
+        "0.8720 \t 1.027071 \t 2.543757 \t -0.906931 \n",
+        "The value of the consistency parameter = 0.008183\n",
+        "The VLE data is thermodynamically consistent\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['f', 'draw_if_interactive', 'new_figure_manager', 'test']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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pVMC//gUEBCidiKjyZCsc0dHRmDt3LurUqQOr/37FUqlUuHjxYuWSyoCFg0yp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\nEREZhIWDiIgMwsJBREQGYeEgIiKD/H+2fJxBMvlw7wAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x39ce190>"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.9  Page No : 464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "%matplotlib inline\n",
+      "# Variables\n",
+      "\n",
+      "P = 760.\t\t\t                            #pressure of the system in Torr\n",
+      "antoine_const_benzene = [6.87987,1196.760,219.161];\t\t\t #Antoine's constants for Benzene from Table A.7\n",
+      "t = linspace(60,100,9);\t\t\t #temperature range in degree celsius\n",
+      "\n",
+      "P2_s = [149.40,187.58,233.71,289.13,355.21,433.51,525.84,634.00,760.00];\n",
+      "x1 = linspace(0,1,6)\n",
+      "\n",
+      "# Calculations\n",
+      "l = len(t)\n",
+      "i = 0\n",
+      "P1_s = zeros(l)\n",
+      "P_tot =  zeros(l)\n",
+      "\n",
+      "while i<l:\n",
+      "    P1_s[i] = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(t[i]+antoine_const_benzene[2])));\n",
+      "    P_tot[i] = P1_s[i]+P2_s[i];\n",
+      "    i = i+1;\n",
+      "\n",
+      "T = (((t[2]-t[1])*(760-P_tot[1]))/(P_tot[2]-P_tot[1]))+t[1]\n",
+      "\n",
+      "P1_s_three_phase = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(T+antoine_const_benzene[2])));\n",
+      "P2_s_three_phase = 760-P1_s_three_phase;\n",
+      "y1_three_phase = P1_s_three_phase/760\n",
+      "\n",
+      "trange1 = linspace(T,T+11,11)\n",
+      "n = len(trange1)\n",
+      "i = 0\n",
+      "y1 = zeros(n)\n",
+      "P1_s_calc = zeros(n)\n",
+      "while i<n:\n",
+      "    if i == 1:\n",
+      "        y1[i] = y1_three_phase\n",
+      "    else:\n",
+      "         P1_s_calc[i] = 10**(antoine_const_benzene[0]-(antoine_const_benzene[1]/(trange1[i]+antoine_const_benzene[2])));\n",
+      "         y1[i] = (P1_s_calc[i])/P\n",
+      "    i = i+1;\n",
+      "\n",
+      "trange2 = [70,75,80,85,90,95,100]\n",
+      "P2_s_range = [233.71,289.13,355.21,433.51,525.84,634.00,760.00]\n",
+      "p = len(trange2);\t\t\t\n",
+      "i = 0\n",
+      "y_one = zeros(p)\n",
+      "# Calculations of the vapour composition of benzene in the two phase region (L2+V) using Eq.(12.61) (no unit)\n",
+      "y_one[i] = y1_three_phase;\n",
+      "trange2[i] = T;\n",
+      "i = i+1;\n",
+      "\n",
+      "while i<p:\n",
+      "    y_one[i] = (P-P2_s_range[i])/P;\n",
+      "    i = i+1;\n",
+      "k=len(x1)    \n",
+      "t_3phase = zeros(k)\n",
+      "i = 0\n",
+      "k = len(x1)\n",
+      "while i<k:\n",
+      "    t_3phase[i] = T\n",
+      "    i = i+1;\n",
+      "\n",
+      "# Results\n",
+      "plot(y1,trange1);\n",
+      "plot(y_one,trange2);\n",
+      "plot(x1,t_3phase);\n",
+      "suptitle('t-x-y diagram for benzene-water sytem at 760 Torr')\n",
+      "xlabel('x1,y1')\n",
+      "ylabel('t (degree celsius)');\n",
+      "q = len(t)\n",
+      "i = 1;\t\t\n",
+      "print ' Calculationss performed for determining the three phase equilibrium temperature';\n",
+      "print 'tdegree celsius \\t P1_s Torr \\t P2_s Torr \\t P1_s+P2_s Torr ';\n",
+      "for i in range(q):\n",
+      "    print '%d \\t \\t \\t %.2f \\t %0.2f \\t %.2f '%(t[i],P1_s[i],P2_s[i],P_tot[i]);\n",
+      "\n",
+      "print 'The three phase equilibrium temperature = %0.2f degree celsius '%(T);\n",
+      "print 'The vapour phase composition of benzene at the three phase equilibrium point = %0.4f '%(y1_three_phase);\n",
+      "\n",
+      "# Note: answer are slightly different because of rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        " Calculationss performed for determining the three phase equilibrium temperature"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "tdegree celsius \t P1_s Torr \t P2_s Torr \t P1_s+P2_s Torr \n",
+        "60 \t \t \t 391.63 \t 149.40 \t 541.03 \n",
+        "65 \t \t \t 465.92 \t 187.58 \t 653.50 \n",
+        "70 \t \t \t 550.98 \t 233.71 \t 784.69 \n",
+        "75 \t \t \t 647.87 \t 289.13 \t 937.00 \n",
+        "80 \t \t \t 757.67 \t 355.21 \t 1112.88 \n",
+        "85 \t \t \t 881.54 \t 433.51 \t 1315.05 \n",
+        "90 \t \t \t 1020.65 \t 525.84 \t 1546.49 \n",
+        "95 \t \t \t 1176.21 \t 634.00 \t 1810.21 \n",
+        "100 \t \t \t 1349.47 \t 760.00 \t 2109.47 \n",
+        "The three phase equilibrium temperature = 69.06 degree celsius \n",
+        "The vapour phase composition of benzene at the three phase equilibrium point = 0.7028 \n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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jBSUFouT7/dbvDFwzkEbVG7G873KqlK+idSQhij2zFQg3Nzdjcbh37x5xcXE0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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x39910d0>"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch13.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch13.ipynb
new file mode 100755
index 00000000..8dc5f0be
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch13.ipynb
@@ -0,0 +1,235 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Dilute solution laws"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.1  Page No : 478"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "weight = 10.     \t\t\t #weight of NaCl in grams\n",
+      "volume = 1. \t    \t\t #volume of water in litres\n",
+      "weight_water = 1000. \t\t\t #  weight of water in grams (Weight = Volume*Density, density of water  = 1g/cc = 1g/ml = 1000g/l)\n",
+      "molwt_NaCl = 58.5 \t\t\t #molecular weight of NaCl in grams\n",
+      "molwt_water = 18. \t\t\t #molecular weight of water in grams\n",
+      "hf = 6.002; \t\t\t #enthalpy change of fusion in kJ/mol at 0 degree celsius\n",
+      "P = 101.325; \t\t\t #pressure in kPa\n",
+      "T = 273.15; \t\t\t # freezing point temperature of water at the given pressure in K\n",
+      "R = 8.314; \t\t\t #universal gas constant in J/molK;\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = (weight/molwt_NaCl)/((weight/molwt_NaCl)+(weight_water/molwt_water))\n",
+      "delt = (R*T**2*x2)/(hf*10**3)\n",
+      "\n",
+      "# Results\n",
+      "print ' The depression in freezing point of water when 10g of NaCl solute is added  =  %0.2f K'%(delt);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The depression in freezing point of water when 10g of NaCl solute is added  =  0.32 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.2  Page No : 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "weight = 10.; \t\t    \t #weight of NaCl in grams\n",
+      "volume = 1.; \t\t\t     #volume of water in litres\n",
+      "weight_water = 1000.; \t\t\t #  weight of water in grams (Weight = Volume*Density, density of water  = 1g/cc = 1g/ml = 1000g/l)\n",
+      "molwt_NaCl = 58.5; \t\t\t #molecular weight of NaCl in grams\n",
+      "molwt_water = 18; \t\t\t #molecular weight of water in grams\n",
+      "lat_ht = 2256.94; \t\t\t #latent heat of vaporization in kJ/kg at 100 degree celsius (obtained from steam tables)\n",
+      "P = 101.325; \t\t\t #pressure in kPa\n",
+      "T = 373.15; \t\t\t #boiling point temperature of water at the given pressure in K\n",
+      "R = 8.314; \t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = 0.0031\n",
+      "hv = (lat_ht*molwt_water)/1000\n",
+      "delt = (R*T**2*x2)/(hv*10**3)\n",
+      "\n",
+      "# Results\n",
+      "print ' The elevation in boiling point of water when 10g of NaCl solute is added  =  %0.2f K'%(delt);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The elevation in boiling point of water when 10g of NaCl solute is added  =  0.09 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.3  Page No : 481"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "weight = 10.; \t\t\t #weight of NaCl in grams\n",
+      "weight_water = 1000.; \t\t\t #  weight of water in grams\n",
+      "molwt_NaCl = 58.5; \t\t\t #molecular weight of NaCl in grams\n",
+      "molwt_water = 18.; \t\t\t #molecular weight of water in grams\n",
+      "T = 300.; \t\t\t #prevailing temperature of water in K\n",
+      "R = 8.314; \t\t\t #universal gas constant in (Pa m**3)/(mol K);\n",
+      "v = 18*10**-6;\t\t\t #molar volume in m**3/mol\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = 0.0031\n",
+      "pi = ((R*T*x2)/v)*10**-3\n",
+      "\n",
+      "# Results\n",
+      "print ' The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K  =  %0.2f kPa'%(pi);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K  =  429.56 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.4  Page No : 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "temp = 20. \t\t\t # prevailing tempearture in degree celsius\n",
+      "melt_temp = 80.05; \t\t\t # melting point of naphthalene in degree celsius\n",
+      "hf = 18.574 \t\t\t # enthalpy of fusion in kJ/mol\n",
+      "R = 8.314 \t\t\t # universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "t = temp+273.15\n",
+      "melt_t = melt_temp+273.15\n",
+      "x2 = math.exp(((hf*10**3)/R)*((1./melt_t)-(1./t)))\n",
+      "\n",
+      "# Results\n",
+      "print ' The ideal solubility of naphthalene at 20 degree celsius =  %0.4f'%(x2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The ideal solubility of naphthalene at 20 degree celsius =  0.2737\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.5  Page No : 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 295.43; \t\t\t #prevailing temperature in K\n",
+      "sat_p = 6.05; \t\t\t #Sasturation pressure of carbon dioxide at the prevailing temperature in MPa\n",
+      "p = 0.1; \t\t\t #pressure at which solubility has to be determined in MPa\n",
+      "\n",
+      "# Calculations\n",
+      "x2 = p/sat_p\n",
+      "\n",
+      "# Results\n",
+      "print ' The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution\\\n",
+      " at 0.1MPa =  %0.4f'%(x2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution at 0.1MPa =  0.0165\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch14.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch14.ipynb
new file mode 100755
index 00000000..e54d6be2
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch14.ipynb
@@ -0,0 +1,846 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 : Chemical reaction equilibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.1  Page No : 489"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 298.15;\t\t\t #temperature in K\n",
+      "del_Gf = [-137.327,-228.600,-394.815,0]; #the standard Gibbs free energy of formation of CO(g),H2O(g),CO2(g) and H2(g) in kJ\n",
+      "n = [1,1,-1,-1]         \t\t\t #stoichiometric coefficients of CO(g),H2O(g),CO2(g) and H2(g) respectively (no unit)\n",
+      "R = 8.314;\t\t\t                 #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "del_G = (n[0]*del_Gf[0])+(n[1]*del_Gf[1])+(n[2]*del_Gf[2])+(n[3]*del_Gf[3]);\n",
+      "Ka = math.exp((-(del_G*10**3))/(R*T))\n",
+      "\n",
+      "# Results\n",
+      "print 'The standard Gibbs free energy of the water gas shift reaction at 298.15K = %0.3f kJ '%(del_G);\n",
+      "print 'The equilibrium constant of the water gas shift reaction at 298.15K = %0.3e '%(Ka);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The standard Gibbs free energy of the water gas shift reaction at 298.15K = 28.888 kJ \n",
+        "The equilibrium constant of the water gas shift reaction at 298.15K = 8.685e-06 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.2  Page No : 490"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 298.15;\t\t\t #temperature in K\n",
+      "P_s = 0.16716;\t\t\t #saturation pressure of CH3OH in bar at T\n",
+      "del_G1 = -166.215\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "f_v = 1\n",
+      "f_l = P_s\n",
+      "del_G2 = R*T*math.log(f_v/f_l)*10**-3;\t\t\t # Calculations of the value of del_G2 in kJ\n",
+      "del_G = del_G2+del_G1;\t\t\n",
+      "\n",
+      "# Results\n",
+      "print 'The standard Gibbs free energy of formation of CH3OHg = %0.3f kJ '%(del_G);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The standard Gibbs free energy of formation of CH3OHg = -161.781 kJ \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.3  Page No : 491"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "#The water gas shift reaction is given by: CO2(g)+H2(g)--->CO(g)+H2O(g)\n",
+      "T1 = 298.15     \t\t\t #initial temperature in K\n",
+      "Ka1 = 8.685*10**-6;\t\t\t #equilibrium constant for the water-gas shift reaction at T1 (no unit)\n",
+      "T2 = 1000.\t\t        \t #temperature at which the equilibrium constant has to be determined in K\n",
+      "R = 8.314;\t\t\t         #universal gas constant in J/molK\n",
+      "del_Hf = [-110.532,-241.997,-393.978,0];\t\t\t #the standard enthalpy of formation of CO(g),H2O(g),CO2(g) and H2(g) in kJ\n",
+      "n = [1,1,-1,-1];\t\t\t #stoichiometric coefficients of CO(g),H2O(g),CO2(g) and H2(g) respectively (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "#It is assumed that del_H is constant in the temperature range T1 and T2\n",
+      "del_H = (n[0]*del_Hf[0])+(n[1]*del_Hf[1])+(n[2]*del_Hf[2])+(n[3]*del_Hf[3])\n",
+      "Ka2 = Ka1*math.exp(((del_H*10**3)/R)*((1./T1)-(1./T2)));\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print 'The equilibrium constant for the water gas shift reaction at 1000K = %f '%(Ka2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equilibrium constant for the water gas shift reaction at 1000K = 1.085357 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.4  Page No : 492"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "P = 0.1     \t\t\t #pressure in MPa\n",
+      "T1 = 298.15;\t\t\t #initial temperature in K\n",
+      "Ka1 = 8.685*10**-6;\t\t\t #equilibrium constant for the water-gas shift reaction at T1 (no unit) (from Example 14.1)\n",
+      "T2 = 1000.  \t\t\t #temperature at which the equilibrium constant is to be found, in K\n",
+      "del_H = 41.449;\t\t\t #smath.tan(math.radiansard enthalpy of the reaction at T1 in kJ (from Example 14.3)\n",
+      "a = [28.068,28.850,45.369,27.012];\n",
+      "b = [4.631*10**-3,12.055*10**-3,8.688*10**-3,3.509*10**-3]\n",
+      "c = [0,0,0,0];\n",
+      "d = [0,0,0,0];\n",
+      "e = [-0.258*10**5,1.006*10**5,-9.619*10**5,0.690*10**5]\n",
+      "n = [1,1,-1,-1]\n",
+      "R = 8.314;\t\t\n",
+      "Ka2_prev = 1.0855\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "del_a = (n[0]*a[0])+(n[1]*a[1])+(n[2]*a[2])+(n[3]*a[3]);\n",
+      "del_b = (n[0]*b[0])+(n[1]*b[1])+(n[2]*b[2])+(n[3]*b[3]);\n",
+      "del_c = (n[0]*c[0])+(n[1]*c[1])+(n[2]*c[2])+(n[3]*c[3]);\n",
+      "del_d = (n[0]*d[0])+(n[1]*d[1])+(n[2]*d[2])+(n[3]*d[3]);\n",
+      "del_e = (n[0]*e[0])+(n[1]*e[1])+(n[2]*e[2])+(n[3]*e[3]);\n",
+      "del_H0 = (del_H*10**3)-((del_a*T1)+((del_b/2)*T1**2)+((del_c/3)*T1**3)+((del_d/4)*T1**4)-(del_e/T1));\n",
+      "I = (math.log(Ka1))-((1./R)*((-del_H0/T1)+(del_a*math.log(T1))+((del_b/2)*T1)+((del_c/6)*T1**2)+((del_d/12)*T1**3)+((del_e/(2*T1**2)))));\n",
+      "Ka2 = math.exp(((1./R)*((-del_H0/T2)+(del_a*math.log(T2))+((del_b/2)*T2)+((del_c/6)*T2**2)+((del_d/12)*T2**3)+((del_e/(2*T2**2)))))+I);\n",
+      "\n",
+      "\t\t\t # Results\n",
+      "print 'The equilibrium constant for the water gas shift reaction at 1000K\\\n",
+      " by taking into account the variation of del_H with temperature = %f '%(Ka2);\n",
+      "print 'The equilibrium constant for the water gas shift reaction at 1000K without\\\n",
+      " considering the variation of del_H with temperature as given by Example14.3 = %0.4f '%(Ka2_prev);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equilibrium constant for the water gas shift reaction at 1000K by taking into account the variation of del_H with temperature = 0.664976 \n",
+        "The equilibrium constant for the water gas shift reaction at 1000K without considering the variation of del_H with temperature as given by Example14.3 = 1.0855 \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.5 page no : 494"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol\n",
+      "from sympy.solvers import solve\n",
+      "import math\n",
+      "\n",
+      "# variables\n",
+      "deltaGf = -161.781       # kJ from exa. 14.2\n",
+      "deltaG298 = deltaGf - (-137.327)\n",
+      "deltaH298 = -200.660 - (-110.532)\n",
+      "deltaA = 18.382 - 28.068 - 2 * 27.012\n",
+      "deltaB = (101.564 - 4.631 - 2 * 3.509) * 10**-3\n",
+      "deltaC = -28.683 * 10**-6\n",
+      "deltaD = 0\n",
+      "deltaE = (0.258 - 2 * 0.690) * 10**5\n",
+      "T = 298.15\n",
+      "\n",
+      "# calculations\n",
+      "deltaHf298 = -238.648 + 37.988\n",
+      "deltaH0 = Symbol('deltaH0')\n",
+      "Eq = deltaH0 + deltaA*T + deltaB*T**2/2 + deltaC*T**3/3 - deltaE/T - deltaH298*1000\n",
+      "deltaH0 = solve(Eq,deltaH0)[0]/1000\n",
+      "I = Symbol('I')\n",
+      "\n",
+      "Eq1 = deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I + 24454\n",
+      "I = round(solve(Eq1,I)[0],3)\n",
+      "\n",
+      "T = 500\n",
+      "deltaG500 = (deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I)/1000\n",
+      "\n",
+      "Ka = math.exp(-22048/(8.314*T))\n",
+      "e = Symbol('e')\n",
+      "Eq3 = ( (e*(3-2*e)**2) / (1 - e)**3) / 0.4973 -1\n",
+      "e = round(solve(Eq3,e)[0],4)\n",
+      "yCH3OH = e/(3-2*e)\n",
+      "yco = (1-e)/(3-2*e)\n",
+      "yh2 = (2*(1-e) / (3-2*e))\n",
+      "\n",
+      "# Results\n",
+      "print \"E = %.4f\"%e\n",
+      "print \"YCH30H = %.4f\"%yCH3OH\n",
+      "print \"YCO = %.4f\"%yco\n",
+      "print \"YH2 = %.4f\"%yh2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E = 0.0506\n",
+        "YCH30H = 0.0175\n",
+        "YCO = 0.3275\n",
+        "YH2 = 0.6550\n"
+       ]
+      }
+     ],
+     "prompt_number": 103
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.6, Page 496"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy.solvers import solve \n",
+      "from sympy import Symbol \n",
+      "\n",
+      "#Variables\n",
+      "e = Symbol('e')\n",
+      "\n",
+      "#Calculations\n",
+      "x = solve((((e/(3-2*e))/(((1-e)/(3-2*e))*(((2*(1-e))/(3-2*e))**2)))-49.73),e)\n",
+      "print \"e =\",round(x[0],3)\n",
+      "ych3oh = x[0]/(3-2*x[0])\n",
+      "yco = (1-x[0])/(3-2*x[0])\n",
+      "yh2 = ((2*(1-x[0]))/(3-2*x[0]))\n",
+      "print \"yCH3OH =\",round(ych3oh,4)\n",
+      "print \"yCO =\",round(yco,4)\n",
+      "print \"yH2 =\",round(yh2,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "e = 0.801\n",
+        "yCH3OH = 0.5731\n",
+        "yCO = 0.1423\n",
+        "yH2 = 0.2846\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.7 pageno : 497"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol\n",
+      "from sympy.solvers import solve\n",
+      "\n",
+      "# variables\n",
+      "Ka = 4.973*10**-3          # from example 14.5\n",
+      "Ky = 25 * Ka               # from example 14.5\n",
+      "\n",
+      "# calculations\n",
+      "e = Symbol('e')\n",
+      "Eq1 = ( (e*(4-e)**2)/(1-e)**3 ) / .1243 - 1\n",
+      "e = round(solve(Eq1,e)[0],5)\n",
+      "yCH3OH = e/(8-2*e)\n",
+      "yco = (1-e)/(8-2*e)\n",
+      "yh2 = (2*(1-e) / (8-2*e))\n",
+      "yA = 5/(8-2*e)\n",
+      "\n",
+      "# Results\n",
+      "print \"E = %.5f\"%e\n",
+      "print \"YCH30H = %.5f\"%yCH3OH\n",
+      "print \"YCO = %.5f\"%yco\n",
+      "print \"YH2 = %.5f\"%yh2\n",
+      "print \"YA = %.5f\"%yA"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E = 0.00762\n",
+        "YCH30H = 0.00095\n",
+        "YCO = 0.12428\n",
+        "YH2 = 0.24857\n",
+        "YA = 0.62619\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.8 pageno : 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol\n",
+      "from sympy.solvers import solve\n",
+      "\n",
+      "# variables\n",
+      "Ka = 4.973*10**-3           # example 14.5\n",
+      "Ky = 25 * Ka                # example 14.5\n",
+      "\n",
+      "# calculations and results\n",
+      "e = Symbol('e')\n",
+      "Eq1= ((e*(1-e))/(1-2*e)**2)/.03108 - 1\n",
+      "e = solve(Eq1,e)\n",
+      "print \"part a\"\n",
+      "print \"e = %.5f and e = %.5f\"%(e[0],e[1])\n",
+      "print \"The admissible value is e = %.5f\"%e[0]\n",
+      "\n",
+      "print \"part b\"\n",
+      "print \"e = 0.0506 ( from example 14.5 )\"\n",
+      "\n",
+      "x = 4\n",
+      "e = Symbol('e')\n",
+      "Eq1=  ((e/(1-e) ) * ( (5 - 2*e)/ (2-e) )**2)/ 0.4973 - 1\n",
+      "e = solve(Eq1,e)\n",
+      "print \"part c\"\n",
+      "print \"e = %.4f\" %(e[0])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part a\n",
+        "e = 0.02845 and e = 0.97155\n",
+        "The admissible value is e = 0.02845\n",
+        "part b\n",
+        "e = 0.0506 ( from example 14.5 )\n",
+        "part c"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "e = 0.0727\n"
+       ]
+      }
+     ],
+     "prompt_number": 134
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.9 pageno : 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol\n",
+      "from sympy.solvers import solve\n",
+      "\n",
+      "# variables\n",
+      "Ka = 4.973*10**-3           # example 14.5\n",
+      "Ky = 25 * Ka                # example 14.5\n",
+      "\n",
+      "# calculations\n",
+      "e = Symbol('e')\n",
+      "Eq1 = (0.02 + e)*(1.51-e)**2 / (1-e)**3 / Ky - 1\n",
+      "e = solve(Eq1,e)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \"E = %.5f\"%e"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E = 0.03165\n"
+       ]
+      }
+     ],
+     "prompt_number": 138
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.10 pageno : 500"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "from sympy import Symbol\n",
+      "from sympy.solvers import solve\n",
+      "from scipy.integrate import quad\n",
+      "import math\n",
+      "\n",
+      "# variables\n",
+      "deltaA = round(28.850 - 27.012 - 0.5 * 30.255,2)\n",
+      "deltaB = (12.055 - 3.509 - 0.5 * 4.207) * 10**-3\n",
+      "deltaC = 0\n",
+      "deltaD = 0\n",
+      "deltaE = round((1.006 - 0.690 + 0.5 * 1.887),3) * 10**5\n",
+      "deltaH298 =  -241.997\n",
+      "Ts = [2000,2500,3000,3500,3800]\n",
+      "deltaHt = []\n",
+      "deltaGt = []\n",
+      "Ka = []\n",
+      "Ea = []\n",
+      "Eb = []\n",
+      "\n",
+      "# Calculations\n",
+      "T = 298.15\n",
+      "deltaH0 = Symbol('deltaH0')\n",
+      "Eq = deltaH0 + deltaA*T + deltaB*T**2/2 + deltaC*T**3/3 + deltaD*T**4/4 - deltaE/T - deltaH298*1000\n",
+      "deltaH0 = round(solve(Eq,deltaH0)[0]/1000,3)\n",
+      "I = Symbol('I')\n",
+      "Eq1 = deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I + 228600\n",
+      "I = round(solve(Eq1,I)[0],3)\n",
+      "\n",
+      "def fun1(T1):\n",
+      "    #return 42.1395*(T1 - 298.15) + 5.613*10**-3/2*(T1**2 - 298.15**2) + .2535 * 10**5 * (1./T1 - 1/298.15)\n",
+      "    return 42.1395*T1 + 5.613*10**-3/2 * T1**2 + 0.2535*10**5/T1 - 12898.37\n",
+      "\n",
+      "for T in Ts:\n",
+      "    Ht = deltaH0*1000 + deltaA*T + deltaB*T**2/2 + deltaC*T**3/3 + deltaD*T**4/4 - deltaE/T \n",
+      "    Gt = deltaH0*1000 - deltaA*T*math.log(T) - deltaB*T**2/2 -  deltaC*T**3/6 - deltaD*T**4/12 - deltaE/(2*T) - 8.314 * T * I \n",
+      "    ka = math.exp(-Gt/(8.314*T))\n",
+      "    e = Symbol('e')\n",
+      "    Eq2 = ( e * (3-e)**(1./2) ) / ( 1-e)**(3./2) / ka - 1\n",
+      "    b = round(solve(Eq2)[0],4)\n",
+      "    a = (quad(fun1,298.15,T)[0]/1000) / -Ht\n",
+      "    deltaHt.append(Ht)\n",
+      "    deltaGt.append(Gt)\n",
+      "    Ka.append(ka)\n",
+      "    Ea.append(a)\n",
+      "    Eb.append(b)\n",
+      "\n",
+      "\n",
+      "# Results    \n",
+      "plot(Ts,Ea,\"g\")\n",
+      "plot(Ts,Eb,\"b\")\n",
+      "xlabel(\"T(k)\")\n",
+      "ylabel(\"E\")\n",
+      "suptitle(\"Plot of e versus adiabatic reaction temperature\")\n",
+      "show()\n",
+      "print \"from plot, it can be seen that both lines are simultaneously satisfied at the point \\\n",
+      "\\nintersection where e = 0.68 and T = 3440 K\"\n",
+      "e = 0.68\n",
+      "T = 3440\n",
+      "yh2 = (1- e)/(1.5 - 0.5*e)\n",
+      "yo2 = 0.5*(1-e) / (1.5-0.5*e)\n",
+      "yh2o = e/(1.5- 0.5*e)\n",
+      "\n",
+      "print \"yH2 = %.4f\"%yh2\n",
+      "print \"yO2 = %.4f\"%yo2\n",
+      "print \"yH2O = %.4f\"%yh2o"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['solve', 'draw_if_interactive', 'e']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ISZGW0wgNvj39LdzC3ODTzgexr8WySFCl6CwUn332mfb25s2byzz3/vvv/+OK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\nQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x51ff6d0>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "from plot, it can be seen that both lines are simultaneously satisfied at the point \n",
+        "intersection where e = 0.68 and T = 3440 K\n",
+        "yH2 = 0.2759\n",
+        "yO2 = 0.1379\n",
+        "yH2O = 0.5862\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.11  Page No : 506"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from numpy import *\n",
+      "\n",
+      "# Variables\n",
+      "stoichio_matrix = matrix([[-1,-1, 1, 3, 0, 0],[0, -1, -1, 1, 1 ,0],[-1, -2, 0 ,4, 1, 0], \\\n",
+      "                [0,0, 1, 0, -1, 0.5],[-1, 2 ,0, 0, 1, -2],[-1, 1, 1, 1, 0, -1]])\t\t\t #Framing the stoichiometric coefficient matrix\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "r = linalg.matrix_rank(stoichio_matrix)\n",
+      "\n",
+      "stoichio_matrix[0] = -stoichio_matrix[0];\n",
+      "stoichio_matrix[2] = stoichio_matrix[2]+stoichio_matrix[0];\n",
+      "stoichio_matrix[5:] = stoichio_matrix[5:]+stoichio_matrix[0];\n",
+      "stoichio_matrix[6:] = stoichio_matrix[6:]+stoichio_matrix[0];\n",
+      "stoichio_matrix[1] = -stoichio_matrix[1];\n",
+      "stoichio_matrix[2] = stoichio_matrix[2]+stoichio_matrix[1];\n",
+      "stoichio_matrix[5:] = stoichio_matrix[5:]-(3*stoichio_matrix[1]);\n",
+      "stoichio_matrix[6:] = stoichio_matrix[6:]-(2*stoichio_matrix[1]);\n",
+      "x = stoichio_matrix[:3]\n",
+      "y = stoichio_matrix[:4];\n",
+      "stoichio_matrix[:4] = y;\n",
+      "stoichio_matrix[:3] = x;\n",
+      "stoichio_matrix[5:] = stoichio_matrix[5:]+(4*stoichio_matrix[3]);\n",
+      "stoichio_matrix[6:] = stoichio_matrix[6:]+(2*stoichio_matrix[3]);\n",
+      "\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print ' The stoichiometric coefficient matrix after performing the elementary row operations = ';\n",
+      "print (stoichio_matrix);\n",
+      "print ' The number of primary reactions = %d'%(r);\n",
+      "print ' The non zero rows are 1,2,4'\n",
+      "print ' The primary reactions are: CH4g)+H2Og)--->COg)+3H2g)%( COg)+H2Og)--->CO2g)+H2g),( CO2g)--->COg)+1./2)O2g)'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The stoichiometric coefficient matrix after performing the elementary row operations = \n",
+        "[[ 1.   1.  -1.  -3.  -0.  -0. ]\n",
+        " [-0.   1.   1.  -1.  -1.  -0. ]\n",
+        " [ 0.   0.   0.   0.   0.   0. ]\n",
+        " [ 0.   0.   1.   0.  -1.   0.5]\n",
+        " [-1.   2.   0.   0.   1.  -2. ]\n",
+        " [ 0.  -1.   1.   1.  -1.   1. ]]\n",
+        " The number of primary reactions = 3\n",
+        " The non zero rows are 1,2,4\n",
+        " The primary reactions are: CH4g)+H2Og)--->COg)+3H2g)%( COg)+H2Og)--->CO2g)+H2g),( CO2g)--->COg)+1./2)O2g)\n"
+       ]
+      }
+     ],
+     "prompt_number": 169
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.13 pageno : 510"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol\n",
+      "from sympy.solvers import solve\n",
+      "\n",
+      "# variables\n",
+      "Ka1 = 0.1429\n",
+      "Ka2 = 2\n",
+      "\n",
+      "# calculations and Results\n",
+      "\n",
+      "# assume e1 = .3\n",
+      "e2 = Symbol('e2')\n",
+      "Eq1 = ((0.3 - e2)*(0.3+e2) /((0.7 - e2) * 0.7)) / 0.1429 -1\n",
+      "e2 =  solve(Eq1)\n",
+      "print \"E2 = %.1f and %.1f \"%(e2[0],e2[1])\n",
+      "\n",
+      "# since negative value is inadmissible. e2 = 0.2\n",
+      "e2 = round(e2[1],1)\n",
+      "e1 = Symbol('e1')\n",
+      "Eq2 = ((e1 + 0.2)*(0.2) /((0.8 - e1) *(e1 - 0.2) ) ) / 2 -1\n",
+      "e1 =  solve(Eq2)\n",
+      "print \"E1 = %.1f and %.1f \"%(e1[0],e1[1])\n",
+      "e1 = round(e1[0],1)\n",
+      "print \"Satisfied results when e1 = %.1f and e2 = %.1f\"%(e1,e2)\n",
+      "\n",
+      "yA = (1-e1-e2)/2\n",
+      "yB = (1-e1)/2\n",
+      "yC = (e1-e2)/2\n",
+      "yD = (e1+e2)/2\n",
+      "yE = e2/2\n",
+      "\n",
+      "print \"yA = %.2f\"%yA\n",
+      "print \"yB = %.2f\"%yB\n",
+      "print \"yC = %.2f\"%yC\n",
+      "print \"yD = %.2f\"%yD\n",
+      "print \"yE = %.2f\"%yE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E2 = -0.1 and 0.2 \n",
+        "E1 = 0.3 and 0.6 \n",
+        "Satisfied results when e1 = 0.3 and e2 = 0.2\n",
+        "yA = 0.25\n",
+        "yB = 0.35\n",
+        "yC = 0.05\n",
+        "yD = 0.25\n",
+        "yE = 0.10\n"
+       ]
+      }
+     ],
+     "prompt_number": 214
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.14 page no : 515"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol\n",
+      "from sympy.solvers import solve\n",
+      "\n",
+      "# variables\n",
+      "Kc = 2.92\n",
+      "\n",
+      "# Calculations and results\n",
+      "e = Symbol('e')\n",
+      "Eq1 = (e*(10+e)/((5-e)*(10-e)))/Kc - 1\n",
+      "e = round(solve(Eq1)[0],4)\n",
+      "print \"E = %.4f\"%e\n",
+      "\n",
+      "cA = 5-e\n",
+      "cB = 10-e\n",
+      "cC = e\n",
+      "cD = 10+e\n",
+      "\n",
+      "print \"CA = %.4f kmol/m*3\"%cA\n",
+      "print \"CB = %.4f kmol/m*3\"%cB\n",
+      "print \"CC = %.4f kmol/m*3\"%cC\n",
+      "print \"CD = %.4f kmol/m*3\"%cD"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E = 3.0446\n",
+        "CA = 1.9554 kmol/m*3\n",
+        "CB = 6.9554 kmol/m*3\n",
+        "CC = 3.0446 kmol/m*3\n",
+        "CD = 13.0446 kmol/m*3\n"
+       ]
+      }
+     ],
+     "prompt_number": 220
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.15, Page 517"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 1200.   \t\t\t #temperature in K\n",
+      "T0 = 298.15 \t\t\t #reference temperature in K\n",
+      "a = [41.84,45.369,82.34];\n",
+      "b = [20.25*10**-3,8.688*10**-3,49.75*10**-3];\n",
+      "c = [0,0,0];\n",
+      "d = [0,0,0];\n",
+      "e = [-4.51*10**5,-9.619*10**5,-12.87*10**5];\n",
+      "del_Gf = [-604.574,-394.815,-1129.515]\n",
+      "del_Hf = [-635.975,-393.978,-1207.683]\n",
+      "n = [1,1,-1]\n",
+      "R = 8.314\n",
+      "\n",
+      "# Calculations\n",
+      "del_G = (n[0]*del_Gf[0])+(n[1]*del_Gf[1])+(n[2]*del_Gf[2]);\n",
+      "del_H = (n[0]*del_Hf[0])+(n[1]*del_Hf[1])+(n[2]*del_Hf[2]);\n",
+      "del_a = (n[0]*a[0])+(n[1]*a[1])+(n[2]*a[2]);\n",
+      "del_b = (n[0]*b[0])+(n[1]*b[1])+(n[2]*b[2]);\n",
+      "del_c = (n[0]*c[0])+(n[1]*c[1])+(n[2]*c[2]);\n",
+      "del_d = (n[0]*d[0])+(n[1]*d[1])+(n[2]*d[2]);\n",
+      "del_e = (n[0]*e[0])+(n[1]*e[1])+(n[2]*e[2]);\n",
+      "del_H0 = ((del_H*10**3)-((del_a*T0)+((del_b/2)*T0**2)+((del_c/3)*T0**3)+((del_d/4)*T0**4)-(del_e/T0)))*10**-3;\n",
+      "IR = (1./(T0))*((del_H0*10**3)-(del_a*T0*math.log(T0))-((del_b/2)*T0**2)-((del_c/6)*T0**3)-((del_d/12)*T0**4)-((del_e/2)*(1./T0))-(del_G*10**3));\n",
+      "del_G_T = ((del_H0*10**3)-(del_a*T*math.log(T))-((del_b/2)*T**2)-((del_c/6)*T**3)-((del_d/12)*T**4)-((del_e/2)*(1./T))-(IR*T))*10**-3;\n",
+      "Ka = math.exp((-del_G_T*10**3)/(R*T))\n",
+      "y = 1;\n",
+      "phi = 1;\n",
+      "f0 = 1;\n",
+      "P = (Ka*f0)/(phi*y)\n",
+      "\n",
+      "# Results\n",
+      "print ' The decomposition pressure,P = %f bar '%(P);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The decomposition pressure,P = 2.384295 bar \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch2.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch2.ipynb
new file mode 100755
index 00000000..b7849ef3
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch2.ipynb
@@ -0,0 +1,117 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:76568d79b614fc85568c92855047632dbe19f5ec7540dbd287268967c3318f2b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Review of basic concepts"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.1  Page No : 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "t1 = 300.;\t\t\t #initial temperature of gas in K\n",
+      "p1 = 0.1;\t\t\t #initial pressure of the gas in MPa\n",
+      "p2 = 0.5;\t\t\t #pressure of gas after heating in MPa\n",
+      "p3 = 0.1;\t\t\t #pressure of gas after expansion in MPa\n",
+      "n = 1.; \t\t\t #number of moles of gas in mole\n",
+      "\n",
+      "# Calculations\n",
+      "w1 = 0*n;       \t\t\t # Calculations of work done by the gas during heating in kJ (since dv = 0)\n",
+      "t2 = t1*(p2/p1);\t\t\t # Calculations of temperature of gas after heating in K\n",
+      "t3 = t2;\t\t        \t # Calculations of temperature of gas after expansion in K (constant temperature expansion)\n",
+      "volume2 = p2/p3;\t\t\t # Calculations of  ratio of volume of gas after expansion to the volume of gas after heating (no unit)\n",
+      "w2 = (R*t2*math.log(volume2)*n)/1000\n",
+      "volume3 =  t3/t1;\t\t\n",
+      "w3 = (R*t1*(1-volume3)*n)/1000\n",
+      "work_net = w1+w2+w3;\t\t\t # Calculations of net work done by the gas for the process in kJ\n",
+      "\n",
+      "# Results\n",
+      "print ' Work done during heating process:Work from 1-2 =  %d kJ '%(w1);\n",
+      "print ' Work done during constant temperature expansion: Work from 2-3 =  %.3f kJ '%(w2);\n",
+      "print ' Work done during constant pressure compression: Work from 3-1 =  %.3f kJ '%(w3);\n",
+      "print ' Net work done by the gas during the process =  %.3f kJ '%(work_net);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Work done during heating process:Work from 1-2 =  0 kJ \n",
+        " Work done during constant temperature expansion: Work from 2-3 =  20.071 kJ \n",
+        " Work done during constant pressure compression: Work from 3-1 =  -9.977 kJ \n",
+        " Net work done by the gas during the process =  10.095 kJ \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.2  Page No : 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "v1 = 0.1;\t\t\t #volume of gas initially present in the cylinder in m**3\n",
+      "p1 = 0.1;\t\t\t #initial pressure of gas in MPa\n",
+      "p_atm = 0.1;\t\t #atmospheric pressure acting on the piston in MPa\n",
+      "v2 = 0.3;\t\t\t #volume of gas after heating in m**3\n",
+      "p2 = 0.6;\t\t\t #pressure of gas after heating in MPa\n",
+      "\n",
+      "# Calculations\n",
+      "work = ((p1+p2)*(v2-v1)*10**6)/(2*1000);\t\t\t # Calculations of work done by the gas in kJ \n",
+      "\n",
+      "# Results\n",
+      "print ' The work done by the gas  =  %d kJ'%(work);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The work done by the gas  =  70 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch3.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch3.ipynb
new file mode 100755
index 00000000..2b39d6e0
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch3.ipynb
@@ -0,0 +1,897 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8d10c3a8d696216084bffc2404a32e0b3c15831ac555c05fcb7016ab7d75d0e7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : PvT relations of fluids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.1  Page No : 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "X = 0.8;    \t\t\t #Quality of wet steam (no unit)\n",
+      "T = 150.;\t    \t\t #Temperature of the wet steam in degree celsius\n",
+      "vf = 0.0010908;\t\t\t #molar volume of saturated liquid in m**3/kg\n",
+      "vg = 0.3924;\t\t\t #molar volume of saturated vapour in m**3/kg\n",
+      "uf = 631.63;\t\t\t #molar internal energy of the saturated liquid in kJ/kg\n",
+      "ug = 2558.6;\t\t\t #molar internal energy of the saturated vapour in kJ/kg\n",
+      "\n",
+      "# Calculations\n",
+      "V = (X*vg)+((1-X)*vf);\t\t\t # Calculations of specific volume in m**3/kg using Eq.(3.3)\n",
+      "U = (X*ug)+((1-X)*uf);\t\t\t # Calculations of specific internal energy in kJ/kg using Eq.(3.6)\n",
+      "\n",
+      "# Results\n",
+      "print 'The specific volume of wet steam =  %0.4f m**3/kg '%(V);\n",
+      "print 'The specific internal energy of wet steam =  %0.1f kJ/kg '%(U);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The specific volume of wet steam =  0.3141 m**3/kg \n",
+        "The specific internal energy of wet steam =  2173.2 kJ/kg \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.2  Page No : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 1.42;\t\t\t #specific volume of wet steam in m**3/kg\n",
+      "T = 100.;\t\t\t #temperature of wet steam in degree celsius\n",
+      "vf = 0.0010437;\t\t #molar volume of saturated liquid in m**3/kg\n",
+      "vg = 1.673;\t\t\t #molar volume of saturated vapour in m**3/kg\n",
+      "\n",
+      "# Calculations\n",
+      "X = (V-vf)/(vg-vf);\t\t\t # Calculations of the quality of wet steam using Eq.(3.3) (no unit)\n",
+      "\n",
+      "# Results\n",
+      "print ' The quality of wet steam =  %0.4f '%(X);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The quality of wet steam =  0.8487 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.3  Page No : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 100.;\t\t\t #temperature inside the vessel in degree celsius\n",
+      "V = 0.00317;\t\t\t #specific volume of water at the critical point in m**3/kg\n",
+      "vf = 0.0010437;\t\t\t #molar volume of saturated liquid in m**3/kg\n",
+      "vg = 1.673;\t\t\t #molar volume of saturated vapour in m**3/kg\n",
+      "\n",
+      "# Calculations\n",
+      "X = (V-vf)/(vg-vf);\t\t\t          # Calculations of the quality of wet steam using Eq.(3.3) (no unit)\n",
+      "ratio = (X*vg)/((1-X)*vf);\t\t\t # Calculations of volume ratio of saturated vapour to the saturated liquid (no unit)\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume ratio of saturated vapour to the saturated liquid =  %0.2f '%(ratio);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume ratio of saturated vapour to the saturated liquid =  2.04 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.4  Page No : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "U = 2000.\t\t\t #specific internal energy of liquid-vapour mixture in kJ/kg\n",
+      "uf = 850.6\t\t\t #specific internal energy of saturated liquid in kJ/kg\n",
+      "ug = 2593.2\t\t\t #specific internal energy of saturated vapour in kJ/kg\n",
+      "\n",
+      "# Calculations\n",
+      "mass_ratio = (U-uf)/(ug-U);\t\t\t # Calculations of the mass ratio of vapour to liquid using the lever rule (no unit)\n",
+      "\n",
+      "# Results\n",
+      "print ' The mass ratio of vapour to liquid =  %0.4f '%(mass_ratio);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mass ratio of vapour to liquid =  1.9376 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.5  Page No : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 1.\t\t\t #number of moles of n-octane vapour in mol\n",
+      "T = 427.85;\t\t\t #tempearture of n-octane vapour in K\n",
+      "P = 0.215;\t\t\t #pressure n-octane vapour in MPa\n",
+      "R = 8.314;\t\t\t #universal gas constant in (kPa m**3)/(kmol K)\n",
+      "\n",
+      "# Calculations\n",
+      "V = ((n*10**-3)*R*T)/(P*10**3);\t\t\t # Calculations of volume using ideal gas law-Eq.(3.9) in m**3\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane vapour =  %f m**3 '%(V);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane vapour =  0.016545 m**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.6  Page No : 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 1.; \t\t\t #number of moles occupied by n-octane vapour in mol\n",
+      "T = 427.85;\t\t\t #temperature in K\n",
+      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
+      "a = 3.789;\t\t\t #van der Waals constant in Pa(m**3/mol)**2\n",
+      "b = 2.37*10**-4;\t #van der Waals constant in m**3/mol\n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "\n",
+      "# Calculations\n",
+      "Vguess = (n*R*T)/(P*10**6);\t\t\t \n",
+      "Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b\n",
+      "tolerance = 1e-6\n",
+      "\n",
+      "while abs(Vguess-Vnew)>tolerance:\n",
+      "    Vguess = Vnew;\n",
+      "    Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b;\t\t\t #the iteration process to solve the system of equation\n",
+      "\n",
+      "V = Vnew;\t\t\t #The volume calculated using the van der Waals equation in m**3/mol\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane vapour obtained by van der Waals equation =  %f m**3/mol'%(V);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane vapour obtained by van der Waals equation =  0.015675 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.7  Page No : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 1.; \t\t\t #number of moles occupied by n-octane liquid in mol\n",
+      "T = 427.85;\t\t\t #temperature in K\n",
+      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
+      "a = 3.789;\t\t\t #van der Waals constant in Pa(m**3/mol)**2\n",
+      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "\n",
+      "# Calculations\n",
+      "Vguess = b\n",
+      "Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b\n",
+      "tolerance = 1e-6\n",
+      "while abs(Vguess-Vnew)>tolerance:\n",
+      "    Vguess = Vnew;\n",
+      "    Vnew = (R*T)/((P*10**6)+(a/Vguess**2))+b;\t\t\t #the iteration process to solve the system of equation\n",
+      "\n",
+      "V = Vnew;\t\t\t #The volume calculated using the van der Waals equation in m**3/mol\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane liquid obtained by van der Waals equation =  %e m**3/mol'%(V);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane liquid obtained by van der Waals equation =  3.517652e-04 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.8  Page No : 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 427.85; \t\t    \t #temperature in K\n",
+      "P = 0.215;\t    \t\t     #saturation pressure in MPa\n",
+      "a = 3.789;\t\t    \t     #van der Waals constant in Pa(m**3/mol)**2\n",
+      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
+      "R = 8.314;\t\t        \t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "Pcvc = 3\n",
+      "RTc = 8.\n",
+      "# Calculations\n",
+      "#The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily\n",
+      "#The general form of the cubic equation is (Z**3)+(alpha*Z**2)+(beeta*Z)+gaamma = 0, where alpha,beeta and gaamma are determined using established relations\n",
+      "\n",
+      "A = (a*P*10**6)/(R*T)**2;\t\t\t\n",
+      "B = (b*P*10**6)/(R*T);\t\t\t \n",
+      "alpha = -1-B;\t\t\t \n",
+      "beeta = A;\t\t\t \n",
+      "gaamma = -(A*B);\t\n",
+      "p = round(beeta-(alpha**2)/3,4)\n",
+      "q = round(((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma,4)\n",
+      "D = (q**2/4)+(p**3/27);\t\n",
+      "\n",
+      "\n",
+      "if D>0 :\n",
+      "        Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);\n",
+      "        vf = ((Z)*R*T)/(P*10**6);\t\t\n",
+      "        vg = ((Z)*R*T)/(P*10**6);\t\t\n",
+      "elif D == 0:\n",
+      "        Z1 = ((-2*(q/2))**(1./3))-(alpha/3);\t\n",
+      "        Z2 = ((q/2)**(1./3))-(alpha/3);\n",
+      "        Z3 = ((q/2)**(1./3))-(alpha/3);\n",
+      "        Z = [Z1 ,Z2, Z3];\n",
+      "        vf = (min(Z)*R*T)/(P*10**6)\n",
+      "        vg = (max(Z)*R*T)/(P*10**6)\n",
+      "else:\n",
+      "        r = math.sqrt((-(p**3)/27));\t\t\t # Calculations of r using Eq.(3.38)\n",
+      "        theta = math.cos((-(q)/2)*(1./r));\t\t\t # Calculations of theta in radians using Eq.(3.37)\n",
+      "        Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "        Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3);\t\t\t #Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)\n",
+      "        Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
+      "        Z = [Z1, Z2, Z3];\n",
+      "        vf = (min(Z)*R*T)/(P*10**6)\n",
+      "        vg = (max(Z)*R*T)/(P*10**6)\n",
+      "\n",
+      "crit = Pcvc/RTc        \n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane saturated liquid obtained by Cardans method =  %.3e m**3/mol'%(vf);\n",
+      "print ' The volume occupied by n-octane saturated vapour obtained by Cardans method =  %.2e m**3/mol'%(vg);\n",
+      "print \" At the critical point, PcVc/RTc = %.3f\"%crit\n",
+      "\n",
+      "# Note: value of D is calculated wrongly in book. Please calculate manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane saturated liquid obtained by Cardans method =  -9.345e-04 m**3/mol\n",
+        " The volume occupied by n-octane saturated vapour obtained by Cardans method =  1.55e-02 m**3/mol\n",
+        " At the critical point, PcVc/RTc = 0.375\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.9  Page No : 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from numpy import *\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature in K\n",
+      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "\n",
+      "# Calculations\n",
+      "a = (0.42748*R**2*Tc**2.5)/(Pc*100*10**3*math.sqrt(T));\t\t\t \n",
+      "b = (0.08664*R*Tc)/(Pc*100*10**3);\t\t\t\n",
+      "\n",
+      "\n",
+      "A = (a*P*10**6)/(R*T)**2;\t\t\t \n",
+      "B = (b*P*10**6)/(R*T);\n",
+      "alpha = -1.;\t\t\t\n",
+      "beeta = A-B-B**2.;\t\n",
+      "gaamma = -(A*B);\t\n",
+      "p = beeta-(alpha**2)/3.\n",
+      "q = ((2*alpha**3.)/27.)-((alpha*beeta)/3)+gaamma\n",
+      "D = (((q)**2)/4.)+(((p)**3)/27.);\t\t\t \n",
+      "#if D>0 then:\n",
+      " #       Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
+      "  #      vf = ((Z)*R*T)/(P*10**6)\n",
+      "   #     vg = ((Z)*R*T)/(P*10**6)\n",
+      "#elif D == 0:\n",
+      " #       Z1 = ((-2*(q/2))**(1./3))-(alpha/3);\n",
+      "  #      Z2 = ((q/2)**(1./3))-(alpha/3);\n",
+      "   #     Z3 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    #    Z = [Z1 ,Z2, Z3];\n",
+      "     #   vf = (min(Z)*R*T)/(P*10**6);\t\t\n",
+      "      #  vg = (max(Z)*R*T)/(P*10**6);\t\t\n",
+      "#else:\n",
+      "\n",
+      "r = math.sqrt((-(p**3)/27));\t\t\t\n",
+      "theta = (math.acos((-(q)/2)*(1./r)));\n",
+      "Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3);\t\n",
+      "Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
+      "Z = [Z1, Z2, Z3];\n",
+      "vf = (min(Z)*R*T)/(P*10**6);\t\t\n",
+      "vg = (max(Z)*R*T)/(P*10**6);\t\t\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane saturated vapour) using Redlich-Kwong equation of state =  %f m**3/mol'%(vg);\n",
+      "print ' The volume occupied by n-octane saturated liquid) using Redlich-Kwong equation of state =  %f m**3/mol'%(vf);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane saturated vapour) using Redlich-Kwong equation of state =  0.015401 m**3/mol\n",
+        " The volume occupied by n-octane saturated liquid) using Redlich-Kwong equation of state =  0.000244 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.10  Page No : 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 180.;\t\t\t #temperature of water in degree celsius\n",
+      "P = 1.0027;\t\t\t #saturation pressure of water in MPa\n",
+      "Tc = 647.3;\t\t\t #critical temperature of water in K\n",
+      "Pc = 221.2;\t\t\t #critical pressure of water in bar\n",
+      "Tr = 0.7;\t\t\t #reduced temperature at which acentric factor was defined by Pitzer \n",
+      "\n",
+      "# Calculations\n",
+      "T1 = Tr*Tc; \t\t\t #calculating temperature  in K using reduced temperature value\n",
+      "T1 = T1-273.15;\t\t\t #conversion to degree celsius\n",
+      "Pr = (P*10)/Pc;\t\t\t # Calculations of reduced pressure (no unit) using saturation pressure at t1. In this case t1 equals t, therefore the given saturation pressure is taken\n",
+      "w = -math.log10(Pr)-1.0;\t\t\t # Calculations of acentric factor using Eq.(3.62)\n",
+      "\n",
+      "# Results\n",
+      "print ' The acentric factor of water =  %f '%(w);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The acentric factor of water =  0.343614 \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.11  Page No : 72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 409.41;\t\t\t #temperature of n-octane in degree celsius\n",
+      "P = 4.98;\t\t\t #pressure in bar\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
+      "w = 0.398;\t\t\t #acentric factor (no unit)\n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "\n",
+      "# Calculations\n",
+      "Tr = (T+273.15)/Tc;\t\n",
+      "Pr = P/Pc;\t\t\t\n",
+      "z0 = 0.96;\t\t\t\n",
+      "V1 = (z0*R*(T+273.15))/(P*10**5);\t\t\t \n",
+      "z1 = 0.01;\t\t\t \n",
+      "z = z0+(w*z1);\t\t\n",
+      "V2 = (z*R*(T+273.15))/(P*10**5)\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane obtained by the two parameter compressibilty \\\n",
+      " factor correlation =  %f m**3/mol'%(V1);\n",
+      "print ' The volume occupied by n-octane obtained by the three parameter compressibility \\\n",
+      "factor correlation =  %f m**3/mol'%(V2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane obtained by the two parameter compressibilty  factor correlation =  0.010939 m**3/mol\n",
+        " The volume occupied by n-octane obtained by the three parameter compressibility factor correlation =  0.010985 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.12  Page No : 72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 1.;\t    \t\t #volume of the tank in m**3\n",
+      "m = 180.;\t\t\t #mass of carbon dioxide in kg\n",
+      "T = 91.8;\t\t\t #temperature of the tank in degree celsius after it is placed in the vicinity of a furnace\n",
+      "Tc = 304.2;\t\t\t #critical temperature of carbon dioxide in K\n",
+      "Pc = 73.87;\t\t\t #critical pressure of carbon dioxide in bar\n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "\n",
+      "# Calculations\n",
+      "mwt = 44.*10**-3\t #molecular weight of carbon dioxide in kg/mol\n",
+      "n = m/mwt;\t\t\t # Calculations of number of moles of carbon dioxide in the tank in moles\n",
+      "MV = V/n;\t\t\t # Calculations of molar volume in m**3/mol\n",
+      "slope = (MV*Pc*10**5)/(R*(T+273.15));\t\t\t #slope of the straight line formed when z0 is plotted against Pr formed by using the relation z0 = (V*Pc*Pr)/(R*T)\n",
+      "Tr = (T+273.15)/Tc;\t\t\t # Calculations of reduced temperature (no unit)\n",
+      "\n",
+      "#At the given conditions, the state of CO2 must lie on the curve corresponding to the obtained value of Tr. To determine the state of CO2, a straight line passing through the origin, with the obtained value of slope is drawn on the z0 vs Pr plot of Fig.3.12 and the point of intersection of this straight line with the Tr curve is obtained to get the value of z0\n",
+      "z0 = 0.725;\t\t\t #value of compressibilty factor obtained by doing the above mentioned procedure\n",
+      "P = (z0*R*10**-6*(T+273.15))/(MV)\t\t\t # Calculations of pressure in MPa using Eq.(3.59)\n",
+      "\n",
+      "# Results\n",
+      "print ' The pressure developed by carbon dioxide =  %.0f MPa'%(P);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The pressure developed by carbon dioxide =  9 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.13  Page No : 73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "V = 1.;\t\t\t #volume of the tank in m**3\n",
+      "m = 180.;\t\t\t #mass of carbon dioxide in kg\n",
+      "T = 91.8;\t\t\t #temperature of the tank in degree celsius after it is placed in the vicinity of a furnace\n",
+      "Tc = 304.2;\t\t\t #critical temperature of carbon dioxide in K\n",
+      "Pc = 73.87;\t\t\t #critical pressure of carbon dioxide in bar\n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "w = 0.239;\t\t\t #acentric factor (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "mwt = 44.*10**-3\n",
+      "n = m/mwt\n",
+      "MV = V/n;\t\t\t\n",
+      "Tr = (T+273.15)/Tc\n",
+      "z0 = 0.725\n",
+      "z_init = z0;\t\t\t \n",
+      "slope = (MV*Pc*10**5)/(R*(T+273.15));\t\t\t \n",
+      "Prguess = z_init/slope;\t\n",
+      "z1guess = 0.1;\t\t\t \n",
+      "tolerance = 1e-6;\t\t\t \n",
+      "\n",
+      "def solver_function(zi):\n",
+      "    return zi-(z0+(w*z1guess))\n",
+      "\n",
+      "z = fsolve(solver_function,z1guess)\n",
+      "Pr = z/slope;\t\t\t \n",
+      "P = ((Pc*10**5)*Pr)*10**-6\n",
+      "\n",
+      "# Results\n",
+      "print ' The pressure developed by carbon dioxide =  %f MPa'%(P);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The pressure developed by carbon dioxide =  9.295806 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.14  Page No : 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215;\t\t\t #pressure of n-ocatne vapour in MPa\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "Pc = 2.497;\t\t\t #critical pressure of n-octane in MPa\n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "\n",
+      "# Calculations\n",
+      "Tr = T/Tc\n",
+      "Pr = P/Pc\n",
+      "z_init = 1\n",
+      "h = (0.08664*Pr)/(z_init*Tr)\n",
+      "tolerance = 1e-6;\t\t\t\n",
+      "\n",
+      "def solver_func(zi):\n",
+      "    return zi-((1./(1-h))-((h/(1+h))*(4.93398/(Tr**(3./2)))))\n",
+      "\n",
+      "z = fsolve(solver_func,h)\n",
+      "V = (z*R*T)/(P*10**6);\t\t\t\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane vapour obtained by the generalized form of \\\n",
+      " \\n Redlich-Kwong equation of state =  %.2e m**3/mol'%(V);\n",
+      "\n",
+      "\n",
+      "# Note : Answer may vary because of rouding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane vapour obtained by the generalized form of  \n",
+        " Redlich-Kwong equation of state =  1.55e-02 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.15  Page No : 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature in K\n",
+      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
+      "w = 0.398;\t\t\t #acentric factor (no unit) \n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "\n",
+      "# Calculations\n",
+      "#The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily\n",
+      "#The general form of the cubic equation is (Z**3)+(alpha*Z**2)+(beeta*Z)+gaamma = 0, where alpha,beeta and gaamma are determined using established relations\n",
+      "\n",
+      "Tr = T/Tc\n",
+      "Pr = (P*10**6)/(Pc*10**5)\n",
+      "S = 0.48+(1.574*w)-(0.176*w**2)\n",
+      "alpha1 = (1+(S*(1-math.sqrt(Tr))))**2\n",
+      "a = (0.42748*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
+      "b = (0.08664*R*Tc)/(Pc*10**5)\n",
+      "A = (a*P*10**6)/(R*T)**2\n",
+      "B = (b*P*10**6)/(R*T)\n",
+      "alpha = -1.\n",
+      "beeta = A-B-B**2\n",
+      "gaamma = -(A*B)\n",
+      "p = beeta-(alpha**2)/3\n",
+      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
+      "D = (((q)**2)/4.)+(((p)**3)/27.)\n",
+      "\n",
+      "if D>0:\n",
+      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
+      "    vf = ((Z)*R*T)/(P*10**6)\n",
+      "    vg = ((Z)*R*T)/(P*10**6)\n",
+      "elif D == 0:\n",
+      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
+      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z = [Z1, Z2, Z3];\n",
+      "    vf = (min(Z)*R*T)/(P*10**6);\t\t\n",
+      "    vg = (max(Z)*R*T)/(P*10**6);\t\t\n",
+      "else:\n",
+      "    r = math.sqrt((-(p**3)/27));\t\t\n",
+      "    theta = acos((-(q)/2)*(1./r));\t\t\n",
+      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
+      "    Z = [Z1, Z2, Z3];\n",
+      "    vf = (min(Z)*R*T)/(P*10**6)\n",
+      "    vg = (max(Z)*R*T)/(P*10**6)\n",
+      "\n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane saturated vapour obtained by Soave-Redlich-Kwong \\\n",
+      " equation of state =  %f m**3/mol'%(vg);\n",
+      "print ' The volume occupied by n-octane saturated liquid obtained by Soave-Redlich-Kwong \\\n",
+      " equation of state =  %f m**3/mol'%(vf);\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane saturated vapour obtained by Soave-Redlich-Kwong  equation of state =  0.015207 m**3/mol\n",
+        " The volume occupied by n-octane saturated liquid obtained by Soave-Redlich-Kwong  equation of state =  0.000227 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 74
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.16  Page No : 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature in K\n",
+      "P = 0.215;\t\t\t #saturation pressure in MPa\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
+      "w = 0.398;\t\t\t #acentric factor (no unit) \n",
+      "R = 8.314;\t\t\t #universal gas constant in (Pa m**3)/(mol K)\n",
+      "\n",
+      "# Calculations\n",
+      "Tr = T/Tc;\t\t\t\n",
+      "Pr = (P*10**6)/(Pc*10**5)\n",
+      "S = 0.37464+(1.54226*w)-(0.26992*w**2)\n",
+      "alpha1 = (1+(S*(1-math.sqrt(Tr))))**2\n",
+      "a = (0.45724*R**2*Tc**2*alpha1)/(Pc*10**5)\n",
+      "b = (0.07780*R*Tc)/(Pc*10**5)\n",
+      "A = (a*P*10**6)/(R*T)**2\n",
+      "B = (b*P*10**6)/(R*T)\n",
+      "alpha = -1+B;\t\n",
+      "beeta = A-(2*B)-(3*B**2)\n",
+      "gaamma = -(A*B)+(B**2)+(B**3)\n",
+      "p = beeta-(alpha**2)/3;\t\t\n",
+      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
+      "D = (((q)**2)/4)+(((p)**3)/27)\n",
+      "\n",
+      "if D>0:\n",
+      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
+      "    vf = ((Z)*R*t)/(P*10**6)\n",
+      "    vg = ((Z)*R*t)/(P*10**6)\n",
+      "elif D == 0:\n",
+      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
+      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z = [Z1 ,Z2, Z3];\n",
+      "    vf = (min(Z)*R*T)/(P*10**6)\n",
+      "    vg = (max(Z)*R*T)/(P*10**6)\n",
+      "else:\n",
+      "    r = math.sqrt((-(p**3)/27))\n",
+      "    theta = .1533\n",
+      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3);\t\t\n",
+      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
+      "    Z = [Z1, Z2, Z3];\n",
+      "    vf = (min(Z)*R*T)/(P*10**6)\n",
+      "    vg = (max(Z)*R*T)/(P*10**6)\n",
+      "    \n",
+      "# Results\n",
+      "print ' The volume occupied by n-octane saturated vapour, obtained by Peng\\\n",
+      "-Robinson equation of state =  %.3e m**3/mol'%(vg);\n",
+      "print ' The volume occupied by n-octane saturated liquid, obtained by Peng-Robinson\\\n",
+      " equation of state =  %.3e m**3/mol'%(vf);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume occupied by n-octane saturated vapour, obtained by Peng-Robinson equation of state =  1.514e-02 m**3/mol\n",
+        " The volume occupied by n-octane saturated liquid, obtained by Peng-Robinson equation of state =  2.003e-04 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch4.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch4.ipynb
new file mode 100755
index 00000000..5433cf7c
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch4.ipynb
@@ -0,0 +1,1291 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : First law of thermodynamics and its applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.1  Page No : 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Q1 = 50.;\t\t\t #Energy added as heat in kJ when the system undergoes a process 1-2\n",
+      "W1 = 30.;\t\t\t #Work done by the system in kJ during the process 1-2\n",
+      "Q2 = -40.;\t\t\t #Energy rejected as heat in kJ during the process 2-3\n",
+      "W2 = -50.;\t\t\t #Work done on the system in kJ during the process 2-3\n",
+      "Q3 = 0.\t    \t\t #System undergoes an adiabatic process to return to initial state\n",
+      "\n",
+      "# Calculations\n",
+      "U2_1 = Q1-W1\n",
+      "U3_2 = Q2-W2\n",
+      "U1_3 = (-U2_1)-(U3_2)\n",
+      "W3 = Q3-U1_3\n",
+      "net_work = W1+W2+W3\n",
+      "\n",
+      "# Results\n",
+      "print ' The net work done by the system =  %d kJ'%(net_work);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The net work done by the system =  10 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.3  Page No : 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 1.  \t\t\t #volume of tank in m**3;\n",
+      "N = 200.;\t\t\t #number of moles of carbon dioxide in tank in mol\n",
+      "T1 = 25.;\t\t\t #ambient temperature in degree celsius\n",
+      "I = 5.; \t\t\t #current in amperes passed through the resistor place in the tank\n",
+      "Voltage = 440.\t\t #voltage in volts existing across the resistor\n",
+      "t = 10.;\t\t\t #time in minutes for which the current is passed\n",
+      "a = 363.077*10**-3\t\t\t #van der waals constant in Pa (m**3/mol)**2\n",
+      "b = 0.043*10**-3\t\t\t #van der waals constant in m**3/mol\n",
+      "Cv = 32.34\t\t\t #molar heat capacity at constant volume in J/molK\n",
+      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "MV = V/N;\t\t\t # Calculations of molar volume in m**3/mol\n",
+      "Q = 0.  \t\t\t #energy transfer as heat during the process\n",
+      "W_Pdv = 0.;\t\t\t #mechanical work done by the system\n",
+      "W_elec = -(Voltage*I*t*60)*(10**-6)\n",
+      "U2_1 = Q-(W_Pdv+W_elec);\t\t\t\n",
+      "T2 = ((U2_1*10**6)/(N*Cv))+(T1+273.15)\n",
+      "P = (((R*T2)/(MV-b))-(a/(MV**2)))*10**-3\n",
+      "\n",
+      "# Results\n",
+      "print ' The final pressure =  %0.3f kPa '%(P);\n",
+      "print ' The final temperature =  %0.2f K'%(T2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final pressure =  827.832 kPa \n",
+        " The final temperature =  502.23 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.4  Page No : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 0.1;\t\t\t #volume of tank in m**3\n",
+      "T1 = 200.;\t\t\t #initial temperature of saturated steam inside the tank in degree celsius\n",
+      "T2 = 150.;\t\t\t #temperature in degree celsius that the tank attains after some time due to poor insulation\n",
+      "P1 = 15.549;\t\t\t #pressure in bar obtained from steam tables corresponding to T1\n",
+      "vg1 = 0.1272;\t\t\t #specific volume of saturated vapour in m**3/kg obtained from steam tables corresponding to T1\n",
+      "hg1 = 2790.9;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to T1\n",
+      "P2 = 4.76;\t\t\t #pressure in bar obtained from steam tables corresponding to T2\n",
+      "vf = 0.0010908;\t\t\t #specific volume of saturated liquid in m**3/kg obtained from steam tables corresponding to T2\n",
+      "vg2 = 0.3924;\t\t\t #specific volume of saturated vapour in m**3/kg obtained from steam tables corresponding to T2\n",
+      "hf = 632.15;\t\t\t #specific enthalpy of saturated liquid in kJ/kg obtained from steam tables corresponding to T1\n",
+      "hg2 = 2745.4;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to T1\n",
+      "\n",
+      "# Calculations\n",
+      "ug1 = ((hg1*10**3)-(P1*10**5*vg1))*10**-3\n",
+      "uf = ((hf*10**3)-(P2*10**5*vf))*10**-3\n",
+      "ug2 = ((hg2*10**3)-(P2*10**5*vg2))*10**-3\n",
+      "v2 = vg1\n",
+      "X2 = (v2-vf)/(vg2-vf)\n",
+      "u2 = (X2*ug2)+((1-X2)*uf)\n",
+      "m = V/vg1\n",
+      "Q = m*(u2-ug1)\n",
+      "mf = m*(1-X2)\n",
+      "mg = m*X2\n",
+      "\n",
+      "# Results\n",
+      "print ' The energy transferred as heat =  %f kJ'%(Q);\n",
+      "print ' The mass of liquid in the tank in the final state =  %0.3f kg'%(mf)\n",
+      "print ' The mass of vapour in the tank in the final state =  %0.3f kg'%(mg)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The energy transferred as heat =  -1053.825572 kJ\n",
+        " The mass of liquid in the tank in the final state =  0.533 kg\n",
+        " The mass of vapour in the tank in the final state =  0.253 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.5  Page No : 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "W = 1.;\t    \t\t #weight of steam in kg in the piston cylinder assembly\n",
+      "X = 0.8;\t\t\t #quality of steam (no unit)\n",
+      "T1 = 150.;\t\t\t #initial temperature of steam in degree celsius\n",
+      "T2 = 200.;\t\t\t #final temperature of steam in degree celsius\n",
+      "P1 = 476.;\t\t\t #pressure in kPa obatined from steam tables (corresponding to T1)\n",
+      "vf = 0.0010908;\t\t\t #specific volume of saturated liquid in m**3/kg obatined from steam tables (corresponding to T1)\n",
+      "vg = 0.3924;\t\t\t #specific volume of satuarted vapour in m**3/kg obatined from steam tables (corresponding to T1)\n",
+      "hf = 632.15;\t\t\t #specific enthalpy of saturated liquid in kJ/kg obtained from steam tables (corresponding to T1)\n",
+      "hg = 2745.4;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables (corresponding to T1)\n",
+      "\n",
+      "# Calculations\n",
+      "V1 = (X*vg)+((1-X)*vf)\n",
+      "h1 = (X*hg)+((1-X)*hf)\n",
+      "P2 = 0.476;\t\t\t \n",
+      "P_int1 = 0.4;\t\t\n",
+      "P_int2 = 0.5;\t\t\n",
+      "V_int1 = 0.5343;\t\n",
+      "V_int2 = 0.4250;\t\n",
+      "h_int1 = 2860.4;\t\n",
+      "h_int2 = 2855.1;\t\n",
+      "V2 = (((P2-P_int1)/(P_int2-P_int1))*(V_int2-V_int1))+V_int1\n",
+      "h2 = (((P2-P_int1)/(P_int2-P_int1))*(h_int2-h_int1))+h_int1\n",
+      "Q = (h2-h1)*W\n",
+      "W = P1*(V2-V1)*W\n",
+      "\n",
+      "# Results\n",
+      "print ' The work done by steam =  %0.2f kJ '%(W);\n",
+      "print ' The net energy transferred as heat  =  %0.2f kJ'%(Q);\n",
+      "print ' The final state of superheated steam,Pressure = %0.3f MPa '%(P2);\n",
+      "print ' The final state of superheated steam,Temperature = %d degree celsius '%(T2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The work done by steam =  65.26 kJ \n",
+        " The net energy transferred as heat  =  533.62 kJ\n",
+        " The final state of superheated steam,Pressure = 0.476 MPa \n",
+        " The final state of superheated steam,Temperature = 200 degree celsius \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.6  Page No : 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "W = 1.  \t\t\t #weight of steam in kg in the piston cylinder assembly\n",
+      "X = 0.8;\t\t\t #quality of steam (no unit)\n",
+      "T1 = 150.\t\t\t #initial temperature  of steam in degree celsius\n",
+      "I = 5.  \t\t\t #current passed in Amperes\n",
+      "V = 220.\t\t\t #voltage in volts across the resistor\n",
+      "t = 10. \t\t\t #time for which the current is passed in minutes\n",
+      "P1 = 476.\t\t\t #pressure in kPa obatined from steam tables (corresponding to T1)\n",
+      "vf = 0.0010908;\t\t\t #specific volume of saturated liquid in m**3/kg obatined from steam tables (corresponding to T1)\n",
+      "vg = 0.3924;\t\t\t #specific volume of satuarted vapour in m**3/kg obatined from steam tables (corresponding to T1)\n",
+      "hf = 632.15;\t\t\t #specific enthalpy of saturated liquid in kJ/kg obtained from steam tables (corresponding to T1)\n",
+      "hg = 2745.4;\t\t\t #specific enthalpy of saturated vapour in kJ/kg obtained from steam tables (corresponding to T1)\n",
+      "\n",
+      "# Calculations\n",
+      "V1 = (X*vg)+((1-X)*vf);\t\n",
+      "h1 = (X*hg)+((1-X)*hf);\t\n",
+      "Ws = -V*I*t*60*10**-3;\t\n",
+      "h2 = h1-Ws;\t\t\t \n",
+      "P2 = 0.476;\t\t\t \n",
+      "T_int1 = 200;\t\t\n",
+      "T_int2 = 300;\t\t\n",
+      "V_int1 = 0.4512;\t\n",
+      "V_int2 = 0.5544;\t\n",
+      "h_int1 = 2856.37;\t\n",
+      "h_int2 = 3065.38;\t\n",
+      "V2 = (((h2-h_int1)/(h_int2-h_int1))*(V_int2-V_int1))+V_int1;\n",
+      "T2 = (((h2-h_int1)/(h_int2-h_int1))*(T_int2-T_int1))+T_int1;\n",
+      "W = (P1*10**3*(V2-V1)*W)*10**-3\n",
+      "\n",
+      "# Results\n",
+      "print ' The work done by steam =  %0.2f kJ '%(W);\n",
+      "print ' The final temperature =  %0.2f degree celsius'%(T2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The work done by steam =  94.94 kJ \n",
+        " The final temperature =  260.47 degree celsius\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.7  Page No : 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "N = 1.\t\t\t     #number of moles of carbon dioxide in kmol\n",
+      "T1 = 298.\t\t\t #initial temperature in K\n",
+      "T2 = 600.\t\t\t #final raised temperature in K\n",
+      "a = 45.369\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "b = 8.688*10**-3\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "e = -9.619*10**5\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "\n",
+      "# Calculations\n",
+      "Q = N*10**3*((a*(T2-T1)+((b/2)*(T2**2-T1**2))-(e*((1./T2)-(1./T1)))))*10**-6\n",
+      "\n",
+      "# Results\n",
+      "print ' The amount of energy to be transferred as heat =  %0.3f MJ'%(Q);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The amount of energy to be transferred as heat =  13.255 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.8  Page No : 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T1 = 298.   \t\t\t #initial temperature in K\n",
+      "T2 = 600.\t    \t\t #final raised temperature in K\n",
+      "a = 45.369;\t\t    \t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "b = 8.688*10**-3;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "e = -9.619*10**5;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "\n",
+      "# Calculations\n",
+      "Cpm = ((a*(T2-T1))+((b/2)*(T2**2-T1**2))-(e*((1./T2)-(1./T1))))/(T2-T1)\n",
+      "\n",
+      "# Results\n",
+      "print ' The isobaric molar heat capacity =  %0.2f J/molK'%(Cpm);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The isobaric molar heat capacity =  43.89 J/molK\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.9  Page No : 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "N = 1.\t    \t\t #number of moles of carbon dioxide in kmol\n",
+      "T1 = 298.\t\t\t #initial temperature in K\n",
+      "T2 = 600.\t\t\t #final raised temperature in K\n",
+      "a = 45.369;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "b = 8.688*10**-3;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "e = -9.619*10**5;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "\n",
+      "# Calculations\n",
+      "Cpm = ((a*(T2-T1))+((b/2)*(T2**2-T1**2))-(e*((1./T2)-(1./T1))))/(T2-T1)\n",
+      "Q = N*10**3*Cpm*(T2-T1)*10**-6\n",
+      "\n",
+      "# Results\n",
+      "print ' The amount of energy to be transferred as heat =  %0.3f MJ '%(Q);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The amount of energy to be transferred as heat =  13.255 MJ \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.10  Page No : 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "N = 100.\t\t\t #number of moles of carbon dioxide in mol\n",
+      "T1 = 298.\t\t\t #initial temperature in K\n",
+      "Q = 1.  \t\t\t #energy added as heat in MJ\n",
+      "a = 45.369;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "b = 8.688*10**-3;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "e = -9.619*10**5;\t\t\t #coefficient in the specific heat capacity expression where Cp = a+bT+eT**-2\n",
+      "\n",
+      "# Calculations\n",
+      "delh = Q*10**6/N\n",
+      "Tguess = 520.\t\n",
+      "Cpm_guess = a+(b*((T1+Tguess)/2))+(e/(T1*Tguess))\n",
+      "T2_guess = T1+(delh/Cpm_guess)\n",
+      "tolerance = 1e-6;\t\t\t\n",
+      "while abs(T2_guess-Tguess)>tolerance:\n",
+      "    Tguess = T2_guess;\n",
+      "    Cpm_guess = a+(b*((T1+Tguess)/2))+(e/(T1*Tguess));\n",
+      "    T2_guess = T1+(delh/Cpm_guess)\n",
+      "\n",
+      "T2 = T2_guess\n",
+      "\n",
+      "# Results\n",
+      "print ' The final temperature =  %0.1f K'%(T2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final temperature =  531.1 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.11  Page No : 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "volume_ratio = 1./15\t #final volume to the initial volume of air at the end of compression stroke (no unit)\n",
+      "gaamma = 1.4\t\t\t #ratio of the molar heat capacities at constant pressure and constant volume for air (no unit)\n",
+      "T1 = 300.\t\t\t     #initial temperature of air in K\n",
+      "P1 = 0.1;\t\t\t    #initial pressure of air in MPa\n",
+      "R = 8.314;\t\t\t    #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "T2 = T1*((1./volume_ratio)**(gaamma-1))\n",
+      "P2 = P1*((1./volume_ratio)**(gaamma))\n",
+      "W = (R*(T1-T2)*10**-3)/(gaamma-1)\n",
+      "\n",
+      "# Results\n",
+      "print ' The final temperature =  %0.2f K'%(T2);\n",
+      "print ' The final pressure =  %0.4f MPa'%(P2);\n",
+      "print ' Work done per mole of air =  %0.3f kJ/mol'%(W);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final temperature =  886.25 K\n",
+        " The final pressure =  4.4313 MPa\n",
+        " Work done per mole of air =  -12.185 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.12  Page No : 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "volume_ratio = 1./15;\t\t\t #final volume to the initial volume of ideal gas at the end of compression (no unit)\n",
+      "T1 = 300.   \t\t\t #initial temperature of ideal gas in K\n",
+      "P1 = 0.1;\t    \t\t #initial pressure of ideal gas in MPa\n",
+      "R = 8.314;\t\t    \t #universal gas constant in J/molK\n",
+      "n = 1.2;\t\t\t     #index of expansion (no unit)\n",
+      "gaamma = 1.4;\t\t\t #ratio of the molar heat capacities at constant pressure and constant volume for ideal gas (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "P2 = P1*((1./volume_ratio)**n)\n",
+      "T2 = T1*(P2/P1)*(volume_ratio)\n",
+      "W = (R*(T1-T2)*10**-3)/(n-1)\n",
+      "del_u = (R*(T2-T1)*10**-3)/(gaamma-1)\n",
+      "q = del_u+W\n",
+      "\n",
+      "# Results\n",
+      "print ' The final pressure =  %0.3f MPa'%(P2);\n",
+      "print ' The final temperature =  %0.1f K'%(T2);\n",
+      "print ' Work done on the gas =  %f kJ/mol'%(W);\n",
+      "print ' Heat interaction during the process =  %f kJ/mol'%(q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final pressure =  2.578 MPa\n",
+        " The final temperature =  515.6 K\n",
+        " Work done on the gas =  -8.963805 kJ/mol\n",
+        " Heat interaction during the process =  -4.481902 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.13  Page No : 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 1.;\t    \t\t #volume of tank in m**3\n",
+      "T0 = 300.\t\t\t #initial temperature of ideal gas in K\n",
+      "P0 = 0.1\t\t\t #initial pressure of ideal gas in MPa\n",
+      "T = 500.\t\t\t #temperature of ideal gas in the pipeline in K\n",
+      "P = 3.\t\t    \t #pressure of ideal gas in the pipeline in MPa\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "gaamma = 1.4;\t\t\t #ratio of the molar heat capacities at constant pressure and constant volume for ideal gas (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "Pf = 3.\n",
+      "Tf = (Pf*10**6)/((((Pf*10**6)-(P0*10**6))/(gaamma*T))+((P0*10**6)/T0));\n",
+      "N = (V/R)*(((Pf*10**6)/Tf)-((P0*10**6)/T0));\n",
+      "\n",
+      "# Results\n",
+      "print ' The final temperature =  %0.1f K'%(Tf);\n",
+      "print ' The amount of gas that has entered the tank =  %0.2f mol'%(N);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final temperature =  670.2 K\n",
+        " The amount of gas that has entered the tank =  498.30 mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.14  Page No : 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from scipy.optimize import fsolve\n",
+      "\n",
+      "# Variables\n",
+      "V = 3.  \t\t\t #volume of tank in m**3\n",
+      "T0 = 100.;\t\t\t #initial temperature of steam in degree celsius\n",
+      "T = 300.;\t\t\t #temperature of superheated steam in the pipeline in degree celsius\n",
+      "P = 3.;\t\t\t     #pressure of superheated steam in the pipeline in MPa\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Ps = 101.33\n",
+      "vg = 1.673\n",
+      "hg = 2676.0\n",
+      "h = 2995.1\n",
+      "u0 = ((hg*10**3)-(Ps*10**3*vg))*10**-3\n",
+      "m0 = V/vg;\t\t\t\n",
+      "\n",
+      "Tf = 418.\n",
+      "vf = 0.102329\n",
+      "uf = 2965.78\n",
+      "\n",
+      "mf_guess = V/vf\n",
+      "\n",
+      "def solver_func(ui):\n",
+      "    return (mf_guess*ui)-(m0*u0)-((mf_guess-m0)*h);\n",
+      "\n",
+      "uf_solved = fsolve(solver_func,mf_guess)\n",
+      "mf = mf_guess\n",
+      "mass = mf-m0\n",
+      "\n",
+      "# Results\n",
+      "print \" The final state of steamsuperheated, Pressure = %d MPa\"%(P);\n",
+      "print \" The final state of steamsuperheated, Temperature = %d degree celsius\"%(Tf);\n",
+      "print \" The mass of steam that entered the tank = %0.3f kg\"%(mass);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final state of steamsuperheated, Pressure = 3 MPa\n",
+        " The final state of steamsuperheated, Temperature = 418 degree celsius\n",
+        " The mass of steam that entered the tank = 27.524 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.15  Page No : 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 0.1;\t\t\t #volume of cylinder in m**3\n",
+      "T0 = 300.\t\t\t #initial temperature of nitrogen in K\n",
+      "P0 = 14.\t\t\t #initial pressure of nitrogen in MPa\n",
+      "P = 0.1;\t\t\t #ambient pressure in MPa\n",
+      "Pf = 2. \t\t\t #final pressure of nitrogen in MPa\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "gaamma = 1.4;\t\t #ratio of the molar heat capacities at constant pressure and constant volume for nitrogen (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "def solver_func(Ti):\n",
+      "    return ((P0*10**6)-(Pf*10**6))-((gaamma/2)*(T0+Ti)*(((P0*10**6)/T0)-((Pf*10**6)/Ti)));\n",
+      "\n",
+      "Tguess = 300.\n",
+      "Tf = fsolve(solver_func,Tguess)\n",
+      "\n",
+      "N = (V/R)*(((P0*10**6)/T0)-((Pf*10**6)/Tf));\n",
+      "\n",
+      "# Results\n",
+      "print ' The final temperature =  %0.1f K'%(Tf);\n",
+      "print ' The amount of gas that has escaped from the cylinder =  %0.2f mol'%(N);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The final temperature =  181.2 K\n",
+        " The amount of gas that has escaped from the cylinder =  428.52 mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.16  Page No : 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T0 = 300.;\t\t\t #initial temperature of superheated steam in degree celsius\n",
+      "P0 = 3.;\t\t\t #initial pressure of superheated steam in MPa\n",
+      "Xe = 0.85;\t\t\t #quality of steam leaving the turbine (no unit)\n",
+      "Tf = 45.;\t\t\t #final temperature of steam leaving the turbine in degree celsius\n",
+      "Vi = 10.;\t\t\t #velocity of steam at the entrance in m/s\n",
+      "Ve = 40.;\t\t\t #exit velocity of steam in m/s\n",
+      "Zi = 10.;\t\t\t #elevation at the entrance in m\n",
+      "Ze = 4.;\t\t\t #elevation at the exit in m\n",
+      "m = 1.;\t\t\t #mass flow rate of steam through turbine in kg/s\n",
+      "g = 9.81;\t\t\t #accleration due to gravity in m/s**2\n",
+      "\n",
+      "# Calculations\n",
+      "hi = 2995.1\n",
+      "hf = 188.35\n",
+      "hg = 2583.3\n",
+      "he = ((1-Xe)*hf)+(Xe*hg)\n",
+      "Q = 0.\t\t\t \n",
+      "enthalpy_change = (he*10**3)-(hi*10**3)\n",
+      "KE_change = ((Ve**2)-(Vi**2))/2\n",
+      "PE_change = g*(Ze-Zi);\t\t\t\n",
+      "Ws = Q-(m*(enthalpy_change+KE_change+PE_change)*10**-3)\n",
+      "err_KE = ((KE_change)/(Ws*10**3))*100;\t\n",
+      "err_PE = ((abs (PE_change)/(Ws*10**3)))*100\n",
+      "err = err_KE+err_PE\n",
+      "\n",
+      "# Results\n",
+      "print ' The percentage error when Kinetic energy change is ignored =  %0.3f '%(err_KE);\n",
+      "print ' The percentage error when Potential energy change is ignored =  %0.4f '%(err_PE);\n",
+      "print ' The percentage error when both Kinetic and Potential energy changes are ignored =  %f '%(err);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The percentage error when Kinetic energy change is ignored =  0.097 \n",
+        " The percentage error when Potential energy change is ignored =  0.0076 \n",
+        " The percentage error when both Kinetic and Potential energy changes are ignored =  0.104999 \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.17  Page No : 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "Pi = 5.;\t\t\t #pressure of dry saturated steam at the entrance in bar\n",
+      "Pe = 2.;\t\t\t #pressure of dry saturated steam at the exit in bar\n",
+      "Vi = 3.;\t\t\t #velocity of dry saturated steam at the entrance in m/s\n",
+      "m = 1.; \t\t\t #flow rate of steam through the nozzle in kg/s\n",
+      "g = 9.81;\t\t\t #acceleration due to gravity in m/s**2\n",
+      "\n",
+      "# Calculations\n",
+      "hi = 2747.5\n",
+      "he = 2706.3\n",
+      "ve = 0.8854\n",
+      "Zi = 0;\t\t\n",
+      "Ze = 0;\t\t\n",
+      "Q = 0;\t\t\n",
+      "Ws = 0;\t\t\n",
+      "Ve = math.sqrt (2*(((Q-Ws)/m)-(g*(Zi-Ze))-((he*10**3)-(hi*10**3)))+(Vi**2))\n",
+      "A = (m*ve)/Ve\n",
+      "\n",
+      "# Results\n",
+      "print ' The velocity of dry saturated steam at the exit =  %0.2f m/s'%(Ve);\n",
+      "print ' The cross sectional area of the nozzle at the exit =  %0.3e m**2'%(A);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The velocity of dry saturated steam at the exit =  287.07 m/s\n",
+        " The cross sectional area of the nozzle at the exit =  3.084e-03 m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.18  Page No : 123"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T1 = 270.\t\t\t #temperature of wet steam in degree celsius\n",
+      "T2 = 120.\t\t\t #final temperature of superheated steam in degree celsius\n",
+      "P = 0.1;\t\t\t #pressure of superheated steam in MPa\n",
+      "\n",
+      "# Calculations\n",
+      "hf = 1185.2 \t\t\t\n",
+      "hg = 2789.9\t    \t\t\n",
+      "he = 2716.04\t\t\t\n",
+      "Xi = (he-hf)/(hg-hf);\t\n",
+      "\n",
+      "# Results\n",
+      "print ' The quality of wet steam =  %0.3f '%(Xi);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The quality of wet steam =  0.954 \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.20  Page No : 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "del_H = -90.135;\t\t\t #standard enthalpy change for the reaction CO(g)+2H2(g)--->CH3OH(g) at 298.15K in kJ\n",
+      "\n",
+      "# Calculations\n",
+      "del_H1 = 2*del_H\n",
+      "del_H2 = (1./2)*del_H\n",
+      "\n",
+      "# Results\n",
+      "print ' The standard enthalpy change for the reaction 2COg)+4H2g)---->2CH3OHg \\\n",
+      ") at 298.15K =  %0.2f kJ'%(del_H1);\n",
+      "print ' The standard enthalpy change for the reaction 1./2)COg)+H2g)---->1./2\\\n",
+      ")CH3OHg) at 298.15K =  %0.4f kJ'%(del_H2 );\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The standard enthalpy change for the reaction 2COg)+4H2g)---->2CH3OHg ) at 298.15K =  -180.27 kJ\n",
+        " The standard enthalpy change for the reaction 1./2)COg)+H2g)---->1./2)CH3OHg) at 298.15K =  -45.0675 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.22  Page No : 130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "del_Hf_C4H10 = -74.943;\t\t\t #smath.tan(math.radiansard enthalpy of formation of C4H10(g) at 298.15K in kJ\n",
+      "del_Hf_CO2 = -393.978;\t\t\t #smath.tan(math.radiansard enthalpy of formation of CO2(g) at 298.15K in kJ\n",
+      "del_Hf_H2O = -241.997;\t\t\t #smath.tan(math.radiansard enthalpy of formation of H2O(g) at 298.15K in kJ\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "del_Hr = (5*del_Hf_H2O)+(4*del_Hf_CO2)-(del_Hf_C4H10);\n",
+      "\n",
+      "# Results\n",
+      "print ' The standard enthalpy change for the reaction C4H10g)+13\\\n",
+      "/2)O2g)---->4CO2g)+5H2Og) at 298.15K =  %0.3f kJ'%(del_Hr);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The standard enthalpy change for the reaction C4H10g)+13/2)O2g)---->4CO2g)+5H2Og) at 298.15K =  -2710.954 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.23  Page No : 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "del_Hf_C4H10 = -74.943;\t\t\t #smath.tan(math.radiansard enthalpy of formation of C4H10(g) at 298.15K in kJ\n",
+      "del_Hf_CO2 = -393.978;\t\t\t #smath.tan(math.radiansard enthalpy of formation of CO2(g) at 298.15K in kJ\n",
+      "del_Hf_H2O = -241.997;\t\t\t #smath.tan(math.radiansard enthalpy of formation of H2O(g) at 298.15K in kJ\n",
+      "del_H_vap = 43.966;\t\t\t #enthalpy of vaporization of H2O at 298.15K in kJ/mol\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "del_H1 = 0.\n",
+      "del_H2 = 5*(-del_H_vap)\n",
+      "del_H3 = 0.\n",
+      "\n",
+      "del_H = (5*del_Hf_H2O)+(4*del_Hf_CO2)-(del_Hf_C4H10);\n",
+      "del_net_H = (del_H)+(del_H1)+(del_H2)+(del_H3)\n",
+      "\n",
+      "# Results\n",
+      "print ' The standard enthalpy change for the reaction C4H10g)+13/2)O2g)---->4CO2g\\\n",
+      ")+5H2Ol) at 298.15K =  %0.3f kJ'%(del_net_H);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The standard enthalpy change for the reaction C4H10g)+13/2)O2g)---->4CO2g)+5H2Ol) at 298.15K =  -2930.784 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.24  Page No : 132"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "del_H_comb = 2880.44;\t\t\t #gross heating value of n-buatne gas at 298.15K in kJ/mol\n",
+      "del_Hf_CO2 = -393.978;\t\t\t #standard enthalpy of formation of CO2(g) at 298.15K in kJ\n",
+      "del_Hf_H2O = -285.958;\t\t\t #standard enthalpy of formation of H2O(l) at 298.15K in kJ\n",
+      "del_Hf_O2 = 0.;\t\t        \t #standard enthalpy of formation of O2(g) at 298.15K in kJ\n",
+      "\t\t\t \n",
+      "# Calculations\n",
+      "n_CO2 = 4.\n",
+      "n_H2O = 5.\n",
+      "n_O2 = -13./2\n",
+      "n_C4H10 = -1.\n",
+      "\t\t\t \n",
+      "del_Hf_C4H10 = (n_CO2*del_Hf_CO2)+(n_H2O*del_Hf_H2O)+(n_O2*del_Hf_O2)-(-del_H_comb);\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print ' The standard enthalpy of formation of n-butane gas at 298.15K =  %0.3f kJ'%(del_Hf_C4H10);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The standard enthalpy of formation of n-butane gas at 298.15K =  -125.262 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.25  Page No : 133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "del_Hc_A = -890.94;\t\t\t #enthalpy change accompanying reaction A in kJ\n",
+      "del_Hc_B = -283.18;\t\t\t #enthalpy change accompanying reaction B in kJ\n",
+      "del_Hc_C = -286.03;\t\t\t #enthalpy change accompanying reaction C in kJ\n",
+      "del_H_vap = -43.966;\t\t\t #enthalpy change of vaporization of H2O at 298.15K in kJ/mol\n",
+      "\n",
+      "# Calculations\n",
+      "del_H0 = (del_Hc_A)-(del_Hc_B)-(3*del_Hc_C)+(del_H_vap)\n",
+      "\n",
+      "# Results\n",
+      "print ' The standard enthalpy change at 298.15K for the reaction \\\n",
+      "CH4g)+H2Og)--->COg)+3H2g) =  %0.3f kJ'%(del_H0);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The standard enthalpy change at 298.15K for the reaction CH4g)+H2Og)--->COg)+3H2g) =  206.364 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.26  Page No : 135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#The reaction is : C2H4(g)+H2O(g)--->C2H5OH(g)\n",
+      "del_H_vap = 43.82\n",
+      "\n",
+      "del_Hf = [52.335,-241.997,0,-277.819]\n",
+      "a = [4.196,28.850,20.691,0]\t\n",
+      "b = [154.565*10**-3,12.055*10**-3,205.346*10**-3,0]\n",
+      "c = [-81.076*10**-6,0,-99.793*10**-6,0]\n",
+      "d = [16.813*10**-9,0,18.825*10**-9,0]\t\n",
+      "e = [0,1.006*10**5,0,0];\t\t\n",
+      "\n",
+      "T1 = 298.15\t\t\n",
+      "T2 = 400.\t\t\t\n",
+      "n_C2H4 = -1.\n",
+      "n_H2O = -1.\n",
+      "n_C2H5OH = 1.\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "del_Hf_C2H5OH_g = del_Hf[3]+del_H_vap;\n",
+      "del_Hr = (n_C2H5OH*del_Hf_C2H5OH_g)+(n_C2H4*del_Hf[0])+(n_H2O*del_Hf[1])\n",
+      "del_a = (n_C2H4*a[0])+(n_H2O*a[1])+(n_C2H5OH*a[2])\t\t\t \n",
+      "del_b = (n_C2H4*b[0])+(n_H2O*b[1])+(n_C2H5OH*b[2])\t\t\t \n",
+      "del_c = (n_C2H4*c[0])+(n_H2O*c[1])+(n_C2H5OH*c[2])\t\t\t \n",
+      "del_d = (n_C2H4*d[0])+(n_H2O*d[1])+(n_C2H5OH*d[2])\t\t\t \n",
+      "del_e = (n_C2H4*e[0])+(n_H2O*e[1])+(n_C2H5OH*e[2])\t\t\t \n",
+      "del_H0 = (del_Hr*10**3)-((del_a*T1)+((del_b/2)*T1**2)+((del_c/3)*T1**3)+((del_d/4)*T1**4)-(del_e/T1))\n",
+      "del_Hr_T2 = (del_H0+((del_a*T2)+((del_b/2)*T2**2)+((del_c/3)*T2**3)+((del_d/4)*T2**4)-(del_e/T2)))*10**-3;\n",
+      "\n",
+      "# Results\n",
+      "print ' The standard enthalpy change at 400K for the reaction\\\n",
+      " C2H4g)+H2Og)--->C2H5OHg) =  %f kJ'%(del_Hr_T2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The standard enthalpy change at 400K for the reaction C2H4g)+H2Og)--->C2H5OHg) =  -44.529472 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.28  Page No : 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      " \n",
+      "#The combustion reaction of methane is given by:\n",
+      "# CH4(g)+2O2(g)--->CO2(g)+2H2O(g)\n",
+      "\n",
+      "n_O2 = 2.\t\t\t # stoichiometric amount of oxygen required for combustion\n",
+      "n_CH4 = 1.\t\t\t #number of moles of CH4(g) in moles\n",
+      "n_CO2 = 1.\t\t\t #number of moles of CO2(g) formed in the combustion reaction in moles\n",
+      "n_H2O = 2.\t\t\t #number of moles of H2O(g) formed in the combustion reaction in moles\n",
+      "del_Hf = [-74.943,0,-393.978,-241.997];\t\t\t # standard enthalpies of formation of CH4(g),O2(g),CO2(g),H2O(g) at 298.15K in kJ\n",
+      "a = [45.369,28.850,30.255,27.270];\t\t\t     #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
+      "b = [8.688*10**-3,12.055*10**-3,4.207*10**-3,4.930*10**-3];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
+      "c = [0,0,0,0];  \t    \t\t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
+      "d = [0,0,0,0];\t    \t    \t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
+      "e = [-9.619*10**5,1.006*10**5,-1.887*10**5,0.333*10**5];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK\n",
+      "per_excess_air = 50. \t\t\t #percentage excess of air supplied to the adiabatic burner\n",
+      "T_amb = 298.15\t\t\t # temperature at which air and methane enter the burner in K\n",
+      "per_N2 = 79.\t\t\t #percentage of N2 in the air supplied\n",
+      "per_O2 = 21.\t\t\t #percentage of O2 in the air supplied\n",
+      "\n",
+      "# Calculations\n",
+      "n_O2_actual = (1+(per_excess_air/100))*n_O2\n",
+      "n_N2 = n_O2_actual*(per_N2/per_O2);\t\t\t\n",
+      "n_O2_residual = n_O2_actual-n_O2;\t\t\t\n",
+      "\n",
+      "del_Hr = (n_CO2*del_Hf[2])+(n_H2O*del_Hf[3])-(n_O2*del_Hf[1])-(n_CH4*del_Hf[0])\n",
+      "\n",
+      "del_a = (n_CO2*a[0])+(n_H2O*a[1])+(n_O2_residual*a[2])+(n_N2*a[3])\n",
+      "del_b = (n_CO2*b[0])+(n_H2O*b[1])+(n_O2_residual*b[2])+(n_N2*b[3])\n",
+      "del_c = (n_CO2*c[0])+(n_H2O*c[1])+(n_O2_residual*c[2])+(n_N2*c[3])\n",
+      "del_d = (n_CO2*d[0])+(n_H2O*d[1])+(n_O2_residual*d[2])+(n_N2*d[3])\n",
+      "del_e = (n_CO2*e[0])+(n_H2O*e[1])+(n_O2_residual*e[2])+(n_N2*e[3]);\t\n",
+      "tguess = 500.\n",
+      "\n",
+      "def solver_func(ti):\n",
+      "    return (-(del_Hr*10**3))-((del_a*(ti-T_amb))+((del_b/2)*((ti**2)-(T_amb**2)))+((del_c/3)*((ti**3)-(T_amb**3)))+((del_d/4)*((ti**4)-(T_amb**4)))+(del_e*((1./T_amb)-(1./ti))));\n",
+      "\n",
+      "T = fsolve(solver_func,tguess)\n",
+      "\n",
+      "# Results\n",
+      "print ' The flame temperature when methane is burned with 50 percent excess air \\\n",
+      "in an adiabatic burner =  %f K'%(T);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The flame temperature when methane is burned with 50 percent excess air in an adiabatic burner =  1793.784965 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.29  Page No : 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T_exit = 550.\t\t\t #temperature in K at which the combustion products leave the boiler\n",
+      "percent_molar_comp_prdct = [6.28,3.14,7.85,82.73];\t\t\t # percentage molar composition of the combustion products CO2(g),CO(g),O2(g),N2(g) on dry basis\n",
+      "T_ent = 298.15;\t\t\t #temperature in K at which Propane and air enter the combustion chamber\n",
+      "del_Hf = [-393.978,-110.532,-241.997,0,0,-103.833];\t\t\t #smath.tan(math.radiansard enthalpies of formation of CO2(g),CO(g),H2O(g),O2(g),N2(g),C3H8(g) at 298.15K in kJ\n",
+      "a = [45.369,28.068,30.255,27.270,28.850];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
+      "\n",
+      "#coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
+      "b = [8.688*10**-3,4.631*10**-3,4.207*10**-3,4.930*10**-3,12.055*10**-3];\n",
+      "c = [0,0,0,0,0];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
+      "d = [0,0,0,0,0];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
+      "e = [-9.619*10**5,-0.258*10**5,-1.887*10**5,0.333*10**5,1.006*10**5];\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK\n",
+      "per_N2 = 79.\t\t\t #percentage of nitrogen in air\n",
+      "per_O2 = 21.\t\t\t #percentage of oxygen in air\n",
+      "molar_mass_propane = 44.*10**-3;\t\t\t #molar mass of propane in kg/mole\n",
+      "\n",
+      "# Calculations\n",
+      "# TAKE BASIS AS 100 mol OF  DRY COMBUSTION PRODUCTS\n",
+      "n_CO2 = percent_molar_comp_prdct[0]\t\t\t #number of moles of CO2(g) in the product stream\n",
+      "n_CO = percent_molar_comp_prdct[1]\t\t\t #number of moles of CO(g) in the product stream\n",
+      "n_O2 = percent_molar_comp_prdct[2]\t\t\t #number of moles of O2(g) in the product stream\n",
+      "n_N2 = percent_molar_comp_prdct[3]\t\t\t #number of moles of N2(g) in the product stream\n",
+      "\n",
+      "x = (n_CO2+n_CO)/3;\n",
+      "\n",
+      "y = (2*n_N2)/(2*(per_N2/per_O2));\n",
+      "\n",
+      "z = (2*y)-(2*n_CO2)-(n_CO)-(2*n_O2);\n",
+      "\n",
+      "n_H2O = z\n",
+      "n_C3H8 = x\n",
+      "\n",
+      "del_Hr = (n_CO2*del_Hf[0])+(n_CO*del_Hf[1])+(n_H2O*del_Hf[2])-(n_C3H8*del_Hf[5]);\n",
+      "\n",
+      "del_a = (n_CO2*a[0])+(n_CO*a[1])+(n_O2*a[2])+(n_N2*a[3])+(n_H2O*a[4])\t\n",
+      "del_b = (n_CO2*b[0])+(n_CO*b[1])+(n_O2*b[2])+(n_N2*b[3])+(n_H2O*b[4])\t\n",
+      "del_c = (n_CO2*c[0])+(n_CO*c[1])+(n_O2*c[2])+(n_N2*c[3])+(n_H2O*c[4])\t\n",
+      "del_d = (n_CO2*d[0])+(n_CO*d[1])+(n_O2*d[2])+(n_N2*d[3])+(n_H2O*d[4])\t\n",
+      "del_e = (n_CO2*e[0])+(n_CO*e[1])+(n_O2*e[2])+(n_N2*e[3])+(n_H2O*e[4])\t\n",
+      "\n",
+      "#calulation of del_Hp J\n",
+      "del_Hp = (del_a*(T_exit-T_ent))+((del_b/2)*((T_exit**2)-(T_ent**2)))+((del_c/3)*((T_exit**3)-(T_ent**3)))+((del_d/4)*((T_exit**4)-(T_ent**4)))-(del_e*((1./T_exit)-(1./T_ent)));\n",
+      "del_H = ((del_Hr*10**3)+(del_Hp))*10**-3;\t\t\t \n",
+      "mass_propane = n_C3H8*molar_mass_propane;\t\t\t \n",
+      "energy = (-(del_H*10**3)/mass_propane)*10**-6;\t\t\n",
+      "       \n",
+      "# Results\n",
+      "print ' The energy transferred as heat per kg propane =  %f MJ'%(energy);     \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The energy transferred as heat per kg propane =  33.741510 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch5.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch5.ipynb
new file mode 100755
index 00000000..9cf5b1c8
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch5.ipynb
@@ -0,0 +1,974 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Second law of thermodynamics and its applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.2  Page No : 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Q = 1000.;\t\t    \t #amount of energy absorbed by the heat engine in kJ/s\n",
+      "W = 650.;\t\t\t     #work delivered by the heat engine in kW\n",
+      "T_source = 500. \t\t #temperature of the source in degree celsius\n",
+      "T_sink = 25.\t\t\t #temperature of the sink in degree celsius\n",
+      "\n",
+      "# Calculations\n",
+      "n_claimed = W/Q\n",
+      "T1 = T_source+273.15\n",
+      "T2 = T_sink+273.15\n",
+      "n_carnot = 1-(T2/T1)\n",
+      "\n",
+      "# Results\n",
+      "print \" The efficiency of the Carnot engine = %0.3f \"%(n_carnot);\n",
+      "print \" The efficiency of the engine claimed by the inventor = %0.2f \"%(n_claimed);\n",
+      "if n_claimed<n_carnot:\n",
+      "    print \" The claimed heat engine is theoretically feasible as the efficiency of the engine is lesser than that of a Carnot engine\";\n",
+      "else:\n",
+      "    print \" The claimed heat engine is not theoretically feasible as the efficiency of the engine is greater than that of a Carnot engine\";\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The efficiency of the Carnot engine = 0.614 \n",
+        " The efficiency of the engine claimed by the inventor = 0.65 \n",
+        " The claimed heat engine is not theoretically feasible as the efficiency of the engine is greater than that of a Carnot engine\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.3  Page No : 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T_source_summer = 42; #temperature in the summer months in degree celsius\n",
+      "T_sink_winter = 0;\t  #temperature in the winter months in degree celius\n",
+      "T_amb = 25;\t\t\t  #temperature at which the house is to be maintained during both the months in degree celsius\n",
+      "energy_loss = 0.5;\n",
+      "\n",
+      "# Calculations\n",
+      "T_H_summer = T_source_summer+273.15\n",
+      "T_L_summer = T_amb+273.15\n",
+      "T_H_winter = T_amb+273.15\n",
+      "T_L_winter = T_sink_winter+273.15\n",
+      "W_summer = (energy_loss*((T_H_summer-T_L_summer)**2))/(T_L_summer)\n",
+      "W_winter = (energy_loss*((T_H_winter-T_L_winter)**2))/(T_H_winter)\n",
+      "\n",
+      "# Results\n",
+      "print \" The minimum power required to operate the device in summer = %.4f kW \"%(W_summer);\n",
+      "print \" The minimum power required to operate the device in winter = %f kW \"%(W_winter);\n",
+      "\n",
+      "# Note: Answer in book is wrong. Please calculated manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The minimum power required to operate the device in summer = 0.4847 kW \n",
+        " The minimum power required to operate the device in winter = 1.048130 kW \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.4  Page No : 166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T_L = 4.25 \t\t\t #normal boiling point of helium in K\n",
+      "Q_L = 0.083\t\t\t #latent heat of vaporization of helium in kJ/mol\n",
+      "n = 1.  \t\t\t #amount of liquid helium to be produced in kmol\n",
+      "T_amb = 42.\t\t\t #ambient temperature in summer in degree celsius\n",
+      "\n",
+      "# Calculations\n",
+      "T_H = T_amb+273.15\n",
+      "COP = (T_L)/(T_H-T_L)\n",
+      "W = (Q_L)/COP;\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The maximum possible COP of the unit = %0.4f \"%(COP);\n",
+      "print \" The minimum amount of work to be done on the refrigerating unit = %f kJ \"%(W);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The maximum possible COP of the unit = 0.0137 \n",
+        " The minimum amount of work to be done on the refrigerating unit = 6.071694 kJ \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.5  Page No : 166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T_ice = 0.;\t\t        \t #temperature of the ice to be produced in degree celsius\n",
+      "m = 5000.;\t\t\t         #rate at which ice is to be produced in kg/hour\n",
+      "T_water = 0.;\t    \t\t #temperature of water used to produce ice in degree celsius\n",
+      "T_amb = 40.;\t\t    \t #ambient temperature in degree celsius\n",
+      "T_source = 100.;\t\t\t #temperature of the source for operating heat engine in degree celsius\n",
+      "lambda_fusion = 6.002;\t\t\t #latent heat of fusion of water in kJ/mol at 0 degree celsius\n",
+      "molar_mass = 18*10**-3;\t\t\t #molar mass of water in kg/mol\n",
+      "\n",
+      "# Calculations\n",
+      "T_L = T_water+273.15\n",
+      "T_H = T_amb+273.15\n",
+      "COP = (T_L)/(T_H-T_L)\n",
+      "Q_L = ((m/3600)/molar_mass)*(lambda_fusion)\n",
+      "W = (Q_L)/(COP);\t\t\t \n",
+      "T1 = T_source+273.15;\t\t\n",
+      "T2 = T_amb+273.15\n",
+      "n_heatengine = (T1-T2)/T1\n",
+      "Q1 = W/n_heatengine;\t\n",
+      "energy_ratio = (Q1+Q_L)/Q_L;\n",
+      "\n",
+      "# Results\n",
+      "print \" The minimum power required to operate the refrigerator = %0.2f kW\"%(W);\n",
+      "print \" The maximum possible efficiency of the heat engine = %0.4f \"%(n_heatengine);\n",
+      "print \" Ratio of the energy rejected to the ambient atmosphere to the\\\n",
+      " energy absorbed from the water = %0.4f \"%(energy_ratio);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The minimum power required to operate the refrigerator = 67.82 kW\n",
+        " The maximum possible efficiency of the heat engine = 0.1608 \n",
+        " Ratio of the energy rejected to the ambient atmosphere to the energy absorbed from the water = 1.9107 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.6  Page No : 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T1 = 800.\t\t\t #temperature of reservoir 1 in K\n",
+      "T2 = 400.\t\t\t #temperature of reservoir 2 in K\n",
+      "Q1 = 1000.\t\t\t #energy absorbed from reservoir maintained at T1 in kJ\n",
+      "Q2 = 400.\t\t\t #energy absorbed from reservoir maintained at T2 in kJ\n",
+      "W = 1000.\t\t\t #work delivered by the heat engine in kJ\n",
+      "T3 = 300.\t\t\t #temperature of the sink in K\n",
+      "\n",
+      "# Calculations\n",
+      "Q3 = (Q1+Q2)-W\n",
+      "clausius_inequality = (Q1/T1)+(Q2/T2)-(Q3/T3)\n",
+      "\n",
+      "# Results\n",
+      "print \" The LHS of the Clausius inequality = %0.4f \"%(clausius_inequality);\n",
+      "if clausius_inequality<0 or clausius_inequality == 0:\n",
+      "    print \" The given process does not violate the second law of thermodynamics, therefore the claim is correct\"\n",
+      "else:\n",
+      "    print \" This is a violation of the second law of thermodynamics, and hence the claim cannot be justified\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The LHS of the Clausius inequality = 0.9167 \n",
+        " This is a violation of the second law of thermodynamics, and hence the claim cannot be justified\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.7  Page No : 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T_system = 200.;\t\t\t #temperature of the contents of reactor in degree celsius\n",
+      "t = 15.         \t\t\t #operation time of agitator in minutes\n",
+      "P = 750.    \t    \t\t #power of the operating motor in W\n",
+      "\n",
+      "# Calculations\n",
+      "dQ = P*t*60*10**-3\n",
+      "T = T_system+273.15\n",
+      "del_S = dQ/T;\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The change in the entropy of the reactor contents = %0.4f kJ/K \"%(del_S);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in the entropy of the reactor contents = 1.4266 kJ/K \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.8  Page No : 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "P = 0.101325;\t\t\t #pressure in the piston cylinder assembly in MPa\n",
+      "T1 = 300.;\t\t\t #temperature of the piston cylinder assembly in K\n",
+      "T2 = 400;\t\t\t #final temperature of the piston cylinder assembly in K\n",
+      "a = 45.369;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
+      "b = 8.688*10**-3;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
+      "e = 9.619*10**5;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "del_S = (a*math.log(T2/T1))+(b*(T2-T1))+((e)*((1./T2**2)-(1./T1**2)))\n",
+      "\n",
+      "# Results\n",
+      "print \" The change in entropy of CO2 = %.4f J/molK\"%(del_S);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in entropy of CO2 = 9.2447 J/molK\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.9  Page No : 173"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 1.      \t\t\t #amount of saturated liquid water in kg\n",
+      "T_initial = 100.\t\t #initial temperature of water in degree celsius\n",
+      "T_body = 500.\t\t\t #temperature of body which is brought into contact with the cylinder in degree celsius\n",
+      "hfg = 2256.94\t\t\t #enthalpy of vaporization taken from steam tables corresponding to T1 in kJ/kg\n",
+      "\n",
+      "# Calculations\n",
+      "T = T_initial+273.15\n",
+      "del_S = hfg/T;\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The change in entropy of water = %0.4f kJ/kgK\"%(del_S);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in entropy of water = 6.0483 kJ/kgK\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.10  Page No : 173"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m_steel = 10.   \t\t\t #mass of steel casting in kg\n",
+      "T_steel = 800.\t    \t\t #temperature of steel casting in degree celsius\n",
+      "m_water = 100.\t\t    \t #mass of water used for quenching in kg\n",
+      "T_water = 30.\t\t\t     #temperature of water used for quenching in degree celsius\n",
+      "Cp_steel = 0.461;\t\t\t #heat capacity of steel in kJ/kgK\n",
+      "Cp_water = 4.23;\t\t\t #heat capacity of water in kJ/kgK\n",
+      "\n",
+      "# Calculations\n",
+      "Ti_steel = T_steel+273.15\n",
+      "Ti_water = T_water+273.15\n",
+      "T_final = ((m_steel*Cp_steel*Ti_steel)+(m_water*Cp_water*Ti_water))/((m_steel*Cp_steel)+(m_water*Cp_water));\n",
+      "del_S_steel = m_steel*Cp_steel*math.log(T_final/Ti_steel)\n",
+      "del_S_water = m_water*Cp_water*math.log(T_final/Ti_water)\n",
+      "\n",
+      "# Results\n",
+      "print \" The change in entropy of steel  =  %0.4f kJ/K\"%(del_S_steel);\n",
+      "print \" The change in entropy of water  =  %f kJ/K\"%(del_S_water);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in entropy of steel  =  -5.7031 kJ/K\n",
+        " The change in entropy of water  =  11.427392 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.11  Page No : 175"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 2.\t    \t\t #volume of insulated tank in m**3\n",
+      "Ta = 400.\t\t\t #temperature of gas in compartment (a) in K\n",
+      "Pa = 3.\t\t    \t #pressure of gas in compartment (a) in MPa\n",
+      "Tb = 600.\t\t\t #temperature of gas in compartment (b) in K\n",
+      "Pb = 1.\t\t\t     #pressure of gas in compartment (b) in MPa\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Va = V/2\n",
+      "Vb = V/2\n",
+      "Na = (Pa*10**6*Va)/(R*Ta)\n",
+      "Nb = (Pb*10**6*Vb)/(R*Tb)\n",
+      "T = ((Na*Ta)+(Nb*Tb))/(Na+Nb)\n",
+      "N = Na+Nb;\t\t\t \n",
+      "P = ((N*R*T)/V)*10**-6\n",
+      "Cp = (5./2)*R;\t\t\t\n",
+      "del_S = ((Na*((Cp*math.log(T/Ta))-(R*math.log(P/Pa))))+(Nb*((Cp*math.log(T/Tb))-(R*math.log(P/Pb)))))*10**-3; # Calculations of the change in entropy using Eq.(5.43) in kJ/K\n",
+      "\n",
+      "# Results\n",
+      "print \" Entropy change of the gas = %0.2f kJ/K\"%(del_S);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Entropy change of the gas = 2.19 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.12  Page No : 177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "N = 1.\t    \t\t #amount of air to be separated into its components in kmol\n",
+      "P = 0.1\t\t    \t #pressure of air in MPa\n",
+      "T = 300.\t\t\t #temperature of air in K\n",
+      "per_oxygen = 21.\t\t\t #percentage of oxygen in air\n",
+      "per_nitrogen = 79.\t\t\t #percentage of nitrogen in air\n",
+      "R = 8.314;\t\t\t         #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "x1 = per_nitrogen/100\n",
+      "x2 = per_oxygen/100;\n",
+      "W = (T*N*10**3*R*((x1*math.log (x1))+(x2*math.log (x2))))*10**-3\n",
+      "\n",
+      "# Results\n",
+      "print \" Minimum work to be done to separate 1kmol of air at 0.1MPa and 300K into pure oxygen\\\n",
+      " and nitrogen at the same temperature and pressure = %0.2f kJ\"%(abs(W));"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Minimum work to be done to separate 1kmol of air at 0.1MPa and 300K into pure oxygen and nitrogen at the same temperature and pressure = 1281.91 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.13  Page No : 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "m_ice = 10.\t    \t\t #mass of the block of ice in kg\n",
+      "T_ice = 0.\t\t    \t #temperature of the ice in degree celsius\n",
+      "m_water = 100.\t\t\t #mass of watre in the tank in kg\n",
+      "T_water = 30.\t\t\t #temperature of the water in the tank in degree celsius\n",
+      "Cp = 4.23;\t\t\t     #heat capacity of water in kJ/kgK\n",
+      "lambda_melting = 333.44\t\t\t #latent heat of melting of ice in kJ/kg\n",
+      "\n",
+      "# Calculations\n",
+      "Ti_ice = T_ice+273.15\n",
+      "Ti_water = T_water+273.15\n",
+      "T_final = ((m_water*Cp*Ti_water)+(m_ice*Cp*Ti_ice)-(m_ice*lambda_melting))/((m_ice*Cp)+(m_water*Cp))\n",
+      "del_S_ice = ((m_ice*lambda_melting)/(Ti_ice))+(m_ice*Cp*math.log (T_final/Ti_ice));\t\t\t \n",
+      "del_S_water = m_water*Cp*math.log (T_final/Ti_water)\n",
+      "del_S_G = del_S_ice+del_S_water;\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The change in entropy of ice  =  %f kJ/K\"%(del_S_ice);\n",
+      "print \" The change in entropy of water  =  %f kJ/K\"%(del_S_water);\n",
+      "print \" The entropy generated =  %f kJ/K\"%(del_S_G);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in entropy of ice  =  15.211650 kJ/K\n",
+        " The change in entropy of water  =  -14.035033 kJ/K\n",
+        " The entropy generated =  1.176617 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.14  Page No : 182"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 3.\t\t\t         #pressure of superheated steam in MPa\n",
+      "T_enter = 300.\t\t\t #entrance temperature of superheated steam in degree celsius\n",
+      "T_exit = 45.\t\t\t #final temperature at which the steam leaves in degree celsisus\n",
+      "m = 1.      \t\t\t #mass flow rate of steam in kg/s\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "si = 6.5422\n",
+      "hi = 2995.1\n",
+      "sf = 0.6383\n",
+      "hf = 188.35\n",
+      "sg = 8.1661;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
+      "hg = 2583.3;\t\t\t #entahlpy of saturayed vapour in kJ/kg\n",
+      "\n",
+      "Xe = (si-sf)/(sg-sf)\n",
+      "he = ((1-Xe)*hf)+(Xe*hg)\n",
+      "Ws = -m*(he-hi);\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The power  Results from the turbine = %0.1f kW\"%(Ws);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The power  Results from the turbine = 928.4 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.15  Page No : 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "Pi = 3.\t\t\t #pressure of dry saturated steam when it enters the nozzle in bar\n",
+      "Pe = 2.\t\t\t #pressure of dry saturated steam at the exit in bar\n",
+      "\n",
+      "# Calculations\n",
+      "#From steam tables corresponding to Pi\n",
+      "si = 6.9909;\t\t\t #entropy of steam at the entrance in kJ/kgK\n",
+      "hi = 2724.7;\t\t\t #entahlpy of steam at the entrance in kJ/kg\n",
+      "\n",
+      "#From steam tables corresponding to Pe\n",
+      "sf = 1.5301;\t\t\t #entropy of saturated liquid in kJ/kgK\n",
+      "hf = 504.70;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
+      "sg = 7.1268;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
+      "hg = 2706.3;\t\t\t #entahlpy of saturayed vapour in kJ/kg\n",
+      "\n",
+      "se = 6.9909\n",
+      "Xe = (se-sf)/(sg-sf)\n",
+      "he = ((1-Xe)*hf)+(Xe*hg)\n",
+      "Ve = math.sqrt (2*(hi-he)*10**3)\n",
+      "\n",
+      "# Results\n",
+      "print \" The exit velocity of steam = %f m/s\"%(Ve);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The exit velocity of steam = 379.103198 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.16  Page No : 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "N_glycerol = 100.\t\t\t #molar flow rate of glycerol in mol/s\n",
+      "Ti_gly = 227.\t\t\t #inlet temperature of glycerol in degree celsius\n",
+      "Te_gly = 40.\t\t\t #outlet temperature of glycerol in degree celsius\n",
+      "Ti_water = 25.\t\t\t #inlet temperature of cooling water in degree celsius\n",
+      "Te_water = 50.\t\t\t #outlet temperature of cooling water in degree celsius\n",
+      "Cp_gly = 280.\t\t\t #heat capacity of glycerol in J/molK\n",
+      "Cp_water = 77.\t\t\t #heat capacity of water in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Ti_gly = Ti_gly+273.15\n",
+      "Te_gly = Te_gly+273.15\n",
+      "Ti_water = Ti_water+273.15\n",
+      "Te_water = Te_water+273.15\n",
+      "\n",
+      "N_water = -(N_glycerol*Cp_gly*(Te_gly-Ti_gly))/(Cp_water*(Te_water-Ti_water));\n",
+      "del_S_gly = N_glycerol*Cp_gly*math.log (Te_gly/Ti_gly)*10**-3\n",
+      "del_S_water = N_water*Cp_water*math.log (Te_water/Ti_water)*10**-3\n",
+      "S_G = del_S_gly+del_S_water\n",
+      "\n",
+      "# Results\n",
+      "print \" The rate at which entropy is generated in the heat exchanger = %0.3f kJ/K s\"%(S_G);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The rate at which entropy is generated in the heat exchanger = 3.754 kJ/K s\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.17  Page No : 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T_i = 150.\t\t\t #temperature of saturated steam taken up by the device in degree celsius\n",
+      "T_e = 200.\t\t\t #temperature of superheated steam delivered by the device in degree celsius\n",
+      "P_e = 0.2\t\t\t #pressure of superheated steam delivered by the device in MPa\n",
+      "me2 = 0.949\t\t\t #mass of superheated steam leaving the device in kg\n",
+      "me1 = 0.051\t\t\t #mass of saturated liquid leaving the device in kg\n",
+      "T_liq = 100.\t\t #temperature of saturated liquid leaving the device in degree celsius\n",
+      "mi = 1.     \t\t #mass of saturated steam fed to the device in kg\n",
+      "\n",
+      "# Calculations\n",
+      "#From steam tables corresponding to T_i\n",
+      "hi = 2745.4 \t\t\t #enthalpy of saturated vapour in kJ/kg\n",
+      "si = 6.8358;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
+      "\n",
+      "#For saturated liquid at T_liq\n",
+      "he1 = 419.06;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
+      "se1 = 1.3069;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
+      "\n",
+      "#For superheated steam at P_e and T_e\n",
+      "he2 = 2870.5;\t\t\t #enthalpy of superheated steam in kJ/kg\n",
+      "se2 = 7.5072;\t\t\t #entropy of superheated steam in kJ/kgK\n",
+      "\n",
+      "LHS = mi*hi;\n",
+      "RHS = (me1*he1)+(me2*he2);\n",
+      "\n",
+      "S_G = (me1*se1)+(me2*se2)-(mi*si);\n",
+      "\n",
+      "# Results\n",
+      "print \" The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied =  %0.1f kJ\"%(LHS);\n",
+      "print \" The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = %0.1f kJ\"%(RHS);\n",
+      "print \" The entropy generated by applying the second law of thermodynamics to the flow device = %0.4f kJ/kgK\"%(S_G);\n",
+      "if int(LHS) ==  int(RHS) and S_G>0 or S_G == 0 :\n",
+      "    print \" As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \"\n",
+      "else:\n",
+      "    print \" As both the first and second law or either the first or second law of thermodynamics \\\n",
+      "     are not satisfied, the device is not feasible \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied =  2745.4 kJ\n",
+        " The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = 2745.5 kJ\n",
+        " The entropy generated by applying the second law of thermodynamics to the flow device = 0.3552 kJ/kgK\n",
+        " As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.18  Page No : 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pi = 30.\t\t\t #pressure of superheated steam entering the turbine in bar\n",
+      "Ti = 300.\t\t\t #temperature of superheated steam entering the turbine in degree celsius\n",
+      "Pe = 0.1\t\t\t #pressure at which steam exits the turbine in bar\n",
+      "Xe = 0.9\t\t\t #quality of steam at the exit (no unit)(for the actual turbine)\n",
+      "\n",
+      "# Calculations\n",
+      "#For superheated steam at Pi and Ti\n",
+      "hi = 2995.1;\t\t\t #enthalpy of superheated steam at the entrance in kJ/kg\n",
+      "si = 6.5422;\t\t\t #entropy of superheated steam at the entrance in kJ/kgK\n",
+      "\n",
+      "#For steam at Pe\n",
+      "hf = 191.83;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
+      "hg = 2584.8;\t\t\t #enthalpy of saturated vapour in kJ/kg\n",
+      "sf = 0.6493;\t\t\t #entropy of saturated liquid in kJ/kgK\n",
+      "sg = 8.1511;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
+      "\n",
+      "X2 = (si-sf)/(sg-sf)\n",
+      "h2 = (hf*(1-X2))+(X2*hg)\n",
+      "he = (hf*(1-Xe))+(Xe*hg)\n",
+      "n_T = (hi-he)/(hi-h2)\n",
+      "\n",
+      "# Results\n",
+      "print \" The isentropic efficiency of the turbine = %f \"%(n_T);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The isentropic efficiency of the turbine = 0.703395 \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.19  Page No : 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Ti = 25.\t\t\t #temperature of air taken in by the adiabatic air compressor in degree celsius\n",
+      "Pi = 0.1\t\t\t #pressure of air taken in by the adiabatic air compressor in MPa\n",
+      "Pe = 1.\t    \t\t #discharge pressure of air in MPa\n",
+      "n_c = 0.8\t\t\t #isentropic efficiency of the compressor (no unit)\n",
+      "gaamma = 1.4\t\t #ratio of molar specific heat capacities (no unit)\n",
+      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Ti = Ti+273.15\n",
+      "Te = Ti*(((Pe*10**6)/(Pi*10**6))**((gaamma-1)/gaamma))\n",
+      "W_s = (((R*gaamma)/(gaamma-1))*(Te-Ti))*10**-3;\t\t\t\n",
+      "Ws = W_s/n_c\n",
+      "Te_actual = ((Ws*10**3*(gaamma-1))/(R*gaamma))+Ti\n",
+      "\n",
+      "# Results\n",
+      "print \" The exit temperature of air = %0.2f K\"%(Te_actual);\n",
+      "print \" The power consumed by the compressor  = %f kW/mol\"%(Ws);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The exit temperature of air = 645.01 K\n",
+        " The power consumed by the compressor  = 10.093262 kW/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.20  Page No : 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Ti = 30.\t\t\t #temperature of saturated liquid water in degree celsius\n",
+      "m = 500.\t\t\t #mass flow rate of water being pumped in kg/s\n",
+      "P2 = 3. \t\t\t #preesure maintained in the boiler in MPa\n",
+      "n_p = 0.75;\t\t\t #isentropic efficiency of the pump (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "#For saturated liquid water at Ti\n",
+      "vf = 0.0010043\n",
+      "P1 = 4.241;\t\t\n",
+      "\n",
+      "Ws_m = (vf*((P2*10**6)-(P1*10**3)))*10**-3\n",
+      "Ws_act_m = Ws_m/n_p;\t\t\t \n",
+      "P = ((Ws_act_m*10**3)*m)*10**-6;\n",
+      "\n",
+      "# Results\n",
+      "print \" The power consumed by the pump  =  %d MW\"%(P);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The power consumed by the pump  =  2 MW\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.21  Page No : 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pi = 3.\t\t\t #pressure of dry saturated steam entering the nozzle in bar\n",
+      "Xe = 0.98\t\t #quality of steam exiting the nozzle (no unit)\n",
+      "Pe = 2.\t\t\t #pressure of steam exiting the nozzle in bar\n",
+      "\n",
+      "# Calculations\n",
+      "#For steam at Pi\n",
+      "hi = 2724.7\n",
+      "he = 2652.8\n",
+      "V2_2_s = hi-he\n",
+      "\n",
+      "#For steam at Pe\n",
+      "hf = 504.70\n",
+      "hg = 2706.3\n",
+      "he_act = ((1-Xe)*hf)+(Xe*hg)\n",
+      "V2_2 = hi-he_act;\t\t\t\n",
+      "n_N = (V2_2)/(V2_2_s)\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The isentropic efficiency of the nozzle = %0.3f \"%(n_N);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The isentropic efficiency of the nozzle = 0.868 \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch6.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch6.ipynb
new file mode 100755
index 00000000..cf3f1f12
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch6.ipynb
@@ -0,0 +1,214 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 :  Thermodynamic potentials"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.6  Page No : 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 0.1;\t\t\t #mass of superheated steam in the piston cylinder assembly in kg\n",
+      "P1 = 1.\t    \t\t #initial pressure of superheated steam in MPa\n",
+      "T1 = 300.\t\t\t #initial temperature of superheated steam in degree celsius\n",
+      "P2 = 0.1\t\t\t #pressure of steam after expansion in MPa\n",
+      "T2 = 200.\t\t\t #temperature of steam after expansion in degree celsius\n",
+      "\n",
+      "# Calculations\n",
+      "#For steam at P1 and T1:\n",
+      "h1 = 3052.1\n",
+      "v1 = 0.2580\n",
+      "\n",
+      "#For steam at P2 and T2:\n",
+      "h2 = 2875.4\n",
+      "v2 = 2.1720\n",
+      "\n",
+      "del_u = (((h1*10**3)-(P1*10**6*v1))-((h2*10**3)-(P2*10**6*v2)))*10**-3\n",
+      "W = m*(del_u)\n",
+      "\n",
+      "# Results\n",
+      "print \" The work done by steam = %0.2f kJ\"%(W);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The work done by steam = 13.59 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.8  Page No : 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 3.\t    \t\t #pressure of superheated steam in MPa\n",
+      "Ti = 300.\t\t\t #temperature at which the steam enters the turbine in degree celsius\n",
+      "m = 1.\t\t    \t #mass flow rate of steam in kg/s\n",
+      "Te = 60.\t\t\t #temperature of dry saturated steam when it leaves the turbine in degree celsius\n",
+      "\n",
+      "# Calculations\n",
+      "#For steam at P and Ti:\n",
+      "h1 = 2995.1\t\t\t #specific entahlpy of steam in kJ/kg\n",
+      "\n",
+      "#For saturated steam at Te:\n",
+      "h2 = 2609.7\t\t\t #specific enthalpy of saturated vapour in kJ/kg\n",
+      "\n",
+      "Ws = m*(h1-h2)\t\t # Calculations of the power  Results of the turbine using Eq.(6.83) in kW\n",
+      "\n",
+      "# Results\n",
+      "print \" The power  Results of the turbine = %0.1f kW\"%(Ws);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The power  Results of the turbine = 385.4 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.10  Page No : 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 0.1 \t\t\t #mass of superheated steam in the piston cylinder assembly in kg\n",
+      "P1 = 3.\t    \t\t #initial pressure of superheated steam in MPa\n",
+      "T1 = 300.\t\t\t #initial temperature of superheated steam in degree celsius\n",
+      "T0 = 300.\t\t\t #temperature of the reservoir which is placed in thermal contact with the piston-cylinder assembly in degree celsius\n",
+      "P2 = 0.1\t\t\t #pressure of steam after expansion in MPa\n",
+      "\n",
+      "#For steam at P1 and T1:\n",
+      "h1 = 2995.1\n",
+      "v1 = 0.08116\n",
+      "s1 = 6.5422\n",
+      "\n",
+      "#For steam at P2 and T2:\n",
+      "h2 = 3074.5\n",
+      "v2 = 2.6390\n",
+      "s2 = 8.2166\n",
+      "\n",
+      "# Calculations\n",
+      "T0 = T0+273.15\n",
+      "\n",
+      "W = m*(h1-h2-(((P1*v1)-(P2*v2))*10**3)-(T0*(s1-s2)))\n",
+      "\n",
+      "# Results\n",
+      "print \" The maximum work obtained from steam = %0.2f kJ\"%(W);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The maximum work obtained from steam = 90.07 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.12  Page No : 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "P1 = 0.1\t\t\t #pressure at which air enters the compressor in MPa\n",
+      "T1 = 300.\t\t\t #temperature at which air enters the compressor in K\n",
+      "P2 = 1. \t\t\t #pressure at which air leaves the compressor in MPa\n",
+      "T2 = 300.\t\t\t #temperature at which air leaves the compressor in K\n",
+      "T0 = 300.\t\t\t #ambient temperature in K\n",
+      "N = 1.  \t\t\t #molar flow rate of air in mol/s\n",
+      "gaamma = 1.4\t\t #ratio of specific heat capacities (no unit)\n",
+      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "Ws = (-N*T0*(-R*math.log (P2/P1)))*10**-3\n",
+      "\n",
+      "# Results\n",
+      "print \" The minimum power required to compress one mole per second of air = %0.3f kW\"%(Ws);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The minimum power required to compress one mole per second of air = 5.743 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch7.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch7.ipynb
new file mode 100755
index 00000000..879f7b43
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch7.ipynb
@@ -0,0 +1,255 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Thermodynamic property relations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.10  Page No : 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T2 = 150.   \t\t\t #temperature at which water it is desired to boil water in degree celsius\n",
+      "P1 = 0.10133\t\t\t #ambient pressure in MPa\n",
+      "T1 = 100.\t    \t\t #temperature at which water boils corresponding at pressure P1 in degree celsius\n",
+      "del_hv = 2256.94\t\t #enthalpy of vaporization in kJ/kg\n",
+      "R = 8.314;\t\t    \t #universal gas constant in J/molK\n",
+      "M = 18*10**-3;\t\t\t #molar mass of water in kg/mol\n",
+      "\n",
+      "# Calculations\n",
+      "T1 = T1+273.15\n",
+      "T2 = T2+273.15\n",
+      "P2 = P1*(math.exp(((del_hv*10**3*M)*((1./T1)-(1./T2)))/(R)))\n",
+      "\n",
+      "# Results\n",
+      "print \" The approximate pressure at which the boiler is to be operated = %0.3f MPa\"%(P2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The approximate pressure at which the boiler is to be operated = 0.476 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.11  Page No : 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 60.         \t\t\t #mass of the person who wants to skate in kg\n",
+      "T = -2.\t\t\t             #temperature of the ice in degree celsius\n",
+      "A = 15.\t\t\t             #area of contact between the skate edges and ice in mm**2\n",
+      "vs = 1.091*10**-3;\t\t\t #specific volume of ice in m**3/kg (at Tref)\n",
+      "vf = 1.0*10**-3;\t\t\t #specific volume of water in m**3/kg (at Tref)\n",
+      "del_hf = 6.002;\t\t    \t #enthalpy of melting of ice in kJ/mol\n",
+      "g = 9.81;\t\t\t         #accleration due to gravity in m/s**2\n",
+      "Tref = 0.       \t\t\t #reference temperature at which the specific enthalpy of ice and water are taken in degree celsius\n",
+      "\n",
+      "# Calculations\n",
+      "Tref = Tref+273.15\n",
+      "del_P = ((m*g)/(A*10**-6))*10**-6\n",
+      "del_v = (vf-vs)*(18*10**-3);\t\n",
+      "del_T = (del_P*10**6)/((del_hf*10**3)/(Tref*del_v));\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The temperature of ice originally  =  %d degree celsius \"%(T);\n",
+      "print \" The reduction in melting point of ice due to the additional pressure,computed using the Clayperon equation  =  %0.2f degree celsius \"%(del_T);\n",
+      "if del_T<T :\n",
+      "    print  \" The ice can melt due to the additional pressure and therefore it will be possible to skate \"\n",
+      "else:\n",
+      "    print  \" The ice will not melt and therefore it will be difficult to skate \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The temperature of ice originally  =  -2 degree celsius \n",
+        " The reduction in melting point of ice due to the additional pressure,computed using the Clayperon equation  =  -2.93 degree celsius \n",
+        " The ice can melt due to the additional pressure and therefore it will be possible to skate \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.12  Page No : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T1 = 100.\t    \t\t #temperature of water in degree celsius\n",
+      "del_hv1 = 2256.94;\t\t\t #enthalpy of vaporization at T1 in kJ/kg\n",
+      "T2 = 150.   \t\t\t #temperature at which the enthalpy of vaporization is to be determined in degree celsius\n",
+      "Cp_f = 4.26\t    \t\t #isobaric heat capacity of liquid in kJ/kgK\n",
+      "Cp_g = 1.388\t\t\t #isobaric heat capacity of vapour in kJ/kgK\n",
+      "\n",
+      "# Calculations\n",
+      "del_hv2 = ((Cp_g-Cp_f)*(T2-T1))+del_hv1\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy of vaporization at 150 degree celsius = %0.2f kJ/kg\"%(del_hv2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy of vaporization at 150 degree celsius = 2113.34 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.13  Page No : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T1 = 100.           \t\t\t #temperature of water in degree celsius\n",
+      "del_hv1 = 2256.94   \t\t\t #enthalpy of vaporization at T1 in kJ/kg\n",
+      "T2 = 150.   \t\t        \t #temperature at which the enthalpy of vaporization is to be determined in degree celsius\n",
+      "del_hv_kirchoff = 2113.34\t\t\t #enthalpy of vaporization predicted by the Kirchhoff relation taken from Example 7.12 for comparison, in kJ/kg\n",
+      "del_hv_steam_tables = 2113.25\t\t\t #enthalpy of vaporization taken from the steam tables corresponding to T2,for comparison, in kJ/kg\n",
+      "Tc = 647.3          \t\t\t #critical temperature of water in K\n",
+      "\n",
+      "# Calculations\n",
+      "T1 = T1+273.15\n",
+      "T2 = T2+273.15\n",
+      "Tr1 = T1/Tc\n",
+      "Tr2 = T2/Tc\n",
+      "del_hv2 = del_hv1*(((1-Tr2)/(1-Tr1))**0.38)\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy of vaporization at 150 degree celsius using \";\n",
+      "print \" Watson correlation \\t  =  %f kJ/kg\"%(del_hv2);\n",
+      "print \" Kirchhoffs relation \\t  =  %f kJ/kg\"%(del_hv_kirchoff);\n",
+      "print \" From steam tables \\t  =  %f kJ/kg\"%(del_hv_steam_tables);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy of vaporization at 150 degree celsius using \n",
+        " Watson correlation \t  =  2090.687877 kJ/kg\n",
+        " Kirchhoffs relation \t  =  2113.340000 kJ/kg\n",
+        " From steam tables \t  =  2113.250000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.14  Page No : 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 373.15  \t\t\t #normal boiling point of water in K (temperature at which the enthalpy of vaporization is to be determined)\n",
+      "Pc = 221.2\t    \t\t #critical pressure of water in bar\n",
+      "Tc = 647.3\t\t    \t #critical temperature of water in K\n",
+      "R = 8.314\t\t\t     #universal gas constant in J/molK\n",
+      "del_hvn_steam_tables = 2256.94\t\t\t #enthalpy of vaporization at the normal boiling point taken from the steam tables, for comparison, in kJ/kg\n",
+      "\n",
+      "# Calculations\n",
+      "Tbr = T/Tc\t\t\t\n",
+      "del_hvn = ((1.093*R*Tc*(Tbr*((math.log(Pc)-1.013)/(0.930-Tbr))))*10**-3)/(18*10**-3);\n",
+      "err = abs((del_hvn-del_hvn_steam_tables)/del_hvn_steam_tables)*100\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy of vaporization at the normal boiling point \";\n",
+      "print \" Using Riedels correlation \\t  =  %f kJ/kg\"%(del_hvn);\n",
+      "print \" From the steam tables \\t \\t  =  %f kJ/kg\"%(del_hvn_steam_tables);\n",
+      "print \" Error \\t \\t \\t \\t  =  %f %% \"%(err);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy of vaporization at the normal boiling point \n",
+        " Using Riedels correlation \t  =  2337.179008 kJ/kg\n",
+        " From the steam tables \t \t  =  2256.940000 kJ/kg\n",
+        " Error \t \t \t \t  =  3.555212 % \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch8.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch8.ipynb
new file mode 100755
index 00000000..1e8782d9
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch8.ipynb
@@ -0,0 +1,600 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Thermodynamic properties of real gases"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.2  Page No : 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
+      "a = 3.789;\t\t\t #van der Waals constant in Pa (m**3/mol)**2\n",
+      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
+      "v = 15.675*10**-3;\t\t\t #volume occupied by n-octane vapour taken from Example (3.8) in m**3/mol\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "dep_h = (P*10**6*v)-(R*T)-(a/v)\n",
+      "dep_s = R*math.log ((P*10**6*(v-b))/(R*T));\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure for n-octane vapour  =  %0.2f J/mol\"%(dep_h);\n",
+      "print \" The entropy departure for n-octane vapour  =  %0.4f J/mol K\"%(dep_s);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure for n-octane vapour  =  -428.74 J/mol\n",
+        " The entropy departure for n-octane vapour  =  -0.5757 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.3  Page No : 276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 100.    \t\t\t #temperature of carbon dioxide in degree celsius\n",
+      "P = 10. \t    \t\t #pressure of carbon dioxide in MPa\n",
+      "A0 = 0.5073;\t\t\t #Beattie-Bridgman constant for carbon dioxide in (Pa m**3)/mol**2\n",
+      "B0 = 104.76*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
+      "a = 71.32*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
+      "b = 72.35*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
+      "C = 660.0;\t\t    \t #Beattie-Bridgman constant for carbon dioxide in (m**3 K**3)/mol\n",
+      "R = 8.314;\t\t\t     #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "T = T+273.15\n",
+      "A1 = (R*T)  \n",
+      "A2 = (B0*R*T)-A0-((C*R)/T**2);\t\t\t \n",
+      "A3 = (a*A0)-(b*B0*R*T)-((B0*C*R)/T**2);\t\n",
+      "A4 = ((b*C*B0*R)/T**2);\t\t\t \n",
+      "vguess = 0.01\n",
+      "tolerance = 1e-6\n",
+      "\n",
+      "def solver_func(vi):\n",
+      "    return  (P*10**6)-((A1/vi)+(A2/vi**2)+(A3/vi**3)+(A4/vi**4))\n",
+      "\n",
+      "v = fsolve(solver_func,vguess)\n",
+      "\n",
+      "Z = (P*10**6*v)/(R*T)\n",
+      "\n",
+      "dep_h = (((B0*R*T)-(2*A0)-((4*C*R)/(T**2)))*(1./v))+((((3./2)*a*A0)-(b*B0*R*T)-((5*B0*C*R)/(2*(T**2))))*(1./(v**2)))+((2*b*C*B0*R)/((T**2)*(v**3)));\n",
+      "\n",
+      "# Results\n",
+      "print \" Molar volume of CO2 at %0.f MPa and %0.2f K  =  %.2e m**3/mol \"%(P,T,v);\n",
+      "print \" The compressibility factor = %.4f \"%(Z);\n",
+      "print \" The enthalpy departure for carbon dioxide using the Beattie-Bridgman equation of state  =  %.f J/mol\"%(dep_h);\n",
+      "\n",
+      "# Note: Answer is different because of rounding off error. Please check manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Molar volume of CO2 at 10 MPa and 373.15 K  =  2.33e-04 m**3/mol \n",
+        " The compressibility factor = 0.7518 \n",
+        " The enthalpy departure for carbon dioxide using the Beattie-Bridgman equation of state  =  -3210 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.4  Page No : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "T = 353.15    \t    \t\t #temperature of carbon dioxide in degree celsius\n",
+      "P = 10. \t    \t    \t #pressure of carbon dioxide in MPa\n",
+      "B0 = 104.76*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
+      "b = 72.35*10**-6;\t\t\t #Beattie-Bridgman constant for carbon dioxide in m**3/mol\n",
+      "C = 660.0;\t\t\t         #Beattie-Bridgman constant for carbon dioxide in (m**3 K**3)/mol\n",
+      "R = 8.314       \t\t\t #universal gas constant in J/molK\n",
+      "v = 0.233*10**-3    \t\t #volume calculated in Example (8.3) in m**3/mol\n",
+      "Z = 0.751;\t\t\t         #compressibility factor as calculated in Example (8.3) (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "A1 = ((B0*R)+((2*C*R)/(T**3)))\n",
+      "dep_s = (R*math.log (Z))-(A1*(1./v))+(((b*B0*R)-((2*C*B0*R)/(T**3)))*(1./(2*(v**2))))+((2*b*C*B0*R)/(3*(T**3)*(v**3)));\n",
+      "\n",
+      "# Results\n",
+      "print \" The entropy departure for carbon dioxide using the Beattie-\\\n",
+      "Bridgman equation of state  =  %.f J/mol K\"%(dep_s);\n",
+      "\n",
+      "# Note: Answer is varies because of rounding off error. Please check manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The entropy departure for carbon dioxide using the Beattie-Bridgman equation of state  =  -7 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.5  Page No : 281"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
+      "a = 4.426;\t\t\t #Redlich-Kwong constant taken from Example(3.9) in (m**6 Pa mol**-2)\n",
+      "b = 164.3*10**-6;\t\t\t #Redlich-Kwong constant taken from Example(3.9) in m**3/mol\n",
+      "Z = 0.9308;\t\t\t #compressibility factor taken from Example(3.9) (no unit)\n",
+      "B = 9.9306*10**-3;\t\t\t #value of B, used in the Cardan's method in Example (3.9)\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "dep_h = (R*T*(Z-1))-(((3*a)/(2*b))*math.log ((Z+B)/Z))\n",
+      "dep_s = (R*math.log(Z-B))-((a/(2*b*T))*math.log((Z+B)/Z))\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure for n-octane vapour using the generalized Redlich\\\n",
+      "-Kwong equation of state  =  %0.2f J/mol\"%(dep_h);\n",
+      "print \" The entropy departure for n-octane vapour using the generalized Redlich\\\n",
+      "-Kwong equation of state  =  %0.4f J/mol K\"%(dep_s);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure for n-octane vapour using the generalized Redlich-Kwong equation of state  =  -674.98 J/mol\n",
+        " The entropy departure for n-octane vapour using the generalized Redlich-Kwong equation of state  =  -1.0195 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.6  Page No : 281"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 427.85\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215\t\t\t #pressure of n-octane vapour in MPa\n",
+      "S = 1.0786\t\t\t #constant used in the SRK equation of state,from Example(3.15)\n",
+      "alpha = 1.3079\t\t #constant used in the SRK equation of state,from Example(3.15)\n",
+      "a = 5.0180\t\t\t #constant used in the SRK equation of state,from Example(3.15) in (m**6 Pa mol**-2)\n",
+      "b = 1.6426*10**-4\t\t\t #constant used in the SRK equation of state,from Example(3.15) in m**3/mol\n",
+      "B = 9.9282*10**-3\t\t\t #factor used in the Cardan's method for solving the SRK equation of state,from Example(3.15) (no unit)\n",
+      "Z = 0.9191;\t\t\t #compressibility factor taken from Example (3.15) (no unit)\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "\n",
+      "# Calculations\n",
+      "da_dT = (-a*S)/(math.sqrt (alpha*T*Tc))\n",
+      "dep_h = (R*T*(Z-1))+((((T*da_dT)-a)/b)*math.log ((Z+B)/Z));\t\t\t \n",
+      "dep_s = (R*math.log (Z-B))+((1./b)*(da_dT)*math.log ((Z+B)/Z));\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure for n-octane vapour using the SRK equation of state  =  %f J/mol\"%(dep_h);\n",
+      "print \" The entropy departure for n-octane vapour using the SRK equation of state  =  %0.4f J/mol K\"%(dep_s);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure for n-octane vapour using the SRK equation of state  =  -884.335509 J/mol\n",
+        " The entropy departure for n-octane vapour using the SRK equation of state  =  -1.4188 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.7  Page No : 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 427.85\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215\t\t\t #pressure of n-octane vapour in MPa\n",
+      "S = 0.9457\t\t\t #constant used in the Peng-Robinson equation of state,from Example(3.16)\n",
+      "alpha = 1.2677\t\t #constant used in the Peng-Robinson equation of state,from Example(3.16)\n",
+      "a = 5.2024\t\t\t #constant used in the Peng-Robinson equation of state,from Example(3.16) in (m**6 Pa mol**-2)\n",
+      "b = 1.4750*10**-4\t #constant used in the Peng-Robinson equation of state,from Example(3.16) in m**3/mol\n",
+      "B = 8.9151*10**-3\t #factor used in the Cardan's method for solving the Peng-Robinson equation of state,from Example(3.16) (no unit)\n",
+      "Z = 0.9151\t\t\t #compressibility factor taken from Example (3.16) (no unit)\n",
+      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
+      "Tc = 569.4\t\t\t #critical temperature of n-octane in K\n",
+      "\n",
+      "# Calculations\n",
+      "da_dT = (-a*S)/(math.sqrt (alpha*T*Tc))\n",
+      "\n",
+      "dep_h = (R*T*(Z-1))+(((((T*da_dT)-a)/(2*math.sqrt(2)*b)))*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
+      "dep_s = (R*math.log (Z-B))+((1./(2*math.sqrt (2)*b))*(da_dT)*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure for n-octane vapour using the Peng-Robinson \\\n",
+      " equation of state  =  %0.1f J/mol\"%(dep_h);\n",
+      "print \" The entropy departure for n-octane vapour using the Peng-Robinson\\\n",
+      " equation of state  =  %0.3f J/mol K\"%(dep_s);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure for n-octane vapour using the Peng-Robinson  equation of state  =  -890.1 J/mol\n",
+        " The entropy departure for n-octane vapour using the Peng-Robinson equation of state  =  -1.398 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.8  Page No : 284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 339.7\t\t\t #temperature of ethylene in K\n",
+      "P = 30.7\t\t\t #pressure of ethylene in bar\n",
+      "Tc = 283.1\t\t\t #critical temperature of ethylene in K\n",
+      "Pc = 51.17\t\t\t #critical pressure of ethylene in bar\n",
+      "w = 0.089\t\t\t #acentric factor (no unit)\n",
+      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Pr = P/Pc\n",
+      "Tr = T/Tc\n",
+      "del_h0 = 0.45\n",
+      "del_h1 = 0.18\n",
+      "del_s0 = 0.26\n",
+      "del_s1 = 0.20\n",
+      "dep_h = ((del_h0)+(w*del_h1))*R*Tc\n",
+      "dep_s = ((del_s0)+(w*del_s1))*R;\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure for ethylene using the Edmister charts  =  %0.3f J/mol\"%(dep_h);\n",
+      "print \" The entropy departure for ethylene using the Edmister charts  =  %0.4f J/mol K\"%(dep_s);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure for ethylene using the Edmister charts  =  1096.868 J/mol\n",
+        " The entropy departure for ethylene using the Edmister charts  =  2.3096 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.9  Page No : 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 339.7;\t\t\t #temperature of ethylene in K\n",
+      "P = 30.7;\t\t\t #pressure of ethylene in bar\n",
+      "Tc = 283.1;\t\t\t #critical temperature of ethylene in K\n",
+      "Pc = 51.17;\t\t\t #critical pressure of ethylene in bar\n",
+      "w = 0.089;\t\t\t #acentric factor (no unit)\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Pr = P/Pc\n",
+      "Tr = T/Tc\n",
+      "del_h0 = 0.474\n",
+      "del_h1 = 0.232\n",
+      "del_s0 = 0.277\n",
+      "del_s1 = 0.220\n",
+      "dep_h = ((del_h0)+(w*del_h1))*R*Tc\n",
+      "dep_s = ((del_s0)+(w*del_s1))*R\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure for ethylene using the Lee-Kesler data  =  %f J/mol\"%(dep_h);\n",
+      "print \" The entropy departure for ethylene using the Lee-Kesler data  =  %f J/mol K\"%(dep_s);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure for ethylene using the Lee-Kesler data  =  1164.249733 J/mol\n",
+        " The entropy departure for ethylene using the Lee-Kesler data  =  2.465766 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.10  Page No : 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 339.7;\t\t\t #temperature of ethylene in K\n",
+      "P = 1.  \t\t\t #pressure of ethylene in bar\n",
+      "Tc = 283.1;\t\t\t #critical temperature of ethylene in K\n",
+      "Pc = 51.17;\t\t\t #critical pressure of ethylene in bar\n",
+      "w = 0.089;\t\t\t #acentric factor (no unit)\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Pr = P/Pc\n",
+      "Tr = T/Tc\n",
+      "dep_h = R*Tc*Pr*((0.083-(1.097/(Tr**1.6)))+(w*(0.139-(0.894/(Tr**4.2)))))\n",
+      "dep_s = -Pr*R*((0.675/(Tr**2.6))+(w*(0.722/(Tr**5.2))));\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure for ethylene using the generalized virial coefficient \\\n",
+      "correlation  =  %f J/mol\"%(dep_h);\n",
+      "print \" The entropy departure for ethylene using the generalized virial coefficient \\\n",
+      "correlation  =  %e J/mol K\"%(dep_s);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure for ethylene using the generalized virial coefficient correlation  =  -35.011078 J/mol\n",
+        " The entropy departure for ethylene using the generalized virial coefficient correlation  =  -7.232682e-02 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.11  Page No : 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "import cmath\n",
+      "\n",
+      "# Variables\n",
+      "T = 427.85\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215\t\t\t #pressure of n-octane vapour in MPa\n",
+      "T_ref = 0.\t\t\t #reference state saturated liquid temperature in degree celsius\n",
+      "h0 = 0. \t\t\t #enthalpy of saturated liquid in J/mol (reference state)\n",
+      "s0 = 0. \t\t\t #entropy of saturated liquid in J/molK (reference state)\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
+      "w = 0.398;\t\t\t #acentric factor (no unit)\n",
+      "NBP = 398.8;\t\t\t #normal boiling point of n-octane (saturated liquid)\n",
+      "Cp = [6.907,741.770*10**-3,-397.204*10**-6,82.629*10**-9,0]\t\t\t #coefficients to compute the isobaric molar heat capacity, where Cp = a+bT+cT**2+dT**3+eT**-2,in J/molK\n",
+      "S = 0.9457;\t\t\t #value of S taken from Example (3.16)\n",
+      "b = 1.4750*10**-4;\t\t\t #value of the Peng-Robinson constant in m**3/mol from Example (3.16)\n",
+      "v = 15.14*10**-3;\t\t\t #volume of saturated vapour in m**3/mol from Example (3.16)\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "P_amb = 101.325;\t\t\t #pressure at which the normal boiling point is computed in kPa\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "#Step a: Vaporization of n-octane at T_ref\n",
+      "T_ref = T_ref+273.15\n",
+      "\n",
+      "del_hv = ((R*math.log ((Pc*10**5)/(P_amb*10**3)))/((1./NBP)-(1./Tc)))*10**-3;\n",
+      "P2 = P_amb* math.exp (((del_hv*10**3)/(R))*((1./NBP)-(1./T_ref)))\n",
+      "Tbr = NBP/Tc\n",
+      "\t\t\t\n",
+      "del_hvn = (1.093*R*Tc*(Tbr*(((math.log (Pc))-1.013)/(0.930-Tbr))))*10**-3;\n",
+      "Tr2 = T_ref/Tc\n",
+      "\n",
+      "del_ha = ((del_hvn*10**3)*(((1-Tr2)/(1-Tbr))**(0.38)))*10**-3;\n",
+      "del_sa = (del_ha*10**3)/T_ref\n",
+      "\n",
+      "alpha = (1+(S*(1-math.sqrt (Tr2))))**2\n",
+      "a = (0.45724*(R**2)*(Tc**2)*alpha)/(Pc*10**5)\n",
+      "\n",
+      "A = (a*P2*10**3)/(R*T_ref)**2\n",
+      "B = (b*P2*10**3)/(R*T_ref);\t\n",
+      "alpha = -1+B;\t\t\t \n",
+      "beeta = A-(2*B)-(3*B**2);\n",
+      "gaamma = -(A*B)+(B**2)+(B**3);\t\t\t \n",
+      "p = beeta-(alpha**2)/3;\t\t\t \n",
+      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
+      "D = (((q)**2)/4)+(((p)**3)/27);\t\t\t\n",
+      "\n",
+      "if D>0:\n",
+      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3)\n",
+      "elif D == 0:\n",
+      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
+      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Za = [Z1 ,Z2, Z3];\n",
+      "    Z = max(Za);\n",
+      "else:\n",
+      "    theta = math.cos((-(q)/2)*(math.sqrt((-27)/(((p)**3)))))\n",
+      "    r = math.sqrt((-(p**3)/27));\t\t\t \n",
+      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Za = [Z1, Z2, Z3];\n",
+      "    Z = max(Za);\n",
+      "da_dT = (-a*S)/(cmath.sqrt(alpha*T_ref*Tc));\t\t\t \n",
+      "\n",
+      "dep_h = (R*T_ref*(Z-1))+(((((T_ref*da_dT)-a)/(2*math.sqrt(2)*b)))*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
+      "dep_s = (R*math.log (Z-B))+((1./(2*math.sqrt (2)*b))*(da_dT)*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
+      "del_hb = -dep_h\n",
+      "del_sb = -dep_s\n",
+      "\n",
+      "del_hc = ((Cp[0]*(T-T_ref))+(((Cp[1])/2)*((T**2)-(T_ref**2)))+(((Cp[2])/3)*((T**3)-(T_ref**3)))+(((Cp[3])/4)*((T**4)-(T_ref**4)))-((Cp[4])*((1./T)-(1./T_ref))))*10**-3;\n",
+      "del_sc = ((Cp[0])*math.log (T/T_ref))+((Cp[1])*(T-T_ref))+(((Cp[2])/2)*((T**2)-(T_ref**2)))+(((Cp[3])/3)*((T**3)-(T_ref**3)))-(((Cp[4])/2)*((1./(T**2))-(1./(T_ref**2))))-(R*math.log((P*10**6)/(P2*10**3)))\t\t\t \n",
+      "\n",
+      "Z = 0.9151\n",
+      "da_dT = (-a*S)/(cmath.sqrt (alpha*T*Tc));\t\t\t \n",
+      "\n",
+      "del_hd = (R*T*(Z-1))+((((T*da_dT)-a)/(2*math.sqrt(2)*b))*math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2))))));\n",
+      "\n",
+      "del_sd = (R*math.log (Z-B))+((1./(2*math.sqrt(2)*b))*(da_dT)*(math.log ((Z+(B*(1+math.sqrt (2))))/(Z+(B*(1-math.sqrt (2)))))));\n",
+      "\n",
+      "h = h0+del_ha+(del_hb*10**-3)+del_hc+(del_hd*10**-3)\n",
+      "s = s0+del_sa+del_sb+del_sc+del_sd;\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation\\\n",
+      " of state  =\",h, \"kJ/mol\"\n",
+      "print \" The entropy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson\\\n",
+      " equation of state  = \",s,\" J/mol K\"\n",
+      "print \" The volume of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson\\\n",
+      " equation of state  =\",v,\" m**3/mol\"\n",
+      "\n",
+      "\n",
+      "#THE VOLUME OF n-OCTANE VAPOUR AS COMPUTED IN EXAMPLE 3.16 IS 15.14*10**-3 m**3/mol AND NOT \n",
+      "#15.41*10**-3 m**3/mol AS PRINTED IN THIS EXAMPLE IN THE TEXTBOOK.\n",
+      "# So ANSWER WOULD BE DIFFERENT\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state  = (76.8343786515+0.000518888144511j) kJ/mol\n",
+        " The entropy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state  =  (207.222784016-0.000740500199878j)  J/mol K\n",
+        " The volume of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state  = 0.01514  m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch9.ipynb b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch9.ipynb
new file mode 100755
index 00000000..9efdc810
--- /dev/null
+++ b/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch9.ipynb
@@ -0,0 +1,1125 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Multicomponent mixtures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.1  Page No : 313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "per_ethanol = 60.\t\t\t #mole percent of ethanol in a ethanol-water system\n",
+      "per_water = 40. \t\t\t #mole percent of water in a ethanol-water system\n",
+      "v1 = 57.5*10**-6\t\t\t #partial molar volume of ethanol in the ethanol-water system in m**3\n",
+      "rho = 849.4;\t    \t\t #density of the mixture in kg/m**3\n",
+      "M_ethanol = 46*10**-3\t\t\t #molar mass of ethanol in kg/mol\n",
+      "M_water = 18*10**-3 \t\t\t #molar mass of ethanol in kg/mol\n",
+      "\n",
+      "# Calculations\n",
+      "X1 = per_ethanol/100\n",
+      "X2 = per_water/100\n",
+      "M = (X1*M_ethanol)+(X2*M_water)\n",
+      "v = M/rho\n",
+      "v2 = (v-(X1*v1))/(X2);\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The partial molar volume of water  =  %f m**3/mol\"%(v2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The partial molar volume of water  =  0.000016 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.2  Page No : 313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 3.\t\t\t            #volume of mixture to be prepared in m**3\n",
+      "per_ethanol = 60.\t\t\t #mole percent of ethanol in a ethanol-water system\n",
+      "per_water = 40.\t\t\t #mole percent of water in a ethanol-water system\n",
+      "v1 = 57.5*10**-6;\t\t\t #partial molar volume of ethanol in the ethanol-water system in m**3/mol\n",
+      "v2 = 16*10**-6\t\t\t #partial molar volume of water in the ethanol-water system in m**3/mol\n",
+      "v1_pure = 57.9*10**-6\t\t\t #molar volume of pure ethanol in m**3/mol\n",
+      "v2_pure = 18*10**-6\t\t\t #molar volume of pure water in m**3/mol\n",
+      "\n",
+      "# Calculations\n",
+      "X1 = per_ethanol/100;\t\t\t # Calculations of the mole fraction of ethanol (no unit)\n",
+      "X2 = per_water/100;\t\t\t    # Calculations of the mole fraction of water (no unit)\n",
+      "v = (X1*v1)+(X2*v2);\t\t\t # Calculations of the molar volume of the solution using Eq.(9.10) in m**3/mol\n",
+      "N = V/v\t            \t\t # Calculations of the mole number of solution required in mol\n",
+      "N1 = N*X1\t\t\t         # Calculations of the mole number of ethanol in solution in mol\n",
+      "N2 = N*X2;\t\t\t         # Calculations of the mole number of water in solution in mol\n",
+      "V1 = N1*v1_pure;\t\t\t # Calculations of the volume of pure ethanol required in m**3\n",
+      "V2 = N2*v2_pure;\t\t\t # Calculations of the volume of pure water required in m**3\n",
+      "\n",
+      "# Results\n",
+      "print \" The volume of pure ethanol required  =  %0.3f m**3\"%(V1);\n",
+      "print \" The volume of pure water required  =  %0.3f m**3\"%(V2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume of pure ethanol required  =  2.548 m**3\n",
+        " The volume of pure water required  =  0.528 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.3  Page No : 318"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
+      "a = 3.789;\t\t\t #van der Waals constant in Pa(m**3/mol)**2\n",
+      "b = 2.37*10**-4;\t\t\t #van der Waals constant in m**3/mol\n",
+      "v = 15.675*10**-3;\t\t\t #molar volume of n-octane saturated vapour taken from Example 3.8 in m**3/mol\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Z = (P*10**6*v)/(R*T)\t\t\t # Calculations of the compressibility factor (no unit)\n",
+      "# Calculations of the fugacity coefficient (f/P) using the expression derived in Example 9.3 (no unit)\n",
+      "phi = math.exp(Z-1-math.log (((P*10**6)*(v-b))/(R*T))-a/(R*T*v));\n",
+      "f = (P*10**6*phi)*10**-6\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of n-octane vapour  =  %0.2f \"%(phi);\n",
+      "print \" The fugacity of n-octane vapour  =  %0.4f MPa\"%(f);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of n-octane vapour  =  0.95 \n",
+        " The fugacity of n-octane vapour  =  0.2043 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.5  Page No : 322"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 427.85;\t\t\t #temperature of n-octane vapour in K\n",
+      "P = 0.215;\t\t\t #pressure of n-octane vapour in MPa\n",
+      "Tc = 569.4;\t\t\t #critical temperature of n-octane in K\n",
+      "Pc = 24.97;\t\t\t #critical pressure of n-octane in bar\n",
+      "w = 0.398;\t\t\t #acentric factor (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "Tr = T/Tc\n",
+      "Pr = (P*10**6)/(Pc*10**5)\n",
+      "log_phi0 = -0.032;\t\n",
+      "log_phi1 = -0.025;\n",
+      "phi = round(log_phi0 + 0.398*log_phi1,3)\n",
+      "phie = 1.1014\n",
+      "f = P*phie\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of n-octane vapour  =  %.3f \"%(phi);\n",
+      "print \" The fugacity of n-octane vapour  =  %.4f MPa\"%(f);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of n-octane vapour  =  -0.042 \n",
+        " The fugacity of n-octane vapour  =  0.2368 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.6  Page No : 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 339.7\t\t\t #temperature of ethylene in K\n",
+      "P = 1.\t    \t\t #pressure of ethylene in bar\n",
+      "Tc = 283.1\t\t\t #critical temperature of ethylene in K\n",
+      "Pc = 51.17\t\t\t #critical pressure of ethylene in bar\n",
+      "w = 0.089\t\t\t #acentric factor (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "Tr = T/Tc\n",
+      "Pr = P/Pc\n",
+      "B0 = 0.083-(0.422/(Tr**1.6))\n",
+      "B1 = 0.139-(0.172/(Tr**4.2))\n",
+      "phi = math.exp((B0+(w*B1))*(Pr/Tr))\n",
+      "f = P*phi\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of ethylene  =  %0.4f \"%(phi);\n",
+      "print \" The fugacity of ethylene  =  %0.4f bar\"%(f);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of ethylene  =  0.9963 \n",
+        " The fugacity of ethylene  =  0.9963 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.7  Page No : 330"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.\t        \t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
+      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
+      "vc = [255.0*10**-6,486.0*10**-6];\t\t\t #critical molar volume of n-butane and n-octane in m**3/mol\n",
+      "Zc = [0.274,0.256];\t        \t\t #compressibility factor of n-butane and n-octane corresponding to Tc,Pc (no unit)\n",
+      "w = [0.199,0.398];\t\t\t         #acentric factor of n-butane and n-octane (no unit)\n",
+      "R = 8.314               \t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "K_12 = 1-((8*((vc[0]*vc[1])**(1./2)))/((((vc[0])**(1./3))+((vc[1])**(1./3)))**3))\n",
+      "Tc_12 = (((Tc[0])*(Tc[1]))**(1./2))*(1-K_12);\t\t\t \n",
+      "w_12 = (w[0]+w[1])/2\n",
+      "Zc_12 = (Zc[0]+Zc[1])/2\n",
+      "vc_12 = ((((vc[0])**(1./3))+((vc[1])**(1./3)))/2)**3\n",
+      "Pc_12 = ((Zc_12*R*Tc_12)/vc_12)*10**-6;\t\t\t \n",
+      "Tr_12 = T/Tc_12\n",
+      "B_12_0 = 0.083-(0.422/(Tr_12**(1.6)));\t\t\t \n",
+      "B_12_1 = 0.139-(0.172/(Tr_12**(4.2)));\t\t\t \n",
+      "B_12 = ((R*Tc_12)/(Pc_12*10**6))*(B_12_0+(w_12*B_12_1))\n",
+      "Tr1 = T/Tc[0]\n",
+      "B_11_0 = 0.083-(0.422/(Tr1**(1.6)));\t\t\t \n",
+      "B_11_1 = 0.139-(0.172/(Tr1**(4.2)));\t\t\t \n",
+      "B_11 = ((R*Tc[0])/(Pc[0]*10**5))*(B_11_0+(w[0]*B_11_1))\n",
+      "Tr2 = T/Tc[1]\n",
+      "B_22_0 = 0.083-(0.422/(Tr2**(1.6)))\n",
+      "B_22_1 = 0.139-(0.172/(Tr2**(4.2)))\n",
+      "B_22 = ((R*Tc[1])/(Pc[1]*10**5))*(B_22_0+(w[1]*B_22_1))\n",
+      "Bm = ((y1**2)*B_11)+((2*y1*y2)*B_12)+((y2**2)*B_22);\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The second virial coefficient for an equimolar mixture of n-butane and n-octane at\\\n",
+      " 600K  =  %.2e m**3/mol\"%(Bm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The second virial coefficient for an equimolar mixture of n-butane and n-octane at 600K  =  -3.09e-04 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.8  Page No : 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "a = [1.3874,3.7890];\t\t\t #van der Waals constant of n-butane and n-octane in Pa(m**3/mol)**2\n",
+      "b = [0.1163*10**-3,0.237*10**-3];\t\t\t #van der Waals constant of n-butane and n-octane in m**3/mol\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "a_m = ((y1**2)*a[0])+((2*y1*y2)*math.sqrt(a[0]*a[1]))+((y2**2)*a[1])\n",
+      "b_m = (y1*b[0])+(y2*b[1])\t\t\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The van der Waals constant for an equimolar mixture of n-butane and n-octane,\\\n",
+      " a_m  =  %0.4f Pam**3/mol)**2\"%(a_m);\n",
+      "print \" The van der Waals constant for an equimolar mixture of n-butane and n-octane,\\\n",
+      " b_m  =  %f m**3/mol\"%(b_m);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The van der Waals constant for an equimolar mixture of n-butane and n-octane, a_m  =  2.4405 Pam**3/mol)**2\n",
+        " The van der Waals constant for an equimolar mixture of n-butane and n-octane, b_m  =  0.000177 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.9  Page No : 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.\t    \t    \t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16.\t    \t    \t     #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
+      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
+      "R = 8.314;\t\t\t        #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "Tr1 = T/Tc[0]\n",
+      "Pr1 = P/Pc[0]\n",
+      "Z1_0 = 0.95\n",
+      "Tr2 = T/Tc[1]\n",
+      "Pr2 = P/Pc[1]\n",
+      "Z2_0 = 0.785\n",
+      "\n",
+      "Zm = (y1*Z1_0)+(y2*Z2_0)\n",
+      "vm = (Zm*R*T)/(P*10**5)\n",
+      "\n",
+      "P1 = y1*P\n",
+      "P2 = y2*P\n",
+      "Pr1 = P1/Pc[0]\n",
+      "Pr2 = P2/Pc[1]\n",
+      "Z1_0 = 0.97\n",
+      "Z2_0 = 0.91\n",
+      "Zm = (y1*Z1_0)+(y2*Z2_0)\n",
+      "vm_dalton = (Zm*R*T)/(P*10**5)\n",
+      "\n",
+      "P1 = ((Z1_0*y1*R*T)/(vm_dalton))*10**-2\n",
+      "P2 = ((Z2_0*y2*R*T)/(vm_dalton))*10**-2\n",
+      "Pr1 = P1/Pc[0]\n",
+      "Pr2 = P2/Pc[1]\n",
+      "Z1_0_new = 0.97\n",
+      "Z2_0_new = 0.91\n",
+      "if Z1_0_new == Z1_0 and Z2_0_new == Z2_0:\n",
+      "    vm_new = vm_dalton;\t\t\t \n",
+      "else:\n",
+      "    Zm = (y1*Z1_0_new)+(y2*Z2_0_new)\n",
+      "    vm_new = (Zm*R*T)/(P*10**5);\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume of an equimolar mixture of n-butane and n-octane found using the\\\n",
+      " Amagats law of additive volumes  =  %0.4e m**3/mol\"%(vm);\n",
+      "print \" The molar volume of an equimolar mixture of n-butane and\\\n",
+      " n-octane found using the Daltons law of additive pressures  =  %0.2e m**3/mol\"%(vm_new);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume of an equimolar mixture of n-butane and n-octane found using the Amagats law of additive volumes  =  2.7046e-03 m**3/mol\n",
+        " The molar volume of an equimolar mixture of n-butane and n-octane found using the Daltons law of additive pressures  =  2.93e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.10  Page No : 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16.\t\t\t     #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
+      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
+      "R = 8.314;\t    \t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "Tcm = (y1*Tc[0])+(y2*Tc[1]);\t\t\t \n",
+      "Pcm = (y1*Pc[0])+(y2*Pc[1])\n",
+      "Trm = T/Tcm;\t\t\t \n",
+      "Prm = P/Pcm;\t\t\t \n",
+      "Zm0 = 0.9;\t\t\t \n",
+      "vm = (Zm0*R*T)/(P*10.**5)\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume of an equimolar mixture of n-butane and n-octane using the\\\n",
+      " pseudocritical properties estimated through Kays rule  =  %0.2e m**3/mol\"%(vm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume of an equimolar mixture of n-butane and n-octane using the pseudocritical properties estimated through Kays rule  =  2.81e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.11  Page No : 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.\t\t\t         #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t         #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
+      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
+      "vc = [255.0*10**-6,486.0*10**-6];\t\t\t #critical molar volume of n-butane and n-octane in m**3/mol\n",
+      "Zc = [0.274,0.256];\t\t\t #compressibility factor of n-butane and n-octane corresponding to Tc,Pc (no unit)\n",
+      "R = 8.314;\t        \t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "Tcm = (y1*Tc[0])+(y2*Tc[1]);\t\t\t \n",
+      "Pcm = ((R*((y1*Zc[0])+(y2*Zc[1]))*Tcm)/((y1*vc[0])+(y2*vc[1])))*10**-5\n",
+      "Trm = T/Tcm\n",
+      "Prm = P/Pcm\n",
+      "Zm0 = 0.89\n",
+      "vm = (Zm0*R*T)/(P*10**5)\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume of an equimolar mixture of n-butane and n-octane at 600K\\\n",
+      " and 16bar estimated using the Prausnitz-Gunn rule  =  %0.2e m**3/mol\"%(vm);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar estimated using the Prausnitz-Gunn rule  =  2.77e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.12  Page No : 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "a_m = 2.4405\t\t\t #van der Waals constant for the mixture as determined in Example 9.8 in Pa(m**3/mol)**2\n",
+      "b_m = 0.1767*10**-3\t\t #van der Waals constant for the mixture as determined in Example 9.8 in m**3/mol\n",
+      "R = 8.314;\t\t\t     #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "A = (a_m*P*10**5)/(R*T)**2\n",
+      "B = (b_m*P*10**5)/(R*T)\n",
+      "alpha = -1-B\n",
+      "beeta = A\n",
+      "gaamma = -(A*B)\n",
+      "p = beeta-((alpha**2)/3)\n",
+      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
+      "D = (((q)**2)/4)+(((p)**3)/27)\n",
+      "\n",
+      "if D>0 :\n",
+      "    Z = (((-(q)/2)+(math.sqrt(D)))**(1./3))+(((-(q)/2)-(math.sqrt(D)))**(1./3))-(alpha/3);\t\n",
+      "elif D == 0:\n",
+      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
+      "    Z2 = ((q/2)**(1./3))-(alpha/3)\n",
+      "    Z3 = ((q/2)**(1./3))-(alpha/3)\n",
+      "    Za = [Z1 ,Z2, Z3];\n",
+      "    Z = max(Za);\n",
+      "else:\n",
+      "    r = math.sqrt((-(p**3)/27));\t\t\t \n",
+      "    theta = amath.cos((-(q)/2)*(1./r));\t\t\n",
+      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Za = [Z1, Z2, Z3];\n",
+      "    Z = max(Za);\n",
+      "\n",
+      "vm = (Z*R*T)/(P*10**5)\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar found \\\n",
+      " using the van der Waals equation of state  =  %e m**3/mol\"%(vm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar found  using the van der Waals equation of state  =  2.780786e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.13  Page No : 336"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.\t\t    \t     #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16.\t\t    \t         #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Bm = -309.*10**-6;\t\t\t #second virial coefficient for the mixture taken from Example(9.7) in m**3/mol\n",
+      "R = 8.314;\t\t\t         #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "Zm = 1+((Bm*P*10**5)/(R*T))\n",
+      "vm = (Zm*R*T)/(P*10**5)\t\n",
+      "\n",
+      "# Results\n",
+      "print \" The molar volume of an equimolar mixture of n-butane and n-octane\\\n",
+      " found using the generalized virial coefficient correlation  =  %0.4e m**3/mol\"%(vm);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The molar volume of an equimolar mixture of n-butane and n-octane found using the generalized virial coefficient correlation  =  2.8088e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.14  Page No : 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "am = 2.4405\t\t\t #van der Waals constant for the mixture taken from Example 9.8 in Pa(m**3/mol)**2\n",
+      "bm = 0.1767*10**-3;\t\t\t #van der Waals constant for the mixture taken from Example 9.8 in m**3/mol\n",
+      "vm = 2.8933*10**-3;\t\t\t #molar volume of the mixture taken from Example 9.12 in m**3/mol\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "dep_h = ((P*10**5*vm)-(R*T)-(am/vm))*10**-3\n",
+      "dep_s = R*(math.log ((P*10**5*(vm-bm))/(R*T)))\n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure of an equimolar mixture of n-butane and n-octane  =  %0.3f kJ/mol\"%(dep_h);\n",
+      "print \" The entropy departure of an equimolar mixture of n-butane and n-octane  =  %0.3f J/mol K\"%(dep_s);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure of an equimolar mixture of n-butane and n-octane  =  -1.203 kJ/mol\n",
+        " The entropy departure of an equimolar mixture of n-butane and n-octane  =  -1.145 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.15  Page No : 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.    \t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t    \t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Tcm = 497.3;\t\t\t #pseudocritical temperature of mixture taken from Example(9.10) in K\n",
+      "Pcm = 31.47;\t\t\t #pseudocritical pressure of mixture taken from Example(9.10) in bar\n",
+      "Trm = 1.21;\t\t    \t #pseudoreduced temperature of the mixture taken from Example(9.10) (no unit)\n",
+      "Prm = 0.51;\t\t\t     #pseudoreduced pressure of the mixture taken from Example(9.10) (no unit)\n",
+      "w_butane = 0.199;\t\t #acentric factor for n-butane (no unit)\n",
+      "w_octane = 0.398;\t\t #acentric factor for n-octane (no unit)\n",
+      "R = 8.314;\t\t    \t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "wm = (y1*w_butane)+(y2*w_octane)\n",
+      "del_h0 = 0.380;\t\t\t \n",
+      "del_h1 = 0.188;\t\t\t \n",
+      "del_s0 = 0.22;\t\t\t \n",
+      "del_s1 = 0.18;\t\t\t \n",
+      "dep_h = ((R*Tcm)*(del_h0+(wm*del_h1)))*10**-3;\t\t\t \n",
+      "dep_s = (R)*(del_s0+(wm*del_s1));\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure of an equimolar mixture of n-butane and n-octane using\\\n",
+      " the generalized compressibility factor correlation  =  %0.3f kJ/mol\"%(dep_h);\n",
+      "print \" The entropy departure of an equimolar mixture of n-butane and n-octane using \\\n",
+      "the generalized compressibility factor correlation  =  %f J/mol K\"%(dep_s);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure of an equimolar mixture of n-butane and n-octane using the generalized compressibility factor correlation  =  1.803 kJ/mol\n",
+        " The entropy departure of an equimolar mixture of n-butane and n-octane using the generalized compressibility factor correlation  =  2.275791 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.16  Page No : 339"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
+      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
+      "w = [0.199,0.398];\t\t\t #acentric factor of n-butane and n-octane (no unit)\n",
+      "Tr1 = 1.411;\t\t\t #reduced temperature of n-butane (no unit) taken from Example (9.7)\n",
+      "Tr2 = 1.054;\t\t\t #reduced temperature of n-octane (no unit) taken from Example (9.7)\n",
+      "Tr_12 = 1.24;\t\t\t #reduced temperature for computing the mixture interaction virial coefficient (no unit) taken from Example(9.7)\n",
+      "Pc_12 = 2.978;\t\t\t #Pc_ij in MPa taken from Example(9.7)\n",
+      "Tc_12 = 483.9;\t\t\t #Tc_ij in K taken from Example(9.7)\n",
+      "w_12 = 0.2985;\t\t\t # w_ij (no unit) taken from Example(9.7)\n",
+      "Bm = -309*10**-6;\t\t\t #second virial coefficient in m**3/mol taken from Example (9.7)\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "dB0_dTr1 = 0.675/(Tr1**2.6);\t\t\t \n",
+      "dB0_dTr2 = 0.675/(Tr2**2.6);\t\t\t \n",
+      "dB1_dTr1 = 0.722/(Tr1**5.2);\t\t\t \n",
+      "dB1_dTr2 = 0.722/(Tr2**5.2);\t\t\t \n",
+      "dB0_dTr12 = 0.675/(Tr_12**2.6);\t\t\t \n",
+      "dB1_dTr12 = 0.722/(Tr_12**5.2);\t\t\t \n",
+      "dB1_dT = (R/(Pc[0]*10**5))*((dB0_dTr1)+(w[0]*(dB1_dTr1)));\t\t\t\n",
+      "dB2_dT = (R/(Pc[1]*10**5))*((dB0_dTr2)+(w[1]*(dB1_dTr2)));\t\t\t\n",
+      "dB12_dT = (R/(Pc_12*10**6))*((dB0_dTr12)+(w_12*(dB1_dTr12)));\t\t\t\n",
+      "dBm_dT = ((y1**2)*(dB1_dT))+((2*y1*y2)*(dB12_dT))+((y2**2)*(dB2_dT));\t\n",
+      "dep_h = ((Bm-(T*dBm_dT))*P*10**5)*10**-3\n",
+      "dep_s = -P*10**5*(dBm_dT);\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The enthalpy departure of an equimolar mixture of n-butane and n-octane\\\n",
+      " using the virial coefficient correlation  =  %f kJ/mol\"%(dep_h);\n",
+      "print \" The entropy departure of an equimolar mixture of n-butane and n-octane\\\n",
+      " using the virial coefficient correlation  =  %0.3f J/mol K\"%(dep_s);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy departure of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  -1.908476 kJ/mol\n",
+        " The entropy departure of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  -2.357 J/mol K\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.17  Page No : 340"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "a_m = 2.4405\t\t\t #van der Waals constant (a_m) in Pa(m**3/mol)**2 taken from Example(9.8)\n",
+      "b_m = 0.1767*10**-3\t\t #van der Waals constant (b_m) in m**3/mol taken from Example(9.8)\n",
+      "Z = 0.928;\t\t\t     #compressibility factor taken from Example(9.12)\n",
+      "vm = 2.8933*10**-3;\t\t #molar volume of the equimolar mixture in m**3/mol taken from Example(9.12)\n",
+      "R = 8.314;\t\t\t     #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "phi = math.exp(Z-1-math.log ((P*10**5*(vm-b_m))/(R*T))-(a_m/(R*T*vm)));\t\t\t\n",
+      "f = phi*P;\t\t\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of an equimolar mixture of n-butane and n-octane\\\n",
+      " using the van der Waals equation of state  =  %0.4f \"%(phi);\n",
+      "print \" The fugacity of an equimolar mixture of n-butane and n-octane using the van der Waals \\\n",
+      "equation of state  =  %0.2f bar\"%(f);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the van der Waals equation of state  =  0.9018 \n",
+        " The fugacity of an equimolar mixture of n-butane and n-octane using the van der Waals equation of state  =  14.43 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.18  Page No : 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16.\t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Tcm = 497.3;\t\t\t #pseudocritical temperature of mixture in K taken from Example(9.10)\n",
+      "Pcm = 31.47;\t\t\t #pseudocritical pressure of mixture in bar taken from Example(9.10)\n",
+      "Trm = 1.21;\t\t\t #pseudoreduced temperature of the mixture (no unit) taken from Example(9.10)\n",
+      "Prm = 0.51;\t\t\t #pseudoreduced pressure of the mixture (no unit) taken from Example(9.10)\n",
+      "w = [0.199,0.398];\t\t\t #acentric factor of n-butane and n-octane (no unit)\n",
+      "\n",
+      "# Calculations\n",
+      "wm = (w[0]+w[1])/2\n",
+      "log_phi0 = -0.042\n",
+      "log_phi1 = 0.01\n",
+      "phi = 10**(log_phi0+(wm*log_phi1));\t\t\n",
+      "f = P*phi;\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of an equimolar mixture of n-butane and n-octane \\\n",
+      "using the pseudocritical constants method  =  %0.3f \"%(phi);\n",
+      "print \" The fugacity of an equimolar mixture of n-butane and n-octane using\\\n",
+      " the pseudocritical constants method  =  %f bar\"%(f);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the pseudocritical constants method  =  0.914 \n",
+        " The fugacity of an equimolar mixture of n-butane and n-octane using the pseudocritical constants method  =  14.625307 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.19  Page No : 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Bm = -309.*10**-6;\t\t\t #second virial coefficient in m**3/mol taken from Example (9.7)\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "phi = math.exp((Bm*P*10**5)/(R*T))\n",
+      "f = phi*P;\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of an equimolar mixture of n-butane and n-octane using\\\n",
+      " the virial coefficient correlation  =  %.3f \"%(phi);\n",
+      "print \" The fugacity of an equimolar mixture of n-butane and n-octane using\\\n",
+      " the virial coefficient correlation  =  %f bar\"%(f);\n",
+      "\n",
+      "# Note : answers in book is wrong please check manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  0.906 \n",
+        " The fugacity of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation  =  14.490290 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.20  Page No : 344"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "Tc = [425.2,569.4];\t\t\t #critical temperature of n-butane and n-octane in K\n",
+      "Pc = [37.97,24.97];\t\t\t #critical pressure of n-butane and n-octane in bar\n",
+      "R = 8.314;\t        \t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "a1 = (0.42748*R**2*Tc[0]**2.5)/(Pc[0]*10**5*math.sqrt(T))\n",
+      "a2 = (0.42748*R**2*Tc[1]**2.5)/(Pc[1]*10**5*math.sqrt(T))\n",
+      "b1 = (0.08664*R*Tc[0])/(Pc[0]*10**5);\t\t\t \n",
+      "b2 = (0.08664*R*Tc[1])/(Pc[1]*10**5);\t\t\t \n",
+      "\n",
+      "a = ((y1**2)*a1)+(2*y1*y2*math.sqrt(a1*a2))+((y2**2)*a2)\n",
+      "b = (y1*b1)+(y2*b2)\n",
+      "\n",
+      "A = (a*P*10**5)/(R*T)**2\n",
+      "B = (b*P*10**5)/(R*T);\t\n",
+      "alpha = -1.\t\t\t \n",
+      "beeta = A-B-B**2\n",
+      "gaamma = -(A*B)\n",
+      "p = beeta-(alpha**2)/3\n",
+      "q = ((2*alpha**3)/27)-((alpha*beeta)/3)+gaamma\n",
+      "D = (((q)**2)/4)+(((p)**3)/27)\n",
+      "\n",
+      "if D>0:\n",
+      "    Z = ((-q/2)+(math.sqrt(D)))**(1./3)+((-q/2)-(math.sqrt(D)))**(1./3)-(alpha/3);\t\t\t #One real root given by  Eq.(3.32)\n",
+      "elif D == 0:\n",
+      "    Z1 = ((-2*(q/2))**(1./3))-(alpha/3)\n",
+      "    Z2 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Z3 = ((q/2)**(1./3))-(alpha/3);\n",
+      "    Za = [Z1, Z2, Z3];\n",
+      "    Z = max(Za);\n",
+      "else:\n",
+      "    r = math.sqrt((-(p**3)/27));\t\t\n",
+      "    theta = amath.cos((-(q)/2)*(1./r));\t\n",
+      "    Z1 = (2*(r**(1./3))*math.cos(theta/3))-(alpha/3);\n",
+      "    Z2 = (2*(r**(1./3))*math.cos(((2*math.pi)+theta)/3))-(alpha/3)\n",
+      "    Z3 = (2*(r**(1./3))*math.cos(((4*math.pi)+theta)/3))-(alpha/3);\n",
+      "    Za = [Z1, Z2, Z3];\n",
+      "    Z = max(Za);\n",
+      "\n",
+      "phi1 = math.exp(((b1/b)*(Z-1))-math.log(Z-B)+((a/(b*R*T))*((b1/b)-(2*math.sqrt(a1/a)))*math.log((Z+B)/Z)));\n",
+      "\n",
+      "phi2 = math.exp(((b2/b)*(Z-1))-math.log(Z-B)+((a/(b*R*T))*((b2/b)-(2*math.sqrt(a2/a)))*math.log((Z+B)/Z)));\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of n-butane in the equimolar mixture using the\\\n",
+      " Redlich-Kwong Equation of state  =  %0.4f \"%(phi1);\n",
+      "print \" The fugacity coefficient of n-octane in the equimolar mixture using the\\\n",
+      " Redlich-Kwong Equation of state  =  %0.4f \"%(phi2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of n-butane in the equimolar mixture using the Redlich-Kwong Equation of state  =  0.9644 \n",
+        " The fugacity coefficient of n-octane in the equimolar mixture using the Redlich-Kwong Equation of state  =  0.8326 \n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.21  Page No : 346"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 600.\t\t\t #temperature of the equimolar n-butane and n-octane mixture in K\n",
+      "P = 16. \t\t\t #pressure of the equimolar n-butane and n-octane mixture in bar\n",
+      "B_11 = -131*10**-6\t\t\t #pure component (n-butane) second virial coefficient in m**3/mol taken from Example(9.7)\n",
+      "B_22 = -577*10**-6\t\t\t #pure component (n-octane) second virial coefficient in m**3/mol taken from Example(9.7)\n",
+      "B_12 = -264*10**-6\t\t\t #mixture interaction virial coefficient in m**3/mol taken from Example(9.7)\n",
+      "Bm = -309*10**-6\t\t\t #second virial coefficient in m**3/mol taken from Example(9.7)\n",
+      "R = 8.314       \t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "y1 = 0.5\n",
+      "y2 = 0.5\n",
+      "Zm = (1./2)*(1+math.sqrt(1+((4*Bm*P*10**5)/(R*T))))\n",
+      "phi1 = math.exp((((2*P*10**5)/(Zm*R*T))*((y1*B_11)+(y2*B_12)))-math.log(Zm))\n",
+      "phi2 = math.exp((((2*P*10**5)/(Zm*R*T))*((y1*B_12)+(y2*B_22)))-math.log(Zm))\n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity coefficient of n-butane in the equimolar mixture using the\\\n",
+      " Virial Equation of state  =  %0.3f \"%(phi1);\n",
+      "print \" The fugacity coefficient of n-octane in the equimolar mixture using the Virial\\\n",
+      " Equation of state  =  %f \"%(phi2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity coefficient of n-butane in the equimolar mixture using the Virial Equation of state  =  0.976 \n",
+        " The fugacity coefficient of n-octane in the equimolar mixture using the Virial Equation of state  =  0.830827 \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.22  Page No : 349"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Variables\n",
+      "T = 427.85  \t\t\t #temperature of n-octane vapour in K\n",
+      "Psat = 0.215;\t\t\t #saturation pressure of n-octane vapour at T in MPa\n",
+      "P = 1.\t\t        \t #pressure at which the fugacity of liquid n-octane is to be determined in MPa\n",
+      "f_sat = 0.2368;\t\t\t #fugacity of n-octane vapour at T and Psat taken from Example(9.5) in MPa\n",
+      "vl = 0.2003*10**-3;\t\t\t #molar volume of n-octane liquid at T and Psat taken from Example(3.16) in m**3/mol\n",
+      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
+      "\n",
+      "# Calculations\n",
+      "f_l = (0.2368*math.exp((vl*(P-Psat)*10**6)/(R*T)));\t\t\t \n",
+      "\n",
+      "# Results\n",
+      "print \" The fugacity of liquid n-octane at 427.85K and 1MPa  =  %0.4f MPa\"%(f_l);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fugacity of liquid n-octane at 427.85K and 1MPa  =  0.2475 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Computer_Programming,_Theory_and_Practice_by_T_Jeyapoovan/README.txt b/Computer_Programming,_Theory_and_Practice_by_T_Jeyapoovan/README.txt
old mode 100644
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diff --git a/DC_Machines_and_Synchronous_Machines/README.txt b/DC_Machines_and_Synchronous_Machines/README.txt
deleted file mode 100755
index 9c84b56e..00000000
--- a/DC_Machines_and_Synchronous_Machines/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Hiren Shah
-Course: mca
-College/Institute/Organization: Financial Technology
-Department/Designation: Developer
-Book Title: DC Machines and Synchronous Machines
-Author: U. A. Bakshi and M. V. Bakshi
-Publisher: Technical Publications, Pune
-Year of publication: 2008
-Isbn: 9788184314830
-Edition: 1
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines/ch1.ipynb b/DC_Machines_and_Synchronous_Machines/ch1.ipynb
deleted file mode 100755
index 04ab6c30..00000000
--- a/DC_Machines_and_Synchronous_Machines/ch1.ipynb
+++ /dev/null
@@ -1,1633 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:28530af987ec737018da049988fa2fb2a82d79acc8dd496873e17355c1e77828"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1 : D.C. Generators"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1 page no : 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "p = 4.\n",
-      "Z = 440.\n",
-      "theta = 0.07    # Wb\n",
-      "N = 900.        # rpm\n",
-      "\n",
-      "# calculations and results\n",
-      "E = theta*N*Z/60.\n",
-      "print \"For lap wound : E = %.f V\"%E\n",
-      "\n",
-      "E = theta*N*Z*4/120.\n",
-      "print \"For wave wound : E = %.f V\"%E\n",
-      "\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For lap wound : E = 462 V\n",
-        "For wave wound : E = 924 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 1.2 page no : 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "phi = 21.*10**-3  \t\t\t#flux produced by each pole in webers\n",
-      "N = 1120.        \t\t\t#Speed of armature in r.p.m\n",
-      "Coils = 42.\n",
-      "turns_per_coil = 8.\n",
-      "Turns = Coils * turns_per_coil\n",
-      "Z = 2*Turns  \t\t\t#Number of armature conductors\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "A1 = Pole  \t\t\t#no of parallel paths for lap winding\n",
-      "E1 = phi*N*Z*Pole/(60*A1)\n",
-      "print 'i)  e.m.f generated is %.3f V'%(E1)\n",
-      "\n",
-      "#Part(ii)\n",
-      "A2 = 2  \t\t\t#wave winding\n",
-      "E2 = E1 \t\t\t#as mentioned in the question\n",
-      "N2 = E2/(phi*Z*Pole/(60*A2)) \t\t\t#E = phi*N*Z*Pole/(60*A)\n",
-      "print 'ii)  For wave-wound armature,above calculated e.m.f is generated at %.0f r.p.m'%(N2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)  e.m.f generated is 263.424 V\n",
-        "ii)  For wave-wound armature,above calculated e.m.f is generated at 560 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3 page no: 27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "coils = 12.\n",
-      "commutator_segments = coils\n",
-      "coil_sides = coils*2\n",
-      "Z = coil_sides  \t\t\t#No of conductors\n",
-      "pole_pitch = Z/Pole\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#for Simplex lap winding\n",
-      "y_f = pole_pitch-1\n",
-      "y_b = pole_pitch+1\n",
-      "\n",
-      "y_c = 1  \t\t\t#Note that it's positive and it's progressive type of Simplex lap winding\n",
-      "\n",
-      "\n",
-      "print 'WINDING TABLE:   1<-    8->    3<-   10->    5<-   12-> 7<-   14->    9<-   16->   11<-   18->13<-   20->   15<-   22->   17<-   24->19<-    2->   21<-    4->   23<-    6-> 1                                                  '\n",
-      "print 'Note that <- indicates back connection with  y_back = %.0f and -> indicates front connection with y_front = %.0f'%(y_b,y_f)\n",
-      "print 'Another form of winding table:'\n",
-      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
-      "\n",
-      "print '       1 to 1+7  =  8                        ->        8 to 8-5  =  3'\n",
-      "print '       3 to 3+7  = 10                        ->       10 to 10-5 =  5'\n",
-      "print '       5 to 5+7  = 12                        ->       12 to 12-5 =  7'\n",
-      "print '       7 to 7+7  = 14                        ->       14 to 14-5 =  9'\n",
-      "print '       9 to 9+7  = 16                        ->       16 to 16-5 = 11'\n",
-      "print '      11 to 11+7 = 18                        ->       18 to 18-5 = 13'\n",
-      "print '      13 to 13+7 = 20                        ->       20 to 20-5 = 15'\n",
-      "print '      15 to 15+7 = 22                        ->       22 to 22-5 = 17'\n",
-      "print '      17 to 17+7 = 24                        ->       24 to 24-5 = 19'\n",
-      "print '      19 to 19+7 = 26 = 26-24 = 2              ->        2 to 26-5 = 21'\n",
-      "print '      21 to 21+7 = 28 = 28-24 = 4              ->        4 to 28-5 = 23'\n",
-      "print '      23 to 23+7 = 30 = 30-24 = 6              ->        6 to 30-5 = 25   =  25-24 = 1'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "WINDING TABLE:   1<-    8->    3<-   10->    5<-   12-> 7<-   14->    9<-   16->   11<-   18->13<-   20->   15<-   22->   17<-   24->19<-    2->   21<-    4->   23<-    6-> 1                                                  \n",
-        "Note that <- indicates back connection with  y_back = 7 and -> indicates front connection with y_front = 5\n",
-        "Another form of winding table:\n",
-        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
-        "       1 to 1+7  =  8                        ->        8 to 8-5  =  3\n",
-        "       3 to 3+7  = 10                        ->       10 to 10-5 =  5\n",
-        "       5 to 5+7  = 12                        ->       12 to 12-5 =  7\n",
-        "       7 to 7+7  = 14                        ->       14 to 14-5 =  9\n",
-        "       9 to 9+7  = 16                        ->       16 to 16-5 = 11\n",
-        "      11 to 11+7 = 18                        ->       18 to 18-5 = 13\n",
-        "      13 to 13+7 = 20                        ->       20 to 20-5 = 15\n",
-        "      15 to 15+7 = 22                        ->       22 to 22-5 = 17\n",
-        "      17 to 17+7 = 24                        ->       24 to 24-5 = 19\n",
-        "      19 to 19+7 = 26 = 26-24 = 2              ->        2 to 26-5 = 21\n",
-        "      21 to 21+7 = 28 = 28-24 = 4              ->        4 to 28-5 = 23\n",
-        "      23 to 23+7 = 30 = 30-24 = 6              ->        6 to 30-5 = 25   =  25-24 = 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4 page no : 32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "Z = 18.               \t\t\t#no of armature conductors\n",
-      "Y_A = (Z+2)/Pole     \t\t\t#For progressive type wave winding,positive sign is used\n",
-      "Y_C = Y_A            \t\t\t#For wave winding\n",
-      "\n",
-      "# Calculations\n",
-      "#Since Y_A = (y_b+y_f)/2, we let y_b = Y_f\n",
-      "y_b = Y_A/2   \t\t\t#say\n",
-      "y_f = y_b\n",
-      "\n",
-      "coils = Z/2\n",
-      "slots = coils\n",
-      "commutator_segments = coils\n",
-      "\n",
-      "print commutator_segments\n",
-      "print 'WINDING TABLE:   1<-    6->  11<-  16->   3<-   8->13<-   18->   5<-  10->  15<-   2-> 7<-   12->  17<-   4->   9<-  14->1                                                  '\n",
-      "\n",
-      "print 'Another form of winding table:'\n",
-      "\n",
-      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
-      "\n",
-      "print '       1 to 1+5   =  6                       ->        6 to 6+5  =  11'\n",
-      "print '      11 to 11+5  = 16                       ->       16 to 16+5 =  21 -18 = 3'\n",
-      "print '       3 to 3+5   =  8                       ->        8 to  8+5 =  13'\n",
-      "print '      13 to 13+5  = 18                       ->       18 to 18+5 =  23 -18 = 5'\n",
-      "print '       5 to 5+5   = 10                       ->       10 to 10+5 =  15'\n",
-      "print '      15 to 15+5  = 20 -18 = 2                 ->        2 to  2+5 =   7'\n",
-      "print '       7 to 7+5   = 12                       ->       12 to 12+5 =  17'\n",
-      "print '      17 to 17+5  = 22 -18 = 4                 ->        4 to  4+5 =   9'\n",
-      "print '       9 to 9+5   = 14                       ->       14 to 14+5 =  19 -18 = 1'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9.0\n",
-        "WINDING TABLE:   1<-    6->  11<-  16->   3<-   8->13<-   18->   5<-  10->  15<-   2-> 7<-   12->  17<-   4->   9<-  14->1                                                  \n",
-        "Another form of winding table:\n",
-        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
-        "       1 to 1+5   =  6                       ->        6 to 6+5  =  11\n",
-        "      11 to 11+5  = 16                       ->       16 to 16+5 =  21 -18 = 3\n",
-        "       3 to 3+5   =  8                       ->        8 to  8+5 =  13\n",
-        "      13 to 13+5  = 18                       ->       18 to 18+5 =  23 -18 = 5\n",
-        "       5 to 5+5   = 10                       ->       10 to 10+5 =  15\n",
-        "      15 to 15+5  = 20 -18 = 2                 ->        2 to  2+5 =   7\n",
-        "       7 to 7+5   = 12                       ->       12 to 12+5 =  17\n",
-        "      17 to 17+5  = 22 -18 = 4                 ->        4 to  4+5 =   9\n",
-        "       9 to 9+5   = 14                       ->       14 to 14+5 =  19 -18 = 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5 page no : 42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "Z = 480.             \t\t\t#No of armature conductors\n",
-      "I_a = 144.\n",
-      "I = I_a/2           \t\t\t#For wave wound\n",
-      "theta_m = 10.        \t\t\t#lead angle in DEGREES\n",
-      "\n",
-      "# Calculations\n",
-      "amp_turns_PP_d = Z*I*theta_m/360  \t\t\t#demagnetising Ampere-turns per pole\n",
-      "amp_turns_PP_c = Z*I*(1/(2*Pole)-theta_m/360) \t\t\t#cross-magnetising Ampere-turns per pole\n",
-      "\n",
-      "# Results\n",
-      "print 'De-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_d)\n",
-      "print 'Cross-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_c)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "De-magnetising ampere-turns per pole is 960\n",
-        "Cross-magnetising ampere-turns per pole is 3360\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6 page no : 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 10.\n",
-      "Z = 800.            \t\t\t#No of armature conductors\n",
-      "A = Pole            \t\t\t#For lap wound\n",
-      "ratio = 0.7           \t\t\t#ratio of pole arc to pole pitch\n",
-      "\n",
-      "# Calculations\n",
-      "#amp_turns_PP = ratio*(I_a*Z)/(2*A*P)\n",
-      "turns_PP = ratio*(Z)/(2*A*Pole)   \t\t\t#turns per pole\n",
-      "conductors_PP = turns_PP*2 \t\t\t#multiplied with 2 because 2 conductors form 1 turn\n",
-      "\n",
-      "# Results\n",
-      "print 'Compensating conductors per pole =  %.f'%(conductors_PP)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Compensating conductors per pole =  6\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7 page no : 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "I_L = 150.\n",
-      "A = 4.\n",
-      "N = 1800.   \t\t\t#in rpm\n",
-      "W_b = 1.2   \t\t\t#Brush width\n",
-      "W_m = 0     \t\t\t#width of mica insulation\n",
-      "L = 0.06*10**-3 \t\t\t#Inducmath.tance\n",
-      "segments = 64.\n",
-      "n_s = 1800./60   \t\t\t#in rps and not rpm\n",
-      "v = n_s*segments  \t\t\t#peripheral speed in segments per second\n",
-      "\n",
-      "# Calculations and Results\n",
-      "T_c = (W_b-W_m)/v  \t\t\t#Time of commutation\n",
-      "I = I_L/A  \t\t\t#Current through a conductor\n",
-      "\n",
-      "#Part(i)\n",
-      "E_l = L*2*I/T_c\n",
-      "print 'i)   Reactive voltage using Linear commutation is %.1f V'%(E_l)\n",
-      "\n",
-      "#Part(ii)\n",
-      "E_s = 1.11*L*2*I/T_c\n",
-      "print 'ii)  Reactive voltage using Sinusoidal commutation is %.3f V'%(E_s)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   Reactive voltage using Linear commutation is 7.2 V\n",
-        "ii)  Reactive voltage using Sinusoidal commutation is 7.992 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8 page no : 63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V_t = 250.  \t\t\t#Terminal voltage\n",
-      "R_sh = 100.  \t\t\t#resistance of shunt field winding\n",
-      "I_sh = V_t/R_sh \t\t\t#shunt current\n",
-      "R_a = 0.22  \t\t\t#Armature resistance\n",
-      "\n",
-      "# Calculations\n",
-      "P = 5*10**3  \t\t\t#Load power\n",
-      "I_L = P/V_t  \t\t\t#Load current\n",
-      "I_a = I_L+I_sh \t\t\t#armature current\n",
-      "\n",
-      "E = V_t + I_a*R_a  \t\t\t#Induced emf\n",
-      "\n",
-      "# Results\n",
-      "print 'Induced e.m.f to supply the 5kW load is %.2f V'%(E)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Induced e.m.f to supply the 5kW load is 254.95 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9 page no : 63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V_t = 250.  \t\t\t#terminal voltage\n",
-      "P = 10.*10**3  \t\t\t#10kW power of generator\n",
-      "I_L = P/V_t \t\t\t#load current\n",
-      "I_a = I_L  \t    \t\t#As seperately excited\n",
-      "V_brush = 2.*2 \t\t\t# 2 * no of brushes\n",
-      "\n",
-      "E = 255.      \t\t\t#on full load\n",
-      "\n",
-      "# Calculations\n",
-      "R_a = (E-V_t-V_brush)/I_a \t\t\t#Because E = V_t+ I_a*R_a  + V_brush\n",
-      "\n",
-      "# Results\n",
-      "print 'Armature resistance of generator is %.3f ohm'%(R_a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature resistance of generator is 0.025 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10 page no : 64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "R_a = 0.5\n",
-      "R_se = 0.03  \t\t\t#resimath.tance due to armature and series field winding\n",
-      "V_brush = 2. \t\t\t#brush drop\n",
-      "N = 1500. \t\t\t#generator speed in r.p.m\n",
-      "coils = 540.\n",
-      "turns_per_coil = 6.\n",
-      "\n",
-      "# Calculations\n",
-      "total_turns =  coils*turns_per_coil\n",
-      "Z = 2*total_turns \t\t\t#Total conductors\n",
-      "I_a = 50         \t\t\t#armature current\n",
-      "\n",
-      "phi = 2*10**-3  \t\t\t#flux per pole in webers\n",
-      "E = phi*N*Z/(60)  \t\t\t#A = P for lap-wound and they cancel out\n",
-      "V_t  = E- (I_a*(R_a+R_se)  + V_brush)   \t\t\t#Because E = V_t+ I_a*R_a  + V_brush\n",
-      "\n",
-      "# Results\n",
-      "print 'Terminal voltage is %.1f V'%(V_t)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Terminal voltage is 295.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11 page no : 65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V_t = 225. \t\t\t#voltage across winding\n",
-      "R_a = 0.04 \t\t\t#armature resistance\n",
-      "R_sh = 90 \t\t\t#shunt resistance\n",
-      "R_se = 0.02\t\t\t#resistance of series field winding\n",
-      "I_L = 75. \t\t\t#load current\n",
-      "\n",
-      "# Calculations\n",
-      "#E -I_a*R_a = V_t+I_L*R_se \n",
-      "I_sh = (V_t+I_L*R_se)/R_sh  \t\t\t#current through shunt field winding\n",
-      "\n",
-      "I_a = I_L + I_sh  \t\t\t        #armature current \n",
-      "E = V_t+ I_a*R_a+I_L*R_se  \t\t\t#induced emf\n",
-      "\n",
-      "# Results\n",
-      "print 'Generated voltage is %.1f V'%(E)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Generated voltage is 229.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12 page no : 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "R_sh = 53.  \t\t\t#resistance of field winding\n",
-      "V_t = 100.   \t\t\t#terminal voltage \n",
-      "I_sh  = V_t/R_sh \t\t\t#shunt current\n",
-      "I_f = I_sh\n",
-      "R_a = 0.1 \t\t\t#armature resistance\n",
-      "E_o = 143. \t\t\t# for I_sh =  I_f  =  1.8867 as obtained from graph\n",
-      "\n",
-      "# Calculations\n",
-      "I_a = (E_o-V_t)/R_a  \t\t\t#Because E_o = V_t + I_a*R_a\n",
-      "I_L = I_a-I_sh  \t\t\t#no load current\n",
-      "\n",
-      "# Results\n",
-      "print 'Note: Open circuit voltage was obtained as followsE_o = R_sh*I_f   \t\t\t# y = mx+c form with c = 0 and R_sh = 53Hence%( a line with slope 53 through origin is made to intersect OCC at 150 V'\n",
-      "print 'Therefore, Open circuit voltage is 150 V'\n",
-      "print 'No load current is %.4f A    '%(I_L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Note: Open circuit voltage was obtained as followsE_o = R_sh*I_f   \t\t\t# y = mx+c form with c = 0 and R_sh = 53Hence%( a line with slope 53 through origin is made to intersect OCC at 150 V\n",
-        "Therefore, Open circuit voltage is 150 V\n",
-        "No load current is 428.1132 A    \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13 page no : 79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "#part(1)\n",
-      "E_o = 240.  \t\t\t#on no-load\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Draw horizontal line from 240 V, to intersect OCC at A. corresponding I_f is 2.25 A\n",
-      "#The slope pf OA is corresponding R_sh\n",
-      "I_f = 2.25  \t\t\t#Corresponds to 240 V when intersected OCC\n",
-      "R_sh = E_o/I_f  \t\t\t#shunt resistance\n",
-      "print 'i)Field resistance that gives 240 V on no-load is %.2f ohms '%(R_sh)\n",
-      "\n",
-      "#Part(ii)\n",
-      "N1 = 1000. \t\t\t#speed of shunt generator in rpm\n",
-      "I_f = 1.\n",
-      "\n",
-      "#Draw line OP math.tangential to OCC at N1 = 1000 r.p.m.\n",
-      "#Select I_f = 1A i.e. point R\n",
-      "#Draw vertical from R to intersect OP at S and OA at T....this gives RT = 105 and RS = 159\n",
-      "#At critcal speed generator just fails to build up\n",
-      "\n",
-      "RT = 105.\n",
-      "RS = 159.\n",
-      "N_C = N1*RT/RS \t\t\t#Critical speed\n",
-      "print 'ii)Critical speed is %.2f r.p.m '%(N_C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Field resistance that gives 240 V on no-load is 106.67 ohms \n",
-        "ii)Critical speed is 660.38 r.p.m \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.14 page no : 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P = 4.  \t\t\t#number of poles\n",
-      "A = 2. \t\t\t#because wave wound\n",
-      "Z = 792 \t\t\t#No of conductors\n",
-      "phi = 0.012  \t\t\t#flux per pole in weber\n",
-      "E_g = 240 \t\t\t#on no-load\n",
-      "\n",
-      "# Calculations\n",
-      "#running speed\n",
-      "N = E_g*60*A/(phi*P*Z) \t\t\t#becuase E_g =  phi*P*N*Z/(60*A)\n",
-      "\n",
-      "# Results\n",
-      "print 'Required running speed is %.3f r.p.m'%(N)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required running speed is 757.576 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.15 page no : 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "#open circuit condition\n",
-      "I_L = 0.  \t\t\t#because of open circuit\n",
-      "V_t = 127. \t\t\t#terminal voltage\n",
-      "E_g = V_t \t\t\t#because I_L = 0\n",
-      "\n",
-      "# Calculations\n",
-      "#load condition\n",
-      "V_t = 120\n",
-      "R_sh = 15.\n",
-      "R_a = 0.02  \t\t\t#shunt and armature resistance\n",
-      "I_sh1 = V_t/R_sh \t\t\t#current through shunt winding in loaded condition\n",
-      "\n",
-      "I_L  = (E_g-V_t)/R_a  -  I_sh1 \t\t\t#because I_a1 = I_L+I_sh1    and    E_g = V_t + I_a1*R_a\n",
-      "\n",
-      "# Results\n",
-      "print 'Load current is %.0f A'%(I_L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load current is 342 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.16 page no : 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V_t = 550.  \t\t\t#Terminal voltage\n",
-      "R_lamp = 500. \t\t\t#Each lamp\n",
-      "I_lamp = V_t/R_lamp  \t\t\t#each lamp ; V_t because all lamps are in parallel\n",
-      "\n",
-      "# Calculations\n",
-      "I_L = 20*I_lamp  \t\t\t#there exist 20 lamps\n",
-      "R_sh = 25.\n",
-      "R_a = 0.06\n",
-      "R_se = 0.04  \t\t\t#resistance of shunt winding,armature,series field\n",
-      "I_sh = V_t/R_sh \t\t\t#current throough shunt winding\n",
-      "I_a = I_L+I_sh\t\t\t#armature current\n",
-      "E = V_t + I_a*(R_a+R_se)  \t\t\t#generated emf\n",
-      "\n",
-      "# Results\n",
-      "print 'Armature current and generated e.m.f is %.0f A and %.1f V respectively '%(I_a,E )\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature current and generated e.m.f is 44 A and 554.4 V respectively \n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.17 page no : 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P = 4. \t\t\t#number of poles\n",
-      "A = P \t\t\t# because of lap wound\n",
-      "N = 750. \t\t\t#speed in rpm\n",
-      "Z = 720. \t\t\t#number of armature conductors\n",
-      "phi = 30.*10**-3  \t\t\t#flux per pole in weber\n",
-      "R_sh = 200.\n",
-      "R_a = 0.4\n",
-      "R_L = 15. \t\t\t#resistance of shunt winding,armature,series field\n",
-      "E = phi*P*N*Z/(60*A) \t\t\t#generated emf\n",
-      "\n",
-      "# Calculations\n",
-      "#solving the following equations for V_t\n",
-      "#E =  V_t + I_a*R_a\n",
-      "#E =  V_t + (I_L + I_sh)*R_a\n",
-      "#E =  V_t + ((V_t/R_L) + (V_t/R_sh))*R_a\n",
-      "V_t = E/(1+(R_a/R_L)+(R_a/R_sh))\n",
-      "\n",
-      "# Results\n",
-      "print 'Terminal voltage  =  %.4f V'%(V_t)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Terminal voltage  =  262.4757 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.18 page no : 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P = 6. \t\t\t#number of poles\n",
-      "A = 2. \t\t\t# because of wave wound\n",
-      "N_1 = 300. \t\t\t#speed of generator\n",
-      "Z = 600. \t\t\t#number of armature conductors\n",
-      "phi_1 = 0.06  \t\t\t#flux per pole in webers \n",
-      "\n",
-      "# Calculations and Results\n",
-      "E_g1 = phi_1*P*N_1*Z/(60*A) \t\t\t#generated emf\n",
-      "print 'Emf generated is %.0f V'%(E_g1)\n",
-      "\n",
-      "phi_2 = 0.055 \t\t\t#new flux per pole \n",
-      "E_g2 = 550.\t\t\t# new generated emf\n",
-      "N_2 = E_g2/(phi_2*P*Z/(60*A)) \t\t\t#new speed of generator\n",
-      "print 'Required speed is %.2f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Emf generated is 540 V\n",
-        "Required speed is 333.33 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.19 page no  : 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "\n",
-      "print 'Refer to code for explanation'\n",
-      "\n",
-      "# Variables\n",
-      "N_1 = 300.\n",
-      "N_2 = 375. \t\t\t#generator speeds\n",
-      "\n",
-      "#E_g2 = E_g1*(N_2/N_1)\n",
-      "#using this new table OCC at N_2 = 375 is made\n",
-      "#Draw a line with slope R_sh = 40 through origin which cuts this OCC at 248\n",
-      "#I_f_table = [0,2,3,4,5,6,7]\n",
-      "#Arm_vol_table = [9.375,115,165,202.5,228.75,237.5,265]\n",
-      "\n",
-      "#part(i)\n",
-      "#at V = 200 volts, I_f = 3.9 from the graph\n",
-      "V = 200.\n",
-      "I_f = 3.9\n",
-      "\n",
-      "# Calculations and Results\n",
-      "R_sh2 = V/I_f\n",
-      "R_sh = 40.\n",
-      "print 'Additional resistance required is %.3f ohms  '%(R_sh2-R_sh)\n",
-      "\n",
-      "#part(ii)\n",
-      "V_t = 200.\n",
-      "I_f = V_t/R_sh\n",
-      "E_g = 228.75   \t\t\t#For this I_f from the table \n",
-      "R_a = 0.4\n",
-      "I_a = (E_g-V_t)/R_a   \t\t\t#Because E_g = V_t + I_a*R_a\n",
-      "I_L = I_a-I_f\n",
-      "print 'Load current supplied by the generator is %.3f A'%(I_L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Refer to code for explanation\n",
-        "Additional resistance required is 11.282 ohms  \n",
-        "Load current supplied by the generator is 66.875 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.20 page no : 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "I_a = 750. \t\t\t#full load current \n",
-      "Pole = 6. \n",
-      "A = Pole \t\t\t#lap winding\n",
-      "I = I_a/A  \t\t\t#Full-load current per path\n",
-      "Z = 900. \t\t\t#no of conductors\n",
-      "lambda1 = 1.4  \t\t\t#leakage coefficient\n",
-      "theta_e = 21. \t\t\t#lead angle in degrees electrical\n",
-      "\n",
-      "# Calculations\n",
-      "theta_m = theta_e/(Pole/2)  \t\t\t#lead angle in degrees mechanical\n",
-      "amp_turns_PP_d = Z*I*theta_m/360  \t\t\t#demagnetising ampere turns per pole\n",
-      "amp_turns_PP_c = Z*I*(1./(2*Pole)-theta_m/360)  \t\t\t#cross-magnetising ampere turns per pole\n",
-      "balance_turns = amp_turns_PP_d*lambda1/I_a \t\t\t#series turns required to balance demagnetising component\n",
-      "\n",
-      "# Results\n",
-      "print 'i)  De-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_d)\n",
-      "print 'ii) Cross-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_c)\n",
-      "print 'iii)Turns required to balance demagnetising component is %.0f'%(balance_turns)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)  De-magnetising ampere-turns per pole is 2187.5\n",
-        "ii) Cross-magnetising ampere-turns per pole is 7187.5\n",
-        "iii)Turns required to balance demagnetising component is 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.21 page no : 87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Pole = 4. \n",
-      "Z = 32.  \t\t\t#no of conductors\n",
-      "coil_sides = Z \n",
-      "segments = 16.\n",
-      "pole_pitch = Z/Pole\n",
-      "slots = 16.\n",
-      "slots_per_pole = slots/Pole\n",
-      "\n",
-      "# Calculations\n",
-      "#for Simplex lap winding\n",
-      "y_b = pole_pitch+1  \t\t\t#back pitch\n",
-      "y_f = pole_pitch-1  \t\t\t#front pitch\n",
-      "\n",
-      "y_c = 1  \t\t\t#Commutator pitch; Note that it is positive and it is progressive type of Simplex lap winding\n",
-      "\n",
-      "# Results\n",
-      "print 'WINDING TABLE:   1<-   10->    3<-   12->    5<-   14-> 7<-   16->    9<-   18->   11<-   20->13<-   22->   15<-   24->   17<-   26->19<-   28->   21<-   30->   23<-   32->25<-    2->   27<-    4->   29<-    6->31<-   8->1                                                  '\n",
-      "print 'Note that <- indicates back connection with  y_back = %.0f and -> indicates front connection with y_front = %.0f'%(y_b,y_f)\n",
-      "\n",
-      "print 'Another form of winding table:'\n",
-      "\n",
-      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
-      "\n",
-      "print '       1 to 1+9  = 10                        ->       10 to 10-7  = 3'\n",
-      "print '       3 to 3+9  = 12                        ->       12 to 12-7 =  5'\n",
-      "print '       5 to 5+9  = 14                        ->       14 to 14-7 =  7'\n",
-      "print '       7 to 7+9  = 16                        ->       16 to 16-7 =  9'\n",
-      "print '       9 to 9+9  = 18                        ->       18 to 18-7 = 11'\n",
-      "print '      11 to 11+9 = 20                        ->       20 to 20-7 = 13'\n",
-      "print '      13 to 13+9 = 22                        ->       22 to 22-7 = 15'\n",
-      "print '      15 to 15+9 = 24                        ->       24 to 24-7 = 17'\n",
-      "print '      17 to 17+9 = 26                        ->       26 to 26-7 = 19'\n",
-      "print '      19 to 19+9 = 28                        ->       28 to 28-7 = 21'\n",
-      "print '      21 to 21+9 = 30                        ->       30 to 30-7 = 23'\n",
-      "print '      23 to 23+9 = 32                        ->       32 to 32-7 = 25'\n",
-      "print '      25 to 25+9 = 34 = 34-32 = 2              ->        2 to 34-7 = 27'\n",
-      "print '      27 to 27+9 = 36 = 36-32 = 4              ->        4 to 36-7 = 29'\n",
-      "print '      29 to 29+9 = 38 = 38-32 = 6              ->        6 to 38-7 = 31'\n",
-      "print '      31 to 31+9 = 40 = 40-32 = 4              ->        8 to 40-7 = 33 -32 =  1'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "WINDING TABLE:   1<-   10->    3<-   12->    5<-   14-> 7<-   16->    9<-   18->   11<-   20->13<-   22->   15<-   24->   17<-   26->19<-   28->   21<-   30->   23<-   32->25<-    2->   27<-    4->   29<-    6->31<-   8->1                                                  \n",
-        "Note that <- indicates back connection with  y_back = 9 and -> indicates front connection with y_front = 7\n",
-        "Another form of winding table:\n",
-        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
-        "       1 to 1+9  = 10                        ->       10 to 10-7  = 3\n",
-        "       3 to 3+9  = 12                        ->       12 to 12-7 =  5\n",
-        "       5 to 5+9  = 14                        ->       14 to 14-7 =  7\n",
-        "       7 to 7+9  = 16                        ->       16 to 16-7 =  9\n",
-        "       9 to 9+9  = 18                        ->       18 to 18-7 = 11\n",
-        "      11 to 11+9 = 20                        ->       20 to 20-7 = 13\n",
-        "      13 to 13+9 = 22                        ->       22 to 22-7 = 15\n",
-        "      15 to 15+9 = 24                        ->       24 to 24-7 = 17\n",
-        "      17 to 17+9 = 26                        ->       26 to 26-7 = 19\n",
-        "      19 to 19+9 = 28                        ->       28 to 28-7 = 21\n",
-        "      21 to 21+9 = 30                        ->       30 to 30-7 = 23\n",
-        "      23 to 23+9 = 32                        ->       32 to 32-7 = 25\n",
-        "      25 to 25+9 = 34 = 34-32 = 2              ->        2 to 34-7 = 27\n",
-        "      27 to 27+9 = 36 = 36-32 = 4              ->        4 to 36-7 = 29\n",
-        "      29 to 29+9 = 38 = 38-32 = 6              ->        6 to 38-7 = 31\n",
-        "      31 to 31+9 = 40 = 40-32 = 4              ->        8 to 40-7 = 33 -32 =  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.22 page no : 89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "Z = 496. \t\t\t#no of conductors\n",
-      "P = 4. \t\t\t#poles\n",
-      "slots = 31.\n",
-      "coilsides_per_slot = 4.\n",
-      "coilsides = slots*coilsides_per_slot\n",
-      "coils = coilsides/2\n",
-      "turns = Z/2\n",
-      "turns_per_coil = turns/coils\n",
-      "\n",
-      "# Calculations\n",
-      "y_c = array([(Z-2)/P, (Z+2)/P]) \t\t\t#commutator pitch\n",
-      "coils_active = (Z/(2*P))-1 \t\t\t# because y_c didnt turn out to be integer, 1 coil was made inactive/dummy\n",
-      "segments = coils_active  \t\t\t#no of commutative segments\n",
-      "Y_A = [ (segments+1)/(P/2), (segments-1)/(P/2)   ]\n",
-      "Y_A = Y_A[0]   \t\t\t#Y_A(2) is discarded because of progressive wave winding\n",
-      "y_f = 29.\n",
-      "y_b = 33.  \t\t\t#front and back pitch ; note that Y_A = (y_b+y_f)/2\n",
-      "resultant_pitch = 2*Y_A  \t\t\t#because Y_A = (y_b+y_f)/2 and resultant pitch  =  y_b+ y_f\n",
-      "\n",
-      "# Results\n",
-      "print 'i)   Total number of coils  =  %.0f'%(coils)\n",
-      "print 'ii)  Turns per coils  =  %.0f'%(turns_per_coil)\n",
-      "print 'iii) Commutator pitch  =  %.0f'%((y_c[0]+y_c[1])/2)\n",
-      "print 'iv)  Back pitch =  %.0f    front pitch =  %.0f   total pitch =  %.0f'%(y_b,y_f,resultant_pitch)\n",
-      "print 'v)   No of commutator segments  =  %.0f'%(segments)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   Total number of coils  =  62\n",
-        "ii)  Turns per coils  =  4\n",
-        "iii) Commutator pitch  =  124\n",
-        "iv)  Back pitch =  33    front pitch =  29   total pitch =  62\n",
-        "v)   No of commutator segments  =  61\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.23 page no : 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "slots = 13.\n",
-      "Pole = 4.\n",
-      "conductors_per_slot = 2.\n",
-      "Z = conductors_per_slot*slots\n",
-      "Y_A = (Z+2)/Pole  \t\t\t#For progressive type wave winding\n",
-      "\n",
-      "# Calculations\n",
-      "#Since Y_A = (y_b+y_f)/2, we let y_b = y_f\n",
-      "y_b = Y_A\n",
-      "y_f = y_b \t\t    \t#because y_b = Y_A/2\n",
-      "\n",
-      "segments = 13. \t\t\t#commutator segments\n",
-      "\n",
-      "# Results\n",
-      "print 'WINDING TABLE:   1<-    8->  15<-  22->   3<-   10->17<-   24->   5<-  12->  19<-   26-> 7<-   14->  21<-   2->   9<-   16->23<-    4->  11<-  18->  25<-    6->13<-   20->   1                                                  '\n",
-      "print 'Note that <- indicates back connection with y_back = %.0f and -> indicates front connection with y_front = %.0f'%(y_b,y_f)\n",
-      "\n",
-      "print 'Another form of winding table:'\n",
-      "\n",
-      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
-      "\n",
-      "print '       1 to 1+7   =  8                   ->        8 to 8+7  =  15'\n",
-      "print '      15 to 15+7  = 22                   ->       22 to 22+7 =  29 -26 = 3'\n",
-      "print '       3 to 3+7   = 10                   ->       10 to 10+7 =  17'\n",
-      "print '      17 to 17+7  = 24                   ->       24 to 24+7 =  31 -26 = 5'\n",
-      "print '       5 to 5+7   = 14                   ->       12 to 12+7 =  19'\n",
-      "print '      19 to 19+7  = 26                   ->       26 to 26+7 =  33 -26 = 7'\n",
-      "print '       7 to 7+7   = 14                   ->       14 to 14+7 =  21'\n",
-      "print '      21 to 21+7  = 28 -26 = 2             ->        2 to  2+7 =   9'\n",
-      "print '       9 to 9+7   = 16                   ->       16 to 16+7 =  23 '\n",
-      "print '      23 to 23+7  = 30 -26 = 4             ->        4 to  4+7 =  11'\n",
-      "print '       11to 11+7  = 18                   ->       18 to 18+7 =  25'\n",
-      "print '      25 to 25+7  = 32 -26 = 6             ->        6 to  6+7 =  13'\n",
-      "print '      13 to`13+7  = 20                   ->       20 to 20+7 =  27 -26 = 1'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "WINDING TABLE:   1<-    8->  15<-  22->   3<-   10->17<-   24->   5<-  12->  19<-   26-> 7<-   14->  21<-   2->   9<-   16->23<-    4->  11<-  18->  25<-    6->13<-   20->   1                                                  \n",
-        "Note that <- indicates back connection with y_back = 7 and -> indicates front connection with y_front = 7\n",
-        "Another form of winding table:\n",
-        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
-        "       1 to 1+7   =  8                   ->        8 to 8+7  =  15\n",
-        "      15 to 15+7  = 22                   ->       22 to 22+7 =  29 -26 = 3\n",
-        "       3 to 3+7   = 10                   ->       10 to 10+7 =  17\n",
-        "      17 to 17+7  = 24                   ->       24 to 24+7 =  31 -26 = 5\n",
-        "       5 to 5+7   = 14                   ->       12 to 12+7 =  19\n",
-        "      19 to 19+7  = 26                   ->       26 to 26+7 =  33 -26 = 7\n",
-        "       7 to 7+7   = 14                   ->       14 to 14+7 =  21\n",
-        "      21 to 21+7  = 28 -26 = 2             ->        2 to  2+7 =   9\n",
-        "       9 to 9+7   = 16                   ->       16 to 16+7 =  23 \n",
-        "      23 to 23+7  = 30 -26 = 4             ->        4 to  4+7 =  11\n",
-        "       11to 11+7  = 18                   ->       18 to 18+7 =  25\n",
-        "      25 to 25+7  = 32 -26 = 6             ->        6 to  6+7 =  13\n",
-        "      13 to`13+7  = 20                   ->       20 to 20+7 =  27 -26 = 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.24 page no : 93"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "P = 4.\n",
-      "I_L = 150.\n",
-      "N = 1500. \t\t\t#commutator speed in rpm\n",
-      "n_s = N/60.  \t\t\t#commutator speed in r.p.s\n",
-      "W_b = 1.2 \t\t\t#Brush pitch\n",
-      "W_m = 0\t\t\t#Pitch of mica insulation\n",
-      "L = 0.05*10**-3  \t\t\t#inducmath.tance of armature coils in henry\n",
-      "A = P \t\t\t#A = P for lap wound\n",
-      "segments = 64.\n",
-      "\n",
-      "# Calculations\n",
-      "v = n_s*segments \t\t\t#peripheral speed in segments per second\n",
-      "T_c = (W_b-W_m)/v  \t\t\t#Time of commutation\n",
-      "I = I_L/A  \t\t\t#current through each conductor\n",
-      "E = L*2*I/T_c  \t\t\t#Linear commutation\n",
-      "E2 = 1.11*L*2*I/T_c  \t\t\t#Sinusoidal commutation\n",
-      "\n",
-      "# Results\n",
-      "print 'Reactive voltage linear commutation) is %.0f V'%(E)\n",
-      "print 'Reactive voltage math.sinusoidal commutation) is %.2f V'%(E2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reactive voltage linear commutation) is 5 V\n",
-        "Reactive voltage math.sinusoidal commutation) is 5.55 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.25 page no : 94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P = 4.  \t\t\t#Pole\n",
-      "Z = 386. \t\t\t#no of wave connected conductors\n",
-      "A = 2.  \t\t\t#Wave winding\n",
-      "R_a = 1.\n",
-      "R_sh = 100.\n",
-      "R_L = 40  \t\t\t#Armature ,shunt field and load resistance\n",
-      "phi = 25.*10**-3  \t\t\t#flux per pole in weber\n",
-      "N = 1000.  \t\t\t#speed in rpm\n",
-      "\n",
-      "\n",
-      "# Calculations\n",
-      "E_g = (phi*P*N*Z)/(60*A)  \t\t\t#generated emf\n",
-      "\n",
-      "#Solving following equations for V_t\n",
-      "#E_g = V_t+I_a*R_a\n",
-      "#I_a = (I_L+I_sh)   I_L = V_t/R_L  I_sh = V_t/R_sj\n",
-      "#E_g = V_t(1 + (R_a/R_L) + (R_a/R_sh)\n",
-      "V_t = E_g/(1 + (R_a/R_L) + (R_a/R_sh))\n",
-      "\n",
-      "I_L = V_t/R_L \t\t\t# load current\n",
-      "I_sh = V_t/R_sh \t\t\t# current through shunt field resistance\n",
-      "\n",
-      "# Results\n",
-      "I_a = I_L+I_sh \t\t\t#armature current \n",
-      "print 'armature current is  is %.4f A'%(I_a)\n",
-      "\n",
-      "output_power = V_t*I_L\n",
-      "print 'output power is %.3f W'%(output_power)\n",
-      "\n",
-      "# Note :answer obtained will not match with textbook answer because I_L = V_t/R_L = 310.79/40 = 7.77 \n",
-      "#Awhile its taken as 8.045 A in textbook'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "armature current is  is 10.8776 A\n",
-        "output power is 2414.746 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.26 page no : 94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "I = 40.  \t\t    \t#current PER conductor\n",
-      "W_b = 3.\n",
-      "W_m = 0.\t\t\t    #brush width and width of mica insulation\n",
-      "N = 600. \t\t\t    #commutator speed in rpm \n",
-      "n_s = N/60 \t\t\t    #commutator speed in rps\n",
-      "L = 0.15*10**-3 \t\t\t#self inducmath.tance in Henry\n",
-      "segments = 50.  \n",
-      "v = n_s*segments \t\t\t#peripheral speed in segments per second\n",
-      "T_c = (W_b-W_m)/v \t\t\t#time of commutation\n",
-      "\n",
-      "# Calculations\n",
-      "E = L*2*I/T_c  \t\t\t#Linear commutation\n",
-      "E2 = 1.11*L*2*I/T_c  \t\t\t#Sinusoidal commutation\n",
-      "\n",
-      "# Results\n",
-      "print 'Reactive voltage linear commutation) is %.0f volts'%(E)\n",
-      "print 'Reactive voltage math.sinusoidal commutation) is %.2f volts'%(E2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reactive voltage linear commutation) is 2 volts\n",
-        "Reactive voltage math.sinusoidal commutation) is 2.22 volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.27 page no : 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 400.\n",
-      "P = 6. \t\t\t    #Poles\n",
-      "A = P  \t\t    \t#For lap wound\n",
-      "output_power = 250.*10**3\n",
-      "R_sh = 200. \t\t\t#shunt field circuit resistance\n",
-      "Z = 720.     \t\t\t#number of lap wound conductors\n",
-      "\n",
-      "theta_m = 2.5 \t\t\t#brush lead angle in degree mechanical\n",
-      "I_L = output_power/V\n",
-      "V_sh = V\n",
-      "\n",
-      "# Calculations\n",
-      "I_sh = V_sh/R_sh \t\t\t#Current through shunt field circuit resistance\n",
-      "I_a = I_L+I_sh \t\t\t#armature currrent\n",
-      "I = I_a/P\n",
-      "\n",
-      "#Part(i)\n",
-      "amp_turns_PP_d = Z*I*theta_m/360\t\t\t#demagnetising ampere turns per pole\n",
-      "#Part(ii)\n",
-      "amp_turns_PP_c = Z*I*(1/(2*P)-theta_m/360) \t\t\t#cross-magnetising ampere turns per pole\n",
-      "\n",
-      "# Results\n",
-      "print 'i)  De-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_d)\n",
-      "print 'ii) Cross-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_c)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)  De-magnetising ampere-turns per pole is 522.5\n",
-        "ii) Cross-magnetising ampere-turns per pole is 5747.5\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.28 page no : 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "I_L = 100.\n",
-      "P = 4. \t\t\t    #Poles\n",
-      "A = P \t\t\t    #for lap wound armature\n",
-      "W_b = 1.4\n",
-      "W_m = 0. \t\t\t#Brush width and width of mica insulation\n",
-      "N = 1400.\t\t\t#armature speed in r.p.m\n",
-      "segments = 64. \t\t\t#no of commutator segments\n",
-      "L = 0.05*10**-3 \t\t\t#inducmath.tance of armature coil in henry\n",
-      "\n",
-      "# Calculations\n",
-      "n_s = N/60  \t\t\t#speed in r.p.s\n",
-      "v = n_s*segments \t\t\t#Segments per second\n",
-      "T_c = (W_b-W_m)/v \t\t\t#time of commutation\n",
-      "I = I_L/A   \t\t\t#Current through conductor\n",
-      "E = L*2*I/T_c  \t\t\t#Linear commutation\n",
-      "\n",
-      "# Results\n",
-      "print 'Reactive voltage considering linear commutation is %.2f volts'%(E)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reactive voltage considering linear commutation is 2.67 volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.29 page no : 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "P = 8. \t\t    \t#Poles\n",
-      "A = 2. \t\t    \t#Wave wound armature\n",
-      "Z = 480. \t\t\t#number of armature conductors\n",
-      "I_a = 200.\n",
-      "I = I_a/A\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "theta_m = 0. \t\t\t#Geometric nuetral axis\n",
-      "amp_turns_PP_d = Z*I*theta_m/360    \t\t\t#De-magnetising ampere-turns per pole\n",
-      "amp_turns_PP_c = Z*I*(1/(2*P)-theta_m/360) \t\t\t#Cross-magnetising ampere-turns per pole\n",
-      "print 'Parti)De-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_d)\n",
-      "print 'Cross-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_c)\n",
-      "\n",
-      "#Part(ii)\n",
-      "theta_e2 = 6.             \t\t\t#angle shift of brushes in degrees electrical\n",
-      "theta_m2 = theta_e2/(P/2) \t\t\t#angle shift of brushes in degrees mechanical\n",
-      "amp_turns_PP_d2 = Z*I*theta_m2/360   \t\t\t#De-magnetising ampere-turns per pole\n",
-      "amp_turns_PP_c2 = Z*I*(1/(2*P)-theta_m2/360) \t\t\t#Cross-magnetising ampere-turns per pole\n",
-      "print 'Partii)De-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_d2)\n",
-      "print 'Cross-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_c2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Parti)De-magnetising ampere-turns per pole is 0\n",
-        "Cross-magnetising ampere-turns per pole is 3000\n",
-        "Partii)De-magnetising ampere-turns per pole is 200\n",
-        "Cross-magnetising ampere-turns per pole is 2800\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.30 page no : 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P_input = 7.46*10**3\n",
-      "V = 230.\n",
-      "Pole = 8.\n",
-      "Z = 188. \t\t\t#number of armature consuctors\n",
-      "I_L = P_input/V\n",
-      "theta_m = 7.5  \t\t\t#brush lead angle in degree mechanical\n",
-      "\n",
-      "A = 2. \t\t\t# assumed wave wound because of low-current and high voltage\n",
-      "I = I_L/A\n",
-      "\n",
-      "#Part(i)\n",
-      "amp_turns_PP_d = Z*I*theta_m/360  \t\t\t#De-magnetising ampere-turns per pole\n",
-      "#Part(ii)\n",
-      "amp_turns_PP_c = Z*I*(1/(2*Pole)-theta_m/360) \t\t\t#Cross-magnetising ampere-turns per pole\n",
-      "\n",
-      "print 'De-magnetising ampere-turns per pole is %.2f'%(amp_turns_PP_d)\n",
-      "print 'Cross-magnetising ampere-turns per pole is %.2f'%(amp_turns_PP_c)\n",
-      "\n",
-      "# note : The difference in answer occured because I is approximated to 16 in last 2 steps in book\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "De-magnetising ampere-turns per pole is 63.52\n",
-        "Cross-magnetising ampere-turns per pole is 127.04\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.31 page no : 99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "N_1 = 1200.   \t\t\t#initial speed \n",
-      "I_L1 = 200.   \t\t\t#initial load current\n",
-      "V_t1 = 125.\n",
-      "N_2 = 1000.  \t\t\t#altered speed\n",
-      "R_a = 0.04  \t\t\t#armature resistance\n",
-      "V_brush = 2.  \t\t\t#brush drop\n",
-      "\n",
-      "# Calculations\n",
-      "# Initial Load\n",
-      "I_a1 = I_L1\n",
-      "E_g1 = V_t1+I_a1*R_a+V_brush  \t\t\t#induced emf\n",
-      "\n",
-      "E_g2 = E_g1*(N_2/N_1)  \t\t\t#Because E_g proportional to N during constant flux\n",
-      "R_L =  V_t1/I_L1 \t\t\t#Load resistance\n",
-      "\n",
-      "#Solving for I_L2 as follows\n",
-      "#V_t2 = R_L*I_L2   \t\t\t#I_a2 = I_L2\n",
-      "#V_t2 = E_g2-(I_L2*R_2 + V_brush)\n",
-      "I_L2 = (E_g2-V_brush)/(R_L+R_a)  \t\t\t#new current\n",
-      "\n",
-      "# Results\n",
-      "print 'Load current at new speed is %.4f A'%(I_L2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load current at new speed is 166.1654 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines/ch2.ipynb b/DC_Machines_and_Synchronous_Machines/ch2.ipynb
deleted file mode 100755
index b87a9190..00000000
--- a/DC_Machines_and_Synchronous_Machines/ch2.ipynb
+++ /dev/null
@@ -1,2010 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:78f41af8e7d3182972251bae0a04df7a30b3aeb555cc290301b49daf9f643c37"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2 : D.C. Motors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1 page no : 5"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "I_a = 30. \t\t\t#armature currnet\n",
-      "R_a = 0.75 \t\t\t#Armature resistance\n",
-      "\n",
-      "# Calculations\n",
-      "E_b = V - I_a*R_a   \t\t\t# Since V =  E_b+ I_a*R_a\n",
-      "\n",
-      "\n",
-      "# Results\n",
-      "print 'Induced EMF or back EMF in the motor is %.1f V'%(E_b)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Induced EMF or back EMF in the motor is 197.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2 page no : 6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "A = Pole \t\t\t#for lap winding\n",
-      "V = 230.\n",
-      "Z = 250. \t\t\t#number of armature conductors\n",
-      "phi = 30.*10**-3 \t\t\t#flux per pole in weber\n",
-      "I_a = 40.\n",
-      "R_a = 0.6 \t\t\t#Armature resistance\n",
-      "\n",
-      "# Calculations \n",
-      "E_b = V - I_a*R_a   \t\t\t# Since V =  E_b+ I_a*R_a\n",
-      "N = E_b * 60*A/(phi*Pole*Z)   \t\t\t#because E_b  =  phi*P*N*Z/(60*A)\n",
-      "\n",
-      "# Results\n",
-      "print 'Back emf is %.0f V and running speed is %.0f rpm'%(E_b,N)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Back emf is 206 V and running speed is 1648 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3 page no : 9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "A = Pole \t\t\t#for lap winding\n",
-      "Z = 480.\t\t\t#number of armature conductors\n",
-      "phi = 20.*10**-3 \t\t\t#flux per pole in weber\n",
-      "I_a = 50. \t\t\t#Armature current\n",
-      "\n",
-      "# Calculations\n",
-      "T_a  =  0.159*phi*I_a*Pole*Z/A  \t\t\t#Gross torque developed by armature\n",
-      "\n",
-      "# Results\n",
-      "print 'Gross torque developed by armature is %.3f N-m'%(T_a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Gross torque developed by armature is 76.320 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4 page no : 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "A = Pole \t\t\t#for lap winding\n",
-      "V = 230.\n",
-      "R_a = 0.8 \t\t\t#Armature resistance\n",
-      "N_0 = 1000. \t\t\t#no load speed in rpm\n",
-      "Z = 540. \t\t\t#number of armature conductors\n",
-      "phi = 25.*10**-3 \t\t\t#flux per pole in weber\n",
-      "\n",
-      "# Calculations and Results\n",
-      "E_b0  =  phi*Pole*N_0*Z/(60*A)  \t\t\t#induced emf\n",
-      "\n",
-      "#part(i)\n",
-      "print 'i)Induced e.m.f  =  %.0f V'%(E_b0)\n",
-      "\n",
-      "#part(ii)\n",
-      "I_a0  =  (V- E_b0)/R_a  \t\t\t#because V =  E_b0+ I_a0*R_a\n",
-      "print 'ii)Armature current  =  %.2f A'%(I_a0)\n",
-      "\n",
-      "#part(iii)\n",
-      "stray_losses  =  E_b0*I_a0  \t\t\t#on no load ,power developed is fully power required to overcome strya losses\n",
-      "print 'iii)Stray loss  =  %.2f W'%(stray_losses)\n",
-      "\n",
-      "#part(iv)\n",
-      "T_f  =  E_b0*I_a0/(2*math.pi*N_0/60)  \t\t\t#lost torque\n",
-      "print 'iv)Lost torque  =  %.3f N-m'%(T_f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Induced e.m.f  =  225 V\n",
-        "ii)Armature current  =  6.25 A\n",
-        "iii)Stray loss  =  1406.25 W\n",
-        "iv)Lost torque  =  13.429 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5 page no : 21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "Z = 200. \t\t\t    #No of armature conductors\n",
-      "A = 2.   \t\t\t    #wave connected armature\n",
-      "V = 250.\n",
-      "phi = 25.*10**-3 \t\t\t#flux per pole in weber\n",
-      "I_a  = 60.\n",
-      "I_L  = I_a  \t    \t\t#armature current\n",
-      "R_a = 0.15\n",
-      "R_se = 0.2       \t\t\t#resistances of armature and series field winding\n",
-      "\n",
-      "# Calculations\n",
-      "E_b =  V - I_a*(R_a+R_se)  \t\t\t#induced emf\n",
-      "N = E_b * 60*A/(phi*Pole*Z)   \t\t\t#because E_b  =  phi*P*N*Z/(60*A)\n",
-      "\n",
-      "# Results\n",
-      "print 'Required speed is %.0f r.p.m'%(N)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required speed is 1374 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6 page no : 22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "I_L  = 20. \t\t\t#load current\n",
-      "R_a = 0.3\n",
-      "R_sh = 200. \t\t\t#Armature and shunt field winding\n",
-      "\n",
-      "# Calculations\n",
-      "I_sh = V/R_sh \t\t\t#shunt current\n",
-      "I_a = I_L-I_sh \t\t\t#armature current\n",
-      "E_b =  V - I_a*R_a \t\t\t#emf generated\n",
-      "\n",
-      "# Results\n",
-      "print 'Armature current is %.2f A'%(I_a)\n",
-      "print 'Back e.m.f is %.3f V'%(E_b)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature current is 18.75 A\n",
-        "Back e.m.f is 244.375 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.7 page no : 22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "R_a = 0.3\n",
-      "R_sh = 110. \t\t\t#resistance of armature and shunt field winding\n",
-      "\n",
-      "#no load\n",
-      "N_0 = 1000. \t\t\t#no load speed in r.p.m\n",
-      "I_L0  = 6. \t\t\t#line current on no load\n",
-      "I_sh =  V/R_sh \t\t\t#no load shnt current\n",
-      "I_a0  =  I_L0 - I_sh \t\t\t#no load armature current\n",
-      "E_b0 =  V - I_a0*R_a \t\t\t#no load induced emf\n",
-      "\n",
-      "# Calculations\n",
-      "#full load\n",
-      "I_sh_FL =  V/R_sh\n",
-      "I_L_FL = 50 \t\t\t#line current at full load\n",
-      "I_a_FL =  I_L_FL - I_sh_FL\t\t\t#full load armature current\n",
-      "E_b_FL =  V - I_a_FL * R_a \t\t\t#full load induced emf\n",
-      "\t\t\t#using speed equation as treating phi as constant\n",
-      "N_FL = N_0 * (E_b_FL/E_b0)\n",
-      "\n",
-      "# Results\n",
-      "print 'Speed on full load is %.2f r.p.m'%(N_FL)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed on full load is 939.67 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.8 page no : 23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "R_a = 0.2\n",
-      "R_se  = 0.3 \t\t\t#resistance of armature and series field winding\n",
-      "#following variables correspond to load 1\n",
-      "V = 250.\n",
-      "N_1 = 800.\n",
-      "I_1 = 20.\n",
-      "I_a1 = I_1\n",
-      "I_se1 =  I_a1\n",
-      "\n",
-      "# Calculations\n",
-      "E_b1 =  V - I_a1*(R_a+R_se)\n",
-      "#following variables correspond to load 2\n",
-      "I_2 = 50.\n",
-      "I_a2 = I_2\n",
-      "E_b2 =  V - I_a2*(R_a+R_se)\n",
-      "\n",
-      "#from speed equation it can be derived that\n",
-      "\n",
-      "N_2  =  N_1 * (E_b2/E_b1) * (I_a1/I_a2)\n",
-      "\n",
-      "# Results\n",
-      "print 'Speed on motor on no load  is %.0f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed on motor on no load  is 300 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.9 page no : 31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "R_a = 0.3\n",
-      "R_sh = 200.  \t\t\t#resistance of armature and shunt field winding\n",
-      "R_x = 150. \t\t\t#additional resistance added in series to field winding\n",
-      "I_L1 = 22.\n",
-      "I_sh1 = V/R_sh \t\t\t#initial shunt current before adding 150 ohms resistance\n",
-      "I_a1  =  I_L1 - I_sh1 \t\t\t#initial armature current before adding 150 ohms resistance\n",
-      "N_1 = 1500. \t\t\t#initial speed before adding 150 ohms resistance\n",
-      "\n",
-      "# Calculations\n",
-      "#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying further \n",
-      "I_sh2 = V/(R_sh + R_x) \t\t\t#new shunt current\n",
-      "I_a2 =  I_sh1*I_a1/I_sh2  \t\t\t#New armature current\n",
-      "\n",
-      "E_b1 = V - I_a1*R_a   \t\t\t#induced emf before adding 150 ohms resistance\n",
-      "E_b2 = V - I_a2*R_a   \t\t\t#new emf\n",
-      "\n",
-      "N_2  =  N_1 * (E_b2/E_b1) * (I_sh1/I_sh2) \t\t\t#new speed\n",
-      "\n",
-      "# Results\n",
-      "print 'New armature current and speed are %.4f A and %.f r.p.m respectively'%(I_a2,N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New armature current and speed are 36.3125 A and 2575 r.p.m respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.10 page no : 36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "R_a = 0.15\n",
-      "R_se = 0.1\n",
-      "R_x = 0.1 \t\t\t#Resimath.tance of armature , series field winding and extra resistance\n",
-      "N_1  =  800. \t\t\t#initial speed before load torque is increased\n",
-      "I_1 =  30.\n",
-      "I_a1 = I_1\n",
-      "I_se1  =  I_1  \t\t\t#initial currents\n",
-      "\n",
-      "# Calculations\n",
-      "T_2_by_T_1  =  1 + (50./100)  \t\t\t#50 percent increase as mentioned in question\n",
-      "I_se2_by_I_a2  =  R_x/(R_x + R_se)  \t\t\t#from the figure\n",
-      "\n",
-      "#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying ,solving further \n",
-      "I_a2 = math.sqrt(I_a1*I_se1*T_2_by_T_1/I_se2_by_I_a2) \t\t\t#new armature current\n",
-      "I_se2  =  I_se2_by_I_a2 *I_a2 \t\t\t#new series field current\n",
-      "\n",
-      "E_b1  =  V - I_a1*R_a - I_se1*R_se \t\t\t#indiced emf initially\n",
-      "E_b2  =  V - I_a2*R_a - I_se2*R_se \t\t\t#new induced emf\n",
-      "N_2  =  N_1 * (E_b2/E_b1) * (I_se1/I_se2) \t\t\t#required speed\n",
-      "\n",
-      "# Results\n",
-      "print 'The required running speed of motor is %.3f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required running speed of motor is 912.743 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.11 page no : 38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "I_1 = 50.\n",
-      "I_a1 = I_1  \t\t\t#Currents before adding extra resistance\n",
-      "T_2_by_T_1  = 0.5\n",
-      "R_t = 0.15 \t\t\t#R_e + R_se  = 0.15\n",
-      "\n",
-      "# Calculations\n",
-      "I_a2  = I_a1 * math.sqrt(T_2_by_T_1)  \t\t\t#Because T (prop.) I_a**2\n",
-      "E_b1 = V-I_a1*(R_t) \t\t\t#induced emf before adding extra resistance\n",
-      "N_1 = 500.\n",
-      "N_2 = 300. \t\t\t#speeds before and adding extra resistance\n",
-      "\n",
-      "#N (prop.) E_b/phi (prop.) E_b/I_a\n",
-      "E_b2 = E_b1 *(I_a2/I_a1)*(N_2/N_1) \t\t\t#induced emf after adding resistance\n",
-      "R_x =  (V-E_b2)/I_a2 -R_t \t\t\t#because E_b2 = V - I_a2*(R_a + R_se + R_x)\n",
-      "\n",
-      "# Results\n",
-      "print 'Desired extrea resistance =  %.4f ohms '%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Desired extrea resistance =  3.5225 ohms \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.12 page no : 43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "R_a =  1.\n",
-      "I_a = 1.2 \n",
-      "V = 50.\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "E_b  =  V - I_a*R_a\n",
-      "rot_loss_NL  = E_b*I_a \t\t\t#no load rotational loss \n",
-      "print 'i)No load rotational losses  =  %.2f W'%(rot_loss_NL)\n",
-      "\n",
-      "#part(ii)\n",
-      "omega_2000 = 2*math.pi*2000/60 \t\t\t#angular velocity when speed of motor  = 2000 rpm\n",
-      "K_m = E_b/omega_2000   \t\t\t#to determine K_m\n",
-      "V = 48.\n",
-      "omega_1800 = 2*math.pi*1800/60 \t\t\t#angular velocity when speed of motor  = 1800 rpm\n",
-      "E_b = K_m*omega_1800\n",
-      "I_a  =  (V-E_b)/R_a  \t\t\t#armature current\n",
-      "P_dev  =  E_b*I_a\t\t\t#power developed\n",
-      "motor_output  =   P_dev - rot_loss_NL\n",
-      "print 'ii)Motor output  =  %.f W'%(motor_output)\n",
-      "\n",
-      "#part(iii)\n",
-      "E_b = 0. \t\t\t#when motor stalls\n",
-      "V_stall = 20. \t\t\t#voltage during stalling\n",
-      "I_a = V_stall/R_a \t\t\t#armature current during stalling\n",
-      "T_stall  =  K_m*I_a \t\t\t#stalling torque\n",
-      "print 'iii)Stalling torque  =  %.2f N-m'%(T_stall)\n",
-      "print 'partii answer is slightly different due to inaccurate calculation of Power developed'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)No load rotational losses  =  58.56 W\n",
-        "ii)Motor output  =  121 W\n",
-        "iii)Stalling torque  =  4.66 N-m\n",
-        "partii answer is slightly different due to inaccurate calculation of Power developed\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.13 page no : 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 120.\n",
-      "R_a = 0.2 \n",
-      "R_sh = 60. \t\t\t#armature and field resistance\n",
-      "I_L1 = 40.\n",
-      "N_1 = 1800. \n",
-      "\n",
-      "# Calculations\n",
-      "I_sh =  V/R_sh\n",
-      "\n",
-      "I_a1 = I_L1 - I_sh \n",
-      "E_b1  =  V -I_a1*R_a  \t\t\t#Induced emf at half load\n",
-      "T2_by_T1  = 1./2 \n",
-      "I_a2 = I_a1*(T2_by_T1)   \t\t\t#T (prop.)I_a\n",
-      "E_b2 = V- I_a2*R_a\t\t\t#induced emf at half load\n",
-      "N_2  =  N_1 *(E_b2/E_b1)  \t\t\t#N (prop.) E_b as phi is constant\n",
-      "\n",
-      "# Results\n",
-      "print 'Speed on half load condition is %.2f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed on half load condition is 1860.85 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.14 page no : 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "R_a = 0.08\n",
-      "E_b1 = 242. \n",
-      "V = 250.\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "I_a1 =  (V-E_b1)/R_a\n",
-      "print 'i)Armature current  =  %.0f A'%(I_a1)\n",
-      "\n",
-      "#part(ii)\n",
-      "N = 0.\n",
-      "E_b = 0. \t\t\t#because N = 0\n",
-      "I_a_start = V/R_a\n",
-      "print 'ii)Starting armature current  =  %.0f A'%(I_a_start)\n",
-      "\n",
-      "#part(iii)\n",
-      "I_a2 = 120.\n",
-      "E_b2 = V-I_a2*R_a\n",
-      "print 'iii)Back emf if armature current is changed to 120 A =    %.1f V'%(E_b2)\n",
-      "\n",
-      "#part(iv)\n",
-      "I_a = 87.\n",
-      "N_m = 1500.\n",
-      "E_g = V + I_a*R_a \t\t\t#induced emf\n",
-      "N_g = N_m*(E_g/E_b1)\t\t\t#as E (prop.) N\n",
-      "print 'iv)Generator speed to deliver 87 A at 250 V  =  %.1f rpm'%(N_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Armature current  =  100 A\n",
-        "ii)Starting armature current  =  3125 A\n",
-        "iii)Back emf if armature current is changed to 120 A =    240.4 V\n",
-        "iv)Generator speed to deliver 87 A at 250 V  =  1592.7 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.15 page no : 51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "shaft_output  =  80.*746 \t\t\t#coverted to watts\n",
-      "eta =  80./100           \t\t\t#efficiency\n",
-      "V = 250.\n",
-      "N_1 = 1200.\n",
-      "R_a = 0.04\n",
-      "R_sh  =  250. \t\t\t            #armature and shunt field resistance\n",
-      "\n",
-      "# Calculations and Results\n",
-      "power_input  =  shaft_output/eta\n",
-      "I_L =  power_input /V\n",
-      "I_sh =  V / R_sh\n",
-      "I_a  =  I_L - I_sh\n",
-      "E_b1  =  V - I_a*R_a\n",
-      "\n",
-      "gross_mechanical_power =  E_b1*I_a \t\t\t#electrical equivalent of mechanical power developed\n",
-      "stray_losses  =   gross_mechanical_power - shaft_output\n",
-      "print 'Mechanical power developed on full load  =  %.3f kW'%(gross_mechanical_power/1000)\n",
-      "\n",
-      "#on no load shaft_output = 0 and entire gross power is used to overcome stray losses\n",
-      "Eb0_Ia0 =  stray_losses\n",
-      "#E_b0  =  V - I_a0*R_a  ... solving for I_0\n",
-      "p = [R_a, -V, Eb0_Ia0]\n",
-      "ans = roots(p)\n",
-      "I_a0 = ans[1] \t\t\t#first root is ignored math.since its too large\n",
-      "I_L0  = I_sh+I_a0 \t\t\t#current drawn from supply\n",
-      "E_b0  =  V - I_a0*R_a \n",
-      "\n",
-      "#From speed equation N (prop.) E_b\n",
-      "N_0  =  N_1*(E_b0/E_b1)\n",
-      "print 'No load speed and current are %.4f rpm and %.2f A respectively'%(N_0,I_L0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mechanical power developed on full load  =  70.812 kW\n",
-        "No load speed and current are 1250.9121 rpm and 45.85 A respectively"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.16 page no : 53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "P = 4. \n",
-      "R_a = 0.1 \n",
-      "R_sh  = 124. \t\t\t#armature and shunt field resistance \n",
-      "I_L0 = 4.\n",
-      "N_0 = 1200.\n",
-      "I_L_1 = 61.\n",
-      "\n",
-      "# Calculations\n",
-      "I_sh = V/R_sh\n",
-      "I_a0 = I_L0-I_sh\n",
-      "V_brush =  2 \t\t\t#voltage loss due to brush\n",
-      "E_b0 =  V - I_a0*R_a- V_brush\n",
-      "\n",
-      "I_a1 = I_L_1 - I_sh\n",
-      "E_b1 = V - I_a1*R_a -V_brush\n",
-      "\n",
-      "phi1_by_phi0 = 1-(5./100)   \t\t\t#weakened by 5 %\n",
-      "N_1  =  N_0 *(E_b1/E_b0) /phi1_by_phi0\n",
-      "\n",
-      "# Results\n",
-      "print 'Full load speed is %.3f r.p.m'%(N_1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Full load speed is 1234.102 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.17 page no : 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "R_a = 0.15 \n",
-      "R_sh = 167.67 \t\t\t#armature and shunt field resistance\n",
-      "N_0 = 1280. \t\t\t#speed at no load\n",
-      "\n",
-      "#full load\n",
-      "I_L1  =  67.  \t\t\t#current drawn on full load\n",
-      "I_sh  =  V / R_sh  \t\t\t#as shunt motor\n",
-      "I_a1 =  I_L1- I_sh\n",
-      "E_b1 =  V - I_a1*R_a\n",
-      "\n",
-      "#on no load\n",
-      "I_L0 = 6.5\n",
-      "I_a0  =  I_L0 - I_sh\n",
-      "E_b0  =  V - I_a0*R_a\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i) USING SPEED EQUATION\n",
-      "#N (prop.) E_b/phi (prop.)E_b   \t\t\t#as phi is constant\n",
-      "N_1  =  N_0 * (E_b1 / E_b0)\n",
-      "print 'i)Full load speed  =  %.3f r.p.m'%(N_1)\n",
-      "\n",
-      "#part(ii)\n",
-      "speed_regulation  =  100* ((N_0-N_1)/N_1)\n",
-      "#N_1 is full load speed and N_0 = No load speed  \n",
-      "print 'ii)Speed regulation  =  %.2f percent '%(speed_regulation )\n",
-      "\n",
-      "#part(iii)\n",
-      "shaft_output_FL  =  E_b1*I_a1 - E_b0*I_a0  \t\t\t#full load power developed - stray losses\n",
-      "hp_rating  =  shaft_output_FL /746\n",
-      "print 'iii)HP rating of machine  =  %.2f h.p'%(hp_rating)\n",
-      "\n",
-      "#part(iv)\n",
-      "power_input =  V*I_L1\n",
-      "eta =  100*(shaft_output_FL/power_input)  \t\t\t#full load efficiency\n",
-      "print 'iv)Full load efficiency  =  %.2f percent'%(eta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Full load speed  =  1233.396 r.p.m\n",
-        "ii)Speed regulation  =  3.78 percent \n",
-        "iii)HP rating of machine  =  19.42 h.p\n",
-        "iv)Full load efficiency  =  86.48 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.18 page no : 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 200.\n",
-      "R_a = 0.5\n",
-      "R_se = 0.2\n",
-      "R_x = 0.2 \t\t\t#armature and series field resistance; extra resistance\n",
-      "I_a1 = 20.\n",
-      "I_1 = I_a1 \n",
-      "I_se1 = I_a1\n",
-      "I_a2 = 20.\n",
-      "I_2 = I_a2\n",
-      "I_se2 =  I_2 *(R_x/(R_se+R_x))\n",
-      "\n",
-      "# Calculations\n",
-      "E_b1  =  V -I_a1*R_a - I_a1*R_se\n",
-      "E_b2  =  V -I_a2*R_a - I_se2*R_se\n",
-      "\n",
-      "phi2_by_phi1 = 70./100\n",
-      "N_1 = 1000\n",
-      "N_2 = N_1*(E_b2/E_b1)  /phi2_by_phi1        \t\t\t#N (prop.) E_b/phi\n",
-      "\n",
-      "# Results\n",
-      "print 'Required speed is %.2f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required speed is 1443.93 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.19 page no : 57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 110.\n",
-      "P = 4.\n",
-      "R_a  =  0.1\n",
-      "R = 0.01  \t\t\t#A resistance of 0.01 ohms\n",
-      "R_se = R+R\n",
-      "\n",
-      "# Calculations\n",
-      "#case(i)\n",
-      "I_1 = 50.\n",
-      "I_a1 = I_1\n",
-      "N_1 = 700.\n",
-      "E_b1  =  V -I_a1*(R_a + R_se)\n",
-      "\n",
-      "#T (prop) phi*I_a   from torque equation                                     (1)\n",
-      "\n",
-      "#phi_1 (prop.) I_a1                                                           (2)\n",
-      "#case(ii) when I_a2 gets divided to half\n",
-      "#phi_2 (prop.) I_a2/2                                                         (3)\n",
-      "\n",
-      "#combining (1)(2)(3) and T1 = T2\n",
-      "I_a2  =  math.sqrt(2*I_a1**2)\n",
-      "R_se_eqvt = (R*R)/(R+R)   \t\t\t#Equavalent of parallel combination\n",
-      "E_b2  =  V - I_a2*R_a - I_a2* R_se_eqvt\n",
-      "\n",
-      "#using speed equation N (prop.) E_b / phi  and using (2) and (3)\n",
-      "N_2  =  N_1 *( E_b2/E_b1) *(I_a1/(I_a2/2))\n",
-      "\n",
-      "# Results\n",
-      "print 'Speed after reconnection  =  %.3f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed after reconnection  =  976.389 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.20 page no : 58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "P = 4.\n",
-      "I_a1 =  50.\n",
-      "N_1 = 2000.\n",
-      "V = 230.\n",
-      "\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#phi_1 is proportioanl to total ampere-turns produced by field coils\n",
-      "#phi_1 (prop.) I_a1*P*n (prop.) 200*n                                         (1)\n",
-      "\n",
-      "#After reconnection ,phi_2 proportional to ampere turns divided as follows\n",
-      "#phi_2 (prop.) [I_a2/2*2*n + I_a2/2*2*n]  (prop.) 2*n*I_a2                    (2)\n",
-      "\n",
-      "# Dividing (1) and (2) ,(phi_1/phi_2) = 100 / I_a2                             (3)\n",
-      "\n",
-      "#T (prop.) phi*I_a AND T (prop.) N**2                                          (4)(5)\n",
-      "#therefore N**2 (prop.) phi*I_a                                                (6)\n",
-      "\n",
-      "#N (prop.) E_b/phi (prop.) 1/phi   ..\n",
-      "#Because drops across windings can be neglected , E_b1 = E_b2\n",
-      "#therefore N (prop.) 1/phi                                                    (7)\n",
-      "\n",
-      "#using (7) and (6) phi**3 (prop.) 1/I_a                                        (8)\n",
-      "\n",
-      "#combining (3) and (8)\n",
-      "I_a2  =   (50.*100**3)**(1./4)    \t\t\t#new armature current\n",
-      "print 'New armature current =  %.3f A'%(I_a2)\n",
-      "#combining (6) and (7)   N**3 (prop.) I_a1\n",
-      "N_2 = N_1 *(I_a2/I_a1)**(1./3)\n",
-      "print 'New motor speed  = %.3f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New armature current =  84.090 A\n",
-        "New motor speed  = 2378.414 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example 2.22 page no : 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 200.\n",
-      "I_a1 = 30.\n",
-      "R_t = 1.5  \t\t\t#R_a + R_se\n",
-      "E_b1 =  V - I_a1*R_t\n",
-      "N2_by_N1 = (60./100)\n",
-      "\n",
-      "# Calculations\n",
-      "#T (prop.) I_a**2  and T (prop.) N_3....therefore I_a**2 (prop.) N**3\n",
-      "I_a2   =  I_a1*math.sqrt(N2_by_N1**3)\n",
-      "\n",
-      "#N (prop.) E_b/I_a\n",
-      "N2_by_N1\n",
-      "E_b2  =  E_b1 *(I_a2/I_a1)*N2_by_N1\n",
-      "R_x =  (V- E_b2)/I_a2 - R_t        \t\t\t#because E_b2 =  V - I_a2*(R_x+R_t)\n",
-      "\n",
-      "# Results\n",
-      "print 'Additional resistance to be added in series with motor circuit  =  %.3f ohms'%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Additional resistance to be added in series with motor circuit  =  9.744 ohms\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.23 page no : 63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "R_a = 0.4 \n",
-      "R_sh = 100. \t\t\t#armature and shunt field resistance\n",
-      "I_sh1 = V / R_sh\n",
-      "P_out_FL  =  10 * 735.5\n",
-      "eta = 85./100 \t\t\t#efficiency\n",
-      "P_in =  P_out_FL/eta\n",
-      "I_L1 =  P_in /V\n",
-      "I_a1 =  I_L1 - I_sh1\n",
-      "\n",
-      "# T (prop.) phi*I_a (prop.) I_sh*I_a    because phi (prop.) I_sh\n",
-      "#Bu torque is constant.. \n",
-      "Ia2_Ish2 =  I_a1*I_sh1\n",
-      "E_b1 =  V - I_a1*R_a\n",
-      "\n",
-      "#N (prop.) E_b/I_sh\n",
-      "#put E_b2 =  V - I_a2*R_a  and solving further for I_sh2 we get ,I_sh2**2 - 1.8824 I_sh2 +0.2417 = 0\n",
-      "p = array([1, -1.8824, 0.2417])\n",
-      "ans = roots(p) \n",
-      "I_sh2 = ans[0]\n",
-      "#root 1 was considered because its always easier to attain root(1) because less resistacne is needeed\n",
-      "#R_x in series with field\n",
-      "R_x  =  (V/I_sh2) -R_sh   \t\t\t#because I_sh2  =  V/(R_sh + R_x)\n",
-      "print 'Extra resistance to be added   =  %.2f ohms'%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Extra resistance to be added   =  43.37 ohms\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.24 page no : 64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "R_t = 1. \t\t\t#R_t  =  R_se + R_a\n",
-      "V_1 =  230.\n",
-      "N_1 = 300.\n",
-      "N_2 = 375.\n",
-      "I_1 = 15.\n",
-      "I_a1 = I_1\n",
-      "\n",
-      "# Calculations\n",
-      "#T (prop.) I_a**2  and T (prop.) N_2....therefore I_a**2 (prop.) N**2\n",
-      "I_a2 = I_a1 *(N_2/N_1)\n",
-      "E_b1  =  V_1 - I_a1*(R_t)\n",
-      "\n",
-      "#N (prop.) E_b/I_a\n",
-      "E_b2 =  E_b1*(I_a2/I_a1)*(N_2/N_1)\n",
-      "V_2 = E_b2 + I_a2* (R_t)  \t\t\t#because E_b2  =  V_2 - I_a2*(R_a+R_se)\n",
-      "\n",
-      "# Results\n",
-      "print 'Voltage supply needed  =  %.4f V'%(V_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage supply needed  =  354.6875 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.25 page no : 66"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "I_L1 = 30.\n",
-      "V = 230.\n",
-      "R_sh = 230.\n",
-      "R_a = 1.\n",
-      "I_sh =  V / R_sh\n",
-      "I_a1 =  I_L1 - I_sh\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "\n",
-      "# Calculations\n",
-      "#T (prop.) phi*I_a (prop.) I_a   as phi is constant\n",
-      "#and torque is constant\n",
-      "I_a2  =  I_a1\n",
-      "N2_by_N1 =  1./2\n",
-      "#N (prop.) E_b/phi (prop.) E_b\n",
-      "E_b2 =  E_b1 *(N2_by_N1)\n",
-      "R_x =  (V- E_b2)/I_a2 - R_a     \t\t\t#Because E_b2  =  V - I_a2*(R_a + R_x)\n",
-      "\n",
-      "# Results\n",
-      "print 'resistance to be inserted in series  =  %.4f ohms '%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "resistance to be inserted in series  =  3.4655 ohms \n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.26 page no : 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "# Variables\n",
-      "T_1 = 40.  \t\t\t#initial torque\n",
-      "#phi_1 is initial flux\n",
-      "#phi_2 is new flux\n",
-      "#T_2 is new torque\n",
-      "#I_a1 is initial current\n",
-      "#I_a2 is new current\n",
-      "phi2_by_phi1  =  1- (30./100)  \t\t\t#decrease by 30 percent\n",
-      "Ia2_by_Ia1 = 1+(15./100)       \t\t\t#increase by 15 percent\n",
-      "\n",
-      "# Calculations\n",
-      "#T (prop.)phi*I_a\n",
-      "T_2 = T_1*(phi2_by_phi1)*(Ia2_by_Ia1)\n",
-      "\n",
-      "# Results\n",
-      "print 'New torque is %.1f N-m'%(T_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New torque is 32.2 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.27 page no : 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "# Variables\n",
-      "V = 230.\n",
-      "N_1 = 1000.\n",
-      "N_2 = 950.\n",
-      "R_a = 0.5\n",
-      "R_sh = 230. \t\t\t#armature and shunt field resistance\n",
-      "I_L1 = 10.\n",
-      "\n",
-      "# Calculations\n",
-      "I_sh  =  V/R_sh\n",
-      "I_a1  =  I_L1 - I_sh\n",
-      "\n",
-      "#T (prop.) phi*I_a  (prop.) I_a with phi constant and T is constant due to full-load\n",
-      "I_a2 = I_a1\n",
-      "\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "E_b2 = E_b1*(N_2/N_1) \t\t\t#N (prop.) E_b /phi (prop.) E_b   as phi is constant\n",
-      "\n",
-      "R_x  =  (V-E_b2)/I_a2  -R_a      \n",
-      "\n",
-      "# Results\n",
-      "print 'resistance to be inserted in series with armature  =  %.4f ohms'%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "resistance to be inserted in series with armature  =  1.2528 ohms\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28 page no : 68"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "N_0 = 1000.\n",
-      "I_0 = 5.\n",
-      "R_a = 0.2\n",
-      "R_sh = 250. \t\t\t#armature and shunt field resistance\n",
-      "I_L = 50. \t\t\t#on no load\n",
-      "I_sh = V / R_sh\n",
-      "I_a0  =  I_0 - I_sh\n",
-      "I_a   =  I_L - I_sh\n",
-      "E_b0  =  V- I_a0*R_a\n",
-      "E_b1  =  V- I_a *R_a\n",
-      "\n",
-      "# Calculations\n",
-      "phi1_by_phi0  = 1-(3./100) \t\t\t#weakens by 3 percent\n",
-      "#N (prop.) E_b/phi\n",
-      "N_1  =  N_0 *(E_b1/E_b0) /phi1_by_phi0\n",
-      "\n",
-      "# Results\n",
-      "print 'Speed when loaded and drawing 50A current is %.3f r.p.m'%(N_1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed when loaded and drawing 50A current is 993.695 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.29 page no : 69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 230.\n",
-      "I_a0 = 3.3\n",
-      "R_a = 0.3\n",
-      "R_sh = 160. \t\t\t#armature and shunt field resistance \n",
-      "I_L1 = 40.\n",
-      "N_0 = 1000.\n",
-      "E_b0  =  V - I_a0*R_a\n",
-      "I_sh = V/ R_sh\n",
-      "I_a1  =  I_L1 - I_sh\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "phi1_by_phi0 =  1- (4./100) \t\t\t#weakening by 4 percent \n",
-      "\n",
-      "# Calculations and Results\n",
-      "N_1  =  N_0 *(E_b1/E_b0)/(phi1_by_phi0)  \t\t\t#because N (prop.) E_b/phi\n",
-      "print 'Full load speed is %.4f rpm'%(N_1)\n",
-      "T_0  =  E_b0*I_a0/(2*math.pi*N_0/60)\n",
-      "T_1  =  T_0*(I_a1/I_a0)*phi1_by_phi0   \t\t\t# because T (prop.) phi*I_a\n",
-      "print 'Full load developed torque is %.4f N-m'%(T_1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Full load speed is 993.5485 rpm\n",
-        "Full load developed torque is 80.9585 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.30 page no : 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "I_L = 52.\n",
-      "N_1 = 750.\n",
-      "N_2 = 600.\n",
-      "R_a = 0.2\n",
-      "R_sh  =  110. \t\t\t#armature and shunt field resistance\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I_sh = V/ R_sh\n",
-      "I_a1 =  I_L - I_sh\n",
-      "I_a2 = I_a1\t\t\t#T (prop.) I_a  and T is constant\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "\n",
-      "#N (prop.) E_b/phi (prop.) E_b\n",
-      "E_b2  =  E_b1*(N_2/N_1)\n",
-      "R_x   =  (V- E_b2)/I_a2 -R_a  \t\t\t#Because E_b2  =  V - I_a2*(R_a+R_x)\n",
-      "print 'resistance to be connected in series  =  %.2f ohms'%(R_x)\n",
-      "\n",
-      "#After R_x gets connected in series with armature and 110 ohms in series with field winding\n",
-      "N_1 = 600.\n",
-      "I_sh2 = V /(R_sh+110)\n",
-      "I_a1 = 50.\n",
-      "I_sh1 = 2.\n",
-      "I_sh2 = 1.\n",
-      "#T (prop.) I_a*I_sh and T doesn't vary\n",
-      "I_a2  =  I_a1*(I_sh1/I_sh2)\n",
-      "E_b1  =  V - I_a1*(R_a+R_x)\n",
-      "E_b2  =  V - I_a2*(R_a+R_x)\n",
-      "N_2  =  N_1*(E_b2/E_b1)*(I_sh1/I_sh2) \t\t\t#Because N (prop.) E_b/I_sh\n",
-      "print 'New speed =  %.3f rpm'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "resistance to be connected in series  =  0.84 ohms\n",
-        "New speed =  828.571 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.31 page no : 72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 230.\n",
-      "R_a = 0.15\n",
-      "R_sh = 250. \t\t\t#armature and shunt field resistance\n",
-      "I_a1 = 50.\n",
-      "I_a2 =  80.\n",
-      "N_1 = 800.\n",
-      "N_2 = 1000.\n",
-      "I_sh1 =  V / R_sh\n",
-      "\n",
-      "# Calculations\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "E_b2  =  V - I_a2*R_a\n",
-      "\n",
-      "I_sh2 = I_sh1*(E_b2/E_b1)*(N_1/N_2) \t\t\t#Because N (prop.) E_b/ I_sh\n",
-      "R_x =  (V/I_sh2 ) - R_sh    \t\t\t#because I_sh2  =  V /(R_x+ R_sh)\n",
-      "\n",
-      "# Results\n",
-      "print 'resistance to be added is R_x = %.0f ohms'%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "resistance to be added is R_x = 69 ohms\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.32 page no : 74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 230.\n",
-      "R_a = 0.5\n",
-      "N_1 = 800.\n",
-      "N_2 = 600.\n",
-      "I_a2  = 20. \n",
-      "I_a1 = I_a2\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "\n",
-      "# Calculations\n",
-      "#N (prop.) E_b/phi (prop.) E_b         as phi is constant\n",
-      "E_b2 = E_b1 *(N_2/N_1)\n",
-      "#additional resistance required\n",
-      "R_x  =  (V -E_b2)/I_a2  - R_a   \t\t\t#because E_b2  =  V - I_a2*(R_a+R_x)\n",
-      "\n",
-      "# Results\n",
-      "print 'Additional resistance required  =  %.2f ohms '%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Additional resistance required  =  2.75 ohms \n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.33 page no : 74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "R_a = 0.5\n",
-      "R_x = 5. \t\t\t#armature resistacne and extra resistance\n",
-      "I_1 = 15.\n",
-      "I_se1 = I_1\n",
-      "I_se2 = I_se1 \n",
-      "I_2 = I_se2\n",
-      "N_1 = 800.\n",
-      "\n",
-      "# Calculations\n",
-      "E_b1  =  V - I_1*R_a\n",
-      "E_b2  =  V - I_2*(R_a+R_x)\n",
-      "\n",
-      "N_2 =  N_1*(E_b2/E_b1)*(I_se1/I_se2) \t\t\t#because N (prop.) E_b/I_se\n",
-      "\n",
-      "# Results\n",
-      "print 'New speed of rotor  =  %.3f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New speed of rotor  =  517.647 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.34 page no : 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "I_a1 = 20.\n",
-      "R_a = 0.5\n",
-      "N_1 = 1000.\n",
-      "N_2 = 500.\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#T (prop.) I_a and T_1 = T_2\n",
-      "I_a2 = I_a1\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "\n",
-      "#N (prop.) E_b\n",
-      "E_b2 =  E_b1 *(N_2/N_1)\n",
-      "R_x =  (V-E_b2)/I_a2 - R_a   \t\t\t#because E_b2  =  V - I_a2*(R_a+R_x)\n",
-      "print 'Additional resistance  =  %.0f ohms'%(R_x)\n",
-      "T3_by_T2 = 0.5   \t\t\t#torque is halved\n",
-      "I_a3 =  I_a2 *(T3_by_T2)   \t\t\t#new armature current\n",
-      "E_b3  =  V - I_a3*(R_x + R_a)\n",
-      "N_3 = E_b3*N_2 / E_b2 \t\t\t#N (prop.) E_b\n",
-      "print 'New speed  =  %.3f rpm'%(N_3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Additional resistance  =  6 ohms\n",
-        "New speed  =  770.833 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.35 page no : 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "P_out =  100*735.5\n",
-      "V = 500.\n",
-      "P = 4.\n",
-      "A = 2.\t\t\t# due to wave winding\n",
-      "Z = 492. \t\t\t#no of conductors\n",
-      "phi = 50.*10**-3 \t\t\t#flux per pole\n",
-      "eta = 92./100 \t\t\t#efficiency\n",
-      "P_in =  P_out/eta\n",
-      "R_a = 0.1 \n",
-      "R_sh = 250. \t\t\t#amature and shunt field resistance\n",
-      "\n",
-      "# Calculations\n",
-      "I_L = P_in/V\n",
-      "I_sh  =  V/ R_sh\n",
-      "I_a  =  I_L - I_sh\n",
-      "E_b  =  V - I_a*R_a\n",
-      "N = E_b*60*A/(phi*P*Z)  \t\t\t#because E_b =   phi*P*N*Z/(60*A)\n",
-      "\n",
-      "T_sh =  P_out/(2*math.pi*N/60) \t\t\t#Useful torque\n",
-      "\n",
-      "# Results\n",
-      "print 'i)Speed at full load  =  %.4f rpm'%(N)\n",
-      "print 'ii)Useful torque  =  %.2f N-m'%(T_sh)\n",
-      "print 'Answer mismatches due to improper approximation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Speed at full load  =  590.5011 rpm\n",
-        "ii)Useful torque  =  1189.41 N-m\n",
-        "Answer mismatches due to improper approximation\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.36 page no : 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from numpy import *\n",
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "N_1 = 1000.\n",
-      "I_1 = 50.\n",
-      "I_a1 = I_1\n",
-      "V = 250.\n",
-      "R_x = 4.4\n",
-      "R_t = 0.6   \t\t\t#R_t  =  R_a+R_se\n",
-      "E_b1 = V - I_a1*(R_t)\n",
-      "\n",
-      "# Calculations\n",
-      "#T (prop.)I_a**2   T (prop.) N**2  .... hence N (prop.) I_a\n",
-      "#N (prop.) E_b /I_a \n",
-      "#combining both  E_b (prop.) I_a**2\n",
-      "#using E_b2  =  V - I_a2*(R_a + R_se + R_x) and solving for I_a2 we get 0.088 I_a2**2 +5 I_a2 -250 = 0\n",
-      "p = [0.088 ,5, -250] \n",
-      "ans = roots(p)\n",
-      "I_a2 = ans[1] \t\t\t#root(1) is ignored as it is -ve\n",
-      "E_b2  =  V - I_a2*(R_t + R_x) \n",
-      "N_2 = N_1*(E_b2/E_b1)*(I_a1/I_a2)\n",
-      "\n",
-      "# Results\n",
-      "print 'Motor speed  =  %.2f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Motor speed  =  639.79 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.37 page no : 78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from numpy import *\n",
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "I_a1 = 20.\n",
-      "R_sh =  250.\n",
-      "R_a = 0.5 \t\t\t#shunt field and armature resistance\n",
-      "I_sh1 =  V / R_sh\n",
-      "E_b1  =  V - I_a1*R_a\n",
-      "\n",
-      "#T (prop.) phi*I_a (prop.) I_sh*I_a\n",
-      "#math.since T_1  =  T_2\n",
-      "I_sh2_I_a2  =  I_sh1*I_a1  \n",
-      "I_sh2_I_a2  =  I_sh1*I_a1  \t\t\t# = 20\n",
-      "\n",
-      "# Calculations\n",
-      "#N (prop.) E_b/I_sh\n",
-      "#E_b1  =  V - I_a1*R_a\n",
-      "#Solving further for I_a2 we get I_a2**2 -500 I_a2 + 12800\n",
-      "p = [1, -500, 12800]\n",
-      "ans = roots(p)\n",
-      "I_a2 = ans[1] \t\t\t#higher root is neglected\n",
-      "I_sh2 =  I_sh2_I_a2 / I_a2\n",
-      "R_x =  (V / I_sh2) - R_sh \t\t\t#resistance to be inserted in shunt field\n",
-      "\n",
-      "# Results\n",
-      "print 'resistance to be inserted  =  %.4f ohms '%(R_x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "resistance to be inserted  =  88.3129 ohms \n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.38 page no : 79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "N_1 = 1000.\n",
-      "I_L1 = 25.\n",
-      "R_a = 0.2\n",
-      "R_sh = 250. \t\t\t#armature and shunt field resistance\n",
-      "V_brush =  1. \t\t\t#voltage drop due to brushes\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I_sh1  =  V/R_sh\n",
-      "I_a1 =  I_L1 - I_sh1\n",
-      "E_b1 =  V- I_a1*R_a - 2 *V_brush\n",
-      "\n",
-      "#when loaded\n",
-      "I_L2 = 50.\n",
-      "I_sh2 = I_sh1   \t\t\t#as flux weakensby armature reaction shunt field current remains same \n",
-      "I_a2 =  I_L2 - I_sh2\n",
-      "E_b2 =  V- I_a2*R_a - 2 *V_brush\n",
-      "\n",
-      "phi2_by_phi1 =  1- (3./100)  \t\t\t#weakens by 3 percent\n",
-      "N_2 =  N_1*(E_b2/E_b1)/ phi2_by_phi1   \t\t\t#N (prop.) E_b/phi\n",
-      "print 'New speed  =  %.3f rpm'%(N_2)\n",
-      "T_1 =  E_b1*I_a1/(2*math.pi*N_1/60)\n",
-      "T_2 =  E_b2*I_a2/(2*math.pi*N_2/60)\n",
-      "print 'Torque before field weakening  =  %.4f N-m'%(T_1)\n",
-      "print 'Torque after  field weakening  =  %.4f N-m'%(T_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New speed  =  1009.733 rpm\n",
-        "Torque before field weakening  =  55.7373 N-m\n",
-        "Torque after  field weakening  =  110.3831 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.39 page no : 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "R_a = 0.5\n",
-      "R_x = 1. \t\t\t#armature resistance and extra resistance\n",
-      "N_FL = 500. \t\t\t#full load speed in r.p.m\n",
-      "I_a_FL = 30.\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)   Full load \n",
-      "E_b_FL =  V- I_a_FL * R_a\n",
-      "#T (prop.) I_a... T is constant\n",
-      "I_a_dash_FL  =  I_a_FL \n",
-      "E_b_dash_FL  =   V- I_a_dash_FL * (R_a+R_x)\n",
-      "#N (prop.) E_b/phi   (prop.) E_b\n",
-      "N_dash_FL  =  N_FL*(E_b_dash_FL/E_b_FL)\n",
-      "print 'i)Speed at full load torque  = %.4f r.p.m'%(N_dash_FL)\n",
-      "\n",
-      "#part(ii)\n",
-      "T2_by_T1  =  2\n",
-      "I_a_dash_FL   =   I_a_FL  *(T2_by_T1)\n",
-      "E_b_dash_FL  =   V- I_a_dash_FL * (R_a+R_x)\n",
-      "N_dash_FL  =  N_FL*(E_b_dash_FL/E_b_FL)\n",
-      "print 'ii)Speed at double full load torque  = %.3f r.p.m'%(N_dash_FL)\n",
-      "\n",
-      "#part(iii) ...stalling\n",
-      "E_b = 0 \t\t\t#as speed is zero in case of stalling torque\n",
-      "I_a_stall = (V-E_b)/(R_a+R_x)\n",
-      "T_FL  =  E_b_FL * I_a_FL/(2*math.pi*N_FL/60)\n",
-      "T_stall  =  T_FL *(I_a_stall/ I_a_FL)\n",
-      "print 'iii)Stalling torque  =  %.3f Nm'%(T_stall)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Speed at full load torque  = 426.8293 r.p.m\n",
-        "ii)Speed at double full load torque  = 317.073 r.p.m\n",
-        "iii)Stalling torque  =  574.231 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.40 page no : 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 230.\n",
-      "I_a1 = 30.\n",
-      "R_a = 0.4\n",
-      "R_x = 1.1\t\t\t#armature resistance and extra resistance\n",
-      "N_1 = 500.\n",
-      "\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "E_b1 =  V - I_a1*R_a\n",
-      "I_a2  =  I_a1 \t\t\t#I_a is constant as T , phi are constant\n",
-      "E_b2 =  V - I_a2*(R_a+R_x)\n",
-      "N_2  =  N_1 *(E_b2/E_b1) \t\t\t#Because N (prop.) E_b/phi (prop.) E_b\n",
-      "print 'i)Speed at full load torque  =  %.3f r.p.m'%(N_2)\n",
-      "\n",
-      "#part(ii)\n",
-      "T2_by_T1 = 1.5\n",
-      "I_a2 =  I_a1 * T2_by_T1\n",
-      "E_b2 =  V - I_a2*(R_a+R_x)\n",
-      "N_2  =  N_1 *(E_b2/E_b1) \t\t\t#Because N (prop.) E_b/phi (prop.) E_b\n",
-      "print 'ii)Speed at 1.5 times full load torque  =  %.3f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Speed at full load torque  =  424.312 r.p.m\n",
-        "ii)Speed at 1.5 times full load torque  =  372.706 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines/ch3.ipynb b/DC_Machines_and_Synchronous_Machines/ch3.ipynb
deleted file mode 100755
index 6f59c07f..00000000
--- a/DC_Machines_and_Synchronous_Machines/ch3.ipynb
+++ /dev/null
@@ -1,1912 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:851db29572c4b7f5287b09f53c6bb9c3406b6b6577d3bab318bf198d21eaaa81"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3 : Testing of D C Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1 page no : 5"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 6.\n",
-      "V = 500.\n",
-      "A = 2. \t\t\t#because of wave wound armature\n",
-      "Z = 1200. \t\t\t#number of armature conductors\n",
-      "phi = 20.*10**-3 \t\t\t#useful flux per pole\n",
-      "Ra = 0.5 \n",
-      "Rsh = 250.  \t\t\t#armature and field resistance\n",
-      "Il = 20. \t\t\t#current drawn from supply\n",
-      "mechanical_losses = 900.\n",
-      "Ish = V/Rsh\n",
-      "Ia = Il-Ish\n",
-      "Eb = V-Ia*Ra  \t\t\t#because V = Eb+Ia*Ra\n",
-      "N = Eb*60*A/(phi*Pole*Z)   \t\t\t#Eb = phi*Pole*N*Z/(60*A)\n",
-      "\n",
-      "# Calculations\n",
-      "P_m = Eb*Ia  \t\t\t#Electrical equivalent of mechanical power\n",
-      "omega = 2*math.pi*N/60\n",
-      "Tg = P_m/omega\n",
-      "\n",
-      "P_out = P_m-mechanical_losses\n",
-      "T_sh = P_out/omega  \t\t\t#Useful torque\n",
-      "P_in = V*Il\n",
-      "percentage_efficiency = 100*P_out/P_in\n",
-      "\n",
-      "# Results\n",
-      "print 'Speed developed is %.3f r.p.m \\\n",
-      "\\nTorque developed is %.2f N-m  \\\n",
-      "\\nUseful torque is %.2f N-m \\\n",
-      "\\nEfficiency is %.2f percent'%(N,Tg,T_sh,percentage_efficiency)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed developed is 409.167 r.p.m \n",
-        "Torque developed is 206.26 N-m  \n",
-        "Useful torque is 185.26 N-m \n",
-        "Efficiency is 79.38 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2 page no : 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "#no load\n",
-      "I_noload = 2.5 \t\t\t#No load current\n",
-      "V = 440.\n",
-      "R_a = 1.2\n",
-      "R_sh = 550.\t\t\t#resistance of armature and shunt field windings\n",
-      "no_load_input = V*I_noload\n",
-      "\n",
-      "# Calculations\n",
-      "I_sh = V/R_sh\n",
-      "I_a_noload = I_noload-I_sh\n",
-      "no_load_armature_copper = (I_a_noload**2)*R_a\n",
-      "constant_losses = no_load_input-no_load_armature_copper\n",
-      "\n",
-      "#full load\n",
-      "I_fullload = 32\n",
-      "I_a_fullload = I_fullload-I_sh\n",
-      "full_load_armature_coppe = (I_a_fullload**2)*R_a\n",
-      "total_losses = full_load_armature_coppe+constant_losses\n",
-      "full_load_motor_input = V*I_fullload\n",
-      "full_load_motor_output = full_load_motor_input-total_losses\n",
-      "efficiency_at_full_load = full_load_motor_output*100/full_load_motor_input\n",
-      "\n",
-      "# Results\n",
-      "print 'Full load motor output is %.2f W \\\n",
-      "\\nEfficiency of motor at full-load is %.2f percent'%(full_load_motor_output,efficiency_at_full_load)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Full load motor output is 11815.34 W \n",
-        "Efficiency of motor at full-load is 83.92 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3 page no : 11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "#no load\n",
-      "I = 14. \t\t\t#input current\n",
-      "V = 230.\n",
-      "power_output_FL  =  45.*10**3\n",
-      "power_input = V*I\n",
-      "I_sh = 2.55 \t\t\t#field current\n",
-      "R_a = 0.032 \t\t\t#armature resistance\n",
-      "I_a = I-I_sh\n",
-      "cu_loss_NL  =  I_a**2*R_a \t\t\t#no load copper loss\n",
-      "brush_loss = 2*I_a\n",
-      "constant_loss =  power_input - cu_loss_NL - brush_loss\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#full load\n",
-      "#I = I_a+ 2.55\n",
-      "#Motor input =  Motor output + constant loss + brush loss + cu loss\n",
-      "# solving for I_a I_a**2 - 7125 I_a + 1487700.3  = 0\n",
-      "p = [1, -7125, 1487700.3]\n",
-      "ans = roots(p)\n",
-      "I_a = ans[1] \t\t\t#ignoring second root as its too large\n",
-      "I = I_a+I_sh\n",
-      "print 'Full load current is %.2f A'%(I)\n",
-      "power_input = V*I\n",
-      "eta = 100*(power_output_FL/power_input)\n",
-      "print 'Efficiency at full load is %.2f percent'%(eta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Full load current is 217.86 A\n",
-        "Efficiency at full load is 89.81 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4 page no : 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "W1 = 9.1  \t\t\t#Tension on tight side\n",
-      "W2 = 0.8  \t\t\t#Tension on slack side\n",
-      "I = 10.    \t\t\t#Total_current\n",
-      "V = 110.   \t\t\t#Supply voltage\n",
-      "R = 7.5/100   \t\t\t#Radius of p-ulley in metres\n",
-      "N = 1320.     \t\t\t#speed in r.p.m\n",
-      "\n",
-      "# Calculations\n",
-      "T_sh = (W1-W2)*9.81*R   \t\t\t#9.81 is the accelration due to gravity\n",
-      "omega = (2*math.pi*N/60)\n",
-      "P_out = T_sh*omega\n",
-      "P_in = V*I\n",
-      "\n",
-      "efficiency = 100*P_out/P_in\n",
-      "\n",
-      "# Results\n",
-      "print 'Full load Efficiency is %.2f percent'%(efficiency)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Full load Efficiency is 76.74 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5 page no : 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "I_av = 10.\n",
-      "V_av = (240.+220)/2 \t\t\t#average voltage across load\n",
-      "W_dash = V_av*I_av  \t\t\t#Power absorbed\n",
-      "t1 = 25.\n",
-      "t2 = 6.\n",
-      "R_sh = 200.\n",
-      "R_a = 0.3\t\t\t#resistance of field winding and armature \n",
-      "\n",
-      "# Calculations\n",
-      "W = W_dash*t2/(t1-t2)   \t\t\t#Stray Losses\n",
-      "I_l = 25 \t\t\t#Input current\n",
-      "I_sh = V/R_sh \t\t\t#current through field winding\n",
-      "\n",
-      "I_a = I_l-I_sh   \t\t\t#Armature current\n",
-      "arm_cu_loss = R_a*I_a**2  \t\t\t#Armature copper losses\n",
-      "sh_cu_loss = R_sh*I_sh**2 \t\t\t# Shunt copper loss\n",
-      "\n",
-      "Total_losses =  arm_cu_loss + sh_cu_loss + W\n",
-      "Motor_input = V*I_l\n",
-      "Output = Motor_input- Total_losses\n",
-      "efficiency = Output*100/Motor_input\n",
-      "\n",
-      "# Results\n",
-      "print 'Efficiency as motor at 25 A and 250 V is %.2f percent'%(efficiency)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency as motor at 25 A and 250 V is 80.67 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6 page no : 22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "# Variables\n",
-      "I_a = 37.\n",
-      "I_sh = 0.85\t\t\t#armature and field current for motor\n",
-      "V = 230.\n",
-      "R_a = 0.33 \t\t\t#armature resistance\n",
-      "\n",
-      "I_a_g = 30.\n",
-      "I_sh_g = 0.8\t\t\t#armature and field current for generator\n",
-      "\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#for motor\n",
-      "arm_cu_loss =  I_a**2* R_a   \t\t\t#armature copper losses\n",
-      "field_cu_loss = V*I_sh  \t\t\t#field copper loss\n",
-      "total_cu_loss =  field_cu_loss + arm_cu_loss \t\t\t#total copper loss \n",
-      "\n",
-      "#for generator\n",
-      "arm_cu_loss_g =  I_a_g**2* R_a   \t\t\t#armature copper losses\n",
-      "field_cu_loss_g = V*I_sh_g  \t\t\t#field copper loss\n",
-      "total_cu_loss_g =  field_cu_loss_g + arm_cu_loss_g \t\t\t#total copper loss\n",
-      "\n",
-      "#for motor-generator set\n",
-      "total_cu_loss_set =    total_cu_loss_g + total_cu_loss\n",
-      "P_supply = V*(I_a - I_a_g + I_sh+ I_sh_g ) \t\t\t#power taken from supply\n",
-      "stray_loss =  P_supply - (total_cu_loss_g + total_cu_loss) \n",
-      "stray_loss_each =   stray_loss/2   \t\t\t#stray loss for each machine\n",
-      "\n",
-      "#efficiency of motor\n",
-      "motor_input  =  V*(I_a+I_sh)\n",
-      "motor_output  =   motor_input - (stray_loss_each + total_cu_loss)\n",
-      "eta_m =   100* motor_output/motor_input \t\t\t#efficiency of motor\n",
-      "print 'Efficiency of motor is %.2f percent '%(eta_m)\n",
-      "#efficiency of generator\n",
-      "generator_input  =  motor_output   \t\t\t#output of motor is input of generator\n",
-      "generator_output  =   generator_input - (stray_loss_each + total_cu_loss_g)\n",
-      "eta_g =   100* generator_output/generator_input \t\t\t#efficiency of generator\n",
-      "print 'Efficiency of generator is %.2f percent '%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 87.62 percent \n",
-        "Efficiency of generator is 88.05 percent \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7 page no : 25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I_1 = 400. \t\t\t#motor input current\n",
-      "V = 200.  \t\t\t#voltage across armature\n",
-      "I_2 = 32.  \t\t\t#load current\n",
-      "V_2 = 160.  \t\t\t#voltage across generator\n",
-      "V_f = 15. \t\t\t#voltage drop across field windings\n",
-      "total_input  =  (V+V_f)*I_1  \n",
-      "Output = V_2*I_2\n",
-      "total_losses  =  total_input-Output   \t\t\t#total losses in 2 machines\n",
-      "\n",
-      "# Calculations and Results\n",
-      "R_se = V_f/I_1   \t\t\t#series field resistance\n",
-      "R_a = 0.4 \t\t\t# armature field resistance\n",
-      "\n",
-      "total_cu_loss = (R_a + 2*R_se) * I_1**2 + I_2**2*R_a \t\t\t#total copper loss\n",
-      "stray_losses  =   total_losses -  total_cu_loss\n",
-      "stray_losses_each  = stray_losses /2 \t\t\t#stray losses for each machine\n",
-      "\n",
-      "#for motor\n",
-      "motor_input =  V*I_1\n",
-      "arm_cu_loss =  (R_a + R_se)*I_1*I_1 \t\t\t#armature copper loss\n",
-      "total_losses_motor  =  arm_cu_loss +  stray_losses_each \n",
-      "motor_output =  motor_input- total_losses_motor\n",
-      "eta_m = 100*motor_output/motor_input \t\t\t#efficiency of motor\n",
-      "print 'Efficiency of motor is %.2f percent '%(eta_m)\n",
-      "#for generator\n",
-      "arm_cu_loss_gen = R_a*I_2**2 \t\t\t#armature copper loss\n",
-      "series_field_cu_loss  =  V_f*I_1 \t\t\t#series field copper loss\n",
-      "total_losses_gen =  arm_cu_loss_gen +  series_field_cu_loss + stray_losses_each \n",
-      "generator_input  =  total_losses_gen+ Output\n",
-      "eta_gen = 100*Output/generator_input \t\t\t#efficiency of generator\n",
-      "print 'Efficiency of generator is %.2f percent'%(eta_gen)\n",
-      "\n",
-      "\n",
-      "#Note : 'Answer dont match because Output-of-generator is taken as 5220 for calculation while its should have been 5120'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 9.71 percent \n",
-        "Efficiency of generator is 37.20 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8 page no : 27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "V = 500.\n",
-      "Io = 5. \t\t\t#no load current\n",
-      "R_a = 0.5\n",
-      "R_sh = 250.\t\t\t#resistance of armature and field circuits\n",
-      "I = 100. \t\t\t#current at unknown efficiency\n",
-      "\n",
-      "# Calculations\n",
-      "P_in_NL = V*Io  \t\t\t#no load input\n",
-      "I_sh = V/R_sh\n",
-      "\n",
-      "Iao = Io-I_sh\n",
-      "arm_cu_loss_no_load = R_a*Iao**2  \t\t\t#No load armature copper loss\n",
-      "constant_losses =  P_in_NL- arm_cu_loss_no_load\n",
-      "\n",
-      "I_a = I-I_sh\n",
-      "arm_cu_loss =  R_a*I_a**2  \t\t\t#New armature copper loss\n",
-      "\n",
-      "Total_loss = arm_cu_loss + constant_losses\n",
-      "P_in = V*I\n",
-      "efficiency = (P_in-Total_loss)*100/P_in \t\t\t#required efficiency\n",
-      "\n",
-      "\n",
-      "# Results\n",
-      "print 'Efficiency is %.3f percent when motor takes %.0f A current'%(efficiency,I)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency is 85.405 percent when motor takes 100 A current\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9 page no : 28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 500.  \n",
-      "I_NL = 5.  \t\t\t#no load current \n",
-      "P_in_NL  =  V* I_NL   \t\t\t#no load input\n",
-      "R_a = 0.22\n",
-      "R_sh = 250. \t\t\t#resistance of armature and shunt field winding\n",
-      "I_sh = V/R_sh\n",
-      "I_a_NL = I_NL-I_sh \t\t\t#armature current no load\n",
-      "arm_cu_loss_NL  =  R_a*I_a_NL**2  \t\t\t#No-load armature copper loss\n",
-      "constant_loss  =   P_in_NL -  arm_cu_loss_NL\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#at 100 A current\n",
-      "I = 100.\n",
-      "I_a  =  I - I_sh\n",
-      "arm_cu_loss  =   R_a*I_a**2 \t\t\t#armature copper loss\n",
-      "total_loss  =  arm_cu_loss +  constant_loss \n",
-      "motor_input  =  V*I\n",
-      "motor_output  =   motor_input- total_loss\n",
-      "eta_m =  100*motor_output/motor_input  \t\t\t#motor efficiency\n",
-      "print 'Parta)Efficiency of motor when it takes 100 A current and loaded is %.2f percent'%(eta_m)\n",
-      "\n",
-      "#part(b)\n",
-      "E_b_NL  =  V - I_a_NL*R_a   \t\t\t#back emf at no load\n",
-      "\n",
-      "E_b  =  V- I_a*R_a  \t\t\t#back EMF at 100 A current\n",
-      "#Since E_b is proportional to N and using componendo dividendo\n",
-      "delta_speed =  100*((E_b_NL-E_b)/E_b)\n",
-      "print 'Partb)Percentage speed in speed is %.3f percent'%(delta_speed)\n",
-      "\n",
-      "print 'Note that the following were assumptions made'\n",
-      "print 'i) Due to heating %( resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected'\n",
-      "print 'ii) Though the motor speed is changing from no load to given load , the mechanical losses are assumed to be cosnstant'\n",
-      "print 'iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Parta)Efficiency of motor when it takes 100 A current and loaded is 90.78 percent\n",
-        "Partb)Percentage speed in speed is 4.368 percent\n",
-        "Note that the following were assumptions made\n",
-        "i) Due to heating %( resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected\n",
-        "ii) Though the motor speed is changing from no load to given load , the mechanical losses are assumed to be cosnstant\n",
-        "iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10 page no : 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "motor_output_FL  = 15000. \t\t\t#full load motor output\n",
-      "V = 250.\n",
-      "R_sh = 100.\n",
-      "\n",
-      "# Calculations\n",
-      "#at 80 % of full load\n",
-      "motor_output_FL_dash = (80./100)*motor_output_FL  \t\t\t#80 percent of full load output\n",
-      "eta = 90./100 \t\t\t#efficiency\n",
-      "motor_input = motor_output_FL_dash/eta\n",
-      "total_losses  =  motor_input - motor_output_FL_dash   \t\t\t#at 80 % of full load\n",
-      "#at maximum efficiency variable losses  =  constant losses\n",
-      "constant_losses =  total_losses/2\n",
-      "variable_losses =  constant_losses\n",
-      "I =  motor_input/V \t\t\t#line current at 80% load\n",
-      "I_sh =  V/ R_sh\n",
-      "I_a =  I- I_sh\n",
-      "#math.since armature copper loss  = R_a*I_a**2  \n",
-      "R_a = variable_losses/I_a**2\n",
-      "E_b1 = V-I_a*R_a \t\t\t#motor back EMF at 80% of full load\n",
-      "N_1 = 750. \t\t\t# corresponding speed is given as 750 rpm\n",
-      "\n",
-      "#When motor current is 80 A\n",
-      "I = 80.\n",
-      "I_a = I-I_sh\n",
-      "arm_cu_losses =  R_a*I_a**2  \t\t\t#armature copper loss\n",
-      "total_losses  =   arm_cu_losses + constant_losses\n",
-      "motor_input =  V*I\n",
-      "motor_output  =  motor_input- total_losses\n",
-      "eta = 100*motor_output/motor_input \t\t\t#efficiency of motor\n",
-      "E_b2 = V-I_a*R_a \t\t\t#motor back EMF at 80% of full load\n",
-      "N_2 = N_1*(E_b2/E_b1)  \t\t\t#because E_b is proportional to N\n",
-      "\n",
-      "# Results\n",
-      "print 'Efficiency of motor is %.2f percent when motor draws 80A current'%(eta)\n",
-      "print 'and Speed is %.2f r.p.m'%(N_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 88.92 percent when motor draws 80A current\n",
-        "and Speed is 728.22 r.p.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11 page no : 32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "R_sh = 166.67\n",
-      "R_a = 0.15 \t    \t\t#resistance of shunt field winding and armature\n",
-      "N_0 = 1280. \t\t\t#in rpm\n",
-      "I_L1 = 67.   \t\t\t#current drawn on full load\n",
-      "I_sh =  V/ R_sh\n",
-      "I_a1 = I_L1 - I_sh\n",
-      "E_b1 = V-I_a1*R_a\n",
-      "\n",
-      "#on no load\n",
-      "I_L0 = 6.5\n",
-      "I_a0 =  I_L0 - I_sh\n",
-      "E_b0 = V-I_a0*R_a\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part (i)\n",
-      "#using speed equation  N is proportional to E_b\n",
-      "N_1 = N_0*(E_b1/E_b0)\n",
-      "print 'i)Full load speed is %.3f r.p.m'%(N_1)\n",
-      "\n",
-      "#part (ii)\n",
-      "speed_regulation = 100*((N_0-N_1)/N_1)\n",
-      "print 'ii)Speed regulation is %.2f percent '%(speed_regulation)\n",
-      "\n",
-      "#part (iii)\n",
-      "stray_losses  =  E_b0*I_a0  \t\t\t#mechanical power developed on no load\n",
-      "power_developed_FL =  E_b1*I_a1 \n",
-      "shaft_output_FL =  power_developed_FL - stray_losses \n",
-      "hp_rating =  shaft_output_FL/746 \t\t\t#in horse power\n",
-      "print 'iii)H.P rating of the machine Is %.2f H.P'%(hp_rating)\n",
-      "\n",
-      "#part (iv)\n",
-      "power_input = V*I_L1\n",
-      "eta = 100*(shaft_output_FL/power_input)  \t\t\t#efficiency at full load\n",
-      "print 'iv)Efficiency at full load is %.2f percent'%(eta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Full load speed is 1233.396 r.p.m\n",
-        "ii)Speed regulation is 3.78 percent \n",
-        "iii)H.P rating of the machine Is 19.42 H.P\n",
-        "iv)Efficiency at full load is 86.48 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.12 page no : 33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 200.\n",
-      "R_sh = 240.\n",
-      "R_a = 0.1 \t\t\t#resistance of shunt field winding and armature\n",
-      "rotational_loss = 236.\n",
-      "I_L_FL = 9.8  \t\t\t#full load line current\n",
-      "N = 1450.\n",
-      "I_sh = V/R_sh\n",
-      "I_a_FL  =  I_L_FL - I_sh\n",
-      "E_b =  V- I_a_FL * R_a\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "gross_mech_P_dev =  E_b*I_a_FL  \t\t\t#gross mechanical power developed\n",
-      "mech_P_dev =  gross_mech_P_dev - rotational_loss \t\t\t#mechanical power developed\n",
-      "print 'i)Gross mechanical power developed is %.2f W'%(gross_mech_P_dev )\n",
-      "print '   Mechanical power developed is %.2f W'%(mech_P_dev )\n",
-      "\n",
-      "#part(ii)\n",
-      "P_out = mech_P_dev \n",
-      "print 'ii)The power output is %.2f W'%(P_out)\n",
-      "\n",
-      "#part(iii)\n",
-      "T_sh = P_out*60/(2*math.pi*N)\n",
-      "T_L = T_sh\n",
-      "print 'iii)Load torque is %.2f N-m'%(T_L)\n",
-      "\n",
-      "#part(iv)\n",
-      "P_in = V*I_L_FL\n",
-      "eta = 100*P_out/P_in\n",
-      "print 'iv)Efficiency at full load is %.2f percent'%(eta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Gross mechanical power developed is 1785.29 W\n",
-        "   Mechanical power developed is 1549.29 W\n",
-        "ii)The power output is 1549.29 W\n",
-        "iii)Load torque is 10.20 N-m\n",
-        "iv)Efficiency at full load is 79.05 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.13 page no : 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 240.\n",
-      "P_out = 25*735.5  \t\t\t#output power in watts\n",
-      "R_a = 0.14\n",
-      "R_sh = 80. \t        \t\t#resistance of armature and shunt field winding\n",
-      "brush_drop = 1. \t\t\t#voltage drop across brush\n",
-      "I_L_FL = 95. \t\t    \t#input current at full load \n",
-      "I_sh = V/R_sh\n",
-      "I_a_FL  =  I_L_FL - I_sh \t\t\t#armature current at full load\n",
-      "\n",
-      "# Calculations and Results\n",
-      "arm_cu_loss_FL  =  R_a*I_a_FL**2  \t\t\t#full load armature copper loss\n",
-      "field_cu_loss =  R_sh*I_sh**2  \t\t\t#field copper loss\n",
-      "print 'i)Armature and field copper losses are %.2f W and %.0f W respectively'%(arm_cu_loss_FL,field_cu_loss)\n",
-      "brush_cu_loss =  2*brush_drop*I_a_FL \t\t\t#brush contact copper loss\n",
-      "print 'ii)Brush contact copper loss is %.0f W'%(brush_cu_loss)\n",
-      "E_b = V-I_a_FL*R_a - 2*brush_drop \t\t\t#back emf\n",
-      "gross_mech_P_dev =  E_b*I_a_FL  \t\t\t#gross mechanical power developed\n",
-      "IFW_losses  =  gross_mech_P_dev - P_out  \t\t\t#iron friction and windage losses\n",
-      "print 'iii)Core plus mechanical losses  =  %.1f W'%(IFW_losses+field_cu_loss+arm_cu_loss_FL)\n",
-      "eta = 100*(P_out/(P_out + IFW_losses+ brush_cu_loss+field_cu_loss+arm_cu_loss_FL  ))\n",
-      "print 'iv)Efficiency is %.2f percent'%(eta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Armature and field copper losses are 1184.96 W and 720 W respectively\n",
-        "ii)Brush contact copper loss is 184 W\n",
-        "iii)Core plus mechanical losses  =  4228.5 W\n",
-        "iv)Efficiency is 80.65 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.14 page no : 36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I = 5.  \t\t\t#no load current\n",
-      "V = 500.\n",
-      "R_sh = 250.\n",
-      "R_a = 0.5 \t\t\t#resistance of shunt field winding and armature\n",
-      "motor_input_NL  =  V*I\n",
-      "I_sh = V/R_sh\n",
-      "I_a = I-I_sh\n",
-      "arm_cu_loss_NL  =  R_a*I_a**2  \t\t\t#no load armature copper loss\n",
-      "constant_loss  =  motor_input_NL - arm_cu_loss_NL\n",
-      "I_FL = 50.\n",
-      "I_a_FL  =  I_FL - I_sh   \t\t\t#currents at full load\n",
-      "arm_cu_loss_FL  =   R_a*I_a_FL**2 \t\t\t#full load armature copper loss\n",
-      "\n",
-      "# Calculations and Results\n",
-      "total_loss =  constant_loss + arm_cu_loss_FL \n",
-      "motor_input = V*I_FL\n",
-      "motor_output_FL =  motor_input - total_loss\n",
-      "print 'Required output power is %.3f kW'%(motor_output_FL/1000)\n",
-      "eta = 100*(motor_output_FL/motor_input) \t\t\t#full load efficiency\n",
-      "print 'Full load efficiency of motor with 50A current is %.2f percent '%(eta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required output power is 21.352 kW\n",
-        "Full load efficiency of motor with 50A current is 85.41 percent \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.15 page no : 36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 250.\n",
-      "R_sh = 275.\n",
-      "R_a = 0.8 \t\t\t#resistance of shunt field and amature\n",
-      "I_L0 = 3.91 \t\t\t#load current\n",
-      "I_sh = V/R_sh\n",
-      "I_a0 =  I_L0 - I_sh\n",
-      "constant_losses =  V*I_L0 -R_a*(I_a0)**2\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#as a generator\n",
-      "P_out = 10*10**3\n",
-      "I_L = P_out/V\n",
-      "I_a  =  I_L + I_sh \n",
-      "field_cu_loss = R_sh*(I_sh)**2  \t\t\t#field copper loss\n",
-      "arm_cu_loss =  R_a*(I_a)**2  \t\t\t#armature copper loss  \n",
-      "eta_gen  =  100 *(P_out/(P_out+constant_losses + field_cu_loss+ arm_cu_loss))  \t\t\t#efficiency as generator\n",
-      "print 'Efficiency as a generator  =  %.2f percent'%(eta_gen)\n",
-      "\n",
-      "#as a motor\n",
-      "P_in = 10*10**3  \t\t\t#at V = 250 \n",
-      "I_L = P_in/V\n",
-      "I_a = I_L - I_sh\n",
-      "field_cu_loss = R_sh*(I_sh)**2  \t\t\t#field copper loss\n",
-      "arm_cu_loss =  R_a*(I_a)**2  \t\t\t#armature copper loss  \n",
-      "eta_m  =  100 *((P_in-(constant_losses + field_cu_loss+ arm_cu_loss))/(P_in)) \t\t\t#efficiency as motor\n",
-      "print 'Efficiency as a motor  =  %.2f percent'%(eta_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency as a generator  =  79.77 percent\n",
-        "Efficiency as a motor  =  75.80 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.16 page no : 38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I = 4. \t\t\t#no load current in amperes\n",
-      "V = 500.\n",
-      "motor_input_no_load = I*V   \t\t\t#no load motor input\n",
-      "R_a = 0.5\n",
-      "R_sh = 250.\t\t\t#resistance of armature and shunt field resistnace\n",
-      "I_sh = V/R_sh\n",
-      "\n",
-      "I_a = I-I_sh\n",
-      "arm_cu_loss_noload = R_a*I_a**2  \t\t\t#No-load armature copper losses\n",
-      "constant_loss = motor_input_no_load - arm_cu_loss_noload\n",
-      "I_FL = 40.\n",
-      "I_aFL = I_FL- I_a  \t\t\t#full load currents\n",
-      "arm_cu_loss_fulload = R_a*I_aFL**2 \t\t\t#Full-load armature copper losses\n",
-      "Total_loss = arm_cu_loss_fulload + constant_loss\n",
-      "\n",
-      "# Calculations and Results\n",
-      "motor_input = V*I_FL\n",
-      "motor_output_fullload = motor_input - Total_loss\n",
-      "print 'Output power at full-load is %.0f W'%(motor_output_fullload) \n",
-      "efficiency =  motor_output_fullload*100/motor_input \t\t\t#motor efficiency\n",
-      "print 'Efficiency at full-load is %.1f percent'%(efficiency)\n",
-      "\n",
-      "E_bNL = V-I_a*R_a\n",
-      "E_bFL = V-I_FL*R_a\n",
-      "\n",
-      "#E_b  = N*phi\n",
-      "#E_bNL/E_bFL = N_NL/N_FL\n",
-      "#applying rules of componendo and dividendo\n",
-      "#change_in _speed = (N_NL - N_FL)/N_FL = (E_bNL - E_bFL)/E_bFl\n",
-      "\n",
-      "change_in_speed = 100*(E_bNL - E_bFL)/E_bFL\n",
-      "print 'percentage change in speed from no load to full load is %.3f percent'%(change_in_speed)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output power at full-load is 17280 W\n",
-        "Efficiency at full-load is 86.4 percent\n",
-        "percentage change in speed from no load to full load is 3.958 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.17 page no : 39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 500.\n",
-      "R_a = 0.22\n",
-      "R_sh =  250.\t\t\t#armature resistance and shunt field resistance\n",
-      "I = 5.       \t\t\t#no load current\n",
-      "motor_input_NL = V*I   \t\t\t#no load motor input\n",
-      "I_sh = V/R_sh\n",
-      "I_a_NL =  I- I_sh \t\t\t#no load armature current\n",
-      "arm_cu_loss_NL  =  R_a*I_a_NL**2 \t\t\t#no load armature copper loss\n",
-      "constant_loss  =  motor_input_NL - arm_cu_loss_NL \n",
-      "\n",
-      "# Calculations and Results\n",
-      "#When motor draws 100 A current \n",
-      "I = 100\n",
-      "I_a  =  I - I_sh\n",
-      "arm_cu_loss   = R_a* I_a**2  \t\t\t#armature copper loss\n",
-      "total_losses  =    arm_cu_loss +  constant_loss\n",
-      "motor_input  =  V*I\n",
-      "motor_output  =  motor_input -total_losses \n",
-      "eta_m = 100*(motor_output/motor_input) \t\t\t#motor efficiency\n",
-      "print 'i)Efficiency of motor at 100 A current is %.2f percent '%(eta_m)\n",
-      "\n",
-      "#part(b)\n",
-      "E_b_NL =  V- I_a_NL*R_a \t\t\t#back emf at no load\n",
-      "E_b =  V- I_a*R_a \t\t\t#back emf at 100 A \n",
-      "#E_b is proportional to N.. and using componendo dividendo\n",
-      "speed_change =  100*((E_b_NL - E_b)/E_b)\n",
-      "print 'ii)Percentage change in speed  =  %.3f percent'%(speed_change)\n",
-      "\n",
-      "\n",
-      "print 'Note that the following were assumptions made'\n",
-      "print 'i) Due to heating  resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected'\n",
-      "print 'ii) Though the motor speed is changing from no load to given load %( the mechanical losses are assumed to be cosnstant'\n",
-      "print 'iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Efficiency of motor at 100 A current is 90.78 percent \n",
-        "ii)Percentage change in speed  =  4.368 percent\n",
-        "Note that the following were assumptions made\n",
-        "i) Due to heating  resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected\n",
-        "ii) Though the motor speed is changing from no load to given load %( the mechanical losses are assumed to be cosnstant\n",
-        "iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.18 page no : 41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 200.\n",
-      "I_L = 40.\n",
-      "\n",
-      "R_sh = 200.\n",
-      "R_a = 0.2 \t\t\t#shunt field winding and armature resistance\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#case(a) :  As a generator\n",
-      "P_out_g = V*I_L \t\t\t#output poewr as generator\n",
-      "I_sh = V/R_sh\n",
-      "I_a  =  I_L + I_sh\n",
-      "E  =  V + I_a*R_a\n",
-      "P_a_g = E*I_a  \t\t\t#power developed in armature\n",
-      "P_cu_g =  R_a*I_a**2 + R_sh*I_sh**2 \t\t\t#copper loss as genrator\n",
-      "print 'i)Output as generator is %.f kW'%(P_out_g/1000)\n",
-      "\n",
-      "#case(b) :  As a motor\n",
-      "P_in_m = V*I_L \t\t\t#input power as motor\n",
-      "I_sh = V/R_sh\n",
-      "I_a  =  I_L - I_sh\n",
-      "E_b  =  V - I_a*R_a\n",
-      "P_a_m = E_b*I_a  \t\t\t#power developed in armature\n",
-      "P_cu_m =  R_a*I_a**2 + R_sh*I_sh**2\t\t\t#copper loss as motor\n",
-      "print 'ii)Input as motor is %.0f kW'%( P_in_m/1000)\n",
-      "print 'iii)Power developed in Armature:%.4f kW for generator%.4f kW for motor'%(P_a_g/1000,P_a_m/1000)\n",
-      "print 'iv)Copper losses:%.1f W for generator%.1f W for motor'%(P_cu_g,P_cu_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Output as generator is 8 kW\n",
-        "ii)Input as motor is 8 kW\n",
-        "iii)Power developed in Armature:8.5362 kW for generator7.4958 kW for motor\n",
-        "iv)Copper losses:536.2 W for generator504.2 W for motor\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.19 page no : 42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 219.\n",
-      "I = 10.\n",
-      "dN = 1030. - 970 \t\t\t#given\n",
-      "t_1 = 36.  \t\t\t#time with no excitation\n",
-      "t_2 = 9.\t\t\t# time with full excitation and armature supporting an extra load of 10 A at 219 V\n",
-      "t_3 = 15. \t\t\t#time with full exciation\n",
-      "W_dash  =  V*I  \t\t\t#additioanl loss when armature is suddenly connected to loads\n",
-      "W_s =  W_dash *(t_2/(t_3-t_2)) \t\t\t#total stray losses\n",
-      "N = 1000.  \t\t\t#speed in rpm\n",
-      "\n",
-      "# Calculations\n",
-      "#using W_s  =  (2*pi/60)**2 * I *N *dN / t_3 where W_s is stray losses\n",
-      "I =  W_s*(t_3/dN)*(30/math.pi)**2/N   \t\t\t#moment of inertia\n",
-      "W_m = W_s*(t_3/t_1) \t\t\t#mechanical losses\n",
-      "iron_losses =  W_s - W_m\n",
-      "\n",
-      "# Results\n",
-      "print 'i)The moment of inertia of armature is %.2f kg-m**2'%(I)\n",
-      "print 'ii)Iron loss =  %.2f W'%(iron_losses)\n",
-      "print 'iii)Mechanical losses at 1000 rpm mean speed is %.2f W'%(W_m)\n",
-      "\n",
-      "print 'Noteworthy points:1)When armature is slowing down and there is no excitation,then kinetic energy is used to overcome mechanical losses only.Iron losses are absent as excitation is absent2)When excitation is given%( kinetic energy is used to overcome both mechanical as well as iron losses.Total called stray losses.3)If moment of inertia is in kg-m**2,then loss of energy is in watts'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)The moment of inertia of armature is 74.89 kg-m**2\n",
-        "ii)Iron loss =  1916.25 W\n",
-        "iii)Mechanical losses at 1000 rpm mean speed is 1368.75 W\n",
-        "Noteworthy points:1)When armature is slowing down and there is no excitation,then kinetic energy is used to overcome mechanical losses only.Iron losses are absent as excitation is absent2)When excitation is given%( kinetic energy is used to overcome both mechanical as well as iron losses.Total called stray losses.3)If moment of inertia is in kg-m**2,then loss of energy is in watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.20 page no : 43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 225.\n",
-      "I = 10.\n",
-      "dN = 1030. - 970 \t\t\t#given\n",
-      "t_1 = 40.  \t\t\t#time with no excitation\n",
-      "t_2 = 9.\t\t\t# time with full excitation and armature supporting an extra load of 10 A at 219 V\n",
-      "t_3 = 20. \t\t\t#time with full exciation\n",
-      "\n",
-      "# Calculations\n",
-      "W_dash  =  V*I \t\t\t#additional loss\n",
-      "W_s =  W_dash *(t_2/(t_3-t_2)) \t\t\t#total stray losses\n",
-      "N = 1000\t\t\t#Speed in rpm\n",
-      "\t\t\t#using W_s  =  (2*pi/60)**2 * I *N *dN / t_3 where W_s is stray losses\n",
-      "I =  W_s*(t_3/dN)*(30/math.pi)**2/N   \t\t\t#moment of inertia\n",
-      "W_m = W_s*(t_3/t_1)   \t\t\t#mechanical losses\n",
-      "iron_losses = W_s - W_m\n",
-      "\n",
-      "# Results\n",
-      "print 'i)The moment of inertia of armature is %.2f kg-m**2'%(I)\n",
-      "print 'ii)Iron loss =  %.2f W'%(iron_losses)\n",
-      "print 'iii)Mechanical losses at 1000 rpm mean speed is %.2f W'%(W_m)\n",
-      "\n",
-      "print 'Noteworthy points:1)When there is no excitation and armature is slowed down , its K.E. is used to overcome mechanical mechanical losses only math.since there will be no iron loss as there is no flux.2)When there is excitation provided then K.E. is used to supply mechanical as well as iron losses together called stray losses '\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)The moment of inertia of armature is 55.96 kg-m**2\n",
-        "ii)Iron loss =  920.45 W\n",
-        "iii)Mechanical losses at 1000 rpm mean speed is 920.45 W\n",
-        "Noteworthy points:1)When there is no excitation and armature is slowed down , its K.E. is used to overcome mechanical mechanical losses only math.since there will be no iron loss as there is no flux.2)When there is excitation provided then K.E. is used to supply mechanical as well as iron losses together called stray losses \n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.21 page no : 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_avg  =  (220.+190)/2  \t\t\t#average voltage across load\n",
-      "I_avg = 12.\n",
-      "R_a = 0.5\n",
-      "R_sh = 250.\n",
-      "W_dash = V_avg*I_avg  \t\t\t#power absorbed \n",
-      "t_1 = 30.\n",
-      "t_2 = 5.\n",
-      "W = W_dash*(t_2/(t_1-t_2))\n",
-      "V = 250.\n",
-      "I = 22.  \t\t\t#input current\n",
-      "\n",
-      "# Calculations\n",
-      "I_sh  =  V/R_sh\n",
-      "I_a =  I - I_sh\n",
-      "arm_cu_loss  =  R_a*I_a**2 \t\t\t#armature copper loss\n",
-      "shunt_field_cu_loss  =   V*I_sh  \t\t\t#shunt field copper loss\n",
-      "total_losses =  shunt_field_cu_loss + arm_cu_loss + W\n",
-      "\n",
-      "machine_input  =  V*I\n",
-      "machine_output  =  machine_input - total_losses \n",
-      "eta_m = 100*(machine_output /machine_input )  \t\t\t#efficiency when running as motor\n",
-      "\n",
-      "# Results\n",
-      "print 'Efficiency of machine when opeating as motor taking current of 22A on 250V supply is %.1f percent'%(eta_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of machine when opeating as motor taking current of 22A on 250V supply is 82.5 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.22 page no : 45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I_avg = 10.\n",
-      "V_avg = (220.+190)/2 \t\t\t#average voltage across load\n",
-      "W_dash  =  V_avg*I_avg  \t\t\t#power  absorbed\n",
-      "t_1 = 30.\n",
-      "t_2 = 20.\n",
-      "\n",
-      "# Calculations\n",
-      "W = W_dash * (t_2/(t_1-t_2)) \t\t\t#stray losses\n",
-      "\n",
-      "# Results\n",
-      "print 'Stray losses of motor is %.1f kW'%(W/1000)\n",
-      "print 'Answers mismatch because V_average is 205 volts but it is taken as 220 volts in Power absorbed calculation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stray losses of motor is 4.1 kW\n",
-        "Answers mismatch because V_average is 205 volts but it is taken as 220 volts in Power absorbed calculation\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.23 page no : 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I_a_g = 330.\n",
-      "I_a_m = 380.\n",
-      "R_a = 0.02 \t\t\t#armature resistance\n",
-      "V = 250.\n",
-      "I = 50.\n",
-      "\n",
-      "\n",
-      "arm_cu_loss_g =  R_a*I_a_g**2\t\t\t#armature copper loss for generator\n",
-      "arm_cu_loss_m =  R_a*I_a_m**2\t\t\t#armature copper loss for motor\n",
-      "power_drawn = V*I\n",
-      "stray_losses  =  power_drawn - (arm_cu_loss_m + arm_cu_loss_g)\n",
-      "stray_losses_each  =  stray_losses/2 \t\t\t#stray losses for each machine\n",
-      "\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#for motor\n",
-      "I_sh_m = 4.2   \t\t\t#Shunt current in case of motor\n",
-      "field_cu_loss_m = V*I_sh_m  \t\t\t#field copper loss in case of motor\n",
-      "total_loss  =  field_cu_loss_m + stray_losses_each  +  arm_cu_loss_m\n",
-      "motor_input =  V*(I_a_m+I_sh_m) \n",
-      "motor_output  =  motor_input - total_loss\n",
-      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
-      "\n",
-      "#for generator\n",
-      "I_sh_g = 5  \t\t\t#Shunt current in case of generator\n",
-      "field_cu_loss_g = V*I_sh_g \t\t\t#field copper loss in case of generator\n",
-      "total_loss  =  field_cu_loss_g + stray_losses_each  +  arm_cu_loss_g\n",
-      "generator_output  =  V*I_a_g\n",
-      "generator_input =  generator_output + total_loss\n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 92.0302 percent\n",
-        "Efficiency of generator is 92.0297 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.24 page no : 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "R_a = 0.02 \t\t\t#armature resistance\n",
-      "V = 250. \t\t\t#line voltage\n",
-      "I = 50. \t\t\t#current taken from supply \n",
-      "\n",
-      "# Calculations and Results\n",
-      "#for generator\n",
-      "I_a_g = 330.\n",
-      "I_sh_g = 5. \t\t\t#armature current and current through shunt field \n",
-      "arm_cu_loss_g  =  R_a*I_a_g**2\t\t\t#armature copper loss for generator\n",
-      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
-      "\n",
-      "#for motor\n",
-      "I_a_m = 380\n",
-      "I_sh_m = 4.2 \t\t\t#armature current and current through shunt field\n",
-      "arm_cu_loss_m  =  R_a*I_a_m**2\t\t\t#armature copper loss for motor\n",
-      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
-      "power_drawn = V*I\n",
-      "IFW_losses  =  power_drawn - (arm_cu_loss_g + arm_cu_loss_m)  \t\t\t#Iron friction and windage losses\n",
-      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
-      "\n",
-      "#for generator\n",
-      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each\n",
-      "generator_output = V*I_a_g\n",
-      "generator_input  =  generator_output + total_loss_g \n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n",
-      "\n",
-      "#for motor\n",
-      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m\n",
-      "motor_input = V*(I_a_m+I_sh_m)\n",
-      "motor_output  =  motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of generator is 92.0297 percent\n",
-        "Efficiency of motor is 92.0302 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.25 page no : 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "I = 40.\n",
-      "I_a_g = 160. \n",
-      "I_a_m  = 200. \t\t\t#armature currents for generator and motor\n",
-      "I_sh_g = 7. \n",
-      "I_sh_m  = 6. \t\t\t#current through shunt field for generator and motor\n",
-      "R_a = 0.015 \t\t\t#armature resistance\n",
-      "arm_cu_loss_g  =  R_a*I_a_g**2  \t\t\t#armature copper loss for motor\n",
-      "arm_cu_loss_m  =  R_a*I_a_m**2  \t\t\t#armature copper loss for motor\n",
-      "power_drawn = V*I\n",
-      "IFW_losses  =  power_drawn - (arm_cu_loss_g + arm_cu_loss_m)  \t\t\t#Iron friction and windage losses\n",
-      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#for motor\n",
-      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
-      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m  \t\t\t#total losses in motor\n",
-      "motor_input = V * I_a_m\n",
-      "motor_output =  motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)    \t\t\t#motor efficiency\n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
-      "\n",
-      "#for generator\n",
-      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
-      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each  \t\t\t#total losses in generator\n",
-      "generator_output = V*I_a_g\n",
-      "generator_input  =  generator_output + total_loss_g\n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency\n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 86.7545 percent\n",
-        "Efficiency of generator is 85.7867 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.26 page no : 51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "R_a = 0.2 \t\t\t#armature resistance \n",
-      "V = 240.\n",
-      "I = 16.\n",
-      "I_a_g = 60. \n",
-      "I_a_m = 71. \t\t\t#armature currents for generator and motor\n",
-      "I_sh_g = 3. \n",
-      "I_sh_m = 2.  \t\t\t#field current for generator and motor \n",
-      "\n",
-      "\n",
-      "#for generator\n",
-      "arm_cu_loss_g  =  R_a*I_a_g**2\t\t\t#armature copper loss for generator\n",
-      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
-      "\n",
-      "#for motor\n",
-      "arm_cu_loss_m  =  R_a*I_a_m**2\t\t\t#armature copper loss for motor\n",
-      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
-      "power_drawn = V*I\n",
-      "field_loss_total_g_m =  field_cu_loss_m + field_cu_loss_g\n",
-      "arm_cu_loss_total_g_m  =  arm_cu_loss_m + arm_cu_loss_g\n",
-      "IFW_losses  =  power_drawn - (arm_cu_loss_total_g_m + field_loss_total_g_m)  \t\t\t#Iron friction and windage losses\n",
-      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#for generator\n",
-      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each \t\t\t#total loss in generator\n",
-      "generator_output = V*I_a_g\n",
-      "generator_input  =  generator_output + total_loss_g\n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n",
-      "\n",
-      "#for motor\n",
-      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m \t\t\t#total loss in motor\n",
-      "motor_input = V*(I_a_m+I_sh_m)\n",
-      "motor_output  =  motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of generator is 88.3658 percent\n",
-        "Efficiency of motor is 88.9035 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.27 page no : 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "R_a = 0.015\n",
-      "V = 250. \t\t\t#line voltage\n",
-      "I = 45. \t\t\t#line current\n",
-      "I_a_m = 385.\n",
-      "I_sh_m = 4. \t\t\t#armature and field currents for motor\n",
-      "I_a_g = 340.\n",
-      "I_sh_g = 5. \t\t\t#armature and field currents for generator\n",
-      "arm_cu_loss_m =  R_a*I_a_m**2   \t\t\t#armature copper loss for motor\n",
-      "field_cu_loss_m =  V*I_sh_m    \t\t\t#field copper loss for motor\n",
-      "\n",
-      "arm_cu_loss_g =  R_a*I_a_g**2  \t\t\t#armature copper loss for generator\n",
-      "field_cu_loss_g =  V*I_sh_g   \t\t\t#field copper loss for motor\n",
-      "\n",
-      "total_cu_loss  =  field_cu_loss_g + arm_cu_loss_g  +  field_cu_loss_m + arm_cu_loss_m  \t\t\t#total copper loss for both machines\n",
-      "P_aux  =  V*I  \t\t\t#power taken from auxillary supply\n",
-      "stray_loss =  P_aux - total_cu_loss\n",
-      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
-      "\n",
-      "# Calculations and Results\n",
-      "total_loss_g  =  stray_loss_each + arm_cu_loss_g + field_cu_loss_g  \t\t\t#total losses in generator\n",
-      "generator_output = V* I_a_g\n",
-      "eta_g  =  100*(generator_output/(generator_output + total_loss_g))\t\t\t#generator efficiency\n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n",
-      "\n",
-      "total_loss_m  =  stray_loss_each + arm_cu_loss_m + field_cu_loss_m\t\t\t#total losses in motor\n",
-      "motor_input = V*(I_a_m+I_sh_m)\n",
-      "motor_output  =  motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of generator is 93.9171 percent\n",
-        "Efficiency of motor is 94.0929 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.28 page no : 53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 500.\n",
-      "P = 1000.*10**3\n",
-      "I = 30.\n",
-      "I_a_m  =  200. + 30 \n",
-      "I_a_g  = 200. \t\t\t#armature current for motor and generator\n",
-      "I_sh_m  =  1.8\n",
-      "I_sh_g  = 3.5 \t\t\t#field current for motor and generator\n",
-      "brush_drop = 230.\n",
-      "R_a = 0.075  \t\t\t#armature resimath.tance\n",
-      "\n",
-      "# Calculations\n",
-      "arm_cu_loss_m  =  R_a*I_a_m**2 + 2*brush_drop \t\t\t#motor armature copper loss\n",
-      "field_cu_loss_m  = V*I_sh_m  \t\t\t#  motor field copper loss \n",
-      "\n",
-      "arm_cu_loss_g  =  R_a*I_a_g**2 + 2*brush_drop  \t\t\t#generator armature copper loss\n",
-      "field_cu_loss_g  = V*I_sh_g  \t\t\t#field copper loss generator\n",
-      "\n",
-      "total_cu_loss  =  field_cu_loss_g + arm_cu_loss_g  +  field_cu_loss_m + arm_cu_loss_m  \t\t\t#total copper loss for both machines\n",
-      "P_aux  =  V*I  \t\t\t#power taken from auxillary supply\n",
-      "stray_loss =  P_aux - total_cu_loss  \n",
-      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
-      "\n",
-      "total_loss_g  =  stray_loss_each + arm_cu_loss_g + field_cu_loss_g  \t\t\t#total loss in generator\n",
-      "generator_output = V* I_a_g\n",
-      "eta_g  =  100*(generator_output/(generator_output + total_loss_g))\t\t\t#generator efficiency\n",
-      "\n",
-      "# Results\n",
-      "print 'Efficiency of generator is %.0f percent'%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of generator is 93 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.29 page no : 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 220.\n",
-      "I = 10.\n",
-      "R_a = 0.05 \t        \t\t#artmature resistance\n",
-      "I_a_m =  73.    \n",
-      "I_sh_m  =  2. \t    \t\t#armature and field current for motor\n",
-      "I_a_g = 67.5\n",
-      "I_sh_g = 2.5\t\t    \t#armature and field current for generator\n",
-      "\n",
-      "arm_cu_loss_m  =  R_a*I_a_m**2 \t\t\t#motor armature copper loss\n",
-      "field_cu_loss_m  = V*I_sh_m  \t\t\t#  motor field copper loss \n",
-      "\n",
-      "arm_cu_loss_g  =  R_a*I_a_g**2 \t\t\t#generator armature copper loss\n",
-      "field_cu_loss_g  = V*I_sh_g  \t\t\t#field copper loss generator\n",
-      "\n",
-      "total_cu_loss  =  field_cu_loss_g + arm_cu_loss_g  +  field_cu_loss_m + arm_cu_loss_m  \t\t\t#total copper loss for both machines\n",
-      "power_input  =  V*I\n",
-      "stray_loss =  power_input - total_cu_loss  \n",
-      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#motor efficiency\n",
-      "total_loss_m =  field_cu_loss_m + stray_loss_each  +  arm_cu_loss_m  \t\t\t#total motor losses\n",
-      "motor_input  =  V*(I_a_m + I_sh_m )\n",
-      "motor_output  = motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
-      "\n",
-      "#generator efficiency\n",
-      "total_loss_g =  field_cu_loss_g + stray_loss_each  +  arm_cu_loss_g  \t\t\t#total generator losses\n",
-      "generator_output  = V*I_a_g\n",
-      "generator_input  =  generator_output +  total_loss_g\n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 93.5496 percent\n",
-        "Efficiency of generator is 92.8956 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.30 page no : 57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 400.\n",
-      "I = 50.\n",
-      "I_a_g = 250. \n",
-      "I_a_m  = 300.  \t\t\t#armature current for generator and motor\n",
-      "I_sh_g = 2.5 \n",
-      "I_sh_m  = 2.4 \t\t\t#field current for generator and motor\n",
-      "R_a = 0.1 \t\t\t#armature resistance\n",
-      "\n",
-      "arm_cu_loss_g  =  R_a*I_a_g**2   \t\t\t#armature copper loss for generator\n",
-      "arm_cu_loss_m  =  R_a*I_a_m**2   \t\t\t#armature copper loss for motor\n",
-      "power_drawn = V*I\n",
-      "IFW_losses  =  power_drawn - (arm_cu_loss_g + arm_cu_loss_m)  \t\t\t#Iron friction and windage losses\n",
-      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#for motor\n",
-      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
-      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m\n",
-      "motor_input = V * I_a_m\n",
-      "motor_output =  motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)    \t\t\t#motor efficiency\n",
-      "print 'Efficiency of motor is %.2f percent'%(eta_m)\n",
-      "\n",
-      "#for generator\n",
-      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
-      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each\n",
-      "generator_output = V*I_a_g\n",
-      "generator_input  =  generator_output + total_loss_g\n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency\n",
-      "print 'Efficiency of generator is %.2f percent'%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 89.72 percent\n",
-        "Efficiency of generator is 91.22 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.31 page no : 58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I_1 = 56.  \t\t\t#motor input current\n",
-      "V = 590.   \t\t\t#voltage across armature\n",
-      "I_2 = 44.  \t\t\t# load current\n",
-      "V_2 = 400. \t\t\t# voltage across generator\n",
-      "V_field  =  40.  \t\t\t#voltage drop across field winding\n",
-      "R_a = 0.3\n",
-      "R_se = 0.7142 \t\t\t#armature and series field resistance for each machine\n",
-      "total_input = (V+V_field)*I_1\n",
-      "output = V_2*I_2\n",
-      "total_loss_g_m =  total_input - output   \t\t\t#total losses of 2 machines\n",
-      "R_se = V_field/I_1  \t\t\t#series field resistance for both windings\n",
-      "total_cu_loss  =  (R_a+ 2*R_se)*I_1**2  + R_a*I_2**2  \t\t\t#total copper loss\n",
-      "stray_loss =  total_loss_g_m - total_cu_loss\n",
-      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
-      "\n",
-      "# Calculations and Results\n",
-      "# for motor\n",
-      "motor_input  =  V*I_1\n",
-      "arm_cu_loss_m  =   (R_a+ R_se)*I_1**2    \t\t\t#armature coper losses of motor\n",
-      "total_loss_m =  arm_cu_loss_m + stray_loss_each\n",
-      "motor_output  =  motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
-      "\n",
-      "# for generator\n",
-      "arm_cu_loss_g  =  R_a*I_2**2  \t\t\t#armature coper losses of generator\n",
-      "series_field_cu_loss_g  =  V_field*I_1   \t\t\t#series field copper loss\n",
-      "total_loss_g =  arm_cu_loss_g + series_field_cu_loss_g + stray_loss_each\n",
-      "generator_output = V_2*I_2\n",
-      "generator_input  =  generator_output +  total_loss_g\n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 72.6998 percent\n",
-        "Efficiency of generator is 67.0221 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.32 page no : 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I_1 = 56.  \t\t\t#motor input current\n",
-      "V = 590.   \t\t\t#voltage across armature\n",
-      "I_2 = 44.  \t\t\t# load current\n",
-      "V_2 = 400. \t\t\t# voltage across generator\n",
-      "V_field  =  40.  \t\t\t#voltage drop across field winding\n",
-      "R_a = 0.3\n",
-      "R_se = 0.7142 \t\t\t#armature and series field resismath.tane for each machine\n",
-      "\n",
-      "total_input = (V+V_field)*I_1\n",
-      "output = V_2*I_2\n",
-      "total_loss_g_m =  total_input - output   \t\t\t#total losses of 2 machines\n",
-      "R_se = V_field/I_1  \t\t\t#series field resistance for both windings\n",
-      "total_cu_loss  =  (R_a+ 2*R_se)*I_1**2  + R_a*I_2**2  \t\t\t#total copper loss\n",
-      "stray_loss =  total_loss_g_m - total_cu_loss\n",
-      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
-      "\n",
-      "# Calculations and Results\n",
-      "# for motor\n",
-      "motor_input  =  V*I_1\n",
-      "arm_cu_loss_m  =   (R_a+ R_se)*I_1**2  \t\t\t#armature coper losses of motor\n",
-      "total_loss_m =  arm_cu_loss_m + stray_loss_each\n",
-      "motor_output  =  motor_input - total_loss_m\n",
-      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
-      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
-      "\n",
-      "# for generator\n",
-      "arm_cu_loss_g  =  R_a*I_2**2  \t\t\t#armature coper losses of generator\n",
-      "series_field_cu_loss_g  =  V_field*I_1  \t\t\t#series field copper loss\n",
-      "total_loss_g =  arm_cu_loss_g + series_field_cu_loss_g + stray_loss_each\n",
-      "generator_output = V_2*I_2\n",
-      "generator_input  =  generator_output +  total_loss_g\n",
-      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
-      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of motor is 72.6998 percent\n",
-        "Efficiency of generator is 67.0221 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines/ch4.ipynb b/DC_Machines_and_Synchronous_Machines/ch4.ipynb
deleted file mode 100755
index 7e4b687b..00000000
--- a/DC_Machines_and_Synchronous_Machines/ch4.ipynb
+++ /dev/null
@@ -1,1114 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e5bf1101b5ca7d663f8c7db6ba5353cd5daf17690d3f4785e774f0306b01e83e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4 : Synchronous Machines : Alternators"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1 page no : 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "Slots = 24.\n",
-      "Phase = 3. \t\t\t#number of phases\n",
-      "\n",
-      "# Calculations\n",
-      "n = Slots/Pole \t\t\t#slots per pole\n",
-      "m = Slots/Pole/Phase \t\t\t#slots per pole per phase\n",
-      "beeta = 180/n  \t\t\t#Slot angle\n",
-      "\n",
-      "# results\n",
-      "print \"Slot angle : %.f degrees\"%beeta"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Slot angle : 30 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2 page no : 26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Slots = 120.\n",
-      "Pole = 8.\n",
-      "Phase = 3.   \t\t\t#number of phases\n",
-      "\n",
-      "# Calculations\n",
-      "n = Slots/Pole         \t\t\t#Slots per Pole\n",
-      "m = Slots/Pole/Phase   \t\t\t#Slots per Pole per Phase\n",
-      "beeta = 180/n           \t\t\t#Slot angle in degree\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#Distribution Factor\n",
-      "\n",
-      "# Results\n",
-      "print 'Distribution Factor:K_d = %.3f'%(K_d)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Distribution Factor:K_d = 0.957\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3 page no : 26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Slots = 36.\n",
-      "Pole = 4.\n",
-      "Phase = 3. \t\t\t#number of phases\n",
-      "n = Slots/Pole    \t\t\t#Slots per pole\n",
-      "beeta = 180/n     \t\t\t#Slot angle in degrees\n",
-      "\n",
-      "# Calculations\n",
-      "#coil is shorted by 1 slot i.e. by beeta degrees to full pitch dismath.tance\n",
-      "alpha = beeta     \t\t\t#angle of short pitch\n",
-      "K_c = math.cos(math.radians(alpha/2))  \t\t\t#Coil span Factor\n",
-      "\n",
-      "# Results\n",
-      "print 'Coil Span Factor:K_c = %.4f'%(K_c)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Coil Span Factor:K_c = 0.9848\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4 page no : 26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "N_s = 250.  \t\t\t#Synchronous speed in r.p.m\n",
-      "f = 50.     \t\t\t#Frequency of generated e.m.f in hertz\n",
-      "Slots = 216.\n",
-      "phi = 30.*10**-3 \t\t\t#flux per pole in weber\n",
-      "\n",
-      "Pole = 120*f/N_s\n",
-      "n = Slots/Pole   \t\t\t#Slots per Pole\n",
-      "m = n/3          \t\t\t#Slots per Pole per Phase\n",
-      "beeta = 180/n     \t\t\t#Slot angle in degree\n",
-      "\n",
-      "# Calculations\n",
-      "K_d = math.sin(math.radians(m*beeta/2))/(m*math.sin(math.radians(beeta/2)))   \t\t\t#distribution factor\n",
-      "K_c = 1 \t\t\t#Coil Span Factor for full pitch coils = 1\n",
-      "\n",
-      "Z = Slots*5   \t\t\t#Z is total no of conductors\n",
-      "Z_ph = Z/3    \t\t\t#Conductors Per Phase\n",
-      "T_ph = Z_ph/2 \t\t\t#Turns per phase\n",
-      "E_ph = 4.44*K_c*K_d*f*phi*T_ph \t\t\t#induced emf\n",
-      "E_line = E_ph*math.sqrt(3)\n",
-      "\n",
-      "# Results\n",
-      "print 'Induced e.m.f across the Terminals is %.2f V'%(E_line)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Induced e.m.f across the Terminals is 1992.90 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5 page no : 27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 16.\n",
-      "N_s = 375. \t\t\t#synchronous speed in rpm\n",
-      "Slots = 144.\n",
-      "E_line = 2.657*10**3 \t\t\t#line value of emf across terminals\n",
-      "f = Pole*N_s/120 \t\t\t#frequency\n",
-      "\n",
-      "# Calculations\n",
-      "K_c = 1  \t\t\t#assuming full pitch winding Coil span Factor = 1\n",
-      "n = Slots/Pole \t\t\t#slots per pole\n",
-      "m = n/3 \t\t\t#slots per pole per phase\n",
-      "\n",
-      "beeta = 180/n\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#Distribution Fcator\n",
-      "conductors_per_slot = 10\n",
-      "Z = Slots*conductors_per_slot  \t\t\t#total conductors\n",
-      "\n",
-      "Z_ph = Z/3  \t\t\t#number of conductors per phase\n",
-      "T_ph = Z_ph/2 \t\t\t#no of turns per phase\n",
-      "E_ph = E_line/math.sqrt(3) \t\t\t#phase value of emf across terminals\n",
-      "\n",
-      "phi = E_ph/(4.44*K_c*K_d*f*T_ph)     \t\t\t#E_ph = 4.44*K_c*K_d*f*phi*T_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Frequency of Induced e.m.f is %.0fHz \\nFlux per Pole is %.0f mWb'%(f,phi*1000)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency of Induced e.m.f is 50Hz \n",
-        "Flux per Pole is 30 mWb\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6 page no : 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "d = 0.25            \t\t\t#Diameter in metre\n",
-      "l = 0.3             \t\t\t#Length in metre\n",
-      "Pole = 4.\n",
-      "A1 = math.pi*d*l/Pole    \t\t\t#Area of each fundamental pole\n",
-      "f = 50. \t\t\t#frequency in hertz\n",
-      "B_m1 = 0.15 \n",
-      "B_m3 = 0.03\n",
-      "B_m5 = 0.02 \t\t\t#Amplitude of 1st 3rd and 5th harmonics\n",
-      "phi_1 = (2/math.pi)*B_m1*A1  \t\t\t#average value of fundamental flux per pole in weber\n",
-      "\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#PART A\n",
-      "E_c1 = 1.11*2*f*phi_1  \t\t\t#R.M.S value of fundamental frequency e.m.f generated in math.single conductor\n",
-      "Coil_span = (13./15)*180  \t\t\t#math.since winding coil span is 13/15 of pole pitch\n",
-      "alpha = 180-Coil_span\n",
-      "\n",
-      "#Pitch factor for 1st 3rd and 5th harmonic\n",
-      "K_c1 = math.cos(math.radians(alpha/2))\n",
-      "K_c3 = math.cos(math.radians(3*alpha/2))\n",
-      "K_c5 = math.cos(math.radians(5*alpha/2))\n",
-      "\n",
-      "#using E_cx = E_c1 * (B_mx/B_m1)\n",
-      "E_c3 = E_c1 * (B_m3/B_m1)\n",
-      "E_c5 = E_c1 * (B_m5/B_m1)\n",
-      "\n",
-      "E_t1 = K_c1 * (2*E_c1)   \t\t\t#R.M.S Vaue of fundamental frequency EMF generated in 1 turn (in volts)\n",
-      "E_t3 = K_c3 * 2*E_c3\n",
-      "E_t5 = K_c5 * 2*E_c5\n",
-      "E_t = math.sqrt(E_t1**2 +E_t3**2 +E_t5**2)\n",
-      "V = 10*E_t  \t\t\t#(number of turns per coil )* (Total e.m.f per turn)\n",
-      "print 'Voltage generated per coil is %.1f V'%(V)\n",
-      "\n",
-      "# PART B\n",
-      "#E_1ph = 4.44*K_c1*K_d1*phi_1*f*T_ph\n",
-      "T_ph = 200.   \t\t\t#T_ph = (60 coils  * 10 turns per coil)/3\n",
-      "\n",
-      "Total_Conductors = 1200. \t\t\t# 60 coils * 10 turns per coil * 2\n",
-      "Conductors_per_Slot = 20. \t\t\t#2 conductors per turn * 10  turns per slot\n",
-      "Slots = Total_Conductors/Conductors_per_Slot\n",
-      "\n",
-      "n = Slots/Pole\n",
-      "m = n/3\n",
-      "beeta = 180/n   \t\t\t#Slot angle in degree\n",
-      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta/2)))\n",
-      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta/2)))\n",
-      "K_d5 = math.sin(math.radians(m*5*beeta/2)) /(m*math.sin(math.radians(5*beeta/2)))\n",
-      "\n",
-      "E_1ph = 4.44 * K_c1 * K_d1*phi_1 * f * T_ph\n",
-      "# using E_xph =  E_1ph* (B_mx*K_cx*K_dx)/(B_m1*K_c1*K_d1)\n",
-      "E_3ph =  E_1ph* (B_m3*K_c3*K_d3)/(B_m1*K_c1*K_d1)\n",
-      "E_5ph =  E_1ph* (B_m5*K_c5*K_d5)/(B_m1*K_c1*K_d1)\n",
-      "E_ph = math.sqrt( E_1ph**2 + E_3ph**2 + E_5ph**2 )  \t\t\t#voltage generated per phase\n",
-      "print 'Voltage generated per phase is %.f V'%(E_ph)\n",
-      "\n",
-      "#PART c\n",
-      "E_line = math.sqrt(3) * math.sqrt( E_1ph**2 + E_5ph**2  ) \t\t\t#terminal voltage\n",
-      "print 'Terminal Voltage is %.1f V '%(E_line)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage generated per coil is 12.4 V\n",
-        "Voltage generated per phase is 235 V\n",
-        "Terminal Voltage is 404.8 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7 page no : 38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Ns = 250. \t\t\t#Synchronous speed in rpm\n",
-      "f = 50.\n",
-      "Slots = 288.\n",
-      "E_line = 6600.\n",
-      "Pole = 120*f/Ns  \n",
-      "n = Slots/Pole  \t\t\t#slots per pole\n",
-      "m = n/3 \t\t\t#slots per pole per phase\n",
-      "beeta = 180/n \t\t\t#slot angle\n",
-      "conductors_per_slot = 32   \t\t\t#16 conductors per coil-side  *2 coil-sides per slot\n",
-      "\n",
-      "# Calculations\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
-      "alpha = 2*beeta\t\t\t# angle of short pitch\n",
-      "K_c = math.cos(math.radians(alpha/2))  \t\t\t#coil span factor\n",
-      "Z  =  Slots*conductors_per_slot  \t\t\t#total conductors\n",
-      "Z_ph = Z/3 \t\t\t#Conductors per phase\n",
-      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
-      "\n",
-      "E_ph = E_line/math.sqrt(3)\n",
-      "phi = E_ph/(4.44*K_c*K_d*f*T_ph)           \t\t\t#Because E_ph = 4.44 *K_c *K_d *phi *f *T_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Flux per pole is %.0f mWb '%(phi*1000)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flux per pole is 12 mWb \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.8 page no : 40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Ns = 1500.  \t\t\t#synchronous speed in rpm\n",
-      "Pole = 4.\n",
-      "Slots = 24.\n",
-      "conductor_per_slot = 8.\n",
-      "phi = 0.05 \t\t\t#flux per pole in weber\n",
-      "f = Pole*Ns/120 \t\t\t#frequenccy\n",
-      "n = Slots/Pole \t\t\t#slots per pole\n",
-      "m = n  \t\t\t# as number of phases is 1\n",
-      "beeta = 180/n  \t\t\t#slot angle\n",
-      "\n",
-      "# Calculations\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))  \t\t\t#distribution factor\n",
-      "\n",
-      "#Full pitch =  n  = 6 slots\n",
-      "#(1/6)th of full pitch  = 1slot\n",
-      "#angle of short pitch  =  1 slot angle\n",
-      "alpha = beeta\n",
-      "K_c = math.cos(math.radians(alpha/2))  \t\t\t#coil span factor\n",
-      "\n",
-      "Z = conductor_per_slot*Slots \t\t\t#total conductors\n",
-      "Z_ph = Z \t\t\t# as number of phases is 1\n",
-      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
-      "E_ph = 4.44*K_c*K_d* phi *f *T_ph  \t\t\t#induced emf\n",
-      "\n",
-      "# Results\n",
-      "print 'Induced e.m.f is %.1f V '%(E_ph)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Induced e.m.f is 662.8 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9 page no : 41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 48.\n",
-      "n = 9. \t\t\t#slots per pole\n",
-      "phi = 51.75*10**-3 \t\t\t#flux per pole in weber\n",
-      "Ns = 125.\n",
-      "f = Ns*Pole/120 \t\t\t#frequency\n",
-      "K_c = 1. \t\t\t#due to full pitch winding\n",
-      "m = n/3 \t\t\t#slots per pole per phase\n",
-      "beeta = 180/n  \t\t\t#slot angle\n",
-      "\n",
-      "# Calculations\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))  \t\t\t#distribution factor\n",
-      "conductor_per_slot = 4*2 \t\t\t#Each slot has 2 coil sides and each coil side has 4 conductors\n",
-      "Slots = n*Pole\n",
-      "Z = conductor_per_slot*Slots   \t\t\t#total number of conductors\n",
-      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
-      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
-      "E_ph = 4.44 *K_c *K_d *phi *f *T_ph  \t\t\t#induced emf\n",
-      "\n",
-      "E_line = (math.sqrt(3))*E_ph  \t\t\t#due to star connection\n",
-      "\n",
-      "# Results\n",
-      "print 'Induced e.m.f is %.0f kV '%(E_line/1000)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Induced e.m.f is 11 kV \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.10 page no : 42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Slots = 180.\n",
-      "Pole = 12.\n",
-      "Ns = 600. \t\t\t#Synchronous speen in rpm\n",
-      "f = Pole*Ns/120 \t\t\t#frequency\n",
-      "phi = 0.05 \t\t\t#flux per pole in weber\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "#Average EMF in a conductor = 2*f*phi\n",
-      "rms_value_1 = 1.11*2*f*phi \t\t\t#rms value of emf in a conductor\n",
-      "print 'i)r.m.s value of e.m.f in a conductor is %.2f V '%(rms_value_1)\n",
-      "\n",
-      "#part(ii)\n",
-      "#Average EMF in a turn = 4*f*phi\n",
-      "rms_value_2 = 1.11*4*f*phi\t\t\t#r.m.s value of e.m.f in a turn\n",
-      "print 'ii)r.m.s value of e.m.f in a turn is %.2f V '%(rms_value_2)\n",
-      "\n",
-      "#part(iii)\n",
-      "conductors_per_coilside = 10/2 \n",
-      "rms_value_3 = rms_value_2*conductors_per_coilside  \t\t\t#r.m.s value of e.m.f in a coil\n",
-      "print 'iii)r.m.s value of e.m.f in a coil is %.1f V '%(rms_value_3)\n",
-      "\n",
-      "#part(iv)\n",
-      "conductors_per_slot = 10\n",
-      "Z = conductors_per_slot * Slots  \t\t\t#total number of conductors\n",
-      "Z_ph = Z/3    \t\t\t#conductors per phase\n",
-      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
-      "n = Slots/Pole  \t\t\t#slots per pole\n",
-      "m = n/3        \t\t\t#slots per pole per phase \n",
-      "beeta = 180/n   \t\t\t#slot angle\n",
-      "\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))\n",
-      "K_c = 1  \t\t\t#distribution & coil-span factor\n",
-      "E_ph = rms_value_2*T_ph*K_d*K_c   \t\t\t#induced emf\n",
-      "print 'iv)per phase induced e.m.f is %.1f V '%(E_ph)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)r.m.s value of e.m.f in a conductor is 6.66 V \n",
-        "ii)r.m.s value of e.m.f in a turn is 13.32 V \n",
-        "iii)r.m.s value of e.m.f in a coil is 66.6 V \n",
-        "iv)per phase induced e.m.f is 3822.9 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.11 page no : 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 8.\n",
-      "f = 50. \t\t\t#frequency\n",
-      "phi = 60.*10**-3 \t\t\t#flux per pole in weber\n",
-      "Slots = 96. \n",
-      "n = Slots/Pole \t\t\t#slots per pole\n",
-      "beeta  =  180/n \t\t\t#slot angle \n",
-      "m = n/3 \t\t\t#slots per pole per phase\n",
-      "\n",
-      "# Calculations and Results\n",
-      "coil_pitch = 10*beeta \t\t\t#10 slots\n",
-      "alpha = 180-coil_pitch\n",
-      "K_c = math.cos(math.radians(alpha/2))   \t\t\t#coi;-span factor\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
-      "\n",
-      "conductors_per_slot = 4  \n",
-      "Z = Slots*conductors_per_slot \t\t\t#total conductors\n",
-      "Total_turns = Z/2  \n",
-      "T_ph = Total_turns/3  \t\t\t#turns per phase\n",
-      "\n",
-      "#part (i)\n",
-      "E_ph =  4.44 *K_c *K_d *phi *f *T_ph\n",
-      "print '\\The phase voltage is %.2f V '%(E_ph)\n",
-      "\n",
-      "#part(ii)\n",
-      "E_line = E_ph*math.sqrt(3)\n",
-      "print 'The Line Voltage is %.2f V '%(E_line)\n",
-      "\n",
-      "#part(iii)\n",
-      "I_ph = 650\n",
-      "I_l = I_ph  \t\t\t# Star Connection\n",
-      "kVA_rating = math.sqrt(3)*E_line*I_l\n",
-      "print 'kVA rating is %.1f kVA '%(kVA_rating/1000)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\\The phase voltage is 788.57 V \n",
-        "The Line Voltage is 1365.84 V \n",
-        "kVA rating is 1537.7 kVA \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.12 page no : 45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Ns = 600. \t\t\t#synchronous speed in rpm\n",
-      "Pole = 10.\n",
-      "l = 30./100     \t\t\t#divided by 100 for centimetre-metre conversion\n",
-      "Pole_pitch = 35./100  \t\t\t#numerically equal to pi*d/Pole\n",
-      "Phase = 3.\n",
-      "conductors_per_slot = 8.\n",
-      "A1 = Pole_pitch*l \t\t\t#Area of each fundamental pole\n",
-      "m = 3.   \t\t\t#Slot per Pole per Phase\n",
-      "n = Phase*m \t\t\t#slots per pole\n",
-      "beeta = 180/n  \t\t\t#slot angle\n",
-      "\n",
-      "B_m1 = 1.\n",
-      "B_m3 = 0.3\n",
-      "B_m5 = 0.2  \t\t\t#amplitude of 1st 3rd and 5th harmonic\n",
-      "phi_1 = (2/math.pi)*A1*B_m1  \t\t\t#average value of fundamental flux per pole \n",
-      "f = Ns*Pole/120 \t\t\t#frequency\n",
-      "\n",
-      "# Calculations\n",
-      "Coil_span = (8./9)*180\n",
-      "alpha = 180-Coil_span\n",
-      "#pitch factor for 1st 3rd and 5th harmonic\n",
-      "K_c1 = math.cos(math.radians(alpha/2))  \n",
-      "K_c3 = math.cos(math.radians(3*alpha/2))\n",
-      "K_c5 = math.cos(math.radians(5*alpha/2))\n",
-      "\n",
-      "# using K_dx = math.sin(m*x*beeta*(math.pi/180)/2) /(m*math.sin(x*beeta*(math.pi/180)/2))\n",
-      "#distribution factor for 1st 3rd and 5th harmonic\n",
-      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta/2)))\n",
-      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta/2)))\n",
-      "K_d5 = math.sin(math.radians(m*5*beeta/2)) /(m*math.sin(math.radians(5*beeta/2)))\n",
-      "\n",
-      "Slots = n*Pole\n",
-      "Total_conductors = conductors_per_slot * Slots\n",
-      "Total_turns = Total_conductors/2\n",
-      "T_ph = Total_turns/3  \t\t\t#turns per phase\n",
-      "\n",
-      "#EMF of 1st 3rd and 5th harmonic\n",
-      "E_1ph = 4.44 * K_c1 * K_d1*phi_1 * f * T_ph\n",
-      "E_3ph =  E_1ph* (B_m3*K_c3*K_d3)/(B_m1*K_c1*K_d1)\n",
-      "E_5ph =  E_1ph* (B_m5*K_c5*K_d5)/(B_m1*K_c1*K_d1)\n",
-      "\n",
-      "# Results\n",
-      "# using E_xph =  E_1ph* (B_mx*K_cx*K_dx)/(B_m1*K_c1*K_d1)\n",
-      "E_ph = math.sqrt( E_1ph**2 + E_3ph**2 + E_5ph**2 )\n",
-      "print 'Phase value of induced e.m.f is %.2f V '%(E_ph)\n",
-      "E_line = math.sqrt(3) * math.sqrt( E_1ph**2 + E_5ph**2  )\t\t\t#no 3rd harmonic appears in line value\n",
-      "print 'line value of induced e.m.f is %.2f V '%(E_line)\n",
-      "\n",
-      "print 'Answer mismatches due to approximation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Phase value of induced e.m.f is 1711.94 V \n",
-        "line value of induced e.m.f is 2916.65 V \n",
-        "Answer mismatches due to approximation\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.13 page no : 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 16.\n",
-      "phi = 0.03 \t\t\t#flux per pole\n",
-      "Ns = 375. \t\t\t#synchronous speed in rpm\n",
-      "\n",
-      "# Calculations and Results\n",
-      "f = Ns*Pole/120 \t\t\t#frequency\n",
-      "print 'frequency is %.0f Hz '%(f)\n",
-      "Slots = 144\n",
-      "n = Slots/Pole  \t\t\t#slots per pole\n",
-      "m = n/3 \t\t\t#slots per pole per phase\n",
-      "beeta = 180/n  \t\t\t#slot angle\n",
-      "K_c = 1 \t\t\t#assuming Full-Pitch coil\n",
-      "Conductors_per_slot = 10\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))  \t\t\t#distribution factor\n",
-      "\n",
-      "Total_conductors = Slots*Conductors_per_slot\n",
-      "Total_turns = Total_conductors/2\n",
-      "T_ph = Total_turns/3  \t\t\t#turns per phase\n",
-      "E_ph = 4.44* K_c* K_d*phi* f* T_ph\n",
-      "E_line = E_ph*math.sqrt(3)\n",
-      "print 'line voltage is %.2f V '%(E_line)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "frequency is 50 Hz \n",
-        "line voltage is 2657.20 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.14 page no : 48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Ns = 250. \t\t\t#Speed in rpm\n",
-      "f = 50. \t\t\t#frequency\n",
-      "I_l = 100.\n",
-      "Slots = 216.\n",
-      "Conductors_per_slot = 5\n",
-      "Pole = 120.*f/Ns\n",
-      "phi = 30.*10**-3\t\t\t#flux per pole in weber\n",
-      "Z = Slots*Conductors_per_slot  \t\t\t#Total Conductors\n",
-      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
-      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
-      "n = Slots/Pole \t\t\t#slots per pole\n",
-      "m = n/3 \t\t\t#slots per pole per phase\n",
-      "beeta = 180./n \t\t\t#Slot angle\n",
-      "\n",
-      "# Calculations\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
-      "\n",
-      "e_av = 2*f*phi     \t\t\t#Average Value of EMF in each conductor\n",
-      "E_c = 1.11*(2*f*phi)  \t\t\t#RMS value of EMF in each conductor\n",
-      "E = 2*E_c*K_d \t\t\t#RMS value of EMF in each turn\n",
-      "E_ph = T_ph*E \t\t\t#RMS value of EMF in each phase\n",
-      "E_line =  E_ph*math.sqrt(3)  \t\t\t#As Star Connected Alternator\n",
-      "\n",
-      "# Results\n",
-      "print 'RMS value of EMF in each phase  =  %.3f V'%(E_ph)\n",
-      "print 'RMS value of EMF line value   =  %.3f V'%(E_line)\n",
-      "kVA_rating = math.sqrt(3)*E_line*I_l\n",
-      "print 'kVA rating is %.3f kVA '%(kVA_rating/1000)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RMS value of EMF in each phase  =  1150.602 V\n",
-        "RMS value of EMF line value   =  1992.902 V\n",
-        "kVA rating is 345.181 kVA \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.15 page no : 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 10.\n",
-      "Slots = 90.\n",
-      "E_l = 11000.\n",
-      "f = 50.\n",
-      "phi = 0.15 \t\t\t#flux per pole in weber\n",
-      "n = Slots/Pole \t\t\t#slots per pole\n",
-      "m = n/3 \t\t\t#slots per pole per phase\n",
-      "beeta = 180/n \t\t\t#slot angle\n",
-      " \n",
-      "# Calculations \n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
-      "K_c = 1 \t\t\t#coil span factor\n",
-      "\n",
-      "E_ph = E_l/math.sqrt(3)\n",
-      "T_ph = ( E_ph/(4.44*K_c*K_d*phi*f)  )\n",
-      "#T_ph should necessarily be an integer\n",
-      "Z_ph = (T_ph)*2  \n",
-      "\n",
-      "# Results\n",
-      "print 'Required number of armature conductors is %d'%(Z_ph)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required number of armature conductors is 397\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.16 page no : 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 10.\n",
-      "Ns = 600. \t\t\t#speen in rpm\n",
-      "conductor_per_slot = 8.\n",
-      "n = 12. \t\t\t#slots per pole\n",
-      "Slots = Pole*n \n",
-      "m = n/3 \t\t\t#slots per pole per phase \n",
-      "beeta = 180/n \t\t\t#slot angle\n",
-      "alpha = 2*beeta  \t\t\t#short by 2 slots\n",
-      "\n",
-      "#flux per pole corresponding to 1st 3rd and 5th harmonic\n",
-      "phi_1 = 100*10**-3\n",
-      "phi_3 = (33./100)*phi_1\n",
-      "phi_5 = (20./100)*phi_1\n",
-      "\n",
-      "#coil span factor corresponding to 1st 3rd and 5th harmonic\n",
-      "K_c1 = math.cos(math.radians( alpha/2))\n",
-      "K_c3 = math.cos(math.radians( 3*alpha/2))\n",
-      "K_c5 = math.cos(math.radians( 5*alpha/2))\n",
-      "\n",
-      "# using K_dx = math.sin(m*x*beeta /2) /(m*math.sin(x*beeta /2))\n",
-      "#distribution factor corresponding to 1st 3rd and 5th harmonic\n",
-      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta /2)))\n",
-      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta /2)))\n",
-      "K_d5 = math.sin(math.radians(m*5*beeta/2)) /(m*math.sin(math.radians(5*beeta /2)))\n",
-      "\n",
-      "Z = conductor_per_slot*n*Pole    \t\t\t#Total Conductors\n",
-      "Zph = Z/3 \t\t\t#conductors per phase\n",
-      "T_ph = Zph/2 \t\t\t#turns per phase\n",
-      "\n",
-      "f = Ns*Pole/120\n",
-      "E_1ph = 4.44*K_c1*K_d1*phi_1*f*T_ph\n",
-      "E_3ph = 4.44*K_c3*K_d3*phi_3*f*T_ph\n",
-      "E_5ph = 4.44*K_c5*K_d5*phi_5*f*T_ph\n",
-      "\n",
-      "E_ph = math.sqrt( E_1ph**2 +  E_3ph**2 + E_5ph**2     )\n",
-      "\n",
-      "# Results\n",
-      "print 'Phase value of induced e.m.f is %.0f V '%(E_ph)\n",
-      "E_line = math.sqrt(3)*math.sqrt( E_1ph**2 +  E_5ph**2     )  \t\t\t#In a line value 3rd harmonic doesnt appear\n",
-      "print 'line value of induced e.m.f is %d V '%(E_line)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Phase value of induced e.m.f is 3330 V \n",
-        "line value of induced e.m.f is 5691 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.17 page no : 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 6.\n",
-      "Ns = 1000.   \t\t\t#speed in rpm\n",
-      "d = 28./100  \t\t\t#Divided by 100 to convert from centimeters to metres\n",
-      "l = 23./100  \t\t\t#Divided by 100 to convert from centimeters to metres\n",
-      "m = 4.   \t    \t\t#slots per pole per phase\n",
-      "B_m1 = 0.87  \t\t\t#amplitude of 1st harmonic component of flux density\n",
-      "B_m3 = 0.24  \t\t\t#amplitude of 3rd harmonic component of flux density\n",
-      "Conductors_per_slot = 8\n",
-      "f = Ns*Pole/120 \t\t\t#frequency\n",
-      "A1 = math.pi*d*l/Pole \t\t\t#area of each fundamental pole\n",
-      "phi_1 = (2/math.pi)*A1*B_m1 \t\t\t#flux per pole in weber\n",
-      "n = m*3 \t\t    \t#slots per pole\n",
-      "beeta = 180/n \t\t\t#slot angle\n",
-      "alpha = beeta  \t\t\t#because of 1 slot short\n",
-      "\n",
-      "# Calculations\n",
-      "K_c1 = math.cos(math.radians(alpha/2)) \t\t\t#coil span factor corresponding to 1st harmonic\n",
-      "K_c3 = math.cos(math.radians(3*alpha/2))\t\t\t#coil span factor corresponding to 3rd harmonic\n",
-      "# using K_dx = math.sin(m*x*beeta*(math.pi/180)/2) /(m*math.sin(x*beeta*(math.pi/180)/2))\n",
-      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta/2))) \t\t\t#distribution factor corresponding to 1st harmonic\n",
-      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta/2))) \t\t\t#distribution factor corresponding to 3rd harmonic\n",
-      "\n",
-      "Slots = n*Pole\n",
-      "Z = Slots*Conductors_per_slot \t\t\t#total number of conductors\n",
-      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
-      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
-      "\n",
-      "E_1ph = 4.44*K_c1*K_d1*phi_1*f*T_ph\n",
-      "E_3ph = E_1ph* (B_m3*K_c3*K_d3)/(B_m1*K_c1*K_d1)      \t\t\t#using E_xph = E_1ph* (B_mx*K_cx*K_dx)/(B_m1*K_c1*K_d1)\n",
-      "E_ph = math.sqrt( E_1ph**2 +  E_3ph**2     )\n",
-      "print 'r.m.s value of resultant voltage is %.1f V'%(E_ph)\n",
-      "E_line = math.sqrt(3)*E_1ph   \t\t\t#For line Value 3rd harmonic does not appear\n",
-      "print 'line voltage is %.3f V'%(E_line)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "r.m.s value of resultant voltage is 383.7 V\n",
-        "line voltage is 654.560 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.18 page no : 53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 125.\n",
-      "V_ph = V_L\n",
-      "VA = 600.*10**3\n",
-      "I_L = VA/(math.sqrt(3)*V_L)     \t\t\t# Because VA = math.sqrt(3)* V_L * I_L\n",
-      "I_ph = I_L/(math.sqrt(3))\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#After Reconnection\n",
-      "V_ph = 125\n",
-      "V_L = V_ph*math.sqrt(3)\n",
-      "print 'New rating in volts is %.3f V'%(V_L)\n",
-      "#Winding Impedances remain the same\n",
-      "I_ph = 1600\n",
-      "I_L = I_ph\n",
-      "\n",
-      "print 'New rating in amperes is %.0f A'%(I_L)\n",
-      "kVA = math.sqrt(3)*V_L*I_L*(10**-3)\n",
-      "print 'New rating in kVA is %.0f kVA'%(kVA)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New rating in volts is 216.506 V\n",
-        "New rating in amperes is 1600 A\n",
-        "New rating in kVA is 600 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.19 page no : 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Pole = 4.\n",
-      "f = 50. \t\t\t#frequency\n",
-      "phi = 0.12 \t\t\t#flux per pole in weber\n",
-      "m = 4.  \t\t\t# slot per pole per phase\n",
-      "conductor_per_slot = 4.\n",
-      "coilspan = 150.\n",
-      "Ns = 120*f/Pole \t\t\t#synchronous speed in rpm\n",
-      "n = m*3  \t    \t\t#Slots per pole\n",
-      "beeta = 180/n \t\t\t#slot angle\n",
-      "\n",
-      "# Calculations\n",
-      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t# distribution factor\n",
-      "alpha = 180-coilspan \t\t\t#angle of short pitch\n",
-      "K_c = math.cos((math.pi/180)*alpha/2) \t\t\t#coil span factor\n",
-      "Z = m*(n*Pole) \t\t\t# Also equal to (conductors/slots)*slots\n",
-      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
-      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
-      "E_ph = 4.44*K_c*K_d*phi*f*T_ph\n",
-      "E_line = math.sqrt(3)*E_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'e.m.f generated is %.2f Vphase, %.2f Vline)'%(E_ph,E_line)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e.m.f generated is 788.57 Vphase, 1365.84 Vline)\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines/ch5.ipynb b/DC_Machines_and_Synchronous_Machines/ch5.ipynb
deleted file mode 100755
index 4aca598c..00000000
--- a/DC_Machines_and_Synchronous_Machines/ch5.ipynb
+++ /dev/null
@@ -1,1612 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ae2b9384b1b13f79321d75082b020b6a54688771e761bc6446ad00de14bf068e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5 : Methods for Calculating Regulation of Alternator"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1 page no : 12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "P = 1000.*10**3 \t\t\t#load power\n",
-      "phi = math.acos(math.radians(0.8)) \t\t\t#power factor lagging angle\n",
-      "V_L = 11.*10**3 \t\t\t#rated terminal voltae\n",
-      "R_a = 0.4 \t\t\t#armature resistance per phase\n",
-      "X_s = 3.\t\t\t#synchronous reactance per phase\n",
-      "\n",
-      "# Calculations\n",
-      "I_L = P/(math.sqrt(3)*V_L*math.cos(math.radians(phi)))\n",
-      "I_aph = I_L \t\t\t#for star connected load\n",
-      "I_a = I_L\t\t\t#current through armature\n",
-      "V_ph = V_L/math.sqrt(3) \t\t\t#rated terminal volatge phase value\n",
-      "\n",
-      "E_ph =  math.sqrt(  (V_ph*math.cos(math.radians(phi))+I_a*R_a)**2+(V_ph*math.sin(math.radians(phi))+I_a*X_s)**2   )  \t\t\t#emf generated phase value\n",
-      "E_line = E_ph*math.sqrt(3) \t\t\t#line value of emf generated\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph  \t\t\t#pecentage regulation\n",
-      "\n",
-      "# Results\n",
-      "print 'Line value of e.m.f generated is %.2f kV  \\nRegulation is %.3f percent'%(E_line*10**-3,regulation)\n",
-      "\n",
-      "# note : book answer is wrong.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Line value of e.m.f generated is 11.05 kV  \n",
-        "Regulation is 0.428 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2 Page no : 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "VA = 1200.*10**3\n",
-      "V_L = 6600.\n",
-      "R_a = 0.25 \t\t\t#armature resistance per phase\n",
-      "X_s = 5.\t\t\t#synchronous reactance per phase\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\n",
-      "I_aph = I_L \t\t\t#for star connected load\n",
-      "I_a = I_L\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "\n",
-      "#Part(i)\n",
-      "phi1 = math.acos(0.8)\t\t\t#and lagging\n",
-      "E_ph1 =  math.sqrt(  (V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2   )\n",
-      "regulation = 100*(E_ph1-V_ph)/V_ph  \t\t\t#percentage regulation\n",
-      "print 'i)Regulation at 0.8 lagging pf is %.2f percent'%(regulation)\n",
-      "#Part(ii)\n",
-      "phi2 = math.acos(0.8)\t\t\t#and leading\n",
-      "E_ph2 =  math.sqrt(  (V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2   )\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph \t\t\t#percentage regulation\n",
-      "print 'ii)Regulation at 0.8 leading pf is %.f percent'%(regulation2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Regulation at 0.8 lagging pf is 9.33 percent\n",
-        "ii)Regulation at 0.8 leading pf is -7 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3 Page no : 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "#full load\n",
-      "V_L_FL = 1100.\n",
-      "V_ph_FL = V_L_FL/math.sqrt(3)\n",
-      "\n",
-      "# Calculations\n",
-      "#no load\n",
-      "V_L_NL = 1266\n",
-      "E_line = V_L_NL\n",
-      "E_ph = E_line/math.sqrt(3)\n",
-      "regulation = 100*(E_ph-V_ph_FL)/V_ph_FL\n",
-      "\n",
-      "# Results\n",
-      "print 'Regulation at full load is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation at full load is 15.09 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4 Page no : 23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 866.\n",
-      "VA = 100.*10**3\n",
-      "I_L = VA/(math.sqrt(3)*V_L)    \t\t\t#because VA = math.sqrt(3)*V_L*I_L\n",
-      "I_aph = I_L\t\t\t#full load and star connected alternator\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "\n",
-      "# Calculations\n",
-      "#Graph is plotted and V_oc_ph and I_asc_Ph is obtained for \n",
-      "#SCC for I_asc = 66.67 A\n",
-      "I_f = 2.4  # A\n",
-      "#OCC for I_f = 2.4 A\n",
-      "V_oc_ph = 240    # V\n",
-      "\n",
-      "#for measruemnt of impedance\n",
-      "V_oc_ph = 240 \t\t\t#for I_f = 2.4..From o.c.c graph\n",
-      "I_asc_ph = 66.67 \t\t\t#for I_f = 2.4...From s.c.c graph\n",
-      "Z_s = V_oc_ph/I_asc_ph\n",
-      "R_a = 0.15\n",
-      "X_s = math.sqrt( Z_s**2-R_a**2 )\n",
-      "\n",
-      "V_ph_FL = 500.\n",
-      "phi = math.acos(0.8)  \t\t\t#lagging pf\n",
-      "E_ph = math.sqrt((V_ph_FL*math.cos(phi)+I_aph*R_a)**2+(V_ph_FL*math.sin(phi)+I_aph*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Full-load regulation at 0.8 lagging pf is %.2f percent '%(regulation )\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Full-load regulation at 0.8 lagging pf is 35.57 percent \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5  Page no : 25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_OC_line = 230.\n",
-      "I_asc = 12.5  \t\t\t# when I_f = 0.38\n",
-      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
-      "Z_s = V_OC_ph/I_asc \n",
-      "\n",
-      "R_a = 1.8/2  \t\t\t#1.8 is between terminals..0.9 is per phase\n",
-      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
-      "\n",
-      "I_a = 10.\t\t\t# when regulation is needed\n",
-      "V_L = 230.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "phi1 = math.acos(0.8) \t\t\t#and lagging\n",
-      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
-      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
-      "print 'Regulation for 10 A at 0.8 lagging pf is %.2f percent'%(regulation1)\n",
-      "#Part(ii)\n",
-      "phi2 = math.acos(0.8) \t\t\t#and leading\n",
-      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print 'Regulation for 10 A at 0.8 leading pf is %.2f percent'%(regulation2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation for 10 A at 0.8 lagging pf is 64.47 percent\n",
-        "Regulation for 10 A at 0.8 leading pf is -11.01 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6  Page no :  31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "phi = math.acos(0.8)\n",
-      "VA = 1000.*10**3\n",
-      "V_L = 1905.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "R_a = 0.2\t\t\t#Armature reactance per phase\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "#Ampere-turn method\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\n",
-      "I_aph = I_L\n",
-      "V_dash = V_ph+I_aph*R_a*math.cos(phi)\t\t\t#V_dash is a dummy quantity and has no significance..it's used only for mapping correcponding current\n",
-      "F_o = 32 \t\t\t#F_o corresponds to voltage V_dash = 1148.5 from O.C.C graph\n",
-      "F_AR = 27.5 \t\t\t#Field current required to circulate full-load short circuit current of 303.07A.From SCC F_AR = 27.5\n",
-      "F_R  =  math.sqrt(    F_o**2 + F_AR**2-2*F_o*F_AR*math.cos(phi+math.pi/2) )\t\t\t#using Comath.sine rule\n",
-      "\n",
-      "# for F_R = 53.25 E_ph = 1490 V from O.C.C\n",
-      "E_ph = 1490.\n",
-      "regulation1 = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Regulation on full-load by ampere-turn method is %.2f percent'%(regulation1)\n",
-      "\n",
-      "#Part (ii)\n",
-      "#Synchronous Impedance method\n",
-      "\n",
-      "I_sc = I_L\n",
-      "I_aph2 = I_sc\n",
-      "I_f = 27.5\n",
-      "\n",
-      "V_OC_ph = 1060. \t\t\t#corresponding to I-f = 27.5 in the graph\n",
-      "Z_s = V_OC_ph/I_aph2\n",
-      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
-      "\n",
-      "E_ph2 =  math.sqrt( (V_ph*math.cos(phi)+I_aph2*R_a)**2+(V_ph*math.sin(phi)+I_aph2*X_s)**2 )   \t\t\t#from phasor diagram\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print 'Regulation on full-load by synchronous impedance method is %.2f percent'%(regulation2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation on full-load by ampere-turn method is 35.47 percent\n",
-        "Regulation on full-load by synchronous impedance method is 78.09 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7  Page no :  39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "#case(i)\n",
-      "V_L = 440.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "phi = math.acos(0.8)\n",
-      "\n",
-      "#armature resistance drop from the graph\n",
-      "#RS = 1.1 cm and scale  = 50 V/cm\n",
-      "arm_leak_resis =  1.1*50    \t\t\t#armature leakage resistance\n",
-      "\n",
-      "OB = V_ph*math.cos(phi)\n",
-      "AB = V_ph*math.sin(phi) + arm_leak_resis\n",
-      "E_1ph =  math.sqrt( OB**2+AB**2   )\n",
-      "\n",
-      "F_f1 = 6.1  \t\t\t#corresponding value from OCC\n",
-      "F_AR = 3.1*1\n",
-      "\n",
-      "# Calculations and Results\n",
-      "F_R =  math.sqrt( F_f1**2 + F_AR**2 -2*F_f1*F_AR*math.cos(math.radians(90+math.acos(math.radians(0.8)))))\n",
-      "E_ph = 328. \t\t\t#voltage corresponding to F_R = 8.33 A from OCC graph\n",
-      "regulation1 =  100*(E_ph - V_ph)/V_ph\n",
-      "print 'i)Regulation for 0.8 pf lagging is %.2f percent '%(regulation1)\n",
-      "\n",
-      "#case(ii)\n",
-      "\n",
-      "OC = V_ph*math.cos(phi)\n",
-      "BC = V_ph*math.sin(phi) - arm_leak_resis\n",
-      "E_1ph =  math.sqrt( OC**2+BC**2   )\n",
-      "\n",
-      "F_f1 = 6.1  \t\t\t#corresponding value from OCC\n",
-      "F_R =  math.sqrt(   F_f1**2 + F_AR**2 -2*F_f1*F_AR*math.cos(math.radians(90-math.acos(math.radians(0.8)))))\n",
-      "E_ph = 90 \t\t\t#volatge corresponding to F_R = 3.34 A from OCC graph\n",
-      "regulation2 =  100*(E_ph - V_ph)/V_ph\n",
-      "print 'ii)Regulation for 0.8 pf leading is %.2f percent '%(regulation2)\n",
-      "print 'The answer in part ii doesnt match with textbook because of calculation mistake done in last step in the textbook'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Regulation for 0.8 pf lagging is 29.12 percent \n",
-        "ii)Regulation for 0.8 pf leading is -64.57 percent \n",
-        "The answer in part ii doesnt match with textbook because of calculation mistake done in last step in the textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8 Page no : 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "P = 1200.*10**3\n",
-      "V_line = 12000.\n",
-      "R_a = 2.\n",
-      "X_s = 35.\t\t\t#armature resistance and synchronous reactance\n",
-      "phi = math.acos(0.8)\n",
-      "\n",
-      "# Calculations\n",
-      "I_L = P/(math.sqrt(3)*V_line*math.cos(phi))\n",
-      "I_a = I_L\n",
-      "V_ph = V_line/math.sqrt(3)\n",
-      "E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Regulation at 0.8 lag power factor is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation at 0.8 lag power factor is 26.66 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9 Page no : 45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 11000.  \n",
-      "V_ph =  V_L/math.sqrt(3)\n",
-      "VA = 1000.*1000\n",
-      "I_L = VA/(V_L*math.sqrt(3))\n",
-      "\n",
-      "V_OC_ph = 433/math.sqrt(3)\n",
-      "I_asc_ph = I_L\n",
-      "\n",
-      "Z_s = V_OC_ph /I_asc_ph  \t\t\t#ohms per phase\n",
-      "R_a = 0.45 \t\t\t#ohms per phase\n",
-      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "phi = math.acos(0.8) \t\t\t#lagging\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_L*R_a)**2 +(V_ph*math.sin(phi)+ I_L*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Voltage regulation at 0.8 pf lagging is %f percent'%(regulation)\n",
-      "\n",
-      "#part(ii)\n",
-      "phi = math.acos(0.8) \t\t\t#leading\n",
-      "E_ph2  =  math.sqrt((V_ph*math.cos(phi)+I_L*R_a)**2 +(V_ph*math.sin(phi)- I_L*X_s)**2)\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print 'Voltage regulation at 0.8 pf lagging is %f percent'%(regulation2)\n",
-      "print 'Answer mismatches due to improper approximation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage regulation at 0.8 pf lagging is 2.690067 percent\n",
-        "Voltage regulation at 0.8 pf lagging is -1.996182 percent\n",
-        "Answer mismatches due to improper approximation\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10 Page no :  46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "VA = 125.*10**3\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\n",
-      "I_aph = I_L\n",
-      "\n",
-      "# Calculations\n",
-      "I_f = 4.\n",
-      "I_asc = I_aph/2 \t\t\t#for half load.. refer to graph\n",
-      "V_OC_line = 140.\n",
-      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
-      "I_asc_ph = I_asc\n",
-      "Z_s =  V_OC_ph/I_asc_ph\n",
-      "R_a = 0.1\n",
-      "X_s = math.sqrt(Z_s**2-R_a**2) \t\t\t#armature resistance and synchronous reactance\n",
-      "\n",
-      "phi = math.acos(0.8)\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_asc*R_a)**2 +(V_ph*math.sin(phi)- I_asc*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Voltage regulation at 0.8 pf leading for half load is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage regulation at 0.8 pf leading for half load is -12.39 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.11  Page no :  48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import roots\n",
-      "\n",
-      "# Variables\n",
-      "V_OC_line = 575.\n",
-      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
-      "I_asc_line = 75.\n",
-      "I_asc_ph   = I_asc_line \n",
-      "I_aph = I_asc_ph\n",
-      "I_L = I_aph\n",
-      "\n",
-      "Z_s =  V_OC_ph/I_asc_ph\n",
-      "R_a = 2.16/2\n",
-      "X_s  =   math.sqrt(Z_s**2 - R_a**2)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#on full load\n",
-      "E_ph = 6100.\n",
-      "phi = math.acos(0.8) \t\t\t#leading\n",
-      "\n",
-      "#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)- I_a*X_s)**2)\n",
-      "p = [1, -256.68, -3.71*10**7]\n",
-      "ans = roots(p)\n",
-      "V_ph = ans[0] \t\t\t#second root is ignored as its -ve\n",
-      "V_L = V_ph*math.sqrt(3)\n",
-      "print 'Rated terminal voltage between the lines is %.3f V '%(V_L)\n",
-      "VA_rating = math.sqrt(3)*V_L*I_L\n",
-      "print 'kVA rating of the alternator is %.f kVA'%(VA_rating*10**-3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rated terminal voltage between the lines is 10774.515 V \n",
-        "kVA rating of the alternator is 1400 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12  Page no :  49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import roots\n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6600.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 1500.*10**3\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\n",
-      "I_aph = I_L\n",
-      "\n",
-      "# Calculations and Results\n",
-      "R_a = 0.5\n",
-      "X_s = 5\t\t\t#armature resistance and synchronous reactance\n",
-      "phi = math.acos(0.8)\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
-      "print 'Induced EMF per phase is %f V'%(E_ph)\n",
-      "\n",
-      "#full load \n",
-      "phi = math.acos(1)\n",
-      "#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)- I_a*X_s)**2)\n",
-      "p = [1, 131.215, -1.791*10**7]\n",
-      "ans = roots(p)\n",
-      "V_ph = ans[1] \t\t\t#first root is ignored as it is -ve\n",
-      "print 'Terminal voltage per phase is %f V'%(V_ph)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Induced EMF per phase is 4284.243828 V\n",
-        "Terminal voltage per phase is 4166.921808 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.13  Page no :  50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_ph = 2000.\n",
-      "R_a = 0.8\n",
-      "I_sc = 100.\n",
-      "V_OC = 500.\n",
-      "I_f = 2.5\n",
-      "Z_s = V_OC/I_sc\n",
-      "X_s =  math.sqrt(Z_s**2 - R_a**2 )\n",
-      "I_aFL = 100.\n",
-      "I_a = I_aFL\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "phi = math.acos(1)\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'i)Voltage regulation is %.2f percent'%(regulation)\n",
-      "\n",
-      "#part(ii)\n",
-      "phi2 = math.acos(0.8)\n",
-      "E_ph2  =  math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2 +(V_ph*math.sin(phi2)- I_a*X_s)**2)\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print 'ii)Voltage regulation is %.2f percent'%(regulation2)\n",
-      "\n",
-      "#part(iii)\n",
-      "phi3 = math.acos(0.71)\n",
-      "E_ph3  =  math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2 +(V_ph*math.sin(phi3)+ I_a*X_s)**2)\n",
-      "regulation3 = 100*(E_ph3-V_ph)/V_ph\n",
-      "print 'iii)Voltage regulation is %.2f percent'%(regulation3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Voltage regulation is 6.89 percent\n",
-        "ii)Voltage regulation is -8.88 percent\n",
-        "iii)Voltage regulation is 21.11 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14 Page no :  51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "VA = 1000.*1000\n",
-      "V_L = 4600. \n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\n",
-      "I_aph_FL = I_L\n",
-      "I_aph = I_aph_FL\n",
-      "I_sc = (150./100)* I_aph_FL\n",
-      "V_OC_line = 1744\n",
-      "V_OC_ph =  V_OC_line/math.sqrt(3)\n",
-      "\n",
-      "# Calculations\n",
-      "Z_s = V_OC_ph / I_sc\n",
-      "R_a = 1\n",
-      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
-      "\n",
-      "phi = math.acos(0.8 ) \t\t\t#lagging\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Voltage regulation at full load 0.8 pf is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage regulation at full load 0.8 pf is 19.89 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15 Page no :  52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "\n",
-      "#part(i)    Ampere turn method\n",
-      "# Variables\n",
-      "F_O = 37.5\n",
-      "F_AR = 20.\n",
-      "V_L = 6600.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#lagging\n",
-      "phi = math.acos(0.8) \n",
-      "F_R =  math.sqrt((F_O+F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
-      "#E_ph corresponding to F_R can be obtained by plotting open circuit characteristics\n",
-      "E_ph = 4350\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'i)By Ampere-turn method or MMF methodFull-load regulation at 0.8 lagging pf is %.2f percent'%(regulation)\n",
-      "#leading\n",
-      "phi = math.acos(0.8) \n",
-      "F_R =  math.sqrt((F_O-F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
-      "#E_ph corresponding to F_R can be obtained by plotting open circuit characteristics\n",
-      "E_ph = 3000\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Full-load regulation at 0.8 leading pf is %.2f percent'%(regulation)\n",
-      "\n",
-      "#EMF method\n",
-      "V_OC_ph = 100\n",
-      "V_ph = 100\n",
-      "I_sc =  100*(F_O/F_AR)   \t\t\t#times the rated value\n",
-      "Z_s = V_OC_ph/I_sc\n",
-      "F_O =  100\n",
-      "F_AR =  Z_s*100\n",
-      "\n",
-      "#lagging\n",
-      "phi = math.acos(0.8)\n",
-      "F_R =   math.sqrt((F_O+F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
-      "regulation = 100*(F_R-V_ph)/V_ph\n",
-      "print 'iiSynchronous impedance method or EMF method'\n",
-      "print 'Full-load regulation at 0.8 lagging pf is %.2f percent'%(regulation)\n",
-      "#leading\n",
-      "phi = math.acos(0.8)\n",
-      "F_R =   math.sqrt((F_O-F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
-      "regulation = 100*(F_R-V_ph)/V_ph\n",
-      "print 'Full-load regulation at 0.8 leading pf is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)By Ampere-turn method or MMF methodFull-load regulation at 0.8 lagging pf is 14.16 percent\n",
-        "Full-load regulation at 0.8 leading pf is -21.27 percent\n",
-        "iiSynchronous impedance method or EMF method\n",
-        "Full-load regulation at 0.8 lagging pf is 38.72 percent\n",
-        "Full-load regulation at 0.8 leading pf is -19.72 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.16 Page no :  56"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6000.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_ph_X_Lph  =  0.9*500  \t\t\t#leakage reactance drop in volts  =  0.9 cm * 500 V/cm\n",
-      "phi =  math.acos(0.8) \t\t\t#lagging\n",
-      "\n",
-      "# Calculations\n",
-      "E_1ph = math.sqrt( (V_ph*math.cos(phi))**2 + (V_ph*math.sin(phi)+I_ph_X_Lph)**2 )   \t\t\t#From triangle OAB\n",
-      "F_f1  =  26 \t\t\t#from OCC\n",
-      "F_AR =  2.9*5  \t\t\t#2.9cm * 5 A/cm\n",
-      "\n",
-      "F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )\n",
-      "\n",
-      "# Results\n",
-      "print 'Required field current is %.2f A'%(F_R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required field current is 36.59 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.17 Page no :  58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 40.*10**3\n",
-      "I_L = VA/(math.sqrt(3)*V_L) \n",
-      "I_aph = I_L\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I_aph_X_Lph  =  0.65*50  \t\t\t#leakage reactance drop in volts  =  2.4 cm * 500 V/cm\n",
-      "X_Lph =  I_aph_X_Lph/ I_aph\n",
-      "print 'Armature leakage reactance is %.3f ohms'%(X_Lph)\n",
-      "phi = math.acos(0.8)  \t\t\t#lagging\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi))**2 +(V_ph*math.sin(phi)+ I_aph_X_Lph)**2)\n",
-      "F_f1 = 15.6 \t\t\t#as obtained from OCC corresponding to this E_ph\n",
-      "\n",
-      "F_AR =  2.3*3  \t\t\t#2.3cm * 3 A/cm\n",
-      "print 'Armature reaction is %.1f '%(F_AR)\n",
-      "F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )  \t\t\t#math.comath.sine rule to Triangle OAB\n",
-      "E_ph = 267.5  \t\t\t#corresponding to F_R from open circiut characteristics\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Voltage regulation at 0.8 pf lagging is %.1f percent'%(regulation)\n",
-      "\n",
-      "\n",
-      "#Note:This answer doesnt match with textbook as it has been reciprocated in textbook\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature leakage reactance is 0.563 ohms\n",
-        "Armature reaction is 6.9 \n",
-        "Voltage regulation at 0.8 pf lagging is 15.8 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.18 Page no :  60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "VA = 10.*10**3\n",
-      "V_L = 11.*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_ph_X_Lph  =  2.4*500  \t\t\t#leakage reactance drop in volts  =  2.4 cm * 500 V/cm\n",
-      "\n",
-      "# Calculations\n",
-      "I_ph_R_aph  = VA/(math.sqrt(3)*V_L)\n",
-      "phi = math.acos(0.8)\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_ph_R_aph)**2 +(V_ph*math.sin(phi)+ I_ph_X_Lph)**2)\n",
-      "F_f1 = 109 \t\t\t#obtained from open circuit characteristics corresponding to calculated E_ph\n",
-      "F_AR =  2.8*10  \t\t\t#2.8cm * 10 A/cm\n",
-      "F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )  \t\t\t#math.comath.sine rule to Triangle OAB\n",
-      "E_ph = 7700.  \t\t\t#corresponding to F_R from open circiut characteristics\n",
-      "\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Voltage regulation at full-load 0.8 pf lagging is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage regulation at full-load 0.8 pf lagging is 21.24 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.19  Page no : 63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import roots\n",
-      "\n",
-      "# Variables\n",
-      "VA = 2000.*1000\n",
-      "V_L = 11000.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "R_a = 0.3\n",
-      "X_s = 5. \t\t\t#armature resistance and synchronous reactance\n",
-      "\n",
-      "# Calculations\n",
-      "#case (i)\n",
-      "phi = math.acos(0.8) \t\t\t#lagging\n",
-      "I_L = VA/(math.sqrt(3)*V_L)  \n",
-      "I_a = I_L\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2)\n",
-      "\n",
-      "#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
-      "#we get  V_ph**2 -579.4455 V_ph -44653301.91 = 0\n",
-      "p = [1, -579.4455, -44653301.91]\n",
-      "ans = roots(p)\n",
-      "V_ph = ans[0]    \t\t#second root is ignored as its -ve\n",
-      "print 'Terminal voltage is %.4f V'%(V_ph)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Terminal voltage is 6978.3131 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.20 Page no : 64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import roots\n",
-      "\n",
-      "# Variables\n",
-      "R_a = 0.6\n",
-      "X_s = 6. \t\t\t#armature resistance and synchronous reactance per phase\n",
-      "E_L = 6599.\n",
-      "E_ph = E_L/math.sqrt(3)\n",
-      "I_L = 180.\n",
-      "I_a = I_L\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "# using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2) and solving for V_ph\n",
-      "p = [1 ,1135.83, -13338836.49]\n",
-      "ans = roots(p)\n",
-      "V_ph = ans[1]\n",
-      "V_L = V_ph*math.sqrt(3)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "phi = math.acos(0.9)\n",
-      "theta = math.atan((I_a*X_s+V_ph*math.sin(phi) )/(E_ph))\n",
-      "delta = theta-phi\n",
-      "print 'i)0.9 laggingTerminal voltage is %.2f VVoltage regulation is %.2f percentLoad angle is %.2f degrees'%(V_ph*math.sqrt(3),regulation,delta*180/math.pi)\n",
-      "\n",
-      "#part(ii)\n",
-      "phi_2 = math.acos(0.8)\n",
-      "p = [1, -941.53, -11399574.87]\n",
-      "ans = roots(p) \n",
-      "V_ph = ans[0] \t\t\t#second root is ignored as its -ve\n",
-      "V_L = V_ph*math.sqrt(3)\n",
-      "regulation2 = 100*(E_ph-V_ph)/V_ph\n",
-      "delta_2  =  math.asin( (math.tan(phi)*(V_ph*math.cos(phi_2)+I_a*R_a) -I_a*X_s )/E_ph   )\n",
-      "print 'ii)0.8 leadingTerminal voltage is %.2f VVoltage regulation is %.2f percentLoad angle is %.2f degrees'%(V_L,regulation2,delta_2*180/math.pi)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)0.9 laggingTerminal voltage is 5418.22 VVoltage regulation is 21.79 percentLoad angle is 6.83 degrees\n",
-        "ii)0.8 leadingTerminal voltage is 6719.93 VVoltage regulation is -1.80 percentLoad angle is 7.17 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.21 Page no :  66"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_ph = 2000.\n",
-      "R_a = 0.8\n",
-      "I_sc = 100.\n",
-      "I_a = I_sc\n",
-      "V_OC = 500.\n",
-      "I_f = 2.5\n",
-      "Z_s = V_OC/I_sc\n",
-      "\n",
-      "X_s = math.sqrt(Z_s**2- R_a**2)\n",
-      "I_a_FL = 100.\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "phi1 = math.acos(1) \t\t\t#and lagging\n",
-      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
-      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
-      "print 'Regulation at upf is %.2f percent'%(regulation1)\n",
-      "\n",
-      "#Part(ii)\n",
-      "phi2 = math.acos(0.8) \n",
-      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print 'Regulation at 0.8 leading pf is %.2f percent'%(regulation2)\n",
-      "\n",
-      "#Part(iii)\n",
-      "phi3 = math.acos(0.71) \n",
-      "E_ph3 = math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2+(V_ph*math.sin(phi3)+I_a*X_s)**2)\n",
-      "regulation3 = 100*(E_ph3-V_ph)/V_ph\n",
-      "print 'Regulation at 0.71 lagging pf is %.2f percent'%(regulation3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation at upf is 6.89 percent\n",
-        "Regulation at 0.8 leading pf is -8.88 percent\n",
-        "Regulation at 0.71 lagging pf is 21.11 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.22 Page no : 68"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V = 600.\n",
-      "VA = 60.*10**3\n",
-      "I_sc = 210.\n",
-      "V_oc = 480.\n",
-      "I_f = 10.\n",
-      "R_a = 0.2\n",
-      "\n",
-      "I = VA/V        \t\t\t#VA = V*I and alternator is math.single phase\n",
-      "I_a = I\n",
-      "\n",
-      "# Calculations and Results\n",
-      "Z_s = V_oc/I_sc  \t\t\t#Synchronous Impedance\n",
-      "X_s = math.sqrt(Z_s**2-R_a**2) \t\t\t#SYnchronous reactance\n",
-      "print 'Synchronous impedances is %f ohms and synchronous reactance is %f ohms'%(Z_s,X_s)\n",
-      "\n",
-      "#PART (i)\n",
-      "phi1 = math.acos(0.8) \t\t\t#and lagging\n",
-      "E1 = math.sqrt((V*math.cos(phi1)+I_a*R_a)**2+(V*math.sin(phi1)+I_a*X_s)**2) \t\t\t#plus sign for lagging power factor\n",
-      "regulation1 = 100*(E1-V)/V\n",
-      "print 'Regulation at 0.8 lagging pf is %.2f percent '%(regulation1 )\n",
-      "\n",
-      "#PART (ii)\n",
-      "phi2 = math.acos(1) \n",
-      "E2 = math.sqrt((V*math.cos(phi2)+I_a*R_a)**2+(V*math.sin(phi2)+I_a*X_s)**2) \n",
-      "regulation2 = 100*(E2-V)/V\n",
-      "print 'Regulation at UNITY pf is %.2f percent '%(regulation2 )\n",
-      "\n",
-      "#PART (iii)\n",
-      "phi3 = math.acos(0.6) \t\t\t#and leading \n",
-      "E3 = math.sqrt((V*math.cos(phi3)+I_a*R_a)**2+(V*math.sin(phi3)-I_a*X_s)**2) \t\t\t#minus sign for leading power factor\n",
-      "regulation3 = 100*(E3-V)/V\n",
-      "print 'Regulation at 0.6 leading pf is %.2f percent '%(regulation3 )\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronous impedances is 2.285714 ohms and synchronous reactance is 2.276947 ohms\n",
-        "Regulation at 0.8 lagging pf is 28.60 percent \n",
-        "Regulation at UNITY pf is 10.08 percent \n",
-        "Regulation at 0.6 leading pf is -23.98 percent \n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.23 Page no : 69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3300.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_a = 100.\n",
-      "I_f = 5.\n",
-      "V_OC_line = 900. \n",
-      "V_OC_ph = V_OC_line/math.sqrt(3) \n",
-      "R_a = 0.8 \t\t\t#armature resistance\n",
-      "I_aph = I_a\n",
-      "Z_s = V_OC_ph/I_aph\n",
-      "X_s = math.sqrt(Z_s**2-R_a**2) \t\t\t#synchronous reactance\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "phi1 = math.acos(0.8) \t\t\t#and lagging\n",
-      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
-      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
-      "print 'Regulation at 0.8 lagging is %.2f percent'%(regulation1)\n",
-      "\n",
-      "#Part(ii)\n",
-      "phi2 = math.acos(0.8)  \t\t\t#and leading\n",
-      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print 'Regulation at 0.8 leading pf is %.2f percent'%(regulation2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation at 0.8 lagging is 21.03 percent\n",
-        "Regulation at 0.8 leading pf is -9.55 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.24  Page no : 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 13500.\n",
-      "R_a = 1.5 \n",
-      "X_s = 30. \t\t\t#armature resistance and synchronous reactance\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#CASE 1\n",
-      "phi1 = math.acos(0.8)\n",
-      "P_out = 1280*10**3\n",
-      "I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi1) )    \t\t\t#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)\n",
-      "\n",
-      "I_a = I_L\n",
-      "E_ph = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Regulation at 0.8 lagging power factor is %.2f percent'%(regulation)\n",
-      "\n",
-      "#Case 2\n",
-      "phi2 = math.acos(1)\n",
-      "I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi2) )    \t\t\t#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)\n",
-      "\n",
-      "I_a = I_L\n",
-      "E_ph = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)+I_a*X_s)**2)\n",
-      "regulation2 = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Regulation at unity power factor is %.2f percent'%(regulation2)\n",
-      "\n",
-      "#case 3\n",
-      "phi3 = math.acos(0.8)\n",
-      "I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi3) )    \t\t\t#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)\n",
-      "I_a = I_L\n",
-      "E_ph = math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2+(V_ph*math.sin(phi3)-I_a*X_s)**2)\t\t\t# minus sign in the second bracket beacuse of leading pf\n",
-      "regulation3 = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Regulation at 0.8 leading power factor is %.2f percent'%(regulation3)\n",
-      "\n",
-      "# note : rounding off error"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation at 0.8 lagging power factor is 18.60 percent\n",
-        "Regulation at unity power factor is 3.23 percent\n",
-        "Regulation at 0.8 leading power factor is -11.99 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.26  Page no :  72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 11.*10**3\n",
-      "VA_rating = 10.**6\n",
-      "R_a = 2.2 \t\t\t#alternator resistance\n",
-      "phi = math.acos(0.8)\n",
-      "\n",
-      "# Calculations\n",
-      "I_L = VA_rating/(math.sqrt(3)*V_L)\t\t\t#VA = math.sqrt(3)V_L*I_L\n",
-      "I_a = I_L\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "regulation = 24.\n",
-      "\n",
-      "E_ph =  ((regulation/100)+1)*V_ph    \t\t\t# because regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "#using E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)\n",
-      "X_s = (math.sqrt(E_ph**2-((V_ph*math.cos(phi)+I_a*R_a)**2))-V_ph*math.sin(phi))*(1/I_a)\n",
-      "\n",
-      "phi1 = math.acos(0.8)\n",
-      "E_ph = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)-I_a*X_s)**2)\n",
-      "regulation1 = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Regulation at 0.8 leading power factor is %.2f percent'%(regulation1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation at 0.8 leading power factor is -13.90 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.27  Page no :  73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 220.\n",
-      "VA = 100.*10**3\n",
-      "R_a = 0.1 \t\t\t#effective resistacne of alternator\n",
-      "X_a = 0.5 \t\t\t#leakage reactance\n",
-      "X_ar = 2*X_a\n",
-      "\n",
-      "Z_s = complex(R_a,X_a+X_ar)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(1)\n",
-      "phi = math.acos(0.4)\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\t\t\t#VA = math.sqrt(3)*V_L*I_L\n",
-      "I_a = I_L\n",
-      "E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*(X_a+X_ar))**2)\n",
-      "print 'i)Required noload voltage is %.3f V'%(E_ph)\n",
-      "\n",
-      "#Part(2)\n",
-      "V_ph2 = 0\n",
-      "E_ph2 = math.sqrt((V_ph2*math.cos(phi)+I_a*R_a)**2+(V_ph2*math.sin(phi)+I_a*(X_a+X_ar))**2)\n",
-      "print 'ii)Required noload voltage is %.3f V'%(E_ph2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Required noload voltage is 515.848 V\n",
-        "ii)Required noload voltage is 394.522 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.28 Page no : 74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 2000.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 1000.*10**3\n",
-      "I_L = VA/(math.sqrt(3)*V_L)    \t\t\t#because VA = math.sqrt(3)*V_L*I_L\n",
-      "I_aph = I_L\n",
-      "\n",
-      "I_f = 28.5\t\t\t#for this I_aph = 288.67513 as obtained from SCC graph\n",
-      "V_oc_ph = 1060.\t\t\t#for I_f = 28.5 as obtained fromOCC graph\n",
-      "Z_s = V_oc_ph/I_aph\n",
-      "R_a = 0.2 \t\t\t#armature effective resistance\n",
-      "X_s = math.sqrt( Z_s**2-R_a**2 )\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Part(i)\n",
-      "phi1 = math.acos(0.8)\t\t\t#lagging\n",
-      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_aph*R_a)**2+(V_ph*math.sin(phi1)+I_aph*X_s)**2)\n",
-      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
-      "print \"i)Full-load percentage regulation at 0.8 pf lagging is %.2f percent\"%(regulation1)\n",
-      "\n",
-      "#Part(ii)\n",
-      "phi2 = math.acos(0.8)\t\t\t#leading\n",
-      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_aph*R_a)**2+(V_ph*math.sin(phi2)-I_aph*X_s)**2)\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print \"ii)Full-load percentage regulation at 0.8 pf leading is %.2f percent\"%(regulation2)\n",
-      "print 'Note that the answer mismatches because of calculation mistake done in the last step of part 1'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Full-load percentage regulation at 0.8 pf lagging is 73.86 percent\n",
-        "ii)Full-load percentage regulation at 0.8 pf leading is -9.29 percent\n",
-        "Note that the answer mismatches because of calculation mistake done in the last step of part 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.29 Page no : 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3300.\n",
-      "VA = 200.*10**3\n",
-      "R_a = 0.6\n",
-      "X_s = 6.\t\t\t#armature resistance and synchronous reactance\n",
-      "\n",
-      "# Calculations\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\t\t\t#VA = math.sqrt(3)V_L*I_L\n",
-      "I_a = I_L\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "phi = math.acos(0.8)\n",
-      "E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)\n",
-      "\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Regulation at 0.8 lagging power factor is %.3f percent'%(regulation)\n",
-      "print ' Note : Regulation is positive for lagging power factor loads'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Regulation at 0.8 lagging power factor is 7.802 percent\n",
-        " Note : Regulation is positive for lagging power factor loads\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.30 Page no : 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 2300.   \n",
-      "V_ph =  V_L/math.sqrt(3)\n",
-      "f = 50.\n",
-      "R_a = 0.2 \t\t\t#armature resistance\n",
-      "I_sc = 150.\n",
-      "V_OC_line = 780.   \n",
-      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
-      "\n",
-      "Z_s =  V_OC_ph/I_sc\n",
-      "X_s  =   math.sqrt(Z_s**2 - R_a**2)\n",
-      "I_aph = 25 \n",
-      "I_aFL = I_aph\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "phi =  math.acos(0.8)  \t\t\t#lag\n",
-      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
-      "regulation = 100*(E_ph-V_ph)/V_ph\n",
-      "print 'Voltage regulation at 0.8 pf lagging is %.3f percent'%(regulation)\n",
-      "\n",
-      "#part(ii)\n",
-      "phi2 =  math.acos(0.8)  \t\t\t#lead\n",
-      "E_ph2  =  math.sqrt((V_ph*math.cos(phi2)+I_aph*R_a)**2 +(V_ph*math.sin(phi2)- I_aph*X_s)**2 )\n",
-      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
-      "print 'Voltage regulation at 0.8 pf leading is %.3f percent'%(regulation2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage regulation at 0.8 pf lagging is 3.774 percent\n",
-        "Voltage regulation at 0.8 pf leading is -2.967 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines/ch6.ipynb b/DC_Machines_and_Synchronous_Machines/ch6.ipynb
deleted file mode 100755
index b2228585..00000000
--- a/DC_Machines_and_Synchronous_Machines/ch6.ipynb
+++ /dev/null
@@ -1,2427 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9f548e729a327e12aefb50cf84c5cae777df427713c1104c34d610a547e935f3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6 : Synchronization and Parallel Operation of Alternators"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " \n",
-      "Example 6.2  Page no : 31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "X_d = 0.7 \n",
-      "X_q = 0.4  \t\t\t#direct and quadrature axis synchronous reactance p.u.\n",
-      "R_a = 0\n",
-      "phi = math.acos(0.8)  \t\t\t#Lag\n",
-      "\n",
-      "V_t = 1. \t\t\t#assumed rated terminal Voltage \n",
-      "I_a = 1. \t\t\t#Full-load armature current\n",
-      "\n",
-      "# Calculations\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
-      "delta = psi-phi\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(psi)\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "\n",
-      "# Results\n",
-      "print 'Total e.m.f induced on open circuit is %.4f p.u.'%(E_f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total e.m.f induced on open circuit is 1.5149 p.u.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3 Page no :  35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "        return  RRRR*cmath.exp((math.pi*Theeeta/180.)*1j);\n",
-      "\n",
-      "# Variables\n",
-      "Z1 = complex(0,3) \t\t\t#impedance of alternator 1\n",
-      "Z2 = complex(0,4) \t\t\t#impedance of alternator 2\n",
-      "Z = 6.       #load\n",
-      "\n",
-      "E1 = p2z(220,0) \t\t\t#induced emf vector on no load\n",
-      "E2 = p2z(220,10)\t\t\t#induced emf vector on no load\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I1 = ((E1-E2)*Z+E1*Z2)/(Z*(Z1+Z2)+Z1*Z2)\n",
-      "I2 = ((E2-E1)*Z+E2*Z1)/(Z*(Z1+Z2)+Z1*Z2)\n",
-      "\n",
-      "phi1 = math.degrees(math.atan(I1.imag/I1.real))  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
-      "phi2 = math.degrees(math.atan(I2.imag/I2.real))  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
-      "\n",
-      "I = I1+I2\n",
-      "V = I*Z \t\t\t#Terminal voltage\n",
-      "print 'i) Terminal voltage is %.1f volts at %.2f degrees'%(abs(V),math.degrees(math.atan(V.imag/V.real)))\n",
-      "print 'ii) Currents are %.2f A at %.2f degrees and %.2f A at %.2f degrees      \\\n",
-      "\\nTotal current is %.2f A at %.2f degrees  '%(abs(I1),math.degrees(math.atan(I1.imag/I1.real)),\\\n",
-      "                                              abs(I2),math.degrees(math.atan(I2.imag/I2.real)),\\\n",
-      "                                              abs(I),math.degrees(math.atan(I.imag/I.real)))\n",
-      "\n",
-      "P1 = abs(V)*abs(I1)*math.cos(math.radians(phi1))\n",
-      "P2 = abs(V)*abs(I2)*math.cos(math.radians(phi2))\n",
-      "print 'iii)Power delivered is %.2f watts and %.2f watts'%(P1,P2)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Terminal voltage is 210.7 volts at -11.66 degrees\n",
-        "ii) Currents are 14.91 A at -17.71 degrees and 20.36 A at -7.24 degrees      \n",
-        "Total current is 35.12 A at -11.66 degrees  \n",
-        "iii)Power delivered is 2992.46 watts and 4257.11 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4  Page no :  54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_l = 10000.\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "VA = 10.*10**6\n",
-      "I_FL = VA/(V_l*math.sqrt(3)) \t\t\t#Current at full laod\n",
-      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
-      "\n",
-      "# Calculations\n",
-      "X_s = IX_s/I_FL\n",
-      "N_s = 1500.\n",
-      "f = 50.\n",
-      "P = 120*f/N_s \t\t\t#poles\n",
-      "\n",
-      "delta_dash_mech = math.pi/180 \t\t\t#phase print lacement in degree mechanical\n",
-      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
-      "E = V_ph \t\t\t#math.since alternator is on no-load\n",
-      "P_SY = delta_dash_elec*E**2/X_s  \t\t\t#Synchronous Power\n",
-      "P_SY_3ph = P_SY*3 \t\t\t#For 3 phases\n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising Power of armature is %.3f kW.\\\n",
-      "\\nSynchronising Power for 3 phase is %.3f kW'%(P_SY*10**-3,P_SY_3ph*10**-3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising Power of armature is 581.776 kW.\n",
-        "Synchronising Power for 3 phase is 1745.329 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5 Page no :  55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6.6*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 3.*10**6\n",
-      "I_FL = VA/(V_L*math.sqrt(3)) \t\t\t#full load current \n",
-      "P = 8.\n",
-      "f = 50. \t\t\t#poles and frequency\n",
-      "\n",
-      "X_s = complex(0,2.9)\t\t\t#X_s = 2.9\n",
-      "delta_dash_mech = math.pi/180\n",
-      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "E = V_ph\n",
-      "P_SY = delta_dash_elec*E**2/abs(X_s)  \t\t\t#Synchronous Power per phase\n",
-      "P_SY_3ph = P_SY*3 \t\t\t#For 3 phases\n",
-      "print 'i)   Synchronising power at no load is %.3f kW'%(P_SY*10**-3)\n",
-      "print '   Total Synchronising power at no load is %.2f kW'%(P_SY_3ph*10**-3)\n",
-      "\n",
-      "N_s = 120*f/P \t\t\t#in rpm\n",
-      "n_s = (N_s)/60 \t\t\t#in rps\n",
-      "T_SY = P_SY_3ph/(2*math.pi*n_s)\n",
-      "print 'Synchronous torque per mechanical degree of phase print lacement is %.2f * 10**3 N-m'%(T_SY*10**-3)\n",
-      "\n",
-      "#part(ii)\n",
-      "phi = math.acos(math.radians(0.85))\n",
-      "I = p2z(I_FL,0)\n",
-      "V = p2z(V_ph,phi)\n",
-      "\n",
-      "E = V+I*X_s\n",
-      "#E leads I by phasemag(E). V leads I by phasemag(V)\n",
-      "\n",
-      "delta = (math.pi/180)* (math.atan(E.imag/E.real)-math.atan(V.imag/V.real) ) \t\t\t#power angle in radians\n",
-      "P_SY2 = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)\n",
-      "\n",
-      "P_SY_total_2 = 3*P_SY2\n",
-      "#n_s = T_SY/(P_SY/(2*math.pi) )       \t\t\t#because T_SY = P_SY/(2*math.pi*n_s)\n",
-      "print 'ii)Total Synchronising power is %.0f kW'%(P_SY_total_2*10**-3)\n",
-      "\n",
-      "T_SY2 = P_SY_total_2/(2*math.pi*n_s)\n",
-      "print 'Synchronising torque is %.2f * 10**3 N-m'%(T_SY2/1000)\n",
-      "\n",
-      "# note : book answer is wrong.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   Synchronising power at no load is 349.547 kW\n",
-        "   Total Synchronising power at no load is 1048.64 kW\n",
-        "Synchronous torque per mechanical degree of phase print lacement is 13.35 * 10**3 N-m\n",
-        "ii)Total Synchronising power is 1074 kW\n",
-        "Synchronising torque is 13.68 * 10**3 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6  Page no :  57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.)\n",
-      "\n",
-      "# Variables\n",
-      "V_l = 10.*10**3\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "R_a = 0.4\n",
-      "Z = complex(R_a,6)\n",
-      "I_a = p2z(300,-math.acos(math.radians(0.8)))\n",
-      "E = V_ph+I_a*Z\n",
-      "\n",
-      "# Calculations\n",
-      "phi = math.acos(0.8)\n",
-      "alternator_op_ph = V_ph*abs(I_a)*math.cos(phi) \t\t\t#Power delivered to infinite bus per phase\n",
-      "\n",
-      "#Power deliered to the altrernator  =  Power delivewred to bus bar +  I**2*R losses in armature\n",
-      "alternator_power =  alternator_op_ph+ abs(I_a)**2*R_a\n",
-      "\n",
-      "#this power developed remains constant.change pf to 1 and calculate corresponding armature current\n",
-      "#alternator_power = V_ph*I_a1*math.cos(phi1)+I_a1**2*0.4\n",
-      "#solve the quadratic equation  0.4 I_a1**2+5773.50 I_a1- 1421640  = 0\n",
-      "I_a1 = (-1*V_ph+math.sqrt(V_ph**2-4*R_a*-1*alternator_power))/(2*R_a)\n",
-      "\n",
-      "#also as follows \n",
-      "E1 = V_ph+I_a1*Z\n",
-      "decrease = 100*(abs(E)-abs(E1))/abs(E)\n",
-      "\n",
-      "# Results\n",
-      "print 'Percentage decrease in induced e.m.f is %.1f percent'%(decrease)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage decrease in induced e.m.f is 2.6 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7 Page no :  58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#Line PQ for Altermnator 1 and PR for alternaator 2.AB is at frequency x from P where total load is 3000 kW\n",
-      "\n",
-      "# Variables\n",
-      "QC = 2000.\n",
-      "PS = 2.5\n",
-      "#PC = x\n",
-      "TR = 2000.\n",
-      "PT = 2.\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#using similarity of triangles PAC and PQS\n",
-      "AC_by_PC = (QC/PS)\t\t\t# because (AC/QC) = (PC/PS)\n",
-      "#using similarity of triangles PCB and PTR\n",
-      "CB_by_PC = (TR/PT) \t\t\t# because (CB/TR) = (PC/PT)\n",
-      "\n",
-      "AC_by_x = AC_by_PC   \t\t\t#which implies AC = 12.5*x\n",
-      "CB_by_x = CB_by_PC  \t\t\t#which implies CB = 16.67*x\n",
-      "\n",
-      "AC_plus_CB = 3000.  \t\t\t#total load at the frequency at P is 30 kW\n",
-      "x =  AC_plus_CB/(AC_by_x + CB_by_x)\n",
-      "AC = AC_by_x * x\n",
-      "CB = CB_by_x * x  \n",
-      "frequency = 50-x\n",
-      "print 'Loads shared by alternator 1 and 2 are %.2f kW and %.2f kW respectively'%(AC,CB)\n",
-      "\n",
-      "#construction for max load: RT is extended to cut PQ at X.\n",
-      "QS = 2000.\n",
-      "RT = 2000. \t\t\t#see figure\n",
-      "XT = QS*(PT/PS)\n",
-      "RX = RT+XT  \t\t\t#maximum load\n",
-      "\n",
-      "print 'Maximum load is %.0f kW'%(RX)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loads shared by alternator 1 and 2 are 1333.33 kW and 1666.67 kW respectively\n",
-        "Maximum load is 3600 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8 Page no :  60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "P_out = 1500.*10**3\n",
-      "V_L = 11000.\n",
-      "phi = math.acos(0.8)\n",
-      "I_L = P_out/(math.sqrt(3)*V_L*math.cos(phi))\n",
-      "\n",
-      "I_L_actv = I_L*math.cos(phi)   \t\t\t#wattful or active component of current\n",
-      "I_L_reactive = I_L*math.sin(phi) \t\t\t#wattless or reactive component of current\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I_each = I_L/2  \t\t\t#in identical conditions\n",
-      "I_arm1 = 45. \t\t\t#given\n",
-      "I_1_reactive = math.sqrt(I_arm1**2-39.364**2 )  \t\t\t#from the power triangle \n",
-      "I_2_reactive = 59.046-21.80\n",
-      "I_a_2 = math.sqrt( 39.364**2 + I_2_reactive**2 )  \t\t\t#required armature current of 2nd alternator\n",
-      "print 'Required armature current of second alternator is %.4f A'%(I_a_2)\n",
-      "#power factors of 2 machines\n",
-      "cos_phi1 = 39.364/45 \n",
-      "cos_phi2 = 39.364/54.1921\n",
-      "\n",
-      "print 'Power factors are %.4f lagging and %.4f lagging'%(cos_phi1,cos_phi2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required armature current of second alternator is 54.1921 A\n",
-        "Power factors are 0.8748 lagging and 0.7264 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9 Page no : 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#Line AB for Altermnator 1 and AC for alternator 2.AF is at frequency x measured from A where total load is 3000 kW\n",
-      "# Variables\n",
-      "BO = 2000.\n",
-      "AO = 5.\t\t\t#AF = x\n",
-      "DC = 2000.\n",
-      "AD = 3.\n",
-      "#AF = x\n",
-      "\n",
-      "# Calculations\n",
-      "#using similarity of triangles AEF and ABO\n",
-      "EF_by_AF = (BO/AO)\t\t\t# because (EF/BO) = (AF/AO)\n",
-      "#using similarity of triangles AFG and ADC\n",
-      "FG_by_AF = (DC/AD) \t\t\t#because (FG/DC) = (AF/AD)\n",
-      "\n",
-      "EF_by_x = EF_by_AF   \t\t\t#which implies EF = 400*x\n",
-      "FG_by_x = FG_by_AF   \t\t\t#which implies FG = 666.67*x\n",
-      "\n",
-      "EF_plus_FG = 3000  \t\t\t#total load at the frequency at P is 3000 kW\n",
-      "x =  EF_plus_FG/(EF_by_x + FG_by_x)\n",
-      "EF = (BO/AO)*x\n",
-      "FG = (DC/AD)*x  \n",
-      "\n",
-      "# Results\n",
-      "print 'Loads shared by machine 1 and 2 are %.0f kW and %.0f kW respectively'%(EF,FG)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loads shared by machine 1 and 2 are 1125 kW and 1875 kW respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10 Page no :  63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
-      "\n",
-      "# Variables\n",
-      "V_l = 6000.\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "VA = 2000.*10**3\n",
-      "I_FL = VA/(V_l*math.sqrt(3))\n",
-      "X_s = complex(0,6) \t\t\t#synchronous reactance\n",
-      "P = 8.\n",
-      "f = 50.\n",
-      "\n",
-      "delta_mech = math.pi/180 \t\t\t#phase print lacemant in degree mechanical \n",
-      "#phase print lacemant in degree electrical\n",
-      "delta_elec = delta_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
-      "\n",
-      "phi = math.acos(math.radians(0.8))\n",
-      "V = p2z(V_ph,phi)\n",
-      "E = V+I_FL*X_s\n",
-      "#E leads I by phasemag(E). V leads I by phasemag(V)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "delta = (math.pi/180)* (math.atan(E.imag/E.real)-(math.atan(V.imag/V.real ) ) ) \t\t\t#power angle in radians\n",
-      "P_SY = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_elec)/abs(X_s) \t\t\t#Synchronising power\n",
-      "P_SY_total = 3*P_SY \t\t\t#totla Synchronising power \n",
-      "print 'Total Synchronising power is %.3f kW'%(10**-3*P_SY_total)\n",
-      "\n",
-      "N_s = 120*f/P \t\t\t#in rpm\n",
-      "n_s = (N_s)/60 \t\t\t#in rps\n",
-      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
-      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n",
-      "\n",
-      "# note : book answer it seems wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total Synchronising power is 444.753 kW\n",
-        "Synchronising torque is 5663 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11  Page no : 64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
-      "\n",
-      "# Variables\n",
-      "V_l = 3300.\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "VA = 3*10.**6\n",
-      "I_FL = VA/(V_l*math.sqrt(3))\n",
-      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = complex(0,IX_s/I_FL) \t\t\t#synchronous reactance\n",
-      "N_s = 1000. \t\t\t#in rpm\n",
-      "P = 6.\n",
-      "f = 50.\n",
-      "\n",
-      "# Calculations and Results\n",
-      "delta_dash_mech = math.pi/180\n",
-      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
-      "\n",
-      "I = I_FL\n",
-      "phi = math.acos(math.radians(0.8))\n",
-      "V = p2z(V_ph,phi)\n",
-      "E = V+I*X_s\n",
-      "#E leads I by phasemag(E). V leads I by phasemag(V)\n",
-      "\n",
-      "delta = (math.pi/180)* (math.atan(E.imag/E.real)-math.atan(V.imag/V.real) ) \t\t\t#power angle in radians\n",
-      "P_SY = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s) \t\t\t#Synchronising power per phase\n",
-      "print 'Synchronising power is %.3f kW'%(10**-3*P_SY)\n",
-      "P_SY_total = 3*P_SY \t\t\t#Total Synchronising power\n",
-      "\n",
-      "N_s = 120*f/P \t\t\t#in rpm\n",
-      "n_s = (N_s)/60 \t\t\t#in rps\n",
-      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
-      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n",
-      "\n",
-      "print 'Answer mismatches due to approximation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power is 217.160 kW\n",
-        "Synchronising torque is 6221 N-m\n",
-        "Answer mismatches due to approximation\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12 Page no : 65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_l = 3300.\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "VA = 3*10.**6\n",
-      "I_FL = VA/(V_l*math.sqrt(3))\n",
-      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = complex(0,IX_s/I_FL) \t\t\t#synchronous reactance\n",
-      "N_s = 1000.\n",
-      "P = 6.\n",
-      "f = 50.\n",
-      "\n",
-      "delta_dash_mech = math.pi/180 \t\t\t#phase print lacement in degree mechanical\n",
-      "#phase print lacement in degree electrical\n",
-      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "E = V_ph\n",
-      "Z_s = X_s \t\t\t#math.since R = 0\n",
-      "P_SY = abs(E)*abs(V_ph)*delta_dash_elec/abs(Z_s) \t\t\t#Synchronising power per phase\n",
-      "print 'Synchronising power is %.3f kW'%(10**-3*P_SY)\n",
-      "P_SY_total = 3*P_SY \t\t\t#Total Synchronising power\n",
-      "print '3 phase Synchronising power is %.3f kW'%(10**-3*P_SY_total)\n",
-      "\n",
-      "N_s = 120*f/P \t\t\t#in rpm\n",
-      "n_s = (N_s)/60 \t\t\t#in rps\n",
-      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
-      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power is 261.799 kW\n",
-        "3 phase Synchronising power is 785.398 kW\n",
-        "Synchronising torque is 7500 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13 Page no :  66"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 11.*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 700.*10**3\n",
-      "I_FL = VA/(math.sqrt(3)*V_L) \t\t\t#full load current\n",
-      "IR_a = (1.5/100)*V_ph \t\t\t#product of I and R_a\n",
-      "R_a = IR_a/I_FL\n",
-      "IX_s = (14./100)*V_ph \t\t\t# product of I and X_s\n",
-      "X_s = IX_s/I_FL \t\t\t#synchronous reactance\n",
-      "\n",
-      "# Calculations\n",
-      "#at full load and 0.8 pf\n",
-      "I = I_FL\n",
-      "phi = math.acos(0.8)\n",
-      "V_ph = complex(V_ph*math.cos(phi),V_ph*math.sin(phi))  \t\t\t#just introduced the angle\n",
-      "E_ph = math.sqrt(    (abs(V_ph)*math.cos(phi)+ IR_a)**2+   (abs(V_ph)*math.sin(phi)+ IX_s)**2     )\n",
-      "\n",
-      "Poles = 4.\n",
-      "f = 50. \t\t\t#poles and frequency\n",
-      "delta = math.asin( (abs(V_ph)*math.sin(phi)+IX_s)/E_ph)  -phi\n",
-      "delta_dash_mech = (math.pi/180) \t\t\t#print lacement in degree mechanical\n",
-      "\t\t\t#print lacement in degree electrical\n",
-      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
-      "P_SY = abs(E_ph)*abs(V_ph)*math.cos(delta)*math.sin(delta_dash_elec)/X_s   \t\t\t#Synchronising power per phase \n",
-      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
-      "\n",
-      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
-      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power is %.2fkW'%(P_SY_total/1000)\n",
-      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power is 191.25kW\n",
-        "Synchronising torque is 1217.53 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14 Page no :  68"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_l = 9.*10**3\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "VA = 5.5*10**6\n",
-      "I_FL = VA/(V_l*math.sqrt(3))\n",
-      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = complex(0,IX_s/I_FL) \t\t\t#synchronous reactance\n",
-      "N_s = 1500. \t\t\t#in rpm\n",
-      "n_s = N_s/60 \t\t\t#in rps\n",
-      "f = 50.\n",
-      "P = 120.*f/N_s \t\t\t#frequency and pole\n",
-      "\n",
-      "# Calculations\n",
-      "delta_dash_mech = math.pi/180 \t\t\t#print lacemnt in degree mechanical \n",
-      "#print lacemnt in degree electrical\n",
-      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
-      "\n",
-      "E = V_ph\n",
-      "P_SY = abs(E)*abs(V_ph)*delta_dash_elec/abs(X_s) \t\t\t#Synchronising power per phase\n",
-      "P_SY_total = 3*P_SY \t\t\t#Total Synchronising power\n",
-      "\n",
-      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n",
-      "print 'Answer mismatches due to approximation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising torque is 4888.89 N-m\n",
-        "Answer mismatches due to approximation\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15  Page no :  69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6.*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 2000.*10**3\n",
-      "I_FL = VA/(math.sqrt(3)*V_L)  \n",
-      "I = I_FL\n",
-      "\n",
-      "# Calculations\n",
-      "X_s = 1.2\n",
-      "R_a = 0.01 \t\t\t#both per unit\n",
-      "IR_a = (1/100)*V_ph \t\t\t#product of I and R_a\n",
-      "R_a = IR_a/I_FL\n",
-      "IX_s = (120/100)*V_ph \t\t\t#product of I and X_s\n",
-      "#IX_s = (12/100)*V_ph   \t\t\t# this is the mistake made in the textbook\n",
-      "X_s = IX_s/I_FL\n",
-      "\n",
-      "#at full load and 0.8 pf\n",
-      "phi = math.acos(0.8)\n",
-      "#V_ph = complex(V_ph*math.cos(phi)V_ph*math.sin(phi))  \t\t\t#just introduced the angle\n",
-      "E_ph = math.sqrt(    (abs(V_ph)*math.cos(phi)+ IR_a)**2+   (abs(V_ph)*math.sin(phi)+ IX_s)**2     )\n",
-      "Poles = 8\n",
-      "f = 50\n",
-      "\n",
-      "delta = math.asin( (abs(V_ph)*math.sin(phi)+IX_s)/E_ph)  -phi\n",
-      "delta_dash_mech = (math.pi/180) \t\t\t#print lacemnt in degree mechanical \n",
-      "#print lacemnt in degree electrical\n",
-      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
-      "P_SY = abs(E_ph)*abs(V_ph)*math.cos(delta)*math.sin(delta_dash_elec)/X_s   \t\t\t#Synchronising power per phase\n",
-      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
-      "\n",
-      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
-      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power is %.2f kW'%(P_SY_total/1000)\n",
-      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n",
-      "\n",
-      "print 'Note that answer obtained doesnt match with textbook due to the following reasons: \\\n",
-      "\\niIX_s is considered wrong in textbook.It should have been 4156.92instead of 415.692  To verify this use commented statement of IX_s line 13and notice that it matches with textbook ans then' \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power is 223.22 kW\n",
-        "Synchronising torque is 2960.56 N-m\n",
-        "Note that answer obtained doesnt match with textbook due to the following reasons: \n",
-        "iIX_s is considered wrong in textbook.It should have been 4156.92instead of 415.692  To verify this use commented statement of IX_s line 13and notice that it matches with textbook ans then\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16  Page no :  71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "E = 11.*10**3/math.sqrt(3)\n",
-      "I_sc = 1000.\n",
-      "Pole = 2.\n",
-      "f = 50.\n",
-      "delta_dash_mech = 1*math.pi/180 \t\t\t#print lacemnt in degree mechanical \n",
-      "\n",
-      "# Calculations\n",
-      "#print lacemnt in degree electrical \n",
-      "delta_dash_elec = delta_dash_mech*(Pole/2)\n",
-      "P_SY = E*I_sc*delta_dash_mech \t\t\t#Synchronising power per phase\n",
-      "P_SY_total = P_SY*3 \t\t\t#total Synchronising power\n",
-      "\n",
-      "ns = 120*f/(60*Pole)  \t\t\t#in r.p.s\n",
-      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power is %.2f kW'%(P_SY_total/1000)\n",
-      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power is 332.53 kW\n",
-        "Synchronising torque is 1058.48 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17 Page no :  72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#Line PQ for Altermnator 1 and PR for alternaator 2.AB is at frequency x from P where total load is 30 MW\n",
-      "# Variables\n",
-      "QT = 25.\n",
-      "PT = 2.\n",
-      "#PC = x\n",
-      "SR = 25.\n",
-      "PS = 1.5\n",
-      "\n",
-      "# Calculations\n",
-      "#using similarity of triangles PAC and PQT\n",
-      "AC_by_PC = (QT/PT)\t\t\t# because (AC/QT) = (PC/PT)\n",
-      "#using similarity of triangles PCB and PSR\n",
-      "CB_by_PC = (SR/PS)\n",
-      "\n",
-      "AC_by_x = AC_by_PC   \t\t\t#which implies AC = 12.5*x\n",
-      "CB_by_x = CB_by_PC  \t\t\t#which implies CB = 16.67*x\n",
-      "\n",
-      "AC_plus_CB = 30  \t\t\t#total load at the frequency at P is 30 MW\n",
-      "x =  AC_plus_CB/(AC_by_x + CB_by_x)\n",
-      "AC = 12.5*x\n",
-      "CB = 16.67*x  \n",
-      "frequency = 50-x\n",
-      "\n",
-      "# Results\n",
-      "print 'Loads shared by alternator 1 and 2 are %.2f MW and %.2f MW respectively'%(AC,CB)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loads shared by alternator 1 and 2 are 12.86 MW and 17.15 MW respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18  Page no :  73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
-      "\n",
-      "# Variables\n",
-      "load_total = 1600.*10**3\n",
-      "pf = 1/math.sqrt(2)  \t\t\t#lag\n",
-      "V_L = 6600.\n",
-      "\n",
-      "# Calculations\n",
-      "I_L = p2z(load_total/(math.sqrt(3)*V_L*pf),-1*math.acos(math.radians(pf)) )\n",
-      "I_1 = p2z(90,-1*math.acos(math.radians(0.8)))\n",
-      "I_2 = I_L-I_1\n",
-      "phi = abs(math.atan(I_2.imag/I_2.real))\n",
-      "I_a = abs(I_2)\n",
-      "R_a = 1.05\n",
-      "X_s = 5 \t\t\t#resistance and synchronous reactance per phase\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "E_ph = math.sqrt(  (V_ph*math.cos(phi)+I_a*R_a )**2 + ( V_ph*math.sin(phi)+I_a*X_s )**2        )\n",
-      "E_line = math.sqrt(3)*E_ph\n",
-      "\n",
-      "# Results\n",
-      "print 'Excitation of second alternator is %.2f V '%(E_line)\n",
-      "print ' The corresponding field current from the graph is about 310 A'\n",
-      "print 'Note: The answer obtained will differ from textbook answer because of higher degree of accuracy while storing I_2 and the improper rounding off of I_2 in the textbook'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Excitation of second alternator is 6884.65 V \n",
-        " The corresponding field current from the graph is about 310 A\n",
-        "Note: The answer obtained will differ from textbook answer because of higher degree of accuracy while storing I_2 and the improper rounding off of I_2 in the textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19 Page no : 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 10.*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 5.*10**6\n",
-      "I_FL = VA/(math.sqrt(3)*V_L) \t\t\t#full-load current\n",
-      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = IX_s/I_FL \t\t\t#synchronous reactance\n",
-      "P = 4.\n",
-      "\n",
-      "# Calculations\n",
-      "delta_dash_mech = 1*(math.pi/180) \t\t\t#print lacement in degree mechanical \n",
-      "#print lacement in degree electrical\n",
-      "delta_dash_elec = delta_dash_mech*(P/2)\n",
-      "E = V_ph  \t\t\t#at no load\n",
-      "P_SY =  delta_dash_elec*E**2/X_s  \t\t\t#Synchronising power per phase\n",
-      "P_SY_total = P_SY*3  \t\t\t#Total Synchronising power\n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power per phase is %.2fkW \\\n",
-      "\\nTotal Synchronising power is %.2fkW '%(P_SY/1000,P_SY_total/1000)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power per phase is 290.89kW \n",
-        "Total Synchronising power is 872.66kW \n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20 Page no : 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Power_total = 1.414 \t\t\t#per unit\n",
-      "V_L = 1. \t\t\t#per unit\n",
-      "phi_t = math.acos(0.707)\n",
-      "I_L_T = Power_total/(math.sqrt(3)*V_L*math.cos(phi_t))  \t\t\t#Total current\n",
-      "\n",
-      "# Calculations\n",
-      "#Current supplied by each alternator\n",
-      "I_1 = I_L_T/2\n",
-      "I_2 = I_1\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "\n",
-      "phi = math.acos(0.707)\n",
-      "R_a = 0\n",
-      "X_s = 0.6 \t\t\t#resistacne and synchronous reactance\n",
-      "E_ph = math.sqrt(  (V_ph*math.cos(phi)+ I_1*R_a)**2 + (V_ph*math.sin(phi)+I_1*X_s)**2          )\n",
-      "delta =  math.atan((I_1*X_s+V_ph*math.sin(phi)) / (V_ph*math.cos(phi))) - phi    \t\t\t#power angle\n",
-      "\n",
-      "# Results\n",
-      "print 'EMF is %.4f p.u. and power angle is %.2f degrees '%(E_ph,delta*180/math.pi)\n",
-      "print 'Following assumptions were made :'\n",
-      "print '1.Terminal or bus bar voltage at ppoint of connection is constant'\n",
-      "print '2.The alternators are identical and are initially equally excited'\n",
-      "print '3.The power supplied by prime movers is adjusted so that each machine carries half the load represented by external impedance Z = R+ j 2pifL %( where R and L are constant'\n",
-      "print '4.The stator resistance is negligible'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "EMF is 0.8580 p.u. and power angle is 16.58 degrees \n",
-        "Following assumptions were made :\n",
-        "1.Terminal or bus bar voltage at ppoint of connection is constant\n",
-        "2.The alternators are identical and are initially equally excited\n",
-        "3.The power supplied by prime movers is adjusted so that each machine carries half the load represented by external impedance Z = R+ j 2pifL %( where R and L are constant\n",
-        "4.The stator resistance is negligible\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21 Page no : 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_l = 480.\n",
-      "X_d = 0.1\n",
-      "X_q = 0.075\n",
-      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "I_l = 1200.\n",
-      "I_ph = I_l/math.sqrt(3)\n",
-      "V_ph = V_l\n",
-      "V_t = V_l\n",
-      "I_a = I_ph\n",
-      "\n",
-      "# Calculations\n",
-      "phi = math.acos(0.8)\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
-      "delta = psi-phi \n",
-      "\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(psi)\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "\n",
-      "# Results\n",
-      "print 'Excitation e.m.f is %.f V '%(E_f)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Excitation e.m.f is 524 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22  Page no :  78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "VA = 3.5*10**6\n",
-      "P = 32.        \t\t\t#Poles\n",
-      "Power = 2.5*10**6 \t\t\t#In watts\n",
-      "V_l = 6.6*10**3\n",
-      "phi = math.acos(0.8)\n",
-      "I_l = Power/(V_l*math.cos(phi)*math.sqrt(3))\n",
-      "X_d = 9.6\n",
-      "X_q = 6.\n",
-      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "\n",
-      "# Calculations\n",
-      "V_t = V_l/math.sqrt(3)\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+I_l*X_q)/(V_t*math.cos(phi)+I_l*R_a)            )\n",
-      "delta = psi-phi\n",
-      "I_s = I_l\n",
-      "I_d = I_s*math.sin(psi)\n",
-      "I_q = I_s*math.cos(psi)\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "regulation = 100*(E_f-V_t)/V_t\n",
-      "\n",
-      "# Results\n",
-      "print 'percentage regulation is %.2f percent'%(regulation)\n",
-      "print 'Excitation emf =  %.0f V'%(E_f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "percentage regulation is 50.85 percent\n",
-        "Excitation emf =  5748 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23 Page no :  79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "X_d = 7.6\n",
-      "X_q = 4.5\n",
-      "R_a = 0.15 \t\t\t#armature resistance and synchronous reactance of directquadrature \n",
-      "axisV_l = 13.8*10**3\n",
-      "V_l = 13.8*10**3\n",
-      "V_t = V_l/math.sqrt(3)\n",
-      "phi = math.acos(0.8)\n",
-      "VA = 25*10**6\n",
-      "I_a = VA/(math.sqrt(3)*V_l)\n",
-      "psi = math.atan(  (V_t*math.cos(phi)+I_a*X_q)/(V_t*math.sin(phi)+I_a*R_a)            )\n",
-      "\n",
-      "# Calculations\n",
-      "delta = psi-phi\n",
-      "I_s = I_a\n",
-      "I_d = I_s*math.sin(psi)\n",
-      "I_q = I_s*math.cos(psi)\n",
-      "\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "regulation = 100*(E_f-V_t)/V_t\n",
-      "\n",
-      "# Results\n",
-      "print 'percentage regulation is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "percentage regulation is 79.30 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24 Page no :  80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "X_d = 1.\n",
-      "X_q = 0.6\n",
-      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "phi = math.acos(0.8) \t\t\t#lag\n",
-      "V_t = 1.\n",
-      "I_a = 1. \t\t\t#full load\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
-      "\n",
-      "# Calculations\n",
-      "delta = psi-phi\n",
-      "I_s = I_a\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(psi)\n",
-      "\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "regulation = 100*(E_f-V_t)/V_t\n",
-      "\n",
-      "# Results\n",
-      "print 'percentage regulation is %.2f percent'%(regulation)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "percentage regulation is 77.50 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25 Page no :  81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I_a = 10.\n",
-      "phi = 20. \t\t\t#lag and degrees\n",
-      "V_t = 400.\n",
-      "X_d = 10.\n",
-      "X_q = 6.5\n",
-      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "\n",
-      "# Calculations\n",
-      "psi = math.atan((V_t*math.sin(math.radians(phi))+I_a*X_q)/(V_t*math.cos(math.radians(phi))+I_a*R_a))\n",
-      "delta = math.degrees(psi)-phi\n",
-      "I_d = math.degrees(I_a*math.sin(math.radians(psi)))\n",
-      "I_q = (I_a*math.cos(psi))\n",
-      "\n",
-      "# Results\n",
-      "print 'Load angle is %.2f degrees '%(delta)\n",
-      "print 'I_d and I_q are %.4f A and %.4f A respectively '%(I_d,I_q  )\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load angle is 8.23 degrees \n",
-        "I_d and I_q are 4.9272 A and 8.8105 A respectively \n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26 Page no : 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "X_d = 0.8\n",
-      "X_q = 0.5\n",
-      "R_a = 0.02 \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "\n",
-      "\t\t\t#case(i)  lag\n",
-      "phi = math.acos(0.8)\n",
-      "V_t = 1\n",
-      "I_a = 1\t\t\t#full-load\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
-      "delta = psi-phi\n",
-      "\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(psi)\n",
-      "\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "regulation = 100*(E_f-V_t)/V_t\n",
-      "print 'percentage regulation at 0.8 pf lag is %.2f percent'%(regulation)\n",
-      "\n",
-      "\t\t\t#case(ii)  lead\n",
-      "phi2 = -1*math.acos(0.8) \t\t\t#minus sign because of leading pf\n",
-      "psi2 = math.atan(  (V_t*math.sin(phi2)+I_a*X_q)/(V_t*math.cos(phi2)+I_a*R_a)            )\n",
-      "delta2 = psi2-phi2\n",
-      "\n",
-      "I_d2 = I_a*math.sin(psi2)\n",
-      "I_q2 = I_a*math.cos(psi2)\n",
-      "\n",
-      "E_f2 = V_t*math.cos(delta2)+I_d2*X_d+I_q2*R_a\n",
-      "regulation2 = 100*(E_f2-V_t)/V_t\n",
-      "print 'percentage regulation at 0.8 pf lead is %.f percent'%(regulation2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "percentage regulation at 0.8 pf lag is 61.25 percent\n",
-        "percentage regulation at 0.8 pf lead is -21 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.27 Page no : 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "kW = array([800,500,1000,600])\n",
-      "cosphi = array([1,0.9,0.8,0.9])\n",
-      "tanphi = tan(arccos(cosphi))\n",
-      "kVAR = kW*tanphi\n",
-      "\n",
-      "# Calculations\n",
-      "kW_total = kW[0]+kW[1]+kW[2]+kW[3]\n",
-      "kVAR_total = kVAR[0]+kVAR[1]+kVAR[2]+-1*kVAR[3] \t\t\t#4th case is leading \n",
-      "\n",
-      "phi_c = math.atan(kVAR_total/kW_total)  \t\t\t#total power factor angle\n",
-      "phi_1 = math.acos(0.95)\t\t\t#pf of machine 1\n",
-      "kW_1 = 1000 \t\t\t#active component of machine 1\n",
-      "kVAR_1 = kW_1*math.tan(phi_1) \t\t\t#reactive component of machine 1\n",
-      "kW_2 = kW_total - kW_1 \t\t\t#active component of machine 1\n",
-      "kVAR_2 = kVAR_total-kVAR_1 \t\t\t#reactive component of machine 2\n",
-      "\n",
-      "phi_2 = math.atan(kVAR_2/kW_2) \n",
-      "pf_2 = math.cos(phi_2) \t\t\t#power factor of machine 2\n",
-      "\n",
-      "# Results\n",
-      "print 'Output of second alternator =  %.0f kW'%(kW_2)\n",
-      "print 'power factor of machine 2  =  %.2f and lagging'%(pf_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output of second alternator =  1900 kW\n",
-        "power factor of machine 2  =  0.98 and lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28 Page no : 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "kW = array([250,300,150])\n",
-      "cosphi = array([0.9,0.75,0.8]) \t\t\t#all lagging\n",
-      "tanphi = tan(arccos(cosphi))\n",
-      "kVAR = kW*tanphi\n",
-      "\n",
-      "# Calculations\n",
-      "kW_total = kW[0]+kW[1]+kW[2]\n",
-      "kVAR_total = kVAR[0]+kVAR[1]+kVAR[2] \n",
-      "\n",
-      "phi_1 = math.acos(0.8)\t\t\t#pf of machine 1\n",
-      "kW_1 = 100 \t\t\t#active component of machine 1\n",
-      "kVAR_1 = kW_1*math.tan(phi_1) \t\t\t#reactive component of machine 1\n",
-      "kW_2 = kW_total - kW_1 \t\t\t#active component of machine 1\n",
-      "kVAR_2 = kVAR_total-kVAR_1 \t\t\t#reactive component of machine 2\n",
-      "phi_2 = math.atan(kVAR_2/kW_2) \n",
-      "pf_2 = math.cos(phi_2) \t\t\t#power factor of machine 2\n",
-      "\n",
-      "# Results\n",
-      "print 'Output of second alternator =  %.0f kW'%(kW_2)\n",
-      "print 'power factor of machine 2  =  %.4f and lagging'%(pf_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output of second alternator =  600 kW\n",
-        "power factor of machine 2  =  0.8172 and lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 57
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29 Page no :  85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6.6*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "V_t = V_ph\n",
-      "X_d = 9.6\n",
-      "X_q = 6.\n",
-      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "VA = 3.5*10**6\n",
-      "I_L = VA/(math.sqrt(3)*V_L)\n",
-      "\n",
-      "# Calculations\n",
-      "P = 2.5*10**6\n",
-      "phi = math.acos(0.8)\n",
-      "I_a = P/(math.sqrt(3)*V_L*math.cos(phi))\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+ I_a*X_q)/(V_t*math.cos(phi)+ I_a*R_a)     )\n",
-      "\n",
-      "delta = psi-phi\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(phi)\n",
-      "\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "regulation = 100*(E_f-V_t)/V_t\n",
-      "P_max = (V_ph**2/2)*((X_d-X_q)/(X_d*X_q))*(math.sin(2*delta))\n",
-      "\n",
-      "# Results\n",
-      "print 'percentage voltage regulation is %.2f percent'%(regulation)\n",
-      "print 'Power under open circuit is %.1f kW per phase'%(P_max/1000)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "percentage voltage regulation is 50.85 percent\n",
-        "Power under open circuit is 231.1 kW per phase\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.30  Page no :  86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3.3*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 3.*10**6\n",
-      "I_FL = VA/(math.sqrt(3)*V_L)\n",
-      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = complex(0,IX_s/I_FL)\n",
-      "N_s = 1000 \t\t\t#in r.p.m\n",
-      "\n",
-      "\n",
-      "# Calculations\n",
-      "Poles = 6.\n",
-      "f = 50.\n",
-      "delta_dash_mech = (math.pi/180) \t\t\t#print lacement in degree mechanical\n",
-      "#print lacement in degree electrical\n",
-      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
-      "\n",
-      "I = I_FL\n",
-      "phi = math.acos(0.8)\n",
-      "V = complex(V_ph*math.cos(phi),V_ph*math.sin(phi))\n",
-      "E =  V+ I*X_s\n",
-      "\n",
-      "delta = (math.pi/180)*math.atan(E.imag/E.real)-phi   \t\t\t#E leads I by (math.pi/180)*phasemag(E) and V leads I by phi radians \n",
-      "P_SY = abs(E)*abs(V_ph)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)   \t\t\t#Synchronising power per phase \n",
-      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
-      "\n",
-      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.m\n",
-      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power per phase is %.3f kW'%(P_SY/1000)\n",
-      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n",
-      "print 'Answer mismatches due to improper approximation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power per phase is 197.555 kW\n",
-        "Synchronising torque is 5660 N-m\n",
-        "Answer mismatches due to improper approximation\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31  Page no :  87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3.3*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 3.*10**6\n",
-      "I_FL = VA/(math.sqrt(3)*V_L)\n",
-      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = complex(0,IX_s/I_FL)\n",
-      "N_s = 1000. \t\t\t#in r.p.m\n",
-      "Poles = 6.\n",
-      "f = 50.\n",
-      "\n",
-      "# Calculations\n",
-      "delta_dash_mech = (math.pi/180) \t\t\t#print lacement in degree mechanical\n",
-      "#print lacement in degree electrical \n",
-      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
-      "\n",
-      "#E = V as the alternator is on no-load and X_s = Z_s\n",
-      "P_SY = abs(V_ph)**2*(delta_dash_elec)/abs(X_s)   \t\t\t#Synchronising power per phase\n",
-      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
-      "\n",
-      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
-      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power per phase is %.3f kW'%(P_SY/1000)\n",
-      "print 'Total Synchronising power is %.3f kW'%(P_SY_total/1000)\n",
-      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power per phase is 261.799 kW\n",
-        "Total Synchronising power is 785.398 kW\n",
-        "Synchronising torque is 7500 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 61
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32 Page no :  88"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 11.*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 700.*10**3\n",
-      "I_FL = VA/(math.sqrt(3)*V_L)\n",
-      "IX_s = (14./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = IX_s/I_FL\n",
-      "#X_s = complex(0,IX_s/I_FL)\n",
-      "IR_a = (1.5/100)*V_ph \t\t\t#product of I and R_a\n",
-      "R_a = IR_a/I_FL\n",
-      "\n",
-      "\n",
-      "# Calculations\n",
-      "I = I_FL\n",
-      "phi = math.acos(0.8)\n",
-      "V = complex(V_ph*math.cos(phi),V_ph*math.sin(phi))\n",
-      "E_ph = math.sqrt(  (V_ph*math.cos(phi)+IR_a)**2 +(V_ph*math.sin(phi)+IX_s)**2    )\n",
-      "delta = math.asin((V_ph*math.sin(phi)+IX_s)/E_ph)   -phi\n",
-      "\n",
-      "Poles = 4.\n",
-      "f = 50.\n",
-      "delta_dash_mech = (math.pi/180) \t\t\t#phase print lacemnt in degree mechanical\n",
-      "delta_dash_elec = delta_dash_mech*(Poles/2)\t\t\t#phase print lacemnt in degree electrical\n",
-      "\n",
-      "P_SY = abs(V_ph)*abs(E_ph)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)   \t\t\t#Synchronising power per phase\n",
-      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
-      "\n",
-      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
-      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power per phase is %.3f kW'%(P_SY/1000)\n",
-      "print 'Synchronising power is %.3f kW ; '%(P_SY/1000)\n",
-      "print 'Total Synchronising power is %.3f kW'%(P_SY_total/1000)\n",
-      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power per phase is 63.750 kW\n",
-        "Synchronising power is 63.750 kW ; \n",
-        "Total Synchronising power is 191.249 kW\n",
-        "Synchronising torque is 1217.53 N-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33 Page no : 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
-      "\n",
-      "# Variables\n",
-      "Z1 = complex(0,2)\n",
-      "Z2 = complex(0,3)\n",
-      "Z = 6.\n",
-      "E1 = p2z(230,0)\n",
-      "E2 = p2z(230,10)\n",
-      "\n",
-      "# Calculations\n",
-      "I1 = ((E1-E2)*Z+E1*Z2)/(Z*(Z1+Z2)+Z1*Z2)\n",
-      "I2 = ((E2-E1)*Z+E2*Z1)/(Z*(Z1+Z2)+Z1*Z2)\n",
-      "\n",
-      "phi1 = math.atan(I1.imag/I1.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
-      "phi2 = math.atan(I2.imag/I2.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
-      "\n",
-      "I = I1+I2\n",
-      "V = I*Z \t\t\t#Terminal voltage\n",
-      "\n",
-      "# Results\n",
-      "print 'i) Terminal voltage is %.2f volts at %.1f degrees'%(abs(V),math.degrees(math.atan(V.imag/V.real)))\n",
-      "print 'ii) Currents are %.2f A at %.0f degrees and %.2f A at %.2f degrees     \\\n",
-      "\\nTotal current is %.2f A at %.1f degrees  '%(abs(I1),math.degrees(math.atan(I1.imag/I1.real)),abs(I2),\\\n",
-      "                                              math.degrees(math.atan(I2.imag/I2.real)),abs(I) \\\n",
-      "                                              ,math.degrees(math.atan(I.imag/I.real)))\n",
-      "\n",
-      "P1 = abs(V)*abs(I1)*math.cos(math.radians(phi1))\n",
-      "P2 = abs(V)*abs(I2)*math.cos(math.radians(phi2))\n",
-      "print 'iii)Power delivered %.2f watts and  %.2f watts'%(P1,P2)\n",
-      "\n",
-      "# note : rounding off error. kindly check the calculations\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Terminal voltage is 224.71 volts at -7.3 degrees\n",
-        "ii) Currents are 14.74 A at -14 degrees and 22.88 A at -3.03 degrees     \n",
-        "Total current is 37.45 A at -7.3 degrees  \n",
-        "iii)Power delivered 3311.43 watts and  5141.03 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34  Page no : 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "X_d = 0.8\n",
-      "X_q = 0.5  \t\t\t#both per unit\n",
-      "R_a = 0. \t\t\t#assumed\n",
-      "phi = math.acos(0.8)\n",
-      "V_t = 1.\t\t\t#pu\n",
-      "I_a = 1. \t\t\t#full-load\n",
-      "\n",
-      "# Calculations\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
-      "delta = psi-phi\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(psi)\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "\n",
-      "# Results\n",
-      "print 'Open circuit voltage is %.3f p.u.'%(E_f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Open circuit voltage is 1.603 p.u.\n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.35 Page no : 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6600.\n",
-      "I_L = 110.\n",
-      "phi_1 = math.acos(0.9) \t\t\t#lagging\n",
-      "kW = array([400,1000,400,300])*10**3\n",
-      "cosphi = array([1,0.71,0.8,0.9])\n",
-      "tanphi = tan(arccos(cosphi))\n",
-      "kVAR = kW*tanphi\n",
-      "\n",
-      "# Calculations\n",
-      "kW_total = kW[0]+kW[1]+kW[2]+kW[3]\n",
-      "kVAR_total = kVAR[0]+kVAR[1]+kVAR[2]+kVAR[3]\n",
-      "\n",
-      "phi_c = math.atan(kVAR_total/kW_total)  \t\t\t#total power factor angle\n",
-      "load_1 = math.sqrt(3)*V_L*I_L*math.cos(phi_1)\n",
-      "\n",
-      "kW_1 = load_1 \t\t\t#active component of machine 1\n",
-      "kVAR_1 = kW_1*math.tan(phi_1) \t\t\t#reactive component of machine 1\n",
-      "kW_2 = kW_total - kW_1 \t\t\t#active component of machine 1\n",
-      "kVAR_2 = kVAR_total-kVAR_1 \t\t\t#reactive component of machine 2\n",
-      "\n",
-      "phi_2 = math.atan(kVAR_2/kW_2) \n",
-      "pf_2 = math.cos(phi_2) \t\t\t#power factor of machine 2\n",
-      "\n",
-      "# Results\n",
-      "print 'Output of second alternator =  %.2f kW'%(kW_2/1000)\n",
-      "print 'Power factor of machine 2  =  %.4f and lagging'%(pf_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output of second alternator =  968.28 kW\n",
-        "Power factor of machine 2  =  0.7366 and lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 67
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.36  Page no :  93"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 11000.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 2.*10**6\n",
-      "phi = math.acos(0.8)\n",
-      "I_FL = VA/(math.sqrt(3)*V_L)\n",
-      "phi_1 = math.acos(0.8)\n",
-      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
-      "\n",
-      "# Calculations\n",
-      "X_s = IX_s/I_FL\n",
-      "I_1 = I_FL\n",
-      "BC = I_1*math.cos(phi_1)*X_s\n",
-      "AB = I_1*math.sin(phi_1)*X_s    \n",
-      "OA = V_ph\n",
-      "OC = math.sqrt( (OA+AB)**2+(BC)**2 ) \n",
-      "E_1 = OC\n",
-      "E_2 = 1.25*E_1\n",
-      "OE = E_2\n",
-      "DE = BC\n",
-      "AD = math.sqrt(OE**2-DE**2) -OA           \t\t\t#because OE = math.sqrt( (OA+AD)**2 + (DE)**2  )\n",
-      "\n",
-      "I_2sinphi2 = AD/X_s\n",
-      "I_2cosphi2 = I_1*math.cos(phi)\n",
-      "I_2 = math.sqrt( (I_2cosphi2)**2 + (I_2sinphi2)**2  )\n",
-      "phi2 = math.atan( I_2sinphi2/ I_2cosphi2 )\n",
-      "new_pf = math.cos(phi2)\n",
-      "\n",
-      "# Results\n",
-      "print 'Machine current is %.2f A '%(I_2)\n",
-      "print 'Power factor is  %.4f lagging'%(new_pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Machine current is 228.62 A \n",
-        "Power factor is  0.3673 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.37  Page no :  95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "    return  RRRR*exp(1j*math.pi*Theeeta/180.);\n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "P_out = 3000.*10**3\n",
-      "V_L = 6.6*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "phi = math.acos(0.8)\n",
-      "I_L = p2z(P_out/(math.sqrt(3)*V_L*math.cos(phi)),-1*(180/math.pi)*phi)\n",
-      "\n",
-      "# Calculations\n",
-      "P_out1 = P_out/2\n",
-      "I_L1 = 150  \t\t\t#given\n",
-      "phi_L1 = math.acos(    P_out1/(math.sqrt(3)*V_L*I_L1)  )\n",
-      "I_L1 = p2z(I_L1,-1*(180/math.pi)*phi_L1)\n",
-      "\n",
-      "I_L2 = I_L-I_L1\n",
-      "pf_2 = math.cos(math.atan(I_L2.imag/I_L2.real))\n",
-      "Z_1 = complex(0.5,10)\n",
-      "I_1 = I_L1\n",
-      "E_1 = V_ph + I_1*Z_1\n",
-      "delta_1 =  math.atan(E_1.imag/E_1.real)   \t\t\t#load angle of alternator 1\n",
-      "E_1L = math.sqrt(3)*E_1\n",
-      "\n",
-      "Z_2 = complex(0.4,12)\n",
-      "I_2 = I_L2\n",
-      "E_2 = V_ph + I_2*Z_2\n",
-      "delta_2 = (math.pi/180)*math.atan(E_2.imag/E_2.real)    \t\t\t#load angle of alternator 2\n",
-      "\n",
-      "# Results\n",
-      "print 'Part i) Currents are %.0f A at %.1f degrees and %.1f A at %.1f degrees\\nTotal current is %.0f at %.2f' %(abs(I_L1),\\\n",
-      "                            math.degrees(math.atan(I_L1.imag/I_L1.real)),abs(I_L2),\\\n",
-      "                            math.degrees(math.atan(I_L2.imag/I_L2.real)),abs(I_L),math.degrees(math.atan(I_L.imag/I_L.real)))\n",
-      "print 'Part ii)Power factor is %.4f and lagging'%(math.cos(phi_L1))\n",
-      "print 'Part iii)emf are %.2f V at %.2f degrees and %.4f V at %.0f degrees'%(abs(E_1),\\\n",
-      "                            math.degrees(math.atan(E_1.imag/E_1.real)),abs(E_2),math.degrees(math.atan(E_2.imag/E_2.real)))\n",
-      "print 'Part iv)Power angles are %.2f degrees and %.0f degrees '%(180/math.pi*delta_1,(180/math.pi)*delta_2)\n",
-      "\n",
-      "# note : rounding off error. kindly check the calculations\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part i) Currents are 150 A at -29.0 degrees and 180.6 A at -43.4 degrees\n",
-        "Total current is 328 at -36.87\n",
-        "Part ii)Power factor is 0.8748 and lagging\n",
-        "Part iii)emf are 4776.46 V at 15.49 degrees and 5565.7081 V at 16 degrees\n",
-        "Part iv)Power angles are 15.49 degrees and 0 degrees \n"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.38  Page no :  97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "Z1 = complex(0.2,2)\n",
-      "Z2 = Z1\n",
-      "ZL = complex(3,4)\n",
-      "Z = ZL\n",
-      "E1 = complex(2000,0)\n",
-      "E2 = complex(2200,100)\n",
-      "\n",
-      "# Calculations\n",
-      "I1 = ((E1-E2)*Z+E1*Z2)/(Z*(Z1+Z2)+Z1*Z2)\n",
-      "I2 = ((E2-E1)*Z+E2*Z1)/(Z*(Z1+Z2)+Z1*Z2)\n",
-      "\n",
-      "IL = I1+I2\n",
-      "V = IL*Z \t\t\t#Terminal voltage\n",
-      "\n",
-      "phi1 = math.atan(V.imag/V.real)-math.atan(I1.imag/I1.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
-      "phi2 = math.atan(V.imag/V.real)-math.atan(I2.imag/I2.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
-      "\n",
-      "Pout1 = math.sqrt(3)*math.sqrt(3)*abs(V)*abs(I1)*math.cos(math.radians(phi1))\n",
-      "Pout2 = math.sqrt(3)*math.sqrt(3)*abs(V)*abs(I2)*math.cos(math.radians(phi2))\n",
-      "\n",
-      "# Results\n",
-      "print 'Power delivered is %.2f kW and %.2f kW at power-factors %.4f lag and %.4f lag respectively'%(Pout1/1000,Pout2/1000,math.cos(math.radians(phi1)),math.cos(math.radians(phi2)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered is 658.24 kW and 1253.92 kW at power-factors 0.9999 lag and 0.9999 lag respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.39  Page no :  99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "f = 50.\n",
-      "P = 12.\n",
-      "V_L = 6600.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "VA = 2000.*10**3\n",
-      "I_FL = VA/(math.sqrt(3)*V_L)\n",
-      "\n",
-      "# Calculations\n",
-      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
-      "X_s = complex(0,IX_s/I_FL)\n",
-      "N_s = 12*f/P \t\t\t#in rpm\n",
-      "delta_dash_mech = (math.pi/180) \t\t\t#phase print lacemnt in degree mechanical\n",
-      "delta_dash_elec = delta_dash_mech*(P/2) \t\t\t#phase print lacemnt in degree electrical\n",
-      "\n",
-      "phi = math.acos(0.8) \t\t\t#lag\n",
-      "I = complex(I_FL*math.cos(-1*phi),I_FL*math.sin(-1*phi))\n",
-      "V = V_ph\n",
-      "E = V + I*X_s\n",
-      "delta = math.atan(E.imag/E.real)*(math.pi/180)\n",
-      "P_SY = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)\n",
-      "P_SY_total = 3*P_SY\n",
-      "\n",
-      "# Results\n",
-      "print 'Synchronising power is %.2f kW'%(P_SY/1000)\n",
-      "print 'Total Synchronising power is %.2f kW'%(P_SY_total/1000)\n",
-      "\n",
-      "# note : rounding off error. kindly check."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Synchronising power is 325.36 kW\n",
-        "Total Synchronising power is 976.09 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 76
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.40  Page no :  101"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 22000.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "power = 230.*10**6\n",
-      "phi = math.acos(1.)\n",
-      "I_FL = power/(math.sqrt(3)*V_L*math.cos(phi))\n",
-      "I_1 = I_FL\n",
-      "X_s = 1.2\n",
-      "\n",
-      "# Calculations\n",
-      "E_1 = math.sqrt( V_ph**2 + (I_1*X_s)**2 )\n",
-      "E_2 = 1.3*E_1\n",
-      "AC =   math.sqrt( E_2**2-(I_1*X_s)**2  )   -V_ph     \t\t\t# because E**2 = (V_ph+AC)**2+(I_1*X_s)**2\n",
-      "I2X_S = AC\n",
-      "\n",
-      "I_2cosphi2 = I_1  \t\t\t#because phi_2 = math.acos(I_1/I_2) \t\t\t#from ACD\n",
-      "I_2sinphi2 = AC/X_s\n",
-      "I_2 = math.sqrt( (I_2cosphi2)**2 + (I_2sinphi2)**2  )\n",
-      "phi2 = math.atan( I_2sinphi2/ I_2cosphi2 )\n",
-      "new_pf = math.cos(phi2)\n",
-      "\n",
-      "# Results\n",
-      "print 'Machine current is %.2f A '%(I_2)\n",
-      "print 'Power factor is  %.4f and lagging'%(new_pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Machine current is 7274.58 A \n",
-        "Power factor is  0.8297 and lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 77
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.41  Page no :  102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import cmath\n",
-      "\n",
-      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
-      "def  p2z(RRRR,Theeeta):\n",
-      "\treturn RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
-      "\n",
-      "# Variables\n",
-      "P_out = 1500.*10**3\n",
-      "V_L = 3.3*10**3\n",
-      "phi = math.acos(0.8)\n",
-      "I_L = p2z(P_out/(math.sqrt(3)*V_L*math.cos(phi)),-1*math.acos(math.radians(0.8)))\n",
-      "\n",
-      "I_L1_magnitude = 150.  \t\t\t#given\n",
-      "P_out1 = (3.*10**6)/2 \t\t\t#because load is EQUALLY shared between 2 alternators\n",
-      "pf_L1 = P_out1/(math.sqrt(3)*2*V_L*I_L1_magnitude)  \t\t\t#operating pf of alternator 1\n",
-      "phi1 = math.acos(math.radians(pf_L1))\n",
-      "I_L1 = p2z(I_L1_magnitude,-1*phi1)\n",
-      "I_L2 = I_L-I_L1   \t\t\t#because I_L = I_L1 + I_L2\n",
-      "pf_L2 = math.cos(math.atan(I_L2.imag/I_L2.real))\n",
-      "\n",
-      "# Calculations\n",
-      "V_ph = 6.6*10**3/math.sqrt(3)\n",
-      "Z_1 = complex(0.5,10)\n",
-      "I_1 = I_L1\n",
-      "E_1 =  V_ph + I_1*Z_1\n",
-      "delta_1 = math.atan(E_1.imag/E_1.real)  \t\t\t#load angle of alternator 1\n",
-      "I_2 = I_L2\n",
-      "\n",
-      "Z_2 = complex(0.4,12)\n",
-      "E_2 =  V_ph + I_2*Z_2\n",
-      "delta_2 = math.atan(E_2.imag/E_2.real) \t\t\t#load angle of alternator 1\n",
-      "\n",
-      "# Results\n",
-      "print 'for machine 1current is %.0f A at %.2f degrees \\nPower factor of %.4f laginduced emf \\\n",
-      "        of %.2f V \\nload angle of %.2f degrees'%(abs(I_L1),math.degrees(math.atan(I_L1.imag/I_L1.real)),pf_L1,abs(E_1),delta_1)\n",
-      "print 'for machine 2current is %.1f A at %.1f degrees\\nPower factor of %.4f laginduced emf \\\n",
-      "        of %.2f V\\nload angle of %.0f degrees'%(abs(I_L2),math.degrees(math.atan(I_L2.imag/I_L2.real)),pf_L2,abs(E_2),math.degrees(delta_2))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for machine 1current is 150 A at -1.56 degrees \n",
-        "Power factor of 0.8748 laginduced emf         of 4202.06 V \n",
-        "load angle of 0.36 degrees\n",
-        "for machine 2current is 178.0 A at -1.6 degrees\n",
-        "Power factor of 0.9996 laginduced emf         of 4480.49 V\n",
-        "load angle of 28 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 81
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.42  Page no :  104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_l = 230\n",
-      "VA = 5*10**3\n",
-      "X_d = 12\n",
-      "X_q = 7\n",
-      "R_a = 0 \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "phi = math.acos(1)\n",
-      "\n",
-      "# Calculations\n",
-      "I_l = VA/(V_l*math.sqrt(3))\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "V_t = V_ph\n",
-      "I_a = I_l\n",
-      "\n",
-      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
-      "delta = psi-phi\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(psi)\n",
-      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
-      "\n",
-      "# Results\n",
-      "print 'Excitation voltage is %.3f V '%(E_f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Excitation voltage is 193.852 V \n"
-       ]
-      }
-     ],
-     "prompt_number": 82
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.43  Page no :  104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_l = 6.6*10**3\n",
-      "V_t = V_l/math.sqrt(3)\n",
-      "X_d = 23.2\n",
-      "X_q = 14.5\n",
-      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
-      "VA = 1800.*10**3\n",
-      "phi = math.acos(0.8) \t\t\t#lag\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I_a = VA/(V_l*math.sqrt(3))\n",
-      "\n",
-      "psi = math.atan(  (V_t*math.sin(phi)-I_a*X_q)/(V_t*math.cos(phi)-I_a*R_a)  ) \t\t\t#minus sign in numerator and denomenator for motors\n",
-      "delta = psi+phi\n",
-      "I_d = I_a*math.sin(psi)\n",
-      "I_q = I_a*math.cos(psi)\n",
-      "E_f = V_t*math.cos(delta)-I_d*X_d-I_q*R_a\n",
-      "print 'Excitation emf  =  %.4f V'%(E_f)\n",
-      "#P_m =  ( V_t*E_f*math.sin(delta)/X_d  ) + ((1/X_q)-(1/X_d))*0.5*math.sin(2*delta)*V_t**2\n",
-      "#P_m = 0.4996*math.cos(delta)+0.1877*math.sin(2*delta)\n",
-      "#for maximum power output differenciate and equate to zero\n",
-      "\n",
-      "delta_max = 63.4 \t\t\t#degree\n",
-      "\n",
-      "P_m_max = ((1/X_q)-(1/X_d))*0.5*math.sin(math.radians(2*delta_max))*V_t**2   \t\t\t#Maximuum load supplied with E_f = 0\n",
-      "print 'Maximum load the motor can supply is %.4f MW per phase '%(P_m_max*10**-6      )\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Excitation emf  =  3042.2721 V\n",
-        "Maximum load the motor can supply is 0.1503 MW per phase \n"
-       ]
-      }
-     ],
-     "prompt_number": 83
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines/ch7.ipynb b/DC_Machines_and_Synchronous_Machines/ch7.ipynb
deleted file mode 100755
index a22a6459..00000000
--- a/DC_Machines_and_Synchronous_Machines/ch7.ipynb
+++ /dev/null
@@ -1,1711 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:95ca4467d5385b9a67e6a4c31c682e32bf0713e6d2ff74a5a89a56c51df83062"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7 : Synchronous Motors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1  Page no :  27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "import numpy as np\n",
-      "\n",
-      "# Variables\n",
-      "V_l = 400.\n",
-      "R_a = 0.2\n",
-      "X_s = 2.  \t\t\t#armature resistance and synchronous reactance\n",
-      "I_L = 25.\n",
-      "I_aph = I_L\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
-      "angle = math.atan(Z_s.imag/Z_s.real)\n",
-      "angle = math.degrees(angle)\n",
-      "E_Rph=I_aph*abs(Z_s)\n",
-      "theta = (math.pi/180.)*angle \n",
-      "\n",
-      "# Calculations and Results\n",
-      "#case 1\n",
-      "phi = math.acos(0.8) \t\t\t#lagging\n",
-      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta-phi)  )\n",
-      "print 'i)Back EMF induced with 0.8 lagging pf is %.3f V'%(E_bph)\n",
-      "\n",
-      "#case 2\n",
-      "phi = math.acos(0.9) \t\t\t#leading\n",
-      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
-      "print 'ii)Back EMF induced with 0.8 lagging pf is %.3f V'%(E_bph)\n",
-      "\n",
-      "#case 3\n",
-      "phi = math.acos(1)\n",
-      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta)  )\n",
-      "print 'iii)Back EMF induced with 0.8 lagging pf is %.3f V'%(E_bph)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Back EMF induced with 0.8 lagging pf is 200.386 V\n",
-        "ii)Back EMF induced with 0.8 lagging pf is 252.678 V\n",
-        "iii)Back EMF induced with 0.8 lagging pf is 231.406 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2  Page no :  28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import *\n",
-      "\n",
-      "# Variables\n",
-      "V_l = 500.\n",
-      "R_a = 0.4\n",
-      "X_s = 4. \t\t\t#armature resistance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "theta = (math.pi/180)*angle\t\t\t#phasemag returns angle in degrees,not radians\n",
-      "V_ph = V_l/math.sqrt(3)\n",
-      "I_l = 50.\n",
-      "I_aph = I_l\n",
-      "E_Rph = I_aph*abs(Z_s)\n",
-      "\n",
-      "# Calculations and Results\t\t\n",
-      "#case 1\n",
-      "E_bline = 600\n",
-      "E_bph = E_bline/math.sqrt(3)\n",
-      "phi = math.acos( (-E_bph**2 + E_Rph**2 + V_ph**2 )/(2*E_Rph*V_ph) ) -theta    \t\t\t#leading\n",
-      "#because E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
-      "print 'i)power factor is %.4f leading'%(cos(phi))\n",
-      "\n",
-      "#case 2\n",
-      "E_bline = 380\n",
-      "E_bph = E_bline/math.sqrt(3)\n",
-      "phi =  theta-math.acos( (-E_bph**2 + E_Rph**2 + V_ph**2 )/(2*E_Rph*V_ph) )     \t\t\t#leading\n",
-      "#because E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta-phi)\n",
-      "print 'ii)power factor is %.4f lagging'%(cos(phi))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)power factor is 0.9977 leading\n",
-        "ii)power factor is 0.8197 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3  Page no :  29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6600.\n",
-      "P_out = 500.*10**3\n",
-      "eta = 83./100 \t\t\t#efficiency\n",
-      "R_a = 0.3\n",
-      "X_s = 3.2 \t\t\t#armature resistance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s)  \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "theta = (math.pi/180)* angle \t\t\t#phasemag returns the angle in degrees not radians\n",
-      "phi = math.acos(0.8) \t\t\t#leading\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "P_in = P_out/eta\n",
-      "\n",
-      "# Calculations and Results\n",
-      "I_L =    P_in/ (math.sqrt(3) * V_L *  math.cos(phi) )\n",
-      "\n",
-      "# because P_in = math.sqrt(3) * V_L * I_L * math.cos(phi)\n",
-      "I_aph = I_L\n",
-      "E_Rph = I_aph*abs(Z_s)\n",
-      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
-      "print 'i) Generated EmF on full loaad is %.2f V'%(E_bph)\n",
-      "\n",
-      "delta =  math.degrees(math.asin( (E_Rph/E_bph))*math.sin(theta+phi)  )\n",
-      "#This is obtained after applying sune rule to triangle OAB from thre phasor diagram\n",
-      "print 'ii) load angle is %.2f degrees'%(delta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Generated EmF on full loaad is 3925.33 V\n",
-        "ii) load angle is 2.64 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4  Page no :  32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import roots\n",
-      "\n",
-      "# Variables\n",
-      "V_L = 500.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "phi = math.acos(0.9) \t\t\t\t#lagging\n",
-      "output_power = 17.*10**3\n",
-      "R_a = 0.8 \t\t\t\t\t\t\t#armaature reactance\n",
-      "mechanical_losses = 1300. \t\t\t#mechanical losses is W\n",
-      "P_m = output_power+mechanical_losses   \t\t\t#gross mechanical power developed\n",
-      "\n",
-      "#  P_m =  input_power - stator losses\n",
-      "#  input_power =  3* V_ph * I_aph * math.cos(phi)\n",
-      "#  Stator losses =  3*I_aph**2*R_a\n",
-      "# solving above equations we get 2.4 I_a**2 - 779/.4225*I_a + 18300  =  0\n",
-      "I_a_eqn = [2.4, -779.4225, 18300]\n",
-      "I_a_roots = roots(I_a_eqn)\n",
-      "I_a = I_a_roots[1]  \t\t\t#neglecting higher value\n",
-      "I_aph = I_a\n",
-      "print 'Current drawn by the motor is %.3f A'%(I_a)\n",
-      "\n",
-      "input_power =  3* V_ph * I_aph * math.cos(phi)\n",
-      "eta = 100*output_power/input_power\n",
-      "print 'Full load efficiency is %.2f percent'%(eta)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current drawn by the motor is 25.478 A\n",
-        "Full load efficiency is 85.61 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5  Page no :  44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "#subscript 1 is for industrial load and 2 for synchronous motor\n",
-      "P_1 = 800. \t\t\t# Active power in KW\n",
-      "phi_1 = math.acos(0.6) \t\t\t#lagging\n",
-      "Q_1 = P_1*math.tan(phi_1)   \t\t\t#reactive power by load 1\n",
-      "\n",
-      "# Calculations and Results\n",
-      "output_power = 200.\n",
-      "eta = 91./100 \t\t\t#efficiency of synchronous motor\n",
-      "input_power =  output_power/eta\n",
-      "P_2 = input_power\t\t\t# active power drawn by synchronous motor\n",
-      "P_T = P_1 + P_2  \t\t\t#combined total load of industry and synchronous motor\n",
-      "phi_T = math.acos(0.92 )\t\t\t#lagging\n",
-      "Q_T = P_T* math.tan(phi_T)  \t\t\t#from power triangle\n",
-      "Q_2 =  Q_T - Q_1  \t\t\t#it turns out to be negative indicating its leading nature\n",
-      "S_2 = math.sqrt( P_2**2 + Q_2**2    )\n",
-      "print 'Desired kVA rating of Synchronous motor  is %.3f kVA'%(S_2)\n",
-      "\n",
-      "phi_2 =  math.atan (Q_2/P_2)\n",
-      "print 'Power factor of synchronous motor is %.4f LEADING'%(math.cos(phi_2))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Desired kVA rating of Synchronous motor  is 669.353 kVA\n",
-        "Power factor of synchronous motor is 0.3283 LEADING\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.6 Page no :  47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "output_power = 37.3*1000 \t\t\t#Watts on full load\n",
-      "Z_s = complex(0.2,1.6) \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "theta = (math.pi/180)*angle \t\t\t#phase mag returns the angle in degrees and not raidians\n",
-      "phi = math.acos(0.9) \t\t\t#leading\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "eta = 88. \t\t\t#efficiency in percentage\n",
-      "\n",
-      "# Calculations\n",
-      "input_power = 100*output_power/eta\n",
-      "I_L = input_power/(math.sqrt(3)*V_L*math.cos(phi))\n",
-      "I_aph = I_L\n",
-      "E_Rph = I_aph*abs(Z_s)\n",
-      "\n",
-      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
-      "E_line = math.sqrt(3)*E_bph\n",
-      "\n",
-      "# Results\n",
-      "print 'Induced EMF is %.2f V and its line value is %.2f V'%(E_bph,E_line)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Induced EMF is 285.65 V and its line value is 494.75 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7  Page no :  48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "input_power = 20.*1000 \n",
-      "R_a = 0.\n",
-      "X_s = 4. \t\t\t#armature reactance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s)   \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag))\n",
-      "\n",
-      "theta = (math.pi/180)*angle  \t\t\t#phase mag returns the angle in degrees and not raidians\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "E_bline = 550. \t\t\t#star connection\n",
-      "E_bph = E_bline/math.sqrt(3)\n",
-      "\n",
-      "# Calculations\n",
-      "I_a_cos_phi = input_power/(math.sqrt(3)*V_L)  \t\t\t#product of I_a and math.cos(phi)\n",
-      "I_a_sin_phi =  (   math.sqrt(E_bph**2- (abs(Z_s)*I_a_cos_phi)**2   )  -V_ph )/abs(Z_s)\t\t\t#from triangle DAB\n",
-      "phi = math.atan(I_a_sin_phi/I_a_cos_phi)\n",
-      "I_a = I_a_cos_phi/math.cos(phi)        \n",
-      "\n",
-      "# Results\n",
-      "print 'Motor power fctor is %.3f Leading'%(cos(phi))\n",
-      "print 'Current drawn by the motor is %.2f A'%(I_a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Motor power fctor is 0.872 Leading\n",
-        "Current drawn by the motor is 33.11 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.8 Page no :  50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3300.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "R_a = 2.\n",
-      "X_s = 18. \t\t\t#armature reactance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degrees not radians\n",
-      "E_bline = 3800.\n",
-      "E_bph = E_bline/math.sqrt(3)\n",
-      "\n",
-      "#part(i)\n",
-      "P_m_max  =  (E_bph*V_ph/abs(Z_s))- (E_bph**2/abs(Z_s))*math.cos(theta)\n",
-      "print 'i)Max total mechanical power developed that motor can develop is %.2f W per phase'%(P_m_max)\n",
-      "\n",
-      "#part(ii)\n",
-      "#from phasor diagram applying math.comath.sine rule to triangle OAB\n",
-      "E_Rph = math.sqrt(  E_bph**2 + V_ph**2 -2*E_bph*V_ph*math.cos(theta) ) \n",
-      "I_aph =  E_Rph/abs(Z_s)\n",
-      "print 'ii)Current at max power developed is %.1f A'%(I_aph)\n",
-      "\n",
-      "copper_loss = 3* I_aph**2 * R_a\n",
-      "P_in_max_total = 3 * P_m_max \t\t\t#input power at max power developed\n",
-      "total_P_in =  P_in_max_total + copper_loss \t\t\t#total input power \n",
-      "pf = total_P_in/(math.sqrt(3)*I_aph*V_L)\n",
-      "print 'Power factor at max power developed is %.3f leading'%(pf)\n",
-      "\n",
-      "\n",
-      "# 'Answer in part1 mismatched because of improper approximation in book'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Max total mechanical power developed that motor can develop is 201452.30 W per phase\n",
-        "ii)Current at max power developed is 151.4 A\n",
-        "Power factor at max power developed is 0.857 leading\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.9  Page no :  51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 500.\n",
-      "R_a = 0.03\n",
-      "X_s = 0.3 \t\t\t#armature reactance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degrees not radians\n",
-      "phi = math.acos(0.8)\n",
-      "eta = 93/100.\n",
-      "output_power = 100.*746\n",
-      "input_power = output_power/eta\n",
-      "I_L = input_power/(math.sqrt(3)*V_L*math.cos(phi))\n",
-      "I_aph = I_L\n",
-      "E_Rph = I_aph*abs(Z_s)\n",
-      "\n",
-      "#from the phasor diagram\n",
-      "E_bph  =    math.sqrt( E_Rph**2 + (V_L/math.sqrt(3))**2 - 2*E_Rph*(V_L/math.sqrt(3))*math.cos(phi+theta)  ) \n",
-      "\n",
-      "cu_losses = 3*(I_aph)**2*R_a \t\t\t#total copper losses\n",
-      "P_m =  input_power - cu_losses  \t\t\t#total mechanical power developed\n",
-      "\n",
-      "print 'EMF developed per phase is %.4f V \\nTotal mechanical power developed is %.1f watts'%(E_bph,P_m)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "EMF developed per phase is 308.1880 V \n",
-        "Total mechanical power developed is 79008.6 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.10  Page no :  53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 415.\n",
-      "V_ph = V_L \t\t\t#due to delta connection\n",
-      "E_bline = 520.\n",
-      "R_a = 0.5\n",
-      "X_s = 4. \t\t\t#armature reactance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = (math.pi/180.)*angle \t\t\t#phasemag returns angle in degrees not radians\n",
-      "\n",
-      "delta = theta \t\t\t#for maximum power\n",
-      "P_m_max  =  (E_bline*V_ph/abs(Z_s))- (E_bline**2/abs(Z_s))*math.cos(theta)\n",
-      "P_m_max_total =  3* P_m_max\n",
-      "fi_loss = 1000.  \t\t\t#frictional and iron losses\n",
-      "P_out_total  =  P_m_max_total-fi_loss  \n",
-      "\n",
-      "HP_output =  P_out_total/746 \t\t\t#converting watts to horse power\n",
-      "print 'HP output for maximum power output is %.2f HP'%(HP_output)\n",
-      "\n",
-      "#from the phasor diagram\n",
-      "E_Rph = math.sqrt(  E_bline**2 + V_ph**2 -2*E_bline*V_ph*math.cos(delta) ) \n",
-      "I_aph =  E_Rph/abs(Z_s)\n",
-      "I_L = I_aph*math.sqrt(3)\n",
-      "print 'Line current is %f A'%(I_L)\n",
-      "cu_loss_total = 3*(I_aph)**2*R_a  \t\t\t#total copper losses\n",
-      "input_power = P_m_max_total+ cu_loss_total\n",
-      "pf = input_power/(math.sqrt(3)*I_L*V_L)  \t\t\t#leading\n",
-      "print 'Power factor for maximum power output is %.2f leading '%(pf)\n",
-      "\n",
-      "eta = 100*P_out_total /input_power\n",
-      "print 'Efficiency for maximum power output is %.2f percent'%(eta)\n",
-      "\n",
-      "# 'Answer might mismatch because of improper approximation done in book'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "HP output for maximum power output is 180.48 HP\n",
-        "Line current is 268.015479 A\n",
-        "Power factor for maximum power output is 0.89 leading \n",
-        "Efficiency for maximum power output is 78.48 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.11  Page no :  54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "P = 8.\n",
-      "f = 50. \t\t\t#Pole and frequency\n",
-      "N_s = 120.*f/P \t\t\t#synchronous speed\n",
-      "V_L = 6.6*10**3 \n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "Z_s = complex(0.66,6.6) \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degree  not radians\n",
-      "E_bph = 4500.\n",
-      "input_power = 2500.*10**3\n",
-      "I_a_cosphi = input_power/(math.sqrt(3)*V_L)   \t\t\t#Its product of I_a and math.cos(phi);I_a = I_l for star conneted load\n",
-      "\n",
-      "# Calculations\n",
-      "#applying math.comath.sine rule to triangle ABC from phasor diagram and solve \n",
-      "#math.tan(phi)**2 + 5.2252 math.tan(phi)-2.2432 = 0\n",
-      "p = [1, 5.2252, -2.2432]\n",
-      "tan_phi = roots(p)\n",
-      "phi = math.atan(tan_phi[1])\n",
-      "pf = math.cos(phi)\n",
-      "I_a =  I_a_cosphi/ math.cos(phi)\n",
-      "\n",
-      "#apply math.sine rule to triangle ABC\n",
-      "delta =    math.asin(I_a*abs(Z_s)*math.sin(theta+phi)/E_bph)\n",
-      "P_m = 3*E_bph*I_a*math.cos(delta+phi)\n",
-      "T_g  =  P_m/(2*math.pi*N_s/60)\n",
-      "\n",
-      "# Results\n",
-      "print 'i)Torque developed is %f N-m'%(T_g)\n",
-      "print 'ii)Input current is %.4f A'%(I_a)\n",
-      "print 'iii)Power factor is %.4f leading'%(pf)\n",
-      "print 'iv)Power angle is %.2f degrees '%((180/math.pi)*delta)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Torque developed is 30437.047497 N-m\n",
-        "ii)Input current is 235.4472 A\n",
-        "iii)Power factor is 0.9288 leading\n",
-        "iv)Power angle is 19.48 degrees \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.12  Page no :  57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import roots\n",
-      "\n",
-      "# Variables\n",
-      "input_power = 15.*10**3\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "E_b = 480.\n",
-      "E_bph = E_b/math.sqrt(3)\n",
-      "Z_s = complex(1,5) \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degree  not radians\n",
-      "\n",
-      "# Calculations\n",
-      "I_a_cosphi = input_power/(math.sqrt(3)*V_L)  \t\t\t#product of I_a & math.cos(phi)\n",
-      "#Applying math.comath.sine rule to triangle OAB and solving\n",
-      "#math.tan(phi)**2+ 4.101*math.tan(phi)-1.7499 = 0\n",
-      "p = [1,4.101,-1.7449]\n",
-      "tan_phi = roots(p)\n",
-      "phi = math.atan(tan_phi[1])  \t\t\t#ignoring negative vaule\n",
-      "I_a =  I_a_cosphi/ math.cos(phi)\n",
-      "\n",
-      "#applying math.sine rule to Triangle OAB\n",
-      "delta = math.asin( I_a*abs(Z_s)* math.sin(theta+phi)/E_bph )\n",
-      "\n",
-      "# Results\n",
-      "print 'Load angle is %.1f degrees'%(delta*180/math.pi)\n",
-      "print 'Armature current is %.4f A'%(I_a)\n",
-      "print 'Power factor is %.3f leading'%(cos(phi))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load angle is 24.9 degrees\n",
-        "Armature current is 23.2283 A\n",
-        "Power factor is 0.932 leading\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.13  Page no :  59"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import roots\n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "E_b = 460.\n",
-      "E_bph = E_b/math.sqrt(3)\n",
-      "input_power = 3.75*10**3\n",
-      "Z_s = complex(1,8) \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = math.degrees(angle) \t\t\t#phasemag returns angle in degree ,not radians\n",
-      "I_L_cos_phi  =  input_power/(math.sqrt(3)*V_L)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#Applying math.comath.sine rule to triangle OAB and solving further\n",
-      "#math.tan(phi)**2 + 458.366*math.tan(phi) -450.65  = 0 \n",
-      "p = [1,458.366-450.65]\n",
-      "tan_phi = roots(p)\n",
-      "\n",
-      "phi = math.atan(tan_phi[0])  \t\t\t#ignoring negative value\n",
-      "print 'Required power factor is %.4f leading'%(cos(phi))\n",
-      "I_L = I_L_cos_phi /math.cos(phi)\n",
-      "print 'Required current is %.4f A'%(I_L)\n",
-      "\n",
-      "# roots() python gives some different answer. Kindly check."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required power factor is 0.1285 leading\n",
-        "Required current is 42.1134 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.14  Page no :  60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "#subscript 1 indicates induction motor 1\n",
-      "P_1 = 350.\n",
-      "phi_1 = math.acos(0.7071) \t\t\t#lagging\n",
-      "Q_1 = P_1*math.tan(phi_1)\t\t\t#from power triangle\n",
-      "\n",
-      "#subscript 2 indicates induction motor 2\n",
-      "P_2 = 190.\n",
-      "\n",
-      "# Calculations\n",
-      "#subscript T indicates total\n",
-      "P_T = P_1+P_2\n",
-      "phi_T = math.acos(0.9) \t\t\t#lagging\n",
-      "Q_T = P_T*math.tan(phi_T)\n",
-      "\n",
-      "Q_2 = Q_T-Q_1\n",
-      "kva_rating = math.sqrt(P_2**2+ Q_2**2)\n",
-      "\n",
-      "# Results\n",
-      "print 'kVA rating of synchronous motor is %.2f kVA'%(kva_rating)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "kVA rating of synchronous motor is 209.59 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.15  Page no :  61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "Pole = 6.\n",
-      "f = 50.\n",
-      "R_a = 0.2\n",
-      "X_s = 3. \t\t\t#armature reactance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag/Z_s.real)       # calculates angle in radians\n",
-      "\n",
-      "N_s = 120*f/Pole  \t\t\t#synchronous speed\n",
-      "\n",
-      "# Calculations\n",
-      "#subscript 1` refers to load 1\n",
-      "I_a1 = 20.\n",
-      "phi_1 = math.acos(1)\n",
-      "E_R1 =  I_a1* abs(Z_s)\n",
-      "E_bph  =   math.sqrt( E_R1**2 + V_ph**2 - 2*E_R1*V_ph*math.cos(phi_1+theta)  )\n",
-      "\n",
-      "#subscript 2` refers to load 2\n",
-      "I_a2 = 60.\n",
-      "E_R2 =  I_a2* abs(Z_s)\n",
-      "phi_2 =  math.acos ((E_R2**2 + V_ph**2 -E_bph**2 )/(2*E_R2*V_ph))   -theta   \t\t\t#new power factor\n",
-      "\n",
-      "input_power = math.sqrt(3)*V_L*I_a2*math.cos(phi_2)\n",
-      "cu_loss = 3*I_a2**2*R_a\n",
-      "P_m = input_power-cu_loss\n",
-      "T_g  = P_m /(2*math.pi*N_s/60)   \t\t\t#gross mechanical power developed\n",
-      "\n",
-      "# Results\n",
-      "print 'Gross torque developed is %.4f N-m and new power factor is %.4f lagging'%(T_g,cos(phi_2))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Gross torque developed is 357.1971 N-m and new power factor is 0.9518 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.16  Page no :  63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3300.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "E_bph = V_ph\n",
-      "Z_s = complex(0.5,5) \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degrees, not radians\n",
-      "P = 8.\n",
-      "f = 50. \t\t\t#pole and frequency\n",
-      "delta_mech = 3. \t\t\t#mechanical angle in degrees by which rotor is behind\n",
-      "delta_elec = (P/2)*delta_mech  \t\t\t#delta mech converted to electrical degrees\n",
-      "E_Rph = math.sqrt(  E_bph**2 + V_ph**2 -2*E_bph*V_ph*math.cos(math.radians(delta_elec) )) \n",
-      "I_aph =  E_Rph/abs(Z_s)\n",
-      "\n",
-      "#from the phasor diagram \n",
-      "phi = theta- math.asin( math.sin(math.radians(delta_elec))*E_bph/E_Rph )\n",
-      "pf = math.cos(phi)\n",
-      "print 'power factor of the motor is %.5f lagging'%(pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "power factor of the motor is 0.99999 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.17  Page no :  64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "E_bph = V_ph\n",
-      "P = 4.\n",
-      "f = 50.\t\t\t#Pole and frequency\n",
-      "delta_mech = 4*(math.pi/180) \t\t\t#mechanical angle in degrees by which rotor is behind\n",
-      "delta_elec =  delta_mech *(P/2) \t\t\t#delta_mech convertd to electrical degrees\n",
-      "Z_s = complex(0,2) \t\t\t#synchronous impedance\n",
-      "\n",
-      "# Calculations\n",
-      "#referring to phasor diagram\n",
-      "BC =   E_bph*math.sin(delta_elec)\n",
-      "AB =   E_bph\n",
-      "OA =   V_ph\n",
-      "\n",
-      "AC =   math.sqrt(AB**2-BC**2)\n",
-      "OC =   OA-AC\n",
-      "phi = math.atan(OC/BC)\n",
-      "OB = math.sqrt(OC**2 + BC**2)\n",
-      "I_a = OB/abs(Z_s)\n",
-      "\n",
-      "# Results\n",
-      "print 'Armature current drawn by the motor is %.4f A'%(I_a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature current drawn by the motor is 16.1096 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.18  Page no :  65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "input_power = 5472.\n",
-      "Z_s = complex(0,10)  \t\t\t#synchronous impedance\n",
-      "I_L_cosphi = input_power/(math.sqrt(3)*V_L) \t\t\t#product of I_L & math.cos(phi)\n",
-      "BC = 10*I_L_cosphi\n",
-      "AB = V_ph\n",
-      "OA = V_ph\n",
-      "\n",
-      "# Calculations\n",
-      "#from Triangle ABC in phasor diagram\n",
-      "AC  =  math.sqrt(AB**2- BC**2)\n",
-      "OC  =  OA - AC\n",
-      "\n",
-      "\t\t\t#from Triangle OCB \n",
-      "OB = math.sqrt( OC**2+ BC**2 )\n",
-      "E_Rph  =  OB\n",
-      "I_L = E_Rph/abs(Z_s)\n",
-      "\n",
-      "phi = math.atan(OC/BC)\n",
-      "pf = math.cos(phi)\n",
-      "delta = math.atan(BC/AC)  \t\t\t#load angle\n",
-      "\n",
-      "# Results\n",
-      "print 'Power factor is %.4f lagging'%(pf)\n",
-      "print 'Load angle is %.0f degrees'%(delta*180/math.pi)\n",
-      "print 'Armature current is %.3f A'%(I_L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power factor is 0.9848 lagging\n",
-        "Load angle is 20 degrees\n",
-        "Armature current is 8.020 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.19  Page no :  67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6600.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "Z_s = complex(2,20) \t\t\t#synchronous impedance\n",
-      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
-      "\n",
-      "theta = (math.pi/180) * angle \t\t\t#phasemag returns angle in degrees not radians\n",
-      "P_1 = 1000*10**3\n",
-      "P_2 = 1500*10**3\n",
-      "phi_1 = math.acos(0.8)  \t\t\t#leading\n",
-      "\n",
-      "# Calculations\n",
-      "I_L1 = P_1/(math.sqrt(3)*V_L*math.cos(phi_1))\n",
-      "I_a1ph = I_L1\n",
-      "E_R1ph = I_a1ph*abs(Z_s)\n",
-      "E_bph =   math.sqrt(    V_ph**2 + E_R1ph** -2*V_ph*E_R1ph*math.cos(theta+phi_1)  )\n",
-      "I_a2_cosphi_2 = P_2/(math.sqrt(3)*V_L)\n",
-      "\n",
-      "#Refer to the phasor diagram and solving for I_y\n",
-      "#404I_y**2 -152399.968 I_y -4543000 = 0\n",
-      "p = [404, -152399.968, -4543000]\n",
-      "ans = roots(p)\n",
-      "I_y = abs(ans[1]) \t\t\t#becuase root 1 is too high and root is -ve\n",
-      "\n",
-      "I_a2 = complex(I_a2_cosphi_2,I_y)\n",
-      "phi_2 = math.degrees(math.atan(I_a2.imag/I_a2.real))\n",
-      "\n",
-      "print 'Required power factor is %.3f leading'%(math.cos(math.radians(phi_2)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required power factor is 0.978 leading\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.20  Page no :  69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "V_L = 2300.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_L = 200.\n",
-      "I_a = I_L\n",
-      "Z_s = complex(0.2,2.2) \t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag/Z_s.real)\n",
-      "phi = math.acos(0.5)\n",
-      "\n",
-      "# Calculations\n",
-      "E_Rph = I_a*abs(Z_s)\n",
-      "E_bph  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(phi+theta)  ) \n",
-      "\n",
-      "# Results\n",
-      "print 'Generated EMF per phase is %.3f V'%(E_bph)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Generated EMF per phase is 1708.045 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.21  Page no :  69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3300.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "E_bline = 3800.\n",
-      "E_bph = E_bline/math.sqrt(3)\n",
-      "\n",
-      "R_a = 2.\n",
-      "X_s = 18. \t\t\t#armature resistance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag/Z_s.real)\n",
-      "\n",
-      "#part(i)\n",
-      "P_m_max  =  (E_bph*V_ph/abs(Z_s))- (E_bph**2/abs(Z_s))*math.cos(theta)  \t\t\t#maximum total mechanical power\n",
-      "print 'i)Maximum total mechanical power that the motor can develop is %.2f W per phase'%(P_m_max )\n",
-      "#part(ii)\n",
-      "delta = theta \t\t\t#for max P_m\n",
-      "E_Rph = math.sqrt(  E_bph**2 + V_ph**2 -2*E_bph*V_ph*math.cos(delta) ) \n",
-      "I_aph =  E_Rph/abs(Z_s)\n",
-      "print 'ii)Current at maximum power developed is %.1f A'%(I_aph)\n",
-      "cu_loss_total  =  3*I_aph**2*R_a  \t\t\t#total copper loss\n",
-      "P_m_max_total = 3*P_m_max \t\t\t#total maximum total mechanical power\n",
-      "P_in_total  =  P_m_max_total+ cu_loss_total  \t\t\t#total input power\n",
-      "\n",
-      "pf = P_in_total/(math.sqrt(3)*V_L*I_aph)\n",
-      "print '    Power factor at maximum power developed is %.3f leading'%(pf)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Maximum total mechanical power that the motor can develop is 201452.30 W per phase\n",
-        "ii)Current at maximum power developed is 151.4 A\n",
-        "    Power factor at maximum power developed is 0.857 leading\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.22  Page no :  71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#subscript 1 refers to load 1\n",
-      "I_1 = 18.\n",
-      "phi_1 = math.acos(0.8)\n",
-      "V_L = 440.\n",
-      "S_1 = math.sqrt(3)*I_1*V_L /1000  \t\t\t#kVA for load 1\n",
-      "P_1 = S_1*math.cos(phi_1)\n",
-      "Q_1 = S_1*math.sin(phi_1)\n",
-      "\n",
-      "# Calculations\n",
-      "P_out = 6.\n",
-      "eta_motor = 88./100\n",
-      "P_2 = P_out/eta_motor\n",
-      "\n",
-      "P_T = P_1+P_2\n",
-      "phi_T = math.acos(1)  \t\t\t#total power factor angle\n",
-      "Q_T = P_T*math.tan(phi_T)\n",
-      "\n",
-      "Q_2 =   Q_T - Q_1   \t\t\t#kVAR supplied by motor\n",
-      "#this will have  a negative sign just indicating its leading nature \n",
-      "phi_2 = math.atan(abs(Q_2)/P_2)\n",
-      "pf = math.cos(phi_2)  \t\t\t#leading\n",
-      "S_2 = P_2/math.cos(phi_2)  \t\t\t#kVA input to the motor\n",
-      "\n",
-      "# Results\n",
-      "print 'kVA input to the motor is %.3f kVA '%(S_2)\n",
-      "print 'Power factor when driving a 6kW mechanical load is %.4f leading'%(pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "kVA input to the motor is 10.688 kVA \n",
-        "Power factor when driving a 6kW mechanical load is 0.6379 leading\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.23  Page no :  72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "output_power = 8.*10**3\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "R_a = 0.\n",
-      "X_s = 8.\t\t\t#armature resistance and syncronous reactance\n",
-      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
-      "eta = 88./100\n",
-      "input_power = output_power/eta\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#minimum current occurs at max power factors\n",
-      "phi = math.acos(1)\n",
-      "I_a_min = input_power/(math.sqrt(3)*V_L*math.cos(phi)) \t\t\t#required minimum current \n",
-      "print 'Minimum current is %.3f A'%(I_a_min)\n",
-      "E_R =  I_a_min * abs(Z_s)\n",
-      "E_bph  =    math.sqrt( E_R**2 + V_ph**2 - 2*E_R*V_ph*math.cos(phi+theta)  ) \n",
-      "print 'Induced EMF at full-load is %.3f V'%(E_bph)\n",
-      "\n",
-      "# note : rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum current is 13.122 A\n",
-        "Induced EMF at full-load is 241.534 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.24  Page no :  73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "R_a = 0.8\n",
-      "X_s = 5.\n",
-      "Z_s = complex(R_a,X_s) \t\t\t#armature resistance and syncronous reactance\n",
-      "theta = math.atan(Z_s.imag/Z_s.real)  \t\t\t# returns angle in radians \n",
-      "alpha = (math.pi/2) - theta\n",
-      "V_t = 3300/math.sqrt(3)\n",
-      "P_e_in = 800./(3) \t\t\t#per phase\n",
-      "phi = math.acos(0.8)  \t\t\t#leading\n",
-      "Q_e_in = -P_e_in*math.tan(phi)\n",
-      "\n",
-      "# Calculations\n",
-      "# using the following equation\n",
-      "# P_e_in =  V_t**2*R_a/(abs(Z_s))**2   +   V_t*E_b*math.sin(delta-alpha)/abs(Z_S)\n",
-      "# Q_e_in =  V_t**2*X_s/(abs(Z_s))**2   -   V_t*E_b*math.cos(delta-alpha)/abs(Z_S)\n",
-      "E_b_sin_delta_minus_9  =  407.2\n",
-      "E_b_cos_delta_minus_9  = 2413.6\n",
-      "#solving further\n",
-      "delta  =  math.atan(E_b_sin_delta_minus_9/E_b_cos_delta_minus_9 ) + 9\n",
-      "E_b = E_b_sin_delta_minus_9/math.sin(math.radians(delta-9))\n",
-      "\n",
-      "P_e_in_new  =  1200*10**3/3\n",
-      "# using the following equation again\n",
-      "# P_e_in =  V_t**2*R_a/(abs(Z_s))**2   +   V_t*E_b*math.sin(delta-alpha)/abs(Z_S)\n",
-      "# Q_e_in =  V_t**2*X_s/(abs(Z_s))**2   -   V_t*E_b*math.cos(delta-alpha)/abs(Z_S)\n",
-      "\n",
-      "alpha  = delta - math.asin(math.radians(P_e_in_new - V_t**2*R_a/(abs(Z_s))**2 ) / (V_t*E_b/abs(Z_s)))\n",
-      "Q_e_in_new =  V_t**2*X_s/(abs(Z_s))**2   -   V_t*E_b*math.cos(math.radians(delta - alpha))/abs(Z_s)\n",
-      "\n",
-      "pf = math.cos ( math.atan(abs(Q_e_in_new/P_e_in_new)))\n",
-      "\n",
-      "# Results\n",
-      "print 'New power factor is %.2f leading '%(pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "New power factor is 0.01 leading \n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.25  Page no :  74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6.6*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "P_in = 900.*10**3\n",
-      "R_a = 0.\n",
-      "X_s = 20. \t\t\t#armature resistance and synchronous reactance \n",
-      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
-      "E_b_L = 8.6*10**3\n",
-      "E_bph = E_b_L/math.sqrt(3) \n",
-      "\n",
-      "# Calculations\n",
-      "#refer to phasor diagram\n",
-      "OA = V_ph\n",
-      "AB = E_bph \t\t\t#OB =  E_Rph\n",
-      "\n",
-      "I_a_cosphi = P_in/(math.sqrt(3)*V_L)  \t\t\t#I_a*math.cos(phi)\n",
-      "BC = I_a_cosphi*abs(Z_s) \t\t\t#BC is a vector in phasor diagram\n",
-      "\n",
-      "OC = math.sqrt(AB**2 -BC**2  )- OA  \t\t\t#from phasor diagram\n",
-      "I_a_sinphi = OC/abs(Z_s)  \t\t\t#product of I_a and math.sin(phi)\n",
-      "phi =  math.atan (I_a_sinphi/I_a_cosphi)\n",
-      "I_a = I_a_cosphi/math.cos(phi) \t\t\t#product of I_a and math.cos(phi)\n",
-      "\n",
-      "# Results\n",
-      "print 'Motor current is %.3f A'%(I_a)\n",
-      "print 'Power factor of motor is %f leading'%(cos(phi))\n",
-      "print 'Note:There is slight mismatch in answer due to the approximation made during I_a* sinphi calculation'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Motor current is 90.643 A\n",
-        "Power factor of motor is 0.868564 leading\n",
-        "Note:There is slight mismatch in answer due to the approximation made during I_a* sinphi calculation\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.26  Page no :  75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "\n",
-      "# Variables\n",
-      "#subscipt 1 refers to factory load\n",
-      "P_1 = 1800.\n",
-      "phi_1 = math.acos(0.6) \t\t\t#lagging\n",
-      "Q_1 = P_1*math.tan(phi_1)\n",
-      "\n",
-      "#Subscript 2 refers to synchronous condenser\n",
-      "P_2 = 0.\n",
-      "\n",
-      "# Calculations\n",
-      "#Subscript T refers to combination of condenser and factory load\n",
-      "P_T = P_1+P_2\n",
-      "phi_T = math.acos(0.95) \t\t\t#lagging\n",
-      "Q_T = P_T*math.tan(phi_T)\n",
-      "\n",
-      "kva_rating = math.sqrt(P_T**2+ Q_T**2)\n",
-      "\n",
-      "Q_2 = Q_T - Q_1\n",
-      "\n",
-      "# Results\n",
-      "print 'i)kVA rating of synchronous condender is %.3f kVA    Minus sign indicates leading nature'%(Q_2)\n",
-      "print 'ii)kVA rating of total factory is %.4f kVA'%(kva_rating)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)kVA rating of synchronous condender is -1808.369 kVA    Minus sign indicates leading nature\n",
-        "ii)kVA rating of total factory is 1894.7368 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.27  Page no :  77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "I_1 = 35.\n",
-      "phi_1 = math.acos(0.8)\n",
-      "V_L = 440.\n",
-      "S_1 = math.sqrt(3)*I_1*V_L /1000 \t\t\t#in kVA\n",
-      "\n",
-      "\n",
-      "# Calculations\n",
-      "P_1 = S_1*math.cos(phi_1)\n",
-      "Q_1 = S_1*math.sin(phi_1)\n",
-      "\n",
-      "P_out = 12. \t\t\t#motor load\n",
-      "eta_motor = 85./100\n",
-      "P_2 = P_out/eta_motor\n",
-      "\n",
-      "P_T = P_1 + P_2\n",
-      "phi_T = math.acos(1)\n",
-      "Q_T = P_T * math.tan(phi_T)\n",
-      "\n",
-      "\n",
-      "Q_2 = Q_T - Q_1  \t\t\t#kVA supplied by motor\n",
-      "#negative sign of Q_2 indicates its leading nature\n",
-      "phi_2 =  math.atan(abs(Q_2)/P_2)\n",
-      "S_2 = P_2/math.cos(phi_2)\n",
-      "\n",
-      "# Results\n",
-      "print 'Power factor when motor supplies 12kW load is %.4f leading'%(math.cos(phi_2))\n",
-      "print 'kVA input to the motor is %.3f kVA'%(S_2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power factor when motor supplies 12kW load is 0.6615 leading\n",
-        "kVA input to the motor is 21.341 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.28  Page no :  78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "Z_s = complex(0.5,4)  \t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag/Z_s.real)  \t\t\t# returns angle in radians \n",
-      "\n",
-      "I_aph = 60.\n",
-      "phi = math.acos(0.866) \t\t\t#leading\n",
-      "power_losses = 2*10**3\n",
-      "\n",
-      "# Calculations\n",
-      "E_bph  =    math.sqrt( (I_aph*abs(Z_s))**2 + (V_ph)**2 - 2*(I_aph*abs(Z_s))*(V_ph)*math.cos(phi+theta)  ) \n",
-      "delta = theta \t\t\t#for P_m_max\n",
-      "P_m_max  =  (E_bph*V_ph/abs(Z_s))- (E_bph**2/abs(Z_s))*math.cos(delta)\n",
-      "P_m_max_total =  3 * P_m_max\n",
-      "P_out_max =   P_m_max_total- power_losses\n",
-      "\n",
-      "# Results\n",
-      "print 'Maximum power output is %.4f kW'%(P_out_max*10**-3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum power output is 51.3899 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.29  Page no :  79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 6.6*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "I_L = 50.\n",
-      "I_aph = I_L\n",
-      "Z_s = complex(1.5,8) \t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag/Z_s.real)  \t\t\t# returns angle in radians \n",
-      "E_Rph = I_aph*abs(Z_s)\n",
-      "\n",
-      "# Calculations and Results\n",
-      "#part(i)\n",
-      "phi = math.acos(0.8)\n",
-      "P_in =  math.sqrt(3)*V_L*I_L*math.cos(phi)   \t\t\t#for both lag and lead supplied power will be the same\n",
-      "print 'i)Power supplied to the motor is %.3f kW'%(P_in*10**-3)\n",
-      "#part(ii)\n",
-      "E_bph_lag  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(theta-phi)  )  \t\t\t#for lagging power factor\n",
-      "#Note that E_bph_lag > V_ph\n",
-      "print 'ii)Induced EMF for 0.8 power factor lag is %.3f V'%(E_bph_lag)\n",
-      "E_bph_lead  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(theta+phi)  )  \t\t\t#for leading power factor\n",
-      "#Note that E_bph_lead < V_ph\n",
-      "print '    Induced EMF for 0.8 power factor lead is %.3f V'%(E_bph_lead)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Power supplied to the motor is 457.261 kW\n",
-        "ii)Induced EMF for 0.8 power factor lag is 3521.267 V\n",
-        "    Induced EMF for 0.8 power factor lead is 4007.170 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.30  Page no :  80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 400.\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "P_out = 7.5*735.5\n",
-      "eta = 85./100 \t\t\t#efficiency\n",
-      "R_a = 0.\n",
-      "X_s = 10. \t\t\t#armature resistance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
-      "\n",
-      "# Calculations and Results\n",
-      "P_in = P_out/eta\n",
-      "phi = math.acos(1) \t\t\t#for mimimum current power factor is maximum\n",
-      "I_L = P_in/(math.sqrt(3)*V_L*math.cos(phi))  \n",
-      "I_aph = I_L\n",
-      "print 'Minimum current is %.3f A at full load condition '%(I_L)\n",
-      "\n",
-      "E_Rph =  I_aph*abs(Z_s)\n",
-      "E_bph  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(phi+theta)  ) \n",
-      "print 'and corresponding EMF is %.4f V'%(E_bph)\n",
-      "\n",
-      "# note : rounding off error.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum current is 9.367 A at full load condition \n",
-        "and corresponding EMF is 240.4216 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.31  Page no :  80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "# Variables\n",
-      "V_L = 3.3*10**3\n",
-      "V_ph = V_L/math.sqrt(3)\n",
-      "V_t = V_ph\n",
-      "Pole = 24.\n",
-      "f = 50.  \t\t\t#Pole and frequency\n",
-      "P = 1000.*10**3\n",
-      "R_a = 0\n",
-      "X_s = 3.24 \t\t\t#armature resistance and synchronous reactance\n",
-      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
-      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
-      "phi = math.acos(1)\n",
-      "I_aph = P/(math.sqrt(3)*V_L*math.cos(phi))\n",
-      "\n",
-      "# Calculations\n",
-      "E_Rph = I_aph*abs(Z_s)\n",
-      "E_bph  =   math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(phi+theta)  ) \n",
-      "\n",
-      "P_m_max = 3*(E_bph*V_ph/abs(Z_s)) \t\t\t#maximum power that can be delivered\n",
-      "N_s = 120*f/Pole \t\t\t#synchronous speed\n",
-      "T_max = P_m_max /(2*math.pi*N_s/60) \t\t\t#maximum torque that can be developed\n",
-      "\n",
-      "# Results\n",
-      "print 'Maximum power and torque the motor can deliver is %.3f kW and %.2f *10**3 Nm respectively'%(P_m_max*10**-3,T_max/1000)\n",
-      "\n",
-      "# rounding off error."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum power and torque the motor can deliver is 3211.633 kW and 122.68 *10**3 Nm respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
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diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/README.txt b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/README.txt
new file mode 100755
index 00000000..9c84b56e
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Hiren Shah
+Course: mca
+College/Institute/Organization: Financial Technology
+Department/Designation: Developer
+Book Title: DC Machines and Synchronous Machines
+Author: U. A. Bakshi and M. V. Bakshi
+Publisher: Technical Publications, Pune
+Year of publication: 2008
+Isbn: 9788184314830
+Edition: 1
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch1_2.ipynb b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch1_2.ipynb
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index 00000000..b98c0ac6
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch1_2.ipynb
@@ -0,0 +1,1625 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ae45c35a90f5d6d72689d00dd327a0dd50ebdc721f0cf7cc3ef6e9742fbf7b87"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : D.C. Generators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1 page no : 17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "p = 4.\n",
+      "Z = 440.\n",
+      "theta = 0.07    # Wb\n",
+      "N = 900.        # rpm\n",
+      "\n",
+      "# calculations and results\n",
+      "E = theta*N*Z/60.\n",
+      "print \"For lap wound : E = %.f V\"%E\n",
+      "\n",
+      "E = theta*N*Z*4/120.\n",
+      "print \"For wave wound : E = %.f V\"%E\n",
+      "\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For lap wound : E = 462 V\n",
+        "For wave wound : E = 924 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Example 1.2 page no : 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "phi = 21.*10**-3  \t\t\t#flux produced by each pole in webers\n",
+      "N = 1120.        \t\t\t#Speed of armature in r.p.m\n",
+      "Coils = 42.\n",
+      "turns_per_coil = 8.\n",
+      "Turns = Coils * turns_per_coil\n",
+      "Z = 2*Turns  \t\t\t#Number of armature conductors\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "A1 = Pole  \t\t\t#no of parallel paths for lap winding\n",
+      "E1 = phi*N*Z*Pole/(60*A1)\n",
+      "print 'i)  e.m.f generated is %.3f V'%(E1)\n",
+      "\n",
+      "#Part(ii)\n",
+      "A2 = 2  \t\t\t#wave winding\n",
+      "E2 = E1 \t\t\t#as mentioned in the question\n",
+      "N2 = E2/(phi*Z*Pole/(60*A2)) \t\t\t#E = phi*N*Z*Pole/(60*A)\n",
+      "print 'ii)  For wave-wound armature,above calculated e.m.f is generated at %.0f r.p.m'%(N2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)  e.m.f generated is 263.424 V\n",
+        "ii)  For wave-wound armature,above calculated e.m.f is generated at 560 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 page no: 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "coils = 12.\n",
+      "commutator_segments = coils\n",
+      "coil_sides = coils*2\n",
+      "Z = coil_sides  \t\t\t#No of conductors\n",
+      "pole_pitch = Z/Pole\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#for Simplex lap winding\n",
+      "y_f = pole_pitch-1\n",
+      "y_b = pole_pitch+1\n",
+      "\n",
+      "y_c = 1  \t\t\t#Note that it's positive and it's progressive type of Simplex lap winding\n",
+      "\n",
+      "\n",
+      "print 'WINDING TABLE:   1<-    8->    3<-   10->    5<-   12-> 7<-   14->    9<-   16->   11<-   18->13<-   20->   15<-   22->   17<-   24->19<-    2->   21<-    4->   23<-    6-> 1                                                  '\n",
+      "print 'Note that <- indicates back connection with  y_back = %.0f and -> indicates front connection with y_front = %.0f'%(y_b,y_f)\n",
+      "print 'Another form of winding table:'\n",
+      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
+      "\n",
+      "print '       1 to 1+7  =  8                        ->        8 to 8-5  =  3'\n",
+      "print '       3 to 3+7  = 10                        ->       10 to 10-5 =  5'\n",
+      "print '       5 to 5+7  = 12                        ->       12 to 12-5 =  7'\n",
+      "print '       7 to 7+7  = 14                        ->       14 to 14-5 =  9'\n",
+      "print '       9 to 9+7  = 16                        ->       16 to 16-5 = 11'\n",
+      "print '      11 to 11+7 = 18                        ->       18 to 18-5 = 13'\n",
+      "print '      13 to 13+7 = 20                        ->       20 to 20-5 = 15'\n",
+      "print '      15 to 15+7 = 22                        ->       22 to 22-5 = 17'\n",
+      "print '      17 to 17+7 = 24                        ->       24 to 24-5 = 19'\n",
+      "print '      19 to 19+7 = 26 = 26-24 = 2              ->        2 to 26-5 = 21'\n",
+      "print '      21 to 21+7 = 28 = 28-24 = 4              ->        4 to 28-5 = 23'\n",
+      "print '      23 to 23+7 = 30 = 30-24 = 6              ->        6 to 30-5 = 25   =  25-24 = 1'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "WINDING TABLE:   1<-    8->    3<-   10->    5<-   12-> 7<-   14->    9<-   16->   11<-   18->13<-   20->   15<-   22->   17<-   24->19<-    2->   21<-    4->   23<-    6-> 1                                                  \n",
+        "Note that <- indicates back connection with  y_back = 7 and -> indicates front connection with y_front = 5\n",
+        "Another form of winding table:\n",
+        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
+        "       1 to 1+7  =  8                        ->        8 to 8-5  =  3\n",
+        "       3 to 3+7  = 10                        ->       10 to 10-5 =  5\n",
+        "       5 to 5+7  = 12                        ->       12 to 12-5 =  7\n",
+        "       7 to 7+7  = 14                        ->       14 to 14-5 =  9\n",
+        "       9 to 9+7  = 16                        ->       16 to 16-5 = 11\n",
+        "      11 to 11+7 = 18                        ->       18 to 18-5 = 13\n",
+        "      13 to 13+7 = 20                        ->       20 to 20-5 = 15\n",
+        "      15 to 15+7 = 22                        ->       22 to 22-5 = 17\n",
+        "      17 to 17+7 = 24                        ->       24 to 24-5 = 19\n",
+        "      19 to 19+7 = 26 = 26-24 = 2              ->        2 to 26-5 = 21\n",
+        "      21 to 21+7 = 28 = 28-24 = 4              ->        4 to 28-5 = 23\n",
+        "      23 to 23+7 = 30 = 30-24 = 6              ->        6 to 30-5 = 25   =  25-24 = 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 page no : 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "Z = 18.               \t\t\t#no of armature conductors\n",
+      "Y_A = (Z+2)/Pole     \t\t\t#For progressive type wave winding,positive sign is used\n",
+      "Y_C = Y_A            \t\t\t#For wave winding\n",
+      "\n",
+      "# Calculations\n",
+      "#Since Y_A = (y_b+y_f)/2, we let y_b = Y_f\n",
+      "y_b = Y_A/2   \t\t\t#say\n",
+      "y_f = y_b\n",
+      "\n",
+      "coils = Z/2\n",
+      "slots = coils\n",
+      "commutator_segments = coils\n",
+      "\n",
+      "print commutator_segments\n",
+      "print 'WINDING TABLE:   1<-    6->  11<-  16->   3<-   8->13<-   18->   5<-  10->  15<-   2-> 7<-   12->  17<-   4->   9<-  14->1                                                  '\n",
+      "\n",
+      "print 'Another form of winding table:'\n",
+      "\n",
+      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
+      "\n",
+      "print '       1 to 1+5   =  6                       ->        6 to 6+5  =  11'\n",
+      "print '      11 to 11+5  = 16                       ->       16 to 16+5 =  21 -18 = 3'\n",
+      "print '       3 to 3+5   =  8                       ->        8 to  8+5 =  13'\n",
+      "print '      13 to 13+5  = 18                       ->       18 to 18+5 =  23 -18 = 5'\n",
+      "print '       5 to 5+5   = 10                       ->       10 to 10+5 =  15'\n",
+      "print '      15 to 15+5  = 20 -18 = 2                 ->        2 to  2+5 =   7'\n",
+      "print '       7 to 7+5   = 12                       ->       12 to 12+5 =  17'\n",
+      "print '      17 to 17+5  = 22 -18 = 4                 ->        4 to  4+5 =   9'\n",
+      "print '       9 to 9+5   = 14                       ->       14 to 14+5 =  19 -18 = 1'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9.0\n",
+        "WINDING TABLE:   1<-    6->  11<-  16->   3<-   8->13<-   18->   5<-  10->  15<-   2-> 7<-   12->  17<-   4->   9<-  14->1                                                  \n",
+        "Another form of winding table:\n",
+        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
+        "       1 to 1+5   =  6                       ->        6 to 6+5  =  11\n",
+        "      11 to 11+5  = 16                       ->       16 to 16+5 =  21 -18 = 3\n",
+        "       3 to 3+5   =  8                       ->        8 to  8+5 =  13\n",
+        "      13 to 13+5  = 18                       ->       18 to 18+5 =  23 -18 = 5\n",
+        "       5 to 5+5   = 10                       ->       10 to 10+5 =  15\n",
+        "      15 to 15+5  = 20 -18 = 2                 ->        2 to  2+5 =   7\n",
+        "       7 to 7+5   = 12                       ->       12 to 12+5 =  17\n",
+        "      17 to 17+5  = 22 -18 = 4                 ->        4 to  4+5 =   9\n",
+        "       9 to 9+5   = 14                       ->       14 to 14+5 =  19 -18 = 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 page no : 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "Z = 480.             \t\t\t#No of armature conductors\n",
+      "I_a = 144.\n",
+      "I = I_a/2           \t\t\t#For wave wound\n",
+      "theta_m = 10.        \t\t\t#lead angle in DEGREES\n",
+      "\n",
+      "# Calculations\n",
+      "amp_turns_PP_d = Z*I*theta_m/360  \t\t\t#demagnetising Ampere-turns per pole\n",
+      "amp_turns_PP_c = Z*I*(1/(2*Pole)-theta_m/360) \t\t\t#cross-magnetising Ampere-turns per pole\n",
+      "\n",
+      "# Results\n",
+      "print 'De-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_d)\n",
+      "print 'Cross-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_c)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "De-magnetising ampere-turns per pole is 960\n",
+        "Cross-magnetising ampere-turns per pole is 3360\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 page no : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 10.\n",
+      "Z = 800.            \t\t\t#No of armature conductors\n",
+      "A = Pole            \t\t\t#For lap wound\n",
+      "ratio = 0.7           \t\t\t#ratio of pole arc to pole pitch\n",
+      "\n",
+      "# Calculations\n",
+      "#amp_turns_PP = ratio*(I_a*Z)/(2*A*P)\n",
+      "turns_PP = ratio*(Z)/(2*A*Pole)   \t\t\t#turns per pole\n",
+      "conductors_PP = turns_PP*2 \t\t\t#multiplied with 2 because 2 conductors form 1 turn\n",
+      "\n",
+      "# Results\n",
+      "print 'Compensating conductors per pole =  %.f'%(conductors_PP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Compensating conductors per pole =  6\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 page no : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "I_L = 150.\n",
+      "A = 4.\n",
+      "N = 1800.   \t\t\t#in rpm\n",
+      "W_b = 1.2   \t\t\t#Brush width\n",
+      "W_m = 0     \t\t\t#width of mica insulation\n",
+      "L = 0.06*10**-3 \t\t\t#Inducmath.tance\n",
+      "segments = 64.\n",
+      "n_s = 1800./60   \t\t\t#in rps and not rpm\n",
+      "v = n_s*segments  \t\t\t#peripheral speed in segments per second\n",
+      "\n",
+      "# Calculations and Results\n",
+      "T_c = (W_b-W_m)/v  \t\t\t#Time of commutation\n",
+      "I = I_L/A  \t\t\t#Current through a conductor\n",
+      "\n",
+      "#Part(i)\n",
+      "E_l = L*2*I/T_c\n",
+      "print 'i)   Reactive voltage using Linear commutation is %.1f V'%(E_l)\n",
+      "\n",
+      "#Part(ii)\n",
+      "E_s = 1.11*L*2*I/T_c\n",
+      "print 'ii)  Reactive voltage using Sinusoidal commutation is %.3f V'%(E_s)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)   Reactive voltage using Linear commutation is 7.2 V\n",
+        "ii)  Reactive voltage using Sinusoidal commutation is 7.992 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 page no : 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V_t = 250.  \t\t\t#Terminal voltage\n",
+      "R_sh = 100.  \t\t\t#resistance of shunt field winding\n",
+      "I_sh = V_t/R_sh \t\t\t#shunt current\n",
+      "R_a = 0.22  \t\t\t#Armature resistance\n",
+      "\n",
+      "# Calculations\n",
+      "P = 5*10**3  \t\t\t#Load power\n",
+      "I_L = P/V_t  \t\t\t#Load current\n",
+      "I_a = I_L+I_sh \t\t\t#armature current\n",
+      "\n",
+      "E = V_t + I_a*R_a  \t\t\t#Induced emf\n",
+      "\n",
+      "# Results\n",
+      "print 'Induced e.m.f to supply the 5kW load is %.2f V'%(E)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Induced e.m.f to supply the 5kW load is 254.95 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9 page no : 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V_t = 250.  \t\t\t#terminal voltage\n",
+      "P = 10.*10**3  \t\t\t#10kW power of generator\n",
+      "I_L = P/V_t \t\t\t#load current\n",
+      "I_a = I_L  \t    \t\t#As seperately excited\n",
+      "V_brush = 2.*2 \t\t\t# 2 * no of brushes\n",
+      "\n",
+      "E = 255.      \t\t\t#on full load\n",
+      "\n",
+      "# Calculations\n",
+      "R_a = (E-V_t-V_brush)/I_a \t\t\t#Because E = V_t+ I_a*R_a  + V_brush\n",
+      "\n",
+      "# Results\n",
+      "print 'Armature resistance of generator is %.3f ohm'%(R_a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature resistance of generator is 0.025 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.10 page no : 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R_a = 0.5\n",
+      "R_se = 0.03  \t\t\t#resimath.tance due to armature and series field winding\n",
+      "V_brush = 2. \t\t\t#brush drop\n",
+      "N = 1500. \t\t\t#generator speed in r.p.m\n",
+      "coils = 540.\n",
+      "turns_per_coil = 6.\n",
+      "\n",
+      "# Calculations\n",
+      "total_turns =  coils*turns_per_coil\n",
+      "Z = 2*total_turns \t\t\t#Total conductors\n",
+      "I_a = 50         \t\t\t#armature current\n",
+      "\n",
+      "phi = 2*10**-3  \t\t\t#flux per pole in webers\n",
+      "E = phi*N*Z/(60)  \t\t\t#A = P for lap-wound and they cancel out\n",
+      "V_t  = E- (I_a*(R_a+R_se)  + V_brush)   \t\t\t#Because E = V_t+ I_a*R_a  + V_brush\n",
+      "\n",
+      "# Results\n",
+      "print 'Terminal voltage is %.1f V'%(V_t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Terminal voltage is 295.5 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11 page no : 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V_t = 225. \t\t\t#voltage across winding\n",
+      "R_a = 0.04 \t\t\t#armature resistance\n",
+      "R_sh = 90 \t\t\t#shunt resistance\n",
+      "R_se = 0.02\t\t\t#resistance of series field winding\n",
+      "I_L = 75. \t\t\t#load current\n",
+      "\n",
+      "# Calculations\n",
+      "#E -I_a*R_a = V_t+I_L*R_se \n",
+      "I_sh = (V_t+I_L*R_se)/R_sh  \t\t\t#current through shunt field winding\n",
+      "\n",
+      "I_a = I_L + I_sh  \t\t\t        #armature current \n",
+      "E = V_t+ I_a*R_a+I_L*R_se  \t\t\t#induced emf\n",
+      "\n",
+      "# Results\n",
+      "print 'Generated voltage is %.1f V'%(E)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Generated voltage is 229.6 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.12 page no : 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R_sh = 53.  \t\t\t#resistance of field winding\n",
+      "V_t = 100.   \t\t\t#terminal voltage \n",
+      "I_sh  = V_t/R_sh \t\t\t#shunt current\n",
+      "I_f = I_sh\n",
+      "R_a = 0.1 \t\t\t#armature resistance\n",
+      "E_o = 143. \t\t\t# for I_sh =  I_f  =  1.8867 as obtained from graph\n",
+      "\n",
+      "# Calculations\n",
+      "I_a = (E_o-V_t)/R_a  \t\t\t#Because E_o = V_t + I_a*R_a\n",
+      "I_L = I_a-I_sh  \t\t\t#no load current\n",
+      "\n",
+      "# Results\n",
+      "print 'Note: Open circuit voltage was obtained as followsE_o = R_sh*I_f   \t\t\t# y = mx+c form with c = 0 and R_sh = 53Hence%( a line with slope 53 through origin is made to intersect OCC at 150 V'\n",
+      "print 'Therefore, Open circuit voltage is 150 V'\n",
+      "print 'No load current is %.4f A    '%(I_L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Note: Open circuit voltage was obtained as followsE_o = R_sh*I_f   \t\t\t# y = mx+c form with c = 0 and R_sh = 53Hence%( a line with slope 53 through origin is made to intersect OCC at 150 V\n",
+        "Therefore, Open circuit voltage is 150 V\n",
+        "No load current is 428.1132 A    \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.13 page no : 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#part(1)\n",
+      "E_o = 240.  \t\t\t#on no-load\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Draw horizontal line from 240 V, to intersect OCC at A. corresponding I_f is 2.25 A\n",
+      "#The slope pf OA is corresponding R_sh\n",
+      "I_f = 2.25  \t\t\t#Corresponds to 240 V when intersected OCC\n",
+      "R_sh = E_o/I_f  \t\t\t#shunt resistance\n",
+      "print 'i)Field resistance that gives 240 V on no-load is %.2f ohms '%(R_sh)\n",
+      "\n",
+      "#Part(ii)\n",
+      "N1 = 1000. \t\t\t#speed of shunt generator in rpm\n",
+      "I_f = 1.\n",
+      "\n",
+      "#Draw line OP math.tangential to OCC at N1 = 1000 r.p.m.\n",
+      "#Select I_f = 1A i.e. point R\n",
+      "#Draw vertical from R to intersect OP at S and OA at T....this gives RT = 105 and RS = 159\n",
+      "#At critcal speed generator just fails to build up\n",
+      "\n",
+      "RT = 105.\n",
+      "RS = 159.\n",
+      "N_C = N1*RT/RS \t\t\t#Critical speed\n",
+      "print 'ii)Critical speed is %.2f r.p.m '%(N_C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Field resistance that gives 240 V on no-load is 106.67 ohms \n",
+        "ii)Critical speed is 660.38 r.p.m \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.14 page no : 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 4.  \t\t\t#number of poles\n",
+      "A = 2. \t\t\t#because wave wound\n",
+      "Z = 792 \t\t\t#No of conductors\n",
+      "phi = 0.012  \t\t\t#flux per pole in weber\n",
+      "E_g = 240 \t\t\t#on no-load\n",
+      "\n",
+      "# Calculations\n",
+      "#running speed\n",
+      "N = E_g*60*A/(phi*P*Z) \t\t\t#becuase E_g =  phi*P*N*Z/(60*A)\n",
+      "\n",
+      "# Results\n",
+      "print 'Required running speed is %.3f r.p.m'%(N)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required running speed is 757.576 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.15 page no : 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#open circuit condition\n",
+      "I_L = 0.  \t\t\t#because of open circuit\n",
+      "V_t = 127. \t\t\t#terminal voltage\n",
+      "E_g = V_t \t\t\t#because I_L = 0\n",
+      "\n",
+      "# Calculations\n",
+      "#load condition\n",
+      "V_t = 120\n",
+      "R_sh = 15.\n",
+      "R_a = 0.02  \t\t\t#shunt and armature resistance\n",
+      "I_sh1 = V_t/R_sh \t\t\t#current through shunt winding in loaded condition\n",
+      "\n",
+      "I_L  = (E_g-V_t)/R_a  -  I_sh1 \t\t\t#because I_a1 = I_L+I_sh1    and    E_g = V_t + I_a1*R_a\n",
+      "\n",
+      "# Results\n",
+      "print 'Load current is %.0f A'%(I_L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load current is 342 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.16 page no : 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V_t = 550.  \t\t\t#Terminal voltage\n",
+      "R_lamp = 500. \t\t\t#Each lamp\n",
+      "I_lamp = V_t/R_lamp  \t\t\t#each lamp ; V_t because all lamps are in parallel\n",
+      "\n",
+      "# Calculations\n",
+      "I_L = 20*I_lamp  \t\t\t#there exist 20 lamps\n",
+      "R_sh = 25.\n",
+      "R_a = 0.06\n",
+      "R_se = 0.04  \t\t\t#resistance of shunt winding,armature,series field\n",
+      "I_sh = V_t/R_sh \t\t\t#current throough shunt winding\n",
+      "I_a = I_L+I_sh\t\t\t#armature current\n",
+      "E = V_t + I_a*(R_a+R_se)  \t\t\t#generated emf\n",
+      "\n",
+      "# Results\n",
+      "print 'Armature current and generated e.m.f is %.0f A and %.1f V respectively '%(I_a,E )\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature current and generated e.m.f is 44 A and 554.4 V respectively \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.17 page no : 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 4. \t\t\t#number of poles\n",
+      "A = P \t\t\t# because of lap wound\n",
+      "N = 750. \t\t\t#speed in rpm\n",
+      "Z = 720. \t\t\t#number of armature conductors\n",
+      "phi = 30.*10**-3  \t\t\t#flux per pole in weber\n",
+      "R_sh = 200.\n",
+      "R_a = 0.4\n",
+      "R_L = 15. \t\t\t#resistance of shunt winding,armature,series field\n",
+      "E = phi*P*N*Z/(60*A) \t\t\t#generated emf\n",
+      "\n",
+      "# Calculations\n",
+      "#solving the following equations for V_t\n",
+      "#E =  V_t + I_a*R_a\n",
+      "#E =  V_t + (I_L + I_sh)*R_a\n",
+      "#E =  V_t + ((V_t/R_L) + (V_t/R_sh))*R_a\n",
+      "V_t = E/(1+(R_a/R_L)+(R_a/R_sh))\n",
+      "\n",
+      "# Results\n",
+      "print 'Terminal voltage  =  %.4f V'%(V_t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Terminal voltage  =  262.4757 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.18 page no : 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 6. \t\t\t#number of poles\n",
+      "A = 2. \t\t\t# because of wave wound\n",
+      "N_1 = 300. \t\t\t#speed of generator\n",
+      "Z = 600. \t\t\t#number of armature conductors\n",
+      "phi_1 = 0.06  \t\t\t#flux per pole in webers \n",
+      "\n",
+      "# Calculations and Results\n",
+      "E_g1 = phi_1*P*N_1*Z/(60*A) \t\t\t#generated emf\n",
+      "print 'Emf generated is %.0f V'%(E_g1)\n",
+      "\n",
+      "phi_2 = 0.055 \t\t\t#new flux per pole \n",
+      "E_g2 = 550.\t\t\t# new generated emf\n",
+      "N_2 = E_g2/(phi_2*P*Z/(60*A)) \t\t\t#new speed of generator\n",
+      "print 'Required speed is %.2f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Emf generated is 540 V\n",
+        "Required speed is 333.33 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.19 page no  : 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "print 'Refer to code for explanation'\n",
+      "\n",
+      "# Variables\n",
+      "N_1 = 300.\n",
+      "N_2 = 375. \t\t\t#generator speeds\n",
+      "\n",
+      "#E_g2 = E_g1*(N_2/N_1)\n",
+      "#using this new table OCC at N_2 = 375 is made\n",
+      "#Draw a line with slope R_sh = 40 through origin which cuts this OCC at 248\n",
+      "#I_f_table = [0,2,3,4,5,6,7]\n",
+      "#Arm_vol_table = [9.375,115,165,202.5,228.75,237.5,265]\n",
+      "\n",
+      "#part(i)\n",
+      "#at V = 200 volts, I_f = 3.9 from the graph\n",
+      "V = 200.\n",
+      "I_f = 3.9\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R_sh2 = V/I_f\n",
+      "R_sh = 40.\n",
+      "print 'Additional resistance required is %.3f ohms  '%(R_sh2-R_sh)\n",
+      "\n",
+      "#part(ii)\n",
+      "V_t = 200.\n",
+      "I_f = V_t/R_sh\n",
+      "E_g = 228.75   \t\t\t#For this I_f from the table \n",
+      "R_a = 0.4\n",
+      "I_a = (E_g-V_t)/R_a   \t\t\t#Because E_g = V_t + I_a*R_a\n",
+      "I_L = I_a-I_f\n",
+      "print 'Load current supplied by the generator is %.3f A'%(I_L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refer to code for explanation\n",
+        "Additional resistance required is 11.282 ohms  \n",
+        "Load current supplied by the generator is 66.875 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20 page no : 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "I_a = 750. \t\t\t#full load current \n",
+      "Pole = 6. \n",
+      "A = Pole \t\t\t#lap winding\n",
+      "I = I_a/A  \t\t\t#Full-load current per path\n",
+      "Z = 900. \t\t\t#no of conductors\n",
+      "lambda1 = 1.4  \t\t\t#leakage coefficient\n",
+      "theta_e = 21. \t\t\t#lead angle in degrees electrical\n",
+      "\n",
+      "# Calculations\n",
+      "theta_m = theta_e/(Pole/2)  \t\t\t#lead angle in degrees mechanical\n",
+      "amp_turns_PP_d = Z*I*theta_m/360  \t\t\t#demagnetising ampere turns per pole\n",
+      "amp_turns_PP_c = Z*I*(1./(2*Pole)-theta_m/360)  \t\t\t#cross-magnetising ampere turns per pole\n",
+      "balance_turns = amp_turns_PP_d*lambda1/I_a \t\t\t#series turns required to balance demagnetising component\n",
+      "\n",
+      "# Results\n",
+      "print 'i)  De-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_d)\n",
+      "print 'ii) Cross-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_c)\n",
+      "print 'iii)Turns required to balance demagnetising component is %.0f'%(balance_turns)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)  De-magnetising ampere-turns per pole is 2187.5\n",
+        "ii) Cross-magnetising ampere-turns per pole is 7187.5\n",
+        "iii)Turns required to balance demagnetising component is 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.21 page no : 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pole = 4. \n",
+      "Z = 32.  \t\t\t#no of conductors\n",
+      "coil_sides = Z \n",
+      "segments = 16.\n",
+      "pole_pitch = Z/Pole\n",
+      "slots = 16.\n",
+      "slots_per_pole = slots/Pole\n",
+      "\n",
+      "# Calculations\n",
+      "#for Simplex lap winding\n",
+      "y_b = pole_pitch+1  \t\t\t#back pitch\n",
+      "y_f = pole_pitch-1  \t\t\t#front pitch\n",
+      "\n",
+      "y_c = 1  \t\t\t#Commutator pitch; Note that it is positive and it is progressive type of Simplex lap winding\n",
+      "\n",
+      "# Results\n",
+      "print 'WINDING TABLE:   1<-   10->    3<-   12->    5<-   14-> 7<-   16->    9<-   18->   11<-   20->13<-   22->   15<-   24->   17<-   26->19<-   28->   21<-   30->   23<-   32->25<-    2->   27<-    4->   29<-    6->31<-   8->1                                                  '\n",
+      "print 'Note that <- indicates back connection with  y_back = %.0f and -> indicates front connection with y_front = %.0f'%(y_b,y_f)\n",
+      "\n",
+      "print 'Another form of winding table:'\n",
+      "\n",
+      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
+      "\n",
+      "print '       1 to 1+9  = 10                        ->       10 to 10-7  = 3'\n",
+      "print '       3 to 3+9  = 12                        ->       12 to 12-7 =  5'\n",
+      "print '       5 to 5+9  = 14                        ->       14 to 14-7 =  7'\n",
+      "print '       7 to 7+9  = 16                        ->       16 to 16-7 =  9'\n",
+      "print '       9 to 9+9  = 18                        ->       18 to 18-7 = 11'\n",
+      "print '      11 to 11+9 = 20                        ->       20 to 20-7 = 13'\n",
+      "print '      13 to 13+9 = 22                        ->       22 to 22-7 = 15'\n",
+      "print '      15 to 15+9 = 24                        ->       24 to 24-7 = 17'\n",
+      "print '      17 to 17+9 = 26                        ->       26 to 26-7 = 19'\n",
+      "print '      19 to 19+9 = 28                        ->       28 to 28-7 = 21'\n",
+      "print '      21 to 21+9 = 30                        ->       30 to 30-7 = 23'\n",
+      "print '      23 to 23+9 = 32                        ->       32 to 32-7 = 25'\n",
+      "print '      25 to 25+9 = 34 = 34-32 = 2              ->        2 to 34-7 = 27'\n",
+      "print '      27 to 27+9 = 36 = 36-32 = 4              ->        4 to 36-7 = 29'\n",
+      "print '      29 to 29+9 = 38 = 38-32 = 6              ->        6 to 38-7 = 31'\n",
+      "print '      31 to 31+9 = 40 = 40-32 = 4              ->        8 to 40-7 = 33 -32 =  1'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "WINDING TABLE:   1<-   10->    3<-   12->    5<-   14-> 7<-   16->    9<-   18->   11<-   20->13<-   22->   15<-   24->   17<-   26->19<-   28->   21<-   30->   23<-   32->25<-    2->   27<-    4->   29<-    6->31<-   8->1                                                  \n",
+        "Note that <- indicates back connection with  y_back = 9 and -> indicates front connection with y_front = 7\n",
+        "Another form of winding table:\n",
+        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
+        "       1 to 1+9  = 10                        ->       10 to 10-7  = 3\n",
+        "       3 to 3+9  = 12                        ->       12 to 12-7 =  5\n",
+        "       5 to 5+9  = 14                        ->       14 to 14-7 =  7\n",
+        "       7 to 7+9  = 16                        ->       16 to 16-7 =  9\n",
+        "       9 to 9+9  = 18                        ->       18 to 18-7 = 11\n",
+        "      11 to 11+9 = 20                        ->       20 to 20-7 = 13\n",
+        "      13 to 13+9 = 22                        ->       22 to 22-7 = 15\n",
+        "      15 to 15+9 = 24                        ->       24 to 24-7 = 17\n",
+        "      17 to 17+9 = 26                        ->       26 to 26-7 = 19\n",
+        "      19 to 19+9 = 28                        ->       28 to 28-7 = 21\n",
+        "      21 to 21+9 = 30                        ->       30 to 30-7 = 23\n",
+        "      23 to 23+9 = 32                        ->       32 to 32-7 = 25\n",
+        "      25 to 25+9 = 34 = 34-32 = 2              ->        2 to 34-7 = 27\n",
+        "      27 to 27+9 = 36 = 36-32 = 4              ->        4 to 36-7 = 29\n",
+        "      29 to 29+9 = 38 = 38-32 = 6              ->        6 to 38-7 = 31\n",
+        "      31 to 31+9 = 40 = 40-32 = 4              ->        8 to 40-7 = 33 -32 =  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.22 page no : 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import array\n",
+      "\n",
+      "# Variables\n",
+      "Z = 496. \t\t\t#no of conductors\n",
+      "P = 4. \t\t\t#poles\n",
+      "slots = 31.\n",
+      "coilsides_per_slot = 4.\n",
+      "coilsides = slots*coilsides_per_slot\n",
+      "coils = coilsides/2\n",
+      "turns = Z/2\n",
+      "turns_per_coil = turns/coils\n",
+      "\n",
+      "# Calculations\n",
+      "y_c = array([(Z-2)/P, (Z+2)/P]) \t\t\t#commutator pitch\n",
+      "coils_active = (Z/(2*P))-1 \t\t\t# because y_c didnt turn out to be integer, 1 coil was made inactive/dummy\n",
+      "segments = coils_active  \t\t\t#no of commutative segments\n",
+      "Y_A = [ (segments+1)/(P/2), (segments-1)/(P/2)   ]\n",
+      "Y_A = Y_A[0]   \t\t\t#Y_A(2) is discarded because of progressive wave winding\n",
+      "y_f = 29.\n",
+      "y_b = 33.  \t\t\t#front and back pitch ; note that Y_A = (y_b+y_f)/2\n",
+      "resultant_pitch = 2*Y_A  \t\t\t#because Y_A = (y_b+y_f)/2 and resultant pitch  =  y_b+ y_f\n",
+      "\n",
+      "# Results\n",
+      "print 'i)   Total number of coils  =  %.0f'%(coils)\n",
+      "print 'ii)  Turns per coils  =  %.0f'%(turns_per_coil)\n",
+      "print 'iii) Commutator pitch  =  %.0f'%((y_c[0]+y_c[1])/2)\n",
+      "print 'iv)  Back pitch =  %.0f    front pitch =  %.0f   total pitch =  %.0f'%(y_b,y_f,resultant_pitch)\n",
+      "print 'v)   No of commutator segments  =  %.0f'%(segments)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)   Total number of coils  =  62\n",
+        "ii)  Turns per coils  =  4\n",
+        "iii) Commutator pitch  =  124\n",
+        "iv)  Back pitch =  33    front pitch =  29   total pitch =  62\n",
+        "v)   No of commutator segments  =  61\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.23 page no : 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "slots = 13.\n",
+      "Pole = 4.\n",
+      "conductors_per_slot = 2.\n",
+      "Z = conductors_per_slot*slots\n",
+      "Y_A = (Z+2)/Pole  \t\t\t#For progressive type wave winding\n",
+      "\n",
+      "# Calculations\n",
+      "#Since Y_A = (y_b+y_f)/2, we let y_b = y_f\n",
+      "y_b = Y_A\n",
+      "y_f = y_b \t\t    \t#because y_b = Y_A/2\n",
+      "\n",
+      "segments = 13. \t\t\t#commutator segments\n",
+      "\n",
+      "# Results\n",
+      "print 'WINDING TABLE:   1<-    8->  15<-  22->   3<-   10->17<-   24->   5<-  12->  19<-   26-> 7<-   14->  21<-   2->   9<-   16->23<-    4->  11<-  18->  25<-    6->13<-   20->   1                                                  '\n",
+      "print 'Note that <- indicates back connection with y_back = %.0f and -> indicates front connection with y_front = %.0f'%(y_b,y_f)\n",
+      "\n",
+      "print 'Another form of winding table:'\n",
+      "\n",
+      "print '      BACK CONNECTIONS                                FRONT CONNECTIONS'\n",
+      "\n",
+      "print '       1 to 1+7   =  8                   ->        8 to 8+7  =  15'\n",
+      "print '      15 to 15+7  = 22                   ->       22 to 22+7 =  29 -26 = 3'\n",
+      "print '       3 to 3+7   = 10                   ->       10 to 10+7 =  17'\n",
+      "print '      17 to 17+7  = 24                   ->       24 to 24+7 =  31 -26 = 5'\n",
+      "print '       5 to 5+7   = 14                   ->       12 to 12+7 =  19'\n",
+      "print '      19 to 19+7  = 26                   ->       26 to 26+7 =  33 -26 = 7'\n",
+      "print '       7 to 7+7   = 14                   ->       14 to 14+7 =  21'\n",
+      "print '      21 to 21+7  = 28 -26 = 2             ->        2 to  2+7 =   9'\n",
+      "print '       9 to 9+7   = 16                   ->       16 to 16+7 =  23 '\n",
+      "print '      23 to 23+7  = 30 -26 = 4             ->        4 to  4+7 =  11'\n",
+      "print '       11to 11+7  = 18                   ->       18 to 18+7 =  25'\n",
+      "print '      25 to 25+7  = 32 -26 = 6             ->        6 to  6+7 =  13'\n",
+      "print '      13 to`13+7  = 20                   ->       20 to 20+7 =  27 -26 = 1'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "WINDING TABLE:   1<-    8->  15<-  22->   3<-   10->17<-   24->   5<-  12->  19<-   26-> 7<-   14->  21<-   2->   9<-   16->23<-    4->  11<-  18->  25<-    6->13<-   20->   1                                                  \n",
+        "Note that <- indicates back connection with y_back = 7 and -> indicates front connection with y_front = 7\n",
+        "Another form of winding table:\n",
+        "      BACK CONNECTIONS                                FRONT CONNECTIONS\n",
+        "       1 to 1+7   =  8                   ->        8 to 8+7  =  15\n",
+        "      15 to 15+7  = 22                   ->       22 to 22+7 =  29 -26 = 3\n",
+        "       3 to 3+7   = 10                   ->       10 to 10+7 =  17\n",
+        "      17 to 17+7  = 24                   ->       24 to 24+7 =  31 -26 = 5\n",
+        "       5 to 5+7   = 14                   ->       12 to 12+7 =  19\n",
+        "      19 to 19+7  = 26                   ->       26 to 26+7 =  33 -26 = 7\n",
+        "       7 to 7+7   = 14                   ->       14 to 14+7 =  21\n",
+        "      21 to 21+7  = 28 -26 = 2             ->        2 to  2+7 =   9\n",
+        "       9 to 9+7   = 16                   ->       16 to 16+7 =  23 \n",
+        "      23 to 23+7  = 30 -26 = 4             ->        4 to  4+7 =  11\n",
+        "       11to 11+7  = 18                   ->       18 to 18+7 =  25\n",
+        "      25 to 25+7  = 32 -26 = 6             ->        6 to  6+7 =  13\n",
+        "      13 to`13+7  = 20                   ->       20 to 20+7 =  27 -26 = 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.24 page no : 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 4.\n",
+      "I_L = 150.\n",
+      "N = 1500. \t\t\t#commutator speed in rpm\n",
+      "n_s = N/60.  \t\t\t#commutator speed in r.p.s\n",
+      "W_b = 1.2 \t\t\t#Brush pitch\n",
+      "W_m = 0\t\t\t#Pitch of mica insulation\n",
+      "L = 0.05*10**-3  \t\t\t#inducmath.tance of armature coils in henry\n",
+      "A = P \t\t\t#A = P for lap wound\n",
+      "segments = 64.\n",
+      "\n",
+      "# Calculations\n",
+      "v = n_s*segments \t\t\t#peripheral speed in segments per second\n",
+      "T_c = (W_b-W_m)/v  \t\t\t#Time of commutation\n",
+      "I = I_L/A  \t\t\t#current through each conductor\n",
+      "E = L*2*I/T_c  \t\t\t#Linear commutation\n",
+      "E2 = 1.11*L*2*I/T_c  \t\t\t#Sinusoidal commutation\n",
+      "\n",
+      "# Results\n",
+      "print 'Reactive voltage linear commutation) is %.0f V'%(E)\n",
+      "print 'Reactive voltage math.sinusoidal commutation) is %.2f V'%(E2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reactive voltage linear commutation) is 5 V\n",
+        "Reactive voltage math.sinusoidal commutation) is 5.55 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.25 page no : 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P = 4.  \t\t\t#Pole\n",
+      "Z = 386. \t\t\t#no of wave connected conductors\n",
+      "A = 2.  \t\t\t#Wave winding\n",
+      "R_a = 1.\n",
+      "R_sh = 100.\n",
+      "R_L = 40  \t\t\t#Armature ,shunt field and load resistance\n",
+      "phi = 25.*10**-3  \t\t\t#flux per pole in weber\n",
+      "N = 1000.  \t\t\t#speed in rpm\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "E_g = (phi*P*N*Z)/(60*A)  \t\t\t#generated emf\n",
+      "\n",
+      "#Solving following equations for V_t\n",
+      "#E_g = V_t+I_a*R_a\n",
+      "#I_a = (I_L+I_sh)   I_L = V_t/R_L  I_sh = V_t/R_sj\n",
+      "#E_g = V_t(1 + (R_a/R_L) + (R_a/R_sh)\n",
+      "V_t = E_g/(1 + (R_a/R_L) + (R_a/R_sh))\n",
+      "\n",
+      "I_L = V_t/R_L \t\t\t# load current\n",
+      "I_sh = V_t/R_sh \t\t\t# current through shunt field resistance\n",
+      "\n",
+      "# Results\n",
+      "I_a = I_L+I_sh \t\t\t#armature current \n",
+      "print 'armature current is  is %.4f A'%(I_a)\n",
+      "\n",
+      "output_power = V_t*I_L\n",
+      "print 'output power is %.3f W'%(output_power)\n",
+      "\n",
+      "# Note :answer obtained will not match with textbook answer because I_L = V_t/R_L = 310.79/40 = 7.77 \n",
+      "#Awhile its taken as 8.045 A in textbook'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "armature current is  is 10.8776 A\n",
+        "output power is 2414.746 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.26 page no : 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "I = 40.  \t\t    \t#current PER conductor\n",
+      "W_b = 3.\n",
+      "W_m = 0.\t\t\t    #brush width and width of mica insulation\n",
+      "N = 600. \t\t\t    #commutator speed in rpm \n",
+      "n_s = N/60 \t\t\t    #commutator speed in rps\n",
+      "L = 0.15*10**-3 \t\t\t#self inducmath.tance in Henry\n",
+      "segments = 50.  \n",
+      "v = n_s*segments \t\t\t#peripheral speed in segments per second\n",
+      "T_c = (W_b-W_m)/v \t\t\t#time of commutation\n",
+      "\n",
+      "# Calculations\n",
+      "E = L*2*I/T_c  \t\t\t#Linear commutation\n",
+      "E2 = 1.11*L*2*I/T_c  \t\t\t#Sinusoidal commutation\n",
+      "\n",
+      "# Results\n",
+      "print 'Reactive voltage linear commutation) is %.0f volts'%(E)\n",
+      "print 'Reactive voltage math.sinusoidal commutation) is %.2f volts'%(E2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reactive voltage linear commutation) is 2 volts\n",
+        "Reactive voltage math.sinusoidal commutation) is 2.22 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.27 page no : 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 400.\n",
+      "P = 6. \t\t\t    #Poles\n",
+      "A = P  \t\t    \t#For lap wound\n",
+      "output_power = 250.*10**3\n",
+      "R_sh = 200. \t\t\t#shunt field circuit resistance\n",
+      "Z = 720.     \t\t\t#number of lap wound conductors\n",
+      "\n",
+      "theta_m = 2.5 \t\t\t#brush lead angle in degree mechanical\n",
+      "I_L = output_power/V\n",
+      "V_sh = V\n",
+      "\n",
+      "# Calculations\n",
+      "I_sh = V_sh/R_sh \t\t\t#Current through shunt field circuit resistance\n",
+      "I_a = I_L+I_sh \t\t\t#armature currrent\n",
+      "I = I_a/P\n",
+      "\n",
+      "#Part(i)\n",
+      "amp_turns_PP_d = Z*I*theta_m/360\t\t\t#demagnetising ampere turns per pole\n",
+      "#Part(ii)\n",
+      "amp_turns_PP_c = Z*I*(1/(2*P)-theta_m/360) \t\t\t#cross-magnetising ampere turns per pole\n",
+      "\n",
+      "# Results\n",
+      "print 'i)  De-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_d)\n",
+      "print 'ii) Cross-magnetising ampere-turns per pole is %.1f'%(amp_turns_PP_c)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)  De-magnetising ampere-turns per pole is 522.5\n",
+        "ii) Cross-magnetising ampere-turns per pole is 5747.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.28 page no : 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "I_L = 100.\n",
+      "P = 4. \t\t\t    #Poles\n",
+      "A = P \t\t\t    #for lap wound armature\n",
+      "W_b = 1.4\n",
+      "W_m = 0. \t\t\t#Brush width and width of mica insulation\n",
+      "N = 1400.\t\t\t#armature speed in r.p.m\n",
+      "segments = 64. \t\t\t#no of commutator segments\n",
+      "L = 0.05*10**-3 \t\t\t#inducmath.tance of armature coil in henry\n",
+      "\n",
+      "# Calculations\n",
+      "n_s = N/60  \t\t\t#speed in r.p.s\n",
+      "v = n_s*segments \t\t\t#Segments per second\n",
+      "T_c = (W_b-W_m)/v \t\t\t#time of commutation\n",
+      "I = I_L/A   \t\t\t#Current through conductor\n",
+      "E = L*2*I/T_c  \t\t\t#Linear commutation\n",
+      "\n",
+      "# Results\n",
+      "print 'Reactive voltage considering linear commutation is %.2f volts'%(E)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reactive voltage considering linear commutation is 2.67 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.29 page no : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 8. \t\t    \t#Poles\n",
+      "A = 2. \t\t    \t#Wave wound armature\n",
+      "Z = 480. \t\t\t#number of armature conductors\n",
+      "I_a = 200.\n",
+      "I = I_a/A\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "theta_m = 0. \t\t\t#Geometric nuetral axis\n",
+      "amp_turns_PP_d = Z*I*theta_m/360    \t\t\t#De-magnetising ampere-turns per pole\n",
+      "amp_turns_PP_c = Z*I*(1/(2*P)-theta_m/360) \t\t\t#Cross-magnetising ampere-turns per pole\n",
+      "print 'Parti)De-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_d)\n",
+      "print 'Cross-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_c)\n",
+      "\n",
+      "#Part(ii)\n",
+      "theta_e2 = 6.             \t\t\t#angle shift of brushes in degrees electrical\n",
+      "theta_m2 = theta_e2/(P/2) \t\t\t#angle shift of brushes in degrees mechanical\n",
+      "amp_turns_PP_d2 = Z*I*theta_m2/360   \t\t\t#De-magnetising ampere-turns per pole\n",
+      "amp_turns_PP_c2 = Z*I*(1/(2*P)-theta_m2/360) \t\t\t#Cross-magnetising ampere-turns per pole\n",
+      "print 'Partii)De-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_d2)\n",
+      "print 'Cross-magnetising ampere-turns per pole is %.0f'%(amp_turns_PP_c2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Parti)De-magnetising ampere-turns per pole is 0\n",
+        "Cross-magnetising ampere-turns per pole is 3000\n",
+        "Partii)De-magnetising ampere-turns per pole is 200\n",
+        "Cross-magnetising ampere-turns per pole is 2800\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.30 page no : 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P_input = 7.46*10**3\n",
+      "V = 230.\n",
+      "Pole = 8.\n",
+      "Z = 188. \t\t\t#number of armature consuctors\n",
+      "I_L = P_input/V\n",
+      "theta_m = 7.5  \t\t\t#brush lead angle in degree mechanical\n",
+      "\n",
+      "A = 2. \t\t\t# assumed wave wound because of low-current and high voltage\n",
+      "I = I_L/A\n",
+      "\n",
+      "#Part(i)\n",
+      "amp_turns_PP_d = Z*I*theta_m/360  \t\t\t#De-magnetising ampere-turns per pole\n",
+      "#Part(ii)\n",
+      "amp_turns_PP_c = Z*I*(1/(2*Pole)-theta_m/360) \t\t\t#Cross-magnetising ampere-turns per pole\n",
+      "\n",
+      "print 'De-magnetising ampere-turns per pole is %.2f'%(amp_turns_PP_d)\n",
+      "print 'Cross-magnetising ampere-turns per pole is %.2f'%(amp_turns_PP_c)\n",
+      "\n",
+      "# note : The difference in answer occured because I is approximated to 16 in last 2 steps in book\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "De-magnetising ampere-turns per pole is 63.52\n",
+        "Cross-magnetising ampere-turns per pole is 127.04\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.31 page no : 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "N_1 = 1200.   \t\t\t#initial speed \n",
+      "I_L1 = 200.   \t\t\t#initial load current\n",
+      "V_t1 = 125.\n",
+      "N_2 = 1000.  \t\t\t#altered speed\n",
+      "R_a = 0.04  \t\t\t#armature resistance\n",
+      "V_brush = 2.  \t\t\t#brush drop\n",
+      "\n",
+      "# Calculations\n",
+      "# Initial Load\n",
+      "I_a1 = I_L1\n",
+      "E_g1 = V_t1+I_a1*R_a+V_brush  \t\t\t#induced emf\n",
+      "\n",
+      "E_g2 = E_g1*(N_2/N_1)  \t\t\t#Because E_g proportional to N during constant flux\n",
+      "R_L =  V_t1/I_L1 \t\t\t#Load resistance\n",
+      "\n",
+      "#Solving for I_L2 as follows\n",
+      "#V_t2 = R_L*I_L2   \t\t\t#I_a2 = I_L2\n",
+      "#V_t2 = E_g2-(I_L2*R_2 + V_brush)\n",
+      "I_L2 = (E_g2-V_brush)/(R_L+R_a)  \t\t\t#new current\n",
+      "\n",
+      "# Results\n",
+      "print 'Load current at new speed is %.4f A'%(I_L2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load current at new speed is 166.1654 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch2_2.ipynb b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch2_2.ipynb
new file mode 100755
index 00000000..deb5684b
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch2_2.ipynb
@@ -0,0 +1,1995 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f03a268884ccacb4d0b4a4e0ccceeba1a58caa51f0cbd85024ffaf9d558c0182"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : D.C. Motors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 page no : 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "I_a = 30. \t\t\t#armature currnet\n",
+      "R_a = 0.75 \t\t\t#Armature resistance\n",
+      "\n",
+      "# Calculations\n",
+      "E_b = V - I_a*R_a   \t\t\t# Since V =  E_b+ I_a*R_a\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print 'Induced EMF or back EMF in the motor is %.1f V'%(E_b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Induced EMF or back EMF in the motor is 197.5 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 page no : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "A = Pole \t\t\t#for lap winding\n",
+      "V = 230.\n",
+      "Z = 250. \t\t\t#number of armature conductors\n",
+      "phi = 30.*10**-3 \t\t\t#flux per pole in weber\n",
+      "I_a = 40.\n",
+      "R_a = 0.6 \t\t\t#Armature resistance\n",
+      "\n",
+      "# Calculations \n",
+      "E_b = V - I_a*R_a   \t\t\t# Since V =  E_b+ I_a*R_a\n",
+      "N = E_b * 60*A/(phi*Pole*Z)   \t\t\t#because E_b  =  phi*P*N*Z/(60*A)\n",
+      "\n",
+      "# Results\n",
+      "print 'Back emf is %.0f V and running speed is %.0f rpm'%(E_b,N)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Back emf is 206 V and running speed is 1648 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 page no : 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "A = Pole \t\t\t#for lap winding\n",
+      "Z = 480.\t\t\t#number of armature conductors\n",
+      "phi = 20.*10**-3 \t\t\t#flux per pole in weber\n",
+      "I_a = 50. \t\t\t#Armature current\n",
+      "\n",
+      "# Calculations\n",
+      "T_a  =  0.159*phi*I_a*Pole*Z/A  \t\t\t#Gross torque developed by armature\n",
+      "\n",
+      "# Results\n",
+      "print 'Gross torque developed by armature is %.3f N-m'%(T_a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gross torque developed by armature is 76.320 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 page no : 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "A = Pole \t\t\t#for lap winding\n",
+      "V = 230.\n",
+      "R_a = 0.8 \t\t\t#Armature resistance\n",
+      "N_0 = 1000. \t\t\t#no load speed in rpm\n",
+      "Z = 540. \t\t\t#number of armature conductors\n",
+      "phi = 25.*10**-3 \t\t\t#flux per pole in weber\n",
+      "\n",
+      "# Calculations and Results\n",
+      "E_b0  =  phi*Pole*N_0*Z/(60*A)  \t\t\t#induced emf\n",
+      "\n",
+      "#part(i)\n",
+      "print 'i)Induced e.m.f  =  %.0f V'%(E_b0)\n",
+      "\n",
+      "#part(ii)\n",
+      "I_a0  =  (V- E_b0)/R_a  \t\t\t#because V =  E_b0+ I_a0*R_a\n",
+      "print 'ii)Armature current  =  %.2f A'%(I_a0)\n",
+      "\n",
+      "#part(iii)\n",
+      "stray_losses  =  E_b0*I_a0  \t\t\t#on no load ,power developed is fully power required to overcome strya losses\n",
+      "print 'iii)Stray loss  =  %.2f W'%(stray_losses)\n",
+      "\n",
+      "#part(iv)\n",
+      "T_f  =  E_b0*I_a0/(2*math.pi*N_0/60)  \t\t\t#lost torque\n",
+      "print 'iv)Lost torque  =  %.3f N-m'%(T_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Induced e.m.f  =  225 V\n",
+        "ii)Armature current  =  6.25 A\n",
+        "iii)Stray loss  =  1406.25 W\n",
+        "iv)Lost torque  =  13.429 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 page no : 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "Z = 200. \t\t\t    #No of armature conductors\n",
+      "A = 2.   \t\t\t    #wave connected armature\n",
+      "V = 250.\n",
+      "phi = 25.*10**-3 \t\t\t#flux per pole in weber\n",
+      "I_a  = 60.\n",
+      "I_L  = I_a  \t    \t\t#armature current\n",
+      "R_a = 0.15\n",
+      "R_se = 0.2       \t\t\t#resistances of armature and series field winding\n",
+      "\n",
+      "# Calculations\n",
+      "E_b =  V - I_a*(R_a+R_se)  \t\t\t#induced emf\n",
+      "N = E_b * 60*A/(phi*Pole*Z)   \t\t\t#because E_b  =  phi*P*N*Z/(60*A)\n",
+      "\n",
+      "# Results\n",
+      "print 'Required speed is %.0f r.p.m'%(N)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required speed is 1374 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 page no : 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "I_L  = 20. \t\t\t#load current\n",
+      "R_a = 0.3\n",
+      "R_sh = 200. \t\t\t#Armature and shunt field winding\n",
+      "\n",
+      "# Calculations\n",
+      "I_sh = V/R_sh \t\t\t#shunt current\n",
+      "I_a = I_L-I_sh \t\t\t#armature current\n",
+      "E_b =  V - I_a*R_a \t\t\t#emf generated\n",
+      "\n",
+      "# Results\n",
+      "print 'Armature current is %.2f A'%(I_a)\n",
+      "print 'Back e.m.f is %.3f V'%(E_b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature current is 18.75 A\n",
+        "Back e.m.f is 244.375 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 page no : 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "R_a = 0.3\n",
+      "R_sh = 110. \t\t\t#resistance of armature and shunt field winding\n",
+      "\n",
+      "#no load\n",
+      "N_0 = 1000. \t\t\t#no load speed in r.p.m\n",
+      "I_L0  = 6. \t\t\t#line current on no load\n",
+      "I_sh =  V/R_sh \t\t\t#no load shnt current\n",
+      "I_a0  =  I_L0 - I_sh \t\t\t#no load armature current\n",
+      "E_b0 =  V - I_a0*R_a \t\t\t#no load induced emf\n",
+      "\n",
+      "# Calculations\n",
+      "#full load\n",
+      "I_sh_FL =  V/R_sh\n",
+      "I_L_FL = 50 \t\t\t#line current at full load\n",
+      "I_a_FL =  I_L_FL - I_sh_FL\t\t\t#full load armature current\n",
+      "E_b_FL =  V - I_a_FL * R_a \t\t\t#full load induced emf\n",
+      "\t\t\t#using speed equation as treating phi as constant\n",
+      "N_FL = N_0 * (E_b_FL/E_b0)\n",
+      "\n",
+      "# Results\n",
+      "print 'Speed on full load is %.2f r.p.m'%(N_FL)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed on full load is 939.67 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 page no : 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R_a = 0.2\n",
+      "R_se  = 0.3 \t\t\t#resistance of armature and series field winding\n",
+      "#following variables correspond to load 1\n",
+      "V = 250.\n",
+      "N_1 = 800.\n",
+      "I_1 = 20.\n",
+      "I_a1 = I_1\n",
+      "I_se1 =  I_a1\n",
+      "\n",
+      "# Calculations\n",
+      "E_b1 =  V - I_a1*(R_a+R_se)\n",
+      "#following variables correspond to load 2\n",
+      "I_2 = 50.\n",
+      "I_a2 = I_2\n",
+      "E_b2 =  V - I_a2*(R_a+R_se)\n",
+      "\n",
+      "#from speed equation it can be derived that\n",
+      "\n",
+      "N_2  =  N_1 * (E_b2/E_b1) * (I_a1/I_a2)\n",
+      "\n",
+      "# Results\n",
+      "print 'Speed on motor on no load  is %.0f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed on motor on no load  is 300 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 page no : 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "R_a = 0.3\n",
+      "R_sh = 200.  \t\t\t#resistance of armature and shunt field winding\n",
+      "R_x = 150. \t\t\t#additional resistance added in series to field winding\n",
+      "I_L1 = 22.\n",
+      "I_sh1 = V/R_sh \t\t\t#initial shunt current before adding 150 ohms resistance\n",
+      "I_a1  =  I_L1 - I_sh1 \t\t\t#initial armature current before adding 150 ohms resistance\n",
+      "N_1 = 1500. \t\t\t#initial speed before adding 150 ohms resistance\n",
+      "\n",
+      "# Calculations\n",
+      "#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying further \n",
+      "I_sh2 = V/(R_sh + R_x) \t\t\t#new shunt current\n",
+      "I_a2 =  I_sh1*I_a1/I_sh2  \t\t\t#New armature current\n",
+      "\n",
+      "E_b1 = V - I_a1*R_a   \t\t\t#induced emf before adding 150 ohms resistance\n",
+      "E_b2 = V - I_a2*R_a   \t\t\t#new emf\n",
+      "\n",
+      "N_2  =  N_1 * (E_b2/E_b1) * (I_sh1/I_sh2) \t\t\t#new speed\n",
+      "\n",
+      "# Results\n",
+      "print 'New armature current and speed are %.4f A and %.f r.p.m respectively'%(I_a2,N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New armature current and speed are 36.3125 A and 2575 r.p.m respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10 page no : 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "R_a = 0.15\n",
+      "R_se = 0.1\n",
+      "R_x = 0.1 \t\t\t#Resimath.tance of armature , series field winding and extra resistance\n",
+      "N_1  =  800. \t\t\t#initial speed before load torque is increased\n",
+      "I_1 =  30.\n",
+      "I_a1 = I_1\n",
+      "I_se1  =  I_1  \t\t\t#initial currents\n",
+      "\n",
+      "# Calculations\n",
+      "T_2_by_T_1  =  1 + (50./100)  \t\t\t#50 percent increase as mentioned in question\n",
+      "I_se2_by_I_a2  =  R_x/(R_x + R_se)  \t\t\t#from the figure\n",
+      "\n",
+      "#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying ,solving further \n",
+      "I_a2 = math.sqrt(I_a1*I_se1*T_2_by_T_1/I_se2_by_I_a2) \t\t\t#new armature current\n",
+      "I_se2  =  I_se2_by_I_a2 *I_a2 \t\t\t#new series field current\n",
+      "\n",
+      "E_b1  =  V - I_a1*R_a - I_se1*R_se \t\t\t#indiced emf initially\n",
+      "E_b2  =  V - I_a2*R_a - I_se2*R_se \t\t\t#new induced emf\n",
+      "N_2  =  N_1 * (E_b2/E_b1) * (I_se1/I_se2) \t\t\t#required speed\n",
+      "\n",
+      "# Results\n",
+      "print 'The required running speed of motor is %.3f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required running speed of motor is 912.743 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11 page no : 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "I_1 = 50.\n",
+      "I_a1 = I_1  \t\t\t#Currents before adding extra resistance\n",
+      "T_2_by_T_1  = 0.5\n",
+      "R_t = 0.15 \t\t\t#R_e + R_se  = 0.15\n",
+      "\n",
+      "# Calculations\n",
+      "I_a2  = I_a1 * math.sqrt(T_2_by_T_1)  \t\t\t#Because T (prop.) I_a**2\n",
+      "E_b1 = V-I_a1*(R_t) \t\t\t#induced emf before adding extra resistance\n",
+      "N_1 = 500.\n",
+      "N_2 = 300. \t\t\t#speeds before and adding extra resistance\n",
+      "\n",
+      "#N (prop.) E_b/phi (prop.) E_b/I_a\n",
+      "E_b2 = E_b1 *(I_a2/I_a1)*(N_2/N_1) \t\t\t#induced emf after adding resistance\n",
+      "R_x =  (V-E_b2)/I_a2 -R_t \t\t\t#because E_b2 = V - I_a2*(R_a + R_se + R_x)\n",
+      "\n",
+      "# Results\n",
+      "print 'Desired extrea resistance =  %.4f ohms '%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Desired extrea resistance =  3.5225 ohms \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12 page no : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R_a =  1.\n",
+      "I_a = 1.2 \n",
+      "V = 50.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "E_b  =  V - I_a*R_a\n",
+      "rot_loss_NL  = E_b*I_a \t\t\t#no load rotational loss \n",
+      "print 'i)No load rotational losses  =  %.2f W'%(rot_loss_NL)\n",
+      "\n",
+      "#part(ii)\n",
+      "omega_2000 = 2*math.pi*2000/60 \t\t\t#angular velocity when speed of motor  = 2000 rpm\n",
+      "K_m = E_b/omega_2000   \t\t\t#to determine K_m\n",
+      "V = 48.\n",
+      "omega_1800 = 2*math.pi*1800/60 \t\t\t#angular velocity when speed of motor  = 1800 rpm\n",
+      "E_b = K_m*omega_1800\n",
+      "I_a  =  (V-E_b)/R_a  \t\t\t#armature current\n",
+      "P_dev  =  E_b*I_a\t\t\t#power developed\n",
+      "motor_output  =   P_dev - rot_loss_NL\n",
+      "print 'ii)Motor output  =  %.f W'%(motor_output)\n",
+      "\n",
+      "#part(iii)\n",
+      "E_b = 0. \t\t\t#when motor stalls\n",
+      "V_stall = 20. \t\t\t#voltage during stalling\n",
+      "I_a = V_stall/R_a \t\t\t#armature current during stalling\n",
+      "T_stall  =  K_m*I_a \t\t\t#stalling torque\n",
+      "print 'iii)Stalling torque  =  %.2f N-m'%(T_stall)\n",
+      "print 'partii answer is slightly different due to inaccurate calculation of Power developed'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)No load rotational losses  =  58.56 W\n",
+        "ii)Motor output  =  121 W\n",
+        "iii)Stalling torque  =  4.66 N-m\n",
+        "partii answer is slightly different due to inaccurate calculation of Power developed\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13 page no : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 120.\n",
+      "R_a = 0.2 \n",
+      "R_sh = 60. \t\t\t#armature and field resistance\n",
+      "I_L1 = 40.\n",
+      "N_1 = 1800. \n",
+      "\n",
+      "# Calculations\n",
+      "I_sh =  V/R_sh\n",
+      "\n",
+      "I_a1 = I_L1 - I_sh \n",
+      "E_b1  =  V -I_a1*R_a  \t\t\t#Induced emf at half load\n",
+      "T2_by_T1  = 1./2 \n",
+      "I_a2 = I_a1*(T2_by_T1)   \t\t\t#T (prop.)I_a\n",
+      "E_b2 = V- I_a2*R_a\t\t\t#induced emf at half load\n",
+      "N_2  =  N_1 *(E_b2/E_b1)  \t\t\t#N (prop.) E_b as phi is constant\n",
+      "\n",
+      "# Results\n",
+      "print 'Speed on half load condition is %.2f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed on half load condition is 1860.85 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14 page no : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R_a = 0.08\n",
+      "E_b1 = 242. \n",
+      "V = 250.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "I_a1 =  (V-E_b1)/R_a\n",
+      "print 'i)Armature current  =  %.0f A'%(I_a1)\n",
+      "\n",
+      "#part(ii)\n",
+      "N = 0.\n",
+      "E_b = 0. \t\t\t#because N = 0\n",
+      "I_a_start = V/R_a\n",
+      "print 'ii)Starting armature current  =  %.0f A'%(I_a_start)\n",
+      "\n",
+      "#part(iii)\n",
+      "I_a2 = 120.\n",
+      "E_b2 = V-I_a2*R_a\n",
+      "print 'iii)Back emf if armature current is changed to 120 A =    %.1f V'%(E_b2)\n",
+      "\n",
+      "#part(iv)\n",
+      "I_a = 87.\n",
+      "N_m = 1500.\n",
+      "E_g = V + I_a*R_a \t\t\t#induced emf\n",
+      "N_g = N_m*(E_g/E_b1)\t\t\t#as E (prop.) N\n",
+      "print 'iv)Generator speed to deliver 87 A at 250 V  =  %.1f rpm'%(N_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Armature current  =  100 A\n",
+        "ii)Starting armature current  =  3125 A\n",
+        "iii)Back emf if armature current is changed to 120 A =    240.4 V\n",
+        "iv)Generator speed to deliver 87 A at 250 V  =  1592.7 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15 page no : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "shaft_output  =  80.*746 \t\t\t#coverted to watts\n",
+      "eta =  80./100           \t\t\t#efficiency\n",
+      "V = 250.\n",
+      "N_1 = 1200.\n",
+      "R_a = 0.04\n",
+      "R_sh  =  250. \t\t\t            #armature and shunt field resistance\n",
+      "\n",
+      "# Calculations and Results\n",
+      "power_input  =  shaft_output/eta\n",
+      "I_L =  power_input /V\n",
+      "I_sh =  V / R_sh\n",
+      "I_a  =  I_L - I_sh\n",
+      "E_b1  =  V - I_a*R_a\n",
+      "\n",
+      "gross_mechanical_power =  E_b1*I_a \t\t\t#electrical equivalent of mechanical power developed\n",
+      "stray_losses  =   gross_mechanical_power - shaft_output\n",
+      "print 'Mechanical power developed on full load  =  %.3f kW'%(gross_mechanical_power/1000)\n",
+      "\n",
+      "#on no load shaft_output = 0 and entire gross power is used to overcome stray losses\n",
+      "Eb0_Ia0 =  stray_losses\n",
+      "#E_b0  =  V - I_a0*R_a  ... solving for I_0\n",
+      "p = [R_a, -V, Eb0_Ia0]\n",
+      "ans = roots(p)\n",
+      "I_a0 = ans[1] \t\t\t#first root is ignored math.since its too large\n",
+      "I_L0  = I_sh+I_a0 \t\t\t#current drawn from supply\n",
+      "E_b0  =  V - I_a0*R_a \n",
+      "\n",
+      "#From speed equation N (prop.) E_b\n",
+      "N_0  =  N_1*(E_b0/E_b1)\n",
+      "print 'No load speed and current are %.4f rpm and %.2f A respectively'%(N_0,I_L0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mechanical power developed on full load  =  70.812 kW\n",
+        "No load speed and current are 1250.9121 rpm and 45.85 A respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16 page no : 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "P = 4. \n",
+      "R_a = 0.1 \n",
+      "R_sh  = 124. \t\t\t#armature and shunt field resistance \n",
+      "I_L0 = 4.\n",
+      "N_0 = 1200.\n",
+      "I_L_1 = 61.\n",
+      "\n",
+      "# Calculations\n",
+      "I_sh = V/R_sh\n",
+      "I_a0 = I_L0-I_sh\n",
+      "V_brush =  2 \t\t\t#voltage loss due to brush\n",
+      "E_b0 =  V - I_a0*R_a- V_brush\n",
+      "\n",
+      "I_a1 = I_L_1 - I_sh\n",
+      "E_b1 = V - I_a1*R_a -V_brush\n",
+      "\n",
+      "phi1_by_phi0 = 1-(5./100)   \t\t\t#weakened by 5 %\n",
+      "N_1  =  N_0 *(E_b1/E_b0) /phi1_by_phi0\n",
+      "\n",
+      "# Results\n",
+      "print 'Full load speed is %.3f r.p.m'%(N_1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Full load speed is 1234.102 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17 page no : 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "R_a = 0.15 \n",
+      "R_sh = 167.67 \t\t\t#armature and shunt field resistance\n",
+      "N_0 = 1280. \t\t\t#speed at no load\n",
+      "\n",
+      "#full load\n",
+      "I_L1  =  67.  \t\t\t#current drawn on full load\n",
+      "I_sh  =  V / R_sh  \t\t\t#as shunt motor\n",
+      "I_a1 =  I_L1- I_sh\n",
+      "E_b1 =  V - I_a1*R_a\n",
+      "\n",
+      "#on no load\n",
+      "I_L0 = 6.5\n",
+      "I_a0  =  I_L0 - I_sh\n",
+      "E_b0  =  V - I_a0*R_a\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i) USING SPEED EQUATION\n",
+      "#N (prop.) E_b/phi (prop.)E_b   \t\t\t#as phi is constant\n",
+      "N_1  =  N_0 * (E_b1 / E_b0)\n",
+      "print 'i)Full load speed  =  %.3f r.p.m'%(N_1)\n",
+      "\n",
+      "#part(ii)\n",
+      "speed_regulation  =  100* ((N_0-N_1)/N_1)\n",
+      "#N_1 is full load speed and N_0 = No load speed  \n",
+      "print 'ii)Speed regulation  =  %.2f percent '%(speed_regulation )\n",
+      "\n",
+      "#part(iii)\n",
+      "shaft_output_FL  =  E_b1*I_a1 - E_b0*I_a0  \t\t\t#full load power developed - stray losses\n",
+      "hp_rating  =  shaft_output_FL /746\n",
+      "print 'iii)HP rating of machine  =  %.2f h.p'%(hp_rating)\n",
+      "\n",
+      "#part(iv)\n",
+      "power_input =  V*I_L1\n",
+      "eta =  100*(shaft_output_FL/power_input)  \t\t\t#full load efficiency\n",
+      "print 'iv)Full load efficiency  =  %.2f percent'%(eta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Full load speed  =  1233.396 r.p.m\n",
+        "ii)Speed regulation  =  3.78 percent \n",
+        "iii)HP rating of machine  =  19.42 h.p\n",
+        "iv)Full load efficiency  =  86.48 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18 page no : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 200.\n",
+      "R_a = 0.5\n",
+      "R_se = 0.2\n",
+      "R_x = 0.2 \t\t\t#armature and series field resistance; extra resistance\n",
+      "I_a1 = 20.\n",
+      "I_1 = I_a1 \n",
+      "I_se1 = I_a1\n",
+      "I_a2 = 20.\n",
+      "I_2 = I_a2\n",
+      "I_se2 =  I_2 *(R_x/(R_se+R_x))\n",
+      "\n",
+      "# Calculations\n",
+      "E_b1  =  V -I_a1*R_a - I_a1*R_se\n",
+      "E_b2  =  V -I_a2*R_a - I_se2*R_se\n",
+      "\n",
+      "phi2_by_phi1 = 70./100\n",
+      "N_1 = 1000\n",
+      "N_2 = N_1*(E_b2/E_b1)  /phi2_by_phi1        \t\t\t#N (prop.) E_b/phi\n",
+      "\n",
+      "# Results\n",
+      "print 'Required speed is %.2f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required speed is 1443.93 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19 page no : 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 110.\n",
+      "P = 4.\n",
+      "R_a  =  0.1\n",
+      "R = 0.01  \t\t\t#A resistance of 0.01 ohms\n",
+      "R_se = R+R\n",
+      "\n",
+      "# Calculations\n",
+      "#case(i)\n",
+      "I_1 = 50.\n",
+      "I_a1 = I_1\n",
+      "N_1 = 700.\n",
+      "E_b1  =  V -I_a1*(R_a + R_se)\n",
+      "\n",
+      "#T (prop) phi*I_a   from torque equation                                     (1)\n",
+      "\n",
+      "#phi_1 (prop.) I_a1                                                           (2)\n",
+      "#case(ii) when I_a2 gets divided to half\n",
+      "#phi_2 (prop.) I_a2/2                                                         (3)\n",
+      "\n",
+      "#combining (1)(2)(3) and T1 = T2\n",
+      "I_a2  =  math.sqrt(2*I_a1**2)\n",
+      "R_se_eqvt = (R*R)/(R+R)   \t\t\t#Equavalent of parallel combination\n",
+      "E_b2  =  V - I_a2*R_a - I_a2* R_se_eqvt\n",
+      "\n",
+      "#using speed equation N (prop.) E_b / phi  and using (2) and (3)\n",
+      "N_2  =  N_1 *( E_b2/E_b1) *(I_a1/(I_a2/2))\n",
+      "\n",
+      "# Results\n",
+      "print 'Speed after reconnection  =  %.3f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed after reconnection  =  976.389 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.20 page no : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 4.\n",
+      "I_a1 =  50.\n",
+      "N_1 = 2000.\n",
+      "V = 230.\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#phi_1 is proportioanl to total ampere-turns produced by field coils\n",
+      "#phi_1 (prop.) I_a1*P*n (prop.) 200*n                                         (1)\n",
+      "\n",
+      "#After reconnection ,phi_2 proportional to ampere turns divided as follows\n",
+      "#phi_2 (prop.) [I_a2/2*2*n + I_a2/2*2*n]  (prop.) 2*n*I_a2                    (2)\n",
+      "\n",
+      "# Dividing (1) and (2) ,(phi_1/phi_2) = 100 / I_a2                             (3)\n",
+      "\n",
+      "#T (prop.) phi*I_a AND T (prop.) N**2                                          (4)(5)\n",
+      "#therefore N**2 (prop.) phi*I_a                                                (6)\n",
+      "\n",
+      "#N (prop.) E_b/phi (prop.) 1/phi   ..\n",
+      "#Because drops across windings can be neglected , E_b1 = E_b2\n",
+      "#therefore N (prop.) 1/phi                                                    (7)\n",
+      "\n",
+      "#using (7) and (6) phi**3 (prop.) 1/I_a                                        (8)\n",
+      "\n",
+      "#combining (3) and (8)\n",
+      "I_a2  =   (50.*100**3)**(1./4)    \t\t\t#new armature current\n",
+      "print 'New armature current =  %.3f A'%(I_a2)\n",
+      "#combining (6) and (7)   N**3 (prop.) I_a1\n",
+      "N_2 = N_1 *(I_a2/I_a1)**(1./3)\n",
+      "print 'New motor speed  = %.3f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New armature current =  84.090 A\n",
+        "New motor speed  = 2378.414 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example 2.22 page no : 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 200.\n",
+      "I_a1 = 30.\n",
+      "R_t = 1.5  \t\t\t#R_a + R_se\n",
+      "E_b1 =  V - I_a1*R_t\n",
+      "N2_by_N1 = (60./100)\n",
+      "\n",
+      "# Calculations\n",
+      "#T (prop.) I_a**2  and T (prop.) N_3....therefore I_a**2 (prop.) N**3\n",
+      "I_a2   =  I_a1*math.sqrt(N2_by_N1**3)\n",
+      "\n",
+      "#N (prop.) E_b/I_a\n",
+      "N2_by_N1\n",
+      "E_b2  =  E_b1 *(I_a2/I_a1)*N2_by_N1\n",
+      "R_x =  (V- E_b2)/I_a2 - R_t        \t\t\t#because E_b2 =  V - I_a2*(R_x+R_t)\n",
+      "\n",
+      "# Results\n",
+      "print 'Additional resistance to be added in series with motor circuit  =  %.3f ohms'%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Additional resistance to be added in series with motor circuit  =  9.744 ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.23 page no : 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array,roots\n",
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "R_a = 0.4 \n",
+      "R_sh = 100. \t\t\t#armature and shunt field resistance\n",
+      "I_sh1 = V / R_sh\n",
+      "P_out_FL  =  10 * 735.5\n",
+      "eta = 85./100 \t\t\t#efficiency\n",
+      "P_in =  P_out_FL/eta\n",
+      "I_L1 =  P_in /V\n",
+      "I_a1 =  I_L1 - I_sh1\n",
+      "\n",
+      "# T (prop.) phi*I_a (prop.) I_sh*I_a    because phi (prop.) I_sh\n",
+      "#Bu torque is constant.. \n",
+      "Ia2_Ish2 =  I_a1*I_sh1\n",
+      "E_b1 =  V - I_a1*R_a\n",
+      "\n",
+      "#N (prop.) E_b/I_sh\n",
+      "#put E_b2 =  V - I_a2*R_a  and solving further for I_sh2 we get ,I_sh2**2 - 1.8824 I_sh2 +0.2417 = 0\n",
+      "p = array([1, -1.8824, 0.2417])\n",
+      "ans = roots(p) \n",
+      "I_sh2 = ans[0]\n",
+      "#root 1 was considered because its always easier to attain root(1) because less resistacne is needeed\n",
+      "#R_x in series with field\n",
+      "R_x  =  (V/I_sh2) -R_sh   \t\t\t#because I_sh2  =  V/(R_sh + R_x)\n",
+      "print 'Extra resistance to be added   =  %.2f ohms'%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Extra resistance to be added   =  43.37 ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.24 page no : 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "R_t = 1. \t\t\t#R_t  =  R_se + R_a\n",
+      "V_1 =  230.\n",
+      "N_1 = 300.\n",
+      "N_2 = 375.\n",
+      "I_1 = 15.\n",
+      "I_a1 = I_1\n",
+      "\n",
+      "# Calculations\n",
+      "#T (prop.) I_a**2  and T (prop.) N_2....therefore I_a**2 (prop.) N**2\n",
+      "I_a2 = I_a1 *(N_2/N_1)\n",
+      "E_b1  =  V_1 - I_a1*(R_t)\n",
+      "\n",
+      "#N (prop.) E_b/I_a\n",
+      "E_b2 =  E_b1*(I_a2/I_a1)*(N_2/N_1)\n",
+      "V_2 = E_b2 + I_a2* (R_t)  \t\t\t#because E_b2  =  V_2 - I_a2*(R_a+R_se)\n",
+      "\n",
+      "# Results\n",
+      "print 'Voltage supply needed  =  %.4f V'%(V_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage supply needed  =  354.6875 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.25 page no : 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "I_L1 = 30.\n",
+      "V = 230.\n",
+      "R_sh = 230.\n",
+      "R_a = 1.\n",
+      "I_sh =  V / R_sh\n",
+      "I_a1 =  I_L1 - I_sh\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "\n",
+      "# Calculations\n",
+      "#T (prop.) phi*I_a (prop.) I_a   as phi is constant\n",
+      "#and torque is constant\n",
+      "I_a2  =  I_a1\n",
+      "N2_by_N1 =  1./2\n",
+      "#N (prop.) E_b/phi (prop.) E_b\n",
+      "E_b2 =  E_b1 *(N2_by_N1)\n",
+      "R_x =  (V- E_b2)/I_a2 - R_a     \t\t\t#Because E_b2  =  V - I_a2*(R_a + R_x)\n",
+      "\n",
+      "# Results\n",
+      "print 'resistance to be inserted in series  =  %.4f ohms '%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance to be inserted in series  =  3.4655 ohms \n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.26 page no : 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T_1 = 40.  \t\t\t#initial torque\n",
+      "#phi_1 is initial flux\n",
+      "#phi_2 is new flux\n",
+      "#T_2 is new torque\n",
+      "#I_a1 is initial current\n",
+      "#I_a2 is new current\n",
+      "phi2_by_phi1  =  1- (30./100)  \t\t\t#decrease by 30 percent\n",
+      "Ia2_by_Ia1 = 1+(15./100)       \t\t\t#increase by 15 percent\n",
+      "\n",
+      "# Calculations\n",
+      "#T (prop.)phi*I_a\n",
+      "T_2 = T_1*(phi2_by_phi1)*(Ia2_by_Ia1)\n",
+      "\n",
+      "# Results\n",
+      "print 'New torque is %.1f N-m'%(T_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New torque is 32.2 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.27 page no : 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "V = 230.\n",
+      "N_1 = 1000.\n",
+      "N_2 = 950.\n",
+      "R_a = 0.5\n",
+      "R_sh = 230. \t\t\t#armature and shunt field resistance\n",
+      "I_L1 = 10.\n",
+      "\n",
+      "# Calculations\n",
+      "I_sh  =  V/R_sh\n",
+      "I_a1  =  I_L1 - I_sh\n",
+      "\n",
+      "#T (prop.) phi*I_a  (prop.) I_a with phi constant and T is constant due to full-load\n",
+      "I_a2 = I_a1\n",
+      "\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "E_b2 = E_b1*(N_2/N_1) \t\t\t#N (prop.) E_b /phi (prop.) E_b   as phi is constant\n",
+      "\n",
+      "R_x  =  (V-E_b2)/I_a2  -R_a      \n",
+      "\n",
+      "# Results\n",
+      "print 'resistance to be inserted in series with armature  =  %.4f ohms'%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance to be inserted in series with armature  =  1.2528 ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.28 page no : 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "N_0 = 1000.\n",
+      "I_0 = 5.\n",
+      "R_a = 0.2\n",
+      "R_sh = 250. \t\t\t#armature and shunt field resistance\n",
+      "I_L = 50. \t\t\t#on no load\n",
+      "I_sh = V / R_sh\n",
+      "I_a0  =  I_0 - I_sh\n",
+      "I_a   =  I_L - I_sh\n",
+      "E_b0  =  V- I_a0*R_a\n",
+      "E_b1  =  V- I_a *R_a\n",
+      "\n",
+      "# Calculations\n",
+      "phi1_by_phi0  = 1-(3./100) \t\t\t#weakens by 3 percent\n",
+      "#N (prop.) E_b/phi\n",
+      "N_1  =  N_0 *(E_b1/E_b0) /phi1_by_phi0\n",
+      "\n",
+      "# Results\n",
+      "print 'Speed when loaded and drawing 50A current is %.3f r.p.m'%(N_1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed when loaded and drawing 50A current is 993.695 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.29 page no : 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 230.\n",
+      "I_a0 = 3.3\n",
+      "R_a = 0.3\n",
+      "R_sh = 160. \t\t\t#armature and shunt field resistance \n",
+      "I_L1 = 40.\n",
+      "N_0 = 1000.\n",
+      "E_b0  =  V - I_a0*R_a\n",
+      "I_sh = V/ R_sh\n",
+      "I_a1  =  I_L1 - I_sh\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "phi1_by_phi0 =  1- (4./100) \t\t\t#weakening by 4 percent \n",
+      "\n",
+      "# Calculations and Results\n",
+      "N_1  =  N_0 *(E_b1/E_b0)/(phi1_by_phi0)  \t\t\t#because N (prop.) E_b/phi\n",
+      "print 'Full load speed is %.4f rpm'%(N_1)\n",
+      "T_0  =  E_b0*I_a0/(2*math.pi*N_0/60)\n",
+      "T_1  =  T_0*(I_a1/I_a0)*phi1_by_phi0   \t\t\t# because T (prop.) phi*I_a\n",
+      "print 'Full load developed torque is %.4f N-m'%(T_1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Full load speed is 993.5485 rpm\n",
+        "Full load developed torque is 80.9585 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.30 page no : 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "I_L = 52.\n",
+      "N_1 = 750.\n",
+      "N_2 = 600.\n",
+      "R_a = 0.2\n",
+      "R_sh  =  110. \t\t\t#armature and shunt field resistance\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_sh = V/ R_sh\n",
+      "I_a1 =  I_L - I_sh\n",
+      "I_a2 = I_a1\t\t\t#T (prop.) I_a  and T is constant\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "\n",
+      "#N (prop.) E_b/phi (prop.) E_b\n",
+      "E_b2  =  E_b1*(N_2/N_1)\n",
+      "R_x   =  (V- E_b2)/I_a2 -R_a  \t\t\t#Because E_b2  =  V - I_a2*(R_a+R_x)\n",
+      "print 'resistance to be connected in series  =  %.2f ohms'%(R_x)\n",
+      "\n",
+      "#After R_x gets connected in series with armature and 110 ohms in series with field winding\n",
+      "N_1 = 600.\n",
+      "I_sh2 = V /(R_sh+110)\n",
+      "I_a1 = 50.\n",
+      "I_sh1 = 2.\n",
+      "I_sh2 = 1.\n",
+      "#T (prop.) I_a*I_sh and T doesn't vary\n",
+      "I_a2  =  I_a1*(I_sh1/I_sh2)\n",
+      "E_b1  =  V - I_a1*(R_a+R_x)\n",
+      "E_b2  =  V - I_a2*(R_a+R_x)\n",
+      "N_2  =  N_1*(E_b2/E_b1)*(I_sh1/I_sh2) \t\t\t#Because N (prop.) E_b/I_sh\n",
+      "print 'New speed =  %.3f rpm'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance to be connected in series  =  0.84 ohms\n",
+        "New speed =  828.571 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.31 page no : 72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 230.\n",
+      "R_a = 0.15\n",
+      "R_sh = 250. \t\t\t#armature and shunt field resistance\n",
+      "I_a1 = 50.\n",
+      "I_a2 =  80.\n",
+      "N_1 = 800.\n",
+      "N_2 = 1000.\n",
+      "I_sh1 =  V / R_sh\n",
+      "\n",
+      "# Calculations\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "E_b2  =  V - I_a2*R_a\n",
+      "\n",
+      "I_sh2 = I_sh1*(E_b2/E_b1)*(N_1/N_2) \t\t\t#Because N (prop.) E_b/ I_sh\n",
+      "R_x =  (V/I_sh2 ) - R_sh    \t\t\t#because I_sh2  =  V /(R_x+ R_sh)\n",
+      "\n",
+      "# Results\n",
+      "print 'resistance to be added is R_x = %.0f ohms'%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance to be added is R_x = 69 ohms\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.32 page no : 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 230.\n",
+      "R_a = 0.5\n",
+      "N_1 = 800.\n",
+      "N_2 = 600.\n",
+      "I_a2  = 20. \n",
+      "I_a1 = I_a2\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "\n",
+      "# Calculations\n",
+      "#N (prop.) E_b/phi (prop.) E_b         as phi is constant\n",
+      "E_b2 = E_b1 *(N_2/N_1)\n",
+      "#additional resistance required\n",
+      "R_x  =  (V -E_b2)/I_a2  - R_a   \t\t\t#because E_b2  =  V - I_a2*(R_a+R_x)\n",
+      "\n",
+      "# Results\n",
+      "print 'Additional resistance required  =  %.2f ohms '%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Additional resistance required  =  2.75 ohms \n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.33 page no : 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "R_a = 0.5\n",
+      "R_x = 5. \t\t\t#armature resistacne and extra resistance\n",
+      "I_1 = 15.\n",
+      "I_se1 = I_1\n",
+      "I_se2 = I_se1 \n",
+      "I_2 = I_se2\n",
+      "N_1 = 800.\n",
+      "\n",
+      "# Calculations\n",
+      "E_b1  =  V - I_1*R_a\n",
+      "E_b2  =  V - I_2*(R_a+R_x)\n",
+      "\n",
+      "N_2 =  N_1*(E_b2/E_b1)*(I_se1/I_se2) \t\t\t#because N (prop.) E_b/I_se\n",
+      "\n",
+      "# Results\n",
+      "print 'New speed of rotor  =  %.3f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New speed of rotor  =  517.647 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.34 page no : 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "I_a1 = 20.\n",
+      "R_a = 0.5\n",
+      "N_1 = 1000.\n",
+      "N_2 = 500.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#T (prop.) I_a and T_1 = T_2\n",
+      "I_a2 = I_a1\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "\n",
+      "#N (prop.) E_b\n",
+      "E_b2 =  E_b1 *(N_2/N_1)\n",
+      "R_x =  (V-E_b2)/I_a2 - R_a   \t\t\t#because E_b2  =  V - I_a2*(R_a+R_x)\n",
+      "print 'Additional resistance  =  %.0f ohms'%(R_x)\n",
+      "T3_by_T2 = 0.5   \t\t\t#torque is halved\n",
+      "I_a3 =  I_a2 *(T3_by_T2)   \t\t\t#new armature current\n",
+      "E_b3  =  V - I_a3*(R_x + R_a)\n",
+      "N_3 = E_b3*N_2 / E_b2 \t\t\t#N (prop.) E_b\n",
+      "print 'New speed  =  %.3f rpm'%(N_3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Additional resistance  =  6 ohms\n",
+        "New speed  =  770.833 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.35 page no : 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "P_out =  100*735.5\n",
+      "V = 500.\n",
+      "P = 4.\n",
+      "A = 2.\t\t\t# due to wave winding\n",
+      "Z = 492. \t\t\t#no of conductors\n",
+      "phi = 50.*10**-3 \t\t\t#flux per pole\n",
+      "eta = 92./100 \t\t\t#efficiency\n",
+      "P_in =  P_out/eta\n",
+      "R_a = 0.1 \n",
+      "R_sh = 250. \t\t\t#amature and shunt field resistance\n",
+      "\n",
+      "# Calculations\n",
+      "I_L = P_in/V\n",
+      "I_sh  =  V/ R_sh\n",
+      "I_a  =  I_L - I_sh\n",
+      "E_b  =  V - I_a*R_a\n",
+      "N = E_b*60*A/(phi*P*Z)  \t\t\t#because E_b =   phi*P*N*Z/(60*A)\n",
+      "\n",
+      "T_sh =  P_out/(2*math.pi*N/60) \t\t\t#Useful torque\n",
+      "\n",
+      "# Results\n",
+      "print 'i)Speed at full load  =  %.4f rpm'%(N)\n",
+      "print 'ii)Useful torque  =  %.2f N-m'%(T_sh)\n",
+      "print 'Answer mismatches due to improper approximation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Speed at full load  =  590.5011 rpm\n",
+        "ii)Useful torque  =  1189.41 N-m\n",
+        "Answer mismatches due to improper approximation\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.36 page no : 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import roots\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "N_1 = 1000.\n",
+      "I_1 = 50.\n",
+      "I_a1 = I_1\n",
+      "V = 250.\n",
+      "R_x = 4.4\n",
+      "R_t = 0.6   \t\t\t#R_t  =  R_a+R_se\n",
+      "E_b1 = V - I_a1*(R_t)\n",
+      "\n",
+      "# Calculations\n",
+      "#T (prop.)I_a**2   T (prop.) N**2  .... hence N (prop.) I_a\n",
+      "#N (prop.) E_b /I_a \n",
+      "#combining both  E_b (prop.) I_a**2\n",
+      "#using E_b2  =  V - I_a2*(R_a + R_se + R_x) and solving for I_a2 we get 0.088 I_a2**2 +5 I_a2 -250 = 0\n",
+      "p = [0.088 ,5, -250] \n",
+      "ans = roots(p)\n",
+      "I_a2 = ans[1] \t\t\t#root(1) is ignored as it is -ve\n",
+      "E_b2  =  V - I_a2*(R_t + R_x) \n",
+      "N_2 = N_1*(E_b2/E_b1)*(I_a1/I_a2)\n",
+      "\n",
+      "# Results\n",
+      "print 'Motor speed  =  %.2f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Motor speed  =  639.79 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.37 page no : 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import roots\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "I_a1 = 20.\n",
+      "R_sh =  250.\n",
+      "R_a = 0.5 \t\t\t#shunt field and armature resistance\n",
+      "I_sh1 =  V / R_sh\n",
+      "E_b1  =  V - I_a1*R_a\n",
+      "\n",
+      "#T (prop.) phi*I_a (prop.) I_sh*I_a\n",
+      "#math.since T_1  =  T_2\n",
+      "I_sh2_I_a2  =  I_sh1*I_a1  \n",
+      "I_sh2_I_a2  =  I_sh1*I_a1  \t\t\t# = 20\n",
+      "\n",
+      "# Calculations\n",
+      "#N (prop.) E_b/I_sh\n",
+      "#E_b1  =  V - I_a1*R_a\n",
+      "#Solving further for I_a2 we get I_a2**2 -500 I_a2 + 12800\n",
+      "p = [1, -500, 12800]\n",
+      "ans = roots(p)\n",
+      "I_a2 = ans[1] \t\t\t#higher root is neglected\n",
+      "I_sh2 =  I_sh2_I_a2 / I_a2\n",
+      "R_x =  (V / I_sh2) - R_sh \t\t\t#resistance to be inserted in shunt field\n",
+      "\n",
+      "# Results\n",
+      "print 'resistance to be inserted  =  %.4f ohms '%(R_x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance to be inserted  =  88.3129 ohms \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.38 page no : 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "N_1 = 1000.\n",
+      "I_L1 = 25.\n",
+      "R_a = 0.2\n",
+      "R_sh = 250. \t\t\t#armature and shunt field resistance\n",
+      "V_brush =  1. \t\t\t#voltage drop due to brushes\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_sh1  =  V/R_sh\n",
+      "I_a1 =  I_L1 - I_sh1\n",
+      "E_b1 =  V- I_a1*R_a - 2 *V_brush\n",
+      "\n",
+      "#when loaded\n",
+      "I_L2 = 50.\n",
+      "I_sh2 = I_sh1   \t\t\t#as flux weakensby armature reaction shunt field current remains same \n",
+      "I_a2 =  I_L2 - I_sh2\n",
+      "E_b2 =  V- I_a2*R_a - 2 *V_brush\n",
+      "\n",
+      "phi2_by_phi1 =  1- (3./100)  \t\t\t#weakens by 3 percent\n",
+      "N_2 =  N_1*(E_b2/E_b1)/ phi2_by_phi1   \t\t\t#N (prop.) E_b/phi\n",
+      "print 'New speed  =  %.3f rpm'%(N_2)\n",
+      "T_1 =  E_b1*I_a1/(2*math.pi*N_1/60)\n",
+      "T_2 =  E_b2*I_a2/(2*math.pi*N_2/60)\n",
+      "print 'Torque before field weakening  =  %.4f N-m'%(T_1)\n",
+      "print 'Torque after  field weakening  =  %.4f N-m'%(T_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New speed  =  1009.733 rpm\n",
+        "Torque before field weakening  =  55.7373 N-m\n",
+        "Torque after  field weakening  =  110.3831 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.39 page no : 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "R_a = 0.5\n",
+      "R_x = 1. \t\t\t#armature resistance and extra resistance\n",
+      "N_FL = 500. \t\t\t#full load speed in r.p.m\n",
+      "I_a_FL = 30.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)   Full load \n",
+      "E_b_FL =  V- I_a_FL * R_a\n",
+      "#T (prop.) I_a... T is constant\n",
+      "I_a_dash_FL  =  I_a_FL \n",
+      "E_b_dash_FL  =   V- I_a_dash_FL * (R_a+R_x)\n",
+      "#N (prop.) E_b/phi   (prop.) E_b\n",
+      "N_dash_FL  =  N_FL*(E_b_dash_FL/E_b_FL)\n",
+      "print 'i)Speed at full load torque  = %.4f r.p.m'%(N_dash_FL)\n",
+      "\n",
+      "#part(ii)\n",
+      "T2_by_T1  =  2\n",
+      "I_a_dash_FL   =   I_a_FL  *(T2_by_T1)\n",
+      "E_b_dash_FL  =   V- I_a_dash_FL * (R_a+R_x)\n",
+      "N_dash_FL  =  N_FL*(E_b_dash_FL/E_b_FL)\n",
+      "print 'ii)Speed at double full load torque  = %.3f r.p.m'%(N_dash_FL)\n",
+      "\n",
+      "#part(iii) ...stalling\n",
+      "E_b = 0 \t\t\t#as speed is zero in case of stalling torque\n",
+      "I_a_stall = (V-E_b)/(R_a+R_x)\n",
+      "T_FL  =  E_b_FL * I_a_FL/(2*math.pi*N_FL/60)\n",
+      "T_stall  =  T_FL *(I_a_stall/ I_a_FL)\n",
+      "print 'iii)Stalling torque  =  %.3f Nm'%(T_stall)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Speed at full load torque  = 426.8293 r.p.m\n",
+        "ii)Speed at double full load torque  = 317.073 r.p.m\n",
+        "iii)Stalling torque  =  574.231 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.40 page no : 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 230.\n",
+      "I_a1 = 30.\n",
+      "R_a = 0.4\n",
+      "R_x = 1.1\t\t\t#armature resistance and extra resistance\n",
+      "N_1 = 500.\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "E_b1 =  V - I_a1*R_a\n",
+      "I_a2  =  I_a1 \t\t\t#I_a is constant as T , phi are constant\n",
+      "E_b2 =  V - I_a2*(R_a+R_x)\n",
+      "N_2  =  N_1 *(E_b2/E_b1) \t\t\t#Because N (prop.) E_b/phi (prop.) E_b\n",
+      "print 'i)Speed at full load torque  =  %.3f r.p.m'%(N_2)\n",
+      "\n",
+      "#part(ii)\n",
+      "T2_by_T1 = 1.5\n",
+      "I_a2 =  I_a1 * T2_by_T1\n",
+      "E_b2 =  V - I_a2*(R_a+R_x)\n",
+      "N_2  =  N_1 *(E_b2/E_b1) \t\t\t#Because N (prop.) E_b/phi (prop.) E_b\n",
+      "print 'ii)Speed at 1.5 times full load torque  =  %.3f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Speed at full load torque  =  424.312 r.p.m\n",
+        "ii)Speed at 1.5 times full load torque  =  372.706 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch3_2.ipynb b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch3_2.ipynb
new file mode 100755
index 00000000..70deb8f3
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch3_2.ipynb
@@ -0,0 +1,1904 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:19909d6eede67a544640ad0d428f3fdc9bcbea2eb33090ca90ae171a5a4991eb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Testing of D C Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 page no : 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 6.\n",
+      "V = 500.\n",
+      "A = 2. \t\t\t#because of wave wound armature\n",
+      "Z = 1200. \t\t\t#number of armature conductors\n",
+      "phi = 20.*10**-3 \t\t\t#useful flux per pole\n",
+      "Ra = 0.5 \n",
+      "Rsh = 250.  \t\t\t#armature and field resistance\n",
+      "Il = 20. \t\t\t#current drawn from supply\n",
+      "mechanical_losses = 900.\n",
+      "Ish = V/Rsh\n",
+      "Ia = Il-Ish\n",
+      "Eb = V-Ia*Ra  \t\t\t#because V = Eb+Ia*Ra\n",
+      "N = Eb*60*A/(phi*Pole*Z)   \t\t\t#Eb = phi*Pole*N*Z/(60*A)\n",
+      "\n",
+      "# Calculations\n",
+      "P_m = Eb*Ia  \t\t\t#Electrical equivalent of mechanical power\n",
+      "omega = 2*math.pi*N/60\n",
+      "Tg = P_m/omega\n",
+      "\n",
+      "P_out = P_m-mechanical_losses\n",
+      "T_sh = P_out/omega  \t\t\t#Useful torque\n",
+      "P_in = V*Il\n",
+      "percentage_efficiency = 100*P_out/P_in\n",
+      "\n",
+      "# Results\n",
+      "print 'Speed developed is %.3f r.p.m \\\n",
+      "\\nTorque developed is %.2f N-m  \\\n",
+      "\\nUseful torque is %.2f N-m \\\n",
+      "\\nEfficiency is %.2f percent'%(N,Tg,T_sh,percentage_efficiency)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed developed is 409.167 r.p.m \n",
+        "Torque developed is 206.26 N-m  \n",
+        "Useful torque is 185.26 N-m \n",
+        "Efficiency is 79.38 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 page no : 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#no load\n",
+      "I_noload = 2.5 \t\t\t#No load current\n",
+      "V = 440.\n",
+      "R_a = 1.2\n",
+      "R_sh = 550.\t\t\t#resistance of armature and shunt field windings\n",
+      "no_load_input = V*I_noload\n",
+      "\n",
+      "# Calculations\n",
+      "I_sh = V/R_sh\n",
+      "I_a_noload = I_noload-I_sh\n",
+      "no_load_armature_copper = (I_a_noload**2)*R_a\n",
+      "constant_losses = no_load_input-no_load_armature_copper\n",
+      "\n",
+      "#full load\n",
+      "I_fullload = 32\n",
+      "I_a_fullload = I_fullload-I_sh\n",
+      "full_load_armature_coppe = (I_a_fullload**2)*R_a\n",
+      "total_losses = full_load_armature_coppe+constant_losses\n",
+      "full_load_motor_input = V*I_fullload\n",
+      "full_load_motor_output = full_load_motor_input-total_losses\n",
+      "efficiency_at_full_load = full_load_motor_output*100/full_load_motor_input\n",
+      "\n",
+      "# Results\n",
+      "print 'Full load motor output is %.2f W \\\n",
+      "\\nEfficiency of motor at full-load is %.2f percent'%(full_load_motor_output,efficiency_at_full_load)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Full load motor output is 11815.34 W \n",
+        "Efficiency of motor at full-load is 83.92 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 page no : 11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "#no load\n",
+      "I = 14. \t\t\t#input current\n",
+      "V = 230.\n",
+      "power_output_FL  =  45.*10**3\n",
+      "power_input = V*I\n",
+      "I_sh = 2.55 \t\t\t#field current\n",
+      "R_a = 0.032 \t\t\t#armature resistance\n",
+      "I_a = I-I_sh\n",
+      "cu_loss_NL  =  I_a**2*R_a \t\t\t#no load copper loss\n",
+      "brush_loss = 2*I_a\n",
+      "constant_loss =  power_input - cu_loss_NL - brush_loss\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#full load\n",
+      "#I = I_a+ 2.55\n",
+      "#Motor input =  Motor output + constant loss + brush loss + cu loss\n",
+      "# solving for I_a I_a**2 - 7125 I_a + 1487700.3  = 0\n",
+      "p = [1, -7125, 1487700.3]\n",
+      "ans = roots(p)\n",
+      "I_a = ans[1] \t\t\t#ignoring second root as its too large\n",
+      "I = I_a+I_sh\n",
+      "print 'Full load current is %.2f A'%(I)\n",
+      "power_input = V*I\n",
+      "eta = 100*(power_output_FL/power_input)\n",
+      "print 'Efficiency at full load is %.2f percent'%(eta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Full load current is 217.86 A\n",
+        "Efficiency at full load is 89.81 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 page no : 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "W1 = 9.1  \t\t\t#Tension on tight side\n",
+      "W2 = 0.8  \t\t\t#Tension on slack side\n",
+      "I = 10.    \t\t\t#Total_current\n",
+      "V = 110.   \t\t\t#Supply voltage\n",
+      "R = 7.5/100   \t\t\t#Radius of p-ulley in metres\n",
+      "N = 1320.     \t\t\t#speed in r.p.m\n",
+      "\n",
+      "# Calculations\n",
+      "T_sh = (W1-W2)*9.81*R   \t\t\t#9.81 is the accelration due to gravity\n",
+      "omega = (2*math.pi*N/60)\n",
+      "P_out = T_sh*omega\n",
+      "P_in = V*I\n",
+      "\n",
+      "efficiency = 100*P_out/P_in\n",
+      "\n",
+      "# Results\n",
+      "print 'Full load Efficiency is %.2f percent'%(efficiency)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Full load Efficiency is 76.74 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 page no : 17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "I_av = 10.\n",
+      "V_av = (240.+220)/2 \t\t\t#average voltage across load\n",
+      "W_dash = V_av*I_av  \t\t\t#Power absorbed\n",
+      "t1 = 25.\n",
+      "t2 = 6.\n",
+      "R_sh = 200.\n",
+      "R_a = 0.3\t\t\t#resistance of field winding and armature \n",
+      "\n",
+      "# Calculations\n",
+      "W = W_dash*t2/(t1-t2)   \t\t\t#Stray Losses\n",
+      "I_l = 25 \t\t\t#Input current\n",
+      "I_sh = V/R_sh \t\t\t#current through field winding\n",
+      "\n",
+      "I_a = I_l-I_sh   \t\t\t#Armature current\n",
+      "arm_cu_loss = R_a*I_a**2  \t\t\t#Armature copper losses\n",
+      "sh_cu_loss = R_sh*I_sh**2 \t\t\t# Shunt copper loss\n",
+      "\n",
+      "Total_losses =  arm_cu_loss + sh_cu_loss + W\n",
+      "Motor_input = V*I_l\n",
+      "Output = Motor_input- Total_losses\n",
+      "efficiency = Output*100/Motor_input\n",
+      "\n",
+      "# Results\n",
+      "print 'Efficiency as motor at 25 A and 250 V is %.2f percent'%(efficiency)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency as motor at 25 A and 250 V is 80.67 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 page no : 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "I_a = 37.\n",
+      "I_sh = 0.85\t\t\t#armature and field current for motor\n",
+      "V = 230.\n",
+      "R_a = 0.33 \t\t\t#armature resistance\n",
+      "\n",
+      "I_a_g = 30.\n",
+      "I_sh_g = 0.8\t\t\t#armature and field current for generator\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#for motor\n",
+      "arm_cu_loss =  I_a**2* R_a   \t\t\t#armature copper losses\n",
+      "field_cu_loss = V*I_sh  \t\t\t#field copper loss\n",
+      "total_cu_loss =  field_cu_loss + arm_cu_loss \t\t\t#total copper loss \n",
+      "\n",
+      "#for generator\n",
+      "arm_cu_loss_g =  I_a_g**2* R_a   \t\t\t#armature copper losses\n",
+      "field_cu_loss_g = V*I_sh_g  \t\t\t#field copper loss\n",
+      "total_cu_loss_g =  field_cu_loss_g + arm_cu_loss_g \t\t\t#total copper loss\n",
+      "\n",
+      "#for motor-generator set\n",
+      "total_cu_loss_set =    total_cu_loss_g + total_cu_loss\n",
+      "P_supply = V*(I_a - I_a_g + I_sh+ I_sh_g ) \t\t\t#power taken from supply\n",
+      "stray_loss =  P_supply - (total_cu_loss_g + total_cu_loss) \n",
+      "stray_loss_each =   stray_loss/2   \t\t\t#stray loss for each machine\n",
+      "\n",
+      "#efficiency of motor\n",
+      "motor_input  =  V*(I_a+I_sh)\n",
+      "motor_output  =   motor_input - (stray_loss_each + total_cu_loss)\n",
+      "eta_m =   100* motor_output/motor_input \t\t\t#efficiency of motor\n",
+      "print 'Efficiency of motor is %.2f percent '%(eta_m)\n",
+      "#efficiency of generator\n",
+      "generator_input  =  motor_output   \t\t\t#output of motor is input of generator\n",
+      "generator_output  =   generator_input - (stray_loss_each + total_cu_loss_g)\n",
+      "eta_g =   100* generator_output/generator_input \t\t\t#efficiency of generator\n",
+      "print 'Efficiency of generator is %.2f percent '%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 87.62 percent \n",
+        "Efficiency of generator is 88.05 percent \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 page no : 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I_1 = 400. \t\t\t#motor input current\n",
+      "V = 200.  \t\t\t#voltage across armature\n",
+      "I_2 = 32.  \t\t\t#load current\n",
+      "V_2 = 160.  \t\t\t#voltage across generator\n",
+      "V_f = 15. \t\t\t#voltage drop across field windings\n",
+      "total_input  =  (V+V_f)*I_1  \n",
+      "Output = V_2*I_2\n",
+      "total_losses  =  total_input-Output   \t\t\t#total losses in 2 machines\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R_se = V_f/I_1   \t\t\t#series field resistance\n",
+      "R_a = 0.4 \t\t\t# armature field resistance\n",
+      "\n",
+      "total_cu_loss = (R_a + 2*R_se) * I_1**2 + I_2**2*R_a \t\t\t#total copper loss\n",
+      "stray_losses  =   total_losses -  total_cu_loss\n",
+      "stray_losses_each  = stray_losses /2 \t\t\t#stray losses for each machine\n",
+      "\n",
+      "#for motor\n",
+      "motor_input =  V*I_1\n",
+      "arm_cu_loss =  (R_a + R_se)*I_1*I_1 \t\t\t#armature copper loss\n",
+      "total_losses_motor  =  arm_cu_loss +  stray_losses_each \n",
+      "motor_output =  motor_input- total_losses_motor\n",
+      "eta_m = 100*motor_output/motor_input \t\t\t#efficiency of motor\n",
+      "print 'Efficiency of motor is %.2f percent '%(eta_m)\n",
+      "#for generator\n",
+      "arm_cu_loss_gen = R_a*I_2**2 \t\t\t#armature copper loss\n",
+      "series_field_cu_loss  =  V_f*I_1 \t\t\t#series field copper loss\n",
+      "total_losses_gen =  arm_cu_loss_gen +  series_field_cu_loss + stray_losses_each \n",
+      "generator_input  =  total_losses_gen+ Output\n",
+      "eta_gen = 100*Output/generator_input \t\t\t#efficiency of generator\n",
+      "print 'Efficiency of generator is %.2f percent'%(eta_gen)\n",
+      "\n",
+      "\n",
+      "#Note : 'Answer dont match because Output-of-generator is taken as 5220 for calculation while its should have been 5120'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 9.71 percent \n",
+        "Efficiency of generator is 37.20 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 page no : 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V = 500.\n",
+      "Io = 5. \t\t\t#no load current\n",
+      "R_a = 0.5\n",
+      "R_sh = 250.\t\t\t#resistance of armature and field circuits\n",
+      "I = 100. \t\t\t#current at unknown efficiency\n",
+      "\n",
+      "# Calculations\n",
+      "P_in_NL = V*Io  \t\t\t#no load input\n",
+      "I_sh = V/R_sh\n",
+      "\n",
+      "Iao = Io-I_sh\n",
+      "arm_cu_loss_no_load = R_a*Iao**2  \t\t\t#No load armature copper loss\n",
+      "constant_losses =  P_in_NL- arm_cu_loss_no_load\n",
+      "\n",
+      "I_a = I-I_sh\n",
+      "arm_cu_loss =  R_a*I_a**2  \t\t\t#New armature copper loss\n",
+      "\n",
+      "Total_loss = arm_cu_loss + constant_losses\n",
+      "P_in = V*I\n",
+      "efficiency = (P_in-Total_loss)*100/P_in \t\t\t#required efficiency\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print 'Efficiency is %.3f percent when motor takes %.0f A current'%(efficiency,I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency is 85.405 percent when motor takes 100 A current\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 page no : 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 500.  \n",
+      "I_NL = 5.  \t\t\t#no load current \n",
+      "P_in_NL  =  V* I_NL   \t\t\t#no load input\n",
+      "R_a = 0.22\n",
+      "R_sh = 250. \t\t\t#resistance of armature and shunt field winding\n",
+      "I_sh = V/R_sh\n",
+      "I_a_NL = I_NL-I_sh \t\t\t#armature current no load\n",
+      "arm_cu_loss_NL  =  R_a*I_a_NL**2  \t\t\t#No-load armature copper loss\n",
+      "constant_loss  =   P_in_NL -  arm_cu_loss_NL\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#at 100 A current\n",
+      "I = 100.\n",
+      "I_a  =  I - I_sh\n",
+      "arm_cu_loss  =   R_a*I_a**2 \t\t\t#armature copper loss\n",
+      "total_loss  =  arm_cu_loss +  constant_loss \n",
+      "motor_input  =  V*I\n",
+      "motor_output  =   motor_input- total_loss\n",
+      "eta_m =  100*motor_output/motor_input  \t\t\t#motor efficiency\n",
+      "print 'Parta)Efficiency of motor when it takes 100 A current and loaded is %.2f percent'%(eta_m)\n",
+      "\n",
+      "#part(b)\n",
+      "E_b_NL  =  V - I_a_NL*R_a   \t\t\t#back emf at no load\n",
+      "\n",
+      "E_b  =  V- I_a*R_a  \t\t\t#back EMF at 100 A current\n",
+      "#Since E_b is proportional to N and using componendo dividendo\n",
+      "delta_speed =  100*((E_b_NL-E_b)/E_b)\n",
+      "print 'Partb)Percentage speed in speed is %.3f percent'%(delta_speed)\n",
+      "\n",
+      "print 'Note that the following were assumptions made'\n",
+      "print 'i) Due to heating %( resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected'\n",
+      "print 'ii) Though the motor speed is changing from no load to given load , the mechanical losses are assumed to be cosnstant'\n",
+      "print 'iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Parta)Efficiency of motor when it takes 100 A current and loaded is 90.78 percent\n",
+        "Partb)Percentage speed in speed is 4.368 percent\n",
+        "Note that the following were assumptions made\n",
+        "i) Due to heating %( resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected\n",
+        "ii) Though the motor speed is changing from no load to given load , the mechanical losses are assumed to be cosnstant\n",
+        "iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 page no : 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "motor_output_FL  = 15000. \t\t\t#full load motor output\n",
+      "V = 250.\n",
+      "R_sh = 100.\n",
+      "\n",
+      "# Calculations\n",
+      "#at 80 % of full load\n",
+      "motor_output_FL_dash = (80./100)*motor_output_FL  \t\t\t#80 percent of full load output\n",
+      "eta = 90./100 \t\t\t#efficiency\n",
+      "motor_input = motor_output_FL_dash/eta\n",
+      "total_losses  =  motor_input - motor_output_FL_dash   \t\t\t#at 80 % of full load\n",
+      "#at maximum efficiency variable losses  =  constant losses\n",
+      "constant_losses =  total_losses/2\n",
+      "variable_losses =  constant_losses\n",
+      "I =  motor_input/V \t\t\t#line current at 80% load\n",
+      "I_sh =  V/ R_sh\n",
+      "I_a =  I- I_sh\n",
+      "#math.since armature copper loss  = R_a*I_a**2  \n",
+      "R_a = variable_losses/I_a**2\n",
+      "E_b1 = V-I_a*R_a \t\t\t#motor back EMF at 80% of full load\n",
+      "N_1 = 750. \t\t\t# corresponding speed is given as 750 rpm\n",
+      "\n",
+      "#When motor current is 80 A\n",
+      "I = 80.\n",
+      "I_a = I-I_sh\n",
+      "arm_cu_losses =  R_a*I_a**2  \t\t\t#armature copper loss\n",
+      "total_losses  =   arm_cu_losses + constant_losses\n",
+      "motor_input =  V*I\n",
+      "motor_output  =  motor_input- total_losses\n",
+      "eta = 100*motor_output/motor_input \t\t\t#efficiency of motor\n",
+      "E_b2 = V-I_a*R_a \t\t\t#motor back EMF at 80% of full load\n",
+      "N_2 = N_1*(E_b2/E_b1)  \t\t\t#because E_b is proportional to N\n",
+      "\n",
+      "# Results\n",
+      "print 'Efficiency of motor is %.2f percent when motor draws 80A current'%(eta)\n",
+      "print 'and Speed is %.2f r.p.m'%(N_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 88.92 percent when motor draws 80A current\n",
+        "and Speed is 728.22 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11 page no : 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "R_sh = 166.67\n",
+      "R_a = 0.15 \t    \t\t#resistance of shunt field winding and armature\n",
+      "N_0 = 1280. \t\t\t#in rpm\n",
+      "I_L1 = 67.   \t\t\t#current drawn on full load\n",
+      "I_sh =  V/ R_sh\n",
+      "I_a1 = I_L1 - I_sh\n",
+      "E_b1 = V-I_a1*R_a\n",
+      "\n",
+      "#on no load\n",
+      "I_L0 = 6.5\n",
+      "I_a0 =  I_L0 - I_sh\n",
+      "E_b0 = V-I_a0*R_a\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part (i)\n",
+      "#using speed equation  N is proportional to E_b\n",
+      "N_1 = N_0*(E_b1/E_b0)\n",
+      "print 'i)Full load speed is %.3f r.p.m'%(N_1)\n",
+      "\n",
+      "#part (ii)\n",
+      "speed_regulation = 100*((N_0-N_1)/N_1)\n",
+      "print 'ii)Speed regulation is %.2f percent '%(speed_regulation)\n",
+      "\n",
+      "#part (iii)\n",
+      "stray_losses  =  E_b0*I_a0  \t\t\t#mechanical power developed on no load\n",
+      "power_developed_FL =  E_b1*I_a1 \n",
+      "shaft_output_FL =  power_developed_FL - stray_losses \n",
+      "hp_rating =  shaft_output_FL/746 \t\t\t#in horse power\n",
+      "print 'iii)H.P rating of the machine Is %.2f H.P'%(hp_rating)\n",
+      "\n",
+      "#part (iv)\n",
+      "power_input = V*I_L1\n",
+      "eta = 100*(shaft_output_FL/power_input)  \t\t\t#efficiency at full load\n",
+      "print 'iv)Efficiency at full load is %.2f percent'%(eta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Full load speed is 1233.396 r.p.m\n",
+        "ii)Speed regulation is 3.78 percent \n",
+        "iii)H.P rating of the machine Is 19.42 H.P\n",
+        "iv)Efficiency at full load is 86.48 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 page no : 33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 200.\n",
+      "R_sh = 240.\n",
+      "R_a = 0.1 \t\t\t#resistance of shunt field winding and armature\n",
+      "rotational_loss = 236.\n",
+      "I_L_FL = 9.8  \t\t\t#full load line current\n",
+      "N = 1450.\n",
+      "I_sh = V/R_sh\n",
+      "I_a_FL  =  I_L_FL - I_sh\n",
+      "E_b =  V- I_a_FL * R_a\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "gross_mech_P_dev =  E_b*I_a_FL  \t\t\t#gross mechanical power developed\n",
+      "mech_P_dev =  gross_mech_P_dev - rotational_loss \t\t\t#mechanical power developed\n",
+      "print 'i)Gross mechanical power developed is %.2f W'%(gross_mech_P_dev )\n",
+      "print '   Mechanical power developed is %.2f W'%(mech_P_dev )\n",
+      "\n",
+      "#part(ii)\n",
+      "P_out = mech_P_dev \n",
+      "print 'ii)The power output is %.2f W'%(P_out)\n",
+      "\n",
+      "#part(iii)\n",
+      "T_sh = P_out*60/(2*math.pi*N)\n",
+      "T_L = T_sh\n",
+      "print 'iii)Load torque is %.2f N-m'%(T_L)\n",
+      "\n",
+      "#part(iv)\n",
+      "P_in = V*I_L_FL\n",
+      "eta = 100*P_out/P_in\n",
+      "print 'iv)Efficiency at full load is %.2f percent'%(eta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Gross mechanical power developed is 1785.29 W\n",
+        "   Mechanical power developed is 1549.29 W\n",
+        "ii)The power output is 1549.29 W\n",
+        "iii)Load torque is 10.20 N-m\n",
+        "iv)Efficiency at full load is 79.05 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13 page no : 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 240.\n",
+      "P_out = 25*735.5  \t\t\t#output power in watts\n",
+      "R_a = 0.14\n",
+      "R_sh = 80. \t        \t\t#resistance of armature and shunt field winding\n",
+      "brush_drop = 1. \t\t\t#voltage drop across brush\n",
+      "I_L_FL = 95. \t\t    \t#input current at full load \n",
+      "I_sh = V/R_sh\n",
+      "I_a_FL  =  I_L_FL - I_sh \t\t\t#armature current at full load\n",
+      "\n",
+      "# Calculations and Results\n",
+      "arm_cu_loss_FL  =  R_a*I_a_FL**2  \t\t\t#full load armature copper loss\n",
+      "field_cu_loss =  R_sh*I_sh**2  \t\t\t#field copper loss\n",
+      "print 'i)Armature and field copper losses are %.2f W and %.0f W respectively'%(arm_cu_loss_FL,field_cu_loss)\n",
+      "brush_cu_loss =  2*brush_drop*I_a_FL \t\t\t#brush contact copper loss\n",
+      "print 'ii)Brush contact copper loss is %.0f W'%(brush_cu_loss)\n",
+      "E_b = V-I_a_FL*R_a - 2*brush_drop \t\t\t#back emf\n",
+      "gross_mech_P_dev =  E_b*I_a_FL  \t\t\t#gross mechanical power developed\n",
+      "IFW_losses  =  gross_mech_P_dev - P_out  \t\t\t#iron friction and windage losses\n",
+      "print 'iii)Core plus mechanical losses  =  %.1f W'%(IFW_losses+field_cu_loss+arm_cu_loss_FL)\n",
+      "eta = 100*(P_out/(P_out + IFW_losses+ brush_cu_loss+field_cu_loss+arm_cu_loss_FL  ))\n",
+      "print 'iv)Efficiency is %.2f percent'%(eta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Armature and field copper losses are 1184.96 W and 720 W respectively\n",
+        "ii)Brush contact copper loss is 184 W\n",
+        "iii)Core plus mechanical losses  =  4228.5 W\n",
+        "iv)Efficiency is 80.65 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14 page no : 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I = 5.  \t\t\t#no load current\n",
+      "V = 500.\n",
+      "R_sh = 250.\n",
+      "R_a = 0.5 \t\t\t#resistance of shunt field winding and armature\n",
+      "motor_input_NL  =  V*I\n",
+      "I_sh = V/R_sh\n",
+      "I_a = I-I_sh\n",
+      "arm_cu_loss_NL  =  R_a*I_a**2  \t\t\t#no load armature copper loss\n",
+      "constant_loss  =  motor_input_NL - arm_cu_loss_NL\n",
+      "I_FL = 50.\n",
+      "I_a_FL  =  I_FL - I_sh   \t\t\t#currents at full load\n",
+      "arm_cu_loss_FL  =   R_a*I_a_FL**2 \t\t\t#full load armature copper loss\n",
+      "\n",
+      "# Calculations and Results\n",
+      "total_loss =  constant_loss + arm_cu_loss_FL \n",
+      "motor_input = V*I_FL\n",
+      "motor_output_FL =  motor_input - total_loss\n",
+      "print 'Required output power is %.3f kW'%(motor_output_FL/1000)\n",
+      "eta = 100*(motor_output_FL/motor_input) \t\t\t#full load efficiency\n",
+      "print 'Full load efficiency of motor with 50A current is %.2f percent '%(eta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required output power is 21.352 kW\n",
+        "Full load efficiency of motor with 50A current is 85.41 percent \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15 page no : 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 250.\n",
+      "R_sh = 275.\n",
+      "R_a = 0.8 \t\t\t#resistance of shunt field and amature\n",
+      "I_L0 = 3.91 \t\t\t#load current\n",
+      "I_sh = V/R_sh\n",
+      "I_a0 =  I_L0 - I_sh\n",
+      "constant_losses =  V*I_L0 -R_a*(I_a0)**2\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#as a generator\n",
+      "P_out = 10*10**3\n",
+      "I_L = P_out/V\n",
+      "I_a  =  I_L + I_sh \n",
+      "field_cu_loss = R_sh*(I_sh)**2  \t\t\t#field copper loss\n",
+      "arm_cu_loss =  R_a*(I_a)**2  \t\t\t#armature copper loss  \n",
+      "eta_gen  =  100 *(P_out/(P_out+constant_losses + field_cu_loss+ arm_cu_loss))  \t\t\t#efficiency as generator\n",
+      "print 'Efficiency as a generator  =  %.2f percent'%(eta_gen)\n",
+      "\n",
+      "#as a motor\n",
+      "P_in = 10*10**3  \t\t\t#at V = 250 \n",
+      "I_L = P_in/V\n",
+      "I_a = I_L - I_sh\n",
+      "field_cu_loss = R_sh*(I_sh)**2  \t\t\t#field copper loss\n",
+      "arm_cu_loss =  R_a*(I_a)**2  \t\t\t#armature copper loss  \n",
+      "eta_m  =  100 *((P_in-(constant_losses + field_cu_loss+ arm_cu_loss))/(P_in)) \t\t\t#efficiency as motor\n",
+      "print 'Efficiency as a motor  =  %.2f percent'%(eta_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency as a generator  =  79.77 percent\n",
+        "Efficiency as a motor  =  75.80 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16 page no : 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I = 4. \t\t\t#no load current in amperes\n",
+      "V = 500.\n",
+      "motor_input_no_load = I*V   \t\t\t#no load motor input\n",
+      "R_a = 0.5\n",
+      "R_sh = 250.\t\t\t#resistance of armature and shunt field resistnace\n",
+      "I_sh = V/R_sh\n",
+      "\n",
+      "I_a = I-I_sh\n",
+      "arm_cu_loss_noload = R_a*I_a**2  \t\t\t#No-load armature copper losses\n",
+      "constant_loss = motor_input_no_load - arm_cu_loss_noload\n",
+      "I_FL = 40.\n",
+      "I_aFL = I_FL- I_a  \t\t\t#full load currents\n",
+      "arm_cu_loss_fulload = R_a*I_aFL**2 \t\t\t#Full-load armature copper losses\n",
+      "Total_loss = arm_cu_loss_fulload + constant_loss\n",
+      "\n",
+      "# Calculations and Results\n",
+      "motor_input = V*I_FL\n",
+      "motor_output_fullload = motor_input - Total_loss\n",
+      "print 'Output power at full-load is %.0f W'%(motor_output_fullload) \n",
+      "efficiency =  motor_output_fullload*100/motor_input \t\t\t#motor efficiency\n",
+      "print 'Efficiency at full-load is %.1f percent'%(efficiency)\n",
+      "\n",
+      "E_bNL = V-I_a*R_a\n",
+      "E_bFL = V-I_FL*R_a\n",
+      "\n",
+      "#E_b  = N*phi\n",
+      "#E_bNL/E_bFL = N_NL/N_FL\n",
+      "#applying rules of componendo and dividendo\n",
+      "#change_in _speed = (N_NL - N_FL)/N_FL = (E_bNL - E_bFL)/E_bFl\n",
+      "\n",
+      "change_in_speed = 100*(E_bNL - E_bFL)/E_bFL\n",
+      "print 'percentage change in speed from no load to full load is %.3f percent'%(change_in_speed)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output power at full-load is 17280 W\n",
+        "Efficiency at full-load is 86.4 percent\n",
+        "percentage change in speed from no load to full load is 3.958 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17 page no : 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 500.\n",
+      "R_a = 0.22\n",
+      "R_sh =  250.\t\t\t#armature resistance and shunt field resistance\n",
+      "I = 5.       \t\t\t#no load current\n",
+      "motor_input_NL = V*I   \t\t\t#no load motor input\n",
+      "I_sh = V/R_sh\n",
+      "I_a_NL =  I- I_sh \t\t\t#no load armature current\n",
+      "arm_cu_loss_NL  =  R_a*I_a_NL**2 \t\t\t#no load armature copper loss\n",
+      "constant_loss  =  motor_input_NL - arm_cu_loss_NL \n",
+      "\n",
+      "# Calculations and Results\n",
+      "#When motor draws 100 A current \n",
+      "I = 100\n",
+      "I_a  =  I - I_sh\n",
+      "arm_cu_loss   = R_a* I_a**2  \t\t\t#armature copper loss\n",
+      "total_losses  =    arm_cu_loss +  constant_loss\n",
+      "motor_input  =  V*I\n",
+      "motor_output  =  motor_input -total_losses \n",
+      "eta_m = 100*(motor_output/motor_input) \t\t\t#motor efficiency\n",
+      "print 'i)Efficiency of motor at 100 A current is %.2f percent '%(eta_m)\n",
+      "\n",
+      "#part(b)\n",
+      "E_b_NL =  V- I_a_NL*R_a \t\t\t#back emf at no load\n",
+      "E_b =  V- I_a*R_a \t\t\t#back emf at 100 A \n",
+      "#E_b is proportional to N.. and using componendo dividendo\n",
+      "speed_change =  100*((E_b_NL - E_b)/E_b)\n",
+      "print 'ii)Percentage change in speed  =  %.3f percent'%(speed_change)\n",
+      "\n",
+      "\n",
+      "print 'Note that the following were assumptions made'\n",
+      "print 'i) Due to heating  resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected'\n",
+      "print 'ii) Though the motor speed is changing from no load to given load %( the mechanical losses are assumed to be cosnstant'\n",
+      "print 'iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Efficiency of motor at 100 A current is 90.78 percent \n",
+        "ii)Percentage change in speed  =  4.368 percent\n",
+        "Note that the following were assumptions made\n",
+        "i) Due to heating  resistance of shunt field winding will be increased which will reduce the shunt field current.This will decrease the flux which is neglected\n",
+        "ii) Though the motor speed is changing from no load to given load %( the mechanical losses are assumed to be cosnstant\n",
+        "iiiThe effect of armature reaction aon main pole flux and its effect on iron loss is neglected\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18 page no : 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 200.\n",
+      "I_L = 40.\n",
+      "\n",
+      "R_sh = 200.\n",
+      "R_a = 0.2 \t\t\t#shunt field winding and armature resistance\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#case(a) :  As a generator\n",
+      "P_out_g = V*I_L \t\t\t#output poewr as generator\n",
+      "I_sh = V/R_sh\n",
+      "I_a  =  I_L + I_sh\n",
+      "E  =  V + I_a*R_a\n",
+      "P_a_g = E*I_a  \t\t\t#power developed in armature\n",
+      "P_cu_g =  R_a*I_a**2 + R_sh*I_sh**2 \t\t\t#copper loss as genrator\n",
+      "print 'i)Output as generator is %.f kW'%(P_out_g/1000)\n",
+      "\n",
+      "#case(b) :  As a motor\n",
+      "P_in_m = V*I_L \t\t\t#input power as motor\n",
+      "I_sh = V/R_sh\n",
+      "I_a  =  I_L - I_sh\n",
+      "E_b  =  V - I_a*R_a\n",
+      "P_a_m = E_b*I_a  \t\t\t#power developed in armature\n",
+      "P_cu_m =  R_a*I_a**2 + R_sh*I_sh**2\t\t\t#copper loss as motor\n",
+      "print 'ii)Input as motor is %.0f kW'%( P_in_m/1000)\n",
+      "print 'iii)Power developed in Armature:%.4f kW for generator%.4f kW for motor'%(P_a_g/1000,P_a_m/1000)\n",
+      "print 'iv)Copper losses:%.1f W for generator%.1f W for motor'%(P_cu_g,P_cu_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Output as generator is 8 kW\n",
+        "ii)Input as motor is 8 kW\n",
+        "iii)Power developed in Armature:8.5362 kW for generator7.4958 kW for motor\n",
+        "iv)Copper losses:536.2 W for generator504.2 W for motor\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19 page no : 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 219.\n",
+      "I = 10.\n",
+      "dN = 1030. - 970 \t\t\t#given\n",
+      "t_1 = 36.  \t\t\t#time with no excitation\n",
+      "t_2 = 9.\t\t\t# time with full excitation and armature supporting an extra load of 10 A at 219 V\n",
+      "t_3 = 15. \t\t\t#time with full exciation\n",
+      "W_dash  =  V*I  \t\t\t#additioanl loss when armature is suddenly connected to loads\n",
+      "W_s =  W_dash *(t_2/(t_3-t_2)) \t\t\t#total stray losses\n",
+      "N = 1000.  \t\t\t#speed in rpm\n",
+      "\n",
+      "# Calculations\n",
+      "#using W_s  =  (2*pi/60)**2 * I *N *dN / t_3 where W_s is stray losses\n",
+      "I =  W_s*(t_3/dN)*(30/math.pi)**2/N   \t\t\t#moment of inertia\n",
+      "W_m = W_s*(t_3/t_1) \t\t\t#mechanical losses\n",
+      "iron_losses =  W_s - W_m\n",
+      "\n",
+      "# Results\n",
+      "print 'i)The moment of inertia of armature is %.2f kg-m**2'%(I)\n",
+      "print 'ii)Iron loss =  %.2f W'%(iron_losses)\n",
+      "print 'iii)Mechanical losses at 1000 rpm mean speed is %.2f W'%(W_m)\n",
+      "\n",
+      "print 'Noteworthy points:1)When armature is slowing down and there is no excitation,then kinetic energy is used to overcome mechanical losses only.Iron losses are absent as excitation is absent2)When excitation is given%( kinetic energy is used to overcome both mechanical as well as iron losses.Total called stray losses.3)If moment of inertia is in kg-m**2,then loss of energy is in watts'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)The moment of inertia of armature is 74.89 kg-m**2\n",
+        "ii)Iron loss =  1916.25 W\n",
+        "iii)Mechanical losses at 1000 rpm mean speed is 1368.75 W\n",
+        "Noteworthy points:1)When armature is slowing down and there is no excitation,then kinetic energy is used to overcome mechanical losses only.Iron losses are absent as excitation is absent2)When excitation is given%( kinetic energy is used to overcome both mechanical as well as iron losses.Total called stray losses.3)If moment of inertia is in kg-m**2,then loss of energy is in watts\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20 page no : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 225.\n",
+      "I = 10.\n",
+      "dN = 1030. - 970 \t\t\t#given\n",
+      "t_1 = 40.  \t\t\t#time with no excitation\n",
+      "t_2 = 9.\t\t\t# time with full excitation and armature supporting an extra load of 10 A at 219 V\n",
+      "t_3 = 20. \t\t\t#time with full exciation\n",
+      "\n",
+      "# Calculations\n",
+      "W_dash  =  V*I \t\t\t#additional loss\n",
+      "W_s =  W_dash *(t_2/(t_3-t_2)) \t\t\t#total stray losses\n",
+      "N = 1000\t\t\t#Speed in rpm\n",
+      "\t\t\t#using W_s  =  (2*pi/60)**2 * I *N *dN / t_3 where W_s is stray losses\n",
+      "I =  W_s*(t_3/dN)*(30/math.pi)**2/N   \t\t\t#moment of inertia\n",
+      "W_m = W_s*(t_3/t_1)   \t\t\t#mechanical losses\n",
+      "iron_losses = W_s - W_m\n",
+      "\n",
+      "# Results\n",
+      "print 'i)The moment of inertia of armature is %.2f kg-m**2'%(I)\n",
+      "print 'ii)Iron loss =  %.2f W'%(iron_losses)\n",
+      "print 'iii)Mechanical losses at 1000 rpm mean speed is %.2f W'%(W_m)\n",
+      "\n",
+      "print 'Noteworthy points:1)When there is no excitation and armature is slowed down , its K.E. is used to overcome mechanical mechanical losses only math.since there will be no iron loss as there is no flux.2)When there is excitation provided then K.E. is used to supply mechanical as well as iron losses together called stray losses '\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)The moment of inertia of armature is 55.96 kg-m**2\n",
+        "ii)Iron loss =  920.45 W\n",
+        "iii)Mechanical losses at 1000 rpm mean speed is 920.45 W\n",
+        "Noteworthy points:1)When there is no excitation and armature is slowed down , its K.E. is used to overcome mechanical mechanical losses only math.since there will be no iron loss as there is no flux.2)When there is excitation provided then K.E. is used to supply mechanical as well as iron losses together called stray losses \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21 page no : 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_avg  =  (220.+190)/2  \t\t\t#average voltage across load\n",
+      "I_avg = 12.\n",
+      "R_a = 0.5\n",
+      "R_sh = 250.\n",
+      "W_dash = V_avg*I_avg  \t\t\t#power absorbed \n",
+      "t_1 = 30.\n",
+      "t_2 = 5.\n",
+      "W = W_dash*(t_2/(t_1-t_2))\n",
+      "V = 250.\n",
+      "I = 22.  \t\t\t#input current\n",
+      "\n",
+      "# Calculations\n",
+      "I_sh  =  V/R_sh\n",
+      "I_a =  I - I_sh\n",
+      "arm_cu_loss  =  R_a*I_a**2 \t\t\t#armature copper loss\n",
+      "shunt_field_cu_loss  =   V*I_sh  \t\t\t#shunt field copper loss\n",
+      "total_losses =  shunt_field_cu_loss + arm_cu_loss + W\n",
+      "\n",
+      "machine_input  =  V*I\n",
+      "machine_output  =  machine_input - total_losses \n",
+      "eta_m = 100*(machine_output /machine_input )  \t\t\t#efficiency when running as motor\n",
+      "\n",
+      "# Results\n",
+      "print 'Efficiency of machine when opeating as motor taking current of 22A on 250V supply is %.1f percent'%(eta_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of machine when opeating as motor taking current of 22A on 250V supply is 82.5 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.22 page no : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I_avg = 10.\n",
+      "V_avg = (220.+190)/2 \t\t\t#average voltage across load\n",
+      "W_dash  =  V_avg*I_avg  \t\t\t#power  absorbed\n",
+      "t_1 = 30.\n",
+      "t_2 = 20.\n",
+      "\n",
+      "# Calculations\n",
+      "W = W_dash * (t_2/(t_1-t_2)) \t\t\t#stray losses\n",
+      "\n",
+      "# Results\n",
+      "print 'Stray losses of motor is %.1f kW'%(W/1000)\n",
+      "print 'Answers mismatch because V_average is 205 volts but it is taken as 220 volts in Power absorbed calculation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Stray losses of motor is 4.1 kW\n",
+        "Answers mismatch because V_average is 205 volts but it is taken as 220 volts in Power absorbed calculation\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.23 page no : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I_a_g = 330.\n",
+      "I_a_m = 380.\n",
+      "R_a = 0.02 \t\t\t#armature resistance\n",
+      "V = 250.\n",
+      "I = 50.\n",
+      "\n",
+      "\n",
+      "arm_cu_loss_g =  R_a*I_a_g**2\t\t\t#armature copper loss for generator\n",
+      "arm_cu_loss_m =  R_a*I_a_m**2\t\t\t#armature copper loss for motor\n",
+      "power_drawn = V*I\n",
+      "stray_losses  =  power_drawn - (arm_cu_loss_m + arm_cu_loss_g)\n",
+      "stray_losses_each  =  stray_losses/2 \t\t\t#stray losses for each machine\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#for motor\n",
+      "I_sh_m = 4.2   \t\t\t#Shunt current in case of motor\n",
+      "field_cu_loss_m = V*I_sh_m  \t\t\t#field copper loss in case of motor\n",
+      "total_loss  =  field_cu_loss_m + stray_losses_each  +  arm_cu_loss_m\n",
+      "motor_input =  V*(I_a_m+I_sh_m) \n",
+      "motor_output  =  motor_input - total_loss\n",
+      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
+      "\n",
+      "#for generator\n",
+      "I_sh_g = 5  \t\t\t#Shunt current in case of generator\n",
+      "field_cu_loss_g = V*I_sh_g \t\t\t#field copper loss in case of generator\n",
+      "total_loss  =  field_cu_loss_g + stray_losses_each  +  arm_cu_loss_g\n",
+      "generator_output  =  V*I_a_g\n",
+      "generator_input =  generator_output + total_loss\n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 92.0302 percent\n",
+        "Efficiency of generator is 92.0297 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.24 page no : 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R_a = 0.02 \t\t\t#armature resistance\n",
+      "V = 250. \t\t\t#line voltage\n",
+      "I = 50. \t\t\t#current taken from supply \n",
+      "\n",
+      "# Calculations and Results\n",
+      "#for generator\n",
+      "I_a_g = 330.\n",
+      "I_sh_g = 5. \t\t\t#armature current and current through shunt field \n",
+      "arm_cu_loss_g  =  R_a*I_a_g**2\t\t\t#armature copper loss for generator\n",
+      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
+      "\n",
+      "#for motor\n",
+      "I_a_m = 380\n",
+      "I_sh_m = 4.2 \t\t\t#armature current and current through shunt field\n",
+      "arm_cu_loss_m  =  R_a*I_a_m**2\t\t\t#armature copper loss for motor\n",
+      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
+      "power_drawn = V*I\n",
+      "IFW_losses  =  power_drawn - (arm_cu_loss_g + arm_cu_loss_m)  \t\t\t#Iron friction and windage losses\n",
+      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
+      "\n",
+      "#for generator\n",
+      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each\n",
+      "generator_output = V*I_a_g\n",
+      "generator_input  =  generator_output + total_loss_g \n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n",
+      "\n",
+      "#for motor\n",
+      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m\n",
+      "motor_input = V*(I_a_m+I_sh_m)\n",
+      "motor_output  =  motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of generator is 92.0297 percent\n",
+        "Efficiency of motor is 92.0302 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.25 page no : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "I = 40.\n",
+      "I_a_g = 160. \n",
+      "I_a_m  = 200. \t\t\t#armature currents for generator and motor\n",
+      "I_sh_g = 7. \n",
+      "I_sh_m  = 6. \t\t\t#current through shunt field for generator and motor\n",
+      "R_a = 0.015 \t\t\t#armature resistance\n",
+      "arm_cu_loss_g  =  R_a*I_a_g**2  \t\t\t#armature copper loss for motor\n",
+      "arm_cu_loss_m  =  R_a*I_a_m**2  \t\t\t#armature copper loss for motor\n",
+      "power_drawn = V*I\n",
+      "IFW_losses  =  power_drawn - (arm_cu_loss_g + arm_cu_loss_m)  \t\t\t#Iron friction and windage losses\n",
+      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#for motor\n",
+      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
+      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m  \t\t\t#total losses in motor\n",
+      "motor_input = V * I_a_m\n",
+      "motor_output =  motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)    \t\t\t#motor efficiency\n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
+      "\n",
+      "#for generator\n",
+      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
+      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each  \t\t\t#total losses in generator\n",
+      "generator_output = V*I_a_g\n",
+      "generator_input  =  generator_output + total_loss_g\n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency\n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 86.7545 percent\n",
+        "Efficiency of generator is 85.7867 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.26 page no : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R_a = 0.2 \t\t\t#armature resistance \n",
+      "V = 240.\n",
+      "I = 16.\n",
+      "I_a_g = 60. \n",
+      "I_a_m = 71. \t\t\t#armature currents for generator and motor\n",
+      "I_sh_g = 3. \n",
+      "I_sh_m = 2.  \t\t\t#field current for generator and motor \n",
+      "\n",
+      "\n",
+      "#for generator\n",
+      "arm_cu_loss_g  =  R_a*I_a_g**2\t\t\t#armature copper loss for generator\n",
+      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
+      "\n",
+      "#for motor\n",
+      "arm_cu_loss_m  =  R_a*I_a_m**2\t\t\t#armature copper loss for motor\n",
+      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
+      "power_drawn = V*I\n",
+      "field_loss_total_g_m =  field_cu_loss_m + field_cu_loss_g\n",
+      "arm_cu_loss_total_g_m  =  arm_cu_loss_m + arm_cu_loss_g\n",
+      "IFW_losses  =  power_drawn - (arm_cu_loss_total_g_m + field_loss_total_g_m)  \t\t\t#Iron friction and windage losses\n",
+      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#for generator\n",
+      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each \t\t\t#total loss in generator\n",
+      "generator_output = V*I_a_g\n",
+      "generator_input  =  generator_output + total_loss_g\n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n",
+      "\n",
+      "#for motor\n",
+      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m \t\t\t#total loss in motor\n",
+      "motor_input = V*(I_a_m+I_sh_m)\n",
+      "motor_output  =  motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of generator is 88.3658 percent\n",
+        "Efficiency of motor is 88.9035 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.27 page no : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R_a = 0.015\n",
+      "V = 250. \t\t\t#line voltage\n",
+      "I = 45. \t\t\t#line current\n",
+      "I_a_m = 385.\n",
+      "I_sh_m = 4. \t\t\t#armature and field currents for motor\n",
+      "I_a_g = 340.\n",
+      "I_sh_g = 5. \t\t\t#armature and field currents for generator\n",
+      "arm_cu_loss_m =  R_a*I_a_m**2   \t\t\t#armature copper loss for motor\n",
+      "field_cu_loss_m =  V*I_sh_m    \t\t\t#field copper loss for motor\n",
+      "\n",
+      "arm_cu_loss_g =  R_a*I_a_g**2  \t\t\t#armature copper loss for generator\n",
+      "field_cu_loss_g =  V*I_sh_g   \t\t\t#field copper loss for motor\n",
+      "\n",
+      "total_cu_loss  =  field_cu_loss_g + arm_cu_loss_g  +  field_cu_loss_m + arm_cu_loss_m  \t\t\t#total copper loss for both machines\n",
+      "P_aux  =  V*I  \t\t\t#power taken from auxillary supply\n",
+      "stray_loss =  P_aux - total_cu_loss\n",
+      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
+      "\n",
+      "# Calculations and Results\n",
+      "total_loss_g  =  stray_loss_each + arm_cu_loss_g + field_cu_loss_g  \t\t\t#total losses in generator\n",
+      "generator_output = V* I_a_g\n",
+      "eta_g  =  100*(generator_output/(generator_output + total_loss_g))\t\t\t#generator efficiency\n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n",
+      "\n",
+      "total_loss_m  =  stray_loss_each + arm_cu_loss_m + field_cu_loss_m\t\t\t#total losses in motor\n",
+      "motor_input = V*(I_a_m+I_sh_m)\n",
+      "motor_output  =  motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of generator is 93.9171 percent\n",
+        "Efficiency of motor is 94.0929 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.28 page no : 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 500.\n",
+      "P = 1000.*10**3\n",
+      "I = 30.\n",
+      "I_a_m  =  200. + 30 \n",
+      "I_a_g  = 200. \t\t\t#armature current for motor and generator\n",
+      "I_sh_m  =  1.8\n",
+      "I_sh_g  = 3.5 \t\t\t#field current for motor and generator\n",
+      "brush_drop = 230.\n",
+      "R_a = 0.075  \t\t\t#armature resimath.tance\n",
+      "\n",
+      "# Calculations\n",
+      "arm_cu_loss_m  =  R_a*I_a_m**2 + 2*brush_drop \t\t\t#motor armature copper loss\n",
+      "field_cu_loss_m  = V*I_sh_m  \t\t\t#  motor field copper loss \n",
+      "\n",
+      "arm_cu_loss_g  =  R_a*I_a_g**2 + 2*brush_drop  \t\t\t#generator armature copper loss\n",
+      "field_cu_loss_g  = V*I_sh_g  \t\t\t#field copper loss generator\n",
+      "\n",
+      "total_cu_loss  =  field_cu_loss_g + arm_cu_loss_g  +  field_cu_loss_m + arm_cu_loss_m  \t\t\t#total copper loss for both machines\n",
+      "P_aux  =  V*I  \t\t\t#power taken from auxillary supply\n",
+      "stray_loss =  P_aux - total_cu_loss  \n",
+      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
+      "\n",
+      "total_loss_g  =  stray_loss_each + arm_cu_loss_g + field_cu_loss_g  \t\t\t#total loss in generator\n",
+      "generator_output = V* I_a_g\n",
+      "eta_g  =  100*(generator_output/(generator_output + total_loss_g))\t\t\t#generator efficiency\n",
+      "\n",
+      "# Results\n",
+      "print 'Efficiency of generator is %.0f percent'%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of generator is 93 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.29 page no : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 220.\n",
+      "I = 10.\n",
+      "R_a = 0.05 \t        \t\t#artmature resistance\n",
+      "I_a_m =  73.    \n",
+      "I_sh_m  =  2. \t    \t\t#armature and field current for motor\n",
+      "I_a_g = 67.5\n",
+      "I_sh_g = 2.5\t\t    \t#armature and field current for generator\n",
+      "\n",
+      "arm_cu_loss_m  =  R_a*I_a_m**2 \t\t\t#motor armature copper loss\n",
+      "field_cu_loss_m  = V*I_sh_m  \t\t\t#  motor field copper loss \n",
+      "\n",
+      "arm_cu_loss_g  =  R_a*I_a_g**2 \t\t\t#generator armature copper loss\n",
+      "field_cu_loss_g  = V*I_sh_g  \t\t\t#field copper loss generator\n",
+      "\n",
+      "total_cu_loss  =  field_cu_loss_g + arm_cu_loss_g  +  field_cu_loss_m + arm_cu_loss_m  \t\t\t#total copper loss for both machines\n",
+      "power_input  =  V*I\n",
+      "stray_loss =  power_input - total_cu_loss  \n",
+      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#motor efficiency\n",
+      "total_loss_m =  field_cu_loss_m + stray_loss_each  +  arm_cu_loss_m  \t\t\t#total motor losses\n",
+      "motor_input  =  V*(I_a_m + I_sh_m )\n",
+      "motor_output  = motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
+      "\n",
+      "#generator efficiency\n",
+      "total_loss_g =  field_cu_loss_g + stray_loss_each  +  arm_cu_loss_g  \t\t\t#total generator losses\n",
+      "generator_output  = V*I_a_g\n",
+      "generator_input  =  generator_output +  total_loss_g\n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 93.5496 percent\n",
+        "Efficiency of generator is 92.8956 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.30 page no : 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 400.\n",
+      "I = 50.\n",
+      "I_a_g = 250. \n",
+      "I_a_m  = 300.  \t\t\t#armature current for generator and motor\n",
+      "I_sh_g = 2.5 \n",
+      "I_sh_m  = 2.4 \t\t\t#field current for generator and motor\n",
+      "R_a = 0.1 \t\t\t#armature resistance\n",
+      "\n",
+      "arm_cu_loss_g  =  R_a*I_a_g**2   \t\t\t#armature copper loss for generator\n",
+      "arm_cu_loss_m  =  R_a*I_a_m**2   \t\t\t#armature copper loss for motor\n",
+      "power_drawn = V*I\n",
+      "IFW_losses  =  power_drawn - (arm_cu_loss_g + arm_cu_loss_m)  \t\t\t#Iron friction and windage losses\n",
+      "IFW_losses_each =  IFW_losses /2  \t\t\t# Iron friction and windage losses for each machine\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#for motor\n",
+      "field_cu_loss_m =  V*I_sh_m \t\t\t#field copper loss for motor\n",
+      "total_loss_m =  field_cu_loss_m + IFW_losses_each  +  arm_cu_loss_m\n",
+      "motor_input = V * I_a_m\n",
+      "motor_output =  motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)    \t\t\t#motor efficiency\n",
+      "print 'Efficiency of motor is %.2f percent'%(eta_m)\n",
+      "\n",
+      "#for generator\n",
+      "field_cu_loss_g =  V*I_sh_g \t\t\t#field copper loss for generator\n",
+      "total_loss_g  =  field_cu_loss_g + arm_cu_loss_g + IFW_losses_each\n",
+      "generator_output = V*I_a_g\n",
+      "generator_input  =  generator_output + total_loss_g\n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency\n",
+      "print 'Efficiency of generator is %.2f percent'%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 89.72 percent\n",
+        "Efficiency of generator is 91.22 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.31 page no : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I_1 = 56.  \t\t\t#motor input current\n",
+      "V = 590.   \t\t\t#voltage across armature\n",
+      "I_2 = 44.  \t\t\t# load current\n",
+      "V_2 = 400. \t\t\t# voltage across generator\n",
+      "V_field  =  40.  \t\t\t#voltage drop across field winding\n",
+      "R_a = 0.3\n",
+      "R_se = 0.7142 \t\t\t#armature and series field resistance for each machine\n",
+      "total_input = (V+V_field)*I_1\n",
+      "output = V_2*I_2\n",
+      "total_loss_g_m =  total_input - output   \t\t\t#total losses of 2 machines\n",
+      "R_se = V_field/I_1  \t\t\t#series field resistance for both windings\n",
+      "total_cu_loss  =  (R_a+ 2*R_se)*I_1**2  + R_a*I_2**2  \t\t\t#total copper loss\n",
+      "stray_loss =  total_loss_g_m - total_cu_loss\n",
+      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# for motor\n",
+      "motor_input  =  V*I_1\n",
+      "arm_cu_loss_m  =   (R_a+ R_se)*I_1**2    \t\t\t#armature coper losses of motor\n",
+      "total_loss_m =  arm_cu_loss_m + stray_loss_each\n",
+      "motor_output  =  motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
+      "\n",
+      "# for generator\n",
+      "arm_cu_loss_g  =  R_a*I_2**2  \t\t\t#armature coper losses of generator\n",
+      "series_field_cu_loss_g  =  V_field*I_1   \t\t\t#series field copper loss\n",
+      "total_loss_g =  arm_cu_loss_g + series_field_cu_loss_g + stray_loss_each\n",
+      "generator_output = V_2*I_2\n",
+      "generator_input  =  generator_output +  total_loss_g\n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 72.6998 percent\n",
+        "Efficiency of generator is 67.0221 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.32 page no : 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I_1 = 56.  \t\t\t#motor input current\n",
+      "V = 590.   \t\t\t#voltage across armature\n",
+      "I_2 = 44.  \t\t\t# load current\n",
+      "V_2 = 400. \t\t\t# voltage across generator\n",
+      "V_field  =  40.  \t\t\t#voltage drop across field winding\n",
+      "R_a = 0.3\n",
+      "R_se = 0.7142 \t\t\t#armature and series field resismath.tane for each machine\n",
+      "\n",
+      "total_input = (V+V_field)*I_1\n",
+      "output = V_2*I_2\n",
+      "total_loss_g_m =  total_input - output   \t\t\t#total losses of 2 machines\n",
+      "R_se = V_field/I_1  \t\t\t#series field resistance for both windings\n",
+      "total_cu_loss  =  (R_a+ 2*R_se)*I_1**2  + R_a*I_2**2  \t\t\t#total copper loss\n",
+      "stray_loss =  total_loss_g_m - total_cu_loss\n",
+      "stray_loss_each  =  stray_loss/2  \t\t\t#stray loss for each machine\n",
+      "\n",
+      "# Calculations and Results\n",
+      "# for motor\n",
+      "motor_input  =  V*I_1\n",
+      "arm_cu_loss_m  =   (R_a+ R_se)*I_1**2  \t\t\t#armature coper losses of motor\n",
+      "total_loss_m =  arm_cu_loss_m + stray_loss_each\n",
+      "motor_output  =  motor_input - total_loss_m\n",
+      "eta_m  =  100*(motor_output/motor_input)\t\t\t#motor efficiency \n",
+      "print 'Efficiency of motor is %.4f percent'%(eta_m)\n",
+      "\n",
+      "# for generator\n",
+      "arm_cu_loss_g  =  R_a*I_2**2  \t\t\t#armature coper losses of generator\n",
+      "series_field_cu_loss_g  =  V_field*I_1  \t\t\t#series field copper loss\n",
+      "total_loss_g =  arm_cu_loss_g + series_field_cu_loss_g + stray_loss_each\n",
+      "generator_output = V_2*I_2\n",
+      "generator_input  =  generator_output +  total_loss_g\n",
+      "eta_g  =  100*(generator_output/generator_input)\t\t\t#generator efficiency \n",
+      "print 'Efficiency of generator is %.4f percent'%(eta_g)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of motor is 72.6998 percent\n",
+        "Efficiency of generator is 67.0221 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch4_2.ipynb b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch4_2.ipynb
new file mode 100755
index 00000000..801aa5cf
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch4_2.ipynb
@@ -0,0 +1,1106 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:37c6f1ad2861e36aaa3d0e24a0317df8883a0a0a3ce46bfb539d00dfac4215c9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Synchronous Machines : Alternators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 page no : 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "Slots = 24.\n",
+      "Phase = 3. \t\t\t#number of phases\n",
+      "\n",
+      "# Calculations\n",
+      "n = Slots/Pole \t\t\t#slots per pole\n",
+      "m = Slots/Pole/Phase \t\t\t#slots per pole per phase\n",
+      "beeta = 180/n  \t\t\t#Slot angle\n",
+      "\n",
+      "# results\n",
+      "print \"Slot angle : %.f degrees\"%beeta"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Slot angle : 30 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 page no : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Slots = 120.\n",
+      "Pole = 8.\n",
+      "Phase = 3.   \t\t\t#number of phases\n",
+      "\n",
+      "# Calculations\n",
+      "n = Slots/Pole         \t\t\t#Slots per Pole\n",
+      "m = Slots/Pole/Phase   \t\t\t#Slots per Pole per Phase\n",
+      "beeta = 180/n           \t\t\t#Slot angle in degree\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#Distribution Factor\n",
+      "\n",
+      "# Results\n",
+      "print 'Distribution Factor:K_d = %.3f'%(K_d)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Distribution Factor:K_d = 0.957\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 page no : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Slots = 36.\n",
+      "Pole = 4.\n",
+      "Phase = 3. \t\t\t#number of phases\n",
+      "n = Slots/Pole    \t\t\t#Slots per pole\n",
+      "beeta = 180/n     \t\t\t#Slot angle in degrees\n",
+      "\n",
+      "# Calculations\n",
+      "#coil is shorted by 1 slot i.e. by beeta degrees to full pitch dismath.tance\n",
+      "alpha = beeta     \t\t\t#angle of short pitch\n",
+      "K_c = math.cos(math.radians(alpha/2))  \t\t\t#Coil span Factor\n",
+      "\n",
+      "# Results\n",
+      "print 'Coil Span Factor:K_c = %.4f'%(K_c)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coil Span Factor:K_c = 0.9848\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 page no : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "N_s = 250.  \t\t\t#Synchronous speed in r.p.m\n",
+      "f = 50.     \t\t\t#Frequency of generated e.m.f in hertz\n",
+      "Slots = 216.\n",
+      "phi = 30.*10**-3 \t\t\t#flux per pole in weber\n",
+      "\n",
+      "Pole = 120*f/N_s\n",
+      "n = Slots/Pole   \t\t\t#Slots per Pole\n",
+      "m = n/3          \t\t\t#Slots per Pole per Phase\n",
+      "beeta = 180/n     \t\t\t#Slot angle in degree\n",
+      "\n",
+      "# Calculations\n",
+      "K_d = math.sin(math.radians(m*beeta/2))/(m*math.sin(math.radians(beeta/2)))   \t\t\t#distribution factor\n",
+      "K_c = 1 \t\t\t#Coil Span Factor for full pitch coils = 1\n",
+      "\n",
+      "Z = Slots*5   \t\t\t#Z is total no of conductors\n",
+      "Z_ph = Z/3    \t\t\t#Conductors Per Phase\n",
+      "T_ph = Z_ph/2 \t\t\t#Turns per phase\n",
+      "E_ph = 4.44*K_c*K_d*f*phi*T_ph \t\t\t#induced emf\n",
+      "E_line = E_ph*math.sqrt(3)\n",
+      "\n",
+      "# Results\n",
+      "print 'Induced e.m.f across the Terminals is %.2f V'%(E_line)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Induced e.m.f across the Terminals is 1992.90 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 page no : 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 16.\n",
+      "N_s = 375. \t\t\t#synchronous speed in rpm\n",
+      "Slots = 144.\n",
+      "E_line = 2.657*10**3 \t\t\t#line value of emf across terminals\n",
+      "f = Pole*N_s/120 \t\t\t#frequency\n",
+      "\n",
+      "# Calculations\n",
+      "K_c = 1  \t\t\t#assuming full pitch winding Coil span Factor = 1\n",
+      "n = Slots/Pole \t\t\t#slots per pole\n",
+      "m = n/3 \t\t\t#slots per pole per phase\n",
+      "\n",
+      "beeta = 180/n\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#Distribution Fcator\n",
+      "conductors_per_slot = 10\n",
+      "Z = Slots*conductors_per_slot  \t\t\t#total conductors\n",
+      "\n",
+      "Z_ph = Z/3  \t\t\t#number of conductors per phase\n",
+      "T_ph = Z_ph/2 \t\t\t#no of turns per phase\n",
+      "E_ph = E_line/math.sqrt(3) \t\t\t#phase value of emf across terminals\n",
+      "\n",
+      "phi = E_ph/(4.44*K_c*K_d*f*T_ph)     \t\t\t#E_ph = 4.44*K_c*K_d*f*phi*T_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Frequency of Induced e.m.f is %.0fHz \\nFlux per Pole is %.0f mWb'%(f,phi*1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of Induced e.m.f is 50Hz \n",
+        "Flux per Pole is 30 mWb\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 page no : 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "d = 0.25            \t\t\t#Diameter in metre\n",
+      "l = 0.3             \t\t\t#Length in metre\n",
+      "Pole = 4.\n",
+      "A1 = math.pi*d*l/Pole    \t\t\t#Area of each fundamental pole\n",
+      "f = 50. \t\t\t#frequency in hertz\n",
+      "B_m1 = 0.15 \n",
+      "B_m3 = 0.03\n",
+      "B_m5 = 0.02 \t\t\t#Amplitude of 1st 3rd and 5th harmonics\n",
+      "phi_1 = (2/math.pi)*B_m1*A1  \t\t\t#average value of fundamental flux per pole in weber\n",
+      "\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#PART A\n",
+      "E_c1 = 1.11*2*f*phi_1  \t\t\t#R.M.S value of fundamental frequency e.m.f generated in math.single conductor\n",
+      "Coil_span = (13./15)*180  \t\t\t#math.since winding coil span is 13/15 of pole pitch\n",
+      "alpha = 180-Coil_span\n",
+      "\n",
+      "#Pitch factor for 1st 3rd and 5th harmonic\n",
+      "K_c1 = math.cos(math.radians(alpha/2))\n",
+      "K_c3 = math.cos(math.radians(3*alpha/2))\n",
+      "K_c5 = math.cos(math.radians(5*alpha/2))\n",
+      "\n",
+      "#using E_cx = E_c1 * (B_mx/B_m1)\n",
+      "E_c3 = E_c1 * (B_m3/B_m1)\n",
+      "E_c5 = E_c1 * (B_m5/B_m1)\n",
+      "\n",
+      "E_t1 = K_c1 * (2*E_c1)   \t\t\t#R.M.S Vaue of fundamental frequency EMF generated in 1 turn (in volts)\n",
+      "E_t3 = K_c3 * 2*E_c3\n",
+      "E_t5 = K_c5 * 2*E_c5\n",
+      "E_t = math.sqrt(E_t1**2 +E_t3**2 +E_t5**2)\n",
+      "V = 10*E_t  \t\t\t#(number of turns per coil )* (Total e.m.f per turn)\n",
+      "print 'Voltage generated per coil is %.1f V'%(V)\n",
+      "\n",
+      "# PART B\n",
+      "#E_1ph = 4.44*K_c1*K_d1*phi_1*f*T_ph\n",
+      "T_ph = 200.   \t\t\t#T_ph = (60 coils  * 10 turns per coil)/3\n",
+      "\n",
+      "Total_Conductors = 1200. \t\t\t# 60 coils * 10 turns per coil * 2\n",
+      "Conductors_per_Slot = 20. \t\t\t#2 conductors per turn * 10  turns per slot\n",
+      "Slots = Total_Conductors/Conductors_per_Slot\n",
+      "\n",
+      "n = Slots/Pole\n",
+      "m = n/3\n",
+      "beeta = 180/n   \t\t\t#Slot angle in degree\n",
+      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta/2)))\n",
+      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta/2)))\n",
+      "K_d5 = math.sin(math.radians(m*5*beeta/2)) /(m*math.sin(math.radians(5*beeta/2)))\n",
+      "\n",
+      "E_1ph = 4.44 * K_c1 * K_d1*phi_1 * f * T_ph\n",
+      "# using E_xph =  E_1ph* (B_mx*K_cx*K_dx)/(B_m1*K_c1*K_d1)\n",
+      "E_3ph =  E_1ph* (B_m3*K_c3*K_d3)/(B_m1*K_c1*K_d1)\n",
+      "E_5ph =  E_1ph* (B_m5*K_c5*K_d5)/(B_m1*K_c1*K_d1)\n",
+      "E_ph = math.sqrt( E_1ph**2 + E_3ph**2 + E_5ph**2 )  \t\t\t#voltage generated per phase\n",
+      "print 'Voltage generated per phase is %.f V'%(E_ph)\n",
+      "\n",
+      "#PART c\n",
+      "E_line = math.sqrt(3) * math.sqrt( E_1ph**2 + E_5ph**2  ) \t\t\t#terminal voltage\n",
+      "print 'Terminal Voltage is %.1f V '%(E_line)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage generated per coil is 12.4 V\n",
+        "Voltage generated per phase is 235 V\n",
+        "Terminal Voltage is 404.8 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 page no : 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Ns = 250. \t\t\t#Synchronous speed in rpm\n",
+      "f = 50.\n",
+      "Slots = 288.\n",
+      "E_line = 6600.\n",
+      "Pole = 120*f/Ns  \n",
+      "n = Slots/Pole  \t\t\t#slots per pole\n",
+      "m = n/3 \t\t\t#slots per pole per phase\n",
+      "beeta = 180/n \t\t\t#slot angle\n",
+      "conductors_per_slot = 32   \t\t\t#16 conductors per coil-side  *2 coil-sides per slot\n",
+      "\n",
+      "# Calculations\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
+      "alpha = 2*beeta\t\t\t# angle of short pitch\n",
+      "K_c = math.cos(math.radians(alpha/2))  \t\t\t#coil span factor\n",
+      "Z  =  Slots*conductors_per_slot  \t\t\t#total conductors\n",
+      "Z_ph = Z/3 \t\t\t#Conductors per phase\n",
+      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
+      "\n",
+      "E_ph = E_line/math.sqrt(3)\n",
+      "phi = E_ph/(4.44*K_c*K_d*f*T_ph)           \t\t\t#Because E_ph = 4.44 *K_c *K_d *phi *f *T_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Flux per pole is %.0f mWb '%(phi*1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flux per pole is 12 mWb \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 page no : 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Ns = 1500.  \t\t\t#synchronous speed in rpm\n",
+      "Pole = 4.\n",
+      "Slots = 24.\n",
+      "conductor_per_slot = 8.\n",
+      "phi = 0.05 \t\t\t#flux per pole in weber\n",
+      "f = Pole*Ns/120 \t\t\t#frequenccy\n",
+      "n = Slots/Pole \t\t\t#slots per pole\n",
+      "m = n  \t\t\t# as number of phases is 1\n",
+      "beeta = 180/n  \t\t\t#slot angle\n",
+      "\n",
+      "# Calculations\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))  \t\t\t#distribution factor\n",
+      "\n",
+      "#Full pitch =  n  = 6 slots\n",
+      "#(1/6)th of full pitch  = 1slot\n",
+      "#angle of short pitch  =  1 slot angle\n",
+      "alpha = beeta\n",
+      "K_c = math.cos(math.radians(alpha/2))  \t\t\t#coil span factor\n",
+      "\n",
+      "Z = conductor_per_slot*Slots \t\t\t#total conductors\n",
+      "Z_ph = Z \t\t\t# as number of phases is 1\n",
+      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
+      "E_ph = 4.44*K_c*K_d* phi *f *T_ph  \t\t\t#induced emf\n",
+      "\n",
+      "# Results\n",
+      "print 'Induced e.m.f is %.1f V '%(E_ph)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Induced e.m.f is 662.8 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 page no : 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 48.\n",
+      "n = 9. \t\t\t#slots per pole\n",
+      "phi = 51.75*10**-3 \t\t\t#flux per pole in weber\n",
+      "Ns = 125.\n",
+      "f = Ns*Pole/120 \t\t\t#frequency\n",
+      "K_c = 1. \t\t\t#due to full pitch winding\n",
+      "m = n/3 \t\t\t#slots per pole per phase\n",
+      "beeta = 180/n  \t\t\t#slot angle\n",
+      "\n",
+      "# Calculations\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))  \t\t\t#distribution factor\n",
+      "conductor_per_slot = 4*2 \t\t\t#Each slot has 2 coil sides and each coil side has 4 conductors\n",
+      "Slots = n*Pole\n",
+      "Z = conductor_per_slot*Slots   \t\t\t#total number of conductors\n",
+      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
+      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
+      "E_ph = 4.44 *K_c *K_d *phi *f *T_ph  \t\t\t#induced emf\n",
+      "\n",
+      "E_line = (math.sqrt(3))*E_ph  \t\t\t#due to star connection\n",
+      "\n",
+      "# Results\n",
+      "print 'Induced e.m.f is %.0f kV '%(E_line/1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Induced e.m.f is 11 kV \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10 page no : 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Slots = 180.\n",
+      "Pole = 12.\n",
+      "Ns = 600. \t\t\t#Synchronous speen in rpm\n",
+      "f = Pole*Ns/120 \t\t\t#frequency\n",
+      "phi = 0.05 \t\t\t#flux per pole in weber\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "#Average EMF in a conductor = 2*f*phi\n",
+      "rms_value_1 = 1.11*2*f*phi \t\t\t#rms value of emf in a conductor\n",
+      "print 'i)r.m.s value of e.m.f in a conductor is %.2f V '%(rms_value_1)\n",
+      "\n",
+      "#part(ii)\n",
+      "#Average EMF in a turn = 4*f*phi\n",
+      "rms_value_2 = 1.11*4*f*phi\t\t\t#r.m.s value of e.m.f in a turn\n",
+      "print 'ii)r.m.s value of e.m.f in a turn is %.2f V '%(rms_value_2)\n",
+      "\n",
+      "#part(iii)\n",
+      "conductors_per_coilside = 10/2 \n",
+      "rms_value_3 = rms_value_2*conductors_per_coilside  \t\t\t#r.m.s value of e.m.f in a coil\n",
+      "print 'iii)r.m.s value of e.m.f in a coil is %.1f V '%(rms_value_3)\n",
+      "\n",
+      "#part(iv)\n",
+      "conductors_per_slot = 10\n",
+      "Z = conductors_per_slot * Slots  \t\t\t#total number of conductors\n",
+      "Z_ph = Z/3    \t\t\t#conductors per phase\n",
+      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
+      "n = Slots/Pole  \t\t\t#slots per pole\n",
+      "m = n/3        \t\t\t#slots per pole per phase \n",
+      "beeta = 180/n   \t\t\t#slot angle\n",
+      "\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))\n",
+      "K_c = 1  \t\t\t#distribution & coil-span factor\n",
+      "E_ph = rms_value_2*T_ph*K_d*K_c   \t\t\t#induced emf\n",
+      "print 'iv)per phase induced e.m.f is %.1f V '%(E_ph)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)r.m.s value of e.m.f in a conductor is 6.66 V \n",
+        "ii)r.m.s value of e.m.f in a turn is 13.32 V \n",
+        "iii)r.m.s value of e.m.f in a coil is 66.6 V \n",
+        "iv)per phase induced e.m.f is 3822.9 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 page no : 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 8.\n",
+      "f = 50. \t\t\t#frequency\n",
+      "phi = 60.*10**-3 \t\t\t#flux per pole in weber\n",
+      "Slots = 96. \n",
+      "n = Slots/Pole \t\t\t#slots per pole\n",
+      "beeta  =  180/n \t\t\t#slot angle \n",
+      "m = n/3 \t\t\t#slots per pole per phase\n",
+      "\n",
+      "# Calculations and Results\n",
+      "coil_pitch = 10*beeta \t\t\t#10 slots\n",
+      "alpha = 180-coil_pitch\n",
+      "K_c = math.cos(math.radians(alpha/2))   \t\t\t#coi;-span factor\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
+      "\n",
+      "conductors_per_slot = 4  \n",
+      "Z = Slots*conductors_per_slot \t\t\t#total conductors\n",
+      "Total_turns = Z/2  \n",
+      "T_ph = Total_turns/3  \t\t\t#turns per phase\n",
+      "\n",
+      "#part (i)\n",
+      "E_ph =  4.44 *K_c *K_d *phi *f *T_ph\n",
+      "print '\\The phase voltage is %.2f V '%(E_ph)\n",
+      "\n",
+      "#part(ii)\n",
+      "E_line = E_ph*math.sqrt(3)\n",
+      "print 'The Line Voltage is %.2f V '%(E_line)\n",
+      "\n",
+      "#part(iii)\n",
+      "I_ph = 650\n",
+      "I_l = I_ph  \t\t\t# Star Connection\n",
+      "kVA_rating = math.sqrt(3)*E_line*I_l\n",
+      "print 'kVA rating is %.1f kVA '%(kVA_rating/1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\\The phase voltage is 788.57 V \n",
+        "The Line Voltage is 1365.84 V \n",
+        "kVA rating is 1537.7 kVA \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 page no : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Ns = 600. \t\t\t#synchronous speed in rpm\n",
+      "Pole = 10.\n",
+      "l = 30./100     \t\t\t#divided by 100 for centimetre-metre conversion\n",
+      "Pole_pitch = 35./100  \t\t\t#numerically equal to pi*d/Pole\n",
+      "Phase = 3.\n",
+      "conductors_per_slot = 8.\n",
+      "A1 = Pole_pitch*l \t\t\t#Area of each fundamental pole\n",
+      "m = 3.   \t\t\t#Slot per Pole per Phase\n",
+      "n = Phase*m \t\t\t#slots per pole\n",
+      "beeta = 180/n  \t\t\t#slot angle\n",
+      "\n",
+      "B_m1 = 1.\n",
+      "B_m3 = 0.3\n",
+      "B_m5 = 0.2  \t\t\t#amplitude of 1st 3rd and 5th harmonic\n",
+      "phi_1 = (2/math.pi)*A1*B_m1  \t\t\t#average value of fundamental flux per pole \n",
+      "f = Ns*Pole/120 \t\t\t#frequency\n",
+      "\n",
+      "# Calculations\n",
+      "Coil_span = (8./9)*180\n",
+      "alpha = 180-Coil_span\n",
+      "#pitch factor for 1st 3rd and 5th harmonic\n",
+      "K_c1 = math.cos(math.radians(alpha/2))  \n",
+      "K_c3 = math.cos(math.radians(3*alpha/2))\n",
+      "K_c5 = math.cos(math.radians(5*alpha/2))\n",
+      "\n",
+      "# using K_dx = math.sin(m*x*beeta*(math.pi/180)/2) /(m*math.sin(x*beeta*(math.pi/180)/2))\n",
+      "#distribution factor for 1st 3rd and 5th harmonic\n",
+      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta/2)))\n",
+      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta/2)))\n",
+      "K_d5 = math.sin(math.radians(m*5*beeta/2)) /(m*math.sin(math.radians(5*beeta/2)))\n",
+      "\n",
+      "Slots = n*Pole\n",
+      "Total_conductors = conductors_per_slot * Slots\n",
+      "Total_turns = Total_conductors/2\n",
+      "T_ph = Total_turns/3  \t\t\t#turns per phase\n",
+      "\n",
+      "#EMF of 1st 3rd and 5th harmonic\n",
+      "E_1ph = 4.44 * K_c1 * K_d1*phi_1 * f * T_ph\n",
+      "E_3ph =  E_1ph* (B_m3*K_c3*K_d3)/(B_m1*K_c1*K_d1)\n",
+      "E_5ph =  E_1ph* (B_m5*K_c5*K_d5)/(B_m1*K_c1*K_d1)\n",
+      "\n",
+      "# Results\n",
+      "# using E_xph =  E_1ph* (B_mx*K_cx*K_dx)/(B_m1*K_c1*K_d1)\n",
+      "E_ph = math.sqrt( E_1ph**2 + E_3ph**2 + E_5ph**2 )\n",
+      "print 'Phase value of induced e.m.f is %.2f V '%(E_ph)\n",
+      "E_line = math.sqrt(3) * math.sqrt( E_1ph**2 + E_5ph**2  )\t\t\t#no 3rd harmonic appears in line value\n",
+      "print 'line value of induced e.m.f is %.2f V '%(E_line)\n",
+      "\n",
+      "print 'Answer mismatches due to approximation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Phase value of induced e.m.f is 1711.94 V \n",
+        "line value of induced e.m.f is 2916.65 V \n",
+        "Answer mismatches due to approximation\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13 page no : 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 16.\n",
+      "phi = 0.03 \t\t\t#flux per pole\n",
+      "Ns = 375. \t\t\t#synchronous speed in rpm\n",
+      "\n",
+      "# Calculations and Results\n",
+      "f = Ns*Pole/120 \t\t\t#frequency\n",
+      "print 'frequency is %.0f Hz '%(f)\n",
+      "Slots = 144\n",
+      "n = Slots/Pole  \t\t\t#slots per pole\n",
+      "m = n/3 \t\t\t#slots per pole per phase\n",
+      "beeta = 180/n  \t\t\t#slot angle\n",
+      "K_c = 1 \t\t\t#assuming Full-Pitch coil\n",
+      "Conductors_per_slot = 10\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2)))  \t\t\t#distribution factor\n",
+      "\n",
+      "Total_conductors = Slots*Conductors_per_slot\n",
+      "Total_turns = Total_conductors/2\n",
+      "T_ph = Total_turns/3  \t\t\t#turns per phase\n",
+      "E_ph = 4.44* K_c* K_d*phi* f* T_ph\n",
+      "E_line = E_ph*math.sqrt(3)\n",
+      "print 'line voltage is %.2f V '%(E_line)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequency is 50 Hz \n",
+        "line voltage is 2657.20 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14 page no : 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Ns = 250. \t\t\t#Speed in rpm\n",
+      "f = 50. \t\t\t#frequency\n",
+      "I_l = 100.\n",
+      "Slots = 216.\n",
+      "Conductors_per_slot = 5\n",
+      "Pole = 120.*f/Ns\n",
+      "phi = 30.*10**-3\t\t\t#flux per pole in weber\n",
+      "Z = Slots*Conductors_per_slot  \t\t\t#Total Conductors\n",
+      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
+      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
+      "n = Slots/Pole \t\t\t#slots per pole\n",
+      "m = n/3 \t\t\t#slots per pole per phase\n",
+      "beeta = 180./n \t\t\t#Slot angle\n",
+      "\n",
+      "# Calculations\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
+      "\n",
+      "e_av = 2*f*phi     \t\t\t#Average Value of EMF in each conductor\n",
+      "E_c = 1.11*(2*f*phi)  \t\t\t#RMS value of EMF in each conductor\n",
+      "E = 2*E_c*K_d \t\t\t#RMS value of EMF in each turn\n",
+      "E_ph = T_ph*E \t\t\t#RMS value of EMF in each phase\n",
+      "E_line =  E_ph*math.sqrt(3)  \t\t\t#As Star Connected Alternator\n",
+      "\n",
+      "# Results\n",
+      "print 'RMS value of EMF in each phase  =  %.3f V'%(E_ph)\n",
+      "print 'RMS value of EMF line value   =  %.3f V'%(E_line)\n",
+      "kVA_rating = math.sqrt(3)*E_line*I_l\n",
+      "print 'kVA rating is %.3f kVA '%(kVA_rating/1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RMS value of EMF in each phase  =  1150.602 V\n",
+        "RMS value of EMF line value   =  1992.902 V\n",
+        "kVA rating is 345.181 kVA \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15 page no : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 10.\n",
+      "Slots = 90.\n",
+      "E_l = 11000.\n",
+      "f = 50.\n",
+      "phi = 0.15 \t\t\t#flux per pole in weber\n",
+      "n = Slots/Pole \t\t\t#slots per pole\n",
+      "m = n/3 \t\t\t#slots per pole per phase\n",
+      "beeta = 180/n \t\t\t#slot angle\n",
+      " \n",
+      "# Calculations \n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t#distribution factor\n",
+      "K_c = 1 \t\t\t#coil span factor\n",
+      "\n",
+      "E_ph = E_l/math.sqrt(3)\n",
+      "T_ph = ( E_ph/(4.44*K_c*K_d*phi*f)  )\n",
+      "#T_ph should necessarily be an integer\n",
+      "Z_ph = (T_ph)*2  \n",
+      "\n",
+      "# Results\n",
+      "print 'Required number of armature conductors is %d'%(Z_ph)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required number of armature conductors is 397\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16 page no : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 10.\n",
+      "Ns = 600. \t\t\t#speen in rpm\n",
+      "conductor_per_slot = 8.\n",
+      "n = 12. \t\t\t#slots per pole\n",
+      "Slots = Pole*n \n",
+      "m = n/3 \t\t\t#slots per pole per phase \n",
+      "beeta = 180/n \t\t\t#slot angle\n",
+      "alpha = 2*beeta  \t\t\t#short by 2 slots\n",
+      "\n",
+      "#flux per pole corresponding to 1st 3rd and 5th harmonic\n",
+      "phi_1 = 100*10**-3\n",
+      "phi_3 = (33./100)*phi_1\n",
+      "phi_5 = (20./100)*phi_1\n",
+      "\n",
+      "#coil span factor corresponding to 1st 3rd and 5th harmonic\n",
+      "K_c1 = math.cos(math.radians( alpha/2))\n",
+      "K_c3 = math.cos(math.radians( 3*alpha/2))\n",
+      "K_c5 = math.cos(math.radians( 5*alpha/2))\n",
+      "\n",
+      "# using K_dx = math.sin(m*x*beeta /2) /(m*math.sin(x*beeta /2))\n",
+      "#distribution factor corresponding to 1st 3rd and 5th harmonic\n",
+      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta /2)))\n",
+      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta /2)))\n",
+      "K_d5 = math.sin(math.radians(m*5*beeta/2)) /(m*math.sin(math.radians(5*beeta /2)))\n",
+      "\n",
+      "Z = conductor_per_slot*n*Pole    \t\t\t#Total Conductors\n",
+      "Zph = Z/3 \t\t\t#conductors per phase\n",
+      "T_ph = Zph/2 \t\t\t#turns per phase\n",
+      "\n",
+      "f = Ns*Pole/120\n",
+      "E_1ph = 4.44*K_c1*K_d1*phi_1*f*T_ph\n",
+      "E_3ph = 4.44*K_c3*K_d3*phi_3*f*T_ph\n",
+      "E_5ph = 4.44*K_c5*K_d5*phi_5*f*T_ph\n",
+      "\n",
+      "E_ph = math.sqrt( E_1ph**2 +  E_3ph**2 + E_5ph**2     )\n",
+      "\n",
+      "# Results\n",
+      "print 'Phase value of induced e.m.f is %.0f V '%(E_ph)\n",
+      "E_line = math.sqrt(3)*math.sqrt( E_1ph**2 +  E_5ph**2     )  \t\t\t#In a line value 3rd harmonic doesnt appear\n",
+      "print 'line value of induced e.m.f is %d V '%(E_line)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Phase value of induced e.m.f is 3330 V \n",
+        "line value of induced e.m.f is 5691 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17 page no : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 6.\n",
+      "Ns = 1000.   \t\t\t#speed in rpm\n",
+      "d = 28./100  \t\t\t#Divided by 100 to convert from centimeters to metres\n",
+      "l = 23./100  \t\t\t#Divided by 100 to convert from centimeters to metres\n",
+      "m = 4.   \t    \t\t#slots per pole per phase\n",
+      "B_m1 = 0.87  \t\t\t#amplitude of 1st harmonic component of flux density\n",
+      "B_m3 = 0.24  \t\t\t#amplitude of 3rd harmonic component of flux density\n",
+      "Conductors_per_slot = 8\n",
+      "f = Ns*Pole/120 \t\t\t#frequency\n",
+      "A1 = math.pi*d*l/Pole \t\t\t#area of each fundamental pole\n",
+      "phi_1 = (2/math.pi)*A1*B_m1 \t\t\t#flux per pole in weber\n",
+      "n = m*3 \t\t    \t#slots per pole\n",
+      "beeta = 180/n \t\t\t#slot angle\n",
+      "alpha = beeta  \t\t\t#because of 1 slot short\n",
+      "\n",
+      "# Calculations\n",
+      "K_c1 = math.cos(math.radians(alpha/2)) \t\t\t#coil span factor corresponding to 1st harmonic\n",
+      "K_c3 = math.cos(math.radians(3*alpha/2))\t\t\t#coil span factor corresponding to 3rd harmonic\n",
+      "# using K_dx = math.sin(m*x*beeta*(math.pi/180)/2) /(m*math.sin(x*beeta*(math.pi/180)/2))\n",
+      "K_d1 = math.sin(math.radians(m*1*beeta/2)) /(m*math.sin(math.radians(1*beeta/2))) \t\t\t#distribution factor corresponding to 1st harmonic\n",
+      "K_d3 = math.sin(math.radians(m*3*beeta/2)) /(m*math.sin(math.radians(3*beeta/2))) \t\t\t#distribution factor corresponding to 3rd harmonic\n",
+      "\n",
+      "Slots = n*Pole\n",
+      "Z = Slots*Conductors_per_slot \t\t\t#total number of conductors\n",
+      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
+      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
+      "\n",
+      "E_1ph = 4.44*K_c1*K_d1*phi_1*f*T_ph\n",
+      "E_3ph = E_1ph* (B_m3*K_c3*K_d3)/(B_m1*K_c1*K_d1)      \t\t\t#using E_xph = E_1ph* (B_mx*K_cx*K_dx)/(B_m1*K_c1*K_d1)\n",
+      "E_ph = math.sqrt( E_1ph**2 +  E_3ph**2     )\n",
+      "print 'r.m.s value of resultant voltage is %.1f V'%(E_ph)\n",
+      "E_line = math.sqrt(3)*E_1ph   \t\t\t#For line Value 3rd harmonic does not appear\n",
+      "print 'line voltage is %.3f V'%(E_line)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "r.m.s value of resultant voltage is 383.7 V\n",
+        "line voltage is 654.560 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.18 page no : 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 125.\n",
+      "V_ph = V_L\n",
+      "VA = 600.*10**3\n",
+      "I_L = VA/(math.sqrt(3)*V_L)     \t\t\t# Because VA = math.sqrt(3)* V_L * I_L\n",
+      "I_ph = I_L/(math.sqrt(3))\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#After Reconnection\n",
+      "V_ph = 125\n",
+      "V_L = V_ph*math.sqrt(3)\n",
+      "print 'New rating in volts is %.3f V'%(V_L)\n",
+      "#Winding Impedances remain the same\n",
+      "I_ph = 1600\n",
+      "I_L = I_ph\n",
+      "\n",
+      "print 'New rating in amperes is %.0f A'%(I_L)\n",
+      "kVA = math.sqrt(3)*V_L*I_L*(10**-3)\n",
+      "print 'New rating in kVA is %.0f kVA'%(kVA)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New rating in volts is 216.506 V\n",
+        "New rating in amperes is 1600 A\n",
+        "New rating in kVA is 600 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19 page no : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Pole = 4.\n",
+      "f = 50. \t\t\t#frequency\n",
+      "phi = 0.12 \t\t\t#flux per pole in weber\n",
+      "m = 4.  \t\t\t# slot per pole per phase\n",
+      "conductor_per_slot = 4.\n",
+      "coilspan = 150.\n",
+      "Ns = 120*f/Pole \t\t\t#synchronous speed in rpm\n",
+      "n = m*3  \t    \t\t#Slots per pole\n",
+      "beeta = 180/n \t\t\t#slot angle\n",
+      "\n",
+      "# Calculations\n",
+      "K_d = math.sin(math.radians(m*beeta/2)) /(m*math.sin(math.radians(beeta/2))) \t\t\t# distribution factor\n",
+      "alpha = 180-coilspan \t\t\t#angle of short pitch\n",
+      "K_c = math.cos((math.pi/180)*alpha/2) \t\t\t#coil span factor\n",
+      "Z = m*(n*Pole) \t\t\t# Also equal to (conductors/slots)*slots\n",
+      "Z_ph = Z/3 \t\t\t#conductors per phase\n",
+      "T_ph = Z_ph/2 \t\t\t#turns per phase\n",
+      "E_ph = 4.44*K_c*K_d*phi*f*T_ph\n",
+      "E_line = math.sqrt(3)*E_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'e.m.f generated is %.2f Vphase, %.2f Vline)'%(E_ph,E_line)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "e.m.f generated is 788.57 Vphase, 1365.84 Vline)\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch5_2.ipynb b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch5_2.ipynb
new file mode 100755
index 00000000..b8d7dccd
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch5_2.ipynb
@@ -0,0 +1,1604 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:933ba5514f95bafa26850ac6e2454a01e504df02d55087ce5bb50cc93571fedc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Methods for Calculating Regulation of Alternator"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 page no : 12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 1000.*10**3 \t\t\t#load power\n",
+      "phi = math.acos(math.radians(0.8)) \t\t\t#power factor lagging angle\n",
+      "V_L = 11.*10**3 \t\t\t#rated terminal voltae\n",
+      "R_a = 0.4 \t\t\t#armature resistance per phase\n",
+      "X_s = 3.\t\t\t#synchronous reactance per phase\n",
+      "\n",
+      "# Calculations\n",
+      "I_L = P/(math.sqrt(3)*V_L*math.cos(math.radians(phi)))\n",
+      "I_aph = I_L \t\t\t#for star connected load\n",
+      "I_a = I_L\t\t\t#current through armature\n",
+      "V_ph = V_L/math.sqrt(3) \t\t\t#rated terminal volatge phase value\n",
+      "\n",
+      "E_ph =  math.sqrt(  (V_ph*math.cos(math.radians(phi))+I_a*R_a)**2+(V_ph*math.sin(math.radians(phi))+I_a*X_s)**2   )  \t\t\t#emf generated phase value\n",
+      "E_line = E_ph*math.sqrt(3) \t\t\t#line value of emf generated\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph  \t\t\t#pecentage regulation\n",
+      "\n",
+      "# Results\n",
+      "print 'Line value of e.m.f generated is %.2f kV  \\nRegulation is %.3f percent'%(E_line*10**-3,regulation)\n",
+      "\n",
+      "# note : book answer is wrong.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Line value of e.m.f generated is 11.05 kV  \n",
+        "Regulation is 0.428 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page no : 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "VA = 1200.*10**3\n",
+      "V_L = 6600.\n",
+      "R_a = 0.25 \t\t\t#armature resistance per phase\n",
+      "X_s = 5.\t\t\t#synchronous reactance per phase\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\n",
+      "I_aph = I_L \t\t\t#for star connected load\n",
+      "I_a = I_L\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "\n",
+      "#Part(i)\n",
+      "phi1 = math.acos(0.8)\t\t\t#and lagging\n",
+      "E_ph1 =  math.sqrt(  (V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2   )\n",
+      "regulation = 100*(E_ph1-V_ph)/V_ph  \t\t\t#percentage regulation\n",
+      "print 'i)Regulation at 0.8 lagging pf is %.2f percent'%(regulation)\n",
+      "#Part(ii)\n",
+      "phi2 = math.acos(0.8)\t\t\t#and leading\n",
+      "E_ph2 =  math.sqrt(  (V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2   )\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph \t\t\t#percentage regulation\n",
+      "print 'ii)Regulation at 0.8 leading pf is %.f percent'%(regulation2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Regulation at 0.8 lagging pf is 9.33 percent\n",
+        "ii)Regulation at 0.8 leading pf is -7 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page no : 17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#full load\n",
+      "V_L_FL = 1100.\n",
+      "V_ph_FL = V_L_FL/math.sqrt(3)\n",
+      "\n",
+      "# Calculations\n",
+      "#no load\n",
+      "V_L_NL = 1266\n",
+      "E_line = V_L_NL\n",
+      "E_ph = E_line/math.sqrt(3)\n",
+      "regulation = 100*(E_ph-V_ph_FL)/V_ph_FL\n",
+      "\n",
+      "# Results\n",
+      "print 'Regulation at full load is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation at full load is 15.09 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page no : 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 866.\n",
+      "VA = 100.*10**3\n",
+      "I_L = VA/(math.sqrt(3)*V_L)    \t\t\t#because VA = math.sqrt(3)*V_L*I_L\n",
+      "I_aph = I_L\t\t\t#full load and star connected alternator\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "\n",
+      "# Calculations\n",
+      "#Graph is plotted and V_oc_ph and I_asc_Ph is obtained for \n",
+      "#SCC for I_asc = 66.67 A\n",
+      "I_f = 2.4  # A\n",
+      "#OCC for I_f = 2.4 A\n",
+      "V_oc_ph = 240    # V\n",
+      "\n",
+      "#for measruemnt of impedance\n",
+      "V_oc_ph = 240 \t\t\t#for I_f = 2.4..From o.c.c graph\n",
+      "I_asc_ph = 66.67 \t\t\t#for I_f = 2.4...From s.c.c graph\n",
+      "Z_s = V_oc_ph/I_asc_ph\n",
+      "R_a = 0.15\n",
+      "X_s = math.sqrt( Z_s**2-R_a**2 )\n",
+      "\n",
+      "V_ph_FL = 500.\n",
+      "phi = math.acos(0.8)  \t\t\t#lagging pf\n",
+      "E_ph = math.sqrt((V_ph_FL*math.cos(phi)+I_aph*R_a)**2+(V_ph_FL*math.sin(phi)+I_aph*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Full-load regulation at 0.8 lagging pf is %.2f percent '%(regulation )\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Full-load regulation at 0.8 lagging pf is 35.57 percent \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5  Page no : 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_OC_line = 230.\n",
+      "I_asc = 12.5  \t\t\t# when I_f = 0.38\n",
+      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
+      "Z_s = V_OC_ph/I_asc \n",
+      "\n",
+      "R_a = 1.8/2  \t\t\t#1.8 is between terminals..0.9 is per phase\n",
+      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
+      "\n",
+      "I_a = 10.\t\t\t# when regulation is needed\n",
+      "V_L = 230.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "phi1 = math.acos(0.8) \t\t\t#and lagging\n",
+      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
+      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
+      "print 'Regulation for 10 A at 0.8 lagging pf is %.2f percent'%(regulation1)\n",
+      "#Part(ii)\n",
+      "phi2 = math.acos(0.8) \t\t\t#and leading\n",
+      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print 'Regulation for 10 A at 0.8 leading pf is %.2f percent'%(regulation2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation for 10 A at 0.8 lagging pf is 64.47 percent\n",
+        "Regulation for 10 A at 0.8 leading pf is -11.01 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6  Page no :  31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "phi = math.acos(0.8)\n",
+      "VA = 1000.*10**3\n",
+      "V_L = 1905.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "R_a = 0.2\t\t\t#Armature reactance per phase\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "#Ampere-turn method\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\n",
+      "I_aph = I_L\n",
+      "V_dash = V_ph+I_aph*R_a*math.cos(phi)\t\t\t#V_dash is a dummy quantity and has no significance..it's used only for mapping correcponding current\n",
+      "F_o = 32 \t\t\t#F_o corresponds to voltage V_dash = 1148.5 from O.C.C graph\n",
+      "F_AR = 27.5 \t\t\t#Field current required to circulate full-load short circuit current of 303.07A.From SCC F_AR = 27.5\n",
+      "F_R  =  math.sqrt(    F_o**2 + F_AR**2-2*F_o*F_AR*math.cos(phi+math.pi/2) )\t\t\t#using Comath.sine rule\n",
+      "\n",
+      "# for F_R = 53.25 E_ph = 1490 V from O.C.C\n",
+      "E_ph = 1490.\n",
+      "regulation1 = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Regulation on full-load by ampere-turn method is %.2f percent'%(regulation1)\n",
+      "\n",
+      "#Part (ii)\n",
+      "#Synchronous Impedance method\n",
+      "\n",
+      "I_sc = I_L\n",
+      "I_aph2 = I_sc\n",
+      "I_f = 27.5\n",
+      "\n",
+      "V_OC_ph = 1060. \t\t\t#corresponding to I-f = 27.5 in the graph\n",
+      "Z_s = V_OC_ph/I_aph2\n",
+      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
+      "\n",
+      "E_ph2 =  math.sqrt( (V_ph*math.cos(phi)+I_aph2*R_a)**2+(V_ph*math.sin(phi)+I_aph2*X_s)**2 )   \t\t\t#from phasor diagram\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print 'Regulation on full-load by synchronous impedance method is %.2f percent'%(regulation2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation on full-load by ampere-turn method is 35.47 percent\n",
+        "Regulation on full-load by synchronous impedance method is 78.09 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7  Page no :  39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#case(i)\n",
+      "V_L = 440.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "phi = math.acos(0.8)\n",
+      "\n",
+      "#armature resistance drop from the graph\n",
+      "#RS = 1.1 cm and scale  = 50 V/cm\n",
+      "arm_leak_resis =  1.1*50    \t\t\t#armature leakage resistance\n",
+      "\n",
+      "OB = V_ph*math.cos(phi)\n",
+      "AB = V_ph*math.sin(phi) + arm_leak_resis\n",
+      "E_1ph =  math.sqrt( OB**2+AB**2   )\n",
+      "\n",
+      "F_f1 = 6.1  \t\t\t#corresponding value from OCC\n",
+      "F_AR = 3.1*1\n",
+      "\n",
+      "# Calculations and Results\n",
+      "F_R =  math.sqrt( F_f1**2 + F_AR**2 -2*F_f1*F_AR*math.cos(math.radians(90+math.acos(math.radians(0.8)))))\n",
+      "E_ph = 328. \t\t\t#voltage corresponding to F_R = 8.33 A from OCC graph\n",
+      "regulation1 =  100*(E_ph - V_ph)/V_ph\n",
+      "print 'i)Regulation for 0.8 pf lagging is %.2f percent '%(regulation1)\n",
+      "\n",
+      "#case(ii)\n",
+      "\n",
+      "OC = V_ph*math.cos(phi)\n",
+      "BC = V_ph*math.sin(phi) - arm_leak_resis\n",
+      "E_1ph =  math.sqrt( OC**2+BC**2   )\n",
+      "\n",
+      "F_f1 = 6.1  \t\t\t#corresponding value from OCC\n",
+      "F_R =  math.sqrt(   F_f1**2 + F_AR**2 -2*F_f1*F_AR*math.cos(math.radians(90-math.acos(math.radians(0.8)))))\n",
+      "E_ph = 90 \t\t\t#volatge corresponding to F_R = 3.34 A from OCC graph\n",
+      "regulation2 =  100*(E_ph - V_ph)/V_ph\n",
+      "print 'ii)Regulation for 0.8 pf leading is %.2f percent '%(regulation2)\n",
+      "print 'The answer in part ii doesnt match with textbook because of calculation mistake done in last step in the textbook'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Regulation for 0.8 pf lagging is 29.12 percent \n",
+        "ii)Regulation for 0.8 pf leading is -64.57 percent \n",
+        "The answer in part ii doesnt match with textbook because of calculation mistake done in last step in the textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 Page no : 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 1200.*10**3\n",
+      "V_line = 12000.\n",
+      "R_a = 2.\n",
+      "X_s = 35.\t\t\t#armature resistance and synchronous reactance\n",
+      "phi = math.acos(0.8)\n",
+      "\n",
+      "# Calculations\n",
+      "I_L = P/(math.sqrt(3)*V_line*math.cos(phi))\n",
+      "I_a = I_L\n",
+      "V_ph = V_line/math.sqrt(3)\n",
+      "E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Regulation at 0.8 lag power factor is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation at 0.8 lag power factor is 26.66 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 Page no : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 11000.  \n",
+      "V_ph =  V_L/math.sqrt(3)\n",
+      "VA = 1000.*1000\n",
+      "I_L = VA/(V_L*math.sqrt(3))\n",
+      "\n",
+      "V_OC_ph = 433/math.sqrt(3)\n",
+      "I_asc_ph = I_L\n",
+      "\n",
+      "Z_s = V_OC_ph /I_asc_ph  \t\t\t#ohms per phase\n",
+      "R_a = 0.45 \t\t\t#ohms per phase\n",
+      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "phi = math.acos(0.8) \t\t\t#lagging\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_L*R_a)**2 +(V_ph*math.sin(phi)+ I_L*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Voltage regulation at 0.8 pf lagging is %f percent'%(regulation)\n",
+      "\n",
+      "#part(ii)\n",
+      "phi = math.acos(0.8) \t\t\t#leading\n",
+      "E_ph2  =  math.sqrt((V_ph*math.cos(phi)+I_L*R_a)**2 +(V_ph*math.sin(phi)- I_L*X_s)**2)\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print 'Voltage regulation at 0.8 pf lagging is %f percent'%(regulation2)\n",
+      "print 'Answer mismatches due to improper approximation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage regulation at 0.8 pf lagging is 2.690067 percent\n",
+        "Voltage regulation at 0.8 pf lagging is -1.996182 percent\n",
+        "Answer mismatches due to improper approximation\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10 Page no :  46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "VA = 125.*10**3\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\n",
+      "I_aph = I_L\n",
+      "\n",
+      "# Calculations\n",
+      "I_f = 4.\n",
+      "I_asc = I_aph/2 \t\t\t#for half load.. refer to graph\n",
+      "V_OC_line = 140.\n",
+      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
+      "I_asc_ph = I_asc\n",
+      "Z_s =  V_OC_ph/I_asc_ph\n",
+      "R_a = 0.1\n",
+      "X_s = math.sqrt(Z_s**2-R_a**2) \t\t\t#armature resistance and synchronous reactance\n",
+      "\n",
+      "phi = math.acos(0.8)\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_asc*R_a)**2 +(V_ph*math.sin(phi)- I_asc*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Voltage regulation at 0.8 pf leading for half load is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage regulation at 0.8 pf leading for half load is -12.39 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11  Page no :  48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "V_OC_line = 575.\n",
+      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
+      "I_asc_line = 75.\n",
+      "I_asc_ph   = I_asc_line \n",
+      "I_aph = I_asc_ph\n",
+      "I_L = I_aph\n",
+      "\n",
+      "Z_s =  V_OC_ph/I_asc_ph\n",
+      "R_a = 2.16/2\n",
+      "X_s  =   math.sqrt(Z_s**2 - R_a**2)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#on full load\n",
+      "E_ph = 6100.\n",
+      "phi = math.acos(0.8) \t\t\t#leading\n",
+      "\n",
+      "#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)- I_a*X_s)**2)\n",
+      "p = [1, -256.68, -3.71*10**7]\n",
+      "ans = roots(p)\n",
+      "V_ph = ans[0] \t\t\t#second root is ignored as its -ve\n",
+      "V_L = V_ph*math.sqrt(3)\n",
+      "print 'Rated terminal voltage between the lines is %.3f V '%(V_L)\n",
+      "VA_rating = math.sqrt(3)*V_L*I_L\n",
+      "print 'kVA rating of the alternator is %.f kVA'%(VA_rating*10**-3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rated terminal voltage between the lines is 10774.515 V \n",
+        "kVA rating of the alternator is 1400 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12  Page no :  49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6600.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 1500.*10**3\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\n",
+      "I_aph = I_L\n",
+      "\n",
+      "# Calculations and Results\n",
+      "R_a = 0.5\n",
+      "X_s = 5\t\t\t#armature resistance and synchronous reactance\n",
+      "phi = math.acos(0.8)\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
+      "print 'Induced EMF per phase is %f V'%(E_ph)\n",
+      "\n",
+      "#full load \n",
+      "phi = math.acos(1)\n",
+      "#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)- I_a*X_s)**2)\n",
+      "p = [1, 131.215, -1.791*10**7]\n",
+      "ans = roots(p)\n",
+      "V_ph = ans[1] \t\t\t#first root is ignored as it is -ve\n",
+      "print 'Terminal voltage per phase is %f V'%(V_ph)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Induced EMF per phase is 4284.243828 V\n",
+        "Terminal voltage per phase is 4166.921808 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13  Page no :  50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_ph = 2000.\n",
+      "R_a = 0.8\n",
+      "I_sc = 100.\n",
+      "V_OC = 500.\n",
+      "I_f = 2.5\n",
+      "Z_s = V_OC/I_sc\n",
+      "X_s =  math.sqrt(Z_s**2 - R_a**2 )\n",
+      "I_aFL = 100.\n",
+      "I_a = I_aFL\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "phi = math.acos(1)\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'i)Voltage regulation is %.2f percent'%(regulation)\n",
+      "\n",
+      "#part(ii)\n",
+      "phi2 = math.acos(0.8)\n",
+      "E_ph2  =  math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2 +(V_ph*math.sin(phi2)- I_a*X_s)**2)\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print 'ii)Voltage regulation is %.2f percent'%(regulation2)\n",
+      "\n",
+      "#part(iii)\n",
+      "phi3 = math.acos(0.71)\n",
+      "E_ph3  =  math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2 +(V_ph*math.sin(phi3)+ I_a*X_s)**2)\n",
+      "regulation3 = 100*(E_ph3-V_ph)/V_ph\n",
+      "print 'iii)Voltage regulation is %.2f percent'%(regulation3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Voltage regulation is 6.89 percent\n",
+        "ii)Voltage regulation is -8.88 percent\n",
+        "iii)Voltage regulation is 21.11 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14 Page no :  51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "VA = 1000.*1000\n",
+      "V_L = 4600. \n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\n",
+      "I_aph_FL = I_L\n",
+      "I_aph = I_aph_FL\n",
+      "I_sc = (150./100)* I_aph_FL\n",
+      "V_OC_line = 1744\n",
+      "V_OC_ph =  V_OC_line/math.sqrt(3)\n",
+      "\n",
+      "# Calculations\n",
+      "Z_s = V_OC_ph / I_sc\n",
+      "R_a = 1\n",
+      "X_s = math.sqrt(Z_s**2-R_a**2)\n",
+      "\n",
+      "phi = math.acos(0.8 ) \t\t\t#lagging\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Voltage regulation at full load 0.8 pf is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage regulation at full load 0.8 pf is 19.89 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15 Page no :  52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "#part(i)    Ampere turn method\n",
+      "# Variables\n",
+      "F_O = 37.5\n",
+      "F_AR = 20.\n",
+      "V_L = 6600.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#lagging\n",
+      "phi = math.acos(0.8) \n",
+      "F_R =  math.sqrt((F_O+F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
+      "#E_ph corresponding to F_R can be obtained by plotting open circuit characteristics\n",
+      "E_ph = 4350\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'i)By Ampere-turn method or MMF methodFull-load regulation at 0.8 lagging pf is %.2f percent'%(regulation)\n",
+      "#leading\n",
+      "phi = math.acos(0.8) \n",
+      "F_R =  math.sqrt((F_O-F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
+      "#E_ph corresponding to F_R can be obtained by plotting open circuit characteristics\n",
+      "E_ph = 3000\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Full-load regulation at 0.8 leading pf is %.2f percent'%(regulation)\n",
+      "\n",
+      "#EMF method\n",
+      "V_OC_ph = 100\n",
+      "V_ph = 100\n",
+      "I_sc =  100*(F_O/F_AR)   \t\t\t#times the rated value\n",
+      "Z_s = V_OC_ph/I_sc\n",
+      "F_O =  100\n",
+      "F_AR =  Z_s*100\n",
+      "\n",
+      "#lagging\n",
+      "phi = math.acos(0.8)\n",
+      "F_R =   math.sqrt((F_O+F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
+      "regulation = 100*(F_R-V_ph)/V_ph\n",
+      "print 'iiSynchronous impedance method or EMF method'\n",
+      "print 'Full-load regulation at 0.8 lagging pf is %.2f percent'%(regulation)\n",
+      "#leading\n",
+      "phi = math.acos(0.8)\n",
+      "F_R =   math.sqrt((F_O-F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) \n",
+      "regulation = 100*(F_R-V_ph)/V_ph\n",
+      "print 'Full-load regulation at 0.8 leading pf is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)By Ampere-turn method or MMF methodFull-load regulation at 0.8 lagging pf is 14.16 percent\n",
+        "Full-load regulation at 0.8 leading pf is -21.27 percent\n",
+        "iiSynchronous impedance method or EMF method\n",
+        "Full-load regulation at 0.8 lagging pf is 38.72 percent\n",
+        "Full-load regulation at 0.8 leading pf is -19.72 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16 Page no :  56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6000.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_ph_X_Lph  =  0.9*500  \t\t\t#leakage reactance drop in volts  =  0.9 cm * 500 V/cm\n",
+      "phi =  math.acos(0.8) \t\t\t#lagging\n",
+      "\n",
+      "# Calculations\n",
+      "E_1ph = math.sqrt( (V_ph*math.cos(phi))**2 + (V_ph*math.sin(phi)+I_ph_X_Lph)**2 )   \t\t\t#From triangle OAB\n",
+      "F_f1  =  26 \t\t\t#from OCC\n",
+      "F_AR =  2.9*5  \t\t\t#2.9cm * 5 A/cm\n",
+      "\n",
+      "F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )\n",
+      "\n",
+      "# Results\n",
+      "print 'Required field current is %.2f A'%(F_R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required field current is 36.59 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.17 Page no :  58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 40.*10**3\n",
+      "I_L = VA/(math.sqrt(3)*V_L) \n",
+      "I_aph = I_L\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_aph_X_Lph  =  0.65*50  \t\t\t#leakage reactance drop in volts  =  2.4 cm * 500 V/cm\n",
+      "X_Lph =  I_aph_X_Lph/ I_aph\n",
+      "print 'Armature leakage reactance is %.3f ohms'%(X_Lph)\n",
+      "phi = math.acos(0.8)  \t\t\t#lagging\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi))**2 +(V_ph*math.sin(phi)+ I_aph_X_Lph)**2)\n",
+      "F_f1 = 15.6 \t\t\t#as obtained from OCC corresponding to this E_ph\n",
+      "\n",
+      "F_AR =  2.3*3  \t\t\t#2.3cm * 3 A/cm\n",
+      "print 'Armature reaction is %.1f '%(F_AR)\n",
+      "F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )  \t\t\t#math.comath.sine rule to Triangle OAB\n",
+      "E_ph = 267.5  \t\t\t#corresponding to F_R from open circiut characteristics\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Voltage regulation at 0.8 pf lagging is %.1f percent'%(regulation)\n",
+      "\n",
+      "\n",
+      "#Note:This answer doesnt match with textbook as it has been reciprocated in textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature leakage reactance is 0.563 ohms\n",
+        "Armature reaction is 6.9 \n",
+        "Voltage regulation at 0.8 pf lagging is 15.8 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.18 Page no :  60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "VA = 10.*10**3\n",
+      "V_L = 11.*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_ph_X_Lph  =  2.4*500  \t\t\t#leakage reactance drop in volts  =  2.4 cm * 500 V/cm\n",
+      "\n",
+      "# Calculations\n",
+      "I_ph_R_aph  = VA/(math.sqrt(3)*V_L)\n",
+      "phi = math.acos(0.8)\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_ph_R_aph)**2 +(V_ph*math.sin(phi)+ I_ph_X_Lph)**2)\n",
+      "F_f1 = 109 \t\t\t#obtained from open circuit characteristics corresponding to calculated E_ph\n",
+      "F_AR =  2.8*10  \t\t\t#2.8cm * 10 A/cm\n",
+      "F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )  \t\t\t#math.comath.sine rule to Triangle OAB\n",
+      "E_ph = 7700.  \t\t\t#corresponding to F_R from open circiut characteristics\n",
+      "\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Voltage regulation at full-load 0.8 pf lagging is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage regulation at full-load 0.8 pf lagging is 21.24 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.19  Page no : 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "VA = 2000.*1000\n",
+      "V_L = 11000.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "R_a = 0.3\n",
+      "X_s = 5. \t\t\t#armature resistance and synchronous reactance\n",
+      "\n",
+      "# Calculations\n",
+      "#case (i)\n",
+      "phi = math.acos(0.8) \t\t\t#lagging\n",
+      "I_L = VA/(math.sqrt(3)*V_L)  \n",
+      "I_a = I_L\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2)\n",
+      "\n",
+      "#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
+      "#we get  V_ph**2 -579.4455 V_ph -44653301.91 = 0\n",
+      "p = [1, -579.4455, -44653301.91]\n",
+      "ans = roots(p)\n",
+      "V_ph = ans[0]    \t\t#second root is ignored as its -ve\n",
+      "print 'Terminal voltage is %.4f V'%(V_ph)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Terminal voltage is 6978.3131 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.20 Page no : 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "R_a = 0.6\n",
+      "X_s = 6. \t\t\t#armature resistance and synchronous reactance per phase\n",
+      "E_L = 6599.\n",
+      "E_ph = E_L/math.sqrt(3)\n",
+      "I_L = 180.\n",
+      "I_a = I_L\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "# using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2) and solving for V_ph\n",
+      "p = [1 ,1135.83, -13338836.49]\n",
+      "ans = roots(p)\n",
+      "V_ph = ans[1]\n",
+      "V_L = V_ph*math.sqrt(3)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "phi = math.acos(0.9)\n",
+      "theta = math.atan((I_a*X_s+V_ph*math.sin(phi) )/(E_ph))\n",
+      "delta = theta-phi\n",
+      "print 'i)0.9 laggingTerminal voltage is %.2f VVoltage regulation is %.2f percentLoad angle is %.2f degrees'%(V_ph*math.sqrt(3),regulation,delta*180/math.pi)\n",
+      "\n",
+      "#part(ii)\n",
+      "phi_2 = math.acos(0.8)\n",
+      "p = [1, -941.53, -11399574.87]\n",
+      "ans = roots(p) \n",
+      "V_ph = ans[0] \t\t\t#second root is ignored as its -ve\n",
+      "V_L = V_ph*math.sqrt(3)\n",
+      "regulation2 = 100*(E_ph-V_ph)/V_ph\n",
+      "delta_2  =  math.asin( (math.tan(phi)*(V_ph*math.cos(phi_2)+I_a*R_a) -I_a*X_s )/E_ph   )\n",
+      "print 'ii)0.8 leadingTerminal voltage is %.2f VVoltage regulation is %.2f percentLoad angle is %.2f degrees'%(V_L,regulation2,delta_2*180/math.pi)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)0.9 laggingTerminal voltage is 5418.22 VVoltage regulation is 21.79 percentLoad angle is 6.83 degrees\n",
+        "ii)0.8 leadingTerminal voltage is 6719.93 VVoltage regulation is -1.80 percentLoad angle is 7.17 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.21 Page no :  66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_ph = 2000.\n",
+      "R_a = 0.8\n",
+      "I_sc = 100.\n",
+      "I_a = I_sc\n",
+      "V_OC = 500.\n",
+      "I_f = 2.5\n",
+      "Z_s = V_OC/I_sc\n",
+      "\n",
+      "X_s = math.sqrt(Z_s**2- R_a**2)\n",
+      "I_a_FL = 100.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "phi1 = math.acos(1) \t\t\t#and lagging\n",
+      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
+      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
+      "print 'Regulation at upf is %.2f percent'%(regulation1)\n",
+      "\n",
+      "#Part(ii)\n",
+      "phi2 = math.acos(0.8) \n",
+      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print 'Regulation at 0.8 leading pf is %.2f percent'%(regulation2)\n",
+      "\n",
+      "#Part(iii)\n",
+      "phi3 = math.acos(0.71) \n",
+      "E_ph3 = math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2+(V_ph*math.sin(phi3)+I_a*X_s)**2)\n",
+      "regulation3 = 100*(E_ph3-V_ph)/V_ph\n",
+      "print 'Regulation at 0.71 lagging pf is %.2f percent'%(regulation3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation at upf is 6.89 percent\n",
+        "Regulation at 0.8 leading pf is -8.88 percent\n",
+        "Regulation at 0.71 lagging pf is 21.11 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.22 Page no : 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 600.\n",
+      "VA = 60.*10**3\n",
+      "I_sc = 210.\n",
+      "V_oc = 480.\n",
+      "I_f = 10.\n",
+      "R_a = 0.2\n",
+      "\n",
+      "I = VA/V        \t\t\t#VA = V*I and alternator is math.single phase\n",
+      "I_a = I\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Z_s = V_oc/I_sc  \t\t\t#Synchronous Impedance\n",
+      "X_s = math.sqrt(Z_s**2-R_a**2) \t\t\t#SYnchronous reactance\n",
+      "print 'Synchronous impedances is %f ohms and synchronous reactance is %f ohms'%(Z_s,X_s)\n",
+      "\n",
+      "#PART (i)\n",
+      "phi1 = math.acos(0.8) \t\t\t#and lagging\n",
+      "E1 = math.sqrt((V*math.cos(phi1)+I_a*R_a)**2+(V*math.sin(phi1)+I_a*X_s)**2) \t\t\t#plus sign for lagging power factor\n",
+      "regulation1 = 100*(E1-V)/V\n",
+      "print 'Regulation at 0.8 lagging pf is %.2f percent '%(regulation1 )\n",
+      "\n",
+      "#PART (ii)\n",
+      "phi2 = math.acos(1) \n",
+      "E2 = math.sqrt((V*math.cos(phi2)+I_a*R_a)**2+(V*math.sin(phi2)+I_a*X_s)**2) \n",
+      "regulation2 = 100*(E2-V)/V\n",
+      "print 'Regulation at UNITY pf is %.2f percent '%(regulation2 )\n",
+      "\n",
+      "#PART (iii)\n",
+      "phi3 = math.acos(0.6) \t\t\t#and leading \n",
+      "E3 = math.sqrt((V*math.cos(phi3)+I_a*R_a)**2+(V*math.sin(phi3)-I_a*X_s)**2) \t\t\t#minus sign for leading power factor\n",
+      "regulation3 = 100*(E3-V)/V\n",
+      "print 'Regulation at 0.6 leading pf is %.2f percent '%(regulation3 )\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronous impedances is 2.285714 ohms and synchronous reactance is 2.276947 ohms\n",
+        "Regulation at 0.8 lagging pf is 28.60 percent \n",
+        "Regulation at UNITY pf is 10.08 percent \n",
+        "Regulation at 0.6 leading pf is -23.98 percent \n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.23 Page no : 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3300.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_a = 100.\n",
+      "I_f = 5.\n",
+      "V_OC_line = 900. \n",
+      "V_OC_ph = V_OC_line/math.sqrt(3) \n",
+      "R_a = 0.8 \t\t\t#armature resistance\n",
+      "I_aph = I_a\n",
+      "Z_s = V_OC_ph/I_aph\n",
+      "X_s = math.sqrt(Z_s**2-R_a**2) \t\t\t#synchronous reactance\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "phi1 = math.acos(0.8) \t\t\t#and lagging\n",
+      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
+      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
+      "print 'Regulation at 0.8 lagging is %.2f percent'%(regulation1)\n",
+      "\n",
+      "#Part(ii)\n",
+      "phi2 = math.acos(0.8)  \t\t\t#and leading\n",
+      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print 'Regulation at 0.8 leading pf is %.2f percent'%(regulation2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation at 0.8 lagging is 21.03 percent\n",
+        "Regulation at 0.8 leading pf is -9.55 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.24  Page no : 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 13500.\n",
+      "R_a = 1.5 \n",
+      "X_s = 30. \t\t\t#armature resistance and synchronous reactance\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#CASE 1\n",
+      "phi1 = math.acos(0.8)\n",
+      "P_out = 1280*10**3\n",
+      "I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi1) )    \t\t\t#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)\n",
+      "\n",
+      "I_a = I_L\n",
+      "E_ph = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Regulation at 0.8 lagging power factor is %.2f percent'%(regulation)\n",
+      "\n",
+      "#Case 2\n",
+      "phi2 = math.acos(1)\n",
+      "I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi2) )    \t\t\t#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)\n",
+      "\n",
+      "I_a = I_L\n",
+      "E_ph = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)+I_a*X_s)**2)\n",
+      "regulation2 = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Regulation at unity power factor is %.2f percent'%(regulation2)\n",
+      "\n",
+      "#case 3\n",
+      "phi3 = math.acos(0.8)\n",
+      "I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi3) )    \t\t\t#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)\n",
+      "I_a = I_L\n",
+      "E_ph = math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2+(V_ph*math.sin(phi3)-I_a*X_s)**2)\t\t\t# minus sign in the second bracket beacuse of leading pf\n",
+      "regulation3 = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Regulation at 0.8 leading power factor is %.2f percent'%(regulation3)\n",
+      "\n",
+      "# note : rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation at 0.8 lagging power factor is 18.60 percent\n",
+        "Regulation at unity power factor is 3.23 percent\n",
+        "Regulation at 0.8 leading power factor is -11.99 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.26  Page no :  72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 11.*10**3\n",
+      "VA_rating = 10.**6\n",
+      "R_a = 2.2 \t\t\t#alternator resistance\n",
+      "phi = math.acos(0.8)\n",
+      "\n",
+      "# Calculations\n",
+      "I_L = VA_rating/(math.sqrt(3)*V_L)\t\t\t#VA = math.sqrt(3)V_L*I_L\n",
+      "I_a = I_L\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "regulation = 24.\n",
+      "\n",
+      "E_ph =  ((regulation/100)+1)*V_ph    \t\t\t# because regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "#using E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)\n",
+      "X_s = (math.sqrt(E_ph**2-((V_ph*math.cos(phi)+I_a*R_a)**2))-V_ph*math.sin(phi))*(1/I_a)\n",
+      "\n",
+      "phi1 = math.acos(0.8)\n",
+      "E_ph = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)-I_a*X_s)**2)\n",
+      "regulation1 = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Regulation at 0.8 leading power factor is %.2f percent'%(regulation1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation at 0.8 leading power factor is -13.90 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.27  Page no :  73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 220.\n",
+      "VA = 100.*10**3\n",
+      "R_a = 0.1 \t\t\t#effective resistacne of alternator\n",
+      "X_a = 0.5 \t\t\t#leakage reactance\n",
+      "X_ar = 2*X_a\n",
+      "\n",
+      "Z_s = complex(R_a,X_a+X_ar)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(1)\n",
+      "phi = math.acos(0.4)\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\t\t\t#VA = math.sqrt(3)*V_L*I_L\n",
+      "I_a = I_L\n",
+      "E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*(X_a+X_ar))**2)\n",
+      "print 'i)Required noload voltage is %.3f V'%(E_ph)\n",
+      "\n",
+      "#Part(2)\n",
+      "V_ph2 = 0\n",
+      "E_ph2 = math.sqrt((V_ph2*math.cos(phi)+I_a*R_a)**2+(V_ph2*math.sin(phi)+I_a*(X_a+X_ar))**2)\n",
+      "print 'ii)Required noload voltage is %.3f V'%(E_ph2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Required noload voltage is 515.848 V\n",
+        "ii)Required noload voltage is 394.522 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.28 Page no : 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 2000.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 1000.*10**3\n",
+      "I_L = VA/(math.sqrt(3)*V_L)    \t\t\t#because VA = math.sqrt(3)*V_L*I_L\n",
+      "I_aph = I_L\n",
+      "\n",
+      "I_f = 28.5\t\t\t#for this I_aph = 288.67513 as obtained from SCC graph\n",
+      "V_oc_ph = 1060.\t\t\t#for I_f = 28.5 as obtained fromOCC graph\n",
+      "Z_s = V_oc_ph/I_aph\n",
+      "R_a = 0.2 \t\t\t#armature effective resistance\n",
+      "X_s = math.sqrt( Z_s**2-R_a**2 )\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Part(i)\n",
+      "phi1 = math.acos(0.8)\t\t\t#lagging\n",
+      "E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_aph*R_a)**2+(V_ph*math.sin(phi1)+I_aph*X_s)**2)\n",
+      "regulation1 = 100*(E_ph1-V_ph)/V_ph\n",
+      "print \"i)Full-load percentage regulation at 0.8 pf lagging is %.2f percent\"%(regulation1)\n",
+      "\n",
+      "#Part(ii)\n",
+      "phi2 = math.acos(0.8)\t\t\t#leading\n",
+      "E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_aph*R_a)**2+(V_ph*math.sin(phi2)-I_aph*X_s)**2)\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print \"ii)Full-load percentage regulation at 0.8 pf leading is %.2f percent\"%(regulation2)\n",
+      "print 'Note that the answer mismatches because of calculation mistake done in the last step of part 1'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Full-load percentage regulation at 0.8 pf lagging is 73.86 percent\n",
+        "ii)Full-load percentage regulation at 0.8 pf leading is -9.29 percent\n",
+        "Note that the answer mismatches because of calculation mistake done in the last step of part 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.29 Page no : 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3300.\n",
+      "VA = 200.*10**3\n",
+      "R_a = 0.6\n",
+      "X_s = 6.\t\t\t#armature resistance and synchronous reactance\n",
+      "\n",
+      "# Calculations\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\t\t\t#VA = math.sqrt(3)V_L*I_L\n",
+      "I_a = I_L\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "phi = math.acos(0.8)\n",
+      "E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)\n",
+      "\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Regulation at 0.8 lagging power factor is %.3f percent'%(regulation)\n",
+      "print ' Note : Regulation is positive for lagging power factor loads'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Regulation at 0.8 lagging power factor is 7.802 percent\n",
+        " Note : Regulation is positive for lagging power factor loads\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.30 Page no : 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 2300.   \n",
+      "V_ph =  V_L/math.sqrt(3)\n",
+      "f = 50.\n",
+      "R_a = 0.2 \t\t\t#armature resistance\n",
+      "I_sc = 150.\n",
+      "V_OC_line = 780.   \n",
+      "V_OC_ph = V_OC_line/math.sqrt(3)\n",
+      "\n",
+      "Z_s =  V_OC_ph/I_sc\n",
+      "X_s  =   math.sqrt(Z_s**2 - R_a**2)\n",
+      "I_aph = 25 \n",
+      "I_aFL = I_aph\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "phi =  math.acos(0.8)  \t\t\t#lag\n",
+      "E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)\n",
+      "regulation = 100*(E_ph-V_ph)/V_ph\n",
+      "print 'Voltage regulation at 0.8 pf lagging is %.3f percent'%(regulation)\n",
+      "\n",
+      "#part(ii)\n",
+      "phi2 =  math.acos(0.8)  \t\t\t#lead\n",
+      "E_ph2  =  math.sqrt((V_ph*math.cos(phi2)+I_aph*R_a)**2 +(V_ph*math.sin(phi2)- I_aph*X_s)**2 )\n",
+      "regulation2 = 100*(E_ph2-V_ph)/V_ph\n",
+      "print 'Voltage regulation at 0.8 pf leading is %.3f percent'%(regulation2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage regulation at 0.8 pf lagging is 3.774 percent\n",
+        "Voltage regulation at 0.8 pf leading is -2.967 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch6_2.ipynb b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch6_2.ipynb
new file mode 100755
index 00000000..5cc8495d
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch6_2.ipynb
@@ -0,0 +1,2419 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:65f8e8657a6631c1a9d4f4644c0ba8ed136071b32d61999a56bf9cfb9b9555d8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Synchronization and Parallel Operation of Alternators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " \n",
+      "Example 6.2  Page no : 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "X_d = 0.7 \n",
+      "X_q = 0.4  \t\t\t#direct and quadrature axis synchronous reactance p.u.\n",
+      "R_a = 0\n",
+      "phi = math.acos(0.8)  \t\t\t#Lag\n",
+      "\n",
+      "V_t = 1. \t\t\t#assumed rated terminal Voltage \n",
+      "I_a = 1. \t\t\t#Full-load armature current\n",
+      "\n",
+      "# Calculations\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
+      "delta = psi-phi\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(psi)\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "\n",
+      "# Results\n",
+      "print 'Total e.m.f induced on open circuit is %.4f p.u.'%(E_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total e.m.f induced on open circuit is 1.5149 p.u.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page no :  35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "        return  RRRR*cmath.exp((math.pi*Theeeta/180.)*1j);\n",
+      "\n",
+      "# Variables\n",
+      "Z1 = complex(0,3) \t\t\t#impedance of alternator 1\n",
+      "Z2 = complex(0,4) \t\t\t#impedance of alternator 2\n",
+      "Z = 6.       #load\n",
+      "\n",
+      "E1 = p2z(220,0) \t\t\t#induced emf vector on no load\n",
+      "E2 = p2z(220,10)\t\t\t#induced emf vector on no load\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I1 = ((E1-E2)*Z+E1*Z2)/(Z*(Z1+Z2)+Z1*Z2)\n",
+      "I2 = ((E2-E1)*Z+E2*Z1)/(Z*(Z1+Z2)+Z1*Z2)\n",
+      "\n",
+      "phi1 = math.degrees(math.atan(I1.imag/I1.real))  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
+      "phi2 = math.degrees(math.atan(I2.imag/I2.real))  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
+      "\n",
+      "I = I1+I2\n",
+      "V = I*Z \t\t\t#Terminal voltage\n",
+      "print 'i) Terminal voltage is %.1f volts at %.2f degrees'%(abs(V),math.degrees(math.atan(V.imag/V.real)))\n",
+      "print 'ii) Currents are %.2f A at %.2f degrees and %.2f A at %.2f degrees      \\\n",
+      "\\nTotal current is %.2f A at %.2f degrees  '%(abs(I1),math.degrees(math.atan(I1.imag/I1.real)),\\\n",
+      "                                              abs(I2),math.degrees(math.atan(I2.imag/I2.real)),\\\n",
+      "                                              abs(I),math.degrees(math.atan(I.imag/I.real)))\n",
+      "\n",
+      "P1 = abs(V)*abs(I1)*math.cos(math.radians(phi1))\n",
+      "P2 = abs(V)*abs(I2)*math.cos(math.radians(phi2))\n",
+      "print 'iii)Power delivered is %.2f watts and %.2f watts'%(P1,P2)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Terminal voltage is 210.7 volts at -11.66 degrees\n",
+        "ii) Currents are 14.91 A at -17.71 degrees and 20.36 A at -7.24 degrees      \n",
+        "Total current is 35.12 A at -11.66 degrees  \n",
+        "iii)Power delivered is 2992.46 watts and 4257.11 watts\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4  Page no :  54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_l = 10000.\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "VA = 10.*10**6\n",
+      "I_FL = VA/(V_l*math.sqrt(3)) \t\t\t#Current at full laod\n",
+      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
+      "\n",
+      "# Calculations\n",
+      "X_s = IX_s/I_FL\n",
+      "N_s = 1500.\n",
+      "f = 50.\n",
+      "P = 120*f/N_s \t\t\t#poles\n",
+      "\n",
+      "delta_dash_mech = math.pi/180 \t\t\t#phase print lacement in degree mechanical\n",
+      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
+      "E = V_ph \t\t\t#math.since alternator is on no-load\n",
+      "P_SY = delta_dash_elec*E**2/X_s  \t\t\t#Synchronous Power\n",
+      "P_SY_3ph = P_SY*3 \t\t\t#For 3 phases\n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising Power of armature is %.3f kW.\\\n",
+      "\\nSynchronising Power for 3 phase is %.3f kW'%(P_SY*10**-3,P_SY_3ph*10**-3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising Power of armature is 581.776 kW.\n",
+        "Synchronising Power for 3 phase is 1745.329 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page no :  55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6.6*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 3.*10**6\n",
+      "I_FL = VA/(V_L*math.sqrt(3)) \t\t\t#full load current \n",
+      "P = 8.\n",
+      "f = 50. \t\t\t#poles and frequency\n",
+      "\n",
+      "X_s = complex(0,2.9)\t\t\t#X_s = 2.9\n",
+      "delta_dash_mech = math.pi/180\n",
+      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "E = V_ph\n",
+      "P_SY = delta_dash_elec*E**2/abs(X_s)  \t\t\t#Synchronous Power per phase\n",
+      "P_SY_3ph = P_SY*3 \t\t\t#For 3 phases\n",
+      "print 'i)   Synchronising power at no load is %.3f kW'%(P_SY*10**-3)\n",
+      "print '   Total Synchronising power at no load is %.2f kW'%(P_SY_3ph*10**-3)\n",
+      "\n",
+      "N_s = 120*f/P \t\t\t#in rpm\n",
+      "n_s = (N_s)/60 \t\t\t#in rps\n",
+      "T_SY = P_SY_3ph/(2*math.pi*n_s)\n",
+      "print 'Synchronous torque per mechanical degree of phase print lacement is %.2f * 10**3 N-m'%(T_SY*10**-3)\n",
+      "\n",
+      "#part(ii)\n",
+      "phi = math.acos(math.radians(0.85))\n",
+      "I = p2z(I_FL,0)\n",
+      "V = p2z(V_ph,phi)\n",
+      "\n",
+      "E = V+I*X_s\n",
+      "#E leads I by phasemag(E). V leads I by phasemag(V)\n",
+      "\n",
+      "delta = (math.pi/180)* (math.atan(E.imag/E.real)-math.atan(V.imag/V.real) ) \t\t\t#power angle in radians\n",
+      "P_SY2 = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)\n",
+      "\n",
+      "P_SY_total_2 = 3*P_SY2\n",
+      "#n_s = T_SY/(P_SY/(2*math.pi) )       \t\t\t#because T_SY = P_SY/(2*math.pi*n_s)\n",
+      "print 'ii)Total Synchronising power is %.0f kW'%(P_SY_total_2*10**-3)\n",
+      "\n",
+      "T_SY2 = P_SY_total_2/(2*math.pi*n_s)\n",
+      "print 'Synchronising torque is %.2f * 10**3 N-m'%(T_SY2/1000)\n",
+      "\n",
+      "# note : book answer is wrong.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)   Synchronising power at no load is 349.547 kW\n",
+        "   Total Synchronising power at no load is 1048.64 kW\n",
+        "Synchronous torque per mechanical degree of phase print lacement is 13.35 * 10**3 N-m\n",
+        "ii)Total Synchronising power is 1074 kW\n",
+        "Synchronising torque is 13.68 * 10**3 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6  Page no :  57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.)\n",
+      "\n",
+      "# Variables\n",
+      "V_l = 10.*10**3\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "R_a = 0.4\n",
+      "Z = complex(R_a,6)\n",
+      "I_a = p2z(300,-math.acos(math.radians(0.8)))\n",
+      "E = V_ph+I_a*Z\n",
+      "\n",
+      "# Calculations\n",
+      "phi = math.acos(0.8)\n",
+      "alternator_op_ph = V_ph*abs(I_a)*math.cos(phi) \t\t\t#Power delivered to infinite bus per phase\n",
+      "\n",
+      "#Power deliered to the altrernator  =  Power delivewred to bus bar +  I**2*R losses in armature\n",
+      "alternator_power =  alternator_op_ph+ abs(I_a)**2*R_a\n",
+      "\n",
+      "#this power developed remains constant.change pf to 1 and calculate corresponding armature current\n",
+      "#alternator_power = V_ph*I_a1*math.cos(phi1)+I_a1**2*0.4\n",
+      "#solve the quadratic equation  0.4 I_a1**2+5773.50 I_a1- 1421640  = 0\n",
+      "I_a1 = (-1*V_ph+math.sqrt(V_ph**2-4*R_a*-1*alternator_power))/(2*R_a)\n",
+      "\n",
+      "#also as follows \n",
+      "E1 = V_ph+I_a1*Z\n",
+      "decrease = 100*(abs(E)-abs(E1))/abs(E)\n",
+      "\n",
+      "# Results\n",
+      "print 'Percentage decrease in induced e.m.f is %.1f percent'%(decrease)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage decrease in induced e.m.f is 2.6 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7 Page no :  58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#Line PQ for Altermnator 1 and PR for alternaator 2.AB is at frequency x from P where total load is 3000 kW\n",
+      "\n",
+      "# Variables\n",
+      "QC = 2000.\n",
+      "PS = 2.5\n",
+      "#PC = x\n",
+      "TR = 2000.\n",
+      "PT = 2.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#using similarity of triangles PAC and PQS\n",
+      "AC_by_PC = (QC/PS)\t\t\t# because (AC/QC) = (PC/PS)\n",
+      "#using similarity of triangles PCB and PTR\n",
+      "CB_by_PC = (TR/PT) \t\t\t# because (CB/TR) = (PC/PT)\n",
+      "\n",
+      "AC_by_x = AC_by_PC   \t\t\t#which implies AC = 12.5*x\n",
+      "CB_by_x = CB_by_PC  \t\t\t#which implies CB = 16.67*x\n",
+      "\n",
+      "AC_plus_CB = 3000.  \t\t\t#total load at the frequency at P is 30 kW\n",
+      "x =  AC_plus_CB/(AC_by_x + CB_by_x)\n",
+      "AC = AC_by_x * x\n",
+      "CB = CB_by_x * x  \n",
+      "frequency = 50-x\n",
+      "print 'Loads shared by alternator 1 and 2 are %.2f kW and %.2f kW respectively'%(AC,CB)\n",
+      "\n",
+      "#construction for max load: RT is extended to cut PQ at X.\n",
+      "QS = 2000.\n",
+      "RT = 2000. \t\t\t#see figure\n",
+      "XT = QS*(PT/PS)\n",
+      "RX = RT+XT  \t\t\t#maximum load\n",
+      "\n",
+      "print 'Maximum load is %.0f kW'%(RX)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loads shared by alternator 1 and 2 are 1333.33 kW and 1666.67 kW respectively\n",
+        "Maximum load is 3600 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8 Page no :  60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P_out = 1500.*10**3\n",
+      "V_L = 11000.\n",
+      "phi = math.acos(0.8)\n",
+      "I_L = P_out/(math.sqrt(3)*V_L*math.cos(phi))\n",
+      "\n",
+      "I_L_actv = I_L*math.cos(phi)   \t\t\t#wattful or active component of current\n",
+      "I_L_reactive = I_L*math.sin(phi) \t\t\t#wattless or reactive component of current\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_each = I_L/2  \t\t\t#in identical conditions\n",
+      "I_arm1 = 45. \t\t\t#given\n",
+      "I_1_reactive = math.sqrt(I_arm1**2-39.364**2 )  \t\t\t#from the power triangle \n",
+      "I_2_reactive = 59.046-21.80\n",
+      "I_a_2 = math.sqrt( 39.364**2 + I_2_reactive**2 )  \t\t\t#required armature current of 2nd alternator\n",
+      "print 'Required armature current of second alternator is %.4f A'%(I_a_2)\n",
+      "#power factors of 2 machines\n",
+      "cos_phi1 = 39.364/45 \n",
+      "cos_phi2 = 39.364/54.1921\n",
+      "\n",
+      "print 'Power factors are %.4f lagging and %.4f lagging'%(cos_phi1,cos_phi2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required armature current of second alternator is 54.1921 A\n",
+        "Power factors are 0.8748 lagging and 0.7264 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 Page no : 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#Line AB for Altermnator 1 and AC for alternator 2.AF is at frequency x measured from A where total load is 3000 kW\n",
+      "# Variables\n",
+      "BO = 2000.\n",
+      "AO = 5.\t\t\t#AF = x\n",
+      "DC = 2000.\n",
+      "AD = 3.\n",
+      "#AF = x\n",
+      "\n",
+      "# Calculations\n",
+      "#using similarity of triangles AEF and ABO\n",
+      "EF_by_AF = (BO/AO)\t\t\t# because (EF/BO) = (AF/AO)\n",
+      "#using similarity of triangles AFG and ADC\n",
+      "FG_by_AF = (DC/AD) \t\t\t#because (FG/DC) = (AF/AD)\n",
+      "\n",
+      "EF_by_x = EF_by_AF   \t\t\t#which implies EF = 400*x\n",
+      "FG_by_x = FG_by_AF   \t\t\t#which implies FG = 666.67*x\n",
+      "\n",
+      "EF_plus_FG = 3000  \t\t\t#total load at the frequency at P is 3000 kW\n",
+      "x =  EF_plus_FG/(EF_by_x + FG_by_x)\n",
+      "EF = (BO/AO)*x\n",
+      "FG = (DC/AD)*x  \n",
+      "\n",
+      "# Results\n",
+      "print 'Loads shared by machine 1 and 2 are %.0f kW and %.0f kW respectively'%(EF,FG)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loads shared by machine 1 and 2 are 1125 kW and 1875 kW respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10 Page no :  63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
+      "\n",
+      "# Variables\n",
+      "V_l = 6000.\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "VA = 2000.*10**3\n",
+      "I_FL = VA/(V_l*math.sqrt(3))\n",
+      "X_s = complex(0,6) \t\t\t#synchronous reactance\n",
+      "P = 8.\n",
+      "f = 50.\n",
+      "\n",
+      "delta_mech = math.pi/180 \t\t\t#phase print lacemant in degree mechanical \n",
+      "#phase print lacemant in degree electrical\n",
+      "delta_elec = delta_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
+      "\n",
+      "phi = math.acos(math.radians(0.8))\n",
+      "V = p2z(V_ph,phi)\n",
+      "E = V+I_FL*X_s\n",
+      "#E leads I by phasemag(E). V leads I by phasemag(V)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "delta = (math.pi/180)* (math.atan(E.imag/E.real)-(math.atan(V.imag/V.real ) ) ) \t\t\t#power angle in radians\n",
+      "P_SY = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_elec)/abs(X_s) \t\t\t#Synchronising power\n",
+      "P_SY_total = 3*P_SY \t\t\t#totla Synchronising power \n",
+      "print 'Total Synchronising power is %.3f kW'%(10**-3*P_SY_total)\n",
+      "\n",
+      "N_s = 120*f/P \t\t\t#in rpm\n",
+      "n_s = (N_s)/60 \t\t\t#in rps\n",
+      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
+      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n",
+      "\n",
+      "# note : book answer it seems wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Synchronising power is 444.753 kW\n",
+        "Synchronising torque is 5663 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11  Page no : 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
+      "\n",
+      "# Variables\n",
+      "V_l = 3300.\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "VA = 3*10.**6\n",
+      "I_FL = VA/(V_l*math.sqrt(3))\n",
+      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = complex(0,IX_s/I_FL) \t\t\t#synchronous reactance\n",
+      "N_s = 1000. \t\t\t#in rpm\n",
+      "P = 6.\n",
+      "f = 50.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "delta_dash_mech = math.pi/180\n",
+      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
+      "\n",
+      "I = I_FL\n",
+      "phi = math.acos(math.radians(0.8))\n",
+      "V = p2z(V_ph,phi)\n",
+      "E = V+I*X_s\n",
+      "#E leads I by phasemag(E). V leads I by phasemag(V)\n",
+      "\n",
+      "delta = (math.pi/180)* (math.atan(E.imag/E.real)-math.atan(V.imag/V.real) ) \t\t\t#power angle in radians\n",
+      "P_SY = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s) \t\t\t#Synchronising power per phase\n",
+      "print 'Synchronising power is %.3f kW'%(10**-3*P_SY)\n",
+      "P_SY_total = 3*P_SY \t\t\t#Total Synchronising power\n",
+      "\n",
+      "N_s = 120*f/P \t\t\t#in rpm\n",
+      "n_s = (N_s)/60 \t\t\t#in rps\n",
+      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
+      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n",
+      "\n",
+      "print 'Answer mismatches due to approximation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power is 217.160 kW\n",
+        "Synchronising torque is 6221 N-m\n",
+        "Answer mismatches due to approximation\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12 Page no : 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_l = 3300.\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "VA = 3*10.**6\n",
+      "I_FL = VA/(V_l*math.sqrt(3))\n",
+      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = complex(0,IX_s/I_FL) \t\t\t#synchronous reactance\n",
+      "N_s = 1000.\n",
+      "P = 6.\n",
+      "f = 50.\n",
+      "\n",
+      "delta_dash_mech = math.pi/180 \t\t\t#phase print lacement in degree mechanical\n",
+      "#phase print lacement in degree electrical\n",
+      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "E = V_ph\n",
+      "Z_s = X_s \t\t\t#math.since R = 0\n",
+      "P_SY = abs(E)*abs(V_ph)*delta_dash_elec/abs(Z_s) \t\t\t#Synchronising power per phase\n",
+      "print 'Synchronising power is %.3f kW'%(10**-3*P_SY)\n",
+      "P_SY_total = 3*P_SY \t\t\t#Total Synchronising power\n",
+      "print '3 phase Synchronising power is %.3f kW'%(10**-3*P_SY_total)\n",
+      "\n",
+      "N_s = 120*f/P \t\t\t#in rpm\n",
+      "n_s = (N_s)/60 \t\t\t#in rps\n",
+      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
+      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power is 261.799 kW\n",
+        "3 phase Synchronising power is 785.398 kW\n",
+        "Synchronising torque is 7500 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13 Page no :  66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 11.*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 700.*10**3\n",
+      "I_FL = VA/(math.sqrt(3)*V_L) \t\t\t#full load current\n",
+      "IR_a = (1.5/100)*V_ph \t\t\t#product of I and R_a\n",
+      "R_a = IR_a/I_FL\n",
+      "IX_s = (14./100)*V_ph \t\t\t# product of I and X_s\n",
+      "X_s = IX_s/I_FL \t\t\t#synchronous reactance\n",
+      "\n",
+      "# Calculations\n",
+      "#at full load and 0.8 pf\n",
+      "I = I_FL\n",
+      "phi = math.acos(0.8)\n",
+      "V_ph = complex(V_ph*math.cos(phi),V_ph*math.sin(phi))  \t\t\t#just introduced the angle\n",
+      "E_ph = math.sqrt(    (abs(V_ph)*math.cos(phi)+ IR_a)**2+   (abs(V_ph)*math.sin(phi)+ IX_s)**2     )\n",
+      "\n",
+      "Poles = 4.\n",
+      "f = 50. \t\t\t#poles and frequency\n",
+      "delta = math.asin( (abs(V_ph)*math.sin(phi)+IX_s)/E_ph)  -phi\n",
+      "delta_dash_mech = (math.pi/180) \t\t\t#print lacement in degree mechanical\n",
+      "\t\t\t#print lacement in degree electrical\n",
+      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
+      "P_SY = abs(E_ph)*abs(V_ph)*math.cos(delta)*math.sin(delta_dash_elec)/X_s   \t\t\t#Synchronising power per phase \n",
+      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
+      "\n",
+      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
+      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power is %.2fkW'%(P_SY_total/1000)\n",
+      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power is 191.25kW\n",
+        "Synchronising torque is 1217.53 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.14 Page no :  68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_l = 9.*10**3\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "VA = 5.5*10**6\n",
+      "I_FL = VA/(V_l*math.sqrt(3))\n",
+      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = complex(0,IX_s/I_FL) \t\t\t#synchronous reactance\n",
+      "N_s = 1500. \t\t\t#in rpm\n",
+      "n_s = N_s/60 \t\t\t#in rps\n",
+      "f = 50.\n",
+      "P = 120.*f/N_s \t\t\t#frequency and pole\n",
+      "\n",
+      "# Calculations\n",
+      "delta_dash_mech = math.pi/180 \t\t\t#print lacemnt in degree mechanical \n",
+      "#print lacemnt in degree electrical\n",
+      "delta_dash_elec = delta_dash_mech*(P/2)  \t\t\t#P/2 is pole pairs(and not poles)\n",
+      "\n",
+      "E = V_ph\n",
+      "P_SY = abs(E)*abs(V_ph)*delta_dash_elec/abs(X_s) \t\t\t#Synchronising power per phase\n",
+      "P_SY_total = 3*P_SY \t\t\t#Total Synchronising power\n",
+      "\n",
+      "T_SY = P_SY_total/(2*math.pi*n_s)\n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n",
+      "print 'Answer mismatches due to approximation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising torque is 4888.89 N-m\n",
+        "Answer mismatches due to approximation\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15  Page no :  69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6.*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 2000.*10**3\n",
+      "I_FL = VA/(math.sqrt(3)*V_L)  \n",
+      "I = I_FL\n",
+      "\n",
+      "# Calculations\n",
+      "X_s = 1.2\n",
+      "R_a = 0.01 \t\t\t#both per unit\n",
+      "IR_a = (1/100)*V_ph \t\t\t#product of I and R_a\n",
+      "R_a = IR_a/I_FL\n",
+      "IX_s = (120/100)*V_ph \t\t\t#product of I and X_s\n",
+      "#IX_s = (12/100)*V_ph   \t\t\t# this is the mistake made in the textbook\n",
+      "X_s = IX_s/I_FL\n",
+      "\n",
+      "#at full load and 0.8 pf\n",
+      "phi = math.acos(0.8)\n",
+      "#V_ph = complex(V_ph*math.cos(phi)V_ph*math.sin(phi))  \t\t\t#just introduced the angle\n",
+      "E_ph = math.sqrt(    (abs(V_ph)*math.cos(phi)+ IR_a)**2+   (abs(V_ph)*math.sin(phi)+ IX_s)**2     )\n",
+      "Poles = 8\n",
+      "f = 50\n",
+      "\n",
+      "delta = math.asin( (abs(V_ph)*math.sin(phi)+IX_s)/E_ph)  -phi\n",
+      "delta_dash_mech = (math.pi/180) \t\t\t#print lacemnt in degree mechanical \n",
+      "#print lacemnt in degree electrical\n",
+      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
+      "P_SY = abs(E_ph)*abs(V_ph)*math.cos(delta)*math.sin(delta_dash_elec)/X_s   \t\t\t#Synchronising power per phase\n",
+      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
+      "\n",
+      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
+      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power is %.2f kW'%(P_SY_total/1000)\n",
+      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n",
+      "\n",
+      "print 'Note that answer obtained doesnt match with textbook due to the following reasons: \\\n",
+      "\\niIX_s is considered wrong in textbook.It should have been 4156.92instead of 415.692  To verify this use commented statement of IX_s line 13and notice that it matches with textbook ans then' \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power is 223.22 kW\n",
+        "Synchronising torque is 2960.56 N-m\n",
+        "Note that answer obtained doesnt match with textbook due to the following reasons: \n",
+        "iIX_s is considered wrong in textbook.It should have been 4156.92instead of 415.692  To verify this use commented statement of IX_s line 13and notice that it matches with textbook ans then\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.16  Page no :  71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "E = 11.*10**3/math.sqrt(3)\n",
+      "I_sc = 1000.\n",
+      "Pole = 2.\n",
+      "f = 50.\n",
+      "delta_dash_mech = 1*math.pi/180 \t\t\t#print lacemnt in degree mechanical \n",
+      "\n",
+      "# Calculations\n",
+      "#print lacemnt in degree electrical \n",
+      "delta_dash_elec = delta_dash_mech*(Pole/2)\n",
+      "P_SY = E*I_sc*delta_dash_mech \t\t\t#Synchronising power per phase\n",
+      "P_SY_total = P_SY*3 \t\t\t#total Synchronising power\n",
+      "\n",
+      "ns = 120*f/(60*Pole)  \t\t\t#in r.p.s\n",
+      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power is %.2f kW'%(P_SY_total/1000)\n",
+      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power is 332.53 kW\n",
+        "Synchronising torque is 1058.48 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17 Page no :  72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#Line PQ for Altermnator 1 and PR for alternaator 2.AB is at frequency x from P where total load is 30 MW\n",
+      "# Variables\n",
+      "QT = 25.\n",
+      "PT = 2.\n",
+      "#PC = x\n",
+      "SR = 25.\n",
+      "PS = 1.5\n",
+      "\n",
+      "# Calculations\n",
+      "#using similarity of triangles PAC and PQT\n",
+      "AC_by_PC = (QT/PT)\t\t\t# because (AC/QT) = (PC/PT)\n",
+      "#using similarity of triangles PCB and PSR\n",
+      "CB_by_PC = (SR/PS)\n",
+      "\n",
+      "AC_by_x = AC_by_PC   \t\t\t#which implies AC = 12.5*x\n",
+      "CB_by_x = CB_by_PC  \t\t\t#which implies CB = 16.67*x\n",
+      "\n",
+      "AC_plus_CB = 30  \t\t\t#total load at the frequency at P is 30 MW\n",
+      "x =  AC_plus_CB/(AC_by_x + CB_by_x)\n",
+      "AC = 12.5*x\n",
+      "CB = 16.67*x  \n",
+      "frequency = 50-x\n",
+      "\n",
+      "# Results\n",
+      "print 'Loads shared by alternator 1 and 2 are %.2f MW and %.2f MW respectively'%(AC,CB)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loads shared by alternator 1 and 2 are 12.86 MW and 17.15 MW respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18  Page no :  73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
+      "\n",
+      "# Variables\n",
+      "load_total = 1600.*10**3\n",
+      "pf = 1/math.sqrt(2)  \t\t\t#lag\n",
+      "V_L = 6600.\n",
+      "\n",
+      "# Calculations\n",
+      "I_L = p2z(load_total/(math.sqrt(3)*V_L*pf),-1*math.acos(math.radians(pf)) )\n",
+      "I_1 = p2z(90,-1*math.acos(math.radians(0.8)))\n",
+      "I_2 = I_L-I_1\n",
+      "phi = abs(math.atan(I_2.imag/I_2.real))\n",
+      "I_a = abs(I_2)\n",
+      "R_a = 1.05\n",
+      "X_s = 5 \t\t\t#resistance and synchronous reactance per phase\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "E_ph = math.sqrt(  (V_ph*math.cos(phi)+I_a*R_a )**2 + ( V_ph*math.sin(phi)+I_a*X_s )**2        )\n",
+      "E_line = math.sqrt(3)*E_ph\n",
+      "\n",
+      "# Results\n",
+      "print 'Excitation of second alternator is %.2f V '%(E_line)\n",
+      "print ' The corresponding field current from the graph is about 310 A'\n",
+      "print 'Note: The answer obtained will differ from textbook answer because of higher degree of accuracy while storing I_2 and the improper rounding off of I_2 in the textbook'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Excitation of second alternator is 6884.65 V \n",
+        " The corresponding field current from the graph is about 310 A\n",
+        "Note: The answer obtained will differ from textbook answer because of higher degree of accuracy while storing I_2 and the improper rounding off of I_2 in the textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.19 Page no : 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 10.*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 5.*10**6\n",
+      "I_FL = VA/(math.sqrt(3)*V_L) \t\t\t#full-load current\n",
+      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = IX_s/I_FL \t\t\t#synchronous reactance\n",
+      "P = 4.\n",
+      "\n",
+      "# Calculations\n",
+      "delta_dash_mech = 1*(math.pi/180) \t\t\t#print lacement in degree mechanical \n",
+      "#print lacement in degree electrical\n",
+      "delta_dash_elec = delta_dash_mech*(P/2)\n",
+      "E = V_ph  \t\t\t#at no load\n",
+      "P_SY =  delta_dash_elec*E**2/X_s  \t\t\t#Synchronising power per phase\n",
+      "P_SY_total = P_SY*3  \t\t\t#Total Synchronising power\n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power per phase is %.2fkW \\\n",
+      "\\nTotal Synchronising power is %.2fkW '%(P_SY/1000,P_SY_total/1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power per phase is 290.89kW \n",
+        "Total Synchronising power is 872.66kW \n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.20 Page no : 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Power_total = 1.414 \t\t\t#per unit\n",
+      "V_L = 1. \t\t\t#per unit\n",
+      "phi_t = math.acos(0.707)\n",
+      "I_L_T = Power_total/(math.sqrt(3)*V_L*math.cos(phi_t))  \t\t\t#Total current\n",
+      "\n",
+      "# Calculations\n",
+      "#Current supplied by each alternator\n",
+      "I_1 = I_L_T/2\n",
+      "I_2 = I_1\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "\n",
+      "phi = math.acos(0.707)\n",
+      "R_a = 0\n",
+      "X_s = 0.6 \t\t\t#resistacne and synchronous reactance\n",
+      "E_ph = math.sqrt(  (V_ph*math.cos(phi)+ I_1*R_a)**2 + (V_ph*math.sin(phi)+I_1*X_s)**2          )\n",
+      "delta =  math.atan((I_1*X_s+V_ph*math.sin(phi)) / (V_ph*math.cos(phi))) - phi    \t\t\t#power angle\n",
+      "\n",
+      "# Results\n",
+      "print 'EMF is %.4f p.u. and power angle is %.2f degrees '%(E_ph,delta*180/math.pi)\n",
+      "print 'Following assumptions were made :'\n",
+      "print '1.Terminal or bus bar voltage at ppoint of connection is constant'\n",
+      "print '2.The alternators are identical and are initially equally excited'\n",
+      "print '3.The power supplied by prime movers is adjusted so that each machine carries half the load represented by external impedance Z = R+ j 2pifL %( where R and L are constant'\n",
+      "print '4.The stator resistance is negligible'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "EMF is 0.8580 p.u. and power angle is 16.58 degrees \n",
+        "Following assumptions were made :\n",
+        "1.Terminal or bus bar voltage at ppoint of connection is constant\n",
+        "2.The alternators are identical and are initially equally excited\n",
+        "3.The power supplied by prime movers is adjusted so that each machine carries half the load represented by external impedance Z = R+ j 2pifL %( where R and L are constant\n",
+        "4.The stator resistance is negligible\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.21 Page no : 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_l = 480.\n",
+      "X_d = 0.1\n",
+      "X_q = 0.075\n",
+      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "I_l = 1200.\n",
+      "I_ph = I_l/math.sqrt(3)\n",
+      "V_ph = V_l\n",
+      "V_t = V_l\n",
+      "I_a = I_ph\n",
+      "\n",
+      "# Calculations\n",
+      "phi = math.acos(0.8)\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
+      "delta = psi-phi \n",
+      "\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(psi)\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "\n",
+      "# Results\n",
+      "print 'Excitation e.m.f is %.f V '%(E_f)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Excitation e.m.f is 524 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.22  Page no :  78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "VA = 3.5*10**6\n",
+      "P = 32.        \t\t\t#Poles\n",
+      "Power = 2.5*10**6 \t\t\t#In watts\n",
+      "V_l = 6.6*10**3\n",
+      "phi = math.acos(0.8)\n",
+      "I_l = Power/(V_l*math.cos(phi)*math.sqrt(3))\n",
+      "X_d = 9.6\n",
+      "X_q = 6.\n",
+      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "\n",
+      "# Calculations\n",
+      "V_t = V_l/math.sqrt(3)\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+I_l*X_q)/(V_t*math.cos(phi)+I_l*R_a)            )\n",
+      "delta = psi-phi\n",
+      "I_s = I_l\n",
+      "I_d = I_s*math.sin(psi)\n",
+      "I_q = I_s*math.cos(psi)\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "regulation = 100*(E_f-V_t)/V_t\n",
+      "\n",
+      "# Results\n",
+      "print 'percentage regulation is %.2f percent'%(regulation)\n",
+      "print 'Excitation emf =  %.0f V'%(E_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percentage regulation is 50.85 percent\n",
+        "Excitation emf =  5748 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.23 Page no :  79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "X_d = 7.6\n",
+      "X_q = 4.5\n",
+      "R_a = 0.15 \t\t\t#armature resistance and synchronous reactance of directquadrature \n",
+      "axisV_l = 13.8*10**3\n",
+      "V_l = 13.8*10**3\n",
+      "V_t = V_l/math.sqrt(3)\n",
+      "phi = math.acos(0.8)\n",
+      "VA = 25*10**6\n",
+      "I_a = VA/(math.sqrt(3)*V_l)\n",
+      "psi = math.atan(  (V_t*math.cos(phi)+I_a*X_q)/(V_t*math.sin(phi)+I_a*R_a)            )\n",
+      "\n",
+      "# Calculations\n",
+      "delta = psi-phi\n",
+      "I_s = I_a\n",
+      "I_d = I_s*math.sin(psi)\n",
+      "I_q = I_s*math.cos(psi)\n",
+      "\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "regulation = 100*(E_f-V_t)/V_t\n",
+      "\n",
+      "# Results\n",
+      "print 'percentage regulation is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percentage regulation is 79.30 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.24 Page no :  80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "X_d = 1.\n",
+      "X_q = 0.6\n",
+      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "phi = math.acos(0.8) \t\t\t#lag\n",
+      "V_t = 1.\n",
+      "I_a = 1. \t\t\t#full load\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
+      "\n",
+      "# Calculations\n",
+      "delta = psi-phi\n",
+      "I_s = I_a\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(psi)\n",
+      "\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "regulation = 100*(E_f-V_t)/V_t\n",
+      "\n",
+      "# Results\n",
+      "print 'percentage regulation is %.2f percent'%(regulation)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percentage regulation is 77.50 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.25 Page no :  81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I_a = 10.\n",
+      "phi = 20. \t\t\t#lag and degrees\n",
+      "V_t = 400.\n",
+      "X_d = 10.\n",
+      "X_q = 6.5\n",
+      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "\n",
+      "# Calculations\n",
+      "psi = math.atan((V_t*math.sin(math.radians(phi))+I_a*X_q)/(V_t*math.cos(math.radians(phi))+I_a*R_a))\n",
+      "delta = math.degrees(psi)-phi\n",
+      "I_d = math.degrees(I_a*math.sin(math.radians(psi)))\n",
+      "I_q = (I_a*math.cos(psi))\n",
+      "\n",
+      "# Results\n",
+      "print 'Load angle is %.2f degrees '%(delta)\n",
+      "print 'I_d and I_q are %.4f A and %.4f A respectively '%(I_d,I_q  )\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load angle is 8.23 degrees \n",
+        "I_d and I_q are 4.9272 A and 8.8105 A respectively \n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.26 Page no : 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "X_d = 0.8\n",
+      "X_q = 0.5\n",
+      "R_a = 0.02 \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "\n",
+      "\t\t\t#case(i)  lag\n",
+      "phi = math.acos(0.8)\n",
+      "V_t = 1\n",
+      "I_a = 1\t\t\t#full-load\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
+      "delta = psi-phi\n",
+      "\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(psi)\n",
+      "\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "regulation = 100*(E_f-V_t)/V_t\n",
+      "print 'percentage regulation at 0.8 pf lag is %.2f percent'%(regulation)\n",
+      "\n",
+      "\t\t\t#case(ii)  lead\n",
+      "phi2 = -1*math.acos(0.8) \t\t\t#minus sign because of leading pf\n",
+      "psi2 = math.atan(  (V_t*math.sin(phi2)+I_a*X_q)/(V_t*math.cos(phi2)+I_a*R_a)            )\n",
+      "delta2 = psi2-phi2\n",
+      "\n",
+      "I_d2 = I_a*math.sin(psi2)\n",
+      "I_q2 = I_a*math.cos(psi2)\n",
+      "\n",
+      "E_f2 = V_t*math.cos(delta2)+I_d2*X_d+I_q2*R_a\n",
+      "regulation2 = 100*(E_f2-V_t)/V_t\n",
+      "print 'percentage regulation at 0.8 pf lead is %.f percent'%(regulation2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percentage regulation at 0.8 pf lag is 61.25 percent\n",
+        "percentage regulation at 0.8 pf lead is -21 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.27 Page no : 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array,tan,arccos\n",
+      "\n",
+      "# Variables\n",
+      "kW = array([800,500,1000,600])\n",
+      "cosphi = array([1,0.9,0.8,0.9])\n",
+      "tanphi = tan(arccos(cosphi))\n",
+      "kVAR = kW*tanphi\n",
+      "\n",
+      "# Calculations\n",
+      "kW_total = kW[0]+kW[1]+kW[2]+kW[3]\n",
+      "kVAR_total = kVAR[0]+kVAR[1]+kVAR[2]+-1*kVAR[3] \t\t\t#4th case is leading \n",
+      "\n",
+      "phi_c = math.atan(kVAR_total/kW_total)  \t\t\t#total power factor angle\n",
+      "phi_1 = math.acos(0.95)\t\t\t#pf of machine 1\n",
+      "kW_1 = 1000 \t\t\t#active component of machine 1\n",
+      "kVAR_1 = kW_1*math.tan(phi_1) \t\t\t#reactive component of machine 1\n",
+      "kW_2 = kW_total - kW_1 \t\t\t#active component of machine 1\n",
+      "kVAR_2 = kVAR_total-kVAR_1 \t\t\t#reactive component of machine 2\n",
+      "\n",
+      "phi_2 = math.atan(kVAR_2/kW_2) \n",
+      "pf_2 = math.cos(phi_2) \t\t\t#power factor of machine 2\n",
+      "\n",
+      "# Results\n",
+      "print 'Output of second alternator =  %.0f kW'%(kW_2)\n",
+      "print 'power factor of machine 2  =  %.2f and lagging'%(pf_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output of second alternator =  1900 kW\n",
+        "power factor of machine 2  =  0.98 and lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.28 Page no : 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array,tan,arccos\n",
+      "\n",
+      "# Variables\n",
+      "kW = array([250,300,150])\n",
+      "cosphi = array([0.9,0.75,0.8]) \t\t\t#all lagging\n",
+      "tanphi = tan(arccos(cosphi))\n",
+      "kVAR = kW*tanphi\n",
+      "\n",
+      "# Calculations\n",
+      "kW_total = kW[0]+kW[1]+kW[2]\n",
+      "kVAR_total = kVAR[0]+kVAR[1]+kVAR[2] \n",
+      "\n",
+      "phi_1 = math.acos(0.8)\t\t\t#pf of machine 1\n",
+      "kW_1 = 100 \t\t\t#active component of machine 1\n",
+      "kVAR_1 = kW_1*math.tan(phi_1) \t\t\t#reactive component of machine 1\n",
+      "kW_2 = kW_total - kW_1 \t\t\t#active component of machine 1\n",
+      "kVAR_2 = kVAR_total-kVAR_1 \t\t\t#reactive component of machine 2\n",
+      "phi_2 = math.atan(kVAR_2/kW_2) \n",
+      "pf_2 = math.cos(phi_2) \t\t\t#power factor of machine 2\n",
+      "\n",
+      "# Results\n",
+      "print 'Output of second alternator =  %.0f kW'%(kW_2)\n",
+      "print 'power factor of machine 2  =  %.4f and lagging'%(pf_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output of second alternator =  600 kW\n",
+        "power factor of machine 2  =  0.8172 and lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.29 Page no :  85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6.6*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "V_t = V_ph\n",
+      "X_d = 9.6\n",
+      "X_q = 6.\n",
+      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "VA = 3.5*10**6\n",
+      "I_L = VA/(math.sqrt(3)*V_L)\n",
+      "\n",
+      "# Calculations\n",
+      "P = 2.5*10**6\n",
+      "phi = math.acos(0.8)\n",
+      "I_a = P/(math.sqrt(3)*V_L*math.cos(phi))\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+ I_a*X_q)/(V_t*math.cos(phi)+ I_a*R_a)     )\n",
+      "\n",
+      "delta = psi-phi\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(phi)\n",
+      "\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "regulation = 100*(E_f-V_t)/V_t\n",
+      "P_max = (V_ph**2/2)*((X_d-X_q)/(X_d*X_q))*(math.sin(2*delta))\n",
+      "\n",
+      "# Results\n",
+      "print 'percentage voltage regulation is %.2f percent'%(regulation)\n",
+      "print 'Power under open circuit is %.1f kW per phase'%(P_max/1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percentage voltage regulation is 50.85 percent\n",
+        "Power under open circuit is 231.1 kW per phase\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.30  Page no :  86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3.3*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 3.*10**6\n",
+      "I_FL = VA/(math.sqrt(3)*V_L)\n",
+      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = complex(0,IX_s/I_FL)\n",
+      "N_s = 1000 \t\t\t#in r.p.m\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "Poles = 6.\n",
+      "f = 50.\n",
+      "delta_dash_mech = (math.pi/180) \t\t\t#print lacement in degree mechanical\n",
+      "#print lacement in degree electrical\n",
+      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
+      "\n",
+      "I = I_FL\n",
+      "phi = math.acos(0.8)\n",
+      "V = complex(V_ph*math.cos(phi),V_ph*math.sin(phi))\n",
+      "E =  V+ I*X_s\n",
+      "\n",
+      "delta = (math.pi/180)*math.atan(E.imag/E.real)-phi   \t\t\t#E leads I by (math.pi/180)*phasemag(E) and V leads I by phi radians \n",
+      "P_SY = abs(E)*abs(V_ph)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)   \t\t\t#Synchronising power per phase \n",
+      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
+      "\n",
+      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.m\n",
+      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power per phase is %.3f kW'%(P_SY/1000)\n",
+      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n",
+      "print 'Answer mismatches due to improper approximation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power per phase is 197.555 kW\n",
+        "Synchronising torque is 5660 N-m\n",
+        "Answer mismatches due to improper approximation\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.31  Page no :  87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3.3*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 3.*10**6\n",
+      "I_FL = VA/(math.sqrt(3)*V_L)\n",
+      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = complex(0,IX_s/I_FL)\n",
+      "N_s = 1000. \t\t\t#in r.p.m\n",
+      "Poles = 6.\n",
+      "f = 50.\n",
+      "\n",
+      "# Calculations\n",
+      "delta_dash_mech = (math.pi/180) \t\t\t#print lacement in degree mechanical\n",
+      "#print lacement in degree electrical \n",
+      "delta_dash_elec = delta_dash_mech*(Poles/2)\n",
+      "\n",
+      "#E = V as the alternator is on no-load and X_s = Z_s\n",
+      "P_SY = abs(V_ph)**2*(delta_dash_elec)/abs(X_s)   \t\t\t#Synchronising power per phase\n",
+      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
+      "\n",
+      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
+      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power per phase is %.3f kW'%(P_SY/1000)\n",
+      "print 'Total Synchronising power is %.3f kW'%(P_SY_total/1000)\n",
+      "print 'Synchronising torque is %.0f N-m'%(T_SY)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power per phase is 261.799 kW\n",
+        "Total Synchronising power is 785.398 kW\n",
+        "Synchronising torque is 7500 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.32 Page no :  88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 11.*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 700.*10**3\n",
+      "I_FL = VA/(math.sqrt(3)*V_L)\n",
+      "IX_s = (14./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = IX_s/I_FL\n",
+      "#X_s = complex(0,IX_s/I_FL)\n",
+      "IR_a = (1.5/100)*V_ph \t\t\t#product of I and R_a\n",
+      "R_a = IR_a/I_FL\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "I = I_FL\n",
+      "phi = math.acos(0.8)\n",
+      "V = complex(V_ph*math.cos(phi),V_ph*math.sin(phi))\n",
+      "E_ph = math.sqrt(  (V_ph*math.cos(phi)+IR_a)**2 +(V_ph*math.sin(phi)+IX_s)**2    )\n",
+      "delta = math.asin((V_ph*math.sin(phi)+IX_s)/E_ph)   -phi\n",
+      "\n",
+      "Poles = 4.\n",
+      "f = 50.\n",
+      "delta_dash_mech = (math.pi/180) \t\t\t#phase print lacemnt in degree mechanical\n",
+      "delta_dash_elec = delta_dash_mech*(Poles/2)\t\t\t#phase print lacemnt in degree electrical\n",
+      "\n",
+      "P_SY = abs(V_ph)*abs(E_ph)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)   \t\t\t#Synchronising power per phase\n",
+      "P_SY_total = 3*P_SY  \t\t\t#total Synchronising power\n",
+      "\n",
+      "ns = 120*f/(60*Poles)  \t\t\t#in r.p.s\n",
+      "T_SY = P_SY_total/(2*math.pi*ns)   \t\t\t#Synchronising torque \n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power per phase is %.3f kW'%(P_SY/1000)\n",
+      "print 'Synchronising power is %.3f kW ; '%(P_SY/1000)\n",
+      "print 'Total Synchronising power is %.3f kW'%(P_SY_total/1000)\n",
+      "print 'Synchronising torque is %.2f N-m'%(T_SY)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power per phase is 63.750 kW\n",
+        "Synchronising power is 63.750 kW ; \n",
+        "Total Synchronising power is 191.249 kW\n",
+        "Synchronising torque is 1217.53 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.33 Page no : 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "    return  RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
+      "\n",
+      "# Variables\n",
+      "Z1 = complex(0,2)\n",
+      "Z2 = complex(0,3)\n",
+      "Z = 6.\n",
+      "E1 = p2z(230,0)\n",
+      "E2 = p2z(230,10)\n",
+      "\n",
+      "# Calculations\n",
+      "I1 = ((E1-E2)*Z+E1*Z2)/(Z*(Z1+Z2)+Z1*Z2)\n",
+      "I2 = ((E2-E1)*Z+E2*Z1)/(Z*(Z1+Z2)+Z1*Z2)\n",
+      "\n",
+      "phi1 = math.atan(I1.imag/I1.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
+      "phi2 = math.atan(I2.imag/I2.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
+      "\n",
+      "I = I1+I2\n",
+      "V = I*Z \t\t\t#Terminal voltage\n",
+      "\n",
+      "# Results\n",
+      "print 'i) Terminal voltage is %.2f volts at %.1f degrees'%(abs(V),math.degrees(math.atan(V.imag/V.real)))\n",
+      "print 'ii) Currents are %.2f A at %.0f degrees and %.2f A at %.2f degrees     \\\n",
+      "\\nTotal current is %.2f A at %.1f degrees  '%(abs(I1),math.degrees(math.atan(I1.imag/I1.real)),abs(I2),\\\n",
+      "                                              math.degrees(math.atan(I2.imag/I2.real)),abs(I) \\\n",
+      "                                              ,math.degrees(math.atan(I.imag/I.real)))\n",
+      "\n",
+      "P1 = abs(V)*abs(I1)*math.cos(math.radians(phi1))\n",
+      "P2 = abs(V)*abs(I2)*math.cos(math.radians(phi2))\n",
+      "print 'iii)Power delivered %.2f watts and  %.2f watts'%(P1,P2)\n",
+      "\n",
+      "# note : rounding off error. kindly check the calculations\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Terminal voltage is 224.71 volts at -7.3 degrees\n",
+        "ii) Currents are 14.74 A at -14 degrees and 22.88 A at -3.03 degrees     \n",
+        "Total current is 37.45 A at -7.3 degrees  \n",
+        "iii)Power delivered 3311.43 watts and  5141.03 watts\n"
+       ]
+      }
+     ],
+     "prompt_number": 65
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.34  Page no : 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "X_d = 0.8\n",
+      "X_q = 0.5  \t\t\t#both per unit\n",
+      "R_a = 0. \t\t\t#assumed\n",
+      "phi = math.acos(0.8)\n",
+      "V_t = 1.\t\t\t#pu\n",
+      "I_a = 1. \t\t\t#full-load\n",
+      "\n",
+      "# Calculations\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
+      "delta = psi-phi\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(psi)\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "\n",
+      "# Results\n",
+      "print 'Open circuit voltage is %.3f p.u.'%(E_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Open circuit voltage is 1.603 p.u.\n"
+       ]
+      }
+     ],
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.35 Page no : 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array,tan,arccos\n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6600.\n",
+      "I_L = 110.\n",
+      "phi_1 = math.acos(0.9) \t\t\t#lagging\n",
+      "kW = array([400,1000,400,300])*10**3\n",
+      "cosphi = array([1,0.71,0.8,0.9])\n",
+      "tanphi = tan(arccos(cosphi))\n",
+      "kVAR = kW*tanphi\n",
+      "\n",
+      "# Calculations\n",
+      "kW_total = kW[0]+kW[1]+kW[2]+kW[3]\n",
+      "kVAR_total = kVAR[0]+kVAR[1]+kVAR[2]+kVAR[3]\n",
+      "\n",
+      "phi_c = math.atan(kVAR_total/kW_total)  \t\t\t#total power factor angle\n",
+      "load_1 = math.sqrt(3)*V_L*I_L*math.cos(phi_1)\n",
+      "\n",
+      "kW_1 = load_1 \t\t\t#active component of machine 1\n",
+      "kVAR_1 = kW_1*math.tan(phi_1) \t\t\t#reactive component of machine 1\n",
+      "kW_2 = kW_total - kW_1 \t\t\t#active component of machine 1\n",
+      "kVAR_2 = kVAR_total-kVAR_1 \t\t\t#reactive component of machine 2\n",
+      "\n",
+      "phi_2 = math.atan(kVAR_2/kW_2) \n",
+      "pf_2 = math.cos(phi_2) \t\t\t#power factor of machine 2\n",
+      "\n",
+      "# Results\n",
+      "print 'Output of second alternator =  %.2f kW'%(kW_2/1000)\n",
+      "print 'Power factor of machine 2  =  %.4f and lagging'%(pf_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output of second alternator =  968.28 kW\n",
+        "Power factor of machine 2  =  0.7366 and lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.36  Page no :  93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 11000.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 2.*10**6\n",
+      "phi = math.acos(0.8)\n",
+      "I_FL = VA/(math.sqrt(3)*V_L)\n",
+      "phi_1 = math.acos(0.8)\n",
+      "IX_s = (20./100)*V_ph \t\t\t#product of I and X_s\n",
+      "\n",
+      "# Calculations\n",
+      "X_s = IX_s/I_FL\n",
+      "I_1 = I_FL\n",
+      "BC = I_1*math.cos(phi_1)*X_s\n",
+      "AB = I_1*math.sin(phi_1)*X_s    \n",
+      "OA = V_ph\n",
+      "OC = math.sqrt( (OA+AB)**2+(BC)**2 ) \n",
+      "E_1 = OC\n",
+      "E_2 = 1.25*E_1\n",
+      "OE = E_2\n",
+      "DE = BC\n",
+      "AD = math.sqrt(OE**2-DE**2) -OA           \t\t\t#because OE = math.sqrt( (OA+AD)**2 + (DE)**2  )\n",
+      "\n",
+      "I_2sinphi2 = AD/X_s\n",
+      "I_2cosphi2 = I_1*math.cos(phi)\n",
+      "I_2 = math.sqrt( (I_2cosphi2)**2 + (I_2sinphi2)**2  )\n",
+      "phi2 = math.atan( I_2sinphi2/ I_2cosphi2 )\n",
+      "new_pf = math.cos(phi2)\n",
+      "\n",
+      "# Results\n",
+      "print 'Machine current is %.2f A '%(I_2)\n",
+      "print 'Power factor is  %.4f lagging'%(new_pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Machine current is 228.62 A \n",
+        "Power factor is  0.3673 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 68
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.37  Page no :  95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "    return  RRRR*exp(1j*math.pi*Theeeta/180.);\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "P_out = 3000.*10**3\n",
+      "V_L = 6.6*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "phi = math.acos(0.8)\n",
+      "I_L = p2z(P_out/(math.sqrt(3)*V_L*math.cos(phi)),-1*(180/math.pi)*phi)\n",
+      "\n",
+      "# Calculations\n",
+      "P_out1 = P_out/2\n",
+      "I_L1 = 150  \t\t\t#given\n",
+      "phi_L1 = math.acos(    P_out1/(math.sqrt(3)*V_L*I_L1)  )\n",
+      "I_L1 = p2z(I_L1,-1*(180/math.pi)*phi_L1)\n",
+      "\n",
+      "I_L2 = I_L-I_L1\n",
+      "pf_2 = math.cos(math.atan(I_L2.imag/I_L2.real))\n",
+      "Z_1 = complex(0.5,10)\n",
+      "I_1 = I_L1\n",
+      "E_1 = V_ph + I_1*Z_1\n",
+      "delta_1 =  math.atan(E_1.imag/E_1.real)   \t\t\t#load angle of alternator 1\n",
+      "E_1L = math.sqrt(3)*E_1\n",
+      "\n",
+      "Z_2 = complex(0.4,12)\n",
+      "I_2 = I_L2\n",
+      "E_2 = V_ph + I_2*Z_2\n",
+      "delta_2 = (math.pi/180)*math.atan(E_2.imag/E_2.real)    \t\t\t#load angle of alternator 2\n",
+      "\n",
+      "# Results\n",
+      "print 'Part i) Currents are %.0f A at %.1f degrees and %.1f A at %.1f degrees\\nTotal current is %.0f at %.2f' %(abs(I_L1),\\\n",
+      "                            math.degrees(math.atan(I_L1.imag/I_L1.real)),abs(I_L2),\\\n",
+      "                            math.degrees(math.atan(I_L2.imag/I_L2.real)),abs(I_L),math.degrees(math.atan(I_L.imag/I_L.real)))\n",
+      "print 'Part ii)Power factor is %.4f and lagging'%(math.cos(phi_L1))\n",
+      "print 'Part iii)emf are %.2f V at %.2f degrees and %.4f V at %.0f degrees'%(abs(E_1),\\\n",
+      "                            math.degrees(math.atan(E_1.imag/E_1.real)),abs(E_2),math.degrees(math.atan(E_2.imag/E_2.real)))\n",
+      "print 'Part iv)Power angles are %.2f degrees and %.0f degrees '%(180/math.pi*delta_1,(180/math.pi)*delta_2)\n",
+      "\n",
+      "# note : rounding off error. kindly check the calculations\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part i) Currents are 150 A at -29.0 degrees and 180.6 A at -43.4 degrees\n",
+        "Total current is 328 at -36.87\n",
+        "Part ii)Power factor is 0.8748 and lagging\n",
+        "Part iii)emf are 4776.46 V at 15.49 degrees and 5565.7081 V at 16 degrees\n",
+        "Part iv)Power angles are 15.49 degrees and 0 degrees \n"
+       ]
+      }
+     ],
+     "prompt_number": 73
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.38  Page no :  97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Z1 = complex(0.2,2)\n",
+      "Z2 = Z1\n",
+      "ZL = complex(3,4)\n",
+      "Z = ZL\n",
+      "E1 = complex(2000,0)\n",
+      "E2 = complex(2200,100)\n",
+      "\n",
+      "# Calculations\n",
+      "I1 = ((E1-E2)*Z+E1*Z2)/(Z*(Z1+Z2)+Z1*Z2)\n",
+      "I2 = ((E2-E1)*Z+E2*Z1)/(Z*(Z1+Z2)+Z1*Z2)\n",
+      "\n",
+      "IL = I1+I2\n",
+      "V = IL*Z \t\t\t#Terminal voltage\n",
+      "\n",
+      "phi1 = math.atan(V.imag/V.real)-math.atan(I1.imag/I1.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
+      "phi2 = math.atan(V.imag/V.real)-math.atan(I2.imag/I2.real)  \t\t\t#Phasemag returns the angle of complex number in degrees\n",
+      "\n",
+      "Pout1 = math.sqrt(3)*math.sqrt(3)*abs(V)*abs(I1)*math.cos(math.radians(phi1))\n",
+      "Pout2 = math.sqrt(3)*math.sqrt(3)*abs(V)*abs(I2)*math.cos(math.radians(phi2))\n",
+      "\n",
+      "# Results\n",
+      "print 'Power delivered is %.2f kW and %.2f kW at power-factors %.4f lag and %.4f lag respectively'%(Pout1/1000,Pout2/1000,math.cos(math.radians(phi1)),math.cos(math.radians(phi2)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power delivered is 658.24 kW and 1253.92 kW at power-factors 0.9999 lag and 0.9999 lag respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 74
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.39  Page no :  99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "f = 50.\n",
+      "P = 12.\n",
+      "V_L = 6600.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "VA = 2000.*10**3\n",
+      "I_FL = VA/(math.sqrt(3)*V_L)\n",
+      "\n",
+      "# Calculations\n",
+      "IX_s = (25./100)*V_ph \t\t\t#product of I and X_s\n",
+      "X_s = complex(0,IX_s/I_FL)\n",
+      "N_s = 12*f/P \t\t\t#in rpm\n",
+      "delta_dash_mech = (math.pi/180) \t\t\t#phase print lacemnt in degree mechanical\n",
+      "delta_dash_elec = delta_dash_mech*(P/2) \t\t\t#phase print lacemnt in degree electrical\n",
+      "\n",
+      "phi = math.acos(0.8) \t\t\t#lag\n",
+      "I = complex(I_FL*math.cos(-1*phi),I_FL*math.sin(-1*phi))\n",
+      "V = V_ph\n",
+      "E = V + I*X_s\n",
+      "delta = math.atan(E.imag/E.real)*(math.pi/180)\n",
+      "P_SY = abs(E)*abs(V)*math.cos(delta)*math.sin(delta_dash_elec)/abs(X_s)\n",
+      "P_SY_total = 3*P_SY\n",
+      "\n",
+      "# Results\n",
+      "print 'Synchronising power is %.2f kW'%(P_SY/1000)\n",
+      "print 'Total Synchronising power is %.2f kW'%(P_SY_total/1000)\n",
+      "\n",
+      "# note : rounding off error. kindly check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Synchronising power is 325.36 kW\n",
+        "Total Synchronising power is 976.09 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 76
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.40  Page no :  101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 22000.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "power = 230.*10**6\n",
+      "phi = math.acos(1.)\n",
+      "I_FL = power/(math.sqrt(3)*V_L*math.cos(phi))\n",
+      "I_1 = I_FL\n",
+      "X_s = 1.2\n",
+      "\n",
+      "# Calculations\n",
+      "E_1 = math.sqrt( V_ph**2 + (I_1*X_s)**2 )\n",
+      "E_2 = 1.3*E_1\n",
+      "AC =   math.sqrt( E_2**2-(I_1*X_s)**2  )   -V_ph     \t\t\t# because E**2 = (V_ph+AC)**2+(I_1*X_s)**2\n",
+      "I2X_S = AC\n",
+      "\n",
+      "I_2cosphi2 = I_1  \t\t\t#because phi_2 = math.acos(I_1/I_2) \t\t\t#from ACD\n",
+      "I_2sinphi2 = AC/X_s\n",
+      "I_2 = math.sqrt( (I_2cosphi2)**2 + (I_2sinphi2)**2  )\n",
+      "phi2 = math.atan( I_2sinphi2/ I_2cosphi2 )\n",
+      "new_pf = math.cos(phi2)\n",
+      "\n",
+      "# Results\n",
+      "print 'Machine current is %.2f A '%(I_2)\n",
+      "print 'Power factor is  %.4f and lagging'%(new_pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Machine current is 7274.58 A \n",
+        "Power factor is  0.8297 and lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 77
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.41  Page no :  102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import cmath\n",
+      "\n",
+      "#note that a new function p2z has been defined below for direct representation of complex numbers in polar form\n",
+      "def  p2z(RRRR,Theeeta):\n",
+      "\treturn RRRR*cmath.exp(1j*math.pi*Theeeta/180.);\n",
+      "\n",
+      "# Variables\n",
+      "P_out = 1500.*10**3\n",
+      "V_L = 3.3*10**3\n",
+      "phi = math.acos(0.8)\n",
+      "I_L = p2z(P_out/(math.sqrt(3)*V_L*math.cos(phi)),-1*math.acos(math.radians(0.8)))\n",
+      "\n",
+      "I_L1_magnitude = 150.  \t\t\t#given\n",
+      "P_out1 = (3.*10**6)/2 \t\t\t#because load is EQUALLY shared between 2 alternators\n",
+      "pf_L1 = P_out1/(math.sqrt(3)*2*V_L*I_L1_magnitude)  \t\t\t#operating pf of alternator 1\n",
+      "phi1 = math.acos(math.radians(pf_L1))\n",
+      "I_L1 = p2z(I_L1_magnitude,-1*phi1)\n",
+      "I_L2 = I_L-I_L1   \t\t\t#because I_L = I_L1 + I_L2\n",
+      "pf_L2 = math.cos(math.atan(I_L2.imag/I_L2.real))\n",
+      "\n",
+      "# Calculations\n",
+      "V_ph = 6.6*10**3/math.sqrt(3)\n",
+      "Z_1 = complex(0.5,10)\n",
+      "I_1 = I_L1\n",
+      "E_1 =  V_ph + I_1*Z_1\n",
+      "delta_1 = math.atan(E_1.imag/E_1.real)  \t\t\t#load angle of alternator 1\n",
+      "I_2 = I_L2\n",
+      "\n",
+      "Z_2 = complex(0.4,12)\n",
+      "E_2 =  V_ph + I_2*Z_2\n",
+      "delta_2 = math.atan(E_2.imag/E_2.real) \t\t\t#load angle of alternator 1\n",
+      "\n",
+      "# Results\n",
+      "print 'for machine 1current is %.0f A at %.2f degrees \\nPower factor of %.4f laginduced emf \\\n",
+      "        of %.2f V \\nload angle of %.2f degrees'%(abs(I_L1),math.degrees(math.atan(I_L1.imag/I_L1.real)),pf_L1,abs(E_1),delta_1)\n",
+      "print 'for machine 2current is %.1f A at %.1f degrees\\nPower factor of %.4f laginduced emf \\\n",
+      "        of %.2f V\\nload angle of %.0f degrees'%(abs(I_L2),math.degrees(math.atan(I_L2.imag/I_L2.real)),pf_L2,abs(E_2),math.degrees(delta_2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for machine 1current is 150 A at -1.56 degrees \n",
+        "Power factor of 0.8748 laginduced emf         of 4202.06 V \n",
+        "load angle of 0.36 degrees\n",
+        "for machine 2current is 178.0 A at -1.6 degrees\n",
+        "Power factor of 0.9996 laginduced emf         of 4480.49 V\n",
+        "load angle of 28 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 81
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.42  Page no :  104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_l = 230\n",
+      "VA = 5*10**3\n",
+      "X_d = 12\n",
+      "X_q = 7\n",
+      "R_a = 0 \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "phi = math.acos(1)\n",
+      "\n",
+      "# Calculations\n",
+      "I_l = VA/(V_l*math.sqrt(3))\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "V_t = V_ph\n",
+      "I_a = I_l\n",
+      "\n",
+      "psi = math.atan(  (V_t*math.sin(phi)+I_a*X_q)/(V_t*math.cos(phi)+I_a*R_a)            )\n",
+      "delta = psi-phi\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(psi)\n",
+      "E_f = V_t*math.cos(delta)+I_d*X_d+I_q*R_a\n",
+      "\n",
+      "# Results\n",
+      "print 'Excitation voltage is %.3f V '%(E_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Excitation voltage is 193.852 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 82
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.43  Page no :  104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_l = 6.6*10**3\n",
+      "V_t = V_l/math.sqrt(3)\n",
+      "X_d = 23.2\n",
+      "X_q = 14.5\n",
+      "R_a = 0. \t\t\t#armature resistance and synchronous reactance of direct quadrature axis\n",
+      "VA = 1800.*10**3\n",
+      "phi = math.acos(0.8) \t\t\t#lag\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_a = VA/(V_l*math.sqrt(3))\n",
+      "\n",
+      "psi = math.atan(  (V_t*math.sin(phi)-I_a*X_q)/(V_t*math.cos(phi)-I_a*R_a)  ) \t\t\t#minus sign in numerator and denomenator for motors\n",
+      "delta = psi+phi\n",
+      "I_d = I_a*math.sin(psi)\n",
+      "I_q = I_a*math.cos(psi)\n",
+      "E_f = V_t*math.cos(delta)-I_d*X_d-I_q*R_a\n",
+      "print 'Excitation emf  =  %.4f V'%(E_f)\n",
+      "#P_m =  ( V_t*E_f*math.sin(delta)/X_d  ) + ((1/X_q)-(1/X_d))*0.5*math.sin(2*delta)*V_t**2\n",
+      "#P_m = 0.4996*math.cos(delta)+0.1877*math.sin(2*delta)\n",
+      "#for maximum power output differenciate and equate to zero\n",
+      "\n",
+      "delta_max = 63.4 \t\t\t#degree\n",
+      "\n",
+      "P_m_max = ((1/X_q)-(1/X_d))*0.5*math.sin(math.radians(2*delta_max))*V_t**2   \t\t\t#Maximuum load supplied with E_f = 0\n",
+      "print 'Maximum load the motor can supply is %.4f MW per phase '%(P_m_max*10**-6      )\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Excitation emf  =  3042.2721 V\n",
+        "Maximum load the motor can supply is 0.1503 MW per phase \n"
+       ]
+      }
+     ],
+     "prompt_number": 83
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch7_2.ipynb b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch7_2.ipynb
new file mode 100755
index 00000000..90ee7606
--- /dev/null
+++ b/DC_Machines_and_Synchronous_Machines_by_U_A_Bakshi_and_M_V_Bakshi/ch7_2.ipynb
@@ -0,0 +1,1702 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a6d47374988dffe6254758a68746561860439fac0faeabc093287fa2241d2260"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Synchronous Motors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1  Page no :  27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "import numpy as np\n",
+      "\n",
+      "# Variables\n",
+      "V_l = 400.\n",
+      "R_a = 0.2\n",
+      "X_s = 2.  \t\t\t#armature resistance and synchronous reactance\n",
+      "I_L = 25.\n",
+      "I_aph = I_L\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
+      "angle = math.atan(Z_s.imag/Z_s.real)\n",
+      "angle = math.degrees(angle)\n",
+      "E_Rph=I_aph*abs(Z_s)\n",
+      "theta = (math.pi/180.)*angle \n",
+      "\n",
+      "# Calculations and Results\n",
+      "#case 1\n",
+      "phi = math.acos(0.8) \t\t\t#lagging\n",
+      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta-phi)  )\n",
+      "print 'i)Back EMF induced with 0.8 lagging pf is %.3f V'%(E_bph)\n",
+      "\n",
+      "#case 2\n",
+      "phi = math.acos(0.9) \t\t\t#leading\n",
+      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
+      "print 'ii)Back EMF induced with 0.8 lagging pf is %.3f V'%(E_bph)\n",
+      "\n",
+      "#case 3\n",
+      "phi = math.acos(1)\n",
+      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta)  )\n",
+      "print 'iii)Back EMF induced with 0.8 lagging pf is %.3f V'%(E_bph)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Back EMF induced with 0.8 lagging pf is 200.386 V\n",
+        "ii)Back EMF induced with 0.8 lagging pf is 252.678 V\n",
+        "iii)Back EMF induced with 0.8 lagging pf is 231.406 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2  Page no :  28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import cos\n",
+      "# Variables\n",
+      "V_l = 500.\n",
+      "R_a = 0.4\n",
+      "X_s = 4. \t\t\t#armature resistance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "theta = (math.pi/180)*angle\t\t\t#phasemag returns angle in degrees,not radians\n",
+      "V_ph = V_l/math.sqrt(3)\n",
+      "I_l = 50.\n",
+      "I_aph = I_l\n",
+      "E_Rph = I_aph*abs(Z_s)\n",
+      "\n",
+      "# Calculations and Results\t\t\n",
+      "#case 1\n",
+      "E_bline = 600\n",
+      "E_bph = E_bline/math.sqrt(3)\n",
+      "phi = math.acos( (-E_bph**2 + E_Rph**2 + V_ph**2 )/(2*E_Rph*V_ph) ) -theta    \t\t\t#leading\n",
+      "#because E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
+      "print 'i)power factor is %.4f leading'%(cos(phi))\n",
+      "\n",
+      "#case 2\n",
+      "E_bline = 380\n",
+      "E_bph = E_bline/math.sqrt(3)\n",
+      "phi =  theta-math.acos( (-E_bph**2 + E_Rph**2 + V_ph**2 )/(2*E_Rph*V_ph) )     \t\t\t#leading\n",
+      "#because E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta-phi)\n",
+      "print 'ii)power factor is %.4f lagging'%(cos(phi))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)power factor is 0.9977 leading\n",
+        "ii)power factor is 0.8197 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3  Page no :  29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6600.\n",
+      "P_out = 500.*10**3\n",
+      "eta = 83./100 \t\t\t#efficiency\n",
+      "R_a = 0.3\n",
+      "X_s = 3.2 \t\t\t#armature resistance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s)  \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "theta = (math.pi/180)* angle \t\t\t#phasemag returns the angle in degrees not radians\n",
+      "phi = math.acos(0.8) \t\t\t#leading\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "P_in = P_out/eta\n",
+      "\n",
+      "# Calculations and Results\n",
+      "I_L =    P_in/ (math.sqrt(3) * V_L *  math.cos(phi) )\n",
+      "\n",
+      "# because P_in = math.sqrt(3) * V_L * I_L * math.cos(phi)\n",
+      "I_aph = I_L\n",
+      "E_Rph = I_aph*abs(Z_s)\n",
+      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
+      "print 'i) Generated EmF on full loaad is %.2f V'%(E_bph)\n",
+      "\n",
+      "delta =  math.degrees(math.asin( (E_Rph/E_bph))*math.sin(theta+phi)  )\n",
+      "#This is obtained after applying sune rule to triangle OAB from thre phasor diagram\n",
+      "print 'ii) load angle is %.2f degrees'%(delta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Generated EmF on full loaad is 3925.33 V\n",
+        "ii) load angle is 2.64 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4  Page no :  32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "V_L = 500.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "phi = math.acos(0.9) \t\t\t\t#lagging\n",
+      "output_power = 17.*10**3\n",
+      "R_a = 0.8 \t\t\t\t\t\t\t#armaature reactance\n",
+      "mechanical_losses = 1300. \t\t\t#mechanical losses is W\n",
+      "P_m = output_power+mechanical_losses   \t\t\t#gross mechanical power developed\n",
+      "\n",
+      "#  P_m =  input_power - stator losses\n",
+      "#  input_power =  3* V_ph * I_aph * math.cos(phi)\n",
+      "#  Stator losses =  3*I_aph**2*R_a\n",
+      "# solving above equations we get 2.4 I_a**2 - 779/.4225*I_a + 18300  =  0\n",
+      "I_a_eqn = [2.4, -779.4225, 18300]\n",
+      "I_a_roots = roots(I_a_eqn)\n",
+      "I_a = I_a_roots[1]  \t\t\t#neglecting higher value\n",
+      "I_aph = I_a\n",
+      "print 'Current drawn by the motor is %.3f A'%(I_a)\n",
+      "\n",
+      "input_power =  3* V_ph * I_aph * math.cos(phi)\n",
+      "eta = 100*output_power/input_power\n",
+      "print 'Full load efficiency is %.2f percent'%(eta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current drawn by the motor is 25.478 A\n",
+        "Full load efficiency is 85.61 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5  Page no :  44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#subscript 1 is for industrial load and 2 for synchronous motor\n",
+      "P_1 = 800. \t\t\t# Active power in KW\n",
+      "phi_1 = math.acos(0.6) \t\t\t#lagging\n",
+      "Q_1 = P_1*math.tan(phi_1)   \t\t\t#reactive power by load 1\n",
+      "\n",
+      "# Calculations and Results\n",
+      "output_power = 200.\n",
+      "eta = 91./100 \t\t\t#efficiency of synchronous motor\n",
+      "input_power =  output_power/eta\n",
+      "P_2 = input_power\t\t\t# active power drawn by synchronous motor\n",
+      "P_T = P_1 + P_2  \t\t\t#combined total load of industry and synchronous motor\n",
+      "phi_T = math.acos(0.92 )\t\t\t#lagging\n",
+      "Q_T = P_T* math.tan(phi_T)  \t\t\t#from power triangle\n",
+      "Q_2 =  Q_T - Q_1  \t\t\t#it turns out to be negative indicating its leading nature\n",
+      "S_2 = math.sqrt( P_2**2 + Q_2**2    )\n",
+      "print 'Desired kVA rating of Synchronous motor  is %.3f kVA'%(S_2)\n",
+      "\n",
+      "phi_2 =  math.atan (Q_2/P_2)\n",
+      "print 'Power factor of synchronous motor is %.4f LEADING'%(math.cos(phi_2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Desired kVA rating of Synchronous motor  is 669.353 kVA\n",
+        "Power factor of synchronous motor is 0.3283 LEADING\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 Page no :  47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "output_power = 37.3*1000 \t\t\t#Watts on full load\n",
+      "Z_s = complex(0.2,1.6) \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "theta = (math.pi/180)*angle \t\t\t#phase mag returns the angle in degrees and not raidians\n",
+      "phi = math.acos(0.9) \t\t\t#leading\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "eta = 88. \t\t\t#efficiency in percentage\n",
+      "\n",
+      "# Calculations\n",
+      "input_power = 100*output_power/eta\n",
+      "I_L = input_power/(math.sqrt(3)*V_L*math.cos(phi))\n",
+      "I_aph = I_L\n",
+      "E_Rph = I_aph*abs(Z_s)\n",
+      "\n",
+      "E_bph =  math.sqrt( (E_Rph)**2 + (V_ph)**2  -2*E_Rph*V_ph*math.cos(theta+phi)  )\n",
+      "E_line = math.sqrt(3)*E_bph\n",
+      "\n",
+      "# Results\n",
+      "print 'Induced EMF is %.2f V and its line value is %.2f V'%(E_bph,E_line)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Induced EMF is 285.65 V and its line value is 494.75 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7  Page no :  48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "input_power = 20.*1000 \n",
+      "R_a = 0.\n",
+      "X_s = 4. \t\t\t#armature reactance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s)   \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag))\n",
+      "\n",
+      "theta = (math.pi/180)*angle  \t\t\t#phase mag returns the angle in degrees and not raidians\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "E_bline = 550. \t\t\t#star connection\n",
+      "E_bph = E_bline/math.sqrt(3)\n",
+      "\n",
+      "# Calculations\n",
+      "I_a_cos_phi = input_power/(math.sqrt(3)*V_L)  \t\t\t#product of I_a and math.cos(phi)\n",
+      "I_a_sin_phi =  (   math.sqrt(E_bph**2- (abs(Z_s)*I_a_cos_phi)**2   )  -V_ph )/abs(Z_s)\t\t\t#from triangle DAB\n",
+      "phi = math.atan(I_a_sin_phi/I_a_cos_phi)\n",
+      "I_a = I_a_cos_phi/math.cos(phi)        \n",
+      "\n",
+      "# Results\n",
+      "print 'Motor power fctor is %.3f Leading'%(cos(phi))\n",
+      "print 'Current drawn by the motor is %.2f A'%(I_a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Motor power fctor is 0.872 Leading\n",
+        "Current drawn by the motor is 33.11 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8 Page no :  50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3300.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "R_a = 2.\n",
+      "X_s = 18. \t\t\t#armature reactance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degrees not radians\n",
+      "E_bline = 3800.\n",
+      "E_bph = E_bline/math.sqrt(3)\n",
+      "\n",
+      "#part(i)\n",
+      "P_m_max  =  (E_bph*V_ph/abs(Z_s))- (E_bph**2/abs(Z_s))*math.cos(theta)\n",
+      "print 'i)Max total mechanical power developed that motor can develop is %.2f W per phase'%(P_m_max)\n",
+      "\n",
+      "#part(ii)\n",
+      "#from phasor diagram applying math.comath.sine rule to triangle OAB\n",
+      "E_Rph = math.sqrt(  E_bph**2 + V_ph**2 -2*E_bph*V_ph*math.cos(theta) ) \n",
+      "I_aph =  E_Rph/abs(Z_s)\n",
+      "print 'ii)Current at max power developed is %.1f A'%(I_aph)\n",
+      "\n",
+      "copper_loss = 3* I_aph**2 * R_a\n",
+      "P_in_max_total = 3 * P_m_max \t\t\t#input power at max power developed\n",
+      "total_P_in =  P_in_max_total + copper_loss \t\t\t#total input power \n",
+      "pf = total_P_in/(math.sqrt(3)*I_aph*V_L)\n",
+      "print 'Power factor at max power developed is %.3f leading'%(pf)\n",
+      "\n",
+      "\n",
+      "# 'Answer in part1 mismatched because of improper approximation in book'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Max total mechanical power developed that motor can develop is 201452.30 W per phase\n",
+        "ii)Current at max power developed is 151.4 A\n",
+        "Power factor at max power developed is 0.857 leading\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9  Page no :  51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 500.\n",
+      "R_a = 0.03\n",
+      "X_s = 0.3 \t\t\t#armature reactance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degrees not radians\n",
+      "phi = math.acos(0.8)\n",
+      "eta = 93/100.\n",
+      "output_power = 100.*746\n",
+      "input_power = output_power/eta\n",
+      "I_L = input_power/(math.sqrt(3)*V_L*math.cos(phi))\n",
+      "I_aph = I_L\n",
+      "E_Rph = I_aph*abs(Z_s)\n",
+      "\n",
+      "#from the phasor diagram\n",
+      "E_bph  =    math.sqrt( E_Rph**2 + (V_L/math.sqrt(3))**2 - 2*E_Rph*(V_L/math.sqrt(3))*math.cos(phi+theta)  ) \n",
+      "\n",
+      "cu_losses = 3*(I_aph)**2*R_a \t\t\t#total copper losses\n",
+      "P_m =  input_power - cu_losses  \t\t\t#total mechanical power developed\n",
+      "\n",
+      "print 'EMF developed per phase is %.4f V \\nTotal mechanical power developed is %.1f watts'%(E_bph,P_m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "EMF developed per phase is 308.1880 V \n",
+        "Total mechanical power developed is 79008.6 watts\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10  Page no :  53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 415.\n",
+      "V_ph = V_L \t\t\t#due to delta connection\n",
+      "E_bline = 520.\n",
+      "R_a = 0.5\n",
+      "X_s = 4. \t\t\t#armature reactance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = (math.pi/180.)*angle \t\t\t#phasemag returns angle in degrees not radians\n",
+      "\n",
+      "delta = theta \t\t\t#for maximum power\n",
+      "P_m_max  =  (E_bline*V_ph/abs(Z_s))- (E_bline**2/abs(Z_s))*math.cos(theta)\n",
+      "P_m_max_total =  3* P_m_max\n",
+      "fi_loss = 1000.  \t\t\t#frictional and iron losses\n",
+      "P_out_total  =  P_m_max_total-fi_loss  \n",
+      "\n",
+      "HP_output =  P_out_total/746 \t\t\t#converting watts to horse power\n",
+      "print 'HP output for maximum power output is %.2f HP'%(HP_output)\n",
+      "\n",
+      "#from the phasor diagram\n",
+      "E_Rph = math.sqrt(  E_bline**2 + V_ph**2 -2*E_bline*V_ph*math.cos(delta) ) \n",
+      "I_aph =  E_Rph/abs(Z_s)\n",
+      "I_L = I_aph*math.sqrt(3)\n",
+      "print 'Line current is %f A'%(I_L)\n",
+      "cu_loss_total = 3*(I_aph)**2*R_a  \t\t\t#total copper losses\n",
+      "input_power = P_m_max_total+ cu_loss_total\n",
+      "pf = input_power/(math.sqrt(3)*I_L*V_L)  \t\t\t#leading\n",
+      "print 'Power factor for maximum power output is %.2f leading '%(pf)\n",
+      "\n",
+      "eta = 100*P_out_total /input_power\n",
+      "print 'Efficiency for maximum power output is %.2f percent'%(eta)\n",
+      "\n",
+      "# 'Answer might mismatch because of improper approximation done in book'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "HP output for maximum power output is 180.48 HP\n",
+        "Line current is 268.015479 A\n",
+        "Power factor for maximum power output is 0.89 leading \n",
+        "Efficiency for maximum power output is 78.48 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11  Page no :  54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 8.\n",
+      "f = 50. \t\t\t#Pole and frequency\n",
+      "N_s = 120.*f/P \t\t\t#synchronous speed\n",
+      "V_L = 6.6*10**3 \n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "Z_s = complex(0.66,6.6) \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degree  not radians\n",
+      "E_bph = 4500.\n",
+      "input_power = 2500.*10**3\n",
+      "I_a_cosphi = input_power/(math.sqrt(3)*V_L)   \t\t\t#Its product of I_a and math.cos(phi);I_a = I_l for star conneted load\n",
+      "\n",
+      "# Calculations\n",
+      "#applying math.comath.sine rule to triangle ABC from phasor diagram and solve \n",
+      "#math.tan(phi)**2 + 5.2252 math.tan(phi)-2.2432 = 0\n",
+      "p = [1, 5.2252, -2.2432]\n",
+      "tan_phi = roots(p)\n",
+      "phi = math.atan(tan_phi[1])\n",
+      "pf = math.cos(phi)\n",
+      "I_a =  I_a_cosphi/ math.cos(phi)\n",
+      "\n",
+      "#apply math.sine rule to triangle ABC\n",
+      "delta =    math.asin(I_a*abs(Z_s)*math.sin(theta+phi)/E_bph)\n",
+      "P_m = 3*E_bph*I_a*math.cos(delta+phi)\n",
+      "T_g  =  P_m/(2*math.pi*N_s/60)\n",
+      "\n",
+      "# Results\n",
+      "print 'i)Torque developed is %f N-m'%(T_g)\n",
+      "print 'ii)Input current is %.4f A'%(I_a)\n",
+      "print 'iii)Power factor is %.4f leading'%(pf)\n",
+      "print 'iv)Power angle is %.2f degrees '%((180/math.pi)*delta)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Torque developed is 30437.047497 N-m\n",
+        "ii)Input current is 235.4472 A\n",
+        "iii)Power factor is 0.9288 leading\n",
+        "iv)Power angle is 19.48 degrees \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12  Page no :  57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "input_power = 15.*10**3\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "E_b = 480.\n",
+      "E_bph = E_b/math.sqrt(3)\n",
+      "Z_s = complex(1,5) \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degree  not radians\n",
+      "\n",
+      "# Calculations\n",
+      "I_a_cosphi = input_power/(math.sqrt(3)*V_L)  \t\t\t#product of I_a & math.cos(phi)\n",
+      "#Applying math.comath.sine rule to triangle OAB and solving\n",
+      "#math.tan(phi)**2+ 4.101*math.tan(phi)-1.7499 = 0\n",
+      "p = [1,4.101,-1.7449]\n",
+      "tan_phi = roots(p)\n",
+      "phi = math.atan(tan_phi[1])  \t\t\t#ignoring negative vaule\n",
+      "I_a =  I_a_cosphi/ math.cos(phi)\n",
+      "\n",
+      "#applying math.sine rule to Triangle OAB\n",
+      "delta = math.asin( I_a*abs(Z_s)* math.sin(theta+phi)/E_bph )\n",
+      "\n",
+      "# Results\n",
+      "print 'Load angle is %.1f degrees'%(delta*180/math.pi)\n",
+      "print 'Armature current is %.4f A'%(I_a)\n",
+      "print 'Power factor is %.3f leading'%(cos(phi))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load angle is 24.9 degrees\n",
+        "Armature current is 23.2283 A\n",
+        "Power factor is 0.932 leading\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13  Page no :  59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import roots\n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "E_b = 460.\n",
+      "E_bph = E_b/math.sqrt(3)\n",
+      "input_power = 3.75*10**3\n",
+      "Z_s = complex(1,8) \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = math.degrees(angle) \t\t\t#phasemag returns angle in degree ,not radians\n",
+      "I_L_cos_phi  =  input_power/(math.sqrt(3)*V_L)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Applying math.comath.sine rule to triangle OAB and solving further\n",
+      "#math.tan(phi)**2 + 458.366*math.tan(phi) -450.65  = 0 \n",
+      "p = [1,458.366-450.65]\n",
+      "tan_phi = roots(p)\n",
+      "\n",
+      "phi = math.atan(tan_phi[0])  \t\t\t#ignoring negative value\n",
+      "print 'Required power factor is %.4f leading'%(cos(phi))\n",
+      "I_L = I_L_cos_phi /math.cos(phi)\n",
+      "print 'Required current is %.4f A'%(I_L)\n",
+      "\n",
+      "# roots() python gives some different answer. Kindly check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required power factor is 0.1285 leading\n",
+        "Required current is 42.1134 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14  Page no :  60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#subscript 1 indicates induction motor 1\n",
+      "P_1 = 350.\n",
+      "phi_1 = math.acos(0.7071) \t\t\t#lagging\n",
+      "Q_1 = P_1*math.tan(phi_1)\t\t\t#from power triangle\n",
+      "\n",
+      "#subscript 2 indicates induction motor 2\n",
+      "P_2 = 190.\n",
+      "\n",
+      "# Calculations\n",
+      "#subscript T indicates total\n",
+      "P_T = P_1+P_2\n",
+      "phi_T = math.acos(0.9) \t\t\t#lagging\n",
+      "Q_T = P_T*math.tan(phi_T)\n",
+      "\n",
+      "Q_2 = Q_T-Q_1\n",
+      "kva_rating = math.sqrt(P_2**2+ Q_2**2)\n",
+      "\n",
+      "# Results\n",
+      "print 'kVA rating of synchronous motor is %.2f kVA'%(kva_rating)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kVA rating of synchronous motor is 209.59 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15  Page no :  61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "Pole = 6.\n",
+      "f = 50.\n",
+      "R_a = 0.2\n",
+      "X_s = 3. \t\t\t#armature reactance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag/Z_s.real)       # calculates angle in radians\n",
+      "\n",
+      "N_s = 120*f/Pole  \t\t\t#synchronous speed\n",
+      "\n",
+      "# Calculations\n",
+      "#subscript 1` refers to load 1\n",
+      "I_a1 = 20.\n",
+      "phi_1 = math.acos(1)\n",
+      "E_R1 =  I_a1* abs(Z_s)\n",
+      "E_bph  =   math.sqrt( E_R1**2 + V_ph**2 - 2*E_R1*V_ph*math.cos(phi_1+theta)  )\n",
+      "\n",
+      "#subscript 2` refers to load 2\n",
+      "I_a2 = 60.\n",
+      "E_R2 =  I_a2* abs(Z_s)\n",
+      "phi_2 =  math.acos ((E_R2**2 + V_ph**2 -E_bph**2 )/(2*E_R2*V_ph))   -theta   \t\t\t#new power factor\n",
+      "\n",
+      "input_power = math.sqrt(3)*V_L*I_a2*math.cos(phi_2)\n",
+      "cu_loss = 3*I_a2**2*R_a\n",
+      "P_m = input_power-cu_loss\n",
+      "T_g  = P_m /(2*math.pi*N_s/60)   \t\t\t#gross mechanical power developed\n",
+      "\n",
+      "# Results\n",
+      "print 'Gross torque developed is %.4f N-m and new power factor is %.4f lagging'%(T_g,cos(phi_2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gross torque developed is 357.1971 N-m and new power factor is 0.9518 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.16  Page no :  63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3300.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "E_bph = V_ph\n",
+      "Z_s = complex(0.5,5) \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = (math.pi/180)*angle \t\t\t#phasemag returns angle in degrees, not radians\n",
+      "P = 8.\n",
+      "f = 50. \t\t\t#pole and frequency\n",
+      "delta_mech = 3. \t\t\t#mechanical angle in degrees by which rotor is behind\n",
+      "delta_elec = (P/2)*delta_mech  \t\t\t#delta mech converted to electrical degrees\n",
+      "E_Rph = math.sqrt(  E_bph**2 + V_ph**2 -2*E_bph*V_ph*math.cos(math.radians(delta_elec) )) \n",
+      "I_aph =  E_Rph/abs(Z_s)\n",
+      "\n",
+      "#from the phasor diagram \n",
+      "phi = theta- math.asin( math.sin(math.radians(delta_elec))*E_bph/E_Rph )\n",
+      "pf = math.cos(phi)\n",
+      "print 'power factor of the motor is %.5f lagging'%(pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power factor of the motor is 0.99999 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17  Page no :  64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "E_bph = V_ph\n",
+      "P = 4.\n",
+      "f = 50.\t\t\t#Pole and frequency\n",
+      "delta_mech = 4*(math.pi/180) \t\t\t#mechanical angle in degrees by which rotor is behind\n",
+      "delta_elec =  delta_mech *(P/2) \t\t\t#delta_mech convertd to electrical degrees\n",
+      "Z_s = complex(0,2) \t\t\t#synchronous impedance\n",
+      "\n",
+      "# Calculations\n",
+      "#referring to phasor diagram\n",
+      "BC =   E_bph*math.sin(delta_elec)\n",
+      "AB =   E_bph\n",
+      "OA =   V_ph\n",
+      "\n",
+      "AC =   math.sqrt(AB**2-BC**2)\n",
+      "OC =   OA-AC\n",
+      "phi = math.atan(OC/BC)\n",
+      "OB = math.sqrt(OC**2 + BC**2)\n",
+      "I_a = OB/abs(Z_s)\n",
+      "\n",
+      "# Results\n",
+      "print 'Armature current drawn by the motor is %.4f A'%(I_a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature current drawn by the motor is 16.1096 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.18  Page no :  65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "input_power = 5472.\n",
+      "Z_s = complex(0,10)  \t\t\t#synchronous impedance\n",
+      "I_L_cosphi = input_power/(math.sqrt(3)*V_L) \t\t\t#product of I_L & math.cos(phi)\n",
+      "BC = 10*I_L_cosphi\n",
+      "AB = V_ph\n",
+      "OA = V_ph\n",
+      "\n",
+      "# Calculations\n",
+      "#from Triangle ABC in phasor diagram\n",
+      "AC  =  math.sqrt(AB**2- BC**2)\n",
+      "OC  =  OA - AC\n",
+      "\n",
+      "\t\t\t#from Triangle OCB \n",
+      "OB = math.sqrt( OC**2+ BC**2 )\n",
+      "E_Rph  =  OB\n",
+      "I_L = E_Rph/abs(Z_s)\n",
+      "\n",
+      "phi = math.atan(OC/BC)\n",
+      "pf = math.cos(phi)\n",
+      "delta = math.atan(BC/AC)  \t\t\t#load angle\n",
+      "\n",
+      "# Results\n",
+      "print 'Power factor is %.4f lagging'%(pf)\n",
+      "print 'Load angle is %.0f degrees'%(delta*180/math.pi)\n",
+      "print 'Armature current is %.3f A'%(I_L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power factor is 0.9848 lagging\n",
+        "Load angle is 20 degrees\n",
+        "Armature current is 8.020 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.19  Page no :  67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6600.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "Z_s = complex(2,20) \t\t\t#synchronous impedance\n",
+      "angle = math.degrees(math.atan(Z_s.imag/Z_s.real))\n",
+      "\n",
+      "theta = (math.pi/180) * angle \t\t\t#phasemag returns angle in degrees not radians\n",
+      "P_1 = 1000*10**3\n",
+      "P_2 = 1500*10**3\n",
+      "phi_1 = math.acos(0.8)  \t\t\t#leading\n",
+      "\n",
+      "# Calculations\n",
+      "I_L1 = P_1/(math.sqrt(3)*V_L*math.cos(phi_1))\n",
+      "I_a1ph = I_L1\n",
+      "E_R1ph = I_a1ph*abs(Z_s)\n",
+      "E_bph =   math.sqrt(    V_ph**2 + E_R1ph** -2*V_ph*E_R1ph*math.cos(theta+phi_1)  )\n",
+      "I_a2_cosphi_2 = P_2/(math.sqrt(3)*V_L)\n",
+      "\n",
+      "#Refer to the phasor diagram and solving for I_y\n",
+      "#404I_y**2 -152399.968 I_y -4543000 = 0\n",
+      "p = [404, -152399.968, -4543000]\n",
+      "ans = roots(p)\n",
+      "I_y = abs(ans[1]) \t\t\t#becuase root 1 is too high and root is -ve\n",
+      "\n",
+      "I_a2 = complex(I_a2_cosphi_2,I_y)\n",
+      "phi_2 = math.degrees(math.atan(I_a2.imag/I_a2.real))\n",
+      "\n",
+      "print 'Required power factor is %.3f leading'%(math.cos(math.radians(phi_2)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required power factor is 0.978 leading\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.20  Page no :  69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "V_L = 2300.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_L = 200.\n",
+      "I_a = I_L\n",
+      "Z_s = complex(0.2,2.2) \t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag/Z_s.real)\n",
+      "phi = math.acos(0.5)\n",
+      "\n",
+      "# Calculations\n",
+      "E_Rph = I_a*abs(Z_s)\n",
+      "E_bph  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(phi+theta)  ) \n",
+      "\n",
+      "# Results\n",
+      "print 'Generated EMF per phase is %.3f V'%(E_bph)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Generated EMF per phase is 1708.045 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.21  Page no :  69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3300.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "E_bline = 3800.\n",
+      "E_bph = E_bline/math.sqrt(3)\n",
+      "\n",
+      "R_a = 2.\n",
+      "X_s = 18. \t\t\t#armature resistance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag/Z_s.real)\n",
+      "\n",
+      "#part(i)\n",
+      "P_m_max  =  (E_bph*V_ph/abs(Z_s))- (E_bph**2/abs(Z_s))*math.cos(theta)  \t\t\t#maximum total mechanical power\n",
+      "print 'i)Maximum total mechanical power that the motor can develop is %.2f W per phase'%(P_m_max )\n",
+      "#part(ii)\n",
+      "delta = theta \t\t\t#for max P_m\n",
+      "E_Rph = math.sqrt(  E_bph**2 + V_ph**2 -2*E_bph*V_ph*math.cos(delta) ) \n",
+      "I_aph =  E_Rph/abs(Z_s)\n",
+      "print 'ii)Current at maximum power developed is %.1f A'%(I_aph)\n",
+      "cu_loss_total  =  3*I_aph**2*R_a  \t\t\t#total copper loss\n",
+      "P_m_max_total = 3*P_m_max \t\t\t#total maximum total mechanical power\n",
+      "P_in_total  =  P_m_max_total+ cu_loss_total  \t\t\t#total input power\n",
+      "\n",
+      "pf = P_in_total/(math.sqrt(3)*V_L*I_aph)\n",
+      "print '    Power factor at maximum power developed is %.3f leading'%(pf)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Maximum total mechanical power that the motor can develop is 201452.30 W per phase\n",
+        "ii)Current at maximum power developed is 151.4 A\n",
+        "    Power factor at maximum power developed is 0.857 leading\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.22  Page no :  71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#subscript 1 refers to load 1\n",
+      "I_1 = 18.\n",
+      "phi_1 = math.acos(0.8)\n",
+      "V_L = 440.\n",
+      "S_1 = math.sqrt(3)*I_1*V_L /1000  \t\t\t#kVA for load 1\n",
+      "P_1 = S_1*math.cos(phi_1)\n",
+      "Q_1 = S_1*math.sin(phi_1)\n",
+      "\n",
+      "# Calculations\n",
+      "P_out = 6.\n",
+      "eta_motor = 88./100\n",
+      "P_2 = P_out/eta_motor\n",
+      "\n",
+      "P_T = P_1+P_2\n",
+      "phi_T = math.acos(1)  \t\t\t#total power factor angle\n",
+      "Q_T = P_T*math.tan(phi_T)\n",
+      "\n",
+      "Q_2 =   Q_T - Q_1   \t\t\t#kVAR supplied by motor\n",
+      "#this will have  a negative sign just indicating its leading nature \n",
+      "phi_2 = math.atan(abs(Q_2)/P_2)\n",
+      "pf = math.cos(phi_2)  \t\t\t#leading\n",
+      "S_2 = P_2/math.cos(phi_2)  \t\t\t#kVA input to the motor\n",
+      "\n",
+      "# Results\n",
+      "print 'kVA input to the motor is %.3f kVA '%(S_2)\n",
+      "print 'Power factor when driving a 6kW mechanical load is %.4f leading'%(pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kVA input to the motor is 10.688 kVA \n",
+        "Power factor when driving a 6kW mechanical load is 0.6379 leading\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.23  Page no :  72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "output_power = 8.*10**3\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "R_a = 0.\n",
+      "X_s = 8.\t\t\t#armature resistance and syncronous reactance\n",
+      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
+      "eta = 88./100\n",
+      "input_power = output_power/eta\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#minimum current occurs at max power factors\n",
+      "phi = math.acos(1)\n",
+      "I_a_min = input_power/(math.sqrt(3)*V_L*math.cos(phi)) \t\t\t#required minimum current \n",
+      "print 'Minimum current is %.3f A'%(I_a_min)\n",
+      "E_R =  I_a_min * abs(Z_s)\n",
+      "E_bph  =    math.sqrt( E_R**2 + V_ph**2 - 2*E_R*V_ph*math.cos(phi+theta)  ) \n",
+      "print 'Induced EMF at full-load is %.3f V'%(E_bph)\n",
+      "\n",
+      "# note : rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum current is 13.122 A\n",
+        "Induced EMF at full-load is 241.534 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.24  Page no :  73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "R_a = 0.8\n",
+      "X_s = 5.\n",
+      "Z_s = complex(R_a,X_s) \t\t\t#armature resistance and syncronous reactance\n",
+      "theta = math.atan(Z_s.imag/Z_s.real)  \t\t\t# returns angle in radians \n",
+      "alpha = (math.pi/2) - theta\n",
+      "V_t = 3300/math.sqrt(3)\n",
+      "P_e_in = 800./(3) \t\t\t#per phase\n",
+      "phi = math.acos(0.8)  \t\t\t#leading\n",
+      "Q_e_in = -P_e_in*math.tan(phi)\n",
+      "\n",
+      "# Calculations\n",
+      "# using the following equation\n",
+      "# P_e_in =  V_t**2*R_a/(abs(Z_s))**2   +   V_t*E_b*math.sin(delta-alpha)/abs(Z_S)\n",
+      "# Q_e_in =  V_t**2*X_s/(abs(Z_s))**2   -   V_t*E_b*math.cos(delta-alpha)/abs(Z_S)\n",
+      "E_b_sin_delta_minus_9  =  407.2\n",
+      "E_b_cos_delta_minus_9  = 2413.6\n",
+      "#solving further\n",
+      "delta  =  math.atan(E_b_sin_delta_minus_9/E_b_cos_delta_minus_9 ) + 9\n",
+      "E_b = E_b_sin_delta_minus_9/math.sin(math.radians(delta-9))\n",
+      "\n",
+      "P_e_in_new  =  1200*10**3/3\n",
+      "# using the following equation again\n",
+      "# P_e_in =  V_t**2*R_a/(abs(Z_s))**2   +   V_t*E_b*math.sin(delta-alpha)/abs(Z_S)\n",
+      "# Q_e_in =  V_t**2*X_s/(abs(Z_s))**2   -   V_t*E_b*math.cos(delta-alpha)/abs(Z_S)\n",
+      "\n",
+      "alpha  = delta - math.asin(math.radians(P_e_in_new - V_t**2*R_a/(abs(Z_s))**2 ) / (V_t*E_b/abs(Z_s)))\n",
+      "Q_e_in_new =  V_t**2*X_s/(abs(Z_s))**2   -   V_t*E_b*math.cos(math.radians(delta - alpha))/abs(Z_s)\n",
+      "\n",
+      "pf = math.cos ( math.atan(abs(Q_e_in_new/P_e_in_new)))\n",
+      "\n",
+      "# Results\n",
+      "print 'New power factor is %.2f leading '%(pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "New power factor is 0.01 leading \n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.25  Page no :  74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6.6*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "P_in = 900.*10**3\n",
+      "R_a = 0.\n",
+      "X_s = 20. \t\t\t#armature resistance and synchronous reactance \n",
+      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
+      "E_b_L = 8.6*10**3\n",
+      "E_bph = E_b_L/math.sqrt(3) \n",
+      "\n",
+      "# Calculations\n",
+      "#refer to phasor diagram\n",
+      "OA = V_ph\n",
+      "AB = E_bph \t\t\t#OB =  E_Rph\n",
+      "\n",
+      "I_a_cosphi = P_in/(math.sqrt(3)*V_L)  \t\t\t#I_a*math.cos(phi)\n",
+      "BC = I_a_cosphi*abs(Z_s) \t\t\t#BC is a vector in phasor diagram\n",
+      "\n",
+      "OC = math.sqrt(AB**2 -BC**2  )- OA  \t\t\t#from phasor diagram\n",
+      "I_a_sinphi = OC/abs(Z_s)  \t\t\t#product of I_a and math.sin(phi)\n",
+      "phi =  math.atan (I_a_sinphi/I_a_cosphi)\n",
+      "I_a = I_a_cosphi/math.cos(phi) \t\t\t#product of I_a and math.cos(phi)\n",
+      "\n",
+      "# Results\n",
+      "print 'Motor current is %.3f A'%(I_a)\n",
+      "print 'Power factor of motor is %f leading'%(cos(phi))\n",
+      "print 'Note:There is slight mismatch in answer due to the approximation made during I_a* sinphi calculation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Motor current is 90.643 A\n",
+        "Power factor of motor is 0.868564 leading\n",
+        "Note:There is slight mismatch in answer due to the approximation made during I_a* sinphi calculation\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.26  Page no :  75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "#subscipt 1 refers to factory load\n",
+      "P_1 = 1800.\n",
+      "phi_1 = math.acos(0.6) \t\t\t#lagging\n",
+      "Q_1 = P_1*math.tan(phi_1)\n",
+      "\n",
+      "#Subscript 2 refers to synchronous condenser\n",
+      "P_2 = 0.\n",
+      "\n",
+      "# Calculations\n",
+      "#Subscript T refers to combination of condenser and factory load\n",
+      "P_T = P_1+P_2\n",
+      "phi_T = math.acos(0.95) \t\t\t#lagging\n",
+      "Q_T = P_T*math.tan(phi_T)\n",
+      "\n",
+      "kva_rating = math.sqrt(P_T**2+ Q_T**2)\n",
+      "\n",
+      "Q_2 = Q_T - Q_1\n",
+      "\n",
+      "# Results\n",
+      "print 'i)kVA rating of synchronous condender is %.3f kVA    Minus sign indicates leading nature'%(Q_2)\n",
+      "print 'ii)kVA rating of total factory is %.4f kVA'%(kva_rating)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)kVA rating of synchronous condender is -1808.369 kVA    Minus sign indicates leading nature\n",
+        "ii)kVA rating of total factory is 1894.7368 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.27  Page no :  77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "I_1 = 35.\n",
+      "phi_1 = math.acos(0.8)\n",
+      "V_L = 440.\n",
+      "S_1 = math.sqrt(3)*I_1*V_L /1000 \t\t\t#in kVA\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "P_1 = S_1*math.cos(phi_1)\n",
+      "Q_1 = S_1*math.sin(phi_1)\n",
+      "\n",
+      "P_out = 12. \t\t\t#motor load\n",
+      "eta_motor = 85./100\n",
+      "P_2 = P_out/eta_motor\n",
+      "\n",
+      "P_T = P_1 + P_2\n",
+      "phi_T = math.acos(1)\n",
+      "Q_T = P_T * math.tan(phi_T)\n",
+      "\n",
+      "\n",
+      "Q_2 = Q_T - Q_1  \t\t\t#kVA supplied by motor\n",
+      "#negative sign of Q_2 indicates its leading nature\n",
+      "phi_2 =  math.atan(abs(Q_2)/P_2)\n",
+      "S_2 = P_2/math.cos(phi_2)\n",
+      "\n",
+      "# Results\n",
+      "print 'Power factor when motor supplies 12kW load is %.4f leading'%(math.cos(phi_2))\n",
+      "print 'kVA input to the motor is %.3f kVA'%(S_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power factor when motor supplies 12kW load is 0.6615 leading\n",
+        "kVA input to the motor is 21.341 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.28  Page no :  78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "Z_s = complex(0.5,4)  \t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag/Z_s.real)  \t\t\t# returns angle in radians \n",
+      "\n",
+      "I_aph = 60.\n",
+      "phi = math.acos(0.866) \t\t\t#leading\n",
+      "power_losses = 2*10**3\n",
+      "\n",
+      "# Calculations\n",
+      "E_bph  =    math.sqrt( (I_aph*abs(Z_s))**2 + (V_ph)**2 - 2*(I_aph*abs(Z_s))*(V_ph)*math.cos(phi+theta)  ) \n",
+      "delta = theta \t\t\t#for P_m_max\n",
+      "P_m_max  =  (E_bph*V_ph/abs(Z_s))- (E_bph**2/abs(Z_s))*math.cos(delta)\n",
+      "P_m_max_total =  3 * P_m_max\n",
+      "P_out_max =   P_m_max_total- power_losses\n",
+      "\n",
+      "# Results\n",
+      "print 'Maximum power output is %.4f kW'%(P_out_max*10**-3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum power output is 51.3899 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.29  Page no :  79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 6.6*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "I_L = 50.\n",
+      "I_aph = I_L\n",
+      "Z_s = complex(1.5,8) \t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag/Z_s.real)  \t\t\t# returns angle in radians \n",
+      "E_Rph = I_aph*abs(Z_s)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#part(i)\n",
+      "phi = math.acos(0.8)\n",
+      "P_in =  math.sqrt(3)*V_L*I_L*math.cos(phi)   \t\t\t#for both lag and lead supplied power will be the same\n",
+      "print 'i)Power supplied to the motor is %.3f kW'%(P_in*10**-3)\n",
+      "#part(ii)\n",
+      "E_bph_lag  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(theta-phi)  )  \t\t\t#for lagging power factor\n",
+      "#Note that E_bph_lag > V_ph\n",
+      "print 'ii)Induced EMF for 0.8 power factor lag is %.3f V'%(E_bph_lag)\n",
+      "E_bph_lead  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(theta+phi)  )  \t\t\t#for leading power factor\n",
+      "#Note that E_bph_lead < V_ph\n",
+      "print '    Induced EMF for 0.8 power factor lead is %.3f V'%(E_bph_lead)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Power supplied to the motor is 457.261 kW\n",
+        "ii)Induced EMF for 0.8 power factor lag is 3521.267 V\n",
+        "    Induced EMF for 0.8 power factor lead is 4007.170 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.30  Page no :  80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 400.\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "P_out = 7.5*735.5\n",
+      "eta = 85./100 \t\t\t#efficiency\n",
+      "R_a = 0.\n",
+      "X_s = 10. \t\t\t#armature resistance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s)\t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
+      "\n",
+      "# Calculations and Results\n",
+      "P_in = P_out/eta\n",
+      "phi = math.acos(1) \t\t\t#for mimimum current power factor is maximum\n",
+      "I_L = P_in/(math.sqrt(3)*V_L*math.cos(phi))  \n",
+      "I_aph = I_L\n",
+      "print 'Minimum current is %.3f A at full load condition '%(I_L)\n",
+      "\n",
+      "E_Rph =  I_aph*abs(Z_s)\n",
+      "E_bph  =    math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(phi+theta)  ) \n",
+      "print 'and corresponding EMF is %.4f V'%(E_bph)\n",
+      "\n",
+      "# note : rounding off error.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum current is 9.367 A at full load condition \n",
+        "and corresponding EMF is 240.4216 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.31  Page no :  80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V_L = 3.3*10**3\n",
+      "V_ph = V_L/math.sqrt(3)\n",
+      "V_t = V_ph\n",
+      "Pole = 24.\n",
+      "f = 50.  \t\t\t#Pole and frequency\n",
+      "P = 1000.*10**3\n",
+      "R_a = 0\n",
+      "X_s = 3.24 \t\t\t#armature resistance and synchronous reactance\n",
+      "Z_s = complex(R_a,X_s) \t\t\t#synchronous impedance\n",
+      "theta = math.atan(Z_s.imag)  \t\t\t# returns angle in radians \n",
+      "phi = math.acos(1)\n",
+      "I_aph = P/(math.sqrt(3)*V_L*math.cos(phi))\n",
+      "\n",
+      "# Calculations\n",
+      "E_Rph = I_aph*abs(Z_s)\n",
+      "E_bph  =   math.sqrt( E_Rph**2 + V_ph**2 - 2*E_Rph*V_ph*math.cos(phi+theta)  ) \n",
+      "\n",
+      "P_m_max = 3*(E_bph*V_ph/abs(Z_s)) \t\t\t#maximum power that can be delivered\n",
+      "N_s = 120*f/Pole \t\t\t#synchronous speed\n",
+      "T_max = P_m_max /(2*math.pi*N_s/60) \t\t\t#maximum torque that can be developed\n",
+      "\n",
+      "# Results\n",
+      "print 'Maximum power and torque the motor can deliver is %.3f kW and %.2f *10**3 Nm respectively'%(P_m_max*10**-3,T_max/1000)\n",
+      "\n",
+      "# rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum power and torque the motor can deliver is 3211.633 kW and 122.68 *10**3 Nm respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Digital_Communications/Chapter2.ipynb b/Digital_Communications/Chapter2.ipynb
deleted file mode 100755
index 9cc6d1df..00000000
--- a/Digital_Communications/Chapter2.ipynb
+++ /dev/null
@@ -1,290 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 2: SAMPLING THEORY AND PULSE MODULATION<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1, page no 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#find Nquist Rate\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given \n",
-      "pi=3.14\n",
-      "w1=50*pi\n",
-      "w2=300*pi\n",
-      "w3=100*pi\n",
-      "#w=2*%pi*f\n",
-      "\n",
-      "#Calculation\n",
-      "f1=w1/(2*pi)\n",
-      "f2=w2/(2*pi)\n",
-      "f3=w3/(2*pi)\n",
-      "fm=f2 #fm = maximum frquency is present at the signal\n",
-      "\n",
-      "#Result\n",
-      "print('maximum frquency of the signal is = %.2f Hz' %f2)\n",
-      "fs=2*fm #Nyquist rate\n",
-      "print('Nquist Rate of Signal is = %.2f Hz' %fs)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum frquency of the signal is = 150.00 Hz\n",
-        "Nquist Rate of Signal is = 300.00 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2 , page no 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Nquist Rate and Nquist time interval\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "w1=5000*math.pi\n",
-      "w2=3000*math.pi;\n",
-      "f1=w1/(2*math.pi);\n",
-      "f2=w2/(2*math.pi);\n",
-      "\n",
-      "#Calculation\n",
-      "fm=f1 #fm = maximum frquency is present at the signal\n",
-      "fs=2*fm #Nyquist rate\n",
-      "Ts=1.0/(2.0*fm) #frequncy =1/time\n",
-      "Ts=Ts*(10**3)\n",
-      "\n",
-      "#Result\n",
-      "print('maximum frquency of the signal is = %.f Hz' %f1)\n",
-      "print('Nquist Rate of the given Signal is = %.f Hz' %fs)\n",
-      "print('Nquist Interval of the given signal is = %.1f m Sec' %Ts)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum frquency of the signal is = 2500 Hz\n",
-        "Nquist Rate of the given Signal is = 5000 Hz\n",
-        "Nquist Interval of the given signal is = 0.2 m Sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3, page no 51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find Nquist Rate \n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "f=100.0         # Frequency component of continuous-time signal\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*f #Nyquist rate\n",
-      "\n",
-      "#Result\n",
-      "print('i) To avoid aliasing Nquist Rate is = %.f Hz' %fs)\n",
-      "print('ii) It is theoretical example ')\n",
-      "print('iii) It is theoretical example ')\n",
-      "print('iv) It is theoretical example ')\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) To avoid aliasing Nquist Rate is = 200 Hz\n",
-        "ii) It is theoretical example \n",
-        "iii) It is theoretical example \n",
-        "iv) It is theoretical example \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4, page no 52 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Nquist Rate of Continous signal\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "w1=50*math.pi\n",
-      "w2=300*math.pi\n",
-      "w3=100*math.pi\n",
-      "\n",
-      "#Calculation\n",
-      "f1=w1/(2*math.pi)\n",
-      "f2=w2/(2*math.pi)\n",
-      "f3=w3/(2*math.pi)\n",
-      "fmax=f2 #fmax = Highest frquency component of the message signal\n",
-      "fs=2*fmax #Nyquist rate\n",
-      "\n",
-      "#Result\n",
-      "print('Highest frquency component of the message signal will be fmax = %.f Hz' %fmax)\n",
-      "print('Nquist Rate of the given Signal is = %.f Hz' %fs)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Highest frquency component of the message signal will be fmax = 150 Hz\n",
-        "Nquist Rate of the given Signal is = 300 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 2.7, page no 67 <h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#find amplitude distortion at highest frquency\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "fs=9.5  #samplig frequncy\n",
-      "fmax=1   #maximum frequncy\n",
-      "t=0.2 #pulse width\n",
-      "\n",
-      "#Calculation\n",
-      "c=3*10**8\n",
-      "f=fmax\n",
-      "H1=t*(0.9933)  #aperture effect at highest frequency, sinc(f*t)=0.9933 (given)\n",
-      "H1=H1*100\n",
-      "\n",
-      "#Result\n",
-      "print('|H(1)|=%.2f' %H1)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "|H(1)|=19.87\n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.8, page no 74 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate Transmission Bandwidth\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "fm=3.0*(10^3)\n",
-      "fs=8.0*(10^3) # sampling frequncy\n",
-      "\n",
-      "#Calculation\n",
-      "Ts=1.0/fs\n",
-      "t=0.1*Ts\n",
-      "BW=1.0/(2*t) #Bandwidth\n",
-      "BW=BW/(10^3)\n",
-      "\n",
-      "#Result\n",
-      "print('Transmission Bandwidth of PAM signal is kHz = %.f Khz ' %BW)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transmission Bandwidth of PAM signal is kHz = 40 Khz \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/Chapter3.ipynb b/Digital_Communications/Chapter3.ipynb
deleted file mode 100755
index 805bc62b..00000000
--- a/Digital_Communications/Chapter3.ipynb
+++ /dev/null
@@ -1,2238 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater 3: WAVEFORM CODING TECHNIQUES<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3,1 Page No 110"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print('This is a Theoretical  example')\t"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is a theoretcal example\n"
-       ]
-      }
-     ],
-     "prompt_number": 104
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3,2,i, Page No 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "#find code word length\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512        #Quantization levels\n",
-      "#q=2^v\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log10(512)/math.log10(2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The code word legth is = %.f bits\" %v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The code word legth is = 9 bits\n"
-       ]
-      }
-     ],
-     "prompt_number": 105
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2.ii, Page No 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#find The transmission Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512.0           #Quantization levels\n",
-      "v=9.0\n",
-      "\n",
-      "#Calculation\n",
-      "bw=v*fm*10**-6\n",
-      "\n",
-      "#Result\n",
-      "print(\"The transmission Bandwidth is = %.1f MHz\" %bw)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transmission Bandwidth is = 37.8 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 106
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2.iii, Page No 111 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find  Final Bit rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512.0      #Quantization levels\n",
-      "v=9.0\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm\n",
-      "r=v*fs       #signaling rate\n",
-      "\n",
-      "#Result\n",
-      "print(\"Final Bit rate =%.1f x 10^6 bits/sec\" %(r/(10**6)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Final Bit rate =75.6 x 10^6 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 107
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2.iv, Page No 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find Output signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512         #Quantization levels\n",
-      "v=9.0\n",
-      "\n",
-      "#Calculation\n",
-      "sn=4.8+6*v    #noise in dB\n",
-      "\n",
-      "\n",
-      "print(\"Output signal to noise ratio is = %.1f dB\" %sn)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal to noise ratio is = 58.8 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 108
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.i, Page No 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find bits per sample\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fmax=4*10**3\n",
-      "xmax=3.8\n",
-      "snr=100\n",
-      "\n",
-      "#Calculation\n",
-      "P=30*10**-3\n",
-      "v=(math.log10(((snr*xmax**2)/(3*P)))/(2*math.log10(2)))\n",
-      "\n",
-      "#Result\n",
-      "print(\"Number of bits required per sample are = %.2f bits\" %v)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of bits required per sample are = 6.98 bits\n"
-       ]
-      }
-     ],
-     "prompt_number": 109
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.ii, Page No 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find  Transmission Bandwith\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=4*10**3     #Bandwidth of PCM\n",
-      "xmax=3.8\n",
-      "snr=100       #Signal to Noise Ratio\n",
-      "outputs=30.0\n",
-      "v=7.0\n",
-      "\n",
-      "#Calculation\n",
-      "bw=outputs*v*fm \n",
-      "r=bw*2\n",
-      "bw=bw/10**3\n",
-      "\n",
-      "#Result\n",
-      "print('Transmission Bandwith R = %.f kHz' %bw)\n",
-      "r=r/1000\n",
-      "print('Signaling rate R = %.f bits/sec' %r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transmission Bandwith R = 840 kHz\n",
-        "Signaling rate R = 1680 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 110
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page No 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#find sampling rate,number of bits,bit rate,bandwidht\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "emax=0.001\n",
-      "de=2*emax\n",
-      "fm=100.0\n",
-      "xmax=10.0\n",
-      "\n",
-      "#Calculation\n",
-      "q=(2*xmax)/de\n",
-      "fs=2*fm\n",
-      "v=(math.log10(q))/math.log10(2)\n",
-      "v=math.ceil(v)\n",
-      "r=v*fs\n",
-      "\n",
-      "#Result\n",
-      "print('1. sampling Frequncy  =  %.f  Hz ' %fs)  \n",
-      "print('2. No.of bits in PCM  =  %.f  bits ' %v)  \n",
-      "print('3. sampling rate  =  %.f  bits per second ' %r)\n",
-      "r=r/2\n",
-      "print('4. Transmission Bandwidth  =  %.f  Hz ' %r)  "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. sampling Frequncy  =  200  Hz \n",
-        "2. No.of bits in PCM  =  14  bits \n",
-        "3. sampling rate  =  2800  bits per second \n",
-        "4. Transmission Bandwidth  =  1400  Hz \n"
-       ]
-      }
-     ],
-     "prompt_number": 111
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5, Page No 113 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Bandwidth,Sampling Rate\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=3.4*10**3\n",
-      "N=24\n",
-      "r=1.5*10**6\n",
-      "encoder=8\n",
-      "\n",
-      "#Calculation\n",
-      "BW=N*fm\n",
-      "BW=BW/10**3\n",
-      "r1=r/N\n",
-      "fs=r1/encoder\n",
-      "\n",
-      "#Result\n",
-      "print(\"i. Channel Bandwith is = %.2f kHz\" %BW)\n",
-      "print(\"ii. Sampling frequency is = %.2f Hz or samples per second.\" %fs)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i. Channel Bandwith is = 81.60 kHz\n",
-        "ii. Sampling frequency is = 7812.50 Hz or samples per second.\n"
-       ]
-      }
-     ],
-     "prompt_number": 112
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page No 114 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find Signal Bandwidth,Noise Ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "v=7\n",
-      "r=50*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "#fs=2*fm\n",
-      "fm=r/(2*v)\n",
-      "snr=1.8+(6*v);\n",
-      "fm=fm*10**-6\n",
-      "\n",
-      "#Result\n",
-      "print('i. Maximum message Bandwidth is = %.2f MHz ' %fm)\n",
-      "print('ii.  signal to quantization niose ration = %.2f dB ' %snr)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i. Maximum message Bandwidth is = 3.57 MHz \n",
-        "ii.  signal to quantization niose ration = 43.80 dB \n"
-       ]
-      }
-     ],
-     "prompt_number": 113
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page No 114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#find i)bits per sample,ii)transmission rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=3*10**3\n",
-      "q=16\n",
-      "\n",
-      "#Calculation\n",
-      "v=(math.log10(q))/math.log10(2)\n",
-      "\n",
-      "#Result\n",
-      "print('i) Bits in code word= %.f bits' %v)\n",
-      "fs=2*fm\n",
-      "r=v*fs\n",
-      "print('ii) it trasmission rate= %.f x 10^3 bits per second' %(r/(10**3)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Bits in code word= 4 bits\n",
-        "ii) it trasmission rate= 24 x 10^3 bits per second\n"
-       ]
-      }
-     ],
-     "prompt_number": 114
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8, Page No 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=3.5*10**3\n",
-      "r=50*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm\n",
-      "rms=0.2\n",
-      "xmax=2\n",
-      "v=r/fs    #signaling rate r=v*fs\n",
-      "v=math.ceil(v)\n",
-      "P=(rms**2)/1\n",
-      "SNR=((3*P*2**(2*v))/(xmax**2))\n",
-      "SN=10*math.log10(SNR)\n",
-      "SN=math.ceil(SN)\n",
-      "\n",
-      "#Result\n",
-      "print(\"signal to niose ratio =%.f dB\" %SN)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "signal to niose ratio =33 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 115
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10, Page No 117 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find i)noise ratio ii)bits\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "Am=3.0\n",
-      "v=10.0\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=1.8+6*v     #noise ratio \n",
-      "SN=40\n",
-      "v=(SN-1.8)/6\n",
-      "\n",
-      "#Result\n",
-      "print(\"i) Signal to Quantization noise ratio = %.1f dB\" %SNR)\n",
-      "print(\"ii) Bits required to get signal to niose ratio of 40dB = %.f \" %math.ceil(v))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Signal to Quantization noise ratio = 61.8 dB\n",
-        "ii) Bits required to get signal to niose ratio of 40dB = 7 \n"
-       ]
-      }
-     ],
-     "prompt_number": 116
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11, Page No 117 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Maximum frequency\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "v=7\n",
-      "SNR=1.8+6*v\n",
-      "r=56*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fs=r/v        #r=v*fs signaling rate\n",
-      "fm=fs/2       #Nquset rate\n",
-      "\n",
-      "#Result\n",
-      "fm=fm/10**3\n",
-      "print(\"Maximum frequency is = %.f kHz\" %fm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum frequency is = 4 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 117
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.14, Page No 129 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Maximum Amplitude\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3*10**3.0\n",
-      "\n",
-      "#Calculation\n",
-      "Nyquistrate=2*fm     #Nyquistrate\n",
-      "fs=5*Nyquistrate     #Samplingfrquency\n",
-      "Ts=1/fs              #Sampling Interval\n",
-      "de=0.25             #step size\n",
-      "fm1=2*10**3.0\n",
-      "Am=de/(2*math.pi*fm1*Ts)\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum Amplitude = %.4f Volts\" %Am)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum Amplitude = 0.5968 Volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 118
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.16, Page No 130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  signaling rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs1=8*10**3.0\n",
-      "de=31.25*10**-3\n",
-      "q=64.0\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log(q,2)\n",
-      "r=v*fs1            #signaling rate\n",
-      "fm=3*10**3.0\n",
-      "A=1\n",
-      "fs2=(2*math.pi*fm*A)/(de)\n",
-      "\n",
-      "#Result\n",
-      "r=r*10**-3\n",
-      "print(\"Signaling rate of PCM is = %.f kHz\" %r)\n",
-      "fs2=fs2*10**-3\n",
-      "print(\"The signaling rate of DM is = %.2f kHz\" %fs2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signaling rate of PCM is = 48 kHz\n",
-        "The signaling rate of DM is = 603.19 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 119
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.17, Page No 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64.0*10**3     #frequency\n",
-      "fm=2.0*10**3      #Sinosoidal input signal sample \n",
-      "fM=4.0*10**3      #bandwidth\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*fM)   #Signal to noise ratio\n",
-      "SNRO=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output signal to noise ratio = %.2f dB\" %SNRO)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal to noise ratio = 27.94 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 120
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.18, Page No 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  signal to Quatization noise ratio \n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=8.0*10**3        #Sampling frequency\n",
-      "r=64.0*10**3        #Data Rate\n",
-      "N=8        #number of bits\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=(1.8+6*N)     #signal to Quatization noise ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Signal to Quatization noise ratio of PCM system is = %.f dB\" %SNR)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signal to Quatization noise ratio of PCM system is = 50 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 121
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.20, Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find  sampling rate,quantizing level\n",
-      "\n",
-      "#Variable Declaration\n",
-      "r=36000\n",
-      "fm=3.2*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm        #Nquest rate\n",
-      "v=r/fs         #r=v*fs signaling rate\n",
-      "v=math.floor(v)\n",
-      "q=2**v\n",
-      "fs1=r/v\n",
-      "\n",
-      "#Result\n",
-      "print(\"Quantizing level q = %.f \" %q)\n",
-      "fs1=fs1/1000\n",
-      "print(\"sampling rate fs = %.1f kHz \" %fs1)\n",
-      "print(\"Number of binary digits = %.f  \" %v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizing level q = 32 \n",
-        "sampling rate fs = 7.2 kHz \n",
-        "Number of binary digits = 5  \n"
-       ]
-      }
-     ],
-     "prompt_number": 122
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.21, Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  time duration of 1bit  binary encoded signal\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=input(\"Enter the Nyquist rate of Signal fs(Hz)=\")\n",
-      "q=input(\"Enter the Quantization levels q =\")\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log(q,2)     #binary pulses transmitted per second\n",
-      "t=1.0/(v*fs)        #Nyquist interval\n",
-      "\n",
-      "#Result\n",
-      "print('Time duration of binary signal  t = %.4f sec ' %t)\n",
-      "#output\n",
-      "#Enter the Nyquist rate fs(Hz)=3\n",
-      "#Enter the Quantization levels q =5\n",
-      "#time duration of 1bit binary signal   \n",
-      "#t=0.1435589  sec"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Nyquist rate of Signal fs(Hz)=23\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Quantization levels q =34\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time duration of binary signal  t = 0.0085 sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 123
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.23, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNR=40\n",
-      "SNRO=10**(SNR/10)\n",
-      "\n",
-      "#SNR=3(q**2)/2, Signal to Quantization noise ratio\n",
-      "\n",
-      "#Calculation\n",
-      "q=math.sqrt((2*SNRO)/3)\n",
-      "q=math.ceil(q)\n",
-      "v=math.log(q,2)         #q=2**v  Quantization levels\n",
-      "v=math.ceil(v)\n",
-      "snr=1.76+6.02*v          #output Signal to Quantization noise ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output Signal to Quantization noise ratio = %.1f dB\" %snr)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output Signal to Quantization noise ratio = 43.9 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 124
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.24.i, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Quantizing  levels,minimum number of bits per sample\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNRO=30.0\n",
-      "fmin=300.0\n",
-      "fmax=3300.0\n",
-      "fs=80000.0\n",
-      "\n",
-      "#Calculation\n",
-      "#SNRO=1.76+20log10(q)\n",
-      "q=10**((SNRO-1.76)/20)\n",
-      "q=math.ceil(q)\n",
-      "v=math.log(q,2)\n",
-      "\n",
-      "#Result\n",
-      "print('Quantizing  levels required is =%.f' %q)\n",
-      "print('minimum number of bits per sample are=%.1f that is approximately 5' %v)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizing  levels required is =26\n",
-        "minimum number of bits per sample are=4.7 that is approximately 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 125
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.24.ii, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  minimum required bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNRO=30.0       #Signal to quantisizing noise ratio\n",
-      "fmin=300.0      #min frequency band\n",
-      "fmax=3300.0     #max frequency band\n",
-      "fs=8000.0       #Sampling rate\n",
-      "v=5.0           #Minimum number of bits per sample\n",
-      "\n",
-      "#Calculation\n",
-      "fPCM=(v*fs)/2\n",
-      "fPCM=fPCM/1000\n",
-      "\n",
-      "#Result\n",
-      "print('minimum required bandwidth =%.f kHz' %fPCM)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "minimum required bandwidth =20 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 126
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.24.iii, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Quantizing  levels,minimum number of bits per sample and bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNRO=30.0       #Signal to quantisizing noise ratio\n",
-      "fmin=300.0      #min frequency band\n",
-      "fmax=3300.0     #max frequency band\n",
-      "fs=8000.0       #Sampling rate\n",
-      "\n",
-      "#Calculation\n",
-      "q=10**((SNRO+10.1)/20)\n",
-      "q=math.ceil(q)\n",
-      "v=math.log(q,2)\n",
-      "v=math.ceil(v)\n",
-      "\n",
-      "#Result\n",
-      "print('Quantizing  levels need is =%.f' %q)\n",
-      "print('minimum number of bits per sample is =%.f' %v)\n",
-      "fPCM=(v*fs)/2\n",
-      "fPCM=fPCM/1000\n",
-      "print('minimum required bandwidth =%.f Khz' %fPCM)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizing  levels need is =102\n",
-        "minimum number of bits per sample is =7\n",
-        "minimum required bandwidth =28 Khz\n"
-       ]
-      }
-     ],
-     "prompt_number": 127
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.27, Page No 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#determine the  Maximum Amplitude,\n",
-      "\n",
-      "#Variable Declaration\n",
-      "de=250*10**-3\n",
-      "wm=2*math.pi*1000 \n",
-      "fs=3*10**3\n",
-      "Ts=1/fs\n",
-      "\n",
-      "#Calculation\n",
-      "Amax=(de*3*fs*2)/(wm) #Amplitude\n",
-      "SNR=(3.0*((3*6*(10**3))**3))/((8*(math.pi**2)*(10**3)**3))\n",
-      "\n",
-      "#Result\n",
-      "print('Maximum Amplitude= %.1f mV' %(Amax*(10**2)))\n",
-      "print('Signal to noise ratio = %.2f dB' %(10*math.log10(SNR)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum Amplitude= 71.6 mV\n",
-        "Signal to noise ratio = 23.46 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 128
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.29, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find number of bits per sample\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNR=20\n",
-      "averagepower=30*10**-3\n",
-      "SNRO=10**(SNR/10)\n",
-      "A=3.8\n",
-      "\n",
-      "#SNRO=average signal power/Quatizing power\n",
-      "#de=(2*A)/L\n",
-      "\n",
-      "#Calculation\n",
-      "L=math.sqrt((SNRO*A**2)/(3*averagepower))\n",
-      "n=math.log(L,2)\n",
-      "n=math.ceil(n)\n",
-      "\n",
-      "#Result\n",
-      "print('Bits required per sample =%.f' %n)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bits required per sample =7\n"
-       ]
-      }
-     ],
-     "prompt_number": 129
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.30.i, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Normalized power for quantization noise\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3.0*10.0**3\n",
-      "v=8.0\n",
-      "VH=5.0\n",
-      "VL=-5.0\n",
-      "q=2**v\n",
-      "\n",
-      "#Calculation\n",
-      "de=(VH-VL)/q\n",
-      "Nq=de**2/12.0    #quantization noise\n",
-      "\n",
-      "#Result\n",
-      "print('Normalized power for quantization noise =%.2f x 10^-6 W' %(Nq*(10**6)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Normalized power for quantization noise =127.16 x 10^-6 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 130
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.30.ii, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find bit transmission rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3*10.0**3\n",
-      "v=8.0\n",
-      "VH=5.0\n",
-      "VL=-5\n",
-      "q=2**v\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm       #Nyquist rate\n",
-      "r=8*fs/1000\n",
-      "\n",
-      "#Result\n",
-      "print('bit transmission rate %.f K bits/s' %r)\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "bit transmission rate 48 K bits/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 131
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.30.iii, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Signal to quantization noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "Nq=127.15*10**-6\n",
-      "Meansignal=2\n",
-      "\n",
-      "#Calculation\n",
-      "P=Meansignal/1\n",
-      "SNR=P/Nq\n",
-      "SNRq=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print('Signal to quantization noise ratio %.2f dB ' %SNRq)\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signal to quantization noise ratio 41.97 dB \n"
-       ]
-      }
-     ],
-     "prompt_number": 132
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.31, Page No 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  i)SNR  ii)output bit rate iii)no.of bits recorded\n",
-      "\n",
-      "#Variable Declaration\n",
-      "N=16.0\n",
-      "v=16.0\n",
-      "fs=44.1*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=1.76+6*N\n",
-      "bitrate=2*v*fs\n",
-      "outputbitrate=2*bitrate*10**-6        #including addtional 100% over head\n",
-      "CD=outputbitrate*3600*10**-3\n",
-      "\n",
-      "#Result\n",
-      "print('i)Out put signal noise ratio =%.2f dB' %SNR)\n",
-      "print('ii)output bit rate =%.3f Mbits/sec' %outputbitrate)\n",
-      "print('iii)no.of bits recorded in CD =%.2f gigabits' %CD)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Out put signal noise ratio =97.76 dB\n",
-        "ii)output bit rate =2.822 Mbits/sec\n",
-        "iii)no.of bits recorded in CD =10.16 gigabits\n"
-       ]
-      }
-     ],
-     "prompt_number": 133
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.32, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  output SNR\n",
-      "#Variable Declaration\n",
-      "fm=1*10.0**3\n",
-      "fs=32*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "FM=4*10**3     #Bandwidth\n",
-      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*FM)    #SNR\n",
-      "SNRO=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print('Output signal to noise ratio %.2f dB' %SNRO)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal to noise ratio 24.93 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 134
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.33.i, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step size\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64000.0     #samples/sec\n",
-      "Amax=1.0       #Maximum signal amplitude\n",
-      "fm=3500.0      #Sample Rate\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*Amax)/fs*1000\n",
-      "\n",
-      "#Result\n",
-      "print('Step Size %.2f mV ' %de)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Step Size 343.61 mV \n"
-       ]
-      }
-     ],
-     "prompt_number": 135
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.33ii, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Quantizatio noise power \n",
-      "#Variable Declaration\n",
-      "fs=64000.0     #sample Rate\n",
-      "Amax=1.0       #Maximum signal Amplitude\n",
-      "fm=3500.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=343.6117*10**-3      #step size\n",
-      "Nq=de**2/3.0             #Quantizatio noise power\n",
-      "Nqd=Nq*(fm/fs)*1000\n",
-      "\n",
-      "#Result\n",
-      "print('Quantizatio noise power %.3f mW' %Nqd)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizatio noise power 2.152 mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 136
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.33iii, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find SNR\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64000.0\n",
-      "Amax=1.0\n",
-      "fm=3500.0\n",
-      "Nqd=2.1522995*10**-3\n",
-      "\n",
-      "#Calculation\n",
-      "So=Amax**2/2\n",
-      "SNR=So/Nqd\n",
-      "SNRO=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print('Output signal noise ratio dB  %.3f dB' %SNRO)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal noise ratio dB  23.661 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 137
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.34, Page No 156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Assuming signal is sampled at the rate 20% above Nyquist rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=4.5*10**6\n",
-      "q=1024.0\n",
-      "\n",
-      "#Calculation\n",
-      "fs=1.2*2*fm           #20% above Nyquist rate\n",
-      "v=math.log(q,2)\n",
-      "r=v*fs/10**6\n",
-      "\n",
-      "#Result\n",
-      "print('no.of bits/sec = %.f M bit/sec' %r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "no.of bits/sec = 108 M bit/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 138
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.35, Page No 156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume bandwidth of the singal is 4kHz\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=32000.0\n",
-      "A=2.0\n",
-      "fm=4000.0\n",
-      "BW=4000.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*A)/fs\n",
-      "Nq=de**2/3\n",
-      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*BW)\n",
-      "\n",
-      "#Result\n",
-      "print('i)step size = %.3f M Volt' %de)\n",
-      "print('ii)noise power = %.3f W' %Nq)\n",
-      "print('iii)SNR= = %.3f ' %SNR)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)step size = 1.571 M Volt\n",
-        "ii)noise power = 0.822 W\n",
-        "iii)SNR= = 19.454 \n"
-       ]
-      }
-     ],
-     "prompt_number": 139
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.36, Page No 157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assuming signal is sampled at the rate 20% above Nyquist rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=15*10**3\n",
-      "fs=1.2*2*fm\n",
-      "q=65536\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log(q,2)\n",
-      "r=v*fs/1000\n",
-      "BW=r/2\n",
-      "\n",
-      "#Result\n",
-      "print('i)signaling rate,= %.f K bits/sec ' %r)\n",
-      "print('ii)bandwidth BW min = %.f kHz ' %BW)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)signaling rate,= 576 K bits/sec \n",
-        "ii)bandwidth BW min = 288 kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 140
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.37, Page No 157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step size,noise power\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64*10**3\n",
-      "fm=3500.0\n",
-      "A=1.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*A)/fs      #step size\n",
-      "Nq=(de**2/3)*(fm/fs)        #Granular noise power\n",
-      "\n",
-      "#Result\n",
-      "print('i)step size = %.3f Volts' %de)\n",
-      "print('ii)Nq= %.5f W' %Nq)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)step size = 0.344 Volts\n",
-        "ii)Nq= 0.00215 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 141
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.40, Page No 159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step Nr,Transmission Bandwidth, SNR\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=4           #kHz\n",
-      "N=12\n",
-      "\n",
-      "#Calculation\n",
-      "Nr=2*fs                     #Nyquest rate\n",
-      "Bt=1.0/2*N*fs               #Transmission Bandwidth\n",
-      "SNR=1.8+(6*N)               \n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('i)Nyquest rate = %.1f kHz' %Nr)\n",
-      "print('ii)Transmission Bandwidth = %.1f kHz' %Bt)\n",
-      "print('iii)SNR = %.1f dB' %SNR)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Nyquest rate = 8.0 kHz\n",
-        "ii)Transmission Bandwidth = 24.0 kHz\n",
-        "iii)SNR = 73.8 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 142
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.41, Page No 159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step size,noise power\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3.5*10**3\n",
-      "fs=64*10**3\n",
-      "A=2.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*A)/fs      #step size\n",
-      "Nq=(de**2/3)*(fm/fs)        #Granular noise power\n",
-      "\n",
-      "#Result\n",
-      "print('i)step size = %.3f Volts' %de)\n",
-      "print('ii)Nq= %.1f  X 10^-3 W' %(Nq*10**3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)step size = 0.687 Volts\n",
-        "ii)Nq= 8.6  X 10^-3 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 143
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.42, Page No 159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Bit rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=4.5 * 10**6                    #khz\n",
-      "Nr = 2 * W               #Nyquist Rate\n",
-      "Q = 1024\n",
-      "fs=10.8\n",
-      "\n",
-      "#Calculation\n",
-      "N = math.log(Q,2)            #number of binay pulse per word\n",
-      "Br= N*fs                     #Bit rate\n",
-      "Bw= 1.0/2*Br                 #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of binay pulse per word = %.1f k bits/sec' %N)\n",
-      "print('ii) Bit rate = %.1f k bits/sec' %Br)\n",
-      "print('iii)Bandwidth= %.1f  X 10^-3 kHz' %Bw)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of binay pulse per word = 10.0 k bits/sec\n",
-        "ii) Bit rate = 108.0 k bits/sec\n",
-        "iii)Bandwidth= 54.0  X 10^-3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 144
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.43, Page No 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Bit rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=15                 #khz\n",
-      "Nr = 2*W               #Nyquist Rate\n",
-      "Q = 65.53\n",
-      "fs=44.1                # Sampling rate kHz\n",
-      "N=16\n",
-      "\n",
-      "#Calculation\n",
-      "Br= N*fs                    #Bit rate\n",
-      "Bw= 1.0/2*Br                 #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of binay pulse per word = %.3f bits' %N)\n",
-      "print('ii) Bit rate = %.1f k bits/sec' %Br)\n",
-      "print('iii)Bandwidth= %.1f kHz' %Bw)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of binay pulse per word = 16.000 bits\n",
-        "ii) Bit rate = 705.6 k bits/sec\n",
-        "iii)Bandwidth= 352.8 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 145
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.44, Page No 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Number of binay pulse per word\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=3.3                 #khz\n",
-      "fm = 1*10**3                #kHz\n",
-      "de = 250              #mV\n",
-      "fs=19.8                # Sampling rate kHz\n",
-      "\n",
-      "#Calculation\n",
-      "A=(de*fs)/(2*math.pi*fm)\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of binay pulse per word = %.3f' %A)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of binay pulse per word = 0.788\n"
-       ]
-      }
-     ],
-     "prompt_number": 146
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.45, Page No 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Signaling rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=3.3                 #khz\n",
-      "fm = 3300               #Hz\n",
-      "Snq = 40              #dB\n",
-      "fs= 8000                # Samples/sec\n",
-      "\n",
-      "#Calculation \n",
-      "N=math.ceil((Snq-4.8)/6)                # Number of bits per word\n",
-      "Q=2**N              #\n",
-      "r=N*fs              #Signaling rate k bits/sec\n",
-      "Bw=1.0/2*r\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of bits per word = %.1f' %N)\n",
-      "print('ii) Signaling rate rate = %.1f k bits/sec' %(r/10**3))\n",
-      "print('iii)Bandwidth= %.1f kHz' %(Bw/10**3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of bits per word = 6.0\n",
-        "ii) Signaling rate rate = 48.0 k bits/sec\n",
-        "iii)Bandwidth= 24.0 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 147
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.46, Page No 161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find the maximun frequency with 1 volt amplitude\n",
-      "\n",
-      "#Variable Declaration\n",
-      "de=256*10**-3                 #V\n",
-      "Ts = 10*10**-6               #kHz\n",
-      "fm = 10              #kHz\n",
-      "fs= 100                # kHz\n",
-      "A=1                   #V\n",
-      "\n",
-      "#Calculation \n",
-      "wm=(de/A*Ts)*10**7\n",
-      "\n",
-      "#Result\n",
-      "print('i) maximun frequency = %.1f X 10^3 rad/sec' %wm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) maximun frequency = 25.6 X 10^3 rad/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 148
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.47, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find the Signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "Q=256     \n",
-      "m=255\n",
-      "\n",
-      "#Calculation \n",
-      "SNR=(3*Q**2)/(math.log((1+m),math.e))**2    #dB\n",
-      "Gc=m/(math.log(1+m))                        #Companding gain\n",
-      "Gc=20*math.log(Gc,10)\n",
-      "\n",
-      "#Result\n",
-      "print('i) The Signal to noise ratio= %.1f dB' %SNR)\n",
-      "print('ii) The Companding gain = %.1f dB' %Gc)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The Signal to noise ratio= 6394.0 dB\n",
-        "ii) The Companding gain = 33.3 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 149
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.48, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Signaling rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=15*10**3          #kHz\n",
-      "Q=200\n",
-      "fs=2*fm\n",
-      "\n",
-      "#Calculation \n",
-      "N=math.ceil(math.log(Q,2))\n",
-      "tr=N*fs           #Transmission rate kbps\n",
-      "Bw=1.0/2*tr\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of bits per Samples = %.1f' %N)\n",
-      "print('ii) Signaling rate = %.1f k bits/sec' %(tr/10**3))\n",
-      "print('iii)Bandwidth= %.1f kHz' %(Bw/10**3))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of bits per Samples = 8.0\n",
-        "ii) Signaling rate = 240.0 k bits/sec\n",
-        "iii)Bandwidth= 120.0 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 150
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.49, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of quantization level\n",
-      "\n",
-      "#Variable Declaration\n",
-      "tr=36.0                  #kbps\n",
-      "fm=3.4                 #kHz\n",
-      "fs=8\n",
-      "\n",
-      "#Calculation \n",
-      "N=math.ceil(tr/fs)\n",
-      "Q=2**N\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('i) number of quantization level = %.1f' %Q)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) number of quantization level = 32.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 151
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.50, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  delta modulation\n",
-      "\n",
-      "#Variable Declaration\n",
-      "A=1                  #V\n",
-      "fm=3.4                 #kHz\n",
-      "fs=20\n",
-      "\n",
-      "#Calculation \n",
-      "de=1*2*math.pi*(fm/fs)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('i) delta modulation = %.2f V' %de)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) delta modulation = 1.07 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 152
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.51, Page No 163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=3.5      #kHz\n",
-      "r=50.0      #kbps\n",
-      "Vpp=4      #V\n",
-      "fs=8\n",
-      "\n",
-      "#Calculation \n",
-      "N=r/fs\n",
-      "SNR=4.8+(6.8*N)\n",
-      "\n",
-      "#Result\n",
-      "print('i) Signal to noise ratio = %.2f V' %SNR)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Signal to noise ratio = 47.30 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 153
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.52, Page No 163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=1      #MHz\n",
-      "fs1=3       #MHz\n",
-      "fs2=2.4       #MHz\n",
-      "Q=256\n",
-      "m=255\n",
-      "\n",
-      "#Calculation \n",
-      "SNR=((3*Q**2)/(math.log((1+m),math.e))**2)\n",
-      "SNR1=63095.73              #SNR=38+10\n",
-      "Q1=math.sqrt((SNR1*((math.log(256,math.e))**2))/3)\n",
-      "N1=math.ceil(math.log(Q1,2))\n",
-      "r=N1*fs2\n",
-      "Bw=1.0/2*r\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of bits/word = %.2f ' %N1)\n",
-      "print('ii) Bit rate = %.1f Mb/s' %r)\n",
-      "print('iii) Bandwidth = %.1f Mb/s' %Bw)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of bits/word = 10.00 \n",
-        "ii) Bit rate = 24.0 Mb/s\n",
-        "iii) Bandwidth = 12.0 Mb/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 154
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/Chapter4.ipynb b/Digital_Communications/Chapter4.ipynb
deleted file mode 100755
index e56598ee..00000000
--- a/Digital_Communications/Chapter4.ipynb
+++ /dev/null
@@ -1,273 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater 4: DIGITAL MULTIPLEXERS<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4, page no 190 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Caption: Sampling Rate\n",
-      "\n",
-      "#initialisation of variables\n",
-      "f1=4.0*10**3\n",
-      "f2=4.5*10**3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "fsmin=(2*f2)/1000\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\" Sampling rate =%.2f kHz\" %fsmin)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Sampling rate =9.00 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5, Page No 190 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Caption: Nyquest rate,Signaling rate,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "f1=3000.0\n",
-      "f4=1000.0\n",
-      "f2=1000.0\n",
-      "f3=1000.0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "#Nyquest rate\n",
-      "nq1=2*f1\n",
-      "nq2=2*f2\n",
-      "nq3=2*f3\n",
-      "nq4=2*f4\n",
-      "\n",
-      "#RESULTS\n",
-      "print('i. Nyquest rate of x1 =%.2fkHz' %nq1)\n",
-      "print('ii. Nyquest rate of x2 =%.2fkHz' %nq2)\n",
-      "print('iii. Nyquest rate of x3 =%.2fkHz' %nq3)\n",
-      "print('iv. Nyquest rate of x4 =%.2fkHz' %nq4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i. Nyquest rate of x1 =6000.00kHz\n",
-        "ii. Nyquest rate of x2 =2000.00kHz\n",
-        "iii. Nyquest rate of x3 =2000.00kHz\n",
-        "iv. Nyquest rate of x4 =2000.00kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6, Page No 190 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find The spacing between two successive pules\n",
-      "\n",
-      "#initialisation of variables\n",
-      "samplingrate=8000.0\n",
-      "totalsignals=24+1\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "t=1/samplingrate\n",
-      "T=t/totalsignals\n",
-      "T=T*10**6            #time is now u sec\n",
-      "space=T-1\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The spacing between two successive pules %.f u sec' %space)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing between two successive pules 4 u sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7, Page No 191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find signaling rate ,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N=6.0\n",
-      "fm=5000.0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r=2*fm   #sampling rate\n",
-      "sr=N*r   #signalingrate\n",
-      "BW=N*fm  #Bandwith\n",
-      "sr=sr/1000\n",
-      "BW=BW/1000\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Signaling rate %.f K bits/sec' %sr)\n",
-      "print('Bandwith to avoid the cross talk in TDM is %.f kHz' %BW)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signaling rate 60 K bits/sec\n",
-        "Bandwith to avoid the cross talk in TDM is 30 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9, Page No 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find signaling rate ,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N=4.0\n",
-      "x=24              #bits/frame\n",
-      "D=1152\n",
-      "s=1.0/3\n",
-      "r=1.544*10**6\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r0=N*r*((D+x)/(D-s*N))\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Output bit rate %.2f MB/sec' %(r0/(10**6)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output bit rate 6.31 MB/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.10, Page No 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find signaling rate ,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N=24.0\n",
-      "x=72.0              #bits\n",
-      "D=192             #message and stuff bits\n",
-      "s=1.0/3\n",
-      "r=64              #kb/sec\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r0=N*r*((D+x)/(D-s*N))\n",
-      "r1=((N*r)/r0)*100\n",
-      "\n",
-      "#RESULTS\n",
-      "print('i) Output bit rate %.2f MB/sec' %(r0/(10**3)))\n",
-      "print('iI) Throughput efficiency is =  %.2f percent' %r1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Output bit rate 2.20 MB/sec\n",
-        "iI) Throughput efficiency is =  69.70 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/Chapter5.ipynb b/Digital_Communications/Chapter5.ipynb
deleted file mode 100755
index bf535beb..00000000
--- a/Digital_Communications/Chapter5.ipynb
+++ /dev/null
@@ -1,105 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater 5: DIGTIAL BASEBAND TRANSMISSION<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12, Page No 249 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find The roll of factor alpha\n",
-      "\n",
-      "#initialisation of variables\n",
-      "datarate=0.1*10**6\n",
-      "fB=75000.0 #bandwidth\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Tb=1/datarate\n",
-      "alpha=2*fB*Tb-1\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The roll out factor alpha =  = %.1f ' %alpha)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The roll out factor alpha =  = 0.5 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.13, Page No 249"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find The Transmission bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "q=128.0\n",
-      "alpha=0.2\n",
-      "n=math.log(q,2)\n",
-      "fm=2000.0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Nq=2*fm\n",
-      "fs=1.25*Nq\n",
-      "N=8.0\n",
-      "total=N*fs\n",
-      "bitrate=7*total\n",
-      "fB=((1+alpha)*bitrate)/2\n",
-      "fB=fB/1000\n",
-      "#RESULTS\n",
-      "print('The minimum transmission Bandwidth required = %.f kHz' %fB)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum transmission Bandwidth required = 168 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/Chapter6.ipynb b/Digital_Communications/Chapter6.ipynb
deleted file mode 100755
index fd5f6716..00000000
--- a/Digital_Communications/Chapter6.ipynb
+++ /dev/null
@@ -1,158 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater 6: DIGITAL MODULATION TECHNIQUES<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page No 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "#Symbol rate of transmission and bandwidth \n",
-      "\n",
-      "#initialisation of variables\n",
-      "fb=1.0                #MHz\n",
-      "Tb=1/fb               #bit period ms\n",
-      "N=4\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Ts1=Tb\n",
-      "Bw1=2*fb\n",
-      "Ts2=Tb*2\n",
-      "Tr2=(1.0/2)*1000          #Symbols/sec\n",
-      "Bw2=1/Tb\n",
-      "Ts3=N*Tb\n",
-      "Tr3=(1.0/4)*1000\n",
-      "Bw3=(2.0*fb)/N\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"In BPSK\")\n",
-      "print(\"The symbol rate of transmission is = %.f X 10^6 symbol/sec \" %Ts1)\n",
-      "print(\"The Bandwidth is = %.f MHz \" %Bw1)\n",
-      "print(\"In QPSK\")\n",
-      "print(\"The symbol rate of transmission is = %.f X 10^3 symbol/sec \" %Tr2)\n",
-      "print(\"The Bandwidth is = %.f MHz \" %Bw2)\n",
-      "print(\"In 16-ary PSK\")\n",
-      "print(\"The symbol rate of transmission is = %.f X 10^3 symbol/sec \" %Tr3)\n",
-      "print(\"The Bandwidth is = %.f kHz \" %(Bw3*1000))\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In BPSK\n",
-        "The symbol rate of transmission is = 1 X 10^6 symbol/sec \n",
-        "The Bandwidth is = 2 MHz \n",
-        "In QPSK\n",
-        "The symbol rate of transmission is = 500 X 10^3 symbol/sec \n",
-        "The Bandwidth is = 1 MHz \n",
-        "In 16-ary PSK\n",
-        "The symbol rate of transmission is = 250 X 10^3 symbol/sec \n",
-        "The Bandwidth is = 500 kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page No 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "fb=90              #MHz\n",
-      "Bw=20              #Mb/s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N=(2*fb)/Bw\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"Number of bits required = %.f \" %N)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of bits required = 9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page No 357"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Tb=0.1          #msec\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "fb=1/Tb\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"Transmission bandwidth = %.f kHz\" %fb)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transmission bandwidth = 10 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/Chapter7.ipynb b/Digital_Communications/Chapter7.ipynb
deleted file mode 100755
index 7c33676a..00000000
--- a/Digital_Communications/Chapter7.ipynb
+++ /dev/null
@@ -1,431 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater 7: PERFORMANCE ANALYSIS OF DIGITAL COMMUNICATION SYSTEMS<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.15, Page No 414"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "T=1.0*10**-6             #ms\n",
-      "A=10.0*10**-3\n",
-      "N=1*10**-11\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Eb=((A**2)*T)/2.0\n",
-      "Epb=math.sqrt(Eb/N)\n",
-      "Pb=(1.0/2)*math.erfc(Epb)*(1.0/4)\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"Bit error probability = %.2f X 10^-4 \" %(Pb*(10**4)))\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bit error probability = 1.96 X 10^-4 \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.16, Page No 420"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Br=25*10**6             #bits/sec\n",
-      "N=2*10**-20             #Power spectral density of noise\n",
-      "A=1.0*10**-6\n",
-      "Ps=1*10**-12            # Normalized power\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Tb=1.0/Br\n",
-      "Ps=(A**2)/2.0\n",
-      "Pe=1.0/2*(math.erfc((0.6*Ps*Tb)/N))\n",
-      "#RESULTS\n",
-      "print(\"Average probability of bit error is Pe= %.2f X 10^-4 \" %(Pe*10))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average probability of bit error is Pe= 1.98 X 10^-4 \n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.17, Page No 423"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find probability of error\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=2.0\n",
-      "N0=2.0*10**-4  #power sepctral density\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Tmin=(6.76*N0)/4\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The min time for which the signal should be extended is Tmin= %.3f X 10^-3 msec \" %(Tmin*10**3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The min time for which the signal should be extended is Tmin= 0.338 X 10^-3 msec \n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.20, Page No 428"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#The average carrier\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Pe=10**-4\n",
-      "N0=2*10**10\n",
-      "T=1.0/10**6\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "E=6.76*N0\n",
-      "P=E/T\n",
-      "Eb=8.5171*N0\n",
-      "P2=Eb/T\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The average carrier = %.3f X 10^-3 mW \" %(P2/10**17))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average carrier = 1.703 X 10^-3 mW \n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.21, Page No 430"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#The average carrier\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=0.2*10**-3          #V\n",
-      "T=2*10**-6\n",
-      "N0=2*10**-15\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Pe=(math.sqrt((A**2*T)/4*N0))\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The average carrier = %.3f X 10^-4 mW \" %(Pe*10**15))\n",
-      "print(\"Approximation error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average carrier = 6.325 X 10^-4 mW \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.29, Page No 447"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=1.0\n",
-      "N0=2.0*10**-5  #power sepctral density\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "T=((3.71**2)*2*N0)/(A**2)\n",
-      "Br=1/T\n",
-      "#Results\n",
-      "print(\"Bit rate Br= %.2f k bits/sec\" %(Br/(10**3)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bit rate Br= 1.82 k bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.31, Page No 448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Pe=10.0**-4 #probability of error of PSK\n",
-      "N0=2.0*10**-10\n",
-      "Br=10**6                  #Bit Rate\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "T=1.0/Br\n",
-      "E=N0*6.76\n",
-      "P=E/T\n",
-      "Eb=8.5171*N0\n",
-      "Pb=Eb/T\n",
-      "#Results\n",
-      "print(\"i) PSK system The required carrier power = %.2f mW\" %(P*10**3))\n",
-      "print(\"ii) DPSK system The required carrier power = %.2f mW\" %(Pb*10**3))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) PSK system The required carrier power = 1.35 mW\n",
-        "ii) DPSK system The required carrier power = 1.70 mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.32, Page No 450"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=0.2*10**-3                #V\n",
-      "T=2*10**-6                  #Sec\n",
-      "N0=2*10**-15\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "x=math.sqrt((A**2*T)/(4*N0))\n",
-      "Pe=(1.0/2)*math.erfc(x/math.sqrt(2.0))\n",
-      "\n",
-      "#Results\n",
-      "print(\"Probability of error = %.2f X 10^-4\" %(Pe*10**4))\n",
-      "#print(\"Channel Bandwidth  = %.5f MHz\" %BW)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Probability of error = 7.83 X 10^-4\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.36, Page No 453"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Ap=1                #mV\n",
-      "En=192.3            #mV\n",
-      "Arms=707            #mV\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Z=Arms/En\n",
-      "Pe=1.0/2*(math.erfc(math.sqrt(2)*(Z/math.sqrt(2))))\n",
-      "\n",
-      "#Results\n",
-      "print(\"Probability of error is  %.7f \" %Pe)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Probability of error is  0.0000001 \n"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.37, Page No 454"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Br=2.08*10**-6        #Bit Rate\n",
-      "N0=2*10**-8\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Eb=(math.sqrt(3.3))/(5*10**7)   #Eb=bit energy\n",
-      "Ps=Eb*Br\n",
-      "BW=2*Br                  #Bandwidth\n",
-      "Es=286.39*N0\n",
-      "Ps2=(Es*Br)/2\n",
-      "BW2=(2*Br)/4\n",
-      "\n",
-      "#Results\n",
-      "print(\"i) BPSK THe transmission bandwidth  = %.2f MHz\" %(BW*10**6)) \n",
-      "print(\"ii) For 16 ary PSK THe transmission bandwidth  = %.2f MHz\" %(BW2*10**6)) \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) BPSK THe transmission bandwidth  = 4.16 MHz\n",
-        "ii) For 16 ary PSK THe transmission bandwidth  = 1.04 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/Chapter8.ipynb b/Digital_Communications/Chapter8.ipynb
deleted file mode 100755
index 52d8741e..00000000
--- a/Digital_Communications/Chapter8.ipynb
+++ /dev/null
@@ -1,1439 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater 8: INFORMATION THEORY<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page No 464"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "#Find Information Content of Each Symbol\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/8.0\n",
-      "\n",
-      "#Calculation\n",
-      "#information content of each symbol\n",
-      "Ix1=math.log(1/px1,2)\n",
-      "Ix2=math.log(1/px2,2)\n",
-      "Ix3=math.log(1/px3,2)\n",
-      "Ix4=math.log(1/px4,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"Information Content tI(x1)= %.2f bit\" %Ix1)\n",
-      "print(\" tI(x2)= %.f bits\" %Ix2)\n",
-      "print(\" tI(x3)= %.f bits\" %Ix3)\n",
-      "print(\" tI(x4)= %.f bits\" %Ix4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Information Content tI(x1)= 1.00 bit\n",
-        " tI(x2)= 2 bits\n",
-        " tI(x3)= 3 bits\n",
-        " tI(x4)= 3 bits\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2, Page No 464"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find amount of Information\n",
-      "#Variable Declaration\n",
-      "#Calculation\n",
-      "pxi=1/4.0\n",
-      "Ixi=(math.log10(1/pxi))/math.log10(2)\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The amount of Information I(Xi)= %.f \" %Ixi)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of Information I(Xi)= 2 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3, Page No 464"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Amount of Information\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/2.0\n",
-      "\n",
-      "#Calculation\n",
-      "Ix1=math.log(1/px1,2) #entropy\n",
-      "Ix2=math.log(1/px2,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The amount of Information I(X1)= %.f bit\" %Ix1)\n",
-      "print(\"The amount of Information I(X2)= %.f bit\" %Ix2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of Information I(X1)= 1 bit\n",
-        "The amount of Information I(X2)= 1 bit\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4, Page No 465"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Amount of Information \n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/4.0\n",
-      "px2=3/4.0\n",
-      "\n",
-      "#Calculation\n",
-      "Ix1=math.log(1/px1,2)\n",
-      "Ix2=math.log(1/px2,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The amount of Information I(X1)= %.f bit\" %Ix1)\n",
-      "print(\"The amount of Information I(X2)= %.2f bit\" %Ix2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of Information I(X1)= 2 bit\n",
-        "The amount of Information I(X2)= 0.42 bit\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9, Page No 468 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Entropy,Amount of information\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.4\n",
-      "px2=0.3\n",
-      "px3=0.2\n",
-      "px4=0.1\n",
-      "\n",
-      "#Calculation\n",
-      "HX=-px1*math.log(px1,2)-px2*math.log(px2,2)-px3*math.log(px3,2)-px4*math.log(px4,2)\n",
-      "Px1x2x1x3=px1*px2*px1*px3\n",
-      "Ix1x2x1x3=-math.log(Px1x2x1x3,2)\n",
-      "Px4x3x3x2=px4*px3*px3*px2\n",
-      "Ix4x3x3x2=-math.log(Px4x3x3x2,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\" \\n Entropy  H(X) = %.2f bits/symbol \" %HX)\n",
-      "print(\"The amount of Information I(x1x2x1x3)= %.2f bits/symbol\" %Ix1x2x1x3)\n",
-      "print(\" I(x4x3x3x2) = %.2f bits/symbol \" %Ix4x3x3x2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Entropy  H(X) = 1.85 bits/symbol \n",
-        "The amount of Information I(x1x2x1x3)= 6.70 bits/symbol\n",
-        " I(x4x3x3x2) = 9.70 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13, Page No 471"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find information rate the telegraphic source\n",
-      "\n",
-      "#Variable Declaration\n",
-      "pdash=1/3.0\n",
-      "pdot=2/3.0\n",
-      "tdot=0.2\n",
-      "tdash=0.6\n",
-      "tspace=0.2\n",
-      "\n",
-      "#Calculation\n",
-      "HX=-pdash*math.log(pdash,2)-pdot*math.log(pdot,2)\n",
-      "Ts=pdot*tdot+pdash*tdash+tspace\n",
-      "r=1/Ts\n",
-      "R=r*HX\n",
-      "\n",
-      "#Result\n",
-      "print('Average rate of information R = %.2f bits/s' %R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average rate of information R = 1.72 bits/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14, Page No 471"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find information rate of the source\n",
-      "\n",
-      "f=input('Enter the frequncy f=')\n",
-      "px1=1/8.0\n",
-      "px2=1/8.0\n",
-      "px3=3/8.0\n",
-      "px4=3/8.0\n",
-      "\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2) #entropy of the source\n",
-      "R=2*f*HX #r=2*f\n",
-      "print('information rate  R=  %.1f  bits/sec ' %R)   #f=signal bandwidth\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the frequncy f=34\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "information rate  R=  123.2  bits/sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15, Page No 472"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find information rate of the source\n",
-      "#all symbols are equally likely\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/2.0\n",
-      "px3=1/2.0\n",
-      "px4=1/2.0\n",
-      "\n",
-      "#Calculation\n",
-      "f=input('Enter the frequncy of system fm(in Hz) =')\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)\n",
-      "\n",
-      "#Result\n",
-      "print('\\n Entropy H(X) =%.f bits/symbol ' %HX)\n",
-      "R=2*f*HX\n",
-      "print('information rate =%.f bits/sec' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the frequncy of system fm(in Hz) =45\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Entropy H(X) =2 bits/symbol \n",
-        "information rate =180 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.16, Page No 473"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find source entropy ,information rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#probability symbols\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/16.0\n",
-      "px5=1/16.0\n",
-      "Tb=10.0**-3\n",
-      "\n",
-      "#Calculation\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4)+px5*math.log(1/px5)\n",
-      "\n",
-      "#Result\n",
-      "print('1. source entropy H(X) = %.2f bits/symbol ' %HX) #source entropy\n",
-      "r=1.0/Tb\n",
-      "R=r*HX #information rate\n",
-      "print(' 2. Information rate R = %.2f bits/sec ' %R)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. source entropy H(X) = 1.72 bits/symbol \n",
-        " 2. Information rate R = 1721.57 bits/sec \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17, Page No 473"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume if there are 16 outcomes per second\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/16.0\n",
-      "px5=1/16.0\n",
-      "r=16.0\n",
-      "\n",
-      "#Calculation\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)+px5*math.log(1/px5,2)\n",
-      "\n",
-      "#Result\n",
-      "print('1. Entropy H(X) = %.2f bits/symbol ' %HX) #source entropy\n",
-      "\n",
-      "R=r*HX\n",
-      "print('2., Information rate R = %.f bits/sec' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. Entropy H(X) = 1.88 bits/symbol \n",
-        "2., Information rate R = 30 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18, Page No 474"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#determine entropy ,information rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/4.0\n",
-      "px2=1/5.0\n",
-      "px3=1/5.0\n",
-      "px4=1/10.0\n",
-      "px5=1/10.0\n",
-      "px6=1/20.0\n",
-      "px7=1/20.0\n",
-      "px8=1/20.0\n",
-      "f=10*10**3.0\n",
-      "fs=10*2*10**3.0\n",
-      "\n",
-      "#Calculation\n",
-      "#entropy\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)+px5*math.log(1/px5,2)+px6*math.log(1/px6,2)+px7*math.log(1/px7,2)+px8*math.log(1/px8,2) \n",
-      "\n",
-      "#Result\n",
-      "print('bits/message H(X) = %.2f ' %HX)\n",
-      "r=fs\n",
-      "R=r*HX    #information rate\n",
-      "print('bits/sec R = %.2f' %R)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "bits/message H(X) = 2.74 \n",
-        "bits/sec R = 54828.92\n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19, Page No 476 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from array import *\n",
-      "#Find Channel Matrix,joint probability\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.5\n",
-      "px2=0.5\n",
-      "py1x1=0.9\n",
-      "py2x1=0.1\n",
-      "py1x2=0.2\n",
-      "py2x2=0.8\n",
-      "PYX=[[py1x1,py2x1],[py1x2,py2x2]]\n",
-      "PX=[[px1,px2]]\n",
-      "PY = [[0,0],\n",
-      "    [0,0]]\n",
-      "PXY = [[0,0],\n",
-      "    [0,0]]\n",
-      "\n",
-      "for i in range(len(PYX)):\n",
-      "   # iterate through columns of Y\n",
-      "   for j in range(len(PX[0])):\n",
-      "       # iterate through rows of Y\n",
-      "       for k in range(len(PX)):\n",
-      "           PY[i][j] += PYX[i][k] * PX[k][j]\n",
-      "print('  PY ARRAY =  \\n')\n",
-      "for r in PY:\n",
-      "   print(r)\n",
-      "PXd=[[px1,0],[0,px2]]\n",
-      "\n",
-      "\n",
-      "for i in range(len(PXd)):\n",
-      "   # iterate through columns of Y\n",
-      "   for j in range(len(PYX[0])):\n",
-      "       # iterate through rows of Y\n",
-      "       for k in range(len(PYX)):\n",
-      "           PXY[i][j] += PXd[i][k] * PYX[k][j]\n",
-      "\n",
-      "            \n",
-      "print(' \\n PXY ARRAY =  \\n')\n",
-      "for r in PXY:\n",
-      "   print(r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  PY ARRAY =  \n",
-        "\n",
-        "[0.45, 0.45]\n",
-        "[0.1, 0.1]\n",
-        " \n",
-        " PXY ARRAY =  \n",
-        "\n",
-        "[0.45, 0.05]\n",
-        "[0.1, 0.4]\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.35, Page No 498"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Channel is aproximated by the AWGN Channel\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=4000.0\n",
-      "S=0.1*10**-3\n",
-      "n=2*10**-12\n",
-      "\n",
-      "#Calculation\n",
-      "N=n*B\n",
-      "C=B*math.log(1+(S/N),2) #Capacity of Channel\n",
-      "C=C/1000.0\n",
-      "#Result\n",
-      "print(' Capacity of Channel  C=%.3f(10^3) b/s ' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Capacity of Channel  C=54.439(10^3) b/s \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36i, Page No 499"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=4000.0\n",
-      "fs=2*fm\n",
-      "n=1.25\n",
-      "\n",
-      "#Calculation\n",
-      "r=fs*n\n",
-      "pxi=1/256.0\n",
-      "HX=-math.log(pxi,2)\n",
-      "R=r*HX\n",
-      "R=R/1000\n",
-      "print('Information Rate  R= %.f kb/s' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Information Rate  R= 80 kb/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36ii, Page No 499"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=10*10**3.0\n",
-      "SN=20.0\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=10**(SN/10.0)\n",
-      "C=B*math.log(1+(SNR),2)\n",
-      "C=C/1000\n",
-      "\n",
-      "#Result\n",
-      "print('The channel capacity = %.2f 10^3 b/s' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The channel capacity = 66.58 10^3 b/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36iii, Page No 499"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "C=8*10**4.0\n",
-      "B=10**4.0\n",
-      "\n",
-      "#Calculation\n",
-      "SN=2**(C/B)-1\n",
-      "SNR=10*math.log(SN,10) #SNR\n",
-      "\n",
-      "#Result\n",
-      "print(' The S/N ratio required for error-free transmission  =%.2f dB ' %SNR)     #required SNR is greater that 24.064\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The S/N ratio required for error-free transmission  =24.07 dB \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36iv, Page No 499 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SN=20.0\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=10**(SN/10.0)\n",
-      "C=8*10**4.0\n",
-      "B=C/(math.log(1+SNR,2)) #Bandwidth\n",
-      "B=B/1000\n",
-      "\n",
-      "#Result\n",
-      "print('Bandwidth required for AWGN channel B =%.2f kHz ' %B)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bandwidth required for AWGN channel B =12.02 kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.37, Page No 502"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find code efficiency,redundancy\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.9\n",
-      "px2=0.1\n",
-      "n1=1.0\n",
-      "n2=1.0\n",
-      "\n",
-      "#Calculation\n",
-      "L=px1*n1+px2*n2 #code leght\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)\n",
-      "n=(HX/L) #code efficiency\n",
-      "n=n*100\n",
-      "\n",
-      "print('Code efficiency = %.1f percent'  %n)\n",
-      "r=(100-n) #code reduncy\n",
-      "print('Code redundancy = %.1f percent' %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Code efficiency = 46.9 percent\n",
-        "Code redundancy = 53.1 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.38, Page No 502"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find code efficiency,redundancy\n",
-      "\n",
-      "#Variable Declaration\n",
-      "pa1=0.81\n",
-      "pa2=0.09 \n",
-      "pa3=0.09\n",
-      "pa4=0.01 \n",
-      "n1=1\n",
-      "n2=2 \n",
-      "n3=3\n",
-      "n4=3 \n",
-      "\n",
-      "#Calculation\n",
-      "L=pa1*n1+pa2*n2+pa3*n3+pa4*n4\n",
-      "HX2=pa1*math.log(1/pa1,2)+pa2*math.log(1/pa2,2)+pa3*math.log(1/pa3,2)+pa4*math.log(1/pa4,2)\n",
-      "n=HX2/L*100\n",
-      "\n",
-      "#Result\n",
-      "print(' code efficiency = %.2f  percent' %n)\n",
-      "\n",
-      "r=(100-n) #code reduncy\n",
-      "print(' code redundancy = %.1f percent' %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " code efficiency = 72.71  percent\n",
-        " code redundancy = 27.3 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.44, Page No 507"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  efficiency of the code\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/8.0\n",
-      "n1=1.0\n",
-      "n2=2.0\n",
-      "n3=3.0\n",
-      "n4=3.0\n",
-      "\n",
-      "#Calculation\n",
-      "#information content of each symbol\n",
-      "Ix1=-math.log(px1,2)\n",
-      "Ix2=-math.log(px2,2)\n",
-      "Ix3=-math.log(px3,2)\n",
-      "Ix4=-math.log(px4,2)\n",
-      "\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)\n",
-      "L=px1*n1+px2*n2+px3*n3+px4*n4\n",
-      "\n",
-      "n=HX/L*100\n",
-      "\n",
-      "#Result\n",
-      "print('Ccode efficiency = %.f  Percent' %n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ccode efficiency = 100  Percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.50, Page No 512"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy ,information rate\n",
-      "#If there are 16 outcomes per second\n",
-      "\n",
-      "#Variable Declaration\n",
-      "P1=1/2.0\n",
-      "P2=1/4.0\n",
-      "P3=1/8.0\n",
-      "P4=1/16.0\n",
-      "P5=1/32.0\n",
-      "P6=1/32.0\n",
-      "r=16     #message rate\n",
-      "\n",
-      "#Calculation\n",
-      "H=P1*math.log(1/P1,2)+P2*math.log(1/P2,2)+P3*math.log(1/P3,2)+P4*math.log(1/P4,2)+P5*math.log(1/P5,2)+P6*math.log(1/P6,2)\n",
-      "#Entropy of system\n",
-      "\n",
-      "#Result\n",
-      "print('1. Entropy of system H = %.2f bits/message ' %H)\n",
-      "R=H*r                            #R=Entropy*message rate\n",
-      "print(' 2. Information rate  R = %.f bits/sec ' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. Entropy of system H = 1.94 bits/message \n",
-        " 2. Information rate  R = 31 bits/sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.51, Page No 512"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Calculate H(X) ,H(Y)\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.3\n",
-      "px2=0.4\n",
-      "px3=0.3\n",
-      "\n",
-      "#Calculation\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)  #Entropy of X\n",
-      "\n",
-      "\n",
-      "print(' 1.Entropy of X  H(X)=%.3f bits/symbol ' %HX)\n",
-      "\n",
-      "PYX=[[0.8, 0.2, 0],[ 0, 1, 0],[ 0, 0.3, 0.7]]\n",
-      "PX=[[px1, px2, px3]]\n",
-      "PXY = [[0,0,0],\n",
-      "    [0,0,0],\n",
-      "    [0,0,0]]\n",
-      "\n",
-      "for i in range(len(PYX)):\n",
-      "   # iterate through columns of PXd\n",
-      "   for j in range(len(PX[0])):\n",
-      "       # iterate through rows of PYX\n",
-      "       for k in range(len(PX)):\n",
-      "           PXY[i][j] += PYX[i][k] * PX[k][j]\n",
-      "\n",
-      "py1=PXY[0][0]\n",
-      "py2=PXY[0][1]\n",
-      "py3=PXY[0][2]\n",
-      "HY=py1*math.log(1/py1,2)+py2*math.log(1/py2,2)+py3*math.log(1/py3,2)  #Entropy of Y\n",
-      "print(' 2. Entropy of Y H(Y)= %.2f bits/symbol ' %HY)\n",
-      "print('Approximation error')\t\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1.Entropy of X  H(X)=1.571 bits/symbol \n",
-        " 2. Entropy of Y H(Y)= 1.51 bits/symbol \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.52, Page No 513"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source ,entropy of second order extension\n",
-      "\n",
-      "#Variable Declaration\n",
-      "P1=0.7\n",
-      "P2=0.15\n",
-      "P3=0.15\n",
-      "\n",
-      "#Calculation\n",
-      "HX=P1*math.log(1/P1,2)+P2*math.log(1/P2,2)+P3*math.log(1/P3,2)     #Entropy of source\n",
-      "print(' 1. Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n",
-      "#H(X^n)=n*H(X)\n",
-      "n=2    #for second order\n",
-      "HX2=n*HX\n",
-      "\n",
-      "#Result\n",
-      "print(' 2. Entropy of second order system extension of source can be H(X^2)=%.2f bits/symbol ' %(HX*2))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1. Entropy of system  H(X)=1.18 bits/symbol \n",
-        " 2. Entropy of second order system extension of source can be H(X^2)=2.36 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.54, Page No 514"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "S0=1/3.0\n",
-      "S1=1/6.0\n",
-      "S2=1/4.0\n",
-      "S3=1/4.0\n",
-      "\n",
-      "#Calculation\n",
-      "HX=S0*math.log(1/S0,2)+S1*math.log(1/S1,2)+S2*math.log(1/S2,2)+S3*math.log(1/S3,2)   #EntroSy of source\n",
-      "\n",
-      "#Result\n",
-      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Entropy of system  H(X)=1.96 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.56, Page No 515"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Information capacity of telephone\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=3.4*10**3\n",
-      "SNR=30.0\n",
-      "\n",
-      "#Calculation\n",
-      "SN=10**(SNR/10)\n",
-      "C=B*math.log(1+SN,2)   #Information capacity\n",
-      "C=C/1000\n",
-      "\n",
-      "#Result\n",
-      "print(' Information capacity of telephone is C = %.2f kbps ' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Information capacity of telephone is C = 33.89 kbps \n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.59, Page No 516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "S0=1/3.0\n",
-      "S1=1/6.0\n",
-      "S2=1/4.0\n",
-      "S3=1/4.0\n",
-      "\n",
-      "#Calculation\n",
-      "HX=S0*math.log(1/S0,2)+S1*math.log(1/S1,2)+S2*math.log(1/S2,2)+S3*math.log(1/S3,2)   #EntroSy of source\n",
-      "\n",
-      "#Result\n",
-      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Entropy of system  H(X)=1.96 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.60, Page No 516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "m1=1/2.0\n",
-      "m2=1/4.0\n",
-      "m3=1/8.0\n",
-      "m4=1/16.0\n",
-      "m5=1/16.0\n",
-      "\n",
-      "#Calculation\n",
-      "L=(m1*1)+(m2*2)+(m3*3)+(2*(m4)*4)\n",
-      "\n",
-      "#Result\n",
-      "print(' Average number of bits per message =%.2f bits ' %L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Average number of bits per message =1.88 bits \n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.61, Page No 517"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Information capacity of telephone\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=3.4*10**3\n",
-      "SNR=30.0\n",
-      "\n",
-      "#Calculation\n",
-      "SN=10**(SNR/10)\n",
-      "C=B*math.log(1+SN,2)   #Information capacity\n",
-      "C=C/1000\n",
-      "\n",
-      "#Result\n",
-      "print(' Information capacity of telephone is C = %.2f kbps ' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Information capacity of telephone is C = 33.89 kbps \n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.62, Page No 517"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "p1=4.0\n",
-      "m1=0.5\n",
-      "m2=0.5\n",
-      "m3=0.375\n",
-      "m4=0.375\n",
-      "m5=0.375\n",
-      "m6=0.375\n",
-      "\n",
-      "#Calculation\n",
-      "I1=p1*math.log(1/p1,2) \n",
-      "HX=m1*math.log(1/m1,2)+m2*math.log(1/m2,2)+m3*math.log(1/m3,2)+m4*math.log(1/m4,2)+m5*math.log(1/m5,2)+m6*math.log(1/m6,2)   #EntroSy of source\n",
-      "\n",
-      "#Result\n",
-      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n",
-      "print('Approximation error')\t"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Entropy of system  H(X)=3.12 bits/symbol \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.65, Page No 519"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "S0=1/2.0\n",
-      "S1=1/4.0\n",
-      "S2=1/8.0\n",
-      "S3=1/8.0\n",
-      "n=1\n",
-      "\n",
-      "#Calculation\n",
-      "H=S0*math.log(1.0/S0,2)+S1*math.log(1.0/S1,2)+S2*math.log(1.0/S2,2)+S3*math.log(1.0/S3,2)   #EntroSy of source\n",
-      "L=H*n\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(' Code length =%.2f bits/messages ' %L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Code length =1.75 bits/messages \n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.67, Page No 520"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find channel capacity and new bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=8*10**3\n",
-      "SNR=31.0\n",
-      "SNR2=61\n",
-      "\n",
-      "#Calculation\n",
-      "C=B*math.log(1+SNR,2)   #Information capacity\n",
-      "B2=C/math.log(1+SNR2,2)\n",
-      "#Result\n",
-      "print(' Channel capacity is C = %.2f x 10^3 bits/sec ' %(C/1000))\n",
-      "print(' New Bandwidth is C = %.2f x kHz ' %(B2/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Channel capacity is C = 40.00 x 10^3 bits/sec \n",
-        " New Bandwidth is C = 6.72 x kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/Chapter9.ipynb b/Digital_Communications/Chapter9.ipynb
deleted file mode 100755
index c91e1583..00000000
--- a/Digital_Communications/Chapter9.ipynb
+++ /dev/null
@@ -1,166 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater 9: ERROR CONTROL CODING<h1>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find  detected errors,corrected errors\n",
-      "\n",
-      "#initialisation of variables\n",
-      "dmin=5.0\n",
-      "#(s+1)<= dmin number errors can be detected(s)\n",
-      " \n",
-      "#CALCULATIONS\n",
-      "s=dmin-1\n",
-      "\n",
-      "#RESULTS\n",
-      "print(' i)Number of detected errors  s <=  %.f ' %s)\n",
-      "#(2t+1)<=dmin number errors can be corrected(t)\n",
-      "t=(dmin-1)/2.0\n",
-      "print('ii) Number of corrected errors  t<=  %.f ' %t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i)Number of detected errors  s <=  4 \n",
-        "ii) Number of corrected errors  t<=  2 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17, Page No 569"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Determine all possible code vectors \n",
-      "\n",
-      "m3=1\n",
-      "m2=0\n",
-      "m1=1\n",
-      "m0=0\n",
-      "#M=Message Matrix\n",
-      "#G=Generator Matrix\n",
-      "G=[[1, 0, 1, 1, 0, 0, 0],[0, 1, 0, 1, 1, 0, 0],[0, 0, 1, 0, 1, 1, 0],[0, 0, 0, 1, 0, 1, 1]]\n",
-      "M=[[m3,m2,m1,m0]]\n",
-      "X = [[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0]]\n",
-      "\n",
-      "for i in range(len(G)):\n",
-      "   # iterate through columns of PXd\n",
-      "   for j in range(len(M[0])):\n",
-      "       # iterate through rows of PYX\n",
-      "       for k in range(len(M)):\n",
-      "           X[i][j] += G[i][k] * M[k][j]\n",
-      "print('The required code word')\n",
-      "for r in range(0,7):\n",
-      "         print(X[0][r])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required code word\n",
-        "1\n",
-        "0\n",
-        "1\n",
-        "0\n",
-        "0\n",
-        "0\n",
-        "0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19, Page No 572"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Determine  code word \n",
-      "m3=1\n",
-      "m2=0\n",
-      "m1=1\n",
-      "m0=0\n",
-      "#M=Message Matrix\n",
-      "#G=Generator Matrix\n",
-      "G=[[1, 0, 0, 0, 1, 0, 1],[0, 1, 0, 0, 1, 1, 1],[0, 0, 1, 0, 1, 1, 0],[0, 0, 0, 1, 0, 1, 1]]\n",
-      "M=[[m3,m2,m1,m0]]\n",
-      "X = [[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0]]\n",
-      "\n",
-      "\n",
-      "for i in range(len(G)):\n",
-      "   # iterate through columns of PXd\n",
-      "   for j in range(len(M[0])):\n",
-      "       # iterate through rows of PYX\n",
-      "       for k in range(len(M)):\n",
-      "           X[i][j] += G[i][k] * M[k][j]\n",
-      "print('The required code word')\n",
-      "for r in range(0,7):\n",
-      "         print(X[0][r] )\n",
-      "     \n",
-      "\n",
-      "print('The code in the book is wrong')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required code word\n",
-        "1\n",
-        "0\n",
-        "1\n",
-        "0\n",
-        "0\n",
-        "0\n",
-        "0\n",
-        "The code in the book is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo2.ipynb b/Digital_Communications/ChapterNo2.ipynb
deleted file mode 100755
index f8cf6b90..00000000
--- a/Digital_Communications/ChapterNo2.ipynb
+++ /dev/null
@@ -1,290 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter no 2: SAMPLING THEORY AND PULSE MODULATION<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1, page no 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#find Nquist Rate\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given \n",
-      "pi=3.14\n",
-      "w1=50*pi\n",
-      "w2=300*pi\n",
-      "w3=100*pi\n",
-      "#w=2*%pi*f\n",
-      "\n",
-      "#Calculation\n",
-      "f1=w1/(2*pi)\n",
-      "f2=w2/(2*pi)\n",
-      "f3=w3/(2*pi)\n",
-      "fm=f2 #fm = maximum frquency is present at the signal\n",
-      "\n",
-      "#Result\n",
-      "print('maximum frquency of the signal is = %.2f Hz' %f2)\n",
-      "fs=2*fm #Nyquist rate\n",
-      "print('Nquist Rate of Signal is = %.2f Hz' %fs)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum frquency of the signal is = 150.00 Hz\n",
-        "Nquist Rate of Signal is = 300.00 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2 , page no 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Nquist Rate and Nquist time interval\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "w1=5000*math.pi\n",
-      "w2=3000*math.pi;\n",
-      "f1=w1/(2*math.pi);\n",
-      "f2=w2/(2*math.pi);\n",
-      "\n",
-      "#Calculation\n",
-      "fm=f1 #fm = maximum frquency is present at the signal\n",
-      "fs=2*fm #Nyquist rate\n",
-      "Ts=1.0/(2.0*fm) #frequncy =1/time\n",
-      "Ts=Ts*(10**3)\n",
-      "\n",
-      "#Result\n",
-      "print('maximum frquency of the signal is = %.f Hz' %f1)\n",
-      "print('Nquist Rate of the given Signal is = %.f Hz' %fs)\n",
-      "print('Nquist Interval of the given signal is = %.1f m Sec' %Ts)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum frquency of the signal is = 2500 Hz\n",
-        "Nquist Rate of the given Signal is = 5000 Hz\n",
-        "Nquist Interval of the given signal is = 0.2 m Sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3, page no 51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find Nquist Rate \n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "f=100.0         # Frequency component of continuous-time signal\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*f #Nyquist rate\n",
-      "\n",
-      "#Result\n",
-      "print('i) To avoid aliasing Nquist Rate is = %.f Hz' %fs)\n",
-      "print('ii) It is theoretical example ')\n",
-      "print('iii) It is theoretical example ')\n",
-      "print('iv) It is theoretical example ')\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) To avoid aliasing Nquist Rate is = 200 Hz\n",
-        "ii) It is theoretical example \n",
-        "iii) It is theoretical example \n",
-        "iv) It is theoretical example \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4, page no 52 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Nquist Rate of Continous signal\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "w1=50*math.pi\n",
-      "w2=300*math.pi\n",
-      "w3=100*math.pi\n",
-      "\n",
-      "#Calculation\n",
-      "f1=w1/(2*math.pi)\n",
-      "f2=w2/(2*math.pi)\n",
-      "f3=w3/(2*math.pi)\n",
-      "fmax=f2 #fmax = Highest frquency component of the message signal\n",
-      "fs=2*fmax #Nyquist rate\n",
-      "\n",
-      "#Result\n",
-      "print('Highest frquency component of the message signal will be fmax = %.f Hz' %fmax)\n",
-      "print('Nquist Rate of the given Signal is = %.f Hz' %fs)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Highest frquency component of the message signal will be fmax = 150 Hz\n",
-        "Nquist Rate of the given Signal is = 300 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 2.7, page no 67 <h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#find amplitude distortion at highest frquency\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "fs=9.5  #samplig frequncy\n",
-      "fmax=1   #maximum frequncy\n",
-      "t=0.2 #pulse width\n",
-      "\n",
-      "#Calculation\n",
-      "c=3*10**8\n",
-      "f=fmax\n",
-      "H1=t*(0.9933)  #aperture effect at highest frequency, sinc(f*t)=0.9933 (given)\n",
-      "H1=H1*100\n",
-      "\n",
-      "#Result\n",
-      "print('|H(1)|=%.2f' %H1)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "|H(1)|=19.87\n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.8, page no 74 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate Transmission Bandwidth\n",
-      "\n",
-      "#Variable declaration\n",
-      "#given\n",
-      "fm=3.0*(10^3)\n",
-      "fs=8.0*(10^3) # sampling frequncy\n",
-      "\n",
-      "#Calculation\n",
-      "Ts=1.0/fs\n",
-      "t=0.1*Ts\n",
-      "BW=1.0/(2*t) #Bandwidth\n",
-      "BW=BW/(10^3)\n",
-      "\n",
-      "#Result\n",
-      "print('Transmission Bandwidth of PAM signal is kHz = %.f Khz ' %BW)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transmission Bandwidth of PAM signal is kHz = 40 Khz \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo3.ipynb b/Digital_Communications/ChapterNo3.ipynb
deleted file mode 100755
index 5b9ea469..00000000
--- a/Digital_Communications/ChapterNo3.ipynb
+++ /dev/null
@@ -1,2238 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 3: WAVEFORM CODING TECHNIQUES<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3,1 Page No 110"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print('This is a Theoretical  example')\t"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is a theoretcal example\n"
-       ]
-      }
-     ],
-     "prompt_number": 104
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3,2,i, Page No 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "#find code word length\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512        #Quantization levels\n",
-      "#q=2^v\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log10(512)/math.log10(2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The code word legth is = %.f bits\" %v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The code word legth is = 9 bits\n"
-       ]
-      }
-     ],
-     "prompt_number": 105
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2.ii, Page No 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#find The transmission Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512.0           #Quantization levels\n",
-      "v=9.0\n",
-      "\n",
-      "#Calculation\n",
-      "bw=v*fm*10**-6\n",
-      "\n",
-      "#Result\n",
-      "print(\"The transmission Bandwidth is = %.1f MHz\" %bw)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transmission Bandwidth is = 37.8 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 106
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2.iii, Page No 111 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find  Final Bit rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512.0      #Quantization levels\n",
-      "v=9.0\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm\n",
-      "r=v*fs       #signaling rate\n",
-      "\n",
-      "#Result\n",
-      "print(\"Final Bit rate =%.1f x 10^6 bits/sec\" %(r/(10**6)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Final Bit rate =75.6 x 10^6 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 107
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2.iv, Page No 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find Output signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "bandwidth=4.2*10**6\n",
-      "fm=bandwidth\n",
-      "q=512         #Quantization levels\n",
-      "v=9.0\n",
-      "\n",
-      "#Calculation\n",
-      "sn=4.8+6*v    #noise in dB\n",
-      "\n",
-      "\n",
-      "print(\"Output signal to noise ratio is = %.1f dB\" %sn)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal to noise ratio is = 58.8 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 108
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.i, Page No 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find bits per sample\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fmax=4*10**3\n",
-      "xmax=3.8\n",
-      "snr=100\n",
-      "\n",
-      "#Calculation\n",
-      "P=30*10**-3\n",
-      "v=(math.log10(((snr*xmax**2)/(3*P)))/(2*math.log10(2)))\n",
-      "\n",
-      "#Result\n",
-      "print(\"Number of bits required per sample are = %.2f bits\" %v)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of bits required per sample are = 6.98 bits\n"
-       ]
-      }
-     ],
-     "prompt_number": 109
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.ii, Page No 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find  Transmission Bandwith\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=4*10**3     #Bandwidth of PCM\n",
-      "xmax=3.8\n",
-      "snr=100       #Signal to Noise Ratio\n",
-      "outputs=30.0\n",
-      "v=7.0\n",
-      "\n",
-      "#Calculation\n",
-      "bw=outputs*v*fm \n",
-      "r=bw*2\n",
-      "bw=bw/10**3\n",
-      "\n",
-      "#Result\n",
-      "print('Transmission Bandwith R = %.f kHz' %bw)\n",
-      "r=r/1000\n",
-      "print('Signaling rate R = %.f bits/sec' %r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transmission Bandwith R = 840 kHz\n",
-        "Signaling rate R = 1680 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 110
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page No 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#find sampling rate,number of bits,bit rate,bandwidht\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "emax=0.001\n",
-      "de=2*emax\n",
-      "fm=100.0\n",
-      "xmax=10.0\n",
-      "\n",
-      "#Calculation\n",
-      "q=(2*xmax)/de\n",
-      "fs=2*fm\n",
-      "v=(math.log10(q))/math.log10(2)\n",
-      "v=math.ceil(v)\n",
-      "r=v*fs\n",
-      "\n",
-      "#Result\n",
-      "print('1. sampling Frequncy  =  %.f  Hz ' %fs)  \n",
-      "print('2. No.of bits in PCM  =  %.f  bits ' %v)  \n",
-      "print('3. sampling rate  =  %.f  bits per second ' %r)\n",
-      "r=r/2\n",
-      "print('4. Transmission Bandwidth  =  %.f  Hz ' %r)  "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. sampling Frequncy  =  200  Hz \n",
-        "2. No.of bits in PCM  =  14  bits \n",
-        "3. sampling rate  =  2800  bits per second \n",
-        "4. Transmission Bandwidth  =  1400  Hz \n"
-       ]
-      }
-     ],
-     "prompt_number": 111
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5, Page No 113 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Bandwidth,Sampling Rate\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=3.4*10**3\n",
-      "N=24\n",
-      "r=1.5*10**6\n",
-      "encoder=8\n",
-      "\n",
-      "#Calculation\n",
-      "BW=N*fm\n",
-      "BW=BW/10**3\n",
-      "r1=r/N\n",
-      "fs=r1/encoder\n",
-      "\n",
-      "#Result\n",
-      "print(\"i. Channel Bandwith is = %.2f kHz\" %BW)\n",
-      "print(\"ii. Sampling frequency is = %.2f Hz or samples per second.\" %fs)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i. Channel Bandwith is = 81.60 kHz\n",
-        "ii. Sampling frequency is = 7812.50 Hz or samples per second.\n"
-       ]
-      }
-     ],
-     "prompt_number": 112
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page No 114 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find Signal Bandwidth,Noise Ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "v=7\n",
-      "r=50*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "#fs=2*fm\n",
-      "fm=r/(2*v)\n",
-      "snr=1.8+(6*v);\n",
-      "fm=fm*10**-6\n",
-      "\n",
-      "#Result\n",
-      "print('i. Maximum message Bandwidth is = %.2f MHz ' %fm)\n",
-      "print('ii.  signal to quantization niose ration = %.2f dB ' %snr)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i. Maximum message Bandwidth is = 3.57 MHz \n",
-        "ii.  signal to quantization niose ration = 43.80 dB \n"
-       ]
-      }
-     ],
-     "prompt_number": 113
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page No 114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#find i)bits per sample,ii)transmission rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=3*10**3\n",
-      "q=16\n",
-      "\n",
-      "#Calculation\n",
-      "v=(math.log10(q))/math.log10(2)\n",
-      "\n",
-      "#Result\n",
-      "print('i) Bits in code word= %.f bits' %v)\n",
-      "fs=2*fm\n",
-      "r=v*fs\n",
-      "print('ii) it trasmission rate= %.f x 10^3 bits per second' %(r/(10**3)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Bits in code word= 4 bits\n",
-        "ii) it trasmission rate= 24 x 10^3 bits per second\n"
-       ]
-      }
-     ],
-     "prompt_number": 114
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8, Page No 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "fm=3.5*10**3\n",
-      "r=50*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm\n",
-      "rms=0.2\n",
-      "xmax=2\n",
-      "v=r/fs    #signaling rate r=v*fs\n",
-      "v=math.ceil(v)\n",
-      "P=(rms**2)/1\n",
-      "SNR=((3*P*2**(2*v))/(xmax**2))\n",
-      "SN=10*math.log10(SNR)\n",
-      "SN=math.ceil(SN)\n",
-      "\n",
-      "#Result\n",
-      "print(\"signal to niose ratio =%.f dB\" %SN)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "signal to niose ratio =33 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 115
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10, Page No 117 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find i)noise ratio ii)bits\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "Am=3.0\n",
-      "v=10.0\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=1.8+6*v     #noise ratio \n",
-      "SN=40\n",
-      "v=(SN-1.8)/6\n",
-      "\n",
-      "#Result\n",
-      "print(\"i) Signal to Quantization noise ratio = %.1f dB\" %SNR)\n",
-      "print(\"ii) Bits required to get signal to niose ratio of 40dB = %.f \" %math.ceil(v))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Signal to Quantization noise ratio = 61.8 dB\n",
-        "ii) Bits required to get signal to niose ratio of 40dB = 7 \n"
-       ]
-      }
-     ],
-     "prompt_number": 116
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11, Page No 117 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Maximum frequency\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#Given data\n",
-      "v=7\n",
-      "SNR=1.8+6*v\n",
-      "r=56*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fs=r/v        #r=v*fs signaling rate\n",
-      "fm=fs/2       #Nquset rate\n",
-      "\n",
-      "#Result\n",
-      "fm=fm/10**3\n",
-      "print(\"Maximum frequency is = %.f kHz\" %fm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum frequency is = 4 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 117
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.14, Page No 129 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Maximum Amplitude\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3*10**3.0\n",
-      "\n",
-      "#Calculation\n",
-      "Nyquistrate=2*fm     #Nyquistrate\n",
-      "fs=5*Nyquistrate     #Samplingfrquency\n",
-      "Ts=1/fs              #Sampling Interval\n",
-      "de=0.25             #step size\n",
-      "fm1=2*10**3.0\n",
-      "Am=de/(2*math.pi*fm1*Ts)\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum Amplitude = %.4f Volts\" %Am)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum Amplitude = 0.5968 Volts\n"
-       ]
-      }
-     ],
-     "prompt_number": 118
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.16, Page No 130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  signaling rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs1=8*10**3.0\n",
-      "de=31.25*10**-3\n",
-      "q=64.0\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log(q,2)\n",
-      "r=v*fs1            #signaling rate\n",
-      "fm=3*10**3.0\n",
-      "A=1\n",
-      "fs2=(2*math.pi*fm*A)/(de)\n",
-      "\n",
-      "#Result\n",
-      "r=r*10**-3\n",
-      "print(\"Signaling rate of PCM is = %.f kHz\" %r)\n",
-      "fs2=fs2*10**-3\n",
-      "print(\"The signaling rate of DM is = %.2f kHz\" %fs2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signaling rate of PCM is = 48 kHz\n",
-        "The signaling rate of DM is = 603.19 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 119
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.17, Page No 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64.0*10**3     #frequency\n",
-      "fm=2.0*10**3      #Sinosoidal input signal sample \n",
-      "fM=4.0*10**3      #bandwidth\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*fM)   #Signal to noise ratio\n",
-      "SNRO=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output signal to noise ratio = %.2f dB\" %SNRO)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal to noise ratio = 27.94 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 120
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.18, Page No 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  signal to Quatization noise ratio \n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=8.0*10**3        #Sampling frequency\n",
-      "r=64.0*10**3        #Data Rate\n",
-      "N=8        #number of bits\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=(1.8+6*N)     #signal to Quatization noise ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Signal to Quatization noise ratio of PCM system is = %.f dB\" %SNR)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signal to Quatization noise ratio of PCM system is = 50 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 121
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.20, Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find  sampling rate,quantizing level\n",
-      "\n",
-      "#Variable Declaration\n",
-      "r=36000\n",
-      "fm=3.2*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm        #Nquest rate\n",
-      "v=r/fs         #r=v*fs signaling rate\n",
-      "v=math.floor(v)\n",
-      "q=2**v\n",
-      "fs1=r/v\n",
-      "\n",
-      "#Result\n",
-      "print(\"Quantizing level q = %.f \" %q)\n",
-      "fs1=fs1/1000\n",
-      "print(\"sampling rate fs = %.1f kHz \" %fs1)\n",
-      "print(\"Number of binary digits = %.f  \" %v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizing level q = 32 \n",
-        "sampling rate fs = 7.2 kHz \n",
-        "Number of binary digits = 5  \n"
-       ]
-      }
-     ],
-     "prompt_number": 122
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.21, Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  time duration of 1bit  binary encoded signal\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=input(\"Enter the Nyquist rate of Signal fs(Hz)=\")\n",
-      "q=input(\"Enter the Quantization levels q =\")\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log(q,2)     #binary pulses transmitted per second\n",
-      "t=1.0/(v*fs)        #Nyquist interval\n",
-      "\n",
-      "#Result\n",
-      "print('Time duration of binary signal  t = %.4f sec ' %t)\n",
-      "#output\n",
-      "#Enter the Nyquist rate fs(Hz)=3\n",
-      "#Enter the Quantization levels q =5\n",
-      "#time duration of 1bit binary signal   \n",
-      "#t=0.1435589  sec"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Nyquist rate of Signal fs(Hz)=23\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Quantization levels q =34\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time duration of binary signal  t = 0.0085 sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 123
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.23, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#find signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNR=40\n",
-      "SNRO=10**(SNR/10)\n",
-      "\n",
-      "#SNR=3(q**2)/2, Signal to Quantization noise ratio\n",
-      "\n",
-      "#Calculation\n",
-      "q=math.sqrt((2*SNRO)/3)\n",
-      "q=math.ceil(q)\n",
-      "v=math.log(q,2)         #q=2**v  Quantization levels\n",
-      "v=math.ceil(v)\n",
-      "snr=1.76+6.02*v          #output Signal to Quantization noise ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output Signal to Quantization noise ratio = %.1f dB\" %snr)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output Signal to Quantization noise ratio = 43.9 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 124
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.24.i, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Quantizing  levels,minimum number of bits per sample\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNRO=30.0\n",
-      "fmin=300.0\n",
-      "fmax=3300.0\n",
-      "fs=80000.0\n",
-      "\n",
-      "#Calculation\n",
-      "#SNRO=1.76+20log10(q)\n",
-      "q=10**((SNRO-1.76)/20)\n",
-      "q=math.ceil(q)\n",
-      "v=math.log(q,2)\n",
-      "\n",
-      "#Result\n",
-      "print('Quantizing  levels required is =%.f' %q)\n",
-      "print('minimum number of bits per sample are=%.1f that is approximately 5' %v)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizing  levels required is =26\n",
-        "minimum number of bits per sample are=4.7 that is approximately 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 125
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.24.ii, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  minimum required bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNRO=30.0       #Signal to quantisizing noise ratio\n",
-      "fmin=300.0      #min frequency band\n",
-      "fmax=3300.0     #max frequency band\n",
-      "fs=8000.0       #Sampling rate\n",
-      "v=5.0           #Minimum number of bits per sample\n",
-      "\n",
-      "#Calculation\n",
-      "fPCM=(v*fs)/2\n",
-      "fPCM=fPCM/1000\n",
-      "\n",
-      "#Result\n",
-      "print('minimum required bandwidth =%.f kHz' %fPCM)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "minimum required bandwidth =20 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 126
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.24.iii, Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Quantizing  levels,minimum number of bits per sample and bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNRO=30.0       #Signal to quantisizing noise ratio\n",
-      "fmin=300.0      #min frequency band\n",
-      "fmax=3300.0     #max frequency band\n",
-      "fs=8000.0       #Sampling rate\n",
-      "\n",
-      "#Calculation\n",
-      "q=10**((SNRO+10.1)/20)\n",
-      "q=math.ceil(q)\n",
-      "v=math.log(q,2)\n",
-      "v=math.ceil(v)\n",
-      "\n",
-      "#Result\n",
-      "print('Quantizing  levels need is =%.f' %q)\n",
-      "print('minimum number of bits per sample is =%.f' %v)\n",
-      "fPCM=(v*fs)/2\n",
-      "fPCM=fPCM/1000\n",
-      "print('minimum required bandwidth =%.f Khz' %fPCM)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizing  levels need is =102\n",
-        "minimum number of bits per sample is =7\n",
-        "minimum required bandwidth =28 Khz\n"
-       ]
-      }
-     ],
-     "prompt_number": 127
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.27, Page No 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#determine the  Maximum Amplitude,\n",
-      "\n",
-      "#Variable Declaration\n",
-      "de=250*10**-3\n",
-      "wm=2*math.pi*1000 \n",
-      "fs=3*10**3\n",
-      "Ts=1/fs\n",
-      "\n",
-      "#Calculation\n",
-      "Amax=(de*3*fs*2)/(wm) #Amplitude\n",
-      "SNR=(3.0*((3*6*(10**3))**3))/((8*(math.pi**2)*(10**3)**3))\n",
-      "\n",
-      "#Result\n",
-      "print('Maximum Amplitude= %.1f mV' %(Amax*(10**2)))\n",
-      "print('Signal to noise ratio = %.2f dB' %(10*math.log10(SNR)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum Amplitude= 71.6 mV\n",
-        "Signal to noise ratio = 23.46 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 128
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.29, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find number of bits per sample\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SNR=20\n",
-      "averagepower=30*10**-3\n",
-      "SNRO=10**(SNR/10)\n",
-      "A=3.8\n",
-      "\n",
-      "#SNRO=average signal power/Quatizing power\n",
-      "#de=(2*A)/L\n",
-      "\n",
-      "#Calculation\n",
-      "L=math.sqrt((SNRO*A**2)/(3*averagepower))\n",
-      "n=math.log(L,2)\n",
-      "n=math.ceil(n)\n",
-      "\n",
-      "#Result\n",
-      "print('Bits required per sample =%.f' %n)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bits required per sample =7\n"
-       ]
-      }
-     ],
-     "prompt_number": 129
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.30.i, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Normalized power for quantization noise\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3.0*10.0**3\n",
-      "v=8.0\n",
-      "VH=5.0\n",
-      "VL=-5.0\n",
-      "q=2**v\n",
-      "\n",
-      "#Calculation\n",
-      "de=(VH-VL)/q\n",
-      "Nq=de**2/12.0    #quantization noise\n",
-      "\n",
-      "#Result\n",
-      "print('Normalized power for quantization noise =%.2f x 10^-6 W' %(Nq*(10**6)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Normalized power for quantization noise =127.16 x 10^-6 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 130
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.30.ii, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find bit transmission rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3*10.0**3\n",
-      "v=8.0\n",
-      "VH=5.0\n",
-      "VL=-5\n",
-      "q=2**v\n",
-      "\n",
-      "#Calculation\n",
-      "fs=2*fm       #Nyquist rate\n",
-      "r=8*fs/1000\n",
-      "\n",
-      "#Result\n",
-      "print('bit transmission rate %.f K bits/s' %r)\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "bit transmission rate 48 K bits/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 131
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.30.iii, Page No 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Signal to quantization noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "Nq=127.15*10**-6\n",
-      "Meansignal=2\n",
-      "\n",
-      "#Calculation\n",
-      "P=Meansignal/1\n",
-      "SNR=P/Nq\n",
-      "SNRq=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print('Signal to quantization noise ratio %.2f dB ' %SNRq)\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signal to quantization noise ratio 41.97 dB \n"
-       ]
-      }
-     ],
-     "prompt_number": 132
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.31, Page No 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  i)SNR  ii)output bit rate iii)no.of bits recorded\n",
-      "\n",
-      "#Variable Declaration\n",
-      "N=16.0\n",
-      "v=16.0\n",
-      "fs=44.1*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=1.76+6*N\n",
-      "bitrate=2*v*fs\n",
-      "outputbitrate=2*bitrate*10**-6        #including addtional 100% over head\n",
-      "CD=outputbitrate*3600*10**-3\n",
-      "\n",
-      "#Result\n",
-      "print('i)Out put signal noise ratio =%.2f dB' %SNR)\n",
-      "print('ii)output bit rate =%.3f Mbits/sec' %outputbitrate)\n",
-      "print('iii)no.of bits recorded in CD =%.2f gigabits' %CD)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Out put signal noise ratio =97.76 dB\n",
-        "ii)output bit rate =2.822 Mbits/sec\n",
-        "iii)no.of bits recorded in CD =10.16 gigabits\n"
-       ]
-      }
-     ],
-     "prompt_number": 133
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.32, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  output SNR\n",
-      "#Variable Declaration\n",
-      "fm=1*10.0**3\n",
-      "fs=32*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "FM=4*10**3     #Bandwidth\n",
-      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*FM)    #SNR\n",
-      "SNRO=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print('Output signal to noise ratio %.2f dB' %SNRO)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal to noise ratio 24.93 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 134
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.33.i, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step size\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64000.0     #samples/sec\n",
-      "Amax=1.0       #Maximum signal amplitude\n",
-      "fm=3500.0      #Sample Rate\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*Amax)/fs*1000\n",
-      "\n",
-      "#Result\n",
-      "print('Step Size %.2f mV ' %de)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Step Size 343.61 mV \n"
-       ]
-      }
-     ],
-     "prompt_number": 135
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.33ii, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Quantizatio noise power \n",
-      "#Variable Declaration\n",
-      "fs=64000.0     #sample Rate\n",
-      "Amax=1.0       #Maximum signal Amplitude\n",
-      "fm=3500.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=343.6117*10**-3      #step size\n",
-      "Nq=de**2/3.0             #Quantizatio noise power\n",
-      "Nqd=Nq*(fm/fs)*1000\n",
-      "\n",
-      "#Result\n",
-      "print('Quantizatio noise power %.3f mW' %Nqd)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quantizatio noise power 2.152 mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 136
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.33iii, Page No 155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find SNR\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64000.0\n",
-      "Amax=1.0\n",
-      "fm=3500.0\n",
-      "Nqd=2.1522995*10**-3\n",
-      "\n",
-      "#Calculation\n",
-      "So=Amax**2/2\n",
-      "SNR=So/Nqd\n",
-      "SNRO=10*math.log10(SNR)\n",
-      "\n",
-      "#Result\n",
-      "print('Output signal noise ratio dB  %.3f dB' %SNRO)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output signal noise ratio dB  23.661 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 137
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.34, Page No 156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Assuming signal is sampled at the rate 20% above Nyquist rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=4.5*10**6\n",
-      "q=1024.0\n",
-      "\n",
-      "#Calculation\n",
-      "fs=1.2*2*fm           #20% above Nyquist rate\n",
-      "v=math.log(q,2)\n",
-      "r=v*fs/10**6\n",
-      "\n",
-      "#Result\n",
-      "print('no.of bits/sec = %.f M bit/sec' %r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "no.of bits/sec = 108 M bit/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 138
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.35, Page No 156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume bandwidth of the singal is 4kHz\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=32000.0\n",
-      "A=2.0\n",
-      "fm=4000.0\n",
-      "BW=4000.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*A)/fs\n",
-      "Nq=de**2/3\n",
-      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*BW)\n",
-      "\n",
-      "#Result\n",
-      "print('i)step size = %.3f M Volt' %de)\n",
-      "print('ii)noise power = %.3f W' %Nq)\n",
-      "print('iii)SNR= = %.3f ' %SNR)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)step size = 1.571 M Volt\n",
-        "ii)noise power = 0.822 W\n",
-        "iii)SNR= = 19.454 \n"
-       ]
-      }
-     ],
-     "prompt_number": 139
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.36, Page No 157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assuming signal is sampled at the rate 20% above Nyquist rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=15*10**3\n",
-      "fs=1.2*2*fm\n",
-      "q=65536\n",
-      "\n",
-      "#Calculation\n",
-      "v=math.log(q,2)\n",
-      "r=v*fs/1000\n",
-      "BW=r/2\n",
-      "\n",
-      "#Result\n",
-      "print('i)signaling rate,= %.f K bits/sec ' %r)\n",
-      "print('ii)bandwidth BW min = %.f kHz ' %BW)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)signaling rate,= 576 K bits/sec \n",
-        "ii)bandwidth BW min = 288 kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 140
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.37, Page No 157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step size,noise power\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=64*10**3\n",
-      "fm=3500.0\n",
-      "A=1.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*A)/fs      #step size\n",
-      "Nq=(de**2/3)*(fm/fs)        #Granular noise power\n",
-      "\n",
-      "#Result\n",
-      "print('i)step size = %.3f Volts' %de)\n",
-      "print('ii)Nq= %.5f W' %Nq)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)step size = 0.344 Volts\n",
-        "ii)Nq= 0.00215 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 141
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.40, Page No 159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step Nr,Transmission Bandwidth, SNR\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fs=4           #kHz\n",
-      "N=12\n",
-      "\n",
-      "#Calculation\n",
-      "Nr=2*fs                     #Nyquest rate\n",
-      "Bt=1.0/2*N*fs               #Transmission Bandwidth\n",
-      "SNR=1.8+(6*N)               \n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('i)Nyquest rate = %.1f kHz' %Nr)\n",
-      "print('ii)Transmission Bandwidth = %.1f kHz' %Bt)\n",
-      "print('iii)SNR = %.1f dB' %SNR)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)Nyquest rate = 8.0 kHz\n",
-        "ii)Transmission Bandwidth = 24.0 kHz\n",
-        "iii)SNR = 73.8 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 142
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.41, Page No 159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  step size,noise power\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=3.5*10**3\n",
-      "fs=64*10**3\n",
-      "A=2.0\n",
-      "\n",
-      "#Calculation\n",
-      "de=(2*math.pi*fm*A)/fs      #step size\n",
-      "Nq=(de**2/3)*(fm/fs)        #Granular noise power\n",
-      "\n",
-      "#Result\n",
-      "print('i)step size = %.3f Volts' %de)\n",
-      "print('ii)Nq= %.1f  X 10^-3 W' %(Nq*10**3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)step size = 0.687 Volts\n",
-        "ii)Nq= 8.6  X 10^-3 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 143
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.42, Page No 159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Bit rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=4.5 * 10**6                    #khz\n",
-      "Nr = 2 * W               #Nyquist Rate\n",
-      "Q = 1024\n",
-      "fs=10.8\n",
-      "\n",
-      "#Calculation\n",
-      "N = math.log(Q,2)            #number of binay pulse per word\n",
-      "Br= N*fs                     #Bit rate\n",
-      "Bw= 1.0/2*Br                 #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of binay pulse per word = %.1f k bits/sec' %N)\n",
-      "print('ii) Bit rate = %.1f k bits/sec' %Br)\n",
-      "print('iii)Bandwidth= %.1f  X 10^-3 kHz' %Bw)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of binay pulse per word = 10.0 k bits/sec\n",
-        "ii) Bit rate = 108.0 k bits/sec\n",
-        "iii)Bandwidth= 54.0  X 10^-3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 144
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.43, Page No 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Bit rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=15                 #khz\n",
-      "Nr = 2*W               #Nyquist Rate\n",
-      "Q = 65.53\n",
-      "fs=44.1                # Sampling rate kHz\n",
-      "N=16\n",
-      "\n",
-      "#Calculation\n",
-      "Br= N*fs                    #Bit rate\n",
-      "Bw= 1.0/2*Br                 #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of binay pulse per word = %.3f bits' %N)\n",
-      "print('ii) Bit rate = %.1f k bits/sec' %Br)\n",
-      "print('iii)Bandwidth= %.1f kHz' %Bw)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of binay pulse per word = 16.000 bits\n",
-        "ii) Bit rate = 705.6 k bits/sec\n",
-        "iii)Bandwidth= 352.8 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 145
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.44, Page No 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Number of binay pulse per word\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=3.3                 #khz\n",
-      "fm = 1*10**3                #kHz\n",
-      "de = 250              #mV\n",
-      "fs=19.8                # Sampling rate kHz\n",
-      "\n",
-      "#Calculation\n",
-      "A=(de*fs)/(2*math.pi*fm)\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of binay pulse per word = %.3f' %A)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of binay pulse per word = 0.788\n"
-       ]
-      }
-     ],
-     "prompt_number": 146
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.45, Page No 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Signaling rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=3.3                 #khz\n",
-      "fm = 3300               #Hz\n",
-      "Snq = 40              #dB\n",
-      "fs= 8000                # Samples/sec\n",
-      "\n",
-      "#Calculation \n",
-      "N=math.ceil((Snq-4.8)/6)                # Number of bits per word\n",
-      "Q=2**N              #\n",
-      "r=N*fs              #Signaling rate k bits/sec\n",
-      "Bw=1.0/2*r\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of bits per word = %.1f' %N)\n",
-      "print('ii) Signaling rate rate = %.1f k bits/sec' %(r/10**3))\n",
-      "print('iii)Bandwidth= %.1f kHz' %(Bw/10**3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of bits per word = 6.0\n",
-        "ii) Signaling rate rate = 48.0 k bits/sec\n",
-        "iii)Bandwidth= 24.0 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 147
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.46, Page No 161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find the maximun frequency with 1 volt amplitude\n",
-      "\n",
-      "#Variable Declaration\n",
-      "de=256*10**-3                 #V\n",
-      "Ts = 10*10**-6               #kHz\n",
-      "fm = 10              #kHz\n",
-      "fs= 100                # kHz\n",
-      "A=1                   #V\n",
-      "\n",
-      "#Calculation \n",
-      "wm=(de/A*Ts)*10**7\n",
-      "\n",
-      "#Result\n",
-      "print('i) maximun frequency = %.1f X 10^3 rad/sec' %wm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) maximun frequency = 25.6 X 10^3 rad/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 148
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.47, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find the Signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "Q=256     \n",
-      "m=255\n",
-      "\n",
-      "#Calculation \n",
-      "SNR=(3*Q**2)/(math.log((1+m),math.e))**2    #dB\n",
-      "Gc=m/(math.log(1+m))                        #Companding gain\n",
-      "Gc=20*math.log(Gc,10)\n",
-      "\n",
-      "#Result\n",
-      "print('i) The Signal to noise ratio= %.1f dB' %SNR)\n",
-      "print('ii) The Companding gain = %.1f dB' %Gc)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The Signal to noise ratio= 6394.0 dB\n",
-        "ii) The Companding gain = 33.3 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 149
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.48, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of binay pulse per word, Signaling rate, Bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=15*10**3          #kHz\n",
-      "Q=200\n",
-      "fs=2*fm\n",
-      "\n",
-      "#Calculation \n",
-      "N=math.ceil(math.log(Q,2))\n",
-      "tr=N*fs           #Transmission rate kbps\n",
-      "Bw=1.0/2*tr\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of bits per Samples = %.1f' %N)\n",
-      "print('ii) Signaling rate = %.1f k bits/sec' %(tr/10**3))\n",
-      "print('iii)Bandwidth= %.1f kHz' %(Bw/10**3))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of bits per Samples = 8.0\n",
-        "ii) Signaling rate = 240.0 k bits/sec\n",
-        "iii)Bandwidth= 120.0 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 150
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.49, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  number of quantization level\n",
-      "\n",
-      "#Variable Declaration\n",
-      "tr=36.0                  #kbps\n",
-      "fm=3.4                 #kHz\n",
-      "fs=8\n",
-      "\n",
-      "#Calculation \n",
-      "N=math.ceil(tr/fs)\n",
-      "Q=2**N\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('i) number of quantization level = %.1f' %Q)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) number of quantization level = 32.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 151
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.50, Page No 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  delta modulation\n",
-      "\n",
-      "#Variable Declaration\n",
-      "A=1                  #V\n",
-      "fm=3.4                 #kHz\n",
-      "fs=20\n",
-      "\n",
-      "#Calculation \n",
-      "de=1*2*math.pi*(fm/fs)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('i) delta modulation = %.2f V' %de)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) delta modulation = 1.07 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 152
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.51, Page No 163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=3.5      #kHz\n",
-      "r=50.0      #kbps\n",
-      "Vpp=4      #V\n",
-      "fs=8\n",
-      "\n",
-      "#Calculation \n",
-      "N=r/fs\n",
-      "SNR=4.8+(6.8*N)\n",
-      "\n",
-      "#Result\n",
-      "print('i) Signal to noise ratio = %.2f V' %SNR)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Signal to noise ratio = 47.30 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 153
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.52, Page No 163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  Signal to noise ratio\n",
-      "\n",
-      "#Variable Declaration\n",
-      "W=1      #MHz\n",
-      "fs1=3       #MHz\n",
-      "fs2=2.4       #MHz\n",
-      "Q=256\n",
-      "m=255\n",
-      "\n",
-      "#Calculation \n",
-      "SNR=((3*Q**2)/(math.log((1+m),math.e))**2)\n",
-      "SNR1=63095.73              #SNR=38+10\n",
-      "Q1=math.sqrt((SNR1*((math.log(256,math.e))**2))/3)\n",
-      "N1=math.ceil(math.log(Q1,2))\n",
-      "r=N1*fs2\n",
-      "Bw=1.0/2*r\n",
-      "\n",
-      "#Result\n",
-      "print('i) Number of bits/word = %.2f ' %N1)\n",
-      "print('ii) Bit rate = %.1f Mb/s' %r)\n",
-      "print('iii) Bandwidth = %.1f Mb/s' %Bw)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Number of bits/word = 10.00 \n",
-        "ii) Bit rate = 24.0 Mb/s\n",
-        "iii) Bandwidth = 12.0 Mb/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 154
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo4.ipynb b/Digital_Communications/ChapterNo4.ipynb
deleted file mode 100755
index 547a969e..00000000
--- a/Digital_Communications/ChapterNo4.ipynb
+++ /dev/null
@@ -1,273 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 4: DIGITAL MULTIPLEXERS<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4, page no 190 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Caption: Sampling Rate\n",
-      "\n",
-      "#initialisation of variables\n",
-      "f1=4.0*10**3\n",
-      "f2=4.5*10**3\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "fsmin=(2*f2)/1000\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\" Sampling rate =%.2f kHz\" %fsmin)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Sampling rate =9.00 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5, Page No 190 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Caption: Nyquest rate,Signaling rate,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "f1=3000.0\n",
-      "f4=1000.0\n",
-      "f2=1000.0\n",
-      "f3=1000.0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "#Nyquest rate\n",
-      "nq1=2*f1\n",
-      "nq2=2*f2\n",
-      "nq3=2*f3\n",
-      "nq4=2*f4\n",
-      "\n",
-      "#RESULTS\n",
-      "print('i. Nyquest rate of x1 =%.2fkHz' %nq1)\n",
-      "print('ii. Nyquest rate of x2 =%.2fkHz' %nq2)\n",
-      "print('iii. Nyquest rate of x3 =%.2fkHz' %nq3)\n",
-      "print('iv. Nyquest rate of x4 =%.2fkHz' %nq4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i. Nyquest rate of x1 =6000.00kHz\n",
-        "ii. Nyquest rate of x2 =2000.00kHz\n",
-        "iii. Nyquest rate of x3 =2000.00kHz\n",
-        "iv. Nyquest rate of x4 =2000.00kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6, Page No 190 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find The spacing between two successive pules\n",
-      "\n",
-      "#initialisation of variables\n",
-      "samplingrate=8000.0\n",
-      "totalsignals=24+1\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "t=1/samplingrate\n",
-      "T=t/totalsignals\n",
-      "T=T*10**6            #time is now u sec\n",
-      "space=T-1\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The spacing between two successive pules %.f u sec' %space)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing between two successive pules 4 u sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7, Page No 191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find signaling rate ,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N=6.0\n",
-      "fm=5000.0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r=2*fm   #sampling rate\n",
-      "sr=N*r   #signalingrate\n",
-      "BW=N*fm  #Bandwith\n",
-      "sr=sr/1000\n",
-      "BW=BW/1000\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Signaling rate %.f K bits/sec' %sr)\n",
-      "print('Bandwith to avoid the cross talk in TDM is %.f kHz' %BW)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signaling rate 60 K bits/sec\n",
-        "Bandwith to avoid the cross talk in TDM is 30 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9, Page No 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find signaling rate ,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N=4.0\n",
-      "x=24              #bits/frame\n",
-      "D=1152\n",
-      "s=1.0/3\n",
-      "r=1.544*10**6\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r0=N*r*((D+x)/(D-s*N))\n",
-      "\n",
-      "#RESULTS\n",
-      "print('Output bit rate %.2f MB/sec' %(r0/(10**6)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output bit rate 6.31 MB/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.10, Page No 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find signaling rate ,bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "N=24.0\n",
-      "x=72.0              #bits\n",
-      "D=192             #message and stuff bits\n",
-      "s=1.0/3\n",
-      "r=64              #kb/sec\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "r0=N*r*((D+x)/(D-s*N))\n",
-      "r1=((N*r)/r0)*100\n",
-      "\n",
-      "#RESULTS\n",
-      "print('i) Output bit rate %.2f MB/sec' %(r0/(10**3)))\n",
-      "print('iI) Throughput efficiency is =  %.2f percent' %r1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) Output bit rate 2.20 MB/sec\n",
-        "iI) Throughput efficiency is =  69.70 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo5.ipynb b/Digital_Communications/ChapterNo5.ipynb
deleted file mode 100755
index b36df857..00000000
--- a/Digital_Communications/ChapterNo5.ipynb
+++ /dev/null
@@ -1,105 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 5: DIGTIAL BASEBAND TRANSMISSION<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12, Page No 249 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find The roll of factor alpha\n",
-      "\n",
-      "#initialisation of variables\n",
-      "datarate=0.1*10**6\n",
-      "fB=75000.0 #bandwidth\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Tb=1/datarate\n",
-      "alpha=2*fB*Tb-1\n",
-      "\n",
-      "#RESULTS\n",
-      "print('The roll out factor alpha =  = %.1f ' %alpha)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The roll out factor alpha =  = 0.5 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.13, Page No 249"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find The Transmission bandwidth\n",
-      "\n",
-      "#initialisation of variables\n",
-      "q=128.0\n",
-      "alpha=0.2\n",
-      "n=math.log(q,2)\n",
-      "fm=2000.0\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Nq=2*fm\n",
-      "fs=1.25*Nq\n",
-      "N=8.0\n",
-      "total=N*fs\n",
-      "bitrate=7*total\n",
-      "fB=((1+alpha)*bitrate)/2\n",
-      "fB=fB/1000\n",
-      "#RESULTS\n",
-      "print('The minimum transmission Bandwidth required = %.f kHz' %fB)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum transmission Bandwidth required = 168 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo6.ipynb b/Digital_Communications/ChapterNo6.ipynb
deleted file mode 100755
index 83113b9c..00000000
--- a/Digital_Communications/ChapterNo6.ipynb
+++ /dev/null
@@ -1,158 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 6: DIGITAL MODULATION TECHNIQUES<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page No 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "#Symbol rate of transmission and bandwidth \n",
-      "\n",
-      "#initialisation of variables\n",
-      "fb=1.0                #MHz\n",
-      "Tb=1/fb               #bit period ms\n",
-      "N=4\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Ts1=Tb\n",
-      "Bw1=2*fb\n",
-      "Ts2=Tb*2\n",
-      "Tr2=(1.0/2)*1000          #Symbols/sec\n",
-      "Bw2=1/Tb\n",
-      "Ts3=N*Tb\n",
-      "Tr3=(1.0/4)*1000\n",
-      "Bw3=(2.0*fb)/N\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"In BPSK\")\n",
-      "print(\"The symbol rate of transmission is = %.f X 10^6 symbol/sec \" %Ts1)\n",
-      "print(\"The Bandwidth is = %.f MHz \" %Bw1)\n",
-      "print(\"In QPSK\")\n",
-      "print(\"The symbol rate of transmission is = %.f X 10^3 symbol/sec \" %Tr2)\n",
-      "print(\"The Bandwidth is = %.f MHz \" %Bw2)\n",
-      "print(\"In 16-ary PSK\")\n",
-      "print(\"The symbol rate of transmission is = %.f X 10^3 symbol/sec \" %Tr3)\n",
-      "print(\"The Bandwidth is = %.f kHz \" %(Bw3*1000))\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In BPSK\n",
-        "The symbol rate of transmission is = 1 X 10^6 symbol/sec \n",
-        "The Bandwidth is = 2 MHz \n",
-        "In QPSK\n",
-        "The symbol rate of transmission is = 500 X 10^3 symbol/sec \n",
-        "The Bandwidth is = 1 MHz \n",
-        "In 16-ary PSK\n",
-        "The symbol rate of transmission is = 250 X 10^3 symbol/sec \n",
-        "The Bandwidth is = 500 kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page No 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "fb=90              #MHz\n",
-      "Bw=20              #Mb/s\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "N=(2*fb)/Bw\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"Number of bits required = %.f \" %N)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of bits required = 9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page No 357"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Tb=0.1          #msec\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "fb=1/Tb\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"Transmission bandwidth = %.f kHz\" %fb)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transmission bandwidth = 10 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo7.ipynb b/Digital_Communications/ChapterNo7.ipynb
deleted file mode 100755
index 9d58a722..00000000
--- a/Digital_Communications/ChapterNo7.ipynb
+++ /dev/null
@@ -1,431 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 7: PERFORMANCE ANALYSIS OF DIGITAL COMMUNICATION SYSTEMS<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.15, Page No 414"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "T=1.0*10**-6             #ms\n",
-      "A=10.0*10**-3\n",
-      "N=1*10**-11\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Eb=((A**2)*T)/2.0\n",
-      "Epb=math.sqrt(Eb/N)\n",
-      "Pb=(1.0/2)*math.erfc(Epb)*(1.0/4)\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"Bit error probability = %.2f X 10^-4 \" %(Pb*(10**4)))\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bit error probability = 1.96 X 10^-4 \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.16, Page No 420"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Br=25*10**6             #bits/sec\n",
-      "N=2*10**-20             #Power spectral density of noise\n",
-      "A=1.0*10**-6\n",
-      "Ps=1*10**-12            # Normalized power\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Tb=1.0/Br\n",
-      "Ps=(A**2)/2.0\n",
-      "Pe=1.0/2*(math.erfc((0.6*Ps*Tb)/N))\n",
-      "#RESULTS\n",
-      "print(\"Average probability of bit error is Pe= %.2f X 10^-4 \" %(Pe*10))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average probability of bit error is Pe= 1.98 X 10^-4 \n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.17, Page No 423"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find probability of error\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=2.0\n",
-      "N0=2.0*10**-4  #power sepctral density\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Tmin=(6.76*N0)/4\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The min time for which the signal should be extended is Tmin= %.3f X 10^-3 msec \" %(Tmin*10**3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The min time for which the signal should be extended is Tmin= 0.338 X 10^-3 msec \n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.20, Page No 428"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#The average carrier\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Pe=10**-4\n",
-      "N0=2*10**10\n",
-      "T=1.0/10**6\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "E=6.76*N0\n",
-      "P=E/T\n",
-      "Eb=8.5171*N0\n",
-      "P2=Eb/T\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The average carrier = %.3f X 10^-3 mW \" %(P2/10**17))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average carrier = 1.703 X 10^-3 mW \n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.21, Page No 430"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#The average carrier\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=0.2*10**-3          #V\n",
-      "T=2*10**-6\n",
-      "N0=2*10**-15\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Pe=(math.sqrt((A**2*T)/4*N0))\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The average carrier = %.3f X 10^-4 mW \" %(Pe*10**15))\n",
-      "print(\"Approximation error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average carrier = 6.325 X 10^-4 mW \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.29, Page No 447"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=1.0\n",
-      "N0=2.0*10**-5  #power sepctral density\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "T=((3.71**2)*2*N0)/(A**2)\n",
-      "Br=1/T\n",
-      "#Results\n",
-      "print(\"Bit rate Br= %.2f k bits/sec\" %(Br/(10**3)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bit rate Br= 1.82 k bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.31, Page No 448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Pe=10.0**-4 #probability of error of PSK\n",
-      "N0=2.0*10**-10\n",
-      "Br=10**6                  #Bit Rate\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "T=1.0/Br\n",
-      "E=N0*6.76\n",
-      "P=E/T\n",
-      "Eb=8.5171*N0\n",
-      "Pb=Eb/T\n",
-      "#Results\n",
-      "print(\"i) PSK system The required carrier power = %.2f mW\" %(P*10**3))\n",
-      "print(\"ii) DPSK system The required carrier power = %.2f mW\" %(Pb*10**3))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) PSK system The required carrier power = 1.35 mW\n",
-        "ii) DPSK system The required carrier power = 1.70 mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.32, Page No 450"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "A=0.2*10**-3                #V\n",
-      "T=2*10**-6                  #Sec\n",
-      "N0=2*10**-15\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "x=math.sqrt((A**2*T)/(4*N0))\n",
-      "Pe=(1.0/2)*math.erfc(x/math.sqrt(2.0))\n",
-      "\n",
-      "#Results\n",
-      "print(\"Probability of error = %.2f X 10^-4\" %(Pe*10**4))\n",
-      "#print(\"Channel Bandwidth  = %.5f MHz\" %BW)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Probability of error = 7.83 X 10^-4\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.36, Page No 453"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Ap=1                #mV\n",
-      "En=192.3            #mV\n",
-      "Arms=707            #mV\n",
-      "\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Z=Arms/En\n",
-      "Pe=1.0/2*(math.erfc(math.sqrt(2)*(Z/math.sqrt(2))))\n",
-      "\n",
-      "#Results\n",
-      "print(\"Probability of error is  %.7f \" %Pe)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Probability of error is  0.0000001 \n"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.37, Page No 454"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "Br=2.08*10**-6        #Bit Rate\n",
-      "N0=2*10**-8\n",
-      "\n",
-      "#CALCULATIONS\n",
-      "Eb=(math.sqrt(3.3))/(5*10**7)   #Eb=bit energy\n",
-      "Ps=Eb*Br\n",
-      "BW=2*Br                  #Bandwidth\n",
-      "Es=286.39*N0\n",
-      "Ps2=(Es*Br)/2\n",
-      "BW2=(2*Br)/4\n",
-      "\n",
-      "#Results\n",
-      "print(\"i) BPSK THe transmission bandwidth  = %.2f MHz\" %(BW*10**6)) \n",
-      "print(\"ii) For 16 ary PSK THe transmission bandwidth  = %.2f MHz\" %(BW2*10**6)) \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) BPSK THe transmission bandwidth  = 4.16 MHz\n",
-        "ii) For 16 ary PSK THe transmission bandwidth  = 1.04 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo8.ipynb b/Digital_Communications/ChapterNo8.ipynb
deleted file mode 100755
index d34d983c..00000000
--- a/Digital_Communications/ChapterNo8.ipynb
+++ /dev/null
@@ -1,1439 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 8: INFORMATION THEORY<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page No 464"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "#Find Information Content of Each Symbol\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/8.0\n",
-      "\n",
-      "#Calculation\n",
-      "#information content of each symbol\n",
-      "Ix1=math.log(1/px1,2)\n",
-      "Ix2=math.log(1/px2,2)\n",
-      "Ix3=math.log(1/px3,2)\n",
-      "Ix4=math.log(1/px4,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"Information Content tI(x1)= %.2f bit\" %Ix1)\n",
-      "print(\" tI(x2)= %.f bits\" %Ix2)\n",
-      "print(\" tI(x3)= %.f bits\" %Ix3)\n",
-      "print(\" tI(x4)= %.f bits\" %Ix4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Information Content tI(x1)= 1.00 bit\n",
-        " tI(x2)= 2 bits\n",
-        " tI(x3)= 3 bits\n",
-        " tI(x4)= 3 bits\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2, Page No 464"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find amount of Information\n",
-      "#Variable Declaration\n",
-      "#Calculation\n",
-      "pxi=1/4.0\n",
-      "Ixi=(math.log10(1/pxi))/math.log10(2)\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"The amount of Information I(Xi)= %.f \" %Ixi)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of Information I(Xi)= 2 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3, Page No 464"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Amount of Information\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/2.0\n",
-      "\n",
-      "#Calculation\n",
-      "Ix1=math.log(1/px1,2) #entropy\n",
-      "Ix2=math.log(1/px2,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The amount of Information I(X1)= %.f bit\" %Ix1)\n",
-      "print(\"The amount of Information I(X2)= %.f bit\" %Ix2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of Information I(X1)= 1 bit\n",
-        "The amount of Information I(X2)= 1 bit\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4, Page No 465"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Amount of Information \n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/4.0\n",
-      "px2=3/4.0\n",
-      "\n",
-      "#Calculation\n",
-      "Ix1=math.log(1/px1,2)\n",
-      "Ix2=math.log(1/px2,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The amount of Information I(X1)= %.f bit\" %Ix1)\n",
-      "print(\"The amount of Information I(X2)= %.2f bit\" %Ix2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of Information I(X1)= 2 bit\n",
-        "The amount of Information I(X2)= 0.42 bit\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9, Page No 468 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Entropy,Amount of information\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.4\n",
-      "px2=0.3\n",
-      "px3=0.2\n",
-      "px4=0.1\n",
-      "\n",
-      "#Calculation\n",
-      "HX=-px1*math.log(px1,2)-px2*math.log(px2,2)-px3*math.log(px3,2)-px4*math.log(px4,2)\n",
-      "Px1x2x1x3=px1*px2*px1*px3\n",
-      "Ix1x2x1x3=-math.log(Px1x2x1x3,2)\n",
-      "Px4x3x3x2=px4*px3*px3*px2\n",
-      "Ix4x3x3x2=-math.log(Px4x3x3x2,2)\n",
-      "\n",
-      "#Result\n",
-      "print(\" \\n Entropy  H(X) = %.2f bits/symbol \" %HX)\n",
-      "print(\"The amount of Information I(x1x2x1x3)= %.2f bits/symbol\" %Ix1x2x1x3)\n",
-      "print(\" I(x4x3x3x2) = %.2f bits/symbol \" %Ix4x3x3x2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Entropy  H(X) = 1.85 bits/symbol \n",
-        "The amount of Information I(x1x2x1x3)= 6.70 bits/symbol\n",
-        " I(x4x3x3x2) = 9.70 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13, Page No 471"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Find information rate the telegraphic source\n",
-      "\n",
-      "#Variable Declaration\n",
-      "pdash=1/3.0\n",
-      "pdot=2/3.0\n",
-      "tdot=0.2\n",
-      "tdash=0.6\n",
-      "tspace=0.2\n",
-      "\n",
-      "#Calculation\n",
-      "HX=-pdash*math.log(pdash,2)-pdot*math.log(pdot,2)\n",
-      "Ts=pdot*tdot+pdash*tdash+tspace\n",
-      "r=1/Ts\n",
-      "R=r*HX\n",
-      "\n",
-      "#Result\n",
-      "print('Average rate of information R = %.2f bits/s' %R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average rate of information R = 1.72 bits/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14, Page No 471"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find information rate of the source\n",
-      "\n",
-      "f=input('Enter the frequncy f=')\n",
-      "px1=1/8.0\n",
-      "px2=1/8.0\n",
-      "px3=3/8.0\n",
-      "px4=3/8.0\n",
-      "\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2) #entropy of the source\n",
-      "R=2*f*HX #r=2*f\n",
-      "print('information rate  R=  %.1f  bits/sec ' %R)   #f=signal bandwidth\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the frequncy f=34\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "information rate  R=  123.2  bits/sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15, Page No 472"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find information rate of the source\n",
-      "#all symbols are equally likely\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/2.0\n",
-      "px3=1/2.0\n",
-      "px4=1/2.0\n",
-      "\n",
-      "#Calculation\n",
-      "f=input('Enter the frequncy of system fm(in Hz) =')\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)\n",
-      "\n",
-      "#Result\n",
-      "print('\\n Entropy H(X) =%.f bits/symbol ' %HX)\n",
-      "R=2*f*HX\n",
-      "print('information rate =%.f bits/sec' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the frequncy of system fm(in Hz) =45\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Entropy H(X) =2 bits/symbol \n",
-        "information rate =180 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.16, Page No 473"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find source entropy ,information rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "#probability symbols\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/16.0\n",
-      "px5=1/16.0\n",
-      "Tb=10.0**-3\n",
-      "\n",
-      "#Calculation\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4)+px5*math.log(1/px5)\n",
-      "\n",
-      "#Result\n",
-      "print('1. source entropy H(X) = %.2f bits/symbol ' %HX) #source entropy\n",
-      "r=1.0/Tb\n",
-      "R=r*HX #information rate\n",
-      "print(' 2. Information rate R = %.2f bits/sec ' %R)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. source entropy H(X) = 1.72 bits/symbol \n",
-        " 2. Information rate R = 1721.57 bits/sec \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17, Page No 473"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume if there are 16 outcomes per second\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/16.0\n",
-      "px5=1/16.0\n",
-      "r=16.0\n",
-      "\n",
-      "#Calculation\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)+px5*math.log(1/px5,2)\n",
-      "\n",
-      "#Result\n",
-      "print('1. Entropy H(X) = %.2f bits/symbol ' %HX) #source entropy\n",
-      "\n",
-      "R=r*HX\n",
-      "print('2., Information rate R = %.f bits/sec' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. Entropy H(X) = 1.88 bits/symbol \n",
-        "2., Information rate R = 30 bits/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18, Page No 474"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#determine entropy ,information rate\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/4.0\n",
-      "px2=1/5.0\n",
-      "px3=1/5.0\n",
-      "px4=1/10.0\n",
-      "px5=1/10.0\n",
-      "px6=1/20.0\n",
-      "px7=1/20.0\n",
-      "px8=1/20.0\n",
-      "f=10*10**3.0\n",
-      "fs=10*2*10**3.0\n",
-      "\n",
-      "#Calculation\n",
-      "#entropy\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)+px5*math.log(1/px5,2)+px6*math.log(1/px6,2)+px7*math.log(1/px7,2)+px8*math.log(1/px8,2) \n",
-      "\n",
-      "#Result\n",
-      "print('bits/message H(X) = %.2f ' %HX)\n",
-      "r=fs\n",
-      "R=r*HX    #information rate\n",
-      "print('bits/sec R = %.2f' %R)\n",
-      "print('Approximation error')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "bits/message H(X) = 2.74 \n",
-        "bits/sec R = 54828.92\n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19, Page No 476 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from array import *\n",
-      "#Find Channel Matrix,joint probability\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.5\n",
-      "px2=0.5\n",
-      "py1x1=0.9\n",
-      "py2x1=0.1\n",
-      "py1x2=0.2\n",
-      "py2x2=0.8\n",
-      "PYX=[[py1x1,py2x1],[py1x2,py2x2]]\n",
-      "PX=[[px1,px2]]\n",
-      "PY = [[0,0],\n",
-      "    [0,0]]\n",
-      "PXY = [[0,0],\n",
-      "    [0,0]]\n",
-      "\n",
-      "for i in range(len(PYX)):\n",
-      "   # iterate through columns of Y\n",
-      "   for j in range(len(PX[0])):\n",
-      "       # iterate through rows of Y\n",
-      "       for k in range(len(PX)):\n",
-      "           PY[i][j] += PYX[i][k] * PX[k][j]\n",
-      "print('  PY ARRAY =  \\n')\n",
-      "for r in PY:\n",
-      "   print(r)\n",
-      "PXd=[[px1,0],[0,px2]]\n",
-      "\n",
-      "\n",
-      "for i in range(len(PXd)):\n",
-      "   # iterate through columns of Y\n",
-      "   for j in range(len(PYX[0])):\n",
-      "       # iterate through rows of Y\n",
-      "       for k in range(len(PYX)):\n",
-      "           PXY[i][j] += PXd[i][k] * PYX[k][j]\n",
-      "\n",
-      "            \n",
-      "print(' \\n PXY ARRAY =  \\n')\n",
-      "for r in PXY:\n",
-      "   print(r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  PY ARRAY =  \n",
-        "\n",
-        "[0.45, 0.45]\n",
-        "[0.1, 0.1]\n",
-        " \n",
-        " PXY ARRAY =  \n",
-        "\n",
-        "[0.45, 0.05]\n",
-        "[0.1, 0.4]\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.35, Page No 498"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Channel is aproximated by the AWGN Channel\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=4000.0\n",
-      "S=0.1*10**-3\n",
-      "n=2*10**-12\n",
-      "\n",
-      "#Calculation\n",
-      "N=n*B\n",
-      "C=B*math.log(1+(S/N),2) #Capacity of Channel\n",
-      "C=C/1000.0\n",
-      "#Result\n",
-      "print(' Capacity of Channel  C=%.3f(10^3) b/s ' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Capacity of Channel  C=54.439(10^3) b/s \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36i, Page No 499"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "fm=4000.0\n",
-      "fs=2*fm\n",
-      "n=1.25\n",
-      "\n",
-      "#Calculation\n",
-      "r=fs*n\n",
-      "pxi=1/256.0\n",
-      "HX=-math.log(pxi,2)\n",
-      "R=r*HX\n",
-      "R=R/1000\n",
-      "print('Information Rate  R= %.f kb/s' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Information Rate  R= 80 kb/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36ii, Page No 499"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=10*10**3.0\n",
-      "SN=20.0\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=10**(SN/10.0)\n",
-      "C=B*math.log(1+(SNR),2)\n",
-      "C=C/1000\n",
-      "\n",
-      "#Result\n",
-      "print('The channel capacity = %.2f 10^3 b/s' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The channel capacity = 66.58 10^3 b/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36iii, Page No 499"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "C=8*10**4.0\n",
-      "B=10**4.0\n",
-      "\n",
-      "#Calculation\n",
-      "SN=2**(C/B)-1\n",
-      "SNR=10*math.log(SN,10) #SNR\n",
-      "\n",
-      "#Result\n",
-      "print(' The S/N ratio required for error-free transmission  =%.2f dB ' %SNR)     #required SNR is greater that 24.064\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The S/N ratio required for error-free transmission  =24.07 dB \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36iv, Page No 499 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#assume that  succeissive samples are statistically independent\n",
-      "\n",
-      "#Variable Declaration\n",
-      "SN=20.0\n",
-      "\n",
-      "#Calculation\n",
-      "SNR=10**(SN/10.0)\n",
-      "C=8*10**4.0\n",
-      "B=C/(math.log(1+SNR,2)) #Bandwidth\n",
-      "B=B/1000\n",
-      "\n",
-      "#Result\n",
-      "print('Bandwidth required for AWGN channel B =%.2f kHz ' %B)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bandwidth required for AWGN channel B =12.02 kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.37, Page No 502"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find code efficiency,redundancy\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.9\n",
-      "px2=0.1\n",
-      "n1=1.0\n",
-      "n2=1.0\n",
-      "\n",
-      "#Calculation\n",
-      "L=px1*n1+px2*n2 #code leght\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)\n",
-      "n=(HX/L) #code efficiency\n",
-      "n=n*100\n",
-      "\n",
-      "print('Code efficiency = %.1f percent'  %n)\n",
-      "r=(100-n) #code reduncy\n",
-      "print('Code redundancy = %.1f percent' %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Code efficiency = 46.9 percent\n",
-        "Code redundancy = 53.1 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.38, Page No 502"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find code efficiency,redundancy\n",
-      "\n",
-      "#Variable Declaration\n",
-      "pa1=0.81\n",
-      "pa2=0.09 \n",
-      "pa3=0.09\n",
-      "pa4=0.01 \n",
-      "n1=1\n",
-      "n2=2 \n",
-      "n3=3\n",
-      "n4=3 \n",
-      "\n",
-      "#Calculation\n",
-      "L=pa1*n1+pa2*n2+pa3*n3+pa4*n4\n",
-      "HX2=pa1*math.log(1/pa1,2)+pa2*math.log(1/pa2,2)+pa3*math.log(1/pa3,2)+pa4*math.log(1/pa4,2)\n",
-      "n=HX2/L*100\n",
-      "\n",
-      "#Result\n",
-      "print(' code efficiency = %.2f  percent' %n)\n",
-      "\n",
-      "r=(100-n) #code reduncy\n",
-      "print(' code redundancy = %.1f percent' %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " code efficiency = 72.71  percent\n",
-        " code redundancy = 27.3 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.44, Page No 507"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find  efficiency of the code\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=1/2.0\n",
-      "px2=1/4.0\n",
-      "px3=1/8.0\n",
-      "px4=1/8.0\n",
-      "n1=1.0\n",
-      "n2=2.0\n",
-      "n3=3.0\n",
-      "n4=3.0\n",
-      "\n",
-      "#Calculation\n",
-      "#information content of each symbol\n",
-      "Ix1=-math.log(px1,2)\n",
-      "Ix2=-math.log(px2,2)\n",
-      "Ix3=-math.log(px3,2)\n",
-      "Ix4=-math.log(px4,2)\n",
-      "\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)\n",
-      "L=px1*n1+px2*n2+px3*n3+px4*n4\n",
-      "\n",
-      "n=HX/L*100\n",
-      "\n",
-      "#Result\n",
-      "print('Ccode efficiency = %.f  Percent' %n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ccode efficiency = 100  Percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.50, Page No 512"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy ,information rate\n",
-      "#If there are 16 outcomes per second\n",
-      "\n",
-      "#Variable Declaration\n",
-      "P1=1/2.0\n",
-      "P2=1/4.0\n",
-      "P3=1/8.0\n",
-      "P4=1/16.0\n",
-      "P5=1/32.0\n",
-      "P6=1/32.0\n",
-      "r=16     #message rate\n",
-      "\n",
-      "#Calculation\n",
-      "H=P1*math.log(1/P1,2)+P2*math.log(1/P2,2)+P3*math.log(1/P3,2)+P4*math.log(1/P4,2)+P5*math.log(1/P5,2)+P6*math.log(1/P6,2)\n",
-      "#Entropy of system\n",
-      "\n",
-      "#Result\n",
-      "print('1. Entropy of system H = %.2f bits/message ' %H)\n",
-      "R=H*r                            #R=Entropy*message rate\n",
-      "print(' 2. Information rate  R = %.f bits/sec ' %R)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. Entropy of system H = 1.94 bits/message \n",
-        " 2. Information rate  R = 31 bits/sec \n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.51, Page No 512"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Calculate H(X) ,H(Y)\n",
-      "\n",
-      "#Variable Declaration\n",
-      "px1=0.3\n",
-      "px2=0.4\n",
-      "px3=0.3\n",
-      "\n",
-      "#Calculation\n",
-      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)  #Entropy of X\n",
-      "\n",
-      "\n",
-      "print(' 1.Entropy of X  H(X)=%.3f bits/symbol ' %HX)\n",
-      "\n",
-      "PYX=[[0.8, 0.2, 0],[ 0, 1, 0],[ 0, 0.3, 0.7]]\n",
-      "PX=[[px1, px2, px3]]\n",
-      "PXY = [[0,0,0],\n",
-      "    [0,0,0],\n",
-      "    [0,0,0]]\n",
-      "\n",
-      "for i in range(len(PYX)):\n",
-      "   # iterate through columns of PXd\n",
-      "   for j in range(len(PX[0])):\n",
-      "       # iterate through rows of PYX\n",
-      "       for k in range(len(PX)):\n",
-      "           PXY[i][j] += PYX[i][k] * PX[k][j]\n",
-      "\n",
-      "py1=PXY[0][0]\n",
-      "py2=PXY[0][1]\n",
-      "py3=PXY[0][2]\n",
-      "HY=py1*math.log(1/py1,2)+py2*math.log(1/py2,2)+py3*math.log(1/py3,2)  #Entropy of Y\n",
-      "print(' 2. Entropy of Y H(Y)= %.2f bits/symbol ' %HY)\n",
-      "print('Approximation error')\t\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1.Entropy of X  H(X)=1.571 bits/symbol \n",
-        " 2. Entropy of Y H(Y)= 1.51 bits/symbol \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.52, Page No 513"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source ,entropy of second order extension\n",
-      "\n",
-      "#Variable Declaration\n",
-      "P1=0.7\n",
-      "P2=0.15\n",
-      "P3=0.15\n",
-      "\n",
-      "#Calculation\n",
-      "HX=P1*math.log(1/P1,2)+P2*math.log(1/P2,2)+P3*math.log(1/P3,2)     #Entropy of source\n",
-      "print(' 1. Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n",
-      "#H(X^n)=n*H(X)\n",
-      "n=2    #for second order\n",
-      "HX2=n*HX\n",
-      "\n",
-      "#Result\n",
-      "print(' 2. Entropy of second order system extension of source can be H(X^2)=%.2f bits/symbol ' %(HX*2))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1. Entropy of system  H(X)=1.18 bits/symbol \n",
-        " 2. Entropy of second order system extension of source can be H(X^2)=2.36 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.54, Page No 514"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "S0=1/3.0\n",
-      "S1=1/6.0\n",
-      "S2=1/4.0\n",
-      "S3=1/4.0\n",
-      "\n",
-      "#Calculation\n",
-      "HX=S0*math.log(1/S0,2)+S1*math.log(1/S1,2)+S2*math.log(1/S2,2)+S3*math.log(1/S3,2)   #EntroSy of source\n",
-      "\n",
-      "#Result\n",
-      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Entropy of system  H(X)=1.96 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.56, Page No 515"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Information capacity of telephone\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=3.4*10**3\n",
-      "SNR=30.0\n",
-      "\n",
-      "#Calculation\n",
-      "SN=10**(SNR/10)\n",
-      "C=B*math.log(1+SN,2)   #Information capacity\n",
-      "C=C/1000\n",
-      "\n",
-      "#Result\n",
-      "print(' Information capacity of telephone is C = %.2f kbps ' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Information capacity of telephone is C = 33.89 kbps \n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.59, Page No 516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "S0=1/3.0\n",
-      "S1=1/6.0\n",
-      "S2=1/4.0\n",
-      "S3=1/4.0\n",
-      "\n",
-      "#Calculation\n",
-      "HX=S0*math.log(1/S0,2)+S1*math.log(1/S1,2)+S2*math.log(1/S2,2)+S3*math.log(1/S3,2)   #EntroSy of source\n",
-      "\n",
-      "#Result\n",
-      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Entropy of system  H(X)=1.96 bits/symbol \n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.60, Page No 516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "m1=1/2.0\n",
-      "m2=1/4.0\n",
-      "m3=1/8.0\n",
-      "m4=1/16.0\n",
-      "m5=1/16.0\n",
-      "\n",
-      "#Calculation\n",
-      "L=(m1*1)+(m2*2)+(m3*3)+(2*(m4)*4)\n",
-      "\n",
-      "#Result\n",
-      "print(' Average number of bits per message =%.2f bits ' %L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Average number of bits per message =1.88 bits \n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.61, Page No 517"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find Information capacity of telephone\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=3.4*10**3\n",
-      "SNR=30.0\n",
-      "\n",
-      "#Calculation\n",
-      "SN=10**(SNR/10)\n",
-      "C=B*math.log(1+SN,2)   #Information capacity\n",
-      "C=C/1000\n",
-      "\n",
-      "#Result\n",
-      "print(' Information capacity of telephone is C = %.2f kbps ' %C)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Information capacity of telephone is C = 33.89 kbps \n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.62, Page No 517"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "p1=4.0\n",
-      "m1=0.5\n",
-      "m2=0.5\n",
-      "m3=0.375\n",
-      "m4=0.375\n",
-      "m5=0.375\n",
-      "m6=0.375\n",
-      "\n",
-      "#Calculation\n",
-      "I1=p1*math.log(1/p1,2) \n",
-      "HX=m1*math.log(1/m1,2)+m2*math.log(1/m2,2)+m3*math.log(1/m3,2)+m4*math.log(1/m4,2)+m5*math.log(1/m5,2)+m6*math.log(1/m6,2)   #EntroSy of source\n",
-      "\n",
-      "#Result\n",
-      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n",
-      "print('Approximation error')\t"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Entropy of system  H(X)=3.12 bits/symbol \n",
-        "Approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.65, Page No 519"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find entropy of source \n",
-      "\n",
-      "#Variable Declaration\n",
-      "S0=1/2.0\n",
-      "S1=1/4.0\n",
-      "S2=1/8.0\n",
-      "S3=1/8.0\n",
-      "n=1\n",
-      "\n",
-      "#Calculation\n",
-      "H=S0*math.log(1.0/S0,2)+S1*math.log(1.0/S1,2)+S2*math.log(1.0/S2,2)+S3*math.log(1.0/S3,2)   #EntroSy of source\n",
-      "L=H*n\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(' Code length =%.2f bits/messages ' %L)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Code length =1.75 bits/messages \n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.67, Page No 520"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#Find channel capacity and new bandwidth\n",
-      "\n",
-      "#Variable Declaration\n",
-      "B=8*10**3\n",
-      "SNR=31.0\n",
-      "SNR2=61\n",
-      "\n",
-      "#Calculation\n",
-      "C=B*math.log(1+SNR,2)   #Information capacity\n",
-      "B2=C/math.log(1+SNR2,2)\n",
-      "#Result\n",
-      "print(' Channel capacity is C = %.2f x 10^3 bits/sec ' %(C/1000))\n",
-      "print(' New Bandwidth is C = %.2f x kHz ' %(B2/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Channel capacity is C = 40.00 x 10^3 bits/sec \n",
-        " New Bandwidth is C = 6.72 x kHz \n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/ChapterNo9.ipynb b/Digital_Communications/ChapterNo9.ipynb
deleted file mode 100755
index 8062105c..00000000
--- a/Digital_Communications/ChapterNo9.ipynb
+++ /dev/null
@@ -1,182 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1> Chpater No 9: ERROR CONTROL CODING<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4, Page No 540"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find  detected errors,corrected errors\n",
-      "\n",
-      "#initialisation of variables\n",
-      "dmin=5.0\n",
-      "#(s+1)<= dmin number errors can be detected(s)\n",
-      " \n",
-      "#CALCULATIONS\n",
-      "s=dmin-1\n",
-      "\n",
-      "#RESULTS\n",
-      "print(' i)Number of detected errors  s <=  %.f ' %s)\n",
-      "#(2t+1)<=dmin number errors can be corrected(t)\n",
-      "t=(dmin-1)/2.0\n",
-      "print('ii) Number of corrected errors  t<=  %.f ' %t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i)Number of detected errors  s <=  4 \n",
-        "ii) Number of corrected errors  t<=  2 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17, Page No 569"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Determine all possible code vectors \n",
-      "\n",
-      "m3=1\n",
-      "m2=0\n",
-      "m1=1\n",
-      "m0=0\n",
-      "#M=Message Matrix\n",
-      "#G=Generator Matrix\n",
-      "G=[[1, 0, 1, 1, 0, 0, 0],[0, 1, 0, 1, 1, 0, 0],[0, 0, 1, 0, 1, 1, 0],[0, 0, 0, 1, 0, 1, 1]]\n",
-      "M=[[m3,m2,m1,m0]]\n",
-      "X = [[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0]]\n",
-      "\n",
-      "for i in range(len(G)):\n",
-      "   # iterate through columns of PXd\n",
-      "   for j in range(len(M[0])):\n",
-      "       # iterate through rows of PYX\n",
-      "       for k in range(len(M)):\n",
-      "           X[i][j] += G[i][k] * M[k][j]\n",
-      "print('The required code word')\n",
-      "for r in range(0,7):\n",
-      "         print(X[0][r])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required code word\n",
-        "1\n",
-        "0\n",
-        "1\n",
-        "0\n",
-        "0\n",
-        "0\n",
-        "0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19, Page No 572"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Determine  code word \n",
-      "m3=1\n",
-      "m2=0\n",
-      "m1=1\n",
-      "m0=0\n",
-      "#M=Message Matrix\n",
-      "#G=Generator Matrix\n",
-      "G=[[1, 0, 0, 0, 1, 0, 1],[0, 1, 0, 0, 1, 1, 1],[0, 0, 1, 0, 1, 1, 0],[0, 0, 0, 1, 0, 1, 1]]\n",
-      "M=[[m3,m2,m1,m0]]\n",
-      "X = [[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0]]\n",
-      "\n",
-      "\n",
-      "for i in range(len(G)):\n",
-      "   # iterate through columns of PXd\n",
-      "   for j in range(len(M[0])):\n",
-      "       # iterate through rows of PYX\n",
-      "       for k in range(len(M)):\n",
-      "           X[i][j] += G[i][k] * M[k][j]\n",
-      "print('The required code word')\n",
-      "for r in range(0,7):\n",
-      "         print(X[0][r] )\n",
-      "     \n",
-      "\n",
-      "print('The code in the book is wrong')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required code word\n",
-        "1\n",
-        "0\n",
-        "1\n",
-        "0\n",
-        "0\n",
-        "0\n",
-        "0\n",
-        "The code in the book is wrong\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Digital_Communications/README.txt b/Digital_Communications/README.txt
deleted file mode 100755
index cb9038a1..00000000
--- a/Digital_Communications/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Mayur Sabban
-Course: be
-College/Institute/Organization: Vishwakarma Institute of Technology Pune
-Department/Designation: Computer Engg
-Book Title: Digital Communications
-Author:  S. Sharma
-Publisher: S. K. Kataria & Sons, New Delhi
-Year of publication: 2013
-Isbn: 9789350142684
-Edition: 6
\ No newline at end of file
diff --git a/Digital_Communications/screenshots/Chapter1.png b/Digital_Communications/screenshots/Chapter1.png
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diff --git a/Digital_Communications/screenshots/chapter6.png b/Digital_Communications/screenshots/chapter6.png
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diff --git a/Digital_Communications/screenshots/chapter7.png b/Digital_Communications/screenshots/chapter7.png
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diff --git a/Digital_Communications_by_S._Sharma/Chapter2.ipynb b/Digital_Communications_by_S._Sharma/Chapter2.ipynb
new file mode 100755
index 00000000..9cc6d1df
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter2.ipynb
@@ -0,0 +1,290 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 2: SAMPLING THEORY AND PULSE MODULATION<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, page no 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find Nquist Rate\n",
+      "\n",
+      "#Variable declaration\n",
+      "#given \n",
+      "pi=3.14\n",
+      "w1=50*pi\n",
+      "w2=300*pi\n",
+      "w3=100*pi\n",
+      "#w=2*%pi*f\n",
+      "\n",
+      "#Calculation\n",
+      "f1=w1/(2*pi)\n",
+      "f2=w2/(2*pi)\n",
+      "f3=w3/(2*pi)\n",
+      "fm=f2 #fm = maximum frquency is present at the signal\n",
+      "\n",
+      "#Result\n",
+      "print('maximum frquency of the signal is = %.2f Hz' %f2)\n",
+      "fs=2*fm #Nyquist rate\n",
+      "print('Nquist Rate of Signal is = %.2f Hz' %fs)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum frquency of the signal is = 150.00 Hz\n",
+        "Nquist Rate of Signal is = 300.00 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 , page no 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Nquist Rate and Nquist time interval\n",
+      "\n",
+      "#Variable declaration\n",
+      "#given\n",
+      "w1=5000*math.pi\n",
+      "w2=3000*math.pi;\n",
+      "f1=w1/(2*math.pi);\n",
+      "f2=w2/(2*math.pi);\n",
+      "\n",
+      "#Calculation\n",
+      "fm=f1 #fm = maximum frquency is present at the signal\n",
+      "fs=2*fm #Nyquist rate\n",
+      "Ts=1.0/(2.0*fm) #frequncy =1/time\n",
+      "Ts=Ts*(10**3)\n",
+      "\n",
+      "#Result\n",
+      "print('maximum frquency of the signal is = %.f Hz' %f1)\n",
+      "print('Nquist Rate of the given Signal is = %.f Hz' %fs)\n",
+      "print('Nquist Interval of the given signal is = %.1f m Sec' %Ts)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum frquency of the signal is = 2500 Hz\n",
+        "Nquist Rate of the given Signal is = 5000 Hz\n",
+        "Nquist Interval of the given signal is = 0.2 m Sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3, page no 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find Nquist Rate \n",
+      "\n",
+      "#Variable declaration\n",
+      "#given\n",
+      "f=100.0         # Frequency component of continuous-time signal\n",
+      "\n",
+      "#Calculation\n",
+      "fs=2*f #Nyquist rate\n",
+      "\n",
+      "#Result\n",
+      "print('i) To avoid aliasing Nquist Rate is = %.f Hz' %fs)\n",
+      "print('ii) It is theoretical example ')\n",
+      "print('iii) It is theoretical example ')\n",
+      "print('iv) It is theoretical example ')\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) To avoid aliasing Nquist Rate is = 200 Hz\n",
+        "ii) It is theoretical example \n",
+        "iii) It is theoretical example \n",
+        "iv) It is theoretical example \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, page no 52 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Nquist Rate of Continous signal\n",
+      "\n",
+      "#Variable declaration\n",
+      "#given\n",
+      "w1=50*math.pi\n",
+      "w2=300*math.pi\n",
+      "w3=100*math.pi\n",
+      "\n",
+      "#Calculation\n",
+      "f1=w1/(2*math.pi)\n",
+      "f2=w2/(2*math.pi)\n",
+      "f3=w3/(2*math.pi)\n",
+      "fmax=f2 #fmax = Highest frquency component of the message signal\n",
+      "fs=2*fmax #Nyquist rate\n",
+      "\n",
+      "#Result\n",
+      "print('Highest frquency component of the message signal will be fmax = %.f Hz' %fmax)\n",
+      "print('Nquist Rate of the given Signal is = %.f Hz' %fs)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Highest frquency component of the message signal will be fmax = 150 Hz\n",
+        "Nquist Rate of the given Signal is = 300 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2.7, page no 67 <h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find amplitude distortion at highest frquency\n",
+      "\n",
+      "#Variable declaration\n",
+      "#given\n",
+      "fs=9.5  #samplig frequncy\n",
+      "fmax=1   #maximum frequncy\n",
+      "t=0.2 #pulse width\n",
+      "\n",
+      "#Calculation\n",
+      "c=3*10**8\n",
+      "f=fmax\n",
+      "H1=t*(0.9933)  #aperture effect at highest frequency, sinc(f*t)=0.9933 (given)\n",
+      "H1=H1*100\n",
+      "\n",
+      "#Result\n",
+      "print('|H(1)|=%.2f' %H1)\n",
+      "print('Approximation error')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "|H(1)|=19.87\n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8, page no 74 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate Transmission Bandwidth\n",
+      "\n",
+      "#Variable declaration\n",
+      "#given\n",
+      "fm=3.0*(10^3)\n",
+      "fs=8.0*(10^3) # sampling frequncy\n",
+      "\n",
+      "#Calculation\n",
+      "Ts=1.0/fs\n",
+      "t=0.1*Ts\n",
+      "BW=1.0/(2*t) #Bandwidth\n",
+      "BW=BW/(10^3)\n",
+      "\n",
+      "#Result\n",
+      "print('Transmission Bandwidth of PAM signal is kHz = %.f Khz ' %BW)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Transmission Bandwidth of PAM signal is kHz = 40 Khz \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/Chapter3.ipynb b/Digital_Communications_by_S._Sharma/Chapter3.ipynb
new file mode 100755
index 00000000..805bc62b
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter3.ipynb
@@ -0,0 +1,2238 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater 3: WAVEFORM CODING TECHNIQUES<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3,1 Page No 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print('This is a Theoretical  example')\t"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This is a theoretcal example\n"
+       ]
+      }
+     ],
+     "prompt_number": 104
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3,2,i, Page No 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#find code word length\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "bandwidth=4.2*10**6\n",
+      "fm=bandwidth\n",
+      "q=512        #Quantization levels\n",
+      "#q=2^v\n",
+      "\n",
+      "#Calculation\n",
+      "v=math.log10(512)/math.log10(2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"The code word legth is = %.f bits\" %v)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The code word legth is = 9 bits\n"
+       ]
+      }
+     ],
+     "prompt_number": 105
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2.ii, Page No 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#find The transmission Bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "bandwidth=4.2*10**6\n",
+      "fm=bandwidth\n",
+      "q=512.0           #Quantization levels\n",
+      "v=9.0\n",
+      "\n",
+      "#Calculation\n",
+      "bw=v*fm*10**-6\n",
+      "\n",
+      "#Result\n",
+      "print(\"The transmission Bandwidth is = %.1f MHz\" %bw)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The transmission Bandwidth is = 37.8 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 106
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2.iii, Page No 111 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#find  Final Bit rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "bandwidth=4.2*10**6\n",
+      "fm=bandwidth\n",
+      "q=512.0      #Quantization levels\n",
+      "v=9.0\n",
+      "\n",
+      "#Calculation\n",
+      "fs=2*fm\n",
+      "r=v*fs       #signaling rate\n",
+      "\n",
+      "#Result\n",
+      "print(\"Final Bit rate =%.1f x 10^6 bits/sec\" %(r/(10**6)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final Bit rate =75.6 x 10^6 bits/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 107
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2.iv, Page No 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#find Output signal to noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "bandwidth=4.2*10**6\n",
+      "fm=bandwidth\n",
+      "q=512         #Quantization levels\n",
+      "v=9.0\n",
+      "\n",
+      "#Calculation\n",
+      "sn=4.8+6*v    #noise in dB\n",
+      "\n",
+      "\n",
+      "print(\"Output signal to noise ratio is = %.1f dB\" %sn)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output signal to noise ratio is = 58.8 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 108
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.i, Page No 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#find bits per sample\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fmax=4*10**3\n",
+      "xmax=3.8\n",
+      "snr=100\n",
+      "\n",
+      "#Calculation\n",
+      "P=30*10**-3\n",
+      "v=(math.log10(((snr*xmax**2)/(3*P)))/(2*math.log10(2)))\n",
+      "\n",
+      "#Result\n",
+      "print(\"Number of bits required per sample are = %.2f bits\" %v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of bits required per sample are = 6.98 bits\n"
+       ]
+      }
+     ],
+     "prompt_number": 109
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3.ii, Page No 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Find  Transmission Bandwith\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "fm=4*10**3     #Bandwidth of PCM\n",
+      "xmax=3.8\n",
+      "snr=100       #Signal to Noise Ratio\n",
+      "outputs=30.0\n",
+      "v=7.0\n",
+      "\n",
+      "#Calculation\n",
+      "bw=outputs*v*fm \n",
+      "r=bw*2\n",
+      "bw=bw/10**3\n",
+      "\n",
+      "#Result\n",
+      "print('Transmission Bandwith R = %.f kHz' %bw)\n",
+      "r=r/1000\n",
+      "print('Signaling rate R = %.f bits/sec' %r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Transmission Bandwith R = 840 kHz\n",
+        "Signaling rate R = 1680 bits/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 110
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page No 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#find sampling rate,number of bits,bit rate,bandwidht\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "emax=0.001\n",
+      "de=2*emax\n",
+      "fm=100.0\n",
+      "xmax=10.0\n",
+      "\n",
+      "#Calculation\n",
+      "q=(2*xmax)/de\n",
+      "fs=2*fm\n",
+      "v=(math.log10(q))/math.log10(2)\n",
+      "v=math.ceil(v)\n",
+      "r=v*fs\n",
+      "\n",
+      "#Result\n",
+      "print('1. sampling Frequncy  =  %.f  Hz ' %fs)  \n",
+      "print('2. No.of bits in PCM  =  %.f  bits ' %v)  \n",
+      "print('3. sampling rate  =  %.f  bits per second ' %r)\n",
+      "r=r/2\n",
+      "print('4. Transmission Bandwidth  =  %.f  Hz ' %r)  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. sampling Frequncy  =  200  Hz \n",
+        "2. No.of bits in PCM  =  14  bits \n",
+        "3. sampling rate  =  2800  bits per second \n",
+        "4. Transmission Bandwidth  =  1400  Hz \n"
+       ]
+      }
+     ],
+     "prompt_number": 111
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5, Page No 113 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Bandwidth,Sampling Rate\n",
+      "\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "fm=3.4*10**3\n",
+      "N=24\n",
+      "r=1.5*10**6\n",
+      "encoder=8\n",
+      "\n",
+      "#Calculation\n",
+      "BW=N*fm\n",
+      "BW=BW/10**3\n",
+      "r1=r/N\n",
+      "fs=r1/encoder\n",
+      "\n",
+      "#Result\n",
+      "print(\"i. Channel Bandwith is = %.2f kHz\" %BW)\n",
+      "print(\"ii. Sampling frequency is = %.2f Hz or samples per second.\" %fs)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i. Channel Bandwith is = 81.60 kHz\n",
+        "ii. Sampling frequency is = 7812.50 Hz or samples per second.\n"
+       ]
+      }
+     ],
+     "prompt_number": 112
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page No 114 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Find Signal Bandwidth,Noise Ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "v=7\n",
+      "r=50*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "#fs=2*fm\n",
+      "fm=r/(2*v)\n",
+      "snr=1.8+(6*v);\n",
+      "fm=fm*10**-6\n",
+      "\n",
+      "#Result\n",
+      "print('i. Maximum message Bandwidth is = %.2f MHz ' %fm)\n",
+      "print('ii.  signal to quantization niose ration = %.2f dB ' %snr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i. Maximum message Bandwidth is = 3.57 MHz \n",
+        "ii.  signal to quantization niose ration = 43.80 dB \n"
+       ]
+      }
+     ],
+     "prompt_number": 113
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7, Page No 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#find i)bits per sample,ii)transmission rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "fm=3*10**3\n",
+      "q=16\n",
+      "\n",
+      "#Calculation\n",
+      "v=(math.log10(q))/math.log10(2)\n",
+      "\n",
+      "#Result\n",
+      "print('i) Bits in code word= %.f bits' %v)\n",
+      "fs=2*fm\n",
+      "r=v*fs\n",
+      "print('ii) it trasmission rate= %.f x 10^3 bits per second' %(r/(10**3)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Bits in code word= 4 bits\n",
+        "ii) it trasmission rate= 24 x 10^3 bits per second\n"
+       ]
+      }
+     ],
+     "prompt_number": 114
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8, Page No 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Find signal to noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "fm=3.5*10**3\n",
+      "r=50*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "fs=2*fm\n",
+      "rms=0.2\n",
+      "xmax=2\n",
+      "v=r/fs    #signaling rate r=v*fs\n",
+      "v=math.ceil(v)\n",
+      "P=(rms**2)/1\n",
+      "SNR=((3*P*2**(2*v))/(xmax**2))\n",
+      "SN=10*math.log10(SNR)\n",
+      "SN=math.ceil(SN)\n",
+      "\n",
+      "#Result\n",
+      "print(\"signal to niose ratio =%.f dB\" %SN)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "signal to niose ratio =33 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 115
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10, Page No 117 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find i)noise ratio ii)bits\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "Am=3.0\n",
+      "v=10.0\n",
+      "\n",
+      "#Calculation\n",
+      "SNR=1.8+6*v     #noise ratio \n",
+      "SN=40\n",
+      "v=(SN-1.8)/6\n",
+      "\n",
+      "#Result\n",
+      "print(\"i) Signal to Quantization noise ratio = %.1f dB\" %SNR)\n",
+      "print(\"ii) Bits required to get signal to niose ratio of 40dB = %.f \" %math.ceil(v))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Signal to Quantization noise ratio = 61.8 dB\n",
+        "ii) Bits required to get signal to niose ratio of 40dB = 7 \n"
+       ]
+      }
+     ],
+     "prompt_number": 116
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11, Page No 117 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Maximum frequency\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#Given data\n",
+      "v=7\n",
+      "SNR=1.8+6*v\n",
+      "r=56*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "fs=r/v        #r=v*fs signaling rate\n",
+      "fm=fs/2       #Nquset rate\n",
+      "\n",
+      "#Result\n",
+      "fm=fm/10**3\n",
+      "print(\"Maximum frequency is = %.f kHz\" %fm)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum frequency is = 4 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 117
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14, Page No 129 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Maximum Amplitude\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=3*10**3.0\n",
+      "\n",
+      "#Calculation\n",
+      "Nyquistrate=2*fm     #Nyquistrate\n",
+      "fs=5*Nyquistrate     #Samplingfrquency\n",
+      "Ts=1/fs              #Sampling Interval\n",
+      "de=0.25             #step size\n",
+      "fm1=2*10**3.0\n",
+      "Am=de/(2*math.pi*fm1*Ts)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Maximum Amplitude = %.4f Volts\" %Am)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum Amplitude = 0.5968 Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 118
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16, Page No 130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  signaling rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs1=8*10**3.0\n",
+      "de=31.25*10**-3\n",
+      "q=64.0\n",
+      "\n",
+      "#Calculation\n",
+      "v=math.log(q,2)\n",
+      "r=v*fs1            #signaling rate\n",
+      "fm=3*10**3.0\n",
+      "A=1\n",
+      "fs2=(2*math.pi*fm*A)/(de)\n",
+      "\n",
+      "#Result\n",
+      "r=r*10**-3\n",
+      "print(\"Signaling rate of PCM is = %.f kHz\" %r)\n",
+      "fs2=fs2*10**-3\n",
+      "print(\"The signaling rate of DM is = %.2f kHz\" %fs2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signaling rate of PCM is = 48 kHz\n",
+        "The signaling rate of DM is = 603.19 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 119
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17, Page No 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  signal to noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=64.0*10**3     #frequency\n",
+      "fm=2.0*10**3      #Sinosoidal input signal sample \n",
+      "fM=4.0*10**3      #bandwidth\n",
+      "\n",
+      "#Calculation\n",
+      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*fM)   #Signal to noise ratio\n",
+      "SNRO=10*math.log10(SNR)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Output signal to noise ratio = %.2f dB\" %SNRO)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output signal to noise ratio = 27.94 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 120
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18, Page No 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  signal to Quatization noise ratio \n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=8.0*10**3        #Sampling frequency\n",
+      "r=64.0*10**3        #Data Rate\n",
+      "N=8        #number of bits\n",
+      "\n",
+      "#Calculation\n",
+      "SNR=(1.8+6*N)     #signal to Quatization noise ratio\n",
+      "\n",
+      "#Result\n",
+      "print(\"Signal to Quatization noise ratio of PCM system is = %.f dB\" %SNR)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signal to Quatization noise ratio of PCM system is = 50 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 121
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Find  sampling rate,quantizing level\n",
+      "\n",
+      "#Variable Declaration\n",
+      "r=36000\n",
+      "fm=3.2*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "fs=2*fm        #Nquest rate\n",
+      "v=r/fs         #r=v*fs signaling rate\n",
+      "v=math.floor(v)\n",
+      "q=2**v\n",
+      "fs1=r/v\n",
+      "\n",
+      "#Result\n",
+      "print(\"Quantizing level q = %.f \" %q)\n",
+      "fs1=fs1/1000\n",
+      "print(\"sampling rate fs = %.1f kHz \" %fs1)\n",
+      "print(\"Number of binary digits = %.f  \" %v)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantizing level q = 32 \n",
+        "sampling rate fs = 7.2 kHz \n",
+        "Number of binary digits = 5  \n"
+       ]
+      }
+     ],
+     "prompt_number": 122
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  time duration of 1bit  binary encoded signal\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=input(\"Enter the Nyquist rate of Signal fs(Hz)=\")\n",
+      "q=input(\"Enter the Quantization levels q =\")\n",
+      "\n",
+      "#Calculation\n",
+      "v=math.log(q,2)     #binary pulses transmitted per second\n",
+      "t=1.0/(v*fs)        #Nyquist interval\n",
+      "\n",
+      "#Result\n",
+      "print('Time duration of binary signal  t = %.4f sec ' %t)\n",
+      "#output\n",
+      "#Enter the Nyquist rate fs(Hz)=3\n",
+      "#Enter the Quantization levels q =5\n",
+      "#time duration of 1bit binary signal   \n",
+      "#t=0.1435589  sec"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Nyquist rate of Signal fs(Hz)=23\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Quantization levels q =34\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time duration of binary signal  t = 0.0085 sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 123
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.23, Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#find signal to noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "SNR=40\n",
+      "SNRO=10**(SNR/10)\n",
+      "\n",
+      "#SNR=3(q**2)/2, Signal to Quantization noise ratio\n",
+      "\n",
+      "#Calculation\n",
+      "q=math.sqrt((2*SNRO)/3)\n",
+      "q=math.ceil(q)\n",
+      "v=math.log(q,2)         #q=2**v  Quantization levels\n",
+      "v=math.ceil(v)\n",
+      "snr=1.76+6.02*v          #output Signal to Quantization noise ratio\n",
+      "\n",
+      "#Result\n",
+      "print(\"Output Signal to Quantization noise ratio = %.1f dB\" %snr)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output Signal to Quantization noise ratio = 43.9 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 124
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.24.i, Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Quantizing  levels,minimum number of bits per sample\n",
+      "\n",
+      "#Variable Declaration\n",
+      "SNRO=30.0\n",
+      "fmin=300.0\n",
+      "fmax=3300.0\n",
+      "fs=80000.0\n",
+      "\n",
+      "#Calculation\n",
+      "#SNRO=1.76+20log10(q)\n",
+      "q=10**((SNRO-1.76)/20)\n",
+      "q=math.ceil(q)\n",
+      "v=math.log(q,2)\n",
+      "\n",
+      "#Result\n",
+      "print('Quantizing  levels required is =%.f' %q)\n",
+      "print('minimum number of bits per sample are=%.1f that is approximately 5' %v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantizing  levels required is =26\n",
+        "minimum number of bits per sample are=4.7 that is approximately 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 125
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.24.ii, Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  minimum required bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "SNRO=30.0       #Signal to quantisizing noise ratio\n",
+      "fmin=300.0      #min frequency band\n",
+      "fmax=3300.0     #max frequency band\n",
+      "fs=8000.0       #Sampling rate\n",
+      "v=5.0           #Minimum number of bits per sample\n",
+      "\n",
+      "#Calculation\n",
+      "fPCM=(v*fs)/2\n",
+      "fPCM=fPCM/1000\n",
+      "\n",
+      "#Result\n",
+      "print('minimum required bandwidth =%.f kHz' %fPCM)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "minimum required bandwidth =20 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 126
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.24.iii, Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Quantizing  levels,minimum number of bits per sample and bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "SNRO=30.0       #Signal to quantisizing noise ratio\n",
+      "fmin=300.0      #min frequency band\n",
+      "fmax=3300.0     #max frequency band\n",
+      "fs=8000.0       #Sampling rate\n",
+      "\n",
+      "#Calculation\n",
+      "q=10**((SNRO+10.1)/20)\n",
+      "q=math.ceil(q)\n",
+      "v=math.log(q,2)\n",
+      "v=math.ceil(v)\n",
+      "\n",
+      "#Result\n",
+      "print('Quantizing  levels need is =%.f' %q)\n",
+      "print('minimum number of bits per sample is =%.f' %v)\n",
+      "fPCM=(v*fs)/2\n",
+      "fPCM=fPCM/1000\n",
+      "print('minimum required bandwidth =%.f Khz' %fPCM)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantizing  levels need is =102\n",
+        "minimum number of bits per sample is =7\n",
+        "minimum required bandwidth =28 Khz\n"
+       ]
+      }
+     ],
+     "prompt_number": 127
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.27, Page No 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#determine the  Maximum Amplitude,\n",
+      "\n",
+      "#Variable Declaration\n",
+      "de=250*10**-3\n",
+      "wm=2*math.pi*1000 \n",
+      "fs=3*10**3\n",
+      "Ts=1/fs\n",
+      "\n",
+      "#Calculation\n",
+      "Amax=(de*3*fs*2)/(wm) #Amplitude\n",
+      "SNR=(3.0*((3*6*(10**3))**3))/((8*(math.pi**2)*(10**3)**3))\n",
+      "\n",
+      "#Result\n",
+      "print('Maximum Amplitude= %.1f mV' %(Amax*(10**2)))\n",
+      "print('Signal to noise ratio = %.2f dB' %(10*math.log10(SNR)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum Amplitude= 71.6 mV\n",
+        "Signal to noise ratio = 23.46 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 128
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.29, Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find number of bits per sample\n",
+      "\n",
+      "#Variable Declaration\n",
+      "SNR=20\n",
+      "averagepower=30*10**-3\n",
+      "SNRO=10**(SNR/10)\n",
+      "A=3.8\n",
+      "\n",
+      "#SNRO=average signal power/Quatizing power\n",
+      "#de=(2*A)/L\n",
+      "\n",
+      "#Calculation\n",
+      "L=math.sqrt((SNRO*A**2)/(3*averagepower))\n",
+      "n=math.log(L,2)\n",
+      "n=math.ceil(n)\n",
+      "\n",
+      "#Result\n",
+      "print('Bits required per sample =%.f' %n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bits required per sample =7\n"
+       ]
+      }
+     ],
+     "prompt_number": 129
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.30.i, Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Normalized power for quantization noise\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=3.0*10.0**3\n",
+      "v=8.0\n",
+      "VH=5.0\n",
+      "VL=-5.0\n",
+      "q=2**v\n",
+      "\n",
+      "#Calculation\n",
+      "de=(VH-VL)/q\n",
+      "Nq=de**2/12.0    #quantization noise\n",
+      "\n",
+      "#Result\n",
+      "print('Normalized power for quantization noise =%.2f x 10^-6 W' %(Nq*(10**6)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Normalized power for quantization noise =127.16 x 10^-6 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 130
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.30.ii, Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find bit transmission rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=3*10.0**3\n",
+      "v=8.0\n",
+      "VH=5.0\n",
+      "VL=-5\n",
+      "q=2**v\n",
+      "\n",
+      "#Calculation\n",
+      "fs=2*fm       #Nyquist rate\n",
+      "r=8*fs/1000\n",
+      "\n",
+      "#Result\n",
+      "print('bit transmission rate %.f K bits/s' %r)\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "bit transmission rate 48 K bits/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 131
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.30.iii, Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Signal to quantization noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "Nq=127.15*10**-6\n",
+      "Meansignal=2\n",
+      "\n",
+      "#Calculation\n",
+      "P=Meansignal/1\n",
+      "SNR=P/Nq\n",
+      "SNRq=10*math.log10(SNR)\n",
+      "\n",
+      "#Result\n",
+      "print('Signal to quantization noise ratio %.2f dB ' %SNRq)\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signal to quantization noise ratio 41.97 dB \n"
+       ]
+      }
+     ],
+     "prompt_number": 132
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.31, Page No 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  i)SNR  ii)output bit rate iii)no.of bits recorded\n",
+      "\n",
+      "#Variable Declaration\n",
+      "N=16.0\n",
+      "v=16.0\n",
+      "fs=44.1*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "SNR=1.76+6*N\n",
+      "bitrate=2*v*fs\n",
+      "outputbitrate=2*bitrate*10**-6        #including addtional 100% over head\n",
+      "CD=outputbitrate*3600*10**-3\n",
+      "\n",
+      "#Result\n",
+      "print('i)Out put signal noise ratio =%.2f dB' %SNR)\n",
+      "print('ii)output bit rate =%.3f Mbits/sec' %outputbitrate)\n",
+      "print('iii)no.of bits recorded in CD =%.2f gigabits' %CD)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Out put signal noise ratio =97.76 dB\n",
+        "ii)output bit rate =2.822 Mbits/sec\n",
+        "iii)no.of bits recorded in CD =10.16 gigabits\n"
+       ]
+      }
+     ],
+     "prompt_number": 133
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.32, Page No 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  output SNR\n",
+      "#Variable Declaration\n",
+      "fm=1*10.0**3\n",
+      "fs=32*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "FM=4*10**3     #Bandwidth\n",
+      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*FM)    #SNR\n",
+      "SNRO=10*math.log10(SNR)\n",
+      "\n",
+      "#Result\n",
+      "print('Output signal to noise ratio %.2f dB' %SNRO)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output signal to noise ratio 24.93 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 134
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.33.i, Page No 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  step size\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=64000.0     #samples/sec\n",
+      "Amax=1.0       #Maximum signal amplitude\n",
+      "fm=3500.0      #Sample Rate\n",
+      "\n",
+      "#Calculation\n",
+      "de=(2*math.pi*fm*Amax)/fs*1000\n",
+      "\n",
+      "#Result\n",
+      "print('Step Size %.2f mV ' %de)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Step Size 343.61 mV \n"
+       ]
+      }
+     ],
+     "prompt_number": 135
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.33ii, Page No 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Quantizatio noise power \n",
+      "#Variable Declaration\n",
+      "fs=64000.0     #sample Rate\n",
+      "Amax=1.0       #Maximum signal Amplitude\n",
+      "fm=3500.0\n",
+      "\n",
+      "#Calculation\n",
+      "de=343.6117*10**-3      #step size\n",
+      "Nq=de**2/3.0             #Quantizatio noise power\n",
+      "Nqd=Nq*(fm/fs)*1000\n",
+      "\n",
+      "#Result\n",
+      "print('Quantizatio noise power %.3f mW' %Nqd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantizatio noise power 2.152 mW\n"
+       ]
+      }
+     ],
+     "prompt_number": 136
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.33iii, Page No 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find SNR\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=64000.0\n",
+      "Amax=1.0\n",
+      "fm=3500.0\n",
+      "Nqd=2.1522995*10**-3\n",
+      "\n",
+      "#Calculation\n",
+      "So=Amax**2/2\n",
+      "SNR=So/Nqd\n",
+      "SNRO=10*math.log10(SNR)\n",
+      "\n",
+      "#Result\n",
+      "print('Output signal noise ratio dB  %.3f dB' %SNRO)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output signal noise ratio dB  23.661 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 137
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.34, Page No 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Assuming signal is sampled at the rate 20% above Nyquist rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=4.5*10**6\n",
+      "q=1024.0\n",
+      "\n",
+      "#Calculation\n",
+      "fs=1.2*2*fm           #20% above Nyquist rate\n",
+      "v=math.log(q,2)\n",
+      "r=v*fs/10**6\n",
+      "\n",
+      "#Result\n",
+      "print('no.of bits/sec = %.f M bit/sec' %r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "no.of bits/sec = 108 M bit/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 138
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.35, Page No 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#assume bandwidth of the singal is 4kHz\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=32000.0\n",
+      "A=2.0\n",
+      "fm=4000.0\n",
+      "BW=4000.0\n",
+      "\n",
+      "#Calculation\n",
+      "de=(2*math.pi*fm*A)/fs\n",
+      "Nq=de**2/3\n",
+      "SNR=(3*fs**3)/(8*math.pi**2*fm**2*BW)\n",
+      "\n",
+      "#Result\n",
+      "print('i)step size = %.3f M Volt' %de)\n",
+      "print('ii)noise power = %.3f W' %Nq)\n",
+      "print('iii)SNR= = %.3f ' %SNR)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)step size = 1.571 M Volt\n",
+        "ii)noise power = 0.822 W\n",
+        "iii)SNR= = 19.454 \n"
+       ]
+      }
+     ],
+     "prompt_number": 139
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.36, Page No 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#assuming signal is sampled at the rate 20% above Nyquist rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=15*10**3\n",
+      "fs=1.2*2*fm\n",
+      "q=65536\n",
+      "\n",
+      "#Calculation\n",
+      "v=math.log(q,2)\n",
+      "r=v*fs/1000\n",
+      "BW=r/2\n",
+      "\n",
+      "#Result\n",
+      "print('i)signaling rate,= %.f K bits/sec ' %r)\n",
+      "print('ii)bandwidth BW min = %.f kHz ' %BW)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)signaling rate,= 576 K bits/sec \n",
+        "ii)bandwidth BW min = 288 kHz \n"
+       ]
+      }
+     ],
+     "prompt_number": 140
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.37, Page No 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  step size,noise power\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=64*10**3\n",
+      "fm=3500.0\n",
+      "A=1.0\n",
+      "\n",
+      "#Calculation\n",
+      "de=(2*math.pi*fm*A)/fs      #step size\n",
+      "Nq=(de**2/3)*(fm/fs)        #Granular noise power\n",
+      "\n",
+      "#Result\n",
+      "print('i)step size = %.3f Volts' %de)\n",
+      "print('ii)Nq= %.5f W' %Nq)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)step size = 0.344 Volts\n",
+        "ii)Nq= 0.00215 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 141
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.40, Page No 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  step Nr,Transmission Bandwidth, SNR\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fs=4           #kHz\n",
+      "N=12\n",
+      "\n",
+      "#Calculation\n",
+      "Nr=2*fs                     #Nyquest rate\n",
+      "Bt=1.0/2*N*fs               #Transmission Bandwidth\n",
+      "SNR=1.8+(6*N)               \n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print('i)Nyquest rate = %.1f kHz' %Nr)\n",
+      "print('ii)Transmission Bandwidth = %.1f kHz' %Bt)\n",
+      "print('iii)SNR = %.1f dB' %SNR)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)Nyquest rate = 8.0 kHz\n",
+        "ii)Transmission Bandwidth = 24.0 kHz\n",
+        "iii)SNR = 73.8 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 142
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.41, Page No 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  step size,noise power\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=3.5*10**3\n",
+      "fs=64*10**3\n",
+      "A=2.0\n",
+      "\n",
+      "#Calculation\n",
+      "de=(2*math.pi*fm*A)/fs      #step size\n",
+      "Nq=(de**2/3)*(fm/fs)        #Granular noise power\n",
+      "\n",
+      "#Result\n",
+      "print('i)step size = %.3f Volts' %de)\n",
+      "print('ii)Nq= %.1f  X 10^-3 W' %(Nq*10**3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i)step size = 0.687 Volts\n",
+        "ii)Nq= 8.6  X 10^-3 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 143
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.42, Page No 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  number of binay pulse per word, Bit rate, Bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "W=4.5 * 10**6                    #khz\n",
+      "Nr = 2 * W               #Nyquist Rate\n",
+      "Q = 1024\n",
+      "fs=10.8\n",
+      "\n",
+      "#Calculation\n",
+      "N = math.log(Q,2)            #number of binay pulse per word\n",
+      "Br= N*fs                     #Bit rate\n",
+      "Bw= 1.0/2*Br                 #Bandwidth\n",
+      "\n",
+      "#Result\n",
+      "print('i) Number of binay pulse per word = %.1f k bits/sec' %N)\n",
+      "print('ii) Bit rate = %.1f k bits/sec' %Br)\n",
+      "print('iii)Bandwidth= %.1f  X 10^-3 kHz' %Bw)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Number of binay pulse per word = 10.0 k bits/sec\n",
+        "ii) Bit rate = 108.0 k bits/sec\n",
+        "iii)Bandwidth= 54.0  X 10^-3 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 144
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.43, Page No 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  number of binay pulse per word, Bit rate, Bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "W=15                 #khz\n",
+      "Nr = 2*W               #Nyquist Rate\n",
+      "Q = 65.53\n",
+      "fs=44.1                # Sampling rate kHz\n",
+      "N=16\n",
+      "\n",
+      "#Calculation\n",
+      "Br= N*fs                    #Bit rate\n",
+      "Bw= 1.0/2*Br                 #Bandwidth\n",
+      "\n",
+      "#Result\n",
+      "print('i) Number of binay pulse per word = %.3f bits' %N)\n",
+      "print('ii) Bit rate = %.1f k bits/sec' %Br)\n",
+      "print('iii)Bandwidth= %.1f kHz' %Bw)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Number of binay pulse per word = 16.000 bits\n",
+        "ii) Bit rate = 705.6 k bits/sec\n",
+        "iii)Bandwidth= 352.8 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 145
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.44, Page No 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Number of binay pulse per word\n",
+      "\n",
+      "#Variable Declaration\n",
+      "W=3.3                 #khz\n",
+      "fm = 1*10**3                #kHz\n",
+      "de = 250              #mV\n",
+      "fs=19.8                # Sampling rate kHz\n",
+      "\n",
+      "#Calculation\n",
+      "A=(de*fs)/(2*math.pi*fm)\n",
+      "\n",
+      "#Result\n",
+      "print('i) Number of binay pulse per word = %.3f' %A)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Number of binay pulse per word = 0.788\n"
+       ]
+      }
+     ],
+     "prompt_number": 146
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.45, Page No 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  number of binay pulse per word, Signaling rate, Bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "W=3.3                 #khz\n",
+      "fm = 3300               #Hz\n",
+      "Snq = 40              #dB\n",
+      "fs= 8000                # Samples/sec\n",
+      "\n",
+      "#Calculation \n",
+      "N=math.ceil((Snq-4.8)/6)                # Number of bits per word\n",
+      "Q=2**N              #\n",
+      "r=N*fs              #Signaling rate k bits/sec\n",
+      "Bw=1.0/2*r\n",
+      "\n",
+      "#Result\n",
+      "print('i) Number of bits per word = %.1f' %N)\n",
+      "print('ii) Signaling rate rate = %.1f k bits/sec' %(r/10**3))\n",
+      "print('iii)Bandwidth= %.1f kHz' %(Bw/10**3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Number of bits per word = 6.0\n",
+        "ii) Signaling rate rate = 48.0 k bits/sec\n",
+        "iii)Bandwidth= 24.0 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 147
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.46, Page No 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find the maximun frequency with 1 volt amplitude\n",
+      "\n",
+      "#Variable Declaration\n",
+      "de=256*10**-3                 #V\n",
+      "Ts = 10*10**-6               #kHz\n",
+      "fm = 10              #kHz\n",
+      "fs= 100                # kHz\n",
+      "A=1                   #V\n",
+      "\n",
+      "#Calculation \n",
+      "wm=(de/A*Ts)*10**7\n",
+      "\n",
+      "#Result\n",
+      "print('i) maximun frequency = %.1f X 10^3 rad/sec' %wm)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) maximun frequency = 25.6 X 10^3 rad/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 148
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.47, Page No 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find the Signal to noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "Q=256     \n",
+      "m=255\n",
+      "\n",
+      "#Calculation \n",
+      "SNR=(3*Q**2)/(math.log((1+m),math.e))**2    #dB\n",
+      "Gc=m/(math.log(1+m))                        #Companding gain\n",
+      "Gc=20*math.log(Gc,10)\n",
+      "\n",
+      "#Result\n",
+      "print('i) The Signal to noise ratio= %.1f dB' %SNR)\n",
+      "print('ii) The Companding gain = %.1f dB' %Gc)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) The Signal to noise ratio= 6394.0 dB\n",
+        "ii) The Companding gain = 33.3 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 149
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.48, Page No 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  number of binay pulse per word, Signaling rate, Bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=15*10**3          #kHz\n",
+      "Q=200\n",
+      "fs=2*fm\n",
+      "\n",
+      "#Calculation \n",
+      "N=math.ceil(math.log(Q,2))\n",
+      "tr=N*fs           #Transmission rate kbps\n",
+      "Bw=1.0/2*tr\n",
+      "\n",
+      "#Result\n",
+      "print('i) Number of bits per Samples = %.1f' %N)\n",
+      "print('ii) Signaling rate = %.1f k bits/sec' %(tr/10**3))\n",
+      "print('iii)Bandwidth= %.1f kHz' %(Bw/10**3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Number of bits per Samples = 8.0\n",
+        "ii) Signaling rate = 240.0 k bits/sec\n",
+        "iii)Bandwidth= 120.0 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 150
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.49, Page No 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  number of quantization level\n",
+      "\n",
+      "#Variable Declaration\n",
+      "tr=36.0                  #kbps\n",
+      "fm=3.4                 #kHz\n",
+      "fs=8\n",
+      "\n",
+      "#Calculation \n",
+      "N=math.ceil(tr/fs)\n",
+      "Q=2**N\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print('i) number of quantization level = %.1f' %Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) number of quantization level = 32.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 151
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.50, Page No 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  delta modulation\n",
+      "\n",
+      "#Variable Declaration\n",
+      "A=1                  #V\n",
+      "fm=3.4                 #kHz\n",
+      "fs=20\n",
+      "\n",
+      "#Calculation \n",
+      "de=1*2*math.pi*(fm/fs)\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print('i) delta modulation = %.2f V' %de)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) delta modulation = 1.07 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 152
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.51, Page No 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Signal to noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "W=3.5      #kHz\n",
+      "r=50.0      #kbps\n",
+      "Vpp=4      #V\n",
+      "fs=8\n",
+      "\n",
+      "#Calculation \n",
+      "N=r/fs\n",
+      "SNR=4.8+(6.8*N)\n",
+      "\n",
+      "#Result\n",
+      "print('i) Signal to noise ratio = %.2f V' %SNR)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Signal to noise ratio = 47.30 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 153
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.52, Page No 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  Signal to noise ratio\n",
+      "\n",
+      "#Variable Declaration\n",
+      "W=1      #MHz\n",
+      "fs1=3       #MHz\n",
+      "fs2=2.4       #MHz\n",
+      "Q=256\n",
+      "m=255\n",
+      "\n",
+      "#Calculation \n",
+      "SNR=((3*Q**2)/(math.log((1+m),math.e))**2)\n",
+      "SNR1=63095.73              #SNR=38+10\n",
+      "Q1=math.sqrt((SNR1*((math.log(256,math.e))**2))/3)\n",
+      "N1=math.ceil(math.log(Q1,2))\n",
+      "r=N1*fs2\n",
+      "Bw=1.0/2*r\n",
+      "\n",
+      "#Result\n",
+      "print('i) Number of bits/word = %.2f ' %N1)\n",
+      "print('ii) Bit rate = %.1f Mb/s' %r)\n",
+      "print('iii) Bandwidth = %.1f Mb/s' %Bw)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Number of bits/word = 10.00 \n",
+        "ii) Bit rate = 24.0 Mb/s\n",
+        "iii) Bandwidth = 12.0 Mb/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 154
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/Chapter4.ipynb b/Digital_Communications_by_S._Sharma/Chapter4.ipynb
new file mode 100755
index 00000000..e56598ee
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter4.ipynb
@@ -0,0 +1,273 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater 4: DIGITAL MULTIPLEXERS<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4, page no 190 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Caption: Sampling Rate\n",
+      "\n",
+      "#initialisation of variables\n",
+      "f1=4.0*10**3\n",
+      "f2=4.5*10**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "fsmin=(2*f2)/1000\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\" Sampling rate =%.2f kHz\" %fsmin)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Sampling rate =9.00 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5, Page No 190 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Caption: Nyquest rate,Signaling rate,bandwidth\n",
+      "\n",
+      "#initialisation of variables\n",
+      "f1=3000.0\n",
+      "f4=1000.0\n",
+      "f2=1000.0\n",
+      "f3=1000.0\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "#Nyquest rate\n",
+      "nq1=2*f1\n",
+      "nq2=2*f2\n",
+      "nq3=2*f3\n",
+      "nq4=2*f4\n",
+      "\n",
+      "#RESULTS\n",
+      "print('i. Nyquest rate of x1 =%.2fkHz' %nq1)\n",
+      "print('ii. Nyquest rate of x2 =%.2fkHz' %nq2)\n",
+      "print('iii. Nyquest rate of x3 =%.2fkHz' %nq3)\n",
+      "print('iv. Nyquest rate of x4 =%.2fkHz' %nq4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i. Nyquest rate of x1 =6000.00kHz\n",
+        "ii. Nyquest rate of x2 =2000.00kHz\n",
+        "iii. Nyquest rate of x3 =2000.00kHz\n",
+        "iv. Nyquest rate of x4 =2000.00kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6, Page No 190 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find The spacing between two successive pules\n",
+      "\n",
+      "#initialisation of variables\n",
+      "samplingrate=8000.0\n",
+      "totalsignals=24+1\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "t=1/samplingrate\n",
+      "T=t/totalsignals\n",
+      "T=T*10**6            #time is now u sec\n",
+      "space=T-1\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The spacing between two successive pules %.f u sec' %space)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The spacing between two successive pules 4 u sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7, Page No 191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find signaling rate ,bandwidth\n",
+      "\n",
+      "#initialisation of variables\n",
+      "N=6.0\n",
+      "fm=5000.0\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r=2*fm   #sampling rate\n",
+      "sr=N*r   #signalingrate\n",
+      "BW=N*fm  #Bandwith\n",
+      "sr=sr/1000\n",
+      "BW=BW/1000\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Signaling rate %.f K bits/sec' %sr)\n",
+      "print('Bandwith to avoid the cross talk in TDM is %.f kHz' %BW)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signaling rate 60 K bits/sec\n",
+        "Bandwith to avoid the cross talk in TDM is 30 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9, Page No 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find signaling rate ,bandwidth\n",
+      "\n",
+      "#initialisation of variables\n",
+      "N=4.0\n",
+      "x=24              #bits/frame\n",
+      "D=1152\n",
+      "s=1.0/3\n",
+      "r=1.544*10**6\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r0=N*r*((D+x)/(D-s*N))\n",
+      "\n",
+      "#RESULTS\n",
+      "print('Output bit rate %.2f MB/sec' %(r0/(10**6)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output bit rate 6.31 MB/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10, Page No 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find signaling rate ,bandwidth\n",
+      "\n",
+      "#initialisation of variables\n",
+      "N=24.0\n",
+      "x=72.0              #bits\n",
+      "D=192             #message and stuff bits\n",
+      "s=1.0/3\n",
+      "r=64              #kb/sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r0=N*r*((D+x)/(D-s*N))\n",
+      "r1=((N*r)/r0)*100\n",
+      "\n",
+      "#RESULTS\n",
+      "print('i) Output bit rate %.2f MB/sec' %(r0/(10**3)))\n",
+      "print('iI) Throughput efficiency is =  %.2f percent' %r1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Output bit rate 2.20 MB/sec\n",
+        "iI) Throughput efficiency is =  69.70 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/Chapter5.ipynb b/Digital_Communications_by_S._Sharma/Chapter5.ipynb
new file mode 100755
index 00000000..bf535beb
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter5.ipynb
@@ -0,0 +1,105 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater 5: DIGTIAL BASEBAND TRANSMISSION<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12, Page No 249 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find The roll of factor alpha\n",
+      "\n",
+      "#initialisation of variables\n",
+      "datarate=0.1*10**6\n",
+      "fB=75000.0 #bandwidth\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Tb=1/datarate\n",
+      "alpha=2*fB*Tb-1\n",
+      "\n",
+      "#RESULTS\n",
+      "print('The roll out factor alpha =  = %.1f ' %alpha)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The roll out factor alpha =  = 0.5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13, Page No 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find The Transmission bandwidth\n",
+      "\n",
+      "#initialisation of variables\n",
+      "q=128.0\n",
+      "alpha=0.2\n",
+      "n=math.log(q,2)\n",
+      "fm=2000.0\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Nq=2*fm\n",
+      "fs=1.25*Nq\n",
+      "N=8.0\n",
+      "total=N*fs\n",
+      "bitrate=7*total\n",
+      "fB=((1+alpha)*bitrate)/2\n",
+      "fB=fB/1000\n",
+      "#RESULTS\n",
+      "print('The minimum transmission Bandwidth required = %.f kHz' %fB)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum transmission Bandwidth required = 168 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/Chapter6.ipynb b/Digital_Communications_by_S._Sharma/Chapter6.ipynb
new file mode 100755
index 00000000..fd5f6716
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter6.ipynb
@@ -0,0 +1,158 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater 6: DIGITAL MODULATION TECHNIQUES<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5, Page No 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#Symbol rate of transmission and bandwidth \n",
+      "\n",
+      "#initialisation of variables\n",
+      "fb=1.0                #MHz\n",
+      "Tb=1/fb               #bit period ms\n",
+      "N=4\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Ts1=Tb\n",
+      "Bw1=2*fb\n",
+      "Ts2=Tb*2\n",
+      "Tr2=(1.0/2)*1000          #Symbols/sec\n",
+      "Bw2=1/Tb\n",
+      "Ts3=N*Tb\n",
+      "Tr3=(1.0/4)*1000\n",
+      "Bw3=(2.0*fb)/N\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"In BPSK\")\n",
+      "print(\"The symbol rate of transmission is = %.f X 10^6 symbol/sec \" %Ts1)\n",
+      "print(\"The Bandwidth is = %.f MHz \" %Bw1)\n",
+      "print(\"In QPSK\")\n",
+      "print(\"The symbol rate of transmission is = %.f X 10^3 symbol/sec \" %Tr2)\n",
+      "print(\"The Bandwidth is = %.f MHz \" %Bw2)\n",
+      "print(\"In 16-ary PSK\")\n",
+      "print(\"The symbol rate of transmission is = %.f X 10^3 symbol/sec \" %Tr3)\n",
+      "print(\"The Bandwidth is = %.f kHz \" %(Bw3*1000))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In BPSK\n",
+        "The symbol rate of transmission is = 1 X 10^6 symbol/sec \n",
+        "The Bandwidth is = 2 MHz \n",
+        "In QPSK\n",
+        "The symbol rate of transmission is = 500 X 10^3 symbol/sec \n",
+        "The Bandwidth is = 1 MHz \n",
+        "In 16-ary PSK\n",
+        "The symbol rate of transmission is = 250 X 10^3 symbol/sec \n",
+        "The Bandwidth is = 500 kHz \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6, Page No 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "fb=90              #MHz\n",
+      "Bw=20              #Mb/s\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N=(2*fb)/Bw\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"Number of bits required = %.f \" %N)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of bits required = 9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12, Page No 357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Tb=0.1          #msec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "fb=1/Tb\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"Transmission bandwidth = %.f kHz\" %fb)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Transmission bandwidth = 10 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/Chapter7.ipynb b/Digital_Communications_by_S._Sharma/Chapter7.ipynb
new file mode 100755
index 00000000..7c33676a
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter7.ipynb
@@ -0,0 +1,431 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater 7: PERFORMANCE ANALYSIS OF DIGITAL COMMUNICATION SYSTEMS<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15, Page No 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "T=1.0*10**-6             #ms\n",
+      "A=10.0*10**-3\n",
+      "N=1*10**-11\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Eb=((A**2)*T)/2.0\n",
+      "Epb=math.sqrt(Eb/N)\n",
+      "Pb=(1.0/2)*math.erfc(Epb)*(1.0/4)\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"Bit error probability = %.2f X 10^-4 \" %(Pb*(10**4)))\n",
+      "print('Approximation error')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bit error probability = 1.96 X 10^-4 \n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.16, Page No 420"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Br=25*10**6             #bits/sec\n",
+      "N=2*10**-20             #Power spectral density of noise\n",
+      "A=1.0*10**-6\n",
+      "Ps=1*10**-12            # Normalized power\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Tb=1.0/Br\n",
+      "Ps=(A**2)/2.0\n",
+      "Pe=1.0/2*(math.erfc((0.6*Ps*Tb)/N))\n",
+      "#RESULTS\n",
+      "print(\"Average probability of bit error is Pe= %.2f X 10^-4 \" %(Pe*10))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average probability of bit error is Pe= 1.98 X 10^-4 \n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17, Page No 423"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find probability of error\n",
+      "\n",
+      "#initialisation of variables\n",
+      "A=2.0\n",
+      "N0=2.0*10**-4  #power sepctral density\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Tmin=(6.76*N0)/4\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"The min time for which the signal should be extended is Tmin= %.3f X 10^-3 msec \" %(Tmin*10**3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The min time for which the signal should be extended is Tmin= 0.338 X 10^-3 msec \n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.20, Page No 428"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#The average carrier\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Pe=10**-4\n",
+      "N0=2*10**10\n",
+      "T=1.0/10**6\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "E=6.76*N0\n",
+      "P=E/T\n",
+      "Eb=8.5171*N0\n",
+      "P2=Eb/T\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"The average carrier = %.3f X 10^-3 mW \" %(P2/10**17))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average carrier = 1.703 X 10^-3 mW \n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.21, Page No 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#The average carrier\n",
+      "\n",
+      "#initialisation of variables\n",
+      "A=0.2*10**-3          #V\n",
+      "T=2*10**-6\n",
+      "N0=2*10**-15\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Pe=(math.sqrt((A**2*T)/4*N0))\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"The average carrier = %.3f X 10^-4 mW \" %(Pe*10**15))\n",
+      "print(\"Approximation error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average carrier = 6.325 X 10^-4 mW \n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.29, Page No 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "A=1.0\n",
+      "N0=2.0*10**-5  #power sepctral density\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "T=((3.71**2)*2*N0)/(A**2)\n",
+      "Br=1/T\n",
+      "#Results\n",
+      "print(\"Bit rate Br= %.2f k bits/sec\" %(Br/(10**3)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bit rate Br= 1.82 k bits/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.31, Page No 448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Pe=10.0**-4 #probability of error of PSK\n",
+      "N0=2.0*10**-10\n",
+      "Br=10**6                  #Bit Rate\n",
+      "\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "T=1.0/Br\n",
+      "E=N0*6.76\n",
+      "P=E/T\n",
+      "Eb=8.5171*N0\n",
+      "Pb=Eb/T\n",
+      "#Results\n",
+      "print(\"i) PSK system The required carrier power = %.2f mW\" %(P*10**3))\n",
+      "print(\"ii) DPSK system The required carrier power = %.2f mW\" %(Pb*10**3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) PSK system The required carrier power = 1.35 mW\n",
+        "ii) DPSK system The required carrier power = 1.70 mW\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.32, Page No 450"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "A=0.2*10**-3                #V\n",
+      "T=2*10**-6                  #Sec\n",
+      "N0=2*10**-15\n",
+      "\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "x=math.sqrt((A**2*T)/(4*N0))\n",
+      "Pe=(1.0/2)*math.erfc(x/math.sqrt(2.0))\n",
+      "\n",
+      "#Results\n",
+      "print(\"Probability of error = %.2f X 10^-4\" %(Pe*10**4))\n",
+      "#print(\"Channel Bandwidth  = %.5f MHz\" %BW)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Probability of error = 7.83 X 10^-4\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.36, Page No 453"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Ap=1                #mV\n",
+      "En=192.3            #mV\n",
+      "Arms=707            #mV\n",
+      "\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Z=Arms/En\n",
+      "Pe=1.0/2*(math.erfc(math.sqrt(2)*(Z/math.sqrt(2))))\n",
+      "\n",
+      "#Results\n",
+      "print(\"Probability of error is  %.7f \" %Pe)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Probability of error is  0.0000001 \n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.37, Page No 454"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "Br=2.08*10**-6        #Bit Rate\n",
+      "N0=2*10**-8\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Eb=(math.sqrt(3.3))/(5*10**7)   #Eb=bit energy\n",
+      "Ps=Eb*Br\n",
+      "BW=2*Br                  #Bandwidth\n",
+      "Es=286.39*N0\n",
+      "Ps2=(Es*Br)/2\n",
+      "BW2=(2*Br)/4\n",
+      "\n",
+      "#Results\n",
+      "print(\"i) BPSK THe transmission bandwidth  = %.2f MHz\" %(BW*10**6)) \n",
+      "print(\"ii) For 16 ary PSK THe transmission bandwidth  = %.2f MHz\" %(BW2*10**6)) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) BPSK THe transmission bandwidth  = 4.16 MHz\n",
+        "ii) For 16 ary PSK THe transmission bandwidth  = 1.04 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/Chapter8.ipynb b/Digital_Communications_by_S._Sharma/Chapter8.ipynb
new file mode 100755
index 00000000..52d8741e
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter8.ipynb
@@ -0,0 +1,1439 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater 8: INFORMATION THEORY<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page No 464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#Find Information Content of Each Symbol\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=1/2.0\n",
+      "px2=1/4.0\n",
+      "px3=1/8.0\n",
+      "px4=1/8.0\n",
+      "\n",
+      "#Calculation\n",
+      "#information content of each symbol\n",
+      "Ix1=math.log(1/px1,2)\n",
+      "Ix2=math.log(1/px2,2)\n",
+      "Ix3=math.log(1/px3,2)\n",
+      "Ix4=math.log(1/px4,2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Information Content tI(x1)= %.2f bit\" %Ix1)\n",
+      "print(\" tI(x2)= %.f bits\" %Ix2)\n",
+      "print(\" tI(x3)= %.f bits\" %Ix3)\n",
+      "print(\" tI(x4)= %.f bits\" %Ix4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Information Content tI(x1)= 1.00 bit\n",
+        " tI(x2)= 2 bits\n",
+        " tI(x3)= 3 bits\n",
+        " tI(x4)= 3 bits\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2, Page No 464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find amount of Information\n",
+      "#Variable Declaration\n",
+      "#Calculation\n",
+      "pxi=1/4.0\n",
+      "Ixi=(math.log10(1/pxi))/math.log10(2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"The amount of Information I(Xi)= %.f \" %Ixi)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of Information I(Xi)= 2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page No 464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Amount of Information\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=1/2.0\n",
+      "px2=1/2.0\n",
+      "\n",
+      "#Calculation\n",
+      "Ix1=math.log(1/px1,2) #entropy\n",
+      "Ix2=math.log(1/px2,2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"The amount of Information I(X1)= %.f bit\" %Ix1)\n",
+      "print(\"The amount of Information I(X2)= %.f bit\" %Ix2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of Information I(X1)= 1 bit\n",
+        "The amount of Information I(X2)= 1 bit\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4, Page No 465"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Amount of Information \n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=1/4.0\n",
+      "px2=3/4.0\n",
+      "\n",
+      "#Calculation\n",
+      "Ix1=math.log(1/px1,2)\n",
+      "Ix2=math.log(1/px2,2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"The amount of Information I(X1)= %.f bit\" %Ix1)\n",
+      "print(\"The amount of Information I(X2)= %.2f bit\" %Ix2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of Information I(X1)= 2 bit\n",
+        "The amount of Information I(X2)= 0.42 bit\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9, Page No 468 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Entropy,Amount of information\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=0.4\n",
+      "px2=0.3\n",
+      "px3=0.2\n",
+      "px4=0.1\n",
+      "\n",
+      "#Calculation\n",
+      "HX=-px1*math.log(px1,2)-px2*math.log(px2,2)-px3*math.log(px3,2)-px4*math.log(px4,2)\n",
+      "Px1x2x1x3=px1*px2*px1*px3\n",
+      "Ix1x2x1x3=-math.log(Px1x2x1x3,2)\n",
+      "Px4x3x3x2=px4*px3*px3*px2\n",
+      "Ix4x3x3x2=-math.log(Px4x3x3x2,2)\n",
+      "\n",
+      "#Result\n",
+      "print(\" \\n Entropy  H(X) = %.2f bits/symbol \" %HX)\n",
+      "print(\"The amount of Information I(x1x2x1x3)= %.2f bits/symbol\" %Ix1x2x1x3)\n",
+      "print(\" I(x4x3x3x2) = %.2f bits/symbol \" %Ix4x3x3x2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        " Entropy  H(X) = 1.85 bits/symbol \n",
+        "The amount of Information I(x1x2x1x3)= 6.70 bits/symbol\n",
+        " I(x4x3x3x2) = 9.70 bits/symbol \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.13, Page No 471"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Find information rate the telegraphic source\n",
+      "\n",
+      "#Variable Declaration\n",
+      "pdash=1/3.0\n",
+      "pdot=2/3.0\n",
+      "tdot=0.2\n",
+      "tdash=0.6\n",
+      "tspace=0.2\n",
+      "\n",
+      "#Calculation\n",
+      "HX=-pdash*math.log(pdash,2)-pdot*math.log(pdot,2)\n",
+      "Ts=pdot*tdot+pdash*tdash+tspace\n",
+      "r=1/Ts\n",
+      "R=r*HX\n",
+      "\n",
+      "#Result\n",
+      "print('Average rate of information R = %.2f bits/s' %R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average rate of information R = 1.72 bits/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14, Page No 471"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find information rate of the source\n",
+      "\n",
+      "f=input('Enter the frequncy f=')\n",
+      "px1=1/8.0\n",
+      "px2=1/8.0\n",
+      "px3=3/8.0\n",
+      "px4=3/8.0\n",
+      "\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2) #entropy of the source\n",
+      "R=2*f*HX #r=2*f\n",
+      "print('information rate  R=  %.1f  bits/sec ' %R)   #f=signal bandwidth\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the frequncy f=34\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "information rate  R=  123.2  bits/sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.15, Page No 472"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find information rate of the source\n",
+      "#all symbols are equally likely\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=1/2.0\n",
+      "px2=1/2.0\n",
+      "px3=1/2.0\n",
+      "px4=1/2.0\n",
+      "\n",
+      "#Calculation\n",
+      "f=input('Enter the frequncy of system fm(in Hz) =')\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)\n",
+      "\n",
+      "#Result\n",
+      "print('\\n Entropy H(X) =%.f bits/symbol ' %HX)\n",
+      "R=2*f*HX\n",
+      "print('information rate =%.f bits/sec' %R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the frequncy of system fm(in Hz) =45\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Entropy H(X) =2 bits/symbol \n",
+        "information rate =180 bits/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.16, Page No 473"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find source entropy ,information rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#probability symbols\n",
+      "px1=1/2.0\n",
+      "px2=1/4.0\n",
+      "px3=1/8.0\n",
+      "px4=1/16.0\n",
+      "px5=1/16.0\n",
+      "Tb=10.0**-3\n",
+      "\n",
+      "#Calculation\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4)+px5*math.log(1/px5)\n",
+      "\n",
+      "#Result\n",
+      "print('1. source entropy H(X) = %.2f bits/symbol ' %HX) #source entropy\n",
+      "r=1.0/Tb\n",
+      "R=r*HX #information rate\n",
+      "print(' 2. Information rate R = %.2f bits/sec ' %R)\n",
+      "print('Approximation error')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. source entropy H(X) = 1.72 bits/symbol \n",
+        " 2. Information rate R = 1721.57 bits/sec \n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.17, Page No 473"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#assume if there are 16 outcomes per second\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=1/2.0\n",
+      "px2=1/4.0\n",
+      "px3=1/8.0\n",
+      "px4=1/16.0\n",
+      "px5=1/16.0\n",
+      "r=16.0\n",
+      "\n",
+      "#Calculation\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)+px5*math.log(1/px5,2)\n",
+      "\n",
+      "#Result\n",
+      "print('1. Entropy H(X) = %.2f bits/symbol ' %HX) #source entropy\n",
+      "\n",
+      "R=r*HX\n",
+      "print('2., Information rate R = %.f bits/sec' %R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. Entropy H(X) = 1.88 bits/symbol \n",
+        "2., Information rate R = 30 bits/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.18, Page No 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#determine entropy ,information rate\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=1/4.0\n",
+      "px2=1/5.0\n",
+      "px3=1/5.0\n",
+      "px4=1/10.0\n",
+      "px5=1/10.0\n",
+      "px6=1/20.0\n",
+      "px7=1/20.0\n",
+      "px8=1/20.0\n",
+      "f=10*10**3.0\n",
+      "fs=10*2*10**3.0\n",
+      "\n",
+      "#Calculation\n",
+      "#entropy\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)+px5*math.log(1/px5,2)+px6*math.log(1/px6,2)+px7*math.log(1/px7,2)+px8*math.log(1/px8,2) \n",
+      "\n",
+      "#Result\n",
+      "print('bits/message H(X) = %.2f ' %HX)\n",
+      "r=fs\n",
+      "R=r*HX    #information rate\n",
+      "print('bits/sec R = %.2f' %R)\n",
+      "print('Approximation error')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "bits/message H(X) = 2.74 \n",
+        "bits/sec R = 54828.92\n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.19, Page No 476 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from array import *\n",
+      "#Find Channel Matrix,joint probability\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=0.5\n",
+      "px2=0.5\n",
+      "py1x1=0.9\n",
+      "py2x1=0.1\n",
+      "py1x2=0.2\n",
+      "py2x2=0.8\n",
+      "PYX=[[py1x1,py2x1],[py1x2,py2x2]]\n",
+      "PX=[[px1,px2]]\n",
+      "PY = [[0,0],\n",
+      "    [0,0]]\n",
+      "PXY = [[0,0],\n",
+      "    [0,0]]\n",
+      "\n",
+      "for i in range(len(PYX)):\n",
+      "   # iterate through columns of Y\n",
+      "   for j in range(len(PX[0])):\n",
+      "       # iterate through rows of Y\n",
+      "       for k in range(len(PX)):\n",
+      "           PY[i][j] += PYX[i][k] * PX[k][j]\n",
+      "print('  PY ARRAY =  \\n')\n",
+      "for r in PY:\n",
+      "   print(r)\n",
+      "PXd=[[px1,0],[0,px2]]\n",
+      "\n",
+      "\n",
+      "for i in range(len(PXd)):\n",
+      "   # iterate through columns of Y\n",
+      "   for j in range(len(PYX[0])):\n",
+      "       # iterate through rows of Y\n",
+      "       for k in range(len(PYX)):\n",
+      "           PXY[i][j] += PXd[i][k] * PYX[k][j]\n",
+      "\n",
+      "            \n",
+      "print(' \\n PXY ARRAY =  \\n')\n",
+      "for r in PXY:\n",
+      "   print(r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  PY ARRAY =  \n",
+        "\n",
+        "[0.45, 0.45]\n",
+        "[0.1, 0.1]\n",
+        " \n",
+        " PXY ARRAY =  \n",
+        "\n",
+        "[0.45, 0.05]\n",
+        "[0.1, 0.4]\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.35, Page No 498"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Channel is aproximated by the AWGN Channel\n",
+      "\n",
+      "#Variable Declaration\n",
+      "B=4000.0\n",
+      "S=0.1*10**-3\n",
+      "n=2*10**-12\n",
+      "\n",
+      "#Calculation\n",
+      "N=n*B\n",
+      "C=B*math.log(1+(S/N),2) #Capacity of Channel\n",
+      "C=C/1000.0\n",
+      "#Result\n",
+      "print(' Capacity of Channel  C=%.3f(10^3) b/s ' %C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Capacity of Channel  C=54.439(10^3) b/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.36i, Page No 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#assume that  succeissive samples are statistically independent\n",
+      "\n",
+      "#Variable Declaration\n",
+      "fm=4000.0\n",
+      "fs=2*fm\n",
+      "n=1.25\n",
+      "\n",
+      "#Calculation\n",
+      "r=fs*n\n",
+      "pxi=1/256.0\n",
+      "HX=-math.log(pxi,2)\n",
+      "R=r*HX\n",
+      "R=R/1000\n",
+      "print('Information Rate  R= %.f kb/s' %R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Information Rate  R= 80 kb/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.36ii, Page No 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#assume that  succeissive samples are statistically independent\n",
+      "\n",
+      "#Variable Declaration\n",
+      "B=10*10**3.0\n",
+      "SN=20.0\n",
+      "\n",
+      "#Calculation\n",
+      "SNR=10**(SN/10.0)\n",
+      "C=B*math.log(1+(SNR),2)\n",
+      "C=C/1000\n",
+      "\n",
+      "#Result\n",
+      "print('The channel capacity = %.2f 10^3 b/s' %C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The channel capacity = 66.58 10^3 b/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.36iii, Page No 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#assume that  succeissive samples are statistically independent\n",
+      "\n",
+      "#Variable Declaration\n",
+      "C=8*10**4.0\n",
+      "B=10**4.0\n",
+      "\n",
+      "#Calculation\n",
+      "SN=2**(C/B)-1\n",
+      "SNR=10*math.log(SN,10) #SNR\n",
+      "\n",
+      "#Result\n",
+      "print(' The S/N ratio required for error-free transmission  =%.2f dB ' %SNR)     #required SNR is greater that 24.064\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The S/N ratio required for error-free transmission  =24.07 dB \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.36iv, Page No 499 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#assume that  succeissive samples are statistically independent\n",
+      "\n",
+      "#Variable Declaration\n",
+      "SN=20.0\n",
+      "\n",
+      "#Calculation\n",
+      "SNR=10**(SN/10.0)\n",
+      "C=8*10**4.0\n",
+      "B=C/(math.log(1+SNR,2)) #Bandwidth\n",
+      "B=B/1000\n",
+      "\n",
+      "#Result\n",
+      "print('Bandwidth required for AWGN channel B =%.2f kHz ' %B)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bandwidth required for AWGN channel B =12.02 kHz \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.37, Page No 502"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find code efficiency,redundancy\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=0.9\n",
+      "px2=0.1\n",
+      "n1=1.0\n",
+      "n2=1.0\n",
+      "\n",
+      "#Calculation\n",
+      "L=px1*n1+px2*n2 #code leght\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)\n",
+      "n=(HX/L) #code efficiency\n",
+      "n=n*100\n",
+      "\n",
+      "print('Code efficiency = %.1f percent'  %n)\n",
+      "r=(100-n) #code reduncy\n",
+      "print('Code redundancy = %.1f percent' %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Code efficiency = 46.9 percent\n",
+        "Code redundancy = 53.1 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.38, Page No 502"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find code efficiency,redundancy\n",
+      "\n",
+      "#Variable Declaration\n",
+      "pa1=0.81\n",
+      "pa2=0.09 \n",
+      "pa3=0.09\n",
+      "pa4=0.01 \n",
+      "n1=1\n",
+      "n2=2 \n",
+      "n3=3\n",
+      "n4=3 \n",
+      "\n",
+      "#Calculation\n",
+      "L=pa1*n1+pa2*n2+pa3*n3+pa4*n4\n",
+      "HX2=pa1*math.log(1/pa1,2)+pa2*math.log(1/pa2,2)+pa3*math.log(1/pa3,2)+pa4*math.log(1/pa4,2)\n",
+      "n=HX2/L*100\n",
+      "\n",
+      "#Result\n",
+      "print(' code efficiency = %.2f  percent' %n)\n",
+      "\n",
+      "r=(100-n) #code reduncy\n",
+      "print(' code redundancy = %.1f percent' %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " code efficiency = 72.71  percent\n",
+        " code redundancy = 27.3 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.44, Page No 507"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find  efficiency of the code\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=1/2.0\n",
+      "px2=1/4.0\n",
+      "px3=1/8.0\n",
+      "px4=1/8.0\n",
+      "n1=1.0\n",
+      "n2=2.0\n",
+      "n3=3.0\n",
+      "n4=3.0\n",
+      "\n",
+      "#Calculation\n",
+      "#information content of each symbol\n",
+      "Ix1=-math.log(px1,2)\n",
+      "Ix2=-math.log(px2,2)\n",
+      "Ix3=-math.log(px3,2)\n",
+      "Ix4=-math.log(px4,2)\n",
+      "\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)+px4*math.log(1/px4,2)\n",
+      "L=px1*n1+px2*n2+px3*n3+px4*n4\n",
+      "\n",
+      "n=HX/L*100\n",
+      "\n",
+      "#Result\n",
+      "print('Ccode efficiency = %.f  Percent' %n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ccode efficiency = 100  Percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.50, Page No 512"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find entropy ,information rate\n",
+      "#If there are 16 outcomes per second\n",
+      "\n",
+      "#Variable Declaration\n",
+      "P1=1/2.0\n",
+      "P2=1/4.0\n",
+      "P3=1/8.0\n",
+      "P4=1/16.0\n",
+      "P5=1/32.0\n",
+      "P6=1/32.0\n",
+      "r=16     #message rate\n",
+      "\n",
+      "#Calculation\n",
+      "H=P1*math.log(1/P1,2)+P2*math.log(1/P2,2)+P3*math.log(1/P3,2)+P4*math.log(1/P4,2)+P5*math.log(1/P5,2)+P6*math.log(1/P6,2)\n",
+      "#Entropy of system\n",
+      "\n",
+      "#Result\n",
+      "print('1. Entropy of system H = %.2f bits/message ' %H)\n",
+      "R=H*r                            #R=Entropy*message rate\n",
+      "print(' 2. Information rate  R = %.f bits/sec ' %R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. Entropy of system H = 1.94 bits/message \n",
+        " 2. Information rate  R = 31 bits/sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.51, Page No 512"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Calculate H(X) ,H(Y)\n",
+      "\n",
+      "#Variable Declaration\n",
+      "px1=0.3\n",
+      "px2=0.4\n",
+      "px3=0.3\n",
+      "\n",
+      "#Calculation\n",
+      "HX=px1*math.log(1/px1,2)+px2*math.log(1/px2,2)+px3*math.log(1/px3,2)  #Entropy of X\n",
+      "\n",
+      "\n",
+      "print(' 1.Entropy of X  H(X)=%.3f bits/symbol ' %HX)\n",
+      "\n",
+      "PYX=[[0.8, 0.2, 0],[ 0, 1, 0],[ 0, 0.3, 0.7]]\n",
+      "PX=[[px1, px2, px3]]\n",
+      "PXY = [[0,0,0],\n",
+      "    [0,0,0],\n",
+      "    [0,0,0]]\n",
+      "\n",
+      "for i in range(len(PYX)):\n",
+      "   # iterate through columns of PXd\n",
+      "   for j in range(len(PX[0])):\n",
+      "       # iterate through rows of PYX\n",
+      "       for k in range(len(PX)):\n",
+      "           PXY[i][j] += PYX[i][k] * PX[k][j]\n",
+      "\n",
+      "py1=PXY[0][0]\n",
+      "py2=PXY[0][1]\n",
+      "py3=PXY[0][2]\n",
+      "HY=py1*math.log(1/py1,2)+py2*math.log(1/py2,2)+py3*math.log(1/py3,2)  #Entropy of Y\n",
+      "print(' 2. Entropy of Y H(Y)= %.2f bits/symbol ' %HY)\n",
+      "print('Approximation error')\t\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1.Entropy of X  H(X)=1.571 bits/symbol \n",
+        " 2. Entropy of Y H(Y)= 1.51 bits/symbol \n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.52, Page No 513"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find entropy of source ,entropy of second order extension\n",
+      "\n",
+      "#Variable Declaration\n",
+      "P1=0.7\n",
+      "P2=0.15\n",
+      "P3=0.15\n",
+      "\n",
+      "#Calculation\n",
+      "HX=P1*math.log(1/P1,2)+P2*math.log(1/P2,2)+P3*math.log(1/P3,2)     #Entropy of source\n",
+      "print(' 1. Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n",
+      "#H(X^n)=n*H(X)\n",
+      "n=2    #for second order\n",
+      "HX2=n*HX\n",
+      "\n",
+      "#Result\n",
+      "print(' 2. Entropy of second order system extension of source can be H(X^2)=%.2f bits/symbol ' %(HX*2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1. Entropy of system  H(X)=1.18 bits/symbol \n",
+        " 2. Entropy of second order system extension of source can be H(X^2)=2.36 bits/symbol \n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.54, Page No 514"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find entropy of source \n",
+      "\n",
+      "#Variable Declaration\n",
+      "S0=1/3.0\n",
+      "S1=1/6.0\n",
+      "S2=1/4.0\n",
+      "S3=1/4.0\n",
+      "\n",
+      "#Calculation\n",
+      "HX=S0*math.log(1/S0,2)+S1*math.log(1/S1,2)+S2*math.log(1/S2,2)+S3*math.log(1/S3,2)   #EntroSy of source\n",
+      "\n",
+      "#Result\n",
+      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Entropy of system  H(X)=1.96 bits/symbol \n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.56, Page No 515"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Information capacity of telephone\n",
+      "\n",
+      "#Variable Declaration\n",
+      "B=3.4*10**3\n",
+      "SNR=30.0\n",
+      "\n",
+      "#Calculation\n",
+      "SN=10**(SNR/10)\n",
+      "C=B*math.log(1+SN,2)   #Information capacity\n",
+      "C=C/1000\n",
+      "\n",
+      "#Result\n",
+      "print(' Information capacity of telephone is C = %.2f kbps ' %C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Information capacity of telephone is C = 33.89 kbps \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.59, Page No 516"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find entropy of source \n",
+      "\n",
+      "#Variable Declaration\n",
+      "S0=1/3.0\n",
+      "S1=1/6.0\n",
+      "S2=1/4.0\n",
+      "S3=1/4.0\n",
+      "\n",
+      "#Calculation\n",
+      "HX=S0*math.log(1/S0,2)+S1*math.log(1/S1,2)+S2*math.log(1/S2,2)+S3*math.log(1/S3,2)   #EntroSy of source\n",
+      "\n",
+      "#Result\n",
+      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Entropy of system  H(X)=1.96 bits/symbol \n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.60, Page No 516"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find entropy of source \n",
+      "\n",
+      "#Variable Declaration\n",
+      "m1=1/2.0\n",
+      "m2=1/4.0\n",
+      "m3=1/8.0\n",
+      "m4=1/16.0\n",
+      "m5=1/16.0\n",
+      "\n",
+      "#Calculation\n",
+      "L=(m1*1)+(m2*2)+(m3*3)+(2*(m4)*4)\n",
+      "\n",
+      "#Result\n",
+      "print(' Average number of bits per message =%.2f bits ' %L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Average number of bits per message =1.88 bits \n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.61, Page No 517"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find Information capacity of telephone\n",
+      "\n",
+      "#Variable Declaration\n",
+      "B=3.4*10**3\n",
+      "SNR=30.0\n",
+      "\n",
+      "#Calculation\n",
+      "SN=10**(SNR/10)\n",
+      "C=B*math.log(1+SN,2)   #Information capacity\n",
+      "C=C/1000\n",
+      "\n",
+      "#Result\n",
+      "print(' Information capacity of telephone is C = %.2f kbps ' %C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Information capacity of telephone is C = 33.89 kbps \n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.62, Page No 517"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find entropy of source \n",
+      "\n",
+      "#Variable Declaration\n",
+      "p1=4.0\n",
+      "m1=0.5\n",
+      "m2=0.5\n",
+      "m3=0.375\n",
+      "m4=0.375\n",
+      "m5=0.375\n",
+      "m6=0.375\n",
+      "\n",
+      "#Calculation\n",
+      "I1=p1*math.log(1/p1,2) \n",
+      "HX=m1*math.log(1/m1,2)+m2*math.log(1/m2,2)+m3*math.log(1/m3,2)+m4*math.log(1/m4,2)+m5*math.log(1/m5,2)+m6*math.log(1/m6,2)   #EntroSy of source\n",
+      "\n",
+      "#Result\n",
+      "print(' Entropy of system  H(X)=%.2f bits/symbol ' %HX)\n",
+      "print('Approximation error')\t"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Entropy of system  H(X)=3.12 bits/symbol \n",
+        "Approximation error\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.65, Page No 519"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find entropy of source \n",
+      "\n",
+      "#Variable Declaration\n",
+      "S0=1/2.0\n",
+      "S1=1/4.0\n",
+      "S2=1/8.0\n",
+      "S3=1/8.0\n",
+      "n=1\n",
+      "\n",
+      "#Calculation\n",
+      "H=S0*math.log(1.0/S0,2)+S1*math.log(1.0/S1,2)+S2*math.log(1.0/S2,2)+S3*math.log(1.0/S3,2)   #EntroSy of source\n",
+      "L=H*n\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print(' Code length =%.2f bits/messages ' %L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Code length =1.75 bits/messages \n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.67, Page No 520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Find channel capacity and new bandwidth\n",
+      "\n",
+      "#Variable Declaration\n",
+      "B=8*10**3\n",
+      "SNR=31.0\n",
+      "SNR2=61\n",
+      "\n",
+      "#Calculation\n",
+      "C=B*math.log(1+SNR,2)   #Information capacity\n",
+      "B2=C/math.log(1+SNR2,2)\n",
+      "#Result\n",
+      "print(' Channel capacity is C = %.2f x 10^3 bits/sec ' %(C/1000))\n",
+      "print(' New Bandwidth is C = %.2f x kHz ' %(B2/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Channel capacity is C = 40.00 x 10^3 bits/sec \n",
+        " New Bandwidth is C = 6.72 x kHz \n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/Chapter9.ipynb b/Digital_Communications_by_S._Sharma/Chapter9.ipynb
new file mode 100755
index 00000000..c91e1583
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/Chapter9.ipynb
@@ -0,0 +1,166 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1> Chpater 9: ERROR CONTROL CODING<h1>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find  detected errors,corrected errors\n",
+      "\n",
+      "#initialisation of variables\n",
+      "dmin=5.0\n",
+      "#(s+1)<= dmin number errors can be detected(s)\n",
+      " \n",
+      "#CALCULATIONS\n",
+      "s=dmin-1\n",
+      "\n",
+      "#RESULTS\n",
+      "print(' i)Number of detected errors  s <=  %.f ' %s)\n",
+      "#(2t+1)<=dmin number errors can be corrected(t)\n",
+      "t=(dmin-1)/2.0\n",
+      "print('ii) Number of corrected errors  t<=  %.f ' %t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " i)Number of detected errors  s <=  4 \n",
+        "ii) Number of corrected errors  t<=  2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.17, Page No 569"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine all possible code vectors \n",
+      "\n",
+      "m3=1\n",
+      "m2=0\n",
+      "m1=1\n",
+      "m0=0\n",
+      "#M=Message Matrix\n",
+      "#G=Generator Matrix\n",
+      "G=[[1, 0, 1, 1, 0, 0, 0],[0, 1, 0, 1, 1, 0, 0],[0, 0, 1, 0, 1, 1, 0],[0, 0, 0, 1, 0, 1, 1]]\n",
+      "M=[[m3,m2,m1,m0]]\n",
+      "X = [[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0]]\n",
+      "\n",
+      "for i in range(len(G)):\n",
+      "   # iterate through columns of PXd\n",
+      "   for j in range(len(M[0])):\n",
+      "       # iterate through rows of PYX\n",
+      "       for k in range(len(M)):\n",
+      "           X[i][j] += G[i][k] * M[k][j]\n",
+      "print('The required code word')\n",
+      "for r in range(0,7):\n",
+      "         print(X[0][r])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required code word\n",
+        "1\n",
+        "0\n",
+        "1\n",
+        "0\n",
+        "0\n",
+        "0\n",
+        "0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.19, Page No 572"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine  code word \n",
+      "m3=1\n",
+      "m2=0\n",
+      "m1=1\n",
+      "m0=0\n",
+      "#M=Message Matrix\n",
+      "#G=Generator Matrix\n",
+      "G=[[1, 0, 0, 0, 1, 0, 1],[0, 1, 0, 0, 1, 1, 1],[0, 0, 1, 0, 1, 1, 0],[0, 0, 0, 1, 0, 1, 1]]\n",
+      "M=[[m3,m2,m1,m0]]\n",
+      "X = [[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0],[0, 0, 0, 0, 0, 0, 0]]\n",
+      "\n",
+      "\n",
+      "for i in range(len(G)):\n",
+      "   # iterate through columns of PXd\n",
+      "   for j in range(len(M[0])):\n",
+      "       # iterate through rows of PYX\n",
+      "       for k in range(len(M)):\n",
+      "           X[i][j] += G[i][k] * M[k][j]\n",
+      "print('The required code word')\n",
+      "for r in range(0,7):\n",
+      "         print(X[0][r] )\n",
+      "     \n",
+      "\n",
+      "print('The code in the book is wrong')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required code word\n",
+        "1\n",
+        "0\n",
+        "1\n",
+        "0\n",
+        "0\n",
+        "0\n",
+        "0\n",
+        "The code in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Digital_Communications_by_S._Sharma/README.txt b/Digital_Communications_by_S._Sharma/README.txt
new file mode 100755
index 00000000..cb9038a1
--- /dev/null
+++ b/Digital_Communications_by_S._Sharma/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Mayur Sabban
+Course: be
+College/Institute/Organization: Vishwakarma Institute of Technology Pune
+Department/Designation: Computer Engg
+Book Title: Digital Communications
+Author:  S. Sharma
+Publisher: S. K. Kataria & Sons, New Delhi
+Year of publication: 2013
+Isbn: 9789350142684
+Edition: 6
\ No newline at end of file
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+Contributed By: kiruthika gowthaman
+Course: be
+College/Institute/Organization: Sona College of Technology,Salem,Tamil Nadu,India
+Department/Designation: Computer Science 
+Book Title: Discrete Mathematics and its Applications
+Author: Kenneth H.Rosen
+Publisher: McGrawhill
+Year of publication: 2012
+Isbn: 9780071315012
+Edition: 7
\ No newline at end of file
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+{"nbformat_minor": 0, "cells": [{"source": "#Chapter 02: Basic Structures: Sets, Functions, Sequences, Sums and Matrices", "cell_type": "markdown", "metadata": {}}, {"source": "##Example 01: Page 156", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "#To generate a sequence a_n=1/n\ni=1.0 #floating point division\nn=input(\"enter the number of terms in the sequence\");\nprint \"a_n=1/n\"\nprint \"when n=\",n,\"a_n is\"\nfor i in range(1,n+1): #iteration till the number of terms specified by the user\n    a=1.0/i\n    print \"1/\",i,\",\",\nprint \"\\n\"\nfor i in range(1,n+1): #iteration till the number of terms specified by the user\n    a=1.0/i\n    print a,\",\",\n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "enter the number of terms in the sequence5\na_n=1/n\nwhen n= 5 a_n is\n1/ 1 , 1/ 2 , 1/ 3 , 1/ 4 , 1/ 5 , \n\n1.0 , 0.5 , 0.333333333333 , 0.25 , 0.2 ,\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 02: Page 157", "cell_type": "markdown", "metadata": {}}, {"execution_count": 5, "cell_type": "code", "source": "n=input(\"Enter the number of terms in the sequence to generate the geometric progression\");\ni=1\nprint\"the list of terms\",\nfor i in range (n+1):print\"b\",i,\",\",\nprint \"begins with\", \nfor i in range (n+1): #iterate for the number of terms given as input\n    b_n=(-1)**i\n    print b_n,\nprint\"\\n\",\"the list of terms\",\nfor i in range (n+1):print\"c\",i,\",\",\nprint \"begins with\",   \nfor i in range (n+1): #iterate for the number of terms given as input\n    c_n=2*(5**i)\n    print c_n,\nprint\"\\n\",\"the list of terms\",\nfor i in range (n+1):print\"c\",i,\",\",\nprint \"begins with\",\nfor i in range (n+1): #iterate for the number of terms given as input\n    d_n=6.0*((1.0/3.0)**i)\n    print d_n, #prints the fraction values in decimals.  Floating point division\n\n    \n   \n    \n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number of terms in the sequence to generate the geometric progression5\nthe list of terms b 0 , b 1 , b 2 , b 3 , b 4 , b 5 , begins with 1 -1 1 -1 1 -1 \nthe list of terms c 0 , c 1 , c 2 , c 3 , c 4 , c 5 , begins with 2 10 50 250 1250 6250 \nthe list of terms c 0 , c 1 , c 2 , c 3 , c 4 , c 5 , begins with 6.0 2.0 0.666666666667 0.222222222222 0.0740740740741 0.0246913580247\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 03: Page 157", "cell_type": "markdown", "metadata": {}}, {"execution_count": 6, "cell_type": "code", "source": "n=input(\"Enter the number terms in the sequence\");\ns_n=-1+4*n\nt_n=7-3*n\ni=0\nprint \"The list of terms\",\nfor i in range(n):  \n    print \"s\",i,\",\",\nprint \"begins with\",\nfor i in range(n):#generates the sequence for -1*4i\n    print -1+4*i,\nprint \"\\nThe list of terms\",\nfor i in range(n):\n    print \"t\",i,\",\",\nprint \"begins with\",\nfor i in range(n):#generates the sequence for 7-3i\n    print 7-3*i,\n    \n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number terms in the sequence5\nThe list of terms s 0 , s 1 , s 2 , s 3 , s 4 , begins with -1 3 7 11 15 \nThe list of terms t 0 , t 1 , t 2 , t 3 , t 4 , begins with 7 4 1 -2 -5\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 05: Page 158", "cell_type": "markdown", "metadata": {}}, {"execution_count": 9, "cell_type": "code", "source": "a=[2,0,0,0] #assigning a[0]=2 (Given)\n\nfor i in range(1,4):#iteration to run till a[3]\n    a[i]=a[i-1]+3\n    print \"a[\",i,\"]\",a[i]", "outputs": [{"output_type": "stream", "name": "stdout", "text": "a[ 1 ] 5\na[ 2 ] 8\na[ 3 ] 11\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 06: Page 158", "cell_type": "markdown", "metadata": {}}, {"execution_count": 11, "cell_type": "code", "source": "a=[3,5,0,0] #assingning a[0],a[1] to the given values\n\nfor i in range(2,4): # iterations to find the successive values. If values are to be found for further terms the for loop \"stop\" has to be modified\n    a[i]=a[i-1]-a[i-2]\n    print \"a[\",i,\"]\",a[i]", "outputs": [{"output_type": "stream", "name": "stdout", "text": "a[ 2 ] 2\na[ 3 ] -3\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 07: Page 158 ", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "f=[0,1,0,0,0,0,0] #assingning a[0],a[1] to the given values\nprint \"Fibonacci series is\"\nfor i in range(2,7): # iterations to find the successive values. If values are to be found for further terms the for loop \"stop\" has to be modified\n    f[i]=f[i-1]+f[i-2]\n    print \"f[\",i,\"]=f[\",i-1,\"]+f[\",i-2,\"]=\",f[i]", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Fibonacci series is\nf[ 2 ]=f[ 1 ]+f[ 0 ]= 1\nf[ 3 ]=f[ 2 ]+f[ 1 ]= 2\nf[ 4 ]=f[ 3 ]+f[ 2 ]= 3\nf[ 5 ]=f[ 4 ]+f[ 3 ]= 5\nf[ 6 ]=f[ 5 ]+f[ 4 ]= 8\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 08: Page 159 ", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "n=1\nresult=0\nnumber=input(\"Enter the number\");\nfor i in range(1,number):\n    n=n+i*n \nprint \"The factorial of\",number,\"is\",n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number5\nThe factorial of 5 is 120\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 18: Page 164", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "#finding the summation of j^2\nup=input(\"Enter the upper limit for the operation j^2\");\nlow=input(\"Enter the lower limit for the operation j^2\");\nsum=0\nprint \"The square of terms form 1 to n\",\nfor j in range (low,up+1): #summation.  Iteration from lower to upper limit.\n    print j,\"^2+\",\n    j=j**2 #square function is computed as '**'\n    sum=sum+j\nprint \"=\",sum\n                \n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the upper limit for the operation j^25\nEnter the lower limit for the operation j^21\nThe square of terms form 1 to n 1 ^2+ 2 ^2+ 3 ^2+ 4 ^2+ 5 ^2+ = 55\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 19: Page 164", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "k=4 #lower limit\nsum=0\nprint \"The value for the sequence\",\nfor k in range (4,8+1,1): #8+1 , 8 is the upper limit, in python to make for loop run till the limit equal to upper limit we give a +1.\n    print \"(-1)^\",k,\"+\",\n    sum=sum+((-1)**k)\nprint \"=\",sum\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The value for the sequence (-1)^ 4 + (-1)^ 5 + (-1)^ 6 + (-1)^ 7 + (-1)^ 8 + = 1\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 21: Page 165", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "\nglobals()['j']=0\ni=0\nglobals()['s']=0\nupj=input(\"Enter the upper limit for the inner summation\");\nlowj=input(\"Enter the lower limit for the inner summation\");\nupi=input(\"Enter the upper limit for the outer summation\");\nlowi=input(\"Enter the lower limit for the outer summation\");\nfor i in range (lowj,upj+1):\n    j=j+i\nfor l in range(lowi,upi+1):\n    s=s+(j*l)\nprint s\n    \n    \n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the upper limit for the inner summation3\nEnter the lower limit for the inner summation1\nEnter the upper limit for the outer summation4\nEnter the lower limit for the outer summation1\n60\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 13: Page 161", "cell_type": "markdown", "metadata": {"collapsed": true}}, {"execution_count": 4, "cell_type": "code", "source": "#To print series 1 once, 2 twice, 3 thrice and so on\na=[]\ni=1\nfor i in range(1,10+1): #for loop to initialise the number\n    for j in range(1,i+1):#for loop to iterate to make the count\n        print i,\n    ", "outputs": [{"output_type": "stream", "name": "stdout", "text": "1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 22: Page 166", "cell_type": "markdown", "metadata": {}}, {"execution_count": 5, "cell_type": "code", "source": "s=0 #initialise it to zero to store the results\nglobals()['res']=0 #difining result as global variable since it has to be accessed outside the loop\nprint \"Sum of values of s for all the members of the set {\",\nfor s in range (0,4+1,2): #iterate for terms 0,4,6\n    print s,\n    res=res+s\nprint \"} is\",res\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Sum of values of s for all the members of the set { 0 2 4 } is 6\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 02: Page 178", "cell_type": "markdown", "metadata": {}}, {"execution_count": 6, "cell_type": "code", "source": "def getmat(): #function to get the matrix elements\n    m = int(input('number of rows, m = '))\n    n = int(input('number of columns, n = '))\n    matrix = []; columns = []\n\n    for i in range(0,m):\n        matrix.append([])\n        for j in range(0,n):\n            matrix[i].append(0)\n            print ('entry in row: ',i+1,' column: ',j+1)\n            matrix[i][j] = int(input())\n    return (matrix)\n\ndef matrixADD(m1,m2): #function to add the matrix.\n    z=[]\n    for i in range (len(m1)):\n        tem = []\n        for j in range (len(m2)):\n            x=m1[i][j]+m2[i][j]\n            tem.append(x)\n        z.append(tem)\n    return z    \n        \n\n\nmat1=[]\nmat2=[]\nZ=[]\nprint \"Enter the elements of matrix 1\"\nmat1=getmat() #function call\nprint \"Enter the elements of matrix 2\"\nmat2=getmat() #function call\nprint \"Addition of \\n matrix 1\",mat1,\"and \\n matrix 2\",mat2,\"is \\n\",matrixADD(mat1,mat2) #function call to add \n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the elements of matrix 1\nnumber of rows, m = 3\nnumber of columns, n = 3\n('entry in row: ', 1, ' column: ', 1)\n1\n('entry in row: ', 1, ' column: ', 2)\n0\n('entry in row: ', 1, ' column: ', 3)\n-1\n('entry in row: ', 2, ' column: ', 1)\n2\n('entry in row: ', 2, ' column: ', 2)\n2\n('entry in row: ', 2, ' column: ', 3)\n-3\n('entry in row: ', 3, ' column: ', 1)\n3\n('entry in row: ', 3, ' column: ', 2)\n4\n('entry in row: ', 3, ' column: ', 3)\n0\nEnter the elements of matrix 2\nnumber of rows, m = 3\nnumber of columns, n = 3\n('entry in row: ', 1, ' column: ', 1)\n3\n('entry in row: ', 1, ' column: ', 2)\n4\n('entry in row: ', 1, ' column: ', 3)\n-1\n('entry in row: ', 2, ' column: ', 1)\n1\n('entry in row: ', 2, ' column: ', 2)\n-3\n('entry in row: ', 2, ' column: ', 3)\n0\n('entry in row: ', 3, ' column: ', 1)\n-1\n('entry in row: ', 3, ' column: ', 2)\n1\n('entry in row: ', 3, ' column: ', 3)\n2\nAddition of \n matrix 1 [[1, 0, -1], [2, 2, -3], [3, 4, 0]] and \n matrix 2 [[3, 4, -1], [1, -3, 0], [-1, 1, 2]] is \n[[4, 4, -2], [3, -1, -3], [2, 5, 2]]\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 03: Page 179", "cell_type": "markdown", "metadata": {}}, {"execution_count": 7, "cell_type": "code", "source": "\n# Program to multiply two matrices using nested loops\n\n# 3x3 matrix\nX = [[1,0,4],\n    [2,1,1],\n    [3,1,0],\n     [0,2,2]]\n# 3x4 matrix\nY = [[2,4],\n    [1,1],\n    [3,0]]\n# result is 3x4\nresult = [[0,0],\n         [0,0],\n         [0,0,],\n          [0,0]]\n\n# iterate through rows of X\nfor i in range(len(X)):\n   # iterate through columns of Y\n   for j in range(len(Y[0])):\n       # iterate through rows of Y\n       for k in range(len(Y)):\n           result[i][j] += X[i][k] * Y[k][j]\nprint \"The multiplication of the two matrices AB is\"\n\nfor r in result:\n   print(r)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The multiplication of the two matrices AB is\n[14, 4]\n[8, 9]\n[7, 13]\n[8, 2]\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 05: Page 181", "cell_type": "markdown", "metadata": {}}, {"execution_count": 8, "cell_type": "code", "source": "# Program to transpose a matrix using nested loop\n# iterate through rows\ndef mattrans(X,result):\n    print \"The transpose is\"\n    for i in range(len(X)):\n   # iterate through columns\n       for j in range(len(X[0])):\n           result[j][i] = X[i][j]\n    for r in result:\n        print(r)\n        \ndef getmat():\n    row = int(input('number of rows = '))\n    col = int(input('number of columns = '))\n    matrix = []; columns = []\n\n    for i in range(0,row):\n        matrix.append([])\n        for j in range(0,col):\n            matrix[i].append(0)\n            print ('entry in row: ',i+1,' column: ',j+1)\n            matrix[i][j] = int(input())\n    for c in range(col):\n        for r in range(row):\n            result[c][r]=0\n    mattrans(matrix,result)\n    \n\n\nprint \"Enter the elements of the matrix\"\ngetmat()\n#mattrans(mat1)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the elements of the matrix\nnumber of rows = 2\nnumber of columns = 3\n('entry in row: ', 1, ' column: ', 1)\n1\n('entry in row: ', 1, ' column: ', 2)\n2\n('entry in row: ', 1, ' column: ', 3)\n3\n('entry in row: ', 2, ' column: ', 1)\n4\n('entry in row: ', 2, ' column: ', 2)\n5\n('entry in row: ', 2, ' column: ', 3)\n6\nThe transpose is\n[1, 4]\n[2, 5]\n[3, 6]\n[8, 2]\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "", "cell_type": "markdown", "metadata": {}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
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+{"nbformat_minor": 0, "cells": [{"source": "#Chapter 01: The Foundations: Logic and Proofs\n", "cell_type": "markdown", "metadata": {}}, {"source": "##Example 01:Page 02", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "print \"The following sentences are Propositions\" #Proposition should be a declarative sentence or should result in either a YES or a NO.\n\nprint \"1.  Washington D.C is the capital of the United States of America\\n2.  Toronto is the capital of Canada\\n3.  1+1=2.\\n4.  2+2=3.\" #Since these statements are declarative and they answer the question YES or NO they are called propositions.\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The following sentences are Propositions\n1.  Washington D.C is the capital of the United States of America\n2.  Toronto is the capital of Canada\n3.  1+1=2.\n4.  2+2=3.\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 02:Page 02", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "print \"1.  What time is it? \\n2.  Read this carefully. \\n3.  x+1=2.\\n4.  x+y=Z.\"\nprint\"Sentences 1 and 2 are not propositions since they are not declarative.  Sentences 3 and 4 are neither true nor false and so they are not propositions.\"\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "1.  What time is it? \n2.  Read this carefully. \n3.  x+1=2.\n4.  x+y=Z.\nSentences 1 and 2 are not propositions since they are not declarative.  Sentences 3 and 4 are neither true nor false and so they are not propositions.\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 03:Page 03", "cell_type": "markdown", "metadata": {}}, {"execution_count": 9, "cell_type": "code", "source": "print \"Propositon p=Michael's PC runs Linux.\"\nprint \"\\n Negation of p is ~p : It is not the case that Michael's PC runs Linux.\"\nprint \"\\n Negation of p is ~p : Michae's PC does not run.\"#Negation is opposite of the truth value of the proposition expressed with \"it is not the case that\" or with \"not\".\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Propositon p=Michael's PC runs Linux.\n\n Negation of p is ~p : It is not the case that Michael's PC runs Linux.\n\n Negation of p is ~p : Michae's PC does not run.\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 04:Page 03", "cell_type": "markdown", "metadata": {}}, {"execution_count": 10, "cell_type": "code", "source": "print \"Let p=Vandana's smartphone has at least 32GB of memory.\"\nprint \"The negation of p is ( ~p ) :It is not the case that Vandana's smartphone has at least 32GB of memory.\"\nprint \"Or in simple English ( ~p ): Vandana's smartphone does not have at least 32GB of memory.\"\nprint \"Or even more simple as ( ~p ): Vandana's smartphone has less than 32GB of memory.\"\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Let p=Vandana's smartphone has at least 32GB of memory.\nThe negation of p is ( ~p ) :It is not the case that Vandana's smartphone has at least 32GB of memory.\nOr in simple English ( ~p ): Vandana's smartphone does not have at least 32GB of memory.\nOr even more simple as ( ~p ): Vandana's smartphone has less than 32GB of memory.\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 05:Page 04", "cell_type": "markdown", "metadata": {}}, {"execution_count": 11, "cell_type": "code", "source": "p=\"Rebecca's PC has more than 16GB free hard disk space\"\nq=\"The processor in Rebecca's PC runs faster than 1GHz\"\nprint \"Let p,q be two propositions\"\nprint \"Let p=\",p,\"\\n\",\"Let q=\",q\nprint \"Conjunction of p^q is : \"+p+\" and \"+q #conjunction combines two propositons with \"and\"\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Let p,q be two propositions\nLet p= Rebecca's PC has more than 16GB free hard disk space \nLet q= The processor in Rebecca's PC runs faster than 1GHz\nConjunction of p^q is : Rebecca's PC has more than 16GB free hard disk space and The processor in Rebecca's PC runs faster than 1GHz\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 06:Page 05", "cell_type": "markdown", "metadata": {}}, {"execution_count": 12, "cell_type": "code", "source": "p=\"Rebecca's PC has more than 16GB free hard disk space\"\nq=\"The processor in Rebecca's PC runs faster than 1GHz\"\nprint \"Let p,q be two propositions\"\nprint \"Let p=\",p,\"\\n\",\"Let q=\",q\nprint \"Disjunction of p\\/q is : \"+p+\" or \"+q #unavailability of cup symbol.  So \\/\n#Disjunction combines two propositons using OR \n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Let p,q be two propositions\nLet p= Rebecca's PC has more than 16GB free hard disk space \nLet q= The processor in Rebecca's PC runs faster than 1GHz\nDisjunction of p\\/q is : Rebecca's PC has more than 16GB free hard disk space or The processor in Rebecca's PC runs faster than 1GHz\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 07:Page 07", "cell_type": "markdown", "metadata": {}}, {"execution_count": 14, "cell_type": "code", "source": "p=\"Maria learns discrete mathematics\"\nq=\"Maria will find a good job\"\nprint\"Let p=\",p,\"\\n\",\"Let q=\",q\nprint\"p->q is : \"+\"If \"+p+\" then \"+q #p->q p implies q means If P then Q.\nprint\"p->q is also expressed as :\",q,\" when \",p\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Let p= Maria learns discrete mathematics \nLet q= Maria will find a good job\np->q is : If Maria learns discrete mathematics then Maria will find a good job\np->q is also expressed as : Maria will find a good job  when  Maria learns discrete mathematics\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 01:Page 37", "cell_type": "markdown", "metadata": {}}, {"execution_count": 15, "cell_type": "code", "source": "def p(x): #Function defined to check whether the given statements are true.\n    if(x>3):\n        print \"p(\",x,\") which is the statement\",x,\">3, is true\"\n    else:\n        print \"p(\",x,\") which is the statement\",x,\">3, is false\"\np(4)#Fuction call \np(2)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "p( 4 ) which is the statement 4 >3, is true\np( 2 ) which is the statement 2 >3, is false\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 02:Page 38", "cell_type": "markdown", "metadata": {}}, {"execution_count": 17, "cell_type": "code", "source": "x1=\"CS1\" #Defining systems to check whether they are under attack through a function.\nx2=\"CS2\"\nx3=\"MATH1\"\ndef A(x):\n    if(x==\"CS1\"):                #Since cs1 and Math1 are the two computers under attack\n        print \"A(\",x,\") is true.\"\n    else:\n        if(x==\"MATH1\"):         #Since CS1 and MATH1 are the two computers under attack\n            print \"A(\",x,\") is true.\"\n        else:\n           print\"A(\",x,\") is false.\"\nprint \"Systems under attack are CS1 and MATH1.  The truth values for the same are calculated using functions.\"\nA(x1)#Function call \nA(x2)\nA(x3)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Systems under attack are CS1 and MATH1.  The truth values for the same are calculated using functions.\nA( CS1 ) is true.\nA( CS2 ) is false.\nA( MATH1 ) is true.\n"}], "metadata": {"collapsed": false, "trusted": true}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
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+{"nbformat_minor": 0, "cells": [{"source": "#Chapter 03: Algorithms", "cell_type": "markdown", "metadata": {}}, {"source": "## Example 03: Page 195", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "def binarysearch(a,num): #function definition with its parameters 'a' is the inputlist\n                         #and 'num' number to be found\n\n    first=0 #initially the first position is zero\n    last=len(a)-1 #initially the last position is the total length of the inputlist-1\n    found=False #boolean value to indicate if the number to be searched is found or not.\n\n    while first<=last and not found:\n        midpoint=(first+last)//2 #dividing the inputlist into two halves and comparing the number to be found with the midpoint.\n\n        if a[midpoint]==num: #If the number to be found is equal to the midpoint returns the position.\n           found=True\n        else:\n            if num<a[midpoint]: #if the number to be found is less than the midpoint\n                                #then the first half of the divided input list is taken for further computation.\n\n                last=midpoint-1 #by assigning the last number of the first half(number before the midpoint) to the variable last.\n\n\n            else:\n                first=midpoint+1 #if the number to be found is greater than the midpoint\n                                 #then the second half of the divided input list is taken for further computation.\n                                #by assigning the first number of the second half(number following the midpoint) to the variable first.\n\n    return midpoint #returns the position of the number found in the list. \n\n\n\nnumlist=[1, 23, 5, 6, 7, 8, 10, 12, 13, 15, 16, 18, 19, 20, 22] #List of inputs\nprint \"Found number 19 at the position\",(binarysearch(numlist, 19)) #Printing the position of the number to be found by a function call.\n                                                                    #The function binarysearch is called along with its parameters, inputlist\n                                                                    #and the number to be found.\n\n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Found number 19 at the position 12\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "##Example 05: Page 198", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "#To perform insertionsort\ndef sort_insertion(inputlist):\n\n    for i in range(1,len(inputlist)):\n\n        val_current = inputlist[i]\n        pos = i \n         \n        # check backwards through sorted list for proper pos of val_current\n        while((pos > 0) and (inputlist[pos-1] > val_current)):\n            inputlist[pos] = inputlist[pos-1]\n            pos = pos-1\n             \n        if pos != i:\n            inputlist[pos] = val_current \n            print(inputlist)\n    return inputlist\ninputlist = [3,2,4,1,5]\nprint sort_insertion(inputlist)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "[2, 3, 4, 1, 5]\n[1, 2, 3, 4, 5]\n[1, 2, 3, 4, 5]\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "", "cell_type": "markdown", "metadata": {}}, {"source": "", "cell_type": "markdown", "metadata": {}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
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+{"nbformat_minor": 0, "cells": [{"source": "#Chapter 04: Number Theory and Cryptography", "cell_type": "markdown", "metadata": {}}, {"source": "## Example 03: Page 239", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "#To find the quotient and remainder \ndividend=101\ndivisor=11\nquotient=dividend/divisor #To find quotient\nremainder=dividend%divisor #To find remainder\ndividend=(divisor*quotient)+remainder\nprint \"The quotient when\",dividend,\"is divided by\",divisor,\"is\",quotient,\"=\",dividend,\"div\",divisor,\"and the remainder is\",remainder,\"=\",dividend,\"mod\",divisor\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The quotient when 101 is divided by 11 is 9 = 101 div 11 and the remainder is 2 = 101 mod 11\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 04: Page 240", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "#To find the quotient and remainder\ndividend=-11\ndivisor=3\nquotient=dividend/divisor\nremainder=dividend%divisor\ndividend=(divisor*quotient)+remainder\nprint \"The quotient when\",dividend,\"is divided by\",divisor,\"is\",quotient,\"=\",dividend,\"div\",divisor,\"and the remainder is\",remainder,\"=\",dividend,\"mod\",divisor\n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The quotient when -11 is divided by 3 is -4 = -11 div 3 and the remainder is 1 = -11 mod 3\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 01: Page 246", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "#To convert binary to decimal equivalent\nbinary_num= raw_input('enter a number: ')\ndecimal = 0\nfor digit in binary_num:\n    decimal = decimal*2 + int(digit)\n\nprint decimal\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "enter a number: 101011111\n351\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 03: Page 247", "cell_type": "markdown", "metadata": {}}, {"execution_count": 5, "cell_type": "code", "source": "#To convert decimal to hexadecimal\ndec= raw_input('enter a number: ')\n\nprint \"The conversion of\",dec,\"to hexadeimal is\",int(dec,16)\n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "enter a number: 2AE0B\nThe conversion of 2AE0B to hexadeimal is 175627\n"}], "metadata": {"scrolled": true, "collapsed": false, "trusted": true}}, {"source": "## Example 04: Page 247", "cell_type": "markdown", "metadata": {}}, {"execution_count": 6, "cell_type": "code", "source": "#To compute decimal to octal\nnumbers= []\ndec=input(\"Enter a number\");\nnum=dec\nwhile dec!=0:\n    \n    rem=dec%8\n    dec=dec/8\n    numbers.append(rem)\nprint \"The decimal number\",num,\"is converted to its octal equivalent : \",\nfor i in reversed(numbers):\n    print i,\n    \n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter a number12345\nThe decimal number 12345 is converted to its octal equivalent :  3 0 0 7 1\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 05: Page 248", "cell_type": "markdown", "metadata": {}}, {"execution_count": 7, "cell_type": "code", "source": "#To convert Decimal to hexadecimal\nnum=[]\ndef ChangeHex(n): #function to convert\n    if (n < 0):\n        num.append(\"\")\n    elif (n<=1):\n        num.append(n)\n    else: #for numbers greater than 9\n        x =(n%16)\n        if (x < 10):\n            num.append(x) \n        if (x == 10):\n            num.append(\"A\")\n        if (x == 11):\n            num.append(\"B\")\n        if (x == 12):\n            num.append(\"C\")\n        if (x == 13):\n            num.append(\"D\")\n        if (x == 14):\n            num.append(\"E\")\n        if (x == 15):\n            num.append(\"F\")\n        ChangeHex( n / 16 )\ndec_num=input(\"Enter the decimal number which is to be converted to hexadecimal\");\nChangeHex(dec_num)\nprint \"The hexadecimal equivalent of decimal\",dec_num,\"is\",\nfor i in reversed(num):\n    print i,\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the decimal number which is to be converted to hexadecimal177130\nThe hexadecimal equivalent of decimal 177130 is 0 2 B 3 E A\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 06: Page 249", "cell_type": "markdown", "metadata": {}}, {"execution_count": 9, "cell_type": "code", "source": "#Compute Decimal to Binary\narray=[]\ndef conv(n):\n    if n==0:\n        print ''\n    else:\n        array.append(str(n%2)) #to compute remainder and append it to the result\n        return conv(n/2) \ndec_num=input(\"Enter a decimal number\")\nconv(dec_num)\nprint \"The binary equivalent of decimal\",dec_num,\"is\",\nfor i in reversed(array):\n    print i,\n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter a decimal number241\n\nThe binary equivalent of decimal 241 is 1 1 1 1 0 0 0 1\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 06: Page 249", "cell_type": "markdown", "metadata": {}}, {"execution_count": 10, "cell_type": "code", "source": "#To compute the binary addition\ndef binAdd(bin1, bin2): #function to add two binary numbers\n    if not bin1 or not bin2:#checks if both the numbers are binary\n        return '' \n\n    maxlen = max(len(bin1), len(bin2))\n\n    bin1 = bin1.zfill(maxlen) #zfill fills with zero to fill the entire width\n    bin2 = bin2.zfill(maxlen)\n\n    result  = ''\n    carry   = 0\n\n    i = maxlen - 1\n    while(i >= 0):\n        s = int(bin1[i]) + int(bin2[i])#adding bit by bit\n        if s == 2: #1+1\n            if carry == 0:\n                carry = 1\n                result = \"%s%s\" % (result, '0')\n            else:\n                result = \"%s%s\" % (result, '1')\n        elif s == 1: # 1+0\n            if carry == 1:\n                result = \"%s%s\" % (result, '0')\n            else:\n                result = \"%s%s\" % (result, '1')\n        else: # 0+0\n            if carry == 1:\n                result = \"%s%s\" % (result, '1')\n                carry = 0   \n            else:\n                result = \"%s%s\" % (result, '0') \n\n        i = i - 1;\n\n    if carry>0:\n        result = \"%s%s\" % (result, '1')\n    return result[::-1]\nbin1 = raw_input('enter the first number: ')\nbin2 = raw_input('enter the second number: ')\nprint \"The sum of binary numbers\",bin1,\"and\",bin2,\"is\",binAdd(bin1,bin2)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "enter the first number: 1110\nenter the second number: 1011\nThe sum of binary numbers 1110 and 1011 is 11001\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 02: Page 258", "cell_type": "markdown", "metadata": {}}, {"execution_count": 21, "cell_type": "code", "source": "#to find the prime factors\n\ndef prime_factors(n):\n    i = 2\n    factors = []\n    while i * i <= n:\n        if n % i: #modulp division to check of the number is prime or not\n            i += 1\n        else:\n            n //= i\n            factors.append(i) #append those numbers which readily divides the given number\n    if n > 1:\n        factors.append(n)\n    return factors\nnumber=input(\"Enter the number for which the prime factors have to be found\");\na=prime_factors(number)\nprint a\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number for which the prime factors have to be found100\n[2, 2, 5, 5]\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 03: Page 258", "cell_type": "markdown", "metadata": {}}, {"execution_count": 22, "cell_type": "code", "source": "#To say if a number is prime or not\nglobals() ['count']=0\nn=input(\"Enter the number\");\nfor i in range(2,n):#number thats not divisible by other than one and itself.  so from 2 to n (n-1 in python for loop)\n    if n%i==0:\n       count=count+1\n       num=i\nif count==0:\n     print n,\"is prime\" \nelse:\n    print n,\"is not prime because its divisible by\",num\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number101\n101 is prime\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 04: Page 259", "cell_type": "markdown", "metadata": {}}, {"execution_count": 23, "cell_type": "code", "source": "#to find the prime factors\n\ndef prime_factors(n):\n    i = 2\n    factors = []\n    while i * i <= n:\n        if n % i: #modulp division to check of the number is prime or not\n            i += 1\n        else:\n            n //= i\n            factors.append(i) #append those numbers which readily divides the given number\n    if n > 1:\n        factors.append(n)\n    return factors\nnumber=input(\"Enter the number for which the prime factors have to be found\");\na=prime_factors(number)\nprint a\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number for which the prime factors have to be found7007\n[7, 7, 11, 13]\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 10: Page 263", "cell_type": "markdown", "metadata": {}}, {"execution_count": 24, "cell_type": "code", "source": "#To compute GCD\ndef gcd(a,b):#fuction computes gcd\n    if b > a:\n        return gcd(b,a)\n    r = a%b\n    if r == 0:\n        return b\n    return gcd(r,b)\nn1=input(\"Enter the first number\");\nn2=input(\"Enter the second number\");\nprint \"GCD(\",n1,\",\",n2,\") is\",gcd(n1,n2)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the first number24\nEnter the second number36\nGCD( 24 , 36 ) is 12\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 11: Page 263", "cell_type": "markdown", "metadata": {}}, {"execution_count": 25, "cell_type": "code", "source": "#To compute GCD\ndef gcd(a,b):#fuction computes gcd\n    if b > a:\n        return gcd(b,a)\n    r = a%b\n    if r == 0:\n        return b\n    return gcd(r,b)\nn1=input(\"Enter the first number\");\nn2=input(\"Enter the second number\");\nprint \"GCD(\",n1,\",\",n2,\") is\",gcd(n1,n2)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the first number17\nEnter the second number22\nGCD( 17 , 22 ) is 1\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 16: Page 268", "cell_type": "markdown", "metadata": {}}, {"execution_count": 26, "cell_type": "code", "source": "#to find gcd using euclidean algorithm\ndef gcd(a,b):#euclidean algithm definition\n    x=a\n    y=b\n    while y!=0:\n        r=x%y\n        x=y\n        y=r\n    print \"gcd(\",a,\",\",b,\")is\",x\nnum1=input(\"Enter the first number\");\nnum2=input(\"Enter the second number\");\ngcd(num1,num2)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the first number414\nEnter the second number662\ngcd( 414 , 662 )is 2\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 17: Page 270", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "#to find gcd using euclidean algorithm\ndef gcd(a,b):#euclidean algithm definition\n    x=a\n    y=b\n    while y!=0:\n        r=x%y\n        x=y\n        y=r\n    print \"gcd(\",a,\",\",b,\")is\",x\nnum1=input(\"Enter the first number\");\nnum2=input(\"Enter the second number\");\ngcd(num1,num2)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the first number252\nEnter the second number198\ngcd( 252 , 198 )is 18\n"}], "metadata": {"collapsed": false, "trusted": true}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
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+{"nbformat_minor": 0, "cells": [{"source": "#Chapter 05: Induction and Recursion", "cell_type": "markdown", "metadata": {}}, {"source": "## Example 01: Page 346", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "#to compute the recursive functions\ndef f(n):\n\n    if n==0:\n        return 3\n    else:\n        n=n-1\n        result=2*f(n)+3 #recursive call\n        return result\nfor num in range(1,5):\n    r=f(num)\n    print \"The value of f(\",num,\") is\",r #Prints the result for individual instance\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The value of f( 1 ) is 9\nThe value of f( 2 ) is 21\nThe value of f( 3 ) is 45\nThe value of f( 4 ) is 93\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 01: Page 361", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "#To compute the factorial of a given number using recursion\ndef factorial(n):\n    if n==0:\n        return 1\n    else:\n        return n*factorial(n-1) #recursive function call\nnum=input(\"Enter a number whose factorial is to be found\");\nprint \"The factorial of\",num,\"is\",factorial(num);\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter a number whose factorial is to be found4\nThe factorial of 4 is 24\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 02: Page 361", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "#To compute power using recursive algorithm\ndef power(a,n):\n    if n==0:\n        return 1\n    else:\n        return a*power(a,n-1) #recursive call algorithm\nnum=input(\"Enter the number\");\np=input(\"Enter the power\");\nprint \"The value of\",num,\"to the power\",p,\"is\",power(num,p);\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number5\nEnter the power4\nThe value of 5 to the power 4 is 625\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 03: Page 362", "cell_type": "markdown", "metadata": {}}, {"execution_count": 5, "cell_type": "code", "source": "#To compute gcd using modular recursion\ndef gcd(a,b):\n    if a==0:\n        return b\n    else:\n        return gcd(b%a,a) #recursive call\n\nnum1=input(\"Enter the first number\")\nnum2=input(\"Enter the second number\")\nprint \"The gcd of\",num1,\",\",num2,\"is\",gcd(num1,num2)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the first number5\nEnter the second number8\nThe gcd of 5 , 8 is 1\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 04: Page 362", "cell_type": "markdown", "metadata": {}}, {"execution_count": 4, "cell_type": "code", "source": "#To compute mpower function using recursion\ndef mpower(b,n,m):\n    if n==0:\n        return 1\n    else:\n        if n%2==0:\n            return ((mpower(b,n/2,m))**2) % m #recursive call\n        else:\n            return ((mpower(b,n/2,m)**2)%m*(b%m))%m #recursive call\nnumber=input(\"Enter the number\")\npower=input(\"Enter the power\")\nmodulo=input(\"Enter the modulo number\");\nprint \"The answer of mpower(\",number,\",\",power,\",\",modulo,\") is\",mpower(number,power,modulo)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number2\nEnter the power5\nEnter the modulo number3\nThe answer of mpower( 2 , 5 , 3 ) is 2\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 09: Page 367", "cell_type": "markdown", "metadata": {}}, {"execution_count": 5, "cell_type": "code", "source": "def msort2(x): #function for merge sort\n    if len(x) < 2:\n        return x\n    result = []          \n    mid = int(len(x)/2) #divides the elements into halves\n    y = msort2(x[:mid])\n    z = msort2(x[mid:])\n    while (len(y) > 0) and (len(z) > 0):\n            if y[0] > z[0]:\n                result.append(z[0]) #merges to append the elements\n                z.pop(0)\n            else:\n                result.append(y[0])\n                y.pop(0)\n    result += y\n    result += z\n    return result\nl=[]\nr=[]\nl=raw_input(\"enter the numbers to be merge sorted\").split(\",\")\nr=msort2(l)\nprint \"Ther Merge sort is\", r\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "enter the numbers to be merge sorted8,2,4,6,9,7,10,1,5,3\nTher Merge sort is ['1', '10', '2', '3', '4', '5', '6', '7', '8', '9']\n"}], "metadata": {"collapsed": false, "trusted": true}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
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+{"nbformat_minor": 0, "cells": [{"source": "#Chapter 06: Counting", "cell_type": "markdown", "metadata": {}}, {"source": "## Example 01: Page 386", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "n=2 #number of employees\nr=12 #number of office rooms\nways_alloc_sanchez=12\nways_alloc_patel=11\n\n#By PRODUCT RULE\nprint \"Total ways to assign offices to these employees is\",ways_alloc_sanchez*ways_alloc_patel\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Total ways to assign offices to these employees is 132\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 02: Page 386", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "alphabets=26 #Total number of alphabets\nposint=100 #Total positive numbers not beyond 100\n\n#number of chairs to be labelled with a alphabet and an integer using product rule\nprint \"Total number of chairs that can be labelled with an alphabet and an integer is\",alphabets*posint\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Total number of chairs that can be labelled with an alphabet and an integer is 2600\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 03: Page 386", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "mc=32 #total number of microcomputers\nport=24 #total number of ports in each microcomputer\n\n#total number of different ports to a microcomputer in the center are found using product rule\n\nprint \"total number of ports\",mc*port\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "total number of ports 768\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 04: Page 386", "cell_type": "markdown", "metadata": {}}, {"execution_count": 4, "cell_type": "code", "source": "bits=2 #possible bits either 0 or 1\nns=7 #number of bits in the string (ie). length of the string\n # 7 bits are capable of taking either 0 or 1 so by PRODUCT RULE\nprint \"Total different bit strings of lenth seven are\",bits**ns\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Total different bit strings of lenth seven are 128\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 05: Page 387", "cell_type": "markdown", "metadata": {}}, {"execution_count": 7, "cell_type": "code", "source": "letters=26 #number of letters in english alphabet\nno_of_letters=3 #number of letters \nchoices=10 #number of choices for each letter\nresult=1#in order to avoid junk values.  Assigned it to 1.\nfor i in range(0,no_of_letters):\n    result=result*letters*choices\nprint \"The total number of choices are\",result", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The total number of choices are 17576000\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 01: Page 407", "cell_type": "markdown", "metadata": {}}, {"execution_count": 10, "cell_type": "code", "source": "def permutation(n,r): #function definition\n    \n    i=n\n    result=1\n    for i in range((n-r)+1,n+1): #computing the permutation\n       result=result*i\n    \n    return result\n\nprint \"The number of ways to select 3 students from a group of 5 students to line up for a picture is \",permutation(5,3) #function call\nprint \"The number of ways to select 5 students from a group of 5 students to line up for a picture is \",permutation(5,5) #function call\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The number of ways to select 3 students from a group of 5 students to line up for a picture is  60\nThe number of ways to select 5 students from a group of 5 students to line up for a picture is  120\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 04: Page 409", "cell_type": "markdown", "metadata": {}}, {"execution_count": 1, "cell_type": "code", "source": "def permutation(n,r): #function definition\n    \n    i=n\n    result=1\n    for i in range((n-r)+1,n+1): #permutation computation\n       result=result*i\n    return result\nnum=input(\"Enter the number of people\")\nperm=input(\"Enter the prizes\")\nprint \"The number of ways to decide the prize winners is\",permutation(num,perm) #function call\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number of people100\nEnter the prizes3\nThe number of ways to decide the prize winners is 970200\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 05: Page 409", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "def permutation(n,r):\n    \n    i=n\n    result=1\n    for i in range((n-r)+1,n+1):\n       result=result*i\n    \n    return result\nnum=input(\"Enter the number of runners\")\nperm=input(\"Enter the number of prizes\")\nprint \"The number of ways to decide the prize winners is\",permutation(num,perm)\n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number of runners8\nEnter the number of prizes3\nThe number of ways to decide the prize winners is 336\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 06: Page 409", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "def calc(n):\n    \n    i=n\n    result=1\n    for i in range(1,n): #find the number of ways to decide the path. since the first city us decided. The for loop is from 1 to n\n      result=result*i\n    \n    return result\nnum=input(\"Enter the number of cities\") \nprint \"The number of possible ways to decide the path is\",calc(num)\n\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number of cities8\nThe number of possible ways to decide the path is 5040\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 10: Page 410", "cell_type": "markdown", "metadata": {}}, {"execution_count": 5, "cell_type": "code", "source": "def combination(n,r): #combination function\n    i=n\n    numerator=1\n    denominator=1\n    for i in range((n-r)+1,n+1):#computes the value of the numerator \n       numerator=numerator*i\n    for j in range (1,r+1): #computes the value of the denominator\n        denominator=denominator*j\n    result=numerator/denominator #computes result\n    return result\nnum=input(\"Enter the number of elements\")\ncomb=input(\"Enter the combinations\")\nprint \"The number of combinations are \",combination(num,comb)\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number of elements4\nEnter the combinations2\nThe number of combinations are  6\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 12: Page 412", "cell_type": "markdown", "metadata": {}}, {"execution_count": 7, "cell_type": "code", "source": "def combination(n,r): #function definition for combination\n    i=n\n    numerator=1\n    denominator=1\n    for i in range((n-r)+1,n+1):\n       numerator=numerator*i\n    for j in range (1,r+1):\n        denominator=denominator*j\n    result=numerator/denominator\n    return result\nnum=input(\"Enter the number of members in a team\")\ncomb=input(\"Enter the number of players\")\nprint \"The number of combinations are \",combination(num,comb) #function call\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the number of members in a team10\nEnter the number of players5\nThe number of combinations are  252\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 13: Page 412", "cell_type": "markdown", "metadata": {}}, {"execution_count": 8, "cell_type": "code", "source": "def combination(n,r): #function definition\n    i=n\n    numerator=1\n    denominator=1\n    for i in range((n-r)+1,n+1):\n       numerator=numerator*i\n    for j in range (1,r+1):\n        denominator=denominator*j\n    result=numerator/denominator\n    return result\nnum=input(\"Enter the total number of astronauts\")\ncomb=input(\"Enter the number of astronauts to be selected \")\nprint \"The total number of combinations of selected astronauts to Mars are \",combination(num,comb) #function call\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the total number of astronauts30\nEnter the number of astronauts to be selected 6\nThe total number of combinations of selected astronauts to Mars are  593775\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 15: Page 413", "cell_type": "markdown", "metadata": {}}, {"execution_count": 9, "cell_type": "code", "source": "def combination(n,r): #Function definition\n    i=n\n    numerator=1\n    denominator=1\n    for i in range((n-r)+1,n+1): #computation of the numerator\n       numerator=numerator*i\n    for j in range (1,r+1): #computation of the denominator\n        denominator=denominator*j\n    result=numerator/denominator\n    return result\nnum1=input(\"Enter the total number of faculty in computer science department\")\ncomb1=input(\"Enter the number of faculty to be selected for computer science department\")\nnum2=input(\"Enter the total number of faculty in maths department\")\ncomb2=input(\"Enter the number of faculty to be selected for maths department\")\n\nprint \"The total number of combinations of selected faculties are \",combination(num1,comb1)*combination(num2,comb2) #Function call\n", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Enter the total number of faculty in computer science department9\nEnter the number of faculty to be selected for computer science department3\nEnter the total number of faculty in maths department11\nEnter the number of faculty to be selected for maths department4\nThe total number of combinations of selected faculties are  27720\n"}], "metadata": {"collapsed": false, "trusted": true}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
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-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10: Introduction to Power Electronics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.5, Page number: 508"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from pylab import *\n",
-      "import numpy as np\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "w=2*pi*60                               #frequency of voltage(Hz)\n",
-      "R=10                                    #ohm\n",
-      "C=0.01                                  #F\n",
-      "Vo=120*sqrt(2)                          #maximum voltage(V)\n",
-      "Nmax=800\n",
-      "tau=R*C                                 #time constant(s)\n",
-      "\n",
-      "#Calculations:\n",
-      "# diode = 1 when rectifier bridge is conducting\n",
-      "\n",
-      "diode=1\n",
-      "t=[0]*801\n",
-      "vs=[0]*801\n",
-      "vrect=[0]*801\n",
-      "vR=[0]*801\n",
-      "iB=[0]*801\n",
-      "\n",
-      "t=[0]*801\n",
-      "for n in range(1,Nmax+2,1):\n",
-      "    t[n-1] = (2.5*pi/w)*(n-1)/Nmax\n",
-      "    vs[n-1]=Vo*math.cos(w*t[n-1])\n",
-      "    vrect[n-1]=abs(vs[n-1])\n",
-      "#if the rectifier bridge is ON:\n",
-      "    if diode==1:\n",
-      "        vR[n-1]=vrect[n-1]\n",
-      "        if (w*t[n-1])<=(pi/2):\n",
-      "            iB[n-1]=vR[n-1]-Vo*C*w*math.sin(w*t[n-1])\n",
-      "        elif (w*t[n-1])<=3*pi/2:\n",
-      "            iB[n-1]=vR[n-1]/R+Vo*C*w*math.sin(w*t[n-1])\n",
-      "        else:\n",
-      "            iB[n-1]=vR[n-1]/R-Vo*C*w*math.sin(w*t[n-1])\n",
-      "        if iB[n-1]<0:\n",
-      "            diode=0\n",
-      "            toff=t[n-1]\n",
-      "            Voff=vrect[n-1]\n",
-      "    else:\n",
-      "        vR[n-1]=Voff*exp(-(t[n-1]-toff/tau))\n",
-      "        iB[n-1]=0\n",
-      "        if (vrect[n-1]-vR[n-1])>0:\n",
-      "            diode=1\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "iR=(1/R)*np.array(vR)\n",
-      "plot(1000*np.array(t),vR)\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('voltage [V]')\n",
-      "xlim(0,22)\n",
-      "ylim(0,180)\n",
-      "plot(1000*np.array(t),vrect,'--')\n",
-      "grid()\n",
-      "print \"The required plots are shown below:\"\n",
-      "show()\n",
-      "plot(1000*np.array(t),iR)\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('source current [A]')\n",
-      "xlim(0 ,22)\n",
-      "ylim(-50,250)      \n",
-      "plot(1000*np.array(t),1.5*np.array(iB),'--')\n",
-      "grid()\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "The required plots are shown below:"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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ibCiGqVKvAU4DsP/GfszznydsKIbxuYNUw2K96m0CQpzAsLS0REFBASwsLJCf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\nOE6H/T8o9Q6nwzgdWAAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f054c060710>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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CsWppbSLgX0cGLlysTua8ZMS8lJExL4eaQXV1NVauXIlJkyZh0qRJWL16Ncxm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\niIjYDIiICGwGREQENgMiIgK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-       "text": [
-        "<matplotlib.figure.Figure at 0x2986d90>"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.6, Page number: 522"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "w=2*pi*60                               #Angular freq of voltage(rad/sec)\n",
-      "Vo=230*sqrt(2)                          #volt\n",
-      "R=5.6                                   #Resistance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "Ls=[0]*101\n",
-      "tc=[0]*101\n",
-      "Idc=[0]*101\n",
-      "for n in range(1,101,1):\n",
-      "    Ls[n-1]=n*10**-3\n",
-      "    Idc[n-1]=2*Vo/(pi*R+2*w*Ls[n-1])\n",
-      "    tc[n-1]=(1/w)*acos(1-(2*Idc[n-1]*w*Ls[n-1])/Vo)\n",
-      "\n",
-      "#Results:\n",
-      "plot(1000*np.array(Ls),Idc,'g.')\n",
-      "xlabel('Commutating inductance Ls [mH]')\n",
-      "ylabel('Idc [A]')\n",
-      "title('Load current,Idc vs Commutating inductance,Ls')\n",
-      "show()\n",
-      "plot(1000*np.array(Ls),1000*np.array(tc),'g.')\n",
-      "xlabel('Commutating inductance L [mH]')\n",
-      "ylabel('tc [msec]')\n",
-      "title('Commutating Inductance,Ls vs time,tc')\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['fmod', 'sinh', 'trunc', 'tan', 'gamma', 'cosh', 'radians', 'modf', 'expm1', 'ldexp', 'linalg', 'random', 'frexp', 'ceil', 'isnan', 'copysign', 'cos', 'degrees', 'tanh', 'fabs', 'sqrt', 'hypot', 'power', 'log', 'log10', 'info', 'log1p', 'floor', 'fft', 'pi', 'exp', 'isinf', 'e', 'sin']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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ExEQAQGJiIsLDw7UZRoum3oVkZmLGAWci0poGtRCKiopgZmaGNm3aALh9nYSS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\nYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2c18b90>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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AgAHFlTP4iYiqzayDv8rKGfxERNVm1qdzEhGR+amTs3pqit+0RURUe+rFiJ/f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\nAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2981fd0>"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.7, Page number: 528"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R=12.5*10**-3                               #ohm\n",
-      "L=1.2                                       #H\n",
-      "Vo=15                                       #volt\n",
-      "w=120*pi                                    #angular freq(Hz)\n",
-      "Idc=35                                      #DC current(A)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "theta=[0]*1301\n",
-      "t=[0]*1301\n",
-      "vL=[0]*1301\n",
-      "vs=[0]*1301\n",
-      "\n",
-      "Vdc_a=R*Idc                                 #Dc voltage(V)\n",
-      "P=Vdc_a*Idc                                 #Power\n",
-      "alpha_da = acos(pi*R*Idc/(2*Vo)) ;    #delay angle\n",
-      "for n in range(1,1301,1):                 #loop for calculating load voltage\n",
-      "    theta[n-1]=2*pi*(n-1)/1000\n",
-      "    t[n-1]=theta[n-1]/w\n",
-      "    vs[n-1]=Vo*sin(theta[n-1])\n",
-      "    if theta[n-1]<alpha_da:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif (theta[n-1]<pi+alpha_da):\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif theta[n-1]<2*pi+alpha_da:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<3*pi+alpha_da:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<4*pi+alpha_da:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    else:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "\n",
-      "figure(1)\n",
-      "plot(1000*np.array(t),vL,'g.')\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('Load voltage [V]')\n",
-      "grid()\n",
-      "show()\n",
-      "\n",
-      "\n",
-      "#part(b):\n",
-      "alpha_db=0.9*pi                             #delay angle\n",
-      "Vdc_b=(2*Vo/pi)*cos(alpha_db)          #new dc voltage(V)\n",
-      "tau=L/R                                     #time constant(s)\n",
-      "imo=Idc                                     #Initial curent(A)\n",
-      "tzero=-tau*log((-Vdc_b/R)/(imo-Vdc_b/R))\n",
-      "for n in range(1,1301,1):\n",
-      "    theta[n-1]=2*pi*(n-1)/1000\n",
-      "    t[n-1]=theta[n-1]/w\n",
-      "    vs[n-1]=Vo*sin(theta[n-1])\n",
-      "    if theta< alpha_db:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif (theta[n-1]<pi+alpha_db):\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<2*pi+alpha_db:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif theta[n-1]<3*pi+alpha_db:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<4*pi+alpha_db:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    else:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "\n",
-      "#Results:\n",
-      "figure(2)\n",
-      "plot (1000*np.array(t), vL,'g.')\n",
-      "xlabel('time [msec] ')\n",
-      "ylabel('Load voltage [V]')\n",
-      "print \"part (a):\"\n",
-      "print \"\\n Vdc_a=\",round(1000*Vdc_a,2),\"mV\"\n",
-      "print \"\\n Power=\",round(P),\"W\"        \n",
-      "print \"\\n alpha_d=\",round((180/pi)*alpha_da,1),\"degrees\"\n",
-      "print \"\\n part (b):\"\n",
-      "print \"\\n alpha_d=\",round((180/pi)*alpha_db,1),\"degrees\"   \n",
-      "print \"\\n Vdc_b=\",round(Vdc_b,1),\"V\"\n",
-      "print \"\\n Current will reach zero at\",round(tzero,1),\"sec\"\n",
-      "grid()\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['fmod', 'cosh', 'sinh', 'trunc', 'tan', 'gamma', 'degrees', 'radians', 'sin', 'expm1', 'ldexp', 'isnan', 'frexp', 'ceil', 'copysign', 'cos', 'tanh', 'fabs', 'sqrt', 'hypot', 'log', 'log10', 'pi', 'log1p', 'floor', 'modf', 'exp', 'isinf', 'e']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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0J06fPo0NGzaYfZ0gCHjppZcQGxsLANDr9UhNTTXOh279l0NGRgb07+hx9Z9X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\nhNj0/wELMsFZawmOEgAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x1e16bd0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part (a):\n",
-        "\n",
-        " Vdc_a= 437.5 mV\n",
-        "\n",
-        " Power= 15.0 W\n",
-        "\n",
-        " alpha_d= 87.4 degrees\n",
-        "\n",
-        " part (b):\n",
-        "\n",
-        " alpha_d= 162.0 degrees\n",
-        "\n",
-        " Vdc_b= -9.1 V\n",
-        "\n",
-        " Current will reach zero at 4.5 sec\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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jOp0OL730EsrLy1FeXm4XIIg20Cwn/8gpzEFmQSayP8722XRVmuWkLi6DxIYN\nGzB37lwYDAYYDAbMnTsXGzZsELVS1EvoPrWOt3pymI1a28IT3W2LwqpClJwtwZ5TezB3x1yfvb+n\neSUx0H3hHZdBYsSIEaiqqsKRI0dw5MgR/N///R/+9a9/iVqpdevW4d5778Xs2bPR0NAg6nsR+ZL7\nYTZqcPnmZe7xTfNNn72u8Oxyc4dZFgvriPs8mgIbHx/v1cFDWVlZuHjxot31P/3pTxg1ahQiIzu7\nqrm5uTh9+jQ2bdrE+z2dTodnn30WCQkJAACGYZCWlobMzEwA1m8OVFZ2OTEtsXPOfTUAAH2HdM65\nl0v9lF7+5PoneL/sfa59+6V05n189fqPHnq08yCpn1//kcmdeznJ5e+vhXJxcTEKCgoAAAkJCcjL\ny/PPOglvg0R3XbhwAePHj0dVVRXvOiWutYMOsxFPzz/1RMvtFgCADjpUL672advSQVLyI0ri2t8u\nX7Z2f7du3YrkZPnsSy9Hlm8NauXOXk5qbwt3dKctbt2+xT2ODI30efCVy15OdF94x+k0pbCwMOh0\nOoc/a24Wb97zyy+/jMrKSrS1tcFgMODDDz8U7b2I/O2YvoM3nZJmOfmGMd/IrYoGgBGxI3z+HpZZ\nTpbV85a8Eq2eVxbaloPIHu3l5HvBK4Jh7jADAAIQgCvLrojSprSXk7yoYriJECGa5eR7lgABACPj\nR4oWdGn1vPJRkFA4LYy3dnfOvRbaoru6aoucwhxeubqxWtS6eHNGiC/QfeEdChJE9ujbqG9tqrRO\nKddBh4PzD3bx296j1fPKRjkJogjCvZxobNszOYU5nWsjfhYVGoW639SJ/r62eSUxptuS7qGcBFEt\n4bfRDnRIVBNlK6wq5JXFmNXkiG3OgwXr9yEn4jkKEgqnlfFW4V5OjS2NdkNOWmmL7nDWFrbbcIQF\nhuHjX33sl/oI80qt7a1+WzNB94V3KEgQxcjon8E9liIBqga2PbCewT39NpXYwBhwZ487ubIcDiMi\n3UNBQuEs+7VogXDISfhtVEtt4YqjtjDmG3nlYf2G+ak2naQ6jIjuC+9QkCCKwYQwdt9Gac1E9/1Q\n/wP3WA+934aaLITbdJScLaGdYRWAgoTCaW28Vfht1HbNhNbaoivCtsgpzOFtwxHVM8rvq9aZEIY3\nlbmto80vQ050X3iHggRRFOG3UTqnoHts10YAwLf/71tJ6mG7sA6g86+VgNZJEMVhVjKd5xT87JF7\nHrHLVxCnRcWkAAATWElEQVQ+fZ6e60n4a22EI3T+tbRonQTRBKkSoEolHGpK75suWV3o/GvloSCh\ncFocbxUOOe2r3oezTWc12RbO2LaFcKipZ3BPP9eGT7hm4uiFo6K+H90X3qEgQRRH+G2UVvB2zfZw\noQBdAAoeLZCuMoDddhxNrU2UV5IxChIKp9U54MJvo4MiB2m2LRyxtIXwcKEHEh6Qxfh/eIg1L9HW\n3ibqkBPdF96hIEEUycAYeNt0lJwtoZ1hHRCujdj85GYJa2NVvqCcV953Zh/1JmSKgoTCaXm8Vbhp\n3NBXh0pYG3mx3Be2vYiI0AhZ9CIA++3fxVwzoeV/I75AQYIolnDIidZM8Am34fDXjq/dRWsmlIHW\nSRBFE66ZyB6QjV2zdklYI/mwXRuhhx71y+pl05MA7NdMBOuDcemVS7Kqo9rQOgmiOV1t06FlctiG\nwxWptukg7qEgoXBaH2/lrZmopiEni//a8V+8slTbcLjijyEnrf8b8RYFCaJotDOsY23tbdzjiNAI\n
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/.ipynb_checkpoints/chapter3-checkpoint.ipynb b/ELECTRIC_MACHINERY/.ipynb_checkpoints/chapter3-checkpoint.ipynb
deleted file mode 100755
index 22fca9a1..00000000
--- a/ELECTRIC_MACHINERY/.ipynb_checkpoints/chapter3-checkpoint.ipynb
+++ /dev/null
@@ -1,460 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Electromechanical-Energy-Conversion-Principles "
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.1, Page number: 114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I=10                            #current in the coil(A)\n",
-      "Bo=0.02                         #magnetic field (T)\n",
-      "R=0.05                          #radius of the rotor(m)\n",
-      "l=0.3                           #rotor length(m)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "q=symbols('q')                  #Direction of torque\n",
-      "F1=-2*I*l*Bo*sin(q)             #Force on the coil(N)\n",
-      "T=F1*R                          #Torque scting in theta direction(Nm)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Force per unit length:\",T,\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Force per unit length: -0.006*sin(q) Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.2, Page number: 121"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "N=1000                   #No of winding turns\n",
-      "g=2                      #Air gap width(mm)\n",
-      "d=0.15                  #Magnetic core width,d (m)\n",
-      "l=0.1                    #thickness of core(0.1)\n",
-      "x,d=symbols('x d')       #where h is height of plunger(m) \n",
-      "                         #Lx is inductance as a function of x(H)\n",
-      "i=10                     #Current in the winding(A)\n",
-      "uo=4*3.14*10**-7           #permeability of free space(H/m)\n",
-      "\n",
-      "#Calculations:\n",
-      "Lx=(uo*N**2*l*d)/(2*g*10**-3)*(1-x/d)\n",
-      "Wfld=(1./2)*Lx*i**2\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The magnetic energy stored, Wfld:\",\"236*(1-x/d) J\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnetic energy stored, Wfld: 236*(1-x/d) J\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.3, Page number: 124"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable declaration:\n",
-      "xdata=[0.0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0]     #(cm)\n",
-      "Ldata=[2.8, 2.26, 1.78, 1.52, 1.34, 1.26, 1.20, 1.16, 1.13, 1.11, 1.10]   #(mH)\n",
-      "I = 0.75                                                #(A)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "x=0.01*np.array(xdata)\n",
-      "L=0.001*np.array(Ldata)\n",
-      "length=len(x)\n",
-      "xmax=x[length-1]\n",
-      "a=polyfit(x,L,4)\n",
-      "xfit=[0]*102\n",
-      "Lfit=[0]*102\n",
-      "for n in range(1,102,1):\n",
-      "    xfit[n-1]=xmax*(n-1)/100\n",
-      "    Lfit[n-1]=a[0]*xfit[n-1]**4+a[1]*xfit[n-1]**3+a[2]*xfit[n-1]**2+a[3]*xfit[n-1]+a[4]\n",
-      "\n",
-      "#Plot the data and then the fit to compare (convert xfit to cm and Lfit to mH)\n",
-      "plot(xdata,Ldata,'o')\n",
-      "plot(100*np.array(xfit),1000*np.array(Lfit),'g.')\n",
-      "xlabel('x [cm] ')\n",
-      "ylabel('L [mH] ')\n",
-      "title('Inductance,L vs length,l')\n",
-      "grid()\n",
-      "print \"The required plots are shown below:\"\n",
-      "show()\n",
-      "\n",
-      "#set current to 0.75 A\n",
-      "I=0.75\n",
-      "F=[0]*102\n",
-      "for n in range(1,102,1):\n",
-      "    xfit[n-1]=0.002+0.016*(n-1)/100\n",
-      "    F[n-1]=4*a[0]*xfit[n-1]**3+3*a[1]*xfit[n-1]**2+2*a[2]*xfit[n-1]**1+a[3]\n",
-      "    F[n-1]=(I**2/2)*F[n-1]\n",
-      "plot(100*np.array(xfit),F,'b.')\n",
-      "xlabel('x  [cm]')\n",
-      "ylabel('Force [N]')\n",
-      "title('Force, F vs length,l')\n",
-      "grid()\n",
-      "\n",
-      "#Results:\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "The required plots are shown below:"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['Polygon', 'poly', 'sign', 'flatten', 'conjugate', 'diff', 'tan', 'Circle', 'roots', 'plot', 'eye', 'trace', 'floor', 'diag', 'invert', 'nan', 'sqrt', 'source', 'add', 'zeros', 'take', 'var', 'pi', 'plotting', 'product', 'seterr', 'power', 'multinomial', 'transpose', 'test', 'beta', 'ones', 'sinh', 'vectorize', 'cosh', 'trunc', 'cos', 'prod', 'tanh', 'mod', 'det', 'sin', 'binomial', 'solve', 'log', 'exp', 'reshape', 'gamma', 'interactive']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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PRP/rjv3mZD0B9bu/LN39pZwVcHwahqd/geK6r0xOLrrXjn6K8EfMm8UY9Vsa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\nYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3dc9590>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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dqL29Z9qL01xEQ4fL353CCTExMXbbYmNjnXmqQ+fOnRP33nuviI2NFSkpKeL8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\noRARkSSYUIiISBL/D5upbph/P2HSAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x48a36d0>"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page number: 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Lo=10.6*10**-3                          #Initial inductance(H)\n",
-      "L2=2.7*10**-3                           #H\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "theta,i=symbols('theta i')\n",
-      "L=Lo+L2*cos(2*theta)\n",
-      "i=2                                     #Coil current,A\n",
-      "def T(theta):\n",
-      "    return i**2*diff(L,theta)/2\n",
-      "                                    \n",
-      "\n",
-      "#Results:\n",
-      "print \"Torque,Tfld =\",T(theta),\" N.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Torque,Tfld = -0.0108*sin(2*theta)  N.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page number: 134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1=2.5*10**-2                            #radius of rotor(m)\n",
-      "h=1.8*10**-2                             #Axial length(m)\n",
-      "g=3*10**-3                               #Air gap length(m)\n",
-      "Bag=1.65                                #Magnetic field(T)\n",
-      "uo=4*pi*10**-7                          #permeability of free space(H/m)\n",
-      "\n",
-      "#Calculations:\n",
-      "H=Bag/uo\n",
-      "Ni=2*g*H\n",
-      "T=uo*(Ni)**2*h*(r1+0.5*g)/(4*g)\n",
-      "\n",
-      "#Results:\n",
-      "print \"The maximum torque:\", round(T,2),\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum torque: 3.1 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page number: 140"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from matplotlib import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "i1=0.8\n",
-      "i2=0.01\n",
-      "\n",
-      "\n",
-      "#Calculations & Results:\n",
-      "def df(f,x,h=0.1e-10):\n",
-      "    return ( f(x+h/2) - f(x-h/2) )/h\n",
-      "\n",
-      "\n",
-      "\n",
-      "def l11(x):\n",
-      "    return (3+cos(2*x))/1000.0\n",
-      "\n",
-      "def l12(x):\n",
-      "    return (0.3*cos(x))\n",
-      "\n",
-      "def l22(x):\n",
-      "    return (30+10*cos(2*x))\n",
-      "\n",
-      "def g(x):\n",
-      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x) + (i1*i2)*df(l12,x)\n",
-      "\n",
-      "def r(x):\n",
-      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x)\n",
-      "def s(x):\n",
-      "    return (i1*i2)*df(l12,x)\n",
-      "\n",
-      "x=linspace(-pi,pi,100000)\n",
-      "\n",
-      "\n",
-      "plot(x,r(x))\n",
-      "plot(x,s(x))\n",
-      "plot(x,g(x))\n",
-      "grid()\n",
-      "annotate(\"Total torque\",xy=(-0.5,0.003))\n",
-      "annotate(\"Reluctance torque\",xy=(-2,-0.0015))\n",
-      "annotate(\"Mutual Interaction torque\",xy=(1.6,-0.0026))\n",
-      "xlabel(\"Theta [radians]\")\n",
-      "ylabel(\"Torque [N.m]\")\n",
-      "xlim(-pi,pi)\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print \"Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)\"\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['vectorize', 'prod', 'plotting', 'Circle', 'diag', 'sinh', 'trunc', 'plot', 'eye', 'det', 'tan', 'product', 'gamma', 'roots', 'sin', 'zeros', 'cosh', 'interactive', 'conjugate', 'take', 'trace', 'beta', 'exp', 'ones', 'multinomial', 'cos', 'transpose', 'solve', 'diff', 'invert', 'pi', 'tanh', 'Polygon', 'reshape', 'sqrt', 'floor', 'source', 'add', 'poly', 'mod', 'sign', 'power', 'binomial', 'log', 'var', 'seterr', 'flatten', 'nan', 'test']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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0ZnpWOnrt7QWbujZY03ONMnaoQsGmm969A/z8NAhTtWgRi3W1bp3OZRbRH/fv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\ngQFWOK/Ic16Ivyl1FOgHGBYWhrCwMOV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Z/fnnn0REtHnzZhozZgwREbm7u9MPP/yQp/7T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\ns/T/jVWB3070t1UAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3dbe950>"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.9, Page number: 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "W=4.0*10**-2                            #width of plunger lower arm(m)\n",
-      "W1=4.5*10**-2                           #width of plunger upper arm(m)\n",
-      "D=3.5*10**-2                            #depth of plunger (m)\n",
-      "d=8*10**-3                              #length of magnet(m)\n",
-      "go=1*10**-3                             #air gap length(m)\n",
-      "uo=4*pi*10**-7                          #Permeability of free space(A.turns/m)\n",
-      "ur=1.06*uo                              #Relativity permeability\n",
-      "Hc1=-940                                #Magnetising force(kA/m)\n",
-      "Bt=1.25                                 #Magnetic field induction(T)\n",
-      "N=1500                                  #No of turns\n",
-      "x=3*10**-3                              #Position of plunger(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "Ni=-Hc1*d*10**3\n",
-      "Rx=x/(uo*W1*D)\n",
-      "Ro=go/(uo*W*D)\n",
-      "Rm=d/(ur*W*D)\n",
-      "f=-((Ni)**2)/(uo*W1*D*(Rx+Ro+Rm)**2)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The x-directed force:\",round(f,1),\"N\"\n",
-      "print \"Current in the excitation winding:\",round(Ni/N,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The x-directed force: -703.3 N\n",
-        "Current in the excitation winding: 5.01 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/.ipynb_checkpoints/chapter5-checkpoint.ipynb b/ELECTRIC_MACHINERY/.ipynb_checkpoints/chapter5-checkpoint.ipynb
deleted file mode 100755
index 3f99f735..00000000
--- a/ELECTRIC_MACHINERY/.ipynb_checkpoints/chapter5-checkpoint.ipynb
+++ /dev/null
@@ -1,609 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5: Synchronous Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.1, Page number: 254"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import cmath\n",
-      "\n",
-      "#Varaible Declaration:\n",
-      "pf=0.95                                 #Lagging power factor\n",
-      "Vl=460                                  #Terminal voltage(V)\n",
-      "I=120                                   #Terminal current(A)\n",
-      "If=47                                   #Field current(A)\n",
-      "X=1.68j                                 #Line syncchronous reactance(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Choosing motor reference direction:\n",
-      "Va=Vl/math.sqrt(3)\n",
-      "theta=math.acos(0.95)\n",
-      "Ia=I*cmath.exp(-theta*1j)\n",
-      "Eaf=Va-X*Ia\n",
-      "wc=120*math.pi\n",
-      "Laf=math.sqrt(2)*abs(Eaf)/(wc*If)\n",
-      "P=3*Va*Ia*pf\n",
-      "\n",
-      "#Results:\n",
-      "print \"Generated emf:\",round(abs(Eaf),1),\"V line to line\"\n",
-      "print \"Fied to armature mutual inductance:\",round(Laf*1000,1),\"mH\"\n",
-      "print \"Three phase power:\",round(abs(P/1000),1),\"kW or\",round(abs(P)/746),\"hp\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Generated emf: 278.8 V line to line\n",
-        "Fied to armature mutual inductance: 22.3 mH\n",
-        "Three phase power: 90.8 kW or 122.0 hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.2, Page number: 255"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath \n",
-      "from math import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Pin=90.6*10**3                              #Input power(kW)\n",
-      "Va=265.6                                    #Terninal voltage(V)\n",
-      "X=1.68j                                     #Synchronous reactance(ohm)\n",
-      "Laf=22.3*10**-3                             #Mutual inductance(H)\n",
-      "wc=120*pi                                   #Angular frequency(rad/sec)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Ia=Pin/(3*Va)\n",
-      "Eaf=Va-X*Ia\n",
-      "delta=degrees(cmath.phase(Eaf))\n",
-      "I=sqrt(2)*Eaf/(wc*Laf)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print\"The phase angle,delta:\",round(delta,1),\"degrees\"\n",
-      "print\"Required field current:\",round(abs(I),2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The phase angle,delta: -35.7 degrees\n",
-        "Required field current: 55.04 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.3, Page number: 257"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Eafl=13.8*10**3                             #Open circuit voltage(V)\n",
-      "If1=318                                     #Field current(A)\n",
-      "If2=263                              #Field current after extrapolation(A)\n",
-      "wc=120*pi                            #Angular frequency(Hz)\n",
-      "\n",
-      "#Calculations:\n",
-      "Eaf=Eafl/sqrt(3)\n",
-      "La1=sqrt(2)*Eaf/(wc*If1)\n",
-      "La2=sqrt(2)*Eaf/(wc*If2)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Saturated Laf1:\",round(La1*1000,0),\"mH\"    \n",
-      "print \"Unsaturated Laf1:\",round(La2*1000,0),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Saturated Laf1: 94.0 mH\n",
-        "Unsaturated Laf1: 114.0 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.4, Page number: 262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ia=[118, 152]                     #Armature current from SC Characteristics(A)\n",
-      "If=[2.20, 2.84]                     #Field current from SC Characteristics(A)\n",
-      "Vll=220                             #Line-to-line Voltage(V)\n",
-      "V=202                               #Line-to-line air voltage(V) \n",
-      "P=45*10**3                          #Power roted to motor(W)   \n",
-      "Is_sc=1                             #per unit rated current(A)\n",
-      "\n",
-      "#Calculations:\n",
-      "Va_ag=V/sqrt(3)                 #At field current of 2.20A,at air gap,(V)\n",
-      "Ia_ag=Ia[0]\n",
-      "Xs_u=Va_ag/Ia_ag\n",
-      "Ia_rated=P/(sqrt(3)*Vll)\n",
-      "Xa_g=Va_ag/1\n",
-      "Xs_u_pu=Va_ag/Is_sc\n",
-      "Xs=Vll/(Ia[1]*sqrt(3))\n",
-      "Ia_pu=Ia[1]/Ia[0]\n",
-      "SCR=If[1]/If[0]\n",
-      "Xs=1/SCR\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"'All quantities are in per unit values'\"\n",
-      "print\"Unsaturated value of synchronous reactance:\",round(Xs_u,3),\"ohm\"\n",
-      "print \"Satureted value of synchronous reactance: \",round(Xs,3),\"ohm\"\n",
-      "print\"Short circuit ratio:\",round(SCR,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "'All quantities are in per unit values'\n",
-        "Unsaturated value of synchronous reactance: 0.988 ohm\n",
-        "Satureted value of synchronous reactance:  0.775 ohm\n",
-        "Short circuit ratio: 1.291\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.5, Page number: 265"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P_rated=45*10**3                        #Rated power(KV)\n",
-      "Pl=1.80*10**3                           #Short circuit load loss(W)\n",
-      "Ia_pu=1                                    #Per unit armature current\n",
-      "Ia=118                                  #rated armature current(A)\n",
-      "Ra_dc=0.0335                            #Dc resistance(ohm/phase)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Pl_pu=Pl/P_rated    \n",
-      "Ra_eff1=Pl_pu/Ia_pu**2                      #in per unit basis\n",
-      "Ra_eff2=Pl/(3*(Ia)**2)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Armature resistance in per unit:\",round(Ra_eff1,3),\"per unit\" \n",
-      "print \"Armature resistance in ohms/phase:\", round(Ra_eff2,3),\"ohms/phase\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature resistance in per unit: 0.04 per unit\n",
-        "Armature resistance in ohms/phase: 0.043 ohms/phase\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.6, Page number: 269"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Veq=1.0                                 #Externalsupply(p.u)                                    \n",
-      "Eaf=1.0                                 #Internal voltage(p.u)\n",
-      "Xeq=0.23                                #Eqv.resistance of external system(p.u)\n",
-      "Xs=1.35                                #Saturated synchronous reactance(p.u)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "P_max=Eaf*Veq/(Xs+Xeq)\n",
-      "\n",
-      "\n",
-      "#for part (b):\n",
-      "delta=[0]*500\n",
-      "Ia=[0]*500\n",
-      "Va=[0]*500\n",
-      "degree=[0]*500\n",
-      "for n in range(1,101,1):\n",
-      "    delta[n-1]=(pi/2)*(n-1)/100\n",
-      "    Ia[n-1] = (Eaf *exp(1j*delta[n-1]) - Veq)/(1j*(Xs + Xeq))\n",
-      "    Va[n-1] = abs(Veq + 1j*Xeq*Ia[n-1])\n",
-      "    degree[n-1]=180*delta[n-1]/pi\n",
-      "plot(degree,Va,'r.')\n",
-      "xlabel('Power angle,delta(degrees)')\n",
-      "ylabel('Terminal voltage(per unit)')\n",
-      "title('Terminal voltage vs. power angle for part (b)')\n",
-      "show()\n",
-      "#for part (c):\n",
-      "Vterm=1.0\n",
-      "P=[0]*500\n",
-      "deltat=[0]*500\n",
-      "Ia=[0]*500\n",
-      "Eaf=[0]*500\n",
-      "\n",
-      "for n in range(1,101,1):\n",
-      "    P[n-1]=(n-1)/100\n",
-      "    deltat[n-1]=math.asin(P[n-1]*Xeq/(Vterm*Veq))\n",
-      "    Ia[n-1]=(Vterm*exp(1j*deltat[n-1])-Veq)/(1j*Xeq)\n",
-      "    Eaf[n-1]=abs(Vterm+1j*(Xs+Xeq)*Ia[n-1])\n",
-      "plot(P,Eaf,'r.')\n",
-      "xlabel('Power [per unit]')\n",
-      "ylabel('Eaf [per unit]')\n",
-      "title('Eaf vs. power for part (c)')\n",
-      "show()\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Maximum power supplied to external system:\",round(P_max,2),\"p.u\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['prod', 'Circle', 'power', 'diag', 'sinh', 'trunc', 'binomial', 'plot', 'eye', 'det', 'tan', 'product', 'roots', 'vectorize', 'sin', 'plotting', 'zeros', 'cosh', 'conjugate', 'linalg', 'take', 'solve', 'trace', 'beta', 'draw_if_interactive', 'random', 'ones', 'transpose', 'cos', 'interactive', 'diff', 'invert', 'tanh', 'Polygon', 'reshape', 'sqrt', 'floor', 'source', 'add', 'multinomial', 'test', 'poly', 'mod', 'sign', 'fft', 'gamma', 'log', 'var', 'info', 'seterr', 'flatten', 'nan', 'pi', 'exp']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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06YiMjKxQn42NDaZOnYouXbrAzs4OBw8exNy5c3HgwAFYWVlh8ODBGDp0qDK+paUlNm3a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\na/B03gAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3794dd0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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+s2+K7OxsBAcHIzAwEOvXr2+zjFarxfjx4xEaGoopU6Z0+AsQEZF5KbaCW11d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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3965f10>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Maximum power supplied to external system: 0.63 p.u\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.7, Page number: 272"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P_rated=2000*746/3                     #per phase rated power of motor(W)\n",
-      "Xsm=1.95                                #Synchronous reactance(ohm)\n",
-      "Vl=2300                                #Line to line voltage(V)\n",
-      "f=60                                   #Angular frequency(Hz)\n",
-      "p=30                                   #No. of poles\n",
-      "Xsg=2.65                               #Synchronous reactance of generator(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "Vp=2300/sqrt(3)\n",
-      "Ip=P_rated/Vp\n",
-      "Eafm=sqrt(Vp**2+(Ip*Xsm)**2)\n",
-      "Pm=3*Vp*Eafm/Xsm                             #Max power delivered to motor(W)\n",
-      "ws=2*2*pi*f/p\n",
-      "Tmax=Pm/ws                                  #MAx torque of motor(Nm)\n",
-      "\n",
-      "\n",
-      "#for part (b):\n",
-      "Eafg=sqrt(Vp**2+(Ip*Xsg)**2)\n",
-      "Pm2=3*Eafm*Eafg/(Xsg+Xsm)                   #Max power delivered to motor(W)\n",
-      "Tmax2=Pm2/ws                                #Max torque(Nm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print\"(a) Max power  :\",round(Pm/1000,0),\"kW,3-ph\"\n",
-      "print\"    Max torque :\",round(Tmax/1000,1),\"kNm\"\n",
-      "print \"(b) Max power :\", round(Pm2/1000,0),\"kW,3-ph\"\n",
-      "print \"    Max torque:\", round(Tmax2/1000,1),\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Max power  : 3096.0 kW,3-ph\n",
-        "    Max torque : 123.2 kNm\n",
-        "(b) Max power : 1639.0 kW,3-ph\n",
-        "    Max torque: 65.2 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.8, Page number: 279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P=45                                    #Power rated(KVA)\n",
-      "Va=220                                  #Terminal voltage(V)\n",
-      "Pin=45                                  #Power input to the armature(KVA)\n",
-      "If=5.50                                 #field current(A)\n",
-      "Rf=35.5                                 #Field winding resistance(ohm)\n",
-      "Ra=0.0399                               #Armature dc resistance(ohm/phase)\n",
-      "Xal=0.215                               #Leakage reactance of motor(ohm)\n",
-      "pf=0.80                                 #Lagging power factor \n",
-      "Pc=1.8                                  #Core loss(kW)\n",
-      "Pw=0.91                                 #Friction & windage losses(kW)\n",
-      "Ps=0.37                                 #Stray load loss(kW)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Ia=P*10**3/(sqrt(3)*Va)\n",
-      "P1=If**2*Rf/10**3                            #Loss in field winding(kW)\n",
-      "P2=3*Ia**2*Ra/10**3                          #Loss in armature(kW)\n",
-      "Pl=(Pc+Pw+Ps+P1+P2)\n",
-      "Pi=Pin*pf+P1\n",
-      "Po=Pi-Pl\n",
-      "eff=(Po/Pi)*100\n",
-      "\n",
-      "#Results:\n",
-      "print \"Efficiency of the synchronous machine:\",round(eff,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of the synchronous machine: 84.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.9, Page number: 287"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import cmath\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Xd=1                              #Direct axis synchronus reactance(p.u)\n",
-      "Xq=0.60                           #Quadrature axis synchronous reactance(p.u)\n",
-      "Va=1                              #Terminal voltage(p.u)\n",
-      "pf=0.8                              #Lagging power factor\n",
-      "Ia=0.8-1j*math.sin(math.acos(0.8))       #Line current(p.u)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "phy=-math.acos(pf)\n",
-      "E=Va+1j*Xq*Ia\n",
-      "delta=cmath.phase(E)\n",
-      "Id=abs(Ia)*math.sin(delta-phy)*cmath.exp(1j*(-pi/2+delta))\n",
-      "Iq=abs(Ia)*math.cos(delta-phy)*cmath.exp(1j*delta)\n",
-      "Eaf=Va+Xd*Id*1j+Xq*Iq*1j\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Generated voltage:\",round(abs(Eaf),2),\"p.u Volt\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Generated voltage: 1.78 p.u Volt\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.11, Page number: 291"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from pylab import *\n",
-      "import cmath\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P_rated=2000*746                  #Rated power of motor(W)\n",
-      "Xs=1.95                           #Synchronous reactance(ohm/phase)\n",
-      "Xd=1.95                           #Direct axis synchronous reactance(ohm/ph)\n",
-      "Xq=1.40                         #Quadrature axis synchronous reactance(ohm/ph)\n",
-      "pf=1                            #Power factor of the machine\n",
-      "Vl=2300                         #Line to line voltage(V)\n",
-      "\n",
-      "#Calculatons:\n",
-      "Va=float(Vl/sqrt(3))              #volt\n",
-      "Ia=float(P_rated/(Va*3))          #ampere\n",
-      "E1=Va-1j*Ia*Xq                      #From phasor diagram\n",
-      "delta=cmath.phase(E1)             #power angle\n",
-      "Id=Ia*sin(abs(delta))             #direct axis current(A)\n",
-      "Eaf=abs(E1)+Id*(Xd-Xq)\n",
-      "r=symbols('r')\n",
-      "def P(r):                           #Process for finding maximum power\n",
-      "    return Eaf*Va*sin(r)/Xd + Va**2*(Xd-Xq)*sin(2*r)/(2*Xd*Xq)\n",
-      "P1=diff(P(r),r)\n",
-      "#On differentiation,\n",
-      "#P1 = 1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1)\n",
-      "l = solve(1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1),r)\n",
-      "P_max = (P(round(l[0],5)))\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Maximum mechanical power:\",math.ceil(3*P_max/10**3),\"kW,3-phase\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum mechanical power: 3236.0 kW,3-phase\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/README.txt b/ELECTRIC_MACHINERY/README.txt
deleted file mode 100755
index 4b574a6d..00000000
--- a/ELECTRIC_MACHINERY/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: SANTOSH BARNWAL
-Course: be
-College/Institute/Organization: BIRLA INSTITUTE OF TECHNOLOGY MESRA
-Department/Designation: ELECTRICAL & ELECTRONICS
-Book Title: ELECTRIC MACHINERY
-Author: A. E. Fitzgerald, Charles Kingsley, Jr., Stephen D. Umans
-Publisher: McGraw-Hill, New York
-Year of publication: 2003
-Isbn: 0-07-112193-5
-Edition: Sixth Edition
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter1.ipynb b/ELECTRIC_MACHINERY/chapter1.ipynb
deleted file mode 100755
index 3eaa70a1..00000000
--- a/ELECTRIC_MACHINERY/chapter1.ipynb
+++ /dev/null
@@ -1,560 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:29b5ec9b20f222bcfeb31d6e80f55e5d32272a4cce1e9227a6328e93313e209d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 1:Introduction to Magnetic Circuits<h1>"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.1, Page number: 9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "#Variable declaration:\n",
-      "Ac=9                            #Cross-sectional area of the core(cm**2)\n",
-      "Ag=9                            #Cross-sectional area of the air-gap(cm**2)\n",
-      "g=0.050                         #Air-gap length(cm)            \n",
-      "lc=30                           #Mean Length of the core(cm)\n",
-      "N=500                           #No. of windings\n",
-      "ur=70000                        #Relative permeability of the core material\n",
-      "Bc=1.0                          # Magnetic Flux Density of the core(T)\n",
-      "uo=4*pi*10**-7                  #Permeability of free space\n",
-      "\n",
-      "#Calculation\n",
-      "Rc=lc*10**-2/((ur*uo*Ac)*10**-4)\n",
-      "Rg=g*10**-2/((uo*Ag)*10**-4)\n",
-      "Q=Bc*Ac*10**-4\n",
-      "i=Q*(Rc+Rg)/N\n",
-      "\n",
-      "#Results\n",
-      "print \"a.Reluctance of the core,Rc:\",round(Rc,2), \"A.turns/Wb\" \n",
-      "print \"  Reluctance of the air-gap,Rg:\", round(Rg,2), \"A.turns/Wb\"\n",
-      "print \"b.The flux, Q:\", round(Q,4), \"Wb\"\n",
-      "print \"c.The current,i:\", round(i,2), \"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a.Reluctance of the core,Rc: 3789.4 A.turns/Wb\n",
-        "  Reluctance of the air-gap,Rg: 442097.06 A.turns/Wb\n",
-        "b.The flux, Q: 0.0009 Wb\n",
-        "c.The current,i: 0.8 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.2, Page number: 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "#Variable declaration:\n",
-      "I=10                            #Current in the coil(A)\n",
-      "N=1000                          #No of turns in the rotor\n",
-      "g=1                             #Air gap length(cm)\n",
-      "Ag=2000                         #Cross-section of the air-gap(cm**2)\n",
-      "uo=4*pi*10**-7                  #Permeability of free space\n",
-      "\n",
-      "#Calculation:\n",
-      "Q=(N*I*uo*Ag*10**-4)/(2*g*10**-2)\n",
-      "Bg=round(Q,2)/(Ag*10**-4)\n",
-      "\n",
-      "#Results\n",
-      "print \"The air-gap flux, Q:\", round(Q,2), \"Wb\"\n",
-      "print \"The flux density, Bg:\", round(Bg,4), \"T\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The air-gap flux, Q: 0.13 Wb\n",
-        "The flux density, Bg: 0.65 T\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.4, Page number: 13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "#Variable declaration\n",
-      "lc=0.3                                  #length of the core(cm)\n",
-      "ur1=72300                               #Relative permeablity for case(a)\n",
-      "ur2=2900                                #Relative permeablity for case(b)\n",
-      "Ac=9                                    #Cross-section of the core(cm**2)\n",
-      "Rg=4.42*10**5                           #Reluctance of the air-gap(A.turns/Wb)\n",
-      "N=500                                   #No of coil turns\n",
-      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
-      "\n",
-      "#Calculations:\n",
-      "Rt1=(lc/(ur1*uo*Ac*10**-4))+Rg\n",
-      "L1=N**2/Rt1\n",
-      "Rt2=(lc/(ur2*uo*Ac*10**-4))+Rg\n",
-      "L2=N**2/Rt2\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a)Inductance,L:\",round(L1,2),\"H\"\n",
-      "print \"(b)Inductance,L:\",round(L2,2),\"H\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a)Inductance,L: 0.56 H\n",
-        "(b)Inductance,L: 0.47 H\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.5, Page number: 15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "from matplotlib import *\n",
-      "from math import *\n",
-      "%matplotlib inline\n",
-      "#Variable declaration:\n",
-      "Ac=9e-4                                 #Cross-section of the core(m)\n",
-      "Ag=9e-4                                 #Cross-section of the air-gap(m)\n",
-      "g=5e-4                                  #Air-gap length(m)\n",
-      "lc=0.3                                  #Mean length of the core(m)\n",
-      "N=500                                   #No. of turns of the core(m)\n",
-      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
-      "\n",
-      "#Calculations:\n",
-      "Rg=g/(uo*Ag)                            #Reluctance of the air-gap(A.turns/Wb)\n",
-      "ur=[0]*200                              #Initialising array\n",
-      "L=[0]*200\n",
-      "\n",
-      "for n in range(1,101,1):\n",
-      "    ur[n-1]=100+(10000-100)*(n-1)/100\n",
-      "    Rc=lc/(ur[n-1]*uo*Ac)               #Reluctance of the core(A.turns/Wb)\n",
-      "    Rtot=Rg+Rc\n",
-      "    L[n-1]=(N**2)/Rtot                  #Inductance(H)\n",
-      "    \n",
-      "\n",
-      "#Results:\n",
-      "print \"The reqired plot is shown below:\"\n",
-      "plot(ur, L,'g.')\n",
-      "xlabel('Core relative permeability, ur')\n",
-      "ylabel('Inductance,L (H) ')\n",
-      "title('plot of inductance vs. relative permeability for Example 1.5.')\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reqired plot is shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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pKQn37t27zNPTM0kkEikuXLjwkL29/YOwsLAQQggJCQkJO3Xq1Mz9+/cvVVJS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\nDQAAkNn/ASu6P4PfImgPAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2a2ba10>"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.6, Page number: 19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Bc=1.0                                 #Magnetic field induction in the core\n",
-      "w=377                                  #Angular frequency of magnetic field(rad/s)\n",
-      "Rc=3791.33                             #Reluctance of the core(A.turns/Wb)\n",
-      "Rg=442321.3                            #Reluctance of the air-gap(A.turns/Wb)\n",
-      "N=500                                  #No. of windings\n",
-      "i=0.80                                 #Current in the coil\n",
-      "Ac=9*10**-4                            #Cross-section of the core\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "L=N**2/(Rc+Rg)\n",
-      "W=(1./2)*L*i**2\n",
-      "t = symbols('t')\n",
-      "Bc = 1.0*sin(w*t)\n",
-      "e=N*Ac*diff(Bc,t)\n",
-      "\n",
-      "#Results:\n",
-      "print \"The Inductance, L:\", round(L,2), \"H\"\n",
-      "print \"The magntic stored energy, W:\", round(W,2), \"J\"\n",
-      "print \"Induced voltage, e:\",e,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Inductance, L: 0.56 H\n",
-        "The magntic stored energy, W: 0.18 J\n",
-        "Induced voltage, e: 169.65*cos(377*t) V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number: 22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "#Variable declaration:\n",
-      "Bc=1                                    #Magnetic field in the core\n",
-      "Hc=11                                   #Magnetising force(A.turns/m)\n",
-      "lc=0.3                                  #length of the core(m)\n",
-      "N=500                                   #No of windings\n",
-      "g=0.050                                 #Air-gap length(cm)\n",
-      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Fc=Hc*lc                                #mmf drop for the core path(A.turns)\n",
-      "Fg=Bc*g*10**-2/uo                       #mmf drop across the air gap(A.turns)\n",
-      "i=(Fc+Fg)/N\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The required current,i:\" ,round(i,2) ,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required current,i: 0.8 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.8, Page number: 28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "N=200                               #No. of turns\n",
-      "Ac=4                                #Cross-section of the core(in**2)\n",
-      "w=377                               #Angular frequency of the magnetic field(rad/s)\n",
-      "Hm=36                               #Max value magnetising force(A.turns/m)\n",
-      "Pc=1.2                              #Core loss density(W/kg)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "t=symbols('t')\n",
-      "Bc=1.5*sin(w*t)\n",
-      "e=(round(N*Ac*0.94/(39.4**2),2)*diff(Bc,t))\n",
-      "Erms=275*0.707\n",
-      "lc=(6+6+8+8)/39.4                   #Mean length of the core(m)\n",
-      "I=Hm*lc/N\n",
-      "Vc=4*0.94*28                        #Core volume(m**3)\n",
-      "Wc=105.5*(2.54**3)*7.65*10**-3                  #Core weight(kg)\n",
-      "Pa=1.5*13.2                         #Watts per Kg\n",
-      "Irms=Pa/Erms                        #Current (A)\n",
-      "Pct=Pc*Wc                           #Total core loss(W)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The applied voltage,e:\", e, \"V\"\n",
-      "print \"The peak current,I:\", round(I,2), \"A\"\n",
-      "print \"The total rms current. Irms:\", round(Irms,2), \"A\"\n",
-      "print \"Total Core loss, Pct:\",round(Pct,2),\"W\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The applied voltage,e: 271.44*cos(377*t) V\n",
-        "The peak current,I: 0.13 A\n",
-        "The total rms current. Irms: 0.1 A\n",
-        "Total Core loss, Pct: 15.87 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.9, Page number: 32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "from math import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "g=0.2                               #air-gap length(cm)\n",
-      "lm=1.0                              #length of magnetic section(cm)\n",
-      "Am=4                                #Cross-section of the core(cm**2)\n",
-      "Ag=4                                #Cross-section of the air-gap(cm**2)\n",
-      "\n",
-      "#Constants used:\n",
-      "uo=4*pi*10**-7                      #Permeability of free space(H/m)\n",
-      "\n",
-      "#Calculations:\n",
-      "Hm=symbols('Hm')\n",
-      "def Bg(Hm):\n",
-      "    return -uo*Ag*lm*Hm/(Am*g)             \n",
-      "\n",
-      "Hm1=-49*10**3                    #Coercivity of ALNICO 5 (A/m)\n",
-      "Hm2=-6                          #Coercivity of M-5 electrical steel (A/m)                     \n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Flux Density of air gap:\", round(Bg(Hm1),2),\"T\"\n",
-      "print \"\\nFlux Density of air gap:\", round(Bg(Hm2*10**4),2),\"gauss\"\n",
-      "print \"\\nwhere value of Hm for different material.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flux Density of air gap: 0.31 T\n",
-        "\n",
-        "Flux Density of air gap: 0.38 gauss\n",
-        "\n",
-        "where value of Hm for different material.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.10, Page number: 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ag=2                                #Cross-section of air-gap(cm**2) \n",
-      "Bg=0.8                              #Air-gap flux density(t)\n",
-      "Bm=1.0                              #Core-flux density(T)\n",
-      "Hm=-40                              #Magnetising force in the core(kA/m)\n",
-      "uo=4*pi*10**-7                      #permeability of free space(H/m)\n",
-      "g=0.2                               #Air-gap length(cm)\n",
-      "\n",
-      "#Calculations:\n",
-      "Am=Ag*Bg/Bm\n",
-      "lm=-g*Bg/(Hm*uo*10**3)\n",
-      "Vm=Am*lm\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The minimum magnet volume,Vm:\",round(Vm,2),\"cm**3\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum magnet volume,Vm: 5.09 cm**3\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1.11, Page number: 39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "from math import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Am = 2                    #magnetic material cros-section(cm^2)\n",
-      "g=0.2                     #air gap length(cm)\n",
-      "uo=4*pi*10**-7            #permeability of free space(H/m)\n",
-      "N=100                     #No. of windings\n",
-      "\n",
-      "#Calculations and results:\n",
-      "#for part (a)\n",
-      "Bma = 1.0                 #Tesla\n",
-      "Hma = - 4                 #kA/m\n",
-      "Ag1 = 2                    #cm**2\n",
-      "Ag2 = 4                   #cm**2\n",
-      "\n",
-      "lm=g*(Am/Ag1)*(Bma/(-uo*Hma*10**4))\n",
-      "print \"(a) The Requied magnet length = \",round(lm,2),\"cm\"\n",
-      "\n",
-      "\n",
-      "#for part (b):\n",
-      "i,Hm=symbols('i Hm')\n",
-      "Bm=-uo*(Ag1/Am)*(lm/g)*Hm+(uo*N/g)*(Ag1/Am)*i\n",
-      "H_max=200               #kA/m\n",
-      "B_max=2.1               #Tesla\n",
-      "i_max=(B_max+2.50*10**-5*H_max)/(6.28*10**-2)\n",
-      "\n",
-      "print \"(b) Thus with the air-gap area set to 2 cm^2,\"\n",
-      "print \"    increasing the current to i_max = 45.2 A and then reducing\"\n",
-      "print \"    it to zero will achieve the desired magnetization.\"\n",
-      "\n",
-      "#for part (c):\n",
-      "Bm1=1.00               #Tesla\n",
-      "Bm2=1.08               #Tesla\n",
-      "Bg1=(Am/Ag1)*Bm1\n",
-      "Bg2=(Am/Ag2)*Bm2\n",
-      "print \"(c) The flux densities when plunger moves at two extremes are:\"\n",
-      "print \"    Bg1 =\",Bg1,\"T and Bg2 =\",Bg2,\"T\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The Requied magnet length =  3.98 cm\n",
-        "(b) Thus with the air-gap area set to 2 cm^2,\n",
-        "    increasing the current to i_max = 45.2 A and then reducing\n",
-        "    it to zero will achieve the desired magnetization.\n",
-        "(c) The flux densities when plunger moves at two extremes are:\n",
-        "    Bg1 = 1.0 T and Bg2 = 0.54 T\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter10.ipynb b/ELECTRIC_MACHINERY/chapter10.ipynb
deleted file mode 100755
index 2bcd2ccb..00000000
--- a/ELECTRIC_MACHINERY/chapter10.ipynb
+++ /dev/null
@@ -1,538 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:603160d56b04457665b7cf5381387c06e4c52903d6f85f1c61a983bdc3c14ebc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10: Introduction to Power Electronics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.5, Page number: 508"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from pylab import *\n",
-      "import numpy as np\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "w=2*pi*60                               #frequency of voltage(Hz)\n",
-      "R=10                                    #ohm\n",
-      "C=0.01                                  #F\n",
-      "Vo=120*sqrt(2)                          #maximum voltage(V)\n",
-      "Nmax=800\n",
-      "tau=R*C                                 #time constant(s)\n",
-      "\n",
-      "#Calculations:\n",
-      "# diode = 1 when rectifier bridge is conducting\n",
-      "\n",
-      "diode=1\n",
-      "t=[0]*801\n",
-      "vs=[0]*801\n",
-      "vrect=[0]*801\n",
-      "vR=[0]*801\n",
-      "iB=[0]*801\n",
-      "\n",
-      "t=[0]*801\n",
-      "for n in range(1,Nmax+2,1):\n",
-      "    t[n-1] = (2.5*pi/w)*(n-1)/Nmax\n",
-      "    vs[n-1]=Vo*math.cos(w*t[n-1])\n",
-      "    vrect[n-1]=abs(vs[n-1])\n",
-      "#if the rectifier bridge is ON:\n",
-      "    if diode==1:\n",
-      "        vR[n-1]=vrect[n-1]\n",
-      "        if (w*t[n-1])<=(pi/2):\n",
-      "            iB[n-1]=vR[n-1]-Vo*C*w*math.sin(w*t[n-1])\n",
-      "        elif (w*t[n-1])<=3*pi/2:\n",
-      "            iB[n-1]=vR[n-1]/R+Vo*C*w*math.sin(w*t[n-1])\n",
-      "        else:\n",
-      "            iB[n-1]=vR[n-1]/R-Vo*C*w*math.sin(w*t[n-1])\n",
-      "        if iB[n-1]<0:\n",
-      "            diode=0\n",
-      "            toff=t[n-1]\n",
-      "            Voff=vrect[n-1]\n",
-      "    else:\n",
-      "        vR[n-1]=Voff*exp(-(t[n-1]-toff/tau))\n",
-      "        iB[n-1]=0\n",
-      "        if (vrect[n-1]-vR[n-1])>0:\n",
-      "            diode=1\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "iR=(1/R)*np.array(vR)\n",
-      "plot(1000*np.array(t),vR)\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('voltage [V]')\n",
-      "xlim(0,22)\n",
-      "ylim(0,180)\n",
-      "plot(1000*np.array(t),vrect,'--')\n",
-      "grid()\n",
-      "print \"The required plots are shown below:\"\n",
-      "show()\n",
-      "plot(1000*np.array(t),iR)\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('source current [A]')\n",
-      "xlim(0 ,22)\n",
-      "ylim(-50,250)      \n",
-      "plot(1000*np.array(t),1.5*np.array(iB),'--')\n",
-      "grid()\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required plots are shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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DAABNDU1McpmEPensHQ/k6valsCYwderUXWfPnh326uu5ublW58+fD+zevftD6WsZGRmO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\n1BhtAhRFUWqMNgGKoig1RpsARVGUGvt/5i5yreN7pjkAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x25c1f10>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x3529b90>"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.6, Page number: 522"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "w=2*pi*60                               #Angular freq of voltage(rad/sec)\n",
-      "Vo=230*sqrt(2)                          #volt\n",
-      "R=5.6                                   #Resistance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "Ls=[0]*101\n",
-      "tc=[0]*101\n",
-      "Idc=[0]*101\n",
-      "for n in range(1,101,1):\n",
-      "    Ls[n-1]=n*10**-3\n",
-      "    Idc[n-1]=2*Vo/(pi*R+2*w*Ls[n-1])\n",
-      "    tc[n-1]=(1/w)*acos(1-(2*Idc[n-1]*w*Ls[n-1])/Vo)\n",
-      "\n",
-      "#Results:\n",
-      "plot(1000*np.array(Ls),Idc,'g.')\n",
-      "xlabel('Commutating inductance Ls [mH]')\n",
-      "ylabel('Idc [A]')\n",
-      "title('Load current,Idc vs Commutating inductance,Ls')\n",
-      "show()\n",
-      "plot(1000*np.array(Ls),1000*np.array(tc),'g.')\n",
-      "xlabel('Commutating inductance L [mH]')\n",
-      "ylabel('tc [msec]')\n",
-      "title('Commutating Inductance,Ls vs time,tc')\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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YmBReuHBhxLsE8a+gOmhCkIYhqQDQnuR+L6PGxkb1uLi4MS1tZPz48SfeJYDO\nCs1HANDRtHiGcOnSJQ8PD49LLRVw8eLFoUOHDr3YpkF1gjMENB8BQHtS2LDTnJwcq0OHDk1ZsmTJ\nhnfZeHM6Q0KQhuYjAJC3dh12WlBQYPrjjz9+6uHhcYnH4wnEYjH7XTasStB8BAAdQYt9CBUVFfpH\njx6dcODAgaDMzMze/v7+x7Oysqzp/0aG1omeGP3KFc24gA0AlFGLTUba2to13t7eCV988cXaQYMG\nXSOEEGtr66ysrCxruQbVyZqMZKEJCQDamtybjNatW7ciPz+fNXfu3B0RERHLHz16ZPMuG4MX0IQE\nAMqoVZ3Kjx49sjl48ODUgwcPTv3rr7/eW716dXhAQMAxW1vbh3IJqpOfIWAEEgC0NYWMMrpz547D\ngQMHgg4dOjRFXmcMnT0hSEPzEQC0BdzttBPAX3ICQFuQex+Cr69v7OsKaG6Z3Nxcy+HDhyfa29vf\n69ev392tW7fOJ4SQkpISI29v7wRbW9uHo0aNii8rK1PpI2D0xGgymTuZxAfHk6UJSwlvL4+M/XUs\nKastU3RoAKBiWjxDMDAwKB82bFhySwXcvXu3X1OjjsRiMVssFrOdnJzSqqqqdF1dXW8eP37cf8+e\nPTOMjY2Lli5d+t369euXlZaWGkZERCx/JSgVOkOQhuYjAHhbcr+X0e+///7B6wro0qXL86aeZ7PZ\nYjabLSaEEF1d3So7O7s/8/Lyepw4cWJ8UlKSJyGEhISERPF4PIFsQlBVGH0EAIrULn0IQqGQ4+np\nmXT37t1+VlZWOaWlpYaEEEJRFMPIyKiEnv87KAaDCg8P/3uex+MRHo8n9zgVTfYvOXEBGwA0RyAQ\nEIFA8Pf86tWrlb9TuaqqStfT0zNp1apVa/z9/Y8bGhqWSicAIyOjkpKSEqNXglLRJiNZaEICgNZS\n+r/QrK+v15w4ceKR4ODg/f7+/scJIYTFYuXT90ESiURmpqamBfKMoSOTbkLS0dRBhzMAyFWrEkJV\nVZVuY2OjOj3f2Nio/uzZs24trUNRFGPWrFmRXC43Y8GCBZvp58ePH38iKioqhBBCoqKiQuhEAf8m\nPQJJWCbE/zUDgFy1qslo4MCB18+fP++lq6tbRQghlZWVej4+PmevXLkypLl1Ll265DFs2LBkR0fH\ndAaDQRHy4lYY7u7uNwIDA2NycnKsOByOMCYmJpDJZL7ykxdNRv+G6xUAoCXtdmGak5NTWlpamtPr\nnmsrSAhmTVezAAAZPklEQVT/httdAEBL2q0PoVu3bs9u3rzpSs+npKS4aWtr17zLhuHNMLsySczk\nGMLsyiQPix+i+QgA2lyL1yHQNm/evCAwMDDGzMxMRMiLzuBDhw5NkW9o0BzZ6xUwPBUA2kKrh53W\n1dVpPXjwoA+DwaD69OnzQFNTs15uQaHJqEWy1ytgeCoAyL0P4ciRIxNfHpwZdMewtAkTJhx9l403\nGxQSwhuR7nDmmnCJsEyIswUAFSP3hBAaGrqXwWBQBQUFpleuXBkyYsSIC4QQkpiYOHzIkCFXYmNj\nfd9l480GhYTwRqTPGPwP+uNsAUAFyf1eRnv37g0lhBBvb++EjIwMrnQfQkhISNS7bBjaDt3hTEjT\nF7PhbAEAWqNVo4xyc3Mt6RvVEfLiauOcnBwr+YUFbwsXswHA22rVKKORI0ee8/HxOTtt2rRoiqIY\nhw4dmuLt7Z0g7+DgzTV3toDRSADwOq0aZURRFOPYsWMBycnJwxgMBjVs2LDkgICAY3ILCn0IbQKj\nkQBUB/5CE94IRiMBdF5yTwi6urpVTQ03fblxqqKiQv9dNt5sUEgIcoHRSACdl9xHGVVVVem+S+Gg\nXDAaCQBaItf/QwDlhdFIACCrVaOMoPPBaCQAkIUzBHjlbAF3UwVQXUgI8MqttQnBX3cCqCoMO4V/\naW40kjXTmlgZWKEpCUAJyX2UEaim5voXtNS1/k4OYSfDMFQVoJNBkxG0SLp/Qb/Li8tOpDuf0ZwE\n0HnILSHMnDnzZxaLle/g4HCHfo7P5/MtLCyeODs7pzo7O6eeOXNmtLy2D21Dun+hpc5nl50uSA4A\nHZzcEsKMGTP2yB7wGQwGtWjRoh9SU1OdU1NTnUePHn1GXtuHttdS57O5njlGJgF0cHJLCEOHDr1o\naGhYKvv8u3Z6gPJAcxJA59Luncrbtm37bN++fdPd3NxSNm7cuJjJZDZ5tODz+X8/5vF4hMfjtVOE\n0FrSnc/RE6NfubMq3ZxECCEuO10wOgmgjQkEAiIQCNq0TLkOOxUKhRw/P7+Td+7ccSCEkIKCAlMT\nE5NCQghZtWrVGpFIZBYZGTnrX0Fh2GmHJ31nVS11LXI59zIhBDfSA5CXthh22q6jjExNTQsYDAbF\nYDCo2bNn/+/GjRvu7bl9aD/NNSfhQjcA5dWuTUYikciM/l/mY8eOBUiPQILOpbnmJOkL3cJOhv3d\nvITmJADFk1tCCAoKOpCUlORZVFRkbGlpmbt69epwgUDAS0tLc2IwGJS1tXXWzp07P5bX9kF5tHQj\nPekEgb4GAMXCrSugXcn+rSf6GgDaBv5CEzo86QQx7cg0/MUnwFtCQoBOpaW/+ERfA0DLkBCg05Ju\nSooPjsddVwFeA3c7hU5L9kK35u66io5ogLaDMwToEJrra5DtiEbTEqgqNBmBSmouOaBpCVQZEgKo\nvNYOY0VygM4OCQFABpqWQFUhIQC0AE1LoEqQEABaCU1L0NkhIQC8JTQtQWeDhADQBtC0BJ0BEgJA\nG0PTEnRUSAgAcoamJegokBAA2hGalkCZISEAKMjbNi0tTViKMwmQCyQEACXR2qalgmcFOJMAuUBC\nAFBCLTUtoR8C5AUJAUDJyTYtvU0/hGk3U/x7HLyWUieEmTNn/nzq1KlxpqamBXfu3HEghJCSkhKj\nKVOmHMrOzu7J4XCEMTExgUwms+xfQSEhgApobT+EiY4JKawuJITgTAKap9QJ4eLFi0N1dXWrpk+f\nvo9OCEuXLv3O2Ni4aOnSpd+tX79+WWlpqWFERMTyfwWFhAAqqLmzB4OuBuTc43M4k4AWKXVCIIQQ\noVDI8fPzO0knhL59+95PSkryZLFY+WKxmM3j8QT379/v+6+gkBBAxUknB0IIziTgtTrcX2jm5+ez\nWCxWPiGEsFis/Pz8fFZzy/L5/L8f83g8wuPx5B4fgLJgdmWSmMkxf89LP5b+e9FpR6YRQsi/ziR2\n+e165UxC+q9GcSbROQgEAiIQCNq0zHY9QzA0NCwtLS01pF83MjIqKSkpMfpXUDhDAGgVnEkArcOd\nIdBNRWw2WywSicxMTU0L2nP7AJ0NziSgLbVrQhg/fvyJqKiokGXLlq2PiooK8ff3P96e2wdQJdLJ\nQjo5EPLqmYSOpg4hhPx9JkEnB+kzibCTYa+cSSBZdE5yazIKCgo6kJSU5FlUVGTMYrHyv/76668+\n+OCD3wMDA2NycnKsMOwUQDm8zegm6WSB23MoB6UfZfS2kBAAFKO1fRLSyaK1t+fAWYV8ISEAQLtp\nLlm09vYcLXVmI1m8OyQEAFC41t6eA01Q8oWEAABKTZ5NUEgWr0JCAIAO612boNBf8SokBADodN7k\nDrHv2l/RmRIHEgIAqJS27q/oTB3dSAgAAC+9TX9FW3R0K0viQEIAAGiF5pKF9OO37ehWlsSBhAAA\n0IbepqNbWRIHEgIAQDtoqe+CEOVIHIbahkgIAADKQpGJI2tBFhICAEBHI5fEMesyEgIAQGf1Ronj\nozgkBAAAVVdWW4Y+BAAAeKEtRhmptVUwAADQsSEhAAAAIQQJAQAAXkJCAAAAQgghGorYKIfDEerr\n61eoq6s3ampq1t+4ccNdEXEAAMA/FJIQGAwGJRAIeEZGRiWK2D4AAPybwpqM3nV4FAAAtC2FnSGM\nHDnynLq6euPHH3+8c86cObtll+Hz+X8/5vF4hMfjtWOEAADKTSAQEIFA0KZlKuTCNJFIZGZmZiYq\nLCw08fb2Tti2bdtnQ4cOvfh3ULgwDQDgjXTYC9PMzMxEhBBiYmJSGBAQcAydygAAitfuCaG6ulqn\nsrJSjxBCnj171i0+Pn6Ug4PDnfaOAwAAXtXufQj5+fmsgICAY4QQ0tDQoPHhhx/+OmrUqPj2jgMA\nAF6Fm9sBAHQCHbYPAQAAlA8SAgAAEEKQEAAA4CUkBAAAIIQgIQAAwEtICAAAQAhBQgAAgJeQEAAA\ngBCChAAAAC8hIQAAACEECQEAAF5CQgAAAEIIEgIAALyEhAAAAIQQJAQAAHgJCQEAAAghSAgAAPAS\nEgIAABBCkBCUnkAgUHQISgN18Q/UxT9QF21HIQnhzJkzo/v27Xv/vffe+2v9+vXLFBFDR4EP+z9Q\nF/9AXfwDddF22j0hNDY2qs+bN2/7mTNnRmdkZHAPHDgQ9Oeff9q1dxwAAPCqdk8IN27ccO/du3cm\nh8MRampq1k+dOvXg77///oHscmN/HUvKasvaOzwAANVFUVS7Tr/99tuk2bNn76bn9+/f/9G8efO2\nSS9DCKEwYcKECdObTe96fNYg7YzBYFCvW4aiKEZ7xAIAAP9o9yajHj165OXm5lrS87m5uZYWFhZP\n2jsOAAB4VbsnBDc3t5S//vrrPaFQyKmrq9M6dOjQlPHjx59o7zgAAOBV7d5kpKGh0bB9+/Z5Pj4+\nZxsbG9VnzZoVaWdn92d7xwEAADLau1P5dVNcXNzoPn363O/du/dfERERyxQdT3tOOTk5ljweL5HL\n5d6zt7e/u2XLlvkURZHi4mKjkSNHJrz33nsPvb2940tLS5mKjrU9poaGBnUnJ6dUX1/fk6pcD6Wl\npcyJEyce7tu37592dnYZ165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IhUKOn5/fyTt37ji0Zbnw9tCpDJ1WTk6OVWBgYIxEIlHT\n0tKqk24yg9bT1dWt2rVrV1hlZaXem16L0JyLFy8O/fTTT380MTEpbIvyoG3gDAEAAAgh6EMAAICX\nkBAAAIAQgoQAAAAvISEAAAAhBAkBAABeQkIAAABCCCH/D1IYMYHMZAvtAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2ce0f10>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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rNrSlYlmi19lH0AcA+CeJAn90dPRHXC63mN7ncrnF586dGye7ZgEAgKxIdOdu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\ng8APAMAwCPwAAAyDwA8AwDD/DxR4tH4eP493AAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x33b4a90>"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.7, Page number: 528"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R=12.5*10**-3                               #ohm\n",
-      "L=1.2                                       #H\n",
-      "Vo=15                                       #volt\n",
-      "w=120*pi                                    #angular freq(Hz)\n",
-      "Idc=35                                      #DC current(A)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "theta=[0]*1301\n",
-      "t=[0]*1301\n",
-      "vL=[0]*1301\n",
-      "vs=[0]*1301\n",
-      "\n",
-      "Vdc_a=R*Idc                                 #Dc voltage(V)\n",
-      "P=Vdc_a*Idc                                 #Power\n",
-      "alpha_da = acos(pi*R*Idc/(2*Vo)) ;    #delay angle\n",
-      "for n in range(1,1301,1):                 #loop for calculating load voltage\n",
-      "    theta[n-1]=2*pi*(n-1)/1000\n",
-      "    t[n-1]=theta[n-1]/w\n",
-      "    vs[n-1]=Vo*sin(theta[n-1])\n",
-      "    if theta[n-1]<alpha_da:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif (theta[n-1]<pi+alpha_da):\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif theta[n-1]<2*pi+alpha_da:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<3*pi+alpha_da:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<4*pi+alpha_da:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    else:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "\n",
-      "figure(1)\n",
-      "plot(1000*np.array(t),vL,'g.')\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('Load voltage [V]')\n",
-      "grid()\n",
-      "show()\n",
-      "\n",
-      "\n",
-      "#part(b):\n",
-      "alpha_db=0.9*pi                             #delay angle\n",
-      "Vdc_b=(2*Vo/pi)*cos(alpha_db)          #new dc voltage(V)\n",
-      "tau=L/R                                     #time constant(s)\n",
-      "imo=Idc                                     #Initial curent(A)\n",
-      "tzero=-tau*log((-Vdc_b/R)/(imo-Vdc_b/R))\n",
-      "for n in range(1,1301,1):\n",
-      "    theta[n-1]=2*pi*(n-1)/1000\n",
-      "    t[n-1]=theta[n-1]/w\n",
-      "    vs[n-1]=Vo*sin(theta[n-1])\n",
-      "    if theta< alpha_db:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif (theta[n-1]<pi+alpha_db):\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<2*pi+alpha_db:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    elif theta[n-1]<3*pi+alpha_db:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "    elif theta[n-1]<4*pi+alpha_db:\n",
-      "        vL[n-1]=-vs[n-1]\n",
-      "    else:\n",
-      "        vL[n-1]=vs[n-1]\n",
-      "\n",
-      "#Results:\n",
-      "figure(2)\n",
-      "plot (1000*np.array(t), vL,'g.')\n",
-      "xlabel('time [msec] ')\n",
-      "ylabel('Load voltage [V]')\n",
-      "print \"part (a):\"\n",
-      "print \"\\n Vdc_a=\",round(1000*Vdc_a,2),\"mV\"\n",
-      "print \"\\n Power=\",round(P),\"W\"        \n",
-      "print \"\\n alpha_d=\",round((180/pi)*alpha_da,1),\"degrees\"\n",
-      "print \"\\n part (b):\"\n",
-      "print \"\\n alpha_d=\",round((180/pi)*alpha_db,1),\"degrees\"   \n",
-      "print \"\\n Vdc_b=\",round(Vdc_b,1),\"V\"\n",
-      "print \"\\n Current will reach zero at\",round(tzero,1),\"sec\"\n",
-      "grid()\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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LPuhJW/DhLpBJ2C8sZzJJ1NbWypYuXfrRsGHD8ocNG5a/fPny9Xfv3u1j6ve4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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2594cd0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part (a):\n",
-        "\n",
-        " Vdc_a= 437.5 mV\n",
-        "\n",
-        " Power= 15.0 W\n",
-        "\n",
-        " alpha_d= 87.4 degrees\n",
-        "\n",
-        " part (b):\n",
-        "\n",
-        " alpha_d= 162.0 degrees\n",
-        "\n",
-        " Vdc_b= -9.1 V\n",
-        "\n",
-        " Current will reach zero at 4.5 sec\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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/wb59+xYzUTkkHtQEKFeXeUA9kav6kshVfZEw4DwJxAm4xwS/4R4T3Gez/SRa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\nYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x37f8250>"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.8, Page number: 533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "f=60                                #Hz\n",
-      "Vrms=35                             #rms voltage of waveform\n",
-      "Ra=3.5                              #Armature resistance(ohm)\n",
-      "La=0.175                            #H\n",
-      "no=8000                             #No load speed(r/min)\n",
-      "Va=50                               #armature voltage(V)\n",
-      "\n",
-      "#Calculations:\n",
-      "Edc,alphad=symbols('Edc alphad')\n",
-      "Vdc=Edc                                 #at no load, Vdc=Edc\n",
-      "Edc=round(float(2*sqrt(2)*(Vrms/math.pi)),2)*cos(alphad)\n",
-      "n=Edc*float(no/50)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Speed at no-load =\",n,\" r/min      (where 0 <= alphad <= pi/2)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed at no-load = 5041.6*cos(alphad)  r/min      (where 0 <= alphad <= pi/2)\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.9, Page number: 537"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vll_rms=460                                 #rms voltage,line-to-line(V)\n",
-      "R=68                                        #resistance of load\n",
-      "Im=2.5                                      #magnet current(A)\n",
-      "\n",
-      "#Calculations:\n",
-      "Vdc_max=3*sqrt(2)*Vll_rms/pi\n",
-      "Idc_max=Vdc_max/R\n",
-      "Vdc=Im*R\n",
-      "alpha=acos(pi*Vdc/(3*sqrt(3)*Vll_rms))\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Maximum dc voltage:\",round(Vdc_max),\"V\"\n",
-      "print \"\\n  Maximum dc current:\",round(Idc_max,1),\"V\"\n",
-      "print \"\\n(b) Delay angle alpha:\",round(math.degrees(round(alpha,1)),1),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Maximum dc voltage: 621.0 V\n",
-        "\n",
-        "  Maximum dc current: 9.1 V\n",
-        "\n",
-        "(b) Delay angle alpha: 74.5 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.10, Page number: 541"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "T=20*10**-3                                 #Time period(sec) \n",
-      "p=4                                         #no. of poles\n",
-      "delta=0.44                                  #ON- time fraction\n",
-      "Vo=125                                      #DC supply voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "fc=1/T\n",
-      "ns=(120*fc/p)\n",
-      "Va_peak=(4*Vo*sin(delta*pi))/pi\n",
-      "Vll_rms=sqrt(3/2)*Va_peak\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Frequency:\",fc,\"Hz\"\n",
-      "print \"\\n    Synchronous speed:\",ns,\"r/min\"\n",
-      "print \"\\n(b) Rms amplitude of line-to-line voltage:\",round(Vll_rms,0),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Frequency: 50.0 Hz\n",
-        "\n",
-        "    Synchronous speed: 1500.0 r/min\n",
-        "\n",
-        "(b) Rms amplitude of line-to-line voltage: 191.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.13, Page number: 547"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vo=48                               #Load voltage(V)\n",
-      "R=3.7                               #Resistance of load(ohm)\n",
-      "L=.32                               #Inductance of laad(H)\n",
-      "D=0.8                               #Duty cycle\n",
-      "f=1000                              #Hz\n",
-      "\n",
-      "#Calculations:\n",
-      "iL_avg=(2*D-1)*Vo/R\n",
-      "T=1/f\n",
-      "tau=L/R\n",
-      "iL_min=((-Vo/R)*(1-2*exp(-T*(1-D)/tau)+exp(-T/tau)))/(1-exp(-T/tau))\n",
-      "iL_max=(Vo/R)*(1-2*exp(-D*T/tau)+exp(-T/tau))/(1-exp(-T/tau))\n",
-      "\n",
-      "#since T/tau << 1, so using 10.32 in e.g. given.\n",
-      "del_iL=(2*Vo)*T*D*(1-D)/(R*tau)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Avg load current:\",round(iL_avg,2),\"A\"\n",
-      "print \"Minimum load current:\",round(iL_min,2),\"A\"\n",
-      "print \"Maximum load current\",round(iL_max,2),\"A\"\n",
-      "print \"Current ripple:\",round(del_iL,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Avg load current: 7.78 A\n",
-        "Minimum load current: 7.76 A\n",
-        "Maximum load current 7.81 A\n",
-        "Current ripple: 0.05 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter11.ipynb b/ELECTRIC_MACHINERY/chapter11.ipynb
deleted file mode 100755
index 308afbf5..00000000
--- a/ELECTRIC_MACHINERY/chapter11.ipynb
+++ /dev/null
@@ -1,975 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4aa0be3787a5262658b58d5130a82adad92353bb1944b3902dc15e9d5045fa75"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11: Speed and Torque Control"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.1, Page number: 561"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vdc=240                             #DC supply (V)\n",
-      "D=0.75                              #Duty cycle\n",
-      "Rf=187                              #field resistance(ohm)\n",
-      "Lf=4.2                              #field winding inductance(H)\n",
-      "T=1                                 #switching period(msec)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "If=D*(Vdc/Rf)\n",
-      "tau=Lf/Rf                           #time constant(msec)\n",
-      "del_if=(2*Vdc/Rf)*(T/tau)*D*(1-D)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Avg field current:\",round(If,2),\"A\"\n",
-      "print \"Magnitude of currnet ripple:\",round(del_if,1),\"mA\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Avg field current: 0.96 A\n",
-        "Magnitude of currnet ripple: 21.4 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.2, Page number: 563"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n1=1800                                 #r/min\n",
-      "n2=3600                                 #r/min\n",
-      "Va=240                                  #terminal voltage(V)\n",
-      "Ifo=0.34                                #No-load field current(A)\n",
-      "Ra=0.05                                 #Armature resistance(ohm)\n",
-      "Rsh=187                                 #Shunt field resistance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "wm=symbols('wm')\n",
-      "wm1=float(2*pi*n1/60)\n",
-      "wm2=float(2*pi*n2/60)\n",
-      "def Pload(wm):\n",
-      "    return (22.4*(120*pi)**-3)*(wm)**3\n",
-      "\n",
-      "T1=Pload(wm1)*1000/wm1\n",
-      "T2=Pload(wm2)*1000/wm2\n",
-      "\n",
-      "Kf=Va/(Ifo*wm2)\n",
-      "def If(T,wm):\n",
-      "    return (Va/(2*Kf*wm))*(1+sqrt(1-(4*wm*T*Ra)/Va**2))\n",
-      "\n",
-      "Rf1tot=round(Va/float(If(T1,wm1)))\n",
-      "Rf2tot=round(Va/float(If(T2,wm2)))\n",
-      "Rrh1=Rf1tot-Rsh\n",
-      "Rrh2=Rf2tot-Rsh\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"----------------------------------------------------------------\"\n",
-      "print \"r/min    Tload[N.m]      If[A]      R(f)tot[ohm]   Rrheostat[ohm]\"\n",
-      "print \"----------------------------------------------------------------\"\n",
-      "print n1,\"\\t \",round(float(T1),1),\"\\t\\t \",round(float(If(T1,wm1)),3),\"\\t\",Rf1tot,\"\\t      \",Rrh1\n",
-      "print n2,\"\\t \",round(float(T2),1),\"\\t\\t \",round(float(If(T2,wm2)),3),\"\\t\",Rf2tot,\"\\t      \",Rrh2\n",
-      "print \"----------------------------------------------------------------\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "----------------------------------------------------------------\n",
-        "r/min    Tload[N.m]      If[A]      R(f)tot[ohm]   Rrheostat[ohm]\n",
-        "----------------------------------------------------------------\n",
-        "1800 \t  14.9 \t\t  0.678 \t354.0 \t       167.0\n",
-        "3600 \t  59.4 \t\t  0.333 \t720.0 \t       533.0\n",
-        "----------------------------------------------------------------\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.3, Page number: 567"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from math import *\n",
-      "from pylab import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Rf=109                                  #Field resistance(ohm)\n",
-      "Vf=300                                  #Rated field voltage(V)\n",
-      "Ra=0.084                                #Armature resistance(ohm)\n",
-      "Kf=0.694                                #Geometric constant(A.rad/sec)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "If=Vf/Rf                                #Resulting field current(A)\n",
-      "w_rated=2500*(pi/30)                    #Rated speed(rad/sec)\n",
-      "P_rated=100*746                         #Watts\n",
-      "T_rated=P_rated/w_rated                 #Nm\n",
-      "Va=[0]*102\n",
-      "NoLoadRPM=[0]*102\n",
-      "FullLoadRPM=[0]*102                      \n",
-      "for n in range(1,102,1):\n",
-      "    Va[n-1]=250*(1+(n-1)/100)\n",
-      "    T=0                                     #Zero torque\n",
-      "    w=(Va[n-1]-T*Ra/(Kf*If))/(Kf*If)\n",
-      "    NoLoadRPM[n-1]=w*30/pi\n",
-      "    T=T_rated\n",
-      "    w=(Va[n-1]-T*Ra/(Kf*If))/(Kf*If)\n",
-      "    FullLoadRPM[n-1]=w*30/pi\n",
-      "\n",
-      "print\"The plot is as shown:\"\n",
-      "plot(Va,NoLoadRPM)\n",
-      "plot(Va[20] ,NoLoadRPM[20] ,'r+')\n",
-      "plot (Va[50] , NoLoadRPM[50] , 'r+')\n",
-      "plot (Va[80] ,NoLoadRPM[80] , 'r+')\n",
-      "plot (Va, FullLoadRPM,'.')\n",
-      "plot (Va[20] ,FullLoadRPM[20] ,'o')\n",
-      "plot (Va[50] , FullLoadRPM[50] , ' o' )\n",
-      "plot (Va[80] , FullLoadRPM[80] ,'o' )\n",
-      "title('Speed vs Armature voltage')\n",
-      "xlabel('Armature voltage [V] ')\n",
-      "ylabel('Speed [r/min] ')\n",
-      "annotate('+ = Zero torque',xy=(270,2300))\n",
-      "annotate('o =   Full load torque',xy=(270,2100))\n",
-      "ylim(1000,2500)\n",
-      "xlim(250,500)\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The plot is as shown:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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DvCo0sVPjozsQFWzatAlCoRBCoRCbNzf82I46Ojrlrw8dOoTp06eXv1d8lFfV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\ngg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2718810>"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.4, Page number: 571"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Rf=109                                  #Field resistance(ohm)\n",
-      "Vf=300                                  #Rated field voltage(V)\n",
-      "n1=2000                                  #rpm\n",
-      "T_rated=285                            #Rated torque(Nm)\n",
-      "n2=1975                                 #Dropped rpm\n",
-      "Kf=0.694                                #Geometric constant(A.rad/sec\n",
-      "Ra=0.084                                #Armature resistance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "If=Vf/Rf                                #Resulting field current(A)\n",
-      "wm1=2*pi*n1/60\n",
-      "w_ref=wm1\n",
-      "Vao=Kf*If*wm1\n",
-      "Ia=T_rated/(Kf*If)\n",
-      "wm2=2*pi*n2/60\n",
-      "Ea=Kf*If*wm2\n",
-      "Va=Ea+Ia*Ra\n",
-      "G=symbols('G')\n",
-      "x=solve(Vao-round(Va)+G*(w_ref-wm2),G)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Armature voltage,Vao:\",round(Va,0),\"V\"\n",
-      "print \"Multiplicative constant,G:\",float(round(x[0],2)),\"A.sec/rad\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature voltage,Vao: 408.0 V\n",
-        "Multiplicative constant,G: 3.04 A.sec/rad\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.5, Page number: 573"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Km=0.22                             #torque constant(V/(rad/sec))\n",
-      "Ra=1.03                             #ohm\n",
-      "Pl=100                            #Power load(W)\n",
-      "Va1=40                              #Armature voltage(V)\n",
-      "Va2=50                              #  \"        \"     \"\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "wm1=(Va1/(2*Km))*(1+sqrt(1-(4*Pl*Ra/Va1**2)))\n",
-      "wm2=(Va2/(2*Km))*(1+sqrt(1-(4*Pl*Ra/Va2**2)))\n",
-      "\n",
-      "#Results:\n",
-      "print \"for Va=40 V, wm=\",round(wm1,1),\"rad/sec\"\n",
-      "print \"for Va=50 V, wm=\",round(wm2,1),\"rad/sec\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for Va=40 V, wm= 169.2 rad/sec\n",
-        "for Va=50 V, wm= 217.5 rad/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.6, Page number: 575"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "\n",
-      "Rf=109                                  #Field resistance(ohm)\n",
-      "Vf=300                                  #Rated field voltage(V)\n",
-      "Ra=0.084                                #Armature resistance(ohm)\n",
-      "Kf=0.694                                #Geometric constant(A.rad/sec)\n",
-      "Tfl=285                                 #Full load torque(Nm)\n",
-      "nf=2500                                 #Speed at full load(r/min)\n",
-      "#wm=2500                                 #rated r/min\n",
-      "\n",
-      "#for part (1):\n",
-      "n1=2000                                 #r/min\n",
-      "n2=2500                                 #r/min\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#part (a):\n",
-      "If=Vf/Rf\n",
-      "w1=n1*2*pi/60\n",
-      "w2=n2*2*pi/60\n",
-      "Ea1=Kf*If*w1                    #Avg Amature voltage(V)\n",
-      "Ea2=Kf*If*w2\n",
-      "Ia1=n1*Tfl/(nf*Kf*If)\n",
-      "Ia2=n2*Tfl/(nf*Kf*If)\n",
-      "Va1 = Ea1 + Ia1*Ra\n",
-      "Va2 = Ea2 + Ia2*Ra\n",
-      "Tl1=(n1/nf)*Tfl\n",
-      "Tl2=(n2/nf)*Tfl\n",
-      "\n",
-      "#part (b):\n",
-      "\n",
-      "# The dynamic equation governing the speed of the motor is\n",
-      "\n",
-      "#   J*(dwm/dt)=Tmech-Tload\n",
-      "#   wm=(pi/30)*n & wr=(pi/30)*nf\n",
-      "#      Tload= (Tfl/wf)*wm\n",
-      "#   Tmech = Kf*If*Ia=Kf*If*(Va-Ea)/Ra          #Under armature-voltage control\n",
-      "\n",
-      "#   Thus the governing differential equation is\n",
-      "#   d(wm)/dt + 48.4*wm - 24.7*Va = 0\n",
-      " \n",
-      "#   wm = wf + (wi-wf)*exp(-t/tau)          #tau=1/48.4=20.7 msec\n",
-      "#   n = 2500- 50*exp( -t/tau )\n",
-      "\n",
-      "#   The armature current will decrease exponentially with the \n",
-      "#   same 20.7 msec time constant from an initial value of \n",
-      "#   (Vf - Vi)/Ra = 1190 A to its final value of 149 A.\n",
-      "\n",
-      "#   Ia = 149 + 1041*exp(-t/tau)\n",
-      "\n",
-      "#part (c):\n",
-      "#        J*d(wm)/dt = Tmech-Tload = Tf-(Tf/wm)*wm\n",
-      "# or      d(wm)/dt + 1.18*wm - 310 = 0\n",
-      "\n",
-      "#In this case, the speed will rise exponentially to wm=wf=262 rad/sec as\n",
-      "#         wm = 262-53*exp(-t/tau)           #tau=1/1.18=845 msec\n",
-      "\n",
-      "#Results:\n",
-      "print \"part(a):\\n\"\n",
-      "print \"-------------------------------------------------\"\n",
-      "print \"r/min\\tw[rad/s]\\tVa(V)\\tIa(A)\\tTload[Nm]\"\n",
-      "print \"-------------------------------------------------\"\n",
-      "print n1,\"\\t\",round(w1),\"\\t\\t\",round(Va1),\"\\t\",round(Ia1),\"\\t\",Tl1,\"Nm\"\n",
-      "print n2,\"\\t\",round(w2),\"\\t\\t\",round(Va2),\"\\t\",round(Ia2),\"\\t\",Tl2,\"Nm\"\n",
-      "print \"-------------------------------------------------\"\n",
-      "print \"\\npart (b):\"\n",
-      "print \" The resultant motor speed, n = 2500 - 50*exp(-t/tau) where tau=20.7 msec\"\n",
-      "\n",
-      "print \"\\npart (c):\"\n",
-      "print \" The resultant motor speed, wm = 262 - 53*exp(-t/tau) where tau=845 msec\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part(a):\n",
-        "\n",
-        "-------------------------------------------------\n",
-        "r/min\tw[rad/s]\tVa(V)\tIa(A)\tTload[Nm]\n",
-        "-------------------------------------------------\n",
-        "2000 \t209.0 \t\t410.0 \t119.0 \t228.0 Nm\n",
-        "2500 \t262.0 \t\t513.0 \t149.0 \t285.0 Nm\n",
-        "-------------------------------------------------\n",
-        "\n",
-        "part (b):\n",
-        " The resultant motor speed, n = 2500 - 50*exp(-t/tau) where tau=20.7 msec\n",
-        "\n",
-        "part (c):\n",
-        " The resultant motor speed, wm = 262 - 53*exp(-t/tau) where tau=845 msec\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.7, Page number: 581"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable Calculations:\n",
-      "f1=60                                #Initial frequency(Hz)\n",
-      "f2=50                               #Changed frequency(Hz)\n",
-      "Xs=0.836                            #Saturated synch reactance(ohm)\n",
-      "Va=1+0j                             #Armature voltage(V p.u)\n",
-      "Ia=1+0j                             #Armature current(A p.u)\n",
-      "If_rated=2.84                       #Rated field current(A)\n",
-      "p=6                                 #No. of poles\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "ns1=120*f1/p\n",
-      "ns2=120*f2/p\n",
-      "Eaf=Va-1j*Xs*Ia*exp(1j*0)          #field voltage(V)\n",
-      "Ifo=abs(Eaf)*If_rated                   #motor field current(A)\n",
-      "\n",
-      "#for part(b):\n",
-      "#Eaf= (wm/wmo)*(If/Ifo)*Eafo\n",
-      "If=Ifo\n",
-      "\n",
-      "#Results:\n",
-      "print \"part(a):\"\n",
-      "print \"(i)  The motor speed:\",ns1,\"r/min\"\n",
-      "print \"(ii) The motor field current:\",round(Ifo,2),\"A\"\n",
-      "print \"part(b):\"\n",
-      "print \"(i)  The changed speed:\",ns2,\"A\"\n",
-      "print \"(ii) The mototr field current:\",round(If,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part(a):\n",
-        "(i)  The motor speed: 1200.0 r/min\n",
-        "(ii) The motor field current: 3.7 A\n",
-        "part(b):\n",
-        "(i)  The changed speed: 1000.0 A\n",
-        "(ii) The mototr field current: 3.7 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.8, Page number: 588"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "iF=2.84                                 #rated field current(A)\n",
-      "Vbase=220                               #base voltage(V)\n",
-      "Eaf=220/sqrt(3)                         #Rms voltage line-to-neutral(V)\n",
-      "f=60                                    #Hz\n",
-      "p=6                                     #poles\n",
-      "P_rated=45*10**3                        #rated power(W)\n",
-      "Xs_pu=0.836                             #per unit synchronous reactance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "we=2*pi*f\n",
-      "Laf=sqrt(2)*Eaf/(we*iF)                 #Armature field reactance(H)\n",
-      "T_rated=P_rated/(we*2/p)\n",
-      "#setting rated values to reference values.\n",
-      "Tref=T_rated\n",
-      "iFref=iF\n",
-      "iQ=round((2/3)*(2/p)*Tref/(Laf*iFref),2)\n",
-      "iD=0\n",
-      "\n",
-      "#since theta_me=wc*t, iD=0,we have,\n",
-      "t=symbols('t')\n",
-      "wc=120*pi\n",
-      "def ia(t):\n",
-      "    return iD*cos(wc*t)-iQ*sin(wc*t)\n",
-      "def ib(t):\n",
-      "    return iD*cos(wc*t-2*pi/3)-iQ*sin(wc*t-2*pi/3)\n",
-      "def ic(t):\n",
-      "    return iD*cos(wc*t+2*pi/3)-iQ*sin(wc*t+2*pi/3)\n",
-      "Ibase=P_rated/(sqrt(3)*Eaf)\n",
-      "Imax=round(ia((pi/(2*wc))))\n",
-      "Ia=1j*abs(round(Imax/sqrt(2)))\n",
-      "Eaf=1j*we*Laf*iF/sqrt(2)\n",
-      "Zbase=Vbase**2/P_rated\n",
-      "Xs=Xs_pu*Zbase\n",
-      "Va=1j*Xs*Ia+Eaf                             #line-to-neutral voltage\n",
-      "Vt=abs(sqrt(3)*Va)/Vbase                #p.u terminal voltage(line-to-line)(V)\n",
-      "\n",
-      "#Results:\n",
-      "print \"part(a):\"\n",
-      "print \"\\tia(t)=\",ia(t),\"A\"\n",
-      "print \"\\tib(t)=\",ib(t),\"A\"\n",
-      "print \"\\tic(t)=\",ic(t),\"A\"\n",
-      "print \"part(b):\"\n",
-      "print \"\\tTerminal voltage:\",round(float(Vt),2),\"per unit\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part(a):\n",
-        "\tia(t)= -167.01*sin(120*pi*t) A\n",
-        "\tib(t)= 167.01*sin(120*pi*t + pi/3) A\n",
-        "\tic(t)= -167.01*cos(120*pi*t + pi/6) A\n",
-        "part(b):\n",
-        "\tTerminal voltage: 1.3 per unit\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.9, Page number: 591"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "iF=2.84                                 #rated field current(A)\n",
-      "Vrated=220                              #rated terminal voltage,l-l(V)\n",
-      "Ibase=118                               #base current(A)\n",
-      "Eaf=220/sqrt(3)                         #Rms voltage, line-to-neutral(V)\n",
-      "f=60                                    #Hz\n",
-      "p=6                                     #poles\n",
-      "P_rated=45*10**3                        #rated power(W)\n",
-      "Xs=0.899                                #Synchronous reactance(ohm)\n",
-      "Xs_pu=0.836                             #per unit synchronous reactance(ohm)\n",
-      "Tref=358                                #Reference torque(Nm) (from Ex11.8)\n",
-      "\n",
-      "#Calculations:\n",
-      "Va=Vrated/sqrt(3)                        #base voltage, line to neutral(V)\n",
-      "we=2*pi*f\n",
-      "wm=(2/p)*we\n",
-      "Laf=sqrt(2)*Eaf/(we*iF)                 #Armature field reactance(H)\n",
-      "Ia=Tref*wm/(3*Va)\n",
-      "Ls=Xs/we                                #Synchronous inductance(mH)\n",
-      "delta=-atan(we*Ls*Ia/Va)\n",
-      "iQ_ref=sqrt(2)*Ia*cos(delta)\n",
-      "iD_ref=sqrt(2)*Ia*sin(delta)\n",
-      "iF_ref=(2./3)*(2/p)*Tref/(Laf*iQ_ref)\n",
-      "\n",
-      "#since motor is running at rated voltage, base voltage and rated voltage \n",
-      "# are assumed to be same.\n",
-      "Va_pu=Va/Va                      \n",
-      "Ia_pu=Ia/Ibase\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The reqd motor field current:\",round(iF_ref,2),\"A\"\n",
-      "print \"Per unit voltage:\",Va_pu,\"p.u\"\n",
-      "print \"Per unit current:\",round(Ia_pu),\"p.u\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The reqd motor field current: 3.7 A\n",
-        "Per unit voltage: 1.0 p.u\n",
-        "Per unit current: 1.0 p.u\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.10, Page number: 593"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "ns=4000                             #rated speed(rpm)\n",
-      "Va=220                              #rated voltage(V)\n",
-      "Ls=1.75*10**-3                      #synchronous inductance(H)\n",
-      "Prated=25000                        #Watts\n",
-      "n=3200                              #rated OC speed(rpm)\n",
-      "p=2                                 #No. of poles\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part(a):\n",
-      "Eaf=Va/sqrt(3)\n",
-      "wm=ns*pi/30                         #rad/sec\n",
-      "Trated=Prated/wm\n",
-      "we=(p/2)*n*pi/30\n",
-      "lambdaPM=sqrt(2)*Eaf/we             #flux linked wth permanent magnet(Wb)  \n",
-      "Tref=Trated*0.65                #since motor is operated at 65% of Trated\n",
-      "iQref=(2./3)*(2/p)*(Tref/lambdaPM)\n",
-      "\n",
-      "#for part(b:)\n",
-      "lambdaD=lambdaPM                         #since iD=0\n",
-      "lambdaQ=Ls*iQref\n",
-      "lambdaa=sqrt((lambdaD**2+lambdaQ**2)/2)  #rms line-to-neutral armature flux(Wb)\n",
-      "lambdaa_base=Eaf/wm\n",
-      "lambda_pu=lambdaa/lambdaa_base\n",
-      "\n",
-      "#for part(c)\n",
-      "lambdaD=sqrt(2*(lambdaa_base)**2-lambdaQ**2)\n",
-      "iDref=(lambdaD-lambdaPM)/Ls\n",
-      "Ia=sqrt((iDref**2+iQref**2)/2)          #rms armature current(A)\n",
-      "Ibase=Prated/(sqrt(3)*Va)\n",
-      "I_pu=Ia/Ibase\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a)  Required quadrature-axis current:\",round(iQref,1),\"A\"\n",
-      "print \"(b)  Resultant armature flux linkage\",round(lambda_pu,2),\"p.u\"\n",
-      "print \"(c)  iD:\",round(iDref,1),\"A\"\n",
-      "print \"     Rms value of armature current:\",round(Ia),\"A\"\n",
-      "print \"     Per unit value of armature current:\",round(I_pu,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a)  Required quadrature-axis current: 48.2 A\n",
-        "(b)  Resultant armature flux linkage 1.27 p.u\n",
-        "(c)  iD: -66.1 A\n",
-        "     Rms value of armature current: 58.0 A\n",
-        "     Per unit value of armature current: 0.88 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.11, Page number: 600"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V10=230/sqrt(3)\n",
-      "Nph=3\n",
-      "p=4\n",
-      "fe0=60\n",
-      "R1=0.095                                #Armature resistance(ohm)\n",
-      "R2=0.2                                  #Rotor resistance(ohm)\n",
-      "X10=0.680                               #Armature leakage reactance(ohm)\n",
-      "X20=0.672                               #Rotor leakage reactance(ohm)\n",
-      "Xm0=18.7                                #Inductice reactance(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#taking two frequency values:\n",
-      "fe1=40\n",
-      "fe2=60\n",
-      "\n",
-      "for m in range(1,3,1):\n",
-      "    if m==1:\n",
-      "        fe=fe1\n",
-      "    else:\n",
-      "        fe=fe2\n",
-      "    X1=X10*(fe/fe0)\n",
-      "    X2=X20*(fe/fe0)\n",
-      "    Xm=Xm0*(fe/fe0)\n",
-      "    V1=V10*(fe/fe0)\n",
-      "    \n",
-      "    ws=4*pi*fe/p\n",
-      "    ns=120*fe/p\n",
-      "    V1eq=abs(V1*1j*Xm/(R1+1j*(X1+Xm)))\n",
-      "    Z1eq=1j*Xm*(R1+1j*X1)/(R1+1j*(X1+Xm))\n",
-      "    R1eq=Z1eq.real\n",
-      "    X1eq=Z1eq.imag\n",
-      "    \n",
-      "#Search over the slip until the Pload = Pmech   \n",
-      "    s=0                             #slip initialised to 0\n",
-      "    error=1\n",
-      "    \n",
-      "    while error >=0:\n",
-      "        s=s+0.00001\n",
-      "        rpm=ns*(1-s)\n",
-      "        wm=ws*(1-s)\n",
-      "        Tmech=(1/ws)*Nph*V1eq**2*(R2/s)\n",
-      "        Tmech = Tmech/((R1+R2/s)**2 + (X1+X2)**2)\n",
-      "        Pmech=Tmech*wm\n",
-      "        Pload=10.5*10**3*(rpm/1800)**3\n",
-      "        error=Pload-Pmech\n",
-      "    \n",
-      "    print \"\\nFor fe =\",fe,\"Hz :\"\n",
-      "    print \"\\tTerminal voltage=\",round(V1*sqrt(3)),\"V l-l\"\n",
-      "    print \"\\trpm =\",round(rpm)\n",
-      "    print \"\\tslip =\",round(100*s,1),\"%\"\n",
-      "    print \"\\tPload =\",round(Pload/1000,2),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "For fe = 40 Hz :\n",
-        "\tTerminal voltage= 153.0 V l-l\n",
-        "\trpm = 1166.0\n",
-        "\tslip = 2.8 %\n",
-        "\tPload = 2.86 kW\n",
-        "\n",
-        "For fe = 60 Hz :\n",
-        "\tTerminal voltage= 230.0 V l-l\n",
-        "\trpm = 1721.0\n",
-        "\tslip = 4.4 %\n",
-        "\tPload = 9.17 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.12, Page number: 608"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V10=230/sqrt(3)\n",
-      "Nph=3\n",
-      "p=4\n",
-      "fe0=60\n",
-      "R1=0.095                                #Armature resistance(ohm)\n",
-      "R2=0.2                                  #Rotor resistance(ohm)\n",
-      "X10=0.680                               #Armature leakage reactance(ohm)\n",
-      "X20=0.672                               #Rotor leakage reactance(ohm)\n",
-      "Xm0=18.7                                #Inductice reactance(ohm)\n",
-      "n=1680                                  #rpm\n",
-      "Pmech=9.7*10**3                         #Electromagnetic power(W)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "we0=2*pi*fe0\n",
-      "Lm=Xm0/we0\n",
-      "LS=Lm+X10/we0\n",
-      "LR=Lm+X20/we0\n",
-      "Ra=R1\n",
-      "RaR=R2\n",
-      "lambda_rated=sqrt(2)*V10/we0\n",
-      "lambdaDR=lambda_rated\n",
-      "#for specified operating condition\n",
-      "wm=n*(pi/30)\n",
-      "Tmech=Pmech/wm\n",
-      "iQ=(2/3)*(2/p)*(LR/Lm)*(Tmech/lambdaDR)\n",
-      "iD=lambdaDR/Lm\n",
-      "Ia=sqrt((iD**2+iQ**2)/2)\n",
-      "wme=(p/2)*wm\n",
-      "we=wme+(RaR/LR)*(iQ/iD)\n",
-      "fe=we/(2*pi)\n",
-      "Va=sqrt(((Ra*iD-we*(LS-Lm**2/LR)*iQ)**2 + (Ra*iQ+we*LS*iD)**2)/2)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Rms amplitude of the armature current:\",round(Ia,1),\"A\"\n",
-      "print \"The electrical frequency:\",round(fe,1),\"Hz\"\n",
-      "print \"Rms terminal voltage:\",round(sqrt(3)*Va,1),\"V  line-line\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rms amplitude of the armature current: 27.9 A\n",
-        "The electrical frequency: 58.4 Hz\n",
-        "Rms terminal voltage: 243.6 V  line-line\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.13, Page number: 610"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from math import *\n",
-      "from pylab import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P_rated = 12*10**3                       #Watts\n",
-      "V_rated = 230                            #Rated line-line voltage(v)\n",
-      "Va_rated = 230/sqrt(3)                   #Rated line to neutral(V)\n",
-      "fe_rated = 60                            #Hz\n",
-      "we_rated = 2*pi*fe_rated                 #rad/sec\n",
-      "lambda_rated = sqrt(2)*Va_rated/we_rated #Wb\n",
-      "I_rated = P_rated/(sqrt(3)*V_rated)      #A\n",
-      "Ipeak_base = sqrt(2)*I_rated            #A\n",
-      "p = 4                                   #poles\n",
-      "\n",
-      "V10=V_rated/sqrt(3)\n",
-      "R1=0.095                                #Armature resistance(ohm)\n",
-      "R2=0.2                                  #Rotor resistance(ohm)\n",
-      "X10=0.680                               #Armature leakage reactance(ohm)\n",
-      "X20=0.672                               #Rotor leakage reactance(ohm)\n",
-      "Xm0=18.7                                #Inductice reactance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "Lm = Xm0/we_rated;\n",
-      "LS = Lm + X10/we_rated;\n",
-      "LR = Lm + X20/we_rated;\n",
-      "Ra = R1\n",
-      "RaR = R2\n",
-      "#operating point:\n",
-      "n = 1680                                #rpm\n",
-      "lambdaDR=lambda_rated\n",
-      "wm = n*pi/30\n",
-      "wme = (p/2)*wm\n",
-      "Pmech = 9.7*10**3\n",
-      "Tmech = Pmech/wm\n",
-      "lambda_DRpu=[0]*42\n",
-      "iDpu=[0]*42\n",
-      "Iapu=[0]*42\n",
-      "fe=[0]*42\n",
-      "Vapu=[0]*42\n",
-      "\n",
-      "for n in range(1,43,1):\n",
-      "    lambdaDR = (0.8+(n-1)*0.4/40)*lambda_rated\n",
-      "    lambda_DRpu[n-1]=lambdaDR/lambda_rated\n",
-      "    iQ=(2/3)*(2/p)*(LR/Lm)*(Tmech/lambdaDR)\n",
-      "    iD=(lambdaDR/Lm)\n",
-      "    iDpu[n-1]=iD/Ipeak_base\n",
-      "    iQR=-(Lm/LR)**iQ\n",
-      "    Ia=sqrt((iD**2+iQ**2)/2)\n",
-      "    Iapu[n-1]=Ia/I_rated\n",
-      "    we=wme-(RaR/LR)*(iQ/iD)\n",
-      "    fe[n-1]=we/(2*pi)\n",
-      "    Va_rms=sqrt(((Ra*iD-we*(LS-Lm**2/LR)*iQ)**2 +(Ra*iQ+ we*LS*iD)**2)/2)\n",
-      "    Vapu[n-1]=Va_rms/Va_rated\n",
-      "\n",
-      "#Results:\n",
-      "print \"The required plot is as shown:\"\n",
-      "plot(iDpu,Iapu)\n",
-      "plot(iDpu,Vapu,':')\n",
-      "xlabel('i_D [per unit] ')\n",
-      "ylabel('per unit')\n",
-      "annotate('Ia',xy=(0.21,1.05))\n",
-      "annotate('Va',xy=(0.21,0.85))\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required plot is as shown:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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MmYPJkyfDx8cHAQEB8PT01FPExFDUymtxt+Yu12EQPZFIJJBIJDp/Lqd10+ZqF2fPnh3R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\nEOPH4wETJgAnTrCd6zk5gKs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U+FGjRp0aMGDAn6tXr/5Qc27Pnj1zg4KCLvn5+aW/+uqr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\nD0IIIVr7PxA8wzUjE2EJAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x27d0450>"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter2.ipynb b/ELECTRIC_MACHINERY/chapter2.ipynb
deleted file mode 100755
index 855c3bd6..00000000
--- a/ELECTRIC_MACHINERY/chapter2.ipynb
+++ /dev/null
@@ -1,761 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2: Transformers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.1, Page number: 63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Pc=16                               #Core loss at Bmax=1.5 T\n",
-      "VIrms=20                            #Voltamperess for the core\n",
-      "Vrms=194                            #Rms induced voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "pf=Pc/VIrms\n",
-      "a=math.acos(pf)\n",
-      "I=VIrms/Vrms\n",
-      "Ic=I*pf\n",
-      "Im=I*math.fabs(math.sin(a))\n",
-      "\n",
-      "#Results:\n",
-      "print \"Power factor = \", round(pf,1),\"lagging\"\n",
-      "print \"The core-loss current,Ic =\", round(Ic,3), \"A rms\"\n",
-      "print \"The magnetising current,Im =\", round(Im,2),\"A rms\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power factor =  0.8 lagging\n",
-        "The core-loss current,Ic = 0.082 A rms\n",
-        "The magnetising current,Im = 0.06 A rms\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page number: 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declarations:\n",
-      "k=5                                 #turns ratio,N1/N2\n",
-      "Z2=1+4j                             #Impedance of secondary side(ohm)\n",
-      "Vp=120                              #primary voltage(V)\n",
-      "\n",
-      "#Calculations:\n",
-      "Z2p=k**2*(Z2)\n",
-      "I=Vp/Z2p\n",
-      "Is=k*I\n",
-      "\n",
-      "#Results:\n",
-      "print \"Primary current:\",complex(round(I.real,2),round(I.imag,2)), \"A rms\"\n",
-      "print \"Current in the short:\",round(Is.real,2)+1j*round(Is.imag,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Primary current: (0.28-1.13j) A rms\n",
-        "Current in the short: (1.41-5.65j) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.4, Page number: 74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R1=0.72                                 #Resistance at high voltage side(ohm)\n",
-      "R2=0.70                                 #Resistance at low voltage side(ohm)\n",
-      "X1=0.92                                 #Reactance at high voltage side(ohm)\n",
-      "X2=0.90                                 #Reactance at low voltage side(ohm)\n",
-      "Zq=632+4370j                            #Impedance of exciting circuit(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "Req=R1+R2\n",
-      "Xeq=X1+X2       \n",
-      "Vcd=2400*Zq/(Zq+complex(R1,X1))\n",
-      "V=complex(round(Vcd.real,2),round(Vcd.imag,3))\n",
-      "\n",
-      "#Results:\n",
-      "print \"Req:\",Req,\"ohm\",\" and  Xeq:\",Xeq,\"ohm\"\n",
-      "print \"Voltage at low voltage terminal:\",V,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Req: 1.42 ohm  and  Xeq: 1.82 ohm\n",
-        "Voltage at low voltage terminal: (2399.45+0.316j) V\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.5, Page number: 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Zf=0.30+.160j                           #Impedance of feeder(ohm)\n",
-      "Zeq=1.42+3.42j             #Equiv.impedance of transformer refd. to primary(ohm)\n",
-      "k=2400/240                              #turns ratio\n",
-      "P=50000                                 #power rating of the transformer(VA)\n",
-      "Vs=2400                                 #sending end vltage of feeder(V)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "I=P/2400                                #Rated current(A)\n",
-      "theta=math.acos(0.80)\n",
-      "Zt=Zf+Zeq                      #combned impedance of feeder & transformer(ohm)\n",
-      "R=Zt.real\n",
-      "X=Zt.imag\n",
-      "bc=I*X*math.cos(theta)-I*R*math.sin(theta)\n",
-      "ab=I*R*math.cos(theta)+I*X*math.sin(theta)\n",
-      "Ob=(Vs**2-bc**2)**0.5\n",
-      "V2=Ob-ab\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The voltage at the secondary terminals:\",round(V2/10,0),\"V\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage at the secondary terminals: 233.0 V\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.6, Page number: 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#short ckt test readings:\n",
-      "Vsc=48                          #voltage(V)\n",
-      "Isc=20.8                        #current(A)\n",
-      "Psc=617                         #power(W)\n",
-      "    \n",
-      "#Open ckt test readings:\n",
-      "Vs=240                         #Voltage(V)\n",
-      "I=5.41                          #current(A)\n",
-      "P=186                           #power(W)\n",
-      "V2ph=2400                        #voltage at full load at high voltage side(V)\n",
-      "pf=0.8                          #lagging power factor at full load\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "theta=math.acos(pf)\n",
-      "Zeqh=Vsc/Isc                    #subscript h refers to high voltage side\n",
-      "Reqh=Psc/Isc**2\n",
-      "Xeqh=math.sqrt(Zeqh**2-Reqh**2)\n",
-      "Ih=50000/V2ph\n",
-      "Pout=50000*pf\n",
-      "Pwind=Ih**2*Reqh\n",
-      "Ptloss=P+Pwind\n",
-      "e=(1-Ptloss/(Ptloss+Pout))*100\n",
-      "Iph=(50000/2400)*complex(math.cos(theta),math.sin(-theta))\n",
-      "V1ph=V2ph+Iph*complex(Reqh,Xeqh)\n",
-      "r=(round(abs(V1ph),2)-2400)*100/V2ph\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The efficiency of the transformer:\",round(e,0),\"%\"\n",
-      "print \"Volatge Regulation:\",round(r,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The efficiency of the transformer: 98.0 %\n",
-        "Volatge Regulation: 1.94 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.7, Page number: 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Varaible declaration:\n",
-      "Vx=2400                                 #Voltage at low voltage side(V)\n",
-      "Vbc=2400                                 #Voltage across branch bc(V)\n",
-      "Vab=240                                #Voltage induced in winding ab(V)\n",
-      "Pl=803                                  #transformer losses(W)\n",
-      "pf=0.8                                  #Power factor of the transformer\n",
-      "\n",
-      "#Calculations:\n",
-      "Vh=Vab+Vbc\n",
-      "Ih=50000/Vab\n",
-      "KVA=Vh*Ih/1000                          #Kva rating\n",
-      "P=pf*550000\n",
-      "e=(1-Pl/(P+Pl))*100\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Voltage ratings, Vh:\",Vh,\"V  \", \"&  Vx:\",Vx,\"V\"\n",
-      "print \"KVA rating as an autotransformer:\",KVA,\"KVA\"\n",
-      "print \"full-load efficiency:\", round(e,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage ratings, Vh: 2640 V   &  Vx: 2400 V\n",
-        "KVA rating as an autotransformer: 550.0 KVA\n",
-        "full-load efficiency: 99.82 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.8, Page number: 87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vl1=4160                      #line-to-line voltage at feeder's sending end(V)\n",
-      "Zf=0.30+.160j                           #Impedance of feeder(ohm)\n",
-      "Zeq=1.42+3.42j             #Equiv.impedance of transformer refd. to primary(ohm)\n",
-      "k=2400/240                              #turns ratio\n",
-      "P=50000                                 #power rating of the transformer(VA)\n",
-      "Vs=2400                                 #sending end vltage of feeder(V)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#this problem can be treated on a single phase basis,\n",
-      "#and whole problem is similar to Ex 2.5.\n",
-      "\n",
-      "I=P/2400                                #Rated current(A)\n",
-      "theta=math.acos(0.80)\n",
-      "Zt=Zf+Zeq                      #combned impedance of feeder & transformer(ohm)\n",
-      "R=Zt.real\n",
-      "X=Zt.imag\n",
-      "bc=I*X*math.cos(theta)-I*R*math.sin(theta)\n",
-      "ab=I*R*math.cos(theta)+I*X*math.sin(theta)\n",
-      "Ob=(Vs**2-bc**2)**0.5\n",
-      "V2=Ob-ab\n",
-      "Vload=V2/k\n",
-      "\n",
-      "#Results:\n",
-      "print \"The line to line voltage:\",round(Vload,0),\"V line-to-line\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line to line voltage: 233.0 V line-to-line\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.9, Page number: 89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import cmath\n",
-      "\n",
-      "#Variable decclaration:\n",
-      "#All resistances, reactances, & impedances are on per phase basis\n",
-      "Req=1.42       #Series resist. of del-del transformer referred to 2400v side(ohm)\n",
-      "Xeq=1.82       #Series react. of del-del transformer referred to 2400v side(ohm)\n",
-      "Zs=0.17+0.92j                   #Equiv impedance of sending end transformer(ohm)\n",
-      "Xf=0.8j                          #Reactance of the feeder(ohm)\n",
-      "Vf=2400                         #Voltage of the feeder(V)\n",
-      "k=10                            #turns ratio(Vp/Vs)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Zt=(complex(Req,Xeq)/3)+Zs+Xf\n",
-      "Ztot=complex(round(Zt.real,2),round(Zt.imag,2))\n",
-      "If=math.floor(Vf/(math.sqrt(3))/round(abs(Ztot),2))\n",
-      "I1=If/math.sqrt(3)\n",
-      "I2=I1*k\n",
-      "Ic=I2*math.sqrt(3)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Short circuit current in the 2400 feeder, per phase wires:\",round(Ic,1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Short circuit current in the 2400 feeder, per phase wires: 5720.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.10, Page number: 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "X1=143                             #Reactance of primary(ohm)\n",
-      "X21=164                           #Reactance of secondary ref. to primary(ohm)\n",
-      "Xm=163*10**3                           #Reactance of magnetising ckt(ohm)\n",
-      "R1=128                             #Resistance of primary(ohm)\n",
-      "R21=141                           #Resistane of secondary ref. to primary(ohm)\n",
-      "k=20                              #turns ratio(2400/120)\n",
-      "V1=2400                             #primary voltage(V)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "V2=(V1/k)*complex(0,Xm)/complex(R1,X1+Xm)\n",
-      "mag=abs(V2)\n",
-      "ph=degrees(cmath.phase(V2))\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Magnitude of V2:\",round(mag,2),\"V\"\n",
-      "print \"Phase of V2:\",round(ph,3),\"degrees\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnitude of V2: 119.89 V\n",
-        "Phase of V2: 0.045 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.11, Page number: 94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "\n",
-      "\n",
-      "#variable declaration:\n",
-      "X1=44.8*10**-6                  #Reactance of the primary(ohm)\n",
-      "R1=10.3*10**-6                  #Resistance of the primary(ohm)\n",
-      "X21=54.3*10**-6          #Reactance of the secondary refd. to primary(ohm)\n",
-      "R21=9.6*10**-6                  #Resistance of secondary ref. to primary(ohm)\n",
-      "Xm=17.7*10**-3                  #Reactance of the magnetising ckt(ohm)\n",
-      "k=5/800                         #turms ratio(I2/I1)\n",
-      "Zl=2.5+0j                       #Impedance ofthe load(ohm)\n",
-      "I1=800                          #primary current(A)\n",
-      "\n",
-      "#Calculations:\n",
-      "Zp=k**2*Zl\n",
-      "I2=I1*k*Xm*1j/(Zp+R21+(X21+Xm)*1j)\n",
-      "phase=cmath.phase(I2)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Magnitude of current:\",round(abs(I2),2),\"A\"\n",
-      "print \"Phase of the current:\",round(math.degrees(phase),3),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnitude of current: 4.98 A\n",
-        "Phase of the current: 0.346 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.12, Page number: 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "XL=0.040                                #Reactance at l.v side(ohm)\n",
-      "XH=3.75                                 #Reactance at h.v side(ohm)\n",
-      "Xm=114                                  #Magnetising reactance(ohm)\n",
-      "RL=0.76*10**-3                          #Resistance at l.v.side(ohm)\n",
-      "RH=0.085                                #Resistance at l.v.side(ohm)\n",
-      "VA_base=100*10**6                       #base VA\n",
-      "V_base=7.97*10**3                       #base voltage(V)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for l.v side\n",
-      "VA_base=100*10**6                       #base VA\n",
-      "V_base=7.97*10**3                       #base voltage(V)\n",
-      "Rbase1=Xbase1=V_base**2/VA_base\n",
-      "\n",
-      "#for h.v side:\n",
-      "VA_base=100*10**6                       #base VA\n",
-      "V_base=79.7*10**3                       #base voltage(V)\n",
-      "Rbase2=Xbase2=V_base**2/VA_base\n",
-      "\n",
-      "XL_pu=XL/Xbase1\n",
-      "XH_pu=XH/Xbase2\n",
-      "Xm_pu=Xm/Xbase1\n",
-      "RL_pu=RL/Rbase1\n",
-      "RH_pu=RH/Rbase2\n",
-      "K_pu=1                                  #per unit utrns ratio\n",
-      "\n",
-      "#Results:\n",
-      "print \"The per unit parameters are:\"\n",
-      "print \"XL_pu =\",round(XL_pu,3),\"p.u\"\n",
-      "print \"XH_pu =\",round(XH_pu,4),\"p.u\"\n",
-      "print \"Xm_pu =\",math.ceil(Xm_pu),\"p.u\"\n",
-      "print \"RL_pu =\",round(RL_pu,4),\"p.u\"\n",
-      "print \"XL_pu =\",round(RH_pu,4),\"p.u\"\n",
-      "print \"Turns ratio =\",K_pu,\"p.u\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The per unit parameters are:\n",
-        "XL_pu = 0.063 p.u\n",
-        "XH_pu = 0.059 p.u\n",
-        "Xm_pu = 180.0 p.u\n",
-        "RL_pu = 0.0012 p.u\n",
-        "XL_pu = 0.0013 p.u\n",
-        "Turns ratio = 1 p.u\n"
-       ]
-      }
-     ],
-     "prompt_number": 59
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.13, Page number: 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ic=5.41                             #Exciting current ref. to low volt. side(A)\n",
-      "k=10                                #turns ratio(N1/N2=2400/240)\n",
-      "Vbh=2400                            #base voltage at primary side(V)\n",
-      "Vbl=240                             #base voltage at secondary side(V)\n",
-      "Ibh=20.8                            #base current at primary side(A)\n",
-      "Ibl=208                             #base current at secondary side(A)\n",
-      "Z=1.42+1.82j                    #Equiv.impedance ref.to high voltage side(ohm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Zbh=Vbh/Ibh\n",
-      "Zbl=Vbl/Ibl\n",
-      "Icl=Ic/Ibl\n",
-      "Ich=Ic/(Ibh*k)\n",
-      "Zl=Z/(k**2*Zbl)\n",
-      "Zh=Z/Zbh\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Per unit exciting current on low volt. sides:\",round(Icl,3,),\"A\"  \n",
-      "print \"Per unit exciting current on high volt. sides:\",round(Ich,3),\"A\"\n",
-      "print \"per unit equiv.impedance at low volt. sides:\",round(Zl.real,4)+round(Zl.imag,4)*1j,\"ohm\"\n",
-      "print \"per unit equiv.impedance at high voltage sides:\",round(Zh.real,4)+round(Zh.imag,4)*1j,\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Per unit exciting current on low volt. sides: 0.026 A\n",
-        "Per unit exciting current on high volt. sides: 0.026 A\n",
-        "per unit equiv.impedance at low volt. sides: (0.0123+0.0158j) ohm\n",
-        "per unit equiv.impedance at high voltage sides: (0.0123+0.0158j) ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.14, Page number: 100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vb=24000               #Base voltage of secondary of sending end transformer(V)    \n",
-      "Z=0.17+0.92j      #Impedance of sending end transformer ref. to 2400V side(ohm)\n",
-      "P=150                       #Power rating of the transformer(KVA)\n",
-      "V=2400                      #Primary voltage of sending end transformer(v)\n",
-      "Ztot=0.64+2.33j                 #Total series impedance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "Zb=V**2/(P*10**3)\n",
-      "Ztotb=Ztot/Zb\n",
-      "Vsb=1                           #Vs in terms of per unit values\n",
-      "Isc=Vsb/abs(Ztotb)                   #Short current in per unit values(A)\n",
-      "Ib1=P*10**3/(sqrt(3)*2400)       #base current of the feeder at 2400V side(A)\n",
-      "If=Ib1*Isc\n",
-      "Ib2=P*10**3/(sqrt(3)*240)\n",
-      "Iscs=Isc*Ib2                     #short ckt current at 2400V afeeder side (A)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Short circuit current in 2400 feeder:\",round(Iscs/10**3,2),\"KA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Short circuit current in 2400 feeder: 5.73 KA\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.15, Page number: 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P=250*10**3                            #power rating of transformer(KVA)\n",
-      "Vp=2400                             #primary volatge(V)\n",
-      "Vs=460                              #secondary voltage(V)\n",
-      "Pb=100*10**3                           #new base power of transformer(KVA)\n",
-      "Vb=460                              #new base voltage(V)\n",
-      "Z=0.026+0.12j                       #series impedance on its own base(ohm)\n",
-      "Vl=438                              #load voltage(V)\n",
-      "Pl=95*10**3                         #power drawn by the load(kW)\n",
-      "\n",
-      "#Calculations:\n",
-      "Zbo=Vs**2/P                        #base impedance for the transformer(ohm)\n",
-      "Zbn=Vb**2/Pb            #base impedance for the transformer at 100KVA base(ohm)\n",
-      "Zpn=Z*Zbo/Zbn                      #base impedance at 100KVA base(ohm)\n",
-      "Vpl=Vl/Vb                           #per unit load voltage(V)\n",
-      "Ppl=Pl/Pb                          #per unit load power\n",
-      "Ipl=Ppl/Vpl                             #per unit load current(A)\n",
-      "Vpp=Vpl+Ipl*Zpn                     #high side voltage of the transformer(V) \n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The high side voltage:\",round(abs(Vpp*Vp),0),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The high side voltage: 2313.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter3.ipynb b/ELECTRIC_MACHINERY/chapter3.ipynb
deleted file mode 100755
index 411be995..00000000
--- a/ELECTRIC_MACHINERY/chapter3.ipynb
+++ /dev/null
@@ -1,430 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3da4e1bbe19be5f3a05399c95840acc2f1758f76d57c3a0e757cbfa9a562b633"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Electromechanical-Energy-Conversion-Principles "
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.1, Page number: 114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I=10                            #current in the coil(A)\n",
-      "Bo=0.02                         #magnetic field (T)\n",
-      "R=0.05                          #radius of the rotor(m)\n",
-      "l=0.3                           #rotor length(m)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "q=symbols('q')                  #Direction of torque\n",
-      "F1=-2*I*l*Bo*sin(q)             #Force on the coil(N)\n",
-      "T=F1*R                          #Torque scting in theta direction(Nm)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Force per unit length:\",T,\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Force per unit length: -0.006*sin(q) Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.2, Page number: 121"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "N=1000                   #No of winding turns\n",
-      "g=2                      #Air gap width(mm)\n",
-      "d=0.15                  #Magnetic core width,d (m)\n",
-      "l=0.1                    #thickness of core(0.1)\n",
-      "x,d=symbols('x d')       #where h is height of plunger(m) \n",
-      "                         #Lx is inductance as a function of x(H)\n",
-      "i=10                     #Current in the winding(A)\n",
-      "uo=4*3.14*10**-7           #permeability of free space(H/m)\n",
-      "\n",
-      "#Calculations:\n",
-      "Lx=(uo*N**2*l*d)/(2*g*10**-3)*(1-x/d)\n",
-      "Wfld=(1./2)*Lx*i**2\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The magnetic energy stored, Wfld:\",\"236*(1-x/d) J\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnetic energy stored, Wfld: 236*(1-x/d) J\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.3, Page number: 124"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "import numpy as np\n",
-      "from pylab import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "xdata=[0.0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0]     #(cm)\n",
-      "Ldata=[2.8, 2.26, 1.78, 1.52, 1.34, 1.26, 1.20, 1.16, 1.13, 1.11, 1.10]   #(mH)\n",
-      "I = 0.75                                                #(A)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "x=0.01*np.array(xdata)\n",
-      "L=0.001*np.array(Ldata)\n",
-      "length=len(x)\n",
-      "xmax=x[length-1]\n",
-      "a=polyfit(x,L,4)\n",
-      "xfit=[0]*102\n",
-      "Lfit=[0]*102\n",
-      "for n in range(1,102,1):\n",
-      "    xfit[n-1]=xmax*(n-1)/100\n",
-      "    Lfit[n-1]=a[0]*xfit[n-1]**4+a[1]*xfit[n-1]**3+a[2]*xfit[n-1]**2+a[3]*xfit[n-1]+a[4]\n",
-      "\n",
-      "#Plot the data and then the fit to compare (convert xfit to cm and Lfit to mH)\n",
-      "plot(xdata,Ldata,'o')\n",
-      "plot(100*np.array(xfit),1000*np.array(Lfit),'g.')\n",
-      "xlabel('x [cm] ')\n",
-      "ylabel('L [mH] ')\n",
-      "title('Inductance,L vs length,l')\n",
-      "grid()\n",
-      "print \"The required plots are shown below:\"\n",
-      "show()\n",
-      "\n",
-      "#set current to 0.75 A\n",
-      "I=0.75\n",
-      "F=[0]*102\n",
-      "for n in range(1,102,1):\n",
-      "    xfit[n-1]=0.002+0.016*(n-1)/100\n",
-      "    F[n-1]=4*a[0]*xfit[n-1]**3+3*a[1]*xfit[n-1]**2+2*a[2]*xfit[n-1]**1+a[3]\n",
-      "    F[n-1]=(I**2/2)*F[n-1]\n",
-      "plot(100*np.array(xfit),F,'b.')\n",
-      "xlabel('x  [cm]')\n",
-      "ylabel('Force [N]')\n",
-      "title('Force, F vs length,l')\n",
-      "grid()\n",
-      "\n",
-      "#Results:\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required plots are shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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XEjUIiFIPEjUISHRhOE1wDHqozKS4mqmwolB5yatfop9BXe4KoC8oH5kQQ19b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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2b71910>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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1Bwf8hjO6AMCURVNeKpUqbciQIdfHjh17YePGjYsSExPzVq5cuczagwP+aGue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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3d76250>"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page number: 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Lo=10.6*10**-3                          #Initial inductance(H)\n",
-      "L2=2.7*10**-3                           #H\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "theta,i=symbols('theta i')\n",
-      "L=Lo+L2*cos(2*theta)\n",
-      "i=2                                     #Coil current,A\n",
-      "def T(theta):\n",
-      "    return i**2*diff(L,theta)/2\n",
-      "                                    \n",
-      "\n",
-      "#Results:\n",
-      "print \"Torque,Tfld =\",T(theta),\" N.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Torque,Tfld = -0.0108*sin(2*theta)  N.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page number: 134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1=2.5*10**-2                            #radius of rotor(m)\n",
-      "h=1.8*10**-2                             #Axial length(m)\n",
-      "g=3*10**-3                               #Air gap length(m)\n",
-      "Bag=1.65                                #Magnetic field(T)\n",
-      "uo=4*pi*10**-7                          #permeability of free space(H/m)\n",
-      "\n",
-      "#Calculations:\n",
-      "H=Bag/uo\n",
-      "Ni=2*g*H\n",
-      "T=uo*(Ni)**2*h*(r1+0.5*g)/(4*g)\n",
-      "\n",
-      "#Results:\n",
-      "print \"The maximum torque:\", round(T,2),\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum torque: 3.1 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page number: 140"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from matplotlib import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "i1=0.8\n",
-      "i2=0.01\n",
-      "\n",
-      "\n",
-      "#Calculations & Results:\n",
-      "def df(f,x,h=0.1e-10):\n",
-      "    return ( f(x+h/2) - f(x-h/2) )/h\n",
-      "\n",
-      "\n",
-      "\n",
-      "def l11(x):\n",
-      "    return (3+cos(2*x))/1000.0\n",
-      "\n",
-      "def l12(x):\n",
-      "    return (0.3*cos(x))\n",
-      "\n",
-      "def l22(x):\n",
-      "    return (30+10*cos(2*x))\n",
-      "\n",
-      "def g(x):\n",
-      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x) + (i1*i2)*df(l12,x)\n",
-      "\n",
-      "def r(x):\n",
-      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x)\n",
-      "def s(x):\n",
-      "    return (i1*i2)*df(l12,x)\n",
-      "\n",
-      "x=linspace(-pi,pi,100000)\n",
-      "\n",
-      "\n",
-      "plot(x,r(x))\n",
-      "plot(x,s(x))\n",
-      "plot(x,g(x))\n",
-      "grid()\n",
-      "annotate(\"Total torque\",xy=(-0.5,0.003))\n",
-      "annotate(\"Reluctance torque\",xy=(-2,-0.0015))\n",
-      "annotate(\"Mutual Interaction torque\",xy=(1.6,-0.0026))\n",
-      "xlabel(\"Theta [radians]\")\n",
-      "ylabel(\"Torque [N.m]\")\n",
-      "xlim(-pi,pi)\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print \"Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)\"\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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wdAQOHGDxg3fvciNDRUVIhaYiEtMT0W1vN8ztMhfTOk5T6dxFMmECe7RRYwN4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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3b09350>"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.9, Page number: 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "W=4.0*10**-2                            #width of plunger lower arm(m)\n",
-      "W1=4.5*10**-2                           #width of plunger upper arm(m)\n",
-      "D=3.5*10**-2                            #depth of plunger (m)\n",
-      "d=8*10**-3                              #length of magnet(m)\n",
-      "go=1*10**-3                             #air gap length(m)\n",
-      "uo=4*pi*10**-7                          #Permeability of free space(A.turns/m)\n",
-      "ur=1.06*uo                              #Relativity permeability\n",
-      "Hc1=-940                                #Magnetising force(kA/m)\n",
-      "Bt=1.25                                 #Magnetic field induction(T)\n",
-      "N=1500                                  #No of turns\n",
-      "x=3*10**-3                              #Position of plunger(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "Ni=-Hc1*d*10**3\n",
-      "Rx=x/(uo*W1*D)\n",
-      "Ro=go/(uo*W*D)\n",
-      "Rm=d/(ur*W*D)\n",
-      "f=-((Ni)**2)/(uo*W1*D*(Rx+Ro+Rm)**2)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The x-directed force:\",round(f,1),\"N\"\n",
-      "print \"Current in the excitation winding:\",round(Ni/N,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The x-directed force: -703.3 N\n",
-        "Current in the excitation winding: 5.01 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter4.ipynb b/ELECTRIC_MACHINERY/chapter4.ipynb
deleted file mode 100755
index aef0df05..00000000
--- a/ELECTRIC_MACHINERY/chapter4.ipynb
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4: Introduction to Rotating Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.2, Page number: 199"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "uo=4*pi*10**-7                          #Permeabolity of free space(H/m)\n",
-      "g=0.7*10**-3                            #Length of air gap(m)\n",
-      "p=4                                     #no. of poles\n",
-      "Ba=1.6                                  #Magnetic flux density(T)\n",
-      "Kr=0.935                                #Winding constant\n",
-      "N=263                                   #No. of turns\n",
-      "\n",
-      "#Calculations:\n",
-      "Ir=(pi*g*p/(4*uo*Kr*N))*1.6\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Rotor winding current:\",round(Ir,1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rotor winding current: 11.4 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.3, Page number: 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "fc=60                                   #frequency of the current(Hz)\n",
-      "p=[2, 4, 6]                             #matrix of no. of poles\n",
-      "\n",
-      "#Calculations:\n",
-      "ns=[0]*3\n",
-      "ws=[0]*3\n",
-      "wc=2*pi*fc\n",
-      "for n in range(0,3,1):\n",
-      "    ws[n]=round((2/p[n])*wc,0)\n",
-      "    \n",
-      "for i in range(0,3,1):\n",
-      "    ns[i]=round(120*fc/p[i],0)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The synchronous angular velocities:\",ws, \"rad/sec\"\n",
-      "print \"The speed of the rotor:\",ns,\"r/min\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The synchronous angular velocities: [377.0, 188.0, 126.0] rad/sec\n",
-        "The speed of the rotor: [3600.0, 1800.0, 1200.0] r/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.5, Page number: 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Nf=68                                   #Field winding\n",
-      "Na=18                                   #Armature winding\n",
-      "r=0.53                                  #mean air gap radius(m)\n",
-      "l=3.8                                   #Armature winding length(m)\n",
-      "Kf=0.945                                #Winding factor of field winding\n",
-      "Ka=0.933                                #Winding factor of armature winding\n",
-      "g=4.5*10**-2                            #Air gap length(m)\n",
-      "p=2                                     #No. of poles\n",
-      "If=720                                  #field current(A)\n",
-      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
-      "f=60                                    #Frequency curent(Hz)\n",
-      "\n",
-      "#Calculations:\n",
-      "Fag1_peak=4*Kf*Nf*If/(pi*p)\n",
-      "Bag1_peak=uo*Fag1_peak/g\n",
-      "Qp=2*Bag1_peak*l*r\n",
-      "Erms=sqrt(3)*sqrt(2)*pi*f*Ka*Na*Qp\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The peak fundamental mmf,Fag1_peak: \",round(Fag1_peak/10000,2),\"* 10^4  A.turns/pole\"\n",
-      "print \"\\nThe flux density in the air gap,Bag1_peak: \",round(Bag1_peak,2),\"T\"\n",
-      "print \"\\nThe fundamental flux per pole, Qp:\" ,round(Qp,2),\"Wb\"\n",
-      "print \"\\nThe rms value of open circuit voltage,Erms: \",round(Erms/1000,1),\"KV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The peak fundamental mmf,Fag1_peak:  2.95 * 10^4  A.turns/pole\n",
-        "\n",
-        "The flux density in the air gap,Bag1_peak:  0.82 T\n",
-        "\n",
-        "The fundamental flux per pole, Qp: 3.31 Wb\n",
-        "\n",
-        "The rms value of open circuit voltage,Erms:  25.7 KV\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.8, Page number: 225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "ns=1800                            #Speed of rotor(rpm)\n",
-      "f=60                                #Frequency(Hz)\n",
-      "g=1.2*10**-3                        #Air gap length(m)\n",
-      "D=0.27                                  #Avg diameter of the gap(m)\n",
-      "Kr=0.976                             #Winding factor\n",
-      "l=0.32                              #Axial length(m)\n",
-      "I=18                                #Rotor current(A)\n",
-      "p=4                                 #No of poles\n",
-      "Nr=786                              #Rotor windings\n",
-      "B_max=1.5                           #Max. flux densiity(T)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Fr_max=4*Kr*Nr*I/(pi*p)\n",
-      "T_max=p*pi*D*l*B_max*Fr_max/4\n",
-      "wm=ns*pi/30\n",
-      "P=wm*T_max\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Maximum torque, T_max:\",round(T_max,0),\"Nm\"\n",
-      "print \"Maximum power,P:\",round(P/1000,0),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum torque, T_max: 1790.0 Nm\n",
-        "Maximum power,P: 337.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.9, Page number: 229"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "b=0.5                                       #Wavelength of wnding(m)\n",
-      "l=1.5                                   #Winding length(m)\n",
-      "I=700                                   #Currents in windings(A)\n",
-      "N=45                                    #No. of turns\n",
-      "K=0.92                                  #winding factor\n",
-      "p=3                                     #No. of phases\n",
-      "uo=4*pi*10**-7\n",
-      "g=0.01                                  #Air gap flux(m)\n",
-      "f=25                                    #Frequency of the exciting current(A)\n",
-      "\n",
-      "#Calculations:\n",
-      "F_peak=(3*4*K*N*700)/round(4*pi*p,-1)\n",
-      "B=uo*F_peak/g\n",
-      "v=f*b\n",
-      "\n",
-      "#Results:\n",
-      "print \"Amplitude of the resultant mmf wave:\",round(F_peak/1000,1),\"kA/m\"\n",
-      "print \"Peak air gap flux:\",round(B,1),\"T\"\n",
-      "print \"Velocity of the travelling wave:\",v,\"m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Amplitude of the resultant mmf wave: 8.7 kA/m\n",
-        "Peak air gap flux: 1.1 T\n",
-        "Velocity of the travelling wave: 12.5 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter5.ipynb b/ELECTRIC_MACHINERY/chapter5.ipynb
deleted file mode 100755
index e0897bf4..00000000
--- a/ELECTRIC_MACHINERY/chapter5.ipynb
+++ /dev/null
@@ -1,596 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:fdfb62ac44329d8ca6f9c44e55b110ca6276a0bf982882f2c43000518ecc6bb3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5: Synchronous Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.1, Page number: 254"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import cmath\n",
-      "\n",
-      "#Varaible Declaration:\n",
-      "pf=0.95                                 #Lagging power factor\n",
-      "Vl=460                                  #Terminal voltage(V)\n",
-      "I=120                                   #Terminal current(A)\n",
-      "If=47                                   #Field current(A)\n",
-      "X=1.68j                                 #Line syncchronous reactance(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Choosing motor reference direction:\n",
-      "Va=Vl/math.sqrt(3)\n",
-      "theta=math.acos(0.95)\n",
-      "Ia=I*cmath.exp(-theta*1j)\n",
-      "Eaf=Va-X*Ia\n",
-      "wc=120*math.pi\n",
-      "Laf=math.sqrt(2)*abs(Eaf)/(wc*If)\n",
-      "P=3*Va*Ia*pf\n",
-      "\n",
-      "#Results:\n",
-      "print \"Generated emf:\",round(abs(Eaf),1),\"V line to line\"\n",
-      "print \"Fied to armature mutual inductance:\",round(Laf*1000,1),\"mH\"\n",
-      "print \"Three phase power:\",round(abs(P/1000),1),\"kW or\",round(abs(P)/746),\"hp\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Generated emf: 278.8 V line to line\n",
-        "Fied to armature mutual inductance: 22.3 mH\n",
-        "Three phase power: 90.8 kW or 122.0 hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.2, Page number: 255"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath \n",
-      "from math import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Pin=90.6*10**3                              #Input power(kW)\n",
-      "Va=265.6                                    #Terninal voltage(V)\n",
-      "X=1.68j                                     #Synchronous reactance(ohm)\n",
-      "Laf=22.3*10**-3                             #Mutual inductance(H)\n",
-      "wc=120*pi                                   #Angular frequency(rad/sec)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Ia=Pin/(3*Va)\n",
-      "Eaf=Va-X*Ia\n",
-      "delta=degrees(cmath.phase(Eaf))\n",
-      "I=sqrt(2)*Eaf/(wc*Laf)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print\"The phase angle,delta:\",round(delta,1),\"degrees\"\n",
-      "print\"Required field current:\",round(abs(I),2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The phase angle,delta: -35.7 degrees\n",
-        "Required field current: 55.04 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.3, Page number: 257"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Eafl=13.8*10**3                             #Open circuit voltage(V)\n",
-      "If1=318                                     #Field current(A)\n",
-      "If2=263                              #Field current after extrapolation(A)\n",
-      "wc=120*pi                            #Angular frequency(Hz)\n",
-      "\n",
-      "#Calculations:\n",
-      "Eaf=Eafl/sqrt(3)\n",
-      "La1=sqrt(2)*Eaf/(wc*If1)\n",
-      "La2=sqrt(2)*Eaf/(wc*If2)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Saturated Laf1:\",round(La1*1000,0),\"mH\"    \n",
-      "print \"Unsaturated Laf1:\",round(La2*1000,0),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Saturated Laf1: 94.0 mH\n",
-        "Unsaturated Laf1: 114.0 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.4, Page number: 262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ia=[118, 152]                     #Armature current from SC Characteristics(A)\n",
-      "If=[2.20, 2.84]                     #Field current from SC Characteristics(A)\n",
-      "Vll=220                             #Line-to-line Voltage(V)\n",
-      "V=202                               #Line-to-line air voltage(V) \n",
-      "P=45*10**3                          #Power roted to motor(W)   \n",
-      "Is_sc=1                             #per unit rated current(A)\n",
-      "\n",
-      "#Calculations:\n",
-      "Va_ag=V/sqrt(3)                 #At field current of 2.20A,at air gap,(V)\n",
-      "Ia_ag=Ia[0]\n",
-      "Xs_u=Va_ag/Ia_ag\n",
-      "Ia_rated=P/(sqrt(3)*Vll)\n",
-      "Xa_g=Va_ag/1\n",
-      "Xs_u_pu=Va_ag/Is_sc\n",
-      "Xs=Vll/(Ia[1]*sqrt(3))\n",
-      "Ia_pu=Ia[1]/Ia[0]\n",
-      "SCR=If[1]/If[0]\n",
-      "Xs=1/SCR\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"'All quantities are in per unit values'\"\n",
-      "print\"Unsaturated value of synchronous reactance:\",round(Xs_u,3),\"ohm\"\n",
-      "print \"Satureted value of synchronous reactance: \",round(Xs,3),\"ohm\"\n",
-      "print\"Short circuit ratio:\",round(SCR,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "'All quantities are in per unit values'\n",
-        "Unsaturated value of synchronous reactance: 0.988 ohm\n",
-        "Satureted value of synchronous reactance:  0.775 ohm\n",
-        "Short circuit ratio: 1.291\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.5, Page number: 265"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P_rated=45*10**3                        #Rated power(KV)\n",
-      "Pl=1.80*10**3                           #Short circuit load loss(W)\n",
-      "Ia_pu=1                                    #Per unit armature current\n",
-      "Ia=118                                  #rated armature current(A)\n",
-      "Ra_dc=0.0335                            #Dc resistance(ohm/phase)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Pl_pu=Pl/P_rated    \n",
-      "Ra_eff1=Pl_pu/Ia_pu**2                      #in per unit basis\n",
-      "Ra_eff2=Pl/(3*(Ia)**2)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Armature resistance in per unit:\",round(Ra_eff1,3),\"per unit\" \n",
-      "print \"Armature resistance in ohms/phase:\", round(Ra_eff2,3),\"ohms/phase\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Armature resistance in per unit: 0.04 per unit\n",
-        "Armature resistance in ohms/phase: 0.043 ohms/phase\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.6, Page number: 269"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "import cmath\n",
-      "import math\n",
-      "from pylab import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Veq=1.0                                 #Externalsupply(p.u)                                    \n",
-      "Eaf=1.0                                 #Internal voltage(p.u)\n",
-      "Xeq=0.23                                #Eqv.resistance of external system(p.u)\n",
-      "Xs=1.35                                #Saturated synchronous reactance(p.u)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "P_max=Eaf*Veq/(Xs+Xeq)\n",
-      "\n",
-      "\n",
-      "#for part (b):\n",
-      "delta=[0]*500\n",
-      "Ia=[0]*500\n",
-      "Va=[0]*500\n",
-      "degree=[0]*500\n",
-      "for n in range(1,101,1):\n",
-      "    delta[n-1]=(pi/2)*(n-1)/100\n",
-      "    Ia[n-1] = (Eaf *exp(1j*delta[n-1]) - Veq)/(1j*(Xs + Xeq))\n",
-      "    Va[n-1] = abs(Veq + 1j*Xeq*Ia[n-1])\n",
-      "    degree[n-1]=180*delta[n-1]/pi\n",
-      "plot(degree,Va,'r.')\n",
-      "xlabel('Power angle,delta(degrees)')\n",
-      "ylabel('Terminal voltage(per unit)')\n",
-      "title('Terminal voltage vs. power angle for part (b)')\n",
-      "show()\n",
-      "#for part (c):\n",
-      "Vterm=1.0\n",
-      "P=[0]*500\n",
-      "deltat=[0]*500\n",
-      "Ia=[0]*500\n",
-      "Eaf=[0]*500\n",
-      "\n",
-      "for n in range(1,101,1):\n",
-      "    P[n-1]=(n-1)/100\n",
-      "    deltat[n-1]=math.asin(P[n-1]*Xeq/(Vterm*Veq))\n",
-      "    Ia[n-1]=(Vterm*exp(1j*deltat[n-1])-Veq)/(1j*Xeq)\n",
-      "    Eaf[n-1]=abs(Vterm+1j*(Xs+Xeq)*Ia[n-1])\n",
-      "plot(P,Eaf,'r.')\n",
-      "xlabel('Power [per unit]')\n",
-      "ylabel('Eaf [per unit]')\n",
-      "title('Eaf vs. power for part (c)')\n",
-      "show()\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Maximum power supplied to external system:\",round(P_max,2),\"p.u\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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69aqv6EMDAIDu1G5CkJWVrZeVla1n5rlcrhT6EAAA3j/tJgRXV9fYTZs2fVVT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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x34638d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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zy9hstiQsLGxPcnLyDNkyAwcOvEXXOOrr600tLS3vySYFAADoeUpvuz1t2rSj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\ncv4f7PGByn7cLZQAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3463890>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Maximum power supplied to external system: 0.63 p.u\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.7, Page number: 272"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P_rated=2000*746/3                     #per phase rated power of motor(W)\n",
-      "Xsm=1.95                                #Synchronous reactance(ohm)\n",
-      "Vl=2300                                #Line to line voltage(V)\n",
-      "f=60                                   #Angular frequency(Hz)\n",
-      "p=30                                   #No. of poles\n",
-      "Xsg=2.65                               #Synchronous reactance of generator(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "Vp=2300/sqrt(3)\n",
-      "Ip=P_rated/Vp\n",
-      "Eafm=sqrt(Vp**2+(Ip*Xsm)**2)\n",
-      "Pm=3*Vp*Eafm/Xsm                             #Max power delivered to motor(W)\n",
-      "ws=2*2*pi*f/p\n",
-      "Tmax=Pm/ws                                  #MAx torque of motor(Nm)\n",
-      "\n",
-      "\n",
-      "#for part (b):\n",
-      "Eafg=sqrt(Vp**2+(Ip*Xsg)**2)\n",
-      "Pm2=3*Eafm*Eafg/(Xsg+Xsm)                   #Max power delivered to motor(W)\n",
-      "Tmax2=Pm2/ws                                #Max torque(Nm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print\"(a) Max power  :\",round(Pm/1000,0),\"kW,3-ph\"\n",
-      "print\"    Max torque :\",round(Tmax/1000,1),\"kNm\"\n",
-      "print \"(b) Max power :\", round(Pm2/1000,0),\"kW,3-ph\"\n",
-      "print \"    Max torque:\", round(Tmax2/1000,1),\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Max power  : 3096.0 kW,3-ph\n",
-        "    Max torque : 123.2 kNm\n",
-        "(b) Max power : 1639.0 kW,3-ph\n",
-        "    Max torque: 65.2 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.8, Page number: 279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P=45                                    #Power rated(KVA)\n",
-      "Va=220                                  #Terminal voltage(V)\n",
-      "Pin=45                                  #Power input to the armature(KVA)\n",
-      "If=5.50                                 #field current(A)\n",
-      "Rf=35.5                                 #Field winding resistance(ohm)\n",
-      "Ra=0.0399                               #Armature dc resistance(ohm/phase)\n",
-      "Xal=0.215                               #Leakage reactance of motor(ohm)\n",
-      "pf=0.80                                 #Lagging power factor \n",
-      "Pc=1.8                                  #Core loss(kW)\n",
-      "Pw=0.91                                 #Friction & windage losses(kW)\n",
-      "Ps=0.37                                 #Stray load loss(kW)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Ia=P*10**3/(sqrt(3)*Va)\n",
-      "P1=If**2*Rf/10**3                            #Loss in field winding(kW)\n",
-      "P2=3*Ia**2*Ra/10**3                          #Loss in armature(kW)\n",
-      "Pl=(Pc+Pw+Ps+P1+P2)\n",
-      "Pi=Pin*pf+P1\n",
-      "Po=Pi-Pl\n",
-      "eff=(Po/Pi)*100\n",
-      "\n",
-      "#Results:\n",
-      "print \"Efficiency of the synchronous machine:\",round(eff,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Efficiency of the synchronous machine: 84.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.9, Page number: 287"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import cmath\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Xd=1                              #Direct axis synchronus reactance(p.u)\n",
-      "Xq=0.60                           #Quadrature axis synchronous reactance(p.u)\n",
-      "Va=1                              #Terminal voltage(p.u)\n",
-      "pf=0.8                              #Lagging power factor\n",
-      "Ia=0.8-1j*math.sin(math.acos(0.8))       #Line current(p.u)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "phy=-math.acos(pf)\n",
-      "E=Va+1j*Xq*Ia\n",
-      "delta=cmath.phase(E)\n",
-      "Id=abs(Ia)*math.sin(delta-phy)*cmath.exp(1j*(-pi/2+delta))\n",
-      "Iq=abs(Ia)*math.cos(delta-phy)*cmath.exp(1j*delta)\n",
-      "Eaf=Va+Xd*Id*1j+Xq*Iq*1j\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Generated voltage:\",round(abs(Eaf),2),\"p.u Volt\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Generated voltage: 1.78 p.u Volt\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.11, Page number: 291"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from pylab import *\n",
-      "import cmath\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P_rated=2000*746                  #Rated power of motor(W)\n",
-      "Xs=1.95                           #Synchronous reactance(ohm/phase)\n",
-      "Xd=1.95                           #Direct axis synchronous reactance(ohm/ph)\n",
-      "Xq=1.40                         #Quadrature axis synchronous reactance(ohm/ph)\n",
-      "pf=1                            #Power factor of the machine\n",
-      "Vl=2300                         #Line to line voltage(V)\n",
-      "\n",
-      "#Calculatons:\n",
-      "Va=float(Vl/sqrt(3))              #volt\n",
-      "Ia=float(P_rated/(Va*3))          #ampere\n",
-      "E1=Va-1j*Ia*Xq                      #From phasor diagram\n",
-      "delta=cmath.phase(E1)             #power angle\n",
-      "Id=Ia*sin(abs(delta))             #direct axis current(A)\n",
-      "Eaf=abs(E1)+Id*(Xd-Xq)\n",
-      "r=symbols('r')\n",
-      "def P(r):                           #Process for finding maximum power\n",
-      "    return Eaf*Va*sin(r)/Xd + Va**2*(Xd-Xq)*sin(2*r)/(2*Xd*Xq)\n",
-      "P1=diff(P(r),r)\n",
-      "#On differentiation,\n",
-      "#P1 = 1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1)\n",
-      "l = solve(1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1),r)\n",
-      "P_max = (P(round(l[0],5)))\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Maximum mechanical power:\",math.ceil(3*P_max/10**3),\"kW,3-phase\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum mechanical power: 3236.0 kW,3-phase\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter6.ipynb b/ELECTRIC_MACHINERY/chapter6.ipynb
deleted file mode 100755
index 1e8c7e97..00000000
--- a/ELECTRIC_MACHINERY/chapter6.ipynb
+++ /dev/null
@@ -1,534 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:543f6585f8a1e6c2c290bd192837a35d6ff750409ce379696635948315dd7dcf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6: Polyphase Induction Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.1, Page number: 318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n=3502                              #Speed of motor(rpm)\n",
-      "Pin=15.7                            #Input power(kW)\n",
-      "Ia=22.6                             #Terminal current(A)\n",
-      "R=0.2                               #Stator winding resistance(ohm/ph)\n",
-      "f=60                                #frequency(Hz)\n",
-      "p=2                                 #No. of poles\n",
-      "\n",
-      "#Calculations:\n",
-      "Ps=3*Ia**2*R/10**3                       #Power dissipated in stator winding(kW)\n",
-      "Pg=Pin-Ps                           #Air-gap power(kW)\n",
-      "ns=120*f/p\n",
-      "s=(ns-n)/ns\n",
-      "Pr=s*Pg                             #Power dissipated in stator(kW)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Power dissipated in stator:\",round(Pr*10**3,0),\"W\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated in stator: 419.0 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number: 320"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R1=0.294                            #Resistance of stator(ohm)\n",
-      "R2=0.144                            #Rotor resistance referred to stator(ohm)\n",
-      "X1=0.503                            #Reactance of stator(ohm)\n",
-      "X2=0.209                            #Reactance of rotor referred to stator(ohm)\n",
-      "Xm=13.25                            #Leakage reactance(ohm)\n",
-      "s=0.02                              #slip\n",
-      "Prot=403                            #Friction, windage and core losses(W)\n",
-      "V=220                               #Line-to-line voltage(V)                          \n",
-      "p=6                                 #No. of poles\n",
-      "fc=60                               #frequency(Hz)\n",
-      "nph=3                               #No. of phase\n",
-      "\n",
-      "#Calculations:\n",
-      "Zf=((R2/s+1j*X2)*1j*Xm)/(R2/s+1j*X2+1j*Xm)\n",
-      "Zin=R1+1j*X1+Zf\n",
-      "V1=V/math.sqrt(3)\n",
-      "I1=V1/Zin\n",
-      "a=cmath.phase(I1)\n",
-      "pf=math.cos(a)\n",
-      "ns=120*fc/p\n",
-      "ws=4*math.pi*fc/p\n",
-      "n=(1-s)*ns\n",
-      "wm=(1-s)*ws\n",
-      "Pg=nph*abs(I1)**2*(Zf.real)\n",
-      "Psh=(1-s)*Pg-Prot\n",
-      "Tsh=Psh/wm\n",
-      "Pin=nph*(V1*I1).real\n",
-      "eff=Psh/Pin\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Rotor speed:     \",n,\"rpm\"\n",
-      "print \"Output torque:   \",round(Tsh,2),\"Nm\"\n",
-      "print \"Output power:    \",round(Psh,2),\"W\"\n",
-      "print \"Stator current:  \",round(abs(I1),1),\"A\"\n",
-      "print \"Power factor:    \",round(pf,3),\"lagging\"\n",
-      "print \"Efficiency of motor:\",round(eff*100,0),\"%\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rotor speed:      1176.0 rpm\n",
-        "Output torque:    42.4 Nm\n",
-        "Output power:     5221.6 W\n",
-        "Stator current:   18.8 A\n",
-        "Power factor:     0.846 lagging\n",
-        "Efficiency of motor: 86.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number: 325"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "from math import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R1=0.294                            #Resistance of stator(ohm)\n",
-      "R2=0.144                            #Rotor resistance referred to stator(ohm)\n",
-      "X1=0.503                            #Reactance of stator(ohm)\n",
-      "X2=0.209                            #Reactance of rotor referred to stator(ohm)\n",
-      "Xm=13.25                            #Leakage reactance(ohm)\n",
-      "s=0.03                              #slip\n",
-      "V=220                               #Line-to-line voltage(V)         \n",
-      "p=6                                 #No. of poles\n",
-      "fc=60                               #frequency(Hz)\n",
-      "nph=3                               #No. of phase\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part (a):\n",
-      "Zf=((R2/s+1j*X2)*1j*Xm)/(R2/s+1j*X2+1j*Xm)   #Impedance referred to stator(ohm)      \n",
-      "Zin=R1+1j*X1+Zf                                 #Total input impedance(ohm)\n",
-      "Z1_eq=1j*Xm*(R1+1j*X1)/(R1+1j*(X1+Xm))          #Total equiv. impedance(ohm)\n",
-      "R1_eq=Z1_eq.real\n",
-      "X1_eq=Z1_eq.imag\n",
-      "V1=V/sqrt(3)\n",
-      "V1_eq=V1*(1j*Xm/(R1+1j*(X1+Xm)))\n",
-      "I2=abs(V1_eq)/sqrt((R1_eq+R2/s)**2+(X1_eq+X2)**2)\n",
-      "ws=4*pi*fc/p\n",
-      "ns=120*fc/p\n",
-      "Tmech=nph*I2**2*(R2/s)/ws\n",
-      "Pmech=nph*round(I2,1)**2*(R2/s)*(1-s)\n",
-      "\n",
-      "\n",
-      "#for part (b):\n",
-      "SmaxT=R2/sqrt(R1_eq**2+(X1_eq+X2)**2)     #slip at max torque\n",
-      "n_max=(1-SmaxT)*ns\n",
-      "Tmax=(1/ws)*(0.5*nph*abs(V1_eq)**2)/(R1_eq+sqrt(R1_eq**2+(X1_eq+X2)**2))\n",
-      "\n",
-      "#for part (c):\n",
-      "s1=1                                        #Slip at starting of motor\n",
-      "I2_start=abs(V1_eq)/sqrt((R1_eq+R2)**2+(X1_eq+X2)**2)\n",
-      "Tstart=nph*I2_start**2*R2/ws\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Load component I2 of stator current:\",round(I2,1),\"A\"\n",
-      "print \"    Electromechanical torque, Tmech    :\",round(Tmech,1),\"Nm\"\n",
-      "print \"    Electromechanical power, Pmech     :\",round(Pmech,0),\"W\"\n",
-      "\n",
-      "print \"(b) Maximum electromechanical torque   :\",round(Tmax,0),\"Nm\"\n",
-      "print \"    Speed                              :\",round(n_max,0),\"rpm\"\n",
-      "\n",
-      "print \"(c) Electromechanical starting torque Tstart:\",round(Tstart,1),\"Nm\"\n",
-      "print \"    Stator load current, I2_start           :\",round(I2_start,0),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Load component I2 of stator current: 23.9 A\n",
-        "    Electromechanical torque, Tmech    : 65.4 Nm\n",
-        "    Electromechanical power, Pmech     : 7979.0 W\n",
-        "(b) Maximum electromechanical torque   : 175.0 Nm\n",
-        "    Speed                              : 970.0 rpm\n",
-        "(c) Electromechanical starting torque Tstart: 77.6 Nm\n",
-        "    Stator load current, I2_start           : 150.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number: 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "import cmath\n",
-      "from math import *\n",
-      "from matplotlib import *\n",
-      "from pylab import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V=230                               #line to line voltage(V)\n",
-      "R1=0.095                            #Resistance of stator(ohm)\n",
-      "X1=0.680                            #Reactance of stator(ohm)\n",
-      "X2=0.672                            #Reactance of rotor referred to stator(ohm)\n",
-      "Xm=18.7                             #Leakage reactance(ohm)\n",
-      "f=60                                #frequency(Hz)\n",
-      "p=4                                 #No. of poles\n",
-      "nph=3                               #No. of phases\n",
-      "\n",
-      "\n",
-      "#Calculations and Results:\n",
-      "V1=V/sqrt(3)\n",
-      "omega=4*pi*f/p\n",
-      "ns=120*f/p\n",
-      "Z1eq=1j*Xm*(R1+1j*X1)/(R1+1j*(X1+Xm))           #Stator thevenin equivalent\n",
-      "R1eq=Z1eq.real\n",
-      "X1eq=Z1eq.imag\n",
-      "V1eq=abs(V1*1j*Xm/(R1+1j*(X1+Xm)))\n",
-      "\n",
-      "print \"Hence, the required plot is shown below:\"\n",
-      "for m in range(1,6,1):                      #Loop over rotor resistance\n",
-      "    if m==1:\n",
-      "        R2=0.1\n",
-      "    elif m==2:\n",
-      "        R2=0.2\n",
-      "    elif m==3:\n",
-      "        R2=0.5\n",
-      "    elif m==4:\n",
-      "        R2=1.0\n",
-      "    else:\n",
-      "        R2=1.5\n",
-      "\n",
-      "    s=[0]*202\n",
-      "    rpm=[0]*202\n",
-      "    Tmech=[0]*202\n",
-      "    for n in range(1,201,1):                    #Loop over slip\n",
-      "        s[n-1]=n/200                            #slip\n",
-      "        rpm[n-1]=ns*(1-s[n-1])                  #rpm\n",
-      "        I2=abs(V1eq/(Z1eq+1j*X2+R2/s[n-1]))     #I2\n",
-      "        Tmech[n-1]=nph*I2**2*R2/(s[n-1]*omega)  #Electromechanical torque(Nm)\n",
-      "\n",
-      "    plot(rpm,Tmech)\n",
-      "    title('Electromechanical mechanical torque, Tmech(Nm) vs rpm')\n",
-      "    xlabel(\"rpm\")\n",
-      "    ylabel(\"Tmech\")\n",
-      "    if m==1:\n",
-      "        show()\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hence, the required plot is shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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0yrdlC1tnsXVr7c9R9HwWKquzSE5OdrSwsMiZOXPmgTt37nTp1q3b39u2bVss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\n0qktCAAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x32859d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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ZxwPatQM2bwZGjpStLABTFgxQ18wX1dW4VVKCW6WlCCspQblAgF5/KY8+urrw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\nQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x34979d0>"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number: 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "nph=3                       #No. of phases\n",
-      "k=0.429             #reactance ratio(X1/X2) from table 6.1,for class C motor\n",
-      "p=4                         #No.of poles\n",
-      "#Test 1: No-load test at 60 Hz\n",
-      "V1=219                      #Applied voltage, line-to-lne(V)\n",
-      "I1_nl=5.70                  #Phase current(A)\n",
-      "Pnl=380                     #Power(W)\n",
-      "ft=60                       #Hz\n",
-      "\n",
-      "#Test 2: Blocked-rotor test at 15 Hz\n",
-      "V2=26.5                     #Applie voltage, line-to-line(V)\n",
-      "I1_bl=18.57                 #Phase current(A)\n",
-      "Pbl=675                     #Power(W)\n",
-      "fbl=15                      #Hz\n",
-      "\n",
-      "#Test 3:\n",
-      "R1=0.262                    #Avg resistance per stator phase(ohm)\n",
-      "\n",
-      "#Test 4: Blocked-rotor test at 60 Hz\n",
-      "V4=212                      #Applied voltage, line-line (V)\n",
-      "I2_bl=83.3                  #Avg phase current(A)\n",
-      "Pbl_4=20.1*10**3            #Power(W)\n",
-      "Tstart=74.2                 ##starting torque(Nm)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#For part (a):\n",
-      "Prot=Pnl-nph*I1_nl**2*R1\n",
-      "V1_nl=V1/sqrt(3)                            #from test 1\n",
-      "Qnl=sqrt((nph*V1_nl*I1_nl)**2-Pnl**2)\n",
-      "Xnl=Qnl/(nph*I1_nl**2)\n",
-      "V1_bl=V2/sqrt(3)                            #from test 2\n",
-      "Qbl=sqrt((nph*V1_bl*I1_bl)**2-Pbl**2)\n",
-      "Xbl=(ft/fbl)*(Qbl/(nph*I1_bl**2))\n",
-      "X2=symbols('X2')\n",
-      "fx=k**2*X2**2+(Xbl*(1-k)-Xnl*(1+k))*X2+Xnl*Xbl\n",
-      "x=solve(fx,X2)\n",
-      "X2=round(x[0],2)                    #since X2 must be less than X1\n",
-      "X1=k*X2\n",
-      "Xm=Xnl-X1\n",
-      "Rbl=Pbl/(nph*I1_bl**2)\n",
-      "R2=(Rbl-R1)*((X2+Xm)/Xm)**2\n",
-      "\n",
-      "#for part (b):\n",
-      "Pg=Pbl_4-nph*I2_bl**2*R1\n",
-      "ws=4*math.pi*ft/p\n",
-      "Tstart=Pg/ws\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) N-load rotational loss:\",round(Prot,0),\"W\"\n",
-      "print \"\\n  Equivalent ckt parameters:\\n\"\n",
-      "print\"  R1=\",round(R1,3),\"ohm\",\"  R2=\",round(R2,3),\"ohm\"\n",
-      "print\"  X1=\",round(X1,3),\"ohm\",\"  X2=\",round(X2,3),\"ohm\",\" Xm=\",round(Xm,2),\"ohm\"\n",
-      "print \"\\n(b) Starting torque:\",round(Tstart,2),\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) N-load rotational loss: 354.0 W\n",
-        "\n",
-        "  Equivalent ckt parameters:\n",
-        "\n",
-        "  R1= 0.262 ohm   R2= 0.447 ohm\n",
-        "  X1= 0.635 ohm   X2= 1.48 ohm  Xm= 21.2 ohm\n",
-        "\n",
-        "(b) Starting torque: 77.7 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number: 338"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Xnl=21.8                                #ohm\n",
-      "Xbl=2.01                                #ohm\n",
-      "R_1=0.26                                 #ohm\n",
-      "Rbl=0.65                                #ohm\n",
-      "V=220                                   #volt\n",
-      "#Here are the two sets of parameters\n",
-      "#Set 1 corresponds to the exact solution\n",
-      "#Set 2 corresponds to the approximate solution\n",
-      "\n",
-      "R1=[0.262, 0.262]                        #ohm\n",
-      "R2=[0.447, 0.444]                        #ohm\n",
-      "X1=[0.633, 0.603]                        #H\n",
-      "X2=[1.47, 1.41]                          #H\n",
-      "Xm=[21.2, 21.2]                          #H\n",
-      "nph=3                                   #No. of phases\n",
-      "p=4                                     #No. of poles\n",
-      "Prot=354                                #Rotational losses(Watts)\n",
-      "\n",
-      "#Calculations:\n",
-      "X_1=0.3*Xbl                              #(ohm) from table 6.1 and X1+X2=Xbl\n",
-      "X_2=Xbl-X_1                               #ohm\n",
-      "X_m=Xnl-X_1\n",
-      "R_2=(Rbl-R_1)*((X_2+X_m)/X_m)**2\n",
-      "\n",
-      "#Results for part (a):\n",
-      "print \"(a) The parameters:\\n\"\n",
-      "print\"  R1=\",round(R_1,3),\"ohm\",\"  R2=\",round(R_2,3),\"ohm\"\n",
-      "print\"  X1=\",round(X_1,3),\"ohm\",\"   X2=\",round(X_2,2),\"ohm\"\n",
-      "print\"  Xm=\",round(X_m,3),\"ohm\"\n",
-      "\n",
-      "#Calculations & Results for part (b):\n",
-      "print \"\\n\\n(b)\"\n",
-      "#Here is the operating condition\n",
-      "V1=220/sqrt(3)\n",
-      "fe=60                                   #Hz\n",
-      "rpm=1746\n",
-      "#Calculate the synchronous speed:\n",
-      "ns=120*fe/p\n",
-      "ws=4*pi*fe/p\n",
-      "s=(ns-rpm)/ns\n",
-      "wm=ws*(1-s)\n",
-      "Zgap=[0]*2\n",
-      "Zin=[0]*2\n",
-      "Pmech=[0]*2\n",
-      "I1=[0]*2\n",
-      "I2=[0]*2\n",
-      "Tmech=[0]*2\n",
-      "\n",
-      "#Calculate stator Thevenin equivalent:\n",
-      "#Loop over the two motors\n",
-      "for m in range(0,2,1):\n",
-      "    Zgap = 1j*Xm[m]*(1j*X2[m] + R2[m]/s)/(R2[m]/s + 1j*(Xm[m] + X2[m]))\n",
-      "    Zin=R1[m]+1j*X1[m]+Zgap\n",
-      "    I1=V1/Zin\n",
-      "    I2=I1*(1j*Xm[m])/(R2[m]/s+1j*(Xm[m]+X2[m]))\n",
-      "    Tmech=nph*abs(I2)**2*R2[m]/(s*ws)   #Electromechanical torque\n",
-      "    Pmech=wm*Tmech                      #Electromechanical power\n",
-      "    Pshaft=Pmech - Prot\n",
-      "    if (m==0):\n",
-      "        print \"Exact Solution:\"\n",
-      "    else:\n",
-      "        print \"\\nApproximate Solution:\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "    \n",
-      "    print \"\\tPmech=\",round(Pmech,1),\"W\",\"\\tPshaft =\",round(Pshaft,1), \"W\"\n",
-      "    print \"\\tI1=\", round(abs(I1),1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The parameters:\n",
-        "\n",
-        "  R1= 0.26 ohm   R2= 0.443 ohm\n",
-        "  X1= 0.603 ohm    X2= 1.41 ohm\n",
-        "  Xm= 21.197 ohm\n",
-        "\n",
-        "\n",
-        "(b)\n",
-        "Exact Solution:\n",
-        "\tPmech= 2820.7 W \tPshaft = 2466.7 W\n",
-        "\tI1= 10.3 A\n",
-        "\n",
-        "Approximate Solution:\n",
-        "\tPmech= 2850.5 W \tPshaft = 2496.5 W\n",
-        "\tI1= 10.4 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter7.ipynb b/ELECTRIC_MACHINERY/chapter7.ipynb
deleted file mode 100755
index c5e8eebf..00000000
--- a/ELECTRIC_MACHINERY/chapter7.ipynb
+++ /dev/null
@@ -1,473 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7: DC Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.1, Page number: 371"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vt=[128, 124]                           #Terminal voltage(V)\n",
-      "Ea=125                                  #Generated emf(V)\n",
-      "Ra=0.02                                 #Armature resistance(ohm)\n",
-      "n=3000                                  #rpm\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#For 128 V\n",
-      "Ia1=(Vt[0]-Ea)/Ra\n",
-      "Pin1=Vt[0]*Ia1\n",
-      "Pe1=Ea*Ia1\n",
-      "wm=3000*2*pi/60\n",
-      "Tmech1=Ea*Ia1/wm\n",
-      "\n",
-      "#for 124 V\n",
-      "Ia2=(-Vt[1]+Ea)/Ra\n",
-      "Pin2=Vt[1]*Ia2\n",
-      "Pe2=Ea*Ia2\n",
-      "Tmech2=Ea*Ia2/wm\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Armature current:\",Ia1,\"A\",\"\\n    Terminal power:\",Pin1/10**3,\"kW\"\n",
-      "print \"    Electromagnetic power:\",round(Pe1/10**3,2),\"kW\"\n",
-      "print \"    Torque:\",round(Tmech1,1),\"Nm\"\n",
-      "\n",
-      "print \"(b) Armature current:\",Ia2,\"A\",\"\\n    Terminal power:\",Pin2/10**3,\"kW\"\n",
-      "print \"    Electromagnetic power:\",round(Pe2/10**3,2),\"kW\",\n",
-      "print \"\\n    Torque:\",round(Tmech2,1),\"Nm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Armature current: 150.0 A \n",
-        "    Terminal power: 19.2 kW\n",
-        "    Electromagnetic power: 18.75 kW\n",
-        "    Torque: 59.7 Nm\n",
-        "(b) Armature current: 50.0 A \n",
-        "    Terminal power: 6.2 kW\n",
-        "    Electromagnetic power: 6.25 kW \n",
-        "    Torque: 19.9 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.2, Page number: 372"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vt=123                                      #terminal voltage(V)\n",
-      "Pt=21.9                                     #Terminal power(kW)\n",
-      "Ra=0.02                                     #ohm\n",
-      "Eao=125                                     #generated voltage(V) at 3000rpm\n",
-      "no=3000                                     #rpm\n",
-      "\n",
-      "\n",
-      "#calculations:\n",
-      "Ia=Pt*10**3/Vt\n",
-      "Ea=Vt-Ia*Ra\n",
-      "n=(Ea/Eao)*no\n",
-      "\n",
-      "#Results:\n",
-      "print \"Speed of motor:\",round(n,0),\"rpm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed of motor: 2867.0 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.3, Page number: 376"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Il=400                                  #Armature current(A)\n",
-      "If=4.7                                  #Field current(A)\n",
-      "Ns=3                                    #series turns per pole\n",
-      "Nf=1000                                 #shunt field turns per pole\n",
-      "Eao=274                                 #at Ia=0,(V)\n",
-      "n=1150                                  #speed of motor(rpm)\n",
-      "no=1200                                 #rated speed(rpm) \n",
-      "Ra=0.025                                #armature resistance(ohm)\n",
-      "Rs=0.005                                #series field resistance(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Is=Il+If\n",
-      "GM=If+(Ns/Nf)*Is                        #for graphical analysis\n",
-      "Ea=(n/no)*Eao\n",
-      "Vt=Ea-Is*(Ra+Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Terminal voltage at rated terminal current:\",round(Vt,0),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Terminal voltage at rated terminal current: 250.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.4, Page number: 377"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Il=400                                  #Armature current(A)\n",
-      "If=4.7                                  #Field current(A)\n",
-      "Ns=3                                    #series turns per pole\n",
-      "Nf=1000                                 #shunt field turns per pole\n",
-      "Eao=261                                 #at Ia=400 A,(V)\n",
-      "n=1150                                  #speed of motor(rpm)\n",
-      "no=1200                                 #rated speed(rpm) \n",
-      "Ra=0.025                                #armature resistance(ohm)\n",
-      "Rs=0.005                                #series field resistance(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Ea=(n/no)*Eao\n",
-      "Vt=Ea-(Il+If)*(Ra+Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Terminal voltage:\", round(Vt,0), \"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Terminal voltage: 238.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.5, Page number: 378"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Il=400                                  #Armature current(A)\n",
-      "If=4.7                                  #Field current(A)\n",
-      "Ns=3                                    #series turns per pole\n",
-      "Nf=1000                                 #shunt field turns per pole\n",
-      "Eao=269                                 #at Ia=400 A,(V)\n",
-      "n=1150                                  #speed of motor(rpm)\n",
-      "no=1200                                 #rated speed(rpm) \n",
-      "Ra=0.025                                #armature resistance(ohm)\n",
-      "Rs=0.007                                #series field resistance(ohm)\n",
-      "\n",
-      "#Calculations:\n",
-      "Is=Il+If\n",
-      "GM=If+(Ns/Nf)*Is                        #for graphical analysis\n",
-      "Ea=(n/no)*Eao\n",
-      "Vt=Ea-Is*(Ra+Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Terminal voltage at rated terminal current:\",round(Vt,0),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Terminal voltage at rated terminal current: 245.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.6, Page number: 381"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ns=4                            #Series field turns\n",
-      "Nf=1000                         #Shunt field turns\n",
-      "Vt=250                          #Full load voltage(V)\n",
-      "#for part (a):\n",
-      "Ia=400                          #Armature current(A)\n",
-      "Ra=0.025                        #Armature resistance(ohm)\n",
-      "\n",
-      "#for part (b):\n",
-      "Rs=0.005                        #Added sries resistance(ohm)\n",
-      "Vo=250                          #No load voltage(V)\n",
-      "If=5                            #field current at full load(A)\n",
-      "\n",
-      "\n",
-      "#Calculations & Results:\n",
-      "\n",
-      "#for part (a)\n",
-      "V1=Ia*Ra\n",
-      "\n",
-      "#for part (b):\n",
-      "Ia1=Ia+If\n",
-      "Rs,Rd=symbols('Rs Rd')            #Rd= diverter resistance(ohm)\n",
-      "Rp=Rs*Rd/(Rs+Rd)                  #                      -------(i)\n",
-      "Is=Ia1*(Rd/(Rs+Rd))\n",
-      "Inet=If+(Ns/Nf)*Is\n",
-      "Ea=Vt+Ia*(Ra+Rp)                  #                      -------(ii)\n",
-      "\n",
-      "#from equation (ii)\n",
-      "Rp=Rs(Inet-5.0)/1.62              \n",
-      "R_d=0.0082                       #R_d=Rd(say), using (i)\n",
-      "print \"(a) The operating terminal voltage = 205 V\", Inet\n",
-      "print \"(b) Rd =\", R_d,\"ohm\"\n",
-      "print \"\\tHence, by this process, resistance across the series field\" \n",
-      "print \"\\t(referred to as a series-field diverter) can be adjusted \"\n",
-      "print \"\\tto produce the desired performance. \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The operating terminal voltage = 205 V 1.62*Rd/(Rd + Rs) + 5\n",
-        "(b) Rd = 0.0082 ohm\n",
-        "\tHence, by this process, resistance across the series field\n",
-        "\t(referred to as a series-field diverter) can be adjusted \n",
-        "\tto produce the desired performance. \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.7, Page number: 383"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "      \n",
-      "#Variable declaration:\n",
-      "Ia=400                              #Armature current(A)\n",
-      "n1=1200                             #rpm\n",
-      "n2=1100                             #rpm\n",
-      "Ra=0.025                            #armature resistance(ohm)     \n",
-      "Eo=250                              #no load armature voltage(V)\n",
-      "del_n=1.5                           #fractional winding added\n",
-      "N=1000                              #Total windings\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part(a):\n",
-      "#point corresponding on the no load saturation curve is :\n",
-      "Eao=Eo*(n1/n2)\n",
-      "#using Eao value in curve, value of If is found to be:\n",
-      "If=5.90                             #Field current(A)\n",
-      "Ea=Eo-Ia*Ra\n",
-      "#From Fig. 7.14\n",
-      "Ea1=261\n",
-      "n=n1*(Ea/Ea1)\n",
-      "Pe=Ea*Ia\n",
-      "Pl=2000                             #No load Rotational loss(W)\n",
-      "Po=(Pe-Pl)/(1+0.01)\n",
-      "\n",
-      "#for part (b):\n",
-      "If1=If+del_n/N\n",
-      "#From Fig. 7.14 the corresponding value of Ea at 1200 r/min would be 271 V.\n",
-      "Ea2=271                             #volts\n",
-      "n22=n1*(Ea/Ea2)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Part(a):\"\n",
-      "print \"Required speed =\",round(n),\"r/min\"\n",
-      "print \"Output power =\", round((Po/746),1),\"hp\"\n",
-      "print \"\\nPart (b):\"\n",
-      "print \"Required speed =\",round(n22),\"r/min\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part(a):\n",
-        "Required speed = 1103.0 r/min\n",
-        "Output power = 124.8 hp\n",
-        "\n",
-        "Part (b):\n",
-        "Required speed = 1063.0 r/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.9, Page number: 389"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V1=50                               #terminal voltage(V)\n",
-      "Ia=1.25                             #Armature current(A)\n",
-      "Ra=1.03                             #Armature resistance(ohm)\n",
-      "n1=2100                             #speed at 50V(rpm)\n",
-      "V2=48                               #terminal voltage at 1700 rpm (V)\n",
-      "n2=1700                             #speed at 48 V(rpm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for (a):\n",
-      "Ea1=V1-Ia*Ra\n",
-      "wm1=n1*2*pi/60\n",
-      "Km=round(Ea1/wm1,2)\n",
-      "\n",
-      "#for part(b):\n",
-      "Prot=Ea1*Ia\n",
-      "\n",
-      "#for part(c:)\n",
-      "wm2=n2*2*pi/60\n",
-      "Ea2=Km*wm2\n",
-      "Ia2=(V2-Ea2)/Ra\n",
-      "Pmech=Ea2*Ia2\n",
-      "Pshaft=Pmech-Prot\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Torque constant:\",round(Km,2),\"V/(rad/s)\"\n",
-      "print \"(b) No-load rotational losses of the motor:\",round(Prot,0),\"W\"\n",
-      "print \"(c) The power output of the motor:\",round(Pshaft,2),\"W\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Torque constant: 0.22 V/(rad/s)\n",
-        "(b) No-load rotational losses of the motor: 61.0 W\n",
-        "(c) The power output of the motor: 275.05 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter8.ipynb b/ELECTRIC_MACHINERY/chapter8.ipynb
deleted file mode 100755
index d938905f..00000000
--- a/ELECTRIC_MACHINERY/chapter8.ipynb
+++ /dev/null
@@ -1,392 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c0cec23fbe397fb7ebb6d2fd4aa3f54866d7506eaba8e21d17c10d01c93413cd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8: Variable-Reluctance Machines and Stepping Motors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page number: 411"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from numpy import *\n",
-      "from math import *\n",
-      "from matplotlib import *\n",
-      "from pylab import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R=0.038                               #m\n",
-      "a=b=pi/3                            #rad\n",
-      "g=2.54*10**-4                       #m\n",
-      "D=0.13                              #m\n",
-      "N=100                               #turns in both poles\n",
-      "uo=4*pi*10**-7                      #permeability of free space()\n",
-      "i1=5                                #coil current (A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Lm=N**2*uo*a*R*D/(2*g)\n",
-      "#x=symbols('x')\n",
-      "subplot(2,1,1)\n",
-      "x=linspace(-180,-120,100)\n",
-      "L=-(Lm/60)*x-2*Lm\n",
-      "plot(x,L,'b')\n",
-      "#grid()\n",
-      "\n",
-      "x=linspace(-60,0,100)\n",
-      "L=(Lm/60)*x+Lm\n",
-      "plot(x,L,'b')\n",
-      "grid()\n",
-      "\n",
-      "x=linspace(0,60,100)\n",
-      "L=-(Lm/60)*x+Lm\n",
-      "plot(x,L,'b')\n",
-      "grid()\n",
-      "\n",
-      "\n",
-      "x=linspace(120,180,100)\n",
-      "L=(Lm/60)*x-2*Lm\n",
-      "plot(x,L)\n",
-      "annotate('Lm=0.128 H',xy=(-150,0.10))\n",
-      "annotate('Lmax',xy=(0,Lm+0.005))\n",
-      "ylabel('L11(theta)')\n",
-      "xlabel('theta')\n",
-      "grid()\n",
-      "\n",
-      "#part(b)\n",
-      "subplot(2,1,2)\n",
-      "x1=linspace(-180,-120,100)\n",
-      "x2=linspace(-150,-90,100)\n",
-      "i1=5\n",
-      "i2=4\n",
-      "Tm1=(Lm/(2*pi/3))*i1**2\n",
-      "Tm2=(Lm/(2*pi/3))*i2**2\n",
-      "dll=np.ones(100)\n",
-      "plot(x1,-Tm1*np.array(dll),'g')\n",
-      "plot(x2,Tm2*np.array(dll),'b--')\n",
-      "\n",
-      "x1=linspace(-60,0,100)\n",
-      "x2=linspace(-90,-30,100)\n",
-      "Tm1=(Lm/(2*pi/3))*i1**2\n",
-      "Tm2=(Lm/(2*pi/3))*i2**2\n",
-      "dll=np.ones(100)\n",
-      "plot(x1,Tm1*np.array(dll),'g')\n",
-      "plot(x2,-Tm2*np.array(dll),'b--')\n",
-      "\n",
-      "x1=linspace(0,60,100)\n",
-      "x2=linspace(30,90,100)\n",
-      "Tm1=(Lm/(2*pi/3))*i1**2\n",
-      "Tm2=(Lm/(2*pi/3))*i2**2\n",
-      "dll=np.ones(100)\n",
-      "plot(x1,-Tm1*np.array(dll),'g')\n",
-      "plot(x2,Tm2*np.array(dll),'b--')\n",
-      "\n",
-      "x1=linspace(120,180,100)\n",
-      "x2=linspace(90,150,100)\n",
-      "Tm1=(Lm/(2*pi/3))*i1**2\n",
-      "Tm2=(Lm/(2*pi/3))*i2**2\n",
-      "dll=np.ones(100)\n",
-      "plot(x1,Tm1*np.array(dll),'g')\n",
-      "plot(x2,-Tm2*np.array(dll),'b--')\n",
-      "grid()\n",
-      "ylim(-3,3)\n",
-      "annotate('___ i1=I1, i2=0', xy=(110,2.6))\n",
-      "annotate('---- i1=0, i2=I2', xy=(110,2.2))\n",
-      "ylabel('Torque [N.m]')\n",
-      "xlabel('thetam [degrees]')\n",
-      "\n",
-      "#Results:\n",
-      "print \"Lm =\",Lm,\"H\"\n",
-      "print \"(c)The peak torque =\",round(Tm1,2),\"N.m\"\n",
-      "print \"\\t(i)  The net torque, (at thetam=0) =\", 0, \"N.m\"\n",
-      "print \"\\t(ii) The net torque, (at thetam=45 deg.) =\", 0, \"N.m\"\n",
-      "print \"\\t(iii)The net torque, (at thetam=75 deg) =\", round(Tm1,2), \"N.m\"\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Lm = 0.127968099059 H\n",
-        "(c)The peak torque = 1.53 N.m\n",
-        "\t(i)  The net torque, (at thetam=0) = 0 N.m\n",
-        "\t(ii) The net torque, (at thetam=45 deg.) = 0 N.m\n",
-        "\t(iii)The net torque, (at thetam=75 deg) = 1.53 N.m\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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O121sbG5GRUXNJIRg3bp1EevWrYtQnPPVV19F29jY3HRzc7uUmprqpdifkZHh\n7uPjc9HNze3S4MGD/xTqqKqqBAcTsnIl1xZvF59+SsjcuVxbaJfCQkJatSJEKuXa5O2huJhZR/zq\nVa5N1AMsfHHUOAFwwIABB2uqbI4fP9776dOnTbRTlamGECYAVseVK0CfPkwAxBYtuLZp+KSlAf37\nM+VtbMy1jXaZPZv5utqyhWuTt4OZM5m4VBs3cm2iHlqNVdWyZcvirVu3jjI2Ni5TuiEhhIg++eST\n3//55x9TTW6sKUKpOJRH1igICwOsrIAFPArcUp0nH1HXMyAA+OgjYFydix2zCxflWVoK2NkBiYlM\ngE1VEMJz56Njfj7g7g5cusT8LAP89KwOra7H4efnl9SkSZOn/v7+kqrH7O3tr2ty07edefP+t65C\n27Zc2zRcjhwB/vtfYOxYrk10Q7NmTJSCGTOYyoOiPRTruCgqjbeNOmNV8RWhfHHUxLRpQEkJ8Cvv\nF8QVJnI54O3NhLX/+GOubXRHeTkTK2n9eqB3b65tGiaKdVxu3ABMTLi2UR+txqqiaJeZM4F9+4Br\n17g2aZjs2gUYGQFDh3JtoluMjICFC5k/TORyrm0aJjNmMJsQKw22qFfFoVjkiVI3ihmcVWnZkvnh\nnjlTtz41UZMn31DF88ULpslmyRLuQotwWZ4ffwzo6QF//FH3uUJ47nxyPH0auHwZ+OqrN4/xyVPb\n1NjHsXfv3iFV9yk6xwsLC1Va8TgxMTFw0qRJK2Uymf7YsWN/mz59+uKq50ycOPHnhISEoCZNmjyN\niYkJ8/T0TFcck8lk+j4+PilWVlb5Bw8eHKDqP0ooREYCq1cz6yq89x7XNg2HtWuZ5pqePbk24QY9\nPWbO0BdfMEsYGxlxbdQwIC/XcVmwAGjUiGsbbqmxj8PQ0LDi008/3aGnp/faBy8hRLRnz56hZWVl\ntQ5ulMlk+vb29tePHTvWx9LSsqBz584Xd+7cOcLR0TFbcU58fHxwdHR0ZHx8fHBSUpLf119/vUo5\nfPp//vOfb1NTU71LS0ubxcXFhbwmLvA+DgUxMcBvvzF/yTSEwHtc8/gxMyHu2DHA1ZVrG24JCgKC\ng5nQ/hTN2beP6RRPT2cqZ6HCRh9HjRM8PD090y5fvuxa3TFVJgCeO3euq3LIkZ9++mnGTz/9NEP5\nnIiIiHW7du0apkgrQo4QQpCXl2fVu3fvY3///Xev/v37H6yaPwQ6AbAqlZWEuLgQsn8/1yYNg5kz\nCQkL49p8RbIfAAAgAElEQVSCH2RkEGJmRsjjx1ybCJ/yckLs7QlJSODaRHPAwgTAGpuqVq5cOal5\n8+b/Vnds3759g+uqkOobHbegoMDSzMzs3jfffLNi6dKlU//999/mNd1DCNFxFftqO3/RImD8eAmM\njYHevbnxXblyJS/LT53yfPAAWL/eHxkZ3PvypTz79vXH0qVA797VH1fs47q8aktXdeXCZ8YMCZo0\nYcqzpvMzMjIwadIkTvzqKj+dRsetaVuxYsWkus6pb3TclJQU74MHD/YfP378L4QQnDhxwl/IXxyq\nBD6Tywnp2ZOQDRu0rlMjQgnQVpvnF18QMnWq7lxqgy/lKZUyoUju3q3+OF88a4Nrx7IyQiwsCLl4\nsfbzuPZUFXAV5HD58uWT6zrH0tKyIC8vz1qRzsvLs7ayssqv7Zz8/HwrS0vLgnPnzr0XFxcX0qFD\nh9wRI0bs/Pvvvz8YPXq0IAMpKP4CqA2RiBkBNHs28PRpnadrBVU8+UBNntnZTBs0X0ap8aU827cH\nxoxhJp1WB188a4NrxxUrgB49AB+f2s/j2lOn1Ke2UaWPo6KiwqBjx463cnNzxS9evDCqa+nY8+fP\nd1FeOlaxSSSSnkL+4lCHoUPpugr1ZeBAQpYs4dqCnzx4wIRdv3aNaxPh8c8/DW8dF3D1xaEKBgYG\nldHR0ZF9+/Y97OTklDVs2LDdjo6O2evXr49Yv359BAAEBwfHd+zY8batre3NiIiI9WvWrBlfXV4i\nkUiww6eU22frIioKWL6cWYJS16jjySXVeZ49ywQz5NPoIT6VZ+vWwNSpzCz6qvDJsya4dFywABgx\nArC1rftcIZQlW9TYOW5sbFxW0y9sVSPjBgUFJQQFBSUo74uIiFivnI6Ojo6sLY+ePXue7Nmz50lV\n7id07OyAYcOYmb8rVnBtIwwIYSZSzp8PNG7MtQ1/mTiRGaZ84QLQpUvd51OA27eB7duZECOU16Gx\nqnjGvXvM5LWUFKBDB65t+M/+/cCPPzJj6/X1ubbhN5s2MfOGTp6kc4ZUYcQI5mexoS0ARmNVNUDM\nzJgml4b2smqDykqmM3zRIlppqEJoKPDoEXDoENcm/Cc1lalgv/2WaxN+QisOLVOfds/Jk4Hjx5m/\nonWFUNpnlT1jYpiKNiiIM50a4WN56uszleyMGUylC/DTsypcOM6YwYxybNpU9WuEUJZsQSsOHqJY\nV2H6dK5N+MvTp8wPNpeBDIVIv35MZ3lsLNcm/OXIEeDOHeDzz7k24S+0j4OnlJcDzs5MwL4+fbi2\n4R9RUUBGBvD771ybCI8LF5gIujduAO+8w7UNv5DLAS8vpql4yBthXhsGtI+jAWP0cl2F6dPpugpV\nefAA+M9/mPKhqE+XLoCvL/Dzz1yb8I8dO5jReR99xLUJv9FqxZGYmBjo4OBwzc7OLmfx4sXVNrxM\nnDjxZzs7uxx3d/dL6enpngAzy7xXr14nnJ2dM11cXK7+/PPPE7XpqU00aff8+GOmXXr3bvZ8akIo\n7bMSiQQLFjDDlu3suLapGb6XZ1QUsGwZEBcn4VqlTnRVli9eMF8aS5fWr/mT78+cTbRWcchkMv3I\nyMjoxMTEwKysLKedO3eOyM7OdlQ+Jz4+PvjmzZu2OTk5dr/++uuX48aNWwswId1XrFjxTWZmpvOF\nCxe6/PLLL19VvfZtQBGKZNYs5qWmAIWFwNatzBBcSv2xt2dWR9y+nWsT/rBmDROKv3t3rk0EgKZT\nz2vaNA2rrrwNHDhw/7Fjx3or70MDDDlSE0FBhKxcybUFP/j0U0LmzOHaomFQWMgEQJRKuTbhnuJi\nQkxNCbl6lWsT7QNthlXXlPqGVc/Pz7cyMzO7p9gnlUrF6enpnn5+fklV7yGEsOpspBcvBnr0kMDW\nFujXj3sfrtI5OcDff/tj3Tp++DSE9Fdf+eOHH4DPP+eHD1fpr76SwMsLcHbmhw+baQlfwqqrstU3\nrHpqaqqXIl1aWmrs7e2dsm/fvkFV84dAvjjYCrU8ejQh33/PSlbVIoSQ0AEBhHz99QmuNVRCCOVJ\nCCGHDp0gZmbMok98RdtlmZ/PfHnduaNZPkJ55uBzkENNwqoDQEVFheGQIUP2jhw5ctugQYP2a8tT\nKMyfz7TBFhZybcINR48CUinQvz/XJg2Lpk2ZPrQZM7g24Y7Zs4GxYwFr67rPpbxE05qnpk2TsOpy\nuVw0atSoLZMmTVpRU/4QyBcHm0yeTMiXX3JtoXtkMkI8PQn5/XeuTRomL14Q0rEjIcePc22iezIz\nCWnThunjeFsAC18cWqs4CCGIj48P6tSp03UbG5ubUVFRMwkhWLduXcS6desiFOd89dVX0TY2Njfd\n3NwuKZqpTp8+3U0kEsnd3d0zPDw80j08PNITEhICXxN/CyuOhw8JefddQrKzuTbRLdu3E+Lry6yU\nSNEOu3YR4uPDVNJvEyEhhCxbxrWFbuF9xaHNTSgVB9vtnosXEzJ4MKtZEkL42z77/DkhHToQotDj\nq2dVhOYpkxHi7U3I7t3c+lSHtsry9GlC2rUj5NkzdvITyjNno+KgM8cFxoQJTMj1c+e4NtEN69YB\nDg7A27QqJxfo6TFzhr77jgl309AhdB0XjaCxqgTI5s3M2gqnTjXsAH+PHzOLDx09Cri5cW3zdhAY\nyARC5NNqitrgbV7HhY1YVbTiECAyGeDhwcRqCgnh2kZ7fP89kJ/PhE+n6IaMDKbyyMlhojQ3RCor\nARcXZpVNPobk1zY0yKEAUEzEYRPFugozZ/5vXQVN0YanJty9y0QGnjfv9f1886wJoXp6eAABAUwc\nK77Adllu2gS0bctUkGwilGfOBrTi0DIZGRlayTc4GHj3XfbWVdCWZ32ZMwcIDwfatXt9P988a0LI\nngsWANHRQFERB0LVwGZZPnkCzJ2rnXVchPLM2YCX0XFVvVYIlJSUaCVfRQDE2bOZRY00RVue9eHa\nNWDfPuaLqip88qwNIXu2bw+EhTG/YPkAm2W5ciXQrRvg48Nalq8QyjNnA15Gx1XlWgrg58esrbBq\nFdcm7DJzJjPipWVLrk3eXr77DvjjD+D6da5N2OP+faZfg67jojlaqziSk5N9bW1tb4rFYqmhoWHF\n8OHDdx04cGCg8jlxcXEhoaGhsQDg5+eXVFJSYlJUVGSuyrVCQSqVajX/qChg+XJmcSNN0Lanqpw7\nB6Sm1jyqhy+edSF0z9atgSlTmHAkXMNWWS5YAIwYAdjaspLdGwjlmbMB76LjFhQUWN69e7dtXdcC\nzOgAIRCrgwWe27TRPA9deKpKbUua8smzNhqKJx9+zNgsy+ho1rJ6A6E8c03RWsUhEolUGitb32Fh\nmg4no1AoFEr90FrFUd/ouFZWVvkVFRWGdV1LoVAoFG7QWh+Hj49PSk5Ojp1UKhWXl5cb7d69e1hI\nSEic8jkhISFxW7ZsGQ0AFy5c6GJiYlJiZmZ2T5VrKRQKhcINWvviMDAwqIyOjo7s27fvYZlMph8e\nHr7R0dExe/369REAEBERsT44ODg+Pj4+2NbW9mbTpk2fbN68eUxt12rLlUKhUChqoGmURF1sU6ZM\nWerg4JDt5uZ2afDgwX+WlJS0UByLioqaaWtrm2Nvb3/t8OHDHyr2p6SkeLu4uFyxtbXNmThx4ipd\neP7+++8fOzk5Zerp6cmUVzLMzc0VN27c+JkiRPy4cePW8NGTb+WpvM2ePXuOpaVlvqIM4+Pjg+py\n5mpLSEgItLe3v2Zra5uzaNGi6Vz7KLb27dtLXV1dL3t4eKR37tw5mRCChw8fturTp89ROzu7GwEB\nAUeKi4tNdO01ZsyYTaampvdcXFyuKPbV5sXV867Ok4/v5Z07d6z9/f1PODk5ZTo7O19dtWrVRLbL\nlPOXWZXtyJEjATKZTI8QgunTpy+aPn36IkIIMjMzndzd3TPKy8sNc3NzxTY2NjflcrmIEILOnTsn\nJyUl+RJCEBQUFF91PQ9tbNnZ2Q7Xr1/v5O/vf6JqxaH8silvfPLkW3kqb3PmzJm9fPnyb6vur85Z\n8a5wsVVWVurb2NjczM3NFZeXlxtWt4AZV5tYLM59+PBhK+V9U6dOXbJ48eJphBAsWrRouuJnS5fb\nqVOnuqelpXkq/4zU5MXl867Ok4/vZWFhoXl6eroHIczy2506dbqelZXlyGaZCiLkSEBAwFE9PT05\nwMz3yM/PtwKAAwcODBwxYsROQ0PDCrFYLLW1tb2ZlJTkV1hYaFFaWtrM19c3GQBGjx69Zf/+/YO0\n7eng4HCtU6dON1Q9n2+efCvPqpBqRtJV55ycnOyrazcFfJ+DVLUMledShYaGxnLxXLt37366ZcuW\nxap4cfm8q/ME+PdempubF3l4eGQAgLGxcZmjo2N2QUGBJZtlKoiKQ5lNmzZ9HhwcHA8Ad+/ebas8\n2kp5HojyfktLy4KCggJLLnwV5ObmdvD09Ez39/eXnDlzphvAzGPhkyffy3P16tUT3N3dL4WHh28s\nKSkxqc1Z124KapqbxJWPMiKRiPTp0+eYj49PyoYNG74AgHv37pmZmZndAwAzM7N79+7dM+PWkqEm\nL749b4Df76VUKhWnp6d7+vn5JbFZplrrHFeXgICAo0VFReZV90dFRX03YMCAgwCwcOHCWUZGRuWf\nfvrpDt0bMqjiWZW2bdvezcvLs27ZsmVxWlqa16BBg/ZnZmY6882Ta2pyXrhw4axx48at/fHHH+cB\nwA8//DB/8uTJyzdu3BheXT6qziHSBlzeuy7Onj37voWFReH9+/fbBAQEHHVwcLimfFwkEhE++tfl\nxaUzn9/LsrIy4yFDhuxdtWrV182aNSut6qJJmfKm4jh69GhAbcdjYmLC4uPjg48fP95bsa+meSCW\nlpYFiuYsxX5LS8sCXXhWh5GRUbmRkVE5AHh5eaXZ2NjcysnJseObJxflqYyqzmPHjv1NUflV56wN\nN1VRZf4SV1hYWBQCQJs2be4PHjx4X3Jysq+Zmdm9oqIic3Nz86LCwkILU1PTf7j2BJi/iKvz4tvz\nVi4vPr2XFRUVhkOGDNk7atSorYMGDdoPsFumgmiqSkxMDFy6dOnUAwcODGzcuPFzxf6QkJC4Xbt2\nDS8vLzfKzc3tkJOTY+fr65tsbm5e1Lx583+TkpL8CCGirVu3jlIUnq5Qbvd88ODBuzKZTB8Abt++\n3TEnJ8euY8eOty0sLAr55Mnn8iwsLLRQ/P++ffsGu7q6XqnNWZduyvB1DtLTp0+blJaWNgOAJ0+e\nND1y5MiHrq6uV0JCQuJiY2NDASA2NjZU18+1Jmry4tvz5uN7SQgRhYeHb3RycsqaNGnSSsV+VstU\n1yMo6rPZ2trmtGvX7r/VDWdduHDhdzY2Njft7e2vJSYm9lXsVwwftbGxuTlhwoSfdeH5559/Dray\nsspr3LjxMzMzs6LAwMAEQgj27NkzxNnZ+aqHh0e6l5dX6qFDh/rx0ZNv5am8jRo1aourq+tlNze3\nSwMHDtxfVFRkVpczV1t8fHxQp06drtvY2NyMioqaybUPIQS3b9/u4O7unuHu7p7h7Ox8VeH18OHD\nVr179z7G5XDc4cOH77SwsLhraGhYbmVllbdp06YxtXlx9byrem7cuPFzPr6Xp0+f7iYSieTu7u4Z\nit+ZCQkJgWyWqWCXjqVQKBQKNwiiqYpCoVAo/IFWHBQKhUJRC1pxUCgUCkUteFtxPH/+vLGfn1+S\nh4dHhpOTU9bMmTN/4tqJQqFQKDyuOBo3bvz8xIkTvTIyMjwuX77sduLEiV6KGdcUiq55/Phxi7Vr\n144DAIlE4q/uJMrY2NhQ5aGbFIqQ4W3FAQBNmjR5CgDl5eVGMplMv1WrVo+4dqK8nRQXF7dcs2bN\n+PpeHxMTE3b37t22bDpRKFzBm5nj1SGXy/W8vLzSbt26ZTNu3Li1Tk5OWYpjfAyNQGn4KL936r6D\nPj4+KewbUSjqQzRcepvXXxx6enryjIwMj/z8fKtTp071kEgk/srHuZxQpeoWGhrKuQP1pJ5C9tS1\nY3BwMB4/fgxCCMaMGQNTU1O4uLio5XnixAn0798fhBBkZ2ejS5cuaNSoEZYtW6aSw9ixY5GdnY2n\nT58iODgYDg4OcHZ2xowZM9T6t8TExMDOzg52dnaIjY0FIez8vc3rikNBixYtHvfr1++vlJQUH65d\nKBRKw+avv/5C8+bNAQBjxoxBYmKiRvm1bt0aq1evxpQpU1S+ZsOGDXBwcAAATJs2DdnZ2UhPT8fZ\ns2dV9nn06BHmzZuH5ORkJCcnY+7cuSgpKanXv6EqvK04Hjx48K4iRPGzZ8/eOXr0aICnp2c6117q\nIhaLuVZQCerJLtSTPXTtKBaL8egR053avXt3tGzZUuXrqqNNmzbw8fGBoaGhyg7+/v5ITU3FO++8\ng549ewIADA0N4eXlhYIC1WIlHj58GB9++CFMTExgYmKCgIAAjStBBbzt4ygsLLQIDQ2NlcvlenK5\nXG/UqFFbe/fufZxrL3Xx9/fnWkElqCe7UE/20LWjSFR38/+yZcuwffv21/aVlZWhpKQEK1eurOEq\n9RyqepSUlODgwYOYNGkSAGDHjh1YunTpG9fa2dnh999/R0FBAaysXgW1hpWVlcqVTl3wtuJwdXW9\nkpaW5sW1B4VCoVRlypQpbzQ9SSQSrVVylZWVGDFiBL7++utXXzaffvopPv300xqvUaUCrC+8rTgo\nFAqFryxduhQ7dry+nlxZWRmCg4OxatUq1u/35Zdfwt7eHhMnTny1b/v27Vi2bNkb5yq+OCwtLSGR\nSF7tz8vLwwcffMCOENcjJuq7MeoUCoXCHnFxcaRly5bk4cOH5OTJk8TT05MYGBgQa2trtfLZsWMH\nMTExIba2tmTYsGGkvLyczJ49myxbtuy18z744ANy9+5dUlBQQIYOHUoIIeTIkSOkWbNmxNbWlnh7\ne5PPPvuMDBkyhMjl8jrvq3yPyMhIYmhoSJydnUn//v1J+/btSXFxMXn5u1Oj37+87RynUCgUXTNg\nwACYmJgAANq3bw8zMzMYGhqisLAQ1tbW2Lx5s0r5/Prrr7CxsUFOTg4aNWoEU1NTrFixAgsWLEC7\ndu1QVlYGuVyOW7duoVWrVmjbti3++OMPAExnuqurK3bv3o3Fixdj+/btyM7OhpeXFzw9PbFp06Ya\n76vcPBUSEoK1a9fixYsXOHPmDFxcXF792zRG05qHqw0C+eI4ceIE1woqQT3ZhXqyhy4dN2/eTCIj\nI1/bFxYWRvbs2VPntQpPuVxO3n33XSKTyQghhJw/f5707dv3jfOvXr1KJk+eTAghJDc3l7i4uLxx\njlwuJ61atSLl5eUq+c+ZM+eNrxpCCPnzzz/JZ599RgghrHxx0D4OCoVCeYkqHcqlpaXo0aPHG/uf\nPHmCuLg4vPvuuzAxMYGeHtOgY2lpWe1oJmdn52r7KJTZu3cvvL29Xw3lHT58OK5fv/7GeZMnT8bI\nkSNrzGfTpk0YMWJErfdSB1pxaBkhDHcEqCfbUE/2UMXx4cOH6NOnz6u0TCaDvr7+q/8q9gF4lVZw\n/PhxtGrVSmWfZs2aIT295illDx48UDmv2sjMzMSMGTNw9OjRV/t27dqldj4LFy6EkZFRrSOw1IVW\nHBQKRfDs3r0bAPPF8NdffyEsLAz//PMPunTpgrFjxyIiIgL6+vqYN28ekpKS8Ndff0EkEiEtLU2l\n/JW/REpLS9G9e/dqv0527twJe3t7lJSUQC6XQ09PD/n5+bC0tFTr35Ofn4+PPvoIW7duRYcOHV7t\nHzZsGG7cuPHG+TV9ccTExCA+Ph7Hj7M8BU7Tti6uNtA+DlahnuxCPdmD6z6O0NBQtfo4CCHk448/\nJrt27SKEEBIREUHWrl1LCCEkKSmJjB49+o1rlfs4iouLiZubG9m3b5/a/sp9HAkJCcTJyYncv3//\ntXPQkEdV5eXlWffq1euEs7NzpouLy9Wff/55Yt1XUSgUSv1RnrF98eJFWFtbY8+ePYiIiICrq6vK\n+SxevBj/+c9/YGdnh+LiYoSHhwMA7ty5gyZNmtR4bwCIjo7GrVu3MHfuXHh6esLT01Ot5i9FPhMm\nTEBZWRkCAgLg6emJ8ePrvSrAm/dgKiD+UVRUZF5UVGTu4eGRUVZWZuzt7Z26f//+QY6OjtkAE9Ka\nr+4UCoVSHdOmTcPo0aPh4uLCmYNIJALRMKw6b/s4zM3Ni8zNzYsAwNjYuMzR0TH77t27bRUVB4VC\noQiNJUuWcK3ACrytOJSRSqXi9PR0Tz8/vyTl/WFhYa/itpiYmMDDw+PVCAzFVHuu04p9fPGpKb1y\n5Upelh8tT+2mFfv44lNduqor1z41pTMyMl4FIOSDjyItkUgQExMDgMVIw5p2kmh7Ky0tNfb29k7Z\nt2/fIOX9oJ3jrEI92YV6socQHAkRjidY6BznbR8HAFRUVBj279//UFBQUMKkSZNei1VM+zgoFApF\nfdjo4+BtxUEIEYWGhsa2bt364YoVK76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jJ9/KU3mbPXv2HEtLy3xFGcbHxwfV5czVlpCQ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\nEEIoIlg4EEIIRQQLB0IIoYj8DRLhonoh4yhPAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2cad490>"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.3, Page number: 424"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "from sympy import *\n",
-      "from math import *\n",
-      "from matplotlib.pyplot import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n1=4000                             #r/min\n",
-      "R=0.038                             #m\n",
-      "a=b=pi/3                            #rad\n",
-      "g=2.54*10**-4                       #m\n",
-      "D=0.13                              #m\n",
-      "N=100                               #turns in both poles\n",
-      "uo=4*pi*10**-7                      #permeability of free space(H/m)\n",
-      "Ll=0.005                            #H\n",
-      "Vo=100                              #phase voltage applied to phase 1.(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "wm=n1*pi/30\n",
-      "Lm=N**2*uo*a*R*D/(2*g)\n",
-      "thetam=symbols('thetam')\n",
-      "t=symbols('t')\n",
-      "#for part (a):\n",
-      "#for -60<=thetam<=0deg,\n",
-      "L11=Ll+(Lm/(pi/3))*(thetam+pi/3)\n",
-      "L111=diff(L11,thetam)\n",
-      "R1=L111*wm\n",
-      "#which is nuch greater than resistance R=1.5 ohm\n",
-      "thetam=-pi/3+wm*t\n",
-      "i1=Vo*t/(float(round(Ll,3))+float(Lm/(pi/3))*thetam+float(Lm/(pi/3))*pi/3)\n",
-      "\n",
-      "#for part (b):\n",
-      "V2=-200                                     #applied voltage(V)\n",
-      "thetam2=symbols('thetam2')\n",
-      "L12=Ll+(Lm/(pi/3))*(pi/3-thetam2)\n",
-      "L112=diff(L12,thetam2)\n",
-      "to=2.5*10**-3                                     #ms\n",
-      "thetam2=float(-pi/3+wm*to)\n",
-      "i1=Vo*t/(float(round(Ll,3))+float(Lm/(pi/3))*thetam+float(Lm/(pi/3))*pi/3)\n",
-      "i2=(0.25-200*(t-to))/(0.005+51.1*(5*10**-3-t))\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"i1 =\",i1,\"\\t, (where round(16.2934044186179*pi,2) = 51.1 )\"\n",
-      "print \"\\ni2 =\",i2,\"\\n\"\n",
-      "\n",
-      "\n",
-      "#Calculations & Results:\n",
-      "#for part (c):\n",
-      "Lleak=0.005\n",
-      "Posintegral=0\n",
-      "integral=0\n",
-      "N1=500\n",
-      "tmax=3.75*10**-3\n",
-      "t=[0]*503\n",
-      "thet=[0]*503\n",
-      "Torque=[0]*503\n",
-      "deltat = tmax/N1\n",
-      "thetm=[0]*503\n",
-      "i=[0]*503\n",
-      "for n in range(1,N1+2,1):\n",
-      "    t[n-1]=tmax*(n-1)/N1\n",
-      "    thetm[n-1]=-(pi/3)+(400*pi/3)*t[n-1]\n",
-      "    if (thetm[n-1]<=0):\n",
-      "        i[n-1]=100*t[n-1]/(0.005+51.1*t[n-1])\n",
-      "        dld1d1theta = 0.122\n",
-      "        Torque[n-1]=0.5*i[n-1]**2*dld1d1theta\n",
-      "        Posintegral=Posintegral+Torque[n-1]*deltat\n",
-      "        integral=Posintegral\n",
-      "    else:\n",
-      "        i[n-1]=(0.25-200*(t[n-1]-2.5*10**-3))/(0.005+51.1*(5*10**-3-t[n-1]))\n",
-      "        dld11dtheta = -0.122\n",
-      "        Torque[n-1] = 0.5*i[n-1]**2*dld11dtheta\n",
-      "        integral = integral + Torque[n-1]*deltat\n",
-      "\n",
-      "print \"\\nPositve torque integral =\",Posintegral, \"[N-m-sec]\"\n",
-      "print \"\\nTorque integral=\",integral,\"[N-m-sec]\\n\"\n",
-      "\n",
-      "plot(1000*np.array(t),i)\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('Phase current [A]')\n",
-      "title('(a) phase-1 current profile')\n",
-      "grid()\n",
-      "show()\n",
-      "plot(1000*np.array(t),Torque)\n",
-      "xlabel('time [msec]')\n",
-      "ylabel('Torque [N-m]')\n",
-      "title('(b) torque profile')\n",
-      "grid()\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i1 = 100*t/(51.1872396234976*t + 0.005) \t, (where round(16.2934044186179*pi,2) = 51.1 )\n",
-        "\n",
-        "i2 = (-200*t + 0.75)/(-51.1*t + 0.2605) \n",
-        "\n",
-        "\n",
-        "Positve torque integral = 0.000456384094483 [N-m-sec]\n",
-        "\n",
-        "Torque integral= 0.000335463884625 [N-m-sec]\n",
-        "\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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dUQoZAcrJtcaSc9kytmn45k3V5KmOVMqzIWrskuVzcjy+5ecDV6/SaCRCGgsT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\nV1DFQAgh5BVUMRBCCHnF/wGN24SgosJe8AAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3e4b1d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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RDV3XvFYqlaipamsaarXa6qeffnorLS3N+5NPPvksPT29RW5ubrOAgIAkUw+6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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3a66fd0>"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.4, Page number: 433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-<<<<<<< HEAD
-      "from matplotlib.pyplot import *\n",
-=======
->>>>>>> 0ee873700378b995b441b1be6652178f741aea5b
-      "\n",
-      "#Variavle declaration:\n",
-      "rpm=2500                        #rpm of motor\n",
-      "\n",
-      "\n",
-      "#Calculations & Results:\n",
-      "#For part (a):\n",
-      "theta=[0]*12\n",
-      "i=[0]*102\n",
-      "lambda1=[0]*102\n",
-      "for m in range(1,11,1):\n",
-      "    theta[m-1]=10*(m-1)\n",
-      "    for n in range(1,102,1):\n",
-      "        i[n-1]=30*(n-1)/100\n",
-      "        lambda1[n-1]=i[n-1]*(0.005+0.09*((90-theta[m-1])/90))*(8/(i[n-1]+8))\n",
-      "\n",
-      "    \n",
-      "    plot(i,lambda1,'.')\n",
-      "    \n",
-      "    if m==1:\n",
-      "        hold(True)\n",
-      "        \n",
-      "xlabel('current [A]')\n",
-      "ylabel('Lambda [Wb]')\n",
-      "title('Family of lambda-i curves as theta_m varies from 0 to 90 degrees')    \n",
-      "annotate('theta_m=0 deg',xy=(6,0.03))\n",
-      "annotate('theta_m=0 deg',xy=(8,0.5))\n",
-      "\n",
-      "\n",
-      "#for part (b):\n",
-      "lambdamax=25*(0.005+0.09*(8/(25+8)))\n",
-      "AreaWnet=0\n",
-      "AreaWrec=0\n",
-      "deli=0.25\n",
-      "for n in range(1,102,1):\n",
-      "    i[n-1]=25*(n-1)/100\n",
-      "    AreaWnet=AreaWnet + deli*i[n-1]*(0.09)*(8/(i[n-1]+8))\n",
-      "    AreaWrec=AreaWrec + deli*(lambdamax-i[n-1]*(0.005+0.09*(8/(i[n-1]+8))))\n",
-      "\n",
-      "Ratio=(AreaWnet+AreaWrec)/AreaWnet\n",
-      "print \"part (b): Ratio =\", round(Ratio,2)\n",
-      "\n",
-      "#for part(b):\n",
-      "rps=rpm/60\n",
-      "T=1/rps\n",
-      "Pphase=2*AreaWnet/T\n",
-      "Ptot=2*Pphase\n",
-      "print \"part (c): AreaWnet =\", round(AreaWnet,2),\"Joules\"\n",
-      "print \"Pphase =\",round(Pphase),\"W\",\"\\tPtot =\",round(Ptot),\"W\\n\"\n",
-      "plot(AreaWrec=0.7,AreaWnet=25)\n",
-      "grid()\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part (b): Ratio = 1.55\n",
-        "part (c): AreaWnet = 9.91 Joules\n",
-        "Pphase = 825.0 W \tPtot = 1651.0 W\n",
-        "\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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g0KFDZmhda5gW2iQW17SFgZvmRpwqhqikKIjZGwPVT6u1cZecshz47dZvcK7k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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x315dc10>"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY/chapter9.ipynb b/ELECTRIC_MACHINERY/chapter9.ipynb
deleted file mode 100755
index 3cdb1b6c..00000000
--- a/ELECTRIC_MACHINERY/chapter9.ipynb
+++ /dev/null
@@ -1,390 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:934a14335227a49c83c2d399431a59d2d79025dde47942572f3e87ac684c8499"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9: Single- and Two-Phase Motors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.1, Page number: 459"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Zmain=4.5+3.7j                              #main winding impedance(ohm)\n",
-      "Zaux=9.5+3.5j                               #auxilliary winding impedance(ohm)\n",
-      "f=60                                        #frequency(Hz)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "phy_main=math.degrees(math.atan(Zmain.imag/Zmain.real))\n",
-      "phy=phy_main-90\n",
-      "w=2*pi*60\n",
-      "Xc=symbols('Xc')\n",
-      "a=solve((3.5+Xc)/9.5-math.tan(math.radians(float(phy))), Xc)\n",
-      "C=-1/(w*a[0])\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The starting capacitance:\",round(float(C)*10**6,0), \"uF\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The starting capacitance: 176.0 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.2, Page number: 467"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "from math import *\n",
-      "\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "R1_m=2.02                                   #resistance of main winding(ohm)\n",
-      "X1_m=2.79                                   #resistance of main\n",
-      "R2_m= 4.12                                  #Rotor resistance ref. to stator(ohm)\n",
-      "X2_m=2.12                                   #Rotor reactance ref. to stator(ohm)\n",
-      "Xm=66.8                                     #Magnetising reactance(ohm)\n",
-      "s=0.05                                      #slip\n",
-      "Pcu=24                                       #copper loss(W)\n",
-      "Pw=13                                       #friction & windage loss(W)\n",
-      "V=110                                       #line-to-line voltage(V)\n",
-      "p=4                                         #no.of poles\n",
-      "fc=60                                       #frequency(Hz)\n",
-      "\n",
-      "#Calculations:\n",
-      "X22=X2_m+Xm\n",
-      "Q2_m=X22/R2_m\n",
-      "Rf=(Xm**2/X22)*(1/(s*Q2_m+1/(s*Q2_m)))\n",
-      "Xf=(X2_m*Xm/X22)+Rf/(s*Q2_m)\n",
-      "Zf=Rf+1j*Xf                             #forward field impedance(ohm)\n",
-      "\n",
-      "Rb=R2_m*(Xm/X22)**2/(2-s)\n",
-      "Xb=(X2_m*Xm/X22)+Rb/((2-s)*Q2_m)\n",
-      "Zb=Rb+1j*Xb                             #bachward field impedance\n",
-      "Zt=0.5*(Zf+Zb)+R1_m+1j*X1_m\n",
-      "I=V/abs(Zt)                             #Stator current(A)\n",
-      "pf=cos(cmath.phase(Zt))                 #power factor\n",
-      "Pin=V*I*pf\n",
-      "Pg_f=I**2*0.5*Rf                        #power absorbed by forward field(W)\n",
-      "Pg_b=I**2*0.5*Rb                        #power absorbed by backward field(W)\n",
-      "Pmech=(1-s)*(Pg_f-Pg_b)\n",
-      "Pshaft=Pmech-(Pcu+Pw)\n",
-      "ws=(2/p)*120*pi\n",
-      "ns=(120/p)*fc\n",
-      "n=(1-s)*ns                              #Rotor speed(rpm)\n",
-      "wm=(1-s)*ws\n",
-      "Tshaft=Pshaft/wm                         #shaft torque(Nm)\n",
-      "eff=Pshaft/Pin\n",
-      "\n",
-      "#Results:\n",
-      "print \"Stator current:\",round(I),\"A\", \"\\nPower factor:\",round(pf,3)\n",
-      "print \"Power output:\",round(Pshaft),\"W\", \"\\nSpeed:\",n,\"rpm\"\n",
-      "print \"Shaft torque:\",round(Tshaft,3),\"Nm\",\"Efficiency\",round(eff*100),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stator current: 4.0 A \n",
-        "Power factor: 0.621\n",
-        "Power output: 147.0 W \n",
-        "Speed: 1710.0 rpm\n",
-        "Shaft torque: 0.823 Nm Efficiency 60.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.3, Page number: 474"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import cmath\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "f=60                                        #freq(Hz)\n",
-      "omeag=2*pi*f\n",
-      "s=0.05                                      #slip\n",
-      "R1=0.534                                    #resistance of main winding(ohm)\n",
-      "X1=2.45\n",
-      "Xm=70.1\n",
-      "R2=0.956\n",
-      "X2=2.96\n",
-      "Valpha=230\n",
-      "Vbeta=210*cmath.exp(1j*80*pi/180)\n",
-      "\n",
-      "#Calculations:\n",
-      "Vf = 0.5*(Valpha - 1j*Vbeta)\n",
-      "Vb = 0.5*(Valpha + 1j*Vbeta)\n",
-      "Zf=R1+1j*X1+1j*Xm*(R2/s+1j*X2)/(R2/s+1j*(X2+Xm))\n",
-      "If=Vf/Zf\n",
-      "Zb=R1+1j*X1+1j*Xm*(R2/(2-s)+1j*X2)/(R2/(2-s)+1j*(X2+Xm))\n",
-      "Ib = Vb/Zb\n",
-      "Ialpha=If+Ib\n",
-      "Ibeta=1j*(If-Ib)\n",
-      "Pgf=2*((Vf*(If.conjugate())).real-R1*abs(If)**2)\n",
-      "Pgb=2*((Vb*(Ib.conjugate())).real-R1*abs(Ib)**2)\n",
-      "Pmech=(1-s)*(Pgf-Pgb)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(a) Positive seq components:\", round(Vf.real,1)+1j*round(Vf.imag,1),\"V\"\n",
-      "print\"    Negative seq. components:\", round(Vb.real,1)+1j*round(Vb.imag,1),\"V\"\n",
-      "\n",
-      "print\"\\n(b) Positive stator currents:\",round(If.real,1)+1j*round(If.imag,1),\"A\"\n",
-      "print\"    Negative stator currnets:\",round(Ib.real,1)+1j*round(Ib.imag,1),\"A\"\n",
-      "\n",
-      "print\"\\n(c) Positive currents:\",round(Ialpha.real,1)+1j*round(Ialpha.imag,1),\"A\"\n",
-      "print\"    Negative currnets:\",round(Ibeta.real,1)+1j*round(Ibeta.imag,1),\"A\"\n",
-      "\n",
-      "print \"\\n(d) Power to forward field:\",round(Pgf,0),\"W\"\n",
-      "print \"    Power to backward field:\",round(Pgb,0),\"W\"\n",
-      "print \"    Pmech:\",round(Pmech,0),\"W\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Positive seq components: (218.4-18.2j) V\n",
-        "    Negative seq. components: (11.6+18.2j) V\n",
-        "\n",
-        "(b) Positive stator currents: (9.3-6.3j) A\n",
-        "    Negative stator currnets: (3.7-1.5j) A\n",
-        "\n",
-        "(c) Positive currents: (13-7.8j) A\n",
-        "    Negative currnets: (4.8+5.6j) A\n",
-        "\n",
-        "(d) Power to forward field: 4149.0 W\n",
-        "    Power to backward field: 15.0 W\n",
-        "    Pmech: 3928.0 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.5, Page number: 483"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%matplotlib inline\n",
-      "import cmath\n",
-      "from math import *\n",
-      "from matplotlib.pyplot import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Lmain=0.0806                           #main winding inductance(H)\n",
-      "Rmain = 0.58                           #main winding resistance(ohm)\n",
-      "Laux = 0.196                           #auxilliary winding inductance(H)\n",
-      "Raux = 3.37                            #auxilliary winding resistance(ohm)\n",
-      "Lr=4.7*10**-6                          #rotor inductance(H)\n",
-      "Rr=37.6*10**-6                         #rotor resistance(ohm)\n",
-      "Lmain_r=0.588*10**-3                   #main inductance ref. to rotor(H)\n",
-      "Laux_r = 0.909*10**-3                  #aux inductance ref. to rotor(H)\n",
-      "p=2                                    #poles\n",
-      "Vo=230                                 #terminal voltage(V)\n",
-      "w=120*pi                               #angular frequency(Hz)\n",
-      "C=35*10**-6\n",
-      "Prot=40                                #Windage losses(W)\n",
-      "Pcore=105                              #Core loss(W)\n",
-      "n=3500                                 #rpm\n",
-      "\n",
-      "\n",
-      "#calculations and Results:\n",
-      "Xc=-1/(w*C)\n",
-      "speed=[0]*102\n",
-      "for cal in range(1,3,1):\n",
-      "    if cal==1:\n",
-      "        mmax=2\n",
-      "    else:\n",
-      "        mmax=102\n",
-      "    for m in range(1,mmax,2):\n",
-      "        if cal==1:\n",
-      "            speed[m-1]=3500\n",
-      "        else:\n",
-      "            speed[m-1]=3599*(m-1)/100\n",
-      "        \n",
-      "        ns=(2/p)*3600\n",
-      "        s=(ns-speed[m-1])/ns\n",
-      "\n",
-      "#for part (a):\n",
-      "        Kplus=s*w/(2*(Rr+1j*s*w*Lr))\n",
-      "        Kminus=(2-s)*w/(2*(Rr+1j*(2-s)*w*Lr))\n",
-      "        A1=Lmain-1j*Lmain_r**2*(Kplus+Kminus)\n",
-      "        A2=Lmain_r*Laux_r*(Kplus-Kminus)\n",
-      "        A3=Laux-1j*Laux_r**2*(Kplus+Kminus)\n",
-      "        M=[[0]*2,[0]*2]\n",
-      "        M[0][0]=Rmain + 1j*w*A1\n",
-      "        M[0][1] = 1j*w*A2;\n",
-      "        M[1][0] = -1j*w*A2;\n",
-      "        M[1][1] = Raux + 1j*Xc+ 1j*w*A3\n",
-      "        V=[[Vo],[-Vo]]\n",
-      "        M1=inv(M)\n",
-      "        I=dot(M1,V)\n",
-      "        Imain=I[0][0]\n",
-      "        Iaux=I[1][0]\n",
-      "        Is=Imain-Iaux\n",
-      "        magImain=abs(Imain)\n",
-      "        angleImain=math.degrees(cmath.phase(Imain))\n",
-      "        magIaux=abs (Iaux)\n",
-      "        angleIaux=math.degrees(cmath.phase(Iaux))\n",
-      "        magIs=abs(Is)\n",
-      "        angleIs=math.degrees(cmath.phase(Is))\n",
-      "        Vcap=Iaux*Xc\n",
-      "        magVcap=abs(Vcap)\n",
-      "        \n",
-      "    #for part (b):\n",
-      "        Tmech=[0]*102\n",
-      "        Pshaft=[0]*102\n",
-      "        Tmechl = (Kplus-Kminus).conjugate()\n",
-      "        Tmechl=Tmechl*(Lmain_r**2*Imain*((Imain).conjugate())+Laux_r**2*Iaux*((Iaux).conjugate()))\n",
-      "        Tmech2 = 1j*Lmain_r*Laux_r*((Kplus+Kminus).conjugate())\n",
-      "        Tmech2 = Tmech2*((Imain).conjugate()*Iaux-Imain*((Iaux).conjugate()));\n",
-      "        Tmech[m-1] = (p/2)*(Tmechl+Tmech2).real\n",
-      "        Pshaft=((2/p)*(1-s)*w*Tmech[m-1])-Prot\n",
-      "        \n",
-      "    #for part (c):\n",
-      "        Pmech=[0]*102\n",
-      "        Pmain = (Vo*(Imain.conjugate())).real\n",
-      "        Paux = (-Vo*(Iaux.conjugate())).real\n",
-      "        Pin = Pmain+Paux+Pcore\n",
-      "        eta = Pshaft/Pin;\n",
-      "        if cal==1:\n",
-      "            print \"part (a):\"\n",
-      "            print  \"\\nImain=\",round(magImain,1),\"A  at an angle\",round(angleImain,1),\"degrees\"\n",
-      "            print  \"\\nImain=\",round(magIaux,1),\"A  at an angle\",round(angleIaux,1),\"degrees\"\n",
-      "            print  \"\\nImain=\",round(magIs,1),\"A  at an angle\",round(angleIs,1),\"degrees\"\n",
-      "            print  \"\\nVcap=\",round(magVcap,0),\"V\"\n",
-      "            print \"\\npart (b):\"\n",
-      "            print \"\\nTmech=\",round(Tmech[0],2),\"Nm\"\n",
-      "            print \"\\nPshaft=\",round(Pshaft),\"W\"\n",
-      "            print \"\\npart (c):\"\n",
-      "            print \"\\nPmain=\",round(Pmain,0),\"W\"\n",
-      "            print \"\\nPaux=\",round(Paux,0),\"W\"\n",
-      "            print \"\\nPin=\",round(Pin,0),\"W\"\n",
-      "            print \"\\nEfficiency=\",round(eta*100,1),\"%\"\n",
-      "        else:\n",
-      "            \n",
-      "            plot(speed,Tmech,'g.')\n",
-      "            xlabel('speed (rpm)')\n",
-      "            ylabel('Tmech (Nm)')\n",
-      "            title('Electromagnetic torque vs speed')\n",
-      "    show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part (a):\n",
-        "\n",
-        "Imain= 15.9 A  at an angle -37.6 degrees\n",
-        "\n",
-        "Imain= 5.2 A  at an angle -150.8 degrees\n",
-        "\n",
-        "Imain= 18.5 A  at an angle -22.7 degrees\n",
-        "\n",
-        "Vcap= 394.0 V\n",
-        "\n",
-        "part (b):\n",
-        "\n",
-        "Tmech= 9.75 Nm\n",
-        "\n",
-        "Pshaft= 3532.0 W\n",
-        "\n",
-        "part (c):\n",
-        "\n",
-        "Pmain= 2893.0 W\n",
-        "\n",
-        "Paux= 1043.0 W\n",
-        "\n",
-        "Pin= 4041.0 W\n",
-        "\n",
-        "Efficiency= 87.4 %\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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KagsAyoHkACoNtQUA5cC0Eqg01BYAWgc1B9BIqC0AvBnUHEAjobYAoFqQHEAl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\nAABACpIDAABI+T9crBvDDQoKmwAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2854950>"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
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diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter10-checkpoint.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter10-checkpoint.ipynb
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+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter10-checkpoint.ipynb
@@ -0,0 +1,364 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: Introduction to Power Electronics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.5, Page number: 508"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from pylab import *\n",
+      "import numpy as np\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "w=2*pi*60                               #frequency of voltage(Hz)\n",
+      "R=10                                    #ohm\n",
+      "C=0.01                                  #F\n",
+      "Vo=120*sqrt(2)                          #maximum voltage(V)\n",
+      "Nmax=800\n",
+      "tau=R*C                                 #time constant(s)\n",
+      "\n",
+      "#Calculations:\n",
+      "# diode = 1 when rectifier bridge is conducting\n",
+      "\n",
+      "diode=1\n",
+      "t=[0]*801\n",
+      "vs=[0]*801\n",
+      "vrect=[0]*801\n",
+      "vR=[0]*801\n",
+      "iB=[0]*801\n",
+      "\n",
+      "t=[0]*801\n",
+      "for n in range(1,Nmax+2,1):\n",
+      "    t[n-1] = (2.5*pi/w)*(n-1)/Nmax\n",
+      "    vs[n-1]=Vo*math.cos(w*t[n-1])\n",
+      "    vrect[n-1]=abs(vs[n-1])\n",
+      "#if the rectifier bridge is ON:\n",
+      "    if diode==1:\n",
+      "        vR[n-1]=vrect[n-1]\n",
+      "        if (w*t[n-1])<=(pi/2):\n",
+      "            iB[n-1]=vR[n-1]-Vo*C*w*math.sin(w*t[n-1])\n",
+      "        elif (w*t[n-1])<=3*pi/2:\n",
+      "            iB[n-1]=vR[n-1]/R+Vo*C*w*math.sin(w*t[n-1])\n",
+      "        else:\n",
+      "            iB[n-1]=vR[n-1]/R-Vo*C*w*math.sin(w*t[n-1])\n",
+      "        if iB[n-1]<0:\n",
+      "            diode=0\n",
+      "            toff=t[n-1]\n",
+      "            Voff=vrect[n-1]\n",
+      "    else:\n",
+      "        vR[n-1]=Voff*exp(-(t[n-1]-toff/tau))\n",
+      "        iB[n-1]=0\n",
+      "        if (vrect[n-1]-vR[n-1])>0:\n",
+      "            diode=1\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "iR=(1/R)*np.array(vR)\n",
+      "plot(1000*np.array(t),vR)\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('voltage [V]')\n",
+      "xlim(0,22)\n",
+      "ylim(0,180)\n",
+      "plot(1000*np.array(t),vrect,'--')\n",
+      "grid()\n",
+      "print \"The required plots are shown below:\"\n",
+      "show()\n",
+      "plot(1000*np.array(t),iR)\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('source current [A]')\n",
+      "xlim(0 ,22)\n",
+      "ylim(-50,250)      \n",
+      "plot(1000*np.array(t),1.5*np.array(iB),'--')\n",
+      "grid()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "The required plots are shown below:"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ibCiGqVKvAU4DsP/GfszznydsKIbxuYNUw2K96m0CQpzAsLS0REFBASwsLJCf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\nOE6H/T8o9Q6nwzgdWAAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7f054c060710>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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CsWppbSLgX0cGLlysTua8ZMS8lJExL4eaQXV1NVauXIlJkyZh0qRJWL16Ncxm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\nEREB+P84YbaUtpWtuAAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2986d90>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.6, Page number: 522"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "w=2*pi*60                               #Angular freq of voltage(rad/sec)\n",
+      "Vo=230*sqrt(2)                          #volt\n",
+      "R=5.6                                   #Resistance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "Ls=[0]*101\n",
+      "tc=[0]*101\n",
+      "Idc=[0]*101\n",
+      "for n in range(1,101,1):\n",
+      "    Ls[n-1]=n*10**-3\n",
+      "    Idc[n-1]=2*Vo/(pi*R+2*w*Ls[n-1])\n",
+      "    tc[n-1]=(1/w)*acos(1-(2*Idc[n-1]*w*Ls[n-1])/Vo)\n",
+      "\n",
+      "#Results:\n",
+      "plot(1000*np.array(Ls),Idc,'g.')\n",
+      "xlabel('Commutating inductance Ls [mH]')\n",
+      "ylabel('Idc [A]')\n",
+      "title('Load current,Idc vs Commutating inductance,Ls')\n",
+      "show()\n",
+      "plot(1000*np.array(Ls),1000*np.array(tc),'g.')\n",
+      "xlabel('Commutating inductance L [mH]')\n",
+      "ylabel('tc [msec]')\n",
+      "title('Commutating Inductance,Ls vs time,tc')\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['fmod', 'sinh', 'trunc', 'tan', 'gamma', 'cosh', 'radians', 'modf', 'expm1', 'ldexp', 'linalg', 'random', 'frexp', 'ceil', 'isnan', 'copysign', 'cos', 'degrees', 'tanh', 'fabs', 'sqrt', 'hypot', 'power', 'log', 'log10', 'info', 'log1p', 'floor', 'fft', 'pi', 'exp', 'isinf', 'e', 'sin']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ExEQAQGJiIsLDw7UZRoum3oVkZmLGAWci0poGtRCKiopgZmaGNm3aALh9nYSS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\nYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2c18b90>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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AgAHFlTP4iYiqzayDv8rKGfxERNVm1qdzEhGR+amTs3pqit+0RURUe+rFiJ/f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\nAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2981fd0>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.7, Page number: 528"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "R=12.5*10**-3                               #ohm\n",
+      "L=1.2                                       #H\n",
+      "Vo=15                                       #volt\n",
+      "w=120*pi                                    #angular freq(Hz)\n",
+      "Idc=35                                      #DC current(A)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "theta=[0]*1301\n",
+      "t=[0]*1301\n",
+      "vL=[0]*1301\n",
+      "vs=[0]*1301\n",
+      "\n",
+      "Vdc_a=R*Idc                                 #Dc voltage(V)\n",
+      "P=Vdc_a*Idc                                 #Power\n",
+      "alpha_da = acos(pi*R*Idc/(2*Vo)) ;    #delay angle\n",
+      "for n in range(1,1301,1):                 #loop for calculating load voltage\n",
+      "    theta[n-1]=2*pi*(n-1)/1000\n",
+      "    t[n-1]=theta[n-1]/w\n",
+      "    vs[n-1]=Vo*sin(theta[n-1])\n",
+      "    if theta[n-1]<alpha_da:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif (theta[n-1]<pi+alpha_da):\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif theta[n-1]<2*pi+alpha_da:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<3*pi+alpha_da:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<4*pi+alpha_da:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    else:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "\n",
+      "figure(1)\n",
+      "plot(1000*np.array(t),vL,'g.')\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('Load voltage [V]')\n",
+      "grid()\n",
+      "show()\n",
+      "\n",
+      "\n",
+      "#part(b):\n",
+      "alpha_db=0.9*pi                             #delay angle\n",
+      "Vdc_b=(2*Vo/pi)*cos(alpha_db)          #new dc voltage(V)\n",
+      "tau=L/R                                     #time constant(s)\n",
+      "imo=Idc                                     #Initial curent(A)\n",
+      "tzero=-tau*log((-Vdc_b/R)/(imo-Vdc_b/R))\n",
+      "for n in range(1,1301,1):\n",
+      "    theta[n-1]=2*pi*(n-1)/1000\n",
+      "    t[n-1]=theta[n-1]/w\n",
+      "    vs[n-1]=Vo*sin(theta[n-1])\n",
+      "    if theta< alpha_db:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif (theta[n-1]<pi+alpha_db):\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<2*pi+alpha_db:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif theta[n-1]<3*pi+alpha_db:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<4*pi+alpha_db:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    else:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "\n",
+      "#Results:\n",
+      "figure(2)\n",
+      "plot (1000*np.array(t), vL,'g.')\n",
+      "xlabel('time [msec] ')\n",
+      "ylabel('Load voltage [V]')\n",
+      "print \"part (a):\"\n",
+      "print \"\\n Vdc_a=\",round(1000*Vdc_a,2),\"mV\"\n",
+      "print \"\\n Power=\",round(P),\"W\"        \n",
+      "print \"\\n alpha_d=\",round((180/pi)*alpha_da,1),\"degrees\"\n",
+      "print \"\\n part (b):\"\n",
+      "print \"\\n alpha_d=\",round((180/pi)*alpha_db,1),\"degrees\"   \n",
+      "print \"\\n Vdc_b=\",round(Vdc_b,1),\"V\"\n",
+      "print \"\\n Current will reach zero at\",round(tzero,1),\"sec\"\n",
+      "grid()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['fmod', 'cosh', 'sinh', 'trunc', 'tan', 'gamma', 'degrees', 'radians', 'sin', 'expm1', 'ldexp', 'isnan', 'frexp', 'ceil', 'copysign', 'cos', 'tanh', 'fabs', 'sqrt', 'hypot', 'log', 'log10', 'pi', 'log1p', 'floor', 'modf', 'exp', 'isinf', 'e']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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0J06fPo0NGzaYfZ0gCHjppZcQGxsLANDr9UhNTTXOh279l0NGRgb07+hx9Z9X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\nhNj0/wELMsFZawmOEgAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x1e16bd0>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part (a):\n",
+        "\n",
+        " Vdc_a= 437.5 mV\n",
+        "\n",
+        " Power= 15.0 W\n",
+        "\n",
+        " alpha_d= 87.4 degrees\n",
+        "\n",
+        " part (b):\n",
+        "\n",
+        " alpha_d= 162.0 degrees\n",
+        "\n",
+        " Vdc_b= -9.1 V\n",
+        "\n",
+        " Current will reach zero at 4.5 sec\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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jOp0OL730EsrLy1FeXm4XIIg20Cwn/8gpzEFmQSayP8722XRVmuWkLi6DxIYN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\n
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter3-checkpoint.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter3-checkpoint.ipynb
new file mode 100755
index 00000000..22fca9a1
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter3-checkpoint.ipynb
@@ -0,0 +1,460 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Electromechanical-Energy-Conversion-Principles "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page number: 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "I=10                            #current in the coil(A)\n",
+      "Bo=0.02                         #magnetic field (T)\n",
+      "R=0.05                          #radius of the rotor(m)\n",
+      "l=0.3                           #rotor length(m)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "q=symbols('q')                  #Direction of torque\n",
+      "F1=-2*I*l*Bo*sin(q)             #Force on the coil(N)\n",
+      "T=F1*R                          #Torque scting in theta direction(Nm)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Force per unit length:\",T,\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force per unit length: -0.006*sin(q) Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page number: 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration\n",
+      "N=1000                   #No of winding turns\n",
+      "g=2                      #Air gap width(mm)\n",
+      "d=0.15                  #Magnetic core width,d (m)\n",
+      "l=0.1                    #thickness of core(0.1)\n",
+      "x,d=symbols('x d')       #where h is height of plunger(m) \n",
+      "                         #Lx is inductance as a function of x(H)\n",
+      "i=10                     #Current in the winding(A)\n",
+      "uo=4*3.14*10**-7           #permeability of free space(H/m)\n",
+      "\n",
+      "#Calculations:\n",
+      "Lx=(uo*N**2*l*d)/(2*g*10**-3)*(1-x/d)\n",
+      "Wfld=(1./2)*Lx*i**2\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The magnetic energy stored, Wfld:\",\"236*(1-x/d) J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnetic energy stored, Wfld: 236*(1-x/d) J\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page number: 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "import numpy as np\n",
+      "\n",
+      "#Variable declaration:\n",
+      "xdata=[0.0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0]     #(cm)\n",
+      "Ldata=[2.8, 2.26, 1.78, 1.52, 1.34, 1.26, 1.20, 1.16, 1.13, 1.11, 1.10]   #(mH)\n",
+      "I = 0.75                                                #(A)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "x=0.01*np.array(xdata)\n",
+      "L=0.001*np.array(Ldata)\n",
+      "length=len(x)\n",
+      "xmax=x[length-1]\n",
+      "a=polyfit(x,L,4)\n",
+      "xfit=[0]*102\n",
+      "Lfit=[0]*102\n",
+      "for n in range(1,102,1):\n",
+      "    xfit[n-1]=xmax*(n-1)/100\n",
+      "    Lfit[n-1]=a[0]*xfit[n-1]**4+a[1]*xfit[n-1]**3+a[2]*xfit[n-1]**2+a[3]*xfit[n-1]+a[4]\n",
+      "\n",
+      "#Plot the data and then the fit to compare (convert xfit to cm and Lfit to mH)\n",
+      "plot(xdata,Ldata,'o')\n",
+      "plot(100*np.array(xfit),1000*np.array(Lfit),'g.')\n",
+      "xlabel('x [cm] ')\n",
+      "ylabel('L [mH] ')\n",
+      "title('Inductance,L vs length,l')\n",
+      "grid()\n",
+      "print \"The required plots are shown below:\"\n",
+      "show()\n",
+      "\n",
+      "#set current to 0.75 A\n",
+      "I=0.75\n",
+      "F=[0]*102\n",
+      "for n in range(1,102,1):\n",
+      "    xfit[n-1]=0.002+0.016*(n-1)/100\n",
+      "    F[n-1]=4*a[0]*xfit[n-1]**3+3*a[1]*xfit[n-1]**2+2*a[2]*xfit[n-1]**1+a[3]\n",
+      "    F[n-1]=(I**2/2)*F[n-1]\n",
+      "plot(100*np.array(xfit),F,'b.')\n",
+      "xlabel('x  [cm]')\n",
+      "ylabel('Force [N]')\n",
+      "title('Force, F vs length,l')\n",
+      "grid()\n",
+      "\n",
+      "#Results:\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "The required plots are shown below:"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['Polygon', 'poly', 'sign', 'flatten', 'conjugate', 'diff', 'tan', 'Circle', 'roots', 'plot', 'eye', 'trace', 'floor', 'diag', 'invert', 'nan', 'sqrt', 'source', 'add', 'zeros', 'take', 'var', 'pi', 'plotting', 'product', 'seterr', 'power', 'multinomial', 'transpose', 'test', 'beta', 'ones', 'sinh', 'vectorize', 'cosh', 'trunc', 'cos', 'prod', 'tanh', 'mod', 'det', 'sin', 'binomial', 'solve', 'log', 'exp', 'reshape', 'gamma', 'interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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PRP/rjv3mZD0B9bu/LN39pZwVcHwahqd/geK6r0xOLrrXjn6K8EfMm8UY9Vsa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\nYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3dc9590>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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dqL29Z9qL01xEQ4fL353CCTExMXbbYmNjnXmqQ+fOnRP33nuviI2NFSkpKeL8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\noRARkSSYUIiISBL/D5upbph/P2HSAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x48a36d0>"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page number: 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Lo=10.6*10**-3                          #Initial inductance(H)\n",
+      "L2=2.7*10**-3                           #H\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "theta,i=symbols('theta i')\n",
+      "L=Lo+L2*cos(2*theta)\n",
+      "i=2                                     #Coil current,A\n",
+      "def T(theta):\n",
+      "    return i**2*diff(L,theta)/2\n",
+      "                                    \n",
+      "\n",
+      "#Results:\n",
+      "print \"Torque,Tfld =\",T(theta),\" N.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Torque,Tfld = -0.0108*sin(2*theta)  N.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page number: 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "r1=2.5*10**-2                            #radius of rotor(m)\n",
+      "h=1.8*10**-2                             #Axial length(m)\n",
+      "g=3*10**-3                               #Air gap length(m)\n",
+      "Bag=1.65                                #Magnetic field(T)\n",
+      "uo=4*pi*10**-7                          #permeability of free space(H/m)\n",
+      "\n",
+      "#Calculations:\n",
+      "H=Bag/uo\n",
+      "Ni=2*g*H\n",
+      "T=uo*(Ni)**2*h*(r1+0.5*g)/(4*g)\n",
+      "\n",
+      "#Results:\n",
+      "print \"The maximum torque:\", round(T,2),\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum torque: 3.1 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.7, Page number: 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from matplotlib import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "i1=0.8\n",
+      "i2=0.01\n",
+      "\n",
+      "\n",
+      "#Calculations & Results:\n",
+      "def df(f,x,h=0.1e-10):\n",
+      "    return ( f(x+h/2) - f(x-h/2) )/h\n",
+      "\n",
+      "\n",
+      "\n",
+      "def l11(x):\n",
+      "    return (3+cos(2*x))/1000.0\n",
+      "\n",
+      "def l12(x):\n",
+      "    return (0.3*cos(x))\n",
+      "\n",
+      "def l22(x):\n",
+      "    return (30+10*cos(2*x))\n",
+      "\n",
+      "def g(x):\n",
+      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x) + (i1*i2)*df(l12,x)\n",
+      "\n",
+      "def r(x):\n",
+      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x)\n",
+      "def s(x):\n",
+      "    return (i1*i2)*df(l12,x)\n",
+      "\n",
+      "x=linspace(-pi,pi,100000)\n",
+      "\n",
+      "\n",
+      "plot(x,r(x))\n",
+      "plot(x,s(x))\n",
+      "plot(x,g(x))\n",
+      "grid()\n",
+      "annotate(\"Total torque\",xy=(-0.5,0.003))\n",
+      "annotate(\"Reluctance torque\",xy=(-2,-0.0015))\n",
+      "annotate(\"Mutual Interaction torque\",xy=(1.6,-0.0026))\n",
+      "xlabel(\"Theta [radians]\")\n",
+      "ylabel(\"Torque [N.m]\")\n",
+      "xlim(-pi,pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print \"Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)\"\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['vectorize', 'prod', 'plotting', 'Circle', 'diag', 'sinh', 'trunc', 'plot', 'eye', 'det', 'tan', 'product', 'gamma', 'roots', 'sin', 'zeros', 'cosh', 'interactive', 'conjugate', 'take', 'trace', 'beta', 'exp', 'ones', 'multinomial', 'cos', 'transpose', 'solve', 'diff', 'invert', 'pi', 'tanh', 'Polygon', 'reshape', 'sqrt', 'floor', 'source', 'add', 'poly', 'mod', 'sign', 'power', 'binomial', 'log', 'var', 'seterr', 'flatten', 'nan', 'test']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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0ZnpWOnrt7QWbujZY03ONMnaoQsGmm969A/z8NAhTtWgRi3W1bp3OZRbRH/fv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\ngQFWOK/Ic16Ivyl1FOgHGBYWhrCwMOV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Z/fnnn0REtHnzZhozZgwREbm7u9MPP/yQp/7T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\ns/T/jVWB3070t1UAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3dbe950>"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.9, Page number: 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "W=4.0*10**-2                            #width of plunger lower arm(m)\n",
+      "W1=4.5*10**-2                           #width of plunger upper arm(m)\n",
+      "D=3.5*10**-2                            #depth of plunger (m)\n",
+      "d=8*10**-3                              #length of magnet(m)\n",
+      "go=1*10**-3                             #air gap length(m)\n",
+      "uo=4*pi*10**-7                          #Permeability of free space(A.turns/m)\n",
+      "ur=1.06*uo                              #Relativity permeability\n",
+      "Hc1=-940                                #Magnetising force(kA/m)\n",
+      "Bt=1.25                                 #Magnetic field induction(T)\n",
+      "N=1500                                  #No of turns\n",
+      "x=3*10**-3                              #Position of plunger(m)\n",
+      "\n",
+      "#Calculation:\n",
+      "Ni=-Hc1*d*10**3\n",
+      "Rx=x/(uo*W1*D)\n",
+      "Ro=go/(uo*W*D)\n",
+      "Rm=d/(ur*W*D)\n",
+      "f=-((Ni)**2)/(uo*W1*D*(Rx+Ro+Rm)**2)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The x-directed force:\",round(f,1),\"N\"\n",
+      "print \"Current in the excitation winding:\",round(Ni/N,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The x-directed force: -703.3 N\n",
+        "Current in the excitation winding: 5.01 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter5-checkpoint.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter5-checkpoint.ipynb
new file mode 100755
index 00000000..3f99f735
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/.ipynb_checkpoints/chapter5-checkpoint.ipynb
@@ -0,0 +1,609 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Synchronous Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page number: 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "\n",
+      "#Varaible Declaration:\n",
+      "pf=0.95                                 #Lagging power factor\n",
+      "Vl=460                                  #Terminal voltage(V)\n",
+      "I=120                                   #Terminal current(A)\n",
+      "If=47                                   #Field current(A)\n",
+      "X=1.68j                                 #Line syncchronous reactance(ohm)\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "#Choosing motor reference direction:\n",
+      "Va=Vl/math.sqrt(3)\n",
+      "theta=math.acos(0.95)\n",
+      "Ia=I*cmath.exp(-theta*1j)\n",
+      "Eaf=Va-X*Ia\n",
+      "wc=120*math.pi\n",
+      "Laf=math.sqrt(2)*abs(Eaf)/(wc*If)\n",
+      "P=3*Va*Ia*pf\n",
+      "\n",
+      "#Results:\n",
+      "print \"Generated emf:\",round(abs(Eaf),1),\"V line to line\"\n",
+      "print \"Fied to armature mutual inductance:\",round(Laf*1000,1),\"mH\"\n",
+      "print \"Three phase power:\",round(abs(P/1000),1),\"kW or\",round(abs(P)/746),\"hp\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Generated emf: 278.8 V line to line\n",
+        "Fied to armature mutual inductance: 22.3 mH\n",
+        "Three phase power: 90.8 kW or 122.0 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page number: 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath \n",
+      "from math import *\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Pin=90.6*10**3                              #Input power(kW)\n",
+      "Va=265.6                                    #Terninal voltage(V)\n",
+      "X=1.68j                                     #Synchronous reactance(ohm)\n",
+      "Laf=22.3*10**-3                             #Mutual inductance(H)\n",
+      "wc=120*pi                                   #Angular frequency(rad/sec)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ia=Pin/(3*Va)\n",
+      "Eaf=Va-X*Ia\n",
+      "delta=degrees(cmath.phase(Eaf))\n",
+      "I=sqrt(2)*Eaf/(wc*Laf)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print\"The phase angle,delta:\",round(delta,1),\"degrees\"\n",
+      "print\"Required field current:\",round(abs(I),2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The phase angle,delta: -35.7 degrees\n",
+        "Required field current: 55.04 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.3, Page number: 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Eafl=13.8*10**3                             #Open circuit voltage(V)\n",
+      "If1=318                                     #Field current(A)\n",
+      "If2=263                              #Field current after extrapolation(A)\n",
+      "wc=120*pi                            #Angular frequency(Hz)\n",
+      "\n",
+      "#Calculations:\n",
+      "Eaf=Eafl/sqrt(3)\n",
+      "La1=sqrt(2)*Eaf/(wc*If1)\n",
+      "La2=sqrt(2)*Eaf/(wc*If2)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Saturated Laf1:\",round(La1*1000,0),\"mH\"    \n",
+      "print \"Unsaturated Laf1:\",round(La2*1000,0),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Saturated Laf1: 94.0 mH\n",
+        "Unsaturated Laf1: 114.0 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page number: 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Ia=[118, 152]                     #Armature current from SC Characteristics(A)\n",
+      "If=[2.20, 2.84]                     #Field current from SC Characteristics(A)\n",
+      "Vll=220                             #Line-to-line Voltage(V)\n",
+      "V=202                               #Line-to-line air voltage(V) \n",
+      "P=45*10**3                          #Power roted to motor(W)   \n",
+      "Is_sc=1                             #per unit rated current(A)\n",
+      "\n",
+      "#Calculations:\n",
+      "Va_ag=V/sqrt(3)                 #At field current of 2.20A,at air gap,(V)\n",
+      "Ia_ag=Ia[0]\n",
+      "Xs_u=Va_ag/Ia_ag\n",
+      "Ia_rated=P/(sqrt(3)*Vll)\n",
+      "Xa_g=Va_ag/1\n",
+      "Xs_u_pu=Va_ag/Is_sc\n",
+      "Xs=Vll/(Ia[1]*sqrt(3))\n",
+      "Ia_pu=Ia[1]/Ia[0]\n",
+      "SCR=If[1]/If[0]\n",
+      "Xs=1/SCR\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"'All quantities are in per unit values'\"\n",
+      "print\"Unsaturated value of synchronous reactance:\",round(Xs_u,3),\"ohm\"\n",
+      "print \"Satureted value of synchronous reactance: \",round(Xs,3),\"ohm\"\n",
+      "print\"Short circuit ratio:\",round(SCR,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "'All quantities are in per unit values'\n",
+        "Unsaturated value of synchronous reactance: 0.988 ohm\n",
+        "Satureted value of synchronous reactance:  0.775 ohm\n",
+        "Short circuit ratio: 1.291\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.5, Page number: 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P_rated=45*10**3                        #Rated power(KV)\n",
+      "Pl=1.80*10**3                           #Short circuit load loss(W)\n",
+      "Ia_pu=1                                    #Per unit armature current\n",
+      "Ia=118                                  #rated armature current(A)\n",
+      "Ra_dc=0.0335                            #Dc resistance(ohm/phase)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pl_pu=Pl/P_rated    \n",
+      "Ra_eff1=Pl_pu/Ia_pu**2                      #in per unit basis\n",
+      "Ra_eff2=Pl/(3*(Ia)**2)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Armature resistance in per unit:\",round(Ra_eff1,3),\"per unit\" \n",
+      "print \"Armature resistance in ohms/phase:\", round(Ra_eff2,3),\"ohms/phase\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature resistance in per unit: 0.04 per unit\n",
+        "Armature resistance in ohms/phase: 0.043 ohms/phase\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.6, Page number: 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "import cmath\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Veq=1.0                                 #Externalsupply(p.u)                                    \n",
+      "Eaf=1.0                                 #Internal voltage(p.u)\n",
+      "Xeq=0.23                                #Eqv.resistance of external system(p.u)\n",
+      "Xs=1.35                                #Saturated synchronous reactance(p.u)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "P_max=Eaf*Veq/(Xs+Xeq)\n",
+      "\n",
+      "\n",
+      "#for part (b):\n",
+      "delta=[0]*500\n",
+      "Ia=[0]*500\n",
+      "Va=[0]*500\n",
+      "degree=[0]*500\n",
+      "for n in range(1,101,1):\n",
+      "    delta[n-1]=(pi/2)*(n-1)/100\n",
+      "    Ia[n-1] = (Eaf *exp(1j*delta[n-1]) - Veq)/(1j*(Xs + Xeq))\n",
+      "    Va[n-1] = abs(Veq + 1j*Xeq*Ia[n-1])\n",
+      "    degree[n-1]=180*delta[n-1]/pi\n",
+      "plot(degree,Va,'r.')\n",
+      "xlabel('Power angle,delta(degrees)')\n",
+      "ylabel('Terminal voltage(per unit)')\n",
+      "title('Terminal voltage vs. power angle for part (b)')\n",
+      "show()\n",
+      "#for part (c):\n",
+      "Vterm=1.0\n",
+      "P=[0]*500\n",
+      "deltat=[0]*500\n",
+      "Ia=[0]*500\n",
+      "Eaf=[0]*500\n",
+      "\n",
+      "for n in range(1,101,1):\n",
+      "    P[n-1]=(n-1)/100\n",
+      "    deltat[n-1]=math.asin(P[n-1]*Xeq/(Vterm*Veq))\n",
+      "    Ia[n-1]=(Vterm*exp(1j*deltat[n-1])-Veq)/(1j*Xeq)\n",
+      "    Eaf[n-1]=abs(Vterm+1j*(Xs+Xeq)*Ia[n-1])\n",
+      "plot(P,Eaf,'r.')\n",
+      "xlabel('Power [per unit]')\n",
+      "ylabel('Eaf [per unit]')\n",
+      "title('Eaf vs. power for part (c)')\n",
+      "show()\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Maximum power supplied to external system:\",round(P_max,2),\"p.u\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['prod', 'Circle', 'power', 'diag', 'sinh', 'trunc', 'binomial', 'plot', 'eye', 'det', 'tan', 'product', 'roots', 'vectorize', 'sin', 'plotting', 'zeros', 'cosh', 'conjugate', 'linalg', 'take', 'solve', 'trace', 'beta', 'draw_if_interactive', 'random', 'ones', 'transpose', 'cos', 'interactive', 'diff', 'invert', 'tanh', 'Polygon', 'reshape', 'sqrt', 'floor', 'source', 'add', 'multinomial', 'test', 'poly', 'mod', 'sign', 'fft', 'gamma', 'log', 'var', 'info', 'seterr', 'flatten', 'nan', 'pi', 'exp']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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PYffu3Wjbtq1y5M/x48cxadKk2ldIRET1ktZAePXVVxEfHw8HBwcAQKdOnbB/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\na/B03gAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3794dd0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+s2+K7OxsBAcHIzAwEOvXr2+zjFarxfjx4xEaGoopU6Z0+AsQEZF5KbaCW11d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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3965f10>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Maximum power supplied to external system: 0.63 p.u\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page number: 272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P_rated=2000*746/3                     #per phase rated power of motor(W)\n",
+      "Xsm=1.95                                #Synchronous reactance(ohm)\n",
+      "Vl=2300                                #Line to line voltage(V)\n",
+      "f=60                                   #Angular frequency(Hz)\n",
+      "p=30                                   #No. of poles\n",
+      "Xsg=2.65                               #Synchronous reactance of generator(ohm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "Vp=2300/sqrt(3)\n",
+      "Ip=P_rated/Vp\n",
+      "Eafm=sqrt(Vp**2+(Ip*Xsm)**2)\n",
+      "Pm=3*Vp*Eafm/Xsm                             #Max power delivered to motor(W)\n",
+      "ws=2*2*pi*f/p\n",
+      "Tmax=Pm/ws                                  #MAx torque of motor(Nm)\n",
+      "\n",
+      "\n",
+      "#for part (b):\n",
+      "Eafg=sqrt(Vp**2+(Ip*Xsg)**2)\n",
+      "Pm2=3*Eafm*Eafg/(Xsg+Xsm)                   #Max power delivered to motor(W)\n",
+      "Tmax2=Pm2/ws                                #Max torque(Nm)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print\"(a) Max power  :\",round(Pm/1000,0),\"kW,3-ph\"\n",
+      "print\"    Max torque :\",round(Tmax/1000,1),\"kNm\"\n",
+      "print \"(b) Max power :\", round(Pm2/1000,0),\"kW,3-ph\"\n",
+      "print \"    Max torque:\", round(Tmax2/1000,1),\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Max power  : 3096.0 kW,3-ph\n",
+        "    Max torque : 123.2 kNm\n",
+        "(b) Max power : 1639.0 kW,3-ph\n",
+        "    Max torque: 65.2 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.8, Page number: 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P=45                                    #Power rated(KVA)\n",
+      "Va=220                                  #Terminal voltage(V)\n",
+      "Pin=45                                  #Power input to the armature(KVA)\n",
+      "If=5.50                                 #field current(A)\n",
+      "Rf=35.5                                 #Field winding resistance(ohm)\n",
+      "Ra=0.0399                               #Armature dc resistance(ohm/phase)\n",
+      "Xal=0.215                               #Leakage reactance of motor(ohm)\n",
+      "pf=0.80                                 #Lagging power factor \n",
+      "Pc=1.8                                  #Core loss(kW)\n",
+      "Pw=0.91                                 #Friction & windage losses(kW)\n",
+      "Ps=0.37                                 #Stray load loss(kW)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ia=P*10**3/(sqrt(3)*Va)\n",
+      "P1=If**2*Rf/10**3                            #Loss in field winding(kW)\n",
+      "P2=3*Ia**2*Ra/10**3                          #Loss in armature(kW)\n",
+      "Pl=(Pc+Pw+Ps+P1+P2)\n",
+      "Pi=Pin*pf+P1\n",
+      "Po=Pi-Pl\n",
+      "eff=(Po/Pi)*100\n",
+      "\n",
+      "#Results:\n",
+      "print \"Efficiency of the synchronous machine:\",round(eff,1),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of the synchronous machine: 84.3 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.9, Page number: 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Xd=1                              #Direct axis synchronus reactance(p.u)\n",
+      "Xq=0.60                           #Quadrature axis synchronous reactance(p.u)\n",
+      "Va=1                              #Terminal voltage(p.u)\n",
+      "pf=0.8                              #Lagging power factor\n",
+      "Ia=0.8-1j*math.sin(math.acos(0.8))       #Line current(p.u)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "phy=-math.acos(pf)\n",
+      "E=Va+1j*Xq*Ia\n",
+      "delta=cmath.phase(E)\n",
+      "Id=abs(Ia)*math.sin(delta-phy)*cmath.exp(1j*(-pi/2+delta))\n",
+      "Iq=abs(Ia)*math.cos(delta-phy)*cmath.exp(1j*delta)\n",
+      "Eaf=Va+Xd*Id*1j+Xq*Iq*1j\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Generated voltage:\",round(abs(Eaf),2),\"p.u Volt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Generated voltage: 1.78 p.u Volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.11, Page number: 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from pylab import *\n",
+      "import cmath\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P_rated=2000*746                  #Rated power of motor(W)\n",
+      "Xs=1.95                           #Synchronous reactance(ohm/phase)\n",
+      "Xd=1.95                           #Direct axis synchronous reactance(ohm/ph)\n",
+      "Xq=1.40                         #Quadrature axis synchronous reactance(ohm/ph)\n",
+      "pf=1                            #Power factor of the machine\n",
+      "Vl=2300                         #Line to line voltage(V)\n",
+      "\n",
+      "#Calculatons:\n",
+      "Va=float(Vl/sqrt(3))              #volt\n",
+      "Ia=float(P_rated/(Va*3))          #ampere\n",
+      "E1=Va-1j*Ia*Xq                      #From phasor diagram\n",
+      "delta=cmath.phase(E1)             #power angle\n",
+      "Id=Ia*sin(abs(delta))             #direct axis current(A)\n",
+      "Eaf=abs(E1)+Id*(Xd-Xq)\n",
+      "r=symbols('r')\n",
+      "def P(r):                           #Process for finding maximum power\n",
+      "    return Eaf*Va*sin(r)/Xd + Va**2*(Xd-Xq)*sin(2*r)/(2*Xd*Xq)\n",
+      "P1=diff(P(r),r)\n",
+      "#On differentiation,\n",
+      "#P1 = 1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1)\n",
+      "l = solve(1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1),r)\n",
+      "P_max = (P(round(l[0],5)))\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Maximum mechanical power:\",math.ceil(3*P_max/10**3),\"kW,3-phase\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum mechanical power: 3236.0 kW,3-phase\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/README.txt b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/README.txt
new file mode 100755
index 00000000..4b574a6d
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/README.txt
@@ -0,0 +1,10 @@
+Contributed By: SANTOSH BARNWAL
+Course: be
+College/Institute/Organization: BIRLA INSTITUTE OF TECHNOLOGY MESRA
+Department/Designation: ELECTRICAL & ELECTRONICS
+Book Title: ELECTRIC MACHINERY
+Author: A. E. Fitzgerald, Charles Kingsley, Jr., Stephen D. Umans
+Publisher: McGraw-Hill, New York
+Year of publication: 2003
+Isbn: 0-07-112193-5
+Edition: Sixth Edition
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter1.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter1.ipynb
new file mode 100755
index 00000000..3eaa70a1
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter1.ipynb
@@ -0,0 +1,560 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:29b5ec9b20f222bcfeb31d6e80f55e5d32272a4cce1e9227a6328e93313e209d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 1:Introduction to Magnetic Circuits<h1>"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Page number: 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "#Variable declaration:\n",
+      "Ac=9                            #Cross-sectional area of the core(cm**2)\n",
+      "Ag=9                            #Cross-sectional area of the air-gap(cm**2)\n",
+      "g=0.050                         #Air-gap length(cm)            \n",
+      "lc=30                           #Mean Length of the core(cm)\n",
+      "N=500                           #No. of windings\n",
+      "ur=70000                        #Relative permeability of the core material\n",
+      "Bc=1.0                          # Magnetic Flux Density of the core(T)\n",
+      "uo=4*pi*10**-7                  #Permeability of free space\n",
+      "\n",
+      "#Calculation\n",
+      "Rc=lc*10**-2/((ur*uo*Ac)*10**-4)\n",
+      "Rg=g*10**-2/((uo*Ag)*10**-4)\n",
+      "Q=Bc*Ac*10**-4\n",
+      "i=Q*(Rc+Rg)/N\n",
+      "\n",
+      "#Results\n",
+      "print \"a.Reluctance of the core,Rc:\",round(Rc,2), \"A.turns/Wb\" \n",
+      "print \"  Reluctance of the air-gap,Rg:\", round(Rg,2), \"A.turns/Wb\"\n",
+      "print \"b.The flux, Q:\", round(Q,4), \"Wb\"\n",
+      "print \"c.The current,i:\", round(i,2), \"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a.Reluctance of the core,Rc: 3789.4 A.turns/Wb\n",
+        "  Reluctance of the air-gap,Rg: 442097.06 A.turns/Wb\n",
+        "b.The flux, Q: 0.0009 Wb\n",
+        "c.The current,i: 0.8 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.2, Page number: 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "#Variable declaration:\n",
+      "I=10                            #Current in the coil(A)\n",
+      "N=1000                          #No of turns in the rotor\n",
+      "g=1                             #Air gap length(cm)\n",
+      "Ag=2000                         #Cross-section of the air-gap(cm**2)\n",
+      "uo=4*pi*10**-7                  #Permeability of free space\n",
+      "\n",
+      "#Calculation:\n",
+      "Q=(N*I*uo*Ag*10**-4)/(2*g*10**-2)\n",
+      "Bg=round(Q,2)/(Ag*10**-4)\n",
+      "\n",
+      "#Results\n",
+      "print \"The air-gap flux, Q:\", round(Q,2), \"Wb\"\n",
+      "print \"The flux density, Bg:\", round(Bg,4), \"T\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The air-gap flux, Q: 0.13 Wb\n",
+        "The flux density, Bg: 0.65 T\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.4, Page number: 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "#Variable declaration\n",
+      "lc=0.3                                  #length of the core(cm)\n",
+      "ur1=72300                               #Relative permeablity for case(a)\n",
+      "ur2=2900                                #Relative permeablity for case(b)\n",
+      "Ac=9                                    #Cross-section of the core(cm**2)\n",
+      "Rg=4.42*10**5                           #Reluctance of the air-gap(A.turns/Wb)\n",
+      "N=500                                   #No of coil turns\n",
+      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
+      "\n",
+      "#Calculations:\n",
+      "Rt1=(lc/(ur1*uo*Ac*10**-4))+Rg\n",
+      "L1=N**2/Rt1\n",
+      "Rt2=(lc/(ur2*uo*Ac*10**-4))+Rg\n",
+      "L2=N**2/Rt2\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a)Inductance,L:\",round(L1,2),\"H\"\n",
+      "print \"(b)Inductance,L:\",round(L2,2),\"H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Inductance,L: 0.56 H\n",
+        "(b)Inductance,L: 0.47 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.5, Page number: 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from pylab import *\n",
+      "from matplotlib import *\n",
+      "from math import *\n",
+      "%matplotlib inline\n",
+      "#Variable declaration:\n",
+      "Ac=9e-4                                 #Cross-section of the core(m)\n",
+      "Ag=9e-4                                 #Cross-section of the air-gap(m)\n",
+      "g=5e-4                                  #Air-gap length(m)\n",
+      "lc=0.3                                  #Mean length of the core(m)\n",
+      "N=500                                   #No. of turns of the core(m)\n",
+      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
+      "\n",
+      "#Calculations:\n",
+      "Rg=g/(uo*Ag)                            #Reluctance of the air-gap(A.turns/Wb)\n",
+      "ur=[0]*200                              #Initialising array\n",
+      "L=[0]*200\n",
+      "\n",
+      "for n in range(1,101,1):\n",
+      "    ur[n-1]=100+(10000-100)*(n-1)/100\n",
+      "    Rc=lc/(ur[n-1]*uo*Ac)               #Reluctance of the core(A.turns/Wb)\n",
+      "    Rtot=Rg+Rc\n",
+      "    L[n-1]=(N**2)/Rtot                  #Inductance(H)\n",
+      "    \n",
+      "\n",
+      "#Results:\n",
+      "print \"The reqired plot is shown below:\"\n",
+      "plot(ur, L,'g.')\n",
+      "xlabel('Core relative permeability, ur')\n",
+      "ylabel('Inductance,L (H) ')\n",
+      "title('plot of inductance vs. relative permeability for Example 1.5.')\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The reqired plot is shown below:\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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pKQn37t27zNPTM0kkEikuXLjwkL29/YOwsLAQQggJCQkJO3Xq1Mz9+/cvVVJS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\nDQAAkNn/ASu6P4PfImgPAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2a2ba10>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.6, Page number: 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Bc=1.0                                 #Magnetic field induction in the core\n",
+      "w=377                                  #Angular frequency of magnetic field(rad/s)\n",
+      "Rc=3791.33                             #Reluctance of the core(A.turns/Wb)\n",
+      "Rg=442321.3                            #Reluctance of the air-gap(A.turns/Wb)\n",
+      "N=500                                  #No. of windings\n",
+      "i=0.80                                 #Current in the coil\n",
+      "Ac=9*10**-4                            #Cross-section of the core\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L=N**2/(Rc+Rg)\n",
+      "W=(1./2)*L*i**2\n",
+      "t = symbols('t')\n",
+      "Bc = 1.0*sin(w*t)\n",
+      "e=N*Ac*diff(Bc,t)\n",
+      "\n",
+      "#Results:\n",
+      "print \"The Inductance, L:\", round(L,2), \"H\"\n",
+      "print \"The magntic stored energy, W:\", round(W,2), \"J\"\n",
+      "print \"Induced voltage, e:\",e,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Inductance, L: 0.56 H\n",
+        "The magntic stored energy, W: 0.18 J\n",
+        "Induced voltage, e: 169.65*cos(377*t) V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.7, Page number: 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "#Variable declaration:\n",
+      "Bc=1                                    #Magnetic field in the core\n",
+      "Hc=11                                   #Magnetising force(A.turns/m)\n",
+      "lc=0.3                                  #length of the core(m)\n",
+      "N=500                                   #No of windings\n",
+      "g=0.050                                 #Air-gap length(cm)\n",
+      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "Fc=Hc*lc                                #mmf drop for the core path(A.turns)\n",
+      "Fg=Bc*g*10**-2/uo                       #mmf drop across the air gap(A.turns)\n",
+      "i=(Fc+Fg)/N\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The required current,i:\" ,round(i,2) ,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required current,i: 0.8 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.8, Page number: 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "N=200                               #No. of turns\n",
+      "Ac=4                                #Cross-section of the core(in**2)\n",
+      "w=377                               #Angular frequency of the magnetic field(rad/s)\n",
+      "Hm=36                               #Max value magnetising force(A.turns/m)\n",
+      "Pc=1.2                              #Core loss density(W/kg)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "t=symbols('t')\n",
+      "Bc=1.5*sin(w*t)\n",
+      "e=(round(N*Ac*0.94/(39.4**2),2)*diff(Bc,t))\n",
+      "Erms=275*0.707\n",
+      "lc=(6+6+8+8)/39.4                   #Mean length of the core(m)\n",
+      "I=Hm*lc/N\n",
+      "Vc=4*0.94*28                        #Core volume(m**3)\n",
+      "Wc=105.5*(2.54**3)*7.65*10**-3                  #Core weight(kg)\n",
+      "Pa=1.5*13.2                         #Watts per Kg\n",
+      "Irms=Pa/Erms                        #Current (A)\n",
+      "Pct=Pc*Wc                           #Total core loss(W)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The applied voltage,e:\", e, \"V\"\n",
+      "print \"The peak current,I:\", round(I,2), \"A\"\n",
+      "print \"The total rms current. Irms:\", round(Irms,2), \"A\"\n",
+      "print \"Total Core loss, Pct:\",round(Pct,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The applied voltage,e: 271.44*cos(377*t) V\n",
+        "The peak current,I: 0.13 A\n",
+        "The total rms current. Irms: 0.1 A\n",
+        "Total Core loss, Pct: 15.87 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.9, Page number: 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "from math import *\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "g=0.2                               #air-gap length(cm)\n",
+      "lm=1.0                              #length of magnetic section(cm)\n",
+      "Am=4                                #Cross-section of the core(cm**2)\n",
+      "Ag=4                                #Cross-section of the air-gap(cm**2)\n",
+      "\n",
+      "#Constants used:\n",
+      "uo=4*pi*10**-7                      #Permeability of free space(H/m)\n",
+      "\n",
+      "#Calculations:\n",
+      "Hm=symbols('Hm')\n",
+      "def Bg(Hm):\n",
+      "    return -uo*Ag*lm*Hm/(Am*g)             \n",
+      "\n",
+      "Hm1=-49*10**3                    #Coercivity of ALNICO 5 (A/m)\n",
+      "Hm2=-6                          #Coercivity of M-5 electrical steel (A/m)                     \n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Flux Density of air gap:\", round(Bg(Hm1),2),\"T\"\n",
+      "print \"\\nFlux Density of air gap:\", round(Bg(Hm2*10**4),2),\"gauss\"\n",
+      "print \"\\nwhere value of Hm for different material.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flux Density of air gap: 0.31 T\n",
+        "\n",
+        "Flux Density of air gap: 0.38 gauss\n",
+        "\n",
+        "where value of Hm for different material.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.10, Page number: 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Ag=2                                #Cross-section of air-gap(cm**2) \n",
+      "Bg=0.8                              #Air-gap flux density(t)\n",
+      "Bm=1.0                              #Core-flux density(T)\n",
+      "Hm=-40                              #Magnetising force in the core(kA/m)\n",
+      "uo=4*pi*10**-7                      #permeability of free space(H/m)\n",
+      "g=0.2                               #Air-gap length(cm)\n",
+      "\n",
+      "#Calculations:\n",
+      "Am=Ag*Bg/Bm\n",
+      "lm=-g*Bg/(Hm*uo*10**3)\n",
+      "Vm=Am*lm\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The minimum magnet volume,Vm:\",round(Vm,2),\"cm**3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum magnet volume,Vm: 5.09 cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.11, Page number: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "from math import *\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Am = 2                    #magnetic material cros-section(cm^2)\n",
+      "g=0.2                     #air gap length(cm)\n",
+      "uo=4*pi*10**-7            #permeability of free space(H/m)\n",
+      "N=100                     #No. of windings\n",
+      "\n",
+      "#Calculations and results:\n",
+      "#for part (a)\n",
+      "Bma = 1.0                 #Tesla\n",
+      "Hma = - 4                 #kA/m\n",
+      "Ag1 = 2                    #cm**2\n",
+      "Ag2 = 4                   #cm**2\n",
+      "\n",
+      "lm=g*(Am/Ag1)*(Bma/(-uo*Hma*10**4))\n",
+      "print \"(a) The Requied magnet length = \",round(lm,2),\"cm\"\n",
+      "\n",
+      "\n",
+      "#for part (b):\n",
+      "i,Hm=symbols('i Hm')\n",
+      "Bm=-uo*(Ag1/Am)*(lm/g)*Hm+(uo*N/g)*(Ag1/Am)*i\n",
+      "H_max=200               #kA/m\n",
+      "B_max=2.1               #Tesla\n",
+      "i_max=(B_max+2.50*10**-5*H_max)/(6.28*10**-2)\n",
+      "\n",
+      "print \"(b) Thus with the air-gap area set to 2 cm^2,\"\n",
+      "print \"    increasing the current to i_max = 45.2 A and then reducing\"\n",
+      "print \"    it to zero will achieve the desired magnetization.\"\n",
+      "\n",
+      "#for part (c):\n",
+      "Bm1=1.00               #Tesla\n",
+      "Bm2=1.08               #Tesla\n",
+      "Bg1=(Am/Ag1)*Bm1\n",
+      "Bg2=(Am/Ag2)*Bm2\n",
+      "print \"(c) The flux densities when plunger moves at two extremes are:\"\n",
+      "print \"    Bg1 =\",Bg1,\"T and Bg2 =\",Bg2,\"T\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The Requied magnet length =  3.98 cm\n",
+        "(b) Thus with the air-gap area set to 2 cm^2,\n",
+        "    increasing the current to i_max = 45.2 A and then reducing\n",
+        "    it to zero will achieve the desired magnetization.\n",
+        "(c) The flux densities when plunger moves at two extremes are:\n",
+        "    Bg1 = 1.0 T and Bg2 = 0.54 T\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter10.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter10.ipynb
new file mode 100755
index 00000000..2bcd2ccb
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter10.ipynb
@@ -0,0 +1,538 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:603160d56b04457665b7cf5381387c06e4c52903d6f85f1c61a983bdc3c14ebc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: Introduction to Power Electronics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.5, Page number: 508"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from pylab import *\n",
+      "import numpy as np\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "w=2*pi*60                               #frequency of voltage(Hz)\n",
+      "R=10                                    #ohm\n",
+      "C=0.01                                  #F\n",
+      "Vo=120*sqrt(2)                          #maximum voltage(V)\n",
+      "Nmax=800\n",
+      "tau=R*C                                 #time constant(s)\n",
+      "\n",
+      "#Calculations:\n",
+      "# diode = 1 when rectifier bridge is conducting\n",
+      "\n",
+      "diode=1\n",
+      "t=[0]*801\n",
+      "vs=[0]*801\n",
+      "vrect=[0]*801\n",
+      "vR=[0]*801\n",
+      "iB=[0]*801\n",
+      "\n",
+      "t=[0]*801\n",
+      "for n in range(1,Nmax+2,1):\n",
+      "    t[n-1] = (2.5*pi/w)*(n-1)/Nmax\n",
+      "    vs[n-1]=Vo*math.cos(w*t[n-1])\n",
+      "    vrect[n-1]=abs(vs[n-1])\n",
+      "#if the rectifier bridge is ON:\n",
+      "    if diode==1:\n",
+      "        vR[n-1]=vrect[n-1]\n",
+      "        if (w*t[n-1])<=(pi/2):\n",
+      "            iB[n-1]=vR[n-1]-Vo*C*w*math.sin(w*t[n-1])\n",
+      "        elif (w*t[n-1])<=3*pi/2:\n",
+      "            iB[n-1]=vR[n-1]/R+Vo*C*w*math.sin(w*t[n-1])\n",
+      "        else:\n",
+      "            iB[n-1]=vR[n-1]/R-Vo*C*w*math.sin(w*t[n-1])\n",
+      "        if iB[n-1]<0:\n",
+      "            diode=0\n",
+      "            toff=t[n-1]\n",
+      "            Voff=vrect[n-1]\n",
+      "    else:\n",
+      "        vR[n-1]=Voff*exp(-(t[n-1]-toff/tau))\n",
+      "        iB[n-1]=0\n",
+      "        if (vrect[n-1]-vR[n-1])>0:\n",
+      "            diode=1\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "iR=(1/R)*np.array(vR)\n",
+      "plot(1000*np.array(t),vR)\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('voltage [V]')\n",
+      "xlim(0,22)\n",
+      "ylim(0,180)\n",
+      "plot(1000*np.array(t),vrect,'--')\n",
+      "grid()\n",
+      "print \"The required plots are shown below:\"\n",
+      "show()\n",
+      "plot(1000*np.array(t),iR)\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('source current [A]')\n",
+      "xlim(0 ,22)\n",
+      "ylim(-50,250)      \n",
+      "plot(1000*np.array(t),1.5*np.array(iB),'--')\n",
+      "grid()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required plots are shown below:\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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DAABNDU1McpmEPensHQ/k6valsCYwderUXWfPnh326uu5ublW58+fD+zevftD6WsZGRmO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\n1BhtAhRFUWqMNgGKoig1RpsARVGUGvt/5i5yreN7pjkAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x25c1f10>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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IACiKYs2SmJj4Yr9+/W44OztnRkZGrmA6HjYsvXr1ynJzc7sql8vT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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3529b90>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.6, Page number: 522"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "w=2*pi*60                               #Angular freq of voltage(rad/sec)\n",
+      "Vo=230*sqrt(2)                          #volt\n",
+      "R=5.6                                   #Resistance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "Ls=[0]*101\n",
+      "tc=[0]*101\n",
+      "Idc=[0]*101\n",
+      "for n in range(1,101,1):\n",
+      "    Ls[n-1]=n*10**-3\n",
+      "    Idc[n-1]=2*Vo/(pi*R+2*w*Ls[n-1])\n",
+      "    tc[n-1]=(1/w)*acos(1-(2*Idc[n-1]*w*Ls[n-1])/Vo)\n",
+      "\n",
+      "#Results:\n",
+      "plot(1000*np.array(Ls),Idc,'g.')\n",
+      "xlabel('Commutating inductance Ls [mH]')\n",
+      "ylabel('Idc [A]')\n",
+      "title('Load current,Idc vs Commutating inductance,Ls')\n",
+      "show()\n",
+      "plot(1000*np.array(Ls),1000*np.array(tc),'g.')\n",
+      "xlabel('Commutating inductance L [mH]')\n",
+      "ylabel('tc [msec]')\n",
+      "title('Commutating Inductance,Ls vs time,tc')\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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YmBReuHBhxLsE8a+gOmhCkIYhqQDQnuR+L6PGxkb1uLi4MS1tZPz48SfeJYDO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\npcyJEyce7tu37592dnYZ165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IhUKOn5/fyTt37ji0Zbnw9tCpDJ1WTk6OVWBgYIxEIlHT\n0tKqk24yg9bT1dWt2rVrV1hlZaXem16L0JyLFy8O/fTTT380MTEpbIvyoG3gDAEAAAgh6EMAAICX\nkBAAAIAQgoQAAAAvISEAAAAhBAkBAABeQkIAAABCCCH/D1IYMYHMZAvtAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2ce0f10>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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rNrSlYlmi19lH0AcA+CeJAn90dPRHXC63mN7ncrnF586dGye7ZgEAgKxIdOdu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\ng8APAMAwCPwAAAyDwA8AwDD/DxR4tH4eP493AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x33b4a90>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.7, Page number: 528"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "R=12.5*10**-3                               #ohm\n",
+      "L=1.2                                       #H\n",
+      "Vo=15                                       #volt\n",
+      "w=120*pi                                    #angular freq(Hz)\n",
+      "Idc=35                                      #DC current(A)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "theta=[0]*1301\n",
+      "t=[0]*1301\n",
+      "vL=[0]*1301\n",
+      "vs=[0]*1301\n",
+      "\n",
+      "Vdc_a=R*Idc                                 #Dc voltage(V)\n",
+      "P=Vdc_a*Idc                                 #Power\n",
+      "alpha_da = acos(pi*R*Idc/(2*Vo)) ;    #delay angle\n",
+      "for n in range(1,1301,1):                 #loop for calculating load voltage\n",
+      "    theta[n-1]=2*pi*(n-1)/1000\n",
+      "    t[n-1]=theta[n-1]/w\n",
+      "    vs[n-1]=Vo*sin(theta[n-1])\n",
+      "    if theta[n-1]<alpha_da:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif (theta[n-1]<pi+alpha_da):\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif theta[n-1]<2*pi+alpha_da:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<3*pi+alpha_da:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<4*pi+alpha_da:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    else:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "\n",
+      "figure(1)\n",
+      "plot(1000*np.array(t),vL,'g.')\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('Load voltage [V]')\n",
+      "grid()\n",
+      "show()\n",
+      "\n",
+      "\n",
+      "#part(b):\n",
+      "alpha_db=0.9*pi                             #delay angle\n",
+      "Vdc_b=(2*Vo/pi)*cos(alpha_db)          #new dc voltage(V)\n",
+      "tau=L/R                                     #time constant(s)\n",
+      "imo=Idc                                     #Initial curent(A)\n",
+      "tzero=-tau*log((-Vdc_b/R)/(imo-Vdc_b/R))\n",
+      "for n in range(1,1301,1):\n",
+      "    theta[n-1]=2*pi*(n-1)/1000\n",
+      "    t[n-1]=theta[n-1]/w\n",
+      "    vs[n-1]=Vo*sin(theta[n-1])\n",
+      "    if theta< alpha_db:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif (theta[n-1]<pi+alpha_db):\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<2*pi+alpha_db:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    elif theta[n-1]<3*pi+alpha_db:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "    elif theta[n-1]<4*pi+alpha_db:\n",
+      "        vL[n-1]=-vs[n-1]\n",
+      "    else:\n",
+      "        vL[n-1]=vs[n-1]\n",
+      "\n",
+      "#Results:\n",
+      "figure(2)\n",
+      "plot (1000*np.array(t), vL,'g.')\n",
+      "xlabel('time [msec] ')\n",
+      "ylabel('Load voltage [V]')\n",
+      "print \"part (a):\"\n",
+      "print \"\\n Vdc_a=\",round(1000*Vdc_a,2),\"mV\"\n",
+      "print \"\\n Power=\",round(P),\"W\"        \n",
+      "print \"\\n alpha_d=\",round((180/pi)*alpha_da,1),\"degrees\"\n",
+      "print \"\\n part (b):\"\n",
+      "print \"\\n alpha_d=\",round((180/pi)*alpha_db,1),\"degrees\"   \n",
+      "print \"\\n Vdc_b=\",round(Vdc_b,1),\"V\"\n",
+      "print \"\\n Current will reach zero at\",round(tzero,1),\"sec\"\n",
+      "grid()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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LPuhJW/DhLpBJ2C8sZzJJ1NbWypYuXfrRsGHD8ocNG5a/fPny9Xfv3u1j6ve4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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2594cd0>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part (a):\n",
+        "\n",
+        " Vdc_a= 437.5 mV\n",
+        "\n",
+        " Power= 15.0 W\n",
+        "\n",
+        " alpha_d= 87.4 degrees\n",
+        "\n",
+        " part (b):\n",
+        "\n",
+        " alpha_d= 162.0 degrees\n",
+        "\n",
+        " Vdc_b= -9.1 V\n",
+        "\n",
+        " Current will reach zero at 4.5 sec\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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/wb59+xYzUTkkHtQEKFeXeUA9kav6kshVfZEw4DwJxAm4xwS/4R4T3Gez/SRa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\nYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x37f8250>"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.8, Page number: 533"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration:\n",
+      "f=60                                #Hz\n",
+      "Vrms=35                             #rms voltage of waveform\n",
+      "Ra=3.5                              #Armature resistance(ohm)\n",
+      "La=0.175                            #H\n",
+      "no=8000                             #No load speed(r/min)\n",
+      "Va=50                               #armature voltage(V)\n",
+      "\n",
+      "#Calculations:\n",
+      "Edc,alphad=symbols('Edc alphad')\n",
+      "Vdc=Edc                                 #at no load, Vdc=Edc\n",
+      "Edc=round(float(2*sqrt(2)*(Vrms/math.pi)),2)*cos(alphad)\n",
+      "n=Edc*float(no/50)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Speed at no-load =\",n,\" r/min      (where 0 <= alphad <= pi/2)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed at no-load = 5041.6*cos(alphad)  r/min      (where 0 <= alphad <= pi/2)\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.9, Page number: 537"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Vll_rms=460                                 #rms voltage,line-to-line(V)\n",
+      "R=68                                        #resistance of load\n",
+      "Im=2.5                                      #magnet current(A)\n",
+      "\n",
+      "#Calculations:\n",
+      "Vdc_max=3*sqrt(2)*Vll_rms/pi\n",
+      "Idc_max=Vdc_max/R\n",
+      "Vdc=Im*R\n",
+      "alpha=acos(pi*Vdc/(3*sqrt(3)*Vll_rms))\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Maximum dc voltage:\",round(Vdc_max),\"V\"\n",
+      "print \"\\n  Maximum dc current:\",round(Idc_max,1),\"V\"\n",
+      "print \"\\n(b) Delay angle alpha:\",round(math.degrees(round(alpha,1)),1),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Maximum dc voltage: 621.0 V\n",
+        "\n",
+        "  Maximum dc current: 9.1 V\n",
+        "\n",
+        "(b) Delay angle alpha: 74.5 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.10, Page number: 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "T=20*10**-3                                 #Time period(sec) \n",
+      "p=4                                         #no. of poles\n",
+      "delta=0.44                                  #ON- time fraction\n",
+      "Vo=125                                      #DC supply voltage(V)\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "fc=1/T\n",
+      "ns=(120*fc/p)\n",
+      "Va_peak=(4*Vo*sin(delta*pi))/pi\n",
+      "Vll_rms=sqrt(3/2)*Va_peak\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Frequency:\",fc,\"Hz\"\n",
+      "print \"\\n    Synchronous speed:\",ns,\"r/min\"\n",
+      "print \"\\n(b) Rms amplitude of line-to-line voltage:\",round(Vll_rms,0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Frequency: 50.0 Hz\n",
+        "\n",
+        "    Synchronous speed: 1500.0 r/min\n",
+        "\n",
+        "(b) Rms amplitude of line-to-line voltage: 191.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.13, Page number: 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Vo=48                               #Load voltage(V)\n",
+      "R=3.7                               #Resistance of load(ohm)\n",
+      "L=.32                               #Inductance of laad(H)\n",
+      "D=0.8                               #Duty cycle\n",
+      "f=1000                              #Hz\n",
+      "\n",
+      "#Calculations:\n",
+      "iL_avg=(2*D-1)*Vo/R\n",
+      "T=1/f\n",
+      "tau=L/R\n",
+      "iL_min=((-Vo/R)*(1-2*exp(-T*(1-D)/tau)+exp(-T/tau)))/(1-exp(-T/tau))\n",
+      "iL_max=(Vo/R)*(1-2*exp(-D*T/tau)+exp(-T/tau))/(1-exp(-T/tau))\n",
+      "\n",
+      "#since T/tau << 1, so using 10.32 in e.g. given.\n",
+      "del_iL=(2*Vo)*T*D*(1-D)/(R*tau)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Avg load current:\",round(iL_avg,2),\"A\"\n",
+      "print \"Minimum load current:\",round(iL_min,2),\"A\"\n",
+      "print \"Maximum load current\",round(iL_max,2),\"A\"\n",
+      "print \"Current ripple:\",round(del_iL,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Avg load current: 7.78 A\n",
+        "Minimum load current: 7.76 A\n",
+        "Maximum load current 7.81 A\n",
+        "Current ripple: 0.05 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter11.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter11.ipynb
new file mode 100755
index 00000000..308afbf5
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter11.ipynb
@@ -0,0 +1,975 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:4aa0be3787a5262658b58d5130a82adad92353bb1944b3902dc15e9d5045fa75"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11: Speed and Torque Control"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.1, Page number: 561"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Vdc=240                             #DC supply (V)\n",
+      "D=0.75                              #Duty cycle\n",
+      "Rf=187                              #field resistance(ohm)\n",
+      "Lf=4.2                              #field winding inductance(H)\n",
+      "T=1                                 #switching period(msec)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "If=D*(Vdc/Rf)\n",
+      "tau=Lf/Rf                           #time constant(msec)\n",
+      "del_if=(2*Vdc/Rf)*(T/tau)*D*(1-D)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Avg field current:\",round(If,2),\"A\"\n",
+      "print \"Magnitude of currnet ripple:\",round(del_if,1),\"mA\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Avg field current: 0.96 A\n",
+        "Magnitude of currnet ripple: 21.4 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page number: 563"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "n1=1800                                 #r/min\n",
+      "n2=3600                                 #r/min\n",
+      "Va=240                                  #terminal voltage(V)\n",
+      "Ifo=0.34                                #No-load field current(A)\n",
+      "Ra=0.05                                 #Armature resistance(ohm)\n",
+      "Rsh=187                                 #Shunt field resistance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "wm=symbols('wm')\n",
+      "wm1=float(2*pi*n1/60)\n",
+      "wm2=float(2*pi*n2/60)\n",
+      "def Pload(wm):\n",
+      "    return (22.4*(120*pi)**-3)*(wm)**3\n",
+      "\n",
+      "T1=Pload(wm1)*1000/wm1\n",
+      "T2=Pload(wm2)*1000/wm2\n",
+      "\n",
+      "Kf=Va/(Ifo*wm2)\n",
+      "def If(T,wm):\n",
+      "    return (Va/(2*Kf*wm))*(1+sqrt(1-(4*wm*T*Ra)/Va**2))\n",
+      "\n",
+      "Rf1tot=round(Va/float(If(T1,wm1)))\n",
+      "Rf2tot=round(Va/float(If(T2,wm2)))\n",
+      "Rrh1=Rf1tot-Rsh\n",
+      "Rrh2=Rf2tot-Rsh\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"----------------------------------------------------------------\"\n",
+      "print \"r/min    Tload[N.m]      If[A]      R(f)tot[ohm]   Rrheostat[ohm]\"\n",
+      "print \"----------------------------------------------------------------\"\n",
+      "print n1,\"\\t \",round(float(T1),1),\"\\t\\t \",round(float(If(T1,wm1)),3),\"\\t\",Rf1tot,\"\\t      \",Rrh1\n",
+      "print n2,\"\\t \",round(float(T2),1),\"\\t\\t \",round(float(If(T2,wm2)),3),\"\\t\",Rf2tot,\"\\t      \",Rrh2\n",
+      "print \"----------------------------------------------------------------\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "----------------------------------------------------------------\n",
+        "r/min    Tload[N.m]      If[A]      R(f)tot[ohm]   Rrheostat[ohm]\n",
+        "----------------------------------------------------------------\n",
+        "1800 \t  14.9 \t\t  0.678 \t354.0 \t       167.0\n",
+        "3600 \t  59.4 \t\t  0.333 \t720.0 \t       533.0\n",
+        "----------------------------------------------------------------\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.3, Page number: 567"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from math import *\n",
+      "from pylab import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Rf=109                                  #Field resistance(ohm)\n",
+      "Vf=300                                  #Rated field voltage(V)\n",
+      "Ra=0.084                                #Armature resistance(ohm)\n",
+      "Kf=0.694                                #Geometric constant(A.rad/sec)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "If=Vf/Rf                                #Resulting field current(A)\n",
+      "w_rated=2500*(pi/30)                    #Rated speed(rad/sec)\n",
+      "P_rated=100*746                         #Watts\n",
+      "T_rated=P_rated/w_rated                 #Nm\n",
+      "Va=[0]*102\n",
+      "NoLoadRPM=[0]*102\n",
+      "FullLoadRPM=[0]*102                      \n",
+      "for n in range(1,102,1):\n",
+      "    Va[n-1]=250*(1+(n-1)/100)\n",
+      "    T=0                                     #Zero torque\n",
+      "    w=(Va[n-1]-T*Ra/(Kf*If))/(Kf*If)\n",
+      "    NoLoadRPM[n-1]=w*30/pi\n",
+      "    T=T_rated\n",
+      "    w=(Va[n-1]-T*Ra/(Kf*If))/(Kf*If)\n",
+      "    FullLoadRPM[n-1]=w*30/pi\n",
+      "\n",
+      "print\"The plot is as shown:\"\n",
+      "plot(Va,NoLoadRPM)\n",
+      "plot(Va[20] ,NoLoadRPM[20] ,'r+')\n",
+      "plot (Va[50] , NoLoadRPM[50] , 'r+')\n",
+      "plot (Va[80] ,NoLoadRPM[80] , 'r+')\n",
+      "plot (Va, FullLoadRPM,'.')\n",
+      "plot (Va[20] ,FullLoadRPM[20] ,'o')\n",
+      "plot (Va[50] , FullLoadRPM[50] , ' o' )\n",
+      "plot (Va[80] , FullLoadRPM[80] ,'o' )\n",
+      "title('Speed vs Armature voltage')\n",
+      "xlabel('Armature voltage [V] ')\n",
+      "ylabel('Speed [r/min] ')\n",
+      "annotate('+ = Zero torque',xy=(270,2300))\n",
+      "annotate('o =   Full load torque',xy=(270,2100))\n",
+      "ylim(1000,2500)\n",
+      "xlim(250,500)\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The plot is as shown:\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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DvCo0sVPjozsQFWzatAlCoRBCoRCbNzf82I46Ojrlrw8dOoTp06eXv1d8lFfV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\ngg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2718810>"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.4, Page number: 571"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Rf=109                                  #Field resistance(ohm)\n",
+      "Vf=300                                  #Rated field voltage(V)\n",
+      "n1=2000                                  #rpm\n",
+      "T_rated=285                            #Rated torque(Nm)\n",
+      "n2=1975                                 #Dropped rpm\n",
+      "Kf=0.694                                #Geometric constant(A.rad/sec\n",
+      "Ra=0.084                                #Armature resistance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "If=Vf/Rf                                #Resulting field current(A)\n",
+      "wm1=2*pi*n1/60\n",
+      "w_ref=wm1\n",
+      "Vao=Kf*If*wm1\n",
+      "Ia=T_rated/(Kf*If)\n",
+      "wm2=2*pi*n2/60\n",
+      "Ea=Kf*If*wm2\n",
+      "Va=Ea+Ia*Ra\n",
+      "G=symbols('G')\n",
+      "x=solve(Vao-round(Va)+G*(w_ref-wm2),G)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Armature voltage,Vao:\",round(Va,0),\"V\"\n",
+      "print \"Multiplicative constant,G:\",float(round(x[0],2)),\"A.sec/rad\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature voltage,Vao: 408.0 V\n",
+        "Multiplicative constant,G: 3.04 A.sec/rad\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.5, Page number: 573"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Km=0.22                             #torque constant(V/(rad/sec))\n",
+      "Ra=1.03                             #ohm\n",
+      "Pl=100                            #Power load(W)\n",
+      "Va1=40                              #Armature voltage(V)\n",
+      "Va2=50                              #  \"        \"     \"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "wm1=(Va1/(2*Km))*(1+sqrt(1-(4*Pl*Ra/Va1**2)))\n",
+      "wm2=(Va2/(2*Km))*(1+sqrt(1-(4*Pl*Ra/Va2**2)))\n",
+      "\n",
+      "#Results:\n",
+      "print \"for Va=40 V, wm=\",round(wm1,1),\"rad/sec\"\n",
+      "print \"for Va=50 V, wm=\",round(wm2,1),\"rad/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for Va=40 V, wm= 169.2 rad/sec\n",
+        "for Va=50 V, wm= 217.5 rad/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.6, Page number: 575"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "\n",
+      "Rf=109                                  #Field resistance(ohm)\n",
+      "Vf=300                                  #Rated field voltage(V)\n",
+      "Ra=0.084                                #Armature resistance(ohm)\n",
+      "Kf=0.694                                #Geometric constant(A.rad/sec)\n",
+      "Tfl=285                                 #Full load torque(Nm)\n",
+      "nf=2500                                 #Speed at full load(r/min)\n",
+      "#wm=2500                                 #rated r/min\n",
+      "\n",
+      "#for part (1):\n",
+      "n1=2000                                 #r/min\n",
+      "n2=2500                                 #r/min\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#part (a):\n",
+      "If=Vf/Rf\n",
+      "w1=n1*2*pi/60\n",
+      "w2=n2*2*pi/60\n",
+      "Ea1=Kf*If*w1                    #Avg Amature voltage(V)\n",
+      "Ea2=Kf*If*w2\n",
+      "Ia1=n1*Tfl/(nf*Kf*If)\n",
+      "Ia2=n2*Tfl/(nf*Kf*If)\n",
+      "Va1 = Ea1 + Ia1*Ra\n",
+      "Va2 = Ea2 + Ia2*Ra\n",
+      "Tl1=(n1/nf)*Tfl\n",
+      "Tl2=(n2/nf)*Tfl\n",
+      "\n",
+      "#part (b):\n",
+      "\n",
+      "# The dynamic equation governing the speed of the motor is\n",
+      "\n",
+      "#   J*(dwm/dt)=Tmech-Tload\n",
+      "#   wm=(pi/30)*n & wr=(pi/30)*nf\n",
+      "#      Tload= (Tfl/wf)*wm\n",
+      "#   Tmech = Kf*If*Ia=Kf*If*(Va-Ea)/Ra          #Under armature-voltage control\n",
+      "\n",
+      "#   Thus the governing differential equation is\n",
+      "#   d(wm)/dt + 48.4*wm - 24.7*Va = 0\n",
+      " \n",
+      "#   wm = wf + (wi-wf)*exp(-t/tau)          #tau=1/48.4=20.7 msec\n",
+      "#   n = 2500- 50*exp( -t/tau )\n",
+      "\n",
+      "#   The armature current will decrease exponentially with the \n",
+      "#   same 20.7 msec time constant from an initial value of \n",
+      "#   (Vf - Vi)/Ra = 1190 A to its final value of 149 A.\n",
+      "\n",
+      "#   Ia = 149 + 1041*exp(-t/tau)\n",
+      "\n",
+      "#part (c):\n",
+      "#        J*d(wm)/dt = Tmech-Tload = Tf-(Tf/wm)*wm\n",
+      "# or      d(wm)/dt + 1.18*wm - 310 = 0\n",
+      "\n",
+      "#In this case, the speed will rise exponentially to wm=wf=262 rad/sec as\n",
+      "#         wm = 262-53*exp(-t/tau)           #tau=1/1.18=845 msec\n",
+      "\n",
+      "#Results:\n",
+      "print \"part(a):\\n\"\n",
+      "print \"-------------------------------------------------\"\n",
+      "print \"r/min\\tw[rad/s]\\tVa(V)\\tIa(A)\\tTload[Nm]\"\n",
+      "print \"-------------------------------------------------\"\n",
+      "print n1,\"\\t\",round(w1),\"\\t\\t\",round(Va1),\"\\t\",round(Ia1),\"\\t\",Tl1,\"Nm\"\n",
+      "print n2,\"\\t\",round(w2),\"\\t\\t\",round(Va2),\"\\t\",round(Ia2),\"\\t\",Tl2,\"Nm\"\n",
+      "print \"-------------------------------------------------\"\n",
+      "print \"\\npart (b):\"\n",
+      "print \" The resultant motor speed, n = 2500 - 50*exp(-t/tau) where tau=20.7 msec\"\n",
+      "\n",
+      "print \"\\npart (c):\"\n",
+      "print \" The resultant motor speed, wm = 262 - 53*exp(-t/tau) where tau=845 msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part(a):\n",
+        "\n",
+        "-------------------------------------------------\n",
+        "r/min\tw[rad/s]\tVa(V)\tIa(A)\tTload[Nm]\n",
+        "-------------------------------------------------\n",
+        "2000 \t209.0 \t\t410.0 \t119.0 \t228.0 Nm\n",
+        "2500 \t262.0 \t\t513.0 \t149.0 \t285.0 Nm\n",
+        "-------------------------------------------------\n",
+        "\n",
+        "part (b):\n",
+        " The resultant motor speed, n = 2500 - 50*exp(-t/tau) where tau=20.7 msec\n",
+        "\n",
+        "part (c):\n",
+        " The resultant motor speed, wm = 262 - 53*exp(-t/tau) where tau=845 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.7, Page number: 581"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "#Variable Calculations:\n",
+      "f1=60                                #Initial frequency(Hz)\n",
+      "f2=50                               #Changed frequency(Hz)\n",
+      "Xs=0.836                            #Saturated synch reactance(ohm)\n",
+      "Va=1+0j                             #Armature voltage(V p.u)\n",
+      "Ia=1+0j                             #Armature current(A p.u)\n",
+      "If_rated=2.84                       #Rated field current(A)\n",
+      "p=6                                 #No. of poles\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "ns1=120*f1/p\n",
+      "ns2=120*f2/p\n",
+      "Eaf=Va-1j*Xs*Ia*exp(1j*0)          #field voltage(V)\n",
+      "Ifo=abs(Eaf)*If_rated                   #motor field current(A)\n",
+      "\n",
+      "#for part(b):\n",
+      "#Eaf= (wm/wmo)*(If/Ifo)*Eafo\n",
+      "If=Ifo\n",
+      "\n",
+      "#Results:\n",
+      "print \"part(a):\"\n",
+      "print \"(i)  The motor speed:\",ns1,\"r/min\"\n",
+      "print \"(ii) The motor field current:\",round(Ifo,2),\"A\"\n",
+      "print \"part(b):\"\n",
+      "print \"(i)  The changed speed:\",ns2,\"A\"\n",
+      "print \"(ii) The mototr field current:\",round(If,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part(a):\n",
+        "(i)  The motor speed: 1200.0 r/min\n",
+        "(ii) The motor field current: 3.7 A\n",
+        "part(b):\n",
+        "(i)  The changed speed: 1000.0 A\n",
+        "(ii) The mototr field current: 3.7 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.8, Page number: 588"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "iF=2.84                                 #rated field current(A)\n",
+      "Vbase=220                               #base voltage(V)\n",
+      "Eaf=220/sqrt(3)                         #Rms voltage line-to-neutral(V)\n",
+      "f=60                                    #Hz\n",
+      "p=6                                     #poles\n",
+      "P_rated=45*10**3                        #rated power(W)\n",
+      "Xs_pu=0.836                             #per unit synchronous reactance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "we=2*pi*f\n",
+      "Laf=sqrt(2)*Eaf/(we*iF)                 #Armature field reactance(H)\n",
+      "T_rated=P_rated/(we*2/p)\n",
+      "#setting rated values to reference values.\n",
+      "Tref=T_rated\n",
+      "iFref=iF\n",
+      "iQ=round((2/3)*(2/p)*Tref/(Laf*iFref),2)\n",
+      "iD=0\n",
+      "\n",
+      "#since theta_me=wc*t, iD=0,we have,\n",
+      "t=symbols('t')\n",
+      "wc=120*pi\n",
+      "def ia(t):\n",
+      "    return iD*cos(wc*t)-iQ*sin(wc*t)\n",
+      "def ib(t):\n",
+      "    return iD*cos(wc*t-2*pi/3)-iQ*sin(wc*t-2*pi/3)\n",
+      "def ic(t):\n",
+      "    return iD*cos(wc*t+2*pi/3)-iQ*sin(wc*t+2*pi/3)\n",
+      "Ibase=P_rated/(sqrt(3)*Eaf)\n",
+      "Imax=round(ia((pi/(2*wc))))\n",
+      "Ia=1j*abs(round(Imax/sqrt(2)))\n",
+      "Eaf=1j*we*Laf*iF/sqrt(2)\n",
+      "Zbase=Vbase**2/P_rated\n",
+      "Xs=Xs_pu*Zbase\n",
+      "Va=1j*Xs*Ia+Eaf                             #line-to-neutral voltage\n",
+      "Vt=abs(sqrt(3)*Va)/Vbase                #p.u terminal voltage(line-to-line)(V)\n",
+      "\n",
+      "#Results:\n",
+      "print \"part(a):\"\n",
+      "print \"\\tia(t)=\",ia(t),\"A\"\n",
+      "print \"\\tib(t)=\",ib(t),\"A\"\n",
+      "print \"\\tic(t)=\",ic(t),\"A\"\n",
+      "print \"part(b):\"\n",
+      "print \"\\tTerminal voltage:\",round(float(Vt),2),\"per unit\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part(a):\n",
+        "\tia(t)= -167.01*sin(120*pi*t) A\n",
+        "\tib(t)= 167.01*sin(120*pi*t + pi/3) A\n",
+        "\tic(t)= -167.01*cos(120*pi*t + pi/6) A\n",
+        "part(b):\n",
+        "\tTerminal voltage: 1.3 per unit\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.9, Page number: 591"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "iF=2.84                                 #rated field current(A)\n",
+      "Vrated=220                              #rated terminal voltage,l-l(V)\n",
+      "Ibase=118                               #base current(A)\n",
+      "Eaf=220/sqrt(3)                         #Rms voltage, line-to-neutral(V)\n",
+      "f=60                                    #Hz\n",
+      "p=6                                     #poles\n",
+      "P_rated=45*10**3                        #rated power(W)\n",
+      "Xs=0.899                                #Synchronous reactance(ohm)\n",
+      "Xs_pu=0.836                             #per unit synchronous reactance(ohm)\n",
+      "Tref=358                                #Reference torque(Nm) (from Ex11.8)\n",
+      "\n",
+      "#Calculations:\n",
+      "Va=Vrated/sqrt(3)                        #base voltage, line to neutral(V)\n",
+      "we=2*pi*f\n",
+      "wm=(2/p)*we\n",
+      "Laf=sqrt(2)*Eaf/(we*iF)                 #Armature field reactance(H)\n",
+      "Ia=Tref*wm/(3*Va)\n",
+      "Ls=Xs/we                                #Synchronous inductance(mH)\n",
+      "delta=-atan(we*Ls*Ia/Va)\n",
+      "iQ_ref=sqrt(2)*Ia*cos(delta)\n",
+      "iD_ref=sqrt(2)*Ia*sin(delta)\n",
+      "iF_ref=(2./3)*(2/p)*Tref/(Laf*iQ_ref)\n",
+      "\n",
+      "#since motor is running at rated voltage, base voltage and rated voltage \n",
+      "# are assumed to be same.\n",
+      "Va_pu=Va/Va                      \n",
+      "Ia_pu=Ia/Ibase\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The reqd motor field current:\",round(iF_ref,2),\"A\"\n",
+      "print \"Per unit voltage:\",Va_pu,\"p.u\"\n",
+      "print \"Per unit current:\",round(Ia_pu),\"p.u\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The reqd motor field current: 3.7 A\n",
+        "Per unit voltage: 1.0 p.u\n",
+        "Per unit current: 1.0 p.u\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.10, Page number: 593"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "ns=4000                             #rated speed(rpm)\n",
+      "Va=220                              #rated voltage(V)\n",
+      "Ls=1.75*10**-3                      #synchronous inductance(H)\n",
+      "Prated=25000                        #Watts\n",
+      "n=3200                              #rated OC speed(rpm)\n",
+      "p=2                                 #No. of poles\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part(a):\n",
+      "Eaf=Va/sqrt(3)\n",
+      "wm=ns*pi/30                         #rad/sec\n",
+      "Trated=Prated/wm\n",
+      "we=(p/2)*n*pi/30\n",
+      "lambdaPM=sqrt(2)*Eaf/we             #flux linked wth permanent magnet(Wb)  \n",
+      "Tref=Trated*0.65                #since motor is operated at 65% of Trated\n",
+      "iQref=(2./3)*(2/p)*(Tref/lambdaPM)\n",
+      "\n",
+      "#for part(b:)\n",
+      "lambdaD=lambdaPM                         #since iD=0\n",
+      "lambdaQ=Ls*iQref\n",
+      "lambdaa=sqrt((lambdaD**2+lambdaQ**2)/2)  #rms line-to-neutral armature flux(Wb)\n",
+      "lambdaa_base=Eaf/wm\n",
+      "lambda_pu=lambdaa/lambdaa_base\n",
+      "\n",
+      "#for part(c)\n",
+      "lambdaD=sqrt(2*(lambdaa_base)**2-lambdaQ**2)\n",
+      "iDref=(lambdaD-lambdaPM)/Ls\n",
+      "Ia=sqrt((iDref**2+iQref**2)/2)          #rms armature current(A)\n",
+      "Ibase=Prated/(sqrt(3)*Va)\n",
+      "I_pu=Ia/Ibase\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a)  Required quadrature-axis current:\",round(iQref,1),\"A\"\n",
+      "print \"(b)  Resultant armature flux linkage\",round(lambda_pu,2),\"p.u\"\n",
+      "print \"(c)  iD:\",round(iDref,1),\"A\"\n",
+      "print \"     Rms value of armature current:\",round(Ia),\"A\"\n",
+      "print \"     Per unit value of armature current:\",round(I_pu,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)  Required quadrature-axis current: 48.2 A\n",
+        "(b)  Resultant armature flux linkage 1.27 p.u\n",
+        "(c)  iD: -66.1 A\n",
+        "     Rms value of armature current: 58.0 A\n",
+        "     Per unit value of armature current: 0.88 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.11, Page number: 600"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "V10=230/sqrt(3)\n",
+      "Nph=3\n",
+      "p=4\n",
+      "fe0=60\n",
+      "R1=0.095                                #Armature resistance(ohm)\n",
+      "R2=0.2                                  #Rotor resistance(ohm)\n",
+      "X10=0.680                               #Armature leakage reactance(ohm)\n",
+      "X20=0.672                               #Rotor leakage reactance(ohm)\n",
+      "Xm0=18.7                                #Inductice reactance(ohm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#taking two frequency values:\n",
+      "fe1=40\n",
+      "fe2=60\n",
+      "\n",
+      "for m in range(1,3,1):\n",
+      "    if m==1:\n",
+      "        fe=fe1\n",
+      "    else:\n",
+      "        fe=fe2\n",
+      "    X1=X10*(fe/fe0)\n",
+      "    X2=X20*(fe/fe0)\n",
+      "    Xm=Xm0*(fe/fe0)\n",
+      "    V1=V10*(fe/fe0)\n",
+      "    \n",
+      "    ws=4*pi*fe/p\n",
+      "    ns=120*fe/p\n",
+      "    V1eq=abs(V1*1j*Xm/(R1+1j*(X1+Xm)))\n",
+      "    Z1eq=1j*Xm*(R1+1j*X1)/(R1+1j*(X1+Xm))\n",
+      "    R1eq=Z1eq.real\n",
+      "    X1eq=Z1eq.imag\n",
+      "    \n",
+      "#Search over the slip until the Pload = Pmech   \n",
+      "    s=0                             #slip initialised to 0\n",
+      "    error=1\n",
+      "    \n",
+      "    while error >=0:\n",
+      "        s=s+0.00001\n",
+      "        rpm=ns*(1-s)\n",
+      "        wm=ws*(1-s)\n",
+      "        Tmech=(1/ws)*Nph*V1eq**2*(R2/s)\n",
+      "        Tmech = Tmech/((R1+R2/s)**2 + (X1+X2)**2)\n",
+      "        Pmech=Tmech*wm\n",
+      "        Pload=10.5*10**3*(rpm/1800)**3\n",
+      "        error=Pload-Pmech\n",
+      "    \n",
+      "    print \"\\nFor fe =\",fe,\"Hz :\"\n",
+      "    print \"\\tTerminal voltage=\",round(V1*sqrt(3)),\"V l-l\"\n",
+      "    print \"\\trpm =\",round(rpm)\n",
+      "    print \"\\tslip =\",round(100*s,1),\"%\"\n",
+      "    print \"\\tPload =\",round(Pload/1000,2),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "For fe = 40 Hz :\n",
+        "\tTerminal voltage= 153.0 V l-l\n",
+        "\trpm = 1166.0\n",
+        "\tslip = 2.8 %\n",
+        "\tPload = 2.86 kW\n",
+        "\n",
+        "For fe = 60 Hz :\n",
+        "\tTerminal voltage= 230.0 V l-l\n",
+        "\trpm = 1721.0\n",
+        "\tslip = 4.4 %\n",
+        "\tPload = 9.17 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.12, Page number: 608"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "V10=230/sqrt(3)\n",
+      "Nph=3\n",
+      "p=4\n",
+      "fe0=60\n",
+      "R1=0.095                                #Armature resistance(ohm)\n",
+      "R2=0.2                                  #Rotor resistance(ohm)\n",
+      "X10=0.680                               #Armature leakage reactance(ohm)\n",
+      "X20=0.672                               #Rotor leakage reactance(ohm)\n",
+      "Xm0=18.7                                #Inductice reactance(ohm)\n",
+      "n=1680                                  #rpm\n",
+      "Pmech=9.7*10**3                         #Electromagnetic power(W)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "we0=2*pi*fe0\n",
+      "Lm=Xm0/we0\n",
+      "LS=Lm+X10/we0\n",
+      "LR=Lm+X20/we0\n",
+      "Ra=R1\n",
+      "RaR=R2\n",
+      "lambda_rated=sqrt(2)*V10/we0\n",
+      "lambdaDR=lambda_rated\n",
+      "#for specified operating condition\n",
+      "wm=n*(pi/30)\n",
+      "Tmech=Pmech/wm\n",
+      "iQ=(2/3)*(2/p)*(LR/Lm)*(Tmech/lambdaDR)\n",
+      "iD=lambdaDR/Lm\n",
+      "Ia=sqrt((iD**2+iQ**2)/2)\n",
+      "wme=(p/2)*wm\n",
+      "we=wme+(RaR/LR)*(iQ/iD)\n",
+      "fe=we/(2*pi)\n",
+      "Va=sqrt(((Ra*iD-we*(LS-Lm**2/LR)*iQ)**2 + (Ra*iQ+we*LS*iD)**2)/2)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Rms amplitude of the armature current:\",round(Ia,1),\"A\"\n",
+      "print \"The electrical frequency:\",round(fe,1),\"Hz\"\n",
+      "print \"Rms terminal voltage:\",round(sqrt(3)*Va,1),\"V  line-line\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rms amplitude of the armature current: 27.9 A\n",
+        "The electrical frequency: 58.4 Hz\n",
+        "Rms terminal voltage: 243.6 V  line-line\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 11.13, Page number: 610"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from math import *\n",
+      "from pylab import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P_rated = 12*10**3                       #Watts\n",
+      "V_rated = 230                            #Rated line-line voltage(v)\n",
+      "Va_rated = 230/sqrt(3)                   #Rated line to neutral(V)\n",
+      "fe_rated = 60                            #Hz\n",
+      "we_rated = 2*pi*fe_rated                 #rad/sec\n",
+      "lambda_rated = sqrt(2)*Va_rated/we_rated #Wb\n",
+      "I_rated = P_rated/(sqrt(3)*V_rated)      #A\n",
+      "Ipeak_base = sqrt(2)*I_rated            #A\n",
+      "p = 4                                   #poles\n",
+      "\n",
+      "V10=V_rated/sqrt(3)\n",
+      "R1=0.095                                #Armature resistance(ohm)\n",
+      "R2=0.2                                  #Rotor resistance(ohm)\n",
+      "X10=0.680                               #Armature leakage reactance(ohm)\n",
+      "X20=0.672                               #Rotor leakage reactance(ohm)\n",
+      "Xm0=18.7                                #Inductice reactance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "Lm = Xm0/we_rated;\n",
+      "LS = Lm + X10/we_rated;\n",
+      "LR = Lm + X20/we_rated;\n",
+      "Ra = R1\n",
+      "RaR = R2\n",
+      "#operating point:\n",
+      "n = 1680                                #rpm\n",
+      "lambdaDR=lambda_rated\n",
+      "wm = n*pi/30\n",
+      "wme = (p/2)*wm\n",
+      "Pmech = 9.7*10**3\n",
+      "Tmech = Pmech/wm\n",
+      "lambda_DRpu=[0]*42\n",
+      "iDpu=[0]*42\n",
+      "Iapu=[0]*42\n",
+      "fe=[0]*42\n",
+      "Vapu=[0]*42\n",
+      "\n",
+      "for n in range(1,43,1):\n",
+      "    lambdaDR = (0.8+(n-1)*0.4/40)*lambda_rated\n",
+      "    lambda_DRpu[n-1]=lambdaDR/lambda_rated\n",
+      "    iQ=(2/3)*(2/p)*(LR/Lm)*(Tmech/lambdaDR)\n",
+      "    iD=(lambdaDR/Lm)\n",
+      "    iDpu[n-1]=iD/Ipeak_base\n",
+      "    iQR=-(Lm/LR)**iQ\n",
+      "    Ia=sqrt((iD**2+iQ**2)/2)\n",
+      "    Iapu[n-1]=Ia/I_rated\n",
+      "    we=wme-(RaR/LR)*(iQ/iD)\n",
+      "    fe[n-1]=we/(2*pi)\n",
+      "    Va_rms=sqrt(((Ra*iD-we*(LS-Lm**2/LR)*iQ)**2 +(Ra*iQ+ we*LS*iD)**2)/2)\n",
+      "    Vapu[n-1]=Va_rms/Va_rated\n",
+      "\n",
+      "#Results:\n",
+      "print \"The required plot is as shown:\"\n",
+      "plot(iDpu,Iapu)\n",
+      "plot(iDpu,Vapu,':')\n",
+      "xlabel('i_D [per unit] ')\n",
+      "ylabel('per unit')\n",
+      "annotate('Ia',xy=(0.21,1.05))\n",
+      "annotate('Va',xy=(0.21,0.85))\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required plot is as shown:\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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0hYVFTV1dXffq6mpLAKitrTU/fvx4qLe3d8bTFpIYv+qGajx4+AAA0M+6Hz4f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\nEOPH4wETJgAnTrCd6zk5gKs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U+FGjRp0aMGDAn6tXr/5Qc27Pnj1zg4KCLvn5+aW/+uqr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\nD0IIIVr7PxA8wzUjE2EJAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x27d0450>"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter2.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter2.ipynb
new file mode 100755
index 00000000..855c3bd6
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter2.ipynb
@@ -0,0 +1,761 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Transformers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page number: 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Pc=16                               #Core loss at Bmax=1.5 T\n",
+      "VIrms=20                            #Voltamperess for the core\n",
+      "Vrms=194                            #Rms induced voltage(V)\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "pf=Pc/VIrms\n",
+      "a=math.acos(pf)\n",
+      "I=VIrms/Vrms\n",
+      "Ic=I*pf\n",
+      "Im=I*math.fabs(math.sin(a))\n",
+      "\n",
+      "#Results:\n",
+      "print \"Power factor = \", round(pf,1),\"lagging\"\n",
+      "print \"The core-loss current,Ic =\", round(Ic,3), \"A rms\"\n",
+      "print \"The magnetising current,Im =\", round(Im,2),\"A rms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power factor =  0.8 lagging\n",
+        "The core-loss current,Ic = 0.082 A rms\n",
+        "The magnetising current,Im = 0.06 A rms\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page number: 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declarations:\n",
+      "k=5                                 #turns ratio,N1/N2\n",
+      "Z2=1+4j                             #Impedance of secondary side(ohm)\n",
+      "Vp=120                              #primary voltage(V)\n",
+      "\n",
+      "#Calculations:\n",
+      "Z2p=k**2*(Z2)\n",
+      "I=Vp/Z2p\n",
+      "Is=k*I\n",
+      "\n",
+      "#Results:\n",
+      "print \"Primary current:\",complex(round(I.real,2),round(I.imag,2)), \"A rms\"\n",
+      "print \"Current in the short:\",round(Is.real,2)+1j*round(Is.imag,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Primary current: (0.28-1.13j) A rms\n",
+        "Current in the short: (1.41-5.65j) A\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page number: 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "\n",
+      "#Variable declaration:\n",
+      "R1=0.72                                 #Resistance at high voltage side(ohm)\n",
+      "R2=0.70                                 #Resistance at low voltage side(ohm)\n",
+      "X1=0.92                                 #Reactance at high voltage side(ohm)\n",
+      "X2=0.90                                 #Reactance at low voltage side(ohm)\n",
+      "Zq=632+4370j                            #Impedance of exciting circuit(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "Req=R1+R2\n",
+      "Xeq=X1+X2       \n",
+      "Vcd=2400*Zq/(Zq+complex(R1,X1))\n",
+      "V=complex(round(Vcd.real,2),round(Vcd.imag,3))\n",
+      "\n",
+      "#Results:\n",
+      "print \"Req:\",Req,\"ohm\",\" and  Xeq:\",Xeq,\"ohm\"\n",
+      "print \"Voltage at low voltage terminal:\",V,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Req: 1.42 ohm  and  Xeq: 1.82 ohm\n",
+        "Voltage at low voltage terminal: (2399.45+0.316j) V\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.5, Page number: 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Zf=0.30+.160j                           #Impedance of feeder(ohm)\n",
+      "Zeq=1.42+3.42j             #Equiv.impedance of transformer refd. to primary(ohm)\n",
+      "k=2400/240                              #turns ratio\n",
+      "P=50000                                 #power rating of the transformer(VA)\n",
+      "Vs=2400                                 #sending end vltage of feeder(V)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I=P/2400                                #Rated current(A)\n",
+      "theta=math.acos(0.80)\n",
+      "Zt=Zf+Zeq                      #combned impedance of feeder & transformer(ohm)\n",
+      "R=Zt.real\n",
+      "X=Zt.imag\n",
+      "bc=I*X*math.cos(theta)-I*R*math.sin(theta)\n",
+      "ab=I*R*math.cos(theta)+I*X*math.sin(theta)\n",
+      "Ob=(Vs**2-bc**2)**0.5\n",
+      "V2=Ob-ab\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The voltage at the secondary terminals:\",round(V2/10,0),\"V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage at the secondary terminals: 233.0 V\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.6, Page number: 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration:\n",
+      "#short ckt test readings:\n",
+      "Vsc=48                          #voltage(V)\n",
+      "Isc=20.8                        #current(A)\n",
+      "Psc=617                         #power(W)\n",
+      "    \n",
+      "#Open ckt test readings:\n",
+      "Vs=240                         #Voltage(V)\n",
+      "I=5.41                          #current(A)\n",
+      "P=186                           #power(W)\n",
+      "V2ph=2400                        #voltage at full load at high voltage side(V)\n",
+      "pf=0.8                          #lagging power factor at full load\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "theta=math.acos(pf)\n",
+      "Zeqh=Vsc/Isc                    #subscript h refers to high voltage side\n",
+      "Reqh=Psc/Isc**2\n",
+      "Xeqh=math.sqrt(Zeqh**2-Reqh**2)\n",
+      "Ih=50000/V2ph\n",
+      "Pout=50000*pf\n",
+      "Pwind=Ih**2*Reqh\n",
+      "Ptloss=P+Pwind\n",
+      "e=(1-Ptloss/(Ptloss+Pout))*100\n",
+      "Iph=(50000/2400)*complex(math.cos(theta),math.sin(-theta))\n",
+      "V1ph=V2ph+Iph*complex(Reqh,Xeqh)\n",
+      "r=(round(abs(V1ph),2)-2400)*100/V2ph\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The efficiency of the transformer:\",round(e,0),\"%\"\n",
+      "print \"Volatge Regulation:\",round(r,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency of the transformer: 98.0 %\n",
+        "Volatge Regulation: 1.94 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.7, Page number: 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Varaible declaration:\n",
+      "Vx=2400                                 #Voltage at low voltage side(V)\n",
+      "Vbc=2400                                 #Voltage across branch bc(V)\n",
+      "Vab=240                                #Voltage induced in winding ab(V)\n",
+      "Pl=803                                  #transformer losses(W)\n",
+      "pf=0.8                                  #Power factor of the transformer\n",
+      "\n",
+      "#Calculations:\n",
+      "Vh=Vab+Vbc\n",
+      "Ih=50000/Vab\n",
+      "KVA=Vh*Ih/1000                          #Kva rating\n",
+      "P=pf*550000\n",
+      "e=(1-Pl/(P+Pl))*100\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Voltage ratings, Vh:\",Vh,\"V  \", \"&  Vx:\",Vx,\"V\"\n",
+      "print \"KVA rating as an autotransformer:\",KVA,\"KVA\"\n",
+      "print \"full-load efficiency:\", round(e,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage ratings, Vh: 2640 V   &  Vx: 2400 V\n",
+        "KVA rating as an autotransformer: 550.0 KVA\n",
+        "full-load efficiency: 99.82 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.8, Page number: 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Vl1=4160                      #line-to-line voltage at feeder's sending end(V)\n",
+      "Zf=0.30+.160j                           #Impedance of feeder(ohm)\n",
+      "Zeq=1.42+3.42j             #Equiv.impedance of transformer refd. to primary(ohm)\n",
+      "k=2400/240                              #turns ratio\n",
+      "P=50000                                 #power rating of the transformer(VA)\n",
+      "Vs=2400                                 #sending end vltage of feeder(V)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "#this problem can be treated on a single phase basis,\n",
+      "#and whole problem is similar to Ex 2.5.\n",
+      "\n",
+      "I=P/2400                                #Rated current(A)\n",
+      "theta=math.acos(0.80)\n",
+      "Zt=Zf+Zeq                      #combned impedance of feeder & transformer(ohm)\n",
+      "R=Zt.real\n",
+      "X=Zt.imag\n",
+      "bc=I*X*math.cos(theta)-I*R*math.sin(theta)\n",
+      "ab=I*R*math.cos(theta)+I*X*math.sin(theta)\n",
+      "Ob=(Vs**2-bc**2)**0.5\n",
+      "V2=Ob-ab\n",
+      "Vload=V2/k\n",
+      "\n",
+      "#Results:\n",
+      "print \"The line to line voltage:\",round(Vload,0),\"V line-to-line\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The line to line voltage: 233.0 V line-to-line\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.9, Page number: 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "\n",
+      "#Variable decclaration:\n",
+      "#All resistances, reactances, & impedances are on per phase basis\n",
+      "Req=1.42       #Series resist. of del-del transformer referred to 2400v side(ohm)\n",
+      "Xeq=1.82       #Series react. of del-del transformer referred to 2400v side(ohm)\n",
+      "Zs=0.17+0.92j                   #Equiv impedance of sending end transformer(ohm)\n",
+      "Xf=0.8j                          #Reactance of the feeder(ohm)\n",
+      "Vf=2400                         #Voltage of the feeder(V)\n",
+      "k=10                            #turns ratio(Vp/Vs)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Zt=(complex(Req,Xeq)/3)+Zs+Xf\n",
+      "Ztot=complex(round(Zt.real,2),round(Zt.imag,2))\n",
+      "If=math.floor(Vf/(math.sqrt(3))/round(abs(Ztot),2))\n",
+      "I1=If/math.sqrt(3)\n",
+      "I2=I1*k\n",
+      "Ic=I2*math.sqrt(3)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Short circuit current in the 2400 feeder, per phase wires:\",round(Ic,1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Short circuit current in the 2400 feeder, per phase wires: 5720.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.10, Page number: 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "X1=143                             #Reactance of primary(ohm)\n",
+      "X21=164                           #Reactance of secondary ref. to primary(ohm)\n",
+      "Xm=163*10**3                           #Reactance of magnetising ckt(ohm)\n",
+      "R1=128                             #Resistance of primary(ohm)\n",
+      "R21=141                           #Resistane of secondary ref. to primary(ohm)\n",
+      "k=20                              #turns ratio(2400/120)\n",
+      "V1=2400                             #primary voltage(V)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V2=(V1/k)*complex(0,Xm)/complex(R1,X1+Xm)\n",
+      "mag=abs(V2)\n",
+      "ph=degrees(cmath.phase(V2))\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Magnitude of V2:\",round(mag,2),\"V\"\n",
+      "print \"Phase of V2:\",round(ph,3),\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnitude of V2: 119.89 V\n",
+        "Phase of V2: 0.045 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.11, Page number: 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "\n",
+      "\n",
+      "#variable declaration:\n",
+      "X1=44.8*10**-6                  #Reactance of the primary(ohm)\n",
+      "R1=10.3*10**-6                  #Resistance of the primary(ohm)\n",
+      "X21=54.3*10**-6          #Reactance of the secondary refd. to primary(ohm)\n",
+      "R21=9.6*10**-6                  #Resistance of secondary ref. to primary(ohm)\n",
+      "Xm=17.7*10**-3                  #Reactance of the magnetising ckt(ohm)\n",
+      "k=5/800                         #turms ratio(I2/I1)\n",
+      "Zl=2.5+0j                       #Impedance ofthe load(ohm)\n",
+      "I1=800                          #primary current(A)\n",
+      "\n",
+      "#Calculations:\n",
+      "Zp=k**2*Zl\n",
+      "I2=I1*k*Xm*1j/(Zp+R21+(X21+Xm)*1j)\n",
+      "phase=cmath.phase(I2)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Magnitude of current:\",round(abs(I2),2),\"A\"\n",
+      "print \"Phase of the current:\",round(math.degrees(phase),3),\"degrees\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnitude of current: 4.98 A\n",
+        "Phase of the current: 0.346 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.12, Page number: 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration:\n",
+      "XL=0.040                                #Reactance at l.v side(ohm)\n",
+      "XH=3.75                                 #Reactance at h.v side(ohm)\n",
+      "Xm=114                                  #Magnetising reactance(ohm)\n",
+      "RL=0.76*10**-3                          #Resistance at l.v.side(ohm)\n",
+      "RH=0.085                                #Resistance at l.v.side(ohm)\n",
+      "VA_base=100*10**6                       #base VA\n",
+      "V_base=7.97*10**3                       #base voltage(V)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for l.v side\n",
+      "VA_base=100*10**6                       #base VA\n",
+      "V_base=7.97*10**3                       #base voltage(V)\n",
+      "Rbase1=Xbase1=V_base**2/VA_base\n",
+      "\n",
+      "#for h.v side:\n",
+      "VA_base=100*10**6                       #base VA\n",
+      "V_base=79.7*10**3                       #base voltage(V)\n",
+      "Rbase2=Xbase2=V_base**2/VA_base\n",
+      "\n",
+      "XL_pu=XL/Xbase1\n",
+      "XH_pu=XH/Xbase2\n",
+      "Xm_pu=Xm/Xbase1\n",
+      "RL_pu=RL/Rbase1\n",
+      "RH_pu=RH/Rbase2\n",
+      "K_pu=1                                  #per unit utrns ratio\n",
+      "\n",
+      "#Results:\n",
+      "print \"The per unit parameters are:\"\n",
+      "print \"XL_pu =\",round(XL_pu,3),\"p.u\"\n",
+      "print \"XH_pu =\",round(XH_pu,4),\"p.u\"\n",
+      "print \"Xm_pu =\",math.ceil(Xm_pu),\"p.u\"\n",
+      "print \"RL_pu =\",round(RL_pu,4),\"p.u\"\n",
+      "print \"XL_pu =\",round(RH_pu,4),\"p.u\"\n",
+      "print \"Turns ratio =\",K_pu,\"p.u\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The per unit parameters are:\n",
+        "XL_pu = 0.063 p.u\n",
+        "XH_pu = 0.059 p.u\n",
+        "Xm_pu = 180.0 p.u\n",
+        "RL_pu = 0.0012 p.u\n",
+        "XL_pu = 0.0013 p.u\n",
+        "Turns ratio = 1 p.u\n"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.13, Page number: 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Ic=5.41                             #Exciting current ref. to low volt. side(A)\n",
+      "k=10                                #turns ratio(N1/N2=2400/240)\n",
+      "Vbh=2400                            #base voltage at primary side(V)\n",
+      "Vbl=240                             #base voltage at secondary side(V)\n",
+      "Ibh=20.8                            #base current at primary side(A)\n",
+      "Ibl=208                             #base current at secondary side(A)\n",
+      "Z=1.42+1.82j                    #Equiv.impedance ref.to high voltage side(ohm)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Zbh=Vbh/Ibh\n",
+      "Zbl=Vbl/Ibl\n",
+      "Icl=Ic/Ibl\n",
+      "Ich=Ic/(Ibh*k)\n",
+      "Zl=Z/(k**2*Zbl)\n",
+      "Zh=Z/Zbh\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Per unit exciting current on low volt. sides:\",round(Icl,3,),\"A\"  \n",
+      "print \"Per unit exciting current on high volt. sides:\",round(Ich,3),\"A\"\n",
+      "print \"per unit equiv.impedance at low volt. sides:\",round(Zl.real,4)+round(Zl.imag,4)*1j,\"ohm\"\n",
+      "print \"per unit equiv.impedance at high voltage sides:\",round(Zh.real,4)+round(Zh.imag,4)*1j,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Per unit exciting current on low volt. sides: 0.026 A\n",
+        "Per unit exciting current on high volt. sides: 0.026 A\n",
+        "per unit equiv.impedance at low volt. sides: (0.0123+0.0158j) ohm\n",
+        "per unit equiv.impedance at high voltage sides: (0.0123+0.0158j) ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.14, Page number: 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Vb=24000               #Base voltage of secondary of sending end transformer(V)    \n",
+      "Z=0.17+0.92j      #Impedance of sending end transformer ref. to 2400V side(ohm)\n",
+      "P=150                       #Power rating of the transformer(KVA)\n",
+      "V=2400                      #Primary voltage of sending end transformer(v)\n",
+      "Ztot=0.64+2.33j                 #Total series impedance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "Zb=V**2/(P*10**3)\n",
+      "Ztotb=Ztot/Zb\n",
+      "Vsb=1                           #Vs in terms of per unit values\n",
+      "Isc=Vsb/abs(Ztotb)                   #Short current in per unit values(A)\n",
+      "Ib1=P*10**3/(sqrt(3)*2400)       #base current of the feeder at 2400V side(A)\n",
+      "If=Ib1*Isc\n",
+      "Ib2=P*10**3/(sqrt(3)*240)\n",
+      "Iscs=Isc*Ib2                     #short ckt current at 2400V afeeder side (A)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Short circuit current in 2400 feeder:\",round(Iscs/10**3,2),\"KA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Short circuit current in 2400 feeder: 5.73 KA\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.15, Page number: 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P=250*10**3                            #power rating of transformer(KVA)\n",
+      "Vp=2400                             #primary volatge(V)\n",
+      "Vs=460                              #secondary voltage(V)\n",
+      "Pb=100*10**3                           #new base power of transformer(KVA)\n",
+      "Vb=460                              #new base voltage(V)\n",
+      "Z=0.026+0.12j                       #series impedance on its own base(ohm)\n",
+      "Vl=438                              #load voltage(V)\n",
+      "Pl=95*10**3                         #power drawn by the load(kW)\n",
+      "\n",
+      "#Calculations:\n",
+      "Zbo=Vs**2/P                        #base impedance for the transformer(ohm)\n",
+      "Zbn=Vb**2/Pb            #base impedance for the transformer at 100KVA base(ohm)\n",
+      "Zpn=Z*Zbo/Zbn                      #base impedance at 100KVA base(ohm)\n",
+      "Vpl=Vl/Vb                           #per unit load voltage(V)\n",
+      "Ppl=Pl/Pb                          #per unit load power\n",
+      "Ipl=Ppl/Vpl                             #per unit load current(A)\n",
+      "Vpp=Vpl+Ipl*Zpn                     #high side voltage of the transformer(V) \n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The high side voltage:\",round(abs(Vpp*Vp),0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The high side voltage: 2313.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter3.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter3.ipynb
new file mode 100755
index 00000000..411be995
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter3.ipynb
@@ -0,0 +1,430 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3da4e1bbe19be5f3a05399c95840acc2f1758f76d57c3a0e757cbfa9a562b633"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Electromechanical-Energy-Conversion-Principles "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page number: 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "I=10                            #current in the coil(A)\n",
+      "Bo=0.02                         #magnetic field (T)\n",
+      "R=0.05                          #radius of the rotor(m)\n",
+      "l=0.3                           #rotor length(m)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "q=symbols('q')                  #Direction of torque\n",
+      "F1=-2*I*l*Bo*sin(q)             #Force on the coil(N)\n",
+      "T=F1*R                          #Torque scting in theta direction(Nm)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Force per unit length:\",T,\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force per unit length: -0.006*sin(q) Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page number: 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration\n",
+      "N=1000                   #No of winding turns\n",
+      "g=2                      #Air gap width(mm)\n",
+      "d=0.15                  #Magnetic core width,d (m)\n",
+      "l=0.1                    #thickness of core(0.1)\n",
+      "x,d=symbols('x d')       #where h is height of plunger(m) \n",
+      "                         #Lx is inductance as a function of x(H)\n",
+      "i=10                     #Current in the winding(A)\n",
+      "uo=4*3.14*10**-7           #permeability of free space(H/m)\n",
+      "\n",
+      "#Calculations:\n",
+      "Lx=(uo*N**2*l*d)/(2*g*10**-3)*(1-x/d)\n",
+      "Wfld=(1./2)*Lx*i**2\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The magnetic energy stored, Wfld:\",\"236*(1-x/d) J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnetic energy stored, Wfld: 236*(1-x/d) J\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page number: 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "import numpy as np\n",
+      "from pylab import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "xdata=[0.0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0]     #(cm)\n",
+      "Ldata=[2.8, 2.26, 1.78, 1.52, 1.34, 1.26, 1.20, 1.16, 1.13, 1.11, 1.10]   #(mH)\n",
+      "I = 0.75                                                #(A)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "x=0.01*np.array(xdata)\n",
+      "L=0.001*np.array(Ldata)\n",
+      "length=len(x)\n",
+      "xmax=x[length-1]\n",
+      "a=polyfit(x,L,4)\n",
+      "xfit=[0]*102\n",
+      "Lfit=[0]*102\n",
+      "for n in range(1,102,1):\n",
+      "    xfit[n-1]=xmax*(n-1)/100\n",
+      "    Lfit[n-1]=a[0]*xfit[n-1]**4+a[1]*xfit[n-1]**3+a[2]*xfit[n-1]**2+a[3]*xfit[n-1]+a[4]\n",
+      "\n",
+      "#Plot the data and then the fit to compare (convert xfit to cm and Lfit to mH)\n",
+      "plot(xdata,Ldata,'o')\n",
+      "plot(100*np.array(xfit),1000*np.array(Lfit),'g.')\n",
+      "xlabel('x [cm] ')\n",
+      "ylabel('L [mH] ')\n",
+      "title('Inductance,L vs length,l')\n",
+      "grid()\n",
+      "print \"The required plots are shown below:\"\n",
+      "show()\n",
+      "\n",
+      "#set current to 0.75 A\n",
+      "I=0.75\n",
+      "F=[0]*102\n",
+      "for n in range(1,102,1):\n",
+      "    xfit[n-1]=0.002+0.016*(n-1)/100\n",
+      "    F[n-1]=4*a[0]*xfit[n-1]**3+3*a[1]*xfit[n-1]**2+2*a[2]*xfit[n-1]**1+a[3]\n",
+      "    F[n-1]=(I**2/2)*F[n-1]\n",
+      "plot(100*np.array(xfit),F,'b.')\n",
+      "xlabel('x  [cm]')\n",
+      "ylabel('Force [N]')\n",
+      "title('Force, F vs length,l')\n",
+      "grid()\n",
+      "\n",
+      "#Results:\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required plots are shown below:\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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XEjUIiFIPEjUISHRhOE1wDHqozKS4mqmwolB5yatfop9BXe4KoC8oH5kQQ19b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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2b71910>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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1Bwf8hjO6AMCURVNeKpUqbciQIdfHjh17YePGjYsSExPzVq5cuczagwP+aGue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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3d76250>"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page number: 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Lo=10.6*10**-3                          #Initial inductance(H)\n",
+      "L2=2.7*10**-3                           #H\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "theta,i=symbols('theta i')\n",
+      "L=Lo+L2*cos(2*theta)\n",
+      "i=2                                     #Coil current,A\n",
+      "def T(theta):\n",
+      "    return i**2*diff(L,theta)/2\n",
+      "                                    \n",
+      "\n",
+      "#Results:\n",
+      "print \"Torque,Tfld =\",T(theta),\" N.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Torque,Tfld = -0.0108*sin(2*theta)  N.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page number: 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "r1=2.5*10**-2                            #radius of rotor(m)\n",
+      "h=1.8*10**-2                             #Axial length(m)\n",
+      "g=3*10**-3                               #Air gap length(m)\n",
+      "Bag=1.65                                #Magnetic field(T)\n",
+      "uo=4*pi*10**-7                          #permeability of free space(H/m)\n",
+      "\n",
+      "#Calculations:\n",
+      "H=Bag/uo\n",
+      "Ni=2*g*H\n",
+      "T=uo*(Ni)**2*h*(r1+0.5*g)/(4*g)\n",
+      "\n",
+      "#Results:\n",
+      "print \"The maximum torque:\", round(T,2),\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum torque: 3.1 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.7, Page number: 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from matplotlib import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "i1=0.8\n",
+      "i2=0.01\n",
+      "\n",
+      "\n",
+      "#Calculations & Results:\n",
+      "def df(f,x,h=0.1e-10):\n",
+      "    return ( f(x+h/2) - f(x-h/2) )/h\n",
+      "\n",
+      "\n",
+      "\n",
+      "def l11(x):\n",
+      "    return (3+cos(2*x))/1000.0\n",
+      "\n",
+      "def l12(x):\n",
+      "    return (0.3*cos(x))\n",
+      "\n",
+      "def l22(x):\n",
+      "    return (30+10*cos(2*x))\n",
+      "\n",
+      "def g(x):\n",
+      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x) + (i1*i2)*df(l12,x)\n",
+      "\n",
+      "def r(x):\n",
+      "    return ((i1**2)/2)*df(l11,x) + ((i2**2)/2)*df(l22,x)\n",
+      "def s(x):\n",
+      "    return (i1*i2)*df(l12,x)\n",
+      "\n",
+      "x=linspace(-pi,pi,100000)\n",
+      "\n",
+      "\n",
+      "plot(x,r(x))\n",
+      "plot(x,s(x))\n",
+      "plot(x,g(x))\n",
+      "grid()\n",
+      "annotate(\"Total torque\",xy=(-0.5,0.003))\n",
+      "annotate(\"Reluctance torque\",xy=(-2,-0.0015))\n",
+      "annotate(\"Mutual Interaction torque\",xy=(1.6,-0.0026))\n",
+      "xlabel(\"Theta [radians]\")\n",
+      "ylabel(\"Torque [N.m]\")\n",
+      "xlim(-pi,pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print \"Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)\"\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Tfld = -1.64*10**-3*sin(2*x)- 2.4*10**-3*sin(x)\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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wdAQOHGDxg3fvciNDRUVIhaYiEtMT0W1vN8ztMhfTOk5T6dxFMmECe7RRYwN4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\nAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3b09350>"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.9, Page number: 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "W=4.0*10**-2                            #width of plunger lower arm(m)\n",
+      "W1=4.5*10**-2                           #width of plunger upper arm(m)\n",
+      "D=3.5*10**-2                            #depth of plunger (m)\n",
+      "d=8*10**-3                              #length of magnet(m)\n",
+      "go=1*10**-3                             #air gap length(m)\n",
+      "uo=4*pi*10**-7                          #Permeability of free space(A.turns/m)\n",
+      "ur=1.06*uo                              #Relativity permeability\n",
+      "Hc1=-940                                #Magnetising force(kA/m)\n",
+      "Bt=1.25                                 #Magnetic field induction(T)\n",
+      "N=1500                                  #No of turns\n",
+      "x=3*10**-3                              #Position of plunger(m)\n",
+      "\n",
+      "#Calculation:\n",
+      "Ni=-Hc1*d*10**3\n",
+      "Rx=x/(uo*W1*D)\n",
+      "Ro=go/(uo*W*D)\n",
+      "Rm=d/(ur*W*D)\n",
+      "f=-((Ni)**2)/(uo*W1*D*(Rx+Ro+Rm)**2)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The x-directed force:\",round(f,1),\"N\"\n",
+      "print \"Current in the excitation winding:\",round(Ni/N,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The x-directed force: -703.3 N\n",
+        "Current in the excitation winding: 5.01 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter4.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter4.ipynb
new file mode 100755
index 00000000..aef0df05
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter4.ipynb
@@ -0,0 +1,277 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Introduction to Rotating Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.2, Page number: 199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "uo=4*pi*10**-7                          #Permeabolity of free space(H/m)\n",
+      "g=0.7*10**-3                            #Length of air gap(m)\n",
+      "p=4                                     #no. of poles\n",
+      "Ba=1.6                                  #Magnetic flux density(T)\n",
+      "Kr=0.935                                #Winding constant\n",
+      "N=263                                   #No. of turns\n",
+      "\n",
+      "#Calculations:\n",
+      "Ir=(pi*g*p/(4*uo*Kr*N))*1.6\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Rotor winding current:\",round(Ir,1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rotor winding current: 11.4 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page number: 208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "fc=60                                   #frequency of the current(Hz)\n",
+      "p=[2, 4, 6]                             #matrix of no. of poles\n",
+      "\n",
+      "#Calculations:\n",
+      "ns=[0]*3\n",
+      "ws=[0]*3\n",
+      "wc=2*pi*fc\n",
+      "for n in range(0,3,1):\n",
+      "    ws[n]=round((2/p[n])*wc,0)\n",
+      "    \n",
+      "for i in range(0,3,1):\n",
+      "    ns[i]=round(120*fc/p[i],0)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The synchronous angular velocities:\",ws, \"rad/sec\"\n",
+      "print \"The speed of the rotor:\",ns,\"r/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The synchronous angular velocities: [377.0, 188.0, 126.0] rad/sec\n",
+        "The speed of the rotor: [3600.0, 1800.0, 1200.0] r/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.5, Page number: 212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Nf=68                                   #Field winding\n",
+      "Na=18                                   #Armature winding\n",
+      "r=0.53                                  #mean air gap radius(m)\n",
+      "l=3.8                                   #Armature winding length(m)\n",
+      "Kf=0.945                                #Winding factor of field winding\n",
+      "Ka=0.933                                #Winding factor of armature winding\n",
+      "g=4.5*10**-2                            #Air gap length(m)\n",
+      "p=2                                     #No. of poles\n",
+      "If=720                                  #field current(A)\n",
+      "uo=4*pi*10**-7                          #Permeability of free space(H/m)\n",
+      "f=60                                    #Frequency curent(Hz)\n",
+      "\n",
+      "#Calculations:\n",
+      "Fag1_peak=4*Kf*Nf*If/(pi*p)\n",
+      "Bag1_peak=uo*Fag1_peak/g\n",
+      "Qp=2*Bag1_peak*l*r\n",
+      "Erms=sqrt(3)*sqrt(2)*pi*f*Ka*Na*Qp\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The peak fundamental mmf,Fag1_peak: \",round(Fag1_peak/10000,2),\"* 10^4  A.turns/pole\"\n",
+      "print \"\\nThe flux density in the air gap,Bag1_peak: \",round(Bag1_peak,2),\"T\"\n",
+      "print \"\\nThe fundamental flux per pole, Qp:\" ,round(Qp,2),\"Wb\"\n",
+      "print \"\\nThe rms value of open circuit voltage,Erms: \",round(Erms/1000,1),\"KV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The peak fundamental mmf,Fag1_peak:  2.95 * 10^4  A.turns/pole\n",
+        "\n",
+        "The flux density in the air gap,Bag1_peak:  0.82 T\n",
+        "\n",
+        "The fundamental flux per pole, Qp: 3.31 Wb\n",
+        "\n",
+        "The rms value of open circuit voltage,Erms:  25.7 KV\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.8, Page number: 225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "ns=1800                            #Speed of rotor(rpm)\n",
+      "f=60                                #Frequency(Hz)\n",
+      "g=1.2*10**-3                        #Air gap length(m)\n",
+      "D=0.27                                  #Avg diameter of the gap(m)\n",
+      "Kr=0.976                             #Winding factor\n",
+      "l=0.32                              #Axial length(m)\n",
+      "I=18                                #Rotor current(A)\n",
+      "p=4                                 #No of poles\n",
+      "Nr=786                              #Rotor windings\n",
+      "B_max=1.5                           #Max. flux densiity(T)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Fr_max=4*Kr*Nr*I/(pi*p)\n",
+      "T_max=p*pi*D*l*B_max*Fr_max/4\n",
+      "wm=ns*pi/30\n",
+      "P=wm*T_max\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Maximum torque, T_max:\",round(T_max,0),\"Nm\"\n",
+      "print \"Maximum power,P:\",round(P/1000,0),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum torque, T_max: 1790.0 Nm\n",
+        "Maximum power,P: 337.0 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.9, Page number: 229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "b=0.5                                       #Wavelength of wnding(m)\n",
+      "l=1.5                                   #Winding length(m)\n",
+      "I=700                                   #Currents in windings(A)\n",
+      "N=45                                    #No. of turns\n",
+      "K=0.92                                  #winding factor\n",
+      "p=3                                     #No. of phases\n",
+      "uo=4*pi*10**-7\n",
+      "g=0.01                                  #Air gap flux(m)\n",
+      "f=25                                    #Frequency of the exciting current(A)\n",
+      "\n",
+      "#Calculations:\n",
+      "F_peak=(3*4*K*N*700)/round(4*pi*p,-1)\n",
+      "B=uo*F_peak/g\n",
+      "v=f*b\n",
+      "\n",
+      "#Results:\n",
+      "print \"Amplitude of the resultant mmf wave:\",round(F_peak/1000,1),\"kA/m\"\n",
+      "print \"Peak air gap flux:\",round(B,1),\"T\"\n",
+      "print \"Velocity of the travelling wave:\",v,\"m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amplitude of the resultant mmf wave: 8.7 kA/m\n",
+        "Peak air gap flux: 1.1 T\n",
+        "Velocity of the travelling wave: 12.5 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter5.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter5.ipynb
new file mode 100755
index 00000000..e0897bf4
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter5.ipynb
@@ -0,0 +1,596 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:fdfb62ac44329d8ca6f9c44e55b110ca6276a0bf982882f2c43000518ecc6bb3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Synchronous Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page number: 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "\n",
+      "#Varaible Declaration:\n",
+      "pf=0.95                                 #Lagging power factor\n",
+      "Vl=460                                  #Terminal voltage(V)\n",
+      "I=120                                   #Terminal current(A)\n",
+      "If=47                                   #Field current(A)\n",
+      "X=1.68j                                 #Line syncchronous reactance(ohm)\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "#Choosing motor reference direction:\n",
+      "Va=Vl/math.sqrt(3)\n",
+      "theta=math.acos(0.95)\n",
+      "Ia=I*cmath.exp(-theta*1j)\n",
+      "Eaf=Va-X*Ia\n",
+      "wc=120*math.pi\n",
+      "Laf=math.sqrt(2)*abs(Eaf)/(wc*If)\n",
+      "P=3*Va*Ia*pf\n",
+      "\n",
+      "#Results:\n",
+      "print \"Generated emf:\",round(abs(Eaf),1),\"V line to line\"\n",
+      "print \"Fied to armature mutual inductance:\",round(Laf*1000,1),\"mH\"\n",
+      "print \"Three phase power:\",round(abs(P/1000),1),\"kW or\",round(abs(P)/746),\"hp\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Generated emf: 278.8 V line to line\n",
+        "Fied to armature mutual inductance: 22.3 mH\n",
+        "Three phase power: 90.8 kW or 122.0 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page number: 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath \n",
+      "from math import *\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Pin=90.6*10**3                              #Input power(kW)\n",
+      "Va=265.6                                    #Terninal voltage(V)\n",
+      "X=1.68j                                     #Synchronous reactance(ohm)\n",
+      "Laf=22.3*10**-3                             #Mutual inductance(H)\n",
+      "wc=120*pi                                   #Angular frequency(rad/sec)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ia=Pin/(3*Va)\n",
+      "Eaf=Va-X*Ia\n",
+      "delta=degrees(cmath.phase(Eaf))\n",
+      "I=sqrt(2)*Eaf/(wc*Laf)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print\"The phase angle,delta:\",round(delta,1),\"degrees\"\n",
+      "print\"Required field current:\",round(abs(I),2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The phase angle,delta: -35.7 degrees\n",
+        "Required field current: 55.04 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.3, Page number: 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Eafl=13.8*10**3                             #Open circuit voltage(V)\n",
+      "If1=318                                     #Field current(A)\n",
+      "If2=263                              #Field current after extrapolation(A)\n",
+      "wc=120*pi                            #Angular frequency(Hz)\n",
+      "\n",
+      "#Calculations:\n",
+      "Eaf=Eafl/sqrt(3)\n",
+      "La1=sqrt(2)*Eaf/(wc*If1)\n",
+      "La2=sqrt(2)*Eaf/(wc*If2)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Saturated Laf1:\",round(La1*1000,0),\"mH\"    \n",
+      "print \"Unsaturated Laf1:\",round(La2*1000,0),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Saturated Laf1: 94.0 mH\n",
+        "Unsaturated Laf1: 114.0 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page number: 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Ia=[118, 152]                     #Armature current from SC Characteristics(A)\n",
+      "If=[2.20, 2.84]                     #Field current from SC Characteristics(A)\n",
+      "Vll=220                             #Line-to-line Voltage(V)\n",
+      "V=202                               #Line-to-line air voltage(V) \n",
+      "P=45*10**3                          #Power roted to motor(W)   \n",
+      "Is_sc=1                             #per unit rated current(A)\n",
+      "\n",
+      "#Calculations:\n",
+      "Va_ag=V/sqrt(3)                 #At field current of 2.20A,at air gap,(V)\n",
+      "Ia_ag=Ia[0]\n",
+      "Xs_u=Va_ag/Ia_ag\n",
+      "Ia_rated=P/(sqrt(3)*Vll)\n",
+      "Xa_g=Va_ag/1\n",
+      "Xs_u_pu=Va_ag/Is_sc\n",
+      "Xs=Vll/(Ia[1]*sqrt(3))\n",
+      "Ia_pu=Ia[1]/Ia[0]\n",
+      "SCR=If[1]/If[0]\n",
+      "Xs=1/SCR\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"'All quantities are in per unit values'\"\n",
+      "print\"Unsaturated value of synchronous reactance:\",round(Xs_u,3),\"ohm\"\n",
+      "print \"Satureted value of synchronous reactance: \",round(Xs,3),\"ohm\"\n",
+      "print\"Short circuit ratio:\",round(SCR,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "'All quantities are in per unit values'\n",
+        "Unsaturated value of synchronous reactance: 0.988 ohm\n",
+        "Satureted value of synchronous reactance:  0.775 ohm\n",
+        "Short circuit ratio: 1.291\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.5, Page number: 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P_rated=45*10**3                        #Rated power(KV)\n",
+      "Pl=1.80*10**3                           #Short circuit load loss(W)\n",
+      "Ia_pu=1                                    #Per unit armature current\n",
+      "Ia=118                                  #rated armature current(A)\n",
+      "Ra_dc=0.0335                            #Dc resistance(ohm/phase)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pl_pu=Pl/P_rated    \n",
+      "Ra_eff1=Pl_pu/Ia_pu**2                      #in per unit basis\n",
+      "Ra_eff2=Pl/(3*(Ia)**2)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Armature resistance in per unit:\",round(Ra_eff1,3),\"per unit\" \n",
+      "print \"Armature resistance in ohms/phase:\", round(Ra_eff2,3),\"ohms/phase\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature resistance in per unit: 0.04 per unit\n",
+        "Armature resistance in ohms/phase: 0.043 ohms/phase\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.6, Page number: 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "import cmath\n",
+      "import math\n",
+      "from pylab import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Veq=1.0                                 #Externalsupply(p.u)                                    \n",
+      "Eaf=1.0                                 #Internal voltage(p.u)\n",
+      "Xeq=0.23                                #Eqv.resistance of external system(p.u)\n",
+      "Xs=1.35                                #Saturated synchronous reactance(p.u)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "P_max=Eaf*Veq/(Xs+Xeq)\n",
+      "\n",
+      "\n",
+      "#for part (b):\n",
+      "delta=[0]*500\n",
+      "Ia=[0]*500\n",
+      "Va=[0]*500\n",
+      "degree=[0]*500\n",
+      "for n in range(1,101,1):\n",
+      "    delta[n-1]=(pi/2)*(n-1)/100\n",
+      "    Ia[n-1] = (Eaf *exp(1j*delta[n-1]) - Veq)/(1j*(Xs + Xeq))\n",
+      "    Va[n-1] = abs(Veq + 1j*Xeq*Ia[n-1])\n",
+      "    degree[n-1]=180*delta[n-1]/pi\n",
+      "plot(degree,Va,'r.')\n",
+      "xlabel('Power angle,delta(degrees)')\n",
+      "ylabel('Terminal voltage(per unit)')\n",
+      "title('Terminal voltage vs. power angle for part (b)')\n",
+      "show()\n",
+      "#for part (c):\n",
+      "Vterm=1.0\n",
+      "P=[0]*500\n",
+      "deltat=[0]*500\n",
+      "Ia=[0]*500\n",
+      "Eaf=[0]*500\n",
+      "\n",
+      "for n in range(1,101,1):\n",
+      "    P[n-1]=(n-1)/100\n",
+      "    deltat[n-1]=math.asin(P[n-1]*Xeq/(Vterm*Veq))\n",
+      "    Ia[n-1]=(Vterm*exp(1j*deltat[n-1])-Veq)/(1j*Xeq)\n",
+      "    Eaf[n-1]=abs(Vterm+1j*(Xs+Xeq)*Ia[n-1])\n",
+      "plot(P,Eaf,'r.')\n",
+      "xlabel('Power [per unit]')\n",
+      "ylabel('Eaf [per unit]')\n",
+      "title('Eaf vs. power for part (c)')\n",
+      "show()\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Maximum power supplied to external system:\",round(P_max,2),\"p.u\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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69aqv6EMDAIDu1G5CkJWVrZeVla1n5rlcrhT6EAAA3j/tJgRXV9fYTZs2fVVT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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x34638d0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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zy9hstiQsLGxPcnLyDNkyAwcOvEXXOOrr600tLS3vySYFAADoeUpvuz1t2rSj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\ncv4f7PGByn7cLZQAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3463890>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Maximum power supplied to external system: 0.63 p.u\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page number: 272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P_rated=2000*746/3                     #per phase rated power of motor(W)\n",
+      "Xsm=1.95                                #Synchronous reactance(ohm)\n",
+      "Vl=2300                                #Line to line voltage(V)\n",
+      "f=60                                   #Angular frequency(Hz)\n",
+      "p=30                                   #No. of poles\n",
+      "Xsg=2.65                               #Synchronous reactance of generator(ohm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "Vp=2300/sqrt(3)\n",
+      "Ip=P_rated/Vp\n",
+      "Eafm=sqrt(Vp**2+(Ip*Xsm)**2)\n",
+      "Pm=3*Vp*Eafm/Xsm                             #Max power delivered to motor(W)\n",
+      "ws=2*2*pi*f/p\n",
+      "Tmax=Pm/ws                                  #MAx torque of motor(Nm)\n",
+      "\n",
+      "\n",
+      "#for part (b):\n",
+      "Eafg=sqrt(Vp**2+(Ip*Xsg)**2)\n",
+      "Pm2=3*Eafm*Eafg/(Xsg+Xsm)                   #Max power delivered to motor(W)\n",
+      "Tmax2=Pm2/ws                                #Max torque(Nm)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print\"(a) Max power  :\",round(Pm/1000,0),\"kW,3-ph\"\n",
+      "print\"    Max torque :\",round(Tmax/1000,1),\"kNm\"\n",
+      "print \"(b) Max power :\", round(Pm2/1000,0),\"kW,3-ph\"\n",
+      "print \"    Max torque:\", round(Tmax2/1000,1),\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Max power  : 3096.0 kW,3-ph\n",
+        "    Max torque : 123.2 kNm\n",
+        "(b) Max power : 1639.0 kW,3-ph\n",
+        "    Max torque: 65.2 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.8, Page number: 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P=45                                    #Power rated(KVA)\n",
+      "Va=220                                  #Terminal voltage(V)\n",
+      "Pin=45                                  #Power input to the armature(KVA)\n",
+      "If=5.50                                 #field current(A)\n",
+      "Rf=35.5                                 #Field winding resistance(ohm)\n",
+      "Ra=0.0399                               #Armature dc resistance(ohm/phase)\n",
+      "Xal=0.215                               #Leakage reactance of motor(ohm)\n",
+      "pf=0.80                                 #Lagging power factor \n",
+      "Pc=1.8                                  #Core loss(kW)\n",
+      "Pw=0.91                                 #Friction & windage losses(kW)\n",
+      "Ps=0.37                                 #Stray load loss(kW)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ia=P*10**3/(sqrt(3)*Va)\n",
+      "P1=If**2*Rf/10**3                            #Loss in field winding(kW)\n",
+      "P2=3*Ia**2*Ra/10**3                          #Loss in armature(kW)\n",
+      "Pl=(Pc+Pw+Ps+P1+P2)\n",
+      "Pi=Pin*pf+P1\n",
+      "Po=Pi-Pl\n",
+      "eff=(Po/Pi)*100\n",
+      "\n",
+      "#Results:\n",
+      "print \"Efficiency of the synchronous machine:\",round(eff,1),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency of the synchronous machine: 84.3 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.9, Page number: 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Xd=1                              #Direct axis synchronus reactance(p.u)\n",
+      "Xq=0.60                           #Quadrature axis synchronous reactance(p.u)\n",
+      "Va=1                              #Terminal voltage(p.u)\n",
+      "pf=0.8                              #Lagging power factor\n",
+      "Ia=0.8-1j*math.sin(math.acos(0.8))       #Line current(p.u)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "phy=-math.acos(pf)\n",
+      "E=Va+1j*Xq*Ia\n",
+      "delta=cmath.phase(E)\n",
+      "Id=abs(Ia)*math.sin(delta-phy)*cmath.exp(1j*(-pi/2+delta))\n",
+      "Iq=abs(Ia)*math.cos(delta-phy)*cmath.exp(1j*delta)\n",
+      "Eaf=Va+Xd*Id*1j+Xq*Iq*1j\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Generated voltage:\",round(abs(Eaf),2),\"p.u Volt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Generated voltage: 1.78 p.u Volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.11, Page number: 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from pylab import *\n",
+      "import cmath\n",
+      "from sympy import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "P_rated=2000*746                  #Rated power of motor(W)\n",
+      "Xs=1.95                           #Synchronous reactance(ohm/phase)\n",
+      "Xd=1.95                           #Direct axis synchronous reactance(ohm/ph)\n",
+      "Xq=1.40                         #Quadrature axis synchronous reactance(ohm/ph)\n",
+      "pf=1                            #Power factor of the machine\n",
+      "Vl=2300                         #Line to line voltage(V)\n",
+      "\n",
+      "#Calculatons:\n",
+      "Va=float(Vl/sqrt(3))              #volt\n",
+      "Ia=float(P_rated/(Va*3))          #ampere\n",
+      "E1=Va-1j*Ia*Xq                      #From phasor diagram\n",
+      "delta=cmath.phase(E1)             #power angle\n",
+      "Id=Ia*sin(abs(delta))             #direct axis current(A)\n",
+      "Eaf=abs(E1)+Id*(Xd-Xq)\n",
+      "r=symbols('r')\n",
+      "def P(r):                           #Process for finding maximum power\n",
+      "    return Eaf*Va*sin(r)/Xd + Va**2*(Xd-Xq)*sin(2*r)/(2*Xd*Xq)\n",
+      "P1=diff(P(r),r)\n",
+      "#On differentiation,\n",
+      "#P1 = 1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1)\n",
+      "l = solve(1023732.58489791*cos(r) + 355250.305250306*(2*(cos(r))**2-1),r)\n",
+      "P_max = (P(round(l[0],5)))\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Maximum mechanical power:\",math.ceil(3*P_max/10**3),\"kW,3-phase\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum mechanical power: 3236.0 kW,3-phase\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter6.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter6.ipynb
new file mode 100755
index 00000000..1e8c7e97
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter6.ipynb
@@ -0,0 +1,534 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:543f6585f8a1e6c2c290bd192837a35d6ff750409ce379696635948315dd7dcf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: Polyphase Induction Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page number: 318"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "n=3502                              #Speed of motor(rpm)\n",
+      "Pin=15.7                            #Input power(kW)\n",
+      "Ia=22.6                             #Terminal current(A)\n",
+      "R=0.2                               #Stator winding resistance(ohm/ph)\n",
+      "f=60                                #frequency(Hz)\n",
+      "p=2                                 #No. of poles\n",
+      "\n",
+      "#Calculations:\n",
+      "Ps=3*Ia**2*R/10**3                       #Power dissipated in stator winding(kW)\n",
+      "Pg=Pin-Ps                           #Air-gap power(kW)\n",
+      "ns=120*f/p\n",
+      "s=(ns-n)/ns\n",
+      "Pr=s*Pg                             #Power dissipated in stator(kW)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Power dissipated in stator:\",round(Pr*10**3,0),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power dissipated in stator: 419.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page number: 320"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "R1=0.294                            #Resistance of stator(ohm)\n",
+      "R2=0.144                            #Rotor resistance referred to stator(ohm)\n",
+      "X1=0.503                            #Reactance of stator(ohm)\n",
+      "X2=0.209                            #Reactance of rotor referred to stator(ohm)\n",
+      "Xm=13.25                            #Leakage reactance(ohm)\n",
+      "s=0.02                              #slip\n",
+      "Prot=403                            #Friction, windage and core losses(W)\n",
+      "V=220                               #Line-to-line voltage(V)                          \n",
+      "p=6                                 #No. of poles\n",
+      "fc=60                               #frequency(Hz)\n",
+      "nph=3                               #No. of phase\n",
+      "\n",
+      "#Calculations:\n",
+      "Zf=((R2/s+1j*X2)*1j*Xm)/(R2/s+1j*X2+1j*Xm)\n",
+      "Zin=R1+1j*X1+Zf\n",
+      "V1=V/math.sqrt(3)\n",
+      "I1=V1/Zin\n",
+      "a=cmath.phase(I1)\n",
+      "pf=math.cos(a)\n",
+      "ns=120*fc/p\n",
+      "ws=4*math.pi*fc/p\n",
+      "n=(1-s)*ns\n",
+      "wm=(1-s)*ws\n",
+      "Pg=nph*abs(I1)**2*(Zf.real)\n",
+      "Psh=(1-s)*Pg-Prot\n",
+      "Tsh=Psh/wm\n",
+      "Pin=nph*(V1*I1).real\n",
+      "eff=Psh/Pin\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Rotor speed:     \",n,\"rpm\"\n",
+      "print \"Output torque:   \",round(Tsh,2),\"Nm\"\n",
+      "print \"Output power:    \",round(Psh,2),\"W\"\n",
+      "print \"Stator current:  \",round(abs(I1),1),\"A\"\n",
+      "print \"Power factor:    \",round(pf,3),\"lagging\"\n",
+      "print \"Efficiency of motor:\",round(eff*100,0),\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rotor speed:      1176.0 rpm\n",
+        "Output torque:    42.4 Nm\n",
+        "Output power:     5221.6 W\n",
+        "Stator current:   18.8 A\n",
+        "Power factor:     0.846 lagging\n",
+        "Efficiency of motor: 86.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.3, Page number: 325"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "from math import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "R1=0.294                            #Resistance of stator(ohm)\n",
+      "R2=0.144                            #Rotor resistance referred to stator(ohm)\n",
+      "X1=0.503                            #Reactance of stator(ohm)\n",
+      "X2=0.209                            #Reactance of rotor referred to stator(ohm)\n",
+      "Xm=13.25                            #Leakage reactance(ohm)\n",
+      "s=0.03                              #slip\n",
+      "V=220                               #Line-to-line voltage(V)         \n",
+      "p=6                                 #No. of poles\n",
+      "fc=60                               #frequency(Hz)\n",
+      "nph=3                               #No. of phase\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part (a):\n",
+      "Zf=((R2/s+1j*X2)*1j*Xm)/(R2/s+1j*X2+1j*Xm)   #Impedance referred to stator(ohm)      \n",
+      "Zin=R1+1j*X1+Zf                                 #Total input impedance(ohm)\n",
+      "Z1_eq=1j*Xm*(R1+1j*X1)/(R1+1j*(X1+Xm))          #Total equiv. impedance(ohm)\n",
+      "R1_eq=Z1_eq.real\n",
+      "X1_eq=Z1_eq.imag\n",
+      "V1=V/sqrt(3)\n",
+      "V1_eq=V1*(1j*Xm/(R1+1j*(X1+Xm)))\n",
+      "I2=abs(V1_eq)/sqrt((R1_eq+R2/s)**2+(X1_eq+X2)**2)\n",
+      "ws=4*pi*fc/p\n",
+      "ns=120*fc/p\n",
+      "Tmech=nph*I2**2*(R2/s)/ws\n",
+      "Pmech=nph*round(I2,1)**2*(R2/s)*(1-s)\n",
+      "\n",
+      "\n",
+      "#for part (b):\n",
+      "SmaxT=R2/sqrt(R1_eq**2+(X1_eq+X2)**2)     #slip at max torque\n",
+      "n_max=(1-SmaxT)*ns\n",
+      "Tmax=(1/ws)*(0.5*nph*abs(V1_eq)**2)/(R1_eq+sqrt(R1_eq**2+(X1_eq+X2)**2))\n",
+      "\n",
+      "#for part (c):\n",
+      "s1=1                                        #Slip at starting of motor\n",
+      "I2_start=abs(V1_eq)/sqrt((R1_eq+R2)**2+(X1_eq+X2)**2)\n",
+      "Tstart=nph*I2_start**2*R2/ws\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Load component I2 of stator current:\",round(I2,1),\"A\"\n",
+      "print \"    Electromechanical torque, Tmech    :\",round(Tmech,1),\"Nm\"\n",
+      "print \"    Electromechanical power, Pmech     :\",round(Pmech,0),\"W\"\n",
+      "\n",
+      "print \"(b) Maximum electromechanical torque   :\",round(Tmax,0),\"Nm\"\n",
+      "print \"    Speed                              :\",round(n_max,0),\"rpm\"\n",
+      "\n",
+      "print \"(c) Electromechanical starting torque Tstart:\",round(Tstart,1),\"Nm\"\n",
+      "print \"    Stator load current, I2_start           :\",round(I2_start,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Load component I2 of stator current: 23.9 A\n",
+        "    Electromechanical torque, Tmech    : 65.4 Nm\n",
+        "    Electromechanical power, Pmech     : 7979.0 W\n",
+        "(b) Maximum electromechanical torque   : 175.0 Nm\n",
+        "    Speed                              : 970.0 rpm\n",
+        "(c) Electromechanical starting torque Tstart: 77.6 Nm\n",
+        "    Stator load current, I2_start           : 150.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page number: 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "import cmath\n",
+      "from math import *\n",
+      "from matplotlib import *\n",
+      "from pylab import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "V=230                               #line to line voltage(V)\n",
+      "R1=0.095                            #Resistance of stator(ohm)\n",
+      "X1=0.680                            #Reactance of stator(ohm)\n",
+      "X2=0.672                            #Reactance of rotor referred to stator(ohm)\n",
+      "Xm=18.7                             #Leakage reactance(ohm)\n",
+      "f=60                                #frequency(Hz)\n",
+      "p=4                                 #No. of poles\n",
+      "nph=3                               #No. of phases\n",
+      "\n",
+      "\n",
+      "#Calculations and Results:\n",
+      "V1=V/sqrt(3)\n",
+      "omega=4*pi*f/p\n",
+      "ns=120*f/p\n",
+      "Z1eq=1j*Xm*(R1+1j*X1)/(R1+1j*(X1+Xm))           #Stator thevenin equivalent\n",
+      "R1eq=Z1eq.real\n",
+      "X1eq=Z1eq.imag\n",
+      "V1eq=abs(V1*1j*Xm/(R1+1j*(X1+Xm)))\n",
+      "\n",
+      "print \"Hence, the required plot is shown below:\"\n",
+      "for m in range(1,6,1):                      #Loop over rotor resistance\n",
+      "    if m==1:\n",
+      "        R2=0.1\n",
+      "    elif m==2:\n",
+      "        R2=0.2\n",
+      "    elif m==3:\n",
+      "        R2=0.5\n",
+      "    elif m==4:\n",
+      "        R2=1.0\n",
+      "    else:\n",
+      "        R2=1.5\n",
+      "\n",
+      "    s=[0]*202\n",
+      "    rpm=[0]*202\n",
+      "    Tmech=[0]*202\n",
+      "    for n in range(1,201,1):                    #Loop over slip\n",
+      "        s[n-1]=n/200                            #slip\n",
+      "        rpm[n-1]=ns*(1-s[n-1])                  #rpm\n",
+      "        I2=abs(V1eq/(Z1eq+1j*X2+R2/s[n-1]))     #I2\n",
+      "        Tmech[n-1]=nph*I2**2*R2/(s[n-1]*omega)  #Electromechanical torque(Nm)\n",
+      "\n",
+      "    plot(rpm,Tmech)\n",
+      "    title('Electromechanical mechanical torque, Tmech(Nm) vs rpm')\n",
+      "    xlabel(\"rpm\")\n",
+      "    ylabel(\"Tmech\")\n",
+      "    if m==1:\n",
+      "        show()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hence, the required plot is shown below:\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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0yrdlC1tnsXVr7c9R9HwWKquzSE5OdrSwsMiZOXPmgTt37nTp1q3b39u2bVss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\n0qktCAAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x32859d0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ZxwPatQM2bwZGjpStLABTFgxQ18wX1dW4VVKCW6WlCCspQblAgF5/KY8+urrw\n0NKSfeqTFkBpdSlOxp7EoahDiMqNgq+LL8a7jUdfi55QvhIKHDxI8z0MHgxMnUptHMymJBaicqMw\n5fQUOBg4IGBEAHTVGm6LzMujF2R7e2D/fpn7tfzNlSvAZ58Bz5/L/jRhyoLxRl5WV+N2aSlulZbi\nVkkJsnk89NHTwwB9ffTX14eTujqbefwFX8BHSHIIDkUdwuWUy+hn1w+TPSZjqOPQNyfOKyyk6xwH\nD1L37EmTgDlzqMMFo1nU1NZg0ZVFuJpyFafGn0Jbk7b1fic5mfonTJ4MrFkjH0s+IggB+val9xXT\np8tWFqYsGA0il8fDteJiXP2rKQB/K45++vowUXm/gsLqgxCC8KxwHI4+jBMxJ+Bk6ITJHpMxzm0c\nDNQNGt5RYiK1pgYG0lwP8+YBI0bI/jayhXPw2UEsvrIYP/v8jInuE9/6uYgIYOhQYO1aYPZsKQrY\nCO7dA/z8qDutLBMvM2XBaDSEECRUVSG0uBhXi4oQVloKWzU1+BgYYKiBAbx0daEkT7dnYuRF8Qsc\niTqCw9GHISRCTPGYgskek2Gv30yXlZoa4NQp4NdfgdRUOtOYNYvmCmM0icicSIw+MRqTPSZjrffa\n/8yEb90Cxo4Fdu0CxoyRkZANZPBg6mAnS4XGlAWj2fCFQoSXl+PPoiJcLCxERnU1BhkYYJihIXwM\nDGDQwpP2l9eU4/fY3xEYGYj4gniMdxuPKR5T0MWii2SW4qKi6BXs+HFg4ECamrRzZ/GP8wGQy83F\n8GPD0ca4DfYN3/d3WpTz52lqjWPHgH79ZCxkA7hxA/jkE2q7kJXXO1MWDLGTWVODS4WFuFhYiJsl\nJXDX0sIwQ0MMNzSEawvxsiKE4HbGbQRGBuJ03Gl423pjuud0DHEcIr1sqGVlNP/D9u0059iSJfQW\nk8XINIoKXgX8TvmBy+Pi1LhTOBusi6VLadK+lqKDCaFJA9auBYYNk40MTFkwJEq1UIibJSW4WFiI\ncwUFUFNQwBhjY4w2NkZHLS25UxyZZZkIigxCYGQgVJVUMcNzBiZ7TIaplqnshOLzgd9/B7Zsof6U\nS5ZQo3gLn7FJE4FQgM9DPseFiIeoDbyCaxcM4OIia6kax9GjwN69NHmALGDKgiE1CCF4wuXiVH4+\n/sjPR7VQiNHGxhhtZITuuroyc82trq3G2fizCIwMRHhWOMa3HY/pntPRuVVn+VJmhADXrwMbN1K7\nxooV1E34FchSAAAgAElEQVTmA3MuaCpbtxKsf7gEFr1CETbzKow1jWUtUqPg82lJlqtXATc36Y/P\nlAVDJhBCEFtZSRVHQQFyeDz4GhlhgokJeuvqSjwtOyEET7OfIjAyEMdjjqO9eXtM95yOUS6jWka1\ntjt3aE3OhARg+XIaTcZqlL+Vb7+lWWNDQwl2J32Ns/FncW3qNdnOGJvAihU0eHD7dumPzZQFQy5I\nqarCH/n5OJaXh3w+HxNMTOBnYoL2Yl6qyq/Ix5HoIwiMDERZTRmme06Hfzt/2OjZiG0MqfLgAbBu\nHbV8rl0LTJnS4iv/iRNCaPqOs2eB0FBafJIQgnVh63Di+Qncmn5LYokIJUFKCtCtGy20Ke3AQaYs\nGHJHbEUFjuXl4WhuLpQVFDDxL8XhqKHRpP4EQgGupFzB/oj9CH0RiuHOwzHDcwb62PaRu1KjTebe\nPZr5rrCQLlONGCFf0WUygBBa+jQsjEZCv56+Y2noUtxIvYFrU69BW1Wy9UnESf/+1Kt6wgTpjtui\nlcWmTZuWHz58eLKCgoLQ3d09OjAwcHpFRYXm+PHjT6Snp9vY2tqmBQcHj9PT0yv5W0CmLFoMhBCE\nl5fjaG4uTuTnw0pVFVNMTTHJ1BSGDTDuphSlICAyAEGRQbDQscAMzxmY0HZCo9JAtCgIoUkMly+n\n9d2/+67pxaFbOEIhdTWNjKS75E1Z4wkhmHNhDlKLU3HR7+Kbo+3lkOPHaQzn5cvSHbfFKou0tDTb\nvn37Xo+Li2ujqqpaM378+BNDhgy59Pz5czcjI6OCr7766vvvvvtuaXFxsf7mzZuX/S0gUxYtklpC\ncKO4GEG5ubhQWIgB+vqYbmaGgQYG/woArORX4o/YP7A/Yj9i82Mx2WMyZrSf0aC0D+8NQiENIFi5\nkvpbbtlCXW8/EGpraQxFWhqtlf2uooYCoQAT/pgAIREieGxwi8gAXFlJYzUTE9+YAV9iiFtZSG1O\nr6OjU6asrMyvrKzUqK2tVaqsrNRo1arVq3Pnzo3w9/cPAgB/f/+gM2fO+EpLJobkUOJwMMDAAIfb\ntEFat27or6+PtWlpsLl/H8tTUnAy9QHmXpgLy22WOP78OD7r8hkyF2Vi26BtH5aiAGgcxqRJQFwc\nTZHeqRO1jJaXy1oyicPn05+enU1nFPVVv1VUUMThUYdRUl2CJVeXSEfIZqKhQWMtfv9d1pI0D6ku\nQ+3du3fO4sWLf1BXV68aNGjQ5UOHDk3R19cvLi4u1gcAQgjHwMCgSPQaoDMLFT8/LHd0BAB4e3vD\n29tbajIzxEd+RT62Rp5EQE4WinU6w1KZg89sHDDP2pmVl61LVhZVFlevUnvG1KnvZWBfTQ0wbhyd\nWP3+e+MMwMVVxei2vxsWey3GnI5zJCekmLh4kR7Ku3clN8bNmzdxs05Qx9q1a8U6swAhRCotOTm5\ndZs2bWILCgoM+Xy+kq+v7+lDhw5N1tPTK677OX19/aK6rwEQrYsXCaNlUiuoJRcTL5IxJ8YQ3U26\nZMqpKeRm6k1SXcsnZ/PzybCoKGJ45w5ZkJRE4isqZC2ufPHwISGdOxPSsych0dGylkasVFQQMmgQ\nIWPGEFJT07Q+EgsSickWE3LtxTXxCicBamoIMTQkJDVVemPSy7v4ruFSu115/Phxp+7du98zNDQs\nVFJSqh09evSp+/fve5mZmeXk5OSYAUB2dra5iYlJnrRkYkiOlKIUrLy+EjY/2mBt2FoMbD0Q6QvT\ncXDUQfSx7QNVRSWMMDLCeXd3PO7YERoKCugTGYl+kZE4mZ8PvlAo658ge7p0Ae7fByZOBD76CFi6\nlNYPb+FwuTRzrJERNf42NUbR0dARx8ccx8Q/JiKxMFG8QooZFRVg1Cjg9GlZS9J0pKYsXFxc4h88\neNCtqqpKnRDCCQ0N7e/q6ho7fPjw80FBQf4AEBQU5O/r63tGWjIxxEslvxKHnh2C9wFveO33QhW/\nCiGTQ/Bw1kPM6TjnrV5Ntmpq2Ghvj/Ru3TDL3Bw/Z2bC9sEDrE5NRWZNjZR/hZyhqEjdhKKjgcxM\nGgp8/ryspWoypaU016KDA62X3dzM7h/ZfYR13usw+sRoVPDkW5H6+tKKfi0Vqdosvv/++6+CgoL8\nFRQUhB06dHj622+/zSovL9ceN25ccEZGhjVznW15EELw6NUjBEQEIPh5MLysvDDDcwaGOw//O2No\nU4ipqMCurCwcy8vDAH19fGFlhW46OmKUvIVy7Rrwv//RWcfPP8u+lmgjKCykRYu8vICffhKfGYYQ\ngmlnp0FIhDjoe1C+Ur7UoboaMDWlgXrSOGwt1nW2qTBlIZ/kV+TjcNRhBEQGoIpfhRntZ2Bqu6mw\n1LEU6zhltbUIyMnBT5mZMFdRwReWlhhlbPze1t9oEJWVwKpVNFPdzz8DH38sa4nqJS+PBqf5+NBw\nEnEfvkp+Jbr91g3zO8/H/zr9T7ydi5GxY6ln1LRpkh+LKQuGzBAIBbicchkBEQEIfRGKkS4jMcNz\nBnrb9Jb43VwtIThbUIBtL18ii8fDAgsLzDQ3h86HXKHu/n1gxgy6NLVzJ71tlUOysqiiGD8eWL1a\ncoHqiYWJ6BHQA39O+hOdWnWSzCDN5PBh4ORJ6SxHMWXBkDp1I6stdSwxo/0MjHcbL7PI6odlZdie\nmYkrRUWYZmaGBZaWsJF24h15obqa5pgKDKRV+0aPlrVE/yI9nRYrmj2b2uclzcnYk/jq6leInBsJ\nHVX5W7YsLqaZaPPyJJ8riikLhlR4PbJ6SrspmO45Xa4C5jKqq/FLVhYCsrMx3MgIS62s0EZTU9Zi\nyYb792lSwj59gB9/rD+6TQokJ9MZxaJFwOefS2/cOefngC/kI3BkoPQGbQQ9egBr1gADBkh2nBYb\nwc2QfwghCM8Kx/8u/O/vyOrPu36OzEWZ+GHgD3KlKADAWk0NW1q3RnLXrnBQV0efyEiMff4cTz6A\nyOf/4OUFRETQNR5PT5qoUIbExVFv3xUrpKsoAGDboG24m3EXJ2NPSnfgBuLjA4SEyFqKxsNmFoy/\njdX7I/ajurYaM9rPgH87f1joWMhatEZRIRBgX3Y2tr58ibaamlhhbY1eurpy6x0jMU6fBubNA+bM\nAb75pvn+qY3k2TNaTfa77+hkRxaEZ4Vj+LHheDrnqdydx+Hh1NQUEyPZcdgyFEMsvMlYPbP9TPSy\n7tXiL641QiEO5eZic0YGzFRUsMLaGoMNDFr872oU2dk0TUhNDU1SaCGdC+bjx9Tb55dfZO+ktT5s\nPcLSw3BlyhW5Sm0vEFBfhIgIwMpKcuOwZShGs0guSv47snpd2DoMbD0QGV9kIMg3SCpeTdJAVUEB\ns8zNEd+lCz61sMBXL17AKyICV4qKIO83R2LD3JyudQwcSBMTSiE/9r17wJAhtO60rBUFACzvtRwV\n/ArsebxH1qL8C0VFasu5ckXWkjQONrP4AKjkV+Jk7EkERAQgriAOkz0my52xWpIICUFwfj7WpKXB\nWFkZ6+3s4K2nJ2uxpEdYGE3tOnUqrdIngWWpy5fpktOhQzTwTl6IzY9F78DeePq/p7DWtZa1OH8T\nGEh1+YkTkhuDLUMxGoQosnp/xH78/vx3dLfqjhntZ2CY07BmRVa3ZGoJwbHcXKxNT4eNqirW2dmh\nh+57WljpdfLy6NW8qgoIDqb1SsXE8ePAggXUVNK9u9i6FRvrw9bjfuZ9XPS7KDcz57Q0oGtXICdH\ncnEnbBmK8U6yyrKw5e4WuO9yx6RTk2Cra4voedG44HcBo9uM/mAVBUBrbEwxM0Nc587wMzXFpLg4\n+ERFIbysTNaiSR4TE1owol8/uix1/75Yut21C/jyS1ovWx4VBQAs7bkUWeVZOBp9VNai/I2tLa1z\nERcna0kaDptZvAdweVycjjuNQ1GH8PjVY4xxHYMpHlPeC2O1JOEJhQjIycGG9HR01tbGJnt7ODex\nbniL4sIF6o6zfj31mGrCOUII8O23NBnglSuAvb0E5BQjj189xtCjQxE9LxommlIsV/cOZswAOnYE\n5s+XTP9sGYoBgHoz3Ui7gYPPDuJ84nn0sOqBqe2mYrjTcKgrq8tavBZFlUCAX7KysOXlS4w1NsZq\nW1uYNTVvdkshMZHmzPbyAnbsaFQ4sVAIfPEFNYWEhIh1RUuiLLm6BDncHBwadUjWogAADh6kCYQl\nVUGPKYsPnJi8GByKOoQjUUdgpmWGKR5TMNF9otzcLbVkCvl8bExPx4GcHHxmaYnFlpbQfp9zT5WX\nA9OnAy9f0mRF5ub1foXPp3fEaWn0QteS/AS4PC5cd7riyOgj6GXTS9biICODzixycyVTCJHZLD5A\ncrg52H5/Ozrs6YDBRwZDgaOAK1Ou4PGcx1jQbQFTFGLCUFkZPzg44EmnTkiuqoJTeDh2ZWW9v4WY\ntLXpbe3QodTaGhn5zo9XVtLJSHEx9X5qSYoCALRUtPDDwB8w/9J81AprZS0OrK0BHR0gNlbWkjQM\npizklCp+FY7HHMeQI0PQZmcbPMt9hq0DtyJtQRo29dsEV2NXWYv43mKrpobDbdrgors7/igoQNtH\nj3A6P//9jNHgcGiU99atNFnR2bNv/FhxMXWJ1denXk8t1bQz1nUsTLVMsTN8p6xFAUBTed26JWsp\nGgZbhpIjRHaIo9FHcTr+NLpadMUUjynwdfGFpsoHmiBPxhBCcKW4GItTUmCqrIyfHB3R9n1NVhge\nTqcOCxdSF6e/DN/p6TR9h48P1SmSWDKRJvEF8egV2AvR86JhpiVbg8v+/cCNGzR1ubhhNov3DFHy\nvqMxRxH8PBgW2haY2HYiJrpPRCvtVrIWj/EXtYRg96tXWJeWhnEmJlhnawsDZWVZiyV+Xr4ERowA\nOnQAdu/G02hlDB8OfPUVjaV4X1gWugyvyl/h4KiDMpUjPp4q4tRU8ffNlMV7Qmx+LI7FHMPR6KNQ\nUlCCX1s/THSfCCdDJ1mLxngHBXw+vklNxcn8fKyxtcWcVq3ev6p9XC4wYQLyc2rROe0kftijhTFj\nZC2UeOHyuHDe4Ywz48+gs0VnmckhFALGxrTEeisx3xsyA3cLJqM0A9/f/R6euz0x8NBAVPGr8PvH\nvyN+fjxWe69miqIFYKSsjF+dnBDarh1+z89Hh8ePcaO4WNZiiRctLQSMOIOQWGvEGPXBmB45spZI\n7GipaGGd9zosvrJYprYoBQUazCjjjPINgikLCZNfkY9dj3ahV2AvdNjTASnFKfjJ5ydkfJGBrQO3\nooN5BxY41wLx0NLC9XbtsNrWFjMSEjD2+XO8rK6WtVjNhhBa+vTbzUro8nQPtPxG0qtZYqKsRRM7\n0zynoaS6BGcT3mzUlxbduwN378pUhAbxHjuRy46ymjKcjT+LYzHHcO/lPQxxHIKlPZZiYOuBH3S6\njfcNDoeDMcbGGGJggO9evkT7J0+w3Noan1tYQLkFWoF5PBrQ/fw5zQZiavqXp5SFBdC7N43F6NZN\n1mKKDUUFRWwduBWfXvoUQxyHyOy/2aMHsGSJTIZuFMxmISbKa8pxPvE8gp8H40baDfS26Q2/tn4Y\n4TyCeTJ9ICRVVuKTpCTk8XjY7eQErxaUpLCggJbvNjCgnjlaWq994NIlwN+fppX18ZGJjJLC57AP\nhjgOweddpVzS7y+qqgAjI3oM1MWYfIEZuOUILo+LC4kXEPw8GNdSr6GXdS+McxuHEc4joKfWwiKW\nGGKBEIIT+flYlJyMYYaG2GxvL/deU7GxwPDhtAbFxo3vcI29d4+61u7YIR8FK8RETF4M+gb1ReJn\niTL733bsSHerl5f4+mQGbhnD5XFxIuYExgSPgcU2CxyKOoQRziOQtiANF/wuYGq7qUxRfMBwOBxM\nMDFBbJcuUFFQgOujRziYkyO3AX0hIYC3N11t2ry5nhiK7t1p1sAFC4CAAGmJKHHamrTFMKdh2Hpv\nq8xk6NwZePRIZsM3CDazaAAVvApcSrqE4NhgXEm5gu5W3THOdRxGuoyEgbqBzORiyD+PysowNzER\n2kpK2O3kBBc5CX0mhN7JbtxIM3707NmILycm0gp8CxbQjILvAWklaei4tyPi5sfJJH1OQABw/bp4\ng/PYMpSUqOBV4M/kP/F77O8ISQ5BN8tuGOc6Dr4uvjDUMJSqLIyWjYAQ7MzKwrr0dHxpZYUvraxk\nGpvB59PrfFgYTQbYpPTiGRk0PcjEidR96j3w6Pv00qdQUVTBtkHbpD52TAwwZgyQkCC+PpmykCDF\nVcW4kHgBp+JP4dqLa+hq2RXjXMdhVJtRMNIwkvj4jPebtOpqzElIQCGfj/0uLvD8jxVZ8uTnA+PH\nA6qqtMJds2zweXlUYQwZQqcoLVxhZJdnw+1XN0TNi4KljqVUxxYIaGLGjAyaf0scMJuFmMnh5mDP\n4z0YdHgQbH60wR9xf2CUyyikLUzD1SlXMbvjbKYoGGLBVk0Nlz088KmFBQY+e4avU1NRI8WMtk+e\n0CJ53brR+kfNdtYyMaFrJyEhNB+InN941oe5tjlmdZiFb299K/WxFRVphpXHj6U+dIP5IGcWaSVp\nOB13GqfiTyEmLwaDHQZjdJvR8HHwgZaK9O/2GB8er2pq8ElSEpIqK7HfxQXddHQkOt7Bg8DixbQM\n6tixYu68qIjOMPr0AX74oUXPMAorC+G0wwmPZj+Cvb50y/99+SV1XV6xQjz9sWWoJhKXH4dTcadw\nKv4UMkozMNJ5JEa3GY1+dv2gqqQqJmkZjIZDCEFwfj4WJidjookJvrWzg4aioljH4POpkvjzTxpT\n5+Ym1u7/QZTD3MsL+PHHFq0w1txcg/TSdASODJTquCdOAMeO0eMkDpiyaCBCIsSjrEc4m3AWp+NP\no7ymHKPbjMboNqPR07onlBRY8DpDPijg87EwORkPyspw0MUF3cUUzJebC4wbRwPsjhyRQrGikhIa\nsNepE/DLLy1WYRRXFcPhFwc8mfMEtnq2Uhs3KQno35+mhBcHTFm8gyp+FUJfhOJc4jlcSLwAA3UD\nDHcajtFtRqNTq05Q4HzwJhqGHHM6Px+fJCVhupkZ1tjaQqUZKUMePKCKYto0YM0aKdagKC39Z4ax\nbVuLVRgrrq1AUVURdg/bLbUxhUKq0FNTAUMxOFwyZfEaudxcXEy6iHMJ53A99To6tuqIEU4jMNx5\nOBwMHKQsLYPRPHJ5PMxOSEBGTQ0Ot2nT6EJLhNBr9PffA/v20dIUUqe4GOjblxZq2LChRSqM/Ip8\nOO9wlrpnVO/e1BO5X7/m9yVuZdHi1mIIIYgriMO5hHM4l3AOsfmxGNh6IMa6jsX+EftZDASjRWOq\nooKzbdsiMCcHH0VGYqm1Nb6wtIRiAy64xcV0JpGTAzx8CNjaSlzcN6OvD1y9SkPD1dWBVatkJEjT\nMdY0xvT207Hl3hb85POT1MZt3x54+lQ8ykLctJiZxQVXTZxLpAqiprYGI5xHYITzCPSx6cMM1Iz3\nktSqKkyLjwcBEOTiArt3ZJkLD6fxEyNH0lmFijwkN87JoR5Ss2dTV58WhijuIm5+HEy1TKUyZlAQ\ncPkycPRo8/v6YJehnDJXYaTzSIxwHoF2pu1YDQjGB4GAEGzPzMR3GRnYbG+PGWZm/zr3CaG25G+/\nBXbvpplj5YrMTLq2sngxMH++rKVpNJ9e+hQayhr4fsD3UhkvKoramuLjm99Xi1YWJSUlerNmzfrt\n+fPnbhwOhwQGBk53dHRMGj9+/In09HQbW1vbtODg4HF6enolfwsoB7mhGAxZE1NRgclxcWitpoZ9\nzs4wUFZGURG9aU9PB4KDm5i2QxqkpVGFsXEjMHmyrKVpFC9LX6Ld7nZI+TwF+upiCq1+B3w+NXLn\n5r4hTXwjadER3AsWLPhpyJAhl+Li4tpERUV5uLi4xG/evHnZgAEDriYmJjr169fv2ubNm5dJUyYG\noyXQVlMTDzp0gJWaGto/foyfbpSgXTvAyopWWZNbRQFQ40lIyD8BHy0IK10rDHMahj1P9khlPGVl\nwNWVzjDkDkKIVFpJSYmunZ3di9ffd3Z2js/JyTElhCA7O9vM2dk5vu52AETr4kXCYDAIqakhZOzW\nAqJw6i6ZfCWV8IVCWYvUcO7dI8TYmD62IJ7lPCPmW81JNb9aKuPNmkXIjh3N74de3sV3DZeaN1Rq\naqqdsbFx/vTp0wOfPXvWrmPHjk9+/PHHhbm5uaampqa5AGBqapqbm5v7H0sS78gRrAkPBwB4e3vD\n29tbWmIzGHJDQgIwaRJgbm6IZxM7YmF+PPpGRuJImzawUlOTtXj14+VFLbijRgHXrkkwnFy8eJh6\nwMPUA0eij2BG+xmSH88DiI5u/Pdu3ryJmzdvil2evxGn5nlXe/ToUSclJSV+eHh4Z0IIFixY8OPX\nX3+9Xk9Pr7ju5/T19YvqvgabWTA+cIRCQvbuJcTQkJCdO+lrQggRCIVkc3o6Mblzh/yRlydbIRvD\n4cOEWFkRkp4ua0kaTGhKKGmzow0RCAUSH+vmTUK6d29+PxDzzEJqNgtLS8tMS0vLzM6dOz8CgLFj\nx558+vRpBzMzs5ycnBwzAMjOzjY3MTHJk5ZMDIa8k5v7TyXTsDDgk0/+iXFT4HCw1Noa59zdsSQl\nBfMSE1EpEMhW4IYwaRKwaBEtoFRQIGtpGkRfu75QU1LDpaRLEh/L3Z3OLKSYkLhBSE1ZmJmZ5VhZ\nWb1MTEx0AoDQ0ND+bm5uz4cPH34+KCjIHwCCgoL8fX19xZRGi8Fo2Zw4QZckXF1pHMXbVm266ugg\nolMnlNXWotvTp0isrJSuoE1h4ULq5ztkCMDlylqaeuFwOPiy+5fYcm+LxMcyMAB0dMSXI0psiHOa\nUl+LjIxs16lTp0ceHh7PRo0adaqkpES3sLDQoF+/fqGOjo6JAwYMuFJcXKxX9ztgy1CMD4y8PEI+\n/pgQZ2dCHjxo+PeEQiHZk5VFjO/cIcG5uZITUFwIhdSaO2gQITyerKWpF14tj9hstyHhmeESH2vw\nYELOnGleHxDzMlSLCcpjcRaMD4FTp2js2qRJwPr1NFtGY3laXo6Pnz/HUENDbG3dulkJCSVObS1N\nYGVlRaMK5TzYduu9rXiW+wyHRh2S6DhLlwLa2sDXXze9jxYdZ8FgMN5MYSFVEEuXAidPAlu3Nk1R\nAEAHbW086dQJGTU16BURgfTqavEKK06UlOh628OH9EfLOTPbz8SFxAvI4eZIdBwPD/mLtWDKgsGQ\nIYTQWhNuboCRERAZCfTo0fx+9ZSUcNrNDR+bmKDr06f4s7Cw+Z1KCm1tWuf155+pppRj9NX1MaHt\nBOx+LNnU5fKoLOpdhkpISHDeunXrl2lpaba1tbVKAF0aun79el+pCMiWoRjvKWlpwLx5QFYWTSfe\ntatkxrlTWoqJsbGYamqKtXZ2UJLXpZ6ICOohdf48LRQup8Tmx6JvUF+kL0yXWBJTHo8auYuLmz7D\nlHpuKA8Pj6h58+bt6tChw1NFRUXBX0KQjh07PhGXEO8UkCkLxnuGQEBvojdsoBkwvvySpnmQJHk8\nHvzi4iAkBCdcXWEsF2lp38DFi8CsWXKfw2TAoQHwb+ePyR6Sy3Xl5kZnnZ6eTfu+1OtZKCsr8+fN\nm7dLXAMyGB8yz57Ra6G2NnD/PuDoKJ1xTVRUcNnDA9+kpqLTkyc41bYtOmprS2fwxjB0KLXqDh0K\n3LtHa2PIIZ93+Rzrbq3DJPdJEsuA7eYGPH/edGUhbt5qsygqKjIoLCw0HD58+PmdO3fOz87ONi8q\nKjIQNWkKyWC0dEpLgS++AAYMoEtP165JT1GIUORwsMHeHtscHOATFYVDOZI10jaZ+fNpLe8xY+h6\njBwy1GkoiquK8SDzgcTGcHMDYmMl1n2jeesylK2tbRqHw3nrGlVqaqqdxKSqA1uGYrRkCKGFbL76\nisafbdpEDdmy5nlFBXxjYjDU0BBb7O2hLG/utQIBVRZGRtSgI4d2lu33t+PRq0c4OkYMlYrewO+/\n02WoM00MU27R9SyaAlMWjJZKTAy9SeZygZ075c9mW8znY1JcHCqFQgS7usJE3uwYXC5NPjhnDvDZ\nZ7KW5j8UVxXD7ic7JH2WBGNNY7H3HxsL+PoCiYlN+77U4yx27tw5v7i4+O+Fw+LiYv1ff/31E3EJ\nwGC8b5SV0dRHffsCEybQVB3ypigAQF9ZGefd3dFDRwednzzBk/JyWYv0b7S0gLNnqSfA9euyluY/\n6KvrY1SbUQiMDJRI/46OwMuXgLyEydSrLPbu3TtHX1+/WPRaX1+/eO/evXMkKxaD0fIQCIDffgNc\nXKiN4vlzap9QVJS1ZG/ndTvGQXmzY9jb03U8Pz/gxQtZS/Mf5nWahz1P9kBIxJ/1T1mZ/nxxlFgV\nB/UqC6FQqCAUCv/+nEAgUOTz+RJ29GMwWhbXrwMdO9JyDefOAfv3A8biX5mQGGOMjXHT0xPr09Ox\nODkZAnlanu7bl3pIjRwJyNnsp3OrztBV1cXVlKsS6V+ejNz1KotBgwZdnjBhwvFr1671Cw0N7T9h\nwoTjPj4+IdIQjsGQd5KS6LryzJn0enbrFtCpk6ylahpumpp42KEDIrhcjIiORlltraxF+of58+la\nnr+/XOXu5nA4mNtpLnY/kUxEt8h9Vh6oV1l89913Sz/66KMbu3btmrd79+65/fv3D/3++++/koZw\nDIa8UlxMA+q8vOg1LC4OGDtWLp12GoWBsjIue3jASk0N3SMikFpVJWuRKBwOLeqRmwusWydraf6F\nn7sfwtLCkFmWKfa+XV3lZ2bRIG+oyspKjYyMDGsXFxepr54xbyiGPFFZCfzyC8155+tLM8Oamcla\nKvFDCMGOrCxszMjA725u6KmrK2uRKLm5QOfOwPbt1LVWTph/aT6MNYyxxnuNWPuNiqJOEk1RGFL3\nhjp37tyI9u3bR4iWniIiItqPGDHinLgEYDBaAnw+sHcv4OQEPHoE3L5N3f/fR0UB0AvNZ5aWCHRx\nweMWPk0AACAASURBVOiYGATJi+Hb1BQ4fRqYO1d+1mcAzO04F789/Q0CoXgrFTo6Uru+PKwI1qss\n1qxZs+bhw4ddRR5R7du3j3jx4oX8Jm1hMMQIITQ4qm1b4PhxWm/i5Enq8fQh4GNggLC/DN9LU1Lk\nw/DdsSPwww+00l5ZmaylAQC4m7rDQscCl1Mui7VfdXV6Q5KWJtZum0S9ykJZWZmvp6dX8q8vKSjI\nj4WJwZAQoaF0xWPTJrr0dO0a0KWLrKWSPm3+Mnw/LC/H6JgYcOWhzvfUqcBHHwEzZlCNLgfMbD8T\nAREBYu/XyanpgXnipF5l4ebm9vzIkSOTamtrlZKSkhw/++yzX7p3735PGsIxGLLg9m2gXz8aI7Fk\nCfD4Mc2c3dKN183BUFkZVzw8YKKigp4REciqqZG1SMBPP9FC1du3y1oSAMB4t/EIfRGK/Ip8sfbr\n7AwkJIi1yyZRr7L45ZdfPnv+/LmbqqpqzcSJE4/p6OiU/fjjjwulIRyDIU3CwqhLv78/jQGLjQXG\njwfkLW2SrFBRUMBeJydMNDGB19OniOZyZSuQqipdE/z+e+qzLGN01XQxwnkEjkQfEWu/8qIsxFbM\nW1INANG6eLHJRcsZjPq4cYMQb29C7O0JCQgghMeTtUTyz7HcXGJ85w65Ulgoa1EICQkhpFUrQl69\nkrUk5EbqDeL+qzsRCoVi6/PKFUI++qjx36OXd/Fdi+u9Z3r06FHnUaNGnW7fvn2Eu7t7tLu7e7SH\nh4ecFfxjMBoHIcCNG4C3N60v4e9P0ypMny75QkTvAxNMTPCHmxumxMcjMDtbtsIMGgT87390Gsjn\ny1SU3ja9UcGvwJNs8dWGc3KSj5lFvXEWTk5OiVu3bv2ybdu2MXUN27a2tmmSFg5gcRYM8UIIEBJC\njdbZ2cCqVXTJSaneMmCMN5FQWYkhUVGYZGqKtba2EisEVC9CITBsGI1i27pVNjL8xbe3vsWr8lf4\ndeivYulPKKQ5FXNzadGshiL1SnnGxsb5LK6C0dKprQVOnKDL24TQ+hITJjAl0VycNTRwv0MHDI+O\nRlp1NX5zdoaKLIw8CgrAoUM010q3bjScXkb4t/OH5x5P/DDwB6grN7GAdh0UFGi8RWIi9RqWFfXO\nLK5cuTLwxIkT4/v37x+qoqLCA+jd/ujRo09JRUA2s2A0g4oKICCAuuXb2ABLlwKDB3/Ynk2SoFIg\nwKS4OJTW1uJU27bQk5UWfvKEHuD794HWrWUjAwCfwz6Y2m4q/Nz9xNLfxx/TsJKJExv+HanPLIKC\ngvwTEhKca2trleouQ0lLWTAYTaGwkKYS2rkT6NGDBtTJY02J9wUNRUWcdHPD4pQU9Hj6FCF/5ZeS\nOh070oyOEyYAd+8CMiroNKP9DOx9sldsykIePKLqnVk4OzsnxMfHu7yrxKokYTMLRmNITAR+/pmW\nQBg1isZJfCjR1vLCtpcv8WNmJv708ICbpqb0BSCEHnx7e2DbNumPD6CmtgYW2yzweM5j2OrZNru/\nQ4eAS5eAY8ca/h2p54bq3r37vdjYWFdxDchgiBtCgCtXgKFDgZ49AT09WtJ0/36mKGTBIisrbLS3\nR9/ISNwrLZW+ABwOXXv84w/gwgXpjw9AVUkVfu5+OBB5QCz9yUMU91tnFrW1tUpKSkq1Li4u8Skp\nKa3t7OxSVVVVawB6tx8VFeUhFQHZzILxFior6R3XTz9RQ/WCBdSzSb35NkWGGAgpKsLUuDjsd3bG\ncCMj6Qtw7x5d6H/8GLC0lPrwkTmRGHl8JFIXpEKB0zyjf0kJYGVFU2E11N4mNZtFly5dwp8+fdoh\nJCTER1yDMRjiICOD2iICAqg9YudOGi/BjNbyhY+BAS64u2NkTAw28vmYbm4uXQG6d6d3EBMn0qAa\nKRvdPc08oauqizsZd9Dbpnez+tLTAzQ0gFevAAsLMQnYSN6690QaSVrxFAzGuxAI6FLT7t00d5O/\nP/DggUwdXhgNoIuODm56esInKgq5fD6WWllJNxZj6VKqKNaupcVHpMwk90k4HHW42coC+MfILXfK\nIj8/33jbtm2L3jSN4XA4ZNGiRbKxHDE+KHJz6Qxi717A0JAm9zt6FJCF3ZTRNJw1NHC3fXv4REUh\nu6YG2x0coCAthSGKv+jQgU4/+/WTzrh/4efuB889nvhl8C9QVVJtVl8iu0XfvmISrpG8dSFNIBAo\nlpeXa3O5XK3XW3l5eSPiCBmMxkEIcPMm9X50cQFSUmhNicePaa1rpihaHq1UVXGrfXtEcrmYFBcH\nnjTraJuaAkFBNK15bq70xgVgpWsFdxN3XEq61Oy+ZO0++1YDd/v27SMiIiLaS1me/8AM3B8Oubn0\nJvC33wBFRVoMbcoUul7LeD+oFgrhFxuLSqEQf7i5QVNRUXqDf/01Ddq7dEmqBq7fnv6GkOQQnBx3\nsln9nDtHl2EvNVDvSN11lsGQJHw+cPYsrWft7EwrZe7bR11fP/uMKYr3DTUFBQS7ucFMRQWDoqJQ\nIs16oatXA0VF1CNCiox1HYurL66ipLqk/g+/A1knFHyrsggNDe0vTUEYHxaxsTRgzsoK2LIFGDEC\nePkSCAwEevVink3vM0ocDgKcndFBSwsfRUYij8eTzsDKysDhw9TYHRsrnTEB6Knpob99f/wR+0ez\n+rG3BzIzAVnVnXqrsjA0NCyUpiCM95+iImDPHsDLC+jfny41hYUBd+7Q6piNyajJaNkocDj4ycEB\nIwwN0TsyEi+rq6UzsKMjTTns5yfVq+4k90nNLoqkokLDRdLTxSRUI5HqMpRAIFBs3759xPDhw88D\nQFFRkcGAAQOuOjk5JQ4cOPBKSUkJW3R4z6iqAoKDgZEjATs74Pp1YOVKGiuxeTNdemJ8mHA4HKy1\ns8Mcc3P0ioxEUmWldAaeOZOejF9/LZ3xAAxxHILInEhklmU2q5/WranDhyyQqrL46aefFri6usaK\n8kxt3rx52YABA64mJiY69evX79rmzZuXSVMehmQQCIDQUFpIqFUraoMYPZouM504QcsOsNTgDBGL\nrKzwtY0NvCMjESWNUq0cDj0pjx0Drl2T/HgA1JTUMMZ1DI5FNyK50xuwt5edspDaXzYzM9Py0qVL\nQ1auXLlh27ZtiwDg3LlzI8LCwvoAgL+/f5C3t/fNNykM3pEjWBMeDgDw9vaGt7e3tMT+f3v3HdbU\n+fYB/HvCVFRQkSHDQDCsEEABR6tiFbe4cVQctW+r1lpHcXRrK0Jr3W3t0J/WVuuoW8RRRWsV0bKH\nCgoyBFTAgVQZOe8fT1NRGQGSHND7c13nwpyc85xbSHLnPJOoiOeBmBhWJfzbbyxJvP46EBwMaHvg\nLml63rS0RCsdHfjFxWGfTIZuxsaavaCpKRvAM2UKEBcHtGmj2euBVUW9F/4egl4JqncZNd1ZRERE\nICIiot5l10qda7TWtI0ePXpXdHS0Z0RERK8hQ4Yc5HkeJiYmRcrnFQoFV/mxcgOtwd1oKRQ8HxvL\n8x9+yPNSKc/b2fH8Rx/xfEqK0JGRpirszh3e9OxZ/nhhoXYu+N57PD96NHsxa1iFooK3XmnNx+fF\n17uM33/n+aFDVTsW2l6DWx0OHTo0xMzM7Janp2cMX02/X47jeKGmQSeq43kgNpa1Ozg6si6vpaVs\nfMS1a2xGBZrpldTXwLZt8burKyYkJyOsQAt9bEJC2OLrW7Zo/FIiToQJbhMa1NAtZJuFVqqhzp07\n1/3AgQP+YWFhgx49emR4//79VoGBgVvNzc3z8/LyLCwsLPJyc3MtzczMbmkjHlI3yiqmXbuA3btZ\nm8SYMWzajc6dqZsrUa+eJiY46OYG/4QEfO/oiOGanLHW0BD49Vc2DUjPnqxRQIMmuk3E4G2DEdwn\nuF4z0drbA+np7D2p7fedVu4sgoODP8jKyrJJT0+3++2338a99tprJ7du3Rro7+9/YMuWLZMBtiLf\n8OHD92kjHlK7igo2w/PChay3YUAA2//bb+ybTWgoW+6YEgXRhC6tWiFMLsf0q1ex45aGv0PK5eyF\nPmUKoOFpSNzM3dDKoBUisyPrdX7LlmzLzVVzYCoQZAS3srpp0aJFIcePH/eTSqVXT548+dqiRYtC\nhIiHMP/8Axw8CLz5Jmugnj6d9e3etQtITWXd0+lOgmhL55YtcUwux5y0NGzNy9PsxebOZV/X16zR\n7HUABLgGYGfSznqfL1SPqFqXVRUazQ2lWXfusMXE9u9nYyA6dWJjIvz9NX5HTohKkh8+hF9cHJba\n2WGaJrvWXbsGdOnC5sB3dtbYZVJup6Dv1r7ImptVr6qoiRPZoNYpU2o+TmuLH5EXE88DKSlsMrID\nB1ivQT8/Ng7ip5/YNOCENCYuRkY45eGBvnFxeKxQYKamFnSQSFgPjcmTWR2shgYDObdzRttmbfFX\n5l/o0aFHnc8XqpGbJhJ8CRQXs8QwfTogFgODBrEX26JFbKbX3bvZ7K6UKEhjJW3eHBEeHvgqKwur\nsrI0d6Hp09nslaGhmrsGgLGuY7EzuX5VUZQsiNoo7x5WrmR3DZaWwNq1rKE6PJz1pvjuO5Y0DA2F\njpYQ1dg3a4bTHh745uZNhGRmauYiHAds3AisXs36iGvIGNcx2J28GxWKijqfK5EA169rIKhaUDXU\nC6K4mK0eeeQI2xQKYOBAYNYsYM8emqSPvBhsDQ1x2sMDfeLioOB5fNChg/ovYmMDrFjBqqOiogCD\nhq1wVxVpWyksWljgz8w/4Sv2rdO5dGdB6qSsjFWrLl3KuodbWgKrVrFG6UOHgIwMtlDKsGGUKMiL\nxcrAAKfc3bElLw/LNTUF66RJbLLBpUs1Uz6AAJf69YoyNwdKSoD79zUQVA2oN1QTwfNsCv4TJ9j2\n558sMfTty7ZXXwWaNxc6SkK05+bjx+gdG4uplpZYZGur/gvk5wPu7sC+fUDXrmov/lrhNXTf1B05\n83KgK6pbJY+bG/Dzz4BnDWuZUm+ol0h2NpsUU5kgmjVjiSEwkM2B1q6d0BESIpz2BgY45eEB39hY\ncAAWqjthmJsD69ez6qjYWPYGVCNJGwlsWtngdMZp9LHvU7dz/62KqilZqBtVQzUiN26wbwvTpgEO\nDoCHB+vi2rMn8NdfrFHrhx/YaGpKFIQ8SRgbc3PxpSYavUePZm/Ezz5Tf9n4d4BePXpFCdFuQXcW\nAuF59uF/+vST7Z9/gF692DZ3LuDiAogonRNSI6tn7jCC1H2HsW4dmxJk9GjA21utRQe4BsD7R2+s\nH7geejp6Kp8nkbAxUtpEyUJLeJ4ttn76NHDmDPvJ80+Sw+LFbBZXmkqDkLpTJozesbHgOA7v29io\nr3AzM9YPfepU4O+/1do7Smwihn1re5zKOIV+kn4qnyeRsF6O2kTJQkNKSoCLF1mPpXPngPPngRYt\nWJXSa6+xNeMlEkoOhKiL9TN3GPPVmTDGj2ezaAYHszevGil7RdU1WWi7Gop6Q6lJdvaTxHDuHJCU\nxHosdO8OvPIK0K0bm5yPEKJZWY8eoXdcHGa2b4956kwYOTms/eLECdZLSk0y72XC83tP5M3PU7kq\nqrSUdYl/8IBN9lkV6g3VCJSVAfHxTyeHkhKWGLp3Z3esnTurvfMEIUQFNoaGOOXu/t8dxlx1JQwr\nKzYNyNSpwIULgJ7qbQw1sTW2hUMbB0RkRMBP4qfSOfr67MvnjRtsZgZtoObTWvA8m5572zZgzhyW\nDExMWG+6hARgwAD2RePWLTZz68KFbMwDJQpChGNjaIhTHh5Yn5ODddnZ6it46lS2fveKFeorE8BI\np5HYc7lujRDaroqiO4tn5OWxEf5RUazN4eJFdrvn48O24GB210Cjoglp3GwNDfGHhwd6xcSgmY4O\n3lTH9OYcB/z4I/sQGD5cbVOZj3AegZ7/64n1A9dDR6Sj0jnaniPqpU4WhYVsudBLl54kh+LiJ4nh\n3XdZTzlzc6EjJYTUh9jQECfc3dE7Lg6GIhEmquPN3KEDmwbkjTeAs2cBHdU+3GsibStFO6N2iMyO\nxCu2r6h0jljMqqG05aVIFjzPliGMjmbJISaG/buwkLVTeXmxLtRffsmm0KAeSoS8ODo2b45jcjn6\n/JswRqtjROv06cCOHWxlvXnzGl4egJHOrCqqLslinxYXon7hkoVysNuziUGhYEPjO3UCxo5l7VQS\nCQ16I+Rl4GJkhCNyOfr/mzCGNHTxFpGIrRbWrRtbOUwsbnCMI51GYviO4VjhtwKcCt9YxWI2Yai2\nNOlk8fgxW7chPv5JYoiJAYyNnySGGTPYv62t6Y6BkJeZR4sWOOjmhiEJCfjV2Rl+bdo0rMCOHdld\nxYwZbF6eBn7AyM3lEHEixObFwtOy9kmfxGK2No22NJlxFlc8BiE+niWGuDj2My2NVRvJ5az7c6dO\nLDGYmgodNSGksTp77x5GJiZit6srepqYNKyw0lLW2P3hh8C4cQ2OLeh4EAx1DfF5789rPVahAIyM\ngDt32M9nqXucRZNJFgaTBsHdnbUxyOVsc3Ghld4IIXX3R1ERxicn46CbG7q0atWwwiIjgREj2Ejc\nBt6tnM86jzcPvomkmUkqHe/kxKb9cHF5/rmXdlDe7dtUjUQIUY8+rVtjs5MT/BMSEC6Xw7MhfeG7\ndmU9ZBYsYO0YDdDFuguK/inC5TuX4WTqVOvxynaLqpKFujWZ5l1KFIQQdRrUti02SKUYlJCApIcP\nG1bYsmXA0aNARESDihFxIoxwHoG9KXtVOl6bjdxNJlkQQoi6jWjXDislEvSLi0NqSUn9C2rVii2U\n9NZbwKNHDYqpLqO5KVkQQoiWjDc3x1I7O/SLj0dWQz7ohw1js4cuW9ageHp26In0onRk3qt9MSdK\nFoQQokXTLC0xy8oKfvHxuF1aWv+C1q0DNmwAEhPrXYSejh6GOg5VqSqKkgUhhGjZfBsbjGnXDgPi\n43GvvLx+hbRvD3z+OauOUijqHYuqVVGULAghRABLxWJ0NzbG0IQElFRU1K+Qt95iPXK+/77ecfhJ\n/BCbF4vbD2/XeJy5OVvToqHt86qgZEEIIf/iOA5rHBzQwdAQo5OSUFqfuwORCPjhB+CTT9iCSfVg\nqGuIvvZ9cTj1cC3xsnkNtXF3QcmCEEIqEXEcNjk6Qo/jMOnyZVTUZ+CyqyubBuTdd+sdh7/UHweu\nHKj1ODs7ShaEECIIPZEIO1xdcau0FDOuXkW9Zrr44APW0H3oUL1iGCwdjD/S/8Cj8pp7aGmr3YKS\nBSGEVMFQJMJ+mQxxxcVYeP163ROGoSHw7bfs7qIeYzhMm5vC3dwdJ9NP1ngcJQtCCBFYS11dHJHL\nEVZYiJDM2sc9PKdvXzYdyBdf1Ov6/o7+2H9lf43HULIghJBGoI2eHo7J5fgpNxff1afBeuVK1uCd\nklLnU/0d/XHwykEo+Oob2ilZEEJII9HewAAn3N0RnJmJ7fn5dTvZ0pL1jJo5k63OVgfStlIYGxrj\n75t/V3vMC5cssrKybHr37n3K1dU1SSaTJa5du3Y2ABQWFrbx8/M7LpVKr/br1+/Y3bt3GzjBPCGE\nqJ9ds2YIc3PDnLQ0HCssrNvJM2cC9+4Bv/xS5+v6O/rjwNXqe0WZmbFxFsXFdS66TrSWLPT09MpW\nrVo1NykpyTUyMrLrN998805KSopzSEjIIj8/v+NXr16V9unT54+QkJBF2oqJEELqwq1FC/wuk2Fi\nSgqi7t9X/URdXTYNyIIFQFFRna5ZWxda5ViLGzfqVGydaS1ZWFhY5Hl4eMQCQIsWLYqdnZ1TcnJy\nrA4cOOA/efLkLQAwefLkLfv27RuurZgIIaSuXjU2xkZHRwxLTMSVuvRy8vEBhg9nq+rVQVfrrsh9\nkIuMuxnVHqONqihBVsrLyMgQ9+rV63RiYqLM1tY2s6ioqDUA8DzPtWnTplD5GGAr5elPmIDFHTsC\nAHx9feHr66v1mAkhpLL/5eZiyY0b+MvTE1YGBqqdVFTEVirav58lDxVN3T8VnhaemN1ldpXPz5wJ\n6OlFoHXriP/2LVmyRK0r5YHnea1uDx48aNGpU6e/9+7dO5zneZiYmBRVfr5169aFlR8D4FscPswT\nQkhjE3LjBu8aFcUXlJaqftLPP/N8p048X16u8il7U/byfbb0qfb50FCenz//6X3s4119n91a7Q1V\nVlamN2rUqN8DAwO3Dh8+fB8AmJub5+fl5VkAQG5urqWZmdktbcZECCH1tcDGBv1at67bxIMTJwIt\nW7IBeyrys/dDVE4U7j66W+XzYjGQnq5ycfWitWTB8zw3bdq0jS4uLslz5sxZrdzv7+9/YMuWLZMB\nYMuWLZOVSYQQQho7juOwQiKBfbNmGJucjDJVJh7kOJYoli4FcnNVuo6RvhF6duiJ8LTwKp/v0AGo\nz5jButBam8XZs2df7dmz5xm5XB7PcRwPAMuXL1/s4+MTFRAQsDMzM9NWLBZn7Ny5M8DExOS/9Mlx\nHN/i8GE8GDRIK3ESQkhdlSkUGJaYCHN9fWxydATHqdBUsHgxa5Xevl2la/zw9w+IyIjAtlHbnnsu\nLw+Qy4FbleplOI5Ta5uFIA3cdUHJghDSFDysqEDfuDj0NDZGqERS+wklJWx22h9/ZNOC1CL3QS5c\nv3VF/vv50NPRe+o5hQJo3hwoLGQ/AfUnCxrBTQghamCko4NDbm44WFCAr7Oyaj+heXNg9Wo20aAK\nS7latrREx7YdcebGmeeeE4kAGxtAlcvWFyULQghRk7Z6ejgql2NNdjZ+zsur/QR/f9Y6vXatSuX7\nS6sfza3pgXmULAghRI1sDA0RLpcj6No1hBUU1HwwxwFr1gAhIcDNm7WW7e/IRnNX1Xxga0vJghBC\nmhQXIyPsk8kw+fJlXKhtWhCplK3bvWBBreXKzGTgwCHxVuJzz9GdBSGENEHdjI3xPycnDE9MRGpt\n04J8+CFw5gzbasBx3H93F8/SdPdZShaEEKIhQ9q2xRKxGAMTEpBfUyO2kRGwYgUwaxZQXl5jmUOl\nQ6tst6A7C0IIacLeat8eE8zMMCQhAcU1jfIeMwZo1w747rsay+vRoQeu3LmCWw+fnuyC2iwIIaSJ\nWyIWw83ICGOTklBe3dg2jgPWrWMju29VP+uRvo4++tj3wZHUI0/tt7FhA8JVnXWkrihZEEKIhnEc\nh++lUigATL96tcreTADYjLSTJrHR3TUY3HEwDqcefmqfvj5gaqpSp6p6oWRBCCFaoCcSYZerK2KL\ni7GkpvqiTz8FwsOByMhqDxnUcRCOXz+Osoqyp/ZrsiqKkgUhhGhJCx0dHHZzw9a8PPxU3SSCrVoB\noaGssbuaOiWLFhaQtJbgr6y/ntqvyUZuShaEEKJF5vr6CJfL8XF6Og5XN2jv9deBZs2AjRurLWew\n9PmqqA4dNDflByULQgjRso7Nm2OfTIYply9XvZY3xwHr1wMffwxUk1AGdxyMw1efTha2tpoba0HJ\nghBCBNClVSts+nct77R//nn+AHd3ICAA+OijKs/3au+Fgn8KkF70ZNUjShaEEPICGmpqiiViMQbE\nx+NWVYP2li4F9u4FoqOfe0rEiTDQYSDCUsP+20fJghBCXlCVB+09fLZBu3VrYNky1thdxSp8z3ah\npWRBCCEvsCViMWRGRgioatDe1KmsV9TWrc+d10/SD2czz6KkjM09ZWLCDr13T/0xUrIghBCBVR60\nN+PZQXsiEVvvYvFi4MGDp84zNjRG5/adcTL95L/laO7ugpIFIYQ0AspBezHFxfji2cESXboA/foB\nX3zx3HnaqoqiZEEIIY1Ei3+XZt2Ul4etz660t3w5G3eRmvrUbmUXWuXdCCULQgh5CVjo6+Owmxve\nv3YNp4qKnjxhaQkEBQHz5z91vJOpE3RFuv8tiETJghBCXhIuRkb4zcUF45KTkfzw4ZMn5swBkpOB\no0f/28Vx3FOjuSlZEELIS6R369ZYIZFgcEIC8pRjMAwMgJUrgblzgbInkwhWbregZEEIIS+ZQAsL\nTLWweHoMxtChbPGKb7757zhfsS/i8+NR+E8hJQtCCHkZfdyhA9yMjDA+ORkVPM/6x65ezQbr3b4N\nADDUNUSvDr1wNO0orKzYIkjqRsmCEEIaMY7j8INUin8UCryXlsZ6PTk7s5lpK80bpayK0tcHRo9W\nfxyULAghpJHTE4mw29UVp+/exarsbLbzs8+A/fuBmBgAbEGk8LRwVCgqsG2b+mOgZEEIIU2Asa4u\nDru5YWV2Nn6/fZvN7bF0KfDeewDPw8bYBlatrHAh54JGrk/JghBCmghbQ0MclMkw4+pVnL93D5g2\njU0BsnMngKrX5lYXShaEENKEeLZsic1OThiZlIS00lJgzRpgwQKgpKTKBZHUhZIFIYQ0MYPatsVn\nYjEGxcejoFs3oGtX4Msv0dW6K7LvZyP7frbar0nJghBCmqC327fHCFNTDEtMxKPQUGDdOuhkZaO/\nQ/+nFkRSF0oWhBDSRC23t4eVvj6mlJRAMXs2sGABBjkM0ki7BSULQghpokQchy3Ozsh+/BgfjBsH\nREZiSG5LnEo/pf5rqb1EQgghWmMoEmGfTIY9d+/i+3XrYLzwE/w06Hu1X4eSRT1FREQIHcJzKCbV\nUEyqa4xxUUzPM9XTQ5hcjs9MTRHm5YWAyPtqv0ajSBbh4eEDnJycLnfs2DE1NDR0odDxqELoF0dV\nKCbVUEyqa4xxUUxVc2jWDHtcXTElMBAxmzapvXzBk0VFRYXOrFmz1oeHhw9ITk522b59+/iUlBRn\noeMihJCmppuxMb5zccHQ0FC1ly14soiKivJxcHBIE4vFGXp6emXjxo37bf/+/cOEjosQQpqiUe3a\nYZenp/oL5nle0G3Xrl2j33zzzR+Vj7du3Tpx1qxZ65SPAfC00UYbbbTVfVPnZ7UuBMZxHF/T8zzP\nc9qKhRBCSNUEr4aysrLKycrKslE+zsrKsrG2tlb/WHVCCCH1Jniy8PLyupSamtoxIyNDXFpalh+R\nTQAACbZJREFUqr9jx46x/v7+B4SOixBCyBOCV0Pp6uqWr1+/flb//v2PVlRU6EybNm2js7NzitBx\nEUIIqUToBu6atiNHjgxwdHS87ODgkBoSErJQW9fNzMy08fX1PeXi4pLk6uqauGbNmtk8z6OgoKBN\n3759j3fs2PGqn5/fsaKiIhPlOcHBwYsdHBxSHR0dLx89erSfpmIrLy/X8fDwiBkyZMjBxhJTUVGR\nyahRo3Y7OTmlODs7J0dGRnYROq7g4ODFLi4uSTKZLGH8+PHbHj16ZKDtmKZOnbrJzMwsXyaTJSj3\n1SeGS5cudZbJZAkODg6ps2fPXqPumN5///2vnJycUuRyedyIESP23L1711ibMVUXl3JbsWLFfI7j\nFAUFBW2E/l3xPI+1a9e+6+TklOLq6pq4YMGCUKFjunDhgo+3t3eUh4dHjJeX18WoqChvTcSk9jep\nurby8nIdiUSSlp6eLi4tLdVzd3ePTU5OdtbGtXNzcy1iYmI8eJ7HgwcPWkil0ivJycnOQUFBX4aG\nhi7geR4hISELFy5cGMLzPJKSklzc3d1jS0tL9dLT08USiSStoqJCpInYvv7663kTJkz4dejQoQd4\nnkdjiGnSpElbNm7c+AbP8ygrK9O9e/eusZBxpaeni+3s7K4/evTIgOd5BAQE7Ni8efNkbcd05syZ\nHtHR0Z6V39h1iUGhUHA8z8Pb2zvqwoULPjzPY+DAgWFHjhwZoM6Yjh075qf8/y5cuDBE2zFVFxfP\nsy9u/fv3DxeLxenKZCHk7+rkyZO9+/bte7y0tFSP53ncunWrndAx9erVKyI8PLw/z/MICwsb6Ovr\ne0oTMan9g0Nd27lz57r1798/XPl4+fLli5YvX75IiFiGDRu27/jx430dHR0v5+XlmfM8SyiOjo6X\neZ5l78p3Pv379w8/f/58V3XHkZWVZd2nT58TJ0+e7K28sxA6prt37xrb2dldf3a/kHEVFBS0kUql\nVwoLC1uXlZXpDhky5OCxY8f8hIgpPT1dXPmNXdcYbt68aenk5JSi3L99+/Zxb7/99gZ1xlR527Nn\nz4jXX3/9F23HVF1co0eP3hUXFyevnCyE/F2NGTNm5x9//PHas8cJGdO4ceO279ixI4DneWzbtm28\npv5+gjdwVycnJ8fKxsYmS/nY2to6Oycnx0rbcWRkZIhjYmI8u3TpciE/P9/c3Nw8HwDMzc3z8/Pz\nzQHg5s2b7Sv34NJUrHPnzl311VdfBYlEIoVyn9Axpaen27Vr1+721KlT/9epU6fo//u///vx4cOH\nRkLG1aZNm8L58+d/bWtrm9m+ffubJiYmd/38/I4L/bsC6v73ena/lZVVjibfB5s2bXpj0KBBYY0h\npv379w+ztrbOlsvl8ZX3CxlXampqxzNnzvTs2rVrpK+vb8SlS5e8hI4pJCRkkfL1HhQU9NXy5csX\nayKmRpssaht/oQ3FxcUtRo0a9fuaNWvea9my5YPKz3Ecx9cUo7rjP3To0BAzM7Nbnp6eMXw1Y0+0\nHRMAlJeX60ZHR3eaOXPmt9HR0Z2MjIwehoSELBIyrmvXrklWr149JyMjQ3zz5s32xcXFLX755ZeJ\nQsZU3TUaw+tcadmyZR/q6+uXTpgwYZvQsZSUlDQPDg7+YMmSJZ8q91X3utem8vJy3aKiotaRkZFd\nv/rqq6CAgICdQsc0bdq0jWvXrp2dmZlpu2rVqrlvvPGG+ieGQiNOFkKPvygrK9MbNWrU74GBgVuH\nDx++D2DfBPPy8iwAIDc319LMzOxWVbFmZ2dbW1lZ5agznnPnznU/cOCAv52dXfr48eO3nzx58rXA\nwMCtQsYEsG8r1tbW2d7e3hcBYPTo0bujo6M7WVhY5AkV16VLl7y6d+9+rm3btgW6urrlI0eO3HP+\n/PluQsakVJe/l7W1dbaVlVVOdna2taZj27x585SwsLBBv/766+vKfULGdO3aNUlGRobY3d09zs7O\nLj07O9u6c+fOf+fn55sLGZe1tXX2yJEj9wCAt7f3RZFIpLhz546pkDFFRUX5jBgxYi/A3n9RUVE+\ngAb+fg2tZ9TUVlZWpmtvb38tPT1d/PjxY31tNnArFAouMDDw5zlz5qyqvD8oKOhLZR3g8uXLFz3b\nEPj48WP969ev29nb219TNiRpYouIiOilbLNoDDH16NHjzJUrV6Q8z+PTTz/9LCgo6Esh44qNjXV3\ndXVNLCkpaaZQKLhJkyZtWb9+/TtCxPRs/XJ9YvDx8bkQGRnZRaFQcOpoTH42piNHjgxwcXFJun37\ntmnl47QZU1VxVd6qauAW4ne1YcOGtz/55JMlPM/jypUrUhsbm0yhY/L09IyOiIjoxfM8Tpw40cfL\ny+uiJmJS+weHOrewsLCBUqn0ikQiSQsODl6srev++eefr3Icp3B3d4/18PCI8fDwiDly5MiAgoKC\nNn369DlRVbfHZcuWfSCRSNIcHR0vK3smaGqLiIjopewN1Rhiio2Ndffy8rpYueul0HGFhoYuUHad\nnTRp0pbS0lI9bcc0bty47ZaWljf19PRKra2tszZt2jS1PjEouzlKJJK0d999d606Y9q4ceMbDg4O\nqba2tjeUr/UZM2Z8q82YKselr6//WPm7qvy8nZ3d9cpdZ7X5u6ocU2lpqd7EiRO3ymSyhE6dOv19\n6tQpXyH/fps2bZp68eJFLx8fnwvu7u6xXbt2PR8dHe2piZg4nm80VaaEEEIaqUbbZkEIIaTxoGRB\nCCGkVpQsCCGE1IqSBSEq4HmeUygU9H4hLy168RNSjYyMDLGjo+OVyZMnb3Fzc0vQ1dUtnzdv3kqZ\nTJbYt2/fE3fu3DEFAF9f34h58+at9Pb2vujs7Jxy8eJF7xEjRuyVSqVXP/7448+F/n8Qog6ULAip\nQVpamsM777zzTWJiogxgA7ESExNlvXr1Oq0cXcxxHG9gYPD44sWL3jNmzPhu2LBh+zds2DA9MTFR\ntnnz5ilFRUWthf1fENJwlCwIqUGHDh1u+Pj4RAGASCRSjB07dgcATJw48ZezZ8++qjxOuWCXTCZL\nlMlkiebm5vn6+vql9vb21zMzM22FiZ4Q9aFkQUgNjIyMHla1n+d5rvK8TgYGBo8BllCU/1Y+rqio\n0NF8pIRoFiULQlSkUChEu3btGgMA27Ztm9CjR48/hY6JEG0RfFlVQhqzyncPRkZGD6Oiony++OKL\nj8zNzfN37NgxtqrjG9NMsoSoC033QYiKWrZs+eDBgwcthY6DECFQNRQhKqI7BvIyozsLQgghtaI7\nC0IIIbWiZEEIIaRWlCwIIYTUipIFIYSQWlGyIIQQUitKFoQQQmr1/0ZwGmquOtPsAAAAAElFTkSu\nQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x34979d0>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.5, Page number: 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from sympy import *\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration:\n",
+      "nph=3                       #No. of phases\n",
+      "k=0.429             #reactance ratio(X1/X2) from table 6.1,for class C motor\n",
+      "p=4                         #No.of poles\n",
+      "#Test 1: No-load test at 60 Hz\n",
+      "V1=219                      #Applied voltage, line-to-lne(V)\n",
+      "I1_nl=5.70                  #Phase current(A)\n",
+      "Pnl=380                     #Power(W)\n",
+      "ft=60                       #Hz\n",
+      "\n",
+      "#Test 2: Blocked-rotor test at 15 Hz\n",
+      "V2=26.5                     #Applie voltage, line-to-line(V)\n",
+      "I1_bl=18.57                 #Phase current(A)\n",
+      "Pbl=675                     #Power(W)\n",
+      "fbl=15                      #Hz\n",
+      "\n",
+      "#Test 3:\n",
+      "R1=0.262                    #Avg resistance per stator phase(ohm)\n",
+      "\n",
+      "#Test 4: Blocked-rotor test at 60 Hz\n",
+      "V4=212                      #Applied voltage, line-line (V)\n",
+      "I2_bl=83.3                  #Avg phase current(A)\n",
+      "Pbl_4=20.1*10**3            #Power(W)\n",
+      "Tstart=74.2                 ##starting torque(Nm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#For part (a):\n",
+      "Prot=Pnl-nph*I1_nl**2*R1\n",
+      "V1_nl=V1/sqrt(3)                            #from test 1\n",
+      "Qnl=sqrt((nph*V1_nl*I1_nl)**2-Pnl**2)\n",
+      "Xnl=Qnl/(nph*I1_nl**2)\n",
+      "V1_bl=V2/sqrt(3)                            #from test 2\n",
+      "Qbl=sqrt((nph*V1_bl*I1_bl)**2-Pbl**2)\n",
+      "Xbl=(ft/fbl)*(Qbl/(nph*I1_bl**2))\n",
+      "X2=symbols('X2')\n",
+      "fx=k**2*X2**2+(Xbl*(1-k)-Xnl*(1+k))*X2+Xnl*Xbl\n",
+      "x=solve(fx,X2)\n",
+      "X2=round(x[0],2)                    #since X2 must be less than X1\n",
+      "X1=k*X2\n",
+      "Xm=Xnl-X1\n",
+      "Rbl=Pbl/(nph*I1_bl**2)\n",
+      "R2=(Rbl-R1)*((X2+Xm)/Xm)**2\n",
+      "\n",
+      "#for part (b):\n",
+      "Pg=Pbl_4-nph*I2_bl**2*R1\n",
+      "ws=4*math.pi*ft/p\n",
+      "Tstart=Pg/ws\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) N-load rotational loss:\",round(Prot,0),\"W\"\n",
+      "print \"\\n  Equivalent ckt parameters:\\n\"\n",
+      "print\"  R1=\",round(R1,3),\"ohm\",\"  R2=\",round(R2,3),\"ohm\"\n",
+      "print\"  X1=\",round(X1,3),\"ohm\",\"  X2=\",round(X2,3),\"ohm\",\" Xm=\",round(Xm,2),\"ohm\"\n",
+      "print \"\\n(b) Starting torque:\",round(Tstart,2),\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) N-load rotational loss: 354.0 W\n",
+        "\n",
+        "  Equivalent ckt parameters:\n",
+        "\n",
+        "  R1= 0.262 ohm   R2= 0.447 ohm\n",
+        "  X1= 0.635 ohm   X2= 1.48 ohm  Xm= 21.2 ohm\n",
+        "\n",
+        "(b) Starting torque: 77.7 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.6, Page number: 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Xnl=21.8                                #ohm\n",
+      "Xbl=2.01                                #ohm\n",
+      "R_1=0.26                                 #ohm\n",
+      "Rbl=0.65                                #ohm\n",
+      "V=220                                   #volt\n",
+      "#Here are the two sets of parameters\n",
+      "#Set 1 corresponds to the exact solution\n",
+      "#Set 2 corresponds to the approximate solution\n",
+      "\n",
+      "R1=[0.262, 0.262]                        #ohm\n",
+      "R2=[0.447, 0.444]                        #ohm\n",
+      "X1=[0.633, 0.603]                        #H\n",
+      "X2=[1.47, 1.41]                          #H\n",
+      "Xm=[21.2, 21.2]                          #H\n",
+      "nph=3                                   #No. of phases\n",
+      "p=4                                     #No. of poles\n",
+      "Prot=354                                #Rotational losses(Watts)\n",
+      "\n",
+      "#Calculations:\n",
+      "X_1=0.3*Xbl                              #(ohm) from table 6.1 and X1+X2=Xbl\n",
+      "X_2=Xbl-X_1                               #ohm\n",
+      "X_m=Xnl-X_1\n",
+      "R_2=(Rbl-R_1)*((X_2+X_m)/X_m)**2\n",
+      "\n",
+      "#Results for part (a):\n",
+      "print \"(a) The parameters:\\n\"\n",
+      "print\"  R1=\",round(R_1,3),\"ohm\",\"  R2=\",round(R_2,3),\"ohm\"\n",
+      "print\"  X1=\",round(X_1,3),\"ohm\",\"   X2=\",round(X_2,2),\"ohm\"\n",
+      "print\"  Xm=\",round(X_m,3),\"ohm\"\n",
+      "\n",
+      "#Calculations & Results for part (b):\n",
+      "print \"\\n\\n(b)\"\n",
+      "#Here is the operating condition\n",
+      "V1=220/sqrt(3)\n",
+      "fe=60                                   #Hz\n",
+      "rpm=1746\n",
+      "#Calculate the synchronous speed:\n",
+      "ns=120*fe/p\n",
+      "ws=4*pi*fe/p\n",
+      "s=(ns-rpm)/ns\n",
+      "wm=ws*(1-s)\n",
+      "Zgap=[0]*2\n",
+      "Zin=[0]*2\n",
+      "Pmech=[0]*2\n",
+      "I1=[0]*2\n",
+      "I2=[0]*2\n",
+      "Tmech=[0]*2\n",
+      "\n",
+      "#Calculate stator Thevenin equivalent:\n",
+      "#Loop over the two motors\n",
+      "for m in range(0,2,1):\n",
+      "    Zgap = 1j*Xm[m]*(1j*X2[m] + R2[m]/s)/(R2[m]/s + 1j*(Xm[m] + X2[m]))\n",
+      "    Zin=R1[m]+1j*X1[m]+Zgap\n",
+      "    I1=V1/Zin\n",
+      "    I2=I1*(1j*Xm[m])/(R2[m]/s+1j*(Xm[m]+X2[m]))\n",
+      "    Tmech=nph*abs(I2)**2*R2[m]/(s*ws)   #Electromechanical torque\n",
+      "    Pmech=wm*Tmech                      #Electromechanical power\n",
+      "    Pshaft=Pmech - Prot\n",
+      "    if (m==0):\n",
+      "        print \"Exact Solution:\"\n",
+      "    else:\n",
+      "        print \"\\nApproximate Solution:\"\n",
+      "\n",
+      "\n",
+      "\n",
+      "    \n",
+      "    print \"\\tPmech=\",round(Pmech,1),\"W\",\"\\tPshaft =\",round(Pshaft,1), \"W\"\n",
+      "    print \"\\tI1=\", round(abs(I1),1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The parameters:\n",
+        "\n",
+        "  R1= 0.26 ohm   R2= 0.443 ohm\n",
+        "  X1= 0.603 ohm    X2= 1.41 ohm\n",
+        "  Xm= 21.197 ohm\n",
+        "\n",
+        "\n",
+        "(b)\n",
+        "Exact Solution:\n",
+        "\tPmech= 2820.7 W \tPshaft = 2466.7 W\n",
+        "\tI1= 10.3 A\n",
+        "\n",
+        "Approximate Solution:\n",
+        "\tPmech= 2850.5 W \tPshaft = 2496.5 W\n",
+        "\tI1= 10.4 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter7.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter7.ipynb
new file mode 100755
index 00000000..c5e8eebf
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter7.ipynb
@@ -0,0 +1,473 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: DC Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.1, Page number: 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Vt=[128, 124]                           #Terminal voltage(V)\n",
+      "Ea=125                                  #Generated emf(V)\n",
+      "Ra=0.02                                 #Armature resistance(ohm)\n",
+      "n=3000                                  #rpm\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#For 128 V\n",
+      "Ia1=(Vt[0]-Ea)/Ra\n",
+      "Pin1=Vt[0]*Ia1\n",
+      "Pe1=Ea*Ia1\n",
+      "wm=3000*2*pi/60\n",
+      "Tmech1=Ea*Ia1/wm\n",
+      "\n",
+      "#for 124 V\n",
+      "Ia2=(-Vt[1]+Ea)/Ra\n",
+      "Pin2=Vt[1]*Ia2\n",
+      "Pe2=Ea*Ia2\n",
+      "Tmech2=Ea*Ia2/wm\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Armature current:\",Ia1,\"A\",\"\\n    Terminal power:\",Pin1/10**3,\"kW\"\n",
+      "print \"    Electromagnetic power:\",round(Pe1/10**3,2),\"kW\"\n",
+      "print \"    Torque:\",round(Tmech1,1),\"Nm\"\n",
+      "\n",
+      "print \"(b) Armature current:\",Ia2,\"A\",\"\\n    Terminal power:\",Pin2/10**3,\"kW\"\n",
+      "print \"    Electromagnetic power:\",round(Pe2/10**3,2),\"kW\",\n",
+      "print \"\\n    Torque:\",round(Tmech2,1),\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Armature current: 150.0 A \n",
+        "    Terminal power: 19.2 kW\n",
+        "    Electromagnetic power: 18.75 kW\n",
+        "    Torque: 59.7 Nm\n",
+        "(b) Armature current: 50.0 A \n",
+        "    Terminal power: 6.2 kW\n",
+        "    Electromagnetic power: 6.25 kW \n",
+        "    Torque: 19.9 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page number: 372"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Vt=123                                      #terminal voltage(V)\n",
+      "Pt=21.9                                     #Terminal power(kW)\n",
+      "Ra=0.02                                     #ohm\n",
+      "Eao=125                                     #generated voltage(V) at 3000rpm\n",
+      "no=3000                                     #rpm\n",
+      "\n",
+      "\n",
+      "#calculations:\n",
+      "Ia=Pt*10**3/Vt\n",
+      "Ea=Vt-Ia*Ra\n",
+      "n=(Ea/Eao)*no\n",
+      "\n",
+      "#Results:\n",
+      "print \"Speed of motor:\",round(n,0),\"rpm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed of motor: 2867.0 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page number: 376"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Il=400                                  #Armature current(A)\n",
+      "If=4.7                                  #Field current(A)\n",
+      "Ns=3                                    #series turns per pole\n",
+      "Nf=1000                                 #shunt field turns per pole\n",
+      "Eao=274                                 #at Ia=0,(V)\n",
+      "n=1150                                  #speed of motor(rpm)\n",
+      "no=1200                                 #rated speed(rpm) \n",
+      "Ra=0.025                                #armature resistance(ohm)\n",
+      "Rs=0.005                                #series field resistance(ohm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Is=Il+If\n",
+      "GM=If+(Ns/Nf)*Is                        #for graphical analysis\n",
+      "Ea=(n/no)*Eao\n",
+      "Vt=Ea-Is*(Ra+Rs)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Terminal voltage at rated terminal current:\",round(Vt,0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Terminal voltage at rated terminal current: 250.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.4, Page number: 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Il=400                                  #Armature current(A)\n",
+      "If=4.7                                  #Field current(A)\n",
+      "Ns=3                                    #series turns per pole\n",
+      "Nf=1000                                 #shunt field turns per pole\n",
+      "Eao=261                                 #at Ia=400 A,(V)\n",
+      "n=1150                                  #speed of motor(rpm)\n",
+      "no=1200                                 #rated speed(rpm) \n",
+      "Ra=0.025                                #armature resistance(ohm)\n",
+      "Rs=0.005                                #series field resistance(ohm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ea=(n/no)*Eao\n",
+      "Vt=Ea-(Il+If)*(Ra+Rs)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Terminal voltage:\", round(Vt,0), \"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Terminal voltage: 238.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page number: 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Il=400                                  #Armature current(A)\n",
+      "If=4.7                                  #Field current(A)\n",
+      "Ns=3                                    #series turns per pole\n",
+      "Nf=1000                                 #shunt field turns per pole\n",
+      "Eao=269                                 #at Ia=400 A,(V)\n",
+      "n=1150                                  #speed of motor(rpm)\n",
+      "no=1200                                 #rated speed(rpm) \n",
+      "Ra=0.025                                #armature resistance(ohm)\n",
+      "Rs=0.007                                #series field resistance(ohm)\n",
+      "\n",
+      "#Calculations:\n",
+      "Is=Il+If\n",
+      "GM=If+(Ns/Nf)*Is                        #for graphical analysis\n",
+      "Ea=(n/no)*Eao\n",
+      "Vt=Ea-Is*(Ra+Rs)\n",
+      "\n",
+      "#Results:\n",
+      "print \"Terminal voltage at rated terminal current:\",round(Vt,0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Terminal voltage at rated terminal current: 245.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.6, Page number: 381"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Ns=4                            #Series field turns\n",
+      "Nf=1000                         #Shunt field turns\n",
+      "Vt=250                          #Full load voltage(V)\n",
+      "#for part (a):\n",
+      "Ia=400                          #Armature current(A)\n",
+      "Ra=0.025                        #Armature resistance(ohm)\n",
+      "\n",
+      "#for part (b):\n",
+      "Rs=0.005                        #Added sries resistance(ohm)\n",
+      "Vo=250                          #No load voltage(V)\n",
+      "If=5                            #field current at full load(A)\n",
+      "\n",
+      "\n",
+      "#Calculations & Results:\n",
+      "\n",
+      "#for part (a)\n",
+      "V1=Ia*Ra\n",
+      "\n",
+      "#for part (b):\n",
+      "Ia1=Ia+If\n",
+      "Rs,Rd=symbols('Rs Rd')            #Rd= diverter resistance(ohm)\n",
+      "Rp=Rs*Rd/(Rs+Rd)                  #                      -------(i)\n",
+      "Is=Ia1*(Rd/(Rs+Rd))\n",
+      "Inet=If+(Ns/Nf)*Is\n",
+      "Ea=Vt+Ia*(Ra+Rp)                  #                      -------(ii)\n",
+      "\n",
+      "#from equation (ii)\n",
+      "Rp=Rs(Inet-5.0)/1.62              \n",
+      "R_d=0.0082                       #R_d=Rd(say), using (i)\n",
+      "print \"(a) The operating terminal voltage = 205 V\", Inet\n",
+      "print \"(b) Rd =\", R_d,\"ohm\"\n",
+      "print \"\\tHence, by this process, resistance across the series field\" \n",
+      "print \"\\t(referred to as a series-field diverter) can be adjusted \"\n",
+      "print \"\\tto produce the desired performance. \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The operating terminal voltage = 205 V 1.62*Rd/(Rd + Rs) + 5\n",
+        "(b) Rd = 0.0082 ohm\n",
+        "\tHence, by this process, resistance across the series field\n",
+        "\t(referred to as a series-field diverter) can be adjusted \n",
+        "\tto produce the desired performance. \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.7, Page number: 383"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "      \n",
+      "#Variable declaration:\n",
+      "Ia=400                              #Armature current(A)\n",
+      "n1=1200                             #rpm\n",
+      "n2=1100                             #rpm\n",
+      "Ra=0.025                            #armature resistance(ohm)     \n",
+      "Eo=250                              #no load armature voltage(V)\n",
+      "del_n=1.5                           #fractional winding added\n",
+      "N=1000                              #Total windings\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for part(a):\n",
+      "#point corresponding on the no load saturation curve is :\n",
+      "Eao=Eo*(n1/n2)\n",
+      "#using Eao value in curve, value of If is found to be:\n",
+      "If=5.90                             #Field current(A)\n",
+      "Ea=Eo-Ia*Ra\n",
+      "#From Fig. 7.14\n",
+      "Ea1=261\n",
+      "n=n1*(Ea/Ea1)\n",
+      "Pe=Ea*Ia\n",
+      "Pl=2000                             #No load Rotational loss(W)\n",
+      "Po=(Pe-Pl)/(1+0.01)\n",
+      "\n",
+      "#for part (b):\n",
+      "If1=If+del_n/N\n",
+      "#From Fig. 7.14 the corresponding value of Ea at 1200 r/min would be 271 V.\n",
+      "Ea2=271                             #volts\n",
+      "n22=n1*(Ea/Ea2)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"Part(a):\"\n",
+      "print \"Required speed =\",round(n),\"r/min\"\n",
+      "print \"Output power =\", round((Po/746),1),\"hp\"\n",
+      "print \"\\nPart (b):\"\n",
+      "print \"Required speed =\",round(n22),\"r/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part(a):\n",
+        "Required speed = 1103.0 r/min\n",
+        "Output power = 124.8 hp\n",
+        "\n",
+        "Part (b):\n",
+        "Required speed = 1063.0 r/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.9, Page number: 389"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "V1=50                               #terminal voltage(V)\n",
+      "Ia=1.25                             #Armature current(A)\n",
+      "Ra=1.03                             #Armature resistance(ohm)\n",
+      "n1=2100                             #speed at 50V(rpm)\n",
+      "V2=48                               #terminal voltage at 1700 rpm (V)\n",
+      "n2=1700                             #speed at 48 V(rpm)\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "#for (a):\n",
+      "Ea1=V1-Ia*Ra\n",
+      "wm1=n1*2*pi/60\n",
+      "Km=round(Ea1/wm1,2)\n",
+      "\n",
+      "#for part(b):\n",
+      "Prot=Ea1*Ia\n",
+      "\n",
+      "#for part(c:)\n",
+      "wm2=n2*2*pi/60\n",
+      "Ea2=Km*wm2\n",
+      "Ia2=(V2-Ea2)/Ra\n",
+      "Pmech=Ea2*Ia2\n",
+      "Pshaft=Pmech-Prot\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Torque constant:\",round(Km,2),\"V/(rad/s)\"\n",
+      "print \"(b) No-load rotational losses of the motor:\",round(Prot,0),\"W\"\n",
+      "print \"(c) The power output of the motor:\",round(Pshaft,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Torque constant: 0.22 V/(rad/s)\n",
+        "(b) No-load rotational losses of the motor: 61.0 W\n",
+        "(c) The power output of the motor: 275.05 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter8.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter8.ipynb
new file mode 100755
index 00000000..785bba7e
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter8.ipynb
@@ -0,0 +1,427 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8: Variable-Reluctance Machines and Stepping Motors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page number: 411"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from numpy import *\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "R=0.038                               #m\n",
+      "a=b=pi/3                            #rad\n",
+      "g=2.54*10**-4                       #m\n",
+      "D=0.13                              #m\n",
+      "N=100                               #turns in both poles\n",
+      "uo=4*pi*10**-7                      #permeability of free space()\n",
+      "i1=5                                #coil current (A)\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "Lm=N**2*uo*a*R*D/(2*g)\n",
+      "#x=symbols('x')\n",
+      "subplot(2,1,1)\n",
+      "x=linspace(-180,-120,100)\n",
+      "L=-(Lm/60)*x-2*Lm\n",
+      "plot(x,L,'b')\n",
+      "#grid()\n",
+      "\n",
+      "x=linspace(-60,0,100)\n",
+      "L=(Lm/60)*x+Lm\n",
+      "plot(x,L,'b')\n",
+      "grid()\n",
+      "\n",
+      "x=linspace(0,60,100)\n",
+      "L=-(Lm/60)*x+Lm\n",
+      "plot(x,L,'b')\n",
+      "grid()\n",
+      "\n",
+      "\n",
+      "x=linspace(120,180,100)\n",
+      "L=(Lm/60)*x-2*Lm\n",
+      "plot(x,L)\n",
+      "annotate('Lm=0.128 H',xy=(-150,0.10))\n",
+      "annotate('Lmax',xy=(0,Lm+0.005))\n",
+      "ylabel('L11(theta)')\n",
+      "xlabel('theta')\n",
+      "grid()\n",
+      "\n",
+      "#part(b)\n",
+      "subplot(2,1,2)\n",
+      "x1=linspace(-180,-120,100)\n",
+      "x2=linspace(-150,-90,100)\n",
+      "i1=5\n",
+      "i2=4\n",
+      "Tm1=(Lm/(2*pi/3))*i1**2\n",
+      "Tm2=(Lm/(2*pi/3))*i2**2\n",
+      "dll=np.ones(100)\n",
+      "plot(x1,-Tm1*np.array(dll),'g')\n",
+      "plot(x2,Tm2*np.array(dll),'b--')\n",
+      "\n",
+      "x1=linspace(-60,0,100)\n",
+      "x2=linspace(-90,-30,100)\n",
+      "Tm1=(Lm/(2*pi/3))*i1**2\n",
+      "Tm2=(Lm/(2*pi/3))*i2**2\n",
+      "dll=np.ones(100)\n",
+      "plot(x1,Tm1*np.array(dll),'g')\n",
+      "plot(x2,-Tm2*np.array(dll),'b--')\n",
+      "\n",
+      "x1=linspace(0,60,100)\n",
+      "x2=linspace(30,90,100)\n",
+      "Tm1=(Lm/(2*pi/3))*i1**2\n",
+      "Tm2=(Lm/(2*pi/3))*i2**2\n",
+      "dll=np.ones(100)\n",
+      "plot(x1,-Tm1*np.array(dll),'g')\n",
+      "plot(x2,Tm2*np.array(dll),'b--')\n",
+      "\n",
+      "x1=linspace(120,180,100)\n",
+      "x2=linspace(90,150,100)\n",
+      "Tm1=(Lm/(2*pi/3))*i1**2\n",
+      "Tm2=(Lm/(2*pi/3))*i2**2\n",
+      "dll=np.ones(100)\n",
+      "plot(x1,Tm1*np.array(dll),'g')\n",
+      "plot(x2,-Tm2*np.array(dll),'b--')\n",
+      "grid()\n",
+      "ylim(-3,3)\n",
+      "annotate('___ i1=I1, i2=0', xy=(110,2.6))\n",
+      "annotate('---- i1=0, i2=I2', xy=(110,2.2))\n",
+      "ylabel('Torque [N.m]')\n",
+      "xlabel('thetam [degrees]')\n",
+      "\n",
+      "#Results:\n",
+      "print \"Lm =\",Lm,\"H\"\n",
+      "print \"(c)The peak torque =\",round(Tm1,2),\"N.m\"\n",
+      "print \"\\t(i)  The net torque, (at thetam=0) =\", 0, \"N.m\"\n",
+      "print \"\\t(ii) The net torque, (at thetam=45 deg.) =\", 0, \"N.m\"\n",
+      "print \"\\t(iii)The net torque, (at thetam=75 deg) =\", round(Tm1,2), \"N.m\"\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "Lm ="
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 0.127968099059 H\n",
+        "(c)The peak torque = 1.53 N.m\n",
+        "\t(i)  The net torque, (at thetam=0) = 0 N.m\n",
+        "\t(ii) The net torque, (at thetam=45 deg.) = 0 N.m\n",
+        "\t(iii)The net torque, (at thetam=75 deg) = 1.53 N.m\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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HDh3Qs2dPkxlKm700dPkq+DoKkTA9e8YMYbq2JSzJOXKkcOX5kSNAixa25ywNKc6ntZSQ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\niIiIL0ocIiLiixKHiIj4osQhIiK+/ADYIdwwaPRWXQAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x1d13bd0>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page number: 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from sympy import *\n",
+      "from math import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "n1=4000                             #r/min\n",
+      "R=0.038                             #m\n",
+      "a=b=pi/3                            #rad\n",
+      "g=2.54*10**-4                       #m\n",
+      "D=0.13                              #m\n",
+      "N=100                               #turns in both poles\n",
+      "uo=4*pi*10**-7                      #permeability of free space(H/m)\n",
+      "Ll=0.005                            #H\n",
+      "Vo=100                              #phase voltage applied to phase 1.(V)\n",
+      "\n",
+      "\n",
+      "#Calculation:\n",
+      "wm=n1*pi/30\n",
+      "Lm=N**2*uo*a*R*D/(2*g)\n",
+      "thetam=symbols('thetam')\n",
+      "t=symbols('t')\n",
+      "#for part (a):\n",
+      "#for -60<=thetam<=0deg,\n",
+      "L11=Ll+(Lm/(pi/3))*(thetam+pi/3)\n",
+      "L111=diff(L11,thetam)\n",
+      "R1=L111*wm\n",
+      "#which is nuch greater than resistance R=1.5 ohm\n",
+      "thetam=-pi/3+wm*t\n",
+      "i1=Vo*t/(float(round(Ll,3))+float(Lm/(pi/3))*thetam+float(Lm/(pi/3))*pi/3)\n",
+      "\n",
+      "#for part (b):\n",
+      "V2=-200                                     #applied voltage(V)\n",
+      "thetam2=symbols('thetam2')\n",
+      "L12=Ll+(Lm/(pi/3))*(pi/3-thetam2)\n",
+      "L112=diff(L12,thetam2)\n",
+      "to=2.5*10**-3                                     #ms\n",
+      "thetam2=float(-pi/3+wm*to)\n",
+      "i1=Vo*t/(float(round(Ll,3))+float(Lm/(pi/3))*thetam+float(Lm/(pi/3))*pi/3)\n",
+      "i2=(0.25-200*(t-to))/(0.005+51.1*(5*10**-3-t))\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"i1 =\",i1,\"\\t, (where round(16.2934044186179*pi,2) = 51.1 )\"\n",
+      "print \"\\ni2 =\",i2,\"\\n\"\n",
+      "\n",
+      "\n",
+      "#Calculations & Results:\n",
+      "#for part (c):\n",
+      "from __future__ import division\n",
+      "from pylab import *\n",
+      "\n",
+      "Lleak=0.005\n",
+      "Posintegral=0\n",
+      "integral=0\n",
+      "N1=500\n",
+      "tmax=3.75*10**-3\n",
+      "t=[0]*503\n",
+      "thet=[0]*503\n",
+      "Torque=[0]*503\n",
+      "deltat = tmax/N1\n",
+      "thetm=[0]*503\n",
+      "i=[0]*503\n",
+      "for n in range(1,N1+2,1):\n",
+      "    t[n-1]=tmax*(n-1)/N1\n",
+      "    thetm[n-1]=-(pi/3)+(400*pi/3)*t[n-1]\n",
+      "    if (thetm[n-1]<=0):\n",
+      "        i[n-1]=100*t[n-1]/(0.005+51.1*t[n-1])\n",
+      "        dld1d1theta = 0.122\n",
+      "        Torque[n-1]=0.5*i[n-1]**2*dld1d1theta\n",
+      "        Posintegral=Posintegral+Torque[n-1]*deltat\n",
+      "        integral=Posintegral\n",
+      "    else:\n",
+      "        i[n-1]=(0.25-200*(t[n-1]-2.5*10**-3))/(0.005+51.1*(5*10**-3-t[n-1]))\n",
+      "        dld11dtheta = -0.122\n",
+      "        Torque[n-1] = 0.5*i[n-1]**2*dld11dtheta\n",
+      "        integral = integral + Torque[n-1]*deltat\n",
+      "\n",
+      "print \"\\nPositve torque integral =\",Posintegral, \"[N-m-sec]\"\n",
+      "print \"\\nTorque integral=\",integral,\"[N-m-sec]\\n\"\n",
+      "\n",
+      "plot(1000*np.array(t),i)\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('Phase current [A]')\n",
+      "title('(a) phase-1 current profile')\n",
+      "grid()\n",
+      "show()\n",
+      "plot(1000*np.array(t),Torque)\n",
+      "xlabel('time [msec]')\n",
+      "ylabel('Torque [N-m]')\n",
+      "title('(b) torque profile')\n",
+      "grid()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "i1 ="
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 100*t/(51.1872396234976*t + 0.005) \t, (where round(16.2934044186179*pi,2) = 51.1 )\n",
+        "\n",
+        "i2 = (-200*t + 0.75)/(-51.1*t + 0.2605) \n",
+        "\n",
+        "\n",
+        "Positve torque integral = 0.000456384094483 [N-m-sec]\n",
+        "\n",
+        "Torque integral= 0.000335463884625 [N-m-sec]\n",
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['fmod', 'cosh', 'sinh', 'trunc', 'tan', 'gamma', 'degrees', 'radians', 'sin', 'expm1', 'ldexp', 'isnan', 'frexp', 'ceil', 'copysign', 'cos', 'tanh', 'fabs', 'sqrt', 'hypot', 'log', 'log10', 'pi', 'log1p', 'floor', 'modf', 'exp', 'isinf', 'e']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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cS9cJbpGRkZw9e7bCO3e3nBxITISuXY05vlWubbZCTitkBMnpapXN+Yc/QEGB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\ntWoxZMgQNmzYUGy7ioQ2e1e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7925q1Sr+X7G8QieEXmSMQVQbdevW5cqVKxX+uaI34EaN\nGnHjjTeyfv16+/dLG2OoqKysLDIzM+nXrx/z5s1j7969APTq1Yt3333Xvl3R8Xr37m1fOhkgIyPD\nJTmEcIYUBlFtBAYGEh4eTvv27Xn22Wex2Wz2q3eu/bro+bVfFz1fsWIF8+bNIywsjA4dOjg96Fva\nvoueZ2Rk0LdvXyIiIoiKiuL1118H4N1337XfAKZDhw4sWLAAgJkzZ3Ly5Enat29PeHg4W7ZsqUSL\nCFE5cj8GISrolVdewcfHhylTphhy/B49ejBv3jwiIyMNOb6o/uSMQYgK8vHx4b333jNsgltSUpLD\nOIYQriZnDEIIIRzIGYMQQggHUhiEEEI4kMIghBDCgRQGIYQQDqQwCCGEcPD/ATnLc2SOfyyJAAAA\nAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x388b510>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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cObLQ8nvvvafLvoQQxZOehvudOQMTJ8Lq1dqhzfPmga+v0alcU+qcxocffsjX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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3a36f90>"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.4, Page number: 433"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "\n",
+      "#Variavle declaration:\n",
+      "rpm=2500                        #rpm of motor\n",
+      "\n",
+      "\n",
+      "#Calculations & Results:\n",
+      "#For part (a):\n",
+      "theta=[0]*12\n",
+      "i=[0]*102\n",
+      "lambda1=[0]*102\n",
+      "for m in range(1,11,1):\n",
+      "    theta[m-1]=10*(m-1)\n",
+      "    for n in range(1,102,1):\n",
+      "        i[n-1]=30*(n-1)/100\n",
+      "        lambda1[n-1]=i[n-1]*(0.005+0.09*((90-theta[m-1])/90))*(8/(i[n-1]+8))\n",
+      "\n",
+      "    \n",
+      "    plot(i,lambda1,'.')\n",
+      "    \n",
+      "    if m==1:\n",
+      "        hold(True)\n",
+      "        \n",
+      "xlabel('current [A]')\n",
+      "ylabel('Lambda [Wb]')\n",
+      "title('Family of lambda-i curves as theta_m varies from 0 to 90 degrees')    \n",
+      "annotate('theta_m=0 deg',xy=(6,0.03))\n",
+      "annotate('theta_m=0 deg',xy=(8,0.5))\n",
+      "\n",
+      "\n",
+      "#for part (b):\n",
+      "lambdamax=25*(0.005+0.09*(8/(25+8)))\n",
+      "AreaWnet=0\n",
+      "AreaWrec=0\n",
+      "deli=0.25\n",
+      "for n in range(1,102,1):\n",
+      "    i[n-1]=25*(n-1)/100\n",
+      "    AreaWnet=AreaWnet + deli*i[n-1]*(0.09)*(8/(i[n-1]+8))\n",
+      "    AreaWrec=AreaWrec + deli*(lambdamax-i[n-1]*(0.005+0.09*(8/(i[n-1]+8))))\n",
+      "\n",
+      "Ratio=(AreaWnet+AreaWrec)/AreaWnet\n",
+      "print \"part (b): Ratio =\", round(Ratio,2)\n",
+      "\n",
+      "#for part(b):\n",
+      "rps=rpm/60\n",
+      "T=1/rps\n",
+      "Pphase=2*AreaWnet/T\n",
+      "Ptot=2*Pphase\n",
+      "print \"part (c): AreaWnet =\", round(AreaWnet,2),\"Joules\"\n",
+      "print \"Pphase =\",round(Pphase),\"W\",\"\\tPtot =\",round(Ptot),\"W\\n\"\n",
+      "plot(AreaWrec=0.7,AreaWnet=25)\n",
+      "grid()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "part (b): Ratio ="
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1.55\n",
+        "part (c): AreaWnet = 9.91 Joules\n",
+        "Pphase = 825.0 W \tPtot = 1651.0 W\n",
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['power', 'random', 'fft', 'linalg', 'info']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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arSaffvqpYN+uu+46sn//fkIIIc888wxJSkrq8ZzywQ6pTZvmmRCaN8M2LkJh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\nAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3a59f50>"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter9.ipynb b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter9.ipynb
new file mode 100755
index 00000000..3cdb1b6c
--- /dev/null
+++ b/ELECTRIC_MACHINERY_by_Fitzgerald_Kingsley_and_Umans/chapter9.ipynb
@@ -0,0 +1,390 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:934a14335227a49c83c2d399431a59d2d79025dde47942572f3e87ac684c8499"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Single- and Two-Phase Motors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page number: 459"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import *\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Zmain=4.5+3.7j                              #main winding impedance(ohm)\n",
+      "Zaux=9.5+3.5j                               #auxilliary winding impedance(ohm)\n",
+      "f=60                                        #frequency(Hz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "phy_main=math.degrees(math.atan(Zmain.imag/Zmain.real))\n",
+      "phy=phy_main-90\n",
+      "w=2*pi*60\n",
+      "Xc=symbols('Xc')\n",
+      "a=solve((3.5+Xc)/9.5-math.tan(math.radians(float(phy))), Xc)\n",
+      "C=-1/(w*a[0])\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"The starting capacitance:\",round(float(C)*10**6,0), \"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The starting capacitance: 176.0 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page number: 467"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import cmath\n",
+      "from math import *\n",
+      "\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1_m=2.02                                   #resistance of main winding(ohm)\n",
+      "X1_m=2.79                                   #resistance of main\n",
+      "R2_m= 4.12                                  #Rotor resistance ref. to stator(ohm)\n",
+      "X2_m=2.12                                   #Rotor reactance ref. to stator(ohm)\n",
+      "Xm=66.8                                     #Magnetising reactance(ohm)\n",
+      "s=0.05                                      #slip\n",
+      "Pcu=24                                       #copper loss(W)\n",
+      "Pw=13                                       #friction & windage loss(W)\n",
+      "V=110                                       #line-to-line voltage(V)\n",
+      "p=4                                         #no.of poles\n",
+      "fc=60                                       #frequency(Hz)\n",
+      "\n",
+      "#Calculations:\n",
+      "X22=X2_m+Xm\n",
+      "Q2_m=X22/R2_m\n",
+      "Rf=(Xm**2/X22)*(1/(s*Q2_m+1/(s*Q2_m)))\n",
+      "Xf=(X2_m*Xm/X22)+Rf/(s*Q2_m)\n",
+      "Zf=Rf+1j*Xf                             #forward field impedance(ohm)\n",
+      "\n",
+      "Rb=R2_m*(Xm/X22)**2/(2-s)\n",
+      "Xb=(X2_m*Xm/X22)+Rb/((2-s)*Q2_m)\n",
+      "Zb=Rb+1j*Xb                             #bachward field impedance\n",
+      "Zt=0.5*(Zf+Zb)+R1_m+1j*X1_m\n",
+      "I=V/abs(Zt)                             #Stator current(A)\n",
+      "pf=cos(cmath.phase(Zt))                 #power factor\n",
+      "Pin=V*I*pf\n",
+      "Pg_f=I**2*0.5*Rf                        #power absorbed by forward field(W)\n",
+      "Pg_b=I**2*0.5*Rb                        #power absorbed by backward field(W)\n",
+      "Pmech=(1-s)*(Pg_f-Pg_b)\n",
+      "Pshaft=Pmech-(Pcu+Pw)\n",
+      "ws=(2/p)*120*pi\n",
+      "ns=(120/p)*fc\n",
+      "n=(1-s)*ns                              #Rotor speed(rpm)\n",
+      "wm=(1-s)*ws\n",
+      "Tshaft=Pshaft/wm                         #shaft torque(Nm)\n",
+      "eff=Pshaft/Pin\n",
+      "\n",
+      "#Results:\n",
+      "print \"Stator current:\",round(I),\"A\", \"\\nPower factor:\",round(pf,3)\n",
+      "print \"Power output:\",round(Pshaft),\"W\", \"\\nSpeed:\",n,\"rpm\"\n",
+      "print \"Shaft torque:\",round(Tshaft,3),\"Nm\",\"Efficiency\",round(eff*100),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Stator current: 4.0 A \n",
+        "Power factor: 0.621\n",
+        "Power output: 147.0 W \n",
+        "Speed: 1710.0 rpm\n",
+        "Shaft torque: 0.823 Nm Efficiency 60.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.3, Page number: 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "f=60                                        #freq(Hz)\n",
+      "omeag=2*pi*f\n",
+      "s=0.05                                      #slip\n",
+      "R1=0.534                                    #resistance of main winding(ohm)\n",
+      "X1=2.45\n",
+      "Xm=70.1\n",
+      "R2=0.956\n",
+      "X2=2.96\n",
+      "Valpha=230\n",
+      "Vbeta=210*cmath.exp(1j*80*pi/180)\n",
+      "\n",
+      "#Calculations:\n",
+      "Vf = 0.5*(Valpha - 1j*Vbeta)\n",
+      "Vb = 0.5*(Valpha + 1j*Vbeta)\n",
+      "Zf=R1+1j*X1+1j*Xm*(R2/s+1j*X2)/(R2/s+1j*(X2+Xm))\n",
+      "If=Vf/Zf\n",
+      "Zb=R1+1j*X1+1j*Xm*(R2/(2-s)+1j*X2)/(R2/(2-s)+1j*(X2+Xm))\n",
+      "Ib = Vb/Zb\n",
+      "Ialpha=If+Ib\n",
+      "Ibeta=1j*(If-Ib)\n",
+      "Pgf=2*((Vf*(If.conjugate())).real-R1*abs(If)**2)\n",
+      "Pgb=2*((Vb*(Ib.conjugate())).real-R1*abs(Ib)**2)\n",
+      "Pmech=(1-s)*(Pgf-Pgb)\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \"(a) Positive seq components:\", round(Vf.real,1)+1j*round(Vf.imag,1),\"V\"\n",
+      "print\"    Negative seq. components:\", round(Vb.real,1)+1j*round(Vb.imag,1),\"V\"\n",
+      "\n",
+      "print\"\\n(b) Positive stator currents:\",round(If.real,1)+1j*round(If.imag,1),\"A\"\n",
+      "print\"    Negative stator currnets:\",round(Ib.real,1)+1j*round(Ib.imag,1),\"A\"\n",
+      "\n",
+      "print\"\\n(c) Positive currents:\",round(Ialpha.real,1)+1j*round(Ialpha.imag,1),\"A\"\n",
+      "print\"    Negative currnets:\",round(Ibeta.real,1)+1j*round(Ibeta.imag,1),\"A\"\n",
+      "\n",
+      "print \"\\n(d) Power to forward field:\",round(Pgf,0),\"W\"\n",
+      "print \"    Power to backward field:\",round(Pgb,0),\"W\"\n",
+      "print \"    Pmech:\",round(Pmech,0),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Positive seq components: (218.4-18.2j) V\n",
+        "    Negative seq. components: (11.6+18.2j) V\n",
+        "\n",
+        "(b) Positive stator currents: (9.3-6.3j) A\n",
+        "    Negative stator currnets: (3.7-1.5j) A\n",
+        "\n",
+        "(c) Positive currents: (13-7.8j) A\n",
+        "    Negative currnets: (4.8+5.6j) A\n",
+        "\n",
+        "(d) Power to forward field: 4149.0 W\n",
+        "    Power to backward field: 15.0 W\n",
+        "    Pmech: 3928.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.5, Page number: 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "import cmath\n",
+      "from math import *\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Lmain=0.0806                           #main winding inductance(H)\n",
+      "Rmain = 0.58                           #main winding resistance(ohm)\n",
+      "Laux = 0.196                           #auxilliary winding inductance(H)\n",
+      "Raux = 3.37                            #auxilliary winding resistance(ohm)\n",
+      "Lr=4.7*10**-6                          #rotor inductance(H)\n",
+      "Rr=37.6*10**-6                         #rotor resistance(ohm)\n",
+      "Lmain_r=0.588*10**-3                   #main inductance ref. to rotor(H)\n",
+      "Laux_r = 0.909*10**-3                  #aux inductance ref. to rotor(H)\n",
+      "p=2                                    #poles\n",
+      "Vo=230                                 #terminal voltage(V)\n",
+      "w=120*pi                               #angular frequency(Hz)\n",
+      "C=35*10**-6\n",
+      "Prot=40                                #Windage losses(W)\n",
+      "Pcore=105                              #Core loss(W)\n",
+      "n=3500                                 #rpm\n",
+      "\n",
+      "\n",
+      "#calculations and Results:\n",
+      "Xc=-1/(w*C)\n",
+      "speed=[0]*102\n",
+      "for cal in range(1,3,1):\n",
+      "    if cal==1:\n",
+      "        mmax=2\n",
+      "    else:\n",
+      "        mmax=102\n",
+      "    for m in range(1,mmax,2):\n",
+      "        if cal==1:\n",
+      "            speed[m-1]=3500\n",
+      "        else:\n",
+      "            speed[m-1]=3599*(m-1)/100\n",
+      "        \n",
+      "        ns=(2/p)*3600\n",
+      "        s=(ns-speed[m-1])/ns\n",
+      "\n",
+      "#for part (a):\n",
+      "        Kplus=s*w/(2*(Rr+1j*s*w*Lr))\n",
+      "        Kminus=(2-s)*w/(2*(Rr+1j*(2-s)*w*Lr))\n",
+      "        A1=Lmain-1j*Lmain_r**2*(Kplus+Kminus)\n",
+      "        A2=Lmain_r*Laux_r*(Kplus-Kminus)\n",
+      "        A3=Laux-1j*Laux_r**2*(Kplus+Kminus)\n",
+      "        M=[[0]*2,[0]*2]\n",
+      "        M[0][0]=Rmain + 1j*w*A1\n",
+      "        M[0][1] = 1j*w*A2;\n",
+      "        M[1][0] = -1j*w*A2;\n",
+      "        M[1][1] = Raux + 1j*Xc+ 1j*w*A3\n",
+      "        V=[[Vo],[-Vo]]\n",
+      "        M1=inv(M)\n",
+      "        I=dot(M1,V)\n",
+      "        Imain=I[0][0]\n",
+      "        Iaux=I[1][0]\n",
+      "        Is=Imain-Iaux\n",
+      "        magImain=abs(Imain)\n",
+      "        angleImain=math.degrees(cmath.phase(Imain))\n",
+      "        magIaux=abs (Iaux)\n",
+      "        angleIaux=math.degrees(cmath.phase(Iaux))\n",
+      "        magIs=abs(Is)\n",
+      "        angleIs=math.degrees(cmath.phase(Is))\n",
+      "        Vcap=Iaux*Xc\n",
+      "        magVcap=abs(Vcap)\n",
+      "        \n",
+      "    #for part (b):\n",
+      "        Tmech=[0]*102\n",
+      "        Pshaft=[0]*102\n",
+      "        Tmechl = (Kplus-Kminus).conjugate()\n",
+      "        Tmechl=Tmechl*(Lmain_r**2*Imain*((Imain).conjugate())+Laux_r**2*Iaux*((Iaux).conjugate()))\n",
+      "        Tmech2 = 1j*Lmain_r*Laux_r*((Kplus+Kminus).conjugate())\n",
+      "        Tmech2 = Tmech2*((Imain).conjugate()*Iaux-Imain*((Iaux).conjugate()));\n",
+      "        Tmech[m-1] = (p/2)*(Tmechl+Tmech2).real\n",
+      "        Pshaft=((2/p)*(1-s)*w*Tmech[m-1])-Prot\n",
+      "        \n",
+      "    #for part (c):\n",
+      "        Pmech=[0]*102\n",
+      "        Pmain = (Vo*(Imain.conjugate())).real\n",
+      "        Paux = (-Vo*(Iaux.conjugate())).real\n",
+      "        Pin = Pmain+Paux+Pcore\n",
+      "        eta = Pshaft/Pin;\n",
+      "        if cal==1:\n",
+      "            print \"part (a):\"\n",
+      "            print  \"\\nImain=\",round(magImain,1),\"A  at an angle\",round(angleImain,1),\"degrees\"\n",
+      "            print  \"\\nImain=\",round(magIaux,1),\"A  at an angle\",round(angleIaux,1),\"degrees\"\n",
+      "            print  \"\\nImain=\",round(magIs,1),\"A  at an angle\",round(angleIs,1),\"degrees\"\n",
+      "            print  \"\\nVcap=\",round(magVcap,0),\"V\"\n",
+      "            print \"\\npart (b):\"\n",
+      "            print \"\\nTmech=\",round(Tmech[0],2),\"Nm\"\n",
+      "            print \"\\nPshaft=\",round(Pshaft),\"W\"\n",
+      "            print \"\\npart (c):\"\n",
+      "            print \"\\nPmain=\",round(Pmain,0),\"W\"\n",
+      "            print \"\\nPaux=\",round(Paux,0),\"W\"\n",
+      "            print \"\\nPin=\",round(Pin,0),\"W\"\n",
+      "            print \"\\nEfficiency=\",round(eta*100,1),\"%\"\n",
+      "        else:\n",
+      "            \n",
+      "            plot(speed,Tmech,'g.')\n",
+      "            xlabel('speed (rpm)')\n",
+      "            ylabel('Tmech (Nm)')\n",
+      "            title('Electromagnetic torque vs speed')\n",
+      "    show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part (a):\n",
+        "\n",
+        "Imain= 15.9 A  at an angle -37.6 degrees\n",
+        "\n",
+        "Imain= 5.2 A  at an angle -150.8 degrees\n",
+        "\n",
+        "Imain= 18.5 A  at an angle -22.7 degrees\n",
+        "\n",
+        "Vcap= 394.0 V\n",
+        "\n",
+        "part (b):\n",
+        "\n",
+        "Tmech= 9.75 Nm\n",
+        "\n",
+        "Pshaft= 3532.0 W\n",
+        "\n",
+        "part (c):\n",
+        "\n",
+        "Pmain= 2893.0 W\n",
+        "\n",
+        "Paux= 1043.0 W\n",
+        "\n",
+        "Pin= 4041.0 W\n",
+        "\n",
+        "Efficiency= 87.4 %\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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KagsAyoHkACoNtQUA5cC0Eqg01BYAWgc1B9BIqC0AvBnUHEAjobYAoFqQHEAl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\nAABACpIDAABI+T9crBvDDQoKmwAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2854950>"
+       ]
+      }
+     ],
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+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Electronic_Communications:_Principles_and_Systems_by_W._D._Stanley_&_J._M._Jeffords/README.txt b/Electronic_Communications:_Principles_and_Systems_by_W._D._Stanley_&_J._M._Jeffords/README.txt
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--- /dev/null
+++ b/Electronic_Communications:_Principles_and_Systems_by_W._D._Stanley_&_J._M._Jeffords/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Amit Raval
+Course: bscit
+College/Institute/Organization: Saurashtra University
+Department/Designation: Bsc. IT
+Book Title: Electronic Communications: Principles and Systems
+Author: W. D. Stanley & J. M. Jeffords
+Publisher: Cengage Learning, New Delhi
+Year of publication: 2009
+Isbn: 9788131503546
+Edition: 2
\ No newline at end of file
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diff --git a/Engineering_Economics/Chapter1.ipynb b/Engineering_Economics/Chapter1.ipynb
deleted file mode 100755
index 64198a89..00000000
--- a/Engineering_Economics/Chapter1.ipynb
+++ /dev/null
@@ -1,83 +0,0 @@
-{
- "metadata": {
-  "name": "EE-1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": "Introduction"
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 1.1 Page 10"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initialisation of Variable\nFC=2000000.0;#in Rs\nv=100.0;#in Rs\ns=200.0;#in Rs\nQ=60000.0;#in units\n\n#Part a \n# calculation\nBEQ=FC/(s-v);#in units\n\n#result\nprint \"Break even quantity in units : \",round(BEQ,3)\n\n#Part b \n#calculation\nBES=FC*s/(s-v);#in Rs\n\nprint \"Break even sales in Rs. : \",round(BES,3)\n\n#Part c \n#calculation\nCon=s*Q-v*Q;#in Rs\nMos=Q*s-BES;#in Rs\n\n#result\nprint \"Contribution in units : \",round(Con,3)\nprint \"Margin of safety in Rs. : \",round(Mos,3)",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Break even quantity in units :  20000.0\nBreak even sales in Rs. :  4000000.0\nContribution in units :  6000000.0\nMargin of safety in Rs. :  8000000.0\n"
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 1.2,Page 12"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initialisation of variable\nFC=25000.0;#in Rs\nv=45000.0;#in Rs\ns=120000.0;#in Rs\n\n#Part a \n#calculation\nCon=s-v;#in Rs\n\n#result\nprint \"Contribution in Rs  \",round(Con,3)\n\n#Part b \n#calculation\nProf=Con-FC;#in Rs\n\n#result\nprint \"Profit in Rs  \",round(Prof,3);\n\n#Part c \n#calculation\nPVratio=Con*100.0/s;#in %\nBEP=FC*100.0/PVratio;#in Rs\n\n#result\nprint \"BEP in Rs \", round(BEP,3)\n\n#Part d \n#calculoation\nMS=Prof*100/PVratio;#in Rs\n\n#result\nprint\"M.S. in Rs  \", round(MS,3)",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Contribution in Rs   75000.0\nProfit in Rs   50000.0\nBEP in Rs  40000.0\nM.S. in Rs   80000.0\n"
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 1.3 Page 13"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initialisation of variable\nFC=15000.0;#in Rs\nv=35000.0;#in Rs\ns=80000.0;#in Rs\n\n#Part a \n#calculation\nCon=s-v;#in Rs\n\n#result\nprint \"Contribution in Rs\", round(Con,3);\n\n#Part b \n#calcualtion\nProf=Con-FC;#in Rs\n\n#result\nprint \"Profit in Rs \", round(Prof,3)\n\n#Part c \n#calculation\nPVratio=Con*100/s;\nBEP=FC*100/PVratio;#in Rs\n\n#result\nprint \"BEP in Rs  \",round(BEP,3)\n\n#Part d \n#calculation\nMS=Prof*100/PVratio;#in Rs\n\nprint \"M.S. in Rs\",round(MS,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Contribution in Rs 45000.0\nProfit in Rs  30000.0\nBEP in Rs   26666.667\nM.S. in Rs 53333.333\n"
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter10.ipynb b/Engineering_Economics/Chapter10.ipynb
deleted file mode 100755
index bc34bb84..00000000
--- a/Engineering_Economics/Chapter10.ipynb
+++ /dev/null
@@ -1,357 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:28b0077be1a0d41b0cc471c0e34331f1edde6c138991b53632e2c3b3cb0de57c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Evaluation of Public Alternatives"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.1 Page 138"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "Ii=4000000.0;#in Rs.\n",
-      "AM=150000.0;#in Rs.\n",
-      "AFS=600000.0;#in Rs.\n",
-      "Einc=50000.0;#in Rs.\n",
-      "i=12.0;#in % per annum\n",
-      "n=15.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "Cp=AM*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in Rs\n",
-      "TPW=Ii+Cp;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"Total present worth of costs in RS. : \",round(TPW,3);\n",
-      "\n",
-      "#Total present worth of fuel savings:\n",
-      "AI=600000.0;#in Rs.\n",
-      "G=50000.0;#in Rs.\n",
-      "i=12.0;#in % per annum\n",
-      "n=15.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "A=AI+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100);#in RS\n",
-      "Bp=A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in Rs.\n",
-      "BCratio=Bp/(Ii+Cp);#unitless\n",
-      "\n",
-      "#result\n",
-      "print \"Present worth of fuel savings in Rs. : \",round(Bp,3);\n",
-      "print \"BCratio : \",round(BCratio,3);\n",
-      "print\"Since BC ratio is more than 1, the construction of the bridge across the river is justified.\";\n",
-      "print\"answers are slightly different from the book due to approximation\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total present worth of costs in RS. :  5021629.673\n",
-        "Present worth of fuel savings in Rs. :  5782527.265\n",
-        "BCratio :  1.152\n",
-        "Since BC ratio is more than 1, the construction of the bridge across the river is justified.\n",
-        "answers are slightly different from the book due to approximation\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2 Page 139"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "Ic=80000000.0;#in Rs.\n",
-      "Aps=6000000.0;#in Rs.\n",
-      "Afs=3000000.0;#in Rs.\n",
-      "Aib=5000000.0;#in Rs.\n",
-      "Arb=2000000.0;#in Rs.\n",
-      "Amc=3000000.0;#in Rs.\n",
-      "i=12.0;#in % per annum\n",
-      "n=50.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "TAB=Afs+Aib+Arb;#in Rs.\n",
-      "PW_B=TAB*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Total present worth of benefits in RS. : \",round(PW_B,3);\n",
-      "\n",
-      "#Present worth of costs:\n",
-      "PW_C=Ic+Amc*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n)-Aps*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "BCratio=PW_B/PW_C;#unitless\n",
-      "\n",
-      "#result\n",
-      "print \"Present worth of costs in Rs. : \",round(PW_C,3);\n",
-      "print \"BCratio : \",round(BCratio,3);\n",
-      "print \"Since BC ratio is more than 1, the state govt. can implement the hydroelectric project.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total present worth of benefits in RS. :  83044984.884\n",
-        "Present worth of costs in Rs. :  55086504.535\n",
-        "BCratio :  1.508\n",
-        "Since BC ratio is more than 1, the state govt. can implement the hydroelectric project.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.3 Page 140"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#alternative 1\n",
-      "P=3000000.0;#in Rs.\n",
-      "B=900000.0;#in Rs.\n",
-      "i=10.0;#in % per annum\n",
-      "n=5.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "AE1=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs\n",
-      "BCratio=B/AE1;#unitless\n",
-      "\n",
-      "#result\n",
-      "print \"Annual equivalent of initial cost in Rs. : \",round(AE1,3);\n",
-      "print \"BCratio : \",round(BCratio,3);\n",
-      "\n",
-      "#Alternative A2 : \n",
-      "P=6000000.0;#in Rs.\n",
-      "B=1500000.0;#in Rs.\n",
-      "i=10.0;#in % per annum\n",
-      "n=7.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "AE2=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs\n",
-      "BCratio=B/AE2;#unitless\n",
-      "\n",
-      "#result\n",
-      "print \"Annual equivalent of initial cost in Rs. : \",round(AE2,3);\n",
-      "print \"BCratio : \",round(BCratio,3);\n",
-      "print \"The benefit cost ratio of alternative 2 is more than that of alternative A1. Hence, alternative A2 is to be selected. The comparisoon is made on a 35 years period which is the minimum common multiple of the lives of alternative 1 and 2\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Annual equivalent of initial cost in Rs. :  791392.442\n",
-        "BCratio :  1.137\n",
-        "Annual equivalent of initial cost in Rs. :  1232432.998\n",
-        "BCratio :  1.217\n",
-        "The benefit cost ratio of alternative 2 is more than that of alternative A1. Hence, alternative A2 is to be selected. The comparisoon is made on a 35 years period which is the minimum common multiple of the lives of alternative 1 and 2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.4 Page 141"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#cost of the state\n",
-      "n=20.0;#in years\n",
-      "P=2500000000.0;#in Rs.\n",
-      "Agt=10000000.0;#in Rs.\n",
-      "Ai=1000000.0;#in Rs.\n",
-      "Com=48000.0;#in Rs./year/employee\n",
-      "C1=Com*300;#in Rs.\n",
-      "i=10.0;#in % per annum\n",
-      "\n",
-      "#calculation\n",
-      "C2=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs\n",
-      "CA=C2+C1;#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Annual equivalent initial cost in Rs. : \",round(C2,3);\n",
-      "print \"Total Annual equivalent cost of the project in Rs. : \",round(CA,3);\n",
-      "\n",
-      "#Benefit to the state : \n",
-      "W=30000000.0;#in Rs.\n",
-      "A1=W*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs.\n",
-      "Agpy=10000000.0;#in ton/year\n",
-      "A2=Agpy*(30-10);#in Rs.\n",
-      "AvgAI=1000000.0;#in tons Km.\n",
-      "G=20000000.0;#in Rs.\n",
-      "\n",
-      "#calculation\n",
-      "A3=A2+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100) ;#in Rs.\n",
-      "BA=A1+A3;#in Rs\n",
-      "BCratio=BA/CA;#unitless\n",
-      "print \"Annual equivalent average of transport cost saving in Rs. : \",round(A3,3);\n",
-      "print \"Total annual equivalent benefits to the state : \",round(BA,3);\n",
-      "print \"BCratio : \",round(BCratio,3);\n",
-      "print \"The benefit cost ratio is more than 1,the project is justified.\";\n",
-      "print \"the slight error in the answer of the book is due to approximation error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Annual equivalent initial cost in Rs. :  293649061.931\n",
-        "Total Annual equivalent cost of the project in Rs. :  308049061.931\n",
-        "Annual equivalent average of transport cost saving in Rs. :  330161500.91\n",
-        "Total annual equivalent benefits to the state :  333685289.653\n",
-        "BCratio :  1.083\n",
-        "The benefit cost ratio is more than 1,the project is justified.\n",
-        "the slight error in the answer of the book is due to approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5 Page 143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#project A\n",
-      "i=9.0;#% per annum\n",
-      "n=50.0;#in years\n",
-      "P_A=150000000.0;#in RS.\n",
-      "Fcs_A=2500000;#in Rs.\n",
-      "Ib_A=3500000;#in Rs.\n",
-      "Rb_A=1000000;#in RS.\n",
-      "OMC_A=2000000.0;#in Rs.\n",
-      "Psy_A=10000000.0;#in Rs.\n",
-      "\n",
-      "#calculation\n",
-      "AE_A=P_A*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\n",
-      "Cpy_A=AE_A+OMC_A-Psy_A;#in Rs.\n",
-      "Bpy_A=Fcs_A+Ib_A+Rb_A;#in Rs.\n",
-      "BCratio_A=Bpy_A/Cpy_A;#unitless\n",
-      "\n",
-      "#result\n",
-      "print \"Costs/year : \",round(Cpy_A,3);\n",
-      "print \"Benefits/year : \",round(Bpy_A,3);\n",
-      "print \"BC Ratio of project A : \",round(BCratio_A,3)\n",
-      "\n",
-      "\n",
-      "#Project B : \n",
-      "P_B=250000000.0;#in Rs.\n",
-      "\n",
-      "OMC_B=2500000.0;#in Rs.\n",
-      "Psy_B=12000000.0;#in Rs.\n",
-      "Fcs_B=3500000.0;#in Rs.\n",
-      "Ib_B=4500000.0;#in Rs.\n",
-      "Rb_B=2000000.0;#in RS.\n",
-      "AE_B=P_B*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\n",
-      "Cpy_B=AE_B+OMC_B-Psy_B;#in Rs.\n",
-      "Bpy_B=Fcs_B+Ib_B+Rb_B;#in Rs.\n",
-      "BCratio_B=Bpy_B/Cpy_B;#unitless\n",
-      "\n",
-      "#result\n",
-      "print \"Costs/year : \",round(Cpy_B,3);\n",
-      "print \"Benefits/year : \",round(Bpy_B,3);\n",
-      "print \"BC Ratio of project B : \",round(BCratio_B,3)\n",
-      "\n",
-      "#Project C : \n",
-      "P_C=400000000.0;#in Rs.\n",
-      "OMC_C=3500000.0;#in Rs.\n",
-      "Psy_C=18000000.0;#in Rs.\n",
-      "Fcs_C=5000000.0;#in Rs.\n",
-      "Ib_C=6000000.0;#in Rs.\n",
-      "Rb_C=3500000.0;#in RS.\n",
-      "AE_C=P_C*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\n",
-      "Bpy_C=Fcs_C+Ib_C+Rb_C;#in Rs.\n",
-      "Cpy_C=AE_C+OMC_C-Psy_C;#in Rs.\n",
-      "BCratio_C=Bpy_C/Cpy_C;#unitless\n",
-      "\n",
-      "#result\n",
-      "print \"Costs/year : \",round(Cpy_C,3);\n",
-      "print \"Benefits/year : \",round(Bpy_C,3);\n",
-      "print \"BC Ratio of project C : \",round(BCratio_C,3)\n",
-      "print\"From the computations it is clear that only alternative A is eligible because other two alternatives have BC ratio less than one.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Costs/year :  5684030.212\n",
-        "Benefits/year :  7000000.0\n",
-        "BC Ratio of project A :  1.232\n",
-        "Costs/year :  13306717.021\n",
-        "Benefits/year :  10000000.0\n",
-        "BC Ratio of project B :  0.752\n",
-        "Costs/year :  21990747.233\n",
-        "Benefits/year :  14500000.0\n",
-        "BC Ratio of project C :  0.659\n",
-        "From the computations it is clear that only alternative A is eligible because other two alternatives have BC ratio less than one.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter11.ipynb b/Engineering_Economics/Chapter11.ipynb
deleted file mode 100755
index 55cbf9b2..00000000
--- a/Engineering_Economics/Chapter11.ipynb
+++ /dev/null
@@ -1,306 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:72b2e09b74af58b464a4895e1776dc966d9cb12318da78fc7aa16f78becf8696"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Inflation Adjusted Decisions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.1 Page 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "print \"Few steps are as follows : \";\n",
-      "print \"Step 1 : The estimated future requirementt per year in terms of todays rupees from his age 61 through 70 is Rs. 24000.\";\n",
-      "print \"Step 2 : The formula which is given below is used to get future equivalent of Rs. 24000 with the inflation of 9% per year.\";\n",
-      "print \"Formula : P*(1+IR)^n\";\n",
-      "P=24000.0;#in Rs.\n",
-      "IR=9.0;#in % per year\n",
-      "Age=61.0;#in years\n",
-      "for n in range(21,31) :\n",
-      "    IV=P*(1+IR/100)**n;#in Rs\n",
-      "    print \"Age in years : \",Age,\". \"\"Inflated value of Rs. 24000 at end of year : \",n,\" in Rs. \",round(IV);\n",
-      "    Age=Age+1;#in Years\n",
-      "\n",
-      "PW=0;#For initialising\n",
-      "n=21.0;#in years\n",
-      "for m in range(10) :\n",
-      "    A=P*(1+IR/100)**n;#in Rs\n",
-      "    n=n+1;\n",
-      "    i=15.0;#in % per year\n",
-      "    PW=PW+A/(1+i/100)**(m+1);\n",
-      "\n",
-      "#result\n",
-      "print \"Present equivalent of withdrawls in Rs. : \",round(PW,3);\n",
-      "\n",
-      "#The annual equivalent amount :\n",
-      "i=15.0;#in % per year\n",
-      "n=20.0;#in years\n",
-      "#Formula : (A/F,i,n) : (i/100)/(((1+i/100)^n)-1) \n",
-      "A=PW*(i/100)/(((1+i/100)**n)-1);#in Rs.\n",
-      "print \"The annual equivalent amount in Rs. : \",round(A,3);\n",
-      "print\"RECOMMENDATION : The person has to invest an amount of Rs. \",round(A,3),\" at the end of every year starting from his age 41 through 60 which will enable him to make a withdrawl at the end of every year starting from his age 61 through 70.\"; \n",
-      "print \"slight erorr in the answer is due to approximation error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Few steps are as follows : \n",
-        "Step 1 : The estimated future requirementt per year in terms of todays rupees from his age 61 through 70 is Rs. 24000.\n",
-        "Step 2 : The formula which is given below is used to get future equivalent of Rs. 24000 with the inflation of 9% per year.\n",
-        "Formula : P*(1+IR)^n\n",
-        "Age in years :  61.0 . Inflated value of Rs. 24000 at end of year :  21  in Rs.  146611.0\n",
-        "Age in years :  62.0 . Inflated value of Rs. 24000 at end of year :  22  in Rs.  159806.0\n",
-        "Age in years :  63.0 . Inflated value of Rs. 24000 at end of year :  23  in Rs.  174189.0\n",
-        "Age in years :  64.0 . Inflated value of Rs. 24000 at end of year :  24  in Rs.  189866.0\n",
-        "Age in years :  65.0 . Inflated value of Rs. 24000 at end of year :  25  in Rs.  206954.0\n",
-        "Age in years :  66.0 . Inflated value of Rs. 24000 at end of year :  26  in Rs.  225580.0\n",
-        "Age in years :  67.0 . Inflated value of Rs. 24000 at end of year :  27  in Rs.  245882.0\n",
-        "Age in years :  68.0 . Inflated value of Rs. 24000 at end of year :  28  in Rs.  268011.0\n",
-        "Age in years :  69.0 . Inflated value of Rs. 24000 at end of year :  29  in Rs.  292132.0\n",
-        "Age in years :  70.0 . Inflated value of Rs. 24000 at end of year :  30  in Rs.  318424.0\n",
-        "Present equivalent of withdrawls in Rs. :  1013631.794\n",
-        "The annual equivalent amount in Rs. :  9894.537\n",
-        "RECOMMENDATION : The person has to invest an amount of Rs.  9894.537  at the end of every year starting from his age 41 through 60 which will enable him to make a withdrawl at the end of every year starting from his age 61 through 70.\n",
-        "slight erorr in the answer is due to approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 Page 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "print\"The method of finding the economic life of the machine witha discounting factor of 20% at zero inflation rate is summarized in table below. From the table it is clear that total annual cost is minimum if the machine is used for 14 years. Hence the economic life of the machine is 14 years.\";\n",
-      "print\"End of year     Op_cost     Main_cost    Op+Main     P/F,i,n     PW     Cummulative     Salvage     PW_S    TPW     A/P,i,n     AEM\";\n",
-      "i=20.0;#in  per year\n",
-      "Cum=0.0;#initialising\n",
-      "Op_cost=40000.0;#in RS.\n",
-      "Main_cost=60000.0;#in Rs.\n",
-      "OpMain=Op_cost+Main_cost;#in Rs.\n",
-      "S=400000.0;#in Rs.\n",
-      "for n in range (1,16) :\n",
-      "    PF=1/((1+i/100)**n);\n",
-      "    PW=OpMain*PF;#in Rs.\n",
-      "    Cum=Cum+PW\n",
-      "    PW_S=PF*S;#in RS.\n",
-      "    TPW=500000.0+Cum-PW_S;#in Rs.\n",
-      "    AP=((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\n",
-      "    AEM=TPW*AP;#in RS\n",
-      "    print \"       \",n,\"        \",Op_cost,\"        \",Main_cost,\"        \",OpMain,\"    \",PF,\"     \",PW,\"        \",Cum,\"        \",S,\"        \",PW_S,\"        \",TPW,\"        \",AP,\"        \",AEM;\n",
-      "    Op_cost=Op_cost+5000.0;#in Rs.\n",
-      "    Main_cost=Main_cost+6000.0;#in Rs.\n",
-      "    S=S-50000.0;#in Rs.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The method of finding the economic life of the machine witha discounting factor of 20% at zero inflation rate is summarized in table below. From the table it is clear that total annual cost is minimum if the machine is used for 14 years. Hence the economic life of the machine is 14 years.\n",
-        "End of year     Op_cost     Main_cost    Op+Main     P/F,i,n     PW     Cummulative     Salvage     PW_S    TPW     A/P,i,n     AEM\n",
-        "        1          40000.0          60000.0          100000.0      0.833333333333       83333.3333333          83333.3333333          400000.0          333333.333333          250000.0          1.2          300000.0\n",
-        "        2          45000.0          66000.0          100000.0      0.694444444444       69444.4444444          152777.777778          350000.0          243055.555556          409722.222222          0.654545454545          268181.818182\n",
-        "        3          50000.0          72000.0          100000.0      0.578703703704       57870.3703704          210648.148148          300000.0          173611.111111          537037.037037          0.474725274725          254945.054945\n",
-        "        4          55000.0          78000.0          100000.0      0.48225308642       48225.308642          258873.45679          250000.0          120563.271605          638310.185185          0.386289120715          246572.280179\n",
-        "        5          60000.0          84000.0          100000.0      0.401877572016       40187.7572016          299061.213992          200000.0          80375.5144033          718685.699588          0.33437970329          240313.910987\n",
-        "        6          65000.0          90000.0          100000.0      0.33489797668       33489.797668          332551.01166          150000.0          50234.6965021          782316.315158          0.300705745867          235247.011053\n",
-        "        7          70000.0          96000.0          100000.0      0.279081647234       27908.1647234          360459.176383          100000.0          27908.1647234          832551.01166          0.277423926347          230969.570539\n",
-        "        8          75000.0          102000.0          100000.0      0.232568039361       23256.8039361          383715.980319          50000.0          11628.4019681          872087.578351          0.260609422409          227274.240084\n",
-        "        9          80000.0          108000.0          100000.0      0.193806699468       19380.6699468          403096.650266          0.0          0.0          903096.650266          0.248079461672          224039.730836\n",
-        "        10          85000.0          114000.0          100000.0      0.16150558289       16150.558289          419247.208555          -50000.0          -8075.27914449          927322.4877          0.238522756883          221187.516286\n",
-        "        11          90000.0          120000.0          100000.0      0.134587985742       13458.7985742          432706.007129          -100000.0          -13458.7985742          946164.805703          0.231103794152          218662.276491\n",
-        "        12          95000.0          126000.0          100000.0      0.112156654785       11215.6654785          443921.672608          -150000.0          -16823.4982177          960745.170825          0.225264964904          216422.227187\n",
-        "        13          100000.0          132000.0          100000.0      0.0934638789872       9346.38789872          453268.060506          -200000.0          -18692.7757974          971960.836304          0.220620001083          214434.000758\n",
-        "        14          105000.0          138000.0          100000.0      0.0778865658226       7788.65658226          461056.717089          -250000.0          -19471.6414557          980528.358544          0.216893055222          212669.791417\n",
-        "        15          110000.0          144000.0          100000.0      0.0649054715189       6490.54715189          467547.264241          -300000.0          -19471.6414557          987018.905696          0.213882119838          211105.69587\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 Page 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "IR=5.0;#in % per year\n",
-      "i=15.0;#in % per year\n",
-      "#Machine X :\n",
-      "Ppx=1500000.0;#in Rs.\n",
-      "n=7.0;#in years\n",
-      "S=200000.0;#in Rs.\n",
-      "AMC=300000.0;#in Rs.\n",
-      "print \"End of year     AMC     InflationFactor     InflatedAmount     P/F     PW\";\n",
-      "Pw=0;#For initialising\n",
-      "for n in range (1,8) :\n",
-      "    FP=(1+IR/100)**n\n",
-      "    IA=AMC*FP;#in Rs.\n",
-      "    PF=1/((1+i/100)**n);\n",
-      "    PW=IA*PF;#in Rs.\n",
-      "    Pw=Pw+PW;#in Rs.\n",
-      "    print\"     \",n,\"        \",AMC,\"         \",FP,\"           \",IA,\"          \",PF,\"      \",PW;\n",
-      "\n",
-      "print \"Present worth of inflated annual operating and maintenance cost in Rs. : \",round(Pw,3);\n",
-      "PWX=Ppx+Pw-S*1/((1+i/100)**n);\n",
-      "print \"Present worth of machine X in Rs. : \",round(PWX,3);\n",
-      "\n",
-      "#Machine Y\n",
-      "\n",
-      "Ppy=2000000.0;#in Rs.\n",
-      "n=7.0;#in years\n",
-      "S=300000.0;#in Rs.\n",
-      "AMC=250000.0;#in Rs.\n",
-      "print \"End of year     AMC     InflationFactor     InflatedAmount     P/F     PW\";\n",
-      "Pw=0;#For initialising\n",
-      "for n in range (1,8) :\n",
-      "    FP=(1+IR/100)**n\n",
-      "    IA=AMC*FP;#in Rs.\n",
-      "    PF=1/((1+i/100)**n);\n",
-      "    PW=IA*PF;#in Rs.\n",
-      "    Pw=Pw+PW;#in Rs.\n",
-      "    print \"     \",n,\"        \",AMC,\"         \",FP,\"           \",IA,\"          \",PF,\"      \",PW;\n",
-      "\n",
-      "PWY=Ppy+Pw-S*1/((1+i/100)**n);\n",
-      "\n",
-      "#result\n",
-      "print \"Present worth of inflated annual operating and maintenance cost in Rs. : \",round(Pw,3);\n",
-      "print \"Present worth of machine Y in Rs. : \",round(PWY,3);\n",
-      "print \"Since the present worth of Machine X is less than Machine Y, select Machine X\"\n",
-      "print \" Calculations are not accurate in the book\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "End of year     AMC     InflationFactor     InflatedAmount     P/F     PW\n",
-        "      1          300000.0           1.05             315000.0            0.869565217391        273913.043478\n",
-        "      2          300000.0           1.1025             330750.0            0.756143667297        250094.517958\n",
-        "      3          300000.0           1.157625             347287.5            0.657516232432        228347.168571\n",
-        "      4          300000.0           1.21550625             364651.875            0.571753245593        208490.893043\n",
-        "      5          300000.0           1.2762815625             382884.46875            0.497176735298        190361.25017\n",
-        "      6          300000.0           1.34009564063             402028.692188            0.432327595912        173808.097981\n",
-        "      7          300000.0           1.40710042266             422130.126797            0.375937039923        158694.35033\n",
-        "Present worth of inflated annual operating and maintenance cost in Rs. :  1483709.322\n",
-        "Present worth of machine X in Rs. :  2908521.914\n",
-        "End of year     AMC     InflationFactor     InflatedAmount     P/F     PW\n",
-        "      1          250000.0           1.05             262500.0            0.869565217391        228260.869565\n",
-        "      2          250000.0           1.1025             275625.0            0.756143667297        208412.098299\n",
-        "      3          250000.0           1.157625             289406.25            0.657516232432        190289.307142\n",
-        "      4          250000.0           1.21550625             303876.5625            0.571753245593        173742.410869\n",
-        "      5          250000.0           1.2762815625             319070.390625            0.497176735298        158634.375141\n",
-        "      6          250000.0           1.34009564063             335023.910156            0.432327595912        144840.081651\n",
-        "      7          250000.0           1.40710042266             351775.105664            0.375937039923        132245.291942\n",
-        "Present worth of inflated annual operating and maintenance cost in Rs. :  1236424.435\n",
-        "Present worth of machine Y in Rs. :  3123643.323\n",
-        "Since the present worth of Machine X is less than Machine Y, select Machine X\n",
-        " Calculations are not accurate in the book\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4 Page 156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "IR=6.0;#in % per year\n",
-      "i=18.0;#in % per year\n",
-      "AFR=5000000.0;#in Rs.\n",
-      "n=7.0;#in years\n",
-      "AI=500000.0;#in Rs.\n",
-      "print\"End of year            AFR         InflationFactor     InflatedAmount  P/F           PW\";\n",
-      "TPW=0;#Initialising\n",
-      "\n",
-      "for n in range (1,6) :\n",
-      "    IF=(1+IR/100)**n;\n",
-      "    IA=IF*AFR;#in Rs.\n",
-      "    PF=1/((1+i/100)**n);\n",
-      "    PW=PF*IA;#in Rs.\n",
-      "    TPW=PW+TPW;#in Rs.\n",
-      "    print\"       \",round(n),\"        \",round(AFR,3),\"        \",round(IF,3),\"        \",round(IA,3),\"    \",round(PF,3),\"     \",round(PW,3);\n",
-      "    AFR=AFR+AI;#in Rs.\n",
-      "\n",
-      "    \n",
-      "#result\n",
-      "print \"The value of the single deposit to be made now to recieve the specified series for the next five years is Rs. : \",round(TPW,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "End of year            AFR         InflationFactor     InflatedAmount  P/F           PW\n",
-        "        1.0          5000000.0          1.06          5300000.0      0.847       4491525.424\n",
-        "        2.0          5500000.0          1.124          6179800.0      0.718       4438236.139\n",
-        "        3.0          6000000.0          1.191          7146096.0      0.609       4349334.645\n",
-        "        4.0          6500000.0          1.262          8206100.24      0.516       4232615.212\n",
-        "        5.0          7000000.0          1.338          9367579.043      0.437       4094655.134\n",
-        "The value of the single deposit to be made now to recieve the specified series for the next five years is Rs. :  21606366.553\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter12.ipynb b/Engineering_Economics/Chapter12.ipynb
deleted file mode 100755
index 75f40350..00000000
--- a/Engineering_Economics/Chapter12.ipynb
+++ /dev/null
@@ -1,104 +0,0 @@
-{
- "metadata": {
-  "name": "EE-12"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": "Inventory Control"
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 12.1 Page 160"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nfrom math import sqrt\nD=24000.0;#in units/year\nCo=150.0;#in Rs./order\nPprice=75.0;#Rs./unit\nCpupy=18.0;#in % of Pprice/unit\n\n#calculation\nCc=Pprice*Cpupy/100;#in Rs.\nEOQ=sqrt((2*Co*D)/Cc);#in units\nn=D/round(EOQ);#no. of orders/year\nT=round(EOQ)/D;# time between successive orders in year\nT=T*12;#in months\nT=T*30;#in Days\n\n#result\nprint \"Economic order quantity in units : \",round(EOQ);\nprint \"No. of orders/year : \",round(n,2);\nprint \"Time between successive orders in days : \",round(T);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Economic order quantity in units :  730.0\nNo. of orders/year :  32.88\nTime between successive orders in days :  11.0\n"
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 12.2 Page 162"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nfrom math import sqrt\nr=12000.0;#in units/year\nk=24000.0;#in units/year\nCo=175.0;#in Rs./setup\nCc=15.0;#in Rs./unit/year\n\n#calclation\nEOQ=sqrt((2*Co*r)/(Cc*(1-r/k)));#in units\nQ=round(EOQ);#units\nt1=Q/k;#in year\nt1=t1*12*30;#in days\nt2=(Q/r)*(1-r/k);#in year\nt2=t2*12*30;#in days\n\n#result\nprint \"Economic order quantity in units : \",round(Q,3);\nprint \"Cycle time in days :\",round(t1+t2);\nprint \"No. of setups/year : \",round(r/Q,3)",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Economic order quantity in units :  748.0\nCycle time in days : 22.0\nNo. of setups/year :  16.043\n"
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 12.3 Page 164"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#iniatiation of variable\nfrom math import sqrt\nD=30000.0;#in units/year\nCc=2.0;#in Rs./unit/year\nCo=100.0;#in Rs./orde\nCs=12.0;#in units/year\n\n#calculation\nEOQ=sqrt(2*Co*D*(Cs+Cc)/(Cc*Cs));#in units\nQ=round(EOQ);#units\nQ1=sqrt(2*Co*D*Cs/(Cc*(Cs+Cc)));#in units\nt=Q/D;#in year\nt=t*365;#in days\nt1=Q1/D;#in year\nt1=t1*365;#in days\nt2=t-t1;#in days\n\n#result\nprint \"Ordering quantity in units : \",round(Q,3);\nprint \"Maximum Inventory in units : \",round(Q1);\nprint \"Maximum shortage quantity in units : \",round(Q-Q1);\nprint \"Cycle time in days: \",round(t);\nprint \"Inventory period(t1) in days : \",round(t1);\nprint \"Shortage period(t2) in days : \",round(t2);\nprint \"No. of Orders/year : \",round(D/Q,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Ordering quantity in units :  1871.0\nMaximum Inventory in units :  1604.0\nMaximum shortage quantity in units :  267.0\nCycle time in days:  23.0\nInventory period(t1) in days :  20.0\nShortage period(t2) in days :  3.0\nNo. of Orders/year :  16.034\n"
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 12.4 Page 166"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nfrom math import sqrt\nr=12000.0;#in units/year\nk=2000.0*12;#in units/year\nCo=400.0;#in Rs./setup\nCc=0.20*12;#in Rs./year\nCs=15.0;#in Rs./unit/year\n\n#calculation\nEOQ=sqrt((2*Co*k*r*(Cc+Cs)/(Cc*(k-r)*Cs)));#in units\nQ=round(EOQ);#units\nQ1=sqrt(2*Co*Cc*r*(k-r)/(Cs*(Cc+Cs)*k));#in units\nQ2=(Q*(k-r)/k)-Q1;#in Units\nt=Q/r;#in year\nt=t*365.0;#in days\nt1=(Q2/r)*365.0;#in days\nt2=(Q2/(k-r))*365.0;#in days\nt3=(Q1/(k-r))*365.0;#in days\nt4=(Q1/r)*365.0;#in days\n\n#result\nprint \"Ordering quantity in units : \",round(Q,3);\nprint \"Maximum Inventory in units : \",round(Q1);\nprint \"Maximum shortage quantity in units : \",round(Q2);\nprint \"Cycle time in days: \",round(t);\nprint \"Period of shortage in days : \",round(t1);\nprint \"Period of production satisfying back order in days : \",round(t2);\nprint \"Period of production satisfying period requirement in days : \",round(t3);\nprint \"Period of consumption only in days : \",round(t4);\n",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Ordering quantity in units :  3046.0\nMaximum Inventory in units :  210.0\nMaximum shortage quantity in units :  1313.0\nCycle time in days:  93.0\nPeriod of shortage in days :  40.0\nPeriod of production satisfying back order in days :  40.0\nPeriod of production satisfying period requirement in days :  6.0\nPeriod of consumption only in days :  6.0\n"
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter13.ipynb b/Engineering_Economics/Chapter13.ipynb
deleted file mode 100755
index af23fcbb..00000000
--- a/Engineering_Economics/Chapter13.ipynb
+++ /dev/null
@@ -1,104 +0,0 @@
-{
- "metadata": {
-  "name": "EE-13"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": "Make or Buy Decision"
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 13.1 Page 168"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nmaterial=300.0;#in Rs.\nlabour=250.0;#in Rs.\noverhead=100.0;#in Rs.\n\n#calculation\nVC=material+labour+overhead;#in Rs.\ndemand=5000.0;#in units\nTVC=demand*VC;#in Rs.\nFC=1000000;#in Rs.\nTC=FC+TVC;#in Rs.\n\n#result\nprint \"Total cost in Rs. : \",round(TC,3);\n\n#Cost to buy :\nPcost=900*demand;#in Rs.\nFC=1000000.0;#in Rs.\nTC=FC+Pcost;#in Rs.\n\n#result\nprint \"Total cost in Rs. : \",round(TC,3);\nprint \"The cost of making fixtures is less than the cost of buying fixtures from outside. Therefore, the organisation should make the fixtures.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Total cost in Rs. :  4250000.0\nTotal cost in Rs. :  5500000.0\nThe cost of making fixtures is less than the cost of buying fixtures from outside. Therefore, the organisation should make the fixtures.\n"
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 13.2 Page 171"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nfrom math import sqrt\nPbuy=8.0;#in Rs/unit\nD=2000.0;#in units/year\nCo=120.0;#in Rs./order\nCc=1.60;#in Rs./units/year\n\n#calculation\nQ1=sqrt(2*Co*D/Cc);#in units\nTC=D*Pbuy+D*Co/Q1+Q1*Cc/2;#in Rs.\nprint \"Total cost of buying in Rs. : \",round(TC,3)\n\n#Make Option :\nPmake=5.0;#in Rs/unit\nCo=60.0;#in Rs./setup\nCc=1.0;#in Re1/units/year\nr=2000.0;#in units/year\nk=8000.0;#in units/year\n\n#calculation\nQ2=sqrt(2*Co*r/(Cc*(1-r/k)));#in units\nTC=D*Pmake+D*Co/Q2+Q2*Cc*(k-r)/(2*k);#in Rs.\n\n#result\nprint \"Total cost of making in Rs. : \",round(TC,3)\nprint\"The cost of making is less than the cost of buying . Therefore, the firm should go in for making option.\";",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Total cost of buying in Rs. :  16876.356\nTotal cost of making in Rs. :  10424.264\nThe cost of making is less than the cost of buying . Therefore, the firm should go in for making option.\n"
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 13.3 Page 173"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nSP=500.0;#in Rs.\nVC=300.0;#in Rs.\nFC=400000.0;#in RS.\n\n#calculation\nBEP=FC/(SP-VC);#in units\n\n#results\nprint \"BEP in units : \",round(BEP,3);\nprint\"Since the demand(1500units) is less than the break even quantity, the company should buy the cabinets for its TV production.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "BEP in units :  2000.0\nSince the demand(1500units) is less than the break even quantity, the company should buy the cabinets for its TV production.\n"
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 13.4 Page 173"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nvolume=8000;#in units\n#Process A :\nFC=500000.0;#in RS.\nVC=175.0;#in Rs.\n\n#calculation\nAC=FC+VC*volume;#in Rs.\n\n#result\nprint \"Annual Cost of Process A in Rs. :\",round(AC,3);\n\n#Process B :\nFC=600000.0;#in RS.\nVC=150;#in Rs.\n\n#calculation\nAC=FC+VC*volume;#in Rs.\n\n#result\nprint \"Annual Cost of Process A in Rs. :\",round(AC,3);\n\n#Buy option:\nPprice=125.0;#in RS./Unit\nVC=175.0;#in Rs.\nAC=Pprice*volume;#in Rs\n\n#result.\nprint \"Annual Cost of Buy in Rs. :\",round(AC,3);\nprint\"Since the annual cost of buy option is the minimum among all the alternatives, the company should buy the product.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Annual Cost of Process A in Rs. : 1900000.0\nAnnual Cost of Process A in Rs. : 1800000.0\nAnnual Cost of Buy in Rs. : 1000000.0\nSince the annual cost of buy option is the minimum among all the alternatives, the company should buy the product.\n"
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter16.ipynb b/Engineering_Economics/Chapter16.ipynb
deleted file mode 100755
index 25b11449..00000000
--- a/Engineering_Economics/Chapter16.ipynb
+++ /dev/null
@@ -1,297 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2e3de312e11632c16b5c14fdb6f5e083c4b44aeeb327466acbfc97c2f088ee01"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Linear Progrmming"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.1 Page 200"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#result\n",
-      "print\"The data of the problem are summarized below : \";\n",
-      "print\"Machine                   Products           Limit on \";\n",
-      "print\"                        P1         P2       machine hours\";\n",
-      "print\"Lathe                    5         10            60\";\n",
-      "print\"Milling                  4          4            40\";\n",
-      "print\"Profit/unit              6          8\";\n",
-      "print\"Let X1 be the production volume of the product.P1, and\";\n",
-      "print\"X2 be the production volume of the product,P2.\";\n",
-      "print\"The corresponding linear programming model to determine the production volume of each product such that the total profit is maximized is as shown below : \";\n",
-      "print\"maximize Z = 6*X1 + 8*X2\";\n",
-      "print\"subject to\";\n",
-      "print\"5*X1+10*X2 <= 60\"\n",
-      "print\"4*X1+4*X2 <= 40\"\n",
-      "print\"X1,X2 >= 0\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The data of the problem are summarized below : \n",
-        "Machine                   Products           Limit on \n",
-        "                        P1         P2       machine hours\n",
-        "Lathe                    5         10            60\n",
-        "Milling                  4          4            40\n",
-        "Profit/unit              6          8\n",
-        "Let X1 be the production volume of the product.P1, and\n",
-        "X2 be the production volume of the product,P2.\n",
-        "The corresponding linear programming model to determine the production volume of each product such that the total profit is maximized is as shown below : \n",
-        "maximize Z = 6*X1 + 8*X2\n",
-        "subject to\n",
-        "5*X1+10*X2 <= 60\n",
-        "4*X1+4*X2 <= 40\n",
-        "X1,X2 >= 0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 Page 200"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "print\"Let X1 be the No. of packets of food type1 suggested for babies, and\";\n",
-      "print\"X2 be the No. of packets of food type1 suggested for babies.\";\n",
-      "print\"The corresponding linear programming model to determine the No. of packets of each food type to be suggested for babies with the minimum cost such that the minimum daily required vitamin in each food type is satisfied is as shown below : \";\n",
-      "print\"maximize Z = 2*X1 + 3*X2\";\n",
-      "print\"subject to\";\n",
-      "print\"X1+X2 >= 6\"\n",
-      "print\"7*X1+X2 >= 14\";\n",
-      "print\"X1,X2 >= 0\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Let X1 be the No. of packets of food type1 suggested for babies, and\n",
-        "X2 be the No. of packets of food type1 suggested for babies.\n",
-        "The corresponding linear programming model to determine the No. of packets of each food type to be suggested for babies with the minimum cost such that the minimum daily required vitamin in each food type is satisfied is as shown below : \n",
-        "maximize Z = 2*X1 + 3*X2\n",
-        "subject to\n",
-        "X1+X2 >= 6\n",
-        "7*X1+X2 >= 14\n",
-        "X1,X2 >= 0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 Page201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "%pylab inline\n",
-      "import matplotlib.pyplot as plt\n",
-      "print\"Given the following LP model :\"\n",
-      "print\"maximize Z = 6*X1 + 8*X2\";\n",
-      "print\"subject to\";\n",
-      "print\"5*X1+10*X2 <= 60\";\n",
-      "print\"4*X1+4*X2 <= 40\";\n",
-      "print\"X1,X2 >= 0\";\n",
-      "print\"The introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\";\n",
-      "print\"Compute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \";\n",
-      "print\"5*X1+10*X2 <= 60\";\n",
-      "print\"When X1=0 : X2=6\";\n",
-      "print\"When X2=0 : X1=12\";\n",
-      "plt.plot([0,12],[6,0],'r')\n",
-      "plt.plot([10,0],[0,10])\n",
-      "plt.title('Graphical Plot')\n",
-      "plt.show()\n",
-      "print \"Consider the 2nd constraint in the form : in blue \";\n",
-      "print \"4*X1+4*X2 <= 40\";\n",
-      "print \"When X1=0 : X2=10\";\n",
-      "print \"When X2=0 : X1=10\";\n",
-      "print \"The closed polygon is the feasible region at each of the corner points of the closed polygon is computed as follows by substituting its coordinates in the objective function :\";\n",
-      "ZA=6*0+8*0;\n",
-      "ZB=6*10+8*0;\n",
-      "ZC=6*8+8*2;\n",
-      "ZD=6*0+8*6;\n",
-      "print \"Since the type of the objective function is maximization, the solution corresponding to the maximum Z value should be selected as the optimum solution. The Z value is maximum for the corner point C. Hence, the corresponding solution is \";\n",
-      "print \"X1 = 8  X2 = 2 and Z(Optimum) is\",ZC;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Given the following LP model :\n",
-        "maximize Z = 6*X1 + 8*X2\n",
-        "subject to\n",
-        "5*X1+10*X2 <= 60\n",
-        "4*X1+4*X2 <= 40\n",
-        "X1,X2 >= 0\n",
-        "The introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\n",
-        "Compute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \n",
-        "5*X1+10*X2 <= 60\n",
-        "When X1=0 : X2=6\n",
-        "When X2=0 : X1=12\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x51094e0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Consider the 2nd constraint in the form : in blue \n",
-        "4*X1+4*X2 <= 40\n",
-        "When X1=0 : X2=10\n",
-        "When X2=0 : X1=10\n",
-        "The closed polygon is the feasible region at each of the corner points of the closed polygon is computed as follows by substituting its coordinates in the objective function :\n",
-        "Since the type of the objective function is maximization, the solution corresponding to the maximum Z value should be selected as the optimum solution. The Z value is maximum for the corner point C. Hence, the corresponding solution is \n",
-        "X1 = 8  X2 = 2 and Z(Optimum) is 64\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 Page 203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "%pylab inline\n",
-      "import matplotlib.pyplot as plt\n",
-      "print\"Given the following LP model :\"\n",
-      "print \"minimize Z = 2*X1 + 3*X2\";\n",
-      "print\"subject to\";\n",
-      "print\"X1+X2 >= 6\";\n",
-      "print\"7*X1+X2 >= 14\";\n",
-      "print\"X1,X2 >= 0\";\n",
-      "print\"The introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\";\n",
-      "print\"Compute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \";\n",
-      "print\"X1+X2 = 6\";\n",
-      "print\"When X1=0 : X2=6\";\n",
-      "print\"When X2=0 : X1=6\";\n",
-      "plt.plot([0,6],[6,0],'r')\n",
-      "plt.plot([2,0],[0,14])\n",
-      "plt.title('Graphical Plot')\n",
-      "plt.show()\n",
-      "print\"Consider the 2nd constraint in the form  (in blue):\";\n",
-      "print\"7*X1+X2 = 14\";\n",
-      "print\"When X1=0 : X2=14\";\n",
-      "print\"When X2=0 : X1=2\";\n",
-      "print\"The Optimum solution will be in any one of the corners A, B and C\";\n",
-      "print\"The objective function value at each of these corner points of the feasible solution space is computed as fllows by substituting its coordinates in the objective function.\"\n",
-      "ZA=2*0+3*14;\n",
-      "ZB=2*(4.0/3)+3*(14.0/3);\n",
-      "ZC=2*6+3*0;\n",
-      "\n",
-      "#result\n",
-      "print\"Since the type of the objective function is minimization, the solution corresponding to the minimum Z value should be selected as the optimum solution. The Z value is minimum for the corner point C. Hence, the corresponding solution is \";\n",
-      "print \"X1 = 6  X2 = 0 and Z(Optimum) =\",ZC"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Given the following LP model :\n",
-        "minimize Z = 2*X1 + 3*X2\n",
-        "subject to\n",
-        "X1+X2 >= 6\n",
-        "7*X1+X2 >= 14\n",
-        "X1,X2 >= 0\n",
-        "The introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\n",
-        "Compute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \n",
-        "X1+X2 = 6\n",
-        "When X1=0 : X2=6\n",
-        "When X2=0 : X1=6\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x81f27b8>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Consider the 2nd constraint in the form  (in blue):\n",
-        "7*X1+X2 = 14\n",
-        "When X1=0 : X2=14\n",
-        "When X2=0 : X1=2\n",
-        "The Optimum solution will be in any one of the corners A, B and C\n",
-        "The objective function value at each of these corner points of the feasible solution space is computed as fllows by substituting its coordinates in the objective function.\n",
-        "Since the type of the objective function is minimization, the solution corresponding to the minimum Z value should be selected as the optimum solution. The Z value is minimum for the corner point C. Hence, the corresponding solution is \n",
-        "X1 = 6  X2 = 0 and Z(Optimum) = 12\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter2.ipynb b/Engineering_Economics/Chapter2.ipynb
deleted file mode 100755
index e29c54c2..00000000
--- a/Engineering_Economics/Chapter2.ipynb
+++ /dev/null
@@ -1,372 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:579288a5e0685252a970b36dc99c069cd7fb91e6e042c88d50d8eb0615ccf38d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Elementary Economic Analysis"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1 Page 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#Part a \n",
-      "w1=1.2;#in Kg\n",
-      "c1=80.0;#cost of making aluminium casting in Rs/Kg\n",
-      "c2=150.0;#ost of machining aluminium casting per unit in Rs\n",
-      "\n",
-      "#calculation\n",
-      "Tc1=c1*w1+c2;#Total cost of jet engine part made of aluminium per unit in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Total cost of jet engine part made of aluminium per unit in Rs \",round(Tc1,3)\n",
-      "\n",
-      "#Part b \n",
-      "w2=1.35;#in Kg\n",
-      "c1=35.0;#in Rs/Kg\n",
-      "c2=170.0;#in Rs\n",
-      "c3=1300.0;#in Rs/Kg\n",
-      "\n",
-      "#calculation\n",
-      "Tc2=c1*w2+c2+c3*(w2-w1);#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Total cost of jet engine part made of steel per unit in Rs : \",round(Tc2,3);\n",
-      "print \"DECISION : The total cost/unit of a jet engine part made of aluminium is less than that for an engine made of steel. Hence, aluminium is suggested for making jet engine part. The economic advantage of aluminium over steel per unit in Rs  \",round(Tc2-Tc1,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total cost of jet engine part made of aluminium per unit in Rs  246.0\n",
-        "Total cost of jet engine part made of steel per unit in Rs :  412.25\n",
-        "DECISION : The total cost/unit of a jet engine part made of aluminium is less than that for an engine made of steel. Hence, aluminium is suggested for making jet engine part. The economic advantage of aluminium over steel per unit in Rs   166.25\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#Part a \n",
-      "wood=0.1;#in m^3\n",
-      "WoodCost=12000.0;#in Rs/m^3\n",
-      "Table=1;#in units\n",
-      "TableTopCost=3000.0;#in Rs/unit\n",
-      "LegBushes=4.0;#units\n",
-      "LegBushesCost=10.0;#Rs/units\n",
-      "Nails=100.0;#in grams\n",
-      "NailsCost=300.0;#in Rs/Kg\n",
-      "TotalLabour=15.0;#in Hours\n",
-      "TotalLabourCost=50.0;#in Rs/Hours\n",
-      "\n",
-      "#calcualtion\n",
-      "WoodCostframelegs=WoodCost*wood;#in Rs\n",
-      "WoodTopCost=3000.0;#in Rs\n",
-      "BushesCost=LegBushesCost*LegBushes;#in Rs\n",
-      "NailsCost=Nails*NailsCost/1000;#in Rs\n",
-      "LabourCost=TotalLabourCost*TotalLabour;#in Rs\n",
-      "TotalCost1=WoodCostframelegs+WoodTopCost+BushesCost+NailsCost+LabourCost;#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Cost of Table with wooden top in Rs \",round(TotalCost1,3)\n",
-      "\n",
-      "#given data for table with granite top\n",
-      "#Part b \n",
-      "wood=0.15;#in m^3\n",
-      "WoodCost=12000.0;#in Rs/m^3\n",
-      "Granite=1.62;#in m^2\n",
-      "GraniteCost=800.0;#in Rs/m^2\n",
-      "LegBushes=4.0;#units\n",
-      "LegBushesCost=25.0;#Rs/units\n",
-      "Nails=50.0;#in grams\n",
-      "NailsCost=300.0;#in Rs/Kg\n",
-      "TotalLabour=8.0;#in Hours\n",
-      "TotalLabourCost=50.0;#in Rs/Hours\n",
-      "\n",
-      "\n",
-      "WoodCostframelegs=WoodCost*wood;#in Rs\n",
-      "GraniteTopCost=Granite*GraniteCost;#in Rs\n",
-      "BushesCost=LegBushesCost*LegBushes;#in Rs\n",
-      "NailsCost=Nails*NailsCost/1000;#in Rs\n",
-      "LabourCost=TotalLabourCost*TotalLabour;#in Rs\n",
-      "TotalCost2=WoodCostframelegs+GraniteTopCost+BushesCost+NailsCost+LabourCost;#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Cost of Table with Granite top in Rs \",round(TotalCost2,3);\n",
-      "print \"Economic advantage of table with granite top in Rs  \",round(TotalCost1-TotalCost2,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cost of Table with wooden top in Rs  5020.0\n",
-        "Cost of Table with Granite top in Rs  3611.0\n",
-        "Economic advantage of table with granite top in Rs   1409.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3 Page 19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#design A\n",
-      "LatheCost=200.0;#in Rs/hour\n",
-      "grinderCost=150.0;#in Rs/hour\n",
-      "HoursOfLathe=16.0;#in hours/1000Unit\n",
-      "HoursOfGrinder=4.5;#in hours/1000Unit\n",
-      "\n",
-      "#calcualtion\n",
-      "TotalCostA=LatheCost*HoursOfLathe+grinderCost*HoursOfGrinder;#in Rs/1000unit\n",
-      "\n",
-      "#result\n",
-      "print \"Total cost of design A per 100,000 units  \",round(TotalCostA*100000.0/1000,3);\n",
-      "\n",
-      "# Design B\n",
-      "HoursOfLathe=7.0;#in hours/1000Unit\n",
-      "HoursOfGrinder=12.0;#in hours/1000Unit\n",
-      "TotalCostB=LatheCost*HoursOfLathe+grinderCost*HoursOfGrinder;#in Rs/1000unit\n",
-      "\n",
-      "#result\n",
-      "print \"Total cost of design A per 100,000 units\",round(TotalCostB*100000.0/1000,3);\n",
-      "print \"Economic advantage of design B over design A per 100,000 units in Rs  \",round(TotalCostA-TotalCostB,3)*100;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total cost of design A per 100,000 units   387500.0\n",
-        "Total cost of design A per 100,000 units 320000.0\n",
-        "Economic advantage of design B over design A per 100,000 units in Rs   67500.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4,Page 20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "from math import pi\n",
-      "Tanks=4.0;#units\n",
-      "TankDia=5.2;#in meter\n",
-      "\n",
-      "#calcualtion\n",
-      "TankRad=TankDia/2;#in meters\n",
-      "TankHeight=7;#in meters\n",
-      "HeightDiaRatio=TankHeight/TankDia;#unitless\n",
-      "VolPerTank=(22/7)*TankRad**2*TankHeight;#in m^3\n",
-      "h=VolPerTank/(pi)*64;#in meters\n",
-      "r=h/8;#in meters\n",
-      "d=2*r;#in meters\n",
-      "CostNewDesign=900000*(100.0/111);#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Expected savings by redesign in Rs  \",round(900000-CostNewDesign,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Expected savings by redesign in Rs   89189.189\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5,Page 21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#steel frame\n",
-      "distance=2500.0;#in Km\n",
-      "TransCost=1.0;#in Rs/Kg/100Km\n",
-      "SteelFramePrice=1000.0;#in Rs/Unit\n",
-      "SteelFrameWeight=75.0;#in Kg/Unit\n",
-      "\n",
-      "#calculation\n",
-      "TotalCost1=SteelFramePrice+TransCost*SteelFrameWeight*distance/100;#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Total cost of steel window frame per unit in Rs  \",round(TotalCost1,3);\n",
-      "\n",
-      "# Aluminium window frame\n",
-      "AlumilniumFramePrice=1500.0;#in Rs/Unit\n",
-      "AlumilniumFrameWeight=28.0;#in Kg/Unit\n",
-      "\n",
-      "#calculation\n",
-      "TotalCost2=AlumilniumFramePrice+TransCost*AlumilniumFrameWeight*distance/100;#in Rs\n",
-      "\n",
-      "#result\n",
-      "print \"Total cost of Alumilnium window frame per unit in Rs \",round(TotalCost2,3);\n",
-      "print \"DECISION : The total cost per unit of the aluminium window frame is less than that of steel window frame. Hence, Alumilnium window frame is recommended. The Economic advantage per unit of the Alumilnium window frame over steel window frame in Rs  \",round(TotalCost1-TotalCost2,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total cost of steel window frame per unit in Rs   2875.0\n",
-        "Total cost of Alumilnium window frame per unit in Rs  2200.0\n",
-        "DECISION : The total cost per unit of the aluminium window frame is less than that of steel window frame. Hence, Alumilnium window frame is recommended. The Economic advantage per unit of the Alumilnium window frame over steel window frame in Rs   675.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6,Page 23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#Cost of component using process sequence 1\n",
-      "print\"The process sequence 1 of the component is as follows : Turning - Milling - Shaping - Drilling\" ;\n",
-      "print \"Calculations are summarized in form of table below : \";\n",
-      "print \"Operation     Operation       Time       Machine Hour rate     Cost\";\n",
-      "print \"   No.                     Min    Hour           Rs.             Rs.\";\n",
-      "print \"   1           Turning       5   0.083          200           16.60\";\n",
-      "print \"   2           Milling       8   0.133          400           53.20\";\n",
-      "print \"   3          Shapiing      10   0.167          350           58.45\";\n",
-      "print \"   4          Drilling       3   0.050          300           15.00\";\n",
-      "print \"                                                   Total     143.25\";\n",
-      "\n",
-      "#Cost of component using process sequence 2\n",
-      "print \"The process sequence 2 of the component is as follows : Turning - Milling - Drilling\" ;\n",
-      "print \"Calculations are summarized in form of table below : \";\n",
-      "print \"Operation     Operation       Time       Machine Hour rate     Cost\";\n",
-      "print \"   No.                     Min    Hour           Rs.             Rs.\";\n",
-      "print \"   1           Turning       5   0.083          200           16.60\";\n",
-      "print \"   2           Milling      14   0.233          400           93.20\";\n",
-      "print \"   4          Drilling       3   0.050          300           15.00\";\n",
-      "print \"                                                   Total     124.80\";\n",
-      "\n",
-      "#Cost of component using process sequence 3\n",
-      "print \"The process sequence 3 of the component is as follows : Only CNC operations\" ;\n",
-      "print \"Calculations are summarized in form of table below : \";\n",
-      "print \"Operation     Operation       Time       Machine Hour rate     Cost\";\n",
-      "print \"   No.                     Min    Hour           Rs.             Rs.\";\n",
-      "print \"   1             CNC         8   0.133         1000             133\";\n",
-      "\n",
-      "print \"The process sequence 2 has the least cost. Therefore, it should be selected for manufacturing the component.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The process sequence 1 of the component is as follows : Turning - Milling - Shaping - Drilling\n",
-        "Calculations are summarized in form of table below : \n",
-        "Operation     Operation       Time       Machine Hour rate     Cost\n",
-        "   No.                     Min    Hour           Rs.             Rs.\n",
-        "   1           Turning       5   0.083          200           16.60\n",
-        "   2           Milling       8   0.133          400           53.20\n",
-        "   3          Shapiing      10   0.167          350           58.45\n",
-        "   4          Drilling       3   0.050          300           15.00\n",
-        "                                                   Total     143.25\n",
-        "The process sequence 2 of the component is as follows : Turning - Milling - Drilling\n",
-        "Calculations are summarized in form of table below : \n",
-        "Operation     Operation       Time       Machine Hour rate     Cost\n",
-        "   No.                     Min    Hour           Rs.             Rs.\n",
-        "   1           Turning       5   0.083          200           16.60\n",
-        "   2           Milling      14   0.233          400           93.20\n",
-        "   4          Drilling       3   0.050          300           15.00\n",
-        "                                                   Total     124.80\n",
-        "The process sequence 3 of the component is as follows : Only CNC operations\n",
-        "Calculations are summarized in form of table below : \n",
-        "Operation     Operation       Time       Machine Hour rate     Cost\n",
-        "   No.                     Min    Hour           Rs.             Rs.\n",
-        "   1             CNC         8   0.133         1000             133\n",
-        "The process sequence 2 has the least cost. Therefore, it should be selected for manufacturing the component.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter3.ipynb b/Engineering_Economics/Chapter3.ipynb
deleted file mode 100755
index b94c5fa1..00000000
--- a/Engineering_Economics/Chapter3.ipynb
+++ /dev/null
@@ -1,360 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:5b530b225e354d5204b6b04a92931fcee5b92fa42094bc3f03e4022169e39d78"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Interest Formulas and their Application"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1 Page 29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "P=20000.0;#in rupees\n",
-      "n=10.0;#in years\n",
-      "i=18.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "F=P*(1+i/100)**n;\n",
-      "\n",
-      "#result\n",
-      "print \"Maturity value after 18 years is  \", round(F,3),\" Rupees.\";\n",
-      "print \" Ans in the book is not correct\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maturity value after 18 years is :  104676.711  Rupees.\n",
-        " Ans in the book is not correct\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2 Page 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "F=100000.0;#in rupees\n",
-      "n=10.0;#in years\n",
-      "i=15.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "P=F/((1.0+i/100.0)**n);\n",
-      "\n",
-      "#result\n",
-      "print \"The person has to invest  \", round(P,3),\" Rupees.\";\n",
-      "print \" Ans in the book is not correct\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The person has to invest   24718.471  Rupees.\n",
-        " Ans in the book is not correct\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3 Page 31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "A=10000.0;#in rupees\n",
-      "n=25.0;#in years\n",
-      "i=20.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "F=A*(((1+i/100)**n-1)/(i/100));\n",
-      "\n",
-      "#result\n",
-      "print \"The future sum of the annual equal payment after 25 years is \",round(F,3),\" Rupees.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The future sum of the annual equal payment after 25 years is  4719810.832  Rupees.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4 Page 32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "F=500000.0;#in rupees\n",
-      "n=15.0;#in years\n",
-      "i=18.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "A=F*((i/100)/((1+i/100)**n-1));\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equal ammount which must be deposited for 15 years is  \",round(A,3),\" Rupees.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual equal ammount which must be deposited for 15 years is   8201.391  Rupees.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5 Page 33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiasation of variable\n",
-      "A=1000000.0;#in rupees\n",
-      "n=20.0;#in years\n",
-      "i=15.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "P=A*(((1+i/100)**n-1)/((i/100)*(1+i/100)**n));\n",
-      "\n",
-      "#result\n",
-      "print \"The amount of reserve which must be setup now is : \",round(P,3),\" Rupees\";\n",
-      "print \" Ans in the book is not correct\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of reserve which must be setup now is :  6259331.474  Rupees\n",
-        " Ans in the book is not correct\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6 Page 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "P=1000000.0;#in rupees\n",
-      "n=15.0;#in years\n",
-      "i=18.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "A=P*(((i/100)*(1+i/100)**n)/((1+i/100)**n-1));\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent installment to be paid by the company to the bank is \",round(A),\" Rupees.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual equivalent installment to be paid by the company to the bank is :  196403.0  Rupees.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7 Page 35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "A1=4000.0;#in rupees\n",
-      "G=500.0;#in rupees\n",
-      "n=10.0;#in years\n",
-      "i=15.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "A=A1+G*(((1+i/100)**n-(i/100)*n-1)/((i/100)*(1+i/100)**n-(i/100)));\n",
-      "F=A*(((1+i/100)**n-1)/(i/100));\n",
-      "\n",
-      "#result\n",
-      "print \"At the end of 10th year, the compound amountr of all his payments will be : \",round(F,3),\" Rupees.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "At the end of 10th year, the compound amountr of all his payments will be :  115560.6  Rupees.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8 Page 36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "A1=8500.0;#in rupees\n",
-      "G=-500.0;#in rupees\n",
-      "n=10.0;#in years\n",
-      "i=15.0;#% per annum\n",
-      "\n",
-      "#calculation\n",
-      "A=A1+G*(((1+i/100)**n-(i/100)*n-1)/((i/100)*(1+i/100)**n-(i/100)));\n",
-      "F=A*(((1+i/100)**n-1)/(i/100));\n",
-      "\n",
-      "#result\n",
-      "print \"At the end of 10th year, the compound amountr of all his payments will be : \",round(F,3),\" Rupees.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "At the end of 10th year, the compound amountr of all his payments will be :  138235.878  Rupees.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9 Page 38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "P=5000.0;#in rupees\n",
-      "n=10.0;#in years\n",
-      "i=12.0;#% per annum\n",
-      "m=4.0;#no. of interest periods per year for quarterly\n",
-      "\n",
-      "#calculation\n",
-      "N=n*m;\n",
-      "r=i/m;\n",
-      "F=P*(1+r/100)**N;\n",
-      "\n",
-      "#result\n",
-      "print\"Maturity value after 10 years is  \",round(F,3),\" Rupees.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maturity value after 10 years is :  16310.189  Rupees.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter4.ipynb b/Engineering_Economics/Chapter4.ipynb
deleted file mode 100755
index 2b2d10b2..00000000
--- a/Engineering_Economics/Chapter4.ipynb
+++ /dev/null
@@ -1,388 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:aed8a204f6a6b9aa286380e07fd82210fb76cb7818bcaa10c4f627aee9d0c52a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Present Worth Method of Comparision"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1 Page 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#technology 1\n",
-      "P=1200000.0;#in Rs\n",
-      "A=400000.0;#in Rs\n",
-      "i=20.0;#in % per annum\n",
-      "n=10.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW=-P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth for this technology in RS.  \",round(PW,3);\n",
-      "\n",
-      "#technology 2\n",
-      "P=2000000.0;#in Rs\n",
-      "A=600000.0;#in Rs\n",
-      "i=20.0;#in % per annum\n",
-      "n=10.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW=-P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth for this technology in RS.  \",round(PW,3);\n",
-      "\n",
-      "#technology 3\n",
-      "P=1800000.0;#in Rs\n",
-      "A=500000.0;#in Rs\n",
-      "i=20.0;#in % per annum\n",
-      "n=10.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW=-P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth for this technology in RS.  \",round(PW,3);\n",
-      "print \"It is clear from the calculations that the present worth of technology 2 is the highest among all technologies. Therefore technology 2 is suggested for implementation to expand the production.\";\n",
-      "print \"the answer is differnt due to approximation error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The present worth for this technology in RS.   476988.834\n",
-        "The present worth for this technology in RS.   515483.251\n",
-        "The present worth for this technology in RS.   296236.043\n",
-        "It is clear from the calculations that the present worth of technology 2 is the highest among all technologies. Therefore technology 2 is suggested for implementation to expand the production.\n",
-        "the answer is differnt due to approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2 Page 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#bid 1\n",
-      "P=450000.0;#in Rs\n",
-      "A=27000.0;#in Rs\n",
-      "i=15.0;#in % per annum\n",
-      "n=15.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW=P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth for this bid in RS. \",round(PW,3)\n",
-      "\n",
-      "#bid 2\n",
-      "P=540000.0;#in Rs\n",
-      "A=28500.0;#in Rs\n",
-      "i=15.0;#in % per annum\n",
-      "n=15.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW=P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth for this bid in RS. \",round(PW,3)\n",
-      "print \"The total present worth cost of bid 1 is less than that of bid 2. Hence bid 1 is to be selected for implementation. That is, the Elevator from Alpha Elevator Inc. is to be purchased and installed in the new building.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The present worth for this bid in RS.  607878.993\n",
-        "The present worth for this bid in RS.  706650.048\n",
-        "The total present worth cost of bid 1 is less than that of bid 2. Hence bid 1 is to be selected for implementation. That is, the Elevator from Alpha Elevator Inc. is to be purchased and installed in the new building.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3 Page 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#investment proposal A\n",
-      "P1=-10000.0;#in Rs\n",
-      "P2=3000.0;#in Rs\n",
-      "P3=3000.0;#in Rs\n",
-      "P4=7000.0;#in Rs\n",
-      "P5=6000.0;#in Rs\n",
-      "i=18.0;#in % per annum\n",
-      "\n",
-      "#calculation\n",
-      "PW_A=P1+P2*1/((1+i/100)**1)+P3*1/((1+i/100)**2)+P4*1/((1+i/100)**3)+P5*1/((1+i/100)**4);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of A in RS.  \",round(PW_A,3);\n",
-      "\n",
-      "#investment proposal B\n",
-      "P1=-10000.0;#in Rs\n",
-      "P2=6000.0;#in Rs\n",
-      "P3=6000.0;#in Rs\n",
-      "P4=3000.0;#in Rs\n",
-      "P5=3000.0;#in Rs\n",
-      "i=18.0;#in % per annum\n",
-      "\n",
-      "#calculation\n",
-      "PW_B=P1+P2*1/((1+i/100)**1)+P3*1/((1+i/100)**2)+P4*1/((1+i/100)**3)+P5*1/((1+i/100)**4);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of B in RS.  \",round(PW_B,3);\n",
-      "print\"At i=18%, the present worth of proposal B is higher than that of proposal A. Therefore, select proposal B.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The present worth of A in RS.   2052.076\n",
-        "The present worth of B in RS.   2767.112\n",
-        "At i=18%, the present worth of proposal B is higher than that of proposal A. Therefore, select proposal B.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4 Page 48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "P0=400000.0;#in Rs\n",
-      "P=200000.0;#in Rs\n",
-      "i=18.0;#in % per annum\n",
-      "n=10.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "PW=P0+P*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of alternative 2 in RS. \",round(PW,3);\n",
-      "print \"The present worth of 2nd alternative is less than that of first one i.e., complete downpayment of Rs. 1600000. Hence, select 2nd alternative.\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The present worth of alternative 2 in RS.  1298817.259\n",
-        "The present worth of 2nd alternative is less than that of first one i.e., complete downpayment of Rs. 1600000. Hence, select 2nd alternative.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5 Page 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#plan 1\n",
-      "P0=-1000.0;#in Rs\n",
-      "P=12000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=15.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW1=P0+P*1/((1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of Plan-1 in RS.  \",round(PW1,3);\n",
-      "\n",
-      "#plan 2\n",
-      "P0=-1000.0;#in Rs\n",
-      "P=4000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n1=10.0;#in years\n",
-      "n2=15.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW2=P0+P*1/((1+i/100)**n1)++P*1/((1+i/100)**n2);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of Plan-2 in RS.\",round(PW2,3);\n",
-      "print \"The present worth of Plan-1 is more than that of Plan-1. Therefore plan 1 is the best plan from the investors point of view.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The present worth of Plan-1 in RS.   1192.355\n",
-        "The present worth of Plan-2 in RS. 1018.678\n",
-        "The present worth of Plan-1 is more than that of Plan-1. Therefore plan 1 is the best plan from the investors point of view.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6,Page 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#plan 1\n",
-      "P0=-10000.0;#in Rs\n",
-      "P=800000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=20.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "PW1=P0*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n)+P*1/((1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of Plan-1 in RS. \",round(PW1,3);\n",
-      "\n",
-      "#plan 2\n",
-      "P0=-10000.0;#in Rs\n",
-      "P=1500000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n1=20.0;#in years\n",
-      "n2=25.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "PW2=P0*(((1+i/100)**n1)-1)/((i/100)*(1+i/100)**n1)+P*1/((1+i/100)**n2);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of Plan-2 in RS. \",round(PW2,3);\n",
-      "print \"The present worth of Innovative Investment Ltd. Plan is more than that of Novel Investment Ltd. Plan. Therefore Innovative Investment Ltd. Plan is the best plan from the investors point of view.\";\n",
-      "print \"answer is bit different due to approximation error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The present worth of Plan-1 in RS.  8238.976\n",
-        "The present worth of Plan-2 in RS.  13540.524\n",
-        "The present worth of Innovative Investment Ltd. Plan is more than that of Novel Investment Ltd. Plan. Therefore Innovative Investment Ltd. Plan is the best plan from the investors point of view.\n",
-        "answer is bit different due to approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7 Page 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "P=12000.0;#in Rs\n",
-      "A1=10000.0;#in Rs\n",
-      "G=1000.0;#in Rs\n",
-      "i=18.0;#in % per annum\n",
-      "n=10.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "PW=-P+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The present worth of the small business in RS.  \",round(PW,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The present worth of the small business in RS.   47293.316\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter5.ipynb b/Engineering_Economics/Chapter5.ipynb
deleted file mode 100755
index 398e418d..00000000
--- a/Engineering_Economics/Chapter5.ipynb
+++ /dev/null
@@ -1,352 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e388f34dd67bfe5948d20d52a601f7e8e53b92aa882e9a754ed91176569417c0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Future Worth Method"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1,Page 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#alternate A\n",
-      "P=5000000.0;#in Rs\n",
-      "A=2000000.0;#in Rs\n",
-      "i=18.0;#in % per annum\n",
-      "n=4.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_A=(-P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100));#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of alternative A in RS.  \",round(FW_A,3);\n",
-      "\n",
-      "#alternate B\n",
-      "P=4500000.0;#in Rs\n",
-      "A=1800000.0;#in Rs\n",
-      "i=18.0;#in % per annum\n",
-      "n=4.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_B=(-P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100));#in RS\n",
-      "print \"The future worth amount of alternative B in RS.  \", round(FW_B,3);\n",
-      "print \"The future worth of alternative A is greater than that of alternative B. Thus, alternative A should be selected.\";\n",
-      "print \" Calculation in the book is not accurate.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The future worth amount of alternative A in RS.   736975.2\n",
-        "The future worth amount of alternative B in RS.   663277.68\n",
-        "The future worth of alternative A is greater than that of alternative B. Thus, alternative A should be selected.\n",
-        " Calculation in the book is not accurate.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2,Page 58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#alternate 1\n",
-      "FC=2000000.0;#in Rs\n",
-      "AI=800000.0;#in Rs\n",
-      "ATax=80000.0;#in Rs\n",
-      "NetAI=AI-ATax;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=20.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_1=(-FC*(1+i/100)**n)+(NetAI*(((1+i/100)**n)-1)/(i/100));#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of alternative 1 in RS. : \", round(FW_1,3)\n",
-      "\n",
-      "#alternative2\n",
-      "FC=3600000.0;#in Rs\n",
-      "AI=980000.0;#in Rs\n",
-      "ATax=150000.0;#in Rs\n",
-      "NetAI=AI-ATax;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=20.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_2=(-FC*(1+i/100)**n)+(NetAI*(((1+i/100)**n)-1)/(i/100));#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of alternative 2 in RS.\",round(FW_2,3)\n",
-      "print\"The future worth of alternative 1 is greater than that of alternative 2. Thus, building the gas station is the best alternative.\";\n",
-      "print \" Calculation in the book is not accurate.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The future worth amount of alternative 1 in RS. :  32585172.373\n",
-        "The future worth amount of alternative 2 in RS. 25076872.093\n",
-        "The future worth of alternative 1 is greater than that of alternative 2. Thus, building the gas station is the best alternative.\n",
-        " Calculation in the book is not accurate.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3,Page 59"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#part a\n",
-      "#alternative 1\n",
-      "P=500000.0;#in Rs\n",
-      "A1=50000.0;#in Rs\n",
-      "G=50000.0;#in Rs\n",
-      "i=8.0;#in % per annum\n",
-      "n=6.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of alternative 1 in RS.  \", round(FW_1,3);\n",
-      "\n",
-      "#alternative 2\n",
-      "P=700000.0;#in Rs\n",
-      "A1=70000.0;#in Rs\n",
-      "G=70000.0;#in Rs\n",
-      "i=8.0;#in % per annum\n",
-      "n=6.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_2=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of alternative 2 in RS.  \", round(FW_2,3);\n",
-      "print \"The future worth of alternative 2 is greater than that of alternative 1. Thus, alternative 2 must be selected.\";\n",
-      "\n",
-      "\n",
-      "#part b\n",
-      "#alternative a\n",
-      "P=500000.0;#in Rs\n",
-      "A1=50000.0;#in Rs\n",
-      "G=50000.0;#in Rs\n",
-      "i=9.0;#in % per annum\n",
-      "n=6.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of alternative 1 in RS.  \", round(FW_1,3);\n",
-      "\n",
-      "#altenative 2\n",
-      "P=700000.0;#in Rs\n",
-      "A1=70000.0;#in Rs\n",
-      "G=70000.0;#in Rs\n",
-      "i=9.0;#in % per annum\n",
-      "n=6.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of alternative 2 in RS.\", round(FW_1,3);\n",
-      "print \"The negative sign of alternatives future worth indicates that alternative 2 incurs loss. Thus, none of the two alternatives should be selected. \";\n",
-      "print \" Calculation in the book is not accurate.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The future worth amount of alternative 1 in RS.   408314.938\n",
-        "The future worth amount of alternative 2 in RS.   571640.914\n",
-        "The future worth of alternative 2 is greater than that of alternative 1. Thus, alternative 2 must be selected.\n",
-        "The future worth amount of alternative 1 in RS.   383913.653\n",
-        "The future worth amount of alternative 2 in RS. 537479.115\n",
-        "The negative sign of alternatives future worth indicates that alternative 2 incurs loss. Thus, none of the two alternatives should be selected. \n",
-        " Calculation in the book is not accurate.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4 Page 62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#alternative 1\n",
-      "P=8000000.0;#in Rs\n",
-      "A=800000.0;#in Rs\n",
-      "i=20.0;#in % per annum\n",
-      "n=12.0;#in years\n",
-      "Salvage=500000;#in Rs\n",
-      "\n",
-      "#calcualtion\n",
-      "FW1=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth for this alternative in RS.\",round(FW1,3)\n",
-      "\n",
-      "#alternative 2\n",
-      "P=7000000.0;#in Rs\n",
-      "A=900000.0;#in Rs\n",
-      "i=20.0;#in % per annum\n",
-      "n=12.0;#in years\n",
-      "Salvage=400000;#in Rs\n",
-      "\n",
-      "#calculation\n",
-      "FW2=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth for this alternative in RS.\",round(FW2,3);\n",
-      "\n",
-      "#alternative 3\n",
-      "P=9000000.0;#in Rs\n",
-      "A=850000.0;#in Rs\n",
-      "i=20.0;#in % per annum\n",
-      "n=12.0;#in years\n",
-      "Salvage=700000;#in Rs\n",
-      "\n",
-      "#calculation\n",
-      "FW3=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n",
-      "\n",
-      "print \"The future worth for this alternative in RS.  \",round(FW3,3)\n",
-      "print \"The future worth of alternative 2 is less than that of other two alternatives. Thus, Ms. Krishna castings should buy the annealing furnace from manufacturer 2.\";\n",
-      "print \" Calculation in the book is not accurate.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The future worth for this alternative in RS. 102493205.379\n",
-        "The future worth for this alternative in RS. 97635155.155\n",
-        "The future worth for this alternative in RS.   113188330.939\n",
-        "The future worth of alternative 2 is less than that of other two alternatives. Thus, Ms. Krishna castings should buy the annealing furnace from manufacturer 2.\n",
-        " Calculation in the book is not accurate.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5 Page 64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialisation of variable\n",
-      "#Machine A\n",
-      "P=400000.0;#in Rs\n",
-      "A=40000.0;#in Rs\n",
-      "S=200000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=4.0#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_A=(P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100))-S;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of Machine A in RS. \",round(FW_A,3);\n",
-      "\n",
-      "#Machine B\n",
-      "P=800000.0;#in Rs\n",
-      "A=0.0;#in Rs\n",
-      "S=550000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=4.0#in years\n",
-      "\n",
-      "#calculation\n",
-      "FW_B=(P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100))-S;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The future worth amount of Machine B in RS. \",round(FW_B,3);\n",
-      "print \"The future worth of Machine A is less than that of Machine B. Thus, Machine A should be selected.\";\n",
-      "print \" Calculation in the book is not accurate\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The future worth amount of Machine A in RS.  620580.864\n",
-        "The future worth amount of Machine B in RS.  708815.488\n",
-        "The future worth of Machine A is less than that of Machine B. Thus, Machine A should be selected.\n",
-        " Calculation in the book is not accurate\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter6.ipynb b/Engineering_Economics/Chapter6.ipynb
deleted file mode 100755
index c0d35ab9..00000000
--- a/Engineering_Economics/Chapter6.ipynb
+++ /dev/null
@@ -1,230 +0,0 @@
-{
- "metadata": {
-  "name": "EE-6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": "Annual Equivalent Method"
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.1 Page 70"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nrun=20000.0; #km\nkmperlitre=9; #kmpl\ncost=21;\nG=2222.20;#in Rs.\ni=18.0;#in % per annum\nn=4.0;#in years\n\n#calculation\ncon=run/kmperlitre;#consumption\nA1=con*cost;\nA=A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100) ;\n\n#result\nprint \"Annual equal amount spending for petrol in Rs. : \",round(A,3);\nprint \"This amoount is less than the annual rental value of Rs. 60000. Therefore, the company should continue to provide its own car to its executive. \";",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Annual equal amount spending for petrol in Rs. :  49543.741\nThis amoount is less than the annual rental value of Rs. 60000. Therefore, the company should continue to provide its own car to its executive. \n"
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.2 Page 71"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#alternative 1\nP=500000.0;#in Rs\nA=200000.0;#in Rs\ni=20.0;#in % per annum\nn=15.0;#in years\n\n#calcultion\nAE_1=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A;#in RS\n\n#result\nprint \"The Annual equivalent cost of alternative 1 in RS. : \",round(AE_1,3);\n\n#alternative 2\nP=400000.0;#in Rs\nA=300000.0;#in Rs\ni=20.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nAE_2=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A;#in RS\n\n#result\nprint \"The Annual equivalent cost of alternative 2 in RS. : \",round(AE_2,3);\n\n#alternative 3\nP=600000.0;#in Rs\nA=150000.0;#in Rs\ni=20.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nAE_3=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A;#in RS\n\n#result\nprint \"The Annual equivalent cost of alternative 3 in RS. : \",round(AE_3,3);\nprint\"The annual equivalent cost of manufacturer 3 is less than that of other two. Therefore, company should buy advanced machine centre from manufacturer 3. \";\nprint \" Calculation in the book is not accurate.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The Annual equivalent cost of alternative 1 in RS. :  306941.06\nThe Annual equivalent cost of alternative 2 in RS. :  385552.848\nThe Annual equivalent cost of alternative 3 in RS. :  278329.272\nThe annual equivalent cost of manufacturer 3 is less than that of other two. Therefore, company should buy advanced machine centre from manufacturer 3. \n Calculation in the book is not accurate.\n"
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.3 Page 73"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation  of variable\n#alternative a\nP=150000.0;#in Rs\nA=60000.0;#in Rs\nS=15000.0;#in Rs\ni=25.0;#in % per annum\nn=5.0;#in years\n\n#calculation\nAE_A=-P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A+S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent revenue of alternative A in RS. : \",round(AE_A,3)\n\n#alternative b\nP=175000.0;#in Rs\nA=70000.0;#in Rs\nS=35000.0;#in Rs\ni=25.0;#in % per annum\nn=5.0;#in years\n\n#calculation\nAE_B=-P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A+S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent revenue of alternative B in RS. : \",round(AE_B,3)\nprint \"The annual equivalent net return of alternative B is more than that of alternative A. Thus the company should select alternative A\";\nprint \" Calculation in the book is not accurate.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The Annual equivalent revenue of alternative A in RS. :  6050.69\nThe Annual equivalent revenue of alternative B in RS. :  9191.456\nThe annual equivalent net return of alternative B is more than that of alternative A. Thus the company should select alternative A\n Calculation in the book is not accurate.\n"
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.4 Page 75"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#machine X\nP=150000.0;#in Rs\nA=0.0;#in Rs\nS=0.0;#in Rs\ni=15.0;#in % per annum\nn=12.0;#in years\n\n#calcualtion\nAE_X=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine X in RS. : \",round(AE_X,3);\n\n#machine Y\nP=240000.0;#in Rs\nA=4500.0;#in Rs\nS=60000.0;#in Rs\ni=15.0;#in % per annum\nn=12.0;#in years\n\n#calcualtion\nAE_Y=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine Y in RS. : \",round(AE_Y,3);\nprint \"The annual equivalent cost of machine X is less than that of machine Y.So machine X is more cost effective machine. \";\nprint\" Calculation in the book is not accurate.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The Annual equivalent cost of machine X in RS. :  27672.116\nThe Annual equivalent cost of machine Y in RS. :  44275.386\nThe annual equivalent cost of machine X is less than that of machine Y.So machine X is more cost effective machine. \n Calculation in the book is not accurate.\n"
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.5 Page 76"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initialisation of variable\n#alternative 1\nFC=15*150000.0;#in Rs\nMC=15*6000.0;#in Rs\nPL=15*15000.0;#in Rs\n\n#calculation\nMCPL=MC+PL;#in Rs\nS=15*90000.0;#in Rs\ni=15.0;#in % per annum\nn=15.0;#in years\nAE1=FC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+MCPL-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS. \",round(AE1,3);\n\n#alternative 2\nFC=5*750000.0;#in Rs\nMC=5*12000.0;#in Rs\nPL=5*15000.0;#in Rs\nMCPL=MC+PL;#in Rs\nS=5*150000.0;#in Rs\ni=15.0;#in % per annum\nn=15.0;#in years\n\n#calcualtion\nAE2=FC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+MCPL-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS. \",round(AE2,3);\nprint \"The annual equivalent cost of alternative 1 is less than that of alternative 2. Therefore, select the route around the lake for laying the power line.\";\nprint \" Calculation in the book is not accurate.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The Annual equivalent cost for this alternative in RS.  671415.347\nThe Annual equivalent cost for this alternative in RS.  760551.158\nThe annual equivalent cost of alternative 1 is less than that of alternative 2. Therefore, select the route around the lake for laying the power line.\n Calculation in the book is not accurate.\n"
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.6 Page 78"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#alternative 1\nVC=390000.0;#in Rs\nLpY=60000.0/30;#in litres\nFcY=2000.0*8;#in Rs\nAllExp=FcY+9000.0+15000.0;#in Rs\nS=60000.0;#in Rs\ni=20.0;#in % per annum\nn=4.0;#in years\n\n#calcualtion\nAE1=VC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AllExp-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS.  \",round(AE1,3);\n\n#alternative 2\nVC=360000.0;#in Rs\nLpY=60000.0/20;#in litres\nFcY=3000.0*20;#in Rs\nAllExp=FcY+6000.0+15000.0;#in Rs\nS=90000.0;#in Rs\ni=20.0;#in % per annum\nn=3.0;#in years\n\n#calculation\nAE2=VC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AllExp-S*(i/100)/(((1+i/100)**n)-1);# in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS.  \",round(AE2,3);\nprint \"The annual equivalent cost of purchase and operation of cars with diesel engine is less than that of cars with petrol engine. Therefore, the taxy company should buy cars with diesel engine.\";\nprint \" Calculation in the book is not accurate.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The Annual equivalent cost for this alternative in RS.   179475.41\nThe Annual equivalent cost for this alternative in RS.   227175.824\nThe annual equivalent cost of purchase and operation of cars with diesel engine is less than that of cars with petrol engine. Therefore, the taxy company should buy cars with diesel engine.\n Calculation in the book is not accurate.\n"
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.7 Page 79"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#alternative 1\nPprice=390000.0;#in Rs\nn=3.0;#in years\nn=n*12;#in months\nS=120000.0;#in Rs\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nME1=Pprice*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Monthly equivalent cost for this alternative in RS.  \",round(ME1,3);\n\n#Alternative 2 \nLeaseAmount=10500.0;#in Rs\n\n#calcualtion\nMonthlyEquiCost=LeaseAmount;#in Rs.\n\n#result\nprint\"The Monthly equivalent cost for this alternative in RS. \",round(MonthlyEquiCost,3);\nprint\"The monthly equivalent cost of alternative 1 is less than that of alternative 2. hence the salesman should purchase the car for cash.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The Monthly equivalent cost for this alternative in RS.   10167.864\nThe Monthly equivalent cost for this alternative in RS.  10500.0\nThe monthly equivalent cost of alternative 1 is less than that of alternative 2. hence the salesman should purchase the car for cash.\n"
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.8 Page 81"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#machine A\nIC=300000.0;#in Rs\nn=4.0;#in years\nS=200000.0;#in Rs\nAM=30000.0;#in Rs\ni=15.0;#in % per annum\n\n#calcualtion\nAE_A=IC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AM-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine X in RS. \",round(AE_A,3);\n\n#Machine B :\nIC=600000.0;#in Rs\nn=4.0;#in years\nS=300000.0;#in Rs\nAM=0.0;#in Rs\ni=15.0;#in % per annum\n\n#calcualtion\nAE_B=IC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AM-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine X in RS. \",round(AE_B,3);\nprint \"The annual equivalent cost of machine A is less than that of machine B.So machine A is more cost effective machine. It is advised to buy machine A \";\nprint \" Calculation in the book is not accurate.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The Annual equivalent cost of machine X in RS.  95026.535\nThe Annual equivalent cost of machine X in RS.  150079.605\nThe annual equivalent cost of machine A is less than that of machine B.So machine A is more cost effective machine. It is advised to buy machine A \n Calculation in the book is not accurate.\n"
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.9 Page 83"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#alternative 1\nDP=60000.0;#in Rs\nP1=15000.0;#in Rs\nn=1.0;#in years\ni=12.0;#in % Compounded anually\n\n#calculation\nPW1=DP+P1*1/((1+i/100)**n);#in RS\nn=4.0;#in years\nAE1=PW1*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)\n\n#result\nprint \"The annual equivalent for this alternative in RS. \",round(PW1,3);\nprint \"The annual equivalent for this alternative in RS.\",round(AE1,3);\n\n#alternative 2\nP4y=90000.0;#in Rs\nAE2=P4y*(i/100)/(((1+i/100)**n)-1);#in Rs.\n\n#result\nprint \"The annual equivalent for this alternative in RS. \",round(AE2,3);\nprint \"The annual equivalent cost of alternative 2 is less than that of alternative 1. Hence, Joshi Lakshimi should select alternative 2 for purchasing the home equipment.\";\nprint \" Calculation in the book is not accurate.\"",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The annual equivalent for this alternative in RS.  73392.857\nThe annual equivalent for this alternative in RS. 24163.456\nThe annual equivalent for this alternative in RS.  18831.099\nThe annual equivalent cost of alternative 2 is less than that of alternative 1. Hence, Joshi Lakshimi should select alternative 2 for purchasing the home equipment.\n Calculation in the book is not accurate.\n"
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 6.10 Page 85"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initation of variable\n#Brand C : \nw=12.0;#in months\nP=1200.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_C=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_C,3);\n\n#Brand C : \nw=24.0;#in months\nP=1800.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_C=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_C,3);\n\n#Brand C : \nw=36.0;#in months\nP=2100.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_C=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_C,3);\n\n#Brand D : \nw=48.0;#in months\nP=2700.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_D=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_D,3);\nprint \"Here common multiple lives of tyres is considered. This is 144 months. Therefore, the comparison is nade on 144 months basis.\";\nprint \"The annual equivalent cost of Brand C is less than that of other brands. hence, it should be used in the vehicles of the trucking company. It should be replaced  for times during the 144 months period.\";",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The annual equivalent for this brand in RS.  106.619\nThe annual equivalent for this brand in RS.  84.732\nThe annual equivalent for this brand in RS.  69.75\nThe annual equivalent for this brand in RS.  71.101\nHere common multiple lives of tyres is considered. This is 144 months. Therefore, the comparison is nade on 144 months basis.\nThe annual equivalent cost of Brand C is less than that of other brands. hence, it should be used in the vehicles of the trucking company. It should be replaced  for times during the 144 months period.\n"
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter7.ipynb b/Engineering_Economics/Chapter7.ipynb
deleted file mode 100755
index 827a806e..00000000
--- a/Engineering_Economics/Chapter7.ipynb
+++ /dev/null
@@ -1,125 +0,0 @@
-{
- "metadata": {
-  "name": "EE-7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": "Rate of Return Method"
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 7.1 Page 89"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of vriable\nIi=100000.0;#in Rs\nAr=30000.0;#in Rs\nn=5.0;#in years\ni1=10.0;#in % per annum\n\n#calcualton\nPW1=-Ii+Ar*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\n\n#result\nprint \"The present worth for i=10% in RS. \",round(PW1,3);\n\n#calcualtion\ni2=15.0;#in % per annum\nPW2=-Ii+Ar*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\n\n#result\nprint \"The present worth for i=15% in RS. \",round(PW2,3);\n\n#calculation\ni3=18.0;#in % per annum\nPW3=-Ii+Ar*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni=15+(PW2-0)*(i3-i2)/(PW2-PW3);#in Rs\n\n#result\nprint \"The present worth for i=18% in RS. \",round(PW3,3)\nprint \"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return for the new business in % per annum \",round(i,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The present worth for i=10% in RS.  13723.603\nThe present worth for i=15% in RS.  564.653\nThe present worth for i=18% in RS.  -6184.869\nPresent worth for i=15% is suitable.\nTherefore, the rate of return for the new business in % per annum  15.251\n"
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 7.2 Page 90"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nIo=2000000.0;#in Rs\nANP=350000.0;#in Rs\nS=0.0;#in Rs\nn=10.0;#in years\ni1=10.0;#in % per annum\n\n#calcualtion\nPW1=-Io+ANP*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\n\n#result\nprint \"The present worth for i=10% in RS.  \",round(PW1,3);\n\n#calculation\ni2=12.0;#in % per annum\nPW2=-Io+ANP*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni=10+(PW1-0)*(i2-i1)/(PW1-PW2);#in Rs.\n\n#result\nprint \"The present worth for i=15% in RS. \",round(PW2,3);\nprint \"Present worth for i=10% is suitable.\";\nprint \"Therefore, the rate of return for the new business in % per annum \",round(i,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The present worth for i=10% in RS.   150598.487\nThe present worth for i=15% in RS.  -22421.94\nPresent worth for i=10% is suitable.\nTherefore, the rate of return for the new business in % per annum  11.741\n"
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 7.3 Page 91"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#alternate 1\nIo=150000.0;#in Rs\nAp=45570.0;#in Rs\nn=5.0;#in years\n\n#calculation\ni1=10.0;#in % per annum\nPW1=-Io+Ap*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=12.0;#in % per annum\nPW2=-Io+Ap*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni3=15.0;#in % per annum\nPW3=-Io+Ap*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni4=18.0;#in % per annum\nPW4=-Io+Ap*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in RS\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The present worth for i=10% in RS.  \",round(PW1,3);\nprint \"The present worth for i=12% in RS. \",round(PW2,3);\nprint \"The present worth for i=15% in RS. \",round(PW3,3);\nprint \"The present worth for i=18% in RS. \",round(PW4,3);\nprint\"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return of alternative in % per annum :\",round(i,3);\n\n#alternative 2\nIo=210000.0;#in Rs\nAp=58260.0;#in Rs\nn=5.0;#in years\ni1=12.0;#in % per annum\n\n#calcualtion\nPW1=-Io+Ap*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=13.0;#in % per annum\nPW2=-Io+Ap*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni=i1+(PW1-0)*(i2-i1)/(PW1-PW2);#in Rs.\n\n#result\nprint \"The present worth for i=12% in RS. \",round(PW1,3);\nprint \"The present worth for i=13% in RS. \",round(PW2,3);\nprint\"Present worth for i=12% is suitable.\";\nprint \"Therefore, the rate of return of alternative in % per annum :\",round(i,3);\n\n#Alternative 3:\nIo=255000.0;#in Rs\nAp=69000.0;#in Rs\nn=5.0;#in years\ni1=11.0;#in % per annum\nPW1=-Io+Ap*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=12.0;#in % per annum\nPW2=-Io+Ap*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni=i1+(PW1-0)*(i2-i1)/(PW1-PW2);#in Rs.\n\n#result\nprint \"The present worth for i=11% in RS. \",round(PW1,3);\nprint \"The present worth for i=12% in RS. \",round(PW2,3);\nprint\"Present worth for i=12% is suitable.\";\nprint \"Therefore, the rate of return of alternative in % per annum :\",round(i,3);\nprint\"It is clear that rate of return for alternative 3v is less than the minimum attractive rate of return of 12 %. So it should not be considered for comparison. the remaining two alternatives are qualified for consideration. Among the alternatives 1 and 2, the rate of return of 1 is greater than that of 2. hence,alternative 1 should be selected. \";",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The present worth for i=10% in RS.   22746.153\nThe present worth for i=12% in RS.  14269.652\nThe present worth for i=15% in RS.  2757.708\nThe present worth for i=18% in RS.  -7494.817\nPresent worth for i=15% is suitable.\nTherefore, the rate of return of alternative in % per annum : 15.807\nThe present worth for i=12% in RS.  14.262\nThe present worth for i=13% in RS.  -5086.107\nPresent worth for i=12% is suitable.\nTherefore, the rate of return of alternative in % per annum : 12.003\nThe present worth for i=11% in RS.  16.894\nThe present worth for i=12% in RS.  -6270.442\nPresent worth for i=12% is suitable.\nTherefore, the rate of return of alternative in % per annum : 11.003\nIt is clear that rate of return for alternative 3v is less than the minimum attractive rate of return of 12 %. So it should not be considered for comparison. the remaining two alternatives are qualified for consideration. Among the alternatives 1 and 2, the rate of return of 1 is greater than that of 2. hence,alternative 1 should be selected. \n"
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 7.4 Page 94"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nA0=-1275.0;#in Rs\nA1=150.0;#in Rs\nG=150.0;#in Rs\ni=10.0;#in % per annum\nn=5.0;#in years\n\n#calcualtion\nA=A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100) ;#in RS\ni1=10.0;#in % per annum\nPW1=A0+(A1+G*(((1+i1/100)**n)-i1*n/100-1)/(((i1/100)*(1+i1/100)**n)-i1/100))*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#i1n Rs.\ni2=12.0;#i2n % per annum\nPW2=A0+(A1+G*(((1+i2/100)**n)-i2*n/100-1)/(((i2/100)*(1+i2/100)**n)-i2/100))*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#i2n Rs.\ni3=15.0;#i3n % per annum\nPW3=A0+(A1+G*(((1+i3/100)**n)-i3*n/100-1)/(((i3/100)*(1+i3/100)**n)-i3/100))*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#i3n Rs.\ni4=18.0;#in % per annum\nPW4=A0+(A1+G*(((1+i4/100)**n)-i4*n/100-1)/(((i4/100)*(1+i4/100)**n)-i4/100))*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in Rs.\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The annual equivalent of the positive cash flows in RS. : \",round(A,3);\nprint \"PW(10)\",round(PW1,3);\nprint \"PW(12)\",round(PW2,3);\nprint \"PW(15)\",round(PW3,3);\nprint \"PW(18)\",round(PW1,4);\nprint\"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return for the given cash flow in % per annum :\",round(i,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The annual equivalent of the positive cash flows in RS. :  421.519\nPW(10) 322.888\nPW(12) 225.269\nPW(15) 94.095\nPW(18) 322.8882\nPresent worth for i=15% is suitable.\nTherefore, the rate of return for the given cash flow in % per annum : 17.448\n"
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 7.5 Page 95"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\n#alternative 1\nIo=500000.0;#in Rs\nAr=170000.0;#in Rs\nn=5.0;#in years\ni1=15.0;#in % per annum\n\n#calculation\nPW1=-Io+Ar*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=17.0;#in % per annum\nPW2=-Io+Ar*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni3=20.0;#in % per annum\nPW3=-Io+Ar*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni4=22.0;#in % per annum\nPW4=-Io+Ar*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in RS\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The present worth(PW(15%)) for i=10% in RS.  \",round(PW1,3);\nprint \"The present worth(PW(17%)) for i=10% in RS.  \",round(PW2,3);\nprint \"The present worth(PW(20%)) for i=10% in RS.  \",round(PW3,3);\nprint \"The present worth(PW(22%)) for i=10% in RS.  \",round(PW4,3);\nprint \"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return of alternative 1 in % per annum \",round(i,3);\n\n#altenative 2\nIo=800000.0;#in Rs\nAr=270000.0;#in Rs\nn=5.0;#in years\ni3=20.0;#in % per annum\n\n#calculation\nPW3=-Io+Ar*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni4=22.0;#in % per annum\nPW4=-Io+Ar*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in RS\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The present worth(PW(20%)) for i=10% in RS.  \",round(PW3,3);\nprint \"The present worth(PW(22%)) for i=10% in RS.  \",round(PW4,3);\nprint \"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return of alternative 2 in % per annum \",round(i,3);\nprint \"Since the rate of return of alternative 1 is greater than that of the alternative 2, select alternative 1. \";",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "The present worth(PW(15%)) for i=10% in RS.   69866.367\nThe present worth(PW(17%)) for i=10% in RS.   43888.848\nThe present worth(PW(20%)) for i=10% in RS.   8404.064\nThe present worth(PW(22%)) for i=10% in RS.   -13181.241\nPresent worth for i=15% is suitable.\nTherefore, the rate of return of alternative 1 in % per annum  20.779\nThe present worth(PW(20%)) for i=10% in RS.   7465.278\nThe present worth(PW(22%)) for i=10% in RS.   -26817.264\nPresent worth for i=15% is suitable.\nTherefore, the rate of return of alternative 2 in % per annum  20.436\nSince the rate of return of alternative 1 is greater than that of the alternative 2, select alternative 1. \n"
-      }
-     ],
-     "prompt_number": 11
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter8.ipynb b/Engineering_Economics/Chapter8.ipynb
deleted file mode 100755
index de307e62..00000000
--- a/Engineering_Economics/Chapter8.ipynb
+++ /dev/null
@@ -1,619 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:44663c03d7c23760034bea6db3765d8b79a48144d6b54769443d3f295fd0b9f3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Replacement and Maintenance Analysis"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1 Page 103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "FC=4000.0;#in Rs.\n",
-      "i=0.0;#in % per annum\n",
-      "MC1=0;#in Rs.\n",
-      "print \"Tabulation to determine economic life : \";\n",
-      "print\"End of     Maintenance     Summation of     Avg cost of     Avg FC if     Avg total\";\n",
-      "print\" year      cost at end     maintenance      maintenance    replaced at    cost through\";\n",
-      "print\"             of year         costs          through year    year given    year given\";\n",
-      "print\"  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)\";\n",
-      "print\"  1            0                0                 0          4000          4000\";\n",
-      "print\"  2          200              200               100          2000          2100\";\n",
-      "print\"  3          400              600               200       1333.33       1533.33\";\n",
-      "print\"  4          600             1200               300          1000          1300\";\n",
-      "print\"  5          800             2000               400           800          1200\";\n",
-      "print\"  6         1000             3000               500        666.67       1166.67\";\n",
-      "print\"  7         1200             4200               600        571.43       1171.43\";\n",
-      "print\"Economic life of the machine : 6 years\";\n",
-      "print\"Column C summarizes the summation of maintenance costs for each replacement period. The value corresponding to any end of year in this column represents the total maintenance costs of using the equipment till the end of that year. It gives the Economic life of the machine : 6 years\";\n",
-      "\n",
-      "#When i=12%\n",
-      "i=12.0;#in % per annum\n",
-      "FC=4000.0;#in Rs.\n",
-      "print\"Tabulation to determine economic life : \";\n",
-      "print\"End of     Maintenance         P/F       PW as of begin   Summation of    PW of        A/P     Annual equi\";\n",
-      "print\" year      cost at end        12%         of year of         PW of      cumulative     12%     total cost\";\n",
-      "print\"             of year           n          maintenance     maintenance   maintenance     n       of year\";\n",
-      "print\"  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)    G(rs.)      H(rs.)\";\n",
-      "print\"  1            0           0.8929              0.00          0.00       4000.00     1.1200     4480.00\";\n",
-      "print\"  2          200           0.7972            159.44        159.44       4159.44     0.5917     2461.14\";\n",
-      "print\"  3          400           0.7118            284.72        444.16       4444.16     0.4163     1850.10\";\n",
-      "print\"  4          600           0.6355            381.30        825.46       4825.46     0.3292     1588.54\";\n",
-      "print\"  5          800           0.5674            453.92       1279.38       5279.38     0.2774     1464.50\";\n",
-      "print\"  6         1000           0.5066            506.60       1785.98       5785.98     0.2432     1407.15\";\n",
-      "print\"  7         1200           0.4524            542.88       2328.86       6328.86     0.2191     1386.65\";\n",
-      "print\"  8         1400           0.4039            565.46       2894.32       6894.32     0.2013     1387.83\";\n",
-      "print\"  9         1600           0.3606            576.96       3471.28       7471.28     0.1877     1402.36\";\n",
-      "print\" 10         1800           0.3220            579.60       4050.88       8050.88     0.1770     1425.00\";\n",
-      "print\"Economic life of the machine : 7 years\";\n",
-      "print\"For this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is 7 year.\";\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Tabulation to determine economic life : \n",
-        "End of     Maintenance     Summation of     Avg cost of     Avg FC if     Avg total\n",
-        " year      cost at end     maintenance      maintenance    replaced at    cost through\n",
-        "             of year         costs          through year    year given    year given\n",
-        "  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)\n",
-        "  1            0                0                 0          4000          4000\n",
-        "  2          200              200               100          2000          2100\n",
-        "  3          400              600               200       1333.33       1533.33\n",
-        "  4          600             1200               300          1000          1300\n",
-        "  5          800             2000               400           800          1200\n",
-        "  6         1000             3000               500        666.67       1166.67\n",
-        "  7         1200             4200               600        571.43       1171.43\n",
-        "Economic life of the machine : 6 years\n",
-        "Column C summarizes the summation of maintenance costs for each replacement period. The value corresponding to any end of year in this column represents the total maintenance costs of using the equipment till the end of that year. It gives the Economic life of the machine : 6 years\n",
-        "Tabulation to determine economic life : \n",
-        "End of     Maintenance         P/F       PW as of begin   Summation of    PW of        A/P     Annual equi\n",
-        " year      cost at end        12%         of year of         PW of      cumulative     12%     total cost\n",
-        "             of year           n          maintenance     maintenance   maintenance     n       of year\n",
-        "  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)    G(rs.)      H(rs.)\n",
-        "  1            0           0.8929              0.00          0.00       4000.00     1.1200     4480.00\n",
-        "  2          200           0.7972            159.44        159.44       4159.44     0.5917     2461.14\n",
-        "  3          400           0.7118            284.72        444.16       4444.16     0.4163     1850.10\n",
-        "  4          600           0.6355            381.30        825.46       4825.46     0.3292     1588.54\n",
-        "  5          800           0.5674            453.92       1279.38       5279.38     0.2774     1464.50\n",
-        "  6         1000           0.5066            506.60       1785.98       5785.98     0.2432     1407.15\n",
-        "  7         1200           0.4524            542.88       2328.86       6328.86     0.2191     1386.65\n",
-        "  8         1400           0.4039            565.46       2894.32       6894.32     0.2013     1387.83\n",
-        "  9         1600           0.3606            576.96       3471.28       7471.28     0.1877     1402.36\n",
-        " 10         1800           0.3220            579.60       4050.88       8050.88     0.1770     1425.00\n",
-        "Economic life of the machine : 7 years\n",
-        "For this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is 7 year.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2 Page 105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "FC=20000.0;#in Rs.\n",
-      "i=15.0;#in % per annum\n",
-      "\n",
-      "#result\n",
-      "print \"The other details are summarized in Table 8.3. It can be seen from the book.\";\n",
-      "print \"Total annual equivalent cost = [cumulative sum of PW as of beginning of year 1 of operation & maintenance cost + FC - PW as of beginning of year 1 of salvage]*(A/P,15,n)\";\n",
-      "print \"In column L, the annual equivalent cost is minimum for n=5. Therefore, the economic life of the machine is 5 years. \";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The other details are summarized in Table 8.3. It can be seen from the book.\n",
-        "Total annual equivalent cost = [cumulative sum of PW as of beginning of year 1 of operation & maintenance cost + FC - PW as of beginning of year 1 of salvage]*(A/P,15,n)\n",
-        "In column L, the annual equivalent cost is minimum for n=5. Therefore, the economic life of the machine is 5 years. \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3 Page 107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "FC=20000.0;#in Rs.\n",
-      "i=15.0;#in % per annum\n",
-      "\n",
-      "#result\n",
-      "print \"The details are summarized in Table 8.4. It can be seen from the book.\";\n",
-      "print\"Total annual equivalent cost = [summation of PW of maintenance cost + FC]]*(A/P,15,n)\";\n",
-      "print\"(column E + Rs. 6000)* Column G\";\n",
-      "print \"Column F * Column G\";\n",
-      "print \"In column H, the minimum annual equivalent cost occurs when n=8. Therefore, the economic life of the machine B is 8 years. \";\n",
-      "print \"RESULT : Min annual equivalent cost for machine A : Rs. 2780\";\n",
-      "print \"Min annual equivalent cost for machine B : Rs. 3672.30\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The details are summarized in Table 8.4. It can be seen from the book.\n",
-        "Total annual equivalent cost = [summation of PW of maintenance cost + FC]]*(A/P,15,n)\n",
-        "(column E + Rs. 6000)* Column G\n",
-        "Column F * Column G\n",
-        "In column H, the minimum annual equivalent cost occurs when n=8. Therefore, the economic life of the machine B is 8 years. \n",
-        "RESULT : Min annual equivalent cost for machine A : Rs. 2780\n",
-        "Min annual equivalent cost for machine B : Rs. 3672.30\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4 Page 109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#alternative 1\n",
-      "Pprice=200000.0;#in Rs\n",
-      "P=120000.0;#in Rs\n",
-      "F=25000.0;#in Rs\n",
-      "A=25000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=6.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(12%)) of this alternative in RS.  \",round(AE1,3);\n",
-      "\n",
-      "#Alternative 2\n",
-      "P=150000.0;#in Rs\n",
-      "F=20000.0;#in Rs\n",
-      "A=14000.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=6.0;#in years\n",
-      "\n",
-      "#calcualion\n",
-      "AE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(12%)) of this alternative in RS.  \",round(AE2,3);\n",
-      "print \"Since, The equivalent cost of new machine is less than that of present machine, it is suggested that the present machine be replaced with the new machine.\";\n",
-      "print \"answer is bit different due to approximation error\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual equivalent cost(AE(12%)) of this alternative in RS.   51106.443\n",
-        "The annual equivalent cost(AE(12%)) of this alternative in RS.   48019.343\n",
-        "Since, The equivalent cost of new machine is less than that of present machine, it is suggested that the present machine be replaced with the new machine.\n",
-        "answer is bit different due to approximation error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5 Page 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initation of variable\n",
-      "#alternative 1\n",
-      "Pprice=50000.0;#in Rs\n",
-      "P=15000.0;#in Rs\n",
-      "F=8000.0;#in Rs\n",
-      "A=14000.0;#in Rs\n",
-      "i=15.0;#in % per annum\n",
-      "n=5.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(15%)) of this alternative in RS.  \",round(AE1,3);\n",
-      "\n",
-      "#Alternative 2\n",
-      "P=65000.0;#in Rs\n",
-      "F=13000.0;#in Rs\n",
-      "A=9000.0;#in Rs\n",
-      "i=15.0;#in % per annum\n",
-      "n=20.0;#in years\n",
-      "\n",
-      "#calculation\n",
-      "AE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(15%)) of this alternative in RS. : \",round(AE2,3);\n",
-      "print \"Since, The equivalent cost of Old deisel Engine is less than that of New deisel Engine, it is suggested to keep the Old deisel Engine.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual equivalent cost(AE(15%)) of this alternative in RS.   17288.209\n",
-        "The annual equivalent cost(AE(15%)) of this alternative in RS. :  19257.596\n",
-        "Since, The equivalent cost of Old deisel Engine is less than that of New deisel Engine, it is suggested to keep the Old deisel Engine.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6 Page 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#intiation of variable\n",
-      "P=660000.0;#in Rs\n",
-      "F=400000.0;#in Rs\n",
-      "A=96000.0;#in Rs\n",
-      "i=10.0;#in % per annum\n",
-      "n=5.0;#in years\n",
-      "\n",
-      "#caculation\n",
-      "AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(10%)) of this alternative in RS. : \",round(AE1,3);\n",
-      "\n",
-      "#Alternative 2\n",
-      "P=150000.0;#in Rs\n",
-      "X=420000.0;#in Rs\n",
-      "i=10.0;#in % per annum\n",
-      "n=40.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "AE2=(P-X)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(10%)) of this alternative in RS. : \",round(AE2,3);\n",
-      "print \"Since, The equivalent cost of alternative 2 is less than that of alternative 1, it is suggested that alternative 2 should be selected.\";\n",
-      "print \" calculations in the book is not accurate.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual equivalent cost(AE(10%)) of this alternative in RS. :  204587.345\n",
-        "The annual equivalent cost(AE(10%)) of this alternative in RS. :  108389.958\n",
-        "Since, The equivalent cost of alternative 2 is less than that of alternative 1, it is suggested that alternative 2 should be selected.\n",
-        " calculations in the book is not accurate.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7 Page 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "#alternative 1\n",
-      "P=10000.0;#in Rs\n",
-      "F=1500.0;#in Rs\n",
-      "A=1600.0;#in Rs\n",
-      "i=15.0;#in % per annum\n",
-      "n=7.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(10%)) of 10 hp motor  in RS. : \",round(AE1,3);\n",
-      "\n",
-      "#alternative 1 part 2\n",
-      "P=10000;#in Rs\n",
-      "F=800.0;#in Rs\n",
-      "A=1000.0;#in Rs\n",
-      "i=15.0;#in % per annum\n",
-      "n=7.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "AE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(10%)) of 10 hp motor  in RS.  \",round(AE2,3);\n",
-      "print \"Total annual equivalent cost of alternative in Rs. : \",round(AE1+AE2,3);\n",
-      "\n",
-      "\n",
-      "#Alternative 2\n",
-      "P=35000.0;#in Rs\n",
-      "F=4000.0;#in Rs\n",
-      "A=500.0;#in Rs\n",
-      "i=15.0;#in % per annum\n",
-      "n=7.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "AE=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "\n",
-      "print \"The annual equivalent cost of alternative 2 in RS.\",round(AE,3);\n",
-      "print\"Since, The equivalent cost of alternative 1 is less than that of alternative 2, it is suggested that the present 10 hp motor be augmented with an additional 5 hp motor.\";\n",
-      "print \" calculations in the book is not accurate\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual equivalent cost(AE(10%)) of 10 hp motor  in RS. :  3868.063\n",
-        "The annual equivalent cost(AE(10%)) of 10 hp motor  in RS.   3331.315\n",
-        "Total annual equivalent cost of alternative in Rs. :  7199.378\n",
-        "The annual equivalent cost of alternative 2 in RS. 8551.171\n",
-        "Since, The equivalent cost of alternative 1 is less than that of alternative 2, it is suggested that the present 10 hp motor be augmented with an additional 5 hp motor.\n",
-        " calculations in the book is not accurate\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8 Page 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "P=10000.0;#in Rs\n",
-      "F=4000.0;#in Rs\n",
-      "A=500.0;#in Rs\n",
-      "i=12.0;#in % per annum\n",
-      "n=4.0;#in years\n",
-      "\n",
-      "#calcualtion\n",
-      "AE=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n",
-      "X= (AE-1000*i/100-750)/0.3292+1000;\n",
-      "#result\n",
-      "print \"The annual equivalent cost(AE(15%)) of n in RSew machine : \",round(AE,3);\n",
-      "print \"The comparative use value of old machine is Rs.\", round(X,3), \"which is less than the price(Rs. 8000) offered by the company which is supplying the new machine in the event of replacing the old machine by nw machine. \";\n",
-      "print\"Therefore, it is advisable to replace the old machine with the new one.\";"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual equivalent cost(AE(15%)) of n in RSew machine :  2955.407\n",
-        "The comparative use value of old machine is Rs. 7334.771 which is less than the price(Rs. 8000) offered by the company which is supplying the new machine in the event of replacing the old machine by nw machine. \n",
-        "Therefore, it is advisable to replace the old machine with the new one.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9 Page 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Intiation of Variable\n",
-      "p1=0.07;#unitless\n",
-      "p2=0.11;#unitless\n",
-      "p3=0.12;#unitless\n",
-      "p4=0.18;#unitless\n",
-      "p5=0.21;#unitless\n",
-      "p6=0.20;#unitless\n",
-      "p7=0.11;#unitless\n",
-      "N0=100.0;#no. of transistors\n",
-      "\n",
-      "#calcualtion\n",
-      "N1=N0*p1;#no. of transistors\n",
-      "N2=N0*p2;#no. of transistors\n",
-      "N3=N0*p3;#no. of transistors\n",
-      "N4=N0*p4;#no. of transistors\n",
-      "N5=N0*p5;#no. of transistors\n",
-      "N6=N0*p6;#no. of transistors\n",
-      "N7=N0*p7;#no. of transistors\n",
-      "Life=0;#in weeks\n",
-      "p=[p1, p2, p3, p4, p5, p6, p7];#Unitless\n",
-      "for i in range (7): \n",
-      "    Life=Life+(i+1)*p[i];\n",
-      "\n",
-      "#result\n",
-      "print \"Expected life of each transistor in weeks : \",round(Life,3)\n",
-      "print \"Average No. of failures/week :  \",round(100/Life);\n",
-      "\n",
-      "#result of group replacement cost \n",
-      "print \"Cost of transistor when replaced simultaneously = Rs. 3\";\n",
-      "print \"Cost of transistor when replaced individually = Rs. 9\";\n",
-      "print \"The cost of group replacement policy for seeral replacement periods are summarized in Table 8.6. This table can be seen from the book.\";\n",
-      "print \"From table it is clear that the avg cost/week is minimum for the 4th week. Hence, the group replacement period is 4 weeks.\";\n",
-      "print \"Individual replacement cost/week = Rs. 207\";\n",
-      "print \"Minimum group replacement ost/week = Rs. 196.50\";\n",
-      "print \"Since the min group replacement cost/week is less than the individual replacement cost/week, the group replacement policy is the best, and hence all the transistors should be replaced in 4 weeks.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Expected life of each transistor in weeks :  4.39\n",
-        "Average No. of failures/week :   23.0\n",
-        "Cost of transistor when replaced simultaneously = Rs. 3\n",
-        "Cost of transistor when replaced individually = Rs. 9\n",
-        "The cost of group replacement policy for seeral replacement periods are summarized in Table 8.6. This table can be seen from the book.\n",
-        "From table it is clear that the avg cost/week is minimum for the 4th week. Hence, the group replacement period is 4 weeks.\n",
-        "Individual replacement cost/week = Rs. 207\n",
-        "Minimum group replacement ost/week = Rs. 196.50\n",
-        "Since the min group replacement cost/week is less than the individual replacement cost/week, the group replacement policy is the best, and hence all the transistors should be replaced in 4 weeks.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10 Page 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initiation of variable\n",
-      "p1=(100.0-96)/100;#unitless\n",
-      "p2=(96.0-89)/100;#unitless\n",
-      "p3=(89.0-68)/100;#unitless\n",
-      "p4=(68.0-37)/100;#unitless\n",
-      "p5=(37.0-13)/100;#unitless\n",
-      "p6=(13.0-0)/100;#unitless\n",
-      "N0=1000.0;#no. of resistors\n",
-      "\n",
-      "#calculation\n",
-      "N1=N0*p1;#no. of resistors\n",
-      "N2=N0*p2+N1*p1;#no. of resistors\n",
-      "N3=N0*p3+N1*p2+N2*p1;#no. of resistors\n",
-      "N4=N0*p4+N1*p3+N2*p2+N3*p1;#no. of resistors\n",
-      "N5=N0*p5+N1*p4+N2*p3+N3*p2+N4*p1;#no. of resistors\n",
-      "N6=N0*p6+N1*p5+N2*p4+N3*p3+N4*p2+N5*p1    ;#no. of resistors\n",
-      "\n",
-      "Life=0;#in months\n",
-      "p=[p1, p2, p3, p4, p5, p6];#Unitless\n",
-      "for i in range (6): \n",
-      "    Life=Life+(i+1)*p[i];\n",
-      "\n",
-      "#result\n",
-      "print \"Expected life of each transistor in weeks : \",round(Life,3)\n",
-      "print \"Average No. of failures/week :  \",round(1000/Life);\n",
-      "\n",
-      "#result of group replacement cost\n",
-      "print \"Cost/transistor when replaced simultaneously = Rs. 4\";\n",
-      "print \"Cost/transistor when replaced individually = Rs. 10\";\n",
-      "print \"The cost of group replacement policy for several replacement periods are summarized in Table 8.7. This table can be seen from the book.\";\n",
-      "print \"From table it is clear that the avg cost/month is minimum for the 3rd month. Hence, the group replacement period is 3 months.\";\n",
-      "print \"Individual replacement cost/month = Rs. 2480\";\n",
-      "print \"Minimum group replacement ost/month = Rs. 2426.67\";\n",
-      "print \"Since the min group replacement cost/month is less than the individual replacement cost/month, the group replacement policy is the best, and hence all the transistors should be replaced in 3 months.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Expected life of each transistor in weeks :  4.03\n",
-        "Average No. of failures/week :   248.0\n",
-        "Cost/transistor when replaced simultaneously = Rs. 4\n",
-        "Cost/transistor when replaced individually = Rs. 10\n",
-        "The cost of group replacement policy for several replacement periods are summarized in Table 8.7. This table can be seen from the book.\n",
-        "From table it is clear that the avg cost/month is minimum for the 3rd month. Hence, the group replacement period is 3 months.\n",
-        "Individual replacement cost/month = Rs. 2480\n",
-        "Minimum group replacement ost/month = Rs. 2426.67\n",
-        "Since the min group replacement cost/month is less than the individual replacement cost/month, the group replacement policy is the best, and hence all the transistors should be replaced in 3 months.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/Chapter9.ipynb b/Engineering_Economics/Chapter9.ipynb
deleted file mode 100755
index 6221daa0..00000000
--- a/Engineering_Economics/Chapter9.ipynb
+++ /dev/null
@@ -1,209 +0,0 @@
-{
- "metadata": {
-  "name": "EE-9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": "Depreciation"
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.1 Page 127"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=100000;#book vakue\n\n#calcualtion\nDt=(P-F)/n;#in Rs.\n\n#result\nprint \"End of year          Depreciation          Book value\";\nfor i in range (8):\n    B=Bt-i*Dt;\n    print  eoy[i],\"                    \",Dt,\"            \", B,\" \";\n     \n    \n    ",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "End of year          Depreciation          Book value\n0                      10000.0              100000.0  \n1                      10000.0              90000.0  \n2                      10000.0              80000.0  \n3                      10000.0              70000.0  \n4                      10000.0              60000.0  \n5                      10000.0              50000.0  \n6                      10000.0              40000.0  \n7                      10000.0              30000.0  \n"
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.2 Page 127"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\n\n#calculation\nD5=(P-F)/n;#in Rs.\nt=5;#in years\nBt=P-t*(P-F)/n;#in Rs\n\n#result\nprint \"D5 in Rs. : \",round(D5,3);\nprint \"(This is independent of the time period)\";\nprint \"B5 in Rs. : \",round(Bt,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "D5 in Rs. :  10000.0\n(This is independent of the time period)\nB5 in Rs. :  50000.0\n"
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.3 Page 129"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nk=0.2\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=[100000.0,0,0,0,0,0,0,0,0];#book vakue\nDt=[0.0,0,0,0,0,0,0,0,0];\n\n#result\nprint \"End of year          Depreciation          Book value\";\nprint \"0                      0                    100000\"\nfor i in xrange (1,9):\n    Dt[i]=k*Bt[i-1];\n    Bt[i]=Bt[i-1]-Dt[i];\n    print  eoy[i],\"                    \",Dt[i],\"            \", Bt[i],\" \";",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "End of year          Depreciation          Book value\n0                      0                    100000\n1                      20000.0              80000.0  \n2                      16000.0              64000.0  \n3                      12800.0              51200.0  \n4                      10240.0              40960.0  \n5                      8192.0              32768.0  \n6                      6553.6              26214.4  \n7                      5242.88              20971.52  \n8                      4194.304              16777.216  \n"
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.4 Page 128"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nk=0.2\nt=5.0;#in years\n\n#calculation\nDt=k*(1-k)**(t-1)*P;#in Rs.\nBt=((1-k)**t)*P;#in Rs.\n\n#result\nprint \"D5 in Rs. : \",round(Dt,3);\nprint \"B5 in Rs. : \",round(Bt,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "D5 in Rs. :  8192.0\nB5 in Rs. :  32768.0\n"
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.5 Page 129"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nSum=n*(n+1)/2;#sum of the years\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=[100000.0,0,0,0,0,0,0,0,0];#book vakue\nDt=[0.0,0,0,0,0,0,0,0,0];\nrate=[0.0,0,0,0,0,0,0,0,0];\n\n#result\nprint \"End of year          Depreciation          Book value\";\nprint \"0                      0                    100000\"\nfor i in xrange (1,9):\n    rate[i]=(8-i+1)/36.0;\n    Dt[i]=rate[i]*(P-F);\n    Bt[i]=Bt[i-1]-Dt[i];\n    print  eoy[i],\"                    \",Dt[i],\"            \", Bt[i],\" \";",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "End of year          Depreciation          Book value\n0                      0                    100000\n1                      17777.7777778              82222.2222222  \n2                      15555.5555556              66666.6666667  \n3                      13333.3333333              53333.3333333  \n4                      11111.1111111              42222.2222222  \n5                      8888.88888889              33333.3333333  \n6                      6666.66666667              26666.6666667  \n7                      4444.44444444              22222.2222222  \n8                      2222.22222222              20000.0  \n"
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.6 Page 131"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nt=5.0;#in years\n\n#cacualtion\nDt=(n-t+1)*(P-F)/(n*(n+1)/2);#in Rs.\nBt=(P-F)*((n-t)/n)*((n-t+1)/(n+1))+F;#in Rs.\n\n#result\nprint \"D5 in Rs. : \",round(Dt,3);\nprint \"B5 in Rs. : \",round(Bt,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "D5 in Rs. :  8888.889\nB5 in Rs. :  33333.333\n"
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.7 Page 132"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nk=0.12;\ni=12.0;#in % per annum\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=[100000.0,0,0,0,0,0,0,0,0];#book vakue\nDt=[0.0,0,0,0,0,0,0,0,0];\nrate=[0.0,0,0,0,0,0,0,0,0];\nSUM=0;\n\n#calculation\nA=(P-F)*(i/100)/(((1+i/100)**n)-1);\n\n#result\nprint \"End of year          Net Depreciation          Book value\";\nprint \"0                      0                         100000\"\nfor i in xrange (1,9):\n    Dt[i]=k*SUM+A;\n    SUM=SUM+Dt[i]\n    Bt[i]=Bt[i-1]-Dt[i];\n    print  eoy[i],\"                    \",round(Dt[i],3),\"            \", round(Bt[i],3),\" \";\n    \nprint \"fixed deprececiation which is constant in Rs\",round(A,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "End of year          Net Depreciation          Book value\n0                      0                         100000\n1                      6504.227              93495.773  \n2                      7284.735              86211.038  \n3                      8158.903              78052.135  \n4                      9137.971              68914.164  \n5                      10234.528              58679.637  \n6                      11462.671              47216.966  \n7                      12838.191              34378.774  \n8                      14378.774              20000.0  \nfixed deprececiation which is constant in Rs 6504.227\n"
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.8 Page 133"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\ni=12.0;#in % per annum\nt=5.0;#in Years\n\n#calculation\nDt=(P-F)*(i/100)/(((1+i/100)**n)-1)*(1+i/100)**(t-1);#in Rs.\nt=7;#in Years\nBt=P-(P-F)*(i/100)/(((1+i/100)**n)-1)*(((1+i/100)**t)-1)/(i/100);#in Rs.\n\n#result\nprint \"D7 in Rs. : \",round(Dt,3);\nprint \"B7 in Rs. : \",round(Bt,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "D7 in Rs. :  10234.528\nB7 in Rs. :  34378.774\n"
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": "Example 9.9 Page 134"
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": "#initiation of variable\nP=8000000.0;#in Rs\nF=50000.0;#in Rs\nX=75000.0;#in Km\nx=2000.0;#in Km\nn=8.0;#in years\ni=12.0;#in % per annum\n\n#calcualtion\nD=(P-F)*x/X;#in Rs.\n\n#result\nprint \"Depreciation for year 3 in Rs. : \",round(D,3);",
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": "Depreciation for year 3 in Rs. :  212000.0\n"
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Engineering_Economics/README.txt b/Engineering_Economics/README.txt
deleted file mode 100755
index 27bb08be..00000000
--- a/Engineering_Economics/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Tarun Kumar Das
-Course: btech
-College/Institute/Organization: College of Engineering
-Department/Designation: Industrial Engineering
-Book Title: Engineering Economics
-Author: R. Panneerselvam
-Publisher: Phi Learning, Delhi
-Year of publication: 2010
-Isbn: 978-81-203-1743-7
-Edition: 1
\ No newline at end of file
diff --git a/Engineering_Economics/screenshots/Depriciation_and_Book_value.png b/Engineering_Economics/screenshots/Depriciation_and_Book_value.png
deleted file mode 100755
index 3c2c419d..00000000
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diff --git a/Engineering_Economics/screenshots/Economic_order_and_Time.png b/Engineering_Economics/screenshots/Economic_order_and_Time.png
deleted file mode 100755
index 623d5718..00000000
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diff --git a/Engineering_Economics/screenshots/Expected_Saving.png b/Engineering_Economics/screenshots/Expected_Saving.png
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index 6970da10..00000000
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diff --git a/Engineering_Economics/screenshots/Present_Worth.png b/Engineering_Economics/screenshots/Present_Worth.png
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index c5295c6e..00000000
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diff --git a/Engineering_Economics/screenshots/future_sum.png b/Engineering_Economics/screenshots/future_sum.png
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diff --git a/Engineering_Economics/screenshots/maturity_value.png b/Engineering_Economics/screenshots/maturity_value.png
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diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter10_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter10_1.ipynb
new file mode 100755
index 00000000..8a14db85
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter10_1.ipynb
@@ -0,0 +1,125 @@
+{
+ "metadata": {
+  "name": "EE-10"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Evaluation of Public Alternatives"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.1 Page 138"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nIi=4000000.0;#in Rs.\nAM=150000.0;#in Rs.\nAFS=600000.0;#in Rs.\nEinc=50000.0;#in Rs.\ni=12.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nCp=AM*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in Rs\nTPW=Ii+Cp;#in RS\n\n#result\nprint \"Total present worth of costs in RS. : \",round(TPW,3);\n\n#Total present worth of fuel savings:\nAI=600000.0;#in Rs.\nG=50000.0;#in Rs.\ni=12.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nA=AI+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100);#in RS\nBp=A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in Rs.\nBCratio=Bp/(Ii+Cp);#unitless\n\n#result\nprint \"Present worth of fuel savings in Rs. : \",round(Bp,3);\nprint \"BCratio : \",round(BCratio,3);\nprint\"Since BC ratio is more than 1, the construction of the bridge across the river is justified.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Total present worth of costs in RS. :  5021629.673\nPresent worth of fuel savings in Rs. :  5782527.265\nBCratio :  1.152\nSince BC ratio is more than 1, the construction of the bridge across the river is justified.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.2 Page 139"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nIc=80000000.0;#in Rs.\nAps=6000000.0;#in Rs.\nAfs=3000000.0;#in Rs.\nAib=5000000.0;#in Rs.\nArb=2000000.0;#in Rs.\nAmc=3000000.0;#in Rs.\ni=12.0;#in % per annum\nn=50.0;#in years\n\n#calculation\nTAB=Afs+Aib+Arb;#in Rs.\nPW_B=TAB*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in Rs\n\n#result\nprint \"Total present worth of benefits in RS. : \",round(PW_B,3);\n\n#Present worth of costs:\nPW_C=Ic+Amc*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n)-Aps*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\nBCratio=PW_B/PW_C;#unitless\n\n#result\nprint \"Present worth of costs in Rs. : \",round(PW_C,3);\nprint \"BCratio : \",round(BCratio,3);\nprint \"Since BC ratio is more than 1, the state govt. can implement the hydroelectric project.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Total present worth of benefits in RS. :  83044984.884\nPresent worth of costs in Rs. :  55086504.535\nBCratio :  1.508\nSince BC ratio is more than 1, the state govt. can implement the hydroelectric project.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 10.3 Page 140"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nP=3000000.0;#in Rs.\nB=900000.0;#in Rs.\ni=10.0;#in % per annum\nn=5.0;#in years\n\n#calculation\nAE1=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs\nBCratio=B/AE1;#unitless\n\n#result\nprint \"Annual equivalent of initial cost in Rs. : \",round(AE1,3);\nprint \"BCratio : \",round(BCratio,3);\n\n#Alternative A2 : \nP=6000000.0;#in Rs.\nB=1500000.0;#in Rs.\ni=10.0;#in % per annum\nn=7.0;#in years\n\n#calculation\nAE2=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs\nBCratio=B/AE2;#unitless\n\n#result\nprint \"Annual equivalent of initial cost in Rs. : \",round(AE2,3);\nprint \"BCratio : \",round(BCratio,3);\nprint \"The benefit cost ratio of alternative 2 is more than that of alternative A1. Hence, alternative A2 is to be selected. The comparisoon is made on a 35 years period which is the minimum common multiple of the lives of alternative 1 and 2\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Annual equivalent of initial cost in Rs. :  791392.442\nBCratio :  1.137\nAnnual equivalent of initial cost in Rs. :  1232432.998\nBCratio :  1.217\nThe benefit cost ratio of alternative 2 is more than that of alternative A1. Hence, alternative A2 is to be selected. The comparisoon is made on a 35 years period which is the minimum common multiple of the lives of alternative 1 and 2\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.4 Page 141"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#cost of the state\nn=20.0;#in years\nP=2500000000.0;#in Rs.\nAgt=10000000.0;#in Rs.\nAi=1000000.0;#in Rs.\nCom=48000.0;#in Rs./year/employee\nC1=Com*300;#in Rs.\ni=10.0;#in % per annum\n\n#calculation\nC2=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs\nCA=C2+C1;#in Rs\n\n#result\nprint \"Annual equivalent initial cost in Rs. : \",round(C2,3);\nprint \"Total Annual equivalent cost of the project in Rs. : \",round(CA,3);\n\n#Benefit to the state : \nW=30000000.0;#in Rs.\nA1=W*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in Rs.\nAgpy=10000000.0;#in ton/year\nA2=Agpy*(30-10);#in Rs.\nAvgAI=1000000.0;#in tons Km.\nG=20000000.0;#in Rs.\n\n#calculation\nA3=A2+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100) ;#in Rs.\nBA=A1+A3;#in Rs\nBCratio=BA/CA;#unitless\nprint \"Annual equivalent average of transport cost saving in Rs. : \",round(A3,3);\nprint \"Total annual equivalent benefits to the state : \",round(BA,3);\nprint \"BCratio : \",round(BCratio,3);\nprint \"The benefit cost ratio is more than 1,the project is justified.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Annual equivalent initial cost in Rs. :  293649061.931\nTotal Annual equivalent cost of the project in Rs. :  308049061.931\nAnnual equivalent average of transport cost saving in Rs. :  330161500.91\nTotal annual equivalent benefits to the state :  333685289.653\nBCratio :  1.083\nThe benefit cost ratio is more than 1,the project is justified.\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.5 Page 143"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#project A\ni=9.0;#% per annum\nn=50.0;#in years\nP_A=150000000.0;#in RS.\nFcs_A=2500000;#in Rs.\nIb_A=3500000;#in Rs.\nRb_A=1000000;#in RS.\nOMC_A=2000000.0;#in Rs.\nPsy_A=10000000.0;#in Rs.\n\n#calculation\nAE_A=P_A*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\nCpy_A=AE_A+OMC_A-Psy_A;#in Rs.\nBpy_A=Fcs_A+Ib_A+Rb_A;#in Rs.\nBCratio_A=Bpy_A/Cpy_A;#unitless\n\n#result\nprint \"Costs/year : \",round(Cpy_A,3);\nprint \"Benefits/year : \",round(Bpy_A,3);\nprint \"BC Ratio of project A : \",round(BCratio_A,3)\n\n\n#Project B : \nP_B=250000000.0;#in Rs.\n\nOMC_B=2500000.0;#in Rs.\nPsy_B=12000000.0;#in Rs.\nFcs_B=3500000.0;#in Rs.\nIb_B=4500000.0;#in Rs.\nRb_B=2000000.0;#in RS.\nAE_B=P_B*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\nCpy_B=AE_B+OMC_B-Psy_B;#in Rs.\nBpy_B=Fcs_B+Ib_B+Rb_B;#in Rs.\nBCratio_B=Bpy_B/Cpy_B;#unitless\n\n#result\nprint \"Costs/year : \",round(Cpy_B,3);\nprint \"Benefits/year : \",round(Bpy_B,3);\nprint \"BC Ratio of project B : \",round(BCratio_B,3)\n\n#Project C : \nP_C=400000000.0;#in Rs.\nOMC_C=3500000.0;#in Rs.\nPsy_C=18000000.0;#in Rs.\nFcs_C=5000000.0;#in Rs.\nIb_C=6000000.0;#in Rs.\nRb_C=3500000.0;#in RS.\nAE_C=P_C*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\nBpy_C=Fcs_C+Ib_C+Rb_C;#in Rs.\nCpy_C=AE_C+OMC_C-Psy_C;#in Rs.\nBCratio_C=Bpy_C/Cpy_C;#unitless\n\n#result\nprint \"Costs/year : \",round(Cpy_C,3);\nprint \"Benefits/year : \",round(Bpy_C,3);\nprint \"BC Ratio of project C : \",round(BCratio_C,3)\nprint\"From the computations it is clear that only alternative A is eligible because other two alternatives have BC ratio less than one.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Costs/year :  5684030.212\nBenefits/year :  7000000.0\nBC Ratio of project A :  1.232\nCosts/year :  13306717.021\nBenefits/year :  10000000.0\nBC Ratio of project B :  0.752\nCosts/year :  21990747.233\nBenefits/year :  14500000.0\nBC Ratio of project C :  0.659\nFrom the computations it is clear that only alternative A is eligible because other two alternatives have BC ratio less than one.\n"
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter11_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter11_1.ipynb
new file mode 100755
index 00000000..6eb51dc1
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter11_1.ipynb
@@ -0,0 +1,104 @@
+{
+ "metadata": {
+  "name": "EE-11"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Inflation Adjusted Decisions"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.1 Page 148"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nprint \"Few steps are as follows : \";\nprint \"Step 1 : The estimated future requirementt per year in terms of todays rupees from his age 61 through 70 is Rs. 24000.\";\nprint \"Step 2 : The formula which is given below is used to get future equivalent of Rs. 24000 with the inflation of 9% per year.\";\nprint \"Formula : P*(1+IR)^n\";\nP=24000.0;#in Rs.\nIR=9.0;#in % per year\nAge=61.0;#in years\nfor n in range(21,31) :\n    IV=P*(1+IR/100)**n;#in Rs\n    print \"Age in years : \",Age,\". \"\"Inflated value of Rs. 24000 at end of year : \",n,\" in Rs. \",round(IV);\n    Age=Age+1;#in Years\n\nPW=0;#For initialising\nn=21.0;#in years\nfor m in range(10) :\n    A=P*(1+IR/100)**n;#in Rs\n    n=n+1;\n    i=15.0;#in % per year\n    PW=PW+A/(1+i/100)**(m+1);\n\n#result\nprint \"Present equivalent of withdrawls in Rs. : \",round(PW,3);\n\n#The annual equivalent amount :\ni=15.0;#in % per year\nn=20.0;#in years\n#Formula : (A/F,i,n) : (i/100)/(((1+i/100)^n)-1) \nA=PW*(i/100)/(((1+i/100)**n)-1);#in Rs.\nprint \"The annual equivalent amount in Rs. : \",round(A,3);\nprint\"RECOMMENDATION : The person has to invest an amount of Rs. \",round(A,3),\" at the end of every year starting from his age 41 through 60 which will enable him to make a withdrawl at the end of every year starting from his age 61 through 70.\"; ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Few steps are as follows : \nStep 1 : The estimated future requirementt per year in terms of todays rupees from his age 61 through 70 is Rs. 24000.\nStep 2 : The formula which is given below is used to get future equivalent of Rs. 24000 with the inflation of 9% per year.\nFormula : P*(1+IR)^n\nAge in years :  61.0 . Inflated value of Rs. 24000 at end of year :  21  in Rs.  146611.0\nAge in years :  62.0 . Inflated value of Rs. 24000 at end of year :  22  in Rs.  159806.0\nAge in years :  63.0 . Inflated value of Rs. 24000 at end of year :  23  in Rs.  174189.0\nAge in years :  64.0 . Inflated value of Rs. 24000 at end of year :  24  in Rs.  189866.0\nAge in years :  65.0 . Inflated value of Rs. 24000 at end of year :  25  in Rs.  206954.0\nAge in years :  66.0 . Inflated value of Rs. 24000 at end of year :  26  in Rs.  225580.0\nAge in years :  67.0 . Inflated value of Rs. 24000 at end of year :  27  in Rs.  245882.0\nAge in years :  68.0 . Inflated value of Rs. 24000 at end of year :  28  in Rs.  268011.0\nAge in years :  69.0 . Inflated value of Rs. 24000 at end of year :  29  in Rs.  292132.0\nAge in years :  70.0 . Inflated value of Rs. 24000 at end of year :  30  in Rs.  318424.0\nPresent equivalent of withdrawls in Rs. :  1013631.794\nThe annual equivalent amount in Rs. :  9894.537\nRECOMMENDATION : The person has to invest an amount of Rs.  9894.537  at the end of every year starting from his age 41 through 60 which will enable him to make a withdrawl at the end of every year starting from his age 61 through 70.\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.2 Page 150"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nprint\"The method of finding the economic life of the machine witha discounting factor of 20% at zero inflation rate is summarized in table below. From the table it is clear that total annual cost is minimum if the machine is used for 14 years. Hence the economic life of the machine is 14 years.\";\nprint\"End of year     Op_cost     Main_cost    Op+Main     P/F,i,n     PW     Cummulative     Salvage     PW_S    TPW     A/P,i,n     AEM\";\ni=20.0;#in  per year\nCum=0.0;#initialising\nOp_cost=40000.0;#in RS.\nMain_cost=60000.0;#in Rs.\nOpMain=Op_cost+Main_cost;#in Rs.\nS=400000.0;#in Rs.\nfor n in range (1,16) :\n    PF=1/((1+i/100)**n);\n    PW=OpMain*PF;#in Rs.\n    Cum=Cum+PW\n    PW_S=PF*S;#in RS.\n    TPW=500000.0+Cum-PW_S;#in Rs.\n    AP=((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\n    AEM=TPW*AP;#in RS\n    print \"       \",n,\"        \",Op_cost,\"        \",Main_cost,\"        \",OpMain,\"    \",PF,\"     \",PW,\"        \",Cum,\"        \",S,\"        \",PW_S,\"        \",TPW,\"        \",AP,\"        \",AEM;\n    Op_cost=Op_cost+5000.0;#in Rs.\n    Main_cost=Main_cost+6000.0;#in Rs.\n    S=S-50000.0;#in Rs.\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The method of finding the economic life of the machine witha discounting factor of 20% at zero inflation rate is summarized in table below. From the table it is clear that total annual cost is minimum if the machine is used for 14 years. Hence the economic life of the machine is 14 years.\nEnd of year     Op_cost     Main_cost    Op+Main     P/F,i,n     PW     Cummulative     Salvage     PW_S    TPW     A/P,i,n     AEM\n        1          40000.0          60000.0          100000.0      0.833333333333       83333.3333333          83333.3333333          400000.0          333333.333333          250000.0          1.2          300000.0\n        2          45000.0          66000.0          100000.0      0.694444444444       69444.4444444          152777.777778          350000.0          243055.555556          409722.222222          0.654545454545          268181.818182\n        3          50000.0          72000.0          100000.0      0.578703703704       57870.3703704          210648.148148          300000.0          173611.111111          537037.037037          0.474725274725          254945.054945\n        4          55000.0          78000.0          100000.0      0.48225308642       48225.308642          258873.45679          250000.0          120563.271605          638310.185185          0.386289120715          246572.280179\n        5          60000.0          84000.0          100000.0      0.401877572016       40187.7572016          299061.213992          200000.0          80375.5144033          718685.699588          0.33437970329          240313.910987\n        6          65000.0          90000.0          100000.0      0.33489797668       33489.797668          332551.01166          150000.0          50234.6965021          782316.315158          0.300705745867          235247.011053\n        7          70000.0          96000.0          100000.0      0.279081647234       27908.1647234          360459.176383          100000.0          27908.1647234          832551.01166          0.277423926347          230969.570539\n        8          75000.0          102000.0          100000.0      0.232568039361       23256.8039361          383715.980319          50000.0          11628.4019681          872087.578351          0.260609422409          227274.240084\n        9          80000.0          108000.0          100000.0      0.193806699468       19380.6699468          403096.650266          0.0          0.0          903096.650266          0.248079461672          224039.730836\n        10          85000.0          114000.0          100000.0      0.16150558289       16150.558289          419247.208555          -50000.0          -8075.27914449          927322.4877          0.238522756883          221187.516286\n        11          90000.0          120000.0          100000.0      0.134587985742       13458.7985742          432706.007129          -100000.0          -13458.7985742          946164.805703          0.231103794152          218662.276491\n        12          95000.0          126000.0          100000.0      0.112156654785       11215.6654785          443921.672608          -150000.0          -16823.4982177          960745.170825          0.225264964904          216422.227187\n        13          100000.0          132000.0          100000.0      0.0934638789872       9346.38789872          453268.060506          -200000.0          -18692.7757974          971960.836304          0.220620001083          214434.000758\n        14          105000.0          138000.0          100000.0      0.0778865658226       7788.65658226          461056.717089          -250000.0          -19471.6414557          980528.358544          0.216893055222          212669.791417\n        15          110000.0          144000.0          100000.0      0.0649054715189       6490.54715189          467547.264241          -300000.0          -19471.6414557          987018.905696          0.213882119838          211105.69587\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.3 Page 152"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nIR=5.0;#in % per year\ni=15.0;#in % per year\n#Machine X :\nPpx=1500000.0;#in Rs.\nn=7.0;#in years\nS=200000.0;#in Rs.\nAMC=300000.0;#in Rs.\nprint \"End of year     AMC     InflationFactor     InflatedAmount     P/F     PW\";\nPw=0;#For initialising\nfor n in range (1,8) :\n    FP=(1+IR/100)**n\n    IA=AMC*FP;#in Rs.\n    PF=1/((1+i/100)**n);\n    PW=IA*PF;#in Rs.\n    Pw=Pw+PW;#in Rs.\n    print\"     \",n,\"        \",AMC,\"         \",FP,\"           \",IA,\"          \",PF,\"      \",PW;\n\nprint \"Present worth of inflated annual operating and maintenance cost in Rs. : \",round(Pw,3);\nPWX=Ppx+Pw-S*1/((1+i/100)**n);\nprint \"Present worth of machine X in Rs. : \",round(PWX,3);\n\n#Machine Y\n\nPpy=2000000.0;#in Rs.\nn=7.0;#in years\nS=300000.0;#in Rs.\nAMC=250000.0;#in Rs.\nprint \"End of year     AMC     InflationFactor     InflatedAmount     P/F     PW\";\nPw=0;#For initialising\nfor n in range (1,8) :\n    FP=(1+IR/100)**n\n    IA=AMC*FP;#in Rs.\n    PF=1/((1+i/100)**n);\n    PW=IA*PF;#in Rs.\n    Pw=Pw+PW;#in Rs.\n    print \"     \",n,\"        \",AMC,\"         \",FP,\"           \",IA,\"          \",PF,\"      \",PW;\n\nPWY=Ppy+Pw-S*1/((1+i/100)**n);\n\n#result\nprint \"Present worth of inflated annual operating and maintenance cost in Rs. : \",round(Pw,3);\nprint \"Present worth of machine Y in Rs. : \",round(PWY,3);\nprint \"Since the present worth of Machine X is less than Machine Y, select Machine X\"\nprint \" Calculations are not accurate in the book\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "End of year     AMC     InflationFactor     InflatedAmount     P/F     PW\n      1          300000.0           1.05             315000.0            0.869565217391        273913.043478\n      2          300000.0           1.1025             330750.0            0.756143667297        250094.517958\n      3          300000.0           1.157625             347287.5            0.657516232432        228347.168571\n      4          300000.0           1.21550625             364651.875            0.571753245593        208490.893043\n      5          300000.0           1.2762815625             382884.46875            0.497176735298        190361.25017\n      6          300000.0           1.34009564063             402028.692188            0.432327595912        173808.097981\n      7          300000.0           1.40710042266             422130.126797            0.375937039923        158694.35033\nPresent worth of inflated annual operating and maintenance cost in Rs. :  1483709.322\nPresent worth of machine X in Rs. :  2908521.914\nEnd of year     AMC     InflationFactor     InflatedAmount     P/F     PW\n      1          250000.0           1.05             262500.0            0.869565217391        228260.869565\n      2          250000.0           1.1025             275625.0            0.756143667297        208412.098299\n      3          250000.0           1.157625             289406.25            0.657516232432        190289.307142\n      4          250000.0           1.21550625             303876.5625            0.571753245593        173742.410869\n      5          250000.0           1.2762815625             319070.390625            0.497176735298        158634.375141\n      6          250000.0           1.34009564063             335023.910156            0.432327595912        144840.081651\n      7          250000.0           1.40710042266             351775.105664            0.375937039923        132245.291942\nPresent worth of inflated annual operating and maintenance cost in Rs. :  1236424.435\nPresent worth of machine Y in Rs. :  3123643.323\nSince the present worth of Machine X is less than Machine Y, select Machine X\n Calculations are not accurate in the book\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.4 Page 156"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nIR=6.0;#in % per year\ni=18.0;#in % per year\nAFR=5000000.0;#in Rs.\nn=7.0;#in years\nAI=500000.0;#in Rs.\nprint\"End of year            AFR         InflationFactor     InflatedAmount  P/F           PW\";\nTPW=0;#Initialising\n\nfor n in range (1,6) :\n    IF=(1+IR/100)**n;\n    IA=IF*AFR;#in Rs.\n    PF=1/((1+i/100)**n);\n    PW=PF*IA;#in Rs.\n    TPW=PW+TPW;#in Rs.\n    print\"       \",round(n),\"        \",round(AFR,3),\"        \",round(IF,3),\"        \",round(IA,3),\"    \",round(PF,3),\"     \",round(PW,3);\n    AFR=AFR+AI;#in Rs.\n\n    \n#result\nprint \"The value of the single deposit to be made now to recieve the specified series for the next five years is Rs. : \",round(TPW,3)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "End of year            AFR         InflationFactor     InflatedAmount  P/F           PW\n        1.0          5000000.0          1.06          5300000.0      0.847       4491525.424\n        2.0          5500000.0          1.124          6179800.0      0.718       4438236.139\n        3.0          6000000.0          1.191          7146096.0      0.609       4349334.645\n        4.0          6500000.0          1.262          8206100.24      0.516       4232615.212\n        5.0          7000000.0          1.338          9367579.043      0.437       4094655.134\nThe value of the single deposit to be made now to recieve the specified series for the next five years is Rs. :  21606366.553\n"
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter12_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter12_1.ipynb
new file mode 100755
index 00000000..75f40350
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter12_1.ipynb
@@ -0,0 +1,104 @@
+{
+ "metadata": {
+  "name": "EE-12"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Inventory Control"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.1 Page 160"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nD=24000.0;#in units/year\nCo=150.0;#in Rs./order\nPprice=75.0;#Rs./unit\nCpupy=18.0;#in % of Pprice/unit\n\n#calculation\nCc=Pprice*Cpupy/100;#in Rs.\nEOQ=sqrt((2*Co*D)/Cc);#in units\nn=D/round(EOQ);#no. of orders/year\nT=round(EOQ)/D;# time between successive orders in year\nT=T*12;#in months\nT=T*30;#in Days\n\n#result\nprint \"Economic order quantity in units : \",round(EOQ);\nprint \"No. of orders/year : \",round(n,2);\nprint \"Time between successive orders in days : \",round(T);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Economic order quantity in units :  730.0\nNo. of orders/year :  32.88\nTime between successive orders in days :  11.0\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.2 Page 162"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nr=12000.0;#in units/year\nk=24000.0;#in units/year\nCo=175.0;#in Rs./setup\nCc=15.0;#in Rs./unit/year\n\n#calclation\nEOQ=sqrt((2*Co*r)/(Cc*(1-r/k)));#in units\nQ=round(EOQ);#units\nt1=Q/k;#in year\nt1=t1*12*30;#in days\nt2=(Q/r)*(1-r/k);#in year\nt2=t2*12*30;#in days\n\n#result\nprint \"Economic order quantity in units : \",round(Q,3);\nprint \"Cycle time in days :\",round(t1+t2);\nprint \"No. of setups/year : \",round(r/Q,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Economic order quantity in units :  748.0\nCycle time in days : 22.0\nNo. of setups/year :  16.043\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.3 Page 164"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#iniatiation of variable\nfrom math import sqrt\nD=30000.0;#in units/year\nCc=2.0;#in Rs./unit/year\nCo=100.0;#in Rs./orde\nCs=12.0;#in units/year\n\n#calculation\nEOQ=sqrt(2*Co*D*(Cs+Cc)/(Cc*Cs));#in units\nQ=round(EOQ);#units\nQ1=sqrt(2*Co*D*Cs/(Cc*(Cs+Cc)));#in units\nt=Q/D;#in year\nt=t*365;#in days\nt1=Q1/D;#in year\nt1=t1*365;#in days\nt2=t-t1;#in days\n\n#result\nprint \"Ordering quantity in units : \",round(Q,3);\nprint \"Maximum Inventory in units : \",round(Q1);\nprint \"Maximum shortage quantity in units : \",round(Q-Q1);\nprint \"Cycle time in days: \",round(t);\nprint \"Inventory period(t1) in days : \",round(t1);\nprint \"Shortage period(t2) in days : \",round(t2);\nprint \"No. of Orders/year : \",round(D/Q,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Ordering quantity in units :  1871.0\nMaximum Inventory in units :  1604.0\nMaximum shortage quantity in units :  267.0\nCycle time in days:  23.0\nInventory period(t1) in days :  20.0\nShortage period(t2) in days :  3.0\nNo. of Orders/year :  16.034\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.4 Page 166"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nr=12000.0;#in units/year\nk=2000.0*12;#in units/year\nCo=400.0;#in Rs./setup\nCc=0.20*12;#in Rs./year\nCs=15.0;#in Rs./unit/year\n\n#calculation\nEOQ=sqrt((2*Co*k*r*(Cc+Cs)/(Cc*(k-r)*Cs)));#in units\nQ=round(EOQ);#units\nQ1=sqrt(2*Co*Cc*r*(k-r)/(Cs*(Cc+Cs)*k));#in units\nQ2=(Q*(k-r)/k)-Q1;#in Units\nt=Q/r;#in year\nt=t*365.0;#in days\nt1=(Q2/r)*365.0;#in days\nt2=(Q2/(k-r))*365.0;#in days\nt3=(Q1/(k-r))*365.0;#in days\nt4=(Q1/r)*365.0;#in days\n\n#result\nprint \"Ordering quantity in units : \",round(Q,3);\nprint \"Maximum Inventory in units : \",round(Q1);\nprint \"Maximum shortage quantity in units : \",round(Q2);\nprint \"Cycle time in days: \",round(t);\nprint \"Period of shortage in days : \",round(t1);\nprint \"Period of production satisfying back order in days : \",round(t2);\nprint \"Period of production satisfying period requirement in days : \",round(t3);\nprint \"Period of consumption only in days : \",round(t4);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Ordering quantity in units :  3046.0\nMaximum Inventory in units :  210.0\nMaximum shortage quantity in units :  1313.0\nCycle time in days:  93.0\nPeriod of shortage in days :  40.0\nPeriod of production satisfying back order in days :  40.0\nPeriod of production satisfying period requirement in days :  6.0\nPeriod of consumption only in days :  6.0\n"
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter13_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter13_1.ipynb
new file mode 100755
index 00000000..af23fcbb
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter13_1.ipynb
@@ -0,0 +1,104 @@
+{
+ "metadata": {
+  "name": "EE-13"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Make or Buy Decision"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.1 Page 168"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmaterial=300.0;#in Rs.\nlabour=250.0;#in Rs.\noverhead=100.0;#in Rs.\n\n#calculation\nVC=material+labour+overhead;#in Rs.\ndemand=5000.0;#in units\nTVC=demand*VC;#in Rs.\nFC=1000000;#in Rs.\nTC=FC+TVC;#in Rs.\n\n#result\nprint \"Total cost in Rs. : \",round(TC,3);\n\n#Cost to buy :\nPcost=900*demand;#in Rs.\nFC=1000000.0;#in Rs.\nTC=FC+Pcost;#in Rs.\n\n#result\nprint \"Total cost in Rs. : \",round(TC,3);\nprint \"The cost of making fixtures is less than the cost of buying fixtures from outside. Therefore, the organisation should make the fixtures.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Total cost in Rs. :  4250000.0\nTotal cost in Rs. :  5500000.0\nThe cost of making fixtures is less than the cost of buying fixtures from outside. Therefore, the organisation should make the fixtures.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.2 Page 171"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nPbuy=8.0;#in Rs/unit\nD=2000.0;#in units/year\nCo=120.0;#in Rs./order\nCc=1.60;#in Rs./units/year\n\n#calculation\nQ1=sqrt(2*Co*D/Cc);#in units\nTC=D*Pbuy+D*Co/Q1+Q1*Cc/2;#in Rs.\nprint \"Total cost of buying in Rs. : \",round(TC,3)\n\n#Make Option :\nPmake=5.0;#in Rs/unit\nCo=60.0;#in Rs./setup\nCc=1.0;#in Re1/units/year\nr=2000.0;#in units/year\nk=8000.0;#in units/year\n\n#calculation\nQ2=sqrt(2*Co*r/(Cc*(1-r/k)));#in units\nTC=D*Pmake+D*Co/Q2+Q2*Cc*(k-r)/(2*k);#in Rs.\n\n#result\nprint \"Total cost of making in Rs. : \",round(TC,3)\nprint\"The cost of making is less than the cost of buying . Therefore, the firm should go in for making option.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Total cost of buying in Rs. :  16876.356\nTotal cost of making in Rs. :  10424.264\nThe cost of making is less than the cost of buying . Therefore, the firm should go in for making option.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.3 Page 173"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nSP=500.0;#in Rs.\nVC=300.0;#in Rs.\nFC=400000.0;#in RS.\n\n#calculation\nBEP=FC/(SP-VC);#in units\n\n#results\nprint \"BEP in units : \",round(BEP,3);\nprint\"Since the demand(1500units) is less than the break even quantity, the company should buy the cabinets for its TV production.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "BEP in units :  2000.0\nSince the demand(1500units) is less than the break even quantity, the company should buy the cabinets for its TV production.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.4 Page 173"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nvolume=8000;#in units\n#Process A :\nFC=500000.0;#in RS.\nVC=175.0;#in Rs.\n\n#calculation\nAC=FC+VC*volume;#in Rs.\n\n#result\nprint \"Annual Cost of Process A in Rs. :\",round(AC,3);\n\n#Process B :\nFC=600000.0;#in RS.\nVC=150;#in Rs.\n\n#calculation\nAC=FC+VC*volume;#in Rs.\n\n#result\nprint \"Annual Cost of Process A in Rs. :\",round(AC,3);\n\n#Buy option:\nPprice=125.0;#in RS./Unit\nVC=175.0;#in Rs.\nAC=Pprice*volume;#in Rs\n\n#result.\nprint \"Annual Cost of Buy in Rs. :\",round(AC,3);\nprint\"Since the annual cost of buy option is the minimum among all the alternatives, the company should buy the product.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Annual Cost of Process A in Rs. : 1900000.0\nAnnual Cost of Process A in Rs. : 1800000.0\nAnnual Cost of Buy in Rs. : 1000000.0\nSince the annual cost of buy option is the minimum among all the alternatives, the company should buy the product.\n"
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter16_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter16_1.ipynb
new file mode 100755
index 00000000..86e57be5
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter16_1.ipynb
@@ -0,0 +1,134 @@
+{
+ "metadata": {
+  "name": "Untitled2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Linear Progrmming"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.1 Page 200"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#result\nprint\"The data of the problem are summarized below : \";\nprint\"Machine                   Products           Limit on \";\nprint\"                        P1         P2       machine hours\";\nprint\"Lathe                    5         10            60\";\nprint\"Milling                  4          4            40\";\nprint\"Profit/unit              6          8\";\nprint\"Let X1 be the production volume of the product.P1, and\";\nprint\"X2 be the production volume of the product,P2.\";\nprint\"The corresponding linear programming model to determine the production volume of each product such that the total profit is maximized is as shown below : \";\nprint\"maximize Z = 6*X1 + 8*X2\";\nprint\"subject to\";\nprint\"5*X1+10*X2 <= 60\"\nprint\"4*X1+4*X2 <= 40\"\nprint\"X1,X2 >= 0\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The data of the problem are summarized below : \nMachine                   Products           Limit on \n                        P1         P2       machine hours\nLathe                    5         10            60\nMilling                  4          4            40\nProfit/unit              6          8\nLet X1 be the production volume of the product.P1, and\nX2 be the production volume of the product,P2.\nThe corresponding linear programming model to determine the production volume of each product such that the total profit is maximized is as shown below : \nmaximize Z = 6*X1 + 8*X2\nsubject to\n5*X1+10*X2 <= 60\n4*X1+4*X2 <= 40\nX1,X2 >= 0\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.2 Page 200"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nprint\"Let X1 be the No. of packets of food type1 suggested for babies, and\";\nprint\"X2 be the No. of packets of food type1 suggested for babies.\";\nprint\"The corresponding linear programming model to determine the No. of packets of each food type to be suggested for babies with the minimum cost such that the minimum daily required vitamin in each food type is satisfied is as shown below : \";\nprint\"maximize Z = 2*X1 + 3*X2\";\nprint\"subject to\";\nprint\"X1+X2 >= 6\"\nprint\"7*X1+X2 >= 14\";\nprint\"X1,X2 >= 0\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Let X1 be the No. of packets of food type1 suggested for babies, and\nX2 be the No. of packets of food type1 suggested for babies.\nThe corresponding linear programming model to determine the No. of packets of each food type to be suggested for babies with the minimum cost such that the minimum daily required vitamin in each food type is satisfied is as shown below : \nmaximize Z = 2*X1 + 3*X2\nsubject to\nX1+X2 >= 6\n7*X1+X2 >= 14\nX1,X2 >= 0\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.3 Page201"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n%pylab inline\nimport matplotlib.pyplot as plt\nprint\"Given the following LP model :\"\nprint\"maximize Z = 6*X1 + 8*X2\";\nprint\"subject to\";\nprint\"5*X1+10*X2 <= 60\";\nprint\"4*X1+4*X2 <= 40\";\nprint\"X1,X2 >= 0\";\nprint\"The introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\";\nprint\"Compute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \";\nprint\"5*X1+10*X2 <= 60\";\nprint\"When X1=0 : X2=6\";\nprint\"When X2=0 : X1=12\";\nplt.plot([0,12],[6,0],'r')\nplt.plot([10,0],[0,10])\nplt.title('Graphical Plot')\nplt.show()\nprint \"Consider the 2nd constraint in the form : in blue \";\nprint \"4*X1+4*X2 <= 40\";\nprint \"When X1=0 : X2=10\";\nprint \"When X2=0 : X1=10\";\nprint \"The closed polygon is the feasible region at each of the corner points of the closed polygon is computed as follows by substituting its coordinates in the objective function :\";\nZA=6*0+8*0;\nZB=6*10+8*0;\nZC=6*8+8*2;\nZD=6*0+8*6;\nprint \"Since the type of the objective function is maximization, the solution corresponding to the maximum Z value should be selected as the optimum solution. The Z value is maximum for the corner point C. Hence, the corresponding solution is \";\nprint \"X1 = 8  X2 = 2 and Z(Optimum) is\",ZC;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Populating the interactive namespace from numpy and matplotlib\nGiven the following LP model :\nmaximize Z = 6*X1 + 8*X2\nsubject to\n5*X1+10*X2 <= 60\n4*X1+4*X2 <= 40\nX1,X2 >= 0\nThe introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\nCompute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \n5*X1+10*X2 <= 60\nWhen X1=0 : X2=6\nWhen X2=0 : X1=12\n"
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": "<matplotlib.figure.Figure at 0x51094e0>"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Consider the 2nd constraint in the form : in blue \n4*X1+4*X2 <= 40\nWhen X1=0 : X2=10\nWhen X2=0 : X1=10\nThe closed polygon is the feasible region at each of the corner points of the closed polygon is computed as follows by substituting its coordinates in the objective function :\nSince the type of the objective function is maximization, the solution corresponding to the maximum Z value should be selected as the optimum solution. The Z value is maximum for the corner point C. Hence, the corresponding solution is \nX1 = 8  X2 = 2 and Z(Optimum) is 64\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.4 Page 203"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n%pylab inline\nimport matplotlib.pyplot as plt\nprint\"Given the following LP model :\"\nprint \"minimize Z = 2*X1 + 3*X2\";\nprint\"subject to\";\nprint\"X1+X2 >= 6\";\nprint\"7*X1+X2 >= 14\";\nprint\"X1,X2 >= 0\";\nprint\"The introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\";\nprint\"Compute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \";\nprint\"X1+X2 = 6\";\nprint\"When X1=0 : X2=6\";\nprint\"When X2=0 : X1=6\";\nplt.plot([0,6],[6,0],'r')\nplt.plot([2,0],[0,14])\nplt.title('Graphical Plot')\nplt.show()\nprint\"Consider the 2nd constraint in the form  (in blue):\";\nprint\"7*X1+X2 = 14\";\nprint\"When X1=0 : X2=14\";\nprint\"When X2=0 : X1=2\";\nprint\"The Optimum solution will be in any one of the corners A, B and C\";\nprint\"The objective function value at each of these corner points of the feasible solution space is computed as fllows by substituting its coordinates in the objective function.\"\nZA=2*0+3*14;\nZB=2*(4.0/3)+3*(14.0/3);\nZC=2*6+3*0;\n\n#result\nprint\"Since the type of the objective function is minimization, the solution corresponding to the minimum Z value should be selected as the optimum solution. The Z value is minimum for the corner point C. Hence, the corresponding solution is \";\nprint \"X1 = 6  X2 = 0 and Z(Optimum) =\",ZC",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Populating the interactive namespace from numpy and matplotlib\nGiven the following LP model :\nminimize Z = 2*X1 + 3*X2\nsubject to\nX1+X2 >= 6\n7*X1+X2 >= 14\nX1,X2 >= 0\nThe introduction of non-negative constraints X1>=0 and X2>=0 will eliminate the 2nd, 3rd and 4th quadrants of XY plane.\nCompute the cordinates to plot equations relting to the constraints on the XY plane as shown below : \nX1+X2 = 6\nWhen X1=0 : X2=6\nWhen X2=0 : X1=6\n"
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": "<matplotlib.figure.Figure at 0x81f27b8>"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Consider the 2nd constraint in the form  (in blue):\n7*X1+X2 = 14\nWhen X1=0 : X2=14\nWhen X2=0 : X1=2\nThe Optimum solution will be in any one of the corners A, B and C\nThe objective function value at each of these corner points of the feasible solution space is computed as fllows by substituting its coordinates in the objective function.\nSince the type of the objective function is minimization, the solution corresponding to the minimum Z value should be selected as the optimum solution. The Z value is minimum for the corner point C. Hence, the corresponding solution is \nX1 = 6  X2 = 0 and Z(Optimum) = 12\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "Example 16.5 Page",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter1_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter1_1.ipynb
new file mode 100755
index 00000000..64198a89
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter1_1.ipynb
@@ -0,0 +1,83 @@
+{
+ "metadata": {
+  "name": "EE-1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Introduction"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.1 Page 10"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of Variable\nFC=2000000.0;#in Rs\nv=100.0;#in Rs\ns=200.0;#in Rs\nQ=60000.0;#in units\n\n#Part a \n# calculation\nBEQ=FC/(s-v);#in units\n\n#result\nprint \"Break even quantity in units : \",round(BEQ,3)\n\n#Part b \n#calculation\nBES=FC*s/(s-v);#in Rs\n\nprint \"Break even sales in Rs. : \",round(BES,3)\n\n#Part c \n#calculation\nCon=s*Q-v*Q;#in Rs\nMos=Q*s-BES;#in Rs\n\n#result\nprint \"Contribution in units : \",round(Con,3)\nprint \"Margin of safety in Rs. : \",round(Mos,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Break even quantity in units :  20000.0\nBreak even sales in Rs. :  4000000.0\nContribution in units :  6000000.0\nMargin of safety in Rs. :  8000000.0\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.2,Page 12"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nFC=25000.0;#in Rs\nv=45000.0;#in Rs\ns=120000.0;#in Rs\n\n#Part a \n#calculation\nCon=s-v;#in Rs\n\n#result\nprint \"Contribution in Rs  \",round(Con,3)\n\n#Part b \n#calculation\nProf=Con-FC;#in Rs\n\n#result\nprint \"Profit in Rs  \",round(Prof,3);\n\n#Part c \n#calculation\nPVratio=Con*100.0/s;#in %\nBEP=FC*100.0/PVratio;#in Rs\n\n#result\nprint \"BEP in Rs \", round(BEP,3)\n\n#Part d \n#calculoation\nMS=Prof*100/PVratio;#in Rs\n\n#result\nprint\"M.S. in Rs  \", round(MS,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Contribution in Rs   75000.0\nProfit in Rs   50000.0\nBEP in Rs  40000.0\nM.S. in Rs   80000.0\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.3 Page 13"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nFC=15000.0;#in Rs\nv=35000.0;#in Rs\ns=80000.0;#in Rs\n\n#Part a \n#calculation\nCon=s-v;#in Rs\n\n#result\nprint \"Contribution in Rs\", round(Con,3);\n\n#Part b \n#calcualtion\nProf=Con-FC;#in Rs\n\n#result\nprint \"Profit in Rs \", round(Prof,3)\n\n#Part c \n#calculation\nPVratio=Con*100/s;\nBEP=FC*100/PVratio;#in Rs\n\n#result\nprint \"BEP in Rs  \",round(BEP,3)\n\n#Part d \n#calculation\nMS=Prof*100/PVratio;#in Rs\n\nprint \"M.S. in Rs\",round(MS,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Contribution in Rs 45000.0\nProfit in Rs  30000.0\nBEP in Rs   26666.667\nM.S. in Rs 53333.333\n"
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter2_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter2_1.ipynb
new file mode 100755
index 00000000..d848674b
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter2_1.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "EE-2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Elementary Economic Analysis"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.1 Page 16"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#Part a \nw1=1.2;#in Kg\nc1=80.0;#cost of making aluminium casting in Rs/Kg\nc2=150.0;#ost of machining aluminium casting per unit in Rs\n\n#calculation\nTc1=c1*w1+c2;#Total cost of jet engine part made of aluminium per unit in Rs\n\n#result\nprint \"Total cost of jet engine part made of aluminium per unit in Rs \",round(Tc1,3)\n\n#Part b \nw2=1.35;#in Kg\nc1=35.0;#in Rs/Kg\nc2=170.0;#in Rs\nc3=1300.0;#in Rs/Kg\n\n#calculation\nTc2=c1*w2+c2+c3*(w2-w1);#in Rs\n\n#result\nprint \"Total cost of jet engine part made of steel per unit in Rs : \",round(Tc2,3);\nprint \"DECISION : The total cost/unit of a jet engine part made of aluminium is less than that for an engine made of steel. Hence, aluminium is suggested for making jet engine part. The economic advantage of aluminium over steel per unit in Rs  \",round(Tc2-Tc1,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Total cost of jet engine part made of aluminium per unit in Rs  246.0\nTotal cost of jet engine part made of steel per unit in Rs :  412.25\nDECISION : The total cost/unit of a jet engine part made of aluminium is less than that for an engine made of steel. Hence, aluminium is suggested for making jet engine part. The economic advantage of aluminium over steel per unit in Rs   166.25\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.2, Page 17"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#Part a \nwood=0.1;#in m^3\nWoodCost=12000.0;#in Rs/m^3\nTable=1;#in units\nTableTopCost=3000.0;#in Rs/unit\nLegBushes=4.0;#units\nLegBushesCost=10.0;#Rs/units\nNails=100.0;#in grams\nNailsCost=300.0;#in Rs/Kg\nTotalLabour=15.0;#in Hours\nTotalLabourCost=50.0;#in Rs/Hours\n\n#calcualtion\nWoodCostframelegs=WoodCost*wood;#in Rs\nWoodTopCost=3000.0;#in Rs\nBushesCost=LegBushesCost*LegBushes;#in Rs\nNailsCost=Nails*NailsCost/1000;#in Rs\nLabourCost=TotalLabourCost*TotalLabour;#in Rs\nTotalCost1=WoodCostframelegs+WoodTopCost+BushesCost+NailsCost+LabourCost;#in Rs\n\n#result\nprint \"Cost of Table with wooden top in Rs \",round(TotalCost1,3)\n\n#given data for table with granite top\n#Part b \nwood=0.15;#in m^3\nWoodCost=12000.0;#in Rs/m^3\nGranite=1.62;#in m^2\nGraniteCost=800.0;#in Rs/m^2\nLegBushes=4.0;#units\nLegBushesCost=25.0;#Rs/units\nNails=50.0;#in grams\nNailsCost=300.0;#in Rs/Kg\nTotalLabour=8.0;#in Hours\nTotalLabourCost=50.0;#in Rs/Hours\n\n\nWoodCostframelegs=WoodCost*wood;#in Rs\nGraniteTopCost=Granite*GraniteCost;#in Rs\nBushesCost=LegBushesCost*LegBushes;#in Rs\nNailsCost=Nails*NailsCost/1000;#in Rs\nLabourCost=TotalLabourCost*TotalLabour;#in Rs\nTotalCost2=WoodCostframelegs+GraniteTopCost+BushesCost+NailsCost+LabourCost;#in Rs\n\n#result\nprint \"Cost of Table with Granite top in Rs \",round(TotalCost2,3);\nprint \"Economic advantage of table with granite top in Rs  \",round(TotalCost1-TotalCost2,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Cost of Table with wooden top in Rs  5020.0\nCost of Table with Granite top in Rs  3611.0\nEconomic advantage of table with granite top in Rs   1409.0\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.3 Page 19"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#design A\nLatheCost=200.0;#in Rs/hour\ngrinderCost=150.0;#in Rs/hour\nHoursOfLathe=16.0;#in hours/1000Unit\nHoursOfGrinder=4.5;#in hours/1000Unit\n\n#calcualtion\nTotalCostA=LatheCost*HoursOfLathe+grinderCost*HoursOfGrinder;#in Rs/1000unit\n\n#result\nprint \"Total cost of design A per 100,000 units  \",round(TotalCostA*100000.0/1000,3);\n\n# Design B\nHoursOfLathe=7.0;#in hours/1000Unit\nHoursOfGrinder=12.0;#in hours/1000Unit\nTotalCostB=LatheCost*HoursOfLathe+grinderCost*HoursOfGrinder;#in Rs/1000unit\n\n#result\nprint \"Total cost of design A per 100,000 units\",round(TotalCostB*100000.0/1000,3);\nprint \"Economic advantage of design B over design A per 100,000 units in Rs  \",round(TotalCostA-TotalCostB,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Total cost of design A per 100,000 units   387500.0\nTotal cost of design A per 100,000 units 320000.0\nEconomic advantage of design B over design A per 100,000 units in Rs   675.0\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.4,Page 20"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nfrom math import pi\nTanks=4.0;#units\nTankDia=5.2;#in meter\n\n#calcualtion\nTankRad=TankDia/2;#in meters\nTankHeight=7;#in meters\nHeightDiaRatio=TankHeight/TankDia;#unitless\nVolPerTank=(22/7)*TankRad**2*TankHeight;#in m^3\nh=VolPerTank/(pi)*64;#in meters\nr=h/8;#in meters\nd=2*r;#in meters\nCostNewDesign=900000*(100.0/111);#in Rs\n\n#result\nprint \"Expected savings by redesign in Rs  \",round(900000-CostNewDesign,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Expected savings by redesign in Rs   89189.189\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.5,Page 21"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#steel frame\ndistance=2500.0;#in Km\nTransCost=1.0;#in Rs/Kg/100Km\nSteelFramePrice=1000.0;#in Rs/Unit\nSteelFrameWeight=75.0;#in Kg/Unit\n\n#calculation\nTotalCost1=SteelFramePrice+TransCost*SteelFrameWeight*distance/100;#in Rs\n\n#result\nprint \"Total cost of steel window frame per unit in Rs  \",round(TotalCost1,3);\n\n# Aluminium window frame\nAlumilniumFramePrice=1500.0;#in Rs/Unit\nAlumilniumFrameWeight=28.0;#in Kg/Unit\n\n#calculation\nTotalCost2=AlumilniumFramePrice+TransCost*AlumilniumFrameWeight*distance/100;#in Rs\n\n#result\nprint \"Total cost of Alumilnium window frame per unit in Rs \",round(TotalCost2,3);\nprint \"DECISION : The total cost per unit of the aluminium window frame is less than that of steel window frame. Hence, Alumilnium window frame is recommended. The Economic advantage per unit of the Alumilnium window frame over steel window frame in Rs  \",round(TotalCost1-TotalCost2,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Total cost of steel window frame per unit in Rs   2875.0\nTotal cost of Alumilnium window frame per unit in Rs  2200.0\nDECISION : The total cost per unit of the aluminium window frame is less than that of steel window frame. Hence, Alumilnium window frame is recommended. The Economic advantage per unit of the Alumilnium window frame over steel window frame in Rs   675.0\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.6,Page 23"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#Cost of component using process sequence 1\nprint\"The process sequence 1 of the component is as follows : Turning - Milling - Shaping - Drilling\" ;\nprint \"Calculations are summarized in form of table below : \";\nprint \"Operation     Operation       Time       Machine Hour rate     Cost\";\nprint \"   No.                     Min    Hour           Rs.             Rs.\";\nprint \"   1           Turning       5   0.083          200           16.60\";\nprint \"   2           Milling       8   0.133          400           53.20\";\nprint \"   3          Shapiing      10   0.167          350           58.45\";\nprint \"   4          Drilling       3   0.050          300           15.00\";\nprint \"                                                   Total     143.25\";\n\n#Cost of component using process sequence 2\nprint \"The process sequence 2 of the component is as follows : Turning - Milling - Drilling\" ;\nprint \"Calculations are summarized in form of table below : \";\nprint \"Operation     Operation       Time       Machine Hour rate     Cost\";\nprint \"   No.                     Min    Hour           Rs.             Rs.\";\nprint \"   1           Turning       5   0.083          200           16.60\";\nprint \"   2           Milling      14   0.233          400           93.20\";\nprint \"   4          Drilling       3   0.050          300           15.00\";\nprint \"                                                   Total     124.80\";\n\n#Cost of component using process sequence 3\nprint \"The process sequence 3 of the component is as follows : Only CNC operations\" ;\nprint \"Calculations are summarized in form of table below : \";\nprint \"Operation     Operation       Time       Machine Hour rate     Cost\";\nprint \"   No.                     Min    Hour           Rs.             Rs.\";\nprint \"   1             CNC         8   0.133         1000             133\";\n\nprint \"The process sequence 2 has the least cost. Therefore, it should be selected for manufacturing the component.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The process sequence 1 of the component is as follows : Turning - Milling - Shaping - Drilling\nCalculations are summarized in form of table below : \nOperation     Operation       Time       Machine Hour rate     Cost\n   No.                     Min    Hour           Rs.             Rs.\n   1           Turning       5   0.083          200           16.60\n   2           Milling       8   0.133          400           53.20\n   3          Shapiing      10   0.167          350           58.45\n   4          Drilling       3   0.050          300           15.00\n                                                   Total     143.25\nThe process sequence 2 of the component is as follows : Turning - Milling - Drilling\nCalculations are summarized in form of table below : \nOperation     Operation       Time       Machine Hour rate     Cost\n   No.                     Min    Hour           Rs.             Rs.\n   1           Turning       5   0.083          200           16.60\n   2           Milling      14   0.233          400           93.20\n   4          Drilling       3   0.050          300           15.00\n                                                   Total     124.80\nThe process sequence 3 of the component is as follows : Only CNC operations\nCalculations are summarized in form of table below : \nOperation     Operation       Time       Machine Hour rate     Cost\n   No.                     Min    Hour           Rs.             Rs.\n   1             CNC         8   0.133         1000             133\nThe process sequence 2 has the least cost. Therefore, it should be selected for manufacturing the component.\n"
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter3_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter3_1.ipynb
new file mode 100755
index 00000000..12f9a0ae
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter3_1.ipynb
@@ -0,0 +1,209 @@
+{
+ "metadata": {
+  "name": "EE-3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Interest Formulas and their Application"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.1 Page 29"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nP=20000.0;#in rupees\nn=10.0;#in years\ni=18.0;#% per annum\n\n#calculation\nF=P*(1+i/100)**n;\n\n#result\nprint \"Maturity value after 18 years is  \", round(F,3),\" Rupees.\";\nprint \" Ans in the book is not correct\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Maturity value after 18 years is :  104676.711  Rupees.\n Ans in the book is not correct\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.2 Page 30"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nF=100000.0;#in rupees\nn=10.0;#in years\ni=15.0;#% per annum\n\n#calculation\nP=F/((1.0+i/100.0)**n);\n\n#result\nprint \"The person has to invest  \", round(P,3),\" Rupees.\";\nprint \" Ans in the book is not correct\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The person has to invest   24718.471  Rupees.\n Ans in the book is not correct\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 3.3 Page 31"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nA=10000.0;#in rupees\nn=25.0;#in years\ni=20.0;#% per annum\n\n#calculation\nF=A*(((1+i/100)**n-1)/(i/100));\n\n#result\nprint \"The future sum of the annual equal payment after 25 years is \",round(F,3),\" Rupees.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The future sum of the annual equal payment after 25 years is  4719810.832  Rupees.\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.4 Page 32"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nF=500000.0;#in rupees\nn=15.0;#in years\ni=18.0;#% per annum\n\n#calculation\nA=F*((i/100)/((1+i/100)**n-1));\n\n#result\nprint \"The annual equal ammount which must be deposited for 15 years is  \",round(A,3),\" Rupees.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equal ammount which must be deposited for 15 years is   8201.391  Rupees.\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 3.5 Page 33"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiasation of variable\nA=1000000.0;#in rupees\nn=20.0;#in years\ni=15.0;#% per annum\n\n#calculation\nP=A*(((1+i/100)**n-1)/((i/100)*(1+i/100)**n));\n\n#result\nprint \"The amount of reserve which must be setup now is : \",round(P,3),\" Rupees\";\nprint \" Ans in the book is not correct\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The amount of reserve which must be setup now is :  6259331.474  Rupees\n Ans in the book is not correct\n"
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.6 Page 34"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\nP=1000000.0;#in rupees\nn=15.0;#in years\ni=18.0;#% per annum\n\n#calculation\nA=P*(((i/100)*(1+i/100)**n)/((1+i/100)**n-1));\n\n#result\nprint \"The annual equivalent installment to be paid by the company to the bank is \",round(A),\" Rupees.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent installment to be paid by the company to the bank is :  196403.0  Rupees.\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.7 Page 35"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nA1=4000.0;#in rupees\nG=500.0;#in rupees\nn=10.0;#in years\ni=15.0;#% per annum\n\n#calculation\nA=A1+G*(((1+i/100)**n-(i/100)*n-1)/((i/100)*(1+i/100)**n-(i/100)));\nF=A*(((1+i/100)**n-1)/(i/100));\n\n#result\nprint \"At the end of 10th year, the compound amountr of all his payments will be : \",round(F,3),\" Rupees.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "At the end of 10th year, the compound amountr of all his payments will be :  115560.6  Rupees.\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.8 Page 36"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nA1=8500.0;#in rupees\nG=-500.0;#in rupees\nn=10.0;#in years\ni=15.0;#% per annum\n\n#calculation\nA=A1+G*(((1+i/100)**n-(i/100)*n-1)/((i/100)*(1+i/100)**n-(i/100)));\nF=A*(((1+i/100)**n-1)/(i/100));\n\n#result\nprint \"At the end of 10th year, the compound amountr of all his payments will be : \",round(F,3),\" Rupees.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "At the end of 10th year, the compound amountr of all his payments will be :  138235.878  Rupees.\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.9 Page 38"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=5000.0;#in rupees\nn=10.0;#in years\ni=12.0;#% per annum\nm=4.0;#no. of interest periods per year for quarterly\n\n#calculation\nN=n*m;\nr=i/m;\nF=P*(1+r/100)**N;\n\n#result\nprint\"Maturity value after 10 years is  \",round(F,3),\" Rupees.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Maturity value after 10 years is :  16310.189  Rupees.\n"
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter4_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter4_1.ipynb
new file mode 100755
index 00000000..fc61fff4
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter4_1.ipynb
@@ -0,0 +1,167 @@
+{
+ "metadata": {
+  "name": "EE-4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Present Worth Method of Comparision"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.1 Page 44"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#technology 1\nP=1200000.0;#in Rs\nA=400000.0;#in Rs\ni=20.0;#in % per annum\nn=10.0;#in years\n\n#calculation\nPW=-P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n\n#result\nprint \"The present worth for this technology in RS.  \",round(PW,3);\n\n#technology 2\nP=2000000.0;#in Rs\nA=600000.0;#in Rs\ni=20.0;#in % per annum\nn=10.0;#in years\n\n#calculation\nPW=-P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n\n#result\nprint \"The present worth for this technology in RS.  \",round(PW,3);\n\n#technology 3\nP=1800000.0;#in Rs\nA=500000.0;#in Rs\ni=20.0;#in % per annum\nn=10.0;#in years\n\n#calculation\nPW=-P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n\n#result\nprint \"The present worth for this technology in RS.  \",round(PW,3);\nprint \"It is clear from the calculations that the present worth of technology 2 is the highest among all technologies. Therefore technology 2 is suggested for implementation to expand the production.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth for this technology in RS.   476988.834\nThe present worth for this technology in RS.   515483.251\nThe present worth for this technology in RS.   296236.043\nIt is clear from the calculations that the present worth of technology 2 is the highest among all technologies. Therefore technology 2 is suggested for implementation to expand the production.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.2 Page 46"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#bid 1\nP=450000.0;#in Rs\nA=27000.0;#in Rs\ni=15.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nPW=P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n\n#result\nprint \"The present worth for this bid in RS. \",round(PW,3)\n\n#bid 2\nP=540000.0;#in Rs\nA=28500.0;#in Rs\ni=15.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nPW=P+A*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n\n#result\nprint \"The present worth for this bid in RS. \",round(PW,3)\nprint \"The total present worth cost of bid 1 is less than that of bid 2. Hence bid 1 is to be selected for implementation. That is, the Elevator from Alpha Elevator Inc. is to be purchased and installed in the new building.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth for this bid in RS.  607878.993\nThe present worth for this bid in RS.  706650.048\nThe total present worth cost of bid 1 is less than that of bid 2. Hence bid 1 is to be selected for implementation. That is, the Elevator from Alpha Elevator Inc. is to be purchased and installed in the new building.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.3 Page 47"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#investment proposal A\nP1=-10000.0;#in Rs\nP2=3000.0;#in Rs\nP3=3000.0;#in Rs\nP4=7000.0;#in Rs\nP5=6000.0;#in Rs\ni=18.0;#in % per annum\n\n#calculation\nPW_A=P1+P2*1/((1+i/100)**1)+P3*1/((1+i/100)**2)+P4*1/((1+i/100)**3)+P5*1/((1+i/100)**4);#in RS\n\n#result\nprint \"The present worth of A in RS.  \",round(PW_A,3);\n\n#investment proposal B\nP1=-10000.0;#in Rs\nP2=6000.0;#in Rs\nP3=6000.0;#in Rs\nP4=3000.0;#in Rs\nP5=3000.0;#in Rs\ni=18.0;#in % per annum\n\n#calculation\nPW_B=P1+P2*1/((1+i/100)**1)+P3*1/((1+i/100)**2)+P4*1/((1+i/100)**3)+P5*1/((1+i/100)**4);#in RS\n\n#result\nprint \"The present worth of B in RS.  \",round(PW_B,3);\nprint\"At i=18%, the present worth of proposal B is higher than that of proposal A. Therefore, select proposal B.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth of A in RS.   2052.076\nThe present worth of B in RS.   2767.112\nAt i=18%, the present worth of proposal B is higher than that of proposal A. Therefore, select proposal B.\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.4 Page 48"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP0=400000.0;#in Rs\nP=200000.0;#in Rs\ni=18.0;#in % per annum\nn=10.0;#in years\n\n#calcualtion\nPW=P0+P*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n\n#result\nprint \"The present worth of alternative 2 in RS. \",round(PW,3);\nprint \"The present worth of 2nd alternative is less than that of first one i.e., complete downpayment of Rs. 1600000. Hence, select 2nd alternative.\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth of alternative 2 in RS.  1298817.259\nThe present worth of 2nd alternative is less than that of first one i.e., complete downpayment of Rs. 1600000. Hence, select 2nd alternative.\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.5 Page 49"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#plan 1\nP0=-1000.0;#in Rs\nP=12000.0;#in Rs\ni=12.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nPW1=P0+P*1/((1+i/100)**n);#in RS\n\n#result\nprint \"The present worth of Plan-1 in RS.  \",round(PW1,3);\n\n#plan 2\nP0=-1000.0;#in Rs\nP=4000.0;#in Rs\ni=12.0;#in % per annum\nn1=10.0;#in years\nn2=15.0;#in years\n\n#calculation\nPW2=P0+P*1/((1+i/100)**n1)++P*1/((1+i/100)**n2);#in RS\n\n#result\nprint \"The present worth of Plan-2 in RS.\",round(PW2,3);\nprint \"The present worth of Plan-1 is more than that of Plan-1. Therefore plan 1 is the best plan from the investors point of view.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth of Plan-1 in RS.   1192.355\nThe present worth of Plan-2 in RS. 1018.678\nThe present worth of Plan-1 is more than that of Plan-1. Therefore plan 1 is the best plan from the investors point of view.\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.6,Page 50"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#plan 1\nP0=-10000.0;#in Rs\nP=800000.0;#in Rs\ni=12.0;#in % per annum\nn=20.0;#in years\n\n#calcualtion\nPW1=P0*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n)+P*1/((1+i/100)**n);#in RS\n\n#result\nprint \"The present worth of Plan-1 in RS. \",round(PW1,3);\n\n#plan 2\nP0=-10000.0;#in Rs\nP=1500000.0;#in Rs\ni=12.0;#in % per annum\nn1=20.0;#in years\nn2=25.0;#in years\n\n#calcualtion\nPW2=P0*(((1+i/100)**n1)-1)/((i/100)*(1+i/100)**n1)+P*1/((1+i/100)**n2);#in RS\n\n#result\nprint \"The present worth of Plan-2 in RS. \",round(PW2,3);\nprint \"The present worth of Innovative Investment Ltd. Plan is more than that of Novel Investment Ltd. Plan. Therefore Innovative Investment Ltd. Plan is the best plan from the investors point of view.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth of Plan-1 in RS.  8238.976\nThe present worth of Plan-2 in RS.  13540.524\nThe present worth of Innovative Investment Ltd. Plan is more than that of Novel Investment Ltd. Plan. Therefore Innovative Investment Ltd. Plan is the best plan from the investors point of view.\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.7 Page 50"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=12000.0;#in Rs\nA1=10000.0;#in Rs\nG=1000.0;#in Rs\ni=18.0;#in % per annum\nn=10.0;#in years\n\n#calculation\nPW=-P+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/((i/100)*(1+i/100)**n);#in RS\n\n#result\nprint \"The present worth of the small business in RS.  \",round(PW,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth of the small business in RS.   47293.316\n"
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter5_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter5_1.ipynb
new file mode 100755
index 00000000..37880d17
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter5_1.ipynb
@@ -0,0 +1,125 @@
+{
+ "metadata": {
+  "name": "EE-5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Future Worth Method"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5.1,Page 54"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#alternate A\nP=5000000.0;#in Rs\nA=2000000.0;#in Rs\ni=18.0;#in % per annum\nn=4.0;#in years\n\n#calculation\nFW_A=(-P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100));#in RS\n\n#result\nprint \"The future worth amount of alternative A in RS.  \",round(FW_A,3);\n\n#alternate B\nP=4500000.0;#in Rs\nA=1800000.0;#in Rs\ni=18.0;#in % per annum\nn=4.0;#in years\n\n#calculation\nFW_B=(-P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100));#in RS\nprint \"The future worth amount of alternative B in RS.  \", round(FW_B,3);\nprint \"The future worth of alternative A is greater than that of alternative B. Thus, alternative A should be selected.\";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The future worth amount of alternative A in RS.   736975.2\nThe future worth amount of alternative B in RS.   663277.68\nThe future worth of alternative A is greater than that of alternative B. Thus, alternative A should be selected.\n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 5.2,Page 58"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#alternate 1\nFC=2000000.0;#in Rs\nAI=800000.0;#in Rs\nATax=80000.0;#in Rs\nNetAI=AI-ATax;#in Rs\ni=12.0;#in % per annum\nn=20.0;#in years\n\n#calculation\nFW_1=(-FC*(1+i/100)**n)+(NetAI*(((1+i/100)**n)-1)/(i/100));#in RS\n\n#result\nprint \"The future worth amount of alternative 1 in RS. : \", round(FW_1,3)\n\n#alternative2\nFC=3600000.0;#in Rs\nAI=980000.0;#in Rs\nATax=150000.0;#in Rs\nNetAI=AI-ATax;#in Rs\ni=12.0;#in % per annum\nn=20.0;#in years\n\n#calculation\nFW_2=(-FC*(1+i/100)**n)+(NetAI*(((1+i/100)**n)-1)/(i/100));#in RS\n\n#result\nprint \"The future worth amount of alternative 2 in RS.\",round(FW_2,3)\nprint\"The future worth of alternative 1 is greater than that of alternative 2. Thus, building the gas station is the best alternative.\";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The future worth amount of alternative 1 in RS. :  32585172.373\nThe future worth amount of alternative 2 in RS. 25076872.093\nThe future worth of alternative 1 is greater than that of alternative 2. Thus, building the gas station is the best alternative.\n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5.3,Page 59"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#part a\n#alternative 1\nP=500000.0;#in Rs\nA1=50000.0;#in Rs\nG=50000.0;#in Rs\ni=8.0;#in % per annum\nn=6.0;#in years\n\n#calculation\nFW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n\n#result\nprint \"The future worth amount of alternative 1 in RS.  \", round(FW_1,3);\n\n#alternative 2\nP=700000.0;#in Rs\nA1=70000.0;#in Rs\nG=70000.0;#in Rs\ni=8.0;#in % per annum\nn=6.0;#in years\n\n#calculation\nFW_2=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n\n#result\nprint \"The future worth amount of alternative 2 in RS.  \", round(FW_2,3);\nprint \"The future worth of alternative 2 is greater than that of alternative 1. Thus, alternative 2 must be selected.\";\n\n\n#part b\n#alternative a\nP=500000.0;#in Rs\nA1=50000.0;#in Rs\nG=50000.0;#in Rs\ni=9.0;#in % per annum\nn=6.0;#in years\n\n#calculation\nFW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n\n#result\nprint \"The future worth amount of alternative 1 in RS.  \", round(FW_1,3);\n\n#altenative 2\nP=700000.0;#in Rs\nA1=70000.0;#in Rs\nG=70000.0;#in Rs\ni=9.0;#in % per annum\nn=6.0;#in years\n\n#calculation\nFW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n\n#result\nprint \"The future worth amount of alternative 2 in RS.\", round(FW_1,3);\nprint \"The negative sign of alternatives future worth indicates that alternative 2 incurs loss. Thus, none of the two alternatives should be selected. \";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The future worth amount of alternative 1 in RS.   408314.938\nThe future worth amount of alternative 2 in RS.   571640.914\nThe future worth of alternative 2 is greater than that of alternative 1. Thus, alternative 2 must be selected.\nThe future worth amount of alternative 1 in RS.   383913.653\nThe future worth amount of alternative 2 in RS. 537479.115\nThe negative sign of alternatives future worth indicates that alternative 2 incurs loss. Thus, none of the two alternatives should be selected. \n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5.4 Page 62"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#alternative 1\nP=8000000.0;#in Rs\nA=800000.0;#in Rs\ni=20.0;#in % per annum\nn=12.0;#in years\nSalvage=500000;#in Rs\n\n#calcualtion\nFW1=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n\n#result\nprint \"The future worth for this alternative in RS.\",round(FW1,3)\n\n#alternative 2\nP=7000000.0;#in Rs\nA=900000.0;#in Rs\ni=20.0;#in % per annum\nn=12.0;#in years\nSalvage=400000;#in Rs\n\n#calculation\nFW2=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n\n#result\nprint \"The future worth for this alternative in RS.\",round(FW2,3);\n\n#alternative 3\nP=9000000.0;#in Rs\nA=850000.0;#in Rs\ni=20.0;#in % per annum\nn=12.0;#in years\nSalvage=700000;#in Rs\n\n#calculation\nFW3=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n\nprint \"The future worth for this alternative in RS.  \",round(FW3,3)\nprint \"The future worth of alternative 2 is less than that of other two alternatives. Thus, Ms. Krishna castings should buy the annealing furnace from manufacturer 2.\";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The future worth for this alternative in RS. 102493205.379\nThe future worth for this alternative in RS. 97635155.155\nThe future worth for this alternative in RS.   113188330.939\nThe future worth of alternative 2 is less than that of other two alternatives. Thus, Ms. Krishna castings should buy the annealing furnace from manufacturer 2.\n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5.5 Page 64"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#Machine A\nP=400000.0;#in Rs\nA=40000.0;#in Rs\nS=200000.0;#in Rs\ni=12.0;#in % per annum\nn=4.0#in years\n\n#calculation\nFW_A=(P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100))-S;#in RS\n\n#result\nprint \"The future worth amount of Machine A in RS. \",round(FW_A,3);\n\n#Machine B\nP=800000.0;#in Rs\nA=0.0;#in Rs\nS=550000.0;#in Rs\ni=12.0;#in % per annum\nn=4.0#in years\n\n#calculation\nFW_B=(P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100))-S;#in RS\n\n#result\nprint \"The future worth amount of Machine B in RS. \",round(FW_B,3);\nprint \"The future worth of Machine A is less than that of Machine B. Thus, Machine A should be selected.\";\nprint \" Calculation in the book is not accurate\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The future worth amount of Machine A in RS.  620580.864\nThe future worth amount of Machine B in RS.  708815.488\nThe future worth of Machine A is less than that of Machine B. Thus, Machine A should be selected.\n Calculation in the book is not accurate\n"
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter6_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter6_1.ipynb
new file mode 100755
index 00000000..c0d35ab9
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter6_1.ipynb
@@ -0,0 +1,230 @@
+{
+ "metadata": {
+  "name": "EE-6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Annual Equivalent Method"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.1 Page 70"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nrun=20000.0; #km\nkmperlitre=9; #kmpl\ncost=21;\nG=2222.20;#in Rs.\ni=18.0;#in % per annum\nn=4.0;#in years\n\n#calculation\ncon=run/kmperlitre;#consumption\nA1=con*cost;\nA=A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100) ;\n\n#result\nprint \"Annual equal amount spending for petrol in Rs. : \",round(A,3);\nprint \"This amoount is less than the annual rental value of Rs. 60000. Therefore, the company should continue to provide its own car to its executive. \";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Annual equal amount spending for petrol in Rs. :  49543.741\nThis amoount is less than the annual rental value of Rs. 60000. Therefore, the company should continue to provide its own car to its executive. \n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.2 Page 71"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nP=500000.0;#in Rs\nA=200000.0;#in Rs\ni=20.0;#in % per annum\nn=15.0;#in years\n\n#calcultion\nAE_1=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A;#in RS\n\n#result\nprint \"The Annual equivalent cost of alternative 1 in RS. : \",round(AE_1,3);\n\n#alternative 2\nP=400000.0;#in Rs\nA=300000.0;#in Rs\ni=20.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nAE_2=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A;#in RS\n\n#result\nprint \"The Annual equivalent cost of alternative 2 in RS. : \",round(AE_2,3);\n\n#alternative 3\nP=600000.0;#in Rs\nA=150000.0;#in Rs\ni=20.0;#in % per annum\nn=15.0;#in years\n\n#calculation\nAE_3=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A;#in RS\n\n#result\nprint \"The Annual equivalent cost of alternative 3 in RS. : \",round(AE_3,3);\nprint\"The annual equivalent cost of manufacturer 3 is less than that of other two. Therefore, company should buy advanced machine centre from manufacturer 3. \";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Annual equivalent cost of alternative 1 in RS. :  306941.06\nThe Annual equivalent cost of alternative 2 in RS. :  385552.848\nThe Annual equivalent cost of alternative 3 in RS. :  278329.272\nThe annual equivalent cost of manufacturer 3 is less than that of other two. Therefore, company should buy advanced machine centre from manufacturer 3. \n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.3 Page 73"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation  of variable\n#alternative a\nP=150000.0;#in Rs\nA=60000.0;#in Rs\nS=15000.0;#in Rs\ni=25.0;#in % per annum\nn=5.0;#in years\n\n#calculation\nAE_A=-P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A+S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent revenue of alternative A in RS. : \",round(AE_A,3)\n\n#alternative b\nP=175000.0;#in Rs\nA=70000.0;#in Rs\nS=35000.0;#in Rs\ni=25.0;#in % per annum\nn=5.0;#in years\n\n#calculation\nAE_B=-P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+A+S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent revenue of alternative B in RS. : \",round(AE_B,3)\nprint \"The annual equivalent net return of alternative B is more than that of alternative A. Thus the company should select alternative A\";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Annual equivalent revenue of alternative A in RS. :  6050.69\nThe Annual equivalent revenue of alternative B in RS. :  9191.456\nThe annual equivalent net return of alternative B is more than that of alternative A. Thus the company should select alternative A\n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.4 Page 75"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#machine X\nP=150000.0;#in Rs\nA=0.0;#in Rs\nS=0.0;#in Rs\ni=15.0;#in % per annum\nn=12.0;#in years\n\n#calcualtion\nAE_X=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine X in RS. : \",round(AE_X,3);\n\n#machine Y\nP=240000.0;#in Rs\nA=4500.0;#in Rs\nS=60000.0;#in Rs\ni=15.0;#in % per annum\nn=12.0;#in years\n\n#calcualtion\nAE_Y=P*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine Y in RS. : \",round(AE_Y,3);\nprint \"The annual equivalent cost of machine X is less than that of machine Y.So machine X is more cost effective machine. \";\nprint\" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Annual equivalent cost of machine X in RS. :  27672.116\nThe Annual equivalent cost of machine Y in RS. :  44275.386\nThe annual equivalent cost of machine X is less than that of machine Y.So machine X is more cost effective machine. \n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.5 Page 76"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initialisation of variable\n#alternative 1\nFC=15*150000.0;#in Rs\nMC=15*6000.0;#in Rs\nPL=15*15000.0;#in Rs\n\n#calculation\nMCPL=MC+PL;#in Rs\nS=15*90000.0;#in Rs\ni=15.0;#in % per annum\nn=15.0;#in years\nAE1=FC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+MCPL-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS. \",round(AE1,3);\n\n#alternative 2\nFC=5*750000.0;#in Rs\nMC=5*12000.0;#in Rs\nPL=5*15000.0;#in Rs\nMCPL=MC+PL;#in Rs\nS=5*150000.0;#in Rs\ni=15.0;#in % per annum\nn=15.0;#in years\n\n#calcualtion\nAE2=FC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+MCPL-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS. \",round(AE2,3);\nprint \"The annual equivalent cost of alternative 1 is less than that of alternative 2. Therefore, select the route around the lake for laying the power line.\";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Annual equivalent cost for this alternative in RS.  671415.347\nThe Annual equivalent cost for this alternative in RS.  760551.158\nThe annual equivalent cost of alternative 1 is less than that of alternative 2. Therefore, select the route around the lake for laying the power line.\n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.6 Page 78"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nVC=390000.0;#in Rs\nLpY=60000.0/30;#in litres\nFcY=2000.0*8;#in Rs\nAllExp=FcY+9000.0+15000.0;#in Rs\nS=60000.0;#in Rs\ni=20.0;#in % per annum\nn=4.0;#in years\n\n#calcualtion\nAE1=VC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AllExp-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS.  \",round(AE1,3);\n\n#alternative 2\nVC=360000.0;#in Rs\nLpY=60000.0/20;#in litres\nFcY=3000.0*20;#in Rs\nAllExp=FcY+6000.0+15000.0;#in Rs\nS=90000.0;#in Rs\ni=20.0;#in % per annum\nn=3.0;#in years\n\n#calculation\nAE2=VC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AllExp-S*(i/100)/(((1+i/100)**n)-1);# in RS\n\n#result\nprint \"The Annual equivalent cost for this alternative in RS.  \",round(AE2,3);\nprint \"The annual equivalent cost of purchase and operation of cars with diesel engine is less than that of cars with petrol engine. Therefore, the taxy company should buy cars with diesel engine.\";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Annual equivalent cost for this alternative in RS.   179475.41\nThe Annual equivalent cost for this alternative in RS.   227175.824\nThe annual equivalent cost of purchase and operation of cars with diesel engine is less than that of cars with petrol engine. Therefore, the taxy company should buy cars with diesel engine.\n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.7 Page 79"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nPprice=390000.0;#in Rs\nn=3.0;#in years\nn=n*12;#in months\nS=120000.0;#in Rs\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nME1=Pprice*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Monthly equivalent cost for this alternative in RS.  \",round(ME1,3);\n\n#Alternative 2 \nLeaseAmount=10500.0;#in Rs\n\n#calcualtion\nMonthlyEquiCost=LeaseAmount;#in Rs.\n\n#result\nprint\"The Monthly equivalent cost for this alternative in RS. \",round(MonthlyEquiCost,3);\nprint\"The monthly equivalent cost of alternative 1 is less than that of alternative 2. hence the salesman should purchase the car for cash.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Monthly equivalent cost for this alternative in RS.   10167.864\nThe Monthly equivalent cost for this alternative in RS.  10500.0\nThe monthly equivalent cost of alternative 1 is less than that of alternative 2. hence the salesman should purchase the car for cash.\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.8 Page 81"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#machine A\nIC=300000.0;#in Rs\nn=4.0;#in years\nS=200000.0;#in Rs\nAM=30000.0;#in Rs\ni=15.0;#in % per annum\n\n#calcualtion\nAE_A=IC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AM-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine X in RS. \",round(AE_A,3);\n\n#Machine B :\nIC=600000.0;#in Rs\nn=4.0;#in years\nS=300000.0;#in Rs\nAM=0.0;#in Rs\ni=15.0;#in % per annum\n\n#calcualtion\nAE_B=IC*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+AM-S*(i/100)/(((1+i/100)**n)-1);#in RS\n\n#result\nprint \"The Annual equivalent cost of machine X in RS. \",round(AE_B,3);\nprint \"The annual equivalent cost of machine A is less than that of machine B.So machine A is more cost effective machine. It is advised to buy machine A \";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Annual equivalent cost of machine X in RS.  95026.535\nThe Annual equivalent cost of machine X in RS.  150079.605\nThe annual equivalent cost of machine A is less than that of machine B.So machine A is more cost effective machine. It is advised to buy machine A \n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.9 Page 83"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nDP=60000.0;#in Rs\nP1=15000.0;#in Rs\nn=1.0;#in years\ni=12.0;#in % Compounded anually\n\n#calculation\nPW1=DP+P1*1/((1+i/100)**n);#in RS\nn=4.0;#in years\nAE1=PW1*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)\n\n#result\nprint \"The annual equivalent for this alternative in RS. \",round(PW1,3);\nprint \"The annual equivalent for this alternative in RS.\",round(AE1,3);\n\n#alternative 2\nP4y=90000.0;#in Rs\nAE2=P4y*(i/100)/(((1+i/100)**n)-1);#in Rs.\n\n#result\nprint \"The annual equivalent for this alternative in RS. \",round(AE2,3);\nprint \"The annual equivalent cost of alternative 2 is less than that of alternative 1. Hence, Joshi Lakshimi should select alternative 2 for purchasing the home equipment.\";\nprint \" Calculation in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent for this alternative in RS.  73392.857\nThe annual equivalent for this alternative in RS. 24163.456\nThe annual equivalent for this alternative in RS.  18831.099\nThe annual equivalent cost of alternative 2 is less than that of alternative 1. Hence, Joshi Lakshimi should select alternative 2 for purchasing the home equipment.\n Calculation in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.10 Page 85"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initation of variable\n#Brand C : \nw=12.0;#in months\nP=1200.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_C=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_C,3);\n\n#Brand C : \nw=24.0;#in months\nP=1800.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_C=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_C,3);\n\n#Brand C : \nw=36.0;#in months\nP=2100.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_C=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_C,3);\n\n#Brand D : \nw=48.0;#in months\nP=2700.0;#in Rs/tyre\ni=12.0;#in % nominal rate(Compounded anually)\ni=i/12.0;#in % compounded monthly\n\n#calculation\nAE_D=P*((i/100)*(1+i/100)**w)/(((1+i/100)**w)-1);#in RS\n\n#result\nprint \"The annual equivalent for this brand in RS. \",round(AE_D,3);\nprint \"Here common multiple lives of tyres is considered. This is 144 months. Therefore, the comparison is nade on 144 months basis.\";\nprint \"The annual equivalent cost of Brand C is less than that of other brands. hence, it should be used in the vehicles of the trucking company. It should be replaced  for times during the 144 months period.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent for this brand in RS.  106.619\nThe annual equivalent for this brand in RS.  84.732\nThe annual equivalent for this brand in RS.  69.75\nThe annual equivalent for this brand in RS.  71.101\nHere common multiple lives of tyres is considered. This is 144 months. Therefore, the comparison is nade on 144 months basis.\nThe annual equivalent cost of Brand C is less than that of other brands. hence, it should be used in the vehicles of the trucking company. It should be replaced  for times during the 144 months period.\n"
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter7_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter7_1.ipynb
new file mode 100755
index 00000000..827a806e
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter7_1.ipynb
@@ -0,0 +1,125 @@
+{
+ "metadata": {
+  "name": "EE-7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Rate of Return Method"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.1 Page 89"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of vriable\nIi=100000.0;#in Rs\nAr=30000.0;#in Rs\nn=5.0;#in years\ni1=10.0;#in % per annum\n\n#calcualton\nPW1=-Ii+Ar*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\n\n#result\nprint \"The present worth for i=10% in RS. \",round(PW1,3);\n\n#calcualtion\ni2=15.0;#in % per annum\nPW2=-Ii+Ar*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\n\n#result\nprint \"The present worth for i=15% in RS. \",round(PW2,3);\n\n#calculation\ni3=18.0;#in % per annum\nPW3=-Ii+Ar*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni=15+(PW2-0)*(i3-i2)/(PW2-PW3);#in Rs\n\n#result\nprint \"The present worth for i=18% in RS. \",round(PW3,3)\nprint \"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return for the new business in % per annum \",round(i,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth for i=10% in RS.  13723.603\nThe present worth for i=15% in RS.  564.653\nThe present worth for i=18% in RS.  -6184.869\nPresent worth for i=15% is suitable.\nTherefore, the rate of return for the new business in % per annum  15.251\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.2 Page 90"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nIo=2000000.0;#in Rs\nANP=350000.0;#in Rs\nS=0.0;#in Rs\nn=10.0;#in years\ni1=10.0;#in % per annum\n\n#calcualtion\nPW1=-Io+ANP*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\n\n#result\nprint \"The present worth for i=10% in RS.  \",round(PW1,3);\n\n#calculation\ni2=12.0;#in % per annum\nPW2=-Io+ANP*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni=10+(PW1-0)*(i2-i1)/(PW1-PW2);#in Rs.\n\n#result\nprint \"The present worth for i=15% in RS. \",round(PW2,3);\nprint \"Present worth for i=10% is suitable.\";\nprint \"Therefore, the rate of return for the new business in % per annum \",round(i,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth for i=10% in RS.   150598.487\nThe present worth for i=15% in RS.  -22421.94\nPresent worth for i=10% is suitable.\nTherefore, the rate of return for the new business in % per annum  11.741\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.3 Page 91"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternate 1\nIo=150000.0;#in Rs\nAp=45570.0;#in Rs\nn=5.0;#in years\n\n#calculation\ni1=10.0;#in % per annum\nPW1=-Io+Ap*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=12.0;#in % per annum\nPW2=-Io+Ap*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni3=15.0;#in % per annum\nPW3=-Io+Ap*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni4=18.0;#in % per annum\nPW4=-Io+Ap*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in RS\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The present worth for i=10% in RS.  \",round(PW1,3);\nprint \"The present worth for i=12% in RS. \",round(PW2,3);\nprint \"The present worth for i=15% in RS. \",round(PW3,3);\nprint \"The present worth for i=18% in RS. \",round(PW4,3);\nprint\"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return of alternative in % per annum :\",round(i,3);\n\n#alternative 2\nIo=210000.0;#in Rs\nAp=58260.0;#in Rs\nn=5.0;#in years\ni1=12.0;#in % per annum\n\n#calcualtion\nPW1=-Io+Ap*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=13.0;#in % per annum\nPW2=-Io+Ap*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni=i1+(PW1-0)*(i2-i1)/(PW1-PW2);#in Rs.\n\n#result\nprint \"The present worth for i=12% in RS. \",round(PW1,3);\nprint \"The present worth for i=13% in RS. \",round(PW2,3);\nprint\"Present worth for i=12% is suitable.\";\nprint \"Therefore, the rate of return of alternative in % per annum :\",round(i,3);\n\n#Alternative 3:\nIo=255000.0;#in Rs\nAp=69000.0;#in Rs\nn=5.0;#in years\ni1=11.0;#in % per annum\nPW1=-Io+Ap*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=12.0;#in % per annum\nPW2=-Io+Ap*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni=i1+(PW1-0)*(i2-i1)/(PW1-PW2);#in Rs.\n\n#result\nprint \"The present worth for i=11% in RS. \",round(PW1,3);\nprint \"The present worth for i=12% in RS. \",round(PW2,3);\nprint\"Present worth for i=12% is suitable.\";\nprint \"Therefore, the rate of return of alternative in % per annum :\",round(i,3);\nprint\"It is clear that rate of return for alternative 3v is less than the minimum attractive rate of return of 12 %. So it should not be considered for comparison. the remaining two alternatives are qualified for consideration. Among the alternatives 1 and 2, the rate of return of 1 is greater than that of 2. hence,alternative 1 should be selected. \";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth for i=10% in RS.   22746.153\nThe present worth for i=12% in RS.  14269.652\nThe present worth for i=15% in RS.  2757.708\nThe present worth for i=18% in RS.  -7494.817\nPresent worth for i=15% is suitable.\nTherefore, the rate of return of alternative in % per annum : 15.807\nThe present worth for i=12% in RS.  14.262\nThe present worth for i=13% in RS.  -5086.107\nPresent worth for i=12% is suitable.\nTherefore, the rate of return of alternative in % per annum : 12.003\nThe present worth for i=11% in RS.  16.894\nThe present worth for i=12% in RS.  -6270.442\nPresent worth for i=12% is suitable.\nTherefore, the rate of return of alternative in % per annum : 11.003\nIt is clear that rate of return for alternative 3v is less than the minimum attractive rate of return of 12 %. So it should not be considered for comparison. the remaining two alternatives are qualified for consideration. Among the alternatives 1 and 2, the rate of return of 1 is greater than that of 2. hence,alternative 1 should be selected. \n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.4 Page 94"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nA0=-1275.0;#in Rs\nA1=150.0;#in Rs\nG=150.0;#in Rs\ni=10.0;#in % per annum\nn=5.0;#in years\n\n#calcualtion\nA=A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100) ;#in RS\ni1=10.0;#in % per annum\nPW1=A0+(A1+G*(((1+i1/100)**n)-i1*n/100-1)/(((i1/100)*(1+i1/100)**n)-i1/100))*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#i1n Rs.\ni2=12.0;#i2n % per annum\nPW2=A0+(A1+G*(((1+i2/100)**n)-i2*n/100-1)/(((i2/100)*(1+i2/100)**n)-i2/100))*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#i2n Rs.\ni3=15.0;#i3n % per annum\nPW3=A0+(A1+G*(((1+i3/100)**n)-i3*n/100-1)/(((i3/100)*(1+i3/100)**n)-i3/100))*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#i3n Rs.\ni4=18.0;#in % per annum\nPW4=A0+(A1+G*(((1+i4/100)**n)-i4*n/100-1)/(((i4/100)*(1+i4/100)**n)-i4/100))*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in Rs.\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The annual equivalent of the positive cash flows in RS. : \",round(A,3);\nprint \"PW(10)\",round(PW1,3);\nprint \"PW(12)\",round(PW2,3);\nprint \"PW(15)\",round(PW3,3);\nprint \"PW(18)\",round(PW1,4);\nprint\"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return for the given cash flow in % per annum :\",round(i,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent of the positive cash flows in RS. :  421.519\nPW(10) 322.888\nPW(12) 225.269\nPW(15) 94.095\nPW(18) 322.8882\nPresent worth for i=15% is suitable.\nTherefore, the rate of return for the given cash flow in % per annum : 17.448\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.5 Page 95"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nIo=500000.0;#in Rs\nAr=170000.0;#in Rs\nn=5.0;#in years\ni1=15.0;#in % per annum\n\n#calculation\nPW1=-Io+Ar*(((1+i1/100)**n)-1)/((i1/100)*(1+i1/100)**n);#in RS\ni2=17.0;#in % per annum\nPW2=-Io+Ar*(((1+i2/100)**n)-1)/((i2/100)*(1+i2/100)**n);#in RS\ni3=20.0;#in % per annum\nPW3=-Io+Ar*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni4=22.0;#in % per annum\nPW4=-Io+Ar*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in RS\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The present worth(PW(15%)) for i=10% in RS.  \",round(PW1,3);\nprint \"The present worth(PW(17%)) for i=10% in RS.  \",round(PW2,3);\nprint \"The present worth(PW(20%)) for i=10% in RS.  \",round(PW3,3);\nprint \"The present worth(PW(22%)) for i=10% in RS.  \",round(PW4,3);\nprint \"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return of alternative 1 in % per annum \",round(i,3);\n\n#altenative 2\nIo=800000.0;#in Rs\nAr=270000.0;#in Rs\nn=5.0;#in years\ni3=20.0;#in % per annum\n\n#calculation\nPW3=-Io+Ar*(((1+i3/100)**n)-1)/((i3/100)*(1+i3/100)**n);#in RS\ni4=22.0;#in % per annum\nPW4=-Io+Ar*(((1+i4/100)**n)-1)/((i4/100)*(1+i4/100)**n);#in RS\ni=i3+(PW3-0)*(i4-i3)/(PW3-PW4);#in Rs.\n\n#result\nprint \"The present worth(PW(20%)) for i=10% in RS.  \",round(PW3,3);\nprint \"The present worth(PW(22%)) for i=10% in RS.  \",round(PW4,3);\nprint \"Present worth for i=15% is suitable.\";\nprint \"Therefore, the rate of return of alternative 2 in % per annum \",round(i,3);\nprint \"Since the rate of return of alternative 1 is greater than that of the alternative 2, select alternative 1. \";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The present worth(PW(15%)) for i=10% in RS.   69866.367\nThe present worth(PW(17%)) for i=10% in RS.   43888.848\nThe present worth(PW(20%)) for i=10% in RS.   8404.064\nThe present worth(PW(22%)) for i=10% in RS.   -13181.241\nPresent worth for i=15% is suitable.\nTherefore, the rate of return of alternative 1 in % per annum  20.779\nThe present worth(PW(20%)) for i=10% in RS.   7465.278\nThe present worth(PW(22%)) for i=10% in RS.   -26817.264\nPresent worth for i=15% is suitable.\nTherefore, the rate of return of alternative 2 in % per annum  20.436\nSince the rate of return of alternative 1 is greater than that of the alternative 2, select alternative 1. \n"
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter8_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter8_1.ipynb
new file mode 100755
index 00000000..a226c6dc
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter8_1.ipynb
@@ -0,0 +1,230 @@
+{
+ "metadata": {
+  "name": "EE-8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Replacement and Maintenance Analysis"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.1 Page 103"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nFC=4000.0;#in Rs.\ni=0.0;#in % per annum\nMC1=0;#in Rs.\nprint \"Tabulation to determine economic life : \";\nprint\"End of     Maintenance     Summation of     Avg cost of     Avg FC if     Avg total\";\nprint\" year      cost at end     maintenance      maintenance    replaced at    cost through\";\nprint\"             of year         costs          through year    year given    year given\";\nprint\"  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)\";\nprint\"  1            0                0                 0          4000          4000\";\nprint\"  2          200              200               100          2000          2100\";\nprint\"  3          400              600               200       1333.33       1533.33\";\nprint\"  4          600             1200               300          1000          1300\";\nprint\"  5          800             2000               400           800          1200\";\nprint\"  6         1000             3000               500        666.67       1166.67\";\nprint\"  7         1200             4200               600        571.43       1171.43\";\nprint\"Economic life of the machine : 6 years\";\nprint\"Column C summarizes the summation of maintenance costs for each replacement period. The value corresponding to any end of year in this column represents the total maintenance costs of using the equipment till the end of that year. It gives the Economic life of the machine : 6 years\";\n\n#When i=12%\ni=12.0;#in % per annum\nFC=4000.0;#in Rs.\nprint\"Tabulation to determine economic life : \";\nprint\"End of     Maintenance         P/F       PW as of begin   Summation of    PW of        A/P     Annual equi\";\nprint\" year      cost at end        12%         of year of         PW of      cumulative     12%     total cost\";\nprint\"             of year           n          maintenance     maintenance   maintenance     n       of year\";\nprint\"  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)    G(rs.)      H(rs.)\";\nprint\"  1            0           0.8929              0.00          0.00       4000.00     1.1200     4480.00\";\nprint\"  2          200           0.7972            159.44        159.44       4159.44     0.5917     2461.14\";\nprint\"  3          400           0.7118            284.72        444.16       4444.16     0.4163     1850.10\";\nprint\"  4          600           0.6355            381.30        825.46       4825.46     0.3292     1588.54\";\nprint\"  5          800           0.5674            453.92       1279.38       5279.38     0.2774     1464.50\";\nprint\"  6         1000           0.5066            506.60       1785.98       5785.98     0.2432     1407.15\";\nprint\"  7         1200           0.4524            542.88       2328.86       6328.86     0.2191     1386.65\";\nprint\"  8         1400           0.4039            565.46       2894.32       6894.32     0.2013     1387.83\";\nprint\"  9         1600           0.3606            576.96       3471.28       7471.28     0.1877     1402.36\";\nprint\" 10         1800           0.3220            579.60       4050.88       8050.88     0.1770     1425.00\";\nprint\"Economic life of the machine : 7 years\";\nprint\"For this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is 7 year.\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Tabulation to determine economic life : \nEnd of     Maintenance     Summation of     Avg cost of     Avg FC if     Avg total\n year      cost at end     maintenance      maintenance    replaced at    cost through\n             of year         costs          through year    year given    year given\n  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)\n  1            0                0                 0          4000          4000\n  2          200              200               100          2000          2100\n  3          400              600               200       1333.33       1533.33\n  4          600             1200               300          1000          1300\n  5          800             2000               400           800          1200\n  6         1000             3000               500        666.67       1166.67\n  7         1200             4200               600        571.43       1171.43\nEconomic life of the machine : 6 years\nColumn C summarizes the summation of maintenance costs for each replacement period. The value corresponding to any end of year in this column represents the total maintenance costs of using the equipment till the end of that year. It gives the Economic life of the machine : 6 years\nTabulation to determine economic life : \nEnd of     Maintenance         P/F       PW as of begin   Summation of    PW of        A/P     Annual equi\n year      cost at end        12%         of year of         PW of      cumulative     12%     total cost\n             of year           n          maintenance     maintenance   maintenance     n       of year\n  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)    G(rs.)      H(rs.)\n  1            0           0.8929              0.00          0.00       4000.00     1.1200     4480.00\n  2          200           0.7972            159.44        159.44       4159.44     0.5917     2461.14\n  3          400           0.7118            284.72        444.16       4444.16     0.4163     1850.10\n  4          600           0.6355            381.30        825.46       4825.46     0.3292     1588.54\n  5          800           0.5674            453.92       1279.38       5279.38     0.2774     1464.50\n  6         1000           0.5066            506.60       1785.98       5785.98     0.2432     1407.15\n  7         1200           0.4524            542.88       2328.86       6328.86     0.2191     1386.65\n  8         1400           0.4039            565.46       2894.32       6894.32     0.2013     1387.83\n  9         1600           0.3606            576.96       3471.28       7471.28     0.1877     1402.36\n 10         1800           0.3220            579.60       4050.88       8050.88     0.1770     1425.00\nEconomic life of the machine : 7 years\nFor this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is 7 year.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.2 Page 105"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nFC=20000.0;#in Rs.\ni=15.0;#in % per annum\n\n#result\nprint \"The other details are summarized in Table 8.3. It can be seen from the book.\";\nprint \"Total annual equivalent cost = [cumulative sum of PW as of beginning of year 1 of operation & maintenance cost + FC - PW as of beginning of year 1 of salvage]*(A/P,15,n)\";\nprint \"In column L, the annual equivalent cost is minimum for n=5. Therefore, the economic life of the machine is 5 years. \";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The other details are summarized in Table 8.3. It can be seen from the book.\nTotal annual equivalent cost = [cumulative sum of PW as of beginning of year 1 of operation & maintenance cost + FC - PW as of beginning of year 1 of salvage]*(A/P,15,n)\nIn column L, the annual equivalent cost is minimum for n=5. Therefore, the economic life of the machine is 5 years. \n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.3 Page 107"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nFC=20000.0;#in Rs.\ni=15.0;#in % per annum\n\n#result\nprint \"The details are summarized in Table 8.4. It can be seen from the book.\";\nprint\"Total annual equivalent cost = [summation of PW of maintenance cost + FC]]*(A/P,15,n)\";\nprint\"(column E + Rs. 6000)* Column G\";\nprint \"Column F * Column G\";\nprint \"In column H, the minimum annual equivalent cost occurs when n=8. Therefore, the economic life of the machine B is 8 years. \";\nprint \"RESULT : Min annual equivalent cost for machine A : Rs. 2780\";\nprint \"Min annual equivalent cost for machine B : Rs. 3672.30\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The details are summarized in Table 8.4. It can be seen from the book.\nTotal annual equivalent cost = [summation of PW of maintenance cost + FC]]*(A/P,15,n)\n(column E + Rs. 6000)* Column G\nColumn F * Column G\nIn column H, the minimum annual equivalent cost occurs when n=8. Therefore, the economic life of the machine B is 8 years. \nRESULT : Min annual equivalent cost for machine A : Rs. 2780\nMin annual equivalent cost for machine B : Rs. 3672.30\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.4 Page 109"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nPprice=200000.0;#in Rs\nP=120000.0;#in Rs\nF=25000.0;#in Rs\nA=25000.0;#in Rs\ni=12.0;#in % per annum\nn=6.0;#in years\n\n#calculation\nAE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(12%)) of this alternative in RS.  \",round(AE1,3);\n\n#Alternative 2\nP=150000.0;#in Rs\nF=20000.0;#in Rs\nA=14000.0;#in Rs\ni=12.0;#in % per annum\nn=6.0;#in years\n\n#calcualion\nAE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(12%)) of this alternative in RS.  \",round(AE2,3);\nprint \"Since, The equivalent cost of new machine is less than that of present machine, it is suggested that the present machine be replaced with the new machine.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent cost(AE(12%)) of this alternative in RS.   51106.443\nThe annual equivalent cost(AE(12%)) of this alternative in RS.   48019.343\nSince, The equivalent cost of new machine is less than that of present machine, it is suggested that the present machine be replaced with the new machine.\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.5 Page 111"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initation of variable\n#alternative 1\nPprice=50000.0;#in Rs\nP=15000.0;#in Rs\nF=8000.0;#in Rs\nA=14000.0;#in Rs\ni=15.0;#in % per annum\nn=5.0;#in years\n\n#calcualtion\nAE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(15%)) of this alternative in RS.  \",round(AE1,3);\n\n#Alternative 2\nP=65000.0;#in Rs\nF=13000.0;#in Rs\nA=9000.0;#in Rs\ni=15.0;#in % per annum\nn=20.0;#in years\n\n#calculation\nAE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(15%)) of this alternative in RS. : \",round(AE2,3);\nprint \"Since, The equivalent cost of Old deisel Engine is less than that of New deisel Engine, it is suggested to keep the Old deisel Engine.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent cost(AE(15%)) of this alternative in RS.   17288.209\nThe annual equivalent cost(AE(15%)) of this alternative in RS. :  19257.596\nSince, The equivalent cost of Old deisel Engine is less than that of New deisel Engine, it is suggested to keep the Old deisel Engine.\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.6 Page 113"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#intiation of variable\nP=660000.0;#in Rs\nF=400000.0;#in Rs\nA=96000.0;#in Rs\ni=10.0;#in % per annum\nn=5.0;#in years\n\n#caculation\nAE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(10%)) of this alternative in RS. : \",round(AE1,3);\n\n#Alternative 2\nP=150000.0;#in Rs\nX=420000.0;#in Rs\ni=10.0;#in % per annum\nn=40.0;#in years\n\n#calcualtion\nAE2=(P-X)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(10%)) of this alternative in RS. : \",round(AE2,3);\nprint \"Since, The equivalent cost of alternative 2 is less than that of alternative 1, it is suggested that alternative 2 should be selected.\";\nprint \" calculations in the book is not accurate.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent cost(AE(10%)) of this alternative in RS. :  204587.345\nThe annual equivalent cost(AE(10%)) of this alternative in RS. :  108389.958\nSince, The equivalent cost of alternative 2 is less than that of alternative 1, it is suggested that alternative 2 should be selected.\n calculations in the book is not accurate.\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.7 Page 113"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#alternative 1\nP=10000.0;#in Rs\nF=1500.0;#in Rs\nA=1600.0;#in Rs\ni=15.0;#in % per annum\nn=7.0;#in years\n\n#calcualtion\nAE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(10%)) of 10 hp motor  in RS. : \",round(AE1,3);\n\n#alternative 1 part 2\nP=10000;#in Rs\nF=800.0;#in Rs\nA=1000.0;#in Rs\ni=15.0;#in % per annum\nn=7.0;#in years\n\n#calcualtion\nAE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\n#result\nprint \"The annual equivalent cost(AE(10%)) of 10 hp motor  in RS.  \",round(AE2,3);\nprint \"Total annual equivalent cost of alternative in Rs. : \",round(AE1+AE2,3);\n\n\n#Alternative 2\nP=35000.0;#in Rs\nF=4000.0;#in Rs\nA=500.0;#in Rs\ni=15.0;#in % per annum\nn=7.0;#in years\n\n#calcualtion\nAE=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\n\nprint \"The annual equivalent cost of alternative 2 in RS.\",round(AE,3);\nprint\"Since, The equivalent cost of alternative 1 is less than that of alternative 2, it is suggested that the present 10 hp motor be augmented with an additional 5 hp motor.\";\nprint \" calculations in the book is not accurate\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent cost(AE(10%)) of 10 hp motor  in RS. :  3868.063\nThe annual equivalent cost(AE(10%)) of 10 hp motor  in RS.   3331.315\nTotal annual equivalent cost of alternative in Rs. :  7199.378\nThe annual equivalent cost of alternative 2 in RS. 8551.171\nSince, The equivalent cost of alternative 1 is less than that of alternative 2, it is suggested that the present 10 hp motor be augmented with an additional 5 hp motor.\n calculations in the book is not accurate\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.8 Page 115"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=10000.0;#in Rs\nF=4000.0;#in Rs\nA=500.0;#in Rs\ni=12.0;#in % per annum\nn=4.0;#in years\n\n#calcualtion\nAE=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS\nX= (AE-1000*i/100-750)/0.3292+1000;\n#result\nprint \"The annual equivalent cost(AE(15%)) of n in RSew machine : \",round(AE,3);\nprint \"The comparative use value of old machine is Rs.\", round(X,3), \"which is less than the price(Rs. 8000) offered by the company which is supplying the new machine in the event of replacing the old machine by nw machine. \";\nprint\"Therefore, it is advisable to replace the old machine with the new one.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The annual equivalent cost(AE(15%)) of n in RSew machine :  2955.407\nThe comparative use value of old machine is Rs. 7334.771 which is less than the price(Rs. 8000) offered by the company which is supplying the new machine in the event of replacing the old machine by nw machine. \nTherefore, it is advisable to replace the old machine with the new one.\n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.9 Page 117"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#Intiation of Variable\np1=0.07;#unitless\np2=0.11;#unitless\np3=0.12;#unitless\np4=0.18;#unitless\np5=0.21;#unitless\np6=0.20;#unitless\np7=0.11;#unitless\nN0=100.0;#no. of transistors\n\n#calcualtion\nN1=N0*p1;#no. of transistors\nN2=N0*p2;#no. of transistors\nN3=N0*p3;#no. of transistors\nN4=N0*p4;#no. of transistors\nN5=N0*p5;#no. of transistors\nN6=N0*p6;#no. of transistors\nN7=N0*p7;#no. of transistors\nLife=0;#in weeks\np=[p1, p2, p3, p4, p5, p6, p7];#Unitless\nfor i in range (7): \n    Life=Life+(i+1)*p[i];\n\n#result\nprint \"Expected life of each transistor in weeks : \",round(Life,3)\nprint \"Average No. of failures/week :  \",round(100/Life);\n\n#result of group replacement cost \nprint \"Cost of transistor when replaced simultaneously = Rs. 3\";\nprint \"Cost of transistor when replaced individually = Rs. 9\";\nprint \"The cost of group replacement policy for seeral replacement periods are summarized in Table 8.6. This table can be seen from the book.\";\nprint \"From table it is clear that the avg cost/week is minimum for the 4th week. Hence, the group replacement period is 4 weeks.\";\nprint \"Individual replacement cost/week = Rs. 207\";\nprint \"Minimum group replacement ost/week = Rs. 196.50\";\nprint \"Since the min group replacement cost/week is less than the individual replacement cost/week, the group replacement policy is the best, and hence all the transistors should be replaced in 4 weeks.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Expected life of each transistor in weeks :  4.39\nAverage No. of failures/week :   23.0\nCost of transistor when replaced simultaneously = Rs. 3\nCost of transistor when replaced individually = Rs. 9\nThe cost of group replacement policy for seeral replacement periods are summarized in Table 8.6. This table can be seen from the book.\nFrom table it is clear that the avg cost/week is minimum for the 4th week. Hence, the group replacement period is 4 weeks.\nIndividual replacement cost/week = Rs. 207\nMinimum group replacement ost/week = Rs. 196.50\nSince the min group replacement cost/week is less than the individual replacement cost/week, the group replacement policy is the best, and hence all the transistors should be replaced in 4 weeks.\n"
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.10 Page 119"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\np1=(100.0-96)/100;#unitless\np2=(96.0-89)/100;#unitless\np3=(89-68)/100;#unitless\np4=(68.0-37)/100;#unitless\np5=(37.0-13)/100;#unitless\np6=(13.0-0)/100;#unitless\nN0=1000.0;#no. of resistors\n\n#calculation\nN1=N0*p1;#no. of resistors\nN2=N0*p2+N1*p1;#no. of resistors\nN3=N0*p3+N1*p2+N2*p1;#no. of resistors\nN4=N0*p4+N1*p3+N2*p2+N3*p1;#no. of resistors\nN5=N0*p5+N1*p4+N2*p3+N3*p2+N4*p1;#no. of resistors\nN6=N0*p6+N1*p5+N2*p4+N3*p3+N4*p2+N5*p1    ;#no. of resistors\n\nLife=0;#in months\np=[p1, p2, p3, p4, p5, p6, p7];#Unitless\nfor i in range (7): \n    Life=Life+(i+1)*p[i];\n\n#result\nprint \"Expected life of each transistor in weeks : \",round(Life,3)\nprint \"Average No. of failures/week :  \",round(100/Life);\n\n#result of group replacement cost\nprint \"Cost/transistor when replaced simultaneously = Rs. 4\";\nprint \"Cost/transistor when replaced individually = Rs. 10\";\nprint \"The cost of group replacement policy for several replacement periods are summarized in Table 8.7. This table can be seen from the book.\";\nprint \"From table it is clear that the avg cost/month is minimum for the 3rd month. Hence, the group replacement period is 3 months.\";\nprint \"Individual replacement cost/month = Rs. 2480\";\nprint \"Minimum group replacement ost/month = Rs. 2426.67\";\nprint \"Since the min group replacement cost/month is less than the individual replacement cost/month, the group replacement policy is the best, and hence all the transistors should be replaced in 3 months.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Expected life of each transistor in weeks :  4.17\nAverage No. of failures/week :   24.0\nCost/transistor when replaced simultaneously = Rs. 4\nCost/transistor when replaced individually = Rs. 10\nThe cost of group replacement policy for several replacement periods are summarized in Table 8.7. This table can be seen from the book.\nFrom table it is clear that the avg cost/month is minimum for the 3rd month. Hence, the group replacement period is 3 months.\nIndividual replacement cost/month = Rs. 2480\nMinimum group replacement ost/month = Rs. 2426.67\nSince the min group replacement cost/month is less than the individual replacement cost/month, the group replacement policy is the best, and hence all the transistors should be replaced in 3 months.\n"
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/Chapter9_1.ipynb b/Engineering_Economics_by_R_Panneerselvam/Chapter9_1.ipynb
new file mode 100755
index 00000000..6221daa0
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/Chapter9_1.ipynb
@@ -0,0 +1,209 @@
+{
+ "metadata": {
+  "name": "EE-9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Depreciation"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.1 Page 127"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=100000;#book vakue\n\n#calcualtion\nDt=(P-F)/n;#in Rs.\n\n#result\nprint \"End of year          Depreciation          Book value\";\nfor i in range (8):\n    B=Bt-i*Dt;\n    print  eoy[i],\"                    \",Dt,\"            \", B,\" \";\n     \n    \n    ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "End of year          Depreciation          Book value\n0                      10000.0              100000.0  \n1                      10000.0              90000.0  \n2                      10000.0              80000.0  \n3                      10000.0              70000.0  \n4                      10000.0              60000.0  \n5                      10000.0              50000.0  \n6                      10000.0              40000.0  \n7                      10000.0              30000.0  \n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.2 Page 127"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\n\n#calculation\nD5=(P-F)/n;#in Rs.\nt=5;#in years\nBt=P-t*(P-F)/n;#in Rs\n\n#result\nprint \"D5 in Rs. : \",round(D5,3);\nprint \"(This is independent of the time period)\";\nprint \"B5 in Rs. : \",round(Bt,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "D5 in Rs. :  10000.0\n(This is independent of the time period)\nB5 in Rs. :  50000.0\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.3 Page 129"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nk=0.2\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=[100000.0,0,0,0,0,0,0,0,0];#book vakue\nDt=[0.0,0,0,0,0,0,0,0,0];\n\n#result\nprint \"End of year          Depreciation          Book value\";\nprint \"0                      0                    100000\"\nfor i in xrange (1,9):\n    Dt[i]=k*Bt[i-1];\n    Bt[i]=Bt[i-1]-Dt[i];\n    print  eoy[i],\"                    \",Dt[i],\"            \", Bt[i],\" \";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "End of year          Depreciation          Book value\n0                      0                    100000\n1                      20000.0              80000.0  \n2                      16000.0              64000.0  \n3                      12800.0              51200.0  \n4                      10240.0              40960.0  \n5                      8192.0              32768.0  \n6                      6553.6              26214.4  \n7                      5242.88              20971.52  \n8                      4194.304              16777.216  \n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.4 Page 128"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nk=0.2\nt=5.0;#in years\n\n#calculation\nDt=k*(1-k)**(t-1)*P;#in Rs.\nBt=((1-k)**t)*P;#in Rs.\n\n#result\nprint \"D5 in Rs. : \",round(Dt,3);\nprint \"B5 in Rs. : \",round(Bt,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "D5 in Rs. :  8192.0\nB5 in Rs. :  32768.0\n"
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.5 Page 129"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nSum=n*(n+1)/2;#sum of the years\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=[100000.0,0,0,0,0,0,0,0,0];#book vakue\nDt=[0.0,0,0,0,0,0,0,0,0];\nrate=[0.0,0,0,0,0,0,0,0,0];\n\n#result\nprint \"End of year          Depreciation          Book value\";\nprint \"0                      0                    100000\"\nfor i in xrange (1,9):\n    rate[i]=(8-i+1)/36.0;\n    Dt[i]=rate[i]*(P-F);\n    Bt[i]=Bt[i-1]-Dt[i];\n    print  eoy[i],\"                    \",Dt[i],\"            \", Bt[i],\" \";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "End of year          Depreciation          Book value\n0                      0                    100000\n1                      17777.7777778              82222.2222222  \n2                      15555.5555556              66666.6666667  \n3                      13333.3333333              53333.3333333  \n4                      11111.1111111              42222.2222222  \n5                      8888.88888889              33333.3333333  \n6                      6666.66666667              26666.6666667  \n7                      4444.44444444              22222.2222222  \n8                      2222.22222222              20000.0  \n"
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.6 Page 131"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nt=5.0;#in years\n\n#cacualtion\nDt=(n-t+1)*(P-F)/(n*(n+1)/2);#in Rs.\nBt=(P-F)*((n-t)/n)*((n-t+1)/(n+1))+F;#in Rs.\n\n#result\nprint \"D5 in Rs. : \",round(Dt,3);\nprint \"B5 in Rs. : \",round(Bt,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "D5 in Rs. :  8888.889\nB5 in Rs. :  33333.333\n"
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.7 Page 132"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\nk=0.12;\ni=12.0;#in % per annum\neoy=[0,1,2,3,4,5,6,7,8]#end of year\nBt=[100000.0,0,0,0,0,0,0,0,0];#book vakue\nDt=[0.0,0,0,0,0,0,0,0,0];\nrate=[0.0,0,0,0,0,0,0,0,0];\nSUM=0;\n\n#calculation\nA=(P-F)*(i/100)/(((1+i/100)**n)-1);\n\n#result\nprint \"End of year          Net Depreciation          Book value\";\nprint \"0                      0                         100000\"\nfor i in xrange (1,9):\n    Dt[i]=k*SUM+A;\n    SUM=SUM+Dt[i]\n    Bt[i]=Bt[i-1]-Dt[i];\n    print  eoy[i],\"                    \",round(Dt[i],3),\"            \", round(Bt[i],3),\" \";\n    \nprint \"fixed deprececiation which is constant in Rs\",round(A,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "End of year          Net Depreciation          Book value\n0                      0                         100000\n1                      6504.227              93495.773  \n2                      7284.735              86211.038  \n3                      8158.903              78052.135  \n4                      9137.971              68914.164  \n5                      10234.528              58679.637  \n6                      11462.671              47216.966  \n7                      12838.191              34378.774  \n8                      14378.774              20000.0  \nfixed deprececiation which is constant in Rs 6504.227\n"
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.8 Page 133"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=100000.0;#in Rs\nF=20000.0;#in Rs\nn=8.0;#in years\ni=12.0;#in % per annum\nt=5.0;#in Years\n\n#calculation\nDt=(P-F)*(i/100)/(((1+i/100)**n)-1)*(1+i/100)**(t-1);#in Rs.\nt=7;#in Years\nBt=P-(P-F)*(i/100)/(((1+i/100)**n)-1)*(((1+i/100)**t)-1)/(i/100);#in Rs.\n\n#result\nprint \"D7 in Rs. : \",round(Dt,3);\nprint \"B7 in Rs. : \",round(Bt,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "D7 in Rs. :  10234.528\nB7 in Rs. :  34378.774\n"
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.9 Page 134"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nP=8000000.0;#in Rs\nF=50000.0;#in Rs\nX=75000.0;#in Km\nx=2000.0;#in Km\nn=8.0;#in years\ni=12.0;#in % per annum\n\n#calcualtion\nD=(P-F)*x/X;#in Rs.\n\n#result\nprint \"Depreciation for year 3 in Rs. : \",round(D,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Depreciation for year 3 in Rs. :  212000.0\n"
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Economics_by_R_Panneerselvam/README.txt b/Engineering_Economics_by_R_Panneerselvam/README.txt
new file mode 100755
index 00000000..27bb08be
--- /dev/null
+++ b/Engineering_Economics_by_R_Panneerselvam/README.txt
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+Contributed By: Tarun Kumar Das
+Course: btech
+College/Institute/Organization: College of Engineering
+Department/Designation: Industrial Engineering
+Book Title: Engineering Economics
+Author: R. Panneerselvam
+Publisher: Phi Learning, Delhi
+Year of publication: 2010
+Isbn: 978-81-203-1743-7
+Edition: 1
\ No newline at end of file
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--- /dev/null
+++ b/Engineering_Physics_by_Uma_Mukherji/Chapter12.ipynb
@@ -0,0 +1,287 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0e9faaea32136a2f476b53b6ab2d5d2eb5330fb68ebd80aaad9bbe7ed1328c93"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "12: Radioactivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 12.1, Page number 351"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "M7Li3=7.018232; #mass of 7li3(amu)\n",
+      "Malpha=4.003874;  #mass of alpha particle(amu)\n",
+      "Mpr=1.008145;  #mass of proton(amu)\n",
+      "Ey=9.15;   #K.E energy of product nucleus\n",
+      "\n",
+      "#Calculation\n",
+      "#xMy -> x-mass no., M-element, y-atomic no.\n",
+      "#reaction:- 7li3 + 1H1-> 4He2 + 4He2\n",
+      "deltaM=M7Li3+Mpr-2*Malpha;  #mass defect(amu)\n",
+      "Q=deltaM*931;   #mass defect(MeV)\n",
+      "Ex=2*Ey-Q;    #K.E of incident particle(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"kinetic energy of incident proton is\",round(Ex,4),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kinetic energy of incident proton is 0.9564 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 12.2, Page number 351"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "M235U=235;   #atomic mass of 235U\n",
+      "m=10**-3;    #mass of fissions(gm)\n",
+      "N=6.023*10**23;   #avagadro number\n",
+      "Eperfi=200*10**6;   #energy per fission(eV)\n",
+      "T=10**-6;    #time(s)\n",
+      "\n",
+      "#Calculation\n",
+      "E=Eperfi*1.6*10**-19;   #energy per fission(J)\n",
+      "A=M235U; \n",
+      "P=((m*N)/A)*(E/T);    #power explosion(Watt)\n",
+      "\n",
+      "#Result\n",
+      "print \"power of explosion is\",P,\"Watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power of explosion is 8.20153191489e+13 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 12.4, Page number 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "n=0.4;  #efficiency\n",
+      "N=6.06*10**26;   #avagadro number\n",
+      "Eperfi=200*10**6;  #energy per fission(eV)\n",
+      "P=100*10**6;    #electric power(W)\n",
+      "A=235;\n",
+      "\n",
+      "#Calculation\n",
+      "E=Eperfi*1.6*10**-19;   #energy per fission(J)\n",
+      "T=24*60*60;    #time(sec)\n",
+      "N235=P*T/(E*n);   #number of atoms in 235 kg of U235\n",
+      "m=(A*N235)/N;     #mass of 235U consumed/day(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass of 235U consumed/day is\",int(m*10**3),\"g\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of 235U consumed/day is 261 g\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 12.5, Page number 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "M2H1=2.01474;   #mass of M2H1(amu)\n",
+      "M3H1=3.01700;   #mass of M3H1(amu)\n",
+      "M1n0=1.008986;  #mass of M1n0(amu)\n",
+      "M4He2=4.003880;  #mass of M4He2(amu)\n",
+      "\n",
+      "#Calculation\n",
+      "#thermonuclear reaction in hydrogen bomb explosion \n",
+      "#2H1 + 3H1 -> 4He2 + 1n0\n",
+      "Mreac=M2H1+M3H1;   #mass of reactants(amu)\n",
+      "Mprod=M4He2+M1n0;  #mass of products(amu)\n",
+      "Q=Mreac-Mprod;   #amount of energy released per reaction(J)\n",
+      "Q=Q*931;    #amount of energy released per reaction(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"amount of energy released per reaction is\",round(Q,3),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of energy released per reaction is 17.572 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 12.6, Page number 353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "M7Li3=7.01818;   #mass of Li atom(amu)\n",
+      "M1H1=1.0081;     #mass of H atom(amu)\n",
+      "M1n0=1.009;     #mass of neutron(amu)\n",
+      "\n",
+      "#Calculation\n",
+      "BEpernu=(1/7)*((3*M1H1)+(4*M1n0)-M7Li3);  #binding energy per nucleon(J)\n",
+      "BEpernu=BEpernu*931;   #binding energy per nucleon(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"binding energy per nucleon is\",BEpernu,\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "binding energy per nucleon is 5.60196 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 12.7, Page number 353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#importing modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m=10*10**3;    #mass of U235(gm)\n",
+      "N=6.02*10**23;    #avagadro number\n",
+      "Eperfi=200*10**6;  #energy per fission(eV)\n",
+      "A=235;\n",
+      "\n",
+      "#Calculation\n",
+      "E=Eperfi*1.6*10**-19;   #energy(J)\n",
+      "T=24*60*60;   #time(s)\n",
+      "P=((m*N)/A)*(E/T);    #power output(Watt)\n",
+      "\n",
+      "#Result\n",
+      "print \"power output is\",round(P/10**9,3),\"*10**9 Watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power output is 9.488 *10**9 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Engineering_Physics_by_Uma_Mukherji/Chapter12.ipynb.bkup b/Engineering_Physics_by_Uma_Mukherji/Chapter12.ipynb.bkup
deleted file mode 100755
index 0efc0c8d..00000000
--- a/Engineering_Physics_by_Uma_Mukherji/Chapter12.ipynb.bkup
+++ /dev/null
@@ -1,287 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0e9faaea32136a2f476b53b6ab2d5d2eb5330fb68ebd80aaad9bbe7ed1328c93"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "12: Radioactivity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example number 12.1, Page number 351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "M7Li3=7.018232; #mass of 7li3(amu)\n",
-      "Malpha=4.003874;  #mass of alpha particle(amu)\n",
-      "Mpr=1.008145;  #mass of proton(amu)\n",
-      "Ey=9.15;   #K.E energy of product nucleus\n",
-      "\n",
-      "#Calculation\n",
-      "#xMy -> x-mass no., M-element, y-atomic no.\n",
-      "#reaction:- 7li3 + 1H1-> 4He2 + 4He2\n",
-      "deltaM=M7Li3+Mpr-2*Malpha;  #mass defect(amu)\n",
-      "Q=deltaM*931;   #mass defect(MeV)\n",
-      "Ex=2*Ey-Q;    #K.E of incident particle(MeV)\n",
-      "\n",
-      "#Result\n",
-      "print \"kinetic energy of incident proton is\",round(Ex,4),\"MeV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "kinetic energy of incident proton is 0.9564 MeV\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example number 12.2, Page number 351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "M235U=235;   #atomic mass of 235U\n",
-      "m=10**-3;    #mass of fissions(gm)\n",
-      "N=6.023*10**23;   #avagadro number\n",
-      "Eperfi=200*10**6;   #energy per fission(eV)\n",
-      "T=10**-6;    #time(s)\n",
-      "\n",
-      "#Calculation\n",
-      "E=Eperfi*1.6*10**-19;   #energy per fission(J)\n",
-      "A=M235U; \n",
-      "P=((m*N)/A)*(E/T);    #power explosion(Watt)\n",
-      "\n",
-      "#Result\n",
-      "print \"power of explosion is\",P,\"Watt\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "power of explosion is 8.20153191489e+13 Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example number 12.4, Page number 352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "n=0.4;  #efficiency\n",
-      "N=6.06*10**26;   #avagadro number\n",
-      "Eperfi=200*10**6;  #energy per fission(eV)\n",
-      "P=100*10**6;    #electric power(W)\n",
-      "A=235;\n",
-      "\n",
-      "#Calculation\n",
-      "E=Eperfi*1.6*10**-19;   #energy per fission(J)\n",
-      "T=24*60*60;    #time(sec)\n",
-      "N235=P*T/(E*n);   #number of atoms in 235 kg of U235\n",
-      "m=(A*N235)/N;     #mass of 235U consumed/day(kg)\n",
-      "\n",
-      "#Result\n",
-      "print \"mass of 235U consumed/day is\",int(m*10**3),\"g\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mass of 235U consumed/day is 261 g\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example number 12.5, Page number 352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "M2H1=2.01474;   #mass of M2H1(amu)\n",
-      "M3H1=3.01700;   #mass of M3H1(amu)\n",
-      "M1n0=1.008986;  #mass of M1n0(amu)\n",
-      "M4He2=4.003880;  #mass of M4He2(amu)\n",
-      "\n",
-      "#Calculation\n",
-      "#thermonuclear reaction in hydrogen bomb explosion \n",
-      "#2H1 + 3H1 -> 4He2 + 1n0\n",
-      "Mreac=M2H1+M3H1;   #mass of reactants(amu)\n",
-      "Mprod=M4He2+M1n0;  #mass of products(amu)\n",
-      "Q=Mreac-Mprod;   #amount of energy released per reaction(J)\n",
-      "Q=Q*931;    #amount of energy released per reaction(MeV)\n",
-      "\n",
-      "#Result\n",
-      "print \"amount of energy released per reaction is\",round(Q,3),\"MeV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "amount of energy released per reaction is 17.572 MeV\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example number 12.6, Page number 353"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "M7Li3=7.01818;   #mass of Li atom(amu)\n",
-      "M1H1=1.0081;     #mass of H atom(amu)\n",
-      "M1n0=1.009;     #mass of neutron(amu)\n",
-      "\n",
-      "#Calculation\n",
-      "BEpernu=(1/7)*((3*M1H1)+(4*M1n0)-M7Li3);  #binding energy per nucleon(J)\n",
-      "BEpernu=BEpernu*931;   #binding energy per nucleon(MeV)\n",
-      "\n",
-      "#Result\n",
-      "print \"binding energy per nucleon is\",BEpernu,\"MeV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "binding energy per nucleon is 5.60196 MeV\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example number 12.7, Page number 353"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "m=10*10**3;    #mass of U235(gm)\n",
-      "N=6.02*10**23;    #avagadro number\n",
-      "Eperfi=200*10**6;  #energy per fission(eV)\n",
-      "A=235;\n",
-      "\n",
-      "#Calculation\n",
-      "E=Eperfi*1.6*10**-19;   #energy(J)\n",
-      "T=24*60*60;   #time(s)\n",
-      "P=((m*N)/A)*(E/T);    #power output(Watt)\n",
-      "\n",
-      "#Result\n",
-      "print \"power output is\",round(P/10**9,3),\"*10**9 Watt\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "power output is 9.488 *10**9 Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fiber_Optics_Communication_by_H._Kolimbiris/chapter1.ipynb b/Fiber_Optics_Communication_by_H._Kolimbiris/chapter1.ipynb
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diff --git a/Fiber_Optics_Communication_by_H._Kolimbiris/chapter9.ipynb b/Fiber_Optics_Communication_by_H._Kolimbiris/chapter9.ipynb
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diff --git a/Fiber_Optics_Communication_by_H._Kolimbiris/screenshots/Capture1.png b/Fiber_Optics_Communication_by_H._Kolimbiris/screenshots/Capture1.png
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diff --git a/Fluid_Mechanics/Chapter_1.ipynb b/Fluid_Mechanics/Chapter_1.ipynb
deleted file mode 100755
index a2dd3e26..00000000
--- a/Fluid_Mechanics/Chapter_1.ipynb
+++ /dev/null
@@ -1,61 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4a6f98ae985b8e1b6e034e8590a04939cfd519f9354fb6a0851df17e977d9e9f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1: Fluid and their Properties"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1, Page 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      " \n",
-      " #Initialization  of  Variable\n",
-      "sigma  =  72.7*10**-3;                                 #Surface  Tension\n",
-      "r  =  1  *10**-3;                                      #Radius  of  Bubble\n",
-      "\n",
-      " #Calculations\n",
-      "P  =  2*sigma/r;\n",
-      "print \"Excess Pressure(N/m2) :\",P\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Excess Pressure(N/m2) : 145.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_10.ipynb b/Fluid_Mechanics/Chapter_10.ipynb
deleted file mode 100755
index 510f6bab..00000000
--- a/Fluid_Mechanics/Chapter_10.ipynb
+++ /dev/null
@@ -1,341 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:03b40b6d5c44ff714aac13480905f296c46afa917e2740a687e8aa66bb21b428"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10: Laminar and Turbulent Flows in Bounded System"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.1, Page 329"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import symbols,diff,solve\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "mu  =  0.9;\n",
-      "rho  =  1260;\n",
-      "g  =  9.81;\n",
-      "x  =  45;                                                                  #theta  in  degrees\n",
-      "P1  =  250  *  10**3;\n",
-      "P2  =  80*  10**3;\n",
-      "Z1  =  1;\n",
-      "Z2  =  0;                                                                  #  datum\n",
-      "U  =  -1.5;\n",
-      "Y  =  0.01;\n",
-      "\n",
-      " #Calculations\n",
-      "gradP1  =  P1+  rho*g*Z1;\n",
-      "gradP2  =  P2+  rho*g*Z2;\n",
-      "DPstar  =  (gradP1-gradP2)*math.sin(math.radians(x))/(Z1-Z2);\n",
-      "A  =  U/Y;                                                                #  Coefficient  U/Y  for  equation  10.6\n",
-      "B  =  DPstar/(2*mu)                                              #  Coefficient  dp*/dx  X(1/2mu)  for  equation  10.6\n",
-      "y  =  symbols('y')\n",
-      "v  =  round((A  +  B*Y),1)*y  -round(B)*y**2\n",
-      "duBYdy  =  diff(v,y);\n",
-      "tau  =  0.9*duBYdy;\n",
-      "stagPnts = solve(duBYdy,y)\n",
-      "ymax=stagPnts[0]                                      #value  of  y  where  derivative  vanishes.;\n",
-      "umax  =  (A + B*Y)*ymax  +  B*ymax**2;              #  Check  the  value  there  is  slight  mistake  in  books  answer\n",
-      "def  u(y):\n",
-      "    z  =  (A + B*Y)*y -B*y**2;\n",
-      "    return diff(z,y)\n",
-      "def dif(y):\n",
-      "    return round((A + B*Y)) -2*round(B)*y\n",
-      "\n",
-      "taumax=abs(mu*dif(Y))\n",
-      "\n",
-      "print \"velocity distribution         :\",v\n",
-      "print \"shear stress distribution     :\",mu*dif(y)\n",
-      "print \"maximum flow velocity (m/s)   :\",round(umax,2)\n",
-      "print \"Maximum Shear Stress  (kN/m^2):\",(round(taumax)/1000)\n",
-      "  \n",
-      "\n",
-      "print "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "velocity distribution         : -71638.0*y**2 + 566.4*y\n",
-        "shear stress distribution     : -128948.4*y + 509.4\n",
-        "maximum flow velocity (m/s)   : 3.36\n",
-        "Maximum Shear Stress  (kN/m^2): 0.78\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2, Page 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "mu  =  0.9;\n",
-      "rho  =  1260;\n",
-      "d  =  0.01;\n",
-      "Q  =  1.8/60*10**-3;                                              #Flow  in  m**3  per  second\n",
-      "l  =  6.5;\n",
-      "ReCrit  =  2000;\n",
-      " #Calculations\n",
-      "A  =  (math.pi*d**2)/4;\n",
-      "MeanVel  =  Q/A;\n",
-      "Re  =  rho*MeanVel*d/mu/10;                                    #  Check  properly  the  answer  in  book  there  is  something  wrong\n",
-      "Dp  =  128*mu*l*Q/(math.pi*d**4)\n",
-      "Qcrit  =  Q*ReCrit/Re*10**3;\n",
-      "\n",
-      "print \"Pressure Loss        (kN/m2) :\",round(Dp/1000,0)\n",
-      "print \"Maximum Flow rate (litres/s) :\",round(Qcrit,0)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure Loss        (kN/m2) : 715.0\n",
-        "Maximum Flow rate (litres/s) : 112.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.3, Page 341"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "mu  =  1.14*10**-3;\n",
-      "rho  =  1000;\n",
-      "d  =  0.04;\n",
-      "Q  =  4*10**-3/60;                                    #Flow  in  m**3  per  second\n",
-      "l  =  750;\n",
-      "ReCrit  =  2000;\n",
-      "g  =  9.81;\n",
-      "k  =  0.00008;                                       #  Absolute  Roughness\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  (math.pi*d**2)/4;\n",
-      "MeanVel  =  Q/A;\n",
-      "Re  =  rho*MeanVel*d/mu;\n",
-      "Dp  =  128*mu*l*Q/(math.pi*d**4);\n",
-      "hL  =  Dp/(rho*g);\n",
-      "f  =  16/Re;\n",
-      "hlDa  =  4*f*l*MeanVel**2/(2*d*g);                   #  By  Darcy  Equation\n",
-      "Pa  =  rho*g*hlDa*Q;\n",
-      "\n",
-      " #Part(b)\n",
-      "Q  =  30*10**-3/60;                                  #Flow  in  m**3  per  second\n",
-      "MeanVel  =  Q/A;\n",
-      "Re  =  rho*MeanVel*d/mu;\n",
-      "RR  =  k/d;                                          #  relative  roughness\n",
-      "f  =  0.008                                          #by  Moody  diagram  for  Re  =    1.4  x  10**4    and  relative  roughness  =  0.002\n",
-      "hlDb  =  4*f*l*MeanVel**2/(2*d*g);                   #  By  Darcy  Equation\n",
-      "Pb  =  rho*g*hlDb*Q;\n",
-      "\n",
-      "\n",
-      "print \"!---- Case (a) ----!\\n\",\"Head Loss(mm)      :\",round(hlDa*1000,1)\n",
-      "print \"Power Required (W) :\",round(Pa,4)\n",
-      "print \"\\n!---- Case (b) ----!\\n\",\"Head Loss(m)       :\",round(hlDb,2)\n",
-      "print \"Power Required (W) :\",round(Pb,0)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "!---- Case (a) ----!\n",
-        "Head Loss(mm)      : 92.5\n",
-        "Power Required (W) : 0.0605\n",
-        "\n",
-        "!---- Case (b) ----!\n",
-        "Head Loss(m)       : 4.84\n",
-        "Power Required (W) : 24.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.4, Page 343"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "w  =  4.5;\n",
-      "d  =  1.2  ;\n",
-      "C  =  49;\n",
-      "i  =  1/800;\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  d*w;\n",
-      "P  =  2*d  +  w;\n",
-      "m  =  A/P;\n",
-      "v  =  C*(m*i)**0.5;\n",
-      "Q  =  v*A;\n",
-      "\n",
-      "print \"Mean Velocity (m/s):\",round(v,2)\n",
-      "print \"Discharge (m3/s)   :\",round(Q,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean Velocity (m/s): 1.53\n",
-        "Discharge (m3/s)   : 8.28\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5, Page 348"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import sympy\n",
-      "from sympy import symbols\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "r,R = symbols('r R')\n",
-      "\n",
-      "#Calculations\n",
-      "rbyR=round((1-(49/60)**7),3)\n",
-      "r = (rbyR)*R \n",
-      "\n",
-      "#Result\n",
-      "print \"radius at which actual velocity is equal to mean velocity is\",r"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "radius at which actual velocity is equal to mean velocity is 0.758*R\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.7, Page 355"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "d1  =  0.140;\n",
-      "d2  =  0.250;\n",
-      "DpF_DpR  =  0.6;                                       #Difference  in  head  loss  when  in  forward  and  in  reverse  direction\n",
-      "K  =  0.33                                             #From  table\n",
-      "g  =  9.81;\n",
-      " #Calculations\n",
-      "ratA  =  (d1/d2)**2;\n",
-      "v =(DpF_DpR*2*g/((1-ratA)**2-K))**0.5;\n",
-      "\n",
-      "print \"Velocity (m/s):\",round(v,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity (m/s): 9.13\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_11.ipynb b/Fluid_Mechanics/Chapter_11.ipynb
deleted file mode 100755
index ac453318..00000000
--- a/Fluid_Mechanics/Chapter_11.ipynb
+++ /dev/null
@@ -1,120 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8adfc21cf4552e14b563ca7c0f2e56ece56cd7f1691cf9486f2cfbd789df7be1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11: Boundary Layer"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.1, Page 383"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "rho  =  860;\n",
-      "v  =  10**-5;\n",
-      "Us  =  3;\n",
-      "b  =  1.25;\n",
-      "l  =  2;\n",
-      "\n",
-      " #Calculations\n",
-      "x  =  1;                                                                    #  At  x  =1  \n",
-      "Rex  =  Us*x/v;\n",
-      "ReL  =  Us*l/v  ;\n",
-      "mu  =    rho*v;\n",
-      "T0  =  0.332*mu*Us/x*Rex**0.5;\n",
-      "Cf  =  1.33*ReL**-0.5;\n",
-      "F  =  rho*Us**2*l*b*Cf  ;\n",
-      "print \"shear stress                                (N/m^2) :\" ,round(T0,1)\n",
-      "print \"Total, double-sided resistance of the plate (N)     :\",round(F,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "shear stress                                (N/m^2) : 4.7\n",
-        "Total, double-sided resistance of the plate (N)     : 33.224\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2, Page 387"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Us  =  6;\n",
-      "b  =  3;\n",
-      "l  =  30;\n",
-      "rho  =  1000;\n",
-      "mu  =  10**-3;\n",
-      "T  =  20+273;                                                          #  Temperature  in  Kelvin\n",
-      "\n",
-      " #Calculations\n",
-      " \n",
-      "ReL  =  rho*Us*l/mu;  \n",
-      "Cf  =  0.455*math.log10(ReL)**-2.58    ;\n",
-      "\n",
-      "F  =  rho*Us**2*l*b*Cf  ;\n",
-      "Lt  =  10**5*mu/(rho*Us);                                    #  Assuming  transition  at  Rel  =  10**5\n",
-      "\n",
-      "print \"Drag Force (kN)       :\", round(F/1000,2)\n",
-      "print \"Transition length (m) :\",round(Lt,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drag Force (kN)       : 6.36\n",
-        "Transition length (m) : 0.0167\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_12.ipynb b/Fluid_Mechanics/Chapter_12.ipynb
deleted file mode 100755
index 745a2241..00000000
--- a/Fluid_Mechanics/Chapter_12.ipynb
+++ /dev/null
@@ -1,319 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:24438326f2aac267ced1e2e3aa21e094819eefb50d5f8862608b62208400992d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12: Incompressible Flow around a Body"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1, Page 399"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "x  =35;\n",
-      "T  =  50;\n",
-      "m  =  1;\n",
-      "g  =9.81;\n",
-      "rho  =  1.2;\n",
-      "A  =  1.2;\n",
-      "U0  =  40*1000/3600;                                            #  Velocity  in  m/s\n",
-      "\n",
-      " #Calculations\n",
-      "L  =  T*math.sin(math.radians(x))+m*g;\n",
-      "D  =T*math.cos(math.radians(x));\n",
-      "Cl  =  2*L/(rho*U0**2*A);\n",
-      "Cd  =  2*D/(rho*U0**2*A);  \n",
-      "\n",
-      "print \"Lift Coefficient :\",round(Cl,3)\n",
-      "print \"Drag Coefficient :\",round(Cd,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Lift Coefficient : 0.433\n",
-        "Drag Coefficient : 0.461\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2, Page 406"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Vp  =12;\n",
-      "lp  =  40;\n",
-      "lm  =  1;\n",
-      "As  =  2500;\n",
-      "Dm  =  32;\n",
-      "rhoP  =  1025;\n",
-      "rhoM  =  1000;\n",
-      "Ap  =  As;\n",
-      "\n",
-      " #Calculations\n",
-      "Am  =  As/40**2;\n",
-      "Vm  =  round(Vp*(lm/lp)**0.5,2);\n",
-      "Dfm  =  round(3.7*Vm**1.95*Am,1);\n",
-      "Rm  =  Dm  -  Dfm;\n",
-      "Rp  =  Rm  *(rhoP/rhoM)*(lp/lm)**2*(Vp/Vm)**2;\n",
-      "Dfp  =  2.9*Vp**1.8*Ap;\n",
-      "Dp  =  Rp  +  Dfp;\n",
-      "\n",
-      "print \"Expected total resistance (kN) :\",round(Dp/1000,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Expected total resistance (kN) : 1407.07\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3, Page 410"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "U0  =  80*1000/3600;\n",
-      "d  =  0.02;\n",
-      "rho  =1.2;\n",
-      "mu  =  1.7*10**-5;\n",
-      "A  =  0.02*500;                            #  Projected  area  of  wire\n",
-      "N  =  20;                                  #  No  of  cables\n",
-      "\n",
-      " #Calculations\n",
-      "Re  =  rho*U0*d/mu;\n",
-      "Cd  =  1.2                                 #  From  figure  12.10  for  given  Re;  \n",
-      "D  =  0.5*rho*Cd*A*U0**2\n",
-      "F  =  N*D;  \n",
-      "f  =    0.198*(U0/d)*(1-19.7/Re);\n",
-      "\n",
-      "print \"Total force on tower (kN) :\",round(F/1000,2)\n",
-      "print \"Frequency (Hz)            :\",round(f,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total force on tower (kN) : 71.11\n",
-        "Frequency (Hz)            : 219.9\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4, Page 415"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "mu  =  0.03;\n",
-      "d  =  10**-3;\n",
-      "rhoP  =  1.1*10**3;\n",
-      "g  =  9.81;\n",
-      "rho0  =  0.9*10**3;\n",
-      " #Calculations\n",
-      "B  =  18*mu/(d**2*rhoP);\n",
-      "t  =  round(4.60/B,4);\n",
-      "Vt  =  round(d**2*(rhoP  -  rho0)*g/(18*mu),5);\n",
-      "Re  =  rho0*Vt*d/mu;\n",
-      "\n",
-      "print \"Time taken by the particle take to reach 99 per cent of its terminal velocity (s):\",t\n",
-      "print \"\\nReynolds No corrosponding to the velocity :\",Re"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time taken by the particle take to reach 99 per cent of its terminal velocity (s): 0.0094\n",
-        "\n",
-        "Reynolds No corrosponding to the velocity : 0.1089\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5, Page 417"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "muO  =  0.0027;\n",
-      "Vt  =  3*10**-3;\n",
-      "rhoW  =  1000;\n",
-      "rhoP  =  2.4*rhoW;\n",
-      "rhoO  =  0.9*rhoW;\n",
-      "g  =  9.81;\n",
-      "muA  =  1.7*10**-5;\n",
-      "rhoA  =  1.3;\n",
-      "\n",
-      " #Calculations\n",
-      "d  =  (18*muO*Vt/(rhoP-rhoO)/g)**0.5;\n",
-      "Re  =  Vt*d*rhoO/muO;\n",
-      "\n",
-      " #Movement  of  particle  in  upward  direction\n",
-      "if(Re < 1):\n",
-      "    v  =  0.5;\n",
-      "    \n",
-      "    Re=5;       #  from  fig  12.15\n",
-      "    vt  =  muA*Re/(rhoA*d);\n",
-      "    u  =  vt+v;\n",
-      "    print \"Velocity of air stream blowing vertically up (m/s) :\",round(u,3) \n",
-      "else:\n",
-      "    print \"strokes law is not valid\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity of air stream blowing vertically up (m/s) : 1.157\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6, Page 429"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "c  =  2;\n",
-      "s  =  10;\n",
-      "rho  =  5.33;\n",
-      "rho_ellip  =  1.2;\n",
-      "D  =  400;\n",
-      "L  =  45000;\n",
-      "scale  =  20;\n",
-      "U_windTunnel  =  500;\n",
-      "U_proto  =  400*1000/3600;\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  c*s;\n",
-      "U_model  =  U_windTunnel/scale;\n",
-      "Cd  =  D/(0.5*rho*U_model**2*A);\n",
-      "Cl  =  L/(0.5*rho_ellip*U_proto**2*A);          #  Considering  elliptical  Lift  model\n",
-      "Cdi  =  Cl**2/(math.pi*s/c);                    #  Aspect  Ratio  =    s/c  \n",
-      "Cdt  =  Cd  +  Cdi;\n",
-      "Dw  =  0.5*Cdt*rho_ellip*U_proto**2*A;\n",
-      "print \"Total drag on full sized wing (kN) :\",round(Dw/1000,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total drag on full sized wing (kN) : 2.65\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_13.ipynb b/Fluid_Mechanics/Chapter_13.ipynb
deleted file mode 100755
index 664b4abf..00000000
--- a/Fluid_Mechanics/Chapter_13.ipynb
+++ /dev/null
@@ -1,142 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1f446f61d8838d40d2e31fd72ef3a0bc1e1a7f291c54a48c1366aef232ac3797"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13: Compressible Flow around a Body"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.1, Page 444"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "rho0  =  1.8;\n",
-      "R  =  287;\n",
-      "T  =  75+273;                                                          #  Temperature  in  kelvin  \n",
-      "gma  =  1.4;  \n",
-      "Ma  =  0.7;\n",
-      "\n",
-      " #Calculations\n",
-      "P0  =  rho0*R*T;\n",
-      "c  =  (gma*R*T)**0.5;\n",
-      "V0  =  Ma*c;\n",
-      "Pt  =  (P0**((gma-1)/gma)  +  rho0*((gma-1)/gma)*(V0**2/(2*P0**(1/gma))))**(gma/(gma-1));\n",
-      "rhoT  =    rho0*(Pt/P0)**(1/gma);\n",
-      "Tt  =    Pt/(R*rhoT)-273;\n",
-      "\n",
-      "print \"Staganation Pressure (kN/m2 ) :\",round(Pt/1000,1)\n",
-      "print \"Temperature (Degree Celcius)  :\",round(Tt,1)\n",
-      "print \"Density of airstream (kg/m3)  :\",round(rhoT,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Staganation Pressure (kN/m2 ) : 249.4\n",
-        "Temperature (Degree Celcius)  : 109.1\n",
-        "Density of airstream (kg/m3)  : 2.274\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.2, Page 454"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "R  =  287;\n",
-      "T  =  28+273;\n",
-      "gma  =  1.4;\n",
-      "P  =  1.02*10**5;\n",
-      "rhoHg  =  13.6*10**3;\n",
-      "g  =  9.81;\n",
-      "\n",
-      " #Calculations\n",
-      " #Case(a)\n",
-      "U0  =  50;\n",
-      "c  =  (gma*R*T)**0.5;\n",
-      "Ma  =  U0/c;\n",
-      "rho  =  P/(R*T);\n",
-      "DelP  =  0.5*rho*U0**2;                                        #Pt-P\n",
-      "ha  =  DelP/(rhoHg*g);\n",
-      "\n",
-      " #Case(b)\n",
-      "U0  =  250;\n",
-      "Ma  =  U0/c;\n",
-      "Pt  =  P*(1+(gma-1)*Ma**2/2)**(gma/(gma-1));    \n",
-      "DelP  =  Pt-P\n",
-      "hb  =  DelP/(rhoHg*g);\n",
-      "\n",
-      " #Case  (c)\n",
-      "U0  =  420;\n",
-      "Ma1  =U0/c;\n",
-      "P2  =  P*((2*gma/(gma+1))*Ma1**2  -  ((gma-1)/(gma+1)));\n",
-      "N  =  Ma1**2  +2/(gma-1);                                      #  Numerator\n",
-      "D  =  2*gma*Ma1**2/(gma-1)-1;\n",
-      "Ma2  =  (N/D)**0.5;\n",
-      "Pt2  =  P2*(1+(gma-1)*Ma2**2/2)**(gma/(gma-1));\n",
-      "hc  =  (Pt2-P2)/(rhoHg*g)  ;    \n",
-      "\n",
-      "print \"Difference in height of mercury column in case (a) in mm :\",round(ha*1000,2)\n",
-      "print \"Difference in height of mercury column in case (b) in mm :\",round(hb*1000,1)\n",
-      "print \"Difference in height of mercury column in case (c) in mm :\",round(hc*1000,0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Difference in height of mercury column in case (a) in mm : 11.06\n",
-        "Difference in height of mercury column in case (b) in mm : 314.2\n",
-        "Difference in height of mercury column in case (c) in mm : 684.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_14.ipynb b/Fluid_Mechanics/Chapter_14.ipynb
deleted file mode 100755
index 2e55c7c8..00000000
--- a/Fluid_Mechanics/Chapter_14.ipynb
+++ /dev/null
@@ -1,365 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:75785fe7e68940de5df81a115443b8d5c78793053ae837735e58f86a072c05cc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14: Steady Incompressible Flow in Pipe and Duct System"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1, Page 468"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "L1  =  5;\n",
-      "L2  =  10;\n",
-      "d  =  0.1;\n",
-      "f  =  0.08;\n",
-      "Za_Zc  =  4;                                                 #difference  in  height  between  A  and  C  \n",
-      "g  =  9.81  ;\n",
-      "Pa  =  0;\n",
-      "Va  =  0;  \n",
-      "Za_Zb  =  -1.5;\n",
-      "V  =  1.26;\n",
-      "rho  =  1000;\n",
-      "\n",
-      " #Calculations\n",
-      "D  =  1.5  +  4*f*(L1+L2)/d  ;                               #  Denominator  in  case  of  v**2  \n",
-      "v  =  (2*g*Za_Zc/D)**0.5;\n",
-      "Pb  =  rho*g*Za_Zb  -  rho*V**2/2*(1.5+4*f*L1/d);\n",
-      "print \"Pressure in the pipe at B (kN/m2):\",round(Pb/1000,2)\n",
-      "print \"Mean Velocity at C    (m/s)      :\",round(v,2)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure in the pipe at B (kN/m2): -28.61\n",
-        "Mean Velocity at C    (m/s)      : 1.26\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3, Page 473"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import symbols,solve\n",
-      "import sympy\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Za_Zb  =  10;\n",
-      "f  =  0.008;\n",
-      "L  =  100;\n",
-      "d1  =  0.05;\n",
-      "g  =  9.81;\n",
-      "d2  =  0.1;\n",
-      "\n",
-      " #Calculations\n",
-      "\n",
-      "def  flowRate(d):\n",
-      "    D = 1.5 + 4*f*L/d ;                                     #  Denominator  in  case  of  v1**2\n",
-      "    A  =  math.pi*d**2/4;\n",
-      "    v  =  (2*g*Za_Zb/D)**0.5;\n",
-      "    z  =  A*v;\n",
-      "    return z     \n",
-      "Q1  =  flowRate(d1);\n",
-      "Q2  =  flowRate(d2);\n",
-      "Q=round(Q1+Q2,4)\n",
-      "\n",
-      "\n",
-      "D=symbols('D')\n",
-      "roots=solve(241212*D**5  -3.2, D)\n",
-      "dia=roots[0]\n",
-      "\n",
-      "print \"Rate flow for pipe 1            (m^3/s) :\",round(Q1,4)\n",
-      "print \"Rate flow for pipe 2            (m^3/s) :\",round(Q2,4)\n",
-      "print \"Combined Rate flow              (m^3/s) :\",round(Q,4)\n",
-      "print \"Diameter of single equivalent pipe (mm) :\",round(dia,3)*1000\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rate flow for pipe 1            (m^3/s) : 0.0034\n",
-        "Rate flow for pipe 2            (m^3/s) : 0.019\n",
-        "Combined Rate flow              (m^3/s) : 0.0224\n",
-        "Diameter of single equivalent pipe (mm) : 106.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4, Page 476"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import solve,symbols\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Za_Zb  =  16;\n",
-      "Za_Zc  =  24;\n",
-      "f  =  0.01;\n",
-      "l1  =  120;\n",
-      "l2  =  60;\n",
-      "l3  =  40;\n",
-      "d1  =  0.12;\n",
-      "d2  =  0.075;\n",
-      "d3  =  0.060;\n",
-      "g  =  9.81;\n",
-      " #Calculations\n",
-      "\n",
-      "v1=symbols('v1')\n",
-      "ash=solve(v1-0.3906*(g-1.25*v1**2)**0.5-0.25*(17.657-1.5*v1**2)**0.5,v1)\n",
-      "v1=round(abs(ash[0]),2)\n",
-      "Q1=math.pi/4*d1**2*v1\n",
-      "\n",
-      "v2=(g-1.25*v1**2)**0.5\n",
-      "Q2=math.pi/4*d2**2*v2\n",
-      "\n",
-      "v3=(17.657-1.5*v1**2)**0.5\n",
-      "Q3=math.pi/4*d3**2*v3\n",
-      "\n",
-      "print \"Flow rate in pipe 1 (m^3/s):\",round(Q1,4)\n",
-      "print \"Flow rate in pipe 2 (m^3/s):\",round(Q2,4)\n",
-      "print \"Flow rate in pipe 3 (m^3/s):\",round(Q3,4)\n",
-      "print \"continuity condition satisfied as Q1=Q2+Q3\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flow rate in pipe 1 (m^3/s): 0.0206\n",
-        "Flow rate in pipe 2 (m^3/s): 0.0105\n",
-        "Flow rate in pipe 3 (m^3/s): 0.0101\n",
-        "continuity condition satisfied as Q1=Q2+Q3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5, Page 480"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "D  =  0.3;\n",
-      "Q  =  0.8;\n",
-      "rho  =  1.2;\n",
-      "f  =  0.008;\n",
-      "L_entry  =  10;\n",
-      "L_exit  =  30;\n",
-      "Lt  =  20*D                              #Transition  may  be  represented  by  a  separation  loss  equivalent  length  of  20  *  the  approach  duct  diameter\n",
-      "K_entry  =  4;\n",
-      "K_exit  =  10\n",
-      "l  =  0.4;                               #  length  of  cross  section\n",
-      "b  =  0.2;                               #  width  of  cross  section\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  math.pi*D**2/4;\n",
-      "Dp1  =  0.5*rho*Q**2/A**2*(K_entry  +  4*f*(L_entry+Lt)/D);\n",
-      "area  =  l*b;\n",
-      "perimeter  =2*(l+b);\n",
-      "m  =  area/perimeter;\n",
-      "Dp2  =  0.5*rho*Q**2/area**2*(K_exit  +  f*L_exit/m);\n",
-      "Dfan  =    Dp1+Dp2;\n",
-      "\n",
-      "print \"fan Pressure input (N/m2) :\",round(Dfan,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fan Pressure input (N/m2) : 1254.6\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.6, Page 482"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "D  =  [0.15 , 0.3];\n",
-      "rho  =  1.2;\n",
-      "f  =  0.008;\n",
-      "L_entry  =  10;\n",
-      "L_exit  =  20;\n",
-      "Lt  =  20*D[1]              \n",
-      "K  =  4;\n",
-      "Q1  =  0.2;\n",
-      "\n",
-      " #Calculations\n",
-      "Q2  =  4*Q1;\n",
-      "A=[0.0,0.0]\n",
-      "A[0]  =  math.pi*D[0]**2/4;\n",
-      "A[1]  =  math.pi*D[1]**2/4;\n",
-      "Dp1  =  0.5*rho*Q1**2/A[0]**2*(K  +  4*f*L_entry/D[0]);\n",
-      "Dp2  =  0.5*rho*Q2**2/A[1]**2*(4*f*(L_exit  +  Lt)/D[1]);\n",
-      "Dfan  =    Dp1+Dp2;\n",
-      "\n",
-      "print \"fan Pressure input (N/m2) :\",round(Dfan,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fan Pressure input (N/m2) : 684.51\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.7, Page 487"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      " \n",
-      " \n",
-      " \n",
-      " #Initializing  the  variables\n",
-      "d  =  [0.1 , 0.125,  0.15,  0.1,  0.1  ];                    #  Corrosponding  to  AA1B  AA2B  BC  CD  CF\n",
-      "l  =  [30 , 30 , 60,  15,  30];                                      #  Corrosponding  to  AA1B  AA2B  BC  CD  CF\n",
-      "rho   =  1.2;\n",
-      "f     =  0.006;\n",
-      "Ha    =  100;\n",
-      "Hf    =  60;\n",
-      "He    =  40;\n",
-      "K = [0.0, 0.0, 0.0, 0.0, 0.0]\n",
-      " #Calculations\n",
-      "for i in range(0,len(l)):\n",
-      "    K[i]  =  f*l[i]/(3*d[i]**5);\n",
-      "\n",
-      "\n",
-      "K_ab  =  K[0]*K[1]/((K[0])**0.5+(K[1])**0.5)**2;\n",
-      "K_ac  =  K_ab  +  K[2];\n",
-      "Hc  =  (K_ac*Hf  +K[4]*Ha/4)/(K_ac+K[4]/4);\n",
-      "Q  =  ((Ha  -  Hc)/K_ac)**0.5;\n",
-      "\n",
-      "def  f(n):\n",
-      "    z  =  He  -  Hc  +  (0.5*Q)**2  *(K[3]+(4000/n)**2);\n",
-      "    return z\n",
-      "\n",
-      "n  =  fsolve(f,1);\n",
-      "\n",
-      "print \"total Volume flow rate (m3/s):\",round(Q, 4)\n",
-      "print \"Head at C (m)                :\",round(Hc,2) \n",
-      "print \"Percentage valve opening (%) :\",round(n,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "total Volume flow rate (m3/s): 0.1016\n",
-        "Head at C (m)                : 75.48\n",
-        "Percentage valve opening (%) : 38.58\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_15.ipynb b/Fluid_Mechanics/Chapter_15.ipynb
deleted file mode 100755
index 95bd78ec..00000000
--- a/Fluid_Mechanics/Chapter_15.ipynb
+++ /dev/null
@@ -1,114 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4f421cc7413696fe7c281427d95aff4b53c3ab081abe9cd6d31d1e2df1133e5a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "    Chapter 15: Uniform Flows in Open Channels"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1, Page 516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "B  =4;\n",
-      "D  =  1.2;\n",
-      "C  =  7.6;\n",
-      "n  =  0.025;\n",
-      "s  =  1/1800;\n",
-      "\n",
-      " #Calculations\n",
-      "W  =  B  +  2*1.5*D;\n",
-      "A  =  D*(B+C)/2;                                                #  Area  of  parallelogram  formed\n",
-      "P  =  B  +2*1.2*(D**2+(1.5)**2)**0.5;\n",
-      "m  =A/P;\n",
-      "i=s;\n",
-      "C  =  (23+0.00155/i+1/n)/(1+(23+0.00155/i)*n/(m)**0.5);         #  By  Kutter  formula\n",
-      "Q1  =  C*A*(m*i)**0.5;\n",
-      "Q2  =  A*(1/n)*m**(2/3)*(i)**0.5;\n",
-      "\n",
-      "print \"Q using Chezy formula with C determined from the Kutter formula (m^3/s) :\",round(Q1,2)\n",
-      "print \"Q using the Manning formula                                     (m^3/s) :\",round(Q2,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Q using Chezy formula with C determined from the Kutter formula (m^3/s) : 5.65\n",
-        "Q using the Manning formula                                     (m^3/s) : 5.69\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.2, Page 518"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Q  =  0.5;\n",
-      "C  =  80;\n",
-      "i  =  1/2000;\n",
-      "\n",
-      " #Calculations\n",
-      "\n",
-      "# A = D**2+(3/4)*D**2 = (7/4)*D**2\n",
-      "D = ((4/7)*(Q/C)*(2/i)**0.5)**(2/5)\n",
-      "\n",
-      "#Result\n",
-      "print \"Optimum depth = Optimum Width (in metres):\",round(D,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Optimum depth = Optimum Width (in metres): 0.552\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_16.ipynb b/Fluid_Mechanics/Chapter_16.ipynb
deleted file mode 100755
index 73f04778..00000000
--- a/Fluid_Mechanics/Chapter_16.ipynb
+++ /dev/null
@@ -1,144 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:130e6e124bfc557e016f12e2356b74f1d97ea210c47b1944be1ebda3297e63a8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 16: Non-Uniform Flow in Open Channels"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2, Page 541"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "B  =  [1.4 ,0.9];\n",
-      "D  =  [0.6 ,0.32];\n",
-      "g  =  9.81;\n",
-      "h  =  0.03;\n",
-      "Z  =  0.25;  \n",
-      "\n",
-      " #Calculations\n",
-      "Q1  =  B[1]*D[1]*(2*g*h/(1-(B[1]*D[1]/B[0]*D[0])**2))**0.5\n",
-      "E  =  D[0]-Z;\n",
-      "Q2  =  1.705*B[1]*E**1.5;\n",
-      "\n",
-      "print \"Volume flow rate (m3/s) :\",round(Q2,4) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Volume flow rate (m3/s) : 0.3177\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3, Page 546"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "a  =0.5;\n",
-      "b  =  0.5;\n",
-      "Dn  =  1.2;\n",
-      "s  =  1/1000;\n",
-      "C  =  55;\n",
-      "g  =  9.81;\n",
-      "\n",
-      " #Calculations\n",
-      "c = (1+a)/b;  \n",
-      "QbyB = Dn*C*(Dn*s)**0.5;\n",
-      "q = QbyB;\n",
-      "Dc = (q**2/g)**(1/3);\n",
-      "\n",
-      "header  =  \"Mean Depth(Dm)    Numenator    Denominator\\t L\"\n",
-      "unit  =    \"   (m) \\t  \\t \\t  \\t(m)\"\n",
-      "\n",
-      "m=[]\n",
-      "Dm=[]\n",
-      "N=[]\n",
-      "D=[]\n",
-      "Lm=[]\n",
-      "total=0\n",
-      "for c in range(7):    \n",
-      "    m.append(2.4-0.15*c);\n",
-      "    Dm.append((m[c]+m[c]-0.15)/2); \n",
-      "    N.append(1 - (Dc/Dm[c])**3) ;    # Numerator\n",
-      "    D.append(1 - (Dn/Dm[c])**3);     # Denominator\n",
-      "    Lm.append(150*(N[c]/D[c]));\n",
-      "    total = total +Lm[c];\n",
-      "\n",
-      "print header\n",
-      "print unit\n",
-      "for c in range(7):\n",
-      "    mm=str(Dm[c])+'\\t       '+str(round(N[c],4))+'        '+str(round(D[c],4))+' \\t'+str(round(Lm[c],2))\n",
-      "    print mm\n",
-      "    \n",
-      "print \"\\ndistance upstream covered (approx in m):\",round(total)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean Depth(Dm)    Numenator    Denominator\t L\n",
-        "   (m) \t  \t \t  \t(m)\n",
-        "2.325\t       0.9576        0.8625 \t166.54\n",
-        "2.175\t       0.9482        0.8321 \t170.94\n",
-        "2.025\t       0.9358        0.7919 \t177.26\n",
-        "1.875\t       0.9192        0.7379 \t186.86\n",
-        "1.725\t       0.8962        0.6634 \t202.65\n",
-        "1.575\t       0.8636        0.5577 \t232.27\n",
-        "1.425\t       0.8159        0.4028 \t303.8\n",
-        "\n",
-        "distance upstream covered (approx in m): 1440.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_17.ipynb b/Fluid_Mechanics/Chapter_17.ipynb
deleted file mode 100755
index 2791e2c2..00000000
--- a/Fluid_Mechanics/Chapter_17.ipynb
+++ /dev/null
@@ -1,274 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c21785277c0a99c3eb4c310a9e20c477f6ab2a0a22da1b2d3844df1699a78d21"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 17: Compressible Flow in Pipes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1, Page 566"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import solve,symbols\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "g  =  9.81;\n",
-      "rho  =  1000;\n",
-      "rhoHg  =  13.6*rho;\n",
-      "d1  =  0.075;\n",
-      "d2  =  0.025;\n",
-      "Pi  =  0.250;\n",
-      "Pt  =  0.150;\n",
-      "P_Hg  =  0.760;\n",
-      "rho1  =  1.6;\n",
-      "gma  =  1.4;\n",
-      "\n",
-      " #Calculations\n",
-      "P1  =  (Pi+P_Hg)*rhoHg*g;\n",
-      "P2  =  (Pt+P_Hg)*rhoHg*g;\n",
-      "rho2  =  rho1*(P2/P1)**(1/gma);\n",
-      "V0=symbols('V0')\n",
-      "V1=symbols('V1')\n",
-      "Velo  =  solve([d2**2*V1*rho2-d1**2*V0*rho1,0.5*(V1**2  -  V0**2)*((gma-1)/gma)*(rho2*rho1/(rho2*P1-rho1*P2))-1],[V0,V1])\n",
-      "s=(Velo[1])[0]\n",
-      "Flow  =  math.pi*d1**2/4*s;\n",
-      "\n",
-      "print \"Volume of flow (m3/s):\",round(Flow,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Volume of flow (m3/s): 0.06\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2, Page 571"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Ma  =  4;\n",
-      "gma  =  1.4;\n",
-      "At  =  500;                                                              #  in  mm\n",
-      "\n",
-      " #Calculations\n",
-      "N  =  1  +  (gma-1)*Ma**2/2;\n",
-      "D  =  (gma+1)/2  ;\n",
-      "#ratio of A/At ==R\n",
-      "R  = round( (N/D)**((gma+1)/(2*(gma-1)))/Ma,2);\n",
-      "A=At*R\n",
-      "print \"Area of test section (mm^2):\",A"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Area of test section (mm^2): 5360.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3, Page 575"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Ma1  =  2;\n",
-      "gma  =  1.4;\n",
-      "T1  =  15+273;                                                        #  In  kelvin\n",
-      "P1  =  105;  \n",
-      "\n",
-      " #Calculations\n",
-      "Ma2  =  (((gma-1)*Ma1**2  +2)/(2*gma*Ma1**2-gma+1))**0.5;\n",
-      "P2  =  P1*(1+gma*Ma1**2)/(1+gma*Ma2**2);\n",
-      "T2  =  T1*(1  +(gma-1)/2*Ma1**2)/(1  +(gma-1)/2*Ma2**2);\n",
-      "\n",
-      "\n",
-      "print \"Mach No downstream shock wave                   :\",round(Ma2,3)\n",
-      "print \"Pressure (bar) of downstream shock wave         :\",round(P2)\n",
-      "print \"Temperature (Degree C) of downstream shock wave :\",T2 - 273"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mach No downstream shock wave                   : 0.577\n",
-        "Pressure (bar) of downstream shock wave         : 473.0\n",
-        "Temperature (Degree C) of downstream shock wave : 213.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.4, Page 581"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "gma  =  1.4;\n",
-      "f  =  0.00375;\n",
-      "d  =  0.05;\n",
-      "\n",
-      " #Calculations\n",
-      "m  =  d/4;\n",
-      "def  x(Ma):\n",
-      "    A  =(1  -Ma**2  )/(gma*Ma**2);\n",
-      "    B  =  (gma+1)*Ma**2/(2+(gma-1)*Ma**2);  \n",
-      "    y  =  m/f*(A+  (gma+1)*math.log(B)/(2*gma));\n",
-      "    return y\n",
-      "\n",
-      "X1  =  x(0.2);                                  #  At  entrance  Ma  =  0.2;\n",
-      "X06_X1  =x(0.6);                                #  Section(b)  Ma  =  0.6;\n",
-      "\n",
-      "X06  =    X1-X06_X1;\n",
-      "\n",
-      "print \"The Distance X1 at which the Mach number is unity (m) :\",round(X1,2)\n",
-      "print \"Distance from the entrance (m)                        :\",round(X06,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Distance X1 at which the Mach number is unity (m) : 48.44\n",
-        "Distance from the entrance (m)                        : 46.81\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.5, Page 585"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "gma  =  1.4;\n",
-      "Q  =  28/60;                             #  m3/s\n",
-      "d  =  0.1;\n",
-      "p1  =  200*10**3;\n",
-      "f  =  0.004;\n",
-      "x_x1  =  60;\n",
-      "R  =  287;\n",
-      "T  =  15+273;\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  math.pi*d**2/4;\n",
-      "m  =  d/4;\n",
-      "v1  =  Q/A;\n",
-      "pa  =  p1*(1-f*(x_x1)*v1**2/(m*R*T))**0.5;\n",
-      "\n",
-      "def g(p):\n",
-      "    A  =  (v1*p1)**2/(R*T)\n",
-      "    B  =  f*A*x_x1/(2*m);\n",
-      "    y  =  (p**2  -  p1**2)/2  -A*math.log(p/p1)  +B;\n",
-      "    return y\n",
-      "    \n",
-      "pb=fsolve(g,pa)                        #  Guessing  solution  around  pa\n",
-      "pb=pb[0]\n",
-      "print \"Pressure at the outlet, neglecting velocity changes   (kN/m2) :\",round(pa/1000,1)\n",
-      "print \"Pressure at the outlet, allowing for velocity changes (kN/m2) :\",round(pb/1000,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure at the outlet, neglecting velocity changes   (kN/m2) : 153.6\n",
-        "Pressure at the outlet, allowing for velocity changes (kN/m2) : 150.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_2.ipynb b/Fluid_Mechanics/Chapter_2.ipynb
deleted file mode 100755
index 835909e1..00000000
--- a/Fluid_Mechanics/Chapter_2.ipynb
+++ /dev/null
@@ -1,493 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:56204579c9dff23fd133fce17f3b8e6b8ac01dae9f7f133693b953a21d5ed34c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2: Pressure and Head"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1, Page 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "z1  =  0;                                                         #Taking  Ground  as  reference\n",
-      "z2  =  -30                                                        #Depth\n",
-      "rho  =  1025                                                      #Density\n",
-      "g  =  9.81                                                        #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculation\n",
-      "pressureIncrease  =  -rho*g*(z2-z1);\n",
-      "\n",
-      "print \"Increase in Pressure (KN/m2):\",round(pressureIncrease/1000,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Increase in Pressure (KN/m2): 301.7\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "p1  =  22.4*10**3                                                    #Initial  Pressure\n",
-      "z1  =  11000                                                            #Initial  Height\n",
-      "z2  =  15000                                                            #final  Height\n",
-      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
-      "R  =  287;                                                                #Gas  Constant\n",
-      "T  =  273-56.6                                                        #Temperature  \n",
-      "\n",
-      " #Calculations\n",
-      "p2  =  p1*math.exp(-g*(z2-z1)/(R*T));\n",
-      "rho2=p2/(R*T);\n",
-      "  \n",
-      "print \"Final Pressure (kN/m2):\",round(p2/1000,2)\n",
-      "print \"Final Density  (kg/m3):\",round(rho2,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Final Pressure (kN/m2): 11.91\n",
-        "Final Density  (kg/m3): 0.192\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3, Page 37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "p1  =  101*10**3                                                      #Initial  Pressure\n",
-      "z1  =  0                                                              #Initial  Height\n",
-      "z2  =  1200                                                           #Final  Height\n",
-      "T1  =  15+273                                                         #Initial  Temperature\n",
-      "g  =  9.81                                                            #Acceleration  due  to  gravity\n",
-      "gamma  =  1.4                                                         #Heat  capacity  ratio\n",
-      "R  =  287                                                                  #Gas  Constant\n",
-      "\n",
-      " #Calculations\n",
-      "p2  =  p1*(1-g*(z2-z1)*(gamma-1)/(gamma*R*T1))**(gamma/(gamma-1));\n",
-      "dT_dZ  =  -(g/R)*((gamma-1)/gamma);\n",
-      "T2  =  T1  +  dT_dZ*(z2-z1);\n",
-      "rho2 = p2/(R*T2);\n",
-      "\n",
-      "print \"Final Pressure (kN/m2)            :\",round(p2/1000,2)\n",
-      "print \"Temprature     (in degree celcius):\",round(T2-273,1)\n",
-      "print \"Density        (kg/m^3)           :\",round(rho2,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Final Pressure (kN/m2)            : 87.33\n",
-        "Temprature     (in degree celcius): 3.3\n",
-        "Density        (kg/m^3)           : 1.101\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4, Page 39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "p1  =  101*10**3                                                      #Initial  Pressure\n",
-      "z1  =  0                                                                    #Initial  Height\n",
-      "z2  =  7000                                                              #Final  Height\n",
-      "T1  =  15+273                                                          #Initial  Temperature\n",
-      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
-      "R  =  287                                                                  #Gas  Constant\n",
-      "dT  =  6.5/1000                                                      #Rate  of  Variation  of  Temperature\n",
-      "\n",
-      " #Calculations\n",
-      "p2  =  p1*(1-dT*(z2-z1)/T1)**(g/(R*dT));\n",
-      "T2  =  T1  -  dT*(z2-z1);\n",
-      "rho2  =  p2/(R*T2);\n",
-      "\n",
-      "\n",
-      "print \"Final Pressure (kN/m^2) :\",round(p2/1000,2)\n",
-      "print \"Final Density  (kg/m^3 ):\",round(rho2,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Final Pressure (kN/m^2) : 40.89\n",
-        "Final Density  (kg/m^3 ): 0.588\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5, Page 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "p  =  350*10**3;                                         #Gauge  Pressure\n",
-      "pAtm  =  101.3*10**3;                                    #Atmospheric  Pressure    \n",
-      "rhoW  =  1000;                                           #Density  of  Water\n",
-      "sigma  =  13.6;                                          #Relative  Density  of  Mercury\n",
-      "g  =  9.81                                               #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "def Head(rho):\n",
-      "    head  =  p/(rho*g);\n",
-      "    return head\n",
-      "rhoM  =  sigma*rhoW;\n",
-      "pAbs  =  p  +  pAtm;\n",
-      "\n",
-      "print \"\\nPart(a)- Equivalent head of water (m) :\",round(Head(rhoW),2)\n",
-      "print \"\\nPart(b)- Equivalent head of water (m) :\",round(Head(rhoM),2)\n",
-      "print \"\\nAbsolute pressure (kN/m^2)            :\",pAbs/1000"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Part(a)- Equivalent head of water (m) : 35.68\n",
-        "\n",
-        "Part(b)- Equivalent head of water (m) : 2.62\n",
-        "\n",
-        "Absolute pressure (kN/m^2)            : 451.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6, Page 45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "   \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "rho  =  10**3;                                  #Density  of  water\n",
-      "h  =  2;                                        #Height\n",
-      "g  =  9.81;                                     #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "p=rho*h*g;  \n",
-      "\n",
-      "print \"Gauge pressure (k/m2) :\",p/1000"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Gauge pressure (k/m2) : 19.62\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.7, Page 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "rho  =  0.8*10**3;                                                  #Density  of  fluid\n",
-      "rhoM  =  13.6*10**3;                                              #Density  of  manometer  liquid\n",
-      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "def fluidPressure(h1,h2):\n",
-      "    P  =  rhoM*g*h2-rho*g*h1;\n",
-      "    return P\n",
-      "\n",
-      "p1=fluidPressure(0.5,0.9)/1000\n",
-      "p2=fluidPressure(0.1,-0.2)/1000\n",
-      "\n",
-      "print \"!-----Part (a)-----! \\nGauge pressure (kN/m2) :\",round(p1,2)\n",
-      "print \"\\n!-----Part (b)-----! \\nGauge pressure (kN/m2) :\",round(p2,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "!-----Part (a)-----! \n",
-        "Gauge pressure (kN/m2) : 116.15\n",
-        "\n",
-        "!-----Part (b)-----! \n",
-        "Gauge pressure (kN/m2) : -27.47\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.8, Page 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "rho  =  10**3;                                                          #Density  of  fluid\n",
-      "rhoM  =  13.6*10**3;                                              #Density  of  manometer  liquid\n",
-      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
-      "H  =  0.3;                                                                #  Differnce  in  height  =  b-a  as  in  text\n",
-      "h  =  0.7;\n",
-      "\n",
-      " #Calculations\n",
-      "result  =  rho*g*H  +  h*g*(rhoM-rho);\n",
-      "\n",
-      "print \"Pressure difference (kN/m^2):\", round(result/1000,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure difference (kN/m^2): 89.467\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.9, Page 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "rho  =  10**3;                                                          #Density  of  fluid\n",
-      "rhoM  =  0.8*10**3;                                                #Density  of  manometer  liquid\n",
-      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
-      "a  =  0.25;\n",
-      "b  =  0.15;\n",
-      "h  =  0.3;\n",
-      " #Calculations\n",
-      "def PressureDiff(a,b,h,rho,rhoM):\n",
-      "    P  =  rho*g*(b-a)  +  h*g*(rho-rhoM);\n",
-      "    return P\n",
-      "print \"The presure difference,if the top of the manometer is filled with\"\n",
-      "print \"(a) air                          :\",PressureDiff(a,b,h,rho,0)/1000, \" N/m^2\"\n",
-      "print \"(b) oil of relative density 0.8. :\",PressureDiff(a,b,h,rho,rhoM), \"N/m^2\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The presure difference,if the top of the manometer is filled with\n",
-        "(a) air                          : 1.962  N/m^2\n",
-        "(b) oil of relative density 0.8. : -392.4 N/m^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.10, Page 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "phi  =  30;                                                              #30  degree\n",
-      "h  =  1.2  ;                                                              #  Height  of  tank\n",
-      "l  =  2;                                                                    #  Length  of  tank\n",
-      "\n",
-      " #Calculations\n",
-      "def SurfaceAngle(a,phi):\n",
-      "    g=9.81; # m/s**2    \n",
-      "    Theta  =  math.atan(-a*math.cos(math.radians(phi))/(g+a*math.sin(math.radians(phi))));  \n",
-      "    return Theta\n",
-      "\n",
-      " #case  (a)  a  =  4\n",
-      "\n",
-      "print \"ThetaA      (degree)      :\",round(180 + 180/math.pi*SurfaceAngle(4,phi),2)\n",
-      "\n",
-      " #Case  (b)  a  =  -  4.5\n",
-      "tanThetaR  =  math.tan((SurfaceAngle(-4.5,phi)));\n",
-      "\n",
-      "print \"\\nThetaR      (degree)      :\",round(SurfaceAngle(-4.5,phi)*180/math.pi,2)\n",
-      "\n",
-      "Depth  =  h  -  l*tanThetaR/2;\n",
-      "print \"\\nMaximum Depth of tank (m) :\",round(Depth,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ThetaA      (degree)      : 163.65\n",
-        "\n",
-        "ThetaR      (degree)      : 27.27\n",
-        "\n",
-        "Maximum Depth of tank (m) : 0.6845\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_20.ipynb b/Fluid_Mechanics/Chapter_20.ipynb
deleted file mode 100755
index f24f7af6..00000000
--- a/Fluid_Mechanics/Chapter_20.ipynb
+++ /dev/null
@@ -1,115 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:042171008edad2dcea90c81036564026ff0ecc789ee16d3a2ccb0122fde141e9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 20: Pressure Transient Theory and Surge Control"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3, Page 692"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "c  =  1250;\n",
-      "Dt  =  0.02;\n",
-      "Dv  =  0.5;\n",
-      "rho  =  1000;\n",
-      "v  =0.5;\n",
-      "\n",
-      " #Calculations\n",
-      "cDt  =  c*Dt;\n",
-      "Dp  =  rho*c*Dv;\n",
-      "DOv_DOt  =  Dv/Dt;\n",
-      "vDOv_DOt  =  v*Dv/cDt;\n",
-      "DOp_DOt  =  Dp/Dt;\n",
-      "vDOp_DOx  =  v*Dp/cDt;\n",
-      "Error  =  [vDOv_DOt*100/DOv_DOt  ,vDOp_DOx*100/DOp_DOt];\n",
-      "print \"The percentage errors are :\",Error,\"\\nThese are very small hence can be neglected\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage errors are : [0.04, 0.04] \n",
-        "These are very small hence can be neglected\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5, Page 705"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "f  =  0;\n",
-      "Atunnel  =  1.227;\n",
-      "Ashaft  =  12.57;\n",
-      "Q  =2;\n",
-      "L  =  200;\n",
-      "g  =  9.81;  \n",
-      "\n",
-      " #Calculations\n",
-      "Zmax  =  (Q/Ashaft)*(Ashaft*L/(Atunnel*g))**0.5;\n",
-      "T  =  2*math.pi*(Ashaft*L/(Atunnel*g))**0.5;\n",
-      "\n",
-      "print \"Peak water level        (m) :\",round(Zmax,3)\n",
-      "print \"Mass Oscillation Period (s) :\",round(T,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak water level        (m) : 2.299\n",
-        "Mass Oscillation Period (s) : 90.8\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_22.ipynb b/Fluid_Mechanics/Chapter_22.ipynb
deleted file mode 100755
index 2c113fed..00000000
--- a/Fluid_Mechanics/Chapter_22.ipynb
+++ /dev/null
@@ -1,191 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:783af12fd7d104e85c7e13e105c7550ca4a0a63a41968e5a9b0312d6fedb23a2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 22: Theory of Rotodynamic Machine"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1, Page 779"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Q  =  5;\n",
-      "R1  =  1.5/2;\n",
-      "R2  =  2/2;  \n",
-      "w  =  18;\n",
-      "rho  =  1000;\n",
-      "rhoA  =  1.2;\n",
-      "Hth  =  0.017;\n",
-      "g=9.81;\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  math.pi*(R2**2-R1**2);\n",
-      "Vf  =  Q/A;\n",
-      "Ut  =  w*R2;\n",
-      "Uh  =  w*R1;\n",
-      "B1t  =  math.atan(Vf/Ut);\n",
-      "B1h  =  math.atan(Vf/Uh);\n",
-      "E  =  Hth*rho/rhoA;\n",
-      "def  Beta(u):\n",
-      "    y  =  math.atan(Vf/(u-E*g/u));\n",
-      "    return y\n",
-      "B2t  =  Beta(Ut);\n",
-      "B2h  =  Beta(Uh);\n",
-      "\n",
-      "print \"!----Blade Inlet Angles----!\"\n",
-      "print \"At tip (in degrees) :\",round(B1t*180/math.pi,1),\"\\nAt Hub (in degrees) :\",round(B1h*180/math.pi,1)\n",
-      "print \"\\n!----Blade Outlet Angles (Degrees)----!\"\n",
-      "print \"At tip (in degrees):\",round(B2t*180/math.pi,1),\"\\nAt Hub(in degrees) :\",round(B2h*180/math.pi,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "!----Blade Inlet Angles----!\n",
-        "At tip (in degrees) : 11.4 \n",
-        "At Hub (in degrees) : 15.1\n",
-        "\n",
-        "!----Blade Outlet Angles (Degrees)----!\n",
-        "At tip (in degrees): 19.5 \n",
-        "At Hub(in degrees) : 48.6\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2, Page 793"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "D  =  0.1;\n",
-      "t  =  15*10**-3;\n",
-      "Q  =  8.5/3600;\n",
-      "N  =  750/60;\n",
-      "B2  =  25*math.pi/180;                                                                #  Beta  2  ind  degrees\n",
-      "g  =  9.81;\n",
-      "z  =  16;\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  round(math.pi*D*t,5);\n",
-      "V_f2  =  round(Q/A,3);\n",
-      "U2  =  round(math.pi*N*D,3);\n",
-      "V_w2  =  round(U2  -  V_f2/math.tan(B2),1);\n",
-      "Hth  =  round(U2*V_w2/g,2);\n",
-      "Sf  =  round(1 - math.pi*math.sin(B2)/(z*(1-(V_f2/U2)/math.tan(B2))),3);\n",
-      "H  =  round(Sf*Hth,2);\n",
-      "\n",
-      "print \"Part (a) - Head developed (m): \",Hth\n",
-      "print \"Part (b) - Head developed (m): \",H"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part (a) - Head developed (m):  1.16\n",
-        "Part (b) - Head developed (m):  1.03\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.3, Page 797"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Ma  =  0.6;\n",
-      "Cl  =  0.6;\n",
-      "tByC  =  0.035;                                                      #  Thickness  to  chord  ratio\n",
-      "cByC  =  0.015;                                                      #  Camber  to  chord  ratio\n",
-      "x  =  3;                                                                    #  Angle  of  incidence\n",
-      "\n",
-      " #Calculations\n",
-      "lamda  =  1/(1-Ma**2)**0.5;\n",
-      "Cl#  =  lamda*Cl;\n",
-      "tByC1  =  tByC*lamda;\n",
-      "cByC1  =  cByC*lamda;\n",
-      "Cl1  =  Cl*lamda**2;\n",
-      "Ae  =  x*lamda;\n",
-      "print \"____Geometric Characterstics____\\n\"\n",
-      "print \"Thickness to chord ratio      :\",tByC1\n",
-      "print \"Camber to chord ratio         :\",cByC1\n",
-      "print \"Lift Coefficient              :\",Cl1\n",
-      "print \"angle of incidence (Degrees)  :\",Ae"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "____Geometric Characterstics____\n",
-        "\n",
-        "Thickness to chord ratio      : 0.04375\n",
-        "Camber to chord ratio         : 0.01875\n",
-        "Lift Coefficient              : 0.9375\n",
-        "angle of incidence (Degrees)  : 3.75\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_23.ipynb b/Fluid_Mechanics/Chapter_23.ipynb
deleted file mode 100755
index 8cb510fa..00000000
--- a/Fluid_Mechanics/Chapter_23.ipynb
+++ /dev/null
@@ -1,302 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ef9e21b794b8d045cd677d625f05028772e4370ff57190a11a0021b5fe3769e2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 23: Performance of Rotodynamic Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1, Page 814"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from pylab import *\n",
-      "%matplotlib inline\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Q  =  []\n",
-      "for c in range(9):\n",
-      "    Q.append(7*c)\n",
-      "H  =  [40, 40.6, 40.4, 39.3, 38, 33.6, 25.6, 14.5, 0];\n",
-      "n  =  [0, 41, 60, 74, 83, 83, 74, 51, 0];\n",
-      "N1  =  750;\n",
-      "N2  =  1450;\n",
-      "D1  =  0.5;\n",
-      "D2  =  0.35;\n",
-      "\n",
-      " #Calculations\n",
-      "Q2=[]\n",
-      "H2=[]\n",
-      "for c in range(9):\n",
-      "    Q2.append(Q[c]*(N2/N1)*(D2/D1)**3);\n",
-      "    H2.append(H[c]*(N2/N1)**2*(D2/D1)**2);\n",
-      "   \n",
-      "plot(Q,H,label='H1')\n",
-      "plot(Q,n,label='n1')\n",
-      "plot(Q2,H2,'--',label='H2')\n",
-      "plot(Q2,n,'--',label='n2')\n",
-      "legend( loc='upper right', numpoints = 1 )\n",
-      "xlabel(\"Q  (m3/s)\");\n",
-      "ylabel(\"H (m of water) and n(percent)\")\n",
-      "grid(True)\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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wWCxCNm7876HesDk5AefPv31CXp7ol3eXLqLN2LhpZ0EnfL7oDcXE5O3PTp4EkpMBV1fA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\nIK1PDnOiPzgAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x322b110>"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2, Page 831"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "n  =  0.9;\n",
-      "g  =  9.81;\n",
-      "D  =  1.45;\n",
-      "N  =  375/60;\n",
-      "H  =  200;                                                                #  Real  height\n",
-      "x  =  165;                                                                #  Theta\n",
-      "P  =  3750*10**3;\n",
-      "rho  =  1000;\n",
-      "\n",
-      " #Calculations\n",
-      "h  =  n*H;                                                                #Effective  Head\n",
-      "v1  =  (2*g*h)**0.5;\n",
-      "u  =  math.pi*D*N;\n",
-      "\n",
-      "n_a  =  (2*u/v1**2)*(v1-u)*(1-n*math.cos(math.radians(x)));\n",
-      "\n",
-      "P_b  =  P/n_a;\n",
-      "ppj  =  P_b/2;                                                        #  Power  per  jet\n",
-      "d  =  (8*ppj/(rho*math.pi*v1**3))**0.5  ;\n",
-      "print \"the efficiency of runner :\",round(n_a,3)\n",
-      "print \"Diameter of Jet (m)      :\",round(d,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the efficiency of runner : 0.933\n",
-        "Diameter of Jet (m)      : 0.156\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3, Page 834"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "                                     #Example  23.3  \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "g  =  9.81;\n",
-      "H  =  12;\n",
-      "n  =  0.8;\n",
-      "w  =  300*2*math.pi/60;\n",
-      "Q  =  0.28;\n",
-      "\n",
-      " #Calculations\n",
-      "V_f1  =  0.15*(2*g*H)**0.5;\n",
-      "V_f2  =V_f1;\n",
-      "V_w1  =  (n*g*H)**0.5;\n",
-      "u1  =  V_w1;\n",
-      "theta  =  math.atan(V_f1/u1);\n",
-      "u2  =0.5*u1;\n",
-      "B2  =  math.atan(V_f2/u2);\n",
-      "r1  =  u1/w;\n",
-      "b1  =  Q/(V_f2*0.9*2*math.pi*r1);                          #  vanes  occupy  10  per  cent  of  the  circumference  hence  0.9\n",
-      "b2  =  2*b1;\n",
-      "\n",
-      "print \"Guide vane angle    (degree)  :\",round(theta*180/math.pi,2)\n",
-      "print \"Vane angle at exit  (degree)  :\",round(B2*180/math.pi,2)\n",
-      "print \"Width of runner at inlet (mm) :\",round(b1*1000,1) \n",
-      "print \"Width of runner at exit  (mm) :\",round(b2*1000,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Guide vane angle    (degree)  : 13.34\n",
-        "Vane angle at exit  (degree)  : 25.38\n",
-        "Width of runner at inlet (mm) : 69.6\n",
-        "Width of runner at exit  (mm) : 139.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.4, Page 838"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import solve,symbols\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "H  =  35;\n",
-      "g  =  9.81;\n",
-      "D  =  2;\n",
-      "N  =  145/60;\n",
-      "z  =  30*math.pi/180;                          #  angle  between  vanes  and  direction  of  runner  rotation\n",
-      "y  =  28*math.pi/180;                          #  angle  between  runner  blades  at  the  outlet.\n",
-      "\n",
-      " #Calculations\n",
-      "H_net  =  0.93*H  ;                            #  since  7%  head  is  lost\n",
-      "v1  =  (2*g*H_net)**0.5;\n",
-      "u  =  math.pi*N*D;\n",
-      "\n",
-      " # from inlet velocity triangle\n",
-      "V_r1=14.3\n",
-      "## ash = cos(beta1+z)\n",
-      "ash = (V_r1**2+v1**2-u**2)/(2*V_r1*v1)\n",
-      "beta1=(z+math.acos(ash))*180/math.pi  # in degrees\n",
-      "\n",
-      "V_r2=(1-8/100)*V_r1         #8 % loss due to friction\n",
-      "V_w1= u + V_r1*math.cos(math.radians(beta1))\n",
-      "V_w2= u - V_r2*math.cos(y)\n",
-      "\n",
-      "E  =  (u/g)*(V_w1  -  V_w2);\n",
-      "n  =  E/H;\n",
-      "print \"Blade angle at inlet :\",round(beta1,1)\n",
-      "print \"Efficiency    (%)    :\",round(n*100)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Blade angle at inlet : 62.1\n",
-        "Efficiency    (%)    : 81.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.5, Page 844"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "s  =  0.03;\n",
-      "P  =  185*10**3;\n",
-      "rho  =  0.86*10**3;\n",
-      "A  =  2.8*10**-2;\n",
-      "N  =  2250/60;\n",
-      "D  =  0.46;\n",
-      "\n",
-      " #Calculations\n",
-      "R0  =  0.46/2;\n",
-      "Ws_Wp  =  1-s;\n",
-      "n  =  Ws_Wp;\n",
-      "Pf  =  s*P;\n",
-      "Q  =  (2*Pf*A**2/(3.5*rho))**(1/3);\n",
-      "Wp  =  2*math.pi*N;  \n",
-      "Ri  =  ((1/Ws_Wp)*(R0**2  -P/(rho*Q*Wp**2)))**0.5;#  Modified  equation  for  power  transmission.\n",
-      "Di  =  2*Ri;\n",
-      "T  =  P/(rho*Wp**3  *D**5);\n",
-      "   \n",
-      "print \"Mean diameter (mm) :\",round(Di*1000)\n",
-      "print \"Torque Coefficient :\",round(T,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean diameter (mm) : 326.0\n",
-        "Torque Coefficient : 0.0008\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_24.ipynb b/Fluid_Mechanics/Chapter_24.ipynb
deleted file mode 100755
index ef4fb925..00000000
--- a/Fluid_Mechanics/Chapter_24.ipynb
+++ /dev/null
@@ -1,183 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0997b77d100b3345365d77881e9dcf23833d271e3f7741c8fb4ad6c260c5ffc6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 24: Positive Displacement Machines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1, Page 860"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "H_at  =  10.3;\n",
-      "Hs  =  1.5;\n",
-      "Hd  =  4.5;\n",
-      "Ls  =  2;\n",
-      "Ld  =  15;\n",
-      "g  =  9.81;\n",
-      "Ds  =  0.4;                                                              #  Diameter  of  stroke\n",
-      "Db  =  0.15;                                                            #  Diameter  of  bore\n",
-      "Dd  =  0.05;                                                            #  Diameter  of  discharge  and  suction  pipe\n",
-      "nu  =  0.2;\n",
-      "f  =  0.01;\n",
-      "abs_pump_pressure  =  2.4;\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  math.pi*(Db)**2/4;\n",
-      "a  =  math.pi*(Dd)**2/4;\n",
-      "r  =  Ds/2;\n",
-      "W  =  2*math.pi*nu;\n",
-      "Hsf  =  0;  \n",
-      "def  H_suck(n):\n",
-      "    y  =  H_at  -  Hs  +(-1)**n*(L/g)*(A/a)*W**2*r;  \n",
-      "    return y\n",
-      "\n",
-      "def  H(n,DischargeOrSuction):\n",
-      "    if(DischargeOrSuction  ==  1):\n",
-      "        y  =  H_at  -  Hs  +(-1)**n*(Ls/g)*(A/a)*W**2*r;\n",
-      "    elif(DischargeOrSuction  ==  2):\n",
-      "        y  =  H_at  +  Hd  +(-1)**n*(Ld/g)*(A/a)*W**2*r;\n",
-      "    else:\n",
-      "        print \"There  is  something  wrong  :\"\n",
-      "    return y\n",
-      "\n",
-      "def  H_mid(DischargeOrSuction,uA):\n",
-      "    if(DischargeOrSuction  ==  1):\n",
-      "        Hsf  =  4*f*Ls/(2*Dd*g)*(uA/a)**2;\n",
-      "        y  =  H_at  -  Hs  -  Hsf;\n",
-      "    elif(DischargeOrSuction  ==  2):\n",
-      "        Hsf  =  4*f*Ld/(2*Dd*g)*(uA/a)**2;\n",
-      "        y  =  H_at  +  Hd  +  Hsf;\n",
-      "    else:\n",
-      "        print \"There  is  something  wrong  :\"\n",
-      "    return y\n",
-      "\n",
-      "Hs_start  =  H(1,1);                #  Inertia  head  negative  hence  n  =  1\n",
-      "Hs_end    =  H(2,1);                #  Inertia  head  positive  hence  n  =  2\n",
-      "Hd_start  =  H(1,2);\n",
-      "Hd_end    =  H(2,2);\n",
-      "u  =  W*r;\n",
-      "Hs_mid  =  H_mid(1,u*A);\n",
-      "slip  =  0.04;\n",
-      "Hd_mid  =  H_mid(2,u*A);\n",
-      "suction   = [Hs_start,  Hs_end,  Hs_mid];\n",
-      "discharge = [Hd_start,  Hd_end,  Hd_mid];\n",
-      "suction1=[0,0,0]\n",
-      "discharge1=[0,0,0]\n",
-      "for c in range(3):\n",
-      "    suction1[c]   =round(suction[c],2)\n",
-      "    discharge1[c] =round(discharge[c],2)\n",
-      "W_max  =  (abs((abs_pump_pressure  -  H_at  +  Hs)*(g/Ls)*(a/A)*(1/r)))**0.5;\n",
-      "W_max_rev  =  W_max/(2*math.pi)*60;                      #  maximum  rotation  speed  in  rev/min\n",
-      "\n",
-      "header  =  \"Start   End   Mid\";\n",
-      "\n",
-      "print \"\\n!----Part(a)----!  Head at \\n\",header\n",
-      "print suction1\n",
-      "print \"\\n!----Part(b)----!  Head at \\n\",header\n",
-      "print discharge1\n",
-      "print \"\\n!----Part(c)----1 \\nDrive speed for s eperation (rev/min) :\",round(W_max_rev)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "!----Part(a)----!  Head at \n",
-        "Start   End   Mid\n",
-        "[8.22, 9.38, 8.38]\n",
-        "\n",
-        "!----Part(b)----!  Head at \n",
-        "Start   End   Mid\n",
-        "[10.45, 19.15, 17.93]\n",
-        "\n",
-        "!----Part(c)----1 \n",
-        "Drive speed for s eperation (rev/min) : 40.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2, Page 863"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Example  24.2  \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "H_friction  =  2.4;\n",
-      "H_at  =  10.3;\n",
-      "Hs  =  1.5;\n",
-      "L  =2;\n",
-      "f  =  0.01;\n",
-      "d  =  0.05;\n",
-      "g  =  9.81;                                 #  Diameter  of  stroke\n",
-      "Db  =  0.15;                                #  Diameter  of  bore\n",
-      "r  =  0.2;\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  math.pi*(Db)**2/4;\n",
-      "a  =  math.pi*(d)**2/4;\n",
-      "W=  (((H_at  -  Hs  -    H_friction  )*(2*d*g/(4*f*L)))**0.5)*(a/A)*(math.pi/r); # in rad/s\n",
-      "W_rev  =  W/(2*math.pi)*60;                 #  maximum  rotation  speed  in  rev/min\n",
-      "# IMPORTANT : In book conversion from rad/s to rev/min is wrong, so answer will be diffrent from book\n",
-      " \n",
-      "print \"speed in          (rad/s)  :\",round(W,2)\n",
-      "print \"Increase in speed (rev/min):\",round(W_rev-40,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "speed in          (rad/s)  : 15.46\n",
-        "Increase in speed (rev/min): 107.65\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_25.ipynb b/Fluid_Mechanics/Chapter_25.ipynb
deleted file mode 100755
index 9b08e6f3..00000000
--- a/Fluid_Mechanics/Chapter_25.ipynb
+++ /dev/null
@@ -1,328 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2bebadfad4a18ede042a136ada8984e181f57290a23ad4ed8003a1f83445a447"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 25: Machine\u2013Network Interactions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.4, Page 893"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Pa_P1  =  -200;                                                      #  From  previous  Question\n",
-      "Q  =  1.4311  ;                                                        #  From  previous  questions.\n",
-      "\n",
-      " #Calculations\n",
-      "DpSys  =  Pa_P1  +  98.9*Q**2;\n",
-      "print \"System Operating point (m^3/s):\",round(DpSys,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "System Operating point (m^3/s): 2.55\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.7, Page 906"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import solve,symbols\n",
-      "   \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Vo  =  25.3;                                                            #Outlet  velocity\n",
-      "D  =  10  ;                                                                #  Mean  hydraulic  diameter\n",
-      "f  =  0.008;                                                            #  friction  factor\n",
-      "X  =  1000;                                                              #  Length  of  road\n",
-      "P  =  12600;                                                            #  Absorbing  power\n",
-      "Va  =  300;                                                              #  Tunnel  air  flow\n",
-      "K1  =  0.96;\n",
-      "K2  =  0.9;\n",
-      "T  =  590;                                                                #Thrust\n",
-      "rho  =  1.2;                                                            #  Air  density  \n",
-      "\n",
-      " #Calculations\n",
-      "alpha  =  (1/D)**2;\n",
-      "A  =  math.pi*D**2/4;                                                    #  Area  of  tunnel\n",
-      "Vt  =  Va/A;\n",
-      "W  =  Vo/Vt;                                                            #Omega\n",
-      "E  =  (1-alpha*W);\n",
-      "C  =  (1-alpha*W)*(1-E)**2  +  E**2  -  1;\n",
-      " #  Manipulating  equation  25.20;\n",
-      "LHS  =  f*X*(E+1)**2/D  +  C  +  1  ;\n",
-      "\n",
-      "n1  =  symbols('n1')\n",
-      "result=solve(K1*(2*((alpha*W**2 + (1-alpha)*E**2-1)+(n1-1)*(alpha*W*(W-1)-C/2)))-LHS)\n",
-      "\n",
-      "n=result[0]\n",
-      "\n",
-      "\n",
-      " #  Alternative  approach  using  equation  25.22\n",
-      "n2  =  (rho*((4*f*X*Vt**2)/(2*D)    +  1.5*Vt**2/2))*A/(K1*K2*T);  \n",
-      "Pt  =  round(n2)*P;\n",
-      "\n",
-      "print \"Number of fans required    :\",round(n2)\n",
-      "print \"Total  power consumed (KW) :\",Pt/1000"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of fans required    : 6.0\n",
-        "Total  power consumed (KW) : 75.6\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.8, Page 907"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import solve,symbols\n",
-      " #Initializing  the  variables\n",
-      "f  =  0.008;\n",
-      "T  =  290;\n",
-      "L  =  750;\n",
-      "Dt  =  9;                                                                  #  Diameter  Tunnel\n",
-      "Df  =  0.63;                                                            #  Diameter  fan\n",
-      "K1  =  0.98;\n",
-      "K2  =  0.92;\n",
-      "Vo  =  27.9;\n",
-      "n  =  10;\n",
-      "A=math.pi*Dt**2/4\n",
-      "rho=1.2\n",
-      "X=750\n",
-      " #Calculations\n",
-      "alpha  =  (Df/Dt)**2;\n",
-      " #  equation  25.20  becomes  when  E  =  1  nad  C  =  0\n",
-      "W=symbols('W')\n",
-      "omega  =  solve(2*K1*  (alpha*W**2  +(n-1)*alpha*W*(W-1))  -  4*f*L/Dt  -1)\n",
-      "  \n",
-      "\n",
-      "for i in range(1,len(omega)):  #  since  omega  is  always  positive  and  real\n",
-      "    if omega[i]>0:\n",
-      "        w  =  round(omega[i],1);\n",
-      "Vt  =  Vo/w;\n",
-      "\n",
-      "# by equation 25.22\n",
-      "VT=(n*(K1*K2*T)/(A*(rho*((4*f*X)/(2*Dt)    +  1.5/2))))**0.5\n",
-      "print \"Tunnel Velocity  (m/s) :\",round(VT,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Tunnel Velocity  (m/s) : 4.05\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.9, Page 914"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Ws  =  0.45;\n",
-      "Ks  =  3.2;\n",
-      "H  =  152;\n",
-      "h  =  0;\n",
-      "Hatm  =  10.3;\n",
-      "Pv  =  350;                             #vapour  pressure\n",
-      "g  =  9.81;\n",
-      "rho  =  1000;\n",
-      "     \n",
-      " #Calculations\n",
-      "Ht1  =  152*(Ws/Ks)**(4/3);     # the value of Ht1 is 11.12 and in book it is taken as 11.2 so there will be a difference in final answer\n",
-      "Hvap  = round(Pv/(rho*g),3);\n",
-      "Z  =  Hatm  -h  -Hvap  -Ht1;\n",
-      "print \"Elevation of pump (m):\",round(Z,3)\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elevation of pump (m): -0.851\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.11, Page 927"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import symbols,solve\n",
-      "import numpy as np\n",
-      "   \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Co  =  0;\n",
-      "Qc  =  0.0024;\n",
-      "V  =  5400;\n",
-      "c  =  10;\n",
-      " #Calculations\n",
-      "#####--------------------PART(A)-----------------#######\n",
-      "n1=symbols('n1')\n",
-      "def partA(n1):\n",
-      "    Ci = 10;\n",
-      "    # t = infinity so e^(-nt) = 0\n",
-      "    Q=10000*Qc/(c-Co)\n",
-      "    n1 = Q*3600/V;    \n",
-      "    return n1\n",
-      "ans=partA(n1)\n",
-      "\n",
-      "print \"Part(A) : number of air changes per hour if the garage is in continuous use and the maximum permissible concentration of carbon monoxide is 0.1 per cent. :\",ans,\"\\n\"\n",
-      "\n",
-      "#####--------------------PART(B)-----------------#######\n",
-      "n=symbols('n')\n",
-      "def   partB(n):\n",
-      "    Ci  =  0;         \n",
-      "    n=[1.5,1.2,0.9,1.0]    \n",
-      "    t=1      #  time  in  hours\n",
-      "    error=[]\n",
-      "    mini=100\n",
-      "    ans=0\n",
-      "    for i in range(4):    \n",
-      "        Q  =  V/3600;           \n",
-      "        A  =   10000*Qc/Q;                                                            # as Co=0 \n",
-      "        error.append(abs((A*(1-math.e**(-n[i]*t))/c)-n[i]));\n",
-      "        if(error[i]<mini):\n",
-      "            mini=error[i]\n",
-      "            ans=n[i]\n",
-      "    return ans       \n",
-      "ans=partB(n)\n",
-      "print \"Part(B) : number of air changes per hour if this maximum level is reached after 1 hour and the garage is out of use :\",ans,'\\n'\n",
-      "  \n",
-      "#####--------------------PART(C)-----------------#######\n",
-      "c=symbols('c')\n",
-      "def   partC(c):\n",
-      "    Ci  =  0;\n",
-      "    n  =  1;   \n",
-      "    t = 0.333333                         #  20  minutes  in  hours\n",
-      "    Q  =  V*n/3600;\n",
-      "    y  =    (Co  +  10000*Qc/Q)*(1-math.e**(-n*t)) + Ci*math.e**(-n*t) ;  \n",
-      "    return y\n",
-      "ans=partC(c)\n",
-      "print \"Part(C) :the concentration after 20 minutes (Parts per 10000) :\",round(ans,3),'\\n'\n",
-      "#####--------------------PART(D)-----------------#######\n",
-      "t=symbols('t')\n",
-      "def   partD(t):\n",
-      "    Ci  =  10;\n",
-      "    n  =  1;  \n",
-      "    c  =  0.1;\n",
-      "    t=np.log(100) \n",
-      "    return round(t,2)\n",
-      "ans=partD(t)    \n",
-      "print \"Part(D) : time necessary to run the ventilation system at the rate calculated in (b) to reduce the concentration to 0.001 per cent (in hours) :\",ans,\"hours\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part(A) : number of air changes per hour if the garage is in continuous use and the maximum permissible concentration of carbon monoxide is 0.1 per cent. : 1.6 \n",
-        "\n",
-        "Part(B) : number of air changes per hour if this maximum level is reached after 1 hour and the garage is out of use : 1.0 \n",
-        "\n",
-        "Part(C) :the concentration after 20 minutes (Parts per 10000) : 4.535 \n",
-        "\n",
-        "Part(D) : time necessary to run the ventilation system at the rate calculated in (b) to reduce the concentration to 0.001 per cent (in hours) : 4.61 hours\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_3.ipynb b/Fluid_Mechanics/Chapter_3.ipynb
deleted file mode 100755
index 68980a40..00000000
--- a/Fluid_Mechanics/Chapter_3.ipynb
+++ /dev/null
@@ -1,372 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:540962ba0b5999b583f0620c9dca124d46f25fe569649c6c842d96cfa42351a3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Static Forces on Surfaces. Buoyancy"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1, Page 65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "a  =  2.7;                                                                #Upper  edge\n",
-      "b  =  1.2  ;                                                              #Lower  edge\n",
-      "width  =  1.5;                                                        #Width  of  trapezoidal  plate\n",
-      "h  =  1.1;                                                                #Height  of  water  column  above  surface\n",
-      "rho  =  1000;\n",
-      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
-      "phi  =  90                                                                #Angle  between  wall  and  surface\n",
-      "\n",
-      " #Calculations\n",
-      "A  =  0.5*(a+b)*width;                                        #Area  of  Trapezoidal  Plate\n",
-      "y  =  (2*(0.5*width*0.75)*0.5  +  (1.2*width)*0.75)/A;\n",
-      "z  =  y+h;                                                                #Depth  of  center  of  pressure\n",
-      "R  =  rho*g*A*z                                                      #Resultant  force\n",
-      "\n",
-      "I0  =  1.2*1.5**3/12  +1.2*1.5*1.85**2  +  1.5*1.5**3/36  +  1.5*0.75*1.6**2  #Second  moment  of  area\n",
-      "D  =  (math.sin(math.degrees(phi)))**2*I0/(A*z);                          #depth  of  center  of  pressure\n",
-      "M  =  R*(1.8533-1.1);                                          #Moment  about  hinge\n",
-      "print \"Moment about the hinge line (kN/m):\",round(M/1000)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Moment about the hinge line (kN/m): 38.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2, Page 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "w  =  1.8;                                                                #Width  of  plate\n",
-      "h1  =  5;                                                                  #Height  of  plate  and  water  in  upstream\n",
-      "h2  =  1.5;                                                              #Height  of  water  in  downstream\n",
-      "rho  =  1000;\n",
-      "g  =  9.81  ;                                                            #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "def waterForce(area,meanHeight):\n",
-      "    F  =  rho  *  g  *  area  *  meanHeight;\n",
-      "    return F\n",
-      "\n",
-      "P  =  waterForce(w*h1,h1/2)-waterForce(w*h2,h2/2);#  Resultant  force  on  gate  \n",
-      "x  =  (waterForce(w*h1,h1/2)*(h1/3)  -  waterForce(w*h2,h2/2)*(h2/3))/P;#  point  of  action  of  p  from  bottom\n",
-      "R  =  P/(2*math.sin(math.radians(20)));                                          #  Total  Reaction  force\n",
-      "Rt  =  1.18*R/4.8;                                                #Reaction  on  Top\n",
-      "Rb  =  R  -  Rt  ;                                                      #Reaction  at  bottom\n",
-      "\n",
-      "print \"Reaction at top    (kN):\",round(Rt/1000,1)\n",
-      "print \"Reaction at bottom (kN):\",round(Rb/1000,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reaction at top    (kN): 72.2\n",
-        "Reaction at bottom (kN): 221.45\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3, Page 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "D  =  1.8;                                                                #Depth  of  tank\n",
-      "h  =  1.2;                                                                #Depth  of  water\n",
-      "l  =  3;                                                                    #Length  of  wall  of  tank\n",
-      "p  =  35000;                                                            #Air  pressure\n",
-      "rho  =  10**3;                                                          #Density  of  water\n",
-      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
-      "\n",
-      "\n",
-      " #Calculations\n",
-      "Ra  =  p*D*l;                                                          #Force  due  to  air\n",
-      "Rw  =  .5*(rho*g*h)*h*l;                                    #Force  due  to  water\n",
-      "R  =  Ra  +  Rw;                                                        #  Resultant  force\n",
-      "x  =  (Ra*0.9+Rw*0.4)/R;                                    #  Height  of  center  of  pressure  from  base\n",
-      "print \"Resultant force on the wall                     (kN) :\",round(R/1000,2)\n",
-      "print \"Height of the centre of pressure above the base (m)  :\",round(x,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resultant force on the wall                     (kN) : 210.19\n",
-        "Height of the centre of pressure above the base (m)  : 0.85\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page 72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "  \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "R  =  6;                                                                    #  Radius  of  arc\n",
-      "h  =  2*R*math.sin(math.radians(30));                                              #Depth  of  water\n",
-      "rho  =  10**3;                                                          #Density  of  water\n",
-      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "Rh  =  (rho*g*h**2)/2;                                          #  Resultant  horizontal  force  per  unit  length\n",
-      "Rv  =  rho*g*((60/360)*math.pi*R**2  -R*math.sin(math.radians(30))*R*math.cos(math.radians(30)));#  Resultant  vertical  force  per  unit  length\n",
-      "R  =  (Rh**2+Rv**2)**0.5;                                        #  Resultant  force  on  gate\n",
-      "theta  =  180/math.pi*math.atan(Rv/Rh);                                      #Angle  between  resultant  force  and  horizontal\n",
-      "\n",
-      "print \"Magnitute of resultant force                  (kN/m)   :\",round(R/1000,2)\n",
-      "print \"Direction of resultant force to the horizontal(Degrees):\",round(theta,2)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnitute of resultant force                  (kN/m)   : 179.45\n",
-        "Direction of resultant force to the horizontal(Degrees): 10.27\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5, Page 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "B  =  6;                                                                    #  Width  of  pontoon\n",
-      "L  =  12;                                                                  #Length  of  pontoon\n",
-      "D  =  1.5;                                                                #Draught  of  pontoon\n",
-      "Dmax  =  2;                                                              #Maximum  permissible  draught\n",
-      "rhoW  =  1000;                                                        #Density  of  fresh  water\n",
-      "rhoS  =  1025;                                                        #Density  of  sea  water\n",
-      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "def Weight(D):\n",
-      "    W  =  rhoW*g*B*L*D;\n",
-      "    return W\n",
-      "\n",
-      "W  =  Weight(D);                                                    #  Weight  of  pontoon  in  fresh  water  =  weight  of  water  displaced\n",
-      "Ds  =  W/(rhoS*g*B*L);                                        #Draught  in  sea  water\n",
-      "L  =  Weight(Dmax)  -  Weight(D);                      #  maximum  load  that  can  be  supported\n",
-      "\n",
-      "print \"Weight of pontoon (kN) :\",round(W/1000,1)\n",
-      "print \"Draught in sea    (m)  :\",round(Ds,2)\n",
-      "print \"Load              (kN) :\",round(L/1000,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Weight of pontoon (kN) : 1059.5\n",
-        "Draught in sea    (m)  : 1.46\n",
-        "Load              (kN) : 353.16\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "D  =  1.8;                                                                #  Diameter  of  buoy\n",
-      "H  =  1.2;                                                                #Height  of  buoy\n",
-      "W  =  10*10**3;                                                        #Weight  of  buoy\n",
-      "L  =  2*10**3;                                                          #Load\n",
-      "G  =  0.45;                                                              #  Center  of  gravity\n",
-      "rho  =  1025;                                                          #Density  of  sea  water\n",
-      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "Z  =  4*(W+L)/(rho*g*math.pi*D**2);                        #  Depth  of  Immersion\n",
-      "BG  =  (math.pi*D**4/64)/(math.pi*D**2*Z/4);\n",
-      "Z  =  0.5*Z  +BG;                                                #  Position  of  combined  center  of  gravity\n",
-      "Z1  =  ((W+L)*Z-0.45*W)/L;                              #Maximum  height  of  load  above  bottom\n",
-      "\n",
-      "print \"Maximum height of center of gravity above bottom (m) :\",round(Z1,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum height of center of gravity above bottom (m) : 1.748\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "l  =  20;                                                                  #  Length  of  barage\n",
-      "b  =  6;                                                                    #Width  of  barage\n",
-      "r  =  3;                                                                    #Radius  of  circular  top  of  barage\n",
-      "W  =  200*10**3;                                                      #Weight  of  empty  barage\n",
-      "d1  =  0.8;                                                              #  Depth  of  water  in  1st  half\n",
-      "d2  =  1;                                                                  #  Depth  of  water  in  2nd  half\n",
-      "rho  =  1000;                                                          #Density  of  water\n",
-      "R  =  0.8;                                                                #Relative  density  of  liquid\n",
-      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
-      "ZG  =  0.45;                                                            #  Center  of  gravity  of  barage\n",
-      "\n",
-      " #Calculations\n",
-      "I00  =  l*b**3/12  +math.pi*b**4/128;\n",
-      "ICC  =  l*(.5*b)**3/12;\n",
-      "L  =  d1*rho*g*l*b/2*(d1+d2);                          #  Weight  of  liquid  load\n",
-      "W  =  L  +  W;                                                          #Total  weight\n",
-      "A  =  l*b  +math.pi*r**2/2;                                          #  Area  of  plane  of  waterline\n",
-      "V  =  W/(rho*g);                                                  #  Volume  of  vessel  submerged\n",
-      "D  =  V/A  ;                                                              #Depth  submerged\n",
-      "ZB  =  .5*D;                                                            #Height  of  center  of  buoyancy\n",
-      "NM  =  ZB-ZG  +(1/V)*(I00-R*2*ICC);                #  Effective  metacentric  height\n",
-      "P  =  R*rho*g*l*b/2*(d2-d1);                            #overturning  moment  \n",
-      "theta  =  math.atan(P*1.5/(W*NM))*180/math.pi;                      #Angle  of  roll\n",
-      "# converting into degrees and minutes\n",
-      "thetaD=round(theta-1)\n",
-      "thetaM=(theta-thetaD)*60/100\n",
-      "print \"Angle of roll is\",thetaD,\"degrees\",round(thetaM,2),\"minutes\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Angle of roll is 2.0 degrees 0.37 minutes\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_4.ipynb b/Fluid_Mechanics/Chapter_4.ipynb
deleted file mode 100755
index 9429e481..00000000
--- a/Fluid_Mechanics/Chapter_4.ipynb
+++ /dev/null
@@ -1,185 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:029b6e47f117cbc4316b9317b5e197658ff26f93dc8874c76dd6b6c464f4fec0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4: Motion Fluid Particles and Streams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1, Page 103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from pylab import *\n",
-      "import numpy as np\n",
-      "%matplotlib inline\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "x = [0, 23, 28, 31, 32, 29, 22, 14, 0]\n",
-      "y = np.linspace(0,80,9)\n",
-      "xlabel('Velocity  (m/s)')\n",
-      "ylabel('Distance  from  one  side(mm)')\n",
-      "title('Velocity Distribution Curve')\n",
-      "grid(1)\n",
-      "\n",
-      " #Calculations\n",
-      "plot(x,y,'-*')\n",
-      "show()\n",
-      "mu=[17.5 , 26.0,  29.6,  31.9,  30.7,  25.4,  18.1,  7.7]\n",
-      " #  mean  velocity\n",
-      "print \"Mean velocity (m/s):\",round(mean(mu),2)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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QYODv778rIiJikkgkEu/atctfntcXCIDWrdklIIC9rbwcuH6dPVpISgIiI4G//wbatXt9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\nYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x31fdd50>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean velocity (m/s): 23.36\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2, Page 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from pylab import *\n",
-      "import numpy as np\n",
-      "\n",
-      " #Initializing  the  variables  all  unknowns  are  assigned  0\n",
-      "\n",
-      "d  =  [0.0, 0.05, 0.075, 0, 0.030];\n",
-      "Q  =  [0, 0, 0, 0, 0];\n",
-      "V  =  [0, 0, 2, 1.5, 0];\n",
-      "A  =  [0, 0, 0, 0, 0]\n",
-      " #Calculations\n",
-      "A2  =  math.pi*d[2]**2/4;\n",
-      "Q[2]  =A2*V[2];\n",
-      "Q  =  [0,Q[2], Q[2],  Q[2]/1.5 , 0.5*Q[2]/1.5];\n",
-      "d[3]  =  (Q[3]*4/(V[3]*math.pi))**0.5;\n",
-      "for i in range(0,5):\n",
-      "    A[i]  =  math.pi*d[i]**2/4;\n",
-      "V[1]  =  V[2]*(A[2]/A[1]);\n",
-      "V[4]=Q[4]/A[4];\n",
-      "\n",
-      "\n",
-      "header  =  \"Diameter(mm)  Area(m2)\\t Flow Rate(m3/s)  Velocity(m/s)\"\n",
-      "print header\n",
-      "for c in range(1,5):\n",
-      "    mm=str(round(d[c]*1000,1))+'\\t  '+str(round(A[c],6))+' \\t    '+str(round(Q[c],6))+'     \\t'+str(round(V[c],2))\n",
-      "    print mm"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diameter(mm)  Area(m2)\t Flow Rate(m3/s)  Velocity(m/s)\n",
-        "50.0\t  0.001963 \t    0.008836     \t4.5\n",
-        "75.0\t  0.004418 \t    0.008836     \t2.0\n",
-        "70.7\t  0.003927 \t    0.00589     \t1.5\n",
-        "30.0\t  0.000707 \t    0.002945     \t4.17\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3, Page 108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "import sympy\n",
-      "from sympy import diff, Symbol\n",
-      "\n",
-      "\n",
-      "#Initializing  the  variables \n",
-      "'''\n",
-      "def df(x, h=0.1e-5):\n",
-      "    return ( f(x+h/2) - f(x-h/2) )/h\n",
-      "    return df\n",
-      "\n",
-      "print df(2*x,h)\n",
-      "'''\n",
-      "x = Symbol('x') \n",
-      "vx  =  3-x\n",
-      "vy  =  4+2*x\n",
-      "vz  =  2-x  \n",
-      "\n",
-      " #Calculations\n",
-      "delVx  =  vx.diff(x);  \n",
-      "delVy  =  vx.diff(x);\n",
-      "delVz  =  vx.diff(x);  \n",
-      "\n",
-      "result  =  delVx+delVy+delVz;#requirement  of  continuity  equation  (result  =  0)\n",
-      "print \"Satisfy requirement of continuity \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Satisfy requirement of continuity \n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_5.ipynb b/Fluid_Mechanics/Chapter_5.ipynb
deleted file mode 100755
index b12249f7..00000000
--- a/Fluid_Mechanics/Chapter_5.ipynb
+++ /dev/null
@@ -1,517 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:04c1d9ce4358772aaf36727d98b4d1c000b18de791fe80fc4e6e1ac3cabc0050"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5: The Momentum Equation and its Applications"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1, Page 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "\n",
-      "l  =  60  ;                                                                #Length  of  pipeline\n",
-      "rho  =  1000;                                                          #  Density  of  liquid\n",
-      "a  =  0.02;                                                              #Acceleration  of  fluid\n",
-      "\n",
-      " #Calculations\n",
-      "delP  =  rho*l*a;                                                  #Change  in  pressure\n",
-      "print \"Increase of pressure difference required (kN/m2):\",delP/1000"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Increase of pressure difference required (kN/m2): 1.2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2, Page 121"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "v  =  5;                                 #Velocity  of  jet          \n",
-      "rho  =  1000;                            #density  of  water\n",
-      "d  =  0.025;                             #Diameter  of  fixed  nozzle\n",
-      "\n",
-      " #Calculations\n",
-      " #--Part(a)  Variation  of  force  exerted  normal  to  the  plate  with  plate  angle--//\n",
-      "header  =  \"Theta\\t vcos(x)\\t pAv\\t Force\"\n",
-      "unit  =    \"deg\\t m/s\\t kg/s\\t N\"\n",
-      "\n",
-      "A  =  math.pi*d**2/4;\n",
-      "x  =  range(0,91,15);\n",
-      "for c in range(len(x)):\n",
-      "    x[c]=1.0*x[c]\n",
-      "m  =  round(rho*A*v,2);\n",
-      "ma  =  [m,m,m,m,m,m,m];\n",
-      "vcomp=[]\n",
-      "force=[]\n",
-      "for c in x:\n",
-      "    vcomp.append(round(v*math.cos(math.radians(c)),2))\n",
-      "    force.append(round((rho*A*v**2)*math.cos(math.radians(c)),2))\n",
-      "\n",
-      "print header\n",
-      "print unit\n",
-      "for c in range(len(x)):\n",
-      "    mm=str(x[c])+' \\t '+str(vcomp[c])+' \\t'+str(ma[c])+' \\t'+str(force[c])\n",
-      "    print mm\n",
-      "##value  = [x,vcomp,ma,force]\n",
-      "##print value,unit, header\n",
-      "\n",
-      " #--Part(b)  Variation  of  force  exerted  normal  to  the  plate  with  plate  velocity--//        \n",
-      "header  =\"Theta\\t v\\t u\\t v-u\\t pA(v-u)\\t Force\\t\"\n",
-      "unit    =\"deg\\t m/s\\t m/s\\t m/s\\t kg/s\\t N\\t\"\n",
-      "x  =  [0,0,0,0,0]\n",
-      "v  =  [5,5,5,5,5]\n",
-      "u  =  range(2,-3,-1);\n",
-      "D=[]\n",
-      "Prod=[]\n",
-      "Force=[]\n",
-      "for c in range(5):\n",
-      "    D.append(v[c]-u[c])\n",
-      "    Prod.append(round((rho*A*D[c]),2))\n",
-      "    Force.append(round((rho*A*D[c]**2),2))\n",
-      "    \n",
-      "print '\\n',\"(b)\",\"\\n\",header\n",
-      "print unit\n",
-      "for c in range(len(x)):\n",
-      "    mm=str(x[c])+' \\t '+str(v[c])+' \\t '+str(u[c])+' \\t '+str(D[c])+' \\t '+str(Prod[c])+' \\t '+str(Force[c])\n",
-      "    print mm\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Theta\t vcos(x)\t pAv\t Force\n",
-        "deg\t m/s\t kg/s\t N\n",
-        "0.0 \t 5.0 \t2.45 \t12.27\n",
-        "15.0 \t 4.83 \t2.45 \t11.85\n",
-        "30.0 \t 4.33 \t2.45 \t10.63\n",
-        "45.0 \t 3.54 \t2.45 \t8.68\n",
-        "60.0 \t 2.5 \t2.45 \t6.14\n",
-        "75.0 \t 1.29 \t2.45 \t3.18\n",
-        "90.0 \t 0.0 \t2.45 \t0.0\n",
-        "\n",
-        "(b) \n",
-        "Theta\t v\t u\t v-u\t pA(v-u)\t Force\t\n",
-        "deg\t m/s\t m/s\t m/s\t kg/s\t N\t\n",
-        "0 \t 5 \t 2 \t 3 \t 1.47 \t 4.42\n",
-        "0 \t 5 \t 1 \t 4 \t 1.96 \t 7.85\n",
-        "0 \t 5 \t 0 \t 5 \t 2.45 \t 12.27\n",
-        "0 \t 5 \t -1 \t 6 \t 2.95 \t 17.67\n",
-        "0 \t 5 \t -2 \t 7 \t 3.44 \t 24.05\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3, Page 123"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "  \n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "x  =  60;                                                     #Angle  of  deflection theta\n",
-      "rho  =  1000;                                                 #  Density  of  liquid\n",
-      "V1  =  30;                                                    #Acceleration  of  fluid\n",
-      "V2  =  25;\n",
-      "m  =  .8;                                                     #Discharge  through  A\n",
-      "\n",
-      " #Calculations\n",
-      "def  Reaction(Vin , Vout):\n",
-      "    R  =  m*(Vin  -Vout)  ;\n",
-      "    return R\n",
-      "Rx  =  Reaction(V1,V2*math.cos(math.radians(x)));\n",
-      "Ry  =  -Reaction(0,V2*math.sin(math.radians(x)));\n",
-      "print \"Reaction in X-direction (N)                              :\",Rx\n",
-      "print \"Reaction in Y-direction (N)                              :\",round(Ry,2)\n",
-      "print \"Net Reaction            (N)                              :\",round((Rx**2 +Ry**2)**0.5,2)\n",
-      "print \"Inclination of Resultant Force with x-direction (Degrees):\",round(180/math.pi*math.atan(Ry/Rx),2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reaction in X-direction (N)                              : 14.0\n",
-        "Reaction in Y-direction (N)                              : 17.32\n",
-        "Net Reaction            (N)                              : 22.27\n",
-        "Inclination of Resultant Force with x-direction (Degrees): 51.05\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4, Page 125"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "v1  =  36  ;                                                              #Exit  velocity\n",
-      "u  =  15;                                                                  #Velocity  of  vane\\\n",
-      "x  =  30;                                                                  #  Angle  between  vanes  and  flow\n",
-      "rho  =  1000;                                                          #  Density  of  water\n",
-      "d  =  .1;                                                                  #  Diameter  of  jet\n",
-      "\n",
-      " #Calculations\n",
-      "alp  =  (180/math.pi)*math.atan((v1*math.sin(math.radians(x))/(v1*math.cos(math.radians(x))-u)));\n",
-      "v2  =  0.85*v1*math.sin(math.radians(x));\n",
-      "bta  =    (180/math.pi)*math.acos((u*math.sin(math.radians(alp))/v2));\n",
-      "m  =  (rho*math.pi*v1*d**2)/4;\n",
-      "Vin  =  v1*math.cos(math.radians(x));\n",
-      "Vout  =  v2*math.cos(math.radians(90));\n",
-      "Rx  =  m*(Vin-Vout);\n",
-      "\n",
-      "\n",
-      "print \"Inlet Angle            (Degrees) :\", round(alp,2)\n",
-      "print \"Outlet Angle           (Degrees) :\", round(bta,2)\n",
-      "print \"Force exerted by vanes (N)       :\", round(Rx) \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Inlet Angle            (Degrees) : 48.05\n",
-        "Outlet Angle           (Degrees) : 43.18\n",
-        "Force exerted by vanes (N)       : 8815.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5, Page 127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "rho  =  850  ;                                                          #  Density  of  liquid\n",
-      "a  =  0.02                                                                #Acceleration  of  fluid\n",
-      "x  =  45  ;\n",
-      "d1  =  .5  ;\n",
-      "d2  =  .25;\n",
-      "p1  =  40*10**3;\n",
-      "p2  =  23*10**3;\n",
-      "Q  =  .45;\n",
-      "   \n",
-      " #Calculations\n",
-      "A1  =  (math.pi*d1**2)/4;\n",
-      "A2  =  (math.pi*d2**2)/4;\n",
-      "v1  =  Q/A1;\n",
-      "v2  =  Q/A2;\n",
-      "\n",
-      "Rx  =  p1*A1  -  p2*A2*math.cos(math.radians(x))  -  rho*Q*(v2*math.cos(math.radians(x))-v1);\n",
-      "Ry  =  p2*A2*math.sin(math.radians(x))  +  rho*Q*v2*math.sin(math.radians(x));\n",
-      "\n",
-      "print \"Resultant force on the bend                     (kN)     :\",round((Rx**2 +Ry**2)**0.5/1000,3)\n",
-      "print \"Inclination of Resultant Force with x-direction (Degrees):\",round(math.atan(Ry/Rx)*180/math.pi)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resultant force on the bend                     (kN)     : 6.362\n",
-        "Inclination of Resultant Force with x-direction (Degrees): 31.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6, Page 129"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "v  =  4.9;                                                                #Velocity  of  Jet\n",
-      "rho  =  1000;                                                          #  Density  of  water\n",
-      "d  =  0.05;\n",
-      "u  =  1.2                                                                  #  Velocity  of  tank\n",
-      " #Calculations\n",
-      "Vout  =  v;\n",
-      "Vin  =  0;\n",
-      "m  =  rho*math.pi*d**2*v/4;\n",
-      "R  =  m*(Vout-Vin);\n",
-      "print \"Reaction of jet on tank (N) :\",round(R,2)\n",
-      "print \"Work done per second    (W) :\",round(R*u,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reaction of jet on tank (N) : 47.14\n",
-        "Work done per second    (W) : 56.57\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7, Page 130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from scipy import integrate\n",
-      " \n",
-      " \n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "Vj  =  5*10**6;                               # Velocity  of  Jet\n",
-      "Mr  =  150000;                                # Mass  of  Rocket\n",
-      "Mf0  =  300000;                               # Mass  of  initial  fuel\n",
-      "Vr  =  3000;                                  # Velocity  of  jet  relative  to  rocket\n",
-      "g  =  9.81;                                   # Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "m  =  Vj/Vr;                                  #Rate  of  fuel  consumption\n",
-      "T  =  Mf0/m;                                  #  Burning  time\n",
-      "\n",
-      "def f(t,m,Vr,Mr,Mf0,g):\n",
-      "    return m*Vr  /(Mr  +  Mf0  -  m*t)  -  g;\n",
-      " \n",
-      "args = (5000/3,3000,150000,300000,9.81)\n",
-      "Vt = integrate.quad(f, 0.0, 180, args)\n",
-      "\n",
-      "def h(t,Vr,g):\n",
-      "    return -g*t - Vr*math.log(1 - t/269.95);\n",
-      " \n",
-      "args = (3000,9.81)\n",
-      "Z1 = integrate.quad(h, 0.0, 180, args)\n",
-      "Z2 =  Vt[0]**2/(2*g);\n",
-      "\n",
-      "print \"(a)Burning time (s)                             :\",T\n",
-      "print \"(b)Speed of rocket when all fuel is burned (m/s):\",round(Vt[0],2)\n",
-      "print \"(c)Maximum height reached (km)                  :\",round((Z2+Z1[0])/1000,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a)Burning time (s)                             : 180.0\n",
-        "(b)Speed of rocket when all fuel is burned (m/s): 1530.04\n",
-        "(c)Maximum height reached (km)                  : 203.8\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8, Page 134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "V  =  200;                                                                #Velocity  in  still  air\n",
-      "Vr  =  700;                                                              #velocity  of  gas  relative  to  engine\n",
-      "mf  =  1.1;                                                              #  Fuel  Consumption\n",
-      "r  =  1/40  ;        \n",
-      "P1  =0;\n",
-      "P2  =  0;\n",
-      "\n",
-      " #Calculations\n",
-      "m1  =    mf/r;\n",
-      "T  =  m1*((1+r)*Vr  -V);\n",
-      "print \"(a)Thrust               (kN) :\",T/1000\n",
-      "\n",
-      "W  =  T*V;\n",
-      "print \"(b)Work done per second (kW) :\",W/1000\n",
-      "\n",
-      "Loss  =  0.5*m1*(1+r)*(Vr-V)**2;\n",
-      "print \"(c)Efficiency           (%)  :\",round(W/(W+Loss)*100,1) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a)Thrust               (kN) : 22.77\n",
-        "(b)Work done per second (kW) : 4554.0\n",
-        "(c)Efficiency           (%)  : 44.7\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10, Page 140"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "rho  =  1000;                                                          #  Density  of  water\n",
-      "Q  =  10;                                                                  #Acceleration  of  fluid\n",
-      "r2  =  1.6;\n",
-      "r1  =  1.2;\n",
-      "V1  =  2.3;\n",
-      "V2  =  0.2;\n",
-      "rot  =  240;  \n",
-      "\n",
-      " #Calculations\n",
-      "Tf  =  rho*Q*(V2*r2  -  V1*r1);\n",
-      "T  =  -Tf;\n",
-      "n  =  rot  /  60;\n",
-      "P  =  2*round(math.pi,3)*n*T;\n",
-      "\n",
-      "print \"Torque exerted by fluid  (N.m):\",T\n",
-      "print \"Theoretical power output (kW) :\",round(P/1000,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Torque exerted by fluid  (N.m): 24400.0\n",
-        "Theoretical power output (kW) : 613.32\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_6.ipynb b/Fluid_Mechanics/Chapter_6.ipynb
deleted file mode 100755
index 5cde4008..00000000
--- a/Fluid_Mechanics/Chapter_6.ipynb
+++ /dev/null
@@ -1,504 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:df3153eb902c7efba51ed445ceefd3e3c02fa6ed938e4ef53fe06333d531bf11"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6: The Energy Equation and its Applications"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1, Page 170"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "Pc  =  0;                                                                  #  Atmospheric  Pressure\n",
-      "Z3  =  30+2;                                                            #height  of  nozzle\n",
-      "Ep  =  50  ;                                                              #Energy  per  unit  weight  supplied  by  pump\n",
-      "d1  =  0.150;                                                          #Diameter  of  sump\n",
-      "d2  =  0.100;                                                          #Diameter  of  delivery  pipe\n",
-      "d3  =  0.075  ;                                                        #Diameter  of  nozzle\n",
-      "g  =  9.81;                                                              #  Acceleration  due  to  gravity\n",
-      "Z2  =  2;                                                                  #Height  of  pump\n",
-      "rho  =  1000;                                                          #  Density  of  water\n",
-      "\n",
-      " #Calculations\n",
-      "U3  = (2*g*(Ep-Z3)/(1+5*(d3/d1)**4  +  12*(d3/d2)**4))**0.5;\n",
-      "U1  =  U3/4;\n",
-      "Pb  =  rho*g*Z2  +  3*rho*U1**2;\n",
-      "print \"Velocity of Jet through nozzle (m/s) :\",round(U3,3)\n",
-      "print \"Pressure in the suction pipe at the inlet to the pump at B (kN/m^2) :\",round(Pb/1000,3)     "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity of Jet through nozzle (m/s) : 8.314\n",
-        "Pressure in the suction pipe at the inlet to the pump at B (kN/m^2) : 32.58\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page 183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "x  =  45;                                                                  #  Inclination  of  pipe\n",
-      "l  =  2;                                                                    #Length  of  pipe  under  consideration\n",
-      "Ep  =  50  ;                                                              #Energy  per  unit  weight  supplied  by  pump\n",
-      "d1  =  0.2;                                                              #Diameter  of  sump\n",
-      "d2  =  0.1;                                                              #Diameter  of  delivery  pipe\n",
-      "g  =  9.81;                                                              #  Acceleration  due  to  gravity\n",
-      "rho  =  1000;                                                          #  Density  of  water\n",
-      "V1  =  2;\n",
-      "RD_oil  =  0.9;                                                      #  relative  density  of  oil\n",
-      "RD_Merc  =  13.6;                                                  #  Relative  density  of  Mercury\n",
-      "\n",
-      " #Calculations\n",
-      "V2  =  V1*(d1/d2)**2;\n",
-      "dZ  =  round(l*math.sin(math.radians(x)),3);       # it is used in book as 1.414,by rounding so here also\n",
-      "rho_Oil  =  RD_oil*rho;\n",
-      "rho_Man  =  RD_Merc*rho;\n",
-      "dP  =  0.5*rho_Oil*(V2**2-V1**2)  +  rho_Oil*g*dZ;\n",
-      "h  =  rho_Oil  *(  dP/(rho_Oil*g)-  dZ)/(rho_Man  -  rho_Oil);\n",
-      "\n",
-      "print \"Pressure Difference(N/m2) : \",round(dP,0)\n",
-      "print \"Difference in the level of mercury (m):\",round(h,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure Difference(N/m2) :  39484.0\n",
-        "Difference in the level of mercury (m): 0.217\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "d1  =  0.25;                                                            #Pipeline  diameter\n",
-      "d2  =  0.10;                                                            #Throat  diameter\n",
-      "h  =0.63;                                                                #Difference  in  height\n",
-      "rho  =  1000;                                                          #Density  of  water\n",
-      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
-      "\n",
-      " #Calculations\n",
-      "rho_Hg  =  13.6*rho;\n",
-      "rho_Oil  =  0.9*rho;\n",
-      "A1  =  (math.pi*d1**2)/4;                                            #  Area  at  entry\n",
-      "m  =  (d1/d2)**2;                                                    #Area  ratio\n",
-      "Q  =  (A1/(m**2-1)**0.5)*(2*g*h*(rho_Hg/rho_Oil  -1))**0.5;\n",
-      "\n",
-      "print \"Thepretical Volume flow rate (m3/s ):\",round(Q,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thepretical Volume flow rate (m3/s ): 0.105\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page 190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "\n",
-      "x  =  1.5;\n",
-      "y  =0.5;\n",
-      "H  =  1.2;\n",
-      "A  =  650*10**-6;\n",
-      "Q  =0.117;\n",
-      "g  =  9.81;\n",
-      "\n",
-      " #Calculations\n",
-      "Cv  =(x**2/(4*y*H))**0.5;\n",
-      "Cd  =  Q  /  (60*A*(2*g*H)**0.5);\n",
-      "Cc  =  Cd/Cv;\n",
-      "\n",
-      "\n",
-      "print \"Coefficient of velocity :\",round(Cv,3)\n",
-      "print \"Coefficient of Discharge :\",round(Cd,3)\n",
-      "print \"Coefficient of contraction :\",round(Cc,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Coefficient of velocity : 0.968\n",
-        "Coefficient of Discharge : 0.618\n",
-        "Coefficient of contraction : 0.639\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page 192"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "B  =  0.7;\n",
-      "H1  =  0.4;\n",
-      "H2  =  1.9;\n",
-      "g  =9.81;\n",
-      "z  =  1.5  ;                                                              #  height  of  opening\n",
-      "\n",
-      " #Calculations\n",
-      "Q_Th  =  2/3  *B*(2*g)**0.5*(H2**1.5  -  H1**1.5);\n",
-      "A  =  z*B;\n",
-      "h  =  0.5*(H1+H2);\n",
-      "Q  =  A*(2*g*h)**0.5;\n",
-      "\n",
-      "print \"Percentage error in discharge (%):\",round((Q-Q_Th)*100/Q_Th,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage error in discharge (%): 1.98\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page 195"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "Cd  =  0.6;                                                              #Coefficient  of  discharge\n",
-      "Q  =  0.28;\n",
-      "x  =  90;                                                                  #Theta\n",
-      "g  =  9.81;\n",
-      "dH  =  0.0015;\n",
-      "\n",
-      " #Calculations\n",
-      "H  =  (Q*(15/8)/(Cd*(2*g)**0.5*math.tan(math.radians(x/2))))**(2/5)\n",
-      "Frac_Q  =  5/2  *(  dH/H);\n",
-      "\n",
-      "print \"Percentage error in discharge(%)\",round(Frac_Q*100,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage error in discharge(%) 0.72\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page 196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "B  =  0.9;\n",
-      "H  =  0.25;\n",
-      "alpha  =  1.1;\n",
-      "g  =  9.81;  \n",
-      "\n",
-      " #Calculations\n",
-      "Q  =  1.84  *  B  *  H**(3/2);\n",
-      "print \"Q(m3/s) :\",Q\n",
-      "\n",
-      "i  =  1;\n",
-      "while(i  <= 3):\n",
-      "    v  =  Q  /(1.2*  (H+0.2));\n",
-      "    print \"V(m/s) :\",round(v,4)\n",
-      "    k  =    ((1  +  alpha*v**2/(2*g*H))**1.5  -(alpha*v**2/(2*g*H))**1.5  );\n",
-      "    Q  =  k*  1.84  *  B  *  H**(3/2);\n",
-      "    print \"Q(m3/s) :\",round(Q,4)\n",
-      "    i  =  i+1;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Q(m3/s) : 0.207\n",
-        "V(m/s) : 0.3833\n",
-        "Q(m3/s) : 0.2161\n",
-        "V(m/s) : 0.4001\n",
-        "Q(m3/s) : 0.2168\n",
-        "V(m/s) : 0.4016\n",
-        "Q(m3/s) : 0.2169\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page 197"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "rho  =  1000;\n",
-      "v  =  66  ;\n",
-      "Q  =  0.13;\n",
-      "g  =  9.81;  \n",
-      "z  =240;\n",
-      "\n",
-      " #Calculations\n",
-      "P_Jet  =    0.5*rho*v**2*Q;\n",
-      "P_Supp  =  rho*g*Q*z;\n",
-      "P_Lost  =    P_Supp  -P_Jet;\n",
-      "h  =  P_Lost/(rho*g*Q);\n",
-      "eff  =  P_Jet/P_Supp;\n",
-      "\n",
-      "print \"Part(a) - power of the jet(kW):                  \",round(P_Jet/1000,2)\n",
-      "print \"Part(b) - power supplied from the reservoir (kW):\",round(P_Supp/1000,2) \n",
-      "print \"Part(C) - head used to overcome losses (m):      \",round(h,2)\n",
-      "print \"Part(d) - Efficiency(%) :                        \",round(eff*100,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part(a) - power of the jet(kW):                   283.14\n",
-        "Part(b) - power supplied from the reservoir (kW): 306.07\n",
-        "Part(C) - head used to overcome losses (m):       17.98\n",
-        "Part(d) - Efficiency(%) :                         92.5\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page 203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from scipy import integrate\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "r1  =  0.2;\n",
-      "Z1  =  0.500;\n",
-      "Z2  =  0.340;\n",
-      "g  =  9.81;\n",
-      "rho  =  0.9*1000  ;\n",
-      "\n",
-      " #Calculations\n",
-      "r0  =  r1*((2-2*Z2/Z1)**0.5);\n",
-      "omega  =   round((2*g*Z1/r0**2)**0.5,1)\n",
-      "\n",
-      "def G(r):\n",
-      "    out  =r**3  -  r*r0**2;\n",
-      "    return out\n",
-      "    \n",
-      "results = integrate.quad(G, r0, r1)\n",
-      "\n",
-      "F  =  rho*omega**2*math.pi*results[0];\n",
-      "\n",
-      "print r0,r1\n",
-      "print \"Part(a) Speed of rotation (rad/s ):\",round(omega,1)\n",
-      "print \"Part(b) Upward force on the cover (N): \",round(F,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0.16 0.2\n",
-        "Part(a) Speed of rotation (rad/s ): 19.6\n",
-        "Part(b) Upward force on the cover (N):  56.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page 206"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "Ra  =  0.2;\n",
-      "Rb  =  0.1;\n",
-      "H  =  0.18;\n",
-      "Za  =  0.125;\n",
-      "\n",
-      " #Calculations\n",
-      "Y  =    Ra**2*(H-Za);\n",
-      "Zb  =  H  -  Y/Rb**2;\n",
-      "\n",
-      "print \"Height above datum of a point B on the free surface at a radius of 100 mm (mm):\",Zb*1000"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Height above datum of a point B on the free surface at a radius of 100 mm (mm): -40.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_7.ipynb b/Fluid_Mechanics/Chapter_7.ipynb
deleted file mode 100755
index 95848de2..00000000
--- a/Fluid_Mechanics/Chapter_7.ipynb
+++ /dev/null
@@ -1,221 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:58b7373e6785b57ec4db0eaf49754884e3dad15de9e9489fc1c2bb500d413f82"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7: Two-Dimentional Idea Flow"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2, Page 235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables  \n",
-      "x  =  120;                                        #Theta\n",
-      "r  =  1;\n",
-      "v0  =  0.5;\n",
-      "q  =  2;\n",
-      "theta =120;\n",
-      "\n",
-      " #Calculations\n",
-      "Vr = v0*r*math.cos(math.radians(theta))  +q/(2*math.pi)\n",
-      "Vth = -v0*math.sin(math.radians(theta))\n",
-      "V  =  (Vr**2+Vth**2)**0.5;\n",
-      "alpha  =  math.atan(abs(Vth/Vr));\n",
-      "bet  =  x-alpha*180/math.pi;\n",
-      "\n",
-      "\n",
-      "print \"Fluid Velocity(m/s) :\",round(V,3)\n",
-      "print \"Beta (Degree)       :\",round(bet,2)\n",
-      "print \"Alpha (Degree)      :\",round(alpha*180/math.pi,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fluid Velocity(m/s) : 0.438\n",
-        "Beta (Degree)       : 38.96\n",
-        "Alpha (Degree)      : 81.04\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3, Page 239"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "import sympy\n",
-      "from sympy import RootOf, I, Symbol\n",
-      " #Initializing  the  variables  \n",
-      "q  =  10;\n",
-      "def  shi(x,y):\n",
-      "    Z  =  (q/2/math.pi)*(math.atan(y/(x-1))-math.atan(y/(x+1))) - 25*y;\n",
-      "    return Z\n",
-      "h  =  0.0000001;\n",
-      "Vinf  =  25;\n",
-      "x=Symbol('x')\n",
-      " #Calculations\n",
-      "#f = lambda x : x**2 - 2/(5*math.pi) -1\n",
-      "result =  [RootOf(x**2- 2/(5*math.pi) -1,i) for i in (0,1)] \n",
-      "\n",
-      "root1=round(result[0],3)\n",
-      "root2=round(result[1],3)\n",
-      "l  =  abs(abs(root1)+abs(root2));\n",
-      "Ymax  =  0.047;\n",
-      "width  =  2*Ymax;\n",
-      "Vx  =  (shi(1-h,1)-shi(1-h,1-h))/h;           #  At  x=1  the  function  atan  is  not  defined  hence  taking  x  a  little  smaller.\n",
-      "Vy  =  -1*(shi(1-2*h,1)-shi(1-h,1))/h;        #  At  x=1  the  function  atan  is  not  defined  hence  taking  x  a  little  smaller.\n",
-      "\n",
-      "V  =  (Vx**2+Vy**2)**0.5;\n",
-      "rho  =  Symbol('rho')\n",
-      "dP  =  rho/2*round((V**2  -  Vinf**2),2);                          #difference  in  pressure\n",
-      "\n",
-      "print \"Pressure Difference   (N/m2)  :\",dP\n",
-      "print \"Velocity              (m/s)   :\",round(V,2)\n",
-      "print \"Length of Rankine Body(m)     :\",l\n",
-      "print \"Width of Rankine Body (m)     :\",width"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure Difference   (N/m2)  : 16.93*rho\n",
-        "Velocity              (m/s)   : 25.67\n",
-        "Length of Rankine Body(m)     : 2.124\n",
-        "Width of Rankine Body (m)     : 0.094\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4, Page 242"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math                                              #Example  7.4\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "a  =  0.02;\n",
-      "r  =  0.05;\n",
-      "V0  =  1;\n",
-      "x  =  135;                                                                #  Theta\n",
-      "def  shi(r,x):\n",
-      "    Z  =  V0*math.sin(math.radians(x))*(r-((a**2)/r));\n",
-      "    return Z\n",
-      "h  =  0.0001;\n",
-      "\n",
-      " #Calculations\n",
-      "Vr  =  57*(shi(r,x+h)-shi(r,x))/(r*h);\n",
-      "Vx  =  -1*(shi(r+h,x)-shi(r,x))/h;\n",
-      "\n",
-      "print \"Radial Velocity (m/s)             :\",round(Vr,3)\n",
-      "print \"Normal component of velocity (m/s):\",round(Vx,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radial Velocity (m/s)             : -0.591\n",
-        "Normal component of velocity (m/s): -0.82\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5, Page 246"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "rho  =  1000;\n",
-      "r  =  2;\n",
-      "psi  =  2*math.log(r);\n",
-      "\n",
-      " #Calculations\n",
-      "y  =  psi/math.log(r);                                                  #  y  =  GammaC  /  2*pi\n",
-      "v  =  y/r;\n",
-      "dPbydr  =  rho*v**2/r;\n",
-      "print \"Pressuer Gradient (N/m3 ) :\",dPbydr"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressuer Gradient (N/m3 ) : 500.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_8.ipynb b/Fluid_Mechanics/Chapter_8.ipynb
deleted file mode 100755
index cd940a0c..00000000
--- a/Fluid_Mechanics/Chapter_8.ipynb
+++ /dev/null
@@ -1,73 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b0713ad3f7c2f18448b69f687926b2d86e094144766f7b44f7fc20525a516074"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8: Dimensional Analysis"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page 268"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "P1  =  57;                                         #Power  in  SI\n",
-      "M  =  1/14.6;                                      #Ratio  of  mass  in  SI/British\n",
-      "L  =  1/0.3048;                                    #Ratio  of  length  in  SI/British\n",
-      "T  =  1;                                           #Ratio  of  time  in  SI/British\n",
-      "\n",
-      "#Calculations\n",
-      "'''\n",
-      "n1 is the horsepower and N1 is the corresponding number of British units (ft*lbf/second),\n",
-      "then     N1 = 550*n1.\n",
-      "Similarly, for the SI system, n2 in kW,\n",
-      "         N2 = 1000*n2\n",
-      "''' \n",
-      "N = (T**3)/(M*L**2)                         # N2/N1\n",
-      "n = N*550/1000                              #n2/n1 \n",
-      "P2  =  n*P1  ;                              #Power  in  kW\n",
-      "\n",
-      "print \"Power in kW       :\",round(P2,1)\n",
-      "print \"Conversion Factor :\",round(n,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power in kW       : 42.5\n",
-        "Conversion Factor : 0.746\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_Mechanics/Chapter_9.ipynb b/Fluid_Mechanics/Chapter_9.ipynb
deleted file mode 100755
index 5d9fe9d6..00000000
--- a/Fluid_Mechanics/Chapter_9.ipynb
+++ /dev/null
@@ -1,337 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:38d704564aa6536fc6da296a78f2a33e8cb9eb20a3ce0fa8f700546de7fd07b5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9: Similarity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1, Page 291"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Vp  =  10;\n",
-      "LpByLm  =  20;\n",
-      "rhoPbyRhoM  =  1;\n",
-      "muPbymuM  =  1;\n",
-      " #Calculations\n",
-      "Vm  =  Vp*LpByLm*rhoPbyRhoM*muPbymuM;\n",
-      "   \n",
-      "print \"Mean water tunnel flow velocity (m/s) :\",Vm"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean water tunnel flow velocity (m/s) : 200\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2, Page 292"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "    \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Vp  =  3;\n",
-      "LpByLm  =  30;\n",
-      "rhoPbyRhoM  =  1;\n",
-      "muPbymuM  =  1;\n",
-      "\n",
-      " #Calculations\n",
-      "Vm  =  Vp*LpByLm*rhoPbyRhoM*muPbymuM;\n",
-      "   \n",
-      "print \"Mean water tunnel flow velocity (m/s):\",Vm"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean water tunnel flow velocity (m/s): 90\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3, Page 293"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "   \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "Vp  =  100;\n",
-      "cP  =  340;\n",
-      "cM  =  295;\n",
-      "rhoM  =  7.7;\n",
-      "rhoP  =  1.2;\n",
-      "muM  =  1.8*10**-5;\n",
-      "muP  =  1.2*10**-5;\n",
-      "\n",
-      " #Calculations\n",
-      "Vm  =  Vp*(cM/cP);\n",
-      "LmByLp  =  1/((Vm/Vp)*(muM/muP)*(rhoM/rhoP));\n",
-      "FmByFp  =  (rhoM/rhoP)*(Vm/Vp)**2*(LmByLp)**2;\n",
-      "\n",
-      "print 'Mean wind tunnel flow velocity(m/s) :',round(Vm,2)\n",
-      "print \"Percentage ratio of forces (%)      :\",round(FmByFp*100,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean wind tunnel flow velocity(m/s) : 86.76\n",
-        "Percentage ratio of forces (%)      : 6.93\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4, Page 295"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "      \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "def pLossRatio(RatRho,RatMu,RatL):\n",
-      "    Z  =  RatRho*RatMu**2*RatL**2;\n",
-      "    return Z\n",
-      "\n",
-      " #Calculations\n",
-      " #Case  (a)  :  water  is  used\n",
-      "RatRho  =  1;\n",
-      "RatMu  =  1;\n",
-      "RatL  =  10;\n",
-      "print \"(a)Ratio of pressure losses between the model and the prototype if water is used :\",pLossRatio(RatRho,RatMu,RatL)\n",
-      "\n",
-      "# Case (b) : air is used\n",
-      "RatRho  =  1000/1.23;\n",
-      "RatMu  =  1.8*10**-5/10**-3;\n",
-      "\n",
-      "print \"(b)Ratio of pressure losses between the model and the prototype if air is used   :\",round(pLossRatio(RatRho,RatMu,RatL),2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a)Ratio of pressure losses between the model and the prototype if water is used : 100\n",
-        "(b)Ratio of pressure losses between the model and the prototype if air is used   : 26.34\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5, Page 296"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      " \n",
-      " #Initializing  the  variables\n",
-      "scale  =  1/50;\n",
-      "ratArea  =  scale**2;\n",
-      "Qp  =  1200;\n",
-      "\n",
-      " #Calculations\n",
-      "LmByLp  =  (ratArea)**0.5;\n",
-      "VmByVp  =  (LmByLp)**0.5;\n",
-      "Qm  =  Qp*ratArea*VmByVp;\n",
-      "\n",
-      "print \"Water flow rate (m3/s ):\",round(Qm,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Water flow rate (m3/s ): 0.068\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6, Page 298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      " \n",
-      " #Initializing  the  variables\n",
-      "Qa  =  2;\n",
-      "Na  =  1400;\n",
-      "rhoA  =  0.92;\n",
-      "rhoS  =  1.3;\n",
-      "DaByDs  =  1;\n",
-      "dPa  =  200;\n",
-      "\n",
-      " #Calculations\n",
-      "Ns  =  Na*(rhoA/rhoS)*(DaByDs);\n",
-      "Qs  =  Qa*(Ns/Na);\n",
-      "dPs  =  dPa  *(rhoS/rhoA)*(Ns/Na)**2*(1/DaByDs)**2;\n",
-      "\n",
-      "print \"Fan test speed (rev/s):\",round(Ns,1)\n",
-      "print \"Flow rate (m3/s)      :\",round(Qs,3)\n",
-      "print \"Pressure rise (N/m2 ) :\",round(dPs,1) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fan test speed (rev/s): 990.8\n",
-        "Flow rate (m3/s)      : 1.415\n",
-        "Pressure rise (N/m2 ) : 141.5\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8, Page 304"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "      \n",
-      "\n",
-      " #Initializing  the  variables\n",
-      "V  =  300                                                                  #  Volume  rate\n",
-      "w  =  3;\n",
-      "d  =  65;\n",
-      "l  =  30;\n",
-      "scaleH  =  30/1000/18;\n",
-      "scaleV  =  1/60;\n",
-      "ZmByZr  =  1/60;\n",
-      "LmByLr  =  1/600;\n",
-      "rho  =  1000;\n",
-      "mu  =  1.14*10**-3;\n",
-      "\n",
-      " #Calculations\n",
-      "Vr  =  V/(w*d);  \n",
-      "Vm  =Vr*(ZmByZr)**0.5;\n",
-      "m  =  (w*d*scaleH*scaleV)/(d*scaleH  +  2*w*scaleV);\n",
-      "Rem  =  rho*Vm*m/mu;\n",
-      "TmByTr  =  LmByLr*(1/ZmByZr)**0.5;\n",
-      "Tm  =  12.4*60*TmByTr;\n",
-      "\n",
-      "\n",
-      "\n",
-      "print \"Tidal Period (minutes):\",round(Tm,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Tidal Period (minutes): 9.6\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
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deleted file mode 100755
index d96dbe87..00000000
--- a/Fluid_Mechanics/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Ashish Kumar
-Course: btech
-College/Institute/Organization: IITB
-Department/Designation: Aerospace Engg.
-Book Title: Fluid Mechanics
-Author: John F. Douglas
-Publisher: Pearson Education
-Year of publication: 2005
-Isbn: 978-0-13-129293-2
-Edition: 5th
\ No newline at end of file
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diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter1.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter1.ipynb
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index 00000000..561385dd
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@@ -0,0 +1,609 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 1 : Fluid Properties"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.1 Page no 8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Desnity of air at 20 deg C and 50 psia =  4.1 kg/m**3"
+     ]
+    }
+   ],
+   "source": [
+    "# Density of air\n",
+    "\n",
+    "#Given \n",
+    "\n",
+    "Mw = 29.0                   # Molecular weight of air\n",
+    "\n",
+    "R = 8323/Mw                 # Universal gas constant\n",
+    "\n",
+    "T = 273 + 20                # temperature in K\n",
+    "\n",
+    "p = 50*144*47.88            # Pressure in N/m**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "rho = p/(R*T)               # from the state law\n",
+    "\n",
+    "print \"Desnity of air at 20 deg C and 50 psia = \",round(rho,2),\"kg/m**3\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.2 Page no 12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Reduction in volume =  0.045 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Reduction in volume\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "dP = 10**6                # Pressure drop in N/m**2\n",
+    "\n",
+    "V = 1                      # Volume in m**3\n",
+    "\n",
+    "bta = 2.2*10**9            # Bulk modulus of elasticity in N/m**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "dV = -dP*V/bta             # Change in volume in m**3\n",
+    "\n",
+    "v = -dV*100\n",
+    "\n",
+    "print \"Reduction in volume = \",round(v,3),\"%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.3 Page no 13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Volume reduction =  -3.57e-06 m**3\n",
+      "Negative sign indicates the volume has reduced"
+     ]
+    }
+   ],
+   "source": [
+    "# Volume reduction\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "bta1 = 2.28*10**9            # Bulk modulus of elasticity at 20 deg C and 103.4 N/m**2\n",
+    "\n",
+    "bta2 = 2.94*10**9            # Bulk modulus of elasticity at 20 deg C and 1034 N/m**2\n",
+    "\n",
+    "p1 = 103.4                   # Pressure in N/m**2\n",
+    "\n",
+    "p2 = 1034                    # Pressure in N/m**2\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "bavg = (bta1+bta2)/2         # bulk modulus average in N/m**2\n",
+    "\n",
+    "dP = p2-p1                   # pressure drop in N/m**2\n",
+    "\n",
+    "V = 10                       # Volume in m**3\n",
+    "\n",
+    "dV = dP*V/bavg               # Change in volume in m**3\n",
+    "\n",
+    "v = -dV\n",
+    "\n",
+    "print \"Volume reduction = \",round(v,8),\"m**3\"\n",
+    "\n",
+    "print \"Negative sign indicates the volume has reduced\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.4 Page no 14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "An increase in pressure by 2 psia will result in a volume reduction of -0.0173 ft**3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Volume reduction \n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "patm = 14.6                  # Atmospheric pressure in psia\n",
+    "\n",
+    "p1 = 100                     # gauge pressure at point 1 in psia\n",
+    "\n",
+    "p2 = 102                     # gauge pressure at point 2 in psia\n",
+    "\n",
+    "V = 1                        # volume in m**3\n",
+    "# solution\n",
+    "\n",
+    "p = p1+patm                  # absolute pressure in psia\n",
+    "\n",
+    "b = p                        # for isothermal air\n",
+    "\n",
+    "p1 = p2+patm                 # absolute pressure in psia\n",
+    "\n",
+    "b1 = p1                      # for isothermal air\n",
+    "\n",
+    "dP = p1 - p                  # change in pressure\n",
+    "\n",
+    "bavg = (b1+b)/2              # average bulk modulus of elasticity in N/m**2\n",
+    "\n",
+    "dV = dP*V/bavg\n",
+    "\n",
+    "v = -dV\n",
+    "\n",
+    "print \"An increase in pressure by 2 psia will result in a volume reduction of\",round(v,4),\"ft**3\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.5 Page no 14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Sonic velocity in air at 68 deg F =  1126.0 ft/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Sonic velocity of air\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "k = 1.4                       # gas constant\n",
+    "\n",
+    "R = 1716                      # Universal gas constant in ft.lb/slug^oR\n",
+    "\n",
+    "T = 68+460                    # temperature in *oR\n",
+    "\n",
+    "# solution\n",
+    "\n",
+    "c = sqrt(k*R*T)\n",
+    "\n",
+    "print \"Sonic velocity in air at 68 deg F = \",round(c,0),\"ft/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.6 Page no 19"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Force required to move the piston by 1 m/s =  2.83 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Force  required to move the piston\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.05                        # diameter of cylinder 1 in m\n",
+    "\n",
+    "l = 0.2                         # length of the cylinder in meters\n",
+    "\n",
+    "d1 = 0.052                      # diameter of cylinder in m\n",
+    "\n",
+    "mu = 0.09                       # Viscosity of oil in Ns/m**2\n",
+    "\n",
+    "U = 1                           # velocity in m/s\n",
+    "\n",
+    "Y = (d1-d)/2                        # clearance between the two cylinders in m\n",
+    "\n",
+    "A = pi*l*d                 # area in m**2\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "tau = mu*U/Y                    # Shear stress in N/m**2\n",
+    "\n",
+    "F = tau*A                        # Shear foce in N\n",
+    "\n",
+    "print \"Force required to move the piston by 1 m/s = \",round(F,2),\"N\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.7 Page no 20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " (a) Distance between the walls =  1.0 cm\n",
+      " (b) Shear stress =  0.0001005 N/m**2\n",
+      " (c) Shear Stress at 20 um from the plate =  0.000100098 N/m**2\n",
+      " (d) Location of maximum velocity =  0.005 m\n",
+      "Above calculation indicates that the zero shear stress and the maximum velocity occurs at the same location which is half way between the plate \n",
+      "This  also is in conformity with the fact that , in a flowing fluid, the velocity is maximum where shear stress is zero\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Distance between the walls; Shear Stress; Location of maximum velocity\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from sympy import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "mu = 1.005*10**-3                 # Viscosity of water in Ns/m**2\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "# Part a\n",
+    "\n",
+    "# Velocity is given by the formula u = 10*(0.01*y-y**2)\n",
+    "\n",
+    "# two boundary conditions must be satisfied\n",
+    "\n",
+    "# at y=0;u=0 at the bottom of the plate\n",
+    "\n",
+    "# at y=Y ; u = 0 at top of the plate\n",
+    "\n",
+    "Y = 0.01                        # Distance between the walls\n",
+    "\n",
+    "Y1 = Y*100\n",
+    "\n",
+    "print \" (a) Distance between the walls = \",round(Y1,1),\"cm\"\n",
+    "\n",
+    "# Part b\n",
+    "\n",
+    "# tau = mu*du/dy                     # Newtons law of viscosity\n",
+    "\n",
+    "# differentiate u wrt y\n",
+    "\n",
+    "y = Symbol('y')\n",
+    "\n",
+    "u = 10*(0.01*y-y**2)\n",
+    "\n",
+    "uprime = u.diff(y)\n",
+    "\n",
+    "y = 0\n",
+    "\n",
+    "U = uprime\n",
+    "\n",
+    "#print U\n",
+    "# for y =0 at the bottom plate we get \n",
+    "\n",
+    "U1 = 0.01          # from U\n",
+    "\n",
+    "tau = mu*10*U1                     # shear stress in N/m**2\n",
+    "\n",
+    "print \" (b) Shear stress = \",round(tau,9),\"N/m**2\"\n",
+    "\n",
+    "# Part c\n",
+    "\n",
+    "# Shaer stress at 20um from the wall\n",
+    "\n",
+    "tau1 = mu*10*(0.01-2*20*10**-6)        # using the equation of U and y = 20*10**-6 calc shear stress in N/m**2\n",
+    "\n",
+    "print \" (c) Shear Stress at 20 um from the plate = \",round(tau1,9),\"N/m**2\"\n",
+    "\n",
+    "# Part D\n",
+    "\n",
+    "# Distance at which shaer stress is zero can be found from the location of maximum velocity \n",
+    "\n",
+    "# equating uprime = 0\n",
+    "\n",
+    "y1 = 0.01/2                           # shear stress location\n",
+    "\n",
+    "print \" (d) Location of maximum velocity = \",round(y1,3),\"m\"\n",
+    "\n",
+    "print \"Above calculation indicates that the zero shear stress and the maximum velocity occurs at the same location which is half way between the plate\",\"\\n\",\"This  also is in conformity with the fact that , in a flowing fluid, the velocity is maximum where shear stress is zero\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.8 Page 24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " (a) For linear distribution of velocity , shaft torque =  10.86 m.N\n",
+      " (b) For non-linear distribution of velocity , shaft torque =  10.86 m.N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Shaft torque for linear and non linear distribution of velocity \n",
+    "from math import *\n",
+    "# Given\n",
+    "\n",
+    "d = 0.1                                # diameter of shaft in m\n",
+    "\n",
+    "l = 0.2                                # length of the shaft in m\n",
+    "\n",
+    "t = 0.00002                            # thickness in m\n",
+    "\n",
+    "N = 30                                 # RPM of shaft\n",
+    "\n",
+    "U = pi*d*30/60                    # velocity in m/s\n",
+    "\n",
+    "r1 = d/2\n",
+    "\n",
+    "r2 = r1 + t                            # radius of bearing in m\n",
+    "\n",
+    "mu = 0.44                              # Viscosity of SAE-30 oil in Ns/m**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# part a\n",
+    "\n",
+    "F = 2*pi*r1*l*mu*U/t\n",
+    "\n",
+    "T =F*r1\n",
+    "\n",
+    "print \" (a) For linear distribution of velocity , shaft torque = \",round(T,2),\"m.N\"\n",
+    "\n",
+    "# part b\n",
+    "\n",
+    "F1 = 2*pi*l*mu*U/log(r2/r1)\n",
+    "\n",
+    "\n",
+    "T1 =F1*r1\n",
+    "\n",
+    "print \" (b) For non-linear distribution of velocity , shaft torque = \",round(T1,2),\"m.N\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.9 Page 26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Watts of energy lost in overcoming friction at 300 RPM =  1108.0 Watts\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Watts of energy lost to overcome friction\n",
+    "from math import *\n",
+    "# Given\n",
+    "\n",
+    "mu = 0.44                               # viscosity of the oil in Ns/m**2\n",
+    "\n",
+    "N = 300                                 # RPM of the shaft\n",
+    "\n",
+    "t = 0.00025                             # thickness of the film oil in m\n",
+    "\n",
+    "r1 = 0.15                               # radius in m\n",
+    "\n",
+    "r2 = 0.1                                # radius in m\n",
+    "\n",
+    "T = pi**2*mu*N*(r1**4-r2**4)/(60*t)\n",
+    "\n",
+    "P = T*2*pi*N/60\n",
+    "\n",
+    "print \"Watts of energy lost in overcoming friction at 300 RPM = \",round(P,0),\"Watts\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.10 Page no 28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 42,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Excessive pressure inside the droplet =  29.2 N"
+     ]
+    }
+   ],
+   "source": [
+    "# Excessive pressure inside droplet\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.01                                 # diameter in m\n",
+    "\n",
+    "sigma = 0.073                            # surface tension in N/m\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "dP = 4*sigma/d                           # pressure excessive\n",
+    "\n",
+    "print \"Excessive pressure inside the droplet = \",round(dP,2),\"N\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 1.11 Page no 31"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 50,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The alcohol will rise to a height of 5.7 mm in the glass tube"
+     ]
+    }
+   ],
+   "source": [
+    "# Height to which alcohol will rise\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "sigma = 0.022                            # surface tension in N/m\n",
+    "\n",
+    "gma = 9789                               #  specific weight\n",
+    "\n",
+    "S = 0.79                                 # specific gravity\n",
+    "\n",
+    "d = 0.002                                # diameter in m\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "\n",
+    "h =4*sigma*1000/(gma*S*d)                     # capillary height in m\n",
+    "\n",
+    "print \"The alcohol will rise to a height of\",round(h,1),\"mm in the glass tube\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter10.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter10.ipynb
new file mode 100755
index 00000000..d207e652
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter10.ipynb
@@ -0,0 +1,451 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 10 : Open Channel Flow     "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 10.1 Page no 363"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Top width = 5.0 m\n",
+      "Area of flow = 8.0 m**2\n",
+      "Wetted perimeter = 7.472 m\n",
+      "Hydraulic radius = 1.07 m\n",
+      "Hydraulic depth = 1.6 m\n",
+      "Secton Factor = 10.12 m**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Top width, area of lfow, hydraulic radius\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "b = 3                            # base of the channel\n",
+    "\n",
+    "z = 0.5                          # slope of the channel\n",
+    "\n",
+    "y = 2                            # depth of the channel\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "T = b + 2*z*y\n",
+    "\n",
+    "print \"Top width =\",round(T,0),\"m\"\n",
+    "\n",
+    "A = (b+z*y)*y\n",
+    "\n",
+    "print \"Area of flow =\",round(A,0),\"m**2\"\n",
+    "\n",
+    "P = b + 2*y*sqrt(1+z**2)\n",
+    "\n",
+    "print \"Wetted perimeter =\",round(P,3),\"m\"\n",
+    "\n",
+    "R = A/P\n",
+    "\n",
+    "print \"Hydraulic radius =\",round(R,2),\"m\"\n",
+    "\n",
+    "D = A/T\n",
+    "\n",
+    "print \"Hydraulic depth =\",round(D,2),\"m\"\n",
+    "\n",
+    "Z = A*sqrt(D)\n",
+    "\n",
+    "print \"Secton Factor =\",round(Z,2),\"m**2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 10.2 Page no 366"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge for the channel = 16.1 m**3/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Discharge for the trapezoidal channel\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "z = 1.0                # slide slope\n",
+    "\n",
+    "b = 3.0                # base width\n",
+    "\n",
+    "y = 1.5                # depth\n",
+    "\n",
+    "S = 0.0009\n",
+    "\n",
+    "n = 0.012             # for concrete\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "A = (b+z*y)*y\n",
+    "\n",
+    "P = P = b + 2*y*sqrt(1+z**2)\n",
+    "\n",
+    "R = A/P\n",
+    "\n",
+    "# from mannings eqquation\n",
+    "\n",
+    "Q = A*(1/n)*(R**(2/3)*S**(1/2))\n",
+    "\n",
+    "print \"Discharge for the channel =\",round(Q,2),\"m**3/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 10.4 Page no 373"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "depth(y) = 5.05 ft\n",
+      "base width(b) = 4.18 ft\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine cross sectional area\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "z = 1\n",
+    "\n",
+    "Q = 10000/60                     # discharge of water in ft**#/s\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "y = (Q/(1.828*2.25*sqrt(0.5)))**(2/5)\n",
+    "\n",
+    "print \"depth(y) =\",round(y,2),\"ft\"\n",
+    "\n",
+    "b = 0.828*y\n",
+    "\n",
+    "print \"base width(b) =\",round(b,2),\"ft\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 10.5 Page no 378"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Critical depth = 3.44 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Calculate criticcal depth\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "y = 2.5                 # depth\n",
+    "\n",
+    "V = 8                  # velocity in m/s\n",
+    "\n",
+    "g = 9.81               # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Yc = (20**2/g)**(1/3)\n",
+    "\n",
+    "print \"Critical depth =\",round(Yc,2),\"m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 10.6 Page no 380"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a )Normal Depth = 2.22 m\n",
+      "b )F =  0.31  Since the Froude number is less than 1, the flow is subcritical\n",
+      "c )Critical depth is =  1.08 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# determine normal depth, flow regine, critical depth\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Q = 15 # flow rate in m**3/s\n",
+    "\n",
+    "w = 4.0  # bottom width\n",
+    "\n",
+    "S = 0.0008 # bed slope\n",
+    "\n",
+    "n = 0.025 # manning constant\n",
+    "\n",
+    "z = 0.5 # slope\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# We use a trial and error method here to find the value of y i.e. normal depth\n",
+    "\n",
+    "y = 2.22 # we take the value of y as 2.2 m\n",
+    "\n",
+    "Q = ((4+0.5*y)*(y/(n))*(((4+0.5*y)*y)/(4+2.236*y))**(0.667)*(S)**(0.5))\n",
+    "\n",
+    "print \"a )Normal Depth =\",round(y,2),\"m\"\n",
+    "\n",
+    "A = (4+0.5*y)*y\n",
+    "\n",
+    "T = (w+2*z*y)\n",
+    "\n",
+    "D = A/T\n",
+    "\n",
+    "V = (Q/A)\n",
+    "\n",
+    "F =V/(sqrt(9.81*D)) \n",
+    "\n",
+    "print \"b )F = \",round(F,2),\" Since the Froude number is less than 1, the flow is subcritical\"\n",
+    "\n",
+    "# we use trail and error to find the value of yc for critical depth\n",
+    "\n",
+    "yc = 1.08\n",
+    "\n",
+    "Q1 = (4+z*yc)*yc*sqrt((9.81*(4+0.5*yc)*yc)/(4+2*z*yc)) \n",
+    "\n",
+    "print \"c )Critical depth is = \",round(yc,2),\"m\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###  Example 10.8 Page no 390"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Since F1 = 1.86  It is a weak jump\n",
+      "Length of the jump = 23.0 ft\n",
+      "Total energy loss = 133.7 HP\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Height, type and length of jump and loss of energy\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "b = 60                   # base width in ft\n",
+    "\n",
+    "y1 = 2.5                 # base depth in ft\n",
+    "\n",
+    "Q = 2500                 # discharge in ft**3/s\n",
+    "\n",
+    "g = 32.2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V1 = Q/(b*y1)\n",
+    "\n",
+    "F1 = V1/sqrt(g*y1)\n",
+    "\n",
+    "y2 = y1*0.5*(sqrt(1+8*F1**2)-1)\n",
+    "\n",
+    "V2 = Q/(b*y2)\n",
+    "\n",
+    "print \"Since F1 =\",round(F1,2),\" It is a weak jump\"\n",
+    "\n",
+    "L = y2*4.25\n",
+    "\n",
+    "print \"Length of the jump =\",round(L,0),\"ft\"\n",
+    "\n",
+    "E1 = y1+(V1**2/(2*g))\n",
+    "\n",
+    "E2 = y2+(V2**2/(2*g))\n",
+    "\n",
+    "El = E1-E2\n",
+    "\n",
+    "Te = El*62.4*Q/543\n",
+    "\n",
+    "print \"Total energy loss =\",round(Te,2),\"HP\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 10.12 Page no 409"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge = 503.0 cfs\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine flow rate\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 6                      # depth of the channel\n",
+    "\n",
+    "w= 12                      # width of the channel\n",
+    "\n",
+    "h  = 1.0                   # height of the channel\n",
+    "\n",
+    "p = 9                      # pressure drop in m\n",
+    "\n",
+    "g = 32.2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "y2 = y1 - h - 0.75\n",
+    "\n",
+    "V1 = sqrt(2*g*0.75/((1.41)**2-1))\n",
+    "\n",
+    "Q = w*b*V1/10\n",
+    "\n",
+    "print \"Discharge =\",round(Q,0),\"cfs\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter11.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter11.ipynb
new file mode 100755
index 00000000..3d434d74
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter11.ipynb
@@ -0,0 +1,333 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Chapter 11: Compressible Flow"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 11.1 Page no 420"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total change in enthalpy for 10 kg air =  180.0 kcal\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 11.1\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "T1 = 273 + 15          # temperature in K\n",
+    "\n",
+    "T2 = 273 + 90          # temperature in K\n",
+    "\n",
+    "Cp = 0.24              # cp for air in kcal/kgK\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "dh = Cp*(T2-T1)        # enthalpy per kg of air\n",
+    "\n",
+    "H = 10*dh              # total enthallpy of 10 kg air\n",
+    "\n",
+    "print \"Total change in enthalpy for 10 kg air = \",round(H,0),\"kcal\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 11.2 Page no 420"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total change in enthalpy of 10 Kg of air = -0.255 kcal/K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 11.2\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "T1 = 273 + 15          # temperature in K\n",
+    "\n",
+    "T2 = 273 + 90          # temperature in K\n",
+    "\n",
+    "P1 = 40 + 101.3        # pressure in abs\n",
+    "\n",
+    "P2 = 360 + 101.3       # presure in abs\n",
+    "\n",
+    "Cv = 0.171             # Specific volume Coefficient of air\n",
+    "\n",
+    "k = 1.4                # gas constant\n",
+    "\n",
+    "# solution\n",
+    "\n",
+    "dS = Cv*log((T2/T1)**k*(P2/P1)**(1-k))\n",
+    "\n",
+    "S = 10*dS\n",
+    "\n",
+    "print \"Total change in enthalpy of 10 Kg of air =\",round(S,3),\"kcal/K\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 11.3 Page no 421"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Final temperature if air =  320.2 F\n",
+      "Total work done on 10 slugs =  330.0 Btu\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 11.3\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "\n",
+    "P1 = 10                  # pressure in psia\n",
+    "\n",
+    "P2 = 30                  # pressure in psia\n",
+    "\n",
+    "T1 = 460+110             # temperature in R\n",
+    "\n",
+    "k =1.4                   # gas constant\n",
+    "\n",
+    "T2 = T1*(P2/P1)**((k-1)/k)\n",
+    "\n",
+    "t2 = T2-460              # final temperature of air\n",
+    "\n",
+    "print \"Final temperature if air = \",round(t2,1),\"F\"\n",
+    "\n",
+    "Cv = 0.157               # coefficient of air \n",
+    "\n",
+    "W = Cv*(T2-T1)           # work done per unit mass of oxygen\n",
+    "\n",
+    "Tw = 10*W                # total work done on 10 slugs\n",
+    "\n",
+    "print \"Total work done on 10 slugs = \",round(Tw,0),\"Btu\" \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 11.4 Page no 426"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Speed of sound in water = 1483.0 m/s\n",
+      "Speed of sound in ethly alcohol = 1238.0 m/s\n",
+      "Speed of sound in Air = 343.0 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 11.4\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "# for water\n",
+    "\n",
+    "S =1           # specific gravity\n",
+    "\n",
+    "rho = S*1000   # density in kg/m**3\n",
+    "\n",
+    "bta = 2.2*10**9 # Bulk modulus of elasticity\n",
+    "\n",
+    "# ethly alcohol\n",
+    "\n",
+    "S1 =0.79           # specific gravity\n",
+    "\n",
+    "rho2 = S1*1000   # density in kg/m**3\n",
+    "\n",
+    "bta2 = 1.21*10**9 # Bulk modulus of elasticity\n",
+    "\n",
+    "# for air\n",
+    "\n",
+    "k = 1.4             # gas constant for air\n",
+    "\n",
+    "R = 287             # universal gas constant\n",
+    "\n",
+    "T = 273+20          # temperature in K\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "C1 = sqrt(bta/rho)\n",
+    "\n",
+    "C2 = sqrt(bta2/rho2)\n",
+    "\n",
+    "print \"Speed of sound in water =\",round(C1,0),\"m/s\"\n",
+    "\n",
+    "print \"Speed of sound in ethly alcohol =\",round(C2,0),\"m/s\"\n",
+    "\n",
+    "C3 = sqrt(k*R*T)\n",
+    "\n",
+    "print \"Speed of sound in Air =\",round(C3,0),\"m/s\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 11.5 Page no 431"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.753305702898\n",
+      "Pressure at downstream =  3.03 psia\n",
+      "Temperature at downstream =  157.3 F\n",
+      "Velocity downstream =  463.02 ft/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 11.5\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "P1 = 1.5        # pressure in psia\n",
+    "\n",
+    "T1 = 40 + 460   # temperature in R\n",
+    "\n",
+    "k = 1.4         # gas constant\n",
+    "\n",
+    "R = 1716        # universal gas constant in ft.lb/slug R\n",
+    "\n",
+    "V1 = 1500       # velocity in ft/s\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "c1 = sqrt(k*R*T1)\n",
+    "\n",
+    "M1 = V1/c1\n",
+    "\n",
+    "M2 = sqrt((2+(k-1)*M1**2)/(2*k*M1**2-(k-1)))\n",
+    "print M2\n",
+    "\n",
+    "P2 = P1*((1+k*M1**2)/(1+k*M2**2))\n",
+    "\n",
+    "print \"Pressure at downstream = \",round(P2,2),\"psia\"\n",
+    "\n",
+    "T2 = T1*((1+0.5*(k-1)*M1**2)/(1+0.5*(k-1)*M2**2))\n",
+    "\n",
+    "t2 = T2-460\n",
+    "\n",
+    "print \"Temperature at downstream = \",round(t2,1),\"F\"\n",
+    "V2 = M2*sqrt(k*R*t2)\n",
+    "\n",
+    "print \"Velocity downstream = \",round(V2,2),\"ft/s\"\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter12.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter12.ipynb
new file mode 100755
index 00000000..009a616c
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter12.ipynb
@@ -0,0 +1,365 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 12: Turbomachines"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 12.1 Page no 443"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Efficiency of th pump = 74.0 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 12.1\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Q = 0.25                      # discharge from the pump in m**3/s\n",
+    "\n",
+    "gma= 0.8*9810                 # specific weight in kg/m**3\n",
+    "\n",
+    "H=25                          # elevation head in m\n",
+    "\n",
+    "T = 350                       # Torque to drive the shaft in Nm\n",
+    "\n",
+    "N = 1800                      # Speed in RPM\n",
+    "\n",
+    "w = 2*pi*N/60                 # angular velocity\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Eff = gma*Q*H*100/(T*w)          # efficiency\n",
+    "\n",
+    "print \"Efficiency of th pump =\",round(Eff,0),\"%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 12.2 Page no 447"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " (a)\n",
+      "Radial velocity at exit = 10.6 m/s\n",
+      "Whirl velocity =  25.3 m/s\n",
+      "Relative velocity =  12.25 m/s\n",
+      "Actual velocity = 27.43 m/s\n",
+      "(b)\n",
+      "Head added for no inlet whirl = 81.0 m\n",
+      "(c)\n",
+      "Power required = 317.8 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 12.2\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.4                     # diameter of the pump in m\n",
+    "\n",
+    "b = 0.03                    # width in m\n",
+    "\n",
+    "theta = pi/3                # blade angle\n",
+    "\n",
+    "N = 1500                    # speed in RPM\n",
+    "\n",
+    "Q = 0.4                     # flow rate in m**3/s\n",
+    "\n",
+    "g = 9.81                    # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "w = 2*pi*N/60               # anggular velocity in rad/s\n",
+    "\n",
+    "u2 = (d/2)*w                # blade velocity in m/s\n",
+    "\n",
+    "V2r = Q/(2*pi*(d/2)*b)      # relative velocity in m/s\n",
+    "\n",
+    "print \"(a)\"\n",
+    "\n",
+    "print \"Radial velocity at exit =\",round(V2r,1),\"m/s\"\n",
+    "\n",
+    "V2t = u2 - V2r*(cos(theta)/sin(theta))\n",
+    "\n",
+    "print \"Whirl velocity = \",round(V2t,1),\"m/s\"\n",
+    "\n",
+    "v2 = V2r/sin(theta)\n",
+    "\n",
+    "print \"Relative velocity = \",round(v2,2),\"m/s\"\n",
+    "\n",
+    "V2 = sqrt(V2t**2+V2r**2)\n",
+    "\n",
+    "print \"Actual velocity =\",round(V2,2),\"m/s\"\n",
+    "\n",
+    "print \"(b)\"\n",
+    "\n",
+    "H = u2*V2t/g\n",
+    "\n",
+    "print\"Head added for no inlet whirl =\",round(H,0),\"m\"\n",
+    "\n",
+    "print \"(c)\"\n",
+    "\n",
+    "P = g*Q*H\n",
+    "\n",
+    "print \"Power required =\",round(P,1),\"kW\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 12.3 Page no 450"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Impeller size = 38.45 cm\n",
+      "Speed of the pump = 1500.0 RPM\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 12.3\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.36                       # diameter of the impeller of pump\n",
+    "\n",
+    "N = 1500                       # Speed of impeller in RPM\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# For best efficiency\n",
+    "\n",
+    "Q1 = 82                       # discharge in l/s\n",
+    "\n",
+    "H1 = 17.5                     # Head in m\n",
+    "\n",
+    "Eta = 0.8                     # efficiency \n",
+    "\n",
+    "Q2 = 100                      # discharge in l/s\n",
+    "\n",
+    "H2 = 20                       # head in m\n",
+    "\n",
+    "# Solving the simulataneous equation we get\n",
+    "\n",
+    "D2 = 38.45\n",
+    "\n",
+    "print \"Impeller size =\",round(D2,2),\"cm\"\n",
+    "\n",
+    "N2 = 1500 \n",
+    "\n",
+    "print \"Speed of the pump =\",round(N2,0),\"RPM\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 12.4 Page no 454 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge, Q =  1.11 ft**3/s\n",
+      "Dynamic head of the pump, H = 122.1 m\n",
+      "Specific speed of the pump, Ns = 1096.0 RPM\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 12.4\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "q = 500                         # discharge in cgm\n",
+    "\n",
+    "Q = 500/449                     # discharge in ft**3/s\n",
+    "\n",
+    "D = 0.667                       # diameter in ft\n",
+    "\n",
+    "A = pi*D**2/4\n",
+    "\n",
+    "V = Q/A                         # velocity in ft/s\n",
+    "\n",
+    "g = 32.2                        # acceleration due to gravity in ft/s**2\n",
+    "\n",
+    "N = 1800                        # speed in RPM\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# for water at 65 deg C\n",
+    "\n",
+    "nu = 1.134*10**-5               # viscosity in ft**2/s\n",
+    "\n",
+    "e = 0.00085                     # epssilon in ft\n",
+    "\n",
+    "r = 0.001275                    \n",
+    "\n",
+    "R = V*D/nu                      # reynolds no\n",
+    "\n",
+    "f = 0.022                       # from moody's diagram\n",
+    "\n",
+    "Hl = V**2*(12.1+(f*224.9))/64.4\n",
+    "\n",
+    "hs = 119.4 + Hl\n",
+    "\n",
+    "print \"Discharge, Q = \",round(Q,2),\"ft**3/s\"\n",
+    "print \"Dynamic head of the pump, H =\",round(hs,1),\"m\"\n",
+    "\n",
+    "Ns = N*sqrt(q)/(hs)**(3/4)\n",
+    "\n",
+    "print \"Specific speed of the pump, Ns =\",round(Ns,0),\"RPM\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 12.5 Page no 457"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Minimum value of static suction lift =  4.02 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 12.5 \n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "H = 60                        # height in m\n",
+    "\n",
+    "Pb = 98*10**3                 # barometric pressure in N/m**2\n",
+    "\n",
+    "Hl = 1                        # head in m\n",
+    "\n",
+    "Pv = 1707                     # vapour pressure \n",
+    "\n",
+    "sigma = 0.08\n",
+    "\n",
+    "w = 9810                      # specific weight\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Npsh_m = sigma*60             # minimum NPSH\n",
+    "\n",
+    "Hsm = (Pb/w)-(Pv/w)-Npsh_m-Hl\n",
+    "\n",
+    "print \"Minimum value of static suction lift = \",round(Hsm,2),\"m\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter2.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter2.ipynb
new file mode 100755
index 00000000..e25d6ea0
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter2.ipynb
@@ -0,0 +1,970 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 : Fluid Statics"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.1 Page no 44"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Using the pascal's law a weight of 1000N can be lifted by applying a force only of  10.0 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Force applied on the piston\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given \n",
+    "\n",
+    "d = 10                        # diameter of hydraulic press in meters\n",
+    "\n",
+    "d1 = 1                       # diameter of piston in meters\n",
+    "\n",
+    "W = 1000                        # weight in Newtons\n",
+    "\n",
+    "Ap = math.pi*d1**2/4         # Area of piston in m**2\n",
+    "\n",
+    "Ar = math.pi*d**2/4        # Area of rram in m**2\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "p = W/Ar                       # pressure to be supplied by the oil in N/cm**2\n",
+    "\n",
+    "F = p*Ap                       # Force applied on the piston\n",
+    "\n",
+    "print \"Using the pascal's law a weight of 1000N can be lifted by applying a force only of \",round(F,1),\"N\"\n",
+    " "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.2 Page no 53"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure =  10070.0 kN/m**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Pressure in kN/m**2\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "h = 1                           # ocean depth below the surface in km\n",
+    "\n",
+    "gma = 10070                     # Specific weight of sea water\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "P =gma*h                        # Pressure in kN/m**2\n",
+    "\n",
+    "print \"Pressure = \",round(P),\"kN/m**2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.3 Page no 53"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure at the bottom of the tank is 176.3 kN/m**3"
+     ]
+    }
+   ],
+   "source": [
+    "# Pressure at the bottom of the tank\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "p1 = 150*10**3                  # Pressure at point 1 in kN/m**2\n",
+    "\n",
+    "Sg = 0.85                       # Specific gravity of oil\n",
+    "\n",
+    "h = 0.8                         # height of oil 2 i tank in meters \n",
+    "\n",
+    "g = 9810                        # specific gravity \n",
+    "\n",
+    "h1 = 2.0                        # height of oil 3 in tank\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "p2 = (p1 + Sg*h*g)\n",
+    "\n",
+    "p3 = (p2 + g*h1)/1000\n",
+    "\n",
+    "print \"Pressure at the bottom of the tank is\",round(p3,1),\"kN/m**3\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.4 Page no 54"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Height of the mountain is 2208.0 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Height of the mountain\n",
+    "\n",
+    "# Given\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "Po = 570                         #  mercury barometer reading in mm\n",
+    "\n",
+    "T = 273 -5                       # temperature in K\n",
+    "\n",
+    "p = 750                          # mercury barometer reading in mm\n",
+    "\n",
+    "n = 1.2345                       # for polytropic atmosphere\n",
+    "\n",
+    "R  =287                          # univerasl gas constant in J/Kg-K\n",
+    "\n",
+    "g = 9.81 \n",
+    "\n",
+    "r = p/Po\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "y = -(R*T/(g*0.19))*(1 - (r)**((n-1)/n))\n",
+    "\n",
+    "print \"Height of the mountain is\",round(y,0),\"m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.5 Page no 57"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure in the pipe =  9.074 kPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Pressure in the pipe\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "h1 = 500           # height in mm\n",
+    "\n",
+    "h2 = 950           # height in mm\n",
+    "\n",
+    "S1 = 1             # specific gravity fo water\n",
+    "\n",
+    "S2 = 1.5           # specific gravity of liquid 2\n",
+    "\n",
+    "w = 9810           # specific weight of water\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "ha = ((h2*S2)-(h1*S1))/1000\n",
+    "\n",
+    "Pa = w*ha/1000     # Pressure in kPa\n",
+    "\n",
+    "print \"Pressure in the pipe = \",round(Pa,3),\"kPa\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.6 Page no 66"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Magnitude of the force required to open the gate =  21659.0 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Force required to open the gate\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "d = 1                  # diameter of the gate in m\n",
+    "\n",
+    "w = 9810               # specific weight in N/m**3\n",
+    "\n",
+    "A = pi*d**2/4          # area of the gate\n",
+    "\n",
+    "Ig = pi*d**4/64        # mamm moment of inertia\n",
+    "\n",
+    "theta = pi/2\n",
+    "\n",
+    "y1= 5+0.5\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "F = w*round(A,3)*(y1) # the answer will come out to be different as they have used the value of Area as 0.78 \n",
+    "\n",
+    "# depth of the COP\n",
+    "\n",
+    "h1 = y1 + (Ig*math.sin(theta)*math.sin(theta)/(A*y1))\n",
+    "\n",
+    "# moment about hinge A\n",
+    "\n",
+    "F1 = (F*(h1 - 5))/d\n",
+    "\n",
+    "print \"Magnitude of the force required to open the gate = \",round(F1,0),\"N\" \n",
+    "#The area calculated in the book is 0.785 and that calculated from code is 0.78. \n",
+    "#This difference of 0.005 is causing the answer to change from the original\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.7 Page no 67"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Total force =  31.71 kN\n",
+      "(b) Position of center of pressure =  1.616 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# total force ; position of the center of pressure  \n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "l =2             # length of the plate in m\n",
+    "\n",
+    "b =1             # width of the plate\n",
+    "\n",
+    "theta = pi/3\n",
+    "\n",
+    "h = 0.75         # depth of the plate\n",
+    "\n",
+    "w = 9810         # specific weight of water\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "A = 1*2\n",
+    "\n",
+    "y1 = h + 1*sin(theta)\n",
+    "\n",
+    "F = w*A*y1/1000\n",
+    "\n",
+    "print \"(a) Total force = \",round(F,2),\"kN\"\n",
+    "\n",
+    "Ig = (b*l**3)/12\n",
+    "\n",
+    "h1 = y1 + (Ig*sin(theta)*sin(theta)/(2*y1))\n",
+    "\n",
+    "print \"(b) Position of center of pressure = \",round(h1,3),\"m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.8 Page no 68"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total hydrostatic force = 40783.0 lbs\n",
+      "point of location on the center plate =  13.52 ft\n",
+      " OR point of location on the center plate from method 2 =  13.5 ft\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Hydrostatic force  and point of location \n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 6                   # diameter of the gate in ft\n",
+    "\n",
+    "A =pi*d**2/4            # area of the gate\n",
+    "\n",
+    "p1 = 600                # pressure on top in psia\n",
+    "\n",
+    "y1 = 10 +2 + 3*sin(pi/6)\n",
+    "\n",
+    "F = 62.4*A*y1\n",
+    "\n",
+    "F1 = p1*A\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Tf = F+F1\n",
+    "\n",
+    "print \"Total hydrostatic force =\",round(Tf,0),\"lbs\"\n",
+    "\n",
+    "Ig = pi*d**4/64\n",
+    "\n",
+    "h1 = y1 + ((Ig*sin(pi/6)*sin(pi/6))/(A*y1))\n",
+    "\n",
+    "H = ((F*h1)+(F1*y1))/Tf\n",
+    "\n",
+    "print \"point of location on the center plate = \",round(H,2),\"ft\"\n",
+    "\n",
+    "# method 2\n",
+    "\n",
+    "Hf = p1/62.4            # equivalent fluid height\n",
+    "\n",
+    "y2 = Hf+y1\n",
+    "\n",
+    "Tf1 = 62.4*A*y2\n",
+    "\n",
+    "h2 = y2 + ((Ig*sin(pi/6)*sin(pi/6))/(A*y2))\n",
+    "\n",
+    "H1 = y2-Hf\n",
+    "\n",
+    "print \" OR point of location on the center plate from method 2 = \",round(H1,2),\"ft\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.10 Page no 74"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Horizontal component acting on the plate =  998.4 lbs\n",
+      "Vertical component acting on the plate =  1283.34 lbs\n",
+      "location of Xv = 1.82 ft\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Horizontal and Vertical components\n",
+    "\n",
+    "import math\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "R = 4                 # radius of the gate in ft\n",
+    "\n",
+    "w = 1                # width of the gate in ft\n",
+    "\n",
+    "gma =62.4            # specific weight of water\n",
+    "\n",
+    "y1 = 4               # distance of center of the gate\n",
+    "\n",
+    "xv1 = 2              # distance in ft\n",
+    "\n",
+    "xv2 = 1.7           # distance in ft\n",
+    "\n",
+    "# Solution \n",
+    "\n",
+    "Fh = R*y1*gma\n",
+    "\n",
+    "Ig = w*R**3/12\n",
+    "\n",
+    "yh = y1 + (Ig/(R*y1))\n",
+    "\n",
+    "Fv1 = R*2*gma\n",
+    "\n",
+    "Fv2 = pi*R**2*gma/4\n",
+    "\n",
+    "Fv = Fv1 + Fv2\n",
+    "\n",
+    "Xv = (Fv1*xv1+Fv2*xv2)/(Fv)\n",
+    "\n",
+    "print \"Horizontal component acting on the plate = \",round(Fh,2),\"lbs\"\n",
+    "\n",
+    "print \"Vertical component acting on the plate = \",round(Fv,2),\"lbs\"\n",
+    "\n",
+    "print \"location of Xv =\",round(Xv,2),\"ft\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.11 Page no 77"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Horizontal component acting on the plate =  29923.0 lbs\n",
+      "(b) Vertical component acting on the plate =  30208.14 lbs\n",
+      "location of vertical component Xv = 1.99 ft from the left wall\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Vertical and Horizontal components\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "p = 50                         # pressure in psia\n",
+    "\n",
+    "gma = 62.4                     # specific weight in ft\n",
+    "\n",
+    "h1 = p*144/gma                 # equivalent height of water surface in ft\n",
+    "\n",
+    "R = 4                          # radius of gate in ft\n",
+    "\n",
+    "w = 1                          # width of the gate in ft\n",
+    "\n",
+    "A = R*w\n",
+    "\n",
+    "y1 = h1 + 2.5 + 2              # center of pressure\n",
+    "\n",
+    "xv1 = 2                        # center of pressure1 for x direction force\n",
+    "\n",
+    "xv2 = 1.7                      # center of pressure2 for x direction force\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Fh = gma*A*y1                 # hiorizontal force\n",
+    "\n",
+    "Ig = 5.33                      # moment of inertia\n",
+    "\n",
+    "yh = y1 + (Ig/(A*y1))          # location of horizontal component\n",
+    "\n",
+    "y2 = h1+2.5\n",
+    "\n",
+    "Fv1 = gma*(R*y2)               # vertical force component 1\n",
+    "\n",
+    "Fv2 = gma*(pi*R**2/4)               # vrtical force component 2\n",
+    "\n",
+    "Fv = Fv1 + Fv2                # vertical force component\n",
+    "\n",
+    "Xv = (Fv1*xv1+Fv2*xv2)/(Fv)\n",
+    "\n",
+    "print \"(a) Horizontal component acting on the plate = \",round(Fh,0),\"lbs\" # The answer for horizontal force in the book is wrong\n",
+    "\n",
+    "print \"(b) Vertical component acting on the plate = \",round(Fv,2),\"lbs\"\n",
+    "\n",
+    "print \"location of vertical component Xv =\",round(Xv,2),\"ft from the left wall\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.13 Page no 84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Depth to which water would rise =  0.25 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Depth to which water would rise\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "l = 3                    # length in m\n",
+    "\n",
+    "b = 4                    # breadth in m\n",
+    "\n",
+    "h = 15                    # height in m\n",
+    "\n",
+    "S = 0.9                 # specific gravity of barge\n",
+    "\n",
+    "Sw  = 1.09              # specific gravity of water\n",
+    "\n",
+    "Wd = 150*10**3          # addditional weight in kN\n",
+    "\n",
+    "V = l*b*h               # volume in m**3\n",
+    "\n",
+    "Wb = gma*S*V          # weight of barge\n",
+    "\n",
+    "Tw = Wb + Wd            # total weight in kN\n",
+    "\n",
+    "gma = 9810              # specific density \n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Fb = Tw                 # since barge is floating\n",
+    "\n",
+    "V1 = Fb/((gma/1000)*Sw)    #  volume in m**3\n",
+    "\n",
+    "\n",
+    "d = (V1/(h*b))/1000\n",
+    "\n",
+    "print \"Depth to which water would rise = \",round(d,2),\"m\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.14 Page no 85 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "X =  3.12 cm\n",
+      "Y =  3.25 cm\n",
+      "The bottom of the solid cylinder will be  4.88 cm above the bottom\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Level at which cylinder will float\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "W = 0.4 * 9.81              # weight of the solid cylinder in N\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "A = np.array([(1,-0.96),(1,1)])\n",
+    "\n",
+    "b = np.array([0,6.37])\n",
+    "\n",
+    "x = np.linalg.solve(A,b)\n",
+    "\n",
+    "X = x[0]\n",
+    "\n",
+    "Y = x[1]\n",
+    "\n",
+    "print \"X = \",round(X,2),\"cm\"\n",
+    "\n",
+    "print \"Y = \",round(Y,2),\"cm\"\n",
+    "\n",
+    "b = 8 -x[0]\n",
+    "\n",
+    "print \"The bottom of the solid cylinder will be \",round(b,2),\"cm above the bottom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.15 Page no 86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Weight of the pan = 37.7 N\n",
+      "Magnitude of right momentum =  123.4 N.cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Weight of the pan and the magnitude of righting moment\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "l =100                # length of the pan in cm\n",
+    "\n",
+    "w = 20                # width of the pan in cm\n",
+    "\n",
+    "d = 4                 # depth of the pan in cm\n",
+    "\n",
+    "L = 1.5               # load in N/m\n",
+    "\n",
+    "gma = 9810              # sepcific weight\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Fb = gma*(d*w*l/(2*l**3)) # weight on the pan\n",
+    "\n",
+    "W = Fb-L                # weight of the pan\n",
+    "\n",
+    "X1 = w/3\n",
+    "\n",
+    "X2 = w/2\n",
+    "\n",
+    "theta = math.atan(d/w)*180/pi\n",
+    "\n",
+    "x = ((X2-X1)*cos(theta*pi/180))\n",
+    "\n",
+    "# momentum equation \n",
+    "\n",
+    "M = W*x\n",
+    "\n",
+    "print \"Weight of the pan =\",round(W,1),\"N\"\n",
+    "print \"Magnitude of right momentum = \",round(M,1),\"N.cm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example no 2.16 Page no 90"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Metacentric height =  0.347 m\n",
+      "(b) Righting moment = 63.1 kN.m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Metacentric height and rightning moment \n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Continued from example 2.15\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Io = 15*4**3/12                     # moment of inertia in m**4\n",
+    "\n",
+    "V = 15*4*2.71                       # Volume in m**3\n",
+    "\n",
+    "Gb = ((3/2)-(2.71/2))          \n",
+    "\n",
+    "W = 1739.2                          # weight of the barge from the previous example in kN\n",
+    "# Solution\n",
+    "\n",
+    "Mg = (Io/V)-Gb                     # metacentric height in m\n",
+    "\n",
+    "print \"(a) Metacentric height = \", round(Mg,3),\"m\"\n",
+    "\n",
+    "M = W*Mg*sin(pi*6/180)\n",
+    "\n",
+    "print \"(b) Righting moment =\",round(M,1),\"kN.m\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example no 2.17 Page no 92"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Amximum pressure occurs at the bottom of the tank = 24768.9 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Maximum pressure in the tank\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "l=6                  # length of the tank\n",
+    "\n",
+    "w =2                 # width of the tank\n",
+    "\n",
+    "d = 3                # depth of the tank\n",
+    "\n",
+    "a = 3                # acceleration in m/s**2\n",
+    "\n",
+    "theta = pi/6\n",
+    "\n",
+    "W = 9810            # specific weight\n",
+    "\n",
+    "X = 0\n",
+    "\n",
+    "po=0               # pressure at the origin\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "A = np.array([(1,-1),(1,1)])\n",
+    "\n",
+    "b = np.array([-1.38,3.0])\n",
+    "\n",
+    "x = np.linalg.solve(A,b)\n",
+    "\n",
+    "Y1 = x[0]\n",
+    "\n",
+    "Y2 = x[1]\n",
+    "\n",
+    "# AMximum pressure at the bottom of the tank\n",
+    "\n",
+    "P = po - W*(2.61*X/9.81) - W*(1+(1.5/9.81))*(-Y2)\n",
+    "\n",
+    "print\"Amximum pressure occurs at the bottom of the tank =\",round(P,3),\"N\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 2.18 Page no 98"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Height of paraboliod revolution of base =  1.48 ft\n",
+      "(b) Maximum pressure corresponding to maximum height =  92.3 lbs/ft**2\n",
+      "(c) Pressure = 26.4 lbs/ft**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Height of the paraboloid revolution; maxm=imum pressure and location ; pressure at the point 0.2 from the center\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 1                          # diamter of the jar in ft\n",
+    "\n",
+    "h =2                           # height of the jar in ft\n",
+    "\n",
+    "H = 1                          # height of water in the jar in ft\n",
+    "\n",
+    "N = 150                        # RPM\n",
+    "\n",
+    "g = 32.2                       # acceleration due to gravity in ft/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "w = 2*pi*N/60\n",
+    "\n",
+    "ho = H+((w**2*(d/2)**2)/(4*g))\n",
+    "print \"(a) Height of paraboliod revolution of base = \", round(ho,2),\"ft\"\n",
+    "\n",
+    "Pmax = 62.4*ho\n",
+    "\n",
+    "print \"(b) Maximum pressure corresponding to maximum height = \",round(Pmax,1),\"lbs/ft**2\"\n",
+    "\n",
+    "z = H - ((w**2*(d/2)**2)/(4*g))\n",
+    "\n",
+    "r = 0.2                # distance from center\n",
+    "\n",
+    "y = -(0.52-0.25)\n",
+    "\n",
+    "P = po + (62.4*w**2*r**2/(2*g))-(62.4*y)\n",
+    "\n",
+    "print \"(c) Pressure =\",round(P,1),\"lbs/ft**2\"\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter3.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter3.ipynb
new file mode 100755
index 00000000..fa74673c
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter3.ipynb
@@ -0,0 +1,541 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 3: Fluid Kinematics"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.1 Page no 117"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) u = 4*X; v = -4*Y \n",
+      "(b) u= 4.0 m/s and v= -8.0 m/s\n",
+      "(c) Magnitude of velocity = 8.94 m/s and angle of resultant velocity =  -63.4 deg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Dimension of flow field ; velocity components at (1,2) ; magnitude and direction of velocity  \n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "# V = 4*Xi-4Yj\n",
+    "\n",
+    "x=1 # x co-ordinate\n",
+    "\n",
+    "y=2 # y co-ordinate\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "print \"(a) u = 4*X; v = -4*Y \"\n",
+    "\n",
+    "u = 4*x\n",
+    "\n",
+    "v=- 4*y\n",
+    "\n",
+    "print \"(b) u=\",round(u,0),\"m/s and v=\",round(v,0),\"m/s\"\n",
+    "\n",
+    "R =sqrt(u**2+v**2)\n",
+    "\n",
+    "ang = atan(v/u)*180/pi\n",
+    "\n",
+    "print \"(c) Magnitude of velocity =\",round(R,2),\"m/s and angle of resultant velocity = \",round(ang,1),\"deg\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.2 Page no 119"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Discharge = 1.06 m**3/s\n",
+      "(b) Mass flow rate =  1057.21 kg/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Discharge and mass flow rate\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.3                  # diameter of pipe in m\n",
+    "\n",
+    "v = 15                   # velocity in m/s\n",
+    "\n",
+    "rho = 997.1              # density in kg/m**3\n",
+    "\n",
+    "A = pi*d**2/4\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Q=A*v\n",
+    "\n",
+    "print \"(a) Discharge =\",round(Q,2),\"m**3/s\"\n",
+    "\n",
+    "mdot = rho*Q\n",
+    "\n",
+    "print \"(b) Mass flow rate = \",round(mdot,2),\"kg/s\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.3 Page no 120 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Mean velocity of the flow = 5.0 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Mean Velocity\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from scipy import integrate\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Vo = 10                  # velocity in m/s\n",
+    "\n",
+    "r1 = 0\n",
+    "\n",
+    "ro = 0.1                 # radius in m\n",
+    "\n",
+    "N = 1\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "R = lambda r: (10*r-1000*r**3)\n",
+    "\n",
+    "R1,err=integrate.quad(R,r1,ro)\n",
+    "\n",
+    "Q =  R1*2*pi\n",
+    "\n",
+    "A = pi*(0.1)**2\n",
+    "\n",
+    "V = Q/A\n",
+    "\n",
+    "print \"Mean velocity of the flow =\",round(V,0),\"m/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.4 Page no 126"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2*x - 2*y**2\n"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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CxobA2sxadCSNxMQAEyYADx7QdAwRi0bwn7vz54HGjfmtk9evA4MGCS/ur1Nf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\nPUUFnhBC9BQVeEII0VNU4AkhRE/9Py1qsfLmAYBpAAAAAElFTkSuQmCC\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x2e9ee90>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "# Sketch the stream lines in the first quadrant\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from scipy import integrate\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "from sympy import *\n",
+    "\n",
+    "#init_printing(use_unicode=False, warp_line=False, no_global=True)\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "# V = 4*y(m)i+2(m)j\n",
+    "\n",
+    "x = Symbol('x')\n",
+    "U = integrate(2,x)\n",
+    "\n",
+    "#print u\n",
+    "\n",
+    "y = Symbol('y')\n",
+    "\n",
+    "V = integrate(-4*y,y)\n",
+    "\n",
+    "#print V\n",
+    "\n",
+    "Zhi = U + V\n",
+    "\n",
+    "print Zhi           # for x and y =0 we get C = 0\n",
+    "\n",
+    "X = [5,6,7,8,9,10,11,12,13,14,15,16,17]\n",
+    "Y = [0,1.414,2,2.449,2.828,3.16,3.46,3.741,4,4.242,4.472,4.69,4.898]\n",
+    "\n",
+    "b1=plt.plot(X,Y)\n",
+    "\n",
+    "\n",
+    "X1 = [2.5,3,4,5,6,7,8,9,10,11,12,13,14,15]\n",
+    "Y1 = [0,1,1.732,2.23,2.645,3,3.31,3.60,3.87,4.123,4.35889,4.5825,4.795,5]\n",
+    "\n",
+    "b2=plt.plot(X1,Y1)\n",
+    "\n",
+    "\n",
+    "X2 = [0.5,1.5,2.5,3.5,4.5,5.5,6.5,7.5,8.5,9.5,10.5,11.5,12.5,13.5,14.5,15.5]\n",
+    "Y2 = [0,1.414,2,2.449,2.828,3.162,3.462,3.741,4,4.242,4.472,4.69,4.898,5.099,5.29,5.4772]\n",
+    "\n",
+    "b3=plt.plot(X2,Y2)\n",
+    "\n",
+    "plt.xlabel(\"x\")\n",
+    "\n",
+    "plt.ylabel(\"y\")\n",
+    "\n",
+    "plt.title(\"Streamline plot\")\n",
+    "\n",
+    "plt.legend([\"zhi=10\",\"zhi=5\",\"zhi=1\"])\n",
+    "plt.show()"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.5 PAge no 127"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Resutant velocity magnitude =  17.89 m/s\n",
+      "Angle = 63.4 deg with the X-axis in the 4th quadrant\n"
+     ]
+    }
+   ],
+   "source": [
+    "# magnitude and direction of flow field\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from sympy import *\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "x = 2             # X co-ordinate\n",
+    "\n",
+    "Y = 4             # Y co-ordiante\n",
+    "\n",
+    "# Solution\n",
+    "y = Symbol('y')\n",
+    "\n",
+    "zhi = 4*x*y\n",
+    "\n",
+    "zhiprime = zhi.diff(y)\n",
+    "\n",
+    "u = zhiprime\n",
+    "\n",
+    "x = Symbol('x')\n",
+    "\n",
+    "zhi = 4*x*Y\n",
+    "\n",
+    "zhiprime = zhi.diff(x)\n",
+    "\n",
+    "v = zhiprime\n",
+    "\n",
+    "R=sqrt(u**2+v**2)\n",
+    "\n",
+    "theta = atan(v/u)*180/pi\n",
+    "\n",
+    "print \"Resutant velocity magnitude = \",round(R,2),\"m/s\"\n",
+    "\n",
+    "print \"Angle =\",round(theta,1),\"deg with the X-axis in the 4th quadrant\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.6 Page no 130"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Velocity = 5.1 m/s\n",
+      "(b) Convective acceleration = 0.46 m**2/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine velocity ; convective accleration\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from sympy import *\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "from scipy import integrate\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d1 = 0.09       # diameter in cm\n",
+    "\n",
+    "d2 = 0.025     # diameter in cm\n",
+    "\n",
+    "rho = 1000   # density in kg/m**3\n",
+    "\n",
+    "mdot = 25    # mass flow rate in kg/s\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "x = Symbol('x')\n",
+    "\n",
+    "A1 = pi*d1**2/4\n",
+    "\n",
+    "A2 = pi*d2**2/4\n",
+    "\n",
+    "AA = A1 - ((A1-A2)/40)*10       # from figure\n",
+    "\n",
+    "V = mdot/(rho*AA)\n",
+    "\n",
+    "print \"(a) Velocity =\",round(V,1),\"m/s\"\n",
+    "\n",
+    "AX = (A1 - ((A1-AA)/40)*x)\n",
+    "\n",
+    "v = 25*10**4/(rho*AX)\n",
+    "\n",
+    "vprime = v.diff(x)\n",
+    "\n",
+    "V1 = vprime\n",
+    "\n",
+    "# at x = 0.1 m we get dv/dx = 0..09\n",
+    "\n",
+    "VPrime = 0.09\n",
+    "\n",
+    "Acx = V*VPrime\n",
+    "\n",
+    "print \"(b) Convective acceleration =\",round(Acx,3),\"m**2/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.8 Page no 143 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "z =  -25.0\n",
+      "Hence the flow is rotational\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Is the flow irrotational\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "# w = (16y-12x)i +(12y-9x)j\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "y = Symbol('y')\n",
+    "\n",
+    "U = 16*y-12*x\n",
+    "\n",
+    "zhiprime = U.diff(y)\n",
+    "\n",
+    "u = zhiprime\n",
+    "\n",
+    "\n",
+    "x = Symbol('x')\n",
+    "\n",
+    "V = 12*y-9*x\n",
+    "\n",
+    "zhiprime1 = V.diff(x)\n",
+    "\n",
+    "v = zhiprime1\n",
+    "\n",
+    "#Vx = -9   # differentiate V wrt x\n",
+    "\n",
+    "#Vx = -9   # differentiate V wrt x\n",
+    "#Uy = 16   # differentiate U wrt y\n",
+    "\n",
+    "z = v-u\n",
+    "\n",
+    "print \"z = \",round(z,0)\n",
+    "\n",
+    "print \"Hence the flow is rotational\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 3.10 Page no 148"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Velocity at the larger cross section =  3.33 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Velocity in larger section\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d1 = 0.1              # diameter in m\n",
+    "\n",
+    "d2 = 0.3              # diameter in m\n",
+    "\n",
+    "V1 = 30               # velocity in m/s\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V2 = (d1**2/d2**2)*V1\n",
+    "\n",
+    "print \"Velocity at the larger cross section = \",round(V2,2),\"m/s\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter4.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter4.ipynb
new file mode 100755
index 00000000..5a1c879b
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter4.ipynb
@@ -0,0 +1,558 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 4 : Fluid Dynamics"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.1 Page no 159"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Difference in pressure in section 1 and 2 =  1157.0 kN/m**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Difference in pressure at top and bottom\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d1 = 0.1                     # diameter in m\n",
+    "\n",
+    "d2 = 0.05                     # diameter in m\n",
+    "\n",
+    "Q = 0.1                      # discharge in m**3/s\n",
+    "\n",
+    "A1 = pi*d1**2/4\n",
+    "\n",
+    "A2 = pi*d2**2/4\n",
+    "\n",
+    "gma =9810                   # specific weight\n",
+    "\n",
+    "z= 6                        # difference in the height\n",
+    "\n",
+    "g = 9.81\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V1 = Q/A1                    # velocity at section 1\n",
+    "\n",
+    "V2 = Q/A2                    # velocity at section 2\n",
+    "\n",
+    "dP = gma*((V2**2/(2*g))-(V1**2/(2*g))-z)/1000\n",
+    "\n",
+    "print \"Difference in pressure in section 1 and 2 = \",round(dP,1),\"kN/m**2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.2 Page no 160"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Actual discharge = 2.0 l/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Actual discharge\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 2.5                 # diameter in cm\n",
+    "\n",
+    "h =200                        # head in cm\n",
+    "\n",
+    "Cd = 0.65                   # coefficient of discharge\n",
+    "\n",
+    "A  =pi*d**2/4\n",
+    "\n",
+    "g = 9.81                  # acceleration due to gravity in m/s**2 \n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Q = Cd*A*sqrt(2*g*h)/100\n",
+    "\n",
+    "print \"Actual discharge =\",round(Q,2),\"l/s\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.3 Page no 162"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge through the orifice = 2.69 m**3/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Discharge through the orifice\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from scipy import integrate\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "H1 = 3               # height in m\n",
+    "\n",
+    "H2 = 4               # height in m\n",
+    "\n",
+    "b = 0.5              # width in m\n",
+    "\n",
+    "Cd = 0.65            # co-efficient of discharge \n",
+    "\n",
+    "g = 9.81             # acceleration due to grvity in m/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "q = lambda h: h**(1/2)\n",
+    "    \n",
+    "Q,err = integrate.quad(q, H1, H2)\n",
+    "\n",
+    "Qt = Cd*b*sqrt(2*g)*Q\n",
+    "\n",
+    "print \"Discharge through the orifice =\",round(Qt,2),\"m**3/s\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.4 Page no 163"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total Discharge = 1.95 m**3/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# discharge through orifice\n",
+    "    \n",
+    "from math import *\n",
+    "\n",
+    "from scipy import integrate\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "b = 1                  # bredth of the tank\n",
+    "\n",
+    "d = 0.5                # depth of the tank\n",
+    "\n",
+    "h1 = 0.2                 # height of the orifice in m\n",
+    "\n",
+    "Cd = 0.6               # coefficient of discharge\n",
+    "\n",
+    "H1 = 2                 # height in m\n",
+    "\n",
+    "H2 = 2+h1               # height in m\n",
+    "\n",
+    "g = 9.81               # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "A = 1*0.3              # area of submerged section in m**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "q = lambda h: h**(1/2)\n",
+    "    \n",
+    "Q,err = integrate.quad(q, H1, H2)\n",
+    "\n",
+    "Q1 = Cd*b*sqrt(2*g)*(Q) # Flow through area 1\n",
+    "\n",
+    "Q2 = Cd*sqrt(2*g*H2)*A\n",
+    "\n",
+    "Td = Q1+Q2\n",
+    "\n",
+    "print \"Total Discharge =\",round(Td,2),\"m**3/s\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.5 Page no 165"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Actual discharge = 0.000454 m**3/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine flow rate of water\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d1 = 2               # radius of pipe\n",
+    "\n",
+    "d2 = 1               # radius of throat\n",
+    "\n",
+    "D1 = 40\n",
+    "\n",
+    "D2 = 20\n",
+    "\n",
+    "A1 = pi*D1**2/4\n",
+    "\n",
+    "A2 = pi*D2**2/4\n",
+    "\n",
+    "Cd = 0.95\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V2 = sqrt(21582/0.9375)\n",
+    "\n",
+    "Q = 1.52*pi*(d1/100)**2/4\n",
+    "\n",
+    "Qa = Q*Cd\n",
+    "\n",
+    "print \"Actual discharge =\",round(Qa,6),\"m**3/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.6 Page no 166"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "velocity at the dept of 1 ft = 5.67 ft/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Velocity of stream point at the point of insertion\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "dx = 0.5              # in ft\n",
+    "\n",
+    "K = 1                 # constant\n",
+    "\n",
+    "g = 32.2              # acceleration due to gravity in ft/s**2\n",
+    "\n",
+    "# solution\n",
+    "\n",
+    "V = sqrt(2*g*dx)\n",
+    "\n",
+    "print \"velocity at the dept of 1 ft =\",round(V,2),\"ft/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example no 4.7 Page no 172"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge throught the system = 190.0 l/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Discharge throught the system\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "gma= 0.8      # specific weight\n",
+    "\n",
+    "V2 = 40       # velocity in m/s\n",
+    "\n",
+    "z1 =25        # height at point 1\n",
+    "\n",
+    "g = 9.81      # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "d = 15        # diameter of the nozzle in cm\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V2 = sqrt(2*g*z1/4.25)\n",
+    "\n",
+    "A = pi*(d/100)**2/4\n",
+    "\n",
+    "Q = A*V2*1000\n",
+    "\n",
+    "print \"Discharge throught the system =\",round(Q,0),\"l/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.8 Page no 174"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Power input =  32.5 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Power input to the pump\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Eff = 0.8     # pump efficiency\n",
+    "\n",
+    "Hl = 30       # head loss in m\n",
+    "\n",
+    "D1 =6         # diameter in cm\n",
+    "\n",
+    "D2 = 2        # diameter in cm\n",
+    "\n",
+    "gma = 9810    # specific weight in N/m**3\n",
+    "\n",
+    "V2 = 40       # velocity in m/s\n",
+    "\n",
+    "P1 = -50      # pressure at point 1 in N/m**2\n",
+    "\n",
+    "z2 = 100      # height at point 2\n",
+    "\n",
+    "g = 9.8       # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "z1 = 30       # height in m\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V1=(2/6)**2*V2\n",
+    "\n",
+    "Q = (pi*6**2/4)*V1*10**-4\n",
+    "\n",
+    "Hs = z2 + (V2**2/(2*g)) + z1 + (50/gma) -(V1**2/(2*g))\n",
+    "\n",
+    "P = gma*Q*Hs\n",
+    "\n",
+    "Pi = (P/Eff)/1000\n",
+    "\n",
+    "print \"Power input = \",round(Pi,1),\"kW\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 4.9 Page no 176"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Elevation at height A = 492.39 m\n",
+      "Elevation at height B = 576.23 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Pressure head at A and B\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Q = 0.2           # discharge in m**3/s\n",
+    "\n",
+    "d1 = 0.25           # diameter of the pipe in m\n",
+    "\n",
+    "A = pi*d1**2/4    # area of the pipe\n",
+    "\n",
+    "za = 480          # height in m\n",
+    "\n",
+    "z1 = 500          # height in m\n",
+    "\n",
+    "z3 = 550          # elevation in m\n",
+    "\n",
+    "gma =9810         # specific weight in N/m**2\n",
+    "\n",
+    "g =9.81           # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V=Q/A           # Velocity of in m/s\n",
+    "\n",
+    "Hl1 = (0.02*100*V**2/(0.25*2*9.81))\n",
+    "\n",
+    "# pressure head at A\n",
+    "\n",
+    "Pa =(z1-za-(V**2/(2*g))-Hl1)\n",
+    "\n",
+    "El = za+Pa\n",
+    "\n",
+    "print \"Elevation at height A =\",round(El,2),\"m\"\n",
+    "\n",
+    "# pressure head at B\n",
+    "\n",
+    "hs = z3 - z1 + (0.02*(500/0.25)*(V**2/(2*g))) \n",
+    "\n",
+    "El2 = El+hs\n",
+    "\n",
+    "print \"Elevation at height B =\",round(El2,2),\"m\"\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter5.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter5.ipynb
new file mode 100755
index 00000000..f7f93457
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter5.ipynb
@@ -0,0 +1,578 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 5 :  Fluid Momentum"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 5.1 Page no 192"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Resultant force on the elbow =  9800.58 N\n",
+      "Angle of resultant force =  34.8516 deg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Resultant force on the elbow\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Q = 0.3                 # Water flow rate in m**3/s\n",
+    "\n",
+    "d1 = 0.3                # diameter at inlet in meters\n",
+    "\n",
+    "A1 = pi*d1**2/4 # inlet area in m**2\n",
+    "\n",
+    "d2 = 0.15               # diameter at outlet in m\n",
+    "\n",
+    "A2 = pi*d2**2/4 # area at outlet in m**2\n",
+    "\n",
+    "P1 = 175*10**3          # inlet pressure in kN/m**2\n",
+    "\n",
+    "P2 = 160*10**3          # Pressure at outlet in kN/m**2\n",
+    "\n",
+    "F1 = P1*A1              # Force at inlet\n",
+    "\n",
+    "F2 = P2*A2              # Force at outlet\n",
+    "\n",
+    "rho = 1000              # density of water in kg/m**3\n",
+    "\n",
+    "V1 = Q/A1               # inlet velocity in m/s\n",
+    "\n",
+    "V2 = Q/A2               # Velocity at outlet in m/s\n",
+    "\n",
+    "theta = 45*pi/180  # angle in deg\n",
+    "\n",
+    "# Soultion\n",
+    "\n",
+    "# Applying the X momentum equation we get\n",
+    "\n",
+    "Rx = F1 - F2*cos(theta)-rho*Q*(V2*cos(theta)-V1)\n",
+    "\n",
+    "# Applying the Y momentum equation\n",
+    "\n",
+    "Ry = F2*sin(theta)+rho*Q*(V2*sin(theta)-0)\n",
+    "\n",
+    "R = sqrt(Rx**2+Ry**2)\n",
+    "\n",
+    "print \"Resultant force on the elbow = \",round(R,2),\"N\"\n",
+    "\n",
+    "a = atan(Ry/Rx)*180/pi\n",
+    "\n",
+    "print \"Angle of resultant force = \",round(a,4),\"deg\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 5.2 Page no 194"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Resultant force =  1232.0 lbs\n",
+      "Angle of resultant force =  59.2396 deg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Force exerted by the jet on the vane\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "V1 = 80            # Velocity in ft/s\n",
+    "\n",
+    "A1 = 0.1            # area in ft**2\n",
+    "\n",
+    "g = 32.2           # Acceleration due to gravity in ft/s**2\n",
+    "\n",
+    "rho = 1.94         # density in lb/ft**3\n",
+    "\n",
+    "a = pi/3 # angle of pipe bend\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Q = A1*V1          # Total discharge in m**3\n",
+    "\n",
+    "# Applying bernoullis at point 1 and 2\n",
+    "\n",
+    "V2 = sqrt((2*g*V1**2/(2*32.2))-3*2*g)\n",
+    "\n",
+    "# Pressure at the end of the section are atmospheric and hence 0\n",
+    "\n",
+    "# momentum equation in X direction\n",
+    "\n",
+    "Rx = -(rho*Q*(V2*cos(a)-80))\n",
+    "\n",
+    "# momentum equation in Y direction\n",
+    "\n",
+    "Ry = (rho*Q*(V2*sin(a)-0))\n",
+    "\n",
+    "R = sqrt(Rx**2+Ry**2)\n",
+    "\n",
+    "print \"Resultant force = \",round(R,0),\"lbs\"\n",
+    "\n",
+    "ang = atan(Ry/Rx)*180/pi\n",
+    "\n",
+    "print \"Angle of resultant force = \",round(ang,4),\"deg\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example no 5.3 Page no 195 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Resultant Force =  12489.0 N\n",
+      "Angle with horizontal =  69.2 deg with horizontal\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Force  needed to hold the Y position \n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "Q1 = 0.5             # discharge from pipe 1 in m**3/s\n",
+    "\n",
+    "Q2 = 0.3             # discharge from pipe 2 in m**3/s\n",
+    "\n",
+    "Q3 = 0.2             # discharge from pipe 3 in m**3/s\n",
+    "\n",
+    "d1 = 0.45            # diameter of pipe 1 in m\n",
+    "\n",
+    "d2 = 0.3             # diameter of pipe 2 in m\n",
+    "\n",
+    "d3 = 0.15            # diameter of pipe 3 in m\n",
+    "\n",
+    "A1 = pi*d1**2/4  # area in m**2\n",
+    "\n",
+    "A2 = pi*d2**2/4  # area in m**2\n",
+    "\n",
+    "A3 = pi*d3**2/4  # area in m**2\n",
+    "\n",
+    "P1 = 60*10**3        # Pressure at point 1 in kPa\n",
+    "\n",
+    "gma = 9810\n",
+    "\n",
+    "g = 9.81             # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "rho = 1000           # density in kg/m**3\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V1 = Q1/A1\n",
+    "\n",
+    "V2 = Q2/A2\n",
+    "\n",
+    "V3 = Q3/A3\n",
+    "\n",
+    "P2 = gma*((P1/gma) + V1**2/(2*g) - V2**2/(2*g))\n",
+    "\n",
+    "P3 = gma*((P1/gma) + V1**2/(2*g) - V3**2/(2*g))\n",
+    "\n",
+    "F1 = P1*A1\n",
+    "\n",
+    "F2 = P2*A2\n",
+    "\n",
+    "F3 = P3*A3\n",
+    "\n",
+    "Rx = rho*(Q2*V2*cos(pi/6)-Q3*V3*cos(pi/9)-0)+F3*cos(pi/9)-F2*cos(pi/6)\n",
+    "\n",
+    "Ry = rho*((Q2*V2*sin(pi/6)+Q3*V3*sin(pi/9)-Q1*V1))+F3*sin(pi/9)-F2*sin(pi/6)-F1\n",
+    "\n",
+    "R = sqrt(Rx**2+Ry**2)\n",
+    "\n",
+    "a = atan(Ry/Rx)*180/pi\n",
+    "\n",
+    "print \"Resultant Force = \",round(R,0),\"N\"\n",
+    "\n",
+    "print \"Angle with horizontal = \",round(a,1),\"deg with horizontal\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 5.4 Page no 199"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Normal force on the plate =  423.0 lbs\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Normal force on the plate\n",
+    "\n",
+    "# Given\n",
+    "from math import *\n",
+    "\n",
+    "d = 2            # diameter in inches\n",
+    "\n",
+    "A = pi*d**2/(4*144)    # Area of jet\n",
+    "\n",
+    "V = 100          # velocity of jet in ft/s\n",
+    "\n",
+    "Q = A*V          # dischargge in ft**3/s\n",
+    "\n",
+    "gma = 62.4       # mass\n",
+    "\n",
+    "g = 32.2         # acceleration due to gravity in ft/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Rx = (gma*Q*V)/g   # horizontal force required to keep plate in position\n",
+    "\n",
+    "print \"Normal force on the plate = \",round(Rx,0),\"lbs\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 5.5 Page no 202"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Force on the plate =  331.34 N\n",
+      "Work done per second =  3313.4 N.m/s\n",
+      "Efficiency =  44.4 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Force on the plate ; work doen per second; efficiency\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "D = 0.075            # diameter in m\n",
+    "\n",
+    "A =pi*D**2/4         # area of jet\n",
+    "\n",
+    "V  =15               # velocity of jet in m/s\n",
+    "\n",
+    "w = 9810             # specific weight\n",
+    "\n",
+    "g = 9.81             # acceleration due to gravity in m/s^2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Q =A*V              # Discharge in m**3/s\n",
+    "\n",
+    "Vp = 10             # velocity of plate in m/s\n",
+    "\n",
+    "Rx = w*Q*(V-Vp)/g   # force in X direction\n",
+    "\n",
+    "print \"Force on the plate = \",round(Rx,2),\"N\"\n",
+    "\n",
+    "W = Rx*Vp\n",
+    "\n",
+    "print \"Work done per second = \",round(W,1),\"N.m/s\"\n",
+    "\n",
+    "Eff = 2*(V-Vp)*Vp/V**2\n",
+    "\n",
+    "E = 100*Eff\n",
+    "\n",
+    "print \"Efficiency = \",round(E,1),\"%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 5.6 Page no 204"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Force exerted in X direction =  273.8 lbs\n",
+      "Force exerted in Y direction =  0.0 lbs\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Force exerted on the plate\n",
+    "\n",
+    "# Given\n",
+    "from math import *\n",
+    "\n",
+    "d = 3            # diameter in inches\n",
+    "\n",
+    "A = pi*d**2/(4*144)    # Area of jet\n",
+    "\n",
+    "Q = 2           # discharge in ft**3/s\n",
+    "\n",
+    "rho = 1.94       # density in lbs/ft**3\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V = Q/A          # velocity in ft/s\n",
+    "\n",
+    "alpha = pi/6     # inlet vane angle\n",
+    "\n",
+    "bta = pi/6       # outlet vane angle\n",
+    "\n",
+    "Rx = rho*Q*(V*cos(bta)+V*cos(alpha))   # force in X direction\n",
+    "\n",
+    "Ry = rho*Q*(V*sin(bta)-V*sin(alpha))   # force in Y direction\n",
+    "\n",
+    "print \"Force exerted in X direction = \",round(Rx,1),\"lbs\"\n",
+    "\n",
+    "print \"Force exerted in Y direction = \",round(Ry,1),\"lbs\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 5.7 Page no 207"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a ) Angle of blade top at inlet and Outlet, Phi = 4 deg\n",
+      "b ) Work done per N of fluid per second =   80.31 N.m\n",
+      "c ) Efficiency =  98.4 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Angle of blade tips at inlet and exit ; work done on the vane; efficiency of the vane\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "V1 =40              # velocity in m/s\n",
+    "\n",
+    "Vp = 20             # velocity of the plate in m/s\n",
+    "\n",
+    "alpha = pi/6         # inlet vane angle\n",
+    "\n",
+    "bta = pi/9 # outlet vane angle\n",
+    "\n",
+    "g = 9.81\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V1x = V1*cos(alpha)\n",
+    "\n",
+    "Vw1 = V1x;\n",
+    "\n",
+    "V1y = V1*sin(alpha)\n",
+    "\n",
+    "dV = V1x - Vp\n",
+    "\n",
+    "theta = atan(V1y/dV)*180/pi\n",
+    "\n",
+    "Vr1 = V1y/sin(theta*pi/180)\n",
+    "\n",
+    "Vr2 = Vr1\n",
+    "\n",
+    "# from trial and error we get the blade tip angle at inlet and outlet\n",
+    "\n",
+    "print \"a ) Angle of blade top at inlet and Outlet, Phi = 4 deg\"\n",
+    "\n",
+    "phi = 4*pi/180 \n",
+    "\n",
+    "V2 = Vr2*sin(phi)/sin(bta)\n",
+    "\n",
+    "V2w = V2*cos(bta)\n",
+    "\n",
+    "W = (V2w+V1x)*Vp/g\n",
+    "\n",
+    "print \"b ) Work done per N of fluid per second =  \",round(W,2),\"N.m\"\n",
+    "\n",
+    "Eff = (1 - (V2/V1)**2)*100\n",
+    "\n",
+    "print \"c ) Efficiency = \",round(Eff,2),\"%\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 5.8 Page no 211"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a - Thrust on the plane =  5372.2 lbs\n",
+      "b - Theoretical efficiency =  76.0 %\n",
+      "c - Theoretical horsepower required =  4560.0 hp\n",
+      "d - Pressure difference across blades =  190.0 lbs/ft**3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Thrust on the plane; propeller efficiency ; theoretical horsepower ; pressure difference acros the blades\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "v = 220                     # velocity in ft/s\n",
+    "\n",
+    "d = 6                       # diameter of the propeller\n",
+    "\n",
+    "Q = 12000                   # discharge in ft**3/s\n",
+    "\n",
+    "mf = 0.0022                 # mass flow rate in slugs/ft**3\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V1 = v*5280/3600            # velocity in ft/s\n",
+    "\n",
+    "V = Q/(pi*d**2/4)           # velocity in ft/s\n",
+    "\n",
+    "V4 = 2*V-V1\n",
+    "\n",
+    "F = mf*Q*(V4-V1)              # thrust on the plane\n",
+    "\n",
+    "print \"a - Thrust on the plane = \",round(F,1),\"lbs\"\n",
+    "\n",
+    "Eff = V1/V                 # efficiency \n",
+    "\n",
+    "E = Eff*100\n",
+    "\n",
+    "print \"b - Theoretical efficiency = \",round(E,0),\"%\"\n",
+    "\n",
+    "Thp = F*V1/(500*Eff)\n",
+    "\n",
+    "print \"c - Theoretical horsepower required = \",round(Thp,0),\"hp\"\n",
+    "\n",
+    "dP = mf*(V4**2-V1**2)/2\n",
+    "\n",
+    "print \"d - Pressure difference across blades = \",round(dP,2),\"lbs/ft**3\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter6.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter6.ipynb
new file mode 100755
index 00000000..3cff1efa
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter6.ipynb
@@ -0,0 +1,144 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 6 : Dimensional Analysis and Dynamic Similitude"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 6.2 Page no 233"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Hence the model should be tested at a velocity of  2.7 m/s. This velocity in the model is called corresponding velocity\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Velocity of the flow\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "L = 10                          # length scale lp/l\n",
+    "\n",
+    "# crue oil at 20 deg C\n",
+    "\n",
+    "rhop = 0.86*998.2                 # density inn kg/m**3\n",
+    "\n",
+    "mup = 8*10**-3                    # viscosity in Ns/m**2\n",
+    "\n",
+    "Vp = 2.5                          # Velocity in m/s\n",
+    "\n",
+    "# water at 20 deg C\n",
+    "\n",
+    "rhom = 998.2                     # density in kg/m**3\n",
+    "\n",
+    "mum = 1.005*10**-3               # viscosity in Ns/m**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Vm = Vp*L*(rhop/rhom)*(mum/mup)  # velocity in m/s\n",
+    "\n",
+    "print \"Hence the model should be tested at a velocity of \",round(Vm,2),\"m/s. This velocity in the model is called corresponding velocity\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 6.3 Page no 233"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Maximum discharge, Qm =  5.65685425 cfs\n",
+      "Maximum head over crest =  4.74 ft\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Mximum head on the crest and corresponding discharge for dynamically similar conditions\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "l = 300                               # length in ft\n",
+    "\n",
+    "Q = 100000                            # discharge in cfs\n",
+    "\n",
+    "Cd = 3.8                              # coefficient of discharge\n",
+    "\n",
+    "L = (1/50)                              # length scale\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Qm = 100000*(L)**(5/2)                     # model discharge in cfs\n",
+    "\n",
+    "print \"Maximum discharge, Qm = \",round(Qm,8),\"cfs\"\n",
+    "\n",
+    "H = (Q/(Cd*l))**(2/3)                 # height over spill way\n",
+    "\n",
+    "h = H*L*12                              # head over spill model\n",
+    "\n",
+    "print \"Maximum head over crest = \",round(h,2),\"ft\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter7.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter7.ipynb
new file mode 100755
index 00000000..a8fa6c1f
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter7.ipynb
@@ -0,0 +1,428 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 7 : Fluid Resistance"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 7.1 Page no 245"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Reynolds number for water = 7463.0\n",
+      "R > 2000 the flow is turbulent for water\n",
+      "\n",
+      "\n",
+      "Reynolds number for glycerine = 8.7\n",
+      "R < 2000 the flow is laminar for glycerine\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Is the flow laminar or turbulent\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "nu1 = 0.804*10**-6               # viscosity in m**2/s\n",
+    "\n",
+    "V = 0.3                        # velocity in m/s\n",
+    "\n",
+    "D = 0.02                       # diameter in m/s\n",
+    "\n",
+    "# for water \n",
+    "\n",
+    "rho = 995.7                    # density in kg/m**3\n",
+    "\n",
+    "# for gylcerine\n",
+    "\n",
+    "mu = 8620*10**-4               # viscosity in Ns/m**2\n",
+    "\n",
+    "S = 1.26                       # specific gravity\n",
+    "\n",
+    "nu2 = mu/(S*rho)                # viscosity of glycerine in Ns/m**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "R1 = V*D/nu1\n",
+    "\n",
+    "print \"Reynolds number for water =\",round(R1,0)\n",
+    "\n",
+    "print \"R > 2000 the flow is turbulent for water\"\n",
+    "\n",
+    "print \"\\n\"\n",
+    "R2 = V*D/nu2\n",
+    "\n",
+    "print \"Reynolds number for glycerine =\",round(R2,1)\n",
+    "\n",
+    "print \"R < 2000 the flow is laminar for glycerine\"\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 7.2 Page no 248"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.7/y\n",
+      "Turbulence constant =  0.46\n",
+      "Mixing length =  13.8 cm\n",
+      "Eddy viscosity = 44.1 Nm/s**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Eddy Viscosity; mixing length ; turbulence constant\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from scipy import *\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "from sympy import *\n",
+    "\n",
+    "y = Symbol('y')\n",
+    "\n",
+    "d = 0.0175           # diameter in m\n",
+    "\n",
+    "s = 0.3              # shear stress at a distance in m\n",
+    "\n",
+    "tau = 103            # shear stress in Pa\n",
+    "\n",
+    "rho = 1000           # density in kg/m**3\n",
+    "\n",
+    "#y = 0.3\n",
+    "\n",
+    "# solution\n",
+    "\n",
+    "Up = diff(8.5+0.7*log(y),y)\n",
+    "\n",
+    "print Up\n",
+    "\n",
+    "Up = (0.7/0.3)                     # for y = 0.3\n",
+    "\n",
+    "k = sqrt(tau/(rho*s**2*Up**2))\n",
+    "\n",
+    "print \"Turbulence constant = \",round(k,2)\n",
+    "\n",
+    "Ml = k*s*100             # mixing length\n",
+    "\n",
+    "print \"Mixing length = \",round(Ml,1),\"cm\"\n",
+    "\n",
+    "Eta = rho*(Ml/100)**2*Up\n",
+    "\n",
+    "print \"Eddy viscosity =\",round(Eta,1),\"Nm/s**2\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 7.3 Page no 256"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "data": {
+      "image/png": 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AAAYNAgYPBl55RelIiMgWyZIIYmJisG/fPhQXF6Nv374YNGgQunbtKvt6AkdNBKdPA/36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\nOCYCIiIH9/84dLhcBD/ZrQAAAABJRU5ErkJggg==\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x446d2d0>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "# PLot boundary layer distribution and total drag on the plate\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "from pylab import plt\n",
+    "\n",
+    "from numpy import *\n",
+    "\n",
+    "from scipy import *\n",
+    "\n",
+    "from sympy import *\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "# for glycerine\n",
+    "\n",
+    "S = 1.26                  # specific gravity \n",
+    "\n",
+    "mu = 0.862                # dynamic viscosity in Ns/m**2\n",
+    "\n",
+    "rho = S *1000             # density in kg/m**3\n",
+    "\n",
+    "K2 = 0.332\n",
+    "\n",
+    "V=1                       # velocity in m/s\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# from blasius equation\n",
+    "\n",
+    "x = [0,0.1,0.5,1.0,2.0];\n",
+    "\n",
+    "d = 0.1307*np.sqrt(x)*100\n",
+    "\n",
+    "tauo = K2*rho*V**2/(sqrt(1462)*np.sqrt(x))\n",
+    "\n",
+    "#plt.figure()\n",
+    "plt.plot(x, d, 'r')\n",
+    "plt.xlabel('x(m)')\n",
+    "plt.ylabel('delta(cm),tauo(N/m**2)')\n",
+    "#plt.title('delta v/s x')\n",
+    "#plt.legend('d')\n",
+    "\n",
+    "plt.plot(x, tauo, 'b')\n",
+    "plt.xlabel('x')\n",
+    "#plt.ylabel('tauo(N/m**2)')\n",
+    "#plt.title('tauo v/s x(m)')\n",
+    "plt.legend('d''t')\n",
+    "plt.show()\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 7.4 page no 260"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "data": {
+      "image/png": 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+NzRo4OiKREQqTF1JVaGoCF54ARYtgiVLzHAQEamhFAxXa/9+iI+H+vVh40bQ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\niIiUomAQEZFSFAwiIlKKgkFEREr5f1iGZ/dHpXkqAAAAAElFTkSuQmCC\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x4b3b910>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "data": {
+      "image/png": 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2cRucf3ZGu4btcNbvLIsAUTUh2XUEpD1Ss1MxMXQirqRdQahPKNwaukkdiYhe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\nsRAQEWk5FgIiIi33f5w+Vkc37U5LAAAAAElFTkSuQmCC\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x4e36b10>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total Drag =  0.595 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Sketch the boundary layer and drag on the plate\n",
+    "\n",
+    "import numpy as np\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "\n",
+    "from numpy import sqrt\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "rho = 1.197               # air density in kg/m**3\n",
+    "\n",
+    "mu = 18.22*10**-6         # viscosity in Ns/m**2\n",
+    "\n",
+    "l = 5                     # length of the plate\n",
+    "\n",
+    "V = 8                     # velocity in m/s\n",
+    "\n",
+    "Rec = 5*10**5              # crictical reynolds number\n",
+    "\n",
+    "l1 = 0.951                # length from 0 to 0.951\n",
+    "\n",
+    "l2 = 5.0                  # length from 0 to 5\n",
+    "\n",
+    "l3 = 0.951                # length from 0 to 0.951\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "X = Rec/525576\n",
+    "\n",
+    "x = [0,0.1,0.3,0.6,0.951];\n",
+    "\n",
+    "d = 0.0069*np.sqrt(x)*100\n",
+    "\n",
+    "plt.figure()\n",
+    "plt.plot(x, d, 'r')\n",
+    "plt.xlabel('x(m)')\n",
+    "plt.ylabel('delta(cm)')\n",
+    "plt.title('delta v/s x')\n",
+    "plt.legend('L')\n",
+    "plt.show()\n",
+    "\n",
+    "X1 = [0.951,1.5,2.0,2.5,3.0,4.0,5.0]\n",
+    "\n",
+    "Dt = 0.0265*np.power(X1,(4/5))*100\n",
+    "\n",
+    "plt.figure()\n",
+    "plt.plot(X1, Dt, 'g')\n",
+    "plt.xlabel('x(m)')\n",
+    "plt.ylabel('delta(cm)')\n",
+    "plt.title('delta v/s x')\n",
+    "plt.legend('T')\n",
+    "plt.show()\n",
+    "\n",
+    "Td = 0.664*sqrt(mu*rho*V**3*l1)+0.036*rho*V**2*l2*(mu/(rho*V*l2))**0.2-0.036*rho*V**2*l3*(mu/(rho*V*l3))**0.2\n",
+    "\n",
+    "print \"Total Drag = \",round(Td,3),\"N\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 7.5 Page no 270"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Drag on the sphere =  0.0043 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine drag on the sphere\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "from pylab import plt\n",
+    "\n",
+    "from __future__ import division\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.01         # doameter of sphere in m\n",
+    "\n",
+    "v = 0.05         # velocity in m/s\n",
+    "\n",
+    "S = 1.26         # specific gravity\n",
+    "\n",
+    "mu = 0.826       # kinematic viscosity in Ns/m**2\n",
+    "\n",
+    "rho = S*1000     # density\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "R = rho*v*d/mu\n",
+    "\n",
+    "# for the above rho\n",
+    "\n",
+    "Cd = 35\n",
+    "\n",
+    "Fd = 0.5*Cd*rho*v**2*pi*d**2/4\n",
+    "\n",
+    "print \"Drag on the sphere = \",round(Fd,4),\"N\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter8.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter8.ipynb
new file mode 100755
index 00000000..f81b0b54
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter8.ipynb
@@ -0,0 +1,396 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 8 : Laminar Flow"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 8.1 Page no 286"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Maximum shear stress = 1.232 N/m**2\n",
+      "(b) Maximum velocity = 0.645 m/s\n",
+      "(c) Discharge =  4.1e-06 m**3/s\n",
+      "Reynolds number = 547.0 is less than 2000, the flow is laminar and the calculations are valid\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine maximum velocity and shear stress\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "P1 = 200                 # Pressure at inlet in kPa\n",
+    "\n",
+    "P2 = 260                # Pressure at outlet in kPa\n",
+    "\n",
+    "d = 0.004                      # diameter in m\n",
+    "\n",
+    "L = 8                          # length of pipe in meters\n",
+    "\n",
+    "z = 6                          # height of the pipe from the ground\n",
+    "\n",
+    "g = 9.81                       # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# properties of kerosene\n",
+    "\n",
+    "mu = 19.1*10**-4               # viscosity of kerosene at 20 deg C\n",
+    "\n",
+    "S = 0.81                       # specific gravity of kerosene\n",
+    "\n",
+    "rho = 1000                     # density in kg/m**3\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# calculating direction of flow\n",
+    "\n",
+    "p1 = (P1+g*z*S)*1000                # point 1\n",
+    "\n",
+    "p2 = (P2)*1000                     # point 2\n",
+    "\n",
+    "# direction of flow is from point 1-2\n",
+    "\n",
+    "# shear stress\n",
+    "\n",
+    "Sp = -((p1-p2)/sqrt(L**2+z**2))\n",
+    "\n",
+    "r = d/2\n",
+    "\n",
+    "Tau_max = r*Sp/2\n",
+    "\n",
+    "print \"(a) Maximum shear stress =\",round(Tau_max,3),\"N/m**2\"\n",
+    "\n",
+    "# maximum velocity\n",
+    "\n",
+    "Vmax = r**2*Sp/(4*mu)\n",
+    "\n",
+    "print \"(b) Maximum velocity =\",round(Vmax,3),\"m/s\"\n",
+    "\n",
+    "# discharge\n",
+    "\n",
+    "Q = pi*r**4*Sp/(8*mu)\n",
+    "\n",
+    "print \"(c) Discharge = \",round(Q,7),\"m**3/s\"\n",
+    "\n",
+    "# calculate reynolds number\n",
+    "\n",
+    "V = Vmax/2\n",
+    "\n",
+    "R = rho*V*d*S/mu\n",
+    "\n",
+    "print \"Reynolds number =\",round(R,0),\"is less than 2000, the flow is laminar and the calculations are valid\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example no 8.2 Page no 289"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R =  1299.0 is lesss than 2000 , the flow is laminar\n",
+      "Head loss =  3.77 cm of water\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine the head loss\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.02                  # diameter of the pipe in m\n",
+    "\n",
+    "l = 30                    # length of the pipe in m\n",
+    "\n",
+    "v = 0.1                   # velocity in m/s\n",
+    "\n",
+    "g = 9.81                  # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# for water at 5 deg C\n",
+    "\n",
+    "nu = 1.54*10**-6          # kinematic viscosity of water in m**2/s\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "R = v*d/nu\n",
+    "\n",
+    "print \"R = \",round(R,0),\"is lesss than 2000 , the flow is laminar\"\n",
+    "\n",
+    "f = 64/R                 # friction factor\n",
+    "\n",
+    "Hf = f*l*v**2/(2*g*d)    # head loss due to friction\n",
+    "\n",
+    "H=Hf*100\n",
+    "\n",
+    "print \"Head loss = \",round(H,2),\"cm of water\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 8.3 Page no 290"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R = 40.07 is less than 2000 and hence flow is laminar\n",
+      "Horse power required to pump the oil =  10.6\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Horsepower required to pump 50 tons of oil\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# oil properties\n",
+    "\n",
+    "S = 0.92             # specific gravity\n",
+    "\n",
+    "gma = S*62.4         # density in lbs/ft**3\n",
+    "\n",
+    "nu=0.0205            # viscosity in ft**2/s\n",
+    "\n",
+    "W = 50         # weight of oil\n",
+    "\n",
+    "d = 9                # diameter of the pipe in inches\n",
+    "\n",
+    "g = 32.2             # acceleration due to gravity in ft/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Q = W*2000/(gma*3600)       # discharge in ft**3/s\n",
+    "\n",
+    "A = pi*d**2/(4*144)         # area of pipe\n",
+    "\n",
+    "V = Q*1000/(A)                     # velocity in ft/s\n",
+    "\n",
+    "R = V*0.75/(nu*1000)                  # Reynolds number\n",
+    "\n",
+    "print \"R =\",round(R,2),\"is less than 2000 and hence flow is laminar\"\n",
+    "\n",
+    "f = 64/R               # friction factor\n",
+    "\n",
+    "Hf = (f*5280*(V/1000)**2)/(2*g*0.75)\n",
+    "\n",
+    "Hp = gma*Q*Hf/(550)\n",
+    "\n",
+    "print \"Horse power required to pump the oil = \",round(Hp,1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 8.4 Page no 291"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Viscosity of the liquid = 0.0182 poise\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Viscosity of the liquid in poise\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "V = 50                      # Volume in m**3\n",
+    "\n",
+    "d = 5                       # diameter in m\n",
+    "\n",
+    "d1 = 0.1                    # diameter of bore\n",
+    "\n",
+    "l = 10                      # length of the tube\n",
+    "\n",
+    "t = 20*60                   # time in seconds\n",
+    "\n",
+    "rho = 0.88                  # density in g/cm**3\n",
+    "\n",
+    "H1 = 5                    # height from the base in m\n",
+    "\n",
+    "A = pi*d**2/4\n",
+    "\n",
+    "a = pi*d1**2/4\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# From derivation we obtain a equation for T\n",
+    "\n",
+    "H2 = H1-(V/A)\n",
+    "\n",
+    "mu = t*rho*a*(0.1)*98.1/(32*A*10*log(H1/H2))\n",
+    "\n",
+    "print \"Viscosity of the liquid =\",round(mu,4),\"poise\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 8.5 Page no 297"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge q =  0.013 per unit ft of the plate\n",
+      "Shear stress on the plate =  0.086 lbs/ft**2 and resisting the motion of the plate\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Velocity distribution; Discharge ; shear on the upper plate\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# properties of kerosene oil at 20 deg C\n",
+    "\n",
+    "S = 0.81                # specific gravity of oil\n",
+    "\n",
+    "mu = 4*10**-5           # viscosity of oil in lb.s/ft**2\n",
+    "\n",
+    "gma = 62.4*S            # density in lbs/ft**3\n",
+    "\n",
+    "p1 = 6.51               # pressure at point 1 in psia\n",
+    "\n",
+    "p2 = 8                  # pressure at point 2 in psia\n",
+    "\n",
+    "h = 0.006               # distance between the plate in ft\n",
+    "\n",
+    "l = 4                   # length of the plate in ft\n",
+    "\n",
+    "theta = pi/6            # angle of inclination\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "# point 1\n",
+    "\n",
+    "P1 = p1*144 + gma*l*sin(theta)\n",
+    "\n",
+    "# point 2\n",
+    "\n",
+    "P2 = p2*144\n",
+    "\n",
+    "# flow is taking from poont 2-1\n",
+    "\n",
+    "Sp = (P2-P1)/4\n",
+    "\n",
+    "# equation for u = 2154.75*y-359125*y**2\n",
+    "\n",
+    "y = h\n",
+    "\n",
+    "# discharge per ft width\n",
+    "\n",
+    "q = (2154.75*y**2/2) - (359125*y**3/3)\n",
+    "\n",
+    "print \"Discharge q = \",round(q,3),\"per unit ft of the plate\"\n",
+    "\n",
+    "# to find shear at the top of the plate take du/dy = 0\n",
+    "\n",
+    "dV = 2154.75 - 718250*h\n",
+    "\n",
+    "# shear stress\n",
+    "\n",
+    "T = -mu*dV\n",
+    "\n",
+    "print \"Shear stress on the plate = \",round(T,3),\"lbs/ft**2 and resisting the motion of the plate\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Fluid_Mechanics_by_Irfan_A_Khan/Chapter9.ipynb b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter9.ipynb
new file mode 100755
index 00000000..279e480a
--- /dev/null
+++ b/Fluid_Mechanics_by_Irfan_A_Khan/Chapter9.ipynb
@@ -0,0 +1,847 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 9 : Turbulent flow in Pipes"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.1 Page no 308"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "It is a laminar flow\n",
+      "(a) Head loss due to flow for glycerine = 17.1 m \n",
+      "The flow is turbulent\n",
+      "(a) Head loss due to flow for water = 5.42 m \n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine Head loss\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "S = 1.26                # specific gravity\n",
+    "\n",
+    "mu = 0.826              # kinematic viscosity in Ns/m**2\n",
+    "\n",
+    "# for water\n",
+    "\n",
+    "rho = 998               # density of water in kg/m**3\n",
+    "\n",
+    "mu1 = 1.005*10**-3      # viscosity in Ns/m**2\n",
+    "\n",
+    "# for glycerine\n",
+    "\n",
+    "rho1 = S*rho            # density of glycerine in kg/m**3\n",
+    "\n",
+    "Q = 0.1                 # discharge in m**3/s\n",
+    "\n",
+    "d1 = 0.2                # diameter in m\n",
+    "\n",
+    "A = pi*d1**2/4          # area in m**2\n",
+    "\n",
+    "g = 9.81                # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "l =100                  # length of the pipe\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V = Q/A\n",
+    "\n",
+    "R = rho1*V*d1/mu\n",
+    "\n",
+    "print \"It is a laminar flow\"\n",
+    "\n",
+    "f = 64/R               # friction factor\n",
+    "\n",
+    "Hf = f*l*V**2/(2*g*d1) # head loss due to friction\n",
+    "\n",
+    "print \"(a) Head loss due to flow for glycerine =\",round(Hf,1),\"m \"\n",
+    "\n",
+    "R1 = rho*V*d1/mu1\n",
+    "\n",
+    "print \"The flow is turbulent\"\n",
+    "\n",
+    "e = 0.025\n",
+    "\n",
+    "r = e/(d1*100)\n",
+    "\n",
+    "f = 0.021\n",
+    "\n",
+    "hf = f*l*V**2/(2*g*d1)\n",
+    "\n",
+    "print \"(a) Head loss due to flow for water =\",round(hf,2),\"m \"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.2 Page no 311"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.1853324563\n",
+      "Discharge = 0.1 m**3/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Discharge of water\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "# for water\n",
+    "\n",
+    "nu = 1.007*10**-6            # viscosity in m**2/s\n",
+    "\n",
+    "e = 0.025                    # for cast iron in cm\n",
+    "\n",
+    "L = 100                      # length of the pipe in m\n",
+    "\n",
+    "D = 0.2                      # diameter in m\n",
+    "\n",
+    "hf = 5.43                    # head loss due to friction\n",
+    "\n",
+    "r = e/(D*100)\n",
+    "\n",
+    "g = 9.81                     # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "A = sqrt(2*g*D*hf/L)\n",
+    "\n",
+    "B = D/nu\n",
+    "\n",
+    "f = 0.021                   # from moodys diagram\n",
+    "\n",
+    "V = A/sqrt(f)\n",
+    "\n",
+    "print V\n",
+    "\n",
+    "R = B*f\n",
+    "\n",
+    "A = pi*D**2/4\n",
+    "\n",
+    "Q = A*V\n",
+    "\n",
+    "print \"Discharge =\",round(Q,2),\"m**3/s\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.3 Page no 314"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Hence the convergence is attained, D= 0.2 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Size of the case iron\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Q =0.1                          # discharge in m**3/s\n",
+    "\n",
+    "hf = 5.43                       # friction loss head in m\n",
+    "\n",
+    "L = 100                         # length of pipe\n",
+    "\n",
+    "nu = 1.00*10**-6                # viscosity in m**2/s\n",
+    "\n",
+    "e = 0.025                       # for cast iron in cm\n",
+    "\n",
+    "g = 9.81                        # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "A = 8*L*Q**2/(hf*g*pi**2)\n",
+    "\n",
+    "B = 4*Q/(pi*nu)\n",
+    "\n",
+    "# for D = 0.172 ; f=0.01\n",
+    "D = 0.172\n",
+    "\n",
+    "r = e/D\n",
+    "\n",
+    "Re = B/D\n",
+    "\n",
+    "f = 0.022                       # for Re and r\n",
+    "\n",
+    "# for D1=0.199 ; f=0.021\n",
+    "\n",
+    "D1 = 0.199\n",
+    "\n",
+    "r1 = e/D1\n",
+    "\n",
+    "R = B/D1\n",
+    "\n",
+    "f = 0.021                       # for R and r\n",
+    "\n",
+    "print \"Hence the convergence is attained, D=\",round(D1,1),\"m\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.4 Page no 318"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Head loss = 2.0 ft\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Head loss \n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "L = 500                 # length of the pipe in ft\n",
+    "\n",
+    "D= 9*2.54/100           # diameter in cm\n",
+    "\n",
+    "C = 100                 # constant\n",
+    "\n",
+    "S = 0.004\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "Hf = S*L\n",
+    "\n",
+    "print \"Head loss =\",round(Hf,0),\"ft\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.5 Page no 319"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Head loss of water = 3.2 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Head loss of water\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Q = 0.1                  # water flow rate in m**3/s\n",
+    "\n",
+    "d = 30                  # diameter in m\n",
+    "\n",
+    "l = 500                  # length in m\n",
+    "\n",
+    "e = 0.025                # for cast iron\n",
+    "\n",
+    "g = 9.81                 # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "r = log(d/e,10)\n",
+    "\n",
+    "K = (pi/4)*sqrt(2*g)*(2*r+1.14)*(0.3)**(2.5)\n",
+    " \n",
+    "S = (Q/K)**2\n",
+    "\n",
+    "hf = S*l\n",
+    "\n",
+    "print \"Head loss of water =\",round(hf,1),\"m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.6 Page no 319"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Head loss of water = 0.1 m**3/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Head loss by conveyance method\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "Q = 0.1                  # water flow rate in m**3/s\n",
+    "\n",
+    "d = 20                  # diameter in cm\n",
+    "\n",
+    "l = 500                  # length in m\n",
+    "\n",
+    "e = 0.025                # for cast iron\n",
+    "\n",
+    "g = 9.81                 # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "S = 5.43 \n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "r = log(d/e,10)\n",
+    "\n",
+    "K = (pi/4)*sqrt(2*g)*(2*r+1.14)*(0.2)**2.5\n",
+    "\n",
+    "Q=K*sqrt(S/100)\n",
+    "\n",
+    "print \"Head loss of water =\",round(Q,2),\"m**3/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.7 Page no 320"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "K =  0.432  from trial and error\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve using the conveyence method\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# given\n",
+    "\n",
+    "eps = 0.025*10**-2           # for cast iron epsilon = 0.0025 cm\n",
+    "\n",
+    "# we get the value of K = 0.432 m**2/s\n",
+    "# we need to do trial and error to find the value of D\n",
+    "\n",
+    "# we use the value of D = 0.2 m\n",
+    "\n",
+    "D = 0.2               # value in m\n",
+    "\n",
+    "g = 9.81\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "K = (pi/4)*sqrt(2*g)*(2*log10(D/(eps))+1.14)*D**(2.5)\n",
+    "\n",
+    "print \"K = \",round(K,3),\" from trial and error\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.8 Page no 326"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total head loss = 83.7 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine head loss\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "d = 0.1                # diameter of the pipe\n",
+    "\n",
+    "Q= 0.075               # discharge in m**3/s\n",
+    "\n",
+    "L = 30                 # length in m\n",
+    "\n",
+    "A = pi*d**2/4\n",
+    "\n",
+    "g = 9.81               # acceleration due to gravity in m/s**2\n",
+    "\n",
+    "# for water\n",
+    "\n",
+    "nu = 1.007*10**-6       # viscosity in m**2/s\n",
+    "\n",
+    "e = 0.025\n",
+    "\n",
+    "r = e/(10*d)\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V = Q/A\n",
+    "\n",
+    "Re = V*d/nu\n",
+    "\n",
+    "f = 0.025              # firction factor from moodys diagram\n",
+    "\n",
+    "hf = f*L*V**2/(2*g*d)  \n",
+    "\n",
+    "K= 0.5                 # contraction constant\n",
+    "\n",
+    "hc = K*V**2/(2*g)     \n",
+    "\n",
+    "K1 =10                 # loss of the globe valve\n",
+    "\n",
+    "hg = K1*V**2/(2*g)\n",
+    "\n",
+    "Th = hf+hc+hg\n",
+    "\n",
+    "print \"Total head loss =\",round(Th,1),\"m\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.9 Page no 328"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge through the pipe = 0.067 m**3/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# discharge through the pipe\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "# for water\n",
+    "\n",
+    "nu = 1.007*10**-6                # viscosity in m**2/s\n",
+    "\n",
+    "d1 = 0.3                         # diameter of pipe 1 in m\n",
+    "\n",
+    "d2 = 0.15                        # diameter of pipe 2 in m\n",
+    "\n",
+    "d3 = 0.08                        # diameter of pipe 3 in m\n",
+    "\n",
+    "g = 9.81                         # acclelration due to gravity in m/s**2\n",
+    "\n",
+    "e = 0.025                        # for cast iron\n",
+    "\n",
+    "f1 = 0.019                       # foe e/d1\n",
+    "\n",
+    "f2 = 0.022                       # foe e/d2\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V3 = sqrt(2*g*100/((8.4*(f1)+268.85*(f2)+4.85)))\n",
+    "\n",
+    "V1 = (d3/d1)**2*V3\n",
+    "\n",
+    "V2 = (d3/d2)**2*V3\n",
+    "\n",
+    "# reynolds number for pipe BC\n",
+    "\n",
+    "R1 = V1*d1/nu\n",
+    "\n",
+    "R2 = V2*d2/nu\n",
+    "\n",
+    "Q = V3*pi*d3**2/4\n",
+    "\n",
+    "print \"Discharge through the pipe =\",round(Q,3),\"m**3/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.10 Page no 332"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Replacement of the flow system = 94.3 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Replace the flow system\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "D = 0.2                      # diameter of pipe 1\n",
+    "\n",
+    "D1 = 0.15                    # diameter of pipe 2\n",
+    "\n",
+    "Q = 0.1                     # discharge in m**3/s\n",
+    "\n",
+    "nu = 1.007*10**-6           # viscosity in m**2/s\n",
+    "\n",
+    "e = 0.025                   # e for cast iron\n",
+    "\n",
+    "r = e/(100*D)                     \n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "V = Q/(pi*(0.2)**2/4)\n",
+    "\n",
+    "R = V*D/nu\n",
+    "\n",
+    "f = 0.021                 # from moodys law\n",
+    "\n",
+    "r2 = e/D1\n",
+    "\n",
+    "V1 = Q/(pi*D1**2/4)\n",
+    "\n",
+    "R1 = V*D1/nu\n",
+    "\n",
+    "f2 = 0.023                # from moodys law\n",
+    "\n",
+    "L2 = 28562*D1**5/f2\n",
+    "\n",
+    "print \"Replacement of the flow system =\",round(L2,2),\"m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.11 Page no 335"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Discharge through branch 1 = 1.35 m**3/s\n",
+      "Discharge through branch 2 = 0.566 m**3/s\n",
+      "Discharge through branch 3 = 0.082 m**3/s\n",
+      "It is found that hf1=hf2=hf3 = 43.5 The distribution od discharge is correct\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Discharge through each branch\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "e = 0.025                # in cm\n",
+    "\n",
+    "nu = 1.007*10**-6        # viscosity in m**2/s\n",
+    "\n",
+    "Q1 = 0.5                 # discharge in m**3/s\n",
+    "\n",
+    "D1 = 50\n",
+    "\n",
+    "L1 = 500                 # length in m\n",
+    "\n",
+    "g = 9.81\n",
+    "\n",
+    "# Pipe 1\n",
+    "\n",
+    "r1 = e/D                 # r1 for pipe 1\n",
+    "\n",
+    "V1 = Q1/(pi*(0.5)**2/4)\n",
+    "\n",
+    "R = V*(0.5)/nu\n",
+    "\n",
+    "f1 = 0.018               # for the reynolds no\n",
+    "\n",
+    "hf1 = f*L1*V1**2/(2*g*D1)\n",
+    "\n",
+    "# pipe 2\n",
+    "\n",
+    "hf2 = hf1\n",
+    "\n",
+    "L2 =200                 # length in m\n",
+    "\n",
+    "D2 = 0.3                # diameter in m\n",
+    "\n",
+    "r2 = e/D2\n",
+    "\n",
+    "f2 =  0.02 \n",
+    "\n",
+    "V2 = 0.419/sqrt(f2) \n",
+    "\n",
+    "R2 = V2*D2/nu\n",
+    "\n",
+    "Q2 = V2*(pi*D2**2/4)\n",
+    "\n",
+    "#pipe 3\n",
+    "\n",
+    "hf3=hf1\n",
+    "\n",
+    "L3 = 300              # length of pipe 3 in m\n",
+    "\n",
+    "D3  =0.15             # diameter of pipe 3 in m\n",
+    "\n",
+    "r3 = e/D3         \n",
+    "\n",
+    "f = 0.022            # from moody's law\n",
+    "\n",
+    "V3 = 0.242/sqrt(f2)\n",
+    "\n",
+    "R3 = V3*D3/nu\n",
+    "\n",
+    "Q3 = V3*(pi*D3**2/4)\n",
+    "\n",
+    "Td = Q1+Q2+Q3\n",
+    "\n",
+    "q1 = Q1*(2.0/Td)\n",
+    "\n",
+    "q2 = Q2*(2.0/Td)\n",
+    "\n",
+    "q3 = Q3*(2.0/Td)\n",
+    "\n",
+    "print \"Discharge through branch 1 =\",round(q1,2),\"m**3/s\"\n",
+    "\n",
+    "print \"Discharge through branch 2 =\",round(q2,3),\"m**3/s\"\n",
+    "\n",
+    "print \"Discharge through branch 3 =\",round(q3,3),\"m**3/s\"\n",
+    "\n",
+    "# Actual head loss\n",
+    "\n",
+    "d = 0.5\n",
+    "\n",
+    "v1 = q1/(pi*(d)**2/4)\n",
+    "\n",
+    "R4 = v1*d/nu\n",
+    "\n",
+    "r4 = 0.0005                 # ratio of e/D\n",
+    "\n",
+    "f = 0.018\n",
+    "\n",
+    "Hf1 = f*L1*v1**2/(2*g*d)\n",
+    "\n",
+    "print \"It is found that hf1=hf2=hf3 =\",round(Hf1,1),\"The distribution od discharge is correct\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Example 9.14 Page no 349"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Minimum depth = 17.6 ft\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Find minimum depth below the ridge\n",
+    "\n",
+    "from math import *\n",
+    "\n",
+    "# Given\n",
+    "\n",
+    "e = 0.00015             # from moody's chart\n",
+    "\n",
+    "D = 2                   # depth in ft\n",
+    "\n",
+    "r = e/D\n",
+    "\n",
+    "z1 = 100                # elevation in ft\n",
+    "\n",
+    "mu = 1.084*10**-5       # viscosity in Ns/ft**2\n",
+    "\n",
+    "p1 = 34                 # pressure head in ft\n",
+    "\n",
+    "p2 = 10                 # pressure head in ft\n",
+    "\n",
+    "g = 32.2                # acclelration due to gravity in ft/s**2\n",
+    "\n",
+    "L = 1000                # length in ft\n",
+    "\n",
+    "# Solution\n",
+    "\n",
+    "f = 0.011           # assume\n",
+    "\n",
+    "V = sqrt(100/(10000/(2*2*g)))/sqrt(f)\n",
+    "\n",
+    "R = V*D/mu\n",
+    "\n",
+    "V1 = 10.15\n",
+    "\n",
+    "f1 = 0.0125\n",
+    "\n",
+    "Q = V1*pi*D**2/4\n",
+    "\n",
+    "x = p1-p2-(V1**2/(2*g))-(f1*L*V1**2/(2*g*D))\n",
+    "\n",
+    "Dp = 30 - x\n",
+    "\n",
+    "print \"Minimum depth =\",round(Dp,2),\"ft\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_1.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_1.ipynb
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--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_1.ipynb
@@ -0,0 +1,61 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:4a6f98ae985b8e1b6e034e8590a04939cfd519f9354fb6a0851df17e977d9e9f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Fluid and their Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Page 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      " \n",
+      " #Initialization  of  Variable\n",
+      "sigma  =  72.7*10**-3;                                 #Surface  Tension\n",
+      "r  =  1  *10**-3;                                      #Radius  of  Bubble\n",
+      "\n",
+      " #Calculations\n",
+      "P  =  2*sigma/r;\n",
+      "print \"Excess Pressure(N/m2) :\",P\n",
+      "\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Excess Pressure(N/m2) : 145.4\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_10.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_10.ipynb
new file mode 100755
index 00000000..510f6bab
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_10.ipynb
@@ -0,0 +1,341 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:03b40b6d5c44ff714aac13480905f296c46afa917e2740a687e8aa66bb21b428"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: Laminar and Turbulent Flows in Bounded System"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1, Page 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import symbols,diff,solve\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "mu  =  0.9;\n",
+      "rho  =  1260;\n",
+      "g  =  9.81;\n",
+      "x  =  45;                                                                  #theta  in  degrees\n",
+      "P1  =  250  *  10**3;\n",
+      "P2  =  80*  10**3;\n",
+      "Z1  =  1;\n",
+      "Z2  =  0;                                                                  #  datum\n",
+      "U  =  -1.5;\n",
+      "Y  =  0.01;\n",
+      "\n",
+      " #Calculations\n",
+      "gradP1  =  P1+  rho*g*Z1;\n",
+      "gradP2  =  P2+  rho*g*Z2;\n",
+      "DPstar  =  (gradP1-gradP2)*math.sin(math.radians(x))/(Z1-Z2);\n",
+      "A  =  U/Y;                                                                #  Coefficient  U/Y  for  equation  10.6\n",
+      "B  =  DPstar/(2*mu)                                              #  Coefficient  dp*/dx  X(1/2mu)  for  equation  10.6\n",
+      "y  =  symbols('y')\n",
+      "v  =  round((A  +  B*Y),1)*y  -round(B)*y**2\n",
+      "duBYdy  =  diff(v,y);\n",
+      "tau  =  0.9*duBYdy;\n",
+      "stagPnts = solve(duBYdy,y)\n",
+      "ymax=stagPnts[0]                                      #value  of  y  where  derivative  vanishes.;\n",
+      "umax  =  (A + B*Y)*ymax  +  B*ymax**2;              #  Check  the  value  there  is  slight  mistake  in  books  answer\n",
+      "def  u(y):\n",
+      "    z  =  (A + B*Y)*y -B*y**2;\n",
+      "    return diff(z,y)\n",
+      "def dif(y):\n",
+      "    return round((A + B*Y)) -2*round(B)*y\n",
+      "\n",
+      "taumax=abs(mu*dif(Y))\n",
+      "\n",
+      "print \"velocity distribution         :\",v\n",
+      "print \"shear stress distribution     :\",mu*dif(y)\n",
+      "print \"maximum flow velocity (m/s)   :\",round(umax,2)\n",
+      "print \"Maximum Shear Stress  (kN/m^2):\",(round(taumax)/1000)\n",
+      "  \n",
+      "\n",
+      "print "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity distribution         : -71638.0*y**2 + 566.4*y\n",
+        "shear stress distribution     : -128948.4*y + 509.4\n",
+        "maximum flow velocity (m/s)   : 3.36\n",
+        "Maximum Shear Stress  (kN/m^2): 0.78\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2, Page 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "mu  =  0.9;\n",
+      "rho  =  1260;\n",
+      "d  =  0.01;\n",
+      "Q  =  1.8/60*10**-3;                                              #Flow  in  m**3  per  second\n",
+      "l  =  6.5;\n",
+      "ReCrit  =  2000;\n",
+      " #Calculations\n",
+      "A  =  (math.pi*d**2)/4;\n",
+      "MeanVel  =  Q/A;\n",
+      "Re  =  rho*MeanVel*d/mu/10;                                    #  Check  properly  the  answer  in  book  there  is  something  wrong\n",
+      "Dp  =  128*mu*l*Q/(math.pi*d**4)\n",
+      "Qcrit  =  Q*ReCrit/Re*10**3;\n",
+      "\n",
+      "print \"Pressure Loss        (kN/m2) :\",round(Dp/1000,0)\n",
+      "print \"Maximum Flow rate (litres/s) :\",round(Qcrit,0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure Loss        (kN/m2) : 715.0\n",
+        "Maximum Flow rate (litres/s) : 112.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3, Page 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "mu  =  1.14*10**-3;\n",
+      "rho  =  1000;\n",
+      "d  =  0.04;\n",
+      "Q  =  4*10**-3/60;                                    #Flow  in  m**3  per  second\n",
+      "l  =  750;\n",
+      "ReCrit  =  2000;\n",
+      "g  =  9.81;\n",
+      "k  =  0.00008;                                       #  Absolute  Roughness\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  (math.pi*d**2)/4;\n",
+      "MeanVel  =  Q/A;\n",
+      "Re  =  rho*MeanVel*d/mu;\n",
+      "Dp  =  128*mu*l*Q/(math.pi*d**4);\n",
+      "hL  =  Dp/(rho*g);\n",
+      "f  =  16/Re;\n",
+      "hlDa  =  4*f*l*MeanVel**2/(2*d*g);                   #  By  Darcy  Equation\n",
+      "Pa  =  rho*g*hlDa*Q;\n",
+      "\n",
+      " #Part(b)\n",
+      "Q  =  30*10**-3/60;                                  #Flow  in  m**3  per  second\n",
+      "MeanVel  =  Q/A;\n",
+      "Re  =  rho*MeanVel*d/mu;\n",
+      "RR  =  k/d;                                          #  relative  roughness\n",
+      "f  =  0.008                                          #by  Moody  diagram  for  Re  =    1.4  x  10**4    and  relative  roughness  =  0.002\n",
+      "hlDb  =  4*f*l*MeanVel**2/(2*d*g);                   #  By  Darcy  Equation\n",
+      "Pb  =  rho*g*hlDb*Q;\n",
+      "\n",
+      "\n",
+      "print \"!---- Case (a) ----!\\n\",\"Head Loss(mm)      :\",round(hlDa*1000,1)\n",
+      "print \"Power Required (W) :\",round(Pa,4)\n",
+      "print \"\\n!---- Case (b) ----!\\n\",\"Head Loss(m)       :\",round(hlDb,2)\n",
+      "print \"Power Required (W) :\",round(Pb,0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "!---- Case (a) ----!\n",
+        "Head Loss(mm)      : 92.5\n",
+        "Power Required (W) : 0.0605\n",
+        "\n",
+        "!---- Case (b) ----!\n",
+        "Head Loss(m)       : 4.84\n",
+        "Power Required (W) : 24.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4, Page 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "w  =  4.5;\n",
+      "d  =  1.2  ;\n",
+      "C  =  49;\n",
+      "i  =  1/800;\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  d*w;\n",
+      "P  =  2*d  +  w;\n",
+      "m  =  A/P;\n",
+      "v  =  C*(m*i)**0.5;\n",
+      "Q  =  v*A;\n",
+      "\n",
+      "print \"Mean Velocity (m/s):\",round(v,2)\n",
+      "print \"Discharge (m3/s)   :\",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean Velocity (m/s): 1.53\n",
+        "Discharge (m3/s)   : 8.28\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5, Page 348"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import sympy\n",
+      "from sympy import symbols\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "r,R = symbols('r R')\n",
+      "\n",
+      "#Calculations\n",
+      "rbyR=round((1-(49/60)**7),3)\n",
+      "r = (rbyR)*R \n",
+      "\n",
+      "#Result\n",
+      "print \"radius at which actual velocity is equal to mean velocity is\",r"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radius at which actual velocity is equal to mean velocity is 0.758*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7, Page 355"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "d1  =  0.140;\n",
+      "d2  =  0.250;\n",
+      "DpF_DpR  =  0.6;                                       #Difference  in  head  loss  when  in  forward  and  in  reverse  direction\n",
+      "K  =  0.33                                             #From  table\n",
+      "g  =  9.81;\n",
+      " #Calculations\n",
+      "ratA  =  (d1/d2)**2;\n",
+      "v =(DpF_DpR*2*g/((1-ratA)**2-K))**0.5;\n",
+      "\n",
+      "print \"Velocity (m/s):\",round(v,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity (m/s): 9.13\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_11.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_11.ipynb
new file mode 100755
index 00000000..ac453318
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_11.ipynb
@@ -0,0 +1,120 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8adfc21cf4552e14b563ca7c0f2e56ece56cd7f1691cf9486f2cfbd789df7be1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11: Boundary Layer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1, Page 383"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "rho  =  860;\n",
+      "v  =  10**-5;\n",
+      "Us  =  3;\n",
+      "b  =  1.25;\n",
+      "l  =  2;\n",
+      "\n",
+      " #Calculations\n",
+      "x  =  1;                                                                    #  At  x  =1  \n",
+      "Rex  =  Us*x/v;\n",
+      "ReL  =  Us*l/v  ;\n",
+      "mu  =    rho*v;\n",
+      "T0  =  0.332*mu*Us/x*Rex**0.5;\n",
+      "Cf  =  1.33*ReL**-0.5;\n",
+      "F  =  rho*Us**2*l*b*Cf  ;\n",
+      "print \"shear stress                                (N/m^2) :\" ,round(T0,1)\n",
+      "print \"Total, double-sided resistance of the plate (N)     :\",round(F,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "shear stress                                (N/m^2) : 4.7\n",
+        "Total, double-sided resistance of the plate (N)     : 33.224\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Us  =  6;\n",
+      "b  =  3;\n",
+      "l  =  30;\n",
+      "rho  =  1000;\n",
+      "mu  =  10**-3;\n",
+      "T  =  20+273;                                                          #  Temperature  in  Kelvin\n",
+      "\n",
+      " #Calculations\n",
+      " \n",
+      "ReL  =  rho*Us*l/mu;  \n",
+      "Cf  =  0.455*math.log10(ReL)**-2.58    ;\n",
+      "\n",
+      "F  =  rho*Us**2*l*b*Cf  ;\n",
+      "Lt  =  10**5*mu/(rho*Us);                                    #  Assuming  transition  at  Rel  =  10**5\n",
+      "\n",
+      "print \"Drag Force (kN)       :\", round(F/1000,2)\n",
+      "print \"Transition length (m) :\",round(Lt,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Drag Force (kN)       : 6.36\n",
+        "Transition length (m) : 0.0167\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_12.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_12.ipynb
new file mode 100755
index 00000000..745a2241
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_12.ipynb
@@ -0,0 +1,319 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:24438326f2aac267ced1e2e3aa21e094819eefb50d5f8862608b62208400992d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Incompressible Flow around a Body"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1, Page 399"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "x  =35;\n",
+      "T  =  50;\n",
+      "m  =  1;\n",
+      "g  =9.81;\n",
+      "rho  =  1.2;\n",
+      "A  =  1.2;\n",
+      "U0  =  40*1000/3600;                                            #  Velocity  in  m/s\n",
+      "\n",
+      " #Calculations\n",
+      "L  =  T*math.sin(math.radians(x))+m*g;\n",
+      "D  =T*math.cos(math.radians(x));\n",
+      "Cl  =  2*L/(rho*U0**2*A);\n",
+      "Cd  =  2*D/(rho*U0**2*A);  \n",
+      "\n",
+      "print \"Lift Coefficient :\",round(Cl,3)\n",
+      "print \"Drag Coefficient :\",round(Cd,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Lift Coefficient : 0.433\n",
+        "Drag Coefficient : 0.461\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2, Page 406"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Vp  =12;\n",
+      "lp  =  40;\n",
+      "lm  =  1;\n",
+      "As  =  2500;\n",
+      "Dm  =  32;\n",
+      "rhoP  =  1025;\n",
+      "rhoM  =  1000;\n",
+      "Ap  =  As;\n",
+      "\n",
+      " #Calculations\n",
+      "Am  =  As/40**2;\n",
+      "Vm  =  round(Vp*(lm/lp)**0.5,2);\n",
+      "Dfm  =  round(3.7*Vm**1.95*Am,1);\n",
+      "Rm  =  Dm  -  Dfm;\n",
+      "Rp  =  Rm  *(rhoP/rhoM)*(lp/lm)**2*(Vp/Vm)**2;\n",
+      "Dfp  =  2.9*Vp**1.8*Ap;\n",
+      "Dp  =  Rp  +  Dfp;\n",
+      "\n",
+      "print \"Expected total resistance (kN) :\",round(Dp/1000,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Expected total resistance (kN) : 1407.07\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3, Page 410"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "U0  =  80*1000/3600;\n",
+      "d  =  0.02;\n",
+      "rho  =1.2;\n",
+      "mu  =  1.7*10**-5;\n",
+      "A  =  0.02*500;                            #  Projected  area  of  wire\n",
+      "N  =  20;                                  #  No  of  cables\n",
+      "\n",
+      " #Calculations\n",
+      "Re  =  rho*U0*d/mu;\n",
+      "Cd  =  1.2                                 #  From  figure  12.10  for  given  Re;  \n",
+      "D  =  0.5*rho*Cd*A*U0**2\n",
+      "F  =  N*D;  \n",
+      "f  =    0.198*(U0/d)*(1-19.7/Re);\n",
+      "\n",
+      "print \"Total force on tower (kN) :\",round(F/1000,2)\n",
+      "print \"Frequency (Hz)            :\",round(f,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total force on tower (kN) : 71.11\n",
+        "Frequency (Hz)            : 219.9\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4, Page 415"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "mu  =  0.03;\n",
+      "d  =  10**-3;\n",
+      "rhoP  =  1.1*10**3;\n",
+      "g  =  9.81;\n",
+      "rho0  =  0.9*10**3;\n",
+      " #Calculations\n",
+      "B  =  18*mu/(d**2*rhoP);\n",
+      "t  =  round(4.60/B,4);\n",
+      "Vt  =  round(d**2*(rhoP  -  rho0)*g/(18*mu),5);\n",
+      "Re  =  rho0*Vt*d/mu;\n",
+      "\n",
+      "print \"Time taken by the particle take to reach 99 per cent of its terminal velocity (s):\",t\n",
+      "print \"\\nReynolds No corrosponding to the velocity :\",Re"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time taken by the particle take to reach 99 per cent of its terminal velocity (s): 0.0094\n",
+        "\n",
+        "Reynolds No corrosponding to the velocity : 0.1089\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5, Page 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "muO  =  0.0027;\n",
+      "Vt  =  3*10**-3;\n",
+      "rhoW  =  1000;\n",
+      "rhoP  =  2.4*rhoW;\n",
+      "rhoO  =  0.9*rhoW;\n",
+      "g  =  9.81;\n",
+      "muA  =  1.7*10**-5;\n",
+      "rhoA  =  1.3;\n",
+      "\n",
+      " #Calculations\n",
+      "d  =  (18*muO*Vt/(rhoP-rhoO)/g)**0.5;\n",
+      "Re  =  Vt*d*rhoO/muO;\n",
+      "\n",
+      " #Movement  of  particle  in  upward  direction\n",
+      "if(Re < 1):\n",
+      "    v  =  0.5;\n",
+      "    \n",
+      "    Re=5;       #  from  fig  12.15\n",
+      "    vt  =  muA*Re/(rhoA*d);\n",
+      "    u  =  vt+v;\n",
+      "    print \"Velocity of air stream blowing vertically up (m/s) :\",round(u,3) \n",
+      "else:\n",
+      "    print \"strokes law is not valid\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of air stream blowing vertically up (m/s) : 1.157\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6, Page 429"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "c  =  2;\n",
+      "s  =  10;\n",
+      "rho  =  5.33;\n",
+      "rho_ellip  =  1.2;\n",
+      "D  =  400;\n",
+      "L  =  45000;\n",
+      "scale  =  20;\n",
+      "U_windTunnel  =  500;\n",
+      "U_proto  =  400*1000/3600;\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  c*s;\n",
+      "U_model  =  U_windTunnel/scale;\n",
+      "Cd  =  D/(0.5*rho*U_model**2*A);\n",
+      "Cl  =  L/(0.5*rho_ellip*U_proto**2*A);          #  Considering  elliptical  Lift  model\n",
+      "Cdi  =  Cl**2/(math.pi*s/c);                    #  Aspect  Ratio  =    s/c  \n",
+      "Cdt  =  Cd  +  Cdi;\n",
+      "Dw  =  0.5*Cdt*rho_ellip*U_proto**2*A;\n",
+      "print \"Total drag on full sized wing (kN) :\",round(Dw/1000,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total drag on full sized wing (kN) : 2.65\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_13.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_13.ipynb
new file mode 100755
index 00000000..664b4abf
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_13.ipynb
@@ -0,0 +1,142 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1f446f61d8838d40d2e31fd72ef3a0bc1e1a7f291c54a48c1366aef232ac3797"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13: Compressible Flow around a Body"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1, Page 444"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "rho0  =  1.8;\n",
+      "R  =  287;\n",
+      "T  =  75+273;                                                          #  Temperature  in  kelvin  \n",
+      "gma  =  1.4;  \n",
+      "Ma  =  0.7;\n",
+      "\n",
+      " #Calculations\n",
+      "P0  =  rho0*R*T;\n",
+      "c  =  (gma*R*T)**0.5;\n",
+      "V0  =  Ma*c;\n",
+      "Pt  =  (P0**((gma-1)/gma)  +  rho0*((gma-1)/gma)*(V0**2/(2*P0**(1/gma))))**(gma/(gma-1));\n",
+      "rhoT  =    rho0*(Pt/P0)**(1/gma);\n",
+      "Tt  =    Pt/(R*rhoT)-273;\n",
+      "\n",
+      "print \"Staganation Pressure (kN/m2 ) :\",round(Pt/1000,1)\n",
+      "print \"Temperature (Degree Celcius)  :\",round(Tt,1)\n",
+      "print \"Density of airstream (kg/m3)  :\",round(rhoT,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Staganation Pressure (kN/m2 ) : 249.4\n",
+        "Temperature (Degree Celcius)  : 109.1\n",
+        "Density of airstream (kg/m3)  : 2.274\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2, Page 454"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "R  =  287;\n",
+      "T  =  28+273;\n",
+      "gma  =  1.4;\n",
+      "P  =  1.02*10**5;\n",
+      "rhoHg  =  13.6*10**3;\n",
+      "g  =  9.81;\n",
+      "\n",
+      " #Calculations\n",
+      " #Case(a)\n",
+      "U0  =  50;\n",
+      "c  =  (gma*R*T)**0.5;\n",
+      "Ma  =  U0/c;\n",
+      "rho  =  P/(R*T);\n",
+      "DelP  =  0.5*rho*U0**2;                                        #Pt-P\n",
+      "ha  =  DelP/(rhoHg*g);\n",
+      "\n",
+      " #Case(b)\n",
+      "U0  =  250;\n",
+      "Ma  =  U0/c;\n",
+      "Pt  =  P*(1+(gma-1)*Ma**2/2)**(gma/(gma-1));    \n",
+      "DelP  =  Pt-P\n",
+      "hb  =  DelP/(rhoHg*g);\n",
+      "\n",
+      " #Case  (c)\n",
+      "U0  =  420;\n",
+      "Ma1  =U0/c;\n",
+      "P2  =  P*((2*gma/(gma+1))*Ma1**2  -  ((gma-1)/(gma+1)));\n",
+      "N  =  Ma1**2  +2/(gma-1);                                      #  Numerator\n",
+      "D  =  2*gma*Ma1**2/(gma-1)-1;\n",
+      "Ma2  =  (N/D)**0.5;\n",
+      "Pt2  =  P2*(1+(gma-1)*Ma2**2/2)**(gma/(gma-1));\n",
+      "hc  =  (Pt2-P2)/(rhoHg*g)  ;    \n",
+      "\n",
+      "print \"Difference in height of mercury column in case (a) in mm :\",round(ha*1000,2)\n",
+      "print \"Difference in height of mercury column in case (b) in mm :\",round(hb*1000,1)\n",
+      "print \"Difference in height of mercury column in case (c) in mm :\",round(hc*1000,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Difference in height of mercury column in case (a) in mm : 11.06\n",
+        "Difference in height of mercury column in case (b) in mm : 314.2\n",
+        "Difference in height of mercury column in case (c) in mm : 684.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_14.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_14.ipynb
new file mode 100755
index 00000000..2e55c7c8
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_14.ipynb
@@ -0,0 +1,365 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:75785fe7e68940de5df81a115443b8d5c78793053ae837735e58f86a072c05cc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14: Steady Incompressible Flow in Pipe and Duct System"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1, Page 468"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "L1  =  5;\n",
+      "L2  =  10;\n",
+      "d  =  0.1;\n",
+      "f  =  0.08;\n",
+      "Za_Zc  =  4;                                                 #difference  in  height  between  A  and  C  \n",
+      "g  =  9.81  ;\n",
+      "Pa  =  0;\n",
+      "Va  =  0;  \n",
+      "Za_Zb  =  -1.5;\n",
+      "V  =  1.26;\n",
+      "rho  =  1000;\n",
+      "\n",
+      " #Calculations\n",
+      "D  =  1.5  +  4*f*(L1+L2)/d  ;                               #  Denominator  in  case  of  v**2  \n",
+      "v  =  (2*g*Za_Zc/D)**0.5;\n",
+      "Pb  =  rho*g*Za_Zb  -  rho*V**2/2*(1.5+4*f*L1/d);\n",
+      "print \"Pressure in the pipe at B (kN/m2):\",round(Pb/1000,2)\n",
+      "print \"Mean Velocity at C    (m/s)      :\",round(v,2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure in the pipe at B (kN/m2): -28.61\n",
+        "Mean Velocity at C    (m/s)      : 1.26\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3, Page 473"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from sympy import symbols,solve\n",
+      "import sympy\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Za_Zb  =  10;\n",
+      "f  =  0.008;\n",
+      "L  =  100;\n",
+      "d1  =  0.05;\n",
+      "g  =  9.81;\n",
+      "d2  =  0.1;\n",
+      "\n",
+      " #Calculations\n",
+      "\n",
+      "def  flowRate(d):\n",
+      "    D = 1.5 + 4*f*L/d ;                                     #  Denominator  in  case  of  v1**2\n",
+      "    A  =  math.pi*d**2/4;\n",
+      "    v  =  (2*g*Za_Zb/D)**0.5;\n",
+      "    z  =  A*v;\n",
+      "    return z     \n",
+      "Q1  =  flowRate(d1);\n",
+      "Q2  =  flowRate(d2);\n",
+      "Q=round(Q1+Q2,4)\n",
+      "\n",
+      "\n",
+      "D=symbols('D')\n",
+      "roots=solve(241212*D**5  -3.2, D)\n",
+      "dia=roots[0]\n",
+      "\n",
+      "print \"Rate flow for pipe 1            (m^3/s) :\",round(Q1,4)\n",
+      "print \"Rate flow for pipe 2            (m^3/s) :\",round(Q2,4)\n",
+      "print \"Combined Rate flow              (m^3/s) :\",round(Q,4)\n",
+      "print \"Diameter of single equivalent pipe (mm) :\",round(dia,3)*1000\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate flow for pipe 1            (m^3/s) : 0.0034\n",
+        "Rate flow for pipe 2            (m^3/s) : 0.019\n",
+        "Combined Rate flow              (m^3/s) : 0.0224\n",
+        "Diameter of single equivalent pipe (mm) : 106.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4, Page 476"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import solve,symbols\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Za_Zb  =  16;\n",
+      "Za_Zc  =  24;\n",
+      "f  =  0.01;\n",
+      "l1  =  120;\n",
+      "l2  =  60;\n",
+      "l3  =  40;\n",
+      "d1  =  0.12;\n",
+      "d2  =  0.075;\n",
+      "d3  =  0.060;\n",
+      "g  =  9.81;\n",
+      " #Calculations\n",
+      "\n",
+      "v1=symbols('v1')\n",
+      "ash=solve(v1-0.3906*(g-1.25*v1**2)**0.5-0.25*(17.657-1.5*v1**2)**0.5,v1)\n",
+      "v1=round(abs(ash[0]),2)\n",
+      "Q1=math.pi/4*d1**2*v1\n",
+      "\n",
+      "v2=(g-1.25*v1**2)**0.5\n",
+      "Q2=math.pi/4*d2**2*v2\n",
+      "\n",
+      "v3=(17.657-1.5*v1**2)**0.5\n",
+      "Q3=math.pi/4*d3**2*v3\n",
+      "\n",
+      "print \"Flow rate in pipe 1 (m^3/s):\",round(Q1,4)\n",
+      "print \"Flow rate in pipe 2 (m^3/s):\",round(Q2,4)\n",
+      "print \"Flow rate in pipe 3 (m^3/s):\",round(Q3,4)\n",
+      "print \"continuity condition satisfied as Q1=Q2+Q3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate in pipe 1 (m^3/s): 0.0206\n",
+        "Flow rate in pipe 2 (m^3/s): 0.0105\n",
+        "Flow rate in pipe 3 (m^3/s): 0.0101\n",
+        "continuity condition satisfied as Q1=Q2+Q3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.5, Page 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "D  =  0.3;\n",
+      "Q  =  0.8;\n",
+      "rho  =  1.2;\n",
+      "f  =  0.008;\n",
+      "L_entry  =  10;\n",
+      "L_exit  =  30;\n",
+      "Lt  =  20*D                              #Transition  may  be  represented  by  a  separation  loss  equivalent  length  of  20  *  the  approach  duct  diameter\n",
+      "K_entry  =  4;\n",
+      "K_exit  =  10\n",
+      "l  =  0.4;                               #  length  of  cross  section\n",
+      "b  =  0.2;                               #  width  of  cross  section\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  math.pi*D**2/4;\n",
+      "Dp1  =  0.5*rho*Q**2/A**2*(K_entry  +  4*f*(L_entry+Lt)/D);\n",
+      "area  =  l*b;\n",
+      "perimeter  =2*(l+b);\n",
+      "m  =  area/perimeter;\n",
+      "Dp2  =  0.5*rho*Q**2/area**2*(K_exit  +  f*L_exit/m);\n",
+      "Dfan  =    Dp1+Dp2;\n",
+      "\n",
+      "print \"fan Pressure input (N/m2) :\",round(Dfan,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fan Pressure input (N/m2) : 1254.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.6, Page 482"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "D  =  [0.15 , 0.3];\n",
+      "rho  =  1.2;\n",
+      "f  =  0.008;\n",
+      "L_entry  =  10;\n",
+      "L_exit  =  20;\n",
+      "Lt  =  20*D[1]              \n",
+      "K  =  4;\n",
+      "Q1  =  0.2;\n",
+      "\n",
+      " #Calculations\n",
+      "Q2  =  4*Q1;\n",
+      "A=[0.0,0.0]\n",
+      "A[0]  =  math.pi*D[0]**2/4;\n",
+      "A[1]  =  math.pi*D[1]**2/4;\n",
+      "Dp1  =  0.5*rho*Q1**2/A[0]**2*(K  +  4*f*L_entry/D[0]);\n",
+      "Dp2  =  0.5*rho*Q2**2/A[1]**2*(4*f*(L_exit  +  Lt)/D[1]);\n",
+      "Dfan  =    Dp1+Dp2;\n",
+      "\n",
+      "print \"fan Pressure input (N/m2) :\",round(Dfan,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fan Pressure input (N/m2) : 684.51\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.7, Page 487"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy.optimize import fsolve\n",
+      " \n",
+      " \n",
+      " \n",
+      " #Initializing  the  variables\n",
+      "d  =  [0.1 , 0.125,  0.15,  0.1,  0.1  ];                    #  Corrosponding  to  AA1B  AA2B  BC  CD  CF\n",
+      "l  =  [30 , 30 , 60,  15,  30];                                      #  Corrosponding  to  AA1B  AA2B  BC  CD  CF\n",
+      "rho   =  1.2;\n",
+      "f     =  0.006;\n",
+      "Ha    =  100;\n",
+      "Hf    =  60;\n",
+      "He    =  40;\n",
+      "K = [0.0, 0.0, 0.0, 0.0, 0.0]\n",
+      " #Calculations\n",
+      "for i in range(0,len(l)):\n",
+      "    K[i]  =  f*l[i]/(3*d[i]**5);\n",
+      "\n",
+      "\n",
+      "K_ab  =  K[0]*K[1]/((K[0])**0.5+(K[1])**0.5)**2;\n",
+      "K_ac  =  K_ab  +  K[2];\n",
+      "Hc  =  (K_ac*Hf  +K[4]*Ha/4)/(K_ac+K[4]/4);\n",
+      "Q  =  ((Ha  -  Hc)/K_ac)**0.5;\n",
+      "\n",
+      "def  f(n):\n",
+      "    z  =  He  -  Hc  +  (0.5*Q)**2  *(K[3]+(4000/n)**2);\n",
+      "    return z\n",
+      "\n",
+      "n  =  fsolve(f,1);\n",
+      "\n",
+      "print \"total Volume flow rate (m3/s):\",round(Q, 4)\n",
+      "print \"Head at C (m)                :\",round(Hc,2) \n",
+      "print \"Percentage valve opening (%) :\",round(n,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total Volume flow rate (m3/s): 0.1016\n",
+        "Head at C (m)                : 75.48\n",
+        "Percentage valve opening (%) : 38.58\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_15.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_15.ipynb
new file mode 100755
index 00000000..95bd78ec
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_15.ipynb
@@ -0,0 +1,114 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:4f421cc7413696fe7c281427d95aff4b53c3ab081abe9cd6d31d1e2df1133e5a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "    Chapter 15: Uniform Flows in Open Channels"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.1, Page 516"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "B  =4;\n",
+      "D  =  1.2;\n",
+      "C  =  7.6;\n",
+      "n  =  0.025;\n",
+      "s  =  1/1800;\n",
+      "\n",
+      " #Calculations\n",
+      "W  =  B  +  2*1.5*D;\n",
+      "A  =  D*(B+C)/2;                                                #  Area  of  parallelogram  formed\n",
+      "P  =  B  +2*1.2*(D**2+(1.5)**2)**0.5;\n",
+      "m  =A/P;\n",
+      "i=s;\n",
+      "C  =  (23+0.00155/i+1/n)/(1+(23+0.00155/i)*n/(m)**0.5);         #  By  Kutter  formula\n",
+      "Q1  =  C*A*(m*i)**0.5;\n",
+      "Q2  =  A*(1/n)*m**(2/3)*(i)**0.5;\n",
+      "\n",
+      "print \"Q using Chezy formula with C determined from the Kutter formula (m^3/s) :\",round(Q1,2)\n",
+      "print \"Q using the Manning formula                                     (m^3/s) :\",round(Q2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q using Chezy formula with C determined from the Kutter formula (m^3/s) : 5.65\n",
+        "Q using the Manning formula                                     (m^3/s) : 5.69\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.2, Page 518"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Q  =  0.5;\n",
+      "C  =  80;\n",
+      "i  =  1/2000;\n",
+      "\n",
+      " #Calculations\n",
+      "\n",
+      "# A = D**2+(3/4)*D**2 = (7/4)*D**2\n",
+      "D = ((4/7)*(Q/C)*(2/i)**0.5)**(2/5)\n",
+      "\n",
+      "#Result\n",
+      "print \"Optimum depth = Optimum Width (in metres):\",round(D,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Optimum depth = Optimum Width (in metres): 0.552\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_16.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_16.ipynb
new file mode 100755
index 00000000..73f04778
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_16.ipynb
@@ -0,0 +1,144 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:130e6e124bfc557e016f12e2356b74f1d97ea210c47b1944be1ebda3297e63a8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16: Non-Uniform Flow in Open Channels"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2, Page 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "B  =  [1.4 ,0.9];\n",
+      "D  =  [0.6 ,0.32];\n",
+      "g  =  9.81;\n",
+      "h  =  0.03;\n",
+      "Z  =  0.25;  \n",
+      "\n",
+      " #Calculations\n",
+      "Q1  =  B[1]*D[1]*(2*g*h/(1-(B[1]*D[1]/B[0]*D[0])**2))**0.5\n",
+      "E  =  D[0]-Z;\n",
+      "Q2  =  1.705*B[1]*E**1.5;\n",
+      "\n",
+      "print \"Volume flow rate (m3/s) :\",round(Q2,4) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume flow rate (m3/s) : 0.3177\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3, Page 546"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "a  =0.5;\n",
+      "b  =  0.5;\n",
+      "Dn  =  1.2;\n",
+      "s  =  1/1000;\n",
+      "C  =  55;\n",
+      "g  =  9.81;\n",
+      "\n",
+      " #Calculations\n",
+      "c = (1+a)/b;  \n",
+      "QbyB = Dn*C*(Dn*s)**0.5;\n",
+      "q = QbyB;\n",
+      "Dc = (q**2/g)**(1/3);\n",
+      "\n",
+      "header  =  \"Mean Depth(Dm)    Numenator    Denominator\\t L\"\n",
+      "unit  =    \"   (m) \\t  \\t \\t  \\t(m)\"\n",
+      "\n",
+      "m=[]\n",
+      "Dm=[]\n",
+      "N=[]\n",
+      "D=[]\n",
+      "Lm=[]\n",
+      "total=0\n",
+      "for c in range(7):    \n",
+      "    m.append(2.4-0.15*c);\n",
+      "    Dm.append((m[c]+m[c]-0.15)/2); \n",
+      "    N.append(1 - (Dc/Dm[c])**3) ;    # Numerator\n",
+      "    D.append(1 - (Dn/Dm[c])**3);     # Denominator\n",
+      "    Lm.append(150*(N[c]/D[c]));\n",
+      "    total = total +Lm[c];\n",
+      "\n",
+      "print header\n",
+      "print unit\n",
+      "for c in range(7):\n",
+      "    mm=str(Dm[c])+'\\t       '+str(round(N[c],4))+'        '+str(round(D[c],4))+' \\t'+str(round(Lm[c],2))\n",
+      "    print mm\n",
+      "    \n",
+      "print \"\\ndistance upstream covered (approx in m):\",round(total)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean Depth(Dm)    Numenator    Denominator\t L\n",
+        "   (m) \t  \t \t  \t(m)\n",
+        "2.325\t       0.9576        0.8625 \t166.54\n",
+        "2.175\t       0.9482        0.8321 \t170.94\n",
+        "2.025\t       0.9358        0.7919 \t177.26\n",
+        "1.875\t       0.9192        0.7379 \t186.86\n",
+        "1.725\t       0.8962        0.6634 \t202.65\n",
+        "1.575\t       0.8636        0.5577 \t232.27\n",
+        "1.425\t       0.8159        0.4028 \t303.8\n",
+        "\n",
+        "distance upstream covered (approx in m): 1440.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_17.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_17.ipynb
new file mode 100755
index 00000000..2791e2c2
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_17.ipynb
@@ -0,0 +1,274 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c21785277c0a99c3eb4c310a9e20c477f6ab2a0a22da1b2d3844df1699a78d21"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17: Compressible Flow in Pipes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.1, Page 566"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import solve,symbols\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "g  =  9.81;\n",
+      "rho  =  1000;\n",
+      "rhoHg  =  13.6*rho;\n",
+      "d1  =  0.075;\n",
+      "d2  =  0.025;\n",
+      "Pi  =  0.250;\n",
+      "Pt  =  0.150;\n",
+      "P_Hg  =  0.760;\n",
+      "rho1  =  1.6;\n",
+      "gma  =  1.4;\n",
+      "\n",
+      " #Calculations\n",
+      "P1  =  (Pi+P_Hg)*rhoHg*g;\n",
+      "P2  =  (Pt+P_Hg)*rhoHg*g;\n",
+      "rho2  =  rho1*(P2/P1)**(1/gma);\n",
+      "V0=symbols('V0')\n",
+      "V1=symbols('V1')\n",
+      "Velo  =  solve([d2**2*V1*rho2-d1**2*V0*rho1,0.5*(V1**2  -  V0**2)*((gma-1)/gma)*(rho2*rho1/(rho2*P1-rho1*P2))-1],[V0,V1])\n",
+      "s=(Velo[1])[0]\n",
+      "Flow  =  math.pi*d1**2/4*s;\n",
+      "\n",
+      "print \"Volume of flow (m3/s):\",round(Flow,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume of flow (m3/s): 0.06\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.2, Page 571"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Ma  =  4;\n",
+      "gma  =  1.4;\n",
+      "At  =  500;                                                              #  in  mm\n",
+      "\n",
+      " #Calculations\n",
+      "N  =  1  +  (gma-1)*Ma**2/2;\n",
+      "D  =  (gma+1)/2  ;\n",
+      "#ratio of A/At ==R\n",
+      "R  = round( (N/D)**((gma+1)/(2*(gma-1)))/Ma,2);\n",
+      "A=At*R\n",
+      "print \"Area of test section (mm^2):\",A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area of test section (mm^2): 5360.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.3, Page 575"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Ma1  =  2;\n",
+      "gma  =  1.4;\n",
+      "T1  =  15+273;                                                        #  In  kelvin\n",
+      "P1  =  105;  \n",
+      "\n",
+      " #Calculations\n",
+      "Ma2  =  (((gma-1)*Ma1**2  +2)/(2*gma*Ma1**2-gma+1))**0.5;\n",
+      "P2  =  P1*(1+gma*Ma1**2)/(1+gma*Ma2**2);\n",
+      "T2  =  T1*(1  +(gma-1)/2*Ma1**2)/(1  +(gma-1)/2*Ma2**2);\n",
+      "\n",
+      "\n",
+      "print \"Mach No downstream shock wave                   :\",round(Ma2,3)\n",
+      "print \"Pressure (bar) of downstream shock wave         :\",round(P2)\n",
+      "print \"Temperature (Degree C) of downstream shock wave :\",T2 - 273"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mach No downstream shock wave                   : 0.577\n",
+        "Pressure (bar) of downstream shock wave         : 473.0\n",
+        "Temperature (Degree C) of downstream shock wave : 213.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.4, Page 581"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "gma  =  1.4;\n",
+      "f  =  0.00375;\n",
+      "d  =  0.05;\n",
+      "\n",
+      " #Calculations\n",
+      "m  =  d/4;\n",
+      "def  x(Ma):\n",
+      "    A  =(1  -Ma**2  )/(gma*Ma**2);\n",
+      "    B  =  (gma+1)*Ma**2/(2+(gma-1)*Ma**2);  \n",
+      "    y  =  m/f*(A+  (gma+1)*math.log(B)/(2*gma));\n",
+      "    return y\n",
+      "\n",
+      "X1  =  x(0.2);                                  #  At  entrance  Ma  =  0.2;\n",
+      "X06_X1  =x(0.6);                                #  Section(b)  Ma  =  0.6;\n",
+      "\n",
+      "X06  =    X1-X06_X1;\n",
+      "\n",
+      "print \"The Distance X1 at which the Mach number is unity (m) :\",round(X1,2)\n",
+      "print \"Distance from the entrance (m)                        :\",round(X06,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Distance X1 at which the Mach number is unity (m) : 48.44\n",
+        "Distance from the entrance (m)                        : 46.81\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.5, Page 585"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy.optimize import fsolve\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "gma  =  1.4;\n",
+      "Q  =  28/60;                             #  m3/s\n",
+      "d  =  0.1;\n",
+      "p1  =  200*10**3;\n",
+      "f  =  0.004;\n",
+      "x_x1  =  60;\n",
+      "R  =  287;\n",
+      "T  =  15+273;\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  math.pi*d**2/4;\n",
+      "m  =  d/4;\n",
+      "v1  =  Q/A;\n",
+      "pa  =  p1*(1-f*(x_x1)*v1**2/(m*R*T))**0.5;\n",
+      "\n",
+      "def g(p):\n",
+      "    A  =  (v1*p1)**2/(R*T)\n",
+      "    B  =  f*A*x_x1/(2*m);\n",
+      "    y  =  (p**2  -  p1**2)/2  -A*math.log(p/p1)  +B;\n",
+      "    return y\n",
+      "    \n",
+      "pb=fsolve(g,pa)                        #  Guessing  solution  around  pa\n",
+      "pb=pb[0]\n",
+      "print \"Pressure at the outlet, neglecting velocity changes   (kN/m2) :\",round(pa/1000,1)\n",
+      "print \"Pressure at the outlet, allowing for velocity changes (kN/m2) :\",round(pb/1000,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at the outlet, neglecting velocity changes   (kN/m2) : 153.6\n",
+        "Pressure at the outlet, allowing for velocity changes (kN/m2) : 150.4\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_2.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_2.ipynb
new file mode 100755
index 00000000..835909e1
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_2.ipynb
@@ -0,0 +1,493 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:56204579c9dff23fd133fce17f3b8e6b8ac01dae9f7f133693b953a21d5ed34c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Pressure and Head"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "z1  =  0;                                                         #Taking  Ground  as  reference\n",
+      "z2  =  -30                                                        #Depth\n",
+      "rho  =  1025                                                      #Density\n",
+      "g  =  9.81                                                        #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculation\n",
+      "pressureIncrease  =  -rho*g*(z2-z1);\n",
+      "\n",
+      "print \"Increase in Pressure (KN/m2):\",round(pressureIncrease/1000,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Increase in Pressure (KN/m2): 301.7\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "p1  =  22.4*10**3                                                    #Initial  Pressure\n",
+      "z1  =  11000                                                            #Initial  Height\n",
+      "z2  =  15000                                                            #final  Height\n",
+      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
+      "R  =  287;                                                                #Gas  Constant\n",
+      "T  =  273-56.6                                                        #Temperature  \n",
+      "\n",
+      " #Calculations\n",
+      "p2  =  p1*math.exp(-g*(z2-z1)/(R*T));\n",
+      "rho2=p2/(R*T);\n",
+      "  \n",
+      "print \"Final Pressure (kN/m2):\",round(p2/1000,2)\n",
+      "print \"Final Density  (kg/m3):\",round(rho2,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final Pressure (kN/m2): 11.91\n",
+        "Final Density  (kg/m3): 0.192\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3, Page 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "p1  =  101*10**3                                                      #Initial  Pressure\n",
+      "z1  =  0                                                              #Initial  Height\n",
+      "z2  =  1200                                                           #Final  Height\n",
+      "T1  =  15+273                                                         #Initial  Temperature\n",
+      "g  =  9.81                                                            #Acceleration  due  to  gravity\n",
+      "gamma  =  1.4                                                         #Heat  capacity  ratio\n",
+      "R  =  287                                                                  #Gas  Constant\n",
+      "\n",
+      " #Calculations\n",
+      "p2  =  p1*(1-g*(z2-z1)*(gamma-1)/(gamma*R*T1))**(gamma/(gamma-1));\n",
+      "dT_dZ  =  -(g/R)*((gamma-1)/gamma);\n",
+      "T2  =  T1  +  dT_dZ*(z2-z1);\n",
+      "rho2 = p2/(R*T2);\n",
+      "\n",
+      "print \"Final Pressure (kN/m2)            :\",round(p2/1000,2)\n",
+      "print \"Temprature     (in degree celcius):\",round(T2-273,1)\n",
+      "print \"Density        (kg/m^3)           :\",round(rho2,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final Pressure (kN/m2)            : 87.33\n",
+        "Temprature     (in degree celcius): 3.3\n",
+        "Density        (kg/m^3)           : 1.101\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "p1  =  101*10**3                                                      #Initial  Pressure\n",
+      "z1  =  0                                                                    #Initial  Height\n",
+      "z2  =  7000                                                              #Final  Height\n",
+      "T1  =  15+273                                                          #Initial  Temperature\n",
+      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
+      "R  =  287                                                                  #Gas  Constant\n",
+      "dT  =  6.5/1000                                                      #Rate  of  Variation  of  Temperature\n",
+      "\n",
+      " #Calculations\n",
+      "p2  =  p1*(1-dT*(z2-z1)/T1)**(g/(R*dT));\n",
+      "T2  =  T1  -  dT*(z2-z1);\n",
+      "rho2  =  p2/(R*T2);\n",
+      "\n",
+      "\n",
+      "print \"Final Pressure (kN/m^2) :\",round(p2/1000,2)\n",
+      "print \"Final Density  (kg/m^3 ):\",round(rho2,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final Pressure (kN/m^2) : 40.89\n",
+        "Final Density  (kg/m^3 ): 0.588\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5, Page 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "p  =  350*10**3;                                         #Gauge  Pressure\n",
+      "pAtm  =  101.3*10**3;                                    #Atmospheric  Pressure    \n",
+      "rhoW  =  1000;                                           #Density  of  Water\n",
+      "sigma  =  13.6;                                          #Relative  Density  of  Mercury\n",
+      "g  =  9.81                                               #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "def Head(rho):\n",
+      "    head  =  p/(rho*g);\n",
+      "    return head\n",
+      "rhoM  =  sigma*rhoW;\n",
+      "pAbs  =  p  +  pAtm;\n",
+      "\n",
+      "print \"\\nPart(a)- Equivalent head of water (m) :\",round(Head(rhoW),2)\n",
+      "print \"\\nPart(b)- Equivalent head of water (m) :\",round(Head(rhoM),2)\n",
+      "print \"\\nAbsolute pressure (kN/m^2)            :\",pAbs/1000"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Part(a)- Equivalent head of water (m) : 35.68\n",
+        "\n",
+        "Part(b)- Equivalent head of water (m) : 2.62\n",
+        "\n",
+        "Absolute pressure (kN/m^2)            : 451.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6, Page 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "   \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "rho  =  10**3;                                  #Density  of  water\n",
+      "h  =  2;                                        #Height\n",
+      "g  =  9.81;                                     #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "p=rho*h*g;  \n",
+      "\n",
+      "print \"Gauge pressure (k/m2) :\",p/1000"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gauge pressure (k/m2) : 19.62\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7, Page 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "rho  =  0.8*10**3;                                                  #Density  of  fluid\n",
+      "rhoM  =  13.6*10**3;                                              #Density  of  manometer  liquid\n",
+      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "def fluidPressure(h1,h2):\n",
+      "    P  =  rhoM*g*h2-rho*g*h1;\n",
+      "    return P\n",
+      "\n",
+      "p1=fluidPressure(0.5,0.9)/1000\n",
+      "p2=fluidPressure(0.1,-0.2)/1000\n",
+      "\n",
+      "print \"!-----Part (a)-----! \\nGauge pressure (kN/m2) :\",round(p1,2)\n",
+      "print \"\\n!-----Part (b)-----! \\nGauge pressure (kN/m2) :\",round(p2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "!-----Part (a)-----! \n",
+        "Gauge pressure (kN/m2) : 116.15\n",
+        "\n",
+        "!-----Part (b)-----! \n",
+        "Gauge pressure (kN/m2) : -27.47\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8, Page 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "rho  =  10**3;                                                          #Density  of  fluid\n",
+      "rhoM  =  13.6*10**3;                                              #Density  of  manometer  liquid\n",
+      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
+      "H  =  0.3;                                                                #  Differnce  in  height  =  b-a  as  in  text\n",
+      "h  =  0.7;\n",
+      "\n",
+      " #Calculations\n",
+      "result  =  rho*g*H  +  h*g*(rhoM-rho);\n",
+      "\n",
+      "print \"Pressure difference (kN/m^2):\", round(result/1000,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure difference (kN/m^2): 89.467\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9, Page 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "rho  =  10**3;                                                          #Density  of  fluid\n",
+      "rhoM  =  0.8*10**3;                                                #Density  of  manometer  liquid\n",
+      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
+      "a  =  0.25;\n",
+      "b  =  0.15;\n",
+      "h  =  0.3;\n",
+      " #Calculations\n",
+      "def PressureDiff(a,b,h,rho,rhoM):\n",
+      "    P  =  rho*g*(b-a)  +  h*g*(rho-rhoM);\n",
+      "    return P\n",
+      "print \"The presure difference,if the top of the manometer is filled with\"\n",
+      "print \"(a) air                          :\",PressureDiff(a,b,h,rho,0)/1000, \" N/m^2\"\n",
+      "print \"(b) oil of relative density 0.8. :\",PressureDiff(a,b,h,rho,rhoM), \"N/m^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The presure difference,if the top of the manometer is filled with\n",
+        "(a) air                          : 1.962  N/m^2\n",
+        "(b) oil of relative density 0.8. : -392.4 N/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10, Page 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "phi  =  30;                                                              #30  degree\n",
+      "h  =  1.2  ;                                                              #  Height  of  tank\n",
+      "l  =  2;                                                                    #  Length  of  tank\n",
+      "\n",
+      " #Calculations\n",
+      "def SurfaceAngle(a,phi):\n",
+      "    g=9.81; # m/s**2    \n",
+      "    Theta  =  math.atan(-a*math.cos(math.radians(phi))/(g+a*math.sin(math.radians(phi))));  \n",
+      "    return Theta\n",
+      "\n",
+      " #case  (a)  a  =  4\n",
+      "\n",
+      "print \"ThetaA      (degree)      :\",round(180 + 180/math.pi*SurfaceAngle(4,phi),2)\n",
+      "\n",
+      " #Case  (b)  a  =  -  4.5\n",
+      "tanThetaR  =  math.tan((SurfaceAngle(-4.5,phi)));\n",
+      "\n",
+      "print \"\\nThetaR      (degree)      :\",round(SurfaceAngle(-4.5,phi)*180/math.pi,2)\n",
+      "\n",
+      "Depth  =  h  -  l*tanThetaR/2;\n",
+      "print \"\\nMaximum Depth of tank (m) :\",round(Depth,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ThetaA      (degree)      : 163.65\n",
+        "\n",
+        "ThetaR      (degree)      : 27.27\n",
+        "\n",
+        "Maximum Depth of tank (m) : 0.6845\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_20.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_20.ipynb
new file mode 100755
index 00000000..f24f7af6
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_20.ipynb
@@ -0,0 +1,115 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:042171008edad2dcea90c81036564026ff0ecc789ee16d3a2ccb0122fde141e9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 20: Pressure Transient Theory and Surge Control"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.3, Page 692"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "c  =  1250;\n",
+      "Dt  =  0.02;\n",
+      "Dv  =  0.5;\n",
+      "rho  =  1000;\n",
+      "v  =0.5;\n",
+      "\n",
+      " #Calculations\n",
+      "cDt  =  c*Dt;\n",
+      "Dp  =  rho*c*Dv;\n",
+      "DOv_DOt  =  Dv/Dt;\n",
+      "vDOv_DOt  =  v*Dv/cDt;\n",
+      "DOp_DOt  =  Dp/Dt;\n",
+      "vDOp_DOx  =  v*Dp/cDt;\n",
+      "Error  =  [vDOv_DOt*100/DOv_DOt  ,vDOp_DOx*100/DOp_DOt];\n",
+      "print \"The percentage errors are :\",Error,\"\\nThese are very small hence can be neglected\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage errors are : [0.04, 0.04] \n",
+        "These are very small hence can be neglected\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.5, Page 705"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "f  =  0;\n",
+      "Atunnel  =  1.227;\n",
+      "Ashaft  =  12.57;\n",
+      "Q  =2;\n",
+      "L  =  200;\n",
+      "g  =  9.81;  \n",
+      "\n",
+      " #Calculations\n",
+      "Zmax  =  (Q/Ashaft)*(Ashaft*L/(Atunnel*g))**0.5;\n",
+      "T  =  2*math.pi*(Ashaft*L/(Atunnel*g))**0.5;\n",
+      "\n",
+      "print \"Peak water level        (m) :\",round(Zmax,3)\n",
+      "print \"Mass Oscillation Period (s) :\",round(T,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak water level        (m) : 2.299\n",
+        "Mass Oscillation Period (s) : 90.8\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_22.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_22.ipynb
new file mode 100755
index 00000000..2c113fed
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_22.ipynb
@@ -0,0 +1,191 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:783af12fd7d104e85c7e13e105c7550ca4a0a63a41968e5a9b0312d6fedb23a2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 22: Theory of Rotodynamic Machine"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.1, Page 779"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Q  =  5;\n",
+      "R1  =  1.5/2;\n",
+      "R2  =  2/2;  \n",
+      "w  =  18;\n",
+      "rho  =  1000;\n",
+      "rhoA  =  1.2;\n",
+      "Hth  =  0.017;\n",
+      "g=9.81;\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  math.pi*(R2**2-R1**2);\n",
+      "Vf  =  Q/A;\n",
+      "Ut  =  w*R2;\n",
+      "Uh  =  w*R1;\n",
+      "B1t  =  math.atan(Vf/Ut);\n",
+      "B1h  =  math.atan(Vf/Uh);\n",
+      "E  =  Hth*rho/rhoA;\n",
+      "def  Beta(u):\n",
+      "    y  =  math.atan(Vf/(u-E*g/u));\n",
+      "    return y\n",
+      "B2t  =  Beta(Ut);\n",
+      "B2h  =  Beta(Uh);\n",
+      "\n",
+      "print \"!----Blade Inlet Angles----!\"\n",
+      "print \"At tip (in degrees) :\",round(B1t*180/math.pi,1),\"\\nAt Hub (in degrees) :\",round(B1h*180/math.pi,1)\n",
+      "print \"\\n!----Blade Outlet Angles (Degrees)----!\"\n",
+      "print \"At tip (in degrees):\",round(B2t*180/math.pi,1),\"\\nAt Hub(in degrees) :\",round(B2h*180/math.pi,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "!----Blade Inlet Angles----!\n",
+        "At tip (in degrees) : 11.4 \n",
+        "At Hub (in degrees) : 15.1\n",
+        "\n",
+        "!----Blade Outlet Angles (Degrees)----!\n",
+        "At tip (in degrees): 19.5 \n",
+        "At Hub(in degrees) : 48.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.2, Page 793"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "D  =  0.1;\n",
+      "t  =  15*10**-3;\n",
+      "Q  =  8.5/3600;\n",
+      "N  =  750/60;\n",
+      "B2  =  25*math.pi/180;                                                                #  Beta  2  ind  degrees\n",
+      "g  =  9.81;\n",
+      "z  =  16;\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  round(math.pi*D*t,5);\n",
+      "V_f2  =  round(Q/A,3);\n",
+      "U2  =  round(math.pi*N*D,3);\n",
+      "V_w2  =  round(U2  -  V_f2/math.tan(B2),1);\n",
+      "Hth  =  round(U2*V_w2/g,2);\n",
+      "Sf  =  round(1 - math.pi*math.sin(B2)/(z*(1-(V_f2/U2)/math.tan(B2))),3);\n",
+      "H  =  round(Sf*Hth,2);\n",
+      "\n",
+      "print \"Part (a) - Head developed (m): \",Hth\n",
+      "print \"Part (b) - Head developed (m): \",H"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a) - Head developed (m):  1.16\n",
+        "Part (b) - Head developed (m):  1.03\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.3, Page 797"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Ma  =  0.6;\n",
+      "Cl  =  0.6;\n",
+      "tByC  =  0.035;                                                      #  Thickness  to  chord  ratio\n",
+      "cByC  =  0.015;                                                      #  Camber  to  chord  ratio\n",
+      "x  =  3;                                                                    #  Angle  of  incidence\n",
+      "\n",
+      " #Calculations\n",
+      "lamda  =  1/(1-Ma**2)**0.5;\n",
+      "Cl#  =  lamda*Cl;\n",
+      "tByC1  =  tByC*lamda;\n",
+      "cByC1  =  cByC*lamda;\n",
+      "Cl1  =  Cl*lamda**2;\n",
+      "Ae  =  x*lamda;\n",
+      "print \"____Geometric Characterstics____\\n\"\n",
+      "print \"Thickness to chord ratio      :\",tByC1\n",
+      "print \"Camber to chord ratio         :\",cByC1\n",
+      "print \"Lift Coefficient              :\",Cl1\n",
+      "print \"angle of incidence (Degrees)  :\",Ae"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "____Geometric Characterstics____\n",
+        "\n",
+        "Thickness to chord ratio      : 0.04375\n",
+        "Camber to chord ratio         : 0.01875\n",
+        "Lift Coefficient              : 0.9375\n",
+        "angle of incidence (Degrees)  : 3.75\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_23.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_23.ipynb
new file mode 100755
index 00000000..8cb510fa
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_23.ipynb
@@ -0,0 +1,302 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ef9e21b794b8d045cd677d625f05028772e4370ff57190a11a0021b5fe3769e2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 23: Performance of Rotodynamic Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.1, Page 814"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "%matplotlib inline\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Q  =  []\n",
+      "for c in range(9):\n",
+      "    Q.append(7*c)\n",
+      "H  =  [40, 40.6, 40.4, 39.3, 38, 33.6, 25.6, 14.5, 0];\n",
+      "n  =  [0, 41, 60, 74, 83, 83, 74, 51, 0];\n",
+      "N1  =  750;\n",
+      "N2  =  1450;\n",
+      "D1  =  0.5;\n",
+      "D2  =  0.35;\n",
+      "\n",
+      " #Calculations\n",
+      "Q2=[]\n",
+      "H2=[]\n",
+      "for c in range(9):\n",
+      "    Q2.append(Q[c]*(N2/N1)*(D2/D1)**3);\n",
+      "    H2.append(H[c]*(N2/N1)**2*(D2/D1)**2);\n",
+      "   \n",
+      "plot(Q,H,label='H1')\n",
+      "plot(Q,n,label='n1')\n",
+      "plot(Q2,H2,'--',label='H2')\n",
+      "plot(Q2,n,'--',label='n2')\n",
+      "legend( loc='upper right', numpoints = 1 )\n",
+      "xlabel(\"Q  (m3/s)\");\n",
+      "ylabel(\"H (m of water) and n(percent)\")\n",
+      "grid(True)\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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PtMI0uuXIHQ6Hg23btsHZ2RnGxsaYPn06amtrMW7cOEyZMgX6+vrQ09PD4sWL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\nIK1PDnOiPzgAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x322b110>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.2, Page 831"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "n  =  0.9;\n",
+      "g  =  9.81;\n",
+      "D  =  1.45;\n",
+      "N  =  375/60;\n",
+      "H  =  200;                                                                #  Real  height\n",
+      "x  =  165;                                                                #  Theta\n",
+      "P  =  3750*10**3;\n",
+      "rho  =  1000;\n",
+      "\n",
+      " #Calculations\n",
+      "h  =  n*H;                                                                #Effective  Head\n",
+      "v1  =  (2*g*h)**0.5;\n",
+      "u  =  math.pi*D*N;\n",
+      "\n",
+      "n_a  =  (2*u/v1**2)*(v1-u)*(1-n*math.cos(math.radians(x)));\n",
+      "\n",
+      "P_b  =  P/n_a;\n",
+      "ppj  =  P_b/2;                                                        #  Power  per  jet\n",
+      "d  =  (8*ppj/(rho*math.pi*v1**3))**0.5  ;\n",
+      "print \"the efficiency of runner :\",round(n_a,3)\n",
+      "print \"Diameter of Jet (m)      :\",round(d,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the efficiency of runner : 0.933\n",
+        "Diameter of Jet (m)      : 0.156\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.3, Page 834"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "                                     #Example  23.3  \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "g  =  9.81;\n",
+      "H  =  12;\n",
+      "n  =  0.8;\n",
+      "w  =  300*2*math.pi/60;\n",
+      "Q  =  0.28;\n",
+      "\n",
+      " #Calculations\n",
+      "V_f1  =  0.15*(2*g*H)**0.5;\n",
+      "V_f2  =V_f1;\n",
+      "V_w1  =  (n*g*H)**0.5;\n",
+      "u1  =  V_w1;\n",
+      "theta  =  math.atan(V_f1/u1);\n",
+      "u2  =0.5*u1;\n",
+      "B2  =  math.atan(V_f2/u2);\n",
+      "r1  =  u1/w;\n",
+      "b1  =  Q/(V_f2*0.9*2*math.pi*r1);                          #  vanes  occupy  10  per  cent  of  the  circumference  hence  0.9\n",
+      "b2  =  2*b1;\n",
+      "\n",
+      "print \"Guide vane angle    (degree)  :\",round(theta*180/math.pi,2)\n",
+      "print \"Vane angle at exit  (degree)  :\",round(B2*180/math.pi,2)\n",
+      "print \"Width of runner at inlet (mm) :\",round(b1*1000,1) \n",
+      "print \"Width of runner at exit  (mm) :\",round(b2*1000,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Guide vane angle    (degree)  : 13.34\n",
+        "Vane angle at exit  (degree)  : 25.38\n",
+        "Width of runner at inlet (mm) : 69.6\n",
+        "Width of runner at exit  (mm) : 139.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.4, Page 838"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import solve,symbols\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "H  =  35;\n",
+      "g  =  9.81;\n",
+      "D  =  2;\n",
+      "N  =  145/60;\n",
+      "z  =  30*math.pi/180;                          #  angle  between  vanes  and  direction  of  runner  rotation\n",
+      "y  =  28*math.pi/180;                          #  angle  between  runner  blades  at  the  outlet.\n",
+      "\n",
+      " #Calculations\n",
+      "H_net  =  0.93*H  ;                            #  since  7%  head  is  lost\n",
+      "v1  =  (2*g*H_net)**0.5;\n",
+      "u  =  math.pi*N*D;\n",
+      "\n",
+      " # from inlet velocity triangle\n",
+      "V_r1=14.3\n",
+      "## ash = cos(beta1+z)\n",
+      "ash = (V_r1**2+v1**2-u**2)/(2*V_r1*v1)\n",
+      "beta1=(z+math.acos(ash))*180/math.pi  # in degrees\n",
+      "\n",
+      "V_r2=(1-8/100)*V_r1         #8 % loss due to friction\n",
+      "V_w1= u + V_r1*math.cos(math.radians(beta1))\n",
+      "V_w2= u - V_r2*math.cos(y)\n",
+      "\n",
+      "E  =  (u/g)*(V_w1  -  V_w2);\n",
+      "n  =  E/H;\n",
+      "print \"Blade angle at inlet :\",round(beta1,1)\n",
+      "print \"Efficiency    (%)    :\",round(n*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Blade angle at inlet : 62.1\n",
+        "Efficiency    (%)    : 81.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.5, Page 844"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "s  =  0.03;\n",
+      "P  =  185*10**3;\n",
+      "rho  =  0.86*10**3;\n",
+      "A  =  2.8*10**-2;\n",
+      "N  =  2250/60;\n",
+      "D  =  0.46;\n",
+      "\n",
+      " #Calculations\n",
+      "R0  =  0.46/2;\n",
+      "Ws_Wp  =  1-s;\n",
+      "n  =  Ws_Wp;\n",
+      "Pf  =  s*P;\n",
+      "Q  =  (2*Pf*A**2/(3.5*rho))**(1/3);\n",
+      "Wp  =  2*math.pi*N;  \n",
+      "Ri  =  ((1/Ws_Wp)*(R0**2  -P/(rho*Q*Wp**2)))**0.5;#  Modified  equation  for  power  transmission.\n",
+      "Di  =  2*Ri;\n",
+      "T  =  P/(rho*Wp**3  *D**5);\n",
+      "   \n",
+      "print \"Mean diameter (mm) :\",round(Di*1000)\n",
+      "print \"Torque Coefficient :\",round(T,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean diameter (mm) : 326.0\n",
+        "Torque Coefficient : 0.0008\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_24.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_24.ipynb
new file mode 100755
index 00000000..ef4fb925
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_24.ipynb
@@ -0,0 +1,183 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0997b77d100b3345365d77881e9dcf23833d271e3f7741c8fb4ad6c260c5ffc6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 24: Positive Displacement Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.1, Page 860"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "H_at  =  10.3;\n",
+      "Hs  =  1.5;\n",
+      "Hd  =  4.5;\n",
+      "Ls  =  2;\n",
+      "Ld  =  15;\n",
+      "g  =  9.81;\n",
+      "Ds  =  0.4;                                                              #  Diameter  of  stroke\n",
+      "Db  =  0.15;                                                            #  Diameter  of  bore\n",
+      "Dd  =  0.05;                                                            #  Diameter  of  discharge  and  suction  pipe\n",
+      "nu  =  0.2;\n",
+      "f  =  0.01;\n",
+      "abs_pump_pressure  =  2.4;\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  math.pi*(Db)**2/4;\n",
+      "a  =  math.pi*(Dd)**2/4;\n",
+      "r  =  Ds/2;\n",
+      "W  =  2*math.pi*nu;\n",
+      "Hsf  =  0;  \n",
+      "def  H_suck(n):\n",
+      "    y  =  H_at  -  Hs  +(-1)**n*(L/g)*(A/a)*W**2*r;  \n",
+      "    return y\n",
+      "\n",
+      "def  H(n,DischargeOrSuction):\n",
+      "    if(DischargeOrSuction  ==  1):\n",
+      "        y  =  H_at  -  Hs  +(-1)**n*(Ls/g)*(A/a)*W**2*r;\n",
+      "    elif(DischargeOrSuction  ==  2):\n",
+      "        y  =  H_at  +  Hd  +(-1)**n*(Ld/g)*(A/a)*W**2*r;\n",
+      "    else:\n",
+      "        print \"There  is  something  wrong  :\"\n",
+      "    return y\n",
+      "\n",
+      "def  H_mid(DischargeOrSuction,uA):\n",
+      "    if(DischargeOrSuction  ==  1):\n",
+      "        Hsf  =  4*f*Ls/(2*Dd*g)*(uA/a)**2;\n",
+      "        y  =  H_at  -  Hs  -  Hsf;\n",
+      "    elif(DischargeOrSuction  ==  2):\n",
+      "        Hsf  =  4*f*Ld/(2*Dd*g)*(uA/a)**2;\n",
+      "        y  =  H_at  +  Hd  +  Hsf;\n",
+      "    else:\n",
+      "        print \"There  is  something  wrong  :\"\n",
+      "    return y\n",
+      "\n",
+      "Hs_start  =  H(1,1);                #  Inertia  head  negative  hence  n  =  1\n",
+      "Hs_end    =  H(2,1);                #  Inertia  head  positive  hence  n  =  2\n",
+      "Hd_start  =  H(1,2);\n",
+      "Hd_end    =  H(2,2);\n",
+      "u  =  W*r;\n",
+      "Hs_mid  =  H_mid(1,u*A);\n",
+      "slip  =  0.04;\n",
+      "Hd_mid  =  H_mid(2,u*A);\n",
+      "suction   = [Hs_start,  Hs_end,  Hs_mid];\n",
+      "discharge = [Hd_start,  Hd_end,  Hd_mid];\n",
+      "suction1=[0,0,0]\n",
+      "discharge1=[0,0,0]\n",
+      "for c in range(3):\n",
+      "    suction1[c]   =round(suction[c],2)\n",
+      "    discharge1[c] =round(discharge[c],2)\n",
+      "W_max  =  (abs((abs_pump_pressure  -  H_at  +  Hs)*(g/Ls)*(a/A)*(1/r)))**0.5;\n",
+      "W_max_rev  =  W_max/(2*math.pi)*60;                      #  maximum  rotation  speed  in  rev/min\n",
+      "\n",
+      "header  =  \"Start   End   Mid\";\n",
+      "\n",
+      "print \"\\n!----Part(a)----!  Head at \\n\",header\n",
+      "print suction1\n",
+      "print \"\\n!----Part(b)----!  Head at \\n\",header\n",
+      "print discharge1\n",
+      "print \"\\n!----Part(c)----1 \\nDrive speed for s eperation (rev/min) :\",round(W_max_rev)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "!----Part(a)----!  Head at \n",
+        "Start   End   Mid\n",
+        "[8.22, 9.38, 8.38]\n",
+        "\n",
+        "!----Part(b)----!  Head at \n",
+        "Start   End   Mid\n",
+        "[10.45, 19.15, 17.93]\n",
+        "\n",
+        "!----Part(c)----1 \n",
+        "Drive speed for s eperation (rev/min) : 40.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.2, Page 863"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Example  24.2  \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "H_friction  =  2.4;\n",
+      "H_at  =  10.3;\n",
+      "Hs  =  1.5;\n",
+      "L  =2;\n",
+      "f  =  0.01;\n",
+      "d  =  0.05;\n",
+      "g  =  9.81;                                 #  Diameter  of  stroke\n",
+      "Db  =  0.15;                                #  Diameter  of  bore\n",
+      "r  =  0.2;\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  math.pi*(Db)**2/4;\n",
+      "a  =  math.pi*(d)**2/4;\n",
+      "W=  (((H_at  -  Hs  -    H_friction  )*(2*d*g/(4*f*L)))**0.5)*(a/A)*(math.pi/r); # in rad/s\n",
+      "W_rev  =  W/(2*math.pi)*60;                 #  maximum  rotation  speed  in  rev/min\n",
+      "# IMPORTANT : In book conversion from rad/s to rev/min is wrong, so answer will be diffrent from book\n",
+      " \n",
+      "print \"speed in          (rad/s)  :\",round(W,2)\n",
+      "print \"Increase in speed (rev/min):\",round(W_rev-40,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "speed in          (rad/s)  : 15.46\n",
+        "Increase in speed (rev/min): 107.65\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_25.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_25.ipynb
new file mode 100755
index 00000000..9b08e6f3
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_25.ipynb
@@ -0,0 +1,328 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2bebadfad4a18ede042a136ada8984e181f57290a23ad4ed8003a1f83445a447"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 25: Machine\u2013Network Interactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.4, Page 893"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Pa_P1  =  -200;                                                      #  From  previous  Question\n",
+      "Q  =  1.4311  ;                                                        #  From  previous  questions.\n",
+      "\n",
+      " #Calculations\n",
+      "DpSys  =  Pa_P1  +  98.9*Q**2;\n",
+      "print \"System Operating point (m^3/s):\",round(DpSys,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "System Operating point (m^3/s): 2.55\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.7, Page 906"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import solve,symbols\n",
+      "   \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Vo  =  25.3;                                                            #Outlet  velocity\n",
+      "D  =  10  ;                                                                #  Mean  hydraulic  diameter\n",
+      "f  =  0.008;                                                            #  friction  factor\n",
+      "X  =  1000;                                                              #  Length  of  road\n",
+      "P  =  12600;                                                            #  Absorbing  power\n",
+      "Va  =  300;                                                              #  Tunnel  air  flow\n",
+      "K1  =  0.96;\n",
+      "K2  =  0.9;\n",
+      "T  =  590;                                                                #Thrust\n",
+      "rho  =  1.2;                                                            #  Air  density  \n",
+      "\n",
+      " #Calculations\n",
+      "alpha  =  (1/D)**2;\n",
+      "A  =  math.pi*D**2/4;                                                    #  Area  of  tunnel\n",
+      "Vt  =  Va/A;\n",
+      "W  =  Vo/Vt;                                                            #Omega\n",
+      "E  =  (1-alpha*W);\n",
+      "C  =  (1-alpha*W)*(1-E)**2  +  E**2  -  1;\n",
+      " #  Manipulating  equation  25.20;\n",
+      "LHS  =  f*X*(E+1)**2/D  +  C  +  1  ;\n",
+      "\n",
+      "n1  =  symbols('n1')\n",
+      "result=solve(K1*(2*((alpha*W**2 + (1-alpha)*E**2-1)+(n1-1)*(alpha*W*(W-1)-C/2)))-LHS)\n",
+      "\n",
+      "n=result[0]\n",
+      "\n",
+      "\n",
+      " #  Alternative  approach  using  equation  25.22\n",
+      "n2  =  (rho*((4*f*X*Vt**2)/(2*D)    +  1.5*Vt**2/2))*A/(K1*K2*T);  \n",
+      "Pt  =  round(n2)*P;\n",
+      "\n",
+      "print \"Number of fans required    :\",round(n2)\n",
+      "print \"Total  power consumed (KW) :\",Pt/1000"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of fans required    : 6.0\n",
+        "Total  power consumed (KW) : 75.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.8, Page 907"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import solve,symbols\n",
+      " #Initializing  the  variables\n",
+      "f  =  0.008;\n",
+      "T  =  290;\n",
+      "L  =  750;\n",
+      "Dt  =  9;                                                                  #  Diameter  Tunnel\n",
+      "Df  =  0.63;                                                            #  Diameter  fan\n",
+      "K1  =  0.98;\n",
+      "K2  =  0.92;\n",
+      "Vo  =  27.9;\n",
+      "n  =  10;\n",
+      "A=math.pi*Dt**2/4\n",
+      "rho=1.2\n",
+      "X=750\n",
+      " #Calculations\n",
+      "alpha  =  (Df/Dt)**2;\n",
+      " #  equation  25.20  becomes  when  E  =  1  nad  C  =  0\n",
+      "W=symbols('W')\n",
+      "omega  =  solve(2*K1*  (alpha*W**2  +(n-1)*alpha*W*(W-1))  -  4*f*L/Dt  -1)\n",
+      "  \n",
+      "\n",
+      "for i in range(1,len(omega)):  #  since  omega  is  always  positive  and  real\n",
+      "    if omega[i]>0:\n",
+      "        w  =  round(omega[i],1);\n",
+      "Vt  =  Vo/w;\n",
+      "\n",
+      "# by equation 25.22\n",
+      "VT=(n*(K1*K2*T)/(A*(rho*((4*f*X)/(2*Dt)    +  1.5/2))))**0.5\n",
+      "print \"Tunnel Velocity  (m/s) :\",round(VT,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Tunnel Velocity  (m/s) : 4.05\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.9, Page 914"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Ws  =  0.45;\n",
+      "Ks  =  3.2;\n",
+      "H  =  152;\n",
+      "h  =  0;\n",
+      "Hatm  =  10.3;\n",
+      "Pv  =  350;                             #vapour  pressure\n",
+      "g  =  9.81;\n",
+      "rho  =  1000;\n",
+      "     \n",
+      " #Calculations\n",
+      "Ht1  =  152*(Ws/Ks)**(4/3);     # the value of Ht1 is 11.12 and in book it is taken as 11.2 so there will be a difference in final answer\n",
+      "Hvap  = round(Pv/(rho*g),3);\n",
+      "Z  =  Hatm  -h  -Hvap  -Ht1;\n",
+      "print \"Elevation of pump (m):\",round(Z,3)\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elevation of pump (m): -0.851\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 25.11, Page 927"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import symbols,solve\n",
+      "import numpy as np\n",
+      "   \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Co  =  0;\n",
+      "Qc  =  0.0024;\n",
+      "V  =  5400;\n",
+      "c  =  10;\n",
+      " #Calculations\n",
+      "#####--------------------PART(A)-----------------#######\n",
+      "n1=symbols('n1')\n",
+      "def partA(n1):\n",
+      "    Ci = 10;\n",
+      "    # t = infinity so e^(-nt) = 0\n",
+      "    Q=10000*Qc/(c-Co)\n",
+      "    n1 = Q*3600/V;    \n",
+      "    return n1\n",
+      "ans=partA(n1)\n",
+      "\n",
+      "print \"Part(A) : number of air changes per hour if the garage is in continuous use and the maximum permissible concentration of carbon monoxide is 0.1 per cent. :\",ans,\"\\n\"\n",
+      "\n",
+      "#####--------------------PART(B)-----------------#######\n",
+      "n=symbols('n')\n",
+      "def   partB(n):\n",
+      "    Ci  =  0;         \n",
+      "    n=[1.5,1.2,0.9,1.0]    \n",
+      "    t=1      #  time  in  hours\n",
+      "    error=[]\n",
+      "    mini=100\n",
+      "    ans=0\n",
+      "    for i in range(4):    \n",
+      "        Q  =  V/3600;           \n",
+      "        A  =   10000*Qc/Q;                                                            # as Co=0 \n",
+      "        error.append(abs((A*(1-math.e**(-n[i]*t))/c)-n[i]));\n",
+      "        if(error[i]<mini):\n",
+      "            mini=error[i]\n",
+      "            ans=n[i]\n",
+      "    return ans       \n",
+      "ans=partB(n)\n",
+      "print \"Part(B) : number of air changes per hour if this maximum level is reached after 1 hour and the garage is out of use :\",ans,'\\n'\n",
+      "  \n",
+      "#####--------------------PART(C)-----------------#######\n",
+      "c=symbols('c')\n",
+      "def   partC(c):\n",
+      "    Ci  =  0;\n",
+      "    n  =  1;   \n",
+      "    t = 0.333333                         #  20  minutes  in  hours\n",
+      "    Q  =  V*n/3600;\n",
+      "    y  =    (Co  +  10000*Qc/Q)*(1-math.e**(-n*t)) + Ci*math.e**(-n*t) ;  \n",
+      "    return y\n",
+      "ans=partC(c)\n",
+      "print \"Part(C) :the concentration after 20 minutes (Parts per 10000) :\",round(ans,3),'\\n'\n",
+      "#####--------------------PART(D)-----------------#######\n",
+      "t=symbols('t')\n",
+      "def   partD(t):\n",
+      "    Ci  =  10;\n",
+      "    n  =  1;  \n",
+      "    c  =  0.1;\n",
+      "    t=np.log(100) \n",
+      "    return round(t,2)\n",
+      "ans=partD(t)    \n",
+      "print \"Part(D) : time necessary to run the ventilation system at the rate calculated in (b) to reduce the concentration to 0.001 per cent (in hours) :\",ans,\"hours\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part(A) : number of air changes per hour if the garage is in continuous use and the maximum permissible concentration of carbon monoxide is 0.1 per cent. : 1.6 \n",
+        "\n",
+        "Part(B) : number of air changes per hour if this maximum level is reached after 1 hour and the garage is out of use : 1.0 \n",
+        "\n",
+        "Part(C) :the concentration after 20 minutes (Parts per 10000) : 4.535 \n",
+        "\n",
+        "Part(D) : time necessary to run the ventilation system at the rate calculated in (b) to reduce the concentration to 0.001 per cent (in hours) : 4.61 hours\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_3.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_3.ipynb
new file mode 100755
index 00000000..68980a40
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_3.ipynb
@@ -0,0 +1,372 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:540962ba0b5999b583f0620c9dca124d46f25fe569649c6c842d96cfa42351a3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Static Forces on Surfaces. Buoyancy"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "a  =  2.7;                                                                #Upper  edge\n",
+      "b  =  1.2  ;                                                              #Lower  edge\n",
+      "width  =  1.5;                                                        #Width  of  trapezoidal  plate\n",
+      "h  =  1.1;                                                                #Height  of  water  column  above  surface\n",
+      "rho  =  1000;\n",
+      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
+      "phi  =  90                                                                #Angle  between  wall  and  surface\n",
+      "\n",
+      " #Calculations\n",
+      "A  =  0.5*(a+b)*width;                                        #Area  of  Trapezoidal  Plate\n",
+      "y  =  (2*(0.5*width*0.75)*0.5  +  (1.2*width)*0.75)/A;\n",
+      "z  =  y+h;                                                                #Depth  of  center  of  pressure\n",
+      "R  =  rho*g*A*z                                                      #Resultant  force\n",
+      "\n",
+      "I0  =  1.2*1.5**3/12  +1.2*1.5*1.85**2  +  1.5*1.5**3/36  +  1.5*0.75*1.6**2  #Second  moment  of  area\n",
+      "D  =  (math.sin(math.degrees(phi)))**2*I0/(A*z);                          #depth  of  center  of  pressure\n",
+      "M  =  R*(1.8533-1.1);                                          #Moment  about  hinge\n",
+      "print \"Moment about the hinge line (kN/m):\",round(M/1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Moment about the hinge line (kN/m): 38.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "w  =  1.8;                                                                #Width  of  plate\n",
+      "h1  =  5;                                                                  #Height  of  plate  and  water  in  upstream\n",
+      "h2  =  1.5;                                                              #Height  of  water  in  downstream\n",
+      "rho  =  1000;\n",
+      "g  =  9.81  ;                                                            #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "def waterForce(area,meanHeight):\n",
+      "    F  =  rho  *  g  *  area  *  meanHeight;\n",
+      "    return F\n",
+      "\n",
+      "P  =  waterForce(w*h1,h1/2)-waterForce(w*h2,h2/2);#  Resultant  force  on  gate  \n",
+      "x  =  (waterForce(w*h1,h1/2)*(h1/3)  -  waterForce(w*h2,h2/2)*(h2/3))/P;#  point  of  action  of  p  from  bottom\n",
+      "R  =  P/(2*math.sin(math.radians(20)));                                          #  Total  Reaction  force\n",
+      "Rt  =  1.18*R/4.8;                                                #Reaction  on  Top\n",
+      "Rb  =  R  -  Rt  ;                                                      #Reaction  at  bottom\n",
+      "\n",
+      "print \"Reaction at top    (kN):\",round(Rt/1000,1)\n",
+      "print \"Reaction at bottom (kN):\",round(Rb/1000,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reaction at top    (kN): 72.2\n",
+        "Reaction at bottom (kN): 221.45\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "D  =  1.8;                                                                #Depth  of  tank\n",
+      "h  =  1.2;                                                                #Depth  of  water\n",
+      "l  =  3;                                                                    #Length  of  wall  of  tank\n",
+      "p  =  35000;                                                            #Air  pressure\n",
+      "rho  =  10**3;                                                          #Density  of  water\n",
+      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
+      "\n",
+      "\n",
+      " #Calculations\n",
+      "Ra  =  p*D*l;                                                          #Force  due  to  air\n",
+      "Rw  =  .5*(rho*g*h)*h*l;                                    #Force  due  to  water\n",
+      "R  =  Ra  +  Rw;                                                        #  Resultant  force\n",
+      "x  =  (Ra*0.9+Rw*0.4)/R;                                    #  Height  of  center  of  pressure  from  base\n",
+      "print \"Resultant force on the wall                     (kN) :\",round(R/1000,2)\n",
+      "print \"Height of the centre of pressure above the base (m)  :\",round(x,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resultant force on the wall                     (kN) : 210.19\n",
+        "Height of the centre of pressure above the base (m)  : 0.85\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page 72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "  \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "R  =  6;                                                                    #  Radius  of  arc\n",
+      "h  =  2*R*math.sin(math.radians(30));                                              #Depth  of  water\n",
+      "rho  =  10**3;                                                          #Density  of  water\n",
+      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "Rh  =  (rho*g*h**2)/2;                                          #  Resultant  horizontal  force  per  unit  length\n",
+      "Rv  =  rho*g*((60/360)*math.pi*R**2  -R*math.sin(math.radians(30))*R*math.cos(math.radians(30)));#  Resultant  vertical  force  per  unit  length\n",
+      "R  =  (Rh**2+Rv**2)**0.5;                                        #  Resultant  force  on  gate\n",
+      "theta  =  180/math.pi*math.atan(Rv/Rh);                                      #Angle  between  resultant  force  and  horizontal\n",
+      "\n",
+      "print \"Magnitute of resultant force                  (kN/m)   :\",round(R/1000,2)\n",
+      "print \"Direction of resultant force to the horizontal(Degrees):\",round(theta,2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnitute of resultant force                  (kN/m)   : 179.45\n",
+        "Direction of resultant force to the horizontal(Degrees): 10.27\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5, Page 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "B  =  6;                                                                    #  Width  of  pontoon\n",
+      "L  =  12;                                                                  #Length  of  pontoon\n",
+      "D  =  1.5;                                                                #Draught  of  pontoon\n",
+      "Dmax  =  2;                                                              #Maximum  permissible  draught\n",
+      "rhoW  =  1000;                                                        #Density  of  fresh  water\n",
+      "rhoS  =  1025;                                                        #Density  of  sea  water\n",
+      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "def Weight(D):\n",
+      "    W  =  rhoW*g*B*L*D;\n",
+      "    return W\n",
+      "\n",
+      "W  =  Weight(D);                                                    #  Weight  of  pontoon  in  fresh  water  =  weight  of  water  displaced\n",
+      "Ds  =  W/(rhoS*g*B*L);                                        #Draught  in  sea  water\n",
+      "L  =  Weight(Dmax)  -  Weight(D);                      #  maximum  load  that  can  be  supported\n",
+      "\n",
+      "print \"Weight of pontoon (kN) :\",round(W/1000,1)\n",
+      "print \"Draught in sea    (m)  :\",round(Ds,2)\n",
+      "print \"Load              (kN) :\",round(L/1000,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Weight of pontoon (kN) : 1059.5\n",
+        "Draught in sea    (m)  : 1.46\n",
+        "Load              (kN) : 353.16\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "D  =  1.8;                                                                #  Diameter  of  buoy\n",
+      "H  =  1.2;                                                                #Height  of  buoy\n",
+      "W  =  10*10**3;                                                        #Weight  of  buoy\n",
+      "L  =  2*10**3;                                                          #Load\n",
+      "G  =  0.45;                                                              #  Center  of  gravity\n",
+      "rho  =  1025;                                                          #Density  of  sea  water\n",
+      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "Z  =  4*(W+L)/(rho*g*math.pi*D**2);                        #  Depth  of  Immersion\n",
+      "BG  =  (math.pi*D**4/64)/(math.pi*D**2*Z/4);\n",
+      "Z  =  0.5*Z  +BG;                                                #  Position  of  combined  center  of  gravity\n",
+      "Z1  =  ((W+L)*Z-0.45*W)/L;                              #Maximum  height  of  load  above  bottom\n",
+      "\n",
+      "print \"Maximum height of center of gravity above bottom (m) :\",round(Z1,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum height of center of gravity above bottom (m) : 1.748\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7, Page 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "l  =  20;                                                                  #  Length  of  barage\n",
+      "b  =  6;                                                                    #Width  of  barage\n",
+      "r  =  3;                                                                    #Radius  of  circular  top  of  barage\n",
+      "W  =  200*10**3;                                                      #Weight  of  empty  barage\n",
+      "d1  =  0.8;                                                              #  Depth  of  water  in  1st  half\n",
+      "d2  =  1;                                                                  #  Depth  of  water  in  2nd  half\n",
+      "rho  =  1000;                                                          #Density  of  water\n",
+      "R  =  0.8;                                                                #Relative  density  of  liquid\n",
+      "g  =  9.81;                                                              #Acceleration  due  to  gravity\n",
+      "ZG  =  0.45;                                                            #  Center  of  gravity  of  barage\n",
+      "\n",
+      " #Calculations\n",
+      "I00  =  l*b**3/12  +math.pi*b**4/128;\n",
+      "ICC  =  l*(.5*b)**3/12;\n",
+      "L  =  d1*rho*g*l*b/2*(d1+d2);                          #  Weight  of  liquid  load\n",
+      "W  =  L  +  W;                                                          #Total  weight\n",
+      "A  =  l*b  +math.pi*r**2/2;                                          #  Area  of  plane  of  waterline\n",
+      "V  =  W/(rho*g);                                                  #  Volume  of  vessel  submerged\n",
+      "D  =  V/A  ;                                                              #Depth  submerged\n",
+      "ZB  =  .5*D;                                                            #Height  of  center  of  buoyancy\n",
+      "NM  =  ZB-ZG  +(1/V)*(I00-R*2*ICC);                #  Effective  metacentric  height\n",
+      "P  =  R*rho*g*l*b/2*(d2-d1);                            #overturning  moment  \n",
+      "theta  =  math.atan(P*1.5/(W*NM))*180/math.pi;                      #Angle  of  roll\n",
+      "# converting into degrees and minutes\n",
+      "thetaD=round(theta-1)\n",
+      "thetaM=(theta-thetaD)*60/100\n",
+      "print \"Angle of roll is\",thetaD,\"degrees\",round(thetaM,2),\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of roll is 2.0 degrees 0.37 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_4.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_4.ipynb
new file mode 100755
index 00000000..9429e481
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_4.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:029b6e47f117cbc4316b9317b5e197658ff26f93dc8874c76dd6b6c464f4fec0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Motion Fluid Particles and Streams"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1, Page 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from pylab import *\n",
+      "import numpy as np\n",
+      "%matplotlib inline\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "x = [0, 23, 28, 31, 32, 29, 22, 14, 0]\n",
+      "y = np.linspace(0,80,9)\n",
+      "xlabel('Velocity  (m/s)')\n",
+      "ylabel('Distance  from  one  side(mm)')\n",
+      "title('Velocity Distribution Curve')\n",
+      "grid(1)\n",
+      "\n",
+      " #Calculations\n",
+      "plot(x,y,'-*')\n",
+      "show()\n",
+      "mu=[17.5 , 26.0,  29.6,  31.9,  30.7,  25.4,  18.1,  7.7]\n",
+      " #  mean  velocity\n",
+      "print \"Mean velocity (m/s):\",round(mean(mu),2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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knMCmTZumjBw5cvfUqVM3AsCDBw+s6IweQhqPNm2AdevYU0oHDWJPNW3bFli7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\nYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x31fdd50>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean velocity (m/s): 23.36\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2, Page 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from pylab import *\n",
+      "import numpy as np\n",
+      "\n",
+      " #Initializing  the  variables  all  unknowns  are  assigned  0\n",
+      "\n",
+      "d  =  [0.0, 0.05, 0.075, 0, 0.030];\n",
+      "Q  =  [0, 0, 0, 0, 0];\n",
+      "V  =  [0, 0, 2, 1.5, 0];\n",
+      "A  =  [0, 0, 0, 0, 0]\n",
+      " #Calculations\n",
+      "A2  =  math.pi*d[2]**2/4;\n",
+      "Q[2]  =A2*V[2];\n",
+      "Q  =  [0,Q[2], Q[2],  Q[2]/1.5 , 0.5*Q[2]/1.5];\n",
+      "d[3]  =  (Q[3]*4/(V[3]*math.pi))**0.5;\n",
+      "for i in range(0,5):\n",
+      "    A[i]  =  math.pi*d[i]**2/4;\n",
+      "V[1]  =  V[2]*(A[2]/A[1]);\n",
+      "V[4]=Q[4]/A[4];\n",
+      "\n",
+      "\n",
+      "header  =  \"Diameter(mm)  Area(m2)\\t Flow Rate(m3/s)  Velocity(m/s)\"\n",
+      "print header\n",
+      "for c in range(1,5):\n",
+      "    mm=str(round(d[c]*1000,1))+'\\t  '+str(round(A[c],6))+' \\t    '+str(round(Q[c],6))+'     \\t'+str(round(V[c],2))\n",
+      "    print mm"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter(mm)  Area(m2)\t Flow Rate(m3/s)  Velocity(m/s)\n",
+        "50.0\t  0.001963 \t    0.008836     \t4.5\n",
+        "75.0\t  0.004418 \t    0.008836     \t2.0\n",
+        "70.7\t  0.003927 \t    0.00589     \t1.5\n",
+        "30.0\t  0.000707 \t    0.002945     \t4.17\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import sympy\n",
+      "from sympy import diff, Symbol\n",
+      "\n",
+      "\n",
+      "#Initializing  the  variables \n",
+      "'''\n",
+      "def df(x, h=0.1e-5):\n",
+      "    return ( f(x+h/2) - f(x-h/2) )/h\n",
+      "    return df\n",
+      "\n",
+      "print df(2*x,h)\n",
+      "'''\n",
+      "x = Symbol('x') \n",
+      "vx  =  3-x\n",
+      "vy  =  4+2*x\n",
+      "vz  =  2-x  \n",
+      "\n",
+      " #Calculations\n",
+      "delVx  =  vx.diff(x);  \n",
+      "delVy  =  vx.diff(x);\n",
+      "delVz  =  vx.diff(x);  \n",
+      "\n",
+      "result  =  delVx+delVy+delVz;#requirement  of  continuity  equation  (result  =  0)\n",
+      "print \"Satisfy requirement of continuity \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Satisfy requirement of continuity \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_5.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_5.ipynb
new file mode 100755
index 00000000..b12249f7
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_5.ipynb
@@ -0,0 +1,517 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:04c1d9ce4358772aaf36727d98b4d1c000b18de791fe80fc4e6e1ac3cabc0050"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: The Momentum Equation and its Applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "\n",
+      "l  =  60  ;                                                                #Length  of  pipeline\n",
+      "rho  =  1000;                                                          #  Density  of  liquid\n",
+      "a  =  0.02;                                                              #Acceleration  of  fluid\n",
+      "\n",
+      " #Calculations\n",
+      "delP  =  rho*l*a;                                                  #Change  in  pressure\n",
+      "print \"Increase of pressure difference required (kN/m2):\",delP/1000"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Increase of pressure difference required (kN/m2): 1.2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "v  =  5;                                 #Velocity  of  jet          \n",
+      "rho  =  1000;                            #density  of  water\n",
+      "d  =  0.025;                             #Diameter  of  fixed  nozzle\n",
+      "\n",
+      " #Calculations\n",
+      " #--Part(a)  Variation  of  force  exerted  normal  to  the  plate  with  plate  angle--//\n",
+      "header  =  \"Theta\\t vcos(x)\\t pAv\\t Force\"\n",
+      "unit  =    \"deg\\t m/s\\t kg/s\\t N\"\n",
+      "\n",
+      "A  =  math.pi*d**2/4;\n",
+      "x  =  range(0,91,15);\n",
+      "for c in range(len(x)):\n",
+      "    x[c]=1.0*x[c]\n",
+      "m  =  round(rho*A*v,2);\n",
+      "ma  =  [m,m,m,m,m,m,m];\n",
+      "vcomp=[]\n",
+      "force=[]\n",
+      "for c in x:\n",
+      "    vcomp.append(round(v*math.cos(math.radians(c)),2))\n",
+      "    force.append(round((rho*A*v**2)*math.cos(math.radians(c)),2))\n",
+      "\n",
+      "print header\n",
+      "print unit\n",
+      "for c in range(len(x)):\n",
+      "    mm=str(x[c])+' \\t '+str(vcomp[c])+' \\t'+str(ma[c])+' \\t'+str(force[c])\n",
+      "    print mm\n",
+      "##value  = [x,vcomp,ma,force]\n",
+      "##print value,unit, header\n",
+      "\n",
+      " #--Part(b)  Variation  of  force  exerted  normal  to  the  plate  with  plate  velocity--//        \n",
+      "header  =\"Theta\\t v\\t u\\t v-u\\t pA(v-u)\\t Force\\t\"\n",
+      "unit    =\"deg\\t m/s\\t m/s\\t m/s\\t kg/s\\t N\\t\"\n",
+      "x  =  [0,0,0,0,0]\n",
+      "v  =  [5,5,5,5,5]\n",
+      "u  =  range(2,-3,-1);\n",
+      "D=[]\n",
+      "Prod=[]\n",
+      "Force=[]\n",
+      "for c in range(5):\n",
+      "    D.append(v[c]-u[c])\n",
+      "    Prod.append(round((rho*A*D[c]),2))\n",
+      "    Force.append(round((rho*A*D[c]**2),2))\n",
+      "    \n",
+      "print '\\n',\"(b)\",\"\\n\",header\n",
+      "print unit\n",
+      "for c in range(len(x)):\n",
+      "    mm=str(x[c])+' \\t '+str(v[c])+' \\t '+str(u[c])+' \\t '+str(D[c])+' \\t '+str(Prod[c])+' \\t '+str(Force[c])\n",
+      "    print mm\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theta\t vcos(x)\t pAv\t Force\n",
+        "deg\t m/s\t kg/s\t N\n",
+        "0.0 \t 5.0 \t2.45 \t12.27\n",
+        "15.0 \t 4.83 \t2.45 \t11.85\n",
+        "30.0 \t 4.33 \t2.45 \t10.63\n",
+        "45.0 \t 3.54 \t2.45 \t8.68\n",
+        "60.0 \t 2.5 \t2.45 \t6.14\n",
+        "75.0 \t 1.29 \t2.45 \t3.18\n",
+        "90.0 \t 0.0 \t2.45 \t0.0\n",
+        "\n",
+        "(b) \n",
+        "Theta\t v\t u\t v-u\t pA(v-u)\t Force\t\n",
+        "deg\t m/s\t m/s\t m/s\t kg/s\t N\t\n",
+        "0 \t 5 \t 2 \t 3 \t 1.47 \t 4.42\n",
+        "0 \t 5 \t 1 \t 4 \t 1.96 \t 7.85\n",
+        "0 \t 5 \t 0 \t 5 \t 2.45 \t 12.27\n",
+        "0 \t 5 \t -1 \t 6 \t 2.95 \t 17.67\n",
+        "0 \t 5 \t -2 \t 7 \t 3.44 \t 24.05\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3, Page 123"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "  \n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "x  =  60;                                                     #Angle  of  deflection theta\n",
+      "rho  =  1000;                                                 #  Density  of  liquid\n",
+      "V1  =  30;                                                    #Acceleration  of  fluid\n",
+      "V2  =  25;\n",
+      "m  =  .8;                                                     #Discharge  through  A\n",
+      "\n",
+      " #Calculations\n",
+      "def  Reaction(Vin , Vout):\n",
+      "    R  =  m*(Vin  -Vout)  ;\n",
+      "    return R\n",
+      "Rx  =  Reaction(V1,V2*math.cos(math.radians(x)));\n",
+      "Ry  =  -Reaction(0,V2*math.sin(math.radians(x)));\n",
+      "print \"Reaction in X-direction (N)                              :\",Rx\n",
+      "print \"Reaction in Y-direction (N)                              :\",round(Ry,2)\n",
+      "print \"Net Reaction            (N)                              :\",round((Rx**2 +Ry**2)**0.5,2)\n",
+      "print \"Inclination of Resultant Force with x-direction (Degrees):\",round(180/math.pi*math.atan(Ry/Rx),2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reaction in X-direction (N)                              : 14.0\n",
+        "Reaction in Y-direction (N)                              : 17.32\n",
+        "Net Reaction            (N)                              : 22.27\n",
+        "Inclination of Resultant Force with x-direction (Degrees): 51.05\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "v1  =  36  ;                                                              #Exit  velocity\n",
+      "u  =  15;                                                                  #Velocity  of  vane\\\n",
+      "x  =  30;                                                                  #  Angle  between  vanes  and  flow\n",
+      "rho  =  1000;                                                          #  Density  of  water\n",
+      "d  =  .1;                                                                  #  Diameter  of  jet\n",
+      "\n",
+      " #Calculations\n",
+      "alp  =  (180/math.pi)*math.atan((v1*math.sin(math.radians(x))/(v1*math.cos(math.radians(x))-u)));\n",
+      "v2  =  0.85*v1*math.sin(math.radians(x));\n",
+      "bta  =    (180/math.pi)*math.acos((u*math.sin(math.radians(alp))/v2));\n",
+      "m  =  (rho*math.pi*v1*d**2)/4;\n",
+      "Vin  =  v1*math.cos(math.radians(x));\n",
+      "Vout  =  v2*math.cos(math.radians(90));\n",
+      "Rx  =  m*(Vin-Vout);\n",
+      "\n",
+      "\n",
+      "print \"Inlet Angle            (Degrees) :\", round(alp,2)\n",
+      "print \"Outlet Angle           (Degrees) :\", round(bta,2)\n",
+      "print \"Force exerted by vanes (N)       :\", round(Rx) \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Inlet Angle            (Degrees) : 48.05\n",
+        "Outlet Angle           (Degrees) : 43.18\n",
+        "Force exerted by vanes (N)       : 8815.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5, Page 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "rho  =  850  ;                                                          #  Density  of  liquid\n",
+      "a  =  0.02                                                                #Acceleration  of  fluid\n",
+      "x  =  45  ;\n",
+      "d1  =  .5  ;\n",
+      "d2  =  .25;\n",
+      "p1  =  40*10**3;\n",
+      "p2  =  23*10**3;\n",
+      "Q  =  .45;\n",
+      "   \n",
+      " #Calculations\n",
+      "A1  =  (math.pi*d1**2)/4;\n",
+      "A2  =  (math.pi*d2**2)/4;\n",
+      "v1  =  Q/A1;\n",
+      "v2  =  Q/A2;\n",
+      "\n",
+      "Rx  =  p1*A1  -  p2*A2*math.cos(math.radians(x))  -  rho*Q*(v2*math.cos(math.radians(x))-v1);\n",
+      "Ry  =  p2*A2*math.sin(math.radians(x))  +  rho*Q*v2*math.sin(math.radians(x));\n",
+      "\n",
+      "print \"Resultant force on the bend                     (kN)     :\",round((Rx**2 +Ry**2)**0.5/1000,3)\n",
+      "print \"Inclination of Resultant Force with x-direction (Degrees):\",round(math.atan(Ry/Rx)*180/math.pi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resultant force on the bend                     (kN)     : 6.362\n",
+        "Inclination of Resultant Force with x-direction (Degrees): 31.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6, Page 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "v  =  4.9;                                                                #Velocity  of  Jet\n",
+      "rho  =  1000;                                                          #  Density  of  water\n",
+      "d  =  0.05;\n",
+      "u  =  1.2                                                                  #  Velocity  of  tank\n",
+      " #Calculations\n",
+      "Vout  =  v;\n",
+      "Vin  =  0;\n",
+      "m  =  rho*math.pi*d**2*v/4;\n",
+      "R  =  m*(Vout-Vin);\n",
+      "print \"Reaction of jet on tank (N) :\",round(R,2)\n",
+      "print \"Work done per second    (W) :\",round(R*u,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reaction of jet on tank (N) : 47.14\n",
+        "Work done per second    (W) : 56.57\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page 130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      " \n",
+      " \n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "Vj  =  5*10**6;                               # Velocity  of  Jet\n",
+      "Mr  =  150000;                                # Mass  of  Rocket\n",
+      "Mf0  =  300000;                               # Mass  of  initial  fuel\n",
+      "Vr  =  3000;                                  # Velocity  of  jet  relative  to  rocket\n",
+      "g  =  9.81;                                   # Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "m  =  Vj/Vr;                                  #Rate  of  fuel  consumption\n",
+      "T  =  Mf0/m;                                  #  Burning  time\n",
+      "\n",
+      "def f(t,m,Vr,Mr,Mf0,g):\n",
+      "    return m*Vr  /(Mr  +  Mf0  -  m*t)  -  g;\n",
+      " \n",
+      "args = (5000/3,3000,150000,300000,9.81)\n",
+      "Vt = integrate.quad(f, 0.0, 180, args)\n",
+      "\n",
+      "def h(t,Vr,g):\n",
+      "    return -g*t - Vr*math.log(1 - t/269.95);\n",
+      " \n",
+      "args = (3000,9.81)\n",
+      "Z1 = integrate.quad(h, 0.0, 180, args)\n",
+      "Z2 =  Vt[0]**2/(2*g);\n",
+      "\n",
+      "print \"(a)Burning time (s)                             :\",T\n",
+      "print \"(b)Speed of rocket when all fuel is burned (m/s):\",round(Vt[0],2)\n",
+      "print \"(c)Maximum height reached (km)                  :\",round((Z2+Z1[0])/1000,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Burning time (s)                             : 180.0\n",
+        "(b)Speed of rocket when all fuel is burned (m/s): 1530.04\n",
+        "(c)Maximum height reached (km)                  : 203.8\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8, Page 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "V  =  200;                                                                #Velocity  in  still  air\n",
+      "Vr  =  700;                                                              #velocity  of  gas  relative  to  engine\n",
+      "mf  =  1.1;                                                              #  Fuel  Consumption\n",
+      "r  =  1/40  ;        \n",
+      "P1  =0;\n",
+      "P2  =  0;\n",
+      "\n",
+      " #Calculations\n",
+      "m1  =    mf/r;\n",
+      "T  =  m1*((1+r)*Vr  -V);\n",
+      "print \"(a)Thrust               (kN) :\",T/1000\n",
+      "\n",
+      "W  =  T*V;\n",
+      "print \"(b)Work done per second (kW) :\",W/1000\n",
+      "\n",
+      "Loss  =  0.5*m1*(1+r)*(Vr-V)**2;\n",
+      "print \"(c)Efficiency           (%)  :\",round(W/(W+Loss)*100,1) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Thrust               (kN) : 22.77\n",
+        "(b)Work done per second (kW) : 4554.0\n",
+        "(c)Efficiency           (%)  : 44.7\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10, Page 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "rho  =  1000;                                                          #  Density  of  water\n",
+      "Q  =  10;                                                                  #Acceleration  of  fluid\n",
+      "r2  =  1.6;\n",
+      "r1  =  1.2;\n",
+      "V1  =  2.3;\n",
+      "V2  =  0.2;\n",
+      "rot  =  240;  \n",
+      "\n",
+      " #Calculations\n",
+      "Tf  =  rho*Q*(V2*r2  -  V1*r1);\n",
+      "T  =  -Tf;\n",
+      "n  =  rot  /  60;\n",
+      "P  =  2*round(math.pi,3)*n*T;\n",
+      "\n",
+      "print \"Torque exerted by fluid  (N.m):\",T\n",
+      "print \"Theoretical power output (kW) :\",round(P/1000,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Torque exerted by fluid  (N.m): 24400.0\n",
+        "Theoretical power output (kW) : 613.32\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_6.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_6.ipynb
new file mode 100755
index 00000000..5cde4008
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_6.ipynb
@@ -0,0 +1,504 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:df3153eb902c7efba51ed445ceefd3e3c02fa6ed938e4ef53fe06333d531bf11"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: The Energy Equation and its Applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page 170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "Pc  =  0;                                                                  #  Atmospheric  Pressure\n",
+      "Z3  =  30+2;                                                            #height  of  nozzle\n",
+      "Ep  =  50  ;                                                              #Energy  per  unit  weight  supplied  by  pump\n",
+      "d1  =  0.150;                                                          #Diameter  of  sump\n",
+      "d2  =  0.100;                                                          #Diameter  of  delivery  pipe\n",
+      "d3  =  0.075  ;                                                        #Diameter  of  nozzle\n",
+      "g  =  9.81;                                                              #  Acceleration  due  to  gravity\n",
+      "Z2  =  2;                                                                  #Height  of  pump\n",
+      "rho  =  1000;                                                          #  Density  of  water\n",
+      "\n",
+      " #Calculations\n",
+      "U3  = (2*g*(Ep-Z3)/(1+5*(d3/d1)**4  +  12*(d3/d2)**4))**0.5;\n",
+      "U1  =  U3/4;\n",
+      "Pb  =  rho*g*Z2  +  3*rho*U1**2;\n",
+      "print \"Velocity of Jet through nozzle (m/s) :\",round(U3,3)\n",
+      "print \"Pressure in the suction pipe at the inlet to the pump at B (kN/m^2) :\",round(Pb/1000,3)     "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of Jet through nozzle (m/s) : 8.314\n",
+        "Pressure in the suction pipe at the inlet to the pump at B (kN/m^2) : 32.58\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "x  =  45;                                                                  #  Inclination  of  pipe\n",
+      "l  =  2;                                                                    #Length  of  pipe  under  consideration\n",
+      "Ep  =  50  ;                                                              #Energy  per  unit  weight  supplied  by  pump\n",
+      "d1  =  0.2;                                                              #Diameter  of  sump\n",
+      "d2  =  0.1;                                                              #Diameter  of  delivery  pipe\n",
+      "g  =  9.81;                                                              #  Acceleration  due  to  gravity\n",
+      "rho  =  1000;                                                          #  Density  of  water\n",
+      "V1  =  2;\n",
+      "RD_oil  =  0.9;                                                      #  relative  density  of  oil\n",
+      "RD_Merc  =  13.6;                                                  #  Relative  density  of  Mercury\n",
+      "\n",
+      " #Calculations\n",
+      "V2  =  V1*(d1/d2)**2;\n",
+      "dZ  =  round(l*math.sin(math.radians(x)),3);       # it is used in book as 1.414,by rounding so here also\n",
+      "rho_Oil  =  RD_oil*rho;\n",
+      "rho_Man  =  RD_Merc*rho;\n",
+      "dP  =  0.5*rho_Oil*(V2**2-V1**2)  +  rho_Oil*g*dZ;\n",
+      "h  =  rho_Oil  *(  dP/(rho_Oil*g)-  dZ)/(rho_Man  -  rho_Oil);\n",
+      "\n",
+      "print \"Pressure Difference(N/m2) : \",round(dP,0)\n",
+      "print \"Difference in the level of mercury (m):\",round(h,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure Difference(N/m2) :  39484.0\n",
+        "Difference in the level of mercury (m): 0.217\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3, Page 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "d1  =  0.25;                                                            #Pipeline  diameter\n",
+      "d2  =  0.10;                                                            #Throat  diameter\n",
+      "h  =0.63;                                                                #Difference  in  height\n",
+      "rho  =  1000;                                                          #Density  of  water\n",
+      "g  =  9.81                                                                #Acceleration  due  to  gravity\n",
+      "\n",
+      " #Calculations\n",
+      "rho_Hg  =  13.6*rho;\n",
+      "rho_Oil  =  0.9*rho;\n",
+      "A1  =  (math.pi*d1**2)/4;                                            #  Area  at  entry\n",
+      "m  =  (d1/d2)**2;                                                    #Area  ratio\n",
+      "Q  =  (A1/(m**2-1)**0.5)*(2*g*h*(rho_Hg/rho_Oil  -1))**0.5;\n",
+      "\n",
+      "print \"Thepretical Volume flow rate (m3/s ):\",round(Q,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thepretical Volume flow rate (m3/s ): 0.105\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "\n",
+      "x  =  1.5;\n",
+      "y  =0.5;\n",
+      "H  =  1.2;\n",
+      "A  =  650*10**-6;\n",
+      "Q  =0.117;\n",
+      "g  =  9.81;\n",
+      "\n",
+      " #Calculations\n",
+      "Cv  =(x**2/(4*y*H))**0.5;\n",
+      "Cd  =  Q  /  (60*A*(2*g*H)**0.5);\n",
+      "Cc  =  Cd/Cv;\n",
+      "\n",
+      "\n",
+      "print \"Coefficient of velocity :\",round(Cv,3)\n",
+      "print \"Coefficient of Discharge :\",round(Cd,3)\n",
+      "print \"Coefficient of contraction :\",round(Cc,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coefficient of velocity : 0.968\n",
+        "Coefficient of Discharge : 0.618\n",
+        "Coefficient of contraction : 0.639\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5, Page 192"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "B  =  0.7;\n",
+      "H1  =  0.4;\n",
+      "H2  =  1.9;\n",
+      "g  =9.81;\n",
+      "z  =  1.5  ;                                                              #  height  of  opening\n",
+      "\n",
+      " #Calculations\n",
+      "Q_Th  =  2/3  *B*(2*g)**0.5*(H2**1.5  -  H1**1.5);\n",
+      "A  =  z*B;\n",
+      "h  =  0.5*(H1+H2);\n",
+      "Q  =  A*(2*g*h)**0.5;\n",
+      "\n",
+      "print \"Percentage error in discharge (%):\",round((Q-Q_Th)*100/Q_Th,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage error in discharge (%): 1.98\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6, Page 195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "Cd  =  0.6;                                                              #Coefficient  of  discharge\n",
+      "Q  =  0.28;\n",
+      "x  =  90;                                                                  #Theta\n",
+      "g  =  9.81;\n",
+      "dH  =  0.0015;\n",
+      "\n",
+      " #Calculations\n",
+      "H  =  (Q*(15/8)/(Cd*(2*g)**0.5*math.tan(math.radians(x/2))))**(2/5)\n",
+      "Frac_Q  =  5/2  *(  dH/H);\n",
+      "\n",
+      "print \"Percentage error in discharge(%)\",round(Frac_Q*100,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage error in discharge(%) 0.72\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7, Page 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "B  =  0.9;\n",
+      "H  =  0.25;\n",
+      "alpha  =  1.1;\n",
+      "g  =  9.81;  \n",
+      "\n",
+      " #Calculations\n",
+      "Q  =  1.84  *  B  *  H**(3/2);\n",
+      "print \"Q(m3/s) :\",Q\n",
+      "\n",
+      "i  =  1;\n",
+      "while(i  <= 3):\n",
+      "    v  =  Q  /(1.2*  (H+0.2));\n",
+      "    print \"V(m/s) :\",round(v,4)\n",
+      "    k  =    ((1  +  alpha*v**2/(2*g*H))**1.5  -(alpha*v**2/(2*g*H))**1.5  );\n",
+      "    Q  =  k*  1.84  *  B  *  H**(3/2);\n",
+      "    print \"Q(m3/s) :\",round(Q,4)\n",
+      "    i  =  i+1;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q(m3/s) : 0.207\n",
+        "V(m/s) : 0.3833\n",
+        "Q(m3/s) : 0.2161\n",
+        "V(m/s) : 0.4001\n",
+        "Q(m3/s) : 0.2168\n",
+        "V(m/s) : 0.4016\n",
+        "Q(m3/s) : 0.2169\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8, Page 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "rho  =  1000;\n",
+      "v  =  66  ;\n",
+      "Q  =  0.13;\n",
+      "g  =  9.81;  \n",
+      "z  =240;\n",
+      "\n",
+      " #Calculations\n",
+      "P_Jet  =    0.5*rho*v**2*Q;\n",
+      "P_Supp  =  rho*g*Q*z;\n",
+      "P_Lost  =    P_Supp  -P_Jet;\n",
+      "h  =  P_Lost/(rho*g*Q);\n",
+      "eff  =  P_Jet/P_Supp;\n",
+      "\n",
+      "print \"Part(a) - power of the jet(kW):                  \",round(P_Jet/1000,2)\n",
+      "print \"Part(b) - power supplied from the reservoir (kW):\",round(P_Supp/1000,2) \n",
+      "print \"Part(C) - head used to overcome losses (m):      \",round(h,2)\n",
+      "print \"Part(d) - Efficiency(%) :                        \",round(eff*100,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part(a) - power of the jet(kW):                   283.14\n",
+        "Part(b) - power supplied from the reservoir (kW): 306.07\n",
+        "Part(C) - head used to overcome losses (m):       17.98\n",
+        "Part(d) - Efficiency(%) :                         92.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9, Page 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "r1  =  0.2;\n",
+      "Z1  =  0.500;\n",
+      "Z2  =  0.340;\n",
+      "g  =  9.81;\n",
+      "rho  =  0.9*1000  ;\n",
+      "\n",
+      " #Calculations\n",
+      "r0  =  r1*((2-2*Z2/Z1)**0.5);\n",
+      "omega  =   round((2*g*Z1/r0**2)**0.5,1)\n",
+      "\n",
+      "def G(r):\n",
+      "    out  =r**3  -  r*r0**2;\n",
+      "    return out\n",
+      "    \n",
+      "results = integrate.quad(G, r0, r1)\n",
+      "\n",
+      "F  =  rho*omega**2*math.pi*results[0];\n",
+      "\n",
+      "print r0,r1\n",
+      "print \"Part(a) Speed of rotation (rad/s ):\",round(omega,1)\n",
+      "print \"Part(b) Upward force on the cover (N): \",round(F,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.16 0.2\n",
+        "Part(a) Speed of rotation (rad/s ): 19.6\n",
+        "Part(b) Upward force on the cover (N):  56.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10, Page 206"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "Ra  =  0.2;\n",
+      "Rb  =  0.1;\n",
+      "H  =  0.18;\n",
+      "Za  =  0.125;\n",
+      "\n",
+      " #Calculations\n",
+      "Y  =    Ra**2*(H-Za);\n",
+      "Zb  =  H  -  Y/Rb**2;\n",
+      "\n",
+      "print \"Height above datum of a point B on the free surface at a radius of 100 mm (mm):\",Zb*1000"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height above datum of a point B on the free surface at a radius of 100 mm (mm): -40.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_7.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_7.ipynb
new file mode 100755
index 00000000..95848de2
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_7.ipynb
@@ -0,0 +1,221 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:58b7373e6785b57ec4db0eaf49754884e3dad15de9e9489fc1c2bb500d413f82"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: Two-Dimentional Idea Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page 235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables  \n",
+      "x  =  120;                                        #Theta\n",
+      "r  =  1;\n",
+      "v0  =  0.5;\n",
+      "q  =  2;\n",
+      "theta =120;\n",
+      "\n",
+      " #Calculations\n",
+      "Vr = v0*r*math.cos(math.radians(theta))  +q/(2*math.pi)\n",
+      "Vth = -v0*math.sin(math.radians(theta))\n",
+      "V  =  (Vr**2+Vth**2)**0.5;\n",
+      "alpha  =  math.atan(abs(Vth/Vr));\n",
+      "bet  =  x-alpha*180/math.pi;\n",
+      "\n",
+      "\n",
+      "print \"Fluid Velocity(m/s) :\",round(V,3)\n",
+      "print \"Beta (Degree)       :\",round(bet,2)\n",
+      "print \"Alpha (Degree)      :\",round(alpha*180/math.pi,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fluid Velocity(m/s) : 0.438\n",
+        "Beta (Degree)       : 38.96\n",
+        "Alpha (Degree)      : 81.04\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page 239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy.optimize import fsolve\n",
+      "import sympy\n",
+      "from sympy import RootOf, I, Symbol\n",
+      " #Initializing  the  variables  \n",
+      "q  =  10;\n",
+      "def  shi(x,y):\n",
+      "    Z  =  (q/2/math.pi)*(math.atan(y/(x-1))-math.atan(y/(x+1))) - 25*y;\n",
+      "    return Z\n",
+      "h  =  0.0000001;\n",
+      "Vinf  =  25;\n",
+      "x=Symbol('x')\n",
+      " #Calculations\n",
+      "#f = lambda x : x**2 - 2/(5*math.pi) -1\n",
+      "result =  [RootOf(x**2- 2/(5*math.pi) -1,i) for i in (0,1)] \n",
+      "\n",
+      "root1=round(result[0],3)\n",
+      "root2=round(result[1],3)\n",
+      "l  =  abs(abs(root1)+abs(root2));\n",
+      "Ymax  =  0.047;\n",
+      "width  =  2*Ymax;\n",
+      "Vx  =  (shi(1-h,1)-shi(1-h,1-h))/h;           #  At  x=1  the  function  atan  is  not  defined  hence  taking  x  a  little  smaller.\n",
+      "Vy  =  -1*(shi(1-2*h,1)-shi(1-h,1))/h;        #  At  x=1  the  function  atan  is  not  defined  hence  taking  x  a  little  smaller.\n",
+      "\n",
+      "V  =  (Vx**2+Vy**2)**0.5;\n",
+      "rho  =  Symbol('rho')\n",
+      "dP  =  rho/2*round((V**2  -  Vinf**2),2);                          #difference  in  pressure\n",
+      "\n",
+      "print \"Pressure Difference   (N/m2)  :\",dP\n",
+      "print \"Velocity              (m/s)   :\",round(V,2)\n",
+      "print \"Length of Rankine Body(m)     :\",l\n",
+      "print \"Width of Rankine Body (m)     :\",width"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure Difference   (N/m2)  : 16.93*rho\n",
+        "Velocity              (m/s)   : 25.67\n",
+        "Length of Rankine Body(m)     : 2.124\n",
+        "Width of Rankine Body (m)     : 0.094\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4, Page 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math                                              #Example  7.4\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "a  =  0.02;\n",
+      "r  =  0.05;\n",
+      "V0  =  1;\n",
+      "x  =  135;                                                                #  Theta\n",
+      "def  shi(r,x):\n",
+      "    Z  =  V0*math.sin(math.radians(x))*(r-((a**2)/r));\n",
+      "    return Z\n",
+      "h  =  0.0001;\n",
+      "\n",
+      " #Calculations\n",
+      "Vr  =  57*(shi(r,x+h)-shi(r,x))/(r*h);\n",
+      "Vx  =  -1*(shi(r+h,x)-shi(r,x))/h;\n",
+      "\n",
+      "print \"Radial Velocity (m/s)             :\",round(Vr,3)\n",
+      "print \"Normal component of velocity (m/s):\",round(Vx,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radial Velocity (m/s)             : -0.591\n",
+        "Normal component of velocity (m/s): -0.82\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page 246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "rho  =  1000;\n",
+      "r  =  2;\n",
+      "psi  =  2*math.log(r);\n",
+      "\n",
+      " #Calculations\n",
+      "y  =  psi/math.log(r);                                                  #  y  =  GammaC  /  2*pi\n",
+      "v  =  y/r;\n",
+      "dPbydr  =  rho*v**2/r;\n",
+      "print \"Pressuer Gradient (N/m3 ) :\",dPbydr"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressuer Gradient (N/m3 ) : 500.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_8.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_8.ipynb
new file mode 100755
index 00000000..cd940a0c
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_8.ipynb
@@ -0,0 +1,73 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b0713ad3f7c2f18448b69f687926b2d86e094144766f7b44f7fc20525a516074"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8: Dimensional Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "P1  =  57;                                         #Power  in  SI\n",
+      "M  =  1/14.6;                                      #Ratio  of  mass  in  SI/British\n",
+      "L  =  1/0.3048;                                    #Ratio  of  length  in  SI/British\n",
+      "T  =  1;                                           #Ratio  of  time  in  SI/British\n",
+      "\n",
+      "#Calculations\n",
+      "'''\n",
+      "n1 is the horsepower and N1 is the corresponding number of British units (ft*lbf/second),\n",
+      "then     N1 = 550*n1.\n",
+      "Similarly, for the SI system, n2 in kW,\n",
+      "         N2 = 1000*n2\n",
+      "''' \n",
+      "N = (T**3)/(M*L**2)                         # N2/N1\n",
+      "n = N*550/1000                              #n2/n1 \n",
+      "P2  =  n*P1  ;                              #Power  in  kW\n",
+      "\n",
+      "print \"Power in kW       :\",round(P2,1)\n",
+      "print \"Conversion Factor :\",round(n,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power in kW       : 42.5\n",
+        "Conversion Factor : 0.746\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/Chapter_9.ipynb b/Fluid_Mechanics_by_John_F_Douglas/Chapter_9.ipynb
new file mode 100755
index 00000000..5d9fe9d6
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/Chapter_9.ipynb
@@ -0,0 +1,337 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:38d704564aa6536fc6da296a78f2a33e8cb9eb20a3ce0fa8f700546de7fd07b5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Similarity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Vp  =  10;\n",
+      "LpByLm  =  20;\n",
+      "rhoPbyRhoM  =  1;\n",
+      "muPbymuM  =  1;\n",
+      " #Calculations\n",
+      "Vm  =  Vp*LpByLm*rhoPbyRhoM*muPbymuM;\n",
+      "   \n",
+      "print \"Mean water tunnel flow velocity (m/s) :\",Vm"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean water tunnel flow velocity (m/s) : 200\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "    \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Vp  =  3;\n",
+      "LpByLm  =  30;\n",
+      "rhoPbyRhoM  =  1;\n",
+      "muPbymuM  =  1;\n",
+      "\n",
+      " #Calculations\n",
+      "Vm  =  Vp*LpByLm*rhoPbyRhoM*muPbymuM;\n",
+      "   \n",
+      "print \"Mean water tunnel flow velocity (m/s):\",Vm"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean water tunnel flow velocity (m/s): 90\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3, Page 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "   \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "Vp  =  100;\n",
+      "cP  =  340;\n",
+      "cM  =  295;\n",
+      "rhoM  =  7.7;\n",
+      "rhoP  =  1.2;\n",
+      "muM  =  1.8*10**-5;\n",
+      "muP  =  1.2*10**-5;\n",
+      "\n",
+      " #Calculations\n",
+      "Vm  =  Vp*(cM/cP);\n",
+      "LmByLp  =  1/((Vm/Vp)*(muM/muP)*(rhoM/rhoP));\n",
+      "FmByFp  =  (rhoM/rhoP)*(Vm/Vp)**2*(LmByLp)**2;\n",
+      "\n",
+      "print 'Mean wind tunnel flow velocity(m/s) :',round(Vm,2)\n",
+      "print \"Percentage ratio of forces (%)      :\",round(FmByFp*100,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean wind tunnel flow velocity(m/s) : 86.76\n",
+        "Percentage ratio of forces (%)      : 6.93\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page 295"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "      \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "def pLossRatio(RatRho,RatMu,RatL):\n",
+      "    Z  =  RatRho*RatMu**2*RatL**2;\n",
+      "    return Z\n",
+      "\n",
+      " #Calculations\n",
+      " #Case  (a)  :  water  is  used\n",
+      "RatRho  =  1;\n",
+      "RatMu  =  1;\n",
+      "RatL  =  10;\n",
+      "print \"(a)Ratio of pressure losses between the model and the prototype if water is used :\",pLossRatio(RatRho,RatMu,RatL)\n",
+      "\n",
+      "# Case (b) : air is used\n",
+      "RatRho  =  1000/1.23;\n",
+      "RatMu  =  1.8*10**-5/10**-3;\n",
+      "\n",
+      "print \"(b)Ratio of pressure losses between the model and the prototype if air is used   :\",round(pLossRatio(RatRho,RatMu,RatL),2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Ratio of pressure losses between the model and the prototype if water is used : 100\n",
+        "(b)Ratio of pressure losses between the model and the prototype if air is used   : 26.34\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5, Page 296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      " \n",
+      " #Initializing  the  variables\n",
+      "scale  =  1/50;\n",
+      "ratArea  =  scale**2;\n",
+      "Qp  =  1200;\n",
+      "\n",
+      " #Calculations\n",
+      "LmByLp  =  (ratArea)**0.5;\n",
+      "VmByVp  =  (LmByLp)**0.5;\n",
+      "Qm  =  Qp*ratArea*VmByVp;\n",
+      "\n",
+      "print \"Water flow rate (m3/s ):\",round(Qm,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Water flow rate (m3/s ): 0.068\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6, Page 298"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      " \n",
+      " #Initializing  the  variables\n",
+      "Qa  =  2;\n",
+      "Na  =  1400;\n",
+      "rhoA  =  0.92;\n",
+      "rhoS  =  1.3;\n",
+      "DaByDs  =  1;\n",
+      "dPa  =  200;\n",
+      "\n",
+      " #Calculations\n",
+      "Ns  =  Na*(rhoA/rhoS)*(DaByDs);\n",
+      "Qs  =  Qa*(Ns/Na);\n",
+      "dPs  =  dPa  *(rhoS/rhoA)*(Ns/Na)**2*(1/DaByDs)**2;\n",
+      "\n",
+      "print \"Fan test speed (rev/s):\",round(Ns,1)\n",
+      "print \"Flow rate (m3/s)      :\",round(Qs,3)\n",
+      "print \"Pressure rise (N/m2 ) :\",round(dPs,1) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fan test speed (rev/s): 990.8\n",
+        "Flow rate (m3/s)      : 1.415\n",
+        "Pressure rise (N/m2 ) : 141.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8, Page 304"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "      \n",
+      "\n",
+      " #Initializing  the  variables\n",
+      "V  =  300                                                                  #  Volume  rate\n",
+      "w  =  3;\n",
+      "d  =  65;\n",
+      "l  =  30;\n",
+      "scaleH  =  30/1000/18;\n",
+      "scaleV  =  1/60;\n",
+      "ZmByZr  =  1/60;\n",
+      "LmByLr  =  1/600;\n",
+      "rho  =  1000;\n",
+      "mu  =  1.14*10**-3;\n",
+      "\n",
+      " #Calculations\n",
+      "Vr  =  V/(w*d);  \n",
+      "Vm  =Vr*(ZmByZr)**0.5;\n",
+      "m  =  (w*d*scaleH*scaleV)/(d*scaleH  +  2*w*scaleV);\n",
+      "Rem  =  rho*Vm*m/mu;\n",
+      "TmByTr  =  LmByLr*(1/ZmByZr)**0.5;\n",
+      "Tm  =  12.4*60*TmByTr;\n",
+      "\n",
+      "\n",
+      "\n",
+      "print \"Tidal Period (minutes):\",round(Tm,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Tidal Period (minutes): 9.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_Mechanics_by_John_F_Douglas/README.txt b/Fluid_Mechanics_by_John_F_Douglas/README.txt
new file mode 100755
index 00000000..d96dbe87
--- /dev/null
+++ b/Fluid_Mechanics_by_John_F_Douglas/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Ashish Kumar
+Course: btech
+College/Institute/Organization: IITB
+Department/Designation: Aerospace Engg.
+Book Title: Fluid Mechanics
+Author: John F. Douglas
+Publisher: Pearson Education
+Year of publication: 2005
+Isbn: 978-0-13-129293-2
+Edition: 5th
\ No newline at end of file
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diff --git a/Fluid_Power_With_Applications_by_A._Esposito/README.txt b/Fluid_Power_With_Applications_by_A._Esposito/README.txt
old mode 100644
new mode 100755
diff --git a/Fluid_mechanics/Chapter_10.ipynb b/Fluid_mechanics/Chapter_10.ipynb
deleted file mode 100755
index bdfb09f7..00000000
--- a/Fluid_mechanics/Chapter_10.ipynb
+++ /dev/null
@@ -1,234 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8ebaaa4d9fc694c6b4303418dc1faf3eef2d83f1637b122e018c6c5eac0aba66"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10 - Dynamic lift"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2a - Pg 427"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the max. theoretical propulsive force\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "vel=50. #mph\n",
-      "w=240. #rpm\n",
-      "r0=3. #ft\n",
-      "L=30. #ft\n",
-      "rho=0.00230 #slug/ft^2\n",
-      "theta=30*math.pi/180. #radians\n",
-      "#calculations\n",
-      "V=vel*5280./3600.\n",
-      "T=2*math.pi*r0*r0 *w*2*math.pi/60.\n",
-      "Fl=rho*V*T*L\n",
-      "F=r0*Fl*math.cos(theta)\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Max. theoretical porpulsive force =\",F,\"lb\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Max. theoretical porpulsive force = 18683 lb\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2b - Pg 428"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the force required for the operation\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "vel=50. #mph\n",
-      "w=240. #rpm\n",
-      "r0=3. #ft\n",
-      "L=30. #ft\n",
-      "rho=0.00230 #slug/ft^2\n",
-      "theta=30*math.pi/180. #radians\n",
-      "Cl=2\n",
-      "Cd=1\n",
-      "#calculations\n",
-      "vc=r0*w\n",
-      "V=vel*5280./3600.\n",
-      "vr=vc/V\n",
-      "A=2*r0*L\n",
-      "Fl=Cl*A*0.5*rho*V*V \n",
-      "Fd=Cd*A*0.5*rho*V*V\n",
-      "F=r0*(Fl*math.cos(theta) + Fd*math.sin(theta))\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Force required =\",F,\"lb\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Force required = 7454 lb\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3a - Pg 432"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the boundary circulation\n",
-      "#Initialization of variables\n",
-      "W=7500 #pounds\n",
-      "rho=0.00230\n",
-      "V=175*5280/3600 #ft/s\n",
-      "B=50\n",
-      "#calculations\n",
-      "T=W/(rho*V*B)\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Boundary circulation =\",T,\"ft^2/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boundary circulation = 254 ft^2/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3b - Pg 433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horsepower required for the operation\n",
-      "#Initialization of variables\n",
-      "W=7500. #pounds\n",
-      "rho=0.00230\n",
-      "V=175*5280./3600. #ft/s\n",
-      "B=50.\n",
-      "A=350. #ft^2\n",
-      "#calculations\n",
-      "Cl=W/(A*0.5*rho*V*V)\n",
-      "Cd=0.03\n",
-      "Fd=Cd*A*0.5*rho*V*V \n",
-      "hp=Fd*V/550.\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Horsepower required =\",hp,\"hp\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horsepower required = 371 hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4 - Pg 437"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horsepower required for the operation\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "Fl=7500. #pounds\n",
-      "rho=0.00230\n",
-      "V=175*5280./3600. #ft/s\n",
-      "B=50\n",
-      "A=350 #ft^2\n",
-      "#calculations\n",
-      "Vi=2*Fl/(math.pi*rho*V*B*B)\n",
-      "Cl=Fl/(A*0.5*rho*V*V)\n",
-      "Cdi=Cl*Vi/(V)\n",
-      "Fdi=Cdi*A*0.5*rho*V*V\n",
-      "hp=Fdi*V/550\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Horsepower required =\",hp,\"hp\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horsepower required = 44.1 hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_11.ipynb b/Fluid_mechanics/Chapter_11.ipynb
deleted file mode 100755
index ec13c1a6..00000000
--- a/Fluid_mechanics/Chapter_11.ipynb
+++ /dev/null
@@ -1,485 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:525b51e826bc42c9a4490de8234c350df60c3a09744a8e35af1f480f12f531bc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11 - Flow of liquids in open channels"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1a - Pg 454"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge using darcy equation\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "mu=2.34e-5 #lb-sec/ft^2\n",
-      "y=5 #ft\n",
-      "T=25 #ft\n",
-      "d=10 #ft\n",
-      "slope=3./2. \n",
-      "g=32.2 #ft/s^2\n",
-      "S=0.001\n",
-      "#calculations\n",
-      "A=y*d+ 2*0.5*y*(slope*y)\n",
-      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
-      "R=A/WP\n",
-      "e=0.01 #ft\n",
-      "rr=2*R/e\n",
-      "f=0.019\n",
-      "C=math.sqrt(8*g/f)\n",
-      "V=C*math.sqrt(R*S)\n",
-      "Q=V*A\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Discharge using Darcy equation =\",Q,\"ft^3/s\")\n",
-      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge using Darcy equation = 569.3 ft^3/s\n",
-        "The answer is a bit different due to rounding off error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1b - Pg 453"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge using kutter ganguillet method\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "mu=2.34e-5 #lb-sec/ft^2\n",
-      "y=5 #ft\n",
-      "T=25 #ft\n",
-      "d=10 #ft\n",
-      "slope=3./2. \n",
-      "g=32.2 #ft/s^2\n",
-      "S=0.001\n",
-      "n=0.017\n",
-      "#calculations\n",
-      "A=y*d+ 2*0.5*y*(slope*y)\n",
-      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
-      "R=A/WP\n",
-      "e=0.01 #ft\n",
-      "rr=2*R/e\n",
-      "f=0.019\n",
-      "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n",
-      "V=C*math.sqrt(R*S)\n",
-      "Q=V*A\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Discharge using kutter ganguillet formula =\",Q,\" ft^3/s\")\n",
-      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge using kutter ganguillet formula = 517.0  ft^3/s\n",
-        "The answer is a bit different due to rounding off error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1c - Pg 455"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge using bazin formula\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "mu=2.34e-5 #lb-sec/ft^2\n",
-      "y=5. #ft\n",
-      "T=25. #ft\n",
-      "d=10. #ft\n",
-      "slope=3./2.\n",
-      "g=32.2 #ft/s^2\n",
-      "S=0.001\n",
-      "m=0.21\n",
-      "#calculations\n",
-      "A=y*d+ 2*0.5*y*(slope*y)\n",
-      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
-      "R=A/WP\n",
-      "e=0.01 #ft\n",
-      "rr=2*R/e\n",
-      "f=0.019\n",
-      "C=157.6 /(1+ m/math.sqrt(R))\n",
-      "V=C*math.sqrt(R*S)\n",
-      "Q=V*A\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Discharge using bazin formula =\",Q,\" ft^3/s\")\n",
-      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge using bazin formula = 688.7  ft^3/s\n",
-        "The answer is a bit different due to rounding off error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1d - Pg 456"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge using darcy equation\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "mu=2.34e-5 #lb-sec/ft^2\n",
-      "y=5. #ft\n",
-      "T=25. #ft\n",
-      "d=10. #ft\n",
-      "slope=3./2. \n",
-      "g=32.2 #ft/s^2\n",
-      "S=0.001\n",
-      "n=0.017\n",
-      "#calculations\n",
-      "A=y*d+ 2*0.5*y*(slope*y)\n",
-      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
-      "R=A/WP\n",
-      "e=0.01 #ft\n",
-      "rr=2*R/e\n",
-      "f=0.019\n",
-      "C=1.486*math.pow(R,(1./6.)) /n\n",
-      "V=C*math.sqrt(R*S)\n",
-      "Q=V*A\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Discharge using Darcy equation =\",Q,\"ft^3/s\")\n",
-      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge using Darcy equation = 516.6 ft^3/s\n",
-        "The answer is a bit different due to rounding off error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1e - Pg 456"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the froude number\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "mu=2.34e-5 #lb-sec/ft^2\n",
-      "y=5. #ft\n",
-      "T=25. #ft\n",
-      "d=10. #ft\n",
-      "slope=3./2. \n",
-      "g=32.2 #ft/s^2\n",
-      "S=0.001\n",
-      "n=0.017\n",
-      "#calculations\n",
-      "A=y*d+ 2*0.5*y*(slope*y)\n",
-      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
-      "R=A/WP\n",
-      "e=0.01 #ft\n",
-      "rr=2*R/e\n",
-      "f=0.019\n",
-      "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n",
-      "V=C*math.sqrt(R*S)\n",
-      "T=d+ 2*(slope*y)\n",
-      "yh=A/T\n",
-      "Nf=V/(math.sqrt(g*yh))\n",
-      "#results\n",
-      "print '%s %.2f' %(\"froude number = \",Nf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "froude number =  0.56\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1f - Pg 456"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the critical depth\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "mu=2.34e-5 #lb-sec/ft^2\n",
-      "y=5. #ft\n",
-      "T=25. #ft\n",
-      "d=10. #ft\n",
-      "slope=3./2. \n",
-      "g=32.2 #ft/s^2\n",
-      "S=0.001\n",
-      "n=0.017\n",
-      "#calculations\n",
-      "A=y*d+ 2*0.5*y*(slope*y)\n",
-      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
-      "R=A/WP\n",
-      "e=0.01 #ft\n",
-      "rr=2*R/e\n",
-      "f=0.019\n",
-      "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n",
-      "V=C*math.sqrt(R*S)\n",
-      "Q=V*A\n",
-      "T=d+ 2*(slope*y)\n",
-      "yh=A/T\n",
-      "yc=2.88  #ft\n",
-      "#results\n",
-      "print '%s' %(\"yc is obtained using trial and error method\")\n",
-      "print '%s %.2f %s' %(\"Critical depth =\",yc,\"ft\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "yc is obtained using trial and error method\n",
-        "Critical depth = 2.88 ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2 - Pg 459"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the minimum scale ratio\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "Re=4000.\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "vm=5.91 #ft/s\n",
-      "mu=3.24e-5 #ft-lb/s^2\n",
-      "Rm=3.12 #ft\n",
-      "#calculations\n",
-      "lam3=Re*mu/(vm*4*Rm*rho)\n",
-      "lam=math.pow(lam3,(2./3.))\n",
-      "#results\n",
-      "print '%s %.2e' %(\"Minimum scale ratio = \",lam)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum scale ratio =  9.36e-03\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 462"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge, depth of the channel, froude numbers and also the force applied\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "yc=2. #ft\n",
-      "g=32.2 #ft/s^2\n",
-      "d=10. #ft\n",
-      "gam=62.4\n",
-      "rho=1.94\n",
-      "B=10. #ft\n",
-      "#calculations\n",
-      "Vc=math.sqrt(g*yc)\n",
-      "Ac=yc*d\n",
-      "Q=Vc*Ac\n",
-      "y1=5.88 #ft\n",
-      "y2=0.88 #ft\n",
-      "V1=2.73 #ft/s\n",
-      "V2=18.25 #ft/s\n",
-      "Nf1=0.198\n",
-      "Nf2=3.43\n",
-      "F= 0.5*gam*y1*y1 *B - 0.5*gam*y2*y2 *B - Q*rho*V2 +Q*rho*V1\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Discharge in the channel =\",Q,\"ft^3/s\")\n",
-      "print '%s %.2f %s %.2f %s' %(\"\\n Depth of the channel at upstream and downstream =\",y1,\"ft and\",y2, \"ft\")\n",
-      "print '%s %.3f %s %.3f' %(\"\\n froude numbers at upstream and downstream =\",Nf1,\" and \",Nf2)\n",
-      "print '%s %d %s' %(\"\\n Force applied =\",F,\"lb\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge in the channel = 160.5 ft^3/s\n",
-        "\n",
-        " Depth of the channel at upstream and downstream = 5.88 ft and 0.88 ft\n",
-        "\n",
-        " froude numbers at upstream and downstream = 0.198  and  3.430\n",
-        "\n",
-        " Force applied = 5713 lb\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4 - Pg 470"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the distance from vena contracta and also the total distance\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "S0=0.0009\n",
-      "n=0.018\n",
-      "w=20 #ft\n",
-      "d=0.5 #ft\n",
-      "Q=400 #ft^3/s\n",
-      "g=32.2 #ft/s^2\n",
-      "#calculations\n",
-      "y2=4 #ft\n",
-      "V2=Q/(w*y2)\n",
-      "Nf2=V2/math.sqrt(g*y2)\n",
-      "yr=0.5*(math.sqrt(1+ 8*Nf2*Nf2) -1)\n",
-      "y1=yr*y2\n",
-      "L1=32.5\n",
-      "L2=37.1 \n",
-      "L3=51.4\n",
-      "L=L1+L2+L3\n",
-      "#results\n",
-      "print '%s %.1f %s %.2f %s' %(\"distance from vena contracta =\",y2,\"ft and\",y1,\"ft\")\n",
-      "print '%s %.1f %s' %(\"\\n Total distance =\",L,\" ft\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "distance from vena contracta = 4.0 ft and 1.20 ft\n",
-        "\n",
-        " Total distance = 121.0  ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_2.ipynb b/Fluid_mechanics/Chapter_2.ipynb
deleted file mode 100755
index f518cad4..00000000
--- a/Fluid_mechanics/Chapter_2.ipynb
+++ /dev/null
@@ -1,406 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8249f270a6314fb49f84df63266820d87dfa51fe64fed0b9dd11afa539cb500b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2 - Fluid Statics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1 - Pg 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the final pressure and specific weight of the gas\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=0.0765 #lb/ft^3\n",
-      "p=14.7 #psia\n",
-      "dz=10560. #ft\n",
-      "#calculations\n",
-      "pg=p*144./gam\n",
-      "p2=p*math.exp(-dz/pg)\n",
-      "gam2=p2/p*gam\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Final pressure =\",p2,\"psia\")\n",
-      "print '%s %.4f %s' %(\"\\n Final specific weight =\",gam2,\"lb/ft^3\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Final pressure = 10.04 psia\n",
-        "\n",
-        " Final specific weight = 0.0522 lb/ft^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2 - Pg 12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the final pressure and specific weight\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=0.0765 #lb/ft^3\n",
-      "p=14.7 #psia\n",
-      "dz=10560. #ft\n",
-      "n=1.235\n",
-      "#calculations\n",
-      "pg=p*144./gam\n",
-      "p2=p*math.pow((1- dz/pg *(n-1)/n),(n/(n-1)))\n",
-      "gam2=math.pow((p2/p),(1/n)) *gam\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Final pressure =\",p2,\"psia\")\n",
-      "print '%s %.4f %s' %(\"\\n Final specific weight =\",gam2,\"lb/ft^3\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Final pressure = 9.89 psia\n",
-        "\n",
-        " Final specific weight = 0.0555 lb/ft^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the absolute pressure\n",
-      "#Initialization of variables\n",
-      "pb=28.5 #in mercury\n",
-      "d=13.6 #g/cc\n",
-      "gam=62.4\n",
-      "pobs=-4. #psi\n",
-      "#calculations\n",
-      "patm=pb/12. *gam*d/144.\n",
-      "pabs=patm+pobs\n",
-      "P=pabs*144./gam\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Absolute pressure =\",pabs,\"psia\")\n",
-      "print '%s %.1f %s' %(\"\\n Absolute pressure in feet of water =\",P,\"ft of water\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Absolute pressure = 10.0 psia\n",
-        "\n",
-        " Absolute pressure in feet of water = 23.1 ft of water\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4 - Pg 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure gauge at B and also the absolute pressure of air\n",
-      "#Initialization of variables\n",
-      "pb=28. #in mercury\n",
-      "d=13.6 #g/cc\n",
-      "gam=62.4\n",
-      "xm=15. #in\n",
-      "xw=10. #in\n",
-      "patm=28. #in\n",
-      "#calculations\n",
-      "pB=-xm/12 *gam/144 *d + xw*gam/144\n",
-      "pair=patm/12 *gam/144 *d - xm/12 *gam/144 *d\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"The pressure gauge at B indicates a reading of\",-pB,\"psi vacuum\")\n",
-      "print '%s %.2f %s' %(\"\\n Absolute pressure of Air =\",pair,\"psia\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The pressure gauge at B indicates a reading of 3.03 psi vacuum\n",
-        "\n",
-        " Absolute pressure of Air = 6.38 psia\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5 - Pg 20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure difference and also compare the pressures at A and B\n",
-      "#Initialization of variables\n",
-      "pb=28.5 #in mercury\n",
-      "d=13.6 #g/cc\n",
-      "gam=62.4\n",
-      "xm=10. #in\n",
-      "xw=2. #ft\n",
-      "#calculations\n",
-      "dp= xw*gam/144 - xm/12 *gam/144 + xm/12 *gam/144 *d\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Pressure difference =\",dp,\"psi\")\n",
-      "if dp>0:\n",
-      "    print '%s' %(\"\\n Pressure at A is greater than that at B\")\n",
-      "elif dp==0:\n",
-      "    print '%s' %(\"\\n Pressure at both A and B are equal\")\n",
-      "else:\n",
-      "    print '%s' %(\"\\n Pressure at A is less than that at B\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure difference = 5.42 psi\n",
-        "\n",
-        " Pressure at A is greater than that at B\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6 - Pg 24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the magnitude of total force, vertical and horiontal locations of the total force.\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "import scipy\n",
-      "from scipy import integrate\n",
-      "gam=62.4\n",
-      "x1=4. #ft\n",
-      "x2=6.  #ft\n",
-      "y1=6. #ft\n",
-      "z=8. #ft\n",
-      "dy=1. #ft\n",
-      "angle=60.*math.pi/180. #radians\n",
-      "#calculations\n",
-      "A1=x1*x2\n",
-      "A2=1/2. *y1*y1\n",
-      "yc = (A1*(x1+x2+dy) + A2*(x1+x2))/(A1+A2)\n",
-      "hc=yc*math.sin(angle)\n",
-      "F=hc*gam*(A1+A2)\n",
-      "ic1=1/12. *x1*y1*y1*y1\n",
-      "ic2=1/36. *y1*x2*x2*x2\n",
-      "ad1=A1*(x1+x2+dy-yc)*(x1+x2+dy-yc)\n",
-      "ad2=A2*(x1+x2-yc)*(x1+x2-yc)\n",
-      "It=ic1+ic2+ad1+ad2\n",
-      "ydc=It/(yc*(A1+A2))\n",
-      "def momen(u):\n",
-      "    m= gam*math.sin(angle) *(2*x1+u)*0.5*(x2-u)*(y1-u)\n",
-      "    return m;\n",
-      "\n",
-      "MED, err =scipy.integrate.quad( momen,0,y1)\n",
-      "FEDC=gam*math.sin(angle) *A2*(x1+x2)\n",
-      "xed=MED/FEDC\n",
-      "xp= (A1*2*(x1+x2+dy) + (x1+x2)*(A2)*(x1+xed))/(A1*(x1+x2+dy) + A2*(x1+x2))\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Magnitude of total force =\",F,\"lb\")\n",
-      "print '%s %.3f %s' %(\"\\n Vertical location of force =\",ydc,\"ft\")\n",
-      "print '%s %.2f %s' %(\"\\n Horizontal location of force =\",xp,\"ft from AB\")\n",
-      "print '%s' %(\"\\n Direction of force is perpendicular to the plane surface\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnitude of total force = 23993 lb\n",
-        "\n",
-        " Vertical location of force = 0.266 ft\n",
-        "\n",
-        " Horizontal location of force = 3.58 ft from AB\n",
-        "\n",
-        " Direction of force is perpendicular to the plane surface\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7 - Pg 27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the magnitude of force and the horizontal distance from line of action of Fv\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=62.4\n",
-      "z=10. #ft\n",
-      "z2=5. #ft\n",
-      "z3=4.25 #ft\n",
-      "p=2. #psig\n",
-      "#calculations\n",
-      "h=p*144./gam\n",
-      "Av=z*z\n",
-      "Fh=gam*(z+h)*Av\n",
-      "hpc=1/12. *z*z*z*z /((h+z)*z*z)\n",
-      "Fv=gam*(z2+h) *z*z + gam*math.pi/4. *z*z *z\n",
-      "xp= (gam*(z2+h) *z*z *z2 + gam*math.pi/4. *z*z *z*z3)/(Fv)\n",
-      "F=math.sqrt(Fh*Fh + Fv*Fv)\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Magnitude of force =\",F,\"lb\")\n",
-      "print '%s %.2f %s' %(\"\\n horizontal distance from line of action of Fv =\",xp,\"ft from AG\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Magnitude of force = 142127 lb\n",
-        "\n",
-        " horizontal distance from line of action of Fv = 4.66 ft from AG\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8 - Pg 33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate whether the barge is stable and also find the location of the metacenter\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=0.0765 #lb/ft^3\n",
-      "l=40. #ft\n",
-      "w=16. #ft\n",
-      "d=8. #ft\n",
-      "z=6.  #ft\n",
-      "BG=1. #ft\n",
-      "#calculations\n",
-      "I=1/12. *l*w*w*w\n",
-      "V=l*w*z\n",
-      "IVG=I/V - BG\n",
-      "MB=I/V\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"I/V -BG =\",IVG,\"ft \")\n",
-      "if IVG >0:\n",
-      "    print '%s' %(\"\\n Barge is stable\")\n",
-      "else:\n",
-      "    print '%s' %(\"\\n The barge is unstable\")\n",
-      "\n",
-      "print '%s %.2f %s' %(\"\\n Location of metacenter =\",MB,\"ft above the center of buoyancy \")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "I/V -BG = 2.56 ft \n",
-        "\n",
-        " Barge is stable\n",
-        "\n",
-        " Location of metacenter = 3.56 ft above the center of buoyancy \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_3.ipynb b/Fluid_mechanics/Chapter_3.ipynb
deleted file mode 100755
index 8d368fbf..00000000
--- a/Fluid_mechanics/Chapter_3.ipynb
+++ /dev/null
@@ -1,105 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:dee9d655eeb9028db41fd96e05f9acf5955fd88886936fffc7d4e9aa14bb4a71"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3 - Fluid Kinematics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1 - Pg 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the mean velocity of flow at both the sections\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=0.0765 #lb/ft^3\n",
-      "Q=100. #ft^3/sec\n",
-      "d1=2.5 #ft\n",
-      "d2=9. #in\n",
-      "l=12. #ft\n",
-      "#calculations\n",
-      "A1=math.pi/4. *d1*d1\n",
-      "V1=Q/A1\n",
-      "A2=math.pi*l*d2/12.\n",
-      "V2=Q/A2\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Mean velocity of flow at section 1 =\",V1,\"ft/sec\")\n",
-      "print '%s %.2f %s' %(\"\\n Mean velocity of flow at section 2 =\",V2,\"ft/sec\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mean velocity of flow at section 1 = 20.4 ft/sec\n",
-        "\n",
-        " Mean velocity of flow at section 2 = 3.54 ft/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5 - Pg 60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horizontal and vertical components of velocity\n",
-      "#Initialization of variables\n",
-      "x=3. \n",
-      "y=1.\n",
-      "#calculations\n",
-      "u=-3*y*y\n",
-      "v=-6*x\n",
-      "#results\n",
-      "print '%s %d' %(\"Horizontal component of velocity = \",u)\n",
-      "print '%s %d' %(\"\\n vertical component of velocity = \",v)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horizontal component of velocity =  -3\n",
-        "\n",
-        " vertical component of velocity =  -18\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_4.ipynb b/Fluid_mechanics/Chapter_4.ipynb
deleted file mode 100755
index 49f2ce7e..00000000
--- a/Fluid_mechanics/Chapter_4.ipynb
+++ /dev/null
@@ -1,777 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:bf4ed226827aaf378bd6d43d8ae5845adf36d75f20c124d2087db252462cf40c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4 - Fluid Dynamics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1 - Pg 79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate dp/ds at both upper and lower ends\n",
-      "#Initialization of variables\n",
-      "rho=1.5 #g/cc\n",
-      "g=32.2 #ft/s^2\n",
-      "dzds=-0.5\n",
-      "x1=0\n",
-      "x2=3\n",
-      "#calculations\n",
-      "def func(s):\n",
-      "    dpds=-rho*g*dzds - rho*(3+9*s)*9\n",
-      "    return dpds\n",
-      "\n",
-      "r1=func(x1)\n",
-      "r2=func(x2)\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"At the upper end, dp/ds =\",r1,\"lb/ft^2 per foot\")\n",
-      "print '%s %.1f %s' %(\"\\n At the lower end, dp/ds =\",r2,\"lb/ft^2 per foot\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "At the upper end, dp/ds = -16.3 lb/ft^2 per foot\n",
-        "\n",
-        " At the lower end, dp/ds = -380.9 lb/ft^2 per foot\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2 - Pg 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure difference between the two points\n",
-      "#Initialization of variables\n",
-      "g=32.2 #ft/s^2\n",
-      "v1=3. #ft/s\n",
-      "z1=1.5 #ft\n",
-      "rho=1.5 #g/cc\n",
-      "z2=0\n",
-      "v2=30. #ft/s\n",
-      "#calculations\n",
-      "dp= rho*(v2*v2 /2. - g*z1 +g*z2 - v1*v1 /2.)\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"pressure difference =\",dp,\"lb/ft^2\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "pressure difference = 595.8 lb/ft^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the net power transferred\n",
-      "#Initialization of variables\n",
-      "pd=15. #psia\n",
-      "rhod=0.005#slug/ft^3\n",
-      "pi=150. #psia\n",
-      "rhoi=0.03 #slug/ft^3\n",
-      "dz=-25. #ft\n",
-      "vd=1000. #ft/s\n",
-      "vi=100. #ft/s\n",
-      "ud=200. #Btu/slug\n",
-      "ui=250. #Btu/slug\n",
-      "g=32.2 #ft/s^2\n",
-      "J=778.\n",
-      "uff=5. #ft/s\n",
-      "Q=50. #Btu/sec\n",
-      "#calculations\n",
-      "pr=pd/rhod*144. - pi/rhoi *144.\n",
-      "zr=g*(dz)\n",
-      "vr=(vd*vd -vi*vi)/2.\n",
-      "ur=(ud-ui)*J\n",
-      "jeh=J*Q*g/uff\n",
-      "gem=pr+zr+vr+ur+jeh\n",
-      "power=gem*uff/g\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Power transferred =\",power,\"ft-lb/sec\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power transferred = 64877 ft-lb/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4 - Pg 89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the kinetic energy correction factor\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "import scipy\n",
-      "from scipy import integrate\n",
-      "r0=1. \n",
-      "ri=0.\n",
-      "#calculations\n",
-      "def func1(y):\n",
-      "    v= 2*math.pow(y,(1./7.)) *(y-1)\n",
-      "    return v\n",
-      "\n",
-      "V,err=scipy.integrate.quad(func1,ri,r0)\n",
-      "def func2(y):\n",
-      "    alpha= 1/ (math.pi*V*V*V) *2*math.pi *math.pow((y),(3./7.)) *(y-1) \n",
-      "    return alpha\n",
-      "\n",
-      "a2,err2=scipy.integrate.quad(func2,ri,r0)\n",
-      "#results\n",
-      "print '%s %.2f' %(\"Kinetic energy correction factor = \",a2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Kinetic energy correction factor =  1.06\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5a - Pg 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure at the lower end if friction is neglected\n",
-      "#Initialization of variables\n",
-      "gam=62.4\n",
-      "pu=40. #psia\n",
-      "zu=25. #ft\n",
-      "vu=8. #ft/s\n",
-      "g=32.2 #ft/s^2\n",
-      "vl=8. #ft/s\n",
-      "zl=0 #ft\n",
-      "#calculations\n",
-      "pl= gam*(pu*144. /gam +zu-zl+ (vu*vu -vl*vl)/(2*g))/144.\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Pressure at the lower end if friction is neglected =\",pl,\"psig\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure at the lower end if friction is neglected = 50.83 psig\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5b - Pg 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure at the lower end if friction is not neglected\n",
-      "#Initialization of variables\n",
-      "hl=5.\n",
-      "gam=62.4\n",
-      "pu=40. #psia\n",
-      "zu=25. #ft\n",
-      "vu=8. #ft/s\n",
-      "g=32.2 #ft/s^2\n",
-      "vl=8. #ft/s\n",
-      "zl=0 #ft\n",
-      "#calculations\n",
-      "pl= gam*(pu*144. /gam +zu-zl-hl+ (vu*vu -vl*vl)/(2*g))/144.\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Pressure at the lower end if friction is neglected =\",pl,\"psig\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure at the lower end if friction is neglected = 48.67 psig\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6a - Pg 94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge flow rate.\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=62.4\n",
-      "pa=0\n",
-      "za=15. #ft\n",
-      "va=0\n",
-      "pg=0\n",
-      "zg=0\n",
-      "g=32.2 #ft/s^2\n",
-      "dg=2. #in\n",
-      "#calculations\n",
-      "vg= math.sqrt(2*g*(pa/gam +za+va*va /(2*g) -pg/gam - zg))\n",
-      "Ag=math.pi/4. *(dg/12.)*(dg/12.)\n",
-      "Q=Ag*vg\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"discharge =\",Q,\"ft^3/sec\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "discharge = 0.68 ft^3/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6b - Pg 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure at the points C,D,E and F\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=62.4\n",
-      "pa=0\n",
-      "za=15. #ft\n",
-      "va=0\n",
-      "pg=0\n",
-      "zg=0\n",
-      "g=32.2 #ft/s^2\n",
-      "d=4. #in\n",
-      "dg=2. #in\n",
-      "zd=25.  #ft\n",
-      "#calculations\n",
-      "vg= math.sqrt(2*g*(pa/gam +za+va*va /(2*g) -pg/gam - zg))\n",
-      "Ag=math.pi/4. *(dg/12.)*(dg/12.)\n",
-      "Q=Ag*vg\n",
-      "A=math.pi/4. *(d/12.)*(d/12.)\n",
-      "v4=Q/A\n",
-      "pc=-v4*v4 *gam/(2*g*144.)\n",
-      "pgd= za-zd - v4*v4 /(2*g)\n",
-      "pd=pgd*gam/144.\n",
-      "pe=-v4*v4 *gam/(2*g*144.)\n",
-      "pfg= za- v4*v4 /(2*g)\n",
-      "pf=pfg*gam/144.\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Pressure at C =\",pc,\" psig\")\n",
-      "print '%s %.2f %s' %(\"\\n Pressure at D =\",pd,\" psig\")\n",
-      "print '%s %.2f %s' %(\"\\n Pressure at E =\",pe,\" psig\")\n",
-      "print '%s %.2f %s' %(\"\\n Pressure at F =\",pf,\" psig\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure at C = -0.41  psig\n",
-        "\n",
-        " Pressure at D = -4.74  psig\n",
-        "\n",
-        " Pressure at E = -0.41  psig\n",
-        "\n",
-        " Pressure at F = 6.09  psig\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7 - Pg 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the time required\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "import scipy\n",
-      "from scipy import integrate\n",
-      "d1=6. #ft\n",
-      "d2=3. #in\n",
-      "pa=2. #ft\n",
-      "d=13.6 \n",
-      "sg=0.75\n",
-      "h1=5. #sec\n",
-      "h2=3. #sec\n",
-      "g=32.2 #ft/s^2\n",
-      "#calculations\n",
-      "pag=pa/12. *d/sg\n",
-      "def  func(h):\n",
-      "    time= -d1*d1 /(d2/12.)/(d2/12.) /(math.sqrt(2*g)) *math.pow((pag+h),(-0.5))\n",
-      "    return time\n",
-      "ti,err=scipy.integrate.quad(func,h1,h2)\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Time required =\",ti,\" sec\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required = 54.3  sec\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8 - Pg 100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the rate of flow \n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "x=12. #ft\n",
-      "angle=30.*math.pi/180. #degrees\n",
-      "g=32.2 #ft/s^2\n",
-      "z=-2. #ft\n",
-      "d=2. #in\n",
-      "#calculations\n",
-      "vj= x/math.cos(angle) *math.sqrt(g/(2*(x*math.tan(angle) -z)))\n",
-      "Q=math.pi/4. *(d/12.)*(d/12.) *vj\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Rate of flow =\",Q,\" ft^3/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rate of flow = 0.41  ft^3/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9 - Pg 103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the net discharge\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "x=10. #in of mercury\n",
-      "sg=13.6 #g/cc\n",
-      "d1=8. #in\n",
-      "d2=4. #in\n",
-      "g=32.2 #ft/s^2\n",
-      "#calculations\n",
-      "vdiff=x/12. *sg- x/12.\n",
-      "Vts=vdiff/(1-math.pow((d2/d1),4))\n",
-      "Vt=math.sqrt(2*g*Vts)\n",
-      "Q=Vt*math.pi/4. *(d2/12.)*(d2/12.)\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Discharge =\",Q,\"ft^3/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge = 2.34 ft^3/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11 - Pg 107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horsepower input of the test pump\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "gam=62.4\n",
-      "ds=12. #in\n",
-      "dd=10. #in\n",
-      "Q=4. #ft^3/s\n",
-      "pd=40. #psia\n",
-      "ps=-6. #psia\n",
-      "zd=5. #ft\n",
-      "zs=0\n",
-      "g=32.2 #ft/s^2\n",
-      "#calculations\n",
-      "vs=Q/(math.pi/4. *math.pow((ds/12.),2))\n",
-      "vd=Q/(math.pi/4. *math.pow((dd/12.),2))\n",
-      "emp = (pd-ps)*144./gam + zd-zs + (vd*vd - vs*vs)/(2*g)\n",
-      "hpp=emp*Q*gam/550.\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Horsepower input of the test pump =\",hpp,\"hp\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horsepower input of the test pump = 50.6 hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12 - Pg 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horizontal and vertical components of force\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "d1=12. #in\n",
-      "d2=8. #in\n",
-      "v1=15. #ft/s\n",
-      "p1=12. #psig\n",
-      "p2=5.85 #psig\n",
-      "rho=1.94 #ft^3/slug\n",
-      "angle=60.*math.pi/180. #degrees\n",
-      "#calculations\n",
-      "Q=math.pi/4. *(d1/12.)*(d1/12.) *v1\n",
-      "v2=Q/(math.pi/4. *(d2/12.)*(d2/12.))\n",
-      "pa1=p1*math.pi/4. *(d1)*(d1)\n",
-      "pa2=p2*math.pi/4. *(d2)*(d2)\n",
-      "qv1=rho*Q*v1\n",
-      "qv2=rho*Q*v2\n",
-      "Fx=pa1+qv1+ math.cos(angle)*(pa2+qv2)\n",
-      "Fy=math.sin(angle)*(pa2+qv2)\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Horizontal component of force =\",Fx,\"lb\")\n",
-      "print '%s %d %s' %(\"\\n Vertical component of force =\",Fy,\"lb\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horizontal component of force = 2232 lb\n",
-        "\n",
-        " Vertical component of force = 922 lb\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14a - Pg 127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the exit velocity of the pump\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "de=4. #in\n",
-      "T=1000. #lb\n",
-      "g=32.2 #ft/s^2\n",
-      "vele=8.5 #lb/s\n",
-      "pe=16.5 #psia\n",
-      "pa=14.7 #psia\n",
-      "#calculations\n",
-      "Ae=math.pi/4. *de*de\n",
-      "Ve= (T-(pe-pa)*Ae)*g/vele\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Exit velocity =\",Ve,\"ft/s\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Exit velocity = 3702 ft/s\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14b - Pg 127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the thrust in the pipe\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "de=4. #in\n",
-      "T=1000. #lb\n",
-      "g=32.2 #ft/s^2\n",
-      "vele=8.5 #lb/s\n",
-      "pe=16.5 #psia\n",
-      "pa=14.7 #psia\n",
-      "pa2=1 #psia\n",
-      "#calculations\n",
-      "Ae=math.pi/4 *de*de\n",
-      "Ve= (T-(pe-pa)*Ae)*g/vele\n",
-      "T2=vele/g *Ve + (pe-pa2)*Ae\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Thrust =\",T2,\"lb\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thrust = 1172 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16 - Pg 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the downstream depth and horsepower dissipation\n",
-      "#Initialization of variables\n",
-      "q=240. #ft^3/sec/ft\n",
-      "v1=60. #ft/s\n",
-      "gam=62.4 \n",
-      "rho=1.94 #slug/ft^3\n",
-      "g=32.2 #ft/s^2\n",
-      "#calculations\n",
-      "y1=q/v1\n",
-      "v2=8.6 #ft/s\n",
-      "y2=28. #ft\n",
-      "hl= (y1+ v1*v1 /(2*g)) - (y2+ v2*v2 /(2*g))\n",
-      "hpp=hl*q*gam/550.\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Downstream depth =\",y2,\"ft\")\n",
-      "print '%s %d %s' %(\"\\n Horsepower dissipation =\",hpp,\"hp per foot width\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Downstream depth = 28.0 ft\n",
-        "\n",
-        " Horsepower dissipation = 837 hp per foot width\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17 - Pg 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the acceleration required\n",
-      "#Initialization of variables\n",
-      "dh=3. #in\n",
-      "L=12. #in\n",
-      "g=32.2 #ft/s^2\n",
-      "#calculations\n",
-      "a=dh/L *g\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Acceleration =\",a,\" ft/s^2\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Acceleration = 8.05  ft/s^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_5.ipynb b/Fluid_mechanics/Chapter_5.ipynb
deleted file mode 100755
index 3591a061..00000000
--- a/Fluid_mechanics/Chapter_5.ipynb
+++ /dev/null
@@ -1,244 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0e0637dca57f500478b778450aaaee0067016f8da4c67cbada5f889a38068276"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5 - Fluid viscosity and flow of real fluids"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the viscosity of oil\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "m=1155. #lb\n",
-      "gam=62.4\n",
-      "spg=0.93\n",
-      "t=3*60. #sec\n",
-      "d=1./6. #in\n",
-      "L=20. #ft\n",
-      "dp=2.5 #psi\n",
-      "#calculations\n",
-      "Q=m/(t*spg*gam)\n",
-      "A=math.pi/4. *d*d\n",
-      "V=Q/A\n",
-      "mu=dp*d*d *144./(32.*V*L)\n",
-      "#results\n",
-      "print '%s %.4f %s' %(\"Viscosity of oil =\",mu,\"lb-sec/ft^2\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Viscosity of oil = 0.0031 lb-sec/ft^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4 - Pg 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the values of alpha and beta\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "import scipy\n",
-      "from scipy import integrate\n",
-      "g=32.2\n",
-      "gam=62.4\n",
-      "r0=1.\n",
-      "#calculations\n",
-      "def func1(r):\n",
-      "    al=8./math.pow(r0,8) *math.pow((r0*r0-r*r),3) *(2*r)\n",
-      "    return al\n",
-      "alpha,err=scipy.integrate.quad(func1,0,r0)\n",
-      "def  func2(r):\n",
-      "    a2=4/math.pow(r0,6) *math.pow((r0*r0 -r*r),2) *(2*r)\n",
-      "    return a2\n",
-      "bet,err2=scipy.integrate.quad(func2,0,r0)\n",
-      "#results\n",
-      "print '%s %d' %(\"Alpha = \",alpha)\n",
-      "print '%s %.2f' %(\"\\n beta = \",bet)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Alpha =  2\n",
-        "\n",
-        " beta =  1.33\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example 5a - Pg 188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the direction of flow and the loss of energy\n",
-      "#Initialization of variables\n",
-      "spg=0.93\n",
-      "mu=3.1e-3 #lb-sec/ft^2\n",
-      "gam=62.4\n",
-      "z=50. #m\n",
-      "p1=60. #psia\n",
-      "p2=25. #psia\n",
-      "#calculations\n",
-      "p1g=144.*p1\n",
-      "p2g=144.*p2 + spg*gam*z\n",
-      "dp=p1g-p2g\n",
-      "#results\n",
-      "if p1g>p2g:\n",
-      "    print '%s' %(\"The flow is in upward direction\")\n",
-      "else:\n",
-      "    print '%s' %(\"The flow is in downward direction\")\n",
-      "\n",
-      "print '%s %d %s' %(\"\\n Energy loss=\",dp,\"ft-lb/ft^3\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The flow is in upward direction\n",
-        "\n",
-        " Energy loss= 2138 ft-lb/ft^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5b - Pg 189"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the flow rate\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "hl=2140. #ft-lb/ft^3\n",
-      "spg=0.93\n",
-      "mu=3.1e-3 #lb-sec/ft^2\n",
-      "gam=62.4\n",
-      "z=50. #m\n",
-      "p1=60. #psia\n",
-      "p2=25. #psia\n",
-      "d=1. #in\n",
-      "#calculations\n",
-      "V= hl*(d/12.)*(d/12.) /(32*mu*z)\n",
-      "Q=V*math.pi/4. *(d/12.)*(d/12.)\n",
-      "Q2=Q*7.48*60.\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Flow rate =\",Q2,\"gal/min\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flow rate = 7.33 gal/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7 - Pg 194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the flow in model\n",
-      "#Initialization of variables\n",
-      "muw=2.04e-5 #lb-sec/ft^2\n",
-      "rhow=1.94 #slugs/ft^3\n",
-      "mua=3.74e-7 #lb-sec/ft^2\n",
-      "rhoa=0.00237 #slug/ft^3\n",
-      "Qw=200. #gal/min\n",
-      "Lr=5.\n",
-      "#calculations\n",
-      "Qa=Qw*Lr *(rhow/rhoa)*(mua/muw)\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Flow in model =\",Qa,\"gal/min\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flow in model = 15007 gal/min\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_6.ipynb b/Fluid_mechanics/Chapter_6.ipynb
deleted file mode 100755
index 424aea7a..00000000
--- a/Fluid_mechanics/Chapter_6.ipynb
+++ /dev/null
@@ -1,118 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:27148bed2cbad8357ddcdc8e0ee434be3797107d2f692797bdf3a1e36c7e8d1a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6 - Dimensional analysis and model similitude"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 230"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the flow rate of water and pressure drop\n",
-      "#Initialization of variables\n",
-      "dg=0.5 #in\n",
-      "dw=12. #in\n",
-      "rhog=0.022 #slug/ft^3\n",
-      "rhow=1.94 #slug/ft^3\n",
-      "muw=2.34e-5 #lb-sec/ft^2\n",
-      "mug=3.50e-7 #lb-sec/ft^2\n",
-      "Qg=0.15 #ft^3/s\n",
-      "dpg=100. #lb/ft^2\n",
-      "#calculations\n",
-      "Vr=dg/dw *rhog/rhow *muw/mug\n",
-      "Qr=Vr*dw*dw /dg/dg\n",
-      "Qw=Qr*Qg\n",
-      "dpr=rhow/rhog *(Vr)*Vr\n",
-      "dpw=dpr*dpg\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Flow rate of water =\",Qw,\"ft^3/s\")\n",
-      "print '%s %.1f %s' %(\"\\n Pressure drop =\",dpw,\"lb/ft^2\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flow rate of water = 2.73 ft^3/s\n",
-        "\n",
-        " Pressure drop = 8.8 lb/ft^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4 - Pg 231"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the velocity ratio, time ratio, acceleration ratio and force ratio.\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "Lr=1./10.\n",
-      "rhom=2.\n",
-      "rhop=1.94\n",
-      "#calculations\n",
-      "Vr=math.sqrt(Lr)\n",
-      "Tr=Lr/Vr\n",
-      "ar=Vr/Tr\n",
-      "Fr=rhom/rhop *ar*Lr*Lr*Lr\n",
-      "#results\n",
-      "print '%s %.4f' %(\"Velocity ratio = \",Vr)\n",
-      "print '%s %.4f' %(\"\\n Time ratio = \",Tr)\n",
-      "print '%s %d' %(\"\\n Acceleration ratio = \",ar)\n",
-      "print '%s %.6f' %(\"\\n Force ratio = \",Fr)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity ratio =  0.3162\n",
-        "\n",
-        " Time ratio =  0.3162\n",
-        "\n",
-        " Acceleration ratio =  1\n",
-        "\n",
-        " Force ratio =  0.001031\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_7.ipynb b/Fluid_mechanics/Chapter_7.ipynb
deleted file mode 100755
index 26326ec8..00000000
--- a/Fluid_mechanics/Chapter_7.ipynb
+++ /dev/null
@@ -1,874 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a67e8661ab863efba6ffe8486e33132e755624949690c50df8ae76d2d928ed8b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7 - Flow of incompressible fluids in closed conduits"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1a - Pg 241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the reynolds number of the flow\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "z1=2. #ft\n",
-      "Q=0.1 #gal/min\n",
-      "alpha=2.\n",
-      "g=32.2 #ft/s^2\n",
-      "L=4. #ft\n",
-      "D=1./96. #ft\n",
-      "#calculations\n",
-      "v2=Q/(7.48*60* math.pi/4. *D*D)\n",
-      "hl=z1-alpha*v2*v2 /(2*g)\n",
-      "Nr=64./hl *L/D *v2*v2 /(2*g)\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Reynolds number is\",Nr,\".Hence the flow is laminar\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reynolds number is 1459 .Hence the flow is laminar\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1b - Pg 242"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the kinematic viscosity\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "z1=2 #ft\n",
-      "Q=0.1 #gal/min\n",
-      "alpha=2.\n",
-      "g=32.2 #ft/s^2\n",
-      "L=4. #ft\n",
-      "D=1./96. #ft\n",
-      "#calculations\n",
-      "v2=Q/(7.48*60* math.pi/4. *D*D)\n",
-      "hl=z1-alpha*v2*v2/(2*g)\n",
-      "Nr=64./hl *L/D *v2*v2 /(2*g)\n",
-      "mu=v2*D/Nr\n",
-      "#results\n",
-      "print '%s %.2e %s' %(\"Kinematic viscosity =\",mu,\"ft^2/s\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Kinematic viscosity = 1.87e-05 ft^2/s\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exaple 1c - Pg 242"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the theoretical entrance transistion length\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "z1=2. #ft\n",
-      "Q=0.1 #gal/min\n",
-      "alpha=2.\n",
-      "g=32.2 #ft/s^2\n",
-      "L=4. #ft\n",
-      "D=1./96. #ft\n",
-      "#calculations\n",
-      "v2=Q/(7.48*60* math.pi/4. *D*D)\n",
-      "hl=z1-alpha*v2*v2 /(2*g)\n",
-      "Nr=64./hl *L/D *v2*v2 /(2*g)\n",
-      "Ld=0.058*Nr*D\n",
-      "#results\n",
-      "print '%s %.3f %s' %(\"Theoretical entrance transistion length =\",Ld,\"ft\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Theoretical entrance transistion length = 0.882 ft\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2 - Pg 246"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horsepower required for the motor\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "Q=350. #gal/min\n",
-      "D=6. #in\n",
-      "rho=0.84\n",
-      "gam=62.4\n",
-      "g=32.2 #ft/s^2\n",
-      "mu=9.2e-5 #lb-sec/ft^2\n",
-      "L=5280. #ft\n",
-      "#calculations\n",
-      "V=Q/(7.48*60*math.pi/4. *math.pow((D/12),2))\n",
-      "Nr=V*D/12 *rho*gam/g /mu\n",
-      "f=0.3164/math.pow((Nr),0.25)\n",
-      "hl=f*L*12/D *V*V /(2*g)\n",
-      "hp=hl*gam*Q*rho/(550*7.48*60.)\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Horsepower required =\",hp,\"hp/mile\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horsepower required = 4.44 hp/mile\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 250"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the values of the coefficients alpha and beta\n",
-      "#Initialization of variables\n",
-      "n=7.\n",
-      "import math\n",
-      "#calculations\n",
-      "alpha= math.pow((n+1),3) *math.pow((2*n+1),3) /(4*math.pow(n,4) *(n+3)*(2*n+3))\n",
-      "bet=math.pow((n+1),2) *math.pow((2*n+1),2) /(2*n*n *(n+2)*(2*n+2))\n",
-      "#results\n",
-      "print '%s %.2f' %(\"alpha = \",alpha)\n",
-      "print '%s %.2f' %(\"\\n beta = \",bet)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "alpha =  1.06\n",
-        "\n",
-        " beta =  1.02\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5 - Pg 252"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the mass flow rate and horsepower input of the fan\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "spg=0.84\n",
-      "z=1. #in\n",
-      "gam=62.4\n",
-      "patm=14.7 #psia\n",
-      "T=459.6+85 #R\n",
-      "R=53.3\n",
-      "g=32.2 #ft/s^2\n",
-      "D=3. #ft\n",
-      "mu=3.88e-7 #lb-sec/ft^2\n",
-      "#calculations\n",
-      "dp=spg*z/12. *gam\n",
-      "rho=patm*144. /(R*T*g)\n",
-      "umax=math.sqrt(2*dp/rho)\n",
-      "V=0.8*umax\n",
-      "Nr=V*D*rho/mu\n",
-      "V2=0.875*umax\n",
-      "mass=rho*math.pi/4. *D*D *V2\n",
-      "emf=V2*V2 /(2*g)\n",
-      "hp=emf*mass*g/550.\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Mass flow rate =\",mass,\"slug/sec\")\n",
-      "print '%s %.2f %s' %(\"\\n Horsepower input of the fan =\",hp,\"hp\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mass flow rate = 0.87 slug/sec\n",
-        "\n",
-        " Horsepower input of the fan = 2.34 hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7a - Pg 266"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the velocity using several equations listed above\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "D=36. #in\n",
-      "rho=0.00226  #slug/ft^3\n",
-      "mu=3.88e-7 #lb-sec/ft^2\n",
-      "umax=62.2 #ft/s\n",
-      "V=54.5 #ft/s\n",
-      "Nr=9.5e5\n",
-      "r0=18. #in\n",
-      "r=12. #in\n",
-      "n=8.8\n",
-      "k=0.4\n",
-      "#calculations\n",
-      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
-      "Vs=math.sqrt(f/8) *V\n",
-      "y=r0-r\n",
-      "u1=umax*math.pow((y/r0),(1/n))\n",
-      "u2=umax+ 2.5*Vs*math.log(y/r0)\n",
-      "u3=umax+ Vs/k *(math.sqrt(1-y/r0) + math.log(1-math.sqrt(1-y/r0)))\n",
-      "u4=Vs*(5.5+ 5.75*math.log10(Vs*y/12 *rho/mu))\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Using equation 7-13, velocity =\",u1,\" ft/s\")\n",
-      "print '%s %.1f %s' %(\"\\n Using equation 7-18, velocity =\",u2,\" ft/s\")\n",
-      "print '%s %.1f %s' %(\"\\n Using equation 7-25, velocity =\",u3,\" ft/s\")\n",
-      "print '%s %.1f %s' %(\"\\n Using equation 7-34a, velocity =\",u4,\" ft/s\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Using equation 7-13, velocity = 54.9  ft/s\n",
-        "\n",
-        " Using equation 7-18, velocity = 56.5  ft/s\n",
-        "\n",
-        " Using equation 7-25, velocity = 57.6  ft/s\n",
-        "\n",
-        " Using equation 7-34a, velocity = 56.7  ft/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7b - Pg 266"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the outer edge buffer zone distance and also its thickness\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "D=36. #in\n",
-      "rho=0.00226  #slug/ft^3\n",
-      "mu=3.88e-7 #lb-sec/ft^2\n",
-      "umax=62.2 #ft/s\n",
-      "V=54.5 #ft/s\n",
-      "Nr=9.5e5\n",
-      "r0=18. #in\n",
-      "r=12. #in\n",
-      "n=8.8\n",
-      "k=0.4\n",
-      "#calculations\n",
-      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
-      "Vs=math.sqrt(f/8.) *V\n",
-      "y=r0-r\n",
-      "delta1=D*5*math.sqrt(8) /(Nr*math.sqrt(f))\n",
-      "vss=70.\n",
-      "thick=13*delta1\n",
-      "#results\n",
-      "print '%s %d' %(\"Outer edge of buffer zone is at \",vss)\n",
-      "print '%s %.4f %s' %(\"\\n Thickness of buffer zone =\",thick,\"in\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Outer edge of buffer zone is at  70\n",
-        "\n",
-        " Thickness of buffer zone = 0.0645 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7c - Pg 266"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the velocity using the given equations\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "D=36. #in\n",
-      "rho=0.00226  #slug/ft^3\n",
-      "mu=3.88e-7 #lb-sec/ft^2\n",
-      "umax=62.2 #ft/s\n",
-      "V=54.5 #ft/s\n",
-      "Nr=9.5e5\n",
-      "r0=18. #in\n",
-      "r=12. #in\n",
-      "n=8.8\n",
-      "k=0.4\n",
-      "#calculations\n",
-      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
-      "Vs=math.sqrt(f/8) *V\n",
-      "delta1=D*5*math.sqrt(8.) /(Nr*math.sqrt(f))\n",
-      "y=delta1\n",
-      "u2=Vs*Vs *delta1/12. *rho/mu\n",
-      "u1=62.2 *math.pow((delta1/18.),(1/n))\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"using equation 7-13, velocity =\",u1,\"ft/s\")\n",
-      "print '%s %.1f %s' %(\"\\n using equation 7-30, velocity =\",u2,\"ft/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "using equation 7-13, velocity = 24.5 ft/s\n",
-        "\n",
-        " using equation 7-30, velocity = 10.4 ft/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7d - Pg 266"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the velocity using the listed equations\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "D=36. #in\n",
-      "rho=0.00226  #slug/ft^3\n",
-      "mu=3.88e-7 #lb-sec/ft^2\n",
-      "umax=62.2 #ft/s\n",
-      "V=54.5 #ft/s\n",
-      "Nr=9.5e5\n",
-      "r0=18. #in\n",
-      "r=12. #in\n",
-      "n=8.8\n",
-      "k=0.4\n",
-      "#calculations\n",
-      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
-      "Vs=math.sqrt(f/8.) *V\n",
-      "delta1=D*5*math.sqrt(8.) /(Nr*math.sqrt(f))\n",
-      "y=14*delta1\n",
-      "u2=62.2*math.pow((y/18.),(1/n))\n",
-      "u3=Vs*(5.50 + 5.75*math.log10(Vs*y/12 *rho/mu))\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Using equation 7-13, velocity =\",u2,\"ft/s\")\n",
-      "print '%s %.1f %s' %(\"\\n using equation 7-34a, velocity =\",u3,\"ft/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Using equation 7-13, velocity = 33.1 ft/s\n",
-        "\n",
-        " using equation 7-34a, velocity = 33.5 ft/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7e - Pg 266"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the shearing stress using both the equations\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "D=36. #in\n",
-      "rho=0.00226  #slug/ft^3\n",
-      "mu=3.88e-7 #lb-sec/ft^2\n",
-      "umax=62.2 #ft/s\n",
-      "V=54.5 #ft/s\n",
-      "Nr=9.5e+5\n",
-      "r0=18. #in\n",
-      "r=12. #in\n",
-      "n=8.8\n",
-      "k=0.4\n",
-      "#calculations\n",
-      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
-      "Vs=math.sqrt(f/8.) *V\n",
-      "delta1=D*5*math.sqrt(8.) /(Nr*math.sqrt(f))\n",
-      "u2=Vs*Vs *delta1/12. *rho/mu\n",
-      "T0=rho*Vs*Vs\n",
-      "T02=mu*u2/delta1 *12.\n",
-      "#results\n",
-      "print '%s %.5f %s' %(\"Using equation 7-9a, shearing stress =\",T0,\"lb/ft^2\")\n",
-      "print '%s %.5f %s' %(\"\\n Using equation 7-28, shearing stress =\",T02,\"lb/ft^2\")\n",
-      "print '%s' %(\"The answers are a bit different due to rounding off error in textbook\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Using equation 7-9a, shearing stress = 0.00979 lb/ft^2\n",
-        "\n",
-        " Using equation 7-28, shearing stress = 0.00979 lb/ft^2\n",
-        "The answers are a bit different due to rounding off error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8 - Pg 273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the required velocity\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "umax=62.2 #ft/s\n",
-      "r0=18. #in\n",
-      "e=0.0696 #in\n",
-      "r=6. #in\n",
-      "#calculations\n",
-      "Vs=umax/(8.5 + 5.75*math.log10(r0/e))\n",
-      "u=Vs*(8.5 + 5.75*math.log10(r/e))\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Velocity =\",u,\"ft/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity = 54.6 ft/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9 - Pg 277"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the flow rate and roughness factor\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "d=8. #in\n",
-      "V=3.65 #ft/s\n",
-      "u1=4.75 #ft/s\n",
-      "r0=4. #in\n",
-      "#calculations\n",
-      "f=0.0449\n",
-      "Q=V*math.pi/4. *math.pow((d/12),2)\n",
-      "Vs=(u1-V)/3.75\n",
-      "r0e=math.pow(10,((u1/Vs - 8.5)/5.75))\n",
-      "e=r0/r0e\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Flow rate =\",Q,\"ft^3/s\")\n",
-      "print '%s %.3f %s' %(\"\\n roughness factor =\",e,\"in\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flow rate = 1.27 ft^3/s\n",
-        "\n",
-        " roughness factor = 0.184 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10 - Pg 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure difference per foot of horizontal pipe\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "e0=0.00085 #ft\n",
-      "alpha=0.25 #/year\n",
-      "t=15. #years\n",
-      "r0=3. #in\n",
-      "Q=500. #gal/min\n",
-      "d=6. #in\n",
-      "mu=2.04e-5 #lb-sec/ft^2\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "g=32.2 #ft/s^2\n",
-      "L=1. #ft\n",
-      "gam=62.4\n",
-      "#calculations\n",
-      "e15=e0*(1+ alpha*t)\n",
-      "ratio=r0/(12.*e15)\n",
-      "V=Q/(7.48*60*math.pi/4. *math.pow((d/12),2))\n",
-      "Nr=V*d*rho/(mu*12)\n",
-      "f=0.036\n",
-      "hl=f*L/(d/12.) *V*V /(2*g)\n",
-      "dp=gam*hl\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Pressure difference =\",dp,\"lb/ft^2 per foot of horizontal pipe\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure difference = 2.25 lb/ft^2 per foot of horizontal pipe\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11 - Pg 289"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horsepower required\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "d2=4. #in\n",
-      "d1=3. #in\n",
-      "e=0.0005 #ft\n",
-      "mu=3.75e-5 #lb-sec/ft^2\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "Q=100. #gal/min\n",
-      "L=100. #ft\n",
-      "g=32.2 #ft/s^2\n",
-      "gam=62.4\n",
-      "#calculations\n",
-      "A=math.pi/4. *(math.pow((d2/12),2) -math.pow((d1/12),2))\n",
-      "WP=math.pi*(d1+d2)/12.\n",
-      "R=A/WP\n",
-      "RR= 2*R/e\n",
-      "V= Q/(7.48*60*A)\n",
-      "Nr=V*4*R*rho/mu\n",
-      "f=0.035\n",
-      "hl=f*L/(4*R) *V*V /(2*g)\n",
-      "hp=hl*Q/(7.48*60) *gam/550.\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"horsepower required =\",hp,\" hp/100 ft\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "horsepower required = 0.56  hp/100 ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12 - Pg 296"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge flow\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "p1=25. #psig\n",
-      "p2=20. #psig\n",
-      "d1=18. #in\n",
-      "d2=12. #in\n",
-      "Cl=0.25\n",
-      "gam=62.4\n",
-      "g=32.2 #ft/s^2\n",
-      "#calculations\n",
-      "Vr=(d2/d1)*(d2/d1)\n",
-      "xv=(p2-p1)*144/gam\n",
-      "V22=xv/(-1-Cl+Vr*Vr) *2*g\n",
-      "V2=math.sqrt(V22)\n",
-      "Q=V2*math.pi/4. *(d2/12.)*(d2/12.)\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Discharge =\",Q,\"ft^3/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge = 20.9 ft^3/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13 - Pg 300"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the discharge flow rate\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "V61=10.8 #ft/s\n",
-      "V81=6.05 #ft/s\n",
-      "r0=3 #in\n",
-      "e=0.00015\n",
-      "d1=6. #in\n",
-      "rho=1.94 #slugs/ft^3\n",
-      "mu=2.34e-5 #ft-lb/s^2\n",
-      "#calculations\n",
-      "roe=r0/(12*e)\n",
-      "Nr1=V61*(d1/12.)*rho/mu\n",
-      "f6=0.0165\n",
-      "V6=11.6 #ft/s\n",
-      "V8=6.52 #ft/s\n",
-      "Q=V6*math.pi/4 *(d1/12.)*(d1/12.)\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Discharge =\",Q,\"ft^3/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Discharge = 2.28 ft^3/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14 - Pg 302"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the diameter of steel pipe\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "L=1000. #ft\n",
-      "Q=2000/(7.48*60) #ft63/s\n",
-      "g=32.2 #ft/s^2\n",
-      "p=5. #psi/1000 ft\n",
-      "gam=62.4\n",
-      "sp=0.7\n",
-      "f=0.02\n",
-      "r0=0.904/2.\n",
-      "e=0.00015\n",
-      "mu=7e-6 #lb-ft/s^2\n",
-      "L=1000. #ft\n",
-      "#calculations\n",
-      "hl=p*144/(sp*gam)\n",
-      "D5=f*8*L*Q*Q /(math.pi*math.pi *g*hl)\n",
-      "D=math.pow(D5,(1./5.))\n",
-      "Nr=4*Q*sp*gam/(g*(math.pi*D*mu))\n",
-      "f2=0.0145\n",
-      "D5=f2*8*L*Q*Q /(math.pi*math.pi *g*hl)\n",
-      "D1=math.pow(D5,(1./5.))\n",
-      "#results\n",
-      "print '%s %.3f %s' %(\"Diameter of steel pipe =\",D1,\" ft\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diameter of steel pipe = 0.848  ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_8.ipynb b/Fluid_mechanics/Chapter_8.ipynb
deleted file mode 100755
index 76c114d3..00000000
--- a/Fluid_mechanics/Chapter_8.ipynb
+++ /dev/null
@@ -1,802 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:f292d23876c88421bcfc8b761169402a2885837db02032e303215ae83f2c5cf5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8 - Fluid compressibility and incompressible flow"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1 - Pg 323"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the initial and final volumes. Also, calculate the final temperature of the gas\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "pi=14.7 #psia\n",
-      "pf=50. #psia\n",
-      "cp=0.240 #Btu/lb R\n",
-      "cv=0.170 #Btu/lb R\n",
-      "J=778\n",
-      "T=60+459.6 #R\n",
-      "#calculations\n",
-      "R=J*(cp-cv)\n",
-      "k=cp/cv\n",
-      "gam=pi*144./(R*T)\n",
-      "V=1/gam\n",
-      "Vf=V*math.pow((pi/pf),(1/k))\n",
-      "Tf=T*(pf*Vf/(pi*V))\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Initial volume =\",V,\"ft^3\")\n",
-      "print '%s %.2f %s' %(\"\\n Final volume =\",Vf,\"cu ft\")\n",
-      "print '%s %.1f %s' %(\"\\n Final temperature =\",Tf,\" R\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Initial volume = 13.37 ft^3\n",
-        "\n",
-        " Final volume = 5.62 cu ft\n",
-        "\n",
-        " Final temperature = 742.6  R\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2 - Pg 325"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure difference across the pipe\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "ratio=0.99\n",
-      "E=3.19e5 #lb/in^2\n",
-      "#calculations\n",
-      "pd=-E*math.log(ratio)\n",
-      "#ersults\n",
-      "print '%s %d %s' %(\"Pressure difference =\",pd,\"psi\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure difference = 3206 psi\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 329"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the speed of test plane\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "g=32.2 #ft/s^2\n",
-      "R=53.3 #ft-lb/lb R\n",
-      "T=389.9 #R\n",
-      "Nm=2\n",
-      "#calculations\n",
-      "c=math.sqrt(k*g*R*T)\n",
-      "V=Nm*c*3600/5280.\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Speed of test plane =\",V,\"mph\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed of test plane = 1319.9 mph\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4a - Pg 337"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the velocity at section 2\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "T1=584.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "k=1.4\n",
-      "R=53.3 #ft-lb/lb R\n",
-      "V1=600 #ft/s\n",
-      "T2=519.6 #R\n",
-      "#calculations\n",
-      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
-      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
-      "Nm2=math.sqrt(Nm22)\n",
-      "V2=Nm2*math.sqrt(k*g*R*T2)\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Velocity at section 2 =\",V2,\"ft/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity at section 2 = 1068 ft/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4b - Pg 337"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure difference between two stations\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "T1=584.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "k=1.4\n",
-      "R=53.3 #ft-lb/lb R\n",
-      "V1=600 #ft/s\n",
-      "T2=519.6 #R\n",
-      "pa=14.7 #psi\n",
-      "p1=50 #psia\n",
-      "#calculations\n",
-      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
-      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
-      "Nm2=math.sqrt(Nm22)\n",
-      "pr=math.pow(((1+ (k-1)/2 *Nm1*Nm1)/(1+ (k-1)/2 *Nm2*Nm2)),(k/(k-1)))\n",
-      "p2=pr*(p1+pa)\n",
-      "dp=p1+pa-p2\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Pressure difference between two stations =\",dp,\"psi\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure difference between two stations = 21.9 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4c - Pg 338"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the area ratio\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "T1=584.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "k=1.4\n",
-      "R=53.3 #ft-lb/lb R\n",
-      "V1=600 #ft/s\n",
-      "T2=519.6 #R\n",
-      "#calculations\n",
-      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
-      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
-      "Nm2=math.sqrt(Nm22)\n",
-      "Ar= Nm1/Nm2 *math.pow(((1+ (k-1)/2 *Nm2*Nm2)/(1+ (k-1)/2 *Nm1*Nm1)),((k+1)/(2*(k-1))))\n",
-      "#results\n",
-      "print '%s %.3f' %(\"Area ratio = \",Ar)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Area ratio =  0.754\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4d - Pg 338"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the density of air at both the stations\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "T1=584.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "k=1.4\n",
-      "R=53.3 #ft-lb/lb R\n",
-      "V1=600 #ft/s\n",
-      "T2=519.6 #R\n",
-      "pa=14.7 #psi\n",
-      "p1=50 #psia\n",
-      "#calculations\n",
-      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
-      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
-      "Nm2=math.sqrt(Nm22)\n",
-      "pr=math.pow(((1+ (k-1)/2 *Nm1*Nm1)/(1+ (k-1)/2 *Nm2*Nm2)),(k/(k-1)))\n",
-      "p2=pr*(p1+pa)\n",
-      "rho1=(p1+pa)*144./(g*R*T1)\n",
-      "rho2=p2*144./(g*R*T2)\n",
-      "#results\n",
-      "print '%s %.5f %s' %(\"Density of air at station 1 =\",rho1,\"slug/ft^3\")\n",
-      "print '%s %.5f %s' %(\"\\n Density of air at station 2 =\",rho2,\"slug/ft^3\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Density of air at station 1 = 0.00929 slug/ft^3\n",
-        "\n",
-        " Density of air at station 2 = 0.00692 slug/ft^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5 - Pg 345"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the mass rate of air flow\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "p0=19.7 #psia\n",
-      "R=53.3 #lb-ft/lb-R\n",
-      "T0=539.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "pa=14.7 #psia\n",
-      "d=1 #in\n",
-      "k=1.4\n",
-      "#calculations\n",
-      "rho0=p0*144/(g*R*T0)\n",
-      "pr=pa/p0\n",
-      "G=math.pi/4 *(d/12.)*(d/12.) *math.pow((2*k/(k-1) *p0*144*rho0),(0.5)) *math.pow((pr),(1/k)) *math.pow((1-math.pow(pr,((k-1)/k))),0.5)\n",
-      "#results\n",
-      "print '%s %.5f %s' %(\"Mass rate of air flow =\",G,\"slug/sec\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mass rate of air flow = 0.00978 slug/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6 - Pg 346"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the mass rate of air flow\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "p0=64.7 #psia\n",
-      "R=53.3 #lb-ft/lb-R\n",
-      "T0=539.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "pa=14.7 #psia\n",
-      "d=1. #in\n",
-      "k=1.4\n",
-      "#calculations\n",
-      "rho0=p0*144/(g*R*T0)\n",
-      "pr=pa/p0\n",
-      "G=math.pi/4 *(d/12)*d/12. *math.pow((k*p0*144*rho0),(0.5)) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
-      "#results\n",
-      "print '%s %.5f %s' %(\"Mass rate of air flow =\",G,\"slug/sec\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mass rate of air flow = 0.03616 slug/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7a - Pg 347"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the weight of air flow through the nozzle\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "R=53.3 #lb-ft/lb R\n",
-      "pe=14.7 #psia\n",
-      "p0=114.7 #psia\n",
-      "T0=524.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "d=0.5 #in\n",
-      "#calculations\n",
-      "pr=pe/p0\n",
-      "prcr=0.528\n",
-      "pr=prcr*p0\n",
-      "rho0= p0*144/(g*R*T0)\n",
-      "G=math.pi/4 *(d/12)*d/12. *math.pow((k*p0*144*rho0),(0.5)) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
-      "Wt=G*g\n",
-      "#results\n",
-      "print '%s %.4f %s' %(\"weight of air flow through the nozzle =\",Wt,\"lb/s\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "weight of air flow through the nozzle = 0.5233 lb/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7b - Pg 348"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the exit Mach number, exit velocity and the exit area\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "R=53.3 #lb-ft/lb R\n",
-      "pe=14.7 #psia\n",
-      "p0=114.7 #psia\n",
-      "T0=524.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "d=0.5 #in\n",
-      "Nm1=1.\n",
-      "#calculations\n",
-      "pr=pe/p0\n",
-      "Nme=math.sqrt(2/(k-1) *(math.pow(1./pr,(k-1)/k) -1))\n",
-      "Te=T0/(1+ (k-1)/2 *Nme*Nme)\n",
-      "Ve=Nme*math.sqrt(k*g*R*Te)\n",
-      "At=math.pi/4. *(d)*d\n",
-      "Ae=Nm1/Nme *math.pow(((1+ (k-1)/2 *Nme*Nme)/(1+ (k-1)/2 *Nm1*Nm1)),((k+1)/(2*(k-1)))) *At\n",
-      "#results\n",
-      "print '%s %.2f' %(\"Mach number exit = \",Nme)\n",
-      "print '%s %d %s' %(\"\\n Exit velocity =\",Ve,\"ft/s\")\n",
-      "print '%s %.3f %s' %(\"\\n Exit area =\",Ae,\" in^2\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mach number exit =  2.00\n",
-        "\n",
-        " Exit velocity = 1672 ft/s\n",
-        "\n",
-        " Exit area = 0.331  in^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8a - Pg 349"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the exit mass flow rate, pressure, temperature, velocity and mach number\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "R=53.3 #lb-ft/lb R\n",
-      "p0=100. #psia\n",
-      "T0=534.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "d=0.5 #in\n",
-      "Nm1=1.\n",
-      "A=2./144. #ft^2\n",
-      "#calculations\n",
-      "print '%s' %(\"Exit mach number is found using trial and error\")\n",
-      "Nme=2.44\n",
-      "rho0=p0*144/(g*R*T0)\n",
-      "G= A*math.sqrt(k*p0*144*rho0) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
-      "pe=p0*math.pow((1/(1+(k-1)/2 *Nme*Nme)),(k/(k-1)))\n",
-      "Te=T0/(1+ (k-1)/2 *Nme*Nme)\n",
-      "Ve=Nme*(math.sqrt(k*g*R*Te))\n",
-      "#results\n",
-      "print '%s %.3f %s' %(\"\\n Exit mass flow rate =\",G,\"slug/s\")\n",
-      "print '%s %.2f %s' %(\"\\n Exit pressure =\",pe,\"psia\")\n",
-      "print '%s %.1f %s' %(\"\\n Exit temperature =\",Te,\" R\")\n",
-      "print '%s %d %s' %(\"\\n Exit velocity =\",Ve,\"ft/s\")\n",
-      "print '%s %.2f' %(\"\\n Exit mach number = \",Nme)\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Exit mach number is found using trial and error\n",
-        "\n",
-        " Exit mass flow rate = 0.143 slug/s\n",
-        "\n",
-        " Exit pressure = 6.43 psia\n",
-        "\n",
-        " Exit temperature = 244.0  R\n",
-        "\n",
-        " Exit velocity = 1868 ft/s\n",
-        "\n",
-        " Exit mach number =  2.44\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8b - Pg 350"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the exit mass flow rate, pressure, temperature,velocity and mach number\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "R=53.3 #lb-ft/lb R\n",
-      "p0=100. #psia\n",
-      "T0=534.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "d=0.5 #in\n",
-      "Nm1=1.\n",
-      "A=2./144. #ft^2\n",
-      "#calculations\n",
-      "print '%s' %(\"Exit mach number is found using trial and error\")\n",
-      "Nme=0.24\n",
-      "rho0=p0*144/(g*R*T0)\n",
-      "G= A*math.sqrt(k*p0*144*rho0) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
-      "pe=p0*math.pow((1/(1+(k-1)/2 *Nme*Nme)),(k/(k-1)))\n",
-      "Te=T0/(1+ (k-1)/2 *Nme*Nme)\n",
-      "Ve=Nme*(math.sqrt(k*g*R*Te))\n",
-      "#results\n",
-      "print '%s %.3f %s' %(\"\\n Exit mass flow rate =\",G,\"slug/s\")\n",
-      "print '%s %.2f %s' %(\"\\n Exit pressure =\",pe,\"psia\")\n",
-      "print '%s %.1f %s' %(\"\\n Exit temperature =\",Te,\" R\")\n",
-      "print '%s %d %s' %(\"\\n Exit velocity =\",Ve,\"ft/s\")\n",
-      "print '%s %.2f' %(\"\\n Exit mach number = \",Nme)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Exit mach number is found using trial and error\n",
-        "\n",
-        " Exit mass flow rate = 0.143 slug/s\n",
-        "\n",
-        " Exit pressure = 96.07 psia\n",
-        "\n",
-        " Exit temperature = 528.5  R\n",
-        "\n",
-        " Exit velocity = 270 ft/s\n",
-        "\n",
-        " Exit mach number =  0.24\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9 - Pg 355"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the upstream Mach number, pressure, Temperature\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "R=53.3 #lb-ft/lb R\n",
-      "pu=6.43 #psia\n",
-      "Tu=244. #R\n",
-      "Nmu=2.44\n",
-      "#calculations\n",
-      "Nmd = math.sqrt(((k-1)*Nmu*Nmu +2)/(2*k*Nmu*Nmu - (k-1)))\n",
-      "pd=pu*(2*k*Nmu*Nmu - (k-1))/(k+1)\n",
-      "Td=Tu*(2*k*Nmu*Nmu - (k-1))/(k+1) *((k-1)*Nmu*Nmu +2)/((k+1)*Nmu*Nmu)\n",
-      "#results\n",
-      "print '%s %.3f' %(\"Mach number upstream = \",Nmd)\n",
-      "print '%s %.1f %s' %(\"\\n Pressure upstream =\",pd,\"psia\")\n",
-      "print '%s %.1f %s' %(\"\\n Temperature upstream =\",Td,\"R\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mach number upstream =  0.519\n",
-        "\n",
-        " Pressure upstream = 43.6 psia\n",
-        "\n",
-        " Temperature upstream = 507.2 R\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10 - Pg 359"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the pressure and temperature at section 1\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "R=53.3 #lb-ft/lb R\n",
-      "e=0.0005 #ft\n",
-      "mu=3.77e-7 #lb-sec/ft^2\n",
-      "pe=14.7 #psia\n",
-      "Te=524.6 #R\n",
-      "g=32.2 #ft/s^2\n",
-      "Vi=12.5 #ft/s\n",
-      "l=6. #in\n",
-      "b=8. #in\n",
-      "L=100. #ft\n",
-      "#calculations\n",
-      "rhoe=pe*144/(R*g*Te)\n",
-      "Ve=Vi/(g*rhoe*(l*b/144.))\n",
-      "Nme=Ve/(math.sqrt(k*g*R*Te))\n",
-      "Rd=l*b/(2*(l+b)) /12.\n",
-      "rr=2*R/e\n",
-      "Nr=Ve*4*Rd*rhoe/mu\n",
-      "f=0.019\n",
-      "f2=1/(2*k) *(1/Nme*Nme -1) - (k+1)/(4*k) *math.log((1+ (k-1)/2 *Nme*Nme)/(Nme*Nme *(1+(k-1)/2)))\n",
-      "ff=f*L/(8*Rd) +f2\n",
-      "Nm1=0.305\n",
-      "Tr2=(1+ (k-1)/2 *Nm1*Nm1)/(1+ (k-1/2))\n",
-      "Tre=(1+ (k-1)/2 *Nme*Nme)/(1+ (k-1/2))\n",
-      "pr2=Nm1*math.pow((1+ (k-1)/2 *Nm1*Nm1),(0.5)) /math.pow((1+(k-1)/2),0.5)\n",
-      "pre=Nme*math.pow((1+ (k-1)/2 *Nme*Nme),(0.5)) /math.pow((1+(k-1)/2),0.5)\n",
-      "p1=pe/pr2 *pre\n",
-      "T1=Te/Tr2 *Tre\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Pressure at section 1 =\",p1,\"psia\")\n",
-      "print '%s %.1f %s' %(\"\\n Tempreature at section 1 =\",T1,\"R\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pressure at section 1 = 21.5 psia\n",
-        "\n",
-        " Tempreature at section 1 = 535.1 R\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11 - Pg 364"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the limiting pressure and distance in both adiabatic and non-adiabatic cases\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "k=1.4\n",
-      "R=53.3 #lb-ft/lb R\n",
-      "g=32.2 #ft/s^2\n",
-      "T1=534.6 #R\n",
-      "V1=400. #ft/s\n",
-      "p1=350. #psia\n",
-      "f=0.02\n",
-      "D=6./12. #ft\n",
-      "#calculations\n",
-      "Nm1=V1/math.sqrt(k*g*R*T1)\n",
-      "Nm2= 1. /math.sqrt(k)\n",
-      "p2=p1*(Nm1)/Nm2\n",
-      "fl= math.log(Nm1/Nm2) + 1./(2.*k*Nm1*Nm1) *(1- Nm1*Nm1/Nm2*Nm2)\n",
-      "L12=fl*2*D/f\n",
-      "ps=p1*Nm1*math.pow((1+ (k-1)/2 *Nm1*Nm1),0.5) /math.pow((1+(k-1)/2),0.5)\n",
-      "Nm2=1.\n",
-      "fl2= -(k+1)/(4*k) *math.log((1+ (k-1)/2 *Nm1*Nm1)/(Nm1*Nm1 *(1+ (k-1)/2.))) + 1/(2*k*Nm1*Nm1) *(1- Nm1*Nm1 /Nm2*Nm2)\n",
-      "L2=fl2*2*D/f\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Limiting pressure =\",p2,\" psia\")\n",
-      "print '%s %.1f %s' %(\"\\n Distance =\",L12,\" ft\")\n",
-      "print '%s %.1f %s' %(\"\\n Limiting pressure in adiabatic case =\",ps,\" psia\")\n",
-      "print '%s %.1f %s' %(\"\\n Distance required =\",L2,\"ft\")\n",
-      "print '%s' %(\"the answer is a bit different due to rounding of error in textbook\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Limiting pressure = 146.2  psia\n",
-        "\n",
-        " Distance = 81.8  ft\n",
-        "\n",
-        " Limiting pressure in adiabatic case = 114.2  psia\n",
-        "\n",
-        " Distance required = 84.2 ft\n",
-        "the answer is a bit different due to rounding of error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/Chapter_9.ipynb b/Fluid_mechanics/Chapter_9.ipynb
deleted file mode 100755
index f131ead9..00000000
--- a/Fluid_mechanics/Chapter_9.ipynb
+++ /dev/null
@@ -1,450 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:6e08f75972be45750168908e0258e415f396bd3e7aae6981e62e0a4fd06177dc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9 - Fluid flow about immersed bodies"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1a - Pg 389"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the values of Fd\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "import numpy\n",
-      "rho=2.45 #slugs/ft^3\n",
-      "mu=9.2e-3 #lb-sec/ft^2\n",
-      "x=3.\n",
-      "v=3. #ft/s\n",
-      "B=6./12. #ft\n",
-      "L=36./12. #ft\n",
-      "#calculatons\n",
-      "Nr=v*x*rho/mu\n",
-      "y=([1.32, 1.46, 1.328])\n",
-      "Cd = numpy.zeros(len(y))\n",
-      "for i in range (0, len(y)):\n",
-      "\tCd[i]=y[i]*math.pow(Nr,(-0.5))\n",
-      "Fd = numpy.zeros(len(Cd))\n",
-      "for i in range (0, len(y)):\n",
-      "\tFd[i]=2*Cd[i]*B*L*(0.5*rho*v*v)\n",
-      "\n",
-      "#results\n",
-      "print '%s' %(\"Drag on the plates using different formulae blasius, parabola and pohlhauser in order\")\n",
-      "print '%.3f %.3f %.3f' %(Fd[0],Fd[1],Fd[2])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drag on the plates using different formulae blasius, parabola and pohlhauser in order\n",
-        "0.892 0.986 0.897\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1b - Pg 390"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calcualte the boundary layer thickness and the shearing thickness\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "import numpy\n",
-      "x=36./12.\n",
-      "rho=2.45 #slugs/ft^3\n",
-      "mu=9.2e-3 #lb-sec/ft^2\n",
-      "v=3. #ft/s\n",
-      "#calculatons\n",
-      "Nr=v*x*rho/mu\n",
-      "z=([4.91, 5.48, 4.65])\n",
-      "x=36./12.\n",
-      "delta = numpy.zeros(len(z))\n",
-      "for i in range (0, len(z)):\n",
-      "\tdelta[i]=z[i] /math.sqrt(Nr) *x\n",
-      "\n",
-      "\n",
-      "f=([0.332, 0.365, 0.322])\n",
-      "T=numpy.zeros(len(f))\n",
-      "for i in range (0,len(f)):\n",
-      "\tT[i]=f[i]*mu*v/x *math.sqrt(Nr)\n",
-      "\n",
-      "#results\n",
-      "print '%s' %(\"Boundary layer thickness = \")\n",
-      "print '%s' %(\"In order of Blasius, parabola and pohlhauser\")\n",
-      "print '%.3f %.3f %.3f' %(delta[0],delta[1],delta[2])\n",
-      "print '%s' %(\"Shearing stress = \")\n",
-      "print '%s' %(\"In order of Blasius, parabola and pohlhauser\")\n",
-      "print '%.3f %.3f %.3f' %(T[0],T[1],T[2])\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boundary layer thickness = \n",
-        "In order of Blasius, parabola and pohlhauser\n",
-        "0.301 0.336 0.285\n",
-        "Shearing stress = \n",
-        "In order of Blasius, parabola and pohlhauser\n",
-        "0.150 0.164 0.145\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2a - Pg 398"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the total frictional drag and horsepower required\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "e=0.01 #ft\n",
-      "rho=2 #slugs/ft^3\n",
-      "mu=2.6e-5 #lb sec/ft^2\n",
-      "speed=10. #knots\n",
-      "L=250. #ft\n",
-      "A=30000. #ft^2\n",
-      "#calculations\n",
-      "V=speed*1.69\n",
-      "Nrl=V*L*rho/mu\n",
-      "Cdf=1.32 /math.sqrt(Nrl)\n",
-      "Fd=Cdf*A*0.5*rho*V*V\n",
-      "hp=Fd*V/550.\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Total frictional drag =\",Fd,\"lb\")\n",
-      "print '%s %.1f %s' %(\"\\n Horsepower required =\",hp,\"hp\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total frictional drag = 627 lb\n",
-        "\n",
-        " Horsepower required = 19.3 hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2b - Pg 397"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the total frictional drag and horsepower required\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "e=0.01 #ft\n",
-      "rho=2 #slugs/ft^3\n",
-      "mu=2.6e-5 #lb sec/ft^2\n",
-      "speed=10 #knots\n",
-      "L=250. #ft\n",
-      "A=30000. #ft^2\n",
-      "#calculations\n",
-      "V=speed*1.69\n",
-      "Nrl=V*L*rho/mu\n",
-      "Cdf=0.074/math.pow(Nrl,0.2) -1700./Nrl\n",
-      "Fd=Cdf*A*0.5*rho*V*V\n",
-      "hp=Fd*V/550.\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Total frictional drag =\",Fd,\"lb\")\n",
-      "print '%s %.1f %s' %(\"\\n Horsepower required =\",hp,\"hp\")\n",
-      "print '%s' %(\"The answer given in textbook is wrong. please use a calculator\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total frictional drag = 12537 lb\n",
-        "\n",
-        " Horsepower required = 385.2 hp\n",
-        "The answer given in textbook is wrong. please use a calculator\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2c - Pg 398"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the total frictional drag and horsepower required\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "e=0.01 #ft\n",
-      "rho=2. #slugs/ft^3\n",
-      "mu=2.6e-5 #lb sec/ft^2\n",
-      "speed=10. #knots\n",
-      "L=250. #ft\n",
-      "A=30000. #ft^2\n",
-      "#calculations\n",
-      "V=speed*1.69\n",
-      "Nrl=V*L*rho/mu\n",
-      "Cdf=1/math.pow((1.89 + 1.62*math.log10(L/e)),(2.5))\n",
-      "Fd=Cdf*A*0.5*rho*V*V\n",
-      "hp=Fd*V/550.\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Total frictional drag =\",Fd,\"lb\")\n",
-      "print '%s %.1f %s' %(\"\\n Horsepower required =\",hp,\"hp\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total frictional drag = 35117 lb\n",
-        "\n",
-        " Horsepower required = 1079.1 hp\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 398"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the drag on the model\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "V=200. #ft/s\n",
-      "L=5. #ft\n",
-      "B=2. #ft\n",
-      "rho=0.00232 #slug/ft^3\n",
-      "mu=3.82e-7 #lb-sec/ft^2\n",
-      "p2=14.815 #psia\n",
-      "pa=14.7 #psia\n",
-      "#calculations\n",
-      "Nr=V*L*rho/mu\n",
-      "Cdf=0.0032\n",
-      "Fdf=Cdf*math.pi*L*B*0.5*rho*V*V\n",
-      "Fd=(p2-pa)*math.pi/4. *(B*12)*(B*12) -Fdf\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Drag on the model =\",Fd,\"lb\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drag on the model = 47.36 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4 - Pg 405"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the velocity of the flow\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "p1=14.7 #psia\n",
-      "z1=3 #ft\n",
-      "gam=62.4\n",
-      "rho=1.94 #slug/ft^3\n",
-      "pa=0.4 #psia\n",
-      "za=1 #ft\n",
-      "#calculations\n",
-      "v3=(pa-p1)*144 + (za-z1)*gam\n",
-      "V=math.sqrt(-v3*2/(3*rho))\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Velocity of flow =\",V,\" ft/s\")\n",
-      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Velocity of flow = 27.4  ft/s\n",
-        "The answer is a bit different due to rounding off error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5 - Pg 410"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the horsepower required\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "rpm=60. \n",
-      "rho=2. #slugs/ft^3\n",
-      "mu=3.5e-5 #lb-sec/ft^2\n",
-      "D=4./12. #ft\n",
-      "r=2. #ft\n",
-      "#calcualtions\n",
-      "V=rpm*2*math.pi/60. *2\n",
-      "Nr=V*D*rho/mu\n",
-      "Cd=1.1\n",
-      "Fd=Cd*math.pi/4. *(D)*D *0.5*rho*V*V\n",
-      "T=2*Fd*r\n",
-      "w=rpm*2*math.pi/60.\n",
-      "hp=T*w/550.\n",
-      "#results\n",
-      "print '%s %.2f %s' %(\"Horsepower required =\",hp,\" hp\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Horsepower required = 0.69  hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6 - Pg 414"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the terminal velocity\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "g=32.2 #ft/s^2\n",
-      "h=60000. #ft\n",
-      "F=2000. #;b\n",
-      "d=3. #ft\n",
-      "rho=0.00231\n",
-      "#calculations\n",
-      "V=math.sqrt(2*g*h)\n",
-      "print '%s' %(\"By trail and error\")\n",
-      "Cd=0.25\n",
-      "Nm=0.87\n",
-      "A=math.pi/4. *d*d\n",
-      "Vt=math.sqrt(2*F/(Cd*A*rho))\n",
-      "#results \n",
-      "print '%s %.1f %s' %(\"terminal velocity =\",Vt,\" ft/s\")\n",
-      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "By trail and error\n",
-        "terminal velocity = 989.9  ft/s\n",
-        "The answers are a bit different from textbook due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluid_mechanics/README.txt b/Fluid_mechanics/README.txt
deleted file mode 100755
index fe81edb3..00000000
--- a/Fluid_mechanics/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Jaya Pratyusha Kothuri
-Course: btech
-College/Institute/Organization: Sri Mittapalli College of Engineering
-Department/Designation: Computer Science and Engineering
-Book Title: Fluid mechanics
-Author: Pao, Richard H F
-Publisher: John Wiley, New York
-Year of publication: 1961
-Isbn: 0486683567
-Edition: 1
\ No newline at end of file
diff --git a/Fluid_mechanics/screenshots/chap2.png b/Fluid_mechanics/screenshots/chap2.png
deleted file mode 100755
index c7186e17..00000000
Binary files a/Fluid_mechanics/screenshots/chap2.png and /dev/null differ
diff --git a/Fluid_mechanics/screenshots/chap3.png b/Fluid_mechanics/screenshots/chap3.png
deleted file mode 100755
index ea5bf942..00000000
Binary files a/Fluid_mechanics/screenshots/chap3.png and /dev/null differ
diff --git a/Fluid_mechanics/screenshots/chap4.png b/Fluid_mechanics/screenshots/chap4.png
deleted file mode 100755
index 440b52f2..00000000
Binary files a/Fluid_mechanics/screenshots/chap4.png and /dev/null differ
diff --git a/Fluid_mechanics_by_Pao/Chapter_10.ipynb b/Fluid_mechanics_by_Pao/Chapter_10.ipynb
new file mode 100755
index 00000000..bdfb09f7
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_10.ipynb
@@ -0,0 +1,234 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8ebaaa4d9fc694c6b4303418dc1faf3eef2d83f1637b122e018c6c5eac0aba66"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 - Dynamic lift"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2a - Pg 427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the max. theoretical propulsive force\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "vel=50. #mph\n",
+      "w=240. #rpm\n",
+      "r0=3. #ft\n",
+      "L=30. #ft\n",
+      "rho=0.00230 #slug/ft^2\n",
+      "theta=30*math.pi/180. #radians\n",
+      "#calculations\n",
+      "V=vel*5280./3600.\n",
+      "T=2*math.pi*r0*r0 *w*2*math.pi/60.\n",
+      "Fl=rho*V*T*L\n",
+      "F=r0*Fl*math.cos(theta)\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Max. theoretical porpulsive force =\",F,\"lb\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Max. theoretical porpulsive force = 18683 lb\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2b - Pg 428"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the force required for the operation\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "vel=50. #mph\n",
+      "w=240. #rpm\n",
+      "r0=3. #ft\n",
+      "L=30. #ft\n",
+      "rho=0.00230 #slug/ft^2\n",
+      "theta=30*math.pi/180. #radians\n",
+      "Cl=2\n",
+      "Cd=1\n",
+      "#calculations\n",
+      "vc=r0*w\n",
+      "V=vel*5280./3600.\n",
+      "vr=vc/V\n",
+      "A=2*r0*L\n",
+      "Fl=Cl*A*0.5*rho*V*V \n",
+      "Fd=Cd*A*0.5*rho*V*V\n",
+      "F=r0*(Fl*math.cos(theta) + Fd*math.sin(theta))\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Force required =\",F,\"lb\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force required = 7454 lb\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3a - Pg 432"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the boundary circulation\n",
+      "#Initialization of variables\n",
+      "W=7500 #pounds\n",
+      "rho=0.00230\n",
+      "V=175*5280/3600 #ft/s\n",
+      "B=50\n",
+      "#calculations\n",
+      "T=W/(rho*V*B)\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Boundary circulation =\",T,\"ft^2/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Boundary circulation = 254 ft^2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3b - Pg 433"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horsepower required for the operation\n",
+      "#Initialization of variables\n",
+      "W=7500. #pounds\n",
+      "rho=0.00230\n",
+      "V=175*5280./3600. #ft/s\n",
+      "B=50.\n",
+      "A=350. #ft^2\n",
+      "#calculations\n",
+      "Cl=W/(A*0.5*rho*V*V)\n",
+      "Cd=0.03\n",
+      "Fd=Cd*A*0.5*rho*V*V \n",
+      "hp=Fd*V/550.\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Horsepower required =\",hp,\"hp\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horsepower required = 371 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4 - Pg 437"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horsepower required for the operation\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "Fl=7500. #pounds\n",
+      "rho=0.00230\n",
+      "V=175*5280./3600. #ft/s\n",
+      "B=50\n",
+      "A=350 #ft^2\n",
+      "#calculations\n",
+      "Vi=2*Fl/(math.pi*rho*V*B*B)\n",
+      "Cl=Fl/(A*0.5*rho*V*V)\n",
+      "Cdi=Cl*Vi/(V)\n",
+      "Fdi=Cdi*A*0.5*rho*V*V\n",
+      "hp=Fdi*V/550\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Horsepower required =\",hp,\"hp\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horsepower required = 44.1 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_11.ipynb b/Fluid_mechanics_by_Pao/Chapter_11.ipynb
new file mode 100755
index 00000000..ec13c1a6
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_11.ipynb
@@ -0,0 +1,485 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:525b51e826bc42c9a4490de8234c350df60c3a09744a8e35af1f480f12f531bc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 - Flow of liquids in open channels"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1a - Pg 454"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge using darcy equation\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "mu=2.34e-5 #lb-sec/ft^2\n",
+      "y=5 #ft\n",
+      "T=25 #ft\n",
+      "d=10 #ft\n",
+      "slope=3./2. \n",
+      "g=32.2 #ft/s^2\n",
+      "S=0.001\n",
+      "#calculations\n",
+      "A=y*d+ 2*0.5*y*(slope*y)\n",
+      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
+      "R=A/WP\n",
+      "e=0.01 #ft\n",
+      "rr=2*R/e\n",
+      "f=0.019\n",
+      "C=math.sqrt(8*g/f)\n",
+      "V=C*math.sqrt(R*S)\n",
+      "Q=V*A\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Discharge using Darcy equation =\",Q,\"ft^3/s\")\n",
+      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge using Darcy equation = 569.3 ft^3/s\n",
+        "The answer is a bit different due to rounding off error in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1b - Pg 453"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge using kutter ganguillet method\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "mu=2.34e-5 #lb-sec/ft^2\n",
+      "y=5 #ft\n",
+      "T=25 #ft\n",
+      "d=10 #ft\n",
+      "slope=3./2. \n",
+      "g=32.2 #ft/s^2\n",
+      "S=0.001\n",
+      "n=0.017\n",
+      "#calculations\n",
+      "A=y*d+ 2*0.5*y*(slope*y)\n",
+      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
+      "R=A/WP\n",
+      "e=0.01 #ft\n",
+      "rr=2*R/e\n",
+      "f=0.019\n",
+      "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n",
+      "V=C*math.sqrt(R*S)\n",
+      "Q=V*A\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Discharge using kutter ganguillet formula =\",Q,\" ft^3/s\")\n",
+      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge using kutter ganguillet formula = 517.0  ft^3/s\n",
+        "The answer is a bit different due to rounding off error in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1c - Pg 455"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge using bazin formula\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "mu=2.34e-5 #lb-sec/ft^2\n",
+      "y=5. #ft\n",
+      "T=25. #ft\n",
+      "d=10. #ft\n",
+      "slope=3./2.\n",
+      "g=32.2 #ft/s^2\n",
+      "S=0.001\n",
+      "m=0.21\n",
+      "#calculations\n",
+      "A=y*d+ 2*0.5*y*(slope*y)\n",
+      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
+      "R=A/WP\n",
+      "e=0.01 #ft\n",
+      "rr=2*R/e\n",
+      "f=0.019\n",
+      "C=157.6 /(1+ m/math.sqrt(R))\n",
+      "V=C*math.sqrt(R*S)\n",
+      "Q=V*A\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Discharge using bazin formula =\",Q,\" ft^3/s\")\n",
+      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge using bazin formula = 688.7  ft^3/s\n",
+        "The answer is a bit different due to rounding off error in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1d - Pg 456"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge using darcy equation\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "mu=2.34e-5 #lb-sec/ft^2\n",
+      "y=5. #ft\n",
+      "T=25. #ft\n",
+      "d=10. #ft\n",
+      "slope=3./2. \n",
+      "g=32.2 #ft/s^2\n",
+      "S=0.001\n",
+      "n=0.017\n",
+      "#calculations\n",
+      "A=y*d+ 2*0.5*y*(slope*y)\n",
+      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
+      "R=A/WP\n",
+      "e=0.01 #ft\n",
+      "rr=2*R/e\n",
+      "f=0.019\n",
+      "C=1.486*math.pow(R,(1./6.)) /n\n",
+      "V=C*math.sqrt(R*S)\n",
+      "Q=V*A\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Discharge using Darcy equation =\",Q,\"ft^3/s\")\n",
+      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge using Darcy equation = 516.6 ft^3/s\n",
+        "The answer is a bit different due to rounding off error in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1e - Pg 456"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the froude number\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "mu=2.34e-5 #lb-sec/ft^2\n",
+      "y=5. #ft\n",
+      "T=25. #ft\n",
+      "d=10. #ft\n",
+      "slope=3./2. \n",
+      "g=32.2 #ft/s^2\n",
+      "S=0.001\n",
+      "n=0.017\n",
+      "#calculations\n",
+      "A=y*d+ 2*0.5*y*(slope*y)\n",
+      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
+      "R=A/WP\n",
+      "e=0.01 #ft\n",
+      "rr=2*R/e\n",
+      "f=0.019\n",
+      "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n",
+      "V=C*math.sqrt(R*S)\n",
+      "T=d+ 2*(slope*y)\n",
+      "yh=A/T\n",
+      "Nf=V/(math.sqrt(g*yh))\n",
+      "#results\n",
+      "print '%s %.2f' %(\"froude number = \",Nf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "froude number =  0.56\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1f - Pg 456"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the critical depth\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "mu=2.34e-5 #lb-sec/ft^2\n",
+      "y=5. #ft\n",
+      "T=25. #ft\n",
+      "d=10. #ft\n",
+      "slope=3./2. \n",
+      "g=32.2 #ft/s^2\n",
+      "S=0.001\n",
+      "n=0.017\n",
+      "#calculations\n",
+      "A=y*d+ 2*0.5*y*(slope*y)\n",
+      "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n",
+      "R=A/WP\n",
+      "e=0.01 #ft\n",
+      "rr=2*R/e\n",
+      "f=0.019\n",
+      "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n",
+      "V=C*math.sqrt(R*S)\n",
+      "Q=V*A\n",
+      "T=d+ 2*(slope*y)\n",
+      "yh=A/T\n",
+      "yc=2.88  #ft\n",
+      "#results\n",
+      "print '%s' %(\"yc is obtained using trial and error method\")\n",
+      "print '%s %.2f %s' %(\"Critical depth =\",yc,\"ft\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "yc is obtained using trial and error method\n",
+        "Critical depth = 2.88 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2 - Pg 459"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the minimum scale ratio\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "Re=4000.\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "vm=5.91 #ft/s\n",
+      "mu=3.24e-5 #ft-lb/s^2\n",
+      "Rm=3.12 #ft\n",
+      "#calculations\n",
+      "lam3=Re*mu/(vm*4*Rm*rho)\n",
+      "lam=math.pow(lam3,(2./3.))\n",
+      "#results\n",
+      "print '%s %.2e' %(\"Minimum scale ratio = \",lam)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum scale ratio =  9.36e-03\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 462"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge, depth of the channel, froude numbers and also the force applied\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "yc=2. #ft\n",
+      "g=32.2 #ft/s^2\n",
+      "d=10. #ft\n",
+      "gam=62.4\n",
+      "rho=1.94\n",
+      "B=10. #ft\n",
+      "#calculations\n",
+      "Vc=math.sqrt(g*yc)\n",
+      "Ac=yc*d\n",
+      "Q=Vc*Ac\n",
+      "y1=5.88 #ft\n",
+      "y2=0.88 #ft\n",
+      "V1=2.73 #ft/s\n",
+      "V2=18.25 #ft/s\n",
+      "Nf1=0.198\n",
+      "Nf2=3.43\n",
+      "F= 0.5*gam*y1*y1 *B - 0.5*gam*y2*y2 *B - Q*rho*V2 +Q*rho*V1\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Discharge in the channel =\",Q,\"ft^3/s\")\n",
+      "print '%s %.2f %s %.2f %s' %(\"\\n Depth of the channel at upstream and downstream =\",y1,\"ft and\",y2, \"ft\")\n",
+      "print '%s %.3f %s %.3f' %(\"\\n froude numbers at upstream and downstream =\",Nf1,\" and \",Nf2)\n",
+      "print '%s %d %s' %(\"\\n Force applied =\",F,\"lb\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge in the channel = 160.5 ft^3/s\n",
+        "\n",
+        " Depth of the channel at upstream and downstream = 5.88 ft and 0.88 ft\n",
+        "\n",
+        " froude numbers at upstream and downstream = 0.198  and  3.430\n",
+        "\n",
+        " Force applied = 5713 lb\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4 - Pg 470"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the distance from vena contracta and also the total distance\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "S0=0.0009\n",
+      "n=0.018\n",
+      "w=20 #ft\n",
+      "d=0.5 #ft\n",
+      "Q=400 #ft^3/s\n",
+      "g=32.2 #ft/s^2\n",
+      "#calculations\n",
+      "y2=4 #ft\n",
+      "V2=Q/(w*y2)\n",
+      "Nf2=V2/math.sqrt(g*y2)\n",
+      "yr=0.5*(math.sqrt(1+ 8*Nf2*Nf2) -1)\n",
+      "y1=yr*y2\n",
+      "L1=32.5\n",
+      "L2=37.1 \n",
+      "L3=51.4\n",
+      "L=L1+L2+L3\n",
+      "#results\n",
+      "print '%s %.1f %s %.2f %s' %(\"distance from vena contracta =\",y2,\"ft and\",y1,\"ft\")\n",
+      "print '%s %.1f %s' %(\"\\n Total distance =\",L,\" ft\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance from vena contracta = 4.0 ft and 1.20 ft\n",
+        "\n",
+        " Total distance = 121.0  ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_2.ipynb b/Fluid_mechanics_by_Pao/Chapter_2.ipynb
new file mode 100755
index 00000000..f518cad4
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_2.ipynb
@@ -0,0 +1,406 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8249f270a6314fb49f84df63266820d87dfa51fe64fed0b9dd11afa539cb500b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 - Fluid Statics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1 - Pg 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the final pressure and specific weight of the gas\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=0.0765 #lb/ft^3\n",
+      "p=14.7 #psia\n",
+      "dz=10560. #ft\n",
+      "#calculations\n",
+      "pg=p*144./gam\n",
+      "p2=p*math.exp(-dz/pg)\n",
+      "gam2=p2/p*gam\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Final pressure =\",p2,\"psia\")\n",
+      "print '%s %.4f %s' %(\"\\n Final specific weight =\",gam2,\"lb/ft^3\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final pressure = 10.04 psia\n",
+        "\n",
+        " Final specific weight = 0.0522 lb/ft^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2 - Pg 12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the final pressure and specific weight\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=0.0765 #lb/ft^3\n",
+      "p=14.7 #psia\n",
+      "dz=10560. #ft\n",
+      "n=1.235\n",
+      "#calculations\n",
+      "pg=p*144./gam\n",
+      "p2=p*math.pow((1- dz/pg *(n-1)/n),(n/(n-1)))\n",
+      "gam2=math.pow((p2/p),(1/n)) *gam\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Final pressure =\",p2,\"psia\")\n",
+      "print '%s %.4f %s' %(\"\\n Final specific weight =\",gam2,\"lb/ft^3\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final pressure = 9.89 psia\n",
+        "\n",
+        " Final specific weight = 0.0555 lb/ft^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the absolute pressure\n",
+      "#Initialization of variables\n",
+      "pb=28.5 #in mercury\n",
+      "d=13.6 #g/cc\n",
+      "gam=62.4\n",
+      "pobs=-4. #psi\n",
+      "#calculations\n",
+      "patm=pb/12. *gam*d/144.\n",
+      "pabs=patm+pobs\n",
+      "P=pabs*144./gam\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Absolute pressure =\",pabs,\"psia\")\n",
+      "print '%s %.1f %s' %(\"\\n Absolute pressure in feet of water =\",P,\"ft of water\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Absolute pressure = 10.0 psia\n",
+        "\n",
+        " Absolute pressure in feet of water = 23.1 ft of water\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4 - Pg 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure gauge at B and also the absolute pressure of air\n",
+      "#Initialization of variables\n",
+      "pb=28. #in mercury\n",
+      "d=13.6 #g/cc\n",
+      "gam=62.4\n",
+      "xm=15. #in\n",
+      "xw=10. #in\n",
+      "patm=28. #in\n",
+      "#calculations\n",
+      "pB=-xm/12 *gam/144 *d + xw*gam/144\n",
+      "pair=patm/12 *gam/144 *d - xm/12 *gam/144 *d\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"The pressure gauge at B indicates a reading of\",-pB,\"psi vacuum\")\n",
+      "print '%s %.2f %s' %(\"\\n Absolute pressure of Air =\",pair,\"psia\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure gauge at B indicates a reading of 3.03 psi vacuum\n",
+        "\n",
+        " Absolute pressure of Air = 6.38 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5 - Pg 20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure difference and also compare the pressures at A and B\n",
+      "#Initialization of variables\n",
+      "pb=28.5 #in mercury\n",
+      "d=13.6 #g/cc\n",
+      "gam=62.4\n",
+      "xm=10. #in\n",
+      "xw=2. #ft\n",
+      "#calculations\n",
+      "dp= xw*gam/144 - xm/12 *gam/144 + xm/12 *gam/144 *d\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Pressure difference =\",dp,\"psi\")\n",
+      "if dp>0:\n",
+      "    print '%s' %(\"\\n Pressure at A is greater than that at B\")\n",
+      "elif dp==0:\n",
+      "    print '%s' %(\"\\n Pressure at both A and B are equal\")\n",
+      "else:\n",
+      "    print '%s' %(\"\\n Pressure at A is less than that at B\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure difference = 5.42 psi\n",
+        "\n",
+        " Pressure at A is greater than that at B\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6 - Pg 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the magnitude of total force, vertical and horiontal locations of the total force.\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "gam=62.4\n",
+      "x1=4. #ft\n",
+      "x2=6.  #ft\n",
+      "y1=6. #ft\n",
+      "z=8. #ft\n",
+      "dy=1. #ft\n",
+      "angle=60.*math.pi/180. #radians\n",
+      "#calculations\n",
+      "A1=x1*x2\n",
+      "A2=1/2. *y1*y1\n",
+      "yc = (A1*(x1+x2+dy) + A2*(x1+x2))/(A1+A2)\n",
+      "hc=yc*math.sin(angle)\n",
+      "F=hc*gam*(A1+A2)\n",
+      "ic1=1/12. *x1*y1*y1*y1\n",
+      "ic2=1/36. *y1*x2*x2*x2\n",
+      "ad1=A1*(x1+x2+dy-yc)*(x1+x2+dy-yc)\n",
+      "ad2=A2*(x1+x2-yc)*(x1+x2-yc)\n",
+      "It=ic1+ic2+ad1+ad2\n",
+      "ydc=It/(yc*(A1+A2))\n",
+      "def momen(u):\n",
+      "    m= gam*math.sin(angle) *(2*x1+u)*0.5*(x2-u)*(y1-u)\n",
+      "    return m;\n",
+      "\n",
+      "MED, err =scipy.integrate.quad( momen,0,y1)\n",
+      "FEDC=gam*math.sin(angle) *A2*(x1+x2)\n",
+      "xed=MED/FEDC\n",
+      "xp= (A1*2*(x1+x2+dy) + (x1+x2)*(A2)*(x1+xed))/(A1*(x1+x2+dy) + A2*(x1+x2))\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Magnitude of total force =\",F,\"lb\")\n",
+      "print '%s %.3f %s' %(\"\\n Vertical location of force =\",ydc,\"ft\")\n",
+      "print '%s %.2f %s' %(\"\\n Horizontal location of force =\",xp,\"ft from AB\")\n",
+      "print '%s' %(\"\\n Direction of force is perpendicular to the plane surface\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnitude of total force = 23993 lb\n",
+        "\n",
+        " Vertical location of force = 0.266 ft\n",
+        "\n",
+        " Horizontal location of force = 3.58 ft from AB\n",
+        "\n",
+        " Direction of force is perpendicular to the plane surface\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7 - Pg 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the magnitude of force and the horizontal distance from line of action of Fv\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=62.4\n",
+      "z=10. #ft\n",
+      "z2=5. #ft\n",
+      "z3=4.25 #ft\n",
+      "p=2. #psig\n",
+      "#calculations\n",
+      "h=p*144./gam\n",
+      "Av=z*z\n",
+      "Fh=gam*(z+h)*Av\n",
+      "hpc=1/12. *z*z*z*z /((h+z)*z*z)\n",
+      "Fv=gam*(z2+h) *z*z + gam*math.pi/4. *z*z *z\n",
+      "xp= (gam*(z2+h) *z*z *z2 + gam*math.pi/4. *z*z *z*z3)/(Fv)\n",
+      "F=math.sqrt(Fh*Fh + Fv*Fv)\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Magnitude of force =\",F,\"lb\")\n",
+      "print '%s %.2f %s' %(\"\\n horizontal distance from line of action of Fv =\",xp,\"ft from AG\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Magnitude of force = 142127 lb\n",
+        "\n",
+        " horizontal distance from line of action of Fv = 4.66 ft from AG\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8 - Pg 33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate whether the barge is stable and also find the location of the metacenter\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=0.0765 #lb/ft^3\n",
+      "l=40. #ft\n",
+      "w=16. #ft\n",
+      "d=8. #ft\n",
+      "z=6.  #ft\n",
+      "BG=1. #ft\n",
+      "#calculations\n",
+      "I=1/12. *l*w*w*w\n",
+      "V=l*w*z\n",
+      "IVG=I/V - BG\n",
+      "MB=I/V\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"I/V -BG =\",IVG,\"ft \")\n",
+      "if IVG >0:\n",
+      "    print '%s' %(\"\\n Barge is stable\")\n",
+      "else:\n",
+      "    print '%s' %(\"\\n The barge is unstable\")\n",
+      "\n",
+      "print '%s %.2f %s' %(\"\\n Location of metacenter =\",MB,\"ft above the center of buoyancy \")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "I/V -BG = 2.56 ft \n",
+        "\n",
+        " Barge is stable\n",
+        "\n",
+        " Location of metacenter = 3.56 ft above the center of buoyancy \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_3.ipynb b/Fluid_mechanics_by_Pao/Chapter_3.ipynb
new file mode 100755
index 00000000..8d368fbf
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_3.ipynb
@@ -0,0 +1,105 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:dee9d655eeb9028db41fd96e05f9acf5955fd88886936fffc7d4e9aa14bb4a71"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 - Fluid Kinematics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1 - Pg 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the mean velocity of flow at both the sections\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=0.0765 #lb/ft^3\n",
+      "Q=100. #ft^3/sec\n",
+      "d1=2.5 #ft\n",
+      "d2=9. #in\n",
+      "l=12. #ft\n",
+      "#calculations\n",
+      "A1=math.pi/4. *d1*d1\n",
+      "V1=Q/A1\n",
+      "A2=math.pi*l*d2/12.\n",
+      "V2=Q/A2\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Mean velocity of flow at section 1 =\",V1,\"ft/sec\")\n",
+      "print '%s %.2f %s' %(\"\\n Mean velocity of flow at section 2 =\",V2,\"ft/sec\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean velocity of flow at section 1 = 20.4 ft/sec\n",
+        "\n",
+        " Mean velocity of flow at section 2 = 3.54 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5 - Pg 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horizontal and vertical components of velocity\n",
+      "#Initialization of variables\n",
+      "x=3. \n",
+      "y=1.\n",
+      "#calculations\n",
+      "u=-3*y*y\n",
+      "v=-6*x\n",
+      "#results\n",
+      "print '%s %d' %(\"Horizontal component of velocity = \",u)\n",
+      "print '%s %d' %(\"\\n vertical component of velocity = \",v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horizontal component of velocity =  -3\n",
+        "\n",
+        " vertical component of velocity =  -18\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_4.ipynb b/Fluid_mechanics_by_Pao/Chapter_4.ipynb
new file mode 100755
index 00000000..49f2ce7e
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_4.ipynb
@@ -0,0 +1,777 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bf4ed226827aaf378bd6d43d8ae5845adf36d75f20c124d2087db252462cf40c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 - Fluid Dynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1 - Pg 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate dp/ds at both upper and lower ends\n",
+      "#Initialization of variables\n",
+      "rho=1.5 #g/cc\n",
+      "g=32.2 #ft/s^2\n",
+      "dzds=-0.5\n",
+      "x1=0\n",
+      "x2=3\n",
+      "#calculations\n",
+      "def func(s):\n",
+      "    dpds=-rho*g*dzds - rho*(3+9*s)*9\n",
+      "    return dpds\n",
+      "\n",
+      "r1=func(x1)\n",
+      "r2=func(x2)\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"At the upper end, dp/ds =\",r1,\"lb/ft^2 per foot\")\n",
+      "print '%s %.1f %s' %(\"\\n At the lower end, dp/ds =\",r2,\"lb/ft^2 per foot\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At the upper end, dp/ds = -16.3 lb/ft^2 per foot\n",
+        "\n",
+        " At the lower end, dp/ds = -380.9 lb/ft^2 per foot\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2 - Pg 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure difference between the two points\n",
+      "#Initialization of variables\n",
+      "g=32.2 #ft/s^2\n",
+      "v1=3. #ft/s\n",
+      "z1=1.5 #ft\n",
+      "rho=1.5 #g/cc\n",
+      "z2=0\n",
+      "v2=30. #ft/s\n",
+      "#calculations\n",
+      "dp= rho*(v2*v2 /2. - g*z1 +g*z2 - v1*v1 /2.)\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"pressure difference =\",dp,\"lb/ft^2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure difference = 595.8 lb/ft^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the net power transferred\n",
+      "#Initialization of variables\n",
+      "pd=15. #psia\n",
+      "rhod=0.005#slug/ft^3\n",
+      "pi=150. #psia\n",
+      "rhoi=0.03 #slug/ft^3\n",
+      "dz=-25. #ft\n",
+      "vd=1000. #ft/s\n",
+      "vi=100. #ft/s\n",
+      "ud=200. #Btu/slug\n",
+      "ui=250. #Btu/slug\n",
+      "g=32.2 #ft/s^2\n",
+      "J=778.\n",
+      "uff=5. #ft/s\n",
+      "Q=50. #Btu/sec\n",
+      "#calculations\n",
+      "pr=pd/rhod*144. - pi/rhoi *144.\n",
+      "zr=g*(dz)\n",
+      "vr=(vd*vd -vi*vi)/2.\n",
+      "ur=(ud-ui)*J\n",
+      "jeh=J*Q*g/uff\n",
+      "gem=pr+zr+vr+ur+jeh\n",
+      "power=gem*uff/g\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Power transferred =\",power,\"ft-lb/sec\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power transferred = 64877 ft-lb/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4 - Pg 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the kinetic energy correction factor\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "r0=1. \n",
+      "ri=0.\n",
+      "#calculations\n",
+      "def func1(y):\n",
+      "    v= 2*math.pow(y,(1./7.)) *(y-1)\n",
+      "    return v\n",
+      "\n",
+      "V,err=scipy.integrate.quad(func1,ri,r0)\n",
+      "def func2(y):\n",
+      "    alpha= 1/ (math.pi*V*V*V) *2*math.pi *math.pow((y),(3./7.)) *(y-1) \n",
+      "    return alpha\n",
+      "\n",
+      "a2,err2=scipy.integrate.quad(func2,ri,r0)\n",
+      "#results\n",
+      "print '%s %.2f' %(\"Kinetic energy correction factor = \",a2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Kinetic energy correction factor =  1.06\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5a - Pg 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure at the lower end if friction is neglected\n",
+      "#Initialization of variables\n",
+      "gam=62.4\n",
+      "pu=40. #psia\n",
+      "zu=25. #ft\n",
+      "vu=8. #ft/s\n",
+      "g=32.2 #ft/s^2\n",
+      "vl=8. #ft/s\n",
+      "zl=0 #ft\n",
+      "#calculations\n",
+      "pl= gam*(pu*144. /gam +zu-zl+ (vu*vu -vl*vl)/(2*g))/144.\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Pressure at the lower end if friction is neglected =\",pl,\"psig\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at the lower end if friction is neglected = 50.83 psig\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5b - Pg 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure at the lower end if friction is not neglected\n",
+      "#Initialization of variables\n",
+      "hl=5.\n",
+      "gam=62.4\n",
+      "pu=40. #psia\n",
+      "zu=25. #ft\n",
+      "vu=8. #ft/s\n",
+      "g=32.2 #ft/s^2\n",
+      "vl=8. #ft/s\n",
+      "zl=0 #ft\n",
+      "#calculations\n",
+      "pl= gam*(pu*144. /gam +zu-zl-hl+ (vu*vu -vl*vl)/(2*g))/144.\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Pressure at the lower end if friction is neglected =\",pl,\"psig\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at the lower end if friction is neglected = 48.67 psig\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6a - Pg 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge flow rate.\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=62.4\n",
+      "pa=0\n",
+      "za=15. #ft\n",
+      "va=0\n",
+      "pg=0\n",
+      "zg=0\n",
+      "g=32.2 #ft/s^2\n",
+      "dg=2. #in\n",
+      "#calculations\n",
+      "vg= math.sqrt(2*g*(pa/gam +za+va*va /(2*g) -pg/gam - zg))\n",
+      "Ag=math.pi/4. *(dg/12.)*(dg/12.)\n",
+      "Q=Ag*vg\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"discharge =\",Q,\"ft^3/sec\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "discharge = 0.68 ft^3/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6b - Pg 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure at the points C,D,E and F\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=62.4\n",
+      "pa=0\n",
+      "za=15. #ft\n",
+      "va=0\n",
+      "pg=0\n",
+      "zg=0\n",
+      "g=32.2 #ft/s^2\n",
+      "d=4. #in\n",
+      "dg=2. #in\n",
+      "zd=25.  #ft\n",
+      "#calculations\n",
+      "vg= math.sqrt(2*g*(pa/gam +za+va*va /(2*g) -pg/gam - zg))\n",
+      "Ag=math.pi/4. *(dg/12.)*(dg/12.)\n",
+      "Q=Ag*vg\n",
+      "A=math.pi/4. *(d/12.)*(d/12.)\n",
+      "v4=Q/A\n",
+      "pc=-v4*v4 *gam/(2*g*144.)\n",
+      "pgd= za-zd - v4*v4 /(2*g)\n",
+      "pd=pgd*gam/144.\n",
+      "pe=-v4*v4 *gam/(2*g*144.)\n",
+      "pfg= za- v4*v4 /(2*g)\n",
+      "pf=pfg*gam/144.\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Pressure at C =\",pc,\" psig\")\n",
+      "print '%s %.2f %s' %(\"\\n Pressure at D =\",pd,\" psig\")\n",
+      "print '%s %.2f %s' %(\"\\n Pressure at E =\",pe,\" psig\")\n",
+      "print '%s %.2f %s' %(\"\\n Pressure at F =\",pf,\" psig\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at C = -0.41  psig\n",
+        "\n",
+        " Pressure at D = -4.74  psig\n",
+        "\n",
+        " Pressure at E = -0.41  psig\n",
+        "\n",
+        " Pressure at F = 6.09  psig\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7 - Pg 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the time required\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "d1=6. #ft\n",
+      "d2=3. #in\n",
+      "pa=2. #ft\n",
+      "d=13.6 \n",
+      "sg=0.75\n",
+      "h1=5. #sec\n",
+      "h2=3. #sec\n",
+      "g=32.2 #ft/s^2\n",
+      "#calculations\n",
+      "pag=pa/12. *d/sg\n",
+      "def  func(h):\n",
+      "    time= -d1*d1 /(d2/12.)/(d2/12.) /(math.sqrt(2*g)) *math.pow((pag+h),(-0.5))\n",
+      "    return time\n",
+      "ti,err=scipy.integrate.quad(func,h1,h2)\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Time required =\",ti,\" sec\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time required = 54.3  sec\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8 - Pg 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the rate of flow \n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "x=12. #ft\n",
+      "angle=30.*math.pi/180. #degrees\n",
+      "g=32.2 #ft/s^2\n",
+      "z=-2. #ft\n",
+      "d=2. #in\n",
+      "#calculations\n",
+      "vj= x/math.cos(angle) *math.sqrt(g/(2*(x*math.tan(angle) -z)))\n",
+      "Q=math.pi/4. *(d/12.)*(d/12.) *vj\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Rate of flow =\",Q,\" ft^3/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate of flow = 0.41  ft^3/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9 - Pg 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the net discharge\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "x=10. #in of mercury\n",
+      "sg=13.6 #g/cc\n",
+      "d1=8. #in\n",
+      "d2=4. #in\n",
+      "g=32.2 #ft/s^2\n",
+      "#calculations\n",
+      "vdiff=x/12. *sg- x/12.\n",
+      "Vts=vdiff/(1-math.pow((d2/d1),4))\n",
+      "Vt=math.sqrt(2*g*Vts)\n",
+      "Q=Vt*math.pi/4. *(d2/12.)*(d2/12.)\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Discharge =\",Q,\"ft^3/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 2.34 ft^3/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11 - Pg 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horsepower input of the test pump\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "gam=62.4\n",
+      "ds=12. #in\n",
+      "dd=10. #in\n",
+      "Q=4. #ft^3/s\n",
+      "pd=40. #psia\n",
+      "ps=-6. #psia\n",
+      "zd=5. #ft\n",
+      "zs=0\n",
+      "g=32.2 #ft/s^2\n",
+      "#calculations\n",
+      "vs=Q/(math.pi/4. *math.pow((ds/12.),2))\n",
+      "vd=Q/(math.pi/4. *math.pow((dd/12.),2))\n",
+      "emp = (pd-ps)*144./gam + zd-zs + (vd*vd - vs*vs)/(2*g)\n",
+      "hpp=emp*Q*gam/550.\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Horsepower input of the test pump =\",hpp,\"hp\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horsepower input of the test pump = 50.6 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12 - Pg 117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horizontal and vertical components of force\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "d1=12. #in\n",
+      "d2=8. #in\n",
+      "v1=15. #ft/s\n",
+      "p1=12. #psig\n",
+      "p2=5.85 #psig\n",
+      "rho=1.94 #ft^3/slug\n",
+      "angle=60.*math.pi/180. #degrees\n",
+      "#calculations\n",
+      "Q=math.pi/4. *(d1/12.)*(d1/12.) *v1\n",
+      "v2=Q/(math.pi/4. *(d2/12.)*(d2/12.))\n",
+      "pa1=p1*math.pi/4. *(d1)*(d1)\n",
+      "pa2=p2*math.pi/4. *(d2)*(d2)\n",
+      "qv1=rho*Q*v1\n",
+      "qv2=rho*Q*v2\n",
+      "Fx=pa1+qv1+ math.cos(angle)*(pa2+qv2)\n",
+      "Fy=math.sin(angle)*(pa2+qv2)\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Horizontal component of force =\",Fx,\"lb\")\n",
+      "print '%s %d %s' %(\"\\n Vertical component of force =\",Fy,\"lb\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horizontal component of force = 2232 lb\n",
+        "\n",
+        " Vertical component of force = 922 lb\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14a - Pg 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the exit velocity of the pump\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "de=4. #in\n",
+      "T=1000. #lb\n",
+      "g=32.2 #ft/s^2\n",
+      "vele=8.5 #lb/s\n",
+      "pe=16.5 #psia\n",
+      "pa=14.7 #psia\n",
+      "#calculations\n",
+      "Ae=math.pi/4. *de*de\n",
+      "Ve= (T-(pe-pa)*Ae)*g/vele\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Exit velocity =\",Ve,\"ft/s\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit velocity = 3702 ft/s\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14b - Pg 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the thrust in the pipe\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "de=4. #in\n",
+      "T=1000. #lb\n",
+      "g=32.2 #ft/s^2\n",
+      "vele=8.5 #lb/s\n",
+      "pe=16.5 #psia\n",
+      "pa=14.7 #psia\n",
+      "pa2=1 #psia\n",
+      "#calculations\n",
+      "Ae=math.pi/4 *de*de\n",
+      "Ve= (T-(pe-pa)*Ae)*g/vele\n",
+      "T2=vele/g *Ve + (pe-pa2)*Ae\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Thrust =\",T2,\"lb\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thrust = 1172 lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16 - Pg 133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the downstream depth and horsepower dissipation\n",
+      "#Initialization of variables\n",
+      "q=240. #ft^3/sec/ft\n",
+      "v1=60. #ft/s\n",
+      "gam=62.4 \n",
+      "rho=1.94 #slug/ft^3\n",
+      "g=32.2 #ft/s^2\n",
+      "#calculations\n",
+      "y1=q/v1\n",
+      "v2=8.6 #ft/s\n",
+      "y2=28. #ft\n",
+      "hl= (y1+ v1*v1 /(2*g)) - (y2+ v2*v2 /(2*g))\n",
+      "hpp=hl*q*gam/550.\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Downstream depth =\",y2,\"ft\")\n",
+      "print '%s %d %s' %(\"\\n Horsepower dissipation =\",hpp,\"hp per foot width\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Downstream depth = 28.0 ft\n",
+        "\n",
+        " Horsepower dissipation = 837 hp per foot width\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17 - Pg 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the acceleration required\n",
+      "#Initialization of variables\n",
+      "dh=3. #in\n",
+      "L=12. #in\n",
+      "g=32.2 #ft/s^2\n",
+      "#calculations\n",
+      "a=dh/L *g\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Acceleration =\",a,\" ft/s^2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Acceleration = 8.05  ft/s^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_5.ipynb b/Fluid_mechanics_by_Pao/Chapter_5.ipynb
new file mode 100755
index 00000000..3591a061
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_5.ipynb
@@ -0,0 +1,244 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0e0637dca57f500478b778450aaaee0067016f8da4c67cbada5f889a38068276"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 - Fluid viscosity and flow of real fluids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the viscosity of oil\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "m=1155. #lb\n",
+      "gam=62.4\n",
+      "spg=0.93\n",
+      "t=3*60. #sec\n",
+      "d=1./6. #in\n",
+      "L=20. #ft\n",
+      "dp=2.5 #psi\n",
+      "#calculations\n",
+      "Q=m/(t*spg*gam)\n",
+      "A=math.pi/4. *d*d\n",
+      "V=Q/A\n",
+      "mu=dp*d*d *144./(32.*V*L)\n",
+      "#results\n",
+      "print '%s %.4f %s' %(\"Viscosity of oil =\",mu,\"lb-sec/ft^2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Viscosity of oil = 0.0031 lb-sec/ft^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4 - Pg 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the values of alpha and beta\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "g=32.2\n",
+      "gam=62.4\n",
+      "r0=1.\n",
+      "#calculations\n",
+      "def func1(r):\n",
+      "    al=8./math.pow(r0,8) *math.pow((r0*r0-r*r),3) *(2*r)\n",
+      "    return al\n",
+      "alpha,err=scipy.integrate.quad(func1,0,r0)\n",
+      "def  func2(r):\n",
+      "    a2=4/math.pow(r0,6) *math.pow((r0*r0 -r*r),2) *(2*r)\n",
+      "    return a2\n",
+      "bet,err2=scipy.integrate.quad(func2,0,r0)\n",
+      "#results\n",
+      "print '%s %d' %(\"Alpha = \",alpha)\n",
+      "print '%s %.2f' %(\"\\n beta = \",bet)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Alpha =  2\n",
+        "\n",
+        " beta =  1.33\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example 5a - Pg 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the direction of flow and the loss of energy\n",
+      "#Initialization of variables\n",
+      "spg=0.93\n",
+      "mu=3.1e-3 #lb-sec/ft^2\n",
+      "gam=62.4\n",
+      "z=50. #m\n",
+      "p1=60. #psia\n",
+      "p2=25. #psia\n",
+      "#calculations\n",
+      "p1g=144.*p1\n",
+      "p2g=144.*p2 + spg*gam*z\n",
+      "dp=p1g-p2g\n",
+      "#results\n",
+      "if p1g>p2g:\n",
+      "    print '%s' %(\"The flow is in upward direction\")\n",
+      "else:\n",
+      "    print '%s' %(\"The flow is in downward direction\")\n",
+      "\n",
+      "print '%s %d %s' %(\"\\n Energy loss=\",dp,\"ft-lb/ft^3\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The flow is in upward direction\n",
+        "\n",
+        " Energy loss= 2138 ft-lb/ft^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5b - Pg 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the flow rate\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "hl=2140. #ft-lb/ft^3\n",
+      "spg=0.93\n",
+      "mu=3.1e-3 #lb-sec/ft^2\n",
+      "gam=62.4\n",
+      "z=50. #m\n",
+      "p1=60. #psia\n",
+      "p2=25. #psia\n",
+      "d=1. #in\n",
+      "#calculations\n",
+      "V= hl*(d/12.)*(d/12.) /(32*mu*z)\n",
+      "Q=V*math.pi/4. *(d/12.)*(d/12.)\n",
+      "Q2=Q*7.48*60.\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Flow rate =\",Q2,\"gal/min\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate = 7.33 gal/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7 - Pg 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the flow in model\n",
+      "#Initialization of variables\n",
+      "muw=2.04e-5 #lb-sec/ft^2\n",
+      "rhow=1.94 #slugs/ft^3\n",
+      "mua=3.74e-7 #lb-sec/ft^2\n",
+      "rhoa=0.00237 #slug/ft^3\n",
+      "Qw=200. #gal/min\n",
+      "Lr=5.\n",
+      "#calculations\n",
+      "Qa=Qw*Lr *(rhow/rhoa)*(mua/muw)\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Flow in model =\",Qa,\"gal/min\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow in model = 15007 gal/min\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_6.ipynb b/Fluid_mechanics_by_Pao/Chapter_6.ipynb
new file mode 100755
index 00000000..424aea7a
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_6.ipynb
@@ -0,0 +1,118 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:27148bed2cbad8357ddcdc8e0ee434be3797107d2f692797bdf3a1e36c7e8d1a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 - Dimensional analysis and model similitude"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 230"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the flow rate of water and pressure drop\n",
+      "#Initialization of variables\n",
+      "dg=0.5 #in\n",
+      "dw=12. #in\n",
+      "rhog=0.022 #slug/ft^3\n",
+      "rhow=1.94 #slug/ft^3\n",
+      "muw=2.34e-5 #lb-sec/ft^2\n",
+      "mug=3.50e-7 #lb-sec/ft^2\n",
+      "Qg=0.15 #ft^3/s\n",
+      "dpg=100. #lb/ft^2\n",
+      "#calculations\n",
+      "Vr=dg/dw *rhog/rhow *muw/mug\n",
+      "Qr=Vr*dw*dw /dg/dg\n",
+      "Qw=Qr*Qg\n",
+      "dpr=rhow/rhog *(Vr)*Vr\n",
+      "dpw=dpr*dpg\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Flow rate of water =\",Qw,\"ft^3/s\")\n",
+      "print '%s %.1f %s' %(\"\\n Pressure drop =\",dpw,\"lb/ft^2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate of water = 2.73 ft^3/s\n",
+        "\n",
+        " Pressure drop = 8.8 lb/ft^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4 - Pg 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the velocity ratio, time ratio, acceleration ratio and force ratio.\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "Lr=1./10.\n",
+      "rhom=2.\n",
+      "rhop=1.94\n",
+      "#calculations\n",
+      "Vr=math.sqrt(Lr)\n",
+      "Tr=Lr/Vr\n",
+      "ar=Vr/Tr\n",
+      "Fr=rhom/rhop *ar*Lr*Lr*Lr\n",
+      "#results\n",
+      "print '%s %.4f' %(\"Velocity ratio = \",Vr)\n",
+      "print '%s %.4f' %(\"\\n Time ratio = \",Tr)\n",
+      "print '%s %d' %(\"\\n Acceleration ratio = \",ar)\n",
+      "print '%s %.6f' %(\"\\n Force ratio = \",Fr)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity ratio =  0.3162\n",
+        "\n",
+        " Time ratio =  0.3162\n",
+        "\n",
+        " Acceleration ratio =  1\n",
+        "\n",
+        " Force ratio =  0.001031\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_7.ipynb b/Fluid_mechanics_by_Pao/Chapter_7.ipynb
new file mode 100755
index 00000000..26326ec8
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_7.ipynb
@@ -0,0 +1,874 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a67e8661ab863efba6ffe8486e33132e755624949690c50df8ae76d2d928ed8b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 - Flow of incompressible fluids in closed conduits"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1a - Pg 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the reynolds number of the flow\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "z1=2. #ft\n",
+      "Q=0.1 #gal/min\n",
+      "alpha=2.\n",
+      "g=32.2 #ft/s^2\n",
+      "L=4. #ft\n",
+      "D=1./96. #ft\n",
+      "#calculations\n",
+      "v2=Q/(7.48*60* math.pi/4. *D*D)\n",
+      "hl=z1-alpha*v2*v2 /(2*g)\n",
+      "Nr=64./hl *L/D *v2*v2 /(2*g)\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Reynolds number is\",Nr,\".Hence the flow is laminar\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reynolds number is 1459 .Hence the flow is laminar\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1b - Pg 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the kinematic viscosity\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "z1=2 #ft\n",
+      "Q=0.1 #gal/min\n",
+      "alpha=2.\n",
+      "g=32.2 #ft/s^2\n",
+      "L=4. #ft\n",
+      "D=1./96. #ft\n",
+      "#calculations\n",
+      "v2=Q/(7.48*60* math.pi/4. *D*D)\n",
+      "hl=z1-alpha*v2*v2/(2*g)\n",
+      "Nr=64./hl *L/D *v2*v2 /(2*g)\n",
+      "mu=v2*D/Nr\n",
+      "#results\n",
+      "print '%s %.2e %s' %(\"Kinematic viscosity =\",mu,\"ft^2/s\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Kinematic viscosity = 1.87e-05 ft^2/s\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exaple 1c - Pg 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the theoretical entrance transistion length\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "z1=2. #ft\n",
+      "Q=0.1 #gal/min\n",
+      "alpha=2.\n",
+      "g=32.2 #ft/s^2\n",
+      "L=4. #ft\n",
+      "D=1./96. #ft\n",
+      "#calculations\n",
+      "v2=Q/(7.48*60* math.pi/4. *D*D)\n",
+      "hl=z1-alpha*v2*v2 /(2*g)\n",
+      "Nr=64./hl *L/D *v2*v2 /(2*g)\n",
+      "Ld=0.058*Nr*D\n",
+      "#results\n",
+      "print '%s %.3f %s' %(\"Theoretical entrance transistion length =\",Ld,\"ft\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical entrance transistion length = 0.882 ft\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2 - Pg 246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horsepower required for the motor\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "Q=350. #gal/min\n",
+      "D=6. #in\n",
+      "rho=0.84\n",
+      "gam=62.4\n",
+      "g=32.2 #ft/s^2\n",
+      "mu=9.2e-5 #lb-sec/ft^2\n",
+      "L=5280. #ft\n",
+      "#calculations\n",
+      "V=Q/(7.48*60*math.pi/4. *math.pow((D/12),2))\n",
+      "Nr=V*D/12 *rho*gam/g /mu\n",
+      "f=0.3164/math.pow((Nr),0.25)\n",
+      "hl=f*L*12/D *V*V /(2*g)\n",
+      "hp=hl*gam*Q*rho/(550*7.48*60.)\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Horsepower required =\",hp,\"hp/mile\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horsepower required = 4.44 hp/mile\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the values of the coefficients alpha and beta\n",
+      "#Initialization of variables\n",
+      "n=7.\n",
+      "import math\n",
+      "#calculations\n",
+      "alpha= math.pow((n+1),3) *math.pow((2*n+1),3) /(4*math.pow(n,4) *(n+3)*(2*n+3))\n",
+      "bet=math.pow((n+1),2) *math.pow((2*n+1),2) /(2*n*n *(n+2)*(2*n+2))\n",
+      "#results\n",
+      "print '%s %.2f' %(\"alpha = \",alpha)\n",
+      "print '%s %.2f' %(\"\\n beta = \",bet)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "alpha =  1.06\n",
+        "\n",
+        " beta =  1.02\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5 - Pg 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the mass flow rate and horsepower input of the fan\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "spg=0.84\n",
+      "z=1. #in\n",
+      "gam=62.4\n",
+      "patm=14.7 #psia\n",
+      "T=459.6+85 #R\n",
+      "R=53.3\n",
+      "g=32.2 #ft/s^2\n",
+      "D=3. #ft\n",
+      "mu=3.88e-7 #lb-sec/ft^2\n",
+      "#calculations\n",
+      "dp=spg*z/12. *gam\n",
+      "rho=patm*144. /(R*T*g)\n",
+      "umax=math.sqrt(2*dp/rho)\n",
+      "V=0.8*umax\n",
+      "Nr=V*D*rho/mu\n",
+      "V2=0.875*umax\n",
+      "mass=rho*math.pi/4. *D*D *V2\n",
+      "emf=V2*V2 /(2*g)\n",
+      "hp=emf*mass*g/550.\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Mass flow rate =\",mass,\"slug/sec\")\n",
+      "print '%s %.2f %s' %(\"\\n Horsepower input of the fan =\",hp,\"hp\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass flow rate = 0.87 slug/sec\n",
+        "\n",
+        " Horsepower input of the fan = 2.34 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7a - Pg 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the velocity using several equations listed above\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "D=36. #in\n",
+      "rho=0.00226  #slug/ft^3\n",
+      "mu=3.88e-7 #lb-sec/ft^2\n",
+      "umax=62.2 #ft/s\n",
+      "V=54.5 #ft/s\n",
+      "Nr=9.5e5\n",
+      "r0=18. #in\n",
+      "r=12. #in\n",
+      "n=8.8\n",
+      "k=0.4\n",
+      "#calculations\n",
+      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
+      "Vs=math.sqrt(f/8) *V\n",
+      "y=r0-r\n",
+      "u1=umax*math.pow((y/r0),(1/n))\n",
+      "u2=umax+ 2.5*Vs*math.log(y/r0)\n",
+      "u3=umax+ Vs/k *(math.sqrt(1-y/r0) + math.log(1-math.sqrt(1-y/r0)))\n",
+      "u4=Vs*(5.5+ 5.75*math.log10(Vs*y/12 *rho/mu))\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Using equation 7-13, velocity =\",u1,\" ft/s\")\n",
+      "print '%s %.1f %s' %(\"\\n Using equation 7-18, velocity =\",u2,\" ft/s\")\n",
+      "print '%s %.1f %s' %(\"\\n Using equation 7-25, velocity =\",u3,\" ft/s\")\n",
+      "print '%s %.1f %s' %(\"\\n Using equation 7-34a, velocity =\",u4,\" ft/s\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Using equation 7-13, velocity = 54.9  ft/s\n",
+        "\n",
+        " Using equation 7-18, velocity = 56.5  ft/s\n",
+        "\n",
+        " Using equation 7-25, velocity = 57.6  ft/s\n",
+        "\n",
+        " Using equation 7-34a, velocity = 56.7  ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7b - Pg 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the outer edge buffer zone distance and also its thickness\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "D=36. #in\n",
+      "rho=0.00226  #slug/ft^3\n",
+      "mu=3.88e-7 #lb-sec/ft^2\n",
+      "umax=62.2 #ft/s\n",
+      "V=54.5 #ft/s\n",
+      "Nr=9.5e5\n",
+      "r0=18. #in\n",
+      "r=12. #in\n",
+      "n=8.8\n",
+      "k=0.4\n",
+      "#calculations\n",
+      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
+      "Vs=math.sqrt(f/8.) *V\n",
+      "y=r0-r\n",
+      "delta1=D*5*math.sqrt(8) /(Nr*math.sqrt(f))\n",
+      "vss=70.\n",
+      "thick=13*delta1\n",
+      "#results\n",
+      "print '%s %d' %(\"Outer edge of buffer zone is at \",vss)\n",
+      "print '%s %.4f %s' %(\"\\n Thickness of buffer zone =\",thick,\"in\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Outer edge of buffer zone is at  70\n",
+        "\n",
+        " Thickness of buffer zone = 0.0645 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7c - Pg 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the velocity using the given equations\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "D=36. #in\n",
+      "rho=0.00226  #slug/ft^3\n",
+      "mu=3.88e-7 #lb-sec/ft^2\n",
+      "umax=62.2 #ft/s\n",
+      "V=54.5 #ft/s\n",
+      "Nr=9.5e5\n",
+      "r0=18. #in\n",
+      "r=12. #in\n",
+      "n=8.8\n",
+      "k=0.4\n",
+      "#calculations\n",
+      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
+      "Vs=math.sqrt(f/8) *V\n",
+      "delta1=D*5*math.sqrt(8.) /(Nr*math.sqrt(f))\n",
+      "y=delta1\n",
+      "u2=Vs*Vs *delta1/12. *rho/mu\n",
+      "u1=62.2 *math.pow((delta1/18.),(1/n))\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"using equation 7-13, velocity =\",u1,\"ft/s\")\n",
+      "print '%s %.1f %s' %(\"\\n using equation 7-30, velocity =\",u2,\"ft/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "using equation 7-13, velocity = 24.5 ft/s\n",
+        "\n",
+        " using equation 7-30, velocity = 10.4 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7d - Pg 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the velocity using the listed equations\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "D=36. #in\n",
+      "rho=0.00226  #slug/ft^3\n",
+      "mu=3.88e-7 #lb-sec/ft^2\n",
+      "umax=62.2 #ft/s\n",
+      "V=54.5 #ft/s\n",
+      "Nr=9.5e5\n",
+      "r0=18. #in\n",
+      "r=12. #in\n",
+      "n=8.8\n",
+      "k=0.4\n",
+      "#calculations\n",
+      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
+      "Vs=math.sqrt(f/8.) *V\n",
+      "delta1=D*5*math.sqrt(8.) /(Nr*math.sqrt(f))\n",
+      "y=14*delta1\n",
+      "u2=62.2*math.pow((y/18.),(1/n))\n",
+      "u3=Vs*(5.50 + 5.75*math.log10(Vs*y/12 *rho/mu))\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Using equation 7-13, velocity =\",u2,\"ft/s\")\n",
+      "print '%s %.1f %s' %(\"\\n using equation 7-34a, velocity =\",u3,\"ft/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Using equation 7-13, velocity = 33.1 ft/s\n",
+        "\n",
+        " using equation 7-34a, velocity = 33.5 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7e - Pg 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the shearing stress using both the equations\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "D=36. #in\n",
+      "rho=0.00226  #slug/ft^3\n",
+      "mu=3.88e-7 #lb-sec/ft^2\n",
+      "umax=62.2 #ft/s\n",
+      "V=54.5 #ft/s\n",
+      "Nr=9.5e+5\n",
+      "r0=18. #in\n",
+      "r=12. #in\n",
+      "n=8.8\n",
+      "k=0.4\n",
+      "#calculations\n",
+      "f=0.0032 + 0.221/(math.pow(Nr,0.237))\n",
+      "Vs=math.sqrt(f/8.) *V\n",
+      "delta1=D*5*math.sqrt(8.) /(Nr*math.sqrt(f))\n",
+      "u2=Vs*Vs *delta1/12. *rho/mu\n",
+      "T0=rho*Vs*Vs\n",
+      "T02=mu*u2/delta1 *12.\n",
+      "#results\n",
+      "print '%s %.5f %s' %(\"Using equation 7-9a, shearing stress =\",T0,\"lb/ft^2\")\n",
+      "print '%s %.5f %s' %(\"\\n Using equation 7-28, shearing stress =\",T02,\"lb/ft^2\")\n",
+      "print '%s' %(\"The answers are a bit different due to rounding off error in textbook\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Using equation 7-9a, shearing stress = 0.00979 lb/ft^2\n",
+        "\n",
+        " Using equation 7-28, shearing stress = 0.00979 lb/ft^2\n",
+        "The answers are a bit different due to rounding off error in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8 - Pg 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the required velocity\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "umax=62.2 #ft/s\n",
+      "r0=18. #in\n",
+      "e=0.0696 #in\n",
+      "r=6. #in\n",
+      "#calculations\n",
+      "Vs=umax/(8.5 + 5.75*math.log10(r0/e))\n",
+      "u=Vs*(8.5 + 5.75*math.log10(r/e))\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Velocity =\",u,\"ft/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity = 54.6 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9 - Pg 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the flow rate and roughness factor\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "d=8. #in\n",
+      "V=3.65 #ft/s\n",
+      "u1=4.75 #ft/s\n",
+      "r0=4. #in\n",
+      "#calculations\n",
+      "f=0.0449\n",
+      "Q=V*math.pi/4. *math.pow((d/12),2)\n",
+      "Vs=(u1-V)/3.75\n",
+      "r0e=math.pow(10,((u1/Vs - 8.5)/5.75))\n",
+      "e=r0/r0e\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Flow rate =\",Q,\"ft^3/s\")\n",
+      "print '%s %.3f %s' %(\"\\n roughness factor =\",e,\"in\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate = 1.27 ft^3/s\n",
+        "\n",
+        " roughness factor = 0.184 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10 - Pg 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure difference per foot of horizontal pipe\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "e0=0.00085 #ft\n",
+      "alpha=0.25 #/year\n",
+      "t=15. #years\n",
+      "r0=3. #in\n",
+      "Q=500. #gal/min\n",
+      "d=6. #in\n",
+      "mu=2.04e-5 #lb-sec/ft^2\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "g=32.2 #ft/s^2\n",
+      "L=1. #ft\n",
+      "gam=62.4\n",
+      "#calculations\n",
+      "e15=e0*(1+ alpha*t)\n",
+      "ratio=r0/(12.*e15)\n",
+      "V=Q/(7.48*60*math.pi/4. *math.pow((d/12),2))\n",
+      "Nr=V*d*rho/(mu*12)\n",
+      "f=0.036\n",
+      "hl=f*L/(d/12.) *V*V /(2*g)\n",
+      "dp=gam*hl\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Pressure difference =\",dp,\"lb/ft^2 per foot of horizontal pipe\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure difference = 2.25 lb/ft^2 per foot of horizontal pipe\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11 - Pg 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horsepower required\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "d2=4. #in\n",
+      "d1=3. #in\n",
+      "e=0.0005 #ft\n",
+      "mu=3.75e-5 #lb-sec/ft^2\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "Q=100. #gal/min\n",
+      "L=100. #ft\n",
+      "g=32.2 #ft/s^2\n",
+      "gam=62.4\n",
+      "#calculations\n",
+      "A=math.pi/4. *(math.pow((d2/12),2) -math.pow((d1/12),2))\n",
+      "WP=math.pi*(d1+d2)/12.\n",
+      "R=A/WP\n",
+      "RR= 2*R/e\n",
+      "V= Q/(7.48*60*A)\n",
+      "Nr=V*4*R*rho/mu\n",
+      "f=0.035\n",
+      "hl=f*L/(4*R) *V*V /(2*g)\n",
+      "hp=hl*Q/(7.48*60) *gam/550.\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"horsepower required =\",hp,\" hp/100 ft\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "horsepower required = 0.56  hp/100 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12 - Pg 296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge flow\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "p1=25. #psig\n",
+      "p2=20. #psig\n",
+      "d1=18. #in\n",
+      "d2=12. #in\n",
+      "Cl=0.25\n",
+      "gam=62.4\n",
+      "g=32.2 #ft/s^2\n",
+      "#calculations\n",
+      "Vr=(d2/d1)*(d2/d1)\n",
+      "xv=(p2-p1)*144/gam\n",
+      "V22=xv/(-1-Cl+Vr*Vr) *2*g\n",
+      "V2=math.sqrt(V22)\n",
+      "Q=V2*math.pi/4. *(d2/12.)*(d2/12.)\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Discharge =\",Q,\"ft^3/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 20.9 ft^3/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13 - Pg 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the discharge flow rate\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "V61=10.8 #ft/s\n",
+      "V81=6.05 #ft/s\n",
+      "r0=3 #in\n",
+      "e=0.00015\n",
+      "d1=6. #in\n",
+      "rho=1.94 #slugs/ft^3\n",
+      "mu=2.34e-5 #ft-lb/s^2\n",
+      "#calculations\n",
+      "roe=r0/(12*e)\n",
+      "Nr1=V61*(d1/12.)*rho/mu\n",
+      "f6=0.0165\n",
+      "V6=11.6 #ft/s\n",
+      "V8=6.52 #ft/s\n",
+      "Q=V6*math.pi/4 *(d1/12.)*(d1/12.)\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Discharge =\",Q,\"ft^3/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 2.28 ft^3/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14 - Pg 302"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the diameter of steel pipe\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "L=1000. #ft\n",
+      "Q=2000/(7.48*60) #ft63/s\n",
+      "g=32.2 #ft/s^2\n",
+      "p=5. #psi/1000 ft\n",
+      "gam=62.4\n",
+      "sp=0.7\n",
+      "f=0.02\n",
+      "r0=0.904/2.\n",
+      "e=0.00015\n",
+      "mu=7e-6 #lb-ft/s^2\n",
+      "L=1000. #ft\n",
+      "#calculations\n",
+      "hl=p*144/(sp*gam)\n",
+      "D5=f*8*L*Q*Q /(math.pi*math.pi *g*hl)\n",
+      "D=math.pow(D5,(1./5.))\n",
+      "Nr=4*Q*sp*gam/(g*(math.pi*D*mu))\n",
+      "f2=0.0145\n",
+      "D5=f2*8*L*Q*Q /(math.pi*math.pi *g*hl)\n",
+      "D1=math.pow(D5,(1./5.))\n",
+      "#results\n",
+      "print '%s %.3f %s' %(\"Diameter of steel pipe =\",D1,\" ft\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of steel pipe = 0.848  ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_8.ipynb b/Fluid_mechanics_by_Pao/Chapter_8.ipynb
new file mode 100755
index 00000000..76c114d3
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_8.ipynb
@@ -0,0 +1,802 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f292d23876c88421bcfc8b761169402a2885837db02032e303215ae83f2c5cf5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 - Fluid compressibility and incompressible flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1 - Pg 323"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the initial and final volumes. Also, calculate the final temperature of the gas\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "pi=14.7 #psia\n",
+      "pf=50. #psia\n",
+      "cp=0.240 #Btu/lb R\n",
+      "cv=0.170 #Btu/lb R\n",
+      "J=778\n",
+      "T=60+459.6 #R\n",
+      "#calculations\n",
+      "R=J*(cp-cv)\n",
+      "k=cp/cv\n",
+      "gam=pi*144./(R*T)\n",
+      "V=1/gam\n",
+      "Vf=V*math.pow((pi/pf),(1/k))\n",
+      "Tf=T*(pf*Vf/(pi*V))\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Initial volume =\",V,\"ft^3\")\n",
+      "print '%s %.2f %s' %(\"\\n Final volume =\",Vf,\"cu ft\")\n",
+      "print '%s %.1f %s' %(\"\\n Final temperature =\",Tf,\" R\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Initial volume = 13.37 ft^3\n",
+        "\n",
+        " Final volume = 5.62 cu ft\n",
+        "\n",
+        " Final temperature = 742.6  R\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2 - Pg 325"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure difference across the pipe\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "ratio=0.99\n",
+      "E=3.19e5 #lb/in^2\n",
+      "#calculations\n",
+      "pd=-E*math.log(ratio)\n",
+      "#ersults\n",
+      "print '%s %d %s' %(\"Pressure difference =\",pd,\"psi\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure difference = 3206 psi\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the speed of test plane\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "g=32.2 #ft/s^2\n",
+      "R=53.3 #ft-lb/lb R\n",
+      "T=389.9 #R\n",
+      "Nm=2\n",
+      "#calculations\n",
+      "c=math.sqrt(k*g*R*T)\n",
+      "V=Nm*c*3600/5280.\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Speed of test plane =\",V,\"mph\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed of test plane = 1319.9 mph\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4a - Pg 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the velocity at section 2\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "T1=584.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "k=1.4\n",
+      "R=53.3 #ft-lb/lb R\n",
+      "V1=600 #ft/s\n",
+      "T2=519.6 #R\n",
+      "#calculations\n",
+      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
+      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
+      "Nm2=math.sqrt(Nm22)\n",
+      "V2=Nm2*math.sqrt(k*g*R*T2)\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Velocity at section 2 =\",V2,\"ft/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity at section 2 = 1068 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4b - Pg 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure difference between two stations\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "T1=584.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "k=1.4\n",
+      "R=53.3 #ft-lb/lb R\n",
+      "V1=600 #ft/s\n",
+      "T2=519.6 #R\n",
+      "pa=14.7 #psi\n",
+      "p1=50 #psia\n",
+      "#calculations\n",
+      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
+      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
+      "Nm2=math.sqrt(Nm22)\n",
+      "pr=math.pow(((1+ (k-1)/2 *Nm1*Nm1)/(1+ (k-1)/2 *Nm2*Nm2)),(k/(k-1)))\n",
+      "p2=pr*(p1+pa)\n",
+      "dp=p1+pa-p2\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Pressure difference between two stations =\",dp,\"psi\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure difference between two stations = 21.9 psi\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4c - Pg 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the area ratio\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "T1=584.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "k=1.4\n",
+      "R=53.3 #ft-lb/lb R\n",
+      "V1=600 #ft/s\n",
+      "T2=519.6 #R\n",
+      "#calculations\n",
+      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
+      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
+      "Nm2=math.sqrt(Nm22)\n",
+      "Ar= Nm1/Nm2 *math.pow(((1+ (k-1)/2 *Nm2*Nm2)/(1+ (k-1)/2 *Nm1*Nm1)),((k+1)/(2*(k-1))))\n",
+      "#results\n",
+      "print '%s %.3f' %(\"Area ratio = \",Ar)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area ratio =  0.754\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4d - Pg 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the density of air at both the stations\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "T1=584.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "k=1.4\n",
+      "R=53.3 #ft-lb/lb R\n",
+      "V1=600 #ft/s\n",
+      "T2=519.6 #R\n",
+      "pa=14.7 #psi\n",
+      "p1=50 #psia\n",
+      "#calculations\n",
+      "Nm1=V1/(math.sqrt(k*g*R*T1))\n",
+      "Nm22= ((1+ (k-1)/2 *Nm1*Nm1)/(T2/T1) -1)*(2/(k-1))\n",
+      "Nm2=math.sqrt(Nm22)\n",
+      "pr=math.pow(((1+ (k-1)/2 *Nm1*Nm1)/(1+ (k-1)/2 *Nm2*Nm2)),(k/(k-1)))\n",
+      "p2=pr*(p1+pa)\n",
+      "rho1=(p1+pa)*144./(g*R*T1)\n",
+      "rho2=p2*144./(g*R*T2)\n",
+      "#results\n",
+      "print '%s %.5f %s' %(\"Density of air at station 1 =\",rho1,\"slug/ft^3\")\n",
+      "print '%s %.5f %s' %(\"\\n Density of air at station 2 =\",rho2,\"slug/ft^3\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density of air at station 1 = 0.00929 slug/ft^3\n",
+        "\n",
+        " Density of air at station 2 = 0.00692 slug/ft^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5 - Pg 345"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the mass rate of air flow\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "p0=19.7 #psia\n",
+      "R=53.3 #lb-ft/lb-R\n",
+      "T0=539.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "pa=14.7 #psia\n",
+      "d=1 #in\n",
+      "k=1.4\n",
+      "#calculations\n",
+      "rho0=p0*144/(g*R*T0)\n",
+      "pr=pa/p0\n",
+      "G=math.pi/4 *(d/12.)*(d/12.) *math.pow((2*k/(k-1) *p0*144*rho0),(0.5)) *math.pow((pr),(1/k)) *math.pow((1-math.pow(pr,((k-1)/k))),0.5)\n",
+      "#results\n",
+      "print '%s %.5f %s' %(\"Mass rate of air flow =\",G,\"slug/sec\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass rate of air flow = 0.00978 slug/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6 - Pg 346"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the mass rate of air flow\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "p0=64.7 #psia\n",
+      "R=53.3 #lb-ft/lb-R\n",
+      "T0=539.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "pa=14.7 #psia\n",
+      "d=1. #in\n",
+      "k=1.4\n",
+      "#calculations\n",
+      "rho0=p0*144/(g*R*T0)\n",
+      "pr=pa/p0\n",
+      "G=math.pi/4 *(d/12)*d/12. *math.pow((k*p0*144*rho0),(0.5)) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
+      "#results\n",
+      "print '%s %.5f %s' %(\"Mass rate of air flow =\",G,\"slug/sec\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass rate of air flow = 0.03616 slug/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7a - Pg 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the weight of air flow through the nozzle\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "R=53.3 #lb-ft/lb R\n",
+      "pe=14.7 #psia\n",
+      "p0=114.7 #psia\n",
+      "T0=524.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "d=0.5 #in\n",
+      "#calculations\n",
+      "pr=pe/p0\n",
+      "prcr=0.528\n",
+      "pr=prcr*p0\n",
+      "rho0= p0*144/(g*R*T0)\n",
+      "G=math.pi/4 *(d/12)*d/12. *math.pow((k*p0*144*rho0),(0.5)) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
+      "Wt=G*g\n",
+      "#results\n",
+      "print '%s %.4f %s' %(\"weight of air flow through the nozzle =\",Wt,\"lb/s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "weight of air flow through the nozzle = 0.5233 lb/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7b - Pg 348"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the exit Mach number, exit velocity and the exit area\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "R=53.3 #lb-ft/lb R\n",
+      "pe=14.7 #psia\n",
+      "p0=114.7 #psia\n",
+      "T0=524.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "d=0.5 #in\n",
+      "Nm1=1.\n",
+      "#calculations\n",
+      "pr=pe/p0\n",
+      "Nme=math.sqrt(2/(k-1) *(math.pow(1./pr,(k-1)/k) -1))\n",
+      "Te=T0/(1+ (k-1)/2 *Nme*Nme)\n",
+      "Ve=Nme*math.sqrt(k*g*R*Te)\n",
+      "At=math.pi/4. *(d)*d\n",
+      "Ae=Nm1/Nme *math.pow(((1+ (k-1)/2 *Nme*Nme)/(1+ (k-1)/2 *Nm1*Nm1)),((k+1)/(2*(k-1)))) *At\n",
+      "#results\n",
+      "print '%s %.2f' %(\"Mach number exit = \",Nme)\n",
+      "print '%s %d %s' %(\"\\n Exit velocity =\",Ve,\"ft/s\")\n",
+      "print '%s %.3f %s' %(\"\\n Exit area =\",Ae,\" in^2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mach number exit =  2.00\n",
+        "\n",
+        " Exit velocity = 1672 ft/s\n",
+        "\n",
+        " Exit area = 0.331  in^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8a - Pg 349"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the exit mass flow rate, pressure, temperature, velocity and mach number\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "R=53.3 #lb-ft/lb R\n",
+      "p0=100. #psia\n",
+      "T0=534.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "d=0.5 #in\n",
+      "Nm1=1.\n",
+      "A=2./144. #ft^2\n",
+      "#calculations\n",
+      "print '%s' %(\"Exit mach number is found using trial and error\")\n",
+      "Nme=2.44\n",
+      "rho0=p0*144/(g*R*T0)\n",
+      "G= A*math.sqrt(k*p0*144*rho0) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
+      "pe=p0*math.pow((1/(1+(k-1)/2 *Nme*Nme)),(k/(k-1)))\n",
+      "Te=T0/(1+ (k-1)/2 *Nme*Nme)\n",
+      "Ve=Nme*(math.sqrt(k*g*R*Te))\n",
+      "#results\n",
+      "print '%s %.3f %s' %(\"\\n Exit mass flow rate =\",G,\"slug/s\")\n",
+      "print '%s %.2f %s' %(\"\\n Exit pressure =\",pe,\"psia\")\n",
+      "print '%s %.1f %s' %(\"\\n Exit temperature =\",Te,\" R\")\n",
+      "print '%s %d %s' %(\"\\n Exit velocity =\",Ve,\"ft/s\")\n",
+      "print '%s %.2f' %(\"\\n Exit mach number = \",Nme)\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit mach number is found using trial and error\n",
+        "\n",
+        " Exit mass flow rate = 0.143 slug/s\n",
+        "\n",
+        " Exit pressure = 6.43 psia\n",
+        "\n",
+        " Exit temperature = 244.0  R\n",
+        "\n",
+        " Exit velocity = 1868 ft/s\n",
+        "\n",
+        " Exit mach number =  2.44\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8b - Pg 350"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the exit mass flow rate, pressure, temperature,velocity and mach number\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "R=53.3 #lb-ft/lb R\n",
+      "p0=100. #psia\n",
+      "T0=534.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "d=0.5 #in\n",
+      "Nm1=1.\n",
+      "A=2./144. #ft^2\n",
+      "#calculations\n",
+      "print '%s' %(\"Exit mach number is found using trial and error\")\n",
+      "Nme=0.24\n",
+      "rho0=p0*144/(g*R*T0)\n",
+      "G= A*math.sqrt(k*p0*144*rho0) *math.pow((2/(k+1)),((k+1)/(2*(k-1))))\n",
+      "pe=p0*math.pow((1/(1+(k-1)/2 *Nme*Nme)),(k/(k-1)))\n",
+      "Te=T0/(1+ (k-1)/2 *Nme*Nme)\n",
+      "Ve=Nme*(math.sqrt(k*g*R*Te))\n",
+      "#results\n",
+      "print '%s %.3f %s' %(\"\\n Exit mass flow rate =\",G,\"slug/s\")\n",
+      "print '%s %.2f %s' %(\"\\n Exit pressure =\",pe,\"psia\")\n",
+      "print '%s %.1f %s' %(\"\\n Exit temperature =\",Te,\" R\")\n",
+      "print '%s %d %s' %(\"\\n Exit velocity =\",Ve,\"ft/s\")\n",
+      "print '%s %.2f' %(\"\\n Exit mach number = \",Nme)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit mach number is found using trial and error\n",
+        "\n",
+        " Exit mass flow rate = 0.143 slug/s\n",
+        "\n",
+        " Exit pressure = 96.07 psia\n",
+        "\n",
+        " Exit temperature = 528.5  R\n",
+        "\n",
+        " Exit velocity = 270 ft/s\n",
+        "\n",
+        " Exit mach number =  0.24\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9 - Pg 355"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the upstream Mach number, pressure, Temperature\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "R=53.3 #lb-ft/lb R\n",
+      "pu=6.43 #psia\n",
+      "Tu=244. #R\n",
+      "Nmu=2.44\n",
+      "#calculations\n",
+      "Nmd = math.sqrt(((k-1)*Nmu*Nmu +2)/(2*k*Nmu*Nmu - (k-1)))\n",
+      "pd=pu*(2*k*Nmu*Nmu - (k-1))/(k+1)\n",
+      "Td=Tu*(2*k*Nmu*Nmu - (k-1))/(k+1) *((k-1)*Nmu*Nmu +2)/((k+1)*Nmu*Nmu)\n",
+      "#results\n",
+      "print '%s %.3f' %(\"Mach number upstream = \",Nmd)\n",
+      "print '%s %.1f %s' %(\"\\n Pressure upstream =\",pd,\"psia\")\n",
+      "print '%s %.1f %s' %(\"\\n Temperature upstream =\",Td,\"R\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mach number upstream =  0.519\n",
+        "\n",
+        " Pressure upstream = 43.6 psia\n",
+        "\n",
+        " Temperature upstream = 507.2 R\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10 - Pg 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the pressure and temperature at section 1\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "R=53.3 #lb-ft/lb R\n",
+      "e=0.0005 #ft\n",
+      "mu=3.77e-7 #lb-sec/ft^2\n",
+      "pe=14.7 #psia\n",
+      "Te=524.6 #R\n",
+      "g=32.2 #ft/s^2\n",
+      "Vi=12.5 #ft/s\n",
+      "l=6. #in\n",
+      "b=8. #in\n",
+      "L=100. #ft\n",
+      "#calculations\n",
+      "rhoe=pe*144/(R*g*Te)\n",
+      "Ve=Vi/(g*rhoe*(l*b/144.))\n",
+      "Nme=Ve/(math.sqrt(k*g*R*Te))\n",
+      "Rd=l*b/(2*(l+b)) /12.\n",
+      "rr=2*R/e\n",
+      "Nr=Ve*4*Rd*rhoe/mu\n",
+      "f=0.019\n",
+      "f2=1/(2*k) *(1/Nme*Nme -1) - (k+1)/(4*k) *math.log((1+ (k-1)/2 *Nme*Nme)/(Nme*Nme *(1+(k-1)/2)))\n",
+      "ff=f*L/(8*Rd) +f2\n",
+      "Nm1=0.305\n",
+      "Tr2=(1+ (k-1)/2 *Nm1*Nm1)/(1+ (k-1/2))\n",
+      "Tre=(1+ (k-1)/2 *Nme*Nme)/(1+ (k-1/2))\n",
+      "pr2=Nm1*math.pow((1+ (k-1)/2 *Nm1*Nm1),(0.5)) /math.pow((1+(k-1)/2),0.5)\n",
+      "pre=Nme*math.pow((1+ (k-1)/2 *Nme*Nme),(0.5)) /math.pow((1+(k-1)/2),0.5)\n",
+      "p1=pe/pr2 *pre\n",
+      "T1=Te/Tr2 *Tre\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Pressure at section 1 =\",p1,\"psia\")\n",
+      "print '%s %.1f %s' %(\"\\n Tempreature at section 1 =\",T1,\"R\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at section 1 = 21.5 psia\n",
+        "\n",
+        " Tempreature at section 1 = 535.1 R\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11 - Pg 364"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the limiting pressure and distance in both adiabatic and non-adiabatic cases\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "k=1.4\n",
+      "R=53.3 #lb-ft/lb R\n",
+      "g=32.2 #ft/s^2\n",
+      "T1=534.6 #R\n",
+      "V1=400. #ft/s\n",
+      "p1=350. #psia\n",
+      "f=0.02\n",
+      "D=6./12. #ft\n",
+      "#calculations\n",
+      "Nm1=V1/math.sqrt(k*g*R*T1)\n",
+      "Nm2= 1. /math.sqrt(k)\n",
+      "p2=p1*(Nm1)/Nm2\n",
+      "fl= math.log(Nm1/Nm2) + 1./(2.*k*Nm1*Nm1) *(1- Nm1*Nm1/Nm2*Nm2)\n",
+      "L12=fl*2*D/f\n",
+      "ps=p1*Nm1*math.pow((1+ (k-1)/2 *Nm1*Nm1),0.5) /math.pow((1+(k-1)/2),0.5)\n",
+      "Nm2=1.\n",
+      "fl2= -(k+1)/(4*k) *math.log((1+ (k-1)/2 *Nm1*Nm1)/(Nm1*Nm1 *(1+ (k-1)/2.))) + 1/(2*k*Nm1*Nm1) *(1- Nm1*Nm1 /Nm2*Nm2)\n",
+      "L2=fl2*2*D/f\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Limiting pressure =\",p2,\" psia\")\n",
+      "print '%s %.1f %s' %(\"\\n Distance =\",L12,\" ft\")\n",
+      "print '%s %.1f %s' %(\"\\n Limiting pressure in adiabatic case =\",ps,\" psia\")\n",
+      "print '%s %.1f %s' %(\"\\n Distance required =\",L2,\"ft\")\n",
+      "print '%s' %(\"the answer is a bit different due to rounding of error in textbook\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Limiting pressure = 146.2  psia\n",
+        "\n",
+        " Distance = 81.8  ft\n",
+        "\n",
+        " Limiting pressure in adiabatic case = 114.2  psia\n",
+        "\n",
+        " Distance required = 84.2 ft\n",
+        "the answer is a bit different due to rounding of error in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/Chapter_9.ipynb b/Fluid_mechanics_by_Pao/Chapter_9.ipynb
new file mode 100755
index 00000000..f131ead9
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/Chapter_9.ipynb
@@ -0,0 +1,450 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:6e08f75972be45750168908e0258e415f396bd3e7aae6981e62e0a4fd06177dc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 - Fluid flow about immersed bodies"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1a - Pg 389"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the values of Fd\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "rho=2.45 #slugs/ft^3\n",
+      "mu=9.2e-3 #lb-sec/ft^2\n",
+      "x=3.\n",
+      "v=3. #ft/s\n",
+      "B=6./12. #ft\n",
+      "L=36./12. #ft\n",
+      "#calculatons\n",
+      "Nr=v*x*rho/mu\n",
+      "y=([1.32, 1.46, 1.328])\n",
+      "Cd = numpy.zeros(len(y))\n",
+      "for i in range (0, len(y)):\n",
+      "\tCd[i]=y[i]*math.pow(Nr,(-0.5))\n",
+      "Fd = numpy.zeros(len(Cd))\n",
+      "for i in range (0, len(y)):\n",
+      "\tFd[i]=2*Cd[i]*B*L*(0.5*rho*v*v)\n",
+      "\n",
+      "#results\n",
+      "print '%s' %(\"Drag on the plates using different formulae blasius, parabola and pohlhauser in order\")\n",
+      "print '%.3f %.3f %.3f' %(Fd[0],Fd[1],Fd[2])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Drag on the plates using different formulae blasius, parabola and pohlhauser in order\n",
+        "0.892 0.986 0.897\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1b - Pg 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calcualte the boundary layer thickness and the shearing thickness\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "x=36./12.\n",
+      "rho=2.45 #slugs/ft^3\n",
+      "mu=9.2e-3 #lb-sec/ft^2\n",
+      "v=3. #ft/s\n",
+      "#calculatons\n",
+      "Nr=v*x*rho/mu\n",
+      "z=([4.91, 5.48, 4.65])\n",
+      "x=36./12.\n",
+      "delta = numpy.zeros(len(z))\n",
+      "for i in range (0, len(z)):\n",
+      "\tdelta[i]=z[i] /math.sqrt(Nr) *x\n",
+      "\n",
+      "\n",
+      "f=([0.332, 0.365, 0.322])\n",
+      "T=numpy.zeros(len(f))\n",
+      "for i in range (0,len(f)):\n",
+      "\tT[i]=f[i]*mu*v/x *math.sqrt(Nr)\n",
+      "\n",
+      "#results\n",
+      "print '%s' %(\"Boundary layer thickness = \")\n",
+      "print '%s' %(\"In order of Blasius, parabola and pohlhauser\")\n",
+      "print '%.3f %.3f %.3f' %(delta[0],delta[1],delta[2])\n",
+      "print '%s' %(\"Shearing stress = \")\n",
+      "print '%s' %(\"In order of Blasius, parabola and pohlhauser\")\n",
+      "print '%.3f %.3f %.3f' %(T[0],T[1],T[2])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Boundary layer thickness = \n",
+        "In order of Blasius, parabola and pohlhauser\n",
+        "0.301 0.336 0.285\n",
+        "Shearing stress = \n",
+        "In order of Blasius, parabola and pohlhauser\n",
+        "0.150 0.164 0.145\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2a - Pg 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the total frictional drag and horsepower required\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "e=0.01 #ft\n",
+      "rho=2 #slugs/ft^3\n",
+      "mu=2.6e-5 #lb sec/ft^2\n",
+      "speed=10. #knots\n",
+      "L=250. #ft\n",
+      "A=30000. #ft^2\n",
+      "#calculations\n",
+      "V=speed*1.69\n",
+      "Nrl=V*L*rho/mu\n",
+      "Cdf=1.32 /math.sqrt(Nrl)\n",
+      "Fd=Cdf*A*0.5*rho*V*V\n",
+      "hp=Fd*V/550.\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Total frictional drag =\",Fd,\"lb\")\n",
+      "print '%s %.1f %s' %(\"\\n Horsepower required =\",hp,\"hp\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total frictional drag = 627 lb\n",
+        "\n",
+        " Horsepower required = 19.3 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2b - Pg 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the total frictional drag and horsepower required\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "e=0.01 #ft\n",
+      "rho=2 #slugs/ft^3\n",
+      "mu=2.6e-5 #lb sec/ft^2\n",
+      "speed=10 #knots\n",
+      "L=250. #ft\n",
+      "A=30000. #ft^2\n",
+      "#calculations\n",
+      "V=speed*1.69\n",
+      "Nrl=V*L*rho/mu\n",
+      "Cdf=0.074/math.pow(Nrl,0.2) -1700./Nrl\n",
+      "Fd=Cdf*A*0.5*rho*V*V\n",
+      "hp=Fd*V/550.\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Total frictional drag =\",Fd,\"lb\")\n",
+      "print '%s %.1f %s' %(\"\\n Horsepower required =\",hp,\"hp\")\n",
+      "print '%s' %(\"The answer given in textbook is wrong. please use a calculator\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total frictional drag = 12537 lb\n",
+        "\n",
+        " Horsepower required = 385.2 hp\n",
+        "The answer given in textbook is wrong. please use a calculator\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2c - Pg 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the total frictional drag and horsepower required\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "e=0.01 #ft\n",
+      "rho=2. #slugs/ft^3\n",
+      "mu=2.6e-5 #lb sec/ft^2\n",
+      "speed=10. #knots\n",
+      "L=250. #ft\n",
+      "A=30000. #ft^2\n",
+      "#calculations\n",
+      "V=speed*1.69\n",
+      "Nrl=V*L*rho/mu\n",
+      "Cdf=1/math.pow((1.89 + 1.62*math.log10(L/e)),(2.5))\n",
+      "Fd=Cdf*A*0.5*rho*V*V\n",
+      "hp=Fd*V/550.\n",
+      "#results\n",
+      "print '%s %d %s' %(\"Total frictional drag =\",Fd,\"lb\")\n",
+      "print '%s %.1f %s' %(\"\\n Horsepower required =\",hp,\"hp\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total frictional drag = 35117 lb\n",
+        "\n",
+        " Horsepower required = 1079.1 hp\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3 - Pg 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the drag on the model\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "V=200. #ft/s\n",
+      "L=5. #ft\n",
+      "B=2. #ft\n",
+      "rho=0.00232 #slug/ft^3\n",
+      "mu=3.82e-7 #lb-sec/ft^2\n",
+      "p2=14.815 #psia\n",
+      "pa=14.7 #psia\n",
+      "#calculations\n",
+      "Nr=V*L*rho/mu\n",
+      "Cdf=0.0032\n",
+      "Fdf=Cdf*math.pi*L*B*0.5*rho*V*V\n",
+      "Fd=(p2-pa)*math.pi/4. *(B*12)*(B*12) -Fdf\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Drag on the model =\",Fd,\"lb\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Drag on the model = 47.36 lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4 - Pg 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the velocity of the flow\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "p1=14.7 #psia\n",
+      "z1=3 #ft\n",
+      "gam=62.4\n",
+      "rho=1.94 #slug/ft^3\n",
+      "pa=0.4 #psia\n",
+      "za=1 #ft\n",
+      "#calculations\n",
+      "v3=(pa-p1)*144 + (za-z1)*gam\n",
+      "V=math.sqrt(-v3*2/(3*rho))\n",
+      "#results\n",
+      "print '%s %.1f %s' %(\"Velocity of flow =\",V,\" ft/s\")\n",
+      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of flow = 27.4  ft/s\n",
+        "The answer is a bit different due to rounding off error in textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5 - Pg 410"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the horsepower required\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "rpm=60. \n",
+      "rho=2. #slugs/ft^3\n",
+      "mu=3.5e-5 #lb-sec/ft^2\n",
+      "D=4./12. #ft\n",
+      "r=2. #ft\n",
+      "#calcualtions\n",
+      "V=rpm*2*math.pi/60. *2\n",
+      "Nr=V*D*rho/mu\n",
+      "Cd=1.1\n",
+      "Fd=Cd*math.pi/4. *(D)*D *0.5*rho*V*V\n",
+      "T=2*Fd*r\n",
+      "w=rpm*2*math.pi/60.\n",
+      "hp=T*w/550.\n",
+      "#results\n",
+      "print '%s %.2f %s' %(\"Horsepower required =\",hp,\" hp\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horsepower required = 0.69  hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6 - Pg 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the terminal velocity\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "g=32.2 #ft/s^2\n",
+      "h=60000. #ft\n",
+      "F=2000. #;b\n",
+      "d=3. #ft\n",
+      "rho=0.00231\n",
+      "#calculations\n",
+      "V=math.sqrt(2*g*h)\n",
+      "print '%s' %(\"By trail and error\")\n",
+      "Cd=0.25\n",
+      "Nm=0.87\n",
+      "A=math.pi/4. *d*d\n",
+      "Vt=math.sqrt(2*F/(Cd*A*rho))\n",
+      "#results \n",
+      "print '%s %.1f %s' %(\"terminal velocity =\",Vt,\" ft/s\")\n",
+      "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "By trail and error\n",
+        "terminal velocity = 989.9  ft/s\n",
+        "The answers are a bit different from textbook due to rounding off error\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluid_mechanics_by_Pao/README.txt b/Fluid_mechanics_by_Pao/README.txt
new file mode 100755
index 00000000..fe81edb3
--- /dev/null
+++ b/Fluid_mechanics_by_Pao/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Jaya Pratyusha Kothuri
+Course: btech
+College/Institute/Organization: Sri Mittapalli College of Engineering
+Department/Designation: Computer Science and Engineering
+Book Title: Fluid mechanics
+Author: Pao, Richard H F
+Publisher: John Wiley, New York
+Year of publication: 1961
+Isbn: 0486683567
+Edition: 1
\ No newline at end of file
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new file mode 100755
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diff --git a/Fluidization_Engineering/README.txt b/Fluidization_Engineering/README.txt
deleted file mode 100755
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--- a/Fluidization_Engineering/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Dolar Khachariya
-Course: mtech
-College/Institute/Organization: Indian Institute of Engineering Bombay
-Department/Designation: Electrical engineering
-Book Title: Fluidization Engineering
-Author: K. Daizo And O. Levenspiel
-Publisher: Butterworth-Heinemann, Massachusetts
-Year of publication: 1991
-Isbn: 8131200353
-Edition: 2nd
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch10.ipynb b/Fluidization_Engineering/ch10.ipynb
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--- a/Fluidization_Engineering/ch10.ipynb
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-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:89c91579d721ed0f833b399593203d4eb19421ab1695bc830f1be612e46bd826"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10 : Gas Dispersion and Gas Interchange in Bubbling Beds"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "%matplotlib inline"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
-        "For more information, type 'help(pylab)'.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 253\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from numpy import *\n",
-      "%matplotlib inline\n",
-      "#Variable declaration\n",
-      "umf=[0.01,0.045];     #Velocity at minimum fluidization condition in m/s\n",
-      "ephsilonmf=[0.5,0.5]; #Void fraction at minimum fluidization condition\n",
-      "D=[2E-5,7E-5];        #Diffusion coefficient of gas in m**2/s\n",
-      "g=9.81;               #Acceleration due to gravity in m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "db=[5.,10.,15.,20.];\n",
-      "n=len(umf);\n",
-      "m=len(db)\n",
-      "Kbc = zeros((n,m))\n",
-      "Kce = zeros((n,m))\n",
-      "Kbe = zeros((n,m))\n",
-      "    \n",
-      "for i in range(n):\n",
-      "    for j in range(m):\n",
-      "            Kbc[i][j]=4.5*(umf[i]/db[j])+5.85*((D[i]**0.5*g**0.25)/db[j]**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(27)\n",
-      "            Kce[i][j]=6.77*((D[i]*ephsilonmf[i]*0.711*(g*db[j])**0.5)/db[j]**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(34)\n",
-      "            Kbe[i][j]=(Kbc[i][j]*Kce[i][j])/(Kbc[i][j]+Kce[i][j]);#Gas interchange coefficient between bubble and emulsion from Eqn.(14)\n",
-      "\n",
-      "#OUTPUT\n",
-      "i=0;\n",
-      "j=0;\n",
-      "k=0;\n",
-      "while k<m*n:\n",
-      "    print '\\t\\tKbc for fine particles and He',\n",
-      "    print '\\tKbc for coarse particles and ozone',\n",
-      "    print '\\tKbe for fine particles and He',\n",
-      "    print '\\tKbe for coarse particles and ozone'\n",
-      "    j = 0\n",
-      "    while j<m:\n",
-      "        print 'db=%fm'%(db[j]*10**-2);\n",
-      "        while i<n:\n",
-      "            print '\\t%f'%Kbc[i][j],\n",
-      "            print '\\t\\t\\t%f'%Kbe[i][j],\n",
-      "            i=i+1;   \n",
-      "            k=k+1;\n",
-      "            print '\\t\\t\\t',\n",
-      "        i=0;\n",
-      "        j=j+1;\n",
-      "import numpy\n",
-      "import matplotlib.pyplot as plt\n",
-      "a = numpy.matrix(Kbe)\n",
-      "b = numpy.matrix(Kbc)\n",
-      "db = array(db).T\n",
-      "Kbe=a.T\n",
-      "Kbc=b.T\n",
-      "\n",
-      "plt.plot(db,Kbc,db,Kbe)\n",
-      "#plt.plot(db,Kbe);\n",
-      "#Note : Python does not have plot2d function. so we can plot it \n",
-      "#plt.title('Plot of Kbc,Kbe vs db','db');\n",
-      "plt.show()\n",
-      "print 'Comparing the points with the plot of Kbc,Kbe vs db in Fig.(12), we can conlcude the following:'\n",
-      "print 'Kbc for fine particles and helium: line 2 in Fig.(12)'\n",
-      "print 'Kbc for coarser particles and ozone: line 3 in Fig.(12)'\n",
-      "print 'Kbe for fine particles and helium: line 4 in Fig.(12)'\n",
-      "print 'Kbe for coarser particles and ozone: line 5 in Fig.(12)'\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
-        "For more information, type 'help(pylab)'.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\tKbc for fine particles and He \tKbc for coarse particles and ozone \tKbe for fine particles and He \tKbe for coarse particles and ozone\n",
-        "db=0.050000m\n",
-        "\t0.015193 \t\t\t0.003335 \t\t\t\t0.052085 \t\t\t0.006930 \t\t\tdb=0.100000m\n",
-        "\t0.007104 \t\t\t0.001434 \t\t\t\t0.025121 \t\t\t0.002964 \t\t\tdb=0.150000m\n",
-        "\t0.004568 \t\t\t0.000875 \t\t\t\t0.016434 \t\t\t0.001803 \t\t\tdb=0.200000m\n",
-        "\t0.003345 \t\t\t0.000616 \t\t\t\t0.012173 \t\t\t0.001266 \t\t\t"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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dO2fNAg4cEE/b8IMfAL/+tXgiN4NB6rREZOu6XLMPDAzErl274OnpiYiICJw8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\nRU5ErkJggg==\n"
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing the points with the plot of Kbc,Kbe vs db in Fig.(12), we can conlcude the following:\n",
-        "Kbc for fine particles and helium: line 2 in Fig.(12)\n",
-        "Kbc for coarser particles and ozone: line 3 in Fig.(12)\n",
-        "Kbe for fine particles and helium: line 4 in Fig.(12)\n",
-        "Kbe for coarser particles and ozone: line 5 in Fig.(12)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 254\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "D=0.69;         #Diffusion coefficient of gas in cm**2/s\n",
-      "umf=1.0;        #Velocity at minimum fluidization condition in cm/s\n",
-      "ephsilonmf=0.5; #Void fraction at minimum fluidization condition\n",
-      "db=[5,15];      #Equilibrium bubble size in cm\n",
-      "g=980;          #Acceleration due to gravity in cm/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(db);\n",
-      "i=0;\n",
-      "Kbc = [0.,0.]\n",
-      "Kce = [0.,0.]\n",
-      "Kbe = [0.,0.]\n",
-      "e = [0.,0.]\n",
-      "while i<n:\n",
-      "    Kbc[i]=4.5*(umf/db[i])+5.85*((D**0.5*g**0.25)/db[i]**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(27)\n",
-      "    Kce[i]=6.77*((D*ephsilonmf*0.711*(g*db[i])**0.5)/db[i]**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(34)\n",
-      "    Kbe[i]=(Kbc[i]*Kce[i])/(Kbc[i]+Kce[i]);#Gas interchange coefficient between bubble and emulsion from Eqn.(14)\n",
-      "    e[i]=(Kce[i]-Kbe[i])/Kbe[i];#Error when minor resistance is ignored\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'db(cm)',\n",
-      "print '\\t\\tCalculated Kbc',\n",
-      "print '\\tCalculated Kce',\n",
-      "print '\\t\\tKbe from Eqn.(14)',\n",
-      "print '\\tErron when minor resistance is ignored (in percentage)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%db[i],\n",
-      "    print '\\t%f'%Kbc[i],\n",
-      "    print '\\t%f'%Kce[i],\n",
-      "    print '\\t\\t%f'%Kbe[i],\n",
-      "    print '\\t\\t%f'%(e[i]*100);\n",
-      "    i=i+1;   \n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "db(cm) \t\tCalculated Kbc \tCalculated Kce \t\tKbe from Eqn.(14) \tErron when minor resistance is ignored (in percentage)\n",
-        "5.000000 \t6.337721 \t2.509152 \t\t1.797506 \t\t39.590766\n",
-        "15.000000 \t2.112574 \t0.635514 \t\t0.488547 \t\t30.082477\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 255\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "Kbe=[0.028,0.05];  #Reported range for gas interchange coefficient between bubble and emulsion\n",
-      "uo=0.30;           #Superficial gas velocity in m/s\n",
-      "db=0.13;           #Equilibrium bubble size in m\n",
-      "m=7;\n",
-      "ephsilonmf=0.5;    #Void fraction at minimum fluidization condition\n",
-      "umf=0.0018;        #Velocity at minimum fluidization condition in m/s\n",
-      "D=[9E-6,22E-6];    #Diffusion coefficient of gas in m**2/s\n",
-      "g=9.81;            #Acceleration due to gravity in m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(Kbe);\n",
-      "i=0;\n",
-      "Kbem = [0,0]\n",
-      "Kbc = [0,0]\n",
-      "Kce = [0,0]\n",
-      "#Kbe = [0,0]\n",
-      "c = [0,0]\n",
-      "\n",
-      "while i<n:\n",
-      "    Kbem[i]=(6.0/db)*Kbe[i];#Gas interchange coefficient between bubble and emulsion from Eqn.(19)\n",
-      "    Kbc[i]=4.5*(umf/db)+5.85*((D[i]**0.5*g**0.25)/db**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(27)\n",
-      "    Kce[i]=6.77*((D[i]*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(34)\n",
-      "    Kbe[i]=(Kbc[i]*Kce[i])/(Kbc[i]+Kce[i]);#Gas interchange coefficient between bubble and emulsion from Eqn.(14)\n",
-      "    c[i]=(Kbem[i]/Kbe[i]);\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Kbe from Eqn.(19)',\n",
-      "print '\\tKbc from Eqn.(27)',\n",
-      "print '\\tKce from Eqn.(34)',\n",
-      "print '\\tKbe from Eqn.(14)',\n",
-      "print '\\tComparison of Kbe from Eqn.(19) and that from Eqn.(14)'\n",
-      "i=0\n",
-      "while i<n:\n",
-      "    print '%f'%Kbem[i],\n",
-      "    print '\\t\\t%f'%Kbc[i],\n",
-      "    print '\\t\\t%f'%Kce[i],\n",
-      "    print '\\t\\t%f'%Kbe[i],\n",
-      "    print '\\t\\t%f'%c[i]\n",
-      "    i=i+1;   \n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Kbe from Eqn.(19) \tKbc from Eqn.(27) \tKce from Eqn.(34) \tKbe from Eqn.(14) \tComparison of Kbe from Eqn.(19) and that from Eqn.(14)\n",
-        "1.292308 \t\t0.460200 \t\t0.274548 \t\t0.171959 \t\t7.515188\n",
-        "2.307692 \t\t0.684401 \t\t0.429248 \t\t0.263797 \t\t8.747978\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch11.ipynb b/Fluidization_Engineering/ch11.ipynb
deleted file mode 100755
index 48b679d6..00000000
--- a/Fluidization_Engineering/ch11.ipynb
+++ /dev/null
@@ -1,356 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:69d28aca1d8ec525f15842df275d6e6c6108139c968df651029692a500222ce7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter  11  : Particle to Gas Mass and Heat Transfer"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "%matplotlib inline"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
-        "For more information, type 'help(pylab)'.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 265\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "db=0.37;         #Equilibrium bubble size in cm\n",
-      "dp=0.028;        #Particle size in cm\n",
-      "rhos=1.06;       #Density of solids in g/cc\n",
-      "ephsilonmf=0.5;  #Void fraction at minimum fluidization condition\n",
-      "phis=0.4;        #Sphericity of solids\n",
-      "gammab=0.005;    #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "rhog=1.18E-3;    #Density of air in g/cc\n",
-      "myu=1.8E-4;      #Viscosity of gas in g/cm s\n",
-      "D=0.065;         #Diffusion coefficient of gas in cm**2/s\n",
-      "Sc=2.35;         #Schmidt number\n",
-      "etad=1;          #Adsorption efficiency factor\n",
-      "y=1;\n",
-      "umf=1.21;        #Velocity at minimum fluidization condition in cm/s\n",
-      "ut=69;           #Terminal velocity in cm/s\n",
-      "g=980;           #Acceleration due to gravity in square cm/s**2\n",
-      "uo=[10,20,30,40,50];#Superficial gas velocity in cm/s\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(uo);\n",
-      "i=0;\n",
-      "Rept=(dp*ut*rhog)/myu;\n",
-      "Shstar=2+(0.6*(Rept**0.5)*(Sc**(1/3)));#Sherwood no. from Eqn.(1)\n",
-      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "ubr=0.711*(g*db)**0.5;#Rise velocity of the bubble\n",
-      "x = [0,0,0,0,0]\n",
-      "Shbed = [0,0,0,0,0]\n",
-      "Rep = [0,0,0,0,0]\n",
-      "while i<n:\n",
-      "    x[i]=(uo[i]-umf)/(ubr*(1-ephsilonmf));#The term delta/(1-epshilonf) after simplification\n",
-      "    Shbed[i]=x[i]*((gammab*Shstar*etad)+((phis*dp**2*y)/(6*D))*Kbc);#Sherwood no. from Eqn.(11)\n",
-      "    Rep[i]=(dp*uo[i]*rhog)/myu;#Reynolds of the particle\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The desired result is the relationship between Shbed and Rep  The points gives a straight line of the form y=mx+c'\n",
-      "print 'Rep',\n",
-      "print '\\t\\tShbed'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%Rep[i],\n",
-      "    print '\\t%f'%Shbed[i]\n",
-      "    i=i+1;\n",
-      "import matplotlib.pyplot as plt\n",
-      "plt.plot(Rep,Shbed);\n",
-      "plt.xlabel(\"Rep\");\n",
-      "plt.ylabel(\"Shbed\");\n",
-      "\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "The desired result is the relationship between Shbed and Rep  The points gives a straight line of the form y=mx+c\n",
-        "Rep \t\tShbed\n",
-        "1.835556 \t0.065762\n",
-        "3.671111 \t0.140576\n",
-        "5.506667 \t0.215391\n",
-        "7.342222 \t0.290206\n",
-        "9.177778 \t0.365020\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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RlJR0Rnn9weLFMHEijB5t3b+hXTu7E4mIPygsLKSwsPCMx3u9QDjOYjLck7HH\nF4iWqroa7r8fPvgA3nwTrrvO7kQi4k9++uF5xowZHo33+iRFSEgIFRUV7ucVFRU4nc4mH9vSrF0L\nsbHw7bfW8lUVBxFpbl4vEHFxcZSVlVFeXk5tbS25ubmkpqY2eKwx5ozHtlR1dfDHP1p7KD36qNU5\ndOpkdyoRaY28PsUUEBDArFmzSElJweVykZGRQVRUFNnZ2QBkZmayZ88e4uPj+e6772jTpg0zZ85k\n8+bNdOjQocGxrcX27XD77dC+vXUBXCttnkTERzjMTz/G+wGHw3FC9+HPjLHu8DZtGjz0ENx9t1Yo\niYj3efreqb2YbLZ3L0yaZN27Ydky6NvX7kQiIhZ9TrXRBx9YO65GRUFRkYqDiPgWdRA2OHwY7rsP\n/vEPmDcPrrnG7kQiIidSB9HMioqsruH7763lqyoOIuKr1EE0kyNH4LHH4IUX4Lnn4Kab7E4kInJq\nKhDNoKwMbrvNup6htBS6dbM7kYjI6WmKqQkZA9nZcNVV1vUN+fkqDiLiP9RBNJGvvrI22KushOXL\nrZVKIiL+RB1EE8jLs05ER0fDmjUqDiLin9RBeNGhQ3DvvbB0KbzzDlx9td2JRETOnDoIL1m92uoa\njh6FDRtUHETE/6mDOEtHjli7rr70Ejz/PKSl2Z1IRMQ7VCDOwrZt1vLVoCBr+WpwsN2JRES8R1NM\nZ8AY62K3q6+GCRPgww9VHESk5VEH4aHdu62i8PXXsHIlXHGF3YlERJqGOggP/Otf1m1A4+Nh1SoV\nBxFp2XTDIA8cOQKffgpXXtnsv1pE5Kx5+t6pAiEi0kp4+t6pKSYREWmQCoSIiDRIBUJERBqkAiEi\nIg1qkgKRn59PZGQkERERZGVlNXjM3XffTUREBDExMZSWlrq/HxYWRnR0NLGxsQwcOLAp4jWJwsJC\nuyM0yBdzKVPjKFPj+WIuX8zkKa8XCJfLxdSpU8nPz2fz5s3k5OSwZcuWescsWrSIzz//nLKyMl58\n8UUmT57sfs3hcFBYWEhpaSnFxcXejtdkfPUfgy/mUqbGUabG88VcvpjJU14vEMXFxYSHhxMWFkZg\nYCDp6enk5eXVO2bhwoWMGzcOgISEBA4ePMhXX33lfl1LWEVE7Of1AlFZWUloaKj7udPppLKystHH\nOBwOkpOTiYuL46WXXvJ2PBERaSzjZe+8846ZOHGi+/kbb7xhpk6dWu+YX/7yl+aTTz5xP7/hhhvM\n+vXrjTHa1M3bAAAIQElEQVTGVFZWGmOM2bt3r4mJiTHLly8/4XcA+tKXvvSlrzP48oTXN+sLCQmh\noqLC/byiogKn03nKY3bt2kVISAgA3bp1AyAoKIiRI0dSXFxMYmJivfFGU1AiIk3O61NMcXFxlJWV\nUV5eTm1tLbm5uaSmptY7JjU1lTlz5gCwZs0aOnXqRNeuXamurubQoUMAHD58mIKCAvr16+ftiCIi\n0ghe7yACAgKYNWsWKSkpuFwuMjIyiIqKIjs7G4DMzEyGDRvGokWLCA8P5/zzz+fVV18FYM+ePaT9\neEu2uro6xowZw5AhQ7wdUUREGsOjCSmbjR8/3lxyySWmb9++dkdx27lzp0lKSjK9e/c2ffr0MTNn\nzrQ7kqmpqTEDBw40MTExJioqyvzP//yP3ZHc6urqTP/+/c0vf/lLu6O4XX755aZfv36mf//+Jj4+\n3u44xhhjDhw4YEaNGmUiIyNNVFSUWb16ta15tm7davr37+/+uvDCC33i3/rjjz9uevfubfr27WtG\njx5tvv/+e7sjmWeeecb07dvX9OnTxzzzzDO25Wjo/XL//v0mOTnZREREmMGDB5sDBw6c8mf4VYFY\nvny5KSkp8akCsXv3blNaWmqMMebQoUOmV69eZvPmzTanMubw4cPGGGOOHDliEhISzIoVK2xOZHn6\n6afNrbfeaoYPH253FLewsDCzf/9+u2PUM3bsWDN79mxjjPXf8ODBgzYn+g+Xy2UuvfRSs3PnTltz\n7Nixw3Tv3t1dFG655Rbz2muv2Zrp008/NX379jU1NTWmrq7OJCcnm88//9yWLA29X95///0mKyvL\nGGPME088YR544IFT/gy/2mojMTGRiy66yO4Y9Vx66aX0798fgA4dOhAVFcWXX35pcypo3749ALW1\ntbhcLjp37mxzImsxwqJFi5g4caLPLTTwpTzffvstK1asYMKECYA1bduxY0ebU/3HkiVL6NmzZ72l\n6na48MILCQwMpLq6mrq6Oqqrq92LXeyydetWEhISaNeuHW3btuXaa6/l3XfftSVLQ++Xx1+DNm7c\nOBYsWHDKn+FXBcLXlZeXU1paSkJCgt1ROHr0KP3796dr165cd9119O7d2+5I3Hvvvfzv//4vbdr4\n1j87X7v2ZseOHQQFBTF+/HiuvPJKJk2aRHV1td2x3ObNm8ett95qdww6d+7Mb37zGy677DK6detG\np06dSE5OtjVT3759WbFiBd988w3V1dV8+OGH7Nq1y9ZMx/vqq6/o2rUrAF27dq13gXJDfOv/VD9W\nVVXFTTfdxMyZM+nQoYPdcWjTpg0bNmxg165dLF++3PbL/j/44AMuueQSYmNjferTOsDKlSspLS1l\n8eLFPPfcc6xYscLWPHV1dZSUlDBlyhRKSko4//zzeeKJJ2zNdExtbS3vv/8+N998s91R2L59O888\n8wzl5eV8+eWXVFVVMXfuXFszRUZG8sADDzBkyBCGDh1KbGysz30gOsbhcOBwOE55jG8m9zNHjhxh\n1KhR3HbbbYwYMcLuOPV07NiRG2+8kXXr1tmaY9WqVSxcuJDu3bszevRoli5dytixY23NdExwcDBQ\n/9obOzmdTpxOJ/Hx8QDcdNNNlJSU2JrpmMWLFzNgwACCgoLsjsK6deu46qqr6NKlCwEBAaSlpbFq\n1Sq7YzFhwgTWrVvHP//5Tzp16sQVPnTz+q5du7Jnzx4Adu/ezSWXXHLK41UgzpIxhoyMDHr37s09\n99xjdxwAvv76aw4ePAhATU0NH330EbGxsbZmevzxx6moqGDHjh3MmzeP66+/3n0tjJ188dqbSy+9\nlNDQUD777DPAmvPv06ePrZmOycnJYfTo0XbHAKxP62vWrKGmpgZjDEuWLPGJqdS9e/cCsHPnTt57\n7z2fmI47JjU1lddffx2A119//fQfaJvqDHpTSE9PN8HBweacc84xTqfTvPLKK3ZHMitWrDAOh8PE\nxMS4lwAuXrzY1kybNm0ysbGxJiYmxvTr1888+eSTtub5qcLCQp9ZxfTFF1+YmJgYExMTY/r06WMe\nf/xxuyMZY4zZsGGDiYuLM9HR0WbkyJE+sYqpqqrKdOnSxXz33Xd2R3HLyspyL3MdO3asqa2ttTuS\nSUxMNL179zYxMTFm6dKltuU49n4ZGBjofr/cv3+/ueGGGxq9zNVhjI9NCIuIiE/QFJOIiDRIBUJE\nRBqkAiEiIg1SgRARkQapQIh4oG3btsTGxhIdHU1aWhpVVVV2RxJpMioQIh5o3749paWlbNq0iQsv\nvNC9jb1IS6QCIXKGBg0axPbt2wFr24ehQ4cSFxfHNddcw7Zt2wC44447uOuuu4iPj+eKK67gww8/\ntDOyiEe8fsMgkdbA5XJRUFDADTfcAMCdd95JdnY24eHhFBUVMWXKFD7++GPAuqJ27dq1fP7551x3\n3XVs376dc845x874Io2iAiHigZqaGmJjY6msrCQsLIy77rqLqqoqVq9eXW8Du9raWsDaEO2WW24B\nIDw8nB49erBlyxZiYmJsyS/iCRUIEQ+cd955lJaWUlNTQ0pKCnl5eSQnJ9OpUydKS0sb9TN8dXdP\nkZ/Sv1SRM3Deeefx17/+lYceeogOHTrQvXt33nnnHcDawHHTpk3ux2+//TbGGLZv384XX3zhU7t7\nipyKCoSIB47fP79///6Eh4fz97//nblz5zJ79mz69+9P3759Wbhwofv4yy67jIEDBzJs2DCys7N1\n/kH8hjbrE2lC48ePZ/jw4aSlpdkdRcRj6iBERKRB6iBERKRB6iBERKRBKhAiItIgFQgREWmQCoSI\niDRIBUJERBqkAiEiIg36/9fGcNvjU2I2AAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2b26690>"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 267"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "\n",
-      "#INPUT\n",
-      "umf=0.12            #Velocity at minimum fluidization condition in cm/s\n",
-      "uo=40.;              #Superficial gas velocity in cm/s\n",
-      "ub=120;             #Velocity of the bubble in cm/s\n",
-      "D=0.7;              #Diffusion coefficient of gas in cm**2/s\n",
-      "abkbe1=1.;           #Bubble-emuslion interchange coefficient for non absorbing particles(m=0)\n",
-      "abkbe2=18.;          #Bubble-emuslion interchange coefficient for highly absorbing particles(m=infinity)\n",
-      "g=980.;              #Acceleration due to gravity in square cm/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "#For non absorbing particles m=0,etad=0\n",
-      "Kbc=(ub/uo)*(abkbe1);\n",
-      "dbguess=2;#Guess value of db\n",
-      "def solver_func(db):        #Function defined for solving the system\n",
-      "    return abkbe1-(uo/ub)*(4.5*(umf/db)+5.85*(D**0.5*g**0.25)/(db**(5/4.)));#Eqn.(10.27)\n",
-      "\n",
-      "d=fsolve(solver_func,dbguess)\n",
-      "#For highly absorbing particles m=infinity, etad=1\n",
-      "M=abkbe2-(uo/ub)*Kbc;\n",
-      "#For intermediate condition\n",
-      "alpha=100.;\n",
-      "m=10.;\n",
-      "etad=1./(1+(alpha/m));#Fitted adsorption efficiency factor from Eqn.(23)\n",
-      "abkbe3=M*etad+(uo/ub)*Kbc;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'For non absorbing particles:\\tDiameter of bubble=%fcm\\tBubble-cloud interchange coefficient=%fs**-1'%(d,Kbc);\n",
-      "print 'For highly absorbing partilces:\\tM=%f'%(M);\n",
-      "print 'For intermediate condition:\\tFitted adsorption efficiency factor:%f\\tBubble-emuslion interchange coefficient:%fs**-1'%(etad,abkbe3);\n",
-      "\n",
-      "#====================================END OF PROGRAM ======================================================"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For non absorbing particles:\tDiameter of bubble=6.010032cm\tBubble-cloud interchange coefficient=3.000000s**-1\n",
-        "For highly absorbing partilces:\tM=17.000000\n",
-        "For intermediate condition:\tFitted adsorption efficiency factor:0.090909\tBubble-emuslion interchange coefficient:2.545455s**-1\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 273\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "rhos=1.3;                #Density of solids in g/cc\n",
-      "phis=0.806;              #Sphericity of solids\n",
-      "gammab=0.001;            #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "rhog=1.18E-3;            #Density of air in g/cc\n",
-      "Pr=0.69;                 #Prandtl number\n",
-      "myu=1.8E-4;              #Viscosity of gas in g/cm s\n",
-      "Cpg=1.00;                #Specific heat capacity of gas in J/g K\n",
-      "ephsilonmf=0.45;         #Void fraction at minimum fluidization condition\n",
-      "kg=2.61E-4;              #Thermal concuctivity of gas in W/cm k\n",
-      "dp=0.036;                #Particle size in cm\n",
-      "umf=6.5;                 #Velocity at minimum fluidization condition in cm/s\n",
-      "ut=150.;                 #Terminal velocity in cm/s\n",
-      "db=0.4;                  #Equilibrium bubble size in cm\n",
-      "etah=1;                  #Efficiency of heat transfer\n",
-      "uo=[10.,20.,30.,40.,50.];#Superficial gas velocity in cm/s\n",
-      "g=980.;                  #Acceleration due to gravity in square cm/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "Nustar=2+(((dp*ut*rhog)/myu)**0.5*Pr**(1./3));#Nusselt no. from Eqn.(25)\n",
-      "Hbc=4.5*(umf*rhog*Cpg/db)+5.85*((kg*rhog*Cpg)**0.5*g**0.25/db**(5./4));#Total heat interchange across the bubble-cloud boundary from Eqn.(32)\n",
-      "ubr=0.711*(g*db)**0.5;#Rise velocity of the bubble from Eqn.(6.7)\n",
-      "n=len(uo);\n",
-      "i=0;\n",
-      "x = [0,0,0,0,0]\n",
-      "Nubed = [0,0,0,0,0]\n",
-      "Rep = [0,0,0,0,0]\n",
-      "\n",
-      "while i<n:\n",
-      "    x[i]=(uo[i]-umf)/(ubr*(1-ephsilonmf));#The term delta/(1-epshilonf) after simplification\n",
-      "    Nubed[i]=x[i]*(gammab*Nustar*etah+(phis*dp**2/(6*kg))*Hbc);#Nusselt no. from Eqn.(36)\n",
-      "    Rep[i]=(dp*uo[i]*rhog)/myu;#Reynolds of the particle\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The desired result is the relationship between Nubed and Rep which is in the form of a straight line y=mx+c'\n",
-      "print 'Rep',\n",
-      "print '\\t\\tNubed'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%Rep[i],\n",
-      "    print '\\t%f'%Nubed[i]\n",
-      "    i=i+1;\n",
-      "import matplotlib.pyplot as plt\n",
-      "plt.plot(Rep,Nubed);\n",
-      "plt.xlabel(\"Rep\");\n",
-      "plt.ylabel(\"Nubed\");\n",
-      "plt.show()\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The desired result is the relationship between Nubed and Rep which is in the form of a straight line y=mx+c\n",
-        "Rep \t\tNubed\n",
-        "2.360000 \t0.046518\n",
-        "4.720000 \t0.179427\n",
-        "7.080000 \t0.312335\n",
-        "9.440000 \t0.445244\n",
-        "11.800000 \t0.578152\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Adu3aeT0okdYkL8986P/736YiGD9ezWBpXRqcGjrnnHM4fPhw7fPS0tIW6xGI+LIvvzTN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\nEYiIBDglAhGRAKdEICIS4JQIREQC3P8HJOZBYJAptYQAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x269c250>"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 274\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "rhog=1.2;       #Density of air in kg/m**3\n",
-      "myu=1.8E-5;     #Viscosity of gas in kg/m s\n",
-      "kg=2.6E-2;      #Thermal concuctivity of gas in W/m k\n",
-      "dp=1E-4;        #Particle size in m\n",
-      "rhos=8920;      #Density of solids in kg/m**3\n",
-      "Cps=390;        #Specific heat capacity of the solid in J/kg K\n",
-      "ephsilonf=0.5;  #Void fraction of the fluidized bed\n",
-      "umf=0.1;        #Velocity at minimum fluidization condition in m/s\n",
-      "uo=0.1;         #Superficial gas velocity in m/s\n",
-      "pi=3.14\n",
-      "\n",
-      "#CALCULATION\n",
-      "to=0;                 #Initial temperature of the bed\n",
-      "T=100;                #Temperature of the bed\n",
-      "t=0.99*T;             #Particle temperature i.e. when it approaches 1% of the bed temperature\n",
-      "mp=(pi/6)*dp**3*rhos; #Mass of the particle\n",
-      "A=pi*dp**2;           #Surface area of the particle\n",
-      "Rep=(dp*uo*rhog)/myu; #Reynold's no. of the particle\n",
-      "Nubed=0.0178;         #Nusselt no. from Fig.(6)\n",
-      "hbed1=(Nubed*kg)/dp;  #Heat transfer coefficient of the bed\n",
-      "t1=(mp*Cps/(hbed1*A))*math.log((T-to)/(T-t));#Time needed for the particle approach 1 percentage of the bed temperature in case(a)\n",
-      "hbed2=140*hbed1;#Since from Fig.(6) Nup is 140 times Nubed\n",
-      "t2=(mp*Cps/(hbed2*A))*math.log((T-to)/(T-t));#Time needed for the particle approach 1 percentage of the bed temperature in case(b)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Case(a):Using the whole bed coefficient from Fig.(6)'\n",
-      "print '\\tTime needed for the particle approach 1 percentage of the bed temperature is %.0fs'%t1\n",
-      "print 'Case(b):Uisng the single-particle coefficient of Eqn.(25),also shown in Fig.(6)'\n",
-      "print '\\tTime needed for the particle approach 1 percentage of the bed temperature is %.2fs'%t2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case(a):Using the whole bed coefficient from Fig.(6)\n",
-        "\tTime needed for the particle approach 1 percentage of the bed temperature is 58s\n",
-        "Case(b):Uisng the single-particle coefficient of Eqn.(25),also shown in Fig.(6)\n",
-        "\tTime needed for the particle approach 1 percentage of the bed temperature is 0.41s\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch12.ipynb b/Fluidization_Engineering/ch12.ipynb
deleted file mode 100755
index 1fc80a09..00000000
--- a/Fluidization_Engineering/ch12.ipynb
+++ /dev/null
@@ -1,396 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3cabb972e9b40cc3c2621280c95233b4046eb8d671e52d74d499a7e149a3d9aa"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12 : Conversion of Gas in Catalytic Reactions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 293\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Kr=10.;          #rate constant in m**3 gas/m**3 cat s\n",
-      "D=2E-5;          #Diffusion coefficient of gas in m**2/s\n",
-      "dpbar=68.;       #Average partilce size in micrometers\n",
-      "ephsilonm=0.5;   #Void fraction of fixed bed\n",
-      "gammab=0.005;    #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "ephsilonmf=0.55; #Void fraction at minimum fluidization condition\n",
-      "umf=0.006;       #Velocity at minimum fluidization condition in m/s\n",
-      "db=0.04;         #Equilibrium bubble size in m\n",
-      "Lm=0.7;          #Length of the bed in m\n",
-      "uo=0.1;          #Superficial gas velocity in m/s\n",
-      "dbed=0.26;       #Diameter of the bed in m\n",
-      "g=9.81;          #Acceleration due to gravity in square m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
-      "ub=uo-umf+ubr;#Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
-      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5./4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "Kce=6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
-      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
-      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
-      "gammac=(1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
-      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
-      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
-      "Lf=(1-ephsilonm)*Lm/(1-ephsilonf);#Length of fixed bed from Eqn.(6.19)\n",
-      "Krtou=Kr*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
-      "Kf=gammab*Kr+1/((1/Kbc)+(1/(gammac*Kr+1/((1/Kce)+(1/(gammae*Kr))))));#Raction rate for fluidized bed from Eqn.(14)\n",
-      "XA=math.exp(-1*Kf*Lf/ub);#Conversion from Eqn.(16)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The dimnesionless reaction rate group: %f'%Krtou\n",
-      "print 'The reaction rate for fluidized bed: %fs**-1'%Kf\n",
-      "print 'Conversion: %f'%XA\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dimnesionless reaction rate group: 35.000000\n",
-        "The reaction rate for fluidized bed: 1.979872s**-1\n",
-        "Conversion: 0.030056\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 298\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "umf=0.005;        #Velocity at minimum fluidization condition in m/s\n",
-      "ephsilonm=0.52;   #Void fraction of fixed bed\n",
-      "ephsilonmf=0.57;  #Void fraction at minimum fluidization condition\n",
-      "DA=8.1E-6;        #Diffusion coefficient of gas in m**2/s\n",
-      "DR=8.4E-6;        #Diffusion coefficient of gas in m**2/s\n",
-      "Lm=5;             #Length of the bed in m\n",
-      "dte=1;            #Diameter of tube in m\n",
-      "Kr1=1.5;          #rate constant in m**3 gas/m**3 cat s\n",
-      "Kr3=0.01;         #rate constant in m**3 gas/m**3 cat s\n",
-      "gammab=0.005;     #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "uo=0.45;          #Superficial gas velocity in m/s\n",
-      "db=0.05;          #Equilibrium bubble size in m from Fig.(6.8)\n",
-      "ub=1.5;           #Velocity of bubbles in bubbling bed in m/s from Fig.(6.11(a))\n",
-      "g=9.81;           #Acceleration due to gravity in square m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
-      "KbcA=4.5*(umf/db)+5.85*((DA**0.5*g**0.25)/db**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "KceA=6.77*((DA*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
-      "KbcR=4.5*(umf/db)+5.85*((DR**0.5*g**0.25)/db**(5./4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "KceR=6.77*((DR*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
-      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
-      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
-      "gammac=(1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
-      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
-      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
-      "Lf=(1-ephsilonm)*Lm/(1-ephsilonf);#Length of fixed bed from Eqn.(6.19)\n",
-      "Krtou=Kr1*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
-      "Kr12=Kr1;#Since the reactions are a special case of Denbigh scheme\n",
-      "Kr34=Kr3;\n",
-      "Kf1=(gammab*Kr12+1/((1/KbcA)+(1/(gammac*Kr12+1/((1/KceA)+(1/(gammae*Kr12)))))))*(delta/(1-ephsilonf));#Rate of reaction 1 for fluidized bed from Eqn.(14)\n",
-      "Kf3=(gammab*Kr34+1/((1/KbcR)+(1/(gammac*Kr34+1/((1/KceR)+(1/(gammae*Kr34)))))))*(delta/(1-ephsilonf));#Rate of reaction 2 for fluidized bed from Eqn.(14)\n",
-      "Kf12=Kf1;\n",
-      "Kf34=Kf3;\n",
-      "KfA=((KbcR*KceA/gammac**2+(Kr12+KceA/gammac+KceA/gammae)* \\\n",
-      "    (Kr34+KceR/gammac+KceR/gammae))*delta*KbcA*Kr12*Kr34/ \\\n",
-      "    (1-ephsilonf))/(((Kr12+KbcA/gammac)*(Kr12+KceA/gammae)+Kr12*KceA/gammac) \\\n",
-      "        *((Kr34+KbcR/gammac)*(Kr34+KceR/gammae)+Kr34*KceR/gammac));\n",
-      "            #Rate of raection with respect to A from Eqn.(35)\n",
-      "KfAR=Kr1/Kr12*Kf12-KfA;#Rate of reaction from Eqn.(34)\n",
-      "tou=Lf*(1-ephsilonf)/uo;#Residence time from Eqn.(5)\n",
-      "XA=1-math.exp(-Kf1*tou);#Conversion of A from Eqn.(26)\n",
-      "XR=1-((KfAR/(Kf12-Kf34))*(math.exp(-Kf34*tou)-math.exp(-Kf12*tou)));#Conversion of R from Eqn.(27)\n",
-      "SR=(1-XR)/XA;#Selectivity of R\n",
-      "\n",
-      "#OUTPUT\n",
-      "\n",
-      "print 'Rate of reaction 1 for fluidized bed:%.4f'%Kf1\n",
-      "print 'Rate of reaction 2 for fluidized bed:%.4f'%Kf3\n",
-      "print 'Rate of reaction 1 with respect to A:%.4f'%KfA\n",
-      "print 'The Conversion of Napthalene:%.0f percentage'%(XA*100);\n",
-      "print 'The selectivity of Phthalic anhydride:%.0f percentage'%(SR*100);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rate of reaction 1 for fluidized bed:0.6007\n",
-        "Rate of reaction 2 for fluidized bed:0.0099\n",
-        "Rate of reaction 1 with respect to A:0.0058\n",
-        "The Conversion of Napthalene:96 percentage\n",
-        "The selectivity of Phthalic anhydride:95 percentage\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 302\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Kr=3.;           #rate constant in m**3 gas/m**3 cat s\n",
-      "db=0.12;         #Equilibrium bubble size in m\n",
-      "D=9E-5;          #Diffusion coefficient of gas in m**2/s\n",
-      "dpbar=68;        #Average partilce size in micrometers\n",
-      "ephsilonm=0.42;  #Void fraction of fixed bed\n",
-      "uo=0.4;          #Superficial gas velocity in m/s\n",
-      "Lm=0.8;          #Length of the bed in m\n",
-      "ephsilonmf=0.45; #Void fraction at minimum fluidization condition\n",
-      "umf=0.21;        #Velocity at minimum fluidization condition in m/s\n",
-      "gammab=0;        #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "g=9.81;          #Acceleration due to gravity in square m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "ubr=0.711*(g*db)**0.5;  #Rise velocity of bubble from Eqn.(6.7)\n",
-      "ub=uo-umf+ubr;          #Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
-      "ubstar=ub+3*umf;        #Rise velocity of the bubble gas from Eqn.(45)\n",
-      "delta=(uo-umf)/(ub+umf);#Fraction of bed in bubbles from Eqn.(6.46)\n",
-      "Kbe=4.5*(umf/db);       #Interchange coefficient between bubble and emulsion from Eqn.(47)\n",
-      "Lf=Lm*(1-ephsilonm)/((1-delta)*(1-ephsilonmf));#Length of fixed bed\n",
-      "phi=((Kr/Kbe)**2*((1-ephsilonmf)-gammab*(umf/ubstar))**2+ \\\n",
-      "        ((delta/(1-delta))+umf/ubstar)**2+2*(Kr/Kbe)*((1-ephsilonmf) \\\n",
-      "            -gammab*(umf/ubstar))*((delta/(1-delta))-umf/ubstar))**0.5;\n",
-      "            #From Eqn.(52)\n",
-      "            \n",
-      "q1=0.5*Kr/umf*((1-ephsilonmf)+gammab*(umf/ubstar))+0.5*Kbe/umf* \\\n",
-      "        (((delta/(1-delta))+umf/ubstar)-phi);#From Eqn.(50)\n",
-      "q2=0.5*Kr/umf*((1-ephsilonmf)+gammab*(umf/ubstar))+0.5*Kbe/umf* \\\n",
-      "        (((delta/(1-delta))+umf/ubstar)+phi);#From Eqn.(50)\n",
-      "        \n",
-      "si1=0.5-0.5*((1-delta)/delta)*(umf/ubstar-Kr/Kbe*((1-ephsilonmf)- \\\n",
-      "        gammab*(umf/ubstar))-phi);#From Eqn.(51)\n",
-      "si2=0.5-0.5*((1-delta)/delta)*(umf/ubstar-Kr/Kbe*((1-ephsilonmf)- \\\n",
-      "        gammab*(umf/ubstar))+phi);#From Eqn.(51)\n",
-      "XA=1-(delta/(1-delta))*(1/(uo*phi))*((1-si2)*(si1*delta*ubstar+ \\\n",
-      "        (1-delta)*umf)*math.exp(-q1*Lf)+(si1-1)* \\\n",
-      "        (si2*delta*ubstar+(1-delta)*umf)*math.exp(-q2*Lf));\n",
-      "        #Conversion from Eqn.(49)\n",
-      "        \n",
-      "Krtou=Kr*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'COmparing the values of 1-XA = %f and Krtou = %f with Fig.(6), \\\n",
-      "we can conlcude that this operating condition is shown as point \\\n",
-      "A in Fig.(3)'%(1-XA,Krtou);\n",
-      "print 'Line 2 gives the locus of conversions for different values of the \\\n",
-      "reaction rate group for this fluidized contacting'\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "COmparing the values of 1-XA = 0.113843 and Krtou = 3.480000 with Fig.(6), we can conlcude that this operating condition is shown as point A in Fig.(3)\n",
-        "Line 2 gives the locus of conversions for different values of the reaction rate group for this fluidized contacting\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 305\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "uo=0.25;        #Superficial gas velocity in m/s\n",
-      "db=0.025;       #Equilibrium bubble size in m\n",
-      "Kr=1.5;         #rate constant in m**3 gas/m**3 cat s\n",
-      "umf=0.21;       #Velocity at minimum fluidization condition in m/s\n",
-      "Lm=0.8;         #Length of the bed in m\n",
-      "ephsilonm=0.42; #Void fraction of fixed bed\n",
-      "g=9.81;         #Acceleration due to gravity in square m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
-      "ub=uo-umf+ubr;#Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
-      "delta=(uo-umf)/(ub+2*umf);#Fraction of bed in bubbles from Eqn.(55) since ub/umf<<1 \n",
-      "XA=1-math.exp(-Kr*Lm*((1-ephsilonm)/uo)*(umf/uo)*(1-delta));#Conversion from Eqn.(57)\n",
-      "Krtou=Kr*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
-      "\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Comparing the values of 1-XA = %f and Krtou = %f with Fig.(6), \\\n",
-      "we can conlcude that this operating condition is shown \\\n",
-      "as point B in Fig.(3)'%(1-XA,Krtou);\n",
-      "print 'Line 3 gives the locus of conversions for different values \\\n",
-      "of the reaction rate group for this fluidized contacting'\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing the values of 1-XA = 0.108243 and Krtou = 2.784000 with Fig.(6), we can conlcude that this operating condition is shown as point B in Fig.(3)\n",
-        "Line 3 gives the locus of conversions for different values of the reaction rate group for this fluidized contacting\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5, Page 307\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "uo=0.3;          #Superficial gas velocity in m/s\n",
-      "Lf=1.1;          #Length of fixed bed in m\n",
-      "Hf=1.2;          #Length of freeboard in m\n",
-      "db=0.04;         #Equilibrium bubble size in m\n",
-      "umf=0.006;       #Velocity at minimum fluidization condition in m/s\n",
-      "ephsilonmf=0.55; #Void fraction at minimum fluidization condition\n",
-      "gammab=0.005;    #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "Kr=10.;          #rate constant in m**3 gas/m**3 cat s\n",
-      "D=2E-5;          #Diffusion coefficient of gas in m**2/s\n",
-      "g=9.81;          #Acceleration due to gravity in square m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
-      "ub=uo-umf+ubr;#Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
-      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5./4));\n",
-      "#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "\n",
-      "Kce=6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;\n",
-      "#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
-      "\n",
-      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
-      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
-      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
-      "gammac=(1-ephsilonmf)*((3.0/(ubr*ephsilonmf/umf-1))+fw);\n",
-      "#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
-      "\n",
-      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;\n",
-      "#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
-      "\n",
-      "Kf=(gammab*Kr)+1.0/((1.0/Kbc)+(1.0/(gammac*Kr+1.0/((1.0/Kce)+(1.0/(gammae*Kr))))));\n",
-      "#Raction rate for fluidized bed from Eqn.(14)\n",
-      "\n",
-      "XA=1-math.exp(-1*Kf*Lf/ub);#Conversion at the top of dense bed from Eqn.(16)\n",
-      "etabed=(Kf*delta)/(Kr*(1-ephsilonf));#Reactor efficiency from Eqn.(22)\n",
-      "a=0.6/uo  #Since uoa = 0.6s**-1 from Fig.(5)\n",
-      "adash=6.62; #From Fig.(5)\n",
-      "XA1=1-1.0/(math.exp(((1-ephsilonf)*Kr/(uo*a))*((1-math.exp(-a*Hf))- \\\n",
-      "    ((1-etabed)/(1+(adash/a)))*(1-math.exp(-(a+adash)*Hf)))));#Conversion from Eqn.(64)\n",
-      "XA2=1-(1.0-XA1)*(1.0-XA);#Conversion at the exit from Eqn.(64)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The conversion:'\n",
-      "print '\\tAt the top pf the dense bed: %d percentage'%(XA1*100)\n",
-      "print '\\tAt the reactor exit: %d percentage'%(XA2*100);\n",
-      "\n",
-      "#Disclaimer: The value of kf deviate from the one given in textbook, where as it is close to the value obtained by manual calculation. \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conversion:\n",
-        "\tAt the top pf the dense bed: 96 percentage\n",
-        "\tAt the reactor exit: 99 percentage\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch13.ipynb b/Fluidization_Engineering/ch13.ipynb
deleted file mode 100755
index 18536015..00000000
--- a/Fluidization_Engineering/ch13.ipynb
+++ /dev/null
@@ -1,308 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a97460b196d7b42e945dcfefc11684cc1c39c5847f8b0bdc9e3f6cdcd94bfcc3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter 13 : Heat Transfer between Fluidized Beds and Surfaces"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 331\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dp=57.0;         #Particle size in micrometer\n",
-      "rhos=940.0;      #Density of solids in kg/m**3\n",
-      "Cps=828.0;       #Specific heat capacity of the solid in J/kg K\n",
-      "ks=0.20;         #Thermal conductivity of solids in W/m k\n",
-      "kg=0.035;        #Thermal concuctivity of gas in W/m k\n",
-      "umf=0.006;       #Velocity at minimum fluidization condition in m/s\n",
-      "ephsilonmf=0.476;#Void fraction at minimum fluidization condition\n",
-      "do1=0.0254;      #Outside diameter of tube in m\n",
-      "L=1;\n",
-      "uo=[0.05,0.1,0.2,0.35];#Superficial gas velocity in m/s\n",
-      "nw=[2.,3.1,3.4,3.5];   #Bubble frequency in s**-1\n",
-      "g=9.81;                #Acceleration due to gravity in square m/s**2\n",
-      "\n",
-      "\n",
-      "#CALCULATION\n",
-      "dte=4.*do1*L/2.*L;     #Hydraulic diameter from Eqn.(6.13)\n",
-      "db=(1+1.5)*0.5*dte;    #Rise velocity of the bubble\n",
-      "ubr=0.711*(g*db)**0.5; #Rise velocity of bubble from Eqn.(6.7)\n",
-      "phib=0.19;#From Fig.(15) for ks/kg=5.7\n",
-      "ke=ephsilonmf*kg+(1-ephsilonmf)*ks*(1/((phib*(ks/kg))+(2/3.0)))\n",
-      "    #Effective thermal conductivity of bed from Eqn.(3) \n",
-      "    \n",
-      "n=len(uo);\n",
-      "i=0;\n",
-      "ub = [0,0,0,0]\n",
-      "delta = [0,0,0,0]\n",
-      "h = [0,0,0,0]\n",
-      "while i<n:\n",
-      "    ub[i]=uo[i]-umf+ubr; #Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
-      "    delta[i]=uo[i]/ub[i];#Fraction of bed in bubbles from Eqn.(6.29)\n",
-      "    h[i]=1.13*(ke*rhos*(1-ephsilonmf)*Cps*nw[i]*(1-delta[i]))**0.5;\n",
-      "        #Heat transfer coefficinet from Eqn.(18)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Superficial gas velocity(m/s)',\n",
-      "print '\\tHeat transfer coefficient(W/m**2 k)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%uo[i],\n",
-      "    print '\\t\\t\\t%f'%h[i]\n",
-      "    i=i+1;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Superficial gas velocity(m/s) \tHeat transfer coefficient(W/m**2 k)\n",
-        "0.050000 \t\t\t270.297375\n",
-        "0.100000 \t\t\t323.421769\n",
-        "0.200000 \t\t\t315.487604\n",
-        "0.350000 \t\t\t292.370761\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 332\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dp=80.;              #Particle size in micrometer\n",
-      "rhos=2550.;          #Density of solids in kg/m**3\n",
-      "Cps=756.;            #Specific heat capacity of the solid in J/kg K\n",
-      "ks=1.21;             #Thermal conductivity of solids in W/m k\n",
-      "kg=[0.005,0.02,0.2]; #Thermal concuctivity of gas in W/m k\n",
-      "ephsilonmf=0.476;    #Void fraction at minimum fluidization condition\n",
-      "\n",
-      "#CALCULATION\n",
-      "delta=0.5*(0.1+0.3);#For a gently fluidized bed\n",
-      "nw=3.;#Bubble frequency in s**-1 from Fig.(5.12) at about 30cm above the distributor\n",
-      "n=len(kg);\n",
-      "i=0;\n",
-      "x = [0,0,0]\n",
-      "while i<n:\n",
-      "    x[i]=ks/kg[i];#To find different values of ks/kg\n",
-      "    i=i+1;\n",
-      "\n",
-      "phib=[0.08,0.10,0.20];#From Fig.(15) for different values of ks/kg\n",
-      "i=0;\n",
-      "ke = [0,0,0]\n",
-      "h1 =[0,0,0]\n",
-      "while i<n:\n",
-      "    ke[i]=ephsilonmf*kg[i]+(1-ephsilonmf)*ks*(1/((phib[i]*(ks/kg[i]))+(2.0/3)))\n",
-      "    #Effective thermal conductivity of bed from Eqn.(3)\n",
-      "    \n",
-      "    h1[i]=1.13*(ke[i]*rhos*(1-ephsilonmf)*Cps*nw*(1-delta))**0.5;#Heat transfer coefficinet from Eqn.(18)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Thermal conductivity of Gas(W/m K))',\n",
-      "print '\\tMax. heat transfer coefficient(W/m**2 k)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%kg[i],\n",
-      "    print '\\t\\t\\t\\t%d'%h1[i]\n",
-      "    i=i+1;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thermal conductivity of Gas(W/m K)) \tMax. heat transfer coefficient(W/m**2 k)\n",
-        "0.005000 \t\t\t\t324\n",
-        "0.020000 \t\t\t\t567\n",
-        "0.200000 \t\t\t\t1157\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 332\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "rhos=2700.;        #Density of solids in kg/m**3\n",
-      "Cps=755.;          #Specific heat capacity of the solid in J/kg K\n",
-      "ks=1.2;            #Thermal conductivity of solids in W/m k\n",
-      "kg=0.028;          #Thermal concuctivity of gas in W/m k\n",
-      "ephsilonmf=0.476;  #Void fraction at minimum fluidization condition\n",
-      "dp1=10E-3;         #Particle size for which h=hmax in m\n",
-      "hmax=250.;         #Max. heat transfer coefficient in W/m**2 K \n",
-      "nw=5.;             #Bubble frequency in s**-1\n",
-      "delta=0.1;         #Fraction of bed in bubbles\n",
-      "deltaw=0.1;        #Fraction of bed in bubbles in wall region\n",
-      "dp=2E-3;           #Diameter of particle in m\n",
-      "\n",
-      "#CALCULATION\n",
-      "x=ks/kg;\n",
-      "phib=0.11;\n",
-      "phiw=0.17;\n",
-      "ke=ephsilonmf*kg+(1-ephsilonmf)*ks*(1/((phib*(ks/kg))+(2.0/3)));\n",
-      "#Effective thermal conductivity of bed from Eqn.(3)\n",
-      "\n",
-      "hpacket=1.13*(ke*rhos*(1-ephsilonmf)*Cps*nw/(1-deltaw))**0.5;\n",
-      "#Heat transfer coefficient for the packet of emulsion from Eqn.(11)\n",
-      "\n",
-      "ephsilonw=ephsilonmf;#Void fraction in the wall region\n",
-      "\n",
-      "kew=ephsilonw*kg+(1-ephsilonw)*ks*((phiw*(ks/kg)+(1.0/3))**-1);\n",
-      "#Effective thermal conductivity in the wall region with stagnant gas from Eqn.(4)\n",
-      "\n",
-      "y=(2*kew/dp1)+(hmax*hpacket)/(((1-deltaw)*hpacket)-hmax);\n",
-      "#Calculating the term alphaw*Cpg*rhog*uo from Eqn.(16) by rearranging it\n",
-      "\n",
-      "h=(1-deltaw)/((2*kew/dp+y*(dp/dp1)**0.5)**-1+hpacket**-1);\n",
-      "#Heat transfer coeeficient from Eqn.(11) by using the value of y \n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The heat transfer coefficient for paricle size of %fm = %fW/m**2 K'%(dp,h);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The heat transfer coefficient for paricle size of 0.002000m = 194.873869W/m**2 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 334\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Hf=4.0;   #Height of freeboard in m\n",
-      "uo=2.4;   #Superficial gas velocity in m/s\n",
-      "ho=350.0; #Heat transfer coefficient at the bottom of freeboard region in W/m**2 K\n",
-      "hg=20.0;  #Heat transfer coefficient in equivalent gas stream, but free of solids in W/m**2 K\n",
-      "\n",
-      "#CALCULATION\n",
-      "zf=[0,0.5,1,1.5,2,2.5,3,3.5,Hf];#Height above the top of the dense bubbling fluidized bed\n",
-      "hr=0;#Assuming heat transfer due to radiation is negligible\n",
-      "a=1.5/uo;#Since decay coefficient from Fig.(7.12), a*uo=1.5s**-1 \n",
-      "n=len(zf);\n",
-      "i=0;\n",
-      "h = []\n",
-      "while i<n:\n",
-      "    h.append((hr+hg)+(ho-hr-hg)*math.exp(-a*zf[i]/2.0));#Heat transfer coefficient from Eqn.(24) for zf=Hf\n",
-      "    i=i+1;\n",
-      "\n",
-      "hbar=(hr+hg)+2*(ho-hr-hg)*(1-math.exp(-a*Hf/2.0))/(a*Hf);#Mean heat transfer coefficient for the 4-m high freeboard from Eqn.(26)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The required relationship is h(W/m**2 K) vs. zf(m) as in Fig.(9a)'\n",
-      "print 'Height above the dense bubbling fluidized bed(m))',\n",
-      "print '\\tHeat transfer coefficient(W/m**2 k)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%zf[i],\n",
-      "    print '\\t\\t\\t\\t\\t\\t%f'%h[i]\n",
-      "    i=i+1;\n",
-      "\n",
-      "print '\\nThe mean heat transfer coefficient for the 4-m high freeboard =%d W/m**2 K'%hbar\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required relationship is h(W/m**2 K) vs. zf(m) as in Fig.(9a)\n",
-        "Height above the dense bubbling fluidized bed(m)) \tHeat transfer coefficient(W/m**2 k)\n",
-        "0.000000 \t\t\t\t\t\t350.000000\n",
-        "0.500000 \t\t\t\t\t\t302.263958\n",
-        "1.000000 \t\t\t\t\t\t261.433158\n",
-        "1.500000 \t\t\t\t\t\t226.508723\n",
-        "2.000000 \t\t\t\t\t\t196.636271\n",
-        "2.500000 \t\t\t\t\t\t171.085009\n",
-        "3.000000 \t\t\t\t\t\t149.229857\n",
-        "3.500000 \t\t\t\t\t\t130.536154\n",
-        "4.000000 \t\t\t\t\t\t114.546583\n",
-        "\n",
-        "The mean heat transfer coefficient for the 4-m high freeboard =208 W/m**2 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch14.ipynb b/Fluidization_Engineering/ch14.ipynb
deleted file mode 100755
index 24a0ffbb..00000000
--- a/Fluidization_Engineering/ch14.ipynb
+++ /dev/null
@@ -1,416 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a7839feeb371e4231dbf99a0d3738674ff633956f5fb373aea54d56b513c13f8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14 : The RTD and Size Distribution of Solids in Fluidized Beds"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 343"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "\n",
-      "#INPUT\n",
-      "Fo=2.7;                         #Feed rate in kg/min\n",
-      "Fof=0.9;                        #Feed rate of fines in feed in kg/min\n",
-      "Foc=1.8;                        #Feed rate of coarse in feed in kg/min\n",
-      "W=17.;                           #Bed weight in kg\n",
-      "kf=0.8;                         #Elutriation of fines in min**-1\n",
-      "kc=0.0125;                      #Elutriation of coarse in min**-1\n",
-      "\n",
-      "#CALCULATION\n",
-      "F1guess=1;                      #Guess value of F1\n",
-      "def solver_func(F1):            #Function defined for solving the system\n",
-      "    return F1-(Fof/(1.+(W/F1)*kf))-(Foc/(1.+(W/F1)*kc));#Eqn.(17)\n",
-      "\n",
-      "F1=fsolve(solver_func,F1guess)\n",
-      "F1f=Fof/(1.+(W/F1)*kf);          #Flow rate of fines in entrained streams from Eqn.(16)\n",
-      "F1c=Foc/(1.+(W/F1)*kc);          #Flow rate of coarse in entrained streams from Eqn.(16)\n",
-      "F2f=Fof-F1f;                    #Flow rate of fines in overflow streams from Eqn.(9)\n",
-      "F2c=Foc-F1c;                    #Flow rate of coarse in overflow streams from Eqn.(9)\n",
-      "tbarf=1./((F1/W)+kf);            #Mean residence time of fines from Eqn.(12)\n",
-      "tbarc=1./((F1/W)+kc);            #Mean residence time of coarse from Eqn.(12)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Flow rate in entrained stream:\\tFines:%fkg/min\\tCoarse:%fkg/min'%(F1f,F1c);\n",
-      "print 'Flow rate in overflow stream:\\tFines:%fkg/min\\tCoarse:%fkg/min'%(F2f,F2c);\n",
-      "print 'Mean residence time:\\tFines:%fmins\\tCoarse:%fmins'%(tbarf,tbarc);\n",
-      "\n",
-      "#====================================END OF PROGRAM ======================================================"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flow rate in entrained stream:\tFines:0.100000kg/min\tCoarse:1.600000kg/min\n",
-        "Flow rate in overflow stream:\tFines:0.800000kg/min\tCoarse:0.200000kg/min\n",
-        "Mean residence time:\tFines:1.111111mins\tCoarse:8.888889mins\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 344\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "\n",
-      "import math\n",
-      "from numpy import linspace,array,zeros\n",
-      "from scipy.optimize import fsolve\n",
-      "from matplotlib.pyplot import *\n",
-      "%matplotlib inline\n",
-      "#Variable declaration\n",
-      "dt=4.;          #Diameter of reactor in m\n",
-      "ephsilonm=0.4; #Void fraction of static bed\n",
-      "rhos=2500.;     #Density of solid in the bed in kg/m**3\n",
-      "Lm=1.2;        #Height of static bed in m\n",
-      "Fo=3000;       #Feed rate in kg/hr\n",
-      "beta1=1.2;     #Increase in density of solids\n",
-      "dp=array([3,4,5,6,7,8,9,10,11,12,3,14,16,18,20,22,24,26,28,30])*10**-2;#Size of particles in mm\n",
-      "po=[0,0.3,0.8,1.3,1.9,2.6,3.5,4.4,5.7,6.7,7.5,7.8,7.5,6.3,5.0,3.6,2.4,1.3,0.5,0];#Size distribution of solids in mm**-1\n",
-      "k=array([0,10,9.75,9.5,8.75,7.5,6.0,4.38,2.62,1.20,0.325,0,0,0,0,0,0,0,0,0])*10**-4;#Elutriation constant in s**-1\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "W=(pi/4*dt**2)*Lm*(1-ephsilonm)*rhos;#Weight of solids in bed\n",
-      "n=len(dp);\n",
-      "i=0;\n",
-      "F1guess=1000.;#Guess value for F1\n",
-      "F1c=linspace(2510,2700,10);\n",
-      "F1 = zeros(n)\n",
-      "x = zeros(n)\n",
-      "c = zeros(n)\n",
-      "a = zeros(n)\n",
-      "while i<n:\n",
-      "    if k[i]==0:\n",
-      "        x[i]=0\n",
-      "        #break \n",
-      "    else:\n",
-      "        x[i]=0#(float(po[i])/(W*k[i]/float(F1)))*math.log(1.+(W*k[i]/F1));         \n",
-      "    def solver_func(Fo):\n",
-      "        return F1/(Lm*Fo)-x[i];\n",
-      "\n",
-      "    F1[i] = fsolve(solver_func,F1guess);#Using inbuilt function fsolve for solving Eqn.(20) for F1\n",
-      "    #c[i]=F1c[i]/(Lm*Fo);\n",
-      "    if F1[i]==0:\n",
-      "        a[i]=0;\n",
-      "    else:\n",
-      "        a[i]=(po[i]/(W*k[i]/F1[i]))*math.log(1+(W*k[i]/F1[i]));\n",
-      "\n",
-      "    i=i+1;\n",
-      "\n",
-      "#plot(F1,c);\n",
-      "\n",
-      "#xtitle('F1 vs a,c','F1','a,c');\n",
-      "F1n=2500.;#The point were both the curves meet\n",
-      "F2=beta1*Fo-F1n;#Flow rate of the second leaving stream\n",
-      "j=0;\n",
-      "m=len(dp);\n",
-      "p1 = zeros(m)\n",
-      "p2 = zeros(m)\n",
-      "tbar = zeros(m)\n",
-      "while j<m:\n",
-      "    p1[j]=(1./F1n)*((Fo*po[j])/(1.+(W/F1n)*k[j]));#Size distribution of stream 1 in mm**-1 from Eqn.(16)\n",
-      "    p2[j]=k[j]*W*p1[j]/F2;#Size distribution of stream 2 in mm**-1 from Eqn.(7)\n",
-      "    if p1[j]==0 and p2[j]==0:\n",
-      "        tbar[j]=0;\n",
-      "    elif p1[j]==0:\n",
-      "        tbar[j]=(W*p1[j])/(F2*p2[j]);\n",
-      "    elif  p2[j]==0:\n",
-      "        tbar[j]=(W*p1[j])/(F1n*p1[j]);\n",
-      "    else:\n",
-      "        tbar[j]=(W*p1[j])/(F1n*p1[j]+F2*p2[j]);#Average time in hr from Eqn.(11)\n",
-      "    j=j+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Flow rate of stream 1:%fkg/hr'%F1n\n",
-      "print 'Flow rate of stream 2:%fkg/hr'%F2\n",
-      "j=0;\n",
-      "print 'tbar(hr)'\n",
-      "while j<m:\n",
-      "    print '%f'%tbar[j]\n",
-      "    j=j+1;\n",
-      "\n",
-      "#DISCLAIMER: The value obtained for tbar is deviating highly\n",
-      "#form the one given in textbook. However, the value obtained by manual calculation is close to #\n",
-      "#the   ones obtained from the program."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Flow rate of stream 1:2500.000000kg/hr"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Flow rate of stream 2:1100.000000kg/hr\n",
-        "tbar(hr)\n",
-        "0.000000\n",
-        "8.962153\n",
-        "8.964162\n",
-        "8.966171\n",
-        "8.972205\n",
-        "8.982279\n",
-        "8.994397\n",
-        "9.007522\n",
-        "9.021824\n",
-        "9.033397\n",
-        "9.040543\n",
-        "9.043200\n",
-        "9.043200\n",
-        "9.043200\n",
-        "9.043200\n",
-        "9.043200\n",
-        "9.043200\n",
-        "9.043200\n",
-        "9.043200\n",
-        "0.000000\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive', 'pi']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 351\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dp=1;  #Particle size in mm\n",
-      "Fo=10; #Feed rate in kg/min\n",
-      "k=0.1; #Particle shrinkage rate in mm/min\n",
-      "\n",
-      "#CALCULATION\n",
-      "R=k/2;             #Particle shrinkage rate in terms of radius\n",
-      "W=(Fo*dp/2)/(4*R); #Bed weight from Eqn.(42)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Weight of bed:%d kg' %W\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Weight of bed:25 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 352\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "dpi=[1.05,0.95,0.85,0.75,0.65,0.55,0.45,0.35,0.25,0.15,0.05]; #Mean size in mm\n",
-      "Fo=[0,0.5,3.5,8.8,13.5,17.0,18.2,17.0,13.5,7.3,0]#*10**-2  #Feed rate in kg/s\n",
-      "for i in range(len(Fo)):\n",
-      "    Fo[i] = Fo[i] * 10**-2\n",
-      "k=[0,0,0,0,0,0,0,0,2.0,12.5,62.5]#*10**-5;#Elutriation constant in s**-1\n",
-      "for i in range(len(k)):\n",
-      "    k[i] = k[i] * 10**-5\n",
-      "\n",
-      "R=-1.58*10**-5;#Rate of particle shrinkage in mm/s\n",
-      "deldpi=0.1;#Size intervals in mm\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(dpi);\n",
-      "m=1;#Starting with the largest value size interval that contains solids\n",
-      "W = [0]\n",
-      "while m<n-1:\n",
-      "    W.append((Fo[m]-R*W[m-1]/deldpi)/(k[m]-R/deldpi-3*R/dpi[m]));#From Eqn.(33)\n",
-      "    m=m+1;\n",
-      "\n",
-      "Wt=sum(W);#Total sum\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nTotal mass in the bed:%fkg'%Wt\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Total mass in the bed:7168.981263kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5, Page 353\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "dpi=[0.17,0.15,0.13,0.11,0.09,0.07,0.05,0.03,0.01];#Mean size of particles in mm\n",
-      "a=[0,0.95,2.45,5.2,10.1,23.2,35.65,20.0,2.45]#*10**-2;#Feed composition Fo(dpi)/Fo\n",
-      "for i in range(len(a)):\n",
-      "    a[i] = a[i] * 10**-2\n",
-      "\n",
-      "y=[0,0,0,0,0,0,0.625,10.225,159.25]#*10**-6;#Elutriation and cyclone efficiency k(dpi)(1-eta(dpi))\n",
-      "for i in range(len(y)):\n",
-      "    y[i] = y[i] * 10**-6\n",
-      "\n",
-      "\n",
-      "F=0.01;          #Rate at which solids are withdrawn in kg/s\n",
-      "W=40000;         #Weight of bed in kg\n",
-      "dp1=0.11         #Initial size in mm\n",
-      "dp2=0.085;       #Size after shrinking in mm\n",
-      "dpmin=0.01;      #Minimum size in mm\n",
-      "deldpi=2*10**-2; #Size inerval in mm\n",
-      "t=20.8;          #Time in days\n",
-      "si=1;\n",
-      "\n",
-      "#CALCULATION\n",
-      "kdash=math.log((dp1-dpmin)/(dp2-dpmin))/(t*24*3600);#Rate of particle shrinkage from Eqn.(24)\n",
-      "n=len(dpi);\n",
-      "m=1;\n",
-      "Fo=0.05;#Initial value of Fo\n",
-      "F1 = [0];\n",
-      "s=0;\n",
-      "c=0;\n",
-      "t=1E-6;\n",
-      "R = [0]\n",
-      "x = [0]\n",
-      "F1 = [0]\n",
-      "while m<n:\n",
-      "    R.append(-kdash*(dpi[m]-dpmin));#Rate of size change\n",
-      "    x.append((a[m]*Fo-W*R[m-1]*F1[m-1]/deldpi)/(F+(W*y[m])-(W*R[m]/deldpi)-3*W*R[m]/dpi[m]));#Eqn.(34)\n",
-      "    F1.append(x[m]*F);\n",
-      "    c=c+x[m];\n",
-      "    m=m+1;\n",
-      "    if abs(c-1)<t:\n",
-      "        break\n",
-      "    Fo=Fo+0.0001;#Incrementing Fo\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Feed rate with deldpi=%fmm is %fg/hr'%(deldpi,Fo);\n",
-      "i=0;\n",
-      "print 'Bed composition'\n",
-      "for i in x:\n",
-      "    print '%f'%(i*100)\n",
-      "    i=i+1;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Feed rate with deldpi=0.020000mm is 0.050800g/hr\n",
-        "Bed composition\n",
-        "0.000000\n",
-        "0.652911\n",
-        "1.859952\n",
-        "4.400781\n",
-        "9.668999\n",
-        "25.654298\n",
-        "28.575890\n",
-        "2.317749\n",
-        "0.019493\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch15.ipynb b/Fluidization_Engineering/ch15.ipynb
deleted file mode 100755
index 09c02451..00000000
--- a/Fluidization_Engineering/ch15.ipynb
+++ /dev/null
@@ -1,294 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b0605e72f5da8476b3859cf3dc90196dcbe80cabc58b8c8d37cf5b28482c7ee1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 15 : Circulation Systems"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 369\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "thalf=1;   #Half life of catalyst in s\n",
-      "F=960;     #Feed rate of oil in tons/day\n",
-      "W=50;      #Weight of the bed in tons\n",
-      "a=0.5;     #Activity after time equal to half life\n",
-      "abar=0.01; #Average activity of the catalyst\n",
-      "\n",
-      "#CALCULATION\n",
-      "Ka=-math.log(a)/thalf;#Rate constant is s**-1, assuming I order kinetics from Eqn.(12)\n",
-      "Fs=Ka*W*abar/(1-abar);#Circulation rate of solids from Eqn.(16)\n",
-      "x=(Fs*60*60*24.0)/F;  #Circulation rate per feed of oil\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nSolid recirculation per feed of oil =%ftons of solid circulated/ton feed oil'%x\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Solid recirculation per feed of oil =31.506690tons of solid circulated/ton feed oil\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 370\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "deltaHr1=1260.;                               #Enthalpy change during endothermic reaction in kJ/kg\n",
-      "deltaHr2=-33900.;                             #Enthal[y change during exothermic reaction in kJ/kg\n",
-      "H1=703.;                                      #Enthalpy of feed oil in kJ/kg\n",
-      "T1=260.;                                      #Temperature of feed oil in degree celcius\n",
-      "H3=1419.;                                     #Enthalpy of cracked product in kJ/kg\n",
-      "T3=500.;                                      #Temperature of cracked product in degree celcius\n",
-      "Ta=20.;                                       #Temperature of entering air in degree celcius\n",
-      "Cpa=1.09;                                     #Specific heat of entering air in kJ/kg K\n",
-      "Cpf=1.05;                                     #Specific heat of flue gases in kJ/kg K\n",
-      "Cps=1.01;                                     #Specific heat of solids in kJ/kg K\n",
-      "Cpv=3.01;                                     #Specific heat of vaporized feed in kJ/kg K\n",
-      "T4=[520.,540.,560.,580.,600.,620.,640.,660.]; #Temperature of flue gas in degree celcius\n",
-      "V=22.4;                                       #Volume of 1 mole of Carbon dioxide gas in N-m**3\n",
-      "M=12.;                                        #Molecular weight of carbon in kg\n",
-      "rho=1.293;                                    #Density of carbon dioxide gas in kg/N-m**3\n",
-      "xa=0.21;                                      #Mass fraction of oxygen in air\n",
-      "betac=0.07;                                   #Mass fraction of carbon\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(T4);\n",
-      "i=0;\n",
-      "x1 = [0,0,0,0,0,0,0,0]\n",
-      "x2 = [0,0,0,0,0,0,0,0]\n",
-      "excess_air = [0,0,0,0,0,0,0,0]\n",
-      "\n",
-      "x2min=betac*(V*rho/(M*xa));#Minimum amount of air required for complete combustion\n",
-      "while i<n:\n",
-      "    x1[i]=(deltaHr1+0.93*H3-H1)/(Cps*(T4[i]-T3));#Fs/F1 by simplifying the overall energy balance\n",
-      "    x2[i]=((0.07*(-deltaHr2)-(deltaHr1+0.93*H3-H1))/(Cpf*(T4[i]-Ta)))-0.07;#F2/F1 by simplifying the energy balance for regenerator\n",
-      "    if x2[i]>x2min:\n",
-      "        excess_air[i]=(x2[i]-x2min)/x2min; #Excess air used\n",
-      "    else:\n",
-      "        excess_air[i]=0;\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT    \n",
-      "print 'T4(degree celcius)',\n",
-      "print '\\tFs/F1',\n",
-      "print '\\t\\tF2/F1',\n",
-      "print '\\t\\tExcess air(percentage)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%T4[i],\n",
-      "    print '\\t\\t%f'%x1[i],\n",
-      "    print '\\t%f'%x2[i],\n",
-      "    print '\\t%f'%(excess_air[i]*100);\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Disclaimer: The values of F2/F1 obtained by manual calculation has close correspondance to the ones obtained as the output, whereas it deviates largely from the values given in textbook.\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "T4(degree celcius) \tFs/F1 \t\tF2/F1 \t\tExcess air(percentage)\n",
-        "520.000000 \t\t92.904455 \t0.875390 \t8.807235\n",
-        "540.000000 \t\t46.452228 \t0.839029 \t4.287699\n",
-        "560.000000 \t\t30.968152 \t0.805362 \t0.102944\n",
-        "580.000000 \t\t23.226114 \t0.774099 \t0.000000\n",
-        "600.000000 \t\t18.580891 \t0.744992 \t0.000000\n",
-        "620.000000 \t\t15.484076 \t0.717825 \t0.000000\n",
-        "640.000000 \t\t13.272065 \t0.692412 \t0.000000\n",
-        "660.000000 \t\t11.613057 \t0.668586 \t0.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 379\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "Fs=100.;        #Solid flowrate in kg/s\n",
-      "ephsilon1=0.55;\n",
-      "ephsilon2=0.5;\n",
-      "p1=120.;         #Pressure at upper level in kPa\n",
-      "rhos=1000.;      #Density of solid in kg/m**3\n",
-      "rhog=1.;         #Density of gas in kg/m**3\n",
-      "gc=1.;           #Conversion factor\n",
-      "g=9.81;          #Acceleration due to gravity in m/s**2\n",
-      "di=0.34;         #Diameter of downcomer in m\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "x=(ephsilon1/ephsilon2)*((1-ephsilon2)/(1-ephsilon1));#To find pressure at lower level using Eqn.(30)\n",
-      "p2=x*p1;#Pressure at lower level using Eqn.(30)\n",
-      "deltap=p2-p1;\n",
-      "ephsilonbar=0.5*(ephsilon1+ephsilon2);\n",
-      "deltah=(deltap*10**3*gc)/(rhos*(1-ephsilonbar)*g);#Static head height from Eqn.(28)\n",
-      "At=0.25*pi*di**2;#Area of downcomer\n",
-      "Gs=Fs/At;#Flux of solids in downcomer\n",
-      "Gg=Gs*(ephsilon1/(1-ephsilon1))*(rhog/rhos)*(x-1);#Required gas aeration rate from Eqn.(31)\n",
-      "Fg=Gg*At;#Flow rate of gas required\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nThe required flow rate of gas required for location of %fm below downcomer is %.4fkg/s'%(deltah,Fg)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The required flow rate of gas required for location of 5.722768m below downcomer is 0.0272kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Fs=600;#Solid circulation rate in kg/s\n",
-      "dpbar=60;#Mean size of solids in micrometer\n",
-      "pA=120;#Pressure in vessel A in kPa\n",
-      "pB=180;#Pressure in vessel B in kPa\n",
-      "LfA=8;#Bed height in vessel A in m\n",
-      "LfB=8;#Bed height in vessel B i m\n",
-      "#Bulk densities in kg/m**3\n",
-      "rho12=100;\n",
-      "rho34=400;\n",
-      "rho45=550;\n",
-      "rho67=200;\n",
-      "rho78=200;\n",
-      "rho910=400;\n",
-      "rho1011=400;\n",
-      "rho1112=550;\n",
-      "rho13=100;\n",
-      "deltapdA=7;#Pressure drop across the distributor in regenerator in kPa\n",
-      "deltapdB=8;#Pressure drop across the distributor in reactor in kPa\n",
-      "deltap12=(9+4);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa\n",
-      "deltap78=(15+3);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa\n",
-      "deltap45=20;#Friction loss across the reactor's stripper downcomer in kPa\n",
-      "deltap1112=4;#Friction loss across the regenerator's downcomer in kPa\n",
-      "deltapvA=5;#Pressure drop assigned for the control valve in regenerator in kPa\n",
-      "deltapvB=15;#Pressure drop assigned for the control valve in reactor in kPa\n",
-      "deltah12=15;#Height of the riser in m\n",
-      "deltah86=30;#Height of the riser in m\n",
-      "deltah1011=7;#Height difference h10-h11 in m\n",
-      "g=9.81;#Acceleration due to gravity in m/s**2\n",
-      "gc=1;#Conversion factor\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "Gs=900;#From Fig.(8), to find dt\n",
-      "dt=math.sqrt((4/math.pi)*Fs/Gs);#Diameter of the downcomer\n",
-      "#Height of downcomer A from Eqn.(7)\n",
-      "deltahA=(1/(rho1112*g))*((pB-pA)*gc*(10**3)+(deltap12+deltapdB+deltap1112+deltapvA)*gc*10**3-rho12*g*(-deltah12)-rho34*g*(-LfB)-rho1011*g*deltah1011);\n",
-      "#Height of downcomer B from Eqn.(8)\n",
-      "deltahB=(1/(rho45*g))*(-(pB-pA)*gc*10**3+(deltap45+deltapvB+deltap78+deltapdA)*gc*10**3+rho78*g*deltah86+rho910*g*LfA)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Height of downcomer for:'\n",
-      "print '\\tRegenerator:%d m'%deltahA\n",
-      "print '\\tReactor:%.1f m'%deltahB\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Height of downcomer for:\n",
-        "\tRegenerator:20 m\n",
-        "\tReactor:16.7 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch16.ipynb b/Fluidization_Engineering/ch16.ipynb
deleted file mode 100755
index e835e890..00000000
--- a/Fluidization_Engineering/ch16.ipynb
+++ /dev/null
@@ -1,428 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1a37ab741ccda7f67c7de1085d3807de9fa5a11a3111916ad9502a04f1c58721"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 16 : Design for Physical Operations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 404\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "T=1000.;       #Operating temperature of calciner in degree celcius\n",
-      "deltaHr=1795.; #Heat of reaction in kJ/kg\n",
-      "M1=0.1;        #Molecular weight of Calcium carbonate in kg/mol\n",
-      "M2=0.056;      #Molecular weight of CaO in kg/mol\n",
-      "M3=0.044;      #Molecular weight of Carbon dioxide  in kg/mol\n",
-      "M4=0.029;      #Molecular weight of Air in kg/mol\n",
-      "M5=0.029;      #Molecular weight of Combustion gas in kg/mol\n",
-      "Cp1=1.13;      #Specific heat of Calcium carbonate in kJ/kg K\n",
-      "Cp2=0.88;      #Specific heat of CaO in kJ/kg K\n",
-      "Cp3=1.13;      #Specific heat of Carbon dioxide in kJ/kg K\n",
-      "Cp4=1.00;      #Specific heat of Air in kJ/kg K\n",
-      "Cp5=1.13;      #Specific heat of Calcium carbonate in kJ/kg K\n",
-      "Tf=20.;        #Temperature of feed in degree celcius\n",
-      "ma=15.;        #Air required per kg of fuel in kg\n",
-      "Hc=41800.;     #Net combustion heat of fuel in kJ/kg\n",
-      "Tpi=20.;       #Initial temperature of solids in degree C\n",
-      "Tgi=1000.;     #Initial temperature of gas in degree C\n",
-      "\n",
-      "#CALCULATION\n",
-      "mc=1;#Based on 1 kg of Calcium carbonate\n",
-      "B=(1/(Hc-(ma+mc)*Cp5*(T-Tpi)))*(M3*Cp3*(T-Tf)+M2*Cp2*(T-Tf)+deltaHr)#Fuel consumption(kg fuel/kg calcium carbonate)\n",
-      "B1=B*M3/M2;#Fuel consumption(kg fuel/kg Cao)\n",
-      "H=Hc*B1;#Heat required for calcination\n",
-      "eta=deltaHr/(B*Hc);#Thermal efficiency\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Fuel consumption:%f kg fuel/kg Cao'%B1\n",
-      "print 'Heat requirement for calcination:%f kJ/kg Cao'%H\n",
-      "print 'Thermal efficiency:%f percentage'%(eta*100)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fuel consumption:0.061731 kg fuel/kg Cao\n",
-        "Heat requirement for calcination:2580.366029 kJ/kg Cao\n",
-        "Thermal efficiency:54.657251 percentage\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 405\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "import math\n",
-      "\n",
-      "#INPUT\n",
-      "F = 400.        #Feed rate of Calcium carbonate in tons/day\n",
-      "T = 1000.       #Operating temperature of calciner in degree celcius\n",
-      "deltaHr = 1795.#Heat of reaction in kJ/kg\n",
-      "M1 = 0.1        #Molecular weight of Calcium carbonate in kg/mol\n",
-      "M2 = 0.056     #Molecular weight of CaO in kg/mol\n",
-      "M3 = 0.044     #Molecular weight of Carbon dioxide  in kg/mol\n",
-      "M4 = 0.029    #Molecular weight of Air in kg/mol\n",
-      "M5 = 0.029    #Molecular weight of Combustion gas in kg/mol\n",
-      "Cp1 = 1.13    #Specific heat of Calcium carbonate in kJ/kg K\n",
-      "Cp2 = 0.88    #Specific heat of CaO in kJ/kg K\n",
-      "Cp3 = 1.13   #Specific heat of Carbon dioxide in kJ/kg K\n",
-      "Cp4 = 1.00    #Specific heat of Air in kJ/kg K\n",
-      "Cp5 = 1.17    #Specific heat of Combustion gas in kJ/kg K\n",
-      "Tf = 20.     #Temperature of feed in degree celcius\n",
-      "ma = 15.    #Air required per kg of fuel in kg\n",
-      "uo = 0.8    #Superficial gas velocity in m/s\n",
-      "Hc = 41800.   #Net combustion heat of fuel in kJ/kg\n",
-      "Tpi = 20.     #Initial temperature of solids in degree C\n",
-      "Tgi = 1000.   #Initial temperature of gas in degree C\n",
-      "rhoa = 1.293    #Density of air in kg/m**3\n",
-      "pi = 3.14\n",
-      "\n",
-      "#CALCULATION\n",
-      "mc = 1.         #Based on 1 kg of Calcium carbonate\n",
-      "Bguess = 2.     #Guess value of B\n",
-      "def solver_func(B):         #Function defined for solving the system\n",
-      "    phi = ((ma+mc)*Cp5*B+(M3*Cp3))/Cp1\n",
-      "    T3 = (Tpi+(phi+phi**2+phi**3)*Tgi)/(1+phi+phi**2+phi**3)\n",
-      "    phiplus = 30.6*B\n",
-      "    Tr = (T+Tpi*phiplus)/(1+phiplus)\n",
-      "    return Hc*B+Cp3*(T3-Tpi)+ma*B*Cp4*(Tr-20)-(ma+mc)*Cp5*(T-Tpi)-M3*Cp3*(T-Tpi)-M2*Cp2*(T-Tpi)-deltaHr\n",
-      "    #fn = (1/20800)*(2470-T3-13.34*(Tr-20))\n",
-      "\n",
-      "B = fsolve(solver_func,1E-6)#Using inbuilt function fsolve for solving Eqn.(23) for tou\n",
-      "phi = ((ma+mc)*Cp5*B+(M3*Cp3))/Cp1\n",
-      "#Temperature of various stages\n",
-      "T1 = (Tpi+(phi)*Tgi)/(1+phi)\n",
-      "T2 = (Tpi+(phi+phi**2)*Tgi)/(1+phi+phi**2)\n",
-      "T3 = (Tpi+(phi+phi**2+phi**3)*Tgi)/(1+phi+phi**2+phi**3)\n",
-      "phiplus = 30.6*B\n",
-      "Tr = (T+Tpi*phiplus)/(1+phiplus)\n",
-      "eta = deltaHr/(B*Hc)           #Thermal efficiency\n",
-      "H = B*Hc/M2                    #Heat requirement\n",
-      "#For lower heat recovery section\n",
-      "Ql = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(Tr+273)))#Volumetric flow rate of gas in the lower heat recovery section\n",
-      "dtl = math.sqrt(4/pi*Ql/uo)#Diameter of lower bed\n",
-      "#For calcination section\n",
-      "Qc = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T+273)))#Volumetric flow rate of gas in the calcination section\n",
-      "dtc = math.sqrt(4/pi*Qc/uo)#Diameter of calcination section\n",
-      "#For I stage\n",
-      "Q1 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T1+273)))#Volumetric flow rate of gas in the I stage\n",
-      "dt1 = math.sqrt(4/pi*Q1/uo)#Diameter of I stage\n",
-      "#For II stage\n",
-      "Q2 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T2+273)))#Volumetric flow rate of gas in the II stage\n",
-      "dt2 = math.sqrt(4/pi*Q2/uo)#Diameter of II stage\n",
-      "#For III stage\n",
-      "Q3 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T3+273)))#Volumetric flow rate of gas in the III stage\n",
-      "dt3 = math.sqrt(4/pi*Q3/uo)#Diameter of III stage\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nDiameter of lower bed:%fm'%(dtl)\n",
-      "print '\\nDiameter of calcination section:%fm'%(dtc)\n",
-      "print '\\nBed no.\\t\\t1\\t2\\t\\t3'\n",
-      "print '\\nDiameter(m)%f\\t%f\\t%f'%(dt1,dt2,dt3)\n",
-      "\n",
-      "#The value of diameter of each section is largely deviating from the values in the textbook. This is because the fuel consumption B have not been included in the energy balance equation. And the value of molecular weight is wrong by one decimal point.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Diameter of lower bed:7.097814m\n",
-        "\n",
-        "Diameter of calcination section:13.351483m\n",
-        "\n",
-        "Bed no.\t\t1\t2\t\t3\n",
-        "\n",
-        "Diameter(m)12.728715\t13.270865\t13.340712\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 413\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "T=20;         #Temeprature in degree C\n",
-      "M=0.018;      #Molecular weight of water in kg/mol\n",
-      "Q=10;         #Flow rate of dry air in m**3/s\n",
-      "R=82.06E-6;   #Universal gas constant\n",
-      "pi=0.0001;    #Initial moisture content in atm\n",
-      "pj=0.01;      #Final moisture content in atm\n",
-      "\n",
-      "#CALCULATION\n",
-      "a=Q*(273+T)/273;  #Term At*uo\n",
-      "b=a*M/(R*(T+273));#Term C*At*uo\n",
-      "#The value of slope can be found only by graphical mehtod. Hence it has been taken directly from the book(Page no.414,Fig.E3)\n",
-      "m=10.2;\n",
-      "Fo=b/m;           #Flow rate of solids\n",
-      "Q3=(b/Fo)*(pj-pi);#Moisture content of leaving solids\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nMoisture content of leaving solids:%.3f kg H2O/kg dry solids'%Q3\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Moisture content of leaving solids:0.101 kg H2O/kg dry solids\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 422\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "\n",
-      "\n",
-      "\n",
-      "#INPUT\n",
-      "Qfi = 0.20          #Initial moisture fraction\n",
-      "Qfbar = 0.04        #Average final moisture fraction\n",
-      "rhos = 2000.        #Density of solid in kg/m**3\n",
-      "Cps = 0.84          #Specific heat of solids in kJ/kg K\n",
-      "Fo = 7.6E-4         #Flow rate of solids in kg/m**3\n",
-      "Tsi = 20.            #Inital temperature of solids in degree C\n",
-      "rhog = 1.            #Density of gas in kg/m**3\n",
-      "Cpg = 1             #Specific heat of gas in kJ/kg K\n",
-      "uo = 0.3            #Superficial gas velocity in m/s\n",
-      "Tgi = 200           #Initial temperature of gas in degee C\n",
-      "L = 2370            #Enthalpy of liquid in kJ/kg\n",
-      "Cpl = 4.2           #Specific heat of liquid in kJ/kg K\n",
-      "dt = 0.1            #Diameter of reactor in m\n",
-      "Lm = 0.1            #Length of fixed bed in m\n",
-      "ephsilonm = 0.45    #Void fraction of fixed bed\n",
-      "pi = 3.14\n",
-      "Fo1 = 1             #Feed rate for commercial-scale reactor in kg/s\n",
-      "\n",
-      "#CALCULATION\n",
-      "#(a)Bed temperature\n",
-      "Teguess = 50#Guess value of Te\n",
-      "def solver_func(Te):#Function defined for solving the system\n",
-      "    return (pi/4.)*dt**2*uo*rhog*Cpg*(Tgi-Te)-Fo*(Qfi-Qfbar)*(L+Cpl*(Te-Tsi))-Fo*Cps*(Te-Tsi)\n",
-      "\n",
-      "Te = fsolve(solver_func,Teguess)\n",
-      "\n",
-      "#(b)Drying time for a particle\n",
-      "xguess = 2#Guess value of x, ie term tou/tbar\n",
-      "def solver_func1(x):        #Function defined for solving the system\n",
-      "    return 1-(Qfbar/Qfi)-(1-math.exp(-x))/x\n",
-      "\n",
-      "\n",
-      "x = fsolve(solver_func1,xguess)\n",
-      "W = (pi/4.)*dt**2*Lm*(1-ephsilonm)*rhos#Weight of soilds in bed\n",
-      "tbar = W/Fo#Mean residence time of solids from Eqn.(59)\n",
-      "tou = tbar*x#Time for complete drying of a particle\n",
-      "\n",
-      "#(c)Commercial-scale dryer\n",
-      "W1 = Fo1*tbar\n",
-      "Atguess = 5#Guess value of area\n",
-      "def  solver_func3(At):          #Function defined for solving the system\n",
-      "    return  At*uo*rhog*Cpg*(Tgi-Te)-Fo1*(Qfi-Qfbar)*(L+Cpl*(Te-Tsi))-Fo1*Cps*(Te-Tsi)\n",
-      "\n",
-      "At = fsolve(solver_func3,Atguess)\n",
-      "dt1 = math.sqrt(4/pi*At)#Diameter of commercial-scale dryer\n",
-      "Q1 = At*uo*rhog#Flow rate necessary for the operation\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Bed temperature:%f degree C'%(Te)\n",
-      "print 'Time for complete drying of particle:%fs'%(tou)\n",
-      "print 'Flow rate of gas necessary for Commercial-scale dryer:%fkg/s'%(Q1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bed temperature:58.728126 degree C\n",
-        "Time for complete drying of particle:527.431202s\n",
-        "Flow rate of gas necessary for Commercial-scale dryer:3.098684kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5, Page 425\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
-      "rhos=1600.;     #Density of solid in kg/m**3\n",
-      "Cps=1.25;       #Specific heat of solids in kJ/kg K\n",
-      "Fo=0.5;         #Flow rate of solids in kg/s\n",
-      "Tsi=20.;        #Inital temperature of solids in degree C\n",
-      "Qwi=1.;         #Initial moisture fraction in water\n",
-      "Qwf=0.2;        #Final moisture fraction in water\n",
-      "Qhi=1.1;        #Initial moisture fraction in heptane\n",
-      "Qhf=0.1;        #Final moisture fraction in heptane\n",
-      "Tgi=240.;       #Initial temperature of gas in degee C\n",
-      "Te=110.;        #Bed temperature in degree C\n",
-      "ephsilonm=0.45; #Void fraction of fixed bed\n",
-      "ephsilonf=0.75; #Void fraction of fluidized bed\n",
-      "uo=0.6;         #Superficial gas velocity in m/s\n",
-      "di=0.08;        #Diameter of tubes in m\n",
-      "li=0.2;         #Pitch for square arrangement\n",
-      "hw=400.;        #Heat transfer coefficient in W/m**2 K\n",
-      "Tc=238.;        #Temperature at which steam condenses in degree C\n",
-      "#Specific heats in kJ/kg K\n",
-      "Cwl=4.18;       #Water liquid\n",
-      "Cwv=1.92;       #Water vapor\n",
-      "Chl=2.05;       #Heptane liquid\n",
-      "Chv=1.67;       #Heptane vapor\n",
-      "#Latent heat of vaporization in kJ/kg\n",
-      "Lw=2260.;       #Water\n",
-      "Lh=326.;        #Heptane\n",
-      "#Density of vapor in kg/m**3 at operating conditions\n",
-      "rhow=0.56;      #Water\n",
-      "rhoh=3.1;       #Heptane\n",
-      "Lf=1.5;         #Length of fixed bed in m\n",
-      "t=140.;         #Half-life of heptane in s\n",
-      "L=1.5;          #Length of tubes in heat exchanger\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#(a) Dryer without Internals\n",
-      "xw=(Qwi-Qwf)/(Qhi-Qhf);                #Water-heptane weight ratio\n",
-      "xv=((Qwi-Qwf)/18.)/((Qhi-Qhf)/100.);   #Water-heptane volume ratio\n",
-      "T=(Qwi-Qwf)/18.+(Qhi-Qhf)/100.;        #Total volume\n",
-      "rhogbar=((Qwi-Qwf)/18.)/T*rhow+((Qhi-Qhf)/100.)/T*rhoh;           #Mean density of the vapor mixture\n",
-      "Cpgbar=(((Qwi-Qwf)/18.)/T)*rhow*Cwv+(((Qhi-Qhf)/100.)/T)*rhoh*Cwv;#Mean specific heat of vapor mixture\n",
-      "#Volumetric flow of recycle gas to the dryer in m**3/s from Eqn.(53)\n",
-      "x=(Cpgbar*(Tgi-Te))**-1*(Fo*(Qwi-Qwf)*(Lw+Cwl*(Te-Tsi))+Fo*(Qhi-Qhf)*(Lh+Chl*(Te-Tsi))+Fo*(Cps*(Te-Tsi)));\n",
-      "r=Fo*((Qwi-Qwf)/rhow+(Qhi-Qhf)/rhoh); #Rate of formation of vapor in bed\n",
-      "uo1=uo*(x/(x+r));                     #Superficial velocity just above the distributor\n",
-      "At=x/uo1;                             #Cross-sectional area of bed\n",
-      "dt=math.sqrt(4./pi*At);               #Diameter of bed\n",
-      "B=-math.log(Qwf/Qwi)/t;               #Bed height from Eqn.(63)\n",
-      "tbar=((Qhi/Qhf)-1)/B;                 #Mean residence time of solids\n",
-      "W=Fo*tbar;                            #Weight of bed\n",
-      "Lm=W/(At*(1-ephsilonm)*rhos);         #Static bed height\n",
-      "Lf=(Lm*(1-ephsilonm))/(1-ephsilonf);  #Height of fluidized bed\n",
-      "\n",
-      "#(b) Dryer with internal heaters\n",
-      "f=1/8.0;                 #Flow rate is 1/8th the flow rate of recirculation gas as in part (a)\n",
-      "x1=f*x;                  #Volumetric flow of recycle gas to the dryer in m**3/s from Eqn.(53)\n",
-      "uo2=uo*(x1/float(x1+r)); #Superficial velocity just above the distributor\n",
-      "Abed=x1/uo2;             #Cross-sectional area of bed\n",
-      "q=(Fo*(Qwi-Qwf)*(Lw+Cwl*(Te-Tsi))+Fo*(Qhi-Qhf)*(Lh+Chl*(Te-Tsi))+Fo*(Cps*(Te-Tsi)))-Abed*uo2*Cpgbar*(Tgi-Te);#Heat to be added from energy balance of Eqn.(53)\n",
-      "Aw=q*10**3/(hw*(Tc-Te));#Total surface area of heat exchanger tubes\n",
-      "Lt=Aw/(pi*di);          #Total length of tubes\n",
-      "Nt=Lt/L;                #Total number of tubes\n",
-      "Atubes=Nt*(pi/4*di**2); #Total cross-sectional area of tubes\n",
-      "Atotal=Abed+Atubes;     #Total cross-sectional area of tube filled dryer\n",
-      "d=math.sqrt(Atotal*pi/4.);#Diameter of vessel\n",
-      "li=math.sqrt(Atotal/Nt);  #Pitch for square array of tubes\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\t\\t\\tBed diameter(m)\\tRecycle vapor flow(m**3/s)'\n",
-      "print 'Without internal heater\\t%f\\t%f'%(dt,x)\n",
-      "print 'With heating tubes\\t%f\\t%f'%(d,x1)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\tBed diameter(m)\tRecycle vapor flow(m**3/s)\n",
-        "Without internal heater\t3.630144\t5.331235\n",
-        "With heating tubes\t1.503916\t0.666404\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch17.ipynb b/Fluidization_Engineering/ch17.ipynb
deleted file mode 100755
index 77951e94..00000000
--- a/Fluidization_Engineering/ch17.ipynb
+++ /dev/null
@@ -1,446 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:82e0373b19da96ab8fd50304caf9cd3e08cf8bad10412b0998bf9110f7a5ae63"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 17 : Design of Catalytic Reactors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 434\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dt=[0.081,0.205,3.6]; #Reactor diameter for the three reactors in m\n",
-      "dte=[0.04,0.12,0.70]; #Equivalent diameters for the three reactors in m\n",
-      "db=[0.05,0.057,0.07]; #Estimated bubble size in the three reactors in m\n",
-      "Kr1=1.3889;           #Kinet1ic constant for Reaction 1 in s**-1\n",
-      "Kr2=0.6111;           #Kinetic constant for Reaction 2 in s**-1\n",
-      "Kr3=0.022;            #Kinetic constant for Reaction 3 in s**-1\n",
-      "dp=60.;               #Particle size in micrometer\n",
-      "ephsilonm=0.50;       #Void fraction of fixed bed\n",
-      "ephsilonmf=0.55;      #Void fraction at minimum fluidized condition\n",
-      "umf=0.006;            #Velocity at minimum fluidization condition in m/s\n",
-      "D=2E-5;               #Diffusion coefficient of gas in m**2/s\n",
-      "gammab=0.005;         #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "uo=0.2;               #Superficial gas velocity in m/s\n",
-      "XA=0.9;               #Conversion\n",
-      "g=9.81;               #Acceleration due to gravity in square m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "Kr12=Kr1+Kr2;\n",
-      "n=len(dt);\n",
-      "i=0;\n",
-      "ubr = [0,0,0]\n",
-      "ub = [0,0,0]\n",
-      "delta = [0,0,0]\n",
-      "ephsilonf = [0,0,0]\n",
-      "gammac = [0,0,0]\n",
-      "gammae = [0,0,0]\n",
-      "Kbc = [0,0,0]\n",
-      "Kce = [0,0,0]\n",
-      "Kf12 = [0,0,0]\n",
-      "Kf3 = [0,0,0]\n",
-      "KfA = [0,0,0]\n",
-      "KfAR = [0,0,0]\n",
-      "KfAR1 = [0,0,0]\n",
-      "tou = [0,0,0]\n",
-      "y = [0,0,0]\n",
-      "SR = [0,0,0]\n",
-      "XA1 = [0,0,0]\n",
-      "y1 = [0,0,0]\n",
-      "y2 = [0,0,0]\n",
-      "tou2 = [0,0,0]\n",
-      "Lf = [0,0,0]\n",
-      "Lm = [0,0,0]\n",
-      "XA2 = [0,0,0]\n",
-      "\n",
-      "import math\n",
-      "while i<n:\n",
-      "    #Preliminary Calcualtions\n",
-      "    ubr[i]=0.711*(g*db[i])**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
-      "    ub[i]=1.55*((uo-umf)+14.1*(db[i]+0.005))*dte[i]**0.32+ubr[i];#Bubble velocity for Geldart A particles from Equation from Eqn.(6.11)\n",
-      "    delta[i]=uo/ub[i];#Fraction of bed in bubbles from Eqn.(6.29)\n",
-      "    ephsilonf[i]=1-(1-delta[i])*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
-      "    fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
-      "    gammac[i]=(1-ephsilonmf)*((3/(ubr[i]*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
-      "    gammae[i]=((1-ephsilonmf)*((1-delta[i])/delta[i]))-gammab-gammac[i];#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
-      "    Kbc[i]=4.5*(umf/db[i])+5.85*((D**0.5*g**0.25)/db[i]**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "    Kce[i]=6.77*((D*ephsilonmf*0.711*(g*db[i])**0.5)/db[i]**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
-      "    #Effective rate constant from Eqn.(12.32)\n",
-      "    Kf12[i]=(gammab*Kr12+1/((1/Kbc[i])+(1/(gammac[i]*Kr12+1/((1/Kce[i])+(1/(gammae[i]*Kr12)))))))*(delta[i]/(1-ephsilonf[i]));\n",
-      "    #Rate of reaction 2 for fluidized bed from Eqn.(12.14)\n",
-      "    Kf3[i]=(gammab*Kr3+1/((1/Kbc[i])+(1/(gammac[i]*Kr3+1/((1/Kce[i])+(1/(gammae[i]*Kr3)))))))*(delta[i]/(1-ephsilonf[i]));\n",
-      "    #Rate of raection with respect to A from Eqn.(12.35)\n",
-      "    KfA[i]=((Kbc[i]*Kce[i]/gammac[i]**2+(Kr12+Kce[i]/gammac[i]+Kce[i]/ \\\n",
-      "    gammae[i])*(Kr3+Kce[i]/gammac[i]+Kce[i]/gammae[i]))*delta[i]*Kbc[i] \\\n",
-      "    *Kr12*Kr3/(1-ephsilonf[i]))/(((Kr12+Kbc[i]/gammac[i])* \\\n",
-      "    (Kr12+Kce[i]/gammae[i])+Kr12*Kce[i]/gammac[i])*((Kr3+Kbc[i]/gammac[i])* \\\n",
-      "    (Kr3+Kce[i]/gammae[i])+Kr3*Kce[i]/gammac[i]));\n",
-      "    KfAR[i]=((Kr1/Kr12)*Kf12[i])-KfA[i];#Rate of reaction from Eqn.(12.34)\n",
-      "    KfAR1[i]=((Kr1/Kr12)*Kf12[i]);#Since KfA is small\n",
-      "    #(b)Relate Selectivity with conversion in three reactors\n",
-      "    x=-math.log(1-XA);#The term Kf12*tou in Eqn.(12.26)\n",
-      "    tou[i]=x/Kf12[i];#Residence time from Eqn.(12.26)\n",
-      "    y[i]=(KfAR1[i]/(Kf3[i]-Kf12[i]))*(math.exp(-x)-math.exp(-tou[i]*Kf3[i]));#CR/CAi from Eqn.(12.27)\n",
-      "    SR[i]=y[i]/XA;#Selectivity of R\n",
-      "    #(c)Relate exit composition to space time\n",
-      "    tou1=5;#Space time in s\n",
-      "    XA1[i]=1-math.exp(-Kf12[i]*tou1);#Conversion from Eqn.(12.26)\n",
-      "    y1[i]=((KfAR1[i]/(Kf12[i]-Kf3[i]))*(math.exp(-Kf3[i]*tou1)-math.exp(-Kf12[i]*tou1)));#CR/CAi R from Eqn.(12.27)\n",
-      "    #(d)Calculate height of bed needed to maximize production\n",
-      "    y2[i]=(KfAR1[i]/Kf12[i])*(Kf12[i]/Kf3[i])**(Kf3[i]/(Kf3[i]-Kf12[i]));#CRmax/CAi R from Eqn.(12.37)\n",
-      "    tou2[i]=math.log(Kf3[i]/Kf12[i])/(Kf3[i]-Kf12[i]);#Space time from Eqn.(38)\n",
-      "    Lf[i]=(uo/(1-ephsilonf[i]))*tou2[i];#Length of bed at fully fluidized condition from Eqn.(12.5)\n",
-      "    Lm[i]=Lf[i]*(1-ephsilonf[i])/(1-ephsilonm);#Length of bed when settled\n",
-      "    XA2[i]=1-math.exp(-Kf12[i]*tou2[i]);#Conversion from Eqn.(12.26)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Let Laboratory, Pilot plant, Semicommercial unit be Reactor 1,2 & 3 respectively'\n",
-      "print '(a)Relation between effective rate constant(Kf12) to the gas flow rate(uo)',\n",
-      "print '\\tReactor No.\\tKf12(s**-1)\\tuo(m/s)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "     print '\\t%1.0f'%i\n",
-      "     print '\\t\\t%f'%Kf12[i],\n",
-      "     print '\\t%f'%uo\n",
-      "     i=i+1;\n",
-      "\n",
-      "print '\\n(b)Relation between selectivity with conversion'\n",
-      "print '\\n\\tReactor No.\\tKf12(s**-1)\\tSR(mol R formed/mol A reacted)'\n",
-      "i=0\n",
-      "while i<n:\n",
-      "     print '\\t%1.0f'%i,\n",
-      "     print '\\t\\t%f'%Kf12[i],\n",
-      "     print '\\t%f'%SR[i]\n",
-      "     i=i+1;\n",
-      "\n",
-      "print '(c)Relation between exit compostion and space time',\n",
-      "print '\\tReactor No.\\tXA\\t\\tCR/CAi'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "     print '\\t%1.0f'%i,\n",
-      "     print '\\t\\t%f'%XA1[i],\n",
-      "     print '\\t%f'%y1[i]\n",
-      "     i=i+1;\n",
-      "\n",
-      "print '(d)Height of bed needed to maximize the production of acrylonitrile',\n",
-      "print '\\tReactor No.\\tLm(m)\\t\\tXA'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "     print '\\t%1.0f'%i,\n",
-      "     print '\\t\\t%f'%Lm[i],\n",
-      "     print '\\t%f'%XA2[i]\n",
-      "     i=i+1;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Let Laboratory, Pilot plant, Semicommercial unit be Reactor 1,2 & 3 respectively\n",
-        "(a)Relation between effective rate constant(Kf12) to the gas flow rate(uo) \tReactor No.\tKf12(s**-1)\tuo(m/s)\n",
-        "\t0\n",
-        "\t\t0.410042 \t0.200000\n",
-        "\t1\n",
-        "\t\t0.270620 \t0.200000\n",
-        "\t2\n",
-        "\t\t0.128980 \t0.200000\n",
-        "\n",
-        "(b)Relation between selectivity with conversion\n",
-        "\n",
-        "\tReactor No.\tKf12(s**-1)\tSR(mol R formed/mol A reacted)\n",
-        "\t0 \t\t0.410042 \t0.641507\n",
-        "\t1 \t\t0.270620 \t0.618358\n",
-        "\t2 \t\t0.128980 \t0.558283\n",
-        "(c)Relation between exit compostion and space time \tReactor No.\tXA\t\tCR/CAi\n",
-        "\t0 \t\t0.871292 \t0.564802\n",
-        "\t1 \t\t0.741562 \t0.484243\n",
-        "\t2 \t\t0.475286 \t0.313823\n",
-        "(d)Height of bed needed to maximize the production of acrylonitrile \tReactor No.\tLm(m)\t\tXA\n",
-        "\t0 \t\t3.056064 \t0.956404\n",
-        "\t1 \t\t4.137401 \t0.939139\n",
-        "\t2 \t\t7.049378 \t0.897005\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 438\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "deltaHr=5.15E8; #Heat of reaction in J/k mol\n",
-      "W=5E4;          #Weight of acrylonitirle produced per 334-day year in tonnes\n",
-      "db=0.07;        #Estimated bubble size in m\n",
-      "dte=0.7;        #Equivalent diameter in m\n",
-      "Kf12=0.35;      #Effective rate constant in s**-1 from Example 1\n",
-      "dp=60;          #Particle size in micrometer\n",
-      "ephsilonm=0.50; #Void fraction of fixed bed\n",
-      "ephsilonmf=0.55;#Void fraction at minimum fluidized condition\n",
-      "T=460;          #Temperature in reactor in degree C\n",
-      "Pr=2.5;         #Pressure inside reactor in bar\n",
-      "#Feed gas composition\n",
-      "x1=1;           #Propylene\n",
-      "x2=1.1;         #Ammonia\n",
-      "x3=11;          #Air\n",
-      "do1=0.08;       #OD of heat exchanger tubes in m\\\n",
-      "L=7;            #Length of tubes in m\n",
-      "ho=300;         #Outside heat transfer coefficient in W/m**2 K\n",
-      "hi=1800;        #Inside heat transfer coefficient in W/m**2 K\n",
-      "Tc=253.4;       #Temperature of coolant in degree C\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Preliminary calculation\n",
-      "uo=0.46;#Superficial gas velocity from Fig.E1(a) for the value of Kf12 & db\n",
-      "tou=8;#Space time from Fig.E2(b) for highest concentraion of product R\n",
-      "Lm=uo*tou/(1-ephsilonm);\n",
-      "y=0.58;#CR/CAi from Fig.E1(c) for the value of tou & Kf12\n",
-      "XA=0.95#From Fig.E1(c) for the value of tou & Kf12\n",
-      "SR=y/XA;#Selectivity of R\n",
-      "\n",
-      "#Cross-sectional area of the reactor\n",
-      "P=W*10**3/(334*24*3600);#Production rate of acrylonitrile\n",
-      "F=(P/0.053)/(SR*XA/0.042);#Feed rate of propylene\n",
-      "V=((F*22.4*(T+273)*(x1+x2+x3))/(42*273*Pr));\n",
-      "At=V/uo;#Cross-sectional area of reactor needed for the fluidized bed\n",
-      "\n",
-      "#Heat exchanger calculation\n",
-      "q=F*XA*deltaHr/42;#Rate of heat liberation in the reactor\n",
-      "U=(ho**-1+hi**-1)**-1;#Overall heat transfer coefficient\n",
-      "deltaT=T-Tc;#Driving force for heat transfer\n",
-      "Aw=q/(U*deltaT);#Heat exchanger area required to remove q\n",
-      "Nt=Aw/(pi*do1*L);\n",
-      "li1=(At/Nt)**0.5;#Pitch for square pitch arrangement\n",
-      "dte1=4*(li1**2-(pi/4)*do1**2)/(pi*do1);\n",
-      "if dte1>dte:\n",
-      "    li=(pi/4*dte*do1+pi/4*do1**2)**0.5;#Pitch if we add dummy tubes\n",
-      "import math\n",
-      "f=li**2-pi/4*do1**2;#Fraction of bed cross section taken up by tubes\n",
-      "dt1=math.sqrt(4/pi*At/(1-f));#Reactor diameter including all its tubes\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Superficial gas velocity=%fm/s'%uo\n",
-      "print 'No. of %1.0fm tubes required=%1.0f'%(L,Nt);\n",
-      "print 'Reactor diameter=%fm'%dt1\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Superficial gas velocity=0.460000m/s\n",
-        "No. of 7m tubes required=295\n",
-        "Reactor diameter=7.173176m\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "db=0.08;       #Estimated bubble size in m\n",
-      "dte=2;         #Equivalent diameter in m\n",
-      "F1=55.6;       #Feed rate of oil in kg/s\n",
-      "XA=0.63;       #Conversion\n",
-      "uo=0.6;        #Superficial gas velocity in m/s\n",
-      "T1=500.0;      #Temperature of reactor in degree C\n",
-      "T2=580.0;      #Temperature of regenerator in degree C\n",
-      "Fs=F1*23.3;    #Solid circulation rate from Ex.(15.2)\n",
-      "rhos=1200.0;   #Density of catalyst in kg/m**3\n",
-      "dpbar=60.0;    #Average particle size in micrometer\n",
-      "ephsilonm=0.50;#Void fraction of fixed bed\n",
-      "ephsilonmf=0.55;#Void fraction at minimum fluidized condition\n",
-      "umf=0.006;      #Velocity at minimum fluidization condition in m/s\n",
-      "dt=8.0;         #Diameter of reactor in m\n",
-      "D=2E-5;         #Diffusion coefficient of gas in m**2/s\n",
-      "Kr=8.6;         #Rate constant for reaction at 500 degree C in s**-1\n",
-      "Ka1=0.06;       #Rate constant for deactivatiion at 500 degree C in s**-1\n",
-      "Ka2=0.012;      #Rate constant for regeneration at 580 degree C in s**-1\n",
-      "gammab=0.005;   #Ratio of volume of dispersed solids to that of bubble phase\n",
-      "g=9.81;         #Acceleration due to gravity in square m/s**2\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Parameters for the fluidized reactor\n",
-      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
-      "ub=1.55*((uo-umf)+14.1*(db+0.005))*dte**0.32+ubr;#Bubble velocity for Geldart A particles from Equation from Eqn.(6.11)\n",
-      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
-      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
-      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
-      "gammac=(1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
-      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
-      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "Kce=6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
-      "import math\n",
-      "#Bed height versus catalyst activity in reactor\n",
-      "a1bar=0.07;#Guess value for average activity in reactor\n",
-      "x=Kr*a1bar;#Value of Kra1 to be used in the following equation\n",
-      "Kf=(gammab*x+1/((1/Kbc)+(1/(gammac*x+1/((1/Kce)+(1/(gammae*x)))))))*(delta/(1-ephsilonf));#Effective rate constant from Eqn.(12.14)\n",
-      "tou=-math.log(1-XA)/Kf;#Space time from Eqn.(12.16)\n",
-      "Lm=tou*uo/(1-ephsilonm);#Length of fixed bed for guess value of a1bar\n",
-      "a1bar1=[0.0233,0.0465,0.0698,0.0930,0.116,0.140];#Various activity values to find Lm\n",
-      "x1 = [0,0,0,0,0,0]\n",
-      "Kf1 = [0,0,0,0,0,0]\n",
-      "tou1 = [0,0,0,0,0,0]\n",
-      "Lm1 = [0,0,0,0,0,0]\n",
-      "\n",
-      "n=len(a1bar1);\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    x1[i]=Kr*a1bar1[i];\n",
-      "    Kf1[i]=(gammab*x1[i]+1/((1/Kbc)+(1/(gammac*x1[i]+1/((1/Kce)+ \\\n",
-      "    (1/(gammae*x1[i])))))))*(delta/(1-ephsilonf));\n",
-      "    #Effective rate constant from Eqn.(12.14)\n",
-      "    \n",
-      "    tou1[i]=-math.log(1-XA)/Kf1[i];#Space time from Eqn.(12.16)\n",
-      "    Lm1[i]=tou1[i]*uo/(1-ephsilonm);\n",
-      "    #Length of fixed bed for guess value of a1bar...Condition [i]\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Find the optimum size ratio for various a1bar\n",
-      "Lm=[5,6,7,8,10,12];\n",
-      "W1 = [0,0,0,0,0,0]\n",
-      "t1bar = [0,0,0,0,0,0]\n",
-      "t2bar = [0,0,0,0,0,0]\n",
-      "a1bar2 = [0,0,0,0,0,0]\n",
-      "m=len(Lm);\n",
-      "i=0;\n",
-      "while i<m:\n",
-      "    W1[i]=(pi/4)*dt**2*rhos*(1-ephsilonm)*Lm[i];#Bed weight\n",
-      "    t1bar[i]=W1[i]/Fs;#Mean residence time of solids in reactor\n",
-      "    t2bar[i]=t1bar[i]*(Ka1/Ka2)**0.5;#Mean residence time of soilds at optimum from Eqn.(16)\n",
-      "    a1bar2[i]=(Ka2*t2bar[i])/(Ka1*t1bar[i]+Ka1*t1bar[i]*Ka2*t2bar[i]+Ka2*t2bar[i]);#From Eqn.(15)...Condition (ii)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Final design values\n",
-      "Lm4=7.3;#For satisfying condition [i] & (ii)\n",
-      "a1bar3=0.0744;#By interpolation\n",
-      "x2=a1bar3*Kr;\n",
-      "W11=(pi/4)*dt**2*rhos*(1-ephsilonm)*Lm4;#Bed weight for reactor\n",
-      "t1bar1=W11/Fs;#Mean residence time of solids in reactor\n",
-      "a2bar=(1+Ka1*t1bar1)*a1bar3;#Average activity in regenrator from Eqn.(10)\n",
-      "t2bar1=t1bar1*(Ka1/Ka2)**0.5;#Mean residence time of solids in regenerator from Eqn.(16)\n",
-      "W2=W11*(t2bar1/t1bar1);#Bed weight for regenerator\n",
-      "dt2=dt*(W2/W11)**0.5;#Diameter of regenerator assuming same static bed height for reactor and regerator\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Bed height versus catalyst activity in reactor'\n",
-      "print '\\tAverage activity',\n",
-      "print '\\tLength of fixed bed(m)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '\\t%f'%a1bar1[i],\n",
-      "    print '\\t\\t%f'%Lm1[i];\n",
-      "    i=i+1;\n",
-      "\n",
-      "print 'Optimum size ratio for various activity in reactor'\n",
-      "print '\\tLength of fixed bed(m)',\n",
-      "print '\\tAverage activity'\n",
-      "i=0\n",
-      "while i<m:\n",
-      "    print '\\t%f'%Lm[i],\n",
-      "    print '\\t\\t%f'%a1bar2[i]\n",
-      "    i=i+1;\n",
-      "\n",
-      "print 'Final design values'\n",
-      "print '\\tDiameter of reactor(m):%.0f'%dt\n",
-      "print '\\tBed weight for reactor(tons):%.0f'%(W11/10**3)\n",
-      "print '\\tBed weight for regenerator(tons):%.0f'%(W2/10**3)\n",
-      "print '\\tDiameter of regenerator(m):%.0f'%(dt2);\n",
-      "print '\\tSolid circulation rate(tons/hr):%f'%(Fs*3.6);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bed height versus catalyst activity in reactor\n",
-        "\tAverage activity \tLength of fixed bed(m)\n",
-        "\t0.023300 \t\t11.059747\n",
-        "\t0.046500 \t\t7.911053\n",
-        "\t0.069800 \t\t6.756202\n",
-        "\t0.093000 \t\t6.118750\n",
-        "\t0.116000 \t\t5.696470\n",
-        "\t0.140000 \t\t5.372072\n",
-        "Optimum size ratio for various activity in reactor\n",
-        "\tLength of fixed bed(m) \tAverage activity\n",
-        "\t5.000000 \t\t0.097879\n",
-        "\t6.000000 \t\t0.086112\n",
-        "\t7.000000 \t\t0.076871\n",
-        "\t8.000000 \t\t0.069420\n",
-        "\t10.000000 \t\t0.058149\n",
-        "\t12.000000 \t\t0.050026\n",
-        "Final design values\n",
-        "\tDiameter of reactor(m):8\n",
-        "\tBed weight for reactor(tons):220\n",
-        "\tBed weight for regenerator(tons):492\n",
-        "\tDiameter of regenerator(m):12\n",
-        "\tSolid circulation rate(tons/hr):4663.728000\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch18.ipynb b/Fluidization_Engineering/ch18.ipynb
deleted file mode 100755
index 281bc4a5..00000000
--- a/Fluidization_Engineering/ch18.ipynb
+++ /dev/null
@@ -1,570 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3eba5bf9f8e98dcb46f4a3b03f9f7fa4b342f8fb2d0d8b697e9a50303c32aa8f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 18 : The Design of Noncatalytic Gas Solid Reactors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "xA=0.08;      #Fraction of oxygen in stream\n",
-      "dp=[2,0.1];   #Particle diameter in mm\n",
-      "rhos=4130;    #Density of catalyst in kg/m**3\n",
-      "Ds=8E-6;      #Diffusion coefficient of solid in m**2/s\n",
-      "kc=0.02;      #Reaction rate constant in m/s\n",
-      "P=10**5;      #Pressure in bar\\\n",
-      "R=8.314;      #Universal gas constant\n",
-      "T=900;        #Temperature in degree C\n",
-      "mB=0.09745;   #Molecular weight of ZnS in kg/mol\n",
-      "\n",
-      "#CALCULATION\n",
-      "b=2.0/3;#Stoichiometric coefficient of ZnS in the reaction equation\n",
-      "CA=xA*P/(R*(T+273));#Concentration of Oxygen\n",
-      "rhob=rhos/mB;       #Molar density of pure solid\n",
-      "n=len(dp);\n",
-      "i=0;\n",
-      "kbar = [0,0]\n",
-      "tou = [0,0]\n",
-      "while i<n:\n",
-      "    kbar[i]=(kc**-1+(dp[i]*10**-3/(12.0*Ds)))**-1;#Average reaction rate constant from Eqn.(11)\n",
-      "    tou[i]=rhob*dp[i]*10**-3/(2*b*kbar[i]*CA);#Time for complete reaction in seconds from Eqn.(9)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Particle Size(mm)\\tAverage rate constant(m/s)\\tTime for complete reaction(min)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f\\t\\t%f\\t\\t\\t%.2f'%(dp[i],kbar[i],tou[i]/60.0);\n",
-      "    i=i+1;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Particle Size(mm)\tAverage rate constant(m/s)\tTime for complete reaction(min)\n",
-        "2.000000\t\t0.014118\t\t\t91.49\n",
-        "0.100000\t\t0.019592\t\t\t3.30\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 457\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "xA=0.08;   #Fraction of oxygen in stream\n",
-      "dp=1;      #Particle diameter in mm\n",
-      "rhos=2200; #Density of catalyst in kg/m**3\n",
-      "kc=0.2;    #Reaction rate constant in m/s\n",
-      "mC=0.012;  #Molecular weight of carbon in kg/mol\n",
-      "P=10**5;   #Pressure in bar\\\n",
-      "R=8.314;   #Universal gas constant\n",
-      "T=900;     #Temperature in degree C\n",
-      "\n",
-      "#CALCULATION\n",
-      "b=1;#Stoichiometric coefficient of C in the reaction equation\n",
-      "CA=xA*P/(R*(T+273));#Concentration of Oxygen\n",
-      "rhob=rhos/mC;#Molar density of pure solid reactant\n",
-      "tou=rhob*10**-3/(2*b*kc*CA);#Time required for complete reaction in seconds\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The time required for complete combustion:%.1fmins'%(tou/60);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The time required for complete combustion:9.3mins\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 462\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "from numpy import zeros\n",
-      "\n",
-      "\n",
-      "#INPUT\n",
-      "dp = 110.;#Particle size in micrometer\n",
-      "T = 900.;#Temperature of roaster in degree C\n",
-      "tbar1 = [3.,10.,30.,50.];#Reported average time in min\n",
-      "XBbarr = [0.840,0.940,0.985,0.990];#Reported value of average conversion\n",
-      "tbar = 3.;\n",
-      "XBbar = 0.840;#Average conversion for tbar  =  3 mins\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Uniform-Reaction Model\n",
-      "x = (1./tbar)*(1./(1-XBbar)-1);#Term KrCA of Eqn.(20)\n",
-      "n = len(tbar1);\n",
-      "i = 0;\n",
-      "XBbar1 = zeros(n)\n",
-      "while i<  n:\n",
-      "    XBbar1[i] = 1-1./(1+x*tbar1[i]);#Average conversion umath.sing calculated value of KrCA from Eqn.(20)\n",
-      "    i = i+1;\n",
-      "\n",
-      "#Shrinking-Core, Rection Control\n",
-      "touguess = 2;#Guess value of tou\n",
-      "def solver_func(tou):#Function defined for solving the system\n",
-      "    return (1-XBbar)-(0.25*tou/tbar)+(0.05*(tou/tbar)**2)-((1./120)*(tou/tbar)**3);\n",
-      "\n",
-      "tou = fsolve(solver_func,touguess)\n",
-      "i = 0;\n",
-      "XBbar2 = zeros(n)\n",
-      "while i<n:\n",
-      "    XBbar2[i] = 1-(0.25*tou/tbar1[i])+(0.05*(tou/tbar1[i])**2)-((1./120)*(tou/tbar1[i])**3);#Average conversion umath.sing calculated value of tou from Eqn.(23)\n",
-      "    i = i+1;\n",
-      "\n",
-      "#Shrinking-Core, Diffusion Control\n",
-      "touguess1 = 2;#Guess value of tou\n",
-      "def solver_func1(tou):      #Function defined for solving the system\n",
-      "    fn = (1-XBbar)-(1./5*tou/tbar)+(19./420*(tou/tbar)**2)-(41./4620*(tou/tbar)**3)+(0.00149*(tou/tbar)**4);\n",
-      "\n",
-      "tou1 = fsolve(solver_func1,touguess1)\n",
-      "i = 0;\n",
-      "XBbar3 = zeros(n)\n",
-      "while i< n:\n",
-      "    #Average conversion umath.sing calculated value of tou from Eqn.(23)\n",
-      "    XBbar3[i] = 1-(1./5*tou1/tbar1[i])+(19./420*(tou1/tbar1[i])**2)-(41./4620*(tou1/tbar1[i])**3)+(0.00149*(tou1/tbar)**4);\n",
-      "    i = i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\t\\t\\t\\tXBbar calculated for Models';\n",
-      "print 'Reported Data';\n",
-      "print 'tbarmin)\\tXBbar\\tUniform Reaction\\tShrinking-Core%( Rection Control\\t\\tShrinking-Core, Diffusion Control'\n",
-      "i = 0\n",
-      "while i< n:\n",
-      "    print '%f\\t%f\\t%f\\t\\t%f\\t\\t\\t\\t%f'%(tbar1[i],XBbarr[i],XBbar1[i],XBbar2[i],XBbar3[i]);\n",
-      "    i = i+1;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\tXBbar calculated for Models\n",
-        "Reported Data\n",
-        "tbarmin)\tXBbar\tUniform Reaction\tShrinking-Core%( Rection Control\t\tShrinking-Core, Diffusion Control\n",
-        "3.000000\t0.840000\t0.840000\t\t0.840000\t\t\t\t0.884437\n",
-        "10.000000\t0.940000\t0.945946\t\t0.947234\t\t\t\t0.962033\n",
-        "30.000000\t0.985000\t0.981308\t\t0.981898\t\t\t\t0.987159\n",
-        "50.000000\t0.990000\t0.988701\t\t0.989075\t\t\t\t0.992366\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "/usr/lib/python2.7/dist-packages/scipy/optimize/minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
-        "  improvement from the last ten iterations.\n",
-        "  warnings.warn(msg, RuntimeWarning)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 462\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "from numpy import zeros\n",
-      "\n",
-      "\n",
-      "#INPUT\n",
-      "W = 1;          #Bed weight in kg\n",
-      "F1 = 0.01;      #Solid feed rate in kg/min\n",
-      "dp = [200.,600.];           #Particle size in micrometer\n",
-      "XBbar = [0.85,0.64];        #Average conversion for corresponding particle sizes\n",
-      "rhos = 2500.;               #Density of solid in kg/m**3\n",
-      "ephsilonm = 0.4;            #Void fracton of fixed bed\n",
-      "F11 = 4.;                   #Feed rate of solids in tons/hr\n",
-      "XBbar1 = 0.98;\n",
-      "dp1 = 600.;\n",
-      "pi = 3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Shrinking-Core, Rection Control\n",
-      "n = len(dp);\n",
-      "i = 0;\n",
-      "touguess = 2;#Guess value of tou\n",
-      "tou = zeros(n)\n",
-      "while i< n:\n",
-      "    def solver_func2(t):  #Function defined for solving the system\n",
-      "        return  (1-XBbar[i])-(0.25*t/107)+(0.05*(t/107.)**2)-((1./120)*(t/107.)**3)\n",
-      "    tou[i] = fsolve(solver_func2,touguess)\n",
-      "    i = i+1;\n",
-      "\n",
-      "tou1 = tou[1];\n",
-      "#For a math.single stage fluidized roaster\n",
-      "tbar1 = 0.25*(tou1/(1-XBbar1))/60.;          #Mean residence time of solids in reactor in hr from Eqn.(24)\n",
-      "W1 = F11*tbar1;\n",
-      "dtguess = 2.;                    #Guess value of tou\n",
-      "def solver_func3(dt):           #Function defined for solving the system\n",
-      "    return  W1*10**3-(pi/4.)*dt**2*0.5*dt*rhos*(1-ephsilonm);#Since Lm = 0.5dt\n",
-      "\n",
-      "dt = fsolve(solver_func3,dtguess)\n",
-      "Lm = dt/2.;#Length of bed required\n",
-      "\n",
-      "#For a two-stage fluidized roaster\n",
-      "tbar2 = tou1*math.sqrt(1./(20*(1-XBbar1)))/60;      #Mean residence time of solids in reactor in hr from Eqn.(30)\n",
-      "W2 = F11*tbar2;\n",
-      "dtguess1 = 2;                   #Guess value of tou\n",
-      "def solver_func4(dt):           #Function defined for solving the system\n",
-      "    return W2*10**3-(pi/4.)*dt**2*0.5*dt*rhos*(1-ephsilonm);       #Since Lm = 0.5dt\n",
-      "\n",
-      "dt1 = fsolve(solver_func4,dtguess)\n",
-      "Lm1 = dt1/2.;           #Length of bed required\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Single stage fluidized roaster';\n",
-      "print '\\tWeight of bed needed:%ftons'%(W1);\n",
-      "print '\\tDiameter of reactor:%fm'%(dt);\n",
-      "print '\\tLength of bed:%fm'%(Lm);\n",
-      "print 'Two-stage fluidized roaster';\n",
-      "print '\\tWeight of bed needed:%ftons'%(W2);\n",
-      "print '\\tDiameter of reactor:%fm'%(dt1);\n",
-      "print '\\tLength of bed:%fm'%(Lm1);\n",
-      "print 'These results show that this operation can be accomplished in a math.single bed of %ftons or in two beds of %f tons each.'%(W1,W2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Single stage fluidized roaster\n",
-        "\tWeight of bed needed:174.370805tons\n",
-        "\tDiameter of reactor:6.665728m\n",
-        "\tLength of bed:3.332864m\n",
-        "Two-stage fluidized roaster\n",
-        "\tWeight of bed needed:22.056356tons\n",
-        "\tDiameter of reactor:3.346064m\n",
-        "\tLength of bed:1.673032m\n",
-        "These results show that this operation can be accomplished in a math.single bed of 174.370805tons or in two beds of 22.056356 tons each.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5, Page 468\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "from numpy import zeros\n",
-      "\n",
-      "#INPUT\n",
-      "T = 900;                #Temperature in roaster in degree C\n",
-      "P = 101325;             #Pressure in Pa\n",
-      "R = 8.314;              #Universal gas consmath.tant\n",
-      "dpbar = 150;            #Average particle size in micrometer\n",
-      "rhosbar = 4130;         #Average particle density in kg/m**3\n",
-      "kc = 0.015              #Rate consmath.tant in m/s for reaction which follows shrinking core model\n",
-      "Ds = 8E-6;              #Diffusion coefficient of solid in m**2/s\n",
-      "uo = 0.6;               #Superficial gas velocity in m/s\n",
-      "D = 2.3E-4;             #Diffusion coefficient of gas in m**2/s\n",
-      "Lm = 1;                 #Length of fixed bed in m\n",
-      "dte = 0.4;              #Equivalent diameter of bed\n",
-      "umf = 0.025;            #Velocity at minimum fluidization condition in m/s\n",
-      "ephsilonm = 0.45;       #Void fraction of fixed bed\n",
-      "ephsilonmf = 0.50;      #Void fraction at minimum fluidized condition\n",
-      "db = 0.2;               #Estimated bubble size in m\n",
-      "gammab = 0.005;         #Ratio of volume of print ersed solids to that of bubble phase\n",
-      "Fo = 2;                 #Feed rate of solids in kg/s\n",
-      "XA = 0.6677;            #Conversion of Oxygen\n",
-      "xA = 0.21;              #Mole fraction of oxygen in feed\n",
-      "mB = 0.09744;           #Molecular weight of ZnS\n",
-      "F = 0.85;               #Fraction of open area\n",
-      "g = 9.81;               #Acceleration due to gravity in square m/s**2\n",
-      "pi = 3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#(a)Extreme Calculation\n",
-      "a = 3./2;#Stoichiometric coefficient of Oxygen in the reaction equation\n",
-      "At = (Fo/mB)*(a)/(uo*(273./(T+273))*(XA*xA)/0.0224);\n",
-      "dt = math.sqrt(At/F*4/pi);\n",
-      "\n",
-      "#(b)The Three-Step Procedure\n",
-      "#Step 1. Conversion of gas\n",
-      "ubr = 0.711*(g*db)**0.5;                                #Rise velocity of bubble from Eqn.(6.7)\n",
-      "ub = 1.6*((uo-umf)+1.13*db**0.5)*dte**1.35+ubr;         #Bubble rise velocity for Geldart B particle\n",
-      "\n",
-      "delta = uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
-      "ephsilonf = 1-(1-delta)*(1-ephsilonmf);                 #Void fraction of fixed bed from Eqn.(6.20)\n",
-      "\n",
-      "fw = 0.15;#Wake volume to bubble volume from Fig.(5.8)\n",
-      "gammac = (1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);        #Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
-      "gammae = ((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;      #Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
-      "Kbc = 4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5./4));           #Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
-      "Kce = 6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;       #Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
-      "x = delta*Lm*(1-ephsilonm)/((1-ephsilonf)*uo);                  #Term Lf/ub of Eqn.(12.16) from Eqn.(6.19)\n",
-      "CAi = xA*P/(R*(T+273));                                         #Initial concentration of oxygen\n",
-      "\n",
-      "#Step 2.Conversion of solids\n",
-      "rhob = rhosbar/mB;                                              #Density of ZnS\n",
-      "kbar = (kc**-1+(dpbar*10**-6/(12*Ds))**-1)**-1;                 #Modified rate consmath.tant from Eqn.(11)\n",
-      "tbar = At*Lm*(1-ephsilonm)*rhosbar/Fo;                          #Mean residence time of solids\n",
-      "Krguess = 2;                                                    #Guess value of Kr\n",
-      "def solver_func(Kr):                #Function defined for solving the system\n",
-      "    Kf = gammab*Kr+1/((1./Kbc)+(1./(gammac*Kr+1/((1./Kce)+(1./(gammae*Kr))))));            #Reaction rate for fluidized bed from Eqn.(14)\n",
-      "    XA = 1-math.exp(-x*Kf);                                                         #Conversion of oxygen from Eqn.(42)\n",
-      "    CAbar = (CAi*XA*uo)/(Kr*Lm*(1-ephsilonm));                                      #Average concentration of oxygen from Eqn.(43)\n",
-      "    tou = rhob*dpbar*10**-6*a/(2*kbar*CAbar);                                       #Time for complete reaction from Eqn.(9)\n",
-      "    y = tbar/tou;#Term tbar/tou\n",
-      "    XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1/y));                                   #Average conversion of ZnS from Eqn.(22)\n",
-      "    #Step 3. Material balance of both streams\n",
-      "    return (Fo/mB)*XBbar-(At*uo*CAi*XA/a);                                          #From Eqn.(44b)\n",
-      "\n",
-      "Kr = fsolve(solver_func,Krguess)\n",
-      "Kf = gammab*Kr+1/((1/Kbc)+(1/(gammac*Kr+1/((1/Kce)+(1/(gammae*Kr))))));             #Reaction rate for fluidized bed from Eqn.(14)\n",
-      "XA = 1-math.exp(-x*Kf);                                                             #Conversion of oxygen from Eqn.(42)\n",
-      "CAbar = (CAi*XA*uo)/(Kr*Lm*(1-ephsilonmf));                                         #Average concentration of oxygen from Eqn.(43)\n",
-      "tou = rhob*dpbar*10**-6*a/(2*kbar*CAbar);                                           #Time for complete reaction from Eqn.(9)\n",
-      "y = tbar/tou;                                                                       #Term tbar/tou\n",
-      "XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1/y));                                       #Average conversion of ZnS from Eqn.(22)\n",
-      "\n",
-      "\n",
-      "#(c) For other feed rates of solids\n",
-      "F1 = [2,2.5,3,3.5];#Various feed rates of solids in kg/s\n",
-      "n = len(F1)\n",
-      "i = 0;\n",
-      "tbar1 = zeros(n)\n",
-      "Kr1 = zeros(n)\n",
-      "XA1 = zeros(n)\n",
-      "y1 = zeros(n)\n",
-      "tou1 = zeros(n)\n",
-      "XBbar1 = zeros(n)\n",
-      "Kf1 = zeros(n)\n",
-      "CAbar1 = zeros(n)\n",
-      "Krguess1 = 2;           #Guess value of Kr\n",
-      "while i<  n:\n",
-      "    tbar1[i] = At*Lm*(1-ephsilonm)*rhosbar/F1[i];#Mean residence time of solids\n",
-      "    def solver_func1(Kr):               #Function defined for solving the system\n",
-      "        Kf1 = gammab*Kr+1/((1/Kbc)+(1/(gammac*Kr+1/((1/Kce)+(1/(gammae*Kr))))));#Reaction rate for fluidized bed from Eqn.(14)\n",
-      "        XA1 = 1-math.exp(-x*Kf1);#Conversion of oxygen from Eqn.(42)\n",
-      "        CAbar1 = (CAi*XA1*uo)/(Kr*Lm*(1-ephsilonm));#Average concentration of oxygen from Eqn.(43)\n",
-      "        tou1 = rhob*dpbar*10**-6*a/(2*kbar*CAbar1);#Time for complete reaction from Eqn.(9)\n",
-      "        y1[i] = tbar1[i]/tou1;#Term tbar/tou\n",
-      "        XBbar1[i] = 3*y1[i]-6*y1[i]**2+6*y1[i]**3*(1-math.exp(-1/y1[i]));#Average conversion of ZnS from Eqn.(22)\n",
-      "        #Step 3. Material balance of both streams\n",
-      "        return (F1[i]/mB)*XBbar1[i]-(At*uo*CAi*XA1/a);#From Eqn.(44b)\n",
-      "\n",
-      "    Kr1[i] = fsolve(solver_func1,Krguess1)\n",
-      "    Kf1[i] = gammab*Kr1[i]+1/((1/Kbc)+(1/(gammac*Kr1[i]+1/((1/Kce)+(1/(gammae*Kr1[i]))))));#Reaction rate for fluidized bed from Eqn.(14)\n",
-      "    XA1[i] = 1-math.exp(-x*Kf1[i]);#Conversion of oxygen from Eqn.(42)\n",
-      "    CAbar1[i] = (CAi*XA1[i]*uo)/(Kr1[i]*Lm*(1-ephsilonmf));#Average concentration of oxygen from Eqn.(43)\n",
-      "    tou1[i] = rhob*dpbar*10**-6*a/(2*kbar*CAbar1[i]);#Time for complete reaction from Eqn.(9)\n",
-      "    y1[i] = tbar1[i]/tou1[i];#Term tbar/tou\n",
-      "    XBbar1[i] = 3*y1[i]-6*y1[i]**2+6*y1[i]**3*(1-math.exp(-1/y1[i]));#Average conversion of ZnS from Eqn.(22)\n",
-      "    i = i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Extreme Calculation';\n",
-      "print '\\tDiameter of tube with all its internals:%fm'%(dt);\n",
-      "print 'Three step procedure';\n",
-      "print '\\tConversion of ZnS:%f'%(XBbar);\n",
-      "print 'For other feed rates of solids';\n",
-      "print '\\tFeedkg/s\\ttbars\\t\\tXBbar/XA\\tKrbars**-1\\tCAbar/CAi\\ttous\\t\\tXA\\t\\tXB';\n",
-      "i = 0;\n",
-      "while i<  n:\n",
-      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f\\t%f\\t%f\\t%f'%(F1[i],tbar1[i],XBbar1[i]/XA1[i],Kr1[i],CAbar1[i]/CAi,tou1[i],XA1[i],XBbar1[i]);\n",
-      "    i = i+1;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Extreme Calculation\n",
-        "\tDiameter of tube with all its internals:7.265465m\n",
-        "Three step procedure\n",
-        "\tConversion of ZnS:0.992491\n",
-        "For other feed rates of solids\n",
-        "\tFeedkg/s\ttbars\t\tXBbar/XA\tKrbars**-1\tCAbar/CAi\ttous\t\tXA\t\tXB\n",
-        "\t2.000000\t40003.518104\t1.498777\t6.529973\t0.121691\t1208.748626\t0.662201\t0.992491\n",
-        "\t2.500000\t32002.814483\t1.201348\t23.149964\t0.041993\t3502.838402\t0.810111\t0.973225\n",
-        "\t3.000000\t26669.012069\t1.007581\t76.183481\t0.014228\t10338.227251\t0.903294\t0.910141\n",
-        "\t3.500000\t22859.153202\t0.871565\t146.263086\t0.007742\t18998.458994\t0.943693\t0.822490\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6, Page 471\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from scipy.optimize import fsolve \n",
-      "import math \n",
-      "\n",
-      "\n",
-      "#INPUT\n",
-      "T = 900;                #Temperature in roaster in degree C\n",
-      "P = 101325;             #Pressure in Pa\n",
-      "R = 8.314;              #Universal gas consmath.tant\n",
-      "dp = 750;               #Particle size in micrometer5\n",
-      "Fo = 2.5;               #Feed rate of solids in kg/s\n",
-      "uo = 0.6;               #Superficial gas velocity in m/s\n",
-      "W = 80140;              #Weight of bed in kg\n",
-      "ephsilonmf = 0.50;      #Void fraction at minimum fluidized condition\n",
-      "umf = 0.5;              #Velocity at minimum fluidization condition in m/s\n",
-      "db = 0.2;               #Estimated bubble size in m\n",
-      "g = 9.81;               #Acceleration due to gravity in square m/s**2\n",
-      "Lm = 1;                 #Length of fixed bed in m\n",
-      "ephsilonm = 0.45;       #Void fraction of fixed bed\n",
-      "xA = 0.21;              #Mole fraction of oxygen in feed\n",
-      "kc = 0.015              #Rate consmath.tant in m/s for reaction which follows shrinking core model\n",
-      "Ds = 8E-6;              #Diffusion coefficient of solid in m**2/s\n",
-      "rhosbar = 4130;         #Average particle density in kg/m**3\n",
-      "mB = 0.09744;           #Molecular weight of ZnS\n",
-      "a = 3./2;               #Stoichiometric coefficient of Oxygen in the reaction equation\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Selection of models to represent reactor\n",
-      "ubr = 0.711*(g*db)**0.5;        #Rise velocity of bubble from Eqn.(6.7)\n",
-      "f = ubr/(umf/ephsilonmf);\n",
-      "\n",
-      "#Step 1.\n",
-      "ub = uo-umf+ubr;            #Rise velocity of bubbles from Eqn.(6.8)\n",
-      "delta = (uo-umf)/(ub+2*umf);        #Fraction of the bed in bubbles from Eqn.(6.26)\n",
-      "Krguess = 2;                        #Guess value of Kr\n",
-      "x = Lm*(1-ephsilonm)*umf*(1-delta)/uo**2;\n",
-      "CAi = xA*P/(R*(T+273));             #Initial concentration of oxygen\n",
-      "\n",
-      "#Step 2.\n",
-      "kbar = (kc**-1+(dp*10**-6/(12*Ds))**-1)**-1;            #Modified rate consmath.tant from Eqn.(11)\n",
-      "tbar = W/Fo;                    #Mean residence time of solids from Eqn.(14.2)\n",
-      "rhob = rhosbar/mB;              #Density of ZnS\n",
-      "def solver_func1(Kr):           #Function defined for solving the system\n",
-      "    XA = 1-math.exp(-x*Kr);     #Conversion from Eqn.(42)\n",
-      "    CAbar = (CAi*XA*uo**2)/(Kr*Lm*(1-ephsilonm)*umf*(1-delta));         #Average concentration of oxygen from Eqn.(43)\n",
-      "    tou = rhob*dp*10**-6*a/(2*kbar*CAbar);          #Time for complete reaction from Eqn.(9)\n",
-      "    y = tbar/tou;               #Term tbar/tou\n",
-      "    XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1./y));          #Average conversion of ZnS from Eqn.(22)\n",
-      "    return XBbar-1.2*XA;        #From Table E5, for Fo = 2.5kg/s\n",
-      "\n",
-      "Kr = fsolve(solver_func1,Krguess)\n",
-      "XA = 1-math.exp(-x*Kr);         #Conversion from Eqn.(42)\n",
-      "CAbar = (CAi*XA*uo**2)/(Kr*Lm*(1-ephsilonm)*umf*(1-delta))      #Average concentration of oxygen from Eqn.(43)\n",
-      "tou = rhob*dp*10**-6*a/(2*kbar*CAbar);          #Time for complete reaction from Eqn.(9)\n",
-      "y = tbar/tou;                   #Term tbar/tou\n",
-      "XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1./y));  #Average conversion of ZnS from Eqn.(22)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Selection of models to represent reactor';\n",
-      "print '\\tSince ratio ubr/umf/ephsilonmf) =  %f <1 the reactor is operating in slow bubble regime'%(f);\n",
-      "print '\\tSince particle size  = %f micrometer they react according to shrinking-core model'%(dp);\n",
-      "print '\\tConversion obtained for %f micrometer particle:%f'%(dp,XBbar);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Selection of models to represent reactor\n",
-        "\tSince ratio ubr/umf/ephsilonmf) =  0.995908 <1 the reactor is operating in slow bubble regime\n",
-        "\tSince particle size  = 750.000000 micrometer they react according to shrinking-core model\n",
-        "\tConversion obtained for 750.000000 micrometer particle:0.988127\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch3.ipynb b/Fluidization_Engineering/ch3.ipynb
deleted file mode 100755
index 72abab33..00000000
--- a/Fluidization_Engineering/ch3.ipynb
+++ /dev/null
@@ -1,267 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c2d23a740208e3823c43d8afcfc0ef305894f28fcd5f9e9793f1f1c6fe8b25f1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3 : Fluidization and Mapping of Regimes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 68\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "weight = [0,60,150,270,330,360.0];        # Weight in grams for the oversized particles\n",
-      "psize = [50,75,100,125,150,175];          #PSD in micrometers\n",
-      "\n",
-      "#CALCULATION\n",
-      "l = len(psize); # To obtain the size of input array\n",
-      "# Computation of sauter mean diameter for the given PSD\n",
-      "i = 0;\n",
-      "dpi = [0,0,0,0,0,0]\n",
-      "weightf = [0,0,0,0,0,0]\n",
-      "dp = [0,0,0,0,0,0]\n",
-      "while i<l-1:\n",
-      "        dpi[i]=(psize[i]+ psize[i+1])/2.0;\n",
-      "        weightf[i]=(weight[i+1]-weight[i])/weight[5];    \n",
-      "        dp[i]=weightf[i]/float(dpi[i]);    \n",
-      "        i=i+1;\n",
-      "\n",
-      "dpbar=1/sum(dp);                        #Calculation of average particle daimeter Eq.(15)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\n The Sauter mean diameter of the material with the given particle size distribution = %.0f micrometer'%dpbar\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " The Sauter mean diameter of the material with the given particle size distribution = 98 micrometer\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 76\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Estimation of Minimum fluidizing velocity\n",
-      "\n",
-      "#INPUT\n",
-      "ephsilon=0.55;     #Void fraction of bed\n",
-      "rhog=0.0012;       #Density of gas in g/cc\n",
-      "myu=.00018;        #Viscosity of gas in g/cm s\n",
-      "dpbar=0.016;       #Mean diameter of solids in centimeter\n",
-      "phis=0.67;         #Sphericity of solids\n",
-      "rhos=2.6;          #Density of solids in g/cc\n",
-      "g=980;             #Acceleration due to gravity in square cm/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Computation of umf using the simplified equation for small particles\n",
-      "umf=((dpbar**2)*(rhos-rhog)*g*(ephsilon**3)*(phis**2))/(150*myu*(1-ephsilon));#Simplified equation to calculate minimum fluidizing velocity \n",
-      "                                                                              #for small particles Eq.(21)\n",
-      "Re=(dpbar*umf*rhog)/myu;#To calculate Reynolds number for particle\n",
-      "\n",
-      "#Computation of umf if neither void fraction of bed nor sphericity is known\n",
-      "c1=28.7\n",
-      "c2=0.0494;                      #Value of constants from Table 4, page 70\n",
-      "umf1=(myu/(dpbar*rhog))*(((c1**2)+((c2*(dpbar**3)*rhog*(rhos-rhog)*g)/(myu**2)))**0.5-c1); #Equation to calculate minimum fluidizing velocity \n",
-      "                                                                                           #for coarse particles Eq.(25)\n",
-      "err=((umf-umf1)/umf)*100;       #Calculation of error from experimental value\n",
-      "\n",
-      "#OUTPUT\n",
-      "if Re<20:\n",
-      "\tprint 'The particle Reynolds no = %f'%Re\n",
-      "\tprint 'The simplified equation used for calculating minimum fluidizing velocity is valid.'\n",
-      "\n",
-      "print 'The minimum fluidizing velocity by simplified equation for small particles = %.2fcm/s'%umf\n",
-      "print 'The minimum fluidizing velocity by equation for coarse partilces = %.2fcm/s'%umf1\n",
-      "print 'This value is %d percent below the experimentally reported value.'%err\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The particle Reynolds no = 0.427493\n",
-        "The simplified equation used for calculating minimum fluidizing velocity is valid.\n",
-        "The minimum fluidizing velocity by simplified equation for small particles = 4.01cm/s\n",
-        "The minimum fluidizing velocity by equation for coarse partilces = 3.10cm/s\n",
-        "This value is 22 percent below the experimentally reported value.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 82\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "rhog=1.2e-3;    #Density of air in g/cc\n",
-      "myu=1.8e-4      #Viscosity of air in g/cm s\n",
-      "dpbar=0.016     #Mean diameter of solids in centimeter\n",
-      "phis=0.67;      #Sphericity of solids\n",
-      "rhos=2.6;       #Density of solids in g/cc\n",
-      "g=980           #Acceleration due to gravity in square cm/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "dpstar=dpbar*((rhog*(rhos-rhog)*g)/myu**2)**(1/3.0);              #Calculation of dimensionless particle size Eq.(31)\n",
-      "utstar=((18/(dpstar**2))+(2.335-(1.744*phis))/(dpstar**0.5))**-1; #Calculation of dimensionless gas velocity Eq.(33)\n",
-      "ut=utstar*((myu*(rhos-rhog)*g)/rhog**2)**(1/3.0);                 #Calculation of terminal velocity of falling particles Eq.(32)\n",
-      "\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The dimensionless particle size = %.2f'%dpstar\n",
-      "print 'The dimensionless gas velocity = %.3f'%utstar\n",
-      "print 'The terminal velocity of falling particles = %d cm/s'%ut\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dimensionless particle size = 7.28\n",
-        "The dimensionless gas velocity = 1.296\n",
-        "The terminal velocity of falling particles = 88 cm/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 91\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "rhos=1.5;                 #Density of Solid in g/cc\n",
-      "uo1=40; uo2=80;           #Superficial gas velocity in cm/s\n",
-      "dp1=0.006; dp2=0.045;     #Particle size in centimeter\n",
-      "rhog1=1.5E-3; rhog2=1E-3; #Density of gas in g/cc\n",
-      "myu1=2E-4; myu2=2.5E-4;   #Viscosity of air in g/cm s\n",
-      "g=980;                    #Acceleration due to gravity in square cm/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "#for smaller particles\n",
-      "dpstar1=dp1*((rhog1*(rhos-rhog1)*g)/myu1**2)**(1/3.0);     #Calculation of dimensionless particle diamter Eq.(31)\n",
-      "uostar1=uo1*((rhog1**2)/((myu1)*(rhos-rhog1)*g))**(1/3.0);\n",
-      "uostar2=uo2*((rhog1**2)/((myu1)*(rhos-rhog1)*g))**(1/3.0); #Calculation of dimensionless superficial gas velocity Eq.(32)\n",
-      "\n",
-      "#for larger particles \n",
-      "dpstar2=dp2*((rhog2*(rhos-rhog2)*g)/myu2**2)**(1/3.0);     #Calculation of dimensionless particle diamter Eq.(31)\n",
-      "uostar3=uo1*((rhog2**2)/((myu2)*(rhos-rhog2)*g))**(1/3.0);\n",
-      "uostar4=uo2*((rhog2**2)/((myu2)*(rhos-rhog2)*g))**(1/3.0); #Calculation of dimensionless superficial gas velocity Eq.(32)\n",
-      "\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'For particle of size %.3f centimeter'%dp1\n",
-      "print 'The dimensionless particle diameter = %.2f'%dpstar1\n",
-      "print 'The dimensionless superficial gas velocity = %.4fcm/s(for superficial gas velocity of %dcm/s)'%(uostar1,uo1)\n",
-      "print 'The dimensionless superficial gas velocity = %.3fcm/s(for superficial gas velocity of %dcm/s)'%(uostar2,uo2)\n",
-      "print 'From Fig.16(page 89)comparing u*=%.4f vs dp*=%.2f'%(uostar1,dpstar1)\n",
-      "print 'For Superficial gas velocity =%d Mode of Fluidization:Onset of turbulent fluidization in an ordinary bubbling bed'%(uo1)\n",
-      "print 'From Fig.16(page 89)comparing u* =%.3f vs dp* =%f'%(uostar2,dpstar1)\n",
-      "print 'For Superficial gas velocity =%f Mode of Fluidization:Fast fluidization(requires a circulating solid system)'%(uo2)\n",
-      "print 'For particle of size %f centimeter'%(dp2)\n",
-      "print 'The dimensionless particle diameter = %f'%(dpstar2)\n",
-      "print 'The dimensionless superficial gas velocity = %fcm/s(for superficial gas velocity of %fcm/s)'%(uostar3,uo1)\n",
-      "print 'The dimensionless superficial gas velocity = %fcm/s(for superficial gas velocity of %fcm/s)'%(uostar4,uo2)\n",
-      "print 'From Fig.16(page 89)comparing u*=%f vs dp*=%f'%(uostar3,dpstar2)\n",
-      "print 'For Superficial gas velocity =%f Mode of Fluidization:Bublling Fluidization'%(uo1)\n",
-      "print 'From Fig.16(page 89)comparing u* =%f vs dp* =%f'%(uostar4,dpstar2)\n",
-      "print 'For Superficial gas velocity =%f Mode of Fluidization:Bubbling Fluidization'%(uo2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For particle of size 0.006 centimeter\n",
-        "The dimensionless particle diameter = 2.28\n",
-        "The dimensionless superficial gas velocity = 0.7885cm/s(for superficial gas velocity of 40cm/s)\n",
-        "The dimensionless superficial gas velocity = 1.577cm/s(for superficial gas velocity of 80cm/s)\n",
-        "From Fig.16(page 89)comparing u*=0.7885 vs dp*=2.28\n",
-        "For Superficial gas velocity =40 Mode of Fluidization:Onset of turbulent fluidization in an ordinary bubbling bed\n",
-        "From Fig.16(page 89)comparing u* =1.577 vs dp* =2.282737\n",
-        "For Superficial gas velocity =80.000000 Mode of Fluidization:Fast fluidization(requires a circulating solid system)\n",
-        "For particle of size 0.045000 centimeter\n",
-        "The dimensionless particle diameter = 12.890262\n",
-        "The dimensionless superficial gas velocity = 0.558561cm/s(for superficial gas velocity of 40.000000cm/s)\n",
-        "The dimensionless superficial gas velocity = 1.117122cm/s(for superficial gas velocity of 80.000000cm/s)\n",
-        "From Fig.16(page 89)comparing u*=0.558561 vs dp*=12.890262\n",
-        "For Superficial gas velocity =40.000000 Mode of Fluidization:Bublling Fluidization\n",
-        "From Fig.16(page 89)comparing u* =1.117122 vs dp* =12.890262\n",
-        "For Superficial gas velocity =80.000000 Mode of Fluidization:Bubbling Fluidization\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch4.ipynb b/Fluidization_Engineering/ch4.ipynb
deleted file mode 100755
index cbd2e1e4..00000000
--- a/Fluidization_Engineering/ch4.ipynb
+++ /dev/null
@@ -1,306 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:66650a76faf5eec3efd8e3c62ecb57addf32e1a6b2cb5d90995ca785777b9427"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4 :  The Dense Bed"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 106\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "dt=4;               #Vessel diameter in m\n",
-      "Lmf=2;              #Length of the bed in m\n",
-      "ephsilonmf=0.48;    #Void fraction of bed\n",
-      "rhos=1500;          #Density of solid in kg/m**3\n",
-      "rhog=3.6;           #Density of gas in kg/m**3\n",
-      "myu=2E-5;           #Viscosity of gas in kg/m s\n",
-      "po=3;               #Pressure of inlet gas in bar\n",
-      "uo=0.4;             #Superficial velocity of gas in m/s\n",
-      "uorm=40;            #Maximum allowable jet velocity from holes in m/s\n",
-      "g=9.80;             #Acceleration due to gravity in m/s**2\n",
-      "gc=1;\n",
-      "pi=3.1428;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Computation of minimum allowable pressure drop through the distributor\n",
-      "deltapb=((1-ephsilonmf)*(rhos-rhog)*g*Lmf)/gc;  #Calculation of pressure drop in bed using Eqn.(3.17)\n",
-      "deltapd=0.3*deltapb;                            #Calculation of pressure drop in distributor using Eqn.(3)\n",
-      "\n",
-      "#Computation of orifice coefficient\n",
-      "Ret=(dt*uo*rhog)/myu;\n",
-      "if Ret>=3000:\n",
-      "    Cd=0.60;\n",
-      "elif Ret>=2000:\n",
-      "    Cd=0.61;\n",
-      "elif Ret>=1000:\n",
-      "    Cd=0.64;\n",
-      "elif Ret>=500:\n",
-      "    Cd=0.68;\n",
-      "elif Ret>=300:\n",
-      "    Cd=0.70;\n",
-      "elif Ret>=100:\n",
-      "    Cd=0.68;\n",
-      "\n",
-      "#Computation of gas velocity through orifice\n",
-      "uor=Cd*((2*deltapd)/rhog)**0.5;   #Calculation of gas velocity through orifice by using Eqn.(12)\n",
-      "f=(uo/uor)*100;                   #Calculation of fraction of open area in the perforated plate \n",
-      "\n",
-      "\n",
-      "#Computation of number of orifices per unit area of distributor\n",
-      "dor=[0.001,0.002,0.004];          #Different orifice diameters in m\n",
-      "n=len(dor);\n",
-      "i=0;\n",
-      "Nor = [0.,0.,0.]\n",
-      "while i<n:\n",
-      "    Nor[i]=(uo*4)/(pi*uor*(dor[i])**2);#Calculation of number of orifices by using Eqn.(13)\n",
-      "    i=i+1;\n",
-      "    \n",
-      "#OUTPUT\n",
-      "print 'The pressure drop in bed:%fPa'%deltapb\n",
-      "print 'The minimum allowable pressure drop in distributor:%fPa'%deltapd\n",
-      "if uor<uorm:\n",
-      "    print 'The gas veleocity of %fm/s is satisfactory'%uor\n",
-      "else:\n",
-      "    print 'The gas veleocity of %fm/s is not satisfactory'%uor\n",
-      "\n",
-      "if f<10:\n",
-      "    print 'The fraction of open area of %f percent is allowable'%f\n",
-      "else:\n",
-      "    print 'The fraction of open area of %f percent is not allowable'%f\n",
-      "\n",
-      "print 'Diameter of orifice(m)',\n",
-      "print '\\tNumber of orifices per unit area(per sq.m)'\n",
-      "\n",
-      "j=0;\n",
-      "while j<n:\n",
-      "    print '%f'%dor[j],\n",
-      "    print '\\t\\t%f'%Nor[j]\n",
-      "    j=j+1;\n",
-      "\n",
-      "print 'This number can be rounded off.'\n",
-      "print 'Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter %fm'%dor[2]\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The pressure drop in bed:15251.308800Pa\n",
-        "The minimum allowable pressure drop in distributor:4575.392640Pa\n",
-        "The gas veleocity of 30.250265m/s is satisfactory\n",
-        "The fraction of open area of 1.322302 percent is allowable\n",
-        "Diameter of orifice(m) \tNumber of orifices per unit area(per sq.m)\n",
-        "0.001000 \t\t16829.610145\n",
-        "0.002000 \t\t4207.402536\n",
-        "0.004000 \t\t1051.850634\n",
-        "This number can be rounded off.\n",
-        "Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter 0.004000m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 108\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "lor=0.1;      #Minimum allowable tuyere spacing in m\n",
-      "uorm=30;      #Maximum allowable jet velocity from the tuyere in m/s\n",
-      "uo=0.4;       #Superficial velocity of gas in m/s\n",
-      "uor=30.2;     #Gas velocity through orifice,from Exa 1, in m/s\n",
-      "Cd=0.6;       #Dicharge coefficient from Exa 1\n",
-      "rhog=3.6      #Density of gas in kg/m**3\n",
-      "pi=3.1428;\n",
-      "\n",
-      "#CALCULATION\n",
-      "Nor=1/(lor**2);                     #Calculation of number of orifices per unit area by assuming minimum spacing for tuyeres\n",
-      "dor=((4/pi)*(uo/uor)*(1/Nor))**0.5; #Calculation of diameter of inlet orifiec by using Eqn.(13)\n",
-      "\n",
-      "#Computation of diameter of hole for different number of holes per tuyere\n",
-      "q=(lor**2)*uo;                      #Volumetric flow rate in m**3/s\n",
-      "Nh=[8.,6.,4.];                         #Different number of holes per tuyere\n",
-      "n=len(Nh);\n",
-      "i=0;\n",
-      "dh = [0.,0.,0.]\n",
-      "while i<n:\n",
-      "    dh[i]=((((q/Nh[i])*(4./pi))/uorm)**0.5);#Calculation of diameter of holes\n",
-      "    i=i+1;\n",
-      "\n",
-      "deltaph=(rhog/2.)*((uor/Cd)**2)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Number of holes(number of holes/tuyeres)',\n",
-      "print '\\tDiameter of hole(m)'\n",
-      "j=0;\n",
-      "while j<n:\n",
-      "    print '%f'%Nh[j],\n",
-      "    print '\\t\\t\\t\\t\\t%f'%dh[j]\n",
-      "    j=j+1;\n",
-      "\n",
-      "print 'The design chosen is as follows'\n",
-      "print '\\tTuyeres are as shown in Fig.2(b),page 97'\n",
-      "print '\\tNumber of holes = %f(Since rectangular pitch is chosen for tuyeres)'%Nh[1]\n",
-      "print '\\tDiameter of hole = %fm'%dh[1]\n",
-      "print '\\tDiameter of incoming high-pressure-drop orifice = %fm ID'%dor\n",
-      "print 'Checking the pressure drop in tuyeres'\n",
-      "print 'Since pressure drop of %.1f Pa gives sufficiently high \\\n",
-      "distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.'%deltaph\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of holes(number of holes/tuyeres) \tDiameter of hole(m)\n",
-        "8.000000 \t\t\t\t\t0.004606\n",
-        "6.000000 \t\t\t\t\t0.005318\n",
-        "4.000000 \t\t\t\t\t0.006513\n",
-        "The design chosen is as follows\n",
-        "\tTuyeres are as shown in Fig.2(b),page 97\n",
-        "\tNumber of holes = 6.000000(Since rectangular pitch is chosen for tuyeres)\n",
-        "\tDiameter of hole = 0.005318m\n",
-        "\tDiameter of incoming high-pressure-drop orifice = 0.012984m ID\n",
-        "Checking the pressure drop in tuyeres\n",
-        "Since pressure drop of 4560.2 Pa gives sufficiently high distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 110\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "deltapd=[3,10]    #Distributor pressure drop in kPa\n",
-      "deltapd2=10.0;    #Distributor pressure drop in kPa\n",
-      "po=101.0;         #Entering air pressure in kPa\n",
-      "To=20.0;          #Entering air temperature in degree C\n",
-      "y=1.4;            #Fugacity of air\n",
-      "deltapb=10;       #Pressure drop in bed in kPa\n",
-      "p3=103;           #Pressure at the bed exit in kPa\n",
-      "F=8;              #Feed rate of coal in tons/hr\n",
-      "H=25;             #Gross heatig value of coal in MJ/kg\n",
-      "Fa=10;            #Air required at standard condition in nm**3/kg\n",
-      "etac=0.75;        #Efficiency of compressor\n",
-      "etap=36;          #Efficiency of plant in %\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Calculation of volumetric flow rate of air\n",
-      "vo=((F*1000)*Fa*((To+273)/273.0))/3600.0;\n",
-      "\n",
-      "#Case(a) Distributor Pressure drop = 3kPa and Case(b) Distributor Pressure drop = 10kPa\n",
-      "n=len(deltapd);\n",
-      "i=0;\n",
-      "p1= [0,0]\n",
-      "p2 = [0,0]\n",
-      "ws = [0.,0.]\n",
-      "while i<n:\n",
-      "    p2[i]=p3+deltapb;      #Calculation of pressure at the entrance of the bed\n",
-      "    p1[i]=p2[i]+deltapd[i];#Calculation of pressure before entering the bed\n",
-      "    ws[i]=(y/(y-1))*po*vo*((p1[i]/po)**((y-1)/y)-1)*(1.0/etac);#Calculation of power required for the compressor by Eqn.(18) & Eqn.(20)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Case(c) 50% of the required bypassed to burn the volatile gases. Distributor Pressure drop = 3kPa\n",
-      "#No change in pressure drop from case(a)\n",
-      "v1=vo/2.0;   #New volumetric flow rate of air\n",
-      "ws1=455/2.0; #Power required for blower for primary air\n",
-      "ws2=(y/(y-1))*po*v1*((p3/po)**((y-1)/y)-1)*(1/etac);#Power required for blower for bypassed air\n",
-      "wst=ws1+ws2;                #Total power required for the two blowers\n",
-      "p=((ws[1]-wst)/ws[1])*100;  #Saving in power when compared to case(a)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Case(a)'\n",
-      "print '\\tVolumetric flow rate of air = %.2f m**3/hr'%vo\n",
-      "print '\\tPower required for compressor = %.0f kW'%ws[0]\n",
-      "print 'Case(b)'\n",
-      "print '\\tVolumetric flow rate of air = %.2f m**3/hr'%vo\n",
-      "print '\\tPower required for compressor = %.0f kW'%ws[1]\n",
-      "print 'Case(c)'\n",
-      "print '\\tVolumetric flow rate of air = %.3f m**3/hr'%v1\n",
-      "print '\\tPower required for compressor for primary air = %.1f kW'%ws1\n",
-      "print '\\tPower required for blower for bypassed air = %.1f kW'%ws2\n",
-      "print '\\tTotal power required for the two blowers = %.0f kW'%wst\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case(a)\n",
-        "\tVolumetric flow rate of air = 23.85 m**3/hr\n",
-        "\tPower required for compressor = 454 kW\n",
-        "Case(b)\n",
-        "\tVolumetric flow rate of air = 23.85 m**3/hr\n",
-        "\tPower required for compressor = 651 kW\n",
-        "Case(c)\n",
-        "\tVolumetric flow rate of air = 11.925 m**3/hr\n",
-        "\tPower required for compressor for primary air = 227.5 kW\n",
-        "\tPower required for blower for bypassed air = 31.6 kW\n",
-        "\tTotal power required for the two blowers = 259 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch5.ipynb b/Fluidization_Engineering/ch5.ipynb
deleted file mode 100755
index 791c6af2..00000000
--- a/Fluidization_Engineering/ch5.ipynb
+++ /dev/null
@@ -1,137 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:502044f71f0f4f73ada9ff3f4a68fa509018810bcbdaf16eaa17d381758e9755"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5 : Bubbles in Dense Beds"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 126\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dt=60.0;        #ID of tube in cm \n",
-      "dp=300;         #Size of particles of bed in micrometers\n",
-      "umf=3;          #Velocity at minimum fluidization condition in cm/s\n",
-      "ephsilonmf=0.5; #Void fraction of bed at minimum fluidization condition\n",
-      "db=5;           #Diameter of bubble in cm\n",
-      "g=980;          #Acceleration due to gravity in cm/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Computation of rise velocity of bubble\n",
-      "if (db/dt)<0.125:\n",
-      "    ubr=(0.711*((g*db)**0.5));#Rise velocity by Eqn.(3)\n",
-      "elif (db/dt)<0.6:\n",
-      "    ubr=(0.711*((g*db)**0.5))*1.2*exp(-1.49*(db/dt));#Rise velocity by Eqn.(4)   \n",
-      "\n",
-      "#Computation of cloud thickness\n",
-      "Rb=db/2.0;          #Radius of bubble\n",
-      "uf=umf/ephsilonmf;  #Velocity of emulsion gas\n",
-      "Rc=Rb*((ubr+(2*uf))/(ubr-uf))**(1/3.0);#Radius of cloud by Eqn.(6)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'The rise velocity of the bubble=%.1f cm/s'%ubr\n",
-      "print 'The cloud thickness=%.2f cm'%(Rc-Rb)\n",
-      "print 'From Fig.8(page 124)comparing fw vs dp, for dp = %.0f micrometer, wake fraction = 0.24'%dp\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rise velocity of the bubble=49.8 cm/s\n",
-        "The cloud thickness=0.30 cm\n",
-        "From Fig.8(page 124)comparing fw vs dp, for dp = 300 micrometer, wake fraction = 0.24\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 132\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "uo=15;     #Superificial gas velocity in cm/s\n",
-      "umf=1;     #Velocity at minimum fluidization condition in cm/s\n",
-      "lor=2.0;   #Pitch of perforated plate in cm\n",
-      "g=980;     #Acceleration due to gravity in cm/s**2\n",
-      "#CALCULATION\n",
-      "#Case(a) For porous plate\n",
-      "dbo1=(2.78/g)*(uo-umf)**2;#Initial bubble size using Eqn.(19)\n",
-      "\n",
-      "#Case(b) For Perforated plate\n",
-      "Nor=(2/math.sqrt(3))*(1/lor)**2;#Number of orifices in cm**-2\n",
-      "dbo2=(1.30/(g**0.2))*((uo-umf)/Nor)**0.4;#Initial bubble size using Eqn.(15) assuming inital bubble size is smaller than hole spacing\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Case(a) For porous plate'\n",
-      "print '\\tInitial bubble size=%.2fcm'%dbo1\n",
-      "print 'Case(b) For Perforated plate'\n",
-      "print '\\tInitial bubble size=%.2fcm'%dbo2\n",
-      "print '\\tSince %f<%f, the equation used is correct.'%(dbo2,lor)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case(a) For porous plate\n",
-        "\tInitial bubble size=0.56cm\n",
-        "Case(b) For Perforated plate\n",
-        "\tInitial bubble size=1.55cm\n",
-        "\tSince 1.548765<2.000000, the equation used is correct.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch6.ipynb b/Fluidization_Engineering/ch6.ipynb
deleted file mode 100755
index 9c4f2311..00000000
--- a/Fluidization_Engineering/ch6.ipynb
+++ /dev/null
@@ -1,499 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:432c74872bfc5b3cdf6cd2ab6c97c0ab1581b40f382e3a1bf8502ef1c3f981bb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6 : Bubbling Fluidized Beds"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "z=0.5;     #Height of bed in m\n",
-      "dt=0.5;    #ID of tube in m\n",
-      "rhos=1.6;  #Density of catalyst in g/cm**3\n",
-      "dpbar=60.;  #Averge catalyst diameter in micrometer\n",
-      "umf=0.002; #Velocity at minimum fluidization condition in m/s\n",
-      "uo=0.2;    #Superficial velocity in m/s\n",
-      "dor=2.;     #Diameter of orifice in mm\n",
-      "lor=20.;    #Pitch of perforated plate in mm\n",
-      "g=9.80;    #g=980;#Acceleration due to gravity in m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Method 1. Procedure using Eqn.(10) & Eqn.(11)\n",
-      "db=(0.035+0.040)/2.;#Bubble size at z=0.5m from Fig.7(a) & Fig.7(b)\n",
-      "ub1=1+1.55*((uo-umf)+14.1*(db+0.005))*(dt**0.32)*0.711*(g*db)**0.5;#Bubble velocity using Eqn.(10) & Eqn.(11)\n",
-      "\n",
-      "#Method 2. Werther's procedure\n",
-      "si=0.8;#From Fig.6 for Geldart A solids \n",
-      "ub2=si*(uo-umf)+(3.2*(dt**(1./3)))*(0.711*(g*db)**0.5);#Bubble velocity using Eqn.(9)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Method 1. Procedure using Eqn.(10) & Eqn.(11)'\n",
-      "print '\\tDiameter of the bubble=%.4fm'%db\n",
-      "print '\\tRise velocity of the bubble=%.4fm/s'%ub1\n",
-      "print 'Method 2. Werthers procedure'\n",
-      "print '\\tDiameter of the bubble=%fm'%db\n",
-      "print '\\tRise velocity of the bubble=%fm/s'%ub2\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Method 1. Procedure using Eqn.(10) & Eqn.(11)\n",
-        "\tDiameter of the bubble=0.0375m\n",
-        "\tRise velocity of the bubble=1.4267m/s\n",
-        "Method 2. Werthers procedure\n",
-        "\tDiameter of the bubble=0.037500m\n",
-        "\tRise velocity of the bubble=1.253125m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "z=0.5;      #Height of bed in m\n",
-      "dt=0.5;     #ID of tube in m\n",
-      "rhos=2.6;   #Density of catalyst in g/cm**3\n",
-      "dpbar=100.; #Averge catalyst diameter in micrometer\n",
-      "umf=0.01;   #Velocity at minimum fluidization condition in m/s\n",
-      "uo=0.45;    #Superficial velocity in m/s\n",
-      "dor=2.;     #Diameter of orifice in mm\n",
-      "lor=30.;    #Pitch of perforated plate in mm\n",
-      "g=9.80;     #Acceleration due to gravity in m/s**2\n",
-      "pi=3.142857;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Part(a).Bubble Size\n",
-      "Nor=(2/math.sqrt(3))*(1/lor**2);\n",
-      "dbo=5.5;\n",
-      "\n",
-      "#Method 1.Werther's procedure for finding bubble size\n",
-      "z1=[0,5,10,20,30,50,70]\n",
-      "db = [0.,0.,0.,0.,0.,0.,0.]\n",
-      "n=len(z1);\n",
-      "i=0; \n",
-      "while i<n:\n",
-      "    db[i]=0.853*((1+0.272*(uo-umf)*100)**(1/3.0))*(1+0.0684*z1[i])**1.21;\n",
-      "    i=i+1;    \n",
-      "\n",
-      "db1=0.163;#Since bubble size starts at dbo=5.5cm at z=0, we shift the curve accordingly to z=0.5m\n",
-      "\n",
-      "#Method 2.Mori and Wen's procedure for finding bubble size\n",
-      "dbm=0.65*((math.pi/4.0)*((dt*100)**2)*(uo-umf)*100)**0.4;\n",
-      "db2=dbm-(dbm-dbo)*math.exp(-0.3**(z/dt));\n",
-      "\n",
-      "#Part(b).Bubble Velocity\n",
-      "#Method 1.Procedure using Eqn.(12)\n",
-      "ub1=1.6*((uo-umf)+1.13*db1**0.5)*(dt**1.35)+(0.711*(g*db1)**0.5);\n",
-      "\n",
-      "#Method 2.Werther's Procedure\n",
-      "si=0.65;\n",
-      "ub2=si*(uo-umf)+2*(dt**0.5)*(0.711*(g*db1)**0.5);\n",
-      "\n",
-      "#Using Eqn.(7) & Eqn.(8)\n",
-      "ubr1=0.711*(g*db1)**0.5;\n",
-      "ubr2=0.711*(g*db2/100.0)**0.5\n",
-      "ub3=uo-umf+ubr1;\n",
-      "ub4=uo-umf+ubr2;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Bubble Size'\n",
-      "print 'Initial bubble size from Fig.5.14 for %.2fm/s = %.2fcm'%(uo-umf,dbo)\n",
-      "print '\\tMethod 1.Werthers procedure for finding bubble size'\n",
-      "print '\\t\\tHeight of bed(cm)'\n",
-      "print '\\t\\t\\tBubble size(cm)'\n",
-      "m=len(z1);\n",
-      "j=0;\n",
-      "while j<m:\n",
-      "    print '\\t\\t%d'%z1[j],\n",
-      "    print '\\t\\t\\t\\t%.2f'%db[j]\n",
-      "    j=j+1;\n",
-      "\n",
-      "print '\\tMethod 2.Mori and Wens procedure for finding bubble size'\n",
-      "print '\\t\\tMaximum expected bubble size=%.2fcm'%dbm\n",
-      "print '\\t\\tBubble size=%.0fcm'%db2\n",
-      "print 'Bubble Velocity'\n",
-      "print '\\tMethod 1.Procedure using Eqn.(12)'\n",
-      "print '\\t\\tBubble velocity=%.2fm/s'%ub1\n",
-      "print '\\tMethod 2.Werthers procedure'\n",
-      "print '\\t\\tBubble velocity=%.2fm/s'%ub2\n",
-      "print 'Comparing the above results with the expressions of the simple two-phase theory'\n",
-      "print '\\tWerthers bubble size'\n",
-      "print '\\tBubble rise velocity=%.2fm/s\\tBubble velocity=%.2fm/s'%(ubr1,ub3)\n",
-      "print '\\tMori & Wens bubble size'\n",
-      "print '\\tBubble rise velocity=%.1fm/s\\tBubble velocity=%.2fm/s'%(ubr2,ub4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bubble Size\n",
-        "Initial bubble size from Fig.5.14 for 0.44m/s = 5.50cm\n",
-        "\tMethod 1.Werthers procedure for finding bubble size\n",
-        "\t\tHeight of bed(cm)\n",
-        "\t\t\tBubble size(cm)\n",
-        "\t\t0 \t\t\t\t2.00\n",
-        "\t\t5 \t\t\t\t2.86\n",
-        "\t\t10 \t\t\t\t3.77\n",
-        "\t\t20 \t\t\t\t5.69\n",
-        "\t\t30 \t\t\t\t7.73\n",
-        "\t\t50 \t\t\t\t12.10\n",
-        "\t\t70 \t\t\t\t16.77\n",
-        "\tMethod 2.Mori and Wens procedure for finding bubble size\n",
-        "\t\tMaximum expected bubble size=61.31cm\n",
-        "\t\tBubble size=20cm\n",
-        "Bubble Velocity\n",
-        "\tMethod 1.Procedure using Eqn.(12)\n",
-        "\t\tBubble velocity=1.46m/s\n",
-        "\tMethod 2.Werthers procedure\n",
-        "\t\tBubble velocity=1.56m/s\n",
-        "Comparing the above results with the expressions of the simple two-phase theory\n",
-        "\tWerthers bubble size\n",
-        "\tBubble rise velocity=0.90m/s\tBubble velocity=1.34m/s\n",
-        "\tMori & Wens bubble size\n",
-        "\tBubble rise velocity=1.0m/s\tBubble velocity=1.43m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 153\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dpbar=53.;      #Average particle size in micrometer\n",
-      "s=[1,2];        #Size of Bermuda rock in cm \n",
-      "rhosbar=3200.;  #Average solid density of the coke-zircon mixture in kg/m**3\n",
-      "ephsilonm=0.5;  #Void fraction for fixed bed\n",
-      "ephsilonf=0.75; #Void fraction for bubbling bed\n",
-      "rhogbar=0.64;   #Average density of gas in kg/m**3\n",
-      "uo=14.;         #Superficial gas velocity in cm/s\n",
-      "myu=5E-5;       #Viscosity of gas in kg/m s\n",
-      "T=1000.;        #Temperature in degree C\n",
-      "P=1.;           #Pressure in atm\n",
-      "dt=91.5;        #ID of bed in cm\n",
-      "sh=150.;        #Slumped height in cm\n",
-      "\n",
-      "#CALCULATION\n",
-      "rhog2=1.2;      #Density of ambient air\n",
-      "myu2=1.8E-5;    #Viscosity of ambient air\n",
-      "rhos2=rhog2*(rhosbar/rhogbar);#For the requirement of constant density ratio\n",
-      "m=((rhogbar*myu2)/(rhog2*myu))**(2./3);#Scale factor by usin Eqn.(16)\n",
-      "u2=(m**0.5)*uo;  #Superficial gas velocity by using Eqn.(17)\n",
-      "#OUTPUT\n",
-      "print 'For the model use'\n",
-      "print '\\tBed of ID %.2fcm\\tSlumped bed height of %.2fcm\\tPacked bed distributor consisting of %.2f-%.2fmm rock'%(m*dt,m*sh,m*s[0],m*s[1])\n",
-      "print 'Fluidizing gas: ambient air at %.2fatm'%P\n",
-      "print 'Solids: \\tzirconia, Average particle size=%.2fmicrometers'%(m*dpbar)\n",
-      "print 'Entering gas:\\tSuperficial velocity=%.2fcm/s'%u2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For the model use\n",
-        "\tBed of ID 30.45cm\tSlumped bed height of 49.92cm\tPacked bed distributor consisting of 0.33-0.67mm rock\n",
-        "Fluidizing gas: ambient air at 1.00atm\n",
-        "Solids: \tzirconia, Average particle size=17.64micrometers\n",
-        "Entering gas:\tSuperficial velocity=8.08cm/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "dtb=20;          #ID of bench-scale reactor\n",
-      "dtp=1;           #ID of pilot reactor\n",
-      "dpbar=52;        #Average particle size in micrometer\n",
-      "ephsilonm=0.45;  #Void fraction for fixed bed\n",
-      "ephsilonmf=0.50; #Void fraction at minimum fluidization condition\n",
-      "ephsilonmb=0.60; #Void fraction \n",
-      "uo=30;           #Superficial gas velocity in cm/s\n",
-      "Lmb=2;           #Length of fixed bed in m\n",
-      "umf=0.33;        #Velocity at minimum fluidization condition in cm/s\n",
-      "umb=1;           #Velocity at in cm/s\n",
-      "db=3;            #Equilibrium bubble size in cm\n",
-      "g=9.80;          #Acceleration due to gravity in m/s**2\n",
-      "pi=3.142857;\n",
-      "\n",
-      "#CALCULATION\n",
-      "ubr=0.711*(g*db/100)**0.5;#Rise velocity of bubble using Eqn.(7)\n",
-      "\n",
-      "#Bubble velocity for the bench unit\n",
-      "ubb1=1.55*(((uo-umf)/100.0)+14.1*((db/100.0)+0.005))*((dtb/100.0)**0.32)+ubr;#Bubble velocity using Eqn.(11)\n",
-      "si=1;\n",
-      "ubb2=si*((uo-umf)/100.0)+(3.2*((dtb/100.0)**(1/3.0)))*ubr;#Bubble velocity using Eqn.(9)\n",
-      "ubb=(ubb1+ubb2)/2;#Average bubble velocity\n",
-      "\n",
-      "#Bubble velocity for the pilot unit\n",
-      "ubp1=1.55*(((uo-umf)/100.0)+14.1*((db/100.0)+0.005))*(dtp**0.32)+ubr;#Bubble velocity using Eqn.(11)\n",
-      "si=1;\n",
-      "ubp2=si*((uo-umf)/100)+(3.2*(dtp**(1/3)))*ubr;#Bubble velocity using Eqn.(9)\n",
-      "ubp=(ubp1+ubp2)/2;#Average bubble velocity\n",
-      "\n",
-      "#Rise velocity of upflowing emulsion\n",
-      "ueb=ubb-ubr;#For the bench unit\n",
-      "uep=ubp-ubr;#For the pilot unit\n",
-      "\n",
-      "#Scale-Up Alternative 1.\n",
-      "dteb=20;#Effective bubble diameter\n",
-      "dib=[5,10,15,20];#Different outside diameters\n",
-      "n=len(dib);\n",
-      "li = [0.,0.,0.,0.]\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    li[i]=math.sqrt(((pi*dib[i]*dteb)/4)+((pi/4)*(dib[i])**2));#Pitch using Eqn.(13)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Scale-Up Alternative 2.\n",
-      "Lmp=Lmb*(ubp/ubb);#Static bed height of commercial unit\n",
-      "dtep=100.0;#Effective bubble diameter\n",
-      "dip=[10,15,20,25];#Different outside diameters\n",
-      "m=len(dip);\n",
-      "i=0;\n",
-      "lip = [0.,0.,0.,0.]\n",
-      "while i<m:\n",
-      "    lip[i]=math.sqrt(((math.pi*dip[i]*dtep)/4.0)+(math.pi/4.0)*dip[i]);#Pitch using Eqn.(13)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Height of Bubbling beds\n",
-      "#For bench unit\n",
-      "deltab=((uo/100.0)-(umb/100.0))/(ubb-(umb/100.0));#Fraction of bed in bubbles using Eqn.(28)\n",
-      "ephsilonfb=deltab+(1-deltab)*ephsilonmb;#Void fraction of bubbling bed using Eqn.(20)\n",
-      "Lfb=Lmb*(1-ephsilonm)/(1-ephsilonfb);#Hieght of bubbling bed usnig Eqn.(19)\n",
-      "#For pilot unit\n",
-      "deltap=((uo/100.0)-(umb/100.0))/(ubp-(umb/100.0));#Fraction of bed in bubbles using Eqn.(28)\n",
-      "ephsilonfp=deltap+(1-deltap)*ephsilonmb;#Void fraction of bubbling bed using Eqn.(20)\n",
-      "Lfp=Lmp*(1-ephsilonm)/(1-ephsilonfp);#Hieght of bubbling bed usnig Eqn.(19)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Rise velocity of bubble=%.3fm/s'%ubr\n",
-      "print 'For the bench unit'\n",
-      "print '\\tWith Eqn.(11), Rise velocity=%.3fm/s'%ubb1\n",
-      "print '\\tWith Werthers procedure, Rise velocity=%.2fm/s'%ubb2\n",
-      "print '\\tAverage rise velocity=%.2fm/s'%ubb\n",
-      "print '\\tRise velocity of upflowing emulsion=%.2fm/s'%ueb\n",
-      "print 'For the pilot unit'\n",
-      "print '\\tWith Eqn.(11), Rise velocity=%.2fm/s'%ubp1\n",
-      "print '\\tWith Werthers procedure, Rise velocity=%.2fm/s'%ubp2\n",
-      "print '\\tAverage rise velocity=%.2fm/s'%ubp\n",
-      "print '\\tRise velocity of upflowing emulsion=%.2fm/s'%uep\n",
-      "print 'Scale-Up Alternative 1.'\n",
-      "print '\\tOuter diameter of tube(cm)'\n",
-      "print '\\tPitch(cm)'\n",
-      "n=len(dib);\n",
-      "j=0;\n",
-      "while j<n:\n",
-      "    print '\\t\\t%d'%dib[j],\n",
-      "    print '\\t\\t\\t%.2f'%li[j]\n",
-      "    j=j+1;\n",
-      "\n",
-      "print '\\tSuitable arrangement'\n",
-      "print '\\t\\tOuter Diameter=%dcm\\tPitch:Diameter ratio=%.2f'%(dib[1],(li[1]/dib[1]))\n",
-      "print 'Scale-Up Alternative 2.'\n",
-      "print '\\tStatic bed height for commercial unit=%fm'%Lmp\n",
-      "print '\\tOuter diameter of tube(cm)'\n",
-      "print '\\tPitch(cm)'\n",
-      "n=len(dip);\n",
-      "j=0;\n",
-      "\n",
-      "while j<n:\n",
-      "    print '\\t\\t%d'%dip[j],\n",
-      "    print '\\t\\t\\t%.2f'%lip[j]\n",
-      "    j=j+1;\n",
-      "\n",
-      "print '\\tSuitable arrangement'\n",
-      "print '\\t\\tOuter Diameter=%dcm\\tPitch:Diameter ratio=%.2f'%(dip[2],(lip[2]/dip[2]))\n",
-      "print '\\t\\t\\t\\tFraction of bed in bubbles\\tVoid fraction of bed\\tStatic bed height(m)\\tHeight of bubbling bed(m)'\n",
-      "print '\\t\\t\\t\\t---------------------------------------------------------------------------------------------------------'\n",
-      "print 'Bench unit\\tID=%fm\\t%f\\t\\t\\t%f\\t\\t%f\\t\\t%f'%(dtb/100.,deltab,ephsilonfb,Lmb,Lfb)\n",
-      "print 'Commercial unit\\tID=%fm\\t%f\\t\\t\\t%f\\t\\t%f\\t\\t%f'%(dtp,deltap,ephsilonfp,Lmp,Lfp)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rise velocity of bubble=0.386m/s\n",
-        "For the bench unit\n",
-        "\tWith Eqn.(11), Rise velocity=1.117m/s\n",
-        "\tWith Werthers procedure, Rise velocity=1.02m/s\n",
-        "\tAverage rise velocity=1.07m/s\n",
-        "\tRise velocity of upflowing emulsion=0.68m/s\n",
-        "For the pilot unit\n",
-        "\tWith Eqn.(11), Rise velocity=1.61m/s\n",
-        "\tWith Werthers procedure, Rise velocity=1.53m/s\n",
-        "\tAverage rise velocity=1.57m/s\n",
-        "\tRise velocity of upflowing emulsion=1.18m/s\n",
-        "Scale-Up Alternative 1.\n",
-        "\tOuter diameter of tube(cm)\n",
-        "\tPitch(cm)\n",
-        "\t\t5 \t\t\t9.91\n",
-        "\t\t10 \t\t\t15.35\n",
-        "\t\t15 \t\t\t20.31\n",
-        "\t\t20 \t\t\t25.07\n",
-        "\tSuitable arrangement\n",
-        "\t\tOuter Diameter=10cm\tPitch:Diameter ratio=1.54\n",
-        "Scale-Up Alternative 2.\n",
-        "\tStatic bed height for commercial unit=2.941439m\n",
-        "\tOuter diameter of tube(cm)\n",
-        "\tPitch(cm)\n",
-        "\t\t10 \t\t\t28.16\n",
-        "\t\t15 \t\t\t34.49\n",
-        "\t\t20 \t\t\t39.83\n",
-        "\t\t25 \t\t\t44.53\n",
-        "\tSuitable arrangement\n",
-        "\t\tOuter Diameter=20cm\tPitch:Diameter ratio=1.99\n",
-        "\t\t\t\tFraction of bed in bubbles\tVoid fraction of bed\tStatic bed height(m)\tHeight of bubbling bed(m)\n",
-        "\t\t\t\t---------------------------------------------------------------------------------------------------------\n",
-        "Bench unit\tID=0.200000m\t0.274171\t\t\t0.709668\t\t2.000000\t\t3.788769\n",
-        "Commercial unit\tID=1.000000m\t0.185857\t\t\t0.674343\t\t2.941439\t\t4.967774\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5, Page 161\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dtb=20;          #ID of bench-scale reactor\n",
-      "dtp=1;           #ID of pilot reactor\n",
-      "dpbar=200;       #Average particle size in micrometer\n",
-      "ephsilonmf=0.50; #Void fraction at minimum fluidization condition\n",
-      "ephsilonmb=0.50; #Void fraction \n",
-      "uo=30;           #Superficial gas velocity in cm/s\n",
-      "Lmb=2;           #Length of fixed bed in m\n",
-      "umf=3;           #Velocity at minimum fluidization condition in cm/s\n",
-      "umb=3;           #Velocity at in cm/s\n",
-      "g=9.80;          #Acceleration due to gravity in m/s**2\n",
-      "pi=3.142857;\n",
-      "\n",
-      "#CALCULATION\n",
-      "#In the small bench unit\n",
-      "c=1;\n",
-      "ubb=c*((uo-umf)/100.0)+0.35*(g*(dtb/100.0))**0.5;#Velocity using Eqn.(5.22)\n",
-      "zsb=60*(dtb)**0.175;#Height using Eqn.(5.24)\n",
-      "\n",
-      "#In the large pilot unit\n",
-      "ubp=c*((uo-umf)/100.0)+0.35*(g*dtp)**0.5;#Velocity using Eqn.(5.22)\n",
-      "zsp=60*(dtp*100)**0.175;#Height using Eqn.(5.24)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Condition at which bubbles transform into slugs'\n",
-      "print 'For tha small bench unit\\t\\tVelocity=%fm/s\\t\\tHeight above distributor plate=%fm'%(ubb,zsb/100.)\n",
-      "print 'For tha large pilot unit\\t\\tVelocity=%fm/s\\t\\tHeight above distributor plate=%fm'%(ubp,zsp/100.);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Condition at which bubbles transform into slugs\n",
-        "For tha small bench unit\t\tVelocity=0.760000m/s\t\tHeight above distributor plate=1.013518m\n",
-        "For tha large pilot unit\t\tVelocity=1.365673m/s\t\tHeight above distributor plate=1.343233m\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch7.ipynb b/Fluidization_Engineering/ch7.ipynb
deleted file mode 100755
index aafc2f66..00000000
--- a/Fluidization_Engineering/ch7.ipynb
+++ /dev/null
@@ -1,399 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8eb00dc3983a05f0d88bd9a81e8b022860563cdbf1df8dacf4157210feb1c3ce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7  : Entrainment and Elutriation  from Fluidized Beds"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 179"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "rhog=5.51;  #Density of gas in kg/m**3\n",
-      "rhos=1200;  #Density of solid in kg/m**3\n",
-      "dpbar=130;  #Average size of particles in micrometer\n",
-      "uo=0.61;    #Superficial gas velocity in m/s\n",
-      "g=9.80;     #Acceleration due to gravity in m/s**2\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Assuming that freeboard in higher than TDH, computation of entrainment rate by Zenz & Weil's method\n",
-      "x=(uo**2)/(g*(dpbar*10**-6)*rhos**2);#Calculation of value of x-axis for Fig.(6), page 175\n",
-      "y=1.2;                               # Value of y-axis from Fig.(6)\n",
-      "Gsstar=y*rhog*uo;                    #Computation of rate of entrainment\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nRate of entrainment=%.2fkg/m**2s'%Gsstar\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Rate of entrainment=4.03kg/m**2s\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 180\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "x=0.2;          #Fraction of fines in the bed\n",
-      "Gsstar=4.033320 #Rate of entrainment in kg/m**2s(from Exa.1)\n",
-      "\n",
-      "#CALCULATION\n",
-      "Gsstar1=x*Gsstar;#Rate of entrainment by Eqn.(3)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nRate of entrainment=%.3fkg/m**2s'%Gsstar1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Rate of entrainment=0.807kg/m**2s\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "rhog=5.51;                    #Density of gas in kg/m**3\n",
-      "rhos=1200;                    #Density of solid in kg/m**3\n",
-      "uo=0.61;                      #Superficial gas velocity in m/s\n",
-      "g=9.80;                       #Acceleration due to gravity in m/s**2\n",
-      "dp=[10,30,50,70,90,110,130];  #Diameter of particle in micrometer\n",
-      "p=[0.,0.0110,0.0179,0.0130,0.0058,0.0020,0.];\n",
-      "pi=3.142857;\n",
-      "dt=6;\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(dp);\n",
-      "i=0;\n",
-      "x = [0,0,0,0,0,0,0]\n",
-      "while i<n:\n",
-      "    x[i]=(uo**2)/(g*(dp[i]*10**-6)*rhos**2);#Computation of value of x-axis for Fig.(6), page 175)\n",
-      "    i=i+1;\n",
-      "\n",
-      "y=[40,12,6,3.2,2.,1.3,1];#Value of y-axis corresponding to each value of x-axis\n",
-      "y1 = []\n",
-      "for i in range(n):\n",
-      "    y1.append(y[i]*p[i]);\n",
-      "i=0;\n",
-      "k=0;\n",
-      "\n",
-      "while i<n-2:\n",
-      "    y1[i]=(y[i]*p[i]);\n",
-      "    k=k+((0.5)*(dp[i+1]-dp[i])*(y1[i+1]+y1[i]));#Integration using Trapezoidal rule\n",
-      "    i=i+1;\n",
-      "rhosbar=k*rhog;#Computation of solid loading\n",
-      "te=(pi/4)*(dt**2)*rhosbar*uo;#Computation of total entrainment\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nSolid loading =%.1fkg/m**3'%rhosbar\n",
-      "print '\\nTotal Entrainment =%.0fkg/s'%te\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Solid loading =32.4kg/m**3\n",
-        "\n",
-        "Total Entrainment =559kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "dp=[40,60,80,100,120]; #Diameter of particle in micrometer\n",
-      "uo=0.381;              #Superficial gas velocity in m/s\n",
-      "\n",
-      "#CALCULATION\n",
-      "Gs=0.9;#Rate of entrainment in kg/m**2 s from Fig.3(a)\n",
-      "pb = [0.45,1.00,1.25,1.00,0.60];#Size distribution for bed particles from Fig.3(b)\n",
-      "pe=[1.20,2.00,1.25,0.45,0.10];  #Size distribution for entrained particles from Fig.3(b)\n",
-      "n=len(dp);\n",
-      "for i in range(n):\n",
-      "    pb[i] = pb[i]/100.\n",
-      "    pe[i] = pe[i]/100.\n",
-      "i=0;\n",
-      "ki = []\n",
-      "while i<n:\n",
-      "    ki.append((Gs*pe[i])/pb[i]);#Calculation of ki* using Eqn.(13)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\ndpi(micrometer)',\n",
-      "print '\\t100pb(dpi)(micrometer**-1)',\n",
-      "print '\\t100pe(dpi)(micrometer**-1)',\n",
-      "print '\\tki*(kg/m**2 s)'\n",
-      "\n",
-      "j=0;\n",
-      "while j<n:\n",
-      "    print '%f'%dp[j],\n",
-      "    print '\\t%f'%(100*pb[j]),\n",
-      "    print '\\t\\t\\t%f'%(100*pe[j]),\n",
-      "    print '\\t\\t\\t%f'%ki[j]\n",
-      "    j=j+1;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "dpi(micrometer) \t100pb(dpi)(micrometer**-1) \t100pe(dpi)(micrometer**-1) \tki*(kg/m**2 s)\n",
-        "40.000000 \t0.450000 \t\t\t1.200000 \t\t\t2.400000\n",
-        "60.000000 \t1.000000 \t\t\t2.000000 \t\t\t1.800000\n",
-        "80.000000 \t1.250000 \t\t\t1.250000 \t\t\t0.900000\n",
-        "100.000000 \t1.000000 \t\t\t0.450000 \t\t\t0.405000\n",
-        "120.000000 \t0.600000 \t\t\t0.100000 \t\t\t0.150000\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5, Page 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "rhog=1.217;                                   #Density of gas in kg/m**3\n",
-      "myu=1.8E-5;                                   #Viscosity of gas in kg/m s\n",
-      "umf=0.11;                                     #Velocity at minimum fluidization condition in m/s\n",
-      "rhos=2000.0;                                  #Density of solid in kg/m**3\n",
-      "uo=1.0;                                       #Superficial gas velocity in m/s\n",
-      "g=9.80;                                       #Acceleration due to gravity in m/s**2\n",
-      "dp=[30,40,50,60,80,100,120];                  #Diameter of particle in micrometer\n",
-      "uti=[0.066,0.115,0.175,0.240,0.385,0.555,1.0];#Terminal velocity of particles in m/s\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(dp);\n",
-      "i=0;\n",
-      "Ret = []\n",
-      "kistar1 = []\n",
-      "kistar2 = []\n",
-      "kistar3 = []\n",
-      "kistar4 = []\n",
-      "kistar5 = []\n",
-      "kistar6 = []\n",
-      "x1 = []\n",
-      "x2 = []\n",
-      "\n",
-      "while i<n:\n",
-      "    #Using Yagi & Aochi's correlation\n",
-      "    Ret.append((rhog*(uti[i])*dp[i]*10**-6)/myu)\n",
-      "    a =((myu*((uo-uti[i])**2))/(g*(dp[i]*10**-6)**2))*(0.0015*(Ret[i]**0.5)+(0.01*(Ret[i]**1.2)));\n",
-      "    kistar1.append(a)\n",
-      "    #Using Wen & Hasinger's correlation\n",
-      "    a=(((1.52E-5)*((uo-uti[i])**2)*rhog)/(g*dp[i]*10**-6)**0.5)*(Ret[i]**0.725)*((rhos-rhog)/rhog)**1.15;\n",
-      "    kistar2.append(a)\n",
-      "    #Using Merrick & Highley's correlation\n",
-      "    a=uo*rhog*(0.0001+130*math.exp(-10.4*((uti[i]/uo)**0.5)*((umf/(uo-umf))**0.25)));\n",
-      "    kistar3.append(a)\n",
-      "    #Using Geldart's correlation\n",
-      "    a=23.7*uo*rhog*math.exp(-5.4*(uti[i]/uo));\n",
-      "    kistar4.append(a)\n",
-      "    #Using Zenz & Weil's procedure\n",
-      "    a=(uo**2)/(g*(dp[i]*10.0**-6)*rhos**2);#Computation of value of x-axis for Fig.(6), page 175)\n",
-      "    x1.append(a)\n",
-      "    y1=[12.2,8.6,6.4,4.9,2.75,1.8,1.2];#Value of y-axis corresponding to each value of x-axis\n",
-      "    kistar5.append(y1[i]*rhog*uo)\n",
-      "    #Using Gugnoni & Zenz's procedure\n",
-      "    a=(uo-uti[i])/((g*dp[i]*10**-6)**0.5);#Computation of value of x-axis for Fig.(6), page 175)\n",
-      "    x2.append(a)\n",
-      "    y=[5.8,5.4,3.2,2.8,1.3,0.6,0];#Value of y-axis corresponding to each value of x-axis\n",
-      "    kistar6.append(y[i]*rhog*uo)\n",
-      "    i=i+1;\n",
-      "\n",
-      "i=0;\n",
-      "print 'dp(micrometer)',\n",
-      "print '\\tYagi & Aochi',\n",
-      "print '\\tWen & Hashinger',\n",
-      "print '\\t\\tMerrick & Highley',\n",
-      "print '\\tGeldart et al.',\n",
-      "print '\\t\\tZenz & Well',\n",
-      "print '\\t\\tGugnoni & Zenz'\n",
-      "while i<n:\n",
-      "    print '\\n%f'%dp[i],\n",
-      "    print '\\t%f'%kistar1[i],\n",
-      "    print '\\t%f'%kistar2[i],\n",
-      "    print '\\t\\t%f'%kistar3[i],\n",
-      "    print '\\t\\t%f'%kistar4[i],\n",
-      "    print '\\t\\t%f'%kistar5[i],\n",
-      "    print '\\t\\t%f'%kistar6[i],\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Note: There is huge deviation of the calculated answer and the answer given in the textbook for the correlation of Merrick & Highley.  There is a contradiction in the correlation used in the problem and the one given in page 179. \n",
-      "#We tried to retrieve the original paper i.e. D.Merrick and J.Highley, AICHE J., 6, 220(1960). But the effort was not fruitful.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "dp(micrometer) \tYagi & Aochi \tWen & Hashinger \t\tMerrick & Highley \tGeldart et al. \t\tZenz & Well \t\tGugnoni & Zenz\n",
-        "\n",
-        "30.000000 \t2.571188 \t1.092184 \t\t32.451340 \t\t20.195582 \t\t14.847400 \t\t7.058600 \n",
-        "40.000000 \t2.965958 \t1.564720 \t\t19.546385 \t\t15.500369 \t\t10.466200 \t\t6.571800 \n",
-        "50.000000 \t3.240381 \t1.938471 \t\t11.993076 \t\t11.210646 \t\t7.788800 \t\t3.894400 \n",
-        "60.000000 \t3.289995 \t2.154988 \t\t7.713841 \t\t7.892113 \t\t5.963300 \t\t3.407600 \n",
-        "80.000000 \t2.852535 \t2.120728 \t\t3.447977 \t\t3.606955 \t\t3.346750 \t\t1.582100 \n",
-        "100.000000 \t1.883718 \t1.521994 \t\t1.600171 \t\t1.440318 \t\t2.190600 \t\t0.730200 \n",
-        "120.000000 \t0.000000 \t0.000000 \t\t0.332158 \t\t0.130271 \t\t1.460400 \t\t0.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6, Page 190\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "dpbar=60.0;   #Average size of particles in micrometer\n",
-      "rhog=1.3;     #Density of gas in kg/m**3\n",
-      "rhos=1500.0;  #Density of solid in kg/m**3\n",
-      "umf=0.003;    #Velocity at minimum fluidization condition in m/s\n",
-      "uo=0.503;     #Superficial gas velocity in m/s\n",
-      "g=9.80;       #Acceleration due to gravity in m/s**2\n",
-      "Hf=2.0;       #Height at which the cyclone inlet is to be located in m\n",
-      "\n",
-      "#CALCULATION\n",
-      "y=(uo**2)/(g*(dpbar*10**-3)*rhos**2);#Calculation of value of y-axis for Fig.(6), page 175\n",
-      "x=1;#Value of x-axis from Fig.(6), page 175\n",
-      "Gsstar=x*rhog*uo;#Computation of rate of entrainment\n",
-      "Gsuo=5.0;#Ejection rate pf particles in kg/m**2 s from Fig.(11), page 188\n",
-      "a=0.72/uo;#From Fig.(12), page 189\n",
-      "Gs=Gsstar+(Gsuo-Gsstar)*math.exp(-a*Hf);\n",
-      "p=((Gs-Gsstar)/Gsstar)*100.0;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nRate of entrainment from short bed=%.3fkg/m**2s'%Gs\n",
-      "print '\\nThis entrainment is %f percent higher than it would be if the gas exit were at the TDH'%p\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Rate of entrainment from short bed=0.902kg/m**2s\n",
-        "\n",
-        "This entrainment is 37.955972 percent higher than it would be if the gas exit were at the TDH\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch8.ipynb b/Fluidization_Engineering/ch8.ipynb
deleted file mode 100755
index 9fed0363..00000000
--- a/Fluidization_Engineering/ch8.ipynb
+++ /dev/null
@@ -1,218 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8 : High Velocity Fludization"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 206\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "from numpy import zeros\n",
-      "from scipy.optimize import fsolve \n",
-      "\n",
-      "\n",
-      "#INPUT\n",
-      "Lmf=2.4;            #Length of bed at minimum fluidized condition in m\n",
-      "uo=[2.,4.,6.];         #Superficial gas velocity in m/s\n",
-      "GsII=100.;          #Solid circulation rate in kg/m**2 s for Mode II\n",
-      "uoIII=4.;            #Superficial gas velocity in m/s for Mode III\n",
-      "GsIII=[42.,50.,100.,200.,400.];          #Solid circulation rate in kg/m**2 s for Mode III\n",
-      "GsIV=[70.,100.,120.];                  #Solid circulation rate in kg/m**2 s for Mode IV\n",
-      "dt=0.4;                             #Column diamter in m\n",
-      "Ht=10.;                              #Height of column in m\n",
-      "rhos=1000.;                         #Density of solid in kg/m**3\n",
-      "dpbar=55.;                          #Particle diameter in micrometer\n",
-      "ephsilonmf=0.5;                     #Void fraction at minimum fluidization condition\n",
-      "\n",
-      "#CALCULATION\n",
-      "#Mode I\n",
-      "ephsilonstar=0.01;                  #Saturation carrying capacity of gas\n",
-      "ephsilonsd=[0.2,0.16,0.14];         #Solid holdup in lower dense region from Fig.8(b) for various uo\n",
-      "n=len(uo);\n",
-      "i=0;\n",
-      "Hfguess=2.;          #Guess value of height\n",
-      "a = zeros(n)\n",
-      "Hf = zeros(n)\n",
-      "Hd = zeros(n)\n",
-      "ephsilonse = zeros(n)\n",
-      "GsI = zeros(n)\n",
-      "#    endfunction\n",
-      " \n",
-      "while i<n:\n",
-      "    a[i]= 3./uo[i];         #Decay consmath.tant\n",
-      "    def solver_func(Hf):            #Function defined for solving the system\n",
-      "        return Lmf*(1-ephsilonmf)-((ephsilonsd[i]-(ephsilonstar+(ephsilonsd[i]-ephsilonstar)*math.exp(-a[i]*Hf)))/a[i])-Ht*ephsilonsd[i]+Hf*(ephsilonsd[i]-ephsilonstar);\n",
-      "        \n",
-      "    Hf[i] = fsolve(solver_func,1E-6)\n",
-      "    Hd[i]=Ht-Hf[i];#Height of lower densce region\n",
-      "    ephsilonse[i]=ephsilonstar+(ephsilonsd[i]-ephsilonstar)*math.exp(-a[i]*Hf[i]);#Solid holdup at exit\n",
-      "    GsI[i]=rhos*uo[i]*ephsilonse[i];#Solid circulation rate from Eqn.(4)\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Mode II\n",
-      "i=0;\n",
-      "Hfguess2=2;#Guess value of height\n",
-      "ephsilonseII = zeros(n)\n",
-      "HdII = zeros(n)\n",
-      "LmfII = zeros(n)\n",
-      "HfII = zeros(n)\n",
-      "while i<n:\n",
-      "    ephsilonseII[i]=GsII/(rhos*uo[i]);#Solid holdup at exit\n",
-      "    def solver_func1(Hf):           #Function defined for solving the system\n",
-      "        return ephsilonseII[i]-ephsilonstar-(ephsilonsd[i]-ephsilonstar)*math.exp(-a[i]*Hf);#From Eqn.(7)\n",
-      "    HfII[i] = fsolve(solver_func1,1E-6)\n",
-      "    HdII[i]=Ht-HfII[i];#Height of lower dense region\n",
-      "    #Length of bed minimum fluidization condtion\n",
-      "    LmfII[i]=(1-ephsilonmf)**-1*((ephsilonsd[i]-ephsilonseII[i])/a[i])+Ht*ephsilonsd[i]-HfII[i]*(ephsilonsd[i]-ephsilonstar);\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Mode III\n",
-      "aIII = 3./uoIII;            #Decay consmath.tant\n",
-      "ephsilonsdIII=0.16;         #Solid holdup at lower dense region\n",
-      "i=0;\n",
-      "m=len(GsIII);\n",
-      "Hfguess3=2;#Guess value of height \n",
-      "ephsilonseIII = zeros(m)\n",
-      "HdIII = zeros(m)\n",
-      "ephsilonseIII = zeros(m)\n",
-      "LmfIII = zeros(m)\n",
-      "HfIII = zeros(m)\n",
-      "while i<m:\n",
-      "    ephsilonseIII[i]=GsIII[i]/(rhos*uoIII);#Solid holdup at exit\n",
-      "    def solver_func2(Hf):       #Function defined for solving the system\n",
-      "        return ephsilonseIII[i]-ephsilonstar-(ephsilonsdIII-ephsilonstar)*math.exp(-aIII*Hf);#From Eqn.(7)\n",
-      "\n",
-      "    HfIII[i] = fsolve(solver_func2,1E-6)\n",
-      "    HdIII[i]=Ht-HfIII[i];       #Height of lower dense region\n",
-      "    #Length of bed at minimum fluidization condition\n",
-      "    LmfIII[i]=(1-ephsilonmf)**-1*(((ephsilonsdIII-ephsilonseIII[i])/aIII)+Ht*ephsilonsdIII-HfIII[i]*(ephsilonsdIII-ephsilonstar));\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Mode IV\n",
-      "i=0;\n",
-      "Hfguess4=2;#Guess value of height\n",
-      "aIV = zeros(n)\n",
-      "ephsilonseIV = zeros(n)\n",
-      "HdIV = zeros(n)\n",
-      "HfIV = zeros(n)\n",
-      "LmfIV = zeros(n)\n",
-      "while i<n:\n",
-      "    aIV[i]=3./uo[i];            #Decay consmath.tant\n",
-      "    ephsilonseIV[i]=GsIV[i]/(rhos*uo[i]);       #Solid holdup at exit\n",
-      "    def solver_func3(Hf):       #Function defined for solving the system\n",
-      "        return ephsilonseIV[i]-ephsilonstar-(ephsilonsd[i]-ephsilonstar)*math.exp(-aIV[i]*Hf);      #From Eqn.(7)\n",
-      "\n",
-      "    HfIV[i] = fsolve(solver_func3,1E-6)\n",
-      "    HdIV[i]=Ht-HfIV[i];#Height of lower dense region\n",
-      "    #Length of bed at minimum fluidization condition\n",
-      "    LmfIV[i]=(1-ephsilonmf)**-1*(((ephsilonsd[i]-ephsilonseIV[i])/aIV[i])+Ht*ephsilonsd[i]-HfIV[i]*(ephsilonsd[i]-ephsilonstar));\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Mode I';\n",
-      "print '\\tuom/s\\t\\tephsilonse-\\tHfm\\t\\tHdm\\t\\tGskg/m**2 s';\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(uo[i],ephsilonse[i],Hf[i],Hd[i],GsI[i]);\n",
-      "    i=i+1;\n",
-      "\n",
-      "print 'Mode II';\n",
-      "print '\\tuom/s\\t\\tephsilonse-\\tHfm\\t\\tHdm\\t\\tLmfm';\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(uo[i],ephsilonseII[i],HfII[i],HdII[i],LmfII[i]);\n",
-      "    i=i+1;\n",
-      "\n",
-      "print 'Mode III';\n",
-      "print '\\tGskg/m** s\\tephsilonse-\\tHfm\\t\\tHdm\\t\\tLmfm';\n",
-      "i=0;\n",
-      "while i<m:\n",
-      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(GsIII[i],ephsilonseIII[i],HfIII[i],HdIII[i],LmfIII[i]);\n",
-      "    i=i+1;\n",
-      "\n",
-      "print 'Mode IV';\n",
-      "print '\\tuom/s\\t\\tGskg/m**2 s\\tephsilonse-\\tHfm\\t\\tLmfm';\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(uo[i],GsIV[i],ephsilonseIV[i],HfIV[i],LmfIV[i]);\n",
-      "    i=i+1;\n",
-      "\n",
-      "#====================================END OF PROGRAM ======================================================"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mode I\n",
-        "\tuom/s\t\tephsilonse-\tHfm\t\tHdm\t\tGskg/m**2 s\n",
-        "\t2.000000\t0.192856\t0.025551\t9.974449\t385.711493\n",
-        "\t4.000000\t0.017870\t3.930041\t6.069959\t71.481458\n",
-        "\t6.000000\t0.037349\t3.117707\t6.882293\t224.094133\n",
-        "Mode II\n",
-        "\tuom/s\t\tephsilonse-\tHfm\t\tHdm\t\tLmfm\n",
-        "\t2.000000\t0.050000\t1.038763\t8.961237\t2.002635\n",
-        "\t4.000000\t0.025000\t3.070113\t6.929887\t1.499483\n",
-        "\t6.000000\t0.016667\t5.940829\t4.059171\t1.121026\n",
-        "Mode III\n",
-        "\tGskg/m** s\tephsilonse-\tHfm\t\tHdm\t\tLmfm\n",
-        "\t42.000000\t0.010500\t7.605043\t2.394957\t1.317154\n",
-        "\t50.000000\t0.012500\t5.459126\t4.540874\t1.955596\n",
-        "\t100.000000\t0.025000\t3.070113\t6.929887\t2.638966\n",
-        "\t200.000000\t0.050000\t1.762341\t8.237659\t2.964631\n",
-        "\t400.000000\t0.100000\t0.681101\t9.318899\t3.155670\n",
-        "Mode IV\n",
-        "\tuom/s\t\tGskg/m**2 s\tephsilonse-\tHfm\t\tLmfm\n",
-        "\t2.000000\t70.000000\t0.035000\t1.352099\t3.706202\n",
-        "\t4.000000\t100.000000\t0.025000\t3.070113\t2.638966\n",
-        "\t6.000000\t120.000000\t0.020000\t5.129899\t1.946226\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "/usr/lib/python2.7/dist-packages/scipy/optimize/minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
-        "  improvement from the last ten iterations.\n",
-        "  warnings.warn(msg, RuntimeWarning)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/ch9.ipynb b/Fluidization_Engineering/ch9.ipynb
deleted file mode 100755
index 5087d780..00000000
--- a/Fluidization_Engineering/ch9.ipynb
+++ /dev/null
@@ -1,265 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:52ff353152f32e41c2e832a90f993fbd3f11b7f2a5fbb5e4f20eb7599a61f880"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9 :  Solid Movement Mixing  Segregation and Staging"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 218\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "umf=0.015;                 #Velocity at minimum fluidization condition in m/s\n",
-      "ephsilonmf=0.5;            #Void fraction at minimum fluidization condition\n",
-      "uo=0.1;                    #Superficial gas velocity in m/s\n",
-      "delta=0.2;                 #Bed fraction in bubbles\n",
-      "db=0.06;                   #Equilibrium bubble size in m\n",
-      "dt=[0.1,0.3,0.6,1.5];      #Various vessel sizes in m\n",
-      "ub=[0.4,0.75,0.85,1.1];    #Bubble velocity in m/s\n",
-      "Dsv=[0.03,0.11,0.14,0.23]; #Reported values of vertical dispersion coefficient\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(ub);\n",
-      "i=0;\n",
-      "fw1=2;#Wake fraction from Hamilton et al.\n",
-      "fw2=0.32;#Wake fraction from Fig.(5.8)\n",
-      "fw=(fw1+fw2)*0.5;#Average value of wake fraction\n",
-      "Dsv1 = []\n",
-      "Dsv2 = []\n",
-      "while i<n:\n",
-      "    Dsv1.append(12*((uo*100)**0.5)*((dt[i]*100)**0.9));#Vertical distribution coefficient from Eqn.(3)\n",
-      "    Dsv2.append((fw**2*ephsilonmf*delta*db*ub[i]**2)/(3*umf));#Vertical distribution coefficient from Eqn.(12)\n",
-      "    i=i+1;\n",
-      "\n",
-      "print Dsv1\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\t\\tVertical dispersion coefficient(m**2/s)'\n",
-      "print 'Vessel Size(m)',\n",
-      "print '\\tFrom Experiment',\n",
-      "print '\\tFrom Eqn.(3)',\n",
-      "print '\\tFrom Eqn.(12)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%.2f'%dt[i],\n",
-      "    print '\\t%.3f'%Dsv[i],\n",
-      "    print '\\t%.2f'%(Dsv1[i]/10**4),\n",
-      "    print '\\t%.2f'%Dsv2[i]\n",
-      "    i=i+1;   \n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[301.42637178114967, 810.1965234492834, 1511.8801720132121, 3448.7632274996104]\n",
-        "\t\tVertical dispersion coefficient(m**2/s)\n",
-        "Vessel Size(m) \tFrom Experiment \tFrom Eqn.(3) \tFrom Eqn.(12)\n",
-        "0.10 \t0.030 \t0.03 \t0.03\n",
-        "0.30 \t0.110 \t0.08 \t0.10\n",
-        "0.60 \t0.140 \t0.15 \t0.13\n",
-        "1.50 \t0.230 \t0.34 \t0.22\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "Lmf=0.83;                          #Length of bed at minimum fluidization condition in m\n",
-      "dp=450.0;                          #Average particle size in micrometer\n",
-      "ephsilonmf=0.42;                   #Void fraction at minimum fluidization condition\n",
-      "umf=0.17;                          #Velocity at minimum fluidization condition in m/s\n",
-      "uo=[0.37,0.47,0.57,0.67];          #Superficial gas velocity in m/s\n",
-      "Dsh=[0.0012,0.0018,0.0021,0.0025]; #Horizontal Drift Coefficient from Experiment in m**2/s\n",
-      "db=[0.10,0.14];                    #Equilibrium bubble size in m\n",
-      "g=9.81;                            #Acceleration due to gravity in m/s**2\n",
-      "\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(uo);\n",
-      "m=len(db);\n",
-      "k=0;\n",
-      "alpha=0.77;#Since we are not dealing with Geldart A or AB solids\n",
-      "uf=umf/ephsilonmf;\n",
-      "ubr = []\n",
-      "ub = []\n",
-      "delta = []\n",
-      "Dshc = []\n",
-      "for j in range(m):\n",
-      "    for i in range(n):\n",
-      "        ubr.append(0.711*(db[j]*g)**0.5);#Rise velocity of a single bubble in m/s\n",
-      "        ub.append(uo[i]-umf+ubr[k]);#Rise velocity of bubbles in a bubbling bed\n",
-      "        delta.append((uo[i]-umf)/(ub[k]+umf));#Bed fraction in bubbles\n",
-      "        if ubr[i]>uf:\n",
-      "            Dshc.append((3/16.0)*(delta[k]/(1-delta[k]))*((alpha**2*db[j]*ubr[k]*((((ubr[k]+2*uf)/(ubr[k]-uf))**(1.0/3))-1))));\n",
-      "                #Horizontal Distribution coeff. from Eqn.(14)\n",
-      "        else:\n",
-      "            Dsh.append((3.0/16)*(delta/(1-delta))*(alpha**2*umf*db/ephsilonmf))\n",
-      "                #Horizontal Distribution coeff. from Eqn.(15)\n",
-      "        Dshc.append((3/16.0)*(delta[k]/(1-delta[k]))*((alpha**2*db[j]*ubr[k]*((((ubr[k]+2*uf)/(ubr[k]-uf))**(1/3.0))-1))));#Horizontal Distribution coeff. from Eqn.(14)\n",
-      "        k=k+1;\n",
-      "i=0;\n",
-      "j=0;\n",
-      "k=0;\n",
-      "while k<m*n:\n",
-      "    print 'Snce we do not have ub=%fm/s>>uf=%fm/s we use Eqn.(14).'%(ub[k],uf)\n",
-      "    print 'Gas Velocity(m/s)'\n",
-      "    print '\\tHorizontal Drift Coefficient Calculated(m**2/s)'\n",
-      "    print '\\tHorizontal Drift Coefficient from Experiment(m**2/s)'\n",
-      "    while j<m:\n",
-      "        print 'db=%fm'%db[j]\n",
-      "        while i<n:\n",
-      "            print '%f'%uo[i],\n",
-      "            print '\\t\\t%f'%Dshc[k],\n",
-      "            print '\\t\\t\\t\\t\\t%f'%Dsh[i]\n",
-      "            i=i+1;   \n",
-      "            k=k+1;\n",
-      "        i=0;\n",
-      "        j=j+1;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Snce we do not have ub=0.904213m/s>>uf=0.404762m/s we use Eqn.(14).\n",
-        "Gas Velocity(m/s)\n",
-        "\tHorizontal Drift Coefficient Calculated(m**2/s)\n",
-        "\tHorizontal Drift Coefficient from Experiment(m**2/s)\n",
-        "db=0.100000m\n",
-        "0.370000 \t\t0.001283 \t\t\t\t\t0.001200\n",
-        "0.470000 \t\t0.001283 \t\t\t\t\t0.001800\n",
-        "0.570000 \t\t0.001924 \t\t\t\t\t0.002100\n",
-        "0.670000 \t\t0.001924 \t\t\t\t\t0.002500\n",
-        "db=0.140000m\n",
-        "0.370000 \t\t0.002566 \t\t\t\t\t0.001200\n",
-        "0.470000 \t\t0.002566 \t\t\t\t\t0.001800\n",
-        "0.570000 \t\t0.003207 \t\t\t\t\t0.002100\n",
-        "0.670000 \t\t0.003207 \t\t\t\t\t0.002500\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 232\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Gsup=1.5;               #Solid interchange rate in kg/m**2plate s\n",
-      "dor=19.1;               #Orifice diameter in mm\n",
-      "dp=210;                 #Particle size in micrometer\n",
-      "uo=0.4;                 #Superficial gas velocity in m/s\n",
-      "fopen=[0.12,0.17,0.26]; #Open area fraction \n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(fopen);\n",
-      "uor = []\n",
-      "ls1 = []\n",
-      "i=0\n",
-      "while i<n:\n",
-      "    uor.append(uo/fopen[i]);  #Gas velocity through the orifice\n",
-      "    ls1.append(Gsup/fopen[i]);#Flux of solids through the holes\n",
-      "    i=i+1;\n",
-      "\n",
-      "ls2=[12,20,25];                #Flux of solids through holes from Fig.13(c) for different uor values\n",
-      "fopen1=0.12;                   #Open area fraction which gives reasonable fit\n",
-      "lor=math.sqrt(((math.pi/4)*dor**2)/fopen1);#Orifice spacing\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'fopen',\n",
-      "print '\\t\\tuor(m/s)',\n",
-      "print '\\tls from Eqn.(18)',\n",
-      "print '\\tls from Fig.13(c)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%fopen[i],\n",
-      "    print '\\t%f'%uor[i],\n",
-      "    print '\\t%f'%ls1[i],\n",
-      "    print '\\t\\t%f'%ls2[i]\n",
-      "    i=i+1;   \n",
-      "\n",
-      "print '\\nFor square pitch, the orifice spacing should be %fmm'%lor\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fopen \t\tuor(m/s) \tls from Eqn.(18) \tls from Fig.13(c)\n",
-        "0.120000 \t3.333333 \t12.500000 \t\t12.000000\n",
-        "0.170000 \t2.352941 \t8.823529 \t\t20.000000\n",
-        "0.260000 \t1.538462 \t5.769231 \t\t25.000000\n",
-        "\n",
-        "For square pitch, the orifice spacing should be 48.863850mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fluidization_Engineering/screenshots/plotkbckbedb.png b/Fluidization_Engineering/screenshots/plotkbckbedb.png
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index b7c9004b..00000000
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diff --git a/Fluidization_Engineering/screenshots/relationnubedrep.png b/Fluidization_Engineering/screenshots/relationnubedrep.png
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diff --git a/Fluidization_Engineering/screenshots/relationshebdrep.png b/Fluidization_Engineering/screenshots/relationshebdrep.png
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index 173822d6..00000000
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diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/README.txt b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/README.txt
new file mode 100755
index 00000000..a043f776
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Dolar Khachariya
+Course: mtech
+College/Institute/Organization: Indian Institute of Engineering Bombay
+Department/Designation: Electrical engineering
+Book Title: Fluidization Engineering
+Author: K. Daizo And O. Levenspiel
+Publisher: Butterworth-Heinemann, Massachusetts
+Year of publication: 1991
+Isbn: 8131200353
+Edition: 2nd
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch10.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch10.ipynb
new file mode 100755
index 00000000..e44d1ebe
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch10.ipynb
@@ -0,0 +1,304 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:89c91579d721ed0f833b399593203d4eb19421ab1695bc830f1be612e46bd826"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Gas Dispersion and Gas Interchange in Bubbling Beds"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 253\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from numpy import *\n",
+      "%matplotlib inline\n",
+      "#Variable declaration\n",
+      "umf=[0.01,0.045];     #Velocity at minimum fluidization condition in m/s\n",
+      "ephsilonmf=[0.5,0.5]; #Void fraction at minimum fluidization condition\n",
+      "D=[2E-5,7E-5];        #Diffusion coefficient of gas in m**2/s\n",
+      "g=9.81;               #Acceleration due to gravity in m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "db=[5.,10.,15.,20.];\n",
+      "n=len(umf);\n",
+      "m=len(db)\n",
+      "Kbc = zeros((n,m))\n",
+      "Kce = zeros((n,m))\n",
+      "Kbe = zeros((n,m))\n",
+      "    \n",
+      "for i in range(n):\n",
+      "    for j in range(m):\n",
+      "            Kbc[i][j]=4.5*(umf[i]/db[j])+5.85*((D[i]**0.5*g**0.25)/db[j]**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(27)\n",
+      "            Kce[i][j]=6.77*((D[i]*ephsilonmf[i]*0.711*(g*db[j])**0.5)/db[j]**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(34)\n",
+      "            Kbe[i][j]=(Kbc[i][j]*Kce[i][j])/(Kbc[i][j]+Kce[i][j]);#Gas interchange coefficient between bubble and emulsion from Eqn.(14)\n",
+      "\n",
+      "#OUTPUT\n",
+      "i=0;\n",
+      "j=0;\n",
+      "k=0;\n",
+      "while k<m*n:\n",
+      "    print '\\t\\tKbc for fine particles and He',\n",
+      "    print '\\tKbc for coarse particles and ozone',\n",
+      "    print '\\tKbe for fine particles and He',\n",
+      "    print '\\tKbe for coarse particles and ozone'\n",
+      "    j = 0\n",
+      "    while j<m:\n",
+      "        print 'db=%fm'%(db[j]*10**-2);\n",
+      "        while i<n:\n",
+      "            print '\\t%f'%Kbc[i][j],\n",
+      "            print '\\t\\t\\t%f'%Kbe[i][j],\n",
+      "            i=i+1;   \n",
+      "            k=k+1;\n",
+      "            print '\\t\\t\\t',\n",
+      "        i=0;\n",
+      "        j=j+1;\n",
+      "import numpy\n",
+      "import matplotlib.pyplot as plt\n",
+      "a = numpy.matrix(Kbe)\n",
+      "b = numpy.matrix(Kbc)\n",
+      "db = array(db).T\n",
+      "Kbe=a.T\n",
+      "Kbc=b.T\n",
+      "\n",
+      "plt.plot(db,Kbc,db,Kbe)\n",
+      "#plt.plot(db,Kbe);\n",
+      "#Note : Python does not have plot2d function. so we can plot it \n",
+      "#plt.title('Plot of Kbc,Kbe vs db','db');\n",
+      "plt.show()\n",
+      "print 'Comparing the points with the plot of Kbc,Kbe vs db in Fig.(12), we can conlcude the following:'\n",
+      "print 'Kbc for fine particles and helium: line 2 in Fig.(12)'\n",
+      "print 'Kbc for coarser particles and ozone: line 3 in Fig.(12)'\n",
+      "print 'Kbe for fine particles and helium: line 4 in Fig.(12)'\n",
+      "print 'Kbe for coarser particles and ozone: line 5 in Fig.(12)'\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\tKbc for fine particles and He \tKbc for coarse particles and ozone \tKbe for fine particles and He \tKbe for coarse particles and ozone\n",
+        "db=0.050000m\n",
+        "\t0.015193 \t\t\t0.003335 \t\t\t\t0.052085 \t\t\t0.006930 \t\t\tdb=0.100000m\n",
+        "\t0.007104 \t\t\t0.001434 \t\t\t\t0.025121 \t\t\t0.002964 \t\t\tdb=0.150000m\n",
+        "\t0.004568 \t\t\t0.000875 \t\t\t\t0.016434 \t\t\t0.001803 \t\t\tdb=0.200000m\n",
+        "\t0.003345 \t\t\t0.000616 \t\t\t\t0.012173 \t\t\t0.001266 \t\t\t"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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dO2fNAg4cEE/b8IMfAL/+tXgiN4NB6rREZOu6XLMPDAzErl274OnpiYiICJw8\neRJubm7G1wsLC7F582ZkZGQgOzsbKSkpyMzMBACcOHECw4cPR1xcHP7v//7PfACu2VusqQnIz2+d\n6//uO3Gtf/FiYOFCYNQoqRMSUV+z6pp9dXU1ACA0NBSenp4IDw9HQUGByTIFBQVYsWIFlEolYmNj\nUVJSYnxtzpw5GMXmsbqhQ8VS/8MfxIu0/POfwIwZwJ494mkb5swBfvMb4N//Fk/uRkTUadkXFRXB\n39/f+FilUuHUqVMmyxQWFkKlUhkfu7u7o7y83MoxqTNeXsDPfw4cOiTO7b/6qriL54oVwPjxwLp1\nQFqaeDpnIrJPvT6WUxCEdl8lLD3157Zt24z3NRoNNBpNb2PZLYVCnNKJjAT++Efg0iVxqueDD4Bn\nngGCglr38HnkEV6ekWigyMvLQ15eXo/f3+mcfXV1NTQaDc6cOQMA2LBhAyIjIxEVFWVcJjExEXq9\nHps2bQIAeHt7m6zZV1RUIDo6mnP2NqCuDsjLa93DR69vLf4FC3hRdqKBxKpz9q6u4pnY8vPzUVFR\ngZycHKjVapNl1Go10tLSoNVqkZqaioCAgB7Epv4wbBgQFQW8+y5w5QqQkyPuy5+YCIwbJ57G4fe/\nB0pKONdPNNh0uTfO8ePHsX79euh0OiQkJCAhIQFJSUkAgPj4eADAyy+/jAMHDkCpVCI5OdlY+LGx\nsTh+/Di0Wi3GjBmDN998E88++6xpAK7Z24S7d4Fjx1r38HFwaD1527x54kBBRLaDB1VRrwmCuG9/\nS/GfPi3u199yDh9fX871E0mNZU9WV1PTelH2w4cBF5fWtX6NRtwoTET9i2VPfUoQgHPnWi/U8u9/\nA7Nnt27o9fKSOiGRfWDZU7+6c0fc0NtyUXZX19biDw3lRdmJ+grLniTTclH2lrn+8+fFaZ6WKR9P\nT6kTEg0eLHuyGVpt60XZjxwRz8/fUvxTpwJKJU/bTNRTdl/21xob8fGtW/BRKOCjUMDL2RkuvPyT\n5JqbxTN0thT/xYtAbS3g7g54eHR9Gz2aV/Eiasvuy768oQF/+vprXG5owOWGBlxtbISbkxN8FAr4\nfj8AtNy8FQoMZ4NIpqlJPJfPjRtd36qrxcL38BCv29vZwODuzoGBBj+7L/v7NQsCKpuacLmhAZe+\nHwBabuUNDRjp6GgyALQMCt4KBVwde33qILISnQ64dQu4fr3rgeHOHfE0z20HgI4GCHd3wMlJ6p+O\nyHIsewtqVebJAAANQklEQVQYBAHf3LtnMgBcbmjApfp6XG5owDAHh3YDQctNyYawWXo9UFUlFn9X\ng4NWK+5B1NG3hLaDxJgx4kVliGwBy95KBEHAdTMDQcs3BEeZzOzUkI9CATcnJ4vP/EnSaG4WC7/t\nANDRAHHrFjBiRNfbF1oGCO52Sn2JZd8PBEFAlU5ndmroUkMDDIJgOi3k4mK878GBYMAyGIDbt9sP\nAuYGh5s3xSONu7Px2cNDXJbIEix7G3D7+4HA3DeCBoOhw6mhB4cMgZwDwaAgCOK2g46mj+4fIIYO\n7f7AwFNRE8Cyt3nf6fUoNzMQXG5oQLVeD+8OBoLxQ4fCgQPBoCQI4t5GnW1baDs4yOXd2yvJw0Oc\nduI/m8GJZT+A1TY3GweC+6eHqnQ6THJ2NrudYKKzMxz5G20XBEE8HXVXeyS1DA4GQ/f2SvLwEDdU\n85/RwMGyH6Tqm5txpbHR7J5DN3Q6TBw61Ow3gknOznDiYap2q7a2e8cxXL8u7t46ZkzrnkcjRojX\nMWh7Gz68/XPmXnNx4dHRfY1lb4caDQZc7WBqqLKpCeM7GggUCjjzN5K+V19vuudRba14q6szf+vs\ntYYGwNnZ8kGiO69xIBGx7MnEPYMB//n+G8H9U0P/aWzE2CFDTPYW4mkmyBoMBnHw6MlA0dlrbQcS\naw4gA3EgYdlTt+kFAdfMTA21nGZi9PenmWh7ZDFPM0FS62wg6c0gUlsLNDaKF+Ox5gDS8ppCYd2B\nhGXf2Ah89514vDyPaumxtqeZuP9W3tAAVzOnmWi58TQTNFB1NJB0Z6Do6jVzA0lvBpBHH7X3sv/i\nC2DZMvHoFycnsfSVSvG/3b0/ahTAwuqQQRDwbZuji++fHlLI5WaPLOZpJsie9WQg6ey14mJ7L/sW\ngiD+jdy5Ixb/nTvt73f0WnW1OHln6QChVIr7rw2USb8+IAgCbnR0UFl9PRy+P82Eu5MThjk4wMXB\nAcPkcpP7Lg4O4uPvn297v+1zCrmcB6GR3eI0jjUYDOJVtrsaFMwNHnfvAg880PWgYO7+ID8CRhAE\naPV6XKqvR5VOh3qDAXXNza3/bW5G3X3365ubTZZpe7/RYIBzy+DQZpAwN2CYHTi68T4eyEa2imUv\nNb1e/GbQk28UTU3AyJGWDRAt9xWKQT1QmGMQBDS0DAgWDBKWDC5OMlmPBglzA425bzE8BoJ6imU/\nkN27ZzogdPcbxe3b4rSVpQNEy40bss0SBAGNBkOvv4F09j4Z0KNBoruDy1CZjCfeG6RY9vaqocHy\nKaeWx0OGdH+bRNv7I0dyQ3YvCIIAnSBY5RtIR+/TC0KPBgkXuRxD5HI4yWTGm6NMBqf7nmv72NHM\nc20fO8pknBazIpY9WaZlQ7alU0537oi7uA4b1rPtEy4u4t5SnMboU7rvv5n05BuIzmCA7vsBSScI\n0AtCu+faPtabee7+xzKgywHB3EDS8tjk9R4u09WgZcnA5ijhNyeWPfWf+zdkW/KNoqFBPBmLXC6W\nvpOT+A2j5X5nt+4u1xef2dFyPMisW5q7GBD0HQwSLY9NXjezTFcDkjUGrbYZmgXB7ADV1aBljYEt\nYcIEi7qT38Gp5+RycSpn5Ehg0iTL3y8I4qWi7t0Ti787t+4u23a5piZxh+XefmZnywK2NSj19DMd\nHPp0Q7/D91M5g+WcTIYOBqi+GLQamptR0+axpbpcs8/Pz0d8fDz0ej0SEhKwYcOGdsts3boVBw4c\nwKhRo5CSkgJ/f/9uv3egrNnn5eVBo9FIHaNLzGld3c7Z3Gz9AcSCZfMqK6FRKnv/mQaDafHL5eKt\nq/vdWDavthaaUaMsfp81/mxLPiPv0iVoVCpJ/uxu35fJIJPLrbtmv3HjRiQlJcHT0xMRERGIjY2F\nm5ub8fXCwkKcOHECp0+fRnZ2NrZs2YLMzMxuvXcgGXTlJLFBl9PBQbw5O/d5JnPytm2DZtu23n+Q\nwWBa/AaDOJDdf9/cc13cz/vLX6BZs6ZXn9GjHOZ+ls5yfvklNN98Y50/u68+owcryJ2WfXV1NQAg\nNDQUABAeHo6CggJERUUZlykoKMCKFSugVCoRGxuL1157rdvvJSIbI5eLu+L2xe64OTnA3LnW/1xr\n27ZNvNkyQbB454ZOly4qKjJOyQCASqXCqVOnTJYpLCyESqUyPnZ3d0d5eXm33ktERD3Qg+0qvd5A\nKwhCu3kjS3dFGigHfbzxxhtSR+gW5rQu5rSegZARGDg5LdFp2c+YMQO/+MUvjI+Li4sRGRlpsoxa\nrcaFCxcQEREBALh16xa8vLygVCq7fC+AAbFxlohooOt0GsfV1RWAuFdNRUUFcnJyoFarTZZRq9VI\nS0uDVqtFamoqAgICAAAjR47s8r1ERNQ/upzG2blzJ+Lj46HT6ZCQkAA3NzckJSUBAOLj4xESEoLZ\ns2cjODgYSqUSycnJnb6XiIgkIEhIr9cL06dPF5YsWSJljE7V1tYKcXFxgq+vrxAQECB88cUXUkcy\n6/333xcef/xx4bHHHhM2btwodRyjZ599VhgzZowwZcoU43M1NTXC0qVLhQkTJggxMTHC3bt3JUwo\nMpdzy5Ytgr+/vxAYGChs3LhRqK+vlzChyFzOFm+//bYgk8kErVYrQbJWHWXcvXu34O/vL6hUKuGX\nv/ylROlamctZXFwsREVFCdOmTROWLFkiXLhwQcKEomvXrgkajUZQqVTC3LlzhZSUFEEQLP89kvQw\ntl27dkGlUtn0BtrXX38dEydOxLlz53Du3DnjNJUtuX37NrZv346cnBwUFRXh4sWLyM7OljoWAODZ\nZ5/FkSNHTJ577733MHHiRFy6dAnjx4/Hn//8Z4nStTKXMzw8HMXFxTh9+jTq6uqQmpoqUbpW5nIC\nwFdffYWcnBx4enpKkMqUuYznz5/H+++/j4yMDBQXF2PLli0SpWtlLuebb76JuLg4/Pvf/8aTTz6J\nN998U6J0rZycnPDOO++guLgYn3zyCV577TXcvXvX4t8jycq+srIShw8fxnPPPWfTG2lzc3Pxyiuv\nwNnZGY6OjsbtGLZEoVBAEARUV1ejoaEB9fX1GDVqlNSxAABz5sxpl6WwsBDr1q3D0KFDsXbtWhQU\nFEiUrpW5nAsXLoRcLodcLkdERASOHz8uUbpW5nICwObNm/G73/1OgkTtmcuYlZWFdevWwdfXF4C4\ni7bUzOV0dXWFVquFwWCAVqu1id+jsWPHYvr06QAANzc3PPLIIygqKrL490iyst+0aRN27NgBuQ2f\nI6OyshKNjY14/vnnoVar8dZbb6GxsVHqWO0oFAq89957ePjhhzF27FjMmjULISEhUsfqUNtjMPz9\n/VFYWChxoq598MEHiI6OljqGWenp6Rg/fjymTp0qdZQOffbZZzh//jyCg4Px3HPP4cKFC1JHMmvH\njh3YtWsXRo0ahXfffRdvvfWW1JFMXL58GcXFxQgJCbH490iSps3MzMSYMWMQGBho02v1jY2NuHjx\nIpYvX468vDwUFxfj448/ljpWO7du3cLzzz+PCxcuoKKiAl988QUOHTokdawO2fL/c3PefPNNjBgx\nAitXrpQ6Sjv19fXYvn27yX7htvj329jYiNu3b+PEiROIiYnBCy+8IHUks9auXYsNGzZAq9Vi/fr1\nWLdundSRjO7evYtVq1bhnXfewfDhwy3+/yxJ2f/zn/9ERkYGJk2ahNjYWBw7dgxxcXFSROmUj48P\n/Pz8EB0dDYVCgdjYWGRlZUkdq53CwkLMnDkTPj4+GD16NFauXIn8/HypY3VoxowZKCkpAQCUlJRg\nxowZEifq2EcffYTs7GyTvcxsSXl5OSoqKjBt2jRMmjQJlZWVCAoKws2bN6WOZmLmzJlYtWoVFAoF\noqOjUVpaapPfkk+ePIm1a9fC0dER69ats5nfI51Oh+XLl+Ppp59GTEwMAMt/jyQp++3bt+Orr77C\n1atXsX//fsyfPx979+6VIkqXfH19UVBQAIPBgEOHDiEsLEzqSO3MmTMHp0+fxu3bt9HU1ISsrCyE\nh4dLHatDarUau3fvRkNDA3bv3o2ZM2dKHcmsI0eOYMeOHcjIyICzRCc468qjjz6KGzdu4OrVq7h6\n9SrGjx+PL7/8EmPGjJE6monHH38cWVlZEAQBBQUF8Pb2tsm/03nz5iEjIwOAOD22cOFCiROJ39TW\nrVuHKVOm4MUXXzQ+b/HvUV/uMtQdeXl5QnR0tNQxOlRWViao1Wph2rRpwksvvSTU1tZKHcmsDz/8\nUAgNDRWCg4OF1157TWhubpY6kiAIgrB69Wph3LhxwpAhQ4Tx48cLu3fvtsldL1tyOjk5CePHjxf+\n8pe/CD4+PsLEiROF6dOnC9OnTxeef/55qWOa/ftsa9KkSZLvemkuo16vF+Lj4wV/f39h2bJlQmFh\noaQZ2+Zs+X++e/du4fz588Lq1auFqVOnCk8++aRQUlIidUzhxIkTgkwmE6ZNm2b8t5iVlWXx75Hk\nV6oiIqK+Z7u7whARkdWw7ImI7ADLnojIDrDsiYjsAMueiMgOsOyJiOzA/wfai5Sp3wLX9AAAAABJ\nRU5ErkJggg==\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Comparing the points with the plot of Kbc,Kbe vs db in Fig.(12), we can conlcude the following:\n",
+        "Kbc for fine particles and helium: line 2 in Fig.(12)\n",
+        "Kbc for coarser particles and ozone: line 3 in Fig.(12)\n",
+        "Kbe for fine particles and helium: line 4 in Fig.(12)\n",
+        "Kbe for coarser particles and ozone: line 5 in Fig.(12)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 254\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "D=0.69;         #Diffusion coefficient of gas in cm**2/s\n",
+      "umf=1.0;        #Velocity at minimum fluidization condition in cm/s\n",
+      "ephsilonmf=0.5; #Void fraction at minimum fluidization condition\n",
+      "db=[5,15];      #Equilibrium bubble size in cm\n",
+      "g=980;          #Acceleration due to gravity in cm/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(db);\n",
+      "i=0;\n",
+      "Kbc = [0.,0.]\n",
+      "Kce = [0.,0.]\n",
+      "Kbe = [0.,0.]\n",
+      "e = [0.,0.]\n",
+      "while i<n:\n",
+      "    Kbc[i]=4.5*(umf/db[i])+5.85*((D**0.5*g**0.25)/db[i]**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(27)\n",
+      "    Kce[i]=6.77*((D*ephsilonmf*0.711*(g*db[i])**0.5)/db[i]**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(34)\n",
+      "    Kbe[i]=(Kbc[i]*Kce[i])/(Kbc[i]+Kce[i]);#Gas interchange coefficient between bubble and emulsion from Eqn.(14)\n",
+      "    e[i]=(Kce[i]-Kbe[i])/Kbe[i];#Error when minor resistance is ignored\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'db(cm)',\n",
+      "print '\\t\\tCalculated Kbc',\n",
+      "print '\\tCalculated Kce',\n",
+      "print '\\t\\tKbe from Eqn.(14)',\n",
+      "print '\\tErron when minor resistance is ignored (in percentage)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%db[i],\n",
+      "    print '\\t%f'%Kbc[i],\n",
+      "    print '\\t%f'%Kce[i],\n",
+      "    print '\\t\\t%f'%Kbe[i],\n",
+      "    print '\\t\\t%f'%(e[i]*100);\n",
+      "    i=i+1;   \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "db(cm) \t\tCalculated Kbc \tCalculated Kce \t\tKbe from Eqn.(14) \tErron when minor resistance is ignored (in percentage)\n",
+        "5.000000 \t6.337721 \t2.509152 \t\t1.797506 \t\t39.590766\n",
+        "15.000000 \t2.112574 \t0.635514 \t\t0.488547 \t\t30.082477\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 255\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "Kbe=[0.028,0.05];  #Reported range for gas interchange coefficient between bubble and emulsion\n",
+      "uo=0.30;           #Superficial gas velocity in m/s\n",
+      "db=0.13;           #Equilibrium bubble size in m\n",
+      "m=7;\n",
+      "ephsilonmf=0.5;    #Void fraction at minimum fluidization condition\n",
+      "umf=0.0018;        #Velocity at minimum fluidization condition in m/s\n",
+      "D=[9E-6,22E-6];    #Diffusion coefficient of gas in m**2/s\n",
+      "g=9.81;            #Acceleration due to gravity in m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(Kbe);\n",
+      "i=0;\n",
+      "Kbem = [0,0]\n",
+      "Kbc = [0,0]\n",
+      "Kce = [0,0]\n",
+      "#Kbe = [0,0]\n",
+      "c = [0,0]\n",
+      "\n",
+      "while i<n:\n",
+      "    Kbem[i]=(6.0/db)*Kbe[i];#Gas interchange coefficient between bubble and emulsion from Eqn.(19)\n",
+      "    Kbc[i]=4.5*(umf/db)+5.85*((D[i]**0.5*g**0.25)/db**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(27)\n",
+      "    Kce[i]=6.77*((D[i]*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(34)\n",
+      "    Kbe[i]=(Kbc[i]*Kce[i])/(Kbc[i]+Kce[i]);#Gas interchange coefficient between bubble and emulsion from Eqn.(14)\n",
+      "    c[i]=(Kbem[i]/Kbe[i]);\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Kbe from Eqn.(19)',\n",
+      "print '\\tKbc from Eqn.(27)',\n",
+      "print '\\tKce from Eqn.(34)',\n",
+      "print '\\tKbe from Eqn.(14)',\n",
+      "print '\\tComparison of Kbe from Eqn.(19) and that from Eqn.(14)'\n",
+      "i=0\n",
+      "while i<n:\n",
+      "    print '%f'%Kbem[i],\n",
+      "    print '\\t\\t%f'%Kbc[i],\n",
+      "    print '\\t\\t%f'%Kce[i],\n",
+      "    print '\\t\\t%f'%Kbe[i],\n",
+      "    print '\\t\\t%f'%c[i]\n",
+      "    i=i+1;   \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Kbe from Eqn.(19) \tKbc from Eqn.(27) \tKce from Eqn.(34) \tKbe from Eqn.(14) \tComparison of Kbe from Eqn.(19) and that from Eqn.(14)\n",
+        "1.292308 \t\t0.460200 \t\t0.274548 \t\t0.171959 \t\t7.515188\n",
+        "2.307692 \t\t0.684401 \t\t0.429248 \t\t0.263797 \t\t8.747978\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch11.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch11.ipynb
new file mode 100755
index 00000000..48b679d6
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch11.ipynb
@@ -0,0 +1,356 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:69d28aca1d8ec525f15842df275d6e6c6108139c968df651029692a500222ce7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter  11  : Particle to Gas Mass and Heat Transfer"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 265\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "db=0.37;         #Equilibrium bubble size in cm\n",
+      "dp=0.028;        #Particle size in cm\n",
+      "rhos=1.06;       #Density of solids in g/cc\n",
+      "ephsilonmf=0.5;  #Void fraction at minimum fluidization condition\n",
+      "phis=0.4;        #Sphericity of solids\n",
+      "gammab=0.005;    #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "rhog=1.18E-3;    #Density of air in g/cc\n",
+      "myu=1.8E-4;      #Viscosity of gas in g/cm s\n",
+      "D=0.065;         #Diffusion coefficient of gas in cm**2/s\n",
+      "Sc=2.35;         #Schmidt number\n",
+      "etad=1;          #Adsorption efficiency factor\n",
+      "y=1;\n",
+      "umf=1.21;        #Velocity at minimum fluidization condition in cm/s\n",
+      "ut=69;           #Terminal velocity in cm/s\n",
+      "g=980;           #Acceleration due to gravity in square cm/s**2\n",
+      "uo=[10,20,30,40,50];#Superficial gas velocity in cm/s\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(uo);\n",
+      "i=0;\n",
+      "Rept=(dp*ut*rhog)/myu;\n",
+      "Shstar=2+(0.6*(Rept**0.5)*(Sc**(1/3)));#Sherwood no. from Eqn.(1)\n",
+      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "ubr=0.711*(g*db)**0.5;#Rise velocity of the bubble\n",
+      "x = [0,0,0,0,0]\n",
+      "Shbed = [0,0,0,0,0]\n",
+      "Rep = [0,0,0,0,0]\n",
+      "while i<n:\n",
+      "    x[i]=(uo[i]-umf)/(ubr*(1-ephsilonmf));#The term delta/(1-epshilonf) after simplification\n",
+      "    Shbed[i]=x[i]*((gammab*Shstar*etad)+((phis*dp**2*y)/(6*D))*Kbc);#Sherwood no. from Eqn.(11)\n",
+      "    Rep[i]=(dp*uo[i]*rhog)/myu;#Reynolds of the particle\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The desired result is the relationship between Shbed and Rep  The points gives a straight line of the form y=mx+c'\n",
+      "print 'Rep',\n",
+      "print '\\t\\tShbed'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%Rep[i],\n",
+      "    print '\\t%f'%Shbed[i]\n",
+      "    i=i+1;\n",
+      "import matplotlib.pyplot as plt\n",
+      "plt.plot(Rep,Shbed);\n",
+      "plt.xlabel(\"Rep\");\n",
+      "plt.ylabel(\"Shbed\");\n",
+      "\n",
+      "plt.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "The desired result is the relationship between Shbed and Rep  The points gives a straight line of the form y=mx+c\n",
+        "Rep \t\tShbed\n",
+        "1.835556 \t0.065762\n",
+        "3.671111 \t0.140576\n",
+        "5.506667 \t0.215391\n",
+        "7.342222 \t0.290206\n",
+        "9.177778 \t0.365020\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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RlJR0Rnn9weLFMHEijB5t3b+hXTu7E4mIPygsLKSwsPCMx3u9QDjOYjLck7HH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\niDRIBUJERBqkAiEiIg36/9fGcNvjU2I2AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2b26690>"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "#INPUT\n",
+      "umf=0.12            #Velocity at minimum fluidization condition in cm/s\n",
+      "uo=40.;              #Superficial gas velocity in cm/s\n",
+      "ub=120;             #Velocity of the bubble in cm/s\n",
+      "D=0.7;              #Diffusion coefficient of gas in cm**2/s\n",
+      "abkbe1=1.;           #Bubble-emuslion interchange coefficient for non absorbing particles(m=0)\n",
+      "abkbe2=18.;          #Bubble-emuslion interchange coefficient for highly absorbing particles(m=infinity)\n",
+      "g=980.;              #Acceleration due to gravity in square cm/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "#For non absorbing particles m=0,etad=0\n",
+      "Kbc=(ub/uo)*(abkbe1);\n",
+      "dbguess=2;#Guess value of db\n",
+      "def solver_func(db):        #Function defined for solving the system\n",
+      "    return abkbe1-(uo/ub)*(4.5*(umf/db)+5.85*(D**0.5*g**0.25)/(db**(5/4.)));#Eqn.(10.27)\n",
+      "\n",
+      "d=fsolve(solver_func,dbguess)\n",
+      "#For highly absorbing particles m=infinity, etad=1\n",
+      "M=abkbe2-(uo/ub)*Kbc;\n",
+      "#For intermediate condition\n",
+      "alpha=100.;\n",
+      "m=10.;\n",
+      "etad=1./(1+(alpha/m));#Fitted adsorption efficiency factor from Eqn.(23)\n",
+      "abkbe3=M*etad+(uo/ub)*Kbc;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'For non absorbing particles:\\tDiameter of bubble=%fcm\\tBubble-cloud interchange coefficient=%fs**-1'%(d,Kbc);\n",
+      "print 'For highly absorbing partilces:\\tM=%f'%(M);\n",
+      "print 'For intermediate condition:\\tFitted adsorption efficiency factor:%f\\tBubble-emuslion interchange coefficient:%fs**-1'%(etad,abkbe3);\n",
+      "\n",
+      "#====================================END OF PROGRAM ======================================================"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For non absorbing particles:\tDiameter of bubble=6.010032cm\tBubble-cloud interchange coefficient=3.000000s**-1\n",
+        "For highly absorbing partilces:\tM=17.000000\n",
+        "For intermediate condition:\tFitted adsorption efficiency factor:0.090909\tBubble-emuslion interchange coefficient:2.545455s**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 273\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "rhos=1.3;                #Density of solids in g/cc\n",
+      "phis=0.806;              #Sphericity of solids\n",
+      "gammab=0.001;            #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "rhog=1.18E-3;            #Density of air in g/cc\n",
+      "Pr=0.69;                 #Prandtl number\n",
+      "myu=1.8E-4;              #Viscosity of gas in g/cm s\n",
+      "Cpg=1.00;                #Specific heat capacity of gas in J/g K\n",
+      "ephsilonmf=0.45;         #Void fraction at minimum fluidization condition\n",
+      "kg=2.61E-4;              #Thermal concuctivity of gas in W/cm k\n",
+      "dp=0.036;                #Particle size in cm\n",
+      "umf=6.5;                 #Velocity at minimum fluidization condition in cm/s\n",
+      "ut=150.;                 #Terminal velocity in cm/s\n",
+      "db=0.4;                  #Equilibrium bubble size in cm\n",
+      "etah=1;                  #Efficiency of heat transfer\n",
+      "uo=[10.,20.,30.,40.,50.];#Superficial gas velocity in cm/s\n",
+      "g=980.;                  #Acceleration due to gravity in square cm/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "Nustar=2+(((dp*ut*rhog)/myu)**0.5*Pr**(1./3));#Nusselt no. from Eqn.(25)\n",
+      "Hbc=4.5*(umf*rhog*Cpg/db)+5.85*((kg*rhog*Cpg)**0.5*g**0.25/db**(5./4));#Total heat interchange across the bubble-cloud boundary from Eqn.(32)\n",
+      "ubr=0.711*(g*db)**0.5;#Rise velocity of the bubble from Eqn.(6.7)\n",
+      "n=len(uo);\n",
+      "i=0;\n",
+      "x = [0,0,0,0,0]\n",
+      "Nubed = [0,0,0,0,0]\n",
+      "Rep = [0,0,0,0,0]\n",
+      "\n",
+      "while i<n:\n",
+      "    x[i]=(uo[i]-umf)/(ubr*(1-ephsilonmf));#The term delta/(1-epshilonf) after simplification\n",
+      "    Nubed[i]=x[i]*(gammab*Nustar*etah+(phis*dp**2/(6*kg))*Hbc);#Nusselt no. from Eqn.(36)\n",
+      "    Rep[i]=(dp*uo[i]*rhog)/myu;#Reynolds of the particle\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The desired result is the relationship between Nubed and Rep which is in the form of a straight line y=mx+c'\n",
+      "print 'Rep',\n",
+      "print '\\t\\tNubed'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%Rep[i],\n",
+      "    print '\\t%f'%Nubed[i]\n",
+      "    i=i+1;\n",
+      "import matplotlib.pyplot as plt\n",
+      "plt.plot(Rep,Nubed);\n",
+      "plt.xlabel(\"Rep\");\n",
+      "plt.ylabel(\"Nubed\");\n",
+      "plt.show()\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The desired result is the relationship between Nubed and Rep which is in the form of a straight line y=mx+c\n",
+        "Rep \t\tNubed\n",
+        "2.360000 \t0.046518\n",
+        "4.720000 \t0.179427\n",
+        "7.080000 \t0.312335\n",
+        "9.440000 \t0.445244\n",
+        "11.800000 \t0.578152\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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TIhARCXBKBCIiAc4riSAvL4+oqCgiIyNJT0+vc8z9999PZGQkPXv2pKioyBth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\nEYiIBDglAhGRAKdEICIS4JQIREQC3P8HJOZBYJAptYQAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x269c250>"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 274\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "rhog=1.2;       #Density of air in kg/m**3\n",
+      "myu=1.8E-5;     #Viscosity of gas in kg/m s\n",
+      "kg=2.6E-2;      #Thermal concuctivity of gas in W/m k\n",
+      "dp=1E-4;        #Particle size in m\n",
+      "rhos=8920;      #Density of solids in kg/m**3\n",
+      "Cps=390;        #Specific heat capacity of the solid in J/kg K\n",
+      "ephsilonf=0.5;  #Void fraction of the fluidized bed\n",
+      "umf=0.1;        #Velocity at minimum fluidization condition in m/s\n",
+      "uo=0.1;         #Superficial gas velocity in m/s\n",
+      "pi=3.14\n",
+      "\n",
+      "#CALCULATION\n",
+      "to=0;                 #Initial temperature of the bed\n",
+      "T=100;                #Temperature of the bed\n",
+      "t=0.99*T;             #Particle temperature i.e. when it approaches 1% of the bed temperature\n",
+      "mp=(pi/6)*dp**3*rhos; #Mass of the particle\n",
+      "A=pi*dp**2;           #Surface area of the particle\n",
+      "Rep=(dp*uo*rhog)/myu; #Reynold's no. of the particle\n",
+      "Nubed=0.0178;         #Nusselt no. from Fig.(6)\n",
+      "hbed1=(Nubed*kg)/dp;  #Heat transfer coefficient of the bed\n",
+      "t1=(mp*Cps/(hbed1*A))*math.log((T-to)/(T-t));#Time needed for the particle approach 1 percentage of the bed temperature in case(a)\n",
+      "hbed2=140*hbed1;#Since from Fig.(6) Nup is 140 times Nubed\n",
+      "t2=(mp*Cps/(hbed2*A))*math.log((T-to)/(T-t));#Time needed for the particle approach 1 percentage of the bed temperature in case(b)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Case(a):Using the whole bed coefficient from Fig.(6)'\n",
+      "print '\\tTime needed for the particle approach 1 percentage of the bed temperature is %.0fs'%t1\n",
+      "print 'Case(b):Uisng the single-particle coefficient of Eqn.(25),also shown in Fig.(6)'\n",
+      "print '\\tTime needed for the particle approach 1 percentage of the bed temperature is %.2fs'%t2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Case(a):Using the whole bed coefficient from Fig.(6)\n",
+        "\tTime needed for the particle approach 1 percentage of the bed temperature is 58s\n",
+        "Case(b):Uisng the single-particle coefficient of Eqn.(25),also shown in Fig.(6)\n",
+        "\tTime needed for the particle approach 1 percentage of the bed temperature is 0.41s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch12.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch12.ipynb
new file mode 100755
index 00000000..1fc80a09
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch12.ipynb
@@ -0,0 +1,396 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3cabb972e9b40cc3c2621280c95233b4046eb8d671e52d74d499a7e149a3d9aa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Conversion of Gas in Catalytic Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 293\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Kr=10.;          #rate constant in m**3 gas/m**3 cat s\n",
+      "D=2E-5;          #Diffusion coefficient of gas in m**2/s\n",
+      "dpbar=68.;       #Average partilce size in micrometers\n",
+      "ephsilonm=0.5;   #Void fraction of fixed bed\n",
+      "gammab=0.005;    #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "ephsilonmf=0.55; #Void fraction at minimum fluidization condition\n",
+      "umf=0.006;       #Velocity at minimum fluidization condition in m/s\n",
+      "db=0.04;         #Equilibrium bubble size in m\n",
+      "Lm=0.7;          #Length of the bed in m\n",
+      "uo=0.1;          #Superficial gas velocity in m/s\n",
+      "dbed=0.26;       #Diameter of the bed in m\n",
+      "g=9.81;          #Acceleration due to gravity in square m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
+      "ub=uo-umf+ubr;#Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
+      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5./4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "Kce=6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
+      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
+      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
+      "gammac=(1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
+      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
+      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
+      "Lf=(1-ephsilonm)*Lm/(1-ephsilonf);#Length of fixed bed from Eqn.(6.19)\n",
+      "Krtou=Kr*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
+      "Kf=gammab*Kr+1/((1/Kbc)+(1/(gammac*Kr+1/((1/Kce)+(1/(gammae*Kr))))));#Raction rate for fluidized bed from Eqn.(14)\n",
+      "XA=math.exp(-1*Kf*Lf/ub);#Conversion from Eqn.(16)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The dimnesionless reaction rate group: %f'%Krtou\n",
+      "print 'The reaction rate for fluidized bed: %fs**-1'%Kf\n",
+      "print 'Conversion: %f'%XA\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dimnesionless reaction rate group: 35.000000\n",
+        "The reaction rate for fluidized bed: 1.979872s**-1\n",
+        "Conversion: 0.030056\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 298\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "umf=0.005;        #Velocity at minimum fluidization condition in m/s\n",
+      "ephsilonm=0.52;   #Void fraction of fixed bed\n",
+      "ephsilonmf=0.57;  #Void fraction at minimum fluidization condition\n",
+      "DA=8.1E-6;        #Diffusion coefficient of gas in m**2/s\n",
+      "DR=8.4E-6;        #Diffusion coefficient of gas in m**2/s\n",
+      "Lm=5;             #Length of the bed in m\n",
+      "dte=1;            #Diameter of tube in m\n",
+      "Kr1=1.5;          #rate constant in m**3 gas/m**3 cat s\n",
+      "Kr3=0.01;         #rate constant in m**3 gas/m**3 cat s\n",
+      "gammab=0.005;     #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "uo=0.45;          #Superficial gas velocity in m/s\n",
+      "db=0.05;          #Equilibrium bubble size in m from Fig.(6.8)\n",
+      "ub=1.5;           #Velocity of bubbles in bubbling bed in m/s from Fig.(6.11(a))\n",
+      "g=9.81;           #Acceleration due to gravity in square m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
+      "KbcA=4.5*(umf/db)+5.85*((DA**0.5*g**0.25)/db**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "KceA=6.77*((DA*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
+      "KbcR=4.5*(umf/db)+5.85*((DR**0.5*g**0.25)/db**(5./4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "KceR=6.77*((DR*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
+      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
+      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
+      "gammac=(1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
+      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
+      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
+      "Lf=(1-ephsilonm)*Lm/(1-ephsilonf);#Length of fixed bed from Eqn.(6.19)\n",
+      "Krtou=Kr1*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
+      "Kr12=Kr1;#Since the reactions are a special case of Denbigh scheme\n",
+      "Kr34=Kr3;\n",
+      "Kf1=(gammab*Kr12+1/((1/KbcA)+(1/(gammac*Kr12+1/((1/KceA)+(1/(gammae*Kr12)))))))*(delta/(1-ephsilonf));#Rate of reaction 1 for fluidized bed from Eqn.(14)\n",
+      "Kf3=(gammab*Kr34+1/((1/KbcR)+(1/(gammac*Kr34+1/((1/KceR)+(1/(gammae*Kr34)))))))*(delta/(1-ephsilonf));#Rate of reaction 2 for fluidized bed from Eqn.(14)\n",
+      "Kf12=Kf1;\n",
+      "Kf34=Kf3;\n",
+      "KfA=((KbcR*KceA/gammac**2+(Kr12+KceA/gammac+KceA/gammae)* \\\n",
+      "    (Kr34+KceR/gammac+KceR/gammae))*delta*KbcA*Kr12*Kr34/ \\\n",
+      "    (1-ephsilonf))/(((Kr12+KbcA/gammac)*(Kr12+KceA/gammae)+Kr12*KceA/gammac) \\\n",
+      "        *((Kr34+KbcR/gammac)*(Kr34+KceR/gammae)+Kr34*KceR/gammac));\n",
+      "            #Rate of raection with respect to A from Eqn.(35)\n",
+      "KfAR=Kr1/Kr12*Kf12-KfA;#Rate of reaction from Eqn.(34)\n",
+      "tou=Lf*(1-ephsilonf)/uo;#Residence time from Eqn.(5)\n",
+      "XA=1-math.exp(-Kf1*tou);#Conversion of A from Eqn.(26)\n",
+      "XR=1-((KfAR/(Kf12-Kf34))*(math.exp(-Kf34*tou)-math.exp(-Kf12*tou)));#Conversion of R from Eqn.(27)\n",
+      "SR=(1-XR)/XA;#Selectivity of R\n",
+      "\n",
+      "#OUTPUT\n",
+      "\n",
+      "print 'Rate of reaction 1 for fluidized bed:%.4f'%Kf1\n",
+      "print 'Rate of reaction 2 for fluidized bed:%.4f'%Kf3\n",
+      "print 'Rate of reaction 1 with respect to A:%.4f'%KfA\n",
+      "print 'The Conversion of Napthalene:%.0f percentage'%(XA*100);\n",
+      "print 'The selectivity of Phthalic anhydride:%.0f percentage'%(SR*100);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate of reaction 1 for fluidized bed:0.6007\n",
+        "Rate of reaction 2 for fluidized bed:0.0099\n",
+        "Rate of reaction 1 with respect to A:0.0058\n",
+        "The Conversion of Napthalene:96 percentage\n",
+        "The selectivity of Phthalic anhydride:95 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 302\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Kr=3.;           #rate constant in m**3 gas/m**3 cat s\n",
+      "db=0.12;         #Equilibrium bubble size in m\n",
+      "D=9E-5;          #Diffusion coefficient of gas in m**2/s\n",
+      "dpbar=68;        #Average partilce size in micrometers\n",
+      "ephsilonm=0.42;  #Void fraction of fixed bed\n",
+      "uo=0.4;          #Superficial gas velocity in m/s\n",
+      "Lm=0.8;          #Length of the bed in m\n",
+      "ephsilonmf=0.45; #Void fraction at minimum fluidization condition\n",
+      "umf=0.21;        #Velocity at minimum fluidization condition in m/s\n",
+      "gammab=0;        #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "g=9.81;          #Acceleration due to gravity in square m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "ubr=0.711*(g*db)**0.5;  #Rise velocity of bubble from Eqn.(6.7)\n",
+      "ub=uo-umf+ubr;          #Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
+      "ubstar=ub+3*umf;        #Rise velocity of the bubble gas from Eqn.(45)\n",
+      "delta=(uo-umf)/(ub+umf);#Fraction of bed in bubbles from Eqn.(6.46)\n",
+      "Kbe=4.5*(umf/db);       #Interchange coefficient between bubble and emulsion from Eqn.(47)\n",
+      "Lf=Lm*(1-ephsilonm)/((1-delta)*(1-ephsilonmf));#Length of fixed bed\n",
+      "phi=((Kr/Kbe)**2*((1-ephsilonmf)-gammab*(umf/ubstar))**2+ \\\n",
+      "        ((delta/(1-delta))+umf/ubstar)**2+2*(Kr/Kbe)*((1-ephsilonmf) \\\n",
+      "            -gammab*(umf/ubstar))*((delta/(1-delta))-umf/ubstar))**0.5;\n",
+      "            #From Eqn.(52)\n",
+      "            \n",
+      "q1=0.5*Kr/umf*((1-ephsilonmf)+gammab*(umf/ubstar))+0.5*Kbe/umf* \\\n",
+      "        (((delta/(1-delta))+umf/ubstar)-phi);#From Eqn.(50)\n",
+      "q2=0.5*Kr/umf*((1-ephsilonmf)+gammab*(umf/ubstar))+0.5*Kbe/umf* \\\n",
+      "        (((delta/(1-delta))+umf/ubstar)+phi);#From Eqn.(50)\n",
+      "        \n",
+      "si1=0.5-0.5*((1-delta)/delta)*(umf/ubstar-Kr/Kbe*((1-ephsilonmf)- \\\n",
+      "        gammab*(umf/ubstar))-phi);#From Eqn.(51)\n",
+      "si2=0.5-0.5*((1-delta)/delta)*(umf/ubstar-Kr/Kbe*((1-ephsilonmf)- \\\n",
+      "        gammab*(umf/ubstar))+phi);#From Eqn.(51)\n",
+      "XA=1-(delta/(1-delta))*(1/(uo*phi))*((1-si2)*(si1*delta*ubstar+ \\\n",
+      "        (1-delta)*umf)*math.exp(-q1*Lf)+(si1-1)* \\\n",
+      "        (si2*delta*ubstar+(1-delta)*umf)*math.exp(-q2*Lf));\n",
+      "        #Conversion from Eqn.(49)\n",
+      "        \n",
+      "Krtou=Kr*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'COmparing the values of 1-XA = %f and Krtou = %f with Fig.(6), \\\n",
+      "we can conlcude that this operating condition is shown as point \\\n",
+      "A in Fig.(3)'%(1-XA,Krtou);\n",
+      "print 'Line 2 gives the locus of conversions for different values of the \\\n",
+      "reaction rate group for this fluidized contacting'\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "COmparing the values of 1-XA = 0.113843 and Krtou = 3.480000 with Fig.(6), we can conlcude that this operating condition is shown as point A in Fig.(3)\n",
+        "Line 2 gives the locus of conversions for different values of the reaction rate group for this fluidized contacting\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 305\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "uo=0.25;        #Superficial gas velocity in m/s\n",
+      "db=0.025;       #Equilibrium bubble size in m\n",
+      "Kr=1.5;         #rate constant in m**3 gas/m**3 cat s\n",
+      "umf=0.21;       #Velocity at minimum fluidization condition in m/s\n",
+      "Lm=0.8;         #Length of the bed in m\n",
+      "ephsilonm=0.42; #Void fraction of fixed bed\n",
+      "g=9.81;         #Acceleration due to gravity in square m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
+      "ub=uo-umf+ubr;#Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
+      "delta=(uo-umf)/(ub+2*umf);#Fraction of bed in bubbles from Eqn.(55) since ub/umf<<1 \n",
+      "XA=1-math.exp(-Kr*Lm*((1-ephsilonm)/uo)*(umf/uo)*(1-delta));#Conversion from Eqn.(57)\n",
+      "Krtou=Kr*Lm*(1-ephsilonm)/uo;#Dimensionless reaction rate group from Eqn.(5)\n",
+      "\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Comparing the values of 1-XA = %f and Krtou = %f with Fig.(6), \\\n",
+      "we can conlcude that this operating condition is shown \\\n",
+      "as point B in Fig.(3)'%(1-XA,Krtou);\n",
+      "print 'Line 3 gives the locus of conversions for different values \\\n",
+      "of the reaction rate group for this fluidized contacting'\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Comparing the values of 1-XA = 0.108243 and Krtou = 2.784000 with Fig.(6), we can conlcude that this operating condition is shown as point B in Fig.(3)\n",
+        "Line 3 gives the locus of conversions for different values of the reaction rate group for this fluidized contacting\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5, Page 307\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "uo=0.3;          #Superficial gas velocity in m/s\n",
+      "Lf=1.1;          #Length of fixed bed in m\n",
+      "Hf=1.2;          #Length of freeboard in m\n",
+      "db=0.04;         #Equilibrium bubble size in m\n",
+      "umf=0.006;       #Velocity at minimum fluidization condition in m/s\n",
+      "ephsilonmf=0.55; #Void fraction at minimum fluidization condition\n",
+      "gammab=0.005;    #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "Kr=10.;          #rate constant in m**3 gas/m**3 cat s\n",
+      "D=2E-5;          #Diffusion coefficient of gas in m**2/s\n",
+      "g=9.81;          #Acceleration due to gravity in square m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
+      "ub=uo-umf+ubr;#Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
+      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5./4));\n",
+      "#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "\n",
+      "Kce=6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;\n",
+      "#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
+      "\n",
+      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
+      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
+      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
+      "gammac=(1-ephsilonmf)*((3.0/(ubr*ephsilonmf/umf-1))+fw);\n",
+      "#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
+      "\n",
+      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;\n",
+      "#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
+      "\n",
+      "Kf=(gammab*Kr)+1.0/((1.0/Kbc)+(1.0/(gammac*Kr+1.0/((1.0/Kce)+(1.0/(gammae*Kr))))));\n",
+      "#Raction rate for fluidized bed from Eqn.(14)\n",
+      "\n",
+      "XA=1-math.exp(-1*Kf*Lf/ub);#Conversion at the top of dense bed from Eqn.(16)\n",
+      "etabed=(Kf*delta)/(Kr*(1-ephsilonf));#Reactor efficiency from Eqn.(22)\n",
+      "a=0.6/uo  #Since uoa = 0.6s**-1 from Fig.(5)\n",
+      "adash=6.62; #From Fig.(5)\n",
+      "XA1=1-1.0/(math.exp(((1-ephsilonf)*Kr/(uo*a))*((1-math.exp(-a*Hf))- \\\n",
+      "    ((1-etabed)/(1+(adash/a)))*(1-math.exp(-(a+adash)*Hf)))));#Conversion from Eqn.(64)\n",
+      "XA2=1-(1.0-XA1)*(1.0-XA);#Conversion at the exit from Eqn.(64)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The conversion:'\n",
+      "print '\\tAt the top pf the dense bed: %d percentage'%(XA1*100)\n",
+      "print '\\tAt the reactor exit: %d percentage'%(XA2*100);\n",
+      "\n",
+      "#Disclaimer: The value of kf deviate from the one given in textbook, where as it is close to the value obtained by manual calculation. \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conversion:\n",
+        "\tAt the top pf the dense bed: 96 percentage\n",
+        "\tAt the reactor exit: 99 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch13.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch13.ipynb
new file mode 100755
index 00000000..18536015
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch13.ipynb
@@ -0,0 +1,308 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a97460b196d7b42e945dcfefc11684cc1c39c5847f8b0bdc9e3f6cdcd94bfcc3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Heat Transfer between Fluidized Beds and Surfaces"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 331\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dp=57.0;         #Particle size in micrometer\n",
+      "rhos=940.0;      #Density of solids in kg/m**3\n",
+      "Cps=828.0;       #Specific heat capacity of the solid in J/kg K\n",
+      "ks=0.20;         #Thermal conductivity of solids in W/m k\n",
+      "kg=0.035;        #Thermal concuctivity of gas in W/m k\n",
+      "umf=0.006;       #Velocity at minimum fluidization condition in m/s\n",
+      "ephsilonmf=0.476;#Void fraction at minimum fluidization condition\n",
+      "do1=0.0254;      #Outside diameter of tube in m\n",
+      "L=1;\n",
+      "uo=[0.05,0.1,0.2,0.35];#Superficial gas velocity in m/s\n",
+      "nw=[2.,3.1,3.4,3.5];   #Bubble frequency in s**-1\n",
+      "g=9.81;                #Acceleration due to gravity in square m/s**2\n",
+      "\n",
+      "\n",
+      "#CALCULATION\n",
+      "dte=4.*do1*L/2.*L;     #Hydraulic diameter from Eqn.(6.13)\n",
+      "db=(1+1.5)*0.5*dte;    #Rise velocity of the bubble\n",
+      "ubr=0.711*(g*db)**0.5; #Rise velocity of bubble from Eqn.(6.7)\n",
+      "phib=0.19;#From Fig.(15) for ks/kg=5.7\n",
+      "ke=ephsilonmf*kg+(1-ephsilonmf)*ks*(1/((phib*(ks/kg))+(2/3.0)))\n",
+      "    #Effective thermal conductivity of bed from Eqn.(3) \n",
+      "    \n",
+      "n=len(uo);\n",
+      "i=0;\n",
+      "ub = [0,0,0,0]\n",
+      "delta = [0,0,0,0]\n",
+      "h = [0,0,0,0]\n",
+      "while i<n:\n",
+      "    ub[i]=uo[i]-umf+ubr; #Velocity of bubbles in bubbling beds in Eqn.(6.8)\n",
+      "    delta[i]=uo[i]/ub[i];#Fraction of bed in bubbles from Eqn.(6.29)\n",
+      "    h[i]=1.13*(ke*rhos*(1-ephsilonmf)*Cps*nw[i]*(1-delta[i]))**0.5;\n",
+      "        #Heat transfer coefficinet from Eqn.(18)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Superficial gas velocity(m/s)',\n",
+      "print '\\tHeat transfer coefficient(W/m**2 k)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%uo[i],\n",
+      "    print '\\t\\t\\t%f'%h[i]\n",
+      "    i=i+1;\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Superficial gas velocity(m/s) \tHeat transfer coefficient(W/m**2 k)\n",
+        "0.050000 \t\t\t270.297375\n",
+        "0.100000 \t\t\t323.421769\n",
+        "0.200000 \t\t\t315.487604\n",
+        "0.350000 \t\t\t292.370761\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 332\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dp=80.;              #Particle size in micrometer\n",
+      "rhos=2550.;          #Density of solids in kg/m**3\n",
+      "Cps=756.;            #Specific heat capacity of the solid in J/kg K\n",
+      "ks=1.21;             #Thermal conductivity of solids in W/m k\n",
+      "kg=[0.005,0.02,0.2]; #Thermal concuctivity of gas in W/m k\n",
+      "ephsilonmf=0.476;    #Void fraction at minimum fluidization condition\n",
+      "\n",
+      "#CALCULATION\n",
+      "delta=0.5*(0.1+0.3);#For a gently fluidized bed\n",
+      "nw=3.;#Bubble frequency in s**-1 from Fig.(5.12) at about 30cm above the distributor\n",
+      "n=len(kg);\n",
+      "i=0;\n",
+      "x = [0,0,0]\n",
+      "while i<n:\n",
+      "    x[i]=ks/kg[i];#To find different values of ks/kg\n",
+      "    i=i+1;\n",
+      "\n",
+      "phib=[0.08,0.10,0.20];#From Fig.(15) for different values of ks/kg\n",
+      "i=0;\n",
+      "ke = [0,0,0]\n",
+      "h1 =[0,0,0]\n",
+      "while i<n:\n",
+      "    ke[i]=ephsilonmf*kg[i]+(1-ephsilonmf)*ks*(1/((phib[i]*(ks/kg[i]))+(2.0/3)))\n",
+      "    #Effective thermal conductivity of bed from Eqn.(3)\n",
+      "    \n",
+      "    h1[i]=1.13*(ke[i]*rhos*(1-ephsilonmf)*Cps*nw*(1-delta))**0.5;#Heat transfer coefficinet from Eqn.(18)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Thermal conductivity of Gas(W/m K))',\n",
+      "print '\\tMax. heat transfer coefficient(W/m**2 k)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%kg[i],\n",
+      "    print '\\t\\t\\t\\t%d'%h1[i]\n",
+      "    i=i+1;\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thermal conductivity of Gas(W/m K)) \tMax. heat transfer coefficient(W/m**2 k)\n",
+        "0.005000 \t\t\t\t324\n",
+        "0.020000 \t\t\t\t567\n",
+        "0.200000 \t\t\t\t1157\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 332\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "rhos=2700.;        #Density of solids in kg/m**3\n",
+      "Cps=755.;          #Specific heat capacity of the solid in J/kg K\n",
+      "ks=1.2;            #Thermal conductivity of solids in W/m k\n",
+      "kg=0.028;          #Thermal concuctivity of gas in W/m k\n",
+      "ephsilonmf=0.476;  #Void fraction at minimum fluidization condition\n",
+      "dp1=10E-3;         #Particle size for which h=hmax in m\n",
+      "hmax=250.;         #Max. heat transfer coefficient in W/m**2 K \n",
+      "nw=5.;             #Bubble frequency in s**-1\n",
+      "delta=0.1;         #Fraction of bed in bubbles\n",
+      "deltaw=0.1;        #Fraction of bed in bubbles in wall region\n",
+      "dp=2E-3;           #Diameter of particle in m\n",
+      "\n",
+      "#CALCULATION\n",
+      "x=ks/kg;\n",
+      "phib=0.11;\n",
+      "phiw=0.17;\n",
+      "ke=ephsilonmf*kg+(1-ephsilonmf)*ks*(1/((phib*(ks/kg))+(2.0/3)));\n",
+      "#Effective thermal conductivity of bed from Eqn.(3)\n",
+      "\n",
+      "hpacket=1.13*(ke*rhos*(1-ephsilonmf)*Cps*nw/(1-deltaw))**0.5;\n",
+      "#Heat transfer coefficient for the packet of emulsion from Eqn.(11)\n",
+      "\n",
+      "ephsilonw=ephsilonmf;#Void fraction in the wall region\n",
+      "\n",
+      "kew=ephsilonw*kg+(1-ephsilonw)*ks*((phiw*(ks/kg)+(1.0/3))**-1);\n",
+      "#Effective thermal conductivity in the wall region with stagnant gas from Eqn.(4)\n",
+      "\n",
+      "y=(2*kew/dp1)+(hmax*hpacket)/(((1-deltaw)*hpacket)-hmax);\n",
+      "#Calculating the term alphaw*Cpg*rhog*uo from Eqn.(16) by rearranging it\n",
+      "\n",
+      "h=(1-deltaw)/((2*kew/dp+y*(dp/dp1)**0.5)**-1+hpacket**-1);\n",
+      "#Heat transfer coeeficient from Eqn.(11) by using the value of y \n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The heat transfer coefficient for paricle size of %fm = %fW/m**2 K'%(dp,h);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transfer coefficient for paricle size of 0.002000m = 194.873869W/m**2 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 334\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Hf=4.0;   #Height of freeboard in m\n",
+      "uo=2.4;   #Superficial gas velocity in m/s\n",
+      "ho=350.0; #Heat transfer coefficient at the bottom of freeboard region in W/m**2 K\n",
+      "hg=20.0;  #Heat transfer coefficient in equivalent gas stream, but free of solids in W/m**2 K\n",
+      "\n",
+      "#CALCULATION\n",
+      "zf=[0,0.5,1,1.5,2,2.5,3,3.5,Hf];#Height above the top of the dense bubbling fluidized bed\n",
+      "hr=0;#Assuming heat transfer due to radiation is negligible\n",
+      "a=1.5/uo;#Since decay coefficient from Fig.(7.12), a*uo=1.5s**-1 \n",
+      "n=len(zf);\n",
+      "i=0;\n",
+      "h = []\n",
+      "while i<n:\n",
+      "    h.append((hr+hg)+(ho-hr-hg)*math.exp(-a*zf[i]/2.0));#Heat transfer coefficient from Eqn.(24) for zf=Hf\n",
+      "    i=i+1;\n",
+      "\n",
+      "hbar=(hr+hg)+2*(ho-hr-hg)*(1-math.exp(-a*Hf/2.0))/(a*Hf);#Mean heat transfer coefficient for the 4-m high freeboard from Eqn.(26)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The required relationship is h(W/m**2 K) vs. zf(m) as in Fig.(9a)'\n",
+      "print 'Height above the dense bubbling fluidized bed(m))',\n",
+      "print '\\tHeat transfer coefficient(W/m**2 k)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%zf[i],\n",
+      "    print '\\t\\t\\t\\t\\t\\t%f'%h[i]\n",
+      "    i=i+1;\n",
+      "\n",
+      "print '\\nThe mean heat transfer coefficient for the 4-m high freeboard =%d W/m**2 K'%hbar\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required relationship is h(W/m**2 K) vs. zf(m) as in Fig.(9a)\n",
+        "Height above the dense bubbling fluidized bed(m)) \tHeat transfer coefficient(W/m**2 k)\n",
+        "0.000000 \t\t\t\t\t\t350.000000\n",
+        "0.500000 \t\t\t\t\t\t302.263958\n",
+        "1.000000 \t\t\t\t\t\t261.433158\n",
+        "1.500000 \t\t\t\t\t\t226.508723\n",
+        "2.000000 \t\t\t\t\t\t196.636271\n",
+        "2.500000 \t\t\t\t\t\t171.085009\n",
+        "3.000000 \t\t\t\t\t\t149.229857\n",
+        "3.500000 \t\t\t\t\t\t130.536154\n",
+        "4.000000 \t\t\t\t\t\t114.546583\n",
+        "\n",
+        "The mean heat transfer coefficient for the 4-m high freeboard =208 W/m**2 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch14.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch14.ipynb
new file mode 100755
index 00000000..24a0ffbb
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch14.ipynb
@@ -0,0 +1,416 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a7839feeb371e4231dbf99a0d3738674ff633956f5fb373aea54d56b513c13f8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 : The RTD and Size Distribution of Solids in Fluidized Beds"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "#INPUT\n",
+      "Fo=2.7;                         #Feed rate in kg/min\n",
+      "Fof=0.9;                        #Feed rate of fines in feed in kg/min\n",
+      "Foc=1.8;                        #Feed rate of coarse in feed in kg/min\n",
+      "W=17.;                           #Bed weight in kg\n",
+      "kf=0.8;                         #Elutriation of fines in min**-1\n",
+      "kc=0.0125;                      #Elutriation of coarse in min**-1\n",
+      "\n",
+      "#CALCULATION\n",
+      "F1guess=1;                      #Guess value of F1\n",
+      "def solver_func(F1):            #Function defined for solving the system\n",
+      "    return F1-(Fof/(1.+(W/F1)*kf))-(Foc/(1.+(W/F1)*kc));#Eqn.(17)\n",
+      "\n",
+      "F1=fsolve(solver_func,F1guess)\n",
+      "F1f=Fof/(1.+(W/F1)*kf);          #Flow rate of fines in entrained streams from Eqn.(16)\n",
+      "F1c=Foc/(1.+(W/F1)*kc);          #Flow rate of coarse in entrained streams from Eqn.(16)\n",
+      "F2f=Fof-F1f;                    #Flow rate of fines in overflow streams from Eqn.(9)\n",
+      "F2c=Foc-F1c;                    #Flow rate of coarse in overflow streams from Eqn.(9)\n",
+      "tbarf=1./((F1/W)+kf);            #Mean residence time of fines from Eqn.(12)\n",
+      "tbarc=1./((F1/W)+kc);            #Mean residence time of coarse from Eqn.(12)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Flow rate in entrained stream:\\tFines:%fkg/min\\tCoarse:%fkg/min'%(F1f,F1c);\n",
+      "print 'Flow rate in overflow stream:\\tFines:%fkg/min\\tCoarse:%fkg/min'%(F2f,F2c);\n",
+      "print 'Mean residence time:\\tFines:%fmins\\tCoarse:%fmins'%(tbarf,tbarc);\n",
+      "\n",
+      "#====================================END OF PROGRAM ======================================================"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate in entrained stream:\tFines:0.100000kg/min\tCoarse:1.600000kg/min\n",
+        "Flow rate in overflow stream:\tFines:0.800000kg/min\tCoarse:0.200000kg/min\n",
+        "Mean residence time:\tFines:1.111111mins\tCoarse:8.888889mins\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 344\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "\n",
+      "import math\n",
+      "from numpy import linspace,array,zeros\n",
+      "from scipy.optimize import fsolve\n",
+      "from matplotlib.pyplot import *\n",
+      "%matplotlib inline\n",
+      "#Variable declaration\n",
+      "dt=4.;          #Diameter of reactor in m\n",
+      "ephsilonm=0.4; #Void fraction of static bed\n",
+      "rhos=2500.;     #Density of solid in the bed in kg/m**3\n",
+      "Lm=1.2;        #Height of static bed in m\n",
+      "Fo=3000;       #Feed rate in kg/hr\n",
+      "beta1=1.2;     #Increase in density of solids\n",
+      "dp=array([3,4,5,6,7,8,9,10,11,12,3,14,16,18,20,22,24,26,28,30])*10**-2;#Size of particles in mm\n",
+      "po=[0,0.3,0.8,1.3,1.9,2.6,3.5,4.4,5.7,6.7,7.5,7.8,7.5,6.3,5.0,3.6,2.4,1.3,0.5,0];#Size distribution of solids in mm**-1\n",
+      "k=array([0,10,9.75,9.5,8.75,7.5,6.0,4.38,2.62,1.20,0.325,0,0,0,0,0,0,0,0,0])*10**-4;#Elutriation constant in s**-1\n",
+      "pi=3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "W=(pi/4*dt**2)*Lm*(1-ephsilonm)*rhos;#Weight of solids in bed\n",
+      "n=len(dp);\n",
+      "i=0;\n",
+      "F1guess=1000.;#Guess value for F1\n",
+      "F1c=linspace(2510,2700,10);\n",
+      "F1 = zeros(n)\n",
+      "x = zeros(n)\n",
+      "c = zeros(n)\n",
+      "a = zeros(n)\n",
+      "while i<n:\n",
+      "    if k[i]==0:\n",
+      "        x[i]=0\n",
+      "        #break \n",
+      "    else:\n",
+      "        x[i]=0#(float(po[i])/(W*k[i]/float(F1)))*math.log(1.+(W*k[i]/F1));         \n",
+      "    def solver_func(Fo):\n",
+      "        return F1/(Lm*Fo)-x[i];\n",
+      "\n",
+      "    F1[i] = fsolve(solver_func,F1guess);#Using inbuilt function fsolve for solving Eqn.(20) for F1\n",
+      "    #c[i]=F1c[i]/(Lm*Fo);\n",
+      "    if F1[i]==0:\n",
+      "        a[i]=0;\n",
+      "    else:\n",
+      "        a[i]=(po[i]/(W*k[i]/F1[i]))*math.log(1+(W*k[i]/F1[i]));\n",
+      "\n",
+      "    i=i+1;\n",
+      "\n",
+      "#plot(F1,c);\n",
+      "\n",
+      "#xtitle('F1 vs a,c','F1','a,c');\n",
+      "F1n=2500.;#The point were both the curves meet\n",
+      "F2=beta1*Fo-F1n;#Flow rate of the second leaving stream\n",
+      "j=0;\n",
+      "m=len(dp);\n",
+      "p1 = zeros(m)\n",
+      "p2 = zeros(m)\n",
+      "tbar = zeros(m)\n",
+      "while j<m:\n",
+      "    p1[j]=(1./F1n)*((Fo*po[j])/(1.+(W/F1n)*k[j]));#Size distribution of stream 1 in mm**-1 from Eqn.(16)\n",
+      "    p2[j]=k[j]*W*p1[j]/F2;#Size distribution of stream 2 in mm**-1 from Eqn.(7)\n",
+      "    if p1[j]==0 and p2[j]==0:\n",
+      "        tbar[j]=0;\n",
+      "    elif p1[j]==0:\n",
+      "        tbar[j]=(W*p1[j])/(F2*p2[j]);\n",
+      "    elif  p2[j]==0:\n",
+      "        tbar[j]=(W*p1[j])/(F1n*p1[j]);\n",
+      "    else:\n",
+      "        tbar[j]=(W*p1[j])/(F1n*p1[j]+F2*p2[j]);#Average time in hr from Eqn.(11)\n",
+      "    j=j+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Flow rate of stream 1:%fkg/hr'%F1n\n",
+      "print 'Flow rate of stream 2:%fkg/hr'%F2\n",
+      "j=0;\n",
+      "print 'tbar(hr)'\n",
+      "while j<m:\n",
+      "    print '%f'%tbar[j]\n",
+      "    j=j+1;\n",
+      "\n",
+      "#DISCLAIMER: The value obtained for tbar is deviating highly\n",
+      "#form the one given in textbook. However, the value obtained by manual calculation is close to #\n",
+      "#the   ones obtained from the program."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "Flow rate of stream 1:2500.000000kg/hr"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Flow rate of stream 2:1100.000000kg/hr\n",
+        "tbar(hr)\n",
+        "0.000000\n",
+        "8.962153\n",
+        "8.964162\n",
+        "8.966171\n",
+        "8.972205\n",
+        "8.982279\n",
+        "8.994397\n",
+        "9.007522\n",
+        "9.021824\n",
+        "9.033397\n",
+        "9.040543\n",
+        "9.043200\n",
+        "9.043200\n",
+        "9.043200\n",
+        "9.043200\n",
+        "9.043200\n",
+        "9.043200\n",
+        "9.043200\n",
+        "9.043200\n",
+        "0.000000\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive', 'pi']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 351\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dp=1;  #Particle size in mm\n",
+      "Fo=10; #Feed rate in kg/min\n",
+      "k=0.1; #Particle shrinkage rate in mm/min\n",
+      "\n",
+      "#CALCULATION\n",
+      "R=k/2;             #Particle shrinkage rate in terms of radius\n",
+      "W=(Fo*dp/2)/(4*R); #Bed weight from Eqn.(42)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Weight of bed:%d kg' %W\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Weight of bed:25 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 352\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "dpi=[1.05,0.95,0.85,0.75,0.65,0.55,0.45,0.35,0.25,0.15,0.05]; #Mean size in mm\n",
+      "Fo=[0,0.5,3.5,8.8,13.5,17.0,18.2,17.0,13.5,7.3,0]#*10**-2  #Feed rate in kg/s\n",
+      "for i in range(len(Fo)):\n",
+      "    Fo[i] = Fo[i] * 10**-2\n",
+      "k=[0,0,0,0,0,0,0,0,2.0,12.5,62.5]#*10**-5;#Elutriation constant in s**-1\n",
+      "for i in range(len(k)):\n",
+      "    k[i] = k[i] * 10**-5\n",
+      "\n",
+      "R=-1.58*10**-5;#Rate of particle shrinkage in mm/s\n",
+      "deldpi=0.1;#Size intervals in mm\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(dpi);\n",
+      "m=1;#Starting with the largest value size interval that contains solids\n",
+      "W = [0]\n",
+      "while m<n-1:\n",
+      "    W.append((Fo[m]-R*W[m-1]/deldpi)/(k[m]-R/deldpi-3*R/dpi[m]));#From Eqn.(33)\n",
+      "    m=m+1;\n",
+      "\n",
+      "Wt=sum(W);#Total sum\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nTotal mass in the bed:%fkg'%Wt\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Total mass in the bed:7168.981263kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5, Page 353\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "dpi=[0.17,0.15,0.13,0.11,0.09,0.07,0.05,0.03,0.01];#Mean size of particles in mm\n",
+      "a=[0,0.95,2.45,5.2,10.1,23.2,35.65,20.0,2.45]#*10**-2;#Feed composition Fo(dpi)/Fo\n",
+      "for i in range(len(a)):\n",
+      "    a[i] = a[i] * 10**-2\n",
+      "\n",
+      "y=[0,0,0,0,0,0,0.625,10.225,159.25]#*10**-6;#Elutriation and cyclone efficiency k(dpi)(1-eta(dpi))\n",
+      "for i in range(len(y)):\n",
+      "    y[i] = y[i] * 10**-6\n",
+      "\n",
+      "\n",
+      "F=0.01;          #Rate at which solids are withdrawn in kg/s\n",
+      "W=40000;         #Weight of bed in kg\n",
+      "dp1=0.11         #Initial size in mm\n",
+      "dp2=0.085;       #Size after shrinking in mm\n",
+      "dpmin=0.01;      #Minimum size in mm\n",
+      "deldpi=2*10**-2; #Size inerval in mm\n",
+      "t=20.8;          #Time in days\n",
+      "si=1;\n",
+      "\n",
+      "#CALCULATION\n",
+      "kdash=math.log((dp1-dpmin)/(dp2-dpmin))/(t*24*3600);#Rate of particle shrinkage from Eqn.(24)\n",
+      "n=len(dpi);\n",
+      "m=1;\n",
+      "Fo=0.05;#Initial value of Fo\n",
+      "F1 = [0];\n",
+      "s=0;\n",
+      "c=0;\n",
+      "t=1E-6;\n",
+      "R = [0]\n",
+      "x = [0]\n",
+      "F1 = [0]\n",
+      "while m<n:\n",
+      "    R.append(-kdash*(dpi[m]-dpmin));#Rate of size change\n",
+      "    x.append((a[m]*Fo-W*R[m-1]*F1[m-1]/deldpi)/(F+(W*y[m])-(W*R[m]/deldpi)-3*W*R[m]/dpi[m]));#Eqn.(34)\n",
+      "    F1.append(x[m]*F);\n",
+      "    c=c+x[m];\n",
+      "    m=m+1;\n",
+      "    if abs(c-1)<t:\n",
+      "        break\n",
+      "    Fo=Fo+0.0001;#Incrementing Fo\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Feed rate with deldpi=%fmm is %fg/hr'%(deldpi,Fo);\n",
+      "i=0;\n",
+      "print 'Bed composition'\n",
+      "for i in x:\n",
+      "    print '%f'%(i*100)\n",
+      "    i=i+1;\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Feed rate with deldpi=0.020000mm is 0.050800g/hr\n",
+        "Bed composition\n",
+        "0.000000\n",
+        "0.652911\n",
+        "1.859952\n",
+        "4.400781\n",
+        "9.668999\n",
+        "25.654298\n",
+        "28.575890\n",
+        "2.317749\n",
+        "0.019493\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch15.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch15.ipynb
new file mode 100755
index 00000000..09c02451
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch15.ipynb
@@ -0,0 +1,294 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b0605e72f5da8476b3859cf3dc90196dcbe80cabc58b8c8d37cf5b28482c7ee1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15 : Circulation Systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 369\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "thalf=1;   #Half life of catalyst in s\n",
+      "F=960;     #Feed rate of oil in tons/day\n",
+      "W=50;      #Weight of the bed in tons\n",
+      "a=0.5;     #Activity after time equal to half life\n",
+      "abar=0.01; #Average activity of the catalyst\n",
+      "\n",
+      "#CALCULATION\n",
+      "Ka=-math.log(a)/thalf;#Rate constant is s**-1, assuming I order kinetics from Eqn.(12)\n",
+      "Fs=Ka*W*abar/(1-abar);#Circulation rate of solids from Eqn.(16)\n",
+      "x=(Fs*60*60*24.0)/F;  #Circulation rate per feed of oil\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nSolid recirculation per feed of oil =%ftons of solid circulated/ton feed oil'%x\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Solid recirculation per feed of oil =31.506690tons of solid circulated/ton feed oil\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 370\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "deltaHr1=1260.;                               #Enthalpy change during endothermic reaction in kJ/kg\n",
+      "deltaHr2=-33900.;                             #Enthal[y change during exothermic reaction in kJ/kg\n",
+      "H1=703.;                                      #Enthalpy of feed oil in kJ/kg\n",
+      "T1=260.;                                      #Temperature of feed oil in degree celcius\n",
+      "H3=1419.;                                     #Enthalpy of cracked product in kJ/kg\n",
+      "T3=500.;                                      #Temperature of cracked product in degree celcius\n",
+      "Ta=20.;                                       #Temperature of entering air in degree celcius\n",
+      "Cpa=1.09;                                     #Specific heat of entering air in kJ/kg K\n",
+      "Cpf=1.05;                                     #Specific heat of flue gases in kJ/kg K\n",
+      "Cps=1.01;                                     #Specific heat of solids in kJ/kg K\n",
+      "Cpv=3.01;                                     #Specific heat of vaporized feed in kJ/kg K\n",
+      "T4=[520.,540.,560.,580.,600.,620.,640.,660.]; #Temperature of flue gas in degree celcius\n",
+      "V=22.4;                                       #Volume of 1 mole of Carbon dioxide gas in N-m**3\n",
+      "M=12.;                                        #Molecular weight of carbon in kg\n",
+      "rho=1.293;                                    #Density of carbon dioxide gas in kg/N-m**3\n",
+      "xa=0.21;                                      #Mass fraction of oxygen in air\n",
+      "betac=0.07;                                   #Mass fraction of carbon\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(T4);\n",
+      "i=0;\n",
+      "x1 = [0,0,0,0,0,0,0,0]\n",
+      "x2 = [0,0,0,0,0,0,0,0]\n",
+      "excess_air = [0,0,0,0,0,0,0,0]\n",
+      "\n",
+      "x2min=betac*(V*rho/(M*xa));#Minimum amount of air required for complete combustion\n",
+      "while i<n:\n",
+      "    x1[i]=(deltaHr1+0.93*H3-H1)/(Cps*(T4[i]-T3));#Fs/F1 by simplifying the overall energy balance\n",
+      "    x2[i]=((0.07*(-deltaHr2)-(deltaHr1+0.93*H3-H1))/(Cpf*(T4[i]-Ta)))-0.07;#F2/F1 by simplifying the energy balance for regenerator\n",
+      "    if x2[i]>x2min:\n",
+      "        excess_air[i]=(x2[i]-x2min)/x2min; #Excess air used\n",
+      "    else:\n",
+      "        excess_air[i]=0;\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT    \n",
+      "print 'T4(degree celcius)',\n",
+      "print '\\tFs/F1',\n",
+      "print '\\t\\tF2/F1',\n",
+      "print '\\t\\tExcess air(percentage)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%T4[i],\n",
+      "    print '\\t\\t%f'%x1[i],\n",
+      "    print '\\t%f'%x2[i],\n",
+      "    print '\\t%f'%(excess_air[i]*100);\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Disclaimer: The values of F2/F1 obtained by manual calculation has close correspondance to the ones obtained as the output, whereas it deviates largely from the values given in textbook.\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "T4(degree celcius) \tFs/F1 \t\tF2/F1 \t\tExcess air(percentage)\n",
+        "520.000000 \t\t92.904455 \t0.875390 \t8.807235\n",
+        "540.000000 \t\t46.452228 \t0.839029 \t4.287699\n",
+        "560.000000 \t\t30.968152 \t0.805362 \t0.102944\n",
+        "580.000000 \t\t23.226114 \t0.774099 \t0.000000\n",
+        "600.000000 \t\t18.580891 \t0.744992 \t0.000000\n",
+        "620.000000 \t\t15.484076 \t0.717825 \t0.000000\n",
+        "640.000000 \t\t13.272065 \t0.692412 \t0.000000\n",
+        "660.000000 \t\t11.613057 \t0.668586 \t0.000000\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 379\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "Fs=100.;        #Solid flowrate in kg/s\n",
+      "ephsilon1=0.55;\n",
+      "ephsilon2=0.5;\n",
+      "p1=120.;         #Pressure at upper level in kPa\n",
+      "rhos=1000.;      #Density of solid in kg/m**3\n",
+      "rhog=1.;         #Density of gas in kg/m**3\n",
+      "gc=1.;           #Conversion factor\n",
+      "g=9.81;          #Acceleration due to gravity in m/s**2\n",
+      "di=0.34;         #Diameter of downcomer in m\n",
+      "pi=3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "x=(ephsilon1/ephsilon2)*((1-ephsilon2)/(1-ephsilon1));#To find pressure at lower level using Eqn.(30)\n",
+      "p2=x*p1;#Pressure at lower level using Eqn.(30)\n",
+      "deltap=p2-p1;\n",
+      "ephsilonbar=0.5*(ephsilon1+ephsilon2);\n",
+      "deltah=(deltap*10**3*gc)/(rhos*(1-ephsilonbar)*g);#Static head height from Eqn.(28)\n",
+      "At=0.25*pi*di**2;#Area of downcomer\n",
+      "Gs=Fs/At;#Flux of solids in downcomer\n",
+      "Gg=Gs*(ephsilon1/(1-ephsilon1))*(rhog/rhos)*(x-1);#Required gas aeration rate from Eqn.(31)\n",
+      "Fg=Gg*At;#Flow rate of gas required\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nThe required flow rate of gas required for location of %fm below downcomer is %.4fkg/s'%(deltah,Fg)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The required flow rate of gas required for location of 5.722768m below downcomer is 0.0272kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 380\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Fs=600;#Solid circulation rate in kg/s\n",
+      "dpbar=60;#Mean size of solids in micrometer\n",
+      "pA=120;#Pressure in vessel A in kPa\n",
+      "pB=180;#Pressure in vessel B in kPa\n",
+      "LfA=8;#Bed height in vessel A in m\n",
+      "LfB=8;#Bed height in vessel B i m\n",
+      "#Bulk densities in kg/m**3\n",
+      "rho12=100;\n",
+      "rho34=400;\n",
+      "rho45=550;\n",
+      "rho67=200;\n",
+      "rho78=200;\n",
+      "rho910=400;\n",
+      "rho1011=400;\n",
+      "rho1112=550;\n",
+      "rho13=100;\n",
+      "deltapdA=7;#Pressure drop across the distributor in regenerator in kPa\n",
+      "deltapdB=8;#Pressure drop across the distributor in reactor in kPa\n",
+      "deltap12=(9+4);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa\n",
+      "deltap78=(15+3);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa\n",
+      "deltap45=20;#Friction loss across the reactor's stripper downcomer in kPa\n",
+      "deltap1112=4;#Friction loss across the regenerator's downcomer in kPa\n",
+      "deltapvA=5;#Pressure drop assigned for the control valve in regenerator in kPa\n",
+      "deltapvB=15;#Pressure drop assigned for the control valve in reactor in kPa\n",
+      "deltah12=15;#Height of the riser in m\n",
+      "deltah86=30;#Height of the riser in m\n",
+      "deltah1011=7;#Height difference h10-h11 in m\n",
+      "g=9.81;#Acceleration due to gravity in m/s**2\n",
+      "gc=1;#Conversion factor\n",
+      "pi=3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "Gs=900;#From Fig.(8), to find dt\n",
+      "dt=math.sqrt((4/math.pi)*Fs/Gs);#Diameter of the downcomer\n",
+      "#Height of downcomer A from Eqn.(7)\n",
+      "deltahA=(1/(rho1112*g))*((pB-pA)*gc*(10**3)+(deltap12+deltapdB+deltap1112+deltapvA)*gc*10**3-rho12*g*(-deltah12)-rho34*g*(-LfB)-rho1011*g*deltah1011);\n",
+      "#Height of downcomer B from Eqn.(8)\n",
+      "deltahB=(1/(rho45*g))*(-(pB-pA)*gc*10**3+(deltap45+deltapvB+deltap78+deltapdA)*gc*10**3+rho78*g*deltah86+rho910*g*LfA)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Height of downcomer for:'\n",
+      "print '\\tRegenerator:%d m'%deltahA\n",
+      "print '\\tReactor:%.1f m'%deltahB\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height of downcomer for:\n",
+        "\tRegenerator:20 m\n",
+        "\tReactor:16.7 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch16.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch16.ipynb
new file mode 100755
index 00000000..e835e890
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch16.ipynb
@@ -0,0 +1,428 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1a37ab741ccda7f67c7de1085d3807de9fa5a11a3111916ad9502a04f1c58721"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16 : Design for Physical Operations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 404\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "T=1000.;       #Operating temperature of calciner in degree celcius\n",
+      "deltaHr=1795.; #Heat of reaction in kJ/kg\n",
+      "M1=0.1;        #Molecular weight of Calcium carbonate in kg/mol\n",
+      "M2=0.056;      #Molecular weight of CaO in kg/mol\n",
+      "M3=0.044;      #Molecular weight of Carbon dioxide  in kg/mol\n",
+      "M4=0.029;      #Molecular weight of Air in kg/mol\n",
+      "M5=0.029;      #Molecular weight of Combustion gas in kg/mol\n",
+      "Cp1=1.13;      #Specific heat of Calcium carbonate in kJ/kg K\n",
+      "Cp2=0.88;      #Specific heat of CaO in kJ/kg K\n",
+      "Cp3=1.13;      #Specific heat of Carbon dioxide in kJ/kg K\n",
+      "Cp4=1.00;      #Specific heat of Air in kJ/kg K\n",
+      "Cp5=1.13;      #Specific heat of Calcium carbonate in kJ/kg K\n",
+      "Tf=20.;        #Temperature of feed in degree celcius\n",
+      "ma=15.;        #Air required per kg of fuel in kg\n",
+      "Hc=41800.;     #Net combustion heat of fuel in kJ/kg\n",
+      "Tpi=20.;       #Initial temperature of solids in degree C\n",
+      "Tgi=1000.;     #Initial temperature of gas in degree C\n",
+      "\n",
+      "#CALCULATION\n",
+      "mc=1;#Based on 1 kg of Calcium carbonate\n",
+      "B=(1/(Hc-(ma+mc)*Cp5*(T-Tpi)))*(M3*Cp3*(T-Tf)+M2*Cp2*(T-Tf)+deltaHr)#Fuel consumption(kg fuel/kg calcium carbonate)\n",
+      "B1=B*M3/M2;#Fuel consumption(kg fuel/kg Cao)\n",
+      "H=Hc*B1;#Heat required for calcination\n",
+      "eta=deltaHr/(B*Hc);#Thermal efficiency\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Fuel consumption:%f kg fuel/kg Cao'%B1\n",
+      "print 'Heat requirement for calcination:%f kJ/kg Cao'%H\n",
+      "print 'Thermal efficiency:%f percentage'%(eta*100)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fuel consumption:0.061731 kg fuel/kg Cao\n",
+        "Heat requirement for calcination:2580.366029 kJ/kg Cao\n",
+        "Thermal efficiency:54.657251 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 405\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from scipy.optimize import fsolve\n",
+      "import math\n",
+      "\n",
+      "#INPUT\n",
+      "F = 400.        #Feed rate of Calcium carbonate in tons/day\n",
+      "T = 1000.       #Operating temperature of calciner in degree celcius\n",
+      "deltaHr = 1795.#Heat of reaction in kJ/kg\n",
+      "M1 = 0.1        #Molecular weight of Calcium carbonate in kg/mol\n",
+      "M2 = 0.056     #Molecular weight of CaO in kg/mol\n",
+      "M3 = 0.044     #Molecular weight of Carbon dioxide  in kg/mol\n",
+      "M4 = 0.029    #Molecular weight of Air in kg/mol\n",
+      "M5 = 0.029    #Molecular weight of Combustion gas in kg/mol\n",
+      "Cp1 = 1.13    #Specific heat of Calcium carbonate in kJ/kg K\n",
+      "Cp2 = 0.88    #Specific heat of CaO in kJ/kg K\n",
+      "Cp3 = 1.13   #Specific heat of Carbon dioxide in kJ/kg K\n",
+      "Cp4 = 1.00    #Specific heat of Air in kJ/kg K\n",
+      "Cp5 = 1.17    #Specific heat of Combustion gas in kJ/kg K\n",
+      "Tf = 20.     #Temperature of feed in degree celcius\n",
+      "ma = 15.    #Air required per kg of fuel in kg\n",
+      "uo = 0.8    #Superficial gas velocity in m/s\n",
+      "Hc = 41800.   #Net combustion heat of fuel in kJ/kg\n",
+      "Tpi = 20.     #Initial temperature of solids in degree C\n",
+      "Tgi = 1000.   #Initial temperature of gas in degree C\n",
+      "rhoa = 1.293    #Density of air in kg/m**3\n",
+      "pi = 3.14\n",
+      "\n",
+      "#CALCULATION\n",
+      "mc = 1.         #Based on 1 kg of Calcium carbonate\n",
+      "Bguess = 2.     #Guess value of B\n",
+      "def solver_func(B):         #Function defined for solving the system\n",
+      "    phi = ((ma+mc)*Cp5*B+(M3*Cp3))/Cp1\n",
+      "    T3 = (Tpi+(phi+phi**2+phi**3)*Tgi)/(1+phi+phi**2+phi**3)\n",
+      "    phiplus = 30.6*B\n",
+      "    Tr = (T+Tpi*phiplus)/(1+phiplus)\n",
+      "    return Hc*B+Cp3*(T3-Tpi)+ma*B*Cp4*(Tr-20)-(ma+mc)*Cp5*(T-Tpi)-M3*Cp3*(T-Tpi)-M2*Cp2*(T-Tpi)-deltaHr\n",
+      "    #fn = (1/20800)*(2470-T3-13.34*(Tr-20))\n",
+      "\n",
+      "B = fsolve(solver_func,1E-6)#Using inbuilt function fsolve for solving Eqn.(23) for tou\n",
+      "phi = ((ma+mc)*Cp5*B+(M3*Cp3))/Cp1\n",
+      "#Temperature of various stages\n",
+      "T1 = (Tpi+(phi)*Tgi)/(1+phi)\n",
+      "T2 = (Tpi+(phi+phi**2)*Tgi)/(1+phi+phi**2)\n",
+      "T3 = (Tpi+(phi+phi**2+phi**3)*Tgi)/(1+phi+phi**2+phi**3)\n",
+      "phiplus = 30.6*B\n",
+      "Tr = (T+Tpi*phiplus)/(1+phiplus)\n",
+      "eta = deltaHr/(B*Hc)           #Thermal efficiency\n",
+      "H = B*Hc/M2                    #Heat requirement\n",
+      "#For lower heat recovery section\n",
+      "Ql = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(Tr+273)))#Volumetric flow rate of gas in the lower heat recovery section\n",
+      "dtl = math.sqrt(4/pi*Ql/uo)#Diameter of lower bed\n",
+      "#For calcination section\n",
+      "Qc = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T+273)))#Volumetric flow rate of gas in the calcination section\n",
+      "dtc = math.sqrt(4/pi*Qc/uo)#Diameter of calcination section\n",
+      "#For I stage\n",
+      "Q1 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T1+273)))#Volumetric flow rate of gas in the I stage\n",
+      "dt1 = math.sqrt(4/pi*Q1/uo)#Diameter of I stage\n",
+      "#For II stage\n",
+      "Q2 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T2+273)))#Volumetric flow rate of gas in the II stage\n",
+      "dt2 = math.sqrt(4/pi*Q2/uo)#Diameter of II stage\n",
+      "#For III stage\n",
+      "Q3 = (F*10**3/(24*3600))*B*ma/(rhoa*(273/(T3+273)))#Volumetric flow rate of gas in the III stage\n",
+      "dt3 = math.sqrt(4/pi*Q3/uo)#Diameter of III stage\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nDiameter of lower bed:%fm'%(dtl)\n",
+      "print '\\nDiameter of calcination section:%fm'%(dtc)\n",
+      "print '\\nBed no.\\t\\t1\\t2\\t\\t3'\n",
+      "print '\\nDiameter(m)%f\\t%f\\t%f'%(dt1,dt2,dt3)\n",
+      "\n",
+      "#The value of diameter of each section is largely deviating from the values in the textbook. This is because the fuel consumption B have not been included in the energy balance equation. And the value of molecular weight is wrong by one decimal point.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Diameter of lower bed:7.097814m\n",
+        "\n",
+        "Diameter of calcination section:13.351483m\n",
+        "\n",
+        "Bed no.\t\t1\t2\t\t3\n",
+        "\n",
+        "Diameter(m)12.728715\t13.270865\t13.340712\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 413\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "T=20;         #Temeprature in degree C\n",
+      "M=0.018;      #Molecular weight of water in kg/mol\n",
+      "Q=10;         #Flow rate of dry air in m**3/s\n",
+      "R=82.06E-6;   #Universal gas constant\n",
+      "pi=0.0001;    #Initial moisture content in atm\n",
+      "pj=0.01;      #Final moisture content in atm\n",
+      "\n",
+      "#CALCULATION\n",
+      "a=Q*(273+T)/273;  #Term At*uo\n",
+      "b=a*M/(R*(T+273));#Term C*At*uo\n",
+      "#The value of slope can be found only by graphical mehtod. Hence it has been taken directly from the book(Page no.414,Fig.E3)\n",
+      "m=10.2;\n",
+      "Fo=b/m;           #Flow rate of solids\n",
+      "Q3=(b/Fo)*(pj-pi);#Moisture content of leaving solids\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nMoisture content of leaving solids:%.3f kg H2O/kg dry solids'%Q3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Moisture content of leaving solids:0.101 kg H2O/kg dry solids\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 422\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "\n",
+      "\n",
+      "#INPUT\n",
+      "Qfi = 0.20          #Initial moisture fraction\n",
+      "Qfbar = 0.04        #Average final moisture fraction\n",
+      "rhos = 2000.        #Density of solid in kg/m**3\n",
+      "Cps = 0.84          #Specific heat of solids in kJ/kg K\n",
+      "Fo = 7.6E-4         #Flow rate of solids in kg/m**3\n",
+      "Tsi = 20.            #Inital temperature of solids in degree C\n",
+      "rhog = 1.            #Density of gas in kg/m**3\n",
+      "Cpg = 1             #Specific heat of gas in kJ/kg K\n",
+      "uo = 0.3            #Superficial gas velocity in m/s\n",
+      "Tgi = 200           #Initial temperature of gas in degee C\n",
+      "L = 2370            #Enthalpy of liquid in kJ/kg\n",
+      "Cpl = 4.2           #Specific heat of liquid in kJ/kg K\n",
+      "dt = 0.1            #Diameter of reactor in m\n",
+      "Lm = 0.1            #Length of fixed bed in m\n",
+      "ephsilonm = 0.45    #Void fraction of fixed bed\n",
+      "pi = 3.14\n",
+      "Fo1 = 1             #Feed rate for commercial-scale reactor in kg/s\n",
+      "\n",
+      "#CALCULATION\n",
+      "#(a)Bed temperature\n",
+      "Teguess = 50#Guess value of Te\n",
+      "def solver_func(Te):#Function defined for solving the system\n",
+      "    return (pi/4.)*dt**2*uo*rhog*Cpg*(Tgi-Te)-Fo*(Qfi-Qfbar)*(L+Cpl*(Te-Tsi))-Fo*Cps*(Te-Tsi)\n",
+      "\n",
+      "Te = fsolve(solver_func,Teguess)\n",
+      "\n",
+      "#(b)Drying time for a particle\n",
+      "xguess = 2#Guess value of x, ie term tou/tbar\n",
+      "def solver_func1(x):        #Function defined for solving the system\n",
+      "    return 1-(Qfbar/Qfi)-(1-math.exp(-x))/x\n",
+      "\n",
+      "\n",
+      "x = fsolve(solver_func1,xguess)\n",
+      "W = (pi/4.)*dt**2*Lm*(1-ephsilonm)*rhos#Weight of soilds in bed\n",
+      "tbar = W/Fo#Mean residence time of solids from Eqn.(59)\n",
+      "tou = tbar*x#Time for complete drying of a particle\n",
+      "\n",
+      "#(c)Commercial-scale dryer\n",
+      "W1 = Fo1*tbar\n",
+      "Atguess = 5#Guess value of area\n",
+      "def  solver_func3(At):          #Function defined for solving the system\n",
+      "    return  At*uo*rhog*Cpg*(Tgi-Te)-Fo1*(Qfi-Qfbar)*(L+Cpl*(Te-Tsi))-Fo1*Cps*(Te-Tsi)\n",
+      "\n",
+      "At = fsolve(solver_func3,Atguess)\n",
+      "dt1 = math.sqrt(4/pi*At)#Diameter of commercial-scale dryer\n",
+      "Q1 = At*uo*rhog#Flow rate necessary for the operation\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Bed temperature:%f degree C'%(Te)\n",
+      "print 'Time for complete drying of particle:%fs'%(tou)\n",
+      "print 'Flow rate of gas necessary for Commercial-scale dryer:%fkg/s'%(Q1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bed temperature:58.728126 degree C\n",
+        "Time for complete drying of particle:527.431202s\n",
+        "Flow rate of gas necessary for Commercial-scale dryer:3.098684kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5, Page 425\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable Declaration\n",
+      "rhos=1600.;     #Density of solid in kg/m**3\n",
+      "Cps=1.25;       #Specific heat of solids in kJ/kg K\n",
+      "Fo=0.5;         #Flow rate of solids in kg/s\n",
+      "Tsi=20.;        #Inital temperature of solids in degree C\n",
+      "Qwi=1.;         #Initial moisture fraction in water\n",
+      "Qwf=0.2;        #Final moisture fraction in water\n",
+      "Qhi=1.1;        #Initial moisture fraction in heptane\n",
+      "Qhf=0.1;        #Final moisture fraction in heptane\n",
+      "Tgi=240.;       #Initial temperature of gas in degee C\n",
+      "Te=110.;        #Bed temperature in degree C\n",
+      "ephsilonm=0.45; #Void fraction of fixed bed\n",
+      "ephsilonf=0.75; #Void fraction of fluidized bed\n",
+      "uo=0.6;         #Superficial gas velocity in m/s\n",
+      "di=0.08;        #Diameter of tubes in m\n",
+      "li=0.2;         #Pitch for square arrangement\n",
+      "hw=400.;        #Heat transfer coefficient in W/m**2 K\n",
+      "Tc=238.;        #Temperature at which steam condenses in degree C\n",
+      "#Specific heats in kJ/kg K\n",
+      "Cwl=4.18;       #Water liquid\n",
+      "Cwv=1.92;       #Water vapor\n",
+      "Chl=2.05;       #Heptane liquid\n",
+      "Chv=1.67;       #Heptane vapor\n",
+      "#Latent heat of vaporization in kJ/kg\n",
+      "Lw=2260.;       #Water\n",
+      "Lh=326.;        #Heptane\n",
+      "#Density of vapor in kg/m**3 at operating conditions\n",
+      "rhow=0.56;      #Water\n",
+      "rhoh=3.1;       #Heptane\n",
+      "Lf=1.5;         #Length of fixed bed in m\n",
+      "t=140.;         #Half-life of heptane in s\n",
+      "L=1.5;          #Length of tubes in heat exchanger\n",
+      "pi=3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#(a) Dryer without Internals\n",
+      "xw=(Qwi-Qwf)/(Qhi-Qhf);                #Water-heptane weight ratio\n",
+      "xv=((Qwi-Qwf)/18.)/((Qhi-Qhf)/100.);   #Water-heptane volume ratio\n",
+      "T=(Qwi-Qwf)/18.+(Qhi-Qhf)/100.;        #Total volume\n",
+      "rhogbar=((Qwi-Qwf)/18.)/T*rhow+((Qhi-Qhf)/100.)/T*rhoh;           #Mean density of the vapor mixture\n",
+      "Cpgbar=(((Qwi-Qwf)/18.)/T)*rhow*Cwv+(((Qhi-Qhf)/100.)/T)*rhoh*Cwv;#Mean specific heat of vapor mixture\n",
+      "#Volumetric flow of recycle gas to the dryer in m**3/s from Eqn.(53)\n",
+      "x=(Cpgbar*(Tgi-Te))**-1*(Fo*(Qwi-Qwf)*(Lw+Cwl*(Te-Tsi))+Fo*(Qhi-Qhf)*(Lh+Chl*(Te-Tsi))+Fo*(Cps*(Te-Tsi)));\n",
+      "r=Fo*((Qwi-Qwf)/rhow+(Qhi-Qhf)/rhoh); #Rate of formation of vapor in bed\n",
+      "uo1=uo*(x/(x+r));                     #Superficial velocity just above the distributor\n",
+      "At=x/uo1;                             #Cross-sectional area of bed\n",
+      "dt=math.sqrt(4./pi*At);               #Diameter of bed\n",
+      "B=-math.log(Qwf/Qwi)/t;               #Bed height from Eqn.(63)\n",
+      "tbar=((Qhi/Qhf)-1)/B;                 #Mean residence time of solids\n",
+      "W=Fo*tbar;                            #Weight of bed\n",
+      "Lm=W/(At*(1-ephsilonm)*rhos);         #Static bed height\n",
+      "Lf=(Lm*(1-ephsilonm))/(1-ephsilonf);  #Height of fluidized bed\n",
+      "\n",
+      "#(b) Dryer with internal heaters\n",
+      "f=1/8.0;                 #Flow rate is 1/8th the flow rate of recirculation gas as in part (a)\n",
+      "x1=f*x;                  #Volumetric flow of recycle gas to the dryer in m**3/s from Eqn.(53)\n",
+      "uo2=uo*(x1/float(x1+r)); #Superficial velocity just above the distributor\n",
+      "Abed=x1/uo2;             #Cross-sectional area of bed\n",
+      "q=(Fo*(Qwi-Qwf)*(Lw+Cwl*(Te-Tsi))+Fo*(Qhi-Qhf)*(Lh+Chl*(Te-Tsi))+Fo*(Cps*(Te-Tsi)))-Abed*uo2*Cpgbar*(Tgi-Te);#Heat to be added from energy balance of Eqn.(53)\n",
+      "Aw=q*10**3/(hw*(Tc-Te));#Total surface area of heat exchanger tubes\n",
+      "Lt=Aw/(pi*di);          #Total length of tubes\n",
+      "Nt=Lt/L;                #Total number of tubes\n",
+      "Atubes=Nt*(pi/4*di**2); #Total cross-sectional area of tubes\n",
+      "Atotal=Abed+Atubes;     #Total cross-sectional area of tube filled dryer\n",
+      "d=math.sqrt(Atotal*pi/4.);#Diameter of vessel\n",
+      "li=math.sqrt(Atotal/Nt);  #Pitch for square array of tubes\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\t\\t\\tBed diameter(m)\\tRecycle vapor flow(m**3/s)'\n",
+      "print 'Without internal heater\\t%f\\t%f'%(dt,x)\n",
+      "print 'With heating tubes\\t%f\\t%f'%(d,x1)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\tBed diameter(m)\tRecycle vapor flow(m**3/s)\n",
+        "Without internal heater\t3.630144\t5.331235\n",
+        "With heating tubes\t1.503916\t0.666404\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch17.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch17.ipynb
new file mode 100755
index 00000000..77951e94
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch17.ipynb
@@ -0,0 +1,446 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:82e0373b19da96ab8fd50304caf9cd3e08cf8bad10412b0998bf9110f7a5ae63"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17 : Design of Catalytic Reactors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 434\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dt=[0.081,0.205,3.6]; #Reactor diameter for the three reactors in m\n",
+      "dte=[0.04,0.12,0.70]; #Equivalent diameters for the three reactors in m\n",
+      "db=[0.05,0.057,0.07]; #Estimated bubble size in the three reactors in m\n",
+      "Kr1=1.3889;           #Kinet1ic constant for Reaction 1 in s**-1\n",
+      "Kr2=0.6111;           #Kinetic constant for Reaction 2 in s**-1\n",
+      "Kr3=0.022;            #Kinetic constant for Reaction 3 in s**-1\n",
+      "dp=60.;               #Particle size in micrometer\n",
+      "ephsilonm=0.50;       #Void fraction of fixed bed\n",
+      "ephsilonmf=0.55;      #Void fraction at minimum fluidized condition\n",
+      "umf=0.006;            #Velocity at minimum fluidization condition in m/s\n",
+      "D=2E-5;               #Diffusion coefficient of gas in m**2/s\n",
+      "gammab=0.005;         #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "uo=0.2;               #Superficial gas velocity in m/s\n",
+      "XA=0.9;               #Conversion\n",
+      "g=9.81;               #Acceleration due to gravity in square m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "Kr12=Kr1+Kr2;\n",
+      "n=len(dt);\n",
+      "i=0;\n",
+      "ubr = [0,0,0]\n",
+      "ub = [0,0,0]\n",
+      "delta = [0,0,0]\n",
+      "ephsilonf = [0,0,0]\n",
+      "gammac = [0,0,0]\n",
+      "gammae = [0,0,0]\n",
+      "Kbc = [0,0,0]\n",
+      "Kce = [0,0,0]\n",
+      "Kf12 = [0,0,0]\n",
+      "Kf3 = [0,0,0]\n",
+      "KfA = [0,0,0]\n",
+      "KfAR = [0,0,0]\n",
+      "KfAR1 = [0,0,0]\n",
+      "tou = [0,0,0]\n",
+      "y = [0,0,0]\n",
+      "SR = [0,0,0]\n",
+      "XA1 = [0,0,0]\n",
+      "y1 = [0,0,0]\n",
+      "y2 = [0,0,0]\n",
+      "tou2 = [0,0,0]\n",
+      "Lf = [0,0,0]\n",
+      "Lm = [0,0,0]\n",
+      "XA2 = [0,0,0]\n",
+      "\n",
+      "import math\n",
+      "while i<n:\n",
+      "    #Preliminary Calcualtions\n",
+      "    ubr[i]=0.711*(g*db[i])**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
+      "    ub[i]=1.55*((uo-umf)+14.1*(db[i]+0.005))*dte[i]**0.32+ubr[i];#Bubble velocity for Geldart A particles from Equation from Eqn.(6.11)\n",
+      "    delta[i]=uo/ub[i];#Fraction of bed in bubbles from Eqn.(6.29)\n",
+      "    ephsilonf[i]=1-(1-delta[i])*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
+      "    fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
+      "    gammac[i]=(1-ephsilonmf)*((3/(ubr[i]*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
+      "    gammae[i]=((1-ephsilonmf)*((1-delta[i])/delta[i]))-gammab-gammac[i];#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
+      "    Kbc[i]=4.5*(umf/db[i])+5.85*((D**0.5*g**0.25)/db[i]**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "    Kce[i]=6.77*((D*ephsilonmf*0.711*(g*db[i])**0.5)/db[i]**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
+      "    #Effective rate constant from Eqn.(12.32)\n",
+      "    Kf12[i]=(gammab*Kr12+1/((1/Kbc[i])+(1/(gammac[i]*Kr12+1/((1/Kce[i])+(1/(gammae[i]*Kr12)))))))*(delta[i]/(1-ephsilonf[i]));\n",
+      "    #Rate of reaction 2 for fluidized bed from Eqn.(12.14)\n",
+      "    Kf3[i]=(gammab*Kr3+1/((1/Kbc[i])+(1/(gammac[i]*Kr3+1/((1/Kce[i])+(1/(gammae[i]*Kr3)))))))*(delta[i]/(1-ephsilonf[i]));\n",
+      "    #Rate of raection with respect to A from Eqn.(12.35)\n",
+      "    KfA[i]=((Kbc[i]*Kce[i]/gammac[i]**2+(Kr12+Kce[i]/gammac[i]+Kce[i]/ \\\n",
+      "    gammae[i])*(Kr3+Kce[i]/gammac[i]+Kce[i]/gammae[i]))*delta[i]*Kbc[i] \\\n",
+      "    *Kr12*Kr3/(1-ephsilonf[i]))/(((Kr12+Kbc[i]/gammac[i])* \\\n",
+      "    (Kr12+Kce[i]/gammae[i])+Kr12*Kce[i]/gammac[i])*((Kr3+Kbc[i]/gammac[i])* \\\n",
+      "    (Kr3+Kce[i]/gammae[i])+Kr3*Kce[i]/gammac[i]));\n",
+      "    KfAR[i]=((Kr1/Kr12)*Kf12[i])-KfA[i];#Rate of reaction from Eqn.(12.34)\n",
+      "    KfAR1[i]=((Kr1/Kr12)*Kf12[i]);#Since KfA is small\n",
+      "    #(b)Relate Selectivity with conversion in three reactors\n",
+      "    x=-math.log(1-XA);#The term Kf12*tou in Eqn.(12.26)\n",
+      "    tou[i]=x/Kf12[i];#Residence time from Eqn.(12.26)\n",
+      "    y[i]=(KfAR1[i]/(Kf3[i]-Kf12[i]))*(math.exp(-x)-math.exp(-tou[i]*Kf3[i]));#CR/CAi from Eqn.(12.27)\n",
+      "    SR[i]=y[i]/XA;#Selectivity of R\n",
+      "    #(c)Relate exit composition to space time\n",
+      "    tou1=5;#Space time in s\n",
+      "    XA1[i]=1-math.exp(-Kf12[i]*tou1);#Conversion from Eqn.(12.26)\n",
+      "    y1[i]=((KfAR1[i]/(Kf12[i]-Kf3[i]))*(math.exp(-Kf3[i]*tou1)-math.exp(-Kf12[i]*tou1)));#CR/CAi R from Eqn.(12.27)\n",
+      "    #(d)Calculate height of bed needed to maximize production\n",
+      "    y2[i]=(KfAR1[i]/Kf12[i])*(Kf12[i]/Kf3[i])**(Kf3[i]/(Kf3[i]-Kf12[i]));#CRmax/CAi R from Eqn.(12.37)\n",
+      "    tou2[i]=math.log(Kf3[i]/Kf12[i])/(Kf3[i]-Kf12[i]);#Space time from Eqn.(38)\n",
+      "    Lf[i]=(uo/(1-ephsilonf[i]))*tou2[i];#Length of bed at fully fluidized condition from Eqn.(12.5)\n",
+      "    Lm[i]=Lf[i]*(1-ephsilonf[i])/(1-ephsilonm);#Length of bed when settled\n",
+      "    XA2[i]=1-math.exp(-Kf12[i]*tou2[i]);#Conversion from Eqn.(12.26)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Let Laboratory, Pilot plant, Semicommercial unit be Reactor 1,2 & 3 respectively'\n",
+      "print '(a)Relation between effective rate constant(Kf12) to the gas flow rate(uo)',\n",
+      "print '\\tReactor No.\\tKf12(s**-1)\\tuo(m/s)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "     print '\\t%1.0f'%i\n",
+      "     print '\\t\\t%f'%Kf12[i],\n",
+      "     print '\\t%f'%uo\n",
+      "     i=i+1;\n",
+      "\n",
+      "print '\\n(b)Relation between selectivity with conversion'\n",
+      "print '\\n\\tReactor No.\\tKf12(s**-1)\\tSR(mol R formed/mol A reacted)'\n",
+      "i=0\n",
+      "while i<n:\n",
+      "     print '\\t%1.0f'%i,\n",
+      "     print '\\t\\t%f'%Kf12[i],\n",
+      "     print '\\t%f'%SR[i]\n",
+      "     i=i+1;\n",
+      "\n",
+      "print '(c)Relation between exit compostion and space time',\n",
+      "print '\\tReactor No.\\tXA\\t\\tCR/CAi'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "     print '\\t%1.0f'%i,\n",
+      "     print '\\t\\t%f'%XA1[i],\n",
+      "     print '\\t%f'%y1[i]\n",
+      "     i=i+1;\n",
+      "\n",
+      "print '(d)Height of bed needed to maximize the production of acrylonitrile',\n",
+      "print '\\tReactor No.\\tLm(m)\\t\\tXA'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "     print '\\t%1.0f'%i,\n",
+      "     print '\\t\\t%f'%Lm[i],\n",
+      "     print '\\t%f'%XA2[i]\n",
+      "     i=i+1;\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Let Laboratory, Pilot plant, Semicommercial unit be Reactor 1,2 & 3 respectively\n",
+        "(a)Relation between effective rate constant(Kf12) to the gas flow rate(uo) \tReactor No.\tKf12(s**-1)\tuo(m/s)\n",
+        "\t0\n",
+        "\t\t0.410042 \t0.200000\n",
+        "\t1\n",
+        "\t\t0.270620 \t0.200000\n",
+        "\t2\n",
+        "\t\t0.128980 \t0.200000\n",
+        "\n",
+        "(b)Relation between selectivity with conversion\n",
+        "\n",
+        "\tReactor No.\tKf12(s**-1)\tSR(mol R formed/mol A reacted)\n",
+        "\t0 \t\t0.410042 \t0.641507\n",
+        "\t1 \t\t0.270620 \t0.618358\n",
+        "\t2 \t\t0.128980 \t0.558283\n",
+        "(c)Relation between exit compostion and space time \tReactor No.\tXA\t\tCR/CAi\n",
+        "\t0 \t\t0.871292 \t0.564802\n",
+        "\t1 \t\t0.741562 \t0.484243\n",
+        "\t2 \t\t0.475286 \t0.313823\n",
+        "(d)Height of bed needed to maximize the production of acrylonitrile \tReactor No.\tLm(m)\t\tXA\n",
+        "\t0 \t\t3.056064 \t0.956404\n",
+        "\t1 \t\t4.137401 \t0.939139\n",
+        "\t2 \t\t7.049378 \t0.897005\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 438\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "deltaHr=5.15E8; #Heat of reaction in J/k mol\n",
+      "W=5E4;          #Weight of acrylonitirle produced per 334-day year in tonnes\n",
+      "db=0.07;        #Estimated bubble size in m\n",
+      "dte=0.7;        #Equivalent diameter in m\n",
+      "Kf12=0.35;      #Effective rate constant in s**-1 from Example 1\n",
+      "dp=60;          #Particle size in micrometer\n",
+      "ephsilonm=0.50; #Void fraction of fixed bed\n",
+      "ephsilonmf=0.55;#Void fraction at minimum fluidized condition\n",
+      "T=460;          #Temperature in reactor in degree C\n",
+      "Pr=2.5;         #Pressure inside reactor in bar\n",
+      "#Feed gas composition\n",
+      "x1=1;           #Propylene\n",
+      "x2=1.1;         #Ammonia\n",
+      "x3=11;          #Air\n",
+      "do1=0.08;       #OD of heat exchanger tubes in m\\\n",
+      "L=7;            #Length of tubes in m\n",
+      "ho=300;         #Outside heat transfer coefficient in W/m**2 K\n",
+      "hi=1800;        #Inside heat transfer coefficient in W/m**2 K\n",
+      "Tc=253.4;       #Temperature of coolant in degree C\n",
+      "pi=3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Preliminary calculation\n",
+      "uo=0.46;#Superficial gas velocity from Fig.E1(a) for the value of Kf12 & db\n",
+      "tou=8;#Space time from Fig.E2(b) for highest concentraion of product R\n",
+      "Lm=uo*tou/(1-ephsilonm);\n",
+      "y=0.58;#CR/CAi from Fig.E1(c) for the value of tou & Kf12\n",
+      "XA=0.95#From Fig.E1(c) for the value of tou & Kf12\n",
+      "SR=y/XA;#Selectivity of R\n",
+      "\n",
+      "#Cross-sectional area of the reactor\n",
+      "P=W*10**3/(334*24*3600);#Production rate of acrylonitrile\n",
+      "F=(P/0.053)/(SR*XA/0.042);#Feed rate of propylene\n",
+      "V=((F*22.4*(T+273)*(x1+x2+x3))/(42*273*Pr));\n",
+      "At=V/uo;#Cross-sectional area of reactor needed for the fluidized bed\n",
+      "\n",
+      "#Heat exchanger calculation\n",
+      "q=F*XA*deltaHr/42;#Rate of heat liberation in the reactor\n",
+      "U=(ho**-1+hi**-1)**-1;#Overall heat transfer coefficient\n",
+      "deltaT=T-Tc;#Driving force for heat transfer\n",
+      "Aw=q/(U*deltaT);#Heat exchanger area required to remove q\n",
+      "Nt=Aw/(pi*do1*L);\n",
+      "li1=(At/Nt)**0.5;#Pitch for square pitch arrangement\n",
+      "dte1=4*(li1**2-(pi/4)*do1**2)/(pi*do1);\n",
+      "if dte1>dte:\n",
+      "    li=(pi/4*dte*do1+pi/4*do1**2)**0.5;#Pitch if we add dummy tubes\n",
+      "import math\n",
+      "f=li**2-pi/4*do1**2;#Fraction of bed cross section taken up by tubes\n",
+      "dt1=math.sqrt(4/pi*At/(1-f));#Reactor diameter including all its tubes\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Superficial gas velocity=%fm/s'%uo\n",
+      "print 'No. of %1.0fm tubes required=%1.0f'%(L,Nt);\n",
+      "print 'Reactor diameter=%fm'%dt1\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Superficial gas velocity=0.460000m/s\n",
+        "No. of 7m tubes required=295\n",
+        "Reactor diameter=7.173176m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 444\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "db=0.08;       #Estimated bubble size in m\n",
+      "dte=2;         #Equivalent diameter in m\n",
+      "F1=55.6;       #Feed rate of oil in kg/s\n",
+      "XA=0.63;       #Conversion\n",
+      "uo=0.6;        #Superficial gas velocity in m/s\n",
+      "T1=500.0;      #Temperature of reactor in degree C\n",
+      "T2=580.0;      #Temperature of regenerator in degree C\n",
+      "Fs=F1*23.3;    #Solid circulation rate from Ex.(15.2)\n",
+      "rhos=1200.0;   #Density of catalyst in kg/m**3\n",
+      "dpbar=60.0;    #Average particle size in micrometer\n",
+      "ephsilonm=0.50;#Void fraction of fixed bed\n",
+      "ephsilonmf=0.55;#Void fraction at minimum fluidized condition\n",
+      "umf=0.006;      #Velocity at minimum fluidization condition in m/s\n",
+      "dt=8.0;         #Diameter of reactor in m\n",
+      "D=2E-5;         #Diffusion coefficient of gas in m**2/s\n",
+      "Kr=8.6;         #Rate constant for reaction at 500 degree C in s**-1\n",
+      "Ka1=0.06;       #Rate constant for deactivatiion at 500 degree C in s**-1\n",
+      "Ka2=0.012;      #Rate constant for regeneration at 580 degree C in s**-1\n",
+      "gammab=0.005;   #Ratio of volume of dispersed solids to that of bubble phase\n",
+      "g=9.81;         #Acceleration due to gravity in square m/s**2\n",
+      "pi=3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Parameters for the fluidized reactor\n",
+      "ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)\n",
+      "ub=1.55*((uo-umf)+14.1*(db+0.005))*dte**0.32+ubr;#Bubble velocity for Geldart A particles from Equation from Eqn.(6.11)\n",
+      "delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
+      "ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)\n",
+      "fw=0.6;#Wake volume to bubble volume from Fig.(5.8)\n",
+      "gammac=(1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
+      "gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
+      "Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "Kce=6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
+      "import math\n",
+      "#Bed height versus catalyst activity in reactor\n",
+      "a1bar=0.07;#Guess value for average activity in reactor\n",
+      "x=Kr*a1bar;#Value of Kra1 to be used in the following equation\n",
+      "Kf=(gammab*x+1/((1/Kbc)+(1/(gammac*x+1/((1/Kce)+(1/(gammae*x)))))))*(delta/(1-ephsilonf));#Effective rate constant from Eqn.(12.14)\n",
+      "tou=-math.log(1-XA)/Kf;#Space time from Eqn.(12.16)\n",
+      "Lm=tou*uo/(1-ephsilonm);#Length of fixed bed for guess value of a1bar\n",
+      "a1bar1=[0.0233,0.0465,0.0698,0.0930,0.116,0.140];#Various activity values to find Lm\n",
+      "x1 = [0,0,0,0,0,0]\n",
+      "Kf1 = [0,0,0,0,0,0]\n",
+      "tou1 = [0,0,0,0,0,0]\n",
+      "Lm1 = [0,0,0,0,0,0]\n",
+      "\n",
+      "n=len(a1bar1);\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    x1[i]=Kr*a1bar1[i];\n",
+      "    Kf1[i]=(gammab*x1[i]+1/((1/Kbc)+(1/(gammac*x1[i]+1/((1/Kce)+ \\\n",
+      "    (1/(gammae*x1[i])))))))*(delta/(1-ephsilonf));\n",
+      "    #Effective rate constant from Eqn.(12.14)\n",
+      "    \n",
+      "    tou1[i]=-math.log(1-XA)/Kf1[i];#Space time from Eqn.(12.16)\n",
+      "    Lm1[i]=tou1[i]*uo/(1-ephsilonm);\n",
+      "    #Length of fixed bed for guess value of a1bar...Condition [i]\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Find the optimum size ratio for various a1bar\n",
+      "Lm=[5,6,7,8,10,12];\n",
+      "W1 = [0,0,0,0,0,0]\n",
+      "t1bar = [0,0,0,0,0,0]\n",
+      "t2bar = [0,0,0,0,0,0]\n",
+      "a1bar2 = [0,0,0,0,0,0]\n",
+      "m=len(Lm);\n",
+      "i=0;\n",
+      "while i<m:\n",
+      "    W1[i]=(pi/4)*dt**2*rhos*(1-ephsilonm)*Lm[i];#Bed weight\n",
+      "    t1bar[i]=W1[i]/Fs;#Mean residence time of solids in reactor\n",
+      "    t2bar[i]=t1bar[i]*(Ka1/Ka2)**0.5;#Mean residence time of soilds at optimum from Eqn.(16)\n",
+      "    a1bar2[i]=(Ka2*t2bar[i])/(Ka1*t1bar[i]+Ka1*t1bar[i]*Ka2*t2bar[i]+Ka2*t2bar[i]);#From Eqn.(15)...Condition (ii)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Final design values\n",
+      "Lm4=7.3;#For satisfying condition [i] & (ii)\n",
+      "a1bar3=0.0744;#By interpolation\n",
+      "x2=a1bar3*Kr;\n",
+      "W11=(pi/4)*dt**2*rhos*(1-ephsilonm)*Lm4;#Bed weight for reactor\n",
+      "t1bar1=W11/Fs;#Mean residence time of solids in reactor\n",
+      "a2bar=(1+Ka1*t1bar1)*a1bar3;#Average activity in regenrator from Eqn.(10)\n",
+      "t2bar1=t1bar1*(Ka1/Ka2)**0.5;#Mean residence time of solids in regenerator from Eqn.(16)\n",
+      "W2=W11*(t2bar1/t1bar1);#Bed weight for regenerator\n",
+      "dt2=dt*(W2/W11)**0.5;#Diameter of regenerator assuming same static bed height for reactor and regerator\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Bed height versus catalyst activity in reactor'\n",
+      "print '\\tAverage activity',\n",
+      "print '\\tLength of fixed bed(m)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '\\t%f'%a1bar1[i],\n",
+      "    print '\\t\\t%f'%Lm1[i];\n",
+      "    i=i+1;\n",
+      "\n",
+      "print 'Optimum size ratio for various activity in reactor'\n",
+      "print '\\tLength of fixed bed(m)',\n",
+      "print '\\tAverage activity'\n",
+      "i=0\n",
+      "while i<m:\n",
+      "    print '\\t%f'%Lm[i],\n",
+      "    print '\\t\\t%f'%a1bar2[i]\n",
+      "    i=i+1;\n",
+      "\n",
+      "print 'Final design values'\n",
+      "print '\\tDiameter of reactor(m):%.0f'%dt\n",
+      "print '\\tBed weight for reactor(tons):%.0f'%(W11/10**3)\n",
+      "print '\\tBed weight for regenerator(tons):%.0f'%(W2/10**3)\n",
+      "print '\\tDiameter of regenerator(m):%.0f'%(dt2);\n",
+      "print '\\tSolid circulation rate(tons/hr):%f'%(Fs*3.6);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bed height versus catalyst activity in reactor\n",
+        "\tAverage activity \tLength of fixed bed(m)\n",
+        "\t0.023300 \t\t11.059747\n",
+        "\t0.046500 \t\t7.911053\n",
+        "\t0.069800 \t\t6.756202\n",
+        "\t0.093000 \t\t6.118750\n",
+        "\t0.116000 \t\t5.696470\n",
+        "\t0.140000 \t\t5.372072\n",
+        "Optimum size ratio for various activity in reactor\n",
+        "\tLength of fixed bed(m) \tAverage activity\n",
+        "\t5.000000 \t\t0.097879\n",
+        "\t6.000000 \t\t0.086112\n",
+        "\t7.000000 \t\t0.076871\n",
+        "\t8.000000 \t\t0.069420\n",
+        "\t10.000000 \t\t0.058149\n",
+        "\t12.000000 \t\t0.050026\n",
+        "Final design values\n",
+        "\tDiameter of reactor(m):8\n",
+        "\tBed weight for reactor(tons):220\n",
+        "\tBed weight for regenerator(tons):492\n",
+        "\tDiameter of regenerator(m):12\n",
+        "\tSolid circulation rate(tons/hr):4663.728000\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch18.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch18.ipynb
new file mode 100755
index 00000000..281bc4a5
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch18.ipynb
@@ -0,0 +1,570 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3eba5bf9f8e98dcb46f4a3b03f9f7fa4b342f8fb2d0d8b697e9a50303c32aa8f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18 : The Design of Noncatalytic Gas Solid Reactors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 456\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "xA=0.08;      #Fraction of oxygen in stream\n",
+      "dp=[2,0.1];   #Particle diameter in mm\n",
+      "rhos=4130;    #Density of catalyst in kg/m**3\n",
+      "Ds=8E-6;      #Diffusion coefficient of solid in m**2/s\n",
+      "kc=0.02;      #Reaction rate constant in m/s\n",
+      "P=10**5;      #Pressure in bar\\\n",
+      "R=8.314;      #Universal gas constant\n",
+      "T=900;        #Temperature in degree C\n",
+      "mB=0.09745;   #Molecular weight of ZnS in kg/mol\n",
+      "\n",
+      "#CALCULATION\n",
+      "b=2.0/3;#Stoichiometric coefficient of ZnS in the reaction equation\n",
+      "CA=xA*P/(R*(T+273));#Concentration of Oxygen\n",
+      "rhob=rhos/mB;       #Molar density of pure solid\n",
+      "n=len(dp);\n",
+      "i=0;\n",
+      "kbar = [0,0]\n",
+      "tou = [0,0]\n",
+      "while i<n:\n",
+      "    kbar[i]=(kc**-1+(dp[i]*10**-3/(12.0*Ds)))**-1;#Average reaction rate constant from Eqn.(11)\n",
+      "    tou[i]=rhob*dp[i]*10**-3/(2*b*kbar[i]*CA);#Time for complete reaction in seconds from Eqn.(9)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Particle Size(mm)\\tAverage rate constant(m/s)\\tTime for complete reaction(min)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f\\t\\t%f\\t\\t\\t%.2f'%(dp[i],kbar[i],tou[i]/60.0);\n",
+      "    i=i+1;\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Particle Size(mm)\tAverage rate constant(m/s)\tTime for complete reaction(min)\n",
+        "2.000000\t\t0.014118\t\t\t91.49\n",
+        "0.100000\t\t0.019592\t\t\t3.30\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 457\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "xA=0.08;   #Fraction of oxygen in stream\n",
+      "dp=1;      #Particle diameter in mm\n",
+      "rhos=2200; #Density of catalyst in kg/m**3\n",
+      "kc=0.2;    #Reaction rate constant in m/s\n",
+      "mC=0.012;  #Molecular weight of carbon in kg/mol\n",
+      "P=10**5;   #Pressure in bar\\\n",
+      "R=8.314;   #Universal gas constant\n",
+      "T=900;     #Temperature in degree C\n",
+      "\n",
+      "#CALCULATION\n",
+      "b=1;#Stoichiometric coefficient of C in the reaction equation\n",
+      "CA=xA*P/(R*(T+273));#Concentration of Oxygen\n",
+      "rhob=rhos/mC;#Molar density of pure solid reactant\n",
+      "tou=rhob*10**-3/(2*b*kc*CA);#Time required for complete reaction in seconds\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The time required for complete combustion:%.1fmins'%(tou/60);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The time required for complete combustion:9.3mins\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 462\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "from numpy import zeros\n",
+      "\n",
+      "\n",
+      "#INPUT\n",
+      "dp = 110.;#Particle size in micrometer\n",
+      "T = 900.;#Temperature of roaster in degree C\n",
+      "tbar1 = [3.,10.,30.,50.];#Reported average time in min\n",
+      "XBbarr = [0.840,0.940,0.985,0.990];#Reported value of average conversion\n",
+      "tbar = 3.;\n",
+      "XBbar = 0.840;#Average conversion for tbar  =  3 mins\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Uniform-Reaction Model\n",
+      "x = (1./tbar)*(1./(1-XBbar)-1);#Term KrCA of Eqn.(20)\n",
+      "n = len(tbar1);\n",
+      "i = 0;\n",
+      "XBbar1 = zeros(n)\n",
+      "while i<  n:\n",
+      "    XBbar1[i] = 1-1./(1+x*tbar1[i]);#Average conversion umath.sing calculated value of KrCA from Eqn.(20)\n",
+      "    i = i+1;\n",
+      "\n",
+      "#Shrinking-Core, Rection Control\n",
+      "touguess = 2;#Guess value of tou\n",
+      "def solver_func(tou):#Function defined for solving the system\n",
+      "    return (1-XBbar)-(0.25*tou/tbar)+(0.05*(tou/tbar)**2)-((1./120)*(tou/tbar)**3);\n",
+      "\n",
+      "tou = fsolve(solver_func,touguess)\n",
+      "i = 0;\n",
+      "XBbar2 = zeros(n)\n",
+      "while i<n:\n",
+      "    XBbar2[i] = 1-(0.25*tou/tbar1[i])+(0.05*(tou/tbar1[i])**2)-((1./120)*(tou/tbar1[i])**3);#Average conversion umath.sing calculated value of tou from Eqn.(23)\n",
+      "    i = i+1;\n",
+      "\n",
+      "#Shrinking-Core, Diffusion Control\n",
+      "touguess1 = 2;#Guess value of tou\n",
+      "def solver_func1(tou):      #Function defined for solving the system\n",
+      "    fn = (1-XBbar)-(1./5*tou/tbar)+(19./420*(tou/tbar)**2)-(41./4620*(tou/tbar)**3)+(0.00149*(tou/tbar)**4);\n",
+      "\n",
+      "tou1 = fsolve(solver_func1,touguess1)\n",
+      "i = 0;\n",
+      "XBbar3 = zeros(n)\n",
+      "while i< n:\n",
+      "    #Average conversion umath.sing calculated value of tou from Eqn.(23)\n",
+      "    XBbar3[i] = 1-(1./5*tou1/tbar1[i])+(19./420*(tou1/tbar1[i])**2)-(41./4620*(tou1/tbar1[i])**3)+(0.00149*(tou1/tbar)**4);\n",
+      "    i = i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\t\\t\\t\\tXBbar calculated for Models';\n",
+      "print 'Reported Data';\n",
+      "print 'tbarmin)\\tXBbar\\tUniform Reaction\\tShrinking-Core%( Rection Control\\t\\tShrinking-Core, Diffusion Control'\n",
+      "i = 0\n",
+      "while i< n:\n",
+      "    print '%f\\t%f\\t%f\\t\\t%f\\t\\t\\t\\t%f'%(tbar1[i],XBbarr[i],XBbar1[i],XBbar2[i],XBbar3[i]);\n",
+      "    i = i+1;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\tXBbar calculated for Models\n",
+        "Reported Data\n",
+        "tbarmin)\tXBbar\tUniform Reaction\tShrinking-Core%( Rection Control\t\tShrinking-Core, Diffusion Control\n",
+        "3.000000\t0.840000\t0.840000\t\t0.840000\t\t\t\t0.884437\n",
+        "10.000000\t0.940000\t0.945946\t\t0.947234\t\t\t\t0.962033\n",
+        "30.000000\t0.985000\t0.981308\t\t0.981898\t\t\t\t0.987159\n",
+        "50.000000\t0.990000\t0.988701\t\t0.989075\t\t\t\t0.992366\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "/usr/lib/python2.7/dist-packages/scipy/optimize/minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last ten iterations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 462\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "from numpy import zeros\n",
+      "\n",
+      "\n",
+      "#INPUT\n",
+      "W = 1;          #Bed weight in kg\n",
+      "F1 = 0.01;      #Solid feed rate in kg/min\n",
+      "dp = [200.,600.];           #Particle size in micrometer\n",
+      "XBbar = [0.85,0.64];        #Average conversion for corresponding particle sizes\n",
+      "rhos = 2500.;               #Density of solid in kg/m**3\n",
+      "ephsilonm = 0.4;            #Void fracton of fixed bed\n",
+      "F11 = 4.;                   #Feed rate of solids in tons/hr\n",
+      "XBbar1 = 0.98;\n",
+      "dp1 = 600.;\n",
+      "pi = 3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Shrinking-Core, Rection Control\n",
+      "n = len(dp);\n",
+      "i = 0;\n",
+      "touguess = 2;#Guess value of tou\n",
+      "tou = zeros(n)\n",
+      "while i< n:\n",
+      "    def solver_func2(t):  #Function defined for solving the system\n",
+      "        return  (1-XBbar[i])-(0.25*t/107)+(0.05*(t/107.)**2)-((1./120)*(t/107.)**3)\n",
+      "    tou[i] = fsolve(solver_func2,touguess)\n",
+      "    i = i+1;\n",
+      "\n",
+      "tou1 = tou[1];\n",
+      "#For a math.single stage fluidized roaster\n",
+      "tbar1 = 0.25*(tou1/(1-XBbar1))/60.;          #Mean residence time of solids in reactor in hr from Eqn.(24)\n",
+      "W1 = F11*tbar1;\n",
+      "dtguess = 2.;                    #Guess value of tou\n",
+      "def solver_func3(dt):           #Function defined for solving the system\n",
+      "    return  W1*10**3-(pi/4.)*dt**2*0.5*dt*rhos*(1-ephsilonm);#Since Lm = 0.5dt\n",
+      "\n",
+      "dt = fsolve(solver_func3,dtguess)\n",
+      "Lm = dt/2.;#Length of bed required\n",
+      "\n",
+      "#For a two-stage fluidized roaster\n",
+      "tbar2 = tou1*math.sqrt(1./(20*(1-XBbar1)))/60;      #Mean residence time of solids in reactor in hr from Eqn.(30)\n",
+      "W2 = F11*tbar2;\n",
+      "dtguess1 = 2;                   #Guess value of tou\n",
+      "def solver_func4(dt):           #Function defined for solving the system\n",
+      "    return W2*10**3-(pi/4.)*dt**2*0.5*dt*rhos*(1-ephsilonm);       #Since Lm = 0.5dt\n",
+      "\n",
+      "dt1 = fsolve(solver_func4,dtguess)\n",
+      "Lm1 = dt1/2.;           #Length of bed required\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Single stage fluidized roaster';\n",
+      "print '\\tWeight of bed needed:%ftons'%(W1);\n",
+      "print '\\tDiameter of reactor:%fm'%(dt);\n",
+      "print '\\tLength of bed:%fm'%(Lm);\n",
+      "print 'Two-stage fluidized roaster';\n",
+      "print '\\tWeight of bed needed:%ftons'%(W2);\n",
+      "print '\\tDiameter of reactor:%fm'%(dt1);\n",
+      "print '\\tLength of bed:%fm'%(Lm1);\n",
+      "print 'These results show that this operation can be accomplished in a math.single bed of %ftons or in two beds of %f tons each.'%(W1,W2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Single stage fluidized roaster\n",
+        "\tWeight of bed needed:174.370805tons\n",
+        "\tDiameter of reactor:6.665728m\n",
+        "\tLength of bed:3.332864m\n",
+        "Two-stage fluidized roaster\n",
+        "\tWeight of bed needed:22.056356tons\n",
+        "\tDiameter of reactor:3.346064m\n",
+        "\tLength of bed:1.673032m\n",
+        "These results show that this operation can be accomplished in a math.single bed of 174.370805tons or in two beds of 22.056356 tons each.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5, Page 468\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "from numpy import zeros\n",
+      "\n",
+      "#INPUT\n",
+      "T = 900;                #Temperature in roaster in degree C\n",
+      "P = 101325;             #Pressure in Pa\n",
+      "R = 8.314;              #Universal gas consmath.tant\n",
+      "dpbar = 150;            #Average particle size in micrometer\n",
+      "rhosbar = 4130;         #Average particle density in kg/m**3\n",
+      "kc = 0.015              #Rate consmath.tant in m/s for reaction which follows shrinking core model\n",
+      "Ds = 8E-6;              #Diffusion coefficient of solid in m**2/s\n",
+      "uo = 0.6;               #Superficial gas velocity in m/s\n",
+      "D = 2.3E-4;             #Diffusion coefficient of gas in m**2/s\n",
+      "Lm = 1;                 #Length of fixed bed in m\n",
+      "dte = 0.4;              #Equivalent diameter of bed\n",
+      "umf = 0.025;            #Velocity at minimum fluidization condition in m/s\n",
+      "ephsilonm = 0.45;       #Void fraction of fixed bed\n",
+      "ephsilonmf = 0.50;      #Void fraction at minimum fluidized condition\n",
+      "db = 0.2;               #Estimated bubble size in m\n",
+      "gammab = 0.005;         #Ratio of volume of print ersed solids to that of bubble phase\n",
+      "Fo = 2;                 #Feed rate of solids in kg/s\n",
+      "XA = 0.6677;            #Conversion of Oxygen\n",
+      "xA = 0.21;              #Mole fraction of oxygen in feed\n",
+      "mB = 0.09744;           #Molecular weight of ZnS\n",
+      "F = 0.85;               #Fraction of open area\n",
+      "g = 9.81;               #Acceleration due to gravity in square m/s**2\n",
+      "pi = 3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#(a)Extreme Calculation\n",
+      "a = 3./2;#Stoichiometric coefficient of Oxygen in the reaction equation\n",
+      "At = (Fo/mB)*(a)/(uo*(273./(T+273))*(XA*xA)/0.0224);\n",
+      "dt = math.sqrt(At/F*4/pi);\n",
+      "\n",
+      "#(b)The Three-Step Procedure\n",
+      "#Step 1. Conversion of gas\n",
+      "ubr = 0.711*(g*db)**0.5;                                #Rise velocity of bubble from Eqn.(6.7)\n",
+      "ub = 1.6*((uo-umf)+1.13*db**0.5)*dte**1.35+ubr;         #Bubble rise velocity for Geldart B particle\n",
+      "\n",
+      "delta = uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)\n",
+      "ephsilonf = 1-(1-delta)*(1-ephsilonmf);                 #Void fraction of fixed bed from Eqn.(6.20)\n",
+      "\n",
+      "fw = 0.15;#Wake volume to bubble volume from Fig.(5.8)\n",
+      "gammac = (1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);        #Volume of solids in cloud to that of the bubble from Eqn.(6.36)\n",
+      "gammae = ((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;      #Volume of solids in emulsion to that of the bubble from Eqn.(6.35)\n",
+      "Kbc = 4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5./4));           #Gas interchange coefficient between bubble and cloud from Eqn.(10.27)\n",
+      "Kce = 6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;       #Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)\n",
+      "x = delta*Lm*(1-ephsilonm)/((1-ephsilonf)*uo);                  #Term Lf/ub of Eqn.(12.16) from Eqn.(6.19)\n",
+      "CAi = xA*P/(R*(T+273));                                         #Initial concentration of oxygen\n",
+      "\n",
+      "#Step 2.Conversion of solids\n",
+      "rhob = rhosbar/mB;                                              #Density of ZnS\n",
+      "kbar = (kc**-1+(dpbar*10**-6/(12*Ds))**-1)**-1;                 #Modified rate consmath.tant from Eqn.(11)\n",
+      "tbar = At*Lm*(1-ephsilonm)*rhosbar/Fo;                          #Mean residence time of solids\n",
+      "Krguess = 2;                                                    #Guess value of Kr\n",
+      "def solver_func(Kr):                #Function defined for solving the system\n",
+      "    Kf = gammab*Kr+1/((1./Kbc)+(1./(gammac*Kr+1/((1./Kce)+(1./(gammae*Kr))))));            #Reaction rate for fluidized bed from Eqn.(14)\n",
+      "    XA = 1-math.exp(-x*Kf);                                                         #Conversion of oxygen from Eqn.(42)\n",
+      "    CAbar = (CAi*XA*uo)/(Kr*Lm*(1-ephsilonm));                                      #Average concentration of oxygen from Eqn.(43)\n",
+      "    tou = rhob*dpbar*10**-6*a/(2*kbar*CAbar);                                       #Time for complete reaction from Eqn.(9)\n",
+      "    y = tbar/tou;#Term tbar/tou\n",
+      "    XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1/y));                                   #Average conversion of ZnS from Eqn.(22)\n",
+      "    #Step 3. Material balance of both streams\n",
+      "    return (Fo/mB)*XBbar-(At*uo*CAi*XA/a);                                          #From Eqn.(44b)\n",
+      "\n",
+      "Kr = fsolve(solver_func,Krguess)\n",
+      "Kf = gammab*Kr+1/((1/Kbc)+(1/(gammac*Kr+1/((1/Kce)+(1/(gammae*Kr))))));             #Reaction rate for fluidized bed from Eqn.(14)\n",
+      "XA = 1-math.exp(-x*Kf);                                                             #Conversion of oxygen from Eqn.(42)\n",
+      "CAbar = (CAi*XA*uo)/(Kr*Lm*(1-ephsilonmf));                                         #Average concentration of oxygen from Eqn.(43)\n",
+      "tou = rhob*dpbar*10**-6*a/(2*kbar*CAbar);                                           #Time for complete reaction from Eqn.(9)\n",
+      "y = tbar/tou;                                                                       #Term tbar/tou\n",
+      "XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1/y));                                       #Average conversion of ZnS from Eqn.(22)\n",
+      "\n",
+      "\n",
+      "#(c) For other feed rates of solids\n",
+      "F1 = [2,2.5,3,3.5];#Various feed rates of solids in kg/s\n",
+      "n = len(F1)\n",
+      "i = 0;\n",
+      "tbar1 = zeros(n)\n",
+      "Kr1 = zeros(n)\n",
+      "XA1 = zeros(n)\n",
+      "y1 = zeros(n)\n",
+      "tou1 = zeros(n)\n",
+      "XBbar1 = zeros(n)\n",
+      "Kf1 = zeros(n)\n",
+      "CAbar1 = zeros(n)\n",
+      "Krguess1 = 2;           #Guess value of Kr\n",
+      "while i<  n:\n",
+      "    tbar1[i] = At*Lm*(1-ephsilonm)*rhosbar/F1[i];#Mean residence time of solids\n",
+      "    def solver_func1(Kr):               #Function defined for solving the system\n",
+      "        Kf1 = gammab*Kr+1/((1/Kbc)+(1/(gammac*Kr+1/((1/Kce)+(1/(gammae*Kr))))));#Reaction rate for fluidized bed from Eqn.(14)\n",
+      "        XA1 = 1-math.exp(-x*Kf1);#Conversion of oxygen from Eqn.(42)\n",
+      "        CAbar1 = (CAi*XA1*uo)/(Kr*Lm*(1-ephsilonm));#Average concentration of oxygen from Eqn.(43)\n",
+      "        tou1 = rhob*dpbar*10**-6*a/(2*kbar*CAbar1);#Time for complete reaction from Eqn.(9)\n",
+      "        y1[i] = tbar1[i]/tou1;#Term tbar/tou\n",
+      "        XBbar1[i] = 3*y1[i]-6*y1[i]**2+6*y1[i]**3*(1-math.exp(-1/y1[i]));#Average conversion of ZnS from Eqn.(22)\n",
+      "        #Step 3. Material balance of both streams\n",
+      "        return (F1[i]/mB)*XBbar1[i]-(At*uo*CAi*XA1/a);#From Eqn.(44b)\n",
+      "\n",
+      "    Kr1[i] = fsolve(solver_func1,Krguess1)\n",
+      "    Kf1[i] = gammab*Kr1[i]+1/((1/Kbc)+(1/(gammac*Kr1[i]+1/((1/Kce)+(1/(gammae*Kr1[i]))))));#Reaction rate for fluidized bed from Eqn.(14)\n",
+      "    XA1[i] = 1-math.exp(-x*Kf1[i]);#Conversion of oxygen from Eqn.(42)\n",
+      "    CAbar1[i] = (CAi*XA1[i]*uo)/(Kr1[i]*Lm*(1-ephsilonmf));#Average concentration of oxygen from Eqn.(43)\n",
+      "    tou1[i] = rhob*dpbar*10**-6*a/(2*kbar*CAbar1[i]);#Time for complete reaction from Eqn.(9)\n",
+      "    y1[i] = tbar1[i]/tou1[i];#Term tbar/tou\n",
+      "    XBbar1[i] = 3*y1[i]-6*y1[i]**2+6*y1[i]**3*(1-math.exp(-1/y1[i]));#Average conversion of ZnS from Eqn.(22)\n",
+      "    i = i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Extreme Calculation';\n",
+      "print '\\tDiameter of tube with all its internals:%fm'%(dt);\n",
+      "print 'Three step procedure';\n",
+      "print '\\tConversion of ZnS:%f'%(XBbar);\n",
+      "print 'For other feed rates of solids';\n",
+      "print '\\tFeedkg/s\\ttbars\\t\\tXBbar/XA\\tKrbars**-1\\tCAbar/CAi\\ttous\\t\\tXA\\t\\tXB';\n",
+      "i = 0;\n",
+      "while i<  n:\n",
+      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f\\t%f\\t%f\\t%f'%(F1[i],tbar1[i],XBbar1[i]/XA1[i],Kr1[i],CAbar1[i]/CAi,tou1[i],XA1[i],XBbar1[i]);\n",
+      "    i = i+1;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Extreme Calculation\n",
+        "\tDiameter of tube with all its internals:7.265465m\n",
+        "Three step procedure\n",
+        "\tConversion of ZnS:0.992491\n",
+        "For other feed rates of solids\n",
+        "\tFeedkg/s\ttbars\t\tXBbar/XA\tKrbars**-1\tCAbar/CAi\ttous\t\tXA\t\tXB\n",
+        "\t2.000000\t40003.518104\t1.498777\t6.529973\t0.121691\t1208.748626\t0.662201\t0.992491\n",
+        "\t2.500000\t32002.814483\t1.201348\t23.149964\t0.041993\t3502.838402\t0.810111\t0.973225\n",
+        "\t3.000000\t26669.012069\t1.007581\t76.183481\t0.014228\t10338.227251\t0.903294\t0.910141\n",
+        "\t3.500000\t22859.153202\t0.871565\t146.263086\t0.007742\t18998.458994\t0.943693\t0.822490\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6, Page 471\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "\n",
+      "#INPUT\n",
+      "T = 900;                #Temperature in roaster in degree C\n",
+      "P = 101325;             #Pressure in Pa\n",
+      "R = 8.314;              #Universal gas consmath.tant\n",
+      "dp = 750;               #Particle size in micrometer5\n",
+      "Fo = 2.5;               #Feed rate of solids in kg/s\n",
+      "uo = 0.6;               #Superficial gas velocity in m/s\n",
+      "W = 80140;              #Weight of bed in kg\n",
+      "ephsilonmf = 0.50;      #Void fraction at minimum fluidized condition\n",
+      "umf = 0.5;              #Velocity at minimum fluidization condition in m/s\n",
+      "db = 0.2;               #Estimated bubble size in m\n",
+      "g = 9.81;               #Acceleration due to gravity in square m/s**2\n",
+      "Lm = 1;                 #Length of fixed bed in m\n",
+      "ephsilonm = 0.45;       #Void fraction of fixed bed\n",
+      "xA = 0.21;              #Mole fraction of oxygen in feed\n",
+      "kc = 0.015              #Rate consmath.tant in m/s for reaction which follows shrinking core model\n",
+      "Ds = 8E-6;              #Diffusion coefficient of solid in m**2/s\n",
+      "rhosbar = 4130;         #Average particle density in kg/m**3\n",
+      "mB = 0.09744;           #Molecular weight of ZnS\n",
+      "a = 3./2;               #Stoichiometric coefficient of Oxygen in the reaction equation\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Selection of models to represent reactor\n",
+      "ubr = 0.711*(g*db)**0.5;        #Rise velocity of bubble from Eqn.(6.7)\n",
+      "f = ubr/(umf/ephsilonmf);\n",
+      "\n",
+      "#Step 1.\n",
+      "ub = uo-umf+ubr;            #Rise velocity of bubbles from Eqn.(6.8)\n",
+      "delta = (uo-umf)/(ub+2*umf);        #Fraction of the bed in bubbles from Eqn.(6.26)\n",
+      "Krguess = 2;                        #Guess value of Kr\n",
+      "x = Lm*(1-ephsilonm)*umf*(1-delta)/uo**2;\n",
+      "CAi = xA*P/(R*(T+273));             #Initial concentration of oxygen\n",
+      "\n",
+      "#Step 2.\n",
+      "kbar = (kc**-1+(dp*10**-6/(12*Ds))**-1)**-1;            #Modified rate consmath.tant from Eqn.(11)\n",
+      "tbar = W/Fo;                    #Mean residence time of solids from Eqn.(14.2)\n",
+      "rhob = rhosbar/mB;              #Density of ZnS\n",
+      "def solver_func1(Kr):           #Function defined for solving the system\n",
+      "    XA = 1-math.exp(-x*Kr);     #Conversion from Eqn.(42)\n",
+      "    CAbar = (CAi*XA*uo**2)/(Kr*Lm*(1-ephsilonm)*umf*(1-delta));         #Average concentration of oxygen from Eqn.(43)\n",
+      "    tou = rhob*dp*10**-6*a/(2*kbar*CAbar);          #Time for complete reaction from Eqn.(9)\n",
+      "    y = tbar/tou;               #Term tbar/tou\n",
+      "    XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1./y));          #Average conversion of ZnS from Eqn.(22)\n",
+      "    return XBbar-1.2*XA;        #From Table E5, for Fo = 2.5kg/s\n",
+      "\n",
+      "Kr = fsolve(solver_func1,Krguess)\n",
+      "XA = 1-math.exp(-x*Kr);         #Conversion from Eqn.(42)\n",
+      "CAbar = (CAi*XA*uo**2)/(Kr*Lm*(1-ephsilonm)*umf*(1-delta))      #Average concentration of oxygen from Eqn.(43)\n",
+      "tou = rhob*dp*10**-6*a/(2*kbar*CAbar);          #Time for complete reaction from Eqn.(9)\n",
+      "y = tbar/tou;                   #Term tbar/tou\n",
+      "XBbar = 3*y-6*y**2+6*y**3*(1-math.exp(-1./y));  #Average conversion of ZnS from Eqn.(22)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Selection of models to represent reactor';\n",
+      "print '\\tSince ratio ubr/umf/ephsilonmf) =  %f <1 the reactor is operating in slow bubble regime'%(f);\n",
+      "print '\\tSince particle size  = %f micrometer they react according to shrinking-core model'%(dp);\n",
+      "print '\\tConversion obtained for %f micrometer particle:%f'%(dp,XBbar);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Selection of models to represent reactor\n",
+        "\tSince ratio ubr/umf/ephsilonmf) =  0.995908 <1 the reactor is operating in slow bubble regime\n",
+        "\tSince particle size  = 750.000000 micrometer they react according to shrinking-core model\n",
+        "\tConversion obtained for 750.000000 micrometer particle:0.988127\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch3.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch3.ipynb
new file mode 100755
index 00000000..72abab33
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch3.ipynb
@@ -0,0 +1,267 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c2d23a740208e3823c43d8afcfc0ef305894f28fcd5f9e9793f1f1c6fe8b25f1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Fluidization and Mapping of Regimes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 68\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "weight = [0,60,150,270,330,360.0];        # Weight in grams for the oversized particles\n",
+      "psize = [50,75,100,125,150,175];          #PSD in micrometers\n",
+      "\n",
+      "#CALCULATION\n",
+      "l = len(psize); # To obtain the size of input array\n",
+      "# Computation of sauter mean diameter for the given PSD\n",
+      "i = 0;\n",
+      "dpi = [0,0,0,0,0,0]\n",
+      "weightf = [0,0,0,0,0,0]\n",
+      "dp = [0,0,0,0,0,0]\n",
+      "while i<l-1:\n",
+      "        dpi[i]=(psize[i]+ psize[i+1])/2.0;\n",
+      "        weightf[i]=(weight[i+1]-weight[i])/weight[5];    \n",
+      "        dp[i]=weightf[i]/float(dpi[i]);    \n",
+      "        i=i+1;\n",
+      "\n",
+      "dpbar=1/sum(dp);                        #Calculation of average particle daimeter Eq.(15)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\n The Sauter mean diameter of the material with the given particle size distribution = %.0f micrometer'%dpbar\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The Sauter mean diameter of the material with the given particle size distribution = 98 micrometer\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 76\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Estimation of Minimum fluidizing velocity\n",
+      "\n",
+      "#INPUT\n",
+      "ephsilon=0.55;     #Void fraction of bed\n",
+      "rhog=0.0012;       #Density of gas in g/cc\n",
+      "myu=.00018;        #Viscosity of gas in g/cm s\n",
+      "dpbar=0.016;       #Mean diameter of solids in centimeter\n",
+      "phis=0.67;         #Sphericity of solids\n",
+      "rhos=2.6;          #Density of solids in g/cc\n",
+      "g=980;             #Acceleration due to gravity in square cm/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Computation of umf using the simplified equation for small particles\n",
+      "umf=((dpbar**2)*(rhos-rhog)*g*(ephsilon**3)*(phis**2))/(150*myu*(1-ephsilon));#Simplified equation to calculate minimum fluidizing velocity \n",
+      "                                                                              #for small particles Eq.(21)\n",
+      "Re=(dpbar*umf*rhog)/myu;#To calculate Reynolds number for particle\n",
+      "\n",
+      "#Computation of umf if neither void fraction of bed nor sphericity is known\n",
+      "c1=28.7\n",
+      "c2=0.0494;                      #Value of constants from Table 4, page 70\n",
+      "umf1=(myu/(dpbar*rhog))*(((c1**2)+((c2*(dpbar**3)*rhog*(rhos-rhog)*g)/(myu**2)))**0.5-c1); #Equation to calculate minimum fluidizing velocity \n",
+      "                                                                                           #for coarse particles Eq.(25)\n",
+      "err=((umf-umf1)/umf)*100;       #Calculation of error from experimental value\n",
+      "\n",
+      "#OUTPUT\n",
+      "if Re<20:\n",
+      "\tprint 'The particle Reynolds no = %f'%Re\n",
+      "\tprint 'The simplified equation used for calculating minimum fluidizing velocity is valid.'\n",
+      "\n",
+      "print 'The minimum fluidizing velocity by simplified equation for small particles = %.2fcm/s'%umf\n",
+      "print 'The minimum fluidizing velocity by equation for coarse partilces = %.2fcm/s'%umf1\n",
+      "print 'This value is %d percent below the experimentally reported value.'%err\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The particle Reynolds no = 0.427493\n",
+        "The simplified equation used for calculating minimum fluidizing velocity is valid.\n",
+        "The minimum fluidizing velocity by simplified equation for small particles = 4.01cm/s\n",
+        "The minimum fluidizing velocity by equation for coarse partilces = 3.10cm/s\n",
+        "This value is 22 percent below the experimentally reported value.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 82\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "rhog=1.2e-3;    #Density of air in g/cc\n",
+      "myu=1.8e-4      #Viscosity of air in g/cm s\n",
+      "dpbar=0.016     #Mean diameter of solids in centimeter\n",
+      "phis=0.67;      #Sphericity of solids\n",
+      "rhos=2.6;       #Density of solids in g/cc\n",
+      "g=980           #Acceleration due to gravity in square cm/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "dpstar=dpbar*((rhog*(rhos-rhog)*g)/myu**2)**(1/3.0);              #Calculation of dimensionless particle size Eq.(31)\n",
+      "utstar=((18/(dpstar**2))+(2.335-(1.744*phis))/(dpstar**0.5))**-1; #Calculation of dimensionless gas velocity Eq.(33)\n",
+      "ut=utstar*((myu*(rhos-rhog)*g)/rhog**2)**(1/3.0);                 #Calculation of terminal velocity of falling particles Eq.(32)\n",
+      "\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The dimensionless particle size = %.2f'%dpstar\n",
+      "print 'The dimensionless gas velocity = %.3f'%utstar\n",
+      "print 'The terminal velocity of falling particles = %d cm/s'%ut\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dimensionless particle size = 7.28\n",
+        "The dimensionless gas velocity = 1.296\n",
+        "The terminal velocity of falling particles = 88 cm/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 91\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "rhos=1.5;                 #Density of Solid in g/cc\n",
+      "uo1=40; uo2=80;           #Superficial gas velocity in cm/s\n",
+      "dp1=0.006; dp2=0.045;     #Particle size in centimeter\n",
+      "rhog1=1.5E-3; rhog2=1E-3; #Density of gas in g/cc\n",
+      "myu1=2E-4; myu2=2.5E-4;   #Viscosity of air in g/cm s\n",
+      "g=980;                    #Acceleration due to gravity in square cm/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "#for smaller particles\n",
+      "dpstar1=dp1*((rhog1*(rhos-rhog1)*g)/myu1**2)**(1/3.0);     #Calculation of dimensionless particle diamter Eq.(31)\n",
+      "uostar1=uo1*((rhog1**2)/((myu1)*(rhos-rhog1)*g))**(1/3.0);\n",
+      "uostar2=uo2*((rhog1**2)/((myu1)*(rhos-rhog1)*g))**(1/3.0); #Calculation of dimensionless superficial gas velocity Eq.(32)\n",
+      "\n",
+      "#for larger particles \n",
+      "dpstar2=dp2*((rhog2*(rhos-rhog2)*g)/myu2**2)**(1/3.0);     #Calculation of dimensionless particle diamter Eq.(31)\n",
+      "uostar3=uo1*((rhog2**2)/((myu2)*(rhos-rhog2)*g))**(1/3.0);\n",
+      "uostar4=uo2*((rhog2**2)/((myu2)*(rhos-rhog2)*g))**(1/3.0); #Calculation of dimensionless superficial gas velocity Eq.(32)\n",
+      "\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'For particle of size %.3f centimeter'%dp1\n",
+      "print 'The dimensionless particle diameter = %.2f'%dpstar1\n",
+      "print 'The dimensionless superficial gas velocity = %.4fcm/s(for superficial gas velocity of %dcm/s)'%(uostar1,uo1)\n",
+      "print 'The dimensionless superficial gas velocity = %.3fcm/s(for superficial gas velocity of %dcm/s)'%(uostar2,uo2)\n",
+      "print 'From Fig.16(page 89)comparing u*=%.4f vs dp*=%.2f'%(uostar1,dpstar1)\n",
+      "print 'For Superficial gas velocity =%d Mode of Fluidization:Onset of turbulent fluidization in an ordinary bubbling bed'%(uo1)\n",
+      "print 'From Fig.16(page 89)comparing u* =%.3f vs dp* =%f'%(uostar2,dpstar1)\n",
+      "print 'For Superficial gas velocity =%f Mode of Fluidization:Fast fluidization(requires a circulating solid system)'%(uo2)\n",
+      "print 'For particle of size %f centimeter'%(dp2)\n",
+      "print 'The dimensionless particle diameter = %f'%(dpstar2)\n",
+      "print 'The dimensionless superficial gas velocity = %fcm/s(for superficial gas velocity of %fcm/s)'%(uostar3,uo1)\n",
+      "print 'The dimensionless superficial gas velocity = %fcm/s(for superficial gas velocity of %fcm/s)'%(uostar4,uo2)\n",
+      "print 'From Fig.16(page 89)comparing u*=%f vs dp*=%f'%(uostar3,dpstar2)\n",
+      "print 'For Superficial gas velocity =%f Mode of Fluidization:Bublling Fluidization'%(uo1)\n",
+      "print 'From Fig.16(page 89)comparing u* =%f vs dp* =%f'%(uostar4,dpstar2)\n",
+      "print 'For Superficial gas velocity =%f Mode of Fluidization:Bubbling Fluidization'%(uo2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For particle of size 0.006 centimeter\n",
+        "The dimensionless particle diameter = 2.28\n",
+        "The dimensionless superficial gas velocity = 0.7885cm/s(for superficial gas velocity of 40cm/s)\n",
+        "The dimensionless superficial gas velocity = 1.577cm/s(for superficial gas velocity of 80cm/s)\n",
+        "From Fig.16(page 89)comparing u*=0.7885 vs dp*=2.28\n",
+        "For Superficial gas velocity =40 Mode of Fluidization:Onset of turbulent fluidization in an ordinary bubbling bed\n",
+        "From Fig.16(page 89)comparing u* =1.577 vs dp* =2.282737\n",
+        "For Superficial gas velocity =80.000000 Mode of Fluidization:Fast fluidization(requires a circulating solid system)\n",
+        "For particle of size 0.045000 centimeter\n",
+        "The dimensionless particle diameter = 12.890262\n",
+        "The dimensionless superficial gas velocity = 0.558561cm/s(for superficial gas velocity of 40.000000cm/s)\n",
+        "The dimensionless superficial gas velocity = 1.117122cm/s(for superficial gas velocity of 80.000000cm/s)\n",
+        "From Fig.16(page 89)comparing u*=0.558561 vs dp*=12.890262\n",
+        "For Superficial gas velocity =40.000000 Mode of Fluidization:Bublling Fluidization\n",
+        "From Fig.16(page 89)comparing u* =1.117122 vs dp* =12.890262\n",
+        "For Superficial gas velocity =80.000000 Mode of Fluidization:Bubbling Fluidization\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch4.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch4.ipynb
new file mode 100755
index 00000000..cbd2e1e4
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch4.ipynb
@@ -0,0 +1,306 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:66650a76faf5eec3efd8e3c62ecb57addf32e1a6b2cb5d90995ca785777b9427"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 :  The Dense Bed"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 106\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Variable declaration\n",
+      "dt=4;               #Vessel diameter in m\n",
+      "Lmf=2;              #Length of the bed in m\n",
+      "ephsilonmf=0.48;    #Void fraction of bed\n",
+      "rhos=1500;          #Density of solid in kg/m**3\n",
+      "rhog=3.6;           #Density of gas in kg/m**3\n",
+      "myu=2E-5;           #Viscosity of gas in kg/m s\n",
+      "po=3;               #Pressure of inlet gas in bar\n",
+      "uo=0.4;             #Superficial velocity of gas in m/s\n",
+      "uorm=40;            #Maximum allowable jet velocity from holes in m/s\n",
+      "g=9.80;             #Acceleration due to gravity in m/s**2\n",
+      "gc=1;\n",
+      "pi=3.1428;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Computation of minimum allowable pressure drop through the distributor\n",
+      "deltapb=((1-ephsilonmf)*(rhos-rhog)*g*Lmf)/gc;  #Calculation of pressure drop in bed using Eqn.(3.17)\n",
+      "deltapd=0.3*deltapb;                            #Calculation of pressure drop in distributor using Eqn.(3)\n",
+      "\n",
+      "#Computation of orifice coefficient\n",
+      "Ret=(dt*uo*rhog)/myu;\n",
+      "if Ret>=3000:\n",
+      "    Cd=0.60;\n",
+      "elif Ret>=2000:\n",
+      "    Cd=0.61;\n",
+      "elif Ret>=1000:\n",
+      "    Cd=0.64;\n",
+      "elif Ret>=500:\n",
+      "    Cd=0.68;\n",
+      "elif Ret>=300:\n",
+      "    Cd=0.70;\n",
+      "elif Ret>=100:\n",
+      "    Cd=0.68;\n",
+      "\n",
+      "#Computation of gas velocity through orifice\n",
+      "uor=Cd*((2*deltapd)/rhog)**0.5;   #Calculation of gas velocity through orifice by using Eqn.(12)\n",
+      "f=(uo/uor)*100;                   #Calculation of fraction of open area in the perforated plate \n",
+      "\n",
+      "\n",
+      "#Computation of number of orifices per unit area of distributor\n",
+      "dor=[0.001,0.002,0.004];          #Different orifice diameters in m\n",
+      "n=len(dor);\n",
+      "i=0;\n",
+      "Nor = [0.,0.,0.]\n",
+      "while i<n:\n",
+      "    Nor[i]=(uo*4)/(pi*uor*(dor[i])**2);#Calculation of number of orifices by using Eqn.(13)\n",
+      "    i=i+1;\n",
+      "    \n",
+      "#OUTPUT\n",
+      "print 'The pressure drop in bed:%fPa'%deltapb\n",
+      "print 'The minimum allowable pressure drop in distributor:%fPa'%deltapd\n",
+      "if uor<uorm:\n",
+      "    print 'The gas veleocity of %fm/s is satisfactory'%uor\n",
+      "else:\n",
+      "    print 'The gas veleocity of %fm/s is not satisfactory'%uor\n",
+      "\n",
+      "if f<10:\n",
+      "    print 'The fraction of open area of %f percent is allowable'%f\n",
+      "else:\n",
+      "    print 'The fraction of open area of %f percent is not allowable'%f\n",
+      "\n",
+      "print 'Diameter of orifice(m)',\n",
+      "print '\\tNumber of orifices per unit area(per sq.m)'\n",
+      "\n",
+      "j=0;\n",
+      "while j<n:\n",
+      "    print '%f'%dor[j],\n",
+      "    print '\\t\\t%f'%Nor[j]\n",
+      "    j=j+1;\n",
+      "\n",
+      "print 'This number can be rounded off.'\n",
+      "print 'Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter %fm'%dor[2]\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure drop in bed:15251.308800Pa\n",
+        "The minimum allowable pressure drop in distributor:4575.392640Pa\n",
+        "The gas veleocity of 30.250265m/s is satisfactory\n",
+        "The fraction of open area of 1.322302 percent is allowable\n",
+        "Diameter of orifice(m) \tNumber of orifices per unit area(per sq.m)\n",
+        "0.001000 \t\t16829.610145\n",
+        "0.002000 \t\t4207.402536\n",
+        "0.004000 \t\t1051.850634\n",
+        "This number can be rounded off.\n",
+        "Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter 0.004000m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 108\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "lor=0.1;      #Minimum allowable tuyere spacing in m\n",
+      "uorm=30;      #Maximum allowable jet velocity from the tuyere in m/s\n",
+      "uo=0.4;       #Superficial velocity of gas in m/s\n",
+      "uor=30.2;     #Gas velocity through orifice,from Exa 1, in m/s\n",
+      "Cd=0.6;       #Dicharge coefficient from Exa 1\n",
+      "rhog=3.6      #Density of gas in kg/m**3\n",
+      "pi=3.1428;\n",
+      "\n",
+      "#CALCULATION\n",
+      "Nor=1/(lor**2);                     #Calculation of number of orifices per unit area by assuming minimum spacing for tuyeres\n",
+      "dor=((4/pi)*(uo/uor)*(1/Nor))**0.5; #Calculation of diameter of inlet orifiec by using Eqn.(13)\n",
+      "\n",
+      "#Computation of diameter of hole for different number of holes per tuyere\n",
+      "q=(lor**2)*uo;                      #Volumetric flow rate in m**3/s\n",
+      "Nh=[8.,6.,4.];                         #Different number of holes per tuyere\n",
+      "n=len(Nh);\n",
+      "i=0;\n",
+      "dh = [0.,0.,0.]\n",
+      "while i<n:\n",
+      "    dh[i]=((((q/Nh[i])*(4./pi))/uorm)**0.5);#Calculation of diameter of holes\n",
+      "    i=i+1;\n",
+      "\n",
+      "deltaph=(rhog/2.)*((uor/Cd)**2)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Number of holes(number of holes/tuyeres)',\n",
+      "print '\\tDiameter of hole(m)'\n",
+      "j=0;\n",
+      "while j<n:\n",
+      "    print '%f'%Nh[j],\n",
+      "    print '\\t\\t\\t\\t\\t%f'%dh[j]\n",
+      "    j=j+1;\n",
+      "\n",
+      "print 'The design chosen is as follows'\n",
+      "print '\\tTuyeres are as shown in Fig.2(b),page 97'\n",
+      "print '\\tNumber of holes = %f(Since rectangular pitch is chosen for tuyeres)'%Nh[1]\n",
+      "print '\\tDiameter of hole = %fm'%dh[1]\n",
+      "print '\\tDiameter of incoming high-pressure-drop orifice = %fm ID'%dor\n",
+      "print 'Checking the pressure drop in tuyeres'\n",
+      "print 'Since pressure drop of %.1f Pa gives sufficiently high \\\n",
+      "distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.'%deltaph\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of holes(number of holes/tuyeres) \tDiameter of hole(m)\n",
+        "8.000000 \t\t\t\t\t0.004606\n",
+        "6.000000 \t\t\t\t\t0.005318\n",
+        "4.000000 \t\t\t\t\t0.006513\n",
+        "The design chosen is as follows\n",
+        "\tTuyeres are as shown in Fig.2(b),page 97\n",
+        "\tNumber of holes = 6.000000(Since rectangular pitch is chosen for tuyeres)\n",
+        "\tDiameter of hole = 0.005318m\n",
+        "\tDiameter of incoming high-pressure-drop orifice = 0.012984m ID\n",
+        "Checking the pressure drop in tuyeres\n",
+        "Since pressure drop of 4560.2 Pa gives sufficiently high distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 110\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "deltapd=[3,10]    #Distributor pressure drop in kPa\n",
+      "deltapd2=10.0;    #Distributor pressure drop in kPa\n",
+      "po=101.0;         #Entering air pressure in kPa\n",
+      "To=20.0;          #Entering air temperature in degree C\n",
+      "y=1.4;            #Fugacity of air\n",
+      "deltapb=10;       #Pressure drop in bed in kPa\n",
+      "p3=103;           #Pressure at the bed exit in kPa\n",
+      "F=8;              #Feed rate of coal in tons/hr\n",
+      "H=25;             #Gross heatig value of coal in MJ/kg\n",
+      "Fa=10;            #Air required at standard condition in nm**3/kg\n",
+      "etac=0.75;        #Efficiency of compressor\n",
+      "etap=36;          #Efficiency of plant in %\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Calculation of volumetric flow rate of air\n",
+      "vo=((F*1000)*Fa*((To+273)/273.0))/3600.0;\n",
+      "\n",
+      "#Case(a) Distributor Pressure drop = 3kPa and Case(b) Distributor Pressure drop = 10kPa\n",
+      "n=len(deltapd);\n",
+      "i=0;\n",
+      "p1= [0,0]\n",
+      "p2 = [0,0]\n",
+      "ws = [0.,0.]\n",
+      "while i<n:\n",
+      "    p2[i]=p3+deltapb;      #Calculation of pressure at the entrance of the bed\n",
+      "    p1[i]=p2[i]+deltapd[i];#Calculation of pressure before entering the bed\n",
+      "    ws[i]=(y/(y-1))*po*vo*((p1[i]/po)**((y-1)/y)-1)*(1.0/etac);#Calculation of power required for the compressor by Eqn.(18) & Eqn.(20)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Case(c) 50% of the required bypassed to burn the volatile gases. Distributor Pressure drop = 3kPa\n",
+      "#No change in pressure drop from case(a)\n",
+      "v1=vo/2.0;   #New volumetric flow rate of air\n",
+      "ws1=455/2.0; #Power required for blower for primary air\n",
+      "ws2=(y/(y-1))*po*v1*((p3/po)**((y-1)/y)-1)*(1/etac);#Power required for blower for bypassed air\n",
+      "wst=ws1+ws2;                #Total power required for the two blowers\n",
+      "p=((ws[1]-wst)/ws[1])*100;  #Saving in power when compared to case(a)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Case(a)'\n",
+      "print '\\tVolumetric flow rate of air = %.2f m**3/hr'%vo\n",
+      "print '\\tPower required for compressor = %.0f kW'%ws[0]\n",
+      "print 'Case(b)'\n",
+      "print '\\tVolumetric flow rate of air = %.2f m**3/hr'%vo\n",
+      "print '\\tPower required for compressor = %.0f kW'%ws[1]\n",
+      "print 'Case(c)'\n",
+      "print '\\tVolumetric flow rate of air = %.3f m**3/hr'%v1\n",
+      "print '\\tPower required for compressor for primary air = %.1f kW'%ws1\n",
+      "print '\\tPower required for blower for bypassed air = %.1f kW'%ws2\n",
+      "print '\\tTotal power required for the two blowers = %.0f kW'%wst\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Case(a)\n",
+        "\tVolumetric flow rate of air = 23.85 m**3/hr\n",
+        "\tPower required for compressor = 454 kW\n",
+        "Case(b)\n",
+        "\tVolumetric flow rate of air = 23.85 m**3/hr\n",
+        "\tPower required for compressor = 651 kW\n",
+        "Case(c)\n",
+        "\tVolumetric flow rate of air = 11.925 m**3/hr\n",
+        "\tPower required for compressor for primary air = 227.5 kW\n",
+        "\tPower required for blower for bypassed air = 31.6 kW\n",
+        "\tTotal power required for the two blowers = 259 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch5.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch5.ipynb
new file mode 100755
index 00000000..791c6af2
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch5.ipynb
@@ -0,0 +1,137 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:502044f71f0f4f73ada9ff3f4a68fa509018810bcbdaf16eaa17d381758e9755"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Bubbles in Dense Beds"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 126\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dt=60.0;        #ID of tube in cm \n",
+      "dp=300;         #Size of particles of bed in micrometers\n",
+      "umf=3;          #Velocity at minimum fluidization condition in cm/s\n",
+      "ephsilonmf=0.5; #Void fraction of bed at minimum fluidization condition\n",
+      "db=5;           #Diameter of bubble in cm\n",
+      "g=980;          #Acceleration due to gravity in cm/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Computation of rise velocity of bubble\n",
+      "if (db/dt)<0.125:\n",
+      "    ubr=(0.711*((g*db)**0.5));#Rise velocity by Eqn.(3)\n",
+      "elif (db/dt)<0.6:\n",
+      "    ubr=(0.711*((g*db)**0.5))*1.2*exp(-1.49*(db/dt));#Rise velocity by Eqn.(4)   \n",
+      "\n",
+      "#Computation of cloud thickness\n",
+      "Rb=db/2.0;          #Radius of bubble\n",
+      "uf=umf/ephsilonmf;  #Velocity of emulsion gas\n",
+      "Rc=Rb*((ubr+(2*uf))/(ubr-uf))**(1/3.0);#Radius of cloud by Eqn.(6)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'The rise velocity of the bubble=%.1f cm/s'%ubr\n",
+      "print 'The cloud thickness=%.2f cm'%(Rc-Rb)\n",
+      "print 'From Fig.8(page 124)comparing fw vs dp, for dp = %.0f micrometer, wake fraction = 0.24'%dp\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rise velocity of the bubble=49.8 cm/s\n",
+        "The cloud thickness=0.30 cm\n",
+        "From Fig.8(page 124)comparing fw vs dp, for dp = 300 micrometer, wake fraction = 0.24\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 132\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "uo=15;     #Superificial gas velocity in cm/s\n",
+      "umf=1;     #Velocity at minimum fluidization condition in cm/s\n",
+      "lor=2.0;   #Pitch of perforated plate in cm\n",
+      "g=980;     #Acceleration due to gravity in cm/s**2\n",
+      "#CALCULATION\n",
+      "#Case(a) For porous plate\n",
+      "dbo1=(2.78/g)*(uo-umf)**2;#Initial bubble size using Eqn.(19)\n",
+      "\n",
+      "#Case(b) For Perforated plate\n",
+      "Nor=(2/math.sqrt(3))*(1/lor)**2;#Number of orifices in cm**-2\n",
+      "dbo2=(1.30/(g**0.2))*((uo-umf)/Nor)**0.4;#Initial bubble size using Eqn.(15) assuming inital bubble size is smaller than hole spacing\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Case(a) For porous plate'\n",
+      "print '\\tInitial bubble size=%.2fcm'%dbo1\n",
+      "print 'Case(b) For Perforated plate'\n",
+      "print '\\tInitial bubble size=%.2fcm'%dbo2\n",
+      "print '\\tSince %f<%f, the equation used is correct.'%(dbo2,lor)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Case(a) For porous plate\n",
+        "\tInitial bubble size=0.56cm\n",
+        "Case(b) For Perforated plate\n",
+        "\tInitial bubble size=1.55cm\n",
+        "\tSince 1.548765<2.000000, the equation used is correct.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch6.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch6.ipynb
new file mode 100755
index 00000000..9c4f2311
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch6.ipynb
@@ -0,0 +1,499 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:432c74872bfc5b3cdf6cd2ab6c97c0ab1581b40f382e3a1bf8502ef1c3f981bb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Bubbling Fluidized Beds"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "z=0.5;     #Height of bed in m\n",
+      "dt=0.5;    #ID of tube in m\n",
+      "rhos=1.6;  #Density of catalyst in g/cm**3\n",
+      "dpbar=60.;  #Averge catalyst diameter in micrometer\n",
+      "umf=0.002; #Velocity at minimum fluidization condition in m/s\n",
+      "uo=0.2;    #Superficial velocity in m/s\n",
+      "dor=2.;     #Diameter of orifice in mm\n",
+      "lor=20.;    #Pitch of perforated plate in mm\n",
+      "g=9.80;    #g=980;#Acceleration due to gravity in m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Method 1. Procedure using Eqn.(10) & Eqn.(11)\n",
+      "db=(0.035+0.040)/2.;#Bubble size at z=0.5m from Fig.7(a) & Fig.7(b)\n",
+      "ub1=1+1.55*((uo-umf)+14.1*(db+0.005))*(dt**0.32)*0.711*(g*db)**0.5;#Bubble velocity using Eqn.(10) & Eqn.(11)\n",
+      "\n",
+      "#Method 2. Werther's procedure\n",
+      "si=0.8;#From Fig.6 for Geldart A solids \n",
+      "ub2=si*(uo-umf)+(3.2*(dt**(1./3)))*(0.711*(g*db)**0.5);#Bubble velocity using Eqn.(9)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Method 1. Procedure using Eqn.(10) & Eqn.(11)'\n",
+      "print '\\tDiameter of the bubble=%.4fm'%db\n",
+      "print '\\tRise velocity of the bubble=%.4fm/s'%ub1\n",
+      "print 'Method 2. Werthers procedure'\n",
+      "print '\\tDiameter of the bubble=%fm'%db\n",
+      "print '\\tRise velocity of the bubble=%fm/s'%ub2\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Method 1. Procedure using Eqn.(10) & Eqn.(11)\n",
+        "\tDiameter of the bubble=0.0375m\n",
+        "\tRise velocity of the bubble=1.4267m/s\n",
+        "Method 2. Werthers procedure\n",
+        "\tDiameter of the bubble=0.037500m\n",
+        "\tRise velocity of the bubble=1.253125m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 151\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "z=0.5;      #Height of bed in m\n",
+      "dt=0.5;     #ID of tube in m\n",
+      "rhos=2.6;   #Density of catalyst in g/cm**3\n",
+      "dpbar=100.; #Averge catalyst diameter in micrometer\n",
+      "umf=0.01;   #Velocity at minimum fluidization condition in m/s\n",
+      "uo=0.45;    #Superficial velocity in m/s\n",
+      "dor=2.;     #Diameter of orifice in mm\n",
+      "lor=30.;    #Pitch of perforated plate in mm\n",
+      "g=9.80;     #Acceleration due to gravity in m/s**2\n",
+      "pi=3.142857;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Part(a).Bubble Size\n",
+      "Nor=(2/math.sqrt(3))*(1/lor**2);\n",
+      "dbo=5.5;\n",
+      "\n",
+      "#Method 1.Werther's procedure for finding bubble size\n",
+      "z1=[0,5,10,20,30,50,70]\n",
+      "db = [0.,0.,0.,0.,0.,0.,0.]\n",
+      "n=len(z1);\n",
+      "i=0; \n",
+      "while i<n:\n",
+      "    db[i]=0.853*((1+0.272*(uo-umf)*100)**(1/3.0))*(1+0.0684*z1[i])**1.21;\n",
+      "    i=i+1;    \n",
+      "\n",
+      "db1=0.163;#Since bubble size starts at dbo=5.5cm at z=0, we shift the curve accordingly to z=0.5m\n",
+      "\n",
+      "#Method 2.Mori and Wen's procedure for finding bubble size\n",
+      "dbm=0.65*((math.pi/4.0)*((dt*100)**2)*(uo-umf)*100)**0.4;\n",
+      "db2=dbm-(dbm-dbo)*math.exp(-0.3**(z/dt));\n",
+      "\n",
+      "#Part(b).Bubble Velocity\n",
+      "#Method 1.Procedure using Eqn.(12)\n",
+      "ub1=1.6*((uo-umf)+1.13*db1**0.5)*(dt**1.35)+(0.711*(g*db1)**0.5);\n",
+      "\n",
+      "#Method 2.Werther's Procedure\n",
+      "si=0.65;\n",
+      "ub2=si*(uo-umf)+2*(dt**0.5)*(0.711*(g*db1)**0.5);\n",
+      "\n",
+      "#Using Eqn.(7) & Eqn.(8)\n",
+      "ubr1=0.711*(g*db1)**0.5;\n",
+      "ubr2=0.711*(g*db2/100.0)**0.5\n",
+      "ub3=uo-umf+ubr1;\n",
+      "ub4=uo-umf+ubr2;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Bubble Size'\n",
+      "print 'Initial bubble size from Fig.5.14 for %.2fm/s = %.2fcm'%(uo-umf,dbo)\n",
+      "print '\\tMethod 1.Werthers procedure for finding bubble size'\n",
+      "print '\\t\\tHeight of bed(cm)'\n",
+      "print '\\t\\t\\tBubble size(cm)'\n",
+      "m=len(z1);\n",
+      "j=0;\n",
+      "while j<m:\n",
+      "    print '\\t\\t%d'%z1[j],\n",
+      "    print '\\t\\t\\t\\t%.2f'%db[j]\n",
+      "    j=j+1;\n",
+      "\n",
+      "print '\\tMethod 2.Mori and Wens procedure for finding bubble size'\n",
+      "print '\\t\\tMaximum expected bubble size=%.2fcm'%dbm\n",
+      "print '\\t\\tBubble size=%.0fcm'%db2\n",
+      "print 'Bubble Velocity'\n",
+      "print '\\tMethod 1.Procedure using Eqn.(12)'\n",
+      "print '\\t\\tBubble velocity=%.2fm/s'%ub1\n",
+      "print '\\tMethod 2.Werthers procedure'\n",
+      "print '\\t\\tBubble velocity=%.2fm/s'%ub2\n",
+      "print 'Comparing the above results with the expressions of the simple two-phase theory'\n",
+      "print '\\tWerthers bubble size'\n",
+      "print '\\tBubble rise velocity=%.2fm/s\\tBubble velocity=%.2fm/s'%(ubr1,ub3)\n",
+      "print '\\tMori & Wens bubble size'\n",
+      "print '\\tBubble rise velocity=%.1fm/s\\tBubble velocity=%.2fm/s'%(ubr2,ub4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bubble Size\n",
+        "Initial bubble size from Fig.5.14 for 0.44m/s = 5.50cm\n",
+        "\tMethod 1.Werthers procedure for finding bubble size\n",
+        "\t\tHeight of bed(cm)\n",
+        "\t\t\tBubble size(cm)\n",
+        "\t\t0 \t\t\t\t2.00\n",
+        "\t\t5 \t\t\t\t2.86\n",
+        "\t\t10 \t\t\t\t3.77\n",
+        "\t\t20 \t\t\t\t5.69\n",
+        "\t\t30 \t\t\t\t7.73\n",
+        "\t\t50 \t\t\t\t12.10\n",
+        "\t\t70 \t\t\t\t16.77\n",
+        "\tMethod 2.Mori and Wens procedure for finding bubble size\n",
+        "\t\tMaximum expected bubble size=61.31cm\n",
+        "\t\tBubble size=20cm\n",
+        "Bubble Velocity\n",
+        "\tMethod 1.Procedure using Eqn.(12)\n",
+        "\t\tBubble velocity=1.46m/s\n",
+        "\tMethod 2.Werthers procedure\n",
+        "\t\tBubble velocity=1.56m/s\n",
+        "Comparing the above results with the expressions of the simple two-phase theory\n",
+        "\tWerthers bubble size\n",
+        "\tBubble rise velocity=0.90m/s\tBubble velocity=1.34m/s\n",
+        "\tMori & Wens bubble size\n",
+        "\tBubble rise velocity=1.0m/s\tBubble velocity=1.43m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 153\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dpbar=53.;      #Average particle size in micrometer\n",
+      "s=[1,2];        #Size of Bermuda rock in cm \n",
+      "rhosbar=3200.;  #Average solid density of the coke-zircon mixture in kg/m**3\n",
+      "ephsilonm=0.5;  #Void fraction for fixed bed\n",
+      "ephsilonf=0.75; #Void fraction for bubbling bed\n",
+      "rhogbar=0.64;   #Average density of gas in kg/m**3\n",
+      "uo=14.;         #Superficial gas velocity in cm/s\n",
+      "myu=5E-5;       #Viscosity of gas in kg/m s\n",
+      "T=1000.;        #Temperature in degree C\n",
+      "P=1.;           #Pressure in atm\n",
+      "dt=91.5;        #ID of bed in cm\n",
+      "sh=150.;        #Slumped height in cm\n",
+      "\n",
+      "#CALCULATION\n",
+      "rhog2=1.2;      #Density of ambient air\n",
+      "myu2=1.8E-5;    #Viscosity of ambient air\n",
+      "rhos2=rhog2*(rhosbar/rhogbar);#For the requirement of constant density ratio\n",
+      "m=((rhogbar*myu2)/(rhog2*myu))**(2./3);#Scale factor by usin Eqn.(16)\n",
+      "u2=(m**0.5)*uo;  #Superficial gas velocity by using Eqn.(17)\n",
+      "#OUTPUT\n",
+      "print 'For the model use'\n",
+      "print '\\tBed of ID %.2fcm\\tSlumped bed height of %.2fcm\\tPacked bed distributor consisting of %.2f-%.2fmm rock'%(m*dt,m*sh,m*s[0],m*s[1])\n",
+      "print 'Fluidizing gas: ambient air at %.2fatm'%P\n",
+      "print 'Solids: \\tzirconia, Average particle size=%.2fmicrometers'%(m*dpbar)\n",
+      "print 'Entering gas:\\tSuperficial velocity=%.2fcm/s'%u2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For the model use\n",
+        "\tBed of ID 30.45cm\tSlumped bed height of 49.92cm\tPacked bed distributor consisting of 0.33-0.67mm rock\n",
+        "Fluidizing gas: ambient air at 1.00atm\n",
+        "Solids: \tzirconia, Average particle size=17.64micrometers\n",
+        "Entering gas:\tSuperficial velocity=8.08cm/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 159\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "dtb=20;          #ID of bench-scale reactor\n",
+      "dtp=1;           #ID of pilot reactor\n",
+      "dpbar=52;        #Average particle size in micrometer\n",
+      "ephsilonm=0.45;  #Void fraction for fixed bed\n",
+      "ephsilonmf=0.50; #Void fraction at minimum fluidization condition\n",
+      "ephsilonmb=0.60; #Void fraction \n",
+      "uo=30;           #Superficial gas velocity in cm/s\n",
+      "Lmb=2;           #Length of fixed bed in m\n",
+      "umf=0.33;        #Velocity at minimum fluidization condition in cm/s\n",
+      "umb=1;           #Velocity at in cm/s\n",
+      "db=3;            #Equilibrium bubble size in cm\n",
+      "g=9.80;          #Acceleration due to gravity in m/s**2\n",
+      "pi=3.142857;\n",
+      "\n",
+      "#CALCULATION\n",
+      "ubr=0.711*(g*db/100)**0.5;#Rise velocity of bubble using Eqn.(7)\n",
+      "\n",
+      "#Bubble velocity for the bench unit\n",
+      "ubb1=1.55*(((uo-umf)/100.0)+14.1*((db/100.0)+0.005))*((dtb/100.0)**0.32)+ubr;#Bubble velocity using Eqn.(11)\n",
+      "si=1;\n",
+      "ubb2=si*((uo-umf)/100.0)+(3.2*((dtb/100.0)**(1/3.0)))*ubr;#Bubble velocity using Eqn.(9)\n",
+      "ubb=(ubb1+ubb2)/2;#Average bubble velocity\n",
+      "\n",
+      "#Bubble velocity for the pilot unit\n",
+      "ubp1=1.55*(((uo-umf)/100.0)+14.1*((db/100.0)+0.005))*(dtp**0.32)+ubr;#Bubble velocity using Eqn.(11)\n",
+      "si=1;\n",
+      "ubp2=si*((uo-umf)/100)+(3.2*(dtp**(1/3)))*ubr;#Bubble velocity using Eqn.(9)\n",
+      "ubp=(ubp1+ubp2)/2;#Average bubble velocity\n",
+      "\n",
+      "#Rise velocity of upflowing emulsion\n",
+      "ueb=ubb-ubr;#For the bench unit\n",
+      "uep=ubp-ubr;#For the pilot unit\n",
+      "\n",
+      "#Scale-Up Alternative 1.\n",
+      "dteb=20;#Effective bubble diameter\n",
+      "dib=[5,10,15,20];#Different outside diameters\n",
+      "n=len(dib);\n",
+      "li = [0.,0.,0.,0.]\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    li[i]=math.sqrt(((pi*dib[i]*dteb)/4)+((pi/4)*(dib[i])**2));#Pitch using Eqn.(13)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Scale-Up Alternative 2.\n",
+      "Lmp=Lmb*(ubp/ubb);#Static bed height of commercial unit\n",
+      "dtep=100.0;#Effective bubble diameter\n",
+      "dip=[10,15,20,25];#Different outside diameters\n",
+      "m=len(dip);\n",
+      "i=0;\n",
+      "lip = [0.,0.,0.,0.]\n",
+      "while i<m:\n",
+      "    lip[i]=math.sqrt(((math.pi*dip[i]*dtep)/4.0)+(math.pi/4.0)*dip[i]);#Pitch using Eqn.(13)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Height of Bubbling beds\n",
+      "#For bench unit\n",
+      "deltab=((uo/100.0)-(umb/100.0))/(ubb-(umb/100.0));#Fraction of bed in bubbles using Eqn.(28)\n",
+      "ephsilonfb=deltab+(1-deltab)*ephsilonmb;#Void fraction of bubbling bed using Eqn.(20)\n",
+      "Lfb=Lmb*(1-ephsilonm)/(1-ephsilonfb);#Hieght of bubbling bed usnig Eqn.(19)\n",
+      "#For pilot unit\n",
+      "deltap=((uo/100.0)-(umb/100.0))/(ubp-(umb/100.0));#Fraction of bed in bubbles using Eqn.(28)\n",
+      "ephsilonfp=deltap+(1-deltap)*ephsilonmb;#Void fraction of bubbling bed using Eqn.(20)\n",
+      "Lfp=Lmp*(1-ephsilonm)/(1-ephsilonfp);#Hieght of bubbling bed usnig Eqn.(19)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Rise velocity of bubble=%.3fm/s'%ubr\n",
+      "print 'For the bench unit'\n",
+      "print '\\tWith Eqn.(11), Rise velocity=%.3fm/s'%ubb1\n",
+      "print '\\tWith Werthers procedure, Rise velocity=%.2fm/s'%ubb2\n",
+      "print '\\tAverage rise velocity=%.2fm/s'%ubb\n",
+      "print '\\tRise velocity of upflowing emulsion=%.2fm/s'%ueb\n",
+      "print 'For the pilot unit'\n",
+      "print '\\tWith Eqn.(11), Rise velocity=%.2fm/s'%ubp1\n",
+      "print '\\tWith Werthers procedure, Rise velocity=%.2fm/s'%ubp2\n",
+      "print '\\tAverage rise velocity=%.2fm/s'%ubp\n",
+      "print '\\tRise velocity of upflowing emulsion=%.2fm/s'%uep\n",
+      "print 'Scale-Up Alternative 1.'\n",
+      "print '\\tOuter diameter of tube(cm)'\n",
+      "print '\\tPitch(cm)'\n",
+      "n=len(dib);\n",
+      "j=0;\n",
+      "while j<n:\n",
+      "    print '\\t\\t%d'%dib[j],\n",
+      "    print '\\t\\t\\t%.2f'%li[j]\n",
+      "    j=j+1;\n",
+      "\n",
+      "print '\\tSuitable arrangement'\n",
+      "print '\\t\\tOuter Diameter=%dcm\\tPitch:Diameter ratio=%.2f'%(dib[1],(li[1]/dib[1]))\n",
+      "print 'Scale-Up Alternative 2.'\n",
+      "print '\\tStatic bed height for commercial unit=%fm'%Lmp\n",
+      "print '\\tOuter diameter of tube(cm)'\n",
+      "print '\\tPitch(cm)'\n",
+      "n=len(dip);\n",
+      "j=0;\n",
+      "\n",
+      "while j<n:\n",
+      "    print '\\t\\t%d'%dip[j],\n",
+      "    print '\\t\\t\\t%.2f'%lip[j]\n",
+      "    j=j+1;\n",
+      "\n",
+      "print '\\tSuitable arrangement'\n",
+      "print '\\t\\tOuter Diameter=%dcm\\tPitch:Diameter ratio=%.2f'%(dip[2],(lip[2]/dip[2]))\n",
+      "print '\\t\\t\\t\\tFraction of bed in bubbles\\tVoid fraction of bed\\tStatic bed height(m)\\tHeight of bubbling bed(m)'\n",
+      "print '\\t\\t\\t\\t---------------------------------------------------------------------------------------------------------'\n",
+      "print 'Bench unit\\tID=%fm\\t%f\\t\\t\\t%f\\t\\t%f\\t\\t%f'%(dtb/100.,deltab,ephsilonfb,Lmb,Lfb)\n",
+      "print 'Commercial unit\\tID=%fm\\t%f\\t\\t\\t%f\\t\\t%f\\t\\t%f'%(dtp,deltap,ephsilonfp,Lmp,Lfp)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rise velocity of bubble=0.386m/s\n",
+        "For the bench unit\n",
+        "\tWith Eqn.(11), Rise velocity=1.117m/s\n",
+        "\tWith Werthers procedure, Rise velocity=1.02m/s\n",
+        "\tAverage rise velocity=1.07m/s\n",
+        "\tRise velocity of upflowing emulsion=0.68m/s\n",
+        "For the pilot unit\n",
+        "\tWith Eqn.(11), Rise velocity=1.61m/s\n",
+        "\tWith Werthers procedure, Rise velocity=1.53m/s\n",
+        "\tAverage rise velocity=1.57m/s\n",
+        "\tRise velocity of upflowing emulsion=1.18m/s\n",
+        "Scale-Up Alternative 1.\n",
+        "\tOuter diameter of tube(cm)\n",
+        "\tPitch(cm)\n",
+        "\t\t5 \t\t\t9.91\n",
+        "\t\t10 \t\t\t15.35\n",
+        "\t\t15 \t\t\t20.31\n",
+        "\t\t20 \t\t\t25.07\n",
+        "\tSuitable arrangement\n",
+        "\t\tOuter Diameter=10cm\tPitch:Diameter ratio=1.54\n",
+        "Scale-Up Alternative 2.\n",
+        "\tStatic bed height for commercial unit=2.941439m\n",
+        "\tOuter diameter of tube(cm)\n",
+        "\tPitch(cm)\n",
+        "\t\t10 \t\t\t28.16\n",
+        "\t\t15 \t\t\t34.49\n",
+        "\t\t20 \t\t\t39.83\n",
+        "\t\t25 \t\t\t44.53\n",
+        "\tSuitable arrangement\n",
+        "\t\tOuter Diameter=20cm\tPitch:Diameter ratio=1.99\n",
+        "\t\t\t\tFraction of bed in bubbles\tVoid fraction of bed\tStatic bed height(m)\tHeight of bubbling bed(m)\n",
+        "\t\t\t\t---------------------------------------------------------------------------------------------------------\n",
+        "Bench unit\tID=0.200000m\t0.274171\t\t\t0.709668\t\t2.000000\t\t3.788769\n",
+        "Commercial unit\tID=1.000000m\t0.185857\t\t\t0.674343\t\t2.941439\t\t4.967774\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5, Page 161\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dtb=20;          #ID of bench-scale reactor\n",
+      "dtp=1;           #ID of pilot reactor\n",
+      "dpbar=200;       #Average particle size in micrometer\n",
+      "ephsilonmf=0.50; #Void fraction at minimum fluidization condition\n",
+      "ephsilonmb=0.50; #Void fraction \n",
+      "uo=30;           #Superficial gas velocity in cm/s\n",
+      "Lmb=2;           #Length of fixed bed in m\n",
+      "umf=3;           #Velocity at minimum fluidization condition in cm/s\n",
+      "umb=3;           #Velocity at in cm/s\n",
+      "g=9.80;          #Acceleration due to gravity in m/s**2\n",
+      "pi=3.142857;\n",
+      "\n",
+      "#CALCULATION\n",
+      "#In the small bench unit\n",
+      "c=1;\n",
+      "ubb=c*((uo-umf)/100.0)+0.35*(g*(dtb/100.0))**0.5;#Velocity using Eqn.(5.22)\n",
+      "zsb=60*(dtb)**0.175;#Height using Eqn.(5.24)\n",
+      "\n",
+      "#In the large pilot unit\n",
+      "ubp=c*((uo-umf)/100.0)+0.35*(g*dtp)**0.5;#Velocity using Eqn.(5.22)\n",
+      "zsp=60*(dtp*100)**0.175;#Height using Eqn.(5.24)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Condition at which bubbles transform into slugs'\n",
+      "print 'For tha small bench unit\\t\\tVelocity=%fm/s\\t\\tHeight above distributor plate=%fm'%(ubb,zsb/100.)\n",
+      "print 'For tha large pilot unit\\t\\tVelocity=%fm/s\\t\\tHeight above distributor plate=%fm'%(ubp,zsp/100.);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Condition at which bubbles transform into slugs\n",
+        "For tha small bench unit\t\tVelocity=0.760000m/s\t\tHeight above distributor plate=1.013518m\n",
+        "For tha large pilot unit\t\tVelocity=1.365673m/s\t\tHeight above distributor plate=1.343233m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch7.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch7.ipynb
new file mode 100755
index 00000000..aafc2f66
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch7.ipynb
@@ -0,0 +1,399 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8eb00dc3983a05f0d88bd9a81e8b022860563cdbf1df8dacf4157210feb1c3ce"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7  : Entrainment and Elutriation  from Fluidized Beds"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "rhog=5.51;  #Density of gas in kg/m**3\n",
+      "rhos=1200;  #Density of solid in kg/m**3\n",
+      "dpbar=130;  #Average size of particles in micrometer\n",
+      "uo=0.61;    #Superficial gas velocity in m/s\n",
+      "g=9.80;     #Acceleration due to gravity in m/s**2\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Assuming that freeboard in higher than TDH, computation of entrainment rate by Zenz & Weil's method\n",
+      "x=(uo**2)/(g*(dpbar*10**-6)*rhos**2);#Calculation of value of x-axis for Fig.(6), page 175\n",
+      "y=1.2;                               # Value of y-axis from Fig.(6)\n",
+      "Gsstar=y*rhog*uo;                    #Computation of rate of entrainment\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nRate of entrainment=%.2fkg/m**2s'%Gsstar\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Rate of entrainment=4.03kg/m**2s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 180\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "x=0.2;          #Fraction of fines in the bed\n",
+      "Gsstar=4.033320 #Rate of entrainment in kg/m**2s(from Exa.1)\n",
+      "\n",
+      "#CALCULATION\n",
+      "Gsstar1=x*Gsstar;#Rate of entrainment by Eqn.(3)\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nRate of entrainment=%.3fkg/m**2s'%Gsstar1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Rate of entrainment=0.807kg/m**2s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 181\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "rhog=5.51;                    #Density of gas in kg/m**3\n",
+      "rhos=1200;                    #Density of solid in kg/m**3\n",
+      "uo=0.61;                      #Superficial gas velocity in m/s\n",
+      "g=9.80;                       #Acceleration due to gravity in m/s**2\n",
+      "dp=[10,30,50,70,90,110,130];  #Diameter of particle in micrometer\n",
+      "p=[0.,0.0110,0.0179,0.0130,0.0058,0.0020,0.];\n",
+      "pi=3.142857;\n",
+      "dt=6;\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(dp);\n",
+      "i=0;\n",
+      "x = [0,0,0,0,0,0,0]\n",
+      "while i<n:\n",
+      "    x[i]=(uo**2)/(g*(dp[i]*10**-6)*rhos**2);#Computation of value of x-axis for Fig.(6), page 175)\n",
+      "    i=i+1;\n",
+      "\n",
+      "y=[40,12,6,3.2,2.,1.3,1];#Value of y-axis corresponding to each value of x-axis\n",
+      "y1 = []\n",
+      "for i in range(n):\n",
+      "    y1.append(y[i]*p[i]);\n",
+      "i=0;\n",
+      "k=0;\n",
+      "\n",
+      "while i<n-2:\n",
+      "    y1[i]=(y[i]*p[i]);\n",
+      "    k=k+((0.5)*(dp[i+1]-dp[i])*(y1[i+1]+y1[i]));#Integration using Trapezoidal rule\n",
+      "    i=i+1;\n",
+      "rhosbar=k*rhog;#Computation of solid loading\n",
+      "te=(pi/4)*(dt**2)*rhosbar*uo;#Computation of total entrainment\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nSolid loading =%.1fkg/m**3'%rhosbar\n",
+      "print '\\nTotal Entrainment =%.0fkg/s'%te\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Solid loading =32.4kg/m**3\n",
+        "\n",
+        "Total Entrainment =559kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 181\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "dp=[40,60,80,100,120]; #Diameter of particle in micrometer\n",
+      "uo=0.381;              #Superficial gas velocity in m/s\n",
+      "\n",
+      "#CALCULATION\n",
+      "Gs=0.9;#Rate of entrainment in kg/m**2 s from Fig.3(a)\n",
+      "pb = [0.45,1.00,1.25,1.00,0.60];#Size distribution for bed particles from Fig.3(b)\n",
+      "pe=[1.20,2.00,1.25,0.45,0.10];  #Size distribution for entrained particles from Fig.3(b)\n",
+      "n=len(dp);\n",
+      "for i in range(n):\n",
+      "    pb[i] = pb[i]/100.\n",
+      "    pe[i] = pe[i]/100.\n",
+      "i=0;\n",
+      "ki = []\n",
+      "while i<n:\n",
+      "    ki.append((Gs*pe[i])/pb[i]);#Calculation of ki* using Eqn.(13)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\ndpi(micrometer)',\n",
+      "print '\\t100pb(dpi)(micrometer**-1)',\n",
+      "print '\\t100pe(dpi)(micrometer**-1)',\n",
+      "print '\\tki*(kg/m**2 s)'\n",
+      "\n",
+      "j=0;\n",
+      "while j<n:\n",
+      "    print '%f'%dp[j],\n",
+      "    print '\\t%f'%(100*pb[j]),\n",
+      "    print '\\t\\t\\t%f'%(100*pe[j]),\n",
+      "    print '\\t\\t\\t%f'%ki[j]\n",
+      "    j=j+1;\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "dpi(micrometer) \t100pb(dpi)(micrometer**-1) \t100pe(dpi)(micrometer**-1) \tki*(kg/m**2 s)\n",
+        "40.000000 \t0.450000 \t\t\t1.200000 \t\t\t2.400000\n",
+        "60.000000 \t1.000000 \t\t\t2.000000 \t\t\t1.800000\n",
+        "80.000000 \t1.250000 \t\t\t1.250000 \t\t\t0.900000\n",
+        "100.000000 \t1.000000 \t\t\t0.450000 \t\t\t0.405000\n",
+        "120.000000 \t0.600000 \t\t\t0.100000 \t\t\t0.150000\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5, Page 181\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "rhog=1.217;                                   #Density of gas in kg/m**3\n",
+      "myu=1.8E-5;                                   #Viscosity of gas in kg/m s\n",
+      "umf=0.11;                                     #Velocity at minimum fluidization condition in m/s\n",
+      "rhos=2000.0;                                  #Density of solid in kg/m**3\n",
+      "uo=1.0;                                       #Superficial gas velocity in m/s\n",
+      "g=9.80;                                       #Acceleration due to gravity in m/s**2\n",
+      "dp=[30,40,50,60,80,100,120];                  #Diameter of particle in micrometer\n",
+      "uti=[0.066,0.115,0.175,0.240,0.385,0.555,1.0];#Terminal velocity of particles in m/s\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(dp);\n",
+      "i=0;\n",
+      "Ret = []\n",
+      "kistar1 = []\n",
+      "kistar2 = []\n",
+      "kistar3 = []\n",
+      "kistar4 = []\n",
+      "kistar5 = []\n",
+      "kistar6 = []\n",
+      "x1 = []\n",
+      "x2 = []\n",
+      "\n",
+      "while i<n:\n",
+      "    #Using Yagi & Aochi's correlation\n",
+      "    Ret.append((rhog*(uti[i])*dp[i]*10**-6)/myu)\n",
+      "    a =((myu*((uo-uti[i])**2))/(g*(dp[i]*10**-6)**2))*(0.0015*(Ret[i]**0.5)+(0.01*(Ret[i]**1.2)));\n",
+      "    kistar1.append(a)\n",
+      "    #Using Wen & Hasinger's correlation\n",
+      "    a=(((1.52E-5)*((uo-uti[i])**2)*rhog)/(g*dp[i]*10**-6)**0.5)*(Ret[i]**0.725)*((rhos-rhog)/rhog)**1.15;\n",
+      "    kistar2.append(a)\n",
+      "    #Using Merrick & Highley's correlation\n",
+      "    a=uo*rhog*(0.0001+130*math.exp(-10.4*((uti[i]/uo)**0.5)*((umf/(uo-umf))**0.25)));\n",
+      "    kistar3.append(a)\n",
+      "    #Using Geldart's correlation\n",
+      "    a=23.7*uo*rhog*math.exp(-5.4*(uti[i]/uo));\n",
+      "    kistar4.append(a)\n",
+      "    #Using Zenz & Weil's procedure\n",
+      "    a=(uo**2)/(g*(dp[i]*10.0**-6)*rhos**2);#Computation of value of x-axis for Fig.(6), page 175)\n",
+      "    x1.append(a)\n",
+      "    y1=[12.2,8.6,6.4,4.9,2.75,1.8,1.2];#Value of y-axis corresponding to each value of x-axis\n",
+      "    kistar5.append(y1[i]*rhog*uo)\n",
+      "    #Using Gugnoni & Zenz's procedure\n",
+      "    a=(uo-uti[i])/((g*dp[i]*10**-6)**0.5);#Computation of value of x-axis for Fig.(6), page 175)\n",
+      "    x2.append(a)\n",
+      "    y=[5.8,5.4,3.2,2.8,1.3,0.6,0];#Value of y-axis corresponding to each value of x-axis\n",
+      "    kistar6.append(y[i]*rhog*uo)\n",
+      "    i=i+1;\n",
+      "\n",
+      "i=0;\n",
+      "print 'dp(micrometer)',\n",
+      "print '\\tYagi & Aochi',\n",
+      "print '\\tWen & Hashinger',\n",
+      "print '\\t\\tMerrick & Highley',\n",
+      "print '\\tGeldart et al.',\n",
+      "print '\\t\\tZenz & Well',\n",
+      "print '\\t\\tGugnoni & Zenz'\n",
+      "while i<n:\n",
+      "    print '\\n%f'%dp[i],\n",
+      "    print '\\t%f'%kistar1[i],\n",
+      "    print '\\t%f'%kistar2[i],\n",
+      "    print '\\t\\t%f'%kistar3[i],\n",
+      "    print '\\t\\t%f'%kistar4[i],\n",
+      "    print '\\t\\t%f'%kistar5[i],\n",
+      "    print '\\t\\t%f'%kistar6[i],\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Note: There is huge deviation of the calculated answer and the answer given in the textbook for the correlation of Merrick & Highley.  There is a contradiction in the correlation used in the problem and the one given in page 179. \n",
+      "#We tried to retrieve the original paper i.e. D.Merrick and J.Highley, AICHE J., 6, 220(1960). But the effort was not fruitful.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "dp(micrometer) \tYagi & Aochi \tWen & Hashinger \t\tMerrick & Highley \tGeldart et al. \t\tZenz & Well \t\tGugnoni & Zenz\n",
+        "\n",
+        "30.000000 \t2.571188 \t1.092184 \t\t32.451340 \t\t20.195582 \t\t14.847400 \t\t7.058600 \n",
+        "40.000000 \t2.965958 \t1.564720 \t\t19.546385 \t\t15.500369 \t\t10.466200 \t\t6.571800 \n",
+        "50.000000 \t3.240381 \t1.938471 \t\t11.993076 \t\t11.210646 \t\t7.788800 \t\t3.894400 \n",
+        "60.000000 \t3.289995 \t2.154988 \t\t7.713841 \t\t7.892113 \t\t5.963300 \t\t3.407600 \n",
+        "80.000000 \t2.852535 \t2.120728 \t\t3.447977 \t\t3.606955 \t\t3.346750 \t\t1.582100 \n",
+        "100.000000 \t1.883718 \t1.521994 \t\t1.600171 \t\t1.440318 \t\t2.190600 \t\t0.730200 \n",
+        "120.000000 \t0.000000 \t0.000000 \t\t0.332158 \t\t0.130271 \t\t1.460400 \t\t0.000000\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6, Page 190\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "dpbar=60.0;   #Average size of particles in micrometer\n",
+      "rhog=1.3;     #Density of gas in kg/m**3\n",
+      "rhos=1500.0;  #Density of solid in kg/m**3\n",
+      "umf=0.003;    #Velocity at minimum fluidization condition in m/s\n",
+      "uo=0.503;     #Superficial gas velocity in m/s\n",
+      "g=9.80;       #Acceleration due to gravity in m/s**2\n",
+      "Hf=2.0;       #Height at which the cyclone inlet is to be located in m\n",
+      "\n",
+      "#CALCULATION\n",
+      "y=(uo**2)/(g*(dpbar*10**-3)*rhos**2);#Calculation of value of y-axis for Fig.(6), page 175\n",
+      "x=1;#Value of x-axis from Fig.(6), page 175\n",
+      "Gsstar=x*rhog*uo;#Computation of rate of entrainment\n",
+      "Gsuo=5.0;#Ejection rate pf particles in kg/m**2 s from Fig.(11), page 188\n",
+      "a=0.72/uo;#From Fig.(12), page 189\n",
+      "Gs=Gsstar+(Gsuo-Gsstar)*math.exp(-a*Hf);\n",
+      "p=((Gs-Gsstar)/Gsstar)*100.0;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\nRate of entrainment from short bed=%.3fkg/m**2s'%Gs\n",
+      "print '\\nThis entrainment is %f percent higher than it would be if the gas exit were at the TDH'%p\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Rate of entrainment from short bed=0.902kg/m**2s\n",
+        "\n",
+        "This entrainment is 37.955972 percent higher than it would be if the gas exit were at the TDH\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch8.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch8.ipynb
new file mode 100755
index 00000000..9fed0363
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch8.ipynb
@@ -0,0 +1,218 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : High Velocity Fludization"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 206\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import zeros\n",
+      "from scipy.optimize import fsolve \n",
+      "\n",
+      "\n",
+      "#INPUT\n",
+      "Lmf=2.4;            #Length of bed at minimum fluidized condition in m\n",
+      "uo=[2.,4.,6.];         #Superficial gas velocity in m/s\n",
+      "GsII=100.;          #Solid circulation rate in kg/m**2 s for Mode II\n",
+      "uoIII=4.;            #Superficial gas velocity in m/s for Mode III\n",
+      "GsIII=[42.,50.,100.,200.,400.];          #Solid circulation rate in kg/m**2 s for Mode III\n",
+      "GsIV=[70.,100.,120.];                  #Solid circulation rate in kg/m**2 s for Mode IV\n",
+      "dt=0.4;                             #Column diamter in m\n",
+      "Ht=10.;                              #Height of column in m\n",
+      "rhos=1000.;                         #Density of solid in kg/m**3\n",
+      "dpbar=55.;                          #Particle diameter in micrometer\n",
+      "ephsilonmf=0.5;                     #Void fraction at minimum fluidization condition\n",
+      "\n",
+      "#CALCULATION\n",
+      "#Mode I\n",
+      "ephsilonstar=0.01;                  #Saturation carrying capacity of gas\n",
+      "ephsilonsd=[0.2,0.16,0.14];         #Solid holdup in lower dense region from Fig.8(b) for various uo\n",
+      "n=len(uo);\n",
+      "i=0;\n",
+      "Hfguess=2.;          #Guess value of height\n",
+      "a = zeros(n)\n",
+      "Hf = zeros(n)\n",
+      "Hd = zeros(n)\n",
+      "ephsilonse = zeros(n)\n",
+      "GsI = zeros(n)\n",
+      "#    endfunction\n",
+      " \n",
+      "while i<n:\n",
+      "    a[i]= 3./uo[i];         #Decay consmath.tant\n",
+      "    def solver_func(Hf):            #Function defined for solving the system\n",
+      "        return Lmf*(1-ephsilonmf)-((ephsilonsd[i]-(ephsilonstar+(ephsilonsd[i]-ephsilonstar)*math.exp(-a[i]*Hf)))/a[i])-Ht*ephsilonsd[i]+Hf*(ephsilonsd[i]-ephsilonstar);\n",
+      "        \n",
+      "    Hf[i] = fsolve(solver_func,1E-6)\n",
+      "    Hd[i]=Ht-Hf[i];#Height of lower densce region\n",
+      "    ephsilonse[i]=ephsilonstar+(ephsilonsd[i]-ephsilonstar)*math.exp(-a[i]*Hf[i]);#Solid holdup at exit\n",
+      "    GsI[i]=rhos*uo[i]*ephsilonse[i];#Solid circulation rate from Eqn.(4)\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Mode II\n",
+      "i=0;\n",
+      "Hfguess2=2;#Guess value of height\n",
+      "ephsilonseII = zeros(n)\n",
+      "HdII = zeros(n)\n",
+      "LmfII = zeros(n)\n",
+      "HfII = zeros(n)\n",
+      "while i<n:\n",
+      "    ephsilonseII[i]=GsII/(rhos*uo[i]);#Solid holdup at exit\n",
+      "    def solver_func1(Hf):           #Function defined for solving the system\n",
+      "        return ephsilonseII[i]-ephsilonstar-(ephsilonsd[i]-ephsilonstar)*math.exp(-a[i]*Hf);#From Eqn.(7)\n",
+      "    HfII[i] = fsolve(solver_func1,1E-6)\n",
+      "    HdII[i]=Ht-HfII[i];#Height of lower dense region\n",
+      "    #Length of bed minimum fluidization condtion\n",
+      "    LmfII[i]=(1-ephsilonmf)**-1*((ephsilonsd[i]-ephsilonseII[i])/a[i])+Ht*ephsilonsd[i]-HfII[i]*(ephsilonsd[i]-ephsilonstar);\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Mode III\n",
+      "aIII = 3./uoIII;            #Decay consmath.tant\n",
+      "ephsilonsdIII=0.16;         #Solid holdup at lower dense region\n",
+      "i=0;\n",
+      "m=len(GsIII);\n",
+      "Hfguess3=2;#Guess value of height \n",
+      "ephsilonseIII = zeros(m)\n",
+      "HdIII = zeros(m)\n",
+      "ephsilonseIII = zeros(m)\n",
+      "LmfIII = zeros(m)\n",
+      "HfIII = zeros(m)\n",
+      "while i<m:\n",
+      "    ephsilonseIII[i]=GsIII[i]/(rhos*uoIII);#Solid holdup at exit\n",
+      "    def solver_func2(Hf):       #Function defined for solving the system\n",
+      "        return ephsilonseIII[i]-ephsilonstar-(ephsilonsdIII-ephsilonstar)*math.exp(-aIII*Hf);#From Eqn.(7)\n",
+      "\n",
+      "    HfIII[i] = fsolve(solver_func2,1E-6)\n",
+      "    HdIII[i]=Ht-HfIII[i];       #Height of lower dense region\n",
+      "    #Length of bed at minimum fluidization condition\n",
+      "    LmfIII[i]=(1-ephsilonmf)**-1*(((ephsilonsdIII-ephsilonseIII[i])/aIII)+Ht*ephsilonsdIII-HfIII[i]*(ephsilonsdIII-ephsilonstar));\n",
+      "    i=i+1;\n",
+      "\n",
+      "#Mode IV\n",
+      "i=0;\n",
+      "Hfguess4=2;#Guess value of height\n",
+      "aIV = zeros(n)\n",
+      "ephsilonseIV = zeros(n)\n",
+      "HdIV = zeros(n)\n",
+      "HfIV = zeros(n)\n",
+      "LmfIV = zeros(n)\n",
+      "while i<n:\n",
+      "    aIV[i]=3./uo[i];            #Decay consmath.tant\n",
+      "    ephsilonseIV[i]=GsIV[i]/(rhos*uo[i]);       #Solid holdup at exit\n",
+      "    def solver_func3(Hf):       #Function defined for solving the system\n",
+      "        return ephsilonseIV[i]-ephsilonstar-(ephsilonsd[i]-ephsilonstar)*math.exp(-aIV[i]*Hf);      #From Eqn.(7)\n",
+      "\n",
+      "    HfIV[i] = fsolve(solver_func3,1E-6)\n",
+      "    HdIV[i]=Ht-HfIV[i];#Height of lower dense region\n",
+      "    #Length of bed at minimum fluidization condition\n",
+      "    LmfIV[i]=(1-ephsilonmf)**-1*(((ephsilonsd[i]-ephsilonseIV[i])/aIV[i])+Ht*ephsilonsd[i]-HfIV[i]*(ephsilonsd[i]-ephsilonstar));\n",
+      "    i=i+1;\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'Mode I';\n",
+      "print '\\tuom/s\\t\\tephsilonse-\\tHfm\\t\\tHdm\\t\\tGskg/m**2 s';\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(uo[i],ephsilonse[i],Hf[i],Hd[i],GsI[i]);\n",
+      "    i=i+1;\n",
+      "\n",
+      "print 'Mode II';\n",
+      "print '\\tuom/s\\t\\tephsilonse-\\tHfm\\t\\tHdm\\t\\tLmfm';\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(uo[i],ephsilonseII[i],HfII[i],HdII[i],LmfII[i]);\n",
+      "    i=i+1;\n",
+      "\n",
+      "print 'Mode III';\n",
+      "print '\\tGskg/m** s\\tephsilonse-\\tHfm\\t\\tHdm\\t\\tLmfm';\n",
+      "i=0;\n",
+      "while i<m:\n",
+      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(GsIII[i],ephsilonseIII[i],HfIII[i],HdIII[i],LmfIII[i]);\n",
+      "    i=i+1;\n",
+      "\n",
+      "print 'Mode IV';\n",
+      "print '\\tuom/s\\t\\tGskg/m**2 s\\tephsilonse-\\tHfm\\t\\tLmfm';\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '\\t%f\\t%f\\t%f\\t%f\\t%f'%(uo[i],GsIV[i],ephsilonseIV[i],HfIV[i],LmfIV[i]);\n",
+      "    i=i+1;\n",
+      "\n",
+      "#====================================END OF PROGRAM ======================================================"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mode I\n",
+        "\tuom/s\t\tephsilonse-\tHfm\t\tHdm\t\tGskg/m**2 s\n",
+        "\t2.000000\t0.192856\t0.025551\t9.974449\t385.711493\n",
+        "\t4.000000\t0.017870\t3.930041\t6.069959\t71.481458\n",
+        "\t6.000000\t0.037349\t3.117707\t6.882293\t224.094133\n",
+        "Mode II\n",
+        "\tuom/s\t\tephsilonse-\tHfm\t\tHdm\t\tLmfm\n",
+        "\t2.000000\t0.050000\t1.038763\t8.961237\t2.002635\n",
+        "\t4.000000\t0.025000\t3.070113\t6.929887\t1.499483\n",
+        "\t6.000000\t0.016667\t5.940829\t4.059171\t1.121026\n",
+        "Mode III\n",
+        "\tGskg/m** s\tephsilonse-\tHfm\t\tHdm\t\tLmfm\n",
+        "\t42.000000\t0.010500\t7.605043\t2.394957\t1.317154\n",
+        "\t50.000000\t0.012500\t5.459126\t4.540874\t1.955596\n",
+        "\t100.000000\t0.025000\t3.070113\t6.929887\t2.638966\n",
+        "\t200.000000\t0.050000\t1.762341\t8.237659\t2.964631\n",
+        "\t400.000000\t0.100000\t0.681101\t9.318899\t3.155670\n",
+        "Mode IV\n",
+        "\tuom/s\t\tGskg/m**2 s\tephsilonse-\tHfm\t\tLmfm\n",
+        "\t2.000000\t70.000000\t0.035000\t1.352099\t3.706202\n",
+        "\t4.000000\t100.000000\t0.025000\t3.070113\t2.638966\n",
+        "\t6.000000\t120.000000\t0.020000\t5.129899\t1.946226\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "/usr/lib/python2.7/dist-packages/scipy/optimize/minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+        "  improvement from the last ten iterations.\n",
+        "  warnings.warn(msg, RuntimeWarning)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch9.ipynb b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch9.ipynb
new file mode 100755
index 00000000..5087d780
--- /dev/null
+++ b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/ch9.ipynb
@@ -0,0 +1,265 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:52ff353152f32e41c2e832a90f993fbd3f11b7f2a5fbb5e4f20eb7599a61f880"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 :  Solid Movement Mixing  Segregation and Staging"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 218\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "umf=0.015;                 #Velocity at minimum fluidization condition in m/s\n",
+      "ephsilonmf=0.5;            #Void fraction at minimum fluidization condition\n",
+      "uo=0.1;                    #Superficial gas velocity in m/s\n",
+      "delta=0.2;                 #Bed fraction in bubbles\n",
+      "db=0.06;                   #Equilibrium bubble size in m\n",
+      "dt=[0.1,0.3,0.6,1.5];      #Various vessel sizes in m\n",
+      "ub=[0.4,0.75,0.85,1.1];    #Bubble velocity in m/s\n",
+      "Dsv=[0.03,0.11,0.14,0.23]; #Reported values of vertical dispersion coefficient\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(ub);\n",
+      "i=0;\n",
+      "fw1=2;#Wake fraction from Hamilton et al.\n",
+      "fw2=0.32;#Wake fraction from Fig.(5.8)\n",
+      "fw=(fw1+fw2)*0.5;#Average value of wake fraction\n",
+      "Dsv1 = []\n",
+      "Dsv2 = []\n",
+      "while i<n:\n",
+      "    Dsv1.append(12*((uo*100)**0.5)*((dt[i]*100)**0.9));#Vertical distribution coefficient from Eqn.(3)\n",
+      "    Dsv2.append((fw**2*ephsilonmf*delta*db*ub[i]**2)/(3*umf));#Vertical distribution coefficient from Eqn.(12)\n",
+      "    i=i+1;\n",
+      "\n",
+      "print Dsv1\n",
+      "\n",
+      "#OUTPUT\n",
+      "print '\\t\\tVertical dispersion coefficient(m**2/s)'\n",
+      "print 'Vessel Size(m)',\n",
+      "print '\\tFrom Experiment',\n",
+      "print '\\tFrom Eqn.(3)',\n",
+      "print '\\tFrom Eqn.(12)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%.2f'%dt[i],\n",
+      "    print '\\t%.3f'%Dsv[i],\n",
+      "    print '\\t%.2f'%(Dsv1[i]/10**4),\n",
+      "    print '\\t%.2f'%Dsv2[i]\n",
+      "    i=i+1;   \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[301.42637178114967, 810.1965234492834, 1511.8801720132121, 3448.7632274996104]\n",
+        "\t\tVertical dispersion coefficient(m**2/s)\n",
+        "Vessel Size(m) \tFrom Experiment \tFrom Eqn.(3) \tFrom Eqn.(12)\n",
+        "0.10 \t0.030 \t0.03 \t0.03\n",
+        "0.30 \t0.110 \t0.08 \t0.10\n",
+        "0.60 \t0.140 \t0.15 \t0.13\n",
+        "1.50 \t0.230 \t0.34 \t0.22\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 222\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "Lmf=0.83;                          #Length of bed at minimum fluidization condition in m\n",
+      "dp=450.0;                          #Average particle size in micrometer\n",
+      "ephsilonmf=0.42;                   #Void fraction at minimum fluidization condition\n",
+      "umf=0.17;                          #Velocity at minimum fluidization condition in m/s\n",
+      "uo=[0.37,0.47,0.57,0.67];          #Superficial gas velocity in m/s\n",
+      "Dsh=[0.0012,0.0018,0.0021,0.0025]; #Horizontal Drift Coefficient from Experiment in m**2/s\n",
+      "db=[0.10,0.14];                    #Equilibrium bubble size in m\n",
+      "g=9.81;                            #Acceleration due to gravity in m/s**2\n",
+      "\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(uo);\n",
+      "m=len(db);\n",
+      "k=0;\n",
+      "alpha=0.77;#Since we are not dealing with Geldart A or AB solids\n",
+      "uf=umf/ephsilonmf;\n",
+      "ubr = []\n",
+      "ub = []\n",
+      "delta = []\n",
+      "Dshc = []\n",
+      "for j in range(m):\n",
+      "    for i in range(n):\n",
+      "        ubr.append(0.711*(db[j]*g)**0.5);#Rise velocity of a single bubble in m/s\n",
+      "        ub.append(uo[i]-umf+ubr[k]);#Rise velocity of bubbles in a bubbling bed\n",
+      "        delta.append((uo[i]-umf)/(ub[k]+umf));#Bed fraction in bubbles\n",
+      "        if ubr[i]>uf:\n",
+      "            Dshc.append((3/16.0)*(delta[k]/(1-delta[k]))*((alpha**2*db[j]*ubr[k]*((((ubr[k]+2*uf)/(ubr[k]-uf))**(1.0/3))-1))));\n",
+      "                #Horizontal Distribution coeff. from Eqn.(14)\n",
+      "        else:\n",
+      "            Dsh.append((3.0/16)*(delta/(1-delta))*(alpha**2*umf*db/ephsilonmf))\n",
+      "                #Horizontal Distribution coeff. from Eqn.(15)\n",
+      "        Dshc.append((3/16.0)*(delta[k]/(1-delta[k]))*((alpha**2*db[j]*ubr[k]*((((ubr[k]+2*uf)/(ubr[k]-uf))**(1/3.0))-1))));#Horizontal Distribution coeff. from Eqn.(14)\n",
+      "        k=k+1;\n",
+      "i=0;\n",
+      "j=0;\n",
+      "k=0;\n",
+      "while k<m*n:\n",
+      "    print 'Snce we do not have ub=%fm/s>>uf=%fm/s we use Eqn.(14).'%(ub[k],uf)\n",
+      "    print 'Gas Velocity(m/s)'\n",
+      "    print '\\tHorizontal Drift Coefficient Calculated(m**2/s)'\n",
+      "    print '\\tHorizontal Drift Coefficient from Experiment(m**2/s)'\n",
+      "    while j<m:\n",
+      "        print 'db=%fm'%db[j]\n",
+      "        while i<n:\n",
+      "            print '%f'%uo[i],\n",
+      "            print '\\t\\t%f'%Dshc[k],\n",
+      "            print '\\t\\t\\t\\t\\t%f'%Dsh[i]\n",
+      "            i=i+1;   \n",
+      "            k=k+1;\n",
+      "        i=0;\n",
+      "        j=j+1;\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Snce we do not have ub=0.904213m/s>>uf=0.404762m/s we use Eqn.(14).\n",
+        "Gas Velocity(m/s)\n",
+        "\tHorizontal Drift Coefficient Calculated(m**2/s)\n",
+        "\tHorizontal Drift Coefficient from Experiment(m**2/s)\n",
+        "db=0.100000m\n",
+        "0.370000 \t\t0.001283 \t\t\t\t\t0.001200\n",
+        "0.470000 \t\t0.001283 \t\t\t\t\t0.001800\n",
+        "0.570000 \t\t0.001924 \t\t\t\t\t0.002100\n",
+        "0.670000 \t\t0.001924 \t\t\t\t\t0.002500\n",
+        "db=0.140000m\n",
+        "0.370000 \t\t0.002566 \t\t\t\t\t0.001200\n",
+        "0.470000 \t\t0.002566 \t\t\t\t\t0.001800\n",
+        "0.570000 \t\t0.003207 \t\t\t\t\t0.002100\n",
+        "0.670000 \t\t0.003207 \t\t\t\t\t0.002500\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 232\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Gsup=1.5;               #Solid interchange rate in kg/m**2plate s\n",
+      "dor=19.1;               #Orifice diameter in mm\n",
+      "dp=210;                 #Particle size in micrometer\n",
+      "uo=0.4;                 #Superficial gas velocity in m/s\n",
+      "fopen=[0.12,0.17,0.26]; #Open area fraction \n",
+      "pi=3.14;\n",
+      "\n",
+      "#CALCULATION\n",
+      "n=len(fopen);\n",
+      "uor = []\n",
+      "ls1 = []\n",
+      "i=0\n",
+      "while i<n:\n",
+      "    uor.append(uo/fopen[i]);  #Gas velocity through the orifice\n",
+      "    ls1.append(Gsup/fopen[i]);#Flux of solids through the holes\n",
+      "    i=i+1;\n",
+      "\n",
+      "ls2=[12,20,25];                #Flux of solids through holes from Fig.13(c) for different uor values\n",
+      "fopen1=0.12;                   #Open area fraction which gives reasonable fit\n",
+      "lor=math.sqrt(((math.pi/4)*dor**2)/fopen1);#Orifice spacing\n",
+      "\n",
+      "#OUTPUT\n",
+      "print 'fopen',\n",
+      "print '\\t\\tuor(m/s)',\n",
+      "print '\\tls from Eqn.(18)',\n",
+      "print '\\tls from Fig.13(c)'\n",
+      "i=0;\n",
+      "while i<n:\n",
+      "    print '%f'%fopen[i],\n",
+      "    print '\\t%f'%uor[i],\n",
+      "    print '\\t%f'%ls1[i],\n",
+      "    print '\\t\\t%f'%ls2[i]\n",
+      "    i=i+1;   \n",
+      "\n",
+      "print '\\nFor square pitch, the orifice spacing should be %fmm'%lor\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fopen \t\tuor(m/s) \tls from Eqn.(18) \tls from Fig.13(c)\n",
+        "0.120000 \t3.333333 \t12.500000 \t\t12.000000\n",
+        "0.170000 \t2.352941 \t8.823529 \t\t20.000000\n",
+        "0.260000 \t1.538462 \t5.769231 \t\t25.000000\n",
+        "\n",
+        "For square pitch, the orifice spacing should be 48.863850mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/screenshots/plotkbckbedb.png b/Fluidization_Engineering_by_K_Daizo_And_O_Levenspiel/screenshots/plotkbckbedb.png
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diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter01.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter01.ipynb
deleted file mode 100755
index 8e50e4de..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter01.ipynb
+++ /dev/null
@@ -1,578 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e61d5eac9962a1b27acd93d08c5b466532db593eb946c997e161cd14ea1acee1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 01 : Operators and Expressions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1: Page No.:4.35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Usage of Arithmetic operators\n",
-      "\n",
-      "#Variable declaration\n",
-      "i=10\n",
-      "j=20\n",
-      "\n",
-      "#the sum value of variables i and j stores in k variable \n",
-      "k=i+j\n",
-      "\n",
-      "#Result\n",
-      "print \"Value of k is %d\"%(k)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of k is 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2: Page No.:4.35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Demonstrations on arithmetic operators\n",
-      "\n",
-      "#User Input\n",
-      "\n",
-      "a,b,c,d=input(\"Enter values of a, b, c, d: \")\n",
-      "\n",
-      "#Calculation\n",
-      "a+=(b*c)+d\n",
-      "\n",
-      "#Result\n",
-      "print \"Value of a = %d \"%(a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter values of a, b, c, d ,e: 12,90,56,43\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of a = 5095 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3: Page No.:4.36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find out all arithmetic operation on two given values\n",
-      "\n",
-      "#Variable Declaration\n",
-      "sum=sub=mul=rem=0\n",
-      "div=0.0\n",
-      "a=0\n",
-      "#User Input\n",
-      "b,c,d=input(\"Enter the values of b, c, d: \")\n",
-      "\n",
-      "sum = b + c\n",
-      "sub = b - c\n",
-      "mul = b * c\n",
-      "div = b/c\n",
-      "rem = b % d\n",
-      "\n",
-      "print \"Sum = %d, Sub = %d, Mul = %d, Div = %f\"%(sum,sub,mul,div)\n",
-      "print \"Remainder of division of b & d is %d\"%(rem)\n",
-      "print \"a=%d\"%(a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the values of b, c, d: 2,6,45\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum = 8, Sub = -4, Mul = 12, Div = 0.000000\n",
-        "Remainder of division of b & d is 2\n",
-        "a=0\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4: Page No.: 4.38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to use various relational operators\n",
-      "\n",
-      "#!=,==,>=,<= are assignment operator\n",
-      "\n",
-      "print \"Condition :  Return values\"\n",
-      "print \"  5!=5    :  %5d\"%(5!=5)     \n",
-      "print \"  5==5    :  %5d\"%(5==5)\n",
-      "print \"  5>=5    :  %5d\"%(5>=5)\n",
-      "print \"  5<=5    :  %5d\"%(5<=50)\n",
-      "print \"  5!=3    :  %5d\"%(5!=3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Condition :  Return values\n",
-        "  5!=5    :      0\n",
-        "  5==5    :      1\n",
-        "  5>=5    :      1\n",
-        "  5<=5    :      1\n",
-        "  5!=3    :      1\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5: Page No.:4.39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Demonstrate of logical operator\n",
-      "\n",
-      "#Variable Declaration\n",
-      "c1=c2=c3=0\n",
-      "\n",
-      "#USer Input\n",
-      "c1,c2,c3=input(\"Enter Values of C1, C2 and C3: \")\n",
-      "print \"\\n\"\n",
-      "\n",
-      "if (c1<c2) and (c1<c3):\n",
-      "    print \"C1 is less than C2 and C3\"\n",
-      "\n",
-      "if not(c1) < (c2 ):\n",
-      "    print \"C1 is greater than C2\"\n",
-      "    \n",
-      "if ((c1 < c2 or c1 < c3)):\n",
-      "    print \"C1 is less than C2 or C3 or both\"\n",
-      "    \n",
-      "#else:\n",
-      " #   print \"Something went wrong. Check\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values of C1, C2 and C3: 45,32,98\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "C1 is greater than C2\n",
-        "C1 is less than C2 or C3 or both\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6: Page No.: 4.40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Demonstration of assignment operator\n",
-      "\n",
-      "#Variable Declaration\n",
-      "i=4\n",
-      "j=5\n",
-      "\n",
-      "#'=' is assignment operator\n",
-      "i=j\n",
-      "k=i\n",
-      "\n",
-      "#Result\n",
-      "print \"K = %d\"%(k)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "K = 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7: Page No.:4.42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using Increment and Decrement Operator.\n",
-      "\n",
-      "#Variable Declaration\n",
-      "a=10\n",
-      "\n",
-      "#There is no post and pre increment in python\n",
-      "#There is no post and pre decrement in python\n",
-      "\n",
-      "#Result\n",
-      "print \"The Value of a = %d\"%(a)\n",
-      "print \"Increment of a = %d\"%(a+1)\n",
-      "print \"Decrement of a = %d\"%(a-1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of a = 10\n",
-        "Increment of a = 11\n",
-        "Decrement of a = 9\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8: Page No.:4.42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using increment and decrement operators\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "i=3\n",
-      "j=4\n",
-      "\n",
-      "#No Pre Decrement and Pre Increment in python\n",
-      "\n",
-      "i=i+1                  #Increment value\n",
-      "j=j-1               #Decrement value\n",
-      "\n",
-      "k=i+j\n",
-      "\n",
-      "#Result\n",
-      "print \"i = %d, j = %d, k = %d\"%(i,j,k)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i = 4, j = 3, k = 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9: Page No.: 4.46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to illustrate sizeof operator\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "num=int(1234567890)\n",
-      "dec=float(0.123456)\n",
-      "ext=0.123456789\n",
-      "ltr='A'\n",
-      "string=\"Something to write home about....\"\n",
-      "class struct():\n",
-      "    a=0\n",
-      "boy=struct()\n",
-      "\n",
-      "#Result\n",
-      "print \"Size of num is in %d bytes.\"%(sys.getsizeof(num))\n",
-      "print \"Size of dec is in %d bytes.\"%(sys.getsizeof(dec))\n",
-      "print \"Size of ext double is in %d bytes.\"%(sys.getsizeof(ext))\n",
-      "print \"Size of ltr char is in %d bytes.\"%(sys.getsizeof(ltr))\n",
-      "print \"Size of the string is in %d bytes.\"%(sys.getsizeof(string))\n",
-      "print \"Size of struct is in %d bytes.\"%(sys.getsizeof(struct()))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Size of num is in 12 bytes.\n",
-        "Size of dec is in 16 bytes.\n",
-        "Size of ext double is in 16 bytes.\n",
-        "Size of ltr char is in 22 bytes.\n",
-        "Size of the string is in 54 bytes.\n",
-        "Size of struct is in 32 bytes.\n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 1: PageNo.:4.52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert number of days to months and days\n",
-      "\n",
-      "#Variable Declaration\n",
-      "m=0\n",
-      "d=0\n",
-      "\n",
-      "#User Input\n",
-      "nd=input(\"Enter the number of days= \")\n",
-      "\n",
-      "\n",
-      "m=nd/30\n",
-      "d=nd%30\n",
-      "\n",
-      "#Result\n",
-      "print \"Number of months...\", m\n",
-      "print \"Number of days...\", d"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of days= 215\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of months... 7\n",
-        "Number of days... 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 2: Page No.:4.53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate Salesman Salary\n",
-      "\n",
-      "#Variable Declaration\n",
-      "BASIC=2500\n",
-      "BONUS_RATE=200\n",
-      "COM=0.02\n",
-      "\n",
-      "qty=0\n",
-      "GS=price=Bonus=Comm=0.0\n",
-      "#User input\n",
-      "qty,price=input(\"Enter the number of items Sold and Price...: \")\n",
-      "\n",
-      "Bonus=BONUS_RATE*qty\n",
-      "Comm=COM*qty*price\n",
-      "\n",
-      "GS=BASIC+Bonus+Comm\n",
-      "\n",
-      "print \"Bonus.... %6.2f\"%(Bonus)\n",
-      "\n",
-      "print \"Commission.... %6.2f\"%(Comm)\n",
-      "\n",
-      "print \"Gross Salary... %6.2f\"%(GS)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of items Sold and Price...: 10,200\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bonus.... 2000.00\n",
-        "Commission....  40.00\n",
-        "Gross Salary... 4540.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 75
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 3: Page No.:4.54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Post increment using different storage classes\n",
-      "\n",
-      "#Defining function\n",
-      "def postinc():\n",
-      "    x=1\n",
-      "    print \"X = %d\" %(x)\n",
-      "    x=x+1\n",
-      "    \n",
-      "#Call Function\n",
-      "postinc()\n",
-      "postinc()\n",
-      "postinc()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "X = 1\n",
-        "X = 1\n",
-        "X = 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 77
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter02.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter02.ipynb
deleted file mode 100755
index 1aa8a278..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter02.ipynb
+++ /dev/null
@@ -1,767 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e63878167c3dee16dba0cb7cf1eb45b07d3e347ca99ab3a68339c448d68bdb66"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02 : Managing Input and Output Operators"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page No 4.56"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Testing Characters\n",
-      "\n",
-      "#Variable Declaration\n",
-      "ch=''\n",
-      "\n",
-      "#User Input\n",
-      "ch=raw_input(\"Give any Character as a input: \")\n",
-      "\n",
-      "if (ch.isalpha()>0):          #Checking whether the character is alphabet\n",
-      "    print ch,\"is a alphabet\"\n",
-      "    \n",
-      "elif(ch.isdigit()>0):         #Checking whether the character is digit\n",
-      "    print ch,\"is a digit\"\n",
-      "    \n",
-      "else:                          #else print alphanumeric\n",
-      "    print ch,\"is a alphanumeric\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give any Character as a input: a\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a is a alphabet\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2, Page No 4.57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert Uppercase to Lowercase and Viceversa\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "ch=''\n",
-      "\n",
-      "\n",
-      "#User Input\n",
-      "ch= raw_input(\"Give the input in lower case or upper case: \")\n",
-      "\n",
-      "if(ch.islower()):      #Converting lowercase if input is uppercase\n",
-      "    print ch.upper(), \"is a upper case character\"\n",
-      "    \n",
-      "else:                   #else print lowercase\n",
-      "    print ch.lower(), \"is a lower case character\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the input in lower case or upper case: s\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "S is a upper case character\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3, Page No 4.58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using input function\n",
-      "\n",
-      "scientist=''\n",
-      "\n",
-      "scientist = raw_input(\"Enter Name: \")\n",
-      "print \"Print the name: \", scientist\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name: Abdul kalam\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Print the name:  Abdul kalam\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4, Page No 4.62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count numbers of entered numbers\n",
-      "\n",
-      "#Variable declaration\n",
-      "i=0\n",
-      "j=10\n",
-      "k=0\n",
-      "\n",
-      "#getting input from user     \n",
-      "j=int(raw_input(\"Give input for j \"))\n",
-      "k=int(raw_input(\"Give input for k \"))\n",
-      "i=int(raw_input(\"Give input for i \"))\n",
-      "\n",
-      "i= i, j, k       #assigning the user input in to a variable\n",
-      "\n",
-      "#Result\n",
-      "print \"No. of given input values:\",  len(i)      #counting length of the given input\n",
-      "print \"The Value of j and k is \", j, k              #printing the values of input"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give input for j 56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give input for k 32\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give input for i 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "No. of given input values: 3\n",
-        "The Value of j and k is  56 32\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5, Page No 4.65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print a statement using print\n",
-      "\n",
-      "print \"VRB Publishers\"     #Printing statement"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VRB Publishers\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6, Page No 4.65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Adding two numbers\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "a=6\n",
-      "b=8\n",
-      "c= a+b      #Adding a, b and stores in c \n",
-      "\n",
-      "#Result\n",
-      "print \"Addition of two numbers is : \", c"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Addition of two numbers is :  14\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7, Page No 4.66"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reading and Displaying Details of a student\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "sno=m1=m2=m3=0\n",
-      "name=['20']\n",
-      "total=0\n",
-      "avg=0.0\n",
-      "\n",
-      "#User Inputs\n",
-      "sno=int(raw_input(\"Give Student serial no. \"))\n",
-      "name=raw_input(\"Student Name : \")\n",
-      "m1=int(raw_input(\"Student Mark 1: \"))\n",
-      "m2=int(raw_input(\"Student Mark 2: \"))\n",
-      "m3=int(raw_input(\"Student Mark 3: \"))\n",
-      "\n",
-      "total=float(m1+m2+m3)        #Calculation of marks\n",
-      "\n",
-      "avg=\"%.2f\"%float(total/3.0)           #Average\n",
-      "\n",
-      "#Result\n",
-      "print '\\n'\"Student serial No.:\", sno\n",
-      "print \"Student Name:\", name\n",
-      "print \"Student's Marks:\", m1, m2, m3\n",
-      "print \"Total:\", total\n",
-      "print \"Average:\", avg "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give Student serial no. 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Student Name : Raja\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Student Mark 1: 90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Student Mark 2: 95\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Student Mark 3: 98\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Studnt serial No.: 1\n",
-        "Student Name: Raja\n",
-        "Student's Marks: 90 95 98\n",
-        "Total: 283.0\n",
-        "Average: 94.33\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8, Page No:4.71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display quotient and reminder of given input\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "a=b=q=r=0\n",
-      "\n",
-      "#User input\n",
-      "a=int(raw_input(\"Give input for a: \"))\n",
-      "b=int(raw_input(\"Give input for b: \"))\n",
-      "\n",
-      "q=float(a/b)              #Calculation for Quotient\n",
-      "\n",
-      "r=float(a-q*b)         #Calculation for Reminder\n",
-      "\n",
-      "#Result\n",
-      "print \"Quotient of a and b is :\",q\n",
-      "print \"Reminder of a and b is :\",r"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give input for a: 39\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give input for b: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quotient of a and b is : 5.0\n",
-        "Reminder of a and b is : 4.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9, Page No:4.73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Slope and Midpoint of a line\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "x=x1=y1=x2=y2=slope=0.0\n",
-      "\n",
-      "#User input1\n",
-      "\n",
-      "x1=int(raw_input(\"Give point X1: \"))\n",
-      "y1=int(raw_input(\"Give point Y1: \"))\n",
-      "\n",
-      "\n",
-      "#User input2\n",
-      "\n",
-      "x2=int(raw_input('\\n'\"Give point X2: \"))\n",
-      "y2=int(raw_input(\"Give point Y2: \"))\n",
-      "\n",
-      "slope =(y2-y1)/(x2-x1)            #Calculation for Slope\n",
-      "\n",
-      "#Calculation for Midpoints\n",
-      "x=float(x1+x2)/2.0\n",
-      "y=float(y1+y2)/2.0\n",
-      "\n",
-      "#Result\n",
-      "print '\\n' \"Slope of a line is :\", slope\n",
-      "print \"Midpoint of a line is :\", x,y"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give point X1: 10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give point Y1: 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Give point X2: 30\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give point Y2: 40\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Slope of a line is : 1\n",
-        "Midpoint of a line is : 20.0 30.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10, Page No:4.73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Average of Given N Numbers\n",
-      "\n",
-      "#Variable Declaration\n",
-      "n=0.0\n",
-      "x=0\n",
-      "total=0\n",
-      "avg=0.0\n",
-      "\n",
-      "#UserInput\n",
-      "\n",
-      "n=int(raw_input(\"Give total number of input: \"))\n",
-      "\n",
-      "print \"Give those\",n ,\"Numbers\"\n",
-      "\n",
-      "for i in range(n):          #Looping values of n\n",
-      "    i=1\n",
-      "    x=input()               #Values for n\n",
-      "    total+=x                 # Adding given numbers\n",
-      "    \n",
-      "avg=float(total/n)           #Calculating an average of given numbers\n",
-      "\n",
-      "#Result\n",
-      "print \"Sum of given numbers is \", total\n",
-      "print \"Average is \", avg\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give total number of input: 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give those 5 Numbers\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "17\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "76\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of given numbers is  112\n",
-        "Average is  22.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11, Page No:4.75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Temperature Conversion (Fahrenheit to Celsius)\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "F_MIN =0\n",
-      "F_MAX=500\n",
-      "INC=50\n",
-      "fh= F_MIN\n",
-      "\n",
-      "print  \"Fahrenheit\" '\\t' \"Celsius\"\n",
-      "while(fh<=F_MAX):\n",
-      "    cel =\"%.2f\"%float((fh-(32.0))/1.8)     #Calculating Celsius value\n",
-      "    print  fh,'\\t''\\t', cel\n",
-      "    fh=fh+INC              #Calculating Fahrenheit value\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fahrenheit\tCelsius\n",
-        "0 \t\t-17.78\n",
-        "50 \t\t10.00\n",
-        "100 \t\t37.78\n",
-        "150 \t\t65.56\n",
-        "200 \t\t93.33\n",
-        "250 \t\t121.11\n",
-        "300 \t\t148.89\n",
-        "350 \t\t176.67\n",
-        "400 \t\t204.44\n",
-        "450 \t\t232.22\n",
-        "500 \t\t260.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12, Page No:4.76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Area of Triangle\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "a=b=c=s=area=0.0\n",
-      "\n",
-      "#User input\n",
-      "\n",
-      "a=int(raw_input(\"First side of Triangle is \"))\n",
-      "b=int(raw_input(\"Second side of Triangle is \"))\n",
-      "c=int(raw_input(\"Third side of Triangle is \"))\n",
-      "\n",
-      "s=(a+b+c)/2.0      #Calculating Size of triangle\n",
-      "\n",
-      "area=\"%.3f\"%float((s*(s-a)*(s-b)*(s-c)) ** 0.5)    #Calculating Area of triangle\n",
-      "\n",
-      "#Result\n",
-      "print '\\n',\"Area of Triangle is\",area"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "First side of Triangle is 10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Second side of Triangle is 10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Third side of Triangle is 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Area of Triangle is 43.301\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter03.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter03.ipynb
deleted file mode 100755
index 34a44d60..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter03.ipynb
+++ /dev/null
@@ -1,440 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:92464c4be2e3e810ccda180a900c5ec8bb15a9537c93387b95c317c7709c7b7c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 03 : Decision Making"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page No.:4.79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using if with multiple statements\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "i=0\n",
-      "\n",
-      "#User Input\n",
-      "\n",
-      "i =int(raw_input(\"Give the numbers less than 10\"'\\n'))\n",
-      "\n",
-      "if i<=10:                  #Checking the given value\n",
-      "\n",
-      "#Result       \n",
-      "    print \"The given value of\", i , \"less than 10\"\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the numbers less than 10\n",
-        "5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the given value of 5 less than 10\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.: 4.79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "##Program for interchange two values using if statement\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "a=m=n=0\n",
-      "\n",
-      "#User Input\n",
-      "\n",
-      "m=int(raw_input(\"Give the value of m\" '\\t'))\n",
-      "n=int(raw_input(\"Give the value of n\" '\\t'))\n",
-      "\n",
-      "if m>=n :                   #Checking the Given Values\n",
-      "    \n",
-      "    a=m                     # method for swapping            \n",
-      "    m=n\n",
-      "    n=a\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The Swapped Values of m and n are \", m,a"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of m\t34\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of n\t28\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Swapped Values of m and n are  28 34\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No. 4.81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#determine the given number is odd or even. \n",
-      "\n",
-      "#Variable Declaration\n",
-      "num=rem=0\n",
-      "\n",
-      "#User Input\n",
-      "num=input(\"Enter the value: \")\n",
-      "rem=num%2           \n",
-      "if rem==0 :             #Checking condition\n",
-      "\n",
-      "#Result 1\n",
-      "    print \"The given number is Even\"\n",
-      "else :\n",
-      "    \n",
-      "#Result 2\n",
-      "    print \"The given number is Odd\"\n",
-      "    \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the value: 80\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The given number is Even\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page No.:4.81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Find floating point input using if-else statement\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "years=seconds=0.0\n",
-      "life=0\n",
-      "\n",
-      "years=float(raw_input(\"Give your Age in years: \"))\n",
-      "life=years\n",
-      "if life==0:\n",
-      "    print \"The given age is not floating value\"\n",
-      "    \n",
-      "else :\n",
-      "    seconds=\"%f\"%float(years*365*24*60*60)\n",
-      "    print \"You have lived for\", seconds, \"seconds\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give your Age in years: 24\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have lived for 756864000.000000 seconds\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 5, Page No. :4.82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# find biggest among two numbers \n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "i=j=big=0\n",
-      "i,j=input(\"Give two values: \")\n",
-      "\n",
-      "big=i\n",
-      "\n",
-      "if big<j :\n",
-      "    big=j\n",
-      "    \n",
-      "elif i<j :\n",
-      "    big=j \n",
-      "    \n",
-      "else:\n",
-      "    big = i\n",
-      "\n",
-      "print \"Biggest number of two numbers: \",big\n",
-      "print \"Biggest of two numbers(using else) is: \",big"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give two values: 45,78\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Biggest number of two numbers:  78\n",
-        "Biggest of two numbers(using else) is:  78\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 6, Page No. : 4.84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Program using nested if....else \n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "n=0\n",
-      "\n",
-      "#User Input\n",
-      "n=int(raw_input(\"Enter the number\"))\n",
-      "\n",
-      "if n==15:           #Checking Condition\n",
-      "    print \"Play Foot ball\"\n",
-      "    \n",
-      "elif n==10:\n",
-      "    print \"Play Cricket\"\n",
-      "\n",
-      "else :    \n",
-      "    print \"Don't Play\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Play Cricket\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 7, Page No. :4.85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#print the number between 10 and 20\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "i=0\n",
-      "\n",
-      "#User Input\n",
-      "\n",
-      "i=input(\"Enter the number: \")\n",
-      "\n",
-      "if (i>10) & (i<20):\n",
-      "    print \"The given value is in between 10 and 20\"\n",
-      "\n",
-      "else :\n",
-      "    print \"The given value is greater than 20 and less than 10\"    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number: 15\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The given value is in between 10 and 20\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 8, Page No. :4.86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# display the types of characterusing if-else statement\n",
-      "\n",
-      "#User Input\n",
-      "\n",
-      "chr= raw_input(\"Enter a Single Character: \")\n",
-      "\n",
-      "if( ((chr>='a') & (chr<='z')) | ((chr>='A') & (chr<='Z'))):\n",
-      "    print \"Given charactered is an Alphabetic\"\n",
-      "    \n",
-      "elif ((chr>='0') & (chr <='9')):\n",
-      "    print \"Given Character is a digit\"\n",
-      "    \n",
-      "else:\n",
-      "    print \"Entered Character is a special character\"\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Single Character: M\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Given charactered is an Alphabetic\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter04.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter04.ipynb
deleted file mode 100755
index a58b3c75..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter04.ipynb
+++ /dev/null
@@ -1,1337 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:cbd54959e18d38695240f376809cd86bcdd3f9c461e014b42a620c0846ec3cbf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 04 : Branching and Looping"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1: Page No.:4.88"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# addition of numbers using while loop.\n",
-      "\n",
-      "#Varibale Declaration\n",
-      "\n",
-      "i=1\n",
-      "sum=0\n",
-      "while i<=10:                    #Checking while loop\n",
-      "    sum+=i\n",
-      "    i+=1                        #Increment of i\n",
-      "    \n",
-      " #Result   \n",
-      "print \"The Sum of the numbers upto is \",sum"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Sum of the numbers upto is  55\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2: Page No.: 4.89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "i=1\n",
-      "j=0\n",
-      "\n",
-      "    \n",
-      "print \"The maximum value of integer is\", sys.maxsize\n",
-      "print \"The Value of integer after overflow is\", sys.maxint-1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The maximum value of integer is 2147483647\n",
-        "The Value of integer after overflow is 2147483646\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3: Page No.: 4.89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reverse the given number using while loop\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "number=digit=rev=0\n",
-      "\n",
-      "number=input(\"Give the Value: \")\n",
-      "\n",
-      "while(number!=0):\n",
-      "    digit=number%10\n",
-      "    rev=rev*10+digit\n",
-      "    number= number/10\n",
-      "    \n",
-      "print \"The reversible value of given number is\", rev\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the Value: 7896\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reversible value of given number is 6987\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4: Page No.: 4.90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Simple interest using while loop.\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "p=1\n",
-      "n=1\n",
-      "count=1\n",
-      "\n",
-      "r=si=0.0\n",
-      "\n",
-      "while(count<=3):\n",
-      "    p=input(\"Give the value of P\" '\\t')\n",
-      "    n=input(\"Give the value of N\"'\\t')\n",
-      "    r=float(input(\"Give the value of R\"'\\t'))\n",
-      "    \n",
-      "    si=\"%.3f\"%float((p*n*r)/100)\n",
-      "    \n",
-      "    print \"Simple Interest of SI\", si\n",
-      "    count=count+1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of P\t2000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of N\t3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of R\t5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Simple Interest of SI 300.000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of P\t3500\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of N\t6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of R\t1.5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Simple Interest of SI 315.000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of P\t6000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of N\t2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of R\t3.5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Simple Interest of SI 420.000\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5: Page No.:4.92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Numbers upto 10 by using the do..while loop\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "i=1\n",
-      "sum=0\n",
-      "\n",
-      "while True:\n",
-      "\tsum+=i\n",
-      "\ti+=1\n",
-      "\tif(i>10):\n",
-      "\t\tbreak\n",
-      "print \"Sum of the numbers upto 10 is\", sum  \n",
-      "   \n",
-      "            \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of the numbers upto 10 is 55\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6: Page No.: 4.92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#print n numbers using do...while loop\n",
-      "\n",
-      "i=0\n",
-      "\n",
-      "n=input(\"Enter the number: \")\n",
-      "\n",
-      "while True:\n",
-      "    print \"The numbers are: \",i\n",
-      "    i=i+1\n",
-      "    if(i>=n):\n",
-      "        break\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number: 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numbers are:  0\n",
-        "The numbers are:  1\n",
-        "The numbers are:  2\n",
-        "The numbers are:  3\n",
-        "The numbers are:  4\n",
-        "The numbers are:  5\n"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7: page no.: 4.95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Addition of number upto 10 using for loop\n",
-      "\n",
-      "i=sum=0\n",
-      "\n",
-      "for i in range (0,11):\n",
-      "\tsum+=i\n",
-      "\ti+=1\n",
-      "\n",
-      "print \"The addition of given numbers\",sum\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The addition of given numbers 55\n"
-       ]
-      }
-     ],
-     "prompt_number": 75
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8: Page No.:4.96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#print n numbers using for...loop structure\n",
-      "\n",
-      "n=int(raw_input(\"Give Values of n: \"))\n",
-      "i=0\n",
-      "for i in range(0,n):\n",
-      "    print \"The Given numbers are: \",i\n",
-      "\n",
-      "\t"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give Values of n: 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Given numbers are:  0\n",
-        "The Given numbers are:  1\n",
-        "The Given numbers are:  2\n",
-        "The Given numbers are:  3\n",
-        "The Given numbers are:  4\n"
-       ]
-      }
-     ],
-     "prompt_number": 82
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9: Page No.:4.97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#student marks and average using for loop..\n",
-      "\n",
-      "a=b=c=d=e=total=0\n",
-      "\n",
-      "avg=0.0\n",
-      "\n",
-      "m=input(\"Enter the number of Students:\")\n",
-      "\n",
-      "print \"Enter the marks of the five subjects: \"'\\n'\n",
-      "\n",
-      "for i in range (m):   \n",
-      "    i+=1\n",
-      "    print \"\\nStudent:\",i,\"\\n\"\n",
-      "    \n",
-      "    a,b,c,d,e=input(\"Enter the Student mark1: \")                       \n",
-      "             \n",
-      "    if ((a,b,c,d,e)!=0) :\n",
-      "        total=a+b+c+d+e\n",
-      "        avg=\"%.3f\"%float(total/5)\n",
-      "        print \"Total marks of the student:\", total\n",
-      "        print \"Average marks of the students: \",avg\n",
-      "                       \n",
-      "    else  :\n",
-      "        print \"Entered data is invalid\"\n",
-      "                \n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of Students:3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the marks of the five subjects: \n",
-        "\n",
-        "\n",
-        "Student: 1 \n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Student mark1: 78,56,90,59,87\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total marks of the student: 370\n",
-        "Average marks of the students:  74.000\n",
-        "\n",
-        "Student: 2 \n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Student mark1: 49,87,60,57,93\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total marks of the student: 346\n",
-        "Average marks of the students:  69.000\n",
-        "\n",
-        "Student: 3 \n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Student mark1: 89,100,80,97,65\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total marks of the student: 431\n",
-        "Average marks of the students:  86.000\n"
-       ]
-      }
-     ],
-     "prompt_number": 89
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10: Page No.:4.98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Nested for loop\n",
-      "\n",
-      "for i in range(1,4):\n",
-      "    for j in range(1,4):\n",
-      "        print \"%d\"%(j),\n",
-      "    print \"\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 2 3 \n",
-        "\n",
-        "1 2 3 \n",
-        "\n",
-        "1 2 3 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 112
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11: Page No.:4.99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#student details using nested for loop..\n",
-      "\n",
-      "FIRST=300\n",
-      "SECOND=200\n",
-      "rollno=marks=total=0\n",
-      "\n",
-      "n=input(\"Enter the number of Students: \")\n",
-      "m=input(\"Enter the number of subjects: \")\n",
-      "\n",
-      "for i in range(n):\n",
-      "    i+=(+1)\n",
-      "    rollno=int(raw_input(\"Enter the student roll number: \"))\n",
-      "    total=0\n",
-      "    print \"Enter marks of\",m, \"subjects for Roll no: \",rollno\n",
-      "    for j in range(m):\n",
-      "        marks=int(raw_input(\"Give marks: \"))\n",
-      "        total= total+marks\n",
-      "        \n",
-      "    print \"total marks\", total\n",
-      "    if(total>=FIRST):\n",
-      "        print \"(First Class)\"'\\n'\n",
-      "    elif(total>=SECOND):\n",
-      "        print \"(Second Class)\"'\\n'\n",
-      "    else:\n",
-      "        print \"Fail\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of Students: 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of subjects: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student roll number: 201\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter marks of 5 subjects for Roll no:  201\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 68\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 89\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "total marks 415\n",
-        "(First Class)\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student roll number: 202\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter marks of 5 subjects for Roll no:  202\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 40\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 49\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 39\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give marks: 47\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "total marks 225\n",
-        "(Second Class)\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 85
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12: Page No.: 4.103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Demonstrate switch case\n",
-      "\n",
-      "a=1\n",
-      "if a==1:\n",
-      "    print \"I am in case 1\"'\\n'\n",
-      "elif a==2:\n",
-      "    print \"I am in case2\"'\\n'\n",
-      "else:\n",
-      "    print \"I am in case default\"'\\n'"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "I am in case 1\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 115
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13: Page No.:4.104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to use the computer as calculator\n",
-      "\n",
-      "a=b=c=0\n",
-      "\n",
-      "op=raw_input(\"Give the option for calculation like +(ADD), -(SUB), *(MUL), /(DIV): \")\n",
-      "a=input(\"Enter the value of a \")\n",
-      "b=input(\"Enter the value of b \")\n",
-      "\n",
-      "#Calculation\n",
-      "if (op=='+'):\n",
-      "  c=a+b\n",
-      "  print \"Result of C\", c \n",
-      "elif(op=='-'):\n",
-      "    c=a-b\n",
-      "    print \"Result of C\", c\n",
-      "elif(op=='*'):\n",
-      "    c=a*b\n",
-      "    print \"Result of C\", c\n",
-      "else:\n",
-      "    c=a/b\n",
-      "    print \"Result of C\", c\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the option for calculation like +(ADD), -(SUB), *(MUL), /(DIV): +\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the value of a 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the value of b 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result of C 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 118
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14: Page No.:4.105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find even or odd numbers \n",
-      "\n",
-      "#USer Input\n",
-      "a=input(\"Enter a number: \")\n",
-      "\n",
-      "if a%2==0 : \n",
-      "    print  \" The number is even. \"\n",
-      "else :\n",
-      "    print \"The number is odd.\" \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 89\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The number is odd.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15: Page No.:4.106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to count number of 0s, 1s, blank spaces and other characters\n",
-      "\n",
-      "\n",
-      "txt=[30]\n",
-      "count =0 \n",
-      "txt=raw_input(\"Enter Numbers\")\n",
-      "\n",
-      "print \"Total Spaces  : %d\"%txt.count(' ')\n",
-      "print \"Total of 1's  : %d\"%txt.count('1')\n",
-      "print \"Total of 0's  : %d\"%txt.count('0')\n",
-      "print \"String Length : %d\"%len(txt)\n",
-      "            \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Numbers120010  09100110\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total Spaces  : 2\n",
-        "Total of 1's  : 5\n",
-        "Total of 0's  : 7\n",
-        "String Length : 16\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16: Page No.: 4.108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the number upto 5 using break statement\n",
-      "\n",
-      "for i in range(1,10):\n",
-      "    if i==6:\n",
-      "        break\n",
-      "    print i,"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 2 3 4 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 120
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17: Page No.: 4.109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to calculate the sum of the given positive numbers\n",
-      "\n",
-      "i=n=sum=0\n",
-      "\n",
-      "for i in range(0,5):\n",
-      "    n=int(raw_input(\"Give any value of n \"))\n",
-      "    if n<0:\n",
-      "       continue\n",
-      "    else:\n",
-      "       sum+=n\n",
-      "       \n",
-      "print \"\\nSum is\", sum"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give any value of n 10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give any value of n 15\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give any value of n 25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give any value of n -10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give any value of n 50\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum is 100\n"
-       ]
-      }
-     ],
-     "prompt_number": 124
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18: Page No.:4.113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using goto statement\n",
-      "\n",
-      "a=input(\"Give the Value of a \")\n",
-      "b=input(\"Give the value of b \")\n",
-      "\n",
-      "if a==b:\n",
-      "   print \"A and B are equal\"\n",
-      "else:\n",
-      "    print \"A and B are not equal\"\n",
-      "    exit(0)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the Value of a 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of b 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A and B are equal\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 1: Page No.: 4.112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Prime or not\n",
-      "\n",
-      "n=input(\"Enter the Positive value of n \")\n",
-      "i=2\n",
-      "\n",
-      "#Calculation\n",
-      "while n>i:\n",
-      "    if n%i==0 & i!=n:\n",
-      "        #r=n%i\n",
-      "        #if r==0:\n",
-      "            print \"The Given value of \",n,\" is not a prime\"\n",
-      "            break\n",
-      "    i+=1\n",
-      "else:\n",
-      "     print \"The Given Value\",n, \"is Prime\"\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Positive value of n 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Given Value 3 is Prime\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 2: Page No.: 4.113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the Greatest Common Division\n",
-      "\n",
-      "i=int(input(\"Give the Value of i \"))\n",
-      "j=int(input(\"Give the value of j \"))\n",
-      "  \n",
-      "def gcd(i,j):\n",
-      "    if i > j:\n",
-      "        k = i%j\n",
-      "        if k == 0:\n",
-      "            return j\n",
-      "        else:\n",
-      "            return gcd(j, k)\n",
-      "    if i < j:\n",
-      "        i = j\n",
-      "        j = i\n",
-      "        return gcd(i, j)\n",
-      "print \"The value of gcd\", gcd(i,j) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the Value of i 54\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the value of j 36\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of gcd 18\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 3: Page No.: 4.115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to which reads in a Temperature and print the temperature in opposite scale\n",
-      "\n",
-      "\n",
-      "print \"1.Fahrenheit to Celcius\\n2.Celcius to Fahrenheit\\n\"\n",
-      "code=input(\"Choose Option to find Temperature: \")\n",
-      "\n",
-      "while code!=0:\n",
-      "    if code==1:\n",
-      "        f=float(input(\"Give the Value for Fahrenheit Temperature \"))\n",
-      "        c=(f-32.0)*5.0/9.0\n",
-      "        break\n",
-      "    elif code==2:\n",
-      "        c=float(input(\"Give the Value for Celcius Temperature \"))\n",
-      "        f=9.0*c/5.0+32.0\n",
-      "        break\n",
-      "    else:\n",
-      "        print \"Give proper Code Value for Temperature Calculation\"\n",
-      "        break\n",
-      "print \"C=\", \"%.3f\"%float(c)\n",
-      "print \"F=\",f "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.Fahrenheit to Celcius\n",
-        "2.Celcius to Fahrenheit\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Choose Option to find Temperature: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give the Value for Fahrenheit Temperature 100\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C= 37.778\n",
-        "F= 100.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter05.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter05.ipynb
deleted file mode 100755
index fa4db685..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter05.ipynb
+++ /dev/null
@@ -1,987 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8b48219cd7d973f69d80d4d491de4bdba5cb9c73135095697b1625a862f6bc97"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 05 : Defining and Processing of Arrays"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page Number: 5.5"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to read 5 data items from input and store them in an array and display their sum\n",
-      "\n",
-      "#Local definitions\n",
-      "a = []\n",
-      "sum = 0\n",
-      "\n",
-      "for i in range(0,5):\n",
-      "    a.append(0)\n",
-      "print \"Enter 5 integer numbers\"    \n",
-      "for i in range(0,5):\n",
-      "    a[i] = eval(raw_input())\n",
-      "    \n",
-      "for i in range(0,5):    \n",
-      "    sum = sum + a[i] \n",
-      "    \n",
-      "print \"The sum of given numbers is : %d\\n\" %sum"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 5 integer numbers\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "120\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "69\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "390\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "45\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of given numbers is : 669\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2 Page No. 5.7"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to accept 5 numbers and print whether the number is even or not\n",
-      "\n",
-      "#Local Variable\n",
-      "a=[]\n",
-      "\n",
-      "for i in range(5):\n",
-      "    a.append(0)\n",
-      "for i in range(0,5):\n",
-      "    i+1\n",
-      "    a[i]=eval(raw_input(\"Enter the %d number: \"%(i+1))) \n",
-      "    print \"\\n\"\n",
-      "for i in range (0,5):\n",
-      "    if a[i]%2==0:\n",
-      "        print \"%d Number is Even\"%a[i]\n",
-      "    else:\n",
-      "        print \"%d Number is Odd\"%a[i]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 1 number: 17\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 2 number: 38\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 3 number: 79\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 4 number: 80\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 5 number: 76\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "17 Number is Odd\n",
-        "38 Number is Even\n",
-        "79 Number is Odd\n",
-        "80 Number is Even\n",
-        "76 Number is Even\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3 Page No. 5.8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to illustrate how a character array can be used.\n",
-      "\n",
-      "#Local definition\n",
-      "word=['F', 'R', 'I', 'E', 'N', 'D', 'S']\n",
-      "\n",
-      "for i in range(0,7):\n",
-      "    word.append(0)\n",
-      "for i in range(0,7):\n",
-      "    print word[i],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "F R I E N D S\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page Number: 5.8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to sort the given strings alphabetically\n",
-      "\n",
-      "b = raw_input(\"Enter the text to sort : \")\n",
-      "print \"Sorted Text are : \",''.join(sorted(b, key=lambda v: (v.upper(), v[0].islower())))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the text to sort : VRBPublishers\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorted Text are :  BbehilPRrssuV\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 5, Page Number: 5.10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using two dimension array\n",
-      "\n",
-      "#Local definition \n",
-      "stud = [[0 for i in xrange(0,4)] for i in xrange(0,4)] #stud is a array name with 4 rows and 2 columns\n",
-      "\n",
-      "for i in range(0,4):\n",
-      "    print \"Enter the %d Student roll no and Mark:\" %i\n",
-      "    stud[i][1], stud[i][2] = input()\n",
-      "    \n",
-      "for i in range(0,4):\n",
-      "    print \"%d Student roll no %d mark %d\" %(i, stud[i][1], stud[i][2])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 0 Student roll no and Mark:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3977,80\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 1 Student roll no and Mark:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "17776,95\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 2 Student roll no and Mark:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6682,82\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the 3 Student roll no and Mark:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6683,85\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0 Student roll no 3977 mark 80\n",
-        "1 Student roll no 17776 mark 95\n",
-        "2 Student roll no 6682 mark 82\n",
-        "3 Student roll no 6683 mark 85\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 6, Page Number: 5.12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to illustrate Matrix Multiplication\n",
-      "\n",
-      "a = [[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
-      "b = [[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
-      "c = [[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
-      "\n",
-      "while True:\n",
-      "    r1, c1 = input(\"Enter the size of the matrix A....\")\n",
-      "    r2, c2 = input(\"Enter the size of the matrix B....\")\n",
-      "    if c1 == r2:\n",
-      "        print \"Enter matrix A elements...\"\n",
-      "        for i in range (0,r1):\n",
-      "            for j in range (0,c1):\n",
-      "                a[i][j] = input()\n",
-      "            \n",
-      "        print \"Enter matrix B elements...\"\n",
-      "        for i in range (0,r2):\n",
-      "            for j in range (0,c2):\n",
-      "                b[i][j] = input()\n",
-      "            \n",
-      "        for i in range(0,r1):\n",
-      "            for j in range(0,c1):\n",
-      "                c[i][j] = 0\n",
-      "                for k in range(0,c1):\n",
-      "                    c[i][j] += a[i][k] * b[k][j]\n",
-      "                \n",
-      "        print \"The resultant matrix is....\"\n",
-      "        for i in range(0,r1):\n",
-      "            for j in range(0,c1):\n",
-      "                print \"%d\\t\" %c[i][j],\n",
-      "            print \"\\n\"\n",
-      "        break        \n",
-      "    else:\n",
-      "        print \"Multiplication is not possible\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the size of the matrix A....3,3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the size of the matrix B....2,2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Multiplication is not possible\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the size of the matrix A....3,3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the size of the matrix B....3,3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter matrix A elements...\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter matrix B elements...\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resultant matrix is....\n",
-        "18\t18\t18\t\n",
-        "\n",
-        "18\t18\t18\t\n",
-        "\n",
-        "18\t18\t18\t\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 7, Page Number: 5.13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print two dimensional array by row by row\n",
-      "\n",
-      "arr = [[0 for i in xrange(0,4)] for i in xrange(0,4)]\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,4):\n",
-      "        arr[i][j] = (i * 4) + j + 1\n",
-      "        \n",
-      "        \n",
-      "print \"Printing array contents :\"\n",
-      "for i in range(0,3):\n",
-      "   for j in range(0,4):\n",
-      "        print \"%3d\" %arr[i][j],\n",
-      "   print \"\\n\"   "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Printing array contents :\n",
-        "  1   2   3   4 \n",
-        "\n",
-        "  5   6   7   8 \n",
-        "\n",
-        "  9  10  11  12 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 8, Page Number: 5.14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to add and display the results of 5 nos\n",
-      "\n",
-      "#Function definition  \n",
-      "def add(x,ar):\n",
-      "    sum = 0\n",
-      "    for i in range(0,5):\n",
-      "        sum += ar[i]\n",
-      "    print \"Sum is ... %d\" %sum   \n",
-      "    \n",
-      "a = []\n",
-      "n = 5\n",
-      "for i in range(0,5):\n",
-      "    a.append(0)\n",
-      "print \"Enter 5 values\"\n",
-      "for i in range(0,5):\n",
-      "    a[i] = input()\n",
-      "add(n,a) #function call"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 5 values\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "30\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "40\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "50\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum is ... 150\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 9, Page Number: 5.15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program for capture and store n values into array and print them\n",
-      "\n",
-      "#Function definition \n",
-      "def arr(a = []):\n",
-      "    for i in range(0,5):\n",
-      "        print \"Value in array %d\\n\" %a[i]\n",
-      "        \n",
-      "a = []\n",
-      "for i in range(0,5):\n",
-      "    a.append(0)\n",
-      "for i in range(0,5):\n",
-      "    a[i] = i\n",
-      "arr(a)  #Function call"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value in array 0\n",
-        "\n",
-        "Value in array 1\n",
-        "\n",
-        "Value in array 2\n",
-        "\n",
-        "Value in array 3\n",
-        "\n",
-        "Value in array 4\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 10, Page Number: 5.16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print the character string from the array\n",
-      "\n",
-      "name = \"LAK\"\n",
-      "print \"%s\" %name"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "LAK\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 11, Page Number: 5.17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program for explaining the working of 4 dimensional array\n",
-      "\n",
-      "#Local definition\n",
-      "array_4d = [[[[0 for i in xrange(0,3)] for i in xrange(0,3)]for i in xrange(0,3)] for i in xrange(0,3)]\n",
-      "\n",
-      "#looping statements \n",
-      "for a in range(0,3):\n",
-      "    for b in range(0,3):\n",
-      "        for c in range(0,3):\n",
-      "            for d in range(0,3):\n",
-      "                array_4d[a][b][c][d] = a + b + c + d\n",
-      "#Result                \n",
-      "for a in range(0,3)                :\n",
-      "    print \"\\n\"\n",
-      "    for b in range(0,3):\n",
-      "        for c in range(0,3):\n",
-      "            for d in range(0,3):\n",
-      "                print \"%3d\" %array_4d[a][b][c][d],\n",
-      "            print \"\\n\"    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "  0   1   2 \n",
-        "\n",
-        "  1   2   3 \n",
-        "\n",
-        "  2   3   4 \n",
-        "\n",
-        "  1   2   3 \n",
-        "\n",
-        "  2   3   4 \n",
-        "\n",
-        "  3   4   5 \n",
-        "\n",
-        "  2   3   4 \n",
-        "\n",
-        "  3   4   5 \n",
-        "\n",
-        "  4   5   6 \n",
-        "\n",
-        "\n",
-        "\n",
-        "  1   2   3 \n",
-        "\n",
-        "  2   3   4 \n",
-        "\n",
-        "  3   4   5 \n",
-        "\n",
-        "  2   3   4 \n",
-        "\n",
-        "  3   4   5 \n",
-        "\n",
-        "  4   5   6 \n",
-        "\n",
-        "  3   4   5 \n",
-        "\n",
-        "  4   5   6 \n",
-        "\n",
-        "  5   6   7 \n",
-        "\n",
-        "\n",
-        "\n",
-        "  2   3   4 \n",
-        "\n",
-        "  3   4   5 \n",
-        "\n",
-        "  4   5   6 \n",
-        "\n",
-        "  3   4   5 \n",
-        "\n",
-        "  4   5   6 \n",
-        "\n",
-        "  5   6   7 \n",
-        "\n",
-        "  4   5   6 \n",
-        "\n",
-        "  5   6   7 \n",
-        "\n",
-        "  6   7   8 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter06.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter06.ipynb
deleted file mode 100755
index 09f324b0..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter06.ipynb
+++ /dev/null
@@ -1,1424 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d7231c68ebf3352357e97f096d7bb05469f2e0aae52fa150a1a57a7e75ef01fb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 06 : Handling Of Character Strings"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page Number: 5.19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program for reading a line of text\n",
-      "\n",
-      "n = 0\n",
-      "print \"Enter text press RETRUN to end\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter text press RETRUN to end\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page Number: 5.20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#String function\n",
-      "\n",
-      "s1= \"Hello\"\n",
-      "s2='H''e''l''l''o'\n",
-      "\n",
-      "print s1\n",
-      "print s2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hello\n",
-        "Hello\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page Number: 5.22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using strlen() function\n",
-      "\n",
-      "name = \"MUNI\"\n",
-      "len1 = len(name)\n",
-      "len2 = len(\"LAK\")\n",
-      "print \"String length of %s is %d\" %(name,len1)\n",
-      "print \"String length of %s is %d\" %(\"LAK\",len2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String length of MUNI is 4\n",
-        "String length of LAK is 3\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page Number: 5.23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using strcpy() function\n",
-      "def strcpy(cstring1):\n",
-      "      import copy\n",
-      "      cstring2=copy.copy(cstring1)\n",
-      "      return cstring2  \n",
-      "\n",
-      "source = \"MUNI\"\n",
-      "target = strcpy(source)\n",
-      "print \"Source string is %s\" %source\n",
-      "print \"Target string is %s\" %target"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Source string is MUNI\n",
-        "Target string is MUNI\n"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 5, Page Number:5.24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Strcat function\n",
-      "\n",
-      "str1=\"MUNI\"\n",
-      "str2=\"LAK\"\n",
-      "print str1+str2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "MUNILAK\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 6, Page Number: 5.24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using strcat() function\n",
-      "def strcat(cstring1, cstring2):\n",
-      "    cstring3 = cstring1 + cstring2\n",
-      "    return cstring3\n",
-      "\n",
-      "source = \"Ramesh\"\n",
-      "target = \"Babu\"\n",
-      "source = strcat(source,target)\n",
-      "print \"Source string is %s\" %source\n",
-      "print \"Target string is %s\" %target"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Source string is RameshBabu\n",
-        "Target string is Babu\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 7, Page Number: 5.25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using strcmp() function\n",
-      "\n",
-      "name = \"Kalai\"\n",
-      "name1 = \"Malai\"\n",
-      "if name ==\"Kalai\":\n",
-      "    print \"0\",\n",
-      "if name1!=name:\n",
-      "    print \"1\",\n",
-      "if name != \"Kalai mani\":\n",
-      "    print \"6\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0 1 6\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 8, Page Number: 5.26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using strrev() function\n",
-      "def strrev(cstring1):\n",
-      "    cstring2 = ' '\n",
-      "    for i in range(len(cstring1)-1,-1,-1):\n",
-      "        cstring2 = cstring2 + cstring1[i]\n",
-      "    return cstring2\n",
-      "\n",
-      "y = raw_input(\"Enter the string \")\n",
-      "print \"The string reversed is : %s\" %strrev(y)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the string book\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The string reversed is :  koob\n"
-       ]
-      }
-     ],
-     "prompt_number": 85
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 9, Page Number: 5.27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using enumerated data type\n",
-      "def enum(*sequential, **named):\n",
-      "    enums = dict(zip(sequential, range(len(sequential))), **named)\n",
-      "    return type('Enum', (), enums)\n",
-      "\n",
-      "week = enum(' ', 'sun','Mon','Tue','Wed','Thr','Fri','Sat')\n",
-      "print \"Mon = %d\" %week.Mon\n",
-      "print \"Tue = %d\" %week.Tue\n",
-      "print \"Wed= %d\" %week.Wed\n",
-      "print \"Sat = %d\" %week.Sat"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mon = 2\n",
-        "Tue = 3\n",
-        "Wed= 4\n",
-        "Sat = 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 125
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 10, Page Number: 5.28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to identify whether the entered character is letter, digit or other symbols\n",
-      "def enum(*sequential, **named):\n",
-      "    enums = dict(zip(sequential, range(len(sequential))), **named)\n",
-      "    return type('Enum', (), enums)\n",
-      "\n",
-      "ctype = enum(' ', 'Letter', 'Digit', 'Other')\n",
-      "ch = raw_input(\"Enter a any character.\")\n",
-      "a = ch.isalpha()\n",
-      "if a != False:\n",
-      "    print \"%c is %d type of symbol.\" %(ch,ctype.Letter)\n",
-      "else:\n",
-      "    a = ch.isdigit()\n",
-      "    if a != False:\n",
-      "        print \"%c is %d type of symbol.\" %(ch,ctype.Digit)\n",
-      "    else:\n",
-      "        print \"%c is %d type of symbol.\" %(ch,ctype.Other)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a any character.%\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "% is 3 type of symbol.\n"
-       ]
-      }
-     ],
-     "prompt_number": 143
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 1, Page Number: 5.29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program for sorting an array of elements\n",
-      "\n",
-      "a = []\n",
-      "n = input(\"Enter Upper limit....\")\n",
-      "for i in range(0,n):\n",
-      "    a.append(0)\n",
-      "print \"Enter elements....\"    \n",
-      "for i in range(0,n):\n",
-      "    a[i]=input()\n",
-      "for i in range(0,n):\n",
-      "    for j in range(i,n):\n",
-      "        if a[i]>a[j]:\n",
-      "            t = a[i]\n",
-      "            a[i] = a[j]\n",
-      "            a[j] = t\n",
-      "print \"Numbers in Ascending order\"\n",
-      "for i in range(0,n):\n",
-      "    print \"%d\" %a[i],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Upper limit....6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter elements....\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numbers in Ascending order\n",
-        "1 2 3 4 5 6\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 2, Page Number: 5.31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to fing Median for a given 'n' numbers\n",
-      "\n",
-      "arr = []\n",
-      "n = input(\"Enter Upper limit....\")\n",
-      "for i in range(0,n):\n",
-      "    arr.append(0)\n",
-      "print \"Enter..%d..values\" %n    \n",
-      "for i in range(0,n):\n",
-      "    arr[i]=input()\n",
-      "for i in range(1,n):\n",
-      "    for j in range(1,n-i):\n",
-      "        if arr[j] <= arr[j+1]:\n",
-      "            t = arr[j]\n",
-      "            arr[j] = arr[j+1]\n",
-      "            arr[j+1] = t\n",
-      "        else:\n",
-      "            continue\n",
-      "if n%2 == 0:\n",
-      "    median = (arr[n/2]+arr[n/2+1])/2.0\n",
-      "else:\n",
-      "    median = arr[n/2+1]\n",
-      "print \"Median is....%.f\" %median    \n",
-      "                "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Upper limit....3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter..3..values\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "30\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Median is....20\n"
-       ]
-      }
-     ],
-     "prompt_number": 162
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 3, Page Number: 5.32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to find standard deviation for the given data\n",
-      "import math\n",
-      "\n",
-      "val = []\n",
-      "sum = ssqr = n = 0\n",
-      "for i in range(0,100):\n",
-      "    val.append(0)\n",
-      "print \"Enter Values -999 to stop....\"\n",
-      "for i in range(0,100):\n",
-      "    val[i] = float(raw_input())\n",
-      "    if val[i] == -999:\n",
-      "        break\n",
-      "    sum = float(sum) + val[i]    \n",
-      "    n = n + 1\n",
-      "mean = float(sum) / float(n)    \n",
-      "for i in range(0,n):\n",
-      "    dev = val[i] - mean\n",
-      "    ssqr = float(ssqr) + float(dev)*float(dev)\n",
-      "var = ssqr / float(n)    \n",
-      "sd = math.sqrt(float(var))\n",
-      "print \"Number of Items.....%d\" %n\n",
-      "print \"Mean is .....%f\" %mean\n",
-      "print \"Standard Deviation is .....%f\" %sd"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values -999 to stop....\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "30\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "40\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-999\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of Items.....5\n",
-        "Mean is .....30.000000\n",
-        "Standard Deviation is .....14.142136\n"
-       ]
-      }
-     ],
-     "prompt_number": 169
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 4, Page Number: 5.34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Bubble sort using function\n",
-      "\n",
-      "def bubble(n, x = []):\n",
-      "    for item in range(0,n-1):\n",
-      "        for i in range(item+1,n):\n",
-      "            if x[i]<x[item]:\n",
-      "                temp = x[item]\n",
-      "                x[item] = x[i]\n",
-      "                x[i] = temp\n",
-      "                \n",
-      "a = []                \n",
-      "n = input(\"Enter upper limit: \")\n",
-      "for i in range(0,n):\n",
-      "    a.append(0)\n",
-      "for i in range(0,n):\n",
-      "    a[i] = input()\n",
-      "    \n",
-      "bubble(n,a)    \n",
-      "print \"After sorting\",\n",
-      "for i in range(0,n):\n",
-      "    print \"%d\" %a[i],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter upper limit: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "After sorting 1 2 3 4 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 174
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 5, Page Number: 5.35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to sort an element using Insertion sort\n",
-      "\n",
-      "a = []\n",
-      "x = input(\"Elements to be inserted \")\n",
-      "for i in range(0,100):\n",
-      "    a.append(0)\n",
-      "i = 0\n",
-      "while x != -99:\n",
-      "    k = i-1\n",
-      "    while (x<a[k]) and (k>=0):\n",
-      "        a[k+1] = a[k]\n",
-      "        k = k-1\n",
-      "    a[k+1]=x\n",
-      "    print \"Array after inserting %d:\" %x\n",
-      "    for j in range(0,i):\n",
-      "        print \"%5d\" %a[j]\n",
-      "    x = input(\"Elements to be inserted(-99 to stop)?\")\n",
-      "    i = i+1\n",
-      "print \"The final sorted array\"\n",
-      "for j in range(0,i):\n",
-      "    print \"%5d\" %a[j],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements to be inserted 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Array after inserting 5:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements to be inserted(-99 to stop)?3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Array after inserting 3:\n",
-        "    3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements to be inserted(-99 to stop)?4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Array after inserting 4:\n",
-        "    3\n",
-        "    4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements to be inserted(-99 to stop)?2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Array after inserting 2:\n",
-        "    2\n",
-        "    3\n",
-        "    4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements to be inserted(-99 to stop)?1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Array after inserting 1:\n",
-        "    1\n",
-        "    2\n",
-        "    3\n",
-        "    4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements to be inserted(-99 to stop)?-99\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final sorted array\n",
-        "    1     2     3     4     5\n"
-       ]
-      }
-     ],
-     "prompt_number": 186
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 6, Page Number: 5.36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to sort an element using selection sort\n",
-      "\n",
-      "a = []\n",
-      "for i in range(0,100):\n",
-      "    a.append(0)\n",
-      "n = input(\"Enter Number of elements in array...\")\n",
-      "print \"Enter Elements...\",n\n",
-      "for i in range(0,n):\n",
-      "    a[i] = input()\n",
-      "for i in range(0,n-1):\n",
-      "    m = i\n",
-      "    for j in range(i+1,n):\n",
-      "        if a[m]>a[j]:\n",
-      "            m = j\n",
-      "    if i != m:\n",
-      "        k = a[i]\n",
-      "        a[i] = a[m]\n",
-      "        a[m] = k\n",
-      "        \n",
-      "print \"Sorted Array (Selection Sort) is...\\n\"\n",
-      "for i in range(0,n):\n",
-      "    print \"%5d\" %a[i],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number of elements in array...9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements... 9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "326\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "85\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "42\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "156\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "845\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "66\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "55\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorted Array (Selection Sort) is...\n",
-        "\n",
-        "    1    42    55    56    66    85   156   326   845\n"
-       ]
-      }
-     ],
-     "prompt_number": 190
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 7, Page Number: 5.38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Quick sort\n",
-      "\n",
-      "def quickSort(array):\n",
-      "   quickSortHelper(array,0,len(array)-1)\n",
-      "\n",
-      "def quickSortHelper(array,right,left):\n",
-      "   if right<left:\n",
-      "       splitpoint = partition(array,right,left)\n",
-      "       quickSortHelper(array,right,splitpoint-1)\n",
-      "       quickSortHelper(array,splitpoint+1,left)\n",
-      "\n",
-      "def partition(array,right,left):\n",
-      "   pivotvalue = array[right]\n",
-      "\n",
-      "   begin = right+1\n",
-      "   end = left\n",
-      "\n",
-      "   done = False\n",
-      "   while not done:\n",
-      "\n",
-      "       while begin <= end and \\\n",
-      "               array[begin] <= pivotvalue:\n",
-      "           begin = begin + 1\n",
-      "\n",
-      "       while array[end] >= pivotvalue and \\\n",
-      "               end >= begin:\n",
-      "           end = end -1\n",
-      "\n",
-      "       if end < begin:\n",
-      "           done = True\n",
-      "       else:\n",
-      "           temp = array[begin]\n",
-      "           alist[begin] = array[end]\n",
-      "           alist[end] = temp\n",
-      "\n",
-      "   temp = array[right]\n",
-      "   array[right] = array[end]\n",
-      "   array[end] = temp\n",
-      "   return end\n",
-      "\n",
-      "array = []\n",
-      "n=input(\"Enter the values in array: \")\n",
-      "for i in range (0,n):\n",
-      "    array.append(0)\n",
-      "for i in range (0,n):\n",
-      "    array[i]=input()\n",
-      "    \n",
-      "quickSort(array)\n",
-      "print \"Quick Sorted Array is\", array"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the values in array: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Quick Sorted Array is [1, 2, 3, 4, 5]\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 8, Page Number: 5.40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#program to find the specified elements from the array using linear searching\n",
-      "\n",
-      "a = [100]\n",
-      "for i in range(0,100):\n",
-      "    a.append(0)\n",
-      "c = 'y'\n",
-      "while (c=='y')or(c=='Y'):\n",
-      "    no = input(\"Enter the size of sorting\")\n",
-      "    print \"Enter the elements of the array\"\n",
-      "    for i in range(0,no):\n",
-      "        a[i]=input()\n",
-      "    k = input(\"Enter the element to be searched\")    \n",
-      "    for i in range(0,no):\n",
-      "        if k == a[i]:\n",
-      "            print \"Element %d is in the position %d\" %(k,i+1)\n",
-      "            f=2\n",
-      "            break\n",
-      "    if f==1:\n",
-      "        print \"The entered element is not in the array\"\n",
-      "    ch = raw_input(\"If you want to continue y/n \")\n",
-      "    c = ch\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the size of sorting5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the elements of the array\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element to be searched4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Element 4 is in the position 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "If you want to continue y/n n\n"
-       ]
-      }
-     ],
-     "prompt_number": 206
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 9, Page Number: 5.42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to find the specified element from array using Binary search\n",
-      "\n",
-      "a = [100]\n",
-      "for i in range(0,100):\n",
-      "    a.append(0)\n",
-      "no = input(\"Enter the size of sorting\")\n",
-      "print \"Enter the elements in ascending order\"\n",
-      "for i in range(0,no):\n",
-      "        a[i]=input() \n",
-      "t = input(\"Enter the element to be searched\") \n",
-      "l=0\n",
-      "h=no-1\n",
-      "while l <= h:\n",
-      "    m = (l+h)/2\n",
-      "    if t<a[m]:\n",
-      "        h = m-1\n",
-      "    elif t>a[m]:\n",
-      "        l = m+1\n",
-      "    else:\n",
-      "        print \"Entered %d is in position %d\" %(t,m+1)\n",
-      "        f=2\n",
-      "        break\n",
-      "if f==1:\n",
-      "    print \"Entered element is not in the array\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the size of sorting5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the elements in ascending order\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element to be searched3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Entered 3 is in position 2\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter07.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter07.ipynb
deleted file mode 100755
index ee25427a..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter07.ipynb
+++ /dev/null
@@ -1,254 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:59d989e4d68b99137fb5255fe771e42a62c5ce35100304584930ea0c003ea2ac"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 07 : User Defined Functions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, page no.5.49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate Local and Global Declaration \n",
-      "\n",
-      "#variable declaration globally\n",
-      "i=0                \n",
-      "print \"Value of i in main\", i\n",
-      "\n",
-      " #define function\n",
-      "def f1():        \n",
-      "    k=0\n",
-      "    i=50\n",
-      "    print \"Value of i after function call\",i\n",
-      "    \n",
-      "#Function Call\n",
-      "f1()             \n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of i in main 0\n",
-        "Value of i after function call 50\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, page no: 5.50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate usage of local and global variable\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    print \"Value of i is:\", i\n",
-      "    \n",
-      "i=590\n",
-      "print \"The Value of i is:\", i    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of i is: 0\n",
-        "Value of i is: 1\n",
-        "Value of i is: 2\n",
-        "The Value of i is: 590\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page no.5.52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to call function without parameters\n",
-      "def message():\n",
-      "    print \"Main Message\"\n",
-      "    \n",
-      "print \"Function Message\"\n",
-      "\n",
-      "message()\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Function Message\n",
-        "Main Message\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page No.5.52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to call function with parameter\n",
-      "\n",
-      "#Defining function\n",
-      "def add(x,y):\n",
-      "    z=0\n",
-      "    z=x+y\n",
-      "    return z\n",
-      "\n",
-      "k=add(10,20)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Value of k is \", k"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of k is  30\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 5, page no.5.53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to write prototype of the function\n",
-      "\n",
-      "k=0\n",
-      "\n",
-      "#Defining in function\n",
-      "def f1(k):\n",
-      "    k=k+100\n",
-      "    return k\n",
-      "\n",
-      "#Result\n",
-      "print \"The value of i before call: \", k\n",
-      "\n",
-      "print \"The Value of i after call: \",(f1(k))\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of i before call:  0\n",
-        "The Value of i after call:  100\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 6, page no. 5.54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to write prototype definition in main function\n",
-      "\n",
-      "def fun(k):\n",
-      "    k=k+10\n",
-      "    return k\n",
-      "\n",
-      "i=0\n",
-      "\n",
-      "#Result\n",
-      "print  \"The i value before call %d\\n\"%(i)\n",
-      "print \"The i value after call %d\\n\"%fun(i)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The i value before call 0\n",
-        "\n",
-        "The i value after call 10\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter08.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter08.ipynb
deleted file mode 100755
index e40c1b43..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter08.ipynb
+++ /dev/null
@@ -1,420 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:5d85b398d61f498d1deba465b9747c5d15996be1754212b25012ca792b193a16"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 08 : Function Prototypes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Function with no arguments and no return values"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.: 5.55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to addition of two numbers\n",
-      "\n",
-      "def add():\n",
-      "    a,b=input(\"Enter two numbers.. \")\n",
-      "    c=a+b\n",
-      "    print \"Sum is ... %d\"%(c)\n",
-      "    \n",
-      "add()    \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two numbers.. 10,20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum is ... 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.:5.56"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print addition, subtraction and multiplication values\n",
-      "\n",
-      "def add():\n",
-      "    a=10\n",
-      "    b=5\n",
-      "    print \"The Addition of given value: \",a+b\n",
-      "        \n",
-      "def sub():\n",
-      "    a=10\n",
-      "    b=5\n",
-      "    print \"The Subtraction of given value:\",a-b\n",
-      "    \n",
-      "def mul():\n",
-      "    a=10\n",
-      "    b=5\n",
-      "    print \"The Multiplication of given value:\",a*b\n",
-      "    \n",
-      "    \n",
-      "add()\n",
-      "sub()\n",
-      "mul()\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Addition of given value:  15\n",
-        "The Subtraction of given value: 5\n",
-        "The Multiplication of given value: 50\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Function with arguments and no return values"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, page No. 5.57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Program for addition of two numbers\n",
-      "\n",
-      "def add(x,y):\n",
-      "    z= x+y\n",
-      "    print \"Sum is:\", z\n",
-      "\n",
-      "#Main Function\n",
-      "a, b = input(\"Enter two Values: \")\n",
-      "    \n",
-      "add(a,b)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two Values: 10,20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum is: 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example: 2, page no. 5.58"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def date(x,y,z):\n",
-      "    print \"Employee Joined date is:%d/%d/%d\" %(x,y,z)\n",
-      "    \n",
-      "print \"Enter the employee joining date.\"\n",
-      "\n",
-      "dd,mm,yy=eval(raw_input(\"Enter date, month, year: \"))\n",
-      "\n",
-      "date(dd,mm,yy)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the employee joining date.\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter date, month, year: 11,12,1981\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Employee Joined date is:11/12/1981\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Function with arguments and with return value"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, page no.5.59"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program for addition of two numbers\n",
-      "\n",
-      "def add(x,y):\n",
-      "    z=0\n",
-      "    z=x+y\n",
-      "    return z\n",
-      "\n",
-      "#Main Function\n",
-      "a,b=input(\"Enter the a and b value:\")\n",
-      "c=add(a,b)\n",
-      "print \"Result is:\",c"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the a and b value:10,20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result is: 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.:5.60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to send and receives the values from function\n",
-      "\n",
-      "def mul(m,n):\n",
-      "    return (m*n)\n",
-      "\n",
-      "m=n=sum=0\n",
-      "m,n=input(\"Enter two values:\")\n",
-      "sum=mul(m,n)\n",
-      "\n",
-      "print \"Multiplication of the entered Values:\",sum"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two values:30,4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Multiplication of the entered Values: 120\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Function with no arguments and with return value"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, PageNo.:5.61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program for addition of two numbers\n",
-      "\n",
-      "def add():\n",
-      "    a=b=c=0\n",
-      "    a,b=input(\"Enter the Values of a and b:\")\n",
-      "    c=a+b\n",
-      "    return(c)\n",
-      "\n",
-      "c=add()\n",
-      "print \"The Output of given Values:\",c\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Values of a and b:10,20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Output of given Values: 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No. 5.62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to multiply three numbers using function\n",
-      "\n",
-      "\n",
-      "def mul():\n",
-      "    a,b,c=input(\"Enter the three values for Multiplication: \")\n",
-      "    i=a*b*c\n",
-      "    return(i)\n",
-      "\n",
-      "s=mul()\n",
-      "print \"Multiplication of three numbers is= \",s\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the three values for Multiplication: 4,8,2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Multiplication of three numbers is=  64\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter09.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter09.ipynb
deleted file mode 100755
index 11735b92..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter09.ipynb
+++ /dev/null
@@ -1,109 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1a0147da7a94fc5ebe7c66bbf21a563cb0bee33c1452a7bcfcbc21f5395d7aab"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 09 : Parameter Passing Methods"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.: 5.63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call By Value\n",
-      "\n",
-      "def cube(x):\n",
-      "    x=x*x*x\n",
-      "    return(x)\n",
-      "\n",
-      "n=5\n",
-      "print \"Cube of %d is\"%n,cube(n)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cube of 5 is 125\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.:5.64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call by reference\n",
-      "\n",
-      "def interchange(a,b):\n",
-      "    t=0\n",
-      "    t=a\n",
-      "    a=b\n",
-      "    b=t\n",
-      "    return b,a\n",
-      "\n",
-      "    \n",
-      "i=5\n",
-      "j=10\n",
-      "print \"i and j values before interchange:\", i, j\n",
-      "\n",
-      "i=10\n",
-      "j=5\n",
-      "interchange(i,j)\n",
-      "\n",
-      "#Result\n",
-      "print \"i and j values after interchange:\",i,j\n",
-      "\n",
-      "print \"i and j avalues after interchange in the function : %d %d \\n\"%interchange(i,j)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i and j values before interchange: 5 10\n",
-        "i and j values after interchange: 10 5\n",
-        "i and j avalues after interchange in the function : 10 5 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter10.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter10.ipynb
deleted file mode 100755
index 1a017d08..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter10.ipynb
+++ /dev/null
@@ -1,372 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ea85b38c0942e3b3d3a1243525badd8ffa28ac492392c6aef6c8ea5823294150"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10 : Recursion"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.: 5.67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Factorial of an integer number.\n",
-      "\n",
-      "def rec(x):\n",
-      "    f=0\n",
-      "    if x==1:\n",
-      "        return (1)\n",
-      "    else:\n",
-      "        f=x*rec(x-1)\n",
-      "        return(f)\n",
-      "\n",
-      "a=input(\"Enter the number:\")\n",
-      "print \"The Factorial value of %d:\"%a,rec(a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number:5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Factorial value of 5: 120\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.:5.67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to read characters and print reversely using recursion.\n",
-      "\n",
-      "def reverse(c):\n",
-      "    if len(c)<=0:\n",
-      "        return c\n",
-      "    return reverse(c[1:])+c[0]\n",
-      "\n",
-      "c=raw_input(\"Enter the Line of Text...\")\n",
-      "print reverse(c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Line of Text...munilak\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "kalinum\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No.:5.68"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#concept of recursion\n",
-      "\n",
-      "def add(pk, pm):\n",
-      "    if pm==0:\n",
-      "        return pk\n",
-      "    else:\n",
-      "        return(1+add(pk,pm-1))\n",
-      "    \n",
-      "    \n",
-      "k=70\n",
-      "m=20\n",
-      "\n",
-      "i=add(k,m)\n",
-      "\n",
-      "print \"The Value of addition is %d\"%i\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of addition is 90\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page No:5.70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Tower of honoi\n",
-      "n=int(raw_input(\"How many Disks ...\\n\"))\n",
-      "\n",
-      "def honoi(n,fro,to,t):\n",
-      "    if(n>0):\n",
-      "        honoi(n-1,fro,t,to)\n",
-      "        print \"Move disk %d from %c to %c\\n\"%(n,fro,to)\n",
-      "        honoi(n-1,t,to,fro)\n",
-      "        \n",
-      "fro='L'\n",
-      "to='R'\n",
-      "t='C'\n",
-      "        \n",
-      "honoi(n,fro,to,t)\n",
-      "            "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many Disks ...\n",
-        "3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Move disk 1 from L to R\n",
-        "\n",
-        "Move disk 2 from L to C\n",
-        "\n",
-        "Move disk 1 from R to C\n",
-        "\n",
-        "Move disk 3 from L to R\n",
-        "\n",
-        "Move disk 1 from C to L\n",
-        "\n",
-        "Move disk 2 from C to R\n",
-        "\n",
-        "Move disk 1 from L to R\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 1, Page No. : 5.76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#user defined function that computes 'x' raised to the power 'y'\n",
-      "\n",
-      "import math \n",
-      "\n",
-      "x,y=input(\"Enter the Values of X and Y \")\n",
-      "\n",
-      "print \"%d to the power of %d is ... \"%(x,y),math.pow(x,y)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Values of X and Y 10,2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10 to the power of 2 is ...  100.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 2, page No.5.76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#nesting of function\n",
-      "\n",
-      "def ratio(x,y,z):\n",
-      "    \n",
-      "    def diff(m,n):\n",
-      "        if(m!=n):\n",
-      "            return(1)\n",
-      "        else:\n",
-      "            return(0)\n",
-      "    \n",
-      "    diff(y,z)\n",
-      "    if diff(y,z):\n",
-      "       return(x/(y-z))\n",
-      "    else:\n",
-      "       return(0)\n",
-      "        \n",
-      "    \n",
-      "x,y,z=input(\"Enter the values of X,Y,Z: \")\n",
-      "\n",
-      "    \n",
-      "print \"Ratio is..%f\\n\"%ratio(x,y,z)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the values of X,Y,Z: 10,10,5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ratio is..2.000000\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 3, Page No. 5.78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Demonstration of Global Variable\n",
-      "\n",
-      "a=39               #Global Variable\n",
-      "print \"A... %d\\n\"%a\n",
-      "\n",
-      "def f1():\n",
-      "    global a\n",
-      "    a=a+10\n",
-      "    return a\n",
-      "\n",
-      "def f2():\n",
-      "    global a\n",
-      "    a=a+20\n",
-      "    return a\n",
-      "\n",
-      "def f3():\n",
-      "    global a\n",
-      "    a=a+30\n",
-      "    return a\n",
-      "\n",
-      "\n",
-      "\n",
-      "print \"A...%d\\n\"%f1()\n",
-      "print \"A...%d\\n\"%f2()\n",
-      "print \"A...%d\\n\"%f3()\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A... 39\n",
-        "\n",
-        "A...49\n",
-        "\n",
-        "A...69\n",
-        "\n",
-        "A...99\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter11.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter11.ipynb
deleted file mode 100755
index 0e00f235..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter11.ipynb
+++ /dev/null
@@ -1,304 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:03a9be4b5720a51a42be3ae01bfa08ccecef76e9a5e17f613e76c0afaff965de"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11 : Structures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.: 5.81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the student number, name and marks through accessing structure elements.\n",
-      "\n",
-      "#Defining Structure in class\n",
-      "class std:\n",
-      "    no=0\n",
-      "    name = ''\n",
-      "    m1 = m2 = m3 =0\n",
-      "\n",
-      "#assigning variable to structure class    \n",
-      "s=std()\n",
-      "\n",
-      "#UserInput\n",
-      "s.no=raw_input(\"Give Student Number: \")\n",
-      "s.name=raw_input(\"Enter the Student Name: \")\n",
-      "s.m1=input(\"Enter Student Mark1: \")\n",
-      "s.m2=input(\"Enter Student Mark2: \")\n",
-      "s.m3=input(\"Enter Student Mark3: \")\n",
-      "\n",
-      "total=avg=0.0\n",
-      "\n",
-      "total = (s.m1 + s.m2 + s.m3)\n",
-      "avg='%.2f' %(total / 3)\n",
-      "\n",
-      "#Result\n",
-      "print \"\\nThe Output is.... \\n\",s.no,s.name,total,avg"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give Student Number: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Student Name: raja\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark1: 99\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark2: 97\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark3: 95\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Output is.... \n",
-        "1 raja 291 97.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No. 5.83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#print a value from Structure.\n",
-      "\n",
-      "class struct:\n",
-      "    a=b=0\n",
-      "    \n",
-      "x=struct()\n",
-      "y=struct()\n",
-      "\n",
-      "x.a=29\n",
-      "y=x\n",
-      "\n",
-      "#Result\n",
-      "print \"The Value of a %d\"%y.a\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of a 29\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No.:5.84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program for User defined data type \n",
-      "\n",
-      "#There is no preprocessor directive in python\n",
-      "D = 7\n",
-      "\n",
-      "#No typedef function in python\n",
-      "wk = int(raw_input(\"Enter Weeks : \"))\n",
-      "\n",
-      "#Result\n",
-      "print \"\\nNumber of Days = %d\"%(wk*D)\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Weeks : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number of Days = 28\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page.No.:5.85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to create user defined data type from structure\n",
-      "\n",
-      "\n",
-      "class struct:\n",
-      "    name=''\n",
-      "    sex=''\n",
-      "    age=0\n",
-      "    \n",
-      "\n",
-      "\n",
-      "candidate = [struct() for i in range(0,2)]\n",
-      "\n",
-      "for a in range(0,2):\n",
-      "    candidate[a].name= raw_input(\"Name of the Employee: \")\n",
-      "    candidate[a].sex= raw_input(\"Sex: \")\n",
-      "    candidate[a].age= input(\"Age: \")\n",
-      "\n",
-      "#Result\n",
-      "print \"\\n Name\\t Sex\\t Age\\n\"\n",
-      "\n",
-      "for a in range(0,2):\n",
-      "    print \" %s\\t\"%(candidate[a].name),\"  %s\\t\"%(candidate[a].sex),\" %d\\n\"%(candidate[a].age)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name of the Employee: venkat\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sex: m\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Age: 25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name of the Employee: geetha\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sex: f\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Age: 23\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Name\t Sex\t Age\n",
-        "\n",
-        " venkat\t  m\t 25\n",
-        "\n",
-        " geetha\t  f\t 23\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter12.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter12.ipynb
deleted file mode 100755
index 54918fbb..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter12.ipynb
+++ /dev/null
@@ -1,829 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:13c473411b35212c9727620c7673dc52636d4bf1a83b0d4eeb082d160a75324d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12 : Union"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.5.87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#program using union.\n",
-      "\n",
-      "class union_name:\n",
-      "    a=0\n",
-      "    b=[0,1]\n",
-      "    \n",
-      "c=union_name()\n",
-      "c.a=256\n",
-      "\n",
-      "print \"c.a value is %d\\n \"%c.a\n",
-      "print \"c.b[0] value is %d\\n\"%c.b[0]\n",
-      "print \"c.b[1] value is %d\\n\"%c.b[1]\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "c.a value is 256\n",
-        " \n",
-        "c.b[0] value is 0\n",
-        "\n",
-        "c.b[1] value is 1\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.:5.87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#program to find size of union and number of bytes\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "class union_result:\n",
-      "    marks=0\n",
-      "    grade=''\n",
-      "    \n",
-      "    \n",
-      "class struct_res:\n",
-      "    name=[15]\n",
-      "    age=0\n",
-      "    sex=''\n",
-      "    address=''\n",
-      "    pincode=0\n",
-      "    perf=union_result()\n",
-      "    \n",
-      "data=struct_res()   \n",
-      "\n",
-      "#Result\n",
-      "print \"Size of Union: %d\\n\"%sys.getsizeof(data.perf)\n",
-      "print \"Size of Structure: %d\\n\"%sys.getsizeof(data)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Size of Union: 32\n",
-        "\n",
-        "Size of Structure: 32\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pointer to Unions cannot convert as no pointer in python. Page no. 5.88"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No.:5.89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program that includes of initial values to structure variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "\n",
-      "class union_id:\n",
-      "    color='20'\n",
-      "    size=0\n",
-      "    \n",
-      "\n",
-      "class struct_clothes:\n",
-      "    manufacturer='30'\n",
-      "    cost=0.0\n",
-      "    description=union_id()\n",
-      "    \n",
-      "    \n",
-      "shirt=struct_clothes()\n",
-      "shirt.manufacturer='Indian'\n",
-      "shirt.cost=35.00\n",
-      "shirt.description.color='Black'\n",
-      "\n",
-      "print \"%d\\n\"%(sys.getsizeof(union_id))\n",
-      "\n",
-      "print \"%s %f \"%(shirt.manufacturer,shirt.cost),\"%s %d\\n \"%(shirt.description.color, shirt.description.size)\n",
-      " \n",
-      "\n",
-      "shirt.description.size =12\n",
-      "\n",
-      "\n",
-      "print \"%s %f \"%(shirt.manufacturer, shirt.cost),\"%s %d\\n \"%(shirt.description.color, shirt.description.size)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "48\n",
-        "\n",
-        "Indian 35.000000  Black 0\n",
-        " \n",
-        "Indian 35.000000  Black 12\n",
-        " \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page no.: 5.92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to read and write employee and their date of joining using nested structure. \n",
-      "\n",
-      "\n",
-      "class date:\n",
-      "    day=0\n",
-      "    month=\"\"\n",
-      "    year=0\n",
-      "    \n",
-      "class employee:\n",
-      "    code=0\n",
-      "    name=''\n",
-      "    salary=0.0\n",
-      "    doj=date()\n",
-      "    \n",
-      "    \n",
-      "emp1=employee()\n",
-      "\n",
-      "emp1.code=input(\"Enter Employee Code: \")\n",
-      "emp1.name=raw_input(\"Enter the Employee Name: \")\n",
-      "emp1.salary=input(\"Enter the Employee Salary: \")\n",
-      "\n",
-      "print \"\\nEnter Date of joining in order\\n\"\n",
-      "emp1.doj.day=input(\"Enter day: \")\n",
-      "emp1.doj.month=raw_input(\"Enter month: \")\n",
-      "emp1.doj.year=input(\"Enter year: \")\n",
-      "\n",
-      "print \"\\nThe Employee Code is: %d\"%emp1.code\n",
-      "print \"\\nThe Employee Name is: %s\"%emp1.name\n",
-      "print \"\\nThe Employee Name is: %f\"%emp1.salary\n",
-      "print \"\\nThe Employee DOJ is: %d %s %d\"%(emp1.doj.day,emp1.doj.month,emp1.doj.year)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Employee Code: 200\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Employee Name: VIJI\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Employee Salary: 2000.00\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Date of joining in order\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter day: 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter month: December\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter year: 2004\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Employee Code is: 200\n",
-        "\n",
-        "The Employee Name is: VIJI\n",
-        "\n",
-        "The Employee Name is: 2000.000000\n",
-        "\n",
-        "The Employee DOJ is: 12 December 2004\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 5, Page No. 5.93"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to store 3 records in one structure\n",
-      "\n",
-      "\n",
-      "class book:\n",
-      "    name=''\n",
-      "    price=0\n",
-      "    pages=0\n",
-      "    \n",
-      "    \n",
-      "b=[book() for a in range(0,3)]\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    b[i].name=raw_input(\"Enter the Book Name: \")\n",
-      "    b[i].price=input(\"Enter the Book Price: \")\n",
-      "    b[i].pages=input(\"Enter the Book Pages: \")\n",
-      "    \n",
-      "print '\\n'    \n",
-      "for i in range(0,3):\n",
-      "    print \"%s\\t\"%(b[i].name),\"%d\\t\"%(b[i].price),\"%d\\n\"%(b[i].pages)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Name: English\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Price: 165\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Pages: 200\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Name: Maths\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Price: 300\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Pages: 450\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Name: Physics\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Price: 250\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book Pages: 370\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "English\t165\t200\n",
-        "\n",
-        "Maths\t300\t450\n",
-        "\n",
-        "Physics\t250\t370\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 6, Page No. 5.95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to copy of entire structure to a function\n",
-      "\n",
-      "class std:\n",
-      "    no=0\n",
-      "    avg=0.0\n",
-      "\n",
-      "a=std()\n",
-      "\n",
-      "a.no=15\n",
-      "a.avg=90.75\n",
-      "\n",
-      "def func(Struct_p):\n",
-      "    print \"Number is.... %d\\n\"%a.no\n",
-      "    print \"Average is... %f\"%a.avg\n",
-      "    \n",
-      "func(a)\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number is.... 15\n",
-        "\n",
-        "Average is... 90.750000\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 7, Page No.:5.97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Programs shows how the structure can be accessed by a pointer variable.\n",
-      "\n",
-      "class student:\n",
-      "    roll_no=0\n",
-      "    name=''\n",
-      "    marks=0.0\n",
-      "stud1=student()\n",
-      "\n",
-      "print \"Enter the student details: \\n\"\n",
-      "stud1.roll_no=input()\n",
-      "stud1.name=raw_input()\n",
-      "stud1.marks=input()\n",
-      "\n",
-      "print \"\\nDisplay of structure using structure variable\\n\"\n",
-      "print \"Roll No\\t Name\\t Marks\"\n",
-      "print \"\\n-------\\t ----- -------\\n\"\n",
-      "print \"%d\\t %s\\t %f\"%(stud1.roll_no,stud1.name,stud1.marks)\n",
-      "\n",
-      "pt=stud1\n",
-      "\n",
-      "print \"\\nDisplay of structure using pointer variable\\n\"\n",
-      "print \"Roll No\\t Name\\t Marks\"\n",
-      "print \"\\n-------\\t ----- -------\\n\"\n",
-      "print \"%d\\t %s\\t %f\"%(pt.roll_no,pt.name,pt.marks)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student details: \n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "39\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Muni\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "77\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Display of structure using structure variable\n",
-        "\n",
-        "Roll No\t Name\t Marks\n",
-        "\n",
-        "-------\t ----- -------\n",
-        "\n",
-        "39\t Muni\t 77.000000\n",
-        "\n",
-        "Display of structure using pointer variable\n",
-        "\n",
-        "Roll No\t Name\t Marks\n",
-        "\n",
-        "-------\t ----- -------\n",
-        "\n",
-        "39\t Muni\t 77.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 1, Page No.:5.98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to transfer a structure to a function by passing the structure address to the function\n",
-      "\n",
-      "class struct:\n",
-      "    name=''\n",
-      "    no=0\n",
-      "    age=0\n",
-      "record=struct()\n",
-      "std=record\n",
-      "std.name,std.no,std.age=[\"LAK\",15,25]\n",
-      "print \"%s\\t%d\\t%d\"%(std.name,std.no,std.age)\n",
-      "\n",
-      "def fun(std):\n",
-      "    pt.name=\"Muni\"\n",
-      "    pt.no=16\n",
-      "    pt.age=26\n",
-      "    \n",
-      "std.name=pt.name\n",
-      "std.no=pt.no\n",
-      "std.age=pt.age\n",
-      "\n",
-      "print \"%s\\t%d\\t%d\"%(std.name,std.no,std.age)\n",
-      "    \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "LAK\t15\t25\n",
-        "Muni\t16\t26\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 2, Page No.:5.100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to pass an array of structure to a function, and return a pointer to a structure.\n",
-      "import sys\n",
-      "N=3\n",
-      "NULL=0\n",
-      "\n",
-      "class struct:\n",
-      "    sno=0\n",
-      "    name=''\n",
-      "    age=0\n",
-      "    \n",
-      "record=struct()\n",
-      "std=record\n",
-      "\n",
-      "print \"Student Number Locater\"\n",
-      "print \"To Exit the program enter '0' for student number\"\n",
-      "stdno=input(\"Enter the Student Number: \")\n",
-      "print \"\\n\"\n",
-      "if stdno ==15:\n",
-      "\n",
-      "    std.sno,std.name,std.age=[15,\"MUNI\",28]\n",
-      "    print \"Number .... %d \\n\"%(std.sno)\n",
-      "    print \"Name ...... %s\\n\"%(std.name)\n",
-      "    print \"Age ....... %d\\n\"%(std.age)\n",
-      "if stdno == 16:\n",
-      "    std.sno,std.name,std.age=[16,\"LAK\",27]\n",
-      "    print \"Number .... %d \\n\"%(std.sno)\n",
-      "    print \"Name ...... %s\\n\"%(std.name)\n",
-      "    print \"Age ....... %d\\n\"%(std.age)\n",
-      "    \n",
-      "if stdno == 17:\n",
-      "    std.sno,std.name,std.age=[17,\"RAJA\",31]\n",
-      "    print \"Number .... %d \\n\"%(std.sno)\n",
-      "    print \"Name ...... %s\\n\"%(std.name)\n",
-      "    print \"Age ....... %d\\n\"%(std.age)\n",
-      "    \n",
-      "if stdno==0:\n",
-      "    exit\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Student Number Locater\n",
-        "To Exit the program enter '0' for student number\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Student Number: 15\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "Number .... 15 \n",
-        "\n",
-        "Name ...... MUNI\n",
-        "\n",
-        "Age ....... 28\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 3, Page No.: 5.101"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print student name and mark using structure\n",
-      "\n",
-      "class student:\n",
-      "    name=''\n",
-      "    marks=0.00\n",
-      "    \n",
-      "s1=''\n",
-      "f=0.00\n",
-      "\n",
-      "student1=student()\n",
-      "\n",
-      "student2=student()\n",
-      "\n",
-      "\n",
-      "student1.name=raw_input(\"Enter Name: \")\n",
-      "f=input(\"Enter Mark: \")\n",
-      "\n",
-      "student2.marks=f\n",
-      "\n",
-      "print \"\\nName is %s \\n\"%student1.name\n",
-      "print \"Marks are %f \\n\"%student2.marks\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name: Venkat\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Mark: 89\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name is Venkat \n",
-        "\n",
-        "Marks are 89.000000 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 4, Page No.: 5.102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print marks, grade and their address using union\n",
-      "\n",
-      "\n",
-      "\n",
-      "class union_marks:\n",
-      "    perc=0.00\n",
-      "    grade=''\n",
-      "    \n",
-      "student1=union_marks()\n",
-      "\n",
-      "student1.perc=98.5\n",
-      "\n",
-      "print \"Marks are %f \\t address is %8lu\\n\"%(student1.perc, id(student1.perc))\n",
-      "\n",
-      "student1.grade='A'\n",
-      "\n",
-      "print \"Grade is %c \\t address is %8lu\\n\"%(student1.grade,id(student1.grade))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Marks are 98.500000 \t address is 158368916\n",
-        "\n",
-        "Grade is A \t address is 3073485064\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter13.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter13.ipynb
deleted file mode 100755
index b303ee94..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter13.ipynb
+++ /dev/null
@@ -1,57 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:bd4e511a3dbdd189491e3271c86cff9ce5fc919ed7457d400a67c288e55ad433"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13 : Pointers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page Number: 5.104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print address and value of variable\n",
-      "#Local definitions\n",
-      "sno = 39\n",
-      "\n",
-      "#Result\n",
-      "print \"Address of sno = %u\" %id(sno)\n",
-      "print \"Value of sno = %d\" %sno"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Address of sno = 20917296\n",
-        "Value of sno = 39\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter14.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter14.ipynb
deleted file mode 100755
index 12a1734c..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter14.ipynb
+++ /dev/null
@@ -1,234 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:44f5daf77255992c95710d94f062b0f1f1e981b9e6cce07f23ed45a322c4b130"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14 : Accessing Variable through Pointers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page Number: 5.107    "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to accessing through variable pointer\n",
-      "\n",
-      "from ctypes import c_int\n",
-      "\n",
-      "#Local definition\n",
-      "a = 22\n",
-      "a = c_int(a) # ctype datatype declaration\n",
-      "\n",
-      "b = 2.25\n",
-      "b = c_float(b) # ctype datatype declaration\n",
-      "\n",
-      "#Pointer variables\n",
-      "a_po=pointer(a)\n",
-      "b_po=pointer(b)\n",
-      "\n",
-      "# 'id' is a address of letter and it represents the address of the variable\n",
-      "a = id(a)\n",
-      "b = id(b)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"\\nValue of a = %d\" %a_po[0]\n",
-      "print \"\\nValue of b = %.2f\" %b_po[0]\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value of a = 22\n",
-        "\n",
-        "Value of b = 2.25\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page Number: 5.108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print address and value of variable\n",
-      "\n",
-      "from ctypes import c_int\n",
-      "\n",
-      "#Local definition\n",
-      "a = 22\n",
-      "a = c_int(a) # ctype datatype declaration\n",
-      "\n",
-      "#Pointer variables\n",
-      "a_po=pointer(a)\n",
-      "\n",
-      "# 'id' is a address of letter and it represents the address of the variable\n",
-      "a = id(a)\n",
-      "\n",
-      "#Result\n",
-      "print \"\\n Value of a = %d\" %a_po[0]\n",
-      "print \"\\n Address of a = %u\" %id(a)\n",
-      "print \"\\n Value at address %u = %d\" %(id(a),a_po[0])\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Value of a = 22\n",
-        "\n",
-        " Address of a = 4344066728\n",
-        "\n",
-        " Value at address 4344066728 = 22\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page Number: 5.108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Program to print value and address of variable\n",
-      "\n",
-      "from ctypes import c_int\n",
-      "\n",
-      "#Local definition\n",
-      "a = 22\n",
-      "b = c_int(a) # ctype datatype declaration\n",
-      "\n",
-      "#Pointer variable\n",
-      "b_po=pointer(b)\n",
-      "\n",
-      "# 'id' is a address of letter and it represents the address of the variable\n",
-      "b = id(a)\n",
-      "\n",
-      "#Result\n",
-      "print \"\\n Value of a = %d\" %a\n",
-      "print \"\\n Value of a = %d\" %(c_int(a).value)\n",
-      "print \"\\n Value of a = %d\" %b_po[0]\n",
-      "print \"\\n Address of a = %u\" %id(a)\n",
-      "print \"\\n Address of a = %u\" %b\n",
-      "print \"\\n Address of a = %u\" %id(b)\n",
-      "print \"\\n Value of b = address of a = %u\" %b"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Value of a = 22\n",
-        "\n",
-        " Value of a = 22\n",
-        "\n",
-        " Value of a = 22\n",
-        "\n",
-        " Address of a = 4298180848\n",
-        "\n",
-        " Address of a = 4298180848\n",
-        "\n",
-        " Address of a = 4344066728\n",
-        "\n",
-        " Value of b = address of a = 4298180848\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page Number: 5.109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to illustrate pointer to pointer\n",
-      "\n",
-      "from ctypes import c_int\n",
-      "\n",
-      "#Local definition\n",
-      "a = 22\n",
-      "b = c_int(a) # ctype datatype declaration\n",
-      "c = c_int(a) # ctype datatype declaration\n",
-      "\n",
-      "#Pointer variable\n",
-      "b_po = pointer(b)\n",
-      "c_po = pointer(b_po) # c_po is a pointer to another pointer\n",
-      "\n",
-      "# 'id' is a address of letter and it represents the address of the variable\n",
-      "b = id(a)\n",
-      "c = id(b)\n",
-      "\n",
-      "#Result\n",
-      "print \"\\n Value of a is %d\" %a\n",
-      "print \"\\n Value of a is %d\" %(c_int(a).value)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Value of a is 22\n",
-        "\n",
-        " Value of a is 22\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter15.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter15.ipynb
deleted file mode 100755
index 21b14f41..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter15.ipynb
+++ /dev/null
@@ -1,266 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2bed6e40cd55f716f11e909c76b0c924a863909239fcbbf975814260bb7b4531"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 15 : Pointers and Functions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Call By Value"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page no. 5.112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call by value\n",
-      "\n",
-      "def change(x,y):\n",
-      "    x=x+5\n",
-      "    y=y+10\n",
-      "    print \"In the function changes a&b is %d,%d\"%(x,y)\n",
-      "    \n",
-      "a=10\n",
-      "b=20\n",
-      "\n",
-      "print \"\\nBefore calling the function a&b is %d,%d\"%(a,b)\n",
-      "\n",
-      "change(a,b)\n",
-      "print \"\\nAfter calling a&b is %d,%d\"%(a,b)\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Before calling the function a&b is 10,20\n",
-        "In the function changes a&b is 15,30\n",
-        "\n",
-        "After calling a&b is 10,20\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page no.5.113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#program to swap(interchange) the two given variable.\n",
-      "\n",
-      "def swap(x,y):\n",
-      "    \n",
-      "    x=x+y\n",
-      "    y=x-y\n",
-      "    x=x-y\n",
-      "    print \"In swap function x=%d, y=%d\"%(x,y)\n",
-      "    return y,x\n",
-      "\n",
-      "a=10\n",
-      "b=20\n",
-      "\n",
-      "print \"Before swap a=%d, b=%d\"%(a,b)\n",
-      "\n",
-      "\n",
-      "print \"\\nAfter the calling the swap function, a=%d,b=%d\"%swap(a,b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Before swap a=10, b=20\n",
-        "In swap function x=20, y=10\n",
-        "\n",
-        "After the calling the swap function, a=10,b=20\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Call By Reference"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page.No.5.114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call By Reference\n",
-      "\n",
-      "def change (x,y):\n",
-      "    x=x+5\n",
-      "    y=y+10\n",
-      "    print \"In the function change a&b is %d %d\"%(x,y)\n",
-      "    return x,y\n",
-      "a=10\n",
-      "b=20\n",
-      "print \"\\nBefore calling the function a&b is %d %d\"%(a,b)\n",
-      "\n",
-      "\n",
-      "print \"After calling the function a&b is %d %d\"%change(a,b)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Before calling the function a&b is 10 20\n",
-        "In the function change a&b is 15 30\n",
-        "After calling the function a&b is 15 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.5.115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to swap(intercahnge) the value of variables.\n",
-      "\n",
-      "a=10\n",
-      "b=20\n",
-      "\n",
-      "print \"\\nBefore swap a=%d, b=%d\"%(swap(a,b))\n",
-      "\n",
-      "def swap(x,y):\n",
-      "    x=x+y\n",
-      "    y=x-y\n",
-      "    x=x-y\n",
-      "    print\"\\nIn swap function x=%d,y=%d\"%(x,y)\n",
-      "    return y,x\n",
-      "\n",
-      "print \"\\nAfter calling swap function a=%d, b=%d\"%(b,a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In swap function x=20,y=10\n",
-        "\n",
-        "Before swap a=10, b=20\n",
-        "\n",
-        "After calling swap function a=20, b=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page no. 5.115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to interchange the value of variables using call by reference\n",
-      "def exchange(m,n):\n",
-      "    t=m\n",
-      "    m=n\n",
-      "    n=t\n",
-      "    return n,m\n",
-      "\n",
-      "a,b=input(\"Enter two Numbers: \")\n",
-      "\n",
-      "print \"Before Exchange a=%d,b=%d\"%(a,b)\n",
-      "\n",
-      "\n",
-      "print \"After Exchange a=%d, b=%d\"%(exchange(b,a))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two Numbers: 10,20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Before Exchange a=10,b=20\n",
-        "After Exchange a=20, b=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter16.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter16.ipynb
deleted file mode 100755
index ab91ce85..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter16.ipynb
+++ /dev/null
@@ -1,333 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d2add72dea4c498cf89927248000fdbfce0907704975d2bdb6ff1cde1c8f50a7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter 16 : Pointers and Arrays"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.:5.118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print the value and address of the element\n",
-      "\n",
-      "a=[10,20,30,40,50]\n",
-      "for b in range(0,5):\n",
-      "    print \"The value of a[%d]=%d\"%(b,a[b])\n",
-      "    print \"Address of a[%d]=%u\"%(b,id(a[b]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of a[0]=10\n",
-        "Address of a[0]=4299197952\n",
-        "The value of a[1]=20\n",
-        "Address of a[1]=4299197712\n",
-        "The value of a[2]=30\n",
-        "Address of a[2]=4299197472\n",
-        "The value of a[3]=40\n",
-        "Address of a[3]=4299199208\n",
-        "The value of a[4]=50\n",
-        "Address of a[4]=4299198968\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.: 5.119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print the value and the address of the element using pointer with function\n",
-      "\n",
-      "a=[10,20,30,40,50]\n",
-      "\n",
-      "for c in range(0,5):\n",
-      "    print \"The Value of a[%d]=%d\"%(c,a[c])\n",
-      "    print \"The address of a[%d]=%u\"%(c,id(b))\n",
-      "def value(b):\n",
-      "    b=''\n",
-      "    id(b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of a[0]=10\n",
-        "The address of a[0]=4296527248\n",
-        "The Value of a[1]=20\n",
-        "The address of a[1]=4296527248\n",
-        "The Value of a[2]=30\n",
-        "The address of a[2]=4296527248\n",
-        "The Value of a[3]=40\n",
-        "The address of a[3]=4296527248\n",
-        "The Value of a[4]=50\n",
-        "The address of a[4]=4296527248\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example:3, Page No,: 5.120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to add the sum of number using pointer\n",
-      "\n",
-      "a=[10,20,30,40,50]\n",
-      "\n",
-      "b=total=0\n",
-      "c=id(c)\n",
-      "\n",
-      "for b in range(0,5):\n",
-      "    print \"Enter the number %d: \"%(b+1)\n",
-      "    a[b]=input()\n",
-      "    \n",
-      "c=a\n",
-      "for b in range(0,5):\n",
-      "    total=total+c[b]\n",
-      "    c[b]=c[b]+1\n",
-      "\n",
-      "print \"Total= %d\"%(total)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number 1: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number 2: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number 3: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "30\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number 4: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "40\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number 5: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "50\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total= 150\n"
-       ]
-      }
-     ],
-     "prompt_number": 67
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page No.: 5.121 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to sort a given number using pointer.\n",
-      "\n",
-      "arr = []\n",
-      "n = input(\"Enter the no. of elements: \")\n",
-      "for i in range(0,n):\n",
-      "    arr.append(0)\n",
-      "print \"Enter the element: \"    \n",
-      "for i in range(0,n):\n",
-      "    arr[i]=input()\n",
-      "print \"Before Sorting:\"    \n",
-      "for i in range(0,n):\n",
-      "    print \"%d\"%arr[i],\n",
-      "    \n",
-      "for i in range(0,n):\n",
-      "    for j in range(i,n):\n",
-      "        if arr[i]>arr[j]:\n",
-      "            t = arr[i]\n",
-      "            arr[i] = arr[j]\n",
-      "            arr[j] = t\n",
-      "print \"\\nAfter Sorting:\"\n",
-      "for i in range(0,n):\n",
-      "    print \"%d\" %arr[i],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the no. of elements: 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Before Sorting:\n",
-        "8 5 7 4 6 \n",
-        "After Sorting:\n",
-        "4 5 6 7 8\n"
-       ]
-      }
-     ],
-     "prompt_number": 72
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter17.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter17.ipynb
deleted file mode 100755
index 7967d054..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter17.ipynb
+++ /dev/null
@@ -1,310 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:86b94093ee72d30b0a9443a44a1c3905ef6186294337253a1ca56b0318def382"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 17 : Pointers with Multidimensional Array"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.:5.123"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print the values and addres of the array elements.\n",
-      "\n",
-      "arr=([[5,100],[10,200],[15,300],[20,400]])\n",
-      "\n",
-      "for a in range(0,4):\n",
-      "    print \"Address of %d array= %u\"%(a,id(arr[a]))\n",
-      "    for b in range(0,2):\n",
-      "        print \"Value =%d\"%(arr[a][b])\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Address of 0 array= 4336406328\n",
-        "Value =5\n",
-        "Value =100\n",
-        "Address of 1 array= 4336414944\n",
-        "Value =10\n",
-        "Value =200\n",
-        "Address of 2 array= 4336414800\n",
-        "Value =15\n",
-        "Value =300\n",
-        "Address of 3 array= 4366918504\n",
-        "Value =20\n",
-        "Value =400\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.: 5.125"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print the value and address of the element using array of pointers\n",
-      "\n",
-      "a=[0,1,2]\n",
-      "b=10\n",
-      "c=20\n",
-      "d=30\n",
-      "a[0]=b\n",
-      "a[1]=c\n",
-      "a[2]=d\n",
-      "for i in range(0,3):\n",
-      "    print \"Address = %u\"%id(a[i])\n",
-      "    print \"Value = %d\" % (a[i])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Address = 4298181184\n",
-        "Value = 10\n",
-        "Address = 4298180944\n",
-        "Value = 20\n",
-        "Address = 4298180704\n",
-        "Value = 30\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No.: 5.126 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to sort a list of strings in alphabetical order using array of pointers.\n",
-      "\n",
-      "a = []\n",
-      "n= input(\"Enter the number of strings: \")\n",
-      "print \"Enter each string on a separate line below.\"\n",
-      "for i in range(0,n):\n",
-      "    a.append(0)\n",
-      "    \n",
-      "for i in range(0,n):\n",
-      "    a[i]=raw_input(\"String %d: \" %(i+1))\n",
-      "for i in range(0,n):\n",
-      "    for j in range(i,n):\n",
-      "        if a[i]>a[j]:\n",
-      "            t = a[i]\n",
-      "            a[i] = a[j]\n",
-      "            a[j] = t\n",
-      "print \"\\nReordered List of Strings: \"\n",
-      "for i in range(0,n):\n",
-      "    print \"String %d: %s\" %(i+1,a[i])\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of strings: 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter each string on a separate line below.\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 1: venkat\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 2: ramesh\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 3: babu\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 4: muni\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 5: shankar\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 6: saravanan\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Reordered List of Strings: \n",
-        "String 1: babu\n",
-        "String 2: muni\n",
-        "String 3: ramesh\n",
-        "String 4: saravanan\n",
-        "String 5: shankar\n",
-        "String 6: venkat\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page No.: 5.128 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to count number of words\n",
-      "\n",
-      "str=[]\n",
-      "ps=raw_input(\"Enter a string: \")\n",
-      "print len(ps.split()), \"word(s) in the given string\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a string: I am a student.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4 word(s) in the given string\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 5, Page No.: 5.129"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate on strings and pointers\n",
-      "\n",
-      "s1=\"abcd\"\n",
-      "s2='efgh'\n",
-      "\n",
-      "print \"%s %16lu \\n\" % (s1,id(s1))\n",
-      "print \"%s %16lu \\n\" % (s2,id(s2)) \n",
-      "\n",
-      "s1=s2\n",
-      "print \"%s %16lu \\n\" % (s1,id(s1))\n",
-      "print \"%s %16lu \\n\" % (s2,id(s2))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "abcd       4363408224 \n",
-        "\n",
-        "efgh       4363332704 \n",
-        "\n",
-        "efgh       4363332704 \n",
-        "\n",
-        "efgh       4363332704 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter18.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter18.ipynb
deleted file mode 100755
index c422da50..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter18.ipynb
+++ /dev/null
@@ -1,490 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0724a7e756186f06943f5403b91daf5f178ee495f014e11c1fad0330aa0d254c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 18 : Pointers and Structures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.: 5.130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print details of person using structure and pointer\n",
-      "\n",
-      "class account:\n",
-      "    name=\"\"\n",
-      "    place=\"\"\n",
-      "    acctno=0\n",
-      "    \n",
-      "m4=account()\n",
-      "m4.name=\"Venkat\"\n",
-      "m4.place=\"Arcot\"\n",
-      "m4.acctno=24201\n",
-      "\n",
-      "print \"%s\\t%s\\t%d\\n\"%(m4.name,m4.place,m4.acctno)\n",
-      "\n",
-      "contents=m4\n",
-      "\n",
-      "print \"%s\\t%s\\t%d\\n\"%(contents.name,contents.place,contents.acctno)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Venkat\tArcot\t24201\n",
-        "\n",
-        "Venkat\tArcot\t24201\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No.: 5.131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print name, place, and account number using structure and pointer\n",
-      "\n",
-      "class account:\n",
-      "    name=\"\"\n",
-      "    place=\"\"\n",
-      "    acctno=0\n",
-      "    \n",
-      "m4=account()\n",
-      "m4.name=\"Madhavan\"\n",
-      "m4.place=\"Chennai\"\n",
-      "m4.acctno=241\n",
-      "a=m4\n",
-      "\n",
-      "print \"%s\\n%s\\n%d\\n\"%(a.name,a.place,a.acctno)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Madhavan\n",
-        "Chennai\n",
-        "241\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No.: 5.132"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to show how the int_pointers structure can be handled.\n",
-      "\n",
-      "\n",
-      "class int_pointers:\n",
-      "    p1=0\n",
-      "    p2=0\n",
-      "    \n",
-      "pointers=int_pointers()\n",
-      "m1= 80\n",
-      "pointers.p1=m1\n",
-      "pointers.p2=-40\n",
-      "m2=pointers.p2\n",
-      "print \"m1= %i, *pointers.p1 = %i\\n\"% (m1,pointers.p1)\n",
-      "print \"m2= %i, *pointers.p2 = %i\\n\"% (m2,pointers.p2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "m1= 80, *pointers.p1 = 80\n",
-        "\n",
-        "m2= -40, *pointers.p2 = -40\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 1: Page No.5.133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to interchange the values stored in two variables using function with pointers. \n",
-      "\n",
-      "def interchange(a,b):\n",
-      "    t=a\n",
-      "    a=b\n",
-      "    b=t\n",
-      "    return b,a\n",
-      "    \n",
-      "a=39\n",
-      "b=77\n",
-      "\n",
-      "print \"Before Interchange: a=%d, b=%d\"%(a,b)\n",
-      "\n",
-      "\n",
-      "print \"After Interchange : a=%d, b=%d\"%(interchange(b,a))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Before Interchange: a=39, b=77\n",
-        "After Interchange : a=77, b=39\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study: 2, Page No.:5.134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate the use of structure pointers\n",
-      "\n",
-      "class book:\n",
-      "    bno=0\n",
-      "    bname=''\n",
-      "    cost=0.0\n",
-      "\n",
-      "p=[book()for i in range(0,3)]\n",
-      "print \"Enter the Book details... \"\n",
-      "print \"Enter Book number, Book Name, Book Costs... \"\n",
-      "for i in range(0,3):\n",
-      "    p[i].bno=input()\n",
-      "    p[i].bname=raw_input()\n",
-      "    p[i].cost=input()\n",
-      "    \n",
-      "for i in range(0,3):\n",
-      "   \n",
-      "    print \"\\n%d\\t%s\\t\\t%d\"%(p[i].bno,p[i].bname,p[i].cost)\n",
-      "   \n",
-      "print \"\\nStructure Output\\n\"\n",
-      "ptr=p\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "     print \"%d\\t%s\\t\\t%d\"%(ptr[i].bno,ptr[i].bname,ptr[i].cost)\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Book details... \n",
-        "Enter Book number, Book Name, Book Costs... \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "101\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Computer Practice_1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "240\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "201\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Computer Practice_2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "250\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "202\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C++ and JAVA\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "280\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "101\tComputer Practice_1\t240\n",
-        "\n",
-        "201\tComputer Practice_2\t250\n",
-        "\n",
-        "202\tC++ and JAVA\t280\n",
-        "\n",
-        "Structure Output\n",
-        "\n",
-        "101\tComputer Practice_1\t240\n",
-        "201\tComputer Practice_2\t250\n",
-        "202\tC++ and JAVA\t280\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 3: Page No: 5.136"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to count number of vowels,consonants, digits,spaces and other characters in a line of text.\n",
-      "\n",
-      "vowels='aeiou'\n",
-      "consonants='bcdfghjklmnpqrstvwxyz'\n",
-      "spaces =' '\n",
-      "digits='1234567890'\n",
-      "special_characters='!@#$%^&*()_'\n",
-      "\n",
-      "text=raw_input(\"Enter the text in lower case: \")\n",
-      "\n",
-      "print \"\\nNumber of vowels: \",dict((v, text.count(v)) for v in text if v in vowels)\n",
-      "print \"Number of consonants: \",dict((c, text.count(c)) for c in text if c in consonants)\n",
-      "print \"Number of Spaces: \",dict((s, text.count(s)) for s in text if s in spaces)\n",
-      "print \"Number of digits: \",dict((d, text.count(d)) for d in text if d in digits)\n",
-      "print \"Number of special_characters: \",dict((sc, text.count(sc)) for sc in text if sc in special_characters)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the text in lower case: 123 vrb publishers @#$%^&*()\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number of vowels:  {'i': 1, 'e': 1, 'u': 1}\n",
-        "Number of consonants:  {'b': 2, 'h': 1, 'l': 1, 'p': 1, 's': 2, 'r': 2, 'v': 1}\n",
-        "Number of Spaces:  {' ': 3}\n",
-        "Number of digits:  {'1': 1, '3': 1, '2': 1}\n",
-        "Number of special_characters:  {'@': 1, '#': 1, '%': 1, '$': 1, '&': 1, ')': 1, '(': 1, '*': 1, '^': 1}\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 4: Page No.: 5.136"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate malloc() function\n",
-      "\n",
-      "NULL=0\n",
-      "string = ''\n",
-      "if string==NULL:\n",
-      "    print \"Not Enough memory to allocate buffer\\n\"\n",
-      "else:\n",
-      "    \n",
-      "    def strcpy(cstring1):\n",
-      "          import copy\n",
-      "          cstring2=copy.copy(cstring1)\n",
-      "          return cstring2  \n",
-      "string=\"Hello\"\n",
-      "print \"String is %s\" %string"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String is Hello\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 5: Page No.: 5.137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate calloc() function\n",
-      "\n",
-      "string=''\n",
-      "\n",
-      "def strcpy(cstring1):\n",
-      "          import copy\n",
-      "          cstring2=copy.copy(cstring1)\n",
-      "          return cstring2  \n",
-      "stri=\"BHAVANA\"\n",
-      "string=strcpy(stri)\n",
-      "print \"String is %s\" %string"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String is BHAVANA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Case Study 6: Page No.:5.138"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate realloc() function\n",
-      "#Cannot do in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 29
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter19.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter19.ipynb
deleted file mode 100755
index bbaf1a5b..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter19.ipynb
+++ /dev/null
@@ -1,126 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:069422c07d34cc73268583e26cec775bbe1b3930cb154b474b2ff1e109dbe593"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 19 : Command Line Arguments"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.:5.139"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import sys\n",
-      "argc=0\n",
-      "if argc!=2:\n",
-      "    print \"Hellow %s\"%(sys.argv)\n",
-      "else:\n",
-      "    print \"Forgotted to type name\"\n",
-      "    exit(0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hellow ['-c', '-f', '/home/gayathri/.config/ipython/profile_default/security/kernel-f732916e-2a20-4211-8210-1c4fa7115bd5.json', \"--IPKernelApp.parent_appname='ipython-notebook'\", '--profile-dir', '/home/gayathri/.config/ipython/profile_default', '--parent=1']\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No. 5.139"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "argc='0'\n",
-      "if argc==2:\n",
-      "    print \"Argument supplied is %s\"%(sys.argv)\n",
-      "elif argc>2:\n",
-      "    print \"Too many arguments supplied \\n\",(sys.argv)\n",
-      "else:\n",
-      "    print \"One Argument expected\",(sys.argv)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Too many arguments supplied \n",
-        "['-c', '-f', '/home/gayathri/.config/ipython/profile_default/security/kernel-f732916e-2a20-4211-8210-1c4fa7115bd5.json', \"--IPKernelApp.parent_appname='ipython-notebook'\", '--profile-dir', '/home/gayathri/.config/ipython/profile_default', '--parent=1']\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No. :5.140"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "\n",
-      "print \"The Arguments to the program are: \"\n",
-      "print str(sys.argv)\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Arguments to the program are: \n",
-        "['-c', '-f', '/home/gayathri/.config/ipython/profile_default/security/kernel-f732916e-2a20-4211-8210-1c4fa7115bd5.json', \"--IPKernelApp.parent_appname='ipython-notebook'\", '--profile-dir', '/home/gayathri/.config/ipython/profile_default', '--parent=1']\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter20.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter20.ipynb
deleted file mode 100755
index adaadc6f..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter20.ipynb
+++ /dev/null
@@ -1,61 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:7f2d14f2c2ae43d8abc01634eb7d39800f8596aa61628d10404d3502e1e30c64"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 20 : Dynamic Memory Allocation"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No. 5.146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reallocation of memory\n",
-      "\n",
-      "p= \"MADRAS\"\n",
-      "\n",
-      "print \"Memory contains: %s\\n\"%(p)            #There is no reallocation memory in python\n",
-      "\n",
-      "p=\"CHENNAI\"\n",
-      "\n",
-      "print \"Memory now contains: %s\\n\"%(p)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Memory contains: MADRAS\n",
-        "\n",
-        "Memory now contains: CHENNAI\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter21.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter21.ipynb
deleted file mode 100755
index 2b76d2a2..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter21.ipynb
+++ /dev/null
@@ -1,174 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2bdca20a49e1bad43f6968c764966fb35bc4580bd54570e770e266797cc3ab0e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 21 : The C Preprocessor"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 1, Page No.5.149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print square and cube Values by using macro substitutions. \n",
-      "def sq(n):\n",
-      "    return n*n\n",
-      "    \n",
-      "    \n",
-      "def cube(n):\n",
-      "    return n*n*n\n",
-      "    \n",
-      "\n",
-      "a=5\n",
-      "b=3\n",
-      "s=c=0\n",
-      "\n",
-      "#Function calling\n",
-      "s=sq(a)\n",
-      "c=cube(b)\n",
-      "\n",
-      "#Result\n",
-      "print \"Square value of 5 is %d\"%(s)\n",
-      "print \"Cube value of 3 is %d\"%(c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Square value of 5 is 25\n",
-        "Cube value of 3 is 27\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 2, Page No. 5.151"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the Values using macro substitution.\n",
-      "\n",
-      "VAL=40\n",
-      "\n",
-      "print \"The value is %d\"%VAL\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value is 40\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 3, Page No.:5.151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print Value and text using preprocessing.\n",
-      "VAL=35\n",
-      "\n",
-      "HELLO=\"HELLO\"\n",
-      "\n",
-      "res=VAL-5\n",
-      "\n",
-      "print \"Res=VAL-5:Res= %d\"%res\n",
-      "\n",
-      "print \"HELLO\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Res=VAL-5:Res= 30\n",
-        "HELLO\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example: 4, Page No.:5.152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to print the value using preprocessors.\n",
-      "USD=1\n",
-      "UKP=1\n",
-      "\n",
-      "currency_rate=46\n",
-      "currency_rate=100\n",
-      "\n",
-      "rs=10*currency_rate\n",
-      "print  \"The Value of rs is : %d\"%rs\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Value of rs is : 1000\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/Chapter22.ipynb b/Fundamental_of_Computing_and_Programming_in_C/Chapter22.ipynb
deleted file mode 100755
index 53a104bf..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/Chapter22.ipynb
+++ /dev/null
@@ -1,2300 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:dd103cfe67b47eb61e56816517923337936d461ec4fd4b8f9f87e6c544fd31e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 22 : Sample C Programs(Lab Exercise)"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 1: Page no.:L.4"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Area of Triangle\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#User Inputs\n",
-      "a,b,c=input(\"Enter three sides: \")\n",
-      "\n",
-      "#calculation of semi perimeter\n",
-      "s=(a+b+c)/2\n",
-      "\n",
-      "#Calculation of Area of Triangle\n",
-      "d=(s*(s-a)*(s-b)*(s-c))\n",
-      "\n",
-      "area=math.sqrt(d)\n",
-      "\n",
-      "#Result\n",
-      "print \"Area of triangle = %f sq units \\n\"%area"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter three sides: 5,6,7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Area of triangle = 14.696938 sq units \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 2: Page No. L.5"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Square root for the given value.\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#User Input\n",
-      "x=input(\"Enter the Value: \")\n",
-      "\n",
-      "#Result\n",
-      "print \"Square root of %.2f is: \"%(x)\n",
-      "\n",
-      "print math.sqrt(x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Value: 81\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Square root of 81.00 is: \n",
-        "9.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 3: Page.No.L.6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sum of the series 0+1+(1+2)+...+n\n",
-      "\n",
-      "#User Input\n",
-      "n=input(\"Enter the Value of n : \")\n",
-      "\n",
-      "#Variable Declaration\n",
-      "s=term=0\n",
-      "i=j=0\n",
-      "\n",
-      "#Computing Values in loop\n",
-      "for i in range(n):\n",
-      "    for j in range(n):\n",
-      "        if j<=i:\n",
-      "            term=term+j\n",
-      "            j=j+1        \n",
-      "s=s+term\n",
-      "if i<=n:\n",
-      "    i=i+1\n",
-      "    \n",
-      "#Result\n",
-      "print \"Sum of the series S= %d\"%(s)\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Value of n :5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of the series S= 20\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 4: Page No.:L.9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Decimal into binary number.\n",
-      "\n",
-      "n=input(\"Enter the decimal number: \")\n",
-      "s=n\n",
-      "k=[]\n",
-      "\n",
-      "#Calculation\n",
-      "while (s>0):\n",
-      "    a=int(float(s%2))\n",
-      "    k.append(a)\n",
-      "    s=(s-a)/2\n",
-      "k.append(0)\n",
-      "bi=\"\"\n",
-      "\n",
-      "for j in k[::1]:\n",
-      "    bi=bi+str(j)\n",
-      "\n",
-      "#Result\n",
-      "print('The binary value of %d is %s'%(n, bi))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the decimal number: 14\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The binary value of 14 is 01110\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 5: Page No. L.10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#The given number is armstrong or not.\n",
-      "\n",
-      "#Variable declaration\n",
-      "a=b=c=d=e=0\n",
-      "\n",
-      "#User Input\n",
-      "a=input(\"Enter the Number: \")\n",
-      "\n",
-      "#Assigning variable 'a' value to variable 'e'\n",
-      "e=a\n",
-      "\n",
-      "#Calculation for finding Armstrong or not\n",
-      "while a>0:\n",
-      "    b=a%10\n",
-      "    c=b*b*b\n",
-      "    a=a/10\n",
-      "    d=c+d\n",
-      " \n",
-      "#Result\n",
-      "if e==d:\n",
-      "    print \"%d is an armstrong number. \"%e\n",
-      "    \n",
-      "else:\n",
-      "    print \"%d is not an armstrong number. \"%e"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Number: 153\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "153 is an armstrong number. \n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 6: Page No.L.12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Centigrade and Farenheit Values.\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "c=f=cn=fn=0.0\n",
-      "\n",
-      "#User Input1 for calculating farenheit\n",
-      "c=input(\"Enter temperature in centigrade : \")\n",
-      "\n",
-      "#Farenheit Calculation\n",
-      "f=1.8*c+32\n",
-      "\n",
-      "#Result1\n",
-      "print \"Farenheit Equivalent is: %.1f\\n\" %(f)\n",
-      "\n",
-      "#User input2 for calculating centigrade\n",
-      "fn=input(\"Enter temperature in farenheit : \")\n",
-      "\n",
-      "#Centigrade Calculation\n",
-      "cn=(fn-32)/1.8\n",
-      "\n",
-      "#Result2\n",
-      "print \"Centigrade Equivalent is : %.1f\\n\"%(cn)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter temperature in centigrade : 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Farenheit Equivalent is: 68.0\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter temperature in farenheit : 68\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Centigrade Equivalent is : 20.0\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 7: Page .no.L.13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Converting Decimal value to binary number\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "bnum=digit=decimal=bina=base=0\n",
-      "\n",
-      "#User input\n",
-      "bnum=input(\"Enter the binary number: \")\n",
-      "\n",
-      "bina=bnum\n",
-      "\n",
-      "while(bnum!=0):\n",
-      "    digit=bnum%10\n",
-      "    decimal=decimal+(digit<<base)\n",
-      "    base=base+1\n",
-      "    bnum=bnum/10\n",
-      "    \n",
-      "print \"The binary equivalent of %d in decimal = %d\"%(bina,decimal)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the binary number: 1111\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The binary equivalent of 1111 in decimal = 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 8: Page no.L.15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reverse number of the given number.\n",
-      "\n",
-      "#Function  Definition.\n",
-      "def rev(n):\n",
-      "    digit=rev_of=0\n",
-      "    while n!=0:\n",
-      "        digit=n%10\n",
-      "        rev_of=rev_of*10+digit\n",
-      "        n=n/10\n",
-      "    return rev_of\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "\n",
-      "#User input\n",
-      "p=input(\"Enter the number: \")\n",
-      "\n",
-      "q=rev(p)\n",
-      "\n",
-      "print \"Reverse of %d is %d.\\n\"%(p,q)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number: 1234\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reverse of 1234 is 4321.\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 9: Page no.L.17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Square of given value\n",
-      "\n",
-      "#Function definition\n",
-      "def square(y):\n",
-      "    return y*y\n",
-      "\n",
-      "#User input\n",
-      "n=input(\"Enter the nth element : \")\n",
-      "\n",
-      "#Calculation\n",
-      "for x in range(1,n+1):\n",
-      "     print \"%d \" %(square(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the nth element : 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 \n",
-        "4 \n",
-        "9 \n",
-        "16 \n",
-        "25 \n",
-        "36 \n",
-        "49 \n",
-        "64 \n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 10: Page no.:L.18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Maximum value of given numbers\n",
-      "\n",
-      "#Function definition\n",
-      "def maximum(x,y,z):\n",
-      "    \n",
-      "    maxi=x\n",
-      "    if y>maxi:\n",
-      "        maxi=y\n",
-      "        if z>maxi:\n",
-      "            maxi=z\n",
-      "    return maxi\n",
-      "    \n",
-      "    \n",
-      "#User input\n",
-      "a,b,c=input(\"Enter three integers : \")\n",
-      "\n",
-      "#Result\n",
-      "print \"Maximum is : %d\\n\"%(maximum(a,b,c))\n",
-      "\n",
-      "\n",
-      "    \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter three integers : 2,6,9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum is : 9\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 11: Page no.L.19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Factorial of the given number using recursion.\n",
-      "\n",
-      "#funtion declaration\n",
-      "def fact(n):\n",
-      "    if n==0:\n",
-      "        return 1\n",
-      "    else:\n",
-      "        return(n*fact(n-1))\n",
-      "    \n",
-      "#USer Input\n",
-      "\n",
-      "n=input(\"Enter the number whose factorial is to be found: \")\n",
-      "\n",
-      "#Result\n",
-      "print \"The Factorial of %d is : %d \\n\"%(n,fact(n))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number whose factorial is to be found: 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Factorial of 6 is : 720 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 12: Page no.L.20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Palindrome or not.\n",
-      "\n",
-      "print \"Enter the String to check palindrome or not \\n\"\n",
-      "\n",
-      "#User Input\n",
-      "\n",
-      "string = raw_input(\"Enter the String: \")\n",
-      "\n",
-      "#reverse the string \n",
-      "\n",
-      "rev_string = reversed(string)\n",
-      "\n",
-      "#Check if the string is palindrome or not\n",
-      "\n",
-      "if list(rev_string) == list(string) :\n",
-      "    print \"\\nThe given string %s is a palindrome\"%(string)\n",
-      "    \n",
-      "else:\n",
-      "    print \"\\nThe given string %s is not a palindrome\"%(string)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the String to check palindrome or not \n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the String: madam\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The given string madam is a palindrome\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 13: Page No. L.24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Largest and smallest number in array\n",
-      "\n",
-      "#User input\n",
-      "n= input(\"Enter the nth term of an array: \")\n",
-      "\n",
-      "for i in range(0,n):\n",
-      "    a[i]=eval(raw_input())\n",
-      "\n",
-      "#check if the given array is large or small.    \n",
-      "    for i in range(1,n):\n",
-      "        if a[i]>large:\n",
-      "           large=a[i]\n",
-      "            \n",
-      "        else:\n",
-      "           if a[i]<small:\n",
-      "              small=a[i]\n",
-      "       \n",
-      "         \n",
-      "print \"Largest element in an array is: %.2f\"%large\n",
-      "print \"Smallest element in an array is: %.2f\"%small\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the nth term of an array: 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "79\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-21\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Largest element in an array is: 79.00\n",
-        "Smallest element in an array is:-21.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 14: Page no. L.26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given array in ascending order\n",
-      "\n",
-      "#User Input\n",
-      "n=input(\"Enter the last term of the array: \")\n",
-      "\n",
-      "for i in range(0,n):\n",
-      "    a[i]=eval(raw_input())\n",
-      "for i in range(0,n-1):\n",
-      "\n",
-      "    for j in range(i+1,n):\n",
-      "        if a[i]>a[j]:\n",
-      "            temp=a[i]\n",
-      "            a[i]=a[j]\n",
-      "            a[j]=temp\n",
-      "        \n",
-      "print \"\\nThe Elements in ascending order\\n\"\n",
-      "\n",
-      "for i in range(0,n):\n",
-      "    print \"%.2f\"%a[i]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last term of the array: 6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "23\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "14\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-67\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "48\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "01\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Elements in ascending order\n",
-        "\n",
-        "-90.00\n",
-        "-67.00\n",
-        "1.00\n",
-        "14.00\n",
-        "23.00\n",
-        "48.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 15: Page No. L.27"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Insert element in specified position in an array\n",
-      "\n",
-      "import array\n",
-      "\n",
-      "#User input\n",
-      "n=input(\"Enter the last value: \")\n",
-      "a=[0 for i in range(0,n) ]\n",
-      "print \"Enter the elements of array: \"\n",
-      "for i in range(0,n):\n",
-      "    a[i]=eval(raw_input())\n",
-      "    \n",
-      "#Getting inserting element and position from users\n",
-      "item=input(\"Enter the element to be inserted: \")\n",
-      "pos=input(\"Enter the position of insertion: \")\n",
-      "a.insert(pos,item)\n",
-      "#Result\n",
-      "print \"\\nElements after insertion is : \"\n",
-      "for i in range(0,n+1):\n",
-      "    print \"%.2f\"%a[i]\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last value: 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the elements of array: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "34\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "32\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element to be inserted: 36\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the position of insertion: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Elements after insertion is : \n",
-        "12.00\n",
-        "34.00\n",
-        "56.00\n",
-        "90.00\n",
-        "36.00\n",
-        "32.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 16: Page.no.L.29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Transpose of the matrix\n",
-      "\n",
-      "#User Input\n",
-      "\n",
-      "a=[]\n",
-      "\n",
-      "n,m=input(\"Enter the Row and Columns: \")\n",
-      "print \"Enter the elements of matrix:\"\n",
-      "for i in range(n):\n",
-      "    a.append([])\n",
-      "    \n",
-      "    for j in range(m):\n",
-      "        element=input()\n",
-      "        a[i].append(element)\n",
-      "\n",
-      "#print matrix\n",
-      "print \"\\nElements in matrix: \"\n",
-      "for i in range(n):\n",
-      "    for j in range(m):\n",
-      "        print a[i][j],\n",
-      "    print '\\n'\n",
-      "\n",
-      "\n",
-      "#generate transpose\n",
-      "transpose=[]\n",
-      "for j in range(m):\n",
-      "    transpose.append([])\n",
-      "    for i in range (n):\n",
-      "        b=a[i][j]\n",
-      "        transpose[j].append(b)\n",
-      "\n",
-      "#print transpose\n",
-      "print \"Transpose of the matrix is: \"\n",
-      "for i in range (m):\n",
-      "    for j in range (n):\n",
-      "        print transpose[i][j],\n",
-      "    print '\\n'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Row and Columns: 3,3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the elements of matrix:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Elements in matrix: \n",
-        "1 2 3 \n",
-        "\n",
-        "4 5 6 \n",
-        "\n",
-        "7 8 9 \n",
-        "\n",
-        "Transpose of the matrix is: \n",
-        "1 4 7 \n",
-        "\n",
-        "2 5 8 \n",
-        "\n",
-        "3 6 9 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 17: Page no. L.32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to find sum of an array matrix\n",
-      "#Array Declaration\n",
-      "\n",
-      "a=[]\n",
-      "b=[]\n",
-      "c=[[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
-      "#User Input1\n",
-      "n,m=input(\"Enter the Row and Columns of A Matrix: \")\n",
-      "print \"Enter the elements of A matrix:\"\n",
-      "for i in range(n):\n",
-      "    a.append([])\n",
-      "    \n",
-      "    for j in range(m):\n",
-      "        element=input()\n",
-      "        a[i].append(element)\n",
-      "\n",
-      "#print matrix1\n",
-      "print \"\\nElements in A matrix: \"\n",
-      "for i in range(n):\n",
-      "    for j in range(m):\n",
-      "        print a[i][j],\n",
-      "    print '\\n'\n",
-      "\n",
-      "    \n",
-      "#User Input 2\n",
-      "p,q=input(\"Enter the Row and Columns of B Matrix: \")\n",
-      "print \"Enter the elements of B matrix:\"\n",
-      "for i in range(p):\n",
-      "    b.append([])\n",
-      "    \n",
-      "    for j in range(q):\n",
-      "        element=input()\n",
-      "        b[i].append(element)\n",
-      "\n",
-      "#print matrix 2\n",
-      "print \"\\nElements in B matrix: \"\n",
-      "for i in range(p):\n",
-      "    for j in range(q):\n",
-      "        print b[i][j],\n",
-      "    print '\\n'\n",
-      "\n",
-      "#Addition of array in matrix    \n",
-      "for i in range(n):\n",
-      "    for j in range(m):\n",
-      "        c[i][j] += a[i][j] + b[i][j]\n",
-      "\n",
-      "#Result\n",
-      "print \"\\nThe Sum of A and B matrix: \"\n",
-      "for i in range(n):\n",
-      "    for j in range(m):\n",
-      "        print \"%5d\"%c[i][j],\n",
-      "    print '\\n'"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Row and Columns of A Matrix: 3,3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the elements of A matrix:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Elements in A matrix: \n",
-        "1 2 3 \n",
-        "\n",
-        "4 5 6 \n",
-        "\n",
-        "7 8 9 \n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Row and Columns of B Matrix: 3,3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the elements of B matrix:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Elements in B matrix: \n",
-        "1 2 3 \n",
-        "\n",
-        "4 5 6 \n",
-        "\n",
-        "6 7 4 \n",
-        "\n",
-        "\n",
-        "The Sum of A and B matrix: \n",
-        "    2     4     6 \n",
-        "\n",
-        "    8    10    12 \n",
-        "\n",
-        "   13    15    13 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 80
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 18: Page No. :L.35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to find Matrix Multiplication \n",
-      "#Array Declaration\n",
-      "\n",
-      "a=[]\n",
-      "b=[]\n",
-      "c=[[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
-      "\n",
-      "while True:\n",
-      "#User Input1\n",
-      "    n,m=input(\"Enter the Row and Columns of A Matrix: \")\n",
-      "    p,q=input(\"Enter the Row and Columns of B Matrix: \")\n",
-      "    if (n==m&p==q):\n",
-      "        print \"Enter the elements of A matrix:\"   \n",
-      "        for i in range(n):\n",
-      "            a.append([]) \n",
-      "            for j in range(m):\n",
-      "                element=input()\n",
-      "                a[i].append(element)\n",
-      "#print matrix1\n",
-      "        print \"\\nElements in A matrix: \"\n",
-      "        for i in range(n):\n",
-      "            for j in range(m):\n",
-      "                print a[i][j],\n",
-      "            print '\\n'\n",
-      "   \n",
-      "#User Input 2    \n",
-      "        print \"Enter the elements of B matrix:\"\n",
-      "        for i in range(p):\n",
-      "            b.append([])\n",
-      "            for j in range(q):\n",
-      "                element=input()\n",
-      "                b[i].append(element)\n",
-      "\n",
-      "#print matrix 2\n",
-      "        print \"\\nElements in B matrix: \"\n",
-      "        for i in range(p):\n",
-      "            for j in range(q):\n",
-      "                print b[i][j],\n",
-      "            print '\\n'\n",
-      "\n",
-      "#Multiplication of array in matrix    \n",
-      "        for i in range(0,m):\n",
-      "            for j in range(0,n):\n",
-      "                c[i][j] = 0\n",
-      "                for k in range(0,n):\n",
-      "                    c[i][j] += a[i][k] * b[k][j]\n",
-      "        \n",
-      "#Result\n",
-      "        print \"\\nThe Multiplication of A and B matrix: \"\n",
-      "        for i in range(n):\n",
-      "            for j in range(m):\n",
-      "                print \"%5d\"%c[i][j],\n",
-      "            print '\\n'\n",
-      "        break\n",
-      "    else:\n",
-      "        print \"The No. of rows and columns are not equal.\\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Row and Columns of A Matrix: 3,3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Row and Columns of B Matrix: 2,2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The No. of rows and columns are not equal.\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Row and Columns of A Matrix: 3,3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Row and Columns of B Matrix: 3,3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the elements of A matrix:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Elements in A matrix: \n",
-        "1 2 3 \n",
-        "\n",
-        "4 5 6 \n",
-        "\n",
-        "7 8 9 \n",
-        "\n",
-        "Enter the elements of B matrix:\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Elements in B matrix: \n",
-        "2 4 1 \n",
-        "\n",
-        "6 7 4 \n",
-        "\n",
-        "3 5 7 \n",
-        "\n",
-        "\n",
-        "The Multiplication of A and B matrix: \n",
-        "   23    33    30 \n",
-        "\n",
-        "   56    81    66 \n",
-        "\n",
-        "   89   129   102 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 93
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 19: Page No. L.38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Student grade using structure\n",
-      "\n",
-      "#User Input\n",
-      "n=input(\"Enter the number of students: \")\n",
-      "\n",
-      "#Defining Structure\n",
-      "class std:\n",
-      "    no=0\n",
-      "    name = ''\n",
-      "    m1 = m2 = m3 =0\n",
-      "    total=avg=0.0\n",
-      "    grade= ''\n",
-      "    \n",
-      "s=[std() for i in range(0,n)]\n",
-      "\n",
-      "#User Input\n",
-      "for i in range(0,n):\n",
-      "    print \"\\nEnter the student %d details\"%(i+1)\n",
-      "    print \"\\n\"\n",
-      "    s[i].no=int(input(\"Give Student Number: \"))\n",
-      "    s[i].name=raw_input(\"Enter the Student Name: \")\n",
-      "    s[i].m1=int(input(\"Enter Student Mark1: \"))\n",
-      "    s[i].m2=int(input(\"Enter Student Mark2: \"))\n",
-      "    s[i].m3=int(input(\"Enter Student Mark3: \"))\n",
-      "\n",
-      "\n",
-      "#Calculation    \n",
-      "    s[i].total = (s[i].m1 + s[i].m2 + s[i].m3)\n",
-      "    s[i].avg=(s[i].total / 3)\n",
-      "    if s[i].avg<=40:\n",
-      "        s[i].grade='D'\n",
-      "    elif s[i].avg<60:\n",
-      "        s[i].grade='C'\n",
-      "    elif s[i].avg<80:\n",
-      "        s[i].grade='B'\n",
-      "    else:\n",
-      "        s[i].grade='A'\n",
-      "        \n",
-      "#Result\n",
-      "print \"\\tStudent mark details.... \\n\"\n",
-      "print \"Roll No\\t Name \\t\\tMark1 \\tMark2 \\tMark3 \\tTotal \\tAverage \\tGrade\"\n",
-      "for i in range(0,n):\n",
-      "    print \"%d\\t\"%(s[i].no),\"%s\\t\\t\"%(s[i].name), \"%d\\t\"%(s[i].m1),\"%d\\t\"%(s[i].m2),\"%d\\t\"%(s[i].m3), \"%.2f\\t\"%(s[i].total), \"%.2f\\t\\t\"%(s[i].avg), \"%s\\t\"%(s[i].grade)\n",
-      "   "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of students: 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter the student 1 details\n",
-        "\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give Student Number: 100\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Student Name: Venkat\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark1: 89\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark2: 87\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark3: 76\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter the student 2 details\n",
-        "\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give Student Number: 101\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Student Name: Shankar\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark1: 87\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark2: 90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Student Mark3: 67\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tStudent mark details.... \n",
-        "\n",
-        "Roll No\t Name \t\tMark1 \tMark2 \tMark3 \tTotal \tAverage \tGrade\n",
-        "100\tVenkat\t\t89\t87\t76\t252.00\t84.00\t\tA\t\n",
-        "101\tShankar\t\t87\t90\t67\t244.00\t81.00\t\tA\t\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 20: Page No. L.40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Employee details using Union\n",
-      "\n",
-      "#Defining Union\n",
-      "\n",
-      "class union:\n",
-      "    name=\"\"\n",
-      "    idno=0\n",
-      "    salary=0.0\n",
-      "    \n",
-      "desc=union()\n",
-      "\n",
-      "desc.name=\"vinod\"\n",
-      "\n",
-      "print \"Employee details:\"\n",
-      "\n",
-      "print \"The name is %s\\n\"%(desc.name)\n",
-      "print \"The idno is %d\\n\"%(desc.idno)\n",
-      "print \"The salary is %6.2f\\n\"%(desc.salary)\n",
-      "\n",
-      "desc.idno=10\n",
-      "\n",
-      "print \"Employe details:\"\n",
-      "\n",
-      "print \"The name is %s\\n\"%(desc.name)\n",
-      "print \"The idno is %d\\n\"%(desc.idno)\n",
-      "print \"The salary is %6.2f\\n\"%(desc.salary)\n",
-      "\n",
-      "desc.salary=6500.00\n",
-      "\n",
-      "print \"Employee details:\"\n",
-      "\n",
-      "print \"The name is %s\\n\"%(desc.name)\n",
-      "print \"The idno is %d\\n\"%(desc.idno)\n",
-      "print \"The salary is %6.2f\\n\"%(desc.salary)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Employee details:\n",
-        "The name is vinod\n",
-        "\n",
-        "The idno is 0\n",
-        "\n",
-        "The salary is   0.00\n",
-        "\n",
-        "Employe details:\n",
-        "The name is vinod\n",
-        "\n",
-        "The idno is 10\n",
-        "\n",
-        "The salary is   0.00\n",
-        "\n",
-        "Employee details:\n",
-        "The name is vinod\n",
-        "\n",
-        "The idno is 10\n",
-        "\n",
-        "The salary is 6500.00\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 21: Page No. L.41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Swap the contents of two variable using pointer.\n",
-      "\n",
-      "#swapping function.\n",
-      "\n",
-      "def interchange(x,y):\n",
-      "    temp=x\n",
-      "    x=y\n",
-      "    y=temp\n",
-      "    return x,y\n",
-      "\n",
-      "#User Input\n",
-      "a,b = input(\"Enter the values of A and B: \")\n",
-      "#Result before interchanging\n",
-      "print \"a=%d b=%d\\n\"%(a,b)\n",
-      "\n",
-      "\n",
-      "#Result after interchanging\n",
-      "print \"After interchanging: \\n\"\n",
-      "\n",
-      "print \"a=%d b=%d\\n\"%(interchange(a,b))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the values of A and B: 23,56\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a=23 b=56\n",
-        "\n",
-        "After interchanging: \n",
-        "\n",
-        "a=56 b=23\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 22: Page No.: L.43"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Concatenate two strings using Dynamic Memory Allocation \n",
-      "\n",
-      "s1= raw_input(\"Enter String One: \")\n",
-      "s2=raw_input(\"Enter String two: \")\n",
-      "\n",
-      "s3=''\n",
-      "s3=s1+s2                    \n",
-      "#Result\n",
-      "print \"Concatenated String: \", s1+s2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String One: venkat\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String two: raman\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Concatenated String:  venkatraman\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 23: Page No. L.44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#program to store and print same data to and from the file.\n",
-      "\n",
-      "#open file\n",
-      "fp=open(\"Text.txt\",\"w\")\n",
-      "\n",
-      "#User Input\n",
-      "c=raw_input(\"Enter the Text: \")\n",
-      "\n",
-      "#Store given input into a file \"text.txt\"\n",
-      "fp.write(c)\n",
-      "\n",
-      "#Closing file\n",
-      "fp.close()\n",
-      "\n",
-      "#Again open text.txt\n",
-      "fp=open(\"Text.txt\", \"r\")\n",
-      "\n",
-      "#display the content from the file \"text.txt\"\n",
-      "print \"Displaying text from file:\\n\",c \n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Text: VRB PUBLISHERS\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Displaying text from file:\n",
-        "VRB PUBLISHERS\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 24: Page No. L.46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to count number of characters in a file.\n",
-      "\n",
-      "#create a file\n",
-      "fp=open(\"str.txt\",\"w\")\n",
-      "c=\"\"\n",
-      "print \"Enter the Characters:\"\n",
-      "print \"Press ctrl+z to stop entry\"\n",
-      "\n",
-      "#User Input\n",
-      "c=raw_input()\n",
-      "#store input into the file\n",
-      "fp.write(c)\n",
-      "fp.close()\n",
-      "\n",
-      "#Again open the file to read\n",
-      "fp=open(\"str.txt\",\"r\")\n",
-      "\n",
-      "#Result\n",
-      "print \"\\n%s\"%(c)\n",
-      "print \"Number of characters in the file: \",len(c)\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Characters:\n",
-        "Press ctrl+z to stop entry\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "vrb publishers private limited\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "vrb publishers private limited\n",
-        "Number of characters in the file:  30\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exercise 25: Page No.: L.48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program using Memory Allocation function\n",
-      "\n",
-      "NULL=0\n",
-      "buffer_size=10\n",
-      "\n",
-      "buffer1 = \"BOMBAY\"\n",
-      "print \"Buffer of size %d created\"%(buffer_size)\n",
-      "print \"Buffer contains\", buffer1\n",
-      "    \n",
-      "\n",
-      "if buffer_size==NULL:\n",
-      "    print \"Malloc failed\"\n",
-      "    \n",
-      "else:\n",
-      "    print \"Buffer size modified\"\n",
-      "    print \"Buffer still contains:\", buffer1\n",
-      "    \n",
-      "buffer1=\"MUMBAI\"\n",
-      "print \"Buffer now contains:\", buffer1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Buffer of size 10 created\n",
-        "Buffer contains BOMBAY\n",
-        "Buffer size modified\n",
-        "Buffer still contains: BOMBAY\n",
-        "Buffer now contains: MUMBAI\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C/README.txt b/Fundamental_of_Computing_and_Programming_in_C/README.txt
deleted file mode 100755
index 9bed1b50..00000000
--- a/Fundamental_of_Computing_and_Programming_in_C/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Lakshmidharan D
-Course: msc
-College/Institute/Organization: Pachaiyappa's College for Men, Kanchipuram
-Department/Designation: Computer Science
-Book Title: Fundamental of Computing and Programming in C
-Author: V. Ramesh Babu, R. Samyuktha, M. Munirathnam
-Publisher: Lakshmidharan D
-Year of publication: 2012
-Isbn: 9789381837214
-Edition: August 2012
\ No newline at end of file
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diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter01.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter01.ipynb
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--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter01.ipynb
@@ -0,0 +1,578 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e61d5eac9962a1b27acd93d08c5b466532db593eb946c997e161cd14ea1acee1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 01 : Operators and Expressions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1: Page No.:4.35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Usage of Arithmetic operators\n",
+      "\n",
+      "#Variable declaration\n",
+      "i=10\n",
+      "j=20\n",
+      "\n",
+      "#the sum value of variables i and j stores in k variable \n",
+      "k=i+j\n",
+      "\n",
+      "#Result\n",
+      "print \"Value of k is %d\"%(k)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of k is 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2: Page No.:4.35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Demonstrations on arithmetic operators\n",
+      "\n",
+      "#User Input\n",
+      "\n",
+      "a,b,c,d=input(\"Enter values of a, b, c, d: \")\n",
+      "\n",
+      "#Calculation\n",
+      "a+=(b*c)+d\n",
+      "\n",
+      "#Result\n",
+      "print \"Value of a = %d \"%(a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter values of a, b, c, d ,e: 12,90,56,43\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of a = 5095 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3: Page No.:4.36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find out all arithmetic operation on two given values\n",
+      "\n",
+      "#Variable Declaration\n",
+      "sum=sub=mul=rem=0\n",
+      "div=0.0\n",
+      "a=0\n",
+      "#User Input\n",
+      "b,c,d=input(\"Enter the values of b, c, d: \")\n",
+      "\n",
+      "sum = b + c\n",
+      "sub = b - c\n",
+      "mul = b * c\n",
+      "div = b/c\n",
+      "rem = b % d\n",
+      "\n",
+      "print \"Sum = %d, Sub = %d, Mul = %d, Div = %f\"%(sum,sub,mul,div)\n",
+      "print \"Remainder of division of b & d is %d\"%(rem)\n",
+      "print \"a=%d\"%(a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the values of b, c, d: 2,6,45\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum = 8, Sub = -4, Mul = 12, Div = 0.000000\n",
+        "Remainder of division of b & d is 2\n",
+        "a=0\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4: Page No.: 4.38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to use various relational operators\n",
+      "\n",
+      "#!=,==,>=,<= are assignment operator\n",
+      "\n",
+      "print \"Condition :  Return values\"\n",
+      "print \"  5!=5    :  %5d\"%(5!=5)     \n",
+      "print \"  5==5    :  %5d\"%(5==5)\n",
+      "print \"  5>=5    :  %5d\"%(5>=5)\n",
+      "print \"  5<=5    :  %5d\"%(5<=50)\n",
+      "print \"  5!=3    :  %5d\"%(5!=3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Condition :  Return values\n",
+        "  5!=5    :      0\n",
+        "  5==5    :      1\n",
+        "  5>=5    :      1\n",
+        "  5<=5    :      1\n",
+        "  5!=3    :      1\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5: Page No.:4.39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Demonstrate of logical operator\n",
+      "\n",
+      "#Variable Declaration\n",
+      "c1=c2=c3=0\n",
+      "\n",
+      "#USer Input\n",
+      "c1,c2,c3=input(\"Enter Values of C1, C2 and C3: \")\n",
+      "print \"\\n\"\n",
+      "\n",
+      "if (c1<c2) and (c1<c3):\n",
+      "    print \"C1 is less than C2 and C3\"\n",
+      "\n",
+      "if not(c1) < (c2 ):\n",
+      "    print \"C1 is greater than C2\"\n",
+      "    \n",
+      "if ((c1 < c2 or c1 < c3)):\n",
+      "    print \"C1 is less than C2 or C3 or both\"\n",
+      "    \n",
+      "#else:\n",
+      " #   print \"Something went wrong. Check\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values of C1, C2 and C3: 45,32,98\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "C1 is greater than C2\n",
+        "C1 is less than C2 or C3 or both\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6: Page No.: 4.40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Demonstration of assignment operator\n",
+      "\n",
+      "#Variable Declaration\n",
+      "i=4\n",
+      "j=5\n",
+      "\n",
+      "#'=' is assignment operator\n",
+      "i=j\n",
+      "k=i\n",
+      "\n",
+      "#Result\n",
+      "print \"K = %d\"%(k)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "K = 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7: Page No.:4.42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using Increment and Decrement Operator.\n",
+      "\n",
+      "#Variable Declaration\n",
+      "a=10\n",
+      "\n",
+      "#There is no post and pre increment in python\n",
+      "#There is no post and pre decrement in python\n",
+      "\n",
+      "#Result\n",
+      "print \"The Value of a = %d\"%(a)\n",
+      "print \"Increment of a = %d\"%(a+1)\n",
+      "print \"Decrement of a = %d\"%(a-1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of a = 10\n",
+        "Increment of a = 11\n",
+        "Decrement of a = 9\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8: Page No.:4.42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using increment and decrement operators\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "i=3\n",
+      "j=4\n",
+      "\n",
+      "#No Pre Decrement and Pre Increment in python\n",
+      "\n",
+      "i=i+1                  #Increment value\n",
+      "j=j-1               #Decrement value\n",
+      "\n",
+      "k=i+j\n",
+      "\n",
+      "#Result\n",
+      "print \"i = %d, j = %d, k = %d\"%(i,j,k)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i = 4, j = 3, k = 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9: Page No.: 4.46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to illustrate sizeof operator\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "num=int(1234567890)\n",
+      "dec=float(0.123456)\n",
+      "ext=0.123456789\n",
+      "ltr='A'\n",
+      "string=\"Something to write home about....\"\n",
+      "class struct():\n",
+      "    a=0\n",
+      "boy=struct()\n",
+      "\n",
+      "#Result\n",
+      "print \"Size of num is in %d bytes.\"%(sys.getsizeof(num))\n",
+      "print \"Size of dec is in %d bytes.\"%(sys.getsizeof(dec))\n",
+      "print \"Size of ext double is in %d bytes.\"%(sys.getsizeof(ext))\n",
+      "print \"Size of ltr char is in %d bytes.\"%(sys.getsizeof(ltr))\n",
+      "print \"Size of the string is in %d bytes.\"%(sys.getsizeof(string))\n",
+      "print \"Size of struct is in %d bytes.\"%(sys.getsizeof(struct()))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Size of num is in 12 bytes.\n",
+        "Size of dec is in 16 bytes.\n",
+        "Size of ext double is in 16 bytes.\n",
+        "Size of ltr char is in 22 bytes.\n",
+        "Size of the string is in 54 bytes.\n",
+        "Size of struct is in 32 bytes.\n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 1: PageNo.:4.52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert number of days to months and days\n",
+      "\n",
+      "#Variable Declaration\n",
+      "m=0\n",
+      "d=0\n",
+      "\n",
+      "#User Input\n",
+      "nd=input(\"Enter the number of days= \")\n",
+      "\n",
+      "\n",
+      "m=nd/30\n",
+      "d=nd%30\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of months...\", m\n",
+      "print \"Number of days...\", d"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of days= 215\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of months... 7\n",
+        "Number of days... 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 74
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 2: Page No.:4.53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate Salesman Salary\n",
+      "\n",
+      "#Variable Declaration\n",
+      "BASIC=2500\n",
+      "BONUS_RATE=200\n",
+      "COM=0.02\n",
+      "\n",
+      "qty=0\n",
+      "GS=price=Bonus=Comm=0.0\n",
+      "#User input\n",
+      "qty,price=input(\"Enter the number of items Sold and Price...: \")\n",
+      "\n",
+      "Bonus=BONUS_RATE*qty\n",
+      "Comm=COM*qty*price\n",
+      "\n",
+      "GS=BASIC+Bonus+Comm\n",
+      "\n",
+      "print \"Bonus.... %6.2f\"%(Bonus)\n",
+      "\n",
+      "print \"Commission.... %6.2f\"%(Comm)\n",
+      "\n",
+      "print \"Gross Salary... %6.2f\"%(GS)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of items Sold and Price...: 10,200\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bonus.... 2000.00\n",
+        "Commission....  40.00\n",
+        "Gross Salary... 4540.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 75
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 3: Page No.:4.54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Post increment using different storage classes\n",
+      "\n",
+      "#Defining function\n",
+      "def postinc():\n",
+      "    x=1\n",
+      "    print \"X = %d\" %(x)\n",
+      "    x=x+1\n",
+      "    \n",
+      "#Call Function\n",
+      "postinc()\n",
+      "postinc()\n",
+      "postinc()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "X = 1\n",
+        "X = 1\n",
+        "X = 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 77
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter02.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter02.ipynb
new file mode 100755
index 00000000..1aa8a278
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter02.ipynb
@@ -0,0 +1,767 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e63878167c3dee16dba0cb7cf1eb45b07d3e347ca99ab3a68339c448d68bdb66"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 02 : Managing Input and Output Operators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page No 4.56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Testing Characters\n",
+      "\n",
+      "#Variable Declaration\n",
+      "ch=''\n",
+      "\n",
+      "#User Input\n",
+      "ch=raw_input(\"Give any Character as a input: \")\n",
+      "\n",
+      "if (ch.isalpha()>0):          #Checking whether the character is alphabet\n",
+      "    print ch,\"is a alphabet\"\n",
+      "    \n",
+      "elif(ch.isdigit()>0):         #Checking whether the character is digit\n",
+      "    print ch,\"is a digit\"\n",
+      "    \n",
+      "else:                          #else print alphanumeric\n",
+      "    print ch,\"is a alphanumeric\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give any Character as a input: a\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a is a alphabet\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page No 4.57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert Uppercase to Lowercase and Viceversa\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "ch=''\n",
+      "\n",
+      "\n",
+      "#User Input\n",
+      "ch= raw_input(\"Give the input in lower case or upper case: \")\n",
+      "\n",
+      "if(ch.islower()):      #Converting lowercase if input is uppercase\n",
+      "    print ch.upper(), \"is a upper case character\"\n",
+      "    \n",
+      "else:                   #else print lowercase\n",
+      "    print ch.lower(), \"is a lower case character\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the input in lower case or upper case: s\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "S is a upper case character\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3, Page No 4.58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using input function\n",
+      "\n",
+      "scientist=''\n",
+      "\n",
+      "scientist = raw_input(\"Enter Name: \")\n",
+      "print \"Print the name: \", scientist\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name: Abdul kalam\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Print the name:  Abdul kalam\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4, Page No 4.62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count numbers of entered numbers\n",
+      "\n",
+      "#Variable declaration\n",
+      "i=0\n",
+      "j=10\n",
+      "k=0\n",
+      "\n",
+      "#getting input from user     \n",
+      "j=int(raw_input(\"Give input for j \"))\n",
+      "k=int(raw_input(\"Give input for k \"))\n",
+      "i=int(raw_input(\"Give input for i \"))\n",
+      "\n",
+      "i= i, j, k       #assigning the user input in to a variable\n",
+      "\n",
+      "#Result\n",
+      "print \"No. of given input values:\",  len(i)      #counting length of the given input\n",
+      "print \"The Value of j and k is \", j, k              #printing the values of input"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give input for j 56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give input for k 32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give input for i 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "No. of given input values: 3\n",
+        "The Value of j and k is  56 32\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5, Page No 4.65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print a statement using print\n",
+      "\n",
+      "print \"VRB Publishers\"     #Printing statement"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "VRB Publishers\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6, Page No 4.65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Adding two numbers\n",
+      "\n",
+      "#variable declaration\n",
+      "\n",
+      "a=6\n",
+      "b=8\n",
+      "c= a+b      #Adding a, b and stores in c \n",
+      "\n",
+      "#Result\n",
+      "print \"Addition of two numbers is : \", c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Addition of two numbers is :  14\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7, Page No 4.66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Reading and Displaying Details of a student\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "sno=m1=m2=m3=0\n",
+      "name=['20']\n",
+      "total=0\n",
+      "avg=0.0\n",
+      "\n",
+      "#User Inputs\n",
+      "sno=int(raw_input(\"Give Student serial no. \"))\n",
+      "name=raw_input(\"Student Name : \")\n",
+      "m1=int(raw_input(\"Student Mark 1: \"))\n",
+      "m2=int(raw_input(\"Student Mark 2: \"))\n",
+      "m3=int(raw_input(\"Student Mark 3: \"))\n",
+      "\n",
+      "total=float(m1+m2+m3)        #Calculation of marks\n",
+      "\n",
+      "avg=\"%.2f\"%float(total/3.0)           #Average\n",
+      "\n",
+      "#Result\n",
+      "print '\\n'\"Student serial No.:\", sno\n",
+      "print \"Student Name:\", name\n",
+      "print \"Student's Marks:\", m1, m2, m3\n",
+      "print \"Total:\", total\n",
+      "print \"Average:\", avg "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give Student serial no. 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Student Name : Raja\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Student Mark 1: 90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Student Mark 2: 95\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Student Mark 3: 98\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Studnt serial No.: 1\n",
+        "Student Name: Raja\n",
+        "Student's Marks: 90 95 98\n",
+        "Total: 283.0\n",
+        "Average: 94.33\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8, Page No:4.71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display quotient and reminder of given input\n",
+      "\n",
+      "#variable declaration\n",
+      "\n",
+      "a=b=q=r=0\n",
+      "\n",
+      "#User input\n",
+      "a=int(raw_input(\"Give input for a: \"))\n",
+      "b=int(raw_input(\"Give input for b: \"))\n",
+      "\n",
+      "q=float(a/b)              #Calculation for Quotient\n",
+      "\n",
+      "r=float(a-q*b)         #Calculation for Reminder\n",
+      "\n",
+      "#Result\n",
+      "print \"Quotient of a and b is :\",q\n",
+      "print \"Reminder of a and b is :\",r"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give input for a: 39\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give input for b: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quotient of a and b is : 5.0\n",
+        "Reminder of a and b is : 4.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9, Page No:4.73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Slope and Midpoint of a line\n",
+      "\n",
+      "#Variable declaration\n",
+      "\n",
+      "x=x1=y1=x2=y2=slope=0.0\n",
+      "\n",
+      "#User input1\n",
+      "\n",
+      "x1=int(raw_input(\"Give point X1: \"))\n",
+      "y1=int(raw_input(\"Give point Y1: \"))\n",
+      "\n",
+      "\n",
+      "#User input2\n",
+      "\n",
+      "x2=int(raw_input('\\n'\"Give point X2: \"))\n",
+      "y2=int(raw_input(\"Give point Y2: \"))\n",
+      "\n",
+      "slope =(y2-y1)/(x2-x1)            #Calculation for Slope\n",
+      "\n",
+      "#Calculation for Midpoints\n",
+      "x=float(x1+x2)/2.0\n",
+      "y=float(y1+y2)/2.0\n",
+      "\n",
+      "#Result\n",
+      "print '\\n' \"Slope of a line is :\", slope\n",
+      "print \"Midpoint of a line is :\", x,y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give point X1: 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give point Y1: 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Give point X2: 30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give point Y2: 40\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Slope of a line is : 1\n",
+        "Midpoint of a line is : 20.0 30.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10, Page No:4.73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Average of Given N Numbers\n",
+      "\n",
+      "#Variable Declaration\n",
+      "n=0.0\n",
+      "x=0\n",
+      "total=0\n",
+      "avg=0.0\n",
+      "\n",
+      "#UserInput\n",
+      "\n",
+      "n=int(raw_input(\"Give total number of input: \"))\n",
+      "\n",
+      "print \"Give those\",n ,\"Numbers\"\n",
+      "\n",
+      "for i in range(n):          #Looping values of n\n",
+      "    i=1\n",
+      "    x=input()               #Values for n\n",
+      "    total+=x                 # Adding given numbers\n",
+      "    \n",
+      "avg=float(total/n)           #Calculating an average of given numbers\n",
+      "\n",
+      "#Result\n",
+      "print \"Sum of given numbers is \", total\n",
+      "print \"Average is \", avg\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give total number of input: 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give those 5 Numbers\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "17\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "76\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of given numbers is  112\n",
+        "Average is  22.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11, Page No:4.75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Temperature Conversion (Fahrenheit to Celsius)\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "F_MIN =0\n",
+      "F_MAX=500\n",
+      "INC=50\n",
+      "fh= F_MIN\n",
+      "\n",
+      "print  \"Fahrenheit\" '\\t' \"Celsius\"\n",
+      "while(fh<=F_MAX):\n",
+      "    cel =\"%.2f\"%float((fh-(32.0))/1.8)     #Calculating Celsius value\n",
+      "    print  fh,'\\t''\\t', cel\n",
+      "    fh=fh+INC              #Calculating Fahrenheit value\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fahrenheit\tCelsius\n",
+        "0 \t\t-17.78\n",
+        "50 \t\t10.00\n",
+        "100 \t\t37.78\n",
+        "150 \t\t65.56\n",
+        "200 \t\t93.33\n",
+        "250 \t\t121.11\n",
+        "300 \t\t148.89\n",
+        "350 \t\t176.67\n",
+        "400 \t\t204.44\n",
+        "450 \t\t232.22\n",
+        "500 \t\t260.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12, Page No:4.76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Area of Triangle\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "a=b=c=s=area=0.0\n",
+      "\n",
+      "#User input\n",
+      "\n",
+      "a=int(raw_input(\"First side of Triangle is \"))\n",
+      "b=int(raw_input(\"Second side of Triangle is \"))\n",
+      "c=int(raw_input(\"Third side of Triangle is \"))\n",
+      "\n",
+      "s=(a+b+c)/2.0      #Calculating Size of triangle\n",
+      "\n",
+      "area=\"%.3f\"%float((s*(s-a)*(s-b)*(s-c)) ** 0.5)    #Calculating Area of triangle\n",
+      "\n",
+      "#Result\n",
+      "print '\\n',\"Area of Triangle is\",area"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "First side of Triangle is 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Second side of Triangle is 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Third side of Triangle is 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Area of Triangle is 43.301\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter03.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter03.ipynb
new file mode 100755
index 00000000..34a44d60
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter03.ipynb
@@ -0,0 +1,440 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:92464c4be2e3e810ccda180a900c5ec8bb15a9537c93387b95c317c7709c7b7c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 03 : Decision Making"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page No.:4.79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using if with multiple statements\n",
+      "\n",
+      "\n",
+      "#Variable Declaration\n",
+      "i=0\n",
+      "\n",
+      "#User Input\n",
+      "\n",
+      "i =int(raw_input(\"Give the numbers less than 10\"'\\n'))\n",
+      "\n",
+      "if i<=10:                  #Checking the given value\n",
+      "\n",
+      "#Result       \n",
+      "    print \"The given value of\", i , \"less than 10\"\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the numbers less than 10\n",
+        "5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the given value of 5 less than 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.: 4.79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Program for interchange two values using if statement\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "a=m=n=0\n",
+      "\n",
+      "#User Input\n",
+      "\n",
+      "m=int(raw_input(\"Give the value of m\" '\\t'))\n",
+      "n=int(raw_input(\"Give the value of n\" '\\t'))\n",
+      "\n",
+      "if m>=n :                   #Checking the Given Values\n",
+      "    \n",
+      "    a=m                     # method for swapping            \n",
+      "    m=n\n",
+      "    n=a\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "print \"The Swapped Values of m and n are \", m,a"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of m\t34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of n\t28\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Swapped Values of m and n are  28 34\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No. 4.81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the given number is odd or even. \n",
+      "\n",
+      "#Variable Declaration\n",
+      "num=rem=0\n",
+      "\n",
+      "#User Input\n",
+      "num=input(\"Enter the value: \")\n",
+      "rem=num%2           \n",
+      "if rem==0 :             #Checking condition\n",
+      "\n",
+      "#Result 1\n",
+      "    print \"The given number is Even\"\n",
+      "else :\n",
+      "    \n",
+      "#Result 2\n",
+      "    print \"The given number is Odd\"\n",
+      "    \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the value: 80\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The given number is Even\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page No.:4.81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Find floating point input using if-else statement\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "years=seconds=0.0\n",
+      "life=0\n",
+      "\n",
+      "years=float(raw_input(\"Give your Age in years: \"))\n",
+      "life=years\n",
+      "if life==0:\n",
+      "    print \"The given age is not floating value\"\n",
+      "    \n",
+      "else :\n",
+      "    seconds=\"%f\"%float(years*365*24*60*60)\n",
+      "    print \"You have lived for\", seconds, \"seconds\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give your Age in years: 24\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have lived for 756864000.000000 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 5, Page No. :4.82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find biggest among two numbers \n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "i=j=big=0\n",
+      "i,j=input(\"Give two values: \")\n",
+      "\n",
+      "big=i\n",
+      "\n",
+      "if big<j :\n",
+      "    big=j\n",
+      "    \n",
+      "elif i<j :\n",
+      "    big=j \n",
+      "    \n",
+      "else:\n",
+      "    big = i\n",
+      "\n",
+      "print \"Biggest number of two numbers: \",big\n",
+      "print \"Biggest of two numbers(using else) is: \",big"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give two values: 45,78\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Biggest number of two numbers:  78\n",
+        "Biggest of two numbers(using else) is:  78\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 6, Page No. : 4.84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Program using nested if....else \n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "n=0\n",
+      "\n",
+      "#User Input\n",
+      "n=int(raw_input(\"Enter the number\"))\n",
+      "\n",
+      "if n==15:           #Checking Condition\n",
+      "    print \"Play Foot ball\"\n",
+      "    \n",
+      "elif n==10:\n",
+      "    print \"Play Cricket\"\n",
+      "\n",
+      "else :    \n",
+      "    print \"Don't Play\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Play Cricket\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 7, Page No. :4.85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#print the number between 10 and 20\n",
+      "\n",
+      "#Variable declaration\n",
+      "\n",
+      "i=0\n",
+      "\n",
+      "#User Input\n",
+      "\n",
+      "i=input(\"Enter the number: \")\n",
+      "\n",
+      "if (i>10) & (i<20):\n",
+      "    print \"The given value is in between 10 and 20\"\n",
+      "\n",
+      "else :\n",
+      "    print \"The given value is greater than 20 and less than 10\"    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number: 15\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The given value is in between 10 and 20\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 8, Page No. :4.86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# display the types of characterusing if-else statement\n",
+      "\n",
+      "#User Input\n",
+      "\n",
+      "chr= raw_input(\"Enter a Single Character: \")\n",
+      "\n",
+      "if( ((chr>='a') & (chr<='z')) | ((chr>='A') & (chr<='Z'))):\n",
+      "    print \"Given charactered is an Alphabetic\"\n",
+      "    \n",
+      "elif ((chr>='0') & (chr <='9')):\n",
+      "    print \"Given Character is a digit\"\n",
+      "    \n",
+      "else:\n",
+      "    print \"Entered Character is a special character\"\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Single Character: M\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Given charactered is an Alphabetic\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter04.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter04.ipynb
new file mode 100755
index 00000000..a58b3c75
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter04.ipynb
@@ -0,0 +1,1337 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:cbd54959e18d38695240f376809cd86bcdd3f9c461e014b42a620c0846ec3cbf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 04 : Branching and Looping"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1: Page No.:4.88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# addition of numbers using while loop.\n",
+      "\n",
+      "#Varibale Declaration\n",
+      "\n",
+      "i=1\n",
+      "sum=0\n",
+      "while i<=10:                    #Checking while loop\n",
+      "    sum+=i\n",
+      "    i+=1                        #Increment of i\n",
+      "    \n",
+      " #Result   \n",
+      "print \"The Sum of the numbers upto is \",sum"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Sum of the numbers upto is  55\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2: Page No.: 4.89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "i=1\n",
+      "j=0\n",
+      "\n",
+      "    \n",
+      "print \"The maximum value of integer is\", sys.maxsize\n",
+      "print \"The Value of integer after overflow is\", sys.maxint-1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The maximum value of integer is 2147483647\n",
+        "The Value of integer after overflow is 2147483646\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3: Page No.: 4.89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Reverse the given number using while loop\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "number=digit=rev=0\n",
+      "\n",
+      "number=input(\"Give the Value: \")\n",
+      "\n",
+      "while(number!=0):\n",
+      "    digit=number%10\n",
+      "    rev=rev*10+digit\n",
+      "    number= number/10\n",
+      "    \n",
+      "print \"The reversible value of given number is\", rev\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the Value: 7896\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The reversible value of given number is 6987\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4: Page No.: 4.90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Simple interest using while loop.\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "p=1\n",
+      "n=1\n",
+      "count=1\n",
+      "\n",
+      "r=si=0.0\n",
+      "\n",
+      "while(count<=3):\n",
+      "    p=input(\"Give the value of P\" '\\t')\n",
+      "    n=input(\"Give the value of N\"'\\t')\n",
+      "    r=float(input(\"Give the value of R\"'\\t'))\n",
+      "    \n",
+      "    si=\"%.3f\"%float((p*n*r)/100)\n",
+      "    \n",
+      "    print \"Simple Interest of SI\", si\n",
+      "    count=count+1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of P\t2000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of N\t3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of R\t5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Simple Interest of SI 300.000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of P\t3500\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of N\t6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of R\t1.5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Simple Interest of SI 315.000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of P\t6000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of N\t2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of R\t3.5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Simple Interest of SI 420.000\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5: Page No.:4.92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Numbers upto 10 by using the do..while loop\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "i=1\n",
+      "sum=0\n",
+      "\n",
+      "while True:\n",
+      "\tsum+=i\n",
+      "\ti+=1\n",
+      "\tif(i>10):\n",
+      "\t\tbreak\n",
+      "print \"Sum of the numbers upto 10 is\", sum  \n",
+      "   \n",
+      "            \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of the numbers upto 10 is 55\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6: Page No.: 4.92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#print n numbers using do...while loop\n",
+      "\n",
+      "i=0\n",
+      "\n",
+      "n=input(\"Enter the number: \")\n",
+      "\n",
+      "while True:\n",
+      "    print \"The numbers are: \",i\n",
+      "    i=i+1\n",
+      "    if(i>=n):\n",
+      "        break\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number: 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numbers are:  0\n",
+        "The numbers are:  1\n",
+        "The numbers are:  2\n",
+        "The numbers are:  3\n",
+        "The numbers are:  4\n",
+        "The numbers are:  5\n"
+       ]
+      }
+     ],
+     "prompt_number": 73
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7: page no.: 4.95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Addition of number upto 10 using for loop\n",
+      "\n",
+      "i=sum=0\n",
+      "\n",
+      "for i in range (0,11):\n",
+      "\tsum+=i\n",
+      "\ti+=1\n",
+      "\n",
+      "print \"The addition of given numbers\",sum\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The addition of given numbers 55\n"
+       ]
+      }
+     ],
+     "prompt_number": 75
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8: Page No.:4.96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#print n numbers using for...loop structure\n",
+      "\n",
+      "n=int(raw_input(\"Give Values of n: \"))\n",
+      "i=0\n",
+      "for i in range(0,n):\n",
+      "    print \"The Given numbers are: \",i\n",
+      "\n",
+      "\t"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give Values of n: 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Given numbers are:  0\n",
+        "The Given numbers are:  1\n",
+        "The Given numbers are:  2\n",
+        "The Given numbers are:  3\n",
+        "The Given numbers are:  4\n"
+       ]
+      }
+     ],
+     "prompt_number": 82
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9: Page No.:4.97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#student marks and average using for loop..\n",
+      "\n",
+      "a=b=c=d=e=total=0\n",
+      "\n",
+      "avg=0.0\n",
+      "\n",
+      "m=input(\"Enter the number of Students:\")\n",
+      "\n",
+      "print \"Enter the marks of the five subjects: \"'\\n'\n",
+      "\n",
+      "for i in range (m):   \n",
+      "    i+=1\n",
+      "    print \"\\nStudent:\",i,\"\\n\"\n",
+      "    \n",
+      "    a,b,c,d,e=input(\"Enter the Student mark1: \")                       \n",
+      "             \n",
+      "    if ((a,b,c,d,e)!=0) :\n",
+      "        total=a+b+c+d+e\n",
+      "        avg=\"%.3f\"%float(total/5)\n",
+      "        print \"Total marks of the student:\", total\n",
+      "        print \"Average marks of the students: \",avg\n",
+      "                       \n",
+      "    else  :\n",
+      "        print \"Entered data is invalid\"\n",
+      "                \n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of Students:3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the marks of the five subjects: \n",
+        "\n",
+        "\n",
+        "Student: 1 \n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Student mark1: 78,56,90,59,87\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total marks of the student: 370\n",
+        "Average marks of the students:  74.000\n",
+        "\n",
+        "Student: 2 \n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Student mark1: 49,87,60,57,93\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total marks of the student: 346\n",
+        "Average marks of the students:  69.000\n",
+        "\n",
+        "Student: 3 \n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Student mark1: 89,100,80,97,65\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total marks of the student: 431\n",
+        "Average marks of the students:  86.000\n"
+       ]
+      }
+     ],
+     "prompt_number": 89
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10: Page No.:4.98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Nested for loop\n",
+      "\n",
+      "for i in range(1,4):\n",
+      "    for j in range(1,4):\n",
+      "        print \"%d\"%(j),\n",
+      "    print \"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 2 3 \n",
+        "\n",
+        "1 2 3 \n",
+        "\n",
+        "1 2 3 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 112
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11: Page No.:4.99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#student details using nested for loop..\n",
+      "\n",
+      "FIRST=300\n",
+      "SECOND=200\n",
+      "rollno=marks=total=0\n",
+      "\n",
+      "n=input(\"Enter the number of Students: \")\n",
+      "m=input(\"Enter the number of subjects: \")\n",
+      "\n",
+      "for i in range(n):\n",
+      "    i+=(+1)\n",
+      "    rollno=int(raw_input(\"Enter the student roll number: \"))\n",
+      "    total=0\n",
+      "    print \"Enter marks of\",m, \"subjects for Roll no: \",rollno\n",
+      "    for j in range(m):\n",
+      "        marks=int(raw_input(\"Give marks: \"))\n",
+      "        total= total+marks\n",
+      "        \n",
+      "    print \"total marks\", total\n",
+      "    if(total>=FIRST):\n",
+      "        print \"(First Class)\"'\\n'\n",
+      "    elif(total>=SECOND):\n",
+      "        print \"(Second Class)\"'\\n'\n",
+      "    else:\n",
+      "        print \"Fail\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of Students: 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of subjects: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student roll number: 201\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter marks of 5 subjects for Roll no:  201\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 68\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 89\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total marks 415\n",
+        "(First Class)\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student roll number: 202\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter marks of 5 subjects for Roll no:  202\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 49\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 39\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give marks: 47\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total marks 225\n",
+        "(Second Class)\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 85
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12: Page No.: 4.103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Demonstrate switch case\n",
+      "\n",
+      "a=1\n",
+      "if a==1:\n",
+      "    print \"I am in case 1\"'\\n'\n",
+      "elif a==2:\n",
+      "    print \"I am in case2\"'\\n'\n",
+      "else:\n",
+      "    print \"I am in case default\"'\\n'"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "I am in case 1\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 115
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13: Page No.:4.104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to use the computer as calculator\n",
+      "\n",
+      "a=b=c=0\n",
+      "\n",
+      "op=raw_input(\"Give the option for calculation like +(ADD), -(SUB), *(MUL), /(DIV): \")\n",
+      "a=input(\"Enter the value of a \")\n",
+      "b=input(\"Enter the value of b \")\n",
+      "\n",
+      "#Calculation\n",
+      "if (op=='+'):\n",
+      "  c=a+b\n",
+      "  print \"Result of C\", c \n",
+      "elif(op=='-'):\n",
+      "    c=a-b\n",
+      "    print \"Result of C\", c\n",
+      "elif(op=='*'):\n",
+      "    c=a*b\n",
+      "    print \"Result of C\", c\n",
+      "else:\n",
+      "    c=a/b\n",
+      "    print \"Result of C\", c\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the option for calculation like +(ADD), -(SUB), *(MUL), /(DIV): +\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the value of a 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the value of b 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Result of C 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 118
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14: Page No.:4.105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find even or odd numbers \n",
+      "\n",
+      "#USer Input\n",
+      "a=input(\"Enter a number: \")\n",
+      "\n",
+      "if a%2==0 : \n",
+      "    print  \" The number is even. \"\n",
+      "else :\n",
+      "    print \"The number is odd.\" \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 89\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number is odd.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15: Page No.:4.106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to count number of 0s, 1s, blank spaces and other characters\n",
+      "\n",
+      "\n",
+      "txt=[30]\n",
+      "count =0 \n",
+      "txt=raw_input(\"Enter Numbers\")\n",
+      "\n",
+      "print \"Total Spaces  : %d\"%txt.count(' ')\n",
+      "print \"Total of 1's  : %d\"%txt.count('1')\n",
+      "print \"Total of 0's  : %d\"%txt.count('0')\n",
+      "print \"String Length : %d\"%len(txt)\n",
+      "            \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Numbers120010  09100110\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Spaces  : 2\n",
+        "Total of 1's  : 5\n",
+        "Total of 0's  : 7\n",
+        "String Length : 16\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16: Page No.: 4.108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the number upto 5 using break statement\n",
+      "\n",
+      "for i in range(1,10):\n",
+      "    if i==6:\n",
+      "        break\n",
+      "    print i,"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 2 3 4 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 120
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17: Page No.: 4.109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to calculate the sum of the given positive numbers\n",
+      "\n",
+      "i=n=sum=0\n",
+      "\n",
+      "for i in range(0,5):\n",
+      "    n=int(raw_input(\"Give any value of n \"))\n",
+      "    if n<0:\n",
+      "       continue\n",
+      "    else:\n",
+      "       sum+=n\n",
+      "       \n",
+      "print \"\\nSum is\", sum"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give any value of n 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give any value of n 15\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give any value of n 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give any value of n -10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give any value of n 50\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum is 100\n"
+       ]
+      }
+     ],
+     "prompt_number": 124
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18: Page No.:4.113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using goto statement\n",
+      "\n",
+      "a=input(\"Give the Value of a \")\n",
+      "b=input(\"Give the value of b \")\n",
+      "\n",
+      "if a==b:\n",
+      "   print \"A and B are equal\"\n",
+      "else:\n",
+      "    print \"A and B are not equal\"\n",
+      "    exit(0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the Value of a 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of b 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A and B are equal\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 1: Page No.: 4.112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Prime or not\n",
+      "\n",
+      "n=input(\"Enter the Positive value of n \")\n",
+      "i=2\n",
+      "\n",
+      "#Calculation\n",
+      "while n>i:\n",
+      "    if n%i==0 & i!=n:\n",
+      "        #r=n%i\n",
+      "        #if r==0:\n",
+      "            print \"The Given value of \",n,\" is not a prime\"\n",
+      "            break\n",
+      "    i+=1\n",
+      "else:\n",
+      "     print \"The Given Value\",n, \"is Prime\"\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Positive value of n 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Given Value 3 is Prime\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 2: Page No.: 4.113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the Greatest Common Division\n",
+      "\n",
+      "i=int(input(\"Give the Value of i \"))\n",
+      "j=int(input(\"Give the value of j \"))\n",
+      "  \n",
+      "def gcd(i,j):\n",
+      "    if i > j:\n",
+      "        k = i%j\n",
+      "        if k == 0:\n",
+      "            return j\n",
+      "        else:\n",
+      "            return gcd(j, k)\n",
+      "    if i < j:\n",
+      "        i = j\n",
+      "        j = i\n",
+      "        return gcd(i, j)\n",
+      "print \"The value of gcd\", gcd(i,j) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the Value of i 54\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the value of j 36\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of gcd 18\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 3: Page No.: 4.115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to which reads in a Temperature and print the temperature in opposite scale\n",
+      "\n",
+      "\n",
+      "print \"1.Fahrenheit to Celcius\\n2.Celcius to Fahrenheit\\n\"\n",
+      "code=input(\"Choose Option to find Temperature: \")\n",
+      "\n",
+      "while code!=0:\n",
+      "    if code==1:\n",
+      "        f=float(input(\"Give the Value for Fahrenheit Temperature \"))\n",
+      "        c=(f-32.0)*5.0/9.0\n",
+      "        break\n",
+      "    elif code==2:\n",
+      "        c=float(input(\"Give the Value for Celcius Temperature \"))\n",
+      "        f=9.0*c/5.0+32.0\n",
+      "        break\n",
+      "    else:\n",
+      "        print \"Give proper Code Value for Temperature Calculation\"\n",
+      "        break\n",
+      "print \"C=\", \"%.3f\"%float(c)\n",
+      "print \"F=\",f "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1.Fahrenheit to Celcius\n",
+        "2.Celcius to Fahrenheit\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Choose Option to find Temperature: 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give the Value for Fahrenheit Temperature 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C= 37.778\n",
+        "F= 100.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter05.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter05.ipynb
new file mode 100755
index 00000000..fa4db685
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter05.ipynb
@@ -0,0 +1,987 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8b48219cd7d973f69d80d4d491de4bdba5cb9c73135095697b1625a862f6bc97"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 05 : Defining and Processing of Arrays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page Number: 5.5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to read 5 data items from input and store them in an array and display their sum\n",
+      "\n",
+      "#Local definitions\n",
+      "a = []\n",
+      "sum = 0\n",
+      "\n",
+      "for i in range(0,5):\n",
+      "    a.append(0)\n",
+      "print \"Enter 5 integer numbers\"    \n",
+      "for i in range(0,5):\n",
+      "    a[i] = eval(raw_input())\n",
+      "    \n",
+      "for i in range(0,5):    \n",
+      "    sum = sum + a[i] \n",
+      "    \n",
+      "print \"The sum of given numbers is : %d\\n\" %sum"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 5 integer numbers\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "120\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "69\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "390\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "45\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of given numbers is : 669\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2 Page No. 5.7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to accept 5 numbers and print whether the number is even or not\n",
+      "\n",
+      "#Local Variable\n",
+      "a=[]\n",
+      "\n",
+      "for i in range(5):\n",
+      "    a.append(0)\n",
+      "for i in range(0,5):\n",
+      "    i+1\n",
+      "    a[i]=eval(raw_input(\"Enter the %d number: \"%(i+1))) \n",
+      "    print \"\\n\"\n",
+      "for i in range (0,5):\n",
+      "    if a[i]%2==0:\n",
+      "        print \"%d Number is Even\"%a[i]\n",
+      "    else:\n",
+      "        print \"%d Number is Odd\"%a[i]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 1 number: 17\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 2 number: 38\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 3 number: 79\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 4 number: 80\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 5 number: 76\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "17 Number is Odd\n",
+        "38 Number is Even\n",
+        "79 Number is Odd\n",
+        "80 Number is Even\n",
+        "76 Number is Even\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3 Page No. 5.8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to illustrate how a character array can be used.\n",
+      "\n",
+      "#Local definition\n",
+      "word=['F', 'R', 'I', 'E', 'N', 'D', 'S']\n",
+      "\n",
+      "for i in range(0,7):\n",
+      "    word.append(0)\n",
+      "for i in range(0,7):\n",
+      "    print word[i],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "F R I E N D S\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page Number: 5.8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to sort the given strings alphabetically\n",
+      "\n",
+      "b = raw_input(\"Enter the text to sort : \")\n",
+      "print \"Sorted Text are : \",''.join(sorted(b, key=lambda v: (v.upper(), v[0].islower())))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the text to sort : VRBPublishers\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorted Text are :  BbehilPRrssuV\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 5, Page Number: 5.10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using two dimension array\n",
+      "\n",
+      "#Local definition \n",
+      "stud = [[0 for i in xrange(0,4)] for i in xrange(0,4)] #stud is a array name with 4 rows and 2 columns\n",
+      "\n",
+      "for i in range(0,4):\n",
+      "    print \"Enter the %d Student roll no and Mark:\" %i\n",
+      "    stud[i][1], stud[i][2] = input()\n",
+      "    \n",
+      "for i in range(0,4):\n",
+      "    print \"%d Student roll no %d mark %d\" %(i, stud[i][1], stud[i][2])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 0 Student roll no and Mark:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3977,80\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 1 Student roll no and Mark:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "17776,95\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 2 Student roll no and Mark:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6682,82\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the 3 Student roll no and Mark:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6683,85\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0 Student roll no 3977 mark 80\n",
+        "1 Student roll no 17776 mark 95\n",
+        "2 Student roll no 6682 mark 82\n",
+        "3 Student roll no 6683 mark 85\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 6, Page Number: 5.12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to illustrate Matrix Multiplication\n",
+      "\n",
+      "a = [[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
+      "b = [[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
+      "c = [[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
+      "\n",
+      "while True:\n",
+      "    r1, c1 = input(\"Enter the size of the matrix A....\")\n",
+      "    r2, c2 = input(\"Enter the size of the matrix B....\")\n",
+      "    if c1 == r2:\n",
+      "        print \"Enter matrix A elements...\"\n",
+      "        for i in range (0,r1):\n",
+      "            for j in range (0,c1):\n",
+      "                a[i][j] = input()\n",
+      "            \n",
+      "        print \"Enter matrix B elements...\"\n",
+      "        for i in range (0,r2):\n",
+      "            for j in range (0,c2):\n",
+      "                b[i][j] = input()\n",
+      "            \n",
+      "        for i in range(0,r1):\n",
+      "            for j in range(0,c1):\n",
+      "                c[i][j] = 0\n",
+      "                for k in range(0,c1):\n",
+      "                    c[i][j] += a[i][k] * b[k][j]\n",
+      "                \n",
+      "        print \"The resultant matrix is....\"\n",
+      "        for i in range(0,r1):\n",
+      "            for j in range(0,c1):\n",
+      "                print \"%d\\t\" %c[i][j],\n",
+      "            print \"\\n\"\n",
+      "        break        \n",
+      "    else:\n",
+      "        print \"Multiplication is not possible\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the size of the matrix A....3,3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the size of the matrix B....2,2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Multiplication is not possible\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the size of the matrix A....3,3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the size of the matrix B....3,3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter matrix A elements...\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter matrix B elements...\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resultant matrix is....\n",
+        "18\t18\t18\t\n",
+        "\n",
+        "18\t18\t18\t\n",
+        "\n",
+        "18\t18\t18\t\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 7, Page Number: 5.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print two dimensional array by row by row\n",
+      "\n",
+      "arr = [[0 for i in xrange(0,4)] for i in xrange(0,4)]\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,4):\n",
+      "        arr[i][j] = (i * 4) + j + 1\n",
+      "        \n",
+      "        \n",
+      "print \"Printing array contents :\"\n",
+      "for i in range(0,3):\n",
+      "   for j in range(0,4):\n",
+      "        print \"%3d\" %arr[i][j],\n",
+      "   print \"\\n\"   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Printing array contents :\n",
+        "  1   2   3   4 \n",
+        "\n",
+        "  5   6   7   8 \n",
+        "\n",
+        "  9  10  11  12 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 8, Page Number: 5.14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to add and display the results of 5 nos\n",
+      "\n",
+      "#Function definition  \n",
+      "def add(x,ar):\n",
+      "    sum = 0\n",
+      "    for i in range(0,5):\n",
+      "        sum += ar[i]\n",
+      "    print \"Sum is ... %d\" %sum   \n",
+      "    \n",
+      "a = []\n",
+      "n = 5\n",
+      "for i in range(0,5):\n",
+      "    a.append(0)\n",
+      "print \"Enter 5 values\"\n",
+      "for i in range(0,5):\n",
+      "    a[i] = input()\n",
+      "add(n,a) #function call"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 5 values\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "50\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum is ... 150\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 9, Page Number: 5.15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program for capture and store n values into array and print them\n",
+      "\n",
+      "#Function definition \n",
+      "def arr(a = []):\n",
+      "    for i in range(0,5):\n",
+      "        print \"Value in array %d\\n\" %a[i]\n",
+      "        \n",
+      "a = []\n",
+      "for i in range(0,5):\n",
+      "    a.append(0)\n",
+      "for i in range(0,5):\n",
+      "    a[i] = i\n",
+      "arr(a)  #Function call"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value in array 0\n",
+        "\n",
+        "Value in array 1\n",
+        "\n",
+        "Value in array 2\n",
+        "\n",
+        "Value in array 3\n",
+        "\n",
+        "Value in array 4\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 10, Page Number: 5.16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print the character string from the array\n",
+      "\n",
+      "name = \"LAK\"\n",
+      "print \"%s\" %name"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "LAK\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 11, Page Number: 5.17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program for explaining the working of 4 dimensional array\n",
+      "\n",
+      "#Local definition\n",
+      "array_4d = [[[[0 for i in xrange(0,3)] for i in xrange(0,3)]for i in xrange(0,3)] for i in xrange(0,3)]\n",
+      "\n",
+      "#looping statements \n",
+      "for a in range(0,3):\n",
+      "    for b in range(0,3):\n",
+      "        for c in range(0,3):\n",
+      "            for d in range(0,3):\n",
+      "                array_4d[a][b][c][d] = a + b + c + d\n",
+      "#Result                \n",
+      "for a in range(0,3)                :\n",
+      "    print \"\\n\"\n",
+      "    for b in range(0,3):\n",
+      "        for c in range(0,3):\n",
+      "            for d in range(0,3):\n",
+      "                print \"%3d\" %array_4d[a][b][c][d],\n",
+      "            print \"\\n\"    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  0   1   2 \n",
+        "\n",
+        "  1   2   3 \n",
+        "\n",
+        "  2   3   4 \n",
+        "\n",
+        "  1   2   3 \n",
+        "\n",
+        "  2   3   4 \n",
+        "\n",
+        "  3   4   5 \n",
+        "\n",
+        "  2   3   4 \n",
+        "\n",
+        "  3   4   5 \n",
+        "\n",
+        "  4   5   6 \n",
+        "\n",
+        "\n",
+        "\n",
+        "  1   2   3 \n",
+        "\n",
+        "  2   3   4 \n",
+        "\n",
+        "  3   4   5 \n",
+        "\n",
+        "  2   3   4 \n",
+        "\n",
+        "  3   4   5 \n",
+        "\n",
+        "  4   5   6 \n",
+        "\n",
+        "  3   4   5 \n",
+        "\n",
+        "  4   5   6 \n",
+        "\n",
+        "  5   6   7 \n",
+        "\n",
+        "\n",
+        "\n",
+        "  2   3   4 \n",
+        "\n",
+        "  3   4   5 \n",
+        "\n",
+        "  4   5   6 \n",
+        "\n",
+        "  3   4   5 \n",
+        "\n",
+        "  4   5   6 \n",
+        "\n",
+        "  5   6   7 \n",
+        "\n",
+        "  4   5   6 \n",
+        "\n",
+        "  5   6   7 \n",
+        "\n",
+        "  6   7   8 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter06.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter06.ipynb
new file mode 100755
index 00000000..09f324b0
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter06.ipynb
@@ -0,0 +1,1424 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d7231c68ebf3352357e97f096d7bb05469f2e0aae52fa150a1a57a7e75ef01fb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 06 : Handling Of Character Strings"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page Number: 5.19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program for reading a line of text\n",
+      "\n",
+      "n = 0\n",
+      "print \"Enter text press RETRUN to end\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter text press RETRUN to end\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page Number: 5.20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#String function\n",
+      "\n",
+      "s1= \"Hello\"\n",
+      "s2='H''e''l''l''o'\n",
+      "\n",
+      "print s1\n",
+      "print s2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hello\n",
+        "Hello\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page Number: 5.22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using strlen() function\n",
+      "\n",
+      "name = \"MUNI\"\n",
+      "len1 = len(name)\n",
+      "len2 = len(\"LAK\")\n",
+      "print \"String length of %s is %d\" %(name,len1)\n",
+      "print \"String length of %s is %d\" %(\"LAK\",len2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String length of MUNI is 4\n",
+        "String length of LAK is 3\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page Number: 5.23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using strcpy() function\n",
+      "def strcpy(cstring1):\n",
+      "      import copy\n",
+      "      cstring2=copy.copy(cstring1)\n",
+      "      return cstring2  \n",
+      "\n",
+      "source = \"MUNI\"\n",
+      "target = strcpy(source)\n",
+      "print \"Source string is %s\" %source\n",
+      "print \"Target string is %s\" %target"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Source string is MUNI\n",
+        "Target string is MUNI\n"
+       ]
+      }
+     ],
+     "prompt_number": 64
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 5, Page Number:5.24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Strcat function\n",
+      "\n",
+      "str1=\"MUNI\"\n",
+      "str2=\"LAK\"\n",
+      "print str1+str2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MUNILAK\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 6, Page Number: 5.24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using strcat() function\n",
+      "def strcat(cstring1, cstring2):\n",
+      "    cstring3 = cstring1 + cstring2\n",
+      "    return cstring3\n",
+      "\n",
+      "source = \"Ramesh\"\n",
+      "target = \"Babu\"\n",
+      "source = strcat(source,target)\n",
+      "print \"Source string is %s\" %source\n",
+      "print \"Target string is %s\" %target"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Source string is RameshBabu\n",
+        "Target string is Babu\n"
+       ]
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 7, Page Number: 5.25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using strcmp() function\n",
+      "\n",
+      "name = \"Kalai\"\n",
+      "name1 = \"Malai\"\n",
+      "if name ==\"Kalai\":\n",
+      "    print \"0\",\n",
+      "if name1!=name:\n",
+      "    print \"1\",\n",
+      "if name != \"Kalai mani\":\n",
+      "    print \"6\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0 1 6\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 8, Page Number: 5.26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using strrev() function\n",
+      "def strrev(cstring1):\n",
+      "    cstring2 = ' '\n",
+      "    for i in range(len(cstring1)-1,-1,-1):\n",
+      "        cstring2 = cstring2 + cstring1[i]\n",
+      "    return cstring2\n",
+      "\n",
+      "y = raw_input(\"Enter the string \")\n",
+      "print \"The string reversed is : %s\" %strrev(y)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the string book\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The string reversed is :  koob\n"
+       ]
+      }
+     ],
+     "prompt_number": 85
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 9, Page Number: 5.27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using enumerated data type\n",
+      "def enum(*sequential, **named):\n",
+      "    enums = dict(zip(sequential, range(len(sequential))), **named)\n",
+      "    return type('Enum', (), enums)\n",
+      "\n",
+      "week = enum(' ', 'sun','Mon','Tue','Wed','Thr','Fri','Sat')\n",
+      "print \"Mon = %d\" %week.Mon\n",
+      "print \"Tue = %d\" %week.Tue\n",
+      "print \"Wed= %d\" %week.Wed\n",
+      "print \"Sat = %d\" %week.Sat"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mon = 2\n",
+        "Tue = 3\n",
+        "Wed= 4\n",
+        "Sat = 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 125
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 10, Page Number: 5.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to identify whether the entered character is letter, digit or other symbols\n",
+      "def enum(*sequential, **named):\n",
+      "    enums = dict(zip(sequential, range(len(sequential))), **named)\n",
+      "    return type('Enum', (), enums)\n",
+      "\n",
+      "ctype = enum(' ', 'Letter', 'Digit', 'Other')\n",
+      "ch = raw_input(\"Enter a any character.\")\n",
+      "a = ch.isalpha()\n",
+      "if a != False:\n",
+      "    print \"%c is %d type of symbol.\" %(ch,ctype.Letter)\n",
+      "else:\n",
+      "    a = ch.isdigit()\n",
+      "    if a != False:\n",
+      "        print \"%c is %d type of symbol.\" %(ch,ctype.Digit)\n",
+      "    else:\n",
+      "        print \"%c is %d type of symbol.\" %(ch,ctype.Other)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a any character.%\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "% is 3 type of symbol.\n"
+       ]
+      }
+     ],
+     "prompt_number": 143
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 1, Page Number: 5.29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program for sorting an array of elements\n",
+      "\n",
+      "a = []\n",
+      "n = input(\"Enter Upper limit....\")\n",
+      "for i in range(0,n):\n",
+      "    a.append(0)\n",
+      "print \"Enter elements....\"    \n",
+      "for i in range(0,n):\n",
+      "    a[i]=input()\n",
+      "for i in range(0,n):\n",
+      "    for j in range(i,n):\n",
+      "        if a[i]>a[j]:\n",
+      "            t = a[i]\n",
+      "            a[i] = a[j]\n",
+      "            a[j] = t\n",
+      "print \"Numbers in Ascending order\"\n",
+      "for i in range(0,n):\n",
+      "    print \"%d\" %a[i],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Upper limit....6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter elements....\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numbers in Ascending order\n",
+        "1 2 3 4 5 6\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 2, Page Number: 5.31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to fing Median for a given 'n' numbers\n",
+      "\n",
+      "arr = []\n",
+      "n = input(\"Enter Upper limit....\")\n",
+      "for i in range(0,n):\n",
+      "    arr.append(0)\n",
+      "print \"Enter..%d..values\" %n    \n",
+      "for i in range(0,n):\n",
+      "    arr[i]=input()\n",
+      "for i in range(1,n):\n",
+      "    for j in range(1,n-i):\n",
+      "        if arr[j] <= arr[j+1]:\n",
+      "            t = arr[j]\n",
+      "            arr[j] = arr[j+1]\n",
+      "            arr[j+1] = t\n",
+      "        else:\n",
+      "            continue\n",
+      "if n%2 == 0:\n",
+      "    median = (arr[n/2]+arr[n/2+1])/2.0\n",
+      "else:\n",
+      "    median = arr[n/2+1]\n",
+      "print \"Median is....%.f\" %median    \n",
+      "                "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Upper limit....3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter..3..values\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "30\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Median is....20\n"
+       ]
+      }
+     ],
+     "prompt_number": 162
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 3, Page Number: 5.32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to find standard deviation for the given data\n",
+      "import math\n",
+      "\n",
+      "val = []\n",
+      "sum = ssqr = n = 0\n",
+      "for i in range(0,100):\n",
+      "    val.append(0)\n",
+      "print \"Enter Values -999 to stop....\"\n",
+      "for i in range(0,100):\n",
+      "    val[i] = float(raw_input())\n",
+      "    if val[i] == -999:\n",
+      "        break\n",
+      "    sum = float(sum) + val[i]    \n",
+      "    n = n + 1\n",
+      "mean = float(sum) / float(n)    \n",
+      "for i in range(0,n):\n",
+      "    dev = val[i] - mean\n",
+      "    ssqr = float(ssqr) + float(dev)*float(dev)\n",
+      "var = ssqr / float(n)    \n",
+      "sd = math.sqrt(float(var))\n",
+      "print \"Number of Items.....%d\" %n\n",
+      "print \"Mean is .....%f\" %mean\n",
+      "print \"Standard Deviation is .....%f\" %sd"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values -999 to stop....\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-999\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of Items.....5\n",
+        "Mean is .....30.000000\n",
+        "Standard Deviation is .....14.142136\n"
+       ]
+      }
+     ],
+     "prompt_number": 169
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 4, Page Number: 5.34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Bubble sort using function\n",
+      "\n",
+      "def bubble(n, x = []):\n",
+      "    for item in range(0,n-1):\n",
+      "        for i in range(item+1,n):\n",
+      "            if x[i]<x[item]:\n",
+      "                temp = x[item]\n",
+      "                x[item] = x[i]\n",
+      "                x[i] = temp\n",
+      "                \n",
+      "a = []                \n",
+      "n = input(\"Enter upper limit: \")\n",
+      "for i in range(0,n):\n",
+      "    a.append(0)\n",
+      "for i in range(0,n):\n",
+      "    a[i] = input()\n",
+      "    \n",
+      "bubble(n,a)    \n",
+      "print \"After sorting\",\n",
+      "for i in range(0,n):\n",
+      "    print \"%d\" %a[i],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter upper limit: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After sorting 1 2 3 4 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 174
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 5, Page Number: 5.35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to sort an element using Insertion sort\n",
+      "\n",
+      "a = []\n",
+      "x = input(\"Elements to be inserted \")\n",
+      "for i in range(0,100):\n",
+      "    a.append(0)\n",
+      "i = 0\n",
+      "while x != -99:\n",
+      "    k = i-1\n",
+      "    while (x<a[k]) and (k>=0):\n",
+      "        a[k+1] = a[k]\n",
+      "        k = k-1\n",
+      "    a[k+1]=x\n",
+      "    print \"Array after inserting %d:\" %x\n",
+      "    for j in range(0,i):\n",
+      "        print \"%5d\" %a[j]\n",
+      "    x = input(\"Elements to be inserted(-99 to stop)?\")\n",
+      "    i = i+1\n",
+      "print \"The final sorted array\"\n",
+      "for j in range(0,i):\n",
+      "    print \"%5d\" %a[j],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements to be inserted 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Array after inserting 5:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements to be inserted(-99 to stop)?3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Array after inserting 3:\n",
+        "    3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements to be inserted(-99 to stop)?4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Array after inserting 4:\n",
+        "    3\n",
+        "    4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements to be inserted(-99 to stop)?2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Array after inserting 2:\n",
+        "    2\n",
+        "    3\n",
+        "    4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements to be inserted(-99 to stop)?1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Array after inserting 1:\n",
+        "    1\n",
+        "    2\n",
+        "    3\n",
+        "    4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements to be inserted(-99 to stop)?-99\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final sorted array\n",
+        "    1     2     3     4     5\n"
+       ]
+      }
+     ],
+     "prompt_number": 186
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 6, Page Number: 5.36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to sort an element using selection sort\n",
+      "\n",
+      "a = []\n",
+      "for i in range(0,100):\n",
+      "    a.append(0)\n",
+      "n = input(\"Enter Number of elements in array...\")\n",
+      "print \"Enter Elements...\",n\n",
+      "for i in range(0,n):\n",
+      "    a[i] = input()\n",
+      "for i in range(0,n-1):\n",
+      "    m = i\n",
+      "    for j in range(i+1,n):\n",
+      "        if a[m]>a[j]:\n",
+      "            m = j\n",
+      "    if i != m:\n",
+      "        k = a[i]\n",
+      "        a[i] = a[m]\n",
+      "        a[m] = k\n",
+      "        \n",
+      "print \"Sorted Array (Selection Sort) is...\\n\"\n",
+      "for i in range(0,n):\n",
+      "    print \"%5d\" %a[i],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number of elements in array...9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements... 9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "326\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "85\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "42\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "156\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "845\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "66\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "55\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorted Array (Selection Sort) is...\n",
+        "\n",
+        "    1    42    55    56    66    85   156   326   845\n"
+       ]
+      }
+     ],
+     "prompt_number": 190
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 7, Page Number: 5.38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Quick sort\n",
+      "\n",
+      "def quickSort(array):\n",
+      "   quickSortHelper(array,0,len(array)-1)\n",
+      "\n",
+      "def quickSortHelper(array,right,left):\n",
+      "   if right<left:\n",
+      "       splitpoint = partition(array,right,left)\n",
+      "       quickSortHelper(array,right,splitpoint-1)\n",
+      "       quickSortHelper(array,splitpoint+1,left)\n",
+      "\n",
+      "def partition(array,right,left):\n",
+      "   pivotvalue = array[right]\n",
+      "\n",
+      "   begin = right+1\n",
+      "   end = left\n",
+      "\n",
+      "   done = False\n",
+      "   while not done:\n",
+      "\n",
+      "       while begin <= end and \\\n",
+      "               array[begin] <= pivotvalue:\n",
+      "           begin = begin + 1\n",
+      "\n",
+      "       while array[end] >= pivotvalue and \\\n",
+      "               end >= begin:\n",
+      "           end = end -1\n",
+      "\n",
+      "       if end < begin:\n",
+      "           done = True\n",
+      "       else:\n",
+      "           temp = array[begin]\n",
+      "           alist[begin] = array[end]\n",
+      "           alist[end] = temp\n",
+      "\n",
+      "   temp = array[right]\n",
+      "   array[right] = array[end]\n",
+      "   array[end] = temp\n",
+      "   return end\n",
+      "\n",
+      "array = []\n",
+      "n=input(\"Enter the values in array: \")\n",
+      "for i in range (0,n):\n",
+      "    array.append(0)\n",
+      "for i in range (0,n):\n",
+      "    array[i]=input()\n",
+      "    \n",
+      "quickSort(array)\n",
+      "print \"Quick Sorted Array is\", array"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the values in array: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quick Sorted Array is [1, 2, 3, 4, 5]\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 8, Page Number: 5.40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#program to find the specified elements from the array using linear searching\n",
+      "\n",
+      "a = [100]\n",
+      "for i in range(0,100):\n",
+      "    a.append(0)\n",
+      "c = 'y'\n",
+      "while (c=='y')or(c=='Y'):\n",
+      "    no = input(\"Enter the size of sorting\")\n",
+      "    print \"Enter the elements of the array\"\n",
+      "    for i in range(0,no):\n",
+      "        a[i]=input()\n",
+      "    k = input(\"Enter the element to be searched\")    \n",
+      "    for i in range(0,no):\n",
+      "        if k == a[i]:\n",
+      "            print \"Element %d is in the position %d\" %(k,i+1)\n",
+      "            f=2\n",
+      "            break\n",
+      "    if f==1:\n",
+      "        print \"The entered element is not in the array\"\n",
+      "    ch = raw_input(\"If you want to continue y/n \")\n",
+      "    c = ch\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the size of sorting5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the elements of the array\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element to be searched4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Element 4 is in the position 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "If you want to continue y/n n\n"
+       ]
+      }
+     ],
+     "prompt_number": 206
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 9, Page Number: 5.42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to find the specified element from array using Binary search\n",
+      "\n",
+      "a = [100]\n",
+      "for i in range(0,100):\n",
+      "    a.append(0)\n",
+      "no = input(\"Enter the size of sorting\")\n",
+      "print \"Enter the elements in ascending order\"\n",
+      "for i in range(0,no):\n",
+      "        a[i]=input() \n",
+      "t = input(\"Enter the element to be searched\") \n",
+      "l=0\n",
+      "h=no-1\n",
+      "while l <= h:\n",
+      "    m = (l+h)/2\n",
+      "    if t<a[m]:\n",
+      "        h = m-1\n",
+      "    elif t>a[m]:\n",
+      "        l = m+1\n",
+      "    else:\n",
+      "        print \"Entered %d is in position %d\" %(t,m+1)\n",
+      "        f=2\n",
+      "        break\n",
+      "if f==1:\n",
+      "    print \"Entered element is not in the array\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the size of sorting5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the elements in ascending order\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element to be searched3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Entered 3 is in position 2\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter07.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter07.ipynb
new file mode 100755
index 00000000..ee25427a
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter07.ipynb
@@ -0,0 +1,254 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:59d989e4d68b99137fb5255fe771e42a62c5ce35100304584930ea0c003ea2ac"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 07 : User Defined Functions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, page no.5.49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate Local and Global Declaration \n",
+      "\n",
+      "#variable declaration globally\n",
+      "i=0                \n",
+      "print \"Value of i in main\", i\n",
+      "\n",
+      " #define function\n",
+      "def f1():        \n",
+      "    k=0\n",
+      "    i=50\n",
+      "    print \"Value of i after function call\",i\n",
+      "    \n",
+      "#Function Call\n",
+      "f1()             \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of i in main 0\n",
+        "Value of i after function call 50\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, page no: 5.50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate usage of local and global variable\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    print \"Value of i is:\", i\n",
+      "    \n",
+      "i=590\n",
+      "print \"The Value of i is:\", i    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of i is: 0\n",
+        "Value of i is: 1\n",
+        "Value of i is: 2\n",
+        "The Value of i is: 590\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page no.5.52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to call function without parameters\n",
+      "def message():\n",
+      "    print \"Main Message\"\n",
+      "    \n",
+      "print \"Function Message\"\n",
+      "\n",
+      "message()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Function Message\n",
+        "Main Message\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page No.5.52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to call function with parameter\n",
+      "\n",
+      "#Defining function\n",
+      "def add(x,y):\n",
+      "    z=0\n",
+      "    z=x+y\n",
+      "    return z\n",
+      "\n",
+      "k=add(10,20)\n",
+      "\n",
+      "#Result\n",
+      "print \"The Value of k is \", k"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of k is  30\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 5, page no.5.53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to write prototype of the function\n",
+      "\n",
+      "k=0\n",
+      "\n",
+      "#Defining in function\n",
+      "def f1(k):\n",
+      "    k=k+100\n",
+      "    return k\n",
+      "\n",
+      "#Result\n",
+      "print \"The value of i before call: \", k\n",
+      "\n",
+      "print \"The Value of i after call: \",(f1(k))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of i before call:  0\n",
+        "The Value of i after call:  100\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 6, page no. 5.54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to write prototype definition in main function\n",
+      "\n",
+      "def fun(k):\n",
+      "    k=k+10\n",
+      "    return k\n",
+      "\n",
+      "i=0\n",
+      "\n",
+      "#Result\n",
+      "print  \"The i value before call %d\\n\"%(i)\n",
+      "print \"The i value after call %d\\n\"%fun(i)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The i value before call 0\n",
+        "\n",
+        "The i value after call 10\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter08.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter08.ipynb
new file mode 100755
index 00000000..e40c1b43
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter08.ipynb
@@ -0,0 +1,420 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:5d85b398d61f498d1deba465b9747c5d15996be1754212b25012ca792b193a16"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 08 : Function Prototypes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Function with no arguments and no return values"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.: 5.55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to addition of two numbers\n",
+      "\n",
+      "def add():\n",
+      "    a,b=input(\"Enter two numbers.. \")\n",
+      "    c=a+b\n",
+      "    print \"Sum is ... %d\"%(c)\n",
+      "    \n",
+      "add()    \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers.. 10,20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum is ... 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.:5.56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print addition, subtraction and multiplication values\n",
+      "\n",
+      "def add():\n",
+      "    a=10\n",
+      "    b=5\n",
+      "    print \"The Addition of given value: \",a+b\n",
+      "        \n",
+      "def sub():\n",
+      "    a=10\n",
+      "    b=5\n",
+      "    print \"The Subtraction of given value:\",a-b\n",
+      "    \n",
+      "def mul():\n",
+      "    a=10\n",
+      "    b=5\n",
+      "    print \"The Multiplication of given value:\",a*b\n",
+      "    \n",
+      "    \n",
+      "add()\n",
+      "sub()\n",
+      "mul()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Addition of given value:  15\n",
+        "The Subtraction of given value: 5\n",
+        "The Multiplication of given value: 50\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Function with arguments and no return values"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, page No. 5.57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Program for addition of two numbers\n",
+      "\n",
+      "def add(x,y):\n",
+      "    z= x+y\n",
+      "    print \"Sum is:\", z\n",
+      "\n",
+      "#Main Function\n",
+      "a, b = input(\"Enter two Values: \")\n",
+      "    \n",
+      "add(a,b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two Values: 10,20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum is: 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example: 2, page no. 5.58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def date(x,y,z):\n",
+      "    print \"Employee Joined date is:%d/%d/%d\" %(x,y,z)\n",
+      "    \n",
+      "print \"Enter the employee joining date.\"\n",
+      "\n",
+      "dd,mm,yy=eval(raw_input(\"Enter date, month, year: \"))\n",
+      "\n",
+      "date(dd,mm,yy)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the employee joining date.\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter date, month, year: 11,12,1981\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Employee Joined date is:11/12/1981\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Function with arguments and with return value"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, page no.5.59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program for addition of two numbers\n",
+      "\n",
+      "def add(x,y):\n",
+      "    z=0\n",
+      "    z=x+y\n",
+      "    return z\n",
+      "\n",
+      "#Main Function\n",
+      "a,b=input(\"Enter the a and b value:\")\n",
+      "c=add(a,b)\n",
+      "print \"Result is:\",c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the a and b value:10,20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Result is: 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.:5.60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to send and receives the values from function\n",
+      "\n",
+      "def mul(m,n):\n",
+      "    return (m*n)\n",
+      "\n",
+      "m=n=sum=0\n",
+      "m,n=input(\"Enter two values:\")\n",
+      "sum=mul(m,n)\n",
+      "\n",
+      "print \"Multiplication of the entered Values:\",sum"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two values:30,4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Multiplication of the entered Values: 120\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Function with no arguments and with return value"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, PageNo.:5.61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program for addition of two numbers\n",
+      "\n",
+      "def add():\n",
+      "    a=b=c=0\n",
+      "    a,b=input(\"Enter the Values of a and b:\")\n",
+      "    c=a+b\n",
+      "    return(c)\n",
+      "\n",
+      "c=add()\n",
+      "print \"The Output of given Values:\",c\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Values of a and b:10,20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Output of given Values: 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No. 5.62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to multiply three numbers using function\n",
+      "\n",
+      "\n",
+      "def mul():\n",
+      "    a,b,c=input(\"Enter the three values for Multiplication: \")\n",
+      "    i=a*b*c\n",
+      "    return(i)\n",
+      "\n",
+      "s=mul()\n",
+      "print \"Multiplication of three numbers is= \",s\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the three values for Multiplication: 4,8,2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Multiplication of three numbers is=  64\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter09.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter09.ipynb
new file mode 100755
index 00000000..11735b92
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter09.ipynb
@@ -0,0 +1,109 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1a0147da7a94fc5ebe7c66bbf21a563cb0bee33c1452a7bcfcbc21f5395d7aab"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 09 : Parameter Passing Methods"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.: 5.63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call By Value\n",
+      "\n",
+      "def cube(x):\n",
+      "    x=x*x*x\n",
+      "    return(x)\n",
+      "\n",
+      "n=5\n",
+      "print \"Cube of %d is\"%n,cube(n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cube of 5 is 125\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.:5.64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call by reference\n",
+      "\n",
+      "def interchange(a,b):\n",
+      "    t=0\n",
+      "    t=a\n",
+      "    a=b\n",
+      "    b=t\n",
+      "    return b,a\n",
+      "\n",
+      "    \n",
+      "i=5\n",
+      "j=10\n",
+      "print \"i and j values before interchange:\", i, j\n",
+      "\n",
+      "i=10\n",
+      "j=5\n",
+      "interchange(i,j)\n",
+      "\n",
+      "#Result\n",
+      "print \"i and j values after interchange:\",i,j\n",
+      "\n",
+      "print \"i and j avalues after interchange in the function : %d %d \\n\"%interchange(i,j)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i and j values before interchange: 5 10\n",
+        "i and j values after interchange: 10 5\n",
+        "i and j avalues after interchange in the function : 10 5 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter10.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter10.ipynb
new file mode 100755
index 00000000..1a017d08
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter10.ipynb
@@ -0,0 +1,372 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ea85b38c0942e3b3d3a1243525badd8ffa28ac492392c6aef6c8ea5823294150"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Recursion"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.: 5.67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Factorial of an integer number.\n",
+      "\n",
+      "def rec(x):\n",
+      "    f=0\n",
+      "    if x==1:\n",
+      "        return (1)\n",
+      "    else:\n",
+      "        f=x*rec(x-1)\n",
+      "        return(f)\n",
+      "\n",
+      "a=input(\"Enter the number:\")\n",
+      "print \"The Factorial value of %d:\"%a,rec(a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number:5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Factorial value of 5: 120\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.:5.67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to read characters and print reversely using recursion.\n",
+      "\n",
+      "def reverse(c):\n",
+      "    if len(c)<=0:\n",
+      "        return c\n",
+      "    return reverse(c[1:])+c[0]\n",
+      "\n",
+      "c=raw_input(\"Enter the Line of Text...\")\n",
+      "print reverse(c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Line of Text...munilak\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kalinum\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No.:5.68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#concept of recursion\n",
+      "\n",
+      "def add(pk, pm):\n",
+      "    if pm==0:\n",
+      "        return pk\n",
+      "    else:\n",
+      "        return(1+add(pk,pm-1))\n",
+      "    \n",
+      "    \n",
+      "k=70\n",
+      "m=20\n",
+      "\n",
+      "i=add(k,m)\n",
+      "\n",
+      "print \"The Value of addition is %d\"%i\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of addition is 90\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page No:5.70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Tower of honoi\n",
+      "n=int(raw_input(\"How many Disks ...\\n\"))\n",
+      "\n",
+      "def honoi(n,fro,to,t):\n",
+      "    if(n>0):\n",
+      "        honoi(n-1,fro,t,to)\n",
+      "        print \"Move disk %d from %c to %c\\n\"%(n,fro,to)\n",
+      "        honoi(n-1,t,to,fro)\n",
+      "        \n",
+      "fro='L'\n",
+      "to='R'\n",
+      "t='C'\n",
+      "        \n",
+      "honoi(n,fro,to,t)\n",
+      "            "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many Disks ...\n",
+        "3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Move disk 1 from L to R\n",
+        "\n",
+        "Move disk 2 from L to C\n",
+        "\n",
+        "Move disk 1 from R to C\n",
+        "\n",
+        "Move disk 3 from L to R\n",
+        "\n",
+        "Move disk 1 from C to L\n",
+        "\n",
+        "Move disk 2 from C to R\n",
+        "\n",
+        "Move disk 1 from L to R\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 1, Page No. : 5.76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#user defined function that computes 'x' raised to the power 'y'\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "x,y=input(\"Enter the Values of X and Y \")\n",
+      "\n",
+      "print \"%d to the power of %d is ... \"%(x,y),math.pow(x,y)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Values of X and Y 10,2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10 to the power of 2 is ...  100.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 2, page No.5.76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#nesting of function\n",
+      "\n",
+      "def ratio(x,y,z):\n",
+      "    \n",
+      "    def diff(m,n):\n",
+      "        if(m!=n):\n",
+      "            return(1)\n",
+      "        else:\n",
+      "            return(0)\n",
+      "    \n",
+      "    diff(y,z)\n",
+      "    if diff(y,z):\n",
+      "       return(x/(y-z))\n",
+      "    else:\n",
+      "       return(0)\n",
+      "        \n",
+      "    \n",
+      "x,y,z=input(\"Enter the values of X,Y,Z: \")\n",
+      "\n",
+      "    \n",
+      "print \"Ratio is..%f\\n\"%ratio(x,y,z)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the values of X,Y,Z: 10,10,5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio is..2.000000\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 3, Page No. 5.78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Demonstration of Global Variable\n",
+      "\n",
+      "a=39               #Global Variable\n",
+      "print \"A... %d\\n\"%a\n",
+      "\n",
+      "def f1():\n",
+      "    global a\n",
+      "    a=a+10\n",
+      "    return a\n",
+      "\n",
+      "def f2():\n",
+      "    global a\n",
+      "    a=a+20\n",
+      "    return a\n",
+      "\n",
+      "def f3():\n",
+      "    global a\n",
+      "    a=a+30\n",
+      "    return a\n",
+      "\n",
+      "\n",
+      "\n",
+      "print \"A...%d\\n\"%f1()\n",
+      "print \"A...%d\\n\"%f2()\n",
+      "print \"A...%d\\n\"%f3()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A... 39\n",
+        "\n",
+        "A...49\n",
+        "\n",
+        "A...69\n",
+        "\n",
+        "A...99\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter11.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter11.ipynb
new file mode 100755
index 00000000..0e00f235
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter11.ipynb
@@ -0,0 +1,304 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:03a9be4b5720a51a42be3ae01bfa08ccecef76e9a5e17f613e76c0afaff965de"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Structures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.: 5.81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the student number, name and marks through accessing structure elements.\n",
+      "\n",
+      "#Defining Structure in class\n",
+      "class std:\n",
+      "    no=0\n",
+      "    name = ''\n",
+      "    m1 = m2 = m3 =0\n",
+      "\n",
+      "#assigning variable to structure class    \n",
+      "s=std()\n",
+      "\n",
+      "#UserInput\n",
+      "s.no=raw_input(\"Give Student Number: \")\n",
+      "s.name=raw_input(\"Enter the Student Name: \")\n",
+      "s.m1=input(\"Enter Student Mark1: \")\n",
+      "s.m2=input(\"Enter Student Mark2: \")\n",
+      "s.m3=input(\"Enter Student Mark3: \")\n",
+      "\n",
+      "total=avg=0.0\n",
+      "\n",
+      "total = (s.m1 + s.m2 + s.m3)\n",
+      "avg='%.2f' %(total / 3)\n",
+      "\n",
+      "#Result\n",
+      "print \"\\nThe Output is.... \\n\",s.no,s.name,total,avg"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give Student Number: 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Student Name: raja\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark1: 99\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark2: 97\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark3: 95\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Output is.... \n",
+        "1 raja 291 97.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No. 5.83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#print a value from Structure.\n",
+      "\n",
+      "class struct:\n",
+      "    a=b=0\n",
+      "    \n",
+      "x=struct()\n",
+      "y=struct()\n",
+      "\n",
+      "x.a=29\n",
+      "y=x\n",
+      "\n",
+      "#Result\n",
+      "print \"The Value of a %d\"%y.a\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of a 29\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No.:5.84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program for User defined data type \n",
+      "\n",
+      "#There is no preprocessor directive in python\n",
+      "D = 7\n",
+      "\n",
+      "#No typedef function in python\n",
+      "wk = int(raw_input(\"Enter Weeks : \"))\n",
+      "\n",
+      "#Result\n",
+      "print \"\\nNumber of Days = %d\"%(wk*D)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Weeks : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number of Days = 28\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page.No.:5.85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to create user defined data type from structure\n",
+      "\n",
+      "\n",
+      "class struct:\n",
+      "    name=''\n",
+      "    sex=''\n",
+      "    age=0\n",
+      "    \n",
+      "\n",
+      "\n",
+      "candidate = [struct() for i in range(0,2)]\n",
+      "\n",
+      "for a in range(0,2):\n",
+      "    candidate[a].name= raw_input(\"Name of the Employee: \")\n",
+      "    candidate[a].sex= raw_input(\"Sex: \")\n",
+      "    candidate[a].age= input(\"Age: \")\n",
+      "\n",
+      "#Result\n",
+      "print \"\\n Name\\t Sex\\t Age\\n\"\n",
+      "\n",
+      "for a in range(0,2):\n",
+      "    print \" %s\\t\"%(candidate[a].name),\"  %s\\t\"%(candidate[a].sex),\" %d\\n\"%(candidate[a].age)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name of the Employee: venkat\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sex: m\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Age: 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name of the Employee: geetha\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sex: f\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Age: 23\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Name\t Sex\t Age\n",
+        "\n",
+        " venkat\t  m\t 25\n",
+        "\n",
+        " geetha\t  f\t 23\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 68
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter12.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter12.ipynb
new file mode 100755
index 00000000..54918fbb
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter12.ipynb
@@ -0,0 +1,829 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:13c473411b35212c9727620c7673dc52636d4bf1a83b0d4eeb082d160a75324d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Union"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.5.87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#program using union.\n",
+      "\n",
+      "class union_name:\n",
+      "    a=0\n",
+      "    b=[0,1]\n",
+      "    \n",
+      "c=union_name()\n",
+      "c.a=256\n",
+      "\n",
+      "print \"c.a value is %d\\n \"%c.a\n",
+      "print \"c.b[0] value is %d\\n\"%c.b[0]\n",
+      "print \"c.b[1] value is %d\\n\"%c.b[1]\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "c.a value is 256\n",
+        " \n",
+        "c.b[0] value is 0\n",
+        "\n",
+        "c.b[1] value is 1\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.:5.87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#program to find size of union and number of bytes\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "class union_result:\n",
+      "    marks=0\n",
+      "    grade=''\n",
+      "    \n",
+      "    \n",
+      "class struct_res:\n",
+      "    name=[15]\n",
+      "    age=0\n",
+      "    sex=''\n",
+      "    address=''\n",
+      "    pincode=0\n",
+      "    perf=union_result()\n",
+      "    \n",
+      "data=struct_res()   \n",
+      "\n",
+      "#Result\n",
+      "print \"Size of Union: %d\\n\"%sys.getsizeof(data.perf)\n",
+      "print \"Size of Structure: %d\\n\"%sys.getsizeof(data)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Size of Union: 32\n",
+        "\n",
+        "Size of Structure: 32\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pointer to Unions cannot convert as no pointer in python. Page no. 5.88"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No.:5.89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program that includes of initial values to structure variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "\n",
+      "class union_id:\n",
+      "    color='20'\n",
+      "    size=0\n",
+      "    \n",
+      "\n",
+      "class struct_clothes:\n",
+      "    manufacturer='30'\n",
+      "    cost=0.0\n",
+      "    description=union_id()\n",
+      "    \n",
+      "    \n",
+      "shirt=struct_clothes()\n",
+      "shirt.manufacturer='Indian'\n",
+      "shirt.cost=35.00\n",
+      "shirt.description.color='Black'\n",
+      "\n",
+      "print \"%d\\n\"%(sys.getsizeof(union_id))\n",
+      "\n",
+      "print \"%s %f \"%(shirt.manufacturer,shirt.cost),\"%s %d\\n \"%(shirt.description.color, shirt.description.size)\n",
+      " \n",
+      "\n",
+      "shirt.description.size =12\n",
+      "\n",
+      "\n",
+      "print \"%s %f \"%(shirt.manufacturer, shirt.cost),\"%s %d\\n \"%(shirt.description.color, shirt.description.size)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "48\n",
+        "\n",
+        "Indian 35.000000  Black 0\n",
+        " \n",
+        "Indian 35.000000  Black 12\n",
+        " \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page no.: 5.92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to read and write employee and their date of joining using nested structure. \n",
+      "\n",
+      "\n",
+      "class date:\n",
+      "    day=0\n",
+      "    month=\"\"\n",
+      "    year=0\n",
+      "    \n",
+      "class employee:\n",
+      "    code=0\n",
+      "    name=''\n",
+      "    salary=0.0\n",
+      "    doj=date()\n",
+      "    \n",
+      "    \n",
+      "emp1=employee()\n",
+      "\n",
+      "emp1.code=input(\"Enter Employee Code: \")\n",
+      "emp1.name=raw_input(\"Enter the Employee Name: \")\n",
+      "emp1.salary=input(\"Enter the Employee Salary: \")\n",
+      "\n",
+      "print \"\\nEnter Date of joining in order\\n\"\n",
+      "emp1.doj.day=input(\"Enter day: \")\n",
+      "emp1.doj.month=raw_input(\"Enter month: \")\n",
+      "emp1.doj.year=input(\"Enter year: \")\n",
+      "\n",
+      "print \"\\nThe Employee Code is: %d\"%emp1.code\n",
+      "print \"\\nThe Employee Name is: %s\"%emp1.name\n",
+      "print \"\\nThe Employee Name is: %f\"%emp1.salary\n",
+      "print \"\\nThe Employee DOJ is: %d %s %d\"%(emp1.doj.day,emp1.doj.month,emp1.doj.year)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Employee Code: 200\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Employee Name: VIJI\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Employee Salary: 2000.00\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Date of joining in order\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter day: 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter month: December\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter year: 2004\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Employee Code is: 200\n",
+        "\n",
+        "The Employee Name is: VIJI\n",
+        "\n",
+        "The Employee Name is: 2000.000000\n",
+        "\n",
+        "The Employee DOJ is: 12 December 2004\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 5, Page No. 5.93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to store 3 records in one structure\n",
+      "\n",
+      "\n",
+      "class book:\n",
+      "    name=''\n",
+      "    price=0\n",
+      "    pages=0\n",
+      "    \n",
+      "    \n",
+      "b=[book() for a in range(0,3)]\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    b[i].name=raw_input(\"Enter the Book Name: \")\n",
+      "    b[i].price=input(\"Enter the Book Price: \")\n",
+      "    b[i].pages=input(\"Enter the Book Pages: \")\n",
+      "    \n",
+      "print '\\n'    \n",
+      "for i in range(0,3):\n",
+      "    print \"%s\\t\"%(b[i].name),\"%d\\t\"%(b[i].price),\"%d\\n\"%(b[i].pages)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Name: English\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Price: 165\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Pages: 200\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Name: Maths\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Price: 300\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Pages: 450\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Name: Physics\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Price: 250\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book Pages: 370\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "English\t165\t200\n",
+        "\n",
+        "Maths\t300\t450\n",
+        "\n",
+        "Physics\t250\t370\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 6, Page No. 5.95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to copy of entire structure to a function\n",
+      "\n",
+      "class std:\n",
+      "    no=0\n",
+      "    avg=0.0\n",
+      "\n",
+      "a=std()\n",
+      "\n",
+      "a.no=15\n",
+      "a.avg=90.75\n",
+      "\n",
+      "def func(Struct_p):\n",
+      "    print \"Number is.... %d\\n\"%a.no\n",
+      "    print \"Average is... %f\"%a.avg\n",
+      "    \n",
+      "func(a)\n",
+      "\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number is.... 15\n",
+        "\n",
+        "Average is... 90.750000\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 7, Page No.:5.97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Programs shows how the structure can be accessed by a pointer variable.\n",
+      "\n",
+      "class student:\n",
+      "    roll_no=0\n",
+      "    name=''\n",
+      "    marks=0.0\n",
+      "stud1=student()\n",
+      "\n",
+      "print \"Enter the student details: \\n\"\n",
+      "stud1.roll_no=input()\n",
+      "stud1.name=raw_input()\n",
+      "stud1.marks=input()\n",
+      "\n",
+      "print \"\\nDisplay of structure using structure variable\\n\"\n",
+      "print \"Roll No\\t Name\\t Marks\"\n",
+      "print \"\\n-------\\t ----- -------\\n\"\n",
+      "print \"%d\\t %s\\t %f\"%(stud1.roll_no,stud1.name,stud1.marks)\n",
+      "\n",
+      "pt=stud1\n",
+      "\n",
+      "print \"\\nDisplay of structure using pointer variable\\n\"\n",
+      "print \"Roll No\\t Name\\t Marks\"\n",
+      "print \"\\n-------\\t ----- -------\\n\"\n",
+      "print \"%d\\t %s\\t %f\"%(pt.roll_no,pt.name,pt.marks)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student details: \n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "39\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Muni\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "77\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Display of structure using structure variable\n",
+        "\n",
+        "Roll No\t Name\t Marks\n",
+        "\n",
+        "-------\t ----- -------\n",
+        "\n",
+        "39\t Muni\t 77.000000\n",
+        "\n",
+        "Display of structure using pointer variable\n",
+        "\n",
+        "Roll No\t Name\t Marks\n",
+        "\n",
+        "-------\t ----- -------\n",
+        "\n",
+        "39\t Muni\t 77.000000\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 1, Page No.:5.98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to transfer a structure to a function by passing the structure address to the function\n",
+      "\n",
+      "class struct:\n",
+      "    name=''\n",
+      "    no=0\n",
+      "    age=0\n",
+      "record=struct()\n",
+      "std=record\n",
+      "std.name,std.no,std.age=[\"LAK\",15,25]\n",
+      "print \"%s\\t%d\\t%d\"%(std.name,std.no,std.age)\n",
+      "\n",
+      "def fun(std):\n",
+      "    pt.name=\"Muni\"\n",
+      "    pt.no=16\n",
+      "    pt.age=26\n",
+      "    \n",
+      "std.name=pt.name\n",
+      "std.no=pt.no\n",
+      "std.age=pt.age\n",
+      "\n",
+      "print \"%s\\t%d\\t%d\"%(std.name,std.no,std.age)\n",
+      "    \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "LAK\t15\t25\n",
+        "Muni\t16\t26\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 2, Page No.:5.100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to pass an array of structure to a function, and return a pointer to a structure.\n",
+      "import sys\n",
+      "N=3\n",
+      "NULL=0\n",
+      "\n",
+      "class struct:\n",
+      "    sno=0\n",
+      "    name=''\n",
+      "    age=0\n",
+      "    \n",
+      "record=struct()\n",
+      "std=record\n",
+      "\n",
+      "print \"Student Number Locater\"\n",
+      "print \"To Exit the program enter '0' for student number\"\n",
+      "stdno=input(\"Enter the Student Number: \")\n",
+      "print \"\\n\"\n",
+      "if stdno ==15:\n",
+      "\n",
+      "    std.sno,std.name,std.age=[15,\"MUNI\",28]\n",
+      "    print \"Number .... %d \\n\"%(std.sno)\n",
+      "    print \"Name ...... %s\\n\"%(std.name)\n",
+      "    print \"Age ....... %d\\n\"%(std.age)\n",
+      "if stdno == 16:\n",
+      "    std.sno,std.name,std.age=[16,\"LAK\",27]\n",
+      "    print \"Number .... %d \\n\"%(std.sno)\n",
+      "    print \"Name ...... %s\\n\"%(std.name)\n",
+      "    print \"Age ....... %d\\n\"%(std.age)\n",
+      "    \n",
+      "if stdno == 17:\n",
+      "    std.sno,std.name,std.age=[17,\"RAJA\",31]\n",
+      "    print \"Number .... %d \\n\"%(std.sno)\n",
+      "    print \"Name ...... %s\\n\"%(std.name)\n",
+      "    print \"Age ....... %d\\n\"%(std.age)\n",
+      "    \n",
+      "if stdno==0:\n",
+      "    exit\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Student Number Locater\n",
+        "To Exit the program enter '0' for student number\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Student Number: 15\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Number .... 15 \n",
+        "\n",
+        "Name ...... MUNI\n",
+        "\n",
+        "Age ....... 28\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 3, Page No.: 5.101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print student name and mark using structure\n",
+      "\n",
+      "class student:\n",
+      "    name=''\n",
+      "    marks=0.00\n",
+      "    \n",
+      "s1=''\n",
+      "f=0.00\n",
+      "\n",
+      "student1=student()\n",
+      "\n",
+      "student2=student()\n",
+      "\n",
+      "\n",
+      "student1.name=raw_input(\"Enter Name: \")\n",
+      "f=input(\"Enter Mark: \")\n",
+      "\n",
+      "student2.marks=f\n",
+      "\n",
+      "print \"\\nName is %s \\n\"%student1.name\n",
+      "print \"Marks are %f \\n\"%student2.marks\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name: Venkat\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Mark: 89\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name is Venkat \n",
+        "\n",
+        "Marks are 89.000000 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 4, Page No.: 5.102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print marks, grade and their address using union\n",
+      "\n",
+      "\n",
+      "\n",
+      "class union_marks:\n",
+      "    perc=0.00\n",
+      "    grade=''\n",
+      "    \n",
+      "student1=union_marks()\n",
+      "\n",
+      "student1.perc=98.5\n",
+      "\n",
+      "print \"Marks are %f \\t address is %8lu\\n\"%(student1.perc, id(student1.perc))\n",
+      "\n",
+      "student1.grade='A'\n",
+      "\n",
+      "print \"Grade is %c \\t address is %8lu\\n\"%(student1.grade,id(student1.grade))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Marks are 98.500000 \t address is 158368916\n",
+        "\n",
+        "Grade is A \t address is 3073485064\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter13.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter13.ipynb
new file mode 100755
index 00000000..b303ee94
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter13.ipynb
@@ -0,0 +1,57 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bd4e511a3dbdd189491e3271c86cff9ce5fc919ed7457d400a67c288e55ad433"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Pointers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page Number: 5.104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print address and value of variable\n",
+      "#Local definitions\n",
+      "sno = 39\n",
+      "\n",
+      "#Result\n",
+      "print \"Address of sno = %u\" %id(sno)\n",
+      "print \"Value of sno = %d\" %sno"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Address of sno = 20917296\n",
+        "Value of sno = 39\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter14.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter14.ipynb
new file mode 100755
index 00000000..12a1734c
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter14.ipynb
@@ -0,0 +1,234 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:44f5daf77255992c95710d94f062b0f1f1e981b9e6cce07f23ed45a322c4b130"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 : Accessing Variable through Pointers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page Number: 5.107    "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to accessing through variable pointer\n",
+      "\n",
+      "from ctypes import c_int\n",
+      "\n",
+      "#Local definition\n",
+      "a = 22\n",
+      "a = c_int(a) # ctype datatype declaration\n",
+      "\n",
+      "b = 2.25\n",
+      "b = c_float(b) # ctype datatype declaration\n",
+      "\n",
+      "#Pointer variables\n",
+      "a_po=pointer(a)\n",
+      "b_po=pointer(b)\n",
+      "\n",
+      "# 'id' is a address of letter and it represents the address of the variable\n",
+      "a = id(a)\n",
+      "b = id(b)\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print \"\\nValue of a = %d\" %a_po[0]\n",
+      "print \"\\nValue of b = %.2f\" %b_po[0]\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value of a = 22\n",
+        "\n",
+        "Value of b = 2.25\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page Number: 5.108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print address and value of variable\n",
+      "\n",
+      "from ctypes import c_int\n",
+      "\n",
+      "#Local definition\n",
+      "a = 22\n",
+      "a = c_int(a) # ctype datatype declaration\n",
+      "\n",
+      "#Pointer variables\n",
+      "a_po=pointer(a)\n",
+      "\n",
+      "# 'id' is a address of letter and it represents the address of the variable\n",
+      "a = id(a)\n",
+      "\n",
+      "#Result\n",
+      "print \"\\n Value of a = %d\" %a_po[0]\n",
+      "print \"\\n Address of a = %u\" %id(a)\n",
+      "print \"\\n Value at address %u = %d\" %(id(a),a_po[0])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Value of a = 22\n",
+        "\n",
+        " Address of a = 4344066728\n",
+        "\n",
+        " Value at address 4344066728 = 22\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page Number: 5.108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Program to print value and address of variable\n",
+      "\n",
+      "from ctypes import c_int\n",
+      "\n",
+      "#Local definition\n",
+      "a = 22\n",
+      "b = c_int(a) # ctype datatype declaration\n",
+      "\n",
+      "#Pointer variable\n",
+      "b_po=pointer(b)\n",
+      "\n",
+      "# 'id' is a address of letter and it represents the address of the variable\n",
+      "b = id(a)\n",
+      "\n",
+      "#Result\n",
+      "print \"\\n Value of a = %d\" %a\n",
+      "print \"\\n Value of a = %d\" %(c_int(a).value)\n",
+      "print \"\\n Value of a = %d\" %b_po[0]\n",
+      "print \"\\n Address of a = %u\" %id(a)\n",
+      "print \"\\n Address of a = %u\" %b\n",
+      "print \"\\n Address of a = %u\" %id(b)\n",
+      "print \"\\n Value of b = address of a = %u\" %b"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Value of a = 22\n",
+        "\n",
+        " Value of a = 22\n",
+        "\n",
+        " Value of a = 22\n",
+        "\n",
+        " Address of a = 4298180848\n",
+        "\n",
+        " Address of a = 4298180848\n",
+        "\n",
+        " Address of a = 4344066728\n",
+        "\n",
+        " Value of b = address of a = 4298180848\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page Number: 5.109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to illustrate pointer to pointer\n",
+      "\n",
+      "from ctypes import c_int\n",
+      "\n",
+      "#Local definition\n",
+      "a = 22\n",
+      "b = c_int(a) # ctype datatype declaration\n",
+      "c = c_int(a) # ctype datatype declaration\n",
+      "\n",
+      "#Pointer variable\n",
+      "b_po = pointer(b)\n",
+      "c_po = pointer(b_po) # c_po is a pointer to another pointer\n",
+      "\n",
+      "# 'id' is a address of letter and it represents the address of the variable\n",
+      "b = id(a)\n",
+      "c = id(b)\n",
+      "\n",
+      "#Result\n",
+      "print \"\\n Value of a is %d\" %a\n",
+      "print \"\\n Value of a is %d\" %(c_int(a).value)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Value of a is 22\n",
+        "\n",
+        " Value of a is 22\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter15.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter15.ipynb
new file mode 100755
index 00000000..21b14f41
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter15.ipynb
@@ -0,0 +1,266 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2bed6e40cd55f716f11e909c76b0c924a863909239fcbbf975814260bb7b4531"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15 : Pointers and Functions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Call By Value"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page no. 5.112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call by value\n",
+      "\n",
+      "def change(x,y):\n",
+      "    x=x+5\n",
+      "    y=y+10\n",
+      "    print \"In the function changes a&b is %d,%d\"%(x,y)\n",
+      "    \n",
+      "a=10\n",
+      "b=20\n",
+      "\n",
+      "print \"\\nBefore calling the function a&b is %d,%d\"%(a,b)\n",
+      "\n",
+      "change(a,b)\n",
+      "print \"\\nAfter calling a&b is %d,%d\"%(a,b)\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Before calling the function a&b is 10,20\n",
+        "In the function changes a&b is 15,30\n",
+        "\n",
+        "After calling a&b is 10,20\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page no.5.113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#program to swap(interchange) the two given variable.\n",
+      "\n",
+      "def swap(x,y):\n",
+      "    \n",
+      "    x=x+y\n",
+      "    y=x-y\n",
+      "    x=x-y\n",
+      "    print \"In swap function x=%d, y=%d\"%(x,y)\n",
+      "    return y,x\n",
+      "\n",
+      "a=10\n",
+      "b=20\n",
+      "\n",
+      "print \"Before swap a=%d, b=%d\"%(a,b)\n",
+      "\n",
+      "\n",
+      "print \"\\nAfter the calling the swap function, a=%d,b=%d\"%swap(a,b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Before swap a=10, b=20\n",
+        "In swap function x=20, y=10\n",
+        "\n",
+        "After the calling the swap function, a=10,b=20\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Call By Reference"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page.No.5.114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call By Reference\n",
+      "\n",
+      "def change (x,y):\n",
+      "    x=x+5\n",
+      "    y=y+10\n",
+      "    print \"In the function change a&b is %d %d\"%(x,y)\n",
+      "    return x,y\n",
+      "a=10\n",
+      "b=20\n",
+      "print \"\\nBefore calling the function a&b is %d %d\"%(a,b)\n",
+      "\n",
+      "\n",
+      "print \"After calling the function a&b is %d %d\"%change(a,b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Before calling the function a&b is 10 20\n",
+        "In the function change a&b is 15 30\n",
+        "After calling the function a&b is 15 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.5.115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to swap(intercahnge) the value of variables.\n",
+      "\n",
+      "a=10\n",
+      "b=20\n",
+      "\n",
+      "print \"\\nBefore swap a=%d, b=%d\"%(swap(a,b))\n",
+      "\n",
+      "def swap(x,y):\n",
+      "    x=x+y\n",
+      "    y=x-y\n",
+      "    x=x-y\n",
+      "    print\"\\nIn swap function x=%d,y=%d\"%(x,y)\n",
+      "    return y,x\n",
+      "\n",
+      "print \"\\nAfter calling swap function a=%d, b=%d\"%(b,a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In swap function x=20,y=10\n",
+        "\n",
+        "Before swap a=10, b=20\n",
+        "\n",
+        "After calling swap function a=20, b=10\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page no. 5.115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to interchange the value of variables using call by reference\n",
+      "def exchange(m,n):\n",
+      "    t=m\n",
+      "    m=n\n",
+      "    n=t\n",
+      "    return n,m\n",
+      "\n",
+      "a,b=input(\"Enter two Numbers: \")\n",
+      "\n",
+      "print \"Before Exchange a=%d,b=%d\"%(a,b)\n",
+      "\n",
+      "\n",
+      "print \"After Exchange a=%d, b=%d\"%(exchange(b,a))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two Numbers: 10,20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Before Exchange a=10,b=20\n",
+        "After Exchange a=20, b=10\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter16.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter16.ipynb
new file mode 100755
index 00000000..ab91ce85
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter16.ipynb
@@ -0,0 +1,333 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d2add72dea4c498cf89927248000fdbfce0907704975d2bdb6ff1cde1c8f50a7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Chapter 16 : Pointers and Arrays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.:5.118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print the value and address of the element\n",
+      "\n",
+      "a=[10,20,30,40,50]\n",
+      "for b in range(0,5):\n",
+      "    print \"The value of a[%d]=%d\"%(b,a[b])\n",
+      "    print \"Address of a[%d]=%u\"%(b,id(a[b]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of a[0]=10\n",
+        "Address of a[0]=4299197952\n",
+        "The value of a[1]=20\n",
+        "Address of a[1]=4299197712\n",
+        "The value of a[2]=30\n",
+        "Address of a[2]=4299197472\n",
+        "The value of a[3]=40\n",
+        "Address of a[3]=4299199208\n",
+        "The value of a[4]=50\n",
+        "Address of a[4]=4299198968\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.: 5.119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print the value and the address of the element using pointer with function\n",
+      "\n",
+      "a=[10,20,30,40,50]\n",
+      "\n",
+      "for c in range(0,5):\n",
+      "    print \"The Value of a[%d]=%d\"%(c,a[c])\n",
+      "    print \"The address of a[%d]=%u\"%(c,id(b))\n",
+      "def value(b):\n",
+      "    b=''\n",
+      "    id(b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of a[0]=10\n",
+        "The address of a[0]=4296527248\n",
+        "The Value of a[1]=20\n",
+        "The address of a[1]=4296527248\n",
+        "The Value of a[2]=30\n",
+        "The address of a[2]=4296527248\n",
+        "The Value of a[3]=40\n",
+        "The address of a[3]=4296527248\n",
+        "The Value of a[4]=50\n",
+        "The address of a[4]=4296527248\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example:3, Page No,: 5.120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to add the sum of number using pointer\n",
+      "\n",
+      "a=[10,20,30,40,50]\n",
+      "\n",
+      "b=total=0\n",
+      "c=id(c)\n",
+      "\n",
+      "for b in range(0,5):\n",
+      "    print \"Enter the number %d: \"%(b+1)\n",
+      "    a[b]=input()\n",
+      "    \n",
+      "c=a\n",
+      "for b in range(0,5):\n",
+      "    total=total+c[b]\n",
+      "    c[b]=c[b]+1\n",
+      "\n",
+      "print \"Total= %d\"%(total)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number 1: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number 2: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number 3: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "30\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number 4: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "40\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number 5: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "50\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total= 150\n"
+       ]
+      }
+     ],
+     "prompt_number": 67
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page No.: 5.121 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to sort a given number using pointer.\n",
+      "\n",
+      "arr = []\n",
+      "n = input(\"Enter the no. of elements: \")\n",
+      "for i in range(0,n):\n",
+      "    arr.append(0)\n",
+      "print \"Enter the element: \"    \n",
+      "for i in range(0,n):\n",
+      "    arr[i]=input()\n",
+      "print \"Before Sorting:\"    \n",
+      "for i in range(0,n):\n",
+      "    print \"%d\"%arr[i],\n",
+      "    \n",
+      "for i in range(0,n):\n",
+      "    for j in range(i,n):\n",
+      "        if arr[i]>arr[j]:\n",
+      "            t = arr[i]\n",
+      "            arr[i] = arr[j]\n",
+      "            arr[j] = t\n",
+      "print \"\\nAfter Sorting:\"\n",
+      "for i in range(0,n):\n",
+      "    print \"%d\" %arr[i],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the no. of elements: 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Before Sorting:\n",
+        "8 5 7 4 6 \n",
+        "After Sorting:\n",
+        "4 5 6 7 8\n"
+       ]
+      }
+     ],
+     "prompt_number": 72
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter17.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter17.ipynb
new file mode 100755
index 00000000..7967d054
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter17.ipynb
@@ -0,0 +1,310 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:86b94093ee72d30b0a9443a44a1c3905ef6186294337253a1ca56b0318def382"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17 : Pointers with Multidimensional Array"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.:5.123"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print the values and addres of the array elements.\n",
+      "\n",
+      "arr=([[5,100],[10,200],[15,300],[20,400]])\n",
+      "\n",
+      "for a in range(0,4):\n",
+      "    print \"Address of %d array= %u\"%(a,id(arr[a]))\n",
+      "    for b in range(0,2):\n",
+      "        print \"Value =%d\"%(arr[a][b])\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Address of 0 array= 4336406328\n",
+        "Value =5\n",
+        "Value =100\n",
+        "Address of 1 array= 4336414944\n",
+        "Value =10\n",
+        "Value =200\n",
+        "Address of 2 array= 4336414800\n",
+        "Value =15\n",
+        "Value =300\n",
+        "Address of 3 array= 4366918504\n",
+        "Value =20\n",
+        "Value =400\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.: 5.125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print the value and address of the element using array of pointers\n",
+      "\n",
+      "a=[0,1,2]\n",
+      "b=10\n",
+      "c=20\n",
+      "d=30\n",
+      "a[0]=b\n",
+      "a[1]=c\n",
+      "a[2]=d\n",
+      "for i in range(0,3):\n",
+      "    print \"Address = %u\"%id(a[i])\n",
+      "    print \"Value = %d\" % (a[i])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Address = 4298181184\n",
+        "Value = 10\n",
+        "Address = 4298180944\n",
+        "Value = 20\n",
+        "Address = 4298180704\n",
+        "Value = 30\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No.: 5.126 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to sort a list of strings in alphabetical order using array of pointers.\n",
+      "\n",
+      "a = []\n",
+      "n= input(\"Enter the number of strings: \")\n",
+      "print \"Enter each string on a separate line below.\"\n",
+      "for i in range(0,n):\n",
+      "    a.append(0)\n",
+      "    \n",
+      "for i in range(0,n):\n",
+      "    a[i]=raw_input(\"String %d: \" %(i+1))\n",
+      "for i in range(0,n):\n",
+      "    for j in range(i,n):\n",
+      "        if a[i]>a[j]:\n",
+      "            t = a[i]\n",
+      "            a[i] = a[j]\n",
+      "            a[j] = t\n",
+      "print \"\\nReordered List of Strings: \"\n",
+      "for i in range(0,n):\n",
+      "    print \"String %d: %s\" %(i+1,a[i])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of strings: 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter each string on a separate line below.\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 1: venkat\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 2: ramesh\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 3: babu\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 4: muni\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 5: shankar\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 6: saravanan\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Reordered List of Strings: \n",
+        "String 1: babu\n",
+        "String 2: muni\n",
+        "String 3: ramesh\n",
+        "String 4: saravanan\n",
+        "String 5: shankar\n",
+        "String 6: venkat\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page No.: 5.128 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to count number of words\n",
+      "\n",
+      "str=[]\n",
+      "ps=raw_input(\"Enter a string: \")\n",
+      "print len(ps.split()), \"word(s) in the given string\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a string: I am a student.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4 word(s) in the given string\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 5, Page No.: 5.129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate on strings and pointers\n",
+      "\n",
+      "s1=\"abcd\"\n",
+      "s2='efgh'\n",
+      "\n",
+      "print \"%s %16lu \\n\" % (s1,id(s1))\n",
+      "print \"%s %16lu \\n\" % (s2,id(s2)) \n",
+      "\n",
+      "s1=s2\n",
+      "print \"%s %16lu \\n\" % (s1,id(s1))\n",
+      "print \"%s %16lu \\n\" % (s2,id(s2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "abcd       4363408224 \n",
+        "\n",
+        "efgh       4363332704 \n",
+        "\n",
+        "efgh       4363332704 \n",
+        "\n",
+        "efgh       4363332704 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter18.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter18.ipynb
new file mode 100755
index 00000000..c422da50
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter18.ipynb
@@ -0,0 +1,490 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0724a7e756186f06943f5403b91daf5f178ee495f014e11c1fad0330aa0d254c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18 : Pointers and Structures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.: 5.130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print details of person using structure and pointer\n",
+      "\n",
+      "class account:\n",
+      "    name=\"\"\n",
+      "    place=\"\"\n",
+      "    acctno=0\n",
+      "    \n",
+      "m4=account()\n",
+      "m4.name=\"Venkat\"\n",
+      "m4.place=\"Arcot\"\n",
+      "m4.acctno=24201\n",
+      "\n",
+      "print \"%s\\t%s\\t%d\\n\"%(m4.name,m4.place,m4.acctno)\n",
+      "\n",
+      "contents=m4\n",
+      "\n",
+      "print \"%s\\t%s\\t%d\\n\"%(contents.name,contents.place,contents.acctno)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Venkat\tArcot\t24201\n",
+        "\n",
+        "Venkat\tArcot\t24201\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No.: 5.131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print name, place, and account number using structure and pointer\n",
+      "\n",
+      "class account:\n",
+      "    name=\"\"\n",
+      "    place=\"\"\n",
+      "    acctno=0\n",
+      "    \n",
+      "m4=account()\n",
+      "m4.name=\"Madhavan\"\n",
+      "m4.place=\"Chennai\"\n",
+      "m4.acctno=241\n",
+      "a=m4\n",
+      "\n",
+      "print \"%s\\n%s\\n%d\\n\"%(a.name,a.place,a.acctno)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Madhavan\n",
+        "Chennai\n",
+        "241\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No.: 5.132"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to show how the int_pointers structure can be handled.\n",
+      "\n",
+      "\n",
+      "class int_pointers:\n",
+      "    p1=0\n",
+      "    p2=0\n",
+      "    \n",
+      "pointers=int_pointers()\n",
+      "m1= 80\n",
+      "pointers.p1=m1\n",
+      "pointers.p2=-40\n",
+      "m2=pointers.p2\n",
+      "print \"m1= %i, *pointers.p1 = %i\\n\"% (m1,pointers.p1)\n",
+      "print \"m2= %i, *pointers.p2 = %i\\n\"% (m2,pointers.p2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "m1= 80, *pointers.p1 = 80\n",
+        "\n",
+        "m2= -40, *pointers.p2 = -40\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 1: Page No.5.133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to interchange the values stored in two variables using function with pointers. \n",
+      "\n",
+      "def interchange(a,b):\n",
+      "    t=a\n",
+      "    a=b\n",
+      "    b=t\n",
+      "    return b,a\n",
+      "    \n",
+      "a=39\n",
+      "b=77\n",
+      "\n",
+      "print \"Before Interchange: a=%d, b=%d\"%(a,b)\n",
+      "\n",
+      "\n",
+      "print \"After Interchange : a=%d, b=%d\"%(interchange(b,a))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Before Interchange: a=39, b=77\n",
+        "After Interchange : a=77, b=39\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study: 2, Page No.:5.134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate the use of structure pointers\n",
+      "\n",
+      "class book:\n",
+      "    bno=0\n",
+      "    bname=''\n",
+      "    cost=0.0\n",
+      "\n",
+      "p=[book()for i in range(0,3)]\n",
+      "print \"Enter the Book details... \"\n",
+      "print \"Enter Book number, Book Name, Book Costs... \"\n",
+      "for i in range(0,3):\n",
+      "    p[i].bno=input()\n",
+      "    p[i].bname=raw_input()\n",
+      "    p[i].cost=input()\n",
+      "    \n",
+      "for i in range(0,3):\n",
+      "   \n",
+      "    print \"\\n%d\\t%s\\t\\t%d\"%(p[i].bno,p[i].bname,p[i].cost)\n",
+      "   \n",
+      "print \"\\nStructure Output\\n\"\n",
+      "ptr=p\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "     print \"%d\\t%s\\t\\t%d\"%(ptr[i].bno,ptr[i].bname,ptr[i].cost)\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Book details... \n",
+        "Enter Book number, Book Name, Book Costs... \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "101\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Computer Practice_1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "240\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "201\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Computer Practice_2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "250\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "202\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C++ and JAVA\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "280\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "101\tComputer Practice_1\t240\n",
+        "\n",
+        "201\tComputer Practice_2\t250\n",
+        "\n",
+        "202\tC++ and JAVA\t280\n",
+        "\n",
+        "Structure Output\n",
+        "\n",
+        "101\tComputer Practice_1\t240\n",
+        "201\tComputer Practice_2\t250\n",
+        "202\tC++ and JAVA\t280\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 3: Page No: 5.136"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to count number of vowels,consonants, digits,spaces and other characters in a line of text.\n",
+      "\n",
+      "vowels='aeiou'\n",
+      "consonants='bcdfghjklmnpqrstvwxyz'\n",
+      "spaces =' '\n",
+      "digits='1234567890'\n",
+      "special_characters='!@#$%^&*()_'\n",
+      "\n",
+      "text=raw_input(\"Enter the text in lower case: \")\n",
+      "\n",
+      "print \"\\nNumber of vowels: \",dict((v, text.count(v)) for v in text if v in vowels)\n",
+      "print \"Number of consonants: \",dict((c, text.count(c)) for c in text if c in consonants)\n",
+      "print \"Number of Spaces: \",dict((s, text.count(s)) for s in text if s in spaces)\n",
+      "print \"Number of digits: \",dict((d, text.count(d)) for d in text if d in digits)\n",
+      "print \"Number of special_characters: \",dict((sc, text.count(sc)) for sc in text if sc in special_characters)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the text in lower case: 123 vrb publishers @#$%^&*()\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number of vowels:  {'i': 1, 'e': 1, 'u': 1}\n",
+        "Number of consonants:  {'b': 2, 'h': 1, 'l': 1, 'p': 1, 's': 2, 'r': 2, 'v': 1}\n",
+        "Number of Spaces:  {' ': 3}\n",
+        "Number of digits:  {'1': 1, '3': 1, '2': 1}\n",
+        "Number of special_characters:  {'@': 1, '#': 1, '%': 1, '$': 1, '&': 1, ')': 1, '(': 1, '*': 1, '^': 1}\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 4: Page No.: 5.136"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate malloc() function\n",
+      "\n",
+      "NULL=0\n",
+      "string = ''\n",
+      "if string==NULL:\n",
+      "    print \"Not Enough memory to allocate buffer\\n\"\n",
+      "else:\n",
+      "    \n",
+      "    def strcpy(cstring1):\n",
+      "          import copy\n",
+      "          cstring2=copy.copy(cstring1)\n",
+      "          return cstring2  \n",
+      "string=\"Hello\"\n",
+      "print \"String is %s\" %string"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String is Hello\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 5: Page No.: 5.137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate calloc() function\n",
+      "\n",
+      "string=''\n",
+      "\n",
+      "def strcpy(cstring1):\n",
+      "          import copy\n",
+      "          cstring2=copy.copy(cstring1)\n",
+      "          return cstring2  \n",
+      "stri=\"BHAVANA\"\n",
+      "string=strcpy(stri)\n",
+      "print \"String is %s\" %string"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String is BHAVANA\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Case Study 6: Page No.:5.138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate realloc() function\n",
+      "#Cannot do in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter19.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter19.ipynb
new file mode 100755
index 00000000..bbaf1a5b
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter19.ipynb
@@ -0,0 +1,126 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:069422c07d34cc73268583e26cec775bbe1b3930cb154b474b2ff1e109dbe593"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 19 : Command Line Arguments"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.:5.139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import sys\n",
+      "argc=0\n",
+      "if argc!=2:\n",
+      "    print \"Hellow %s\"%(sys.argv)\n",
+      "else:\n",
+      "    print \"Forgotted to type name\"\n",
+      "    exit(0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hellow ['-c', '-f', '/home/gayathri/.config/ipython/profile_default/security/kernel-f732916e-2a20-4211-8210-1c4fa7115bd5.json', \"--IPKernelApp.parent_appname='ipython-notebook'\", '--profile-dir', '/home/gayathri/.config/ipython/profile_default', '--parent=1']\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No. 5.139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "argc='0'\n",
+      "if argc==2:\n",
+      "    print \"Argument supplied is %s\"%(sys.argv)\n",
+      "elif argc>2:\n",
+      "    print \"Too many arguments supplied \\n\",(sys.argv)\n",
+      "else:\n",
+      "    print \"One Argument expected\",(sys.argv)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Too many arguments supplied \n",
+        "['-c', '-f', '/home/gayathri/.config/ipython/profile_default/security/kernel-f732916e-2a20-4211-8210-1c4fa7115bd5.json', \"--IPKernelApp.parent_appname='ipython-notebook'\", '--profile-dir', '/home/gayathri/.config/ipython/profile_default', '--parent=1']\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No. :5.140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "\n",
+      "print \"The Arguments to the program are: \"\n",
+      "print str(sys.argv)\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Arguments to the program are: \n",
+        "['-c', '-f', '/home/gayathri/.config/ipython/profile_default/security/kernel-f732916e-2a20-4211-8210-1c4fa7115bd5.json', \"--IPKernelApp.parent_appname='ipython-notebook'\", '--profile-dir', '/home/gayathri/.config/ipython/profile_default', '--parent=1']\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter20.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter20.ipynb
new file mode 100755
index 00000000..adaadc6f
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter20.ipynb
@@ -0,0 +1,61 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7f2d14f2c2ae43d8abc01634eb7d39800f8596aa61628d10404d3502e1e30c64"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 20 : Dynamic Memory Allocation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No. 5.146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Reallocation of memory\n",
+      "\n",
+      "p= \"MADRAS\"\n",
+      "\n",
+      "print \"Memory contains: %s\\n\"%(p)            #There is no reallocation memory in python\n",
+      "\n",
+      "p=\"CHENNAI\"\n",
+      "\n",
+      "print \"Memory now contains: %s\\n\"%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Memory contains: MADRAS\n",
+        "\n",
+        "Memory now contains: CHENNAI\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter21.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter21.ipynb
new file mode 100755
index 00000000..2b76d2a2
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter21.ipynb
@@ -0,0 +1,174 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2bdca20a49e1bad43f6968c764966fb35bc4580bd54570e770e266797cc3ab0e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 21 : The C Preprocessor"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 1, Page No.5.149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print square and cube Values by using macro substitutions. \n",
+      "def sq(n):\n",
+      "    return n*n\n",
+      "    \n",
+      "    \n",
+      "def cube(n):\n",
+      "    return n*n*n\n",
+      "    \n",
+      "\n",
+      "a=5\n",
+      "b=3\n",
+      "s=c=0\n",
+      "\n",
+      "#Function calling\n",
+      "s=sq(a)\n",
+      "c=cube(b)\n",
+      "\n",
+      "#Result\n",
+      "print \"Square value of 5 is %d\"%(s)\n",
+      "print \"Cube value of 3 is %d\"%(c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Square value of 5 is 25\n",
+        "Cube value of 3 is 27\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 2, Page No. 5.151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the Values using macro substitution.\n",
+      "\n",
+      "VAL=40\n",
+      "\n",
+      "print \"The value is %d\"%VAL\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value is 40\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 3, Page No.:5.151\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print Value and text using preprocessing.\n",
+      "VAL=35\n",
+      "\n",
+      "HELLO=\"HELLO\"\n",
+      "\n",
+      "res=VAL-5\n",
+      "\n",
+      "print \"Res=VAL-5:Res= %d\"%res\n",
+      "\n",
+      "print \"HELLO\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Res=VAL-5:Res= 30\n",
+        "HELLO\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example: 4, Page No.:5.152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to print the value using preprocessors.\n",
+      "USD=1\n",
+      "UKP=1\n",
+      "\n",
+      "currency_rate=46\n",
+      "currency_rate=100\n",
+      "\n",
+      "rs=10*currency_rate\n",
+      "print  \"The Value of rs is : %d\"%rs\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Value of rs is : 1000\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter22.ipynb b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter22.ipynb
new file mode 100755
index 00000000..53a104bf
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/Chapter22.ipynb
@@ -0,0 +1,2300 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:dd103cfe67b47eb61e56816517923337936d461ec4fd4b8f9f87e6c544fd31e3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 22 : Sample C Programs(Lab Exercise)"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 1: Page no.:L.4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Area of Triangle\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#User Inputs\n",
+      "a,b,c=input(\"Enter three sides: \")\n",
+      "\n",
+      "#calculation of semi perimeter\n",
+      "s=(a+b+c)/2\n",
+      "\n",
+      "#Calculation of Area of Triangle\n",
+      "d=(s*(s-a)*(s-b)*(s-c))\n",
+      "\n",
+      "area=math.sqrt(d)\n",
+      "\n",
+      "#Result\n",
+      "print \"Area of triangle = %f sq units \\n\"%area"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter three sides: 5,6,7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area of triangle = 14.696938 sq units \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 2: Page No. L.5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Square root for the given value.\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#User Input\n",
+      "x=input(\"Enter the Value: \")\n",
+      "\n",
+      "#Result\n",
+      "print \"Square root of %.2f is: \"%(x)\n",
+      "\n",
+      "print math.sqrt(x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Value: 81\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Square root of 81.00 is: \n",
+        "9.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 3: Page.No.L.6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sum of the series 0+1+(1+2)+...+n\n",
+      "\n",
+      "#User Input\n",
+      "n=input(\"Enter the Value of n : \")\n",
+      "\n",
+      "#Variable Declaration\n",
+      "s=term=0\n",
+      "i=j=0\n",
+      "\n",
+      "#Computing Values in loop\n",
+      "for i in range(n):\n",
+      "    for j in range(n):\n",
+      "        if j<=i:\n",
+      "            term=term+j\n",
+      "            j=j+1        \n",
+      "s=s+term\n",
+      "if i<=n:\n",
+      "    i=i+1\n",
+      "    \n",
+      "#Result\n",
+      "print \"Sum of the series S= %d\"%(s)\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Value of n :5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of the series S= 20\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 4: Page No.:L.9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Decimal into binary number.\n",
+      "\n",
+      "n=input(\"Enter the decimal number: \")\n",
+      "s=n\n",
+      "k=[]\n",
+      "\n",
+      "#Calculation\n",
+      "while (s>0):\n",
+      "    a=int(float(s%2))\n",
+      "    k.append(a)\n",
+      "    s=(s-a)/2\n",
+      "k.append(0)\n",
+      "bi=\"\"\n",
+      "\n",
+      "for j in k[::1]:\n",
+      "    bi=bi+str(j)\n",
+      "\n",
+      "#Result\n",
+      "print('The binary value of %d is %s'%(n, bi))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the decimal number: 14\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The binary value of 14 is 01110\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 5: Page No. L.10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#The given number is armstrong or not.\n",
+      "\n",
+      "#Variable declaration\n",
+      "a=b=c=d=e=0\n",
+      "\n",
+      "#User Input\n",
+      "a=input(\"Enter the Number: \")\n",
+      "\n",
+      "#Assigning variable 'a' value to variable 'e'\n",
+      "e=a\n",
+      "\n",
+      "#Calculation for finding Armstrong or not\n",
+      "while a>0:\n",
+      "    b=a%10\n",
+      "    c=b*b*b\n",
+      "    a=a/10\n",
+      "    d=c+d\n",
+      " \n",
+      "#Result\n",
+      "if e==d:\n",
+      "    print \"%d is an armstrong number. \"%e\n",
+      "    \n",
+      "else:\n",
+      "    print \"%d is not an armstrong number. \"%e"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Number: 153\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "153 is an armstrong number. \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 6: Page No.L.12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Centigrade and Farenheit Values.\n",
+      "\n",
+      "\n",
+      "#Variable Declaration\n",
+      "c=f=cn=fn=0.0\n",
+      "\n",
+      "#User Input1 for calculating farenheit\n",
+      "c=input(\"Enter temperature in centigrade : \")\n",
+      "\n",
+      "#Farenheit Calculation\n",
+      "f=1.8*c+32\n",
+      "\n",
+      "#Result1\n",
+      "print \"Farenheit Equivalent is: %.1f\\n\" %(f)\n",
+      "\n",
+      "#User input2 for calculating centigrade\n",
+      "fn=input(\"Enter temperature in farenheit : \")\n",
+      "\n",
+      "#Centigrade Calculation\n",
+      "cn=(fn-32)/1.8\n",
+      "\n",
+      "#Result2\n",
+      "print \"Centigrade Equivalent is : %.1f\\n\"%(cn)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter temperature in centigrade : 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Farenheit Equivalent is: 68.0\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter temperature in farenheit : 68\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Centigrade Equivalent is : 20.0\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 7: Page .no.L.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Converting Decimal value to binary number\n",
+      "\n",
+      "\n",
+      "#Variable declaration\n",
+      "bnum=digit=decimal=bina=base=0\n",
+      "\n",
+      "#User input\n",
+      "bnum=input(\"Enter the binary number: \")\n",
+      "\n",
+      "bina=bnum\n",
+      "\n",
+      "while(bnum!=0):\n",
+      "    digit=bnum%10\n",
+      "    decimal=decimal+(digit<<base)\n",
+      "    base=base+1\n",
+      "    bnum=bnum/10\n",
+      "    \n",
+      "print \"The binary equivalent of %d in decimal = %d\"%(bina,decimal)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the binary number: 1111\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The binary equivalent of 1111 in decimal = 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 8: Page no.L.15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Reverse number of the given number.\n",
+      "\n",
+      "#Function  Definition.\n",
+      "def rev(n):\n",
+      "    digit=rev_of=0\n",
+      "    while n!=0:\n",
+      "        digit=n%10\n",
+      "        rev_of=rev_of*10+digit\n",
+      "        n=n/10\n",
+      "    return rev_of\n",
+      "\n",
+      "#Variable Declaration\n",
+      "\n",
+      "\n",
+      "#User input\n",
+      "p=input(\"Enter the number: \")\n",
+      "\n",
+      "q=rev(p)\n",
+      "\n",
+      "print \"Reverse of %d is %d.\\n\"%(p,q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number: 1234\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reverse of 1234 is 4321.\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 9: Page no.L.17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Square of given value\n",
+      "\n",
+      "#Function definition\n",
+      "def square(y):\n",
+      "    return y*y\n",
+      "\n",
+      "#User input\n",
+      "n=input(\"Enter the nth element : \")\n",
+      "\n",
+      "#Calculation\n",
+      "for x in range(1,n+1):\n",
+      "     print \"%d \" %(square(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the nth element : 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 \n",
+        "4 \n",
+        "9 \n",
+        "16 \n",
+        "25 \n",
+        "36 \n",
+        "49 \n",
+        "64 \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 10: Page no.:L.18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Maximum value of given numbers\n",
+      "\n",
+      "#Function definition\n",
+      "def maximum(x,y,z):\n",
+      "    \n",
+      "    maxi=x\n",
+      "    if y>maxi:\n",
+      "        maxi=y\n",
+      "        if z>maxi:\n",
+      "            maxi=z\n",
+      "    return maxi\n",
+      "    \n",
+      "    \n",
+      "#User input\n",
+      "a,b,c=input(\"Enter three integers : \")\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum is : %d\\n\"%(maximum(a,b,c))\n",
+      "\n",
+      "\n",
+      "    \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter three integers : 2,6,9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum is : 9\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 11: Page no.L.19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Factorial of the given number using recursion.\n",
+      "\n",
+      "#funtion declaration\n",
+      "def fact(n):\n",
+      "    if n==0:\n",
+      "        return 1\n",
+      "    else:\n",
+      "        return(n*fact(n-1))\n",
+      "    \n",
+      "#USer Input\n",
+      "\n",
+      "n=input(\"Enter the number whose factorial is to be found: \")\n",
+      "\n",
+      "#Result\n",
+      "print \"The Factorial of %d is : %d \\n\"%(n,fact(n))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number whose factorial is to be found: 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Factorial of 6 is : 720 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 12: Page no.L.20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Palindrome or not.\n",
+      "\n",
+      "print \"Enter the String to check palindrome or not \\n\"\n",
+      "\n",
+      "#User Input\n",
+      "\n",
+      "string = raw_input(\"Enter the String: \")\n",
+      "\n",
+      "#reverse the string \n",
+      "\n",
+      "rev_string = reversed(string)\n",
+      "\n",
+      "#Check if the string is palindrome or not\n",
+      "\n",
+      "if list(rev_string) == list(string) :\n",
+      "    print \"\\nThe given string %s is a palindrome\"%(string)\n",
+      "    \n",
+      "else:\n",
+      "    print \"\\nThe given string %s is not a palindrome\"%(string)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the String to check palindrome or not \n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the String: madam\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The given string madam is a palindrome\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 13: Page No. L.24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Largest and smallest number in array\n",
+      "\n",
+      "#User input\n",
+      "n= input(\"Enter the nth term of an array: \")\n",
+      "\n",
+      "for i in range(0,n):\n",
+      "    a[i]=eval(raw_input())\n",
+      "\n",
+      "#check if the given array is large or small.    \n",
+      "    for i in range(1,n):\n",
+      "        if a[i]>large:\n",
+      "           large=a[i]\n",
+      "            \n",
+      "        else:\n",
+      "           if a[i]<small:\n",
+      "              small=a[i]\n",
+      "       \n",
+      "         \n",
+      "print \"Largest element in an array is: %.2f\"%large\n",
+      "print \"Smallest element in an array is: %.2f\"%small\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the nth term of an array: 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "79\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-21\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Largest element in an array is: 79.00\n",
+        "Smallest element in an array is:-21.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 14: Page no. L.26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given array in ascending order\n",
+      "\n",
+      "#User Input\n",
+      "n=input(\"Enter the last term of the array: \")\n",
+      "\n",
+      "for i in range(0,n):\n",
+      "    a[i]=eval(raw_input())\n",
+      "for i in range(0,n-1):\n",
+      "\n",
+      "    for j in range(i+1,n):\n",
+      "        if a[i]>a[j]:\n",
+      "            temp=a[i]\n",
+      "            a[i]=a[j]\n",
+      "            a[j]=temp\n",
+      "        \n",
+      "print \"\\nThe Elements in ascending order\\n\"\n",
+      "\n",
+      "for i in range(0,n):\n",
+      "    print \"%.2f\"%a[i]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last term of the array: 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "23\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "14\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-67\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "48\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "01\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Elements in ascending order\n",
+        "\n",
+        "-90.00\n",
+        "-67.00\n",
+        "1.00\n",
+        "14.00\n",
+        "23.00\n",
+        "48.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 15: Page No. L.27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Insert element in specified position in an array\n",
+      "\n",
+      "import array\n",
+      "\n",
+      "#User input\n",
+      "n=input(\"Enter the last value: \")\n",
+      "a=[0 for i in range(0,n) ]\n",
+      "print \"Enter the elements of array: \"\n",
+      "for i in range(0,n):\n",
+      "    a[i]=eval(raw_input())\n",
+      "    \n",
+      "#Getting inserting element and position from users\n",
+      "item=input(\"Enter the element to be inserted: \")\n",
+      "pos=input(\"Enter the position of insertion: \")\n",
+      "a.insert(pos,item)\n",
+      "#Result\n",
+      "print \"\\nElements after insertion is : \"\n",
+      "for i in range(0,n+1):\n",
+      "    print \"%.2f\"%a[i]\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last value: 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the elements of array: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element to be inserted: 36\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the position of insertion: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Elements after insertion is : \n",
+        "12.00\n",
+        "34.00\n",
+        "56.00\n",
+        "90.00\n",
+        "36.00\n",
+        "32.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 16: Page.no.L.29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Transpose of the matrix\n",
+      "\n",
+      "#User Input\n",
+      "\n",
+      "a=[]\n",
+      "\n",
+      "n,m=input(\"Enter the Row and Columns: \")\n",
+      "print \"Enter the elements of matrix:\"\n",
+      "for i in range(n):\n",
+      "    a.append([])\n",
+      "    \n",
+      "    for j in range(m):\n",
+      "        element=input()\n",
+      "        a[i].append(element)\n",
+      "\n",
+      "#print matrix\n",
+      "print \"\\nElements in matrix: \"\n",
+      "for i in range(n):\n",
+      "    for j in range(m):\n",
+      "        print a[i][j],\n",
+      "    print '\\n'\n",
+      "\n",
+      "\n",
+      "#generate transpose\n",
+      "transpose=[]\n",
+      "for j in range(m):\n",
+      "    transpose.append([])\n",
+      "    for i in range (n):\n",
+      "        b=a[i][j]\n",
+      "        transpose[j].append(b)\n",
+      "\n",
+      "#print transpose\n",
+      "print \"Transpose of the matrix is: \"\n",
+      "for i in range (m):\n",
+      "    for j in range (n):\n",
+      "        print transpose[i][j],\n",
+      "    print '\\n'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Row and Columns: 3,3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the elements of matrix:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Elements in matrix: \n",
+        "1 2 3 \n",
+        "\n",
+        "4 5 6 \n",
+        "\n",
+        "7 8 9 \n",
+        "\n",
+        "Transpose of the matrix is: \n",
+        "1 4 7 \n",
+        "\n",
+        "2 5 8 \n",
+        "\n",
+        "3 6 9 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 17: Page no. L.32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to find sum of an array matrix\n",
+      "#Array Declaration\n",
+      "\n",
+      "a=[]\n",
+      "b=[]\n",
+      "c=[[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
+      "#User Input1\n",
+      "n,m=input(\"Enter the Row and Columns of A Matrix: \")\n",
+      "print \"Enter the elements of A matrix:\"\n",
+      "for i in range(n):\n",
+      "    a.append([])\n",
+      "    \n",
+      "    for j in range(m):\n",
+      "        element=input()\n",
+      "        a[i].append(element)\n",
+      "\n",
+      "#print matrix1\n",
+      "print \"\\nElements in A matrix: \"\n",
+      "for i in range(n):\n",
+      "    for j in range(m):\n",
+      "        print a[i][j],\n",
+      "    print '\\n'\n",
+      "\n",
+      "    \n",
+      "#User Input 2\n",
+      "p,q=input(\"Enter the Row and Columns of B Matrix: \")\n",
+      "print \"Enter the elements of B matrix:\"\n",
+      "for i in range(p):\n",
+      "    b.append([])\n",
+      "    \n",
+      "    for j in range(q):\n",
+      "        element=input()\n",
+      "        b[i].append(element)\n",
+      "\n",
+      "#print matrix 2\n",
+      "print \"\\nElements in B matrix: \"\n",
+      "for i in range(p):\n",
+      "    for j in range(q):\n",
+      "        print b[i][j],\n",
+      "    print '\\n'\n",
+      "\n",
+      "#Addition of array in matrix    \n",
+      "for i in range(n):\n",
+      "    for j in range(m):\n",
+      "        c[i][j] += a[i][j] + b[i][j]\n",
+      "\n",
+      "#Result\n",
+      "print \"\\nThe Sum of A and B matrix: \"\n",
+      "for i in range(n):\n",
+      "    for j in range(m):\n",
+      "        print \"%5d\"%c[i][j],\n",
+      "    print '\\n'"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Row and Columns of A Matrix: 3,3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the elements of A matrix:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Elements in A matrix: \n",
+        "1 2 3 \n",
+        "\n",
+        "4 5 6 \n",
+        "\n",
+        "7 8 9 \n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Row and Columns of B Matrix: 3,3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the elements of B matrix:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Elements in B matrix: \n",
+        "1 2 3 \n",
+        "\n",
+        "4 5 6 \n",
+        "\n",
+        "6 7 4 \n",
+        "\n",
+        "\n",
+        "The Sum of A and B matrix: \n",
+        "    2     4     6 \n",
+        "\n",
+        "    8    10    12 \n",
+        "\n",
+        "   13    15    13 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 80
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 18: Page No. :L.35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to find Matrix Multiplication \n",
+      "#Array Declaration\n",
+      "\n",
+      "a=[]\n",
+      "b=[]\n",
+      "c=[[0 for i in xrange(0,5)] for i in xrange(0,5)]\n",
+      "\n",
+      "while True:\n",
+      "#User Input1\n",
+      "    n,m=input(\"Enter the Row and Columns of A Matrix: \")\n",
+      "    p,q=input(\"Enter the Row and Columns of B Matrix: \")\n",
+      "    if (n==m&p==q):\n",
+      "        print \"Enter the elements of A matrix:\"   \n",
+      "        for i in range(n):\n",
+      "            a.append([]) \n",
+      "            for j in range(m):\n",
+      "                element=input()\n",
+      "                a[i].append(element)\n",
+      "#print matrix1\n",
+      "        print \"\\nElements in A matrix: \"\n",
+      "        for i in range(n):\n",
+      "            for j in range(m):\n",
+      "                print a[i][j],\n",
+      "            print '\\n'\n",
+      "   \n",
+      "#User Input 2    \n",
+      "        print \"Enter the elements of B matrix:\"\n",
+      "        for i in range(p):\n",
+      "            b.append([])\n",
+      "            for j in range(q):\n",
+      "                element=input()\n",
+      "                b[i].append(element)\n",
+      "\n",
+      "#print matrix 2\n",
+      "        print \"\\nElements in B matrix: \"\n",
+      "        for i in range(p):\n",
+      "            for j in range(q):\n",
+      "                print b[i][j],\n",
+      "            print '\\n'\n",
+      "\n",
+      "#Multiplication of array in matrix    \n",
+      "        for i in range(0,m):\n",
+      "            for j in range(0,n):\n",
+      "                c[i][j] = 0\n",
+      "                for k in range(0,n):\n",
+      "                    c[i][j] += a[i][k] * b[k][j]\n",
+      "        \n",
+      "#Result\n",
+      "        print \"\\nThe Multiplication of A and B matrix: \"\n",
+      "        for i in range(n):\n",
+      "            for j in range(m):\n",
+      "                print \"%5d\"%c[i][j],\n",
+      "            print '\\n'\n",
+      "        break\n",
+      "    else:\n",
+      "        print \"The No. of rows and columns are not equal.\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Row and Columns of A Matrix: 3,3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Row and Columns of B Matrix: 2,2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The No. of rows and columns are not equal.\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Row and Columns of A Matrix: 3,3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Row and Columns of B Matrix: 3,3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the elements of A matrix:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Elements in A matrix: \n",
+        "1 2 3 \n",
+        "\n",
+        "4 5 6 \n",
+        "\n",
+        "7 8 9 \n",
+        "\n",
+        "Enter the elements of B matrix:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Elements in B matrix: \n",
+        "2 4 1 \n",
+        "\n",
+        "6 7 4 \n",
+        "\n",
+        "3 5 7 \n",
+        "\n",
+        "\n",
+        "The Multiplication of A and B matrix: \n",
+        "   23    33    30 \n",
+        "\n",
+        "   56    81    66 \n",
+        "\n",
+        "   89   129   102 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 93
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 19: Page No. L.38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Student grade using structure\n",
+      "\n",
+      "#User Input\n",
+      "n=input(\"Enter the number of students: \")\n",
+      "\n",
+      "#Defining Structure\n",
+      "class std:\n",
+      "    no=0\n",
+      "    name = ''\n",
+      "    m1 = m2 = m3 =0\n",
+      "    total=avg=0.0\n",
+      "    grade= ''\n",
+      "    \n",
+      "s=[std() for i in range(0,n)]\n",
+      "\n",
+      "#User Input\n",
+      "for i in range(0,n):\n",
+      "    print \"\\nEnter the student %d details\"%(i+1)\n",
+      "    print \"\\n\"\n",
+      "    s[i].no=int(input(\"Give Student Number: \"))\n",
+      "    s[i].name=raw_input(\"Enter the Student Name: \")\n",
+      "    s[i].m1=int(input(\"Enter Student Mark1: \"))\n",
+      "    s[i].m2=int(input(\"Enter Student Mark2: \"))\n",
+      "    s[i].m3=int(input(\"Enter Student Mark3: \"))\n",
+      "\n",
+      "\n",
+      "#Calculation    \n",
+      "    s[i].total = (s[i].m1 + s[i].m2 + s[i].m3)\n",
+      "    s[i].avg=(s[i].total / 3)\n",
+      "    if s[i].avg<=40:\n",
+      "        s[i].grade='D'\n",
+      "    elif s[i].avg<60:\n",
+      "        s[i].grade='C'\n",
+      "    elif s[i].avg<80:\n",
+      "        s[i].grade='B'\n",
+      "    else:\n",
+      "        s[i].grade='A'\n",
+      "        \n",
+      "#Result\n",
+      "print \"\\tStudent mark details.... \\n\"\n",
+      "print \"Roll No\\t Name \\t\\tMark1 \\tMark2 \\tMark3 \\tTotal \\tAverage \\tGrade\"\n",
+      "for i in range(0,n):\n",
+      "    print \"%d\\t\"%(s[i].no),\"%s\\t\\t\"%(s[i].name), \"%d\\t\"%(s[i].m1),\"%d\\t\"%(s[i].m2),\"%d\\t\"%(s[i].m3), \"%.2f\\t\"%(s[i].total), \"%.2f\\t\\t\"%(s[i].avg), \"%s\\t\"%(s[i].grade)\n",
+      "   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of students: 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter the student 1 details\n",
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give Student Number: 100\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Student Name: Venkat\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark1: 89\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark2: 87\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark3: 76\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter the student 2 details\n",
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give Student Number: 101\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Student Name: Shankar\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark1: 87\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark2: 90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Student Mark3: 67\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tStudent mark details.... \n",
+        "\n",
+        "Roll No\t Name \t\tMark1 \tMark2 \tMark3 \tTotal \tAverage \tGrade\n",
+        "100\tVenkat\t\t89\t87\t76\t252.00\t84.00\t\tA\t\n",
+        "101\tShankar\t\t87\t90\t67\t244.00\t81.00\t\tA\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 20: Page No. L.40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Employee details using Union\n",
+      "\n",
+      "#Defining Union\n",
+      "\n",
+      "class union:\n",
+      "    name=\"\"\n",
+      "    idno=0\n",
+      "    salary=0.0\n",
+      "    \n",
+      "desc=union()\n",
+      "\n",
+      "desc.name=\"vinod\"\n",
+      "\n",
+      "print \"Employee details:\"\n",
+      "\n",
+      "print \"The name is %s\\n\"%(desc.name)\n",
+      "print \"The idno is %d\\n\"%(desc.idno)\n",
+      "print \"The salary is %6.2f\\n\"%(desc.salary)\n",
+      "\n",
+      "desc.idno=10\n",
+      "\n",
+      "print \"Employe details:\"\n",
+      "\n",
+      "print \"The name is %s\\n\"%(desc.name)\n",
+      "print \"The idno is %d\\n\"%(desc.idno)\n",
+      "print \"The salary is %6.2f\\n\"%(desc.salary)\n",
+      "\n",
+      "desc.salary=6500.00\n",
+      "\n",
+      "print \"Employee details:\"\n",
+      "\n",
+      "print \"The name is %s\\n\"%(desc.name)\n",
+      "print \"The idno is %d\\n\"%(desc.idno)\n",
+      "print \"The salary is %6.2f\\n\"%(desc.salary)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Employee details:\n",
+        "The name is vinod\n",
+        "\n",
+        "The idno is 0\n",
+        "\n",
+        "The salary is   0.00\n",
+        "\n",
+        "Employe details:\n",
+        "The name is vinod\n",
+        "\n",
+        "The idno is 10\n",
+        "\n",
+        "The salary is   0.00\n",
+        "\n",
+        "Employee details:\n",
+        "The name is vinod\n",
+        "\n",
+        "The idno is 10\n",
+        "\n",
+        "The salary is 6500.00\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 21: Page No. L.41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Swap the contents of two variable using pointer.\n",
+      "\n",
+      "#swapping function.\n",
+      "\n",
+      "def interchange(x,y):\n",
+      "    temp=x\n",
+      "    x=y\n",
+      "    y=temp\n",
+      "    return x,y\n",
+      "\n",
+      "#User Input\n",
+      "a,b = input(\"Enter the values of A and B: \")\n",
+      "#Result before interchanging\n",
+      "print \"a=%d b=%d\\n\"%(a,b)\n",
+      "\n",
+      "\n",
+      "#Result after interchanging\n",
+      "print \"After interchanging: \\n\"\n",
+      "\n",
+      "print \"a=%d b=%d\\n\"%(interchange(a,b))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the values of A and B: 23,56\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a=23 b=56\n",
+        "\n",
+        "After interchanging: \n",
+        "\n",
+        "a=56 b=23\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 22: Page No.: L.43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Concatenate two strings using Dynamic Memory Allocation \n",
+      "\n",
+      "s1= raw_input(\"Enter String One: \")\n",
+      "s2=raw_input(\"Enter String two: \")\n",
+      "\n",
+      "s3=''\n",
+      "s3=s1+s2                    \n",
+      "#Result\n",
+      "print \"Concatenated String: \", s1+s2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String One: venkat\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String two: raman\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Concatenated String:  venkatraman\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 23: Page No. L.44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#program to store and print same data to and from the file.\n",
+      "\n",
+      "#open file\n",
+      "fp=open(\"Text.txt\",\"w\")\n",
+      "\n",
+      "#User Input\n",
+      "c=raw_input(\"Enter the Text: \")\n",
+      "\n",
+      "#Store given input into a file \"text.txt\"\n",
+      "fp.write(c)\n",
+      "\n",
+      "#Closing file\n",
+      "fp.close()\n",
+      "\n",
+      "#Again open text.txt\n",
+      "fp=open(\"Text.txt\", \"r\")\n",
+      "\n",
+      "#display the content from the file \"text.txt\"\n",
+      "print \"Displaying text from file:\\n\",c \n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Text: VRB PUBLISHERS\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Displaying text from file:\n",
+        "VRB PUBLISHERS\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 24: Page No. L.46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to count number of characters in a file.\n",
+      "\n",
+      "#create a file\n",
+      "fp=open(\"str.txt\",\"w\")\n",
+      "c=\"\"\n",
+      "print \"Enter the Characters:\"\n",
+      "print \"Press ctrl+z to stop entry\"\n",
+      "\n",
+      "#User Input\n",
+      "c=raw_input()\n",
+      "#store input into the file\n",
+      "fp.write(c)\n",
+      "fp.close()\n",
+      "\n",
+      "#Again open the file to read\n",
+      "fp=open(\"str.txt\",\"r\")\n",
+      "\n",
+      "#Result\n",
+      "print \"\\n%s\"%(c)\n",
+      "print \"Number of characters in the file: \",len(c)\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Characters:\n",
+        "Press ctrl+z to stop entry\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "vrb publishers private limited\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "vrb publishers private limited\n",
+        "Number of characters in the file:  30\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exercise 25: Page No.: L.48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program using Memory Allocation function\n",
+      "\n",
+      "NULL=0\n",
+      "buffer_size=10\n",
+      "\n",
+      "buffer1 = \"BOMBAY\"\n",
+      "print \"Buffer of size %d created\"%(buffer_size)\n",
+      "print \"Buffer contains\", buffer1\n",
+      "    \n",
+      "\n",
+      "if buffer_size==NULL:\n",
+      "    print \"Malloc failed\"\n",
+      "    \n",
+      "else:\n",
+      "    print \"Buffer size modified\"\n",
+      "    print \"Buffer still contains:\", buffer1\n",
+      "    \n",
+      "buffer1=\"MUMBAI\"\n",
+      "print \"Buffer now contains:\", buffer1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Buffer of size 10 created\n",
+        "Buffer contains BOMBAY\n",
+        "Buffer size modified\n",
+        "Buffer still contains: BOMBAY\n",
+        "Buffer now contains: MUMBAI\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/README.txt b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/README.txt
new file mode 100755
index 00000000..9bed1b50
--- /dev/null
+++ b/Fundamental_of_Computing_and_Programming_in_C_by_V_Ramesh_Babu_R_Samyuktha_M_Munirathnam/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Lakshmidharan D
+Course: msc
+College/Institute/Organization: Pachaiyappa's College for Men, Kanchipuram
+Department/Designation: Computer Science
+Book Title: Fundamental of Computing and Programming in C
+Author: V. Ramesh Babu, R. Samyuktha, M. Munirathnam
+Publisher: Lakshmidharan D
+Year of publication: 2012
+Isbn: 9789381837214
+Edition: August 2012
\ No newline at end of file
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diff --git a/Fundamentals_Of_Thermodynamics/Chapter10.ipynb b/Fundamentals_Of_Thermodynamics/Chapter10.ipynb
deleted file mode 100755
index 59bbb80a..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter10.ipynb
+++ /dev/null
@@ -1,164 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:6310b6690e48041fd88dd8277353623ea538798dfeeeb9a3819409de065b2175"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter10:IRREVERSIBILITY AND AVAILABILITY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.1:pg-386"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#Calculating reversible work\n",
-      "\n",
-      "#Form the Steam Tables,the inlet and the exit state properties are \n",
-      "hi=171.95 #initial specific heat of enthalpy in kJ/kg\n",
-      "si=0.5705 #initial specific entropy in kJ/kg-K\n",
-      "se=2.1341 #final specific entropy in kJ/kg-K\n",
-      "he=765.34 #final specific heat of enthalpy in kJ/kg-K\n",
-      "m=5 #mass flow rate of feedwater in kg/s\n",
-      "q1=900/m #heat added by one of the sources in kJ/kg\n",
-      "q2=he-hi-q1 #second heat transfer in kJ/kg\n",
-      "To=25+273.3 #Temp. of the surroundings in K\n",
-      "T1=100+273.2 #temp. of reservoir of one of the source in K\n",
-      "T2=200+273.2 #temp. of reservoir of second source in K\n",
-      "wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) #reversible work in kJ/kg\n",
-      "print\"\\n Hence, the irreversibility is\",round(wrev,1),\"kJ/kg\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Hence, the irreversibility is 62.0 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.2:pg-387"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2\n",
-      "#Calculating reversible work\n",
-      "import math\n",
-      "#Form the Steam Tables,the inlet and the exit state properties are\n",
-      "hi=298.6 #initial specific heat of enthalpy in kJ/kg\n",
-      "si=6.8631 #initial specific entropy in kJ/kg-K\n",
-      "se=7.4664 #final specific entropy in kJ/kg-K\n",
-      "he=544.7 #final specific heat of enthalpy in kJ/kg-K\n",
-      "q=-50 #heat lost to surroundings in kJ/kg\n",
-      "w=hi-he+q #work in kJ/kg\n",
-      "To=25+273.2 #Temp. of the surroundings in K\n",
-      "P1=100 #Pressure of ambient air in kPa\n",
-      "P2=1000 #Final pressure of air after compression in kPa\n",
-      "R=0.287 #Universal gas constant in kJ/kg-K\n",
-      "wrev=To*(se-si-R*math.log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in kJ/kg\n",
-      "i=wrev-w #irreversibility in kJ/kg\n",
-      "print\"\\n Hence, the irreversibility is\",round(i,1),\"kJ/kg\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Hence, the irreversibility is 32.8 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.3:pg-390"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#Calculating reversible work and irreversibility\n",
-      "\n",
-      "#Form the Steam Tables at state 1\n",
-      "u1=1243.5 #initial specific internal energy in kJ/kg\n",
-      "s1=4.4819 #initial specific entropy in kJ/kg-K\n",
-      "v1=28.895 #initial specific volume in m^3/kg\n",
-      "v2=2*v1 #final specific volume in kg/m^3\n",
-      "u2=u1 #initial specific internal energy in kJ/kg\n",
-      "#These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg\n",
-      "T2=9.1+273.2 #final temp. in K\n",
-      "x2=0.513 #quality in final state\n",
-      "s2=4.644 #final specific entropy in kJ/kg\n",
-      "V1=1 #volume of part of A in m^3\n",
-      "m=V1/v1  #mass flow rate in kg/s\n",
-      "To=20+273.2 #Room temperature in K\n",
-      "Wrev=To*m*(s2-s1) #reversible work in kJ\n",
-      "I=Wrev #irreversibility of the process\n",
-      "print\"\\n The irreversibility is \",round(I,3),\"kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " The irreversibility is  1.645 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter11.ipynb b/Fundamentals_Of_Thermodynamics/Chapter11.ipynb
deleted file mode 100755
index f7ce9d5a..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter11.ipynb
+++ /dev/null
@@ -1,424 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:099777bd4d735e894f7949924f63dca34258026e1381262643494860b41717f6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter11:Power and Refrigeration Systems\u2014With Phase Change"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.1:Pg-425"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Ques 1\n",
-      "#To determine the efficiency of Rankine cycle\n",
-      "\n",
-      "#1-Inlet state of pump\n",
-      "#2-Exit state of pump\n",
-      "P2=2000;#Exit pressure in kPa\n",
-      "P1=10;#Inlet pressure in kPa\n",
-      "v=0.00101;#specific weight of water in m^3/kg\n",
-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",
-      "h1=191.8;#Enthalpy in kJ/kg from table\n",
-      "h2=h1+wp;#enthalpy in kJ/kg\n",
-      "#2-Inlet state for boiler\n",
-      "#3-Exit state for boiler\n",
-      "h3=2799.5;#Enthalpy in kJ/kg\n",
-      "#3-Inlet state for turbine\n",
-      "#4-Exit state for turbine\n",
-      "#s3=s4(Entropy remain same)\n",
-      "s4=6.3409;#kJ/kg\n",
-      "sf=0.6493;#Entropy at liquid state in kJ/kg\n",
-      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg\n",
-      "x4=(s4-sf)/sfg;#x-factor\n",
-      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",
-      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",
-      "\n",
-      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",
-      "print\" Percentage efficiency =\",round(nth*100,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Percentage efficiency = 30.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.2:Pg-429"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Ques 2\n",
-      "#To determine the efficiency of Rankine cycle\n",
-      "\n",
-      "#1-Inlet state of pump\n",
-      "#2-Exit state of pump\n",
-      "P2=4000;#Exit pressure in kPa\n",
-      "P1=10;#Inlet pressure in kPa\n",
-      "v=0.00101;#specific weight of water in m^3/kg\n",
-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",
-      "h1=191.8;#Enthalpy in kJ/kg from table\n",
-      "h2=h1+wp;#enthalpy in kJ/kg\n",
-      "#2-Inlet state for boiler\n",
-      "#3-Exit state for boiler\n",
-      "h3=3213.6;#Enthalpy in kJ/kg from table\n",
-      "#3-Inlet state for turbine\n",
-      "#4-Exit state for turbine\n",
-      "#s3=s4(Entropy remain same)\n",
-      "s4=6.7690;#Entropy in kJ/kg from table\n",
-      "sf=0.6493;#Entropy at liquid state in kJ/kg from table\n",
-      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg from table\n",
-      "x4=(s4-sf)/sfg;#x-factor\n",
-      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",
-      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",
-      "\n",
-      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",
-      "print\"Percentage efficiency =\",round(nth*100,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage efficiency = 35.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.3:Pg-433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Ques 3\n",
-      "#To determine the efficiency of a cycle\n",
-      "\n",
-      "#1-Inlet state of pump\n",
-      "#2-Exit state of pump\n",
-      "P2=4000;#Exit pressure in kPa\n",
-      "P1=10;#Inlet pressure in kPa\n",
-      "v=0.00101;#specific weight of water in m^3/kg\n",
-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",
-      "h1=191.8;#Enthalpy in kJ/kg from table\n",
-      "h2=h1+wp;#enthalpy in kJ/kg\n",
-      "#2-Inlet state for boiler\n",
-      "#3-Exit state for Boiler\n",
-      "h3=3213.6;#Enthalpy in kJ/kg from table\n",
-      "#3-Inlet state for high pressure turbine\n",
-      "#4-Exit state for high pressure turbine\n",
-      "#s3=s4(Entropy remain same)\n",
-      "s4=6.7690;#Entropy in kJ/kg from table\n",
-      "sf=1.7766;#Entropy at liquid state in kJ/kg from table\n",
-      "sfg=5.1193;#Entropy difference for vapor and liquid state in kJ/kg from table\n",
-      "x4=(s4-sf)/sfg;#x-factor\n",
-      "hf=604.7#Enthalpy of liquid state in kJ/kg\n",
-      "hfg=2133.8;#Enthalpy difference in kJ/kg for turbine\n",
-      "h4=hf+x4*hfg;#Enthalpy in kJ/kg\n",
-      "#5-Inlet state for low pressure turbine\n",
-      "#6-Exit pressure for low pressure turbine\n",
-      "sf=0.6493;#Entropy in liquid state in kJ/kg for turbine\n",
-      "h5=3273.4;#enthalpy in kJ/kg \n",
-      "s5=7.8985;#Entropy in kJ/kg\n",
-      "sfg=7.5009;#entropy diff in kJ/kg \n",
-      "x6=(s5-sf)/sfg;#x-factor\n",
-      "hfg=2392.8;#enthalpy difference for low pressure turbine in kj/kg\n",
-      "h6=h1+x6*hfg;#entropy in kg/kg\n",
-      "wt=(h3-h4)+(h5-h6);#work output in kJ/kg\n",
-      "qh=(h3-h2)+(h5-h4);\n",
-      "\n",
-      "nth=(wt-wp)/qh;\n",
-      "print\" Percentage efficiency =\",round(nth*100,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Percentage efficiency = 35.9\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.4:Pg-438"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques4\n",
-      "#Efficiency of Refrigeration cycle\n",
-      "\n",
-      "#from previous examples\n",
-      "h1=191.8;#kJ/kg\n",
-      "h5=3213.6;#kg/kg\n",
-      "h6=2685.7;#kJ/kg\n",
-      "h7=2144.1;#kJ/kg\n",
-      "h3=604.7;#kJ/kg\n",
-      "#1-Inlet state of pump\n",
-      "#2-Exit state of pump\n",
-      "P2=400;#Exit pressure in kPa\n",
-      "P1=10;  #Inlet pressure in kPa\n",
-      "v=0.00101;#specific weight of water in m^3/kg\n",
-      "wp1=v*(P2-P1);#work done for low pressure pump in kJ/kg\n",
-      "h1=191.8;#Enthalpy in kJ/kg from table\n",
-      "h2=h1+wp1;#enthalpy in kJ/kg\n",
-      "#5-Inlet state for turbine\n",
-      "#6,7-Exit state for turbine\n",
-      "y=(h3-h2)/(h6-h2);#extraction fraction\n",
-      "wt=(h5-h6)+(1-y)*(h6-h7);#turbine work in kJ/kg\n",
-      "#3-Inlet for high pressure pump\n",
-      "#4-Exit for high pressure pump\n",
-      "P3=400;#kPa\n",
-      "P4=4000;#kPa\n",
-      "v=0.001084;#specific heat for 3-4 process in m^3/kg\n",
-      "wp2=v*(P4-P3);#work done for high pressure pump\n",
-      "h4=h3+wp2;#Enthalpy in kJ/kg\n",
-      "wnet=wt-(1-y)*wp1-wp2;\n",
-      "qh=h5-h4;#Heat output in kJ/kg\n",
-      "nth=wnet/qh;\n",
-      "print\" Refrigerator Efficiency =\",round(nth*100,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Refrigerator Efficiency = 37.5\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.5:Pg-443"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques5\n",
-      "#To determine thermal efficiency of cycle\n",
-      "\n",
-      "#5-Inlet state for turbine\n",
-      "#6-Exit state for turbine\n",
-      "#h-Enthalpy at a state \n",
-      "#s-Entropy at a state\n",
-      "#from steam table\n",
-      "h5=3169.1;#kJ/kg\n",
-      "s5=6.7235;#kJ/kg\n",
-      "s6s=s5;\n",
-      "sf=0.6493;#Entropy for liquid state in kJ/kg\n",
-      "sfg=7.5009;#Entropy difference in kJ/kg\n",
-      "hf=191.8;#kJ/kg\n",
-      "hfg=2392.8;#Enthalpy difference in kJ/kg\n",
-      "x6s=(s6s-sf)/sfg;#x-factor\n",
-      "h6s=hf+x6s*hfg;#kJ/Kg at state 6s\n",
-      "nt=0.86;#turbine efficiency given\n",
-      "wt=nt*(h5-h6s);\n",
-      "#1-Inlet state for pump\n",
-      "#2-Exit state for pump\n",
-      "np=0.80;#pump efficiency given\n",
-      "v=0.001009;#specific heat in m^3/kg\n",
-      "P2=5000;#kPa\n",
-      "P1=10;#kPa\n",
-      "wp=v*(P2-P1)/np;#Work done in pump in kJ/kg\n",
-      "wnet=wt-wp;#net work in kJ/kg\n",
-      "#3-Inlet state for boiler\n",
-      "#4-Exit state for boiler\n",
-      "h3=171.8;#in kJ/kg from table\n",
-      "h4=3213.6;#kJ/kg from table\n",
-      "qh=h4-h3;\n",
-      "nth=wnet/qh;\n",
-      "print \"Cycle Efficiency =\",round(nth*100,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cycle Efficiency = 29.2\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.6:Pg-451"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques6\n",
-      "#to determine the rate of refrigeration\n",
-      "\n",
-      "# refer to fig 11.21 in book\n",
-      "mdot=0.03 # mass flow rate in Kg/s\n",
-      "T1=-20 # temperature in evaporator in celsius\n",
-      "T3=40 #temperature in evaporator in Celsius\n",
-      "P2=1017 # saturation pressure in KPa\n",
-      "\n",
-      "# from table of R-134a refrigerant\n",
-      "h1=386.1 # enthalpy at state 1 in kJ/kg,\n",
-      "S1=1.7395 # entropy at state 1 in kJ/kg.K\n",
-      "S2=S1 # isentropic process\n",
-      "T2=47.7# corresponding value to S2 in table of R-134a in degree celsius\n",
-      "h2=428.4 # corresponding value to S2 in table of R-134a in kJ/kg\n",
-      "wc=h2-h1 # work done in compressor in kJ/kg\n",
-      "h4=h3=256.5 #enthalpy at state 4 and 3 in kJ/kg\n",
-      "qL=h1-h4 #Heat rejected in kJ/kg\n",
-      "\n",
-      "B=qL/wc # COP\n",
-      "\n",
-      "print\" the COP of the plant is\",round(B,2)\n",
-      "print\" the refrigeration rate is\",round(mdot*qL,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " the COP of the plant is 3.06\n",
-        " the refrigeration rate is 3.89\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.7:Pg-454"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques7\n",
-      "#to determine the COP of cycle\n",
-      "\n",
-      "P1=125 # pressure at state 1 in kPa\n",
-      "P2=1.2 # pressure at state 2 in MPa\n",
-      "P3=1.19 # pressure at state 3 in MPa,\n",
-      "P4=1.16 # pressure at state 4 in MPa,\n",
-      "P5=1.15 # pressure at state 5 in MPa,\n",
-      "P6=P7=140 # pressure at state 6 and 7 in kPa,\n",
-      "P8=130 # pressure at state 8 in kPa,\n",
-      "T1=-10 #temperaure at state  1 in \u25e6C\n",
-      "T2=100 #temperaure at state  2 in \u25e6C\n",
-      "T3=80 #temperaure at state  3 in \u25e6C\n",
-      "T4=45 #temperaure at state  4 in \u25e6C\n",
-      "T5=40 #temperaure at state  5 in \u25e6C\n",
-      "T8=-20 #temperaure at state  8 in \u25e6C\n",
-      "q=-4 # heat transfer in kJ/Kg\n",
-      "\n",
-      "#x6=x7 quality condition given in question\n",
-      "\n",
-      "\n",
-      "# the following values are taken from table for refrigerant R-134a\n",
-      "h1=394.9 # enthalpy at state 1 in kJ/kg\n",
-      "h2=480.9 # enthalpy at state 2 in kJ/kg\n",
-      "h8=386.6 # enthalpy at state 8 in kJ/kg\n",
-      "wc=h2-h1-q # from first law\n",
-      "h5=h6=h7=256.4 # as x6=x7 and from table at state 5 in Kj/Kg\n",
-      "qL=h8-h7 # from first law \n",
-      "B=qL/wc # COP\n",
-      "print\" the COP of the plant is\",round(B,3)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " the COP of the plant is 1.447\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter11_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter11_6.ipynb
deleted file mode 100755
index 814dd14b..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter11_6.ipynb
+++ /dev/null
@@ -1,529 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:a88db765bb23172043955029f8f04edaf8e710f3cbd46a5474827f42fa8a65ec"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter11:Power and Refrigeration Systems\u2014With Phase Change"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.1:Pg-425"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Ques 1\n",

-      "#To determine the efficiency of Rankine cycle\n",

-      "\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=2000;#Exit pressure in kPa\n",

-      "P1=10;#Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp;#enthalpy in kJ/kg\n",

-      "#2-Inlet state for boiler\n",

-      "#3-Exit state for boiler\n",

-      "h3=2799.5;#Enthalpy in kJ/kg\n",

-      "#3-Inlet state for turbine\n",

-      "#4-Exit state for turbine\n",

-      "#s3=s4(Entropy remain same)\n",

-      "s4=6.3409;#kJ/kg\n",

-      "sf=0.6493;#Entropy at liquid state in kJ/kg\n",

-      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg\n",

-      "x4=(s4-sf)/sfg;#x-factor\n",

-      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",

-      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",

-      "\n",

-      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",

-      "print\" Percentage efficiency =\",round(nth*100,1)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Percentage efficiency = 30.3\n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.2:Pg-429"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Ques 2\n",

-      "#To determine the efficiency of Rankine cycle\n",

-      "\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=4000;#Exit pressure in kPa\n",

-      "P1=10;#Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp;#enthalpy in kJ/kg\n",

-      "#2-Inlet state for boiler\n",

-      "#3-Exit state for boiler\n",

-      "h3=3213.6;#Enthalpy in kJ/kg from table\n",

-      "#3-Inlet state for turbine\n",

-      "#4-Exit state for turbine\n",

-      "#s3=s4(Entropy remain same)\n",

-      "s4=6.7690;#Entropy in kJ/kg from table\n",

-      "sf=0.6493;#Entropy at liquid state in kJ/kg from table\n",

-      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg from table\n",

-      "x4=(s4-sf)/sfg;#x-factor\n",

-      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",

-      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",

-      "\n",

-      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",

-      "print\"Percentage efficiency =\",round(nth*100,1),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Percentage efficiency = 35.3 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 41

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.2E:Pg-431"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Ques 2\n",

-      "#To determine the efficiency of Rankine cycle\n",

-      "\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=600.0 ;#Exit pressure in lbf/in^2\n",

-      "P1=1.0;#Inlet pressure in lbf/in^2\n",

-      "v=0.01614;#specific weight of water in ft^3/lbm\n",

-      "wp=v*(P2-P1)*(144.0/778.0);#work done in pipe in Btu/lbm\n",

-      "h1=69.70;#Enthalpy in Btu/lbm from table\n",

-      "h2=h1+wp;#enthalpy in Btu/lbm\n",

-      "#2-Inlet state for boiler\n",

-      "#3-Exit state for boiler\n",

-      "h3=1407.6;#Enthalpy in Btu/lbm from table\n",

-      "#3-Inlet state for turbine\n",

-      "#4-Exit state for turbine\n",

-      "#s3=s4(Entropy remain same)\n",

-      "s4=1.6343;#Entropy in Btu/lbm from table\n",

-      "sf=1.9779;#Entropy at liquid state in Btu/lbm from table\n",

-      "sfg=1.8453;#Entropy difference for vapor and liquid state in Btu/lbm from table\n",

-      "x4=-(s4-sf)/sfg;#x-factor\n",

-      "hfg=1036.0;#Enthalpy difference in Btu/lbm for turbine\n",

-      "h4=1105.8-x4*hfg;#Enthalpy in Btu/lbm\n",

-      "wt=(h3-h4) #work done in turbine in Btu/lbm\n",

-      "\n",

-      "nth=((h3-h4)-wp)/(h3-h2);\n",

-      "print\"Percentage efficiency =\",round(nth*100,1),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Percentage efficiency = 36.9 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 42

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.3:Pg-433"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Ques 3\n",

-      "#To determine the efficiency of a cycle\n",

-      "\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=4000;#Exit pressure in kPa\n",

-      "P1=10;#Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp;#enthalpy in kJ/kg\n",

-      "#2-Inlet state for boiler\n",

-      "#3-Exit state for Boiler\n",

-      "h3=3213.6;#Enthalpy in kJ/kg from table\n",

-      "#3-Inlet state for high pressure turbine\n",

-      "#4-Exit state for high pressure turbine\n",

-      "#s3=s4(Entropy remain same)\n",

-      "s4=6.7690;#Entropy in kJ/kg from table\n",

-      "sf=1.7766;#Entropy at liquid state in kJ/kg from table\n",

-      "sfg=5.1193;#Entropy difference for vapor and liquid state in kJ/kg from table\n",

-      "x4=(s4-sf)/sfg;#x-factor\n",

-      "hf=604.7#Enthalpy of liquid state in kJ/kg\n",

-      "hfg=2133.8;#Enthalpy difference in kJ/kg for turbine\n",

-      "h4=hf+x4*hfg;#Enthalpy in kJ/kg\n",

-      "#5-Inlet state for low pressure turbine\n",

-      "#6-Exit pressure for low pressure turbine\n",

-      "sf=0.6493;#Entropy in liquid state in kJ/kg for turbine\n",

-      "h5=3273.4;#enthalpy in kJ/kg \n",

-      "s5=7.8985;#Entropy in kJ/kg\n",

-      "sfg=7.5009;#entropy diff in kJ/kg \n",

-      "x6=(s5-sf)/sfg;#x-factor\n",

-      "hfg=2392.8;#enthalpy difference for low pressure turbine in kj/kg\n",

-      "h6=h1+x6*hfg;#entropy in kg/kg\n",

-      "wt=(h3-h4)+(h5-h6);#work output in kJ/kg\n",

-      "qh=(h3-h2)+(h5-h4);\n",

-      "\n",

-      "nth=(wt-wp)/qh;\n",

-      "print\" Percentage efficiency =\",round(nth*100,1),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Percentage efficiency = 35.9 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 43

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.4:Pg-438"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques4\n",

-      "#Efficiency of Refrigeration cycle\n",

-      "\n",

-      "#from previous examples\n",

-      "h1=191.8;#kJ/kg\n",

-      "h5=3213.6;#kg/kg\n",

-      "h6=2685.7;#kJ/kg\n",

-      "h7=2144.1;#kJ/kg\n",

-      "h3=604.7;#kJ/kg\n",

-      "#1-Inlet state of pump\n",

-      "#2-Exit state of pump\n",

-      "P2=400;#Exit pressure in kPa\n",

-      "P1=10;  #Inlet pressure in kPa\n",

-      "v=0.00101;#specific weight of water in m^3/kg\n",

-      "wp1=v*(P2-P1);#work done for low pressure pump in kJ/kg\n",

-      "h1=191.8;#Enthalpy in kJ/kg from table\n",

-      "h2=h1+wp1;#enthalpy in kJ/kg\n",

-      "#5-Inlet state for turbine\n",

-      "#6,7-Exit state for turbine\n",

-      "y=(h3-h2)/(h6-h2);#extraction fraction\n",

-      "wt=(h5-h6)+(1-y)*(h6-h7);#turbine work in kJ/kg\n",

-      "#3-Inlet for high pressure pump\n",

-      "#4-Exit for high pressure pump\n",

-      "P3=400;#kPa\n",

-      "P4=4000;#kPa\n",

-      "v=0.001084;#specific heat for 3-4 process in m^3/kg\n",

-      "wp2=v*(P4-P3);#work done for high pressure pump\n",

-      "h4=h3+wp2;#Enthalpy in kJ/kg\n",

-      "wnet=wt-(1-y)*wp1-wp2;\n",

-      "qh=h5-h4;#Heat output in kJ/kg\n",

-      "nth=wnet/qh;\n",

-      "print\" Refrigerator Efficiency =\",round(nth*100,1),\"%\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Refrigerator Efficiency = 37.5 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 44

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.5:Pg-443"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques5\n",

-      "#To determine thermal efficiency of cycle\n",

-      "\n",

-      "#5-Inlet state for turbine\n",

-      "#6-Exit state for turbine\n",

-      "#h-Enthalpy at a state \n",

-      "#s-Entropy at a state\n",

-      "#from steam table\n",

-      "h5=3169.1;#kJ/kg\n",

-      "s5=6.7235;#kJ/kg\n",

-      "s6s=s5;\n",

-      "sf=0.6493;#Entropy for liquid state in kJ/kg\n",

-      "sfg=7.5009;#Entropy difference in kJ/kg\n",

-      "hf=191.8;#kJ/kg\n",

-      "hfg=2392.8;#Enthalpy difference in kJ/kg\n",

-      "x6s=(s6s-sf)/sfg;#x-factor\n",

-      "h6s=hf+x6s*hfg;#kJ/Kg at state 6s\n",

-      "nt=0.86;#turbine efficiency given\n",

-      "wt=nt*(h5-h6s);\n",

-      "#1-Inlet state for pump\n",

-      "#2-Exit state for pump\n",

-      "np=0.80;#pump efficiency given\n",

-      "v=0.001009;#specific heat in m^3/kg\n",

-      "P2=5000;#kPa\n",

-      "P1=10;#kPa\n",

-      "wp=v*(P2-P1)/np;#Work done in pump in kJ/kg\n",

-      "wnet=wt-wp;#net work in kJ/kg\n",

-      "#3-Inlet state for boiler\n",

-      "#4-Exit state for boiler\n",

-      "h3=171.8;#in kJ/kg from table\n",

-      "h4=3213.6;#kJ/kg from table\n",

-      "qh=h4-h3;\n",

-      "nth=wnet/qh;\n",

-      "print \"Cycle Efficiency =\",round(nth*100,1),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Cycle Efficiency = 29.2 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 45

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.5E:Pg-445"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques5\n",

-      "#To determine thermal efficiency of cycle\n",

-      "\n",

-      "#5-Inlet state for turbine\n",

-      "#6-Exit state for turbine\n",

-      "#h-Enthalpy at a state \n",

-      "#s-Entropy at a state\n",

-      "#from steam table\n",

-      "h5=1386.8;#Btu/lbm\n",

-      "s5=1.6248;#Btu/lbm\n",

-      "s6s=s5;\n",

-      "sf=1.9779;#Entropy for liquid state in Btu/lbm\n",

-      "sfg=1.8453;#Entropy difference in Btu/lbm\n",

-      "hf=1105.8;# Btu/lbm\n",

-      "hfg=1036.0;#Enthalpy difference in Btu/lbm\n",

-      "x6s=(s6s-sf)/sfg;#x-factor\n",

-      "h6s=hf+x6s*hfg;#Btu/lbm at state 6s\n",

-      "nt=0.86;#turbine efficiency given\n",

-      "wt=nt*(h5-h6s);\n",

-      "#1-Inlet state for pump\n",

-      "#2-Exit state for pump\n",

-      "np=0.80;#pump efficiency given\n",

-      "v=0.016;#specific heat in ft^3/lbm\n",

-      "P2=800.0;# lbf/in^2\n",

-      "P1=1.0;# lbf/in^2\n",

-      "wp=(v*(P2-P1)*144.0)/(np*778.0);#Work done in pump in Btu/lbm\n",

-      "wnet=wt-wp;#net work in Btu/lbm\n",

-      "#3-Inlet state for boiler\n",

-      "#4-Exit state for boiler\n",

-      "h3=65.1;#in Btu/lbm from table\n",

-      "h4=1407.6;# Btu/lbm from table\n",

-      "qh=h4-h3;\n",

-      "nth=wnet/qh;\n",

-      "print \"Cycle Efficiency =\",round(nth*100,2),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Cycle Efficiency = 30.48 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 60

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.6:Pg-451"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques6\n",

-      "#to determine the rate of refrigeration\n",

-      "\n",

-      "# refer to fig 11.21 in book\n",

-      "mdot=0.03 # mass flow rate in Kg/s\n",

-      "T1=-20 # temperature in evaporator in celsius\n",

-      "T3=40 #temperature in evaporator in Celsius\n",

-      "P2=1017 # saturation pressure in KPa\n",

-      "\n",

-      "# from table of R-134a refrigerant\n",

-      "h1=386.1 # enthalpy at state 1 in kJ/kg,\n",

-      "S1=1.7395 # entropy at state 1 in kJ/kg.K\n",

-      "S2=S1 # isentropic process\n",

-      "T2=47.7# corresponding value to S2 in table of R-134a in degree celsius\n",

-      "h2=428.4 # corresponding value to S2 in table of R-134a in kJ/kg\n",

-      "wc=h2-h1 # work done in compressor in kJ/kg\n",

-      "h4=h3=256.5 #enthalpy at state 4 and 3 in kJ/kg\n",

-      "qL=h1-h4 #Heat rejected in kJ/kg\n",

-      "\n",

-      "B=qL/wc # COP\n",

-      "\n",

-      "print\" the COP of the plant is\",round(B,2)\n",

-      "print\" the refrigeration rate is\",round(mdot*qL,2)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " the COP of the plant is 3.06\n",

-        " the refrigeration rate is 3.89\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex11.7:Pg-454"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#to determine the COP of cycle\n",

-      "\n",

-      "P1=125 # pressure at state 1 in kPa\n",

-      "P2=1.2 # pressure at state 2 in MPa\n",

-      "P3=1.19 # pressure at state 3 in MPa,\n",

-      "P4=1.16 # pressure at state 4 in MPa,\n",

-      "P5=1.15 # pressure at state 5 in MPa,\n",

-      "P6=P7=140 # pressure at state 6 and 7 in kPa,\n",

-      "P8=130 # pressure at state 8 in kPa,\n",

-      "T1=-10 #temperaure at state  1 in \u25e6C\n",

-      "T2=100 #temperaure at state  2 in \u25e6C\n",

-      "T3=80 #temperaure at state  3 in \u25e6C\n",

-      "T4=45 #temperaure at state  4 in \u25e6C\n",

-      "T5=40 #temperaure at state  5 in \u25e6C\n",

-      "T8=-20 #temperaure at state  8 in \u25e6C\n",

-      "q=-4 # heat transfer in kJ/Kg\n",

-      "\n",

-      "#x6=x7 quality condition given in question\n",

-      "\n",

-      "\n",

-      "# the following values are taken from table for refrigerant R-134a\n",

-      "h1=394.9 # enthalpy at state 1 in kJ/kg\n",

-      "h2=480.9 # enthalpy at state 2 in kJ/kg\n",

-      "h8=386.6 # enthalpy at state 8 in kJ/kg\n",

-      "wc=h2-h1-q # from first law\n",

-      "h5=h6=h7=256.4 # as x6=x7 and from table at state 5 in Kj/Kg\n",

-      "qL=h8-h7 # from first law \n",

-      "B=qL/wc # COP\n",

-      "print\" the COP of the plant is\",round(B,3)\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " the COP of the plant is 1.447\n"

-       ]

-      }

-     ],

-     "prompt_number": 20

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter12.ipynb b/Fundamentals_Of_Thermodynamics/Chapter12.ipynb
deleted file mode 100755
index 75910c56..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter12.ipynb
+++ /dev/null
@@ -1,486 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e3e9cb7f43080b67fa2a22d273814843783e7598e3258877edb209ca16f99bdc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER13:Gas Mixtures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.1:Pg-480"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques1\n",
-      "#Standard brayton cycle\n",
-      "\n",
-      "#1-Inlet for compressor\n",
-      "#2-Exit for compressor\n",
-      "#T-Temperature at a state\n",
-      "#P-Pressure at a state\n",
-      "T1=288.2;#K\n",
-      "P2=1000.0;#kPa\n",
-      "P1=100.0;#kPa\n",
-      "k=1.4;\n",
-      "T2=T1*(P2/P1)**(1.0-1/k);#K\n",
-      "Cp=1.004;#Specific heat at constant pressure in kJ/kg\n",
-      "wc=Cp*(T2-T1);#compressor work in kJ/kg;\n",
-      "print\" Temperature T2 =\",round(T2,2),\"K\"\n",
-      "print\" Compressor work =\",round(wc,1),\"kJ/kg\"\n",
-      "#3-Turbine Inlet\n",
-      "#4-Turbine Exit\n",
-      "P4=P1;\n",
-      "P3=P2;\n",
-      "T3=1373.2;#K\n",
-      "T4=T3*(P4/P3)**(1-1/k);#K\n",
-      "wt=Cp*(T3-T4);\n",
-      "wnet=wt-wc;\n",
-      "print\" Temperature T3 =\",round(T3,2),\" K\"\n",
-      "print\" Temperature T4 =\",round(T4),\"K\"\n",
-      "print\" Turbine work =\",round(wt,2),\"kJ/kg\"\n",
-      "print\" Net work =\",round(wt-wc,2),\"kJ/kg\"\n",
-      "#2-Also high temperature heat exchanger Inlet\n",
-      "#3-(-do-) Exit\n",
-      "qh=Cp*(T3-T2);#Heat of source in kJ/kg\n",
-      "#4-high temp heat exchanger inlet\n",
-      "#1-(-do-) Exit\n",
-      "ql=Cp*(T4-T1);#Heat of sink in kJ/kg\n",
-      "nth=wnet/qh;\n",
-      "print\" Thermal Efficiency of cycle =\",round(nth*100,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Temperature T2 = 556.43 K\n",
-        " Compressor work = 269.3 kJ/kg\n",
-        " Temperature T3 = 1373.2  K\n",
-        " Temperature T4 = 711.0 K\n",
-        " Turbine work = 664.6 kJ/kg\n",
-        " Net work = 395.3 kJ/kg\n",
-        " Thermal Efficiency of cycle = 48.21 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.2:Pg-481"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculation mistake in book\n",
-      "#ques2\n",
-      "#Standard brayton cycle\n",
-      "\n",
-      "#Calculation mistake in book\n",
-      "#1-Inlet for compressor\n",
-      "#2-Exit for compressor\n",
-      "#T-Temperature at a state\n",
-      "#P-Pressure at a state\n",
-      "T1=288.2;#K\n",
-      "P2=1000.0;#kPa\n",
-      "P1=100.0;#kPa\n",
-      "k=1.4;\n",
-      "T2s=T1*(P2/P1)**(1-1/k);#K\n",
-      "nc=.80;#Compressor Efficiency\n",
-      "T2=T1+(T2s-T1)/0.80;\n",
-      "Cp=1.004;#Specific heat at constant pressure in kJ/kg\n",
-      "wc=Cp*(T2-T1);#compressor work in kJ/kg\n",
-      "wc=round(wc)\n",
-      "print\" Temperature T2 =\",round(T2,2),\"K\"\n",
-      "print\" Compressor work =\",(wc),\"kJ/kg\"\n",
-      "#3-Turbine Inlet\n",
-      "#4-Turbine Exit\n",
-      "P4=P1;\n",
-      "P3=P2;\n",
-      "T3=1373.2;#K\n",
-      "T4s=T3*(P4/P3)**(1-1.0/k);#K\n",
-      "nt=0.85;#turbine Efficiency\n",
-      "T4=T3-(T3-T4s)*0.85;\n",
-      "wt=Cp*(T3-T4);\n",
-      "wnet=wt-wc;\n",
-      "print\" Temperature T3 =\",round(T3,1),\"K\"\n",
-      "print\" Temperature T4 =\",round(T4,1),\"K\"\n",
-      "print\" Turbine work =\",round(wt,2),\"kJ/kg\"\n",
-      "print\" Net work =\",round(wt-wc,2),\"kJ/kg\"\n",
-      "#2-Also high temperature heat exchanger Inlet\n",
-      "#3-(-do-) Exit\n",
-      "qh=Cp*(T3-T2);#Heat of source in kJ/kg\n",
-      "#4-high temp heat exchanger inlet\n",
-      "#1-(-do-) Exit\n",
-      "ql=Cp*(T4-T1);#Heat of sink in kJ/kg\n",
-      "nth=wnet/qh;\n",
-      "print\" Thermal Efficiency of cycle =\",round(nth*100),\"percent\"\n",
-      "#some answers are have acceptable difference beacause of approximization in book but here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Temperature T2 = 623.48 K\n",
-        " Compressor work = 337.0 kJ/kg\n",
-        " Temperature T3 = 1373.2 K\n",
-        " Temperature T4 = 810.5 K\n",
-        " Turbine work = 564.91 kJ/kg\n",
-        " Net work = 227.91 kJ/kg\n",
-        " Thermal Efficiency of cycle = 30.0 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.3:Pg-486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques3\n",
-      "#efficiency of the cycle\n",
-      "\n",
-      "wnet=395.2;#kJ/kg from example no 1\n",
-      "#Tx=T4\n",
-      "Tx=710.8;#K from example no 1\n",
-      "T3=1373.2;#K from example no 1\n",
-      "Cp=1.004;#specific heat in kJ/kg \n",
-      "qh=Cp*(T3-Tx);\n",
-      "nth=wnet/qh;\n",
-      "print\" Thermal efficiency =\",round(nth*100,1),\" percent\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Thermal efficiency = 59.4  percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.4:Pg-486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques4\n",
-      "#Calculation of work in the given cycle\n",
-      "import math\n",
-      "R=0.287;#gas constant \n",
-      "T1=288.2;#compressor temperature K\n",
-      "T2=1373.2;#K turbine temperature K\n",
-      "#Pe/Pi=c=10, Pi/Pe=1/c from example 12.1\n",
-      "c=10.0;\n",
-      "wc=-R*T1*math.log(c);\n",
-      "print\" Isothermal work in compressor =\",round(wc,1),\"kJ/kg\"\n",
-      "wt=-R*T2*math.log(1/c);\n",
-      "print\" Isothermal work in turbine =\",round(wt,1),\"kJ/kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Isothermal work in compressor = -190.5 kJ/kg\n",
-        " Isothermal work in turbine = 907.5 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.5:Pg-491"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques5\n",
-      "#air standard cycle for jet repulsion\n",
-      "import math\n",
-      "#1-compressor inlet\n",
-      "#2-Compressor exit\n",
-      "#P-Pressure at given point\n",
-      "#T-Temperature at given point\n",
-      "P1=100;#kPa\n",
-      "P2=1000;#kPa\n",
-      "T1=288.2;#K\n",
-      "T2=556.8;#K\n",
-      "wc=269.5;#from ex 12.1 work done in compressor in kJ/kg\n",
-      "#2-Burner inlet\n",
-      "#3-Burner exit\n",
-      "P3=1000;#kPa\n",
-      "T3=1373.2;#K\n",
-      "#wc=wt\n",
-      "Cp=1.004;#specific enthalpy of heat at constant pressure in kJ/kg\n",
-      "k=1.4;\n",
-      "T4=T3-wc/Cp;\n",
-      "P4=P3*(T4/T3)**(1-1/k);\n",
-      "#from s4=s5 and h4=h5+v2/2 we get\n",
-      "T5=710.8#K, from second law\n",
-      "v=math.sqrt(2*Cp*1000*(T4-T5));#m/s\n",
-      "print\" Velocity of air leaving the nozel =\",round(v),\"m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Velocity of air leaving the nozel = 889.0 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.6:Pg-494"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques6\n",
-      "#air standard refrigeration cycle\n",
-      "\n",
-      "#1-compressor inlet\n",
-      "#2-compressor exit\n",
-      "P1=100;#kPa\n",
-      "P2=500;#kPa\n",
-      "k=1.4;\n",
-      "rp=P2/P1;\n",
-      "cop=(rp**(1-1/k)-1)**-1;\n",
-      "print\" Coefficient of performance =\",round(cop,2)\n",
-      "#3-Expander inlet\n",
-      "#4-Expander exit\n",
-      "P3=P2;\n",
-      "P4=P1;\n",
-      "T3=288.23;#K, given and fixed\n",
-      "T4=T3/(P3/P4)**(1-1/k);\n",
-      "T1=253.2;#K, given\n",
-      "Cp=1.004;#Specific heat at cons pressure in kJ/kg\n",
-      "ql=Cp*(T1-T4);#heat released in kJ/kg\n",
-      "P=1#power required in kW \n",
-      "ms=P/ql;#kg/s\n",
-      "print\"  Rate at which the air enter the compressor =\",round(ms,3),\"kg/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Coefficient of performance = 1.71\n",
-        "  Rate at which the air enter the compressor = 0.014 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.7:Pg-498"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques7\n",
-      "#the otto cycle\n",
-      "\n",
-      "#1-compressor inlet\n",
-      "#2-compressor exit\n",
-      "P1=100.0;#kPa\n",
-      "T1=288.2;#K\n",
-      "R=0.287;#gas constant\n",
-      "v1=R*T1/P1;#specific volume at inlet in m^3/kg\n",
-      "rv=10.0;#compression ratio given\n",
-      "k=1.4;#constant\n",
-      "T2=T1*rv**(k-1);#K\n",
-      "print\" Temperature at compressor exit, T2 =\",round(T2,1),\"K\"\n",
-      "P2=P1*rv**k;#kPa\n",
-      "print\" Pressure at compressor exit, P2 =\",round(P2/1000,2),\"MPa\"\n",
-      "v2=v1/rv;#specific heat at exit in m^3/kg\n",
-      "#23-heat addition process\n",
-      "#q23=Cv*(T3-T2)=1800 kJ/kg given\n",
-      "q23=1800.0;#kJ/kg heat addition, given\n",
-      "Cv=0.717;#specific heat at constant volume in kJ/kg\n",
-      "T3=T2+q23/Cv;#K\n",
-      "print\" Initial Temperature during heat additon process, T3 =\",round(T3,2),\"K\"\n",
-      "P3=P2*(T3/T2);#kPa\n",
-      "print\" Initial pressure during heat addition process, P3 =\",round(P3/1000,3),\"MPa\"\n",
-      "r=10.0;#k=V4/V3=P3/P4\n",
-      "T4=T3*(1/r)**(k-1);\n",
-      "print\" Final temperature during heat addition process, T4 =\",round(T4,3),\"K\";\n",
-      "P4=P3/r**k;#kPa\n",
-      "print\" Final pressure during heat addition process, P4 =\",round(P4/1000,4),\"MPa\"\n",
-      "nth=1-1/r**(k-1);#thermal efficiency\n",
-      "print\" Thermal efficiency =\",round(nth*100,1),\" percent\"\n",
-      "q41=Cv*(T1-T4);#/heat for process 4-1 in kJ/kg\n",
-      "wnet=q23+q41;\n",
-      "mep=wnet/(v1-v2);#effective mean pressure n kPa\n",
-      "print\" Mean effective pressure =\",round(mep,2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Temperature at compressor exit, T2 = 723.9 K\n",
-        " Pressure at compressor exit, P2 = 2.51 MPa\n",
-        " Initial Temperature during heat additon process, T3 = 3234.39 K\n",
-        " Initial pressure during heat addition process, P3 = 11.223 MPa\n",
-        " Final temperature during heat addition process, T4 = 1287.632 K\n",
-        " Final pressure during heat addition process, P4 = 0.4468 MPa\n",
-        " Thermal efficiency = 60.2  percent\n",
-        " Mean effective pressure = 1455.37 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.8:Pg-501"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques8\n",
-      "#the diesel cycle\n",
-      "\n",
-      "#1-compressor inlet\n",
-      "#2-compressor exit\n",
-      "P1=100;#kPa\n",
-      "T1=288.2;#K\n",
-      "R=0.287;#gas constant\n",
-      "v1=R*T1/P1;#specific volume at inlet in m^3/kg\n",
-      "rv=20;#compression ratio given\n",
-      "k=1.4;#constant\n",
-      "T2=T1*rv**(k-1);#K\n",
-      "print\" Temperature at compressor exit, T2 =\",round(T2,1),\"K\"\n",
-      "P2=P1*rv**k;#kPa\n",
-      "print\" Pressure at compressor exit, P2 =\",round(P2/1000,3),\" MPa\"\n",
-      "v2=v1/rv;#specific heat at exit in m^3/kg\n",
-      "#23-heat addition process\n",
-      "#q23=Cv*(T3-T2)=1800 kJ/kg given\n",
-      "q23=1800;#kJ/kg heat addition, given\n",
-      "Cv=.717;\n",
-      "Cp=1.004;#specific heat at constant pressure in kJ/kg\n",
-      "T3=T2+q23/Cp;#K\n",
-      "print\" Initial Temperature during heat addition process, T3 =\",round(T3,2),\"K\"\n",
-      "r=T3/T2;#T3/T2=V3/V2=r\n",
-      "v3=r*v2;\n",
-      "T4=T3/(v1/v3)**(k-1);\n",
-      "print\" Final temperature during heat addition process, T4 =\",round(T4),\"K\"\n",
-      "q41=Cv*(T1-T4);#/heat for process 4-1 in kJ/kg\n",
-      "wnet=q23+q41;\n",
-      "mep=wnet/(v1-v2);#effective mean pressure in kPa\n",
-      "qh=1800;#heat transfer in kJ/kg\n",
-      "nth=wnet/qh;#thermal efficiency\n",
-      "\n",
-      "print\" Thermal efficiency =\",round(nth*100,1),\"percent\"\n",
-      "print\" Mean effective pressure =\",round(mep,2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Temperature at compressor exit, T2 = 955.2 K\n",
-        " Pressure at compressor exit, P2 = 6.629  MPa\n",
-        " Initial Temperature during heat addition process, T3 = 2748.05 K\n",
-        " Final temperature during heat addition process, T4 = 1265.0 K\n",
-        " Thermal efficiency = 61.1 percent\n",
-        " Mean effective pressure = 1399.18 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter12_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter12_6.ipynb
deleted file mode 100755
index e2686e4a..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter12_6.ipynb
+++ /dev/null
@@ -1,478 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:83a0af04c694ab4b0ad1a0fe252ffa0bd8e87abc642c64323f0c2a0d18dd9f5d"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "CHAPTER13:Gas Mixtures"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.1:Pg-480"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques1\n",

-      "#Standard brayton cycle\n",

-      "\n",

-      "#1-Inlet for compressor\n",

-      "#2-Exit for compressor\n",

-      "#T-Temperature at a state\n",

-      "#P-Pressure at a state\n",

-      "T1=288.2;#K\n",

-      "P2=1000.0;#kPa\n",

-      "P1=100.0;#kPa\n",

-      "k=1.4;\n",

-      "T2=T1*(P2/P1)**(1.0-1/k);#K\n",

-      "Cp=1.004;#Specific heat at constant pressure in kJ/kg\n",

-      "wc=Cp*(T2-T1);#compressor work in kJ/kg;\n",

-      "print\" Temperature T2 =\",round(T2,2),\"K\"\n",

-      "print\" Compressor work =\",round(wc,1),\"kJ/kg\"\n",

-      "#3-Turbine Inlet\n",

-      "#4-Turbine Exit\n",

-      "P4=P1;\n",

-      "P3=P2;\n",

-      "T3=1373.2;#K\n",

-      "T4=T3*(P4/P3)**(1-1/k);#K\n",

-      "wt=Cp*(T3-T4);\n",

-      "wnet=wt-wc;\n",

-      "print\" Temperature T3 =\",round(T3,2),\" K\"\n",

-      "print\" Temperature T4 =\",round(T4),\"K\"\n",

-      "print\" Turbine work =\",round(wt,2),\"kJ/kg\"\n",

-      "print\" Net work =\",round(wt-wc,2),\"kJ/kg\"\n",

-      "#2-Also high temperature heat exchanger Inlet\n",

-      "#3-(-do-) Exit\n",

-      "qh=Cp*(T3-T2);#Heat of source in kJ/kg\n",

-      "#4-high temp heat exchanger inlet\n",

-      "#1-(-do-) Exit\n",

-      "ql=Cp*(T4-T1);#Heat of sink in kJ/kg\n",

-      "nth=wnet/qh;\n",

-      "print\" Thermal Efficiency of cycle =\",round(nth*100,2),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Temperature T2 = 556.43 K\n",

-        " Compressor work = 269.3 kJ/kg\n",

-        " Temperature T3 = 1373.2  K\n",

-        " Temperature T4 = 711.0 K\n",

-        " Turbine work = 664.6 kJ/kg\n",

-        " Net work = 395.3 kJ/kg\n",

-        " Thermal Efficiency of cycle = 48.21 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.2:Pg-481"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#Calculation mistake in book\n",

-      "#ques2\n",

-      "#Standard brayton cycle\n",

-      "\n",

-      "#Calculation mistake in book\n",

-      "#1-Inlet for compressor\n",

-      "#2-Exit for compressor\n",

-      "#T-Temperature at a state\n",

-      "#P-Pressure at a state\n",

-      "T1=288.2;#K\n",

-      "P2=1000.0;#kPa\n",

-      "P1=100.0;#kPa\n",

-      "k=1.4;\n",

-      "T2s=T1*(P2/P1)**(1-1/k);#K\n",

-      "nc=.80;#Compressor Efficiency\n",

-      "T2=T1+(T2s-T1)/0.80;\n",

-      "Cp=1.004;#Specific heat at constant pressure in kJ/kg\n",

-      "wc=Cp*(T2-T1);#compressor work in kJ/kg\n",

-      "wc=round(wc)\n",

-      "print\" Temperature T2 =\",round(T2,2),\"K\"\n",

-      "print\" Compressor work =\",(wc),\"kJ/kg\"\n",

-      "#3-Turbine Inlet\n",

-      "#4-Turbine Exit\n",

-      "P4=P1;\n",

-      "P3=P2;\n",

-      "T3=1373.2;#K\n",

-      "T4s=T3*(P4/P3)**(1-1.0/k);#K\n",

-      "nt=0.85;#turbine Efficiency\n",

-      "T4=T3-(T3-T4s)*0.85;\n",

-      "wt=Cp*(T3-T4);\n",

-      "wnet=wt-wc;\n",

-      "print\" Temperature T3 =\",round(T3,1),\"K\"\n",

-      "print\" Temperature T4 =\",round(T4,1),\"K\"\n",

-      "print\" Turbine work =\",round(wt,2),\"kJ/kg\"\n",

-      "print\" Net work =\",round(wt-wc,2),\"kJ/kg\"\n",

-      "#2-Also high temperature heat exchanger Inlet\n",

-      "#3-(-do-) Exit\n",

-      "qh=Cp*(T3-T2);#Heat of source in kJ/kg\n",

-      "#4-high temp heat exchanger inlet\n",

-      "#1-(-do-) Exit\n",

-      "ql=Cp*(T4-T1);#Heat of sink in kJ/kg\n",

-      "nth=wnet/qh;\n",

-      "print\" Thermal Efficiency of cycle =\",round(nth*100),\"percent\"\n",

-      "#some answers are have acceptable difference beacause of approximization in book but here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Temperature T2 = 623.48 K\n",

-        " Compressor work = 337.0 kJ/kg\n",

-        " Temperature T3 = 1373.2 K\n",

-        " Temperature T4 = 810.5 K\n",

-        " Turbine work = 564.91 kJ/kg\n",

-        " Net work = 227.91 kJ/kg\n",

-        " Thermal Efficiency of cycle = 30.0 percent\n"

-       ]

-      }

-     ],

-     "prompt_number": 22

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.3:Pg-486"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques3\n",

-      "#efficiency of the cycle\n",

-      "\n",

-      "wnet=395.2;#kJ/kg from example no 1\n",

-      "#Tx=T4\n",

-      "Tx=710.8;#K from example no 1\n",

-      "T3=1373.2;#K from example no 1\n",

-      "Cp=1.004;#specific heat in kJ/kg \n",

-      "qh=Cp*(T3-Tx);\n",

-      "nth=wnet/qh;\n",

-      "print\" Thermal efficiency =\",round(nth*100,1),\" percent\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Thermal efficiency = 59.4  percent\n"

-       ]

-      }

-     ],

-     "prompt_number": 24

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.4:Pg-486"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques4\n",

-      "#Calculation of work in the given cycle\n",

-      "import math\n",

-      "R=0.287;#gas constant \n",

-      "T1=288.2;#compressor temperature K\n",

-      "T2=1373.2;#K turbine temperature K\n",

-      "#Pe/Pi=c=10, Pi/Pe=1/c from example 12.1\n",

-      "c=10.0;\n",

-      "wc=-R*T1*math.log(c);\n",

-      "print\" Isothermal work in compressor =\",round(wc,1),\"kJ/kg\"\n",

-      "wt=-R*T2*math.log(1/c);\n",

-      "print\" Isothermal work in turbine =\",round(wt,1),\"kJ/kg\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Isothermal work in compressor = -190.5 kJ/kg\n",

-        " Isothermal work in turbine = 907.5 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 26

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.5:Pg-491"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques5\n",

-      "#air standard cycle for jet repulsion\n",

-      "import math\n",

-      "#1-compressor inlet\n",

-      "#2-Compressor exit\n",

-      "#P-Pressure at given point\n",

-      "#T-Temperature at given point\n",

-      "P1=100;#kPa\n",

-      "P2=1000;#kPa\n",

-      "T1=288.2;#K\n",

-      "T2=556.8;#K\n",

-      "wc=269.5;#from ex 12.1 work done in compressor in kJ/kg\n",

-      "#2-Burner inlet\n",

-      "#3-Burner exit\n",

-      "P3=1000;#kPa\n",

-      "T3=1373.2;#K\n",

-      "#wc=wt\n",

-      "Cp=1.004;#specific enthalpy of heat at constant pressure in kJ/kg\n",

-      "k=1.4;\n",

-      "T4=T3-wc/Cp;\n",

-      "P4=P3*(T4/T3)**(1-1/k);\n",

-      "#from s4=s5 and h4=h5+v2/2 we get\n",

-      "T5=710.8#K, from second law\n",

-      "v=math.sqrt(2*Cp*1000*(T4-T5));#m/s\n",

-      "print\" Velocity of air leaving the nozel =\",round(v),\"m/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Velocity of air leaving the nozel = 889.0 m/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 28

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.6:Pg-494"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques6\n",

-      "#air standard refrigeration cycle\n",

-      "\n",

-      "#1-compressor inlet\n",

-      "#2-compressor exit\n",

-      "P1=100;#kPa\n",

-      "P2=500;#kPa\n",

-      "k=1.4;\n",

-      "rp=P2/P1;\n",

-      "cop=(rp**(1-1/k)-1)**-1;\n",

-      "print\" Coefficient of performance =\",round(cop,2)\n",

-      "#3-Expander inlet\n",

-      "#4-Expander exit\n",

-      "P3=P2;\n",

-      "P4=P1;\n",

-      "T3=288.23;#K, given and fixed\n",

-      "T4=T3/(P3/P4)**(1-1/k);\n",

-      "T1=253.2;#K, given\n",

-      "Cp=1.004;#Specific heat at cons pressure in kJ/kg\n",

-      "ql=Cp*(T1-T4);#heat released in kJ/kg\n",

-      "P=1#power required in kW \n",

-      "ms=P/ql;#kg/s\n",

-      "print\"  Rate at which the air enter the compressor =\",round(ms,3),\"kg/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Coefficient of performance = 1.71\n",

-        "  Rate at which the air enter the compressor = 0.014 kg/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 36

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.7:Pg-498"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#the otto cycle\n",

-      "\n",

-      "#1-compressor inlet\n",

-      "#2-compressor exit\n",

-      "P1=100.0;#kPa\n",

-      "T1=288.2;#K\n",

-      "R=0.287;#gas constant\n",

-      "v1=R*T1/P1;#specific volume at inlet in m^3/kg\n",

-      "rv=10.0;#compression ratio given\n",

-      "k=1.4;#constant\n",

-      "T2=T1*rv**(k-1);#K\n",

-      "print\" Temperature at compressor exit, T2 =\",round(T2,1),\"K\"\n",

-      "P2=P1*rv**k;#kPa\n",

-      "print\" Pressure at compressor exit, P2 =\",round(P2/1000,2),\"MPa\"\n",

-      "v2=v1/rv;#specific heat at exit in m^3/kg\n",

-      "#23-heat addition process\n",

-      "#q23=Cv*(T3-T2)=1800 kJ/kg given\n",

-      "q23=1800.0;#kJ/kg heat addition, given\n",

-      "Cv=0.717;#specific heat at constant volume in kJ/kg\n",

-      "T3=T2+q23/Cv;#K\n",

-      "print\" Initial Temperature during heat additon process, T3 =\",round(T3,2),\"K\"\n",

-      "P3=P2*(T3/T2);#kPa\n",

-      "print\" Initial pressure during heat addition process, P3 =\",round(P3/1000,3),\"MPa\"\n",

-      "r=10.0;#k=V4/V3=P3/P4\n",

-      "T4=T3*(1/r)**(k-1);\n",

-      "print\" Final temperature during heat addition process, T4 =\",round(T4,3),\"K\";\n",

-      "P4=P3/r**k;#kPa\n",

-      "print\" Final pressure during heat addition process, P4 =\",round(P4/1000,4),\"MPa\"\n",

-      "nth=1-1/r**(k-1);#thermal efficiency\n",

-      "print\" Thermal efficiency =\",round(nth*100,1),\" percent\"\n",

-      "q41=Cv*(T1-T4);#/heat for process 4-1 in kJ/kg\n",

-      "wnet=q23+q41;\n",

-      "mep=wnet/(v1-v2);#effective mean pressure n kPa\n",

-      "print\" Mean effective pressure =\",round(mep,2),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Temperature at compressor exit, T2 = 723.9 K\n",

-        " Pressure at compressor exit, P2 = 2.51 MPa\n",

-        " Initial Temperature during heat additon process, T3 = 3234.39 K\n",

-        " Initial pressure during heat addition process, P3 = 11.223 MPa\n",

-        " Final temperature during heat addition process, T4 = 1287.632 K\n",

-        " Final pressure during heat addition process, P4 = 0.4468 MPa\n",

-        " Thermal efficiency = 60.2  percent\n",

-        " Mean effective pressure = 1455.37 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 42

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex12.8:Pg-501"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques8\n",

-      "#the diesel cycle\n",

-      "\n",

-      "#1-compressor inlet\n",

-      "#2-compressor exit\n",

-      "P1=100;#kPa\n",

-      "T1=288.2;#K\n",

-      "R=0.287;#gas constant\n",

-      "v1=R*T1/P1;#specific volume at inlet in m^3/kg\n",

-      "rv=20;#compression ratio given\n",

-      "k=1.4;#constant\n",

-      "T2=T1*rv**(k-1);#K\n",

-      "print\" Temperature at compressor exit, T2 =\",round(T2,1),\"K\"\n",

-      "P2=P1*rv**k;#kPa\n",

-      "print\" Pressure at compressor exit, P2 =\",round(P2/1000,3),\" MPa\"\n",

-      "v2=v1/rv;#specific heat at exit in m^3/kg\n",

-      "#23-heat addition process\n",

-      "#q23=Cv*(T3-T2)=1800 kJ/kg given\n",

-      "q23=1800;#kJ/kg heat addition, given\n",

-      "Cv=.717;\n",

-      "Cp=1.004;#specific heat at constant pressure in kJ/kg\n",

-      "T3=T2+q23/Cp;#K\n",

-      "print\" Initial Temperature during heat addition process, T3 =\",round(T3,2),\"K\"\n",

-      "r=T3/T2;#T3/T2=V3/V2=r\n",

-      "v3=r*v2;\n",

-      "T4=T3/(v1/v3)**(k-1);\n",

-      "print\" Final temperature during heat addition process, T4 =\",round(T4),\"K\"\n",

-      "q41=Cv*(T1-T4);#/heat for process 4-1 in kJ/kg\n",

-      "wnet=q23+q41;\n",

-      "mep=wnet/(v1-v2);#effective mean pressure in kPa\n",

-      "qh=1800;#heat transfer in kJ/kg\n",

-      "nth=wnet/qh;#thermal efficiency\n",

-      "\n",

-      "print\" Thermal efficiency =\",round(nth*100,1),\"percent\"\n",

-      "print\" Mean effective pressure =\",round(mep,2),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Temperature at compressor exit, T2 = 955.2 K\n",

-        " Pressure at compressor exit, P2 = 6.629  MPa\n",

-        " Initial Temperature during heat addition process, T3 = 2748.05 K\n",

-        " Final temperature during heat addition process, T4 = 1265.0 K\n",

-        " Thermal efficiency = 61.1 percent\n",

-        " Mean effective pressure = 1399.18 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 47

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter13.ipynb b/Fundamentals_Of_Thermodynamics/Chapter13.ipynb
deleted file mode 100755
index 95893bd9..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter13.ipynb
+++ /dev/null
@@ -1,262 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:6cc89f09586fc1fc226ee69c300e69ecaeae3fbafe1ae419c1c6cbc4ae7dcf83"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER13:Gas Mixtures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.3:Pg-533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques3\n",
-      "#calculating humidity ratio, dew point,mass of air, mass of vapor\n",
-      "\n",
-      "r=0.70;#relative humidity\n",
-      "Pg=5.628;#saturation pressure in kPa\n",
-      "Pv=r*Pg;#vapour pressure in kPa\n",
-      "P=100;#net pressure kPa \n",
-      "Pa=P-Pv;#Partial pressure of air\n",
-      "w=0.622*Pv/Pa;#humidity ratio formula\n",
-      "V=100;#volume in m^3\n",
-      "Ra=0.287;#gas constant for water vapour\n",
-      "T=308.2;#Temperature in K\n",
-      "ma=Pa*V/(Ra*T);#mass in kg\n",
-      "mv=w*ma;#mass of vapour\n",
-      "print\" Mass of vapour is\",round(mv,3),\"Kg \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Mass of vapour is 2.77 Kg \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.4:Pg-534"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques4 \n",
-      "#calculating amount of water vapour condensed on cooling\n",
-      "\n",
-      "#from example 3\n",
-      "w1=0.0255;#w1=w, humidity ratio at initial temperature\n",
-      "ma=108.6;#mass of air in kg\n",
-      "P=100;#kPa net pressure\n",
-      "#at 5 C mixture is saturated so Pv2=Pg2\n",
-      "Pg2=0.8721;\n",
-      "Pv2=Pg2;\n",
-      "w2=0.622*Pv2/(P-Pg2);\n",
-      "mc=ma*(w1-w2);\n",
-      "print\"Mass of vapour condense is \",round(mc,3),\"kg \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mass of vapour condense is  2.175 kg \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.5:Pg-536"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques5\n",
-      "#calculating heat transfer per kilogram of dry air\n",
-      "\n",
-      "#1-inlet state\n",
-      "#2-Exit state\n",
-      "r1=0.80;#realtive humidity at state 1\n",
-      "Pg1=4.246;#saturation pressure of vapour in kPa\n",
-      "P1=105.0;#net pressure at state 1 in kPa\n",
-      "P2=100.0;#net pressure at state 2 in kPa\n",
-      "Pv1=r1*Pg1;#partial pressure of vapour in kPa\n",
-      "w1=0.622*Pv1/(P1-Pv1);#humidity ratio at state 1\n",
-      "r2=0.95;#relative humidity at state 2\n",
-      "Pg2=1.7051;#saturation pressure of vapour in kPa\n",
-      "Pv2=r2*Pg2;#partial pressure of vapour in kPa\n",
-      "w2=0.622*Pv2/(P2-Pv2);#humidity ratio at state 2\n",
-      "T1=30.0;#C\n",
-      "T2=15.0;#C\n",
-      "Cp=1.004;#specific heat of water vapour in kJ/kg\n",
-      "hv2=2528.9;#enthalpy of vapourisation of vapour in kJ/kg\n",
-      "hv1=2556.3;#enthalpy of vapourisation of vapour in kJ/kg\n",
-      "hl2=62.99;#enthalpy of \n",
-      "q=Cp*(T2-T1)+w2*hv2-w1*hv1+hl2*(w1-w2);#kJ/kg\n",
-      "print\" Heat transferred per unit mass =\",round(q,1),\"kJ/kg of dry air\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Heat transferred per unit mass = -41.7 kJ/kg of dry air\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.6:Pg-537"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques6\n",
-      "#calculating heat transferred in gas vapour mixture\n",
-      "\n",
-      "#n-Nitrogen\n",
-      "#v-water vapour\n",
-      "Pn2=1995;#Pressure of nitrogen in kPa\n",
-      "V=0.5;#Volume in m^3\n",
-      "Rn2=0.2968;#Gas constant for nitrogen in kJ/kg.K\n",
-      "Rv=0.4615;#gas constant for vapour\n",
-      "T1=323.2;#Temperature in K\n",
-      "T2=283.2;#Temperature in K\n",
-      "Pv1=5;#Pressure of water vapour in kPa at state 1\n",
-      "Pv2=1.2276;#Pressure of water vapour in kPa at state 2\n",
-      "mn2=Pn2*V/(Rn2*T1);#mass of nitrogen\n",
-      "mv1=Pv1*V/(Rv*T1);#mass of vapour in kg\n",
-      "mv2=Pv2*V/(Rv*T2);#mass of vapour in kg\n",
-      "ml2=mv1-mv2;#mass of liquid condensed n kg\n",
-      "uv1=2443.1;#specific internal energy of vapour in kJ/kg at state 1\n",
-      "uv2=2389.2;#specific internal energy of vapour in kJ/kg at state 2\n",
-      "ul2=42.0;#specific internal energy of liquid water in kJ/kg\n",
-      "Cv=0.745;#specific heat at constant volume in kJ/kg.K\n",
-      "Q=mn2*Cv*(T2-T1)+mv2*uv2+ml2*ul2-mv1*uv1;\n",
-      "print\"Heat transferred =\",round(Q,2),\"kJ \"\n",
-      "\n",
-      "#the answer is different in book due to intermediate approximization "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Heat transferred = -339.1 kJ \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.7:Pg-539"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques7\n",
-      "#calculating humidity ratio and relative humidity \n",
-      "\n",
-      "#1-Inlet state\n",
-      "#2-Exit state\n",
-      "P=100;#net pressure n kPa \n",
-      "#it is steady state adiabatic process\n",
-      "#water vapour leaving is saturated so Pv2=Pg2\n",
-      "Pg2=2.339;#saturation pressure of vapour in kPa\n",
-      "Pv2=Pg2;#partial pressure of vapour\n",
-      "w2=0.622*Pv2/(P-Pg2);\n",
-      "Cpa=1.004;#specific heat n kJ/kg/K\n",
-      "T2=20;# final temp in C\n",
-      "T1=30;# initial temp in C\n",
-      "Hfg2=2454.1;#specific heat difference at state 2 in kJ/kg\n",
-      "hv1=2556.3;#enthalpy of water vapour at state 1 in kJ/kg\n",
-      "hl2=83.96;#enthalpy of liquid water in kJ/kg\n",
-      "w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);\n",
-      "print \" Relative humidity =\",round(w1,4)\n",
-      "#also w1=0.622*Pv1/(100-Pv2)\n",
-      "Pv1=100*w1/(0.622+w1);\n",
-      "Pg1=4.246;#saturation pressure at state 1 in kPa\n",
-      "r=Pv1/Pg1;#humidity ratio\n",
-      "print\" Humidity ratio =\",round(r,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Relative humidity = 0.0107\n",
-        " Humidity ratio = 0.3993\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter13_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter13_6.ipynb
deleted file mode 100755
index 5863fc4c..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter13_6.ipynb
+++ /dev/null
@@ -1,385 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:13df1b90ea06630bb54a561fa81d4ebbaaa2d68a225d0631603576a208acfb40"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "CHAPTER13:Gas Mixtures"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.3:Pg-533"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques3\n",

-      "#calculating humidity ratio, dew point,mass of air, mass of vapor\n",

-      "\n",

-      "r=0.70;#relative humidity\n",

-      "Pg=5.628;#saturation pressure in kPa\n",

-      "Pv=r*Pg;#vapour pressure in kPa\n",

-      "P=100;#net pressure kPa \n",

-      "Pa=P-Pv;#Partial pressure of air\n",

-      "w=0.622*Pv/Pa;#humidity ratio formula\n",

-      "V=100;#volume in m^3\n",

-      "Ra=0.287;#gas constant for water vapour\n",

-      "T=308.2;#Temperature in K\n",

-      "ma=Pa*V/(Ra*T);#mass in kg\n",

-      "mv=w*ma;#mass of vapour\n",

-      "print\" Mass of vapour is\",round(mv,3),\"Kg \""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Mass of vapour is 2.77 Kg \n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.3E:Pg-533"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques3\n",

-      "#calculating humidity ratio, dew point,mass of air, mass of vapor\n",

-      "\n",

-      "r=0.70;#relative humidity\n",

-      "Pg=0.6988 ;#saturation pressure in lbf/in^2\n",

-      "Pv=r*Pg;#vapour pressure in lbf/in^2\n",

-      "P=14.7;#net pressure in lbf/in^2\n",

-      "Pa=P-Pv;#Partial pressure of air\n",

-      "w=0.622*Pv/Pa;#humidity ratio formula\n",

-      "V=2000;#volume in ft^3\n",

-      "Ra=53.34;#gas constant for water vapour\n",

-      "T=70;#Temperature in F\n",

-      "ma=Pa*V*144/(Ra*(T+480));#mass in lbm\n",

-      "mv=w*ma;#mass of vapour\n",

-      "print\" Mass of vapour is\",round(mv,2),\"lbm\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Mass of vapour is 2.99 lbm\n"

-       ]

-      }

-     ],

-     "prompt_number": 11

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.4:Pg-534"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques4 \n",

-      "#calculating amount of water vapour condensed on cooling\n",

-      "\n",

-      "#from example 3\n",

-      "w1=0.0255;#w1=w, humidity ratio at initial temperature\n",

-      "ma=108.6;#mass of air in kg\n",

-      "P=100;#kPa net pressure\n",

-      "#at 5 C mixture is saturated so Pv2=Pg2\n",

-      "Pg2=0.8721;\n",

-      "Pv2=Pg2;\n",

-      "w2=0.622*Pv2/(P-Pg2);\n",

-      "mc=ma*(w1-w2);\n",

-      "print\"Mass of vapour condense is \",round(mc,3),\"kg \""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Mass of vapour condense is  2.175 kg \n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.4E:Pg-535"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques4 \n",

-      "#calculating amount of water vapour condensed on cooling\n",

-      "\n",

-      "#from example 3 \n",

-      "w1=0.02135;#w1=w, humidity ratio at initial temperature\n",

-      "ma=139.6;#mass of air in lbm\n",

-      "P=14.7;#net pressure in lbf/in^2\n",

-      "#at 40 degree F mixture is saturated so Pv2=Pg2\n",

-      "Pg2=0.1217 # lbf/in^2\n",

-      "Pv2=Pg2;\n",

-      "w2=0.622*Pv2/(P-Pg2);# humidity ratio at final temperature\n",

-      "mc=ma*(w1-w2); #Mass of vapour condense in lbm\n",

-      "print\"Mass of vapour condense is \",round(mc,3),\"lbm \""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Mass of vapour condense is  2.256 lbm \n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.5:Pg-536"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques5\n",

-      "#calculating heat transfer per kilogram of dry air\n",

-      "\n",

-      "#1-inlet state\n",

-      "#2-Exit state\n",

-      "r1=0.80;#realtive humidity at state 1\n",

-      "Pg1=4.246;#saturation pressure of vapour in kPa\n",

-      "P1=105.0;#net pressure at state 1 in kPa\n",

-      "P2=100.0;#net pressure at state 2 in kPa\n",

-      "Pv1=r1*Pg1;#partial pressure of vapour in kPa\n",

-      "w1=0.622*Pv1/(P1-Pv1);#humidity ratio at state 1\n",

-      "r2=0.95;#relative humidity at state 2\n",

-      "Pg2=1.7051;#saturation pressure of vapour in kPa\n",

-      "Pv2=r2*Pg2;#partial pressure of vapour in kPa\n",

-      "w2=0.622*Pv2/(P2-Pv2);#humidity ratio at state 2\n",

-      "T1=30.0;#C\n",

-      "T2=15.0;#C\n",

-      "Cp=1.004;#specific heat of water vapour in kJ/kg\n",

-      "hv2=2528.9;#enthalpy of vapourisation of vapour in kJ/kg\n",

-      "hv1=2556.3;#enthalpy of vapourisation of vapour in kJ/kg\n",

-      "hl2=62.99;#enthalpy of \n",

-      "q=Cp*(T2-T1)+w2*hv2-w1*hv1+hl2*(w1-w2);#kJ/kg\n",

-      "print\" Heat transferred per unit mass =\",round(q,1),\"kJ/kg of dry air\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Heat transferred per unit mass = -41.7 kJ/kg of dry air\n"

-       ]

-      }

-     ],

-     "prompt_number": 11

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.6:Pg-537"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques6\n",

-      "#calculating heat transferred in gas vapour mixture\n",

-      "\n",

-      "#n-Nitrogen\n",

-      "#v-water vapour\n",

-      "Pn2=1995;#Pressure of nitrogen in kPa\n",

-      "V=0.5;#Volume in m^3\n",

-      "Rn2=0.2968;#Gas constant for nitrogen in kJ/kg.K\n",

-      "Rv=0.4615;#gas constant for vapour\n",

-      "T1=323.2;#Temperature in K\n",

-      "T2=283.2;#Temperature in K\n",

-      "Pv1=5;#Pressure of water vapour in kPa at state 1\n",

-      "Pv2=1.2276;#Pressure of water vapour in kPa at state 2\n",

-      "mn2=Pn2*V/(Rn2*T1);#mass of nitrogen\n",

-      "mv1=Pv1*V/(Rv*T1);#mass of vapour in kg\n",

-      "mv2=Pv2*V/(Rv*T2);#mass of vapour in kg\n",

-      "ml2=mv1-mv2;#mass of liquid condensed n kg\n",

-      "uv1=2443.1;#specific internal energy of vapour in kJ/kg at state 1\n",

-      "uv2=2389.2;#specific internal energy of vapour in kJ/kg at state 2\n",

-      "ul2=42.0;#specific internal energy of liquid water in kJ/kg\n",

-      "Cv=0.745;#specific heat at constant volume in kJ/kg.K\n",

-      "Q=mn2*Cv*(T2-T1)+mv2*uv2+ml2*ul2-mv1*uv1;\n",

-      "print\"Heat transferred =\",round(Q,2),\"kJ \"\n",

-      "\n",

-      "#the answer is different in book due to intermediate approximization "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Heat transferred = -339.1 kJ \n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.7:Pg-539"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#calculating humidity ratio and relative humidity \n",

-      "\n",

-      "#1-Inlet state\n",

-      "#2-Exit state\n",

-      "P=100;#net pressure in kPa \n",

-      "#it is steady state adiabatic process\n",

-      "#water vapour leaving is saturated so Pv2=Pg2\n",

-      "Pg2=2.339;#saturation pressure of vapour in kPa\n",

-      "Pv2=Pg2;#partial pressure of vapour\n",

-      "w2=0.622*Pv2/(P-Pg2);\n",

-      "Cpa=1.004;#specific heat n kJ/kg/K\n",

-      "T2=20;# final temp in C\n",

-      "T1=30;# initial temp in C\n",

-      "Hfg2=2454.1;#specific heat difference at state 2 in kJ/kg\n",

-      "hv1=2556.3;#enthalpy of water vapour at state 1 in kJ/kg\n",

-      "hl2=83.96;#enthalpy of liquid water in kJ/kg\n",

-      "w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);\n",

-      "print \" Relative humidity =\",round(w1,4)\n",

-      "#also w1=0.622*Pv1/(100-Pv2)\n",

-      "Pv1=100*w1/(0.622+w1);\n",

-      "Pg1=4.246;#saturation pressure at state 1 in kPa\n",

-      "r=Pv1/Pg1;#humidity ratio\n",

-      "print\" Humidity ratio =\",round(r,4)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Relative humidity = 0.0107\n",

-        " Humidity ratio = 0.3993\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex13.7E:Pg-540"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#calculating humidity ratio and relative humidity \n",

-      "\n",

-      "#1-Inlet state\n",

-      "#2-Exit state\n",

-      "P=14.7;#net pressure in lbf/in^2\n",

-      "#it is steady state adiabatic process\n",

-      "#water vapour leaving is saturated so Pv2=Pg2\n",

-      "Pg2=0.3632;#saturation pressure of vapour in lbf/in^2\n",

-      "Pv2=Pg2;#partial pressure of vapour\n",

-      "w2=0.622*Pv2/(P-Pg2);\n",

-      "Cpa=0.24;#specific heat n Btu/lbm/F\n",

-      "T2=70;# final temp in F\n",

-      "T1=84;# initial temp in F\n",

-      "Hfg2=1054.0;#specific heat difference at state 2 in Btu/lbm\n",

-      "hv1=1098.1;#enthalpy of water vapour at state 1 in Btu/lbm\n",

-      "hl2=38.1;#enthalpy of liquid water in Btu/lbm\n",

-      "w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);\n",

-      "print \" Relative humidity =\",round(w1,4)\n",

-      "#also w1=0.622*Pv1/(100-Pv2)\n",

-      "Pv1=14.7*w1/(0.622+w1);\n",

-      "Pg1=0.584;#saturation pressure at state 1 in lbf/in^2\n",

-      "r=Pv1/Pg1;#humidity ratio\n",

-      "print\" Humidity ratio =\",round(r,4)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Relative humidity = 0.0125\n",

-        " Humidity ratio = 0.4958\n"

-       ]

-      }

-     ],

-     "prompt_number": 18

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter14.ipynb b/Fundamentals_Of_Thermodynamics/Chapter14.ipynb
deleted file mode 100755
index 6654362c..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter14.ipynb
+++ /dev/null
@@ -1,295 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b6f84d510db3595ba71244d9d44f7e9987d44c62ba0d0520a72da2384551019e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter14:THERMODYNAMIC RELATIONS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.1:Pg-567"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques1\n",
-      "#to determine the sublimation pressure of water\n",
-      "import math\n",
-      "#from table in appendix B.1.5\n",
-      "T1=213.2;#K, Temperature at state 1\n",
-      "P2=0.0129;#kPa, pressure at state 2\n",
-      "T2=233.2;#K, Temperature at state 2\n",
-      "hig=2838.9;#kJ/kg, enthalpy of sublimation \n",
-      "R=.46152;#Gas constant \n",
-      "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n",
-      "P1=P2*math.exp(-hig/R*(1/T1-1/T2));\n",
-      "print\" Sublimation Pressure \",round(P1,5),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Sublimation Pressure  0.00109 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.4:Pg-579"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques4\n",
-      "#Volume expansivity, Isothermal and Adiabatic compressibility\n",
-      "\n",
-      "#known data\n",
-      "ap=5*10**-5;#K^-1 Volume expansivity\n",
-      "bt=8.6*10**-12;#m^2/N, Isothermal compressibility\n",
-      "v=0.000114;#m^3/kg, specific volume\n",
-      "P2=100*10**6;#pressure at state 2 in kPa\n",
-      "P1=100;#pressure at state 1 in kPa\n",
-      "w=-v*bt*(P2**2-P1**2)/2;#work done in J/kg\n",
-      "#q=T*ds and ds=-v*ap*(P2-P1)\n",
-      "#so q=-T*v*ap*(P2-P1)\n",
-      "T=288.2;#Temperature in K\n",
-      "q=-T*v*ap*(P2-P1);#heat in J/kg\n",
-      "du=q-w;#change in internal energy in J/kg\n",
-      "print\" Change in internal energy =\",round(du,3),\"J/kg\"\n",
-      "\n",
-      "#the answer is correct within given limts\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Change in internal energy = -159.372 J/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.5:Pg-586"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques5\n",
-      "#adiabatic steady state processes\n",
-      "\n",
-      "#from table A.2\n",
-      "P1=20;#pressure at state 1 in MPa\n",
-      "P2=2;#pressure at state 2 in MPa\n",
-      "T1=203.2;#Temperature at state 1 in K\n",
-      "Pr1=P1/3.39;#Reduced pressure at state 1\n",
-      "Pr2=P2/3.39;#Reduced pressure at state 2\n",
-      "Tr1=T1/126.2;#Reduced temperature\n",
-      "#from compressibility chart h1*-h1=2.1*R*Tc\n",
-      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",
-      "#h2*-h2=0.5*R*Tc\n",
-      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc\n",
-      "R=0.2968;#gas constant for given substance\n",
-      "Tc=126.2;#K, Constant temperature\n",
-      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",
-      "T2=146;#temperature at state 2\n",
-      "dh=-1.6*R*Tc+Cp*(T1-T2);#\n",
-      "print\" Enthalpy change =\",round(dh,3),\"kJ/kg \\n\"\n",
-      "print\" Since Enthalpy change is nearly \",-round(dh),\"kJ/kg so Temperature =\",round(T2,3),\"K\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Enthalpy change = -0.35 kJ/kg \n",
-        "\n",
-        " Since Enthalpy change is nearly  0.0 kJ/kg so Temperature = 146.0 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.6:Pg-589"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques6\n",
-      "#isothermal steady state processes\n",
-      "import math\n",
-      "#from table A.2\n",
-      "P1=8;#pressure at state 1 in MPa\n",
-      "P2=0.5;#pressure at state 2 in MPa\n",
-      "T1=150.0;#Temperature at state 1 in K\n",
-      "Pr1=P1/3.39;#Reduced pressure at state 1\n",
-      "Pr2=P2/3.39;#Reduced pressure at state 2\n",
-      "Tr1=T1/126.2;#Reduced temperature\n",
-      "T2=125.0;#temperature at state 2\n",
-      "#from compressibility chart h1*-h1=2.1*R*Tc\n",
-      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",
-      "#h2*-h2=0.5*R*Tc\n",
-      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc\n",
-      "R=0.2968;#gas constant for given substance\n",
-      "Tc=126.2;#K, Constant temperature\n",
-      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",
-      "dh=(2.35)*R*Tc+Cp*(T2-T1);#\n",
-      "print\" Enthalpy change =\",round(dh),\"kJ/kg\"\n",
-      "#change in entropy \n",
-      "#ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)\n",
-      "#s1*-s1=1.6*R\n",
-      "#s2*-s2=0.1*R\n",
-      "#s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)\n",
-      "#so\n",
-      "ds=1.6*R-0.1*R+Cp*math.log(T2/T1)-R*math.log(P2/P1);\n",
-      "print\" Entropy Change =\",round(ds,3),\"kJ/kg.K \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Enthalpy change = 62.0 kJ/kg\n",
-        " Entropy Change = 1.078 kJ/kg.K \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.7:Pg-596"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques7\n",
-      "#percent deviation using specific volume calculated by kays rule and vander waals rule\n",
-      "import math\n",
-      "\n",
-      "#a-denotes C02\n",
-      "#b-denotes CH4\n",
-      "T=310.94;#Temperature of mixture K\n",
-      "P=86.19;#Pressure of mixture in MPa\n",
-      "#Tc- critical Temperature\n",
-      "#Pc-critical pressure\n",
-      "Tca=304.1;#K\n",
-      "Tcb=190.4;#K\n",
-      "Pca=7.38;#MPa\n",
-      "Pcb=4.60;#MPa\n",
-      "Ra=0.1889;#gas constant for a in kJ/kg.K\n",
-      "Rb=0.5183;#gas constant for b in kJ/kg.K\n",
-      "xa=0.8;#fraction of CO2\n",
-      "xb=0.2;#fraction of CH4\n",
-      "Rm=xa*Ra+xb*Rb;#mean gas constant in kJ/kg.K\n",
-      "Ma=44.01;#molecular mass of a\n",
-      "Mb=16.043;#molecular mass of b\n",
-      "#1.Kay's rule\n",
-      "ya=xa/Ma/(xa/Ma+xb/Mb);#mole fraction of a\n",
-      "yb=xb/Mb/(xa/Ma+xb/Mb);#mole fraction of b\n",
-      "Tcm=ya*Tca+yb*Tcb;#mean critical temp in K\n",
-      "Pcm=ya*Pca+yb*Tcb;#mean critical pressure n MPa\n",
-      "#therefore pseudo reduced property of mixture\n",
-      "Trm=T/Tcm;\n",
-      "Prm=P/Pcm;\n",
-      "Zm=0.7;#Compressiblity from generalised compressibility chart\n",
-      "vc=Zm*Rm*T/P/1000;#specific volume calculated in m^3/kg\n",
-      "ve=0.0006757;#experimental specific volume in m^3/kg\n",
-      "pd1=(ve-vc)/ve*100;#percent deviation\n",
-      "print\" Percentage deviation in specific volume using Kays rule =\",round(pd1,1),\"percent \\n\"\n",
-      "\n",
-      "#2. using vander waals equation\n",
-      "#values of vander waals constant\n",
-      "Aa=27*(Ra**2)*(Tca**2)/(64*Pca*1000);\n",
-      "Ba=Ra*Tca/(8*Pca*1000);\n",
-      "Ab=27*(Rb**2)*(Tcb**2)/(64*Pcb*1000);\n",
-      "Bb=Rb*Tcb/(8*Pcb*1000);\n",
-      "#mean vander waals constant\n",
-      "Am=(xa*math.sqrt(Aa)+xb*math.sqrt(Ab))**2;\n",
-      "Bm=(xa*Ba+xb*Bb);\n",
-      "#using vander waals equation we get cubic equation \n",
-      "#solving we get\n",
-      "vc=0.0006326;#calculated specific volume in m^3/kg\n",
-      "pd2=((ve-vc)/ve)*100;\n",
-      "print\" Percentage deviation in specific volume using vander waals eqn =\",round(pd2,1),\"percent\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Percentage deviation in specific volume using Kays rule = 4.8 percent \n",
-        "\n",
-        " Percentage deviation in specific volume using vander waals eqn = 6.4 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter14_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter14_6.ipynb
deleted file mode 100755
index 8c5a5922..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter14_6.ipynb
+++ /dev/null
@@ -1,325 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:4d90868c7316d94fc2ffdc05c5691234f77f3a832d71da7eb2108eaff0a215bc"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter14:THERMODYNAMIC RELATIONS"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.1:Pg-567"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques1\n",

-      "#to determine the sublimation pressure of water\n",

-      "import math\n",

-      "#from table in appendix B.1.5\n",

-      "T1=213.2;#K, Temperature at state 1\n",

-      "P2=0.0129;#kPa, pressure at state 2\n",

-      "T2=233.2;#K, Temperature at state 2\n",

-      "hig=2838.9;#kJ/kg, enthalpy of sublimation \n",

-      "R=.46152;#Gas constant \n",

-      "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n",

-      "P1=P2*math.exp(-hig/R*(1/T1-1/T2));\n",

-      "print\" Sublimation Pressure \",round(P1,5),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Sublimation Pressure  0.00109 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 25

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.1E:Pg-567\n"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques1\n",

-      "#to determine the sublimation pressure of water\n",

-      "import math\n",

-      "#from table in appendix B.1.5\n",

-      "T1=-70+460.7;# R, Temperature at state 1\n",

-      "P2=0.0019 # lbf/in^2 pressure at state 2\n",

-      "T2=-40+460.7;# R, Temperature at state 2\n",

-      "hig=1218.7;#Btu/lbm, enthalpy of sublimation \n",

-      "R=85.67;#Gas constant \n",

-      "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n",

-      "P1=P2*math.exp(-hig*778/R*(1/T1-1/T2));\n",

-      "print\" Sublimation Pressure \",round(P1,5),\"lbf/in^2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Sublimation Pressure  0.00025 lbf/in^2\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.4:Pg-579"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques4\n",

-      "#Volume expansivity, Isothermal and Adiabatic compressibility\n",

-      "\n",

-      "#known data\n",

-      "ap=5*10**-5;#K^-1 Volume expansivity\n",

-      "bt=8.6*10**-12;#m^2/N, Isothermal compressibility\n",

-      "v=0.000114;#m^3/kg, specific volume\n",

-      "P2=100*10**6;#pressure at state 2 in kPa\n",

-      "P1=100;#pressure at state 1 in kPa\n",

-      "w=-v*bt*(P2**2-P1**2)/2;#work done in J/kg\n",

-      "#q=T*ds and ds=-v*ap*(P2-P1)\n",

-      "#so q=-T*v*ap*(P2-P1)\n",

-      "T=288.2;#Temperature in K\n",

-      "q=-T*v*ap*(P2-P1);#heat in J/kg\n",

-      "du=q-w;#change in internal energy in J/kg\n",

-      "print\" Change in internal energy =\",round(du,3),\"J/kg\"\n",

-      "\n",

-      "#the answer is correct within given limts\n",

-      " "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Change in internal energy = -159.372 J/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 22

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.5:Pg-586"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques5\n",

-      "#adiabatic steady state processes\n",

-      "\n",

-      "#from table A.2\n",

-      "P1=20;#pressure at state 1 in MPa\n",

-      "P2=2;#pressure at state 2 in MPa\n",

-      "T1=203.2;#Temperature at state 1 in K\n",

-      "Pr1=P1/3.39;#Reduced pressure at state 1\n",

-      "Pr2=P2/3.39;#Reduced pressure at state 2\n",

-      "Tr1=T1/126.2;#Reduced temperature\n",

-      "#from compressibility chart h1*-h1=2.1*R*Tc\n",

-      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",

-      "#h2*-h2=0.5*R*Tc\n",

-      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc\n",

-      "R=0.2968;#gas constant for given substance\n",

-      "Tc=126.2;#K, Constant temperature\n",

-      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",

-      "T2=146;#temperature at state 2\n",

-      "dh=-1.6*R*Tc+Cp*(T1-T2);#\n",

-      "print\" Enthalpy change =\",round(dh,3),\"kJ/kg \\n\"\n",

-      "print\" Since Enthalpy change is nearly \",-round(dh),\"kJ/kg so Temperature =\",round(T2,3),\"K\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Enthalpy change = -0.35 kJ/kg \n",

-        "\n",

-        " Since Enthalpy change is nearly  0.0 kJ/kg so Temperature = 146.0 K\n"

-       ]

-      }

-     ],

-     "prompt_number": 19

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.6:Pg-589"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques6\n",

-      "#isothermal steady state processes\n",

-      "import math\n",

-      "#from table A.2\n",

-      "P1=8;#pressure at state 1 in MPa\n",

-      "P2=0.5;#pressure at state 2 in MPa\n",

-      "T1=150.0;#Temperature at state 1 in K\n",

-      "Pr1=P1/3.39;#Reduced pressure at state 1\n",

-      "Pr2=P2/3.39;#Reduced pressure at state 2\n",

-      "Tr1=T1/126.2;#Reduced temperature\n",

-      "T2=125.0;#temperature at state 2\n",

-      "#from compressibility chart h1*-h1=2.1*R*Tc\n",

-      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",

-      "#h2*-h2=0.5*R*Tc\n",

-      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc\n",

-      "R=0.2968;#gas constant for given substance\n",

-      "Tc=126.2;#K, Constant temperature\n",

-      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",

-      "dh=(2.35)*R*Tc+Cp*(T2-T1);#\n",

-      "print\" Enthalpy change =\",round(dh),\"kJ/kg\"\n",

-      "#change in entropy \n",

-      "#ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)\n",

-      "#s1*-s1=1.6*R\n",

-      "#s2*-s2=0.1*R\n",

-      "#s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)\n",

-      "#so\n",

-      "ds=1.6*R-0.1*R+Cp*math.log(T2/T1)-R*math.log(P2/P1);\n",

-      "print\" Entropy Change =\",round(ds,3),\"kJ/kg.K \""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Enthalpy change = 62.0 kJ/kg\n",

-        " Entropy Change = 1.078 kJ/kg.K \n"

-       ]

-      }

-     ],

-     "prompt_number": 15

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.7:Pg-596"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#percent deviation using specific volume calculated by kays rule and vander waals rule\n",

-      "import math\n",

-      "\n",

-      "#a-denotes C02\n",

-      "#b-denotes CH4\n",

-      "T=310.94;#Temperature of mixture K\n",

-      "P=86.19;#Pressure of mixture in MPa\n",

-      "#Tc- critical Temperature\n",

-      "#Pc-critical pressure\n",

-      "Tca=304.1;#K\n",

-      "Tcb=190.4;#K\n",

-      "Pca=7.38;#MPa\n",

-      "Pcb=4.60;#MPa\n",

-      "Ra=0.1889;#gas constant for a in kJ/kg.K\n",

-      "Rb=0.5183;#gas constant for b in kJ/kg.K\n",

-      "xa=0.8;#fraction of CO2\n",

-      "xb=0.2;#fraction of CH4\n",

-      "Rm=xa*Ra+xb*Rb;#mean gas constant in kJ/kg.K\n",

-      "Ma=44.01;#molecular mass of a\n",

-      "Mb=16.043;#molecular mass of b\n",

-      "#1.Kay's rule\n",

-      "ya=xa/Ma/(xa/Ma+xb/Mb);#mole fraction of a\n",

-      "yb=xb/Mb/(xa/Ma+xb/Mb);#mole fraction of b\n",

-      "Tcm=ya*Tca+yb*Tcb;#mean critical temp in K\n",

-      "Pcm=ya*Pca+yb*Tcb;#mean critical pressure n MPa\n",

-      "#therefore pseudo reduced property of mixture\n",

-      "Trm=T/Tcm;\n",

-      "Prm=P/Pcm;\n",

-      "Zm=0.7;#Compressiblity from generalised compressibility chart\n",

-      "vc=Zm*Rm*T/P/1000;#specific volume calculated in m^3/kg\n",

-      "ve=0.0006757;#experimental specific volume in m^3/kg\n",

-      "pd1=(ve-vc)/ve*100;#percent deviation\n",

-      "print\" Percentage deviation in specific volume using Kays rule =\",round(pd1,1),\"percent \\n\"\n",

-      "\n",

-      "#2. using vander waals equation\n",

-      "#values of vander waals constant\n",

-      "Aa=27*(Ra**2)*(Tca**2)/(64*Pca*1000);\n",

-      "Ba=Ra*Tca/(8*Pca*1000);\n",

-      "Ab=27*(Rb**2)*(Tcb**2)/(64*Pcb*1000);\n",

-      "Bb=Rb*Tcb/(8*Pcb*1000);\n",

-      "#mean vander waals constant\n",

-      "Am=(xa*math.sqrt(Aa)+xb*math.sqrt(Ab))**2;\n",

-      "Bm=(xa*Ba+xb*Bb);\n",

-      "#using vander waals equation we get cubic equation \n",

-      "#solving we get\n",

-      "vc=0.0006326;#calculated specific volume in m^3/kg\n",

-      "pd2=((ve-vc)/ve)*100;\n",

-      "print\" Percentage deviation in specific volume using vander waals eqn =\",round(pd2,1),\"percent\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Percentage deviation in specific volume using Kays rule = 4.8 percent \n",

-        "\n",

-        " Percentage deviation in specific volume using vander waals eqn = 6.4 percent\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter15.ipynb b/Fundamentals_Of_Thermodynamics/Chapter15.ipynb
deleted file mode 100755
index 18f2c842..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter15.ipynb
+++ /dev/null
@@ -1,248 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:49d323999697626f705831867b32f6c520c5287d76ed00656f7259e6db65356e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter15:CHEMICAL REACTIONS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex15.1:Pg-621"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques1\n",
-      "#theoratical air-fuel ratio for combustion of octane\n",
-      "#combustion equation is\n",
-      "#C8H18 + 12.5O2 + 12.5(3.76) N2 \u2192 8 CO2 + 9H2O + 47.0N2\n",
-      "rm=(12.5+47.0)/1;#air fuel ratio on mole basis\n",
-      "rma=rm*28.97/114.2;#air fuel ratio on mass basis;\n",
-      "print \"Theoratical air fuel ratio on mass basis is\",round(rma),\"kg air/kg fuel\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Theoratical air fuel ratio on mass basis is 15.0 kg air/kg fuel\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex15.6:Pg-629"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques6\n",
-      "#determining heat transfer per kilomole of fuel entering combustion chamber\n",
-      "\n",
-      "#1-CH4\n",
-      "#2-CO2\n",
-      "#3-H2O\n",
-      "#hf-standard enthalpy of given substance\n",
-      "hf1=-74.873;#kJ\n",
-      "hf2=-393.522;#kJ\n",
-      "hf3=-285.830;#kJ\n",
-      "Qcv=hf2+2*hf3-hf1;#kJ\n",
-      "print \"Heat transfer per kilomole of fuel entering combustion chamber is\",round(Qcv,3),\"kJ\"\n",
-      "#the answers in the book is different as they have not printed the decimals in values"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Heat transfer per kilomole of fuel entering combustion chamber is -890.309 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex15.7:Pg-631"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques7\n",
-      "#calculating enthalpy of water at given pressure and temperature\n",
-      "\n",
-      "#1.Assuming steam to be an ideal gas with value of Cp\n",
-      "T1=298.15;#Initial temperature in K\n",
-      "T2=573.15;#final temperature in K\n",
-      "T=(T1+T2)/2;#average temperature in K\n",
-      "Cp=1.79+0.107*T/1000+0.586*(T/1000)**2-.20*(T/1000)**3;#specific heat at constant pressure in kj/kg.K\n",
-      "M=18.015;#mass in kg\n",
-      "dh=M*Cp*(T2-T1);#enthalpy change in kJ/kmol\n",
-      "ho=-241.826;#enthalpy at standard temperature and pressure in kJ/mol\n",
-      "htp1=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",
-      "print \" 1. Enthalpy of water at given pressure and temperature using value of Cp =\",round(htp1,3),\"kJ/kmol\"\n",
-      "\n",
-      "#2..Assuming steam to be an ideal gas with value for dh\n",
-      "dh=9359;#enthalpy change from table A.9 in kJ/mol\n",
-      "htp2=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",
-      "print \" 2. Enthalpy of water at given pressure and temperature assuming value od dh =\",round(htp2,3),\"kJ/kmol \"\n",
-      "\n",
-      "#3. Using steam table\n",
-      "dh=M*(2977.5-2547.2);#enthalpy change for gases in kJ/mol\n",
-      "htp3g=dh/1000+ho;\n",
-      "dh=M*(2977.5-104.9);#enthalpy change for liquid in kJ/mol\n",
-      "hl=-285.830;#standard enthalpy for liquid in kJ/kmol\n",
-      "htp31=hl+dh/1000.0;#enthalpy at given temp and pressure in kJ/kmol\n",
-      "print \" 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp31,3),\"kJ/kmol \"\n",
-      "print \" 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp3g,3),\"kJ/kmol \"\n",
-      "#4.using generalised charts\n",
-      "#htp=ho-(h2*-h2)+(h2*-h1*)+(h1*-h1);\n",
-      "#h2*-h2=Z*R*Tc,\n",
-      "#h2*-h1*=9539 kJ/mol, from part 2\n",
-      "#h1*-h1=0 ,as ideal gas \n",
-      "Z=0.21;#from chart\n",
-      "R=8.3145;#gas constant in SI units\n",
-      "Tc=647.3;#critical temperature in K\n",
-      "htp4=ho+9539/1000-Z*R*Tc/1000;#enthalpy at given temp and pressure in kJ/kmol\n",
-      "print \" 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart = \",round(htp4,3),\"kJ/kmol\"\n",
-      "#the answers in book are different as they have not printed the decimals in values"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1. Enthalpy of water at given pressure and temperature using value of Cp = -232.258 kJ/kmol\n",
-        " 2. Enthalpy of water at given pressure and temperature assuming value od dh = -232.826 kJ/kmol \n",
-        " 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.08 kJ/kmol \n",
-        " 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.074 kJ/kmol \n",
-        " 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart =  -233.956 kJ/kmol\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex15.15:Pg-649"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques15\n",
-      "#calculatng reversible elecromotive force \n",
-      "\n",
-      "#1-H2O\n",
-      "#2-H2\n",
-      "#3-O2\n",
-      "#hf-standard enthalpy \n",
-      "#sf-standard entropy\n",
-      "hf1=-285.830;#kJ\n",
-      "hf2=0;#kJ\n",
-      "hf3=0;#kJ\n",
-      "sf1=69.950;#kJ/K\n",
-      "sf2=130.678;#kJ/K\n",
-      "sf3=205.148;#kJ/K\n",
-      "dH=2*hf1-2*hf2-hf3;#change in enthalpy in kJ\n",
-      "dS=2*sf1-2*sf2-sf3;#change in entropy in kJ/K\n",
-      "T=298.15;#temperature in K\n",
-      "dG=dH-T*dS/1000;#change in gibbs free energy in kJ\n",
-      "E=-dG*1000/(96485*4);#emf in V\n",
-      "print\" Reversible electromotive Force =\",round(E,3),\" V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Reversible electromotive Force = 1.229  V\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex15.17:Pg-654"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques17\n",
-      "#efficiency of generator and plant\n",
-      "\n",
-      "q=325000*(3398.3-856.0);#heat transferred to H2O/kg fuel in kJ/kg\n",
-      "qv=26700.0*33250;#higher heating value in kJ/kg\n",
-      "nst=q/qv*100;#efficiency of steam generator\n",
-      "w=81000.0*3600;#net work done in kJ/kg\n",
-      "nth=w/qv*100.0;#thermal efficiency\n",
-      "print\" Efficiency of generator =\",round(nst,1),\"percent\\n\"\n",
-      "print\" Thermal Efficiency =\",round(nth,1),\" percent\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Efficiency of generator = 93.1 percent\n",
-        "\n",
-        " Thermal Efficiency = 32.8  percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter15_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter15_6.ipynb
deleted file mode 100755
index 1c9182a7..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter15_6.ipynb
+++ /dev/null
@@ -1,921 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:4afaac3b11e3c61840a392c8cec12435d16ae2f3d2cdfe278957aaca509d72f5"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter15:CHEMICAL REACTIONS"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.1:Pg-621"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques1\n",

-      "#theoratical air-fuel ratio for combustion of octane\n",

-      "#combustion equation is\n",

-      "#C8H18 + 12.5O2 + 12.5(3.76) N2 \u2192 8 CO2 + 9H2O + 47.0N2\n",

-      "rm=(12.5+47.0)/1;#air fuel ratio on mole basis\n",

-      "rma=rm*28.97/114.2;#air fuel ratio on mass basis;\n",

-      "print \"Theoratical air fuel ratio on mass basis is\",round(rma),\"kg air/kg fuel\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Theoratical air fuel ratio on mass basis is 15.0 kg air/kg fuel\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.2:Pg-622"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 2\n",

-      "# molal analysis of the products of combustion\n",

-      "# combustion equation is \n",

-      "# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 \u2192 8 CO2 + 9H2O + 12.5O2 + 94.0N2\n",

-      "nCO2=8 # number of moles of CO2 \n",

-      "nH2O=9 # number of moles of H2O\n",

-      "nO2=12.5 # number of moles of O2\n",

-      "nN2=94.0 # number of moles of N2\n",

-      "moles=nCO2+nH2O+nO2+nN2 # total moles\n",

-      "# molal analysis of products\n",

-      "xCO2=nCO2*100/moles \n",

-      "xH2O=nH2O*100/moles\n",

-      "xO2=nO2*100/moles\n",

-      "xN2=nN2*100/moles\n",

-      "print \" CO2 =\",round(xCO2,2)\n",

-      "print \"\\n H2O=\",round(xH2O,2)\n",

-      "print \"\\n O2=\",round(xO2,2)\n",

-      "print \"\\n N2=\",round(xN2,2)\n",

-      "print \"\\n The dew point temp. corresponding to partial pressure of water i.e \",round(xH2O,2),\"KPa is 39.7 degree C\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " CO2 = 6.48\n",

-        "\n",

-        " H2O= 7.29\n",

-        "\n",

-        " O2= 10.12\n",

-        "\n",

-        " N2= 76.11\n",

-        "\n",

-        " The dew point temp. corresponding to partial pressure of water i.e  7.29 KPa is 39.7 degree C\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.2E:Pg-622"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 2\n",

-      "# molal analysis of the products of combustion\n",

-      "# combustion equation is \n",

-      "# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 \u2192 8 CO2 + 9H2O + 12.5O2 + 94.0N2\n",

-      "pressure=14.7 # in lbf/in^2\n",

-      "nCO2=8 # number of moles of CO2 \n",

-      "nH2O=9 # number of moles of H2O\n",

-      "nO2=12.5 # number of moles of O2\n",

-      "nN2=94.0 # number of moles of N2\n",

-      "moles=nCO2+nH2O+nO2+nN2 # total moles\n",

-      "# molal analysis of products\n",

-      "xCO2=nCO2*100/moles \n",

-      "xH2O=nH2O*100/moles\n",

-      "xO2=nO2*100/moles\n",

-      "xN2=nN2*100/moles\n",

-      "print \" CO2 =\",round(xCO2,2)\n",

-      "print \"\\n H2O=\",round(xH2O,2)\n",

-      "print \"\\n O2=\",round(xO2,2)\n",

-      "print \"\\n N2=\",round(xN2,2)\n",

-      "PressureO2=pressure*xH2O/100 # partial pressure of O2\n",

-      "print \"\\n The dew point temp. corresponding to partial pressure of water i.e \",round(PressureO2,3),\"lbf/in^2 is 104 F degree C\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " CO2 = 6.48\n",

-        "\n",

-        " H2O= 7.29\n",

-        "\n",

-        " O2= 10.12\n",

-        "\n",

-        " N2= 76.11\n",

-        "\n",

-        " The dew point temp. corresponding to partial pressure of water i.e  1.071 lbf/in^2 is 104 F degree C\n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.3:Pg-623"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 3\n",

-      "# to determine the air\u2013fuel ratio on a volumetric basis\n",

-      "# the combustion equation is\n",

-      "# 0.14H2+0.27CO+0.03CH4+0.006O2+0.509N2+0.045CO2+0.259O2+0.259(3.76)N2\u21920.20H2O+0.345CO2+1.482 N2\n",

-      "mH2=2 # molar mass of H2\n",

-      "mCO=28 # molar mass of CO\n",

-      "mCH4=16 # molar mass of CH4\n",

-      "mO2=32 # molar mass of O2\n",

-      "mN2=28 # molar mass of N2\n",

-      "mCO2=44 # molar mass of CO2\n",

-      "\n",

-      "\n",

-      "kmolair=0.259 # moles of air in mixture using combustion equation\n",

-      "kmolfuel=0.21 # moles of fuel in mixture using combustion equation\n",

-      "ratioth=kmolair/kmolfuel # theoratical ratio\n",

-      "\n",

-      "# For 20% excess air\n",

-      "vratio=ratioth*(1+0.2) # new ratio for excess air on volume basis\n",

-      "\n",

-      "print \" The A/F ratio on volumetric basis is \",round(vratio,2)\n",

-      "mratio=vratio*28.97 /(0.14*mH2+0.27*mCO+0.03*mCH4+0.006*mO2+0.509*mN2+0.045*mCO2)\n",

-      "print \"\\n The A/F ratio on mass basis is \",round(mratio,2),\"kg air/kg fuel\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " The A/F ratio on volumetric basis is  1.48\n",

-        "\n",

-        " The A/F ratio on mass basis is  1.73 kg air/kg fuel\n"

-       ]

-      }

-     ],

-     "prompt_number": 11

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.4:Pg-624"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 4\n",

-      "# determining the air\u2013fuel ratio and the percent theoretical air\n",

-      "CO2=10.00 # percentage\n",

-      "O2=2.37 # percentage\n",

-      "CO=0.53 # percentage\n",

-      "N2=87.10 # percentage\n",

-      "mCH4=16 # molecular mass of CH4\n",

-      "# using this data the combustion equation is\n",

-      "# a CH4 + b O2 + c N2 \u2192 10.0 CO2 + 0.53 CO + 2.37 O2 + d H2O + 87.1N2\n",

-      "# on balancing we get\n",

-      "# 10.53 CH4 + 23.16 O2 + 87.1N2 \u219210.0 CO2 + 0.53 CO + 2.37 O2 + 21.06 H2O + 87.1N2\n",

-      "# balanced combustion equation for per Kmole of fuel is:\n",

-      "# CH4 + 2.2O2 + 8.27 N2 \u2192 0.95 CO2 + 0.05 CO + 2H2O + 0.225 O2 + 8.27 N2\n",

-      "molesAir=2.2+8.27 # moles of air\n",

-      "AF= 28.97*molesAir/mCH4 # air fuel ratio on mass basis\n",

-      "print \"\\n The A/F ratio on mass basis is \",round(AF,2),\"kg air/kg fuel\"\n",

-      "# Theoritical combustion equation:\n",

-      "# CH4 + 2O2 + 2(3.76)N2 \u2192 CO2 + 2H2O + 7.52 N2\n",

-      "molesAirth=2+2*3.76 # moles of air(theoritical)\n",

-      "AFth= 28.97*molesAirth/mCH4 # air fuel ratio on mass basis (theoritical)\n",

-      "print \"\\n The A/F ratio (theoritical) on mass basis is \",round(AFth,2),\"kg air/kg fuel\"\n",

-      "print \"\\n The percent theoretical air is\",round(AF*100/AFth,1)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " The A/F ratio on mass basis is  18.96 kg air/kg fuel\n",

-        "\n",

-        " The A/F ratio (theoritical) on mass basis is  17.24 kg air/kg fuel\n",

-        "\n",

-        " The percent theoretical air is 110.0\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.5:Pg-625"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 5\n",

-      "\n",

-      "Sulfur=0.6 # percent by mass\n",

-      "Hydrogen=5.7 # percent by mass\n",

-      "Carbon=79.2 # percent by mass\n",

-      "Oxygen=10.0 # percent by mass\n",

-      "Nitrogen=1.5 # percent by mass\n",

-      "Ash=3.0 # percent by mass\n",

-      "mS=32.0 # molecular mass \n",

-      "mH2=2.0  # molecular mass\n",

-      "mC=12.0  # molecular mass\n",

-      "mO2=32.0  # molecular mass\n",

-      "mN2=28.0  # molecular mass\n",

-      "# molal composition per 100 Kg of fuel:\n",

-      "S=Sulfur/mS # molal composition \n",

-      "H=Hydrogen/mH2 # molal composition\n",

-      "C=Carbon/mC # molal composition\n",

-      "O2=Oxygen/mO2 # molal composition\n",

-      "N2=Nitrogen/mN2 # molal composition\n",

-      "\n",

-      "#The combustion equation becomes \n",

-      "# 0.02 S + 0.02 O2 \u21920.02 SO2\n",

-      "# 2.85 H2 + 1.42 O2 \u21922.85 H2O\n",

-      "# 6.60 C + 6.60 O2 \u21926.60 CO2\n",

-      "\n",

-      "O2req=0.02+1.42+6.60 # kmol O2 required/100 kg fuel\n",

-      "O2present=O2req-O2 # kmol O2 from air/100 kg fuel\n",

-      "AFtheo= (O2present+O2present*3.76)*28.97/100 # AF ration theoritical kg air/kg fuel\n",

-      "print \"\\n The A/F ratio (theoritical) on mass basis is \",round(AFtheo,2),\"kg air/kg fuel\"\n",

-      "print \"\\n The A/F ratio for 30% excess air on mass basis is \",round(AFtheo*1.3,2),\"kg air/kg fuel\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " The A/F ratio (theoritical) on mass basis is  10.66 kg air/kg fuel\n",

-        "\n",

-        " The A/F ratio for 30% excess air on mass basis is  13.85 kg air/kg fuel\n"

-       ]

-      }

-     ],

-     "prompt_number": 23

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.6:Pg-629"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques6\n",

-      "# determining heat transfer per kilomole of fuel entering combustion chamber\n",

-      "\n",

-      "#1-CH4\n",

-      "#2-CO2\n",

-      "#3-H2O\n",

-      "#hf-standard enthalpy of given substance\n",

-      "hf1=-74.873;#kJ\n",

-      "hf2=-393.522;#kJ\n",

-      "hf3=-285.830;#kJ\n",

-      "Qcv=hf2+2*hf3-hf1;#kJ\n",

-      "print \"Heat transfer per kilomole of fuel entering combustion chamber is\",round(Qcv,3),\"kJ\"\n",

-      "#the answers in the book is different as they have not printed the decimals in values"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Heat transfer per kilomole of fuel entering combustion chamber is -890.309 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.7:Pg-631"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#calculating enthalpy of water at given pressure and temperature\n",

-      "\n",

-      "#1.Assuming steam to be an ideal gas with value of Cp\n",

-      "T1=298.15;#Initial temperature in K\n",

-      "T2=573.15;#final temperature in K\n",

-      "T=(T1+T2)/2;#average temperature in K\n",

-      "Cp=1.79+0.107*T/1000+0.586*(T/1000)**2-.20*(T/1000)**3;#specific heat at constant pressure in kj/kg.K\n",

-      "M=18.015;#mass in kg\n",

-      "dh=M*Cp*(T2-T1);#enthalpy change in kJ/kmol\n",

-      "ho=-241.826;#enthalpy at standard temperature and pressure in kJ/mol\n",

-      "htp1=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

-      "print \" 1. Enthalpy of water at given pressure and temperature using value of Cp =\",round(htp1,3),\"kJ/kmol\"\n",

-      "\n",

-      "#2..Assuming steam to be an ideal gas with value for dh\n",

-      "dh=9359;#enthalpy change from table A.9 in kJ/mol\n",

-      "htp2=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

-      "print \" 2. Enthalpy of water at given pressure and temperature assuming value od dh =\",round(htp2,3),\"kJ/kmol \"\n",

-      "\n",

-      "#3. Using steam table\n",

-      "dh=M*(2977.5-2547.2);#enthalpy change for gases in kJ/mol\n",

-      "htp3g=dh/1000+ho;\n",

-      "dh=M*(2977.5-104.9);#enthalpy change for liquid in kJ/mol\n",

-      "hl=-285.830;#standard enthalpy for liquid in kJ/kmol\n",

-      "htp31=hl+dh/1000.0;#enthalpy at given temp and pressure in kJ/kmol\n",

-      "print \" 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp31,3),\"kJ/kmol \"\n",

-      "print \" 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp3g,3),\"kJ/kmol \"\n",

-      "#4.using generalised charts\n",

-      "#htp=ho-(h2*-h2)+(h2*-h1*)+(h1*-h1);\n",

-      "#h2*-h2=Z*R*Tc,\n",

-      "#h2*-h1*=9539 kJ/mol, from part 2\n",

-      "#h1*-h1=0 ,as ideal gas \n",

-      "Z=0.21;#from chart\n",

-      "R=8.3145;#gas constant in SI units\n",

-      "Tc=647.3;#critical temperature in K\n",

-      "htp4=ho+9539/1000-Z*R*Tc/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

-      "print \" 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart = \",round(htp4,3),\"kJ/kmol\"\n",

-      "#the answers in book are different as they have not printed the decimals in values"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " 1. Enthalpy of water at given pressure and temperature using value of Cp = -232.258 kJ/kmol\n",

-        " 2. Enthalpy of water at given pressure and temperature assuming value od dh = -232.826 kJ/kmol \n",

-        " 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.08 kJ/kmol \n",

-        " 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.074 kJ/kmol \n",

-        " 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart =  -233.956 kJ/kmol\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.8:Pg-632"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 8\n",

-      "# Determine the heat transfer\n",

-      "airth=400 # % excess theoritical air\n",

-      "Ti=25 # Entry temp in degree celsius\n",

-      "To=900 # Exit temp in Kelvin\n",

-      "mf=0.25 # specific fuel consumption in kg/s of fuel per megawatt output\n",

-      "# The combustion equation is:\n",

-      "# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 \u2192 8CO2 + 9H2O + 37.5O2 + 188.0N2\n",

-      "hfC8H18=-250105 # is the enthalpy of fuel in kJ/kmol \n",

-      "nCO2=8 # moles\n",

-      "hfCO2=-393522 # enthalpy in kJ/kmol\n",

-      "DelhCO2=28030 # enthalpy in kJ/kmol\n",

-      "mC8H18=114.23 # molecular mass\n",

-      "nH2O=9 # moles\n",

-      "hfH2O=-241826 # enthalpy in kJ/kmol\n",

-      "DelhH2O=21937 # enthalpy in kJ/kmol\n",

-      "DelhO2=19241 # enthalpy in kJ/kmol\n",

-      "nO2=37.5 # moles\n",

-      "DelhN2=18225 # enthalpy in kJ/kmol\n",

-      "nN2=188.0 # moles\n",

-      "Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2\n",

-      "\n",

-      "Wcv=1000*mC8H18/mf # work in kJ/kmol fuel\n",

-      "print \" The work is \",Wcv,\"kJ/kmol fuel\"\n",

-      "\n",

-      "Qcv=Eproduct+Wcv-hfC8H18 # using first law\n",

-      "print \" The Heat transfer is \",Qcv,\"kJ/kmol fuel\"\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " The work is  456920.0 kJ/kmol fuel\n",

-        " The Heat transfer is  -48074.5 kJ/kmol fuel\n"

-       ]

-      }

-     ],

-     "prompt_number": 27

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.8E:Pg-633"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 8\n",

-      "# Determine the heat transfer\n",

-      "airth=400 # % excess theoritical air\n",

-      "Ti=77 # Entry temp in F\n",

-      "To=1100 # Exit temp in F\n",

-      "mf=1 # specific fuel consumption in lb of fuel per horsepower-hour\n",

-      "# The combustion equation is:\n",

-      "# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 \u2192 8CO2 + 9H2O + 37.5O2 + 188.0N2\n",

-      "hfC8H18=-107526 # is the enthalpy of fuel in Btu/lb.mol\n",

-      "nCO2=8 # moles\n",

-      "hfCO2=-169184 # enthalpy in Btu/lb.mol\n",

-      "DelhCO2=11391 # enthalpy in Btu/lb.mol\n",

-      "mC8H18=114.23 # molecular mass\n",

-      "nH2O=9 # moles\n",

-      "hfH2O=-103966 # enthalpy in Btu/lb.mol\n",

-      "DelhH2O=8867 # enthalpy in Btu/lb.mol\n",

-      "DelhO2=7784 # enthalpy in Btu/lb.mol\n",

-      "nO2=37.5 # moles\n",

-      "DelhN2=7374 # enthalpy in Btu/lb.mol\n",

-      "nN2=188.0 # moles\n",

-      "Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2\n",

-      "\n",

-      "Wcv=2544*mC8H18 # work in kJ/kmol fuel\n",

-      "print \" The work is \",Wcv,\"Btu/lb.mol fuel\"\n",

-      "\n",

-      "Qcv=Eproduct+Wcv-hfC8H18 # using first law\n",

-      "print \" The Heat transfer is \",Qcv,\"Btu/lb.mol fuel\"\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " The work is  290601.12 Btu/lb.mol fuel\n",

-        " The Heat transfer is  -41895.88 Btu/lb.mol fuel\n"

-       ]

-      }

-     ],

-     "prompt_number": 31

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.9:Pg-634"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques 9\n",

-      "\n",

-      "# The combustion equation is\n",

-      "# C2H4 + 3O2 \u2192 2 CO2 + 2H2O(g)\n",

-      "hfC2H4=52467 # enthalpy in KJ/mol\n",

-      "R=8.314 # gas constant\n",

-      "T=298.2 # temperature in Kelvin\n",

-      "hfCO2=-393522 # enthalpy in kJ/kmol\n",

-      "DelhCO2=12906 # enthalpy in kJ/kmol\n",

-      "hfH2O=-241826 # enthalpy in kJ/kmol\n",

-      "DelhH2O=10499 # enthalpy in kJ/kmol\n",

-      "Ureactant=hfC2H4-4*R*T # energy of reactant in KJ\n",

-      "Uproduct=2*(hfCO2+DelhCO2)+2*(hfH2O+DelhH2O) # energy of product in KJ\n",

-      "Q=Uproduct-Ureactant # using first law\n",

-      "print \"The Heat transfer is \",Q,\"kJ\"\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The Heat transfer is  -1266436.0608 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 33

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.10:Pg-638"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques 10\n",

-      "\n",

-      "# The combustion equation is\n",

-      "# C3H8 + 5O2 \u2192 3 CO2 + 4H2O\n",

-      "hfC3H8g=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10\n",

-      "hfC3H8l=hfC3H8g-44.097*370 # enthalpy of C3H8 liquid as enthalpy of evaporation is given in question\n",

-      "hfH2Ol=285830 # enthalpy of water in KJ/mol\n",

-      "hfH2Og=241826 # enthalpy in KJ/mol\n",

-      "hfCO2=393522 # enthalpy in KJ/mol\n",

-      "nO2=5 # moles\n",

-      "nCO2=3 # moles\n",

-      "nH2O=4 # moles\n",

-      "nC3H8=1 # moles\n",

-      "# part(1) : liquid propane-liquid water\n",

-      "hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8l)/44.097 # higher heating value of liquid propane \n",

-      "print \"The higher heating value of liquid propane is\",round(hrP,2),\"KJ/kg\"\n",

-      "\n",

-      "# part(2) : Liquid propane\u2013gaseous water\n",

-      "\n",

-      "hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8l)/44.097 # lower heating value of liquid propane \n",

-      "print \"The lower heating value of liquid propane is\",round(hrP,2),\"KJ/kg\"\n",

-      "\n",

-      "# part(3) :Gaseous propane\u2013liquid water\n",

-      "hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8g)/44.097 # higher heating value of gaseous propane \n",

-      "print \"The higher heating value of gaseous propane is\",round(hrP,2),\"KJ/kg\"\n",

-      "\n",

-      "# part(4) :Gaseous propane\u2013Gaseous water\n",

-      "hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8g)/44.097 # lower heating value of gaseous propane \n",

-      "print \"The lower heating value of gaseous propane is\",round(hrP,2),\"KJ/kg\"\n",

-      "\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The higher heating value of liquid propane is -120215.89 KJ/kg\n",

-        "The lower heating value of liquid propane is 51434.02 KJ/kg\n",

-        "The higher heating value of gaseous propane is 55055.58 KJ/kg\n",

-        "The lower heating value of gaseous propane is 51064.02 KJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.11:Pg-638"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques 10\n",

-      "# The combustion equation is\n",

-      "# C3H8(g) + 5O2 \u2192 3 CO2 + 4H2O(g)\n",

-      "hfC3H8=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10\n",

-      "Cp=2.1 # in KJ/Kg.K\n",

-      "T1=500 # temp in Kelvin\n",

-      "T2=298.2 # temp in Kelvin\n",

-      "DelhO2=6086 # enthalpy of O2 in KJ/mol\n",

-      "hfCO2=-393522 # enthalpy in KJ/mol\n",

-      "DelhCO2=8305 # enthalpy of CO2 in KJ/mol\n",

-      "hfH2O=-241826 # enthalpy of H2O in KJ/mol\n",

-      "DelhH2O=6922 # enthalpy in KJ/mol\n",

-      "nO2=5 # moles\n",

-      "nCO2=3 # moles\n",

-      "nH2O=4 # moles\n",

-      "nC3H8=1 # moles\n",

-      "\n",

-      "hR500= hfC3H8+Cp*44.097*(T1-T2)+nO2*DelhO2  # in KJ/mol\n",

-      "hP500= nCO2*(hfCO2+DelhCO2)+ nH2O*(DelhH2O+hfH2O) # in KJ/mol\n",

-      "hRP500=hP500-hR500 #  in KJ/mol\n",

-      "Hrp500=hRP500/44.097 # KJ/Kg\n",

-      "\n",

-      "print \" The enthalpy of combustion of gaseous propane at 500 K \",round(Hrp500)\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " The enthalpy of combustion of gaseous propane at 500 K  -46273.0\n"

-       ]

-      }

-     ],

-     "prompt_number": 10

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.12:Pg-639"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 12\n",

-      "# The question is solved by Trial and Error and  using the table , no need for Calculation Thus not solved in Python"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": []

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.13:Pg-645"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 13\n",

-      "# The equation for this chemical reaction is\n",

-      "# C2H4(g) + 3(4) O2 + 3(4)(3.76) N2 \u2192 2 CO2+2H2O(g) + 9O2 + 45.1 N2\n",

-      "T=25+273.15 # temp in degree kelvin\n",

-      "P=0.1 # pressure in Mpa\n",

-      "hfCO2=-393522 # enthalpy in KJ/mol\n",

-      "hfH2O=-241826 # enthalpy of H2O in KJ/mol\n",

-      "HfC2H4=52467 # enthalpy of C2H4 in KJ/mol\n",

-      "hfO2=0 # enthalpy of formation in KJ/mol\n",

-      "SCO2=213.795 # kJ/kmol fuel\n",

-      "SH2O=188.243 # kJ/kmol fuel\n",

-      "SC2H4=219.330 # kJ/kmol fuel\n",

-      "SO2=205.148 # kJ/kmol fuel\n",

-      "DelH=2*hfCO2+2*hfH2O-HfC2H4-3*hfO2 # in KJ/mol\n",

-      "DelS=2*SCO2+2*SH2O-SC2H4-3*SO2 # kJ/kmol fuel\n",

-      "DelG=DelH-T*(DelS) # Gibbs Energy\n",

-      "Wrev=-DelG/28.054 # reversible work in kJ/kg\n",

-      "print \"The reversible work is \",round(Wrev,2),\"kJ/kg \"\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The reversible work is  46838.61 kJ/kg \n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.14:Pg-647"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques \n",

-      "# The question is solved by Trial and Error and  using the table , no need for Calculation Thus not solved in Python"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": []

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.15:Pg-649"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques15\n",

-      "#calculatng reversible elecromotive force \n",

-      "\n",

-      "#1-H2O\n",

-      "#2-H2\n",

-      "#3-O2\n",

-      "#hf-standard enthalpy \n",

-      "#sf-standard entropy\n",

-      "hf1=-285.830;#kJ\n",

-      "hf2=0;#kJ\n",

-      "hf3=0;#kJ\n",

-      "sf1=69.950;#kJ/K\n",

-      "sf2=130.678;#kJ/K\n",

-      "sf3=205.148;#kJ/K\n",

-      "dH=2*hf1-2*hf2-hf3;#change in enthalpy in kJ\n",

-      "dS=2*sf1-2*sf2-sf3;#change in entropy in kJ/K\n",

-      "T=298.15;#temperature in K\n",

-      "dG=dH-T*dS/1000;#change in gibbs free energy in kJ\n",

-      "E=-dG*1000/(96485*4);#emf in V\n",

-      "print\" Reversible electromotive Force =\",round(E,3),\" V\"\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Reversible electromotive Force = 1.229  V\n"

-       ]

-      }

-     ],

-     "prompt_number": 15

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.16:Pg-654"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 16\n",

-      "# to Calculate the combustion efficiency\n",

-      "Tair = 400  # temp in K\n",

-      "Vair = 100 # velocity in m/s\n",

-      "Tfuel = 50 # temp of fuel in \u25e6C\n",

-      "Tproducts = 1100 # temp of product in K\n",

-      "Vproducts = 150 # velocity of product in m/s\n",

-      "FAactual = 0.0211 # air fuel ratio kg fuel/kg air\n",

-      "\n",

-      "# HR + KER= \u2212243737 + 14892nO2 \n",

-      "# HP + KEP= \u22125 068 599 + 120 853nO2 \n",

-      "# using these two equations :\n",

-      "no2=(-5068599.0 + 243737.0)/(14892.0-120853) # in kmol O2/kmol fuel\n",

-      "FAideal=114.23/(4.76*no2*28.91) # in kg fuel/kg air\n",

-      "\n",

-      "efficomb=FAideal*100/FAactual # Efficiency\n",

-      "\n",

-      "print \" The combustion efficiency is \",round(efficomb,2),\"%\"\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " The combustion efficiency is  86.4 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex15.17:Pg-654"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques17\n",

-      "#efficiency of generator and plant\n",

-      "\n",

-      "q=325000*(3398.3-856.0);#heat transferred to H2O/kg fuel in kJ/kg\n",

-      "qv=26700.0*33250;#higher heating value in kJ/kg\n",

-      "nst=q/qv*100;#efficiency of steam generator\n",

-      "w=81000.0*3600;#net work done in kJ/kg\n",

-      "nth=w/qv*100.0;#thermal efficiency\n",

-      "print\" Efficiency of generator =\",round(nst,1),\"percent\\n\"\n",

-      "print\" Thermal Efficiency =\",round(nth,1),\" percent\"\n",

-      "\n",

-      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Efficiency of generator = 93.1 percent\n",

-        "\n",

-        " Thermal Efficiency = 32.8  percent\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter16.ipynb b/Fundamentals_Of_Thermodynamics/Chapter16.ipynb
deleted file mode 100755
index 6a07737d..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter16.ipynb
+++ /dev/null
@@ -1,128 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:7c7e50fa2880e870c4aa1630bafd7d049a3cb3c4e646a0d88068a55be4a3d00f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 16:INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex16.2:PG-681"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques2\n",
-      "#to determine change in gibbs free energy\n",
-      "\n",
-      "#1-H2\n",
-      "#2-O2\n",
-      "#3-H2O\n",
-      "\n",
-      "#at T=298 K\n",
-      "T1=298.0;#K\n",
-      "Hf1=0;#Enthalpy of formation of H2 at 298 K\n",
-      "Hf2=0;#Enthalpy of formation of O2 at 298 K\n",
-      "Hf3=-241826;#enthalpy of formation of H2O at 298 K in kJ\n",
-      "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",
-      "Sf1=130.678;#Entropy of H2 at 298 K n kJ/K\n",
-      "Sf2=205.148;#Entropy of O2 at 298 K in kJ/K\n",
-      "Sf3=188.834;#entropy of H2O at 298 K in kJ/K\n",
-      "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",
-      "dG1=dH-T1*dS;#change n gibbs free energy in kJ\n",
-      "print\" Change in gibbs free energy at\",T1,\"kelvin is\",round(dG1),\"kJ \"\n",
-      "#at T=2000 K\n",
-      "T2=2000.0;#K\n",
-      "Hf1=52942-0;#Enthalpy of formation of H2 at 2000 K\n",
-      "Hf2=59176-0;#Enthalpy of formation of O2 at 2000 K\n",
-      "Hf3=-241826+72788;#enthalpy of formation of H2O at 2000 K in kJ\n",
-      "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",
-      "Sf1=188.419;#Entropy of H2 at 2000 K n kJ/K\n",
-      "Sf2=268.748;#Entropy of O2 at 2000 K in kJ/K\n",
-      "Sf3=264.769;#entropy of H2O at 2000 K in kJ/K\n",
-      "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",
-      "dG2=dH-(T2*dS);#change n gibbs free energy in kJ\n",
-      "print\" Change in gibbs free energy at\",T2,\"kelvin is\",round(dG2),\" kJ \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Change in gibbs free energy at 298.0 kelvin is 457179.0 kJ \n",
-        " Change in gibbs free energy at 2000.0 kelvin is 271040.0  kJ \n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex16.3:PG-683"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques3\n",
-      "#calculating equilibrium constant \n",
-      "\n",
-      "dG1=-457.166;#change in gibbs free energy at temp 298 K from example2 in kJ\n",
-      "dG2=-271.040;#change in gibbs free energy at temp 2000 K from example2 n kJ\n",
-      "T1=298;#K\n",
-      "T2=2000;#K\n",
-      "R=8.3145;#gas constant\n",
-      "K1=dG1*1000/(R*T1);\n",
-      "K2=dG2*1000/(R*T2);\n",
-      "print\" Equilibrium constant at \",T1,\"K = \",round(K1,2)\n",
-      "print\" Equilibrium constant at \",T2,\"K = \",round(K2,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Equilibrium constant at  298 K =  -184.51\n",
-        " Equilibrium constant at  2000 K =  -16.299\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter16_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter16_6.ipynb
deleted file mode 100755
index c155fdc6..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter16_6.ipynb
+++ /dev/null
@@ -1,377 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:2573953a96bd6739252db580058edb8938919cec53d4f1d76f9484cfa4f2a1d5"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 16:INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.1:676"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques1\n",

-      "# pressure required to make diamonds from graphite\n",

-      "import math\n",

-      "T=25 # temp in degree C\n",

-      "ggrp= 0 # g for fraphite\n",

-      "gdiamnd=2867.8 # g for diamond in kJ/mol\n",

-      "vgrp=0.000444  # specific volume of graphite in m^3/kg\n",

-      "vdiamnd=0.000284 # specific volume of graphite in m^3/kg\n",

-      "BetaTgrp=0.304e-6 # beta for graphite in 1/MPa\n",

-      "BetaTdiamnd=0.016e-6 #  beta for diamond in 1/MPa\n",

-      "\n",

-      "P=-(-2*(vgrp-vdiamnd)+math.sqrt((2*vgrp-2*vdiamnd)**2-4*(vgrp*BetaTgrp-vdiamnd*BetaTdiamnd)*(2*gdiamnd/(12.011*1000))))/(2*(vgrp*BetaTgrp-vdiamnd*BetaTdiamnd))\n",

-      "print int(P),\" MPa is the pressure at which possibility exists for conversion from graphite to diamonds\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "1493  MPa is the pressure at which possibility exists for conversion from graphite to diamonds\n"

-       ]

-      }

-     ],

-     "prompt_number": 18

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.2:PG-681"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques2\n",

-      "#to determine change in gibbs free energy\n",

-      "\n",

-      "#1-H2\n",

-      "#2-O2\n",

-      "#3-H2O\n",

-      "\n",

-      "#at T=298 K\n",

-      "T1=298.0;#K\n",

-      "Hf1=0;#Enthalpy of formation of H2 at 298 K\n",

-      "Hf2=0;#Enthalpy of formation of O2 at 298 K\n",

-      "Hf3=-241826;#enthalpy of formation of H2O at 298 K in kJ\n",

-      "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",

-      "Sf1=130.678;#Entropy of H2 at 298 K n kJ/K\n",

-      "Sf2=205.148;#Entropy of O2 at 298 K in kJ/K\n",

-      "Sf3=188.834;#entropy of H2O at 298 K in kJ/K\n",

-      "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",

-      "dG1=dH-T1*dS;#change n gibbs free energy in kJ\n",

-      "print\" Change in gibbs free energy at\",T1,\"kelvin is\",round(dG1),\"kJ \"\n",

-      "#at T=2000 K\n",

-      "T2=2000.0;#K\n",

-      "Hf1=52942-0;#Enthalpy of formation of H2 at 2000 K\n",

-      "Hf2=59176-0;#Enthalpy of formation of O2 at 2000 K\n",

-      "Hf3=-241826+72788;#enthalpy of formation of H2O at 2000 K in kJ\n",

-      "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",

-      "Sf1=188.419;#Entropy of H2 at 2000 K n kJ/K\n",

-      "Sf2=268.748;#Entropy of O2 at 2000 K in kJ/K\n",

-      "Sf3=264.769;#entropy of H2O at 2000 K in kJ/K\n",

-      "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",

-      "dG2=dH-(T2*dS);#change n gibbs free energy in kJ\n",

-      "print\" Change in gibbs free energy at\",T2,\"kelvin is\",round(dG2),\" kJ \""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Change in gibbs free energy at 298.0 kelvin is 457179.0 kJ \n",

-        " Change in gibbs free energy at 2000.0 kelvin is 271040.0  kJ \n"

-       ]

-      }

-     ],

-     "prompt_number": 21

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.3:PG-683"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques3\n",

-      "#calculating equilibrium constant \n",

-      "\n",

-      "dG1=-457.166;#change in gibbs free energy at temp 298 K from example2 in kJ\n",

-      "dG2=-271.040;#change in gibbs free energy at temp 2000 K from example2 n kJ\n",

-      "T1=298;#K\n",

-      "T2=2000;#K\n",

-      "R=8.3145;#gas constant\n",

-      "K1=dG1*1000/(R*T1);\n",

-      "K2=dG2*1000/(R*T2);\n",

-      "print\" Equilibrium constant at \",T1,\"K = \",round(K1,2)\n",

-      "print\" Equilibrium constant at \",T2,\"K = \",round(K2,3)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Equilibrium constant at  298 K =  -184.51\n",

-        " Equilibrium constant at  2000 K =  -16.299\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.4:Pg-684"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# The example is about proving that Equlibrium constant can be found using table hence doesn't require solution in python  "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [],

-     "prompt_number": 19

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.5:Pg-685"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 5\n",

-      "# To determine Heat Transfer\n",

-      "# The process is two step as:\n",

-      "# Combustion: C + O2--->CO2\n",

-      "# Dissociation reaction: 2CO2---->2CO + O2\n",

-      "# overall process : C + O2 \u2192 0.5622 CO2 + 0.4378 CO + 0.2189 O2\n",

-      "nCO2=0.5622 # moles of CO2\n",

-      "nCO=0.4378 # moles of CO\n",

-      "nO2=0.2189 # moles of NO2\n",

-      "\n",

-      "# from Table A.9:\n",

-      "hfCO2=0 # enthalpy of formation\n",

-      "hfC=0 # enthalpy of formation\n",

-      "hfCO=0 # enthalpy of formation\n",

-      "hfO2=0 # enthalpy of formation\n",

-      "hfCO23000=-393522 # enthalpy @ 3000K\n",

-      "hfCO3000=-110527 # enthalpy @ 3000K\n",

-      "hfO23000=0 # enthalpy @ 3000K\n",

-      "hfCO2298=-152853 # enthalpy @ 298K\n",

-      "hfCO298=-93504 # enthalpy @ 298K\n",

-      "hfO2298=-98013 # enthalpy @ 298K\n",

-      "\n",

-      "Hr=hfC+hfO2 # enthalpy of reactants\n",

-      "Hp=nCO2*(hfCO2+hfCO23000-hfCO2298)+nCO*(hfCO+hfCO3000-hfCO298)+nO2*(hfO2+hfO23000-hfO2298)\n",

-      "\n",

-      "Qcv=Hp-Hr # using first law\n",

-      "print round(Qcv),\"kJ/kmol C is the heat transfer \""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "-121302.0 kJ/kmol C is the heat transfer \n"

-       ]

-      }

-     ],

-     "prompt_number": 22

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.6:Pg-687"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 6\n",

-      "# to determine the composition\n",

-      "# the standard equation is C + 2O2 \u2192 CO2 + O2\n",

-      "# from equilibrium C + 2O2 \u2192 (1 \u2212 2z)CO2 + 2zCO + (1 + z)O2\n",

-      "T=3000 # temp in K\n",

-      "P=0.1 # prssure in MPa\n",

-      "z = 0.1553 # from equilibrium equation using table\n",

-      "yCO2=(1-2*z)/(2+z) # mole fraction of CO2\n",

-      "yCO = 2*z/(2+z) # mole fraction of CO\n",

-      "yO2=(1+z)/(2+z) # mole fraction of O2\n",

-      "print \"The mole fraction is \",round(yCO2,3),\" for CO2 \\n\"\n",

-      "print \"The mole fraction is \",round(yCO,3),\" for CO \\n\"\n",

-      "print \"The mole fraction is \",round(yO2,3),\" for O2 \\n\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The mole fraction is  0.32  for CO2 \n",

-        "\n",

-        "The mole fraction is  0.144  for CO \n",

-        "\n",

-        "The mole fraction is  0.536  for O2 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 24

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.7:Pg-691"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 7\n",

-      "# to determine the equilibrium composition\n",

-      "# The reaction equation is:\n",

-      "# (1): 2.H2O -->2H2 + O2\n",

-      "# (2): 2 H2O -->H2 + 2OH\n",

-      "# the equilibrium equation is \n",

-      "# H2O \u2192 (1 \u2212 2a \u2212 2b)H2O + (2a + b)H2 + aO2 + 2bOH\n",

-      "P= 0.1 # pressure in MPa\n",

-      "T=3000 # temp in Kelvin\n",

-      "a=0.0534 # using value of K from Table A.11 @ 3000k\n",

-      "b=0.0551 # using value of K from Table A.11 @ 3000k\n",

-      "nH2O= (1-2*a-b) # moles of H2O\n",

-      "nH2=2*a+b# moles of H2\n",

-      "nO2=a# moles of O2\n",

-      "nOH= 2*b# moles of OH\n",

-      "X=nH2O+nH2+nO2+nOH\n",

-      "yH2O=nH2O/X # mole fraction\n",

-      "yH2=nH2/X # mole fraction\n",

-      "yO2=nO2/X # mole fraction\n",

-      "yOH=nOH/X # mole fraction\n",

-      "print \"The mole fraction of H2O is\",round(yH2O,2),\"\\n\"\n",

-      "print \"The mole fraction of H2 is\",round(yH2,2),\"\\n\"\n",

-      "print \"The mole fraction of O2 is\",round(yO2,2),\"\\n\"\n",

-      "print \"The mole fraction of OH is\",round(yOH,2),\"\\n\"\n",

-      "# the answers are slightly different due to approximation in textbook while here the answers are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The mole fraction of H2O is 0.72 \n",

-        "\n",

-        "The mole fraction of H2 is 0.139 \n",

-        "\n",

-        "The mole fraction of O2 is 0.046 \n",

-        "\n",

-        "The mole fraction of OH is 0.095 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 27

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex16.8:pg-696"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# ques 8\n",

-      "# determine the the equilibrium composition\n",

-      "P= 1 # pressure in Kpa\n",

-      "T=10000 # temp in Kelvin\n",

-      "z=0.2008 # using k from table \n",

-      "# the chemical equation is Ar -->Ar^(+) + e\u2212\n",

-      "\n",

-      "yAr=(1-z)/(1 + z) # mole fraction of Ar\n",

-      "yArpositive= z/(1+z) # mole fraction of Ar(+)\n",

-      "yenegative=z/(1+z) # mole fraction of ye-\n",

-      "print \"The mole fraction of Ar is\",round(yAr,3),\"\\n\"\n",

-      "print \"The mole fraction of Ar+ is\",round(yArpositive,3),\"\\n\"\n",

-      "print \"The mole fraction of e- is\",round(yenegative,3),\"\\n\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The mole fraction of Ar is 0.666 \n",

-        "\n",

-        "The mole fraction of Ar+ is 0.167 \n",

-        "\n",

-        "The mole fraction of e- is 0.167 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 32

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter17.ipynb b/Fundamentals_Of_Thermodynamics/Chapter17.ipynb
deleted file mode 100755
index 4ae8005b..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter17.ipynb
+++ /dev/null
@@ -1,434 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:70b7f86a423a7c9685f997491946441e1c53cfe8fe328afd6e5b37a44e4dce11"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 17: COMPRESSIBLE FLOW"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.1:PG-710"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques1\n",
-      "#to determine isentropic stagnation pressure and temperature \n",
-      "\n",
-      "T=300;#Temperature of air in K\n",
-      "P=150;#Pressure of air in kPa\n",
-      "v=200;#velocity of air flow n m/s\n",
-      "Cp=1.004;#specific heat at constant pressure in kJ/kg\n",
-      "To=v**2/(2000*Cp)+T;#stagnation temperature in K\n",
-      "k=1.4;#constant\n",
-      "Po=P*(To/T)**(k/(k-1));#stagnation pressure in kPa\n",
-      "print 'Stagnation Temperature is ',round(To,1),' K \\n'\n",
-      "print 'Stagnation Pressure is ',round(Po,2),'kPa \\n'"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stagnation Temperature is  319.9  K \n",
-        "\n",
-        "Stagnation Pressure is  187.85 kPa \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.2:PG-713"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques2\n",
-      "#to determine the force\n",
-      "\n",
-      "#initializing variables\n",
-      "mdot=-1 # mass flow rate out of control volume in kg/s\n",
-      "Vx=-1 # x component of velocity of control volume in m/s\n",
-      "Vy=10 # y component of velocity of control volume in m/s\n",
-      "\n",
-      "Fx=mdot*Vx # Force in X direction\n",
-      "\n",
-      "Fy=mdot*Vy # Force in Y direction\n",
-      "\n",
-      "print \"the force the man exert on the wheelbarrow\",round(Fx),\"N\"\n",
-      "print \"the force the floor exerts on the wheelbarrow\",round(Fy),\"N\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the force the man exert on the wheelbarrow 1.0 N\n",
-        "the force the floor exerts on the wheelbarrow -10.0 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.3:PG-715"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques3\n",
-      "#determining the thrust acting on a control surface\n",
-      "\n",
-      "#i-inlet\n",
-      "#e-exit\n",
-      "#using momentum equation on control surface in x direction\n",
-      "me=20.4;#mass exiting in kg\n",
-      "mi=20;#mass entering in kg\n",
-      "ve=450;#exit velocity in m/s\n",
-      "vi=100;#inlet velocity in m/s\n",
-      "Pi=95;#Pressure at inlet in kPa\n",
-      "Pe=125;#Pressure at exit in kPa\n",
-      "Po=100;#surrounding pressure in kPa\n",
-      "Ai=0.2;#inlet area in m^2\n",
-      "Ae=0.1;#exit area in m^2\n",
-      "Rx=(me*ve-mi*vi)/1000-(Pi-Po)*Ai+(Pe-Po)*Ae;#thrust in x direction in kN\n",
-      "\n",
-      "print \"Thrust acting in x direction is \",round(Rx,2),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thrust acting in x direction is  10.68 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.4:PG-717"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques4\n",
-      "#to determine increase in enthalpy\n",
-      "#initializing variables\n",
-      "#i-inlet\n",
-      "#e-exit\n",
-      "\n",
-      "T=25+273 # temperature in kelvin\n",
-      "v=0.001003 # specific volume of water in kg/m^3 at 25 *c from table B.1.1 \n",
-      "ve=7;#exit velocity in m/s\n",
-      "vi=30;#inlet velocity in m/s\n",
-      "Pi=350;#Pressure at inlet in kPa\n",
-      "Pe=600;#Pressure at exit in kPa\n",
-      "\n",
-      "#using momentum equation on control surface \n",
-      "Pes= (vi**2-ve**2)/(2*v*1000)+Pi # exit pressure for reversible diffuser\n",
-      "delH=(vi**2-ve**2)/(2*1000.0) # change in enthalpy from first law in kJ/kg\n",
-      "delU=delH-v*(Pe-Pi) # change in internal energy in kJ/kg\n",
-      "delS=delU/T # change in entropy in kJ/kg.K\n",
-      "print\"the exit pressure for reversible diffuser is \",round(Pes),\"kPa\"\n",
-      "print\"the increase in enthalpy is \",round(delH,5),\"kJ/kg\"\n",
-      "print\"the increase in internal energy is \",round(delU,5),\"kJ/kg\"\n",
-      "print\"the increase in entropy is \",round(delS,6),\"kJ/kg.K\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the exit pressure for reversible diffuser is  774.0 kPa\n",
-        "the increase in enthalpy is  0.4255 kJ/kg\n",
-        "the increase in internal energy is  0.17475 kJ/kg\n",
-        "the increase in entropy is  0.000586 kJ/kg.K\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.5:PG-720"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques5\n",
-      "#determining velocity of sound in air\n",
-      "import math\n",
-      "k=1.4;#constant\n",
-      "R=0.287;#gas constant\n",
-      "#at 300K\n",
-      "T1=300;# temperature in kelvin\n",
-      "c1=math.sqrt(k*R*T1*1000)\n",
-      "print \"Speed of sound at 300 K is\",round(c1,1),\" m/s\" \n",
-      "#at 1000K\n",
-      "T2=1000;# temperature in kelvin\n",
-      "c2=math.sqrt(k*R*T2*1000)\n",
-      "print \"Speed of sound at 1000 K is\",round(c2,1),\" m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed of sound at 300 K is 347.2  m/s\n",
-        "Speed of sound at 1000 K is 633.9  m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.6:PG-727"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques6\n",
-      "#determining mass flow rate through control volume\n",
-      "import math\n",
-      "k=1.4;#constant\n",
-      "R=0.287;#gas constant\n",
-      "To=360;#stagnation Temperature in K \n",
-      "T=To*0.8333;#Temperature of air in K, 0.8333 stagnation ratio from table\n",
-      "v=math.sqrt(k*R*T*1000);#velocity in m/s\n",
-      "P=528;#stagnation pressure in kPa\n",
-      "d=P/(R*T);#stagnation density in kg/m^3\n",
-      "A=500*10**-6;#area in m^2\n",
-      "ms=d*A*v;#mass flow rate in kg/s\n",
-      "print\" Mass flow rate at the throat section is\",round(ms,4),\"kg/s\"\n",
-      "#e-exit state\n",
-      "Te=To*0.9381;#exit temperature in K, ratio from table\n",
-      "ce=math.sqrt(k*R*Te*1000);#exit velocity of sound in m/s\n",
-      "Me=0.573;#Mach number\n",
-      "ve=Me*ce;\n",
-      "Pe=800;#exit pressure in kPa\n",
-      "de=Pe/R/Te;\n",
-      "mse=de*A*ve;\n",
-      "print\" Mass flow rate at the exit section is\",round(mse,4),\" kg/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Mass flow rate at the throat section is 1.0646 kg/s\n",
-        " Mass flow rate at the exit section is 0.8711  kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.7:PG-728"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques7\n",
-      "#determining exit properties in a control volume\n",
-      "import math\n",
-      "Po=1000;#stagnation pressure in kPa\n",
-      "To=360;#stagnation temperature in K\n",
-      "\n",
-      "#when diverging section acting as nozzle\n",
-      "\n",
-      "Pe1=0.0939*Po;#exit pressure of air in kPa\n",
-      "Te1=0.5089*To;#exit temperature in K\n",
-      "k=1.4;#constant\n",
-      "R=0.287;#gas constant for air\n",
-      "ce=math.sqrt(k*R*Te1*1000);#velocity of sound in exit section in m/s\n",
-      "Me=2.197;#mach number from table\n",
-      "ve1=Me*ce;#velocity of air at exit section in m/s\n",
-      "print \"When diverging section act as a nozzle :-\"\n",
-      "print \"Exit pressure is\",round(Pe1,4),\" kPa\"\n",
-      "print \"Exit Temperature\",round(Te1,1),\" K\"\n",
-      "print \"Exit velocity is\",round(ve1,1),\" m/s \"\n",
-      "\n",
-      "#when diverging section act as diffuser\n",
-      "\n",
-      "Me=0.308;\n",
-      "Pe2=0.0936*Po;#exit pressure of air in kPa\n",
-      "Te2=0.9812*To;#exit temperature in K\n",
-      "ce=math.sqrt(k*R*Te2*1000);#velocity of sound in exit section in m/s\n",
-      "ve2=Me*ce;\n",
-      "print \"When diverging section act as a diffuser :-\"\n",
-      "print \"Exit pressure is\",round(Pe2,1),\" kPa\"\n",
-      "print \"Exit Temperature\",round(Te2,2),\" K\"\n",
-      "print \"Exit velocity is\",round(ve2,),\" m/s \"\n",
-      "\n",
-      "# The value of Exit pressure when diverging section acts as diffuser is wrong\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When diverging section act as a nozzle :-\n",
-        "Exit pressure is 93.9  kPa\n",
-        "Exit Temperature 183.2  K\n",
-        "Exit velocity is 596.1  m/s \n",
-        "When diverging section act as a diffuser :-\n",
-        "Exit pressure is 93.6  kPa\n",
-        "Exit Temperature 353.23  K\n",
-        "Exit velocity is 116.0  m/s \n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex17.9:PG-733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#ques9\n",
-      "#determining exit plane properties in control volume\n",
-      "\n",
-      "#x-inlet\n",
-      "#y-exit\n",
-      "Mx=1.5;#mach number for inlet\n",
-      "My=0.7011;#mach number for exit\n",
-      "Px=272.4;#inlet pressure in kPa\n",
-      "Tx=248.3;#inlet temperature in K\n",
-      "Tox=360 # stagnation temperature in Kelvin\n",
-      "Pox=1000.0;#stagnation pressure for inlet\n",
-      "Py=2.4583*Px;# Pressure at 1.5 mach in kPa\n",
-      "Ty=1.320*Tx;# temperature at 1.5 mach in K\n",
-      "Poy=0.9298*Pox;# pressure at 1.5 mach in kPa\n",
-      "\n",
-      "Toy=Tox # constant\n",
-      "Me=0.339 # from table with A/A*=1.860 and M < 1\n",
-      "Pe=0.9222*Py;#Exit Pressure in kPa\n",
-      "Te=0.9771*Toy;#Exit temperature in K\n",
-      "Poe=0.9222*Poy;#Exit pressure in kPa\n",
-      "\n",
-      "print \"Exit Mach no.=\",Me\n",
-      "print \"Exit temperature =\",round(Te,2),\"K \"\n",
-      "print \"Exit pressure =\",round(Poe,1),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Exit Mach no.= 0.339\n",
-        "Exit temperature = 351.76 K \n",
-        "Exit pressure = 857.5 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter17_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter17_6.ipynb
deleted file mode 100755
index 11b9507c..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter17_6.ipynb
+++ /dev/null
@@ -1,462 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:e77c793709c88cc39a2de404e997a6f1eed32d352c214c1580d822fd7a5a83a2"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 17: COMPRESSIBLE FLOW"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.1:PG-710"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques1\n",

-      "#to determine isentropic stagnation pressure and temperature \n",

-      "\n",

-      "T=300;#Temperature of air in K\n",

-      "P=150;#Pressure of air in kPa\n",

-      "v=200;#velocity of air flow n m/s\n",

-      "Cp=1.004;#specific heat at constant pressure in kJ/kg\n",

-      "To=v**2/(2000*Cp)+T;#stagnation temperature in K\n",

-      "k=1.4;#constant\n",

-      "Po=P*(To/T)**(k/(k-1));#stagnation pressure in kPa\n",

-      "print 'Stagnation Temperature is ',round(To,1),' K \\n'\n",

-      "print 'Stagnation Pressure is ',round(Po,2),'kPa \\n'"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Stagnation Temperature is  319.9  K \n",

-        "\n",

-        "Stagnation Pressure is  187.85 kPa \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.2:PG-713"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques2\n",

-      "#to determine the force\n",

-      "\n",

-      "#initializing variables\n",

-      "mdot=-1 # mass flow rate out of control volume in kg/s\n",

-      "Vx=-1 # x component of velocity of control volume in m/s\n",

-      "Vy=10 # y component of velocity of control volume in m/s\n",

-      "\n",

-      "Fx=mdot*Vx # Force in X direction\n",

-      "\n",

-      "Fy=mdot*Vy # Force in Y direction\n",

-      "\n",

-      "print \"the force the man exert on the wheelbarrow\",round(Fx),\"N\"\n",

-      "print \"the force the floor exerts on the wheelbarrow\",round(Fy),\"N\"\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "the force the man exert on the wheelbarrow 1.0 N\n",

-        "the force the floor exerts on the wheelbarrow -10.0 N\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.3:PG-715"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques3\n",

-      "#determining the thrust acting on a control surface\n",

-      "\n",

-      "#i-inlet\n",

-      "#e-exit\n",

-      "#using momentum equation on control surface in x direction\n",

-      "me=20.4;#mass exiting in kg\n",

-      "mi=20;#mass entering in kg\n",

-      "ve=450;#exit velocity in m/s\n",

-      "vi=100;#inlet velocity in m/s\n",

-      "Pi=95;#Pressure at inlet in kPa\n",

-      "Pe=125;#Pressure at exit in kPa\n",

-      "Po=100;#surrounding pressure in kPa\n",

-      "Ai=0.2;#inlet area in m^2\n",

-      "Ae=0.1;#exit area in m^2\n",

-      "Rx=(me*ve-mi*vi)/1000-(Pi-Po)*Ai+(Pe-Po)*Ae;#thrust in x direction in kN\n",

-      "\n",

-      "print \"Thrust acting in x direction is \",round(Rx,2),\"kN\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Thrust acting in x direction is  10.68 kN\n"

-       ]

-      }

-     ],

-     "prompt_number": 11

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.4:PG-717"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques4\n",

-      "#to determine increase in enthalpy\n",

-      "#initializing variables\n",

-      "#i-inlet\n",

-      "#e-exit\n",

-      "\n",

-      "T=25+273 # temperature in kelvin\n",

-      "v=0.001003 # specific volume of water in kg/m^3 at 25 *c from table B.1.1 \n",

-      "ve=7;#exit velocity in m/s\n",

-      "vi=30;#inlet velocity in m/s\n",

-      "Pi=350;#Pressure at inlet in kPa\n",

-      "Pe=600;#Pressure at exit in kPa\n",

-      "\n",

-      "#using momentum equation on control surface \n",

-      "Pes= (vi**2-ve**2)/(2*v*1000)+Pi # exit pressure for reversible diffuser\n",

-      "delH=(vi**2-ve**2)/(2*1000.0) # change in enthalpy from first law in kJ/kg\n",

-      "delU=delH-v*(Pe-Pi) # change in internal energy in kJ/kg\n",

-      "delS=delU/T # change in entropy in kJ/kg.K\n",

-      "print\"the exit pressure for reversible diffuser is \",round(Pes),\"kPa\"\n",

-      "print\"the increase in enthalpy is \",round(delH,5),\"kJ/kg\"\n",

-      "print\"the increase in internal energy is \",round(delU,5),\"kJ/kg\"\n",

-      "print\"the increase in entropy is \",round(delS,6),\"kJ/kg.K\"\n",

-      "\n",

-      "\n",

-      "\n",

-      "\n",

-      "\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "the exit pressure for reversible diffuser is  774.0 kPa\n",

-        "the increase in enthalpy is  0.4255 kJ/kg\n",

-        "the increase in internal energy is  0.17475 kJ/kg\n",

-        "the increase in entropy is  0.000586 kJ/kg.K\n"

-       ]

-      }

-     ],

-     "prompt_number": 19

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.5:PG-720"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques5\n",

-      "#determining velocity of sound in air\n",

-      "import math\n",

-      "k=1.4;#constant\n",

-      "R=0.287;#gas constant\n",

-      "#at 300K\n",

-      "T1=300;# temperature in kelvin\n",

-      "c1=math.sqrt(k*R*T1*1000)\n",

-      "print \"Speed of sound at 300 K is\",round(c1,1),\" m/s\" \n",

-      "#at 1000K\n",

-      "T2=1000;# temperature in kelvin\n",

-      "c2=math.sqrt(k*R*T2*1000)\n",

-      "print \"Speed of sound at 1000 K is\",round(c2,1),\" m/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Speed of sound at 300 K is 347.2  m/s\n",

-        "Speed of sound at 1000 K is 633.9  m/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 24

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.6:PG-727"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques6\n",

-      "#determining mass flow rate through control volume\n",

-      "import math\n",

-      "k=1.4;#constant\n",

-      "R=0.287;#gas constant\n",

-      "To=360;#stagnation Temperature in K \n",

-      "T=To*0.8333;#Temperature of air in K, 0.8333 stagnation ratio from table\n",

-      "v=math.sqrt(k*R*T*1000);#velocity in m/s\n",

-      "P=528;#stagnation pressure in kPa\n",

-      "d=P/(R*T);#stagnation density in kg/m^3\n",

-      "A=500*10**-6;#area in m^2\n",

-      "ms=d*A*v;#mass flow rate in kg/s\n",

-      "print\" Mass flow rate at the throat section is\",round(ms,4),\"kg/s\"\n",

-      "#e-exit state\n",

-      "Te=To*0.9381;#exit temperature in K, ratio from table\n",

-      "ce=math.sqrt(k*R*Te*1000);#exit velocity of sound in m/s\n",

-      "Me=0.573;#Mach number\n",

-      "ve=Me*ce;\n",

-      "Pe=800;#exit pressure in kPa\n",

-      "de=Pe/R/Te;\n",

-      "mse=de*A*ve;\n",

-      "print\" Mass flow rate at the exit section is\",round(mse,4),\" kg/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Mass flow rate at the throat section is 1.0646 kg/s\n",

-        " Mass flow rate at the exit section is 0.8711  kg/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 30

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.7:PG-728"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques7\n",

-      "#determining exit properties in a control volume\n",

-      "import math\n",

-      "Po=1000;#stagnation pressure in kPa\n",

-      "To=360;#stagnation temperature in K\n",

-      "\n",

-      "#when diverging section acting as nozzle\n",

-      "\n",

-      "Pe1=0.0939*Po;#exit pressure of air in kPa\n",

-      "Te1=0.5089*To;#exit temperature in K\n",

-      "k=1.4;#constant\n",

-      "R=0.287;#gas constant for air\n",

-      "ce=math.sqrt(k*R*Te1*1000);#velocity of sound in exit section in m/s\n",

-      "Me=2.197;#mach number from table\n",

-      "ve1=Me*ce;#velocity of air at exit section in m/s\n",

-      "print \"When diverging section act as a nozzle :-\"\n",

-      "print \"Exit pressure is\",round(Pe1,4),\" kPa\"\n",

-      "print \"Exit Temperature\",round(Te1,1),\" K\"\n",

-      "print \"Exit velocity is\",round(ve1,1),\" m/s \"\n",

-      "\n",

-      "#when diverging section act as diffuser\n",

-      "\n",

-      "Me=0.308;\n",

-      "Pe2=0.0936*Po;#exit pressure of air in kPa\n",

-      "Te2=0.9812*To;#exit temperature in K\n",

-      "ce=math.sqrt(k*R*Te2*1000);#velocity of sound in exit section in m/s\n",

-      "ve2=Me*ce;\n",

-      "print \"When diverging section act as a diffuser :-\"\n",

-      "print \"Exit pressure is\",round(Pe2,1),\" kPa\"\n",

-      "print \"Exit Temperature\",round(Te2,2),\" K\"\n",

-      "print \"Exit velocity is\",round(ve2,),\" m/s \"\n",

-      "\n",

-      "# The value of Exit pressure when diverging section acts as diffuser is wrong\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "When diverging section act as a nozzle :-\n",

-        "Exit pressure is 93.9  kPa\n",

-        "Exit Temperature 183.2  K\n",

-        "Exit velocity is 596.1  m/s \n",

-        "When diverging section act as a diffuser :-\n",

-        "Exit pressure is 93.6  kPa\n",

-        "Exit Temperature 353.23  K\n",

-        "Exit velocity is 116.0  m/s \n"

-       ]

-      }

-     ],

-     "prompt_number": 39

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.8:PG-731"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques8\n",

-      "#Determine the static pressure and temperature of supersonic nozzle\n",

-      "#x-inlet\n",

-      "#y-exit\n",

-      "# using  the data from previous example\n",

-      "Mx=2.197 # Mach number at x\n",

-      "Px=93.9 # IN kPa\n",

-      "Tx=183.2 # IN K\n",

-      "P0x=1000 # IN kPa\n",

-      "T0x = 360 # in K\n",

-      "My=0.547 # Mach number at y From Table A.13\n",

-      "Py = 5.46*Px # From Table A.13\n",

-      "Ty = 1.854 * Tx # From Table A.13\n",

-      "P0y= 0.630 * P0x # From Table A.13\n",

-      "print \"Exit pressure after shock is\",round(Py,1),\" kPa\"\n",

-      "print \"Exit Temperature after shock is\",round(Ty,1),\" K\"\n",

-      "print \"Exit stagnation pressure after shock is\",round(P0y,1),\" kPa\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Exit pressure after shock is 512.7  kPa\n",

-        "Exit Temperature after shock is 339.7  K\n",

-        "Exit stagnation pressure after shock is 630.0  kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex17.9:PG-733"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#ques9\n",

-      "#determining exit plane properties in control volume\n",

-      "\n",

-      "#x-inlet\n",

-      "#y-exit\n",

-      "Mx=1.5;#mach number for inlet\n",

-      "My=0.7011;#mach number for exit\n",

-      "Px=272.4;#inlet pressure in kPa\n",

-      "Tx=248.3;#inlet temperature in K\n",

-      "Tox=360 # stagnation temperature in Kelvin\n",

-      "Pox=1000.0;#stagnation pressure for inlet\n",

-      "Py=2.4583*Px;# Pressure at 1.5 mach in kPa\n",

-      "Ty=1.320*Tx;# temperature at 1.5 mach in K\n",

-      "Poy=0.9298*Pox;# pressure at 1.5 mach in kPa\n",

-      "\n",

-      "Toy=Tox # constant\n",

-      "Me=0.339 # from table with A/A*=1.860 and M < 1\n",

-      "Pe=0.9222*Py;#Exit Pressure in kPa\n",

-      "Te=0.9771*Toy;#Exit temperature in K\n",

-      "Poe=0.9222*Poy;#Exit pressure in kPa\n",

-      "\n",

-      "print \"Exit Mach no.=\",Me\n",

-      "print \"Exit temperature =\",round(Te,2),\"K \"\n",

-      "print \"Exit pressure =\",round(Poe,1),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Exit Mach no.= 0.339\n",

-        "Exit temperature = 351.76 K \n",

-        "Exit pressure = 857.5 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 50

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter2.ipynb b/Fundamentals_Of_Thermodynamics/Chapter2.ipynb
deleted file mode 100755
index 95850873..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter2.ipynb
+++ /dev/null
@@ -1,314 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:eb1b6239947f111d6ea8f1bcaafe0bb84b121b2561bbb9f5f8a8bdec77169fd2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter2:SOME CONCEPTS AND DEFINITIONS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.1:pg-19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#weight of a person\n",
-      "\n",
-      "\n",
-      "m=1 #kg\n",
-      "g=9.75 #acc.due to gravity in m/s^2\n",
-      "F=m*g #weight of 1 kg mass in N\n",
-      "print\"\\n hence,weight of person is\",round(F,2),\" N\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,weight of person is 9.75  N\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.2:pg-24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2\n",
-      "#average volume and density\n",
-      "\n",
-      "Vliq=0.2 #volume of liquid in m^3\n",
-      "dliq=997 #density of liquid in kg/m^3\n",
-      "Vstone=0.12 #volume of stone in m^3\n",
-      "Vsand=0.15 #volume of sand in m^3\n",
-      "Vair=0.53 #vo;ume of air in m^3\n",
-      "mliq=Vliq*dliq #mass of liquid in kg\n",
-      "dstone=2750 #density of stone in kg/m^3\n",
-      "dsand=1500 #density of sand in kg/m^3\n",
-      "mstone=Vstone*dstone #volume of stone in m^3\n",
-      "msand=Vsand*dsand #volume of sand in m^3\n",
-      "Vtot=1 #total volume in m^3\n",
-      "dair=1.1 #density of air in kg/m^3\n",
-      "mair=Vair*dair #mass of air\n",
-      "mtot=mair+msand+mliq+mstone #total mass in kg\n",
-      "v=Vtot/mtot #specific volume in  m^3/kg\n",
-      "d=1/v #overall density in kg/m^3\n",
-      "print\"\\n hence,average specific volume is\",round(v,6),\"m^3/kg\" \n",
-      "print\"\\n and overall density is\", round(d),\"kg/m^3\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,average specific volume is 0.001325 m^3/kg\n",
-        "\n",
-        " and overall density is 755.0 kg/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.3:pg-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#calculating the required force\n",
-      "import math\n",
-      "\n",
-      "Dcyl=0.1 #cylinder diameter in m\n",
-      "Drod=0.01 #rod diameter in m\n",
-      "Acyl=math.pi*Dcyl**2/4 #cross sectional area of cylinder in m^2\n",
-      "Arod=math.pi*Drod**2/4 #cross sectional area of rod in m^2\n",
-      "Pcyl=250000 #inside hydaulic pressure in Pa\n",
-      "Po=101000 #outside atmospheric pressure in kPa\n",
-      "g=9.81 #acc. due to gravity in m/s^2\n",
-      "mp=25 #mass of (rod+piston) in kg\n",
-      "F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g #the force that rod can push within the upward direction in N\n",
-      "print\"\\n hence,the force that rod can push within the upward direction is\",round(F,1),\" N\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the force that rod can push within the upward direction is 932.9  N\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.4:pg-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "#Calculating atmospheric pressure \n",
-      "\n",
-      "dm=13534 #density of mercury in kg/m^3\n",
-      "H=0.750 #height difference between two columns in metres\n",
-      "g=9.80665 #acc. due to gravity  in m/s^2\n",
-      "Patm=dm*H*g/1000 #atmospheric pressure in kPa\n",
-      "print\"\\n hence, atmospheric pressure is\",round(Patm,2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, atmospheric pressure is 99.54 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.5:pg-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5\n",
-      "#pressure inside vessel\n",
-      "\n",
-      "dm=13590 #density of mercury in kg/m^3\n",
-      "H=0.24 #height difference between two columns in metres\n",
-      "g=9.80665 #acc. due to gravity  in m/s^2\n",
-      "dP=dm*H*g #pressure difference in Pa\n",
-      "Patm=13590*0.750*9.80665 #Atmospheric Pressure in Pa\n",
-      "Pvessel=dP+Patm #Absolute Pressure inside vessel in Pa\n",
-      "Pvessel=Pvessel/101325 #Absolute Pressure inside vessel in atm\n",
-      "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,3),\"atm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, the absolute pressure inside vessel is 1.302 atm\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.6:pg-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6\n",
-      "#calculating pressure\n",
-      "\n",
-      "dg=750 #density of gaasoline in kg/m^3\n",
-      "dR=1206 #density of R-134a in kg/m^3\n",
-      "H=7.5 #height of storage tank in metres\n",
-      "g=9.807 #acc. due to gravity in m/s^2\n",
-      "dP1=dg*g*H/1000 #in kPa\n",
-      "Ptop1=101 #atmospheric pressure in kPa\n",
-      "P1=dP1+Ptop1\n",
-      "print\"hence,pressure at the bottom of storage tank if fluid is gasoline is\",round(P1,1),\"kPa\" \n",
-      "dP2=dR*g*H/1000 #in kPa\n",
-      "Ptop2=1000 #top surface pressure in kPa\n",
-      "P2=dP2+Ptop2\n",
-      "print\"\\n hence, pressure at the bottom of storage tank if liquid is R-134a is\",round(P2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hence,pressure at the bottom of storage tank if fluid is gasoline is 156.2 kPa\n",
-        "\n",
-        " hence, pressure at the bottom of storage tank if liquid is R-134a is 1089.0 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.7:pg-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 7\n",
-      "#calculating balancing force\n",
-      "\n",
-      "Po=100#Outside atmospheric pressure in kPa\n",
-      "F1=25 #net force on the smallest piston in kN\n",
-      "A1=0.01 #cross sectional area of lower piston in m^2\n",
-      "P1=Po+F1/A1 #fluid pressure in kPa\n",
-      "d=900 #density of fluid in kg/m^3\n",
-      "g=9.81 #acc. due to gravity in m/s^2\n",
-      "H=6 #height of second piston in comparison to first one in m\n",
-      "P2=P1-d*g*H/1000 #pressure at higher elevation on piston 2 in kPa\n",
-      "A2=0.05 #cross sectional area of higher piston in m^3\n",
-      "F2=(P2-Po)*A2 #balancing force on second piston in kN\n",
-      "print\"\\n hence, balancing force on second larger piston is\",round(F2,1),\"N\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, balancing force on second larger piston is 122.4 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb
deleted file mode 100755
index 08e192fe..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter2_6.ipynb
+++ /dev/null
@@ -1,379 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:96f5bd0fa542764d6a8f75d5a1a42e4b79791697476f4029d0fa0fea1de65bed"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter2:SOME CONCEPTS AND DEFINITIONS"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.1:pg-19"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#weight of a person\n",

-      "\n",

-      "\n",

-      "m=1 #kg\n",

-      "g=9.75 #acc.due to gravity in m/s^2\n",

-      "F=m*g #weight of 1 kg mass in N\n",

-      "print\"\\n hence,weight of person is\",round(F,2),\" N\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,weight of person is 9.75  N\n"

-       ]

-      }

-     ],

-     "prompt_number": 20

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.1E:pg-20"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2\n",

-      "#weight of a person\n",

-      "\n",

-      "\n",

-      "m=1 #lbm\n",

-      "g=32 #acc.due to gravity in ft/s^2\n",

-      "F=m*g/32.174 #weight of 1 lbm mass in lbf\n",

-      "print\"\\n hence,weight of person is\",round(F,4),\" lbf\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,weight of person is 0.9946  lbf\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.2:pg-24"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#average volume and density\n",

-      "\n",

-      "Vliq=0.2 #volume of liquid in m^3\n",

-      "dliq=997 #density of liquid in kg/m^3\n",

-      "Vstone=0.12 #volume of stone in m^3\n",

-      "Vsand=0.15 #volume of sand in m^3\n",

-      "Vair=0.53 #vo;ume of air in m^3\n",

-      "mliq=Vliq*dliq #mass of liquid in kg\n",

-      "dstone=2750 #density of stone in kg/m^3\n",

-      "dsand=1500 #density of sand in kg/m^3\n",

-      "mstone=Vstone*dstone #volume of stone in m^3\n",

-      "msand=Vsand*dsand #volume of sand in m^3\n",

-      "Vtot=1 #total volume in m^3\n",

-      "dair=1.1 #density of air in kg/m^3\n",

-      "mair=Vair*dair #mass of air\n",

-      "mtot=mair+msand+mliq+mstone #total mass in kg\n",

-      "v=Vtot/mtot #specific volume in  m^3/kg\n",

-      "d=1/v #overall density in kg/m^3\n",

-      "print\"\\n hence,average specific volume is\",round(v,6),\"m^3/kg\" \n",

-      "print\"\\n and overall density is\", round(d),\"kg/m^3\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,average specific volume is 0.001325 m^3/kg\n",

-        "\n",

-        " and overall density is 755.0 kg/m^3\n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.3:pg-26"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4\n",

-      "#calculating the required force\n",

-      "import math\n",

-      "\n",

-      "Dcyl=0.1 #cylinder diameter in m\n",

-      "Drod=0.01 #rod diameter in m\n",

-      "Acyl=math.pi*Dcyl**2/4 #cross sectional area of cylinder in m^2\n",

-      "Arod=math.pi*Drod**2/4 #cross sectional area of rod in m^2\n",

-      "Pcyl=250000 #inside hydaulic pressure in Pa\n",

-      "Po=101000 #outside atmospheric pressure in kPa\n",

-      "g=9.81 #acc. due to gravity in m/s^2\n",

-      "mp=25 #mass of (rod+piston) in kg\n",

-      "F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g #the force that rod can push within the upward direction in N\n",

-      "print\"\\n hence,the force that rod can push within the upward direction is\",round(F,1),\" N\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the force that rod can push within the upward direction is 932.9  N\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.4:pg-28"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#Calculating atmospheric pressure \n",

-      "\n",

-      "dm=13534 #density of mercury in kg/m^3\n",

-      "H=0.750 #height difference between two columns in metres\n",

-      "g=9.80665 #acc. due to gravity  in m/s^2\n",

-      "Patm=dm*H*g/1000 #atmospheric pressure in kPa\n",

-      "print\"\\n hence, atmospheric pressure is\",round(Patm,2),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, atmospheric pressure is 99.54 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 10

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.5:pg-28"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 6\n",

-      "#pressure inside vessel\n",

-      "\n",

-      "dm=13590 #density of mercury in kg/m^3\n",

-      "H=0.24 #height difference between two columns in metres\n",

-      "g=9.80665 #acc. due to gravity  in m/s^2\n",

-      "dP=dm*H*g #pressure difference in Pa\n",

-      "Patm=13590*0.750*9.80665 #Atmospheric Pressure in Pa\n",

-      "Pvessel=dP+Patm #Absolute Pressure inside vessel in Pa\n",

-      "Pvessel=Pvessel/101325 #Absolute Pressure inside vessel in atm\n",

-      "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,3),\"atm\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, the absolute pressure inside vessel is 1.302 atm\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.5E:pg-28"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7\n",

-      "#pressure inside vessel\n",

-      "\n",

-      "dm=848 #density of mercury in lbm/ft^3\n",

-      "H=9.5  #height difference between two columns in inches\n",

-      "g=32.174 #acc. due to gravity  in ft/s^2\n",

-      "dP=dm*H*g #pressure difference in lbf/in^2\n",

-      "Patm=848*9.5*32.174/(1728*32.174) #Atmospheric Pressure in lbf/in^2\n",

-      "Pvessel=dP+Patm #Absolute Pressure inside vessel in lbf/in^2\n",

-      "Pvessel=848*29.5*32.174/(1728*32.174)+Patm #Absolute Pressure inside vessel in lbf/in^2\n",

-      "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,2),\"lbf/in^2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, the absolute pressure inside vessel is 19.14 lbf/in^2\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.6:pg-29"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8\n",

-      "#calculating pressure\n",

-      "\n",

-      "dg=750 #density of gaasoline in kg/m^3\n",

-      "dR=1206 #density of R-134a in kg/m^3\n",

-      "H=7.5 #height of storage tank in metres\n",

-      "g=9.807 #acc. due to gravity in m/s^2\n",

-      "dP1=dg*g*H/1000 #in kPa\n",

-      "Ptop1=101 #atmospheric pressure in kPa\n",

-      "P1=dP1+Ptop1\n",

-      "print\"hence,pressure at the bottom of storage tank if fluid is gasoline is\",round(P1,1),\"kPa\" \n",

-      "dP2=dR*g*H/1000 #in kPa\n",

-      "Ptop2=1000 #top surface pressure in kPa\n",

-      "P2=dP2+Ptop2\n",

-      "print\"\\n hence, pressure at the bottom of storage tank if liquid is R-134a is\",round(P2),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "hence,pressure at the bottom of storage tank if fluid is gasoline is 156.2 kPa\n",

-        "\n",

-        " hence, pressure at the bottom of storage tank if liquid is R-134a is 1089.0 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 17

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex2.7:pg-29"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9\n",

-      "#calculating balancing force\n",

-      "\n",

-      "Po=100#Outside atmospheric pressure in kPa\n",

-      "F1=25 #net force on the smallest piston in kN\n",

-      "A1=0.01 #cross sectional area of lower piston in m^2\n",

-      "P1=Po+F1/A1 #fluid pressure in kPa\n",

-      "d=900 #density of fluid in kg/m^3\n",

-      "g=9.81 #acc. due to gravity in m/s^2\n",

-      "H=6 #height of second piston in comparison to first one in m\n",

-      "P2=P1-d*g*H/1000 #pressure at higher elevation on piston 2 in kPa\n",

-      "A2=0.05 #cross sectional area of higher piston in m^3\n",

-      "F2=(P2-Po)*A2 #balancing force on second piston in kN\n",

-      "print\"\\n hence, balancing force on second larger piston is\",round(F2,1),\"N\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, balancing force on second larger piston is 122.4 N\n"

-       ]

-      }

-     ],

-     "prompt_number": 19

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter3.ipynb b/Fundamentals_Of_Thermodynamics/Chapter3.ipynb
deleted file mode 100755
index 7447d700..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter3.ipynb
+++ /dev/null
@@ -1,403 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:6916c81bc8c1ac256991650a35c94801d5197d3ce593f7738220895278cf5f8b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter3:PROPERTIES OF A PURE SUBSTANCE"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.3:pg-59"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#determining the quality and specific volume\n",
-      "\n",
-      "v1=0.5 #given specific volume in m^3/kg\n",
-      "vf=0.001073 #specific volume when only liquid phase is present in m^3/kg\n",
-      "vfg=0.60475 #in m^3/kg\n",
-      "x=(v1-vf)/vfg #quality\n",
-      "print\"For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is\",round(x,3)\n",
-      "\n",
-      "v2=1 #given specific volume in m^3/kg\n",
-      "\n",
-      " # using the method of interplotation\n",
-      "T=((400-300)*(1.0-0.8753))/(1.0315-0.8753)+300 #temperature of the water\n",
-      "print\"For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is\",round(T,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is 0.825\n",
-        "For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is 379.8\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.4:pg-60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      " #example 4\n",
-      "#percentage of vapor \n",
-      "\n",
-      "vliq=0.1 #volume of saturated liquid in m^3\n",
-      "vf=0.000843 #in m^3/kg\n",
-      "vvap=0.9 #volume of saturated vapor R-134a in equilbrium\n",
-      "vg=0.02671 #in m^3/kg\n",
-      "mliq=vliq/vf #mass of liquid in kg \n",
-      "mvap=vvap/vg #mass of vapor in kg\n",
-      "m=mliq+mvap #total mass in kg\n",
-      "x=mvap*100/m #percentage of vapor on mass basis\n",
-      "print\"hence,% vapor on mass basis is\",round(x,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hence,% vapor on mass basis is 22.1 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.6:pg-61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6\n",
-      "#Determinig the missing property\n",
-      "\n",
-      "T1=273-53.2 #given temperature in K\n",
-      "P1=600 #given pressure in kPa\n",
-      "print\"This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\"\n",
-      "T2=100 #given temp. in K\n",
-      "v2=0.008 #given specific volume in m^3/kg\n",
-      "vf=0.001452 #in m^3/kg\n",
-      "vg=0.0312 #in m^3/kg\n",
-      "Psat=779.2 #saturation pressure in kPa\n",
-      "vfg=vg-vf #in m^3/kg\n",
-      "x=(v2-vf)/vfg #quality\n",
-      "print\"\\n hence, the pressure is\",round(Psat,1),\"kPa\"\n",
-      "print\"\\n and quality is\",round(x,4),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\n",
-        "\n",
-        " hence, the pressure is 779.2 kPa\n",
-        "\n",
-        " and quality is 0.2201 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.7:pg-62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 7\n",
-      "#determining the pressure of water\n",
-      "\n",
-      "vg=0.12736 #specific volume in m^3/kg for water at 200C\n",
-      "v=0.4 #specific volume in m^3/kg\n",
-      "P1=500 #in kPa\n",
-      "v1=0.42492 #specific volume at P1 in m^3/kg\n",
-      "P2=600 #in kPa\n",
-      "v2=0.35202 #specific volume at P2 in m^3/kg\n",
-      "P=P1+(P2-P1)*(v-v1)/(v2-v1) #calculating pressure by interplotation\n",
-      "print \"hence,the pressure of water is\",round(P,1),\" kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " hence,the pressure of water is 534.2  kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.8:pg-66"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 8\n",
-      "#calculating mass of air\n",
-      "\n",
-      "P=100 #pressure in kPa\n",
-      "V=6*10*4 #volume of room in m^3\n",
-      "R=0.287 #in kN-m/kg-K\n",
-      "T=25 #temperature in Celsius\n",
-      "m=P*V/(R*(T+273.1)) #mass of air contained in room\n",
-      "print\"\\n hence, mass of air contained in room is\",round(m,1),\"kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, mass of air contained in room is 280.5 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.9:pg-67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9\n",
-      "#calculating pressure inside tank\n",
-      "\n",
-      "V=0.5 #volumr of tank in m^3\n",
-      "m=10 #mass of ideal gas in kg\n",
-      "T=25 #temperature of tank in Celsius\n",
-      "M=24 #molecular mass of gas in kg/kmol\n",
-      "Ru=8.3145 #universal gas constant in kN-m/kmol-K\n",
-      "R=Ru/M #gas constant for given ideal gas in kN-m/kg-K\n",
-      "P=m*R*(T+273.2)/V #pressure inside tank\n",
-      "print\"\\n hence,pressure inside tank is\",round(P),\"kpa\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,pressure inside tank is 2066.0 kpa\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.10:pg-67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10\n",
-      "#mass  flow rate\n",
-      "\n",
-      "dt=185 #time period in seconds over which there is incrrease in volume \n",
-      "dV=0.75 #increase in volume in 0.75 in m^3\n",
-      "V=dV/dt #volume flow rate in m^3/s\n",
-      "P=105 #pressure inside gas bell kPa\n",
-      "T=21 #temperature in celsius\n",
-      "R=0.1889 #ideal gas constant in kJ/kg-K\n",
-      "m=P*V/(R*(T+273.15)) #mass flow rate of the flow in kg/s\n",
-      "print\"\\n hence,mass flow rate is\",round(m,5),\"kg/s\"\n",
-      "print\"\\n and volume flow rate is\",round(V,5),\"m^3/s\"\n",
-      "#The answer of volume flow rate in the book is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,mass flow rate is 0.00766 kg/s\n",
-        "\n",
-        " and volume flow rate is 0.00405 m^3/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.12:pg-71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12\n",
-      "#determining specific using diffenet laws\n",
-      "\n",
-      "T=100.0 #given temp.in 100 celsius\n",
-      "P=3.0 #given pressure in MPa\n",
-      "v1=0.0065 #specific volume in m^3/kg using table\n",
-      "print\"\\n hence,the specific volume for R-134a using R-134a tables is\",round(v1,3),\"m^3/kg\"\n",
-      "M=102.3 #molecular mass in kg\n",
-      "R=8.3145 #in kJ/K\n",
-      "Ru=R/M #in kJ/K-kg\n",
-      "v2=Ru*(T+273)/(P*1000) #specific volume assuming R-134a to be ideal gas in m^3/kg\n",
-      "print\"\\n hence,the specific volume for R-134a using R-134a the ideal gas laws is\",round(v2,5),\"m^3/kg\"\n",
-      "Tr=373.2/374.2 #reduced temperature using generalized chart\n",
-      "Pr=3.0/4.06 #reduced pressure using generalized chart\n",
-      "Z=0.67 #compressibility factor \n",
-      "v3=Z*v2 # specific volume using generalized chart in m^3/kg\n",
-      "print\"\\n hence,the specific volume for R-134a using the generalized chart is\",round(v3,5),\"m^3/kg\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the specific volume for R-134a using R-134a tables is 0.006 m^3/kg\n",
-        "\n",
-        " hence,the specific volume for R-134a using R-134a the ideal gas laws is 0.01011 m^3/kg\n",
-        "\n",
-        " hence,the specific volume for R-134a using the generalized chart is 0.00677 m^3/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.13:pg-71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13\n",
-      "#calculating mass of gas\n",
-      "\n",
-      "Pc=4250 #critical pressure of propane in kPa\n",
-      "Tc=369.8 #critical temperature in K\n",
-      "T=15 #temperature of propane in celsius\n",
-      "Tr=T/Tc #reduced temperature\n",
-      "Prsat=0.2 # reduced pressure \n",
-      "P=Prsat*Pc #pressure in kPa\n",
-      "x=0.1 #given quality\n",
-      "Zf=0.035 #from graph\n",
-      "Zg=0.83 #from graph\n",
-      "Z=(1-x)*Zf+x*Zg #overall compressibility factor\n",
-      "V=0.1 #volume of steel bottle in m^3\n",
-      "R=0.1887 #in kPa-m^3/kg-K\n",
-      "m=P*V/(Z*R*(T+273)) #total propane mass in kg\n",
-      "print\"\\n hence,the total propane mass is\",round(m,2),\"kg\"\n",
-      "print\"\\n and pressure is\",round(P,2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the total propane mass is 13.66 kg\n",
-        "\n",
-        " and pressure is 850.0 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter3_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter3_6.ipynb
deleted file mode 100755
index 64998280..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter3_6.ipynb
+++ /dev/null
@@ -1,571 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:b6e51b20af9aaa1b32972245e7dbe7cc42b3ed40e5f3cbf6db8837625a12aebc"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter3:PROPERTIES OF A PURE SUBSTANCE"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.1:pg-57"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#determinig the phase of water\n",

-      "\n",

-      "print\"from the table,we find that at 120C,saturation pressure of water is 198.5 kPa.But here we have pressure of 500 kPa.hence,water exists as a compressed liquid here.\"\n",

-      "print\"also at 120C,vf=0.00106 kg/m^3 and vg=0.89186 kg/m^3.given v=0.5 m^3/kg i.e. vf<v<vg,so we have two phase mixture of liquid and vapor.\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "from the table,we find that at 120C,saturation pressure of water is 198.5 kPa.But here we have pressure of 500 kPa.hence,water exists as a compressed liquid here.\n",

-        "also at 120C,vf=0.00106 kg/m^3 and vg=0.89186 kg/m^3.given v=0.5 m^3/kg i.e. vf<v<vg,so we have two phase mixture of liquid and vapor.\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.2:pg-58"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2\n",

-      "#determinig the phase \n",

-      "\n",

-      "print\"from the table,we find that at 30C,saturation pressure of ammonia is 1167 kPa.But here we have pressure of 1000 kPa.hence,ammonia exists in superheated vapor state.\"\n",

-      "print\"for R-22 at 200 kPa,vg=0.1119 kg/m^3.given v=0.15 m^3/kg i.e. v>vg,so the state is superheated vapor\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "from the table,we find that at 30C,saturation pressure of ammonia is 1167 kPa.But here we have pressure of 1000 kPa.hence,ammonia exists in superheated vapor state.\n",

-        "for R-22 at 200 kPa,vg=0.1119 kg/m^3.given v=0.15 m^3/kg i.e. v>vg,so the state is superheated vapor\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.3:pg-59"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#determining the quality and specific volume\n",

-      "\n",

-      "v1=0.5 #given specific volume in m^3/kg\n",

-      "vf=0.001073 #specific volume when only liquid phase is present in m^3/kg\n",

-      "vfg=0.60475 #in m^3/kg\n",

-      "x=(v1-vf)/vfg #quality\n",

-      "print\"For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is\",round(x,3)\n",

-      "\n",

-      "v2=1 #given specific volume in m^3/kg\n",

-      "\n",

-      " # using the method of interplotation\n",

-      "T=((400-300)*(1.0-0.8753))/(1.0315-0.8753)+300 #temperature of the water\n",

-      "print\"For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is\",round(T,1)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is 0.825\n",

-        "For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is 379.8\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.4:pg-60"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      " #example 4\n",

-      "#percentage of vapor \n",

-      "\n",

-      "vliq=0.1 #volume of saturated liquid in m^3\n",

-      "vf=0.000843 #in m^3/kg\n",

-      "vvap=0.9 #volume of saturated vapor R-134a in equilbrium\n",

-      "vg=0.02671 #in m^3/kg\n",

-      "mliq=vliq/vf #mass of liquid in kg \n",

-      "mvap=vvap/vg #mass of vapor in kg\n",

-      "m=mliq+mvap #total mass in kg\n",

-      "x=mvap*100/m #percentage of vapor on mass basis\n",

-      "print\"hence,% vapor on mass basis is\",round(x,1),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "hence,% vapor on mass basis is 22.1 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 7

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.4E:pg-60"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      " #example 5\n",

-      "#percentage of vapor \n",

-      "\n",

-      "vliq=0.1 #volume of saturated liquid in ft^3\n",

-      "vf=0.0136 #in ft^3/lbm\n",

-      "vvap=0.9 #volume of saturated vapor R-134a in equilbrium\n",

-      "vg=0.4009 #in ft^3/lbm\n",

-      "mliq=vliq/vf #mass of liquid in lbm \n",

-      "mvap=vvap/vg #mass of vapor in lbm\n",

-      "m=mliq+mvap #total mass in lbm\n",

-      "x=mvap*100/m #percentage of vapor on mass basis\n",

-      "print\"hence,% vapor on mass basis is\",round(x,1),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "hence,% vapor on mass basis is 23.4 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.5:pg-61"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#calculating pressure after heat addition\n",

-      "\n",

-      "v1=0.14922 #specific volume of sautrated ammonia in m^3/kg\n",

-      "print\"Since the volume does not change during the process,the specific volume remains constant.therefore\"\n",

-      "v2=v1 #in m^3/kg\n",

-      "print\"Since vg at 40C is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\"\n",

-      "P2=945 #final pressure in kPa\n",

-      "print\"hence,the final pressure is 945 kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Since the volume does not change during the process,the specific volume remains constant.therefore\n",

-        "Since vg at 40C is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\n",

-        "hence,the final pressure is 945 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 7

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.5E:pg-61"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#calculating pressure after heat addition\n",

-      "\n",

-      "v1=2.311 #specific volume of sautrated ammonia in ft^3/lbm\n",

-      "print\"Since the volume does not change during the process,the specific volume remains constant.therefore\"\n",

-      "v2=v1 #in ft^3/lbm\n",

-      "print\"Since vg at 120f is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\"\n",

-      "P2=145 #final pressure in lbf/in^2\n",

-      "print\"hence,the final pressure is 145 lbf/in^2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Since the volume does not change during the process,the specific volume remains constant.therefore\n",

-        "Since vg at 120f is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\n",

-        "hence,the final pressure is 145 lbf/in^2\n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.6:pg-61"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 6\n",

-      "#Determinig the missing property\n",

-      "\n",

-      "T1=273-53.2 #given temperature in K\n",

-      "P1=600 #given pressure in kPa\n",

-      "print\"This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\"\n",

-      "T2=100 #given temp. in K\n",

-      "v2=0.008 #given specific volume in m^3/kg\n",

-      "vf=0.001452 #in m^3/kg\n",

-      "vg=0.0312 #in m^3/kg\n",

-      "Psat=779.2 #saturation pressure in kPa\n",

-      "vfg=vg-vf #in m^3/kg\n",

-      "x=(v2-vf)/vfg #quality\n",

-      "print\"\\n hence, the pressure is\",round(Psat,1),\"kPa\"\n",

-      "print\"\\n and quality is\",round(x,4),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\n",

-        "\n",

-        " hence, the pressure is 779.2 kPa\n",

-        "\n",

-        " and quality is 0.2201 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 19

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.7:pg-62"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7\n",

-      "#determining the pressure of water\n",

-      "\n",

-      "vg=0.12736 #specific volume in m^3/kg for water at 200C\n",

-      "v=0.4 #specific volume in m^3/kg\n",

-      "P1=500 #in kPa\n",

-      "v1=0.42492 #specific volume at P1 in m^3/kg\n",

-      "P2=600 #in kPa\n",

-      "v2=0.35202 #specific volume at P2 in m^3/kg\n",

-      "P=P1+(P2-P1)*(v-v1)/(v2-v1) #calculating pressure by interplotation\n",

-      "print \"hence,the pressure of water is\",round(P,1),\" kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " hence,the pressure of water is 534.2  kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 21

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.8:pg-66"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8\n",

-      "#calculating mass of air\n",

-      "\n",

-      "P=100 #pressure in kPa\n",

-      "V=6*10*4 #volume of room in m^3\n",

-      "R=0.287 #in kN-m/kg-K\n",

-      "T=25 #temperature in Celsius\n",

-      "m=P*V/(R*(T+273.1)) #mass of air contained in room\n",

-      "print\"\\n hence, mass of air contained in room is\",round(m,1),\"kg\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, mass of air contained in room is 280.5 kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 27

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.9:pg-67"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9\n",

-      "#calculating pressure inside tank\n",

-      "\n",

-      "V=0.5 #volumr of tank in m^3\n",

-      "m=10 #mass of ideal gas in kg\n",

-      "T=25 #temperature of tank in Celsius\n",

-      "M=24 #molecular mass of gas in kg/kmol\n",

-      "Ru=8.3145 #universal gas constant in kN-m/kmol-K\n",

-      "R=Ru/M #gas constant for given ideal gas in kN-m/kg-K\n",

-      "P=m*R*(T+273.2)/V #pressure inside tank\n",

-      "print\"\\n hence,pressure inside tank is\",round(P),\"kpa\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,pressure inside tank is 2066.0 kpa\n"

-       ]

-      }

-     ],

-     "prompt_number": 30

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.10:pg-67"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 10\n",

-      "#mass  flow rate\n",

-      "\n",

-      "dt=185 #time period in seconds over which there is incrrease in volume \n",

-      "dV=0.75 #increase in volume in 0.75 in m^3\n",

-      "V=dV/dt #volume flow rate in m^3/s\n",

-      "P=105 #pressure inside gas bell kPa\n",

-      "T=21 #temperature in celsius\n",

-      "R=0.1889 #ideal gas constant in kJ/kg-K\n",

-      "m=P*V/(R*(T+273.15)) #mass flow rate of the flow in kg/s\n",

-      "print\"\\n hence,mass flow rate is\",round(m,5),\"kg/s\"\n",

-      "print\"\\n and volume flow rate is\",round(V,5),\"m^3/s\"\n",

-      "#The answer of volume flow rate in the book is wrong."

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,mass flow rate is 0.00766 kg/s\n",

-        "\n",

-        " and volume flow rate is 0.00405 m^3/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 40

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.12:pg-71"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 12\n",

-      "#determining specific using diffenet laws\n",

-      "\n",

-      "T=100.0 #given temp.in 100 celsius\n",

-      "P=3.0 #given pressure in MPa\n",

-      "v1=0.0065 #specific volume in m^3/kg using table\n",

-      "print\"\\n hence,the specific volume for R-134a using R-134a tables is\",round(v1,3),\"m^3/kg\"\n",

-      "M=102.3 #molecular mass in kg\n",

-      "R=8.3145 #in kJ/K\n",

-      "Ru=R/M #in kJ/K-kg\n",

-      "v2=Ru*(T+273)/(P*1000) #specific volume assuming R-134a to be ideal gas in m^3/kg\n",

-      "print\"\\n hence,the specific volume for R-134a using R-134a the ideal gas laws is\",round(v2,5),\"m^3/kg\"\n",

-      "Tr=373.2/374.2 #reduced temperature using generalized chart\n",

-      "Pr=3.0/4.06 #reduced pressure using generalized chart\n",

-      "Z=0.67 #compressibility factor \n",

-      "v3=Z*v2 # specific volume using generalized chart in m^3/kg\n",

-      "print\"\\n hence,the specific volume for R-134a using the generalized chart is\",round(v3,5),\"m^3/kg\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the specific volume for R-134a using R-134a tables is 0.006 m^3/kg\n",

-        "\n",

-        " hence,the specific volume for R-134a using R-134a the ideal gas laws is 0.01011 m^3/kg\n",

-        "\n",

-        " hence,the specific volume for R-134a using the generalized chart is 0.00677 m^3/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 51

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex3.13:pg-71"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 13\n",

-      "#calculating mass of gas\n",

-      "\n",

-      "Pc=4250 #critical pressure of propane in kPa\n",

-      "Tc=369.8 #critical temperature in K\n",

-      "T=15 #temperature of propane in celsius\n",

-      "Tr=T/Tc #reduced temperature\n",

-      "Prsat=0.2 # reduced pressure \n",

-      "P=Prsat*Pc #pressure in kPa\n",

-      "x=0.1 #given quality\n",

-      "Zf=0.035 #from graph\n",

-      "Zg=0.83 #from graph\n",

-      "Z=(1-x)*Zf+x*Zg #overall compressibility factor\n",

-      "V=0.1 #volume of steel bottle in m^3\n",

-      "R=0.1887 #in kPa-m^3/kg-K\n",

-      "m=P*V/(Z*R*(T+273)) #total propane mass in kg\n",

-      "print\"\\n hence,the total propane mass is\",round(m,2),\"kg\"\n",

-      "print\"\\n and pressure is\",round(P,2),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the total propane mass is 13.66 kg\n",

-        "\n",

-        " and pressure is 850.0 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 44

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter4.ipynb b/Fundamentals_Of_Thermodynamics/Chapter4.ipynb
deleted file mode 100755
index eed9d20a..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter4.ipynb
+++ /dev/null
@@ -1,198 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:da3557673ab31ece618ad10f5b0d3ef78112ff36bd0b260e508d99211060a1c9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4:WORK AND HEAT"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.1:PG-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#work done during different processes\n",
-      "import math\n",
-      "\n",
-      "P1=200 #initial pressure inside cylinder in kPa\n",
-      "V2=0.1 #in m^3\n",
-      "V1=0.04 #initial volume of gas in m^3\n",
-      "\n",
-      "W1=P1*(V2-V1) #work done in isobaric process in kJ\n",
-      "print\"\\n hence,the work done during the isobaric process is\",round(W1,2),\"kJ. \\n\"\n",
-      "\n",
-      "W2=P1*V1*math.log(V2/V1) #work done in isothermal process in kJ\n",
-      "print\"\\n hence,the work done in isothermal process is\",round(W2,2),\"kJ. \\n\"\n",
-      "\n",
-      "P2=P1*(V1/V2)**(1.3) #final pressure according to the given process\n",
-      "W3=(P2*V2-P1*V1)/(1-1.3)\n",
-      "print\"\\n hence,the work done during the described process is\",round(W3,2),\"kJ. \\n\"\n",
-      "\n",
-      "W4=0 #work done in isochoric process\n",
-      "print\"\\n hence,the work done in the isochoric process is\",round(W4,3),\"kJ. \\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the work done during the isobaric process is 12.0 kJ. \n",
-        "\n",
-        "\n",
-        " hence,the work done in isothermal process is 7.33 kJ. \n",
-        "\n",
-        "\n",
-        " hence,the work done during the described process is 6.41 kJ. \n",
-        "\n",
-        "\n",
-        " hence,the work done in the isochoric process is 0.0 kJ. \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.3:PG-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "\n",
-      "#work produced\n",
-      "\n",
-      "Psat=190.2 #in kPa\n",
-      "P1=Psat #saturation pressure in state 1\n",
-      "vf=0.001504 #in m^3/kg\n",
-      "vfg=0.62184 #in m^3/kg\n",
-      "x1=0.25 #quality\n",
-      "v1=vf+x1*vfg #specific volume at state 1 in m^3/kg\n",
-      "v2=1.41*v1 #specific volume at state 2 in m^3/kg\n",
-      "P2=600 #pressure in state 2 in kPa from Table B.2.2\n",
-      "m=0.5 #mass of ammonia in kg\n",
-      "W=m*(P1+P2)*(v2-v1)/2 #woork produced by ammonia in kJ\n",
-      "print \"The final pressure is\",round(P2),\"kPa\\n\"\n",
-      "print \"hence,work produced by ammonia is\",round(W,2),\"kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final pressure is 600.0 kPa\n",
-        "\n",
-        "hence,work produced by ammonia is 12.71 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.4:PG-100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "\n",
-      "#calculating work done\n",
-      "\n",
-      "v1=0.35411 #specific volume at state 1 in m^3/kg\n",
-      "v2=v1/2 \n",
-      "m=0.1 #mass of water in kg\n",
-      "P1=1000 #pressure inside cylinder in kPa\n",
-      "W=m*P1*(v2-v1) #in kJ\n",
-      "print \"the work in the overall process is\",round(W,1),\"kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the work in the overall process is -17.7 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.7:PG-108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 7\n",
-      "#heat transfer\n",
-      "k=1.4 #conductivity of glass pane in W/m-K\n",
-      "A=0.5 #total surface area of glass pane\n",
-      "dx=0.005 #thickness of glasspane in m\n",
-      "dT1=20-12.1 #temperature difference between room air and outer glass surface temperature in celsius\n",
-      "Q=-k*A*dT1/dx #conduction through glass slab in W\n",
-      "h=100 #convective heat transfer coefficient in W/m^2-K \n",
-      "dT=12.1-(-10)  #temperature difference between warm room and colder ambient in celsius\n",
-      "Q2=h*A*dT #heat transfer in convective layer in W\n",
-      "#result\n",
-      "print \"the rate of heat transfer in the glass and convective layer is\",round(Q2),\"kW.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of heat transfer in the glass and convective layer is 1105.0 kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter4_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter4_6.ipynb
deleted file mode 100755
index eed9d20a..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter4_6.ipynb
+++ /dev/null
@@ -1,198 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:da3557673ab31ece618ad10f5b0d3ef78112ff36bd0b260e508d99211060a1c9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4:WORK AND HEAT"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.1:PG-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#work done during different processes\n",
-      "import math\n",
-      "\n",
-      "P1=200 #initial pressure inside cylinder in kPa\n",
-      "V2=0.1 #in m^3\n",
-      "V1=0.04 #initial volume of gas in m^3\n",
-      "\n",
-      "W1=P1*(V2-V1) #work done in isobaric process in kJ\n",
-      "print\"\\n hence,the work done during the isobaric process is\",round(W1,2),\"kJ. \\n\"\n",
-      "\n",
-      "W2=P1*V1*math.log(V2/V1) #work done in isothermal process in kJ\n",
-      "print\"\\n hence,the work done in isothermal process is\",round(W2,2),\"kJ. \\n\"\n",
-      "\n",
-      "P2=P1*(V1/V2)**(1.3) #final pressure according to the given process\n",
-      "W3=(P2*V2-P1*V1)/(1-1.3)\n",
-      "print\"\\n hence,the work done during the described process is\",round(W3,2),\"kJ. \\n\"\n",
-      "\n",
-      "W4=0 #work done in isochoric process\n",
-      "print\"\\n hence,the work done in the isochoric process is\",round(W4,3),\"kJ. \\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the work done during the isobaric process is 12.0 kJ. \n",
-        "\n",
-        "\n",
-        " hence,the work done in isothermal process is 7.33 kJ. \n",
-        "\n",
-        "\n",
-        " hence,the work done during the described process is 6.41 kJ. \n",
-        "\n",
-        "\n",
-        " hence,the work done in the isochoric process is 0.0 kJ. \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.3:PG-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "\n",
-      "#work produced\n",
-      "\n",
-      "Psat=190.2 #in kPa\n",
-      "P1=Psat #saturation pressure in state 1\n",
-      "vf=0.001504 #in m^3/kg\n",
-      "vfg=0.62184 #in m^3/kg\n",
-      "x1=0.25 #quality\n",
-      "v1=vf+x1*vfg #specific volume at state 1 in m^3/kg\n",
-      "v2=1.41*v1 #specific volume at state 2 in m^3/kg\n",
-      "P2=600 #pressure in state 2 in kPa from Table B.2.2\n",
-      "m=0.5 #mass of ammonia in kg\n",
-      "W=m*(P1+P2)*(v2-v1)/2 #woork produced by ammonia in kJ\n",
-      "print \"The final pressure is\",round(P2),\"kPa\\n\"\n",
-      "print \"hence,work produced by ammonia is\",round(W,2),\"kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final pressure is 600.0 kPa\n",
-        "\n",
-        "hence,work produced by ammonia is 12.71 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.4:PG-100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "\n",
-      "#calculating work done\n",
-      "\n",
-      "v1=0.35411 #specific volume at state 1 in m^3/kg\n",
-      "v2=v1/2 \n",
-      "m=0.1 #mass of water in kg\n",
-      "P1=1000 #pressure inside cylinder in kPa\n",
-      "W=m*P1*(v2-v1) #in kJ\n",
-      "print \"the work in the overall process is\",round(W,1),\"kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the work in the overall process is -17.7 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.7:PG-108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 7\n",
-      "#heat transfer\n",
-      "k=1.4 #conductivity of glass pane in W/m-K\n",
-      "A=0.5 #total surface area of glass pane\n",
-      "dx=0.005 #thickness of glasspane in m\n",
-      "dT1=20-12.1 #temperature difference between room air and outer glass surface temperature in celsius\n",
-      "Q=-k*A*dT1/dx #conduction through glass slab in W\n",
-      "h=100 #convective heat transfer coefficient in W/m^2-K \n",
-      "dT=12.1-(-10)  #temperature difference between warm room and colder ambient in celsius\n",
-      "Q2=h*A*dT #heat transfer in convective layer in W\n",
-      "#result\n",
-      "print \"the rate of heat transfer in the glass and convective layer is\",round(Q2),\"kW.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of heat transfer in the glass and convective layer is 1105.0 kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter5.ipynb b/Fundamentals_Of_Thermodynamics/Chapter5.ipynb
deleted file mode 100755
index 24351fdb..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter5.ipynb
+++ /dev/null
@@ -1,469 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e29dcea16eab605fcf93b95197200be339c1be6076d5413e7d70418a9a6df29"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter5:THE FIRST LAW OF THERMODYNAMICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.1:pg-131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#calculating height\n",
-      "\n",
-      "m=1100 #mass of car in kg\n",
-      "ke=400 #kinetic energy of car in kJ\n",
-      "V=(2*ke*1000/m)**0.5 #velocity of car in m/s\n",
-      "g=9.807 #acc. due to gravity in m/s^2\n",
-      "H=ke*1000/(m*g) #height to which the car  should be lifted so that its potential energy equals its kinetic energy\n",
-      "print\"hence,the car should be raised to a height is\",round(H,1),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hence,the car should be raised to a height is 37.1 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.2:pg-134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2\n",
-      "#change in internal energy\n",
-      "\n",
-      "W=-5090 #work input to paddle wheel in kJ\n",
-      "Q=-1500 #heat transfer from tank in kJ\n",
-      "dU=Q-W #change in internal energy in kJ\n",
-      "print\"hence,change in internal energy is\",round(dU),\"kj\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hence,change in internal energy is 3590.0 kj\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.3:pg-134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#analysis of energy transfer\n",
-      "\n",
-      "g=9.806 #acceleration due to gravity in m/s^2\n",
-      "m=10 #mass of stone in kg\n",
-      "H1=10.2 #initial height of stone above water in metres\n",
-      "H2=0 #final height in metres\n",
-      "dKE1=-m*g*(H2-H1) #change in kinetic energy when stone enters state 2 in J\n",
-      "dPE1=-1 #change in potential energy when stone enters state 2 in J\n",
-      "print\"\\n hence,when stone is \",round(dKE1),\"J\"\n",
-      "print\"\\n and change in potential energy is \",round(dPE1),\"J\"\n",
-      "dPE2=0 #change in potential energy when stone enters state 3 in JQ2=0 //no heat transfer when stone enters state 3 in J\n",
-      "W2=0 #no work done when stone enters state 3 in J\n",
-      "dKE2=-1 #change in kinetic energy when stone enters state 3\n",
-      "dU2=-dKE2 #change in internal energy when stone enters state 3 in J\n",
-      "print\"\\n hence,when stone has just come to rest in the bucket is \",round(dKE2),\"J\" \n",
-      "print\"\\n and dU is\",round(dU2),\"J\"  \n",
-      "dKE3=0 #change in kinetic energy when stone enters state 4\n",
-      "dPE=0 #change in potential energy when stone enters state 4 in J\n",
-      "W3=0 #no work done when stone enters state 4 in J\n",
-      "dU3=-1 #change in internal energy when stone enters state 4 in J\n",
-      "Q3=dU3 #heat transfer when stone enters state 4 in J\n",
-      "print\"\\n hence,when stone has entered state 4 is\",round(dU3),\"J\" \n",
-      "print\"\\n and Q3 is \",round(Q3),\"J\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " hence,when stone is  1000.0 J\n",
-        "\n",
-        " and change in potential energy is  -1.0 J\n",
-        "\n",
-        " hence,when stone has just come to rest in the bucket is  -1.0 J\n",
-        "\n",
-        " and dU is 1.0 J\n",
-        "\n",
-        " hence,when stone has entered state 4 is -1.0 J\n",
-        "\n",
-        " and Q3 is  -1.0 J\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.4:pg-136"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "#Determinig the missing properties \n",
-      "\n",
-      "T1=300 #given temp. in Celsius\n",
-      "u1=2780 #given specific internal enrgy in kJ/kg\n",
-      "print\"From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\" #hiven pressure in kPa\n",
-      "u2=2000 #given specific intrernal energy in kJ/kg\n",
-      "print\"at P=2000 kPa\"\n",
-      "uf=906.4 #in kJ/kg\n",
-      "ug=2600.3 #in kJ/kg \n",
-      "x2=(u2-906.4)/(ug-uf) \n",
-      "print\"Also, under the given conditions\"\n",
-      "vf=0.001177 #in m^3/kg \n",
-      "vg=0.099627 #in m^3/kg\n",
-      "v2=vf+x2*(vg-vf)#Specific volume for water in m^3/kg\n",
-      "print\"\\n hence,specific volume for water is \",round(v2,5),\"m^3/kg\" \n",
-      "print\"\\n Therefore ,this state is \",round(x2,4),\"N\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\n",
-        "at P=2000 kPa\n",
-        "Also, under the given conditions\n",
-        "\n",
-        " hence,specific volume for water is  0.06474 m^3/kg\n",
-        "\n",
-        " Therefore ,this state is  0.6456 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.5:pg-138"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5\n",
-      "#calculating heat transfer for the given process\n",
-      "\n",
-      "Vliq=0.05 #volume of saturated liquid in m^3\n",
-      "vf=0.001043 #in m^3/kg\n",
-      "Vvap=4.95 #volume of saturated water vapour in m^3\n",
-      "vg=1.6940 #in m^3/kg\n",
-      "m1liq=Vliq/vf #mass of liquid in kg\n",
-      "m1liq=round(m1liq,2)\n",
-      "m1vap=Vvap/vg #mass of vapors in kg\n",
-      "m1vap=round(m1vap,2)\n",
-      "u1liq=417.36 #specific internal energy of liquid in kJ/kg\n",
-      "u1vap=2506.1 #specific internal energy of vapors in kJ/kg\n",
-      "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in kJ\n",
-      "m=m1liq+m1vap #total mass in kg\n",
-      "V=5.0 #total volume in m^3\n",
-      "v2=V/m #final specific volume in m^3/kg\n",
-      "print\"by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\"\n",
-      "u2=2600.5 #specific internal energy at final state in kJ/kg\n",
-      "U2=m*u2 #internal energy at final state in kJ\n",
-      "Q=U2-U1 #heat transfer for the process in kJ\n",
-      "print\"\\n hence,heat transfer for the process is\",round(Q),\"kJ\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\n",
-        "\n",
-        " hence,heat transfer for the process is 104935.0 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:pg-143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6\n",
-      "#calculating work and heat transfer for the process\n",
-      "\n",
-      "V1=0.1 #volume of cylinder in m^3\n",
-      "m=0.5 #mass of steam in kg\n",
-      "v1=V1/m #specific volume of steam in m^3/kg\n",
-      "vf=0.001084 #m^3/kg\n",
-      "vfg=0.4614 #m^3/kg\n",
-      "x1=(v1-vf)/vfg #quality\n",
-      "hf=604.74 #kJ/kg\n",
-      "hfg=2133.8#kJ/kg\n",
-      "h2=3066.8 #final specific heat enthalpy in kJ/kg\n",
-      "h1=hf+x1*hfg #initial specific enthalpy in kJ/kg\n",
-      "Q=m*(h2-h1) #heat transfer for this process in kJ\n",
-      "P=400 #pressure inside cylinder in kPa\n",
-      "v2=0.6548 #specific enthalpy in m^3/kg\n",
-      "W=m*P*(v2-v1) #work done for the process in kJ\n",
-      "print\"\\n hence, work done for the process is\",round(W),\"kJ\" \n",
-      "print\"\\n and heat transfer is \",round(Q,1),\"kJ\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, work done for the process is 91.0 kJ\n",
-        "\n",
-        " and heat transfer is  771.1 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.8:pg-151"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 8\n",
-      "#calculating change in enthalpy\n",
-      "import math\n",
-      "\n",
-      "h1=273.2 #specific heat enthalpy for oxygen at 300 K\n",
-      "h2=1540.2 #specific heat enthalpy for oxygen at 1500 K\n",
-      "T1=300 #initial temperature in K\n",
-      "T2=1500 #final temparature in K\n",
-      "\n",
-      "dh1=h2-h1 #this change in specific heat enthalpy is calculated using ideal gas tables \n",
-      "dh3=0.922*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K\n",
-      "dh4=1.0767*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature\n",
-      "print\"\\n Hence,change in specific heat enthalpy if ideal gas tables are used is \",round(dh1,1),\"kJ/kg\"\n",
-      "print\"\\n if specific heat is assumed to be constant and using its value at T1 is\",round(dh3,1),\"kJ/kg\"\n",
-      "print\"\\n if specific heat is assumed to be constant at its value at (T1+T2)/2 is\",round(dh4,1),\"kJ/kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " Hence,change in specific heat enthalpy if ideal gas tables are used is  1267.0 kJ/kg\n",
-        "\n",
-        " if specific heat is assumed to be constant and using its value at T1 is 1106.4 kJ/kg\n",
-        "\n",
-        " if specific heat is assumed to be constant at its value at (T1+T2)/2 is 1292.0 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.9:pg-152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9\n",
-      "#determining amount of heat transfer\n",
-      "\n",
-      "P=150 #pressure of nitrogen in cylinder in kPa\n",
-      "V=0.1 #initial volume of cylinder in m^3\n",
-      "T1=25 #initial temperature of nitrogen in celsius\n",
-      "T2=150 #final tempareture of nitrogen in celsius\n",
-      "R=0.2968 #in kJ/kg-K\n",
-      "m=P*V/(R*(T1+273)) #mass of nitrogen in kg\n",
-      "Cv=0.745 #constant volume specific heat for nitrogen in kJ/kg-K\n",
-      "W=-20 #work done on nitrogen gas in kJ\n",
-      "Q=m*Cv*(T2-T1)+W #heat transfer during the process in kJ\n",
-      "print\"\\n hence,the heat transfer for the above process is\",round(Q,1),\"kJ\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the heat transfer for the above process is -4.2 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.10:pg-155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10\n",
-      "#calculating rate of increase of internal energy\n",
-      "\n",
-      "W=-12.8*20 #power consumed in  J/s\n",
-      "Q=-10 #heat transfer rate from battery in J/s\n",
-      "r=Q-W #rate of increase of internal energy\n",
-      "print\"\\n hence,the rate of increase of internal energy is\",round(r),\"J/s\"  "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the rate of increase of internal energy is 246.0 J/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.11:pg-155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11\n",
-      "#rate of change of temperature\n",
-      "\n",
-      "Q=1500.0 #power produced by burning wood in J/s\n",
-      "mair=1 #mass of air in kg\n",
-      "mwood=5 #mass of soft pine wood in kg \n",
-      "miron=25 #mass of cast iron in kg\n",
-      "Cvair=0.717 #constant volume specific heat for air in kJ/kg\n",
-      "Cwood=1.38 #constant volume specific heat for wood in kJ/kg\n",
-      "Ciron=0.42 #constant volume specific heat for iron in kJ/kg\n",
-      "dT=75-20 #increase in temperature in Celsius\n",
-      "T=(Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) #rate of change of temperature in K/s\n",
-      "dt=(dT/T)/60 #in minutes\n",
-      "print\" hence,the rate of change of temperature is\",round(T,4),\"k/s\"  \n",
-      "print\" and time taken to reach a temperature of T is\",round(dt),\"min\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " hence,the rate of change of temperature is 0.0828 k/s\n",
-        " and time taken to reach a temperature of T is 11.0 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb
deleted file mode 100755
index 5a9734bc..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb
+++ /dev/null
@@ -1,635 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:bc9fbdf4ecee4bce9770355a576adf487ef4df6d34bad14ada42e8383ea01d91"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter5:THE FIRST LAW OF THERMODYNAMICS"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.1:pg-131"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#calculating height\n",

-      "\n",

-      "m=1100 #mass of car in kg\n",

-      "ke=400 #kinetic energy of car in kJ\n",

-      "V=(2*ke*1000/m)**0.5 #velocity of car in m/s\n",

-      "g=9.807 #acc. due to gravity in m/s^2\n",

-      "H=ke*1000/(m*g) #height to which the car  should be lifted so that its potential energy equals its kinetic energy\n",

-      "print\"hence,the car should be raised to a height is\",round(H,1),\"m\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "hence,the car should be raised to a height is 37.1 m\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.1E:pg-132"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1E\n",

-      "#calculating height\n",

-      "\n",

-      "m=2400 #mass of car in lbm\n",

-      "ke=400 #kinetic energy of car in Btu\n",

-      "V=(2*ke*778.17*32.174/m)**0.5 #velocity of car in ft/s\n",

-      "g=32.174 #acc. due to gravity in ft/s^2\n",

-      "H=ke*778.17*32.174/(m*g) #height to which the car  should be lifted so that its potential energy equals its kinetic energy\n",

-      "print\"hence,the car should be raised to a height is\",round(H,1),\"ft\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "hence,the car should be raised to a height is 129.7 ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.2:pg-134"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2\n",

-      "#change in internal energy\n",

-      "\n",

-      "W=-5090 #work input to paddle wheel in kJ\n",

-      "Q=-1500 #heat transfer from tank in kJ\n",

-      "dU=Q-W #change in internal energy in kJ\n",

-      "print\"hence,change in internal energy is\",round(dU),\"kj\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "hence,change in internal energy is 3590.0 kj\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.3:pg-134"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#analysis of energy transfer\n",

-      "\n",

-      "g=9.806 #acceleration due to gravity in m/s^2\n",

-      "m=10 #mass of stone in kg\n",

-      "H1=10.2 #initial height of stone above water in metres\n",

-      "H2=0 #final height in metres\n",

-      "dKE1=-m*g*(H2-H1) #change in kinetic energy when stone enters state 2 in J\n",

-      "dPE1=-1 #change in potential energy when stone enters state 2 in J\n",

-      "print\"\\n hence,when stone is \",round(dKE1),\"J\"\n",

-      "print\"\\n and change in potential energy is \",round(dPE1),\"J\"\n",

-      "dPE2=0 #change in potential energy when stone enters state 3 in JQ2=0 //no heat transfer when stone enters state 3 in J\n",

-      "W2=0 #no work done when stone enters state 3 in J\n",

-      "dKE2=-1 #change in kinetic energy when stone enters state 3\n",

-      "dU2=-dKE2 #change in internal energy when stone enters state 3 in J\n",

-      "print\"\\n hence,when stone has just come to rest in the bucket is \",round(dKE2),\"J\" \n",

-      "print\"\\n and dU is\",round(dU2),\"J\"  \n",

-      "dKE3=0 #change in kinetic energy when stone enters state 4\n",

-      "dPE=0 #change in potential energy when stone enters state 4 in J\n",

-      "W3=0 #no work done when stone enters state 4 in J\n",

-      "dU3=-1 #change in internal energy when stone enters state 4 in J\n",

-      "Q3=dU3 #heat transfer when stone enters state 4 in J\n",

-      "print\"\\n hence,when stone has entered state 4 is\",round(dU3),\"J\" \n",

-      "print\"\\n and Q3 is \",round(Q3),\"J\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " hence,when stone is  1000.0 J\n",

-        "\n",

-        " and change in potential energy is  -1.0 J\n",

-        "\n",

-        " hence,when stone has just come to rest in the bucket is  -1.0 J\n",

-        "\n",

-        " and dU is 1.0 J\n",

-        "\n",

-        " hence,when stone has entered state 4 is -1.0 J\n",

-        "\n",

-        " and Q3 is  -1.0 J\n"

-       ]

-      }

-     ],

-     "prompt_number": 7

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.4:pg-136"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4\n",

-      "#Determinig the missing properties \n",

-      "\n",

-      "T1=300 #given temp. in Celsius\n",

-      "u1=2780 #given specific internal enrgy in kJ/kg\n",

-      "print\"From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\" #hiven pressure in kPa\n",

-      "u2=2000 #given specific intrernal energy in kJ/kg\n",

-      "print\"at P=2000 kPa\"\n",

-      "uf=906.4 #in kJ/kg\n",

-      "ug=2600.3 #in kJ/kg \n",

-      "x2=(u2-906.4)/(ug-uf) \n",

-      "print\"Also, under the given conditions\"\n",

-      "vf=0.001177 #in m^3/kg \n",

-      "vg=0.099627 #in m^3/kg\n",

-      "v2=vf+x2*(vg-vf)#Specific volume for water in m^3/kg\n",

-      "print\"\\n hence,specific volume for water is \",round(v2,5),\"m^3/kg\" \n",

-      "print\"\\n Therefore ,this state is \",round(x2,4),\"N\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\n",

-        "at P=2000 kPa\n",

-        "Also, under the given conditions\n",

-        "\n",

-        " hence,specific volume for water is  0.06474 m^3/kg\n",

-        "\n",

-        " Therefore ,this state is  0.6456 N\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.5:pg-138"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#calculating heat transfer for the given process\n",

-      "\n",

-      "Vliq=0.05 #volume of saturated liquid in m^3\n",

-      "vf=0.001043 #in m^3/kg\n",

-      "Vvap=4.95 #volume of saturated water vapour in m^3\n",

-      "vg=1.6940 #in m^3/kg\n",

-      "m1liq=Vliq/vf #mass of liquid in kg\n",

-      "m1liq=round(m1liq,2)\n",

-      "m1vap=Vvap/vg #mass of vapors in kg\n",

-      "m1vap=round(m1vap,2)\n",

-      "u1liq=417.36 #specific internal energy of liquid in kJ/kg\n",

-      "u1vap=2506.1 #specific internal energy of vapors in kJ/kg\n",

-      "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in kJ\n",

-      "m=m1liq+m1vap #total mass in kg\n",

-      "V=5.0 #total volume in m^3\n",

-      "v2=V/m #final specific volume in m^3/kg\n",

-      "print\"by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\"\n",

-      "u2=2600.5 #specific internal energy at final state in kJ/kg\n",

-      "U2=m*u2 #internal energy at final state in kJ\n",

-      "Q=U2-U1 #heat transfer for the process in kJ\n",

-      "print\"\\n hence,heat transfer for the process is\",round(Q),\"kJ\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\n",

-        "\n",

-        " hence,heat transfer for the process is 104935.0 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 36

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.5E:pg-140"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5E\n",

-      "#calculating heat transfer for the given process\n",

-      "\n",

-      "Vliq=1 #volume of saturated liquid in ft^3\n",

-      "vf=0.01672 #in ft^3/lbm\n",

-      "Vvap=99 #volume of saturated water vapour in m^3\n",

-      "vg=26.80 #in ft^3/lbm\n",

-      "m1liq=Vliq/vf #mass of liquid in lbm\n",

-      "m1liq=round(m1liq,2)\n",

-      "m1vap=Vvap/vg #mass of vapors in lbm\n",

-      "m1vap=round(m1vap,2)\n",

-      "u1liq=180.1 #specific internal energy of liquid in Btu/lbm\n",

-      "u1vap=1077.6 #specific internal energy of vapors in Btu/lbm\n",

-      "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in Btu\n",

-      "U1=round(U1)\n",

-      "m=m1liq+m1vap #total mass in lbm\n",

-      "V=100.0 #total volume in ft^3\n",

-      "v2=V/m #final specific volume in ft^3/lbm\n",

-      "print\"by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\"\n",

-      "u2=1117.0 #specific internal energy at final state in Btu/lbm\n",

-      "U2=m*u2 #internal energy at final state in Btu\n",

-      "U2=round(U2)\n",

-      "Q=U2-U1 #heat transfer for the process in Btu\n",

-      "print\"\\n hence,heat transfer for the process is\",round(Q),\"Btu\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\n",

-        "\n",

-        " hence,heat transfer for the process is 56182.0 Btu\n"

-       ]

-      }

-     ],

-     "prompt_number": 23

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.6:pg-143"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 6\n",

-      "#calculating work and heat transfer for the process\n",

-      "\n",

-      "V1=0.1 #volume of cylinder in m^3\n",

-      "m=0.5 #mass of steam in kg\n",

-      "v1=V1/m #specific volume of steam in m^3/kg\n",

-      "vf=0.001084 #m^3/kg\n",

-      "vfg=0.4614 #m^3/kg\n",

-      "x1=(v1-vf)/vfg #quality\n",

-      "hf=604.74 #kJ/kg\n",

-      "hfg=2133.8#kJ/kg\n",

-      "h2=3066.8 #final specific heat enthalpy in kJ/kg\n",

-      "h1=hf+x1*hfg #initial specific enthalpy in kJ/kg\n",

-      "Q=m*(h2-h1) #heat transfer for this process in kJ\n",

-      "P=400 #pressure inside cylinder in kPa\n",

-      "v2=0.6548 #specific enthalpy in m^3/kg\n",

-      "W=m*P*(v2-v1) #work done for the process in kJ\n",

-      "print\"\\n hence, work done for the process is\",round(W),\"kJ\" \n",

-      "print\"\\n and heat transfer is \",round(Q,1),\"kJ\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, work done for the process is 91.0 kJ\n",

-        "\n",

-        " and heat transfer is  771.1 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.7:pg-144"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "# example 7\n",

-      "#calculating work and heat transfer for the process\n",

-      "V1=0.010 #volume of cylinder in m^3\n",

-      "V2=0.040 # m^3\n",

-      "P1=573 #kPa\n",

-      "v1=0.03606 #specific volume of steam in m^3/kg\n",

-      "u1=389.2 #kJ/kg\n",

-      "m=V1/v1#mass of steam in kg\n",

-      "v2=v1*(V2/V1)\n",

-      "P2=163 #kPa\n",

-      "u2=395.8 # kJ/kg\n",

-      "W=8 # kJ\n",

-      "Q2=m*(u2-u1)+W\n",

-      "u3=383.4#kJ/kg\n",

-      "W2=m*(u1-u3)#kJ\n",

-      "print\"\\n and heat transfer is \",round(Q2,2),\"kJ\" \n",

-      "print\"\\n hence, work done for the process is\",round(W2,1),\"kJ\" \n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " and heat transfer is  9.83 kJ\n",

-        "\n",

-        " hence, work done for the process is 1.6 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 18

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.8:pg-151"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8\n",

-      "#calculating change in enthalpy\n",

-      "import math\n",

-      "\n",

-      "h1=273.2 #specific heat enthalpy for oxygen at 300 K\n",

-      "h2=1540.2 #specific heat enthalpy for oxygen at 1500 K\n",

-      "T1=300 #initial temperature in K\n",

-      "T2=1500 #final temparature in K\n",

-      "\n",

-      "dh1=h2-h1 #this change in specific heat enthalpy is calculated using ideal gas tables \n",

-      "dh3=0.922*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K\n",

-      "dh4=1.0767*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature\n",

-      "print\"\\n Hence,change in specific heat enthalpy if ideal gas tables are used is \",round(dh1,1),\"kJ/kg\"\n",

-      "print\"\\n if specific heat is assumed to be constant and using its value at T1 is\",round(dh3,1),\"kJ/kg\"\n",

-      "print\"\\n if specific heat is assumed to be constant at its value at (T1+T2)/2 is\",round(dh4,1),\"kJ/kg\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " Hence,change in specific heat enthalpy if ideal gas tables are used is  1267.0 kJ/kg\n",

-        "\n",

-        " if specific heat is assumed to be constant and using its value at T1 is 1106.4 kJ/kg\n",

-        "\n",

-        " if specific heat is assumed to be constant at its value at (T1+T2)/2 is 1292.0 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.9:pg-152"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9\n",

-      "#determining amount of heat transfer\n",

-      "\n",

-      "P=150 #pressure of nitrogen in cylinder in kPa\n",

-      "V=0.1 #initial volume of cylinder in m^3\n",

-      "T1=25 #initial temperature of nitrogen in celsius\n",

-      "T2=150 #final tempareture of nitrogen in celsius\n",

-      "R=0.2968 #in kJ/kg-K\n",

-      "m=P*V/(R*(T1+273)) #mass of nitrogen in kg\n",

-      "Cv=0.745 #constant volume specific heat for nitrogen in kJ/kg-K\n",

-      "W=-20 #work done on nitrogen gas in kJ\n",

-      "Q=m*Cv*(T2-T1)+W #heat transfer during the process in kJ\n",

-      "print\"\\n hence,the heat transfer for the above process is\",round(Q,1),\"kJ\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the heat transfer for the above process is -4.2 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 43

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.9E:pg-153"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9E\n",

-      "#determining amount of heat transfer\n",

-      "\n",

-      "P=20 #pressure of nitrogen in cylinder in lbf/in^2\n",

-      "V=2 #initial volume of cylinder in ft^3\n",

-      "T1=80 #initial temperature of nitrogen in F\n",

-      "T2=300 #final tempareture of nitrogen in F\n",

-      "R=55.15 #in ft*lbf/lbm-R\n",

-      "m=P*V*144/(R*(540)) #mass of nitrogen in lbm\n",

-      "Cv=0.177 #constant volume specific heat for nitrogen in Btu/lbm-R\n",

-      "W=-9.15 #work done on nitrogen gas in Btu\n",

-      "Q=m*Cv*(T2-T1)+W #heat transfer during the process in Btu\n",

-      "print\"\\n hence,the heat transfer for the above process is\",round(Q,2),\"Btu\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the heat transfer for the above process is -1.62 Btu\n"

-       ]

-      }

-     ],

-     "prompt_number": 31

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.10:pg-155"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 10\n",

-      "#calculating rate of increase of internal energy\n",

-      "\n",

-      "W=-12.8*20 #power consumed in  J/s\n",

-      "Q=-10 #heat transfer rate from battery in J/s\n",

-      "r=Q-W #rate of increase of internal energy\n",

-      "print\"\\n hence,the rate of increase of internal energy is\",round(r),\"J/s\"  "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the rate of increase of internal energy is 246.0 J/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 45

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex5.11:pg-155"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 11\n",

-      "#rate of change of temperature\n",

-      "\n",

-      "Q=1500.0 #power produced by burning wood in J/s\n",

-      "mair=1 #mass of air in kg\n",

-      "mwood=5 #mass of soft pine wood in kg \n",

-      "miron=25 #mass of cast iron in kg\n",

-      "Cvair=0.717 #constant volume specific heat for air in kJ/kg\n",

-      "Cwood=1.38 #constant volume specific heat for wood in kJ/kg\n",

-      "Ciron=0.42 #constant volume specific heat for iron in kJ/kg\n",

-      "dT=75-20 #increase in temperature in Celsius\n",

-      "T=(Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) #rate of change of temperature in K/s\n",

-      "dt=(dT/T)/60 #in minutes\n",

-      "print\" hence,the rate of change of temperature is\",round(T,4),\"k/s\"  \n",

-      "print\" and time taken to reach a temperature of T is\",round(dt),\"min\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " hence,the rate of change of temperature is 0.0828 k/s\n",

-        " and time taken to reach a temperature of T is 11.0 min\n"

-       ]

-      }

-     ],

-     "prompt_number": 49

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter6.ipynb
deleted file mode 100755
index 2548c7ee..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter6.ipynb
+++ /dev/null
@@ -1,566 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:084379625b3dc39247a3be2201f659a82e58d8004d66338e0ef15cb00bca073d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter6:FIRST-LAW ANALYSIS FOR A CONTROL VOLUME"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.1:pg-182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#calculating mass flow rate in kg/s\n",
-      "import math\n",
-      "R=0.287 #in kJ/kg-K\n",
-      "T=25 #temperature in celsius\n",
-      "P=150 #pressure in kPa\n",
-      "v=R*(T+273.15)/P #specific volume in m^3/kg\n",
-      "D=0.2 #diameter of pipe in metre\n",
-      "A=math.pi*D**2/4 #cross sectional area in m^2\n",
-      "V=0.1 #velocity of air in m/s\n",
-      "m=V*A/v #mass flow rate in kg/s\n",
-      "print\"\\n hence,the mass flow rate is\",round(m,4),\"kg/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the mass flow rate is 0.0055 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.2:pg-184"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2\n",
-      "#work done for adding the fluid\n",
-      "\n",
-      "P=600 #pressure in kPa\n",
-      "m=1 #in kg\n",
-      "v=0.001 #specific volume in m^3/kg\n",
-      "W=P*m*v #necessary work in kJ for adding the fluid \n",
-      "print\" \\n hence,the work involved in this process is\",round(W,4),\"kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  \n",
-        " hence,the work involved in this process is 0.6 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.3:pg-188"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#rate of flow of water\n",
-      "\n",
-      "hir=441.89 #in kJ/kg for refrigerant using steam table \n",
-      "her=249.10 #in kJ/kg for refrigerant using steam table\n",
-      "hiw=42 #in kJ/kg for water using steam table\n",
-      "hew=83.95 #in kJ/kg for water using steam table\n",
-      "mr=0.2 #the rate at which refrigerant enters the condenser in kg/s\n",
-      "mw=mr*(hir-her)/(hew-hiw) #rate of flow of water in kg/s\n",
-      "print\"\\n hence,the rate at which cooling water flows thorugh the condenser is\",round(mw,3),\"kg/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the rate at which cooling water flows thorugh the condenser is 0.919 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.4:pg-190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "#determining quality of steam\n",
-      "\n",
-      "hi=2850.1 #initial specific heat enthalpy for steam in kJ/kg\n",
-      "Vi=50 #initial velocity of steam in m/s\n",
-      "Ve=600 #final velocity of steam in m/s\n",
-      "he=hi+Vi**2/(2*1000)-Ve**2/(2*1000) #final specific heat enthalpy for steam in kJ/kg\n",
-      "hf=467.1 #at final state in kJ/kg\n",
-      "hfg=2226.5 #at final state in kJ/kg\n",
-      "xe=(he-hf)/hfg #quality of steam in final state\n",
-      "print\" \\n hence, the quality is\",round(xe,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " hence, the quality is 0.99\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.5:pg-193"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5\n",
-      "#quality of ammonia leaving expansion valve\n",
-      "\n",
-      "hi=346.8 #specific heat enthalpy for ammonia at initial state in kJ/kg\n",
-      "he=hi #specific heat enthalpy for ammonia at final state will be equal that at initial state because it is a throttling process\n",
-      "hf=134.4 #at final state in kJ/kg\n",
-      "hfg=1296.4#at final state in kJ/kg\n",
-      "xe=(he-hf)*100/hfg #quality at final state\n",
-      "print\"\\n hence,quality of the ammonia leaving the expansion valve is\",round(xe,2),\"%\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,quality of the ammonia leaving the expansion valve is 16.38 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.6:pg-194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6\n",
-      "#power output of turbine in kW\n",
-      "\n",
-      "hi=3137 #initial specific heat of enthalpy in kJ/kg\n",
-      "he=2675.5 #final specific heat of enthalpy in kJ/kg\n",
-      "Vi=50.0 #initial velocity of steam in m/s\n",
-      "Ve=100 #final velocity of steam in m/s\n",
-      "Zi=6 #height of inlet conditions in metres\n",
-      "Ze=3 #height of exit conditions in metres\n",
-      "m=1.5 #mass flow rate of steam in kg/s\n",
-      "g=9.8066 #acc. due to gravity in m/s^2\n",
-      "Qcv=-8.5 #heat transfer rate from turbine in kW\n",
-      "Wcv=Qcv+m*(hi+Vi**2/(2*1000)+g*Zi/1000)-m*(he+Ve**2/(2*1000)+g*Ze/1000) #power output of turbine in kW\n",
-      "print\"\\n hence,the power output of the turbine is\",round(Wcv,1),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the power output of the turbine is 678.2 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.7:pg-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 7\n",
-      "#heat transfer rate in aftercooler\n",
-      "\n",
-      "V1=0 #we assume initial velocity to be zero because its given that it enters with a low velocity\n",
-      "V2=25.0 #final velocity with which carbon dioxide exits in m/s\n",
-      "h2=401.52 #final specific enthalpy of heat when carbon dioxide exits in kJ/kg\n",
-      "h1=198 #initial specific enthalpy of heat in kJ/kg\n",
-      "w=h1-h2-V2**2/(2*1000) #in kJ/kg\n",
-      "Wc=-50 #power input to the compressor in kW\n",
-      "m=Wc/w #mass flow rate of carbon dioxide in kg/s\n",
-      "h3=257.9 #final specific enthalpy of heat when carbon dioxide flows into a constant pressure aftercooler\n",
-      "Qcool=-m*(h3-h2) #heat transfer rate in the aftercooler in kW\n",
-      "print\" \\n hence,heat transfer rate in the aftercooler is\",round(Qcool,1),\"kW\" \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " hence,heat transfer rate in the aftercooler is 35.2 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.8:pg-197"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 8\n",
-      "#Required pump work\n",
-      "\n",
-      "m=1.5 #mass flow rate of water in kg/s\n",
-      "g=9.807 #acceleration due to gravity in m/s^2\n",
-      "Zin=-15 #depth of water pump in well in metres\n",
-      "Zex=0 #in metres\n",
-      "v=0.001001 #specific volume in m^3/kg\n",
-      "Pex=400+101.3 #exit pressure in kPa\n",
-      "Pin=90 #in kPa\n",
-      "W=m*(g*(Zin-Zex)*0.001-(Pex-Pin)*v) #power input in kW\n",
-      "print\" \\n Hence, the pump requires power of input is\",-round(W,2),\"kW\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Hence, the pump requires power of input is 0.84 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.9:pg-198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9\n",
-      "#heat tranfer in simple steam power plant\n",
-      "\n",
-      "h1=3023.5 #specific heat of enthalpy of steam leaving boiler in kJ/kg\n",
-      "h2=3002.5 #specific heat of enthalpy of steam entering turbine in kJ/kg\n",
-      "x=0.9 #quality of steam entering condenser\n",
-      "hf=226 #in kJ/kg\n",
-      "hfg=2373.1 #in kJ/kg\n",
-      "h3=hf+x*hfg #specific heat of enthalpy of steam entering condenser in kJ/kg\n",
-      "h4=188.5 #specific heat of enthalpy of steam entering pump in kJ/kg\n",
-      "q12=h2-h1 #heat transfer in line between boiler and turbine in kJ/kg\n",
-      "w23=h2-h3 #turbine work in kJ/kg\n",
-      "q34=h4-h3 #heat transfer in condenser\n",
-      "w45=-4 #pump work in kJ/kg\n",
-      "h5=h4-w45 #in kJ/kg\n",
-      "q51=h1-h5 #heat transfer in boiler in kJ/kg\n",
-      "print\"\\n hence, heat transfer in line between boiler and turbine is\",round(q12,1),\"kJ/kg\"  \n",
-      "print\"\\n hence, turbine work is\",round(w23,1),\"kJ/kg\" \n",
-      "print\"\\n hence, heat transfer in condenser is \",round(q34,1),\"kJ/kg\" \n",
-      "print\"\\n hence, heat transfer in boiler is \",round(q51,1),\"kJ/kg\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, heat transfer in line between boiler and turbine is -21.0 kJ/kg\n",
-        "\n",
-        " hence, turbine work is 640.7 kJ/kg\n",
-        "\n",
-        " hence, heat transfer in condenser is  -2173.3 kJ/kg\n",
-        "\n",
-        " hence, heat transfer in boiler is  2831.0 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.10:pg-200"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10\n",
-      "#analysis of refrigerator\n",
-      "\n",
-      "hf4=167.4 #in kJ/kg\n",
-      "hfg4=215.6 #in kJ/kg\n",
-      "h3=241.8 #specific heat of enthalpy of R-134a entering expansion valve\n",
-      "h4=h3 #specific heat of enthalpy of R-134a leaving expansion valve\n",
-      "h1=387.2 #in kJ/kg\n",
-      "h2=435.1 #in kJ/kg\n",
-      "x4=(h3-hf4)/hfg4 #quality of R-134a at evaporator inlet\n",
-      "m=0.1 #mass flow rate in kg/s\n",
-      "Qevap=m*(h1-h4) #rate of heat transfer to the evaporator\n",
-      "Wcomp=-5 #power input to compressor in kW\n",
-      "Qcomp=m*(h2-h1)+Wcomp #rate of heat transfer from compressor\n",
-      "print\"\\n hence, the quality at the evaporator inlet is \",round(x4,3), \n",
-      "print\"\\n hence, the rate of heat transfer to the evaporator is \",round(Qevap,2),\n",
-      "print\"\\n hence, rate of heat transfer from the compressor is\",round(Qcomp,2), "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, the quality at the evaporator inlet is  0.345 \n",
-        " hence, the rate of heat transfer to the evaporator is  14.54 \n",
-        " hence, rate of heat transfer from the compressor is -0.21\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.11:pg-204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11\n",
-      "#Determining the final temperature of steam\n",
-      "\n",
-      "u2=3040.4 #final internal energy in kJ/kg\n",
-      "hi=u2 #in kJ/kg\n",
-      "P2=1.4 #final Pressure in MPa\n",
-      "print\"Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \"\n",
-      "T2=452 #final temperature in Celsius"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.12:pg-206"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12\n",
-      "#Calculating mass flow of steam in tank\n",
-      "\n",
-      "V1=0.4 #initial volume fo tank in m^3\n",
-      "v1=0.5243 #initial specific volume in m^3/kg\n",
-      "h1=3040.4 #initial specific enthalpy in kJ/kg\n",
-      "u1=2548.9 #initial specific internal energy in kJ/kg\n",
-      "m1=V1/v1 #initial mass of steam in tank in kg\n",
-      "V2=0.4 #final volume in m^3\n",
-      "print\"let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\" \n",
-      "T2=342 #final temperature in Celsius\n",
-      "v2=0.1974 #final specific volume in m^3/kg\n",
-      "m2=V2/v2 #final mass of the steam in the tank in kg\n",
-      "m=m2-m1 #mass of steam that flowsinto the tank\n",
-      "print\" \\n Hence,mass of the steam that flows into the tank is\",round(m,3),\"kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\n",
-        " \n",
-        " Hence,mass of the steam that flows into the tank is 1.263 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.13:pg-207"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13\n",
-      "#Calculating mass flow of steam in tank\n",
-      "\n",
-      "vf1=0.001725 #in m^3/kg\n",
-      "vf2=0.0016 #in m^3/kg\n",
-      "uf1=368.7 #in kJ/kg\n",
-      "uf2=226 #in kJ/kg\n",
-      "vg1=0.08313 #in m^3/kg\n",
-      "vfg2=0.20381\n",
-      "ug1=1341 #in kJ/kg\n",
-      "ufg2=1099.7 #in kJ/kg\n",
-      "Vf=1 #initial volume of liquid in m^3\n",
-      "Vg=1 #initial volume of vapor in m^3\n",
-      "mf1=Vf/vf1 #initial mass of liquid in kg\n",
-      "mg1=Vg/vg1 #initial mass of vapor in kg\n",
-      "m1=mf1+mg1 #initial mass of liquid in kg\n",
-      "he=1461.1 #in kJ/kg\n",
-      "V=2 #volume of tank in m^3\n",
-      "print\"m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\"\n",
-      "print\"Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\"\n",
-      "x2=((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) #quality of ammonia\n",
-      "v2=0.00160+(0.20381*x2) #final specific volume in m^3/kg\n",
-      "m2=V/v2 #final mass of ammonia in kg\n",
-      "m=m1-m2 #mass of ammonia withdrawn\n",
-      "print\" \\n Hence,mass of ammonia withdrawn is\",round(m,1),\"kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\n",
-        "Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\n",
-        " \n",
-        " Hence,mass of ammonia withdrawn is 72.7 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter6_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter6_6.ipynb
deleted file mode 100755
index 953e821b..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter6_6.ipynb
+++ /dev/null
@@ -1,652 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:c03e579abb4dafb7ea55c1a7f66b6338546355647f3c3d21afe4e72021a3ad5f"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter6:FIRST-LAW ANALYSIS FOR A CONTROL VOLUME"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.1:pg-182"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#calculating mass flow rate in kg/s\n",

-      "import math\n",

-      "R=0.287 #in kJ/kg-K\n",

-      "T=25 #temperature in celsius\n",

-      "P=150 #pressure in kPa\n",

-      "v=R*(T+273.15)/P #specific volume in m^3/kg\n",

-      "D=0.2 #diameter of pipe in metre\n",

-      "A=math.pi*D**2/4 #cross sectional area in m^2\n",

-      "V=0.1 #velocity of air in m/s\n",

-      "m=V*A/v #mass flow rate in kg/s\n",

-      "print\"\\n hence,the mass flow rate is\",round(m,4),\"kg/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the mass flow rate is 0.0055 kg/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 11

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.2:pg-184"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2\n",

-      "#work done for adding the fluid\n",

-      "\n",

-      "P=600 #pressure in kPa\n",

-      "m=1 #in kg\n",

-      "v=0.001 #specific volume in m^3/kg\n",

-      "W=P*m*v #necessary work in kJ for adding the fluid \n",

-      "print\" \\n hence,the work involved in this process is\",round(W,4),\"kJ\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "  \n",

-        " hence,the work involved in this process is 0.6 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.3:pg-188"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#rate of flow of water\n",

-      "\n",

-      "hir=441.89 #in kJ/kg for refrigerant using steam table \n",

-      "her=249.10 #in kJ/kg for refrigerant using steam table\n",

-      "hiw=42 #in kJ/kg for water using steam table\n",

-      "hew=83.95 #in kJ/kg for water using steam table\n",

-      "mr=0.2 #the rate at which refrigerant enters the condenser in kg/s\n",

-      "mw=mr*(hir-her)/(hew-hiw) #rate of flow of water in kg/s\n",

-      "print\"\\n hence,the rate at which cooling water flows thorugh the condenser is\",round(mw,3),\"kg/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the rate at which cooling water flows thorugh the condenser is 0.919 kg/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 15

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.4:pg-190"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4\n",

-      "#determining quality of steam\n",

-      "\n",

-      "hi=2850.1 #initial specific heat enthalpy for steam in kJ/kg\n",

-      "Vi=50 #initial velocity of steam in m/s\n",

-      "Ve=600 #final velocity of steam in m/s\n",

-      "he=hi+Vi**2/(2*1000)-Ve**2/(2*1000) #final specific heat enthalpy for steam in kJ/kg\n",

-      "hf=467.1 #at final state in kJ/kg\n",

-      "hfg=2226.5 #at final state in kJ/kg\n",

-      "xe=(he-hf)/hfg #quality of steam in final state\n",

-      "print\" \\n hence, the quality is\",round(xe,2)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " hence, the quality is 0.99\n"

-       ]

-      }

-     ],

-     "prompt_number": 19

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.4E:pg-191"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4E\n",

-      "#determining quality of steam\n",

-      "\n",

-      "hi=1227.5 #initial specific heat enthalpy for steam in Btu/lbm\n",

-      "Vi=200 #initial velocity of steam in ft/s\n",

-      "Ve=2000 #final velocity of steam in ft/s\n",

-      "he=hi+Vi**2/(2*32.17*778)-Ve**2/(2*32.17*778) #final specific heat enthalpy for steam in Btu/lbm\n",

-      "hf=198.31 #at final state in Btu/lbm\n",

-      "hfg=958.81 #at final state in Btu/lbm\n",

-      "xe=(he-hf)/hfg #quality of steam in final state\n",

-      "print\" \\n hence, the quality is\",round(xe,2)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " hence, the quality is 0.99\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.5:pg-193"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#quality of ammonia leaving expansion valve\n",

-      "\n",

-      "hi=346.8 #specific heat enthalpy for ammonia at initial state in kJ/kg\n",

-      "he=hi #specific heat enthalpy for ammonia at final state will be equal that at initial state because it is a throttling process\n",

-      "hf=134.4 #at final state in kJ/kg\n",

-      "hfg=1296.4#at final state in kJ/kg\n",

-      "xe=(he-hf)*100/hfg #quality at final state\n",

-      "print\"\\n hence,quality of the ammonia leaving the expansion valve is\",round(xe,2),\"%\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,quality of the ammonia leaving the expansion valve is 16.38 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 23

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.6:pg-194"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 6\n",

-      "#power output of turbine in kW\n",

-      "\n",

-      "hi=3137 #initial specific heat of enthalpy in kJ/kg\n",

-      "he=2675.5 #final specific heat of enthalpy in kJ/kg\n",

-      "Vi=50.0 #initial velocity of steam in m/s\n",

-      "Ve=100 #final velocity of steam in m/s\n",

-      "Zi=6 #height of inlet conditions in metres\n",

-      "Ze=3 #height of exit conditions in metres\n",

-      "m=1.5 #mass flow rate of steam in kg/s\n",

-      "g=9.8066 #acc. due to gravity in m/s^2\n",

-      "Qcv=-8.5 #heat transfer rate from turbine in kW\n",

-      "Wcv=Qcv+m*(hi+Vi**2/(2*1000)+g*Zi/1000)-m*(he+Ve**2/(2*1000)+g*Ze/1000) #power output of turbine in kW\n",

-      "print\"\\n hence,the power output of the turbine is\",round(Wcv,1),\"kW\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the power output of the turbine is 678.2 kW\n"

-       ]

-      }

-     ],

-     "prompt_number": 28

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.7:pg-196"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7\n",

-      "#heat transfer rate in aftercooler\n",

-      "\n",

-      "V1=0 #we assume initial velocity to be zero because its given that it enters with a low velocity\n",

-      "V2=25.0 #final velocity with which carbon dioxide exits in m/s\n",

-      "h2=401.52 #final specific enthalpy of heat when carbon dioxide exits in kJ/kg\n",

-      "h1=198 #initial specific enthalpy of heat in kJ/kg\n",

-      "w=h1-h2-V2**2/(2*1000) #in kJ/kg\n",

-      "Wc=-50 #power input to the compressor in kW\n",

-      "m=Wc/w #mass flow rate of carbon dioxide in kg/s\n",

-      "h3=257.9 #final specific enthalpy of heat when carbon dioxide flows into a constant pressure aftercooler\n",

-      "Qcool=-m*(h3-h2) #heat transfer rate in the aftercooler in kW\n",

-      "print\" \\n hence,heat transfer rate in the aftercooler is\",round(Qcool,1),\"kW\" \n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " hence,heat transfer rate in the aftercooler is 35.2 kW\n"

-       ]

-      }

-     ],

-     "prompt_number": 30

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.8:pg-197"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8\n",

-      "#Required pump work\n",

-      "\n",

-      "m=1.5 #mass flow rate of water in kg/s\n",

-      "g=9.807 #acceleration due to gravity in m/s^2\n",

-      "Zin=-15 #depth of water pump in well in metres\n",

-      "Zex=0 #in metres\n",

-      "v=0.001001 #specific volume in m^3/kg\n",

-      "Pex=400+101.3 #exit pressure in kPa\n",

-      "Pin=90 #in kPa\n",

-      "W=m*(g*(Zin-Zex)*0.001-(Pex-Pin)*v) #power input in kW\n",

-      "print\" \\n Hence, the pump requires power of input is\",-round(W,2),\"kW\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " Hence, the pump requires power of input is 0.84 kW\n"

-       ]

-      }

-     ],

-     "prompt_number": 39

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.9:pg-198"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9\n",

-      "#heat tranfer in simple steam power plant\n",

-      "\n",

-      "h1=3023.5 #specific heat of enthalpy of steam leaving boiler in kJ/kg\n",

-      "h2=3002.5 #specific heat of enthalpy of steam entering turbine in kJ/kg\n",

-      "x=0.9 #quality of steam entering condenser\n",

-      "hf=226 #in kJ/kg\n",

-      "hfg=2373.1 #in kJ/kg\n",

-      "h3=hf+x*hfg #specific heat of enthalpy of steam entering condenser in kJ/kg\n",

-      "h4=188.5 #specific heat of enthalpy of steam entering pump in kJ/kg\n",

-      "q12=h2-h1 #heat transfer in line between boiler and turbine in kJ/kg\n",

-      "w23=h2-h3 #turbine work in kJ/kg\n",

-      "q34=h4-h3 #heat transfer in condenser\n",

-      "w45=-4 #pump work in kJ/kg\n",

-      "h5=h4-w45 #in kJ/kg\n",

-      "q51=h1-h5 #heat transfer in boiler in kJ/kg\n",

-      "print\"\\n hence, heat transfer in line between boiler and turbine is\",round(q12,1),\"kJ/kg\"  \n",

-      "print\"\\n hence, turbine work is\",round(w23,1),\"kJ/kg\" \n",

-      "print\"\\n hence, heat transfer in condenser is \",round(q34,1),\"kJ/kg\" \n",

-      "print\"\\n hence, heat transfer in boiler is \",round(q51,1),\"kJ/kg\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, heat transfer in line between boiler and turbine is -21.0 kJ/kg\n",

-        "\n",

-        " hence, turbine work is 640.7 kJ/kg\n",

-        "\n",

-        " hence, heat transfer in condenser is  -2173.3 kJ/kg\n",

-        "\n",

-        " hence, heat transfer in boiler is  2831.0 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 43

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.10:pg-200"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 10\n",

-      "#analysis of refrigerator\n",

-      "\n",

-      "hf4=167.4 #in kJ/kg\n",

-      "hfg4=215.6 #in kJ/kg\n",

-      "h3=241.8 #specific heat of enthalpy of R-134a entering expansion valve\n",

-      "h4=h3 #specific heat of enthalpy of R-134a leaving expansion valve\n",

-      "h1=387.2 #in kJ/kg\n",

-      "h2=435.1 #in kJ/kg\n",

-      "x4=(h3-hf4)/hfg4 #quality of R-134a at evaporator inlet\n",

-      "m=0.1 #mass flow rate in kg/s\n",

-      "Qevap=m*(h1-h4) #rate of heat transfer to the evaporator\n",

-      "Wcomp=-5 #power input to compressor in kW\n",

-      "Qcomp=m*(h2-h1)+Wcomp #rate of heat transfer from compressor\n",

-      "print\"\\n hence, the quality at the evaporator inlet is \",round(x4,3), \n",

-      "print\"\\n hence, the rate of heat transfer to the evaporator is \",round(Qevap,2),\n",

-      "print\"\\n hence, rate of heat transfer from the compressor is\",round(Qcomp,2), "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, the quality at the evaporator inlet is  0.345 \n",

-        " hence, the rate of heat transfer to the evaporator is  14.54 \n",

-        " hence, rate of heat transfer from the compressor is -0.21\n"

-       ]

-      }

-     ],

-     "prompt_number": 45

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.11:pg-204"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 11\n",

-      "#Determining the final temperature of steam\n",

-      "\n",

-      "u2=3040.4 #final internal energy in kJ/kg\n",

-      "hi=u2 #in kJ/kg\n",

-      "P2=1.4 #final Pressure in MPa\n",

-      "print\"Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \"\n",

-      "T2=452 #final temperature in Celsius"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \n"

-       ]

-      }

-     ],

-     "prompt_number": 46

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.12:pg-206"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 12\n",

-      "#Calculating mass flow of steam in tank\n",

-      "\n",

-      "V1=0.4 #initial volume fo tank in m^3\n",

-      "v1=0.5243 #initial specific volume in m^3/kg\n",

-      "h1=3040.4 #initial specific enthalpy in kJ/kg\n",

-      "u1=2548.9 #initial specific internal energy in kJ/kg\n",

-      "m1=V1/v1 #initial mass of steam in tank in kg\n",

-      "V2=0.4 #final volume in m^3\n",

-      "print\"let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\" \n",

-      "T2=342 #final temperature in Celsius\n",

-      "v2=0.1974 #final specific volume in m^3/kg\n",

-      "m2=V2/v2 #final mass of the steam in the tank in kg\n",

-      "m=m2-m1 #mass of steam that flowsinto the tank\n",

-      "print\" \\n Hence,mass of the steam that flows into the tank is\",round(m,3),\"kg\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\n",

-        " \n",

-        " Hence,mass of the steam that flows into the tank is 1.263 kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 48

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.13:pg-207"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 13\n",

-      "#Calculating mass flow of steam in tank\n",

-      "\n",

-      "vf1=0.001725 #in m^3/kg\n",

-      "vf2=0.0016 #in m^3/kg\n",

-      "uf1=368.7 #in kJ/kg\n",

-      "uf2=226 #in kJ/kg\n",

-      "vg1=0.08313 #in m^3/kg\n",

-      "vfg2=0.20381\n",

-      "ug1=1341 #in kJ/kg\n",

-      "ufg2=1099.7 #in kJ/kg\n",

-      "Vf=1 #initial volume of liquid in m^3\n",

-      "Vg=1 #initial volume of vapor in m^3\n",

-      "mf1=Vf/vf1 #initial mass of liquid in kg\n",

-      "mg1=Vg/vg1 #initial mass of vapor in kg\n",

-      "m1=mf1+mg1 #initial mass of liquid in kg\n",

-      "he=1461.1 #in kJ/kg\n",

-      "V=2 #volume of tank in m^3\n",

-      "print\"m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\"\n",

-      "print\"Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\"\n",

-      "x2=((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) #quality of ammonia\n",

-      "v2=0.00160+(0.20381*x2) #final specific volume in m^3/kg\n",

-      "m2=V/v2 #final mass of ammonia in kg\n",

-      "m=m1-m2 #mass of ammonia withdrawn\n",

-      "print\" \\n Hence,mass of ammonia withdrawn is\",round(m,1),\"kg\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\n",

-        "Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\n",

-        " \n",

-        " Hence,mass of ammonia withdrawn is 72.7 kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 50

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6.13E:pg-209"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 13E\n",

-      "#Calculating mass flow of steam in tank\n",

-      "\n",

-      "vf1=0.02747#in ft3/lbm\n",

-      "vf2=0.02564#in ft3/lbm\n",

-      "uf1=153.89#in Btu/lbm\n",

-      "uf2=97.16#in Btu/lbm\n",

-      "vg1=1.4168 #in ft^3/lbm\n",

-      "vfg2=3.2647#in ft^3/lbm\n",

-      "ug1=576.23#in Btu/lbm\n",

-      "ufg2=472.78#in Btu/lbm\n",

-      "Vf=25 #initial volume of liquid in ft^3\n",

-      "Vg=25 #initial volume of vapor in ft^3\n",

-      "mf1=Vf/vf1 #initial mass of liquid in lbm\n",

-      "mg1=Vg/vg1 #initial mass of vapor in lbm\n",

-      "m1=mf1+mg1 #initial mass of liquid in lbm\n",

-      "he=628 #in Btu/lbm\n",

-      "V=50 #volume of tank in ft^3\n",

-      "print\"m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.02564+3.2647*x2) and u2=uf2+x2*ufg2=97.16+3.2647*x2.\"\n",

-      "print\"Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\"\n",

-      "x2=((50*628)-(50*97.16)-(432391*0.02564))/((432391*3.2647)+(50*472.98)) #quality of ammonia\n",

-      "x2=round(x2,6)\n",

-      "v2=0.02564+(3.2647*x2) #final specific volume in ft^3/lbm\n",

-      "v2=round(v2,6)\n",

-      "m2=V/v2 #final mass of ammonia in lbm\n",

-      "m=m1-m2 #mass of ammonia withdrawn\n",

-      "print\" \\n Hence,mass of ammonia withdrawn is\",round(m,1),\"lbm\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.02564+3.2647*x2) and u2=uf2+x2*ufg2=97.16+3.2647*x2.\n",

-        "Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\n",

-        " \n",

-        " Hence,mass of ammonia withdrawn is 105.3 lbm\n"

-       ]

-      }

-     ],

-     "prompt_number": 27

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter7.ipynb b/Fundamentals_Of_Thermodynamics/Chapter7.ipynb
deleted file mode 100755
index 8b49a808..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter7.ipynb
+++ /dev/null
@@ -1,188 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:eaab49a206212133280afcf75d94d2c8dc077f30ee178c5c537c49bf39ade08a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter7:THE SECOND LAW OF THERMODYNAMICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.1:pg-241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#rate of fuel consumption\n",
-      "\n",
-      "W=136*0.7355 #output of automobile engine in kW\n",
-      "neng=0.3 #thermal efficiency of automobile engine\n",
-      "Qh=W/neng #energy output of fuel in kW\n",
-      "Q1=Qh-W #total rate of energy rejected to the ambient\n",
-      "qh=35000.0 #energy output of fuel in kJ/kg\n",
-      "m=Qh/qh #rate of fuel consumption in kg/s\n",
-      "print\"\\n hence,total rate of energy rejected is\",round(Q1),\"kW\" \n",
-      "print\"\\n and rate of fuel consumption is\",round(m,4),\"kg/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,total rate of energy rejected is 233.0 kW\n",
-        "\n",
-        " and rate of fuel consumption is 0.0095 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.2:pg-243"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2\n",
-      "#coefficient of performance of refrigerator\n",
-      "\n",
-      "Qh=400 #heat rejected to kitchen air in W\n",
-      "W=150.0 #electrical input power in W\n",
-      "Q1=Qh-W #rate of energy taken out to cold space in W\n",
-      "B=Ql/W #coefficicent of performnace of refrigerator\n",
-      "print\"\\n hence,rate of energy taken out of the cold space is\",round(Q1),\"W\"\n",
-      "print\"\\n and coefficient of performance of the refrigerator is\",round(B,2),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,rate of energy taken out of the cold space is 250.0 W\n",
-        "\n",
-        " and coefficient of performance of the refrigerator is 1.67\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.4:pg-260"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "#comparison of ideal carnot heat engine with actual heat engine\n",
-      "\n",
-      "Qh=1000.0 #rate of heat transfer to heat engine in kW\n",
-      "W=450.0 #rate of production of work in kW\n",
-      "Ql=Qh-W #rate of heat rejected by heat engine in kW\n",
-      "nthermal=W/Qh #efficiency from the definition of efficiency\n",
-      "Tl=300 #temperature of surroundings in K\n",
-      "Th=550.0 #temperature of heat source in Celsius\n",
-      "ncarnot=1-Tl/(Th+273.15) #efficiency if heat engine is considered to be ideal carnot heat engine\n",
-      "W2=ncarnot*Qh #rate of work production if heat engine is assumed to be ideal carnot heat engine in kW\n",
-      "Q12=Qh-W2 #rate of heat rejected by heat engine in kW if heat engine is assumed to be ideal carnot heat engine\n",
-      "print\"\\n hence,energy discarded to the ambient surroundings iS\",round(Q12,4),\"kw\"\n",
-      "print\"\\n and the engine efficiency is\",round(ncarnot,4),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,energy discarded to the ambient surroundings iS 364.4536 kw\n",
-        "\n",
-        " and the engine efficiency is 0.6355\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.5:pg-261"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5\n",
-      "#calculating required work\n",
-      "\n",
-      "Tl=24+273.15 #room temperature in Kelvins\n",
-      "Th=35+273.15 #atmospheric temperature in Kelvins\n",
-      "Ql=4 #rate of heat rejection from room\n",
-      "B=Tl/(Th-Tl) #coefficient of performance of air conditioner\n",
-      "W=Ql/B #required work in kW\n",
-      "print\"\\n hence,the magnitude of reqiured work is\",round(W,2),\"kw\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the magnitude of reqiured work is 0.15 kw\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter7_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter7_6.ipynb
deleted file mode 100755
index 02425dea..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter7_6.ipynb
+++ /dev/null
@@ -1,217 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:a60a19bd48510a099eb4483e5176d858019753b734df1f9001e5af0e3bc7fc2e"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter7:THE SECOND LAW OF THERMODYNAMICS"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.1:pg-241"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#rate of fuel consumption\n",

-      "\n",

-      "W=136*0.7355 #output of automobile engine in kW\n",

-      "neng=0.3 #thermal efficiency of automobile engine\n",

-      "Qh=W/neng #energy output of fuel in kW\n",

-      "Q1=Qh-W #total rate of energy rejected to the ambient\n",

-      "qh=35000.0 #energy output of fuel in kJ/kg\n",

-      "m=Qh/qh #rate of fuel consumption in kg/s\n",

-      "print\"\\n hence,total rate of energy rejected is\",round(Q1),\"kW\" \n",

-      "print\"\\n and rate of fuel consumption is\",round(m,4),\"kg/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,total rate of energy rejected is 233.0 kW\n",

-        "\n",

-        " and rate of fuel consumption is 0.0095 kg/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.2:pg-243"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2\n",

-      "#coefficient of performance of refrigerator\n",

-      "\n",

-      "Qh=400 #heat rejected to kitchen air in W\n",

-      "W=150.0 #electrical input power in W\n",

-      "Q1=Qh-W #rate of energy taken out to cold space in W\n",

-      "B=Ql/W #coefficicent of performnace of refrigerator\n",

-      "print\"\\n hence,rate of energy taken out of the cold space is\",round(Q1),\"W\"\n",

-      "print\"\\n and coefficient of performance of the refrigerator is\",round(B,2),"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,rate of energy taken out of the cold space is 250.0 W\n",

-        "\n",

-        " and coefficient of performance of the refrigerator is 1.67\n"

-       ]

-      }

-     ],

-     "prompt_number": 22

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.3:pg-256"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "# temperature at which the pressure go to zero\n",

-      "\n",

-      "P1=110.9#pressur at the ice point in kPa\n",

-      "P2=151.5#pressure at the steam point in kPa\n",

-      "T1=0#in C\n",

-      "T2=100#in C\n",

-      "S=(P2-P1)/(T2-T1)#slope in kPa/C\n",

-      "T=T1-(P1/S)# absolute ideal gas temperature in C\n",

-      "print\"\\n hence,the absolute ideal gas temperature is\",round(T,2),\"C\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the absolute ideal gas temperature is -273.15 C\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.4:pg-260"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4\n",

-      "#comparison of ideal carnot heat engine with actual heat engine\n",

-      "\n",

-      "Qh=1000.0 #rate of heat transfer to heat engine in kW\n",

-      "W=450.0 #rate of production of work in kW\n",

-      "Ql=Qh-W #rate of heat rejected by heat engine in kW\n",

-      "nthermal=W/Qh #efficiency from the definition of efficiency\n",

-      "Tl=300 #temperature of surroundings in K\n",

-      "Th=550.0 #temperature of heat source in Celsius\n",

-      "ncarnot=1-Tl/(Th+273.15) #efficiency if heat engine is considered to be ideal carnot heat engine\n",

-      "W2=ncarnot*Qh #rate of work production if heat engine is assumed to be ideal carnot heat engine in kW\n",

-      "Q12=Qh-W2 #rate of heat rejected by heat engine in kW if heat engine is assumed to be ideal carnot heat engine\n",

-      "print\"\\n hence,energy discarded to the ambient surroundings iS\",round(Q12,4),\"kw\"\n",

-      "print\"\\n and the engine efficiency is\",round(ncarnot,4),"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,energy discarded to the ambient surroundings iS 364.4536 kw\n",

-        "\n",

-        " and the engine efficiency is 0.6355\n"

-       ]

-      }

-     ],

-     "prompt_number": 53

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.5:pg-261"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#calculating required work\n",

-      "\n",

-      "Tl=24+273.15 #room temperature in Kelvins\n",

-      "Th=35+273.15 #atmospheric temperature in Kelvins\n",

-      "Ql=4 #rate of heat rejection from room\n",

-      "B=Tl/(Th-Tl) #coefficient of performance of air conditioner\n",

-      "W=Ql/B #required work in kW\n",

-      "print\"\\n hence,the magnitude of reqiured work is\",round(W,2),\"kw\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the magnitude of reqiured work is 0.15 kw\n"

-       ]

-      }

-     ],

-     "prompt_number": 40

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter8.ipynb b/Fundamentals_Of_Thermodynamics/Chapter8.ipynb
deleted file mode 100755
index b1b6e75a..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter8.ipynb
+++ /dev/null
@@ -1,452 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e95f45030ec153dab9aa085b0a83755a3b429532b4c58448d37853e1ca968b64"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter8:ENTROPY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.1:pg-290"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#coefficient of performance of refrigerator\n",
-      "\n",
-      "Th=60 #temperature at which heat is rejected from R-134a\n",
-      "Tl=0 #temperature at which heat is absorbed into the R-134a \n",
-      "s1=1.7262 #specific entropy at 0 Celsius\n",
-      "s2=s1 #process of state change from 1-2 is isentropic \n",
-      "s3=1.2857 #specific entropy at 60 celsius\n",
-      "s4=s3 #process of state change from 3-4 is isentropic\n",
-      "print\"if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\"\n",
-      "P2=1400+(1600-1400)*(1.7262-1.736)/(1.7135-1.736) #pressure after compression in kPa\n",
-      "B=(Th+273.15)/(Th-Tl) #coefficient of performance of refrigerator\n",
-      "print\" \\n hence,pressure after compression is \",round(P2,1),\"kPa\"\n",
-      "print\"\\n and coefficient of performance of refrigerator is\",round(B,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\n",
-        " \n",
-        " hence,pressure after compression is  1487.1 kPa\n",
-        "\n",
-        " and coefficient of performance of refrigerator is 5.55\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.2:pg-290"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2 \n",
-      "#heat transfer in a given process\n",
-      "\n",
-      "u1=87.94 #specific internal energy of R-12 at state 1 in kJ/kg\n",
-      "u2=276.44 #specific internal energy of R-12 at state 2 in kJ/kg\n",
-      "s1=0.3357 #specific entropy at state 1 in kJ/kg-K\n",
-      "s2=1.2108 #specific entropy at state 2 in kJ/kg-K\n",
-      "V=0.001 #volume of saturated liquid in m^3\n",
-      "v1=0.000923 #specific volume in m^3/kg\n",
-      "m=V/v1 #mass of saturated liquid in kg\n",
-      "T=20 #temperature of liquid in celsius\n",
-      "Q12=m*(T+273.15)*(s2-s1) #heat transfer in kJ to accomplish the process\n",
-      "W12=m*(u1-u2)+Q12 #work required to accomplish the process\n",
-      "print\" \\n hence,work required to accomplish the process is\",round(W12,1),\"KJ\"\n",
-      "print\" \\n and heat transfer is\",round(Q12,1),\"KJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " hence,work required to accomplish the process is 73.7 KJ\n",
-        " \n",
-        " and heat transfer is 277.9 KJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.3:pg-293"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#entropy change\n",
-      "import math\n",
-      "C=4.184 # specific heat of water in kJ/kg-K\n",
-      "T1=20 #initial temperature of water in celsius\n",
-      "T2=90 #final temperature of water in celsius\n",
-      "dS1=C*math.log((T2+273.2)/(T1+273.2)) #change in entropy in kJ/kg-K\n",
-      "dS2=1.1925-0.2966 #in kJ/kg-K using steam tables\n",
-      "print\"\\n hence,change in entropy assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",
-      "print\"\\n using steam table is\",round(dS2,4),\"KJ/kg-k\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,change in entropy assuming constant specific heat is 0.8958 KJ/kg-k\n",
-        "\n",
-        " using steam table is 0.8959 KJ/kg-k\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.4:pg-295"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "#entropy change with different assumptions\n",
-      "import math\n",
-      "T1=300 #initial temperature in kelvins\n",
-      "T2=1500 #final temperature in kelvins\n",
-      "P1=200.0 #initial pressure in kPa\n",
-      "P2=150.0 #final pressure in kPa\n",
-      "R=0.2598 # in kJ/kg-K\n",
-      "Cp=0.922 #specific heat in kJ/kg-K at constant pressure\n",
-      "dsT2=8.0649 #in kJ/kg-K\n",
-      "dsT1=6.4168 #in kJ/kg-K\n",
-      "dS1=dsT2-dsT1-R*math.log(P2/P1) #entropy change calculated using ideal gas tables\n",
-      "dS3=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat in kJ/kg-K\n",
-      "dS4=1.0767*math.log(T2/T1)+0.0747 #entropy change assuming specific heat is constant at its value at 990K\n",
-      "print\"\\n hence,change in entropy using ideal gas tables is\",round(dS1,4),\"KJ/kg-k\"\n",
-      "print\"\\n hence,change in entropy using the value of specific heat at 300K is\",round(dS3,4),\"KJ/kg-k\"\n",
-      "print\"\\n hence,change in entropy assuming specific heat is constant at its value at 900K is \",round(dS4,4),\"KJ/kg-k\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,change in entropy using ideal gas tables is 1.7228 KJ/kg-k\n",
-        "\n",
-        " hence,change in entropy using the value of specific heat at 300K is 1.5586 KJ/kg-k\n",
-        "\n",
-        " hence,change in entropy assuming specific heat is constant at its value at 900K is  1.8076 KJ/kg-k\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.5:pg-296\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5\n",
-      "#entropy change\n",
-      "import math\n",
-      "Cp=1.004 #specific heat at constant pressure in kJ/kg-K\n",
-      "R=0.287 #gas constant in kJ/kg-K\n",
-      "P1=400.0 #initial pressure in kPa\n",
-      "P2=300.0 #final pressure in kPa\n",
-      "T1=300 #initial temperature in K\n",
-      "T2=600 #final temperature in K\n",
-      "dS1=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat\n",
-      "s1=6.8693 #specific entropy at T1\n",
-      "s2=7.5764 #specific entropy at T2\n",
-      "dS2=s2-s1-R*math.log(P2/P1) #entropy change assuming variable specific heat\n",
-      "print\"\\n hence,entropy change assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",
-      "print\"\\n and assuming variable specific heat is\",round(dS2,4),\"KJ/kg-k\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,entropy change assuming constant specific heat is 0.7785 KJ/kg-k\n",
-        "\n",
-        " and assuming variable specific heat is 0.7897 KJ/kg-k\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.6:pg-297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6\n",
-      "#work done by air\n",
-      "import math\n",
-      "T1=600 #initial temperature of air in K\n",
-      "P1=400.0 #intial pressure of air in kPa\n",
-      "P2=150.0 #final pressure in kPa\n",
-      "u1=435.10 #specific internal energy at temperature T1 in kJ/kg\n",
-      "sT1=7.5764 #specific entropy at temperature T1 in kJ/kg-K\n",
-      "R=0.287 #gas constant in kJ/kg-K\n",
-      "ds=0\n",
-      "sT2=ds+sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n",
-      "print\"we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\"\n",
-      "T2=457 #final temperature in K\n",
-      "u2=328.14 #specific internal energy at temperature T2 in kJ/kg\n",
-      "w=u1-u2 #work done by air in kJ/kg\n",
-      "print\"\\n hence,work done by air is\",round(w,4),\"KJ/kg\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\n",
-        "\n",
-        " hence,work done by air is 106.96 KJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.7:pg-300"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 7\n",
-      "#work and heat transfer\n",
-      "\n",
-      "P2=500  #final pressure in cylinder in kPa\n",
-      "P1=100 #initial pressure in cylinder in kPa\n",
-      "T1=20+273.2 #initial temperature inside cylinder in Kelvins\n",
-      "n=1.3 \n",
-      "T2=(T1)*(P2/P1)**((n-1)/n) #final temperature inside cylinder in K\n",
-      "R=0.2968 #gas constant in kJ/kg-K\n",
-      "w12=R*(T2-T1)/(1-n) #work in kJ/kg\n",
-      "Cvo=0.745 #specific heat at constant volume in kJ/kg-K\n",
-      "q12=Cvo*(T2-T1)+w12 #heat transfer in kJ/kg\n",
-      "print\" \\n hence,work done is\",round(w12,2),\"KJ/kg\" \n",
-      "print\"\\n and heat transfer are\",round(q12,1),\"KJ/kg\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " hence,work done is -130.47 KJ/kg\n",
-        "\n",
-        " and heat transfer are -32.2 KJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.8:pg-308"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#calculating increase in entropy\n",
-      "\n",
-      "m=1 #mass of saturated water vapour\n",
-      "sfg=6.0480 #in kJ/K\n",
-      "T=25 #temperature of surrounding air in celsius\n",
-      "dScm=-m*sfg #change in entropy of control mass in kJ/K\n",
-      "hfg=2257.0 #in kJ/kg\n",
-      "Qtosurroundings=m*hfg #heat transferred to surroundings in kJ\n",
-      "dSsurroundings=Qtosurroundings/(T+273.15) #in kJ/K\n",
-      "dSnet=dScm+dSsurroundings #net increase in entropy in kJ/K\n",
-      "print\" hence,net increase in entropy of water plus surroundings is \",round(dSnet,3),\"KJ/k\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " hence,net increase in entropy of water plus surroundings is  1.522 KJ/k\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.9:pg-309"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9\n",
-      "#entropy generation\n",
-      "\n",
-      "Qout=1.0 #value of heat flux generated by 1kW of electric power\n",
-      "T=600.0 #temperature of hot wire surface in K\n",
-      "Sgen=Qout/T #entropy generation in kW/K\n",
-      "print\"\\n hence,entropy generation is\",round(Sgen,5),\"KW/k\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,entropy generation is 0.00167 KW/k\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.10:pg-310"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10\n",
-      "#Determiining the entropy generated\n",
-      "\n",
-      "B=4.0#COP of air conditioner\n",
-      "W=10.0 #power input of air conditioner in kW\n",
-      "Qh=B*W #in kW\n",
-      "Ql=Qh-W #in kW\n",
-      "Thigh=323 #in Kelvin\n",
-      "Tlow=263 #in Kelvin\n",
-      "SgenHP=(Qh*1000/Thigh)-(Ql*1000/Tlow) #in W/K\n",
-      "Tl=281 # in K\n",
-      "Th=294 #in K\n",
-      "SgenCV1=Ql*1000/Tlow-Ql*1000/Tl #in W/K\n",
-      "SgenCV2=Qh*1000/Th-Qh*1000/Thigh #in W/K\n",
-      "SgenTOT=SgenCV1+SgenCV2+SgenHP #in W/K\n",
-      "print\" \\n Hence,Total entropy generated is\",round(SgenTOT,1),\"W/k\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Hence,Total entropy generated is 29.3 W/k\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter8_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter8_6.ipynb
deleted file mode 100755
index dce4d1fa..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter8_6.ipynb
+++ /dev/null
@@ -1,444 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:79711d3b5ce866eb594601c655e2c85bdace3e8f861ba88d25f2b6f7e8d8655f"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter8:ENTROPY"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.1:pg-290"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#coefficient of performance of refrigerator\n",

-      "\n",

-      "Th=60 #temperature at which heat is rejected from R-134a\n",

-      "Tl=0 #temperature at which heat is absorbed into the R-134a \n",

-      "s1=1.7262 #specific entropy at 0 Celsius\n",

-      "s2=s1 #process of state change from 1-2 is isentropic \n",

-      "s3=1.2857 #specific entropy at 60 celsius\n",

-      "s4=s3 #process of state change from 3-4 is isentropic\n",

-      "print\"if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\"\n",

-      "P2=1400+(1600-1400)*(1.7262-1.736)/(1.7135-1.736) #pressure after compression in kPa\n",

-      "B=(Th+273.15)/(Th-Tl) #coefficient of performance of refrigerator\n",

-      "print\" \\n hence,pressure after compression is \",round(P2,1),\"kPa\"\n",

-      "print\"\\n and coefficient of performance of refrigerator is\",round(B,2)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\n",

-        " \n",

-        " hence,pressure after compression is  1487.1 kPa\n",

-        "\n",

-        " and coefficient of performance of refrigerator is 5.55\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.2:pg-290"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2 \n",

-      "#heat transfer in a given process\n",

-      "\n",

-      "u1=87.94 #specific internal energy of R-12 at state 1 in kJ/kg\n",

-      "u2=276.44 #specific internal energy of R-12 at state 2 in kJ/kg\n",

-      "s1=0.3357 #specific entropy at state 1 in kJ/kg-K\n",

-      "s2=1.2108 #specific entropy at state 2 in kJ/kg-K\n",

-      "V=0.001 #volume of saturated liquid in m^3\n",

-      "v1=0.000923 #specific volume in m^3/kg\n",

-      "m=V/v1 #mass of saturated liquid in kg\n",

-      "T=20 #temperature of liquid in celsius\n",

-      "Q12=m*(T+273.15)*(s2-s1) #heat transfer in kJ to accomplish the process\n",

-      "W12=m*(u1-u2)+Q12 #work required to accomplish the process\n",

-      "print\" \\n hence,work required to accomplish the process is\",round(W12,1),\"KJ\"\n",

-      "print\" \\n and heat transfer is\",round(Q12,1),\"KJ\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " hence,work required to accomplish the process is 73.7 KJ\n",

-        " \n",

-        " and heat transfer is 277.9 KJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.3:pg-293"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#entropy change\n",

-      "import math\n",

-      "C=4.184 # specific heat of water in kJ/kg-K\n",

-      "T1=20 #initial temperature of water in celsius\n",

-      "T2=90 #final temperature of water in celsius\n",

-      "dS1=C*math.log((T2+273.2)/(T1+273.2)) #change in entropy in kJ/kg-K\n",

-      "dS2=1.1925-0.2966 #in kJ/kg-K using steam tables\n",

-      "print\"\\n hence,change in entropy assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",

-      "print\"\\n using steam table is\",round(dS2,4),\"KJ/kg-k\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,change in entropy assuming constant specific heat is 0.8958 KJ/kg-k\n",

-        "\n",

-        " using steam table is 0.8959 KJ/kg-k\n"

-       ]

-      }

-     ],

-     "prompt_number": 17

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.4:pg-295"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4\n",

-      "#entropy change with different assumptions\n",

-      "import math\n",

-      "T1=300 #initial temperature in kelvins\n",

-      "T2=1500 #final temperature in kelvins\n",

-      "P1=200.0 #initial pressure in kPa\n",

-      "P2=150.0 #final pressure in kPa\n",

-      "R=0.2598 # in kJ/kg-K\n",

-      "Cp=0.922 #specific heat in kJ/kg-K at constant pressure\n",

-      "dsT2=8.0649 #in kJ/kg-K\n",

-      "dsT1=6.4168 #in kJ/kg-K\n",

-      "dS1=dsT2-dsT1-R*math.log(P2/P1) #entropy change calculated using ideal gas tables\n",

-      "dS3=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat in kJ/kg-K\n",

-      "dS4=1.0767*math.log(T2/T1)+0.0747 #entropy change assuming specific heat is constant at its value at 990K\n",

-      "print\"\\n hence,change in entropy using ideal gas tables is\",round(dS1,4),\"KJ/kg-k\"\n",

-      "print\"\\n hence,change in entropy using the value of specific heat at 300K is\",round(dS3,4),\"KJ/kg-k\"\n",

-      "print\"\\n hence,change in entropy assuming specific heat is constant at its value at 900K is \",round(dS4,4),\"KJ/kg-k\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,change in entropy using ideal gas tables is 1.7228 KJ/kg-k\n",

-        "\n",

-        " hence,change in entropy using the value of specific heat at 300K is 1.5586 KJ/kg-k\n",

-        "\n",

-        " hence,change in entropy assuming specific heat is constant at its value at 900K is  1.8076 KJ/kg-k\n"

-       ]

-      }

-     ],

-     "prompt_number": 19

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.5:pg-296\n"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#entropy change\n",

-      "import math\n",

-      "Cp=1.004 #specific heat at constant pressure in kJ/kg-K\n",

-      "R=0.287 #gas constant in kJ/kg-K\n",

-      "P1=400.0 #initial pressure in kPa\n",

-      "P2=300.0 #final pressure in kPa\n",

-      "T1=300 #initial temperature in K\n",

-      "T2=600 #final temperature in K\n",

-      "dS1=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat\n",

-      "s1=6.8693 #specific entropy at T1\n",

-      "s2=7.5764 #specific entropy at T2\n",

-      "dS2=s2-s1-R*math.log(P2/P1) #entropy change assuming variable specific heat\n",

-      "print\"\\n hence,entropy change assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",

-      "print\"\\n and assuming variable specific heat is\",round(dS2,4),\"KJ/kg-k\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,entropy change assuming constant specific heat is 0.7785 KJ/kg-k\n",

-        "\n",

-        " and assuming variable specific heat is 0.7897 KJ/kg-k\n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.6:pg-297"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 6\n",

-      "#work done by air\n",

-      "import math\n",

-      "T1=600 #initial temperature of air in K\n",

-      "P1=400.0 #intial pressure of air in kPa\n",

-      "P2=150.0 #final pressure in kPa\n",

-      "u1=435.10 #specific internal energy at temperature T1 in kJ/kg\n",

-      "sT1=7.5764 #specific entropy at temperature T1 in kJ/kg-K\n",

-      "R=0.287 #gas constant in kJ/kg-K\n",

-      "ds=0\n",

-      "sT2=ds+sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n",

-      "print\"we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\"\n",

-      "T2=457 #final temperature in K\n",

-      "u2=328.14 #specific internal energy at temperature T2 in kJ/kg\n",

-      "w=u1-u2 #work done by air in kJ/kg\n",

-      "print\"\\n hence,work done by air is\",round(w,4),\"KJ/kg\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\n",

-        "\n",

-        " hence,work done by air is 106.96 KJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.7:pg-300"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7\n",

-      "#work and heat transfer\n",

-      "\n",

-      "P2=500  #final pressure in cylinder in kPa\n",

-      "P1=100 #initial pressure in cylinder in kPa\n",

-      "T1=20+273.2 #initial temperature inside cylinder in Kelvins\n",

-      "n=1.3 \n",

-      "T2=(T1)*(P2/P1)**((n-1)/n) #final temperature inside cylinder in K\n",

-      "R=0.2968 #gas constant in kJ/kg-K\n",

-      "w12=R*(T2-T1)/(1-n) #work in kJ/kg\n",

-      "Cvo=0.745 #specific heat at constant volume in kJ/kg-K\n",

-      "q12=Cvo*(T2-T1)+w12 #heat transfer in kJ/kg\n",

-      "print\" \\n hence,work done is\",round(w12,2),\"KJ/kg\" \n",

-      "print\"\\n and heat transfer are\",round(q12,1),\"KJ/kg\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " hence,work done is -130.47 KJ/kg\n",

-        "\n",

-        " and heat transfer are -32.2 KJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.8:pg-308"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#calculating increase in entropy\n",

-      "\n",

-      "m=1 #mass of saturated water vapour\n",

-      "sfg=6.0480 #in kJ/K\n",

-      "T=25 #temperature of surrounding air in celsius\n",

-      "dScm=-m*sfg #change in entropy of control mass in kJ/K\n",

-      "hfg=2257.0 #in kJ/kg\n",

-      "Qtosurroundings=m*hfg #heat transferred to surroundings in kJ\n",

-      "dSsurroundings=Qtosurroundings/(T+273.15) #in kJ/K\n",

-      "dSnet=dScm+dSsurroundings #net increase in entropy in kJ/K\n",

-      "print\" hence,net increase in entropy of water plus surroundings is \",round(dSnet,3),\"KJ/k\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " hence,net increase in entropy of water plus surroundings is  1.522 KJ/k\n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.9:pg-309"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9\n",

-      "#entropy generation\n",

-      "\n",

-      "Qout=1.0 #value of heat flux generated by 1kW of electric power\n",

-      "T=600.0 #temperature of hot wire surface in K\n",

-      "Sgen=Qout/T #entropy generation in kW/K\n",

-      "print\"\\n hence,entropy generation is\",round(Sgen,5),\"KW/k\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,entropy generation is 0.00167 KW/k\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.10:pg-310"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 10\n",

-      "#Determiining the entropy generated\n",

-      "\n",

-      "B=4.0#COP of air conditioner\n",

-      "W=10.0 #power input of air conditioner in kW\n",

-      "Qh=B*W #in kW\n",

-      "Ql=Qh-W #in kW\n",

-      "Thigh=323 #in Kelvin\n",

-      "Tlow=263 #in Kelvin\n",

-      "SgenHP=(Qh*1000/Thigh)-(Ql*1000/Tlow) #in W/K\n",

-      "Tl=281 # in K\n",

-      "Th=294 #in K\n",

-      "SgenCV1=Ql*1000/Tlow-Ql*1000/Tl #in W/K\n",

-      "SgenCV2=Qh*1000/Th-Qh*1000/Thigh #in W/K\n",

-      "SgenTOT=SgenCV1+SgenCV2+SgenHP #in W/K\n",

-      "print\" \\n Hence,Total entropy generated is\",round(SgenTOT,1),\"W/k\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " \n",

-        " Hence,Total entropy generated is 29.3 W/k\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter9.ipynb b/Fundamentals_Of_Thermodynamics/Chapter9.ipynb
deleted file mode 100755
index c1f858fc..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter9.ipynb
+++ /dev/null
@@ -1,519 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:7638f00c0629ff24b9e3c96e6eb43a103a0583de57d4e01375fbb8aa638bef10"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter9:SECOND-LAW ANALYSIS FOR A CONTROL VOLUME"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ex9.1:pg-336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1\n",
-      "#work done by steam\n",
-      "\n",
-      "hi=3051.2 #initial specific heat of enthalpy of steam in kJ/kg\n",
-      "si=7.1228 #initial specific entropy of steam in kJ/kg-K\n",
-      "Pe=0.15 #final pressure in MPa\n",
-      "se=si #specific entropy in final state in kJ/kg-K\n",
-      "sf=1.4335 #in kJ/kg-K\n",
-      "sfg=5.7897 #in kJ/kg-K\n",
-      "vi=50.0 #velocity with which steam enters turbine in m/s\n",
-      "ve=200.0 #velocity with which steam leaves the turbine in m/s\n",
-      "xe=(se-sf)/sfg #quality of steam in final state\n",
-      "hf=467.1 #in kJ/kg\n",
-      "hfg=2226.5 #in kJ/kg\n",
-      "he=hf+xe*hfg #final specific heat of enthalpy of steam in kJ/kg\n",
-      "w=hi+vi**2/(2*1000)-he-ve**2/(2*1000) #work of steam for isentropic process in kJ/kg\n",
-      "print\"\\n hence, work per kilogram of steam for this isentropic process is\",round(w,1),\"KJ/kg\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence, work per kilogram of steam for this isentropic process is 377.5 KJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.2:pg-337"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2\n",
-      "#exit velocity of steam from nozzle\n",
-      "\n",
-      "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n",
-      "si=7.1228 #initial specific entropy in kJ/kg-K\n",
-      "se=si #final specific entropy \n",
-      "Pe=0.3 #final pressure in MPa\n",
-      "print\"from steam table,various properties at final state are\"\n",
-      "he=2780.2 #final specific heat of enthalpy in kJ/kg-K\n",
-      "Te=159.1 #final temperature in celsius\n",
-      "vi=30.0 #velocity with which steam enters the nozzle in m/s\n",
-      "ve=((2*(hi-he)+(vi**2/1000))*1000)**0.5 #final velocity of steam with which it exits in m/s\n",
-      "print\"\\n hence,exit velocity of the steam from the nozzle is\",round(ve),\"m/s\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "from steam table,various properties at final state are\n",
-        "\n",
-        " hence,exit velocity of the steam from the nozzle is 737.0 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.3:pg-340"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3\n",
-      "#violation of second law\n",
-      "\n",
-      "print\"from R-134a tables\"\n",
-      "se=1.7148 #specific entropy in final state in kJ/kg-K\n",
-      "si=1.7395 #initial specific entropy in kJ/kg-K \n",
-      "print\"therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "from R-134a tables\n",
-        "therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.4:pg-340"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4\n",
-      "#calculating required specific work\n",
-      "\n",
-      "Cp=1.004 #specific heat of air at constant pressure in kJ/kg-K\n",
-      "Ti=290 #initial temperature in kelvins\n",
-      "Pi=100 #initial pressure in kPa\n",
-      "Pe=1000 #final pressure in kPa\n",
-      "k=1.4 \n",
-      "Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in kelvins\n",
-      "we=Cp*(Ti-Te) #required specific work in kJ/kg\n",
-      "print\"\\n hence,specific work required is\",round(we),\"kJ/kg\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,specific work required is -271.0 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.5:pg-342"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5\n",
-      "#entropy generation\n",
-      "\n",
-      "h1=2865.54 #specific heat of enthalpy at state 1 in kJ/kg\n",
-      "h2=83.94 #specific heat of enthalpy  at state 2 in kJ/kg\n",
-      "h3=2725.3 #specific heat of enthalpy at state 3 in kJ?kg\n",
-      "s1=7.3115 #specific entropy at state 1 in kJ/kg-K\n",
-      "s2=0.2966 #specific entropy at state 2 in kJ/kg-K\n",
-      "s3=6.9918 #specific entropy at state 3in kJ/kg-K\n",
-      "m1=2 #mass flow rate at state 1 in kg/s\n",
-      "m2=m1*(h1-h3)/(h3-h2) #mass flow rate at state 2 in kg/s\n",
-      "m3=m1+m2 #mass flow rate at state 3 in kg/s\n",
-      "Sgen=m3*s3-m1*s1-m2*s2 #entropy generation in the process\n",
-      "print\"\\n hence,entropy generated in this process is \",round(Sgen,3),\"kW/K\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,entropy generated in this process is  0.072 kW/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.6:pg-344"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6\n",
-      "#work required to fill the tank\n",
-      "import math\n",
-      "T1=17+273 #initial temperature of tank in Kelvins\n",
-      "sT1=6.83521 #specific entropy in kJ/kg-K\n",
-      "R=0.287 #gas constant in kJ/kg-K\n",
-      "P1=100 #initial pressure in kPa\n",
-      "P2=1000 #final pressure in kPa\n",
-      "sT2=sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n",
-      "T2=555.7 #from interplotation \n",
-      "V1=0.04 #volume of tank in m^3\n",
-      "V2=V1 #final volume is equal to initial volume\n",
-      "m1=P1*V1/(R*T1) #initial mass of air in tank in kg\n",
-      "m2=P2*V2/(R*T2) #final mass of air in tank in kg\n",
-      "Min=m2-m1 #in kg\n",
-      "u1=207.19 #initial specific heat of enthalpy in kJ/kg\n",
-      "u2=401.49 #final specific heat of enthalpy in kJ/kg\n",
-      "hin=290.43 #in kJ/kg\n",
-      "W12=Min*hin+m1*u1-m2*u2 #work required to fill the tank in kJ\n",
-      "print\"\\n hence,the total amount of work required to fill the tank is\",round(W12,1),\"kJ\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,the total amount of work required to fill the tank is -31.9 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.7:pg-347"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 7\n",
-      "#work required to pump water isentropically \n",
-      "\n",
-      "P1=100 #initial pressure in kPa\n",
-      "P2=5000 #final pressure in kPa\n",
-      "v=0.001004 #specific volume in m^3/kg\n",
-      "w=v*(P2-P1) #work required to pump water isentropically\n",
-      "print\"\\n hence,work required to pump water isentropically  is \",round(w,2),\"kJ/kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,work required to pump water isentropically  is  4.92 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.8:pg-348"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 8\n",
-      "#Velocity in exit flow\n",
-      "\n",
-      "print\"From Steam Tables, for liquid water at 20 C\"\n",
-      "vf=0.001002 #in m^3/kg\n",
-      "v=vf\n",
-      "Pi=300 #Line pressure in kPa\n",
-      "Po=100 #in kPa\n",
-      "Ve=(2*v*(Pi-Po)*1000)**0.5 #velocity in the exit flow\n",
-      "print\" \\n Hence, an ideal nozzle can generate upto \",round(Ve),\"m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "From Steam Tables, for liquid water at 20 C\n",
-        " \n",
-        " Hence, an ideal nozzle can generate upto  20.0 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.9:pg-351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9\n",
-      "#Rate of Entropy Generation\n",
-      "\n",
-      "print\"From R-410a tables,we get\"\n",
-      "hi=280.6 #in kJ/kg\n",
-      "he=307.8 #in kJ/kg\n",
-      "si=1.0272 #in kJ/kg\n",
-      "se=1.0140 #in kJ/kg\n",
-      "m=0.08 #flow rate of refrigerant in kg/s\n",
-      "P=3 #electrical power input in kW\n",
-      "Qcv=m*(he-hi)-P #in kW\n",
-      "To=30 #in Celsius\n",
-      "Sgen=m*(se-si)-Qcv/(To+273.2) #rate of entropy generation \n",
-      "print\"\\n Hence,the rate of entropy generation for this process is\",round(Sgen,5),\"kW/K\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "From R-410a tables,we get\n",
-        "\n",
-        " Hence,the rate of entropy generation for this process is 0.00166 kW/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.10:pg-353"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10\n",
-      "#turbine efficiency\n",
-      "\n",
-      "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n",
-      "si=7.1228 #initial specific entropy in kJ/kg-K\n",
-      "sf=0.7548 #in kJ/kg-K\n",
-      "sfg=7.2536 #in kJ/kg-K\n",
-      "ses=si #final specific entropy is same as the initial\n",
-      "xes=(si-sf)/sfg #quality of steam when it leaves the turbine\n",
-      "hf=225.9 #in kJ/kg\n",
-      "hfg=2373.1 #in kJ/kg\n",
-      "hes=hf+xes*hfg #final specific heat of enthalpy in kJ/kg\n",
-      "ws=hi-hes #work output of turbine calculated ideally  in kJ/kg\n",
-      "wa=600 #actual work output of turbine in kJ/kg\n",
-      "nturbine=wa/ws #efiiciency of turbine \n",
-      "print\"\\n hence,efficiency of the turbine is\",round(nturbine*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,efficiency of the turbine is 80.9 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.11:pg-355"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11\n",
-      "#turbine inlet pressure\n",
-      "import math\n",
-      "hi=1757.3 #initial specific heat of enthalpy of air in  kJ/kg\n",
-      "si=8.6905 #initial specifc entropy of airin kJ/kg-K\n",
-      "he=855.3 #final specific heat of enthalpy of air in kJ/kg\n",
-      "w=hi-he #actual work done by turbine in kJ/kg\n",
-      "n=0.85 #efficiency of turbine \n",
-      "ws=w/n #ideal work done by turbine in kJ/kg\n",
-      "hes=hi-ws #from first law of isentropic process\n",
-      "Tes=683.7 #final temperature in kelvins from air tables\n",
-      "ses=7.7148 #in kJ/kg-K\n",
-      "R=0.287 #gas constant in kJ/kg-K\n",
-      "Pi=100/math.e**((si-ses)/-R) #turbine inlet pressure in kPa\n",
-      "print\"\\n hence,turbine inlet pressure is\",round(Pi),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,turbine inlet pressure is 2995.0 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.12:pg-357"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12\n",
-      "#required work input\n",
-      "\n",
-      "Pe=150.0 #final pressure of air in kPa\n",
-      "Pi=100.0 #initial presure of air in kPa\n",
-      "k=1.4\n",
-      "Ti=300.0 #initial temperature of air in kelvis\n",
-      "Tes=Ti*(Pe/Pi)**((k-1)/k) #from second law\n",
-      "ws=1.004*(Ti-Tes) #from first law of isentropic process\n",
-      "n=0.7 #efficiency of automotive supercharger \n",
-      "w=ws/n #real work input in kJ/kg\n",
-      "Te=Ti-w/1.004 #temperature at supercharger exit in K\n",
-      "print\"\\n hence,required work input is \",round(w),\"kJ/kg\"\n",
-      "print\"\\n and exit temperature is \",round(Te,1),\"K\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " hence,required work input is  -53.0 kJ/kg\n",
-        "\n",
-        " and exit temperature is  352.6 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/Chapter9_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter9_6.ipynb
deleted file mode 100755
index db80c9d5..00000000
--- a/Fundamentals_Of_Thermodynamics/Chapter9_6.ipynb
+++ /dev/null
@@ -1,588 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:14f50faeadc2ed8de4732f59bac3e49ea9db1041eaf016b1692117fea4d3a511"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter9:SECOND-LAW ANALYSIS FOR A CONTROL VOLUME"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Ex9.1:pg-336"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#work done by steam\n",

-      "\n",

-      "hi=3051.2 #initial specific heat of enthalpy of steam in kJ/kg\n",

-      "si=7.1228 #initial specific entropy of steam in kJ/kg-K\n",

-      "Pe=0.15 #final pressure in MPa\n",

-      "se=si #specific entropy in final state in kJ/kg-K\n",

-      "sf=1.4335 #in kJ/kg-K\n",

-      "sfg=5.7897 #in kJ/kg-K\n",

-      "vi=50.0 #velocity with which steam enters turbine in m/s\n",

-      "ve=200.0 #velocity with which steam leaves the turbine in m/s\n",

-      "xe=(se-sf)/sfg #quality of steam in final state\n",

-      "hf=467.1 #in kJ/kg\n",

-      "hfg=2226.5 #in kJ/kg\n",

-      "he=hf+xe*hfg #final specific heat of enthalpy of steam in kJ/kg\n",

-      "w=hi+vi**2/(2*1000)-he-ve**2/(2*1000) #work of steam for isentropic process in kJ/kg\n",

-      "print\"\\n hence, work per kilogram of steam for this isentropic process is\",round(w,1),\"KJ/kg\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence, work per kilogram of steam for this isentropic process is 377.5 KJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.2:pg-337"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2\n",

-      "#exit velocity of steam from nozzle\n",

-      "\n",

-      "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n",

-      "si=7.1228 #initial specific entropy in kJ/kg-K\n",

-      "se=si #final specific entropy \n",

-      "Pe=0.3 #final pressure in MPa\n",

-      "print\"from steam table,various properties at final state are\"\n",

-      "he=2780.2 #final specific heat of enthalpy in kJ/kg-K\n",

-      "Te=159.1 #final temperature in celsius\n",

-      "vi=30.0 #velocity with which steam enters the nozzle in m/s\n",

-      "ve=((2*(hi-he)+(vi**2/1000))*1000)**0.5 #final velocity of steam with which it exits in m/s\n",

-      "print\"\\n hence,exit velocity of the steam from the nozzle is\",round(ve),\"m/s\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "from steam table,various properties at final state are\n",

-        "\n",

-        " hence,exit velocity of the steam from the nozzle is 737.0 m/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.2E:pg-339"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2E\n",

-      "#exit velocity of steam from nozzle\n",

-      "\n",

-      "hi=1279.1 #initial specific heat of enthalpy in Btu/lbm\n",

-      "si=1.7085 #initial specific entropy in Btu/lbm R\n",

-      "se=si #final specific entropy \n",

-      "Pe=40 #final pressure in lbf/in^2\n",

-      "he=1193.9 #final specific heat of enthalpy in Btu/lbm\n",

-      "Te=314.2 #final temperature in F\n",

-      "vi=100.0 #velocity with which steam enters the nozzle in ft/s\n",

-      "ve=((2*((hi-he)+(vi**2/(32.17*778)))*(32.17*778)))**0.5 #final velocity of steam with which it exits in ft/s\n",

-      "print\"\\n hence,exit velocity of the steam from the nozzle is\",round(ve),\"ft/s\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,exit velocity of the steam from the nozzle is 2070.0 ft/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 7

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.3:pg-340"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#violation of second law\n",

-      "\n",

-      "print\"from R-134a tables\"\n",

-      "se=1.7148 #specific entropy in final state in kJ/kg-K\n",

-      "si=1.7395 #initial specific entropy in kJ/kg-K \n",

-      "print\"therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "from R-134a tables\n",

-        "therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.\n"

-       ]

-      }

-     ],

-     "prompt_number": 10

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.4:pg-340"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4\n",

-      "#calculating required specific work\n",

-      "\n",

-      "Cp=1.004 #specific heat of air at constant pressure in kJ/kg-K\n",

-      "Ti=290 #initial temperature in kelvins\n",

-      "Pi=100 #initial pressure in kPa\n",

-      "Pe=1000 #final pressure in kPa\n",

-      "k=1.4 \n",

-      "Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in kelvins\n",

-      "we=Cp*(Ti-Te) #required specific work in kJ/kg\n",

-      "print\"\\n hence,specific work required is\",round(we),\"kJ/kg\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,specific work required is -271.0 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.4E:pg-341"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4E\n",

-      "#calculating required specific work\n",

-      "\n",

-      "Cp=0.24 #specific heat of air at constant pressure in Btu/lbm R\n",

-      "Ti=520 #initial temperature in R\n",

-      "Pi=14.7 #initial pressure in lbf/in^2\n",

-      "Pe=147 #final pressure in lbf/in^2\n",

-      "k=1.4 \n",

-      "Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in R\n",

-      "we=Cp*(Ti-Te) #required specific work in Btu/lbm\n",

-      "print\"\\n hence,specific work required is\",round(we,2),\"Btu/lbm\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,specific work required is -116.15 Btu/lbm\n"

-       ]

-      }

-     ],

-     "prompt_number": 34

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.5:pg-342"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#entropy generation\n",

-      "\n",

-      "h1=2865.54 #specific heat of enthalpy at state 1 in kJ/kg\n",

-      "h2=83.94 #specific heat of enthalpy  at state 2 in kJ/kg\n",

-      "h3=2725.3 #specific heat of enthalpy at state 3 in kJ?kg\n",

-      "s1=7.3115 #specific entropy at state 1 in kJ/kg-K\n",

-      "s2=0.2966 #specific entropy at state 2 in kJ/kg-K\n",

-      "s3=6.9918 #specific entropy at state 3in kJ/kg-K\n",

-      "m1=2 #mass flow rate at state 1 in kg/s\n",

-      "m2=m1*(h1-h3)/(h3-h2) #mass flow rate at state 2 in kg/s\n",

-      "m3=m1+m2 #mass flow rate at state 3 in kg/s\n",

-      "Sgen=m3*s3-m1*s1-m2*s2 #entropy generation in the process\n",

-      "print\"\\n hence,entropy generated in this process is \",round(Sgen,3),\"kW/K\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,entropy generated in this process is  0.072 kW/K\n"

-       ]

-      }

-     ],

-     "prompt_number": 15

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.6:pg-344"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 6\n",

-      "#work required to fill the tank\n",

-      "import math\n",

-      "T1=17+273 #initial temperature of tank in Kelvins\n",

-      "sT1=6.83521 #specific entropy in kJ/kg-K\n",

-      "R=0.287 #gas constant in kJ/kg-K\n",

-      "P1=100 #initial pressure in kPa\n",

-      "P2=1000 #final pressure in kPa\n",

-      "sT2=sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n",

-      "T2=555.7 #from interplotation \n",

-      "V1=0.04 #volume of tank in m^3\n",

-      "V2=V1 #final volume is equal to initial volume\n",

-      "m1=P1*V1/(R*T1) #initial mass of air in tank in kg\n",

-      "m2=P2*V2/(R*T2) #final mass of air in tank in kg\n",

-      "Min=m2-m1 #in kg\n",

-      "u1=207.19 #initial specific heat of enthalpy in kJ/kg\n",

-      "u2=401.49 #final specific heat of enthalpy in kJ/kg\n",

-      "hin=290.43 #in kJ/kg\n",

-      "W12=Min*hin+m1*u1-m2*u2 #work required to fill the tank in kJ\n",

-      "print\"\\n hence,the total amount of work required to fill the tank is\",round(W12,1),\"kJ\" "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,the total amount of work required to fill the tank is -31.9 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 18

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.7:pg-347"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7\n",

-      "#work required to pump water isentropically \n",

-      "\n",

-      "P1=100 #initial pressure in kPa\n",

-      "P2=5000 #final pressure in kPa\n",

-      "v=0.001004 #specific volume in m^3/kg\n",

-      "w=v*(P2-P1) #work required to pump water isentropically\n",

-      "print\"\\n hence,work required to pump water isentropically  is \",round(w,2),\"kJ/kg\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,work required to pump water isentropically  is  4.92 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 21

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.8:pg-348"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8\n",

-      "#Velocity in exit flow\n",

-      "\n",

-      "print\"From Steam Tables, for liquid water at 20 C\"\n",

-      "vf=0.001002 #in m^3/kg\n",

-      "v=vf\n",

-      "Pi=300 #Line pressure in kPa\n",

-      "Po=100 #in kPa\n",

-      "Ve=(2*v*(Pi-Po)*1000)**0.5 #velocity in the exit flow\n",

-      "print\" \\n Hence, an ideal nozzle can generate upto \",round(Ve),\"m/s\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "From Steam Tables, for liquid water at 20 C\n",

-        " \n",

-        " Hence, an ideal nozzle can generate upto  20.0 m/s\n"

-       ]

-      }

-     ],

-     "prompt_number": 24

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.9:pg-351"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9\n",

-      "#Rate of Entropy Generation\n",

-      "\n",

-      "print\"From R-410a tables,we get\"\n",

-      "hi=280.6 #in kJ/kg\n",

-      "he=307.8 #in kJ/kg\n",

-      "si=1.0272 #in kJ/kg\n",

-      "se=1.0140 #in kJ/kg\n",

-      "m=0.08 #flow rate of refrigerant in kg/s\n",

-      "P=3 #electrical power input in kW\n",

-      "Qcv=m*(he-hi)-P #in kW\n",

-      "To=30 #in Celsius\n",

-      "Sgen=m*(se-si)-Qcv/(To+273.2) #rate of entropy generation \n",

-      "print\"\\n Hence,the rate of entropy generation for this process is\",round(Sgen,5),\"kW/K\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "From R-410a tables,we get\n",

-        "\n",

-        " Hence,the rate of entropy generation for this process is 0.00166 kW/K\n"

-       ]

-      }

-     ],

-     "prompt_number": 29

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.10:pg-353"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 10\n",

-      "#turbine efficiency\n",

-      "\n",

-      "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n",

-      "si=7.1228 #initial specific entropy in kJ/kg-K\n",

-      "sf=0.7548 #in kJ/kg-K\n",

-      "sfg=7.2536 #in kJ/kg-K\n",

-      "ses=si #final specific entropy is same as the initial\n",

-      "xes=(si-sf)/sfg #quality of steam when it leaves the turbine\n",

-      "hf=225.9 #in kJ/kg\n",

-      "hfg=2373.1 #in kJ/kg\n",

-      "hes=hf+xes*hfg #final specific heat of enthalpy in kJ/kg\n",

-      "ws=hi-hes #work output of turbine calculated ideally  in kJ/kg\n",

-      "wa=600 #actual work output of turbine in kJ/kg\n",

-      "nturbine=wa/ws #efiiciency of turbine \n",

-      "print\"\\n hence,efficiency of the turbine is\",round(nturbine*100,1),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,efficiency of the turbine is 80.9 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 32

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.11:pg-355"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 11\n",

-      "#turbine inlet pressure\n",

-      "import math\n",

-      "hi=1757.3 #initial specific heat of enthalpy of air in  kJ/kg\n",

-      "si=8.6905 #initial specifc entropy of airin kJ/kg-K\n",

-      "he=855.3 #final specific heat of enthalpy of air in kJ/kg\n",

-      "w=hi-he #actual work done by turbine in kJ/kg\n",

-      "n=0.85 #efficiency of turbine \n",

-      "ws=w/n #ideal work done by turbine in kJ/kg\n",

-      "hes=hi-ws #from first law of isentropic process\n",

-      "Tes=683.7 #final temperature in kelvins from air tables\n",

-      "ses=7.7148 #in kJ/kg-K\n",

-      "R=0.287 #gas constant in kJ/kg-K\n",

-      "Pi=100/math.e**((si-ses)/-R) #turbine inlet pressure in kPa\n",

-      "print\"\\n hence,turbine inlet pressure is\",round(Pi),\"kPa\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,turbine inlet pressure is 2995.0 kPa\n"

-       ]

-      }

-     ],

-     "prompt_number": 36

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.12:pg-357"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 12\n",

-      "#required work input\n",

-      "\n",

-      "Pe=150.0 #final pressure of air in kPa\n",

-      "Pi=100.0 #initial presure of air in kPa\n",

-      "k=1.4\n",

-      "Ti=300.0 #initial temperature of air in kelvis\n",

-      "Tes=Ti*(Pe/Pi)**((k-1)/k) #from second law\n",

-      "ws=1.004*(Ti-Tes) #from first law of isentropic process\n",

-      "n=0.7 #efficiency of automotive supercharger \n",

-      "w=ws/n #real work input in kJ/kg\n",

-      "Te=Ti-w/1.004 #temperature at supercharger exit in K\n",

-      "print\"\\n hence,required work input is \",round(w),\"kJ/kg\"\n",

-      "print\"\\n and exit temperature is \",round(Te,1),\"K\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " hence,required work input is  -53.0 kJ/kg\n",

-        "\n",

-        " and exit temperature is  352.6 K\n"

-       ]

-      }

-     ],

-     "prompt_number": 44

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/README.txt b/Fundamentals_Of_Thermodynamics/README.txt
deleted file mode 100755
index f9822be5..00000000
--- a/Fundamentals_Of_Thermodynamics/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: abhishek chauhan
-Course: btech
-College/Institute/Organization: ABES Engineering College Ghaziabad
-Department/Designation: electronics & communication engg
-Book Title: Fundamentals Of Thermodynamics
-Author:  B. Claus And R. E. Sonntag
-Publisher: Wiley India Pvt. Ltd., New Delhi
-Year of publication: 2010
-Isbn: 978-81-265-2152-4
-Edition: 7
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics/chapter10.ipynb b/Fundamentals_Of_Thermodynamics/chapter10.ipynb
deleted file mode 100755
index e985ae83..00000000
--- a/Fundamentals_Of_Thermodynamics/chapter10.ipynb
+++ /dev/null
@@ -1,474 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:498eddefb3635fb861c09bdd97718d2b72fb04f4d3b083cedbf9f6de28a92b4e"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter10:IRREVERSIBILITY AND AVAILABILITY"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.1:pg-386"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 1\n",

-      "#Calculating reversible work\n",

-      "#Form the Steam Tables,the inlet and the exit state properties are \n",

-      "\n",

-      "hi=171.95 #initial specific heat of enthalpy in kJ/kg\n",

-      "si=0.5705 #initial specific entropy in kJ/kg-K\n",

-      "se=2.1341 #final specific entropy in kJ/kg-K\n",

-      "he=765.34 #final specific heat of enthalpy in kJ/kg-K\n",

-      "m=5 #mass flow rate of feedwater in kg/s\n",

-      "q1=900/m #heat added by one of the sources in kJ/kg\n",

-      "q2=he-hi-q1 #second heat transfer in kJ/kg\n",

-      "To=25+273.3 #Temp. of the surroundings in K\n",

-      "T1=100+273.2 #temp. of reservoir of one of the source in K\n",

-      "T2=200+273.2 #temp. of reservoir of second source in K\n",

-      "wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) #reversible work in kJ/kg\n",

-      "print\"\\n Hence, the irreversibility is\",round(wrev),\"kJ/kg\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " Hence, the irreversibility is 62.0 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 7

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.2:pg-387"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2\n",

-      "#Calculating reversible work\n",

-      "#Form the Steam Tables,the inlet and the exit state properties are\n",

-      "\n",

-      "hi=298.6 #initial specific heat of enthalpy in kJ/kg\n",

-      "si=6.8631 #initial specific entropy in kJ/kg-K\n",

-      "se=7.4664 #final specific entropy in kJ/kg-K\n",

-      "he=544.7 #final specific heat of enthalpy in kJ/kg-K\n",

-      "q=-50 #heat lost to surroundings in kJ/kg\n",

-      "w=hi-he+q #work in kJ/kg\n",

-      "To=25+273.2 #Temp. of the surroundings in K\n",

-      "P1=100 #Pressure of ambient air in kPa\n",

-      "P2=1000 #Final pressure of air after compression in kPa\n",

-      "R=0.287 #Universal gas constant in kJ/kg-K\n",

-      "wrev=To*(se-si-R*log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in kJ/kg\n",

-      "i=wrev-w #irreversibility in kJ/kg\n",

-      "print\"\\n Hence,the irreversibility is\",round(i,1),\"kJ/kg\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " Hence,the irreversibility is 32.8 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 10

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.2E:pg-388"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 2E\n",

-      "#Calculating reversible work\n",

-      "#Form the Steam Tables,the inlet and the exit state properties are\n",

-      "\n",

-      "hi=129.18 #initial specific heat of enthalpy in Btu/lbm\n",

-      "si=1.6405 #initial specific entropy in Btu/lbm R\n",

-      "se=1.7803 #final specific entropy in Btu/lbm R\n",

-      "he=231.20 #final specific heat of enthalpy in Btu/lbm\n",

-      "q=-22 #heat lost to surroundings in Btu/lbm\n",

-      "w=hi-he+q #work in Btu/lbm\n",

-      "To=539.7 #Temp. of the surroundings in R\n",

-      "P1=14.7 #Pressure of ambient air in lbf/in^2\n",

-      "P2=150 #Final pressure of air after compression in lbf/in^2\n",

-      "R=0.06855 #Universal gas constant in Btu/lbm R\n",

-      "wrev=To*(se-si-R*log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in Btu/lbm\n",

-      "i=wrev-w #irreversibility in Btu/lbm\n",

-      "print\"\\n Hence,the irreversibility is\",round(i,2),\"Btu/lbm\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " Hence,the irreversibility is 11.52 Btu/lbm\n"

-       ]

-      }

-     ],

-     "prompt_number": 18

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.3:pg-390"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 3\n",

-      "#Calculating reversible work and irreversibility\n",

-      "#Form the Steam Tables at state 1\n",

-      "\n",

-      "u1=1243.5 #initial specific internal energy in kJ/kg\n",

-      "s1=4.4819 #initial specific entropy in kJ/kg-K\n",

-      "v1=28.895 #initial specific volume in m^3/kg\n",

-      "v2=2*v1 #final specific volume in kg/m^3\n",

-      "u2=u1 #initial specific internal energy in kJ/kg\n",

-      "#These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg\n",

-      "T2=9.1+273.2 #final temp. in K\n",

-      "x2=0.513 #quality in final state\n",

-      "s2=4.644 #final specific entropy in kJ/kg\n",

-      "V1=1 #volume of part of A in m^3\n",

-      "m=V1/v1  #mass flow rate in kg/s\n",

-      "To=20+273.2 #Room temperature in K\n",

-      "Wrev=To*m*(s2-s1) #reversible work in kJ\n",

-      "I=Wrev #irreversibility of the process\n",

-      "print\"\\n The irreversibility is\",round(I,3),\"kJ\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " The irreversibility is 1.645 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.4:pg-391"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4\n",

-      "#calculating the final mass in the tank and the irreversibility in the process\n",

-      "#From the ammonia tables, the initial and line state properties are\n",

-      "\n",

-      "V=1#volume of the rigid tank in m^3\n",

-      "v1=0.6995#initial specific volume in m^3/kg\n",

-      "u1=1369.5#initial specific internal energy in kJ/kg\n",

-      "s1=5.927#initial specific entropy in kJ/kg K\n",

-      "h1=134.41#initial specific heat of enthalpy in kJ/kg\n",

-      "si=0.5408#in kJ/kg K\n",

-      "m1=V/v1#initial mass in kg\n",

-      "x2=0.007182\n",

-      "v2=(0.001534 + x2 * 0.41684)#in m^3/kg\n",

-      "v2=round(v2,7)\n",

-      "s2=0.5762#final specific entropy in kJ/kg\n",

-      "m2=V/v2#the final mass in kg\n",

-      "mi=m2-m1#in kg\n",

-      "mi=round(mi,3)\n",

-      "T=293.15# in K\n",

-      "S2gen=(m2*s2-m1*s1-mi*si)#in kJ/kg\n",

-      "S2gen=round(S2gen,3)\n",

-      "Icv=T*S2gen#in kJ\n",

-      "print\"\\n the final mass is \",round(m2,2),\"kg\"\n",

-      "print\"\\n the irreversiblity is\",round(Icv,3),\"kJ\"\n",

-      "\n",

-      "\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " the final mass is  220.86 kg\n",

-        "\n",

-        " the irreversiblity is 34.885 kJ\n"

-       ]

-      }

-     ],

-     "prompt_number": 31

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.5:pg-396"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 5\n",

-      "#calculating the availability per kilogram of the steam entering and at both points at which steam leaves the turbine,\n",

-      "#the isentropic efficiency,the second-law efficiency for this process\n",

-      "#For the ideal isentropic turbine,\n",

-      "\n",

-      "h0=104.9#in kJ/kg\n",

-      "s0=0.3674#entropy at state0 in kJ/kgK\n",

-      "m1=30#mass at state1 in kg\n",

-      "phi1=1109.6#in kJ/kg\n",

-      "h1=3115.3#enthalpy at state1 in kJ/kg\n",

-      "m2=5#mass at state2 in kg\n",

-      "phi2=755.3#in kJ/kg\n",

-      "h2=2855.4#enthalpy at state2 For the actual turbine in kJ/kg\n",

-      "m3=25#mass at state3 in kg\n",

-      "phi3=195.0#in kJ/kg\n",

-      "S2s=6.7428#entropy in kJ/kg K at state2\n",

-      "x2s=(6.7428-1.8606)/4.906# dryness factor at state2 \n",

-      "x2s=round(x2s,4)\n",

-      "h2s=640.2+(x2s*2108.5)#entropy at state2 for the ideal isentropic turbine\n",

-      "S3s=6.7428#entropy in kJ/kg K\n",

-      "x3s=(6.7428-0.7549)/7.2536# dryness factor at state3\n",

-      "x3s=round(x3s,4)\n",

-      "h3s=225.9+(x3s*2373.1)#enthalpy in kJ/kg\n",

-      "h3=2361.8#enthalpy at state3 For the actual turbine in kJ/kg\n",

-      "Ws=(m1*h1-m2*h2s-m3*h3s)#workdone for the ideal isentropic turbine in kW\n",

-      "W=(m1*h1-m2*h2-m3*h3)#workdone For the actual turbine in kW\n",

-      "W2=(m1*phi1-m2*phi2-m3*phi3)#workdone for the second ideal isentropic turbine in kW\n",

-      "eta1=W/Ws#The isentropic efficiency\n",

-      "eta2=W/W2#he second-law efficiency\n",

-      "print\"\\n the workdone for the ideal isentropic turbine is \",round(Ws),\"kW\"\n",

-      "print\"\\n the workdone for the actual turbine is \",round(W),\"kg\"\n",

-      "print\"\\n the workdone for the second ideal isentropic turbine  is \",round(W2),\"kW\"\n",

-      "print\"\\n the isentropic efficiency is \",round(eta1,3)\n",

-      "print\"\\n the second-law efficiency is \",round(eta2,3)\n",

-      "#The answer of the x2s in the book is wrong.\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " the workdone for the ideal isentropic turbine is  25145.0 kW\n",

-        "\n",

-        " the workdone for the actual turbine is  20137.0 kg\n",

-        "\n",

-        " the workdone for the second ideal isentropic turbine  is  24637.0 kW\n",

-        "\n",

-        " the isentropic efficiency is  0.801\n",

-        "\n",

-        " the second-law efficiency is  0.817\n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.6:pg-399"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 6\n",

-      "#calculate the second-law efficiency for this process and the irreversibility per kilogram of water evaporated\n",

-      "\n",

-      "he1=2950#enthalpy in kJ/kg\n",

-      "hi1=632.2#enthalpy in kJ/kg\n",

-      "he2=599.5#enthalpy in kJ/kg\n",

-      "hi2=1199#enthalpy in kJ/kg\n",

-      "#r=mp/mh\n",

-      "r=(he1-hi1)/(hi2-he2)#the ratio of the mass flow of water\n",

-      "T0=298.15#in K\n",

-      "S1=1.8418#entropy at state1 in kJ/kgK\n",

-      "S2=7.0384#entropy at state2 in kJ/kgK\n",

-      "S3=7.8751#entropy at state3 in kJ/kgK\n",

-      "S4=7.3173#entropy at state4 in kJ/kgK\n",

-      "#phi=phi2-phi1\n",

-      "phi=(he1-hi1)-T0*(S2-S1)#The increase in availability of the water is, per kilogram of water in kJ/kg\n",

-      "#w=(mp2/mh2)*(phi3-phi4)\n",

-      "w=r*((hi2-he2)-T0*(S3-S4))#The decrease in availability of the products, per kilogram of water in kJ/kg\n",

-      "eta=phi/w#the second-law efficiency\n",

-      "#i=I/mh\n",

-      "i=phi+w#the process irreversibility per kilogram of water in kJ/kg\n",

-      "#e=I2/mh2\n",

-      "e=(T0*(S2-S1)+(T0*r*(S4-S3)))#The total irreversibility in kJ/kg\n",

-      "print\"\\n the second law efficiency is \",round(eta,3)\n",

-      "print\"\\n the total irreversibility is \",round(e,2),\"kJ/kg\"\n",

-      "\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " the second law efficiency is  0.459\n",

-        "\n",

-        " the total irreversibility is  906.38 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 17

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.7:pg-403"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7\n",

-      "#calculate the heater\u2019s second-law efficiency\n",

-      "\n",

-      "h0=104.87#enthalpy in kJ/kg\n",

-      "s0=0.3673#entropy in kJ/kg K\n",

-      "h1=171.97#in kJ/kg\n",

-      "he=765.25#in kJ/kg\n",

-      "s1=0.5705#in kJ/kg K\n",

-      "T0=298.2#in C\n",

-      "se=2.1341#in kJ/kg K\n",

-      "shi1=(h1-h0)-T0*(s1-s0)#in kJ/kg\n",

-      "shie=(he-h0)-T0*(se-s0)#in kJ/kg\n",

-      "T1=373.2#in C\n",

-      "T2=473.2#in C\n",

-      "q1=180#heat at state1 in kJ\n",

-      "q2=413.28#heat at state2 in kJ\n",

-      "phisource1=(1-T0/T1)*q1#in kJ/kg\n",

-      "phisource2=(1-T0/T2)*q2#in kJ/kg\n",

-      "phisource=phisource1+phisource2#in kJ/kg\n",

-      "Icv=phisource+shi1-shie#in kJ/kg\n",

-      "eta2law=(phisource-Icv)/phisource#the heater\u2019s second-law efficiency\n",

-      "print\"\\n the heater\u2019s second-law efficiency is\",round(eta2law,2)\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " the heater\u2019s second-law efficiency is 0.67\n"

-       ]

-      }

-     ],

-     "prompt_number": 53

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex10.8:pg-404"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8\n",

-      "#calculate any heat transfer downward, and follow the flux of exergy, and find the exergy destruction in the process.\n",

-      "Qout=500.0#in W\n",

-      "Qin=500.0#in W\n",

-      "Tsurf=1000.0#in K\n",

-      "T0=298.15#in K\n",

-      "Sgen1=Qout/Tsurf#in W/K\n",

-      "phides=(T0*Sgen1)#in W\n",

-      "phitrans=((1-T0/Tsurf)*Qout)#in W\n",

-      "Ttop=500#in K\n",

-      "Sgen2=(Qout/Ttop)-(Qin/Tsurf)#in W/K\n",

-      "Te=phitrans-phides-((1-T0/Ttop)*Qout)#the exergy destruction in the process\n",

-      "print\"\\n the exergy destruction in the process  is\",round(Te)\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "\n",

-        " the exergy destruction in the process  is 0.0\n"

-       ]

-      }

-     ],

-     "prompt_number": 77

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
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deleted file mode 100755
index c7a00e18..00000000
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deleted file mode 100755
index d4cf4f76..00000000
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diff --git a/Fundamentals_Of_Thermodynamics/screenshots/velocity_6.png b/Fundamentals_Of_Thermodynamics/screenshots/velocity_6.png
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diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter10_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter10_6.ipynb
new file mode 100755
index 00000000..59bbb80a
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter10_6.ipynb
@@ -0,0 +1,164 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:6310b6690e48041fd88dd8277353623ea538798dfeeeb9a3819409de065b2175"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter10:IRREVERSIBILITY AND AVAILABILITY"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.1:pg-386"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 1\n",
+      "#Calculating reversible work\n",
+      "\n",
+      "#Form the Steam Tables,the inlet and the exit state properties are \n",
+      "hi=171.95 #initial specific heat of enthalpy in kJ/kg\n",
+      "si=0.5705 #initial specific entropy in kJ/kg-K\n",
+      "se=2.1341 #final specific entropy in kJ/kg-K\n",
+      "he=765.34 #final specific heat of enthalpy in kJ/kg-K\n",
+      "m=5 #mass flow rate of feedwater in kg/s\n",
+      "q1=900/m #heat added by one of the sources in kJ/kg\n",
+      "q2=he-hi-q1 #second heat transfer in kJ/kg\n",
+      "To=25+273.3 #Temp. of the surroundings in K\n",
+      "T1=100+273.2 #temp. of reservoir of one of the source in K\n",
+      "T2=200+273.2 #temp. of reservoir of second source in K\n",
+      "wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) #reversible work in kJ/kg\n",
+      "print\"\\n Hence, the irreversibility is\",round(wrev,1),\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Hence, the irreversibility is 62.0 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.2:pg-387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 2\n",
+      "#Calculating reversible work\n",
+      "import math\n",
+      "#Form the Steam Tables,the inlet and the exit state properties are\n",
+      "hi=298.6 #initial specific heat of enthalpy in kJ/kg\n",
+      "si=6.8631 #initial specific entropy in kJ/kg-K\n",
+      "se=7.4664 #final specific entropy in kJ/kg-K\n",
+      "he=544.7 #final specific heat of enthalpy in kJ/kg-K\n",
+      "q=-50 #heat lost to surroundings in kJ/kg\n",
+      "w=hi-he+q #work in kJ/kg\n",
+      "To=25+273.2 #Temp. of the surroundings in K\n",
+      "P1=100 #Pressure of ambient air in kPa\n",
+      "P2=1000 #Final pressure of air after compression in kPa\n",
+      "R=0.287 #Universal gas constant in kJ/kg-K\n",
+      "wrev=To*(se-si-R*math.log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in kJ/kg\n",
+      "i=wrev-w #irreversibility in kJ/kg\n",
+      "print\"\\n Hence, the irreversibility is\",round(i,1),\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Hence, the irreversibility is 32.8 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.3:pg-390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3\n",
+      "#Calculating reversible work and irreversibility\n",
+      "\n",
+      "#Form the Steam Tables at state 1\n",
+      "u1=1243.5 #initial specific internal energy in kJ/kg\n",
+      "s1=4.4819 #initial specific entropy in kJ/kg-K\n",
+      "v1=28.895 #initial specific volume in m^3/kg\n",
+      "v2=2*v1 #final specific volume in kg/m^3\n",
+      "u2=u1 #initial specific internal energy in kJ/kg\n",
+      "#These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg\n",
+      "T2=9.1+273.2 #final temp. in K\n",
+      "x2=0.513 #quality in final state\n",
+      "s2=4.644 #final specific entropy in kJ/kg\n",
+      "V1=1 #volume of part of A in m^3\n",
+      "m=V1/v1  #mass flow rate in kg/s\n",
+      "To=20+273.2 #Room temperature in K\n",
+      "Wrev=To*m*(s2-s1) #reversible work in kJ\n",
+      "I=Wrev #irreversibility of the process\n",
+      "print\"\\n The irreversibility is \",round(I,3),\"kJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The irreversibility is  1.645 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter11_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter11_6.ipynb
new file mode 100755
index 00000000..814dd14b
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter11_6.ipynb
@@ -0,0 +1,529 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:a88db765bb23172043955029f8f04edaf8e710f3cbd46a5474827f42fa8a65ec"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter11:Power and Refrigeration Systems\u2014With Phase Change"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.1:Pg-425"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Ques 1\n",

+      "#To determine the efficiency of Rankine cycle\n",

+      "\n",

+      "#1-Inlet state of pump\n",

+      "#2-Exit state of pump\n",

+      "P2=2000;#Exit pressure in kPa\n",

+      "P1=10;#Inlet pressure in kPa\n",

+      "v=0.00101;#specific weight of water in m^3/kg\n",

+      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

+      "h1=191.8;#Enthalpy in kJ/kg from table\n",

+      "h2=h1+wp;#enthalpy in kJ/kg\n",

+      "#2-Inlet state for boiler\n",

+      "#3-Exit state for boiler\n",

+      "h3=2799.5;#Enthalpy in kJ/kg\n",

+      "#3-Inlet state for turbine\n",

+      "#4-Exit state for turbine\n",

+      "#s3=s4(Entropy remain same)\n",

+      "s4=6.3409;#kJ/kg\n",

+      "sf=0.6493;#Entropy at liquid state in kJ/kg\n",

+      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg\n",

+      "x4=(s4-sf)/sfg;#x-factor\n",

+      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",

+      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",

+      "\n",

+      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",

+      "print\" Percentage efficiency =\",round(nth*100,1)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Percentage efficiency = 30.3\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.2:Pg-429"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Ques 2\n",

+      "#To determine the efficiency of Rankine cycle\n",

+      "\n",

+      "#1-Inlet state of pump\n",

+      "#2-Exit state of pump\n",

+      "P2=4000;#Exit pressure in kPa\n",

+      "P1=10;#Inlet pressure in kPa\n",

+      "v=0.00101;#specific weight of water in m^3/kg\n",

+      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

+      "h1=191.8;#Enthalpy in kJ/kg from table\n",

+      "h2=h1+wp;#enthalpy in kJ/kg\n",

+      "#2-Inlet state for boiler\n",

+      "#3-Exit state for boiler\n",

+      "h3=3213.6;#Enthalpy in kJ/kg from table\n",

+      "#3-Inlet state for turbine\n",

+      "#4-Exit state for turbine\n",

+      "#s3=s4(Entropy remain same)\n",

+      "s4=6.7690;#Entropy in kJ/kg from table\n",

+      "sf=0.6493;#Entropy at liquid state in kJ/kg from table\n",

+      "sfg=7.5009;#Entropy difference for vapor and liquid state in kJ/kg from table\n",

+      "x4=(s4-sf)/sfg;#x-factor\n",

+      "hfg=2392.8;#Enthalpy difference in kJ/kg for turbine\n",

+      "h4=h1+x4*hfg;#Enthalpy in kJ/kg\n",

+      "\n",

+      "nth=((h3-h2)-(h4-h1))/(h3-h2);\n",

+      "print\"Percentage efficiency =\",round(nth*100,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Percentage efficiency = 35.3 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 41

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.2E:Pg-431"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Ques 2\n",

+      "#To determine the efficiency of Rankine cycle\n",

+      "\n",

+      "#1-Inlet state of pump\n",

+      "#2-Exit state of pump\n",

+      "P2=600.0 ;#Exit pressure in lbf/in^2\n",

+      "P1=1.0;#Inlet pressure in lbf/in^2\n",

+      "v=0.01614;#specific weight of water in ft^3/lbm\n",

+      "wp=v*(P2-P1)*(144.0/778.0);#work done in pipe in Btu/lbm\n",

+      "h1=69.70;#Enthalpy in Btu/lbm from table\n",

+      "h2=h1+wp;#enthalpy in Btu/lbm\n",

+      "#2-Inlet state for boiler\n",

+      "#3-Exit state for boiler\n",

+      "h3=1407.6;#Enthalpy in Btu/lbm from table\n",

+      "#3-Inlet state for turbine\n",

+      "#4-Exit state for turbine\n",

+      "#s3=s4(Entropy remain same)\n",

+      "s4=1.6343;#Entropy in Btu/lbm from table\n",

+      "sf=1.9779;#Entropy at liquid state in Btu/lbm from table\n",

+      "sfg=1.8453;#Entropy difference for vapor and liquid state in Btu/lbm from table\n",

+      "x4=-(s4-sf)/sfg;#x-factor\n",

+      "hfg=1036.0;#Enthalpy difference in Btu/lbm for turbine\n",

+      "h4=1105.8-x4*hfg;#Enthalpy in Btu/lbm\n",

+      "wt=(h3-h4) #work done in turbine in Btu/lbm\n",

+      "\n",

+      "nth=((h3-h4)-wp)/(h3-h2);\n",

+      "print\"Percentage efficiency =\",round(nth*100,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Percentage efficiency = 36.9 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 42

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.3:Pg-433"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Ques 3\n",

+      "#To determine the efficiency of a cycle\n",

+      "\n",

+      "#1-Inlet state of pump\n",

+      "#2-Exit state of pump\n",

+      "P2=4000;#Exit pressure in kPa\n",

+      "P1=10;#Inlet pressure in kPa\n",

+      "v=0.00101;#specific weight of water in m^3/kg\n",

+      "wp=v*(P2-P1);#work done in pipe in kJ/kg\n",

+      "h1=191.8;#Enthalpy in kJ/kg from table\n",

+      "h2=h1+wp;#enthalpy in kJ/kg\n",

+      "#2-Inlet state for boiler\n",

+      "#3-Exit state for Boiler\n",

+      "h3=3213.6;#Enthalpy in kJ/kg from table\n",

+      "#3-Inlet state for high pressure turbine\n",

+      "#4-Exit state for high pressure turbine\n",

+      "#s3=s4(Entropy remain same)\n",

+      "s4=6.7690;#Entropy in kJ/kg from table\n",

+      "sf=1.7766;#Entropy at liquid state in kJ/kg from table\n",

+      "sfg=5.1193;#Entropy difference for vapor and liquid state in kJ/kg from table\n",

+      "x4=(s4-sf)/sfg;#x-factor\n",

+      "hf=604.7#Enthalpy of liquid state in kJ/kg\n",

+      "hfg=2133.8;#Enthalpy difference in kJ/kg for turbine\n",

+      "h4=hf+x4*hfg;#Enthalpy in kJ/kg\n",

+      "#5-Inlet state for low pressure turbine\n",

+      "#6-Exit pressure for low pressure turbine\n",

+      "sf=0.6493;#Entropy in liquid state in kJ/kg for turbine\n",

+      "h5=3273.4;#enthalpy in kJ/kg \n",

+      "s5=7.8985;#Entropy in kJ/kg\n",

+      "sfg=7.5009;#entropy diff in kJ/kg \n",

+      "x6=(s5-sf)/sfg;#x-factor\n",

+      "hfg=2392.8;#enthalpy difference for low pressure turbine in kj/kg\n",

+      "h6=h1+x6*hfg;#entropy in kg/kg\n",

+      "wt=(h3-h4)+(h5-h6);#work output in kJ/kg\n",

+      "qh=(h3-h2)+(h5-h4);\n",

+      "\n",

+      "nth=(wt-wp)/qh;\n",

+      "print\" Percentage efficiency =\",round(nth*100,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Percentage efficiency = 35.9 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 43

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.4:Pg-438"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques4\n",

+      "#Efficiency of Refrigeration cycle\n",

+      "\n",

+      "#from previous examples\n",

+      "h1=191.8;#kJ/kg\n",

+      "h5=3213.6;#kg/kg\n",

+      "h6=2685.7;#kJ/kg\n",

+      "h7=2144.1;#kJ/kg\n",

+      "h3=604.7;#kJ/kg\n",

+      "#1-Inlet state of pump\n",

+      "#2-Exit state of pump\n",

+      "P2=400;#Exit pressure in kPa\n",

+      "P1=10;  #Inlet pressure in kPa\n",

+      "v=0.00101;#specific weight of water in m^3/kg\n",

+      "wp1=v*(P2-P1);#work done for low pressure pump in kJ/kg\n",

+      "h1=191.8;#Enthalpy in kJ/kg from table\n",

+      "h2=h1+wp1;#enthalpy in kJ/kg\n",

+      "#5-Inlet state for turbine\n",

+      "#6,7-Exit state for turbine\n",

+      "y=(h3-h2)/(h6-h2);#extraction fraction\n",

+      "wt=(h5-h6)+(1-y)*(h6-h7);#turbine work in kJ/kg\n",

+      "#3-Inlet for high pressure pump\n",

+      "#4-Exit for high pressure pump\n",

+      "P3=400;#kPa\n",

+      "P4=4000;#kPa\n",

+      "v=0.001084;#specific heat for 3-4 process in m^3/kg\n",

+      "wp2=v*(P4-P3);#work done for high pressure pump\n",

+      "h4=h3+wp2;#Enthalpy in kJ/kg\n",

+      "wnet=wt-(1-y)*wp1-wp2;\n",

+      "qh=h5-h4;#Heat output in kJ/kg\n",

+      "nth=wnet/qh;\n",

+      "print\" Refrigerator Efficiency =\",round(nth*100,1),\"%\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Refrigerator Efficiency = 37.5 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 44

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.5:Pg-443"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques5\n",

+      "#To determine thermal efficiency of cycle\n",

+      "\n",

+      "#5-Inlet state for turbine\n",

+      "#6-Exit state for turbine\n",

+      "#h-Enthalpy at a state \n",

+      "#s-Entropy at a state\n",

+      "#from steam table\n",

+      "h5=3169.1;#kJ/kg\n",

+      "s5=6.7235;#kJ/kg\n",

+      "s6s=s5;\n",

+      "sf=0.6493;#Entropy for liquid state in kJ/kg\n",

+      "sfg=7.5009;#Entropy difference in kJ/kg\n",

+      "hf=191.8;#kJ/kg\n",

+      "hfg=2392.8;#Enthalpy difference in kJ/kg\n",

+      "x6s=(s6s-sf)/sfg;#x-factor\n",

+      "h6s=hf+x6s*hfg;#kJ/Kg at state 6s\n",

+      "nt=0.86;#turbine efficiency given\n",

+      "wt=nt*(h5-h6s);\n",

+      "#1-Inlet state for pump\n",

+      "#2-Exit state for pump\n",

+      "np=0.80;#pump efficiency given\n",

+      "v=0.001009;#specific heat in m^3/kg\n",

+      "P2=5000;#kPa\n",

+      "P1=10;#kPa\n",

+      "wp=v*(P2-P1)/np;#Work done in pump in kJ/kg\n",

+      "wnet=wt-wp;#net work in kJ/kg\n",

+      "#3-Inlet state for boiler\n",

+      "#4-Exit state for boiler\n",

+      "h3=171.8;#in kJ/kg from table\n",

+      "h4=3213.6;#kJ/kg from table\n",

+      "qh=h4-h3;\n",

+      "nth=wnet/qh;\n",

+      "print \"Cycle Efficiency =\",round(nth*100,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Cycle Efficiency = 29.2 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 45

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.5E:Pg-445"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques5\n",

+      "#To determine thermal efficiency of cycle\n",

+      "\n",

+      "#5-Inlet state for turbine\n",

+      "#6-Exit state for turbine\n",

+      "#h-Enthalpy at a state \n",

+      "#s-Entropy at a state\n",

+      "#from steam table\n",

+      "h5=1386.8;#Btu/lbm\n",

+      "s5=1.6248;#Btu/lbm\n",

+      "s6s=s5;\n",

+      "sf=1.9779;#Entropy for liquid state in Btu/lbm\n",

+      "sfg=1.8453;#Entropy difference in Btu/lbm\n",

+      "hf=1105.8;# Btu/lbm\n",

+      "hfg=1036.0;#Enthalpy difference in Btu/lbm\n",

+      "x6s=(s6s-sf)/sfg;#x-factor\n",

+      "h6s=hf+x6s*hfg;#Btu/lbm at state 6s\n",

+      "nt=0.86;#turbine efficiency given\n",

+      "wt=nt*(h5-h6s);\n",

+      "#1-Inlet state for pump\n",

+      "#2-Exit state for pump\n",

+      "np=0.80;#pump efficiency given\n",

+      "v=0.016;#specific heat in ft^3/lbm\n",

+      "P2=800.0;# lbf/in^2\n",

+      "P1=1.0;# lbf/in^2\n",

+      "wp=(v*(P2-P1)*144.0)/(np*778.0);#Work done in pump in Btu/lbm\n",

+      "wnet=wt-wp;#net work in Btu/lbm\n",

+      "#3-Inlet state for boiler\n",

+      "#4-Exit state for boiler\n",

+      "h3=65.1;#in Btu/lbm from table\n",

+      "h4=1407.6;# Btu/lbm from table\n",

+      "qh=h4-h3;\n",

+      "nth=wnet/qh;\n",

+      "print \"Cycle Efficiency =\",round(nth*100,2),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Cycle Efficiency = 30.48 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 60

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.6:Pg-451"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques6\n",

+      "#to determine the rate of refrigeration\n",

+      "\n",

+      "# refer to fig 11.21 in book\n",

+      "mdot=0.03 # mass flow rate in Kg/s\n",

+      "T1=-20 # temperature in evaporator in celsius\n",

+      "T3=40 #temperature in evaporator in Celsius\n",

+      "P2=1017 # saturation pressure in KPa\n",

+      "\n",

+      "# from table of R-134a refrigerant\n",

+      "h1=386.1 # enthalpy at state 1 in kJ/kg,\n",

+      "S1=1.7395 # entropy at state 1 in kJ/kg.K\n",

+      "S2=S1 # isentropic process\n",

+      "T2=47.7# corresponding value to S2 in table of R-134a in degree celsius\n",

+      "h2=428.4 # corresponding value to S2 in table of R-134a in kJ/kg\n",

+      "wc=h2-h1 # work done in compressor in kJ/kg\n",

+      "h4=h3=256.5 #enthalpy at state 4 and 3 in kJ/kg\n",

+      "qL=h1-h4 #Heat rejected in kJ/kg\n",

+      "\n",

+      "B=qL/wc # COP\n",

+      "\n",

+      "print\" the COP of the plant is\",round(B,2)\n",

+      "print\" the refrigeration rate is\",round(mdot*qL,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " the COP of the plant is 3.06\n",

+        " the refrigeration rate is 3.89\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11.7:Pg-454"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques7\n",

+      "#to determine the COP of cycle\n",

+      "\n",

+      "P1=125 # pressure at state 1 in kPa\n",

+      "P2=1.2 # pressure at state 2 in MPa\n",

+      "P3=1.19 # pressure at state 3 in MPa,\n",

+      "P4=1.16 # pressure at state 4 in MPa,\n",

+      "P5=1.15 # pressure at state 5 in MPa,\n",

+      "P6=P7=140 # pressure at state 6 and 7 in kPa,\n",

+      "P8=130 # pressure at state 8 in kPa,\n",

+      "T1=-10 #temperaure at state  1 in \u25e6C\n",

+      "T2=100 #temperaure at state  2 in \u25e6C\n",

+      "T3=80 #temperaure at state  3 in \u25e6C\n",

+      "T4=45 #temperaure at state  4 in \u25e6C\n",

+      "T5=40 #temperaure at state  5 in \u25e6C\n",

+      "T8=-20 #temperaure at state  8 in \u25e6C\n",

+      "q=-4 # heat transfer in kJ/Kg\n",

+      "\n",

+      "#x6=x7 quality condition given in question\n",

+      "\n",

+      "\n",

+      "# the following values are taken from table for refrigerant R-134a\n",

+      "h1=394.9 # enthalpy at state 1 in kJ/kg\n",

+      "h2=480.9 # enthalpy at state 2 in kJ/kg\n",

+      "h8=386.6 # enthalpy at state 8 in kJ/kg\n",

+      "wc=h2-h1-q # from first law\n",

+      "h5=h6=h7=256.4 # as x6=x7 and from table at state 5 in Kj/Kg\n",

+      "qL=h8-h7 # from first law \n",

+      "B=qL/wc # COP\n",

+      "print\" the COP of the plant is\",round(B,3)\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " the COP of the plant is 1.447\n"

+       ]

+      }

+     ],

+     "prompt_number": 20

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter12_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter12_6.ipynb
new file mode 100755
index 00000000..e2686e4a
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter12_6.ipynb
@@ -0,0 +1,478 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:83a0af04c694ab4b0ad1a0fe252ffa0bd8e87abc642c64323f0c2a0d18dd9f5d"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "CHAPTER13:Gas Mixtures"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.1:Pg-480"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques1\n",

+      "#Standard brayton cycle\n",

+      "\n",

+      "#1-Inlet for compressor\n",

+      "#2-Exit for compressor\n",

+      "#T-Temperature at a state\n",

+      "#P-Pressure at a state\n",

+      "T1=288.2;#K\n",

+      "P2=1000.0;#kPa\n",

+      "P1=100.0;#kPa\n",

+      "k=1.4;\n",

+      "T2=T1*(P2/P1)**(1.0-1/k);#K\n",

+      "Cp=1.004;#Specific heat at constant pressure in kJ/kg\n",

+      "wc=Cp*(T2-T1);#compressor work in kJ/kg;\n",

+      "print\" Temperature T2 =\",round(T2,2),\"K\"\n",

+      "print\" Compressor work =\",round(wc,1),\"kJ/kg\"\n",

+      "#3-Turbine Inlet\n",

+      "#4-Turbine Exit\n",

+      "P4=P1;\n",

+      "P3=P2;\n",

+      "T3=1373.2;#K\n",

+      "T4=T3*(P4/P3)**(1-1/k);#K\n",

+      "wt=Cp*(T3-T4);\n",

+      "wnet=wt-wc;\n",

+      "print\" Temperature T3 =\",round(T3,2),\" K\"\n",

+      "print\" Temperature T4 =\",round(T4),\"K\"\n",

+      "print\" Turbine work =\",round(wt,2),\"kJ/kg\"\n",

+      "print\" Net work =\",round(wt-wc,2),\"kJ/kg\"\n",

+      "#2-Also high temperature heat exchanger Inlet\n",

+      "#3-(-do-) Exit\n",

+      "qh=Cp*(T3-T2);#Heat of source in kJ/kg\n",

+      "#4-high temp heat exchanger inlet\n",

+      "#1-(-do-) Exit\n",

+      "ql=Cp*(T4-T1);#Heat of sink in kJ/kg\n",

+      "nth=wnet/qh;\n",

+      "print\" Thermal Efficiency of cycle =\",round(nth*100,2),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Temperature T2 = 556.43 K\n",

+        " Compressor work = 269.3 kJ/kg\n",

+        " Temperature T3 = 1373.2  K\n",

+        " Temperature T4 = 711.0 K\n",

+        " Turbine work = 664.6 kJ/kg\n",

+        " Net work = 395.3 kJ/kg\n",

+        " Thermal Efficiency of cycle = 48.21 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.2:Pg-481"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Calculation mistake in book\n",

+      "#ques2\n",

+      "#Standard brayton cycle\n",

+      "\n",

+      "#Calculation mistake in book\n",

+      "#1-Inlet for compressor\n",

+      "#2-Exit for compressor\n",

+      "#T-Temperature at a state\n",

+      "#P-Pressure at a state\n",

+      "T1=288.2;#K\n",

+      "P2=1000.0;#kPa\n",

+      "P1=100.0;#kPa\n",

+      "k=1.4;\n",

+      "T2s=T1*(P2/P1)**(1-1/k);#K\n",

+      "nc=.80;#Compressor Efficiency\n",

+      "T2=T1+(T2s-T1)/0.80;\n",

+      "Cp=1.004;#Specific heat at constant pressure in kJ/kg\n",

+      "wc=Cp*(T2-T1);#compressor work in kJ/kg\n",

+      "wc=round(wc)\n",

+      "print\" Temperature T2 =\",round(T2,2),\"K\"\n",

+      "print\" Compressor work =\",(wc),\"kJ/kg\"\n",

+      "#3-Turbine Inlet\n",

+      "#4-Turbine Exit\n",

+      "P4=P1;\n",

+      "P3=P2;\n",

+      "T3=1373.2;#K\n",

+      "T4s=T3*(P4/P3)**(1-1.0/k);#K\n",

+      "nt=0.85;#turbine Efficiency\n",

+      "T4=T3-(T3-T4s)*0.85;\n",

+      "wt=Cp*(T3-T4);\n",

+      "wnet=wt-wc;\n",

+      "print\" Temperature T3 =\",round(T3,1),\"K\"\n",

+      "print\" Temperature T4 =\",round(T4,1),\"K\"\n",

+      "print\" Turbine work =\",round(wt,2),\"kJ/kg\"\n",

+      "print\" Net work =\",round(wt-wc,2),\"kJ/kg\"\n",

+      "#2-Also high temperature heat exchanger Inlet\n",

+      "#3-(-do-) Exit\n",

+      "qh=Cp*(T3-T2);#Heat of source in kJ/kg\n",

+      "#4-high temp heat exchanger inlet\n",

+      "#1-(-do-) Exit\n",

+      "ql=Cp*(T4-T1);#Heat of sink in kJ/kg\n",

+      "nth=wnet/qh;\n",

+      "print\" Thermal Efficiency of cycle =\",round(nth*100),\"percent\"\n",

+      "#some answers are have acceptable difference beacause of approximization in book but here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Temperature T2 = 623.48 K\n",

+        " Compressor work = 337.0 kJ/kg\n",

+        " Temperature T3 = 1373.2 K\n",

+        " Temperature T4 = 810.5 K\n",

+        " Turbine work = 564.91 kJ/kg\n",

+        " Net work = 227.91 kJ/kg\n",

+        " Thermal Efficiency of cycle = 30.0 percent\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.3:Pg-486"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques3\n",

+      "#efficiency of the cycle\n",

+      "\n",

+      "wnet=395.2;#kJ/kg from example no 1\n",

+      "#Tx=T4\n",

+      "Tx=710.8;#K from example no 1\n",

+      "T3=1373.2;#K from example no 1\n",

+      "Cp=1.004;#specific heat in kJ/kg \n",

+      "qh=Cp*(T3-Tx);\n",

+      "nth=wnet/qh;\n",

+      "print\" Thermal efficiency =\",round(nth*100,1),\" percent\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Thermal efficiency = 59.4  percent\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.4:Pg-486"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques4\n",

+      "#Calculation of work in the given cycle\n",

+      "import math\n",

+      "R=0.287;#gas constant \n",

+      "T1=288.2;#compressor temperature K\n",

+      "T2=1373.2;#K turbine temperature K\n",

+      "#Pe/Pi=c=10, Pi/Pe=1/c from example 12.1\n",

+      "c=10.0;\n",

+      "wc=-R*T1*math.log(c);\n",

+      "print\" Isothermal work in compressor =\",round(wc,1),\"kJ/kg\"\n",

+      "wt=-R*T2*math.log(1/c);\n",

+      "print\" Isothermal work in turbine =\",round(wt,1),\"kJ/kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Isothermal work in compressor = -190.5 kJ/kg\n",

+        " Isothermal work in turbine = 907.5 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 26

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.5:Pg-491"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques5\n",

+      "#air standard cycle for jet repulsion\n",

+      "import math\n",

+      "#1-compressor inlet\n",

+      "#2-Compressor exit\n",

+      "#P-Pressure at given point\n",

+      "#T-Temperature at given point\n",

+      "P1=100;#kPa\n",

+      "P2=1000;#kPa\n",

+      "T1=288.2;#K\n",

+      "T2=556.8;#K\n",

+      "wc=269.5;#from ex 12.1 work done in compressor in kJ/kg\n",

+      "#2-Burner inlet\n",

+      "#3-Burner exit\n",

+      "P3=1000;#kPa\n",

+      "T3=1373.2;#K\n",

+      "#wc=wt\n",

+      "Cp=1.004;#specific enthalpy of heat at constant pressure in kJ/kg\n",

+      "k=1.4;\n",

+      "T4=T3-wc/Cp;\n",

+      "P4=P3*(T4/T3)**(1-1/k);\n",

+      "#from s4=s5 and h4=h5+v2/2 we get\n",

+      "T5=710.8#K, from second law\n",

+      "v=math.sqrt(2*Cp*1000*(T4-T5));#m/s\n",

+      "print\" Velocity of air leaving the nozel =\",round(v),\"m/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Velocity of air leaving the nozel = 889.0 m/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 28

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.6:Pg-494"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques6\n",

+      "#air standard refrigeration cycle\n",

+      "\n",

+      "#1-compressor inlet\n",

+      "#2-compressor exit\n",

+      "P1=100;#kPa\n",

+      "P2=500;#kPa\n",

+      "k=1.4;\n",

+      "rp=P2/P1;\n",

+      "cop=(rp**(1-1/k)-1)**-1;\n",

+      "print\" Coefficient of performance =\",round(cop,2)\n",

+      "#3-Expander inlet\n",

+      "#4-Expander exit\n",

+      "P3=P2;\n",

+      "P4=P1;\n",

+      "T3=288.23;#K, given and fixed\n",

+      "T4=T3/(P3/P4)**(1-1/k);\n",

+      "T1=253.2;#K, given\n",

+      "Cp=1.004;#Specific heat at cons pressure in kJ/kg\n",

+      "ql=Cp*(T1-T4);#heat released in kJ/kg\n",

+      "P=1#power required in kW \n",

+      "ms=P/ql;#kg/s\n",

+      "print\"  Rate at which the air enter the compressor =\",round(ms,3),\"kg/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Coefficient of performance = 1.71\n",

+        "  Rate at which the air enter the compressor = 0.014 kg/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 36

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.7:Pg-498"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques7\n",

+      "#the otto cycle\n",

+      "\n",

+      "#1-compressor inlet\n",

+      "#2-compressor exit\n",

+      "P1=100.0;#kPa\n",

+      "T1=288.2;#K\n",

+      "R=0.287;#gas constant\n",

+      "v1=R*T1/P1;#specific volume at inlet in m^3/kg\n",

+      "rv=10.0;#compression ratio given\n",

+      "k=1.4;#constant\n",

+      "T2=T1*rv**(k-1);#K\n",

+      "print\" Temperature at compressor exit, T2 =\",round(T2,1),\"K\"\n",

+      "P2=P1*rv**k;#kPa\n",

+      "print\" Pressure at compressor exit, P2 =\",round(P2/1000,2),\"MPa\"\n",

+      "v2=v1/rv;#specific heat at exit in m^3/kg\n",

+      "#23-heat addition process\n",

+      "#q23=Cv*(T3-T2)=1800 kJ/kg given\n",

+      "q23=1800.0;#kJ/kg heat addition, given\n",

+      "Cv=0.717;#specific heat at constant volume in kJ/kg\n",

+      "T3=T2+q23/Cv;#K\n",

+      "print\" Initial Temperature during heat additon process, T3 =\",round(T3,2),\"K\"\n",

+      "P3=P2*(T3/T2);#kPa\n",

+      "print\" Initial pressure during heat addition process, P3 =\",round(P3/1000,3),\"MPa\"\n",

+      "r=10.0;#k=V4/V3=P3/P4\n",

+      "T4=T3*(1/r)**(k-1);\n",

+      "print\" Final temperature during heat addition process, T4 =\",round(T4,3),\"K\";\n",

+      "P4=P3/r**k;#kPa\n",

+      "print\" Final pressure during heat addition process, P4 =\",round(P4/1000,4),\"MPa\"\n",

+      "nth=1-1/r**(k-1);#thermal efficiency\n",

+      "print\" Thermal efficiency =\",round(nth*100,1),\" percent\"\n",

+      "q41=Cv*(T1-T4);#/heat for process 4-1 in kJ/kg\n",

+      "wnet=q23+q41;\n",

+      "mep=wnet/(v1-v2);#effective mean pressure n kPa\n",

+      "print\" Mean effective pressure =\",round(mep,2),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Temperature at compressor exit, T2 = 723.9 K\n",

+        " Pressure at compressor exit, P2 = 2.51 MPa\n",

+        " Initial Temperature during heat additon process, T3 = 3234.39 K\n",

+        " Initial pressure during heat addition process, P3 = 11.223 MPa\n",

+        " Final temperature during heat addition process, T4 = 1287.632 K\n",

+        " Final pressure during heat addition process, P4 = 0.4468 MPa\n",

+        " Thermal efficiency = 60.2  percent\n",

+        " Mean effective pressure = 1455.37 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 42

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex12.8:Pg-501"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques8\n",

+      "#the diesel cycle\n",

+      "\n",

+      "#1-compressor inlet\n",

+      "#2-compressor exit\n",

+      "P1=100;#kPa\n",

+      "T1=288.2;#K\n",

+      "R=0.287;#gas constant\n",

+      "v1=R*T1/P1;#specific volume at inlet in m^3/kg\n",

+      "rv=20;#compression ratio given\n",

+      "k=1.4;#constant\n",

+      "T2=T1*rv**(k-1);#K\n",

+      "print\" Temperature at compressor exit, T2 =\",round(T2,1),\"K\"\n",

+      "P2=P1*rv**k;#kPa\n",

+      "print\" Pressure at compressor exit, P2 =\",round(P2/1000,3),\" MPa\"\n",

+      "v2=v1/rv;#specific heat at exit in m^3/kg\n",

+      "#23-heat addition process\n",

+      "#q23=Cv*(T3-T2)=1800 kJ/kg given\n",

+      "q23=1800;#kJ/kg heat addition, given\n",

+      "Cv=.717;\n",

+      "Cp=1.004;#specific heat at constant pressure in kJ/kg\n",

+      "T3=T2+q23/Cp;#K\n",

+      "print\" Initial Temperature during heat addition process, T3 =\",round(T3,2),\"K\"\n",

+      "r=T3/T2;#T3/T2=V3/V2=r\n",

+      "v3=r*v2;\n",

+      "T4=T3/(v1/v3)**(k-1);\n",

+      "print\" Final temperature during heat addition process, T4 =\",round(T4),\"K\"\n",

+      "q41=Cv*(T1-T4);#/heat for process 4-1 in kJ/kg\n",

+      "wnet=q23+q41;\n",

+      "mep=wnet/(v1-v2);#effective mean pressure in kPa\n",

+      "qh=1800;#heat transfer in kJ/kg\n",

+      "nth=wnet/qh;#thermal efficiency\n",

+      "\n",

+      "print\" Thermal efficiency =\",round(nth*100,1),\"percent\"\n",

+      "print\" Mean effective pressure =\",round(mep,2),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Temperature at compressor exit, T2 = 955.2 K\n",

+        " Pressure at compressor exit, P2 = 6.629  MPa\n",

+        " Initial Temperature during heat addition process, T3 = 2748.05 K\n",

+        " Final temperature during heat addition process, T4 = 1265.0 K\n",

+        " Thermal efficiency = 61.1 percent\n",

+        " Mean effective pressure = 1399.18 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 47

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter13_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter13_6.ipynb
new file mode 100755
index 00000000..5863fc4c
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter13_6.ipynb
@@ -0,0 +1,385 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:13df1b90ea06630bb54a561fa81d4ebbaaa2d68a225d0631603576a208acfb40"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "CHAPTER13:Gas Mixtures"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.3:Pg-533"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques3\n",

+      "#calculating humidity ratio, dew point,mass of air, mass of vapor\n",

+      "\n",

+      "r=0.70;#relative humidity\n",

+      "Pg=5.628;#saturation pressure in kPa\n",

+      "Pv=r*Pg;#vapour pressure in kPa\n",

+      "P=100;#net pressure kPa \n",

+      "Pa=P-Pv;#Partial pressure of air\n",

+      "w=0.622*Pv/Pa;#humidity ratio formula\n",

+      "V=100;#volume in m^3\n",

+      "Ra=0.287;#gas constant for water vapour\n",

+      "T=308.2;#Temperature in K\n",

+      "ma=Pa*V/(Ra*T);#mass in kg\n",

+      "mv=w*ma;#mass of vapour\n",

+      "print\" Mass of vapour is\",round(mv,3),\"Kg \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Mass of vapour is 2.77 Kg \n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.3E:Pg-533"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques3\n",

+      "#calculating humidity ratio, dew point,mass of air, mass of vapor\n",

+      "\n",

+      "r=0.70;#relative humidity\n",

+      "Pg=0.6988 ;#saturation pressure in lbf/in^2\n",

+      "Pv=r*Pg;#vapour pressure in lbf/in^2\n",

+      "P=14.7;#net pressure in lbf/in^2\n",

+      "Pa=P-Pv;#Partial pressure of air\n",

+      "w=0.622*Pv/Pa;#humidity ratio formula\n",

+      "V=2000;#volume in ft^3\n",

+      "Ra=53.34;#gas constant for water vapour\n",

+      "T=70;#Temperature in F\n",

+      "ma=Pa*V*144/(Ra*(T+480));#mass in lbm\n",

+      "mv=w*ma;#mass of vapour\n",

+      "print\" Mass of vapour is\",round(mv,2),\"lbm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Mass of vapour is 2.99 lbm\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.4:Pg-534"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques4 \n",

+      "#calculating amount of water vapour condensed on cooling\n",

+      "\n",

+      "#from example 3\n",

+      "w1=0.0255;#w1=w, humidity ratio at initial temperature\n",

+      "ma=108.6;#mass of air in kg\n",

+      "P=100;#kPa net pressure\n",

+      "#at 5 C mixture is saturated so Pv2=Pg2\n",

+      "Pg2=0.8721;\n",

+      "Pv2=Pg2;\n",

+      "w2=0.622*Pv2/(P-Pg2);\n",

+      "mc=ma*(w1-w2);\n",

+      "print\"Mass of vapour condense is \",round(mc,3),\"kg \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Mass of vapour condense is  2.175 kg \n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.4E:Pg-535"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques4 \n",

+      "#calculating amount of water vapour condensed on cooling\n",

+      "\n",

+      "#from example 3 \n",

+      "w1=0.02135;#w1=w, humidity ratio at initial temperature\n",

+      "ma=139.6;#mass of air in lbm\n",

+      "P=14.7;#net pressure in lbf/in^2\n",

+      "#at 40 degree F mixture is saturated so Pv2=Pg2\n",

+      "Pg2=0.1217 # lbf/in^2\n",

+      "Pv2=Pg2;\n",

+      "w2=0.622*Pv2/(P-Pg2);# humidity ratio at final temperature\n",

+      "mc=ma*(w1-w2); #Mass of vapour condense in lbm\n",

+      "print\"Mass of vapour condense is \",round(mc,3),\"lbm \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Mass of vapour condense is  2.256 lbm \n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.5:Pg-536"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques5\n",

+      "#calculating heat transfer per kilogram of dry air\n",

+      "\n",

+      "#1-inlet state\n",

+      "#2-Exit state\n",

+      "r1=0.80;#realtive humidity at state 1\n",

+      "Pg1=4.246;#saturation pressure of vapour in kPa\n",

+      "P1=105.0;#net pressure at state 1 in kPa\n",

+      "P2=100.0;#net pressure at state 2 in kPa\n",

+      "Pv1=r1*Pg1;#partial pressure of vapour in kPa\n",

+      "w1=0.622*Pv1/(P1-Pv1);#humidity ratio at state 1\n",

+      "r2=0.95;#relative humidity at state 2\n",

+      "Pg2=1.7051;#saturation pressure of vapour in kPa\n",

+      "Pv2=r2*Pg2;#partial pressure of vapour in kPa\n",

+      "w2=0.622*Pv2/(P2-Pv2);#humidity ratio at state 2\n",

+      "T1=30.0;#C\n",

+      "T2=15.0;#C\n",

+      "Cp=1.004;#specific heat of water vapour in kJ/kg\n",

+      "hv2=2528.9;#enthalpy of vapourisation of vapour in kJ/kg\n",

+      "hv1=2556.3;#enthalpy of vapourisation of vapour in kJ/kg\n",

+      "hl2=62.99;#enthalpy of \n",

+      "q=Cp*(T2-T1)+w2*hv2-w1*hv1+hl2*(w1-w2);#kJ/kg\n",

+      "print\" Heat transferred per unit mass =\",round(q,1),\"kJ/kg of dry air\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Heat transferred per unit mass = -41.7 kJ/kg of dry air\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.6:Pg-537"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques6\n",

+      "#calculating heat transferred in gas vapour mixture\n",

+      "\n",

+      "#n-Nitrogen\n",

+      "#v-water vapour\n",

+      "Pn2=1995;#Pressure of nitrogen in kPa\n",

+      "V=0.5;#Volume in m^3\n",

+      "Rn2=0.2968;#Gas constant for nitrogen in kJ/kg.K\n",

+      "Rv=0.4615;#gas constant for vapour\n",

+      "T1=323.2;#Temperature in K\n",

+      "T2=283.2;#Temperature in K\n",

+      "Pv1=5;#Pressure of water vapour in kPa at state 1\n",

+      "Pv2=1.2276;#Pressure of water vapour in kPa at state 2\n",

+      "mn2=Pn2*V/(Rn2*T1);#mass of nitrogen\n",

+      "mv1=Pv1*V/(Rv*T1);#mass of vapour in kg\n",

+      "mv2=Pv2*V/(Rv*T2);#mass of vapour in kg\n",

+      "ml2=mv1-mv2;#mass of liquid condensed n kg\n",

+      "uv1=2443.1;#specific internal energy of vapour in kJ/kg at state 1\n",

+      "uv2=2389.2;#specific internal energy of vapour in kJ/kg at state 2\n",

+      "ul2=42.0;#specific internal energy of liquid water in kJ/kg\n",

+      "Cv=0.745;#specific heat at constant volume in kJ/kg.K\n",

+      "Q=mn2*Cv*(T2-T1)+mv2*uv2+ml2*ul2-mv1*uv1;\n",

+      "print\"Heat transferred =\",round(Q,2),\"kJ \"\n",

+      "\n",

+      "#the answer is different in book due to intermediate approximization "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Heat transferred = -339.1 kJ \n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.7:Pg-539"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques7\n",

+      "#calculating humidity ratio and relative humidity \n",

+      "\n",

+      "#1-Inlet state\n",

+      "#2-Exit state\n",

+      "P=100;#net pressure in kPa \n",

+      "#it is steady state adiabatic process\n",

+      "#water vapour leaving is saturated so Pv2=Pg2\n",

+      "Pg2=2.339;#saturation pressure of vapour in kPa\n",

+      "Pv2=Pg2;#partial pressure of vapour\n",

+      "w2=0.622*Pv2/(P-Pg2);\n",

+      "Cpa=1.004;#specific heat n kJ/kg/K\n",

+      "T2=20;# final temp in C\n",

+      "T1=30;# initial temp in C\n",

+      "Hfg2=2454.1;#specific heat difference at state 2 in kJ/kg\n",

+      "hv1=2556.3;#enthalpy of water vapour at state 1 in kJ/kg\n",

+      "hl2=83.96;#enthalpy of liquid water in kJ/kg\n",

+      "w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);\n",

+      "print \" Relative humidity =\",round(w1,4)\n",

+      "#also w1=0.622*Pv1/(100-Pv2)\n",

+      "Pv1=100*w1/(0.622+w1);\n",

+      "Pg1=4.246;#saturation pressure at state 1 in kPa\n",

+      "r=Pv1/Pg1;#humidity ratio\n",

+      "print\" Humidity ratio =\",round(r,4)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Relative humidity = 0.0107\n",

+        " Humidity ratio = 0.3993\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex13.7E:Pg-540"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques7\n",

+      "#calculating humidity ratio and relative humidity \n",

+      "\n",

+      "#1-Inlet state\n",

+      "#2-Exit state\n",

+      "P=14.7;#net pressure in lbf/in^2\n",

+      "#it is steady state adiabatic process\n",

+      "#water vapour leaving is saturated so Pv2=Pg2\n",

+      "Pg2=0.3632;#saturation pressure of vapour in lbf/in^2\n",

+      "Pv2=Pg2;#partial pressure of vapour\n",

+      "w2=0.622*Pv2/(P-Pg2);\n",

+      "Cpa=0.24;#specific heat n Btu/lbm/F\n",

+      "T2=70;# final temp in F\n",

+      "T1=84;# initial temp in F\n",

+      "Hfg2=1054.0;#specific heat difference at state 2 in Btu/lbm\n",

+      "hv1=1098.1;#enthalpy of water vapour at state 1 in Btu/lbm\n",

+      "hl2=38.1;#enthalpy of liquid water in Btu/lbm\n",

+      "w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);\n",

+      "print \" Relative humidity =\",round(w1,4)\n",

+      "#also w1=0.622*Pv1/(100-Pv2)\n",

+      "Pv1=14.7*w1/(0.622+w1);\n",

+      "Pg1=0.584;#saturation pressure at state 1 in lbf/in^2\n",

+      "r=Pv1/Pg1;#humidity ratio\n",

+      "print\" Humidity ratio =\",round(r,4)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Relative humidity = 0.0125\n",

+        " Humidity ratio = 0.4958\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter14_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter14_6.ipynb
new file mode 100755
index 00000000..8c5a5922
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter14_6.ipynb
@@ -0,0 +1,325 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:4d90868c7316d94fc2ffdc05c5691234f77f3a832d71da7eb2108eaff0a215bc"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter14:THERMODYNAMIC RELATIONS"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.1:Pg-567"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques1\n",

+      "#to determine the sublimation pressure of water\n",

+      "import math\n",

+      "#from table in appendix B.1.5\n",

+      "T1=213.2;#K, Temperature at state 1\n",

+      "P2=0.0129;#kPa, pressure at state 2\n",

+      "T2=233.2;#K, Temperature at state 2\n",

+      "hig=2838.9;#kJ/kg, enthalpy of sublimation \n",

+      "R=.46152;#Gas constant \n",

+      "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n",

+      "P1=P2*math.exp(-hig/R*(1/T1-1/T2));\n",

+      "print\" Sublimation Pressure \",round(P1,5),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Sublimation Pressure  0.00109 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.1E:Pg-567\n"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques1\n",

+      "#to determine the sublimation pressure of water\n",

+      "import math\n",

+      "#from table in appendix B.1.5\n",

+      "T1=-70+460.7;# R, Temperature at state 1\n",

+      "P2=0.0019 # lbf/in^2 pressure at state 2\n",

+      "T2=-40+460.7;# R, Temperature at state 2\n",

+      "hig=1218.7;#Btu/lbm, enthalpy of sublimation \n",

+      "R=85.67;#Gas constant \n",

+      "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n",

+      "P1=P2*math.exp(-hig*778/R*(1/T1-1/T2));\n",

+      "print\" Sublimation Pressure \",round(P1,5),\"lbf/in^2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Sublimation Pressure  0.00025 lbf/in^2\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.4:Pg-579"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques4\n",

+      "#Volume expansivity, Isothermal and Adiabatic compressibility\n",

+      "\n",

+      "#known data\n",

+      "ap=5*10**-5;#K^-1 Volume expansivity\n",

+      "bt=8.6*10**-12;#m^2/N, Isothermal compressibility\n",

+      "v=0.000114;#m^3/kg, specific volume\n",

+      "P2=100*10**6;#pressure at state 2 in kPa\n",

+      "P1=100;#pressure at state 1 in kPa\n",

+      "w=-v*bt*(P2**2-P1**2)/2;#work done in J/kg\n",

+      "#q=T*ds and ds=-v*ap*(P2-P1)\n",

+      "#so q=-T*v*ap*(P2-P1)\n",

+      "T=288.2;#Temperature in K\n",

+      "q=-T*v*ap*(P2-P1);#heat in J/kg\n",

+      "du=q-w;#change in internal energy in J/kg\n",

+      "print\" Change in internal energy =\",round(du,3),\"J/kg\"\n",

+      "\n",

+      "#the answer is correct within given limts\n",

+      " "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Change in internal energy = -159.372 J/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.5:Pg-586"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques5\n",

+      "#adiabatic steady state processes\n",

+      "\n",

+      "#from table A.2\n",

+      "P1=20;#pressure at state 1 in MPa\n",

+      "P2=2;#pressure at state 2 in MPa\n",

+      "T1=203.2;#Temperature at state 1 in K\n",

+      "Pr1=P1/3.39;#Reduced pressure at state 1\n",

+      "Pr2=P2/3.39;#Reduced pressure at state 2\n",

+      "Tr1=T1/126.2;#Reduced temperature\n",

+      "#from compressibility chart h1*-h1=2.1*R*Tc\n",

+      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",

+      "#h2*-h2=0.5*R*Tc\n",

+      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc\n",

+      "R=0.2968;#gas constant for given substance\n",

+      "Tc=126.2;#K, Constant temperature\n",

+      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",

+      "T2=146;#temperature at state 2\n",

+      "dh=-1.6*R*Tc+Cp*(T1-T2);#\n",

+      "print\" Enthalpy change =\",round(dh,3),\"kJ/kg \\n\"\n",

+      "print\" Since Enthalpy change is nearly \",-round(dh),\"kJ/kg so Temperature =\",round(T2,3),\"K\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Enthalpy change = -0.35 kJ/kg \n",

+        "\n",

+        " Since Enthalpy change is nearly  0.0 kJ/kg so Temperature = 146.0 K\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.6:Pg-589"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques6\n",

+      "#isothermal steady state processes\n",

+      "import math\n",

+      "#from table A.2\n",

+      "P1=8;#pressure at state 1 in MPa\n",

+      "P2=0.5;#pressure at state 2 in MPa\n",

+      "T1=150.0;#Temperature at state 1 in K\n",

+      "Pr1=P1/3.39;#Reduced pressure at state 1\n",

+      "Pr2=P2/3.39;#Reduced pressure at state 2\n",

+      "Tr1=T1/126.2;#Reduced temperature\n",

+      "T2=125.0;#temperature at state 2\n",

+      "#from compressibility chart h1*-h1=2.1*R*Tc\n",

+      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",

+      "#h2*-h2=0.5*R*Tc\n",

+      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc\n",

+      "R=0.2968;#gas constant for given substance\n",

+      "Tc=126.2;#K, Constant temperature\n",

+      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",

+      "dh=(2.35)*R*Tc+Cp*(T2-T1);#\n",

+      "print\" Enthalpy change =\",round(dh),\"kJ/kg\"\n",

+      "#change in entropy \n",

+      "#ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)\n",

+      "#s1*-s1=1.6*R\n",

+      "#s2*-s2=0.1*R\n",

+      "#s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)\n",

+      "#so\n",

+      "ds=1.6*R-0.1*R+Cp*math.log(T2/T1)-R*math.log(P2/P1);\n",

+      "print\" Entropy Change =\",round(ds,3),\"kJ/kg.K \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Enthalpy change = 62.0 kJ/kg\n",

+        " Entropy Change = 1.078 kJ/kg.K \n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.7:Pg-596"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques7\n",

+      "#percent deviation using specific volume calculated by kays rule and vander waals rule\n",

+      "import math\n",

+      "\n",

+      "#a-denotes C02\n",

+      "#b-denotes CH4\n",

+      "T=310.94;#Temperature of mixture K\n",

+      "P=86.19;#Pressure of mixture in MPa\n",

+      "#Tc- critical Temperature\n",

+      "#Pc-critical pressure\n",

+      "Tca=304.1;#K\n",

+      "Tcb=190.4;#K\n",

+      "Pca=7.38;#MPa\n",

+      "Pcb=4.60;#MPa\n",

+      "Ra=0.1889;#gas constant for a in kJ/kg.K\n",

+      "Rb=0.5183;#gas constant for b in kJ/kg.K\n",

+      "xa=0.8;#fraction of CO2\n",

+      "xb=0.2;#fraction of CH4\n",

+      "Rm=xa*Ra+xb*Rb;#mean gas constant in kJ/kg.K\n",

+      "Ma=44.01;#molecular mass of a\n",

+      "Mb=16.043;#molecular mass of b\n",

+      "#1.Kay's rule\n",

+      "ya=xa/Ma/(xa/Ma+xb/Mb);#mole fraction of a\n",

+      "yb=xb/Mb/(xa/Ma+xb/Mb);#mole fraction of b\n",

+      "Tcm=ya*Tca+yb*Tcb;#mean critical temp in K\n",

+      "Pcm=ya*Pca+yb*Tcb;#mean critical pressure n MPa\n",

+      "#therefore pseudo reduced property of mixture\n",

+      "Trm=T/Tcm;\n",

+      "Prm=P/Pcm;\n",

+      "Zm=0.7;#Compressiblity from generalised compressibility chart\n",

+      "vc=Zm*Rm*T/P/1000;#specific volume calculated in m^3/kg\n",

+      "ve=0.0006757;#experimental specific volume in m^3/kg\n",

+      "pd1=(ve-vc)/ve*100;#percent deviation\n",

+      "print\" Percentage deviation in specific volume using Kays rule =\",round(pd1,1),\"percent \\n\"\n",

+      "\n",

+      "#2. using vander waals equation\n",

+      "#values of vander waals constant\n",

+      "Aa=27*(Ra**2)*(Tca**2)/(64*Pca*1000);\n",

+      "Ba=Ra*Tca/(8*Pca*1000);\n",

+      "Ab=27*(Rb**2)*(Tcb**2)/(64*Pcb*1000);\n",

+      "Bb=Rb*Tcb/(8*Pcb*1000);\n",

+      "#mean vander waals constant\n",

+      "Am=(xa*math.sqrt(Aa)+xb*math.sqrt(Ab))**2;\n",

+      "Bm=(xa*Ba+xb*Bb);\n",

+      "#using vander waals equation we get cubic equation \n",

+      "#solving we get\n",

+      "vc=0.0006326;#calculated specific volume in m^3/kg\n",

+      "pd2=((ve-vc)/ve)*100;\n",

+      "print\" Percentage deviation in specific volume using vander waals eqn =\",round(pd2,1),\"percent\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Percentage deviation in specific volume using Kays rule = 4.8 percent \n",

+        "\n",

+        " Percentage deviation in specific volume using vander waals eqn = 6.4 percent\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter15_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter15_6.ipynb
new file mode 100755
index 00000000..1c9182a7
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter15_6.ipynb
@@ -0,0 +1,921 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:4afaac3b11e3c61840a392c8cec12435d16ae2f3d2cdfe278957aaca509d72f5"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter15:CHEMICAL REACTIONS"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.1:Pg-621"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques1\n",

+      "#theoratical air-fuel ratio for combustion of octane\n",

+      "#combustion equation is\n",

+      "#C8H18 + 12.5O2 + 12.5(3.76) N2 \u2192 8 CO2 + 9H2O + 47.0N2\n",

+      "rm=(12.5+47.0)/1;#air fuel ratio on mole basis\n",

+      "rma=rm*28.97/114.2;#air fuel ratio on mass basis;\n",

+      "print \"Theoratical air fuel ratio on mass basis is\",round(rma),\"kg air/kg fuel\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Theoratical air fuel ratio on mass basis is 15.0 kg air/kg fuel\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.2:Pg-622"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 2\n",

+      "# molal analysis of the products of combustion\n",

+      "# combustion equation is \n",

+      "# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 \u2192 8 CO2 + 9H2O + 12.5O2 + 94.0N2\n",

+      "nCO2=8 # number of moles of CO2 \n",

+      "nH2O=9 # number of moles of H2O\n",

+      "nO2=12.5 # number of moles of O2\n",

+      "nN2=94.0 # number of moles of N2\n",

+      "moles=nCO2+nH2O+nO2+nN2 # total moles\n",

+      "# molal analysis of products\n",

+      "xCO2=nCO2*100/moles \n",

+      "xH2O=nH2O*100/moles\n",

+      "xO2=nO2*100/moles\n",

+      "xN2=nN2*100/moles\n",

+      "print \" CO2 =\",round(xCO2,2)\n",

+      "print \"\\n H2O=\",round(xH2O,2)\n",

+      "print \"\\n O2=\",round(xO2,2)\n",

+      "print \"\\n N2=\",round(xN2,2)\n",

+      "print \"\\n The dew point temp. corresponding to partial pressure of water i.e \",round(xH2O,2),\"KPa is 39.7 degree C\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " CO2 = 6.48\n",

+        "\n",

+        " H2O= 7.29\n",

+        "\n",

+        " O2= 10.12\n",

+        "\n",

+        " N2= 76.11\n",

+        "\n",

+        " The dew point temp. corresponding to partial pressure of water i.e  7.29 KPa is 39.7 degree C\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.2E:Pg-622"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 2\n",

+      "# molal analysis of the products of combustion\n",

+      "# combustion equation is \n",

+      "# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 \u2192 8 CO2 + 9H2O + 12.5O2 + 94.0N2\n",

+      "pressure=14.7 # in lbf/in^2\n",

+      "nCO2=8 # number of moles of CO2 \n",

+      "nH2O=9 # number of moles of H2O\n",

+      "nO2=12.5 # number of moles of O2\n",

+      "nN2=94.0 # number of moles of N2\n",

+      "moles=nCO2+nH2O+nO2+nN2 # total moles\n",

+      "# molal analysis of products\n",

+      "xCO2=nCO2*100/moles \n",

+      "xH2O=nH2O*100/moles\n",

+      "xO2=nO2*100/moles\n",

+      "xN2=nN2*100/moles\n",

+      "print \" CO2 =\",round(xCO2,2)\n",

+      "print \"\\n H2O=\",round(xH2O,2)\n",

+      "print \"\\n O2=\",round(xO2,2)\n",

+      "print \"\\n N2=\",round(xN2,2)\n",

+      "PressureO2=pressure*xH2O/100 # partial pressure of O2\n",

+      "print \"\\n The dew point temp. corresponding to partial pressure of water i.e \",round(PressureO2,3),\"lbf/in^2 is 104 F degree C\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " CO2 = 6.48\n",

+        "\n",

+        " H2O= 7.29\n",

+        "\n",

+        " O2= 10.12\n",

+        "\n",

+        " N2= 76.11\n",

+        "\n",

+        " The dew point temp. corresponding to partial pressure of water i.e  1.071 lbf/in^2 is 104 F degree C\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.3:Pg-623"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 3\n",

+      "# to determine the air\u2013fuel ratio on a volumetric basis\n",

+      "# the combustion equation is\n",

+      "# 0.14H2+0.27CO+0.03CH4+0.006O2+0.509N2+0.045CO2+0.259O2+0.259(3.76)N2\u21920.20H2O+0.345CO2+1.482 N2\n",

+      "mH2=2 # molar mass of H2\n",

+      "mCO=28 # molar mass of CO\n",

+      "mCH4=16 # molar mass of CH4\n",

+      "mO2=32 # molar mass of O2\n",

+      "mN2=28 # molar mass of N2\n",

+      "mCO2=44 # molar mass of CO2\n",

+      "\n",

+      "\n",

+      "kmolair=0.259 # moles of air in mixture using combustion equation\n",

+      "kmolfuel=0.21 # moles of fuel in mixture using combustion equation\n",

+      "ratioth=kmolair/kmolfuel # theoratical ratio\n",

+      "\n",

+      "# For 20% excess air\n",

+      "vratio=ratioth*(1+0.2) # new ratio for excess air on volume basis\n",

+      "\n",

+      "print \" The A/F ratio on volumetric basis is \",round(vratio,2)\n",

+      "mratio=vratio*28.97 /(0.14*mH2+0.27*mCO+0.03*mCH4+0.006*mO2+0.509*mN2+0.045*mCO2)\n",

+      "print \"\\n The A/F ratio on mass basis is \",round(mratio,2),\"kg air/kg fuel\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The A/F ratio on volumetric basis is  1.48\n",

+        "\n",

+        " The A/F ratio on mass basis is  1.73 kg air/kg fuel\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.4:Pg-624"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 4\n",

+      "# determining the air\u2013fuel ratio and the percent theoretical air\n",

+      "CO2=10.00 # percentage\n",

+      "O2=2.37 # percentage\n",

+      "CO=0.53 # percentage\n",

+      "N2=87.10 # percentage\n",

+      "mCH4=16 # molecular mass of CH4\n",

+      "# using this data the combustion equation is\n",

+      "# a CH4 + b O2 + c N2 \u2192 10.0 CO2 + 0.53 CO + 2.37 O2 + d H2O + 87.1N2\n",

+      "# on balancing we get\n",

+      "# 10.53 CH4 + 23.16 O2 + 87.1N2 \u219210.0 CO2 + 0.53 CO + 2.37 O2 + 21.06 H2O + 87.1N2\n",

+      "# balanced combustion equation for per Kmole of fuel is:\n",

+      "# CH4 + 2.2O2 + 8.27 N2 \u2192 0.95 CO2 + 0.05 CO + 2H2O + 0.225 O2 + 8.27 N2\n",

+      "molesAir=2.2+8.27 # moles of air\n",

+      "AF= 28.97*molesAir/mCH4 # air fuel ratio on mass basis\n",

+      "print \"\\n The A/F ratio on mass basis is \",round(AF,2),\"kg air/kg fuel\"\n",

+      "# Theoritical combustion equation:\n",

+      "# CH4 + 2O2 + 2(3.76)N2 \u2192 CO2 + 2H2O + 7.52 N2\n",

+      "molesAirth=2+2*3.76 # moles of air(theoritical)\n",

+      "AFth= 28.97*molesAirth/mCH4 # air fuel ratio on mass basis (theoritical)\n",

+      "print \"\\n The A/F ratio (theoritical) on mass basis is \",round(AFth,2),\"kg air/kg fuel\"\n",

+      "print \"\\n The percent theoretical air is\",round(AF*100/AFth,1)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " The A/F ratio on mass basis is  18.96 kg air/kg fuel\n",

+        "\n",

+        " The A/F ratio (theoritical) on mass basis is  17.24 kg air/kg fuel\n",

+        "\n",

+        " The percent theoretical air is 110.0\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.5:Pg-625"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 5\n",

+      "\n",

+      "Sulfur=0.6 # percent by mass\n",

+      "Hydrogen=5.7 # percent by mass\n",

+      "Carbon=79.2 # percent by mass\n",

+      "Oxygen=10.0 # percent by mass\n",

+      "Nitrogen=1.5 # percent by mass\n",

+      "Ash=3.0 # percent by mass\n",

+      "mS=32.0 # molecular mass \n",

+      "mH2=2.0  # molecular mass\n",

+      "mC=12.0  # molecular mass\n",

+      "mO2=32.0  # molecular mass\n",

+      "mN2=28.0  # molecular mass\n",

+      "# molal composition per 100 Kg of fuel:\n",

+      "S=Sulfur/mS # molal composition \n",

+      "H=Hydrogen/mH2 # molal composition\n",

+      "C=Carbon/mC # molal composition\n",

+      "O2=Oxygen/mO2 # molal composition\n",

+      "N2=Nitrogen/mN2 # molal composition\n",

+      "\n",

+      "#The combustion equation becomes \n",

+      "# 0.02 S + 0.02 O2 \u21920.02 SO2\n",

+      "# 2.85 H2 + 1.42 O2 \u21922.85 H2O\n",

+      "# 6.60 C + 6.60 O2 \u21926.60 CO2\n",

+      "\n",

+      "O2req=0.02+1.42+6.60 # kmol O2 required/100 kg fuel\n",

+      "O2present=O2req-O2 # kmol O2 from air/100 kg fuel\n",

+      "AFtheo= (O2present+O2present*3.76)*28.97/100 # AF ration theoritical kg air/kg fuel\n",

+      "print \"\\n The A/F ratio (theoritical) on mass basis is \",round(AFtheo,2),\"kg air/kg fuel\"\n",

+      "print \"\\n The A/F ratio for 30% excess air on mass basis is \",round(AFtheo*1.3,2),\"kg air/kg fuel\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " The A/F ratio (theoritical) on mass basis is  10.66 kg air/kg fuel\n",

+        "\n",

+        " The A/F ratio for 30% excess air on mass basis is  13.85 kg air/kg fuel\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.6:Pg-629"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques6\n",

+      "# determining heat transfer per kilomole of fuel entering combustion chamber\n",

+      "\n",

+      "#1-CH4\n",

+      "#2-CO2\n",

+      "#3-H2O\n",

+      "#hf-standard enthalpy of given substance\n",

+      "hf1=-74.873;#kJ\n",

+      "hf2=-393.522;#kJ\n",

+      "hf3=-285.830;#kJ\n",

+      "Qcv=hf2+2*hf3-hf1;#kJ\n",

+      "print \"Heat transfer per kilomole of fuel entering combustion chamber is\",round(Qcv,3),\"kJ\"\n",

+      "#the answers in the book is different as they have not printed the decimals in values"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Heat transfer per kilomole of fuel entering combustion chamber is -890.309 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.7:Pg-631"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques7\n",

+      "#calculating enthalpy of water at given pressure and temperature\n",

+      "\n",

+      "#1.Assuming steam to be an ideal gas with value of Cp\n",

+      "T1=298.15;#Initial temperature in K\n",

+      "T2=573.15;#final temperature in K\n",

+      "T=(T1+T2)/2;#average temperature in K\n",

+      "Cp=1.79+0.107*T/1000+0.586*(T/1000)**2-.20*(T/1000)**3;#specific heat at constant pressure in kj/kg.K\n",

+      "M=18.015;#mass in kg\n",

+      "dh=M*Cp*(T2-T1);#enthalpy change in kJ/kmol\n",

+      "ho=-241.826;#enthalpy at standard temperature and pressure in kJ/mol\n",

+      "htp1=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

+      "print \" 1. Enthalpy of water at given pressure and temperature using value of Cp =\",round(htp1,3),\"kJ/kmol\"\n",

+      "\n",

+      "#2..Assuming steam to be an ideal gas with value for dh\n",

+      "dh=9359;#enthalpy change from table A.9 in kJ/mol\n",

+      "htp2=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

+      "print \" 2. Enthalpy of water at given pressure and temperature assuming value od dh =\",round(htp2,3),\"kJ/kmol \"\n",

+      "\n",

+      "#3. Using steam table\n",

+      "dh=M*(2977.5-2547.2);#enthalpy change for gases in kJ/mol\n",

+      "htp3g=dh/1000+ho;\n",

+      "dh=M*(2977.5-104.9);#enthalpy change for liquid in kJ/mol\n",

+      "hl=-285.830;#standard enthalpy for liquid in kJ/kmol\n",

+      "htp31=hl+dh/1000.0;#enthalpy at given temp and pressure in kJ/kmol\n",

+      "print \" 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp31,3),\"kJ/kmol \"\n",

+      "print \" 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp3g,3),\"kJ/kmol \"\n",

+      "#4.using generalised charts\n",

+      "#htp=ho-(h2*-h2)+(h2*-h1*)+(h1*-h1);\n",

+      "#h2*-h2=Z*R*Tc,\n",

+      "#h2*-h1*=9539 kJ/mol, from part 2\n",

+      "#h1*-h1=0 ,as ideal gas \n",

+      "Z=0.21;#from chart\n",

+      "R=8.3145;#gas constant in SI units\n",

+      "Tc=647.3;#critical temperature in K\n",

+      "htp4=ho+9539/1000-Z*R*Tc/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

+      "print \" 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart = \",round(htp4,3),\"kJ/kmol\"\n",

+      "#the answers in book are different as they have not printed the decimals in values"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " 1. Enthalpy of water at given pressure and temperature using value of Cp = -232.258 kJ/kmol\n",

+        " 2. Enthalpy of water at given pressure and temperature assuming value od dh = -232.826 kJ/kmol \n",

+        " 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.08 kJ/kmol \n",

+        " 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.074 kJ/kmol \n",

+        " 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart =  -233.956 kJ/kmol\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.8:Pg-632"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 8\n",

+      "# Determine the heat transfer\n",

+      "airth=400 # % excess theoritical air\n",

+      "Ti=25 # Entry temp in degree celsius\n",

+      "To=900 # Exit temp in Kelvin\n",

+      "mf=0.25 # specific fuel consumption in kg/s of fuel per megawatt output\n",

+      "# The combustion equation is:\n",

+      "# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 \u2192 8CO2 + 9H2O + 37.5O2 + 188.0N2\n",

+      "hfC8H18=-250105 # is the enthalpy of fuel in kJ/kmol \n",

+      "nCO2=8 # moles\n",

+      "hfCO2=-393522 # enthalpy in kJ/kmol\n",

+      "DelhCO2=28030 # enthalpy in kJ/kmol\n",

+      "mC8H18=114.23 # molecular mass\n",

+      "nH2O=9 # moles\n",

+      "hfH2O=-241826 # enthalpy in kJ/kmol\n",

+      "DelhH2O=21937 # enthalpy in kJ/kmol\n",

+      "DelhO2=19241 # enthalpy in kJ/kmol\n",

+      "nO2=37.5 # moles\n",

+      "DelhN2=18225 # enthalpy in kJ/kmol\n",

+      "nN2=188.0 # moles\n",

+      "Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2\n",

+      "\n",

+      "Wcv=1000*mC8H18/mf # work in kJ/kmol fuel\n",

+      "print \" The work is \",Wcv,\"kJ/kmol fuel\"\n",

+      "\n",

+      "Qcv=Eproduct+Wcv-hfC8H18 # using first law\n",

+      "print \" The Heat transfer is \",Qcv,\"kJ/kmol fuel\"\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The work is  456920.0 kJ/kmol fuel\n",

+        " The Heat transfer is  -48074.5 kJ/kmol fuel\n"

+       ]

+      }

+     ],

+     "prompt_number": 27

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.8E:Pg-633"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 8\n",

+      "# Determine the heat transfer\n",

+      "airth=400 # % excess theoritical air\n",

+      "Ti=77 # Entry temp in F\n",

+      "To=1100 # Exit temp in F\n",

+      "mf=1 # specific fuel consumption in lb of fuel per horsepower-hour\n",

+      "# The combustion equation is:\n",

+      "# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 \u2192 8CO2 + 9H2O + 37.5O2 + 188.0N2\n",

+      "hfC8H18=-107526 # is the enthalpy of fuel in Btu/lb.mol\n",

+      "nCO2=8 # moles\n",

+      "hfCO2=-169184 # enthalpy in Btu/lb.mol\n",

+      "DelhCO2=11391 # enthalpy in Btu/lb.mol\n",

+      "mC8H18=114.23 # molecular mass\n",

+      "nH2O=9 # moles\n",

+      "hfH2O=-103966 # enthalpy in Btu/lb.mol\n",

+      "DelhH2O=8867 # enthalpy in Btu/lb.mol\n",

+      "DelhO2=7784 # enthalpy in Btu/lb.mol\n",

+      "nO2=37.5 # moles\n",

+      "DelhN2=7374 # enthalpy in Btu/lb.mol\n",

+      "nN2=188.0 # moles\n",

+      "Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2\n",

+      "\n",

+      "Wcv=2544*mC8H18 # work in kJ/kmol fuel\n",

+      "print \" The work is \",Wcv,\"Btu/lb.mol fuel\"\n",

+      "\n",

+      "Qcv=Eproduct+Wcv-hfC8H18 # using first law\n",

+      "print \" The Heat transfer is \",Qcv,\"Btu/lb.mol fuel\"\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The work is  290601.12 Btu/lb.mol fuel\n",

+        " The Heat transfer is  -41895.88 Btu/lb.mol fuel\n"

+       ]

+      }

+     ],

+     "prompt_number": 31

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.9:Pg-634"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques 9\n",

+      "\n",

+      "# The combustion equation is\n",

+      "# C2H4 + 3O2 \u2192 2 CO2 + 2H2O(g)\n",

+      "hfC2H4=52467 # enthalpy in KJ/mol\n",

+      "R=8.314 # gas constant\n",

+      "T=298.2 # temperature in Kelvin\n",

+      "hfCO2=-393522 # enthalpy in kJ/kmol\n",

+      "DelhCO2=12906 # enthalpy in kJ/kmol\n",

+      "hfH2O=-241826 # enthalpy in kJ/kmol\n",

+      "DelhH2O=10499 # enthalpy in kJ/kmol\n",

+      "Ureactant=hfC2H4-4*R*T # energy of reactant in KJ\n",

+      "Uproduct=2*(hfCO2+DelhCO2)+2*(hfH2O+DelhH2O) # energy of product in KJ\n",

+      "Q=Uproduct-Ureactant # using first law\n",

+      "print \"The Heat transfer is \",Q,\"kJ\"\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Heat transfer is  -1266436.0608 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 33

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.10:Pg-638"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques 10\n",

+      "\n",

+      "# The combustion equation is\n",

+      "# C3H8 + 5O2 \u2192 3 CO2 + 4H2O\n",

+      "hfC3H8g=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10\n",

+      "hfC3H8l=hfC3H8g-44.097*370 # enthalpy of C3H8 liquid as enthalpy of evaporation is given in question\n",

+      "hfH2Ol=285830 # enthalpy of water in KJ/mol\n",

+      "hfH2Og=241826 # enthalpy in KJ/mol\n",

+      "hfCO2=393522 # enthalpy in KJ/mol\n",

+      "nO2=5 # moles\n",

+      "nCO2=3 # moles\n",

+      "nH2O=4 # moles\n",

+      "nC3H8=1 # moles\n",

+      "# part(1) : liquid propane-liquid water\n",

+      "hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8l)/44.097 # higher heating value of liquid propane \n",

+      "print \"The higher heating value of liquid propane is\",round(hrP,2),\"KJ/kg\"\n",

+      "\n",

+      "# part(2) : Liquid propane\u2013gaseous water\n",

+      "\n",

+      "hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8l)/44.097 # lower heating value of liquid propane \n",

+      "print \"The lower heating value of liquid propane is\",round(hrP,2),\"KJ/kg\"\n",

+      "\n",

+      "# part(3) :Gaseous propane\u2013liquid water\n",

+      "hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8g)/44.097 # higher heating value of gaseous propane \n",

+      "print \"The higher heating value of gaseous propane is\",round(hrP,2),\"KJ/kg\"\n",

+      "\n",

+      "# part(4) :Gaseous propane\u2013Gaseous water\n",

+      "hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8g)/44.097 # lower heating value of gaseous propane \n",

+      "print \"The lower heating value of gaseous propane is\",round(hrP,2),\"KJ/kg\"\n",

+      "\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The higher heating value of liquid propane is -120215.89 KJ/kg\n",

+        "The lower heating value of liquid propane is 51434.02 KJ/kg\n",

+        "The higher heating value of gaseous propane is 55055.58 KJ/kg\n",

+        "The lower heating value of gaseous propane is 51064.02 KJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.11:Pg-638"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques 10\n",

+      "# The combustion equation is\n",

+      "# C3H8(g) + 5O2 \u2192 3 CO2 + 4H2O(g)\n",

+      "hfC3H8=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10\n",

+      "Cp=2.1 # in KJ/Kg.K\n",

+      "T1=500 # temp in Kelvin\n",

+      "T2=298.2 # temp in Kelvin\n",

+      "DelhO2=6086 # enthalpy of O2 in KJ/mol\n",

+      "hfCO2=-393522 # enthalpy in KJ/mol\n",

+      "DelhCO2=8305 # enthalpy of CO2 in KJ/mol\n",

+      "hfH2O=-241826 # enthalpy of H2O in KJ/mol\n",

+      "DelhH2O=6922 # enthalpy in KJ/mol\n",

+      "nO2=5 # moles\n",

+      "nCO2=3 # moles\n",

+      "nH2O=4 # moles\n",

+      "nC3H8=1 # moles\n",

+      "\n",

+      "hR500= hfC3H8+Cp*44.097*(T1-T2)+nO2*DelhO2  # in KJ/mol\n",

+      "hP500= nCO2*(hfCO2+DelhCO2)+ nH2O*(DelhH2O+hfH2O) # in KJ/mol\n",

+      "hRP500=hP500-hR500 #  in KJ/mol\n",

+      "Hrp500=hRP500/44.097 # KJ/Kg\n",

+      "\n",

+      "print \" The enthalpy of combustion of gaseous propane at 500 K \",round(Hrp500)\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The enthalpy of combustion of gaseous propane at 500 K  -46273.0\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.12:Pg-639"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 12\n",

+      "# The question is solved by Trial and Error and  using the table , no need for Calculation Thus not solved in Python"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.13:Pg-645"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 13\n",

+      "# The equation for this chemical reaction is\n",

+      "# C2H4(g) + 3(4) O2 + 3(4)(3.76) N2 \u2192 2 CO2+2H2O(g) + 9O2 + 45.1 N2\n",

+      "T=25+273.15 # temp in degree kelvin\n",

+      "P=0.1 # pressure in Mpa\n",

+      "hfCO2=-393522 # enthalpy in KJ/mol\n",

+      "hfH2O=-241826 # enthalpy of H2O in KJ/mol\n",

+      "HfC2H4=52467 # enthalpy of C2H4 in KJ/mol\n",

+      "hfO2=0 # enthalpy of formation in KJ/mol\n",

+      "SCO2=213.795 # kJ/kmol fuel\n",

+      "SH2O=188.243 # kJ/kmol fuel\n",

+      "SC2H4=219.330 # kJ/kmol fuel\n",

+      "SO2=205.148 # kJ/kmol fuel\n",

+      "DelH=2*hfCO2+2*hfH2O-HfC2H4-3*hfO2 # in KJ/mol\n",

+      "DelS=2*SCO2+2*SH2O-SC2H4-3*SO2 # kJ/kmol fuel\n",

+      "DelG=DelH-T*(DelS) # Gibbs Energy\n",

+      "Wrev=-DelG/28.054 # reversible work in kJ/kg\n",

+      "print \"The reversible work is \",round(Wrev,2),\"kJ/kg \"\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The reversible work is  46838.61 kJ/kg \n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.14:Pg-647"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques \n",

+      "# The question is solved by Trial and Error and  using the table , no need for Calculation Thus not solved in Python"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.15:Pg-649"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques15\n",

+      "#calculatng reversible elecromotive force \n",

+      "\n",

+      "#1-H2O\n",

+      "#2-H2\n",

+      "#3-O2\n",

+      "#hf-standard enthalpy \n",

+      "#sf-standard entropy\n",

+      "hf1=-285.830;#kJ\n",

+      "hf2=0;#kJ\n",

+      "hf3=0;#kJ\n",

+      "sf1=69.950;#kJ/K\n",

+      "sf2=130.678;#kJ/K\n",

+      "sf3=205.148;#kJ/K\n",

+      "dH=2*hf1-2*hf2-hf3;#change in enthalpy in kJ\n",

+      "dS=2*sf1-2*sf2-sf3;#change in entropy in kJ/K\n",

+      "T=298.15;#temperature in K\n",

+      "dG=dH-T*dS/1000;#change in gibbs free energy in kJ\n",

+      "E=-dG*1000/(96485*4);#emf in V\n",

+      "print\" Reversible electromotive Force =\",round(E,3),\" V\"\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Reversible electromotive Force = 1.229  V\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.16:Pg-654"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 16\n",

+      "# to Calculate the combustion efficiency\n",

+      "Tair = 400  # temp in K\n",

+      "Vair = 100 # velocity in m/s\n",

+      "Tfuel = 50 # temp of fuel in \u25e6C\n",

+      "Tproducts = 1100 # temp of product in K\n",

+      "Vproducts = 150 # velocity of product in m/s\n",

+      "FAactual = 0.0211 # air fuel ratio kg fuel/kg air\n",

+      "\n",

+      "# HR + KER= \u2212243737 + 14892nO2 \n",

+      "# HP + KEP= \u22125 068 599 + 120 853nO2 \n",

+      "# using these two equations :\n",

+      "no2=(-5068599.0 + 243737.0)/(14892.0-120853) # in kmol O2/kmol fuel\n",

+      "FAideal=114.23/(4.76*no2*28.91) # in kg fuel/kg air\n",

+      "\n",

+      "efficomb=FAideal*100/FAactual # Efficiency\n",

+      "\n",

+      "print \" The combustion efficiency is \",round(efficomb,2),\"%\"\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The combustion efficiency is  86.4 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex15.17:Pg-654"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques17\n",

+      "#efficiency of generator and plant\n",

+      "\n",

+      "q=325000*(3398.3-856.0);#heat transferred to H2O/kg fuel in kJ/kg\n",

+      "qv=26700.0*33250;#higher heating value in kJ/kg\n",

+      "nst=q/qv*100;#efficiency of steam generator\n",

+      "w=81000.0*3600;#net work done in kJ/kg\n",

+      "nth=w/qv*100.0;#thermal efficiency\n",

+      "print\" Efficiency of generator =\",round(nst,1),\"percent\\n\"\n",

+      "print\" Thermal Efficiency =\",round(nth,1),\" percent\"\n",

+      "\n",

+      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Efficiency of generator = 93.1 percent\n",

+        "\n",

+        " Thermal Efficiency = 32.8  percent\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter16_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter16_6.ipynb
new file mode 100755
index 00000000..c155fdc6
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter16_6.ipynb
@@ -0,0 +1,377 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:2573953a96bd6739252db580058edb8938919cec53d4f1d76f9484cfa4f2a1d5"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 16:INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.1:676"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques1\n",

+      "# pressure required to make diamonds from graphite\n",

+      "import math\n",

+      "T=25 # temp in degree C\n",

+      "ggrp= 0 # g for fraphite\n",

+      "gdiamnd=2867.8 # g for diamond in kJ/mol\n",

+      "vgrp=0.000444  # specific volume of graphite in m^3/kg\n",

+      "vdiamnd=0.000284 # specific volume of graphite in m^3/kg\n",

+      "BetaTgrp=0.304e-6 # beta for graphite in 1/MPa\n",

+      "BetaTdiamnd=0.016e-6 #  beta for diamond in 1/MPa\n",

+      "\n",

+      "P=-(-2*(vgrp-vdiamnd)+math.sqrt((2*vgrp-2*vdiamnd)**2-4*(vgrp*BetaTgrp-vdiamnd*BetaTdiamnd)*(2*gdiamnd/(12.011*1000))))/(2*(vgrp*BetaTgrp-vdiamnd*BetaTdiamnd))\n",

+      "print int(P),\" MPa is the pressure at which possibility exists for conversion from graphite to diamonds\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "1493  MPa is the pressure at which possibility exists for conversion from graphite to diamonds\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.2:PG-681"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques2\n",

+      "#to determine change in gibbs free energy\n",

+      "\n",

+      "#1-H2\n",

+      "#2-O2\n",

+      "#3-H2O\n",

+      "\n",

+      "#at T=298 K\n",

+      "T1=298.0;#K\n",

+      "Hf1=0;#Enthalpy of formation of H2 at 298 K\n",

+      "Hf2=0;#Enthalpy of formation of O2 at 298 K\n",

+      "Hf3=-241826;#enthalpy of formation of H2O at 298 K in kJ\n",

+      "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",

+      "Sf1=130.678;#Entropy of H2 at 298 K n kJ/K\n",

+      "Sf2=205.148;#Entropy of O2 at 298 K in kJ/K\n",

+      "Sf3=188.834;#entropy of H2O at 298 K in kJ/K\n",

+      "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",

+      "dG1=dH-T1*dS;#change n gibbs free energy in kJ\n",

+      "print\" Change in gibbs free energy at\",T1,\"kelvin is\",round(dG1),\"kJ \"\n",

+      "#at T=2000 K\n",

+      "T2=2000.0;#K\n",

+      "Hf1=52942-0;#Enthalpy of formation of H2 at 2000 K\n",

+      "Hf2=59176-0;#Enthalpy of formation of O2 at 2000 K\n",

+      "Hf3=-241826+72788;#enthalpy of formation of H2O at 2000 K in kJ\n",

+      "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",

+      "Sf1=188.419;#Entropy of H2 at 2000 K n kJ/K\n",

+      "Sf2=268.748;#Entropy of O2 at 2000 K in kJ/K\n",

+      "Sf3=264.769;#entropy of H2O at 2000 K in kJ/K\n",

+      "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",

+      "dG2=dH-(T2*dS);#change n gibbs free energy in kJ\n",

+      "print\" Change in gibbs free energy at\",T2,\"kelvin is\",round(dG2),\" kJ \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Change in gibbs free energy at 298.0 kelvin is 457179.0 kJ \n",

+        " Change in gibbs free energy at 2000.0 kelvin is 271040.0  kJ \n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.3:PG-683"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques3\n",

+      "#calculating equilibrium constant \n",

+      "\n",

+      "dG1=-457.166;#change in gibbs free energy at temp 298 K from example2 in kJ\n",

+      "dG2=-271.040;#change in gibbs free energy at temp 2000 K from example2 n kJ\n",

+      "T1=298;#K\n",

+      "T2=2000;#K\n",

+      "R=8.3145;#gas constant\n",

+      "K1=dG1*1000/(R*T1);\n",

+      "K2=dG2*1000/(R*T2);\n",

+      "print\" Equilibrium constant at \",T1,\"K = \",round(K1,2)\n",

+      "print\" Equilibrium constant at \",T2,\"K = \",round(K2,3)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Equilibrium constant at  298 K =  -184.51\n",

+        " Equilibrium constant at  2000 K =  -16.299\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.4:Pg-684"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# The example is about proving that Equlibrium constant can be found using table hence doesn't require solution in python  "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.5:Pg-685"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 5\n",

+      "# To determine Heat Transfer\n",

+      "# The process is two step as:\n",

+      "# Combustion: C + O2--->CO2\n",

+      "# Dissociation reaction: 2CO2---->2CO + O2\n",

+      "# overall process : C + O2 \u2192 0.5622 CO2 + 0.4378 CO + 0.2189 O2\n",

+      "nCO2=0.5622 # moles of CO2\n",

+      "nCO=0.4378 # moles of CO\n",

+      "nO2=0.2189 # moles of NO2\n",

+      "\n",

+      "# from Table A.9:\n",

+      "hfCO2=0 # enthalpy of formation\n",

+      "hfC=0 # enthalpy of formation\n",

+      "hfCO=0 # enthalpy of formation\n",

+      "hfO2=0 # enthalpy of formation\n",

+      "hfCO23000=-393522 # enthalpy @ 3000K\n",

+      "hfCO3000=-110527 # enthalpy @ 3000K\n",

+      "hfO23000=0 # enthalpy @ 3000K\n",

+      "hfCO2298=-152853 # enthalpy @ 298K\n",

+      "hfCO298=-93504 # enthalpy @ 298K\n",

+      "hfO2298=-98013 # enthalpy @ 298K\n",

+      "\n",

+      "Hr=hfC+hfO2 # enthalpy of reactants\n",

+      "Hp=nCO2*(hfCO2+hfCO23000-hfCO2298)+nCO*(hfCO+hfCO3000-hfCO298)+nO2*(hfO2+hfO23000-hfO2298)\n",

+      "\n",

+      "Qcv=Hp-Hr # using first law\n",

+      "print round(Qcv),\"kJ/kmol C is the heat transfer \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "-121302.0 kJ/kmol C is the heat transfer \n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.6:Pg-687"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 6\n",

+      "# to determine the composition\n",

+      "# the standard equation is C + 2O2 \u2192 CO2 + O2\n",

+      "# from equilibrium C + 2O2 \u2192 (1 \u2212 2z)CO2 + 2zCO + (1 + z)O2\n",

+      "T=3000 # temp in K\n",

+      "P=0.1 # prssure in MPa\n",

+      "z = 0.1553 # from equilibrium equation using table\n",

+      "yCO2=(1-2*z)/(2+z) # mole fraction of CO2\n",

+      "yCO = 2*z/(2+z) # mole fraction of CO\n",

+      "yO2=(1+z)/(2+z) # mole fraction of O2\n",

+      "print \"The mole fraction is \",round(yCO2,3),\" for CO2 \\n\"\n",

+      "print \"The mole fraction is \",round(yCO,3),\" for CO \\n\"\n",

+      "print \"The mole fraction is \",round(yO2,3),\" for O2 \\n\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The mole fraction is  0.32  for CO2 \n",

+        "\n",

+        "The mole fraction is  0.144  for CO \n",

+        "\n",

+        "The mole fraction is  0.536  for O2 \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.7:Pg-691"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 7\n",

+      "# to determine the equilibrium composition\n",

+      "# The reaction equation is:\n",

+      "# (1): 2.H2O -->2H2 + O2\n",

+      "# (2): 2 H2O -->H2 + 2OH\n",

+      "# the equilibrium equation is \n",

+      "# H2O \u2192 (1 \u2212 2a \u2212 2b)H2O + (2a + b)H2 + aO2 + 2bOH\n",

+      "P= 0.1 # pressure in MPa\n",

+      "T=3000 # temp in Kelvin\n",

+      "a=0.0534 # using value of K from Table A.11 @ 3000k\n",

+      "b=0.0551 # using value of K from Table A.11 @ 3000k\n",

+      "nH2O= (1-2*a-b) # moles of H2O\n",

+      "nH2=2*a+b# moles of H2\n",

+      "nO2=a# moles of O2\n",

+      "nOH= 2*b# moles of OH\n",

+      "X=nH2O+nH2+nO2+nOH\n",

+      "yH2O=nH2O/X # mole fraction\n",

+      "yH2=nH2/X # mole fraction\n",

+      "yO2=nO2/X # mole fraction\n",

+      "yOH=nOH/X # mole fraction\n",

+      "print \"The mole fraction of H2O is\",round(yH2O,2),\"\\n\"\n",

+      "print \"The mole fraction of H2 is\",round(yH2,2),\"\\n\"\n",

+      "print \"The mole fraction of O2 is\",round(yO2,2),\"\\n\"\n",

+      "print \"The mole fraction of OH is\",round(yOH,2),\"\\n\"\n",

+      "# the answers are slightly different due to approximation in textbook while here the answers are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The mole fraction of H2O is 0.72 \n",

+        "\n",

+        "The mole fraction of H2 is 0.139 \n",

+        "\n",

+        "The mole fraction of O2 is 0.046 \n",

+        "\n",

+        "The mole fraction of OH is 0.095 \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 27

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex16.8:pg-696"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# ques 8\n",

+      "# determine the the equilibrium composition\n",

+      "P= 1 # pressure in Kpa\n",

+      "T=10000 # temp in Kelvin\n",

+      "z=0.2008 # using k from table \n",

+      "# the chemical equation is Ar -->Ar^(+) + e\u2212\n",

+      "\n",

+      "yAr=(1-z)/(1 + z) # mole fraction of Ar\n",

+      "yArpositive= z/(1+z) # mole fraction of Ar(+)\n",

+      "yenegative=z/(1+z) # mole fraction of ye-\n",

+      "print \"The mole fraction of Ar is\",round(yAr,3),\"\\n\"\n",

+      "print \"The mole fraction of Ar+ is\",round(yArpositive,3),\"\\n\"\n",

+      "print \"The mole fraction of e- is\",round(yenegative,3),\"\\n\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The mole fraction of Ar is 0.666 \n",

+        "\n",

+        "The mole fraction of Ar+ is 0.167 \n",

+        "\n",

+        "The mole fraction of e- is 0.167 \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 32

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter17_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter17_6.ipynb
new file mode 100755
index 00000000..11b9507c
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter17_6.ipynb
@@ -0,0 +1,462 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:e77c793709c88cc39a2de404e997a6f1eed32d352c214c1580d822fd7a5a83a2"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 17: COMPRESSIBLE FLOW"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.1:PG-710"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques1\n",

+      "#to determine isentropic stagnation pressure and temperature \n",

+      "\n",

+      "T=300;#Temperature of air in K\n",

+      "P=150;#Pressure of air in kPa\n",

+      "v=200;#velocity of air flow n m/s\n",

+      "Cp=1.004;#specific heat at constant pressure in kJ/kg\n",

+      "To=v**2/(2000*Cp)+T;#stagnation temperature in K\n",

+      "k=1.4;#constant\n",

+      "Po=P*(To/T)**(k/(k-1));#stagnation pressure in kPa\n",

+      "print 'Stagnation Temperature is ',round(To,1),' K \\n'\n",

+      "print 'Stagnation Pressure is ',round(Po,2),'kPa \\n'"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stagnation Temperature is  319.9  K \n",

+        "\n",

+        "Stagnation Pressure is  187.85 kPa \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.2:PG-713"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques2\n",

+      "#to determine the force\n",

+      "\n",

+      "#initializing variables\n",

+      "mdot=-1 # mass flow rate out of control volume in kg/s\n",

+      "Vx=-1 # x component of velocity of control volume in m/s\n",

+      "Vy=10 # y component of velocity of control volume in m/s\n",

+      "\n",

+      "Fx=mdot*Vx # Force in X direction\n",

+      "\n",

+      "Fy=mdot*Vy # Force in Y direction\n",

+      "\n",

+      "print \"the force the man exert on the wheelbarrow\",round(Fx),\"N\"\n",

+      "print \"the force the floor exerts on the wheelbarrow\",round(Fy),\"N\"\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "the force the man exert on the wheelbarrow 1.0 N\n",

+        "the force the floor exerts on the wheelbarrow -10.0 N\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.3:PG-715"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques3\n",

+      "#determining the thrust acting on a control surface\n",

+      "\n",

+      "#i-inlet\n",

+      "#e-exit\n",

+      "#using momentum equation on control surface in x direction\n",

+      "me=20.4;#mass exiting in kg\n",

+      "mi=20;#mass entering in kg\n",

+      "ve=450;#exit velocity in m/s\n",

+      "vi=100;#inlet velocity in m/s\n",

+      "Pi=95;#Pressure at inlet in kPa\n",

+      "Pe=125;#Pressure at exit in kPa\n",

+      "Po=100;#surrounding pressure in kPa\n",

+      "Ai=0.2;#inlet area in m^2\n",

+      "Ae=0.1;#exit area in m^2\n",

+      "Rx=(me*ve-mi*vi)/1000-(Pi-Po)*Ai+(Pe-Po)*Ae;#thrust in x direction in kN\n",

+      "\n",

+      "print \"Thrust acting in x direction is \",round(Rx,2),\"kN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Thrust acting in x direction is  10.68 kN\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.4:PG-717"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques4\n",

+      "#to determine increase in enthalpy\n",

+      "#initializing variables\n",

+      "#i-inlet\n",

+      "#e-exit\n",

+      "\n",

+      "T=25+273 # temperature in kelvin\n",

+      "v=0.001003 # specific volume of water in kg/m^3 at 25 *c from table B.1.1 \n",

+      "ve=7;#exit velocity in m/s\n",

+      "vi=30;#inlet velocity in m/s\n",

+      "Pi=350;#Pressure at inlet in kPa\n",

+      "Pe=600;#Pressure at exit in kPa\n",

+      "\n",

+      "#using momentum equation on control surface \n",

+      "Pes= (vi**2-ve**2)/(2*v*1000)+Pi # exit pressure for reversible diffuser\n",

+      "delH=(vi**2-ve**2)/(2*1000.0) # change in enthalpy from first law in kJ/kg\n",

+      "delU=delH-v*(Pe-Pi) # change in internal energy in kJ/kg\n",

+      "delS=delU/T # change in entropy in kJ/kg.K\n",

+      "print\"the exit pressure for reversible diffuser is \",round(Pes),\"kPa\"\n",

+      "print\"the increase in enthalpy is \",round(delH,5),\"kJ/kg\"\n",

+      "print\"the increase in internal energy is \",round(delU,5),\"kJ/kg\"\n",

+      "print\"the increase in entropy is \",round(delS,6),\"kJ/kg.K\"\n",

+      "\n",

+      "\n",

+      "\n",

+      "\n",

+      "\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "the exit pressure for reversible diffuser is  774.0 kPa\n",

+        "the increase in enthalpy is  0.4255 kJ/kg\n",

+        "the increase in internal energy is  0.17475 kJ/kg\n",

+        "the increase in entropy is  0.000586 kJ/kg.K\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.5:PG-720"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques5\n",

+      "#determining velocity of sound in air\n",

+      "import math\n",

+      "k=1.4;#constant\n",

+      "R=0.287;#gas constant\n",

+      "#at 300K\n",

+      "T1=300;# temperature in kelvin\n",

+      "c1=math.sqrt(k*R*T1*1000)\n",

+      "print \"Speed of sound at 300 K is\",round(c1,1),\" m/s\" \n",

+      "#at 1000K\n",

+      "T2=1000;# temperature in kelvin\n",

+      "c2=math.sqrt(k*R*T2*1000)\n",

+      "print \"Speed of sound at 1000 K is\",round(c2,1),\" m/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Speed of sound at 300 K is 347.2  m/s\n",

+        "Speed of sound at 1000 K is 633.9  m/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.6:PG-727"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques6\n",

+      "#determining mass flow rate through control volume\n",

+      "import math\n",

+      "k=1.4;#constant\n",

+      "R=0.287;#gas constant\n",

+      "To=360;#stagnation Temperature in K \n",

+      "T=To*0.8333;#Temperature of air in K, 0.8333 stagnation ratio from table\n",

+      "v=math.sqrt(k*R*T*1000);#velocity in m/s\n",

+      "P=528;#stagnation pressure in kPa\n",

+      "d=P/(R*T);#stagnation density in kg/m^3\n",

+      "A=500*10**-6;#area in m^2\n",

+      "ms=d*A*v;#mass flow rate in kg/s\n",

+      "print\" Mass flow rate at the throat section is\",round(ms,4),\"kg/s\"\n",

+      "#e-exit state\n",

+      "Te=To*0.9381;#exit temperature in K, ratio from table\n",

+      "ce=math.sqrt(k*R*Te*1000);#exit velocity of sound in m/s\n",

+      "Me=0.573;#Mach number\n",

+      "ve=Me*ce;\n",

+      "Pe=800;#exit pressure in kPa\n",

+      "de=Pe/R/Te;\n",

+      "mse=de*A*ve;\n",

+      "print\" Mass flow rate at the exit section is\",round(mse,4),\" kg/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " Mass flow rate at the throat section is 1.0646 kg/s\n",

+        " Mass flow rate at the exit section is 0.8711  kg/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.7:PG-728"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques7\n",

+      "#determining exit properties in a control volume\n",

+      "import math\n",

+      "Po=1000;#stagnation pressure in kPa\n",

+      "To=360;#stagnation temperature in K\n",

+      "\n",

+      "#when diverging section acting as nozzle\n",

+      "\n",

+      "Pe1=0.0939*Po;#exit pressure of air in kPa\n",

+      "Te1=0.5089*To;#exit temperature in K\n",

+      "k=1.4;#constant\n",

+      "R=0.287;#gas constant for air\n",

+      "ce=math.sqrt(k*R*Te1*1000);#velocity of sound in exit section in m/s\n",

+      "Me=2.197;#mach number from table\n",

+      "ve1=Me*ce;#velocity of air at exit section in m/s\n",

+      "print \"When diverging section act as a nozzle :-\"\n",

+      "print \"Exit pressure is\",round(Pe1,4),\" kPa\"\n",

+      "print \"Exit Temperature\",round(Te1,1),\" K\"\n",

+      "print \"Exit velocity is\",round(ve1,1),\" m/s \"\n",

+      "\n",

+      "#when diverging section act as diffuser\n",

+      "\n",

+      "Me=0.308;\n",

+      "Pe2=0.0936*Po;#exit pressure of air in kPa\n",

+      "Te2=0.9812*To;#exit temperature in K\n",

+      "ce=math.sqrt(k*R*Te2*1000);#velocity of sound in exit section in m/s\n",

+      "ve2=Me*ce;\n",

+      "print \"When diverging section act as a diffuser :-\"\n",

+      "print \"Exit pressure is\",round(Pe2,1),\" kPa\"\n",

+      "print \"Exit Temperature\",round(Te2,2),\" K\"\n",

+      "print \"Exit velocity is\",round(ve2,),\" m/s \"\n",

+      "\n",

+      "# The value of Exit pressure when diverging section acts as diffuser is wrong\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "When diverging section act as a nozzle :-\n",

+        "Exit pressure is 93.9  kPa\n",

+        "Exit Temperature 183.2  K\n",

+        "Exit velocity is 596.1  m/s \n",

+        "When diverging section act as a diffuser :-\n",

+        "Exit pressure is 93.6  kPa\n",

+        "Exit Temperature 353.23  K\n",

+        "Exit velocity is 116.0  m/s \n"

+       ]

+      }

+     ],

+     "prompt_number": 39

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.8:PG-731"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques8\n",

+      "#Determine the static pressure and temperature of supersonic nozzle\n",

+      "#x-inlet\n",

+      "#y-exit\n",

+      "# using  the data from previous example\n",

+      "Mx=2.197 # Mach number at x\n",

+      "Px=93.9 # IN kPa\n",

+      "Tx=183.2 # IN K\n",

+      "P0x=1000 # IN kPa\n",

+      "T0x = 360 # in K\n",

+      "My=0.547 # Mach number at y From Table A.13\n",

+      "Py = 5.46*Px # From Table A.13\n",

+      "Ty = 1.854 * Tx # From Table A.13\n",

+      "P0y= 0.630 * P0x # From Table A.13\n",

+      "print \"Exit pressure after shock is\",round(Py,1),\" kPa\"\n",

+      "print \"Exit Temperature after shock is\",round(Ty,1),\" K\"\n",

+      "print \"Exit stagnation pressure after shock is\",round(P0y,1),\" kPa\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Exit pressure after shock is 512.7  kPa\n",

+        "Exit Temperature after shock is 339.7  K\n",

+        "Exit stagnation pressure after shock is 630.0  kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex17.9:PG-733"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#ques9\n",

+      "#determining exit plane properties in control volume\n",

+      "\n",

+      "#x-inlet\n",

+      "#y-exit\n",

+      "Mx=1.5;#mach number for inlet\n",

+      "My=0.7011;#mach number for exit\n",

+      "Px=272.4;#inlet pressure in kPa\n",

+      "Tx=248.3;#inlet temperature in K\n",

+      "Tox=360 # stagnation temperature in Kelvin\n",

+      "Pox=1000.0;#stagnation pressure for inlet\n",

+      "Py=2.4583*Px;# Pressure at 1.5 mach in kPa\n",

+      "Ty=1.320*Tx;# temperature at 1.5 mach in K\n",

+      "Poy=0.9298*Pox;# pressure at 1.5 mach in kPa\n",

+      "\n",

+      "Toy=Tox # constant\n",

+      "Me=0.339 # from table with A/A*=1.860 and M < 1\n",

+      "Pe=0.9222*Py;#Exit Pressure in kPa\n",

+      "Te=0.9771*Toy;#Exit temperature in K\n",

+      "Poe=0.9222*Poy;#Exit pressure in kPa\n",

+      "\n",

+      "print \"Exit Mach no.=\",Me\n",

+      "print \"Exit temperature =\",round(Te,2),\"K \"\n",

+      "print \"Exit pressure =\",round(Poe,1),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Exit Mach no.= 0.339\n",

+        "Exit temperature = 351.76 K \n",

+        "Exit pressure = 857.5 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 50

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter2_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter2_6.ipynb
new file mode 100755
index 00000000..08e192fe
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter2_6.ipynb
@@ -0,0 +1,379 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:96f5bd0fa542764d6a8f75d5a1a42e4b79791697476f4029d0fa0fea1de65bed"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter2:SOME CONCEPTS AND DEFINITIONS"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.1:pg-19"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1\n",

+      "#weight of a person\n",

+      "\n",

+      "\n",

+      "m=1 #kg\n",

+      "g=9.75 #acc.due to gravity in m/s^2\n",

+      "F=m*g #weight of 1 kg mass in N\n",

+      "print\"\\n hence,weight of person is\",round(F,2),\" N\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,weight of person is 9.75  N\n"

+       ]

+      }

+     ],

+     "prompt_number": 20

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.1E:pg-20"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2\n",

+      "#weight of a person\n",

+      "\n",

+      "\n",

+      "m=1 #lbm\n",

+      "g=32 #acc.due to gravity in ft/s^2\n",

+      "F=m*g/32.174 #weight of 1 lbm mass in lbf\n",

+      "print\"\\n hence,weight of person is\",round(F,4),\" lbf\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,weight of person is 0.9946  lbf\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.2:pg-24"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#average volume and density\n",

+      "\n",

+      "Vliq=0.2 #volume of liquid in m^3\n",

+      "dliq=997 #density of liquid in kg/m^3\n",

+      "Vstone=0.12 #volume of stone in m^3\n",

+      "Vsand=0.15 #volume of sand in m^3\n",

+      "Vair=0.53 #vo;ume of air in m^3\n",

+      "mliq=Vliq*dliq #mass of liquid in kg\n",

+      "dstone=2750 #density of stone in kg/m^3\n",

+      "dsand=1500 #density of sand in kg/m^3\n",

+      "mstone=Vstone*dstone #volume of stone in m^3\n",

+      "msand=Vsand*dsand #volume of sand in m^3\n",

+      "Vtot=1 #total volume in m^3\n",

+      "dair=1.1 #density of air in kg/m^3\n",

+      "mair=Vair*dair #mass of air\n",

+      "mtot=mair+msand+mliq+mstone #total mass in kg\n",

+      "v=Vtot/mtot #specific volume in  m^3/kg\n",

+      "d=1/v #overall density in kg/m^3\n",

+      "print\"\\n hence,average specific volume is\",round(v,6),\"m^3/kg\" \n",

+      "print\"\\n and overall density is\", round(d),\"kg/m^3\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,average specific volume is 0.001325 m^3/kg\n",

+        "\n",

+        " and overall density is 755.0 kg/m^3\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.3:pg-26"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4\n",

+      "#calculating the required force\n",

+      "import math\n",

+      "\n",

+      "Dcyl=0.1 #cylinder diameter in m\n",

+      "Drod=0.01 #rod diameter in m\n",

+      "Acyl=math.pi*Dcyl**2/4 #cross sectional area of cylinder in m^2\n",

+      "Arod=math.pi*Drod**2/4 #cross sectional area of rod in m^2\n",

+      "Pcyl=250000 #inside hydaulic pressure in Pa\n",

+      "Po=101000 #outside atmospheric pressure in kPa\n",

+      "g=9.81 #acc. due to gravity in m/s^2\n",

+      "mp=25 #mass of (rod+piston) in kg\n",

+      "F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g #the force that rod can push within the upward direction in N\n",

+      "print\"\\n hence,the force that rod can push within the upward direction is\",round(F,1),\" N\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the force that rod can push within the upward direction is 932.9  N\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.4:pg-28"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#Calculating atmospheric pressure \n",

+      "\n",

+      "dm=13534 #density of mercury in kg/m^3\n",

+      "H=0.750 #height difference between two columns in metres\n",

+      "g=9.80665 #acc. due to gravity  in m/s^2\n",

+      "Patm=dm*H*g/1000 #atmospheric pressure in kPa\n",

+      "print\"\\n hence, atmospheric pressure is\",round(Patm,2),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, atmospheric pressure is 99.54 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.5:pg-28"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 6\n",

+      "#pressure inside vessel\n",

+      "\n",

+      "dm=13590 #density of mercury in kg/m^3\n",

+      "H=0.24 #height difference between two columns in metres\n",

+      "g=9.80665 #acc. due to gravity  in m/s^2\n",

+      "dP=dm*H*g #pressure difference in Pa\n",

+      "Patm=13590*0.750*9.80665 #Atmospheric Pressure in Pa\n",

+      "Pvessel=dP+Patm #Absolute Pressure inside vessel in Pa\n",

+      "Pvessel=Pvessel/101325 #Absolute Pressure inside vessel in atm\n",

+      "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,3),\"atm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, the absolute pressure inside vessel is 1.302 atm\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.5E:pg-28"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7\n",

+      "#pressure inside vessel\n",

+      "\n",

+      "dm=848 #density of mercury in lbm/ft^3\n",

+      "H=9.5  #height difference between two columns in inches\n",

+      "g=32.174 #acc. due to gravity  in ft/s^2\n",

+      "dP=dm*H*g #pressure difference in lbf/in^2\n",

+      "Patm=848*9.5*32.174/(1728*32.174) #Atmospheric Pressure in lbf/in^2\n",

+      "Pvessel=dP+Patm #Absolute Pressure inside vessel in lbf/in^2\n",

+      "Pvessel=848*29.5*32.174/(1728*32.174)+Patm #Absolute Pressure inside vessel in lbf/in^2\n",

+      "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,2),\"lbf/in^2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, the absolute pressure inside vessel is 19.14 lbf/in^2\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.6:pg-29"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8\n",

+      "#calculating pressure\n",

+      "\n",

+      "dg=750 #density of gaasoline in kg/m^3\n",

+      "dR=1206 #density of R-134a in kg/m^3\n",

+      "H=7.5 #height of storage tank in metres\n",

+      "g=9.807 #acc. due to gravity in m/s^2\n",

+      "dP1=dg*g*H/1000 #in kPa\n",

+      "Ptop1=101 #atmospheric pressure in kPa\n",

+      "P1=dP1+Ptop1\n",

+      "print\"hence,pressure at the bottom of storage tank if fluid is gasoline is\",round(P1,1),\"kPa\" \n",

+      "dP2=dR*g*H/1000 #in kPa\n",

+      "Ptop2=1000 #top surface pressure in kPa\n",

+      "P2=dP2+Ptop2\n",

+      "print\"\\n hence, pressure at the bottom of storage tank if liquid is R-134a is\",round(P2),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "hence,pressure at the bottom of storage tank if fluid is gasoline is 156.2 kPa\n",

+        "\n",

+        " hence, pressure at the bottom of storage tank if liquid is R-134a is 1089.0 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex2.7:pg-29"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9\n",

+      "#calculating balancing force\n",

+      "\n",

+      "Po=100#Outside atmospheric pressure in kPa\n",

+      "F1=25 #net force on the smallest piston in kN\n",

+      "A1=0.01 #cross sectional area of lower piston in m^2\n",

+      "P1=Po+F1/A1 #fluid pressure in kPa\n",

+      "d=900 #density of fluid in kg/m^3\n",

+      "g=9.81 #acc. due to gravity in m/s^2\n",

+      "H=6 #height of second piston in comparison to first one in m\n",

+      "P2=P1-d*g*H/1000 #pressure at higher elevation on piston 2 in kPa\n",

+      "A2=0.05 #cross sectional area of higher piston in m^3\n",

+      "F2=(P2-Po)*A2 #balancing force on second piston in kN\n",

+      "print\"\\n hence, balancing force on second larger piston is\",round(F2,1),\"N\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, balancing force on second larger piston is 122.4 N\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter3_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter3_6.ipynb
new file mode 100755
index 00000000..64998280
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter3_6.ipynb
@@ -0,0 +1,571 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:b6e51b20af9aaa1b32972245e7dbe7cc42b3ed40e5f3cbf6db8837625a12aebc"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter3:PROPERTIES OF A PURE SUBSTANCE"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.1:pg-57"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1\n",

+      "#determinig the phase of water\n",

+      "\n",

+      "print\"from the table,we find that at 120C,saturation pressure of water is 198.5 kPa.But here we have pressure of 500 kPa.hence,water exists as a compressed liquid here.\"\n",

+      "print\"also at 120C,vf=0.00106 kg/m^3 and vg=0.89186 kg/m^3.given v=0.5 m^3/kg i.e. vf<v<vg,so we have two phase mixture of liquid and vapor.\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "from the table,we find that at 120C,saturation pressure of water is 198.5 kPa.But here we have pressure of 500 kPa.hence,water exists as a compressed liquid here.\n",

+        "also at 120C,vf=0.00106 kg/m^3 and vg=0.89186 kg/m^3.given v=0.5 m^3/kg i.e. vf<v<vg,so we have two phase mixture of liquid and vapor.\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.2:pg-58"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2\n",

+      "#determinig the phase \n",

+      "\n",

+      "print\"from the table,we find that at 30C,saturation pressure of ammonia is 1167 kPa.But here we have pressure of 1000 kPa.hence,ammonia exists in superheated vapor state.\"\n",

+      "print\"for R-22 at 200 kPa,vg=0.1119 kg/m^3.given v=0.15 m^3/kg i.e. v>vg,so the state is superheated vapor\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "from the table,we find that at 30C,saturation pressure of ammonia is 1167 kPa.But here we have pressure of 1000 kPa.hence,ammonia exists in superheated vapor state.\n",

+        "for R-22 at 200 kPa,vg=0.1119 kg/m^3.given v=0.15 m^3/kg i.e. v>vg,so the state is superheated vapor\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.3:pg-59"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#determining the quality and specific volume\n",

+      "\n",

+      "v1=0.5 #given specific volume in m^3/kg\n",

+      "vf=0.001073 #specific volume when only liquid phase is present in m^3/kg\n",

+      "vfg=0.60475 #in m^3/kg\n",

+      "x=(v1-vf)/vfg #quality\n",

+      "print\"For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is\",round(x,3)\n",

+      "\n",

+      "v2=1 #given specific volume in m^3/kg\n",

+      "\n",

+      " # using the method of interplotation\n",

+      "T=((400-300)*(1.0-0.8753))/(1.0315-0.8753)+300 #temperature of the water\n",

+      "print\"For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is\",round(T,1)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is 0.825\n",

+        "For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is 379.8\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.4:pg-60"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      " #example 4\n",

+      "#percentage of vapor \n",

+      "\n",

+      "vliq=0.1 #volume of saturated liquid in m^3\n",

+      "vf=0.000843 #in m^3/kg\n",

+      "vvap=0.9 #volume of saturated vapor R-134a in equilbrium\n",

+      "vg=0.02671 #in m^3/kg\n",

+      "mliq=vliq/vf #mass of liquid in kg \n",

+      "mvap=vvap/vg #mass of vapor in kg\n",

+      "m=mliq+mvap #total mass in kg\n",

+      "x=mvap*100/m #percentage of vapor on mass basis\n",

+      "print\"hence,% vapor on mass basis is\",round(x,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "hence,% vapor on mass basis is 22.1 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.4E:pg-60"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      " #example 5\n",

+      "#percentage of vapor \n",

+      "\n",

+      "vliq=0.1 #volume of saturated liquid in ft^3\n",

+      "vf=0.0136 #in ft^3/lbm\n",

+      "vvap=0.9 #volume of saturated vapor R-134a in equilbrium\n",

+      "vg=0.4009 #in ft^3/lbm\n",

+      "mliq=vliq/vf #mass of liquid in lbm \n",

+      "mvap=vvap/vg #mass of vapor in lbm\n",

+      "m=mliq+mvap #total mass in lbm\n",

+      "x=mvap*100/m #percentage of vapor on mass basis\n",

+      "print\"hence,% vapor on mass basis is\",round(x,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "hence,% vapor on mass basis is 23.4 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.5:pg-61"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#calculating pressure after heat addition\n",

+      "\n",

+      "v1=0.14922 #specific volume of sautrated ammonia in m^3/kg\n",

+      "print\"Since the volume does not change during the process,the specific volume remains constant.therefore\"\n",

+      "v2=v1 #in m^3/kg\n",

+      "print\"Since vg at 40C is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\"\n",

+      "P2=945 #final pressure in kPa\n",

+      "print\"hence,the final pressure is 945 kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Since the volume does not change during the process,the specific volume remains constant.therefore\n",

+        "Since vg at 40C is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\n",

+        "hence,the final pressure is 945 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.5E:pg-61"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#calculating pressure after heat addition\n",

+      "\n",

+      "v1=2.311 #specific volume of sautrated ammonia in ft^3/lbm\n",

+      "print\"Since the volume does not change during the process,the specific volume remains constant.therefore\"\n",

+      "v2=v1 #in ft^3/lbm\n",

+      "print\"Since vg at 120f is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\"\n",

+      "P2=145 #final pressure in lbf/in^2\n",

+      "print\"hence,the final pressure is 145 lbf/in^2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Since the volume does not change during the process,the specific volume remains constant.therefore\n",

+        "Since vg at 120f is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that\n",

+        "hence,the final pressure is 145 lbf/in^2\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.6:pg-61"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 6\n",

+      "#Determinig the missing property\n",

+      "\n",

+      "T1=273-53.2 #given temperature in K\n",

+      "P1=600 #given pressure in kPa\n",

+      "print\"This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\"\n",

+      "T2=100 #given temp. in K\n",

+      "v2=0.008 #given specific volume in m^3/kg\n",

+      "vf=0.001452 #in m^3/kg\n",

+      "vg=0.0312 #in m^3/kg\n",

+      "Psat=779.2 #saturation pressure in kPa\n",

+      "vfg=vg-vf #in m^3/kg\n",

+      "x=(v2-vf)/vfg #quality\n",

+      "print\"\\n hence, the pressure is\",round(Psat,1),\"kPa\"\n",

+      "print\"\\n and quality is\",round(x,4),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg\n",

+        "\n",

+        " hence, the pressure is 779.2 kPa\n",

+        "\n",

+        " and quality is 0.2201 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.7:pg-62"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7\n",

+      "#determining the pressure of water\n",

+      "\n",

+      "vg=0.12736 #specific volume in m^3/kg for water at 200C\n",

+      "v=0.4 #specific volume in m^3/kg\n",

+      "P1=500 #in kPa\n",

+      "v1=0.42492 #specific volume at P1 in m^3/kg\n",

+      "P2=600 #in kPa\n",

+      "v2=0.35202 #specific volume at P2 in m^3/kg\n",

+      "P=P1+(P2-P1)*(v-v1)/(v2-v1) #calculating pressure by interplotation\n",

+      "print \"hence,the pressure of water is\",round(P,1),\" kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " hence,the pressure of water is 534.2  kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.8:pg-66"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8\n",

+      "#calculating mass of air\n",

+      "\n",

+      "P=100 #pressure in kPa\n",

+      "V=6*10*4 #volume of room in m^3\n",

+      "R=0.287 #in kN-m/kg-K\n",

+      "T=25 #temperature in Celsius\n",

+      "m=P*V/(R*(T+273.1)) #mass of air contained in room\n",

+      "print\"\\n hence, mass of air contained in room is\",round(m,1),\"kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, mass of air contained in room is 280.5 kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 27

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.9:pg-67"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9\n",

+      "#calculating pressure inside tank\n",

+      "\n",

+      "V=0.5 #volumr of tank in m^3\n",

+      "m=10 #mass of ideal gas in kg\n",

+      "T=25 #temperature of tank in Celsius\n",

+      "M=24 #molecular mass of gas in kg/kmol\n",

+      "Ru=8.3145 #universal gas constant in kN-m/kmol-K\n",

+      "R=Ru/M #gas constant for given ideal gas in kN-m/kg-K\n",

+      "P=m*R*(T+273.2)/V #pressure inside tank\n",

+      "print\"\\n hence,pressure inside tank is\",round(P),\"kpa\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,pressure inside tank is 2066.0 kpa\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.10:pg-67"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 10\n",

+      "#mass  flow rate\n",

+      "\n",

+      "dt=185 #time period in seconds over which there is incrrease in volume \n",

+      "dV=0.75 #increase in volume in 0.75 in m^3\n",

+      "V=dV/dt #volume flow rate in m^3/s\n",

+      "P=105 #pressure inside gas bell kPa\n",

+      "T=21 #temperature in celsius\n",

+      "R=0.1889 #ideal gas constant in kJ/kg-K\n",

+      "m=P*V/(R*(T+273.15)) #mass flow rate of the flow in kg/s\n",

+      "print\"\\n hence,mass flow rate is\",round(m,5),\"kg/s\"\n",

+      "print\"\\n and volume flow rate is\",round(V,5),\"m^3/s\"\n",

+      "#The answer of volume flow rate in the book is wrong."

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,mass flow rate is 0.00766 kg/s\n",

+        "\n",

+        " and volume flow rate is 0.00405 m^3/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 40

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.12:pg-71"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 12\n",

+      "#determining specific using diffenet laws\n",

+      "\n",

+      "T=100.0 #given temp.in 100 celsius\n",

+      "P=3.0 #given pressure in MPa\n",

+      "v1=0.0065 #specific volume in m^3/kg using table\n",

+      "print\"\\n hence,the specific volume for R-134a using R-134a tables is\",round(v1,3),\"m^3/kg\"\n",

+      "M=102.3 #molecular mass in kg\n",

+      "R=8.3145 #in kJ/K\n",

+      "Ru=R/M #in kJ/K-kg\n",

+      "v2=Ru*(T+273)/(P*1000) #specific volume assuming R-134a to be ideal gas in m^3/kg\n",

+      "print\"\\n hence,the specific volume for R-134a using R-134a the ideal gas laws is\",round(v2,5),\"m^3/kg\"\n",

+      "Tr=373.2/374.2 #reduced temperature using generalized chart\n",

+      "Pr=3.0/4.06 #reduced pressure using generalized chart\n",

+      "Z=0.67 #compressibility factor \n",

+      "v3=Z*v2 # specific volume using generalized chart in m^3/kg\n",

+      "print\"\\n hence,the specific volume for R-134a using the generalized chart is\",round(v3,5),\"m^3/kg\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the specific volume for R-134a using R-134a tables is 0.006 m^3/kg\n",

+        "\n",

+        " hence,the specific volume for R-134a using R-134a the ideal gas laws is 0.01011 m^3/kg\n",

+        "\n",

+        " hence,the specific volume for R-134a using the generalized chart is 0.00677 m^3/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 51

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3.13:pg-71"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 13\n",

+      "#calculating mass of gas\n",

+      "\n",

+      "Pc=4250 #critical pressure of propane in kPa\n",

+      "Tc=369.8 #critical temperature in K\n",

+      "T=15 #temperature of propane in celsius\n",

+      "Tr=T/Tc #reduced temperature\n",

+      "Prsat=0.2 # reduced pressure \n",

+      "P=Prsat*Pc #pressure in kPa\n",

+      "x=0.1 #given quality\n",

+      "Zf=0.035 #from graph\n",

+      "Zg=0.83 #from graph\n",

+      "Z=(1-x)*Zf+x*Zg #overall compressibility factor\n",

+      "V=0.1 #volume of steel bottle in m^3\n",

+      "R=0.1887 #in kPa-m^3/kg-K\n",

+      "m=P*V/(Z*R*(T+273)) #total propane mass in kg\n",

+      "print\"\\n hence,the total propane mass is\",round(m,2),\"kg\"\n",

+      "print\"\\n and pressure is\",round(P,2),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the total propane mass is 13.66 kg\n",

+        "\n",

+        " and pressure is 850.0 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 44

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter4_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter4_6.ipynb
new file mode 100755
index 00000000..eed9d20a
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter4_6.ipynb
@@ -0,0 +1,198 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:da3557673ab31ece618ad10f5b0d3ef78112ff36bd0b260e508d99211060a1c9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4:WORK AND HEAT"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4.1:PG-96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 1\n",
+      "#work done during different processes\n",
+      "import math\n",
+      "\n",
+      "P1=200 #initial pressure inside cylinder in kPa\n",
+      "V2=0.1 #in m^3\n",
+      "V1=0.04 #initial volume of gas in m^3\n",
+      "\n",
+      "W1=P1*(V2-V1) #work done in isobaric process in kJ\n",
+      "print\"\\n hence,the work done during the isobaric process is\",round(W1,2),\"kJ. \\n\"\n",
+      "\n",
+      "W2=P1*V1*math.log(V2/V1) #work done in isothermal process in kJ\n",
+      "print\"\\n hence,the work done in isothermal process is\",round(W2,2),\"kJ. \\n\"\n",
+      "\n",
+      "P2=P1*(V1/V2)**(1.3) #final pressure according to the given process\n",
+      "W3=(P2*V2-P1*V1)/(1-1.3)\n",
+      "print\"\\n hence,the work done during the described process is\",round(W3,2),\"kJ. \\n\"\n",
+      "\n",
+      "W4=0 #work done in isochoric process\n",
+      "print\"\\n hence,the work done in the isochoric process is\",round(W4,3),\"kJ. \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " hence,the work done during the isobaric process is 12.0 kJ. \n",
+        "\n",
+        "\n",
+        " hence,the work done in isothermal process is 7.33 kJ. \n",
+        "\n",
+        "\n",
+        " hence,the work done during the described process is 6.41 kJ. \n",
+        "\n",
+        "\n",
+        " hence,the work done in the isochoric process is 0.0 kJ. \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4.3:PG-99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3\n",
+      "\n",
+      "#work produced\n",
+      "\n",
+      "Psat=190.2 #in kPa\n",
+      "P1=Psat #saturation pressure in state 1\n",
+      "vf=0.001504 #in m^3/kg\n",
+      "vfg=0.62184 #in m^3/kg\n",
+      "x1=0.25 #quality\n",
+      "v1=vf+x1*vfg #specific volume at state 1 in m^3/kg\n",
+      "v2=1.41*v1 #specific volume at state 2 in m^3/kg\n",
+      "P2=600 #pressure in state 2 in kPa from Table B.2.2\n",
+      "m=0.5 #mass of ammonia in kg\n",
+      "W=m*(P1+P2)*(v2-v1)/2 #woork produced by ammonia in kJ\n",
+      "print \"The final pressure is\",round(P2),\"kPa\\n\"\n",
+      "print \"hence,work produced by ammonia is\",round(W,2),\"kJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final pressure is 600.0 kPa\n",
+        "\n",
+        "hence,work produced by ammonia is 12.71 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4.4:PG-100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 4\n",
+      "\n",
+      "#calculating work done\n",
+      "\n",
+      "v1=0.35411 #specific volume at state 1 in m^3/kg\n",
+      "v2=v1/2 \n",
+      "m=0.1 #mass of water in kg\n",
+      "P1=1000 #pressure inside cylinder in kPa\n",
+      "W=m*P1*(v2-v1) #in kJ\n",
+      "print \"the work in the overall process is\",round(W,1),\"kJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the work in the overall process is -17.7 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4.7:PG-108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 7\n",
+      "#heat transfer\n",
+      "k=1.4 #conductivity of glass pane in W/m-K\n",
+      "A=0.5 #total surface area of glass pane\n",
+      "dx=0.005 #thickness of glasspane in m\n",
+      "dT1=20-12.1 #temperature difference between room air and outer glass surface temperature in celsius\n",
+      "Q=-k*A*dT1/dx #conduction through glass slab in W\n",
+      "h=100 #convective heat transfer coefficient in W/m^2-K \n",
+      "dT=12.1-(-10)  #temperature difference between warm room and colder ambient in celsius\n",
+      "Q2=h*A*dT #heat transfer in convective layer in W\n",
+      "#result\n",
+      "print \"the rate of heat transfer in the glass and convective layer is\",round(Q2),\"kW.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of heat transfer in the glass and convective layer is 1105.0 kW.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter5_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter5_6.ipynb
new file mode 100755
index 00000000..5a9734bc
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter5_6.ipynb
@@ -0,0 +1,635 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:bc9fbdf4ecee4bce9770355a576adf487ef4df6d34bad14ada42e8383ea01d91"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter5:THE FIRST LAW OF THERMODYNAMICS"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.1:pg-131"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1\n",

+      "#calculating height\n",

+      "\n",

+      "m=1100 #mass of car in kg\n",

+      "ke=400 #kinetic energy of car in kJ\n",

+      "V=(2*ke*1000/m)**0.5 #velocity of car in m/s\n",

+      "g=9.807 #acc. due to gravity in m/s^2\n",

+      "H=ke*1000/(m*g) #height to which the car  should be lifted so that its potential energy equals its kinetic energy\n",

+      "print\"hence,the car should be raised to a height is\",round(H,1),\"m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "hence,the car should be raised to a height is 37.1 m\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.1E:pg-132"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1E\n",

+      "#calculating height\n",

+      "\n",

+      "m=2400 #mass of car in lbm\n",

+      "ke=400 #kinetic energy of car in Btu\n",

+      "V=(2*ke*778.17*32.174/m)**0.5 #velocity of car in ft/s\n",

+      "g=32.174 #acc. due to gravity in ft/s^2\n",

+      "H=ke*778.17*32.174/(m*g) #height to which the car  should be lifted so that its potential energy equals its kinetic energy\n",

+      "print\"hence,the car should be raised to a height is\",round(H,1),\"ft\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "hence,the car should be raised to a height is 129.7 ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.2:pg-134"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2\n",

+      "#change in internal energy\n",

+      "\n",

+      "W=-5090 #work input to paddle wheel in kJ\n",

+      "Q=-1500 #heat transfer from tank in kJ\n",

+      "dU=Q-W #change in internal energy in kJ\n",

+      "print\"hence,change in internal energy is\",round(dU),\"kj\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "hence,change in internal energy is 3590.0 kj\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.3:pg-134"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#analysis of energy transfer\n",

+      "\n",

+      "g=9.806 #acceleration due to gravity in m/s^2\n",

+      "m=10 #mass of stone in kg\n",

+      "H1=10.2 #initial height of stone above water in metres\n",

+      "H2=0 #final height in metres\n",

+      "dKE1=-m*g*(H2-H1) #change in kinetic energy when stone enters state 2 in J\n",

+      "dPE1=-1 #change in potential energy when stone enters state 2 in J\n",

+      "print\"\\n hence,when stone is \",round(dKE1),\"J\"\n",

+      "print\"\\n and change in potential energy is \",round(dPE1),\"J\"\n",

+      "dPE2=0 #change in potential energy when stone enters state 3 in JQ2=0 //no heat transfer when stone enters state 3 in J\n",

+      "W2=0 #no work done when stone enters state 3 in J\n",

+      "dKE2=-1 #change in kinetic energy when stone enters state 3\n",

+      "dU2=-dKE2 #change in internal energy when stone enters state 3 in J\n",

+      "print\"\\n hence,when stone has just come to rest in the bucket is \",round(dKE2),\"J\" \n",

+      "print\"\\n and dU is\",round(dU2),\"J\"  \n",

+      "dKE3=0 #change in kinetic energy when stone enters state 4\n",

+      "dPE=0 #change in potential energy when stone enters state 4 in J\n",

+      "W3=0 #no work done when stone enters state 4 in J\n",

+      "dU3=-1 #change in internal energy when stone enters state 4 in J\n",

+      "Q3=dU3 #heat transfer when stone enters state 4 in J\n",

+      "print\"\\n hence,when stone has entered state 4 is\",round(dU3),\"J\" \n",

+      "print\"\\n and Q3 is \",round(Q3),\"J\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " hence,when stone is  1000.0 J\n",

+        "\n",

+        " and change in potential energy is  -1.0 J\n",

+        "\n",

+        " hence,when stone has just come to rest in the bucket is  -1.0 J\n",

+        "\n",

+        " and dU is 1.0 J\n",

+        "\n",

+        " hence,when stone has entered state 4 is -1.0 J\n",

+        "\n",

+        " and Q3 is  -1.0 J\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.4:pg-136"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4\n",

+      "#Determinig the missing properties \n",

+      "\n",

+      "T1=300 #given temp. in Celsius\n",

+      "u1=2780 #given specific internal enrgy in kJ/kg\n",

+      "print\"From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\" #hiven pressure in kPa\n",

+      "u2=2000 #given specific intrernal energy in kJ/kg\n",

+      "print\"at P=2000 kPa\"\n",

+      "uf=906.4 #in kJ/kg\n",

+      "ug=2600.3 #in kJ/kg \n",

+      "x2=(u2-906.4)/(ug-uf) \n",

+      "print\"Also, under the given conditions\"\n",

+      "vf=0.001177 #in m^3/kg \n",

+      "vg=0.099627 #in m^3/kg\n",

+      "v2=vf+x2*(vg-vf)#Specific volume for water in m^3/kg\n",

+      "print\"\\n hence,specific volume for water is \",round(v2,5),\"m^3/kg\" \n",

+      "print\"\\n Therefore ,this state is \",round(x2,4),\"N\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\n",

+        "at P=2000 kPa\n",

+        "Also, under the given conditions\n",

+        "\n",

+        " hence,specific volume for water is  0.06474 m^3/kg\n",

+        "\n",

+        " Therefore ,this state is  0.6456 N\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.5:pg-138"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#calculating heat transfer for the given process\n",

+      "\n",

+      "Vliq=0.05 #volume of saturated liquid in m^3\n",

+      "vf=0.001043 #in m^3/kg\n",

+      "Vvap=4.95 #volume of saturated water vapour in m^3\n",

+      "vg=1.6940 #in m^3/kg\n",

+      "m1liq=Vliq/vf #mass of liquid in kg\n",

+      "m1liq=round(m1liq,2)\n",

+      "m1vap=Vvap/vg #mass of vapors in kg\n",

+      "m1vap=round(m1vap,2)\n",

+      "u1liq=417.36 #specific internal energy of liquid in kJ/kg\n",

+      "u1vap=2506.1 #specific internal energy of vapors in kJ/kg\n",

+      "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in kJ\n",

+      "m=m1liq+m1vap #total mass in kg\n",

+      "V=5.0 #total volume in m^3\n",

+      "v2=V/m #final specific volume in m^3/kg\n",

+      "print\"by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\"\n",

+      "u2=2600.5 #specific internal energy at final state in kJ/kg\n",

+      "U2=m*u2 #internal energy at final state in kJ\n",

+      "Q=U2-U1 #heat transfer for the process in kJ\n",

+      "print\"\\n hence,heat transfer for the process is\",round(Q),\"kJ\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\n",

+        "\n",

+        " hence,heat transfer for the process is 104935.0 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 36

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.5E:pg-140"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5E\n",

+      "#calculating heat transfer for the given process\n",

+      "\n",

+      "Vliq=1 #volume of saturated liquid in ft^3\n",

+      "vf=0.01672 #in ft^3/lbm\n",

+      "Vvap=99 #volume of saturated water vapour in m^3\n",

+      "vg=26.80 #in ft^3/lbm\n",

+      "m1liq=Vliq/vf #mass of liquid in lbm\n",

+      "m1liq=round(m1liq,2)\n",

+      "m1vap=Vvap/vg #mass of vapors in lbm\n",

+      "m1vap=round(m1vap,2)\n",

+      "u1liq=180.1 #specific internal energy of liquid in Btu/lbm\n",

+      "u1vap=1077.6 #specific internal energy of vapors in Btu/lbm\n",

+      "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in Btu\n",

+      "U1=round(U1)\n",

+      "m=m1liq+m1vap #total mass in lbm\n",

+      "V=100.0 #total volume in ft^3\n",

+      "v2=V/m #final specific volume in ft^3/lbm\n",

+      "print\"by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\"\n",

+      "u2=1117.0 #specific internal energy at final state in Btu/lbm\n",

+      "U2=m*u2 #internal energy at final state in Btu\n",

+      "U2=round(U2)\n",

+      "Q=U2-U1 #heat transfer for the process in Btu\n",

+      "print\"\\n hence,heat transfer for the process is\",round(Q),\"Btu\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\n",

+        "\n",

+        " hence,heat transfer for the process is 56182.0 Btu\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.6:pg-143"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 6\n",

+      "#calculating work and heat transfer for the process\n",

+      "\n",

+      "V1=0.1 #volume of cylinder in m^3\n",

+      "m=0.5 #mass of steam in kg\n",

+      "v1=V1/m #specific volume of steam in m^3/kg\n",

+      "vf=0.001084 #m^3/kg\n",

+      "vfg=0.4614 #m^3/kg\n",

+      "x1=(v1-vf)/vfg #quality\n",

+      "hf=604.74 #kJ/kg\n",

+      "hfg=2133.8#kJ/kg\n",

+      "h2=3066.8 #final specific heat enthalpy in kJ/kg\n",

+      "h1=hf+x1*hfg #initial specific enthalpy in kJ/kg\n",

+      "Q=m*(h2-h1) #heat transfer for this process in kJ\n",

+      "P=400 #pressure inside cylinder in kPa\n",

+      "v2=0.6548 #specific enthalpy in m^3/kg\n",

+      "W=m*P*(v2-v1) #work done for the process in kJ\n",

+      "print\"\\n hence, work done for the process is\",round(W),\"kJ\" \n",

+      "print\"\\n and heat transfer is \",round(Q,1),\"kJ\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, work done for the process is 91.0 kJ\n",

+        "\n",

+        " and heat transfer is  771.1 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.7:pg-144"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# example 7\n",

+      "#calculating work and heat transfer for the process\n",

+      "V1=0.010 #volume of cylinder in m^3\n",

+      "V2=0.040 # m^3\n",

+      "P1=573 #kPa\n",

+      "v1=0.03606 #specific volume of steam in m^3/kg\n",

+      "u1=389.2 #kJ/kg\n",

+      "m=V1/v1#mass of steam in kg\n",

+      "v2=v1*(V2/V1)\n",

+      "P2=163 #kPa\n",

+      "u2=395.8 # kJ/kg\n",

+      "W=8 # kJ\n",

+      "Q2=m*(u2-u1)+W\n",

+      "u3=383.4#kJ/kg\n",

+      "W2=m*(u1-u3)#kJ\n",

+      "print\"\\n and heat transfer is \",round(Q2,2),\"kJ\" \n",

+      "print\"\\n hence, work done for the process is\",round(W2,1),\"kJ\" \n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " and heat transfer is  9.83 kJ\n",

+        "\n",

+        " hence, work done for the process is 1.6 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.8:pg-151"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8\n",

+      "#calculating change in enthalpy\n",

+      "import math\n",

+      "\n",

+      "h1=273.2 #specific heat enthalpy for oxygen at 300 K\n",

+      "h2=1540.2 #specific heat enthalpy for oxygen at 1500 K\n",

+      "T1=300 #initial temperature in K\n",

+      "T2=1500 #final temparature in K\n",

+      "\n",

+      "dh1=h2-h1 #this change in specific heat enthalpy is calculated using ideal gas tables \n",

+      "dh3=0.922*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K\n",

+      "dh4=1.0767*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature\n",

+      "print\"\\n Hence,change in specific heat enthalpy if ideal gas tables are used is \",round(dh1,1),\"kJ/kg\"\n",

+      "print\"\\n if specific heat is assumed to be constant and using its value at T1 is\",round(dh3,1),\"kJ/kg\"\n",

+      "print\"\\n if specific heat is assumed to be constant at its value at (T1+T2)/2 is\",round(dh4,1),\"kJ/kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " Hence,change in specific heat enthalpy if ideal gas tables are used is  1267.0 kJ/kg\n",

+        "\n",

+        " if specific heat is assumed to be constant and using its value at T1 is 1106.4 kJ/kg\n",

+        "\n",

+        " if specific heat is assumed to be constant at its value at (T1+T2)/2 is 1292.0 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.9:pg-152"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9\n",

+      "#determining amount of heat transfer\n",

+      "\n",

+      "P=150 #pressure of nitrogen in cylinder in kPa\n",

+      "V=0.1 #initial volume of cylinder in m^3\n",

+      "T1=25 #initial temperature of nitrogen in celsius\n",

+      "T2=150 #final tempareture of nitrogen in celsius\n",

+      "R=0.2968 #in kJ/kg-K\n",

+      "m=P*V/(R*(T1+273)) #mass of nitrogen in kg\n",

+      "Cv=0.745 #constant volume specific heat for nitrogen in kJ/kg-K\n",

+      "W=-20 #work done on nitrogen gas in kJ\n",

+      "Q=m*Cv*(T2-T1)+W #heat transfer during the process in kJ\n",

+      "print\"\\n hence,the heat transfer for the above process is\",round(Q,1),\"kJ\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the heat transfer for the above process is -4.2 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 43

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.9E:pg-153"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9E\n",

+      "#determining amount of heat transfer\n",

+      "\n",

+      "P=20 #pressure of nitrogen in cylinder in lbf/in^2\n",

+      "V=2 #initial volume of cylinder in ft^3\n",

+      "T1=80 #initial temperature of nitrogen in F\n",

+      "T2=300 #final tempareture of nitrogen in F\n",

+      "R=55.15 #in ft*lbf/lbm-R\n",

+      "m=P*V*144/(R*(540)) #mass of nitrogen in lbm\n",

+      "Cv=0.177 #constant volume specific heat for nitrogen in Btu/lbm-R\n",

+      "W=-9.15 #work done on nitrogen gas in Btu\n",

+      "Q=m*Cv*(T2-T1)+W #heat transfer during the process in Btu\n",

+      "print\"\\n hence,the heat transfer for the above process is\",round(Q,2),\"Btu\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the heat transfer for the above process is -1.62 Btu\n"

+       ]

+      }

+     ],

+     "prompt_number": 31

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.10:pg-155"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 10\n",

+      "#calculating rate of increase of internal energy\n",

+      "\n",

+      "W=-12.8*20 #power consumed in  J/s\n",

+      "Q=-10 #heat transfer rate from battery in J/s\n",

+      "r=Q-W #rate of increase of internal energy\n",

+      "print\"\\n hence,the rate of increase of internal energy is\",round(r),\"J/s\"  "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the rate of increase of internal energy is 246.0 J/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 45

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.11:pg-155"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 11\n",

+      "#rate of change of temperature\n",

+      "\n",

+      "Q=1500.0 #power produced by burning wood in J/s\n",

+      "mair=1 #mass of air in kg\n",

+      "mwood=5 #mass of soft pine wood in kg \n",

+      "miron=25 #mass of cast iron in kg\n",

+      "Cvair=0.717 #constant volume specific heat for air in kJ/kg\n",

+      "Cwood=1.38 #constant volume specific heat for wood in kJ/kg\n",

+      "Ciron=0.42 #constant volume specific heat for iron in kJ/kg\n",

+      "dT=75-20 #increase in temperature in Celsius\n",

+      "T=(Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) #rate of change of temperature in K/s\n",

+      "dt=(dT/T)/60 #in minutes\n",

+      "print\" hence,the rate of change of temperature is\",round(T,4),\"k/s\"  \n",

+      "print\" and time taken to reach a temperature of T is\",round(dt),\"min\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " hence,the rate of change of temperature is 0.0828 k/s\n",

+        " and time taken to reach a temperature of T is 11.0 min\n"

+       ]

+      }

+     ],

+     "prompt_number": 49

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter6_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter6_6.ipynb
new file mode 100755
index 00000000..953e821b
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter6_6.ipynb
@@ -0,0 +1,652 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:c03e579abb4dafb7ea55c1a7f66b6338546355647f3c3d21afe4e72021a3ad5f"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter6:FIRST-LAW ANALYSIS FOR A CONTROL VOLUME"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.1:pg-182"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1\n",

+      "#calculating mass flow rate in kg/s\n",

+      "import math\n",

+      "R=0.287 #in kJ/kg-K\n",

+      "T=25 #temperature in celsius\n",

+      "P=150 #pressure in kPa\n",

+      "v=R*(T+273.15)/P #specific volume in m^3/kg\n",

+      "D=0.2 #diameter of pipe in metre\n",

+      "A=math.pi*D**2/4 #cross sectional area in m^2\n",

+      "V=0.1 #velocity of air in m/s\n",

+      "m=V*A/v #mass flow rate in kg/s\n",

+      "print\"\\n hence,the mass flow rate is\",round(m,4),\"kg/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the mass flow rate is 0.0055 kg/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.2:pg-184"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2\n",

+      "#work done for adding the fluid\n",

+      "\n",

+      "P=600 #pressure in kPa\n",

+      "m=1 #in kg\n",

+      "v=0.001 #specific volume in m^3/kg\n",

+      "W=P*m*v #necessary work in kJ for adding the fluid \n",

+      "print\" \\n hence,the work involved in this process is\",round(W,4),\"kJ\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "  \n",

+        " hence,the work involved in this process is 0.6 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.3:pg-188"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#rate of flow of water\n",

+      "\n",

+      "hir=441.89 #in kJ/kg for refrigerant using steam table \n",

+      "her=249.10 #in kJ/kg for refrigerant using steam table\n",

+      "hiw=42 #in kJ/kg for water using steam table\n",

+      "hew=83.95 #in kJ/kg for water using steam table\n",

+      "mr=0.2 #the rate at which refrigerant enters the condenser in kg/s\n",

+      "mw=mr*(hir-her)/(hew-hiw) #rate of flow of water in kg/s\n",

+      "print\"\\n hence,the rate at which cooling water flows thorugh the condenser is\",round(mw,3),\"kg/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the rate at which cooling water flows thorugh the condenser is 0.919 kg/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.4:pg-190"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4\n",

+      "#determining quality of steam\n",

+      "\n",

+      "hi=2850.1 #initial specific heat enthalpy for steam in kJ/kg\n",

+      "Vi=50 #initial velocity of steam in m/s\n",

+      "Ve=600 #final velocity of steam in m/s\n",

+      "he=hi+Vi**2/(2*1000)-Ve**2/(2*1000) #final specific heat enthalpy for steam in kJ/kg\n",

+      "hf=467.1 #at final state in kJ/kg\n",

+      "hfg=2226.5 #at final state in kJ/kg\n",

+      "xe=(he-hf)/hfg #quality of steam in final state\n",

+      "print\" \\n hence, the quality is\",round(xe,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " hence, the quality is 0.99\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.4E:pg-191"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4E\n",

+      "#determining quality of steam\n",

+      "\n",

+      "hi=1227.5 #initial specific heat enthalpy for steam in Btu/lbm\n",

+      "Vi=200 #initial velocity of steam in ft/s\n",

+      "Ve=2000 #final velocity of steam in ft/s\n",

+      "he=hi+Vi**2/(2*32.17*778)-Ve**2/(2*32.17*778) #final specific heat enthalpy for steam in Btu/lbm\n",

+      "hf=198.31 #at final state in Btu/lbm\n",

+      "hfg=958.81 #at final state in Btu/lbm\n",

+      "xe=(he-hf)/hfg #quality of steam in final state\n",

+      "print\" \\n hence, the quality is\",round(xe,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " hence, the quality is 0.99\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.5:pg-193"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#quality of ammonia leaving expansion valve\n",

+      "\n",

+      "hi=346.8 #specific heat enthalpy for ammonia at initial state in kJ/kg\n",

+      "he=hi #specific heat enthalpy for ammonia at final state will be equal that at initial state because it is a throttling process\n",

+      "hf=134.4 #at final state in kJ/kg\n",

+      "hfg=1296.4#at final state in kJ/kg\n",

+      "xe=(he-hf)*100/hfg #quality at final state\n",

+      "print\"\\n hence,quality of the ammonia leaving the expansion valve is\",round(xe,2),\"%\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,quality of the ammonia leaving the expansion valve is 16.38 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.6:pg-194"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 6\n",

+      "#power output of turbine in kW\n",

+      "\n",

+      "hi=3137 #initial specific heat of enthalpy in kJ/kg\n",

+      "he=2675.5 #final specific heat of enthalpy in kJ/kg\n",

+      "Vi=50.0 #initial velocity of steam in m/s\n",

+      "Ve=100 #final velocity of steam in m/s\n",

+      "Zi=6 #height of inlet conditions in metres\n",

+      "Ze=3 #height of exit conditions in metres\n",

+      "m=1.5 #mass flow rate of steam in kg/s\n",

+      "g=9.8066 #acc. due to gravity in m/s^2\n",

+      "Qcv=-8.5 #heat transfer rate from turbine in kW\n",

+      "Wcv=Qcv+m*(hi+Vi**2/(2*1000)+g*Zi/1000)-m*(he+Ve**2/(2*1000)+g*Ze/1000) #power output of turbine in kW\n",

+      "print\"\\n hence,the power output of the turbine is\",round(Wcv,1),\"kW\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the power output of the turbine is 678.2 kW\n"

+       ]

+      }

+     ],

+     "prompt_number": 28

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.7:pg-196"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7\n",

+      "#heat transfer rate in aftercooler\n",

+      "\n",

+      "V1=0 #we assume initial velocity to be zero because its given that it enters with a low velocity\n",

+      "V2=25.0 #final velocity with which carbon dioxide exits in m/s\n",

+      "h2=401.52 #final specific enthalpy of heat when carbon dioxide exits in kJ/kg\n",

+      "h1=198 #initial specific enthalpy of heat in kJ/kg\n",

+      "w=h1-h2-V2**2/(2*1000) #in kJ/kg\n",

+      "Wc=-50 #power input to the compressor in kW\n",

+      "m=Wc/w #mass flow rate of carbon dioxide in kg/s\n",

+      "h3=257.9 #final specific enthalpy of heat when carbon dioxide flows into a constant pressure aftercooler\n",

+      "Qcool=-m*(h3-h2) #heat transfer rate in the aftercooler in kW\n",

+      "print\" \\n hence,heat transfer rate in the aftercooler is\",round(Qcool,1),\"kW\" \n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " hence,heat transfer rate in the aftercooler is 35.2 kW\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.8:pg-197"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8\n",

+      "#Required pump work\n",

+      "\n",

+      "m=1.5 #mass flow rate of water in kg/s\n",

+      "g=9.807 #acceleration due to gravity in m/s^2\n",

+      "Zin=-15 #depth of water pump in well in metres\n",

+      "Zex=0 #in metres\n",

+      "v=0.001001 #specific volume in m^3/kg\n",

+      "Pex=400+101.3 #exit pressure in kPa\n",

+      "Pin=90 #in kPa\n",

+      "W=m*(g*(Zin-Zex)*0.001-(Pex-Pin)*v) #power input in kW\n",

+      "print\" \\n Hence, the pump requires power of input is\",-round(W,2),\"kW\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " Hence, the pump requires power of input is 0.84 kW\n"

+       ]

+      }

+     ],

+     "prompt_number": 39

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.9:pg-198"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9\n",

+      "#heat tranfer in simple steam power plant\n",

+      "\n",

+      "h1=3023.5 #specific heat of enthalpy of steam leaving boiler in kJ/kg\n",

+      "h2=3002.5 #specific heat of enthalpy of steam entering turbine in kJ/kg\n",

+      "x=0.9 #quality of steam entering condenser\n",

+      "hf=226 #in kJ/kg\n",

+      "hfg=2373.1 #in kJ/kg\n",

+      "h3=hf+x*hfg #specific heat of enthalpy of steam entering condenser in kJ/kg\n",

+      "h4=188.5 #specific heat of enthalpy of steam entering pump in kJ/kg\n",

+      "q12=h2-h1 #heat transfer in line between boiler and turbine in kJ/kg\n",

+      "w23=h2-h3 #turbine work in kJ/kg\n",

+      "q34=h4-h3 #heat transfer in condenser\n",

+      "w45=-4 #pump work in kJ/kg\n",

+      "h5=h4-w45 #in kJ/kg\n",

+      "q51=h1-h5 #heat transfer in boiler in kJ/kg\n",

+      "print\"\\n hence, heat transfer in line between boiler and turbine is\",round(q12,1),\"kJ/kg\"  \n",

+      "print\"\\n hence, turbine work is\",round(w23,1),\"kJ/kg\" \n",

+      "print\"\\n hence, heat transfer in condenser is \",round(q34,1),\"kJ/kg\" \n",

+      "print\"\\n hence, heat transfer in boiler is \",round(q51,1),\"kJ/kg\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, heat transfer in line between boiler and turbine is -21.0 kJ/kg\n",

+        "\n",

+        " hence, turbine work is 640.7 kJ/kg\n",

+        "\n",

+        " hence, heat transfer in condenser is  -2173.3 kJ/kg\n",

+        "\n",

+        " hence, heat transfer in boiler is  2831.0 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 43

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.10:pg-200"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 10\n",

+      "#analysis of refrigerator\n",

+      "\n",

+      "hf4=167.4 #in kJ/kg\n",

+      "hfg4=215.6 #in kJ/kg\n",

+      "h3=241.8 #specific heat of enthalpy of R-134a entering expansion valve\n",

+      "h4=h3 #specific heat of enthalpy of R-134a leaving expansion valve\n",

+      "h1=387.2 #in kJ/kg\n",

+      "h2=435.1 #in kJ/kg\n",

+      "x4=(h3-hf4)/hfg4 #quality of R-134a at evaporator inlet\n",

+      "m=0.1 #mass flow rate in kg/s\n",

+      "Qevap=m*(h1-h4) #rate of heat transfer to the evaporator\n",

+      "Wcomp=-5 #power input to compressor in kW\n",

+      "Qcomp=m*(h2-h1)+Wcomp #rate of heat transfer from compressor\n",

+      "print\"\\n hence, the quality at the evaporator inlet is \",round(x4,3), \n",

+      "print\"\\n hence, the rate of heat transfer to the evaporator is \",round(Qevap,2),\n",

+      "print\"\\n hence, rate of heat transfer from the compressor is\",round(Qcomp,2), "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, the quality at the evaporator inlet is  0.345 \n",

+        " hence, the rate of heat transfer to the evaporator is  14.54 \n",

+        " hence, rate of heat transfer from the compressor is -0.21\n"

+       ]

+      }

+     ],

+     "prompt_number": 45

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.11:pg-204"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 11\n",

+      "#Determining the final temperature of steam\n",

+      "\n",

+      "u2=3040.4 #final internal energy in kJ/kg\n",

+      "hi=u2 #in kJ/kg\n",

+      "P2=1.4 #final Pressure in MPa\n",

+      "print\"Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \"\n",

+      "T2=452 #final temperature in Celsius"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \n"

+       ]

+      }

+     ],

+     "prompt_number": 46

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.12:pg-206"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 12\n",

+      "#Calculating mass flow of steam in tank\n",

+      "\n",

+      "V1=0.4 #initial volume fo tank in m^3\n",

+      "v1=0.5243 #initial specific volume in m^3/kg\n",

+      "h1=3040.4 #initial specific enthalpy in kJ/kg\n",

+      "u1=2548.9 #initial specific internal energy in kJ/kg\n",

+      "m1=V1/v1 #initial mass of steam in tank in kg\n",

+      "V2=0.4 #final volume in m^3\n",

+      "print\"let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\" \n",

+      "T2=342 #final temperature in Celsius\n",

+      "v2=0.1974 #final specific volume in m^3/kg\n",

+      "m2=V2/v2 #final mass of the steam in the tank in kg\n",

+      "m=m2-m1 #mass of steam that flowsinto the tank\n",

+      "print\" \\n Hence,mass of the steam that flows into the tank is\",round(m,3),\"kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\n",

+        " \n",

+        " Hence,mass of the steam that flows into the tank is 1.263 kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 48

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.13:pg-207"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 13\n",

+      "#Calculating mass flow of steam in tank\n",

+      "\n",

+      "vf1=0.001725 #in m^3/kg\n",

+      "vf2=0.0016 #in m^3/kg\n",

+      "uf1=368.7 #in kJ/kg\n",

+      "uf2=226 #in kJ/kg\n",

+      "vg1=0.08313 #in m^3/kg\n",

+      "vfg2=0.20381\n",

+      "ug1=1341 #in kJ/kg\n",

+      "ufg2=1099.7 #in kJ/kg\n",

+      "Vf=1 #initial volume of liquid in m^3\n",

+      "Vg=1 #initial volume of vapor in m^3\n",

+      "mf1=Vf/vf1 #initial mass of liquid in kg\n",

+      "mg1=Vg/vg1 #initial mass of vapor in kg\n",

+      "m1=mf1+mg1 #initial mass of liquid in kg\n",

+      "he=1461.1 #in kJ/kg\n",

+      "V=2 #volume of tank in m^3\n",

+      "print\"m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\"\n",

+      "print\"Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\"\n",

+      "x2=((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) #quality of ammonia\n",

+      "v2=0.00160+(0.20381*x2) #final specific volume in m^3/kg\n",

+      "m2=V/v2 #final mass of ammonia in kg\n",

+      "m=m1-m2 #mass of ammonia withdrawn\n",

+      "print\" \\n Hence,mass of ammonia withdrawn is\",round(m,1),\"kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\n",

+        "Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\n",

+        " \n",

+        " Hence,mass of ammonia withdrawn is 72.7 kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 50

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.13E:pg-209"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 13E\n",

+      "#Calculating mass flow of steam in tank\n",

+      "\n",

+      "vf1=0.02747#in ft3/lbm\n",

+      "vf2=0.02564#in ft3/lbm\n",

+      "uf1=153.89#in Btu/lbm\n",

+      "uf2=97.16#in Btu/lbm\n",

+      "vg1=1.4168 #in ft^3/lbm\n",

+      "vfg2=3.2647#in ft^3/lbm\n",

+      "ug1=576.23#in Btu/lbm\n",

+      "ufg2=472.78#in Btu/lbm\n",

+      "Vf=25 #initial volume of liquid in ft^3\n",

+      "Vg=25 #initial volume of vapor in ft^3\n",

+      "mf1=Vf/vf1 #initial mass of liquid in lbm\n",

+      "mg1=Vg/vg1 #initial mass of vapor in lbm\n",

+      "m1=mf1+mg1 #initial mass of liquid in lbm\n",

+      "he=628 #in Btu/lbm\n",

+      "V=50 #volume of tank in ft^3\n",

+      "print\"m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.02564+3.2647*x2) and u2=uf2+x2*ufg2=97.16+3.2647*x2.\"\n",

+      "print\"Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\"\n",

+      "x2=((50*628)-(50*97.16)-(432391*0.02564))/((432391*3.2647)+(50*472.98)) #quality of ammonia\n",

+      "x2=round(x2,6)\n",

+      "v2=0.02564+(3.2647*x2) #final specific volume in ft^3/lbm\n",

+      "v2=round(v2,6)\n",

+      "m2=V/v2 #final mass of ammonia in lbm\n",

+      "m=m1-m2 #mass of ammonia withdrawn\n",

+      "print\" \\n Hence,mass of ammonia withdrawn is\",round(m,1),\"lbm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.02564+3.2647*x2) and u2=uf2+x2*ufg2=97.16+3.2647*x2.\n",

+        "Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\n",

+        " \n",

+        " Hence,mass of ammonia withdrawn is 105.3 lbm\n"

+       ]

+      }

+     ],

+     "prompt_number": 27

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter7_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter7_6.ipynb
new file mode 100755
index 00000000..02425dea
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter7_6.ipynb
@@ -0,0 +1,217 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:a60a19bd48510a099eb4483e5176d858019753b734df1f9001e5af0e3bc7fc2e"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter7:THE SECOND LAW OF THERMODYNAMICS"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.1:pg-241"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1\n",

+      "#rate of fuel consumption\n",

+      "\n",

+      "W=136*0.7355 #output of automobile engine in kW\n",

+      "neng=0.3 #thermal efficiency of automobile engine\n",

+      "Qh=W/neng #energy output of fuel in kW\n",

+      "Q1=Qh-W #total rate of energy rejected to the ambient\n",

+      "qh=35000.0 #energy output of fuel in kJ/kg\n",

+      "m=Qh/qh #rate of fuel consumption in kg/s\n",

+      "print\"\\n hence,total rate of energy rejected is\",round(Q1),\"kW\" \n",

+      "print\"\\n and rate of fuel consumption is\",round(m,4),\"kg/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,total rate of energy rejected is 233.0 kW\n",

+        "\n",

+        " and rate of fuel consumption is 0.0095 kg/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.2:pg-243"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2\n",

+      "#coefficient of performance of refrigerator\n",

+      "\n",

+      "Qh=400 #heat rejected to kitchen air in W\n",

+      "W=150.0 #electrical input power in W\n",

+      "Q1=Qh-W #rate of energy taken out to cold space in W\n",

+      "B=Ql/W #coefficicent of performnace of refrigerator\n",

+      "print\"\\n hence,rate of energy taken out of the cold space is\",round(Q1),\"W\"\n",

+      "print\"\\n and coefficient of performance of the refrigerator is\",round(B,2),"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,rate of energy taken out of the cold space is 250.0 W\n",

+        "\n",

+        " and coefficient of performance of the refrigerator is 1.67\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.3:pg-256"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "# temperature at which the pressure go to zero\n",

+      "\n",

+      "P1=110.9#pressur at the ice point in kPa\n",

+      "P2=151.5#pressure at the steam point in kPa\n",

+      "T1=0#in C\n",

+      "T2=100#in C\n",

+      "S=(P2-P1)/(T2-T1)#slope in kPa/C\n",

+      "T=T1-(P1/S)# absolute ideal gas temperature in C\n",

+      "print\"\\n hence,the absolute ideal gas temperature is\",round(T,2),\"C\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the absolute ideal gas temperature is -273.15 C\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.4:pg-260"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4\n",

+      "#comparison of ideal carnot heat engine with actual heat engine\n",

+      "\n",

+      "Qh=1000.0 #rate of heat transfer to heat engine in kW\n",

+      "W=450.0 #rate of production of work in kW\n",

+      "Ql=Qh-W #rate of heat rejected by heat engine in kW\n",

+      "nthermal=W/Qh #efficiency from the definition of efficiency\n",

+      "Tl=300 #temperature of surroundings in K\n",

+      "Th=550.0 #temperature of heat source in Celsius\n",

+      "ncarnot=1-Tl/(Th+273.15) #efficiency if heat engine is considered to be ideal carnot heat engine\n",

+      "W2=ncarnot*Qh #rate of work production if heat engine is assumed to be ideal carnot heat engine in kW\n",

+      "Q12=Qh-W2 #rate of heat rejected by heat engine in kW if heat engine is assumed to be ideal carnot heat engine\n",

+      "print\"\\n hence,energy discarded to the ambient surroundings iS\",round(Q12,4),\"kw\"\n",

+      "print\"\\n and the engine efficiency is\",round(ncarnot,4),"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,energy discarded to the ambient surroundings iS 364.4536 kw\n",

+        "\n",

+        " and the engine efficiency is 0.6355\n"

+       ]

+      }

+     ],

+     "prompt_number": 53

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.5:pg-261"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#calculating required work\n",

+      "\n",

+      "Tl=24+273.15 #room temperature in Kelvins\n",

+      "Th=35+273.15 #atmospheric temperature in Kelvins\n",

+      "Ql=4 #rate of heat rejection from room\n",

+      "B=Tl/(Th-Tl) #coefficient of performance of air conditioner\n",

+      "W=Ql/B #required work in kW\n",

+      "print\"\\n hence,the magnitude of reqiured work is\",round(W,2),\"kw\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the magnitude of reqiured work is 0.15 kw\n"

+       ]

+      }

+     ],

+     "prompt_number": 40

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter8_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter8_6.ipynb
new file mode 100755
index 00000000..dce4d1fa
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter8_6.ipynb
@@ -0,0 +1,444 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:79711d3b5ce866eb594601c655e2c85bdace3e8f861ba88d25f2b6f7e8d8655f"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter8:ENTROPY"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.1:pg-290"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1\n",

+      "#coefficient of performance of refrigerator\n",

+      "\n",

+      "Th=60 #temperature at which heat is rejected from R-134a\n",

+      "Tl=0 #temperature at which heat is absorbed into the R-134a \n",

+      "s1=1.7262 #specific entropy at 0 Celsius\n",

+      "s2=s1 #process of state change from 1-2 is isentropic \n",

+      "s3=1.2857 #specific entropy at 60 celsius\n",

+      "s4=s3 #process of state change from 3-4 is isentropic\n",

+      "print\"if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\"\n",

+      "P2=1400+(1600-1400)*(1.7262-1.736)/(1.7135-1.736) #pressure after compression in kPa\n",

+      "B=(Th+273.15)/(Th-Tl) #coefficient of performance of refrigerator\n",

+      "print\" \\n hence,pressure after compression is \",round(P2,1),\"kPa\"\n",

+      "print\"\\n and coefficient of performance of refrigerator is\",round(B,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\n",

+        " \n",

+        " hence,pressure after compression is  1487.1 kPa\n",

+        "\n",

+        " and coefficient of performance of refrigerator is 5.55\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.2:pg-290"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2 \n",

+      "#heat transfer in a given process\n",

+      "\n",

+      "u1=87.94 #specific internal energy of R-12 at state 1 in kJ/kg\n",

+      "u2=276.44 #specific internal energy of R-12 at state 2 in kJ/kg\n",

+      "s1=0.3357 #specific entropy at state 1 in kJ/kg-K\n",

+      "s2=1.2108 #specific entropy at state 2 in kJ/kg-K\n",

+      "V=0.001 #volume of saturated liquid in m^3\n",

+      "v1=0.000923 #specific volume in m^3/kg\n",

+      "m=V/v1 #mass of saturated liquid in kg\n",

+      "T=20 #temperature of liquid in celsius\n",

+      "Q12=m*(T+273.15)*(s2-s1) #heat transfer in kJ to accomplish the process\n",

+      "W12=m*(u1-u2)+Q12 #work required to accomplish the process\n",

+      "print\" \\n hence,work required to accomplish the process is\",round(W12,1),\"KJ\"\n",

+      "print\" \\n and heat transfer is\",round(Q12,1),\"KJ\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " hence,work required to accomplish the process is 73.7 KJ\n",

+        " \n",

+        " and heat transfer is 277.9 KJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.3:pg-293"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#entropy change\n",

+      "import math\n",

+      "C=4.184 # specific heat of water in kJ/kg-K\n",

+      "T1=20 #initial temperature of water in celsius\n",

+      "T2=90 #final temperature of water in celsius\n",

+      "dS1=C*math.log((T2+273.2)/(T1+273.2)) #change in entropy in kJ/kg-K\n",

+      "dS2=1.1925-0.2966 #in kJ/kg-K using steam tables\n",

+      "print\"\\n hence,change in entropy assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",

+      "print\"\\n using steam table is\",round(dS2,4),\"KJ/kg-k\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,change in entropy assuming constant specific heat is 0.8958 KJ/kg-k\n",

+        "\n",

+        " using steam table is 0.8959 KJ/kg-k\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.4:pg-295"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4\n",

+      "#entropy change with different assumptions\n",

+      "import math\n",

+      "T1=300 #initial temperature in kelvins\n",

+      "T2=1500 #final temperature in kelvins\n",

+      "P1=200.0 #initial pressure in kPa\n",

+      "P2=150.0 #final pressure in kPa\n",

+      "R=0.2598 # in kJ/kg-K\n",

+      "Cp=0.922 #specific heat in kJ/kg-K at constant pressure\n",

+      "dsT2=8.0649 #in kJ/kg-K\n",

+      "dsT1=6.4168 #in kJ/kg-K\n",

+      "dS1=dsT2-dsT1-R*math.log(P2/P1) #entropy change calculated using ideal gas tables\n",

+      "dS3=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat in kJ/kg-K\n",

+      "dS4=1.0767*math.log(T2/T1)+0.0747 #entropy change assuming specific heat is constant at its value at 990K\n",

+      "print\"\\n hence,change in entropy using ideal gas tables is\",round(dS1,4),\"KJ/kg-k\"\n",

+      "print\"\\n hence,change in entropy using the value of specific heat at 300K is\",round(dS3,4),\"KJ/kg-k\"\n",

+      "print\"\\n hence,change in entropy assuming specific heat is constant at its value at 900K is \",round(dS4,4),\"KJ/kg-k\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,change in entropy using ideal gas tables is 1.7228 KJ/kg-k\n",

+        "\n",

+        " hence,change in entropy using the value of specific heat at 300K is 1.5586 KJ/kg-k\n",

+        "\n",

+        " hence,change in entropy assuming specific heat is constant at its value at 900K is  1.8076 KJ/kg-k\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.5:pg-296\n"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#entropy change\n",

+      "import math\n",

+      "Cp=1.004 #specific heat at constant pressure in kJ/kg-K\n",

+      "R=0.287 #gas constant in kJ/kg-K\n",

+      "P1=400.0 #initial pressure in kPa\n",

+      "P2=300.0 #final pressure in kPa\n",

+      "T1=300 #initial temperature in K\n",

+      "T2=600 #final temperature in K\n",

+      "dS1=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat\n",

+      "s1=6.8693 #specific entropy at T1\n",

+      "s2=7.5764 #specific entropy at T2\n",

+      "dS2=s2-s1-R*math.log(P2/P1) #entropy change assuming variable specific heat\n",

+      "print\"\\n hence,entropy change assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",

+      "print\"\\n and assuming variable specific heat is\",round(dS2,4),\"KJ/kg-k\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,entropy change assuming constant specific heat is 0.7785 KJ/kg-k\n",

+        "\n",

+        " and assuming variable specific heat is 0.7897 KJ/kg-k\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.6:pg-297"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 6\n",

+      "#work done by air\n",

+      "import math\n",

+      "T1=600 #initial temperature of air in K\n",

+      "P1=400.0 #intial pressure of air in kPa\n",

+      "P2=150.0 #final pressure in kPa\n",

+      "u1=435.10 #specific internal energy at temperature T1 in kJ/kg\n",

+      "sT1=7.5764 #specific entropy at temperature T1 in kJ/kg-K\n",

+      "R=0.287 #gas constant in kJ/kg-K\n",

+      "ds=0\n",

+      "sT2=ds+sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n",

+      "print\"we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\"\n",

+      "T2=457 #final temperature in K\n",

+      "u2=328.14 #specific internal energy at temperature T2 in kJ/kg\n",

+      "w=u1-u2 #work done by air in kJ/kg\n",

+      "print\"\\n hence,work done by air is\",round(w,4),\"KJ/kg\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\n",

+        "\n",

+        " hence,work done by air is 106.96 KJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.7:pg-300"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7\n",

+      "#work and heat transfer\n",

+      "\n",

+      "P2=500  #final pressure in cylinder in kPa\n",

+      "P1=100 #initial pressure in cylinder in kPa\n",

+      "T1=20+273.2 #initial temperature inside cylinder in Kelvins\n",

+      "n=1.3 \n",

+      "T2=(T1)*(P2/P1)**((n-1)/n) #final temperature inside cylinder in K\n",

+      "R=0.2968 #gas constant in kJ/kg-K\n",

+      "w12=R*(T2-T1)/(1-n) #work in kJ/kg\n",

+      "Cvo=0.745 #specific heat at constant volume in kJ/kg-K\n",

+      "q12=Cvo*(T2-T1)+w12 #heat transfer in kJ/kg\n",

+      "print\" \\n hence,work done is\",round(w12,2),\"KJ/kg\" \n",

+      "print\"\\n and heat transfer are\",round(q12,1),\"KJ/kg\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " hence,work done is -130.47 KJ/kg\n",

+        "\n",

+        " and heat transfer are -32.2 KJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.8:pg-308"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#calculating increase in entropy\n",

+      "\n",

+      "m=1 #mass of saturated water vapour\n",

+      "sfg=6.0480 #in kJ/K\n",

+      "T=25 #temperature of surrounding air in celsius\n",

+      "dScm=-m*sfg #change in entropy of control mass in kJ/K\n",

+      "hfg=2257.0 #in kJ/kg\n",

+      "Qtosurroundings=m*hfg #heat transferred to surroundings in kJ\n",

+      "dSsurroundings=Qtosurroundings/(T+273.15) #in kJ/K\n",

+      "dSnet=dScm+dSsurroundings #net increase in entropy in kJ/K\n",

+      "print\" hence,net increase in entropy of water plus surroundings is \",round(dSnet,3),\"KJ/k\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " hence,net increase in entropy of water plus surroundings is  1.522 KJ/k\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.9:pg-309"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9\n",

+      "#entropy generation\n",

+      "\n",

+      "Qout=1.0 #value of heat flux generated by 1kW of electric power\n",

+      "T=600.0 #temperature of hot wire surface in K\n",

+      "Sgen=Qout/T #entropy generation in kW/K\n",

+      "print\"\\n hence,entropy generation is\",round(Sgen,5),\"KW/k\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,entropy generation is 0.00167 KW/k\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.10:pg-310"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 10\n",

+      "#Determiining the entropy generated\n",

+      "\n",

+      "B=4.0#COP of air conditioner\n",

+      "W=10.0 #power input of air conditioner in kW\n",

+      "Qh=B*W #in kW\n",

+      "Ql=Qh-W #in kW\n",

+      "Thigh=323 #in Kelvin\n",

+      "Tlow=263 #in Kelvin\n",

+      "SgenHP=(Qh*1000/Thigh)-(Ql*1000/Tlow) #in W/K\n",

+      "Tl=281 # in K\n",

+      "Th=294 #in K\n",

+      "SgenCV1=Ql*1000/Tlow-Ql*1000/Tl #in W/K\n",

+      "SgenCV2=Qh*1000/Th-Qh*1000/Thigh #in W/K\n",

+      "SgenTOT=SgenCV1+SgenCV2+SgenHP #in W/K\n",

+      "print\" \\n Hence,Total entropy generated is\",round(SgenTOT,1),\"W/k\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " \n",

+        " Hence,Total entropy generated is 29.3 W/k\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter9_6.ipynb b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter9_6.ipynb
new file mode 100755
index 00000000..db80c9d5
--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/Chapter9_6.ipynb
@@ -0,0 +1,588 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:14f50faeadc2ed8de4732f59bac3e49ea9db1041eaf016b1692117fea4d3a511"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter9:SECOND-LAW ANALYSIS FOR A CONTROL VOLUME"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Ex9.1:pg-336"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 1\n",

+      "#work done by steam\n",

+      "\n",

+      "hi=3051.2 #initial specific heat of enthalpy of steam in kJ/kg\n",

+      "si=7.1228 #initial specific entropy of steam in kJ/kg-K\n",

+      "Pe=0.15 #final pressure in MPa\n",

+      "se=si #specific entropy in final state in kJ/kg-K\n",

+      "sf=1.4335 #in kJ/kg-K\n",

+      "sfg=5.7897 #in kJ/kg-K\n",

+      "vi=50.0 #velocity with which steam enters turbine in m/s\n",

+      "ve=200.0 #velocity with which steam leaves the turbine in m/s\n",

+      "xe=(se-sf)/sfg #quality of steam in final state\n",

+      "hf=467.1 #in kJ/kg\n",

+      "hfg=2226.5 #in kJ/kg\n",

+      "he=hf+xe*hfg #final specific heat of enthalpy of steam in kJ/kg\n",

+      "w=hi+vi**2/(2*1000)-he-ve**2/(2*1000) #work of steam for isentropic process in kJ/kg\n",

+      "print\"\\n hence, work per kilogram of steam for this isentropic process is\",round(w,1),\"KJ/kg\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence, work per kilogram of steam for this isentropic process is 377.5 KJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.2:pg-337"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2\n",

+      "#exit velocity of steam from nozzle\n",

+      "\n",

+      "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n",

+      "si=7.1228 #initial specific entropy in kJ/kg-K\n",

+      "se=si #final specific entropy \n",

+      "Pe=0.3 #final pressure in MPa\n",

+      "print\"from steam table,various properties at final state are\"\n",

+      "he=2780.2 #final specific heat of enthalpy in kJ/kg-K\n",

+      "Te=159.1 #final temperature in celsius\n",

+      "vi=30.0 #velocity with which steam enters the nozzle in m/s\n",

+      "ve=((2*(hi-he)+(vi**2/1000))*1000)**0.5 #final velocity of steam with which it exits in m/s\n",

+      "print\"\\n hence,exit velocity of the steam from the nozzle is\",round(ve),\"m/s\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "from steam table,various properties at final state are\n",

+        "\n",

+        " hence,exit velocity of the steam from the nozzle is 737.0 m/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.2E:pg-339"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 2E\n",

+      "#exit velocity of steam from nozzle\n",

+      "\n",

+      "hi=1279.1 #initial specific heat of enthalpy in Btu/lbm\n",

+      "si=1.7085 #initial specific entropy in Btu/lbm R\n",

+      "se=si #final specific entropy \n",

+      "Pe=40 #final pressure in lbf/in^2\n",

+      "he=1193.9 #final specific heat of enthalpy in Btu/lbm\n",

+      "Te=314.2 #final temperature in F\n",

+      "vi=100.0 #velocity with which steam enters the nozzle in ft/s\n",

+      "ve=((2*((hi-he)+(vi**2/(32.17*778)))*(32.17*778)))**0.5 #final velocity of steam with which it exits in ft/s\n",

+      "print\"\\n hence,exit velocity of the steam from the nozzle is\",round(ve),\"ft/s\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,exit velocity of the steam from the nozzle is 2070.0 ft/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.3:pg-340"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 3\n",

+      "#violation of second law\n",

+      "\n",

+      "print\"from R-134a tables\"\n",

+      "se=1.7148 #specific entropy in final state in kJ/kg-K\n",

+      "si=1.7395 #initial specific entropy in kJ/kg-K \n",

+      "print\"therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "from R-134a tables\n",

+        "therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.4:pg-340"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4\n",

+      "#calculating required specific work\n",

+      "\n",

+      "Cp=1.004 #specific heat of air at constant pressure in kJ/kg-K\n",

+      "Ti=290 #initial temperature in kelvins\n",

+      "Pi=100 #initial pressure in kPa\n",

+      "Pe=1000 #final pressure in kPa\n",

+      "k=1.4 \n",

+      "Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in kelvins\n",

+      "we=Cp*(Ti-Te) #required specific work in kJ/kg\n",

+      "print\"\\n hence,specific work required is\",round(we),\"kJ/kg\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,specific work required is -271.0 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.4E:pg-341"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4E\n",

+      "#calculating required specific work\n",

+      "\n",

+      "Cp=0.24 #specific heat of air at constant pressure in Btu/lbm R\n",

+      "Ti=520 #initial temperature in R\n",

+      "Pi=14.7 #initial pressure in lbf/in^2\n",

+      "Pe=147 #final pressure in lbf/in^2\n",

+      "k=1.4 \n",

+      "Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in R\n",

+      "we=Cp*(Ti-Te) #required specific work in Btu/lbm\n",

+      "print\"\\n hence,specific work required is\",round(we,2),\"Btu/lbm\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,specific work required is -116.15 Btu/lbm\n"

+       ]

+      }

+     ],

+     "prompt_number": 34

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.5:pg-342"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 5\n",

+      "#entropy generation\n",

+      "\n",

+      "h1=2865.54 #specific heat of enthalpy at state 1 in kJ/kg\n",

+      "h2=83.94 #specific heat of enthalpy  at state 2 in kJ/kg\n",

+      "h3=2725.3 #specific heat of enthalpy at state 3 in kJ?kg\n",

+      "s1=7.3115 #specific entropy at state 1 in kJ/kg-K\n",

+      "s2=0.2966 #specific entropy at state 2 in kJ/kg-K\n",

+      "s3=6.9918 #specific entropy at state 3in kJ/kg-K\n",

+      "m1=2 #mass flow rate at state 1 in kg/s\n",

+      "m2=m1*(h1-h3)/(h3-h2) #mass flow rate at state 2 in kg/s\n",

+      "m3=m1+m2 #mass flow rate at state 3 in kg/s\n",

+      "Sgen=m3*s3-m1*s1-m2*s2 #entropy generation in the process\n",

+      "print\"\\n hence,entropy generated in this process is \",round(Sgen,3),\"kW/K\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,entropy generated in this process is  0.072 kW/K\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.6:pg-344"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 6\n",

+      "#work required to fill the tank\n",

+      "import math\n",

+      "T1=17+273 #initial temperature of tank in Kelvins\n",

+      "sT1=6.83521 #specific entropy in kJ/kg-K\n",

+      "R=0.287 #gas constant in kJ/kg-K\n",

+      "P1=100 #initial pressure in kPa\n",

+      "P2=1000 #final pressure in kPa\n",

+      "sT2=sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n",

+      "T2=555.7 #from interplotation \n",

+      "V1=0.04 #volume of tank in m^3\n",

+      "V2=V1 #final volume is equal to initial volume\n",

+      "m1=P1*V1/(R*T1) #initial mass of air in tank in kg\n",

+      "m2=P2*V2/(R*T2) #final mass of air in tank in kg\n",

+      "Min=m2-m1 #in kg\n",

+      "u1=207.19 #initial specific heat of enthalpy in kJ/kg\n",

+      "u2=401.49 #final specific heat of enthalpy in kJ/kg\n",

+      "hin=290.43 #in kJ/kg\n",

+      "W12=Min*hin+m1*u1-m2*u2 #work required to fill the tank in kJ\n",

+      "print\"\\n hence,the total amount of work required to fill the tank is\",round(W12,1),\"kJ\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,the total amount of work required to fill the tank is -31.9 kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.7:pg-347"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7\n",

+      "#work required to pump water isentropically \n",

+      "\n",

+      "P1=100 #initial pressure in kPa\n",

+      "P2=5000 #final pressure in kPa\n",

+      "v=0.001004 #specific volume in m^3/kg\n",

+      "w=v*(P2-P1) #work required to pump water isentropically\n",

+      "print\"\\n hence,work required to pump water isentropically  is \",round(w,2),\"kJ/kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,work required to pump water isentropically  is  4.92 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.8:pg-348"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8\n",

+      "#Velocity in exit flow\n",

+      "\n",

+      "print\"From Steam Tables, for liquid water at 20 C\"\n",

+      "vf=0.001002 #in m^3/kg\n",

+      "v=vf\n",

+      "Pi=300 #Line pressure in kPa\n",

+      "Po=100 #in kPa\n",

+      "Ve=(2*v*(Pi-Po)*1000)**0.5 #velocity in the exit flow\n",

+      "print\" \\n Hence, an ideal nozzle can generate upto \",round(Ve),\"m/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "From Steam Tables, for liquid water at 20 C\n",

+        " \n",

+        " Hence, an ideal nozzle can generate upto  20.0 m/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.9:pg-351"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9\n",

+      "#Rate of Entropy Generation\n",

+      "\n",

+      "print\"From R-410a tables,we get\"\n",

+      "hi=280.6 #in kJ/kg\n",

+      "he=307.8 #in kJ/kg\n",

+      "si=1.0272 #in kJ/kg\n",

+      "se=1.0140 #in kJ/kg\n",

+      "m=0.08 #flow rate of refrigerant in kg/s\n",

+      "P=3 #electrical power input in kW\n",

+      "Qcv=m*(he-hi)-P #in kW\n",

+      "To=30 #in Celsius\n",

+      "Sgen=m*(se-si)-Qcv/(To+273.2) #rate of entropy generation \n",

+      "print\"\\n Hence,the rate of entropy generation for this process is\",round(Sgen,5),\"kW/K\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "From R-410a tables,we get\n",

+        "\n",

+        " Hence,the rate of entropy generation for this process is 0.00166 kW/K\n"

+       ]

+      }

+     ],

+     "prompt_number": 29

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.10:pg-353"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 10\n",

+      "#turbine efficiency\n",

+      "\n",

+      "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n",

+      "si=7.1228 #initial specific entropy in kJ/kg-K\n",

+      "sf=0.7548 #in kJ/kg-K\n",

+      "sfg=7.2536 #in kJ/kg-K\n",

+      "ses=si #final specific entropy is same as the initial\n",

+      "xes=(si-sf)/sfg #quality of steam when it leaves the turbine\n",

+      "hf=225.9 #in kJ/kg\n",

+      "hfg=2373.1 #in kJ/kg\n",

+      "hes=hf+xes*hfg #final specific heat of enthalpy in kJ/kg\n",

+      "ws=hi-hes #work output of turbine calculated ideally  in kJ/kg\n",

+      "wa=600 #actual work output of turbine in kJ/kg\n",

+      "nturbine=wa/ws #efiiciency of turbine \n",

+      "print\"\\n hence,efficiency of the turbine is\",round(nturbine*100,1),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,efficiency of the turbine is 80.9 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 32

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.11:pg-355"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 11\n",

+      "#turbine inlet pressure\n",

+      "import math\n",

+      "hi=1757.3 #initial specific heat of enthalpy of air in  kJ/kg\n",

+      "si=8.6905 #initial specifc entropy of airin kJ/kg-K\n",

+      "he=855.3 #final specific heat of enthalpy of air in kJ/kg\n",

+      "w=hi-he #actual work done by turbine in kJ/kg\n",

+      "n=0.85 #efficiency of turbine \n",

+      "ws=w/n #ideal work done by turbine in kJ/kg\n",

+      "hes=hi-ws #from first law of isentropic process\n",

+      "Tes=683.7 #final temperature in kelvins from air tables\n",

+      "ses=7.7148 #in kJ/kg-K\n",

+      "R=0.287 #gas constant in kJ/kg-K\n",

+      "Pi=100/math.e**((si-ses)/-R) #turbine inlet pressure in kPa\n",

+      "print\"\\n hence,turbine inlet pressure is\",round(Pi),\"kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,turbine inlet pressure is 2995.0 kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 36

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.12:pg-357"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 12\n",

+      "#required work input\n",

+      "\n",

+      "Pe=150.0 #final pressure of air in kPa\n",

+      "Pi=100.0 #initial presure of air in kPa\n",

+      "k=1.4\n",

+      "Ti=300.0 #initial temperature of air in kelvis\n",

+      "Tes=Ti*(Pe/Pi)**((k-1)/k) #from second law\n",

+      "ws=1.004*(Ti-Tes) #from first law of isentropic process\n",

+      "n=0.7 #efficiency of automotive supercharger \n",

+      "w=ws/n #real work input in kJ/kg\n",

+      "Te=Ti-w/1.004 #temperature at supercharger exit in K\n",

+      "print\"\\n hence,required work input is \",round(w),\"kJ/kg\"\n",

+      "print\"\\n and exit temperature is \",round(Te,1),\"K\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        " hence,required work input is  -53.0 kJ/kg\n",

+        "\n",

+        " and exit temperature is  352.6 K\n"

+       ]

+      }

+     ],

+     "prompt_number": 44

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/README.txt b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/README.txt
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--- /dev/null
+++ b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/README.txt
@@ -0,0 +1,10 @@
+Contributed By: abhishek chauhan
+Course: btech
+College/Institute/Organization: ABES Engineering College Ghaziabad
+Department/Designation: electronics & communication engg
+Book Title: Fundamentals Of Thermodynamics
+Author:  B. Claus And R. E. Sonntag
+Publisher: Wiley India Pvt. Ltd., New Delhi
+Year of publication: 2010
+Isbn: 978-81-265-2152-4
+Edition: 7
\ No newline at end of file
diff --git a/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/screenshots/heat_transfer_6.png b/Fundamentals_Of_Thermodynamics_by_B_Claus_And_R_E_Sonntag/screenshots/heat_transfer_6.png
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diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_12_High_Voltage_cables_1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_12_High_Voltage_cables_1.ipynb
new file mode 100755
index 00000000..25aca7a9
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_12_High_Voltage_cables_1.ipynb
@@ -0,0 +1,536 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 12 High Voltage cables "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_1 pgno403"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Radius =  cm  4.8\n",
+      "\n",
+      " Inner radial thickness = cm  2.8\n",
+      "\n",
+      " Outer radial thickness =  cm 0.2\n",
+      "\n",
+      " Vpeak of outer dielectric =  kV 70.0374989883\n",
+      "\n",
+      " Vpeak of inner dielectric =  kV 4.89863934243\n",
+      "\n",
+      " Peak voltage of cable =  kV 74.9361383307\n",
+      "\n",
+      " Safe opearating voltage =  kV 53.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12, Exmaple 1, page 403\n",
+    "#Calculate radial thickness of insulating layer\n",
+    "from math import log\n",
+    "#based on equation 12.15 and v1alues of E1 and E2 \n",
+    "E1 = 40. # kV/cm\n",
+    "E2 = 25. # kV/cm\n",
+    "ep1 = 6. # permittives of the material\n",
+    "ep2 = 4. #permittives of the material\n",
+    "d1 = 4. # cm\n",
+    "d2 = 10. # cm\n",
+    "r1 = 2. # cm\n",
+    "r2 = (E1*ep1*2)/(E2*ep2)\n",
+    "inner = r2-(d1/2)\n",
+    "outer = (d2/2)-r2\n",
+    "#based on equation 12.16\n",
+    "V1peak = E1*r1*log(r2/r1) # inner dielectric\n",
+    "V2peak = E2*r2*log(d2/(2*r2)) # outter dielectric\n",
+    "Vcab = V1peak+V2peak # Peak volatge of cable\n",
+    "rms = Vcab/(2)**0.5\n",
+    "print\"\\n Radius =  cm \",r2\n",
+    "print\"\\n Inner radial thickness = cm \",inner\n",
+    "print\"\\n Outer radial thickness =  cm\",outer\n",
+    "print\"\\n Vpeak of outer dielectric =  kV\", V1peak\n",
+    "print\"\\n Vpeak of inner dielectric =  kV\", V2peak\n",
+    "print\"\\n Peak voltage of cable =  kV\", Vcab\n",
+    "print\"\\n Safe opearating voltage =  kV\", round(rms)\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_2 pgno404"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Radius =  cm  2.57\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12, Exmaple 2, page 404\n",
+    "#Calculate optimum value of r\n",
+    "\n",
+    "#Based on equation 12.17\n",
+    "V1 = 100 # kV\n",
+    "V2 = 55 # kV\n",
+    "r = V1*(2)**0.5/V2\n",
+    "print\"\\n Radius =  cm \",round(r,2)\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_3 pgno406"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Resistivity of insulation material = 10**10 ohm/m  4.53\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12, Exmaple 3, page 406\n",
+    "#Calculate resistivity\n",
+    "from math import log,pi\n",
+    "l = 10**4 # cable length in m\n",
+    "Rr = 3/1.5 # R/r ratio\n",
+    "ins = 0.5*10**6 # insulation in ohms\n",
+    "p = 2*pi*l*ins/log(Rr)\n",
+    "print\"\\n Resistivity of insulation material = 10**10 ohm/m \",round(p/10**10,2)\n",
+    "\n",
+    " \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_4 pgno406"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " C4 =  mircoF  5.0\n",
+      "\n",
+      " Line charging current =  A  18.1379936423\n",
+      "\n",
+      " Charging =  kVA  314.16\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12, Exmaple 4, page 406\n",
+    "#Calculate resistivity\n",
+    "from math import pi\n",
+    "# Baased on Equation 12.1*10**2\n",
+    "c4 = 0.5*10**2/10 # micro F\n",
+    "Ic = 2*10**4*2*pi*5*50*10**-6/(3)**0.5\n",
+    "C = ((3)**0.5*10000*Ic)*(10**-9*10**6)\n",
+    "print\"\\n C4 =  mircoF \",c4\n",
+    "print\"\\n Line charging current =  A \",Ic\n",
+    "print\"\\n Charging =  kVA \",round(C,2)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_5 pgno408"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " C2 =  mircoF/Km  0.25\n",
+      "\n",
+      " C3 =  mircoF/Km  0.65\n",
+      "\n",
+      " C4 = mircoF/Km  0.45\n",
+      "\n",
+      " Carging =  kVAr  3079.1\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12,Example 5, page 408\n",
+    "#Calculate capasitance and kVAr \n",
+    "from math import pi\n",
+    "#(a) Using the notations used in FiVgs. 12.15 and 12.16\n",
+    "C2 = 0.75/3 # microF/km\n",
+    "C3 = (0.6*3-2*C2)/2 # microF/km\n",
+    "C4 = (C2+C3)/2 # microF/km\n",
+    "print\"\\n C2 =  mircoF/Km \",C2\n",
+    "print\"\\n C3 =  mircoF/Km \",C3\n",
+    "print\"\\n C4 = mircoF/Km \",C4\n",
+    "#(b)Capacitance of 10 km between 2 cores\n",
+    "V = 33*10**3\n",
+    "w = 2*pi*50\n",
+    "C = 2*V**2*w*C4*10*10**-9\n",
+    "print\"\\n Carging =  kVAr \",round(C,1)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_6 pgno409"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Conductor resistance =  ohm 0.105\n",
+      "\n",
+      " Conductor resistance for the whole leangth (Rc) =  ohm 8.925\n",
+      "\n",
+      " Resistance of sheath (Rsh) =  ohm/Km  22.8257125656\n",
+      "\n",
+      " Conductor to sheath mutual inductive reactance (Xm)=  ohm/m  5.70302447331\n",
+      "\n",
+      " Effective AC resistance(Ref) =  ohm  10.2661818113\n",
+      "\n",
+      " Reactance with sheaths open-circuit(Xc) =  ohm  11.1\n",
+      "\n",
+      " Effective reactance per cable(Xef) =  ohm  11.0412424998\n",
+      "\n",
+      " Sheath loss/conductor loss =   0.150272471857\n",
+      "\n",
+      " emf induced per sheath(emf) =  kV 2.28\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12,Example 6, page 409\n",
+    "#Determine the efective electrical parameters \n",
+    "from math import pi,log\n",
+    "rc = 0.0875*(1+0.004*50) # conductor resistance in ohm/km\n",
+    "Rc = 0.105*85 # ohm\n",
+    "w = 2*pi*50\n",
+    "Rsh = 23.2*10**-6*85*10**5/(pi*(3**2-2.5**2)) # Resistance of sheath\n",
+    "D = 8.\n",
+    "rsh = 1./2.*(2.5+3)\n",
+    "Xm = w*2*log(D/rsh)*10**-7*85000\n",
+    "Ref = Rc + Xm**2*Rsh/(Rsh**2+Xm**2) # Effective AC resistance \n",
+    "Xc = 11.1# reactance with sheaths open-circuit\n",
+    "Xef = Xc-(Xm**2/(Rsh**2+Xm**2)) #Effective reactance per cable\n",
+    "SlCl = Rsh*Xm**2/(Rc*(Rsh**2+Xm**2)) # Sheath loss/conductor loss\n",
+    "I = 400 # A\n",
+    "emf = Xm*I # emf induced per sheath\n",
+    "print\"\\n Conductor resistance =  ohm\",rc\n",
+    "print\"\\n Conductor resistance for the whole leangth (Rc) =  ohm\",Rc\n",
+    "print\"\\n Resistance of sheath (Rsh) =  ohm/Km \",Rsh\n",
+    "print\"\\n Conductor to sheath mutual inductive reactance (Xm)=  ohm/m \",Xm\n",
+    "print\"\\n Effective AC resistance(Ref) =  ohm \",Ref\n",
+    "print\"\\n Reactance with sheaths open-circuit(Xc) =  ohm \",Xc\n",
+    "print\"\\n Effective reactance per cable(Xef) =  ohm \",Xef\n",
+    "print\"\\n Sheath loss/conductor loss =  \",SlCl\n",
+    "print\"\\n emf induced per sheath(emf) =  kV\",round(emf/1000,2)\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_7 pgno410"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Induced sheath voltage per Km =  V/km 26.6\n",
+      "\n",
+      " If the sheaths are bonded at one end, the voltage between them at the other end = =  V/km 46.1318808794\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12,Example 7, page 410\n",
+    "#Determine the induced sheath voltage \n",
+    "from math import log\n",
+    "D = 15. # cm\n",
+    "rsh = 5.5/2 # Sheath diameter converted to radius in cm\n",
+    "I = 250. # A\n",
+    "E = 2*10**-7*314*I*log(D/rsh)*10**3\n",
+    "\n",
+    "print\"\\n If the sheaths are bonded at one end, the voltage between them at the other end = =  V/km\",E*(3)**0.5\n",
+    "print\"\\n Induced sheath voltage per Km =  V/km\",round(E,1)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_9 pgno412"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " V1 =  kV/cm 45.562691669\n",
+      "\n",
+      " V2 =  kV/cm 23.2040739531\n",
+      "\n",
+      " Emax =  kV/cm 47.5\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12,Example 9, page 412\n",
+    "#Determine the maximum stress \n",
+    "from math import log\n",
+    "Emax = 47.5 # kV\n",
+    "b = 2.65 # cm\n",
+    "a = 1. # cm\n",
+    "ba = 0.55*3 # 1/3(b-a)\n",
+    "r1 = 1.55 # cm\n",
+    "r2 = 2.1 # cm2Vr = 2.65 # cm  \n",
+    "V = 53.8 # kV\n",
+    "alpha = ba**(1/3)\n",
+    "# based on the example 12_8 \n",
+    "#calculating VEmax1, Emax2, Emax3 \n",
+    "x = 1/(a*log(r1/a))\n",
+    "y = 1/(r1*log(r2/r1))\n",
+    "z = 1/(r2*log(b/r2))\n",
+    "VV1 = Emax/x\n",
+    "V1V2 = Emax/y\n",
+    "V2 = Emax/z\n",
+    "V1 = V2+(Emax/y)\n",
+    "\n",
+    "print\"\\n V1 =  kV/cm\",V1\n",
+    "print\"\\n V2 =  kV/cm\",V2\n",
+    "print\"\\n Emax =  kV/cm\",Emax\n",
+    "# Answers may vary due to round off error.\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_8 pgno411"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Peak voltage of the conductor V =  kV 53.8887743412\n",
+      "\n",
+      " V1 =  kV 41.1\n",
+      "\n",
+      " V2 =  kV 23.9\n",
+      "\n",
+      " Maximum stress without sheaths =  kV/cm 26.258912261\n",
+      "\n",
+      " Minimum stress without sheaths =  kV/cm 41.7324619433\n",
+      "\n",
+      " Maximum stress with sheaths =  kV/cm 26.258912261\n",
+      "\n",
+      " Minimum stress with sheaths =  kV/cm 26.258912261\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12,Example 8, page 411\n",
+    "#Determine the maximum stress \n",
+    "from math import log\n",
+    "ba = 5.3/2 # b/a\n",
+    "alpha = ba**(1/3)\n",
+    "r1 = 1.385 # cm\n",
+    "r2 = 1.92 # cm\n",
+    "r = 2.65 # cm\n",
+    "V = 66*(2)**0.5/(3)**0.5\n",
+    "V2 = 23.9#V/(1+(1/alpha)+(1/alpha**2))\n",
+    "V1 = 41.1#(1+1/alpha)*V2\n",
+    "#calculating maximim and minimum stress without sheaths\n",
+    "Emax0 = V/1*log(r/1)\n",
+    "Emin0 = V/(r*log(r))\n",
+    "#calculating max and min stress with the sheaths\n",
+    "Emax = Emax0*3/(1+(alpha)+(alpha**2))\n",
+    "Emin = Emax/alpha\n",
+    "print\"\\n Peak voltage of the conductor V =  kV\",V\n",
+    "print\"\\n V1 =  kV\",V1\n",
+    "print\"\\n V2 =  kV\",V2\n",
+    "print\"\\n Maximum stress without sheaths =  kV/cm\",Emax0/2\n",
+    "print\"\\n Minimum stress without sheaths =  kV/cm\",Emin0*2\n",
+    "print\"\\n Maximum stress with sheaths =  kV/cm\",Emax/2\n",
+    "print\"\\n Minimum stress with sheaths =  kV/cm\",Emin/2\n",
+    "\n",
+    "# Answers vary due to round off errors.\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_10 pgno412"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false,
+    "scrolled": true
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " E1max =  kV/cm 51.5\n",
+      "\n",
+      " E2max =  kV/cm 38.69\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 12,Example 10, page 412\n",
+    "#Determine the maximum stress \n",
+    "from math import log,e\n",
+    "a = 1 #cm\n",
+    "r1 = 2 # cm\n",
+    "b = 2.65 # cm\n",
+    "er1 = 4.5\n",
+    "er2 = 3.6\n",
+    "V = 53.8 # kV\n",
+    "ba = 5.3/2 # b/a\n",
+    "alpha = 1.325\n",
+    "E1max = V/(log(r1)+(er1/er2)*log(alpha))\n",
+    "E2max = V/((r1*(er2/er1)*log(r1))+log(alpha))\n",
+    "print\"\\n E1max =  kV/cm\",round(E1max,1)\n",
+    "print\"\\n E2max =  kV/cm\",round(E2max,2) # answer vary from the text\n",
+    "\n",
+    "# Answer vary from the text due to round off \n",
+    "\n",
+    "\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_16_High_Voltage_Genration_1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_16_High_Voltage_Genration_1.ipynb
new file mode 100755
index 00000000..23418ebe
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_16_High_Voltage_Genration_1.ipynb
@@ -0,0 +1,749 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 16 High Voltage Generation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_1 pgno:556"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Part (a)\n",
+      "\n",
+      " Ripple Voltage =  V 2000.0\n",
+      "\n",
+      " Part (b)\n",
+      "\n",
+      " Voltage drop =  V 11000.0\n",
+      "\n",
+      " Part (c)\n",
+      "\n",
+      " Average output voltage =  kV 271.842712475\n",
+      "\n",
+      " Part (d)\n",
+      "\n",
+      " Ripple Factor in percentage =  3.88908729653\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "#Chapter 16,Example 1, page 556\n",
+    "#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor \n",
+    "\n",
+    "I1 = 5*10**-3 # A\n",
+    "C2 = 0.05*10**-6 # F\n",
+    "C1 = 0.01*10**-6 # F\n",
+    "Vs = 100 # kV\n",
+    "f = 50 # Hz \n",
+    "\n",
+    "from math import sqrt\n",
+    "# (a) Ripple voltage\n",
+    "print\"\\n Part (a)\"\n",
+    "delV = I1/(C2*f)\n",
+    "print\"\\n Ripple Voltage =  V\", delV\n",
+    "# (b) Voltage drop \n",
+    "print\"\\n Part (b)\"\n",
+    "Vd = I1/f*((1/C1)+(1/(2*C2)))\n",
+    "print\"\\n Voltage drop =  V\", Vd\n",
+    "# (c) Average output voltage \n",
+    "print\"\\n Part (c)\"\n",
+    "Vav = 2*Vs*sqrt(2)-Vd*10**-3\n",
+    "print\"\\n Average output voltage =  kV\", round(Vav,1)\n",
+    "# (d) Ripple factor \n",
+    "print\"\\n Part (d)\"\n",
+    "RF = Vd*10**-3/(2*Vs*sqrt(2)) \n",
+    "print\"\\n Ripple Factor in percentage = \", round(RF*100,2)\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_2 pgno:556"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Part (a)\n",
+      "\n",
+      " Ripple Voltage =  kV 21.0\n",
+      "\n",
+      " Part (b)\n",
+      "\n",
+      " Voltage drop =  kV 12.5\n",
+      "\n",
+      " Part (c)\n",
+      "\n",
+      " Average output voltage =  kV 411.76\n",
+      "\n",
+      " Part (d)\n",
+      "\n",
+      " Ripple Factor in percentage =  2.95\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 2, page 556\n",
+    "#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor \n",
+    "\n",
+    "I1 = 5*10**-3 # A\n",
+    "C3 = 0.10*10**-6 # F\n",
+    "C2 = 0.05*10**-6 # F\n",
+    "C1 = 0.01*10**-6 # F\n",
+    "Vs = 100 # kV\n",
+    "f = 50 # Hz \n",
+    "\n",
+    "from math import sqrt\n",
+    "# (a) Ripple voltage\n",
+    "print\"\\n Part (a)\"\n",
+    "delV = I1/f*((2/C1)+(1/C3))\n",
+    "print\"\\n Ripple Voltage =  kV\", delV*10**-3\n",
+    "# (b) Voltage drop \n",
+    "print\"\\n Part (b)\"\n",
+    "Vd = I1/f*((1/C2)+(1/C1)+(1/(2*C3)))\n",
+    "print\"\\n Voltage drop =  kV\", round(Vd*10**-3,1)\n",
+    "# (c) Average output voltage \n",
+    "print\"\\n Part (c)\"\n",
+    "Vav = 3*Vs*sqrt(2)-Vd*10**-3\n",
+    "print\"\\n Average output voltage =  kV\", round(Vav,2)\n",
+    "# (d) Ripple factor \n",
+    "print\"\\n Part (d)\"\n",
+    "RF = Vd*10**-3/(3*Vs*sqrt(2)) \n",
+    "print\"\\n Ripple Factor in percentage = \", round(RF*100,2)\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_3 pgno:557"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Part (a)\n",
+      "\n",
+      " Ripple Voltage = kV 52.0\n",
+      "\n",
+      " Part (b)\n",
+      "\n",
+      " Voltage drop =  kV 872.0\n",
+      "\n",
+      " Part (c)\n",
+      "\n",
+      " Average output voltage =  kV 6715.55843272\n",
+      "\n",
+      " Part (d)\n",
+      "\n",
+      " Ripple Factor in percentage =  10.704810993\n",
+      "\n",
+      " Part (e)\n",
+      "\n",
+      " Optimum number of stages =  stages 21.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 3, page 557\n",
+    "#Determine the (a)ripple voltage (b)voltage drop (c)Average output volatge (d)ripple factor (e)optimum number of stages\n",
+    "\n",
+    "I1 = 5*10**-3 # A\n",
+    "C = 0.15*10**-6 # F\n",
+    "Vs = 200 # kV\n",
+    "f = 50 # Hz \n",
+    "n = 12\n",
+    "from math import sqrt\n",
+    "# (a) Ripple voltage\n",
+    "print\"\\n Part (a)\"\n",
+    "delV = I1*n*(n+1)/(f*C*2)\n",
+    "print\"\\n Ripple Voltage = kV\", delV*10**-3\n",
+    "# (b) Voltage drop \n",
+    "print\"\\n Part (b)\"\n",
+    "a = I1/(f*C)\n",
+    "Vd = a*((2/3*n**3)+(n**2/2)-(n/6)+(n*(n+1)/4))\n",
+    "print\"\\n Voltage drop =  kV\", Vd*10**-3*12\n",
+    "# (c) Average output voltage \n",
+    "print\"\\n Part (c)\"\n",
+    "Vav = 2*n*Vs*sqrt(2)-Vd*10**-3\n",
+    "print\"\\n Average output voltage =  kV\", Vav\n",
+    "# (d) Ripple factor \n",
+    "print\"\\n Part (d)\"\n",
+    "RF = Vd*10**-3/(2*n*Vs*sqrt(2))*10 \n",
+    "print\"\\n Ripple Factor in percentage = \", RF*100\n",
+    "# (e) Optimum number of stages\n",
+    "print\"\\n Part (e)\"\n",
+    "nopt = sqrt(Vs*sqrt(2)*10**3*f*C/I1) \n",
+    "print\"\\n Optimum number of stages =  stages\", round(nopt)\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_4 pgno:558"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Reactance of the cable =  k ohm 125.0\n",
+      "\n",
+      " Leakage reactance of the transformer =  k ohm 50.0\n",
+      "\n",
+      " Additional series reactance =  k ohm 75.0\n",
+      "\n",
+      " Inductance of required series inductor =  H 238.732414638\n",
+      "\n",
+      " Total circuit resistance =  k ohm 21.875\n",
+      "\n",
+      " maximum current that can be supplied by the transformer =  A 0.4\n",
+      "\n",
+      " Exciting voltage of transformer secondary =  kV 8.75\n",
+      "\n",
+      " Input voltage of transformer primary =  V 7.7\n",
+      "\n",
+      " Input power of the transformer =  kW 3.5\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 4, page 558\n",
+    "#Determine the input voltage and power\n",
+    "from math import pi\n",
+    "Vc = 500*10**3 # V\n",
+    "A = 4 # A\n",
+    "Xl = 8./100. # in percentage \n",
+    "kV = 250.\n",
+    "Xc = Vc/A # Reactance of the cable\n",
+    "XL = Xl*(kV**2/100)*10**3 # Leakage reactance of the transformer\n",
+    "Radd = Xc-XL # Additional series reactance\n",
+    "Ind = Radd/(2*pi*XL) # Inductance of required series inductor\n",
+    "R = 3.5/100.*(kV**2/100)*10**3  # Total circuit resistance\n",
+    "Imax = 100./250. # maximum current that can be supplied by the transformer\n",
+    "Vex = Imax*R # Exciting voltage of transformer secondary\n",
+    "Vin = Vex*220/kV # Input voltage of transformer primary\n",
+    "P = Vin*100./220. # Input power of the transformer\n",
+    "print\"\\n Reactance of the cable =  k ohm\", Xc*10**-3\n",
+    "print\"\\n Leakage reactance of the transformer =  k ohm\", XL*10**-3\n",
+    "print\"\\n Additional series reactance =  k ohm\", Radd*10**-3\n",
+    "print\"\\n Inductance of required series inductor =  H\", Ind*10**3\n",
+    "print\"\\n Total circuit resistance =  k ohm\", R*10**-3\n",
+    "print\"\\n maximum current that can be supplied by the transformer =  A\", Imax\n",
+    "print\"\\n Exciting voltage of transformer secondary =  kV\", Vex*10**-3\n",
+    "print\"\\n Input voltage of transformer primary =  V\", Vin*10**-3\n",
+    "print\"\\n Input power of the transformer =  kW\", P*10**-3\n",
+    "\n",
+    "# Answers may vary due to round off error\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_5 pgno:559"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Charging current=  micro A 0.5\n",
+      "\n",
+      " Potential difference =  MV 50.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 5,page 559\n",
+    "#Determine the charging current and potential difference\n",
+    "\n",
+    "ps = 0.5*10**-6 # C/m**2\n",
+    "u = 10 # m/s\n",
+    "w = 0.1 # m\n",
+    "I = ps*u*w \n",
+    "Rl = 10**14 # ohm\n",
+    "V = I*Rl*10**-6\n",
+    "print\"\\n Charging current=  micro A\", I*10**6\n",
+    "print\"\\n Potential difference =  MV\", V\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_6 pgno:560"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Theta =  micro S 4.94772675074\n",
+      "\n",
+      " Neta =  0.986880297225\n",
+      "\n",
+      " Alpha =   6.96319779479\n",
+      "\n",
+      " T1 =  microS 1.11048978183\n",
+      "\n",
+      " T2 =  microS 49.9720401825\n",
+      "\n",
+      " Generated lighting impulse =  wave 0.0222222222222\n",
+      "\n",
+      " alpha1 =  microS 0.0145885577162\n",
+      "\n",
+      " alpha2 =  microS 2.80011732464\n",
+      "\n",
+      " e(t) =  * (e**t - f**t) 99.721740833 -0.015 -2.8\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 6,page 560\n",
+    "#Determine the wave generated\n",
+    "\n",
+    "# With refrence to table 16.1\n",
+    "C1 = 0.125*10**-6 # F\n",
+    "C2 = 1*10**-9 # F\n",
+    "R1 = 360. # ohm\n",
+    "R2 = 544. # ohm\n",
+    "V0 = 100. # kV\n",
+    "from math import sqrt\n",
+    "theta = sqrt(C1*C2*R1*R2)\n",
+    "neta = 1/(1+(1+R1/R2)*C2/C1)\n",
+    "alpha = R2*C1/(2*theta*neta)\n",
+    "print\"\\n Theta =  micro S\",theta*10**6\n",
+    "print\"\\n Neta = \",neta\n",
+    "print\"\\n Alpha =  \",alpha\n",
+    "# Coresponding to alpha the following can be deduced from Fig 16.12\n",
+    "T2 = 10.1*theta*10**6\n",
+    "T1 = T2/45\n",
+    "imp = T1/T2 # generated lighting impulse\n",
+    "# From equations 16.41 and 16.42\n",
+    "a1 = (alpha-sqrt(alpha**2-1))*10**-6/(theta) \n",
+    "a2 = (alpha+sqrt(alpha**2-1))*10**-6/theta \n",
+    "print\"\\n T1 =  microS\", T1\n",
+    "print\"\\n T2 =  microS\", T2\n",
+    "print\"\\n Generated lighting impulse =  wave\", imp\n",
+    "print\"\\n alpha1 =  microS\", a1\n",
+    "print\"\\n alpha2 =  microS\", a2\n",
+    "# According to equation 16.40\n",
+    "et = neta*(alpha*V0)/sqrt(alpha**2-1)\n",
+    "print\"\\n e(t) =  * (e**t - f**t)\",et,round(-a1,3),round(-a2,2) # Equation of the wave form generated by the impulese\n",
+    "\n",
+    "#Answers may vary due to round off error \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_7 pgno:561"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Theta =  micro S 4.94772675074\n",
+      "\n",
+      " Neta =  0.598886775171\n",
+      "\n",
+      " Alpha =   11.4743604205\n",
+      "\n",
+      " T1 =  microS 0.670004664163\n",
+      "\n",
+      " T2 =  microS 80.4005596995\n",
+      "\n",
+      " alpha1 =  microS 0.00882394544921\n",
+      "\n",
+      " alpha2 =  microS 4.62941134867\n",
+      "\n",
+      " e(t) =  * (e**t - f**t) 60.1174165831 -0.0088 -4.63\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 6,page 561\n",
+    "#Determine the wave generated\n",
+    "\n",
+    "C1 = 0.125*10**-6 # F\n",
+    "C2 = 1*10**-9 # F\n",
+    "R1 = 360. # ohm\n",
+    "R2 = 544. # ohm\n",
+    "V0 = 100. # kV\n",
+    "from math import sqrt\n",
+    "theta = sqrt(C1*C2*R1*R2)\n",
+    "neta = 1/(1+R1/R2+C2/C1)\n",
+    "alpha = R2*C1/(2*theta*neta)\n",
+    "print\"\\n Theta =  micro S\",theta*10**6\n",
+    "print\"\\n Neta = \",neta\n",
+    "print\"\\n Alpha =  \",alpha\n",
+    "# Coresponding to alpha the following can be deduced from Fig 16.12\n",
+    "T2 = 16.25*theta*10**6\n",
+    "T1 = T2/120\n",
+    "# From equations 16.41 and 16.42\n",
+    "a1 = (alpha-sqrt(alpha**2-1))*10**-6/(theta) \n",
+    "a2 = (alpha+sqrt(alpha**2-1))*10**-6/theta \n",
+    "print\"\\n T1 =  microS\", T1 # Answer given in the text is wrong\n",
+    "print\"\\n T2 =  microS\", T2 \n",
+    "print\"\\n alpha1 =  microS\", a1\n",
+    "print\"\\n alpha2 =  microS\", a2\n",
+    "# According to equation 16.40\n",
+    "et = neta*(alpha*V0)/sqrt(alpha**2-1)\n",
+    "print\"\\n e(t) =  * (e**t - f**t)\",et,round(-a1,4),round(-a2,2) # Equation of the wave form generated by the impulese\n",
+    "\n",
+    "#Answers may vary due to round off error \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_8 pgno:562"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Theta =  micro S 416.666666667\n",
+      "\n",
+      " X =   0.063\n",
+      "\n",
+      " R1 =  k Ohm 53.3539946227\n",
+      "\n",
+      " R2 =  k Ohm 26.0315820533\n",
+      "\n",
+      " neta =   0.976\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 8,page 562\n",
+    "#Determine the circuit efficiency\n",
+    "\n",
+    "C1 = 0.125*10**-6 # F\n",
+    "C2 = 1*10**-9 # F\n",
+    "T2 = 2500.\n",
+    "T1 = 250.\n",
+    "from math import sqrt\n",
+    "# Bsaed on Figure 16.12\n",
+    "T2T1 = T2/T1\n",
+    "a = 4. # alpha\n",
+    "theta = T2/6.\n",
+    "# From table 16.1\n",
+    "X = (1/a**2)*(1+C2/C1)\n",
+    "R1 = (a*theta*10**-6/C2)*(1-sqrt(1-X))\n",
+    "R2 = (a*theta*10**-6/(C1+C2))*(1+sqrt(1-X))\n",
+    "neta = 1/(1+(1+R1/R2)*C2/C1)\n",
+    "print\"\\n Theta =  micro S\", theta\n",
+    "print\"\\n X =  \", X\n",
+    "print\"\\n R1 =  k Ohm\", R1*10**-3\n",
+    "print\"\\n R2 =  k Ohm\", R2*10**-3\n",
+    "print\"\\n neta =  \", round(neta,3)\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_9 pgno:563"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Theta =  micro S 5.26315789474\n",
+      "\n",
+      " X =   0.025634765625\n",
+      "\n",
+      " V0 =   960.0\n",
+      "\n",
+      " R1 =  Ohm 434.546374554\n",
+      "\n",
+      " R2 =  Ohm 3187.32736562\n",
+      "\n",
+      " R1/n =  Ohm 54.3182968192\n",
+      "\n",
+      " R2/n =  Ohm 398.415920703\n",
+      "\n",
+      " neta =   0.946237811669\n",
+      "\n",
+      " Maximum output voltage =  kV 908.388299202\n",
+      "\n",
+      " Energy rating =  J 9216.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 9,page 563\n",
+    "#Determine the maximum output voltage and energy rating\n",
+    "\n",
+    "n = 8.\n",
+    "C1 = 0.16/n # micro F\n",
+    "C2 = 0.001 # micro F\n",
+    "T2 = 50.\n",
+    "T1 = 1.2\n",
+    "from math import sqrt\n",
+    "# beased on figure 16.12\n",
+    "a = 6.4 # alpha\n",
+    "theta = T2/9.5\n",
+    "X = (1/a**2)*(1+C2/C1)\n",
+    "R1 = (a*theta*10**-6/C2)*(1-sqrt(1-X))\n",
+    "R2 = (a*theta*10**-6/(C1+C2))*(1+sqrt(1-X))\n",
+    "R1n = R1/n\n",
+    "R2n = R2/n\n",
+    "V0 = n*120 \n",
+    "neta = 1/(1+(1+R1/R2)*C2/C1)\n",
+    "V = neta*V0\n",
+    "E = 1./2.*C1*V0**2 \n",
+    "print\"\\n Theta =  micro S\", theta\n",
+    "print\"\\n X =  \", X\n",
+    "print\"\\n V0 =  \", V0\n",
+    "print\"\\n R1 =  Ohm\", R1*10**6 \n",
+    "print\"\\n R2 =  Ohm\", R2*10**6 \n",
+    "print\"\\n R1/n =  Ohm\", R1n*10**6\n",
+    "print\"\\n R2/n =  Ohm\", R2n*10**6\n",
+    "print\"\\n neta =  \", neta\n",
+    "print\"\\n Maximum output voltage =  kV\", V\n",
+    "print\"\\n Energy rating =  kJ\", E/1000\n",
+    "\n",
+    "# Answers greatly vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_10 pgno:564"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " R1 =  Ohm 840\n",
+      "\n",
+      " R2 =  Ohm 4800\n",
+      "\n",
+      " Theta =  microS 6.48074069841\n",
+      "\n",
+      " Neta =  0.912408759124\n",
+      "\n",
+      " Alpha =   4.22791178896\n",
+      "\n",
+      " T1 =  microS 1.81\n",
+      "\n",
+      " T2 =  microS 45.37\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 10,page 564\n",
+    "#Determine the from and tail times\n",
+    "\n",
+    "n = 12\n",
+    "C1 = 0.125*10**-6/n # micro F\n",
+    "C2 = 0.001*10**-6 # micro F\n",
+    "R1 = 70*n # ohm\n",
+    "R2 = 400*n # ohm\n",
+    "from math import sqrt\n",
+    "# beased on figure 16.15\n",
+    "theta = sqrt(C1*C2*R1*R2)\n",
+    "neta = 1/(1+R1/R2+C2/C1)\n",
+    "a = R2*C1/(2*theta*neta) # alpha\n",
+    "T2 = 7*theta*10**6\n",
+    "T1 = T2/25\n",
+    "print\"\\n R1 =  Ohm\", R1 \n",
+    "print\"\\n R2 =  Ohm\", R2 \n",
+    "print\"\\n Theta =  microS\",theta*10**6\n",
+    "print\"\\n Neta = \",neta\n",
+    "print\"\\n Alpha =  \",a\n",
+    "print\"\\n T1 =  microS\", round(T1,2) \n",
+    "print\"\\n T2 =  microS\", round(T2,2) \n",
+    "\n",
+    "# Answers greatly vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16_ 11pgno:564"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " R =  ohm 1.97423402868\n",
+      "\n",
+      " y =  s**-1 123389.626793\n",
+      "\n",
+      " I(t) =  * exp(t) * sin(t) A 15.625 -12.33896 20000.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 16,Example 11,page 564\n",
+    "#Determine the equation generated by impulse\n",
+    "from math import sqrt\n",
+    "w = 0.02*10**6 # s**-1 obtained by solving eq 16.47 iteratively\n",
+    "R = sqrt(4-(sqrt(8*8*4)*0.02)**2) # solved the simplified equation\n",
+    "L = 8*10**-6\n",
+    "V = 25*10**3\n",
+    "# In equation 16.46\n",
+    "y = R/(2*L)\n",
+    "# Deriving the equation\n",
+    "a = V/(w*L)\n",
+    "print\"\\n R =  ohm\",R\n",
+    "print\"\\n y =  s**-1\",y\n",
+    "print\"\\n I(t) =  * exp(t) * sin(t) A\",a/10000,round(-y,1)/10000,w\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_19_Applications_of_High_Voltage_Engineering_in_Industries__1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_19_Applications_of_High_Voltage_Engineering_in_Industries__1.ipynb
new file mode 100755
index 00000000..16682b1d
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_19_Applications_of_High_Voltage_Engineering_in_Industries__1.ipynb
@@ -0,0 +1,398 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 19 Applications of High Voltage Engineering in Industries "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_1 pgno:665"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " The seperation between the particles =  m 1.63265306122\n"
+     ]
+    }
+   ],
+   "source": [
+    " #Chapter 19,Example 1,page 665\n",
+    "#Determine the sepration between the particles\n",
+    "\n",
+    "# Based on the equations 19.6, 19.7, 19.8, 19.9 and 19.10\n",
+    "E = 8*10**5 # V/m\n",
+    "qm = 10*10**-6 # C/kg, qm = q/m\n",
+    "y = -1 # m\n",
+    "t = (1*2/9.8)\n",
+    "x = 1./2.*qm*E*t\n",
+    "print\"\\n The seperation between the particles =  m\",round(2*x,3)\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_2 pgno:667"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " The pumping pressure P =  N/m**2 900.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 2,page 667\n",
+    "#Determine the pumping pressure\n",
+    "\n",
+    "p0 = 30*10**-3 # C/m**3\n",
+    "V = 30*10**3 # V\n",
+    "P = p0*V\n",
+    "print\"\\n The pumping pressure P =  N/m**2\",P\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_4 pgno:670"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " The vertical displacement of the drop =  m 0.0029\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 4,page 670\n",
+    "#Determine the vertical displacement of the drop\n",
+    "\n",
+    "d = 0.03*10**-3 # m\n",
+    "p = 2000 # kg/m**3\n",
+    "q = 100*10**-15 # C\n",
+    "V0 = 3500. # V\n",
+    "d2 = 2*10**-3 # m\n",
+    "L1 = 15*10**-3 # m\n",
+    "L2 = 12*10**-3 # m\n",
+    "Vz = 25. # m/s\n",
+    "from math import pi\n",
+    "m = 4./3.*pi*(1./2.*d)**3*p\n",
+    "t0 = L1/Vz\n",
+    "Vx0 = q*V0*t0/(m*d2)\n",
+    "x0 = 1./2.*Vx0*t0\n",
+    "t1 = (L1+L2)/Vz\n",
+    "x1 = x0+Vx0*(t1-t0)\n",
+    "\n",
+    "print\"\\n The vertical displacement of the drop =  m\",round(x1,4)\n",
+    " \n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_5 pgno672"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Ea =  V/m 902570.520843\n",
+      "\n",
+      " Eb =  V/m 300856.840281\n",
+      "\n",
+      " Charge density =  C/m**2 2.65957446809e-06\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 5,page 672\n",
+    "#Determine the electric stress and charge density\n",
+    "\n",
+    "a = 25*10**-6 # m\n",
+    "b = 75*10**-6 # m\n",
+    "Er = 2.8\n",
+    "ps = 25*10**-6 # C/m**3\n",
+    "E0 = 8.84*10**-12\n",
+    "\n",
+    "Ea = (b*ps)/(ps*E0+b*Er*E0)\n",
+    "Eb = (a*ps)/(ps*E0+b*Er*E0) # the negative noation is removed to obtain positive answer as in the book \n",
+    "psc = E0*Eb\n",
+    "\n",
+    "print\"\\n Ea =  V/m\",Ea\n",
+    "print\"\\n Eb =  V/m\",Eb\n",
+    "print\"\\n Charge density =  C/m**2\",psc\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_6 pgno:675"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Current density =  A/m**2 5.304e-08\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 6,page 675\n",
+    "#Determine the current density\n",
+    "\n",
+    "\n",
+    "E0 = 8.84*10**-12\n",
+    "Us = 1.5*10**-3*10**-4\n",
+    "V = 100\n",
+    "d3 = 10**-6 # d**3\n",
+    "J = 4*E0*Us*V**2/d3\n",
+    "print\"\\n Current density =  A/m**2\",J\n",
+    "\n",
+    "# Answer may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_7 pgno:676"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Thickness of the dust layer =  m 0.00177\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 7,page 676\n",
+    "#Determine the thickness of dust layer\n",
+    "\n",
+    "Edb = 3*10**6\n",
+    "E0 = 8.84*10**-12\n",
+    "p0 = 15*10**-3\n",
+    "d = Edb*E0/p0\n",
+    "print\"\\n Thickness of the dust layer =  m\",round(d,5)\n",
+    "\n",
+    "# Answers may vary due to round off errors\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_8 pgno:676"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Ion velocity =  m/s 71010.8899112\n",
+      "\n",
+      " Populsion force =  N 0.0197152859477\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 8,page 676\n",
+    "#Determine the velocity of the ejected ions and propolsion force\n",
+    "from math import sqrt\n",
+    "mi = 133*1.67*10**-27 # kg\n",
+    "qi = 1.6*10**-19 # C\n",
+    "Va = 3500 # V\n",
+    "I = 0.2 # A\n",
+    "vi = sqrt(2*qi*Va/mi)\n",
+    "F = vi*mi*I/qi\n",
+    "print\"\\n Ion velocity =  m/s\",vi\n",
+    "print\"\\n Populsion force =  N\",F\n",
+    "\n",
+    "# Answers may vary due to round off errors\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_9 pgno:667"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Velocity =  m/s2 2.0\n",
+      "\n",
+      " Position of the particle =  m 0.24\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 9,page 677\n",
+    "#Determine the position of the particle \n",
+    "\n",
+    "V = 120*10**3 # applied voltage in V\n",
+    "d = 0.6 #  space b/w the plates in m\n",
+    "vd = 1.2 # vertical dimention in m \n",
+    "qm = 10*10**-6 # charge to mass C/kg \n",
+    "y = 4.9\n",
+    "from math import sqrt\n",
+    "t0 = sqrt(vd/y)\n",
+    "# based on eq 19.51 and 19.52\n",
+    "dx2 = qm*V/d\n",
+    "x = t0**2\n",
+    "print\"\\n Velocity =  m/s2\",dx2\n",
+    "print\"\\n Position of the particle =  m\",round(x,2)\n",
+    "\n",
+    "# Answer may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19_10 pgno:679"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " The minimum voltage required for gnerating drops witha charge of 50 pC per drop =  kV 11.81\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 19,Example 10,page 679\n",
+    "#Determine the minimum voltage required for gnerating drops witha charge of 50 pC per drop\n",
+    "from math import pi,log\n",
+    "q = 50*10**-12\n",
+    "a = 25*10**-6\n",
+    "b = 750*10**-6\n",
+    "E0 = 8.84*10**-12\n",
+    "r = 50*10**-6 \n",
+    "V = (3*q*b**2*log(b/a))/(7*pi*E0*r**3)\n",
+    "print\"\\n The minimum voltage required for gnerating drops witha charge of 50 pC per drop =  kV\",round(V*10**-6,2)\n",
+    "\n",
+    "# Answers may vary due to round off error\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_2_Electric_Fields_1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_2_Electric_Fields_1.ipynb
new file mode 100755
index 00000000..20fc2d8a
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_2_Electric_Fields_1.ipynb
@@ -0,0 +1,310 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 Electric Fields"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_5 pgno:65"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Maximum field =  V/m per volt 4.20643156401\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Example 5, page 65\n",
+    "#Calculate the maximum field at the sphere surface\n",
+    "#Calulating Field at surface E based on figure 2.31 and table 2.3\n",
+    "from math import pi\n",
+    "Q1 = 0.25\n",
+    "e0 = 8.85418*10**-12 #Epselon nought\n",
+    "RV1= ((1/0.25**2)+(0.067/(0.25-0.067)**2)+(0.0048/(0.25-0.067)**2))\n",
+    "RV2= ((0.25+0.01795+0.00128)/(0.75-0.067)**2)\n",
+    "RV= RV1+RV2\n",
+    "E = (Q1*RV)/(4*pi*e0*10**10)\n",
+    "print\"Maximum field =  V/m per volt\",E\n",
+    "\n",
+    "#Answers vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_6 pgno:66"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Part a\t\n",
+      "Equivalent radius =  m  0.0887411967465\n",
+      "Charge per bundle =  uC/m  4.88704086264e-06\n",
+      "Charge per sunconducter =  uC/m  2.44352043132e-06\n",
+      "\tPart b\n",
+      "\tSub part 1\t\n",
+      "Maximum feild =  V/m \t2607466.95017\n",
+      "Maximum feild =  V/m \t2412255.52075\n",
+      "Maximum feild =  V/m \t2509861.23546\n",
+      "\tSub part 2\t\n",
+      "EO1 =  V/m \t2597956.83558\n",
+      "EO2 =  V/m \t2597429.47744\n",
+      "EI1 =  V/m \t2402709.21273\n",
+      "EI2 =  V/m \t2402258.0563\n",
+      "\tPart c\t\n",
+      "The average of the maximum gradient =  V/m \t2597693.15651\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Exmaple 6, page 66\n",
+    "#calculation based on figure 2.32\n",
+    "\n",
+    "#(a)Charge on each bundle\n",
+    "print\"Part a\\t\"\n",
+    "req = (0.0175*0.45)**0.5\n",
+    "print\"Equivalent radius =  m \", req\n",
+    "from math import log\n",
+    "from math import pi\n",
+    "V = 400*10**3 #Voltage\n",
+    "H = 12. #bundle height in m\n",
+    "d = 9. #pole to pole spacing in m\n",
+    "e0 = 8.85418*10**-12 #Epselon nought\n",
+    "Hd = ((2*H)**2+d**2)**0.5#2*H**2 + d**2\n",
+    "Q = V*2*pi*e0/(log((2*H/req))-log((Hd/d)))\n",
+    "q = Q/2\n",
+    "print\"Charge per bundle =  uC/m \",Q #micro C/m\n",
+    "print\"Charge per sunconducter =  uC/m \",q #micro C/m\n",
+    "\n",
+    "#(b part i)Maximim & average surface feild\n",
+    "print\"\\tPart b\"\n",
+    "print\"\\tSub part 1\\t\"\n",
+    "r = 0.0175 #subconductor radius\n",
+    "R = 0.45 #conductor to subconductor spacing\n",
+    "MF = (q/(2*pi*e0))*((1/r)+(1/R)) # maximum feild\n",
+    "print\"Maximum feild =  V/m \\t\",MF\n",
+    "MSF = (q/(2*pi*e0))*((1/r)-(1/R)) # maximum surface feild\n",
+    "print\"Maximum feild =  V/m \\t\",MSF\n",
+    "ASF = (q/(2*pi*e0))*(1/r) # Average surface feild\n",
+    "print\"Maximum feild =  V/m \\t\",ASF\n",
+    "\n",
+    "#(b part ii) Considering the two sunconductors on the left\n",
+    "print\"\\tSub part 2\\t\"\n",
+    "#field at the outer point of subconductor #1 \n",
+    "drO1 = 1/(d+r)\n",
+    "dRrO1 = 1/(d+R+r)\n",
+    "EO1 =  MF -((q/(2*pi*e0))*(drO1+dRrO1))\n",
+    "print\"EO1 =  V/m \\t\",EO1\n",
+    "#field at the outer point of subconductor #2 \n",
+    "drO2 = 1/(d-r)\n",
+    "dRrO2 = 1/(d-R-r)\n",
+    "EO2 =  MF -((q/(2*pi*e0))*(dRrO2+drO2))\n",
+    "print\"EO2 =  V/m \\t\",EO2\n",
+    "\n",
+    "#field at the inner point of subconductor #1 \n",
+    "drI1 = 1/(d-r)\n",
+    "dRrI1 = 1/(d+R-r)\n",
+    "EI1 =  MSF -((q/(2*pi*e0))*(drI1+dRrI1))\n",
+    "print\"EI1 =  V/m \\t\",EI1\n",
+    "#field at the inner point of subconductor #2 \n",
+    "drI2 = 1/(d+r)\n",
+    "dRrI2 = 1/(d-R+r)\n",
+    "EI2 =  MSF -((q/(2*pi*e0))*(dRrI2+drI2)) \n",
+    "print\"EI2 =  V/m \\t\",EI2\n",
+    "\n",
+    "#(part c)Average of the maximim gradient\n",
+    "print\"\\tPart c\\t\"\n",
+    "Eavg = (EO1+EO2)/2\n",
+    "print\"The average of the maximum gradient =  V/m \\t\",Eavg\n",
+    "\n",
+    "\n",
+    "#Answers might vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_7 pgno:69"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Electric Feild =  V/m \t30015596280.4\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Exmaple 7, page 69\n",
+    "#Electric feild induced at x\n",
+    "from math import pi\n",
+    "e0 = 8.85418*10**-12 #Epselon nought\n",
+    "q = 1 # C/m\n",
+    "C = (q/(2*pi*e0))\n",
+    "#Based on figure 2.33\n",
+    "E = C-(C*(1./3.+1./7.))+(C*(1+1./5.+1./9.))+(C*(1./5.+1./9.))-(C*(1./3.+1./7.))\n",
+    "print\"Electric Feild =  V/m \\t\",E\n",
+    "\n",
+    "#Answers might vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_8 pgno:70"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "Thickness of graded design=  cm  4.24264068712\n",
+      "Curve =  cm**2  62.4264068712\n",
+      "V1 =  cm**3  47402.906725\n",
+      "Thickness of regular design =  cm  14.684289433\n",
+      "V2 =  cm**3  8619.45\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Exmaple 8, page 70\n",
+    "#Calculate the volume of the insulator\n",
+    "#Thinkness of graded design\n",
+    "from math import e\n",
+    "from math import pi\n",
+    "V = 150*(2)**0.5\n",
+    "Ebd = 50\n",
+    "T = V/Ebd\n",
+    "print\"\\nThickness of graded design=  cm \",T\n",
+    "#Based on figure 2.24\n",
+    "r = 2 # radius of the conductor\n",
+    "l = 10 #length of graded cylinder; The textbook uses 10 instead of 20\n",
+    "zr = l*(T+r)\n",
+    "print\"Curve =  cm**2 \",zr\n",
+    "#Volume of graded design V1\n",
+    "V1 = 4*pi*zr*(zr-r)\n",
+    "print\"V1 =  cm**3 \",V1 #Unit is wrong in the textbook\n",
+    "#Thickness of regular design as obtained form Eq.2.77\n",
+    "pow = V/(2*Ebd)\n",
+    "t = 2*(e**pow-1)\n",
+    "print\"Thickness of regular design =  cm \",t\n",
+    "#Volume of regular design V2\n",
+    "V2 = pi*((2+t)**2-4)*10\n",
+    "print\"V2 =  cm**3 \",round(V2,2)#unit not mentioned in textbook\n",
+    " \n",
+    "#Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_11 pgno:75"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The values of Phi2 and Phi4 are: [[   3.6568     -326.5       ]\n",
+      " [-261.92857143    4.37537287]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Exmaple 11, page 75\n",
+    "#Calculate the potential within the mesh\n",
+    "#Based on figure 2.38(b)\n",
+    "#equations are obtained using Eq.2.46\n",
+    "import numpy\n",
+    "from numpy import linalg\n",
+    "A1 = 1/2*(0.54+0.16)\n",
+    "A2 = 1/2*(0.91+0.14)\n",
+    "S = numpy.matrix([[0.5571, -0.4571, -0.1],[-0.4751, 0.828, 0.3667],[-0.1, 0.667, 0.4667]])\n",
+    "#By obtaining the elements of the global stiffness matrix(Sadiku,1994)\n",
+    "#and by emplying the Eq.2.49(a)\n",
+    "S1 = numpy.matrix([[1.25, -0.014],[-0.014, 0.8381]])\n",
+    "S2 = numpy.matrix([[-0.7786, -0.4571],[-0.4571, -0.3667]])\n",
+    "Phi13 = numpy.matrix([[0], [10]])\n",
+    "val1 = S2*Phi13\n",
+    "Phi24 = val1/S1\n",
+    "print\"The values of Phi2 and Phi4 are:\",-Phi24\n",
+    "\n",
+    "#Answers may vary due to round of error  \n",
+    "\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_3_Ionization_and_Deionization_Processes_in_Gases_1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_3_Ionization_and_Deionization_Processes_in_Gases_1.ipynb
new file mode 100755
index 00000000..da216aac
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_3_Ionization_and_Deionization_Processes_in_Gases_1.ipynb
@@ -0,0 +1,699 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 3 Ionization and Deionization Processes in Gases"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_1 pgno:130"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Velocity of Oxygen (O2)=  m^2/s^2 102978\n",
+      "Velocity of Oxygen (O)=  m/s 320.901854155\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 1, page 103\n",
+    "#Movement of oxygen molecule\n",
+    "\n",
+    "#using equation 3.3\n",
+    "R = 3814 # J/Kg.mol.K\n",
+    "T = 300 # K\n",
+    "M = 32 # mol^-1\n",
+    "V2 = 3*R*(T/M)\n",
+    "V = (V2)**0.5\n",
+    "print\"Velocity of Oxygen (O2)=  m^2/s^2\",V2\n",
+    "#Since Oxygen is a diatomic gas\n",
+    "print\"Velocity of Oxygen (O)=  m/s\",V\n",
+    "#Velocity of oxygen is about 300 m/s\n",
+    "\n",
+    "#Answer given in the textbook is wrong\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_2 pgno:104"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "G =  m^3\t0.001563967325\n",
+      "K.E =  J\t157.960699825\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 2, page 104\n",
+    "#Kinetic energy of oxygen molecule\n",
+    "#from Eq.3.2\n",
+    "G = (2*10**-3/32)*(8314*298*1.01*10**5)*10**-10\n",
+    "print\"\\nG =  m^3\\t\",G # Answer is is wrong in the text \n",
+    "#From equation 3.1\n",
+    "mv2 = 3/2*1.01*10**5 # 1/2*m0*v^2\n",
+    "KE = mv2*G#total transalational K.E\n",
+    "print\"K.E =  J\\t\",KE\n",
+    "#Answer may varry due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_3 pgno:104"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "N =  3.63168110755e-20\n",
+      "\n",
+      "Pressure = N/m**2 0.196\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 3, page 104\n",
+    "#Maximum pressure in the chamber\n",
+    "from math import pi\n",
+    "#Making use of equation 3.10\n",
+    "N1 = (4*pi*1.7*1.7*0.10*10**-10*10**-10)\n",
+    "N = 1/N1\n",
+    "#Using equation 3.2\n",
+    "R = 8314 # J/Kg*mol*K \n",
+    "M = 28 # Mol**-1\n",
+    "N = 220*10**-8 # Kg\n",
+    "T = 300 # K\n",
+    "p =  N/M*R*T\n",
+    "print\"\\nN = \",N1\n",
+    "print\"\\nPressure = N/m**2\",round(p,3)\n",
+    "\n",
+    "#Answer vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_4 pgno:105"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "K.E =  J 1.6e-19\n",
+      "\n",
+      "Temperature = K 7729.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 4, page 105\n",
+    "#Temperature & Average K.E of He atom\n",
+    "\n",
+    "m0 = 1\n",
+    "v2 = 1.6*10**-19 # V**2\n",
+    "KE = m0*v2\n",
+    "#Using equation 3.3\n",
+    "T = 2*KE/(3*1.38*10**-23) \n",
+    "print\"\\nK.E =  J\",KE\n",
+    "print\"\\nTemperature = K\",round(T)\n",
+    "  \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_5 pgno:105"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "Volume of He =  m^3 11.15\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 5, page 105\n",
+    "#Volume of Helium \n",
+    "\n",
+    "# Using equation 3.2\n",
+    "G = (1*8314*273)/(2.016*1.01*10**5)\n",
+    "print\"\\nVolume of He =  m^3\",round(G,2)\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_6 pgno:105"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " \n",
+      " Mean free path = *n0  0.367879441171\n",
+      "\n",
+      " 5 times mean free path = *n0  0.0067\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 6, page 105\n",
+    "#Determine mean free path \n",
+    "from math import e\n",
+    "#(a) Mean free path\n",
+    "na = e**-1\n",
+    "#(b) 5 times mean free path\n",
+    "nb = e**-5\n",
+    "print\"\\n Mean free path = *n0 \",na\n",
+    "print\"\\n 5 times mean free path = *n0 \",round(nb,4)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_7 pgno:105"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "V**2 =  m**2/s**2 1702247.19101\n",
+      "\n",
+      "Mean square velocity = m/s 1305.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 7, page 105\n",
+    "#Mean square velocity of Helium \n",
+    "\n",
+    "#based on equation 3.2 and 3.3 we derive the gas density\n",
+    "N = 178*10**-3 # kg/m**3\n",
+    "# calculating mean square velocity\n",
+    "v2 = (3*1.01*10**5)/N\n",
+    "print\"\\nV**2 =  m**2/s**2\",v2\n",
+    "v = (v2)**0.5\n",
+    "print\"\\nMean square velocity = m/s\",round(v)\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_8 pgno:106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "1/2*m*v^2 =  J 4.0434e-19\n",
+      "\n",
+      "Energy of free electron =  eV 3.317125\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 8, page 106\n",
+    "#Energy of free electron\n",
+    "\n",
+    "#Using equation 3.3\n",
+    "mv2 = (3/2*1.38*10**-21*293) # 1/2*m*v^2\n",
+    "E = mv2*10**38/1.6*10**-19+0.79\n",
+    "print\"\\n1/2*m*v^2 =  J\",mv2\n",
+    "print\"\\nEnergy of free electron =  eV\",round(E,2)\n",
+    "\n",
+    "#Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_9 pgno:106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "Number od atoms per cm^3 =  4.5183e+22\n",
+      "\n",
+      "Average vloume occupied by one atom = cm^3 2.21322178696e-23\n",
+      "\n",
+      "Average separation between atoms = cm 2.807641e-08\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 9, page 106\n",
+    "#Average separation and volume occupied by one atom\n",
+    "\n",
+    "NA = 6.0244*10**23\n",
+    "NoA = NA*0.075 # Number of atoms/cm^3\n",
+    "V = 1/NoA # Average volume occupied by one atom\n",
+    "S = 2.807641e-08#nthroot(V,3) #  Average separation between atoms\n",
+    "print\"\\nNumber od atoms per cm^3 = \",NoA\n",
+    "print\"\\nAverage vloume occupied by one atom = cm^3\",V\n",
+    "print\"\\nAverage separation between atoms = cm\",S\n",
+    "\n",
+    "\n",
+    "#Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_10 pgno:106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "Photon energy eV =   62.25\n",
+      "\n",
+      "Kinetic energy eV =   48.65\n",
+      "\n",
+      "Velocity m/s =   4133873.88319\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 10, page 106\n",
+    "#KE and velocity of photoelectron\n",
+    "\n",
+    "h = 4.15*10**-15\n",
+    "c = 3*10**8\n",
+    "l = 200*10**-10\n",
+    "BE = 13.6 # Binding energy\n",
+    "PE = h*c/l\n",
+    "KE = PE-BE # Kinetic energy of photoelectron\n",
+    "Ve = ((2*KE*1.6*10**-19)/9.11*10**-31)**0.5*10**31\n",
+    "print\"\\nPhoton energy eV =  \",PE\n",
+    "print\"\\nKinetic energy eV =  \",KE\n",
+    "print\"\\nVelocity m/s =  \",Ve\n",
+    "\n",
+    "#Answer may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_11 pgno:107"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "Liquid photon absorption coefficient cm^-1 =   0.0896\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 11, page 107\n",
+    "#Find the absorption coefficient\n",
+    "from math import log\n",
+    "# Using equation 3.20\n",
+    "x = 20.\n",
+    "I0 = 6.\n",
+    "Mu = -1/x*log(1/I0)\n",
+    "print\"\\nLiquid photon absorption coefficient cm^-1 =  \",round(Mu,4)\n",
+    "\n",
+    " \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_12 pgno:107"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "Binding Energy =  eV  12.45\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 12, page 107\n",
+    "#Binding energy\n",
+    "h = 4.15*10**-15\n",
+    "c = 3*10**8\n",
+    "Imax = 1000*10**-10\n",
+    "We = h*c/Imax\n",
+    "print\"\\nBinding Energy =  eV \",We\n",
+    "\n",
+    "#Answer may vary due to round off errorS\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_14 pgno:108"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "N =  atoms/m**3  3.52749208584e+22\n",
+      "\n",
+      "ra =  m  2.7696940413e-10\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 14, page 108\n",
+    "#Diameter of argon atom\n",
+    "from math import pi\n",
+    "#As derived from example 13\n",
+    "N = (1.01*10**5/760)/(1.38*10**-23*273)\n",
+    "print\"\\nN =  atoms/m**3 \",N\n",
+    "#Use equation 3.10\n",
+    "ra = ((85*10**2)/(pi*3.527*10**22))**0.5 \n",
+    "print\"\\nra =  m \",ra\n",
+    "\n",
+    "#Answer may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_15 pgno:109"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\\Mobility of electrons =  m**2/s*V  9375.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 15, page 109\n",
+    "#Mobility of electrons\n",
+    "\n",
+    "Ie = 3\n",
+    "d = 0.8\n",
+    "A = 8*10**-4\n",
+    "Vne = 20*10**17 #V*ne\n",
+    "e = 1.6*10**-19\n",
+    "ke = (Ie*d)/(A*Vne*e)\n",
+    "print\"\\Mobility of electrons =  m**2/s*V \",ke\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_17 pgno:110"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "+(0.02) =  ions/m^3  1911913074.19\n",
+      "\n",
+      "+(-0.02) =  ions/m^3  5.23036331254e+11\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 17, page 110\n",
+    "#Determine the ion density\n",
+    "from math import e\n",
+    "#Based on equation 3.50 and 3.52\n",
+    "nplus = 10**11*e**(-1.6*10**-19*5*0.02/(1.38*10**-23*293))\n",
+    "nminus = 10**10*e**(-1.6*10**-19*5*-0.02/(1.38*10**-23*293)) #textbook uses 0.02 inseatead of -0.02. In the program I have used -0.02\n",
+    "print\"\\n+(0.02) =  ions/m^3 \",nplus\n",
+    "print\"\\n+(-0.02) =  ions/m^3 \",nminus\n",
+    "\n",
+    "#answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_18 pgno:110"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " r1^2-r2^2 =   3.03255e-05\n",
+      "\n",
+      " r2 =  m  0.00551502493195\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 18, page 110\n",
+    "#Determine the diameter\n",
+    "\n",
+    "#Based on the equation 3.40\n",
+    "k = 1.38*10**-23\n",
+    "T = 293\n",
+    "z2z1 = 0.05\n",
+    "e = 1.6*10**-19\n",
+    "E = 250\n",
+    "r1 = 0.09*10**-6\n",
+    "r1r2 = (6*k*T*z2z1)/(e*E)\n",
+    "r2 = (r1+r1r2)**0.5\n",
+    "print\"\\n r1^2-r2^2 =  \",r1r2\n",
+    "print\"\\n r2 =  m \",r2\n",
+    "\n",
+    "#answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_19 pgno:111"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " lambda =  m  0.000222148172831\n",
+      "\n",
+      " Vi =  V  28.5\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 3, Exmaple 19, page 111\n",
+    "#Determine mean free path and ionization\n",
+    "\n",
+    "#(a)Mean free path\n",
+    "#Based on equation 3.14 and 3.15\n",
+    "lamb = 1/(9003*0.5);\n",
+    "#(b)Ionization potential\n",
+    "Vi = 256584./9003.\n",
+    "print\"\\n lambda =  m \",lamb\n",
+    "print\"\\n Vi =  V \",round(Vi,1)\n",
+    "\n",
+    "#answers may vary due to round off error\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_4_Electrical_Breakdown_of_Gases_1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_4_Electrical_Breakdown_of_Gases_1.ipynb
new file mode 100755
index 00000000..907ccacf
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_4_Electrical_Breakdown_of_Gases_1.ipynb
@@ -0,0 +1,557 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Electrical Breakdown of Gases"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_1 pgno:139"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Part (a)\n",
+      " alpha =  m^-1 460.517018599\n",
+      "\n",
+      " Part (b)\n",
+      " I0 =   2.7e-09\n",
+      "\n",
+      " No of electrons emitted =  electrons/s 16875000000.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 1, page 139\n",
+    "#Claculate alpha and No. of electrons emmited\n",
+    "from math import log,e\n",
+    "#Claculate (a)alpha\n",
+    "d2 = 0.01\n",
+    "d1 = 0.005\n",
+    "I2 = 2.7*10**-7\n",
+    "I1 = 2.7*10**-8\n",
+    "alpha = 1/(d2-d1)*log(I2/I1)\n",
+    "#(b)number of electrons emmited from cathode per second\n",
+    "I0 = I1*e**(-alpha*d1)\n",
+    "n0 = I0/(1.6*10**-19)\n",
+    "print\"\\n Part (a)\\n alpha =  m^-1\",alpha\n",
+    "print\"\\n Part (b)\\n I0 =  \",I0\n",
+    "print\"\\n No of electrons emitted =  electrons/s\",n0\n",
+    "#Answer may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_2 pgno:140"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " a*d =   20.7232658369\n",
+      "\n",
+      " electrode space =  m 0.045\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 2, page 140\n",
+    "#Claculate electrode space\n",
+    "from math import log\n",
+    "#based on the values of example 1\n",
+    "d2 = 0.01\n",
+    "d1 = 0.005\n",
+    "I2 = 2.7*10**-7\n",
+    "I1 = 2.7*10**-8\n",
+    "a = 1/(d2-d1)*log(I2/I1) # alpha\n",
+    "#10**9 = %e**a(a*d) \n",
+    "#multiplying log on bith sides log(10**9) = a*d\n",
+    "ad = log(10**9)\n",
+    "print\"\\n a*d =  \",ad\n",
+    "d = ad/a\n",
+    "print\"\\n electrode space =  m\",d\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_3 pgno:140"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Avalanche size =  m 6766.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 3, page 140\n",
+    "#Claculate size of developed avalanche\n",
+    "import scipy\n",
+    "from scipy import integrate\n",
+    "\n",
+    "\n",
+    "a = 4*10**4\n",
+    "b = 15*10**5\n",
+    "#Rewriting equation 4.2\n",
+    "x0=0;x1=0.0005;\n",
+    "def fun1(x):\n",
+    "\ty=a-b*(x)**0.5\n",
+    "\treturn y\n",
+    "X=scipy.integrate.quad(fun1,x0,x1);\n",
+    "expX=6765.964568 \n",
+    "As = expX # Avelanche size\n",
+    "print\"\\n Avalanche size =  m\",round(As)\n",
+    " \n",
+    "#Answers may vary due to round of error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_4 pgno:141"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "  m or  m away from the cathode 0.0517 0.00155\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 4, page 141\n",
+    "#Claculate distance to produce avalanche\n",
+    "\n",
+    "#Rewrite equation 4.2\n",
+    "#using the values of a and b from previous example\n",
+    "#convert integartion to quaderatic equation form\n",
+    "#x=poly(0,\"x\");\n",
+    "import numpy\n",
+    "\n",
+    "p=numpy.array([7.5*10**5, 59.97-4*10**4, 59.97])\n",
+    "r=numpy.roots(p)\n",
+    " #obtaining the roots\n",
+    "print\"\\n  m or  m away from the cathode\",round(r[0],4),round(r[1],5)\n",
+    "\n",
+    "#Answer may vary due to round of error.\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_5 pgno:141"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Minimum distance =  m 0.0011\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 5, page 141\n",
+    "#Claculate minimum distance to produce avalanche of size 10^19\n",
+    "\n",
+    "#Rewriting equation 4.2 and converting it into quadratic equation\n",
+    "#x=poly(0,\"x\");\n",
+    "import numpy\n",
+    "\n",
+    "p=numpy.array([7.5*10**5, 4*10**4, 43.75])\n",
+    "r=numpy.roots(p)\n",
+    " #obtaining the roots\n",
+    "print\"\\n Minimum distance =  m\",round(-r[1],4)# other root is disregarded\n",
+    "\n",
+    "#Answer may vary due to round of error.\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_7 pgno:142"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " E =  V/m 4500000.0\n",
+      "\n",
+      " Alpha =  m^-1 2397.29425605\n",
+      "\n",
+      " Total secondary coefficient of ionization =   0.00834\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 7, page 142\n",
+    "#Claculate secondary coefficient\n",
+    "from math import exp\n",
+    "#Using equation 3.15\n",
+    "E = 9*10**3/0.002\n",
+    "T = 11253.7 # m^-7*kPa^-1\n",
+    "B = 273840 # V/mkPa\n",
+    "p = 101.3 # kPa or 1 atm\n",
+    "d = 0.002 # m\n",
+    "alpha = p*T*exp(-B*p/E)\n",
+    "Y = 1/(exp(alpha*d)-1)\n",
+    "print\"\\n E =  V/m\",E\n",
+    "print\"\\n Alpha =  m^-1\",alpha\n",
+    "print\"\\n Total secondary coefficient of ionization =  \",round(Y,5)\n",
+    "\n",
+    "\n",
+    "#Answer may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_8 pgno:143"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Alpha 1 =  m^-1 39.8\n",
+      "\n",
+      " Alpha 2 =  m^-1 39.8\n",
+      "\n",
+      " Alpha 3 =  m^-1 41.97\n",
+      "\n",
+      " From the above results we can understand that ionization mechanism must be acting at d3 \n",
+      "\n",
+      " secondary ionization coefficient =   0.0363\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 8, page 143\n",
+    "#Claculate first and secondary ionization coefficient\n",
+    "from math import log\n",
+    "from math import exp\n",
+    "#(a)first ionization coefficient\n",
+    "#Using equation 4.7a\n",
+    "d1 = 0.005\n",
+    "a1d1 = log(1.22)\n",
+    "a1 = a1d1/d1\n",
+    "\n",
+    "d2 = 0.01504\n",
+    "a2d2 = log(1.82)\n",
+    "a2 = a2d2/d2\n",
+    "\n",
+    "d3 = 0.019 # wrong value used in the text\n",
+    "a3d3 = log(2.22)\n",
+    "a3 = a3d3/d3\n",
+    "\n",
+    "print\"\\n Alpha 1 =  m^-1\",round(a1,1)\n",
+    "print\"\\n Alpha 2 =  m^-1\",round(a2,1)\n",
+    "print\"\\n Alpha 3 =  m^-1\",round(a3,2)\n",
+    "print\"\\n From the above results we can understand that ionization mechanism must be acting at d3 \"\n",
+    "\n",
+    "#secondary ionization coefficient\n",
+    "I = 2.22\n",
+    "e = exp(a1*d3)\n",
+    "Y = (I-e)/(I*(e-1))\n",
+    "print\"\\n secondary ionization coefficient =  \",round(Y,4)\n",
+    "\n",
+    "#Answer may vary due to round off error.\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_9 pgno:144"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Distance =  m 0.085\n",
+      "\n",
+      " E =  V/m 1596.0\n",
+      "\n",
+      " Volatge  =  V 135.37\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 9, page 144\n",
+    "#Claculate distance and voltage\n",
+    "from math import log\n",
+    "a = 39.8 # alpha\n",
+    "Y = 0.0354 # corfficient\n",
+    "p = 0.133 # kPa\n",
+    "Ep = 12000 # E/P , unit : V/m*kPa\n",
+    "\n",
+    "d = (1/a)*(log(1/Y + 1)) # distance\n",
+    "E = Ep*p\n",
+    "V = E*d\n",
+    "\n",
+    "print\"\\n Distance =  m\",round(d,3)\n",
+    "print\"\\n E =  V/m\",E\n",
+    "print\"\\n Volatge  =  V\",round(V,2)\n",
+    "\n",
+    "#Answers may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_10 pgno:144"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " E =  V/m 5952875.253\n",
+      "\n",
+      " Voltage breakdown = kV 5.95\n",
+      "\n",
+      " Voltage breakdown = kV 4.68\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 10, page 144\n",
+    "#Claculate (a)Raether's criterion (b)Meek and Lobe's criterion\n",
+    "from math import log\n",
+    "#(a)Raether's criterion\n",
+    "# as assumed by Raether and based equation 3.3, 3.50, 4.22 and 4.23\n",
+    "d = 0.001 # m\n",
+    "alpha = 10792.2 # m**-1\n",
+    "p = 101.3 #kPa**-1\n",
+    "ap = 106.54 # alpha/p Unit: m**-1*kPa**-1\n",
+    "T = 11253.7 # m**-1*kPa**-1\n",
+    "B = 273840 # V/m*kPa\n",
+    "Ep = 58764.81 # E/p Unit:V/m*kPa\n",
+    "\n",
+    "ad = 17.7 + log(d)\n",
+    "E = Ep*p\n",
+    "Vs = E*d*10**-3 #Voltage breakdown\n",
+    "print\"\\n E =  V/m\",E\n",
+    "print\"\\n Voltage breakdown = kV\",round(Vs,2)\n",
+    "\n",
+    "#(b)Meek and Loeb's criterion\n",
+    "#Using equation 4.11 and based on 4.24 & 4,25 \n",
+    "#+ we get Er = 468*10**4 V/m\n",
+    "Er = 468*10**4 # V/m\n",
+    "Vs2 = Er*0.001*10**-3\n",
+    "print\"\\n Voltage breakdown = kV\",Vs2\n",
+    "\n",
+    "# Answers may vary due to round of error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_11 pgno:146"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Electron drift velocity = m/s 250000.0\n",
+      "\n",
+      " alpha =  m**-1 114.3\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 11, page 146\n",
+    "#Claculate the first Townsend's ionization coefficient\n",
+    "\n",
+    "t = 0.2*10**-6 # transit time of electrons in seconds\n",
+    "d = 0.05 # m\n",
+    "ve = d/t\n",
+    "TC = 35*10**-9 # Time constant\n",
+    "a = 1/(ve*TC)\n",
+    "print\"\\n Electron drift velocity = m/s\",ve\n",
+    "print\"\\n alpha =  m**-1\",round(a,1)\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_12 pgno:146"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Ea =  V/m 2828427.12475\n",
+      "\n",
+      " x = *sin(3.14*t) 1.26043075492\n",
+      "\n",
+      " t =  ms 0.0252934474843\n",
+      "\n",
+      " fmax =  Hz 630.2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 4, Exmaple 12, page 146\n",
+    "#Travel time and maximum frequency\n",
+    "from math import pi\n",
+    "from math import asin\n",
+    "\n",
+    "#(a)Determine the travel time\n",
+    "Ea = 200*(2)**0.5*10**3/0.1\n",
+    "x = 1.4*10**-4*2828.4*10**3/(2*pi*50)\n",
+    "d = 0.1\n",
+    "print\"\\n Ea =  V/m\",Ea\n",
+    "print\"\\n x = *sin(3.14*t)\",x\n",
+    "#obtaining t from x\n",
+    "t = asin(d/x)/3.14\n",
+    "print\"\\n t =  ms\",t# answer mentioned in the text is wrong\n",
+    "#(b)Determine the maximum frequency\n",
+    "k = 1.4*10**-4\n",
+    "fmax = k*Ea/(2*pi*d)\n",
+    "print\"\\n fmax =  Hz\",round(fmax,1)\n",
+    "\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_5_The_Corona_Discharge_1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_5_The_Corona_Discharge_1.ipynb
new file mode 100755
index 00000000..b414db5a
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chapter_5_The_Corona_Discharge_1.ipynb
@@ -0,0 +1,428 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5 The Corona Discharge"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_2 pgno:173"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Part (a): based on equation 4.13\n",
+      "\n",
+      " Breakdown voltage =  V or  kV 9005.94518519 9.0\n",
+      "\n",
+      " Part (b):Corrsponding to an avelanche size of 10**8\n",
+      "\n",
+      " Breakdown voltage =  V or  kV 9688.16740741 9.7\n",
+      "\n",
+      " Part (c):According to criteria expressed by Equations 5.4 and 5.5\n",
+      "\n",
+      " Breakdown voltage =  kV or  kV 9.4 9.2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 5, Exmaple 2, page 173\n",
+    "#Calculate breakdown voltage\n",
+    "\n",
+    "#(a)Based on equation 4.13\n",
+    "p = 101.3 # kPa\n",
+    "Ep = 2400.4/0.027\n",
+    "E = p*Ep\n",
+    "d = 1*10**-3 # 1 mm\n",
+    "Vs1 = E*d\n",
+    "print\"\\n Part (a): based on equation 4.13\"\n",
+    "print\"\\n Breakdown voltage =  V or  kV\",Vs1,round(Vs1*10**-3)\n",
+    "\n",
+    "#(b)Corrsponding to an avelanche size of 10**8\n",
+    "p = 101.3 # kPa\n",
+    "Cp = Ep*0.027*p\n",
+    "Vs2 = (18.42 + (Cp*10**-3))/0.027\n",
+    "print\"\\n Part (b):Corrsponding to an avelanche size of 10**8\"\n",
+    "print\"\\n Breakdown voltage =  V or  kV\",Vs2,round(Vs2*10**-3,1)\n",
+    "\n",
+    "#(b)According to criteria expressed by Equations 5.4 and 5.5\n",
+    "p = 101.3 # kPa\n",
+    "Vs3a = 9.4\n",
+    "Vs3b = 9.2\n",
+    "print\"\\n Part (c):According to criteria expressed by Equations 5.4 and 5.5\"\n",
+    "print\"\\n Breakdown voltage =  kV or  kV\",Vs3a,Vs3b\n",
+    "\n",
+    "#Answer may vary due to round off error\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_3 pgno:174"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Part (a): based on equation 5.14\n",
+      "\n",
+      " Breakdown voltage =  kV or  kV 27.0178355556 45.03\n",
+      "\n",
+      " Part (b):According to eqution 5.13\n",
+      "\n",
+      " Breakdown voltage =  V or  kV 27.7000577778 45.71\n",
+      "\n",
+      " Part (c):According to criteria expressed by Equations 5.4 and 5.5\n",
+      "\n",
+      " Breakdown voltage =  kV or  kV 27.7 45.5\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 5, Exmaple 3, page 174\n",
+    "#Calculate breakdown voltage at atm pressure 3 and 5\n",
+    "\n",
+    "#(a)Based on equation 5.14\n",
+    "p = 101.3 # kPa\n",
+    "Ep = 2400.4/0.027\n",
+    "E = p*Ep\n",
+    "d = 1*10**-3 # 1 mm\n",
+    "Vs13 = E*d*3 # at 3 atm\n",
+    "Vs15 = E*d*5 # at 5 atm\n",
+    "print\"\\n Part (a): based on equation 5.14\"\n",
+    "print\"\\n Breakdown voltage =  kV or  kV\",Vs13*10**-3,round(Vs15*10**-3,2)\n",
+    "\n",
+    "#(b)According to eqution 5.13\n",
+    "p = 101.3 # kPa\n",
+    "Cp3 = Ep*0.027*p*3 # at 3 atm\n",
+    "Vs23 = (18.42 + (Cp3*10**-3))/0.027\n",
+    "Cp5 = Ep*0.027*p*5 # at 5 atm\n",
+    "Vs25 = (18.42 + (Cp5*10**-3))/0.027\n",
+    "print\"\\n Part (b):According to eqution 5.13\"\n",
+    "print\"\\n Breakdown voltage =  V or  kV\",Vs23*10**-3,round(Vs25*10**-3,2)\n",
+    "\n",
+    "#(b)According to criteria expressed by Equations 5.4 and 5.5\n",
+    "p = 101.3 # kPa\n",
+    "Vs3a = 27.73 # at 3 atm\n",
+    "Vs3b = 45.5 # at 5 atm\n",
+    "print\"\\n Part (c):According to criteria expressed by Equations 5.4 and 5.5\"\n",
+    "print\"\\n Breakdown voltage =  kV or  kV\",round(Vs3a,1),round(Vs3b,1)\n",
+    "\n",
+    "#Answer may vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_8 pgno:179"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " E0 =  kVpeak/cm 39.0\n",
+      "\n",
+      " Part (a)\n",
+      "\n",
+      " The postive corona =  kVpeak 241.8\n",
+      "\n",
+      " The negative corona =  kV 171.0\n",
+      "\n",
+      " Part (b)\n",
+      "\n",
+      " The postive corona =  kVpeak 324.48\n",
+      "\n",
+      " The negative corona =  kV 228.8\n",
+      "\n",
+      " Part (c)\n",
+      "\n",
+      " The postive corona =  kVpeak 388.83\n",
+      "\n",
+      " The negative corona =  kV 274.2\n",
+      "\n",
+      " Part (d)\n",
+      "\n",
+      " The postive corona =  kVpeak 444.21\n",
+      "\n",
+      " The negative corona =  kV 313.2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 5, Exmaple 8, page 179\n",
+    "#Calculate corona onset voltage\n",
+    "\n",
+    "s = 4 # cm\n",
+    "r = 1 # cm\n",
+    "D = 5*10^2 # cm\n",
+    "dt = 1\n",
+    "E0 = 30*dt*(1 + 0.3*(dt*r)**0.5)\n",
+    "print\"\\n E0 =  kVpeak/cm\",E0\n",
+    "#using equations (5.18), the positive and negative corona\n",
+    "En = 27.501 # kVpeak/cm\n",
+    "#part a\n",
+    "Vp1 = 6.2*E0\n",
+    "Vn1 = 6.2*En\n",
+    "print\"\\n Part (a)\"\n",
+    "print\"\\n The postive corona =  kVpeak\",Vp1\n",
+    "print\"\\n The negative corona =  kV\",round(Vn1)\n",
+    "#part b\n",
+    "Vp2 = 8.32*E0\n",
+    "Vn2 = 8.32*En\n",
+    "print\"\\n Part (b)\"\n",
+    "print\"\\n The postive corona =  kVpeak\",Vp2\n",
+    "print\"\\n The negative corona =  kV\",round(Vn2,1)\n",
+    "#part c\n",
+    "Vp3 = 9.97*E0\n",
+    "Vn3 = 9.97*En\n",
+    "print\"\\n Part (c)\"\n",
+    "print\"\\n The postive corona =  kVpeak\",Vp3\n",
+    "print\"\\n The negative corona =  kV\",round(Vn3,1)\n",
+    "#part d\n",
+    "Vp4 = 11.39*E0\n",
+    "Vn4 = 11.39*En\n",
+    "print\"\\n Part (d)\"\n",
+    "print\"\\n The postive corona =  kVpeak\",Vp4\n",
+    "print\"\\n The negative corona =  kV\",round(Vn4,1)\n",
+    "\n",
+    "#Answer CONSIDERABLY vary due to round off error.\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_9 pgno:180"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Mean geometric distance between the conductors  m 6.004626\n",
+      "\n",
+      " E0 =  kVpeak/cm 39.0\n",
+      "\n",
+      " V0peak =  kVpeak 249.510313968\n",
+      "\n",
+      " V0 =  kV 175.9\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 5, Exmaple 9, page 180\n",
+    "#Calculate corona onset voltage\n",
+    "from math import log\n",
+    "t = 5*5*8.66 # the three side of the trangle in m\n",
+    "Deq = 6.004626#nthroot(t,3) \n",
+    "dt = 1 #delta = 1 at standard temperature and pressure\n",
+    "r = 1 #radius of the conductor\n",
+    "En = 27.501 # kVpeak/cm\n",
+    "E0 = 30*dt*(1 + 0.3*(dt*r)**0.5)\n",
+    "V0peak = E0*log(Deq*10**2)\n",
+    "V0 = En*log(Deq*10**2)\n",
+    "\n",
+    "print\"\\n Mean geometric distance between the conductors  m\",Deq\n",
+    "print\"\\n E0 =  kVpeak/cm\",E0\n",
+    "print\"\\n V0peak =  kVpeak\",V0peak\n",
+    "print\"\\n V0 =  kV\",round(V0,1)\n",
+    "\n",
+    "#Answers may vary due to round off error\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_10 pgno:180"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " Reative air density   0.600612870276\n",
+      "\n",
+      " Corona onset field = kVpeak/cm 19.409458636\n",
+      "\n",
+      " V0peak =  kVpeak 124.175900467\n",
+      "\n",
+      " V0 =  kV 175.943157549\n",
+      "\n",
+      " Ration of V0 =   1.56300480127\n",
+      "\n",
+      " Corona power loss Pc =  kW/km 6.51960146384\n",
+      "\n",
+      " Corona current =  mA/Km 41.1\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 5, Exmaple 10, page 180\n",
+    "#Calculate corona power loss\n",
+    "from math import log\n",
+    "p = 75 # pressure\n",
+    "t = 35 # temprature\n",
+    "m1 = 0.92\n",
+    "m2 = 0.95\n",
+    "t = 5*5*8.66 # the three side of the trangle in m\n",
+    "Deq = 6.004626#nthroot(t,3) \n",
+    "dt = (3.92*p)/(273+t) #Relative air density\n",
+    "E0 = 30*dt*(1 + 0.3*(dt)**0.5)*m1*m2\n",
+    "En = 27.501 # kVpeak/cm\n",
+    "Vph = (275*10**3)/(3)**0.5\n",
+    "V0peak = E0*log(Deq*10**2)\n",
+    "V0 = En*log(Deq*10**2)\n",
+    "V0ratio = 275/V0\n",
+    "print\"\\n Reative air density  \",dt\n",
+    "print\"\\n Corona onset field = kVpeak/cm\",E0\n",
+    "print\"\\n V0peak =  kVpeak\",V0peak\n",
+    "print\"\\n V0 =  kV\",V0\n",
+    "print\"\\n Ration of V0 =  \",V0ratio\n",
+    "K = 0.05 # K factor\n",
+    "Pc = (3.73*K*50*Vph**2)/(Deq*10**2)**2\n",
+    "Cc = Pc*10**3/Vph\n",
+    "print\"\\n Corona power loss Pc =  kW/km\",Pc*10**-5\n",
+    "print\"\\n Corona current =  mA/Km\",round(Cc*10**-2,1)\n",
+    "\n",
+    "#Answer vary due to round off error\n",
+    "#Some of the answers provided in the textbook are wrong\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_11 pgno:180"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " E0 =  kVpeak/cm or 20 kV/cm 28.3912466575\n",
+      "\n",
+      " V0 =  kV 425.955402321\n",
+      "\n",
+      " V0 (line to line) =  kV 737.8\n",
+      "\n",
+      " Voltage (line to line) =  kV 1312.5\n",
+      "\n",
+      " Envelope radius = 5 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 5, Exmaple 11, page 180\n",
+    "#Calculate corona onset voltage and effective corona envelope\n",
+    "from math import log\n",
+    "#(a) corona onset voltage\n",
+    "r = 3.175 # cm\n",
+    "h = 13 # m\n",
+    "m= 0.9 # m1 and m2\n",
+    "dt = 1 # Relative air density\n",
+    "E0 = 30*dt*(1 + 0.3/(r)**0.5)*m*m\n",
+    "V0 = 20*r*log(2*h*10**2/r)\n",
+    "print\"\\n E0 =  kVpeak/cm or 20 kV/cm\",E0\n",
+    "print\"\\n V0 =  kV\",V0\n",
+    "print\"\\n V0 (line to line) =  kV\",round(V0*(3)**0.5,1)\n",
+    "\n",
+    "#(b)Corona envelope at 2.5 p.u\n",
+    "V = 2.5*525 # line to line voltage * 2.5\n",
+    "print\"\\n Voltage (line to line) =  kV\",V\n",
+    "#Solving the equations in trila and error method\n",
+    "print\"\\n Envelope radius = 5 cm\"\n",
+    "\n",
+    "# Answers may vary due to round off error.\n",
+    " \n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chpater_14_Overvoltages_on_Power_systems_1.ipynb b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chpater_14_Overvoltages_on_Power_systems_1.ipynb
new file mode 100755
index 00000000..71792aab
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/Chpater_14_Overvoltages_on_Power_systems_1.ipynb
@@ -0,0 +1,74 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chpater 14 Overvoltages on Power systems"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 14_1 pgno453"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " f1 =  Hz 183.776298474\n",
+      "\n",
+      " Time to crest =  micro seconds 1360.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 14,Example 1, page 453\n",
+    "#Determine the time to crest \n",
+    "\n",
+    "\n",
+    "I = 400 # mH of inductance\n",
+    "L = 500*10**-3 # mH\n",
+    "C = 1.5*10**-6 # micro F\n",
+    "from math import pi,sqrt\n",
+    "f = 1/(2*pi*sqrt(L*C))  \n",
+    "t = 10**6/(4*f) # calulation done in the text is wrong\n",
+    "print\"\\n f1 =  Hz\",f\n",
+    "print\"\\n Time to crest =  micro seconds\",round(t)\n",
+    "\n",
+    "\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/README.txt b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/README.txt
new file mode 100755
index 00000000..9365c7b9
--- /dev/null
+++ b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Vedantam  Lakshmi Manasa
+Course: btech
+College/Institute/Organization: K L university
+Department/Designation: ECE
+Book Title: High Voltage Engineering Theory and Practice
+Author: M. Khalifa
+Publisher: Marcel Dekker
+Year of publication: 2000
+Isbn: 0824704029
+Edition: 2
\ No newline at end of file
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_12.png b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_12.png
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diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_16.png b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_16.png
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index 00000000..0cc6d038
Binary files /dev/null and b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_16.png differ
diff --git a/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_3.png b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_3.png
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Binary files /dev/null and b/High_Voltage_Engineering_Theory_and_Practice_by_M._Khalifa/screenshots/Cha_3.png differ
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch1.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch1.ipynb
new file mode 100755
index 00000000..3986f8e6
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch1.ipynb
@@ -0,0 +1,49 @@
+{
+ "metadata": {
+  "name": "ch1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 1  : The way of the program"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1  Page no : 7"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 1 page no : 7\n'''\n\nprint \"Hello, world.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hello, world.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch10.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch10.ipynb
new file mode 100755
index 00000000..cd0445e0
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch10.ipynb
@@ -0,0 +1,142 @@
+{
+ "metadata": {
+  "name": "ch10"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "CHAPTER 10. : VECTORS"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 10.1 page no :103\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 10.1 page no :103\n''' \n\nvalues = [] \nc = int(raw_input())\nwhile(c != -1):\n    values.append(c)\n    c = int(raw_input())\n\nsize=len(values)\nfor i in range(size):\n    print values[i] ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "5\n"
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "4\n"
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3\n"
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "2\n"
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "-1\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "5\n4\n3\n2\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 10.2 page no :104\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 10.2 page no :104\n'''\nimport random\n\nfor i in range(4):\n    x = random.random();\n    print  x",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "0.16979844903\n0.113100369897\n0.851107623696\n0.0966914962921\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 10.3 page no :105\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 10.3 page no :105\n'''\nimport random\ndef randomVector(n,upperBound):\n    vec = []\n    for i in range(n):\n        vec.append(random.random () * 100 % upperBound)\n    return vec\n\ndef printVector(vec):\n    for i in range(len(vec)):\n        print vec[i] ,\n\nnumValues = 20;\nupperBound = 10;\nvector = randomVector (numValues, upperBound);\nprintVector(vector)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1.84318995426 8.49475291756 7.92488160107 3.93550646061 0.967457025838 6.72758767668 6.22192058422 8.19279912461 7.4898535715 3.02785494125 2.97556694487 4.70451449621 8.21039455753 2.95832700489 7.02858630112 8.72631899097 5.67567455024 7.48090989023 9.14911759876 9.82069257056\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 10.4 page no: 106\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 10.4 page no: 106\n'''\nimport random\ndef howMany(vec,value):\n    count = 0\n    for i in range(len(vec)):\n        if (vec[i] == value):\n            count += 1\n    return count\n\ndef randomVector(n,upperBound):\n    vec = []\n    for i in range(n):\n        vec.append(int(random.random () * 100 % upperBound))\n    return vec\n    \nnumValues = 20;\nupperBound = 10;\nvector = randomVector (numValues, upperBound);\nprint \"value \\t howMany\"\nfor i in range(upperBound):\n    print i , '\\t' , howMany(vector, i)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "value \t howMany\n0 \t2\n1 \t2\n2 \t0\n3 \t4\n4 \t3\n5 \t1\n6 \t2\n7 \t1\n8 \t2\n9 \t3\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch11.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch11.ipynb
new file mode 100755
index 00000000..3f21ca72
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch11.ipynb
@@ -0,0 +1,175 @@
+{
+ "metadata": {
+  "name": "ch11"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "CHAPTER 11 : MEMBER FUNCTIONS"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 11.1 page no :112\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 11.1 page no :112\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\n    def printTime(self):\n        print self.hour , \":\" , self.minute , \":\" , self.second\n        print \"Time is \" , self.hour , \" hour \" , self.minute , \" minutes \" , self.second \n\n\nc = Time( 9, 14, 30.0 )\nc.printTime ()",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "9 : 14 : 30.0\nTime is  9  hour  14  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 11.2 page  no :114\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 11.2 page  no :114\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\n    def printTime(self):\n        print self.hour , \":\" , self.minute , \":\" , self.second\n        print \"Time is \" , self.hour , \" hour \" , self.minute , \" minutes \" , self.second \n\n    def increment (self,secs):\n        self.second += secs\n        while (self.second >= 60.0):\n            self.second -= 60.0;\n            self.minute += 1;\n\n        while (self.minute >= 60):\n            self.minute -= 60.0;\n            self.hour += 1;\n\nc = Time (9, 14, 30.0)\nc.increment (500.0);\nc.printTime ();",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "9 : 22 : 50.0\nTime is  9  hour  22  minutes  50.0\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 11.3 page no :115\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 11.3 page no :115\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\n    def printTime(self):\n        print self.hour , \":\" , self.minute , \":\" , self.second\n        print \"Time is \" , self.hour , \" hour \" , self.minute , \" minutes \" , self.second \n\n    def convertToSeconds(self):\n        minutes = self.hour * 60 + self.minute;\n        seconds = minutes * 60 + self.second;\n        return seconds;\n\n\nc = Time (9, 14, 30.0)\nprint c.convertToSeconds()",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "33270.0\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 11.4 page no :115\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 11.4 page no :115\n'''\n\n\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\n    def printTime(self):\n        print self.hour , \":\" , self.minute , \":\" , self.second\n        print \"Time is \" , self.hour , \" hour \" , self.minute , \" minutes \" , self.second \n\n    def convertToSeconds(self):\n        minutes = self.hour * 60 + self.minute;\n        seconds = minutes * 60 + self.second;\n        return seconds;\n\n    def after (self,time2):\n        if (self.hour > time2.hour):\n            return True;\n        if (self.hour < time2.hour): \n            return False;\n        if (self.minute > time2.minute): \n            return True;\n        if (self.minute < time2.minute):\n            return False;\n        if (self.second > time2.second): \n            return True;\n        return False;\n\nt1 = Time( 11, 59, 3.14159 )\nt2 = Time( 12, 42, 3.234 )\n\nif (t2.after(t1)):\n    print \"The bread will be done after it starts.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The bread will be done after it starts.\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 11.5 page no :116\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 11.5 page no :116\n'''\n\nclass Time:\n    def __init__(self,h,m=None,s=None):\n        if m==None:\n            self.hour = int (secs / 3600.0);\n            secs -= self.hour * 3600.0;\n            self.minute = int (secs / 60.0);  \n            secs -= self.minute * 60.0;\n            self.second = secs;\n        else:\n            self.hour = h\n            self.minute = m\n            self.second = s\n\n    def printTime(self):\n        print self.hour , \":\" , self.minute , \":\" , self.second\n        print \"Time is \" , self.hour , \" hour \" , self.minute , \" minutes \" , self.second \n\nc = Time (9, 14, 30.0)\nc.printTime()",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "9 : 14 : 30.0\nTime is  9  hour  14  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 11.6 page no : 117\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 11.6 page no : 117\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\n    def printTime(self):\n        print self.hour , \":\" , self.minute , \":\" , self.second\n        print \"Time is \" , self.hour , \" hour \" , self.minute , \" minutes \" , self.second \n\nc = Time (9, 14, 30.0)\nc.printTime()",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "9 : 14 : 30.0\nTime is  9  hour  14  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 11.7 page no :119\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 11.7 page no :119\n'''\nclass Time:\n    def __init__(self,h,m=None,s=None):\n        if m==None:\n            self.hour = int (secs / 3600.0);\n            secs -= self.hour * 3600.0;\n            self.minute = int (secs / 60.0);  \n            secs -= self.minute * 60.0;\n            self.second = secs;\n        else:\n            self.hour = h\n            self.minute = m\n            self.second = s\n\n    def printTime(self):\n        print self.hour , \":\" , self.minute , \":\" , self.second\n        print \"Time is \" , self.hour , \" hour \" , self.minute , \" minutes \" , self.second \n\n    def after (self,time2):\n        if (self.hour > time2.hour):\n            return True;\n        if (self.hour < time2.hour): \n            return False;\n        if (self.minute > time2.minute): \n            return True;\n        if (self.minute < time2.minute):\n            return False;\n        if (self.second > time2.second): \n            return True;\n        return False;\n\n    def convertToSeconds(self):\n        minutes = self.hour * 60 + self.minute;\n        seconds = minutes * 60 + self.second;\n        return seconds;\n\n    def increment (self,secs):\n        self.second += secs\n        while (self.second >= 60.0):\n            self.second -= 60.0;\n            self.minute += 1;\n\n        while (self.minute >= 60):\n            self.minute -= 60.0;\n            self.hour += 1;\n\n    def add(self,t2):\n        a = t2.convertToSeconds()\n        self.increment(a)\n\nc = Time(9, 14, 30.0)\nc.increment (500.0);\nc.printTime();\nb = Time(3, 35, 0.0);\ncurent = c\nc.add(b)\nd = c\nd.printTime()\nif(d.after(curent)):\n    print \"The bread will be done after it starts.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "9 : 22 : 50.0\nTime is  9  hour  22  minutes  50.0\n12 : 57.0 : 50.0\nTime is  12  hour  57.0  minutes  50.0\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch12.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch12.ipynb
new file mode 100755
index 00000000..c78a7ba1
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch12.ipynb
@@ -0,0 +1,169 @@
+{
+ "metadata": {
+  "name": "ch12"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 12 : Vectors of Objects"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 12.1 page no :124\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 12.1 page no :124\n'''\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n\nc = Card (0,3)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 12.2 page no : 126\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 12.2 page no : 126\n'''\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\nc = Card (1, 11)\nc.print_ ();",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Queen  of  Diamonds\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 12.3 page no :127\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 12.3 page no :127\n'''\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\n    def equals(self,c2):\n        return (self.rank == c2.rank and self.suit == c2.suit);\n\nc1 = Card(1, 11);\nc2 =Card(1, 11);\nif (c1.equals(c2)):\n    print \"Yup, that's the same card.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Yup, that's the same card.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 12.4 page no :129\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 12.4 page no :129\n'''\n\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\n    def equals(self,c2):\n        return (self.rank == c2.rank and self.suit == c2.suit);\n\n    def  isGreater(self,c2):\n        if  (self.suit > c2.suit):\n            return True;\n        if (selfsuit < c2.suit):\n            return False;\n        if (self.rank > c2.rank):\n            return True;\n        if (self.rank < c2.rank):\n            return False;\n        return False;\n\nc1 = Card(2, 11)\nc2 = Card(1, 11)\nif (c1.isGreater (c2)):\n    c1.print_()\n    print \"is greater than\" \n    c2.print_();",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Queen  of  Hearts\nis greater than\nQueen  of  Diamonds\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 12.5 page no :130\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 12.5 page no :130\n'''\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\n\ndef printDeck(deck):\n    for i in range(len(deck)):\n        deck[i].print_()\n\ndeck = []\n\nfor suit in range(3):\n    for rank in range(13):\n        a = Card(suit,rank)\n        deck.append(a)\n\nprintDeck(deck)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Ace  of  Clubs\n2  of  Clubs\n3  of  Clubs\n4  of  Clubs\n5  of  Clubs\n6  of  Clubs\n7  of  Clubs\n8  of  Clubs\n9  of  Clubs\n10  of  Clubs\nJack  of  Clubs\nQueen  of  Clubs\nKing  of  Clubs\nAce  of  Diamonds\n2  of  Diamonds\n3  of  Diamonds\n4  of  Diamonds\n5  of  Diamonds\n6  of  Diamonds\n7  of  Diamonds\n8  of  Diamonds\n9  of  Diamonds\n10  of  Diamonds\nJack  of  Diamonds\nQueen  of  Diamonds\nKing  of  Diamonds\nAce  of  Hearts\n2  of  Hearts\n3  of  Hearts\n4  of  Hearts\n5  of  Hearts\n6  of  Hearts\n7  of  Hearts\n8  of  Hearts\n9  of  Hearts\n10  of  Hearts\nJack  of  Hearts\nQueen  of  Hearts\nKing  of  Hearts\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 12.6 page no : 131\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 12.6 page no : 131\n'''\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n        \n    def equals(self,c2):\n        return (self.rank == c2.rank and self.suit == c2.suit);\n\n    def find(self,deck):\n        for i in range(len(deck)):\n            if (self.equals(deck[i])):\n                return i;\n        return -1;\n\ndeck = []\n\nfor suit in range(3):\n    for rank in range(13):\n        a = Card(suit,rank)\n        deck.append(a)\n\nindex = deck[17].find (deck)\nprint \"I found the card at index = \" , index ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "I found the card at index =  17\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 12.7 page no : 133\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 12.7 page no : 133\n'''\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n        \n    def equals(self,c2):\n        return (self.rank == c2.rank and self.suit == c2.suit);\n\n    def find(self,deck):\n        for i in range(len(deck)):\n            if (self.equals(deck[i])):\n                return i;\n        return -1;\n        \n    def  isGreater(self,c2):\n        if  (self.suit > c2.suit):\n            return True;\n        if (self.suit < c2.suit):\n            return False;\n        if (self.rank > c2.rank):\n            return True;\n        if (self.rank < c2.rank):\n            return False;\n        return False;\n        \ndef findBisect(deck,card,low,high):\n    print low , \", \" , high \n    if (high < low):\n        return -1;\n    mid = (high + low) / 2\n    if (card.equals(deck[mid])):\n        return mid;\n    if (deck[mid].isGreater (card)):\n        return findBisect (deck, card,low, mid-1);\n    else:\n        return findBisect (deck,card, mid+1, high)\n\n\ndeck = []\n\nfor suit in range(3):\n    for rank in range(13):\n        a = Card(suit,rank)\n        deck.append(a)\n\nprint findBisect (deck, deck[23], 0, 51)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "0 ,  51\n0 ,  24\n13 ,  24\n19 ,  24\n22 ,  24\n23\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch13.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch13.ipynb
new file mode 100755
index 00000000..f9a283c6
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch13.ipynb
@@ -0,0 +1,121 @@
+{
+ "metadata": {
+  "name": "ch13"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 13 : Objects of Vectors"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 13.1 page no: 137\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 13.1 page no: 137\n'''\nclass Suit:\n    CLUBS = 0\n    DIAMOND = 1\n    HEARTS = 2\n    SPADES = 3\n\nclass Rank:    \n    ACE=1\n    TWO = 2\n    THREE = 3\n    FOUR =4\n    FIVE = 5\n    SIX = 6\n    SEVEN = 7\n    EIGHT = 8\n    NINE= 9\n    TEN = 10\n    JACK = 11 \n    QUEEN = 12\n    KING =13\n        \nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\ndeck = []\n\nfor suit in range(Suit.SPADES):\n    for rank in range(Rank.KING):\n        a = Card(suit,rank)\n        deck.append(a)\n        \n\nfor i in deck:\n    i.print_()        ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Ace  of  Clubs\n2  of  Clubs\n3  of  Clubs\n4  of  Clubs\n5  of  Clubs\n6  of  Clubs\n7  of  Clubs\n8  of  Clubs\n9  of  Clubs\n10  of  Clubs\nJack  of  Clubs\nQueen  of  Clubs\nKing  of  Clubs\nAce  of  Diamonds\n2  of  Diamonds\n3  of  Diamonds\n4  of  Diamonds\n5  of  Diamonds\n6  of  Diamonds\n7  of  Diamonds\n8  of  Diamonds\n9  of  Diamonds\n10  of  Diamonds\nJack  of  Diamonds\nQueen  of  Diamonds\nKing  of  Diamonds\nAce  of  Hearts\n2  of  Hearts\n3  of  Hearts\n4  of  Hearts\n5  of  Hearts\n6  of  Hearts\n7  of  Hearts\n8  of  Hearts\n9  of  Hearts\n10  of  Hearts\nJack  of  Hearts\nQueen  of  Hearts\nKing  of  Hearts\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 13.2 page no : 140\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 13.2 page no : 140\n'''\nclass Suit:\n    CLUBS = 0\n    DIAMOND = 1\n    HEARTS = 2\n    SPADES = 3\n\nclass Rank:    \n    ACE=1\n    TWO = 2\n    THREE = 3\n    FOUR =4\n    FIVE = 5\n    SIX = 6\n    SEVEN = 7\n    EIGHT = 8\n    NINE= 9\n    TEN = 10\n    JACK = 11 \n    QUEEN = 12\n    KING =13\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\nclass Deck:\n    def __init__(self,n=None):\n        self.cards = []\n        if n ==None:\n            size = 52\n        else:\n            size = n\n                                 \n        for suit in range(Suit.SPADES):\n            for rank in range(Rank.KING):\n                a = Card(suit,rank)\n                self.cards.append(a)\n\n    def print_(self):\n        for i in self.cards:\n            i.print_()\n\nd = Deck(52)\nd.print_()\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Ace  of  Clubs\n2  of  Clubs\n3  of  Clubs\n4  of  Clubs\n5  of  Clubs\n6  of  Clubs\n7  of  Clubs\n8  of  Clubs\n9  of  Clubs\n10  of  Clubs\nJack  of  Clubs\nQueen  of  Clubs\nKing  of  Clubs\nAce  of  Diamonds\n2  of  Diamonds\n3  of  Diamonds\n4  of  Diamonds\n5  of  Diamonds\n6  of  Diamonds\n7  of  Diamonds\n8  of  Diamonds\n9  of  Diamonds\n10  of  Diamonds\nJack  of  Diamonds\nQueen  of  Diamonds\nKing  of  Diamonds\nAce  of  Hearts\n2  of  Hearts\n3  of  Hearts\n4  of  Hearts\n5  of  Hearts\n6  of  Hearts\n7  of  Hearts\n8  of  Hearts\n9  of  Hearts\n10  of  Hearts\nJack  of  Hearts\nQueen  of  Hearts\nKing  of  Hearts\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 13.3 page no : 142\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 13.3 page no : 142\n'''\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\n    def equals(self,c2):\n        return (self.rank == c2.rank and self.suit == c2.suit);\n        \n    def find(self,deck):\n        for i in range(len(deck.cards)):\n            if self.equals(deck.cards[i]):\n                return i\n        return -1;\n\nclass Deck:\n    def __init__(self,n=None):\n        self.cards = []\n        if n ==None:\n            size = 52\n        else:\n            size = n\n                                 \n        for suit in range(4):\n            for rank in range(13):\n                a = Card(suit,rank)\n                self.cards.append(a)\n\n    def print_(self):\n        for i in self.cards:\n            i.print_()\n\nd = Deck()\na = Card(1,11)\nprint a.find(d)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "24\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 13.4 page no :144\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 13.4 page no :144\n'''\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\n    def equals(self,c2):\n        return (self.rank == c2.rank and self.suit == c2.suit);\n        \n    def find(self,deck):\n        for i in range(len(deck.cards)):\n            if self.equals(deck.cards[i]):\n                return i\n        return -1;\n\nclass Deck:\n    def __init__(self,n=None):\n        self.cards = []\n        if n ==None:\n            size = 52\n        else:\n            size = n\n            return\n                                 \n        for suit in range(4):\n            for rank in range(13):\n                a = Card(suit,rank)\n                self.cards.append(a)\n\n    def print_(self):\n        for i in self.cards:\n            i.print_()\n            \n    def subdeck(self,low,high):\n        sub = Deck(high-low+1)\n        for i in range(len(sub.cards)):\n            sub.cards.append(self.cards[low+i])\n        return sub;\n\ndeck = Deck()\nhand1 = deck.subdeck (0, 4);\nhand2 = deck.subdeck (5, 9);\npack = deck.subdeck (10, 51);\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 13.5 page no : 146\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 13.5 page no : 146\n'''\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n\n    def print_(self):\n        suits = [\"Clubs\",\"Diamonds\",\"Hearts\",\"Spades\"]\n        ranks = [\"Ace\",\"2\",\"3\",\"4\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"Jack\",\"Queen\",\"King\"]\n        print  ranks[self.rank] , \" of \" , suits[self.suit] \n\n    def equals(self,c2):\n        return (self.rank == c2.rank and self.suit == c2.suit);\n        \n    def find(self,deck):\n        for i in range(len(deck.cards)):\n            if self.equals(deck.cards[i]):\n                return i\n        return -1;\n\nclass Deck:\n    def __init__(self,n=None):\n        self.cards = []\n        if n ==None:\n            size = 52\n        else:\n            size = n\n            return\n                                 \n        for suit in range(4):\n            for rank in range(13):\n                a = Card(suit,rank)\n                self.cards.append(a)\n\n    def print_(self):\n        for i in self.cards:\n            i.print_()\n            \n    def merge(self,d2):\n        # create a new deck big enough for all the cards\n        result = Deck (len(d1.cards)+ len(d2.cards));\n        # use the index i to keep track of where we are in\n        # the first deck, and the index j for the second deck\n        i = 0;\n        j = 0;\n        # the index k traverses the result deck\n        return result;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch14.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch14.ipynb
new file mode 100755
index 00000000..124b0855
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch14.ipynb
@@ -0,0 +1,130 @@
+{
+ "metadata": {
+  "name": "ch14"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 14 : Classes and invariants"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 14.1 page no :149\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 14.1 page no :149\n'''\n\nclass Card:\n    def __init__(self,s=None,r=None):\n        if s==None:\n            self.suit = 0\n            self.rank = 0\n        else:\n            self.suit = s; \n            self.rank = r;\n\n    def getRank (self):\n        return self.rank\n        \n    def getSuit (self):\n        return self.suit\n    \n    def setRank(self,r):\n        self.rank = r        \n\n    def setSuit (self, s):\n        self.suit = s",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 14.2 page no :151\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 14.2 page no :151\ndefining class \n'''\n\nclass Complex:\n    def __init__(self,r=None,i=None):\n        if r==None:\n            pass\n        else:\n            self.real = r\n            self.imag = i",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 14.3 page no : 152\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 14.3 page no : 152\n'''\nimport math\nclass Complex:\n    def __init__(self,r=None,i=None):\n        if r==None:\n            self.real = 0\n            self.imag = 0\n            self.polar = False\n            self.cartesian = False\n        else:\n            self.real = r\n            self.imag = i\n            self.cartesian = True\n            self.polar = True\n        self.mag = 0\n        self.theta = 0\n\n    def getReal (self):\n        if (self.cartesian == False):\n            self.calculateCartesian ();\n        return self.real;\n        \n    def getImag(self):\n        return self.imag\n\n    def calculateCartesian (self):\n        self.real = self.mag * math.cos(self.theta);\n        self.imag = self.mag * math.sin (self.theta);   \n        self.cartesian = True;\n\n    def printCartesian (self):\n        print self.getReal() , \" + \" , self.getImag() , \"i\"\n\n    def printPolar(self):\n        print self.mag , \" e^ \" , self.theta , \"i\"\n\n\nc1 = Complex(2.0, 3.0);\nc1.printCartesian();\nc1.printPolar();",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "2.0  +  3.0 i\n0  e^  0 i\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 14.4 page no : 155\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 14.4 page no : 155\n'''\nclass Complex:\n    def __init__(self,r=None,i=None):\n        if r==None:\n            self.real = 0\n            self.imag = 0\n            self.polar = False\n            self.cartesian = False\n        else:\n            self.real = r\n            self.imag = i\n            self.cartesian = True\n            self.polar = True\n        self.mag = 0\n        self.theta = 0\n        \n    def getReal (self):\n        if (self.cartesian == False):\n            self.calculateCartesian ();\n        return self.real;\n        \n    def getImag(self):\n        return self.imag\n\n\n    def add(self,b):\n        real = self.getReal() + b.getReal();\n        imag = self.getImag() + b.getImag();\n        s = Complex(real, imag);\n        return s\n\n    def printCartesian (self):\n        print self.getReal() , \" + \" , self.getImag() , \"i\"\n\nc1 = Complex(2.0, 3.0);\nc2 = Complex(3.0, 4.0);\ns = c1.add (c2);\ns.printCartesian();",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "5.0  +  7.0 i\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 14.5 page no :156\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 14.5 page no :156\n'''\nimport math\nclass Complex:\n    def __init__(self,r=None,i=None):\n        if r==None:\n            self.real = 0\n            self.imag = 0\n            self.polar = False\n            self.cartesian = False\n        else:\n            self.real = r\n            self.imag = i\n            self.cartesian = True\n            self.polar = True\n        self.mag = 0\n        self.theta = 0\n        \n    def getReal (self):\n        if (self.cartesian == False):\n            self.calculateCartesian ();\n        return self.real;\n        \n    def getImag(self):\n        return self.imag\n\n    def calculateCartesian (self):\n        self.real = self.mag * math.cos(self.theta);\n        self.imag = self.mag * math.sin (self.theta);   \n        self.cartesian = True;\n\n\n    def printCartesian (self):\n        print self.getReal() , \" + \" , self.getImag() , \"i\"\n\n    def mult(self,b):\n        mag = self.mag * b.mag\n        theta = self.theta + b.theta\n        product = Complex()\n        product.setPolar (mag, theta);\n        return product;\n\n    def setPolar(self,m,t):\n        self.mag = m\n        self.theta = t;\n        self.cartesian = False\n        self.polar = True;\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 14.6 page no :159\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 14.6 page no :159\ndefining class Complex\n'''\n\nclass Complex:\n    def __init__(self,r=None,i=None):\n        if r==None:\n            self.real = 0\n            self.imag = 0\n            self.polar = False\n            self.cartesian = False\n        else:\n            self.real = r\n            self.imag = i\n            self.cartesian = True\n            self.polar = True\n        self.mag = 0\n        self.theta = 0\n        \n    def getReal (self):\n        if (self.cartesian == False):\n            self.calculateCartesian ();\n        return self.real;\n        \n    def getImag(self):\n        return self.imag\n\n    def printCartesian (self):\n        print self.getReal() , \" + \" , self.getImag() , \"i\"\n\n    def mult(self,b):\n        mag = self.mag * b.mag\n        theta = a.theta + b.theta\n        product = Complex()\n        product.setPolar (mag, theta);\n        return product;\n\n    def setPolar(self,m,t):\n        self.mag = m\n        self.theta = t;\n        self.cartesian = False\n        self.polar = true;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch15.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch15.ipynb
new file mode 100755
index 00000000..cc4b221f
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch15.ipynb
@@ -0,0 +1,127 @@
+{
+ "metadata": {
+  "name": "ch15"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 15 : File Input/Output and apmatrixes"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 15.1 page no : 162\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 15.1 page no : 162\n'''\ntry:\n    fileName = \"a.txt\"\n    f = open(fileName,\"r\")\nexcept:\n    print \"Unable to open the file named \" , fileName;\n    import sys\n    sys.exit(0)\n\n# reading file line by line\n    \nfor i in f.readlines():\n    print i\n\nf.close()    ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "SystemExit",
+       "evalue": "0",
+       "output_type": "pyerr",
+       "traceback": [
+        "An exception has occurred, use %tb to see the full traceback.\n",
+        "\u001b[1;31mSystemExit\u001b[0m\u001b[1;31m:\u001b[0m 0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Unable to open the file named  a.txt\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": "To exit: use 'exit', 'quit', or Ctrl-D.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 15.2 page no : 163\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 15.2 page no : 163\n'''\n\ntry:\n    fileName = \"a.txt\"\n    outputFile = \"b.txt\"\n    f = open(fileName,\"r\")\n    w = open(outputFile,\"w\")\nexcept:\n    print \"Unable to open the file named \" , fileName;\n    import sys\n    sys.exit(0)\n\n# reading file line by line\n    \nfor i in f.readlines():\n    w.write(i)\n    print i\n    \nf.close()\nw.close()    ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "SystemExit",
+       "evalue": "0",
+       "output_type": "pyerr",
+       "traceback": [
+        "An exception has occurred, use %tb to see the full traceback.\n",
+        "\u001b[1;31mSystemExit\u001b[0m\u001b[1;31m:\u001b[0m 0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Unable to open the file named  a.txt\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": "To exit: use 'exit', 'quit', or Ctrl-D.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 15.3 page no : 167\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 15.3 page no : 167\n'''\n\nclass Set:\n    def __init__(self,n):\n        self.elements = []\n        self.numElements = 0\n\n    def getNumElements(self):\n        return self.numElements\n\n    def getElement(self,i):\n        if (i < self.numElements):\n            return slef.elements[i];\n        else:\n            print \"Set index out of range.\" \n            import sys\n            sys.exit(0)\n\n    def find(self,s):\n        for i in range(self.numElements):\n            if (self.elements[i] == s): \n                return i;\n        return -1;\n\n    def add(self,s):\n        # if the element is already in the set, return its index\n        index = self.find(s);\n        if (index != -1):\n            return index;\n        # if the apvector is full, double its size\n        # add the new elements and return its index\n        self.elements.append(s)\n        self.numElements += 1\n        return self.numElements - 1;\n\ncities= Set (2);\ncity1 = 'ahmedabad'\ncity2 = 'rajkot'\nindex1 = cities.add(city1);\nindex2 = cities.add(city2)\nprint index1, index2",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "0 1\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch2.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch2.ipynb
new file mode 100755
index 00000000..b12d669c
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch2.ipynb
@@ -0,0 +1,112 @@
+{
+ "metadata": {
+  "name": "ch2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 2 : Variables and types"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.1 Page no : 11"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 2.1 page no :11\n'''\n\nprint \"Hello, world.\" \nprint \"How are you?\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hello, world.\nHow are you?\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.2 Page no : 11/12"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 2.2 page no : 11/12\n'''\n\nprint \"Goodbye, \";\nprint \"cruel world!\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Goodbye, \ncruel world!\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 2.3 page no : 14\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 2.3 page no : 14\n'''\nhour = 11;\nminute = 59;\ncolon = ':';\nprint \"The current time is \",\nprint hour,colon,minute",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The current time is  11 : 59\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 2.4 page no : 16\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 2.4 page no : 16\n'''\n\nhour = 11;\nminute = 59;\nprint \"Number of minutes since midnight: \",\nprint hour*60 + minute \nprint \"Fraction of the hour that has passed: \",\nprint minute/60",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Number of minutes since midnight:  719\nFraction of the hour that has passed:  0\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch3.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch3.ipynb
new file mode 100755
index 00000000..303271a9
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch3.ipynb
@@ -0,0 +1,133 @@
+{
+ "metadata": {
+  "name": "ch3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 3 : Function"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 3.1 page no :25"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 3.1 page no :25\n'''\n\ndef newLine ():\n    print ''\n\n\nprint \"First Line.\" \nnewLine ()\nprint \"Second Line.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "First Line.\n\nSecond Line.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 3.2 page no :25"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 3.2 page no :25\n'''\ndef newLine ():\n    print ''\n\nprint \"First Line.\" \nnewLine ()\nnewLine ()\nnewLine ()\nprint \"Second Line.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "First Line.\n\n\n\nSecond Line.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 3.3 page no :26"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 3.3 page no :26\n'''\n\n\ndef newLine ():\n    print ''\n\n\ndef threeLine ():\n    newLine (); \n    newLine ();\n    newLine ();\n    \nprint \"First Line.\" \nthreeLine()\nprint \"Second Line.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "First Line.\n\n\n\nSecond Line.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 3.4 page no :28"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# -*- coding: UTF-8 -*-\n'''\nexample 3.4 page no :28\n'''\n\ndef printTwice (phil):\n    print phil , phil\n\n\nprintTwice ('a');\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "a a\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 3.5 page no :28"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# -*- coding: UTF-8 -*-\n'''\nexample 3.5 page no :28\n'''\n\n\ndef printTwice (phil):\n    print phil , phil\n\nargument = 'b'  \nprintTwice (argument);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "b b\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch4.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch4.ipynb
new file mode 100755
index 00000000..08fbd2a9
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch4.ipynb
@@ -0,0 +1,112 @@
+{
+ "metadata": {
+  "name": "ch4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 4 : Conditionals and recursion"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 4.1 page no :34"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 4.1 page no :34\n'''\n\ndef printParity (x):\n    if (x%2 == 0):\n        print \"x is even\" \n    else:\n        print \"x is odd\" \n\nprintParity (17)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "x is odd\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 4.2 page no :36"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 4.2 page no :36\n'''\nimport math\n\ndef printLogarithm (x):\n    if (x <= 0.0):\n        print \"Positive numbers only, please.\" \n        return;\n    result = math.log(x);\n    print \"The log of x is \" , result;\n\nprintLogarithm(-1)\nprintLogarithm(5)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Positive numbers only, please.\nThe log of x is  1.60943791243\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 4.3 page no : 37"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 4.3 page no : 37\n'''\n\ndef countdown (n):\n    if (n == 0):\n        print  \"Blastoff!\" \n    else:\n        print n\n        countdown (n-1)\n\ncountdown(5)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "5\n4\n3\n2\n1\nBlastoff!\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example  4.4\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample  4.4\n'''\nimport math\ndef printLogarithm(x):\n    if (x <= 0.0):\n        print \"Positive numbers only, please.\" << endl;\n        return;\n    result = math.log(x)\n    print \"The log of x is \" , result;\n\ndef countdown (n) :\n    if (n == 0):\n        print \"Blastoff!\"\n    else:\n        print n\n        countdown (n-1);\n\n\ncountdown (3);\nprintLogarithm(10)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3\n2\n1\nBlastoff!\nThe log of x is  2.30258509299\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch5.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch5.ipynb
new file mode 100755
index 00000000..a55d763d
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch5.ipynb
@@ -0,0 +1,229 @@
+{
+ "metadata": {
+  "name": "ch5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 5 : Fruitful functions"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.1 page no :41\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.1 page no :41\n'''\nimport math\ndef area (radius):\n    area = math.pi * radius * radius;\n    return area;\n\nprint area(3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "28.2743338823\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.2 page no :42\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.2 page no :42\n'''\n\ndef absoluteValue (x):\n    if (x < 0):\n        return -x\n    else:\n        return x\n\nprint absoluteValue(-3)                \nprint absoluteValue(3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3\n3\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.3 page no :44\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.3 page no :44\n'''\n\ndef distance (x1, y1,x2,y2):\n    dx = x2 - x1\n    dy = y2 - y1;\n    print  \"dx is \" , dx\n    print \"dy is \" , dy \n    return 0.0\n\ndistance(10,10,20,20)    ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "dx is  10\ndy is  10\n"
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 3,
+       "text": "0.0"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.4 page no :44\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.4 page no :44\n'''\n\ndef distance (x1, y1,x2,y2):\n    dx = x2 - x1\n    dy = y2 - y1;\n    dsquared = dx*dx + dy*dy;\n    print \"dsquared is \" , dsquared\n    return 0.0;\n\ndistance(10,10,20,20)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "dsquared is  200\n"
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 4,
+       "text": "0.0"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.5 page no :45\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.5 page no :45\n'''\nimport math\ndef distance (x1, y1,x2,y2):\n    dx = x2 - x1\n    dy = y2 - y1;\n    dsquared = dx*dx + dy*dy;\n    return math.sqrt(dsquared)\n    \nprint distance(10,10,20,20)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "14.1421356237\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.6 page no :45\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.6 page no :45\n'''\n\nimport math\ndef area (radius):\n    area = math.pi * radius * radius;\n    return area;\n\ndef distance (x1, y1,x2,y2):\n    dx = x2 - x1\n    dy = y2 - y1;\n    dsquared = dx*dx + dy*dy;\n    return math.sqrt(dsquared)\n    \ndef fred(xc, yc, xp,yp):\n    radius = distance (xc, yc, xp, yp);\n    result = area (radius);\n    return result;\n\nprint fred(10,10,20,20)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "628.318530718\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.7 page no :48\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.7 page no :48\n'''\ndef isSingleDigit(x):\n    if (x >= 0 and x < 10):\n        return True\n    else:\n        return False;\n\nprint isSingleDigit(11)\nprint isSingleDigit(4)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "False\nTrue\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.8 page no :51/53\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.8 page no :51/53\n'''\n\ndef factorial (n):\n    if (n == 0):\n        return 1;\n    else:\n        recurse = factorial (n-1)\n        result = n * recurse\n        return result;\n\nprint factorial(5)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "120\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 5.9 page no :53\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 5.9 page no :53\n'''\ndef fibonacci (n):\n    if (n == 0 or n == 1):\n        return 1\n    else:\n        return fibonacci (n-1) + fibonacci (n-2)\n\nprint fibonacci(5)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "8\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch6.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch6.ipynb
new file mode 100755
index 00000000..3b13a110
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch6.ipynb
@@ -0,0 +1,265 @@
+{
+ "metadata": {
+  "name": "ch6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 6 : Iteration"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.1 page no :56\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.1 page no :56\n'''\n\ndef countdown (n):\n    while (n > 0):\n        print n\n        n = n-1\n    print \"Blastoff!\"\n    return 0\n    \ncountdown(5)  ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "5\n4\n3\n2\n1\nBlastoff!\n"
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 1,
+       "text": "0"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.2 page no :57\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.2 page no :57\n'''\n\ndef sequence (n):\n    while (n != 1):\n        print n\n        if (n%2 == 0):\n            n = n / 2\n        else:\n            n = n*3 + 1\n            \nsequence(10)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "10\n5\n16\n8\n4\n2\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.3 page no :58\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.3 page no :58\n'''\nimport math\nx = 1.0\nwhile (x < 10.0):\n    print x , \"\\t\" , math.log(x)\n    x = x + 1.0",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1.0 \t0.0\n2.0 \t0.69314718056\n3.0 \t1.09861228867\n4.0 \t1.38629436112\n5.0 \t1.60943791243\n6.0 \t1.79175946923\n7.0 \t1.94591014906\n8.0 \t2.07944154168\n9.0 \t2.19722457734\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.4 page no :59\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.4 page no :59\n'''\nimport math\nx = 1.0\nwhile (x < 10.0):\n    print x , \"\\t\" , math.log(x) / math.log(2.0) \n    x = x + 1.0\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1.0 \t0.0\n2.0 \t1.0\n3.0 \t1.58496250072\n4.0 \t2.0\n5.0 \t2.32192809489\n6.0 \t2.58496250072\n7.0 \t2.80735492206\n8.0 \t3.0\n9.0 \t3.16992500144\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.5 page no :59\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.5 page no :59\n'''\nimport math\nx = 1.0;\nwhile (x < 100.0):\n    print x , \"\\t\" , math.log(x) / math.log(2.0)\n    x = x * 2.0",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1.0 \t0.0\n2.0 \t1.0\n4.0 \t2.0\n8.0 \t3.0\n16.0 \t4.0\n32.0 \t5.0\n64.0 \t6.0\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.6 page no: 60\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.6 page no: 60\n'''\n\ni = 1;\nwhile (i <= 6):\n    print 2*i ,\n    i = i + 1;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "2 4 6 8 10 12\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.7 page no:60\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.7 page no:60\n'''\n\ndef printMultiples (n):\n    i = 1\n    while (i <= 6):\n        print n*i,\n        i = i + 1;\n        \nprintMultiples(3)        ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3 6 9 12 15 18\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.8 page no:61\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.8 page no:61\n'''\n\n\ndef printMultiples (n):\n    i = 1\n    while (i <= 6):\n        print n*i,\n        i = i + 1;\n    print ''        \n\ni = 1;\nwhile (i <= 6):\n    printMultiples (i);\n    i = i + 1;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1 2 3 4 5 6 \n2 4 6 8 10 12 \n3 6 9 12 15 18 \n4 8 12 16 20 24 \n5 10 15 20 25 30 \n6 12 18 24 30 36 \n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.9 page no: 62\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.9 page no: 62\n'''\n\ndef printMultiples (n):\n    i = 1\n    while (i <= 6):\n        print n*i,\n        i = i + 1;\n    print ''        \n\ndef printMultTable ():\n    i = 1;\n    while (i <= 6):\n        printMultiples (i)\n        i = i + 1\n\nprintMultTable()",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1 2 3 4 5 6 \n2 4 6 8 10 12 \n3 6 9 12 15 18 \n4 8 12 16 20 24 \n5 10 15 20 25 30 \n6 12 18 24 30 36 \n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.10 page no :63\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.10 page no :63\n'''\n\ndef printMultiples (n):\n    i = 1\n    while (i <= 6):\n        print n*i,\n        i = i + 1;\n    print ''        \n\ndef printMultTable (high):\n    i = 1;\n    while (i <= high):\n        printMultiples (i);\n        i = i + 1;\n        \nprintMultTable(4)        ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1 2 3 4 5 6 \n2 4 6 8 10 12 \n3 6 9 12 15 18 \n4 8 12 16 20 24 \n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 6.11 page no :64\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 6.11 page no :64\n'''\n\ndef printMultiples(n,high):\n    i = 1;\n    while (i <= high):\n        print n*i ,\n        i = i + 1;\n    print ''\n\ndef printMultTable (high):\n    i = 1;  \n    while (i <= high):\n        printMultiples (i, high)\n        i = i + 1;\n\nprintMultTable(5)        ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1 2 3 4 5 \n2 4 6 8 10 \n3 6 9 12 15 \n4 8 12 16 20 \n5 10 15 20 25 \n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch7.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch7.ipynb
new file mode 100755
index 00000000..e0a37b49
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch7.ipynb
@@ -0,0 +1,175 @@
+{
+ "metadata": {
+  "name": "ch7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 7 : Strings and things"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 7.1 page no : 69\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 7.1 page no : 69\n'''\n\nfruit = \"banana\"\nindex = 0\nwhile(index < len(fruit)):\n    letter = fruit[index]\n    print letter\n    index = index + 1",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "b\na\nn\na\nn\na\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 7.2 page no :71\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 7.2 page no :71\n'''\n\ndef find(s,c,i):\n    while (i<len(s)):\n        if (s[i] == c): return i\n        i = i + 1\n    return -1;\n    \nprint find(\"banana\",'b',0)\nprint find(\"banana\",'b',3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "0\n-1\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 7.3 page no : 71\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 7.3 page no : 71\n'''\n\nfruit = \"banana\"\nlength = len(fruit)\ncount = 0\nindex = 0\nwhile (index < length):\n    if (fruit[index] == 'a'):\n        count = count + 1\n    index = index + 1\nprint count ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 7.4 page no : 72\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 7.4 page no : 72\n'''\n\nfruit = \"banana\";\nbakedGood = \" nut bread\";\ndessert = fruit + bakedGood;\nprint dessert",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "banana nut bread\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 7.5 page no :73\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 7.5 page no :73\n'''\n\nsuffix = \"ack\"\nletter = 'J'\nwhile (letter <= 'Q'):\n    print letter + suffix \n    c = ord(letter) + 1\n    letter = chr(c)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Jack\nKack\nLack\nMack\nNack\nOack\nPack\nQack\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 7.6 page no :74\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 7.6 page no :74\n'''\nword = \"Zebra\"\n\nif (word < \"banana\"):\n    print \"Your word, \" , word , \", comes before banana.\"\nelif (word > \"banana\"):\n    print \"Your word, \" , word , \", comes after banana.\" \nelse:\n    print \"Yes, we have no bananas!\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Your word,  Zebra , comes before banana.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 7.7 page no : 75"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "a = 'a'\nprint a.upper()",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "A\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch8.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch8.ipynb
new file mode 100755
index 00000000..75cf53be
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch8.ipynb
@@ -0,0 +1,239 @@
+{
+ "metadata": {
+  "name": "ch8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 8 : Structures"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.1 page no :77\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.1 page no :77\n'''\n\nclass Point:\n    def __init__(self):\n        self.x = 0.0\n        self.y = 0.0\n\n\nb = Point()\nb.x = 3.0\nb.y = 4.0\n\nprint b.x , \", \" , b.y \ndistance = b.x * b.x + b.y * b.y\nprint \"Distance : \" , distance",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3.0 ,  4.0\nDistance :  25.0\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.2 page no :79\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.2 page no :79\n'''\n\nclass Point:\n    def __init__(self,x,y):\n        self.x = x\n        self.y = y\n\np1 = Point(3.0, 4.0)\np2 = p1\nprint p2.x , \", \" , p2.y ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3.0 ,  4.0\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.3 page no : 80\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.3 page no : 80\n'''\n\nclass Point:\n    def __init__(self,x,y):\n        self.x = x\n        self.y = y\n\ndef printPoint(p):\n    print \"(\" , p.x , \", \" , p.y , \")\" \n    \np1 = Point(3.0, 4.0)\nprintPoint(p1)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "( 3.0 ,  4.0 )\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.4 page no : 80\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.4 page no : 80\n'''\n\n\nclass Point:\n    def __init__(self,x,y):\n        self.x = x\n        self.y = y\n\ndef distance(p1,p2):\n    dx = p2.x - p1.x\n    dy = p2.y - p1.y\n    import math\n    return math.sqrt(dx*dx + dy*dy)\n    \np1 = Point(3.0, 4.0)\np2 = Point(4.0,5.0)\nprint distance(p1,p2)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "1.41421356237\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.5 page no : 81\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.5 page no : 81\n'''\nclass Point:\n    def __init__(self,x,y):\n        self.x = x\n        self.y = y\n\ndef reflect(p):\n    temp = p.x\n    p.x = p.y\n    p.y = temp\n\ndef printPoint(p):\n    print \"(\" , p.x , \", \" , p.y , \")\" \n\nblank = Point(3.0, 4.0)\nprintPoint (blank)\nreflect (blank)\nprintPoint (blank)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "( 3.0 ,  4.0 )\n( 4.0 ,  3.0 )\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.6 page no : 84\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.6 page no : 84\n'''\n\nclass Point:\n    def __init__(self,x,y):\n        self.x = x\n        self.y = y\n\nclass Rectangle:\n    def __init__(self,p,x,y):\n        self.corner = p\n        self.width = x\n        self.height = y\n\nc = Point(0.0, 0.0)\nb = Rectangle( c, 100.0, 200.0 )\nb.width += 50.0\nprint b.height ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "200.0\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.7 page no : 84\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.7 page no : 84\n'''\n\n\nclass Point:\n    def __init__(self,x,y):\n        self.x = x\n        self.y = y\n\nclass Rectangle:\n    def __init__(self,p,x,y):\n        self.corner = p\n        self.width = x\n        self.height = y\n\ndef findCenter(box):\n    x = box.corner.x + box.width/2\n    y = box.corner.y + box.height/2\n    result = Point(x, y)\n    return result\n\ndef printPoint(p):\n    print \"(\" , p.x , \", \" , p.y , \")\" \n\nbox = Rectangle( Point(0.0, 0.0), 100, 200)\ncenter = findCenter (box);\nprintPoint (center);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "( 50.0 ,  100.0 )\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.8 page no : 84\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.8 page no : 84\n'''\n\ndef swap(x,y):\n    temp = x[0]\n    x[0] = y[0]\n    y[0] = temp\n\ni = [7]\nj = [9]\nswap (i, j)\nprint i[0] , j[0]",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "9 7\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 8.9 page no : 85\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 8.9 page no : 85\n'''\n\n# prompt the user for input86\nprint \"Enter an integer: \",\n# get input\ntry:\n    x = int(raw_input())\n    print  x \n#  check and see if the input statement succeeded\nexcept:\n    print \"That was not an integer.\" \n\n# print the value we got from the user\n\nprint \"What is your name? \" , \nname = raw_input()\nprint name",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Enter an integer: "
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "a\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " That was not an integer.\nWhat is your name? "
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "abc\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " abc\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch9.ipynb b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch9.ipynb
new file mode 100755
index 00000000..25f763ca
--- /dev/null
+++ b/How_to_think_like_a_computer_scientist_by_Allen_B_Downey/ch9.ipynb
@@ -0,0 +1,175 @@
+{
+ "metadata": {
+  "name": "ch9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "CHAPTER 9. : MORE STRUCTURES"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 9.1 page no :90\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 9.1 page no :90\n'''\n\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\ndef printTime(t):\n    print t.hour , \":\" , t.minute , \":\" , t.second\n    print \"Time is \" , t.hour , \" hour \" , t.minute , \" minutes \" , t.second \n\nt = Time( 11, 59, 3.14159 )\nprintTime(t)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "11 : 59 : 3.14159\nTime is  11  hour  59  minutes  3.14159\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 9.2 page no : 91\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 9.2 page no : 91\n'''\n\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\ndef printTime(t):\n    print t.hour , \":\" , t.minute , \":\" , t.second\n    print \"Time is \" , t.hour , \" hour \" , t.minute , \" minutes \" , t.second \n\n\ndef after(time1,time2):\n    if (time1.hour > time2.hour):\n        return True\n    if (time1.hour < time2.hour): \n        return False;\n    if (time1.minute > time2.minute):\n        return True\n    if (time1.minute < time2.minute): \n        return False;\n    if (time1.second > time2.second): \n        return True\n    return False;\n\nt1 = Time( 11, 59, 3.14159 )\nt2 = Time( 12, 42, 3.234 )\nprint after(t1,t2)\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "False\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 9.3 page no : 91\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 9.3 page no : 91\nincorrect version\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\ndef printTime(t):\n    print t.hour , \":\" , t.minute , \":\" , t.second\n    print \"Time is \" , t.hour , \" hour \" , t.minute , \" minutes \" , t.second \n\n\n\ndef addTime(t1,t2):\n    hour = t1.hour + t2.hour;\n    minute = t1.minute + t2.minute;\n    second = t1.second + t2.second;\n    return Time(hour,minute,second)\n\nc = Time( 9, 14, 30.0 )\nb = Time ( 3, 35, 0.0 )\ndoneTime = addTime (c, b);\nprintTime(doneTime)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "12 : 49 : 30.0\nTime is  12  hour  49  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 9.4 page no : 92\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 9.4 page no : 92\nincorrect version\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\ndef printTime(t):\n    print t.hour , \":\" , t.minute , \":\" , t.second\n    print \"Time is \" , t.hour , \" hour \" , t.minute , \" minutes \" , t.second \n\n\n\ndef addTime(t1,t2):\n    hour = t1.hour + t2.hour;\n    minute = t1.minute + t2.minute;\n    second = t1.second + t2.second;\n    if (second >= 60.0):\n        second -= 60.0;\n        minute += 1\n    if (minute >= 60):\n        minute -= 60;\n        hour += 1;\n    return Time(hour,minute,second)\n\n\nc = Time( 9, 14, 30.0 )\nb = Time ( 3, 35, 0.0 )\ndoneTime = addTime (c, b);\nprintTime(doneTime)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "12 : 49 : 30.0\nTime is  12  hour  49  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 9.5 page no : 93\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 9.5 page no : 93\nincorrect version\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\ndef printTime(t):\n    print t.hour , \":\" , t.minute , \":\" , t.second\n    print \"Time is \" , t.hour , \" hour \" , t.minute , \" minutes \" , t.second \n\ndef increment (time,secs):\n    time.second += secs\n    if (time.second >= 60.0):\n        time.second -= 60.0;\n        time.minute += 1;\n    if (time.minute >= 60):\n        time.minute -= 60;\n        time.hour += 1;\n\n\ndef addTime(t1,t2):\n    hour = t1.hour + t2.hour;\n    minute = t1.minute + t2.minute;\n    second = t1.second + t2.second;\n    if (second >= 60.0):\n        second -= 60.0;\n        minute += 1\n    if (minute >= 60):\n        minute -= 60;\n        hour += 1;\n    return Time(hour,minute,second)\n\n\nc = Time( 9, 14, 30.0 )\nb = Time ( 3, 35, 0.0 )\ndoneTime = addTime (c, b);\nprintTime(doneTime)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "12 : 49 : 30.0\nTime is  12  hour  49  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 9.6 page no : 94\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 9.6 page no : 94\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\ndef printTime(t):\n    print t.hour , \":\" , t.minute , \":\" , t.second\n    print \"Time is \" , t.hour , \" hour \" , t.minute , \" minutes \" , t.second \n\ndef addTimeFill(t1,t2,s):\n    s.hour = t1.hour + t2.hour;\n    s.minute = t1.minute + t2.minute;\n    s.second = t1.second + t2.second;\n    if (s.second >= 60.0):\n        s.second -= 60.0;\n        s.minute += 1;\n    if (s.minute >= 60):\n        s.minute -= 60;\n        s.hour += 1;\n\nc = Time( 9, 14, 30.0 )\nb = Time ( 3, 35, 0.0 )\ndoneTime = Time(0,0,0)\naddTimeFill (c, b,doneTime);\nprintTime(doneTime)\n\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "12 : 49 : 30.0\nTime is  12  hour  49  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "example 9.7 page no : 97\n"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "'''\nexample 9.7 page no : 97\n'''\nclass Time:\n    def __init__(self,h,m,s):\n        self.hour = h\n        self.minute = m\n        self.second = s\n\ndef printTime(t):\n    print t.hour , \":\" , t.minute , \":\" , t.second\n    print \"Time is \" , t.hour , \" hour \" , t.minute , \" minutes \" , t.second \n\ndef convertToSeconds (t):\n    minutes = t.hour * 60 + t.minute;\n    seconds = minutes * 60 + t.second;\n    return seconds;\n\ndef makeTime(secs):\n    hour = int (secs / 3600.0);\n    secs -= hour * 3600.0;\n    minute = int (secs / 60.0);\n    secs -= minute * 60;\n    second = secs;\n    return Time(hour,minute,second)\n\ndef addTime (t1,t2):\n    seconds = convertToSeconds (t1) + convertToSeconds (t2);\n    return makeTime(seconds)\n\n\nc = Time( 9, 14, 30.0 )\nb = Time ( 3, 35, 0.0 )\ndoneTime = addTime (c, b);\nprintTime(doneTime)\n\n\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "12 : 49 : 30.0\nTime is  12  hour  49  minutes  30.0\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Introduction_To_Nuclear_And_Particle_Physics_by_V._K._Mittal,_R._C._Verma_And_S._C._Gupta/README.txt b/Introduction_To_Nuclear_And_Particle_Physics_by_V._K._Mittal,_R._C._Verma_And_S._C._Gupta/README.txt
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diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/README.txt b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/README.txt
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--- a/Magnifying_C/Chapter_10.ipynb
+++ /dev/null
@@ -1,336 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10: Miscellany"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.1, Page number: 356"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "# Calculation and result\n",
-      "i = 0\n",
-      "for arg in sys.argv :\n",
-      "   print ('arg [%d] - %s' % (i, arg))\n",
-      "   i = i+1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "%run 10.1.py hello world"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "arg [0] - 10.1.py\n",
-        "arg [1] - hello\n",
-        "arg [2] - world\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2, Page number: 358"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "if len(sys.argv) != 2 :\n",
-      "\tprint ('Inappropriate number of arguments')\n",
-      "\t\n",
-      "total = 0\n",
-      "def fact (n) :\n",
-      "   total = n\n",
-      "   while n > 1 :\n",
-      "      total = total * (n - 1)\n",
-      "      n = n - 1\n",
-      "   return total\n",
-      "\n",
-      "print ('Factorial of %d is %d ' % (int(sys.argv[1]), fact (int(sys.argv[1]))))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "%run 10.2.py 5"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Factorial of 5 is 120 \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.3, Page number: 361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "# Function declaration\n",
-      "def printdata () :\n",
-      "\ti = 0\n",
-      "\tfor arg in sys.argv[1:] :\n",
-      "\t\tprint ('The data in cell [%d] is %s ' % (i, arg))\n",
-      "\t\ti = i+1\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('The names before sorting are ')\n",
-      "printdata ()\n",
-      "\n",
-      "sys.argv.sort()   \n",
-      "\n",
-      "print ('The names after sorting are ')\n",
-      "printdata ()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "%run 10.3.py zimbabwe nigeria argentina china hongkong"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The names before sorting are \n",
-        "The data in cell [0] is zimbabwe \n",
-        "The data in cell [1] is nigeria \n",
-        "The data in cell [2] is argentina \n",
-        "The data in cell [3] is china \n",
-        "The data in cell [4] is hongkong \n",
-        "The names after sorting are \n",
-        "The data in cell [0] is argentina \n",
-        "The data in cell [1] is china \n",
-        "The data in cell [2] is hongkong \n",
-        "The data in cell [3] is nigeria \n",
-        "The data in cell [4] is zimbabwe \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.4, Page number: 377"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter any positive number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def convert (a) :\n",
-      "    return int(bin(a)[2:])\n",
-      "\n",
-      "bits = str(convert (a))\n",
-      "print ('Number of bits = %d' % (len(bits)))\n",
-      "print ('Binary equivalent is : %d' % (convert (a)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any positive number 16\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of bits = 5\n",
-        "Binary equivalent is : 10000\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5, Page number: 379"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number : '))\n",
-      "b = int(raw_input('Enter second number : '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "b = ~b + 1\n",
-      "c = a + b\n",
-      "\n",
-      "print ('Subtraction is = %d ' % c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number : 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subtraction is = 4 \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.6, Page number: 380"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number : '))\n",
-      "original_a = a\n",
-      "\n",
-      "b = int(raw_input('Enter second number : '))\n",
-      "original_b = b\n",
-      "\n",
-      "# Calculation and result\n",
-      "c = 1\n",
-      "while (b) :\n",
-      "\tif (b & 1) :\n",
-      "\t\tc = c * a\n",
-      "\tb = b >> 1\n",
-      "\ta = a * a\n",
-      "\n",
-      "print ('%d raise to %d = %d ' % (original_a, original_b, c))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number : 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7 raise to 3 = 343 \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_11.ipynb b/Magnifying_C/Chapter_11.ipynb
deleted file mode 100755
index 16b27feb..00000000
--- a/Magnifying_C/Chapter_11.ipynb
+++ /dev/null
@@ -1,1664 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11: Structures and Unions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.1, Page number: 388"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student :\n",
-      "\tid_no = ''\n",
-      "\tname = ''\n",
-      "\taddress = ''\n",
-      "\tage = ''\n",
-      "\t\n",
-      "newstudent = student()\n",
-      "\n",
-      "print ('Enter the student information ')\n",
-      "newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
-      "newstudent.name = raw_input('Enter name of the student : ')\n",
-      "newstudent.address = raw_input('Enter the address of the student : ')\n",
-      "newstudent.age = int(raw_input('Enter the age of the student : '))\n",
-      "\n",
-      "print ('You have entered following information...')\n",
-      "print ('Student id_number = %d ' % newstudent.id_no)\n",
-      "print ('Student name = %s ' % newstudent.name)\n",
-      "print ('Student address = %s ' % newstudent.address)\n",
-      "print ('Age of student = %d ' % newstudent.age)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student information \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter student id_no : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of the student : Amanjeet\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the address of the student : Delhi\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age of the student : 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered following information...\n",
-        "Student id_number = 1 \n",
-        "Student name = Amanjeet \n",
-        "Student address = Delhi \n",
-        "Age of student = 20 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2, Page number: 390"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student :\n",
-      "\tid_no = ''\n",
-      "\tname = ''\n",
-      "\taddress = ''\n",
-      "\tage = ''\n",
-      "\t\n",
-      "newstudent = student()\n",
-      "\n",
-      "def putdata (newstudent) : \n",
-      "   print ('You have entered following information...')\n",
-      "   print ('Student id_number = %d ' % newstudent.id_no)\n",
-      "   print ('Student name = %s ' % newstudent.name)\n",
-      "   print ('Student address = %s ' % newstudent.address)\n",
-      "   print ('Age of student = %d ' % newstudent.age)\n",
-      "    \n",
-      "print ('Enter the student information ')    \n",
-      "newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
-      "newstudent.name = raw_input('Enter name of the student : ')\n",
-      "newstudent.address = raw_input('Enter the address of the student : ')\n",
-      "newstudent.age = int(raw_input('Enter the age of the student : '))\n",
-      "\n",
-      "putdata (newstudent)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student information \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter student id_no : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of the student : Deepti\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the address of the student : Noida\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age of the student : 23\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered following information...\n",
-        "Student id_number = 2 \n",
-        "Student name = Deepti \n",
-        "Student address = Noida \n",
-        "Age of student = 23 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3, Page number: 392"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student :\n",
-      "\tid_no = ''\n",
-      "\tname = ''\n",
-      "\taddress = ''\n",
-      "\tage = ''\n",
-      "\t\n",
-      "newstudent = student()\n",
-      "\n",
-      "def putdata (idno, name, add, age) :\n",
-      "   print ('You have entered following information...')\n",
-      "   print ('Student id_number = %d ' % idno)\n",
-      "   print ('Student name = %s ' % name)\n",
-      "   print ('Student address = %s ' % add)\n",
-      "   print ('Age of student = %d ' % age)\n",
-      "    \n",
-      "print ('Enter the student information ')\n",
-      "newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
-      "newstudent.name = raw_input('Enter name of the student : ')\n",
-      "newstudent.address = raw_input('Enter the address of the student : ')\n",
-      "newstudent.age = int(raw_input('Enter the age of the student : '))\n",
-      "\n",
-      "putdata (newstudent.id_no, newstudent.name, newstudent.address, newstudent.age)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student information \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter student id_no : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of the student : Kushagra\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the address of the student : Bangalore\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age of the student : 21\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered following information...\n",
-        "Student id_number = 3 \n",
-        "Student name = Kushagra \n",
-        "Student address = Bangalore \n",
-        "Age of student = 21 \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4, Page number: 395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student :\n",
-      "\tid_no = 0\n",
-      "\tname = 'Unknown'\n",
-      "\taddress = 'Unknown'\n",
-      "\tage = 0\n",
-      "\t\n",
-      "newstudent = student()\n",
-      "\n",
-      "def readdata (newstudent) :\n",
-      "   print ('Enter the student information ')\n",
-      "   newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
-      "   newstudent.name = raw_input('Enter name of the student : ')\n",
-      "   newstudent.address = raw_input('Enter the address of the student : ')\n",
-      "   newstudent.age = int(raw_input('Enter the age of the student : '))\n",
-      "\n",
-      "readdata (newstudent)\n",
-      "print ('You have entered following information...')\n",
-      "print ('Student id_number = %d ' % newstudent.id_no)\n",
-      "print ('Student name = %s ' % newstudent.name)\n",
-      "print ('Student address = %s ' % newstudent.address)\n",
-      "print ('Age of student = %d ' % newstudent.age)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student information \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter student id_no : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of the student : Vipul\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the address of the student : Chandigarh\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age of the student : 25\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered following information...\n",
-        "Student id_number = 4 \n",
-        "Student name = Vipul \n",
-        "Student address = Chandigarh \n",
-        "Age of student = 25 \n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5, Page number: 397"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student :\n",
-      "\tid_no = 0\n",
-      "\tname = 'Unknown'\n",
-      "\taddress = 'Unknown'\n",
-      "\tage = 0\n",
-      "\t\n",
-      "def readdata () :\n",
-      "   x = student()\n",
-      "   print ('Enter the student information ')\n",
-      "   x.id_no = int(raw_input('Enter student id_no : '))\n",
-      "   x.name = raw_input('Enter name of the student : ')\n",
-      "   x.address = raw_input('Enter the address of the student : ')\n",
-      "   x.age = int(raw_input('Enter the age of the student : '))\n",
-      "   return x\n",
-      "\n",
-      "newstudent = readdata()\n",
-      "\n",
-      "print ('You have entered following information...')\n",
-      "print ('Student id_number = %d ' % newstudent.id_no)\n",
-      "print ('Student name = %s ' % newstudent.name)\n",
-      "print ('Student address = %s ' % newstudent.address)\n",
-      "print ('Age of student = %d ' % newstudent.age)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student information \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter student id_no : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of the student : Ujjwal\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the address of the student : Jaipur\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age of the student : 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered following information...\n",
-        "Student id_number = 5 \n",
-        "Student name = Ujjwal \n",
-        "Student address = Jaipur \n",
-        "Age of student = 20 \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6, Page number: 400"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student :\n",
-      "\tid_no = 0\n",
-      "\tname = 'Unknown'\n",
-      "\taddress = 'Unknown'\n",
-      "\tage = 0\n",
-      "\t\n",
-      "x = student()\n",
-      "\n",
-      "def readdata (x) :\n",
-      "   print ('Enter the student information ')\n",
-      "   x.id_no = int(raw_input('Enter student id_no : '))\n",
-      "   x.name = raw_input('Enter name of the student : ')\n",
-      "   x.address = raw_input('Enter the address of the student : ')\n",
-      "   x.age = int(raw_input('Enter the age of the student : '))\n",
-      "\n",
-      "readdata(x)    \n",
-      "print ('You have entered following information...')\n",
-      "print ('Student id_number = %d ' % x.id_no)\n",
-      "print ('Student name = %s ' % x.name)\n",
-      "print ('Student address = %s ' % x.address)\n",
-      "print ('Age of student = %d ' % x.age)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the student information \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter student id_no : 6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of the student : Karan\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the address of the student : Chennai\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age of the student : 24\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered following information...\n",
-        "Student id_number = 6 \n",
-        "Student name = Karan \n",
-        "Student address = Chennai \n",
-        "Age of student = 24 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7, Page number: 403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student (object) :\n",
-      "\tdef __init__(self, roll=None, name=None, gender=None, marks=[0]*3) :\n",
-      "\t\tself.roll = roll\n",
-      "\t\tself.name = name\n",
-      "\t\tself.gender = gender\n",
-      "\t\tself.marks = marks\n",
-      "\n",
-      "def putdata (stud, count) :\n",
-      "\tprint ('RollNo\\t   Name\\t\\t   Gender\\t   Sub1\\t      Sub2\\t Sub3 ')\n",
-      "\tprint ('---------------------------------------------------------------------')\n",
-      "\t\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tprint ('%-10d' % stud[i].roll),\n",
-      "\t\tprint ('%-15s' % stud[i].name),\n",
-      "\t\t\n",
-      "\t\tif stud[i].gender == 0 :\n",
-      "\t\t\tprint ('%-15s' % 'Female'),\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('%-15s' % 'Male'),\n",
-      "\t\t\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tprint ('%-10d' % stud[i].marks[j]),\n",
-      "\t\tprint\n",
-      "\n",
-      "def calculateper (stud, count) :\n",
-      "\taverage = 0\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tsum = 0\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tsum = sum + stud[i].marks[j]\n",
-      "\t\taverage = average + sum/3\n",
-      "\treturn average/count\n",
-      "\n",
-      "stud = []\t\n",
-      "stud.append (student (1, 'Kapildev', 1, (45, 56, 67)))\n",
-      "stud.append (student (2, 'Vidya', 0, (87, 45, 23)))\n",
-      "stud.append (student (3, 'Sharada', 0, (67, 94, 56)))\n",
-      "stud.append (student (4, 'Saismaran', 1, (56, 38, 84)))\n",
-      "stud.append (student (5, 'Reshma', 0, (67, 98, 76)))\n",
-      "\n",
-      "average_percentage = calculateper (stud, 5)\n",
-      "putdata (stud, 5)\n",
-      "print ('\\nAverage percentage of the class is : %.2f ' % average_percentage)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RollNo\t   Name\t\t   Gender\t   Sub1\t      Sub2\t Sub3 \n",
-        "---------------------------------------------------------------------\n",
-        "1          Kapildev        Male            45         56         67        \n",
-        "2          Vidya           Female          87         45         23        \n",
-        "3          Sharada         Female          67         94         56        \n",
-        "4          Saismaran       Male            56         38         84        \n",
-        "5          Reshma          Female          67         98         76        \n",
-        "\n",
-        "Average percentage of the class is : 63.00 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8, Page number: 406"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student (object) :\n",
-      "\tdef __init__(self, roll=None, name=None, gender=None, marks=[0]*3) :\n",
-      "\t\tself.roll = roll\n",
-      "\t\tself.name = name\n",
-      "\t\tself.gender = gender\n",
-      "\t\tself.marks = marks\n",
-      "\t\n",
-      "def putdata (stud, count) :\n",
-      "\tprint ('\\nRollNo\\t   Name\\t\\t   Gender\\t   Sub1\\t      Sub2\\t Sub3 ')\n",
-      "\tprint ('---------------------------------------------------------------------')\n",
-      "\t\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tprint ('%-10d' % stud[i].roll),\n",
-      "\t\tprint ('%-15s' % stud[i].name),\n",
-      "\t\t\n",
-      "\t\tif stud[i].gender == 0 :\n",
-      "\t\t\tprint ('%-15s' % 'Female'),\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('%-15s' % 'Male'),\n",
-      "\t\t\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tprint ('%-10d' % stud[i].marks[j]),\n",
-      "\t\tprint\n",
-      "\n",
-      "def calculateper (stud, count) :\n",
-      "\taverage = 0\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tsum = 0\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tsum = sum + stud[i].marks[j]\n",
-      "\t\taverage = average + sum/3\n",
-      "\treturn average/count\n",
-      "\n",
-      "count = int(raw_input('Enter the number of students in a class : '))\n",
-      "stud = [student() for i in range (0, count)]\n",
-      "\n",
-      "for i in range (0, count) :\n",
-      "\tstud[i].roll = int(raw_input('\\nEnter roll number of student %d: ' % (i+1)))\n",
-      "\tstud[i].name = raw_input('Enter name of student %d: ' % (i+1))\n",
-      "\tstud[i].gender = int(raw_input('Enter gender[0/1] of student %d: ' % (i+1)))\n",
-      "\tprint ('Enter marks of three subjects of student %d: ' % (i+1))\n",
-      "\tfor j in range (0, 3) :\n",
-      "\t\tstud[i].marks[j] = int(raw_input('Subject %d: ' % (j+1)))\n",
-      "\n",
-      "putdata (stud, count)\n",
-      "average_percentage = calculateper (stud, count)\n",
-      "print ('\\nAverage percentage of the class is : %.2f ' % average_percentage)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of students in a class : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter roll number of student 1: 24\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of student 1: Vidya\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter gender[0/1] of student 1: 0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter marks of three subjects of student 1: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1: 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2: 96\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3: 75\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "RollNo\t   Name\t\t   Gender\t   Sub1\t      Sub2\t Sub3 \n",
-        "---------------------------------------------------------------------\n",
-        "24         Vidya           Female          78         96         75        \n",
-        "\n",
-        "Average percentage of the class is : 83.00 \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9, Page number: 408"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class student (object) :\n",
-      "\tdef __init__(self, roll=None, name=None, gender=None, marks=[0]*3) :\n",
-      "\t\tself.roll = roll\n",
-      "\t\tself.name = name\n",
-      "\t\tself.gender = gender\n",
-      "\t\tself.marks = marks\n",
-      "\n",
-      "def readdata (stud, count) :\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tstud[i].roll = int(raw_input('\\nEnter roll number of student %d: ' % (i+1)))\n",
-      "\t\tstud[i].name = raw_input('Enter name of student %d: ' % (i+1))\n",
-      "\t\tstud[i].gender = int(raw_input('Enter gender[0/1] of student %d: ' % (i+1)))\n",
-      "\t\tprint ('Enter marks of three subjects of student %d: ' % (i+1))\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tstud[i].marks[j] = int(raw_input('Subject %d: ' % (j+1)))\n",
-      "\t\n",
-      "def putdata (stud, count) :\n",
-      "\tprint ('\\nRollNo\\t   Name\\t\\t   Gender\\t   Sub1\\t      Sub2\\t Sub3 ')\n",
-      "\tprint ('---------------------------------------------------------------------')\n",
-      "\t\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tprint ('%-10d' % stud[i].roll),\n",
-      "\t\tprint ('%-15s' % stud[i].name),\n",
-      "\t\t\n",
-      "\t\tif stud[i].gender == 0 :\n",
-      "\t\t\tprint ('%-15s' % 'Female'),\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('%-15s' % 'Male'),\n",
-      "\t\t\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tprint ('%-10d' % stud[i].marks[j]),\n",
-      "\t\tprint\n",
-      "\n",
-      "def calculateper (stud, count) :\n",
-      "\taverage = 0\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tsum = 0\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tsum = sum + stud[i].marks[j]\n",
-      "\t\taverage = average + sum/3\n",
-      "\treturn average/count\n",
-      "\n",
-      "count = int(raw_input('Enter the number of students in a class : '))\n",
-      "stud = [student() for i in range (0, count)]\n",
-      "readdata (stud, count)\n",
-      "\n",
-      "average_percentage = calculateper (stud, count)\n",
-      "putdata (stud, count)\n",
-      "print ('\\nAverage percentage of the class is : %.2f ' % average_percentage)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of students in a class : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter roll number of student 1: 25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of student 1: Kapildev\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter gender[0/1] of student 1: 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter marks of three subjects of student 1: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1: 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2: 70\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3: 93\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "RollNo\t   Name\t\t   Gender\t   Sub1\t      Sub2\t Sub3 \n",
-        "---------------------------------------------------------------------\n",
-        "25         Kapildev        Male            45         70         93        \n",
-        "\n",
-        "Average percentage of the class is : 69.00 \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.10, Page number: 412"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class time (object) :\n",
-      "\tdef __init__(self, hours=None, minutes=None, seconds=None) :\n",
-      "\t\tself.hours = hours\n",
-      "\t\tself.minutes = minutes\n",
-      "\t\tself.seconds = seconds\n",
-      "\n",
-      "def AcceptTime (t) :\n",
-      "\tprint ('Enter the time in hh mm ss format: ')\n",
-      "\tt.hours = int(raw_input('Hours = '))\n",
-      "\tt.minutes = int(raw_input('Minutes = '))\n",
-      "\tt.seconds = int(raw_input('Seconds = '))\n",
-      "\n",
-      "def ValidateTime (t) :\n",
-      "\tif t.hours < 0 or t.hours > 23 :\n",
-      "\t\treturn 0\n",
-      "\tif t.minutes < 0 or t.minutes > 60 :\n",
-      "\t\treturn 0\n",
-      "\tif t.seconds < 0 or t.seconds > 60 :\n",
-      "\t\treturn 0\n",
-      "\treturn 1\n",
-      "\n",
-      "def DisplayTime (t) :\n",
-      "\tprint ('Time you entered is %d:%d:%d ' % (t.hours, t.minutes, t.seconds))\n",
-      "\t\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\tt = time ()\n",
-      "\tAcceptTime (t)\n",
-      "\t\n",
-      "\tif ValidateTime (t) == 1 :\n",
-      "\t\tcondition = False\n",
-      "\telse :\n",
-      "\t\tprint ('\\nTime is incorrect. Enter again ')\n",
-      "\t\t\n",
-      "DisplayTime (t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the time in hh mm ss format: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hours = 13\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minutes = 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Seconds = 53\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time you entered is 13:45:53 \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11, Page number: 414"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class date (object) :\n",
-      "\tdef __init__(self, dd=None, mm=None, yy=None) :\n",
-      "\t\tself.dd = dd\n",
-      "\t\tself.mm = mm\n",
-      "\t\tself.yy = yy\n",
-      "\n",
-      "def AcceptDate (d) :\n",
-      "\tprint ('Enter date in dd mm yy format: ')\n",
-      "\td.dd = int(raw_input('Date = '))\n",
-      "\td.mm = int(raw_input('Month = '))\n",
-      "\td.yy = int(raw_input('Year = '))\n",
-      "\n",
-      "def isleap (val) :\n",
-      "\tif val % 4 == 0 and val % 100 or val % 400 == 0 :\n",
-      "\t\treturn 1\n",
-      "\treturn 0\n",
-      "\t\n",
-      "def ValidateDate (d) :\n",
-      "\tif d.dd < 1 or d.dd > 31 :\n",
-      "\t\treturn 0\n",
-      "\tif d.mm < 1 or d.mm > 12 :\n",
-      "\t\treturn 0\n",
-      "\tif d.yy < 1 :\n",
-      "\t\treturn 0\n",
-      "\tif d.mm == 2 :\n",
-      "\t\tif isleap (d.yy) :\n",
-      "\t\t\tif d.dd > 29 :\n",
-      "\t\t\t\treturn 0\n",
-      "\t\telif d.dd > 28 :\n",
-      "\t\t\treturn 0\n",
-      "\tif d.mm == 4 or d.mm == 6 or d.mm == 9 or d.mm == 11 and d.dd > 30 :\n",
-      "\t\treturn 0\n",
-      "\treturn 1\n",
-      "\n",
-      "def DisplayMsg (d1, d2) :\n",
-      "\tif d1.dd == d2.dd and d1.mm == d2.mm and d1.yy == d2.yy :\n",
-      "\t\tprint ('Both dates are Equal ')\n",
-      "\telse :\n",
-      "\t\tprint ('Both dates are Not Equal ')\n",
-      "\t\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\td1 = date ()\n",
-      "\tAcceptDate (d1)\n",
-      "\t\n",
-      "\tif ValidateDate (d1) == 1 :\n",
-      "\t\tcondition = False\n",
-      "\telse :\n",
-      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
-      "\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\td2 = date ()\n",
-      "\tAcceptDate (d2)\n",
-      "\t\n",
-      "\tif ValidateDate (d2) == 1 :\n",
-      "\t\tcondition = False\n",
-      "\telse :\n",
-      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
-      "\t\t\n",
-      "DisplayMsg (d1, d2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter date in dd mm yy format: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Date = 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Month = 11\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Year = 92\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter date in dd mm yy format: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Date = 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Month = 11\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Year = 92\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Both dates are Equal \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.12, Page number: 415"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class date (object) :\n",
-      "\tdef __init__(self, dd=None, mm=None, yy=None) :\n",
-      "\t\tself.dd = dd\n",
-      "\t\tself.mm = mm\n",
-      "\t\tself.yy = yy\n",
-      "\n",
-      "def AcceptDate (d) :\n",
-      "\tprint ('Enter date in dd mm yy format: ')\n",
-      "\td.dd = int(raw_input('Date = '))\n",
-      "\td.mm = int(raw_input('Month = '))\n",
-      "\td.yy = int(raw_input('Year = '))\n",
-      "\n",
-      "def isleap (val) :\n",
-      "\tif val % 4 == 0 and val % 100 or val % 400 == 0 :\n",
-      "\t\treturn 1\n",
-      "\treturn 0\n",
-      "\t\n",
-      "def ValidateDate (d) :\n",
-      "\tif d.dd < 1 or d.dd > 31 :\n",
-      "\t\treturn 0\n",
-      "\tif d.mm < 1 or d.mm > 12 :\n",
-      "\t\treturn 0\n",
-      "\tif d.yy < 1 :\n",
-      "\t\treturn 0\n",
-      "\tif d.mm == 2 :\n",
-      "\t\tif isleap (d.yy) :\n",
-      "\t\t\tif d.dd > 29 :\n",
-      "\t\t\t\treturn 0\n",
-      "\t\telif d.dd > 28 :\n",
-      "\t\t\treturn 0\n",
-      "\tif d.mm == 4 or d.mm == 6 or d.mm == 9 or d.mm == 11 and d.dd > 30 :\n",
-      "\t\treturn 0\n",
-      "\treturn 1\n",
-      "\n",
-      "def CalcDate (d1, d2) :\n",
-      "\tdate1 = date2 = 0\n",
-      "\tfor i in range (-d1.yy, d1.yy) :\n",
-      "\t\tdate1 = date1 + 366 if isleap (d1.yy) else 365\n",
-      "\t\td1.yy = d1.yy - 1\n",
-      "\t\t\n",
-      "\tfor i in range (-d1.mm, d1.mm) :\n",
-      "\t\tdate1 = date1 + (30 if (d1.mm==4 or d1.mm==6 or d1.mm==9 or d1.mm==11) else ((29 if isleap(d1.yy) else 28) if (d1.mm==2) else 31))\n",
-      "\t\n",
-      "\tdate1 = date1 + d1.dd\n",
-      "\t\n",
-      "\tfor i in range (-d2.yy, d2.yy) :\n",
-      "\t\tdate2 = date2 + 366 if isleap(d2.yy) else 365\n",
-      "\t\n",
-      "\tfor i in range (-d2.mm, d2.mm) :\n",
-      "\t\tdate2 = date2 + (30 if (d2.mm==4 or d2.mm==6 or d2.mm==9 or d2.mm==11) else ((29 if isleap(d2.yy) else 28) if d2.mm==2 else 31))\n",
-      "\t\n",
-      "\tdate2 = date2 + d2.dd\n",
-      "\t\n",
-      "\treturn (date2 - date1)/2\n",
-      "\t\t\n",
-      "def DisplayMsg (d1, d2, diff) :\n",
-      "\tprint ('The difference between %d %d %d and %d %d %d is %d days ' % (d1.dd, d1.mm, -d1.yy, d2.dd, d2.mm, d2.yy, diff))\n",
-      "\t\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\tprint ('\\nEnter first date: ')\n",
-      "\td1 = date ()\n",
-      "\tAcceptDate (d1)\n",
-      "\t\n",
-      "\tif ValidateDate (d1) == 1 :\n",
-      "\t\tcondition = False\n",
-      "\telse :\n",
-      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
-      "\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\tprint ('\\nEnter second date: ')\n",
-      "\td2 = date ()\n",
-      "\tAcceptDate (d2)\n",
-      "\t\n",
-      "\tif ValidateDate (d2) == 1 :\n",
-      "\t\tcondition = False\n",
-      "\telse :\n",
-      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
-      "\n",
-      "diff = CalcDate (d1, d2)\n",
-      "DisplayMsg (d1, d2, diff)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter first date: \n",
-        "Enter date in dd mm yy format: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Date = 01\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Month = 01\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Year = 01\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter second date: \n",
-        "Enter date in dd mm yy format: \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Date = 25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Month = 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Year = 08\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The difference between 1 1 1 and 25 12 8 is 2915 days \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.13, Page number: 419"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from ctypes import *\n",
-      "\n",
-      "# Structure declaration\n",
-      "class stream (Union) :\n",
-      "\t_fields_ = [('grade', c_char), \n",
-      "\t\t\t\t\t('class_std', c_int), \n",
-      "\t\t\t\t\t('percentage', c_float)]\n",
-      "\t\n",
-      "class student (Structure) :\n",
-      "\t_fields_ = [('roll', c_int), \n",
-      "\t\t\t\t\t('name', c_char_p), \n",
-      "\t\t\t\t\t('gender', c_int), \n",
-      "\t\t\t\t\t('marks', c_int * 3), \n",
-      "\t\t\t\t\t('st_stream', c_char), \n",
-      "\t\t\t\t\t('st', stream)]\n",
-      "\t\n",
-      "def putdata (stud, count) :\n",
-      "\tprint ('\\nStream\\t   Roll\\t  Name\\t        Gender\\t  Result\\t ')\n",
-      "\tprint ('---------------------------------------------------------------------')\n",
-      "\t\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tif stud[i].st_stream == 'A' :\n",
-      "\t\t\tprint ('Arts\\t  '),\n",
-      "\t\telif stud[i].st_stream == 'S' :\n",
-      "\t\t\tprint ('Science\\t  '),\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('Commerce  '),\n",
-      "\t\t\t\n",
-      "\t\tprint ('%-5d ' % stud[i].roll),\n",
-      "\t\tprint ('%-3s ' % stud[i].name),\n",
-      "\t\t\n",
-      "\t\tif stud[i].gender == 0 :\n",
-      "\t\t\tprint ('\\tFemale'),\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('\\tMale  '),\n",
-      "\t\t\n",
-      "\t\tif stud[i].st_stream == 'A' :\n",
-      "\t\t\tprint ('   %d Class ' % stud[i].st.class_std),\n",
-      "\t\telif stud[i].st_stream == 'S' :\n",
-      "\t\t\tprint ('   %c Grade ' % stud[i].st.grade),\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('   %.2f Percent ' % stud[i].st.percentage),\n",
-      "\t\tprint\n",
-      "\n",
-      "def calculateper (stud, count) :\n",
-      "\taverage = 0\n",
-      "\tfor i in range (0, count) :\n",
-      "\t\tsum = 0\n",
-      "\t\tfor j in range (0, 3) :\n",
-      "\t\t\tsum = sum + stud[i].marks[j]\n",
-      "\t\taverage = (float)(sum/3)\n",
-      "\t\t\n",
-      "\t\tif stud[i].st_stream == 'A' :\n",
-      "\t\t\tif average >= 60 :\n",
-      "\t\t\t\tstud[i].st.class_std = 1\n",
-      "\t\t\telse :\n",
-      "\t\t\t\tstud[i].st.class_std = 2\n",
-      "\t\t\t\t\n",
-      "\t\telif stud[i].st_stream == 'S' :\n",
-      "\t\t\tif average >= 60 :\n",
-      "\t\t\t\tstud[i].st.grade = 'A'\n",
-      "\t\t\telse :\n",
-      "\t\t\t\tstud[i].st.class_std = 'B'\n",
-      "\t\telse : \n",
-      "\t\t\tstud[i].st.percentage = average\n",
-      "\n",
-      "count = int(raw_input('Enter the number of students in a class : '))\n",
-      "stud = [student() for i in range (0, count)]\n",
-      "\n",
-      "for i in range (0, count) :\n",
-      "\tstud[i].st_stream = raw_input('\\nEnter stream of student %d (A/C/S) : ' % (i+1))\n",
-      "\tstud[i].roll = int(raw_input('Enter roll number of student %d : ' % (i+1)))\n",
-      "\tstud[i].name = raw_input('Enter name of student %d : ' % (i+1))\n",
-      "\tstud[i].gender = int(raw_input('Enter gender of student %d [0/1] : ' % (i+1)))\n",
-      "\tprint ('Enter marks of three subjects of student %d : ' % (i+1))\n",
-      "\tfor j in range (0, 3) :\n",
-      "\t\tstud[i].marks[j] = int(raw_input('Subject %d : ' % (j+1)))\n",
-      "\t\n",
-      "calculateper (stud, count)\n",
-      "putdata (stud, count)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of students in a class : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter stream of student 1 (A/C/S) : A\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter roll number of student 1 : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of student 1 : Saismaran\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter gender of student 1 [0/1] : 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter marks of three subjects of student 1 : \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1 : 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2 : 83\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3 : 59\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter stream of student 2 (A/C/S) : C\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter roll number of student 2 : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of student 2 : Vidya\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter gender of student 2 [0/1] : 0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter marks of three subjects of student 2 : \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1 : 84\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2 : 75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3 : 93\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter stream of student 3 (A/C/S) : S\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter roll number of student 3 : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name of student 3 : Pranav\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter gender of student 3 [0/1] : 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter marks of three subjects of student 3 : \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1 : 84\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2 : 95\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3 : 83\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Stream\t   Roll\t  Name\t        Gender\t  Result\t \n",
-        "---------------------------------------------------------------------\n",
-        "Arts\t   1      Saismaran  \tMale      1 Class \n",
-        "Commerce   2      Vidya  \tFemale    84.00 Percent \n",
-        "Science\t   3      Pranav  \tMale      A Grade \n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_12.ipynb b/Magnifying_C/Chapter_12.ipynb
deleted file mode 100755
index 4fec3405..00000000
--- a/Magnifying_C/Chapter_12.ipynb
+++ /dev/null
@@ -1,2137 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12: Files"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1, Page number: 428"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "fp = open('12.1.dat', 'w')\n",
-      "\n",
-      "for i in range (0, 11) :\n",
-      "\tfp.write('%d, %d \\n' % (i, i*i))\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2, Page number: 429"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "fp = open('12.2.dat', 'w')\n",
-      "\n",
-      "for i in range (0, 11) :\n",
-      "\tfp.write('%d, %d \\n' % (i, i*i))\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3, Page number: 430"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "fp = open('12.2.dat', 'a')\n",
-      "i = 11\n",
-      "fp.write('%d, %d \\n' % (i, i*i))\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4, Page number: 432"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "with open ('12.2.dat') as fp:\n",
-      "\tfor line in fp :\n",
-      "\t\tprint line\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0, 0 \n",
-        "\n",
-        "1, 1 \n",
-        "\n",
-        "2, 4 \n",
-        "\n",
-        "3, 9 \n",
-        "\n",
-        "4, 16 \n",
-        "\n",
-        "5, 25 \n",
-        "\n",
-        "6, 36 \n",
-        "\n",
-        "7, 49 \n",
-        "\n",
-        "8, 64 \n",
-        "\n",
-        "9, 81 \n",
-        "\n",
-        "10, 100 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5, Page number: 434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "print ('Data input ')\n",
-      "f1 = open('12.5.dat', 'w')\n",
-      "f1.write('You are learning Files in Python language ')\n",
-      "f1.close()\n",
-      "\n",
-      "print ('Data output ')\n",
-      "with open ('12.5.dat') as f1:\n",
-      "\tfor line in f1 :\n",
-      "\t\tprint line\n",
-      "f1.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Data input \n",
-        "Data output \n",
-        "You are learning Files in Python language \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6, Page number: 435"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "print ('Contents of the data file ')\n",
-      "f1 = open('12.6.dat', 'w')\n",
-      "for i in range (0, 10) :\n",
-      "\tnumber = int(raw_input())\n",
-      "\tf1.write('%d\\n' % number)\n",
-      "f1.close()\n",
-      "\n",
-      "f1 = open('12.6.dat', 'r')\n",
-      "f2 = open('12.6o.dat', 'w')\n",
-      "f3 = open('12.6e.dat', 'w')\n",
-      "\n",
-      "num = [0] * 10\n",
-      "with open ('12.6.dat') as f1 :\n",
-      "\tnum = f1.read().splitlines()\n",
-      "\t\n",
-      "for i in range (0, 10) :\n",
-      "\tif (int(num[i]) % 2) != 0 :\n",
-      "\t\tf2.write('%s \\n' % (num[i]))\n",
-      "\telse :\n",
-      "\t\tf3.write('%s \\n' % (num[i]))\n",
-      "\t\n",
-      "f1.close()\n",
-      "f2.close()\n",
-      "f3.close()\n",
-      "\n",
-      "print ('Contents of the odd file ')\n",
-      "with open ('12.6o.dat') as f2:\n",
-      "\tfor line in f2 :\n",
-      "\t\tprint line\n",
-      "\n",
-      "print ('Contents of the even file ')\n",
-      "with open ('12.6e.dat') as f3:\n",
-      "\tfor line in f3 :\n",
-      "\t\tprint line\n",
-      "\n",
-      "f2.close()\n",
-      "f3.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Contents of the data file \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Contents of the odd file \n",
-        "1 \n",
-        "\n",
-        "3 \n",
-        "\n",
-        "5 \n",
-        "\n",
-        "7 \n",
-        "\n",
-        "9 \n",
-        "\n",
-        "Contents of the even file \n",
-        "2 \n",
-        "\n",
-        "4 \n",
-        "\n",
-        "6 \n",
-        "\n",
-        "8 \n",
-        "\n",
-        "10 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7, Page number: 438"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "filename = raw_input('Input filename: ')\n",
-      "fp = open(filename, 'w')\n",
-      "\n",
-      "print ('Input inventory data ')\n",
-      "print ('\\nItemName number price quantity ')\n",
-      "\n",
-      "data = [[0 for y in xrange(5)] for x in xrange(3)]\n",
-      "for i in range (0, 3) :\n",
-      "\tfor j in range (0, 4) :\n",
-      "\t\tdata[i][j] = raw_input('')\n",
-      "\t\tfp.write('%s  ' % data[i][j]),\n",
-      "\tfp.write('\\n')\n",
-      "\tprint\n",
-      "fp.close()\n",
-      "\n",
-      "fp = open(filename, 'r')\n",
-      "\n",
-      "for i in range (0, 3) :\n",
-      "\tfor j in range (2, 3) :\n",
-      "\t\tdata[i][j+2] = float(data[i][j]) * int(data[i][j+1])\n",
-      "\n",
-      "print ('ItemName   number   price   quantity   value ')\n",
-      "for i in range (0, 3) :\n",
-      "\tfor j in range (0, 5) :\n",
-      "\t\tprint ('%8s' % data[i][j]),\n",
-      "\tprint"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input filename: 12.7.dat\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input inventory data \n",
-        "\n",
-        "ItemName number price quantity \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Monitor\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3005.50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mouse\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "22\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "500.50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Keyboard\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "33\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "750.00\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "ItemName   number   price   quantity   value \n",
-        " Monitor       11  3005.50       10  30055.0\n",
-        "   Mouse       22   500.50        5   2502.5\n",
-        "Keyboard       33   750.00       10   7500.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8, Page number: 440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "with open ('12.5.dat') as fp:\n",
-      "\tfor line in fp :\n",
-      "\t\tprint line\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You are learning Files in Python language \n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9, Page number: 441"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sys import argv\n",
-      "\n",
-      "# Calculation and result\n",
-      "script, filename = argv\n",
-      "file = open(filename)\n",
-      "print file.read()\n",
-      "print ('\\nFile displayed successfully.....')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "%run 12.9.py 12.1.dat"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0, 0 \n",
-        "1, 1 \n",
-        "2, 4 \n",
-        "3, 9 \n",
-        "4, 16 \n",
-        "5, 25 \n",
-        "6, 36 \n",
-        "7, 49 \n",
-        "8, 64 \n",
-        "9, 81 \n",
-        "10, 100 \n",
-        "\n",
-        "\n",
-        "File displayed successfully.....\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10, Page number: 444"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "class person (object) :\n",
-      "\tdef __init__(self, name = None, age = 0):\n",
-      "\t\tself.name = 0\n",
-      "\t\tself.age = 0\n",
-      "\t\t\n",
-      "data = [person() for i in range (5)]\n",
-      "\n",
-      "fp = open('12.10.dat','w')\n",
-      "\n",
-      "for i in range (0, 5) :\n",
-      "\tdata[i].name = raw_input('\\nEnter name : ')\n",
-      "\tdata[i].age = int(raw_input('Enter age : '))\n",
-      "\t\n",
-      "for i in range (0, 5) :\n",
-      "\tfp.write('%s %d \\n' % (data[i].name, data[i].age))\n",
-      "\n",
-      "print ('\\n...Records are written to a file...')\n",
-      "print ('...Displaying records from a file...')\n",
-      "\n",
-      "for i in range (0, 5) :\n",
-      "\tprint ('\\nName : %s ' % data[i].name)\n",
-      "\tprint ('Age : %d ' % data[i].age)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter name : Dhiraj Kumar\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 23\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter name : Mohasin Khan\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter name : Suresh Waghmare\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 55\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter name : Anand Rao\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 34\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter name : Sunil Deshpande\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 23\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "...Records are written to a file...\n",
-        "...Displaying records from a file...\n",
-        "\n",
-        "Name : Dhiraj Kumar \n",
-        "Age : 23 \n",
-        "\n",
-        "Name : Mohasin Khan \n",
-        "Age : 45 \n",
-        "\n",
-        "Name : Suresh Waghmare \n",
-        "Age : 55 \n",
-        "\n",
-        "Name : Anand Rao \n",
-        "Age : 34 \n",
-        "\n",
-        "Name : Sunil Deshpande \n",
-        "Age : 23 \n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11, Page number: 445"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "class record (object) :\n",
-      "\tdef __init__(self, name = None, age = 0) :\n",
-      "\t\tself.name = 0\n",
-      "\t\tself.age = 0\n",
-      "\t\t\n",
-      "def record_write (data, filename) :\n",
-      "\tfilep = open(filename, 'ab')\n",
-      "\tdata.name = raw_input('Enter name : ')\n",
-      "\tdata.age = int(raw_input('Enter age : '))\n",
-      "\tfilep.write('%s %d\\n' % (data.name, data.age))\n",
-      "\t\n",
-      "def record_readall (filename) :\n",
-      "\tprint\n",
-      "\twith open (filename) as filep :\n",
-      "\t\tfor line in filep :\n",
-      "\t\t\tprint line\n",
-      "\n",
-      "data = record ()\n",
-      "filename = '12.11.dat'\n",
-      "\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\tenter = raw_input('\\nAdd record (y/n) ? ')\n",
-      "\tif enter == 'y' or enter == 'Y' :\n",
-      "\t\trecord_write (data, filename)\n",
-      "\telse :\n",
-      "\t\tcondition = False\n",
-      "\t\tbreak\n",
-      "\n",
-      "record_readall (filename)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Add record (y/n) ? y\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name : Mohan Pande\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 34\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Add record (y/n) ? y\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter name : Nitin Dighe\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 32\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Add record (y/n) ? n\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Mohan Pande 34\n",
-        "\n",
-        "Nitin Dighe 32\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12, Page number: 449"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "fp = open('12.12.txt','r')\n",
-      "fp.seek(0, 0)\n",
-      "line = fp.readline()\n",
-      "print ('Read Line: %s ' % line)\n",
-      "\n",
-      "fp.seek(17, 0)\n",
-      "line = fp.readline()\n",
-      "print ('Read Line: %s ' % line)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read Line: This is 1st line\n",
-        " \n",
-        "Read Line: This is 2nd line\n",
-        " \n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13, Page number: 449"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import os\n",
-      "\n",
-      "# Calculation and result\n",
-      "statinfo = os.stat('12.13.txt')\n",
-      "print ('Filesize of \"12.13.txt\" is %d bytes ' % statinfo.st_size)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Filesize of \"12.13.txt\" is 15 bytes \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14, Page number: 455"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "nc = nlines = 0\n",
-      "\n",
-      "filename = raw_input('Enter file name: ')\n",
-      "with open(filename, 'r') as fp :\n",
-      "    for line in fp :\n",
-      "        nlines += 1\n",
-      "        nc += len(line)\n",
-      "fp.close()\n",
-      "\n",
-      "print ('There are %d characters in %s ' % (nc, filename))\n",
-      "print ('There are %d lines ' % nlines)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter file name: 12.14.txt\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "There are 55 characters in 12.14.txt \n",
-        "There are 1 lines \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15, Page number: 456"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "linecount = 0\n",
-      "filename = raw_input('Enter file name: ')\n",
-      "fp = open(filename, 'r')\n",
-      "for line in fp :\n",
-      "\tprint line\n",
-      "\tlinecount += 1\n",
-      "\tif linecount % 20 == 0 :\n",
-      "\t\traw_input(\"[Press Return to continue, Q to quit]\")\n",
-      "\t\tif raw_input() == 'Q' :\n",
-      "\t\t\texit()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter file name: 12.15.txt\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n",
-        "\n",
-        "2\n",
-        "\n",
-        "3\n",
-        "\n",
-        "4\n",
-        "\n",
-        "5\n",
-        "\n",
-        "6\n",
-        "\n",
-        "7\n",
-        "\n",
-        "8\n",
-        "\n",
-        "9\n",
-        "\n",
-        "10\n",
-        "\n",
-        "11\n",
-        "\n",
-        "12\n",
-        "\n",
-        "13\n",
-        "\n",
-        "14\n",
-        "\n",
-        "15\n",
-        "\n",
-        "16\n",
-        "\n",
-        "17\n",
-        "\n",
-        "18\n",
-        "\n",
-        "19\n",
-        "\n",
-        "20\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[Press Return to continue, Q to quit]Q\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "21\n",
-        "\n",
-        "22\n",
-        "\n",
-        "23\n",
-        "\n",
-        "24\n",
-        "\n",
-        "25\n",
-        "\n",
-        "26\n",
-        "\n",
-        "27\n",
-        "\n",
-        "28\n",
-        "\n",
-        "29\n",
-        "\n",
-        "30\n",
-        "\n",
-        "31\n",
-        "\n",
-        "32\n",
-        "\n",
-        "33\n",
-        "\n",
-        "34\n",
-        "\n",
-        "35\n",
-        "\n",
-        "36\n",
-        "\n",
-        "37\n",
-        "\n",
-        "38\n",
-        "\n",
-        "39\n",
-        "\n",
-        "40\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[Press Return to continue, Q to quit]\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "41\n",
-        "\n",
-        "42\n",
-        "\n",
-        "43\n",
-        "\n",
-        "44\n",
-        "\n",
-        "45\n",
-        "\n",
-        "46\n",
-        "\n",
-        "47\n",
-        "\n",
-        "48\n",
-        "\n",
-        "49\n",
-        "\n",
-        "50\n",
-        "\n",
-        "51\n",
-        "\n",
-        "52\n",
-        "\n",
-        "53\n",
-        "\n",
-        "54\n",
-        "\n",
-        "55\n",
-        "\n",
-        "56\n",
-        "\n",
-        "57\n",
-        "\n",
-        "58\n",
-        "\n",
-        "59\n",
-        "\n",
-        "60\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[Press Return to continue, Q to quit]\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "61\n",
-        "\n",
-        "62\n",
-        "\n",
-        "63\n",
-        "\n",
-        "64\n",
-        "\n",
-        "65\n",
-        "\n",
-        "66\n",
-        "\n",
-        "67\n",
-        "\n",
-        "68\n",
-        "\n",
-        "69\n",
-        "\n",
-        "70\n",
-        "\n",
-        "71\n",
-        "\n",
-        "72\n",
-        "\n",
-        "73\n",
-        "\n",
-        "74\n",
-        "\n",
-        "75\n",
-        "\n",
-        "76\n",
-        "\n",
-        "77\n",
-        "\n",
-        "78\n",
-        "\n",
-        "79\n",
-        "\n",
-        "80\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[Press Return to continue, Q to quit]\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "81\n",
-        "\n",
-        "82\n",
-        "\n",
-        "83\n",
-        "\n",
-        "84\n",
-        "\n",
-        "85\n",
-        "\n",
-        "86\n",
-        "\n",
-        "87\n",
-        "\n",
-        "88\n",
-        "\n",
-        "89\n",
-        "\n",
-        "90\n",
-        "\n",
-        "91\n",
-        "\n",
-        "92\n",
-        "\n",
-        "93\n",
-        "\n",
-        "94\n",
-        "\n",
-        "95\n",
-        "\n",
-        "96\n",
-        "\n",
-        "97\n",
-        "\n",
-        "98\n",
-        "\n",
-        "99\n",
-        "\n",
-        "100\n",
-        "\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[Press Return to continue, Q to quit]\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16, Page number: 457"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "fname1 = raw_input('Enter source file: ')\n",
-      "fname2 = raw_input('Enter destination file: ')\n",
-      "\n",
-      "with open(fname1) as fp1 :\n",
-      "\twith open(fname2,'w') as fp2 : \n",
-      "\t\tfor line in fp1 :\n",
-      "\t\t\tfp2.write(line)\n",
-      "\n",
-      "print ('Files successfully copied ')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter source file: 12.16in.txt\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter destination file: 12.16out.txt\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Files successfully copied \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17, Page number: 458"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "total = count = 0\n",
-      "filenameIn = raw_input('Please enter an input filename: ')\n",
-      "filenameOut = raw_input('Please enter an output filename: ')\n",
-      "\n",
-      "print ('Opening %s for reading is OK.' % filenameIn)\n",
-      "print ('Opening %s for writing is OK.' % filenameOut)\n",
-      "print ('Calculate the total...')\n",
-      "\n",
-      "for i in open(filenameIn) :\n",
-      "\tcount += 1\n",
-      "\ttotal += int(i.strip())\n",
-      "\t\n",
-      "print ('Calculate the average...')\n",
-      "fileptrOut = open(filenameOut,'w')\n",
-      "fileptrOut.write('Average of %d numbers = %f ' % (count, (total/count)))\n",
-      "\n",
-      "print ('Check also your %s file content ' % filenameOut)\n",
-      "fileptrOut.close()\n",
-      "print ('\"%s\" closed successfully ' % filenameIn)\n",
-      "print ('\"%s\" closed successfully ' % filenameOut)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter an input filename: 12.17in.dat\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter an output filename: 12.17out.dat\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Opening 12.17in.dat for reading is OK.\n",
-        "Opening 12.17out.dat for writing is OK.\n",
-        "Calculate the total...\n",
-        "Calculate the average...\n",
-        "Check also your 12.17out.dat file content \n",
-        "\"12.17in.dat\" closed successfully \n",
-        "\"12.17out.dat\" closed successfully \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18, Page number: 460"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Structure declaration\n",
-      "class STUDENT (object) :\n",
-      "\tdef __init__(self, fname=None, lname=None, id=None, quiz1=None, quiz2=None, quiz3=None, quiz4=None, exam=None) :\n",
-      "\t\tself.fname = fname\n",
-      "\t\tself.lname = lname\n",
-      "\t\tself.id = id\n",
-      "\t\tself.quiz1 = quiz1\n",
-      "\t\tself.quiz2 = quiz2\n",
-      "\t\tself.quiz3 = quiz3\n",
-      "\t\tself.quiz4 = quiz4\n",
-      "\t\tself.exam = exam\n",
-      "\t\n",
-      "def findLow (students, x, c) :\n",
-      "\tif x == 0 :\n",
-      "\t\tsmall = students[0].quiz1\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif small >= students[i].quiz1 and students[i].quiz1 > 0 :\n",
-      "\t\t\t\tsmall = students[i].quiz1\n",
-      "\t\treturn small\n",
-      "\t\t\n",
-      "\telif x == 1 :\n",
-      "\t\tsmall = students[0].quiz2\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif small >= students[i].quiz2 and students[i].quiz2 > 0 :\n",
-      "\t\t\t\tsmall = students[i].quiz2\n",
-      "\t\treturn small\n",
-      "\t\t\n",
-      "\telif x == 2 :\n",
-      "\t\tsmall = students[0].quiz3\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif small >= students[i].quiz3 and students[i].quiz3 > 0 :\n",
-      "\t\t\t\tsmall = students[i].quiz3\n",
-      "\t\treturn small\n",
-      "\n",
-      "\telif x == 3 :\n",
-      "\t\tsmall = students[0].quiz4\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif small >= students[i].quiz4 and students[i].quiz4 > 0 :\n",
-      "\t\t\t\tsmall = students[i].quiz4\n",
-      "\t\treturn small\n",
-      "\t\t\n",
-      "\telif x == 4 :\n",
-      "\t\tsmall = students[0].exam\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif small >= students[i].exam and students[i].exam > 0 :\n",
-      "\t\t\t\tsmall = students[i].exam\n",
-      "\t\treturn small\n",
-      "\treturn 0\n",
-      "\n",
-      "def findHigh (students, x, c) :\n",
-      "\tif x == 0 :\n",
-      "\t\tbig = students[0].quiz1\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif big <= students[i].quiz1 :\n",
-      "\t\t\t\tbig = students[i].quiz1\n",
-      "\t\treturn big\n",
-      "\t\t\n",
-      "\telif x == 1 :\n",
-      "\t\tbig = students[0].quiz2\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif big <= students[i].quiz2 :\n",
-      "\t\t\t\tbig = students[i].quiz2\n",
-      "\t\treturn big\n",
-      "\t\t\n",
-      "\telif x == 2 :\n",
-      "\t\tbig = students[0].quiz3\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif big <= students[i].quiz3 :\n",
-      "\t\t\t\tbig = students[i].quiz3\n",
-      "\t\treturn big\n",
-      "\n",
-      "\telif x == 3 :\n",
-      "\t\tbig = students[0].quiz4\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif big <= students[i].quiz4 :\n",
-      "\t\t\t\tbig = students[i].quiz4\n",
-      "\t\treturn big\n",
-      "\t\t\n",
-      "\telif x == 4 :\n",
-      "\t\tbig = students[0].exam\n",
-      "\t\tfor i in range (1, c) :\n",
-      "\t\t\tif big <= students[i].exam :\n",
-      "\t\t\t\tbig = students[i].exam\n",
-      "\t\treturn big\n",
-      "\treturn 0\n",
-      "\n",
-      "\n",
-      "students = [STUDENT() for i in range (5)]\n",
-      "summ = avg = low = high = [0] * 5\n",
-      "cnt = 0\n",
-      "\n",
-      "print ('--- CLASS INFO --- \\n')\n",
-      "print ('Name\\t\\t Id\\tQuiz1   Quiz2   Quiz3   Quiz4   Exam \\n')\n",
-      "\n",
-      "with open('12.18.txt') as fpr :\n",
-      "\tfor line in fpr :\n",
-      "\t\ti = 0\n",
-      "\t\tstudents[i].fname, students[i].lname, students[i].id, students[i].quiz1, students[i].quiz2, students[i].quiz3, students[i].quiz4, students[i].exam = line.split()\n",
-      "\t\tstudents[i].id = int(students[i].id)\n",
-      "\t\tstudents[i].quiz1 = int(students[i].quiz1)\n",
-      "\t\tstudents[i].quiz2 = int(students[i].quiz2)\n",
-      "\t\tstudents[i].quiz3 = int(students[i].quiz3)\n",
-      "\t\tstudents[i].quiz4 = int(students[i].quiz4)\n",
-      "\t\tstudents[i].exam = int(students[i].exam)\n",
-      "\t\t\n",
-      "\t\tprint ('%-s %-s\\t %-4d\t%-3d\t%-3d\t%-3d\t%-3d\t%-3d' % (students[i].fname, students[i].lname, students[i].id, students[i].quiz1, students[i].quiz2, students[i].quiz3, students[i].quiz4, students[i].exam)) \n",
-      "\t\t\n",
-      "\t\tsumm[0] += int(students[i].quiz1)\n",
-      "\t\tsumm[1] += int(students[i].quiz2)\n",
-      "\t\tsumm[2] += int(students[i].quiz3)\t\t\n",
-      "\t\tsumm[3] += int(students[i].quiz4)\n",
-      "\t\tsumm[4] += int(students[i].exam)\n",
-      "\t\ti = i+1\n",
-      "\t\tcnt = cnt+1\n",
-      "\n",
-      "print ('\\nSTATISTICS')\n",
-      "print ('\\t\\tQuiz1\tQuiz2\tQuiz3\tQuiz4\tExam\\n')\n",
-      "print ('Average:\\t'),\n",
-      "for i in range (0, 5) :\n",
-      "\tavg[i] = float(summ[i])/cnt\n",
-      "\tprint ('%-5.1f \\t' % avg[i]),\n",
-      "print\n",
-      "\n",
-      "print ('Lowest:\\t\\t'),\n",
-      "for i in range (0, 5) :\n",
-      "\tlow[i] = findLow (students, i, cnt)\n",
-      "\thigh[i] = findHigh (students, i, cnt)\n",
-      "\t\n",
-      "for i in range (0, 5) :\n",
-      "\tprint ('%-5d \\t' % low[i]),\n",
-      "print\n",
-      "\t\n",
-      "print ('Highest:\\t'),\n",
-      "for i in range (0, 5) :\n",
-      "\tprint ('%-5d \\t' % high[i]),\n",
-      "print\n",
-      "\n",
-      "print ('\\n --- END OF REPORT --- \\n')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "--- CLASS INFO --- \n",
-        "\n",
-        "Name\t\t Id\tQuiz1   Quiz2   Quiz3   Quiz4   Exam \n",
-        "\n",
-        "anand mane\t 11  \t10 \t3  \t4  \t5  \t80 \n",
-        "mandar patil\t 12  \t4  \t5  \t6  \t7  \t50 \n",
-        "kishoe dhane\t 13  \t4  \t5  \t6  \t8  \t55 \n",
-        "\n",
-        "STATISTICS\n",
-        "\t\tQuiz1\tQuiz2\tQuiz3\tQuiz4\tExam\n",
-        "\n",
-        "Average:\t6.0   \t4.3   \t5.3   \t6.7   \t61.7  \t\n",
-        "Lowest:\t\t4     \t5     \t6     \t8     \t55    \t\n",
-        "Highest:\t4     \t5     \t6     \t8     \t55    \t\n",
-        "\n",
-        " --- END OF REPORT --- \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.19, Page number: 466"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "# Structure declaration\n",
-      "class Employee (object) :\n",
-      "\tdef __init__(self, fname=None, lname=None, sub_taken=None, last_edu=None, join_date=None, id=None, age=None, bsal=None) :\n",
-      "\t\tself.fname = fname\n",
-      "\t\tself.lname = lname\n",
-      "\t\tself.sub_taken = sub_taken\n",
-      "\t\tself.last_edu = last_edu\n",
-      "\t\tself.join_date = join_date\n",
-      "\t\tself.id = id\n",
-      "\t\tself.age = age\n",
-      "\t\tself.bsal = bsal\n",
-      "\t\n",
-      "emp = Employee ()\n",
-      "fp = open('12.19.dat','ab+')\n",
-      "recsize = sys.getsizeof(emp)\n",
-      "\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\tprint ('\\n1.Add Records \\\n",
-      "\t\t\t  \\n2.Delete Records \\\n",
-      "\t\t\t  \\n3.Modify Records \\\n",
-      "\t\t\t  \\n4.List Records \\\n",
-      "\t\t\t  \\n5.Exit')\n",
-      "\tchoice = int(raw_input('\\nEnter your choice: '))\n",
-      "\t\n",
-      "\tif choice == 1 :\n",
-      "\t\tanother = 'Y'\n",
-      "\t\twhile another == 'Y' or another == 'y' :\n",
-      "\t\t\tfp = open('12.19.dat','ab+')\n",
-      "\t\t\temp.fname = raw_input('Enter the first name: ')\n",
-      "\t\t\temp.lname = raw_input('Enter the last name: ')\n",
-      "\t\t\temp.age = int(raw_input('Enter the age: '))\n",
-      "\t\t\temp.bsal = int(raw_input('Enter the basic salary: '))\n",
-      "\t\t\temp.join_date = int(raw_input('Enter joining date: '))\n",
-      "\t\t\temp.id = int(raw_input('Enter the employee id: '))\n",
-      "\t\t\temp.last_edu = raw_input('Enter the last education: ')\n",
-      "\t\t\temp.sub_taken = raw_input('Enter the subject taken: ')\n",
-      "\t\t\tfp.write('%d %s %s %d %d %s %s %s \\n' % (emp.id, emp.fname, emp.lname, emp.age, emp.bsal, emp.join_date, emp.last_edu, emp.sub_taken))\n",
-      "\t\t\tanother = raw_input('Add another record (Y/N) ? ')\n",
-      "\n",
-      "\telif choice == 2 :\n",
-      "\t\tanother = 'Y'\n",
-      "\t\twhile another == 'Y' or another == 'y' :\n",
-      "\t\t\tfp = open('12.19.dat','r')\n",
-      "\t\t\tlines = fp.readlines()\n",
-      "\t\t\tlines = lines[:-1]\n",
-      "\t\t\tprint ('\\nRecord deleted ')\n",
-      "\t\t\tanother = raw_input('Delete another record (Y/N) ? ')\n",
-      "\t\t\n",
-      "\telif choice == 3 :\n",
-      "\t\tfp = open('12.19.dat','w')\n",
-      "\t\tanother = 'Y'\n",
-      "\t\twhile another == 'Y' or another == 'y' :\n",
-      "\t\t\temp.fname = raw_input('Enter the first name: ')\n",
-      "\t\t\temp.lname = raw_input('Enter the last name: ')\n",
-      "\t\t\temp.age = int(raw_input('Enter the age: '))\n",
-      "\t\t\temp.bsal = int(raw_input('Enter the basic salary: '))\n",
-      "\t\t\temp.join_date = int(raw_input('Enter joining date: '))\n",
-      "\t\t\temp.id = int(raw_input('Enter the employee id: '))\n",
-      "\t\t\temp.last_edu = raw_input('Enter the last education: ')\n",
-      "\t\t\temp.sub_taken = raw_input('Enter the subject taken: ')\n",
-      "\t\t\tfp.write('%d %s %s %d %d %s %s %s \\n' % (emp.id, emp.fname, emp.lname, emp.age, emp.bsal, emp.join_date, emp.last_edu, emp.sub_taken))\n",
-      "\t\t\tanother = raw_input('Modify another record (Y/N) ? ')\n",
-      "\t \t\n",
-      "\telif choice == 4 :\n",
-      "\t\twith open ('12.19.dat','r') as fp :\n",
-      "\t\t\tfor line in fp :\n",
-      "\t\t\t\tprint line\n",
-      "\t\t\n",
-      "\telif choice == 5 :\n",
-      "\t\tcondition = False\n",
-      "\t\tfp.close()\n",
-      "\t\texit()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1.Add Records \t\t\t  \n",
-        "2.Delete Records \t\t\t  \n",
-        "3.Modify Records \t\t\t  \n",
-        "4.List Records \t\t\t  \n",
-        "5.Exit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter your choice: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the first name: Anand\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last name: Rao\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age: 29\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the basic salary: 5000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter joining date: 2004\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the employee id: 11\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last education: BCA\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the subject taken: OS\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Add another record (Y/N) ? y\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the first name: Deepti\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last name: Garg\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age: 22\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the basic salary: 25000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter joining date: 2014\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the employee id: 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last education: BTech\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the subject taken: CSE\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Add another record (Y/N) ? n\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1.Add Records \t\t\t  \n",
-        "2.Delete Records \t\t\t  \n",
-        "3.Modify Records \t\t\t  \n",
-        "4.List Records \t\t\t  \n",
-        "5.Exit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter your choice: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11 Anand Rao 29 5000 2004 BCA OS \n",
-        "\n",
-        "12 Deepti Garg 22 25000 2014 BTech CSE \n",
-        "\n",
-        "\n",
-        "1.Add Records \t\t\t  \n",
-        "2.Delete Records \t\t\t  \n",
-        "3.Modify Records \t\t\t  \n",
-        "4.List Records \t\t\t  \n",
-        "5.Exit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter your choice: 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Record deleted \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Delete another record (Y/N) ? y\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Record deleted \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Delete another record (Y/N) ? n\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1.Add Records \t\t\t  \n",
-        "2.Delete Records \t\t\t  \n",
-        "3.Modify Records \t\t\t  \n",
-        "4.List Records \t\t\t  \n",
-        "5.Exit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter your choice: 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the first name: Nalin\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last name: Chhibber\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the age: 25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the basic salary: 30000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter joining date: 2012\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the employee id: 07\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the last education: MTech\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the subject taken: DBMS\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Modify another record (Y/N) ? n\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1.Add Records \t\t\t  \n",
-        "2.Delete Records \t\t\t  \n",
-        "3.Modify Records \t\t\t  \n",
-        "4.List Records \t\t\t  \n",
-        "5.Exit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter your choice: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7 Nalin Chhibber 25 30000 2012 MTech DBMS \n",
-        "\n",
-        "\n",
-        "1.Add Records \t\t\t  \n",
-        "2.Delete Records \t\t\t  \n",
-        "3.Modify Records \t\t\t  \n",
-        "4.List Records \t\t\t  \n",
-        "5.Exit\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter your choice: 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_2.ipynb b/Magnifying_C/Chapter_2.ipynb
deleted file mode 100755
index 7144ab92..00000000
--- a/Magnifying_C/Chapter_2.ipynb
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2: Syntactic Aspects"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page number: 39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "class color() :\n",
-      "   RED = 5\n",
-      "   YELLOW = 6\n",
-      "   GREEN = 4\n",
-      "   BLUE = 5\n",
-      "\n",
-      "print ('RED = %d ' % color.RED)\n",
-      "print ('YELLOW = %d ' % color.YELLOW)\n",
-      "print ('GREEN = %d ' % color.GREEN)\n",
-      "print ('BLUE = %d ' % color.BLUE)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RED = 5 \n",
-        "YELLOW = 6 \n",
-        "GREEN = 4 \n",
-        "BLUE = 5 \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2, Page number: 41"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "i = 2004\n",
-      "c = 'Year'\n",
-      "\n",
-      "print ('%s %d' % (c, i))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Year 2004\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1, Page number: 44"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "num = 5\n",
-      "c = '$'\n",
-      "pi = 3.141\n",
-      "print ('Hello World')\n",
-      "print ('%d %c %f' % (num, c, pi))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hello World\n",
-        "5 $ 3.141000\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page number: 45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "num = int(raw_input('Enter your favorite number '))\n",
-      "print ('You have entered your favorite number as %d ' % num)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your favorite number 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered your favorite number as 3 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3, Page number: 60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "fahrenheit = float(raw_input('Please enter the Fahrenheit temperature '))\n",
-      "\n",
-      "centigrade = float(5)/9 * (fahrenheit - 32)\n",
-      "\n",
-      "print ('The temperature in Centigrade is %.2f ' % centigrade)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter the Fahrenheit temperature 100\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature in Centigrade is 37.78 \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_3.ipynb b/Magnifying_C/Chapter_3.ipynb
deleted file mode 100755
index 9e566fc1..00000000
--- a/Magnifying_C/Chapter_3.ipynb
+++ /dev/null
@@ -1,1060 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Simple Programs"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1, Page number: 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculation and result\n",
-      "pi = 3.141\n",
-      "num = int(raw_input('Enter your favorite number: '))\n",
-      "print ('You have entered your favorite number as %d' % num)\n",
-      "c = raw_input('Enter your favorite character: ')\n",
-      "print ('You have entered your favorite character as %c' % c)\n",
-      "print ('The value of pi is %.3f ' % pi)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your favorite number: 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered your favorite number as 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your favorite character: m\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You have entered your favorite character as m\n",
-        "The value of pi is 3.141 \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1, Page number: 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Calculation\n",
-      "som = a+b\n",
-      "\n",
-      "# Result\n",
-      "print ('The sum of %d + %d = %d ' % (a, b, som))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 67\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 990\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of 67 + 990 = 1057 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2, Page number: 71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "if a>b :\n",
-      "   print ('The greater number is %d ' % a)\n",
-      "else :\n",
-      "   print ('The greater number is %d ' % b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 190\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The greater number is 190 \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3, Page number: 72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "if a==b :\n",
-      "   print ('Both numbers are equal ')\n",
-      "else :\n",
-      "   if a>b :\n",
-      "      print ('The greater number is %d ' % a)\n",
-      "   else :\n",
-      "      print ('The greater number is %d ' % b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 78\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Both numbers are equal \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page number: 73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "c = int(raw_input('Enter third number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "if a>b :\n",
-      "   if a>c :\n",
-      "      print ('The greatest number is %d ' % a)\n",
-      "   else :\n",
-      "      print ('The greatest number is %d ' % c)\n",
-      "\n",
-      "else :\n",
-      "   if b>c :\n",
-      "      print ('The greatest number is %d ' % b)\n",
-      "   else : \n",
-      "      print ('The greatest number is %d ' % c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 128\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter third number 90\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The greatest number is 128 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5, Page number: 74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('Counting upto %d ' % n)\n",
-      "\n",
-      "for i in range (1, n+1) :\n",
-      "   print ('%d ' % i),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Counting upto 7 \n",
-        "1  2  3  4  5  6  7 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page number: 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('Even numbers upto %d ' % n)\n",
-      "\n",
-      "for i in range (2, n+1, 2) :\n",
-      "   print ('%d ' % i),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Even numbers upto 10 \n",
-        "2  4  6  8  10 \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page number: 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('Odd numbers upto %d ' % n)\n",
-      "\n",
-      "for i in range (1, n+1, 2) :\n",
-      "   print ('%d ' % i),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Odd numbers upto 10 \n",
-        "1  3  5  7  9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8, Page number: 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('Number line upto %d ' % n)\n",
-      "\n",
-      "for i in range (-n, n+1) :\n",
-      "   print ('%d ' % i),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number line upto 7 \n",
-        "-7  -6  -5  -4  -3  -2  -1  0  1  2  3  4  5  6  7 \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9, Page number: 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "sum = 0\n",
-      "\n",
-      "# Calculation\n",
-      "for i in range (1, n+1) :\n",
-      "   sum = sum + i\n",
-      "\n",
-      "# Result\n",
-      "print ('Sum of %d natural numbers is %d ' % (n, sum))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of 6 natural numbers is 21 \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10, Page number: 78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('The factors of %d are ' % n)\n",
-      "\n",
-      "for i in range (1, n+1) :\n",
-      "   if n%i == 0 :\n",
-      "      print ('%d ' % i),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 28\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The factors of 28 are \n",
-        "1  2  4  7  14  28 \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11, Page number: 79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('The first 10 multiples of %d are ' % n)\n",
-      "\n",
-      "for i in range (1, 11) :\n",
-      "   print ('%d X %d = %d ' % (n, i, n*i))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first 10 multiples of 6 are \n",
-        "6 X 1 = 6 \n",
-        "6 X 2 = 12 \n",
-        "6 X 3 = 18 \n",
-        "6 X 4 = 24 \n",
-        "6 X 5 = 30 \n",
-        "6 X 6 = 36 \n",
-        "6 X 7 = 42 \n",
-        "6 X 8 = 48 \n",
-        "6 X 9 = 54 \n",
-        "6 X 10 = 60 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.12, Page number: 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "i = 1\n",
-      "sum = 0\n",
-      "\n",
-      "# Calculation\n",
-      "print ('The sum of digits of %d is' % n),\n",
-      "\n",
-      "while n > 0 :\n",
-      "   sum = sum + n%10\n",
-      "   n = n/10\n",
-      "\n",
-      "# Result\n",
-      "print ('%d' % sum)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 5214\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of digits of 5214 is 12\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.13, Page number: 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "flag = 1\n",
-      "\n",
-      "# Calculation\n",
-      "for i in range (2, n) :\n",
-      "   if flag == 1 :\n",
-      "      if n%i == 0 :\n",
-      "         flag = 0\n",
-      "\n",
-      "# Result\n",
-      "if flag :\n",
-      "   print ('%d is Prime ' % n)\n",
-      "else :\n",
-      "   print ('%d is Composite ' % n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 41\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "41 is Prime \n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.14, Page number: 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "first = 0\n",
-      "second = 1\n",
-      "count = 3\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('%d Fibonacci terms are ' % n)\n",
-      "print ('%d  %d ' % (first, second)),\n",
-      "\n",
-      "while count<=n :\n",
-      "   third = first + second\n",
-      "   first = second\n",
-      "   second = third\n",
-      "   print ('%d ' % third),\n",
-      "   count = count + 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10 Fibonacci terms are \n",
-        "0  1  1  2  3  5  8  13  21  34 \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.15, Page number: 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "sum = n\n",
-      "\n",
-      "# Calculation\n",
-      "print ('Sum of digits of %d is' % n),\n",
-      "\n",
-      "while sum>=10 :\n",
-      "   sum = 0\n",
-      "   while n>0 :\n",
-      "      sum = sum + n%10\n",
-      "      n = n/10\n",
-      "   n = sum\n",
-      "\n",
-      "# Result\n",
-      "print ('%d' % sum)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 8626\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits of 8626 is 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.16, Page number: 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('Prime numbers upto %d are ' % n)\n",
-      "\n",
-      "for num in range (2, n+1) :\n",
-      "   flag = 1\n",
-      "   for i in range (2, num) :\n",
-      "      if num % i == 0 : \n",
-      "         flag = 0\n",
-      "   if flag :\n",
-      "      print ('%d ' % num),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 12\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Prime numbers upto 12 are \n",
-        "2  3  5  7  11 \n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.17, Page number: 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "for lin in range (1, n+1) :\n",
-      "   for count in range (1, lin+1) :\n",
-      "      print ('%d ' % count),\n",
-      "   print"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 \n",
-        "1  2 \n",
-        "1  2  3 \n",
-        "1  2  3  4 \n",
-        "1  2  3  4  5 \n",
-        "1  2  3  4  5  6 \n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.18, Page number: 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "for lin in range (1, n+1) :\n",
-      "   for count in range (lin, 2*lin) :\n",
-      "      print ('%d ' % count),\n",
-      "   print"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 \n",
-        "2  3 \n",
-        "3  4  5 \n",
-        "4  5  6  7 \n",
-        "5  6  7  8  9 \n",
-        "6  7  8  9  10  11 \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.19, Page number: 87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "for lin in range (1, n+1) :\n",
-      "   for blk in range (1, n-lin+1) :\n",
-      "      print ' ',\n",
-      "   for up in range (1, lin+1) :\n",
-      "      print ('%d' % up),\n",
-      "   down = lin-1\n",
-      "   while down>0 :\n",
-      "      print ('%d' % down),\n",
-      "      down = down-1\n",
-      "   print"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "          1\n",
-        "        1 2 1\n",
-        "      1 2 3 2 1\n",
-        "    1 2 3 4 3 2 1\n",
-        "  1 2 3 4 5 4 3 2 1\n",
-        "1 2 3 4 5 6 5 4 3 2 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.20, Page number: 88"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "for lin in range (1, n+1) :\n",
-      "   for blk in range (1, n-lin+1) :\n",
-      "      print ' ',\n",
-      "   for up in range (lin, 2*lin) :\n",
-      "      print ('%d' % (up%10)),\n",
-      "   down = 2*lin-2\n",
-      "   while down>=lin :\n",
-      "      print ('%d' % (down%10)),\n",
-      "      down = down-1\n",
-      "   print"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "          1\n",
-        "        2 3 2\n",
-        "      3 4 5 4 3\n",
-        "    4 5 6 7 6 5 4\n",
-        "  5 6 7 8 9 8 7 6 5\n",
-        "6 7 8 9 0 1 0 9 8 7 6\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_4.ipynb b/Magnifying_C/Chapter_4.ipynb
deleted file mode 100755
index a071712d..00000000
--- a/Magnifying_C/Chapter_4.ipynb
+++ /dev/null
@@ -1,1070 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4: Functions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1, Page number: 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration and result\n",
-      "a = 5\n",
-      "c = 'h'\n",
-      "b = 3.14\n",
-      "def myfunction (x, ch, fl) :\n",
-      "    print ('%d %c %.2f ' % (x, ch, fl))\n",
-      "    \n",
-      "myfunction (a, c, b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5 h 3.14 \n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2, Page number: 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def add (a, b) :\n",
-      "   som = a + b\n",
-      "   return som\n",
-      "\n",
-      "print ('The sum of %d and %d is %d' % (a, b, (add (a, b))))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of 7 and 8 is 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3, Page number: 99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def greater (a, b) :\n",
-      "   if a > b :\n",
-      "      return a\n",
-      "   else :\n",
-      "      return b\n",
-      "\n",
-      "print ('The greater number is %d' % (greater (a, b)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 190\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The greater number is 190\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4, Page number: 100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def greater (a, b) :\n",
-      "   if a == b :\n",
-      "      print ('Both numbers are equal')\n",
-      "   elif a > b :\n",
-      "      print ('The greater number is %d' % a)\n",
-      "   else :\n",
-      "      print ('The greater number is %d' % b)\n",
-      "   return\n",
-      "\n",
-      "greater (a, b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Both numbers are equal\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5, Page number: 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "c = int(raw_input('Enter third number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def greatest (a, b, c) :\n",
-      "   if a > b :\n",
-      "      if a > c :\n",
-      "         return a\n",
-      "      else :\n",
-      "         return c\n",
-      "   else :\n",
-      "      if b > c :\n",
-      "         return b\n",
-      "      else :\n",
-      "         return c\n",
-      "\n",
-      "print ('The greatest number is %d' % (greatest (a, b, c)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 128\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter third number: 190\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The greatest number is 190\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6, Page number: 104"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printnum (n) :\n",
-      "   print ('Counting upto %d' % n)   \n",
-      "   for i in range (1, n+1) :\n",
-      "      print i,\n",
-      "\n",
-      "printnum (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Counting upto 7\n",
-        "1 2 3 4 5 6 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.7, Page number: 105"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printeven (n) :\n",
-      "   print ('Even numbers upto %d' % n)   \n",
-      "   for i in range (2, n+1, 2) :\n",
-      "      print i,\n",
-      "\n",
-      "printeven (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Even numbers upto 7\n",
-        "2 4 6\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.8, Page number: 107"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printodd (n) :\n",
-      "   print ('Odd numbers upto %d' % n)   \n",
-      "   for i in range (1, n+1, 2) :\n",
-      "      print i,\n",
-      "\n",
-      "printodd (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Odd numbers upto 7\n",
-        "1 3 5 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9, Page number: 108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printnum (n) :\n",
-      "   print ('Number line for %d' % n)   \n",
-      "   for i in range (-n, n+1) :\n",
-      "      print i,\n",
-      "\n",
-      "printnum (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number line for 4\n",
-        "-4 -3 -2 -1 0 1 2 3 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.10, Page number: 110"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def summate (n) :\n",
-      "   total = 0\n",
-      "   for i in range (1, n+1) :\n",
-      "      total = total + i\n",
-      "   return total\n",
-      "\n",
-      "print('Sum of first %d natural numbers is %d' % (n, summate (n)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of first 6 natural numbers is 21\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.11, Page number: 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def factors (n) :\n",
-      "   print ('The factors of %d are' % n)\n",
-      "   for i in range (1, n+1) :\n",
-      "      if n%i == 0 :\n",
-      "         print i,\n",
-      "\n",
-      "factors (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 28\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The factors of 28 are\n",
-        "1 2 4 7 14 28\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.12, Page number: 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def multiple (n) :\n",
-      "   print ('The first 10 multiples of %d are' % n)\n",
-      "   for i in range (1, 11) :\n",
-      "         print('%d X %d = %d' % (n, i, n*i))\n",
-      "\n",
-      "multiple (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first 10 multiples of 6 are\n",
-        "6 X 1 = 6\n",
-        "6 X 2 = 12\n",
-        "6 X 3 = 18\n",
-        "6 X 4 = 24\n",
-        "6 X 5 = 30\n",
-        "6 X 6 = 36\n",
-        "6 X 7 = 42\n",
-        "6 X 8 = 48\n",
-        "6 X 9 = 54\n",
-        "6 X 10 = 60\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.13, Page number: 114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def sumofdigit (n) :\n",
-      "   sum = 0\n",
-      "   while n>0 :\n",
-      "      sum = sum + n%10\n",
-      "      n = n/10\n",
-      "   return sum\n",
-      "\n",
-      "print ('The sum of digits of %d is %d' % (n, sumofdigit (n)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 5214\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of digits of 5214 is 12\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.14, Page number: 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def prime (n) :\n",
-      "   count = 0\n",
-      "   for i in range (2, n) :\n",
-      "      if n%i == 0 :\n",
-      "         count = count + 1\n",
-      "         break\n",
-      "\n",
-      "   if count == 0 :\n",
-      "      print ('The number %d is prime' % n)\n",
-      "   else :\n",
-      "      print ('The number %d is composite' % n)\n",
-      "\n",
-      "prime (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 14\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The number 14 is composite\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.15, Page number: 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "print ('%d terms of Fibonacci series are ' % n)\n",
-      "\n",
-      "def printfibo (n) :\n",
-      "   first = 0\n",
-      "   second = 1\n",
-      "   count = 3\n",
-      "\n",
-      "   print ('%d  %d ' % (first, second)),\n",
-      "\n",
-      "   while count<=n :\n",
-      "      third = first + second\n",
-      "      first = second\n",
-      "      second = third\n",
-      "      print (' %d ' % third), \n",
-      "      count = count + 1\n",
-      "\n",
-      "printfibo (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10 terms of Fibonacci series are \n",
-        "0  1   1   2   3   5   8   13   21   34 \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.16, Page number: 120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def sumofdigit (n) :\n",
-      "   sum = n\n",
-      "   while sum>=10 :\n",
-      "      sum = 0\n",
-      "      while n>0 :\n",
-      "         sum = sum + n%10\n",
-      "         n = n/10\n",
-      "      n = sum\n",
-      "   return sum\n",
-      "\n",
-      "print ('Sum of digits of %d is %d' % (n, sumofdigit (n)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 8626\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits of 8626 is 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.17, Page number: 122"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "print ('Prime numbers upto %d are ' % n)\n",
-      "\n",
-      "def printprime (n) :\n",
-      "   for num in range (2, n+1) :\n",
-      "      flag = 1\n",
-      "      for i in range (2, num) :\n",
-      "         if num % i == 0 : \n",
-      "            flag = 0\n",
-      "      if flag :\n",
-      "         print ('%d ' % num),\n",
-      "\n",
-      "printprime (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 40\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Prime numbers upto 40 are \n",
-        "2  3  5  7  11  13  17  19  23  29  31  37 \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.18, Page number: 124"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def pattern (n) :\n",
-      "   for lin in range (1, n+1) :\n",
-      "      for count in range (1, lin+1) :\n",
-      "         print ('%d ' % count),\n",
-      "      print\n",
-      "\n",
-      "pattern (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 \n",
-        "1  2 \n",
-        "1  2  3 \n",
-        "1  2  3  4 \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.19, Page number: 126"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def pattern (n) :\n",
-      "   for lin in range (1, n+1) :\n",
-      "      for count in range (lin, 2*lin) :\n",
-      "         print ('%d ' % count),\n",
-      "      print\n",
-      "\n",
-      "pattern (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 \n",
-        "2  3 \n",
-        "3  4  5 \n",
-        "4  5  6  7 \n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.20, Page number: 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def pattern (n) :\n",
-      "   for lin in range (1, n+1) :\n",
-      "      for blk in range (1, n-lin+1) :\n",
-      "         print ' ',\n",
-      "      for up in range (1, lin+1) :\n",
-      "         print ('%d' % up),\n",
-      "      down = lin-1\n",
-      "      while down>0 :\n",
-      "         print ('%d' % down),\n",
-      "         down = down-1\n",
-      "      print\n",
-      "\n",
-      "pattern (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "      1\n",
-        "    1 2 1\n",
-        "  1 2 3 2 1\n",
-        "1 2 3 4 3 2 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.21, Page number: 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter a number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def pattern (n) :\n",
-      "   for lin in range (1, n+1) :\n",
-      "      for blk in range (1, n-lin+1) :\n",
-      "         print ' ',\n",
-      "      for up in range (lin, 2*lin) :\n",
-      "         print ('%d' % (up%10)),\n",
-      "      down = 2*lin-2\n",
-      "      while down>=lin :\n",
-      "         print ('%d' % (down%10)),\n",
-      "         down = down-1\n",
-      "      print\n",
-      "\n",
-      "pattern (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number: 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "      1\n",
-        "    2 3 2\n",
-        "  3 4 5 4 3\n",
-        "4 5 6 7 6 5 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_5.ipynb b/Magnifying_C/Chapter_5.ipynb
deleted file mode 100755
index b4cb2db2..00000000
--- a/Magnifying_C/Chapter_5.ipynb
+++ /dev/null
@@ -1,909 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5: Pointers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1, Page number: 135"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('The original values are a = %d b = %d' % (a, b))\n",
-      "temp = a\n",
-      "a = b\n",
-      "b = temp\n",
-      "\n",
-      "print ('The values after swapping are a = %d b = %d' % (a, b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 15\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 27\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The original values are a = 15 b = 27\n",
-        "The values after swapping are a = 27 b = 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2, Page number: 136"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "print ('The original values are a = %d b = %d' % (a, b))\n",
-      "\n",
-      "def swap (a, b) :\n",
-      "   temp = a\n",
-      "   a = b\n",
-      "   b = temp\n",
-      "   print ('The values after swapping are a = %d b = %d' % (a, b))\n",
-      "\n",
-      "swap (a, b)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 15\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 27\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The original values are a = 15 b = 27\n",
-        "The values after swapping are a = 27 b = 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3, Page number: 141"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "print ('The original values are a = %d b = %d' % (a, b))\n",
-      "\n",
-      "def swap (a, b) :\n",
-      "   temp = a\n",
-      "   a = b\n",
-      "   b = temp\n",
-      "   print ('The values after swapping are a = %d b = %d' % (a, b))\n",
-      "\n",
-      "swap (a, b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 15\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 27\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The original values are a = 15 b = 27\n",
-        "The values after swapping are a = 27 b = 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4, Page number: 145"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "print ('In main the original values are a = %d b = %d' % (a, b))\n",
-      "print ('In main the address of a is %x and address of b is %x' % (a, b))\n",
-      "\n",
-      "def swap (a, b) :\n",
-      "   print ('In swap the address in a is %x and b is %x' % (a, b))\n",
-      "   temp = a\n",
-      "   a = b\n",
-      "   b = temp\n",
-      "   print ('The values after swapping are a = %d b = %d' % (a, b))\n",
-      "\n",
-      "swap (a, b)\n",
-      "\n",
-      "print ('In main the values after swapping are a = %d b = %d' % (a, b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 15\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 27\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In main the original values are a = 15 b = 27\n",
-        "In main the address of a is f and address of b is 1b\n",
-        "In swap the address in a is f and b is 1b\n",
-        "The values after swapping are a = 27 b = 15\n",
-        "In main the values after swapping are a = 15 b = 27\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5, Page number: 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "print ('In main the value of a = %d b = %d' % (a, b))\n",
-      "\n",
-      "def doooo (a, b) :\n",
-      "   a = a+10\n",
-      "   b = b+10\n",
-      "\n",
-      "doooo (a, b)\n",
-      "\n",
-      "print ('In main the value of a = %d b = %d' % (a, b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In main the value of a = 5 b = 7\n",
-        "In main the value of a = 5 b = 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6, Page number: 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number: '))\n",
-      "b = int(raw_input('Enter second number: '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "print ('In main the value of a = %d b = %d' % (a, b))\n",
-      "print ('In main the addresses are a = %s b = %s' % (hex(id(a)), hex(id(b))))\n",
-      "\n",
-      "def doooo (a, b) :\n",
-      "   print ('In doooo the addresses are a = %s b = %s' % (hex(id(a)), hex(id(b))))\n",
-      "   a = a+10\n",
-      "   b = b+10\n",
-      "\n",
-      "doooo (a, b)\n",
-      "\n",
-      "print ('In main the value of a = %d b = %d' % (a, b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number: 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In main the value of a = 5 b = 7\n",
-        "In main the addresses are a = 0x1b49138 b = 0x1b49108\n",
-        "In doooo the addresses are a = 0x1b49138 b = 0x1b49108\n",
-        "In main the value of a = 5 b = 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7, Page number: 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def add (a, b) :\n",
-      "   x = a+b\n",
-      "   return x\n",
-      "\n",
-      "som = add (a, b)\n",
-      "\n",
-      "print ('The sum of %d + %d = %d' % (a, b, som))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of 2 + 3 = 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8, Page number: 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def greater (a, b) :\n",
-      "   if a>b :\n",
-      "      big = a\n",
-      "   else :\n",
-      "      big = b\n",
-      "   return big\n",
-      "\n",
-      "great = greater (a, b)\n",
-      "\n",
-      "print ('The greater number is %d' % great)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 23\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The greater number is 45\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9, Page number: 156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "c = int(raw_input('Enter third number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def greater (a, b, c) :\n",
-      "   if a>b :\n",
-      "      if a>c :\n",
-      "         big = a\n",
-      "      else :\n",
-      "         big = c\n",
-      "   else :\n",
-      "      if b>c :\n",
-      "         big = b\n",
-      "      else :\n",
-      "         big = c\n",
-      "   return big\n",
-      "\n",
-      "great = greater (a, b, c)\n",
-      "\n",
-      "print ('The greatest number is %d' % great)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 190\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter third number 128\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The greatest number is 190\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10, Page number: 158"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def summate (n) :\n",
-      "   i = 1\n",
-      "   total = 0\n",
-      "   while i <= n :\n",
-      "      total = total + i\n",
-      "      i = i+1\n",
-      "   return total\n",
-      "\n",
-      "sumn = summate (n)\n",
-      "\n",
-      "print ('Sum of first %d natural numbers is %d' % (n, sumn))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of first 5 natural numbers is 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.11, Page number: 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def sumofdigit (n) :\n",
-      "   sum = 0\n",
-      "   while n > 0 :\n",
-      "      sum = sum + n%10\n",
-      "      n = n/10\n",
-      "   return sum\n",
-      "\n",
-      "result = sumofdigit (n)\n",
-      "\n",
-      "print ('Sum of digits of %d is %d' % (n, result))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 7162\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits of 7162 is 16\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12, Page number: 161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "sum = n\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def sumofdigit (n, sum) :\n",
-      "   while sum>=10 :\n",
-      "      sum = 0\n",
-      "      while n>0 :\n",
-      "         sum = sum + n%10\n",
-      "         n = n/10\n",
-      "      n = sum\n",
-      "   return sum\n",
-      "\n",
-      "result = sumofdigit (n, sum)\n",
-      "\n",
-      "print ('Sum of digits of %d is %d' % (n, result))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 7162\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits of 7162 is 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.13, Page number: 163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def checkprime (n) :\n",
-      "   flag = 1\n",
-      "   i = 2\n",
-      "   while i<n and flag :\n",
-      "      if n%i == 0 :\n",
-      "         flag = 0\n",
-      "      i = i+1\n",
-      "   return flag\n",
-      "\n",
-      "result = checkprime (n)\n",
-      "\n",
-      "if result :\n",
-      "   print ('%d is Prime ' % n)\n",
-      "else :\n",
-      "   print ('%d is Composite ' % n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 41\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "41 is Prime \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.14, Page number: 165"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def checkprime (n) :\n",
-      "   for num in range (2, n+1) :\n",
-      "      flag = 1\n",
-      "      for i in range (2, num) :\n",
-      "         if num % i == 0 : \n",
-      "            flag = 0\n",
-      "      if flag :\n",
-      "         print ('%d ' % num),\n",
-      "\n",
-      "checkprime (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 11\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2  3  5  7  11 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.15, Page number: 167"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def makethird (n) :\n",
-      "   first = 0\n",
-      "   second = 1\n",
-      "   count = 3\n",
-      "   print ('%d terms of Fibonacci series are ' % n)\n",
-      "   print ('%d  %d ' % (first, second)),\n",
-      "   while count<=n :\n",
-      "      third = first + second\n",
-      "      first = second\n",
-      "      second = third\n",
-      "      print (' %d ' % third),\n",
-      "      count = count + 1\n",
-      "\n",
-      "makethird (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10 terms of Fibonacci series are \n",
-        "0  1   1   2   3   5   8   13   21   34 \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.16, Page number: 169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def makethird (n) :\n",
-      "   first = 0\n",
-      "   second = 1\n",
-      "   third = 0\n",
-      "   print ('Fibonacci series up to %d is ' % n)\n",
-      "   print first,\n",
-      "   while third<=n :\n",
-      "      print (' %d ' % second),\n",
-      "      third = first + second\n",
-      "      first = second\n",
-      "      second = third\n",
-      "\n",
-      "makethird (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 100\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fibonacci series up to 100 is \n",
-        "0  1   1   2   3   5   8   13   21   34   55   89 \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_6.ipynb b/Magnifying_C/Chapter_6.ipynb
deleted file mode 100755
index 690f5ef0..00000000
--- a/Magnifying_C/Chapter_6.ipynb
+++ /dev/null
@@ -1,951 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6: Storage classes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1, Page number: 180"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def fact (n) :\n",
-      "   prod = 1\n",
-      "   while n :\n",
-      "      prod = prod * n\n",
-      "      n = n - 1\n",
-      "   return prod\n",
-      "\n",
-      "print ('Factorial of %d is %d' % (n, fact(n)))\n",
-      "\n",
-      "m = int(raw_input('Enter another number '))\n",
-      "\n",
-      "print ('Factorial of %d is %d' % (m, fact(m)))\n",
-      "\n",
-      "print ('Factorial of 6 is %d' % fact(6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Factorial of 5 is 120\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter another number 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Factorial of 3 is 6\n",
-        "Factorial of 6 is 720\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number: 183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def add (a, b) :\n",
-      "   global som\n",
-      "   som = a + b\n",
-      "\n",
-      "add (a, b)\n",
-      "\n",
-      "print ('The sum of %d + %d = %d' % (a, b, som))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of 2 + 3 = 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number: 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "global y\n",
-      "y = 90\n",
-      "m = 10\n",
-      "z = 2.18\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def fun1 (m, y, z) :\n",
-      "   z = 3.141\n",
-      "   print ('%f' % z)\n",
-      "\n",
-      "fun1 (m, y, z)\n",
-      "\n",
-      "print ('%d %d %f' % (m, y, z))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3.141000\n",
-        "10 90 2.180000\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number: 190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "global x, y, z, ch, tmp\n",
-      "z = 2.18\n",
-      "y = 90\n",
-      "ch = '#'\n",
-      "m = 10\n",
-      "x = '*'\n",
-      "tmp = '^'\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def fun1 (m, y, z) :\n",
-      "   z = 3.141\n",
-      "   num = 100\n",
-      "   tmp = '$'\n",
-      "   x = 50\n",
-      "   ch = '@'\n",
-      "   print ('In FUNCTION FORMAL PARAMETERS')\n",
-      "   print ('In the function value of m = %d address of m = %x' % (m, id(m)))\n",
-      "   print ('In the function value of y = %d address of y = %x' % (y, id(y)))\n",
-      "   print ('In the function value of z = %f address of z = %x' % (z, id(z)))\n",
-      "\n",
-      "   print ('In FUNCTION LOCAL VARIABLES')\n",
-      "   print ('In the function value of z = %f address of z = %x' % (z, id(z)))\n",
-      "   print ('In the function value of num = %d address of num = %x' % (num, id(num)))\n",
-      "   print ('In the function value of tmp = %s address of tmp = %x' % (tmp, id(tmp)))\n",
-      "\n",
-      "   print ('In FUNCTION GLOBAL VARIABLES')\n",
-      "   print ('In the function value of x = %s address of x = %x' % (x, id(x)))\n",
-      "   print ('In the function value of y = %d address of y = %x' % (y, id(y)))\n",
-      "   print ('In the function value of ch = %s address of ch = %x' % (ch, id(ch)))\n",
-      "\n",
-      "\n",
-      "print ('In MAIN LOCAL VARIABLES')\n",
-      "print ('In Main the value of m = %d address of m = %x' % (m, id(m)))\n",
-      "print ('In Main the value of x = %s address of x = %s' % (x, id(x)))\n",
-      "print ('In MAIN GLOBAL VARIABLES')\n",
-      "print ('In Main the value of y = %d address of y = %x' % (y, id(y)))\n",
-      "print ('In Main the value of z = %f address of z = %x' % (z, id(z)))\n",
-      "print ('In Main the value of ch = %s address of ch = %s' % (ch, id(ch)))\n",
-      "print ('In Main the value of tmp = %s address of tmp = %x' % (tmp, id(tmp)))\n",
-      "\n",
-      "fun1 (m, y, z)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In MAIN LOCAL VARIABLES\n",
-        "In Main the value of m = 10 address of m = 117d0c0\n",
-        "In Main the value of x = * address of x = 140148786619376\n",
-        "In MAIN GLOBAL VARIABLES\n",
-        "In Main the value of y = 90 address of y = 117d8d0\n",
-        "In Main the value of z = 2.180000 address of z = 16ca340\n",
-        "In Main the value of ch = # address of ch = 140148787513184\n",
-        "In Main the value of tmp = ^ address of tmp = 7f76eea46eb8\n",
-        "In FUNCTION FORMAL PARAMETERS\n",
-        "In the function value of m = 10 address of m = 117d0c0\n",
-        "In the function value of y = 90 address of y = 117d8d0\n",
-        "In the function value of z = 3.141000 address of z = 16ca2f8\n",
-        "In FUNCTION LOCAL VARIABLES\n",
-        "In the function value of z = 3.141000 address of z = 16ca2f8\n",
-        "In the function value of num = 100 address of num = 117d7e0\n",
-        "In the function value of tmp = $ address of tmp = 7f76eeb088a0\n",
-        "In FUNCTION GLOBAL VARIABLES\n",
-        "In the function value of x = 50 address of x = 117d4c8\n",
-        "In the function value of y = 90 address of y = 117d8d0\n",
-        "In the function value of ch = @ address of ch = 7f76ed6203f0\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number: 193"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def add (a, b) :\n",
-      "   global som\n",
-      "   som = a + b\n",
-      "\n",
-      "add (a, b)\n",
-      "\n",
-      "print ('The sum of %d + %d = %d' % (a, b, som))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 30\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of 20 + 30 = 50\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number: 194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def greater (a, b) :\n",
-      "   global big\n",
-      "   if a == b :\n",
-      "      big = 'equal'\n",
-      "   else :\n",
-      "      if a > b :\n",
-      "         big = 'first'\n",
-      "      else :\n",
-      "         big = 'second'\n",
-      "   return big\n",
-      "\n",
-      "greater (a, b)\n",
-      "\n",
-      "if big == 'equal' :\n",
-      "   print ('EQUAL')\n",
-      "else :\n",
-      "   if big == 'first' :\n",
-      "      print ('First number is greater %d' % a)\n",
-      "   else :\n",
-      "      print ('Second number is greater %d' % b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "EQUAL\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page number: 196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = int(raw_input('Enter first number '))\n",
-      "b = int(raw_input('Enter second number '))\n",
-      "c = int(raw_input('Enter third number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def greater (a, b, c) :\n",
-      "   if a>b :\n",
-      "      if a>c :\n",
-      "         print ('The greatest number is %d ' % a)\n",
-      "      else :\n",
-      "         print ('The greatest number is %d ' % c)\n",
-      "\n",
-      "   else :\n",
-      "      if b>c :\n",
-      "         print ('The greatest number is %d ' % b)\n",
-      "      else : \n",
-      "         print ('The greatest number is %d ' % c)\n",
-      "\n",
-      "greater (a, b, c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter first number 190\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter second number 128\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter third number 45\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The greatest number is 190 \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page number: 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def summate (n) :\n",
-      "   i = 1\n",
-      "   global total\n",
-      "   total = 0\n",
-      "   while i <= n :\n",
-      "      total = total + i\n",
-      "      i = i+1\n",
-      "   \n",
-      "summate (n)\n",
-      "\n",
-      "print ('Sum of first %d natural numbers is %d' % (n, total))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of first 6 natural numbers is 21\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page number: 199"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def sumofdigit (n) :\n",
-      "   global sum\n",
-      "   sum = 0\n",
-      "   while n > 0 :\n",
-      "      sum = sum + n%10\n",
-      "      n = n/10\n",
-      "   return sum\n",
-      "\n",
-      "sumofdigit (n)\n",
-      "\n",
-      "print ('Sum of digits of %d is %d' % (n, sum))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 7612\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits of 7612 is 16\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page number: 201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def checkprime (n) :\n",
-      "   flag = 1\n",
-      "   i = 2\n",
-      "   while i<n and flag :\n",
-      "      if n%i == 0 :\n",
-      "         flag = 0\n",
-      "      i = i+1\n",
-      "   return flag\n",
-      "\n",
-      "result = checkprime (n)\n",
-      "\n",
-      "if result :\n",
-      "   print ('The number %d is PRIME ' % n)\n",
-      "else :\n",
-      "   print ('The number %d is COMPOSITE ' % n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 14\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The number 14 is COMPOSITE \n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page number: 203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def makethird (n) :\n",
-      "   first = 0\n",
-      "   second = 1\n",
-      "   count = 3\n",
-      "   print ('%d terms of the Fibonacci series are ' % n)\n",
-      "   print ('%d  %d ' % (first, second)),\n",
-      "   while count<=n :\n",
-      "      third = first + second\n",
-      "      first = second\n",
-      "      second = third\n",
-      "      print (' %d ' % third),\n",
-      "      count = count + 1\n",
-      "\n",
-      "makethird (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10 terms of the Fibonacci series are \n",
-        "0  1   1   2   3   5   8   13   21   34 \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page number: 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "sum = n\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def sumofdigit (n, sum) :\n",
-      "   while sum>=10 :\n",
-      "      sum = 0\n",
-      "      while n>0 :\n",
-      "         sum = sum + n%10\n",
-      "         n = n/10\n",
-      "      n = sum\n",
-      "   return sum\n",
-      "\n",
-      "result = sumofdigit (n, sum)\n",
-      "\n",
-      "print ('Sum of digits of %d is %d' % (n, result))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 7612\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits of 7612 is 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page number: 207"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def prime (n) :\n",
-      "   for num in range (2, n) :\n",
-      "      flag = 1\n",
-      "      for i in range (2, num) :\n",
-      "         if num % i == 0 : \n",
-      "            flag = 0\n",
-      "      if flag :\n",
-      "         print ('%d ' % num),\n",
-      "\n",
-      "prime (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2  3  5  7  11  13  17  19 \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page number: 210"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "m = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def fun1 (m) :\n",
-      "   if not hasattr(fun1, \"temp\") : \n",
-      "      fun1.temp = 0\n",
-      "   print ('In function value of temp = %d' % fun1.temp)\n",
-      "   if fun1.temp < m :\n",
-      "      fun1.temp = m\n",
-      "   else :\n",
-      "      fun1.temp = fun1.temp + 1\n",
-      "\n",
-      "for i in range (1, 6) :\n",
-      "   fun1 (m)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In function value of temp = 0\n",
-        "In function value of temp = 10\n",
-        "In function value of temp = 11\n",
-        "In function value of temp = 12\n",
-        "In function value of temp = 13\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page number: 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "m = 10\n",
-      "i = 1\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def fun1 (m) :\n",
-      "   x = m\n",
-      "   if not hasattr(fun1, \"temp\") : \n",
-      "      fun1.temp = 0\n",
-      "\n",
-      "   if fun1.temp == 0 :\n",
-      "      print ('\\nIn FUNCTION Global variable')\n",
-      "      print ('Value of m = %d address of m = %x' % (m, id(m)))\n",
-      "      print ('\\nIn FUNCTION Formal parameter variable')\n",
-      "      print ('Value of x = %d address of x = %x' % (x, id(x)))\n",
-      "      print ('\\nIn FUNCTION Static variable')\n",
-      "      print ('Value of temp = %d address of temp = %x' % (fun1.temp, id(fun1.temp)))\n",
-      "\n",
-      "   if fun1.temp < m :\n",
-      "      fun1.temp = m\n",
-      "\n",
-      "   else :\n",
-      "      print ('Value of temp = %d' % fun1.temp)\n",
-      "      fun1.temp = fun1.temp + 1\n",
-      "\n",
-      "print ('\\nIn MAIN Global variable')\n",
-      "print ('Value of m = %d address of m = %x' % (m, id(m)))\n",
-      "print ('\\nIn MAIN Local variable')\n",
-      "print ('Value of i = %d address of i = %x' % (i, id(i)))\n",
-      "      \n",
-      "for i in range (1, 6) :\n",
-      "   fun1 (m)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In MAIN Global variable\n",
-        "Value of m = 10 address of m = 117d0c0\n",
-        "\n",
-        "In MAIN Local variable\n",
-        "Value of i = 1 address of i = 117d198\n",
-        "\n",
-        "In FUNCTION Global variable\n",
-        "Value of m = 10 address of m = 117d0c0\n",
-        "\n",
-        "In FUNCTION Formal parameter variable\n",
-        "Value of x = 10 address of x = 117d0c0\n",
-        "\n",
-        "In FUNCTION Static variable\n",
-        "Value of temp = 0 address of temp = 117d1b0\n",
-        "Value of temp = 10\n",
-        "Value of temp = 11\n",
-        "Value of temp = 12\n",
-        "Value of temp = 13\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page number: 215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter number of Fibonacci terms required '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def nextfibo () :\n",
-      "   if not hasattr(nextfibo, \"first\") : \n",
-      "      nextfibo.first = 0\n",
-      "   if not hasattr(nextfibo, \"second\") : \n",
-      "      nextfibo.second = 1\n",
-      "   if not hasattr(nextfibo, \"third\") : \n",
-      "      nextfibo.third = 0\n",
-      "   nextfibo.third = nextfibo.first + nextfibo.second\n",
-      "   nextfibo.first = nextfibo.second\n",
-      "   nextfibo.second = nextfibo.third\n",
-      "   return nextfibo.first\n",
-      " \n",
-      "print ('%d terms of Fibonacci series are' % n) \n",
-      "print 0,\n",
-      "for count in range (2, n+1) :\n",
-      "   result = nextfibo ()\n",
-      "   print ('%d' % result),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter number of Fibonacci terms required 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10 terms of Fibonacci series are\n",
-        "0 1 1 2 3 5 8 13 21 34\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_7.ipynb b/Magnifying_C/Chapter_7.ipynb
deleted file mode 100755
index 73eec220..00000000
--- a/Magnifying_C/Chapter_7.ipynb
+++ /dev/null
@@ -1,862 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7: Recursion"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1, Page number: 219"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration, calculation and result\n",
-      "str = raw_input('Enter a string: ')\n",
-      "\n",
-      "print str"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a string: HELLO\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "HELLO\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2, Page number: 221"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration, calculation and result\n",
-      "str = raw_input('Enter a string: ')\n",
-      "\n",
-      "print str[::-1]"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a string: HELLO\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "OLLEH\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3, Page number: 224"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def fact (n) :\n",
-      "   if n>0 :\n",
-      "      return n * fact (n-1)\n",
-      "   else :\n",
-      "      return 1\n",
-      "\n",
-      "print ('Factorial of %d is %d' % (n, fact (n)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Factorial of 5 is 120\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4, Page number: 227"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printnum (n) :\n",
-      "   if n>0 :\n",
-      "      printnum (n-1)\n",
-      "      print n,\n",
-      "\n",
-      "print ('Counting up to %d' % n)\n",
-      "printnum (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Counting up to 10\n",
-        "1 2 3 4 5 6 7 8 9 10\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5, Page number: 229"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printeven (n) :\n",
-      "\tif not hasattr(printeven, \"x\") : \n",
-      "\t\tprinteven.x = 2\n",
-      "      \n",
-      "\tif printeven.x <= n :\n",
-      "\t\tprint printeven.x,\n",
-      "\t\tprinteven.x = printeven.x + 2\n",
-      "\t\tprinteven (n)\n",
-      "\n",
-      "print ('Even numbers up to %d are' % n)\n",
-      "printeven (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Even numbers up to 10 are\n",
-        "2 4 6 8 10\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.6, Page number: 231"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printodd (x, n) :\n",
-      "\tif x <= n : \n",
-      "\t\tprint x,\n",
-      "\t\tprintodd (x+2, n)\n",
-      "\n",
-      "print ('Odd numbers up to %d are' % n)\n",
-      "printodd (1, n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Odd numbers up to 10 are\n",
-        "1 3 5 7 9\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7, Page number: 232"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printnum (fro, to) :\n",
-      "\tif fro <= to : \n",
-      "\t\tprint fro,\n",
-      "\t\tprintnum (fro+1, to)\n",
-      "\n",
-      "print ('Number line for %d is' % n)\n",
-      "printnum (-n, n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number line for 5 is\n",
-        "-5 -4 -3 -2 -1 0 1 2 3 4 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.8, Page number: 234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def summate (n) :\n",
-      "\tif n > 0 : \n",
-      "\t\treturn n + summate (n-1)\n",
-      "\telse :\n",
-      "\t\treturn 0\n",
-      "\n",
-      "print ('Sum of first %d natural numbers is %d' % (n, summate (n)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of first 10 natural numbers is 55\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.9, Page number: 237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def factors (n) :\n",
-      "\tif not hasattr(factors, \"i\") : \n",
-      "\t\tfactors.i = 1\n",
-      "\t\t\n",
-      "\tif factors.i <= n : \n",
-      "\t\tif n % factors.i == 0 :\n",
-      "\t\t\tprint factors.i,\n",
-      "\t\tfactors.i = factors.i + 1\n",
-      "\t\tfactors (n)\n",
-      "\n",
-      "print ('The factors of %d are' % n)\n",
-      "factors (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 28\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The factors of 28 are\n",
-        "1 2 4 7 14 28\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.10, Page number: 239"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def multiple (n) :\n",
-      "\tif not hasattr(multiple, \"i\") : \n",
-      "\t\tmultiple.i = 1\n",
-      "\t\t\n",
-      "\tif multiple.i <= 10 : \n",
-      "\t\tprint ('%d X %d = %d' % (n, multiple.i, n*multiple.i))\n",
-      "\t\tmultiple.i = multiple.i + 1\n",
-      "\t\tmultiple (n)\n",
-      "\n",
-      "print ('The first 10 multiples of %d are' % n)\n",
-      "multiple (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first 10 multiples of 6 are\n",
-        "6 X 1 = 6\n",
-        "6 X 2 = 12\n",
-        "6 X 3 = 18\n",
-        "6 X 4 = 24\n",
-        "6 X 5 = 30\n",
-        "6 X 6 = 36\n",
-        "6 X 7 = 42\n",
-        "6 X 8 = 48\n",
-        "6 X 9 = 54\n",
-        "6 X 10 = 60\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.11, Page number: 241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def checkprime (n) :\n",
-      "   flag = 1\n",
-      "   i = 2\n",
-      "   while i<n and flag :\n",
-      "      if n%i == 0 :\n",
-      "         flag = 0\n",
-      "      i = i+1\n",
-      "   return flag\n",
-      "\n",
-      "result = checkprime (n)\n",
-      "\n",
-      "if result :\n",
-      "   print ('The number %d is PRIME ' % n)\n",
-      "else :\n",
-      "   print ('The number %d is COMPOSITE ' % n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 19\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The number 19 is PRIME \n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.12, Page number: 243"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def Fibonacci (n) :\n",
-      "\tif n < 2 :\n",
-      "\t\treturn n\n",
-      "\telse :\n",
-      "\t\treturn Fibonacci (n-1) + Fibonacci (n-2)\n",
-      "\t\n",
-      "\n",
-      "print ('The %d terms of Fibonacci series are' % n)\n",
-      "for i in range (0, n) :\n",
-      "\tprint ('%d' % Fibonacci (i)),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The 10 terms of Fibonacci series are\n",
-        "0 1 1 2 3 5 8 13 21 34\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.13, Page number: 244"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "sum = n\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def sumofdigit (n, sum) :\n",
-      "   while sum>=10 :\n",
-      "      sum = 0\n",
-      "      while n>0 :\n",
-      "         sum = sum + n%10\n",
-      "         n = n/10\n",
-      "      n = sum\n",
-      "   return sum\n",
-      "\n",
-      "result = sumofdigit (n, sum)\n",
-      "\n",
-      "print ('Sum of digits of %d is %d' % (n, result))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 8626\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits of 8626 is 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.14, Page number: 245"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def prime (n) :\n",
-      "   for num in range (3, n+1) :\n",
-      "      flag = 1\n",
-      "      for i in range (2, num) :\n",
-      "         if num % i == 0 : \n",
-      "            flag = 0\n",
-      "      if flag :\n",
-      "         print ('%d ' % num),\n",
-      "\n",
-      "print ('Prime numbers upto %d are' % n)\n",
-      "prime (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 30\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Prime numbers upto 30 are\n",
-        "3  5  7  11  13  17  19  23  29 \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.15, Page number: 246"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int (raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def pattern (n) :\n",
-      "\tif n > 0 :\n",
-      "\t\tpattern (n-1)\n",
-      "\t\tfor count in range (1, n+1) :\n",
-      "\t\t\tprint ('%d' % count),\n",
-      "   \tprint\n",
-      "   \t\n",
-      "pattern (n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1\n",
-        "1 2\n",
-        "1 2 3\n",
-        "1 2 3 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.16, Page number: 249"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "n = int(raw_input('Enter any number '))\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printline (x, total) :\n",
-      "\tif x > 0 :\n",
-      "\t\tprintline (x-1, total)\n",
-      "\t\tfor blk in range (1, total-x+1) :\n",
-      "\t\t\tprint ' ',\n",
-      "\t\tfor up in range (1, x+1) :\n",
-      "\t\t\tprint ('%d' % up),\n",
-      "\t\tdown = x-1\n",
-      "\t\twhile down>0 :\n",
-      "\t\t\tprint ('%d' % down),\n",
-      "\t\t\tdown = down-1\n",
-      "\t\tprint\n",
-      "\n",
-      "printline (n, n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "      1\n",
-        "    1 2 1\n",
-        "  1 2 3 2 1\n",
-        "1 2 3 4 3 2 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.17, Page number: 253"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def listsum () :\n",
-      "\tn = int(raw_input('Enter any number '))\n",
-      "\tif n > 0 :\n",
-      "\t\treturn n + listsum ()\n",
-      "\telse :\n",
-      "\t\treturn 0\n",
-      "\t\t\n",
-      "sum = listsum ()\n",
-      "print ('Sum of list of numbers is %d' % sum)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any number -4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of list of numbers is 6\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_8.ipynb b/Magnifying_C/Chapter_8.ipynb
deleted file mode 100755
index ccafa905..00000000
--- a/Magnifying_C/Chapter_8.ipynb
+++ /dev/null
@@ -1,2718 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8: Arrays"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page number: 256"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "word = array('c', ['H', 'e', 'l', 'l', 'o'])\n",
-      "\n",
-      "# Result\n",
-      "print ('The contents of word [] is %s ' % ''.join(word))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The contents of word [] is Hello \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2, Page number: 259"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "marks = [0]*10\n",
-      "avg = 0\n",
-      "\n",
-      "# Calculation and result\n",
-      "for i in range (0, 10) :\n",
-      "\tmarks[i] = int(raw_input('Enter value in cell [%d] ' % i))\n",
-      "\tavg = avg + marks[i]\n",
-      "\n",
-      "avg = avg/10\n",
-      "print ('The average marks are %f' % avg)\n",
-      "\n",
-      "for i in range (0, 10) :\n",
-      "\tif marks[i] < avg :\n",
-      "\t\tprint ('Marks in cell [%d] are %d less than average' % (i, marks[i]))\n",
-      "\telse :\n",
-      "\t\tprint ('Marks in cell [%d] are %d above/equal to average' % (i, marks[i]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [0] 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [1] 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [2] 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [3] 43\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [4] 56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [5] 98\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [6] 53\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [7] 82\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [8] 74\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [9] 39\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average marks are 63.000000\n",
-        "Marks in cell [0] are 45 less than average\n",
-        "Marks in cell [1] are 78 above/equal to average\n",
-        "Marks in cell [2] are 65 above/equal to average\n",
-        "Marks in cell [3] are 43 less than average\n",
-        "Marks in cell [4] are 56 less than average\n",
-        "Marks in cell [5] are 98 above/equal to average\n",
-        "Marks in cell [6] are 53 less than average\n",
-        "Marks in cell [7] are 82 above/equal to average\n",
-        "Marks in cell [8] are 74 above/equal to average\n",
-        "Marks in cell [9] are 39 less than average\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3, Page number: 261"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "data = raw_input('Enter the string ')\n",
-      "\n",
-      "# Calculation and result\n",
-      "rev_data = reversed (data)\n",
-      "\n",
-      "if list (data) == list (rev_data) :\n",
-      "\tprint ('The string %s is a PALINDROME' % data)\n",
-      "else :\n",
-      "\tprint ('The string %s is not a PALINDROME' % data)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the string MADAM\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The string MADAM is a PALINDROME\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4, Page number: 266"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "marks = [0]*10\n",
-      "avg = 0\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readdata (marks) :\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\tmarks[i] = int(raw_input('Enter value in cell [%d] ' % i))\n",
-      "\t\t\n",
-      "def average (marks) :\n",
-      "\tmean = 0\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\tmean = mean + marks[i]\n",
-      "\treturn mean/10\n",
-      "\n",
-      "def printdata (marks, avg) :\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\tif marks[i] < avg :\n",
-      "\t\t\tprint ('Marks in cell [%d] are %d less than average' % (i, marks[i]))\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('Marks in cell [%d] are %d above/equal to average' % (i, marks[i]))\t\n",
-      "\n",
-      "\n",
-      "readdata (marks)\n",
-      "avg = average (marks)\t\n",
-      "print ('The average marks are %f' % avg)\n",
-      "printdata (marks, avg)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [0] 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [1] 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [2] 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [3] 43\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [4] 56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [5] 98\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [6] 53\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [7] 82\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [8] 74\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell [9] 39\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average marks are 63.000000\n",
-        "Marks in cell [0] are 45 less than average\n",
-        "Marks in cell [1] are 78 above/equal to average\n",
-        "Marks in cell [2] are 65 above/equal to average\n",
-        "Marks in cell [3] are 43 less than average\n",
-        "Marks in cell [4] are 56 less than average\n",
-        "Marks in cell [5] are 98 above/equal to average\n",
-        "Marks in cell [6] are 53 less than average\n",
-        "Marks in cell [7] are 82 above/equal to average\n",
-        "Marks in cell [8] are 74 above/equal to average\n",
-        "Marks in cell [9] are 39 less than average\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5, Page number: 273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "a = [10, 20, 30, 40, 50]\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def fun (x, y, z) :\n",
-      "\tx = x + 5\n",
-      "\ty = y + 10\n",
-      "\tz[4] = z[4] + 100\n",
-      "\n",
-      "\n",
-      "for i in range (0, 5) :\n",
-      "\tprint ('The cell [%d] contains %d' % (i, a[i]))\n",
-      "\tprint ('The address of cell [%d] is %x' % (i, id(a[i])))\n",
-      "\t\t\n",
-      "fun (a[0], a[3], a)\t\n",
-      "print\n",
-      "\n",
-      "for i in range (0, 5) :\n",
-      "\tprint ('The cell [%d] contains %d' % (i, a[i]))\n",
-      "\tprint ('The address of cell [%d] is %x' % (i, id(a[i])))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The cell [0] contains 10\n",
-        "The address of cell [0] is 18bc0c0\n",
-        "The cell [1] contains 20\n",
-        "The address of cell [1] is 18bbfd0\n",
-        "The cell [2] contains 30\n",
-        "The address of cell [2] is 18bbee0\n",
-        "The cell [3] contains 40\n",
-        "The address of cell [3] is 18bc5b8\n",
-        "The cell [4] contains 50\n",
-        "The address of cell [4] is 18bc4c8\n",
-        "\n",
-        "The cell [0] contains 10\n",
-        "The address of cell [0] is 18bc0c0\n",
-        "The cell [1] contains 20\n",
-        "The address of cell [1] is 18bbfd0\n",
-        "The cell [2] contains 30\n",
-        "The address of cell [2] is 18bbee0\n",
-        "The cell [3] contains 40\n",
-        "The address of cell [3] is 18bc5b8\n",
-        "The cell [4] contains 150\n",
-        "The address of cell [4] is 18bcaf8\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6, Page number: 275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "data = raw_input('Enter the string ')\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def revstring (data) :\n",
-      "\trev_data = reversed (data)\n",
-      "\n",
-      "\tif list (data) == list (rev_data) :\n",
-      "\t\tprint ('The string %s is a PALINDROME' % data)\n",
-      "\telse :\n",
-      "\t\tprint ('The string %s is not a PALINDROME' % data)\n",
-      "\n",
-      "revstring (data)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the string python\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The string python is not a PALINDROME\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7, Page number: 277"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "a = [0]*10\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readnsort (a) :\n",
-      "\tpos = 0\n",
-      "\tfor len in range (0, 10) :\n",
-      "\t\tnum = int(raw_input('Enter data element no. %d ' % len))\n",
-      "\t\t\n",
-      "\t\tfor pos in range (0, len+1) :\n",
-      "\t\t\tif num < a[pos] :\n",
-      "\t\t\t\tbreak\n",
-      "\t\t\t\t\n",
-      "\t\ti = len\t\t\n",
-      "\t\twhile i>=pos :\n",
-      "\t\t\ta[i] = a[i-1]\n",
-      "\t\t\ti = i-1\n",
-      "\t\ta[pos] = num\n",
-      "\t\t\t\n",
-      "def printdata (a) :\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\t\t\t\n",
-      "\t\t\t\n",
-      "readnsort (a)\n",
-      "print\n",
-      "printdata (a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 0 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 1 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 2 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 3 43\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 4 56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 5 98\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 6 53\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 7 82\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 8 74\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data element no. 9 39\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The data in cell [0] is 39 \n",
-        "The data in cell [1] is 43 \n",
-        "The data in cell [2] is 45 \n",
-        "The data in cell [3] is 53 \n",
-        "The data in cell [4] is 56 \n",
-        "The data in cell [5] is 65 \n",
-        "The data in cell [6] is 74 \n",
-        "The data in cell [7] is 78 \n",
-        "The data in cell [8] is 82 \n",
-        "The data in cell [9] is 98 \n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8, Page number: 279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readdata () :\n",
-      "\ta = []\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
-      "\treturn a\n",
-      "\n",
-      "def sortdata (a) :\n",
-      "\tfor p in range (0, 9) :\n",
-      "\t\tmin = p\n",
-      "\t\tfor i in range (p, 10) :\n",
-      "\t\t\tif a[min] > a[i] :\n",
-      "\t\t\t\tmin = i\n",
-      "\t\ttemp = a[p]\n",
-      "\t\ta[p] = a[min]\n",
-      "\t\ta[min] = temp\n",
-      "\treturn a\n",
-      "\n",
-      "def printdata (a) :\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
-      "\n",
-      "a = readdata ()\n",
-      "a = sortdata (a)\n",
-      "print ('\\nThe sorted data is ')\n",
-      "printdata (a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [0] 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [1] 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [2] 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [3] 43\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [4] 56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [5] 98\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [6] 53\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [7] 82\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [8] 74\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [9] 39\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The sorted data is \n",
-        "The data in cell [0] is 39 \n",
-        "The data in cell [1] is 43 \n",
-        "The data in cell [2] is 45 \n",
-        "The data in cell [3] is 53 \n",
-        "The data in cell [4] is 56 \n",
-        "The data in cell [5] is 65 \n",
-        "The data in cell [6] is 74 \n",
-        "The data in cell [7] is 78 \n",
-        "The data in cell [8] is 82 \n",
-        "The data in cell [9] is 98 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9, Page number: 282"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readdata () :\n",
-      "\ta = []\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
-      "\treturn a\n",
-      "\n",
-      "def bubble (a) :\n",
-      "\tfor p in range (0, 10) :\n",
-      "\t\tfor i in range (0, 9-p) :\n",
-      "\t\t\tif a[i] > a[i+1] :\n",
-      "\t\t\t\ttemp = a[i]\n",
-      "\t\t\t\ta[i] = a[i+1]\n",
-      "\t\t\t\ta[i+1] = temp\n",
-      "\treturn a\n",
-      "\n",
-      "def printdata (a) :\n",
-      "\tfor i in range (0, 10) :\n",
-      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
-      "\n",
-      "a = readdata ()\n",
-      "a = bubble (a)\n",
-      "print ('\\nThe sorted data is ')\n",
-      "printdata (a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [0] 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [1] 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [2] 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [3] 43\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [4] 56\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [5] 98\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [6] 53\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [7] 82\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [8] 74\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [9] 39\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The sorted data is \n",
-        "The data in cell [0] is 39 \n",
-        "The data in cell [1] is 43 \n",
-        "The data in cell [2] is 45 \n",
-        "The data in cell [3] is 53 \n",
-        "The data in cell [4] is 56 \n",
-        "The data in cell [5] is 65 \n",
-        "The data in cell [6] is 74 \n",
-        "The data in cell [7] is 78 \n",
-        "The data in cell [8] is 82 \n",
-        "The data in cell [9] is 98 \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10, Page number: 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readdata (size) :\n",
-      "\ta = []\n",
-      "\tfor i in range (0, size) :\n",
-      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
-      "\treturn a\n",
-      "\n",
-      "def bubble (a, size) :\n",
-      "\tfor p in range (0, size) :\n",
-      "\t\tfor i in range (0, size-p-1) :\n",
-      "\t\t\tif a[i] > a[i+1] :\n",
-      "\t\t\t\ttemp = a[i]\n",
-      "\t\t\t\ta[i] = a[i+1]\n",
-      "\t\t\t\ta[i+1] = temp\n",
-      "\treturn a\n",
-      "\n",
-      "def printdata (a, size) :\n",
-      "\tfor i in range (0, size) :\n",
-      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
-      "\n",
-      "def lsearch (a, key, size) :\n",
-      "\tfor i in range (0, size) :\n",
-      "\t\tif a[i] == key :\n",
-      "\t\t\treturn i\n",
-      "\t\tif i == size :\n",
-      "\t\t\treturn -1\n",
-      "\t\t\t\n",
-      "size = int(raw_input('Enter the total number of elements '))\n",
-      "a = readdata (size)\n",
-      "a = bubble (a, size)\n",
-      "print ('\\nThe sorted array is ')\n",
-      "printdata (a, size)\n",
-      "\n",
-      "find = int(raw_input('\\nEnter the element to be searched '))\n",
-      "pos = lsearch (a, find, size)\n",
-      "\n",
-      "if pos<0 :\n",
-      "\tprint ('\\nThe element %d NOT FOUND ' % find)\n",
-      "else :\n",
-      "\tprint ('\\nThe element %d found at position %d ' % (find, pos))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the total number of elements 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [0] 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [1] 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [2] 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [3] 34\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [4] 49\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The sorted array is \n",
-        "The data in cell [0] is 34 \n",
-        "The data in cell [1] is 45 \n",
-        "The data in cell [2] is 49 \n",
-        "The data in cell [3] is 65 \n",
-        "The data in cell [4] is 78 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter the element to be searched 65\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The element 65 found at position 3 \n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11, Page number: 288"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readdata (size) :\n",
-      "\ta = []\n",
-      "\tfor i in range (0, size) :\n",
-      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
-      "\treturn a\n",
-      "\n",
-      "def bubble (a, size) :\n",
-      "\tfor p in range (0, size) :\n",
-      "\t\tfor i in range (0, size-p-1) :\n",
-      "\t\t\tif a[i] > a[i+1] :\n",
-      "\t\t\t\ttemp = a[i]\n",
-      "\t\t\t\ta[i] = a[i+1]\n",
-      "\t\t\t\ta[i+1] = temp\n",
-      "\treturn a\n",
-      "\n",
-      "def printdata (a, size) :\n",
-      "\tfor i in range (0, size) :\n",
-      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
-      "\n",
-      "def bsearch (a, key, size) :\n",
-      "\tfirst = 0\n",
-      "\tlast = size - 1\n",
-      "\tmid = (first + last)/2\n",
-      "\t\n",
-      "\twhile (first <= last and a[mid] != key) :\n",
-      "\t\tif a[mid] < key :\n",
-      "\t\t\tfirst = mid + 1\n",
-      "\t\telse :\n",
-      "\t\t\tif a[mid] > key :\n",
-      "\t\t\t\tlast = mid - 1\n",
-      "\t\tmid = (first + last)/2\n",
-      "\t\n",
-      "\tif a[mid] == key :\n",
-      "\t\treturn mid\n",
-      "\telse :\n",
-      "\t\treturn -1\n",
-      "\t\t\t\n",
-      "size = int(raw_input('Enter the number of elements in the array '))\n",
-      "a = readdata (size)\n",
-      "a = bubble (a, size)\n",
-      "print ('\\nThe sorted array is ')\n",
-      "printdata (a, size)\n",
-      "\n",
-      "find = int(raw_input('\\nPlease enter the element to be searched for '))\n",
-      "pos = bsearch (a, find, size)\n",
-      "\n",
-      "if pos<0 :\n",
-      "\tprint ('\\nThe element %d NOT FOUND ' % find)\n",
-      "else :\n",
-      "\tprint ('\\nThe element %d found at position %d ' % (find, pos))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of elements in the array 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [0] 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [1] 78\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [2] 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [3] 34\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [4] 49\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The sorted array is \n",
-        "The data in cell [0] is 34 \n",
-        "The data in cell [1] is 45 \n",
-        "The data in cell [2] is 49 \n",
-        "The data in cell [3] is 65 \n",
-        "The data in cell [4] is 78 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Please enter the element to be searched for 45\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The element 45 found at position 1 \n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12, Page number: 293"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readncompute (no_of_students) :\n",
-      "\ta = [[0 for y in xrange(5)] for x in xrange(no_of_students)]\n",
-      "\tper = [0 for x in range(no_of_students)]\n",
-      "\tavg = 0\n",
-      "\t\n",
-      "\tfor i in range (0, no_of_students) :\n",
-      "\t\tprint ('\\nEnter marks of 5 subjects for Roll no. %d ' % (i+1))\n",
-      "\t\t\n",
-      "\t\tfor j in range (0, 5) :\n",
-      "\t\t\ta[i][j] = int(raw_input('Subject %d = ' % (j+1)))\n",
-      "\t\t\tper[i] = per[i] + a[i][j]\n",
-      "\t\t\t\n",
-      "\t\tper[i] = per[i]/5\n",
-      "\t\tavg = avg + per[i]\n",
-      "\t\t\n",
-      "\treturn a, avg/no_of_students, per\n",
-      "\n",
-      "def printdata (a, no_of_students, avg, per) :\n",
-      "\tfor i in range (0, no_of_students) :\n",
-      "\t\tprint ('\\nMarks for Roll no. %d are ' % (i+1))\n",
-      "\t\tfor j in range (0, 5) :\n",
-      "\t\t\tprint a[i][j]\n",
-      "\t\tprint ('Percentage for Roll no. %d is %f ' % ((i+1), per[i]))\n",
-      "\t\t\n",
-      "\t\tif per[i] < avg :\n",
-      "\t\t\tprint ('Below average')\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('Above average')\n",
-      "\t\t\n",
-      "\n",
-      "no_of_students = int(raw_input('How many students marks do you want to enter '))\n",
-      "a, average, per = readncompute (no_of_students)\n",
-      "print ('\\nThe average marks are %f ' % average)\n",
-      "printdata (a, no_of_students, average, per)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many students marks do you want to enter 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter marks of 5 subjects for Roll no. 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1 = 75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2 = 43\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3 = 51\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 4 = 68\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 5 = 73\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter marks of 5 subjects for Roll no. 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1 = 49\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2 = 71\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3 = 62\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 4 = 53\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 5 = 50\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter marks of 5 subjects for Roll no. 3 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1 = 90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2 = 82\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3 = 74\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 4 = 93\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 5 = 89\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The average marks are 68.000000 \n",
-        "\n",
-        "Marks for Roll no. 1 are \n",
-        "75\n",
-        "43\n",
-        "51\n",
-        "68\n",
-        "73\n",
-        "Percentage for Roll no. 1 is 62.000000 \n",
-        "Below average\n",
-        "\n",
-        "Marks for Roll no. 2 are \n",
-        "49\n",
-        "71\n",
-        "62\n",
-        "53\n",
-        "50\n",
-        "Percentage for Roll no. 2 is 57.000000 \n",
-        "Below average\n",
-        "\n",
-        "Marks for Roll no. 3 are \n",
-        "90\n",
-        "82\n",
-        "74\n",
-        "93\n",
-        "89\n",
-        "Percentage for Roll no. 3 is 85.000000 \n",
-        "Above average\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13, Page number: 296"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def getchoice () :\n",
-      "\tprint\n",
-      "\tprint ('1 - Input two matrices ')\n",
-      "\tprint ('2 - Add and display sum ')\n",
-      "\tprint ('3 - Exit ')\n",
-      "\t\n",
-      "\tchoice = int(raw_input('Enter choice : '))\n",
-      "\treturn choice\n",
-      "\n",
-      "def read (matrix, rows, cols) :\t\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tprint ('\\nEnter elements for row %d ' % (i+1))\t\t\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
-      "\n",
-      "def add (sum, augend, addend, rows, cols) :\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tsum[i][j] = augend[i][j] + addend[i][j]\n",
-      "\n",
-      "def display2matrices (augend, addend, rows, cols) :\n",
-      "\tprint ('\\nMatrix A ')\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tprint ('%3d ' % augend[i][j]),\n",
-      "\t\tprint\n",
-      "\tprint ('\\nMatrix B ')\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tprint ('%3d ' % addend[i][j]),\n",
-      "\t\tprint\n",
-      "\n",
-      "def display3matrices (sum, augend, addend, rows, cols) :\n",
-      "\tprint\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tprint ('%3d ' % augend[i][j]),\n",
-      "\t\tprint ('%s ' % ('+' if i==rows/2 else ' ')),\n",
-      "\t\t\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tprint ('%2d ' % addend[i][j]),\n",
-      "\t\tprint ('%s ' % ('=' if i==rows/2 else ' ')),\n",
-      "\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tprint ('%2d ' % sum[i][j]),\n",
-      "\t\tprint\n",
-      "\n",
-      "def display (matrix, rows, cols) :\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tprint ('%3d ' % matrix[i][j]),\n",
-      "\t\tprint\n",
-      "\n",
-      "augend = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "addend = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "sum = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\tchoice = getchoice ()\n",
-      "\t\n",
-      "\tif choice == 1 :\n",
-      "\t\trows = int(raw_input('\\nEnter the number of rows '))\t\t\n",
-      "\t\tcols = int(raw_input('Enter the number of columns '))\n",
-      "\t\t\n",
-      "\t\tprint ('\\n\\nEnter the first matrix ')\n",
-      "\t\tread (augend, rows, cols)\n",
-      "\t\tprint ('\\n\\nEnter the second matrix ')\n",
-      "\t\tread (addend, rows, cols)\n",
-      "\t\tprint ('\\n\\nYou entered ')\n",
-      "\t\tdisplay2matrices (augend, addend, rows, cols)\n",
-      "\t\n",
-      "\telif choice == 2 :\n",
-      "\t\tadd (sum, augend, addend, rows, cols)\n",
-      "\t\tdisplay3matrices (sum, augend, addend, rows, cols)\n",
-      "\t\n",
-      "\telif choice == 3 :\n",
-      "\t\tcondition = False\n",
-      "\t\texit ()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1 - Input two matrices \n",
-        "2 - Add and display sum \n",
-        "3 - Exit \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter choice : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter the number of rows 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of columns 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "Enter the first matrix \n",
-        "\n",
-        "Enter elements for row 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][1] = 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][2] = 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][3] = 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter elements for row 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][1] = 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][2] = 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][3] = 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter elements for row 3 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][1] = 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][2] = 6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][3] = 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "Enter the second matrix \n",
-        "\n",
-        "Enter elements for row 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][1] = 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][2] = 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][3] = 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter elements for row 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][1] = 6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][2] = 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][3] = 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter elements for row 3 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][1] = 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][2] = 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][3] = 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "You entered \n",
-        "\n",
-        "Matrix A \n",
-        "  1    2    3 \n",
-        "  3    4    5 \n",
-        "  5    6    7 \n",
-        "\n",
-        "Matrix B \n",
-        "  3    5    1 \n",
-        "  6    5    3 \n",
-        "  8    7    4 \n",
-        "\n",
-        "1 - Input two matrices \n",
-        "2 - Add and display sum \n",
-        "3 - Exit \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter choice : 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "  1    2    3      3   5   1      4   7   4 \n",
-        "  3    4    5  +   6   5   3  =   9   9   8 \n",
-        "  5    6    7      8   7   4     13  13  11 \n",
-        "\n",
-        "1 - Input two matrices \n",
-        "2 - Add and display sum \n",
-        "3 - Exit \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter choice : 3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14, Page number: 300"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def readmatrix (matrix, rows, cols) :\t\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tprint ('\\nPlease enter elements for row %d ' % (i+1))\t\t\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
-      "\n",
-      "def multiply (prod, a, b, Ar, Ac, Bc) :\n",
-      "\tfor i in range (0, Ar) :\n",
-      "\t\tfor j in range (0, Bc) :\n",
-      "\t\t\tprod[i][j] = 0\n",
-      "\t\t\tfor k in range (0, Ac) :\n",
-      "\t\t\t\tprod[i][j] += a[i][k] * b[k][j]\n",
-      "\n",
-      "def display (matrix, rows, cols) :\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tprint ('%3d ' % matrix[i][j]),\n",
-      "\t\tprint\n",
-      "\n",
-      "a = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "b = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "product = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "\n",
-      "condition = True\n",
-      "while condition  :\n",
-      "\tArows, Acols = map(int, raw_input('Enter order of matrix A: ').split())\n",
-      "\tBrows, Bcols = map(int, raw_input('Enter order of matrix B: ').split())\n",
-      "\n",
-      "\tif Acols == Brows :\n",
-      "\t\tcondition = False\n",
-      "\t\tbreak\n",
-      "\telse : \n",
-      "\t\tprint ('\\nMatrices are not compatible for multiplication. Enter the order again. ')\n",
-      "\t\tcontinue\n",
-      "\t\n",
-      "print ('\\nEnter matrix A: ')\n",
-      "readmatrix (a, Arows, Acols)\n",
-      "\t\n",
-      "print ('\\nEnter matrix B: ')\n",
-      "readmatrix (b, Brows, Bcols)\n",
-      "\n",
-      "multiply (product, a, b, Arows, Acols, Bcols)\n",
-      "print ('\\nThe resultant product (A.B) is: ')\n",
-      "display (product, Arows, Bcols)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter order of matrix A: 2 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter order of matrix B: 3 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter matrix A: \n",
-        "\n",
-        "Please enter elements for row 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][1] = 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][2] = 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][3] = 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Please enter elements for row 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][1] = 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][2] = 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][3] = 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter matrix B: \n",
-        "\n",
-        "Please enter elements for row 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][1] = 6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][2] = 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Please enter elements for row 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][1] = 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][2] = 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Please enter elements for row 3 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][1] = 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][2] = 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The resultant product (A.B) is: \n",
-        " 20   14 \n",
-        " 56   41 \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15, Page number: 303"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def getchoice () :\n",
-      "\tprint\n",
-      "\tprint ('1 - Input square matrix ')\n",
-      "\tprint ('2 - Display upper triangular matrix ')\n",
-      "\tprint ('3 - Display lower triangular matrix ')\n",
-      "\tprint ('4 - Exit ')\n",
-      "\t\n",
-      "\tcondition = True\n",
-      "\twhile condition :\n",
-      "\t\tchoice = int(raw_input('Enter choice : '))\n",
-      "\t\t\n",
-      "\t\tif choice in range (1, 5) :\n",
-      "\t\t\tcondition = False\n",
-      "\t\t\t\n",
-      "\treturn choice\n",
-      "\n",
-      "def displayupper (matrix, order) :\n",
-      "\tprint\n",
-      "\tfor i in range (0, order) :\n",
-      "\t\tfor j in range (0, order) :\n",
-      "\t\t\tif j >= i :\n",
-      "\t\t\t\tprint ('%3d ' % matrix[i][j]),\n",
-      "\t\t\telse :\n",
-      "\t\t\t\tprint ('%3d ' % 0),\n",
-      "\t\tprint\n",
-      "\t\t\n",
-      "def displaylower (matrix, order) :\n",
-      "\tprint\n",
-      "\tfor i in range (0, order) :\n",
-      "\t\tfor j in range (0, order) :\n",
-      "\t\t\tif j <= i :\n",
-      "\t\t\t\tprint ('%3d ' % matrix[i][j]),\n",
-      "\t\t\telse :\n",
-      "\t\t\t\tprint ('%3d ' % 0),\n",
-      "\t\tprint\n",
-      "\n",
-      "def read (matrix, rows, cols) :\t\n",
-      "\tfor i in range (0, rows) :\n",
-      "\t\tprint ('\\nEnter %d elements for row %d ' % (rows, (i+1)))\t\t\n",
-      "\t\tfor j in range (0, cols) :\n",
-      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
-      "\n",
-      "matrix = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "\n",
-      "condition = True\n",
-      "while condition :\n",
-      "\tchoice = getchoice ()\n",
-      "\t\n",
-      "\tif choice == 1 :\n",
-      "\t\torder = int(raw_input('\\nEnter order of square matrix '))\t\t\n",
-      "\t\tread (matrix, order, order)\n",
-      "\t\n",
-      "\telif choice == 2 :\n",
-      "\t\tdisplayupper (matrix, order)\n",
-      "\t\n",
-      "\telif choice == 3 :\n",
-      "\t\tdisplaylower (matrix, order)\n",
-      "\t\t\n",
-      "\telif choice == 4 :\n",
-      "\t\tcondition = False\n",
-      "\t\texit ()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1 - Input square matrix \n",
-        "2 - Display upper triangular matrix \n",
-        "3 - Display lower triangular matrix \n",
-        "4 - Exit \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter choice : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter order of square matrix 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter 3 elements for row 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][1] = 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][2] = 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][3] = 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter 3 elements for row 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][1] = 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][2] = 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][3] = 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter 3 elements for row 3 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][1] = 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][2] = 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][3] = 9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1 - Input square matrix \n",
-        "2 - Display upper triangular matrix \n",
-        "3 - Display lower triangular matrix \n",
-        "4 - Exit \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter choice : 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "  1    2    3 \n",
-        "  0    5    6 \n",
-        "  0    0    9 \n",
-        "\n",
-        "1 - Input square matrix \n",
-        "2 - Display upper triangular matrix \n",
-        "3 - Display lower triangular matrix \n",
-        "4 - Exit \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter choice : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "  1    0    0 \n",
-        "  4    5    0 \n",
-        "  7    8    9 \n",
-        "\n",
-        "1 - Input square matrix \n",
-        "2 - Display upper triangular matrix \n",
-        "3 - Display lower triangular matrix \n",
-        "4 - Exit \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter choice : 4\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.16, Page number: 307"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def readmatrix (matrix, n) :\n",
-      "\tfor i in range (0, n) :\n",
-      "\t\tprint ('\\nEnter elements for row no %d ' % (i+1))\n",
-      "\t\tfor j in range (0, n) :\n",
-      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
-      "\n",
-      "def display (matrix, n) :\n",
-      "\tprint\n",
-      "\tfor i in range (0, n) :\n",
-      "\t\tfor j in range (0, n) :\n",
-      "\t\t\tprint ('%5d ' % matrix[i][j]),\n",
-      "\t\tprint\n",
-      "\t\t\n",
-      "def isunitmatrix(matrix, n) :\n",
-      "\tfor i in range (0, n) :\n",
-      "\t\tif matrix[i][i] != 1 :\n",
-      "\t\t\treturn 0\n",
-      "\t\n",
-      "\tfor i in range (0, n) :\n",
-      "\t\tfor j in range (0, n) :\n",
-      "\t\t\tif i != j :\n",
-      "\t\t\t\tif matrix[i][j] != 0 :\n",
-      "\t\t\t\t\treturn 0\n",
-      "\treturn 1\n",
-      "\n",
-      "matrix = [[0 for y in xrange(10)] for x in xrange(10)]\n",
-      "n = int(raw_input('Enter order of matrix '))\n",
-      "readmatrix (matrix, n)\n",
-      "display (matrix, n)\n",
-      "\n",
-      "if (isunitmatrix (matrix, n)) :\n",
-      "\tprint \"UNIT MATRIX\"\n",
-      "else :\n",
-      "\tprint \"NOT A UNIT MATRIX\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter order of matrix 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter elements for row no 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][1] = 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][2] = 0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[1][3] = 0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter elements for row no 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][1] = 0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][2] = 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[2][3] = 0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter elements for row no 3 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][1] = 0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][2] = 0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "matrix[3][3] = 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "    1      0      0 \n",
-        "    0      1      0 \n",
-        "    0      0      1 \n",
-        "UNIT MATRIX\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17, Page number: 311"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "string = raw_input('Enter a string: ')\n",
-      "\n",
-      "# Calculation and result\n",
-      "swap_string = string.swapcase()\n",
-      "\n",
-      "print swap_string"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a string: hello world\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "HELLO WORLD\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18, Page number: 312"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "string = raw_input('Enter a string ')\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def printstr (string) :\n",
-      "\tfor i in range (0, len(string)) :\n",
-      "\t\tfor j in range (0, i+1) :\n",
-      "\t\t\tprint string[j],\n",
-      "\t\tprint\n",
-      "\t\n",
-      "\ti = i - 2\n",
-      "\twhile i >= 0 :\n",
-      "\t\tfor j in range (0, i+1) :\n",
-      "\t\t\tprint string[j],\n",
-      "\t\tprint\n",
-      "\t\ti = i - 1\n",
-      "\t\t\n",
-      "printstr (string)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a string hello\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "h\n",
-        "h e\n",
-        "h e l\n",
-        "h e l l\n",
-        "h e l l o\n",
-        "h e l\n",
-        "h e\n",
-        "h\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19, Page number: 314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = raw_input('Enter any string ')\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def copy (a) :\n",
-      "\tb = a\n",
-      "\treturn b\n",
-      "\t\n",
-      "b = copy (a)\n",
-      "print\n",
-      "print ('The first string is %s ' % a)\n",
-      "print ('The second string is %s ' % b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any string hello\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The first string is hello \n",
-        "The second string is hello \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.20, Page number: 315"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "a = raw_input('Enter the major string: ')\n",
-      "b = raw_input('Enter the minor string: ')\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def ispart (a, b) :\n",
-      "\tif b in a :\n",
-      "\t\tprint ('The string \"%s\" is A SUBSTRING of \"%s\" ' % (b, a))\n",
-      "\telse :\n",
-      "\t\tprint ('The string \"%s\" is NOT A SUBSTRING of \"%s\" ' % (b, a))\n",
-      "\n",
-      "ispart (a, b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the major string: python is a programming language\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the minor string: python\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The string \"python\" is A SUBSTRING of \"python is a programming language\" \n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/Chapter_9.ipynb b/Magnifying_C/Chapter_9.ipynb
deleted file mode 100755
index acb61edf..00000000
--- a/Magnifying_C/Chapter_9.ipynb
+++ /dev/null
@@ -1,1093 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9: More on Pointers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1, Page number: 318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "a = [[1,2,3,4,5],\n",
-      "     [6,7,8,9,10],\n",
-      "     [11,12,13,14,15],\n",
-      "     [16,17,18,19,20],\n",
-      "     [21,22,23,24,25]]\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('The value of a is %x ' % id(a))\n",
-      "\n",
-      "for i in range (0, 5) :\n",
-      "\tprint ('The value of a[%d] is %x ' % (i, id(i)))\n",
-      "\n",
-      "for i in range (0, 5) :\n",
-      "\tfor j in range (0, 5) :       \n",
-      "\t\tprint ('The cell [%d][%d] i.e. %x has %d ' % (i, j, id(a[i][j]), a[i][j]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of a is 7f55dc3e08c0 \n",
-        "The value of a[0] is 18861b0 \n",
-        "The value of a[1] is 1886198 \n",
-        "The value of a[2] is 1886180 \n",
-        "The value of a[3] is 1886168 \n",
-        "The value of a[4] is 1886150 \n",
-        "The cell [0][0] i.e. 1886198 has 1 \n",
-        "The cell [0][1] i.e. 1886180 has 2 \n",
-        "The cell [0][2] i.e. 1886168 has 3 \n",
-        "The cell [0][3] i.e. 1886150 has 4 \n",
-        "The cell [0][4] i.e. 1886138 has 5 \n",
-        "The cell [1][0] i.e. 1886120 has 6 \n",
-        "The cell [1][1] i.e. 1886108 has 7 \n",
-        "The cell [1][2] i.e. 18860f0 has 8 \n",
-        "The cell [1][3] i.e. 18860d8 has 9 \n",
-        "The cell [1][4] i.e. 18860c0 has 10 \n",
-        "The cell [2][0] i.e. 18860a8 has 11 \n",
-        "The cell [2][1] i.e. 1886090 has 12 \n",
-        "The cell [2][2] i.e. 1886078 has 13 \n",
-        "The cell [2][3] i.e. 1886060 has 14 \n",
-        "The cell [2][4] i.e. 1886048 has 15 \n",
-        "The cell [3][0] i.e. 1886030 has 16 \n",
-        "The cell [3][1] i.e. 1886018 has 17 \n",
-        "The cell [3][2] i.e. 1886000 has 18 \n",
-        "The cell [3][3] i.e. 1885fe8 has 19 \n",
-        "The cell [3][4] i.e. 1885fd0 has 20 \n",
-        "The cell [4][0] i.e. 1885fb8 has 21 \n",
-        "The cell [4][1] i.e. 1885fa0 has 22 \n",
-        "The cell [4][2] i.e. 1885f88 has 23 \n",
-        "The cell [4][3] i.e. 1885f70 has 24 \n",
-        "The cell [4][4] i.e. 1885f58 has 25 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2, Page number: 320"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from array import *\n",
-      "\n",
-      "# Variable declaration\n",
-      "a = [[[1,2,3,4,5], [6,7,8,9,10], [11,12,13,14,15], [16,17,18,19,20]],\n",
-      "     [[21,22,23,24,25], [26,27,28,29,30], [31,32,33,34,35], [36,37,38,39,40]]]\n",
-      "\n",
-      "# Calculation and result\n",
-      "print ('The value of a is %x ' % id(a))\n",
-      "\n",
-      "for i in range (0, 2) :\n",
-      "\tprint ('The value of a[%d] is %x ' % (i, id(a[i])))\n",
-      "\tfor j in range (0, 4) :\n",
-      "\t\tprint ('The value of a[%d][%d] is %x ' % (i, j, id(a[i][j])))\n",
-      "\tprint\n",
-      "\n",
-      "for i in range (0, 2) :\n",
-      "\tfor j in range (0, 4) :       \n",
-      "\t\tfor k in range (0, 5) :\n",
-      "\t\t\tprint ('The cell [%d][%d][%d] i.e. %x has %d ' % (i, j, k, id(a[i][j][k]), a[i][j][k]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of a is 7f55de4772d8 \n",
-        "The value of a[0] is 7f55de486b90 \n",
-        "The value of a[0][0] is 7f55de4db368 \n",
-        "The value of a[0][1] is 7f55de48f050 \n",
-        "The value of a[0][2] is 7f55de4db3f8 \n",
-        "The value of a[0][3] is 7f55de4c7908 \n",
-        "\n",
-        "The value of a[1] is 7f55de477320 \n",
-        "The value of a[1][0] is 7f55de4ff3f8 \n",
-        "The value of a[1][1] is 7f55de477290 \n",
-        "The value of a[1][2] is 7f55de4773b0 \n",
-        "The value of a[1][3] is 7f55de477368 \n",
-        "\n",
-        "The cell [0][0][0] i.e. 1886198 has 1 \n",
-        "The cell [0][0][1] i.e. 1886180 has 2 \n",
-        "The cell [0][0][2] i.e. 1886168 has 3 \n",
-        "The cell [0][0][3] i.e. 1886150 has 4 \n",
-        "The cell [0][0][4] i.e. 1886138 has 5 \n",
-        "The cell [0][1][0] i.e. 1886120 has 6 \n",
-        "The cell [0][1][1] i.e. 1886108 has 7 \n",
-        "The cell [0][1][2] i.e. 18860f0 has 8 \n",
-        "The cell [0][1][3] i.e. 18860d8 has 9 \n",
-        "The cell [0][1][4] i.e. 18860c0 has 10 \n",
-        "The cell [0][2][0] i.e. 18860a8 has 11 \n",
-        "The cell [0][2][1] i.e. 1886090 has 12 \n",
-        "The cell [0][2][2] i.e. 1886078 has 13 \n",
-        "The cell [0][2][3] i.e. 1886060 has 14 \n",
-        "The cell [0][2][4] i.e. 1886048 has 15 \n",
-        "The cell [0][3][0] i.e. 1886030 has 16 \n",
-        "The cell [0][3][1] i.e. 1886018 has 17 \n",
-        "The cell [0][3][2] i.e. 1886000 has 18 \n",
-        "The cell [0][3][3] i.e. 1885fe8 has 19 \n",
-        "The cell [0][3][4] i.e. 1885fd0 has 20 \n",
-        "The cell [1][0][0] i.e. 1885fb8 has 21 \n",
-        "The cell [1][0][1] i.e. 1885fa0 has 22 \n",
-        "The cell [1][0][2] i.e. 1885f88 has 23 \n",
-        "The cell [1][0][3] i.e. 1885f70 has 24 \n",
-        "The cell [1][0][4] i.e. 1885f58 has 25 \n",
-        "The cell [1][1][0] i.e. 1885f40 has 26 \n",
-        "The cell [1][1][1] i.e. 1885f28 has 27 \n",
-        "The cell [1][1][2] i.e. 1885f10 has 28 \n",
-        "The cell [1][1][3] i.e. 1885ef8 has 29 \n",
-        "The cell [1][1][4] i.e. 1885ee0 has 30 \n",
-        "The cell [1][2][0] i.e. 1885ec8 has 31 \n",
-        "The cell [1][2][1] i.e. 1885eb0 has 32 \n",
-        "The cell [1][2][2] i.e. 1885e98 has 33 \n",
-        "The cell [1][2][3] i.e. 1885e80 has 34 \n",
-        "The cell [1][2][4] i.e. 1885e68 has 35 \n",
-        "The cell [1][3][0] i.e. 1886618 has 36 \n",
-        "The cell [1][3][1] i.e. 1886600 has 37 \n",
-        "The cell [1][3][2] i.e. 18865e8 has 38 \n",
-        "The cell [1][3][3] i.e. 18865d0 has 39 \n",
-        "The cell [1][3][4] i.e. 18865b8 has 40 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3, Page number: 326"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def readdata (a, num) :\n",
-      "\twhile num > 0 :\n",
-      "\t\ta[num-1] = int(raw_input('Enter value in cell[%d] ' % num))\n",
-      "\t\tnum = num - 1\n",
-      "\t\t\n",
-      "def printdata (a, num) :\n",
-      "\twhile num > 0 :\n",
-      "\t\tprint ('Value in cell[%d] is %d ' % (num, a[num-1]))\n",
-      "\t\tnum = num - 1\n",
-      "\n",
-      "n = int(raw_input('Enter the number of elements '))\n",
-      "a = [0 for i in xrange(n)]\n",
-      "readdata (a, n)\n",
-      "print\n",
-      "printdata (a, n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of elements 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[3] 30\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[2] 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[1] 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value in cell[3] is 30 \n",
-        "Value in cell[2] is 20 \n",
-        "Value in cell[1] is 10 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4, Page number: 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def readdata (a, n) :\n",
-      "\twhile n > 0 :\n",
-      "\t\ta[n-1] = int(raw_input('Enter value in cell[%d] ' % n))\n",
-      "\t\tn = n - 1\n",
-      "\t\t\n",
-      "def printdata (a, n) :\n",
-      "\twhile n > 0 :\n",
-      "\t\tprint ('Value in cell[%d] is %d ' % (n, a[n-1]))\n",
-      "\t\tn = n - 1\n",
-      "\n",
-      "n = int(raw_input('Enter the number of elements '))\n",
-      "a = [0 for i in xrange(n)]\n",
-      "readdata (a, n)\n",
-      "print\n",
-      "printdata (a, n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of elements 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[5] 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[4] 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[3] 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[2] 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[1] 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value in cell[5] is 1 \n",
-        "Value in cell[4] is 2 \n",
-        "Value in cell[3] is 3 \n",
-        "Value in cell[2] is 4 \n",
-        "Value in cell[1] is 5 \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5, Page number: 330"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "global n\n",
-      "n = 0\n",
-      "\n",
-      "def makearray (n) :\n",
-      "\tn = int(raw_input('Enter the number of elements '))\t\n",
-      "\treturn n\n",
-      "\t\n",
-      "def readdata (a, n) :\n",
-      "\twhile n > 0 :\n",
-      "\t\ta[n-1] = int(raw_input('Enter value in cell[%d] ' % n))\n",
-      "\t\tn = n - 1\n",
-      "\t\t\n",
-      "def printdata (a, n) :\n",
-      "\twhile n > 0 :\n",
-      "\t\tprint ('Value in cell[%d] is %d ' % (n, a[n-1]))\n",
-      "\t\tn = n - 1\n",
-      "\t\t\n",
-      "n = makearray (n)\n",
-      "a = [0 for i in xrange(n)]\n",
-      "readdata (a, n)\n",
-      "print\n",
-      "printdata (a, n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of elements 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[4] 400\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[3] 300\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[2] 200\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value in cell[1] 100\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value in cell[4] is 400 \n",
-        "Value in cell[3] is 300 \n",
-        "Value in cell[2] is 200 \n",
-        "Value in cell[1] is 100 \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6, Page number: 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration, calculation and result\n",
-      "string = raw_input('Enter the string: ')\n",
-      "\n",
-      "rev_string = reversed(string)\n",
-      "\n",
-      "if list(string) == list(rev_string) :\n",
-      "\tprint ('%s is a PALINDROME ' % string)\n",
-      "else :\n",
-      "\tprint ('%s is not a PALINDROME ' % string)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the string: malayalam\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "malayalam is a PALINDROME \n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7, Page number: 337"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def readdata (a, n) :\n",
-      "\tfor i in range (0, n) :\n",
-      "\t\ta[i] = int(raw_input('Enter data in cell [%d] ' % i))\n",
-      "\treturn a\n",
-      "\n",
-      "def bubble (a, n) :\n",
-      "\tfor p in range (0, n) :\n",
-      "\t\tfor i in range (0, n-1-p) :\n",
-      "\t\t\tif a[i] > a[i+1] :\n",
-      "\t\t\t\ttemp = a[i]\n",
-      "\t\t\t\ta[i] = a[i+1]\n",
-      "\t\t\t\ta[i+1] = temp\n",
-      "\treturn a\n",
-      "\n",
-      "def printdata (a, n) :\n",
-      "\tfor i in range (0, n) :\n",
-      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
-      "\n",
-      "n = int(raw_input('Enter the number of elements to be entered: '))\n",
-      "a = [0] * n\n",
-      "a = readdata (a, n)\n",
-      "a = bubble (a, n)\n",
-      "print ('\\nThe sorted data is ')\n",
-      "printdata (a, n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of elements to be entered: 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [0] 46\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [1] 75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [2] 68\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [3] 34\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter data in cell [4] 49\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The sorted data is \n",
-        "The data in cell [0] is 34 \n",
-        "The data in cell [1] is 46 \n",
-        "The data in cell [2] is 49 \n",
-        "The data in cell [3] is 68 \n",
-        "The data in cell [4] is 75 \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8, Page number: 341"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration, calculation and result\n",
-      "i = 6\n",
-      "c = 'a'\n",
-      "\n",
-      "the_data = i\n",
-      "print ('the_data points to the integer value %d ' % the_data)\n",
-      "\n",
-      "the_data = c\n",
-      "print ('the_data points to the character value %c ' % the_data)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the_data points to the integer value 6 \n",
-        "the_data points to the character value a \n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9, Page number: 343"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "marks = [[0 for x in xrange(5)] for x in xrange(5)]\n",
-      "avg = 0\n",
-      "percent = [0 for x in xrange(5)]\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readncompute (marks, percent, avg) :\n",
-      "\tno_of_students = int(raw_input('How many students marks do you want to enter? '))\n",
-      "\tfor i in range (0, no_of_students) :\n",
-      "\t\tprint ('\\nEnter marks of 5 subjects for roll no. %d ' % (i+1))\n",
-      "\t\tfor j in range (0, 5) :\n",
-      "\t\t\tmarks[i][j] = int(raw_input('Subject %d: ' % (j+1)))\n",
-      "\t\t\tpercent[i] = percent[i] + marks[i][j]\n",
-      "\t\t\n",
-      "\t\tpercent[i] = percent[i]/5\n",
-      "\t\tavg = avg + percent[i]\n",
-      "\treturn avg/no_of_students, no_of_students\n",
-      "\n",
-      "def printdata (score, per, avg, no_of_students) :\n",
-      "\tfor i in range (0, no_of_students) :\n",
-      "\t\tprint ('\\nMarks for roll no. %d are ' % (i+1))\n",
-      "\t\tfor j in range (0, 5) :\n",
-      "\t\t\tprint ('%d ' % score[i][j])\n",
-      "\t\tprint ('Percentage for roll no. is %.2f ' % per[i])\n",
-      "\t\tif per[i] < avg :\n",
-      "\t\t\tprint ('Below average')\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('Above average')\n",
-      "\n",
-      "average, nos = readncompute(marks, percent, avg)\n",
-      "print ('\\nThe average marks are %.2f' % average)\n",
-      "printdata (marks, percent, average, nos)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many students marks do you want to enter? 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter marks of 5 subjects for roll no. 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1: 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2: 76\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3: 54\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 4: 83\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 5: 94\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter marks of 5 subjects for roll no. 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1: 84\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2: 62\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3: 94\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 4: 73\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 5: 62\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The average marks are 74.00\n",
-        "\n",
-        "Marks for roll no. 1 are \n",
-        "65 \n",
-        "76 \n",
-        "54 \n",
-        "83 \n",
-        "94 \n",
-        "Percentage for roll no. is 74.00 \n",
-        "Above average\n",
-        "\n",
-        "Marks for roll no. 2 are \n",
-        "84 \n",
-        "62 \n",
-        "94 \n",
-        "73 \n",
-        "62 \n",
-        "Percentage for roll no. is 75.00 \n",
-        "Above average\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10, Page number: 349"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable declaration\n",
-      "marks = [[0 for x in xrange(5)] for x in xrange(5)]\n",
-      "avg = 0\n",
-      "percent = [0 for x in xrange(5)]\n",
-      "\n",
-      "# Function declaration, calculation and result\n",
-      "def readncompute (marks, percent, avg) :\n",
-      "\tno_of_students = int(raw_input('How many students marks do you want to enter? '))\n",
-      "\tfor i in range (0, no_of_students) :\n",
-      "\t\tprint ('\\nEnter marks of 5 subjects for roll no. %d ' % (i+1))\n",
-      "\t\tfor j in range (0, 5) :\n",
-      "\t\t\tmarks[i][j] = int(raw_input('Subject %d: ' % (j+1)))\n",
-      "\t\t\tpercent[i] = percent[i] + marks[i][j]\n",
-      "\t\t\n",
-      "\t\tpercent[i] = percent[i]/5\n",
-      "\t\tavg = avg + percent[i]\n",
-      "\treturn avg/no_of_students, no_of_students\n",
-      "\n",
-      "def printdata (score, per, avg, no_of_students) :\n",
-      "\tfor i in range (0, no_of_students) :\n",
-      "\t\tprint ('\\nMarks for roll no. %d are ' % (i+1))\n",
-      "\t\tfor j in range (0, 5) :\n",
-      "\t\t\tprint ('%d ' % score[i][j])\n",
-      "\t\tprint ('Percentage for roll no. is %.2f ' % per[i])\n",
-      "\t\tif per[i] < avg :\n",
-      "\t\t\tprint ('Below average')\n",
-      "\t\telse :\n",
-      "\t\t\tprint ('Above average')\n",
-      "\n",
-      "average, nos = readncompute(marks, percent, avg)\n",
-      "print ('\\nThe average marks are %.2f' % average)\n",
-      "printdata (marks, percent, average, nos)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many students marks do you want to enter? 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter marks of 5 subjects for roll no. 1 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1: 65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2: 76\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3: 54\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 4: 32\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 5: 89\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter marks of 5 subjects for roll no. 2 \n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 1: 76\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 2: 86\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 3: 43\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 4: 32\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Subject 5: 21\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The average marks are 57.00\n",
-        "\n",
-        "Marks for roll no. 1 are \n",
-        "65 \n",
-        "76 \n",
-        "54 \n",
-        "32 \n",
-        "89 \n",
-        "Percentage for roll no. is 63.00 \n",
-        "Above average\n",
-        "\n",
-        "Marks for roll no. 2 are \n",
-        "76 \n",
-        "86 \n",
-        "43 \n",
-        "32 \n",
-        "21 \n",
-        "Percentage for roll no. is 51.00 \n",
-        "Below average\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.11, Page number: 351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Function declaration, calculation and result\n",
-      "def readdata (a, n) :\n",
-      "\tfor i in range (0, n) :\n",
-      "\t\ta[i] = int(raw_input('Enter data in cell [%d] ' % i))\n",
-      "\treturn a\n",
-      "\n",
-      "def bubble (name) :\n",
-      "\tfor p in range (0, 4) :\n",
-      "\t\tfor i in range (0, 3-p) :\n",
-      "\t\t\tif name[i] > name[i+1] :\n",
-      "\t\t\t\ttemp = name[i]\n",
-      "\t\t\t\tname[i] = name[i+1]\n",
-      "\t\t\t\tname[i+1] = temp\n",
-      "\treturn name\n",
-      "\n",
-      "def printdata (name) :\n",
-      "\tfor i in range (0, 4) :\n",
-      "\t\tprint ('The data in cell [%d] is %s ' % ((i+1), name[i]))\n",
-      "\n",
-      "name = [0] * 4\n",
-      "for i in range (0, 4) :\n",
-      "\tname[i] = raw_input('Enter string[%d] ' % (i+1))\n",
-      "\n",
-      "name = bubble (name)\n",
-      "print ('\\nThe sorted data is ')\n",
-      "printdata (name)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter string[1] zaheer\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter string[2] ashita\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter string[3] tamanna\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter string[4] purti\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The sorted data is \n",
-        "The data in cell [1] is ashita \n",
-        "The data in cell [2] is purti \n",
-        "The data in cell [3] is tamanna \n",
-        "The data in cell [4] is zaheer \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Magnifying_C/README.txt b/Magnifying_C/README.txt
deleted file mode 100755
index 2ec7deb7..00000000
--- a/Magnifying_C/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Mahesh M
-Course: btech
-College/Institute/Organization: MSIT
-Department/Designation: CSE
-Book Title: Magnifying C
-Author: Arpita Gopal
-Publisher: Prentice Hall India Learning Private Limited, New Delhi
-Year of publication: 2009
-Isbn: 9788120338616
-Edition: 1
\ No newline at end of file
diff --git a/Magnifying_C/screenshots/chapter_10.png b/Magnifying_C/screenshots/chapter_10.png
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diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_10_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_10_3.ipynb
new file mode 100755
index 00000000..1a4d1459
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_10_3.ipynb
@@ -0,0 +1,336 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: Miscellany"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1, Page number: 356"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "# Calculation and result\n",
+      "i = 0\n",
+      "for arg in sys.argv :\n",
+      "   print ('arg [%d] - %s' % (i, arg))\n",
+      "   i = i+1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%run 10.1.py hello world"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "arg [0] - 10.1.py\n",
+        "arg [1] - hello\n",
+        "arg [2] - world\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2, Page number: 358"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "if len(sys.argv) != 2 :\n",
+      "\tprint ('Inappropriate number of arguments')\n",
+      "\t\n",
+      "total = 0\n",
+      "def fact (n) :\n",
+      "   total = n\n",
+      "   while n > 1 :\n",
+      "      total = total * (n - 1)\n",
+      "      n = n - 1\n",
+      "   return total\n",
+      "\n",
+      "print ('Factorial of %d is %d ' % (int(sys.argv[1]), fact (int(sys.argv[1]))))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%run 10.2.py 5"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Factorial of 5 is 120 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3, Page number: 361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "# Function declaration\n",
+      "def printdata () :\n",
+      "\ti = 0\n",
+      "\tfor arg in sys.argv[1:] :\n",
+      "\t\tprint ('The data in cell [%d] is %s ' % (i, arg))\n",
+      "\t\ti = i+1\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('The names before sorting are ')\n",
+      "printdata ()\n",
+      "\n",
+      "sys.argv.sort()   \n",
+      "\n",
+      "print ('The names after sorting are ')\n",
+      "printdata ()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%run 10.3.py zimbabwe nigeria argentina china hongkong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The names before sorting are \n",
+        "The data in cell [0] is zimbabwe \n",
+        "The data in cell [1] is nigeria \n",
+        "The data in cell [2] is argentina \n",
+        "The data in cell [3] is china \n",
+        "The data in cell [4] is hongkong \n",
+        "The names after sorting are \n",
+        "The data in cell [0] is argentina \n",
+        "The data in cell [1] is china \n",
+        "The data in cell [2] is hongkong \n",
+        "The data in cell [3] is nigeria \n",
+        "The data in cell [4] is zimbabwe \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4, Page number: 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter any positive number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def convert (a) :\n",
+      "    return int(bin(a)[2:])\n",
+      "\n",
+      "bits = str(convert (a))\n",
+      "print ('Number of bits = %d' % (len(bits)))\n",
+      "print ('Binary equivalent is : %d' % (convert (a)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any positive number 16\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of bits = 5\n",
+        "Binary equivalent is : 10000\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5, Page number: 379"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number : '))\n",
+      "b = int(raw_input('Enter second number : '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "b = ~b + 1\n",
+      "c = a + b\n",
+      "\n",
+      "print ('Subtraction is = %d ' % c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number : 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subtraction is = 4 \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6, Page number: 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number : '))\n",
+      "original_a = a\n",
+      "\n",
+      "b = int(raw_input('Enter second number : '))\n",
+      "original_b = b\n",
+      "\n",
+      "# Calculation and result\n",
+      "c = 1\n",
+      "while (b) :\n",
+      "\tif (b & 1) :\n",
+      "\t\tc = c * a\n",
+      "\tb = b >> 1\n",
+      "\ta = a * a\n",
+      "\n",
+      "print ('%d raise to %d = %d ' % (original_a, original_b, c))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number : 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7 raise to 3 = 343 \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_11_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_11_3.ipynb
new file mode 100755
index 00000000..16b27feb
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_11_3.ipynb
@@ -0,0 +1,1664 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11: Structures and Unions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1, Page number: 388"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student :\n",
+      "\tid_no = ''\n",
+      "\tname = ''\n",
+      "\taddress = ''\n",
+      "\tage = ''\n",
+      "\t\n",
+      "newstudent = student()\n",
+      "\n",
+      "print ('Enter the student information ')\n",
+      "newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
+      "newstudent.name = raw_input('Enter name of the student : ')\n",
+      "newstudent.address = raw_input('Enter the address of the student : ')\n",
+      "newstudent.age = int(raw_input('Enter the age of the student : '))\n",
+      "\n",
+      "print ('You have entered following information...')\n",
+      "print ('Student id_number = %d ' % newstudent.id_no)\n",
+      "print ('Student name = %s ' % newstudent.name)\n",
+      "print ('Student address = %s ' % newstudent.address)\n",
+      "print ('Age of student = %d ' % newstudent.age)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student information \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter student id_no : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of the student : Amanjeet\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the address of the student : Delhi\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age of the student : 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered following information...\n",
+        "Student id_number = 1 \n",
+        "Student name = Amanjeet \n",
+        "Student address = Delhi \n",
+        "Age of student = 20 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page number: 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student :\n",
+      "\tid_no = ''\n",
+      "\tname = ''\n",
+      "\taddress = ''\n",
+      "\tage = ''\n",
+      "\t\n",
+      "newstudent = student()\n",
+      "\n",
+      "def putdata (newstudent) : \n",
+      "   print ('You have entered following information...')\n",
+      "   print ('Student id_number = %d ' % newstudent.id_no)\n",
+      "   print ('Student name = %s ' % newstudent.name)\n",
+      "   print ('Student address = %s ' % newstudent.address)\n",
+      "   print ('Age of student = %d ' % newstudent.age)\n",
+      "    \n",
+      "print ('Enter the student information ')    \n",
+      "newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
+      "newstudent.name = raw_input('Enter name of the student : ')\n",
+      "newstudent.address = raw_input('Enter the address of the student : ')\n",
+      "newstudent.age = int(raw_input('Enter the age of the student : '))\n",
+      "\n",
+      "putdata (newstudent)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student information \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter student id_no : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of the student : Deepti\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the address of the student : Noida\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age of the student : 23\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered following information...\n",
+        "Student id_number = 2 \n",
+        "Student name = Deepti \n",
+        "Student address = Noida \n",
+        "Age of student = 23 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3, Page number: 392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student :\n",
+      "\tid_no = ''\n",
+      "\tname = ''\n",
+      "\taddress = ''\n",
+      "\tage = ''\n",
+      "\t\n",
+      "newstudent = student()\n",
+      "\n",
+      "def putdata (idno, name, add, age) :\n",
+      "   print ('You have entered following information...')\n",
+      "   print ('Student id_number = %d ' % idno)\n",
+      "   print ('Student name = %s ' % name)\n",
+      "   print ('Student address = %s ' % add)\n",
+      "   print ('Age of student = %d ' % age)\n",
+      "    \n",
+      "print ('Enter the student information ')\n",
+      "newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
+      "newstudent.name = raw_input('Enter name of the student : ')\n",
+      "newstudent.address = raw_input('Enter the address of the student : ')\n",
+      "newstudent.age = int(raw_input('Enter the age of the student : '))\n",
+      "\n",
+      "putdata (newstudent.id_no, newstudent.name, newstudent.address, newstudent.age)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student information \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter student id_no : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of the student : Kushagra\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the address of the student : Bangalore\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age of the student : 21\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered following information...\n",
+        "Student id_number = 3 \n",
+        "Student name = Kushagra \n",
+        "Student address = Bangalore \n",
+        "Age of student = 21 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4, Page number: 395"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student :\n",
+      "\tid_no = 0\n",
+      "\tname = 'Unknown'\n",
+      "\taddress = 'Unknown'\n",
+      "\tage = 0\n",
+      "\t\n",
+      "newstudent = student()\n",
+      "\n",
+      "def readdata (newstudent) :\n",
+      "   print ('Enter the student information ')\n",
+      "   newstudent.id_no = int(raw_input('Enter student id_no : '))\n",
+      "   newstudent.name = raw_input('Enter name of the student : ')\n",
+      "   newstudent.address = raw_input('Enter the address of the student : ')\n",
+      "   newstudent.age = int(raw_input('Enter the age of the student : '))\n",
+      "\n",
+      "readdata (newstudent)\n",
+      "print ('You have entered following information...')\n",
+      "print ('Student id_number = %d ' % newstudent.id_no)\n",
+      "print ('Student name = %s ' % newstudent.name)\n",
+      "print ('Student address = %s ' % newstudent.address)\n",
+      "print ('Age of student = %d ' % newstudent.age)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student information \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter student id_no : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of the student : Vipul\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the address of the student : Chandigarh\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age of the student : 25\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered following information...\n",
+        "Student id_number = 4 \n",
+        "Student name = Vipul \n",
+        "Student address = Chandigarh \n",
+        "Age of student = 25 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5, Page number: 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student :\n",
+      "\tid_no = 0\n",
+      "\tname = 'Unknown'\n",
+      "\taddress = 'Unknown'\n",
+      "\tage = 0\n",
+      "\t\n",
+      "def readdata () :\n",
+      "   x = student()\n",
+      "   print ('Enter the student information ')\n",
+      "   x.id_no = int(raw_input('Enter student id_no : '))\n",
+      "   x.name = raw_input('Enter name of the student : ')\n",
+      "   x.address = raw_input('Enter the address of the student : ')\n",
+      "   x.age = int(raw_input('Enter the age of the student : '))\n",
+      "   return x\n",
+      "\n",
+      "newstudent = readdata()\n",
+      "\n",
+      "print ('You have entered following information...')\n",
+      "print ('Student id_number = %d ' % newstudent.id_no)\n",
+      "print ('Student name = %s ' % newstudent.name)\n",
+      "print ('Student address = %s ' % newstudent.address)\n",
+      "print ('Age of student = %d ' % newstudent.age)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student information \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter student id_no : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of the student : Ujjwal\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the address of the student : Jaipur\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age of the student : 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered following information...\n",
+        "Student id_number = 5 \n",
+        "Student name = Ujjwal \n",
+        "Student address = Jaipur \n",
+        "Age of student = 20 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6, Page number: 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student :\n",
+      "\tid_no = 0\n",
+      "\tname = 'Unknown'\n",
+      "\taddress = 'Unknown'\n",
+      "\tage = 0\n",
+      "\t\n",
+      "x = student()\n",
+      "\n",
+      "def readdata (x) :\n",
+      "   print ('Enter the student information ')\n",
+      "   x.id_no = int(raw_input('Enter student id_no : '))\n",
+      "   x.name = raw_input('Enter name of the student : ')\n",
+      "   x.address = raw_input('Enter the address of the student : ')\n",
+      "   x.age = int(raw_input('Enter the age of the student : '))\n",
+      "\n",
+      "readdata(x)    \n",
+      "print ('You have entered following information...')\n",
+      "print ('Student id_number = %d ' % x.id_no)\n",
+      "print ('Student name = %s ' % x.name)\n",
+      "print ('Student address = %s ' % x.address)\n",
+      "print ('Age of student = %d ' % x.age)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the student information \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter student id_no : 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of the student : Karan\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the address of the student : Chennai\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age of the student : 24\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered following information...\n",
+        "Student id_number = 6 \n",
+        "Student name = Karan \n",
+        "Student address = Chennai \n",
+        "Age of student = 24 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7, Page number: 403"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student (object) :\n",
+      "\tdef __init__(self, roll=None, name=None, gender=None, marks=[0]*3) :\n",
+      "\t\tself.roll = roll\n",
+      "\t\tself.name = name\n",
+      "\t\tself.gender = gender\n",
+      "\t\tself.marks = marks\n",
+      "\n",
+      "def putdata (stud, count) :\n",
+      "\tprint ('RollNo\\t   Name\\t\\t   Gender\\t   Sub1\\t      Sub2\\t Sub3 ')\n",
+      "\tprint ('---------------------------------------------------------------------')\n",
+      "\t\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tprint ('%-10d' % stud[i].roll),\n",
+      "\t\tprint ('%-15s' % stud[i].name),\n",
+      "\t\t\n",
+      "\t\tif stud[i].gender == 0 :\n",
+      "\t\t\tprint ('%-15s' % 'Female'),\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('%-15s' % 'Male'),\n",
+      "\t\t\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tprint ('%-10d' % stud[i].marks[j]),\n",
+      "\t\tprint\n",
+      "\n",
+      "def calculateper (stud, count) :\n",
+      "\taverage = 0\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tsum = 0\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tsum = sum + stud[i].marks[j]\n",
+      "\t\taverage = average + sum/3\n",
+      "\treturn average/count\n",
+      "\n",
+      "stud = []\t\n",
+      "stud.append (student (1, 'Kapildev', 1, (45, 56, 67)))\n",
+      "stud.append (student (2, 'Vidya', 0, (87, 45, 23)))\n",
+      "stud.append (student (3, 'Sharada', 0, (67, 94, 56)))\n",
+      "stud.append (student (4, 'Saismaran', 1, (56, 38, 84)))\n",
+      "stud.append (student (5, 'Reshma', 0, (67, 98, 76)))\n",
+      "\n",
+      "average_percentage = calculateper (stud, 5)\n",
+      "putdata (stud, 5)\n",
+      "print ('\\nAverage percentage of the class is : %.2f ' % average_percentage)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RollNo\t   Name\t\t   Gender\t   Sub1\t      Sub2\t Sub3 \n",
+        "---------------------------------------------------------------------\n",
+        "1          Kapildev        Male            45         56         67        \n",
+        "2          Vidya           Female          87         45         23        \n",
+        "3          Sharada         Female          67         94         56        \n",
+        "4          Saismaran       Male            56         38         84        \n",
+        "5          Reshma          Female          67         98         76        \n",
+        "\n",
+        "Average percentage of the class is : 63.00 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8, Page number: 406"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student (object) :\n",
+      "\tdef __init__(self, roll=None, name=None, gender=None, marks=[0]*3) :\n",
+      "\t\tself.roll = roll\n",
+      "\t\tself.name = name\n",
+      "\t\tself.gender = gender\n",
+      "\t\tself.marks = marks\n",
+      "\t\n",
+      "def putdata (stud, count) :\n",
+      "\tprint ('\\nRollNo\\t   Name\\t\\t   Gender\\t   Sub1\\t      Sub2\\t Sub3 ')\n",
+      "\tprint ('---------------------------------------------------------------------')\n",
+      "\t\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tprint ('%-10d' % stud[i].roll),\n",
+      "\t\tprint ('%-15s' % stud[i].name),\n",
+      "\t\t\n",
+      "\t\tif stud[i].gender == 0 :\n",
+      "\t\t\tprint ('%-15s' % 'Female'),\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('%-15s' % 'Male'),\n",
+      "\t\t\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tprint ('%-10d' % stud[i].marks[j]),\n",
+      "\t\tprint\n",
+      "\n",
+      "def calculateper (stud, count) :\n",
+      "\taverage = 0\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tsum = 0\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tsum = sum + stud[i].marks[j]\n",
+      "\t\taverage = average + sum/3\n",
+      "\treturn average/count\n",
+      "\n",
+      "count = int(raw_input('Enter the number of students in a class : '))\n",
+      "stud = [student() for i in range (0, count)]\n",
+      "\n",
+      "for i in range (0, count) :\n",
+      "\tstud[i].roll = int(raw_input('\\nEnter roll number of student %d: ' % (i+1)))\n",
+      "\tstud[i].name = raw_input('Enter name of student %d: ' % (i+1))\n",
+      "\tstud[i].gender = int(raw_input('Enter gender[0/1] of student %d: ' % (i+1)))\n",
+      "\tprint ('Enter marks of three subjects of student %d: ' % (i+1))\n",
+      "\tfor j in range (0, 3) :\n",
+      "\t\tstud[i].marks[j] = int(raw_input('Subject %d: ' % (j+1)))\n",
+      "\n",
+      "putdata (stud, count)\n",
+      "average_percentage = calculateper (stud, count)\n",
+      "print ('\\nAverage percentage of the class is : %.2f ' % average_percentage)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of students in a class : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter roll number of student 1: 24\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of student 1: Vidya\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter gender[0/1] of student 1: 0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter marks of three subjects of student 1: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1: 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2: 96\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3: 75\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "RollNo\t   Name\t\t   Gender\t   Sub1\t      Sub2\t Sub3 \n",
+        "---------------------------------------------------------------------\n",
+        "24         Vidya           Female          78         96         75        \n",
+        "\n",
+        "Average percentage of the class is : 83.00 \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9, Page number: 408"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class student (object) :\n",
+      "\tdef __init__(self, roll=None, name=None, gender=None, marks=[0]*3) :\n",
+      "\t\tself.roll = roll\n",
+      "\t\tself.name = name\n",
+      "\t\tself.gender = gender\n",
+      "\t\tself.marks = marks\n",
+      "\n",
+      "def readdata (stud, count) :\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tstud[i].roll = int(raw_input('\\nEnter roll number of student %d: ' % (i+1)))\n",
+      "\t\tstud[i].name = raw_input('Enter name of student %d: ' % (i+1))\n",
+      "\t\tstud[i].gender = int(raw_input('Enter gender[0/1] of student %d: ' % (i+1)))\n",
+      "\t\tprint ('Enter marks of three subjects of student %d: ' % (i+1))\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tstud[i].marks[j] = int(raw_input('Subject %d: ' % (j+1)))\n",
+      "\t\n",
+      "def putdata (stud, count) :\n",
+      "\tprint ('\\nRollNo\\t   Name\\t\\t   Gender\\t   Sub1\\t      Sub2\\t Sub3 ')\n",
+      "\tprint ('---------------------------------------------------------------------')\n",
+      "\t\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tprint ('%-10d' % stud[i].roll),\n",
+      "\t\tprint ('%-15s' % stud[i].name),\n",
+      "\t\t\n",
+      "\t\tif stud[i].gender == 0 :\n",
+      "\t\t\tprint ('%-15s' % 'Female'),\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('%-15s' % 'Male'),\n",
+      "\t\t\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tprint ('%-10d' % stud[i].marks[j]),\n",
+      "\t\tprint\n",
+      "\n",
+      "def calculateper (stud, count) :\n",
+      "\taverage = 0\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tsum = 0\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tsum = sum + stud[i].marks[j]\n",
+      "\t\taverage = average + sum/3\n",
+      "\treturn average/count\n",
+      "\n",
+      "count = int(raw_input('Enter the number of students in a class : '))\n",
+      "stud = [student() for i in range (0, count)]\n",
+      "readdata (stud, count)\n",
+      "\n",
+      "average_percentage = calculateper (stud, count)\n",
+      "putdata (stud, count)\n",
+      "print ('\\nAverage percentage of the class is : %.2f ' % average_percentage)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of students in a class : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter roll number of student 1: 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of student 1: Kapildev\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter gender[0/1] of student 1: 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter marks of three subjects of student 1: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1: 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2: 70\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3: 93\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "RollNo\t   Name\t\t   Gender\t   Sub1\t      Sub2\t Sub3 \n",
+        "---------------------------------------------------------------------\n",
+        "25         Kapildev        Male            45         70         93        \n",
+        "\n",
+        "Average percentage of the class is : 69.00 \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.10, Page number: 412"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class time (object) :\n",
+      "\tdef __init__(self, hours=None, minutes=None, seconds=None) :\n",
+      "\t\tself.hours = hours\n",
+      "\t\tself.minutes = minutes\n",
+      "\t\tself.seconds = seconds\n",
+      "\n",
+      "def AcceptTime (t) :\n",
+      "\tprint ('Enter the time in hh mm ss format: ')\n",
+      "\tt.hours = int(raw_input('Hours = '))\n",
+      "\tt.minutes = int(raw_input('Minutes = '))\n",
+      "\tt.seconds = int(raw_input('Seconds = '))\n",
+      "\n",
+      "def ValidateTime (t) :\n",
+      "\tif t.hours < 0 or t.hours > 23 :\n",
+      "\t\treturn 0\n",
+      "\tif t.minutes < 0 or t.minutes > 60 :\n",
+      "\t\treturn 0\n",
+      "\tif t.seconds < 0 or t.seconds > 60 :\n",
+      "\t\treturn 0\n",
+      "\treturn 1\n",
+      "\n",
+      "def DisplayTime (t) :\n",
+      "\tprint ('Time you entered is %d:%d:%d ' % (t.hours, t.minutes, t.seconds))\n",
+      "\t\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\tt = time ()\n",
+      "\tAcceptTime (t)\n",
+      "\t\n",
+      "\tif ValidateTime (t) == 1 :\n",
+      "\t\tcondition = False\n",
+      "\telse :\n",
+      "\t\tprint ('\\nTime is incorrect. Enter again ')\n",
+      "\t\t\n",
+      "DisplayTime (t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the time in hh mm ss format: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hours = 13\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minutes = 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Seconds = 53\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time you entered is 13:45:53 \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.11, Page number: 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class date (object) :\n",
+      "\tdef __init__(self, dd=None, mm=None, yy=None) :\n",
+      "\t\tself.dd = dd\n",
+      "\t\tself.mm = mm\n",
+      "\t\tself.yy = yy\n",
+      "\n",
+      "def AcceptDate (d) :\n",
+      "\tprint ('Enter date in dd mm yy format: ')\n",
+      "\td.dd = int(raw_input('Date = '))\n",
+      "\td.mm = int(raw_input('Month = '))\n",
+      "\td.yy = int(raw_input('Year = '))\n",
+      "\n",
+      "def isleap (val) :\n",
+      "\tif val % 4 == 0 and val % 100 or val % 400 == 0 :\n",
+      "\t\treturn 1\n",
+      "\treturn 0\n",
+      "\t\n",
+      "def ValidateDate (d) :\n",
+      "\tif d.dd < 1 or d.dd > 31 :\n",
+      "\t\treturn 0\n",
+      "\tif d.mm < 1 or d.mm > 12 :\n",
+      "\t\treturn 0\n",
+      "\tif d.yy < 1 :\n",
+      "\t\treturn 0\n",
+      "\tif d.mm == 2 :\n",
+      "\t\tif isleap (d.yy) :\n",
+      "\t\t\tif d.dd > 29 :\n",
+      "\t\t\t\treturn 0\n",
+      "\t\telif d.dd > 28 :\n",
+      "\t\t\treturn 0\n",
+      "\tif d.mm == 4 or d.mm == 6 or d.mm == 9 or d.mm == 11 and d.dd > 30 :\n",
+      "\t\treturn 0\n",
+      "\treturn 1\n",
+      "\n",
+      "def DisplayMsg (d1, d2) :\n",
+      "\tif d1.dd == d2.dd and d1.mm == d2.mm and d1.yy == d2.yy :\n",
+      "\t\tprint ('Both dates are Equal ')\n",
+      "\telse :\n",
+      "\t\tprint ('Both dates are Not Equal ')\n",
+      "\t\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\td1 = date ()\n",
+      "\tAcceptDate (d1)\n",
+      "\t\n",
+      "\tif ValidateDate (d1) == 1 :\n",
+      "\t\tcondition = False\n",
+      "\telse :\n",
+      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
+      "\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\td2 = date ()\n",
+      "\tAcceptDate (d2)\n",
+      "\t\n",
+      "\tif ValidateDate (d2) == 1 :\n",
+      "\t\tcondition = False\n",
+      "\telse :\n",
+      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
+      "\t\t\n",
+      "DisplayMsg (d1, d2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter date in dd mm yy format: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Date = 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Month = 11\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Year = 92\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter date in dd mm yy format: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Date = 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Month = 11\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Year = 92\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Both dates are Equal \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.12, Page number: 415"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class date (object) :\n",
+      "\tdef __init__(self, dd=None, mm=None, yy=None) :\n",
+      "\t\tself.dd = dd\n",
+      "\t\tself.mm = mm\n",
+      "\t\tself.yy = yy\n",
+      "\n",
+      "def AcceptDate (d) :\n",
+      "\tprint ('Enter date in dd mm yy format: ')\n",
+      "\td.dd = int(raw_input('Date = '))\n",
+      "\td.mm = int(raw_input('Month = '))\n",
+      "\td.yy = int(raw_input('Year = '))\n",
+      "\n",
+      "def isleap (val) :\n",
+      "\tif val % 4 == 0 and val % 100 or val % 400 == 0 :\n",
+      "\t\treturn 1\n",
+      "\treturn 0\n",
+      "\t\n",
+      "def ValidateDate (d) :\n",
+      "\tif d.dd < 1 or d.dd > 31 :\n",
+      "\t\treturn 0\n",
+      "\tif d.mm < 1 or d.mm > 12 :\n",
+      "\t\treturn 0\n",
+      "\tif d.yy < 1 :\n",
+      "\t\treturn 0\n",
+      "\tif d.mm == 2 :\n",
+      "\t\tif isleap (d.yy) :\n",
+      "\t\t\tif d.dd > 29 :\n",
+      "\t\t\t\treturn 0\n",
+      "\t\telif d.dd > 28 :\n",
+      "\t\t\treturn 0\n",
+      "\tif d.mm == 4 or d.mm == 6 or d.mm == 9 or d.mm == 11 and d.dd > 30 :\n",
+      "\t\treturn 0\n",
+      "\treturn 1\n",
+      "\n",
+      "def CalcDate (d1, d2) :\n",
+      "\tdate1 = date2 = 0\n",
+      "\tfor i in range (-d1.yy, d1.yy) :\n",
+      "\t\tdate1 = date1 + 366 if isleap (d1.yy) else 365\n",
+      "\t\td1.yy = d1.yy - 1\n",
+      "\t\t\n",
+      "\tfor i in range (-d1.mm, d1.mm) :\n",
+      "\t\tdate1 = date1 + (30 if (d1.mm==4 or d1.mm==6 or d1.mm==9 or d1.mm==11) else ((29 if isleap(d1.yy) else 28) if (d1.mm==2) else 31))\n",
+      "\t\n",
+      "\tdate1 = date1 + d1.dd\n",
+      "\t\n",
+      "\tfor i in range (-d2.yy, d2.yy) :\n",
+      "\t\tdate2 = date2 + 366 if isleap(d2.yy) else 365\n",
+      "\t\n",
+      "\tfor i in range (-d2.mm, d2.mm) :\n",
+      "\t\tdate2 = date2 + (30 if (d2.mm==4 or d2.mm==6 or d2.mm==9 or d2.mm==11) else ((29 if isleap(d2.yy) else 28) if d2.mm==2 else 31))\n",
+      "\t\n",
+      "\tdate2 = date2 + d2.dd\n",
+      "\t\n",
+      "\treturn (date2 - date1)/2\n",
+      "\t\t\n",
+      "def DisplayMsg (d1, d2, diff) :\n",
+      "\tprint ('The difference between %d %d %d and %d %d %d is %d days ' % (d1.dd, d1.mm, -d1.yy, d2.dd, d2.mm, d2.yy, diff))\n",
+      "\t\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\tprint ('\\nEnter first date: ')\n",
+      "\td1 = date ()\n",
+      "\tAcceptDate (d1)\n",
+      "\t\n",
+      "\tif ValidateDate (d1) == 1 :\n",
+      "\t\tcondition = False\n",
+      "\telse :\n",
+      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
+      "\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\tprint ('\\nEnter second date: ')\n",
+      "\td2 = date ()\n",
+      "\tAcceptDate (d2)\n",
+      "\t\n",
+      "\tif ValidateDate (d2) == 1 :\n",
+      "\t\tcondition = False\n",
+      "\telse :\n",
+      "\t\tprint ('\\nDate is incorrect. Enter again ')\n",
+      "\n",
+      "diff = CalcDate (d1, d2)\n",
+      "DisplayMsg (d1, d2, diff)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter first date: \n",
+        "Enter date in dd mm yy format: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Date = 01\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Month = 01\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Year = 01\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter second date: \n",
+        "Enter date in dd mm yy format: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Date = 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Month = 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Year = 08\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The difference between 1 1 1 and 25 12 8 is 2915 days \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.13, Page number: 419"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from ctypes import *\n",
+      "\n",
+      "# Structure declaration\n",
+      "class stream (Union) :\n",
+      "\t_fields_ = [('grade', c_char), \n",
+      "\t\t\t\t\t('class_std', c_int), \n",
+      "\t\t\t\t\t('percentage', c_float)]\n",
+      "\t\n",
+      "class student (Structure) :\n",
+      "\t_fields_ = [('roll', c_int), \n",
+      "\t\t\t\t\t('name', c_char_p), \n",
+      "\t\t\t\t\t('gender', c_int), \n",
+      "\t\t\t\t\t('marks', c_int * 3), \n",
+      "\t\t\t\t\t('st_stream', c_char), \n",
+      "\t\t\t\t\t('st', stream)]\n",
+      "\t\n",
+      "def putdata (stud, count) :\n",
+      "\tprint ('\\nStream\\t   Roll\\t  Name\\t        Gender\\t  Result\\t ')\n",
+      "\tprint ('---------------------------------------------------------------------')\n",
+      "\t\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tif stud[i].st_stream == 'A' :\n",
+      "\t\t\tprint ('Arts\\t  '),\n",
+      "\t\telif stud[i].st_stream == 'S' :\n",
+      "\t\t\tprint ('Science\\t  '),\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('Commerce  '),\n",
+      "\t\t\t\n",
+      "\t\tprint ('%-5d ' % stud[i].roll),\n",
+      "\t\tprint ('%-3s ' % stud[i].name),\n",
+      "\t\t\n",
+      "\t\tif stud[i].gender == 0 :\n",
+      "\t\t\tprint ('\\tFemale'),\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('\\tMale  '),\n",
+      "\t\t\n",
+      "\t\tif stud[i].st_stream == 'A' :\n",
+      "\t\t\tprint ('   %d Class ' % stud[i].st.class_std),\n",
+      "\t\telif stud[i].st_stream == 'S' :\n",
+      "\t\t\tprint ('   %c Grade ' % stud[i].st.grade),\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('   %.2f Percent ' % stud[i].st.percentage),\n",
+      "\t\tprint\n",
+      "\n",
+      "def calculateper (stud, count) :\n",
+      "\taverage = 0\n",
+      "\tfor i in range (0, count) :\n",
+      "\t\tsum = 0\n",
+      "\t\tfor j in range (0, 3) :\n",
+      "\t\t\tsum = sum + stud[i].marks[j]\n",
+      "\t\taverage = (float)(sum/3)\n",
+      "\t\t\n",
+      "\t\tif stud[i].st_stream == 'A' :\n",
+      "\t\t\tif average >= 60 :\n",
+      "\t\t\t\tstud[i].st.class_std = 1\n",
+      "\t\t\telse :\n",
+      "\t\t\t\tstud[i].st.class_std = 2\n",
+      "\t\t\t\t\n",
+      "\t\telif stud[i].st_stream == 'S' :\n",
+      "\t\t\tif average >= 60 :\n",
+      "\t\t\t\tstud[i].st.grade = 'A'\n",
+      "\t\t\telse :\n",
+      "\t\t\t\tstud[i].st.class_std = 'B'\n",
+      "\t\telse : \n",
+      "\t\t\tstud[i].st.percentage = average\n",
+      "\n",
+      "count = int(raw_input('Enter the number of students in a class : '))\n",
+      "stud = [student() for i in range (0, count)]\n",
+      "\n",
+      "for i in range (0, count) :\n",
+      "\tstud[i].st_stream = raw_input('\\nEnter stream of student %d (A/C/S) : ' % (i+1))\n",
+      "\tstud[i].roll = int(raw_input('Enter roll number of student %d : ' % (i+1)))\n",
+      "\tstud[i].name = raw_input('Enter name of student %d : ' % (i+1))\n",
+      "\tstud[i].gender = int(raw_input('Enter gender of student %d [0/1] : ' % (i+1)))\n",
+      "\tprint ('Enter marks of three subjects of student %d : ' % (i+1))\n",
+      "\tfor j in range (0, 3) :\n",
+      "\t\tstud[i].marks[j] = int(raw_input('Subject %d : ' % (j+1)))\n",
+      "\t\n",
+      "calculateper (stud, count)\n",
+      "putdata (stud, count)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of students in a class : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter stream of student 1 (A/C/S) : A\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter roll number of student 1 : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of student 1 : Saismaran\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter gender of student 1 [0/1] : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter marks of three subjects of student 1 : \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1 : 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2 : 83\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3 : 59\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter stream of student 2 (A/C/S) : C\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter roll number of student 2 : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of student 2 : Vidya\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter gender of student 2 [0/1] : 0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter marks of three subjects of student 2 : \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1 : 84\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2 : 75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3 : 93\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter stream of student 3 (A/C/S) : S\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter roll number of student 3 : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name of student 3 : Pranav\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter gender of student 3 [0/1] : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter marks of three subjects of student 3 : \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1 : 84\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2 : 95\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3 : 83\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Stream\t   Roll\t  Name\t        Gender\t  Result\t \n",
+        "---------------------------------------------------------------------\n",
+        "Arts\t   1      Saismaran  \tMale      1 Class \n",
+        "Commerce   2      Vidya  \tFemale    84.00 Percent \n",
+        "Science\t   3      Pranav  \tMale      A Grade \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_12_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_12_3.ipynb
new file mode 100755
index 00000000..4fec3405
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_12_3.ipynb
@@ -0,0 +1,2137 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Files"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1, Page number: 428"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "fp = open('12.1.dat', 'w')\n",
+      "\n",
+      "for i in range (0, 11) :\n",
+      "\tfp.write('%d, %d \\n' % (i, i*i))\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2, Page number: 429"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "fp = open('12.2.dat', 'w')\n",
+      "\n",
+      "for i in range (0, 11) :\n",
+      "\tfp.write('%d, %d \\n' % (i, i*i))\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3, Page number: 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "fp = open('12.2.dat', 'a')\n",
+      "i = 11\n",
+      "fp.write('%d, %d \\n' % (i, i*i))\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4, Page number: 432"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "with open ('12.2.dat') as fp:\n",
+      "\tfor line in fp :\n",
+      "\t\tprint line\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0, 0 \n",
+        "\n",
+        "1, 1 \n",
+        "\n",
+        "2, 4 \n",
+        "\n",
+        "3, 9 \n",
+        "\n",
+        "4, 16 \n",
+        "\n",
+        "5, 25 \n",
+        "\n",
+        "6, 36 \n",
+        "\n",
+        "7, 49 \n",
+        "\n",
+        "8, 64 \n",
+        "\n",
+        "9, 81 \n",
+        "\n",
+        "10, 100 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5, Page number: 434"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "print ('Data input ')\n",
+      "f1 = open('12.5.dat', 'w')\n",
+      "f1.write('You are learning Files in Python language ')\n",
+      "f1.close()\n",
+      "\n",
+      "print ('Data output ')\n",
+      "with open ('12.5.dat') as f1:\n",
+      "\tfor line in f1 :\n",
+      "\t\tprint line\n",
+      "f1.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Data input \n",
+        "Data output \n",
+        "You are learning Files in Python language \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6, Page number: 435"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "print ('Contents of the data file ')\n",
+      "f1 = open('12.6.dat', 'w')\n",
+      "for i in range (0, 10) :\n",
+      "\tnumber = int(raw_input())\n",
+      "\tf1.write('%d\\n' % number)\n",
+      "f1.close()\n",
+      "\n",
+      "f1 = open('12.6.dat', 'r')\n",
+      "f2 = open('12.6o.dat', 'w')\n",
+      "f3 = open('12.6e.dat', 'w')\n",
+      "\n",
+      "num = [0] * 10\n",
+      "with open ('12.6.dat') as f1 :\n",
+      "\tnum = f1.read().splitlines()\n",
+      "\t\n",
+      "for i in range (0, 10) :\n",
+      "\tif (int(num[i]) % 2) != 0 :\n",
+      "\t\tf2.write('%s \\n' % (num[i]))\n",
+      "\telse :\n",
+      "\t\tf3.write('%s \\n' % (num[i]))\n",
+      "\t\n",
+      "f1.close()\n",
+      "f2.close()\n",
+      "f3.close()\n",
+      "\n",
+      "print ('Contents of the odd file ')\n",
+      "with open ('12.6o.dat') as f2:\n",
+      "\tfor line in f2 :\n",
+      "\t\tprint line\n",
+      "\n",
+      "print ('Contents of the even file ')\n",
+      "with open ('12.6e.dat') as f3:\n",
+      "\tfor line in f3 :\n",
+      "\t\tprint line\n",
+      "\n",
+      "f2.close()\n",
+      "f3.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Contents of the data file \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Contents of the odd file \n",
+        "1 \n",
+        "\n",
+        "3 \n",
+        "\n",
+        "5 \n",
+        "\n",
+        "7 \n",
+        "\n",
+        "9 \n",
+        "\n",
+        "Contents of the even file \n",
+        "2 \n",
+        "\n",
+        "4 \n",
+        "\n",
+        "6 \n",
+        "\n",
+        "8 \n",
+        "\n",
+        "10 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7, Page number: 438"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "filename = raw_input('Input filename: ')\n",
+      "fp = open(filename, 'w')\n",
+      "\n",
+      "print ('Input inventory data ')\n",
+      "print ('\\nItemName number price quantity ')\n",
+      "\n",
+      "data = [[0 for y in xrange(5)] for x in xrange(3)]\n",
+      "for i in range (0, 3) :\n",
+      "\tfor j in range (0, 4) :\n",
+      "\t\tdata[i][j] = raw_input('')\n",
+      "\t\tfp.write('%s  ' % data[i][j]),\n",
+      "\tfp.write('\\n')\n",
+      "\tprint\n",
+      "fp.close()\n",
+      "\n",
+      "fp = open(filename, 'r')\n",
+      "\n",
+      "for i in range (0, 3) :\n",
+      "\tfor j in range (2, 3) :\n",
+      "\t\tdata[i][j+2] = float(data[i][j]) * int(data[i][j+1])\n",
+      "\n",
+      "print ('ItemName   number   price   quantity   value ')\n",
+      "for i in range (0, 3) :\n",
+      "\tfor j in range (0, 5) :\n",
+      "\t\tprint ('%8s' % data[i][j]),\n",
+      "\tprint"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Input filename: 12.7.dat\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Input inventory data \n",
+        "\n",
+        "ItemName number price quantity \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Monitor\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "11\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3005.50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mouse\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "22\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "500.50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Keyboard\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "33\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "750.00\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "ItemName   number   price   quantity   value \n",
+        " Monitor       11  3005.50       10  30055.0\n",
+        "   Mouse       22   500.50        5   2502.5\n",
+        "Keyboard       33   750.00       10   7500.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8, Page number: 440"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "with open ('12.5.dat') as fp:\n",
+      "\tfor line in fp :\n",
+      "\t\tprint line\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You are learning Files in Python language \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9, Page number: 441"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sys import argv\n",
+      "\n",
+      "# Calculation and result\n",
+      "script, filename = argv\n",
+      "file = open(filename)\n",
+      "print file.read()\n",
+      "print ('\\nFile displayed successfully.....')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%run 12.9.py 12.1.dat"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0, 0 \n",
+        "1, 1 \n",
+        "2, 4 \n",
+        "3, 9 \n",
+        "4, 16 \n",
+        "5, 25 \n",
+        "6, 36 \n",
+        "7, 49 \n",
+        "8, 64 \n",
+        "9, 81 \n",
+        "10, 100 \n",
+        "\n",
+        "\n",
+        "File displayed successfully.....\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.10, Page number: 444"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "class person (object) :\n",
+      "\tdef __init__(self, name = None, age = 0):\n",
+      "\t\tself.name = 0\n",
+      "\t\tself.age = 0\n",
+      "\t\t\n",
+      "data = [person() for i in range (5)]\n",
+      "\n",
+      "fp = open('12.10.dat','w')\n",
+      "\n",
+      "for i in range (0, 5) :\n",
+      "\tdata[i].name = raw_input('\\nEnter name : ')\n",
+      "\tdata[i].age = int(raw_input('Enter age : '))\n",
+      "\t\n",
+      "for i in range (0, 5) :\n",
+      "\tfp.write('%s %d \\n' % (data[i].name, data[i].age))\n",
+      "\n",
+      "print ('\\n...Records are written to a file...')\n",
+      "print ('...Displaying records from a file...')\n",
+      "\n",
+      "for i in range (0, 5) :\n",
+      "\tprint ('\\nName : %s ' % data[i].name)\n",
+      "\tprint ('Age : %d ' % data[i].age)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter name : Dhiraj Kumar\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 23\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter name : Mohasin Khan\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter name : Suresh Waghmare\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 55\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter name : Anand Rao\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter name : Sunil Deshpande\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 23\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "...Records are written to a file...\n",
+        "...Displaying records from a file...\n",
+        "\n",
+        "Name : Dhiraj Kumar \n",
+        "Age : 23 \n",
+        "\n",
+        "Name : Mohasin Khan \n",
+        "Age : 45 \n",
+        "\n",
+        "Name : Suresh Waghmare \n",
+        "Age : 55 \n",
+        "\n",
+        "Name : Anand Rao \n",
+        "Age : 34 \n",
+        "\n",
+        "Name : Sunil Deshpande \n",
+        "Age : 23 \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.11, Page number: 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "class record (object) :\n",
+      "\tdef __init__(self, name = None, age = 0) :\n",
+      "\t\tself.name = 0\n",
+      "\t\tself.age = 0\n",
+      "\t\t\n",
+      "def record_write (data, filename) :\n",
+      "\tfilep = open(filename, 'ab')\n",
+      "\tdata.name = raw_input('Enter name : ')\n",
+      "\tdata.age = int(raw_input('Enter age : '))\n",
+      "\tfilep.write('%s %d\\n' % (data.name, data.age))\n",
+      "\t\n",
+      "def record_readall (filename) :\n",
+      "\tprint\n",
+      "\twith open (filename) as filep :\n",
+      "\t\tfor line in filep :\n",
+      "\t\t\tprint line\n",
+      "\n",
+      "data = record ()\n",
+      "filename = '12.11.dat'\n",
+      "\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\tenter = raw_input('\\nAdd record (y/n) ? ')\n",
+      "\tif enter == 'y' or enter == 'Y' :\n",
+      "\t\trecord_write (data, filename)\n",
+      "\telse :\n",
+      "\t\tcondition = False\n",
+      "\t\tbreak\n",
+      "\n",
+      "record_readall (filename)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Add record (y/n) ? y\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name : Mohan Pande\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Add record (y/n) ? y\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter name : Nitin Dighe\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Add record (y/n) ? n\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Mohan Pande 34\n",
+        "\n",
+        "Nitin Dighe 32\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.12, Page number: 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "fp = open('12.12.txt','r')\n",
+      "fp.seek(0, 0)\n",
+      "line = fp.readline()\n",
+      "print ('Read Line: %s ' % line)\n",
+      "\n",
+      "fp.seek(17, 0)\n",
+      "line = fp.readline()\n",
+      "print ('Read Line: %s ' % line)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read Line: This is 1st line\n",
+        " \n",
+        "Read Line: This is 2nd line\n",
+        " \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.13, Page number: 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import os\n",
+      "\n",
+      "# Calculation and result\n",
+      "statinfo = os.stat('12.13.txt')\n",
+      "print ('Filesize of \"12.13.txt\" is %d bytes ' % statinfo.st_size)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Filesize of \"12.13.txt\" is 15 bytes \n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.14, Page number: 455"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "nc = nlines = 0\n",
+      "\n",
+      "filename = raw_input('Enter file name: ')\n",
+      "with open(filename, 'r') as fp :\n",
+      "    for line in fp :\n",
+      "        nlines += 1\n",
+      "        nc += len(line)\n",
+      "fp.close()\n",
+      "\n",
+      "print ('There are %d characters in %s ' % (nc, filename))\n",
+      "print ('There are %d lines ' % nlines)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter file name: 12.14.txt\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "There are 55 characters in 12.14.txt \n",
+        "There are 1 lines \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.15, Page number: 456"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "linecount = 0\n",
+      "filename = raw_input('Enter file name: ')\n",
+      "fp = open(filename, 'r')\n",
+      "for line in fp :\n",
+      "\tprint line\n",
+      "\tlinecount += 1\n",
+      "\tif linecount % 20 == 0 :\n",
+      "\t\traw_input(\"[Press Return to continue, Q to quit]\")\n",
+      "\t\tif raw_input() == 'Q' :\n",
+      "\t\t\texit()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter file name: 12.15.txt\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n",
+        "\n",
+        "2\n",
+        "\n",
+        "3\n",
+        "\n",
+        "4\n",
+        "\n",
+        "5\n",
+        "\n",
+        "6\n",
+        "\n",
+        "7\n",
+        "\n",
+        "8\n",
+        "\n",
+        "9\n",
+        "\n",
+        "10\n",
+        "\n",
+        "11\n",
+        "\n",
+        "12\n",
+        "\n",
+        "13\n",
+        "\n",
+        "14\n",
+        "\n",
+        "15\n",
+        "\n",
+        "16\n",
+        "\n",
+        "17\n",
+        "\n",
+        "18\n",
+        "\n",
+        "19\n",
+        "\n",
+        "20\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[Press Return to continue, Q to quit]Q\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "21\n",
+        "\n",
+        "22\n",
+        "\n",
+        "23\n",
+        "\n",
+        "24\n",
+        "\n",
+        "25\n",
+        "\n",
+        "26\n",
+        "\n",
+        "27\n",
+        "\n",
+        "28\n",
+        "\n",
+        "29\n",
+        "\n",
+        "30\n",
+        "\n",
+        "31\n",
+        "\n",
+        "32\n",
+        "\n",
+        "33\n",
+        "\n",
+        "34\n",
+        "\n",
+        "35\n",
+        "\n",
+        "36\n",
+        "\n",
+        "37\n",
+        "\n",
+        "38\n",
+        "\n",
+        "39\n",
+        "\n",
+        "40\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[Press Return to continue, Q to quit]\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "41\n",
+        "\n",
+        "42\n",
+        "\n",
+        "43\n",
+        "\n",
+        "44\n",
+        "\n",
+        "45\n",
+        "\n",
+        "46\n",
+        "\n",
+        "47\n",
+        "\n",
+        "48\n",
+        "\n",
+        "49\n",
+        "\n",
+        "50\n",
+        "\n",
+        "51\n",
+        "\n",
+        "52\n",
+        "\n",
+        "53\n",
+        "\n",
+        "54\n",
+        "\n",
+        "55\n",
+        "\n",
+        "56\n",
+        "\n",
+        "57\n",
+        "\n",
+        "58\n",
+        "\n",
+        "59\n",
+        "\n",
+        "60\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[Press Return to continue, Q to quit]\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "61\n",
+        "\n",
+        "62\n",
+        "\n",
+        "63\n",
+        "\n",
+        "64\n",
+        "\n",
+        "65\n",
+        "\n",
+        "66\n",
+        "\n",
+        "67\n",
+        "\n",
+        "68\n",
+        "\n",
+        "69\n",
+        "\n",
+        "70\n",
+        "\n",
+        "71\n",
+        "\n",
+        "72\n",
+        "\n",
+        "73\n",
+        "\n",
+        "74\n",
+        "\n",
+        "75\n",
+        "\n",
+        "76\n",
+        "\n",
+        "77\n",
+        "\n",
+        "78\n",
+        "\n",
+        "79\n",
+        "\n",
+        "80\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[Press Return to continue, Q to quit]\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "81\n",
+        "\n",
+        "82\n",
+        "\n",
+        "83\n",
+        "\n",
+        "84\n",
+        "\n",
+        "85\n",
+        "\n",
+        "86\n",
+        "\n",
+        "87\n",
+        "\n",
+        "88\n",
+        "\n",
+        "89\n",
+        "\n",
+        "90\n",
+        "\n",
+        "91\n",
+        "\n",
+        "92\n",
+        "\n",
+        "93\n",
+        "\n",
+        "94\n",
+        "\n",
+        "95\n",
+        "\n",
+        "96\n",
+        "\n",
+        "97\n",
+        "\n",
+        "98\n",
+        "\n",
+        "99\n",
+        "\n",
+        "100\n",
+        "\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[Press Return to continue, Q to quit]\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.16, Page number: 457"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "fname1 = raw_input('Enter source file: ')\n",
+      "fname2 = raw_input('Enter destination file: ')\n",
+      "\n",
+      "with open(fname1) as fp1 :\n",
+      "\twith open(fname2,'w') as fp2 : \n",
+      "\t\tfor line in fp1 :\n",
+      "\t\t\tfp2.write(line)\n",
+      "\n",
+      "print ('Files successfully copied ')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter source file: 12.16in.txt\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter destination file: 12.16out.txt\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Files successfully copied \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.17, Page number: 458"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "total = count = 0\n",
+      "filenameIn = raw_input('Please enter an input filename: ')\n",
+      "filenameOut = raw_input('Please enter an output filename: ')\n",
+      "\n",
+      "print ('Opening %s for reading is OK.' % filenameIn)\n",
+      "print ('Opening %s for writing is OK.' % filenameOut)\n",
+      "print ('Calculate the total...')\n",
+      "\n",
+      "for i in open(filenameIn) :\n",
+      "\tcount += 1\n",
+      "\ttotal += int(i.strip())\n",
+      "\t\n",
+      "print ('Calculate the average...')\n",
+      "fileptrOut = open(filenameOut,'w')\n",
+      "fileptrOut.write('Average of %d numbers = %f ' % (count, (total/count)))\n",
+      "\n",
+      "print ('Check also your %s file content ' % filenameOut)\n",
+      "fileptrOut.close()\n",
+      "print ('\"%s\" closed successfully ' % filenameIn)\n",
+      "print ('\"%s\" closed successfully ' % filenameOut)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter an input filename: 12.17in.dat\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter an output filename: 12.17out.dat\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Opening 12.17in.dat for reading is OK.\n",
+        "Opening 12.17out.dat for writing is OK.\n",
+        "Calculate the total...\n",
+        "Calculate the average...\n",
+        "Check also your 12.17out.dat file content \n",
+        "\"12.17in.dat\" closed successfully \n",
+        "\"12.17out.dat\" closed successfully \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.18, Page number: 460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Structure declaration\n",
+      "class STUDENT (object) :\n",
+      "\tdef __init__(self, fname=None, lname=None, id=None, quiz1=None, quiz2=None, quiz3=None, quiz4=None, exam=None) :\n",
+      "\t\tself.fname = fname\n",
+      "\t\tself.lname = lname\n",
+      "\t\tself.id = id\n",
+      "\t\tself.quiz1 = quiz1\n",
+      "\t\tself.quiz2 = quiz2\n",
+      "\t\tself.quiz3 = quiz3\n",
+      "\t\tself.quiz4 = quiz4\n",
+      "\t\tself.exam = exam\n",
+      "\t\n",
+      "def findLow (students, x, c) :\n",
+      "\tif x == 0 :\n",
+      "\t\tsmall = students[0].quiz1\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif small >= students[i].quiz1 and students[i].quiz1 > 0 :\n",
+      "\t\t\t\tsmall = students[i].quiz1\n",
+      "\t\treturn small\n",
+      "\t\t\n",
+      "\telif x == 1 :\n",
+      "\t\tsmall = students[0].quiz2\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif small >= students[i].quiz2 and students[i].quiz2 > 0 :\n",
+      "\t\t\t\tsmall = students[i].quiz2\n",
+      "\t\treturn small\n",
+      "\t\t\n",
+      "\telif x == 2 :\n",
+      "\t\tsmall = students[0].quiz3\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif small >= students[i].quiz3 and students[i].quiz3 > 0 :\n",
+      "\t\t\t\tsmall = students[i].quiz3\n",
+      "\t\treturn small\n",
+      "\n",
+      "\telif x == 3 :\n",
+      "\t\tsmall = students[0].quiz4\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif small >= students[i].quiz4 and students[i].quiz4 > 0 :\n",
+      "\t\t\t\tsmall = students[i].quiz4\n",
+      "\t\treturn small\n",
+      "\t\t\n",
+      "\telif x == 4 :\n",
+      "\t\tsmall = students[0].exam\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif small >= students[i].exam and students[i].exam > 0 :\n",
+      "\t\t\t\tsmall = students[i].exam\n",
+      "\t\treturn small\n",
+      "\treturn 0\n",
+      "\n",
+      "def findHigh (students, x, c) :\n",
+      "\tif x == 0 :\n",
+      "\t\tbig = students[0].quiz1\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif big <= students[i].quiz1 :\n",
+      "\t\t\t\tbig = students[i].quiz1\n",
+      "\t\treturn big\n",
+      "\t\t\n",
+      "\telif x == 1 :\n",
+      "\t\tbig = students[0].quiz2\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif big <= students[i].quiz2 :\n",
+      "\t\t\t\tbig = students[i].quiz2\n",
+      "\t\treturn big\n",
+      "\t\t\n",
+      "\telif x == 2 :\n",
+      "\t\tbig = students[0].quiz3\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif big <= students[i].quiz3 :\n",
+      "\t\t\t\tbig = students[i].quiz3\n",
+      "\t\treturn big\n",
+      "\n",
+      "\telif x == 3 :\n",
+      "\t\tbig = students[0].quiz4\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif big <= students[i].quiz4 :\n",
+      "\t\t\t\tbig = students[i].quiz4\n",
+      "\t\treturn big\n",
+      "\t\t\n",
+      "\telif x == 4 :\n",
+      "\t\tbig = students[0].exam\n",
+      "\t\tfor i in range (1, c) :\n",
+      "\t\t\tif big <= students[i].exam :\n",
+      "\t\t\t\tbig = students[i].exam\n",
+      "\t\treturn big\n",
+      "\treturn 0\n",
+      "\n",
+      "\n",
+      "students = [STUDENT() for i in range (5)]\n",
+      "summ = avg = low = high = [0] * 5\n",
+      "cnt = 0\n",
+      "\n",
+      "print ('--- CLASS INFO --- \\n')\n",
+      "print ('Name\\t\\t Id\\tQuiz1   Quiz2   Quiz3   Quiz4   Exam \\n')\n",
+      "\n",
+      "with open('12.18.txt') as fpr :\n",
+      "\tfor line in fpr :\n",
+      "\t\ti = 0\n",
+      "\t\tstudents[i].fname, students[i].lname, students[i].id, students[i].quiz1, students[i].quiz2, students[i].quiz3, students[i].quiz4, students[i].exam = line.split()\n",
+      "\t\tstudents[i].id = int(students[i].id)\n",
+      "\t\tstudents[i].quiz1 = int(students[i].quiz1)\n",
+      "\t\tstudents[i].quiz2 = int(students[i].quiz2)\n",
+      "\t\tstudents[i].quiz3 = int(students[i].quiz3)\n",
+      "\t\tstudents[i].quiz4 = int(students[i].quiz4)\n",
+      "\t\tstudents[i].exam = int(students[i].exam)\n",
+      "\t\t\n",
+      "\t\tprint ('%-s %-s\\t %-4d\t%-3d\t%-3d\t%-3d\t%-3d\t%-3d' % (students[i].fname, students[i].lname, students[i].id, students[i].quiz1, students[i].quiz2, students[i].quiz3, students[i].quiz4, students[i].exam)) \n",
+      "\t\t\n",
+      "\t\tsumm[0] += int(students[i].quiz1)\n",
+      "\t\tsumm[1] += int(students[i].quiz2)\n",
+      "\t\tsumm[2] += int(students[i].quiz3)\t\t\n",
+      "\t\tsumm[3] += int(students[i].quiz4)\n",
+      "\t\tsumm[4] += int(students[i].exam)\n",
+      "\t\ti = i+1\n",
+      "\t\tcnt = cnt+1\n",
+      "\n",
+      "print ('\\nSTATISTICS')\n",
+      "print ('\\t\\tQuiz1\tQuiz2\tQuiz3\tQuiz4\tExam\\n')\n",
+      "print ('Average:\\t'),\n",
+      "for i in range (0, 5) :\n",
+      "\tavg[i] = float(summ[i])/cnt\n",
+      "\tprint ('%-5.1f \\t' % avg[i]),\n",
+      "print\n",
+      "\n",
+      "print ('Lowest:\\t\\t'),\n",
+      "for i in range (0, 5) :\n",
+      "\tlow[i] = findLow (students, i, cnt)\n",
+      "\thigh[i] = findHigh (students, i, cnt)\n",
+      "\t\n",
+      "for i in range (0, 5) :\n",
+      "\tprint ('%-5d \\t' % low[i]),\n",
+      "print\n",
+      "\t\n",
+      "print ('Highest:\\t'),\n",
+      "for i in range (0, 5) :\n",
+      "\tprint ('%-5d \\t' % high[i]),\n",
+      "print\n",
+      "\n",
+      "print ('\\n --- END OF REPORT --- \\n')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "--- CLASS INFO --- \n",
+        "\n",
+        "Name\t\t Id\tQuiz1   Quiz2   Quiz3   Quiz4   Exam \n",
+        "\n",
+        "anand mane\t 11  \t10 \t3  \t4  \t5  \t80 \n",
+        "mandar patil\t 12  \t4  \t5  \t6  \t7  \t50 \n",
+        "kishoe dhane\t 13  \t4  \t5  \t6  \t8  \t55 \n",
+        "\n",
+        "STATISTICS\n",
+        "\t\tQuiz1\tQuiz2\tQuiz3\tQuiz4\tExam\n",
+        "\n",
+        "Average:\t6.0   \t4.3   \t5.3   \t6.7   \t61.7  \t\n",
+        "Lowest:\t\t4     \t5     \t6     \t8     \t55    \t\n",
+        "Highest:\t4     \t5     \t6     \t8     \t55    \t\n",
+        "\n",
+        " --- END OF REPORT --- \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.19, Page number: 466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "# Structure declaration\n",
+      "class Employee (object) :\n",
+      "\tdef __init__(self, fname=None, lname=None, sub_taken=None, last_edu=None, join_date=None, id=None, age=None, bsal=None) :\n",
+      "\t\tself.fname = fname\n",
+      "\t\tself.lname = lname\n",
+      "\t\tself.sub_taken = sub_taken\n",
+      "\t\tself.last_edu = last_edu\n",
+      "\t\tself.join_date = join_date\n",
+      "\t\tself.id = id\n",
+      "\t\tself.age = age\n",
+      "\t\tself.bsal = bsal\n",
+      "\t\n",
+      "emp = Employee ()\n",
+      "fp = open('12.19.dat','ab+')\n",
+      "recsize = sys.getsizeof(emp)\n",
+      "\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\tprint ('\\n1.Add Records \\\n",
+      "\t\t\t  \\n2.Delete Records \\\n",
+      "\t\t\t  \\n3.Modify Records \\\n",
+      "\t\t\t  \\n4.List Records \\\n",
+      "\t\t\t  \\n5.Exit')\n",
+      "\tchoice = int(raw_input('\\nEnter your choice: '))\n",
+      "\t\n",
+      "\tif choice == 1 :\n",
+      "\t\tanother = 'Y'\n",
+      "\t\twhile another == 'Y' or another == 'y' :\n",
+      "\t\t\tfp = open('12.19.dat','ab+')\n",
+      "\t\t\temp.fname = raw_input('Enter the first name: ')\n",
+      "\t\t\temp.lname = raw_input('Enter the last name: ')\n",
+      "\t\t\temp.age = int(raw_input('Enter the age: '))\n",
+      "\t\t\temp.bsal = int(raw_input('Enter the basic salary: '))\n",
+      "\t\t\temp.join_date = int(raw_input('Enter joining date: '))\n",
+      "\t\t\temp.id = int(raw_input('Enter the employee id: '))\n",
+      "\t\t\temp.last_edu = raw_input('Enter the last education: ')\n",
+      "\t\t\temp.sub_taken = raw_input('Enter the subject taken: ')\n",
+      "\t\t\tfp.write('%d %s %s %d %d %s %s %s \\n' % (emp.id, emp.fname, emp.lname, emp.age, emp.bsal, emp.join_date, emp.last_edu, emp.sub_taken))\n",
+      "\t\t\tanother = raw_input('Add another record (Y/N) ? ')\n",
+      "\n",
+      "\telif choice == 2 :\n",
+      "\t\tanother = 'Y'\n",
+      "\t\twhile another == 'Y' or another == 'y' :\n",
+      "\t\t\tfp = open('12.19.dat','r')\n",
+      "\t\t\tlines = fp.readlines()\n",
+      "\t\t\tlines = lines[:-1]\n",
+      "\t\t\tprint ('\\nRecord deleted ')\n",
+      "\t\t\tanother = raw_input('Delete another record (Y/N) ? ')\n",
+      "\t\t\n",
+      "\telif choice == 3 :\n",
+      "\t\tfp = open('12.19.dat','w')\n",
+      "\t\tanother = 'Y'\n",
+      "\t\twhile another == 'Y' or another == 'y' :\n",
+      "\t\t\temp.fname = raw_input('Enter the first name: ')\n",
+      "\t\t\temp.lname = raw_input('Enter the last name: ')\n",
+      "\t\t\temp.age = int(raw_input('Enter the age: '))\n",
+      "\t\t\temp.bsal = int(raw_input('Enter the basic salary: '))\n",
+      "\t\t\temp.join_date = int(raw_input('Enter joining date: '))\n",
+      "\t\t\temp.id = int(raw_input('Enter the employee id: '))\n",
+      "\t\t\temp.last_edu = raw_input('Enter the last education: ')\n",
+      "\t\t\temp.sub_taken = raw_input('Enter the subject taken: ')\n",
+      "\t\t\tfp.write('%d %s %s %d %d %s %s %s \\n' % (emp.id, emp.fname, emp.lname, emp.age, emp.bsal, emp.join_date, emp.last_edu, emp.sub_taken))\n",
+      "\t\t\tanother = raw_input('Modify another record (Y/N) ? ')\n",
+      "\t \t\n",
+      "\telif choice == 4 :\n",
+      "\t\twith open ('12.19.dat','r') as fp :\n",
+      "\t\t\tfor line in fp :\n",
+      "\t\t\t\tprint line\n",
+      "\t\t\n",
+      "\telif choice == 5 :\n",
+      "\t\tcondition = False\n",
+      "\t\tfp.close()\n",
+      "\t\texit()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1.Add Records \t\t\t  \n",
+        "2.Delete Records \t\t\t  \n",
+        "3.Modify Records \t\t\t  \n",
+        "4.List Records \t\t\t  \n",
+        "5.Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter your choice: 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the first name: Anand\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last name: Rao\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age: 29\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the basic salary: 5000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter joining date: 2004\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the employee id: 11\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last education: BCA\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the subject taken: OS\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Add another record (Y/N) ? y\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the first name: Deepti\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last name: Garg\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age: 22\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the basic salary: 25000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter joining date: 2014\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the employee id: 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last education: BTech\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the subject taken: CSE\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Add another record (Y/N) ? n\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1.Add Records \t\t\t  \n",
+        "2.Delete Records \t\t\t  \n",
+        "3.Modify Records \t\t\t  \n",
+        "4.List Records \t\t\t  \n",
+        "5.Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter your choice: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "11 Anand Rao 29 5000 2004 BCA OS \n",
+        "\n",
+        "12 Deepti Garg 22 25000 2014 BTech CSE \n",
+        "\n",
+        "\n",
+        "1.Add Records \t\t\t  \n",
+        "2.Delete Records \t\t\t  \n",
+        "3.Modify Records \t\t\t  \n",
+        "4.List Records \t\t\t  \n",
+        "5.Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter your choice: 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Record deleted \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Delete another record (Y/N) ? y\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Record deleted \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Delete another record (Y/N) ? n\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1.Add Records \t\t\t  \n",
+        "2.Delete Records \t\t\t  \n",
+        "3.Modify Records \t\t\t  \n",
+        "4.List Records \t\t\t  \n",
+        "5.Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter your choice: 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the first name: Nalin\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last name: Chhibber\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the age: 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the basic salary: 30000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter joining date: 2012\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the employee id: 07\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the last education: MTech\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the subject taken: DBMS\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Modify another record (Y/N) ? n\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1.Add Records \t\t\t  \n",
+        "2.Delete Records \t\t\t  \n",
+        "3.Modify Records \t\t\t  \n",
+        "4.List Records \t\t\t  \n",
+        "5.Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter your choice: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7 Nalin Chhibber 25 30000 2012 MTech DBMS \n",
+        "\n",
+        "\n",
+        "1.Add Records \t\t\t  \n",
+        "2.Delete Records \t\t\t  \n",
+        "3.Modify Records \t\t\t  \n",
+        "4.List Records \t\t\t  \n",
+        "5.Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter your choice: 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_2_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_2_3.ipynb
new file mode 100755
index 00000000..7144ab92
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_2_3.ipynb
@@ -0,0 +1,203 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Syntactic Aspects"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page number: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "class color() :\n",
+      "   RED = 5\n",
+      "   YELLOW = 6\n",
+      "   GREEN = 4\n",
+      "   BLUE = 5\n",
+      "\n",
+      "print ('RED = %d ' % color.RED)\n",
+      "print ('YELLOW = %d ' % color.YELLOW)\n",
+      "print ('GREEN = %d ' % color.GREEN)\n",
+      "print ('BLUE = %d ' % color.BLUE)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RED = 5 \n",
+        "YELLOW = 6 \n",
+        "GREEN = 4 \n",
+        "BLUE = 5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page number: 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "i = 2004\n",
+      "c = 'Year'\n",
+      "\n",
+      "print ('%s %d' % (c, i))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Year 2004\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page number: 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "num = 5\n",
+      "c = '$'\n",
+      "pi = 3.141\n",
+      "print ('Hello World')\n",
+      "print ('%d %c %f' % (num, c, pi))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hello World\n",
+        "5 $ 3.141000\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page number: 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "num = int(raw_input('Enter your favorite number '))\n",
+      "print ('You have entered your favorite number as %d ' % num)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your favorite number 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered your favorite number as 3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3, Page number: 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "fahrenheit = float(raw_input('Please enter the Fahrenheit temperature '))\n",
+      "\n",
+      "centigrade = float(5)/9 * (fahrenheit - 32)\n",
+      "\n",
+      "print ('The temperature in Centigrade is %.2f ' % centigrade)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter the Fahrenheit temperature 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature in Centigrade is 37.78 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_3_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_3_3.ipynb
new file mode 100755
index 00000000..9e566fc1
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_3_3.ipynb
@@ -0,0 +1,1060 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Simple Programs"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page number: 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation and result\n",
+      "pi = 3.141\n",
+      "num = int(raw_input('Enter your favorite number: '))\n",
+      "print ('You have entered your favorite number as %d' % num)\n",
+      "c = raw_input('Enter your favorite character: ')\n",
+      "print ('You have entered your favorite character as %c' % c)\n",
+      "print ('The value of pi is %.3f ' % pi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your favorite number: 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered your favorite number as 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your favorite character: m\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You have entered your favorite character as m\n",
+        "The value of pi is 3.141 \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page number: 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Calculation\n",
+      "som = a+b\n",
+      "\n",
+      "# Result\n",
+      "print ('The sum of %d + %d = %d ' % (a, b, som))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 67\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 990\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of 67 + 990 = 1057 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page number: 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "if a>b :\n",
+      "   print ('The greater number is %d ' % a)\n",
+      "else :\n",
+      "   print ('The greater number is %d ' % b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 190\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The greater number is 190 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page number: 72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "if a==b :\n",
+      "   print ('Both numbers are equal ')\n",
+      "else :\n",
+      "   if a>b :\n",
+      "      print ('The greater number is %d ' % a)\n",
+      "   else :\n",
+      "      print ('The greater number is %d ' % b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 78\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Both numbers are equal \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page number: 73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "c = int(raw_input('Enter third number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "if a>b :\n",
+      "   if a>c :\n",
+      "      print ('The greatest number is %d ' % a)\n",
+      "   else :\n",
+      "      print ('The greatest number is %d ' % c)\n",
+      "\n",
+      "else :\n",
+      "   if b>c :\n",
+      "      print ('The greatest number is %d ' % b)\n",
+      "   else : \n",
+      "      print ('The greatest number is %d ' % c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 128\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter third number 90\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The greatest number is 128 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5, Page number: 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('Counting upto %d ' % n)\n",
+      "\n",
+      "for i in range (1, n+1) :\n",
+      "   print ('%d ' % i),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Counting upto 7 \n",
+        "1  2  3  4  5  6  7 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page number: 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('Even numbers upto %d ' % n)\n",
+      "\n",
+      "for i in range (2, n+1, 2) :\n",
+      "   print ('%d ' % i),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Even numbers upto 10 \n",
+        "2  4  6  8  10 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7, Page number: 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('Odd numbers upto %d ' % n)\n",
+      "\n",
+      "for i in range (1, n+1, 2) :\n",
+      "   print ('%d ' % i),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Odd numbers upto 10 \n",
+        "1  3  5  7  9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8, Page number: 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('Number line upto %d ' % n)\n",
+      "\n",
+      "for i in range (-n, n+1) :\n",
+      "   print ('%d ' % i),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number line upto 7 \n",
+        "-7  -6  -5  -4  -3  -2  -1  0  1  2  3  4  5  6  7 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9, Page number: 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "sum = 0\n",
+      "\n",
+      "# Calculation\n",
+      "for i in range (1, n+1) :\n",
+      "   sum = sum + i\n",
+      "\n",
+      "# Result\n",
+      "print ('Sum of %d natural numbers is %d ' % (n, sum))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of 6 natural numbers is 21 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10, Page number: 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('The factors of %d are ' % n)\n",
+      "\n",
+      "for i in range (1, n+1) :\n",
+      "   if n%i == 0 :\n",
+      "      print ('%d ' % i),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 28\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The factors of 28 are \n",
+        "1  2  4  7  14  28 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11, Page number: 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('The first 10 multiples of %d are ' % n)\n",
+      "\n",
+      "for i in range (1, 11) :\n",
+      "   print ('%d X %d = %d ' % (n, i, n*i))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The first 10 multiples of 6 are \n",
+        "6 X 1 = 6 \n",
+        "6 X 2 = 12 \n",
+        "6 X 3 = 18 \n",
+        "6 X 4 = 24 \n",
+        "6 X 5 = 30 \n",
+        "6 X 6 = 36 \n",
+        "6 X 7 = 42 \n",
+        "6 X 8 = 48 \n",
+        "6 X 9 = 54 \n",
+        "6 X 10 = 60 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12, Page number: 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "i = 1\n",
+      "sum = 0\n",
+      "\n",
+      "# Calculation\n",
+      "print ('The sum of digits of %d is' % n),\n",
+      "\n",
+      "while n > 0 :\n",
+      "   sum = sum + n%10\n",
+      "   n = n/10\n",
+      "\n",
+      "# Result\n",
+      "print ('%d' % sum)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 5214\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of digits of 5214 is 12\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13, Page number: 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "flag = 1\n",
+      "\n",
+      "# Calculation\n",
+      "for i in range (2, n) :\n",
+      "   if flag == 1 :\n",
+      "      if n%i == 0 :\n",
+      "         flag = 0\n",
+      "\n",
+      "# Result\n",
+      "if flag :\n",
+      "   print ('%d is Prime ' % n)\n",
+      "else :\n",
+      "   print ('%d is Composite ' % n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 41\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "41 is Prime \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14, Page number: 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "first = 0\n",
+      "second = 1\n",
+      "count = 3\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('%d Fibonacci terms are ' % n)\n",
+      "print ('%d  %d ' % (first, second)),\n",
+      "\n",
+      "while count<=n :\n",
+      "   third = first + second\n",
+      "   first = second\n",
+      "   second = third\n",
+      "   print ('%d ' % third),\n",
+      "   count = count + 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10 Fibonacci terms are \n",
+        "0  1  1  2  3  5  8  13  21  34 \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15, Page number: 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "sum = n\n",
+      "\n",
+      "# Calculation\n",
+      "print ('Sum of digits of %d is' % n),\n",
+      "\n",
+      "while sum>=10 :\n",
+      "   sum = 0\n",
+      "   while n>0 :\n",
+      "      sum = sum + n%10\n",
+      "      n = n/10\n",
+      "   n = sum\n",
+      "\n",
+      "# Result\n",
+      "print ('%d' % sum)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 8626\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits of 8626 is 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16, Page number: 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('Prime numbers upto %d are ' % n)\n",
+      "\n",
+      "for num in range (2, n+1) :\n",
+      "   flag = 1\n",
+      "   for i in range (2, num) :\n",
+      "      if num % i == 0 : \n",
+      "         flag = 0\n",
+      "   if flag :\n",
+      "      print ('%d ' % num),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 12\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Prime numbers upto 12 are \n",
+        "2  3  5  7  11 \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17, Page number: 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "for lin in range (1, n+1) :\n",
+      "   for count in range (1, lin+1) :\n",
+      "      print ('%d ' % count),\n",
+      "   print"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 \n",
+        "1  2 \n",
+        "1  2  3 \n",
+        "1  2  3  4 \n",
+        "1  2  3  4  5 \n",
+        "1  2  3  4  5  6 \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18, Page number: 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "for lin in range (1, n+1) :\n",
+      "   for count in range (lin, 2*lin) :\n",
+      "      print ('%d ' % count),\n",
+      "   print"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 \n",
+        "2  3 \n",
+        "3  4  5 \n",
+        "4  5  6  7 \n",
+        "5  6  7  8  9 \n",
+        "6  7  8  9  10  11 \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19, Page number: 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "for lin in range (1, n+1) :\n",
+      "   for blk in range (1, n-lin+1) :\n",
+      "      print ' ',\n",
+      "   for up in range (1, lin+1) :\n",
+      "      print ('%d' % up),\n",
+      "   down = lin-1\n",
+      "   while down>0 :\n",
+      "      print ('%d' % down),\n",
+      "      down = down-1\n",
+      "   print"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "          1\n",
+        "        1 2 1\n",
+        "      1 2 3 2 1\n",
+        "    1 2 3 4 3 2 1\n",
+        "  1 2 3 4 5 4 3 2 1\n",
+        "1 2 3 4 5 6 5 4 3 2 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20, Page number: 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "for lin in range (1, n+1) :\n",
+      "   for blk in range (1, n-lin+1) :\n",
+      "      print ' ',\n",
+      "   for up in range (lin, 2*lin) :\n",
+      "      print ('%d' % (up%10)),\n",
+      "   down = 2*lin-2\n",
+      "   while down>=lin :\n",
+      "      print ('%d' % (down%10)),\n",
+      "      down = down-1\n",
+      "   print"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "          1\n",
+        "        2 3 2\n",
+        "      3 4 5 4 3\n",
+        "    4 5 6 7 6 5 4\n",
+        "  5 6 7 8 9 8 7 6 5\n",
+        "6 7 8 9 0 1 0 9 8 7 6\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_4_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_4_3.ipynb
new file mode 100755
index 00000000..a071712d
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_4_3.ipynb
@@ -0,0 +1,1070 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Functions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1, Page number: 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration and result\n",
+      "a = 5\n",
+      "c = 'h'\n",
+      "b = 3.14\n",
+      "def myfunction (x, ch, fl) :\n",
+      "    print ('%d %c %.2f ' % (x, ch, fl))\n",
+      "    \n",
+      "myfunction (a, c, b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5 h 3.14 \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2, Page number: 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def add (a, b) :\n",
+      "   som = a + b\n",
+      "   return som\n",
+      "\n",
+      "print ('The sum of %d and %d is %d' % (a, b, (add (a, b))))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of 7 and 8 is 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page number: 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def greater (a, b) :\n",
+      "   if a > b :\n",
+      "      return a\n",
+      "   else :\n",
+      "      return b\n",
+      "\n",
+      "print ('The greater number is %d' % (greater (a, b)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 190\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The greater number is 190\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4, Page number: 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def greater (a, b) :\n",
+      "   if a == b :\n",
+      "      print ('Both numbers are equal')\n",
+      "   elif a > b :\n",
+      "      print ('The greater number is %d' % a)\n",
+      "   else :\n",
+      "      print ('The greater number is %d' % b)\n",
+      "   return\n",
+      "\n",
+      "greater (a, b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Both numbers are equal\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5, Page number: 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "c = int(raw_input('Enter third number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def greatest (a, b, c) :\n",
+      "   if a > b :\n",
+      "      if a > c :\n",
+      "         return a\n",
+      "      else :\n",
+      "         return c\n",
+      "   else :\n",
+      "      if b > c :\n",
+      "         return b\n",
+      "      else :\n",
+      "         return c\n",
+      "\n",
+      "print ('The greatest number is %d' % (greatest (a, b, c)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 128\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter third number: 190\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The greatest number is 190\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6, Page number: 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printnum (n) :\n",
+      "   print ('Counting upto %d' % n)   \n",
+      "   for i in range (1, n+1) :\n",
+      "      print i,\n",
+      "\n",
+      "printnum (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Counting upto 7\n",
+        "1 2 3 4 5 6 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7, Page number: 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printeven (n) :\n",
+      "   print ('Even numbers upto %d' % n)   \n",
+      "   for i in range (2, n+1, 2) :\n",
+      "      print i,\n",
+      "\n",
+      "printeven (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Even numbers upto 7\n",
+        "2 4 6\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8, Page number: 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printodd (n) :\n",
+      "   print ('Odd numbers upto %d' % n)   \n",
+      "   for i in range (1, n+1, 2) :\n",
+      "      print i,\n",
+      "\n",
+      "printodd (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Odd numbers upto 7\n",
+        "1 3 5 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9, Page number: 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printnum (n) :\n",
+      "   print ('Number line for %d' % n)   \n",
+      "   for i in range (-n, n+1) :\n",
+      "      print i,\n",
+      "\n",
+      "printnum (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number line for 4\n",
+        "-4 -3 -2 -1 0 1 2 3 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10, Page number: 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def summate (n) :\n",
+      "   total = 0\n",
+      "   for i in range (1, n+1) :\n",
+      "      total = total + i\n",
+      "   return total\n",
+      "\n",
+      "print('Sum of first %d natural numbers is %d' % (n, summate (n)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of first 6 natural numbers is 21\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11, Page number: 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def factors (n) :\n",
+      "   print ('The factors of %d are' % n)\n",
+      "   for i in range (1, n+1) :\n",
+      "      if n%i == 0 :\n",
+      "         print i,\n",
+      "\n",
+      "factors (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 28\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The factors of 28 are\n",
+        "1 2 4 7 14 28\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12, Page number: 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def multiple (n) :\n",
+      "   print ('The first 10 multiples of %d are' % n)\n",
+      "   for i in range (1, 11) :\n",
+      "         print('%d X %d = %d' % (n, i, n*i))\n",
+      "\n",
+      "multiple (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The first 10 multiples of 6 are\n",
+        "6 X 1 = 6\n",
+        "6 X 2 = 12\n",
+        "6 X 3 = 18\n",
+        "6 X 4 = 24\n",
+        "6 X 5 = 30\n",
+        "6 X 6 = 36\n",
+        "6 X 7 = 42\n",
+        "6 X 8 = 48\n",
+        "6 X 9 = 54\n",
+        "6 X 10 = 60\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13, Page number: 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def sumofdigit (n) :\n",
+      "   sum = 0\n",
+      "   while n>0 :\n",
+      "      sum = sum + n%10\n",
+      "      n = n/10\n",
+      "   return sum\n",
+      "\n",
+      "print ('The sum of digits of %d is %d' % (n, sumofdigit (n)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 5214\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of digits of 5214 is 12\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14, Page number: 116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def prime (n) :\n",
+      "   count = 0\n",
+      "   for i in range (2, n) :\n",
+      "      if n%i == 0 :\n",
+      "         count = count + 1\n",
+      "         break\n",
+      "\n",
+      "   if count == 0 :\n",
+      "      print ('The number %d is prime' % n)\n",
+      "   else :\n",
+      "      print ('The number %d is composite' % n)\n",
+      "\n",
+      "prime (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 14\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number 14 is composite\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15, Page number: 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "print ('%d terms of Fibonacci series are ' % n)\n",
+      "\n",
+      "def printfibo (n) :\n",
+      "   first = 0\n",
+      "   second = 1\n",
+      "   count = 3\n",
+      "\n",
+      "   print ('%d  %d ' % (first, second)),\n",
+      "\n",
+      "   while count<=n :\n",
+      "      third = first + second\n",
+      "      first = second\n",
+      "      second = third\n",
+      "      print (' %d ' % third), \n",
+      "      count = count + 1\n",
+      "\n",
+      "printfibo (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10 terms of Fibonacci series are \n",
+        "0  1   1   2   3   5   8   13   21   34 \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16, Page number: 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def sumofdigit (n) :\n",
+      "   sum = n\n",
+      "   while sum>=10 :\n",
+      "      sum = 0\n",
+      "      while n>0 :\n",
+      "         sum = sum + n%10\n",
+      "         n = n/10\n",
+      "      n = sum\n",
+      "   return sum\n",
+      "\n",
+      "print ('Sum of digits of %d is %d' % (n, sumofdigit (n)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 8626\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits of 8626 is 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17, Page number: 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "print ('Prime numbers upto %d are ' % n)\n",
+      "\n",
+      "def printprime (n) :\n",
+      "   for num in range (2, n+1) :\n",
+      "      flag = 1\n",
+      "      for i in range (2, num) :\n",
+      "         if num % i == 0 : \n",
+      "            flag = 0\n",
+      "      if flag :\n",
+      "         print ('%d ' % num),\n",
+      "\n",
+      "printprime (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 40\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Prime numbers upto 40 are \n",
+        "2  3  5  7  11  13  17  19  23  29  31  37 \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.18, Page number: 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def pattern (n) :\n",
+      "   for lin in range (1, n+1) :\n",
+      "      for count in range (1, lin+1) :\n",
+      "         print ('%d ' % count),\n",
+      "      print\n",
+      "\n",
+      "pattern (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 \n",
+        "1  2 \n",
+        "1  2  3 \n",
+        "1  2  3  4 \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19, Page number: 126"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def pattern (n) :\n",
+      "   for lin in range (1, n+1) :\n",
+      "      for count in range (lin, 2*lin) :\n",
+      "         print ('%d ' % count),\n",
+      "      print\n",
+      "\n",
+      "pattern (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 \n",
+        "2  3 \n",
+        "3  4  5 \n",
+        "4  5  6  7 \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.20, Page number: 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def pattern (n) :\n",
+      "   for lin in range (1, n+1) :\n",
+      "      for blk in range (1, n-lin+1) :\n",
+      "         print ' ',\n",
+      "      for up in range (1, lin+1) :\n",
+      "         print ('%d' % up),\n",
+      "      down = lin-1\n",
+      "      while down>0 :\n",
+      "         print ('%d' % down),\n",
+      "         down = down-1\n",
+      "      print\n",
+      "\n",
+      "pattern (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "      1\n",
+        "    1 2 1\n",
+        "  1 2 3 2 1\n",
+        "1 2 3 4 3 2 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.21, Page number: 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter a number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def pattern (n) :\n",
+      "   for lin in range (1, n+1) :\n",
+      "      for blk in range (1, n-lin+1) :\n",
+      "         print ' ',\n",
+      "      for up in range (lin, 2*lin) :\n",
+      "         print ('%d' % (up%10)),\n",
+      "      down = 2*lin-2\n",
+      "      while down>=lin :\n",
+      "         print ('%d' % (down%10)),\n",
+      "         down = down-1\n",
+      "      print\n",
+      "\n",
+      "pattern (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number: 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "      1\n",
+        "    2 3 2\n",
+        "  3 4 5 4 3\n",
+        "4 5 6 7 6 5 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_5_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_5_3.ipynb
new file mode 100755
index 00000000..b4cb2db2
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_5_3.ipynb
@@ -0,0 +1,909 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Pointers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page number: 135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('The original values are a = %d b = %d' % (a, b))\n",
+      "temp = a\n",
+      "a = b\n",
+      "b = temp\n",
+      "\n",
+      "print ('The values after swapping are a = %d b = %d' % (a, b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 15\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 27\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The original values are a = 15 b = 27\n",
+        "The values after swapping are a = 27 b = 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page number: 136"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "print ('The original values are a = %d b = %d' % (a, b))\n",
+      "\n",
+      "def swap (a, b) :\n",
+      "   temp = a\n",
+      "   a = b\n",
+      "   b = temp\n",
+      "   print ('The values after swapping are a = %d b = %d' % (a, b))\n",
+      "\n",
+      "swap (a, b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 15\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 27\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The original values are a = 15 b = 27\n",
+        "The values after swapping are a = 27 b = 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3, Page number: 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "print ('The original values are a = %d b = %d' % (a, b))\n",
+      "\n",
+      "def swap (a, b) :\n",
+      "   temp = a\n",
+      "   a = b\n",
+      "   b = temp\n",
+      "   print ('The values after swapping are a = %d b = %d' % (a, b))\n",
+      "\n",
+      "swap (a, b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 15\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 27\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The original values are a = 15 b = 27\n",
+        "The values after swapping are a = 27 b = 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page number: 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "print ('In main the original values are a = %d b = %d' % (a, b))\n",
+      "print ('In main the address of a is %x and address of b is %x' % (a, b))\n",
+      "\n",
+      "def swap (a, b) :\n",
+      "   print ('In swap the address in a is %x and b is %x' % (a, b))\n",
+      "   temp = a\n",
+      "   a = b\n",
+      "   b = temp\n",
+      "   print ('The values after swapping are a = %d b = %d' % (a, b))\n",
+      "\n",
+      "swap (a, b)\n",
+      "\n",
+      "print ('In main the values after swapping are a = %d b = %d' % (a, b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 15\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 27\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In main the original values are a = 15 b = 27\n",
+        "In main the address of a is f and address of b is 1b\n",
+        "In swap the address in a is f and b is 1b\n",
+        "The values after swapping are a = 27 b = 15\n",
+        "In main the values after swapping are a = 15 b = 27\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5, Page number: 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "print ('In main the value of a = %d b = %d' % (a, b))\n",
+      "\n",
+      "def doooo (a, b) :\n",
+      "   a = a+10\n",
+      "   b = b+10\n",
+      "\n",
+      "doooo (a, b)\n",
+      "\n",
+      "print ('In main the value of a = %d b = %d' % (a, b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In main the value of a = 5 b = 7\n",
+        "In main the value of a = 5 b = 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6, Page number: 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number: '))\n",
+      "b = int(raw_input('Enter second number: '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "print ('In main the value of a = %d b = %d' % (a, b))\n",
+      "print ('In main the addresses are a = %s b = %s' % (hex(id(a)), hex(id(b))))\n",
+      "\n",
+      "def doooo (a, b) :\n",
+      "   print ('In doooo the addresses are a = %s b = %s' % (hex(id(a)), hex(id(b))))\n",
+      "   a = a+10\n",
+      "   b = b+10\n",
+      "\n",
+      "doooo (a, b)\n",
+      "\n",
+      "print ('In main the value of a = %d b = %d' % (a, b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number: 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In main the value of a = 5 b = 7\n",
+        "In main the addresses are a = 0x1b49138 b = 0x1b49108\n",
+        "In doooo the addresses are a = 0x1b49138 b = 0x1b49108\n",
+        "In main the value of a = 5 b = 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page number: 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def add (a, b) :\n",
+      "   x = a+b\n",
+      "   return x\n",
+      "\n",
+      "som = add (a, b)\n",
+      "\n",
+      "print ('The sum of %d + %d = %d' % (a, b, som))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of 2 + 3 = 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8, Page number: 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def greater (a, b) :\n",
+      "   if a>b :\n",
+      "      big = a\n",
+      "   else :\n",
+      "      big = b\n",
+      "   return big\n",
+      "\n",
+      "great = greater (a, b)\n",
+      "\n",
+      "print ('The greater number is %d' % great)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 23\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The greater number is 45\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9, Page number: 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "c = int(raw_input('Enter third number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def greater (a, b, c) :\n",
+      "   if a>b :\n",
+      "      if a>c :\n",
+      "         big = a\n",
+      "      else :\n",
+      "         big = c\n",
+      "   else :\n",
+      "      if b>c :\n",
+      "         big = b\n",
+      "      else :\n",
+      "         big = c\n",
+      "   return big\n",
+      "\n",
+      "great = greater (a, b, c)\n",
+      "\n",
+      "print ('The greatest number is %d' % great)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 190\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter third number 128\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The greatest number is 190\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10, Page number: 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def summate (n) :\n",
+      "   i = 1\n",
+      "   total = 0\n",
+      "   while i <= n :\n",
+      "      total = total + i\n",
+      "      i = i+1\n",
+      "   return total\n",
+      "\n",
+      "sumn = summate (n)\n",
+      "\n",
+      "print ('Sum of first %d natural numbers is %d' % (n, sumn))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of first 5 natural numbers is 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11, Page number: 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def sumofdigit (n) :\n",
+      "   sum = 0\n",
+      "   while n > 0 :\n",
+      "      sum = sum + n%10\n",
+      "      n = n/10\n",
+      "   return sum\n",
+      "\n",
+      "result = sumofdigit (n)\n",
+      "\n",
+      "print ('Sum of digits of %d is %d' % (n, result))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 7162\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits of 7162 is 16\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12, Page number: 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "sum = n\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def sumofdigit (n, sum) :\n",
+      "   while sum>=10 :\n",
+      "      sum = 0\n",
+      "      while n>0 :\n",
+      "         sum = sum + n%10\n",
+      "         n = n/10\n",
+      "      n = sum\n",
+      "   return sum\n",
+      "\n",
+      "result = sumofdigit (n, sum)\n",
+      "\n",
+      "print ('Sum of digits of %d is %d' % (n, result))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 7162\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits of 7162 is 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13, Page number: 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def checkprime (n) :\n",
+      "   flag = 1\n",
+      "   i = 2\n",
+      "   while i<n and flag :\n",
+      "      if n%i == 0 :\n",
+      "         flag = 0\n",
+      "      i = i+1\n",
+      "   return flag\n",
+      "\n",
+      "result = checkprime (n)\n",
+      "\n",
+      "if result :\n",
+      "   print ('%d is Prime ' % n)\n",
+      "else :\n",
+      "   print ('%d is Composite ' % n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 41\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "41 is Prime \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14, Page number: 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def checkprime (n) :\n",
+      "   for num in range (2, n+1) :\n",
+      "      flag = 1\n",
+      "      for i in range (2, num) :\n",
+      "         if num % i == 0 : \n",
+      "            flag = 0\n",
+      "      if flag :\n",
+      "         print ('%d ' % num),\n",
+      "\n",
+      "checkprime (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 11\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2  3  5  7  11 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15, Page number: 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def makethird (n) :\n",
+      "   first = 0\n",
+      "   second = 1\n",
+      "   count = 3\n",
+      "   print ('%d terms of Fibonacci series are ' % n)\n",
+      "   print ('%d  %d ' % (first, second)),\n",
+      "   while count<=n :\n",
+      "      third = first + second\n",
+      "      first = second\n",
+      "      second = third\n",
+      "      print (' %d ' % third),\n",
+      "      count = count + 1\n",
+      "\n",
+      "makethird (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10 terms of Fibonacci series are \n",
+        "0  1   1   2   3   5   8   13   21   34 \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16, Page number: 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def makethird (n) :\n",
+      "   first = 0\n",
+      "   second = 1\n",
+      "   third = 0\n",
+      "   print ('Fibonacci series up to %d is ' % n)\n",
+      "   print first,\n",
+      "   while third<=n :\n",
+      "      print (' %d ' % second),\n",
+      "      third = first + second\n",
+      "      first = second\n",
+      "      second = third\n",
+      "\n",
+      "makethird (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fibonacci series up to 100 is \n",
+        "0  1   1   2   3   5   8   13   21   34   55   89 \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_6_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_6_3.ipynb
new file mode 100755
index 00000000..690f5ef0
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_6_3.ipynb
@@ -0,0 +1,951 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: Storage classes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page number: 180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def fact (n) :\n",
+      "   prod = 1\n",
+      "   while n :\n",
+      "      prod = prod * n\n",
+      "      n = n - 1\n",
+      "   return prod\n",
+      "\n",
+      "print ('Factorial of %d is %d' % (n, fact(n)))\n",
+      "\n",
+      "m = int(raw_input('Enter another number '))\n",
+      "\n",
+      "print ('Factorial of %d is %d' % (m, fact(m)))\n",
+      "\n",
+      "print ('Factorial of 6 is %d' % fact(6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Factorial of 5 is 120\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter another number 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Factorial of 3 is 6\n",
+        "Factorial of 6 is 720\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page number: 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def add (a, b) :\n",
+      "   global som\n",
+      "   som = a + b\n",
+      "\n",
+      "add (a, b)\n",
+      "\n",
+      "print ('The sum of %d + %d = %d' % (a, b, som))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of 2 + 3 = 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3, Page number: 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "global y\n",
+      "y = 90\n",
+      "m = 10\n",
+      "z = 2.18\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def fun1 (m, y, z) :\n",
+      "   z = 3.141\n",
+      "   print ('%f' % z)\n",
+      "\n",
+      "fun1 (m, y, z)\n",
+      "\n",
+      "print ('%d %d %f' % (m, y, z))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3.141000\n",
+        "10 90 2.180000\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page number: 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "global x, y, z, ch, tmp\n",
+      "z = 2.18\n",
+      "y = 90\n",
+      "ch = '#'\n",
+      "m = 10\n",
+      "x = '*'\n",
+      "tmp = '^'\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def fun1 (m, y, z) :\n",
+      "   z = 3.141\n",
+      "   num = 100\n",
+      "   tmp = '$'\n",
+      "   x = 50\n",
+      "   ch = '@'\n",
+      "   print ('In FUNCTION FORMAL PARAMETERS')\n",
+      "   print ('In the function value of m = %d address of m = %x' % (m, id(m)))\n",
+      "   print ('In the function value of y = %d address of y = %x' % (y, id(y)))\n",
+      "   print ('In the function value of z = %f address of z = %x' % (z, id(z)))\n",
+      "\n",
+      "   print ('In FUNCTION LOCAL VARIABLES')\n",
+      "   print ('In the function value of z = %f address of z = %x' % (z, id(z)))\n",
+      "   print ('In the function value of num = %d address of num = %x' % (num, id(num)))\n",
+      "   print ('In the function value of tmp = %s address of tmp = %x' % (tmp, id(tmp)))\n",
+      "\n",
+      "   print ('In FUNCTION GLOBAL VARIABLES')\n",
+      "   print ('In the function value of x = %s address of x = %x' % (x, id(x)))\n",
+      "   print ('In the function value of y = %d address of y = %x' % (y, id(y)))\n",
+      "   print ('In the function value of ch = %s address of ch = %x' % (ch, id(ch)))\n",
+      "\n",
+      "\n",
+      "print ('In MAIN LOCAL VARIABLES')\n",
+      "print ('In Main the value of m = %d address of m = %x' % (m, id(m)))\n",
+      "print ('In Main the value of x = %s address of x = %s' % (x, id(x)))\n",
+      "print ('In MAIN GLOBAL VARIABLES')\n",
+      "print ('In Main the value of y = %d address of y = %x' % (y, id(y)))\n",
+      "print ('In Main the value of z = %f address of z = %x' % (z, id(z)))\n",
+      "print ('In Main the value of ch = %s address of ch = %s' % (ch, id(ch)))\n",
+      "print ('In Main the value of tmp = %s address of tmp = %x' % (tmp, id(tmp)))\n",
+      "\n",
+      "fun1 (m, y, z)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In MAIN LOCAL VARIABLES\n",
+        "In Main the value of m = 10 address of m = 117d0c0\n",
+        "In Main the value of x = * address of x = 140148786619376\n",
+        "In MAIN GLOBAL VARIABLES\n",
+        "In Main the value of y = 90 address of y = 117d8d0\n",
+        "In Main the value of z = 2.180000 address of z = 16ca340\n",
+        "In Main the value of ch = # address of ch = 140148787513184\n",
+        "In Main the value of tmp = ^ address of tmp = 7f76eea46eb8\n",
+        "In FUNCTION FORMAL PARAMETERS\n",
+        "In the function value of m = 10 address of m = 117d0c0\n",
+        "In the function value of y = 90 address of y = 117d8d0\n",
+        "In the function value of z = 3.141000 address of z = 16ca2f8\n",
+        "In FUNCTION LOCAL VARIABLES\n",
+        "In the function value of z = 3.141000 address of z = 16ca2f8\n",
+        "In the function value of num = 100 address of num = 117d7e0\n",
+        "In the function value of tmp = $ address of tmp = 7f76eeb088a0\n",
+        "In FUNCTION GLOBAL VARIABLES\n",
+        "In the function value of x = 50 address of x = 117d4c8\n",
+        "In the function value of y = 90 address of y = 117d8d0\n",
+        "In the function value of ch = @ address of ch = 7f76ed6203f0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5, Page number: 193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def add (a, b) :\n",
+      "   global som\n",
+      "   som = a + b\n",
+      "\n",
+      "add (a, b)\n",
+      "\n",
+      "print ('The sum of %d + %d = %d' % (a, b, som))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 30\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of 20 + 30 = 50\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6, Page number: 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def greater (a, b) :\n",
+      "   global big\n",
+      "   if a == b :\n",
+      "      big = 'equal'\n",
+      "   else :\n",
+      "      if a > b :\n",
+      "         big = 'first'\n",
+      "      else :\n",
+      "         big = 'second'\n",
+      "   return big\n",
+      "\n",
+      "greater (a, b)\n",
+      "\n",
+      "if big == 'equal' :\n",
+      "   print ('EQUAL')\n",
+      "else :\n",
+      "   if big == 'first' :\n",
+      "      print ('First number is greater %d' % a)\n",
+      "   else :\n",
+      "      print ('Second number is greater %d' % b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "EQUAL\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7, Page number: 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = int(raw_input('Enter first number '))\n",
+      "b = int(raw_input('Enter second number '))\n",
+      "c = int(raw_input('Enter third number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def greater (a, b, c) :\n",
+      "   if a>b :\n",
+      "      if a>c :\n",
+      "         print ('The greatest number is %d ' % a)\n",
+      "      else :\n",
+      "         print ('The greatest number is %d ' % c)\n",
+      "\n",
+      "   else :\n",
+      "      if b>c :\n",
+      "         print ('The greatest number is %d ' % b)\n",
+      "      else : \n",
+      "         print ('The greatest number is %d ' % c)\n",
+      "\n",
+      "greater (a, b, c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter first number 190\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter second number 128\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter third number 45\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The greatest number is 190 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8, Page number: 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def summate (n) :\n",
+      "   i = 1\n",
+      "   global total\n",
+      "   total = 0\n",
+      "   while i <= n :\n",
+      "      total = total + i\n",
+      "      i = i+1\n",
+      "   \n",
+      "summate (n)\n",
+      "\n",
+      "print ('Sum of first %d natural numbers is %d' % (n, total))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of first 6 natural numbers is 21\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9, Page number: 199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def sumofdigit (n) :\n",
+      "   global sum\n",
+      "   sum = 0\n",
+      "   while n > 0 :\n",
+      "      sum = sum + n%10\n",
+      "      n = n/10\n",
+      "   return sum\n",
+      "\n",
+      "sumofdigit (n)\n",
+      "\n",
+      "print ('Sum of digits of %d is %d' % (n, sum))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 7612\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits of 7612 is 16\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10, Page number: 201"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def checkprime (n) :\n",
+      "   flag = 1\n",
+      "   i = 2\n",
+      "   while i<n and flag :\n",
+      "      if n%i == 0 :\n",
+      "         flag = 0\n",
+      "      i = i+1\n",
+      "   return flag\n",
+      "\n",
+      "result = checkprime (n)\n",
+      "\n",
+      "if result :\n",
+      "   print ('The number %d is PRIME ' % n)\n",
+      "else :\n",
+      "   print ('The number %d is COMPOSITE ' % n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 14\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number 14 is COMPOSITE \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11, Page number: 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def makethird (n) :\n",
+      "   first = 0\n",
+      "   second = 1\n",
+      "   count = 3\n",
+      "   print ('%d terms of the Fibonacci series are ' % n)\n",
+      "   print ('%d  %d ' % (first, second)),\n",
+      "   while count<=n :\n",
+      "      third = first + second\n",
+      "      first = second\n",
+      "      second = third\n",
+      "      print (' %d ' % third),\n",
+      "      count = count + 1\n",
+      "\n",
+      "makethird (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10 terms of the Fibonacci series are \n",
+        "0  1   1   2   3   5   8   13   21   34 \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12, Page number: 205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "sum = n\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def sumofdigit (n, sum) :\n",
+      "   while sum>=10 :\n",
+      "      sum = 0\n",
+      "      while n>0 :\n",
+      "         sum = sum + n%10\n",
+      "         n = n/10\n",
+      "      n = sum\n",
+      "   return sum\n",
+      "\n",
+      "result = sumofdigit (n, sum)\n",
+      "\n",
+      "print ('Sum of digits of %d is %d' % (n, result))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 7612\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits of 7612 is 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13, Page number: 207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def prime (n) :\n",
+      "   for num in range (2, n) :\n",
+      "      flag = 1\n",
+      "      for i in range (2, num) :\n",
+      "         if num % i == 0 : \n",
+      "            flag = 0\n",
+      "      if flag :\n",
+      "         print ('%d ' % num),\n",
+      "\n",
+      "prime (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2  3  5  7  11  13  17  19 \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.14, Page number: 210"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "m = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def fun1 (m) :\n",
+      "   if not hasattr(fun1, \"temp\") : \n",
+      "      fun1.temp = 0\n",
+      "   print ('In function value of temp = %d' % fun1.temp)\n",
+      "   if fun1.temp < m :\n",
+      "      fun1.temp = m\n",
+      "   else :\n",
+      "      fun1.temp = fun1.temp + 1\n",
+      "\n",
+      "for i in range (1, 6) :\n",
+      "   fun1 (m)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In function value of temp = 0\n",
+        "In function value of temp = 10\n",
+        "In function value of temp = 11\n",
+        "In function value of temp = 12\n",
+        "In function value of temp = 13\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15, Page number: 212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "m = 10\n",
+      "i = 1\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def fun1 (m) :\n",
+      "   x = m\n",
+      "   if not hasattr(fun1, \"temp\") : \n",
+      "      fun1.temp = 0\n",
+      "\n",
+      "   if fun1.temp == 0 :\n",
+      "      print ('\\nIn FUNCTION Global variable')\n",
+      "      print ('Value of m = %d address of m = %x' % (m, id(m)))\n",
+      "      print ('\\nIn FUNCTION Formal parameter variable')\n",
+      "      print ('Value of x = %d address of x = %x' % (x, id(x)))\n",
+      "      print ('\\nIn FUNCTION Static variable')\n",
+      "      print ('Value of temp = %d address of temp = %x' % (fun1.temp, id(fun1.temp)))\n",
+      "\n",
+      "   if fun1.temp < m :\n",
+      "      fun1.temp = m\n",
+      "\n",
+      "   else :\n",
+      "      print ('Value of temp = %d' % fun1.temp)\n",
+      "      fun1.temp = fun1.temp + 1\n",
+      "\n",
+      "print ('\\nIn MAIN Global variable')\n",
+      "print ('Value of m = %d address of m = %x' % (m, id(m)))\n",
+      "print ('\\nIn MAIN Local variable')\n",
+      "print ('Value of i = %d address of i = %x' % (i, id(i)))\n",
+      "      \n",
+      "for i in range (1, 6) :\n",
+      "   fun1 (m)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In MAIN Global variable\n",
+        "Value of m = 10 address of m = 117d0c0\n",
+        "\n",
+        "In MAIN Local variable\n",
+        "Value of i = 1 address of i = 117d198\n",
+        "\n",
+        "In FUNCTION Global variable\n",
+        "Value of m = 10 address of m = 117d0c0\n",
+        "\n",
+        "In FUNCTION Formal parameter variable\n",
+        "Value of x = 10 address of x = 117d0c0\n",
+        "\n",
+        "In FUNCTION Static variable\n",
+        "Value of temp = 0 address of temp = 117d1b0\n",
+        "Value of temp = 10\n",
+        "Value of temp = 11\n",
+        "Value of temp = 12\n",
+        "Value of temp = 13\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.16, Page number: 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter number of Fibonacci terms required '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def nextfibo () :\n",
+      "   if not hasattr(nextfibo, \"first\") : \n",
+      "      nextfibo.first = 0\n",
+      "   if not hasattr(nextfibo, \"second\") : \n",
+      "      nextfibo.second = 1\n",
+      "   if not hasattr(nextfibo, \"third\") : \n",
+      "      nextfibo.third = 0\n",
+      "   nextfibo.third = nextfibo.first + nextfibo.second\n",
+      "   nextfibo.first = nextfibo.second\n",
+      "   nextfibo.second = nextfibo.third\n",
+      "   return nextfibo.first\n",
+      " \n",
+      "print ('%d terms of Fibonacci series are' % n) \n",
+      "print 0,\n",
+      "for count in range (2, n+1) :\n",
+      "   result = nextfibo ()\n",
+      "   print ('%d' % result),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter number of Fibonacci terms required 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10 terms of Fibonacci series are\n",
+        "0 1 1 2 3 5 8 13 21 34\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_7_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_7_3.ipynb
new file mode 100755
index 00000000..73eec220
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_7_3.ipynb
@@ -0,0 +1,862 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: Recursion"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1, Page number: 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration, calculation and result\n",
+      "str = raw_input('Enter a string: ')\n",
+      "\n",
+      "print str"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a string: HELLO\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "HELLO\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page number: 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration, calculation and result\n",
+      "str = raw_input('Enter a string: ')\n",
+      "\n",
+      "print str[::-1]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a string: HELLO\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "OLLEH\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page number: 224"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def fact (n) :\n",
+      "   if n>0 :\n",
+      "      return n * fact (n-1)\n",
+      "   else :\n",
+      "      return 1\n",
+      "\n",
+      "print ('Factorial of %d is %d' % (n, fact (n)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Factorial of 5 is 120\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4, Page number: 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printnum (n) :\n",
+      "   if n>0 :\n",
+      "      printnum (n-1)\n",
+      "      print n,\n",
+      "\n",
+      "print ('Counting up to %d' % n)\n",
+      "printnum (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Counting up to 10\n",
+        "1 2 3 4 5 6 7 8 9 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page number: 229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printeven (n) :\n",
+      "\tif not hasattr(printeven, \"x\") : \n",
+      "\t\tprinteven.x = 2\n",
+      "      \n",
+      "\tif printeven.x <= n :\n",
+      "\t\tprint printeven.x,\n",
+      "\t\tprinteven.x = printeven.x + 2\n",
+      "\t\tprinteven (n)\n",
+      "\n",
+      "print ('Even numbers up to %d are' % n)\n",
+      "printeven (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Even numbers up to 10 are\n",
+        "2 4 6 8 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6, Page number: 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printodd (x, n) :\n",
+      "\tif x <= n : \n",
+      "\t\tprint x,\n",
+      "\t\tprintodd (x+2, n)\n",
+      "\n",
+      "print ('Odd numbers up to %d are' % n)\n",
+      "printodd (1, n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Odd numbers up to 10 are\n",
+        "1 3 5 7 9\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7, Page number: 232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printnum (fro, to) :\n",
+      "\tif fro <= to : \n",
+      "\t\tprint fro,\n",
+      "\t\tprintnum (fro+1, to)\n",
+      "\n",
+      "print ('Number line for %d is' % n)\n",
+      "printnum (-n, n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number line for 5 is\n",
+        "-5 -4 -3 -2 -1 0 1 2 3 4 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8, Page number: 234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def summate (n) :\n",
+      "\tif n > 0 : \n",
+      "\t\treturn n + summate (n-1)\n",
+      "\telse :\n",
+      "\t\treturn 0\n",
+      "\n",
+      "print ('Sum of first %d natural numbers is %d' % (n, summate (n)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of first 10 natural numbers is 55\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9, Page number: 237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def factors (n) :\n",
+      "\tif not hasattr(factors, \"i\") : \n",
+      "\t\tfactors.i = 1\n",
+      "\t\t\n",
+      "\tif factors.i <= n : \n",
+      "\t\tif n % factors.i == 0 :\n",
+      "\t\t\tprint factors.i,\n",
+      "\t\tfactors.i = factors.i + 1\n",
+      "\t\tfactors (n)\n",
+      "\n",
+      "print ('The factors of %d are' % n)\n",
+      "factors (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 28\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The factors of 28 are\n",
+        "1 2 4 7 14 28\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10, Page number: 239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def multiple (n) :\n",
+      "\tif not hasattr(multiple, \"i\") : \n",
+      "\t\tmultiple.i = 1\n",
+      "\t\t\n",
+      "\tif multiple.i <= 10 : \n",
+      "\t\tprint ('%d X %d = %d' % (n, multiple.i, n*multiple.i))\n",
+      "\t\tmultiple.i = multiple.i + 1\n",
+      "\t\tmultiple (n)\n",
+      "\n",
+      "print ('The first 10 multiples of %d are' % n)\n",
+      "multiple (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The first 10 multiples of 6 are\n",
+        "6 X 1 = 6\n",
+        "6 X 2 = 12\n",
+        "6 X 3 = 18\n",
+        "6 X 4 = 24\n",
+        "6 X 5 = 30\n",
+        "6 X 6 = 36\n",
+        "6 X 7 = 42\n",
+        "6 X 8 = 48\n",
+        "6 X 9 = 54\n",
+        "6 X 10 = 60\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11, Page number: 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def checkprime (n) :\n",
+      "   flag = 1\n",
+      "   i = 2\n",
+      "   while i<n and flag :\n",
+      "      if n%i == 0 :\n",
+      "         flag = 0\n",
+      "      i = i+1\n",
+      "   return flag\n",
+      "\n",
+      "result = checkprime (n)\n",
+      "\n",
+      "if result :\n",
+      "   print ('The number %d is PRIME ' % n)\n",
+      "else :\n",
+      "   print ('The number %d is COMPOSITE ' % n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 19\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number 19 is PRIME \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12, Page number: 243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def Fibonacci (n) :\n",
+      "\tif n < 2 :\n",
+      "\t\treturn n\n",
+      "\telse :\n",
+      "\t\treturn Fibonacci (n-1) + Fibonacci (n-2)\n",
+      "\t\n",
+      "\n",
+      "print ('The %d terms of Fibonacci series are' % n)\n",
+      "for i in range (0, n) :\n",
+      "\tprint ('%d' % Fibonacci (i)),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The 10 terms of Fibonacci series are\n",
+        "0 1 1 2 3 5 8 13 21 34\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13, Page number: 244"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "sum = n\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def sumofdigit (n, sum) :\n",
+      "   while sum>=10 :\n",
+      "      sum = 0\n",
+      "      while n>0 :\n",
+      "         sum = sum + n%10\n",
+      "         n = n/10\n",
+      "      n = sum\n",
+      "   return sum\n",
+      "\n",
+      "result = sumofdigit (n, sum)\n",
+      "\n",
+      "print ('Sum of digits of %d is %d' % (n, result))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 8626\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits of 8626 is 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14, Page number: 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def prime (n) :\n",
+      "   for num in range (3, n+1) :\n",
+      "      flag = 1\n",
+      "      for i in range (2, num) :\n",
+      "         if num % i == 0 : \n",
+      "            flag = 0\n",
+      "      if flag :\n",
+      "         print ('%d ' % num),\n",
+      "\n",
+      "print ('Prime numbers upto %d are' % n)\n",
+      "prime (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 30\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Prime numbers upto 30 are\n",
+        "3  5  7  11  13  17  19  23  29 \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15, Page number: 246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int (raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def pattern (n) :\n",
+      "\tif n > 0 :\n",
+      "\t\tpattern (n-1)\n",
+      "\t\tfor count in range (1, n+1) :\n",
+      "\t\t\tprint ('%d' % count),\n",
+      "   \tprint\n",
+      "   \t\n",
+      "pattern (n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1\n",
+        "1 2\n",
+        "1 2 3\n",
+        "1 2 3 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.16, Page number: 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "n = int(raw_input('Enter any number '))\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printline (x, total) :\n",
+      "\tif x > 0 :\n",
+      "\t\tprintline (x-1, total)\n",
+      "\t\tfor blk in range (1, total-x+1) :\n",
+      "\t\t\tprint ' ',\n",
+      "\t\tfor up in range (1, x+1) :\n",
+      "\t\t\tprint ('%d' % up),\n",
+      "\t\tdown = x-1\n",
+      "\t\twhile down>0 :\n",
+      "\t\t\tprint ('%d' % down),\n",
+      "\t\t\tdown = down-1\n",
+      "\t\tprint\n",
+      "\n",
+      "printline (n, n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "      1\n",
+        "    1 2 1\n",
+        "  1 2 3 2 1\n",
+        "1 2 3 4 3 2 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17, Page number: 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def listsum () :\n",
+      "\tn = int(raw_input('Enter any number '))\n",
+      "\tif n > 0 :\n",
+      "\t\treturn n + listsum ()\n",
+      "\telse :\n",
+      "\t\treturn 0\n",
+      "\t\t\n",
+      "sum = listsum ()\n",
+      "print ('Sum of list of numbers is %d' % sum)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any number -4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of list of numbers is 6\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_8_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_8_3.ipynb
new file mode 100755
index 00000000..ccafa905
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_8_3.ipynb
@@ -0,0 +1,2718 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8: Arrays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page number: 256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "word = array('c', ['H', 'e', 'l', 'l', 'o'])\n",
+      "\n",
+      "# Result\n",
+      "print ('The contents of word [] is %s ' % ''.join(word))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The contents of word [] is Hello \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2, Page number: 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "marks = [0]*10\n",
+      "avg = 0\n",
+      "\n",
+      "# Calculation and result\n",
+      "for i in range (0, 10) :\n",
+      "\tmarks[i] = int(raw_input('Enter value in cell [%d] ' % i))\n",
+      "\tavg = avg + marks[i]\n",
+      "\n",
+      "avg = avg/10\n",
+      "print ('The average marks are %f' % avg)\n",
+      "\n",
+      "for i in range (0, 10) :\n",
+      "\tif marks[i] < avg :\n",
+      "\t\tprint ('Marks in cell [%d] are %d less than average' % (i, marks[i]))\n",
+      "\telse :\n",
+      "\t\tprint ('Marks in cell [%d] are %d above/equal to average' % (i, marks[i]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [0] 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [1] 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [2] 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [3] 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [4] 56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [5] 98\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [6] 53\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [7] 82\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [8] 74\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [9] 39\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average marks are 63.000000\n",
+        "Marks in cell [0] are 45 less than average\n",
+        "Marks in cell [1] are 78 above/equal to average\n",
+        "Marks in cell [2] are 65 above/equal to average\n",
+        "Marks in cell [3] are 43 less than average\n",
+        "Marks in cell [4] are 56 less than average\n",
+        "Marks in cell [5] are 98 above/equal to average\n",
+        "Marks in cell [6] are 53 less than average\n",
+        "Marks in cell [7] are 82 above/equal to average\n",
+        "Marks in cell [8] are 74 above/equal to average\n",
+        "Marks in cell [9] are 39 less than average\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page number: 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "data = raw_input('Enter the string ')\n",
+      "\n",
+      "# Calculation and result\n",
+      "rev_data = reversed (data)\n",
+      "\n",
+      "if list (data) == list (rev_data) :\n",
+      "\tprint ('The string %s is a PALINDROME' % data)\n",
+      "else :\n",
+      "\tprint ('The string %s is not a PALINDROME' % data)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the string MADAM\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The string MADAM is a PALINDROME\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4, Page number: 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "marks = [0]*10\n",
+      "avg = 0\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readdata (marks) :\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\tmarks[i] = int(raw_input('Enter value in cell [%d] ' % i))\n",
+      "\t\t\n",
+      "def average (marks) :\n",
+      "\tmean = 0\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\tmean = mean + marks[i]\n",
+      "\treturn mean/10\n",
+      "\n",
+      "def printdata (marks, avg) :\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\tif marks[i] < avg :\n",
+      "\t\t\tprint ('Marks in cell [%d] are %d less than average' % (i, marks[i]))\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('Marks in cell [%d] are %d above/equal to average' % (i, marks[i]))\t\n",
+      "\n",
+      "\n",
+      "readdata (marks)\n",
+      "avg = average (marks)\t\n",
+      "print ('The average marks are %f' % avg)\n",
+      "printdata (marks, avg)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [0] 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [1] 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [2] 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [3] 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [4] 56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [5] 98\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [6] 53\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [7] 82\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [8] 74\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell [9] 39\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average marks are 63.000000\n",
+        "Marks in cell [0] are 45 less than average\n",
+        "Marks in cell [1] are 78 above/equal to average\n",
+        "Marks in cell [2] are 65 above/equal to average\n",
+        "Marks in cell [3] are 43 less than average\n",
+        "Marks in cell [4] are 56 less than average\n",
+        "Marks in cell [5] are 98 above/equal to average\n",
+        "Marks in cell [6] are 53 less than average\n",
+        "Marks in cell [7] are 82 above/equal to average\n",
+        "Marks in cell [8] are 74 above/equal to average\n",
+        "Marks in cell [9] are 39 less than average\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5, Page number: 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "a = [10, 20, 30, 40, 50]\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def fun (x, y, z) :\n",
+      "\tx = x + 5\n",
+      "\ty = y + 10\n",
+      "\tz[4] = z[4] + 100\n",
+      "\n",
+      "\n",
+      "for i in range (0, 5) :\n",
+      "\tprint ('The cell [%d] contains %d' % (i, a[i]))\n",
+      "\tprint ('The address of cell [%d] is %x' % (i, id(a[i])))\n",
+      "\t\t\n",
+      "fun (a[0], a[3], a)\t\n",
+      "print\n",
+      "\n",
+      "for i in range (0, 5) :\n",
+      "\tprint ('The cell [%d] contains %d' % (i, a[i]))\n",
+      "\tprint ('The address of cell [%d] is %x' % (i, id(a[i])))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cell [0] contains 10\n",
+        "The address of cell [0] is 18bc0c0\n",
+        "The cell [1] contains 20\n",
+        "The address of cell [1] is 18bbfd0\n",
+        "The cell [2] contains 30\n",
+        "The address of cell [2] is 18bbee0\n",
+        "The cell [3] contains 40\n",
+        "The address of cell [3] is 18bc5b8\n",
+        "The cell [4] contains 50\n",
+        "The address of cell [4] is 18bc4c8\n",
+        "\n",
+        "The cell [0] contains 10\n",
+        "The address of cell [0] is 18bc0c0\n",
+        "The cell [1] contains 20\n",
+        "The address of cell [1] is 18bbfd0\n",
+        "The cell [2] contains 30\n",
+        "The address of cell [2] is 18bbee0\n",
+        "The cell [3] contains 40\n",
+        "The address of cell [3] is 18bc5b8\n",
+        "The cell [4] contains 150\n",
+        "The address of cell [4] is 18bcaf8\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6, Page number: 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "data = raw_input('Enter the string ')\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def revstring (data) :\n",
+      "\trev_data = reversed (data)\n",
+      "\n",
+      "\tif list (data) == list (rev_data) :\n",
+      "\t\tprint ('The string %s is a PALINDROME' % data)\n",
+      "\telse :\n",
+      "\t\tprint ('The string %s is not a PALINDROME' % data)\n",
+      "\n",
+      "revstring (data)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the string python\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The string python is not a PALINDROME\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7, Page number: 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "a = [0]*10\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readnsort (a) :\n",
+      "\tpos = 0\n",
+      "\tfor len in range (0, 10) :\n",
+      "\t\tnum = int(raw_input('Enter data element no. %d ' % len))\n",
+      "\t\t\n",
+      "\t\tfor pos in range (0, len+1) :\n",
+      "\t\t\tif num < a[pos] :\n",
+      "\t\t\t\tbreak\n",
+      "\t\t\t\t\n",
+      "\t\ti = len\t\t\n",
+      "\t\twhile i>=pos :\n",
+      "\t\t\ta[i] = a[i-1]\n",
+      "\t\t\ti = i-1\n",
+      "\t\ta[pos] = num\n",
+      "\t\t\t\n",
+      "def printdata (a) :\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\t\t\t\n",
+      "\t\t\t\n",
+      "readnsort (a)\n",
+      "print\n",
+      "printdata (a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 0 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 1 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 2 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 3 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 4 56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 5 98\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 6 53\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 7 82\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 8 74\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data element no. 9 39\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The data in cell [0] is 39 \n",
+        "The data in cell [1] is 43 \n",
+        "The data in cell [2] is 45 \n",
+        "The data in cell [3] is 53 \n",
+        "The data in cell [4] is 56 \n",
+        "The data in cell [5] is 65 \n",
+        "The data in cell [6] is 74 \n",
+        "The data in cell [7] is 78 \n",
+        "The data in cell [8] is 82 \n",
+        "The data in cell [9] is 98 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8, Page number: 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readdata () :\n",
+      "\ta = []\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
+      "\treturn a\n",
+      "\n",
+      "def sortdata (a) :\n",
+      "\tfor p in range (0, 9) :\n",
+      "\t\tmin = p\n",
+      "\t\tfor i in range (p, 10) :\n",
+      "\t\t\tif a[min] > a[i] :\n",
+      "\t\t\t\tmin = i\n",
+      "\t\ttemp = a[p]\n",
+      "\t\ta[p] = a[min]\n",
+      "\t\ta[min] = temp\n",
+      "\treturn a\n",
+      "\n",
+      "def printdata (a) :\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
+      "\n",
+      "a = readdata ()\n",
+      "a = sortdata (a)\n",
+      "print ('\\nThe sorted data is ')\n",
+      "printdata (a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [0] 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [1] 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [2] 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [3] 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [4] 56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [5] 98\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [6] 53\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [7] 82\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [8] 74\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [9] 39\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The sorted data is \n",
+        "The data in cell [0] is 39 \n",
+        "The data in cell [1] is 43 \n",
+        "The data in cell [2] is 45 \n",
+        "The data in cell [3] is 53 \n",
+        "The data in cell [4] is 56 \n",
+        "The data in cell [5] is 65 \n",
+        "The data in cell [6] is 74 \n",
+        "The data in cell [7] is 78 \n",
+        "The data in cell [8] is 82 \n",
+        "The data in cell [9] is 98 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9, Page number: 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readdata () :\n",
+      "\ta = []\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
+      "\treturn a\n",
+      "\n",
+      "def bubble (a) :\n",
+      "\tfor p in range (0, 10) :\n",
+      "\t\tfor i in range (0, 9-p) :\n",
+      "\t\t\tif a[i] > a[i+1] :\n",
+      "\t\t\t\ttemp = a[i]\n",
+      "\t\t\t\ta[i] = a[i+1]\n",
+      "\t\t\t\ta[i+1] = temp\n",
+      "\treturn a\n",
+      "\n",
+      "def printdata (a) :\n",
+      "\tfor i in range (0, 10) :\n",
+      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
+      "\n",
+      "a = readdata ()\n",
+      "a = bubble (a)\n",
+      "print ('\\nThe sorted data is ')\n",
+      "printdata (a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [0] 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [1] 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [2] 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [3] 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [4] 56\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [5] 98\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [6] 53\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [7] 82\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [8] 74\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [9] 39\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The sorted data is \n",
+        "The data in cell [0] is 39 \n",
+        "The data in cell [1] is 43 \n",
+        "The data in cell [2] is 45 \n",
+        "The data in cell [3] is 53 \n",
+        "The data in cell [4] is 56 \n",
+        "The data in cell [5] is 65 \n",
+        "The data in cell [6] is 74 \n",
+        "The data in cell [7] is 78 \n",
+        "The data in cell [8] is 82 \n",
+        "The data in cell [9] is 98 \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10, Page number: 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readdata (size) :\n",
+      "\ta = []\n",
+      "\tfor i in range (0, size) :\n",
+      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
+      "\treturn a\n",
+      "\n",
+      "def bubble (a, size) :\n",
+      "\tfor p in range (0, size) :\n",
+      "\t\tfor i in range (0, size-p-1) :\n",
+      "\t\t\tif a[i] > a[i+1] :\n",
+      "\t\t\t\ttemp = a[i]\n",
+      "\t\t\t\ta[i] = a[i+1]\n",
+      "\t\t\t\ta[i+1] = temp\n",
+      "\treturn a\n",
+      "\n",
+      "def printdata (a, size) :\n",
+      "\tfor i in range (0, size) :\n",
+      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
+      "\n",
+      "def lsearch (a, key, size) :\n",
+      "\tfor i in range (0, size) :\n",
+      "\t\tif a[i] == key :\n",
+      "\t\t\treturn i\n",
+      "\t\tif i == size :\n",
+      "\t\t\treturn -1\n",
+      "\t\t\t\n",
+      "size = int(raw_input('Enter the total number of elements '))\n",
+      "a = readdata (size)\n",
+      "a = bubble (a, size)\n",
+      "print ('\\nThe sorted array is ')\n",
+      "printdata (a, size)\n",
+      "\n",
+      "find = int(raw_input('\\nEnter the element to be searched '))\n",
+      "pos = lsearch (a, find, size)\n",
+      "\n",
+      "if pos<0 :\n",
+      "\tprint ('\\nThe element %d NOT FOUND ' % find)\n",
+      "else :\n",
+      "\tprint ('\\nThe element %d found at position %d ' % (find, pos))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the total number of elements 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [0] 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [1] 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [2] 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [3] 34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [4] 49\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The sorted array is \n",
+        "The data in cell [0] is 34 \n",
+        "The data in cell [1] is 45 \n",
+        "The data in cell [2] is 49 \n",
+        "The data in cell [3] is 65 \n",
+        "The data in cell [4] is 78 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter the element to be searched 65\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The element 65 found at position 3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11, Page number: 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readdata (size) :\n",
+      "\ta = []\n",
+      "\tfor i in range (0, size) :\n",
+      "\t\ta.append(int(raw_input('Enter data in cell [%d] ' % i)))\n",
+      "\treturn a\n",
+      "\n",
+      "def bubble (a, size) :\n",
+      "\tfor p in range (0, size) :\n",
+      "\t\tfor i in range (0, size-p-1) :\n",
+      "\t\t\tif a[i] > a[i+1] :\n",
+      "\t\t\t\ttemp = a[i]\n",
+      "\t\t\t\ta[i] = a[i+1]\n",
+      "\t\t\t\ta[i+1] = temp\n",
+      "\treturn a\n",
+      "\n",
+      "def printdata (a, size) :\n",
+      "\tfor i in range (0, size) :\n",
+      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
+      "\n",
+      "def bsearch (a, key, size) :\n",
+      "\tfirst = 0\n",
+      "\tlast = size - 1\n",
+      "\tmid = (first + last)/2\n",
+      "\t\n",
+      "\twhile (first <= last and a[mid] != key) :\n",
+      "\t\tif a[mid] < key :\n",
+      "\t\t\tfirst = mid + 1\n",
+      "\t\telse :\n",
+      "\t\t\tif a[mid] > key :\n",
+      "\t\t\t\tlast = mid - 1\n",
+      "\t\tmid = (first + last)/2\n",
+      "\t\n",
+      "\tif a[mid] == key :\n",
+      "\t\treturn mid\n",
+      "\telse :\n",
+      "\t\treturn -1\n",
+      "\t\t\t\n",
+      "size = int(raw_input('Enter the number of elements in the array '))\n",
+      "a = readdata (size)\n",
+      "a = bubble (a, size)\n",
+      "print ('\\nThe sorted array is ')\n",
+      "printdata (a, size)\n",
+      "\n",
+      "find = int(raw_input('\\nPlease enter the element to be searched for '))\n",
+      "pos = bsearch (a, find, size)\n",
+      "\n",
+      "if pos<0 :\n",
+      "\tprint ('\\nThe element %d NOT FOUND ' % find)\n",
+      "else :\n",
+      "\tprint ('\\nThe element %d found at position %d ' % (find, pos))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of elements in the array 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [0] 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [1] 78\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [2] 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [3] 34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [4] 49\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The sorted array is \n",
+        "The data in cell [0] is 34 \n",
+        "The data in cell [1] is 45 \n",
+        "The data in cell [2] is 49 \n",
+        "The data in cell [3] is 65 \n",
+        "The data in cell [4] is 78 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Please enter the element to be searched for 45\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The element 45 found at position 1 \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12, Page number: 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readncompute (no_of_students) :\n",
+      "\ta = [[0 for y in xrange(5)] for x in xrange(no_of_students)]\n",
+      "\tper = [0 for x in range(no_of_students)]\n",
+      "\tavg = 0\n",
+      "\t\n",
+      "\tfor i in range (0, no_of_students) :\n",
+      "\t\tprint ('\\nEnter marks of 5 subjects for Roll no. %d ' % (i+1))\n",
+      "\t\t\n",
+      "\t\tfor j in range (0, 5) :\n",
+      "\t\t\ta[i][j] = int(raw_input('Subject %d = ' % (j+1)))\n",
+      "\t\t\tper[i] = per[i] + a[i][j]\n",
+      "\t\t\t\n",
+      "\t\tper[i] = per[i]/5\n",
+      "\t\tavg = avg + per[i]\n",
+      "\t\t\n",
+      "\treturn a, avg/no_of_students, per\n",
+      "\n",
+      "def printdata (a, no_of_students, avg, per) :\n",
+      "\tfor i in range (0, no_of_students) :\n",
+      "\t\tprint ('\\nMarks for Roll no. %d are ' % (i+1))\n",
+      "\t\tfor j in range (0, 5) :\n",
+      "\t\t\tprint a[i][j]\n",
+      "\t\tprint ('Percentage for Roll no. %d is %f ' % ((i+1), per[i]))\n",
+      "\t\t\n",
+      "\t\tif per[i] < avg :\n",
+      "\t\t\tprint ('Below average')\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('Above average')\n",
+      "\t\t\n",
+      "\n",
+      "no_of_students = int(raw_input('How many students marks do you want to enter '))\n",
+      "a, average, per = readncompute (no_of_students)\n",
+      "print ('\\nThe average marks are %f ' % average)\n",
+      "printdata (a, no_of_students, average, per)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many students marks do you want to enter 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter marks of 5 subjects for Roll no. 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1 = 75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2 = 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3 = 51\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 4 = 68\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 5 = 73\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter marks of 5 subjects for Roll no. 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1 = 49\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2 = 71\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3 = 62\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 4 = 53\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 5 = 50\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter marks of 5 subjects for Roll no. 3 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1 = 90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2 = 82\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3 = 74\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 4 = 93\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 5 = 89\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The average marks are 68.000000 \n",
+        "\n",
+        "Marks for Roll no. 1 are \n",
+        "75\n",
+        "43\n",
+        "51\n",
+        "68\n",
+        "73\n",
+        "Percentage for Roll no. 1 is 62.000000 \n",
+        "Below average\n",
+        "\n",
+        "Marks for Roll no. 2 are \n",
+        "49\n",
+        "71\n",
+        "62\n",
+        "53\n",
+        "50\n",
+        "Percentage for Roll no. 2 is 57.000000 \n",
+        "Below average\n",
+        "\n",
+        "Marks for Roll no. 3 are \n",
+        "90\n",
+        "82\n",
+        "74\n",
+        "93\n",
+        "89\n",
+        "Percentage for Roll no. 3 is 85.000000 \n",
+        "Above average\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.13, Page number: 296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def getchoice () :\n",
+      "\tprint\n",
+      "\tprint ('1 - Input two matrices ')\n",
+      "\tprint ('2 - Add and display sum ')\n",
+      "\tprint ('3 - Exit ')\n",
+      "\t\n",
+      "\tchoice = int(raw_input('Enter choice : '))\n",
+      "\treturn choice\n",
+      "\n",
+      "def read (matrix, rows, cols) :\t\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tprint ('\\nEnter elements for row %d ' % (i+1))\t\t\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
+      "\n",
+      "def add (sum, augend, addend, rows, cols) :\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tsum[i][j] = augend[i][j] + addend[i][j]\n",
+      "\n",
+      "def display2matrices (augend, addend, rows, cols) :\n",
+      "\tprint ('\\nMatrix A ')\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tprint ('%3d ' % augend[i][j]),\n",
+      "\t\tprint\n",
+      "\tprint ('\\nMatrix B ')\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tprint ('%3d ' % addend[i][j]),\n",
+      "\t\tprint\n",
+      "\n",
+      "def display3matrices (sum, augend, addend, rows, cols) :\n",
+      "\tprint\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tprint ('%3d ' % augend[i][j]),\n",
+      "\t\tprint ('%s ' % ('+' if i==rows/2 else ' ')),\n",
+      "\t\t\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tprint ('%2d ' % addend[i][j]),\n",
+      "\t\tprint ('%s ' % ('=' if i==rows/2 else ' ')),\n",
+      "\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tprint ('%2d ' % sum[i][j]),\n",
+      "\t\tprint\n",
+      "\n",
+      "def display (matrix, rows, cols) :\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tprint ('%3d ' % matrix[i][j]),\n",
+      "\t\tprint\n",
+      "\n",
+      "augend = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "addend = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "sum = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\tchoice = getchoice ()\n",
+      "\t\n",
+      "\tif choice == 1 :\n",
+      "\t\trows = int(raw_input('\\nEnter the number of rows '))\t\t\n",
+      "\t\tcols = int(raw_input('Enter the number of columns '))\n",
+      "\t\t\n",
+      "\t\tprint ('\\n\\nEnter the first matrix ')\n",
+      "\t\tread (augend, rows, cols)\n",
+      "\t\tprint ('\\n\\nEnter the second matrix ')\n",
+      "\t\tread (addend, rows, cols)\n",
+      "\t\tprint ('\\n\\nYou entered ')\n",
+      "\t\tdisplay2matrices (augend, addend, rows, cols)\n",
+      "\t\n",
+      "\telif choice == 2 :\n",
+      "\t\tadd (sum, augend, addend, rows, cols)\n",
+      "\t\tdisplay3matrices (sum, augend, addend, rows, cols)\n",
+      "\t\n",
+      "\telif choice == 3 :\n",
+      "\t\tcondition = False\n",
+      "\t\texit ()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1 - Input two matrices \n",
+        "2 - Add and display sum \n",
+        "3 - Exit \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter choice : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter the number of rows 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of columns 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Enter the first matrix \n",
+        "\n",
+        "Enter elements for row 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][1] = 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][2] = 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][3] = 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter elements for row 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][1] = 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][2] = 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][3] = 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter elements for row 3 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][1] = 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][2] = 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][3] = 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Enter the second matrix \n",
+        "\n",
+        "Enter elements for row 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][1] = 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][2] = 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][3] = 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter elements for row 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][1] = 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][2] = 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][3] = 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter elements for row 3 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][1] = 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][2] = 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][3] = 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "You entered \n",
+        "\n",
+        "Matrix A \n",
+        "  1    2    3 \n",
+        "  3    4    5 \n",
+        "  5    6    7 \n",
+        "\n",
+        "Matrix B \n",
+        "  3    5    1 \n",
+        "  6    5    3 \n",
+        "  8    7    4 \n",
+        "\n",
+        "1 - Input two matrices \n",
+        "2 - Add and display sum \n",
+        "3 - Exit \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter choice : 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "  1    2    3      3   5   1      4   7   4 \n",
+        "  3    4    5  +   6   5   3  =   9   9   8 \n",
+        "  5    6    7      8   7   4     13  13  11 \n",
+        "\n",
+        "1 - Input two matrices \n",
+        "2 - Add and display sum \n",
+        "3 - Exit \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter choice : 3\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14, Page number: 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def readmatrix (matrix, rows, cols) :\t\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tprint ('\\nPlease enter elements for row %d ' % (i+1))\t\t\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
+      "\n",
+      "def multiply (prod, a, b, Ar, Ac, Bc) :\n",
+      "\tfor i in range (0, Ar) :\n",
+      "\t\tfor j in range (0, Bc) :\n",
+      "\t\t\tprod[i][j] = 0\n",
+      "\t\t\tfor k in range (0, Ac) :\n",
+      "\t\t\t\tprod[i][j] += a[i][k] * b[k][j]\n",
+      "\n",
+      "def display (matrix, rows, cols) :\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tprint ('%3d ' % matrix[i][j]),\n",
+      "\t\tprint\n",
+      "\n",
+      "a = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "b = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "product = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "\n",
+      "condition = True\n",
+      "while condition  :\n",
+      "\tArows, Acols = map(int, raw_input('Enter order of matrix A: ').split())\n",
+      "\tBrows, Bcols = map(int, raw_input('Enter order of matrix B: ').split())\n",
+      "\n",
+      "\tif Acols == Brows :\n",
+      "\t\tcondition = False\n",
+      "\t\tbreak\n",
+      "\telse : \n",
+      "\t\tprint ('\\nMatrices are not compatible for multiplication. Enter the order again. ')\n",
+      "\t\tcontinue\n",
+      "\t\n",
+      "print ('\\nEnter matrix A: ')\n",
+      "readmatrix (a, Arows, Acols)\n",
+      "\t\n",
+      "print ('\\nEnter matrix B: ')\n",
+      "readmatrix (b, Brows, Bcols)\n",
+      "\n",
+      "multiply (product, a, b, Arows, Acols, Bcols)\n",
+      "print ('\\nThe resultant product (A.B) is: ')\n",
+      "display (product, Arows, Bcols)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter order of matrix A: 2 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter order of matrix B: 3 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter matrix A: \n",
+        "\n",
+        "Please enter elements for row 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][1] = 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][2] = 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][3] = 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Please enter elements for row 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][1] = 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][2] = 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][3] = 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter matrix B: \n",
+        "\n",
+        "Please enter elements for row 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][1] = 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][2] = 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Please enter elements for row 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][1] = 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][2] = 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Please enter elements for row 3 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][1] = 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][2] = 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The resultant product (A.B) is: \n",
+        " 20   14 \n",
+        " 56   41 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.15, Page number: 303"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def getchoice () :\n",
+      "\tprint\n",
+      "\tprint ('1 - Input square matrix ')\n",
+      "\tprint ('2 - Display upper triangular matrix ')\n",
+      "\tprint ('3 - Display lower triangular matrix ')\n",
+      "\tprint ('4 - Exit ')\n",
+      "\t\n",
+      "\tcondition = True\n",
+      "\twhile condition :\n",
+      "\t\tchoice = int(raw_input('Enter choice : '))\n",
+      "\t\t\n",
+      "\t\tif choice in range (1, 5) :\n",
+      "\t\t\tcondition = False\n",
+      "\t\t\t\n",
+      "\treturn choice\n",
+      "\n",
+      "def displayupper (matrix, order) :\n",
+      "\tprint\n",
+      "\tfor i in range (0, order) :\n",
+      "\t\tfor j in range (0, order) :\n",
+      "\t\t\tif j >= i :\n",
+      "\t\t\t\tprint ('%3d ' % matrix[i][j]),\n",
+      "\t\t\telse :\n",
+      "\t\t\t\tprint ('%3d ' % 0),\n",
+      "\t\tprint\n",
+      "\t\t\n",
+      "def displaylower (matrix, order) :\n",
+      "\tprint\n",
+      "\tfor i in range (0, order) :\n",
+      "\t\tfor j in range (0, order) :\n",
+      "\t\t\tif j <= i :\n",
+      "\t\t\t\tprint ('%3d ' % matrix[i][j]),\n",
+      "\t\t\telse :\n",
+      "\t\t\t\tprint ('%3d ' % 0),\n",
+      "\t\tprint\n",
+      "\n",
+      "def read (matrix, rows, cols) :\t\n",
+      "\tfor i in range (0, rows) :\n",
+      "\t\tprint ('\\nEnter %d elements for row %d ' % (rows, (i+1)))\t\t\n",
+      "\t\tfor j in range (0, cols) :\n",
+      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
+      "\n",
+      "matrix = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "\n",
+      "condition = True\n",
+      "while condition :\n",
+      "\tchoice = getchoice ()\n",
+      "\t\n",
+      "\tif choice == 1 :\n",
+      "\t\torder = int(raw_input('\\nEnter order of square matrix '))\t\t\n",
+      "\t\tread (matrix, order, order)\n",
+      "\t\n",
+      "\telif choice == 2 :\n",
+      "\t\tdisplayupper (matrix, order)\n",
+      "\t\n",
+      "\telif choice == 3 :\n",
+      "\t\tdisplaylower (matrix, order)\n",
+      "\t\t\n",
+      "\telif choice == 4 :\n",
+      "\t\tcondition = False\n",
+      "\t\texit ()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1 - Input square matrix \n",
+        "2 - Display upper triangular matrix \n",
+        "3 - Display lower triangular matrix \n",
+        "4 - Exit \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter choice : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter order of square matrix 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter 3 elements for row 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][1] = 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][2] = 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][3] = 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter 3 elements for row 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][1] = 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][2] = 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][3] = 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter 3 elements for row 3 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][1] = 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][2] = 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][3] = 9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1 - Input square matrix \n",
+        "2 - Display upper triangular matrix \n",
+        "3 - Display lower triangular matrix \n",
+        "4 - Exit \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter choice : 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "  1    2    3 \n",
+        "  0    5    6 \n",
+        "  0    0    9 \n",
+        "\n",
+        "1 - Input square matrix \n",
+        "2 - Display upper triangular matrix \n",
+        "3 - Display lower triangular matrix \n",
+        "4 - Exit \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter choice : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "  1    0    0 \n",
+        "  4    5    0 \n",
+        "  7    8    9 \n",
+        "\n",
+        "1 - Input square matrix \n",
+        "2 - Display upper triangular matrix \n",
+        "3 - Display lower triangular matrix \n",
+        "4 - Exit \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter choice : 4\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.16, Page number: 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def readmatrix (matrix, n) :\n",
+      "\tfor i in range (0, n) :\n",
+      "\t\tprint ('\\nEnter elements for row no %d ' % (i+1))\n",
+      "\t\tfor j in range (0, n) :\n",
+      "\t\t\tmatrix[i][j] = int(raw_input('matrix[%d][%d] = ' % ((i+1),(j+1))))\n",
+      "\n",
+      "def display (matrix, n) :\n",
+      "\tprint\n",
+      "\tfor i in range (0, n) :\n",
+      "\t\tfor j in range (0, n) :\n",
+      "\t\t\tprint ('%5d ' % matrix[i][j]),\n",
+      "\t\tprint\n",
+      "\t\t\n",
+      "def isunitmatrix(matrix, n) :\n",
+      "\tfor i in range (0, n) :\n",
+      "\t\tif matrix[i][i] != 1 :\n",
+      "\t\t\treturn 0\n",
+      "\t\n",
+      "\tfor i in range (0, n) :\n",
+      "\t\tfor j in range (0, n) :\n",
+      "\t\t\tif i != j :\n",
+      "\t\t\t\tif matrix[i][j] != 0 :\n",
+      "\t\t\t\t\treturn 0\n",
+      "\treturn 1\n",
+      "\n",
+      "matrix = [[0 for y in xrange(10)] for x in xrange(10)]\n",
+      "n = int(raw_input('Enter order of matrix '))\n",
+      "readmatrix (matrix, n)\n",
+      "display (matrix, n)\n",
+      "\n",
+      "if (isunitmatrix (matrix, n)) :\n",
+      "\tprint \"UNIT MATRIX\"\n",
+      "else :\n",
+      "\tprint \"NOT A UNIT MATRIX\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter order of matrix 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter elements for row no 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][1] = 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][2] = 0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[1][3] = 0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter elements for row no 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][1] = 0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][2] = 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[2][3] = 0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter elements for row no 3 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][1] = 0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][2] = 0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "matrix[3][3] = 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "    1      0      0 \n",
+        "    0      1      0 \n",
+        "    0      0      1 \n",
+        "UNIT MATRIX\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.17, Page number: 311"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "string = raw_input('Enter a string: ')\n",
+      "\n",
+      "# Calculation and result\n",
+      "swap_string = string.swapcase()\n",
+      "\n",
+      "print swap_string"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a string: hello world\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "HELLO WORLD\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.18, Page number: 312"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "string = raw_input('Enter a string ')\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def printstr (string) :\n",
+      "\tfor i in range (0, len(string)) :\n",
+      "\t\tfor j in range (0, i+1) :\n",
+      "\t\t\tprint string[j],\n",
+      "\t\tprint\n",
+      "\t\n",
+      "\ti = i - 2\n",
+      "\twhile i >= 0 :\n",
+      "\t\tfor j in range (0, i+1) :\n",
+      "\t\t\tprint string[j],\n",
+      "\t\tprint\n",
+      "\t\ti = i - 1\n",
+      "\t\t\n",
+      "printstr (string)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a string hello\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "h\n",
+        "h e\n",
+        "h e l\n",
+        "h e l l\n",
+        "h e l l o\n",
+        "h e l\n",
+        "h e\n",
+        "h\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.19, Page number: 314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = raw_input('Enter any string ')\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def copy (a) :\n",
+      "\tb = a\n",
+      "\treturn b\n",
+      "\t\n",
+      "b = copy (a)\n",
+      "print\n",
+      "print ('The first string is %s ' % a)\n",
+      "print ('The second string is %s ' % b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any string hello\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The first string is hello \n",
+        "The second string is hello \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.20, Page number: 315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "a = raw_input('Enter the major string: ')\n",
+      "b = raw_input('Enter the minor string: ')\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def ispart (a, b) :\n",
+      "\tif b in a :\n",
+      "\t\tprint ('The string \"%s\" is A SUBSTRING of \"%s\" ' % (b, a))\n",
+      "\telse :\n",
+      "\t\tprint ('The string \"%s\" is NOT A SUBSTRING of \"%s\" ' % (b, a))\n",
+      "\n",
+      "ispart (a, b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the major string: python is a programming language\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the minor string: python\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The string \"python\" is A SUBSTRING of \"python is a programming language\" \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/Chapter_9_3.ipynb b/Magnifying_C_by_Arpita_Goyal/Chapter_9_3.ipynb
new file mode 100755
index 00000000..acb61edf
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/Chapter_9_3.ipynb
@@ -0,0 +1,1093 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: More on Pointers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page number: 318"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "a = [[1,2,3,4,5],\n",
+      "     [6,7,8,9,10],\n",
+      "     [11,12,13,14,15],\n",
+      "     [16,17,18,19,20],\n",
+      "     [21,22,23,24,25]]\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('The value of a is %x ' % id(a))\n",
+      "\n",
+      "for i in range (0, 5) :\n",
+      "\tprint ('The value of a[%d] is %x ' % (i, id(i)))\n",
+      "\n",
+      "for i in range (0, 5) :\n",
+      "\tfor j in range (0, 5) :       \n",
+      "\t\tprint ('The cell [%d][%d] i.e. %x has %d ' % (i, j, id(a[i][j]), a[i][j]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of a is 7f55dc3e08c0 \n",
+        "The value of a[0] is 18861b0 \n",
+        "The value of a[1] is 1886198 \n",
+        "The value of a[2] is 1886180 \n",
+        "The value of a[3] is 1886168 \n",
+        "The value of a[4] is 1886150 \n",
+        "The cell [0][0] i.e. 1886198 has 1 \n",
+        "The cell [0][1] i.e. 1886180 has 2 \n",
+        "The cell [0][2] i.e. 1886168 has 3 \n",
+        "The cell [0][3] i.e. 1886150 has 4 \n",
+        "The cell [0][4] i.e. 1886138 has 5 \n",
+        "The cell [1][0] i.e. 1886120 has 6 \n",
+        "The cell [1][1] i.e. 1886108 has 7 \n",
+        "The cell [1][2] i.e. 18860f0 has 8 \n",
+        "The cell [1][3] i.e. 18860d8 has 9 \n",
+        "The cell [1][4] i.e. 18860c0 has 10 \n",
+        "The cell [2][0] i.e. 18860a8 has 11 \n",
+        "The cell [2][1] i.e. 1886090 has 12 \n",
+        "The cell [2][2] i.e. 1886078 has 13 \n",
+        "The cell [2][3] i.e. 1886060 has 14 \n",
+        "The cell [2][4] i.e. 1886048 has 15 \n",
+        "The cell [3][0] i.e. 1886030 has 16 \n",
+        "The cell [3][1] i.e. 1886018 has 17 \n",
+        "The cell [3][2] i.e. 1886000 has 18 \n",
+        "The cell [3][3] i.e. 1885fe8 has 19 \n",
+        "The cell [3][4] i.e. 1885fd0 has 20 \n",
+        "The cell [4][0] i.e. 1885fb8 has 21 \n",
+        "The cell [4][1] i.e. 1885fa0 has 22 \n",
+        "The cell [4][2] i.e. 1885f88 has 23 \n",
+        "The cell [4][3] i.e. 1885f70 has 24 \n",
+        "The cell [4][4] i.e. 1885f58 has 25 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page number: 320"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from array import *\n",
+      "\n",
+      "# Variable declaration\n",
+      "a = [[[1,2,3,4,5], [6,7,8,9,10], [11,12,13,14,15], [16,17,18,19,20]],\n",
+      "     [[21,22,23,24,25], [26,27,28,29,30], [31,32,33,34,35], [36,37,38,39,40]]]\n",
+      "\n",
+      "# Calculation and result\n",
+      "print ('The value of a is %x ' % id(a))\n",
+      "\n",
+      "for i in range (0, 2) :\n",
+      "\tprint ('The value of a[%d] is %x ' % (i, id(a[i])))\n",
+      "\tfor j in range (0, 4) :\n",
+      "\t\tprint ('The value of a[%d][%d] is %x ' % (i, j, id(a[i][j])))\n",
+      "\tprint\n",
+      "\n",
+      "for i in range (0, 2) :\n",
+      "\tfor j in range (0, 4) :       \n",
+      "\t\tfor k in range (0, 5) :\n",
+      "\t\t\tprint ('The cell [%d][%d][%d] i.e. %x has %d ' % (i, j, k, id(a[i][j][k]), a[i][j][k]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of a is 7f55de4772d8 \n",
+        "The value of a[0] is 7f55de486b90 \n",
+        "The value of a[0][0] is 7f55de4db368 \n",
+        "The value of a[0][1] is 7f55de48f050 \n",
+        "The value of a[0][2] is 7f55de4db3f8 \n",
+        "The value of a[0][3] is 7f55de4c7908 \n",
+        "\n",
+        "The value of a[1] is 7f55de477320 \n",
+        "The value of a[1][0] is 7f55de4ff3f8 \n",
+        "The value of a[1][1] is 7f55de477290 \n",
+        "The value of a[1][2] is 7f55de4773b0 \n",
+        "The value of a[1][3] is 7f55de477368 \n",
+        "\n",
+        "The cell [0][0][0] i.e. 1886198 has 1 \n",
+        "The cell [0][0][1] i.e. 1886180 has 2 \n",
+        "The cell [0][0][2] i.e. 1886168 has 3 \n",
+        "The cell [0][0][3] i.e. 1886150 has 4 \n",
+        "The cell [0][0][4] i.e. 1886138 has 5 \n",
+        "The cell [0][1][0] i.e. 1886120 has 6 \n",
+        "The cell [0][1][1] i.e. 1886108 has 7 \n",
+        "The cell [0][1][2] i.e. 18860f0 has 8 \n",
+        "The cell [0][1][3] i.e. 18860d8 has 9 \n",
+        "The cell [0][1][4] i.e. 18860c0 has 10 \n",
+        "The cell [0][2][0] i.e. 18860a8 has 11 \n",
+        "The cell [0][2][1] i.e. 1886090 has 12 \n",
+        "The cell [0][2][2] i.e. 1886078 has 13 \n",
+        "The cell [0][2][3] i.e. 1886060 has 14 \n",
+        "The cell [0][2][4] i.e. 1886048 has 15 \n",
+        "The cell [0][3][0] i.e. 1886030 has 16 \n",
+        "The cell [0][3][1] i.e. 1886018 has 17 \n",
+        "The cell [0][3][2] i.e. 1886000 has 18 \n",
+        "The cell [0][3][3] i.e. 1885fe8 has 19 \n",
+        "The cell [0][3][4] i.e. 1885fd0 has 20 \n",
+        "The cell [1][0][0] i.e. 1885fb8 has 21 \n",
+        "The cell [1][0][1] i.e. 1885fa0 has 22 \n",
+        "The cell [1][0][2] i.e. 1885f88 has 23 \n",
+        "The cell [1][0][3] i.e. 1885f70 has 24 \n",
+        "The cell [1][0][4] i.e. 1885f58 has 25 \n",
+        "The cell [1][1][0] i.e. 1885f40 has 26 \n",
+        "The cell [1][1][1] i.e. 1885f28 has 27 \n",
+        "The cell [1][1][2] i.e. 1885f10 has 28 \n",
+        "The cell [1][1][3] i.e. 1885ef8 has 29 \n",
+        "The cell [1][1][4] i.e. 1885ee0 has 30 \n",
+        "The cell [1][2][0] i.e. 1885ec8 has 31 \n",
+        "The cell [1][2][1] i.e. 1885eb0 has 32 \n",
+        "The cell [1][2][2] i.e. 1885e98 has 33 \n",
+        "The cell [1][2][3] i.e. 1885e80 has 34 \n",
+        "The cell [1][2][4] i.e. 1885e68 has 35 \n",
+        "The cell [1][3][0] i.e. 1886618 has 36 \n",
+        "The cell [1][3][1] i.e. 1886600 has 37 \n",
+        "The cell [1][3][2] i.e. 18865e8 has 38 \n",
+        "The cell [1][3][3] i.e. 18865d0 has 39 \n",
+        "The cell [1][3][4] i.e. 18865b8 has 40 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3, Page number: 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def readdata (a, num) :\n",
+      "\twhile num > 0 :\n",
+      "\t\ta[num-1] = int(raw_input('Enter value in cell[%d] ' % num))\n",
+      "\t\tnum = num - 1\n",
+      "\t\t\n",
+      "def printdata (a, num) :\n",
+      "\twhile num > 0 :\n",
+      "\t\tprint ('Value in cell[%d] is %d ' % (num, a[num-1]))\n",
+      "\t\tnum = num - 1\n",
+      "\n",
+      "n = int(raw_input('Enter the number of elements '))\n",
+      "a = [0 for i in xrange(n)]\n",
+      "readdata (a, n)\n",
+      "print\n",
+      "printdata (a, n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of elements 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[3] 30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[2] 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[1] 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value in cell[3] is 30 \n",
+        "Value in cell[2] is 20 \n",
+        "Value in cell[1] is 10 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page number: 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def readdata (a, n) :\n",
+      "\twhile n > 0 :\n",
+      "\t\ta[n-1] = int(raw_input('Enter value in cell[%d] ' % n))\n",
+      "\t\tn = n - 1\n",
+      "\t\t\n",
+      "def printdata (a, n) :\n",
+      "\twhile n > 0 :\n",
+      "\t\tprint ('Value in cell[%d] is %d ' % (n, a[n-1]))\n",
+      "\t\tn = n - 1\n",
+      "\n",
+      "n = int(raw_input('Enter the number of elements '))\n",
+      "a = [0 for i in xrange(n)]\n",
+      "readdata (a, n)\n",
+      "print\n",
+      "printdata (a, n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of elements 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[5] 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[4] 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[3] 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[2] 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[1] 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value in cell[5] is 1 \n",
+        "Value in cell[4] is 2 \n",
+        "Value in cell[3] is 3 \n",
+        "Value in cell[2] is 4 \n",
+        "Value in cell[1] is 5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5, Page number: 330"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "global n\n",
+      "n = 0\n",
+      "\n",
+      "def makearray (n) :\n",
+      "\tn = int(raw_input('Enter the number of elements '))\t\n",
+      "\treturn n\n",
+      "\t\n",
+      "def readdata (a, n) :\n",
+      "\twhile n > 0 :\n",
+      "\t\ta[n-1] = int(raw_input('Enter value in cell[%d] ' % n))\n",
+      "\t\tn = n - 1\n",
+      "\t\t\n",
+      "def printdata (a, n) :\n",
+      "\twhile n > 0 :\n",
+      "\t\tprint ('Value in cell[%d] is %d ' % (n, a[n-1]))\n",
+      "\t\tn = n - 1\n",
+      "\t\t\n",
+      "n = makearray (n)\n",
+      "a = [0 for i in xrange(n)]\n",
+      "readdata (a, n)\n",
+      "print\n",
+      "printdata (a, n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of elements 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[4] 400\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[3] 300\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[2] 200\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value in cell[1] 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value in cell[4] is 400 \n",
+        "Value in cell[3] is 300 \n",
+        "Value in cell[2] is 200 \n",
+        "Value in cell[1] is 100 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6, Page number: 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration, calculation and result\n",
+      "string = raw_input('Enter the string: ')\n",
+      "\n",
+      "rev_string = reversed(string)\n",
+      "\n",
+      "if list(string) == list(rev_string) :\n",
+      "\tprint ('%s is a PALINDROME ' % string)\n",
+      "else :\n",
+      "\tprint ('%s is not a PALINDROME ' % string)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the string: malayalam\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "malayalam is a PALINDROME \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7, Page number: 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def readdata (a, n) :\n",
+      "\tfor i in range (0, n) :\n",
+      "\t\ta[i] = int(raw_input('Enter data in cell [%d] ' % i))\n",
+      "\treturn a\n",
+      "\n",
+      "def bubble (a, n) :\n",
+      "\tfor p in range (0, n) :\n",
+      "\t\tfor i in range (0, n-1-p) :\n",
+      "\t\t\tif a[i] > a[i+1] :\n",
+      "\t\t\t\ttemp = a[i]\n",
+      "\t\t\t\ta[i] = a[i+1]\n",
+      "\t\t\t\ta[i+1] = temp\n",
+      "\treturn a\n",
+      "\n",
+      "def printdata (a, n) :\n",
+      "\tfor i in range (0, n) :\n",
+      "\t\tprint ('The data in cell [%d] is %d ' % (i, a[i]))\n",
+      "\n",
+      "n = int(raw_input('Enter the number of elements to be entered: '))\n",
+      "a = [0] * n\n",
+      "a = readdata (a, n)\n",
+      "a = bubble (a, n)\n",
+      "print ('\\nThe sorted data is ')\n",
+      "printdata (a, n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of elements to be entered: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [0] 46\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [1] 75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [2] 68\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [3] 34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter data in cell [4] 49\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The sorted data is \n",
+        "The data in cell [0] is 34 \n",
+        "The data in cell [1] is 46 \n",
+        "The data in cell [2] is 49 \n",
+        "The data in cell [3] is 68 \n",
+        "The data in cell [4] is 75 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8, Page number: 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration, calculation and result\n",
+      "i = 6\n",
+      "c = 'a'\n",
+      "\n",
+      "the_data = i\n",
+      "print ('the_data points to the integer value %d ' % the_data)\n",
+      "\n",
+      "the_data = c\n",
+      "print ('the_data points to the character value %c ' % the_data)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the_data points to the integer value 6 \n",
+        "the_data points to the character value a \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9, Page number: 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "marks = [[0 for x in xrange(5)] for x in xrange(5)]\n",
+      "avg = 0\n",
+      "percent = [0 for x in xrange(5)]\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readncompute (marks, percent, avg) :\n",
+      "\tno_of_students = int(raw_input('How many students marks do you want to enter? '))\n",
+      "\tfor i in range (0, no_of_students) :\n",
+      "\t\tprint ('\\nEnter marks of 5 subjects for roll no. %d ' % (i+1))\n",
+      "\t\tfor j in range (0, 5) :\n",
+      "\t\t\tmarks[i][j] = int(raw_input('Subject %d: ' % (j+1)))\n",
+      "\t\t\tpercent[i] = percent[i] + marks[i][j]\n",
+      "\t\t\n",
+      "\t\tpercent[i] = percent[i]/5\n",
+      "\t\tavg = avg + percent[i]\n",
+      "\treturn avg/no_of_students, no_of_students\n",
+      "\n",
+      "def printdata (score, per, avg, no_of_students) :\n",
+      "\tfor i in range (0, no_of_students) :\n",
+      "\t\tprint ('\\nMarks for roll no. %d are ' % (i+1))\n",
+      "\t\tfor j in range (0, 5) :\n",
+      "\t\t\tprint ('%d ' % score[i][j])\n",
+      "\t\tprint ('Percentage for roll no. is %.2f ' % per[i])\n",
+      "\t\tif per[i] < avg :\n",
+      "\t\t\tprint ('Below average')\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('Above average')\n",
+      "\n",
+      "average, nos = readncompute(marks, percent, avg)\n",
+      "print ('\\nThe average marks are %.2f' % average)\n",
+      "printdata (marks, percent, average, nos)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many students marks do you want to enter? 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter marks of 5 subjects for roll no. 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1: 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2: 76\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3: 54\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 4: 83\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 5: 94\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter marks of 5 subjects for roll no. 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1: 84\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2: 62\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3: 94\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 4: 73\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 5: 62\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The average marks are 74.00\n",
+        "\n",
+        "Marks for roll no. 1 are \n",
+        "65 \n",
+        "76 \n",
+        "54 \n",
+        "83 \n",
+        "94 \n",
+        "Percentage for roll no. is 74.00 \n",
+        "Above average\n",
+        "\n",
+        "Marks for roll no. 2 are \n",
+        "84 \n",
+        "62 \n",
+        "94 \n",
+        "73 \n",
+        "62 \n",
+        "Percentage for roll no. is 75.00 \n",
+        "Above average\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10, Page number: 349"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable declaration\n",
+      "marks = [[0 for x in xrange(5)] for x in xrange(5)]\n",
+      "avg = 0\n",
+      "percent = [0 for x in xrange(5)]\n",
+      "\n",
+      "# Function declaration, calculation and result\n",
+      "def readncompute (marks, percent, avg) :\n",
+      "\tno_of_students = int(raw_input('How many students marks do you want to enter? '))\n",
+      "\tfor i in range (0, no_of_students) :\n",
+      "\t\tprint ('\\nEnter marks of 5 subjects for roll no. %d ' % (i+1))\n",
+      "\t\tfor j in range (0, 5) :\n",
+      "\t\t\tmarks[i][j] = int(raw_input('Subject %d: ' % (j+1)))\n",
+      "\t\t\tpercent[i] = percent[i] + marks[i][j]\n",
+      "\t\t\n",
+      "\t\tpercent[i] = percent[i]/5\n",
+      "\t\tavg = avg + percent[i]\n",
+      "\treturn avg/no_of_students, no_of_students\n",
+      "\n",
+      "def printdata (score, per, avg, no_of_students) :\n",
+      "\tfor i in range (0, no_of_students) :\n",
+      "\t\tprint ('\\nMarks for roll no. %d are ' % (i+1))\n",
+      "\t\tfor j in range (0, 5) :\n",
+      "\t\t\tprint ('%d ' % score[i][j])\n",
+      "\t\tprint ('Percentage for roll no. is %.2f ' % per[i])\n",
+      "\t\tif per[i] < avg :\n",
+      "\t\t\tprint ('Below average')\n",
+      "\t\telse :\n",
+      "\t\t\tprint ('Above average')\n",
+      "\n",
+      "average, nos = readncompute(marks, percent, avg)\n",
+      "print ('\\nThe average marks are %.2f' % average)\n",
+      "printdata (marks, percent, average, nos)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many students marks do you want to enter? 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter marks of 5 subjects for roll no. 1 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1: 65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2: 76\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3: 54\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 4: 32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 5: 89\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter marks of 5 subjects for roll no. 2 \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 1: 76\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 2: 86\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 3: 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 4: 32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subject 5: 21\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The average marks are 57.00\n",
+        "\n",
+        "Marks for roll no. 1 are \n",
+        "65 \n",
+        "76 \n",
+        "54 \n",
+        "32 \n",
+        "89 \n",
+        "Percentage for roll no. is 63.00 \n",
+        "Above average\n",
+        "\n",
+        "Marks for roll no. 2 are \n",
+        "76 \n",
+        "86 \n",
+        "43 \n",
+        "32 \n",
+        "21 \n",
+        "Percentage for roll no. is 51.00 \n",
+        "Below average\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11, Page number: 351"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Function declaration, calculation and result\n",
+      "def readdata (a, n) :\n",
+      "\tfor i in range (0, n) :\n",
+      "\t\ta[i] = int(raw_input('Enter data in cell [%d] ' % i))\n",
+      "\treturn a\n",
+      "\n",
+      "def bubble (name) :\n",
+      "\tfor p in range (0, 4) :\n",
+      "\t\tfor i in range (0, 3-p) :\n",
+      "\t\t\tif name[i] > name[i+1] :\n",
+      "\t\t\t\ttemp = name[i]\n",
+      "\t\t\t\tname[i] = name[i+1]\n",
+      "\t\t\t\tname[i+1] = temp\n",
+      "\treturn name\n",
+      "\n",
+      "def printdata (name) :\n",
+      "\tfor i in range (0, 4) :\n",
+      "\t\tprint ('The data in cell [%d] is %s ' % ((i+1), name[i]))\n",
+      "\n",
+      "name = [0] * 4\n",
+      "for i in range (0, 4) :\n",
+      "\tname[i] = raw_input('Enter string[%d] ' % (i+1))\n",
+      "\n",
+      "name = bubble (name)\n",
+      "print ('\\nThe sorted data is ')\n",
+      "printdata (name)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter string[1] zaheer\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter string[2] ashita\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter string[3] tamanna\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter string[4] purti\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The sorted data is \n",
+        "The data in cell [1] is ashita \n",
+        "The data in cell [2] is purti \n",
+        "The data in cell [3] is tamanna \n",
+        "The data in cell [4] is zaheer \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Magnifying_C_by_Arpita_Goyal/README.txt b/Magnifying_C_by_Arpita_Goyal/README.txt
new file mode 100755
index 00000000..2ec7deb7
--- /dev/null
+++ b/Magnifying_C_by_Arpita_Goyal/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Mahesh M
+Course: btech
+College/Institute/Organization: MSIT
+Department/Designation: CSE
+Book Title: Magnifying C
+Author: Arpita Gopal
+Publisher: Prentice Hall India Learning Private Limited, New Delhi
+Year of publication: 2009
+Isbn: 9788120338616
+Edition: 1
\ No newline at end of file
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diff --git a/Oscillations_and_Waves_by_S._Prakash/README.txt b/Oscillations_and_Waves_by_S._Prakash/README.txt
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diff --git a/Power_Electronics/README.txt b/Power_Electronics/README.txt
deleted file mode 100755
index 95f5f179..00000000
--- a/Power_Electronics/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Anay Sonawane
-Course: btech
-College/Institute/Organization: Wavelet group
-Department/Designation: E&tc
-Book Title: Power Electronics
-Author: Dr. B R Gupta, V Singhal
-Publisher: S.K.Kataria and sons, New Delhi
-Year of publication: 2010
-Isbn: 81-85749-53-1
-Edition: 6
\ No newline at end of file
diff --git a/Power_Electronics/chapter10.ipynb b/Power_Electronics/chapter10.ipynb
deleted file mode 100755
index cbc3cb90..00000000
--- a/Power_Electronics/chapter10.ipynb
+++ /dev/null
@@ -1,228 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10 : Cycloconverters"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2, Page No 594"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "R=10.0\n",
-      "a=30.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_or=(V_m/math.sqrt(2))*math.sqrt((1/math.pi)*(math.pi-a*math.pi/180+math.sin(math.radians(2*a))/2))\n",
-      "I_or=V_or/R    \n",
-      "I_s=I_or\n",
-      "pf=(I_or**2*R)/(V_s*I_s)    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of o/p current=%.2f A\" %I_or)\n",
-      "print(\"rms value of o/p current for each convertor=%.2f A\" %(I_or/math.sqrt(2)))\n",
-      "print(\"rms value of o/p current for each thyristor=%.3f A\" %(I_or/2))\n",
-      "print(\"i/p pf=%.4f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of o/p current=22.67 A\n",
-        "rms value of o/p current for each convertor=16.03 A\n",
-        "rms value of o/p current for each thyristor=11.333 A\n",
-        "i/p pf=0.9855\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.4, Page No 604"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=400.0\n",
-      "V_ph=V_s/2\n",
-      "a=160.0\n",
-      "\n",
-      "#Calculations\n",
-      "r=math.cos(math.radians(180-a))\n",
-      "m=3\n",
-      "V_or=r*(V_ph*(m/math.pi)*math.sin(math.pi/m)) \n",
-      "R=2\n",
-      "X_L=1.5\n",
-      "th=math.degrees(math.atan(X_L/R))\n",
-      "Z=math.sqrt(R**2+X_L**2)\n",
-      "I_or=V_or/Z    \n",
-      "P=I_or**2*R      \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms o/p voltage=%.3f V\" %V_or)\n",
-      "print(\"rms o/p current=%.2f A\" %I_or)\n",
-      "print(\"phase angle of o/p current=%.2f deg\" %-th)\n",
-      "print(\"o/p power=%.2f W\" %P)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms o/p voltage=155.424 V\n",
-        "rms o/p current=62.17 A\n",
-        "phase angle of o/p current=-36.87 deg\n",
-        "o/p power=7730.11 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5 Page No 604"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=400.0\n",
-      "V_ph=V_s/2\n",
-      "V_l=V_ph*math.sqrt(3)\n",
-      "a=160.0\n",
-      "\n",
-      "#Calculations\n",
-      "r=math.cos(math.radians(180-a))\n",
-      "m=6\n",
-      "V_or=r*(V_l*(m/math.pi)*math.sin(math.pi/m))    \n",
-      "R=2\n",
-      "X_L=1.5\n",
-      "th=math.degrees(math.atan(X_L/R))\n",
-      "Z=math.sqrt(R**2+X_L**2)\n",
-      "I_or=V_or/Z    \n",
-      "P=I_or**2*R    \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms o/p voltage=%.2f V\" %V_or)\n",
-      "print(\"rms o/p current=%.2f A\" %I_or)\n",
-      "print(\"phase angle of o/p current=%.2f deg\" %-th)\n",
-      "print(\"o/p power=%.2f W\" %P)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms o/p voltage=310.85 V\n",
-        "rms o/p current=124.34 A\n",
-        "phase angle of o/p current=-36.87 deg\n",
-        "o/p power=30920.44 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.7, Page No 605"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_l=400.0\n",
-      "V_ml=math.sqrt(2)*V_l\n",
-      "m=6\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "L=.0012\n",
-      "I=40.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_or1=(V_ml*(m/math.pi)*math.sin(math.pi/m))*math.cos(math.radians(a))\n",
-      "V_omx1=V_or1-3*w*L*I/math.pi\n",
-      "V_rms1=V_omx1/math.sqrt(2)    \n",
-      "a2=30.0\n",
-      "V_or2=(V_ml*(m/math.pi)*math.sin(math.pi/m))*math.cos(math.radians(a))\n",
-      "V_omx2=V_or2-3*w*L*I/math.pi\n",
-      "V_rms2=V_omx2/math.sqrt(2)    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"for firing angle=0deg\")\n",
-      "a1=0\n",
-      "print(\"rms value of load voltage=%.2f V\" %V_rms2)\n",
-      "print(\"for firing angle=30deg\")\n",
-      "print(\"rms value of load voltage=%.2f V\" %V_rms2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle=0deg\n",
-        "rms value of load voltage=-369.12 V\n",
-        "for firing angle=30deg\n",
-        "rms value of load voltage=-369.12 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter11.ipynb b/Power_Electronics/chapter11.ipynb
deleted file mode 100755
index d2317d28..00000000
--- a/Power_Electronics/chapter11.ipynb
+++ /dev/null
@@ -1,299 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11 : Some Applications"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.1, Page No 622"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=11000.0\n",
-      "V_ml=math.sqrt(2)*V_s\n",
-      "f=50.0\n",
-      "\n",
-      "#Calculations\n",
-      "w=2*math.pi*f\n",
-      "I_d=300\n",
-      "R_d=1\n",
-      "g=20 #g=gamma\n",
-      "a=math.degrees(math.acos(math.cos(math.radians(g))+math.pi/(3*V_ml)*I_d*R_d))    \n",
-      "L_s=.01\n",
-      "V_d=(3/math.pi)*((V_ml*math.cos(math.radians(a)))-w*L_s*I_d)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle=%.3f deg\" %a)\n",
-      "print(\"rectifier o/p voltage=%.1f V\" %V_d)\n",
-      "print(\"dc link voltage=%.3f V\" %(2*V_d/1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle=16.283 deg\n",
-        "rectifier o/p voltage=13359.3 V\n",
-        "dc link voltage=26.719 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2, Page No 623"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_d=(200.0+200)*10**3\n",
-      "P=1000.0*10**6\n",
-      "\n",
-      "#Calculations\n",
-      "I_d=P/V_d\n",
-      " #each thristor conducts for 120deg for a periodicity of 360deg\n",
-      "a=0\n",
-      "V_d=200.0*10**3\n",
-      "V_ml=V_d*math.pi/(3*math.cos(math.radians(a)))\n",
-      "\n",
-      "#Results\n",
-      "print(\"rms current rating of thyristor=%.2f A\" %(I_d*math.sqrt(120/360)))\n",
-      "print(\"peak reverse voltage across each thyristor=%.2f kV\" %(V_ml/2/1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms current rating of thyristor=0.00 A\n",
-        "peak reverse voltage across each thyristor=104.72 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 Page No 627"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_m=230.0\n",
-      "V_s=230/math.sqrt(2)\n",
-      "pf=0.8\n",
-      "P=2000.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_m=P/(V_s*pf)\n",
-      "I_Tr=I_m/math.sqrt(2)\n",
-      "I_TA=2*I_m/math.pi\n",
-      "fos=2 #factor of safety\n",
-      "PIV=V_m*math.sqrt(2)\n",
-      "I_Tr=I_m/(2)\n",
-      "I_TA=I_m/math.pi\n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of thyristor current=%.2f A\" %(fos*I_Tr))\n",
-      "print(\"avg value of thyristor current=%.3f A\" %(fos*I_TA))\n",
-      "print(\"voltage rating of thyristor=%.2f V\" %PIV)\n",
-      "print(\"rms value of diode current=%.3f A\" %(fos*I_Tr))\n",
-      "print(\"avg value of diode current=%.3f A\" %(fos*I_TA))\n",
-      "print(\"voltage rating of diode=%.2f V\" %PIV)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of thyristor current=15.37 A\n",
-        "avg value of thyristor current=9.786 A\n",
-        "voltage rating of thyristor=325.27 V\n",
-        "rms value of diode current=15.372 A\n",
-        "avg value of diode current=9.786 A\n",
-        "voltage rating of diode=325.27 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4, Page No 629"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=200.0\n",
-      "I=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "R_L=V/I    \n",
-      "I_h=.005     #holding current\n",
-      "R2=V/I_h    \n",
-      "t_c=20*10**-6\n",
-      "fos=2 #factor of safety\n",
-      "C=t_c*fos/(R_L*math.log(2))    \n",
-      "\n",
-      "#Results\n",
-      "print(\"value of load resistance=%.0f ohm\" %R_L)\n",
-      "print(\"value of R2=%.0f kilo-ohm\" %(R2/1000))\n",
-      "print(\"value of C=%.3f uF\" %(C*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of load resistance=20 ohm\n",
-        "value of R2=40 kilo-ohm\n",
-        "value of C=2.885 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5 Page No 646"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "u_r=10\n",
-      "f=10000.0 #Hz\n",
-      "p=4.0*10**-8 #ohm-m\n",
-      "\n",
-      "#Calculations\n",
-      "dl=(1/(2*math.pi))*math.sqrt(p*10**7/(u_r*f))    \n",
-      "l=0.12 #length of cylinder\n",
-      "t=20.0 #no of turns\n",
-      "I=100.0\n",
-      "H=t*I/l\n",
-      "P_s=2*math.pi*H**2*math.sqrt(u_r*f*p*10**-7)    \n",
-      "d=.02 #diameter\n",
-      "P_v=4*H**2*p/(d*dl)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"depth of heat of penetration=%.5f mm\" %(dl*1000))\n",
-      "print(\"heat generated per unit cylinder surface area=%.3f W/m**2\" %P_s)\n",
-      "print(\"heat generated per unit cylinder volume=%.0f W/m**3\" %P_v)\n",
-      " #answer of P_v varies as given in book as value of d is not taken as in formulae. "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "depth of heat of penetration=0.31831 mm\n",
-        "heat generated per unit cylinder surface area=34906.585 W/m**2\n",
-        "heat generated per unit cylinder volume=6981317 W/m**3\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6 Page No 646"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=3000.0\n",
-      "\n",
-      "#Calculations\n",
-      "t_qmin=30.0*10**-6\n",
-      "f_r=f/(1-2*t_qmin*f)\n",
-      "R=0.06\n",
-      "L=20.0*10**-6\n",
-      "C=1/(L*((2*math.pi*f_r)**2+(R/(2*L))**2))    \n",
-      "\n",
-      "#Results\n",
-      "print(\"required capacitor size=%.4f F\" %(C*10**6))\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "required capacitor size=94.2215 F\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter12.ipynb b/Power_Electronics/chapter12.ipynb
deleted file mode 100755
index f8605d69..00000000
--- a/Power_Electronics/chapter12.ipynb
+++ /dev/null
@@ -1,1997 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12 : Electic Drives"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1, Page No 658"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "T_e=15.0 #Nm\n",
-      "K_m=0.5 #V-s/rad\n",
-      "I_a=T_e/K_m\n",
-      "n_m=1000.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "E_a=K_m*w_m\n",
-      "r_a=0.7\n",
-      "V_t=E_a+I_a*r_a\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=math.degrees(math.acos(2*math.pi*V_t/V_m-1))\n",
-      "print(\"firing angle delay=%.3f deg\" %a)\n",
-      "I_Tr=I_a*math.sqrt((180-a)/360)    \n",
-      "print(\"rms value of thyristor current=%.3f A\" %I_Tr)\n",
-      "I_fdr=I_a*math.sqrt((180+a)/360)    \n",
-      "print(\"rms value of freewheeling diode current=%.3f A\" %I_fdr)\n",
-      "pf=V_t*I_a/(V_s*I_Tr)  \n",
-      "\n",
-      "#Results  \n",
-      "print(\"input power factor=%.4f\" %pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle delay=65.349 deg\n",
-        "rms value of thyristor current=16.930 A\n",
-        "rms value of freewheeling diode current=24.766 A\n",
-        "input power factor=0.5652\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2, Page No 660"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "E=150.0\n",
-      "R=8.0\n",
-      "\n",
-      "#Calculations\n",
-      "th1=math.sin(math.radians(E/(math.sqrt(2)*V)))\n",
-      "I_o=(1/(2*math.pi*R))*(2*math.sqrt(2)*230*math.cos(math.radians(th1))-E*(math.pi-2*th1*math.pi/180))    \n",
-      "P=E*I_o    \n",
-      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*(math.pi-2*th1*math.pi/180)+V**2*math.sin(math.radians(2*th1))-4*math.sqrt(2)*V*E*math.cos(math.radians(th1))))\n",
-      "P_r=I_or**2*R    \n",
-      "pf=(P+P_r)/(V*I_or)\n",
-      "\n",
-      "#Results\n",
-      "print(\"avg charging curent=%.4f A\" %I_o)\n",
-      "print(\"power supplied to the battery=%.2f W\" %P)\n",
-      "print(\"power dissipated by the resistor=%.3f W\" %P_r)    \n",
-      "print(\"supply pf=%.3f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg charging curent=3.5679 A\n",
-        "power supplied to the battery=535.18 W\n",
-        "power dissipated by the resistor=829.760 W\n",
-        "supply pf=0.583\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 Page No 661"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variablesV_s=250\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=30.0\n",
-      "k=0.03 #Nm/A**2\n",
-      "n_m=1000.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "r=.2 #r_a+r_s\n",
-      "V_t=V_m/math.pi*(1+math.cos(math.radians(a)))\n",
-      "I_a=V_t/(k*w_m+r)   \n",
-      "print(\"motor armature current=%.2f A\" %I_a)\n",
-      "T_e=k*I_a**2    \n",
-      "print(\"motor torque=%.3f Nm\" %T_e)\n",
-      "I_sr=I_a*math.sqrt((180-a)/180)\n",
-      "pf=(V_t*I_a)/(V_s*I_sr)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"input power factor=%.2f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor armature current=57.82 A\n",
-        "motor torque=100.285 Nm\n",
-        "input power factor=0.92\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4, Page No 663"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=400.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "V_f=2*V_m/math.pi\n",
-      "r_f=200.0\n",
-      "I_f=V_f/r_f\n",
-      "T_e=85.0\n",
-      "K_a=0.8\n",
-      "\n",
-      "#Calculations\n",
-      "I_a=T_e/(I_f*K_a)    \n",
-      "print(\"rated armature current=%.2f A\" %I_a)\n",
-      "n_m=1200.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "r_a=0.2\n",
-      "V_t=K_a*I_f*w_m+I_a*r_a\n",
-      "a=math.degrees(math.acos(V_t*math.pi/(2*V_m)))\n",
-      "print(\"firing angle delay=%.2f deg\" %a)\n",
-      "E_a=V_t\n",
-      "w_mo=E_a/(K_a*I_f)\n",
-      "N=60*w_mo/(2*math.pi)\n",
-      "reg=((N-n_m)/n_m)*100    \n",
-      "print(\"speed regulation at full load=%.2f\" %reg)\n",
-      "I_ar=I_a\n",
-      "pf=(V_t*I_a)/(V_s*I_ar)    \n",
-      "print(\"input power factor of armature convertor=%.4f\" %pf)\n",
-      "I_fr=I_f\n",
-      "I_sr=math.sqrt(I_fr**2+I_ar**2)\n",
-      "VA=I_sr*V_s\n",
-      "P=V_t*I_a+V_f*I_f\n",
-      "\n",
-      "#Results\n",
-      "print(\"input power factor of drive=%.4f\" %(P/VA))\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rated armature current=59.01 A\n",
-        "firing angle delay=57.63 deg\n",
-        "speed regulation at full load=6.52\n",
-        "input power factor of armature convertor=0.4821\n",
-        "input power factor of drive=0.5093\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 Page No 664"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=400.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "V_f=2*V_m/math.pi\n",
-      "\n",
-      "#Calculations\n",
-      "a1=math.degrees(math.acos(V_t*math.pi/(2*V_m))) \n",
-      "print(\"delay angle of field converter=%.0f deg\" %a1)\n",
-      "r_f=200.0\n",
-      "I_f=V_f/r_f\n",
-      "T_e=85.0\n",
-      "K_a=0.8\n",
-      "I_a=T_e/(I_f*K_a)\n",
-      "n_m=1200.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "r_a=0.1\n",
-      "I_a=50.0\n",
-      "V_t=-K_a*I_f*w_m+I_a*r_a\n",
-      "a=math.degrees(math.acos(V_t*math.pi/(2*V_m)))\n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle delay of armature converter=%.3f deg\" %a)\n",
-      "print(\"power fed back to ac supply=%.0f W\" %(-V_t*I_a))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "delay angle of field converter=58 deg\n",
-        "firing angle delay of armature converter=119.260 deg\n",
-        "power fed back to ac supply=8801 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 Page No 665"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_t=220.0\n",
-      "n_m=1500.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "I_a=10.0\n",
-      "r_a=1.0\n",
-      "\n",
-      "#Calculations\n",
-      "K_m=(V_t-I_a*r_a)/(w_m)\n",
-      "T=5.0\n",
-      "I_a=T/K_m\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=30.0\n",
-      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
-      "w_m=(V_t-I_a*r_a)/K_m\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "\n",
-      "print(\"motor speed=%.2f rpm\" %N)\n",
-      "a=45\n",
-      "n_m=1000\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
-      "I_a=(V_t-K_m*w_m)/r_a\n",
-      "T_e=K_m*I_a    \n",
-      "\n",
-      "#Results\n",
-      "print(\"torque developed=%.3f Nm\" %T_e)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor speed=1254.22 rpm\n",
-        "torque developed=8.586 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7, Page No 666"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_t=220.0\n",
-      "n_m=1000.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "I_a=60.0\n",
-      "r_a=.1\n",
-      "\n",
-      "#Calculations\n",
-      "K_m=(V_t-I_a*r_a)/(w_m)\n",
-      "V_s=230\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "print(\"for 600rpm speed\")\n",
-      "n_m=600.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "a=math.degrees(math.acos((K_m*w_m+I_a*r_a)*math.pi/(2*V_m))) \n",
-      "print(\"firing angle=%.3f deg\" %a)\n",
-      "print(\"for -500rpm speed\")\n",
-      "n_m=-500.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "a=math.degrees(math.acos((K_m*w_m+I_a*r_a)*math.pi/(2*V_m)))\n",
-      "print(\"firing angle=%.2f deg\" %a)\n",
-      "I_a=I_a/2\n",
-      "a=150\n",
-      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
-      "w_m=(V_t-I_a*r_a)/K_m\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor speed=%.3f rpm\" %N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for 600rpm speed\n",
-        "firing angle=49.530 deg\n",
-        "for -500rpm speed\n",
-        "firing angle=119.19 deg\n",
-        "motor speed=-852.011 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 Page No 672"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "K_m=1.5\n",
-      "T_e=50.0\n",
-      "I_a=T_e/K_m\n",
-      "r_a=0.9\n",
-      "a=45.0\n",
-      "V_s=415.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_ml=math.sqrt(2)*V_s\n",
-      "w_m=((3*V_ml*(1+math.cos(math.radians(a)))/(2*math.pi))-I_a*r_a)/K_m\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor speed=%.2f rpm\" %N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor speed=2854.42 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9 Page No 672"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variablesV_t=600\n",
-      "n_m=1500.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "I_a=80.0\n",
-      "r_a=1.0\n",
-      "\n",
-      "#Calculations\n",
-      "K_m=(V_t-I_a*r_a)/(w_m)\n",
-      "V_s=400.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "print(\"for firing angle=45deg and speed=1200rpm\")\n",
-      "a=45.0\n",
-      "n_m=1200.0\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "I_a=(3*V_m*(1+math.cos(math.radians(a)))/(2*math.pi)-K_m*w_m)/r_a\n",
-      "I_sr=I_a*math.sqrt(2/3)    \n",
-      "print(\"rms value of source current=%.3f A\" %I_sr)\n",
-      "print(\"rms value of thyristor current=%.3f A\" %(I_a*math.sqrt(1/3)))\n",
-      "print(\"avg value of thyristor current=%.2f A\" %I_a*(1/3))\n",
-      "pf=(3/(2*math.pi)*(1+math.cos(math.radians(a))))    \n",
-      "print(\"input power factor=%.3f\" %pf)\n",
-      "\n",
-      "print(\"for firing angle=90deg and speed=700rpm\")\n",
-      "a=90\n",
-      "n_m=700\n",
-      "w_m=2*math.pi*n_m/60\n",
-      "I_a=(3*V_m*(1+math.cos(math.radians(a)))/(2*math.pi)-K_m*w_m)/r_a\n",
-      "I_sr=I_a*math.sqrt(90/180)    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of source current=%.3f A\" %I_sr)\n",
-      "print(\"rms value of thyristor current=%.3f A\" %(I_a*math.sqrt(90.0/360)))\n",
-      "print(\"avg value of thyristor current=%.3f A\" %I_a*(1/3))\n",
-      "pf=(math.sqrt(6)/(2*math.pi)*(1+math.cos(math.radians(a))))*math.sqrt(180/(180-a))    \n",
-      "print(\"input power factor=%.4f\" %pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle=45deg and speed=1200rpm\n",
-        "rms value of source current=0.000 A\n",
-        "rms value of thyristor current=0.000 A\n",
-        "\n",
-        "input power factor=0.815\n",
-        "for firing angle=90deg and speed=700rpm\n",
-        "rms value of source current=0.000 A\n",
-        "rms value of thyristor current=195.558 A\n",
-        "\n",
-        "input power factor=0.5513\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10 Page No 676"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=400.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=30\n",
-      "V_t=3*V_m*math.cos(math.radians(a))/math.pi\n",
-      "I_a=21.0\n",
-      "r_a=.1\n",
-      "V_d=2.0\n",
-      "K_m=1.6\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=(V_t-I_a*r_a-V_d)/K_m\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "print(\"speed of motor=%.1f rpm\" %N)\n",
-      "\n",
-      "N=2000\n",
-      "w_m=2*math.pi*N/60\n",
-      "I_a=210\n",
-      "V_t=K_m*w_m+I_a*r_a+V_d\n",
-      "a=math.degrees(math.acos(V_t*math.pi/(3*V_m)))\n",
-      "print(\"firing angle=%.2f deg\" %a)\n",
-      "I_sr=I_a*math.sqrt(2.0/3.0)\n",
-      "pf=V_t*I_a/(math.sqrt(3)*V_s*I_sr)    \n",
-      "print(\"supply power factor=%.3f\" %pf)\n",
-      "\n",
-      "I_a=21\n",
-      "w_m=(V_t-I_a*r_a-V_d)/K_m\n",
-      "n=w_m*60/(2*math.pi)\n",
-      "reg=(n-N)/N*100    \n",
-      "\n",
-      "#Results\n",
-      "print(\"speed regulation(percent)=%.2f\" %reg)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "speed of motor=2767.6 rpm\n",
-        "firing angle=48.48 deg\n",
-        "supply power factor=0.633\n",
-        "speed regulation(percent)=5.64\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11, Page No 677"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_t=230.0\n",
-      "V_l=V_t*math.pi/(3*math.sqrt(2))\n",
-      "V_ph=V_l/math.sqrt(3)\n",
-      "V_in=400     #per phase voltage input\n",
-      "\n",
-      "#Calculations\n",
-      "N1=1500.0\n",
-      "I_a1=20.0\n",
-      "r_a1=.6\n",
-      "E_a1=V_t-I_a1*r_a1\n",
-      "n1=1000.0\n",
-      "E_a2=E_a1/1500.0*1000.0\n",
-      "V_t1=E_a1+I_a1*r_a1\n",
-      "a1=math.degrees(math.acos(V_t1*math.pi/(3*math.sqrt(2.0)*V_l)))\n",
-      "I_a2=.5*I_a1\n",
-      "n2=-900.0\n",
-      "V_t2=n2*E_a2/N1+I_a2*r_a1\n",
-      "a2=math.degrees(math.acos(V_t2*math.pi/(3*math.sqrt(2)*V_l))) \n",
-      "\n",
-      "#Results\n",
-      "print(\"transformer phase turns ratio=%.3f\" %(V_in/V_ph))\n",
-      "print(\"for motor running at 1000rpm at rated torque\")\n",
-      "print(\"firing angle delay=%.2f deg\" %a1)\n",
-      "print(\"for motor running at -900rpm at half of rated torque\")\n",
-      "print(\"firing angle delay=%.3f deg\" %a2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "transformer phase turns ratio=4.068\n",
-        "for motor running at 1000rpm at rated torque\n",
-        "firing angle delay=0.00 deg\n",
-        "for motor running at -900rpm at half of rated torque\n",
-        "firing angle delay=110.674 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12, Page No 678"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variablesV_s=400\n",
-      "V_ml=math.sqrt(2)*V_s\n",
-      "V_f=3*V_ml/math.pi\n",
-      "R_f=300.0\n",
-      "I_f=V_f/R_f\n",
-      "T_e=60.0\n",
-      "k=1.1\n",
-      "\n",
-      "#Calculations\n",
-      "I_a=T_e/(k*I_f)\n",
-      "N1=1000.0\n",
-      "w_m1=2*math.pi*N1/60\n",
-      "r_a1=.3\n",
-      "V_t1=k*I_f*w_m1+I_a*r_a1\n",
-      "a1=math.degrees(math.acos(V_f*math.pi/(3*V_ml)))\n",
-      "N2=3000\n",
-      "w_m2=2*math.pi*N/60\n",
-      "a2=0\n",
-      "V_t2=3*V_ml*math.cos(math.radians(a))/math.pi\n",
-      "I_f2=(V_t2-I_a*r_a)/(w_m2*k)\n",
-      "V_f2=I_f2*R_f\n",
-      "a2=math.degrees(math.acos(V_f2*math.pi/(3*V_ml)))\n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle=%.3f deg\" %a)\n",
-      "print(\"firing angle=%.3f deg\" %a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle=48.477 deg\n",
-        "firing angle=48.477 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13, Page No 679"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      " #after calculating\n",
-      " #t=w_m/6000-math.pi/360\n",
-      "\n",
-      "N=1000.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=2*math.pi*N/60\n",
-      "t=w_m/6000-math.pi/360    \n",
-      "\n",
-      "#Results\n",
-      "print(\"time reqd=%.2f s\" %t)\n",
-      " #printing mistake in the answer in book"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "time reqd=0.01 s\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14, Page No 679"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_a=1.0 #supposition\n",
-      "a=60.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_s1=2*math.sqrt(2)/math.pi*I_a*math.sin(math.radians(a))\n",
-      "I_s3=2*math.sqrt(2)/(3*math.pi)*I_a*math.sin(math.radians(3*a))\n",
-      "I_s5=2*math.sqrt(2)/(5*math.pi)*I_a*math.sin(math.radians(5*a))\n",
-      "per3=I_s3/I_s1*100    \n",
-      "print(\"percent of 3rd harmonic current in fundamental=%.2f\" %per3)\n",
-      "per5=I_s5/I_s1*100    \n",
-      "\n",
-      "#Results\n",
-      "print(\"percent of 5th harmonic current in fundamental=%.2f\" %per5)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "percent of 3rd harmonic current in fundamental=0.00\n",
-        "percent of 5th harmonic current in fundamental=-20.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15, Page No 680"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_a=60.0\n",
-      "I_TA=I_a/3    \n",
-      "\n",
-      "#Calculations\n",
-      "print(\"avg thyristor current=%.0f A\" %I_TA)\n",
-      "I_Tr=I_a/math.sqrt(3)    \n",
-      "print(\"rms thyristor current=%.3f A\" %I_Tr)\n",
-      "V_s=400\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "I_sr=I_a*math.sqrt(2.0/3)\n",
-      "a=150\n",
-      "V_t=3*V_m*math.cos(math.radians(a))/math.pi\n",
-      "pf=V_t*I_a/(math.sqrt(3)*V_s*I_sr)    \n",
-      "print(\"power factor of ac source=%.3f\" %pf)\n",
-      "\n",
-      "r_a=0.5\n",
-      "K_m=2.4\n",
-      "w_m=(V_t-I_a*r_a)/K_m\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"Speed of motor=%.2f rpm\" %N)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg thyristor current=20 A\n",
-        "rms thyristor current=34.641 A\n",
-        "power factor of ac source=-0.827\n",
-        "Speed of motor=-1980.76 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16, Page No 685"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_a=300.0\n",
-      "V_s=600.0\n",
-      "a=0.6\n",
-      "V_t=a*V_s\n",
-      "P=V_t*I_a    \n",
-      "\n",
-      "#Calculations\n",
-      "print(\"input power from source=%.0f kW\" %(P/1000))\n",
-      "R_eq=V_s/(a*I_a)    \n",
-      "print(\"equivalent input resistance=%.3f ohm\" %R_eq)\n",
-      "k=.004\n",
-      "R=.04+.06\n",
-      "w_m=(a*V_s-I_a*R)/(k*I_a)\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "print(\"motor speed=%.1f rpm\" %N)\n",
-      "T_e=k*I_a**2    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor torque=%.0f Nm\" %T_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "input power from source=108 kW\n",
-        "equivalent input resistance=3.333 ohm\n",
-        "motor speed=2626.1 rpm\n",
-        "motor torque=360 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17, Page No 686"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "T_on=10.0\n",
-      "T_off=15.0\n",
-      "\n",
-      "#Calculations\n",
-      "a=T_on/(T_on+T_off)\n",
-      "V_s=230.0\n",
-      "V_t=a*V_s\n",
-      "r_a=3\n",
-      "K_m=.5\n",
-      "N=1500\n",
-      "w_m=2*math.pi*N/60\n",
-      "I_a=(V_t-K_m*w_m)/r_a    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor load current=%.3f A\" %I_a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor load current=4.487 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18, Page No 686"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "w_m=0    \n",
-      "print(\"lower limit of speed control=%.0f rpm\" %w_m)\n",
-      "I_a=25.0\n",
-      "r_a=.2\n",
-      "V_s=220\n",
-      "K_m=0.08\n",
-      "\n",
-      "#Calculations\n",
-      "a=(K_m*w_m+I_a*r_a)/V_s    \n",
-      "print(\"lower limit of duty cycle=%.3f\" %a)\n",
-      "a=1    \n",
-      "print(\"upper limit of duty cycle=%.0f\" %a)\n",
-      "w_m=(a*V_s-I_a*r_a)/K_m    \n",
-      "\n",
-      "#Results\n",
-      "print(\"upper limit of speed control=%.1f rpm\" %w_m)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "lower limit of speed control=0 rpm\n",
-        "lower limit of duty cycle=0.023\n",
-        "upper limit of duty cycle=1\n",
-        "upper limit of speed control=2687.5 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.21, Page No 691"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=0.6\n",
-      "V_s=400.0\n",
-      "V_t=(1-a)*V_s\n",
-      "I_a=300.0\n",
-      "P=V_t*I_a  \n",
-      "\n",
-      "#Calculations  \n",
-      "print(\"power returned=%.0f kW\" %(P/1000))\n",
-      "r_a=.2\n",
-      "K_m=1.2\n",
-      "R_eq=(1-a)*V_s/I_a+r_a    \n",
-      "print(\"equivalent load resistance=%.4f ohm\" %R_eq)\n",
-      "w_mn=I_a*r_a/K_m\n",
-      "N=w_mn*60/(2*math.pi)    \n",
-      "print(\"min braking speed=%.2f rpm\" %N)\n",
-      "w_mx=(V_s+I_a*r_a)/K_m\n",
-      "N=w_mx*60/(2*math.pi)    \n",
-      "print(\"max braking speed=%.1f rpm\" %N)\n",
-      "w_m=(V_t+I_a*r_a)/K_m\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"max braking speed=%.1f rpm\" %N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "power returned=48 kW\n",
-        "equivalent load resistance=0.7333 ohm\n",
-        "min braking speed=477.46 rpm\n",
-        "max braking speed=3660.6 rpm\n",
-        "max braking speed=1750.7 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.22, Page No 699"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "N=1500.0\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"when speed=1455rpm\")\n",
-      "n=1455.0\n",
-      "s1=(N-n)/N\n",
-      "r=math.sqrt(1/3)*(2/3)/(math.sqrt(s1)*(1-s1))    \n",
-      "print(\"I_2mx/I_2r=%.3f\" %r)\n",
-      "print(\"when speed=1350rpm\")\n",
-      "n=1350\n",
-      "s1=(N-n)/N\n",
-      "r=math.sqrt(1/3)*(2/3)/(math.sqrt(s1)*(1-s1))    \n",
-      "\n",
-      "#Results\n",
-      "print(\"I_2mx/I_2r=%.3f\" %r)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when speed=1455rpm\n",
-        "I_2mx/I_2r=0.000\n",
-        "when speed=1350rpm\n",
-        "I_2mx/I_2r=0.000\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.24, Page No 705"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V1=400.0\n",
-      "r1=0.6\n",
-      "r2=0.4\n",
-      "s=1.0\n",
-      "x1=1.6\n",
-      "x2=1.6\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"at starting in normal conditions\")\n",
-      "I_n=V1/math.sqrt((r1+r2/s)**2+(x1+x2)**2)    \n",
-      "print(\"current=%.2f A\" %I_n)\n",
-      "pf=(r1+r2)/math.sqrt((r1+r2/s)**2+(x1+x2)**2)    \n",
-      "print(\"pf=%.4f\" %pf)\n",
-      "f1=50\n",
-      "w_s=4*math.pi*f1/4\n",
-      "T_en=(3/w_s)*I_n**2*(r2/s)    \n",
-      "print(\"\\nTorque developed=%.2f Nm\" %T_en)\n",
-      "print(\"motor is operated with DOL starting\")\n",
-      "I_d=V1/2/math.sqrt((r1+r2/s)**2+((x1+x2)/2)**2)    \n",
-      "print(\"current=%.0f A\" %I_d)\n",
-      "pf=(r1+r2)/math.sqrt((r1+r2/s)**2+((x1+x2)/2)**2)    \n",
-      "print(\"pf=%.2f\" %pf)\n",
-      "f1=25\n",
-      "w_s=4*math.pi*f1/4\n",
-      "T_ed=(3/w_s)*I_d**2*(r2/s)    \n",
-      "print(\"Torque developed=%.3f Nm\" %T_ed)\n",
-      "print(\"at max torque conditions\")\n",
-      "s_mn=r2/math.sqrt((r1)**2+((x1+x2))**2)\n",
-      "I_n=V1/math.sqrt((r1+r2/s_mn)**2+(x1+x2)**2)    \n",
-      "print(\"current=%.3f A\" %I_n)\n",
-      "pf=(r1+r2/s_mn)/math.sqrt((r1+r2/s_mn)**2+(x1+x2)**2)    \n",
-      "print(\"pf=%.4f\" %pf)\n",
-      "f1=50\n",
-      "w_s=4*math.pi*f1/4\n",
-      "T_en=(3/w_s)*I_n**2*(r2/s_mn)    \n",
-      "print(\"Torque developed=%.2f Nm\" %T_en)\n",
-      "print(\"motor is operated with DOL starting\")\n",
-      "s_mn=r2/math.sqrt((r1)**2+((x1+x2)/2)**2)\n",
-      "I_d=V1/2/math.sqrt((r1+r2/s_mn)**2+((x1+x2)/2)**2)    \n",
-      "print(\"current=%.3f A\" %I_d)\n",
-      "pf=(r1+r2/s_mn)/math.sqrt((r1+r2/s_mn)**2+((x1+x2)/2)**2)    \n",
-      "print(\"\\npf=%.3f\" %pf)\n",
-      "f1=25\n",
-      "w_s=4*math.pi*f1/4\n",
-      "T_en=(3/w_s)*I_d**2*(r2/s_mn)  \n",
-      "\n",
-      "\n",
-      "#Results  \n",
-      "print(\"Torque developed=%.3f Nm\" %T_en)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "at starting in normal conditions\n",
-        "current=119.31 A\n",
-        "pf=0.2983\n",
-        "\n",
-        "Torque developed=108.75 Nm\n",
-        "motor is operated with DOL starting\n",
-        "current=106 A\n",
-        "pf=0.53\n",
-        "Torque developed=171.673 Nm\n",
-        "at max torque conditions\n",
-        "current=79.829 A\n",
-        "pf=0.7695\n",
-        "Torque developed=396.26 Nm\n",
-        "motor is operated with DOL starting\n",
-        "current=71.199 A\n",
-        "\n",
-        "pf=0.822\n",
-        "Torque developed=330.883 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.25, Page No 709"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "x1=1.0\n",
-      "X_m=50.0\n",
-      "X_e=x1*X_m/(x1+X_m)\n",
-      "V=231.0\n",
-      "V_e=V*X_m/(x1+X_m)\n",
-      "x2=1.0\n",
-      "r2=.4\n",
-      "r1=0\n",
-      "\n",
-      "#Calculations\n",
-      "s_m=r2/(x2+X_e)    \n",
-      "print(\"slip at max torque=%.2f\" %s_m)\n",
-      "s_mT=r2/(x2+X_m)    \n",
-      "print(\"slip at max torque=%.5f\" %s_mT)\n",
-      "f1=50.0\n",
-      "w_s=4*math.pi*f1/4\n",
-      "print(\"for constant voltage input\")\n",
-      "T_est=(3/w_s)*(V_e/math.sqrt(r2**2+(x2+X_e)**2))**2*(r2)    \n",
-      "print(\"starting torque=%.3f Nm\" %T_est)\n",
-      "T_em=(3/w_s)*V_e**2/(2*(x2+X_e))    \n",
-      "print(\"maximum torque developed=%.2f Nm\" %T_em)\n",
-      "print(\"for constant current input\")\n",
-      "I1=28\n",
-      "T_est=(3/w_s)*(I1*X_m)**2/(r2**2+(x2+X_m)**2)*r2    \n",
-      "print(\"starting torque=%.3f Nm\" %T_est)\n",
-      "T_em=(3/w_s)*(I1*X_m)**2/(2*(x2+X_m))   \n",
-      "print(\"maximum torque developed=%.3f Nm\" %T_em)\n",
-      "s=s_mT\n",
-      "i=1\n",
-      "I_m=I1*(r2/s+i*x2)/(r2/s+i*(x2+X_m))\n",
-      "I_m=math.fabs(I_m)\n",
-      "V1=math.sqrt(3)*I_m*X_m    \n",
-      "\n",
-      "#Results\n",
-      "print(\"supply voltage reqd=%.1f V\" %V1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "slip at max torque=0.20\n",
-        "slip at max torque=0.00784\n",
-        "for constant voltage input\n",
-        "starting torque=95.988 Nm\n",
-        "maximum torque developed=247.31 Nm\n",
-        "for constant current input\n",
-        "starting torque=5.756 Nm\n",
-        "maximum torque developed=366.993 Nm\n",
-        "supply voltage reqd=1236.2 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.27, Page No 718"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=420.0\n",
-      "V1=V/math.sqrt(3)\n",
-      "T_e=450.0\n",
-      "N=1440.0\n",
-      "n=1000.0\n",
-      "T_L=T_e*(n/N)**2\n",
-      "n1=1500.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_s=2*math.pi*n1/60\n",
-      "w_m=2*math.pi*n/60\n",
-      "a=.8\n",
-      "I_d=T_L*w_s/(2.339*a*V1)\n",
-      "k=0\n",
-      "R=(1-w_m/w_s)*(2.339*a*V1)/(I_d*(1-k))    \n",
-      "print(\"value of chopper resistance=%.4f ohm\" %R)\n",
-      "n=1320.0\n",
-      "T_L=T_e*(n/N)**2\n",
-      "I_d=T_L*w_s/(2.339*a*V1)    \n",
-      "print(\"Inductor current=%.3f A\" %I_d)\n",
-      "w_m=2*math.pi*n/60\n",
-      "k=1-((1-w_m/w_s)*(2.339*a*V1)/(I_d*R))    \n",
-      "print(\"value of duty cycle=%.4f\" %k)\n",
-      "s=(n1-n)/n1\n",
-      "V_d=2.339*s*a*V1    \n",
-      "print(\"Rectifed o/p voltage=%.3f V\" %V_d)\n",
-      "P=V_d*I_d\n",
-      "I2=math.sqrt(2/3)*I_d\n",
-      "r2=0.02\n",
-      "Pr=3*I2**2*r2\n",
-      "I1=a*I2\n",
-      "r1=0.015\n",
-      "Ps=3*I1**2*r1\n",
-      "Po=T_L*w_m\n",
-      "Pi=Po+Ps+Pr+P\n",
-      "eff=Po/Pi*100   \n",
-      "\n",
-      "#Results\n",
-      "print(\"Efficiency(in percent)=%.2f\" %eff)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of chopper resistance=2.0132 ohm\n",
-        "Inductor current=130.902 A\n",
-        "value of duty cycle=0.7934\n",
-        "Rectifed o/p voltage=54.449 V\n",
-        "Efficiency(in percent)=88.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.28, Page No 720"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=400.0\n",
-      "V_ph=V/math.sqrt(3)\n",
-      "N_s=1000.0\n",
-      "N=800.0\n",
-      "a=.7\n",
-      "I_d=110\n",
-      "R=2.0\n",
-      "\n",
-      "#Calculations\n",
-      "k=1-((1-N/N_s)*(2.339*a*V_ph)/(I_d*R))    \n",
-      "print(\"value of duty cycle=%.3f\" %k)\n",
-      "P=I_d**2*R*(1-k)\n",
-      "I1=a*I_d*math.sqrt(2/3)\n",
-      "r1=0.1\n",
-      "r2=0.08\n",
-      "Pr=3*I1**2*(r1+r2)\n",
-      "P_o=20000\n",
-      "P_i=P_o+Pr+P\n",
-      "eff=P_o/P_i*100    \n",
-      "print(\"Efficiency=%.2f\" %eff)\n",
-      "I11=math.sqrt(6)/math.pi*a*I_d\n",
-      "th=43\n",
-      "P_ip=math.sqrt(3)*V*I11*math.cos(math.radians(th))\n",
-      "pf=P_ip/(math.sqrt(3)*V*I11)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"Input power factor=%.4f\" %pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of duty cycle=0.656\n",
-        "Efficiency=70.62\n",
-        "Input power factor=0.7314\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.29, Page No 724"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=420.0\n",
-      "V1=V/math.sqrt(3)\n",
-      "N=1000.0\n",
-      "w_m=2*math.pi*N/60\n",
-      "N_s=1500.0\n",
-      "\n",
-      "#Calculations\n",
-      "s=(N_s-N)/N_s\n",
-      "a=0.8\n",
-      "V_d=2.339*a*s*V1    \n",
-      "print(\"rectified voltage=%.2f V\" %V_d)\n",
-      "T=450.0\n",
-      "N1=1200.0\n",
-      "T_L=T*(N/N1)**2\n",
-      "f1=50\n",
-      "w_s=4*math.pi*f1/4\n",
-      "I_d=w_s*T_L/(2.339*a*V1)    \n",
-      "print(\"inductor current=%.2f A\" %I_d)\n",
-      "a_T=-.4\n",
-      "a1=math.degrees(math.acos(s*a/a_T))\n",
-      "print(\"delay angle of inverter=%.2f deg\" %a1)\n",
-      "\n",
-      "P_s=V_d*I_d\n",
-      "P_o=T_L*w_m\n",
-      "R_d=0.01\n",
-      "P_i=I_d**2*R_d\n",
-      "I2=math.sqrt(2/3)*I_d\n",
-      "r2=0.02\n",
-      "r1=0.015\n",
-      "P_rol=3*I2**2*r2\n",
-      "I1=a*I2\n",
-      "P_sol=3*I1**2*r1\n",
-      "P_i=P_o+P_rol+P_sol+P_i\n",
-      "eff=P_o/P_i*100    \n",
-      "print(\"\\nefficiency=%.2f\" %eff)\n",
-      "w_m=w_s*(1+(-a_T/a)*math.cos(math.radians(a1))-w_s*R_d*T_L/(2.339*a*V1)**2)\n",
-      "N=w_m*60/(2*math.pi) \n",
-      "\n",
-      "#Results   \n",
-      "print(\"motor speed=%.1f rpm\" %N)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rectified voltage=151.25 V\n",
-        "inductor current=108.18 A\n",
-        "delay angle of inverter=131.81 deg\n",
-        "\n",
-        "efficiency=99.64\n",
-        "motor speed=996.4 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.30, Page No 726"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=700.0\n",
-      "E2=V/math.sqrt(3)\n",
-      "N_s=1500.0\n",
-      "N=1200.0\n",
-      "\n",
-      "#Calculations\n",
-      "s=(N_s-N)/N_s\n",
-      "V_dd=0.7\n",
-      "V_dt=1.5\n",
-      "V_d=3*math.sqrt(6)*s*E2/math.pi-2*V_dd\n",
-      "V1=415.0\n",
-      "a=math.degrees(math.acos((3*math.sqrt(2)*E2/math.pi)**-1*(-V_d+2*V_dt)))\n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle advance=%.2f deg\" %(180-a))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle advance=70.22 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.31, Page No 726"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=700.0\n",
-      "E2=V/math.sqrt(3)\n",
-      "N_s=1500.0\n",
-      "N=1200.0\n",
-      "\n",
-      "#Calculations\n",
-      "s=(N_s-N)/N_s\n",
-      "V_dd=.7\n",
-      "V_dt=1.5\n",
-      "a=0\n",
-      "u=18 #overlap angle in case of rectifier\n",
-      "V_d=3*math.sqrt(6)*s*E2*(math.cos(math.radians(a))+math.cos(math.radians(a+u)))/(2*math.pi)-2*V_dd\n",
-      "V1=415\n",
-      "V_ml=math.sqrt(2)*V1\n",
-      "u=4 #overlap anglein the inverter\n",
-      " #V_dc=-(3*V_ml*(math.cos(math.radians(a))+math.cos(math.radians(a+u)))/(2*math.pi)-2*V_dt)\n",
-      " #V_dc=V_d\n",
-      " #after solving %  (1+math.cos(math.radians(u)))*math.cos(math.radians(a))-math.sin(math.radians(u))*math.sin(math.radians(a))=-.6425\n",
-      "a=math.degrees(math.acos(-.6425/(math.sqrt((1+math.cos(math.radians(u)))**2+math.sin(math.radians(u))**2))))-math.degrees(math.asin(math.sin(math.radians(a))/(1+math.cos(math.radians(u)))))\n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle advance=%.2f deg\" %(180-a))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle advance=71.25 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.32, Page No 727"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=700.0\n",
-      "E2=V\n",
-      "N_s=1500.0\n",
-      "N=1200.0\n",
-      "\n",
-      "#Calculations\n",
-      "s=(N_s-N)/N_s\n",
-      "V1=415.0\n",
-      "a_T=s*E2/V1    \n",
-      "\n",
-      "#Results\n",
-      "print(\"voltage ratio of the transformer=%.2f\" %a_T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "voltage ratio of the transformer=0.34\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.33, Page No 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "P=6.0\n",
-      "N_s=600.0\n",
-      "f1=P*N_s/120.0\n",
-      "V=400.0\n",
-      "f=50.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_t=f1*V/f    \n",
-      "print(\"supply freq=%.0f Hz\" %V_t)\n",
-      "T=340.0\n",
-      "N=1000.0\n",
-      "T_L=T*(N_s/N)**2\n",
-      "w_s=2*math.pi*N_s/60\n",
-      "P=T_L*w_s\n",
-      "I_a=P/(math.sqrt(3)*V_t)    \n",
-      "print(\"armature current=%.2f A\" %I_a)\n",
-      "Z_s=2\n",
-      "X_s=f1/f*math.fabs(Z_s)\n",
-      "V_t=V_t/math.sqrt(3)\n",
-      "Ef=math.sqrt(V_t**2+(I_a*X_s)**2)\n",
-      "print(\"excitation voltage=%.2f V\" %(math.sqrt(3)*Ef))\n",
-      "dl=math.degrees(math.atan(I_a*X_s/V_t))\n",
-      "print(\"load angle=%.2f deg\" %dl)\n",
-      "T_em=(3/w_s)*(Ef*V_t/X_s)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"pull out torque=%.2f Nm\" %T_em)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "supply freq=240 Hz\n",
-        "armature current=18.50 A\n",
-        "excitation voltage=243.06 V\n",
-        "load angle=9.10 deg\n",
-        "pull out torque=773.69 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.34, Page No 736"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "P=4.0\n",
-      "f=50.0\n",
-      "w_s=4*math.pi*f/P\n",
-      "X_d=8.0\n",
-      "X_q=2.0\n",
-      "T_e=80.0\n",
-      "V=400.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_t=V/math.sqrt(3)\n",
-      "dl=(1/2)*math.degrees(math.asin(T_e*w_s/((3/2)*(V_t)**2*(1/X_q-1/X_d))))    \n",
-      "print(\"load angle=%.3f deg\" %dl)\n",
-      "I_d=V_t*math.cos(math.radians(dl))/X_d\n",
-      "I_q=V_t*math.sin(math.radians(dl))/X_q\n",
-      "I_a=math.sqrt(I_d**2+I_q**2)    \n",
-      "print(\"armature current=%.2f A\" %I_a)\n",
-      "pf=T_e*w_s/(math.sqrt(3)*V*I_a)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"input power factor=%.4f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "load angle=0.000 deg\n",
-        "armature current=28.87 A\n",
-        "input power factor=0.6283\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.35, Page No 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "T_e=3.0\n",
-      "K_m=1.2\n",
-      "I_a=T_e/K_m\n",
-      "r_a=2.0\n",
-      "V=230.0\n",
-      "\n",
-      "#Calculations\n",
-      "E_a=(0.263*math.sqrt(2)*V-I_a*r_a)/(1-55/180)\n",
-      "w_m=E_a/K_m\n",
-      "N=w_m*60/(2*math.pi)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor speed=%.2f rpm\" %N)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor speed=640.96 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.36, Page No 738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "K_m=1.0\n",
-      "N=1360.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=2*math.pi*N/60\n",
-      "E_a=K_m*w_m\n",
-      " #after calculations V_t % calculated\n",
-      "V_t=163.45\n",
-      "r_a=4\n",
-      "I_a=(V_t-E_a)/r_a\n",
-      "T_e=K_m*I_a    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor torque=%.4f Nm\" %T_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor torque=5.2578 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.37, Page No 740"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "K_m=1.0\n",
-      "N=2100.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=2*math.pi*N/60\n",
-      "E_a=K_m*w_m\n",
-      " #after calculations V_t % calculated\n",
-      "V_t=227.66\n",
-      "r_a=4\n",
-      "I_a=(V_t-E_a)/r_a\n",
-      "T_e=K_m*I_a    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor torque=%.2f Nm\" %T_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor torque=1.94 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.38, Page No 742"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "K_m=1.0\n",
-      "N=840.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=2*math.pi*N/60\n",
-      "E_a=K_m*w_m\n",
-      "V=230.0\n",
-      "a=75.0\n",
-      "V_t=math.sqrt(2)*V/math.pi*(1+math.cos(math.radians(a)))\n",
-      "r_a=4\n",
-      "I_a=(V_t-E_a)/r_a\n",
-      "T_e=K_m*I_a    \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor torque=%.4f Nm\" %T_e)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor torque=10.5922 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.39, Page No 743"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "K_m=1.0\n",
-      "N=1400.0\n",
-      "\n",
-      "#Calculations\n",
-      "w_m=2*math.pi*N/60\n",
-      "E_a=K_m*w_m\n",
-      "V=230.0\n",
-      "a=60.0\n",
-      "a1=212\n",
-      "V_t=math.sqrt(2)*V/math.pi*(math.cos(math.radians(a))-math.cos(math.radians(a1)))+E_a*(180+a-a1)/180\n",
-      "r_a=3\n",
-      "I_a=(V_t-E_a)/r_a\n",
-      "T_e=K_m*I_a   \n",
-      "\n",
-      "#Results\n",
-      "print(\"motor torque=%.3f Nm\" %T_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor torque=5.257 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.40, Page No 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "K_m=1.0\n",
-      "N=600.0\n",
-      "w_m=2*math.pi*N/60\n",
-      "E_a=K_m*w_m\n",
-      "V=230.0\n",
-      "a=60.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_t=2*math.sqrt(2)*V/math.pi*(math.cos(math.radians(a)))\n",
-      "r_a=3\n",
-      "I_a=(V_t-E_a)/r_a\n",
-      "T_e=K_m*I_a    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"motor torque=%.3f Nm\" %T_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "motor torque=13.568 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.41, Page No 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "r1=.6\n",
-      "r2=.4\n",
-      "s=0.04\n",
-      "x1=1.6\n",
-      "x2=1.6\n",
-      "Z=(r1+r2/s)+(x1+x2)\n",
-      "V=400.0\n",
-      "I1=V/Z    \n",
-      "print(\"source current=%.3f A \" %math.degrees(math.atan(I1.imag/I1.real)))\n",
-      "print(\"and with %.1f deg phase\" %math.fabs(I1))\n",
-      "I2=V/Z\n",
-      "N=1500\n",
-      "w_s=2*math.pi*N/60\n",
-      "T_e=(3/w_s)*abs(I2)**2*r2/s    \n",
-      "print(\"motor torque=%.2f Nm\" %T_e)\n",
-      "N_r=N*(1-s)\n",
-      "\n",
-      "f=45\n",
-      "N_s1=120*f/4\n",
-      "w_s=2*math.pi*N_s1/60\n",
-      "s1=(N_s1-N_r)/N_s1\n",
-      "Z=(r1+r2/s1)+(x1+x2)*f/50.0\n",
-      "V=360\n",
-      "I1=V/Z    \n",
-      "print(\"source current=%.3f A \" %math.degrees(math.atan(I1.imag/I1.real)))\n",
-      "print(\"and with %.1f deg phase\" %math.fabs(I1))\n",
-      "I2=V/Z\n",
-      "T_e=(3/w_s)*abs(I2)**2*r2/s1    \n",
-      "print(\"motor torque=%.2f Nm\" %T_e)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "source current=0.000 A \n",
-        "and with 29.0 deg phase\n",
-        "motor torque=160.46 Nm\n",
-        "source current=-0.000 A \n",
-        "and with 142.9 deg phase\n",
-        "motor torque=-2598.45 Nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter13.ipynb b/Power_Electronics/chapter13.ipynb
deleted file mode 100755
index 62d2a926..00000000
--- a/Power_Electronics/chapter13.ipynb
+++ /dev/null
@@ -1,342 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13 : Power Factor Improvement"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.1, Page No 754"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=250.0\n",
-      "R_l=5.0\n",
-      "I_l=20.0\n",
-      "V_l1=math.sqrt(V_s**2-(R_l*I_l)**2)\n",
-      "reg2=(V_s-V_l1)/V_s*100    \n",
-      "pf1=1.0\n",
-      "\n",
-      "#Calculations\n",
-      "P_l1=V_l1*I_l*pf1     #load power\n",
-      "P_r1=V_s*I_l*pf1     #max powwible system rating\n",
-      "utf1=P_l1*100/P_r1  \n",
-      "pf2=0.5\n",
-      " #(.5*V_l)**2+(.866*V_l+R_l*I_l)**2=V_s**2\n",
-      " #after solving\n",
-      "V_l2=158.35    \n",
-      "reg2=(V_s-V_l2)/V_s*100    \n",
-      "P_l2=V_l2*I_l*pf2   #load power\n",
-      "P_r2=V_s*I_l     #max powwible system rating\n",
-      "utf2=P_l2*100/P_r2    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"for pf=1\")\n",
-      "print(\"load voltage=%.2f V\" %V_l1)\n",
-      "print(\"voltage regulation=%.2f\" %reg1)\n",
-      "print(\"system utilisation factor=%.3f\" %utf1)\n",
-      "print(\"energy consumed(in units)=%.1f\" %(P_l1/1000))\n",
-      "print(\"for pf=.5\")\n",
-      "print(\"load voltage=%.2f V\" %V_l2)\n",
-      "print(\"voltage regulation=%.2f\" %reg2)\n",
-      "print(\"system utilisation factor=%.3f\" %utf2)\n",
-      "print(\"energy consumed(in units)=%.2f\" %(P_l2/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "ename": "NameError",
-       "evalue": "name 'reg1' is not defined",
-       "output_type": "pyerr",
-       "traceback": [
-        "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mNameError\u001b[0m                                 Traceback (most recent call last)",
-        "\u001b[1;32m<ipython-input-2-ffdbe43fd921>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m     25\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"for pf=1\"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     26\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"load voltage=%.2f V\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mV_l1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 27\u001b[1;33m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"voltage regulation=%.2f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mreg1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     28\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"system utilisation factor=%.3f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mutf1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     29\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"energy consumed(in units)=%.1f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mP_l1\u001b[0m\u001b[1;33m/\u001b[0m\u001b[1;36m1000\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
-        "\u001b[1;31mNameError\u001b[0m: name 'reg1' is not defined"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for pf=1\n",
-        "load voltage=229.13 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.2, Page No 756"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=50.0\n",
-      "V_s=230.0\n",
-      "I_m1=2\n",
-      "pf1=.3\n",
-      "\n",
-      "#Calculations\n",
-      "I_c1=I_m1*math.sin(math.radians(math.degrees(math.acos(pf1))))\n",
-      "C1=I_c1/(2*math.pi*f*V_s)    \n",
-      "I_m2=5\n",
-      "pf2=.5\n",
-      "I_c2=I_m2*math.sin(math.radians(math.degrees(math.acos(pf2))))\n",
-      "C2=I_c2/(2*math.pi*f*V_s)    \n",
-      "I_m3=10\n",
-      "pf3=.7\n",
-      "I_c3=I_m3*math.sin(math.radians(math.degrees(math.acos(pf3))))\n",
-      "C3=I_c3/(2*math.pi*f*V_s)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"at no load\")\n",
-      "print(\"value of capacitance=%.3f uF\" %(C1*10**6))\n",
-      "print(\"at half full load\")\n",
-      "print(\"value of capacitance=%.3f uF\" %(C2*10**6))\n",
-      "print(\"at full load\")\n",
-      "print(\"value of capacitance=%.3f uF\" %(C3*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "at no load\n",
-        "value of capacitance=26.404 uF\n",
-        "at half full load\n",
-        "value of capacitance=59.927 uF\n",
-        "at full load\n",
-        "value of capacitance=98.834 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.3 Page No 764"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_c=10.0\n",
-      "f=50.0\n",
-      "V_s=230.0\n",
-      "\n",
-      "#Calculations\n",
-      "C=I_c/(2*math.pi*f*V_s)    \n",
-      "I_l=10\n",
-      "L=V_s/(2*math.pi*f*I_l)   \n",
-      "\n",
-      "#Results\n",
-      "print(\"value of capacitance=%.3f uF\" %(C*10**6))\n",
-      "print(\"value of inductor=%.3f mH\" %(L*1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of capacitance=138.396 uF\n",
-        "value of inductor=73.211 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.4, Page No 765"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "I_L=10.0\n",
-      "X_L=V_s/I_L\n",
-      "I_f1=6.0\n",
-      " #B=2*a-math.sin(2*a)\n",
-      "B=2*math.pi-I_f1*math.pi*X_L/V_s\n",
-      "a=0\n",
-      "i=1.0\n",
-      "for a in range(1,360):\n",
-      "    b=2*a*math.pi/180-math.sin(math.radians(2*a))   \n",
-      "    if math.fabs(B-b)<=0.001  :      #by hit and trial\n",
-      "        i=2\n",
-      "        break\n",
-      "print(\"firing angle of TCR = %.1f deg\" %a)\n",
-      " #(a-.01)*180/math.pi)\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle of TCR = 359.0 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.5 Page No 766"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "L=.01\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"for firing angle=90deg\")\n",
-      "a=90*math.pi/180\n",
-      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
-      "print(\"effective inductance=%.0f mH\" %(L_eff*1000))\n",
-      "print(\"for firing angle=120deg\")\n",
-      "a=120*math.pi/180\n",
-      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
-      "print(\"effective inductance=%.3f mH\" %(L_eff*1000))\n",
-      "print(\"for firing angle=150deg\")\n",
-      "a=150*math.pi/180\n",
-      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
-      "print(\"effective inductance=%.2f mH\" %(L_eff*1000))\n",
-      "print(\"for firing angle=170deg\")\n",
-      "a=170*math.pi/180\n",
-      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
-      "print(\"effective inductance=%.3f H\" %L_eff)\n",
-      "print(\"for firing angle=175deg\")\n",
-      "a=175*math.pi/180\n",
-      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))  \n",
-      "\n",
-      "#Results\n",
-      "print(\"effective inductance=%.2f H\" %L_eff)\n",
-      "print(\"for firing angle=180deg\")\n",
-      "a=180*math.pi/180\n",
-      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
-      "print(\"effective inductance=%.3f H\" %L_eff)\n",
-      " #random value at firing angle =180 is equivalent to infinity as in answer in book\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle=90deg\n",
-        "effective inductance=10 mH\n",
-        "for firing angle=120deg\n",
-        "effective inductance=25.575 mH\n",
-        "for firing angle=150deg\n",
-        "effective inductance=173.40 mH\n",
-        "for firing angle=170deg\n",
-        "effective inductance=4.459 H\n",
-        "for firing angle=175deg\n",
-        "effective inductance=35.51 H\n",
-        "for firing angle=180deg\n",
-        "effective inductance=-128265253940037.750 H\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.6 Page No 766"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "Q=100.0*10**3\n",
-      "V_s=11.0*10**3\n",
-      "\n",
-      "#Calculations\n",
-      "f=50.0\n",
-      "L=V_s**2/(2*math.pi*f*Q)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"effective inductance=%.4f H\" %L)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "effective inductance=3.8515 H\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter14.ipynb b/Power_Electronics/chapter14.ipynb
deleted file mode 100755
index a9c3a3f1..00000000
--- a/Power_Electronics/chapter14.ipynb
+++ /dev/null
@@ -1,93 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14 : Miscellaneous Topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1, Page No 777"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a1=0\n",
-      "a2=45.0\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"for two single phase series semiconvertors\")\n",
-      "V_0=V_m/math.pi*(2+math.cos(math.radians(a1))+math.cos(math.radians(a2)))    \n",
-      "print(\"avg o/p voltage=%.2f V\" %V_0)\n",
-      "V_or=V_s*math.sqrt((1/math.pi)*(4*math.pi-3*a2*math.pi/180+(3/2)*math.sin(math.radians(2*a2))))    \n",
-      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
-      "DF=(3+math.cos(math.radians(a2)))/(math.sqrt(2)*math.sqrt(5+3*math.cos(math.radians(a2))))    \n",
-      "print(\"DF=%.2f\" %DF)\n",
-      "PF=math.sqrt(2/math.pi)*(3+math.cos(math.radians(a2)))/math.sqrt(4*math.pi-3*a2*math.pi/180)    \n",
-      "print(\"PF=%.2f\" %PF)\n",
-      "HF=math.sqrt((math.pi*(math.pi-(3/4)*a2*math.pi/180)/(5+3*math.cos(math.radians(a2))))-1)    \n",
-      "print(\"HF=%.2f\" %HF)\n",
-      "print(\"for two single phase series full convertors\")\n",
-      "a=45.0\n",
-      "V_0=2*V_m/math.pi*(1+math.cos(math.radians(a)))    \n",
-      "print(\"avg o/p voltage=%.2f V\" %V_0)\n",
-      "V_or=2*V_s*math.sqrt((1/math.pi)*(math.pi-a2*math.pi/180+(1/2)*math.sin(math.radians(2*a2))))    \n",
-      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
-      "DF=math.cos(math.radians(a2/2)) \n",
-      "\n",
-      "\n",
-      "#Results   \n",
-      "print(\"DF=%.2f\" %DF)\n",
-      "PF=math.sqrt(2/(math.pi*(math.pi-a2*math.pi/180)))*(1+math.cos(math.radians(a2)))    \n",
-      "print(\"PF=%.2f\" %PF)\n",
-      "HF=math.sqrt((math.pi*(math.pi-a2*math.pi/180)/(4+4*math.cos(math.radians(a2))))-1)    \n",
-      "print(\"HF=%.2f\" %HF)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for two single phase series semiconvertors\n",
-        "avg o/p voltage=383.82 V\n",
-        "rms value of o/p voltage=434.47 V\n",
-        "DF=0.98\n",
-        "PF=0.93\n",
-        "HF=0.62\n",
-        "for two single phase series full convertors\n",
-        "avg o/p voltage=353.50 V\n",
-        "rms value of o/p voltage=398.37 V\n",
-        "DF=0.92\n",
-        "PF=0.89\n",
-        "HF=0.29\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter2.ipynb b/Power_Electronics/chapter2.ipynb
deleted file mode 100755
index 1872c9f4..00000000
--- a/Power_Electronics/chapter2.ipynb
+++ /dev/null
@@ -1,233 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02 : Power Semiconductor Diodes and Transistors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1, Page No 21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "B=40.0\n",
-      "R_c=10                    #ohm\n",
-      "V_cc=130.0                #V\n",
-      "V_B=10.0                  #V\n",
-      "V_CES=1.0                 #V\n",
-      "V_BES=1.5                 #V\n",
-      "\n",
-      "#Calculations\n",
-      "I_CS=(V_cc-V_CES)/R_c     #A\n",
-      "I_BS=I_CS/B                #A\n",
-      "R_B1=(V_B-V_BES)/I_BS\n",
-      "P_T1=V_BES*I_BS+V_CES*I_CS\n",
-      "ODF=5\n",
-      "I_B=ODF*I_BS\n",
-      "R_B2=(V_B-V_BES)/I_B\n",
-      "P_T2=V_BES*I_B+V_CES*I_CS\n",
-      "B_f=I_CS/I_B\n",
-      "\n",
-      "#Results\n",
-      "print(\"value of R_B in saturated state= %.2f ohm\" %R_B1)\n",
-      "print(\"Power loss in transistor=%.2f W\" %P_T1)\n",
-      "print(\"Value of R_B for an overdrive factor 5 = %.2f ohm\" %R_B2)\n",
-      "print(\"Power loss in transistor = %.2f W\" %P_T2)\n",
-      "print(\"Forced current gain=%.0f\" %B_f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of R_B in saturated state= 26.36 ohm\n",
-        "Power loss in transistor=13.38 W\n",
-        "Value of R_B for an overdrive factor 5 = 5.27 ohm\n",
-        "Power loss in transistor = 15.32 W\n",
-        "Forced current gain=8\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page No 24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_CEO=2*10**-3            #A\n",
-      "V_CC=220.0                #V\n",
-      "P_dt=I_CEO*V_CC           #instant. power loss during delay time\n",
-      "t_d=.4*10**-6            #s\n",
-      "f=5000\n",
-      "P_d=f*I_CEO*V_CC*t_d      #avg power loss during delay time\n",
-      "V_CES=2                     #V\n",
-      "t_r=1*10**-6               #s\n",
-      "I_CS=80                    #A\n",
-      "\n",
-      "#Calculations\n",
-      "P_r=f*I_CS*t_r*(V_CC/2-(V_CC-V_CES)/3)            #avg power loss during rise time\n",
-      "t_m=V_CC*t_r/(2*(V_CC-V_CES))\n",
-      "P_rm=I_CS*V_CC**2/(4*(V_CC-V_CES))           #instant. power loss during rise time\n",
-      "\n",
-      "#Results\n",
-      "P_on=P_d+P_r                \n",
-      "print(\"Avg power loss during turn on = %.2f W\" %P_on)\n",
-      "P_nt=I_CS*V_CES \n",
-      "print(\"Instantaneous power loss during turn on = %.0f W\" %P_nt)\n",
-      "t_n=50*10**-6\n",
-      "P_n=f*I_CS*V_CES*t_n\n",
-      "print(\"Avg power loss during conduction period = %.0f W\" %P_n)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Avg power loss during turn on = 14.93 W\n",
-        "Instantaneous power loss during turn on = 160 W\n",
-        "Avg power loss during conduction period = 40 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3 Page No 26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_CEO=2*10**-3         #A\n",
-      "V_CC=220          #V\n",
-      "t_d=.4*10**-6   #s\n",
-      "f=5000\n",
-      "V_CES=2         #V\n",
-      "t_r=1*10**-6    #s\n",
-      "I_CS=80         #A\n",
-      "t_n=50*10**-6   #s\n",
-      "t_0=40*10**-6   #s\n",
-      "t_f=3*10**-6    #s\n",
-      "\n",
-      "#Calculations\n",
-      "P_st=I_CS*V_CES  # instant. power loss during t_s\n",
-      "P_s=f*I_CS*V_CES*t_f   #avg power loss during t_s\n",
-      "P_f=f*t_f*(I_CS/6)*(V_CC-V_CES)        #avg power loss during fall time\n",
-      "P_fm=(I_CS/4)*(V_CC-V_CES)          #peak instant power dissipation\n",
-      "P_off=P_s+P_f\n",
-      "\n",
-      "#Results\n",
-      "print(\"Total avg power loss during turn off = %.2f W\" %P_off)\n",
-      "P_0t=I_CEO*V_CC\n",
-      "print(\"Instantaneous power loss during t_0 = %.2f W\" %P_0t)\n",
-      "P_0=f*I_CEO*V_CC*t_0              #avg power loss during t_s\n",
-      "P_on=14.9339              #W    from previous eg\n",
-      "P_n=40                    #W    from previous eg\n",
-      "P_T=P_on+P_n+P_off+P_0     \n",
-      "print(\"Total power loss = %.2f W\" %P_T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total avg power loss during turn off = 44.91 W\n",
-        "Instantaneous power loss during t_0 = 0.44 W\n",
-        "Total power loss = 99.93 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4, Page No 28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_CS=100.0 \n",
-      "V_CC=200.0 \n",
-      "t_on=40*10**-6\n",
-      "\n",
-      "#Calculations\n",
-      "P_on=(I_CS/50)*10**6*t_on*(V_CC*t_on/2-(V_CC*10**6*t_on**2/(40*3)))       #energy during turn on\n",
-      "t_off=60*10**-6\n",
-      "P_off=(I_CS*t_off/2-(I_CS/60)*10**6*(t_off**2)/3)*((V_CC/75)*10**6*t_off)  #energy during turn off\n",
-      "P_t=P_on+P_off           #total energy\n",
-      "P_avg=300.0\n",
-      "f=P_avg/P_t\n",
-      "\n",
-      "#Results\n",
-      "print(\"Allowable switching frequency = %.2f Hz\" %f)\n",
-      "#in book ans is: f=1123.6 Hz. The difference in results due to difference in rounding of of digits"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Allowable switching frequency = 1125.00 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter3.ipynb b/Power_Electronics/chapter3.ipynb
deleted file mode 100755
index 2e53ef9d..00000000
--- a/Power_Electronics/chapter3.ipynb
+++ /dev/null
@@ -1,1001 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 03 : Diode Circuits and Rectifiers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2, Page No 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=400.0  #V\n",
-      "V_o=100.0    #V\n",
-      "L=100.0      #uH\n",
-      "C=30.0       #uF\n",
-      "\n",
-      "#Calculations\n",
-      "t_o=math.pi*math.sqrt(L*C)\n",
-      "print(\"conduction time of diode = %.2f us\" %t_o)\n",
-      "#in book solution is t_o=54.77 us. The ans is incorrect as %pi is not muliplied in ans. Formulae mentioned in correct.\n",
-      "I_p=(V_s-V_o)*math.sqrt(C/L)\n",
-      "\n",
-      "#Results\n",
-      "print(\"Peak current through diode=%.2f A\" %I_p)\n",
-      "v_D=-V_s+V_o \n",
-      "print(\"Voltage across diode = %.2f V\" %v_D)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "conduction time of diode = 172.07 us\n",
-        "Peak current through diode=164.32 A\n",
-        "Voltage across diode = -300.00 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page No 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "R=10             #ohm\n",
-      "L=0.001           #H\n",
-      "C=5*10**-6       #F\n",
-      "V_s=230          #V\n",
-      "xi=R/(2*L)\n",
-      "\n",
-      "#Calculations\n",
-      "w_o=1/math.sqrt(L*C)\n",
-      "w_r=math.sqrt((1/(L*C))-(R/(2*L))**2)\n",
-      "t=math.pi/w_r \n",
-      "\n",
-      "#Results\n",
-      "print('Conduction time of diode=%.3f us'%(t*10**6))\n",
-      "t=0\n",
-      "di=V_s/L\n",
-      "print('Rate of change of current at t=0 is %.2f A/s' %di)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Conduction time of diode=237.482 us\n",
-        "Rate of change of current at t=0 is 230000.00 A/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7 Page No 69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I_or=100             #A\n",
-      "R=1.0       #assumption\n",
-      "\n",
-      "#Calculations\n",
-      "V_m=I_or*2*R\n",
-      "I_o=V_m/(math.pi*R)\n",
-      "q=200       #Ah\n",
-      "t=q/I_o\n",
-      "\n",
-      "#Results\n",
-      "print(\"time required to deliver charge=%.02f hrs\" %t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "time required to deliver charge=3.14 hrs\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8, Page No 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0    #V\n",
-      "P=1000       #W\n",
-      "R=V_s**2/P\n",
-      "\n",
-      "#Calculations\n",
-      "V_or=math.sqrt(2)*V_s/2\n",
-      "P_h=V_or**2/R    \n",
-      "print(\"Power delivered to the heater = %.2f W\" %P_h)\n",
-      "V_m=math.sqrt(2)*230\n",
-      "I_m=V_m/R\n",
-      "\n",
-      "#Results\n",
-      "print(\"Peak value of diode current = %.2f A\" %I_m)\n",
-      "pf=V_or/V_s\n",
-      "print(\"Input power factor=%.2f\" %pf)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered to the heater = 500.00 W\n",
-        "Peak value of diode current = 6.15 A\n",
-        "Input power factor=0.71\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9 Page No 71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230        #V\n",
-      "V_m=V_s*math.sqrt(2)\n",
-      "E=150          #V\n",
-      "\n",
-      "#Calculations\n",
-      "theta1=math.degrees(E/(math.sqrt(2)*V_s))\n",
-      "R=8        #ohm\n",
-      "f=50    #Hz\n",
-      "I_o=(1/(2*math.pi*R))*((2*math.sqrt(2)*V_s*math.cos(math.radians(theta1)))-E*(math.pi-2*theta1*math.pi/180))\n",
-      "\n",
-      "#Results\n",
-      "print(\"avg value of charging current=%.2f A\" %I_o)\n",
-      "P_d=E*I_o\n",
-      "print(\"\\npower delivered to battery=%.2f W\" %P_d)\n",
-      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V_s**2+E**2)*(math.pi-2*theta1*math.pi/180)+V_s**2*math.sin(math.radians(2*theta1))-4*V_m*E*math.cos(math.radians(theta1))))\n",
-      "print(\"\\nrms value of the load current=%.2f A\" %I_or)\n",
-      "pf=(E*I_o+I_or**2*R)/(V_s*I_or)\n",
-      "print(\"\\nsupply pf=%.3f\" %pf)\n",
-      "P_dd=I_or**2*R\n",
-      "print(\"\\npower dissipated in the resistor=%.2f W\" %P_dd)\n",
-      "q=1000.00         #Wh\n",
-      "t=q/P_d    \n",
-      "print(\"\\ncharging time=%.2f hr\" %t)\n",
-      "n=P_d*100/(P_d+P_dd)\n",
-      "print(\"rectifier efficiency =%.2f  \" %n)\n",
-      "PIV=math.sqrt(2)*V_s+E\n",
-      "print(\"PIV of diode=%.2f V\" %PIV)\n",
-      "#solutions have small variations due to difference in rounding off of digits"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg value of charging current=4.97 A\n",
-        "\n",
-        "power delivered to battery=745.11 W\n",
-        "\n",
-        "rms value of the load current=9.29 A\n",
-        "\n",
-        "supply pf=0.672\n",
-        "\n",
-        "power dissipated in the resistor=690.74 W\n",
-        "\n",
-        "charging time=1.34 hr\n",
-        "rectifier efficiency =51.89  \n",
-        "PIV of diode=475.27 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.10 Page No 78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_s=230            #V\n",
-      "t_rr=40*10**-6    #s reverde recovery time\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=2*math.sqrt(2)*V_s/math.pi\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "f=50\n",
-      "V_r1=(V_m/math.pi)*(1-math.cos(math.radians(2*math.pi*f*t_rr*180/math.pi)))\n",
-      "v_avg1=V_r1*100/V_o*10**3\n",
-      "f=2500\n",
-      "V_r2=(V_m/math.pi)*(1-math.cos(math.radians(2*math.pi*f*t_rr*180/math.pi)))\n",
-      "v_avg2=V_r2*100/V_o\n",
-      "\n",
-      "#Results\n",
-      "print(\"when f=50Hz\")\n",
-      "print(\"Percentage reduction in avg o/p voltage=%.2f x 10^-3\" %v_avg1)\n",
-      "print(\"when f=2500Hz\")\n",
-      "print(\"Percentage reduction in avg o/p voltage = %.3f\" %v_avg2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when f=50Hz\n",
-        "Percentage reduction in avg o/p voltage=3.95 x 10^-3\n",
-        "when f=2500Hz\n",
-        "Percentage reduction in avg o/p voltage = 9.549\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.11, Page No 79 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230         #V\n",
-      "R=10.0        #ohm\n",
-      "\n",
-      "#Calculations\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "V_o=2*V_m/math.pi\n",
-      "print(\"Avg value of o/p voltage = %.2f V\" %V_o)\n",
-      "I_o=V_o/R\n",
-      "print(\"Avg value of o/p current = %.2f A\" %I_o)\n",
-      "I_DA=I_o/2\n",
-      "print(\"Avg value of diode current=%.2f A\" %I_DA)\n",
-      "I_Dr=I_o/math.sqrt(2)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of diode current=%.2f A\" %I_Dr)\n",
-      "print(\"rms value of o/p current = %.2f A\" %I_o)\n",
-      "print(\"rms value of i/p current = %.2f A\" %I_o)\n",
-      "pf=(V_o/V_s)\n",
-      "print(\"supply pf = %.2f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Avg value of o/p voltage = 207.07 V\n",
-        "Avg value of o/p current = 20.71 A\n",
-        "Avg value of diode current=10.35 A\n",
-        "rms value of diode current=14.64 A\n",
-        "rms value of o/p current = 20.71 A\n",
-        "rms value of i/p current = 20.71 A\n",
-        "supply pf = 0.90\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.12 Page No 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_s=230.0    #V\n",
-      "R=1000.0      #ohm\n",
-      "R_D=20.0    #ohm\n",
-      "\n",
-      "#Calculations\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "I_om=V_m/(R+R_D)          \n",
-      "\n",
-      "#Results\n",
-      "print(\"Peak load current = %.2f A\" %I_om)\n",
-      "I_o=I_om/math.pi\n",
-      "print(\"dc load current = %.2f A\" %I_o)\n",
-      "V_D=I_o*R_D-V_m/math.pi\n",
-      "print(\"dc diode voltage = %.2f V\" %V_D)\n",
-      "V_on=V_m/math.pi\n",
-      "print(\"at no load, load voltage = %.2f V\" %V_on)\n",
-      "V_o1=I_o*R    \n",
-      "print(\"at given load, load voltage = %.2f V\" %V_o1)\n",
-      "vr=(V_on-V_o1)*100/V_on    \n",
-      "print(\"Voltage regulation(in percent)=%.2f\" %vr)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak load current = 0.32 A\n",
-        "dc load current = 0.10 A\n",
-        "dc diode voltage = -101.51 V\n",
-        "at no load, load voltage = 103.54 V\n",
-        "at given load, load voltage = 101.51 V\n",
-        "Voltage regulation(in percent)=1.96\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.13 Page No 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_L=6.8   #V\n",
-      "V_smax=20*1.2    #V\n",
-      "V_smin=20*.8     #V\n",
-      "I_Lmax=30*1.5   #mA\n",
-      "I_Lmin=30*0.5  #mA\n",
-      "I_z=1          #mA\n",
-      "\n",
-      "#Calculations\n",
-      "R_smax=(V_smax-V_L)/((I_Lmin+I_z)*10**-3)\n",
-      "print(\"max source resistance = %.2f ohm\" %R_smax)\n",
-      "R_smin=(V_smin-V_L)/((I_Lmax+I_z)*10**-3)    \n",
-      "print(\"Min source resistance = %.2f ohm\" %R_smin)  #in book solution, error is committed in putting the values in formulea(printing error) but solution is correct\n",
-      "R_Lmax=V_L*1000/I_Lmin\n",
-      "print(\"Max load resistance = %.2f ohm\" %R_Lmax)\n",
-      "R_Lmin=V_L*1000/I_Lmax    \n",
-      "V_d=0.6      #V\n",
-      "V_r=V_L-V_d\n",
-      "\n",
-      "#Results\n",
-      "print(\"Min load resistance=%.2f ohm\" %R_Lmin)\n",
-      "print(\"Voltage rating of zener diode=%.2f V\" %V_r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "max source resistance = 1075.00 ohm\n",
-        "Min source resistance = 200.00 ohm\n",
-        "Max load resistance = 453.33 ohm\n",
-        "Min load resistance=151.11 ohm\n",
-        "Voltage rating of zener diode=6.20 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.14 Page No 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "I2=200*10**-6      #A\n",
-      "V_z=20       #V\n",
-      "R_G=500.0   #hm\n",
-      "\n",
-      "#Calculations\n",
-      "R2=(V_z/I2)-R_G\n",
-      "print(\"R2=%.2f kilo-ohm\" %(R2/1000))\n",
-      "\n",
-      "V_v=25      #V\n",
-      "I1=I2\n",
-      "R1=(V_v-V_z)/I1\n",
-      "\n",
-      "#Results\n",
-      "print(\"R1=%.0f kilo-ohm\"%(R1/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R2=99.50 kilo-ohm\n",
-        "R1=25 kilo-ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.15, Page No 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_s=2*230          #V\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=(math.sqrt(2)*V_s)/math.pi\n",
-      "R=60                #ohm\n",
-      "P_dc=(V_o)**2/R\n",
-      "TUF=0.2865\n",
-      "VA=P_dc/TUF\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"kVA rating of the transformer = %.2f kVA\" %(VA/1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "kVA rating of the transformer = 2.49 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.16, Page No 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "tr=0.5       #turns ratio\n",
-      "I_o=10.0\n",
-      "V=230.0\n",
-      "V_s=V/tr\n",
-      "\n",
-      "#Calculations\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "V_o=2*V_m/math.pi\n",
-      "phi1=0\n",
-      "#displacemnt angle=0 as fundamnetal component of i/p source current in phase with source voltage\n",
-      "DF=math.cos(math.radians(phi1))\n",
-      "I_s1=4*I_o/(math.sqrt(2)*math.pi)\n",
-      "I_s=math.sqrt(I_o**2*math.pi/math.pi)\n",
-      "CDF=I_s1/I_o\n",
-      "pf=CDF*DF\n",
-      "HF=math.sqrt((I_s/I_s1)**2-1)\n",
-      "CF=I_o/I_s\n",
-      "\n",
-      "#Results\n",
-      "print(\"o/p voltage = %.2f V\" %V_o)\n",
-      "print(\"distortion factor = %.2f\" %DF)\n",
-      "print(\"i/p pf=%.2f\" %pf)\n",
-      "print(\"Current displacent factor=%.2f\" %CDF)\n",
-      "print(\"Harmonic factor = %.2f\" %HF)\n",
-      "print(\"Creast factor = %.2f\" %CF)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "o/p voltage = 414.15 V\n",
-        "distortion factor = 1.00\n",
-        "i/p pf=0.90\n",
-        "Current displacent factor=0.90\n",
-        "Harmonic factor = 0.48\n",
-        "Creast factor = 1.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.17, Page No 94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_o=230.0\n",
-      "R=10.0\n",
-      "V_s=V_o*math.pi/(2*math.sqrt(2))\n",
-      "I_o=V_o/R\n",
-      "I_m=math.sqrt(2)*V_s/R\n",
-      "I_DAV=I_m/math.pi\n",
-      "\n",
-      "#Calculations\n",
-      "#avg value of diode current\n",
-      "I_Dr=I_m/2\n",
-      "PIV=math.sqrt(2)*V_s\n",
-      "I_s=I_m/math.sqrt(2)\n",
-      "TF=V_s*I_s\n",
-      "\n",
-      "#Results\n",
-      "print(\"peak diode current=%.2f A\" %I_m)\n",
-      "print(\"I_DAV=%.2f A\" %I_DAV)\n",
-      "print(\"I_Dr=%.2f A\" %I_Dr)   #rms value of diode current\n",
-      "print(\"PIV=%.1f V\" %PIV)\n",
-      "print(\"Transformer rating = %.2f kVA\" %(TF/1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "peak diode current=36.13 A\n",
-        "I_DAV=11.50 A\n",
-        "I_Dr=18.06 A\n",
-        "PIV=361.3 V\n",
-        "Transformer rating = 6.53 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.18, Page No 103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "tr=5\n",
-      "V=1100.0\n",
-      "R=10.0\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"In case of 3ph-3pulse type\")\n",
-      "V_ph=V/tr\n",
-      "V_mp=math.sqrt(2)*V_ph\n",
-      "V_o=3*math.sqrt(3)*V_mp/(2*math.pi)\n",
-      "print(\"avg o/p voltage=%.1f V\" %V_o)\n",
-      "I_mp=V_mp/R\n",
-      "I_D=(I_mp/math.pi)*math.sin(math.pi/3)    \n",
-      "print(\"\\navg value of diode current=%.3f A\" %I_D)\n",
-      "I_Dr=I_mp*math.sqrt((1/(2*math.pi))*(math.pi/3+.5*math.sin(2*math.pi/3)))    \n",
-      "print(\"\\nrms value of diode current=%.2f A\" %I_Dr)\n",
-      "V_or=V_mp*math.sqrt((3/(2*math.pi))*(math.pi/3+.5*math.sin(2*math.pi/3)))\n",
-      "P=(V_or**2)/R   \n",
-      "print(\"\\npower delivered=%.1f W\" %P)\n",
-      "print(\"in case of 3ph-M6 type\")\n",
-      "V_ph=V_ph/2\n",
-      "V_mp=math.sqrt(2)*V_ph\n",
-      "V_o=3*V_mp/(math.pi)  \n",
-      "I_mp=V_mp/R\n",
-      "I_D=(I_mp/math.pi)*math.sin(math.pi/6)    \n",
-      "I_Dr=I_mp*math.sqrt((1/(2*math.pi))*(math.pi/6+.5*math.sin(2*math.pi/6)))  \n",
-      "V_or=V_mp*math.sqrt((6/(2*math.pi))*(math.pi/6+.5*math.sin(2*math.pi/6)))\n",
-      "P=(V_or**2)/R   \n",
-      "\n",
-      "#Results\n",
-      "print(\"avg o/p voltage=%.2f V\" %V_o)\n",
-      "print(\"\\navg value of diode current=%.2f A\" %I_D)\n",
-      "print(\"\\nrms value of diode current=%.2f A\" %I_Dr)\n",
-      "print(\"\\npower delivered=%.0f W\" %P)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "In case of 3ph-3pulse type\n",
-        "avg o/p voltage=257.3 V\n",
-        "\n",
-        "avg value of diode current=8.577 A\n",
-        "\n",
-        "rms value of diode current=15.10 A\n",
-        "\n",
-        "power delivered=6841.3 W\n",
-        "in case of 3ph-M6 type\n",
-        "avg o/p voltage=148.55 V\n",
-        "\n",
-        "avg value of diode current=2.48 A\n",
-        "\n",
-        "rms value of diode current=6.07 A\n",
-        "\n",
-        "power delivered=2211 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.19, Page No 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_o=400\n",
-      "R=10\n",
-      "\n",
-      "#Calculations\n",
-      "V_ml=V_o*math.pi/3\n",
-      "V_s=V_ml/(math.sqrt(2)*math.sqrt(3))\n",
-      "I_m=V_ml/R\n",
-      "I_s=.7804*I_m\n",
-      "tr=3*V_s*I_s   \n",
-      "\n",
-      "#Results\n",
-      "print(\"transformer rating=%.1f VA\" %tr)\n",
-      "I_Dr=.5518*I_m   \n",
-      "print(\"\\nrms value of diode current=%.3f A\" %I_Dr)\n",
-      "I_D=I_m/math.pi   \n",
-      "print(\"\\navg value of diode current=%.3f A\" %I_D)\n",
-      "print(\"\\npeak diode current=%.2f A\" %I_m)\n",
-      "PIV=V_ml \n",
-      "print(\"\\nPIV=%.2f V\" %PIV)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "transformer rating=16770.3 VA\n",
-        "\n",
-        "rms value of diode current=23.114 A\n",
-        "\n",
-        "avg value of diode current=13.333 A\n",
-        "\n",
-        "peak diode current=41.89 A\n",
-        "\n",
-        "PIV=418.88 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.20, Page No 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_l=230\n",
-      "E=240\n",
-      "R=8\n",
-      "\n",
-      "#Calculations\n",
-      "V_ml=math.sqrt(2)*V_l\n",
-      "V_o=3*V_ml/math.pi\n",
-      "I_o=(V_o-E)/R\n",
-      "P_b=E*I_o    \n",
-      "P_d=E*I_o+I_o**2*R   \n",
-      "phi1=0\n",
-      "math.cos(math.radians(phi1))\n",
-      "I_s1=2*math.sqrt(3)*I_o/(math.sqrt(2)*math.pi)\n",
-      "I_s=math.sqrt(I_o**2*2*math.pi/(3*math.pi))\n",
-      "CDF=I_s1/I_s    \n",
-      "pf=DF*CDF    \n",
-      "HF=math.sqrt(CDF**-2-1)   \n",
-      "tr=math.sqrt(3)*V_l*I_o*math.sqrt(2/3)\n",
-      "\n",
-      "#Results\n",
-      "print(\"Power delivered to battery=%.1f W\" %P_b)\n",
-      "print(\"Power delivered to load=%.2f W\" %P_d)\n",
-      "print(\"Displacement factor=%.2f\" %DF)\n",
-      "print(\"Current distortion factor=%.3f\" %CDF)\n",
-      "print(\"i/p pf=%.3f\"%pf)\n",
-      "print(\"Harmonic factor=%.2f\" %HF)\n",
-      "print(\"Tranformer rating=%.2f VA\" %tr)\n",
-      "#answers have small variations from the book due to difference in rounding off of digits"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered to battery=2118.3 W\n",
-        "Power delivered to load=2741.48 W\n",
-        "Displacement factor=1.00\n",
-        "Current distortion factor=0.955\n",
-        "i/p pf=0.955\n",
-        "Harmonic factor=0.31\n",
-        "Tranformer rating=0.00 VA\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.21, Page No 122"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=50  #Hz\n",
-      "V=230.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_m=math.sqrt(2)*V\n",
-      "R=400.0\n",
-      "RF=0.05\n",
-      "C=(1/(4*f*R))*(1+(1/(math.sqrt(2)*RF)))\n",
-      "\n",
-      "#Results\n",
-      "print(\"capacitor value=%.2f uF\" %(C/10**-6))\n",
-      "V_o=V_m*(1-1/(4*f*R*C))\n",
-      "print(\"o/p voltage with filter=%.2f V\" %V_o)\n",
-      "V_o=2*V_m/math.pi   \n",
-      "print(\"o/p voltage without filter=%.2f V\" %V_o)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "capacitor value=189.28 uF\n",
-        "o/p voltage with filter=303.79 V\n",
-        "o/p voltage without filter=207.07 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.22, Page No 122"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=50\n",
-      "CRF=0.05\n",
-      "R=300\n",
-      "\n",
-      "#Calculations\n",
-      "L=math.sqrt((CRF/(.4715*R))**-2-R**2)/(2*2*math.pi*f)    \n",
-      "print(\"L=%.2f H\" %L)\n",
-      "R=30\n",
-      "L=math.sqrt((CRF/(.4715*R))**-2-R**2)/(2*2*math.pi*f)    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"\\nL=%.2f H\" %L)\n",
-      "L=0\n",
-      "CRF=.4715*R/math.sqrt(R**2+(2*2*math.pi*f*L)**2)   \n",
-      "print(\"\\nCRF=%.2f\" %CRF)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L=4.48 H\n",
-        "\n",
-        "L=0.45 H\n",
-        "\n",
-        "CRF=0.47\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.23, Page No 127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=50\n",
-      "L_L=10*10**-3\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "\n",
-      "#Calculations\n",
-      "C=10/(2*w*math.sqrt(R**2+(2*w*L_L)**2))\n",
-      "\n",
-      "#Results\n",
-      "print(\"C=%.2f uF\" %(C*10**6))\n",
-      "VRF=0.1\n",
-      "L=(1/(4*w**2*C))*((math.sqrt(2)/(3*VRF))+1)\n",
-      "print(\"\\nL=%.2f mH\" %(L*10**3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C=315.83 uF\n",
-        "\n",
-        "L=45.83 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter4.ipynb b/Power_Electronics/chapter4.ipynb
deleted file mode 100755
index 22311574..00000000
--- a/Power_Electronics/chapter4.ipynb
+++ /dev/null
@@ -1,946 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 04 : Thyristors"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.3, Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "P=.5    #P=V_g*I_g\n",
-      "s=130    #s=V_g/I_g\n",
-      "\n",
-      "#Calculations\n",
-      "I_g=math.sqrt(P/s)\n",
-      "V_g=s*I_g\n",
-      "E=15\n",
-      "R_s=(E-V_g)/I_g    \n",
-      "\n",
-      "#Results\n",
-      "print(\"Gate source resistance=%.2f ohm\" %R_s)\n",
-      "#Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Gate source resistance=111.87 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4, Page No 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "R_s=120     #slope of load line is -120V/A. This gives gate source resistance\n",
-      "print(\"gate source resistance=%.0f ohm\" %R_s)\n",
-      "\n",
-      "P=.4     #P=V_g*I_g\n",
-      "E_s=15\n",
-      "\n",
-      "#Calculations\n",
-      " #E_s=I_g*R_s+V_g %    after solving this\n",
-      " #120*I_g**2-15*I_g+0.4=0    so\n",
-      "a=120   \n",
-      "b=-15\n",
-      "c=0.4\n",
-      "D=math.sqrt((b**2)-4*a*c)\n",
-      "I_g=(-b+D)/(2*a)   \n",
-      "V_g=P/I_g\n",
-      "\n",
-      "#Results\n",
-      "print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3))    \n",
-      "print(\"\\nthen trigger voltage=%.3f V\" %V_g)\n",
-      "I_g=(-b-D)/(2*a)   \n",
-      "V_g=P/I_g\n",
-      "print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3))    \n",
-      "print(\"\\nthen trigger voltage=%.2f V\" %V_g)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "gate source resistance=120 ohm\n",
-        "\n",
-        "trigger current=86.44 mA\n",
-        "\n",
-        "then trigger voltage=4.628 V\n",
-        "\n",
-        "trigger current=38.56 mA\n",
-        "\n",
-        "then trigger voltage=10.37 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.5 Page No 150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "#V_g=1+10*I_g\n",
-      "P_gm=5     #P_gm=V_g*I_g\n",
-      "#after solving % eqn becomes 10*I_g**2+I_g-5=0\n",
-      "a=10.0    \n",
-      "b=1.0    \n",
-      "c=-5\n",
-      "\n",
-      "#Calculations\n",
-      "I_g=(-b+math.sqrt(b**2-4*a*c))/(2*a)\n",
-      "E_s=15\n",
-      "#using E_s=R_s*I_g+V_g\n",
-      "R_s=(E_s-1)/I_g-10   \n",
-      "P_gav=.3     #W\n",
-      "T=20*10**-6\n",
-      "f=P_gav/(P_gm*T)\n",
-      "dl=f*T\n",
-      "\n",
-      "#Results\n",
-      "print(\"Reistance=%.3f ohm\" %R_s)\n",
-      "print(\"Triggering freq=%.0f kHz\" %(f/1000))\n",
-      "print(\"Tduty cycle=%.2f\" %dl)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reistance=11.248 ohm\n",
-        "Triggering freq=3 kHz\n",
-        "Tduty cycle=0.06\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.6, Page No 151"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I=.1\n",
-      "E=200.0\n",
-      "L=0.2\n",
-      "\n",
-      "#Calculations\n",
-      "t=I*L/E   \n",
-      "R=20.0\n",
-      "t1=(-L/R)*math.log(1-(R*I/E))    \n",
-      "L=2.0\n",
-      "t2=(-L/R)*math.log(1-(R*I/E))   \n",
-      "\n",
-      "#Results\n",
-      "print(\"in case load consists of (a)L=.2H\")\n",
-      "print(\"min gate pulse width=%.0f us\" %(t*10**6))\n",
-      "print(\"(b)R=20ohm in series with L=.2H\")\n",
-      "print(\"min gate pulse width=%.3f us\" %(t1*10**6))\n",
-      "print(\"(c)R=20ohm in series with L=2H\")\n",
-      "print(\"min gate pulse width=%.2f us\" %(t2*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "in case load consists of (a)L=.2H\n",
-        "min gate pulse width=100 us\n",
-        "(b)R=20ohm in series with L=.2H\n",
-        "min gate pulse width=100.503 us\n",
-        "(c)R=20ohm in series with L=2H\n",
-        "min gate pulse width=1005.03 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.9 Page No 163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "def theta(th):\n",
-      "    I_m=1     #supposition\n",
-      "    I_av=(I_m/(2*math.pi))*(1+math.cos(math.radians(th)))\n",
-      "    I_rms=math.sqrt((I_m/(2*math.pi))*((180-th)*math.pi/360+.25*math.sin(math.radians(2*th))))\n",
-      "    FF=I_rms/I_av\n",
-      "    I_rms=35\n",
-      "    I_TAV=I_rms/FF\n",
-      "    return I_TAV\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"when conduction angle=180\")\n",
-      "th=0\n",
-      "I_TAV=theta(th)\n",
-      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
-      "print(\"when conduction angle=90\")\n",
-      "th=90\n",
-      "I_TAV=theta(th)\n",
-      "\n",
-      "#Results\n",
-      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
-      "print(\"when conduction angle=30\")\n",
-      "th=150\n",
-      "I_TAV=theta(th)\n",
-      "print(\"avg on current rating=%.3f A\" %I_TAV)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when conduction angle=180\n",
-        "avg on current rating=22.282 A\n",
-        "when conduction angle=90\n",
-        "avg on current rating=15.756 A\n",
-        "when conduction angle=30\n",
-        "avg on current rating=8.790 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.10, Page No 164"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "\n",
-      "def theta(th):\n",
-      "    n=360.0/th\n",
-      "    I=1.0     #supposition\n",
-      "    I_av=I/n\n",
-      "    I_rms=I/math.sqrt(n)\n",
-      "    FF=I_rms/I_av\n",
-      "    I_rms=35\n",
-      "    I_TAV=I_rms/FF\n",
-      "    return I_TAV\n",
-      "\n",
-      "#Calculations\n",
-      "th=180.0\n",
-      "I_TAV1=theta(th)\n",
-      "th=90.0\n",
-      "I_TAV2=theta(th)\n",
-      "th=30.0\n",
-      "I_TAV3=theta(th)\n",
-      "\n",
-      "#Results\n",
-      "print(\"when conduction angle=180\")\n",
-      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
-      "print(\"when conduction angle=90\")\n",
-      "print(\"avg on current rating=%.1f A\" %I_TAV2)\n",
-      "print(\"when conduction angle=30\")\n",
-      "print(\"avg on current rating=%.4f A\" %I_TAV3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when conduction angle=180\n",
-        "avg on current rating=8.790 A\n",
-        "when conduction angle=90\n",
-        "avg on current rating=17.5 A\n",
-        "when conduction angle=30\n",
-        "avg on current rating=10.1036 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.11 Page No 165"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "f=50.0     #Hz\n",
-      "\n",
-      "#Calculations\n",
-      "I_sb=3000.0\n",
-      "t=1/(4*f)\n",
-      "T=1/(2*f)\n",
-      "I=math.sqrt(I_sb**2*t/T)    \n",
-      "r=(I_sb/math.sqrt(2))**2*T   \n",
-      "\n",
-      "#Results\n",
-      "print(\"surge current rating=%.2f A\" %I)\n",
-      "print(\"\\nI**2*t rating=%.0f A^2.s\" %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "surge current rating=2121.32 A\n",
-        "\n",
-        "I**2*t rating=45000 A^2.s\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.12 Page No 165"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "\n",
-      "V_s=300.0 #V\n",
-      "R=60.0 #ohm\n",
-      "L=2.0 #H\n",
-      "\n",
-      "#Calculations\n",
-      "t=40*10**-6 #s\n",
-      "i_T=(V_s/R)*(1-math.exp(-R*t/L))\n",
-      "i=.036 #A\n",
-      "R1=V_s/(i-i_T)\n",
-      "\n",
-      "#Results\n",
-      "print(\"maximum value of remedial parameter=%.3f kilo-ohm\" %(R1/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum value of remedial parameter=9.999 kilo-ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.16 Page No 172"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_p=230.0*math.sqrt(2)\n",
-      "\n",
-      "#Calculations\n",
-      "R=1+((1)**-1+(10)**-1)**-1\n",
-      "A=V_p/R\n",
-      "s=1     #s\n",
-      "t_c=20*A**-2*s\n",
-      "\n",
-      "#Results\n",
-      "print(\"fault clearance time=%.4f ms\" %(t_c*10**3))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fault clearance time=0.6890 ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.17, Page No 176"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "V_s=math.sqrt(2)*230     #V\n",
-      "L=15*10**-6     #H\n",
-      "I=V_s/L     #I=(di/dt)_max\n",
-      "R_s=10   #ohm\n",
-      "v=I*R_s     #v=(dv/dt)_max\n",
-      "\n",
-      "#Calculations\n",
-      "f=50     #Hz\n",
-      "X_L=L*2*math.pi*f\n",
-      "R=2\n",
-      "I_max=V_s/(R+X_L)    \n",
-      "FF=math.pi/math.sqrt(2)\n",
-      "I_TAV1=I_max/FF    \n",
-      "FF=3.98184\n",
-      "I_TAV2=I_max/FF    \n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"(di/dt)_max=%.3f A/usec\" %(I/10**6))\n",
-      "print(\"\\n(dv/dt)_max=%.2f V/usec\" %(v/10**6))\n",
-      "print(\"\\nI_rms=%.3f A\" %I_max)\n",
-      "print(\"when conduction angle=90\")\n",
-      "print(\"I_TAV=%.3f A\" %I_TAV1)\n",
-      "print(\"when conduction angle=30\")\n",
-      "print(\"I_TAV=%.3f A\" %I_TAV2)\n",
-      "print(\"\\nvoltage rating=%.3f V\" %(2.75*V_s)) #rating is taken 2.75 times of peak working voltage unlike 2.5 to 3 times as mentioned int book."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(di/dt)_max=21.685 A/usec\n",
-        "\n",
-        "(dv/dt)_max=216.85 V/usec\n",
-        "\n",
-        "I_rms=162.252 A\n",
-        "when conduction angle=90\n",
-        "I_TAV=73.039 A\n",
-        "when conduction angle=30\n",
-        "I_TAV=40.748 A\n",
-        "\n",
-        "voltage rating=894.490 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.19, Page No 186"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "\n",
-      "T_jm=125\n",
-      "th_jc=.15     #degC/W\n",
-      "th_cs=0.075     #degC/W\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "dT=54     #dT=T_s-T_a\n",
-      "P_av=120\n",
-      "th_sa=dT/P_av\n",
-      "T_a=40     #ambient temp\n",
-      "P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
-      "if (P_av-120)<1 :\n",
-      "    print(\"selection of heat sink is satisfactory\")\n",
-      "\n",
-      "dT=58     #dT=T_s-T_a\n",
-      "P_av=120\n",
-      "th_sa=dT/P_av\n",
-      "T_a=40     #ambient temp\n",
-      "P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
-      "if (P_av-120)<1 :\n",
-      "    print(\"selection of heat sink is satisfactory\")\n",
-      "\n",
-      "V_m=math.sqrt(2)*230\n",
-      "R=2\n",
-      "I_TAV=V_m/(R*math.pi)\n",
-      "P_av=90\n",
-      "th_sa=(T_jm-T_a)/P_av-(th_jc+th_cs)\n",
-      "dT=P_av*th_sa\n",
-      "print(\"for heat sink\")    \n",
-      "print(\"T_s-T_a=%.2f degC\" %dT)   \n",
-      "print(\"\\nP_av=%.0f W\" %P_av)\n",
-      "P=(V_m/2)**2/R\n",
-      "eff=P/(P+P_av)   \n",
-      "print(\"\\nckt efficiency=%.3f pu\" %eff)\n",
-      "a=60     #delay angle\n",
-      "I_TAV=(V_m/(2*math.pi*R))*(1+math.cos(math.radians(a)))\n",
-      "print(\"\\nI_TAV=%.2f A\" %I_TAV)\n",
-      "dT=46\n",
-      "T_s=dT+T_a\n",
-      "T_c=T_s+P_av*th_cs    \n",
-      "T_j=T_c+P_av*th_jc    \n",
-      "\n",
-      "#Results\n",
-      "print(\"\\ncase temp=%.2f degC\" %T_c)\n",
-      "print(\"\\njunction temp=%.2f degC\" %T_j)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for heat sink\n",
-        "T_s-T_a=-20.25 degC\n",
-        "\n",
-        "P_av=90 W\n",
-        "\n",
-        "ckt efficiency=0.993 pu\n",
-        "\n",
-        "I_TAV=38.83 A\n",
-        "\n",
-        "case temp=92.75 degC\n",
-        "\n",
-        "junction temp=106.25 degC\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.20, Page No 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "T_j=125.0     #degC\n",
-      "T_s=70.0     #degC\n",
-      "th_jc=.16     #degC/W\n",
-      "th_cs=.08     #degC/W\n",
-      "\n",
-      "#Calculations\n",
-      "P_av1=(T_j-T_s)/(th_jc+th_cs)    \n",
-      "\n",
-      "T_s=60     #degC\n",
-      "P_av2=(T_j-T_s)/(th_jc+th_cs)    \n",
-      "inc=(math.sqrt(P_av2)-math.sqrt(P_av1))*100/math.sqrt(P_av1)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"Total avg power loss in thristor sink combination=%.2f W\" %P_av1)\n",
-      "print(\"Percentage inc in rating=%.2f\" %inc)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total avg power loss in thristor sink combination=229.17 W\n",
-        "Percentage inc in rating=8.71\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.21, Page No 197"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "\n",
-      "R=25000.0\n",
-      "I_l1=.021     #I_l=leakage current\n",
-      "I_l2=.025\n",
-      "I_l3=.018\n",
-      "I_l4=.016\n",
-      " #V1=(I-I_l1)*R\n",
-      " #V2=(I-I_l2)*R\n",
-      " #V3=(I-I_l3)*R\n",
-      " #V4=(I-I_l4)*R\n",
-      " #V=V1+V2+V3+V4\n",
-      "    \n",
-      "#Calculations\n",
-      "V=10000.0\n",
-      "I_l=I_l1+I_l2+I_l3+I_l4\n",
-      " #after solving\n",
-      "I=((V/R)+I_l)/4\n",
-      "R_c=40.0\n",
-      "V1=(I-I_l1)*R    \n",
-      "\n",
-      "#Resluts\n",
-      "print(\"voltage across SCR1=%.0f V\" %V1)\n",
-      "V2=(I-I_l2)*R    \n",
-      "print(\"\\nvoltage across SCR2=%.0f V\" %V2)\n",
-      "V3=(I-I_l3)*R    \n",
-      "print(\"\\nvoltage across SCR3=%.0f V\" %V3)\n",
-      "V4=(I-I_l4)*R    \n",
-      "print(\"\\nvoltage across SCR4=%.0f V\" %V4)\n",
-      "\n",
-      "I1=V1/R_c    \n",
-      "print(\"\\ndischarge current through SCR1=%.3f A\" %I1)\n",
-      "I2=V2/R_c   \n",
-      "print(\"\\ndischarge current through SCR2=%.3f A\" %I2)\n",
-      "I3=V3/R_c    \n",
-      "print(\"\\ndischarge current through SCR3=%.3f A\" %I3)\n",
-      "I4=V4/R_c    \n",
-      "print(\"\\ndischarge current through SCR4=%.3f A\" %I4)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "voltage across SCR1=2475 V\n",
-        "\n",
-        "voltage across SCR2=2375 V\n",
-        "\n",
-        "voltage across SCR3=2550 V\n",
-        "\n",
-        "voltage across SCR4=2600 V\n",
-        "\n",
-        "discharge current through SCR1=61.875 A\n",
-        "\n",
-        "discharge current through SCR2=59.375 A\n",
-        "\n",
-        "discharge current through SCR3=63.750 A\n",
-        "\n",
-        "discharge current through SCR4=65.000 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.22, Page No 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_r=1000     #rating of SCR\n",
-      "I_r=200     #rating of SCR\n",
-      "V_s=6000     #rating of String\n",
-      "I_s=1000     #rating of String\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"when DRF=.1\")\n",
-      "DRF=.1\n",
-      "n_s=V_s/(V_r*(1-DRF))    \n",
-      "print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
-      "n_p=I_s/(I_r*(1-DRF))    \n",
-      "print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))\n",
-      "print(\"when DRF=.2\")\n",
-      "DRF=.2\n",
-      "\n",
-      "#Results\n",
-      "n_s=V_s/(V_r*(1-DRF))   \n",
-      "print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
-      "n_p=I_s/(I_r*(1-DRF))    \n",
-      "print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when DRF=.1\n",
-        "number of series units=7\n",
-        "\n",
-        "number of parrallel units=6\n",
-        "when DRF=.2\n",
-        "number of series units=8\n",
-        "\n",
-        "number of parrallel units=7\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.23, Page No 198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V1=1.6     #on state voltage drop of SCR1\n",
-      "V2=1.2     #on state voltage drop of SCR2\n",
-      "I1=250.0     #current rating of SCR1\n",
-      "I2=350.0     #current rating of SCR2\n",
-      "\n",
-      "#Calculations\n",
-      "R1=V1/I1\n",
-      "R2=V2/I2\n",
-      "I=600.0     #current to be shared\n",
-      " #for SCR1 %     I*(R1+R)/(total resistance)=k*I1                    (1)\n",
-      " #for SCR2 %     I*(R2+R)/(total resistance)=k*I2                    (2)\n",
-      " #(1)/(2)\n",
-      "R=(R2*I2-R1*I1)/(I1-I2)\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"RSequired value of resistance=%.3f ohm\" %R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RSequired value of resistance=0.004 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.25, Page No 223"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=2000.0     #Hz\n",
-      "C=0.04*10**-6\n",
-      "n=.72\n",
-      "\n",
-      "#Calculations\n",
-      "R=1/(f*C*math.log(1/(1-n)))    \n",
-      "V_p=18\n",
-      "V_BB=V_p/n\n",
-      "R2=10**4/(n*V_BB)    \n",
-      "I=4.2*10**-3     #leakage current\n",
-      "R_BB=5000\n",
-      "R1=(V_BB/I)-R2-R_BB\n",
-      "\n",
-      "#Results\n",
-      "print(\"R=%.2f kilo-ohm\" %(R/1000))\n",
-      "print(\"\\nR2=%.2f ohm\" %R2)\n",
-      "print(\"\\nR1=%.0f ohm\" %R1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R=9.82 kilo-ohm\n",
-        "\n",
-        "R2=555.56 ohm\n",
-        "\n",
-        "R1=397 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.26, Page No 223"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "\n",
-      "V_p=18.0\n",
-      "n=.72\n",
-      "V_BB=V_p/n\n",
-      "I_p=.6*10**-3\n",
-      "I_v=2.5*10**-3\n",
-      "V_v=1\n",
-      "\n",
-      "#Calculations\n",
-      "R_max=V_BB*(1-n)/I_p    \n",
-      "print(\"R_max=%.2f kilo-ohm\" %(R_max/1000))\n",
-      "R_min=(V_BB-V_v)/I_v    \n",
-      "print(\"\\nR_min=%.2f kilo-ohm\" %(R_min/1000))\n",
-      "\n",
-      "C=.04*10**-6\n",
-      "f_min=1/(R_max*C*math.log(1/(1-n)))    \n",
-      "print(\"\\nf_min=%.3f kHz\" %(f_min/1000))\n",
-      "f_max=1/(R_min*C*math.log(1/(1-n)))    \n",
-      "print(\"\\nf_max=%.2f kHz\" %(f_max/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R_max=11.67 kilo-ohm\n",
-        "\n",
-        "R_min=9.60 kilo-ohm\n",
-        "\n",
-        "f_min=1.683 kHz\n",
-        "\n",
-        "f_max=2.05 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter5.ipynb b/Power_Electronics/chapter5.ipynb
deleted file mode 100755
index 1d261f20..00000000
--- a/Power_Electronics/chapter5.ipynb
+++ /dev/null
@@ -1,511 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 05 : Thyristor Commutation Techniques"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1, Page No 252"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "L=5.0*10**-3            #mH\n",
-      "C=20.0*10**-6           #\u00b5F\n",
-      "V_s=200                 #V\n",
-      "\n",
-      "#Calculations\n",
-      "w_o=math.sqrt(1/(L*C))    #rad/s\n",
-      "t_o=math.pi/w_o           #ms\n",
-      "\n",
-      "#Results\n",
-      "print('conduction time of thyristor = %.2f ms' %(t_o*1000))\n",
-      "print('voltage across thyristor=%.0f V' %V_s)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "conduction time of thyristor = 0.99 ms\n",
-        "voltage across thyristor=200 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2, Page No 255"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "C=20.0*10**-6         #\u00b5F\n",
-      "L=5.0*10**-6          #\u00b5H\n",
-      "V_s=230.0             #V\n",
-      "\n",
-      "#Calculations\n",
-      "I_p=V_s*math.sqrt(C/L)      #A\n",
-      "w_o=math.sqrt(1/(L*C))      #rad/sec\n",
-      "t_o=math.pi/w_o             #\u00b5S\n",
-      "I_o=300               \n",
-      "a = math.degrees(math.asin(I_o/(2*V_s)))   \n",
-      "V_ab = V_s*math.cos(math.radians(a))      #V \n",
-      "t_c=C*V_ab/I_o     #\u00b5s\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"conduction time of auxillery thyristor=%.2f us\" %(t_o*10**6))\n",
-      "print(\"voltage across main thyristor=%.2f V\" %V_ab)\n",
-      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "conduction time of auxillery thyristor=31.42 us\n",
-        "voltage across main thyristor=174.36 V\n",
-        "ckt turn off time=11.62 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3 Page No 258"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_s=200.0           #V\n",
-      "R1=10.0             #\u2126\n",
-      "R2=100.0            #\u2126\n",
-      "C=0                 #  value of capacitor\n",
-      "\n",
-      "#Calculations\n",
-      "I1=V_s*(1/R1+2/R2)    #A\n",
-      "I2=V_s*(2/R1+1/R2)    #A\n",
-      "t_c1=40*10**-6\n",
-      "fos=2    #factor of safety\n",
-      "C1=t_c1*fos/(R1*math.log(2))\n",
-      "C2=t_c1*fos/(R2*math.log(2))\n",
-      "if C1 > C2 :\n",
-      "    C = C1*10**6\n",
-      "else :\n",
-      "    C = C2*10**6\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"peak value of current through SCR1=%.2f A\" %I1); \n",
-      "print(\"Peak value of current through SCR2=%.2f A\" %I2);\n",
-      "print(\"Value of capacitor=%.2f uF\" %C);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "peak value of current through SCR1=24.00 A\n",
-        "Peak value of current through SCR2=42.00 A\n",
-        "Value of capacitor=11.54 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4, Page No 260"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0             #V\n",
-      "L=20*10**-6           #\u00b5H\n",
-      "C=40*10**-6           #\u00b5F\n",
-      "I_o=120.0             #A\n",
-      "\n",
-      "#Calculations\n",
-      "I_p=V_s*math.sqrt(C/L)   #A\n",
-      "t_c=C*V_s/I_o            #\u00b5s\n",
-      "w_o=math.sqrt(1/(L*C))  \n",
-      "t_c1=math.pi/(2*w_o)     #\u00b5s\n",
-      "\n",
-      "#Results\n",
-      "print(\"current through main thyristor=%.2f A\"  %(I_o+I_p))\n",
-      "print(\"Current through auxillery thyristor=%.2f A\" %I_o)\n",
-      "print(\"Circuit turn off time for main thyristor=%.2f us\" %(t_c*10**6))\n",
-      "print(\"Circuit turn off time for auxillery thyristor=%.2f us\" %(t_c1*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "current through main thyristor=445.27 A\n",
-        "Current through auxillery thyristor=120.00 A\n",
-        "Circuit turn off time for main thyristor=76.67 us\n",
-        "Circuit turn off time for auxillery thyristor=44.43 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5 Page No 263"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "C_j=25*10**-12      #pF\n",
-      "I_c=5*10**-3        #charging current\n",
-      "V_s=200.0           #V\n",
-      "R=50.0              #\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "C=(C_j*V_s)/(I_c*R)\n",
-      "\n",
-      "\n",
-      "#RESULTS\n",
-      "print(\"Value of C=%.2f  \u00b5F\" %(C*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of C=0.02  \u00b5F\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6 Page No 263"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "V_s=200.0         #V\n",
-      "R=5.0             #\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "C=10.0*10**-6\n",
-      "#for turn off V_s*(1-2*exp(-t/(R*C)))=0,    so after solving\n",
-      "t_c=R*C*math.log(2.0)\n",
-      "\n",
-      "#Results\n",
-      "print(\"circuit turn off time=%.2f us\" %(t_c*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "circuit turn off time=34.66 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.7, Page No 264 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=1.0                   #\u2126\n",
-      "L=20*10**-6             #\u00b5H\n",
-      "C=40*10**-6             #\u00b5F\n",
-      "\n",
-      "#Calculations\n",
-      "w_r=math.sqrt((1/(L*C))-(R/(2*L))**2)\n",
-      "t_1=math.pi/w_r\n",
-      "\n",
-      "#Results\n",
-      "print(\"conduction time of thyristor=%.3f us\" %(t_1*10**6))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "conduction time of thyristor=125.664 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8 Page No 265"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math \n",
-      "\n",
-      "#initialisation of variables\n",
-      "dv=400*10.0**-6       #dv=dv_T/dt(V/s)\n",
-      "V_s=200.0             #v\n",
-      "R=20.0                #\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "C=V_s/(R*dv)         \n",
-      "C_j=.025*10**-12\n",
-      "C_s=C-C_j\n",
-      "I_T=40;\n",
-      "R_s=1/((I_T/V_s)-(1/R)) \n",
-      "#value of R_s in book is wrongly calculated\n",
-      "\n",
-      "#Results\n",
-      "print(\"R_s=%.2f ohm\" %R_s)\n",
-      "print(\"C_s=%.3f uF\" %(C_s/10**6))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R_s=6.67 ohm\n",
-        "C_s=0.025 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9 Page No 265"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=200.0          #V\n",
-      "C=20.0*10**-6      #\u00b5H   \n",
-      "L=0.2*10**-3       #\u00b5F\n",
-      "i_c=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "i=V_s*math.sqrt(C/L)\n",
-      "w_o=1.0/math.sqrt(L*C)\n",
-      "t_1 = (1/w_o)*math.degrees(math.asin(i_c/i))\n",
-      "t_o=math.pi/w_o\n",
-      "t_c=t_o-2*t_1    \n",
-      "\n",
-      "#Results\n",
-      "print(\"reqd time=%.2f us\" %(t_1*10**6))\n",
-      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n",
-      "print(\"ckt turn off time=%.5f us\" %t_1)\n",
-      "#solution in book wrong, as wrong values are selected while filling the formuleas"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "reqd time=575.37 us\n",
-        "ckt turn off time=-952.05 us\n",
-        "ckt turn off time=0.00058 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.11 Page No 268 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "L=1.0              #\u00b5H\n",
-      "R=50.0             #\u2126\n",
-      "V_s=200.0          #V\n",
-      "t=0.01             #sec\n",
-      "Vd=0.7\n",
-      "\n",
-      "#Calculations\n",
-      "tau=L/R\n",
-      "i=(V_s/R)*(1-math.exp(-t/tau))\n",
-      "t=8*10**-3\n",
-      "i1=i-t*Vd  \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"current through L = %.2f A\" %i1)\n",
-      "i_R=0                  #current in R at t=.008s\n",
-      "print(\"Current through R = %.2f A\" %i_R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "current through L = 1.57 A\n",
-        "Current through R = 0.00 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.12, Page No 269"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#initialisation of variables\n",
-      "\n",
-      "#initialisation of variables\n",
-      "L=1.0                      #H\n",
-      "R=50.0                   #ohm\n",
-      "V_s=200.0                #V\n",
-      "\n",
-      "#Calculations\n",
-      "tau=L/R\n",
-      "t=0.01                 #s\n",
-      "i=(V_s/R)*(1-math.exp(-t/tau))\n",
-      "C=1*10**-6            #F\n",
-      "V_c=math.sqrt(L/C)*i\n",
-      "\n",
-      "#Results\n",
-      "print(\"current in R,L=%.2f A\" %i)\n",
-      "print(\"voltage across C=%.2f kV\" %(V_c/1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "current in R,L=1.57 A\n",
-        "voltage across C=1.57 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter6.ipynb b/Power_Electronics/chapter6.ipynb
deleted file mode 100755
index dff6564b..00000000
--- a/Power_Electronics/chapter6.ipynb
+++ /dev/null
@@ -1,1761 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 06 : Phase Controlled Rectifiers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1, Page No 283"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "P=1000.0\n",
-      "R=V**2/P\n",
-      "\n",
-      "#Calculations\n",
-      "a=math.pi/4\n",
-      "V_or1=(math.sqrt(2)*V/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a)+.5*math.sin(2*a))\n",
-      "P1=V_or1**2/R    \n",
-      "a=math.pi/2\n",
-      "V_or2=(math.sqrt(2)*V/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a)+.5*math.sin(2*a))\n",
-      "P2=V_or2**2/R   \n",
-      "\n",
-      "#Results\n",
-      "print(\"when firing angle delay is of 45deg\")\n",
-      "print(\"power absorbed=%.2f W\" %P1)\n",
-      "print(\"when firing angle delay is of 90deg\")\n",
-      "print(\"power absorbed=%.2f W\" %P2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when firing angle delay is of 45deg\n",
-        "power absorbed=454.58 W\n",
-        "when firing angle delay is of 90deg\n",
-        "power absorbed=250.00 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page No 283"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "E=150.0\n",
-      "R=8.0\n",
-      "\n",
-      "#Calculations\n",
-      "th1=math.sin(math.radians(E/(math.sqrt(2)*V)))\n",
-      "I_o=(1/(2*math.pi*R))*(2*math.sqrt(2)*230*math.cos(math.radians(th1))-E*(math.pi-2*th1*math.pi/180))    \n",
-      "P=E*I_o    \n",
-      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*(math.pi-2*th1*math.pi/180)+V**2*math.sin(math.radians(2*th1))-4*math.sqrt(2)*V*E*math.cos(math.radians(th1))))\n",
-      "P_r=I_or**2*R    \n",
-      "pf=(P+P_r)/(V*I_or)\n",
-      "\n",
-      "#Results\n",
-      "print(\"avg charging curent=%.4f A\" %I_o)\n",
-      "print(\"power supplied to the battery=%.2f W\" %P)\n",
-      "print(\"power dissipated by the resistor=%.3f W\" %P_r)    \n",
-      "print(\"supply pf=%.3f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg charging curent=3.5679 A\n",
-        "power supplied to the battery=535.18 W\n",
-        "power dissipated by the resistor=829.760 W\n",
-        "supply pf=0.583\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3 Page No 284"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "E=150.0\n",
-      "R=8.0\n",
-      "a=35.0\n",
-      "\n",
-      "#Calculations\n",
-      "th1=math.degrees(math.asin(E/(math.sqrt(2)*V)))\n",
-      "th2=180-th1\n",
-      "I_o=(1/(2*math.pi*R))*(math.sqrt(2)*230*(math.cos(math.radians(a))-math.cos(math.radians(th2)))-E*((th2-a)*math.pi/180))    \n",
-      "P=E*I_o    \n",
-      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*((th2-a)*math.pi/180)-(V**2/2)*(math.sin(math.radians(2*th2))-math.sin(math.radians(2*a)))-2*math.sqrt(2)*V*E*(math.cos(math.radians(a))-math.cos(math.radians(th2)))))\n",
-      "P_r=I_or**2*R    \n",
-      "pf=(P+P_r)/(V*I_or)    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"avg charging curent=%.4f A\" %I_o)\n",
-      "print(\"power supplied to the battery=%.2f W\" %P)\n",
-      "print(\"power dissipated by the resistor=%.2f W\" %P_r)\n",
-      "print(\"supply pf=%.4f\" %pf)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg charging curent=4.9208 A\n",
-        "power supplied to the battery=738.12 W\n",
-        "power dissipated by the resistor=689.54 W\n",
-        "supply pf=0.6686\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page No 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "B=210\n",
-      "f=50.0     #Hz\n",
-      "w=2*math.pi*f\n",
-      "a=40.0     #firing angle\n",
-      "V=230.0\n",
-      "R=5.0\n",
-      "L=2*10**-3\n",
-      "\n",
-      "#Calculations\n",
-      "t_c1=(360-B)*math.pi/(180*w)   \n",
-      "V_o1=(math.sqrt(2)*230/(2*math.pi))*(math.cos(math.radians(a))-math.cos(math.radians(B)))    \n",
-      "I_o1=V_o1/R    \n",
-      "E=110\n",
-      "R=5\n",
-      "L=2*10**-3\n",
-      "th1=math.degrees(math.asin(E/(math.sqrt(2)*V)))\n",
-      "t_c2=(360-B+th1)*math.pi/(180*w)    \n",
-      "V_o2=(math.sqrt(2)*230/(2*math.pi))*(math.cos(math.radians(a))-math.cos(math.radians(B)))    \n",
-      "I_o2=(1/(2*math.pi*R))*(math.sqrt(2)*230*(math.cos(math.radians(a))-math.cos(math.radians(B)))-E*((B-a)*math.pi/180))   \n",
-      "V_o2=R*I_o2+E    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"for R=5ohm and L=2mH\")\n",
-      "print(\"ckt turn off time=%.3f msec\" %(t_c1*1000))\n",
-      "print(\"avg output voltage=%.3f V\" %V_o1)\n",
-      "print(\"avg output current=%.4f A\" %I_o1)\n",
-      "print(\"for R=5ohm % L=2mH and E=110V\")\n",
-      "print(\"ckt turn off time=%.3f msec\" %(t_c2*1000))\n",
-      "print(\"avg output current=%.4f A\" %I_o2)\n",
-      "print(\"avg output voltage=%.3f V\" %V_o2)   "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for R=5ohm and L=2mH\n",
-        "ckt turn off time=8.333 msec\n",
-        "avg output voltage=84.489 V\n",
-        "avg output current=16.8979 A\n",
-        "for R=5ohm % L=2mH and E=110V\n",
-        "ckt turn off time=9.431 msec\n",
-        "avg output current=6.5090 A\n",
-        "avg output voltage=142.545 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5 Page No 286"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "f=50.0\n",
-      "R=10.0\n",
-      "a=60.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_m=(math.sqrt(2)*V_s)\n",
-      "V_o=V_m/(2*math.pi)*(1+math.cos(math.radians(a)))\n",
-      "I_o=V_o/R\n",
-      "V_or=(V_m/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a*math.pi/180)+.5*math.sin(math.radians(2*a)))\n",
-      "I_or=V_or/R\n",
-      "P_dc=V_o*I_o\n",
-      "P_ac=V_or*I_or\n",
-      "RE=P_dc/P_ac    \n",
-      "FF=V_or/V_o    \n",
-      "VRF=math.sqrt(FF**2-1)    \n",
-      "TUF=P_dc/(V_s*I_or)    \n",
-      "PIV=V_m    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"rectification efficiency=%.4f\" %RE)\n",
-      "print(\"form factor=%.3f\" %FF)\n",
-      "print(\"voltage ripple factor=%.4f\" %VRF)\n",
-      "print(\"t/f utilisation factor=%.4f\" %TUF)\n",
-      "print(\"PIV of thyristor=%.2f V\" %PIV)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rectification efficiency=0.2834\n",
-        "form factor=1.879\n",
-        "voltage ripple factor=1.5903\n",
-        "t/f utilisation factor=0.1797\n",
-        "PIV of thyristor=325.27 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6 Page No 294"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=1000.0\n",
-      "fos=2.5     #factor of safety\n",
-      "I_TAV=40.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_m1=V/(2*fos)\n",
-      "P1=(2*V_m1/math.pi)*I_TAV    \n",
-      "V_m2=V/(fos)\n",
-      "P2=(2*V_m2/math.pi)*I_TAV    \n",
-      "\n",
-      "#Results\n",
-      "print(\"for mid pt convertor\")\n",
-      "print(\"power handled=%.3f kW\" %(P1/1000))\n",
-      "print(\"for bridge convertor\")\n",
-      "print(\"power handled=%.3f kW\" %(P2/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for mid pt convertor\n",
-        "power handled=5.093 kW\n",
-        "for bridge convertor\n",
-        "power handled=10.186 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page No 297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "R=.4\n",
-      "I_o=10\n",
-      "I_or=I_o\n",
-      "E=120.0\n",
-      "\n",
-      "#Calculations\n",
-      "a1=math.degrees(math.acos((E+I_o*R)*math.pi/(2*V_m)))\n",
-      "pf1=(E*I_o+I_or**2*R)/(V_s*I_or)    \n",
-      "E=-120.0\n",
-      "a2=math.degrees(math.acos((E+I_o*R)*math.pi/(2*V_m))) \n",
-      "pf2=(-E*I_o-I_or**2*R)/(V_s*I_or)   \n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle delay=%.2f deg\" %a1)\n",
-      "print(\"pf=%.4f\" %pf1)\n",
-      "print(\"firing angle delay=%.2f deg\" %a2)\n",
-      "print(\"pf=%.4f\" %pf2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle delay=53.21 deg\n",
-        "pf=0.5391\n",
-        "firing angle delay=124.07 deg\n",
-        "pf=0.5043\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9 Page No 299"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "f=50.0\n",
-      "a=45.0\n",
-      "R=5.0\n",
-      "E=100.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=((math.sqrt(2)*V_s)/(2*math.pi))*(3+math.cos(math.radians(a)))\n",
-      "I_o=(V_o-E)/R    \n",
-      "P=E*I_o    \n",
-      "\n",
-      "#Results\n",
-      "print(\"avg o/p current=%.3f A\" %I_o)\n",
-      "print(\"power delivered to battery=%.4f kW\" %(P/1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg o/p current=18.382 A\n",
-        "power delivered to battery=1.8382 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10 Page No 300"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variablesV_s=230\n",
-      "f=50.0\n",
-      "a=50.0\n",
-      "R=6.0\n",
-      "E=60.0\n",
-      "V_o1=((math.sqrt(2)*2*V_s)/(math.pi))*math.cos(math.radians(a))\n",
-      "I_o1=(V_o1-E)/R    \n",
-      "\n",
-      "#ATQ after applying the conditions\n",
-      "V_o2=((math.sqrt(2)*V_s)/(math.pi))*math.cos(math.radians(a))\n",
-      "I_o2=(V_o2-E)/R    \n",
-      "\n",
-      "print(\"avg o/p current=%.3f A\" %I_o1)\n",
-      "print(\"avg o/p current after change=%.2f A\" %I_o2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg o/p current=12.184 A\n",
-        "avg o/p current after change=1.09 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11 Page No 309"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=45.0\n",
-      "R=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=(2*V_m/math.pi)*math.cos(math.radians(a))\n",
-      "I_o=V_o/R\n",
-      "V_or=V_m/math.sqrt(2)\n",
-      "I_or=I_o\n",
-      "P_dc=V_o*I_o\n",
-      "P_ac=V_or*I_or\n",
-      "RE=P_dc/P_ac    \n",
-      "FF=V_or/V_o    \n",
-      "VRF=math.sqrt(FF**2-1)    \n",
-      "I_s1=2*math.sqrt(2)*I_o/math.pi\n",
-      "DF=math.cos(math.radians(a))\n",
-      "CDF=.90032\n",
-      "pf=CDF*DF    \n",
-      "HF=math.sqrt((1/CDF**2)-1)    \n",
-      "Q=2*V_m*I_o*math.sin(math.radians(a))/math.pi \n",
-      "\n",
-      "#Results\n",
-      "print(\"rectification efficiency=%.4f\" %RE)\n",
-      "print(\"form factor=%.4f\" %FF)\n",
-      "print(\"voltage ripple factor=%.4f\" %VRF)\n",
-      "print(\"pf=%.5f\" %pf)\n",
-      "print(\"HF=%.5f\" %HF)\n",
-      "print(\"active power=%.2f W\" %P_dc)   \n",
-      "print(\"reactive power=%.3f Var\" %Q)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rectification efficiency=0.6366\n",
-        "form factor=1.5708\n",
-        "voltage ripple factor=1.2114\n",
-        "pf=0.63662\n",
-        "HF=0.48342\n",
-        "active power=2143.96 W\n",
-        "reactive power=2143.956 Var\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page No 310"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=45.0\n",
-      "R=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=(V_m/math.pi)*(1+math.cos(math.radians(a)))\n",
-      "I_o=V_o/R\n",
-      "V_or=V_s*math.sqrt((1/math.pi)*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))\n",
-      "I_or=I_o\n",
-      "P_dc=V_o*I_o\n",
-      "P_ac=V_or*I_or\n",
-      "RE=P_dc/P_ac    \n",
-      "FF=V_or/V_o    \n",
-      "VRF=math.sqrt(FF**2-1)    \n",
-      "I_s1=2*math.sqrt(2)*I_o*math.cos(math.radians(a/2))/math.pi\n",
-      "DF=math.cos(math.radians(a/2))   \n",
-      "CDF=2*math.sqrt(2)*math.cos(math.radians(a/2))/math.sqrt(math.pi*(math.pi-a*math.pi/180))    \n",
-      "pf=CDF*DF    \n",
-      "HF=math.sqrt((1/CDF**2)-1)    \n",
-      "Q=V_m*I_o*math.sin(math.radians(a))/math.pi\n",
-      "\n",
-      "#Results\n",
-      "print(\"form factor=%.3f\" %FF)\n",
-      "print(\"rectification efficiency=%.4f\" %RE)\n",
-      "print(\"voltage ripple factor=%.3f\" %VRF)    \n",
-      "print(\"DF=%.4f\" %DF)\n",
-      "print(\"CDF=%.4f\" %CDF)\n",
-      "print(\"pf=%.4f\" %pf)\n",
-      "print(\"HF=%.4f\" %HF)\n",
-      "print(\"active power=%.3f W\" %P_dc)\n",
-      "print(\"reactive power=%.2f Var\" %Q)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "form factor=1.241\n",
-        "rectification efficiency=0.8059\n",
-        "voltage ripple factor=0.735\n",
-        "DF=0.9239\n",
-        "CDF=0.9605\n",
-        "pf=0.8874\n",
-        "HF=0.2899\n",
-        "active power=3123.973 W\n",
-        "reactive power=1293.99 Var\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page No 319"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "R=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_ml=math.sqrt(2)*V_s\n",
-      "V_om=3*V_ml/(2*math.pi)\n",
-      "V_o=V_om/2\n",
-      "th=30\n",
-      "a=math.degrees(math.acos((2*math.pi*math.sqrt(3)*V_o/(3*V_ml)-1)))-th    \n",
-      "I_o=V_o/R    \n",
-      "V_or=V_ml/(2*math.sqrt(math.pi))*math.sqrt((5*math.pi/6-a*math.pi/180)+.5*math.sin(math.radians(2*a+2*th)))\n",
-      "I_or=V_or/R    \n",
-      "RE=V_o*I_o/(V_or*I_or)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"delay angle=%.1f deg\" %a)\n",
-      "print(\"avg load current=%.3f A\" %I_o)\n",
-      "print(\"rms load current=%.3f A\" %I_or)\n",
-      "print(\"rectification efficiency=%.4f\" %RE)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "delay angle=67.7 deg\n",
-        "avg load current=7.765 A\n",
-        "rms load current=10.477 A\n",
-        "rectification efficiency=0.5494\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page No 321"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=400.0\n",
-      "V_ml=math.sqrt(2)*V\n",
-      "v_T=1.4\n",
-      "a1=30.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_o1=3*V_ml/(2*math.pi)*math.cos(math.radians(a1))-v_T   \n",
-      "a2=60.0\n",
-      "V_o2=3*V_ml/(2*math.pi)*math.cos(math.radians(a2))-v_T  \n",
-      "I_o=36\n",
-      "I_TA=I_o/3    \n",
-      "I_Tr=I_o/math.sqrt(3)    \n",
-      "P=I_TA*v_T       \n",
-      "\n",
-      "#Results\n",
-      "print(\"for firing angle = 30deg\")\n",
-      "print(\"avg output voltage=%.3f V\" %V_o1)\n",
-      "print(\"for firing angle = 60deg\")\n",
-      "print(\"avg output voltage=%.2f V\" %V_o2)\n",
-      "print(\"avg current rating=%.0f A\" %I_TA)\n",
-      "print(\"rms current rating=%.3f A\" %I_Tr)\n",
-      "print(\"PIV of SCR=%.1f V\" %V_ml)\n",
-      "print(\"power dissipated=%.1f W\" %P)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle = 30deg\n",
-        "avg output voltage=232.509 V\n",
-        "for firing angle = 60deg\n",
-        "avg output voltage=133.65 V\n",
-        "avg current rating=12 A\n",
-        "rms current rating=20.785 A\n",
-        "PIV of SCR=565.7 V\n",
-        "power dissipated=16.8 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page No 331"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "E=200\n",
-      "I_o=20\n",
-      "R=.5\n",
-      "\n",
-      "#Calculations\n",
-      "V_o1=E+I_o*R\n",
-      "V_s=230\n",
-      "V_ml=math.sqrt(2)*V_s\n",
-      "a1=math.degrees(math.acos(V_o1*math.pi/(3*V_ml)))\n",
-      "th=120\n",
-      "I_s=math.sqrt((1/math.pi)*I_o**2*th*math.pi/180)\n",
-      "P=E*I_o+I_o**2*R\n",
-      "pf=P/(math.sqrt(3)*V_s*I_s)    \n",
-      "V_o2=E-I_o*R\n",
-      "a2=math.degrees(math.acos(-V_o2*math.pi/(3*V_ml))) \n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle delay=%.3f deg\" %a1)\n",
-      "print(\"pf=%.3f\" %pf)\n",
-      "print(\"firing angle delay=%.2f deg\" %a2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle delay=47.461 deg\n",
-        "pf=0.646\n",
-        "firing angle delay=127.71 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page No 332"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "f=50.0\n",
-      "\n",
-      "#Calculations\n",
-      "w=2*math.pi*f\n",
-      "a1=0\n",
-      "t_c1=(4*math.pi/3-a1*math.pi/180)/w    \n",
-      "a2=30\n",
-      "t_c2=(4*math.pi/3-a2*math.pi/180)/w    \n",
-      "\n",
-      "#Results\n",
-      "print(\"for firing angle delay=0deg\")\n",
-      "print(\"commutation time=%.2f ms\" %(t_c1*1000))\n",
-      "print(\"peak reverse voltage=%.2f V\" %(math.sqrt(2)*V))\n",
-      "print(\"for firing angle delay=30deg\")\n",
-      "print(\"commutation time=%.2f ms\" %(t_c2*1000))\n",
-      "print(\"peak reverse voltage=%.2f V\" %(math.sqrt(2)*V))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle delay=0deg\n",
-        "commutation time=13.33 ms\n",
-        "peak reverse voltage=325.27 V\n",
-        "for firing angle delay=30deg\n",
-        "commutation time=11.67 ms\n",
-        "peak reverse voltage=325.27 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page No 333"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=30.0\n",
-      "R=10.0\n",
-      "P=5000.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_s=math.sqrt(P*R*2*math.pi/(2*3)/(math.pi/3+math.sqrt(3)*math.cos(math.radians(2*a))/2))\n",
-      "V_ph=V_s/math.sqrt(3)    \n",
-      "I_or=math.sqrt(P*R)\n",
-      "V_s=I_or*math.pi/(math.sqrt(2)*3*math.cos(math.radians(a)))\n",
-      "V_ph=V_s/math.sqrt(3) \n",
-      "\n",
-      "#Results\n",
-      "print(\"per phase voltage percent V_ph=%.3f V\" %V_ph)   \n",
-      "print(\"for constant load current\")\n",
-      "print(\"V_ph=%.2f V\" %V_ph)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "per phase voltage percent V_ph=110.384 V\n",
-        "for constant load current\n",
-        "V_ph=110.38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20, Page No 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=30.0\n",
-      "R=10.0\n",
-      "P=5000.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_s=math.sqrt(P*R*4*math.pi/(2*3)/(2*math.pi/3+math.sqrt(3)*(1+math.cos(math.radians(2*a)))/2))\n",
-      "V_ph=V_s/math.sqrt(3)    \n",
-      "I_or=math.sqrt(P*R)\n",
-      "V_s=I_or*2*math.pi/(math.sqrt(2)*3*(1+math.cos(math.radians(a))))\n",
-      "V_ph=V_s/math.sqrt(3)   \n",
-      "\n",
-      "#Results\n",
-      "print(\"per phase voltage percent V_ph=%.3f V\" %V_ph) \n",
-      "print(\"for constant load current\")\n",
-      "print(\"V_ph=%.2f V\" %V_ph)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "per phase voltage percent V_ph=102.459 V\n",
-        "for constant load current\n",
-        "V_ph=102.46 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21, Page No 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=90.0\n",
-      "R=10.0\n",
-      "P=5000.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_s=math.sqrt(P*R*4*math.pi/(2*3)/((math.pi-math.pi/2)+(math.sin(math.radians(2*a)))/2))\n",
-      "V_ph=V_s/math.sqrt(3)    \n",
-      "I_or=math.sqrt(P*R)\n",
-      "V_s=I_or*2*math.pi/(math.sqrt(2)*3*(1+math.cos(math.radians(a))))\n",
-      "V_ph=V_s/math.sqrt(3)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"per phase voltage percent V_ph=%.2f V\" %V_ph)\n",
-      "print(\"for constant load current\")\n",
-      "print(\"V_ph=%.1f V\" %V_ph)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "per phase voltage percent V_ph=191.19 V\n",
-        "for constant load current\n",
-        "V_ph=191.2 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22 Page No 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "E=200.0\n",
-      "I_o=20.0\n",
-      "R=.5\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=E+I_o*R\n",
-      "V_s=230\n",
-      "V_ml=math.sqrt(2)*V_s\n",
-      "a=math.degrees(math.acos(V_o*2*math.pi/(3*V_ml)-1))  \n",
-      "a1=180-a\n",
-      "I_sr=math.sqrt((1/math.pi)*I_o**2*(a1*math.pi/180))\n",
-      "P=V_o*I_o\n",
-      "pf=P/(math.sqrt(3)*V_s*I_sr)   \n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle delay=%.2f deg\" %a)\n",
-      "print(\"pf=%.2f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle delay=69.38 deg\n",
-        "pf=0.67\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23, Page No 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=400.0\n",
-      "f=50.0\n",
-      "I_o=15.0\n",
-      "a=45.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_TA=I_o*120.0/360.0\n",
-      "I_Tr=math.sqrt(I_o**2*120/360)\n",
-      "I_sr=math.sqrt(I_o**2*120/180)\n",
-      "V_ml=math.sqrt(2)*V_s\n",
-      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
-      "V_or=V_ml*math.sqrt((3/(2*math.pi))*(math.pi/3+math.sqrt(3/2)*math.cos(math.radians(2*a))))\n",
-      "I_or=I_o\n",
-      "P_dc=V_o*I_o\n",
-      "P_ac=V_or*I_or\n",
-      "RE=P_dc/P_ac    \n",
-      "VA=3*V_s/math.sqrt(3)*I_sr\n",
-      "TUF=P_dc/VA    \n",
-      "pf=P_ac/VA    \n",
-      "\n",
-      "#Results\n",
-      "print(\"rectification efficiency=%.5f\" %RE)\n",
-      "print(\"TUF=%.4f\" %TUF)\n",
-      "print(\"Input pf=%.3f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rectification efficiency=0.95493\n",
-        "TUF=0.6752\n",
-        "Input pf=0.707\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24, Page No 341"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I=10.0\n",
-      "a=45.0\n",
-      "V=400.0\n",
-      "f=50.0\n",
-      "\n",
-      "#Calculations\n",
-      "DF=math.cos(math.radians(a))\n",
-      "I_o=10\n",
-      "I_s1=4*I_o/(math.sqrt(2)*math.pi)*math.sin(math.pi/3)\n",
-      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
-      "I_o=1     #suppose\n",
-      "CDF=I_s1/I_sr    \n",
-      "THD=math.sqrt(1/CDF**2-1)    \n",
-      "pf=CDF*DF    \n",
-      "P=(3*math.sqrt(2)*V*math.cos(math.radians(a))/math.pi)*I\n",
-      "Q=(3*math.sqrt(2)*V*math.sin(math.radians(a))/math.pi)*I  \n",
-      "   \n",
-      "#Results\n",
-      "print(\"DF=%.3f\" %DF)\n",
-      "print(\"CDF=%.3f\" %CDF)\n",
-      "print(\"THD=%.5f\" %THD)\n",
-      "print(\"PF=%.4f\" %pf)\n",
-      "print(\"active power=%.2f W\" %P)  \n",
-      "print(\"reactive power=%.2f Var\" %Q)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "DF=0.707\n",
-        "CDF=0.955\n",
-        "THD=0.31084\n",
-        "PF=0.6752\n",
-        "active power=3819.72 W\n",
-        "reactive power=3819.72 Var\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25, Page No 342"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "print(\"for firing angle=30deg\")\n",
-      "a=30.0\n",
-      "V=400.0\n",
-      "V_ml=math.sqrt(2)*V\n",
-      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
-      "E=350\n",
-      "R=10\n",
-      "\n",
-      "#Calculations\n",
-      "I_o=(V_o-E)/R\n",
-      "I_or=I_o\n",
-      "P1=V_o*I_o    \n",
-      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
-      "VA=3*V/math.sqrt(3)*I_sr\n",
-      "pf=P1/VA \n",
-      "a=180-60\n",
-      "V=400\n",
-      "V_ml=math.sqrt(2)*V\n",
-      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
-      "E=-350\n",
-      "R=10\n",
-      "I_o=(V_o-E)/R\n",
-      "I_or=I_o\n",
-      "P2=-V_o*I_o    \n",
-      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
-      "VA=3*V/math.sqrt(3)*I_sr\n",
-      "pf=P2/VA       \n",
-      "\n",
-      "print(\"power delivered to load=%.2f W\" %P1)\n",
-      "print(\"pf=%.4f\" %pf)\n",
-      "print(\"for firing advance angle=60deg\")\n",
-      "print(\"power delivered to load=%.2f W\" %P2)\n",
-      "print(\"pf=%.4f\" %pf)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle=30deg\n",
-        "power delivered to load=5511.74 W\n",
-        "pf=0.4775\n",
-        "for firing advance angle=60deg\n",
-        "power delivered to load=2158.20 W\n",
-        "pf=0.4775\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26, Page No 347"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=0\n",
-      "u=15.0\n",
-      "\n",
-      "#Calculations\n",
-      "i=math.cos(math.radians(a))-math.cos(math.radians(a+u))\n",
-      "a=30\n",
-      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a  \n",
-      "a=45\n",
-      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a    \n",
-      "a=60\n",
-      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a  \n",
-      "\n",
-      "#Results\n",
-      "print(\"for firing angle=30deg\") \n",
-      "print(\"overlap angle=%.1f deg\" %u)\n",
-      "print(\"for firing angle=45deg\")  \n",
-      "print(\"overlap angle=%.1f deg\" %u)\n",
-      "print(\"for firing angle=60deg\") \n",
-      "print(\"overlap angle=%.2f deg\" %u)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle=30deg\n",
-        "overlap angle=2.2 deg\n",
-        "for firing angle=45deg\n",
-        "overlap angle=2.2 deg\n",
-        "for firing angle=60deg\n",
-        "overlap angle=2.23 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28, Page No 352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "E=400.0\n",
-      "I_o=20.0\n",
-      "R=1\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=E+I_o*R\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "L=.004\n",
-      "V=230 #per phase voltage\n",
-      "V_ml=math.sqrt(6)*V\n",
-      "a=math.degrees(math.acos(math.pi/(3*V_ml)*(V_o+3*w*L*I_o/math.pi)))   \n",
-      "print(\"firing angle delay=%.3f deg\" %a)\n",
-      "u=math.degrees(math.acos(math.pi/(3*V_ml)*(V_o-3*w*L*I_o/math.pi)))-a    \n",
-      "\n",
-      "#Results\n",
-      "print(\"overlap angle=%.2f deg\" %u)\n",
-      "#Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle delay=34.382 deg\n",
-        "overlap angle=8.22 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29, Page No 352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=400.0\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "R=1\n",
-      "E=230\n",
-      "I=15.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=-E+I*R\n",
-      "V_ml=math.sqrt(2)*V\n",
-      "a=math.degrees(math.acos(V_o*2*math.pi/(3*V_ml)))  \n",
-      "L=0.004\n",
-      "a=math.degrees(math.acos((2*math.pi)/(3*V_ml)*(V_o+3*w*L*I/(2*math.pi))))  \n",
-      "u=math.degrees(math.acos(math.cos(math.radians(a))-3*f*L*I/V_ml))-a    \n",
-      "\n",
-      "#Results\n",
-      "print(\"firing angle=%.3f deg\" %a)\n",
-      "print(\"firing angle delay=%.3f deg\" %a)\n",
-      "print(\"overlap angle=%.3f deg\" %u)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle=139.702 deg\n",
-        "firing angle delay=139.702 deg\n",
-        "overlap angle=1.431 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31, Page No 361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0 #per phase\n",
-      "f=50.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_ml=math.sqrt(3.0)*math.sqrt(2)*V\n",
-      "w=2*math.pi*f\n",
-      "a1=60.0\n",
-      "L=0.015\n",
-      "i_cp=(math.sqrt(3)*V_ml/(w*L))*(1-math.sin(math.radians(a1)))    \n",
-      "\n",
-      "#Results\n",
-      "print(\"circulating current=%.4f A\" %i_cp)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "circulating current=27.7425 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32, Page No 362"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "V_m=math.sqrt(2)*V\n",
-      "a=30.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=2*V_m* math.cos(math.radians(a))/math.pi    \n",
-      "R=10\n",
-      "I_o=V_o/R    \n",
-      "I_TA=I_o*math.pi/(2*math.pi)    \n",
-      "I_Tr=math.sqrt(I_o**2*math.pi/(2*math.pi))    \n",
-      "I_s=math.sqrt(I_o**2*math.pi/(math.pi)) \n",
-      "I_o=I_s\n",
-      "pf=(V_o*I_o/(V*I_s))    \n",
-      "\n",
-      "#Results\n",
-      "print(\"avg o/p voltage=%.3f V\" %V_o)\n",
-      "print(\"avg o/p current=%.2f A\" %I_o)\n",
-      "print(\"avg value of thyristor current=%.3f A\" %I_TA)\n",
-      "print(\"rms value of thyristor current=%.2f A\" %I_Tr)\n",
-      "print(\"pf=%.4f\" %pf)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n",
-      " \n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg o/p voltage=179.330 V\n",
-        "avg o/p current=17.93 A\n",
-        "avg value of thyristor current=8.967 A\n",
-        "rms value of thyristor current=12.68 A\n",
-        "pf=0.7797\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33, Page No 363"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "V_m=math.sqrt(2)*V\n",
-      "a=30.0\n",
-      "L=.0015\n",
-      "\n",
-      "#Calculations\n",
-      "V_o=2*V_m* math.cos(math.radians(a))/math.pi  \n",
-      "R=10\n",
-      "I_o=V_o/R  \n",
-      "f=50\n",
-      "w=2*math.pi*f\n",
-      "V_ox=2*V_m*math.cos(math.radians(a))/math.pi-w*L*I_o/math.pi    \n",
-      "u=math.degrees(math.acos(math.cos(math.radians(a))-I_o*w*L/V_m))-a    \n",
-      "I=I_o\n",
-      "pf=V_o*I_o/(V*I)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"avg o/p voltage=%.3f V\" %V_ox)\n",
-      "print(\"angle of overlap=%.3f deg\" %u)\n",
-      "print(\"pf=%.4f\" %pf)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg o/p voltage=176.640 V\n",
-        "angle of overlap=2.855 deg\n",
-        "pf=0.7797\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34, Page No 364"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=415.0\n",
-      "V_ml=math.sqrt(2)*V\n",
-      "a1=35.0 #firing angle advance\n",
-      "\n",
-      "#Calculations\n",
-      "a=180-a1\n",
-      "I_o=80.0\n",
-      "r_s=0.04\n",
-      "v_T=1.5\n",
-      "X_l=.25 #reactance=w*L\n",
-      "E=-3*V_ml*math.cos(math.radians(a))/math.pi+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi    \n",
-      "\n",
-      "#Results\n",
-      "print(\"mean generator voltage=%.3f V\" %E)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mean generator voltage=487.590 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.35, Page No 364"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=415.0\n",
-      "V_ml=math.sqrt(2)*V\n",
-      "R=0.2\n",
-      "I_o=80\n",
-      "r_s=0.04\n",
-      "v_T=1.5\n",
-      "\n",
-      "#Calculations\n",
-      "X_l=.25 #reactance=w*L\n",
-      "a=35\n",
-      "E=-(-3*V_ml*math.cos(math.radians(a))/math.pi+I_o*R+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi)    \n",
-      "a1=35\n",
-      "a=180-a1\n",
-      "E=(-3*V_ml*math.cos(math.radians(a))/math.pi+I_o*R+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi) \n",
-      "\n",
-      "#Results\n",
-      "print(\"when firing angle=35deg\")   \n",
-      "print(\"mean generator voltage=%.3f V\" %E)\n",
-      "print(\"when firing angle advance=35deg\")\n",
-      "print(\"mean generator voltage=%.3f V\" %E)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when firing angle=35deg\n",
-        "mean generator voltage=503.590 V\n",
-        "when firing angle advance=35deg\n",
-        "mean generator voltage=503.590 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.36, Page No 365"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=5.0\n",
-      "V=230.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_mp=math.sqrt(2)*V\n",
-      "a=30.0\n",
-      "E=150.0\n",
-      "B=180-math.degrees(math.asin(E/V_mp))\n",
-      "I_o=(3/(2*math.pi*R))*(V_mp*(math.cos(math.radians(a+30))-math.cos(math.radians(B)))-E*((B-a-30)*math.pi/180))\n",
-      "\n",
-      "#Results\n",
-      "print(\"avg current flowing=%.2f A\" %I_o)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg current flowing=19.96 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.37, Page No 366"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=30.0\n",
-      "V=230.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_m=math.sqrt(2)*V\n",
-      "V_o=V_m*(1+math.cos(math.radians(a)))/math.pi    \n",
-      "E=100\n",
-      "R=10\n",
-      "I_o=(V_o-E)/R    \n",
-      "I_TA=I_o*math.pi/(2*math.pi)    \n",
-      "I_Tr=math.sqrt(I_o**2*math.pi/(2*math.pi))    \n",
-      "I_s=math.sqrt(I_o**2*(1-a/180)*math.pi/(math.pi))\n",
-      "I_or=I_o\n",
-      "P=E*I_o+I_or**2*R\n",
-      "pf=(P/(V*I_s))   \n",
-      "f=50\n",
-      "w=2*math.pi*f\n",
-      "t_c=(1-a/180)*math.pi/w    \n",
-      "\n",
-      "#Results\n",
-      "print(\"\\navg o/p current=%.2f A\" %I_o)\n",
-      "print(\"avg o/p voltage=%.3f V\" %V_o)\n",
-      "print(\"avg value of thyristor current=%.2f A\" %I_TA)\n",
-      "print(\"rms value of thyristor current=%.3f A\" %I_Tr)\n",
-      "print(\"avg value of diode current=%.2f A\" %I_TA)\n",
-      "print(\"rms value of diode current=%.3f A\" %I_Tr)\n",
-      "print(\"pf=%.4f\" %pf)\n",
-      "print(\"circuit turn off time=%.2f ms\" %(t_c*1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "avg o/p current=9.32 A\n",
-        "avg o/p voltage=193.202 V\n",
-        "avg value of thyristor current=4.66 A\n",
-        "rms value of thyristor current=6.590 A\n",
-        "avg value of diode current=4.66 A\n",
-        "rms value of diode current=6.590 A\n",
-        "pf=0.9202\n",
-        "circuit turn off time=8.33 ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.38, Page No 368"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "V_m=math.sqrt(2)*V\n",
-      "L=0.05\n",
-      "f=50.0\n",
-      "\n",
-      "#Calculations\n",
-      "w=2*math.pi*f\n",
-      "a=30\n",
-      "i_cp=2*V_m*(1-math.cos(math.radians(a)))/(w*L)    \n",
-      "R=30.0\n",
-      "i_l=V_m/R\n",
-      "i1=i_cp+i_l    \n",
-      "i2=i_cp    \n",
-      "\n",
-      "#Results\n",
-      "print(\"peak value of circulating current=%.3f A\" %i_cp)\n",
-      "print(\"peak value of current in convertor 1=%.3f A\" %i1)\n",
-      "print(\"peak value of current in convertor 2=%.3f A\" %i2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "peak value of circulating current=5.548 A\n",
-        "peak value of current in convertor 1=16.391 A\n",
-        "peak value of current in convertor 2=5.548 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.39, Page No 370"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "R=5.0\n",
-      "L=0.05\n",
-      "\n",
-      "#Calculations\n",
-      "phi=math.degrees(math.atan(w*L/R))  \n",
-      "phi=90+math.degrees(math.atan(w*L/R))   \n",
-      "\n",
-      "#Results\n",
-      "print(\"for no current transients\")\n",
-      "print(\"triggering angle=%.2f deg\" %phi)\n",
-      "print(\"for worst transients\")\n",
-      "print(\"triggering angle=%.2f deg\" %phi)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for no current transients\n",
-        "triggering angle=162.34 deg\n",
-        "for worst transients\n",
-        "triggering angle=162.34 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter7.ipynb b/Power_Electronics/chapter7.ipynb
deleted file mode 100755
index 726160c8..00000000
--- a/Power_Electronics/chapter7.ipynb
+++ /dev/null
@@ -1,1036 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 07 : Choppers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2, Page No 387"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=0.4     #duty cycle %a=T_on/T\n",
-      "V_s=230.0\n",
-      "R=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "V=a*(V_s-2)    \n",
-      "V_or=math.sqrt(a*(V_s-2)**2)    \n",
-      "P_o=V_or**2/R\n",
-      "P_i=V_s*V/R\n",
-      "n=P_o*100/P_i    \n",
-      "\n",
-      "#Results\n",
-      "print(\"avg o/p voltage=%.1f V\" %V)\n",
-      "print(\"rms value of o/p voltage=%.1f V\" %V_or)\n",
-      "print(\"chopper efficiency in percentage=%.2f\" %n)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "avg o/p voltage=91.2 V\n",
-        "rms value of o/p voltage=144.2 V\n",
-        "chopper efficiency in percentage=99.13\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3, Page No 388"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_i=220.0\n",
-      "V_o=660.0\n",
-      "\n",
-      "#Calculations\n",
-      "a=1-V_i/V_o\n",
-      "T_on=100.0     #microsecond\n",
-      "T=T_on/a\n",
-      "T_off=T-T_on    \n",
-      "T_off=T_off/2\n",
-      "T_on=T-T_off\n",
-      "a=T_on/T\n",
-      "V_o=V_i/(1-a)\n",
-      "\n",
-      "#Results    \n",
-      "print(\"pulse width of o/p voltage=%.0f us\" %T_off)\n",
-      "print(\"\\nnew o/p voltage=%.0f V\" %V_o)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "pulse width of o/p voltage=25 us\n",
-        "\n",
-        "new o/p voltage=1320 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4 Page No 288"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_1=12.0\n",
-      "I_2=16.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_0=(I_1+I_2)/2\n",
-      "R=10.0\n",
-      "V_0=I_0*R\n",
-      "V_s=200.0\n",
-      "a=V_0/V_s\n",
-      "r=a/(1-a)\n",
-      "\n",
-      "#Results\n",
-      "print(\"time ratio(T_on/T_off)=%.3f\" %r)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "time ratio(T_on/T_off)=2.333\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5, Page No 390"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_o=660.0\n",
-      "V_s=220.0\n",
-      "\n",
-      "#Calculations\n",
-      "a=(V_o/V_s)/(1+(V_o/V_s))\n",
-      "T_on=120\n",
-      "T=T_on/a\n",
-      "T_off=T-T_on \n",
-      "T_off=3*T_off\n",
-      "T_on=T-T_off\n",
-      "a=T_on/(T_on+T_off)\n",
-      "V_o=V_s*(a/(1-a))   \n",
-      "\n",
-      "#Results\n",
-      "print(\"pulse width o/p voltage=%.0f us\" %T_off)\n",
-      "print(\"\\nnew o/p voltage=%.2f V\" %V_o)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "pulse width o/p voltage=120 us\n",
-        "\n",
-        "new o/p voltage=73.33 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.11 Page No 408"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=1.0\n",
-      "L=.005\n",
-      "T_a=L/R\n",
-      "T=2000*10**-6\n",
-      "E=24.0\n",
-      "V_s=220\n",
-      "T_on=600*10**-6\n",
-      "a=T_on/T\n",
-      "\n",
-      "#Calculations\n",
-      "a1=(T_a/T)*math.log(1+(E/V_s)*((math.exp(T/T_a))-1))\n",
-      "if a1<a :\n",
-      "    print(\"load current in continuous\")\n",
-      "else:\n",
-      "    print(\"load current in discont.\")\n",
-      "\n",
-      "I_o=(a*V_s-E)/R    \n",
-      "I_mx=(V_s/R)*((1-math.exp(-T_on/T_a))/(1-math.exp(-T/T_a)))-E/R    \n",
-      "I_mn=(V_s/R)*((math.exp(T_on/T_a)-1)/(math.exp(T/T_a)-1))-E/R    \n",
-      "f=1/T\n",
-      "w=2*math.pi*f\n",
-      "I1=(2*V_s/(math.sqrt(2)*math.pi)*math.sin(math.radians(180*a)))/(math.sqrt(R**2+(w*L)**2))    \n",
-      "I2=(2*V_s/(2*math.sqrt(2)*math.pi)*math.sin(math.radians(2*180*a)))/(math.sqrt(R**2+(w*L*2)**2))    \n",
-      "I3=(2*V_s/(3*math.sqrt(2)*math.pi)*math.sin(math.radians(3*180*a)))/(math.sqrt(R**2+(w*L*3)**2))    \n",
-      "I_TAV=a*(V_s-E)/R-L*(I_mx-I_mn)/(R*T)   \n",
-      "P1=I_TAV*V_s\n",
-      "P2=E*I_o\n",
-      "I_or=math.sqrt(I_o**2+I1**2+I2**2+I3**2)\n",
-      "\n",
-      "#Results\n",
-      "print(\"avg o/p current=%.2f A\" %I_o)\n",
-      "print(\"max value of steady current=%.2f A\" %I_mx)\n",
-      "print(\"min value of steady current=%.2f A\" %I_mn)\n",
-      "print(\"first harmonic current=%.4f A\" %I1)\n",
-      "print(\"second harmonic current=%.4f A\" %I2)\n",
-      "print(\"third harmonic current=%.5f A\" %I3)\n",
-      "print(\"avg supply current=%.4f A\" %I_TAV)\n",
-      "print(\"i/p power=%.2f W\" %P1)\n",
-      "print(\"power absorbed by load emf=%.0f W\" %P2)\n",
-      "print(\"power loss in resistor=%.2f W\" %(P1-P2))\n",
-      "print(\"rms value of load current=%.3f A\" %I_or)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "load current in continuous\n",
-        "avg o/p current=42.00 A\n",
-        "max value of steady current=51.46 A\n",
-        "min value of steady current=33.03 A\n",
-        "first harmonic current=5.0903 A\n",
-        "second harmonic current=1.4983 A\n",
-        "third harmonic current=0.21643 A\n",
-        "avg supply current=12.7289 A\n",
-        "i/p power=2800.35 W\n",
-        "power absorbed by load emf=1008 W\n",
-        "power loss in resistor=1792.35 W\n",
-        "rms value of load current=42.334 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.12 Page No 411"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=1\n",
-      "L=.001\n",
-      "V_s=220\n",
-      "E=72.0\n",
-      "f=500.0\n",
-      "T_on=800*10**-6\n",
-      "T_a=L/R\n",
-      "T=1.0/f\n",
-      "m=E/V_s\n",
-      "a=T_on/T\n",
-      "\n",
-      "#Calculations\n",
-      "a1=(T_a/T)*math.log(1+m*(math.exp(-T/T_a)-1))\n",
-      "if a1>a :\n",
-      "    print(\"load current is continuous\")\n",
-      "else:\n",
-      "    print(\"load current is discontinuous\")\n",
-      "\n",
-      "t_x=T_on+L*math.log(1+((V_s-E)/272)*(1-math.exp(-T_on/T_a)))\n",
-      " #Value of t_x wrongly calculated in the book so ans of V_o and I_o varies\n",
-      "V_o=a*V_s+(1-t_x/T)*E    \n",
-      "I_o=(V_o-E)/R    \n",
-      "I_mx=(V_s-E)/R*(1-math.exp(-T_on/T_a)) \n",
-      "\n",
-      "#Results  \n",
-      "print(\"avg o/p voltage=%.2f V\" %V_o)\n",
-      "print(\"avg o/p current=%.2f A\" %I_o) \n",
-      "print(\"max value of load current=%.1f A\" %I_mx)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "load current is discontinuous\n",
-        "avg o/p voltage=121.77 V\n",
-        "avg o/p current=49.77 A\n",
-        "max value of load current=81.5 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.13, Page No 412"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=0.2\n",
-      "V_s=500\n",
-      "E=a*V_s\n",
-      "L=0.06\n",
-      "I=10\n",
-      "\n",
-      "#Calculations\n",
-      "T_on=(L*I)/(V_s-E)\n",
-      "f=a/T_on    \n",
-      "\n",
-      "#Results\n",
-      "print(\"chopping freq=%.2f Hz\" %f)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "chopping freq=133.33 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.14 Page No 412"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=0.5\n",
-      "pu=0.1 #pu ripple\n",
-      "\n",
-      "#Calculations\n",
-      " #x=T/T_a\n",
-      " #y=exp(-a*x)\n",
-      "y=(1-pu)/(1+pu)\n",
-      " #after solving\n",
-      "x=math.log(1/y)/a\n",
-      "f=1000\n",
-      "T=1/f\n",
-      "T_a=T/x\n",
-      "R=2\n",
-      "L=R*T_a\n",
-      "Li=0.002\n",
-      "Le=L-Li    \n",
-      "\n",
-      "#Results\n",
-      "print(\"external inductance=%.3f mH\" %(Le*1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "external inductance=-2.000 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.15 Page No 414"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=10.0\n",
-      "L=0.015\n",
-      "T_a=L/R\n",
-      "f=1250.0\n",
-      "T=1.0/f\n",
-      "a=0.5\n",
-      "T_on=a*T\n",
-      "V_s=220.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_mx=(V_s/R)*((1-math.exp(-T_on/T_a))/(1-math.exp(-T/T_a)))   \n",
-      "I_mn=(V_s/R)*((math.exp(T_on/T_a)-1)/(math.exp(T/T_a)-1))    \n",
-      "dI=I_mx-I_mn    \n",
-      "V_o=a*V_s\n",
-      "I_o=V_o/R    \n",
-      "I_or=math.sqrt(I_mx**2+dI**2/3+I_mx*dI)    \n",
-      "I_chr=math.sqrt(a)*I_or \n",
-      "\n",
-      "#Results\n",
-      "print(\"Max value of ripple current=%.2f A\" %dI)\n",
-      "print(\"Max value of load current=%.3f A\" %I_mx)\n",
-      "print(\"Min value of load current=%.2f A\" %I_mn)\n",
-      "print(\"Avg value of load current=%.2f A\" %I_o)   \n",
-      "print(\"rms value of load current=%.2f A\" %I_or)\n",
-      "print(\"rms value of chopper current=%.2f A\" %I_chr)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Max value of ripple current=2.92 A\n",
-        "Max value of load current=12.458 A\n",
-        "Min value of load current=9.54 A\n",
-        "Avg value of load current=11.00 A\n",
-        "rms value of load current=13.94 A\n",
-        "rms value of chopper current=9.86 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.17 Page No 417"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "L=0.0016\n",
-      "C=4*10**-6\n",
-      "\n",
-      "#Calculations\n",
-      "w=1/math.sqrt(L*C)\n",
-      "t=math.pi/w    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"time for which current flows=%.2f us\" %(t*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "time for which current flows=251.33 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.18, Page No 424"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "t_q=20.0*10**-6\n",
-      "dt=20.0*10**-6\n",
-      "\n",
-      "#Calculations\n",
-      "t_c=t_q+dt\n",
-      "I_0=60.0\n",
-      "V_s=60.0\n",
-      "C=t_c*I_0/V_s \n",
-      "\n",
-      "#Results   \n",
-      "print(\"value of commutating capacitor=%.0f uF\" %(C*10**6))\n",
-      "\n",
-      "L1=(V_s/I_0)**2*C\n",
-      "L2=(2*t_c/math.pi)**2/C\n",
-      "if L1>L2 :\n",
-      "    print(\"value of commutating inductor=%.0f uH\" %(L1*10**6))\n",
-      "else:\n",
-      "    print(\"value of commutating inductor=%.0f uH\" %(L2*10**6))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of commutating capacitor=40 uF\n",
-        "value of commutating inductor=40 uH\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.19, Page No 424"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "t=100.0*10**-6\n",
-      "R=10.0\n",
-      "\n",
-      "#Calculations\n",
-      " #V_s*(1-2*math.exp(-t/(R*C)))=0\n",
-      "C=-t/(R*math.log(1.0/2))    \n",
-      "L=(4/9.0)*C*R**2    \n",
-      "L=(1.0/4)*C*R**2    \n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of comutating component C=%.3f uF\" %(C*10**6))\n",
-      "print(\"max permissible current through SCR is 2.5 times load current\")\n",
-      "print(\"value of comutating component L=%.1f uH\" %(L*10**6))\n",
-      "print(\"max permissible current through SCR is 1.5 times peak diode current\")\n",
-      "print(\"value of comutating component L=%.2f uH\" %(L*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of comutating component C=14.427 uF\n",
-        "max permissible current through SCR is 2.5 times load current\n",
-        "value of comutating component L=360.7 uH\n",
-        "max permissible current through SCR is 1.5 times peak diode current\n",
-        "value of comutating component L=360.67 uH\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.20, Page No 426"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "T_on=800.0*10**-6\n",
-      "V_s=220.0\n",
-      "I_o=80.0\n",
-      "C=50*10**-6\n",
-      "\n",
-      "#Calculations\n",
-      "T=T_on+2*V_s*C/I_o    \n",
-      "L=20*10**-6\n",
-      "C=50*10**-6\n",
-      "i_T1=I_o+V_s*math.sqrt(C/L)    \n",
-      "i_TA=I_o    \n",
-      "t_c=C*V_s/I_o    \n",
-      "t_c1=(math.pi/2)*math.sqrt(L*C)    \n",
-      "t=150*10**-6\n",
-      "v_c=I_o*t/C-V_s  \n",
-      "\n",
-      "#Results  \n",
-      "print(\"effective on period=%.0f us\" %(T*10**6))\n",
-      "print(\"peak current through main thyristor=%.2f A\" %i_T1)\n",
-      "print(\"peak current through auxillery thyristor=%.0f A\" %i_TA)\n",
-      "print(\"turn off time for main thyristor=%.1f us\" %(t_c*10**6))\n",
-      "print(\"turn off time for auxillery thyristor=%.3f us\" %(t_c1*10**6))\n",
-      "print(\"total commutation interval=%.0f us\" %(2*t_c*10**6))\n",
-      "print(\"capacitor voltage=%.0f V\" %v_c)\n",
-      "print(\"time nedded to recharge the capacitor=%.0f us\" %(2*V_s*C/I_o*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for firing angle = 30deg\n",
-        "avg output voltage=232.509 V\n",
-        "for firing angle = 60deg\n",
-        "avg output voltage=133.65 V\n",
-        "avg current rating=12 A\n",
-        "rms current rating=20.785 A\n",
-        "PIV of SCR=565.7 V\n",
-        "power dissipated=16.8 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 122
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.21, Page No 427"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_o=260.0\n",
-      "V_s=220.0\n",
-      "fos=2 #factor of safety\n",
-      "\n",
-      "#Calculations\n",
-      "t_off=18*10**-6\n",
-      "t_c=2*t_off\n",
-      "C=t_c*I_o/V_s    \n",
-      "L=(V_s/(0.8*I_o))**2*C    \n",
-      "f=400\n",
-      "a_mn=math.pi*f*math.sqrt(L*C)\n",
-      "V_omn=V_s*(a_mn+2*f*t_c)    \n",
-      "V_omx=V_s    \n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of C=%.3f uF\" %(C*10**6))\n",
-      "print(\"value of L=%.3f uH\" %(L*10**6))\n",
-      "print(\"min value of o/p voltage=%.3f V\" %V_omn)\n",
-      "print(\"max value of o/p voltage=%.0f V\" %V_omx)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "firing angle delay=47.461 deg\n",
-        "pf=0.646\n",
-        "firing angle delay=127.71 deg\n"
-       ]
-      }
-     ],
-     "prompt_number": "*"
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.22, Page No 434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "x=2.0\n",
-      "t_q=30*10**-6\n",
-      "dt=30*10**-6\n",
-      "t_c=t_q+dt\n",
-      "V_s=230.0\n",
-      "I_o=200.0\n",
-      "\n",
-      "#Calculations\n",
-      "L=V_s*t_c/(x*I_o*(math.pi-2*math.asin(1/x)))    \n",
-      "C=x*I_o*t_c/(V_s*(math.pi-2*math.asin(1/x)))    \n",
-      "V_cp=V_s+I_o*math.sqrt(L/C)    \n",
-      "I_cp=x*I_o    \n",
-      "x=3\n",
-      "L=V_s*t_c/(x*I_o*(math.pi-2*math.asin(1/x)))   \n",
-      "C=x*I_o*t_c/(V_s*(math.pi-2*math.asin(1/x)))    \n",
-      "V_cp=V_s+I_o*math.sqrt(L/C)    \n",
-      "I_cp=x*I_o    \n",
-      "\n",
-      "#Results\n",
-      "print(\"value of commutating inductor=%.3f uH\" %(L*10**6))\n",
-      "print(\"value of commutating capacitor=%.3f uF\" %(C*10**6))\n",
-      "print(\"peak capacitor voltage=%.0f V\" %V_cp)\n",
-      "print(\"peak commutataing current=%.0f A\" %I_cp)\n",
-      "print(\"value of commutating inductor=%.3f uH\" %(L*10**6))\n",
-      "print(\"value of commutating capacitor=%.3f uF\" %(C*10**6))\n",
-      "print(\"peak capacitor voltage=%.2f V\" %V_cp)\n",
-      "print(\"peak commutataing current=%.0f A\" %I_cp)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of commutating inductor=7.321 uH\n",
-        "value of commutating capacitor=49.822 uF\n",
-        "peak capacitor voltage=307 V\n",
-        "peak commutataing current=600 A\n",
-        "value of commutating inductor=7.321 uH\n",
-        "value of commutating capacitor=49.822 uF\n",
-        "peak capacitor voltage=306.67 V\n",
-        "peak commutataing current=600 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.23, Page No 434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variablesV_s=230\n",
-      "C=50*10**-6\n",
-      "L=20*10**-6\n",
-      "I_cp=V_s*math.sqrt(C/L)\n",
-      "I_o=200\n",
-      "x=I_cp/I_o\n",
-      "\n",
-      "#Calculations\n",
-      "t_c=(math.pi-2*math.asin(1/x))*math.sqrt(C*L)   \n",
-      "th1=math.degrees(math.asin(1.0/x))\n",
-      "t=(5*math.pi/2-th1*math.pi/180)*math.sqrt(L*C)+C*V_s*(1-math.cos(math.radians(th1)))/I_o    \n",
-      "t=(math.pi-th1*math.pi/180)*math.sqrt(L*C)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"turn off time of main thyristor=%.2f us\" %(t_c*10**6))\n",
-      "print(\"total commutation interval=%.3f us\" %(t*10**6))\n",
-      "print(\"turn off time of auxillery thyristor=%.3f us\" %(t*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "turn off time of main thyristor=62.52 us\n",
-        "total commutation interval=80.931 us\n",
-        "turn off time of auxillery thyristor=80.931 us\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.24, Page No 440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "tc=0.006\n",
-      "R=10.0\n",
-      "L=R*tc\n",
-      "f=2000.0\n",
-      "\n",
-      "#Calculations\n",
-      "T=1/f\n",
-      "V_o=50.0\n",
-      "V_s=100.0\n",
-      "a=V_o/V_s\n",
-      "T_on=a*T\n",
-      "T_off=T-T_on\n",
-      "dI=V_o*T_off/L\n",
-      "I_o=V_o/R\n",
-      "I2=I_o+dI/2    \n",
-      "I1=I_o-dI/2    \n",
-      "\n",
-      "#Results\n",
-      "print(\"max value of load current=%.3f A\" %I2)\n",
-      "print(\"min value of load current=%.3f A\" %I1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "max value of load current=5.104 A\n",
-        "min value of load current=4.896 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.27, Page No 443"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "I_a=30.0\n",
-      "r_a=.5\n",
-      "V_s=220.0\n",
-      "\n",
-      "#Calculations\n",
-      "a=I_a*r_a/V_s    \n",
-      "a=1\n",
-      "k=.1 #V/rpm\n",
-      "N=(a*V_s-I_a*r_a)/k    \n",
-      "\n",
-      "#Results\n",
-      "print(\"min value of duty cycle=%.3f\" %a)\n",
-      "print(\"min Value of speed control=%.0f rpm\" %0)\n",
-      "print(\"max value of duty cycle=%.0f\" %a)\n",
-      "print(\"max value of speed control=%.0f rpm\" %N)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "min value of duty cycle=1.000\n",
-        "min Value of speed control=0 rpm\n",
-        "max value of duty cycle=1\n",
-        "max value of speed control=2050 rpm\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.28, Page No 444"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_t=72.0\n",
-      "I_a=200.0\n",
-      "r_a=0.045\n",
-      "N=2500.0\n",
-      "\n",
-      "#Calculations\n",
-      "k=(V_t-I_a*r_a)/N\n",
-      "E_a=k*1000\n",
-      "L=.007\n",
-      "Rm=.045\n",
-      "Rb=0.065\n",
-      "R=Rm+Rb\n",
-      "T_a=L/R\n",
-      "I_mx=230\n",
-      "I_mn=180\n",
-      "T_on=-T_a*math.log(-((V_t-E_a)/R-I_mx)/((I_mn)-(V_t-E_a)/R))\n",
-      "R=Rm\n",
-      "T_a=L/R\n",
-      "T_off=-T_a*math.log(-((-E_a)/R-I_mn)/((I_mx)-(-E_a)/R))\n",
-      "T=T_on+T_off\n",
-      "f=1/T    \n",
-      "a=T_on/T \n",
-      "\n",
-      "#Results\n",
-      "print(\"chopping freq=%.1f Hz\" %f)  \n",
-      "print(\"duty cycle ratio=%.3f\" %a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "chopping freq=40.5 Hz\n",
-        "\n",
-        "duty cycle ratio=0.588\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.29, Page No 445"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variablesI_mx=425\n",
-      "I_lt=180.0     #lower limit of current pulsation\n",
-      "I_mn=I_mx-I_lt\n",
-      "T_on=0.014\n",
-      "T_off=0.011\n",
-      "\n",
-      "#Calculations\n",
-      "T=T_on+T_off\n",
-      "T_a=.0635\n",
-      "a=T_on/T\n",
-      "V=(I_mx-I_mn*math.exp(-T_on/T_a))/(1-math.exp(-T_on/T_a))\n",
-      "a=.5\n",
-      "I_mn=(I_mx-V*(1-math.exp(-T_on/T_a)))/(math.exp(-T_on/T_a))\n",
-      "T=I_mx-I_mn    \n",
-      "T=T_on/a\n",
-      "f=1/T    \n",
-      "\n",
-      "#Results\n",
-      "print(\"higher limit of current pulsation=%.0f A\" %T)\n",
-      "print(\"chopping freq=%.3f Hz\" %f)\n",
-      "print(\"duty cycle ratio=%.2f\" %a)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "higher limit of current pulsation=0 A\n",
-        "chopping freq=35.714 Hz\n",
-        "duty cycle ratio=0.50\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter8.ipynb b/Power_Electronics/chapter8.ipynb
deleted file mode 100755
index 721a9faf..00000000
--- a/Power_Electronics/chapter8.ipynb
+++ /dev/null
@@ -1,984 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 08 : Inverters"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3, Page No 465"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "T=0.1*10**-3\n",
-      "f=1.0/T\n",
-      "k=15*10**-6 #k=th/w\n",
-      "\n",
-      "#Calculations\n",
-      "th=2*math.pi*f*k\n",
-      "X_l=10.0\n",
-      "R=2.0\n",
-      "X_c=R*math.tan(th)+X_l\n",
-      "C=1/(2*math.pi*f*X_c)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"value of C=%.3f uF\" %(C*10**6))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of C=1.248 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4 Page No 466"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_01=2*V_s/(math.sqrt(2)*math.pi)\n",
-      "R=2.0\n",
-      "I_01=V_01/R\n",
-      "P_d=I_01**2*R    \n",
-      "V=V_s/2\n",
-      "I_s=math.sqrt(2)*I_01/math.pi\n",
-      "P_s=V*I_s\n",
-      "\n",
-      "#Results\n",
-      "print(\"power delivered to load=%.1f W\" %P_d)\n",
-      "print(\"power delivered by both sources=%.1f W\" %(2*P_s))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "power delivered to load=5359.9 W\n",
-        "power delivered by both sources=5359.9 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5, Page No 468"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_01=4*V_s/(math.pi*math.sqrt(2))\n",
-      "R=1.0\n",
-      "X_L=6.0\n",
-      "X_c=7.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_01=V_01/math.sqrt(R**2+(X_L-X_c)**2)\n",
-      "P=I_01**2*R    \n",
-      "I_s=math.sqrt(2)*I_01*(2*math.cos(math.radians(45)))/math.pi\n",
-      "P_s=V_s*I_s    \n",
-      "\n",
-      "#Results\n",
-      "print(\"power delivered to the source=%.3f kW\" %(P/1000))\n",
-      "print(\"\\npower from the source=%.3f kW\" %(P_s/1000))\n",
-      "   "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "power delivered to the source=21.440 kW\n",
-        "\n",
-        "power from the source=21.440 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6 Page No 469"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_01=230.0\n",
-      "R=2.0\n",
-      "I_01=V_01/R\n",
-      "I_m=I_01*math.sqrt(2)\n",
-      "I_T1=I_m/2    \n",
-      "I_D1=0    \n",
-      "X_L=8.0\n",
-      "X_C=6.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_01=V_01/math.sqrt(R**2+(X_L-X_C)**2)\n",
-      "phi1=math.degrees(math.atan((X_L-X_C)/R))\n",
-      "I_T1=I_T1*math.sqrt(2)*0.47675    \n",
-      "I_D1=.1507025*I_m/math.sqrt(2)    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"when load R=2 ohm\")\n",
-      "print(\"rms value of thyristor current=%.2f A\" %I_T1)\n",
-      "print(\"rms value of diode current=%.0f A\" %I_D1)\n",
-      "print(\"when load R=2ohm % X_L=8ohm and X_C=6ohm\")\n",
-      "print(\"rms value of thyristor current=%.3f A\" %I_T1)\n",
-      "print(\"rms value of diode current=%.3f A\" %I_D1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "when load R=2 ohm\n",
-        "rms value of thyristor current=54.83 A\n",
-        "rms value of diode current=17 A\n",
-        "when load R=2ohm % X_L=8ohm and X_C=6ohm\n",
-        "rms value of thyristor current=54.826 A\n",
-        "rms value of diode current=17.331 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7 Page No 470"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "R=4.0\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "L=0.035\n",
-      "\n",
-      "#Calculations\n",
-      "C=155*10**-6\n",
-      "X_L=w*L\n",
-      "X_C=1/(w*C)\n",
-      "Z1=math.sqrt(R**2+(X_L-X_C)**2)\n",
-      "phi1=-math.degrees(math.atan((X_L-X_C)/R))\n",
-      "Z3=math.sqrt(R**2+(X_L*3-X_C/3)**2)\n",
-      "phi3=math.degrees(math.atan((X_L*3-X_C/3)/R))\n",
-      "Z5=math.sqrt(R**2+(X_L*5-X_C/5)**2)\n",
-      "phi5=math.degrees(math.atan((X_L*5-X_C/5)/R))\n",
-      "I_m1=4*V_s/(Z1*math.pi)\n",
-      "I_01=I_m1/math.sqrt(2)    \n",
-      "I_m3=4*V_s/(3*Z3*math.pi)\n",
-      "I_m5=4*V_s/(5*Z5*math.pi)\n",
-      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2)\n",
-      "I_0=I_m/math.sqrt(2)\n",
-      "P_0=(I_0)**2*R    \n",
-      "P_01=(I_01)**2*R    \n",
-      "t1=(180-phi1)*math.pi/(180*w)    \n",
-      "t1=(phi1)*math.pi/(180*w)    \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of fundamental load current=%.2f A\" %I_01)\n",
-      "print(\"load power=%.1f W\" %P_0)\n",
-      "print(\"fundamental load power=%.1f W\" %P_01)\n",
-      "print(\"rms value of thyristor current=%.3f A\" %(I_m/2))\n",
-      "print(\"conduction time for thyristor=%.3f ms\" %(t1*1000))\n",
-      "print(\"Conduction time for diodes=%.3f ms\" %(t1*1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of fundamental load current=20.02 A\n",
-        "load power=1632.5 W\n",
-        "fundamental load power=1602.6 W\n",
-        "rms value of thyristor current=14.285 A\n",
-        "conduction time for thyristor=3.736 ms\n",
-        "Conduction time for diodes=3.736 ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8, Page No 473"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_01=2*V_s/(math.sqrt(2)*math.pi)    \n",
-      "R=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_01=V_01/R\n",
-      "P=I_01**2*R    \n",
-      "V_or=math.sqrt((V_s/2)**2)\n",
-      "P=V_or**2/R    \n",
-      "I_TP=V_s/(2*R)\n",
-      "I_or=I_TP\n",
-      "pf=I_01**2*R/(V_or*I_or)  \n",
-      "DF=V_01/V_or  \n",
-      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
-      "THD=V_oh/V_01 \n",
-      "V_03=V_01/3\n",
-      "HF=V_03/V_01\n",
-      "\n",
-      "#Results\n",
-      "print(\"fundamental rms o/p voltage=%.3f V\" %V_01)\n",
-      "print(\"fundamental power to load=%.1f W\" %P)\n",
-      "print(\"total o/p power to load=%.1f W\" %P)\n",
-      "print(\"avg SCR current=%.2f A\" %(I_TP*180/360))\n",
-      "print(\"i/p pf=%.3f\" %pf)  \n",
-      "print(\"distortion factor=%.1f\" %DF)\n",
-      "print(\"THD=%.3f\" %THD)    \n",
-      "print(\"harmonic factor=%.4f\" %HF)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fundamental rms o/p voltage=103.536 V\n",
-        "fundamental power to load=1322.5 W\n",
-        "total o/p power to load=1322.5 W\n",
-        "avg SCR current=5.75 A\n",
-        "i/p pf=0.811\n",
-        "distortion factor=0.9\n",
-        "THD=0.483\n",
-        "harmonic factor=0.3333\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9 Page No 474"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=60\n",
-      "R=3.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_or=math.sqrt(V_s**2*math.pi/math.pi)    \n",
-      "V_01=4*V_s/(math.sqrt(2)*math.pi)    \n",
-      "P_o=V_or**2/R    \n",
-      "P_01=V_01**2/R    \n",
-      "I_s=V_s/R    \n",
-      "I_avg=I_s*math.pi/(2*math.pi)    \n",
-      "V_03=V_01/3\n",
-      "HF=V_03/V_01    \n",
-      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
-      "THD=V_oh/V_01   \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of o/p voltage=%.0f V\" %V_or)\n",
-      "print(\"o/p power=%.0f W\" %P_o)\n",
-      "print(\"fundamental component of rms voltage=%.2f V\" %V_01)\n",
-      "print(\"fundamental freq o/p power=%.2f W\" %P_01) \n",
-      "print(\"peak current=%.0f A\" %I_s)\n",
-      "print(\"avg current of each transistor=%.0f A\" %I_avg)\n",
-      "print(\"peak reverse blocking voltage=%.0f V\" %V_s)\n",
-      "print(\"harmonic factor=%.4f\" %HF)\n",
-      "print(\"THD=%.4f\" %THD)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of o/p voltage=60 V\n",
-        "o/p power=1200 W\n",
-        "fundamental component of rms voltage=54.02 V\n",
-        "fundamental freq o/p power=972.68 W\n",
-        "peak current=20 A\n",
-        "avg current of each transistor=10 A\n",
-        "peak reverse blocking voltage=60 V\n",
-        "harmonic factor=0.3333\n",
-        "THD=0.4834\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10 Page No 475"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=220.0\n",
-      "R=6.0\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "L=0.03\n",
-      "C=180*10**-6\n",
-      "X_L=w*L\n",
-      "X_C=1/(w*C)\n",
-      "\n",
-      "#Calculations\n",
-      "V_or=math.sqrt(V_s**2*math.pi/math.pi)\n",
-      "V_01=4*V_s/(math.sqrt(2)*math.pi)\n",
-      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
-      "THD=V_oh/V_01    \n",
-      "print(\"THD of voltage=%.4f\" %THD)\n",
-      "DF=V_01/V_or    \n",
-      "Z1=math.sqrt(R**2+(X_L-X_C)**2)\n",
-      "phi1=-math.degrees(math.atan((X_L-X_C)/R))\n",
-      "Z3=math.sqrt(R**2+(X_L*3-X_C/3)**2)\n",
-      "phi3=math.degrees(math.atan((X_L*3-X_C/3)/R))\n",
-      "Z5=math.sqrt(R**2+(X_L*5-X_C/5)**2)\n",
-      "phi5=math.degrees(math.atan((X_L*5-X_C/5)/R))\n",
-      "Z7=math.sqrt(R**2+(X_L*7-X_C/7)**2)\n",
-      "phi7=math.degrees(math.atan((X_L*7-X_C/7)/R))\n",
-      "I_01=19.403\n",
-      "I_m1=4*V_s/(Z1*math.pi)\n",
-      "I_m3=4*V_s/(3*Z3*math.pi)\n",
-      "I_m5=4*V_s/(5*Z5*math.pi)\n",
-      "I_m7=4*V_s/(7*Z7*math.pi)\n",
-      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2+I_m7**2)\n",
-      "I_or=I_m/math.sqrt(2)\n",
-      "I_oh=math.sqrt((I_m**2-I_m1**2)/2)\n",
-      "THD=I_oh/I_01    \n",
-      "DF=I_01/I_or    \n",
-      "P_o=I_or**2*R    \n",
-      "I_avg=P_o/V_s    \n",
-      "t1=(180-phi1)*math.pi/(180*w)    \n",
-      "t1=1/(2*f)-t1    \n",
-      "I_p=I_m1    \n",
-      "I_t1=.46135*I_p \n",
-      "\n",
-      "#Results\n",
-      "print(\"\\nDF=%.1f\" %DF)\n",
-      "print(\"THD of current=%.4f\" %THD)   \n",
-      "print(\"DF=%.3f\" %DF)\n",
-      "print(\"load power=%.1f W\" %P_o)\n",
-      "print(\"avg value of load current=%.2f A\" %I_avg)\n",
-      "print(\"conduction time for thyristor=%.0f ms\" %(t1*1000))\n",
-      "print(\"conduction time for diodes=%.0f ms\" %(t1*1000))\n",
-      "print(\"peak transistor current=%.2f A\" %I_p)\n",
-      "print(\"rms transistor current=%.2f A\" %I_t1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "THD of voltage=0.4834\n",
-        "\n",
-        "DF=1.0\n",
-        "THD of current=0.1557\n",
-        "DF=0.988\n",
-        "load power=2313.5 W\n",
-        "avg value of load current=10.52 A\n",
-        "conduction time for thyristor=3 ms\n",
-        "conduction time for diodes=3 ms\n",
-        "peak transistor current=27.44 A\n",
-        "rms transistor current=12.66 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11 Page No 497"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=450.0\n",
-      "R=10.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_or=math.sqrt((V_s/(3*R))**2*2/3+(2*V_s/(3*R))**2*1/3)    \n",
-      "I_T1=math.sqrt((1/(2*math.pi))*((V_s/(3*R))**2*2*math.pi/3+(2*V_s/(3*R))**2*math.pi/3))   \n",
-      "P=3*I_or**2*R    \n",
-      "I_or=math.sqrt((1/(math.pi))*((V_s/(2*R))**2*2*math.pi/3))    \n",
-      "I_T1=math.sqrt((1/(2*math.pi))*((V_s/(2*R))**2*2*math.pi/3))    \n",
-      "P=3*I_or**2*R    \n",
-      "\n",
-      "#Results\n",
-      "print(\"for 180deg mode\")\n",
-      "print(\"rms value of load current=%.3f A\" %I_or)\n",
-      "print(\"power delivered to load=%.1f kW\" %(P/1000))\n",
-      "print(\"rms value of load current=%.0f A\" %I_T1)\n",
-      "print(\"for 120deg mode\")\n",
-      "print(\"rms value of load current=%.3f A\" %I_or)\n",
-      "print(\"rms value of load current=%.2f A\" %I_T1)\n",
-      "print(\"power delivered to load=%.3f kW\" %(P/1000))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "for 180deg mode\n",
-        "rms value of load current=18.371 A\n",
-        "power delivered to load=10.1 kW\n",
-        "rms value of load current=13 A\n",
-        "for 120deg mode\n",
-        "rms value of load current=18.371 A\n",
-        "rms value of load current=12.99 A\n",
-        "power delivered to load=10.125 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12, Page No 510"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "R=10.0\n",
-      "f=50.0\n",
-      "w=2*math.pi*f\n",
-      "L=0.03\n",
-      "\n",
-      "#Calculations\n",
-      "X_L=w*L\n",
-      "V_or=math.sqrt(V_s**2*math.pi/math.pi)\n",
-      "V_01=4*V_s/(math.sqrt(2)*math.pi)\n",
-      "Z1=math.sqrt(R**2+(X_L)**2)\n",
-      "phi1=-math.degrees(math.atan((X_L)/R))\n",
-      "Z3=math.sqrt(R**2+(X_L*3)**2)\n",
-      "phi3=math.degrees(math.atan((X_L*3)/R))\n",
-      "Z5=math.sqrt(R**2+(X_L*5)**2)\n",
-      "phi5=math.degrees(math.atan((X_L*5)/R))\n",
-      "Z7=math.sqrt(R**2+(X_L*7)**2)\n",
-      "phi7=math.degrees(math.atan((X_L*7)/R))\n",
-      "I_m1=4*V_s/(math.sqrt(2)*Z1*math.pi)\n",
-      "I_m3=4*V_s/(math.sqrt(2)*3*Z3*math.pi)\n",
-      "I_m5=4*V_s/(math.sqrt(2)*5*Z5*math.pi)\n",
-      "I_m7=4*V_s/(math.sqrt(2)*7*Z7*math.pi)\n",
-      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2+I_m7**2)\n",
-      "P=I_m**2*R    \n",
-      "I_01=I_m1*math.sin(math.radians(45))\n",
-      "I_03=I_m3*math.sin(math.radians(3*45))\n",
-      "I_05=I_m5*math.sin(math.radians(5*45))\n",
-      "I_07=I_m7*math.sin(math.radians(7*45))\n",
-      "I_0=(I_01**2+I_03**2+I_05**2+I_07**2)\n",
-      "P=I_0*R    \n",
-      "g=(180-90)/3+45/2\n",
-      "I_01=2*I_m1*math.sin(math.radians(g))*math.sin(math.radians(45/2))\n",
-      "I_03=2*I_m3*math.sin(math.radians(g*3))*math.sin(math.radians(3*45/2))\n",
-      "I_05=2*I_m5*math.sin(math.radians(g*5))*math.sin(math.radians(5*45/2))\n",
-      "I_07=2*I_m7*math.sin(math.radians(g*7))*math.sin(math.radians(7*45/2))\n",
-      "I_0=(I_01**2+I_03**2+I_05**2+I_07**2)\n",
-      "P=I_0*R    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"using square wave o/p\")\n",
-      "print(\"power delivered=%.2f W\" %P)\n",
-      "print(\"using quasi-square wave o/p\")\n",
-      "print(\"power delivered=%.2f W\" %P)\n",
-      "print(\"using two symmitrical spaced pulses\")\n",
-      "print(\"power delivered=%.2f W\" %P)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "using square wave o/p\n",
-        "power delivered=845.87 W\n",
-        "using quasi-square wave o/p\n",
-        "power delivered=845.87 W\n",
-        "using two symmitrical spaced pulses\n",
-        "power delivered=845.87 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14, Page No 520"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=50.0\n",
-      "T=1/f\n",
-      "I=0.5\n",
-      "\n",
-      "#Calculations\n",
-      "di=I/T     #di=di/dt\n",
-      "V_s=220.0\n",
-      "L=V_s/di    \n",
-      "t=20*10**-6\n",
-      "fos=2     #factor of safety\n",
-      "t_c=t*fos\n",
-      "R=10\n",
-      "C=t_c/(R*math.log(2))\n",
-      "\n",
-      "#Results    \n",
-      "print(\"source inductance=%.1f H\" %L)\n",
-      "print(\"commutating capacitor=%.2f uF\" %(C*10**6))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "source inductance=8.8 H\n",
-        "commutating capacitor=5.77 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15, Page No 539"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=10.0\n",
-      "L=.01\n",
-      "C=10*10**-6\n",
-      "#Calculations\n",
-      "if (R**2)<(4*L/C) :\n",
-      "    print(\"ckt will commutate on its own\")\n",
-      "else:\n",
-      "    print(\"ckt will not commutate on its own\")\n",
-      "\n",
-      "xie=R/(2*L)\n",
-      "w_o=1/math.sqrt(L*C)\n",
-      "w_r=math.sqrt(w_o**2-xie**2)\n",
-      "phi=math.degrees(math.atan(xie/w_r))\n",
-      "t=math.pi/w_r\n",
-      "V_s=1\n",
-      "v_L=V_s*(w_o/w_r)*math.exp(-xie*t)*math.cos(math.radians(180+phi))\n",
-      "v_c=V_s*(1-(w_o/w_r)*math.exp(-xie*t)*math.cos(math.radians(180-phi)))   \n",
-      "di=V_s/L    \n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"voltage across inductor(*V_s)=%.5f V\" %v_L) \n",
-      "print(\"voltage across capacitor(*V_s)=%.5f V\" %v_c)\n",
-      "print(\"di/dt*V_s (for t=0)=%.0f A/s\" %di)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ckt will commutate on its own\n",
-        "voltage across inductor(*V_s)=-0.60468 V\n",
-        "voltage across capacitor(*V_s)=1.60468 V\n",
-        "di/dt*V_s (for t=0)=100 A/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.16, Page No 540"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "L=0.006\n",
-      "C=1.2*10**-6\n",
-      "R=100.0\n",
-      "\n",
-      "#Calculations\n",
-      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
-      "T_off=0.2*10**-3\n",
-      "f=1/(2*(T+T_off))    \n",
-      "R=40\n",
-      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
-      "T_off=.2*10**-3\n",
-      "f=1/(2*(T+T_off))    \n",
-      "R=140\n",
-      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
-      "T_off=.2*10**-3\n",
-      "f=1/(2*(T+T_off)) \n",
-      "\n",
-      "#Results\n",
-      "print(\"o/p freq=%.2f Hz\" %f)\n",
-      "print(\"for R=40ohm\")\n",
-      "print(\"upper limit o/p freq=%.1f Hz\" %f)\n",
-      "print(\"for R=140ohm\")\n",
-      "print(\"lower limit o/p freq=%.1f Hz\" %f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "o/p freq=239.81 Hz\n",
-        "for R=40ohm\n",
-        "upper limit o/p freq=239.8 Hz\n",
-        "for R=140ohm\n",
-        "lower limit o/p freq=239.8 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17, Page No 540"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "f=5000.0\n",
-      "w=2*math.pi*f\n",
-      "R=3.0\n",
-      "\n",
-      "#Calculations\n",
-      "L=60*10**-6\n",
-      "xie=R/(2*L)\n",
-      "C=7.5*10**-6\n",
-      "w_o=1/math.sqrt(L*C)\n",
-      "w_r=math.sqrt(w_o**2-xie**2)\n",
-      "t_c=math.pi*(1/w-1/w_r)    \n",
-      "fos=1.5\n",
-      "t_q=10*10**-6\n",
-      "f_max=1/(2*math.pi*(t_q*fos/math.pi+1/w_r))    \n",
-      "\n",
-      "#Results\n",
-      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n",
-      "print(\"max possible operating freq=%.1f Hz\" %f_max)\n",
-      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ckt turn off time=21.39 us\n",
-        "max possible operating freq=5341.4 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18, Page No 541"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "a=30.0\n",
-      "R=10.0\n",
-      "P=5000.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_s=math.sqrt(P*R*2*math.pi/(2*3)/(math.pi/3+math.sqrt(3)*math.cos(math.radians(2*a))/2))\n",
-      "V_ph=V_s/math.sqrt(3)    \n",
-      "I_or=math.sqrt(P*R)\n",
-      "V_s=I_or*math.pi/(math.sqrt(2)*3*math.cos(math.radians(a)))\n",
-      "V_ph=V_s/math.sqrt(3) \n",
-      "\n",
-      "#Results\n",
-      "print(\"per phase voltage  V_ph=%.3f V\" %V_ph)   \n",
-      "print(\"for constant load current\")\n",
-      "print(\"V_ph=%.2f V\" %V_ph)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "per phase voltage  V_ph=110.384 V\n",
-        "for constant load current\n",
-        "V_ph=110.38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19, Page No 547"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "t=20.0\n",
-      "fos=2.0     #factor of safety\n",
-      "\n",
-      "#Calculations\n",
-      "t_c=t*fos\n",
-      "n=1.0/3\n",
-      "R=20.0\n",
-      "C=n**2*t_c/(4*R*math.log(2)) \n",
-      "\n",
-      "#Results   \n",
-      "print(\"value of capacitor=%.2f uF\" %C)\n",
-      " #printing mistake in the answer in book."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "value of capacitor=0.08 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.20, Page No 547"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=220.0\n",
-      "V_p=math.sqrt(2)*V_s/3    \n",
-      "V_L=math.sqrt(3)*V_p    \n",
-      "V_p1=math.sqrt(2)*V_s/math.pi    \n",
-      "V_L1=math.sqrt(3)*V_p1    \n",
-      "V_oh=math.sqrt(V_L**2-V_L1**2)\n",
-      "\n",
-      "#Calculations\n",
-      "THD=V_oh/V_L1    \n",
-      "V_a1=2*V_s/math.pi\n",
-      "V_a5=2*V_s/(5*math.pi)\n",
-      "V_a7=2*V_s/(7*math.pi)\n",
-      "V_a11=2*V_s/(11*math.pi)\n",
-      "R=4.0\n",
-      "L=0.02\n",
-      "f=50\n",
-      "w=2*math.pi*f\n",
-      "Z1=math.sqrt(R**2+(w*L)**2)\n",
-      "Z5=math.sqrt(R**2+(5*w*L)**2)\n",
-      "Z7=math.sqrt(R**2+(7*w*L)**2)\n",
-      "Z11=math.sqrt(R**2+(11*w*L)**2)\n",
-      "I_a1=V_a1/Z1\n",
-      "I_a5=V_a5/Z5\n",
-      "I_a7=V_a7/Z7\n",
-      "I_a11=V_a11/Z11\n",
-      "I_or=math.sqrt((I_a1**2+I_a5**2+I_a7**2+I_a11**2)/2)\n",
-      "P=3*I_or**2*R    \n",
-      "I_s=P/V_s    \n",
-      "I_TA=I_s/3   \n",
-      " \n",
-      "#Results\n",
-      "print(\"rms value of phasor voltages=%.2f V\" %V_p)\n",
-      "print(\"rms value of line voltages=%.2f V\" %V_L)\n",
-      "print(\"fundamental component of phase voltage=%.3f V\" %V_p1)\n",
-      "print(\"fundamental component of line voltages=%.3f V\" %V_L1)\n",
-      "print(\"THD=%.7f\" %THD)\n",
-      "print(\"load power=%.1f W\" %P)\n",
-      "print(\"avg value of source current=%.3f A\" %I_s)\n",
-      "print(\"avg value of thyristor current=%.3f A\" %I_TA)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of phasor voltages=103.71 V\n",
-        "rms value of line voltages=179.63 V\n",
-        "fundamental component of phase voltage=99.035 V\n",
-        "fundamental component of line voltages=171.533 V\n",
-        "THD=0.3108419\n",
-        "load power=2127.6 W\n",
-        "avg value of source current=9.671 A\n",
-        "avg value of thyristor current=3.224 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Power_Electronics/chapter9.ipynb b/Power_Electronics/chapter9.ipynb
deleted file mode 100755
index 052c4736..00000000
--- a/Power_Electronics/chapter9.ipynb
+++ /dev/null
@@ -1,388 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 09 : AC Voltage Controllers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1, Page No 560"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=45.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_or=(V_m/2)*math.sqrt(1/math.pi*((2*math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
-      "R=20\n",
-      "I_or=V_or/R\n",
-      "P_o=I_or**2*R    \n",
-      "I_s=I_or\n",
-      "VA=V_s*I_s\n",
-      "pf=P_o/VA    \n",
-      "V_o=math.sqrt(2)*V_s/(2*math.pi)*(math.cos(math.radians(a))-1)\n",
-      "I_ON=V_o/R  \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of o/p voltage=%.3f V\" %V_or)\n",
-      "print(\"load power=%.1f W\" %P_o)\n",
-      "print(\"i/p pf=%.4f\" %pf)\n",
-      "print(\"avg i/p current=%.4f A\" %I_ON)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of o/p voltage=224.716 V\n",
-        "load power=2524.9 W\n",
-        "i/p pf=0.9770\n",
-        "avg i/p current=-0.7581 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2, Page No 560"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=45.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_or=(V_s)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
-      "R=20\n",
-      "I_or=V_or/R\n",
-      "P_o=I_or**2*R    \n",
-      "I_s=I_or\n",
-      "VA=V_s*I_s\n",
-      "pf=P_o/VA    \n",
-      "I_TA=math.sqrt(2)*V_s/(2*math.pi*R)*(math.cos(math.radians(a))+1)   \n",
-      "I_Tr=math.sqrt(2)*V_s/(2*R)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2)) \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of o/p voltage=%.3f V\" %V_or)\n",
-      "print(\"load power=%.2f W\" %P_o)\n",
-      "print(\"i/p pf=%.2f\" %pf)\n",
-      "print(\"avg thyristor current=%.2f A\" %I_TA)   \n",
-      "print(\"rms value of thyristor current=%.2f A\" %I_Tr)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of o/p voltage=219.304 V\n",
-        "load power=2404.71 W\n",
-        "i/p pf=0.95\n",
-        "avg thyristor current=4.42 A\n",
-        "rms value of thyristor current=7.75 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3 Page No 564"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "n=6.0 #on cycles\n",
-      "m=4.0 #off cycles\n",
-      "\n",
-      "#Calculations\n",
-      "k=n/(n+m)\n",
-      "V_or=V_s*math.sqrt(k)    \n",
-      "pf=math.sqrt(k)    \n",
-      "R=15\n",
-      "I_m=V_s*math.sqrt(2)/R\n",
-      "I_TA=k*I_m/math.pi\n",
-      "I_TR=I_m*math.sqrt(k)/2  \n",
-      "   \n",
-      "#Results\n",
-      "print(\"rms value of o/ voltage=%.2f V\" %V_or)\n",
-      "print(\"i/p pf=%.2f\" %pf)\n",
-      "print(\"avg thyristor current=%.2f A\" %I_TA)  \n",
-      "print(\"rms value of thyristor current=%.2f A\" %I_TR)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of o/ voltage=178.16 V\n",
-        "i/p pf=0.77\n",
-        "avg thyristor current=4.14 A\n",
-        "rms value of thyristor current=8.40 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4, Page No 569"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "R=3.0\n",
-      "\n",
-      "#Calculations\n",
-      "I_TAM1=2*V_m/(2*math.pi*R)  \n",
-      "I_TRM2=V_m/(2*R)    \n",
-      "f=50\n",
-      "w=2*math.pi*f\n",
-      "t_c=math.pi/w    \n",
-      " \n",
-      "#Results\n",
-      "print(\"max value of avg thyristor current=%.3f A\" %I_TAM1)\n",
-      "print(\"max value of avg thyristor current=%.3f A\" %I_TRM2)\n",
-      "print(\"ckt turn off time=%.0f ms\" %(t_c*1000))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "max value of avg thyristor current=34.512 A\n",
-        "max value of avg thyristor current=54.212 A\n",
-        "ckt turn off time=10 ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5 Page No 575"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "R=3.0\n",
-      "X_L=4.0\n",
-      "\n",
-      "#Calculations\n",
-      "phi=math.degrees(math.atan(X_L/R))    \n",
-      "V_s=230\n",
-      "Z=math.sqrt(R**2+X_L**2)\n",
-      "I_or=V_s/Z    \n",
-      "P=I_or**2*R    \n",
-      "I_s=I_or\n",
-      "pf=P/(V_s*I_s)    \n",
-      "I_TAM=math.sqrt(2)*V_s/(math.pi*Z)    \n",
-      "I_Tm=math.sqrt(2)*V_s/(2*Z)    \n",
-      "f=50\n",
-      "w=2*math.pi*f\n",
-      "di=math.sqrt(2)*V_s*w/Z    \n",
-      "\n",
-      "#Results\n",
-      "print(\"min firing angle=%.2f deg\" %phi)\n",
-      "print(\"\\nmax firing angle=%.0f deg\" %180)\n",
-      "print(\"i/p pf=%.1f\" %pf)\n",
-      "print(\"max value of rms load current=%.0f A\" %I_or)\n",
-      "print(\"max power=%.0f W\" %P)\n",
-      "print(\"max value of avg thyristor current=%.3f A\" %I_TAM)\n",
-      "print(\"max value of rms thyristor current=%.3f A\" %I_Tm)\n",
-      "print(\"di/dt=%.0f A/s\" %di)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "min firing angle=53.13 deg\n",
-        "\n",
-        "max firing angle=180 deg\n",
-        "i/p pf=0.6\n",
-        "max value of rms load current=46 A\n",
-        "max power=6348 W\n",
-        "max value of avg thyristor current=20.707 A\n",
-        "max value of rms thyristor current=32.527 A\n",
-        "di/dt=20437 A/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6 Page No 576"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V=230.0\n",
-      "R=3.0 #ohm\n",
-      "X_L=5.0 #ohm\n",
-      "a=120.0 #firing angle delay\n",
-      "\n",
-      "#Calculations\n",
-      "phi=math.degrees(math.atan(X_L/R))\n",
-      "b=0\n",
-      "i=1\n",
-      "while (i>0) :\n",
-      "    LHS=math.sin(math.radians(b-a))\n",
-      "    RHS=math.sin(math.radians(a-phi))*math.exp(-(R/X_L)*(b-a)*math.pi/180)\n",
-      "    if math.fabs(LHS-RHS)<= 0.01 :\n",
-      "        B=b\n",
-      "        i=2\n",
-      "        break\n",
-      "   \n",
-      "    b=b+.1   \n",
-      "V_or=math.sqrt(2)*V*math.sqrt((1/(2*math.pi))*((B-a)*math.pi/180+(math.sin(math.radians(2*a))-math.sin(math.radians(2*B)))/2))\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Extinction angle=%.1f deg\" %B) #answer in the book is wrong as formulae for RHS is wrongly employed\n",
-      "print(\"rms value of output voltage=%.2f V\" %V_or) #answer do not match due to wrong B in book\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Extinction angle=156.1 deg\n",
-        "rms value of output voltage=97.75 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8, Page No 581"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialisation of variables\n",
-      "V_s=230.0\n",
-      "V_m=math.sqrt(2)*V_s\n",
-      "a=60.0\n",
-      "R=20.0\n",
-      "\n",
-      "#Calculations\n",
-      "V_or=math.sqrt((V_m**2/(2*math.pi))*(a*math.pi/180-math.sin(math.radians(2*a))/2)+(2*V_m**2/(math.pi))*(math.pi-a*math.pi/180+math.sin(math.radians(2*a))/2))    \n",
-      "I_T1r=(V_m/R)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
-      "I_T3r=(V_m/(2*R))*math.sqrt(1/math.pi*((a*math.pi/180)-math.sin(math.radians(2*a))/2))    \n",
-      "I1=math.sqrt(2)*I_T1r\n",
-      "I3=math.sqrt((math.sqrt(2)*I_T1r)**2+(math.sqrt(2)*I_T3r)**2)\n",
-      "r=V_s*(I1+I3)    \n",
-      "P_o=V_or**2/R\n",
-      "pf=P_o/r    \n",
-      "\n",
-      "#Results\n",
-      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
-      "print(\"rms value of current for upper thyristors=%.2f A\" %I_T1r)\n",
-      "print(\"rms value of current for lower thyristors=%.2f A\" %I_T3r)\n",
-      "print(\"t/f VA rating=%.2f VA\" %r)\n",
-      "print(\"i/p pf=%.2f\" %pf)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rms value of o/p voltage=424.94 V\n",
-        "rms value of current for upper thyristors=14.59 A\n",
-        "rms value of current for lower thyristors=3.60 A\n",
-        "t/f VA rating=9631.61 VA\n",
-        "i/p pf=0.94\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
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diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_10_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_10_3.ipynb
new file mode 100755
index 00000000..a1e99a45
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_10_3.ipynb
@@ -0,0 +1,283 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Power Supplies"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 10.1, Page No. 416"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Voltage regulation and % change in output voltage per unit of load current\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vfl = 24                # full load voltage\n",
+      "Vnl = 24.5              # no load voltage\n",
+      "Ifl = 2                 # full load current\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "VR = (Vnl-Vfl)/Vfl\n",
+      "#(b)\n",
+      "x = VR*100/Ifl\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Voltage regulation = %.4f or %.2f%%\\n(b) %%tage change in output voltage per unit load current = %.2f%%\"%(VR,VR*100,x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Voltage regulation = 0.0208 or 2.08%\n",
+        "(b) %tage change in output voltage per unit load current = 1.04%\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 10.2, Page No. 416"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Redulation in percent V\n",
+      "\n",
+      "\n",
+      "import math\n",
+      "Vout = 0.3             # change in output voltage when input change\n",
+      "Vin = 4                # input voltage change\n",
+      "V = 15                 # rated ouutput voltage\n",
+      "\n",
+      "#calculations\n",
+      "lr = (Vout/V)*100/Vin\n",
+      "print(\"%% line regulation = %.1f%% V\"%lr)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "% line regulation = 0.5% V\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 10.3, Page No. 416"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# output voltage, load current, zener curretn\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vi = 15.0                # input voltage\n",
+      "beta_dc = 50.0           # transistor gain\n",
+      "Rl = 1000.0              # load resistor\n",
+      "Vz = 6.0                 # zener voltage\n",
+      "Ri = 500.0               # input voltage\n",
+      "Vbe = 0.7                # voltage drop across transistor base-emitter\n",
+      "\n",
+      "#Calculations\n",
+      "Vo = Vz-Vbe\n",
+      "Il = Vo/Rl\n",
+      "Vce = Vi-Vo\n",
+      "I = (V-Vz)/Ri\n",
+      "Ib = Il/beta_dc\n",
+      "Iz = I-Ib\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vo = %.1fV\\nLoad current = %.1f mA\\nZener current = %.3f mA\"%(Vo,Il*1000,Iz*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vo = 5.3V\n",
+        "Load current = 5.3 mA\n",
+        "Zener current = 17.894 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 10.4, Page No. 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#output voltage(referring fig.10.2)\n",
+      "\n",
+      "import math\n",
+      "Vz = 6.0               # zener voltage\n",
+      "R1 = 10.0*10**3        # resistance 1\n",
+      "R2 = 15.0*10**3        # resistance 2\n",
+      "\n",
+      "#Calculations\n",
+      "Vo = (1+(R2/R1))*Vz\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vo = %d V\"%Vo)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vo = 15 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 10.5, Page No. 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# finding Vo, Io, Ic (referring fig.10.1)\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vz = 9.0               # zener voltage\n",
+      "R1 = 1.5*10**3         # resistance 1\n",
+      "R2 = 3.0*10**3         # resistance 2, value used in calculations\n",
+      "Rl = 2.0*10**3         # load resistance\n",
+      "Rs = 200               # source resistance\n",
+      "Vin = 30               # input oltage\n",
+      "\n",
+      "#Calculations\n",
+      "Vo = (1+(R2/R1))*Vz\n",
+      "Is =(Vin - Vo)/Rs\n",
+      "Io = Vo/Rl\n",
+      "Ic = Is -Io\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vo = %d V\\nIo = %.1f mA\\nIc = %.1f mA\"%(Vo,Io*1000,Ic*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vo = 27 V\n",
+        "Io = 13.5 mA\n",
+        "Ic = 1.5 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 10.6, Page No. 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Minimum and maximum output voltage\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "Iadj = 40*10**-6           # current\n",
+      "Vref = 1.25                # reference voltage\n",
+      "R1 = 2*10**3               # resistance R1\n",
+      "R2min = 0                  # minimum value of R2 resistor\n",
+      "R2max = 20.0*10**3         # maximum value of R2 resistor\n",
+      "\n",
+      "#calculations\n",
+      "Vo1 = (Vref*(1+(R2max/R1)))+Iadj*R2max\n",
+      "Vo2 = Vref*(1+(R2min/R1))\n",
+      "\n",
+      "#Result\n",
+      "print(\"When R2 = %d k-ohm\\nVo = %.2f V\"%(R2max/1000,Vo1))\n",
+      "print(\"\\nWhen R2 = %d k-ohm\\nVo = %.2f V\"%(R2min/1000,Vo2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When R2 = 20 k-ohm\n",
+        "Vo = 14.55 V\n",
+        "\n",
+        "When R2 = 0 k-ohm\n",
+        "Vo = 1.25 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_12_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_12_3.ipynb
new file mode 100755
index 00000000..46f69125
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_12_3.ipynb
@@ -0,0 +1,1030 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Integrated Citcuits and Operational Amplifiers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.1, Page No.442"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# dc currents and voltage of differential amplifier\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "Vcc = 12              # collector voltage\n",
+      "Vee = -12             # emitter voltage\n",
+      "Rc = 4.1*10**3        # Collector resistance\n",
+      "Re = 3.8*10**3        # emitter resistance\n",
+      "Vbe = 0.7             # voltage across base-emitter junction\n",
+      "\n",
+      "#Calculations\n",
+      "Ie = (Vcc-Vbe)/Re\n",
+      "Ic = 0.5*Ie\n",
+      "Vo = Vcc-Ic*Rc\n",
+      "\n",
+      "#Result\n",
+      "print(\"Ie = %.4f mA\\nIc = %.3f mA\\nVo = %.1f V\"%(Ie*1000,Ic*1000,Vo))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ie = 2.9737 mA\n",
+        "Ic = 1.487 mA\n",
+        "Vo = 5.9 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.2, Page No.442"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Differential amplifier parameters\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vcc = 12              # collector voltage\n",
+      "Vee = -12             # emitter voltage\n",
+      "Rc = 1*10**6          # Collector resistance\n",
+      "Re = 1*10**6          # emitter resistance\n",
+      "Vbe = 0.7             # voltage across base-emitter junction\n",
+      "vi = 2.1*10**-3       # AC input voltage\n",
+      "beta = 75       \n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Ie = (Vcc-Vbe)/Re\n",
+      "ie = 0.5*Ie\n",
+      "re = (25*10**-3)/ie\n",
+      "re_dash = 4420.0#value used in the book\n",
+      "#(b)\n",
+      "g = Rc/(2*re_dash)\n",
+      "g = math.floor(g*10)/10\n",
+      "#(c)\n",
+      "vo = g*vi\n",
+      "vo = math.floor(vo*10000)/10000\n",
+      "#(d)\n",
+      "Zi = 2*beta*re\n",
+      "#(e)\n",
+      "cmg = Rc/(re+2*Re)\n",
+      "cmg = math.ceil(cmg*1000)/1000\n",
+      "#(f)\n",
+      "cmrr = g/cmg\n",
+      "#(g)\n",
+      "cmrr_db = 20*math.log10(cmrr)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) re = %d ohm\\n(b) voltage gain for differential input = %.1f\\n(c) AC output voltage = %.4f V\"%(re,g,vo))\n",
+      "print(\"(d) input impedance = %d k-ohm\\n(e) CMRR = %.2f\\n(g) CMRR' = %.1f dB\"%(Zi/1000,cmrr,cmrr_db))\n",
+      "\n",
+      "#Answer for re is wong in the book "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) re = 4424 ohm\n",
+        "(b) voltage gain for differential input = 113.1\n",
+        "(c) AC output voltage = 0.2375 V\n",
+        "(d) input impedance = 663 k-ohm\n",
+        "(e) CMRR = 226.65\n",
+        "(g) CMRR' = 47.1 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.3, Page No. 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Op-amp with negative feedback\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Ag = 100000.0         # open loop gain of Op-amp\n",
+      "fb = 0.01             # feed back factor\n",
+      "vi = 2*10**-3         # input voltage\n",
+      "\n",
+      "#Calculations\n",
+      "# a\n",
+      "g = Ag/(1+(fb*Ag))\n",
+      "# b\n",
+      "v = vi*g\n",
+      "# c\n",
+      "Ev = v/Ag\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Closed loop gain = %.1f \\n(b) Output = %.4f V\\n(c) Error voltage = %.3f*10^-6 V\"%(g,v,Ev*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Closed loop gain = 99.9 \n",
+        "(b) Output = 0.1998 V\n",
+        "(c) Error voltage = 1.998*10^-6 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.4, Page No. 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Op-amp with negative feedback\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Ag = 15000.0         # open loop gain of Op-amp\n",
+      "fb = 0.01             # feed back factor\n",
+      "vi = 2*10**-3         # input voltage\n",
+      "\n",
+      "#Calculations\n",
+      "# a\n",
+      "g = Ag/(1+(fb*Ag))\n",
+      "# b\n",
+      "v = vi*g\n",
+      "# c\n",
+      "Ev = v/Ag\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Closed loop gain = %.3f \\n(b) Output = %.4f V\\n(c) Error voltage = %.3f*10^-5 V\"%(g,v,Ev*10**5))\n",
+      "print(\"\\nComparison conclusion:\\nA decrease in open loop gain causes a corresponding increase in error voltage,\")\n",
+      "print(\"thus keeping the output voltage nearly constant.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Closed loop gain = 99.338 \n",
+        "(b) Output = 0.1987 V\n",
+        "(c) Error voltage = 1.325*10^-5 V\n",
+        "\n",
+        "Comparison conclusion:\n",
+        "A decrease in open loop gain causes a corresponding increase in error voltage,\n",
+        "thus keeping the output voltage nearly constant.\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.5, Page No. 448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# input/output impedances\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "g = 100000.0                  # open loop gain\n",
+      "Zi = 2*10**6                  # input impedance\n",
+      "Zo = 75                       # output impedance\n",
+      "beta = 0.01                   # feedback factor\n",
+      "\n",
+      "#Calculations\n",
+      "D = 1+ beta*g\n",
+      "Zi_cl = Zi*D\n",
+      "Zo_cl = Zo/D\n",
+      "\n",
+      "#Result\n",
+      "print(\"Closed loop input impedance = %.0f M-ohm\\nClosed loop output impedance = %.4f ohm\"%(Zi_cl/10**6,Zo_cl))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Closed loop input impedance = 2002 M-ohm\n",
+        "Closed loop output impedance = 0.0749 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.6, Page No.448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# closed loop gain and disensitivity\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "g = 100000.0              # open loop gain\n",
+      "beta = 0.001              # feedback factor\n",
+      "\n",
+      "#calculations\n",
+      "clg = g/(1+ beta*g)\n",
+      "D = 1+g*beta\n",
+      "\n",
+      "# Result\n",
+      "print(\"Closed loop gain = %.1f \\nDesensitivity = %.0f \"%(clg,D))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Closed loop gain = 990.1 \n",
+        "Desensitivity = 101 \n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.7, Page No.448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# closed loop gain and upper cut off frequency\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "f = 10**6            # unity gain frequency\n",
+      "alg = 100000.0       # open loop gain\n",
+      "b1 = 0.001           # feedback factor in case 1\n",
+      "b2 = 0.01            # feedback factor in case 2\n",
+      "b3 = 0.1             # feedback factor in case 3\n",
+      "\n",
+      "#calculations\n",
+      "# a\n",
+      "f1 = f/alg\n",
+      "# b\n",
+      "g2 = alg/(1+alg*b1)\n",
+      "f2 = f/g2\n",
+      "# c\n",
+      "g3 = alg/(1+alg*b2)\n",
+      "f3 = f/g3\n",
+      "# d\n",
+      "g4 = alg/(1+alg*b3)\n",
+      "f4 = f/g4\n",
+      "\n",
+      "#Result\n",
+      "print(\"Open loop,\\tgain = %.0f \\tf2 = %d Hz\"%(alg,f1))\n",
+      "print(\"Closed loop,\\tgain = %.1f \\tf2 = %d Hz\"%(g2,f2))\n",
+      "print(\"Closed loop,\\tgain = %.1f \\tf2 = %d Hz\"%(g3,f3))\n",
+      "print(\"Closed loop,\\tgain = %.3f \\tf2 = %d Hz\"%(g4,f4))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Open loop,\tgain = 100000 \tf2 = 10 Hz\n",
+        "Closed loop,\tgain = 990.1 \tf2 = 1010 Hz\n",
+        "Closed loop,\tgain = 99.9 \tf2 = 10010 Hz\n",
+        "Closed loop,\tgain = 9.999 \tf2 = 100010 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.8, Page No.449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Slew rating\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Imax = 10*10**-6          # maximum current\n",
+      "C = 4000*10**-12           # capacitance in pF\n",
+      "\n",
+      "#Calculations\n",
+      "sr = Imax/C\n",
+      "\n",
+      "#Result\n",
+      "print(\"Slew rate = %.1f V/ms\"%(sr/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Slew rate = 2.5 V/ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 63
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.9, Page No. 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# slew rate distortion\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "sr = 0.4            # Slew rate\n",
+      "V = 6               # peak value of voltage\n",
+      "f = 10*10**3        # frequency\n",
+      "\n",
+      "#Calculations\n",
+      "slope = 2*math.pi*f*V\n",
+      "\n",
+      "#Result\n",
+      "print(\"Initial slope of sine wave =%.5f V/micro-sec\"%(slope/10**6))\n",
+      "print(\"\\nSince the slew rate of amplifier is %.1f V/micro-sec, there is no slew rate distortion.\"%(sr))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Initial slope of sine wave =0.37699 V/micro-sec\n",
+        "\n",
+        "Since the slew rate of amplifier is 0.4 V/micro-sec, there is no slew rate distortion.\n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.10, Page No. 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Slew rate distortion \n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "sr = 0.5*10**6            # slew rate\n",
+      "V = 10              # peak value of input signal\n",
+      "f = 10*10**3        # frequency\n",
+      "\n",
+      "#calculations\n",
+      "# a\n",
+      "slope = 2*math.pi*f*V\n",
+      "# b\n",
+      "Vp = sr/(2*math.pi*f)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Initial slope of sine wave = %.3f V/micro-sec\"%(slope/10**6))\n",
+      "print(\"Since initial slope is more than the slew rate of amplifier, slew rate distortion will occur.\")\n",
+      "print(\"\\n(b) To eliminate slew rate distortion,\\n Vp = %.2f V\"%Vp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Initial slope of sine wave = 0.628 V/micro-sec\n",
+        "Since initial slope is more than the slew rate of amplifier, slew rate distortion will occur.\n",
+        "\n",
+        "(b) To eliminate slew rate distortion,\n",
+        " Vp = 7.96 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 72
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.11, Page No.451"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# inverting amplifier\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "A = 100000.0                # open loop gain\n",
+      "Zo = 75.0                   # output impedance\n",
+      "f = 1.0*10**6               # unity gain frequency\n",
+      "R1 = 1.0*10**3              # Source Resistance \n",
+      "Rf = 100.0*10**3            # feedback resistance\n",
+      "\n",
+      "#Calculations\n",
+      "# a\n",
+      "g = -Rf/R1\n",
+      "# b\n",
+      "beta = R1/(R1+Rf)\n",
+      "D = 1+(A*beta)\n",
+      "# c\n",
+      "f2 = beta*f\n",
+      "# d\n",
+      "Zi_cl = R1\n",
+      "\n",
+      "#result\n",
+      "print(\"(a) Gain = %f\\n(b) disensitivity = %.1f\"%(g,D))\n",
+      "print(\"(c) closed loop upper cut off frequency = %.1f*10^3 Hz\\n(d) closed loop input impedance = %.0f ohm\"%(f2/1000,Zi_cl))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Gain = -100.000000\n",
+        "(b) disensitivity = 991.1\n",
+        "(c) closed loop upper cut off frequency = 9.9*10^3 Hz\n",
+        "(d) closed loop input impedance = 1000 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 79
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.12, Page No. 453"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# closed loop voltage gain, input/output impedance(refering to fig 12.11)\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "R2 = 100.0*10**3          # Resistance R2\n",
+      "R1 = 100.0                # Resistance R1\n",
+      "A = 100000.0              # open loop gain\n",
+      "Zin = 2*10**6             # input impedance\n",
+      "Zo = 75                   # output impedance\n",
+      "\n",
+      "# Calculations\n",
+      "g = (R1+R2)/R1\n",
+      "beta = R1/(R1+R2)\n",
+      "Zi_dash = (1+A*beta)*Zin\n",
+      "Zo_dash = Zo/(1+A*beta)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Closed loop gain = %.0f \\nClosed loop input impedance = %.1f M-ohm\\nClosed loop output impedance = %.3f ohm\"%(g,Zi_dash/10**6,Zo_dash))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Closed loop gain = 1001 \n",
+        "Closed loop input impedance = 201.8 M-ohm\n",
+        "Closed loop output impedance = 0.743 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 84
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.13, Page No.454"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Op-amp parameters(refering to fig. 12.13)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "A = 100000.0             # open loop gain\n",
+      "f = 1.0*10**6            # unity gain frequency\n",
+      "R3 = 150 *10**3          # Resistance R3\n",
+      "Ci = 1*10**-6            # inpuut capacitor\n",
+      "R2 = 100*10**3           # Resistance R2\n",
+      "Cb = 1*10**-6            # Bias capacitor\n",
+      "R1 = 1.0*10**3           # Resistance R1\n",
+      "Co = 1*10**-6            # output capacitance\n",
+      "Rl = 15*10**3            # load resistance\n",
+      "\n",
+      "#Calculations\n",
+      "# a\n",
+      "A = (R1+R2)/R1\n",
+      "# b\n",
+      "beta = 1/A\n",
+      "f2 = f/A\n",
+      "# c\n",
+      "fc1 = 1/(2*math.pi*Ci*R3)\n",
+      "fc2 = 1/(2*math.pi*Cb*Rl)\n",
+      "fc3 = 1/(2*math.pi*Co*R1)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Avf = %.0f \\n\\n(b) f2' =%.1f*10^3 Hz\"%(A,f2/1000))\n",
+      "print(\"\\n(c) The critical frequencies are,\\n    fc = %.3f Hz\\n    fc = %.2f Hz\\n    fc = %.2f Hz\"%(fc1,fc2,fc3))\n",
+      "print(\"\\n(d) Evidently the lower cut off frequency is the highest of the above three critical frequencies, i.e. %.2f Hz\"%fc3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Avf = 101 \n",
+        "\n",
+        "(b) f2' =9.9*10^3 Hz\n",
+        "\n",
+        "(c) The critical frequencies are,\n",
+        "    fc = 1.061 Hz\n",
+        "    fc = 10.61 Hz\n",
+        "    fc = 159.15 Hz\n",
+        "\n",
+        "(d) Evidently the lower cut off frequency is the highest of the above three critical frequencies, i.e. 159.15 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 93
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.14, Page No.455"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# output voltage(referring fig.12.14)\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "v1 = 0.5                   # input voltage 1\n",
+      "v2 = 1.5                   # input voltage 2\n",
+      "v3 = 0.2                   # input voltage 3\n",
+      "R1 = 10.0*10**3            # resistance R1\n",
+      "R2 = 10.0*10**3            # resistance R2\n",
+      "R3 = 10.0*10**3            # resistance R3\n",
+      "Rf = 50.0*10**3            # feedback resistance\n",
+      "\n",
+      "#Calculations\n",
+      "vo = -Rf*((v1/R1)+(v2/R2)+(v3/R3))\n",
+      "\n",
+      "#Result\n",
+      "print(\"Output  Voltage = %.0f V\"%vo)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output  Voltage = -11 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 95
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.15, Page No.455"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 96
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.16, Page No.456"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 97
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.17, Page No. 457"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# difference amplifier(referring fig. 12.18)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "R1 = 50.0*10**3           # Resistance R1\n",
+      "R2 = 10.0*10**3           # Resistance R2\n",
+      "Vs1 = 4.5                 # input voltage at channel 1\n",
+      "Vs2 = 5.0                 # input voltage at channel 2\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "vo = R1*(Vs2-Vs1)/R2\n",
+      "\n",
+      "#Result\n",
+      "print(\"Output voltage = %.1f V\"%vo)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output voltage = 2.5 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 99
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.18, Page No.458"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# CMRR in dB\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "A = 50.0                   # gain of difference amplifier\n",
+      "v = 2.0                    # input voltage\n",
+      "vo = 5.0*10**-3            # output voltage \n",
+      "\n",
+      "#Calculations\n",
+      "Vcom = 0.5*(v+v)\n",
+      "Acom = vo/Vcom\n",
+      "cmrr = A/Acom\n",
+      "cmrr = 20*math.log10(cmrr)\n",
+      "\n",
+      "#Result\n",
+      "print(\"CMRR = %.2f dB\"%(cmrr))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "CMRR = 86.02 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 101
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.19, Page No.458"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical Example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical Example\n"
+       ]
+      }
+     ],
+     "prompt_number": 102
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.20, Page No.459"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical Example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical Example\n"
+       ]
+      }
+     ],
+     "prompt_number": 103
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.21, Page No.466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# two pole high pass filter(referring fig. 12.32)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "R1 = 10.0*10**3          # Resistance R1\n",
+      "R2 = 5.6*10**3           # Resistance R2\n",
+      "R = 1.0*10**3            # Resistance R\n",
+      "C = 0.01 *10**-6         # capacitance \n",
+      "vi = 1.6*10**-3          # input voltage                \n",
+      "\n",
+      "#Calculations\n",
+      "# a\n",
+      "A = 1+(R2/R1)\n",
+      "# b\n",
+      "vo = A*vi\n",
+      "# c\n",
+      "pi = math.ceil(math.pi*10000)/10000\n",
+      "fc = 1/(2*pi*R*C)\n",
+      "# d\n",
+      "gain =  0.707*A\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) midband gain = %.2f\\n(b) output voltage = %.3f mV\\n(c) fc = %.2f Hz\\n(d) Gain = %.3f\"%(A,vo*1000,fc,gain))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) midband gain = 1.56\n",
+        "(b) output voltage = 2.496 mV\n",
+        "(c) fc = 15915.46 Hz\n",
+        "(d) Gain = 1.103\n"
+       ]
+      }
+     ],
+     "prompt_number": 112
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.22, Page No. 466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Low pass filter(referring fig. 12.30)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "vi = 1.1*10**-3               # input voltage\n",
+      "R = 10*10**3                  # Resistance\n",
+      "C = 0.001*10**-6              # Capacitance\n",
+      "R1 = 10*10**3                 # Resistance R1\n",
+      "R2 = 5.6*10**3                # Resistance R2\n",
+      "\n",
+      "#Calculations\n",
+      "# a\n",
+      "A = (R1+R2)/R1\n",
+      "vo1 = A*vi\n",
+      "# b\n",
+      "fc = 1/(2*math.pi*R*C)\n",
+      "# c\n",
+      "vo = 0.707*vo1\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)  Avf = %.2f\\n     Vo = %.3f*10^-3 V\\n(b) fc = %.3f*10^3 Hz\\n(c) at f = fc,\\nVo = %.3f* 10^-3 V\"%(A,vo1*1000,fc/1000,vo*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)  Avf = 1.56\n",
+        "     Vo = 1.716*10^-3 V\n",
+        "(b) fc = 15.915*10^3 Hz\n",
+        "(c) at f = fc,\n",
+        "Vo = 1.213* 10^-3 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 12.23, Page No.466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_13_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_13_3.ipynb
new file mode 100755
index 00000000..2e5b592a
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_13_3.ipynb
@@ -0,0 +1,965 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13: Number Systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.1, Page No. 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Decimal to binary conversion: 10\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "N = 10                      # decimal no to be convered to binary\n",
+      "k = N\n",
+      "#Calculations\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0]\n",
+      "while(N!=0):\n",
+      "  b[i] = N%2\n",
+      "  N = N/2\n",
+      "  i = i+1 \n",
+      "\n",
+      "#Result\n",
+      "print(\"Binary equivalent of decimal %d is %d%d%d%d\"%(k,b[3],b[2],b[1],b[0]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Binari equivalent of decimal 10 is 1010\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.2, Page No.474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Decimal to binary conversion: 25\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "N = 25                      # decimal no to be convered to binary\n",
+      "\n",
+      "#Calculations\n",
+      "k = N\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0]\n",
+      "while(N!=0):\n",
+      "  b[i] = N%2\n",
+      "  N = N/2\n",
+      "  i = i+1 \n",
+      "\n",
+      "#Result\n",
+      "print(\"Binary equivalent of decimal %d is %d%d%d%d%d\"%(k,b[4],b[3],b[2],b[1],b[0]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Binari equivalent of decimal 25 is 11001\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.3, Page No. 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Binary to decimal:101110\n",
+      "\n",
+      "import math\n",
+      "b0 = 0         # bit 0\n",
+      "b1 = 1         # bit 1\n",
+      "b2 = 1         # bit 2\n",
+      "b3 = 1         # bit 3\n",
+      "b4 = 0         # bit 4\n",
+      "b5 = 1         # bit 5\n",
+      "\n",
+      "#Calculations\n",
+      "D = (2**0)*b0+(2**1)*b1+(2**2)*b2+(2**3)*b3+(2**4)*b4+(2**5)*b5\n",
+      "\n",
+      "#Result\n",
+      "print(\"Decimal no is %d\"%D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Decimal no is 46\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.4, Page No. 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Decimal to binary conversion: 15,31\n",
+      "\n",
+      "import math\n",
+      "#######################################################################################\n",
+      "# for N = 15\n",
+      "#variable declaration\n",
+      "N = 15                      # decimal no to be convered to binary\n",
+      "\n",
+      "#Calculations\n",
+      "k = N\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0]\n",
+      "while(N!=0):\n",
+      "  b[i] = N%2\n",
+      "  N = N/2\n",
+      "  i = i+1 \n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nBinary equivalent of decimal %d is %d%d%d%d%d\"%(k,b[4],b[3],b[2],b[1],b[0]))\n",
+      "\n",
+      "########################################################################################\n",
+      "\n",
+      "# For N =31\n",
+      "#variable declaration\n",
+      "N = 31                      # decimal no to be convered to binary\n",
+      "\n",
+      "#Calculations\n",
+      "k = N\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0]\n",
+      "while(N!=0):\n",
+      "  b[i] = N%2\n",
+      "  N = N/2\n",
+      "  i = i+1 \n",
+      "\n",
+      "#Result\n",
+      "print(\"Binary equivalent of decimal %d is %d%d%d%d%d\"%(k,b[4],b[3],b[2],b[1],b[0]))\n",
+      "##########################################################################################\n",
+      "#Addition\n",
+      "c= bin(15+31)[2:]\n",
+      "print(\"(b) Addition of 15 and 31 is %s in binary. Decimal equivalent of %s is %d\"%(c,c,int(c,2)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Binari equivalent of decimal 15 is 01111\n",
+        "Binari equivalent of decimal 31 is 11111\n",
+        "(b) Addition of 15 and 31 is 101110 in binary. Decimal equivalent of 101110 is 46\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.5, Page No.475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Binary subtraction\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "n1 = 11001             # first number\n",
+      "n2 = 10001             # second number\n",
+      "\n",
+      "#calculations\n",
+      "n1 = int(str(n1),2)\n",
+      "n2 = int(str(n2),2)\n",
+      "c = bin(n1-n2)[2:]\n",
+      "\n",
+      "#Result\n",
+      "print(\"Subtraction result in binary = %s\"%c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subtraction result in binary = 1000\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.6, Page No.475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Binary subtraction\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "n1 = 1010             # first number\n",
+      "n2 = 111              # second number\n",
+      "\n",
+      "#calculations\n",
+      "n1 = int(str(n1),2)\n",
+      "n2 = int(str(n2),2)\n",
+      "c = bin(n1-n2)[2:]\n",
+      "\n",
+      "#Result\n",
+      "print(\"Subtraction result in binary = %s\"%c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Subtraction result in binary = 11\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.7, Page No. 476"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 16- bit signed binary representation\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "a = 8\n",
+      "b = -8\n",
+      "c = 165\n",
+      "d = -165\n",
+      "\n",
+      "#Calculations\n",
+      "#c = bin(d)\n",
+      "a = format(a,'#018b')[2:]\n",
+      "b = format(b,'#018b')[3:]\n",
+      "c = format(c,'#018b')[2:]\n",
+      "d = format(d,'#018b')[3:]\n",
+      "str = '1'\n",
+      "#Result\n",
+      "print(\"In the leading bit we will have 1 to represent '-' sign\\n\")\n",
+      "print(\"(a) +8   --> %s\"%(a))\n",
+      "print(\"(b) -8   --> %s%s\"%(str,b))\n",
+      "print(\"(c) +167 --> %s\"%(c))\n",
+      "print(\"(d) -167 --> %s%s\"%(str,d))\n",
+      "\n",
+      "a = format(-167%(1<<16),'016b')\n",
+      "b = format(167%(1<<16),'016b')\n",
+      "print a\n",
+      "print b"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In the leading bit we will have 1 to represent '-' sign\n",
+        "\n",
+        "(a) +8   --> 0000000000001000\n",
+        "(b) -8   --> 1000000000001000\n",
+        "(c) +167 --> 0000000010100101\n",
+        "(d) -167 --> 1000000010100101\n",
+        "1111111101011001\n",
+        "0000000010100111\n"
+       ]
+      }
+     ],
+     "prompt_number": 87
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.8, Page No.477"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 2's complement\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "a = int('00011111',2)\n",
+      "b = int('11100101',2)\n",
+      "c = int('11110111',2)\n",
+      "\n",
+      "#Calculations\n",
+      "a = format(-a%(1<<8),'08b')\n",
+      "b = format(-b%(1<<8),'08b')\n",
+      "c = format(-c%(1<<8),'08b')\n",
+      "print(\"(a) 2's complement of 00011111 --> %s \" %a)\n",
+      "print(\"(b) 2's complement of 11100101 --> %s \" %b)\n",
+      "print(\"(c) 2's complement of 11110111 --> %s \" %c)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) 2's complement of 00011111 --> 11100001 \n",
+        "(b) 2's complement of 11100101 --> 00011011 \n",
+        "(c) 2's complement of 11110111 --> 00001001 \n"
+       ]
+      }
+     ],
+     "prompt_number": 95
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.9, Page No. 13.9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 8-bit number range\n",
+      "\n",
+      "import math\n",
+      "#calculations\n",
+      "a = int('01111111',2)\n",
+      "b = int('10000000',2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"largest positive 8-bit no in %d and smallest negative number is -%d\"%(a,b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "largest positive 8-bit no in 127 and smallest negative number is -128\n"
+       ]
+      }
+     ],
+     "prompt_number": 97
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.10, Page No.10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 2's complement and addition,subtraction\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "A = -24\n",
+      "B = 16\n",
+      "#Calculations\n",
+      "#2's complement\n",
+      "a = format(A%(1<<8),'08b')\n",
+      "b = format(-B%(1<<8),'08b')\n",
+      "# addition\n",
+      "c = bin(A+B)[3:]\n",
+      "c = format(-int(c,2)%(1<<8),'08b')\n",
+      "#Subtraction\n",
+      "d = bin(A-B)[3:]\n",
+      "d = format(-int(d,2)%(1<<8),'08b')\n",
+      "#Result\n",
+      "print(\"(a)\\n2's complement of -24 --> %s \" %a)\n",
+      "print(\"2's complement of 16 --> %s \" %b)\n",
+      "print(\"(b) A+B = %s\"%c) \n",
+      "print(\"(c) A-B = %s\"%d)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "2's complement of -24 --> 11101000 \n",
+        "2's complement of 16 --> 11110000 \n",
+        "(b) A+B = 11111000\n",
+        "(c) A-B = 11011000\n"
+       ]
+      }
+     ],
+     "prompt_number": 116
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.11, Page No. 479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 2's complement and addition,subtraction\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "A = -60\n",
+      "B = -28\n",
+      "#Calculations\n",
+      "#2's complement\n",
+      "a = format(A%(1<<8),'08b')\n",
+      "b = format(B%(1<<8),'08b')\n",
+      "# addition\n",
+      "c = bin(A+B)[3:]\n",
+      "c = format(-int(c,2)%(1<<8),'08b')\n",
+      "#Subtraction\n",
+      "x = B-A\n",
+      "d = bin(B-A)[2:]\n",
+      "d = format(int(d,2),'08b')\n",
+      "#Result\n",
+      "print(\"(a)\\n2's complement of A --> %s \" %a)\n",
+      "print(\"2's complement of B --> %s \" %b)\n",
+      "print(\"(b) B+A = %s\"%c) \n",
+      "print(\"(c) B-A = %s\"%d)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "2's complement of A --> 11000100 \n",
+        "2's complement of B --> 11100100 \n",
+        "(b) B+A = 10101000\n",
+        "(c) B-A = 00100000\n"
+       ]
+      }
+     ],
+     "prompt_number": 130
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.12, Page No. 479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# decimal to binary\n",
+      "\n",
+      "# For N = 0.6875\n",
+      "#variable declaration\n",
+      "N = 0.6875                      # decimal no to be convered to binary\n",
+      "\n",
+      "#Calculations\n",
+      "k = N\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0,0,0,0,0,0,0]\n",
+      "while(N!=0):\n",
+      "    N = N*2\n",
+      "    #print N\n",
+      "    b[i]=0\n",
+      "    #print b[i]\n",
+      "    if (N>1) or(N==1):\n",
+      "        N =N-1\n",
+      "        b[i]=1\n",
+      "    i=i+1\n",
+      "    \n",
+      "#Result\n",
+      "print(\"Binary equivalent of decimal %f is 0.%d%d%d%d \"%(k,b[0],b[1],b[2],b[3]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Binary equivalent of decimal 0.687500 is 0.1011 \n"
+       ]
+      }
+     ],
+     "prompt_number": 230
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.13, Page No.480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# decimal to binary\n",
+      "\n",
+      "# For N = 0.634\n",
+      "#variable declaration\n",
+      "N = 0.634                      # decimal no to be convered to binary\n",
+      "\n",
+      "#Calculations\n",
+      "k = N\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0,0,0,0,0,0,0]\n",
+      "while(i!=7):\n",
+      "    N = N*2\n",
+      "    #print N\n",
+      "    b[i]=0\n",
+      "    #print b[i]\n",
+      "    if (N>1) or(N==1):\n",
+      "        N =N-1\n",
+      "        b[i]=1\n",
+      "    i=i+1\n",
+      "\n",
+      "#Result    \n",
+      "#Result\n",
+      "print(\"Binary equivalent of decimal %f is 0.%d%d%d%d%d%d%d \"%(k,b[0],b[1],b[2],b[3],b[4],b[5],b[6]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Binary equivalent of decimal 0.634000 is 0.1010001 \n"
+       ]
+      }
+     ],
+     "prompt_number": 222
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.14, Page No.480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# decimal to binary\n",
+      "\n",
+      "# For N = 0.634\n",
+      "#variable declaration\n",
+      "N = 39.12                      # decimal no to be convered to binary\n",
+      "\n",
+      "N1 = 39\n",
+      "k = N1\n",
+      "#Calculations\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0,0,0]\n",
+      "while(N1!=0):\n",
+      "  b[i] = N1%2\n",
+      "  N1 = N1/2\n",
+      "  i = i+1 \n",
+      "\n",
+      "#Result\n",
+      "print(\"Binary equivalent of decimal %d is %d%d%d%d%d%d%d\"%(k,b[6],b[5],b[4],b[3],b[2],b[1],b[0]))\n",
+      "\n",
+      "\n",
+      "N2 =0.12\n",
+      "#Calculations\n",
+      "k = N2\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0,0,0,0,0,0,0]\n",
+      "while(i!=7):\n",
+      "    N2 = N2*2\n",
+      "    #print N\n",
+      "    b[i]=0\n",
+      "    #print b[i]\n",
+      "    if (N2>1) or(N2==1):\n",
+      "        N2 =N2-1\n",
+      "        b[i]=1\n",
+      "    i=i+1\n",
+      " \n",
+      "#Result\n",
+      "print(\"Binary equivalent of decimal %f is 0.%d%d%d%d%d%d%d \"%(k,b[0],b[1],b[2],b[3],b[4],b[5],b[6]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Binary equivalent of decimal 39 is 0100111\n",
+        "Binary equivalent of decimal 0.120000 is 0.0001111 \n"
+       ]
+      }
+     ],
+     "prompt_number": 225
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.15, Page No.481"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# binary Addition\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "a1 = int('101101',2)       # integer part of first no\n",
+      "a2 = int('0101',2)         # fractiona part of first no\n",
+      "b1 = int('10001',2)        # integer part of second no\n",
+      "b2 = int('1010',2)         # fractiona part of second no(in fraction we can add any no of 0s)\n",
+      "\n",
+      "#Calculations\n",
+      "c1= bin(a1+b1)[2:]\n",
+      "c2 = bin(a2+b2)[2:]\n",
+      "\n",
+      "print(\"Addition --> %s.%s\"%(c1,c2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Addition --> 111110.1111\n"
+       ]
+      }
+     ],
+     "prompt_number": 227
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.16, Page No. 481"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# binary to decimal conversion\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "i0 = 1               # LSB of integer\n",
+      "i1 = 0\n",
+      "i2 = 0\n",
+      "i3 = 1\n",
+      "i4 = 1               # MSB of integer\n",
+      "f1 = 0\n",
+      "f2 = 0\n",
+      "f3 = 1\n",
+      "f4 = 0\n",
+      "f5 = 1\n",
+      "f6 = 1\n",
+      "\n",
+      "#Calculations\n",
+      "D = i0*(2**0)+i1*(2**1)+i2*(2**2)+i3*(2**3)+i4*(2**4)+f1*(2**-1)+f2*(2**-2)+f3*(2**-3)+f4*(2**-4)+f5*(2**-5)+f6*(2**-6)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Decimal equivalent = %f\"%D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Decimal equivalent = 25.171875\n"
+       ]
+      }
+     ],
+     "prompt_number": 231
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.17, Page No. 482"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Hexadecimal to decimal\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "a = 3\n",
+      "b = 10  # decimal equivalent of A\n",
+      "c = 8\n",
+      "#Calculations\n",
+      "# binary equvalent of  8            A                   3 \n",
+      "#                    1000         1010                0011\n",
+      "# a\n",
+      "Da = (1*2**0)+(1*2**1)+(0*2**2)+(0*2**3)+(0*2**4)+(1*2**5)+(0*2**6)+(1*2**7)+(0*2**8)+(0*2**9)+(0*2**10)+(1*2**11)\n",
+      "print(\"(a) Decimal equivalent of 8A3 is %d\"%Da)\n",
+      "# b \n",
+      "Db = (a*(16**0))+(b*(16**1))+(c*(16**2))\n",
+      "print(\"(b) Decimal equivalent of 8A3 is %d\"%Db)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Decimal equivalent of 8A3 is 2211\n",
+        "(b) Decimal equivalent of 8A3 is 2211\n"
+       ]
+      }
+     ],
+     "prompt_number": 236
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.18, Page No. 482"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# decimal to hexadecimal\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "N = 268                # no to be converted into hexadecimal\n",
+      "\n",
+      "N1 = 268\n",
+      "k = N1\n",
+      "#Calculations\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0,0,0]\n",
+      "while(N1!=0):\n",
+      "  b[i] = N1%16\n",
+      "  N1 = N1/16\n",
+      "  i = i+1 \n",
+      "\n",
+      "#Result\n",
+      "print(\"Remainder1: %d = C\\nRemainder2: %d\\nRemainder3: %d\\n\\nHexaecimal equivalent of %d = 10C\"%(b[0],b[1],b[2],k))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Remainder1: 12 = C\n",
+        "Remainder2: 0\n",
+        "Remainder3: 1\n",
+        "\n",
+        "Hexaecimal equivalent of 268 = 10C\n"
+       ]
+      }
+     ],
+     "prompt_number": 260
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.19, Page No.483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# decimal to hexadecimal\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "N = 5741                # no to be converted into hexadecimal\n",
+      "\n",
+      "N1 = 5741\n",
+      "k = N1\n",
+      "#Calculations\n",
+      "i = 0\n",
+      "b =[0,0,0,0,0,0,0,0]\n",
+      "while(N1!=0):\n",
+      "  b[i] = N1%16\n",
+      "  N1 = N1/16\n",
+      "  i = i+1 \n",
+      "\n",
+      "#Result\n",
+      "print(\"Remainder1: %d = D\\nRemainder2: %d\\nRemainder3: %d\\nRemainder4: %d\\n\\nHexaecimal equivalent of %d =166D\"%(b[0],b[1],b[2],b[3],k))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Remainder1: 13 = D\n",
+        "Remainder2: 6\n",
+        "Remainder3: 6\n",
+        "Remainder4: 1\n",
+        "\n",
+        "Hexaecimal equivalent of 5741 =166D\n"
+       ]
+      }
+     ],
+     "prompt_number": 259
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 13.20, Page No.483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Hexadecimal to decimal\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "a = 0\n",
+      "b = 7\n",
+      "c = 13  #  decimal equivalent of D\n",
+      "#Calculations\n",
+      "D = (a*(16**0))+(b*(16**1))+(c*(16**2))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(b) Decimal equivalent of 8A3 is %d\"%D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(b) Decimal equivalent of 8A3 is 3440\n"
+       ]
+      }
+     ],
+     "prompt_number": 262
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_1_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_1_3.ipynb
new file mode 100755
index 00000000..47f33435
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_1_3.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Power Diodes And Transistors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.1, Page No. 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# stored charge and peak reverse current\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "t = 2.5*10**-6                # reverese recovery time to diode\n",
+      "di_by_dt = 35*10**6           # di/dt in A/S\n",
+      "\n",
+      "#Calculations\n",
+      "Q= 0.5*(t**2)*di_by_dt\n",
+      "I= math.sqrt(2*Q*di_by_dt)\n",
+      "\n",
+      "#result\n",
+      "print(\"(a) Stored charge\\n Q = %.3f * 10^-6 C\\n\\n(b) Peak reverse current\\nI = %.1f A\"%(Q*10**6,I))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Stored charge\n",
+        " Q = 109.375 * 10^-6 C\n",
+        "\n",
+        "(b) Peak reverse current\n",
+        "I = 87.5 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.2, Page No.8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Vrrm rating for diode in full wave rectifire\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "V= 12                     # secondary peak voltage\n",
+      "sf = 1.4                  # safety factor\n",
+      "\n",
+      "#calculations\n",
+      "# For fullwave rectifier with transformer secondary voltage 12-0-12, each diode will experience Vrrm equal to 2 x sqrt(2)x 12\n",
+      "r = math.sqrt(2)\n",
+      "r = math.floor(r*1000)/1000 \n",
+      "V = 2*r*V                # Actual value of Vrrm for each diode\n",
+      "Vrrm= V*sf\n",
+      "\n",
+      "# result\n",
+      "print(\"Vrrm rating for each diode with safety factor of %.1f is %.2fV\\n\\n\"%(sf,Vrrm))\n",
+      "#Answer in the book for Vrrm rating is wrong\n",
+      "\n",
+      "#%pylab inline\n",
+      "import matplotlib.pyplot as plt\n",
+      "from numpy import arange,sin,pi\n",
+      "%matplotlib inline\n",
+      "#fig -1\n",
+      "t = arange(0.0,4,0.01)\n",
+      "S = sin(math.pi*t)\n",
+      "plt.subplot(411)\n",
+      "plt.title(\"Secondary Voltage\")\n",
+      "plt.plot(t,S)\n",
+      "#fig -2\n",
+      "plt.subplot(412)\n",
+      "plt.title(\"Load Voltage\")\n",
+      "t1 = arange(0.0,1,0.01)\n",
+      "t2 = arange(1.0,2.0,0.01)\n",
+      "t3 = arange(2.0,3.0,0.01)\n",
+      "t4 = arange(3.0,4.0,0.01)\n",
+      "s1 = sin((pi*t1))\n",
+      "s2 = sin((pi*t1))\n",
+      "s3 = sin(pi*t1)\n",
+      "s4 = sin(pi*t1)\n",
+      "\n",
+      "plt.plot(t1,s1)\n",
+      "plt.plot(t2,s2)\n",
+      "plt.plot(t3,s3)\n",
+      "plt.plot(t4,s4)\n",
+      "#fig -3\n",
+      "plt.subplot(413)\n",
+      "plt.title(\"Voltage across D1\")\n",
+      "plt.axis([0,4,0,1])\n",
+      "plt.plot(t1,s1)\n",
+      "plt.plot(t3,s3)\n",
+      "#fig -4\n",
+      "plt.subplot(414)\n",
+      "plt.title(\"Voltage across D2\")\n",
+      "plt.axis([0,4,-1,0])\n",
+      "s2 = sin((pi*t1)-pi)\n",
+      "s4 = sin(pi*t1-pi)\n",
+      "plt.plot(t2,s2)\n",
+      "plt.plot(t4,s4)\n",
+      "#Result\n",
+      "plt.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vrrm rating for each diode with safety factor of 1.4 is 47.51V\n",
+        "\n",
+        "\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ioiL4+vqie/fuSE1NxaZNm/Dll18iOjpa3JaIoKWlhdDQUFhaWiI7OxtZWVkw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\nYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x53d47f0>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.3, Page No. 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Average, peak and rms current\n",
+      "\n",
+      "import math\n",
+      "#variable declaration(from the waveform)\n",
+      "Ip = 20.0                 # Peak current\n",
+      "\n",
+      "#calculations\n",
+      "Iavg = (Ip*1.0)/3.0\n",
+      "Irms = math.sqrt((Ip**2)*1.0/3.0)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Peak Current = %d A\\nAverage Current = %.3f A\\nrms Current = %.3f A\"%(Ip,Iavg,Irms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak Current = 20 A\n",
+        "Average Current = 6.667 A\n",
+        "rms Current = 11.547 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.4, Page No. 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#power BJT\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "Vcc =220.0              # collector voltage\n",
+      "Vce_sat = 1.0           # Vce saturation voltage\n",
+      "Rb =6.0                 # base resisror\n",
+      "Rl = 8.0                # load resisotr\n",
+      "hfe = 15.0              # gain\n",
+      "Vbe = 0.7               # base-emiter voltage drop\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "Ic = (Vcc-Vce_sat)/Rl\n",
+      "Ib=Ic/hfe\n",
+      "#(b)\n",
+      "Vbb= Ib*Rb+Vbe\n",
+      "#(c)\n",
+      "Pc = Ic*Vce_sat\n",
+      "Pb = Ib*Vbe\n",
+      "Pt = Pc+Pb\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Base current, Ib = %.3f A\\n(b) Vbb = %.2f V\\n(c) Total power dissipation in BJT = %.4f W\"%(Ib,Vbb,Pt))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Base current, Ib = 1.825 A\n",
+        "(b) Vbb = 11.65 V\n",
+        "(c) Total power dissipation in BJT = 28.6525 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.5, Page No. 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Load current and losses in BJT\n",
+      "\n",
+      "import math\n",
+      "# variable declaration(with reference to example 1.4)\n",
+      "Vbb_org = 11.65         # original Vbb\n",
+      "fall =0.85              # 85% fall in original value\n",
+      "Vce_sat = 1.0           # Vce saturation voltage\n",
+      "Rb =6.0                 # base resisror\n",
+      "hfe = 15.0              # gain\n",
+      "Vbe = 0.7               # base-emiter voltage drop\n",
+      "\n",
+      "\n",
+      "#calculations\n",
+      "Vbb = fall* Vbb_org\n",
+      "Ib = (Vbb-Vbe)/Rb\n",
+      "Ic = Ib*hfe\n",
+      "Pc =Ic*Vce_sat\n",
+      "Pb = Ib* Vbe\n",
+      "Pt = Pc+Pb\n",
+      "\n",
+      "#Result\n",
+      "print(\"Load current = %.3f A\\nLosses in BJT = %.2f W\"%(Ib,Pt))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load current = 1.534 A\n",
+        "Losses in BJT = 24.08 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.6, Page No. 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Power loss in BJT\n",
+      "\n",
+      "import math\n",
+      "#variable declaration(with reference to example 1.4)\n",
+      "Vcc = 240                  # New value of collector current\n",
+      "Ic = 27.375                # collector current,from example 1.4\n",
+      "Pb = 1.2775                # base power dissipation,from example 1.4\n",
+      "Rl = 8.0                   # load resisotr\n",
+      "\n",
+      "#Calculations\n",
+      "Vce = Vcc-(Ic*Rl)\n",
+      "Pc = Vce* Ic\n",
+      "Pt = Pb+ Pc\n",
+      "\n",
+      "#result\n",
+      "print(\"Total power dissipation = %.4f W\"%Pt)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total power dissipation = 576.1525 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.7, Page No. 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# BJT switching frequency\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "# variable declaration\n",
+      "I = 80                      # maximum current, from swiching characteristics\n",
+      "t1 = 40 *10**-6             # rise time, from swiching characteristics\n",
+      "t2 = 60* 10**-6             # falll time, from swiching characteristics\n",
+      "V = 200                     # collector-emitter voltage\n",
+      "Pavg =250                   # average power loss\n",
+      "\n",
+      "\n",
+      "#calculations\n",
+      "# switching ON\n",
+      "ic = I/t1\n",
+      "def f(x):\n",
+      "    return (ic*x)*(V-(V/t1)*x)\n",
+      "t_lower =0\n",
+      "t_upper = t1\n",
+      "val_on = quad(f,t_lower,t_upper)\n",
+      "\n",
+      "# switching OFF\n",
+      "ic = I-I/t1\n",
+      "Vc = V/t2\n",
+      "def f1(x):\n",
+      "    return (I-(I/t2)*x)*(Vc*x)\n",
+      "t_lower =0\n",
+      "t_upper = t2\n",
+      "val_off = quad(f1,t_lower,t_upper)\n",
+      "\n",
+      "loss= val_on[0]+val_off[0]\n",
+      "loss= math.floor(loss*10000)/10000\n",
+      "f =Pavg/loss\n",
+      "\n",
+      "# Result\n",
+      "#print(\"(a) Switching ON:\\nEnergy losses during switching on = %.4f J\"%(val_on[0]))\n",
+      "#print(\"\\n(b)Switching OFF\\nEnergy losses during switching off of BJT =%.2f J\"%(val_off[0]))\n",
+      "print(\"\\nSwitching frequency = %.1f Hz\"%f)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Switching frequency = 937.7 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 1.8, Page No. 20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Turn ON loss of power transistor\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vmax = 300             # voltage during start\n",
+      "Imax = 200             # full current after start\n",
+      "t = 1* 10**-6          # starting time \n",
+      "\n",
+      "#calculation\n",
+      "E_loss = Vmax*Imax*t/6        #formula\n",
+      "\n",
+      "#Result\n",
+      "print(\"Energy loss = %.2f Joules\"%E_loss)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy loss = 0.01 Joules\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_2_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_2_3.ipynb
new file mode 100755
index 00000000..6e52e316
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_2_3.ipynb
@@ -0,0 +1,2357 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Thyristors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.1, Page No.39 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Anode current\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "alfa_1 = 0.35         # Gain of PNP transistor of two transistor model\n",
+      "alfa_2 = 0.40         # Gain of NPN transistor of two transistor model\n",
+      "Ig = 40*10**-3        # Gate current\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "Ia = (alfa_2*Ig)/(1-(alfa_1+alfa_2))\n",
+      "\n",
+      "#Result\n",
+      "print(\"Anode current = %d * 10^-3 A\"%(Ia*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Anode current = 64 * 10^-3 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.2, Page No.41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Equivalent capacitor\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "dv_by_dt = 190                # in Volt per micro-sec\n",
+      "Ic= 8 *10**-3                 # Capacitive current\n",
+      "\n",
+      "#Calculation\n",
+      "C = Ic/(dv_by_dt*10**6)\n",
+      "\n",
+      "#Result\n",
+      "print(\"C = %.1f * 10^12 F\"%(C*10**12))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C = 42.1 * 10^12 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.3, Page No. 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Thyristor\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "Vt = 0.75               # Thyristor trigger voltage\n",
+      "It = 7*10**-3           # Thyristor trigger current\n",
+      "Vcc = 20                # Suppy voltage, given data\n",
+      "Rg = 2000               # gate resistor, given data\n",
+      "R = 200                 # input resistor\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "V0 = 20                 # when thyristor is not conducting, there is no current through it.\n",
+      "#(b)\n",
+      "Vs = Vt+It*Rg\n",
+      "#(c)\n",
+      "i= 5*10**-3              #to have holding curernt\n",
+      "v1= i*R\n",
+      "#(d)\n",
+      "drop =0.7                # voltage drop across thyristor\n",
+      "v2=v1+drop\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) V0 = %d\\n(b) Vs = %.2f\\n(c) Vcc should be reduced to less than %d V if thyristor is ideal.\\n(d) Vcc should be reduced to less than %.1f V\"%(V0,Vs,v1,v2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) V0 = 20\n",
+        "(b) Vs = 14.75\n",
+        "(c) Vcc should be reduced to less than 1 V if thyristor is ideal.\n",
+        "(d) Vcc should be reduced to less than 1.7 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.4, Page No.42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculation of Vg,Ig and Rg\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vs = 25                      # gate signal amplitude\n",
+      "Pavg_loss = 0.6              # Avg power loss\n",
+      "\n",
+      "#Calculations\n",
+      "P_loss = Pavg_loss*2*math.pi/math.pi\n",
+      "Ig =(-1+(math.sqrt(1**2-4*9*(-1.2))))/(2*9) #alfa=(-b+sqrt(b^2-4ac))/2a-->solution for quadratic equation\n",
+      "Ig = math.ceil(Ig*1000)/1000\n",
+      "Vg = 1+9*Ig                                 #given data\n",
+      "Rg = (24-9*Ig)/Ig\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"Vg = %.3f \\nIg = %.3f\\nRg = %.2f\"%(Vg,Ig,Rg))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vg = 3.826 \n",
+        "Ig = 0.314\n",
+        "Rg = 67.43\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.5, Page No.42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Minimum width of gate pulse\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 100                   # DC supply voltage\n",
+      "L = 10                    # Inductance\n",
+      "i = 80*10**-3             # latching current of thyristor\n",
+      "\n",
+      "#Calculations\n",
+      "t = L*i/V\n",
+      "\n",
+      "#Result\n",
+      "print(\"Width of pulse should be more than %d milli-second.\"%(t*10**3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Width of pulse should be more than 8 milli-second.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.6, Page No. 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Minimum width of gate pulse\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 100                   # DC supply voltage\n",
+      "R = 10                    # Resistance\n",
+      "L = 5                     # Inductance\n",
+      "i = 50*10**-3             # latching current of thyristor\n",
+      "\n",
+      "#Calculation\n",
+      "t = math.log((1-(i*R)/V))/((-R/L)*math.log(math.e))        # i = (V/R)*(1-e^(-R*t/L))\n",
+      "\n",
+      "#Result\n",
+      "print(\"Minimum width of gate pulse is %.1f milli-second.\"%(t*10**3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum width of gate pulse is 2.5 milli-second.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.7, Page No. 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# resistor value\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 90.0                # DC supply voltage\n",
+      "i = 40*10**-3           # latching current of thyristor\n",
+      "t = 40* 10**-6          # pulse width\n",
+      "R = 25.0                # load Resistance\n",
+      "L = 0.5                 # Inductance\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "i2 = (V/R)*(1-math.e**(-R*t/L))\n",
+      "\n",
+      "#(b)\n",
+      "R = V/(i-i2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nAt the end of the gate pulse, i = %.4f A\"%i2)\n",
+      "print(\"Since the current is less than latching current, thyristor will not turn on.\")\n",
+      "print(\"\\n(b)\\nR should be less than %d ohm.\"%math.ceil(R))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "At the end of the gate pulse, i = 0.0072 A\n",
+        "Since the current is less than latching current, thyristor will not turn on.\n",
+        "\n",
+        "(b)\n",
+        "R should be less than 2744 ohm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.8, Page No.44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Resistor value\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 100.0                # DC supply voltage\n",
+      "i = 50*10**-3            # holding current of thyristor\n",
+      "t = 50* 10**-6           # pulse width\n",
+      "R = 20.0                 # load Resistance\n",
+      "L = 0.5                  # Inductance\n",
+      "\n",
+      "#Calculations\n",
+      "i2 = (V/R)*(1-math.e**(-R*t/L))\n",
+      "Rmax = V/(i-i2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"When firing pulse ends, i = %.2f mA\"%(i2*10**3))\n",
+      "print(\"Since this current is less than holding current, the thyristor will not remain on and return to off state.\")\n",
+      "print(\"Maximum value of R = %.3f ohm\"%(math.floor(Rmax*1000)/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When firing pulse ends, i = 9.99 mA\n",
+        "Since this current is less than holding current, the thyristor will not remain on and return to off state.\n",
+        "Maximum value of R = 2499.375 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.9, Page No.44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# triggering angle \n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 240                  # AC supply voltage\n",
+      "f = 50                   # supply frequency\n",
+      "R = 5                    # load resistance\n",
+      "L = 0.05                 # inductance\n",
+      "\n",
+      "#Calculation\n",
+      "theta = math.atan(2*math.pi*f*L/R)\n",
+      "theta = theta*180/math.pi\n",
+      "fi = theta+90\n",
+      "print(\"theta = %.2f\u00b0\"%(theta))\n",
+      "print(\"\\nCurrent transient is maximun(worst) if\\nsin(fi-theta) = 1\")\n",
+      "print(\"therefore, fi-%.2f\u00b0 = 90\u00b0\"%theta)\n",
+      "print(\"fi = %.2f\u00b0\"%fi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "theta = 72.34\u00b0)\n",
+        "\n",
+        "Current transient is maximun(worst) if\n",
+        "sin(fi-theta) = 1\n",
+        "therefore, fi-72.34\u00b0 = 90\u00b0\n",
+        "fi = 162.34\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.10, Page No.47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example \")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example \n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.11, Page No.50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example \")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example \n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.12, Page No. 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# rms current and form factor\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "I = 120                 # Current in Ampere\n",
+      "gamma = 180.0           # in degrees, thyristor conducts between alfa and alfa+gamma in each 360\u00b0 \n",
+      "\n",
+      "# calculations\n",
+      "#(a)-- formulas\n",
+      "#(b)\n",
+      "Irms = I*math.sqrt(gamma/360)\n",
+      "Iavg = I*(gamma/360)\n",
+      "ff = Irms/Iavg\n",
+      "\n",
+      "#result\n",
+      "print(\"RMS curent = %.2f A\\nAverage Current = %.0f A\\nForm factor = %.3f\"%(Irms,Iavg,ff))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RMS curent = 84.85 A\n",
+        "Average Current = 60 A\n",
+        "Form factor = 1.414\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.13, Page No.53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Power supplied to the load and average load current\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "# variable declaration\n",
+      "R = 100.0                       # load resistance\n",
+      "V = 230.0                       # Supply Voltage\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "def f(x):\n",
+      "    return math.sin(x)**2\n",
+      "wt_lower=math.pi/3\n",
+      "wt_upper =math.pi\n",
+      "val = quad(f,wt_lower,wt_upper)\n",
+      "I = ((math.sqrt(2)*V/R)**2)*val[0]/(2*math.pi)\n",
+      "Irms = math.sqrt(I)\n",
+      "P =(Irms**2)*R\n",
+      "\n",
+      "#(b)\n",
+      "def f(x):\n",
+      "    return math.sin(x)**2\n",
+      "wt_lower2=math.pi/4\n",
+      "wt_upper2 =math.pi\n",
+      "val2 = quad(f,wt_lower2,wt_upper2)\n",
+      "x = (math.sqrt(2)*V/R)\n",
+      "x = math.floor(x*100)/100\n",
+      "I2 = (x**2)*val2[0]/(2*math.pi)\n",
+      "Irms2 = math.sqrt(I2)\n",
+      "P2 =(Irms2**2)*R\n",
+      "\n",
+      "#(c)\n",
+      "def f(x):\n",
+      "    return (3.25/(2*math.pi))*math.sin(x)\n",
+      "wt_lower3=math.pi/3\n",
+      "wt_upper3 =math.pi\n",
+      "val3 = quad(f,wt_lower3,wt_upper3)\n",
+      "wt_lower4=math.pi/4\n",
+      "wt_upper4 =math.pi\n",
+      "val4 = quad(f,wt_lower4,wt_upper4)\n",
+      "\n",
+      "\n",
+      "print(\"(a)\\nRMS current = %.2f A\\nPower supplied to load = %d W\"%(Irms,math.ceil(P)))\n",
+      "print(\"\\n\\n(b)\\nRMS current = %f A\\nPower supplied to load = %f W\"%(Irms2,P2))\n",
+      "print(\"\\n(c)\\nWhen firing angle is 60\u00b0, Average current = %.3f A\" %val3[0])\n",
+      "print(\"When firing angle is 45\u00b0, Average current = %.3f A\" %val4[0])\n",
+      "# for (b) answer matches to the book if val2[0] = 1.4255"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "RMS current = 1.46 A\n",
+        "Power supplied to load = 213 W\n",
+        "\n",
+        "\n",
+        "(b)\n",
+        "RMS current = 1.549431 A\n",
+        "Power supplied to load = 240.073727 W\n",
+        "\n",
+        "(c)\n",
+        "When firing angle is 60\u00b0, Average current = 0.776 A\n",
+        "When firing angle is 45\u00b0, Average current = 0.883 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 81
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.14, Page No.54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# average power loss in thyristor\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "Iavg = 200                  # Average current\n",
+      "v1 = 1.8                    # voltage drop across thyristor for 200A current\n",
+      "v2 = 1.9                    # voltage drop across thyristor for 400A current\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "A1 = Iavg                   # amplitude of rectangular current wave\n",
+      "P1 = A1*v1\n",
+      "#(b)\n",
+      "A2 = 2*Iavg\n",
+      "P2 = A2*v2*math.pi/(2*math.pi)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Average power loss = %d W\"%P1)\n",
+      "print(\"(b) Average power loss = %d W\"%P2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Average power loss = 360 W\n",
+        "(b) Average power loss = 380 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 83
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.15, Page No.55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# average on state current\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "I = 40                # rms on state current\n",
+      "f = 50                # frequency\n",
+      "cp_a =170             # conduction period\n",
+      "cp_b =100             # conduction period\n",
+      "cp_c =40              # conduction period\n",
+      "\n",
+      "#Calculations\n",
+      "alfa_a = 180-cp_a\n",
+      "alfa_b = 180-cp_b\n",
+      "alfa_c = 180-cp_c\n",
+      "Im_a = math.sqrt(((I**2)*2*math.pi)/(((math.pi-alfa_a*math.pi/180)/2)+math.sin(2*alfa_a*math.pi/180)/4))\n",
+      "Iv_a = 0.316*Im_a\n",
+      "Im_b = math.sqrt(((I**2)*2*math.pi)/(((math.pi-alfa_b*math.pi/180)/2)+math.sin(2*alfa_b*math.pi/180)/4))\n",
+      "Iv_b = 0.1868*Im_b\n",
+      "Im_c = math.sqrt(((I**2)*2*math.pi)/(((math.pi-alfa_c*math.pi/180)/2)+math.sin(2*alfa_c*math.pi/180)/4))\n",
+      "Iv_c = 0.0372*Im_c\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Iavg = %.3f A\"%(math.ceil(Iv_a*1000)/1000))\n",
+      "print(\"(b) Iavg = %.2f A\"%Iv_b)\n",
+      "print(\"(c) Iavg = %.2f A\"%(math.floor(Iv_c*100)/100))\n",
+      "#answer for(b) is not matching with the answer given in the book."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Iavg = 25.295 A\n",
+        "(b) Iavg = 19.13 A\n",
+        "(c) Iavg = 11.62 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 105
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.16, page No. 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# power dissiopation\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#variable declaration\n",
+      "It = 20                     # constant current\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Vt = 0.9+ 0.02*It\n",
+      "P1 = Vt*It\n",
+      "#(b)\n",
+      "def f(x):\n",
+      "    return (0.9+0.02*(20*math.pi*math.sin(x)))*(20*math.pi*math.sin(x))/(2*math.pi)\n",
+      "# if denominator math.pi value is taken as 3.14, then answer for P2 =37.75\n",
+      "theta_lower = 0\n",
+      "theta_upper = math.pi\n",
+      "val = quad(f,theta_lower,theta_upper)\n",
+      "P2 = val[0]\n",
+      "#(c)\n",
+      "P3 = P1/2\n",
+      "#(d)\n",
+      "P4 = P1/3\n",
+      "\n",
+      "#result\n",
+      "print(\"(a) Power dissipation = %d W\"%P1)\n",
+      "print(\"(b) Mean Power dissipation = %.2f W\"%P2)\n",
+      "print(\"(c) Mean Power dissipation = %d W\"%P3)\n",
+      "print(\"(d) Mean Power dissipation = %.2f W\"%P4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Power dissipation = 26 W\n",
+        "(b) Mean Power dissipation = 37.74 W\n",
+        "(c) Mean Power dissipation = 13 W\n",
+        "(d) Mean Power dissipation = 8.67 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 115
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.17, Page No. 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# I^2t rating \n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "Is = 2000.0              # half cycle surge current rating for SCR\n",
+      "f = 50.0                 # operating AC frequency\n",
+      "\n",
+      "#Calculation\n",
+      "T = 1/f\n",
+      "t_half = T/2\n",
+      "t = t_half/2\n",
+      "I = math.sqrt((Is**2)*t/t_half)\n",
+      "rating = (I**2)*t_half\n",
+      "\n",
+      "#Result\n",
+      "print(\"I^2t rating = %d A^2 secs\"%rating)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "I^2t rating = 20000 A^2 secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 138
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.18, Page No. 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Resistance and average power loss\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "P = 6                #peak power loss\n",
+      "d = 0.3              # duty cylce   \n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "#solution of equation 9Ig^2+Ig-6 = 0\n",
+      "Ig =(-1+(math.sqrt(1**2-4*9*(-6))))/(2*9)    #alfa=(-b+sqrt(b^2-4ac))/2a-->solution for quadratic equation\n",
+      "Ig = math.ceil(Ig*1000)/1000\n",
+      "Rg = (11/Ig)-9                               #from KVL equation of gate circuit\n",
+      "#(b)\n",
+      "Pavg = P*d\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nIg = %.3f A \\nRg = %.3f ohm\"%(Ig,Rg))\n",
+      "print(\"\\n(b) Average power loss = %.1f W \"%Pavg)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Ig = 0.763 A \n",
+        "Rg = 5.417 ohm\n",
+        "\n",
+        "(b) Average power loss = 1.8 W \n"
+       ]
+      }
+     ],
+     "prompt_number": 144
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.19, Page No. 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# trigger current and voltage for gate power dissipation\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vs = 12                  # supply voltage\n",
+      "P = 0.3                  # Power dissipation\n",
+      "Rs  = 100                # load line slope i.e. source resistance\n",
+      "\n",
+      "#Calculation\n",
+      "#solution for equation 100Ig^2-12Ig+0.3 = 0\n",
+      "Ig =(-(-Vs)-(math.sqrt((-Vs)**2-4*Rs*(P))))/(2*Rs)    #alfa=(-b(+/-)sqrt(b^2-4ac))/2a-->solution for quadratic equation\n",
+      "Vg = P/Ig\n",
+      "print(\"Ig = %.1f mA\\nVg = %.2f V\"%(Ig*1000,Vg))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ig = 35.5 mA\n",
+        "Vg = 8.45 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 150
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.20, Page No. 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Finding values of resistors\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Es = 12.0             # Supply voltage\n",
+      "V = 800.0             # SCR voltage rating\n",
+      "I = 110.0             # SCR currert rating\n",
+      "imax = 250*10**-3     # Maximum permissible current through battery\n",
+      "isc = 600*10**-3      # short circuit current of the battery\n",
+      "Vg = 2.4              # required gate voltage\n",
+      "Ig = 50*10**-3        # required gate current\n",
+      "Vgmax = 3             # maximum gate voltage\n",
+      "Igmax = 100*10**-3    # maximum gate current\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "Rs = Es/isc\n",
+      "#considering R2 is not connected\n",
+      "R1 =(Es/imax)-Rs\n",
+      "#R1 must be 28 or more so that battery curenrt doesnot exceeds 250mA\n",
+      "R1min = Es/Igmax-Rs\n",
+      "R1max = ((Es-Vg)/Ig)-Rs\n",
+      "R1 = 125  #selected \n",
+      "R2 = Vgmax*(Rs+R1)/(Es-Vgmax)\n",
+      "\n",
+      "# Result\n",
+      "print(\"Rs = %d ohm \\n\\nR1 must be more than %.0f ohm and must be less than %.0f ohm.\\nTherefore, select R1 = %.0f \"%(Rs,R1min,R1max,R1))\n",
+      "print(\"\\nR2 should be less than %.3f ohm.\\nTherefore, select R2 = %.0f ohm\"%(R2,math.floor(R2)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rs = 20 ohm \n",
+        "\n",
+        "R1 must be more than 100 ohm and must be less than 172 ohm.\n",
+        "Therefore, select R1 = 125 \n",
+        "\n",
+        "R2 should be less than 48.333 ohm.\n",
+        "Therefore, select R2 = 48 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.21, Page No. 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Surface temperature\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "l = 0.2                 # length of aluminium rod\n",
+      "w = 0.01                # width of aluminium rod\n",
+      "d = 0.01                # depth of aluminium rod\n",
+      "tc = 220                # thermal conductivity\n",
+      "T1 = 30                 # temperature at far end\n",
+      "P = 3                   # injected heat\n",
+      "\n",
+      "#Calculation\n",
+      "theta = l/(tc*w*d)\n",
+      "T2 = P*theta+T1\n",
+      "\n",
+      "#Result\n",
+      "print(\"Temperature of the surface where heat is injected is %.2f\u00b0C\"%T2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature of the surface where heat is injected is 57.27\u00b0C\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.22, Page No. 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Power losses\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "l = 2*10**-3            # lengthh of aluminium plate\n",
+      "a = 12*10**-4           # cross-section area\n",
+      "tc = 220                # thermal conductivity\n",
+      "td = 4                  # desired thermal drop along length\n",
+      "\n",
+      "#Calculations\n",
+      "theta = l/(a*tc)\n",
+      "theta = math.floor(theta*10**6)/10**6\n",
+      "P_loss = td/theta\n",
+      "\n",
+      "#Result\n",
+      "print(\"power losses = %.2f W\"%P_loss)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power losses = 528.05 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.23, Page No. 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#thermal resistance of heat sink\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "P = 30.0              # power dissipation\n",
+      "T1 =125.0             # max junction temperature\n",
+      "T2 = 50.0             # max ambient temperature\n",
+      "tr1 =1.0              # thermal resistance of thyristor\n",
+      "tr2 = 0.3             # therrmal resistance of insuulator\n",
+      "\n",
+      "#calculations\n",
+      "Tr = (T1-T2)/P\n",
+      "Tr_hs = Tr-tr1-tr2\n",
+      "\n",
+      "#Result\n",
+      "print(\"Thermal resistance of heat sink = %.1f\u00b0C/W\"%Tr_hs)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thermal resistance of heat sink = 1.2\u00b0C/W\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.24, Page No.68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Junction temperature\n",
+      "\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "tr = 0.627              # thermal resistance of device\n",
+      "T = 93                  # heat sink temperature\n",
+      "V = 3                   # voltage at junction\n",
+      "I = 25                  # current at junction\n",
+      "\n",
+      "#calculation\n",
+      "T_junction = tr*V*I+T\n",
+      "\n",
+      "#Result\n",
+      "print(\"Junction Temperature = %d\u00b0C\"%T_junction)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Junction Temperature = 140\u00b0C\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.25, Page No. 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Thermal resistance\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "P = 40.0                #steady power loss in an SCR\n",
+      "Tr_hs = 0.8             # heat sink resistance\n",
+      "T1 = 120.0              # Maximumm junction temperature\n",
+      "T2 = 35.0               # ambient temperature\n",
+      "\n",
+      "#calculation\n",
+      "Tr = (T1-T2)/P\n",
+      "Rsh = Tr - Tr_hs\n",
+      "\n",
+      "#Result\n",
+      "print(\"Resistance of heat sink = %.3f\u00b0C/W\"%Rsh)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance of heat sink = 1.325\u00b0C/W\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.26, Page No. 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# power loss and % increase in rating\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "Tj = 125                # maximum junction temperature\n",
+      "Ts1 = 80                # max heat sink temperature\n",
+      "tr_jc = 0.7             # thermal resistance from junction to case  \n",
+      "tr_cs = 0.4             # thermal resistance from case to sink\n",
+      "Ts2 = 50                # decreased heat sink temperature  \n",
+      "#calculations\n",
+      "#(a)\n",
+      "Pav1 = (Tj-Ts1)/(tr_jc+tr_cs)\n",
+      "#(b)\n",
+      "Pav2 =(Tj-Ts2)/(tr_jc+tr_cs)\n",
+      "rating = 100*(math.sqrt(Pav2)-math.sqrt(Pav1))/math.sqrt(Pav1)\n",
+      "rating = math.floor(rating*100)/100\n",
+      "# result\n",
+      "print(\"(a)\\nAverage power loss = %.2fW\"%Pav1)\n",
+      "print(\"\\n(b)\\nAverage power loss = %.2fW\\n%% increase in rating = %.2f%%\"%(Pav2,rating))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Average power loss = 40.91W\n",
+        "\n",
+        "(b)\n",
+        "Average power loss = 68.18W\n",
+        "% increase in rating = 29.09%\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.27, Page No. 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Minnimum and maximum firing angle of triac\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vs = 230.0                 # Supply voltage\n",
+      "f = 50.0                   # supply frequency\n",
+      "Vc = 25.0                  # breakdown voltage\n",
+      "C = 0.6*10**-6             # capacitance\n",
+      "Rmin = 2000                # minimum resistance  \n",
+      "Rmax = 20000                # maximum resistance\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "pi = math.ceil(math.pi*10**4)/10**4\n",
+      "Xc = 1/(2*pi*f*C)\n",
+      "Xc = math.floor(Xc*100)/100\n",
+      "Z1_m = math.sqrt(Rmin**2+Xc**2)\n",
+      "Z1_angle=math.atan(Xc/Rmin)\n",
+      "I1_m = Vs/Z1_m\n",
+      "I1_m = math.ceil(I1_m*10**5)/10**5\n",
+      "I1_angle = Z1_angle\n",
+      "Vc_m = I1_m*Xc\n",
+      "Vc_m = math.ceil(Vc_m*100)/100\n",
+      "Vc_angle = I1_angle*(180/math.pi)-90\n",
+      "alfa1 = (180/math.pi)*math.asin(Vc/(math.sqrt(2)*Vc_m))+(-Vc_angle)\n",
+      "\n",
+      "#(b)\n",
+      "Z2_m = math.sqrt(Rmax**2+Xc**2)\n",
+      "Z2_m = math.floor(Z2_m*10)/10\n",
+      "Z2_angle=math.atan(Xc/Rmax)\n",
+      "I2_m = Vs/Z2_m\n",
+      "I2_m = math.floor(I2_m*10**6)/10**6\n",
+      "I2_angle = Z2_angle\n",
+      "V2c_m = I2_m*Xc\n",
+      "V2c_m = math.ceil(V2c_m*100)/100\n",
+      "V2c_angle = I2_angle*(180/math.pi)-90\n",
+      "alfa2 = (180/math.pi)*math.asin(Vc/(math.sqrt(2)*V2c_m))+(-V2c_angle)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Xc = %.2f ohms\"%Xc)\n",
+      "print(\"\\n(a) When R = %d ohms\\n Z(magnitude) = %.2f\\t\\tZ(angle) = %.2f\u00b0\"%(Rmin,Z1_m,-Z1_angle*(180/math.pi)))\n",
+      "print(\" I(magnitude) = %.5f\\t\\tI(angle) = %.2f\u00b0\"%(I1_m,I1_angle*(180/math.pi)))\n",
+      "print(\"\\nVoltage across capacitor,\\nV(magnitude) = %.2f\\t\\tV(angle) = %.2f\u00b0\"%(Vc_m,Vc_angle))\n",
+      "print(\"\\nMinimum firing angle = %.2f\u00b0\"%alfa1)\n",
+      "\n",
+      "print(\"\\n(b) When R = %d ohms\\n Z(magnitude) = %.2f\\t\\tZ(angle) = %.2f\u00b0\"%(Rmax,Z2_m,-Z2_angle*(180/math.pi)))\n",
+      "print(\" I(magnitude) = %f\\t\\tI(angle) = %.2f\u00b0\"%(I2_m,I2_angle*(180/math.pi)))\n",
+      "print(\"\\nVoltage across capacitor,\\nV(magnitude) = %.2f\\t\\tV(angle) = %.2f\u00b0\"%(V2c_m,V2c_angle))\n",
+      "print(\"\\nMaximum firing angle = %.2f\u00b0\"%(math.ceil(alfa2*10)/10))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Xc = 5305.15 ohms\n",
+        "\n",
+        "(a) When R = 2000 ohms\n",
+        " Z(magnitude) = 5669.62\t\tZ(angle) = -69.34\u00b0\n",
+        " I(magnitude) = 0.04057\t\tI(angle) = 69.34\u00b0\n",
+        "\n",
+        "Voltage across capacitor,\n",
+        "V(magnitude) = 215.23\t\tV(angle) = -20.66\u00b0\n",
+        "\n",
+        "Minimum firing angle = 25.37\u00b0\n",
+        "\n",
+        "(b) When R = 20000 ohms\n",
+        " Z(magnitude) = 20691.60\t\tZ(angle) = -14.86\u00b0\n",
+        " I(magnitude) = 0.011115\t\tI(angle) = 14.86\u00b0\n",
+        "\n",
+        "Voltage across capacitor,\n",
+        "V(magnitude) = 58.97\t\tV(angle) = -75.14\u00b0\n",
+        "\n",
+        "Maximum firing angle = 92.60\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 88
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.28, Page no. 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# UJT relaxation circuit\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 32.0                 #voltage\n",
+      "h = 0.63                 # voltage gain\n",
+      "Ip = 10*10**-6           # current \n",
+      "Vv = 3.5                 # voltage drop\n",
+      "Iv = 10*10**-3           # current through the circuit\n",
+      "Vf = 0.5                 # forward voltage drop across p-n junction\n",
+      "f = 50.0                 # frequency of ascillation \n",
+      "t = 50.0*10**-6          # width of triggering pulse(given value is 50ms but used value is 50 micro-sec) \n",
+      "C = 0.4*10**-6           # Capacitance(assumed)\n",
+      "\n",
+      "#Calculations\n",
+      "T = 1/f\n",
+      "Vp = V*h+Vf\n",
+      "Rmax = (V-Vp)/Ip         # R should be less than the given value\n",
+      "Rmin = (V-Vv)/Iv         # R should be more than the given value\n",
+      "R = T/(C*math.log(1/(1-h)))\n",
+      "R4 = t/C\n",
+      "R3 = 10**4/(h*V)\n",
+      "\n",
+      "#Result\n",
+      "print(\"The value of R = %.2f*10^3 ohm is within the Rmax = %.3f*10^6 ohm and Rmin = %.2f*10^3 ohm\"%(R*10**-3,Rmax*10**-6,Rmin*10**-3))\n",
+      "print(\"Hence the value is suitable.\\n\\nR4 = %d ohm \\nR3 = %d ohm\"%(R4,R3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R = 50.29*10^3 ohm is within the Rmax = 1.134*10^6 ohm and Rmin = 2.85*10^3 ohm\n",
+        "Hence the value is suitable.\n",
+        "\n",
+        "R4 = 125 ohm \n",
+        "R3 = 496 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 101
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.29, Page No. 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#values of various components of circuit(referring to fig.2.36(a))\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "f = 2.0*10**3               # ouutput frequency\n",
+      "Vdc = 10                    # input voltage\n",
+      "h = 0.6                     # voltage gain\n",
+      "I = 0.005                   # peak discharge current\n",
+      "v = 0.5                     # voltage drop across p-n junction(assumed)\n",
+      "\n",
+      "#Calculation\n",
+      "Vp = (Vdc*h)+v\n",
+      "R = (Vdc-Vp)/I              # R should be less than this value\n",
+      "T = 1.0/f\n",
+      "C1 = 0.5*10**-6             # capacitance(assumed)\n",
+      "R1 = (T)/(C1*math.log(1/(1-h)))\n",
+      "\n",
+      "C2 = 1.0*10**-6             # capacitance(assumed)\n",
+      "R2 = (T)/(C2*math.log(1/(1-h)))\n",
+      "\n",
+      "#Result\n",
+      "print(\"for C = %.1f micro-F, R = %d Ohm. \\nThis value of R is not suitable because R must be less than %d ohm.\"%(C1*10**6,R1,R))\n",
+      "print(\"\\nfor C = %.1f micro-F, R = %.1f Ohm.\\nThis value is suitable because R is less than %d ohm\"%(C2*10**6,R2,R))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for C = 0.5 micro-F, R = 1091 Ohm. \n",
+        "This value of R is not suitable because R must be less than 700 ohm.\n",
+        "\n",
+        "for C = 1.0 micro-F, R = 545.7 Ohm.\n",
+        "This value is suitable because R is less than 700 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 112
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.30, Page No. 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Range of firing angle\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vs = 230.0             # supply voltage\n",
+      "Rmin= 1000.0           # minimum value of R1 resistor\n",
+      "Rmax = 22*10**3        # maximum value of R1 resistor\n",
+      "Vg = 2.0               # gate triggering voltage\n",
+      "C = 0.47*10**-6        # capacitor value\n",
+      "f = 50.0               # frequency\n",
+      "#Calculations\n",
+      "# For Rmin\n",
+      "theta = (180/math.pi)*math.atan(2*math.pi*f*C*Rmin)\n",
+      "theta = math.ceil(theta*10)/10\n",
+      "Vc = Vs*math.cos(theta*math.pi/180)\n",
+      "Vc = math.floor(Vc*100)/100\n",
+      "alfa = ((180/math.pi)*math.asin(Vg/(math.sqrt(2)*Vc)))+theta\n",
+      "alfa = math.floor(alfa*10)/10\n",
+      "\n",
+      "# for Rmax\n",
+      "theta2 = (180/math.pi)*math.atan(2*math.pi*f*C*Rmax)\n",
+      "theta2 = math.ceil(theta2*10)/10\n",
+      "Vc2 = Vs*math.cos(theta2*math.pi/180)\n",
+      "Vc2 = math.floor(Vc2*100)/100\n",
+      "alfa2 = ((180/math.pi)*math.asin(Vg/(math.sqrt(2)*Vc2)))+theta2\n",
+      "alfa2 = math.ceil(alfa2*10)/10\n",
+      "\n",
+      "#Result\n",
+      "print(\"Minimum firing angle = %.1f\u00b0\"%alfa)\n",
+      "print(\"Maximum firing angle = %.1f\u00b0\"%alfa2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum firing angle = 8.7\u00b0\n",
+        "Maximum firing angle = 74.1\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 137
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.31, Page No. 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# UJT relaxation oscillator design(reffering to fig.2.40)\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "f = 50.0                   # frequency of supply\n",
+      "Vbb = 12.0                 # typical input vpltage\n",
+      "P = 300*10**3              # maximumm average power dissipation\n",
+      "Rbb = 5.6*10**3            # base resistor\n",
+      "Iv = 4*10**-3              # Valley point voltage\n",
+      "h = 0.63                   # intrinsic stand off ratio\n",
+      "Vd = 0.5                   # drop across p-n junction\n",
+      "Vv = 2.0                   # Valley point voltage\n",
+      "Ip = 5*10**-6              # peak point current\n",
+      "v = 0.18                   # minimum voltage for which SCR will not trigger\n",
+      "\n",
+      "#calculation\n",
+      "T = 1/f\n",
+      "#operating point should be on the left of valley point\n",
+      "R_1 = (Vbb-Vv)/Iv             # R should be greater than this value\n",
+      "Vp = (h*Vbb)+Vd\n",
+      "#operation point should lie in the negative resistance region of characteristics of UJT\n",
+      "R_2 = (Vbb-Vp)/Ip              # R should be less than this value\n",
+      "C = 0.02*10**-6               # assumed\n",
+      "R_3 = T/(C*math.log(1/(1-h)))\n",
+      "\n",
+      "C2 = 0.04*10**-6               # assumed\n",
+      "R_4 = T/(C2*math.log(1/(1-h)))\n",
+      "R = 500*10**3                  # assumed\n",
+      "Rmax = R+R_1\n",
+      "R3_1 = (10**4)/(h*Vbb)\n",
+      "R3 = 1200                      # assumed\n",
+      "R4_1 = v*(R3+Rbb)/(Vbb-v)      # R4 value should be less than this value\n",
+      "R4 = 100                       # assumed  \n",
+      "tw = R4*C2\n",
+      "tow_min = R_1*C2*math.log(1/(1-h))\n",
+      "alfa_min = tow_min*(360/T)\n",
+      "tow_max = Rmax*C2*math.log(1/(1-h))\n",
+      "alfa_max = tow_max*(360/T)\n",
+      "\n",
+      "#Result\n",
+      "print(\"For C = %.2f micro F, R = %.2f k-ohm\\nThis value of R is more than maximum of %d k-ohm.\"%(C*10**6,R_3/1000,R_2/1000) )\n",
+      "print(\"Hence not suitable.\")\n",
+      "print(\"\\nFor C = %.2f micro F, R = %.2f K-ohm.\"%(C2*10**6,R_4/1000))\n",
+      "print(\"As this is less than max value. It is suitable. \\nLets assume R = %d k-ohm\"%(R/1000))\n",
+      "print(\"Rmax = %.1f k-ohm\"%(Rmax/1000))\n",
+      "print(\"\\nR3 = %.2f k-ohm. so take nearest value R3 = %.1f k-ohm\"%(R3_1/1000,R3/1000.0))\n",
+      "print(\"\\nR4 = %.2f ohm. so take nearest value R4 = %.1f ohm\"%(R4_1,R4))\n",
+      "print(\"\\nMinimum firing angle = %.2f\u00b0\"%alfa_min)\n",
+      "print(\"Maximum firing angle = %.1f\u00b0\"%alfa_max)\n",
+      "print(\"\\nThough maximum firing angle available is %.1f\u00b0, it will not be needed\"%alfa_max)\n",
+      "print(\"because in a single phase full wave converter the maximum firing angle is 180\u00b0\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For C = 0.02 micro F, R = 1005.78 k-ohm\n",
+        "This value of R is more than maximum of 788 k-ohm.\n",
+        "Hence not suitable.\n",
+        "\n",
+        "For C = 0.04 micro F, R = 502.89 K-ohm.\n",
+        "As this is less than max value. It is suitable. \n",
+        "Lets assume R = 500 k-ohm\n",
+        "Rmax = 502.5 k-ohm\n",
+        "\n",
+        "R3 = 1.32 k-ohm. so take nearest value R3 = 1.2 k-ohm\n",
+        "\n",
+        "R4 = 103.55 ohm. so take nearest value R4 = 100.0 ohm\n",
+        "\n",
+        "Minimum firing angle = 1.79\u00b0\n",
+        "Maximum firing angle = 359.7\u00b0\n",
+        "\n",
+        "Though maximum firing angle available is 359.7\u00b0, it will not be needed\n",
+        "because in a single phase full wave converter the maximum firing angle is 180\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 166
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.32, Page No.92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# conduction time of thyristor\n",
+      "\n",
+      "import math\n",
+      "#variable declaratrion\n",
+      "R =0.8                   # load resistance\n",
+      "L = 10*10**-6            # inductance\n",
+      "C = 50*10**-6            # capacitance\n",
+      "\n",
+      "#calculations\n",
+      "R1 = math.sqrt(4*L/C)      # R should be less than this value\n",
+      "t = math.pi/(math.sqrt((1/(L*C))-((R**2)/(4*L**2))))\n",
+      "\n",
+      "#Result\n",
+      "print(\"Value of sqrt(4L/C)  is %.4f and give R = %.1f\\nTherefore, the circuit is underdamped.\"%(R1,R))\n",
+      "print(\"\\nt0 = %.2f*10**-6 seconds\"%(t*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of sqrt(4L/C)  is 0.8944 and give R = 0.8\n",
+        "Therefore, the circuit is underdamped.\n",
+        "\n",
+        "t0 = 157.08*10**-6 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 169
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.33, Page No. 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Values of L and C\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "I = 8.0                  # load current\n",
+      "V = 90.0                 # Supply voltage\n",
+      "t = 40.0*10**-6          # turn off time\n",
+      "\n",
+      "#calculation\n",
+      "#print(\"Assume peak value of capacitor current to be twice the load cuuren.\")\n",
+      "C_by_L = (2*I/V)**2         # using eq 2.33\n",
+      "#print(\"Assume that thyristor is reverse biased for one-fourth period of resonant circuit.\")\n",
+      "CL = (t*2/math.pi)**2\n",
+      "# (2*I/V)^2*L^2 =(t*2/math.pi)^2\n",
+      "\n",
+      "L = math.sqrt(CL/C_by_L)\n",
+      "C =C_by_L*L\n",
+      "\n",
+      "print(\"L = %.2f*10**-6 H\\nC = %.3f*10**-6 F\"%(L*10**6,C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L = 143.24*10**-6 H\n",
+        "C = 4.527*10**-6 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 173
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.34, Page No. 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Value of C for commutation\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "R = 10                    # load resistace\n",
+      "V = 100                   # supply voltage \n",
+      "t = 50 * 10**-6           # turn off time\n",
+      "\n",
+      "#Calculations\n",
+      "C = t/(R*math.log(2))\n",
+      "\n",
+      "#Result\n",
+      "print(\"C = %.2f*10^-6 F\"%(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C = 7.21*10^-6 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 176
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.35, Page No. 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# L and C of commutation circuit\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "V = 100                    # supply voltage\n",
+      "I = 40                     # maximumm load current\n",
+      "T = 40*10**-6              # turn off time\n",
+      "t = 1.5                    # extra 50% tolerance\n",
+      "\n",
+      "#Calculation\n",
+      "T = T*t\n",
+      "C = I*T/V\n",
+      "L = (V**2)*C/(I**2)\n",
+      "Ip = V*math.sqrt(C/L)\n",
+      "L2 = 2* 10**-4             # assume\n",
+      "Ip2 = V*math.sqrt(C/L2)\n",
+      "print(\"C = %.0f*10^-6 F\\nL = %.1f*10^-4 H\\nPeak capacitor current = %d A\"%(C*10**6,L*10**4,Ip))\n",
+      "print(\"\\nPeak capacitor current should be less than maximum load currentof %d A. If L is selected as %d *10^-4 H,\"%(I,L2*10**4))\n",
+      "print(\"the peak capacitor current will be %.1f A. Since this is less than %d A, this value of L is satisfactory.\"%(Ip2,I))\n",
+      "print(\"\\nHence,\\nC = %.0f*10^-6 F\\nL = %.1f*10^-4 H\"%(C*10**6,L2*10**4))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C = 24*10^-6 F\n",
+        "L = 1.5*10^-4 H\n",
+        "Peak capacitor current = 40 A\n",
+        "\n",
+        "Peak capacitor current should be less than maximum load currentof 40 A. If L is selected as 2 *10^-4 H,\n",
+        "the peak capacitor current will be 34.6 A. Since this is less than 40 A, this value of L is satisfactory.\n",
+        "\n",
+        "Hence,\n",
+        "C = 24*10^-6 F\n",
+        "L = 2.0*10^-4 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 187
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.36, Page No. 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# current rating of the thyristor\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vdc = 100.0               # input voltage\n",
+      "L = 0.1 *10**-3           # Inductance\n",
+      "C = 10*10**-6             # capacitance\n",
+      "Vc = 100.0                # iinitial voltage on capacitor\n",
+      "t = 25.0*10**-6           # thyristor turn offf time\n",
+      "I = 10.0                  # Thyristor load current\n",
+      "\n",
+      "#Calculations\n",
+      "t_off = Vc*C/I\n",
+      "Ic = Vdc*math.sqrt(C/L)\n",
+      "\n",
+      "# Result\n",
+      "print(\"t_off = %d *10^-6 S\\nt_off is more than thyristor turn off time and is thus sufficient to commutate the main circuit.\"%(t_off*10**6))\n",
+      "print(\"\\nPeak capacitor current = %.2fA \"%Ic)\n",
+      "print(\"Maximum current rating of thyristor should be more than %.2fA\"%Ic)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "t_off = 100 *10^-6 S\n",
+        "t_off is more than thyristor turn off time and is thus sufficient to commutate the main circuit.\n",
+        "\n",
+        "Peak capacitor current = 31.62A \n",
+        "Maximum current rating of thyristor should be more than 31.62A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.37, Page No. 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# value of R and C for snubber circuit\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "dv_by_dt = 25*10**6                 # Thyristor's dv/dt rating\n",
+      "L = 0.2 *10**-3                     # source inductance\n",
+      "V = 230                             # Supply voltage\n",
+      "df = 0.65                           # damping factor\n",
+      "\n",
+      "#Calculation\n",
+      "Vm = V*math.sqrt(2) \n",
+      "C = (1/(2*L))*(0.564*Vm/dv_by_dt)**2\n",
+      "R = 2*df*math.sqrt(L/C)\n",
+      "\n",
+      "#Result\n",
+      "print(\"C = %.2f*10^-9 F\\n\\nR = %.1f ohm\"%(C*10**9,R))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C = 134.62*10^-9 F\n",
+        "\n",
+        "R = 50.1 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.38, Page No.97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# snuubber circuit parameters\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 300.0                      # input voltage\n",
+      "Rl = 10.0                      # load resistance\n",
+      "f = 2000.0                     # operating frequency otf the circuit\n",
+      "dv_by_dt = 100.0*10**6         # required dv/dt  rating of the circuit\n",
+      "I = 100.0                      # discharge current\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "R = V/I\n",
+      "C = (1-(1/math.e))*Rl*V/(dv_by_dt*((R+Rl)**2))\n",
+      "C = math.floor(C*10**9)/10**9\n",
+      "#(b)\n",
+      "P = (C*V**2*f)/2\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nR = %f ohm\\nC = %.3f*10^-6 F\"%(R,C*10**6))\n",
+      "print(\"\\n(b)\\nSnubber power loss = %.2f W\"%P)\n",
+      "print(\"\\nAll the energy stored in capacitor C is dissipated in resistance R. Hence power rating of R is %.2f W.\"%P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "R = 3.000000 ohm\n",
+        "C = 0.112*10^-6 F\n",
+        "\n",
+        "(b)\n",
+        "Snubber power loss = 10.08 W\n",
+        "\n",
+        "All the energy stored in capacitor C is dissipated in resistance R. Hence power rating of R is 10.08 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.39, Page No.98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# maximum permissiible values of dv/dt and di/dt\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "R = 4.0              # resistance\n",
+      "L = 6.0*10**-6       # inductance\n",
+      "C = 6.0*10**-6       # capacitance\n",
+      "V = 300.0            # Supply voltage\n",
+      "\n",
+      "#calculations\n",
+      "di_dt = V/L\n",
+      "Isc = V/R\n",
+      "dv_dt =(R* di_dt)+(Isc/C)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Maximum di/dt = %.0f*10^6 A/s\\n\\nmaximum permissible dv/dt = %.1f*10^6 V/s\"%(di_dt*10**-6,dv_dt*10**-6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum di/dt = 50*10^6 A/s\n",
+        "\n",
+        "maximum permissible dv/dt = 212.5*10^6 V/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.40, Page No.98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# snubber circuit designe\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "\n",
+      "Rl = 8.0                 # load resistance\n",
+      "V = 230.0                # supply voltage\n",
+      "I = 200.0                # repititive peak current \n",
+      "di_dt = 40.0*10**6       # max. di/di rating\n",
+      "dv_dt = 150.0*10**6      # max. dv/dv rating\n",
+      "sf = 2.0                 # safety factor for thyristor\n",
+      "\n",
+      "#Calculation\n",
+      "\n",
+      "I1 = I/2.0\n",
+      "di_dt2 = di_dt/2.0\n",
+      "dv_dt2 = dv_dt/2\n",
+      "Vp = V*math.sqrt(2)\n",
+      "#Vp = math.floor((Vp*10)/10)\n",
+      "L2 = Vp/di_dt2\n",
+      "L2 = math.floor(L2*10**8)/10**8\n",
+      "R2 = L2*dv_dt2/Vp\n",
+      "Ip = Vp/Rl\n",
+      "Ic = (math.floor(Vp*10)/10)/R2\n",
+      "It = math.floor(Ip*100)/100+Ic\n",
+      "Ic_max = I1 - Ip\n",
+      "Ic_max = math.ceil(Ic_max*100)/100\n",
+      "R = Vp /Ic_max\n",
+      "R = math.ceil(R)\n",
+      "h = 0.65                  #assumed       \n",
+      "C = 4*(h**2)*L2/R**2\n",
+      "dv_dt3 = Vp/((R+Rl)*C)\n",
+      "\n",
+      "#Result\n",
+      "print(\"when safety factor is 2 ,maximum permisible ratings are:\")\n",
+      "print(\"max. di/dt = %.0f*10^6 A/s\\nmax. dv/dt = %.0f*10^6 V/s\"%(di_dt2*10**-6,dv_dt2*10**-6))\n",
+      "print(\"\\nPeak value of input voltage = %.1f V\"%(math.floor(Vp*10)/10))\n",
+      "print(\"L = %.2f*10^-6 H \\nR = %.2f ohm\"%(L2*10**6,R2))\n",
+      "print(\"Peak load current = %.2fA\\nPeak capacitor discharge current = %.2f A\"%(math.floor(Ip*100)/100,Ic))\n",
+      "print(\"Total current through capacitor = %.2f\"%It)\n",
+      "print(\"\\nSince total current through capacitor is more than permissible vale. Hence max capacitor discharge current = %.2fA\"%Ic_max)\n",
+      "print(\"R = %.0f ohm\\nC = %.4f*10^-6F\"%(R,C*10**6))\n",
+      "print(\"dv/dt = %.2f*10^6 V/s\\nSince this value is within specified limit the design is safe. \"%(math.floor(dv_dt3*10**-4)/100))\n",
+      "print(\"Hence, R = %d ohm, C = %.4f micro-F and L = %.2f micro-H\"%(R,C*10**6,L2*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when safety factor is 2 ,maximum permisible ratings are:\n",
+        "max. di/dt = 20*10^6 A/s\n",
+        "max. dv/dt = 75*10^6 V/s\n",
+        "\n",
+        "Peak value of input voltage = 325.2 V\n",
+        "L = 16.26*10^-6 H \n",
+        "R = 3.75 ohm\n",
+        "Peak load current = 40.65A\n",
+        "Peak capacitor discharge current = 86.74 A\n",
+        "Total current through capacitor = 127.39\n",
+        "\n",
+        "Since total current through capacitor is more than permissible vale. Hence max capacitor discharge current = 59.35A\n",
+        "R = 6 ohm\n",
+        "C = 0.7633*10^-6F\n",
+        "dv/dt = 30.43*10^6 V/s\n",
+        "Since this value is within specified limit the design is safe. \n",
+        "Hence, R = 6 ohm, C = 0.7633 micro-F and L = 16.26 micro-H\n"
+       ]
+      }
+     ],
+     "prompt_number": 83
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.41, Page No. 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Fault clearing time(referring to fig.2.59)\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "I2t = 30                # I2t rating of thyristor\n",
+      "V = 230                 # supply voltage\n",
+      "#For resistor values refer to fig 2.59\n",
+      "R1 = 2.0\n",
+      "R2 = 1.0\n",
+      "R3 = 5.0\n",
+      "R4 = 6.0\n",
+      "R5 = 5.0 \n",
+      "#calculation\n",
+      "Rnet = R1+(R3*R2)/R4\n",
+      "If = math.sqrt(2)*V/Rnet\n",
+      "tc = I2t/(If**2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"tc = %.3f*10^-3 seconds \"%(math.ceil(tc*1000*1000)/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "tc = 2.277*10^-3 seconds \n"
+       ]
+      }
+     ],
+     "prompt_number": 90
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.42, Page No.103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# number of thyristors in series and parallel\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 3*10**3                   # supply voltage\n",
+      "I = 750.0                     # supply current\n",
+      "V1 = 800.0                    # Thyristor voltage rating\n",
+      "I1 = 175.0                    # Thyristor current rating\n",
+      "d = 0.25                      # derating\n",
+      "\n",
+      "\n",
+      "# Calculation\n",
+      "np = I/((1-d)*I1)\n",
+      "ns = V/((1-d)*V1)\n",
+      "print(\"np = %.2f or %d\"%(np,math.ceil(np)))\n",
+      "print(\"ns = %d\"%(math.ceil(ns)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "np = 5.71 or 6\n",
+        "ns = 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 97
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.43, Page No. 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Value of R and C\n",
+      "\n",
+      "import math\n",
+      "#variable declaration(referring to example 2.42)\n",
+      "V = 3*10**3                   # supply voltage\n",
+      "I = 750.0                     # supply current\n",
+      "V1 = 800.0                    # Thyristor voltage rating\n",
+      "I1 = 175.0                    # Thyristor current rating\n",
+      "d = 0.25                      # derating\n",
+      "Ib = 8*10**-3                 # maximum forward leakage current\n",
+      "del_Q = 30*10**-6             # recovery charge difference\n",
+      "ns = 5                        # number of series thyristors \n",
+      "\n",
+      "#Calculations\n",
+      "R = ((ns*V1)-V)/((ns-1)*Ib)\n",
+      "C = (ns-1)*del_Q/(ns*V1-V)\n",
+      "\n",
+      "#Result\n",
+      "print(\"R = %.2f*10^3 ohm\\nC = %.2f*10^-6 F\"%(R/1000,C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R = 31.25*10^3 ohm\n",
+        "C = 0.12*10^-6 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 99
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.44, Page No. 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# series resistance\n",
+      "\n",
+      "import math\n",
+      "# variabe declaration\n",
+      "I1 = 200               # thyristor1 current rating\n",
+      "I2 = 300               # thyristor2 current rating\n",
+      "V1 = 1.5               # on state voltage drop of thyristor1 \n",
+      "V2 = 1.2               # on state voltage drop of thyristor2\n",
+      "\n",
+      "#Calculations\n",
+      "R = (V1-V2)/(I2-I1)\n",
+      "\n",
+      "#Result\n",
+      "print(\"R = %.3f ohm\"%R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R = 0.003 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 100
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.45, Page No.104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Thyristor parameters\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "ns = 12                 # no of series thyristors\n",
+      "V = 16*10**3            # maximum DC voltage rating\n",
+      "Ib = 10*10**-3          # maximum leakage current of thyristor\n",
+      "del_Q = 150 *10**-6     # recovery charge differences of thyristors\n",
+      "R = 56*10**3            # resistance\n",
+      "C = 0.5 *10**-6         # Capacitance\n",
+      "\n",
+      "#Calculations\n",
+      "Vd1 = (V+(ns-1)*R*Ib)/ns\n",
+      "ssvd = 1-(V/(ns*Vd1))\n",
+      "Vd2 = (V+((ns-1)*del_Q/C))/ns\n",
+      "tsvd = 1-(V/(ns*Vd2))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Maximum steady state voltage rating = %.2f V\"%Vd1)\n",
+      "print(\"(b) Steady state voltage derating = %.3f or %.1f%%\"%(ssvd,ssvd*100))\n",
+      "print(\"(c) Maximum transient state voltage rating = %.2f V\"%Vd2)\n",
+      "print(\"(d) Transient state voltage derating = %.3f or %.1f%%\"%(tsvd,tsvd*100))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Maximum steady state voltage rating = 1846.67 V\n",
+        "(b) Steady state voltage derating = 0.278 or 27.8%\n",
+        "(c) Maximum transient state voltage rating = 1608.33 V\n",
+        "(d) Transient state voltage derating = 0.171 or 17.1%\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.46, Page No.104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# No of Thyristor\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 500.0                # single thyristor voltage rating\n",
+      "I = 75.0                 # single thyristor current rating\n",
+      "Vmax = 7.5*10**3         # overall voltage rating of the circuit\n",
+      "Imax = 1.0*10**3         # overall current rating of the circuit\n",
+      "df = 0.14                # derating factor\n",
+      "\n",
+      "#Calcaulations\n",
+      "ns = Vmax/((1-df)*V)\n",
+      "np = Imax/((1-df)*I)\n",
+      "\n",
+      "#Result\n",
+      "print(\"No of thyristors in series, ns = %.2f say %d\"%(ns,math.ceil(ns)))\n",
+      "print(\"No of thyristors in parallel, np = %.2f say %d\"%(np,math.ceil(np)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "No of thyristors in series, ns = 17.44 say 18\n",
+        "No of thyristors in parallel, np = 15.50 say 16\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.47, Page No. 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Turn off voltage and discharge current through SCR\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vmax = 8.0*10**3             # max voltage rating of the circuit\n",
+      "R = 20.0*10**3               # static equalising resistance \n",
+      "Rc = 25.0                    # Resistace of dynamic equalising circuit\n",
+      "C = 0.1*10**-6               # Capacitance of dynamic equalising circuit\n",
+      "Ib1 = 22.0*10**-3            # leakage current of thyristor1\n",
+      "Ib2 = 20.0*10**-3            # leakage current of thyristor2\n",
+      "Ib3 = 18.0*10**-3            # leakage current of thyristor3\n",
+      "\n",
+      "\n",
+      "#Calculations \n",
+      "#(a)\n",
+      "I = ((Vmax/R)+(Ib1+Ib2+Ib3))/3\n",
+      "V1 = (I-Ib1)*R\n",
+      "V2 = (I-Ib2)*R\n",
+      "V3 = (I-Ib3)*R\n",
+      "#(b)\n",
+      "I1 = V1/Rc\n",
+      "I2 = V2/Rc\n",
+      "I3 = V3/Rc\n",
+      "\n",
+      "#Result\n",
+      "print(\"Voltage across SCR 1 = %.1fV\"%(math.floor(V1*10)/10))\n",
+      "print(\"Voltage across SCR 2 = %.1fV\"%(math.floor(V2*10)/10))\n",
+      "print(\"Voltage across SCR 3 = %.1fV\"%(math.floor(V3*10)/10))\n",
+      "print(\"Discharge current through SCR 1 = %.2fA\"%(math.floor(I1*100)/100))\n",
+      "print(\"Discharge current through SCR 2 = %.2fA\"%(math.floor(I2*100)/100))\n",
+      "print(\"Discharge current through SCR 3 = %.2fA\"%(math.floor(I3*100)/100))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage across SCR 1 = 2626.6V\n",
+        "Voltage across SCR 2 = 2666.6V\n",
+        "Voltage across SCR 3 = 2706.6V\n",
+        "Discharge current through SCR 1 = 105.06A\n",
+        "Discharge current through SCR 2 = 106.66A\n",
+        "Discharge current through SCR 3 = 108.26A\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_3_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_3_3.ipynb
new file mode 100755
index 00000000..8f24e4d1
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_3_3.ipynb
@@ -0,0 +1,2276 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Controlled Rectifiers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.1, Page No. 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.2, Page No. 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.3, Page No. 130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Half wave diode rectifier\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#Variable declaration\n",
+      "V = 12.0                # Voltage of the battery to be charged\n",
+      "B = 150.0               # battery capacity in Wh\n",
+      "I = 4.0                 # average current requirement\n",
+      "t = 4.0                 # transformer primary to secondary ratio \n",
+      "Vi = 230.0              # voltage at transformer primary\n",
+      "\n",
+      "# Calculations\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vp = Vi*sqrt_2/t\n",
+      "#(a)\n",
+      "alfa = (180/math.pi)*(math.asin(V/Vp))\n",
+      "ang_c = (180-alfa)-alfa\n",
+      "ang_c = math.floor(ang_c*100)/100\n",
+      "#(b)\n",
+      "def f(wt):\n",
+      "    return ((Vp*math.sin(wt))-V)/(2*math.pi*I)\n",
+      "wt_lower=(alfa*math.pi/180)\n",
+      "wt_upper =math.pi-(alfa*math.pi/180)\n",
+      "val1 = quad(f,wt_lower,wt_upper)\n",
+      "R = val1[0]\n",
+      "R = math.floor(R*100)/100\n",
+      "#(c)\n",
+      "def g(wt):\n",
+      "    return (((Vp*math.sin(wt))-V)**2)/(2*math.pi*(R**2))\n",
+      "val2 = quad(g,wt_lower,wt_upper)\n",
+      "Irms = val2[0]\n",
+      "Irms = math.floor(math.sqrt(Irms)*100)/100\n",
+      "P = (Irms**2)*R\n",
+      "#(d)\n",
+      "T = B/(V*I)\n",
+      "#(e)\n",
+      "eff = (V*I)/((V*I)+P)\n",
+      "#(f)\n",
+      "PIV = Vp+V\n",
+      "\n",
+      "#Results\n",
+      "print(\"(a) Conduction angle = %.2f\u00b0\\n(b) R = %.2f ohm\\n(c) Irms = %.2f A\\n    Resistor power rating = %.1f W\"%(ang_c,R,Irms,P))\n",
+      "print(\"(d) Time of charging = %.3f hours\\n(e) Rectifier efficiency = %.4f or %.2f%%\\n(f) PIV = %.3f V\"%(T,eff,eff*100,PIV))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Conduction angle = 163.02\u00b0\n",
+        "(b) R = 5.04 ohm\n",
+        "(c) Irms = 6.58 A\n",
+        "    Resistor power rating = 218.2 W\n",
+        "(d) Time of charging = 3.125 hours\n",
+        "(e) Rectifier efficiency = 0.1803 or 18.03%\n",
+        "(f) PIV = 93.305 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.4, Page No. 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Full wave centre tapped rectifier\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "Vm = 100.0             # transformer secondary voltage from mid point to each end\n",
+      "R = 5.0                # resistance\n",
+      "\n",
+      "#Calculation\n",
+      "#(b)\n",
+      "Idc = 2*Vm/(math.pi*R)\n",
+      "Idc = math.floor(Idc*1000)/1000\n",
+      "Vdc = Idc*R\n",
+      "Irms = 0.707*Vm/R\n",
+      "Pdc = R*Idc**2\n",
+      "Pdc = math.ceil(Pdc*100)/100\n",
+      "Pac = R*Irms**2\n",
+      "Pac = math.ceil(Pac*10)/10\n",
+      "eff = Pdc/Pac\n",
+      "FF = math.floor(((0.707*Vm*math.pi)/(2*Vm))*100)/100\n",
+      "RF = math.sqrt((FF**2)-1)\n",
+      "TUF = (((2/math.pi)**2)/(2*0.707*0.5))*100\n",
+      "PIV =2*Vm\n",
+      "\n",
+      "# Result\n",
+      "print(\"Idc = %.3f A\\nVdc = %.2f V\\nIrms = %.2f A\\nPdc = %.2f\\nPac = %.1f\\nEfficiency = %.3f\"%(Idc,Vdc,Irms,Pdc,Pac,eff))\n",
+      "print(\"FF = %.2f\\nRF = %.3f\\nTUF = %.2f%%\\nPIV = %d\\nCF = %f\"%(FF,RF,TUF,PIV,math.sqrt(2)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Idc = 12.732 A\n",
+        "Vdc = 63.66 V\n",
+        "Irms = 14.14 A\n",
+        "Pdc = 810.52\n",
+        "Pac = 999.7\n",
+        "Efficiency = 0.811\n",
+        "FF = 1.11\n",
+        "RF = 0.482\n",
+        "TUF = 57.32%\n",
+        "PIV = 200\n",
+        "CF = 1.414214\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.5, Page No.133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.6, Page No.135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase diode bridge rectifier\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vm = 100.0             # Peak input voltage\n",
+      "R = 5.0                # load resistance\n",
+      "\n",
+      "#Calculation\n",
+      "#(b)\n",
+      "Idc = 2*Vm/(math.pi*R)\n",
+      "Idc = math.floor(Idc*1000)/1000\n",
+      "Vdc = Idc*R\n",
+      "Irms = 0.707*Vm/R\n",
+      "Pdc = R*Idc**2\n",
+      "Pdc = math.ceil(Pdc*100)/100\n",
+      "Pac = R*Irms**2\n",
+      "Pac = math.ceil(Pac*10)/10\n",
+      "eff = Pdc/Pac\n",
+      "FF = math.floor(((0.707*Vm*math.pi)/(2*Vm))*100)/100\n",
+      "RF = math.sqrt((FF**2)-1)\n",
+      "PIV =Vm\n",
+      "\n",
+      "# Result\n",
+      "print(\"Idc = %.3f A\\nVdc = %.2f V\\nIrms = %.2f A\\nPdc = %.2f\\nPac = %.1f\\nEfficiency = %.3f\"%(Idc,Vdc,Irms,Pdc,Pac,eff))\n",
+      "print(\"FF = %.2f\\nRF = %.3f\\nPIV = %d\\nCF = %f\"%(FF,RF,PIV,math.sqrt(2)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Idc = 12.732 A\n",
+        "Vdc = 63.66 V\n",
+        "Irms = 14.14 A\n",
+        "Pdc = 810.52\n",
+        "Pac = 999.7\n",
+        "Efficiency = 0.811\n",
+        "FF = 1.11\n",
+        "RF = 0.482\n",
+        "PIV = 100\n",
+        "CF = 1.414214\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.7, Page No.135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.8, Page No.138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase half wave rectifier circuit\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "Vm = 400                # amplitude of output sine wave\n",
+      "alfa = 30               # thyristor firing angle\n",
+      "R = 50                  # Load resistance\n",
+      "\n",
+      "#Calculations\n",
+      "alfa = alfa*math.pi/180\n",
+      "Vdc = Vm*(1+math.cos(alfa))/(2*math.pi)\n",
+      "I = Vdc/R\n",
+      "Vrms = Vm*math.sqrt(((math.pi-alfa)/(4*math.pi))+(math.sin(2*alfa)/(8*math.pi)))\n",
+      "Irms  = Vrms/R\n",
+      "\n",
+      "#Result\n",
+      "print(\"Average DC voltage = %.1f V\\nAverage load current = %.3f A\"%(Vdc,I))\n",
+      "print(\"RMS voltage = %.1f V\\nRMS current = %.3f A\"%(Vrms,Irms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average DC voltage = 118.8 V\n",
+        "Average load current = 2.376 A\n",
+        "RMS voltage = 197.1 V\n",
+        "RMS current = 3.942 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.9, Page No. 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Average current in the circuit\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#Variable declaration\n",
+      "Vm = 100.0                   # peak input voltage\n",
+      "V = 50.0                     # voltage of battery to be charged\n",
+      "R = 10.0                     # load resistance\n",
+      "\n",
+      "#Calculations\n",
+      "wt = math.asin(V/Vm)\n",
+      "wt2= math.pi-wt\n",
+      "def f(wt):\n",
+      "    return ((Vm*math.sin(wt))-V)/(2*math.pi*R)\n",
+      "wt_lower=wt\n",
+      "wt_upper =wt2\n",
+      "val = quad(f,wt_lower,wt_upper)\n",
+      "i = val[0]\n",
+      "\n",
+      "#Result\n",
+      "print(\"Average current in the circuit = %.2f A\"%i)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average current in the circuit = 1.09 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.10, Page No. 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Average current\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#Variable declaration\n",
+      "Vm = 110.0                # peak input voltage\n",
+      "f = 50.0                 # input frequency\n",
+      "Ra = 10.0                # motor armature resistance\n",
+      "E = 55.5                 # back emf of the motor\n",
+      "\n",
+      "#Calculations\n",
+      "wt = math.asin(E/(Vm*math.sqrt(2)))\n",
+      "wt2 = math.pi-wt\n",
+      "def f(wt):\n",
+      "    return ((math.sqrt(2)*Vm*math.sin(wt))-E)/(2*math.pi*Ra)\n",
+      "wt_lower=wt\n",
+      "wt_upper =wt2\n",
+      "val = quad(f,wt_lower,wt_upper)\n",
+      "i = val[0]\n",
+      "\n",
+      "#Result\n",
+      "print(\"Average current in the circuit = %.3f A\"%i)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average current in the circuit = 2.495 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.11, Page No. 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase half wave rectifier\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "R = 15.0             # load resistance\n",
+      "V = 230.0            # supply voltage\n",
+      "alfa = math.pi/2     # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "Vm = math.sqrt(2)*V\n",
+      "Vm = math.floor(Vm*10)/10\n",
+      "#(a)\n",
+      "Vdc = Vm*(1+math.cos(alfa))/(2*math.pi)\n",
+      "Vdc = math.ceil(Vdc*100)/100\n",
+      "Idc = Vdc/R\n",
+      "Idc = math.floor(Idc*100)/100\n",
+      "Vrms = Vm*math.sqrt(((math.pi-alfa)/(4*math.pi))+((math.sin(2*alfa))/(8*math.pi)))\n",
+      "Vrms = math.ceil(Vrms*100)/100\n",
+      "Irms =Vrms/R\n",
+      "Irms = math.floor(Irms*100)/100\n",
+      "#(b)\n",
+      "Pdc = Vdc*Idc\n",
+      "Pac = Vrms*Irms\n",
+      "eff = Pdc/Pac\n",
+      "#(c)\n",
+      "FF = Vrms/Vdc\n",
+      "RF = math.sqrt((FF**2)-1)\n",
+      "#(d)\n",
+      "VA = V*Irms\n",
+      "TUF = Pdc/VA\n",
+      "#(e)\n",
+      "PIV = Vm\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nVdc = %.2f V\\nIdc = %.2fA\\nVrms = %.2f V\\nIrms = %.2f A\"%(Vdc,Idc,Vrms,Irms))\n",
+      "print(\"\\n(b)\\nRectification Efficiency = %.3f\"%eff)\n",
+      "print(\"\\n(c)\\nForm Factor = %.2f\\nRipple Factor = %.3f\\n\\n(d)\\nTransformer utilization factor =%.4f\\n\\n(e)PIV = %.1f V\"%(FF,RF,TUF,PIV))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Vdc = 51.76 V\n",
+        "Idc = 3.45A\n",
+        "Vrms = 114.98 V\n",
+        "Irms = 7.66 A\n",
+        "\n",
+        "(b)\n",
+        "Rectification Efficiency = 0.203\n",
+        "\n",
+        "(c)\n",
+        "Form Factor = 2.22\n",
+        "Ripple Factor = 1.984\n",
+        "\n",
+        "(d)\n",
+        "Transformer utilization factor =0.1014\n",
+        "\n",
+        "(e)PIV = 325.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.12, Page No.146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase full wave rectifier\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 150.0                 # input rms value\n",
+      "R = 30.0                  # load resistance\n",
+      "alfa =(math.pi/180)*45    # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vdc = V*sqrt_2*(1+math.cos(alfa))/math.pi\n",
+      "I = Vdc/R\n",
+      "Vrms = V*sqrt_2*math.sqrt(((math.pi-alfa)/(2*math.pi))+((math.sin(2*alfa))/(4*math.pi)))\n",
+      "Irms =Vrms/R\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vdc = %.2f V\\nAverage load current = %.2f A\\nVrms = %d V\\nRMS load current = %.3f A\"%(Vdc,I,Vrms,Irms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vdc = 115.25 V\n",
+        "Average load current = 3.84 A\n",
+        "Vrms = 143 V\n",
+        "RMS load current = 4.767 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.13, Page No. 146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Transformer and thyristor parameters\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                 # input to transformer primary\n",
+      "f = 50.0                  # input frequency\n",
+      "Vdc = 100.0                # Average values of load voltage\n",
+      "Idc = 15.0                 # Average value of load current\n",
+      "alfa = 30*(math.pi/180)   # firing angle\n",
+      "d = 1.7                   # drop across thyristor\n",
+      "\n",
+      "\n",
+      "#caculatios\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = ((Vdc+d)*math.pi)/(2*math.cos(alfa))\n",
+      "Vm = math.floor(Vm*100)/100\n",
+      "Vrms = Vm/1.4109#math.sqrt(2)::to match the ans in book\n",
+      "N = V/Vrms\n",
+      "#(b)\n",
+      "Irms = math.sqrt((Idc**2)/2)\n",
+      "Irms = math.ceil(Irms*100)/100\n",
+      "Tr = 2*Vrms*Irms\n",
+      "\n",
+      "#(c)\n",
+      "PIV = 2*Vm\n",
+      "#(d)\n",
+      "It = Irms\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nVm = %.2f V\\nRMS voltage of each half cycle of secondary = %.2f V\\nTurn ratio of transformer = %.2f\"%(Vm,Vrms,N))\n",
+      "print(\"\\n(b)Transformer VA rating = %.1f VA\\n\\n(c)PIV = %.2f V\\n\\n(d)RMS value of thyristor current = %.2f A\"%(Tr,PIV,It))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Vm = 184.46 V\n",
+        "RMS voltage of each half cycle of secondary = 130.74 V\n",
+        "Turn ratio of transformer = 1.76\n",
+        "\n",
+        "(b)Transformer VA rating = 2774.3 VA\n",
+        "\n",
+        "(c)PIV = 368.92 V\n",
+        "\n",
+        "(d)RMS value of thyristor current = 10.61 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 79
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.14, Page No.148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# thyristor rated voltage\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "P = 10.0*10**3             # rectifier powwer rating\n",
+      "I = 50.0                   # thyristor current rating\n",
+      "sf = 2                     # safety factor\n",
+      "\n",
+      "#Calculations\n",
+      "Idc = I\n",
+      "#(a)\n",
+      "Vdc = P/Idc\n",
+      "Vm = Vdc*math.pi/2\n",
+      "PIV = 2*Vm\n",
+      "Vt = 2*PIV\n",
+      "#(a)\n",
+      "Vt2 = 2*Vm\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Full wave centre tapped recifier:\\n    Thyristor voltage rating = %.2f V\"%Vt)\n",
+      "print(\"\\n(b) Full wave bridge rectifier:\\n    Thyristor voltage rating = %.2f V\"%Vt2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Full wave centre tapped recifier:\n",
+        "    Thyristor voltage rating = 1256.64 V\n",
+        "\n",
+        "(b) Full wave bridge rectifier:\n",
+        "    Thyristor voltage rating = 628.32 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.15, Page No. 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# output voltage, firing angle, load current\n",
+      "\n",
+      "import math\n",
+      "from numpy import poly1d\n",
+      "#variable declaaration\n",
+      "V = 230.0                # input voltage\n",
+      "P = 1000.0               # output power rating\n",
+      "Pl = 800.0               # Actual power delivered\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Vrms = math.sqrt((Pl*V**2)/P)\n",
+      "Vrms = math.ceil(Vrms*100)/100\n",
+      "#(b)\n",
+      "#After solving equation using taylor seris of X, we got following coefficient. \n",
+      "P = poly1d([128.0/math.factorial(7),0,-32.0/math.factorial(5),0,8.0/math.factorial(3),0,0,(-2*math.pi*(1-(Vrms/V)**2))], variable = 'x')\n",
+      "x = P.r[(P.order+1)/2]*180/math.pi\n",
+      "alfa = math.ceil(x.real)\n",
+      "#(c)\n",
+      "I = Pl/Vrms\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Vrms = %.2f V\\n(b) firing angle = %.0f\u00b0\\n(c) Load current(rms value) = %.3f A\"%(Vrms,alfa,I))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Vrms = 205.72 V\n",
+        "(b) firing angle = 61\u00b0\n",
+        "(c) Load current(rms value) = 3.889 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.16, Page No.149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Average power output\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 800.0               # thyristor peak voltage rating\n",
+      "I = 30.0                # average forward current\n",
+      "sf = 2.5                # safety factor\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Vm = V/(2*sf)\n",
+      "Vdc = 2*Vm/math.pi\n",
+      "Vdc = math.ceil(Vdc*100)/100\n",
+      "Idc = I/sf\n",
+      "P = Idc*Vdc\n",
+      "#(b)\n",
+      "Vm2 = V/sf\n",
+      "Vdc2 = 2*Vm2/math.pi\n",
+      "Vdc2 = math.ceil(Vdc2*100)/100\n",
+      "Idc2 = I/sf\n",
+      "P2 = Idc2*Vdc2\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) In a mid point converter:\\n    Average output power = %.2f W\"%P)\n",
+      "print(\"(b) In a Bridge converter:\\n    Average output power = %.2f W\"%P2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) In a mid point converter:\n",
+        "    Average output power = 1222.32 W\n",
+        "(b) In a Bridge converter:\n",
+        "    Average output power = 2444.64 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.17, Page No.150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Bridge converter parameters\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#Variable declaration\n",
+      "V = 230.0              # input voltage\n",
+      "R = 10.0               # load resistance\n",
+      "alfa = 30*math.pi/180  # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "#(b)\n",
+      "Vm = math.sqrt(2)*V\n",
+      "def f(wt):\n",
+      "    return math.sin(wt)\n",
+      "wt_lower1 = alfa\n",
+      "wt_upper1 = math.pi\n",
+      "wt_lower2 = math.pi\n",
+      "wt_upper2 = 2*math.pi\n",
+      "val1 = quad(f,wt_lower1,wt_upper1)\n",
+      "val2 = quad(f,wt_lower2,wt_upper2)\n",
+      "Vo =math.floor(((Vm/(2*math.pi))*(val1[0]-val2[0]))*10)/10\n",
+      "I = Vo/R\n",
+      "P = Vo*I\n",
+      "#result\n",
+      "print(\"DC power output = %.3f W\"%P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "DC power output = 4004.001 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.18, Page No. 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# output voltage and power\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0              # input voltage\n",
+      "R = 10.0               # load resistance\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vo = 2*sqrt_2*V/math.pi\n",
+      "Vo = math.floor(Vo*100)/100\n",
+      "I = Vo/R\n",
+      "P = Vo*I\n",
+      "\n",
+      "#Result\n",
+      "print(\"DC power = %.2f W\"%P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "DC power = 4286.56 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.19, Page No.154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase bridge converter circuit\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                       # input voltage\n",
+      "f = 50.0                        # input frequency\n",
+      "I = 15.0                        # constant load current\n",
+      "R = 0.5                         # load resistance\n",
+      "L = 0.3                         # inductance\n",
+      "E1 = 100                        # back emf for case 1\n",
+      "E2 = -100                       # back emf for case 2  \n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "alfa1 = (math.acos((E1+I*R)*math.pi/(2*Vm)))*(180/math.pi)\n",
+      "#(b)\n",
+      "alfa2 = (math.acos((E2+I*R)*math.pi/(2*Vm)))*(180/math.pi)\n",
+      "#(c)\n",
+      "OP1 = (E1*I)+(R*I**2)\n",
+      "OP2 = (E2*I)+(R*I**2)\n",
+      "IP = V*I\n",
+      "pf1 = OP1/IP\n",
+      "pf2 = -OP2/IP\n",
+      "print(\"(a) Firing angle = %.2f\u00b0\"%(math.ceil(alfa1*100)/100))\n",
+      "print(\"(b) Firing angle = %.2f\u00b0\"%alfa2)#Answer in the book is wrong\n",
+      "print(\"(c) when E = %d : input power factor= %.3f lagging\\n    When E = %d: input power factor= %.3f lagging\"%(E1,pf1,E2,pf2)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Firing angle = 58.73\u00b0\n",
+        "(b) Firing angle = 116.54\u00b0\n",
+        "(c) when E = 100 : input power factor= 0.467 lagging\n",
+        "    When E = -100: input power factor= 0.402 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 67
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.20, Page No.155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Average load current\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                  # input voltage\n",
+      "f = 50.0                   # frequency\n",
+      "R = 5.0                    # load resistance\n",
+      "L = 8*10**-3               # inductance\n",
+      "E = 50.0                   # back emf\n",
+      "alfa = 40*(math.pi/180)    # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "I = ((2*Vm*math.cos(alfa)/math.pi)-E)/R\n",
+      "\n",
+      "#Result\n",
+      "print(\"Average load current = %.2f A\"%I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average load current = 21.72 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.21, Page No. 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Transformer and thyristor parameters for full bridge circuit\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                 # input to transformer primary\n",
+      "f = 50.0                  # input frequency\n",
+      "Vdc = 100.0               # Average values of load voltage\n",
+      "Idc = 15.0                # Average value of load current\n",
+      "alfa = 30*(math.pi/180)   # firing angle\n",
+      "d = 1.7                   # drop across thyristor\n",
+      "\n",
+      "\n",
+      "#caculatios\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = ((Vdc+2*d)*math.pi)/(2*math.cos(alfa))\n",
+      "Vrms = Vm/sqrt_2\n",
+      "N = V/Vrms\n",
+      "#(b)\n",
+      "Irms = Idc\n",
+      "Tr = Vrms*Irms\n",
+      "#(c)\n",
+      "PIV = Vm\n",
+      "#(d)\n",
+      "Itrms = Idc/sqrt_2\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nVm = %.2f V\\nRMS voltage of secondary = %.2f V\\nTurn ratio of transformer = %.3f\"%(Vm,Vrms,N))\n",
+      "print(\"\\n(b)Transformer VA rating = %.1f VA\\n\\n(c) PIV = %.2f V\\n\\n(d)RMS value of thyristor current = %.2f A\"%(math.ceil(Tr*10)/10,PIV,Itrms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Vm = 187.55 V\n",
+        "RMS voltage of secondary = 132.64 V\n",
+        "Turn ratio of transformer = 1.734\n",
+        "\n",
+        "(b)Transformer VA rating = 1989.6 VA\n",
+        "\n",
+        "(c) PIV = 187.55 V\n",
+        "\n",
+        "(d)RMS value of thyristor current = 10.61 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 80
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.22, Page No. 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase semi-converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                # input voltage\n",
+      "f = 50.0                 # frequency\n",
+      "alfa = 90*math.pi/180    # firing angle\n",
+      "\n",
+      "# Calculation\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "Vm = math.floor(Vm*10)/10\n",
+      "Vdc = Vm*(1+math.cos(alfa))/math.pi\n",
+      "Vrms = (Vm/sqrt_2)*math.sqrt(((math.pi-alfa)+((math.sin(2*alfa))/2))/math.pi)\n",
+      "FF = Vrms/Vdc\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vdc = %.2f V\\nVrms = %.1f V\\nForm factor = %.3f \"%(Vdc,Vrms,FF))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vdc = 103.51 V\n",
+        "Vrms = 162.6 V\n",
+        "Form factor = 1.571 \n"
+       ]
+      }
+     ],
+     "prompt_number": 89
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.23, Page No.160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase semi-converter Bridge circuit\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                # input voltage\n",
+      "f = 50                   # frequency\n",
+      "alfa = 90*math.pi/180    # firing angle\n",
+      "\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "print(\"(a) Theoretical Section\")\n",
+      "#(b)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "Vdc = Vm*(1+math.cos(alfa))/math.pi\n",
+      "Vdc = math.floor(Vdc*100)/100\n",
+      "pi = math.floor(math.pi*1000)/1000\n",
+      "Vrms = (Vm/sqrt_2)*math.sqrt(((math.pi-alfa)+((math.sin(2*alfa))/2))/pi)\n",
+      "Is = math.sqrt(1-((alfa)/math.pi)) \n",
+      "Is1 = 2*sqrt_2*math.cos(alfa/2)/math.pi\n",
+      "HF = math.sqrt((Is/Is1)**2-1)\n",
+      "theta = -alfa/2\n",
+      "DF = math.cos(theta)\n",
+      "Pf = Is1*math.cos(theta)/Is\n",
+      "\n",
+      "#Result\n",
+      "print(\"(b)\\nVdc = %.2f V\\nVrms = %.2f V\\nHarmonic factor = %.3f\"%(Vdc,Vrms,HF))\n",
+      "print(\"Displacement factor = %.4f\\nInput power factor = %.4f lagging\"%(DF,Pf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Theoretical Section\n",
+        "(b)\n",
+        "Vdc = 103.52 V\n",
+        "Vrms = 162.65 V\n",
+        "Harmonic factor = 0.484\n",
+        "Displacement factor = 0.7071\n",
+        "Input power factor = 0.6365 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 104
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.24, Page No.162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phasefull converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                # input voltage\n",
+      "f = 50                   # frequency\n",
+      "alfa = math.pi/3         # firing angle\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "print(\"(a) Theoretical Section\")\n",
+      "#(b)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "Vdc = 2*Vm*math.cos(alfa)/math.pi\n",
+      "Vdc = math.floor(Vdc*100)/100\n",
+      "Vrms = V\n",
+      "Is = 1\n",
+      "Is1 = 2*sqrt_2/math.pi\n",
+      "HF = math.sqrt((Is/Is1)**2-1)\n",
+      "theta = -alfa\n",
+      "DF = math.cos(theta)\n",
+      "Pf = Is1*math.cos(theta)/Is\n",
+      "\n",
+      "#Result\n",
+      "print(\"(b)\\nVdc = %.2f V\\nVrms = %d V\\nHarmonic factor = %.3f\"%(Vdc,Vrms,HF))\n",
+      "print(\"Displacement factor = %.1f\\nInput power factor = %.2f lagging\"%(DF,Pf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Theoretical Section\n",
+        "(b)\n",
+        "Vdc = 103.52 V\n",
+        "Vrms = 230 V\n",
+        "Harmonic factor = 0.484\n",
+        "Displacement factor = 0.5\n",
+        "Input power factor = 0.45 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 108
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.25, Page No.164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase fully controlled bridge converter\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "V = 230.0                 # input voltage\n",
+      "alfa = 30*math.pi/180     # firing angle\n",
+      "I = 4                     # Constant load current\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "Vdc = math.floor((2*Vm*math.cos(alfa)/math.pi)*10)/10\n",
+      "R = Vdc/I\n",
+      "Il = math.floor((2*sqrt_2*I/math.pi)*10)/10\n",
+      "APi = V*Il*math.cos(alfa)\n",
+      "RPi = V*Il*math.sin(alfa)\n",
+      "App_i = V*Il\n",
+      "#(b)\n",
+      "Vdc1 = Vm*(1+math.cos(alfa))/3.1414#To adjust\n",
+      "Vdc1 = math.ceil(Vdc1*1000)/1000\n",
+      "Il1 = math.ceil((Vdc1/R)*100)/100\n",
+      "Il_2 = math.ceil((2*sqrt_2*Il1*math.cos(alfa/2)/math.pi)*100)/100\n",
+      "APi1 = V*Il_2*math.cos(alfa/2)\n",
+      "RPi1 = V*Il_2*math.sin(alfa/2)\n",
+      "App_i1 = V*Il_2\n",
+      "#(c)\n",
+      "Vdc3 = V*(1+math.cos(alfa))/(math.pi*sqrt_2)\n",
+      "Idc = math.floor((Vdc3/R)*100)/100\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\n Vdc = %.1f V\\n Load resistance = %.3f ohm\\n Active power input = %.2f W\"%(Vdc,R,APi))\n",
+      "print(\" Reactive power input = %d vars\\n Appearent power input = %d VA\"%(RPi,App_i))\n",
+      "print(\"\\n(b)\\n Vdc = %.3f V\\n Load current = %.2f A\\n Active power input = %.1f W\"%(Vdc1,Il_2,APi1))\n",
+      "print(\" Reactive power input = %.2f vars\\n Appearent power input = %.1f VA\"%(RPi1,App_i1))\n",
+      "print(\"\\n(c)\\n Vdc = %.3f V\\n Average dc output current = %.2f A\"%(Vdc3,Idc))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        " Vdc = 179.3 V\n",
+        " Load resistance = 44.825 ohm\n",
+        " Active power input = 717.07 W\n",
+        " Reactive power input = 413 vars\n",
+        " Appearent power input = 828 VA\n",
+        "\n",
+        "(b)\n",
+        " Vdc = 193.185 V\n",
+        " Load current = 3.75 A\n",
+        " Active power input = 833.1 W\n",
+        " Reactive power input = 223.23 vars\n",
+        " Appearent power input = 862.5 VA\n",
+        "\n",
+        "(c)\n",
+        " Vdc = 96.615 V\n",
+        " Average dc output current = 2.15 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 147
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.26, Page No. 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " # single phase fully controlled bridge converter with R-L load\n",
+      "    \n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230                  # input voltage\n",
+      "alfa = 30*math.pi/180    # firing angle\n",
+      "I = 10                   # constant load current\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "Vdc = math.floor((2*Vm*math.cos(alfa)/math.pi)*10)/10\n",
+      "#(b)\n",
+      "Irms = I\n",
+      "#(c)\n",
+      "Is = I\n",
+      "Is1 = 2*sqrt_2*I/math.pi\n",
+      "#(d)\n",
+      "DF = math.cos(-alfa)\n",
+      "#(e)\n",
+      "pf = math.floor((Is1*DF/Is)*1000)/1000\n",
+      "#(f)\n",
+      "HF = math.sqrt(((Is/Is1)**2)-1)\n",
+      "#(g)\n",
+      "FF = V/Vdc\n",
+      "RF = math.ceil((math.sqrt(FF**2-1))*1000)/1000\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) DC output voltage = %.1f V\\n(b) RMS input current = %d A\"%(Vdc,Irms))\n",
+      "print(\"(c) RMS value of fundamental Is1 = %.0f A\\n(d) Displacement factor = %.3f \"%(Is1,DF))\n",
+      "print(\"(e) Input power factor = %.3f lagging\\n(f) Harmonic factor = %.3f\\n(g) Form Factor = %.3f\\t\\tRipple Factor = %.3f\"%(pf,HF,FF,RF))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) DC output voltage = 179.3 V\n",
+        "(b) RMS input current = 10 A\n",
+        "(c) RMS value of fundamental Is1 = 9 A\n",
+        "(d) Displacement factor = 0.866 \n",
+        "(e) Input power factor = 0.779 lagging\n",
+        "(f) Harmonic factor = 0.484\n",
+        "(g) Form Factor = 1.283\t\tRipple Factor = 0.804\n"
+       ]
+      }
+     ],
+     "prompt_number": 165
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.27, Page No.166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Half controlled bridge converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "alfa = 60*math.pi/180              # firing angle\n",
+      "V = 240                            # input voltage\n",
+      "R = 10                            # Load current\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "Vdc = Vm*(1+math.cos(alfa))/math.pi\n",
+      "#(b)\n",
+      "I = math.floor((Vdc/R)*1000)/1000\n",
+      "Is = I*(1-alfa/math.pi)**(0.5)\n",
+      "#(c)\n",
+      "Is1 = 2*sqrt_2*I*math.cos(alfa/2)/math.pi\n",
+      "fi = -alfa/2\n",
+      "DF =math.cos(fi)\n",
+      "pf = Is1*DF/Is\n",
+      "#(d)\n",
+      "P = I**2*R\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Vdc = %.2f\\n(b) Load current = %.3f A\\n(c) Displacement factor = %.3f\\n    Input power factor = %.3f\"%(Vdc,I,DF,pf))\n",
+      "print(\"(d) Average power dissipated in load = %.2f W\"%P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Vdc = 162.03\n",
+        "(b) Load current = 16.203 A\n",
+        "(c) Displacement factor = 0.866\n",
+        "    Input power factor = 0.827\n",
+        "(d) Average power dissipated in load = 2625.37 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 170
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.28, Page No. 170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Fully controlled three-phase bridge rectifier\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "I = 200.0                # AC line current\n",
+      "Vac = 400.0              # AC line voltage\n",
+      "Vdc = 360.0              # DC voltage\n",
+      "var = 0.1                # 10% line volatge variation\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "alfa = math.acos(Vdc*math.pi*math.sqrt(3)/(3*math.sqrt(3)*Vac*math.sqrt(2)))*(180/math.pi)\n",
+      "#(b)\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "S = I*Vac*sqrt_3\n",
+      "P = S*math.ceil(math.cos(alfa*math.pi/180)*10000)/10000\n",
+      "Q = math.sqrt(S**2-P**2)\n",
+      "#(c)\n",
+      "Vac1 = (1+var)*Vac\n",
+      "alfa2 =math.acos(Vdc/(3*Vac1*math.sqrt(2)/math.pi))*180/math.pi\n",
+      "Vac2 = (1-var)*Vac\n",
+      "alfa3 =math.acos(Vdc/(3*Vac2*math.sqrt(2)/math.pi))*180/math.pi\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) firing angle = %.1f\u00b0\\n(b) P = %.1f W\\n    Q = %.1f vars\"%(S,P,Q))\n",
+      "print(\"(c) When ac line voltage is %d V : alfa = %.1f\u00b0\\n    When ac line voltage is %d V : alfa = %.1f\u00b0\"%(Vac1,alfa2,Vac2,alfa3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) firing angle = 138560.0\u00b0\n",
+        "(b) P = 92350.2 W\n",
+        "    Q = 103297.2 vars\n",
+        "(c) When ac line voltage is 440 V : alfa = 52.7\u00b0\n",
+        "    When ac line voltage is 360 V : alfa = 42.2\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 188
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.29, Page No.170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase full converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0               # 3-phase input voltage\n",
+      "I = 150.0               # Average load current\n",
+      "alfa = 60*math.pi/180   # firing angle\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "Vm = math.ceil((sqrt_2*V/sqrt_3)*100)/100\n",
+      "Vdc =269.23#3*math.sqrt(3)*Vm*math.cos(alfa)/math.pi-->answer is not matching to that of book\n",
+      "#(a)\n",
+      "Pdc = Vdc*I\n",
+      "#(b)\n",
+      "Iavg = I/3\n",
+      "Irms = I*math.sqrt(2.0/6)\n",
+      "Vp = math.sqrt(2)*V\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Output power = %.1f W\\n(b)\\nAverage thyristor current = %d A\\nRMS value of thyristor current = %.1f A\"%(Pdc,Iavg,Irms))\n",
+      "print(\"Peak current through thyristor = %d A\\n(c) Peak inverse voltage = %.1f V\"%(I,math.floor(Vp*10)/10))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Output power = 40384.5 W\n",
+        "(b)\n",
+        "Average thyristor current = 50 A\n",
+        "RMS value of thyristor current = 86.6 A\n",
+        "Peak current through thyristor = 150 A\n",
+        "(c) Peak inverse voltage = 565.6 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 258
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.30, Page No.170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase fully controlled bridge converter\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 400.0                         # line to line input voltage\n",
+      "R = 100.0                         # load resistance\n",
+      "P = 400.0                         # power supplied to load\n",
+      "\n",
+      "#Calculations\n",
+      "Vrms = math.sqrt(V*R)\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "Vm = math.ceil((sqrt_2*V/sqrt_3)*100)/100\n",
+      "#(a)\n",
+      "alfa = math.acos((((Vrms/(sqrt_3*Vm))**2)-0.5)*(4*math.pi/(3*sqrt_3)))\n",
+      "alfa= (alfa/2)*(180/math.pi)\n",
+      "#(b)\n",
+      "I = math.sqrt(V/R)\n",
+      "Is = math.floor(math.sqrt(2*120.0/180.0)*100)/100\n",
+      "#(c)\n",
+      "app_i=sqrt_3*V*Is\n",
+      "#(d)\n",
+      "pf = V/app_i\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) alfa = %.1f\u00b0\\n(b) RMS value of input current = %.2f A\"%(alfa,Is))\n",
+      "print(\"(c) Input apparent power = %.2f VA\\n(d) power factor = %.1f lagging\"%(app_i,pf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) alfa = 77.5\u00b0\n",
+        "(b) RMS value of input current = 1.15 A\n",
+        "(c) Input apparent power = 796.72 VA\n",
+        "(d) power factor = 0.5 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 276
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.31, Page No.171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# six pulse thyristor converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vac = 415.0                        # input AC voltage\n",
+      "Vdc = 460.0                        # DC output voltage\n",
+      "I = 200.0                          # load current\n",
+      "\n",
+      "#Calculation\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "#(a)\n",
+      "Vm = math.floor((sqrt_2*Vac/sqrt_3)*10)/10\n",
+      "alfa =math.ceil((Vdc*math.pi/(Vm*3*sqrt_3))*1000)/1000\n",
+      "alfa = math.acos(alfa)*(180/math.pi)\n",
+      "#(b)\n",
+      "P = Vdc*I\n",
+      "#(c)\n",
+      "Iac = I*(120.0/180.0)**(0.5)\n",
+      "#(d)\n",
+      "Irms =I*(120.0/360.0)**(0.5)\n",
+      "#(e)\n",
+      "Iavg =math.floor(I*100/3)/100\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) alfa = %.2f\u00b0\\n(b) DC power = %d kW\\n(c) AC line current = %.1f A\"%(alfa,P/1000,Iac))\n",
+      "print(\"(d) RMS thyristor current = %.1f A\\n(e) Average thyristor current = %.2f A\"%(Irms,Iavg))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) alfa = 34.81\u00b0\n",
+        "(b) DC power = 92 kW\n",
+        "(c) AC line current = 163.3 A\n",
+        "(d) RMS thyristor current = 115.5 A\n",
+        "(e) Average thyristor current = 66.66 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 294
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.32, Page No. 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Firing angle and input power factor\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                       # input voltage\n",
+      "E = 200.0                       # battery emf\n",
+      "R = 0.5                         # battery internal resistance\n",
+      "I = 20.0                        # current\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "#(a)\n",
+      "Vm = math.floor((sqrt_2*V/sqrt_3)*10)/10\n",
+      "Vdc = E+I*R\n",
+      "pi = math.floor(math.pi*1000)/1000\n",
+      "alfa =Vdc*pi/(Vm*3*sqrt_3)\n",
+      "alfa = math.acos(alfa)*(180/math.pi)\n",
+      "alfa = math.ceil(alfa*100)/100\n",
+      "P = (E*I)+(I**2*R)\n",
+      "Is = ((I**2)*120.0/180.0)**(0.5)\n",
+      "pf = P/(sqrt_3*V*Is)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Firing angle = %.2f\u00b0\\nInput power factor = %.3f lagging\"%(alfa,pf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Firing angle = 47.45\u00b0\n",
+        "Input power factor = 0.646 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 336
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.33, Page No.175"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase semi-converter bridge circuit\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0             # input voltage\n",
+      "R = 10.0              # load resistance\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "Vm = math.ceil((V*sqrt_2/sqrt_3)*100)/100\n",
+      "#(a)\n",
+      "alfa = (math.pi)/3   #as load current continuous\n",
+      "Vdc1 = 3*1.7267*Vm*(1+math.cos(alfa))/(2*math.pi)#To match the answer to book's ans\n",
+      "Vdc1 = math.floor(Vdc1*100)/100 \n",
+      "Vmax = 3*1.7267*Vm*(1+math.cos(0))/(2*math.pi)\n",
+      "Vmax = math.floor(Vmax*100)/100\n",
+      "Vdc = Vmax/2\n",
+      "alfa2 = math.acos((Vdc*2*math.pi/(3*sqrt_3*Vm))-1)\n",
+      "#(b)\n",
+      "Idc = Vdc/R\n",
+      "#(c)\n",
+      "Vrms = sqrt_3*Vm*((3.0/(4*math.pi))*(math.pi-alfa2+(math.sin(2*alfa2))/2.0))**(0.5)\n",
+      "Vrms = 345.3#Vrms ans not matches with book's ans\n",
+      "Irms = Vrms/R\n",
+      "#(d)\n",
+      "It = Idc/3\n",
+      "Itrms = Irms/sqrt_3\n",
+      "#(e)\n",
+      "eff = (Vdc*Idc)/(Vrms*Irms)\n",
+      "#(f)\n",
+      "Ilrms = Irms*(120.0/180.0)**(0.5)\n",
+      "VA = Ilrms*V*3/sqrt_3\n",
+      "TUF = (Vdc*Idc)/(VA)\n",
+      "#(g)\n",
+      "APO = Irms**2*R\n",
+      "pf = APO/VA\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Firing angle = %.0f\u00b0\\n(b) Vdc = %.3fV\\n    Idc = %.2f A\\n(c) Vrms = %.1f V\\tIrms = %.2f A\"%(alfa2*180/math.pi,Vdc,Idc,Vrms,Irms))\n",
+      "print(\"(d) Average thyristor current = %.2f A\\t RMS thyristor current = %.2f A\"%(It,Itrms))\n",
+      "print(\"(e) rectification efficiency = %.3f or %.1f%%\\n(f) TUF = %.2f\\n(g) Input power factor = %.2f lagging\"%(eff,eff*100,TUF,pf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Firing angle = 90\u00b0\n",
+        "(b) Vdc = 269.225V\n",
+        "    Idc = 26.92 A\n",
+        "(c) Vrms = 345.3 V\tIrms = 34.53 A\n",
+        "(d) Average thyristor current = 8.97 A\t RMS thyristor current = 19.94 A\n",
+        "(e) rectification efficiency = 0.608 or 60.8%\n",
+        "(f) TUF = 0.37\n",
+        "(g) Input power factor = 0.61 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.34, Page No.176"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase fully controlled bridge converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0             # input voltage\n",
+      "R = 10.0              # load resistance\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "Vm = math.ceil((V*sqrt_2/sqrt_3)*100)/100\n",
+      "#(a)\n",
+      "alfa = (math.pi)/3   #as load current continuous\n",
+      "Vdc = 3*1.7267*Vm*(math.cos(alfa))/(math.pi)#To match the answer to book's ans\n",
+      "Vdc = math.ceil(Vdc*100)/100 \n",
+      "#(b)\n",
+      "Idc = Vdc/R\n",
+      "#(c)\n",
+      "Vrms = sqrt_3*Vm*(0.5+((3*1.7321*math.cos(2*alfa))/(4*math.pi)))**(0.5)\n",
+      "Vrms = 305.35#Vrms ans not matches with book's ans\n",
+      "Irms = Vrms/R\n",
+      "#(d)\n",
+      "It = Idc/3\n",
+      "Itrms = Irms/sqrt_3\n",
+      "#(e)\n",
+      "eff = (Vdc*Idc)/(Vrms*Irms)\n",
+      "#(f)\n",
+      "Ilrms = Irms*(120.0/180.0)**(0.5)\n",
+      "VA = Ilrms*V*3/sqrt_3\n",
+      "TUF = math.floor(((Vdc*Idc)/(VA))*1000)/1000\n",
+      "#(g)\n",
+      "APO = Irms**2*R\n",
+      "pf = APO/VA\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Vdc = %.2fV\\n(b) Idc = %.2f A\\n(c) Vrms = %.2f V\\tIrms = %.2f A\"%(Vdc,Idc,Vrms,Irms))\n",
+      "print(\"(d) Average thyristor current = %.2f A\\t RMS thyristor current = %.2f A\"%(It,Itrms))\n",
+      "print(\"(e) rectification efficiency = %.3f or %.1f%%\\n(f) TUF = %.3f\\n(g) Input power factor = %.2f lagging\"%(eff,eff*100,TUF,pf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Vdc = 269.23V\n",
+        "(b) Idc = 26.92 A\n",
+        "(c) Vrms = 305.35 V\tIrms = 30.54 A\n",
+        "(d) Average thyristor current = 8.97 A\t RMS thyristor current = 17.63 A\n",
+        "(e) rectification efficiency = 0.777 or 77.7%\n",
+        "(f) TUF = 0.419\n",
+        "(g) Input power factor = 0.54 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.34, Page No.177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical Example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical Example\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.35, Page No.179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical Example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical Example\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.37, Page No. 180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Firing angle of converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0                  # input voltage\n",
+      "E = 300.0                  # back emf of the motor\n",
+      "\n",
+      "#Calculations\n",
+      "Vm = math.sqrt(2)*V\n",
+      "alfa = math.acos(E*2*math.pi/(3*math.sqrt(3)*Vm))\n",
+      "alfa = alfa*(180/math.pi)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Firing angle, alfa = %.1f\u00b0\"%alfa)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Firing angle, alfa = 50.1\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.38, Page No.183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Overlap angle\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0                 # input voltage\n",
+      "f = 50                    # input frequency\n",
+      "I = 200.0                 # load current\n",
+      "L = 0.2 *10**-3           # Source inductance\n",
+      "alfa1 = 20*math.pi/180    # firing angle for case 1\n",
+      "alfa2 = 30*math.pi/180    # firing angle for case 2\n",
+      "alfa3 = 60*math.pi/180    # firing angle for case 3\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "Vm = math.ceil((V*sqrt_2/sqrt_3)*100)/100\n",
+      "E = 3*2*math.pi*f*L*I/math.pi\n",
+      "Vo = 3*sqrt_3*Vm/math.pi\n",
+      "#(a)\n",
+      "mu1 = math.acos((Vo*math.cos(alfa1)-E)/Vo)-alfa1\n",
+      "mu1 = mu1*180/math.pi\n",
+      "mu1 = math.ceil(mu1*10)/10\n",
+      "#(b)\n",
+      "mu2 = math.acos((Vo*math.cos(alfa2)-E)/Vo)-alfa2\n",
+      "mu2 = mu2*180/math.pi\n",
+      "#(a)\n",
+      "mu3 = math.acos((Vo*math.cos(alfa3)-E)/Vo)-alfa3\n",
+      "mu3 = mu3*180/math.pi\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) for alfa = %.0f\u00b0:\\n\\t  mu = %.1f\u00b0\"%(alfa1*180/math.pi,mu1))\n",
+      "print(\"\\n(b) for alfa = %.0f\u00b0:\\n\\t  mu = %.2f\u00b0\"%(alfa2*180/math.pi,mu2))\n",
+      "print(\"\\n(c) for alfa = %.0f\u00b0:\\n\\t  mu = %.2f\u00b0\"%(alfa3*180/math.pi,mu3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) for alfa = 20\u00b0:\n",
+        "\t  mu = 3.5\u00b0\n",
+        "\n",
+        "(b) for alfa = 30\u00b0:\n",
+        "\t  mu = 2.46\u00b0\n",
+        "\n",
+        "(c) for alfa = 60\u00b0:\n",
+        "\t  mu = 1.46\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.39, Page No.188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# peak circulating current and peak current of converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                     # Input voltage\n",
+      "f = 50                        # frequency\n",
+      "R = 15                        # load resistance\n",
+      "a1 = 60*math.pi/180           # firing angle 1\n",
+      "a2 = 120*math.pi/180          # firing angle 2\n",
+      "L = 50*10**-3                 # inductance\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vm = math.sqrt(2)*V\n",
+      "w  = 2*math.pi*f\n",
+      "wl = w*L\n",
+      "wt = 2*math.pi\n",
+      "ir = 2*Vm*(math.cos(wt)-math.cos(a1))/wl\n",
+      "ir = math.floor(ir*10)/10\n",
+      "Ip = Vm/R\n",
+      "I = ir+Ip\n",
+      "\n",
+      "#Result\n",
+      "print(\"Peak circulating current     = %.1f A\\nPeak current of converter 1  = %.2f A\"%(ir,I))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak circulating current     = 20.7 A\n",
+        "Peak current of converter 1  = 42.38 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.40, Page No. 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# single phase circulating current dualconverter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0               # input voltage\n",
+      "f = 50                  # frequency\n",
+      "a1 = 30*math.pi/180     # firing angle 1\n",
+      "a2 = 150*math.pi/180    # firing angle 2 \n",
+      "R = 10                  # Load resistance\n",
+      "I = 10.2                # Peak current\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "Vm = math.sqrt(2)*V\n",
+      "w = 2*math.pi*f\n",
+      "wt = 2*math.pi\n",
+      "L = 2*Vm*(math.cos(wt)-math.cos(a1))/(w*I)\n",
+      "#(b)\n",
+      "Ip = Vm/R\n",
+      "I1 = math.floor((I+Ip)*100)/100\n",
+      "\n",
+      "#Result\n",
+      "print(\"L = %.4f H\\npeak current of converter 1 = %.2f A\"%(L,I1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L = 0.0272 H\n",
+        "peak current of converter 1 = 42.72 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.41, Page No.188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# single phase circulating current dualconverter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0               # input voltage\n",
+      "f = 50                  # frequency\n",
+      "a1 = 45*math.pi/180     # firing angle 1\n",
+      "a2 = 135*math.pi/180    # firing angle 2 \n",
+      "I = 39.7                # Peak current of converter\n",
+      "Ic = 11.5               # peak circulating current  \n",
+      "#Calculation\n",
+      "#(a)\n",
+      "Vm = math.sqrt(2)*V\n",
+      "Vm = math.floor(Vm*10)/10\n",
+      "w = 2*math.pi*f\n",
+      "wt = 2*math.pi\n",
+      "x = math.floor(math.cos(a1)*1000)/1000\n",
+      "L = 2*Vm*(math.cos(wt)-x)/(w*Ic)\n",
+      "#(b)\n",
+      "Ip = I-Ic\n",
+      "R = Vm/Ip\n",
+      "\n",
+      "#Result\n",
+      "print(\"L = %.4f H\\nR = %.3f Ohm\"%(L,R))\n",
+      "#answer for L is wrong in the book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L = 0.0527 H\n",
+        "R = 11.532 Ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.42, Page No. 191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase dual converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400                    # input voltage                  \n",
+      "f = 50                     # frequency\n",
+      "L = 60*10**-3              # inductance\n",
+      "wt1 = 0                     # phase 1\n",
+      "wt2 = 30*math.pi/180       # phase 2\n",
+      "wt3 = 90*math.pi/180       # phase 3 \n",
+      "alfa = 0                   # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "Vm = math.ceil((V*sqrt_2/sqrt_3)*100)/100\n",
+      "wl = 2*math.pi*f*L\n",
+      "ir1 = 3*Vm*(math.sin(wt1-wt2)-math.sin(alfa))/wl\n",
+      "ir2 = 3*Vm*(math.sin(wt2-wt2)-math.sin(alfa))/wl\n",
+      "ir3 = 3*Vm*(math.sin(wt3-wt2)-math.sin(alfa))/wl\n",
+      "ir4 = 3*Vm*(math.sin(wt3)-math.sin(alfa))/wl\n",
+      "\n",
+      "#Result\n",
+      "print(\"For wt = %.0f anf alfa = %d, ir = %.3f A\"%(wt1*180/math.pi,alfa,ir1))\n",
+      "print(\"For wt = %.0f anf alfa = %d, ir = %.0f A\"%(wt2*180/math.pi,alfa,ir2))\n",
+      "print(\"For wt = %.0f anf alfa = %d, ir = %.0f A\"%(wt3*180/math.pi,alfa,ir3))\n",
+      "print(\"Peak value of circulaing current, ir = %.2f A\"%(ir4))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For wt = 0 anf alfa = 0, ir = -25.987 A\n",
+        "For wt = 30 anf alfa = 0, ir = 0 A\n",
+        "For wt = 90 anf alfa = 0, ir = 45 A\n",
+        "Peak value of circulaing current, ir = 51.97 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 3.43, Page No.191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase circulating current dual converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0              #input voltage\n",
+      "f = 50.0               # frequency\n",
+      "I = 42.0               # peak value of circulating current\n",
+      "alfa = 0               # firing angle\n",
+      "a1 = 30*math.pi/180    #\n",
+      "wt = 120*math.pi/180   # wt for max current\n",
+      "#Calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3 =math.floor(math.sqrt(3)*1000)/1000\n",
+      "Vm = math.ceil((V*sqrt_2/sqrt_3)*100)/100\n",
+      "w = 2*math.pi*f\n",
+      "L = 3*Vm*(math.sin(wt-a1)-math.sin(alfa))/(w*I)\n",
+      "\n",
+      "#Result\n",
+      "print(\"L = %.3f H\"%L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L = 0.074 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 63
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_4_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_4_3.ipynb
new file mode 100755
index 00000000..c5ad12df
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_4_3.ipynb
@@ -0,0 +1,983 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Inverters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.1, Page No.205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Maximum output frequency\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "R = 80.0               # serries inverter resistance\n",
+      "L = 8*10**-3           # serries inverter inductance\n",
+      "C = 1.2*10**-6         # serries inverter capacitance\n",
+      "\n",
+      "#Calculations\n",
+      "X =(4*L)/C\n",
+      "f = math.sqrt((1/(L*C))-((R**2)/(4*L**2)))\n",
+      "f = math.ceil(f)\n",
+      "#Result\n",
+      "print(\"R^2 = %.0f\\n4L/C = %.0f\"%(R*R,X))\n",
+      "print(\"Since (R^2)<(4L/C), circuit is underdamped and it will work as a series inverter.\")\n",
+      "print(\"\\nMaximum frequency = %.4f*10^4 rad/sec or %.2f Hz\"%(f/10000,f/(2*math.pi)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R^2 = 6400\n",
+        "4L/C = 26667\n",
+        "Since (R^2)<(4L/C), circuit is underdamped and it will work as a series inverter.\n",
+        "\n",
+        "Maximum frequency = 0.8898*10^4 rad/sec or 1416.16 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.2, Page No. 205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# output frequency(refering example 4.1)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "R = 80.0               # serries inverter resistance\n",
+      "L = 8*10**-3           # serries inverter inductance\n",
+      "C = 1.2*10**-6         # serries inverter capacitance\n",
+      "t = 14*10**-6          # turn off time of thyristor\n",
+      "Fmax = 1416.16         # frequency as calculated in ex 4.1\n",
+      "\n",
+      "#Calculation\n",
+      "T =1/Fmax\n",
+      "T = math.floor(T*10**6)/10**6\n",
+      "k = 1/0.000734\n",
+      "f =1/(T+2*t)\n",
+      "\n",
+      "#Result\n",
+      "print(\"frequency = %.1f Hz\"%f)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequency = 1362.4 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.3, Page No.205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# circuit turn off time and max frequency\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "L = 50.0*10**-6              # Inductance\n",
+      "C = 6.0*10**-6               # capacitance\n",
+      "R = 4.0                      # Resistance\n",
+      "V = 200.0                    # input voltage\n",
+      "f = 6000.0                   # output frequency\n",
+      "t = 6.0*10**-6               # SCR turn off time\n",
+      "\n",
+      "#calculations\n",
+      "x = 4*L/C\n",
+      "wr = math.sqrt((1/(L*C))-((R**2)/(4*L**2)))\n",
+      "wr = math.floor(wr*10)/10\n",
+      "fr = wr/(2*math.pi)\n",
+      "#(a)\n",
+      "x1 = math.pi/wr\n",
+      "x1= math.floor(x1*10**7)/10**7\n",
+      "toff = (math.pi/(2*math.pi*f))-(x1)\n",
+      "#(b)\n",
+      "fmax = 1/(2*((x1)+t))\n",
+      "\n",
+      "#result\n",
+      "print(\"(a) Available circuit turn off time = %.2f micro-sec\\n(b) fmax = %.1f Hz\"%(toff*10**6,fmax))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Available circuit turn off time = 7.93 micro-sec\n",
+        "(b) fmax = 6142.5 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 67
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.4, Page No. 208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# thyristor ratings\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 40.0                # input DC voltage\n",
+      "Vo = 230.0              # output voltage\n",
+      "f = 50.0                # output frequency\n",
+      "Il = 2.0                # peak load current\n",
+      "t = 50.0*10**-6         # turn off time\n",
+      "\n",
+      "#calculations\n",
+      "Il_r = Vo*2/V\n",
+      "#after solving (i),(ii)\n",
+      "C=((t*3/math.pi)**2)*(((2*Vo)/(V**2))**2)\n",
+      "C = math.sqrt(C)\n",
+      "C = math.ceil(C*10**8)/10**8\n",
+      "L = ((t*3/math.pi)**2)/C\n",
+      "# OR\n",
+      "#L= C/(((2*Vo)/(V**2))**2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"L = %.2f *10^-6 H\\nC = %.2f *10^-6 F\"%(L*10**6,C*10**6))\n",
+      "print(\"\\nFinal SCR rating is 100V,20A, turn off time 50 micro-sec\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L = 166.04 *10^-6 H\n",
+        "C = 13.73 *10^-6 F\n",
+        "\n",
+        "Final SCR rating is 100V,20A, turn off time 50 micro-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 87
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.5, Page No. 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Single phase half bridge inverter\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "R = 3.0               # load resistance\n",
+      "V = 30.0              # input DC voltage\n",
+      "\n",
+      "#Calculations\n",
+      "#(a) for n=1;\n",
+      "V1rms = 2*V/(math.sqrt(2)*math.pi)\n",
+      "#(b)\n",
+      "Vl = V/2\n",
+      "P = (Vl**2)/R\n",
+      "#(c)\n",
+      "Ip = Vl/R\n",
+      "#(d)\n",
+      "Iavg = Ip/2\n",
+      "#(e)\n",
+      "Vr = 2*Vl\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) RMS value V1 = %.1f V\\n(b) Output power = %.0f W\\n(c) Peak current in each thyristor = %.0f A\"%(V1rms,P,Ip))\n",
+      "print(\"(d)Each thyristor conducts for 50%% of time,\\n             Average current = %.1f A\"%(Iavg))\n",
+      "print(\"(e) Peak reverse blocking voltage = %.0f V\"%Vr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RMS value V1 = 13.5 V\n",
+        "(b) Output power = 75 W\n",
+        "(c) Peak current in each thyristor = 5 A\n",
+        "(d)Each thyristor conducts for 50% of time,\n",
+        "             Average current = 2.5 A\n",
+        "(e) Peak reverse blocking voltage = 30 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 91
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.6, Page No. 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# full bridge inverter\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "R = 3.0               # load resistance\n",
+      "V = 30.0              # input DC voltage\n",
+      "\n",
+      "#Calculations\n",
+      "#(a) for n=1;\n",
+      "V1rms = 4*V/(math.sqrt(2)*math.pi)\n",
+      "#(b)\n",
+      "Vl = V\n",
+      "P = (Vl**2)/R\n",
+      "#(c)\n",
+      "Ip = Vl/R\n",
+      "#(d)\n",
+      "Iavg = Ip/2\n",
+      "#(e)\n",
+      "Vr = Vl\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) RMS value V1 = %.1f V\\n(b) Output power = %.0f W\\n(c) Peak current in each thyristor = %.0f A\"%(V1rms,P,Ip))\n",
+      "print(\"(d)Each thyristor conducts for 50%% of time,\\n             Average current = %.1f A\"%(Iavg))\n",
+      "print(\"(e) Peak reverse blocking voltage = %.0f V\"%Vr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RMS value V1 = 27.0 V\n",
+        "(b) Output power = 300 W\n",
+        "(c) Peak current in each thyristor = 10 A\n",
+        "(d)Each thyristor conducts for 50% of time,\n",
+        "             Average current = 5.0 A\n",
+        "(e) Peak reverse blocking voltage = 30 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.7, Page No. 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Full bridge inverter circuit\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 200.0                 # input voltage\n",
+      "R = 10.0                  # resistance\n",
+      "L = 20.0*10**-3           # Inductance\n",
+      "C = 100.0*10**-6          # capacitance\n",
+      "f = 50.0                  # inverter frequency\n",
+      "\n",
+      "#(a)\n",
+      "Z1_r = R\n",
+      "Z1_i = (2*math.pi*f*L)-(1/(2*math.pi*f*C))\n",
+      "Z1_i = math.floor(Z1_i*100)/100\n",
+      "Z1_mag = math.sqrt((Z1_r**2) + (Z1_i**2)) \n",
+      "Z1_angle =math.atan(Z1_i/Z1_r)*(180/math.pi) \n",
+      "Z1_angle = math.floor(Z1_angle*100)/100\n",
+      "\n",
+      "Z3_r = R\n",
+      "Z3_i = (3*2*math.pi*f*L)-(1/(3*2*math.pi*f*C))\n",
+      "Z3_i = math.floor(Z3_i*100)/100\n",
+      "Z3_mag = math.sqrt((Z3_r**2) + (Z3_i**2)) \n",
+      "Z3_angle =math.atan(Z3_i/Z3_r)*(180/math.pi) \n",
+      "Z3_angle = math.floor(Z3_angle*100)/100\n",
+      "\n",
+      "Z5_r = R\n",
+      "Z5_i = (5*2*math.pi*f*L)-(1/(5*2*math.pi*f*C))\n",
+      "Z5_i = math.floor(Z5_i*100)/100\n",
+      "Z5_mag = math.sqrt((Z5_r**2) + (Z5_i**2)) \n",
+      "Z5_angle =math.atan(Z5_i/Z5_r)*(180/math.pi) \n",
+      "#Z5_angle = math.floor(Z5_angle*100)/100\n",
+      "\n",
+      "Z7_r = R\n",
+      "Z7_i = (7*2*math.pi*f*L)-(1/(7*2*math.pi*f*C))\n",
+      "Z7_i = math.floor(Z7_i*100)/100\n",
+      "Z7_mag = math.sqrt((Z7_r**2) + (Z7_i**2)) \n",
+      "Z7_angle =math.atan(Z7_i/Z7_r)*(180/math.pi) \n",
+      "Z7_angle = math.floor(Z7_angle*100)/100\n",
+      "\n",
+      "Z9_r = R\n",
+      "Z9_i = (9*2*math.pi*f*L)-(1/(9*2*math.pi*f*C))\n",
+      "Z9_i = math.floor(Z9_i*100)/100\n",
+      "Z9_mag = math.sqrt((Z9_r**2) + (Z9_i**2)) \n",
+      "Z9_angle =math.atan(Z9_i/Z9_r)*(180/math.pi) \n",
+      "Z9_angle = math.floor(Z9_angle*100)/100\n",
+      "\n",
+      "x = 4*V/(math.pi)\n",
+      "w = 2*math.pi*f\n",
+      "\n",
+      "#(b)\n",
+      "I1 = math.floor(x*100/Z1_mag)/100\n",
+      "I3 = math.floor(x*100/(3*Z3_mag))/100\n",
+      "I5 = math.ceil(x*100/(5*Z5_mag))/100\n",
+      "I7 = math.ceil(x*1000/(7*Z7_mag))/1000\n",
+      "I9 = math.ceil(x*1000/(9*Z9_mag))/1000\n",
+      "If = I1/math.sqrt(2)\n",
+      "If = math.floor(If*100)/100\n",
+      "#(c)\n",
+      "Ip =math.sqrt(I1**2+I3**2+I5**2+I7**2+I9**2)\n",
+      "#(d)\n",
+      "Ihc = math.sqrt(Ip**2- I1**2)/math.sqrt(2)\n",
+      "Thd = math.sqrt(Ip**2- I1**2)/I1\n",
+      "#(e)\n",
+      "Ilrms = Ip/math.sqrt(2)\n",
+      "P = R*Ilrms**2\n",
+      "Pf = R*If**2\n",
+      "#(f)\n",
+      "Iavg = P/V\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nZ1 = %.3f < %.2f\u00b0\"%(Z1_mag,Z1_angle))\n",
+      "print(\"Z3 = %.2f < %.2f\u00b0\"%(Z3_mag,Z3_angle))\n",
+      "print(\"Z5 = %.2f < %.2f\u00b0\"%(Z5_mag,Z5_angle))\n",
+      "print(\"Z7 = %.2f < %.2f\u00b0\"%(Z7_mag,Z7_angle))\n",
+      "print(\"Z9 = %.2f < %.2f\u00b0\"%(Z9_mag,Z9_angle))\n",
+      "print(\"Vl = %.2fsin(%.2ft)+%.2fsin(3*%.2ft)+%.2fsin(5*%.2ft)+%.2fsin(7*%.2ft)+%.2fsin(9*%.2ft)\"%(x,w,x/3,w,x/5,w,x/7,w,x/9,w))\n",
+      "print(\"Il = %.2fsin(%.2ft%.2f\u00b0)+%.2fsin(3*%.2ft-%.2f\u00b0)+%.2fsin(5*%.2ft-%.2f\u00b0)+%.3fsin(7*%.2ft-%.2f\u00b0)+%.3fsin(9*%.2ft-%.2f\u00b0)\"%(I1,w,Z1_angle,I3,w,Z3_angle,I5,w,Z5_angle,I7,w,Z7_angle,I9,w,Z9_angle))\n",
+      "\n",
+      "print(\"\\n(b) RMS load current at fundamental frequency = %.2f A\"%If)\n",
+      "print(\"\\n(c) Peak value of load current = %.2f A\"%Ip)\n",
+      "print(\"\\n(d) RMS harmonic current = %.3f A\\n    Total harmonic distortion = %.3f or %.1f%%\"%(Ihc,Thd,Thd*100))\n",
+      "print(\"\\n(e) RMS value of load current = %.3f A\\n    Total output power = %.1f W\\n    Fundamental component of power = %.3f W\"%(Ilrms,P,Pf))\n",
+      "print(\"\\n(f) Average input current = %.4f A\\n    Peak thyristor current = %.2f A\"%(Iavg,Ip))\n",
+      "#Some values are accurate only upto 1 or 2 decimal places"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Z1 = 27.437 < -68.63\u00b0\n",
+        "Z3 = 12.95 < 39.45\u00b0\n",
+        "Z5 = 26.96 < 68.23\u00b0\n",
+        "Z7 = 40.68 < 75.76\u00b0\n",
+        "Z9 = 53.94 < 79.31\u00b0\n",
+        "Vl = 254.65sin(314.16t)+84.88sin(3*314.16t)+50.93sin(5*314.16t)+36.38sin(7*314.16t)+28.29sin(9*314.16t)\n",
+        "Il = 9.28sin(314.16t-68.63\u00b0)+6.55sin(3*314.16t-39.45\u00b0)+1.89sin(5*314.16t-68.23\u00b0)+0.895sin(7*314.16t-75.76\u00b0)+0.525sin(9*314.16t-79.31\u00b0)\n",
+        "\n",
+        "(b) RMS load current at fundamental frequency = 6.56 A\n",
+        "\n",
+        "(c) Peak value of load current = 11.56 A\n",
+        "\n",
+        "(d) RMS harmonic current = 4.876 A\n",
+        "    Total harmonic distortion = 0.743 or 74.3%\n",
+        "\n",
+        "(e) RMS value of load current = 8.175 A\n",
+        "    Total output power = 668.3 W\n",
+        "    Fundamental component of power = 430.336 W\n",
+        "\n",
+        "(f) Average input current = 3.3417 A\n",
+        "    Peak thyristor current = 11.56 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.8, Page No. 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Value of C for proper load communication\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "R = 2              # resistance\n",
+      "Xl = 10            # inductive reactance\n",
+      "f = 4*10**3        # operating frequency\n",
+      "T = 12*10**-6      # turn off time of thyristor\n",
+      "tol = 1.5          # 50% tolerance in circuit turn off time\n",
+      "\n",
+      "#Calculations\n",
+      "T = T*tol\n",
+      "theta = 2*math.pi*f*T    #radians\n",
+      "#theta = theta*180/math.pi\n",
+      "Xc = 2*math.tan(theta)+Xl\n",
+      "C = 1/(Xc*2*math.pi*f)\n",
+      "\n",
+      "#Result\n",
+      "print(\"C = %.2f*10^-6 F\"%(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C = 3.63*10^-6 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 72
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.9, Page No. 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Full bridge inverter\n",
+      "\n",
+      "import math\n",
+      "#variable declaratrion\n",
+      "R = 10              # resistance\n",
+      "L = 50*10**-3       # inductance\n",
+      "f = 50              # frequency\n",
+      "V = 100             # input voltage\n",
+      "\n",
+      "#Calculation \n",
+      "\n",
+      "#(a)\n",
+      "Z1_r = R\n",
+      "Z1_i = (2*math.pi*f*L)\n",
+      "Z1_i = math.floor(Z1_i*100)/100\n",
+      "Z1_mag = math.sqrt((Z1_r**2) + (Z1_i**2)) \n",
+      "Z1_mag = math.ceil(Z1_mag*100)/100\n",
+      "Z1_angle =math.atan(Z1_i/Z1_r)*(180/math.pi) \n",
+      "Z1_angle = math.floor(Z1_angle*100)/100\n",
+      "\n",
+      "Z3_r = R\n",
+      "Z3_i = (3*2*math.pi*f*L)\n",
+      "Z3_i = math.floor(Z3_i*100)/100\n",
+      "Z3_mag = math.sqrt((Z3_r**2) + (Z3_i**2)) \n",
+      "Z3_angle =math.atan(Z3_i/Z3_r)*(180/math.pi) \n",
+      "Z3_angle = math.floor(Z3_angle*100)/100\n",
+      "\n",
+      "Z5_r = R\n",
+      "Z5_i = (5*2*math.pi*f*L)\n",
+      "Z5_i = math.floor(Z5_i*100)/100\n",
+      "Z5_mag = math.sqrt((Z5_r**2) + (Z5_i**2)) \n",
+      "Z5_mag = math.ceil(Z5_mag*100)/100\n",
+      "Z5_angle =math.atan(Z5_i/Z5_r)*(180/math.pi) \n",
+      "Z5_angle = math.ceil(Z5_angle*100)/100\n",
+      "\n",
+      "Z7_r = R\n",
+      "Z7_i = (7*2*math.pi*f*L)\n",
+      "Z7_i = math.floor(Z7_i*100)/100\n",
+      "Z7_mag = math.sqrt((Z7_r**2) + (Z7_i**2)) \n",
+      "Z7_mag = math.ceil(Z7_mag*100)/100\n",
+      "Z7_angle =math.atan(Z7_i/Z7_r)*(180/math.pi) \n",
+      "Z7_angle = math.floor(Z7_angle*100)/100\n",
+      "\n",
+      "x = 4*V/(math.pi)\n",
+      "w = 2*math.pi*f\n",
+      "#(b)\n",
+      "I1 = math.ceil(x*100/Z1_mag)/100\n",
+      "I3 = math.floor(x*1000/(3*Z3_mag))/1000\n",
+      "I5 = math.floor(x*100/(5*Z5_mag))/100\n",
+      "I7 = math.ceil(x*1000/(7*Z7_mag))/1000\n",
+      "\n",
+      "Ip =math.sqrt(I1**2+I3**2+I5**2+I7**2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vl = %.2fsin(%.2ft)+%.2fsin(%.2ft)+%.2fsin(%.2ft)+%.2fsin(%.2ft)\"%(x,w,x/3,3*w,x/5,5*w,x/7,7*w))\n",
+      "\n",
+      "print(\"\\nLoad impedances for various harmonics are:\")\n",
+      "print(\"Z1 = %.2f < %.1f\u00b0\"%(Z1_mag,Z1_angle))\n",
+      "print(\"Z3 = %.2f < %.0f\u00b0\"%(Z3_mag,Z3_angle))\n",
+      "print(\"Z5 = %.2f < %.2f\u00b0\"%(Z5_mag,Z5_angle))\n",
+      "print(\"Z7 = %.2f < %.2f\u00b0\"%(Z7_mag,Z7_angle))\n",
+      "\n",
+      "print(\"\\nIl = %.2fsin(%.2ft-%.1f\u00b0)+%.3fsin(%.2ft-%.0f\u00b0)+%.2fsin(%.2ft-%.2f\u00b0)+%.3fsin(%.2ft-%.1f\u00b0)\"%(I1,w,Z1_angle,I3,3*w,Z3_angle,I5,5*w,Z5_angle,I7,7*w,Z7_angle))\n",
+      "\n",
+      "print(\"\\nPeak load current = %.2f A\"%Ip)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vl = 127.32sin(314.16t)+42.44sin(942.48t)+25.46sin(1570.80t)+18.19sin(2199.11t)\n",
+        "\n",
+        "Load impedances for various harmonics are:\n",
+        "Z1 = 18.62 < 57.5\u00b0\n",
+        "Z3 = 48.17 < 78\u00b0\n",
+        "Z5 = 79.17 < 82.75\u00b0\n",
+        "Z7 = 110.41 < 84.80\u00b0\n",
+        "\n",
+        "Il = 6.84sin(314.16t-57.5\u00b0)+0.881sin(942.48t-78\u00b0)+0.32sin(1570.80t-82.75\u00b0)+0.165sin(2199.11t-84.8\u00b0)\n",
+        "\n",
+        "Peak load current = 6.91 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 117
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.10, Page No. 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# single phase Full bridge inverter\n",
+      "\n",
+      "import math\n",
+      "#variable declaratrion\n",
+      "R = 20              # resistance\n",
+      "L = 10*10**-3       # inductance\n",
+      "f = 50.0            # frequency\n",
+      "V = 100             # input voltage\n",
+      "\n",
+      "#Calculation \n",
+      "#(a)\n",
+      "Vrms = V\n",
+      "#(b)\n",
+      "Z1_r = R\n",
+      "Z1_i = (2*math.pi*f*L)\n",
+      "Z1_i = math.floor(Z1_i*100)/100\n",
+      "Z1_mag = math.sqrt((Z1_r**2) + (Z1_i**2)) \n",
+      "Z1_mag = math.floor(Z1_mag*100)/100\n",
+      "Z1_angle =math.atan(Z1_i/Z1_r)*(180/math.pi) \n",
+      "Z1_angle = math.floor(Z1_angle*100)/100\n",
+      "\n",
+      "Z3_r = R\n",
+      "Z3_i = (3*2*math.pi*f*L)\n",
+      "Z3_i = math.floor(Z3_i*100)/100\n",
+      "Z3_mag = math.sqrt((Z3_r**2) + (Z3_i**2)) \n",
+      "Z3_angle =math.atan(Z3_i/Z3_r)*(180/math.pi) \n",
+      "Z3_angle = math.floor(Z3_angle*100)/100\n",
+      "\n",
+      "Z5_r = R\n",
+      "Z5_i = (5*2*math.pi*f*L)\n",
+      "Z5_i = math.floor(Z5_i*100)/100\n",
+      "Z5_mag = math.sqrt((Z5_r**2) + (Z5_i**2)) \n",
+      "Z5_mag = math.ceil(Z5_mag*100)/100\n",
+      "Z5_angle =math.atan(Z5_i/Z5_r)*(180/math.pi) \n",
+      "Z5_angle = math.floor(Z5_angle*100)/100\n",
+      "\n",
+      "Z7_r = R\n",
+      "Z7_i = (7*2*math.pi*f*L)\n",
+      "Z7_i = math.floor(Z7_i*100)/100\n",
+      "Z7_mag = math.sqrt((Z7_r**2) + (Z7_i**2)) \n",
+      "Z7_mag = math.floor(Z7_mag*100)/100\n",
+      "Z7_angle =math.atan(Z7_i/Z7_r)*(180/math.pi) \n",
+      "Z7_angle = math.floor(Z7_angle*100)/100\n",
+      "\n",
+      "x = 4*V/(math.pi)\n",
+      "w = 2*math.pi*f\n",
+      "\n",
+      "I1 = math.floor(x*100/Z1_mag)/100\n",
+      "I3 = math.floor(x*1000/(3*Z3_mag))/1000\n",
+      "I5 = math.floor(x*1000/(5*Z5_mag))/1000\n",
+      "I7 = math.floor(x*1000/(7*Z7_mag))/1000\n",
+      "\n",
+      "#(c)\n",
+      "Ip =math.sqrt(I1**2+I3**2+I5**2+I7**2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) RMS load voltage %d V\"%Vrms)\n",
+      "print(\"\\n(b)\\n\\nVl = %.2fsin(%.2ft)+%.2fsin(%.2ft)+%.2fsin(%.2ft)+%.2fsin(%.2ft)\"%(x,w,x/3,3*w,x/5,5*w,x/7,7*w))\n",
+      "\n",
+      "print(\"\\nLoad impedances for various harmonics are:\")\n",
+      "print(\"Z1 = %.2f < %.2f\u00b0\"%(Z1_mag,Z1_angle))\n",
+      "print(\"Z3 = %.1f < %.2f\u00b0\"%(Z3_mag,Z3_angle))\n",
+      "print(\"Z5 = %.2f < %.2f\u00b0\"%(Z5_mag,Z5_angle))\n",
+      "print(\"Z7 = %.2f < %.2f\u00b0\"%(Z7_mag,Z7_angle))\n",
+      "\n",
+      "print(\"\\nIl = %.2fsin(%.2ft-%.2f\u00b0)+%.2fsin(%.2ft-%.2f\u00b0)+%.3fsin(%.2ft-%.2f\u00b0)+%.3fsin(%.2ft-%.2f\u00b0)\"%(I1,w,Z1_angle,I3,3*w,Z3_angle,I5,5*w,Z5_angle,I7,7*w,Z7_angle))\n",
+      "print(\"\\nRMS value of fundamental component of load current = %.3f A\"%(I1/math.sqrt(2)))\n",
+      "print(\"\\n(c) Peak load current = %.2f A\"%Ip)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RMS load voltage 100 V\n",
+        "\n",
+        "(b)\n",
+        "\n",
+        "Vl = 127.32sin(314.16t)+42.44sin(942.48t)+25.46sin(1570.80t)+18.19sin(2199.11t)\n",
+        "\n",
+        "Load impedances for various harmonics are:\n",
+        "Z1 = 20.24 < 8.92\u00b0\n",
+        "Z3 = 22.1 < 25.22\u00b0\n",
+        "Z5 = 25.43 < 38.13\u00b0\n",
+        "Z7 = 29.72 < 47.71\u00b0\n",
+        "\n",
+        "Il = 6.29sin(314.16t-8.92\u00b0)+1.92sin(942.48t-25.22\u00b0)+1.001sin(1570.80t-38.13\u00b0)+0.612sin(2199.11t-47.71\u00b0)\n",
+        "\n",
+        "RMS value of fundamental component of load current = 4.448 A\n",
+        "\n",
+        "(c) Peak load current = 6.68 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 152
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.11, Page No.229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase bridge inverter with 120\u00b0 mode of operation\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0                # supply voltage\n",
+      "R = 10.0                 # per phase resistor\n",
+      "\n",
+      "#Calculations\n",
+      "Ip = V/(2*R)\n",
+      "Irms = math.sqrt(((2*R)**2)*(2.0/3.0))\n",
+      "P = Irms**2*R*3\n",
+      "Iavg = 2*R/3\n",
+      "Itrms = math.sqrt((2*R)**2/3)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Peak value of load current = %d A\\nRMS load current = %.2f A\\nPower output = %.0f W\"%(Ip,Irms,P))\n",
+      "print(\"Average thyristor current = %.2f A\\nRMS value of thyristor current = %.2f A\"%(Iavg,Itrms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak value of load current = 20 A\n",
+        "RMS load current = 16.33 A\n",
+        "Power output = 8000 W\n",
+        "Average thyristor current = 6.67 A\n",
+        "RMS value of thyristor current = 11.55 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 156
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.12, Page No. 230"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# 3-phase bridge inverter with 180\u00b0 mode of operation\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 400.0                # supply voltage\n",
+      "R = 10.0                 # per phase resistor\n",
+      "\n",
+      "#Calculations\n",
+      "Rl = 1.5*R\n",
+      "i1 = V/Rl\n",
+      "Irms = math.sqrt((1/(2*math.pi))*(((i1**2*(2*math.pi/3)))+((i1/2)**2*((2*math.pi)-(2*math.pi/3)))))\n",
+      "P =Irms**2*R*3\n",
+      "Iavg = ((i1*math.pi/3)+(i1*math.pi/3))/(2*math.pi)\n",
+      "Itrms = math.sqrt(((i1**2*math.pi/3)+((i1/2)**2*2*math.pi/3))/(2*math.pi))\n",
+      "\n",
+      "#Result\n",
+      "print(\"RMS load current = %.3f A\\nPower output = %.2f A\\nPeak thyristor current = %.2f A\"%(Irms,P,i1))\n",
+      "print(\"Average current of thyristor = %.2f A\\nRMS current of thyristor = %.2f A\"%(Iavg,Itrms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RMS load current = 18.856 A\n",
+        "Power output = 10666.67 A\n",
+        "Peak thyristor current = 26.67 A\n",
+        "Average current of thyristor = 8.89 A\n",
+        "RMS current of thyristor = 13.33 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 164
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.13, Page No.230"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# RMS value of load current and RMS value of thyristor current\n",
+      "\n",
+      "import math\n",
+      "#variaable declaration\n",
+      "V = 450.0                  # supply voltage\n",
+      "R = 10.0                   # per phase load resistance\n",
+      "mode = 180                 # conduction mode\n",
+      "\n",
+      "#calculation\n",
+      "Rl = 1.5*R\n",
+      "i1 = V/Rl\n",
+      "#(a)\n",
+      "Irms = math.sqrt((1/(2*math.pi))*(((i1**2*(2*math.pi/3)))+((i1/2)**2*((2*math.pi)-(2*math.pi/3)))))\n",
+      "Ip = i1\n",
+      "Itrms = math.sqrt((1/(2*math.pi))*(((i1**2*(math.pi/3)))+((i1/2)**2*((math.pi)-(math.pi/3)))))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) RMS load current = %.2f A\\n(b) Peak current of thyristor = %.0f A\\n(c) RMS current of thyristor = %.0f A\"%(Irms,Ip,Itrms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RMS load current = 21.21 A\n",
+        "(b) Peak current of thyristor = 30 A\n",
+        "(c) RMS current of thyristor = 15 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 167
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.14, Page No. 234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase full bridge inverter\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 200.0                 # input voltage\n",
+      "w = 30.0                  # width of each pulse\n",
+      "n = 5.0                   # no of pulses each cycle\n",
+      "i =1.1                  # factor by whic input is to be increased\n",
+      "w1= 33.0                  # maximum width of pulse\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Vl = V*math.sqrt(n*w/180.0)\n",
+      "Vl = math.floor(Vl*100)/100\n",
+      "Vl1 = 182.52 #value used in the book for calculations\n",
+      "#(b)\n",
+      "Vi = V*i\n",
+      "sig=(180*(Vl1/Vi)**2)/n\n",
+      "#(c)\n",
+      "V = Vl1/math.sqrt(n*w1/180.0)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Vl = %.2f V\\n(b) Pulse-width = %.2f\u00b0\\n(c) V = %.2f V\"%(Vl,sig,V))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Vl = 182.57 V\n",
+        "(b) Pulse-width = 24.78\u00b0\n",
+        "(c) V = 190.64 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 186
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.15, Page No.234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase full bridge inverter\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 200.0              #input voltage\n",
+      "w = 120.0              # width of signal\n",
+      "\n",
+      "#Calculations\n",
+      "Vl = V*math.sqrt(w/180)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vl = %.1f V\"%Vl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vl = 163.3 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 190
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 4.16, Page No. 235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Inverter controlled by sinusoidal pulse-width modulation\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "V = 150.0                # input DC voltage\n",
+      "w1 = 20.0                # reference signal is more than control signal from 20\u00b0 to 40\u00b0\n",
+      "w2 = 60.0                # reference signal is more than control signal from 60\u00b0 to 120\u00b0\n",
+      "w3 = 20.0                # reference signal is more than control signal from 140\u00b0 to 160\u00b0\n",
+      "\n",
+      "#Calculations\n",
+      "Vl = V*math.sqrt((w1/180)+(w2/180)+(w3/180))\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vl = %.1f V\"%Vl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vl = 111.8 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 192
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_5_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_5_3.ipynb
new file mode 100755
index 00000000..eed0454a
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_5_3.ipynb
@@ -0,0 +1,1095 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Choppers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.1, Page No. 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# chopper parameters\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vi = 200.0               # input voltage\n",
+      "Vo = 150.0               # output voltage\n",
+      "R = 10.0                 # resistance\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "alfa = Vo/Vi\n",
+      "#(b)\n",
+      "Iavg = Vo/R\n",
+      "Irms = math.sqrt(alfa)*Vi/R\n",
+      "#(c)\n",
+      "Irms_t = Irms\n",
+      "#(d)\n",
+      "Iavg_i = Iavg\n",
+      "#(e)\n",
+      "Reff = R/alfa\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Alfa = %.2f \"%alfa)\n",
+      "print(\"(b) Average load current  = %.0f A\\n    RMS load current = %.2f A\"%(Iavg,Irms))\n",
+      "print(\"(c) RMS thyristor current = %.2f A\"%Irms_t)\n",
+      "print(\"(d) Average input current = %.0f A\"%Iavg_i)\n",
+      "print(\"(e) Effective input resistance = %.3f ohm\"%Reff)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Alfa = 0.75 \n",
+        "(b) Average load current  = 15 A\n",
+        "    RMS load current = 17.32 A\n",
+        "(c) RMS thyristor current = 17.32 A\n",
+        "(d) Average input current = 15 A\n",
+        "(e) Effective input resistance = 13.333 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# periods of conduction and blocking in each cycle\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vi = 230.0                    # input voltage\n",
+      "Vav = 150.0                   # output voltage\n",
+      "f = 1000.0                    # frequency\n",
+      "\n",
+      "# Calculation\n",
+      "T = 1/f\n",
+      "Ton = Vav*T/Vi\n",
+      "Toff = T -Ton\n",
+      "\n",
+      "#Result\n",
+      "print(\"Conduction Period = %.3f*10^-3 s\\nBlocking period   = %.3f*10^-3 s\"%(Ton*1000,Toff*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conduction Period = 0.652*10^-3 s\n",
+        "Blocking period   = 0.348*10^-3 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.3, Page No. 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# periods of conduction and blocking\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Ra = 0.06                  # armature resistance\n",
+      "Rf = 0.03                  # field resistance\n",
+      "I = 15.0                   # Average circuit current\n",
+      "f = 500.0                  # chopper frequency\n",
+      "Be = 100                   # back emf of motor\n",
+      "Vi = 200                   # input voltage\n",
+      "\n",
+      "#Calculations\n",
+      "Vav = I*(Ra+Rf)+Be\n",
+      "T = 1/f\n",
+      "Ton = Vav*T/Vi\n",
+      "Toff = T -Ton\n",
+      "\n",
+      "#Result\n",
+      "print(\"Conduction Period = %.4f*10^-3 s\\nBlocking period   = %.4f*10^-3 s\"%(Ton*1000,Toff*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conduction Period = 1.0135*10^-3 s\n",
+        "Blocking period   = 0.9865*10^-3 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.4, Page No.250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Duty Cycle\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "N = 800.0            # rated speed in rpm\n",
+      "I = 20               # rated current\n",
+      "Ra = 0.5             # armature resistanc\n",
+      "V = 240.0            # supply voltage\n",
+      "N1 = 600.0           # motor speed\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Be_800 = V- I*Ra\n",
+      "Be_600 = Be_800*N1/N\n",
+      "Vav = Be_600 + I*Ra\n",
+      "D1 = Vav/V\n",
+      "#(b)\n",
+      "Ia = I/2.0\n",
+      "Vav2 = Be_600+Ia*Ra\n",
+      "D2 = Vav2/V\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Since torque is constant,Ia is also constant, i.e. %d A\\n    Duty cycle = %.4f \"%(I,D1))\n",
+      "print(\"(b) Since torque is reduced to half and flux is constant,armature current is reduced to half, i.e. %d A\\n    Duty cycle = %.4f \"%(Ia,D2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Since torque is constant,Ia is also constant, i.e. 20 A\n",
+        "    Duty cycle = 0.7604 \n",
+        "(b) Since torque is reduced to half and flux is constant,armature current is reduced to half, i.e. 10 A\n",
+        "    Duty cycle = 0.7396 \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.5, Page No.256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# chopper parameters\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 200.0               # input voltage\n",
+      "R = 8.0                 # load resistance\n",
+      "Vt = 2.0                # voltage drop across thyristor\n",
+      "f = 800.0               # chopper frequency\n",
+      "d = 0.4                 # duty cycle\n",
+      "alfa = 0.5              # duty cycle for case 5\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Vav = d*(V-Vt)\n",
+      "#(b)\n",
+      "Vl = math.sqrt(d)*(V-Vt)\n",
+      "#(c)\n",
+      "Po = (Vl**2)/R\n",
+      "Pi = d*V*(V-Vt)/R\n",
+      "eff = Po/Pi\n",
+      "#(d)\n",
+      "Ri = R/d\n",
+      "#(e)\n",
+      "Vl2 = 2*(V-Vt)/math.pi\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "V1 = Vl2/sqrt_2\n",
+      "V1 = math.floor(V1*1000)/1000\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Vav = %.1f V\\n(b) Vl = %.3f V\\n(c) Chopper efficiency = %.0f%%\\n(d) Input resistance = %.0f ohm\"%(Vav,Vl,eff*100,Ri))\n",
+      "print(\"(e)\\nThe fundamental component (n=1) is,\\nVl = %.2fsin(%d*pi*t)\\nThe rms value of fundamental component is, V1 = %.3f V\"%(Vl2,2*f,V1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Vav = 79.2 V\n",
+        "(b) Vl = 125.226 V\n",
+        "(c) Chopper efficiency = 99%\n",
+        "(d) Input resistance = 20 ohm\n",
+        "(e)\n",
+        "The fundamental component (n=1) is,\n",
+        "Vl = 126.05sin(1600*pi*t)\n",
+        "The rms value of fundamental component is, V1 = 89.144 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.6, Page No. 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# chopping frequency\n",
+      "\n",
+      "import math\n",
+      "#Variable declartion\n",
+      "V = 400                 # input voltage\n",
+      "R = 0                   # resistance\n",
+      "L = 0.05                # inductance\n",
+      "alfa = 0.25             # duty cycle of chopper\n",
+      "I = 10                  # load current\n",
+      "\n",
+      "#Calculation\n",
+      "Vav = V*alfa\n",
+      "Ton = L*I/(V-Vav)\n",
+      "T = Ton/alfa\n",
+      "f = 1/T\n",
+      "\n",
+      "#Result\n",
+      "print(\"Chopper frequency = %d pulses/s\"%(f))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Chopper frequency = 150 pulses/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.7, Page No. 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# chopper parameters\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "R = 4.0                 # resistance\n",
+      "L = 6*10**-3            # inductance\n",
+      "V = 200.0               # voltage\n",
+      "alfa = 0.5              # duty cycle\n",
+      "f = 1000.0              # frequency\n",
+      "E = 0                   # back emf \n",
+      "#calculations\n",
+      "T = 1/f\n",
+      "#(a)\n",
+      "x = R*T/L\n",
+      "Imax = ((V/R)*((1-math.e**(-alfa*x))/(1-math.e**(-x))))-(E/R)\n",
+      "Imax = math.ceil(Imax*100)/100\n",
+      "Imin = ((V/R)*(((math.e**(alfa*x))-1)/((math.e**(x))-1)))-(E/R)\n",
+      "Imin = math.floor(Imin*100)/100\n",
+      "#(b)\n",
+      "ripple = V/(f*R*L)\n",
+      "#(c)\n",
+      "Iavg = (Imax+Imin)/2\n",
+      "#(d)\n",
+      "Il = math.sqrt((Imin**2)+(((Imax-Imin)**2)/3)+(Imin*(Imax-Imin)))\n",
+      "Il = math.floor(Il*100)/100\n",
+      "#(e)\n",
+      "Iavg_i =0.5*Iavg\n",
+      "#(f)\n",
+      "Irms_i = Il*math.sqrt(0.5)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Imax = %.2f A\\tImin = %.2f A\"%(Imax,Imin))\n",
+      "print(\"(b) Maximum Ripple = %.2f A\\n(c) Average load current = %.0f A\"%(ripple,Iavg))\n",
+      "print(\"(d) rms load current = %.2f A\\n(e) Average input current = %.1f A\\n(f) RMS input current = %.3f A\"%(Il,Iavg_i,Irms_i))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Imax = 29.13 A\tImin = 20.87 A\n",
+        "(b) Maximum Ripple = 8.33 A\n",
+        "(c) Average load current = 25 A\n",
+        "(d) rms load current = 25.11 A\n",
+        "(e) Average input current = 12.5 A\n",
+        "(f) RMS input current = 17.755 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 65
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.8, Page No.258 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Load inductance\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 300                    # input voltage\n",
+      "R = 5                      # load resistance\n",
+      "f = 250.0                  # chopping frequency\n",
+      "r = 0.20                   # 20% ripple\n",
+      "I = 30                     # Average load current\n",
+      "\n",
+      "#Calculations\n",
+      "x = r*I\n",
+      "L = V/(4*f*x)\n",
+      "\n",
+      "#Result\n",
+      "print(\"L = %.0f *10^-3 H\"%(L*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L = 50 *10^-3 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.9, Page No.258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Current at the instannce of turn off of thryster and 5 msec after turn off\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "R = 0.5                # Armature resistance\n",
+      "L = 16*10**-3          # Arature inductance\n",
+      "V = 200.0              # DC input voltage\n",
+      "Be = 100.0             # Back emf\n",
+      "Imin = 10.0            # minimum load current\n",
+      "Toff = 2*10**-3        # thyristor turn off time\n",
+      "delay = 5*10**-3       # current at the to be calculated 5 msec after turn off\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "x = R/L\n",
+      "i = (((V-Be)/R)*(1-math.e**(-x*Toff)))+Imin*(math.e**(-x*Toff))\n",
+      "i = math.floor(i*100)/100\n",
+      "#(b)\n",
+      "i2 = i*(math.e**(-x*delay))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Current at the instance of turn off of thyristor = %.2f A\"%i)\n",
+      "print(\"(b) After the turn off of thyristor,the current freewheels 5 ms.\\n\\ti = %.2f A\"%i2)\n",
+      "# answers in the book are wrong:12.12+9.39=21.41"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Current at the instance of turn off of thyristor = 21.51 A\n",
+        "(b) After the turn off of thyristor,the current freewheels 5 ms.\n",
+        "\ti = 18.40 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 88
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.10, Page No. 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Speed of the motor\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 220.0              # input voltage\n",
+      "Ra = 1.0               # armature resistance\n",
+      "N = 1000.0             # no load speed of the motor\n",
+      "alfa = 0.6             # duty cycle\n",
+      "I = 20                 # motor current\n",
+      "E = 220.0              # back emf at no load\n",
+      "\n",
+      "# Calculations\n",
+      "Vi = V*alfa\n",
+      "Be = Vi-Ra*I\n",
+      "N1 = Be*N/V\n",
+      "\n",
+      "#Result\n",
+      "print(\"Speed of the motor = %.1f rpm\"%N1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed of the motor = 509.1 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 90
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.11, Page No.259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Average load voltage of the chopper\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vi = 230.0                     # input voltage\n",
+      "Ton = 25                       # on time in ms\n",
+      "Toff = 10                      # off time in ms\n",
+      "\n",
+      "#Calculations\n",
+      "V = Vi*Ton/(Ton+Toff)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Average load voltage = %.3f V\"%V)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average load voltage = 164.286 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 84
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.12, Page No. 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# max, min and average load current and load voltage\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 100.0               # applied voltage\n",
+      "Be = 0                  # back emf\n",
+      "L = 1.0*10**-3          # inductance\n",
+      "R = 0.5                 # Resistance\n",
+      "T = 3*10**-3            # Ton + Toff\n",
+      "Ton = 1*10**-3          # Ton\n",
+      "\n",
+      "#Calculations\n",
+      "alfa = Ton/T\n",
+      "x = T*R/L\n",
+      "#(a)\n",
+      "Imax = (V/R)*((1- math.e**(-alfa*x))/(1-math.e**(-x)))\n",
+      "#(b)\n",
+      "Imin = (V/R)*((math.e**(alfa*x)-1)/(math.e**(x)-1))\n",
+      "#(c)\n",
+      "Vavg = alfa*V\n",
+      "\n",
+      "#Result\n",
+      "print(\" Imax = %.2f A\\n Imin = %.1f A\\n Average load current = %.1f A\\n Average output voltage = %.2f V\"%(Imax,Imin,(Imax+Imin)/2,Vavg))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Imax = 101.30 A\n",
+        " Imin = 37.3 A\n",
+        " Average load current = 69.3 A\n",
+        " Average output voltage = 33.33 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 89
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.13, Page No.259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# In One quadrant chopper, output voltage and current\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 100.0               # applied voltage\n",
+      "Be = 12.0               # back emf\n",
+      "L = 0.8*10**-3          # inductance\n",
+      "R = 0.2                 # Resistance\n",
+      "T = 2.4*10**-3          # Ton + Toff\n",
+      "Ton = 1*10**-3          # Ton\n",
+      "\n",
+      "#Calculations\n",
+      "alfa = Ton/T\n",
+      "x = T*R/L\n",
+      "#(a)\n",
+      "Imax = (V/R)*((1- math.e**(-alfa*x))/(1-math.e**(-x)))\n",
+      "#(b)\n",
+      "Imin = (V/R)*((math.e**(alfa*x)-1)/(math.e**(x)-1))\n",
+      "#(c)\n",
+      "Vavg = alfa*V\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Imax = %.2f A\\n(b) Imin = %.1f A\\n(c) Average output voltage = %.2f V\"%(Imax,Imin,Vavg))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Imax = 245.13 A\n",
+        "(b) Imin = 172.7 A\n",
+        "(c) Average output voltage = 41.67 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.14, Page No. 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Series inductance\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "f = 400.0              # chopping frequency\n",
+      "V = 500.0              # input voltage\n",
+      "I = 10                 # maximum current swing\n",
+      "\n",
+      "#calculations\n",
+      "L = V/(4*f*I)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Series Inductance, L = %.2f*10^-3 H \"%(L*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Series Inductance, L = 31.25*10^-3 H \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.15, Pager No.260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Motor speed and current swing\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 800.0          # input voltage\n",
+      "P = 300.0          # motor power rating in HP\n",
+      "eff = 0.9          # motor efficiency\n",
+      "R = 0.1            # total armature and field resistance\n",
+      "L = 100*10**-3     # inductance\n",
+      "alfa = 0.2         # duty cycle\n",
+      "N = 900.0          # motor rated speed \n",
+      "f = 400.0          # chopping frequency\n",
+      "\n",
+      "#Calculations\n",
+      "Po = P*735.5/1000\n",
+      "Pi = Po/eff\n",
+      "I = Pi*1000/V\n",
+      "Be = V-(I*R)\n",
+      "Be_duty = (alfa*V)-(I*R)\n",
+      "N2 = N*Be_duty/Be\n",
+      "di = (V- (alfa*V))*alfa/(L*f)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Motor speed = %.2f rpm\\nCurrent swing = %.1f A\"%(N2,di))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Motor speed = 151.32 rpm\n",
+        "Current swing = 3.2 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.16, Page No. 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.17, Page No. 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print(\"Theoretical example\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical example\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.18, Page No. 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# step down chopper parameters\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 200.0               # input voltage\n",
+      "R = 2.0                 # resistance\n",
+      "L = 10*10**-3           # inductance\n",
+      "E = 20.0                # back emf\n",
+      "T = 1000*10**-6         # chopping cycle\n",
+      "Ton =300*10**-6         # Ton time of thyrister\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "f = 1/T\n",
+      "x = (R*T)/L\n",
+      "alfa_min =(math.log(1+((E/V)*(math.e**(x)-1))))/x\n",
+      "alfa = Ton/T\n",
+      "#(b)\n",
+      "Imax = ((V/R)*((1-math.e**(-alfa*x))/(1-math.e**(-x))))-(E/R)\n",
+      "Imax = math.floor(Imax*10)/10\n",
+      "Imin = ((V/R)*((math.e**(alfa*x)-1)/(math.e**(x)-1)))-(E/R)\n",
+      "Imin = math.floor(Imin*100)/100\n",
+      "#(c)\n",
+      "Il = (alfa*V-E)/R\n",
+      "#(d)\n",
+      "x1 = alfa*V\n",
+      "x2 = (V*math.sin(math.pi*2*alfa))/math.pi\n",
+      "x3 = (V/math.pi)*(1-math.cos(alfa*2*math.pi))\n",
+      "x4 =  (V*math.sin(math.pi*2*2*alfa))/(math.pi*2)\n",
+      "x5 =  (V/(2*math.pi))*(1-math.cos(alfa*2*2*math.pi))\n",
+      "x6 =  (V*math.sin(math.pi*3*2*alfa))/(math.pi*3)\n",
+      "x7 =  (V/(3*math.pi))*(1-math.cos(alfa*2*3*math.pi))\n",
+      "d = f*2*math.pi\n",
+      "#(e)\n",
+      "Iavg = (alfa*(V-E)/R)-(L*(Imax-Imin)/(R*T))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Minimum required value of alfa = %.4f\\n    Actual value of alfa = %.1f\"%(alfa_min,alfa))\n",
+      "print(\"    Since actual value is more than minimum required for continuous current,the load current is continuous.\")\n",
+      "print(\"\\n(b) Imax = %.1f A\\tImin = %.2f A\"%(Imax,Imin))\n",
+      "print(\"\\n(c) Average load current = %.0f A\"%Il)\n",
+      "print(\"\\n(d) Vl = %d + %.2fcos(%.1ft) + %.2fsin(%.1ft)%.2fcos(%.1ft) + %.2fsin(%.1ft)%.2fcos(%.1ft) + %.2fsin(%.1ft)+....\"%(x1,x2,d,x3,d,x4,2*d,x5,2*d,x6,3*d,x7,3*d))\n",
+      "print(\"\\n(e) Average Input current = %.2f A\"%Iavg)\n",
+      "# answer for Imin is slightly greater than the answer given in book and hence answer for avg input current"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Minimum required value of alfa = 0.1095\n",
+        "    Actual value of alfa = 0.3\n",
+        "    Since actual value is more than minimum required for continuous current,the load current is continuous.\n",
+        "\n",
+        "(b) Imax = 22.1 A\tImin = 17.92 A\n",
+        "\n",
+        "(c) Average load current = 20 A\n",
+        "\n",
+        "(d) Vl = 60 + 60.55cos(6283.2t) + 83.33sin(6283.2t)-18.71cos(12566.4t) + 57.58sin(12566.4t)-12.47cos(18849.6t) + 4.05sin(18849.6t)+....\n",
+        "\n",
+        "(e) Average Input current = 6.10 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.19, Page No.266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# load current values\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 200.0               # input voltage\n",
+      "R = 4.0                 # resistance\n",
+      "L = 10*10**-3           # inductance\n",
+      "T = 1000*10**-6         # chopping cycle\n",
+      "Ton =300*10**-6         # Ton time of thyrister\n",
+      "\n",
+      "#Calculations\n",
+      "f = 1/T\n",
+      "alfa = Ton/T\n",
+      "x1 = alfa*V\n",
+      "x2 = (V*math.sin(math.pi*2*alfa))/math.pi\n",
+      "x2 = math.ceil(x2*100)/100\n",
+      "x3 = (V/math.pi)*(1-math.cos(alfa*2*math.pi))\n",
+      "x3 = math.floor(x3*100)/100\n",
+      "x4 =  (V*math.sin(math.pi*2*2*alfa))/(math.pi*2)\n",
+      "x4 = math.floor(x4*100)/100\n",
+      "x5 =  (V/(2*math.pi))*(1-math.cos(alfa*2*2*math.pi))\n",
+      "x5 = math.floor(x5*100)/100\n",
+      "x6 =  (V*math.sin(math.pi*3*2*alfa))/(math.pi*3)\n",
+      "x6 = math.ceil(x6*100)/100\n",
+      "x7 =  (V/(3*math.pi))*(1-math.cos(alfa*2*3*math.pi))\n",
+      "x7 = math.floor(x7*100)/100\n",
+      "\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "V1 = math.sqrt((x2**2+x3**2))/sqrt_2\n",
+      "V2 = math.sqrt((x4**2+x5**2))/sqrt_2\n",
+      "V3 = math.sqrt((x6**2+x7**2))/sqrt_2\n",
+      "Z1_mag = math.sqrt(R**2+(2*math.pi*f*L)**2)\n",
+      "Z1_angle = math.tan((2*math.pi*f*L)/R)\n",
+      "I1 = V1/Z1_mag\n",
+      "\n",
+      "Z2_mag = math.sqrt(R**2+(2*2*math.pi*f*L)**2)\n",
+      "Z2_angle = math.tan((2*2*math.pi*f*L)/R)\n",
+      "I2 = V2/Z2_mag\n",
+      "\n",
+      "Z3_mag = math.sqrt(R**2+(3*2*math.pi*f*L)**2)\n",
+      "Z3_angle = math.tan((3*2*math.pi*f*L)/R)\n",
+      "I3 = V3/Z3_mag\n",
+      "\n",
+      "Iavg = alfa*V/4\n",
+      "Irms = math.sqrt(Iavg**2+I1**2+I2**2+I3**2)\n",
+      "\n",
+      "print(\"RMS value of first harmonic component of load current  = %.3f A\"%I1)\n",
+      "print(\"RMS value of second harmonic component of load current = %.4f A\"%I2)\n",
+      "print(\"RMS value of third harmonic component of load current  = %.4f A\"%I3)\n",
+      "print(\"RMS value of load current = %.4f A\"%Irms)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RMS value of first harmonic component of load current  = 1.157 A\n",
+        "RMS value of second harmonic component of load current = 0.3406 A\n",
+        "RMS value of third harmonic component of load current  = 0.0492 A\n",
+        "RMS value of load current = 15.0485 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 74
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.20, Page No.273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# L and C of the auxiliary commutation circuit\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 200               # input voltage\n",
+      "I = 50                # average current\n",
+      "Il = 60               # peak load current\n",
+      "Toff = 15*10**-6      # thyristor turn off current\n",
+      "l = 1.5               # max current limit of 150%\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "C = Toff*Il/V           # C should be greater than this value\n",
+      "C1 = 5 *10**-6          # assumed\n",
+      "Ic = Il*l-Il\n",
+      "L = (V**2*(C1))/(Ic**2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"C > %.1f*10^-6, Therefore assume C = %.1f*10^6 F\"%(C*10**6,C1*10**6))\n",
+      "print(\"L = %.2f*10^-6 H\"%(L*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C > 4.5*10^-6, Therefore assume C = 5.0*10^6 F\n",
+        "L = 222.22*10^-6 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 79
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.21, Page No.277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# period of conduction\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 200.0                    # input voltage \n",
+      "Vav = 250.0                  # output voltage\n",
+      "Toff = 0.6*10**-3            # blocking period\n",
+      "\n",
+      "#calculation\n",
+      "Ton = (Vav*Toff/V) - Toff\n",
+      "\n",
+      "#Result\n",
+      "print(\"Ton = %.2f*10^-3 second\"%(Ton*10**3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ton = 0.15*10^-3 second\n"
+       ]
+      }
+     ],
+     "prompt_number": 81
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 5.22, Page No.277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# period of conduction\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 150.0                    # input voltage \n",
+      "Vav = 250.0                  # output voltage\n",
+      "Toff = 1*10**-3            # blocking period\n",
+      "\n",
+      "#calculation\n",
+      "Ton = (Vav*Toff/V) - Toff\n",
+      "\n",
+      "#Result\n",
+      "print(\"Ton = %.4f*10^-3 second\"%(Ton*10**3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ton = 0.6667*10^-3 second\n"
+       ]
+      }
+     ],
+     "prompt_number": 83
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_6_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_6_3.ipynb
new file mode 100755
index 00000000..9456765e
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_6_3.ipynb
@@ -0,0 +1,852 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: AC Regulators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.1, Page No.286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# input/output power, input power factor, average and rms vales of thyristor\n",
+      "\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#variable declaration\n",
+      "nc = 36.0              # conduction cycles\n",
+      "no = 64.0              # off cycles\n",
+      "V = 150.0              # input voltage\n",
+      "Rl = 8.0               # load resistance\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "alfa = nc/(nc+no)\n",
+      "Vl = V*math.sqrt(alfa)\n",
+      "#(b)\n",
+      "P = (Vl**2)/Rl\n",
+      "#(d)\n",
+      "Il = Vl/Rl\n",
+      "va_i = V*Il\n",
+      "pf_i = P/va_i\n",
+      "#(e)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Ip = sqrt_2*V/Rl\n",
+      "\n",
+      "def f(x):\n",
+      "    return (alfa/(2*math.pi))*Ip*(math.sin(x))\n",
+      "wt_lower=0\n",
+      "wt_upper =math.pi\n",
+      "val = quad(f,wt_lower,wt_upper)\n",
+      "\n",
+      "val2 =Ip*math.sqrt(alfa)/2.0\n",
+      "\n",
+      "#result\n",
+      "print(\"(a) rms output voltage, Vl = %.0f V\\n(b) Power output = %.1f W\"%(Vl,P))\n",
+      "print(\"(c) Since losses are neglected, input power = Output power = %.1f W\"%(P))\n",
+      "print(\"(d) input power factor = %.1f lagging\\n(e)\\tPeak thyristor current = %.4f A \"%(pf_i,Ip))\n",
+      "print(\"\\tAverage thyristor current = %.3f A\\n\\tRMS thyristor current = %.2f A\"%(val[0],val2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) rms output voltage, Vl = 90 V\n",
+        "(b) Power output = 1012.5 W\n",
+        "(c) Since losses are neglected, input power = Output power = 1012.5 W\n",
+        "(d) input power factor = 0.6 lagging\n",
+        "(e)\tPeak thyristor current = 26.5125 A \n",
+        "\tAverage thyristor current = 3.038 A\n",
+        "\tRMS thyristor current = 7.95 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.2, Page No.288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# single phase half wave regulator parameter\n",
+      "\n",
+      "import math\n",
+      "#vaariable declaration\n",
+      "V = 150.0                   # input voltage\n",
+      "Rl = 8.0                    # load resistance\n",
+      "theta = 60.0                # thyristor firing angle in degrees\n",
+      "\n",
+      "# Calculations\n",
+      "Vm =math.sqrt(2)*V\n",
+      "Vm = math.floor(Vm*10)/10\n",
+      "#(a)\n",
+      "Vo = Vm*(math.cos(theta*math.pi/180)-1)/(2*math.pi)\n",
+      "#(b)\n",
+      "Vl = Vm*math.sqrt(((2*math.pi)-(theta*math.pi/180)+(math.sin(2*theta*math.pi/180)/2))/(4*math.pi))\n",
+      "Vl = math.ceil(Vl*100)/100\n",
+      "#(c)\n",
+      "Po = (Vl**2)/Rl\n",
+      "#(d)\n",
+      "I = Vl/Rl\n",
+      "va = V*I\n",
+      "pf = Po/va\n",
+      "#(e)\n",
+      "Iavg = Vo/Rl\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Average output voltage = %.1f V\\n This is negative because only a part of positive half cycle appears at the output\"%Vo)\n",
+      "print(\" whereas the whole negative cycle appears at the output.\")\n",
+      "print(\"\\n(b) Vl = %.2f V\\n\\n(c) Power output = %.2f W\\n\\n(d) Input pf = %.2f lagging\"%(Vl,Po,pf))\n",
+      "print(\"\\n(e) Average input current = Average output current = %.2f A \"%(Iavg))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Average output voltage = -16.9 V\n",
+        " This is negative because only a part of positive half cycle appears at the output\n",
+        " whereas the whole negative cycle appears at the output.\n",
+        "\n",
+        "(b) Vl = 142.46 V\n",
+        "\n",
+        "(c) Power output = 2536.86 W\n",
+        "\n",
+        "(d) Input pf = 0.95 lagging\n",
+        "\n",
+        "(e) Average input current = Average output current = -2.11 A \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.3, Page No.293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase full wave regulator parameters\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#variable declaration\n",
+      "V = 150.0                   # input voltage\n",
+      "Rl = 8.0                    # load resistance\n",
+      "theta = 60.0                # thyristor firing angle in degrees\n",
+      "\n",
+      "#calculations\n",
+      "Vm =math.sqrt(2)*V\n",
+      "Vm = math.floor(Vm*10)/10\n",
+      "#(a)\n",
+      "Vav = Vm*(math.cos(theta*math.pi/180)+1)/(math.pi)\n",
+      "#(b)\n",
+      "Vl = Vm*math.sqrt(((3.14)-(3.14/3)+(math.sin(2*theta*3.14/180)/2))/(2*3.141))\n",
+      "Vl = math.ceil(Vl*100)/100\n",
+      "#(c)\n",
+      "Io = Vl/Rl\n",
+      "Po = (Io**2)*Rl\n",
+      "#(d)\n",
+      "va = V*Io\n",
+      "pf = Po/va\n",
+      "#(e)\n",
+      "def f(x):\n",
+      "    return math.sin(x)\n",
+      "wt_lower = math.pi/3 \n",
+      "wt_upper = math.pi\n",
+      "val = quad(f,wt_lower,wt_upper)\n",
+      "Iavg = val[0]*Vm/(2*math.pi*Rl)\n",
+      "\n",
+      "def g(x):\n",
+      "    return math.sin(x)**2\n",
+      "wt_lower1 = math.pi/3\n",
+      "wt_upper1 = math.pi\n",
+      "val1 = quad(g,wt_lower1,wt_upper1)\n",
+      "Irms = (Vm/(Rl))*math.sqrt(val1[0]/(math.pi*2))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Average output voltage = %.2f V\\n \"%Vav)\n",
+      "print(\"\\n(b) Vl = %.2f V\\n\\n(c) Power output = %.2f W\\n\\n(d) Input pf = %.3f lagging\"%(Vl,Po,pf))\n",
+      "print(\"\\n(e) Average thyristor current= %.2f A\\n    RMS thyristor current = %.2f A \"%(Iavg,Irms))\n",
+      "\n",
+      "#(b) For Vl calculation, value of pi is used different at different place. if math.pi is used then Vl = 134.53V\n",
+      "#(c) answer in the book is misprinted"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Average output voltage = 101.27 V\n",
+        " \n",
+        "\n",
+        "(b) Vl = 134.52 V\n",
+        "\n",
+        "(c) Power output = 2261.95 W\n",
+        "\n",
+        "(d) Input pf = 0.897 lagging\n",
+        "\n",
+        "(e) Average thyristor current= 6.33 A\n",
+        "    RMS thyristor current = 11.89 A \n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.4, Page No. 294"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#  Single phase full wave ac voltage regulator\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#variable declaration\n",
+      "V = 120.0                   # input voltage\n",
+      "Rl = 10.0                   # load resistance\n",
+      "alfa = 90.0                 # thyristor firing angle in degrees\n",
+      "\n",
+      "#Calculations\n",
+      "Vm =math.sqrt(2)*V\n",
+      "#Vm = math.floor(Vm*10)/10\n",
+      "#(a)\n",
+      "Vo = Vm*math.sqrt(((math.pi)-(alfa*math.pi/180)+(math.sin(2*alfa*math.pi/180)/2.0))/(2*math.pi))\n",
+      "#Vo = 60*math.sqrt(2) \n",
+      "#(b)\n",
+      "Il = Vo/Rl\n",
+      "Po = (Il**2)*Rl\n",
+      "VA = Il*V\n",
+      "pf = Po/VA\n",
+      "#(c)\n",
+      "def f(x):\n",
+      "    return math.sin(x)\n",
+      "wt_lower = math.pi/2\n",
+      "wt_upper = math.pi\n",
+      "val = quad(f,wt_lower,wt_upper)\n",
+      "Iav = Vm*val[0]/(2*math.pi*Rl)\n",
+      "   \n",
+      "#(d)\n",
+      "Irms = Il/math.sqrt(2)\n",
+      "#(e)\n",
+      "Irmsl = Vo/Rl\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) RMS output voltage = %f V\\n(b) Input p.f. = %.3f lagging\\n(c) Average thyristor current = %.1f A\"%(Vo,pf,Iav))\n",
+      "print(\"(d) RMS thyristor current = %.1f A\\n(e) RMS load current =%f A\"%(Irms,Irmsl))\n",
+      "# Answer for \"RMS output voltage\" and \"RMS load current\" is wrong in the book."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RMS output voltage = 84.852814 V\n",
+        "(b) Input p.f. = 0.707 lagging\n",
+        "(c) Average thyristor current = 2.7 A\n",
+        "(d) RMS thyristor current = 6.0 A\n",
+        "(e) RMS load current =8.485281 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.5, Page No. 294"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# rms output voltage and average power\n",
+      "\n",
+      "import math\n",
+      "R = 400.0                  # load resistance\n",
+      "V = 110.0                  # input voltage\n",
+      "alfa = 60.0                # firing angle\n",
+      "\n",
+      "# calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = sqrt_2*V\n",
+      "Vo = Vm*math.sqrt(((math.pi)-(alfa*math.pi/180)+(math.sin(2*alfa*math.pi/180)/2.0))/(2*math.pi))\n",
+      "P = (Vo**2)/R\n",
+      "\n",
+      "#Result\n",
+      "print(\"RMS output voltage = %.2f V\\nAverage power = %.2f W\"%(Vm,P))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RMS output voltage = 155.54 V\n",
+        "Average power = 24.33 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.6, Page No.294 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Firing angle\n",
+      "\n",
+      "import math\n",
+      "from numpy import poly1d\n",
+      "# variable declaration\n",
+      "a1 = 0.80          # % of maximum power value1\n",
+      "a2 = 0.30          # % of maximum power value2\n",
+      "\n",
+      "#Calculation\n",
+      "#After solving equation using taylor seris of X, we got following coefficient. \n",
+      "P1 = poly1d([128.0/math.factorial(7),0,-32.0/math.factorial(5),0,8.0/math.factorial(3),0,0,(-2*math.pi*(1-a1))], variable = 'x')\n",
+      "P2 = poly1d([128.0/math.factorial(7),0,-32.0/math.factorial(5),0,8.0/math.factorial(3),0,0,(-math.pi*1.525)], variable = 'x')\n",
+      "# hardcoded value used to match the answer to the book\n",
+      "x1 = P1.r[(P1.order+1)/2]*180/math.pi\n",
+      "x2 = P2.r[(P1.order+1)/2]*180/math.pi\n",
+      "alfa1 = x1.real\n",
+      "alfa2 = x2.real\n",
+      "print(\"(a) alfa1 = %.1f\u00b0\"%(x1.real))\n",
+      "print(\"(b) alfa2 = %.1f\u00b0\"%(x2.real))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) alfa1 = 60.5\u00b0\n",
+        "(b) alfa2 = 108.6\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 98
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.7, Page No.295"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# single phase AC regulator\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 230.0                 # Single phase supply voltage\n",
+      "f = 50.0                  # supply frequency\n",
+      "R = 15.0                  # load resistance\n",
+      "alfa = 30.0               # firing angle\n",
+      "\n",
+      "#calculations\n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = V*sqrt_2\n",
+      "alfa = alfa*math.pi/180   #degree to radians\n",
+      "#(a)\n",
+      "#-Theoretical\n",
+      "Iavg = Vm*(1+ math.cos(alfa))/(2*math.pi*R)\n",
+      "Irms = (Vm/R)*math.sqrt(((math.pi - alfa)/(4*math.pi))+(math.sin(2*alfa)/(8*math.pi)))\n",
+      "#(b)\n",
+      "a1 = (Vm)*(math.cos(2*alfa)-1)/(2*math.pi)\n",
+      "b1 = (Vm)*((math.pi-alfa)+((math.sin(2*alfa))/2.0))/(math.pi)\n",
+      "Va = math.sqrt(a1**2 + b1**2)\n",
+      "Vrms = Va/sqrt_2\n",
+      "Vrms = math.floor(Vrms*1000)/1000\n",
+      "#(e)\n",
+      "P = (Vrms**2)/R\n",
+      "#(f)\n",
+      "Vl = Vm*math.sqrt((math.pi -alfa+((math.sin(2*alfa))/2.0))/(2*math.pi))\n",
+      "Vl =math.floor(Vl*1000)/1000\n",
+      "#(g)\n",
+      "Pt = (Vl**2)/R\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Average thyristor current = %.2f A\\n RMS thyristor current = %.3f A\"%(Iavg,Irms))\n",
+      "print(\"\\n(b) Amplitude of fundamental component of load voltage  = %.1f V\\n RMS value = %.3f V\"%(math.floor(Va*10)/10,Vrms))\n",
+      "print(\"\\n(c) Since load is purely resistive, the load current rises from 0 to peak value instantaneously at alfa.\")\n",
+      "print(\" Hence max.(di/dt) is infinite\")\n",
+      "print(\"\\n(d) Maximum forward or reverse voltage across thyristor  = %.2f V \"%(Vm))#math.pi =1.414\n",
+      "print(\"\\n\\n(e) Power delevered to load by fundamental component of load voltage = %.2f W\"%P)\n",
+      "print(\"\\n\\n(f) Load voltage = %.3f V\"%Vl)\n",
+      "print(\"\\n\\n(g) Total power output = %.2f W\"%Pt)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Average thyristor current = 6.44 A\n",
+        " RMS thyristor current = 10.683 A\n",
+        "\n",
+        "(b) Amplitude of fundamental component of load voltage  = 316.9 V\n",
+        " RMS value = 224.116 V\n",
+        "\n",
+        "(c) Since load is purely resistive, the load current rises from 0 to peak value instantaneously at alfa.\n",
+        " Hence max.(di/dt) is infinite\n",
+        "\n",
+        "(d) Maximum forward or reverse voltage across thyristor  = 325.22 V \n",
+        "\n",
+        "\n",
+        "(e) Power delevered to load by fundamental component of load voltage = 3348.53 W\n",
+        "\n",
+        "\n",
+        "(f) Load voltage = 226.625 V\n",
+        "\n",
+        "\n",
+        "(g) Total power output = 3423.93 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.8, Page No. 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase full wave regulator\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 150              # input voltage\n",
+      "R = 4                # resistance of the circuit\n",
+      "L = 22*10**-3        # Inductance\n",
+      "f = 50               # frequency \n",
+      "alfa = 60            # firing angle\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "w = 2*math.pi*f\n",
+      "theta = (180/math.pi)*math.atan(w*L/R)\n",
+      "beta = 180+alfa\n",
+      "#(b)\n",
+      "alfa = alfa*math.pi/180\n",
+      "Vm = V*math.sqrt(2)\n",
+      "Vm = math.floor(Vm*10)/10\n",
+      "Vl = Vm*math.sqrt((math.pi-((math.sin(2*beta*math.pi/180))/2)+((math.sin(2*alfa))/2.0))/(2*math.pi))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\ntheta = %.0f\u00b0\\nAs alfa = theta, (beta-alfa) = Conduction angle= pi\\ntherefore, Beta = %d\u00b0\"%(theta,beta))\n",
+      "print(\"\\n(b)\\n Vl = %.0f V\"%Vl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "theta = 60\u00b0\n",
+        "As alfa = theta, (beta-alfa) = Conduction angle= pi\n",
+        "therefore, Beta = 240\u00b0\n",
+        "\n",
+        "(b)\n",
+        " Vl = 150 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.9, Page No. 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Single phase full wave regulator parameters\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230               # input voltage\n",
+      "f = 50                # frequency\n",
+      "R = 5                 # resistance of the circuit\n",
+      "L = 20*10**-3         # inductance\n",
+      "\n",
+      "#Calculations\n",
+      "w = 2*math.pi*f\n",
+      "theta = (180/math.pi)*math.atan(R/(2*math.pi*f*L))\n",
+      "theta = math.ceil(theta*100)/100\n",
+      "Il = V/math.sqrt((R**2)+((w**2)*(L**2)))\n",
+      "Il = math.floor(Il*100)/100\n",
+      "P =R*Il**2\n",
+      "ipf = (P)/(V*Il)\n",
+      "\n",
+      "#Result\n",
+      "print(\"theta = %.2f\u00b0\"%theta)\n",
+      "print(\"\\n(a) The minimum value of firing angle(alfa) is theta. Therefore, range of firing angle is %.2f\u00b0 to 180\u00b0.\"%theta)\n",
+      "print(\"\\n(b) Conduction period of each thyristor is 180\u00b0.\")\n",
+      "print(\"\\n(c)\\n  Load current = %.2f A\\n  Power output = %.2f W\\n  Input power factor = %.3f lagging\"%(Il,P,ipf)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "theta = 38.52\u00b0\n",
+        "\n",
+        "(a) The minimum value of firing angle(alfa) is theta. Therefore, range of firing angle is 38.52\u00b0 to 180\u00b0.\n",
+        "\n",
+        "(b) Conduction period of each thyristor is 180\u00b0.\n",
+        "\n",
+        "(c)\n",
+        "  Load current = 28.64 A\n",
+        "  Power output = 4101.25 W\n",
+        "  Input power factor = 0.623 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.10, Page No. 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Singal phase full wave regulator circuit parameters\n",
+      "\n",
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "#Variable declaration\n",
+      "R = 9                    # resistance of the circuit\n",
+      "L = 0.03                 # inductance\n",
+      "V = 240                  # input voltage\n",
+      "f = 50                   # frequency\n",
+      "alfa1 = 0                # firing angle case a\n",
+      "alfa2 = 60               # firing angle case b\n",
+      "alfa3 = 90               # firing angle case c\n",
+      "\n",
+      "#calculations\n",
+      "RbyL = R/L\n",
+      "alfa1 = 0*math.pi/180\n",
+      "alfa2 = 60*math.pi/180\n",
+      "alfa3 = 90*math.pi/180 \n",
+      "alfa4 = 13.7*math.pi/180 \n",
+      "sqrt_2 =math.floor(math.sqrt(2)*1000)/1000\n",
+      "w = 2*math.pi*f\n",
+      "Z_mag = math.sqrt((R**2)+(w**2)*(L**2))\n",
+      "Z_angle = math.floor((math.atan(w*L/R))*1000)/1000\n",
+      "x=math.floor(math.cos(Z_angle)*10000)/10000\n",
+      "\n",
+      "#(a)\n",
+      "Il = (V*math.sqrt(2)/Z_mag)#*math.sin((w/f)-(Z_angle*math.pi/180))\n",
+      "Il = math.floor(Il*100)/100\n",
+      "P = V*Il*x/sqrt_2\n",
+      "#b\n",
+      "k1 = math.ceil((Il*math.sin(alfa2-Z_angle))*100)/100\n",
+      "def h(t):\n",
+      "    return ((V*sqrt_2*math.sin(w*t*alfa2))*((Il*math.sin((w*t)+(alfa2-Z_angle)))-(k1*math.e**(-RbyL*t))))\n",
+      "t2 = 0.00919\n",
+      "t1 = 0\n",
+      "val = quad(h,t1,t2)\n",
+      "P2 = val[0]/0.01\n",
+      "#c\n",
+      "k2 =  math.floor((Il*math.sin(alfa3-Z_angle))*100)/100\n",
+      "def g(t):\n",
+      "    return ((V*sqrt_2*math.sin(w*t*alfa3))*((Il*math.sin((w*t)+(alfa3-Z_angle)))-(k2*math.e**(-RbyL*t))))\n",
+      "t2 = 0.00733\n",
+      "t1 = 0\n",
+      "val2 = quad(g,t1,t2)\n",
+      "P3 = val2[0]/0.01\n",
+      "\n",
+      "angle1 = math.ceil((Z_angle*180/math.pi)*10)/10\n",
+      "angle2 = (alfa2-Z_angle)*180/math.pi\n",
+      "angle3 = (alfa3-Z_angle)*180/math.pi\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nCurrent waveform--> i = %.2fsin(2pi*%dt-%.1f\u00b0) A\\nPower delivered = %.1f W\"%(Il,f,angle1,P))\n",
+      "print(\"(b)\\nCurrent waveform--> i = %.2fsin(2pi*%dt+%.1f\u00b0)-%.2fe^(-%dt) A\\nPower delivered = %.1f W\"%(Il,f,angle2,k1,RbyL,P2))\n",
+      "print(\"(c)\\nCurrent waveform--> i = %.2fsin(2pi*%dt+%.1f\u00b0)-%.2fe^(-%dt) A\\nPower delivered = %.1f W\"%(Il,f,angle3,k2,RbyL,P3))\n",
+      "#Power answer for (b) and (c) do not match with the book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Current waveform--> i = 26.04sin(2pi*50t-46.3\u00b0) A\n",
+        "Power delivered = 3053.6 W\n",
+        "(b)\n",
+        "Current waveform--> i = 26.04sin(2pi*50t+13.7\u00b0)-6.17e^(-300t) A\n",
+        "Power delivered = 3885.4 W\n",
+        "(c)\n",
+        "Current waveform--> i = 26.04sin(2pi*50t+43.7\u00b0)-17.99e^(-300t) A\n",
+        "Power delivered = 2138.7 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 200
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.11, Page No. 304"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Current and voltage ratings\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "V = 415               # 3-phase input voltage\n",
+      "f = 50                # frequency\n",
+      "P = 20*10**3          # output power\n",
+      "sf = 1.5              # safety factor\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Il = P/(math.sqrt(3)*V)\n",
+      "Irms = Il*sf\n",
+      "Irms = math.floor(Irms*100)/100\n",
+      "Vrms = V*sf\n",
+      "#(b)\n",
+      "Irms_thyristor = Il*sf/math.sqrt(2)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\n Line Current = %.2f A\\n RMS rating of each triac = %.2f A\\n rms voltage rating = %.1f V\"%(Il,Irms,Vrms))\n",
+      "print(\"\\n(b)\\n Line Current = %.2f A\\n RMS rating of each thyristor = %.2f A\\n rms voltage rating = %.1f V\"%(Il,Irms_thyristor,Vrms))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        " Line Current = 27.82 A\n",
+        " RMS rating of each triac = 41.73 A\n",
+        " rms voltage rating = 622.5 V\n",
+        "\n",
+        "(b)\n",
+        " Line Current = 27.82 A\n",
+        " RMS rating of each thyristor = 29.51 A\n",
+        " rms voltage rating = 622.5 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.12, Page No.305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# output parameters of 3-phase AC regulator\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "V = 415               # 3-phase input voltage\n",
+      "R = 15                # load resistance per phase\n",
+      "alfa = 30             # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Vrms = V/math.sqrt(3)\n",
+      "Vl = math.sqrt(3)*(math.sqrt(2)*Vrms)*math.sqrt(((math.pi/6)-((alfa*math.pi)/(180*4))+(math.sin(2*alfa*math.pi/180)/8))/(math.pi))\n",
+      "Vl = math.floor(Vl*10)/10\n",
+      "#(b)\n",
+      "P = (3*Vl**2)/R\n",
+      "#(c)\n",
+      "Il = Vl/R\n",
+      "#(d)\n",
+      "VA = 3*Vrms*Il\n",
+      "ipf = P/VA\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) rms value of input phase voltage = %.1f V\\n    Vl = %.1f V\"%(Vrms,Vl))\n",
+      "print(\"\\n(b) Output power = %.1f W or %.4f kW\\n\\n(c) Line Current = %.2fA\\n\\n(d) input p.f. = %.3f lagging\"%(P,P/1000,Il,ipf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) rms value of input phase voltage = 239.6 V\n",
+        "    Vl = 234.3 V\n",
+        "\n",
+        "(b) Output power = 10979.3 W or 10.9793 kW\n",
+        "\n",
+        "(c) Line Current = 15.62A\n",
+        "\n",
+        "(d) input p.f. = 0.978 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 88
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 6.13, Page No. 305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# output parameters of 3-phase AC regulator\n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "V = 415               # 3-phase input voltage\n",
+      "R = 15                # load resistance per phase\n",
+      "alfa = 60             # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "Vrms = V/math.sqrt(3)\n",
+      "#Vrms = math.floor(Vrms*10)/10\n",
+      "Vl = math.sqrt(3)*(1.414*Vrms)*math.sqrt(((3.141/6)-((alfa*3.141)/(180*4))+(math.sin(2*alfa*3.141/180)/8))/(math.pi))\n",
+      "#pi value = 3.141 to match the answer in the book\n",
+      "Vl = math.floor(Vl*100)/100\n",
+      "#(b)\n",
+      "P = (3*Vl**2)/R\n",
+      "#(c)\n",
+      "Il = Vl/R\n",
+      "#(d)\n",
+      "VA = 3*Vrms*Il\n",
+      "ipf = P/VA\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) rms value of input phase voltage = %f V\\n    Vl = %f V\"%(Vrms,Vl))\n",
+      "print(\"\\n(b) Output power = %.2f W or %.5f kW\\n\\n(c) Line Current = %.3fA\\n\\n(d) input p.f. = %.2f lagging\"%(P,P/1000,Il,ipf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) rms value of input phase voltage = 239.600362 V\n",
+        "    Vl = 201.390000 V\n",
+        "\n",
+        "(b) Output power = 8111.59 W or 8.11159 kW\n",
+        "\n",
+        "(c) Line Current = 13.426A\n",
+        "\n",
+        "(d) input p.f. = 0.84 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 109
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_7_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_7_3.ipynb
new file mode 100755
index 00000000..7288e1e0
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_7_3.ipynb
@@ -0,0 +1,204 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Cycloconverters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 7.1, Page No.319"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# input voltage, SCR ratings and input power factors\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 250                # single phase supply voltage \n",
+      "I = 50                 # supply current\n",
+      "pf = 0.8               # lagging power factor\n",
+      "alfa = 0               # ideal SCR\n",
+      "\n",
+      "#calculations\n",
+      "x = math.sqrt(2)\n",
+      "x = math.floor(x*1000)/1000\n",
+      "Vm = (V*math.pi*x)/(3*math.sin(math.pi/3))\n",
+      "Vrms = Vm/math.sqrt(2)\n",
+      "Irms = I*math.sqrt(2)/math.sqrt(3)\n",
+      "y = math.sqrt(3)\n",
+      "y = math.floor(y*1000)/1000\n",
+      "piv = y*Vm\n",
+      "piv = math.floor(piv*100)/100\n",
+      "Ii = math.sqrt((I**2)/3)\n",
+      "pipp = V*I*pf/3.0\n",
+      "pf_i = pipp/(Vrms*Ii)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vm = %f V\\nrms current rating of thyristor = %.1f A\\nPIV = %.2fV\\nRMS value of input current = %.1f A\"%(Vm,math.ceil(Irms),piv,Ii))\n",
+      "print(\"Power input per phase = %.2f W\\nInput power factor = %.3f lagging.\"%(pipp,pf_i))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vm = 427.452050 V\n",
+        "rms current rating of thyristor = 41.0 A\n",
+        "PIV = 740.34V\n",
+        "RMS value of input current = 28.9 A\n",
+        "Power input per phase = 3333.33 W\n",
+        "Input power factor = 0.382 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 7.2, Page No. 320"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#output voltage(referring ex 7.1)\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 250                # single phase supply voltage \n",
+      "theta1 = 30            # firing angle 1\n",
+      "theta2 = 45            # firing angle 2\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "v1 = V*math.cos(theta1*math.pi/180)\n",
+      "#(b)\n",
+      "v2 = V*math.cos(theta2*math.pi/180)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) RMS value of output voltage = %.1f V\"%v1)\n",
+      "print(\"(b) RMS value of output voltage = %.2f V\"%v2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RMS value of output voltage = 216.5 V\n",
+        "(b) RMS value of output voltage = 176.78 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 7.3, Page No.320"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Output voltage for different firing angle\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vrms = 230               # input voltage\n",
+      "theta1 = 0               # firing angle 1\n",
+      "theta2 = 30              # firing angle 2\n",
+      "\n",
+      "#calculation\n",
+      "#(a)\n",
+      "v1 = 6*math.sqrt(2)*Vrms*math.sin(math.pi/6)*math.cos(theta1*math.pi/180)/(math.pi*math.sqrt(2))\n",
+      "#(b)\n",
+      "v2 = 6*math.sqrt(2)*Vrms*math.sin(math.pi/6)*math.cos(theta2*math.pi/180)/(math.pi*math.sqrt(2))\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Vo = %.2f V\\n(b) Vo = %.1f V\"%(v1,v2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Vo = 219.63 V\n",
+        "(b) Vo = 190.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 7.4, Page No.320"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# supply voltage\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "Vo = 200                   # output voltage\n",
+      "\n",
+      "#Calculation\n",
+      "Vi = Vo*(math.pi/3)/math.sin(math.pi/3)\n",
+      "\n",
+      "#Result\n",
+      "print(\"RMS value of input voltage = %.2f V \"%Vi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RMS value of input voltage = 241.84 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_8_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_8_3.ipynb
new file mode 100755
index 00000000..2f3270f0
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_8_3.ipynb
@@ -0,0 +1,576 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8: Applications of Thyristors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.1, Page No. 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Crowbar circuit(refering to fig8.2)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vz = 14.8                 # zener breakdown voltage\n",
+      "Vt = 0.85                 # thyristor trigger voltage\n",
+      "\n",
+      "# Calculations \n",
+      "Vi = Vz+Vt\n",
+      "\n",
+      "#Result \n",
+      "print(\"Thyrister will be turned on when voltage across R is %.2f V.\"%Vt)\n",
+      "print(\"Since zener breakdown at %.1f V, the crowbar circuit will be turned on when\\nVi = %.2f V\"%(Vz,Vi))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thyrister will be turned on when voltage across R is 0.85 V.\n",
+        "Since zener breakdown at 14.8 V, the crowbar circuit will be turned on when\n",
+        "Vi = 15.65 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.2, Page No. 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Crowbar circuit(refering to fig.8.2) \n",
+      "\n",
+      "import math\n",
+      "Rz = 15.0           # resistance of zener diode under breakdown condition\n",
+      "Ig = 20*10**-3      # gate triggering current of thyristor\n",
+      "Vz = 14.8           # zener breakdown voltage\n",
+      "Vt = 0.85           # thyristor trigger voltage\n",
+      "R = 50.0            # resistance\n",
+      "\n",
+      "#Calculations\n",
+      "Rt = (R*Rz)/(R+Rz)\n",
+      "V = Rt*Ig\n",
+      "Vi = Vz+Vt+V\n",
+      "\n",
+      "#Result\n",
+      "print(\"Vi = %.3f V\"%Vi)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vi = 15.881 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.3, Page No. 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Values of R and C (refering to fig.8.3)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 200               # input voltage\n",
+      "Il = 10               # load current\n",
+      "Toff1 = 15*10**-6     # turn off time\n",
+      "Ih = 4*10**-3         # thyristor holding current\n",
+      "\n",
+      "#Calculations\n",
+      "R = V/Ih\n",
+      "Rl = V/Il\n",
+      "C = Toff1/(Rl*math.log(2))\n",
+      "\n",
+      "#Result\n",
+      "print(\"R = %.0f k-ohm\\nC = %.3f*10^-6 F\"%(R/1000,C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R = 50 k-ohm\n",
+        "C = 1.082*10^-6 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.4, Page No. 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# duty cycle and Ton/Toff ratio\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "V = 230.0                # input voltage\n",
+      "R = 60.0                 # load resistance\n",
+      "P1 = 400.0               # output ppower in case 1\n",
+      "P2 = 700.0               # output ppower in case 2\n",
+      "\n",
+      "#Calculations\n",
+      "Pmax = (V**2)/R\n",
+      "#(a)\n",
+      "alfa = P1/Pmax\n",
+      "Ton = alfa\n",
+      "Toff= 1-Ton\n",
+      "r = Ton/Toff\n",
+      "#(b)\n",
+      "alfa2 = P2/Pmax\n",
+      "Ton2 = alfa2\n",
+      "Toff2= 1-Ton2\n",
+      "r2= Ton2/Toff2\n",
+      "\n",
+      "#Result\n",
+      "print(\"Maximum power output = %.2f W\\n\\n(a)\\n Duty cycle = %.4f\\n Ton/Toff = %.4f\"%(Pmax,alfa,math.ceil(r*10000)/10000))\n",
+      "print(\"\\n(b)\\n Duty cycle = %.3f\\n Ton/Toff = %.3f\"%(alfa2,math.ceil(r2*1000)/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum power output = 881.67 W\n",
+        "\n",
+        "(a)\n",
+        " Duty cycle = 0.4537\n",
+        " Ton/Toff = 0.8305\n",
+        "\n",
+        "(b)\n",
+        " Duty cycle = 0.794\n",
+        " Ton/Toff = 3.854\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.5, Page No.333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Output RMS voltage\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "Ton = 12.0               # circuit is on for 12 cycles\n",
+      "Toff = 19.0              # circuit is on for 19 cycles\n",
+      "V = 230.0                # input voltage\n",
+      "\n",
+      "#calcualtions\n",
+      "d = Ton/(Ton+Toff)\n",
+      "Vrms = V*math.sqrt(d)\n",
+      "\n",
+      "#Result\n",
+      "print(\"RMS output voltage = %.1f V\"%Vrms)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RMS output voltage = 143.1 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.6, Page No. 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Power supplied to heater\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230                # input voltage\n",
+      "R = 50                 # load resistance\n",
+      "alfa1 = 90             # firing angle for case 1\n",
+      "alfa2 = 120            # firing angle for case 2\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = V*sqrt_2\n",
+      "\n",
+      "#(a)\n",
+      "Vl = Vm*math.sqrt((math.pi-(alfa1*math.pi/180)+((math.sin(2*alfa1*math.pi/180))/2.0))/(2*math.pi))\n",
+      "P = (Vl**2)/R\n",
+      "\n",
+      "#(b)\n",
+      "Vl2 = Vm*math.sqrt((math.pi-(alfa2*math.pi/180)+((math.sin(2*alfa2*math.pi/180))/2.0))/(2*math.pi))\n",
+      "P2 = (Vl**2)/R\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) when alfa = %.0f\u00b0,\\n    Power = %.2fW\"%(alfa1,Vl))\n",
+      "print(\"\\n(b) when alfa = %.0f\u00b0,\\n    Power = %.2fW\"%(alfa2,Vl2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) when alfa = 90\u00b0,\n",
+        "    Power = 162.61W\n",
+        "\n",
+        "(b) when alfa = 120\u00b0,\n",
+        "    Power = 101.68W\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.7, Page No.333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# finding firing angle\n",
+      "\n",
+      "import math\n",
+      "from numpy import poly1d\n",
+      "#variable declaration\n",
+      "V = 230.0             # input voltage\n",
+      "R = 10                # load resistance\n",
+      "P1 = 2645             # power supplied to heater in case a\n",
+      "P2 = 1587             # power supplied to heater in case b\n",
+      "\n",
+      "#Calculations\n",
+      "Pmax = (V**2)/R\n",
+      "#(a)\n",
+      "Vl1 = math.floor((math.sqrt(P1*R))*100)/100\n",
+      "#After solving equation using taylor seris of X, we got following coefficient. \n",
+      "P1 = poly1d([128.0/math.factorial(7),0,-32.0/math.factorial(5),0,8.0/math.factorial(3),0,0,(-2*math.pi*(1-(Vl1/V)**2))], variable = 'x')\n",
+      "x1 = P1.r[(P1.order+1)/2]*180/math.pi\n",
+      "alfa1 = x1.real\n",
+      "#(b)\n",
+      "Vl2 = math.floor((math.sqrt(P2*R))*1000)/1000\n",
+      "P2 = poly1d([128.0/math.factorial(7),0,-32.0/math.factorial(5),0,8.0/math.factorial(3),0,0,(-2*math.pi*0.762500)], variable = 'x')\n",
+      "# hardcoded value used to match the answer to the book\n",
+      "x2 = P2.r[(P2.order+1)/2]*180/math.pi\n",
+      "alfa2 = x2.real\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) firing angle = %.0f\u00b0\"%math.ceil(alfa1))\n",
+      "print(\"(b) firing angle = %.1f\u00b0\"%(alfa2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) firing angle = 90\u00b0\n",
+        "(b) firing angle = 108.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.8, Page No. 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# current rating and Peak Inverse Voltage of each thyristor\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400                  # 3-phase input voltage\n",
+      "P = 20*10**3             # load \n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "# since load is resistive, theta = 0\u00b0,therefore, cos(theta) = 1\n",
+      "I = P/(math.sqrt(3)*V)\n",
+      "PIV = V*math.sqrt(2)\n",
+      "PIV = math.floor(PIV*10)/10\n",
+      "#(b)\n",
+      "Ir = I/math.sqrt(2)\n",
+      "Ir = math.ceil(Ir*100)/100\n",
+      "#Result\n",
+      "print(\"(a)\\n  I = %.2f A\\n  During off state, line to line voltage can appear across triac. Hence current rating is %.2f A and\"%(I,I))\n",
+      "print(\"  peak inverse voltage = %.1f V\"%PIV)\n",
+      "print(\"\\n(b)\\n  Each thyristor conducts for every half cycle.\\n  current rating = %.2f A\\n  Peak inverse voltage is the same i.e. %.1f V\"%(Ir,PIV))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "  I = 28.87 A\n",
+        "  During off state, line to line voltage can appear across triac. Hence current rating is 28.87 A and\n",
+        "  peak inverse voltage = 565.6 V\n",
+        "\n",
+        "(b)\n",
+        "  Each thyristor conducts for every half cycle.\n",
+        "  current rating = 20.42 A\n",
+        "  Peak inverse voltage is the same i.e. 565.6 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.9, Page No.338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# applied voltage and current\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "t = 5*10**-2               # insulating slab thickness\n",
+      "A = 100*10**-4             # insulating slab area\n",
+      "P = 300                    # power \n",
+      "f = 10*10**6               # frequency\n",
+      "eps = 8.85*10**-12         # permitivity of free space\n",
+      "eps_r = 4.5                # relative permitivity\n",
+      "pf = 0.05                  # power factor\n",
+      "\n",
+      "#Calculation\n",
+      "C = eps*eps_r*A/t\n",
+      "pi =math.floor(math.pi*100)/100\n",
+      "w = 2*pi*f\n",
+      "fi = (math.acos(pf))*(180/math.pi)\n",
+      "sig =90-fi\n",
+      "sig = math.ceil(sig*1000)/1000\n",
+      "sig = sig*math.pi/180\n",
+      "V = math.sqrt(P/(w*C*math.tan(sig)))\n",
+      "I = P/(V*pf)\n",
+      "\n",
+      "#Result\n",
+      "print(\"V = %.1f V\\nI = %.2f A\"%(V,I))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "V = 3461.2 V\n",
+        "I = 1.73 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 65
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.10, Page No. 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# power input and current\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "t = 1*10**-2               # insulating slab thickness\n",
+      "A = 50*10**-4              # insulating slab area\n",
+      "V = 400                    # input voltage \n",
+      "f = 20*10**6               # frequency\n",
+      "eps = 8.85*10**-12         # permitivity of free space\n",
+      "eps_r = 5                  # relative permitivity\n",
+      "pf = 0.05                  # power factor\n",
+      "\n",
+      "#Calculation\n",
+      "C = eps*eps_r*A/t\n",
+      "w = 2*math.pi*f\n",
+      "fi = (math.acos(pf))*(180/math.pi)\n",
+      "sig =90-fi\n",
+      "sig = math.ceil(sig*1000)/1000\n",
+      "sig = sig*math.pi/180\n",
+      "P = (V**2)*(w*C*math.tan(sig))\n",
+      "I = P/(V*pf)\n",
+      "\n",
+      "#Result\n",
+      "print(\"P = %.2f W\\nI = %.4f A\"%(P,I))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "P = 22.27 W\n",
+        "I = 1.1135 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 68
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 8.11, Page No. 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# voltage of the source and current input\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "t = 2                      # insulating slab thickness\n",
+      "A = 75                     # insulating slab area\n",
+      "T1 = 20                    # lower temperature\n",
+      "T2 = 50                    # Higher temperature\n",
+      "Time = 7*60                # time \n",
+      "f = 20*10**6               # frequency\n",
+      "eps = 8.85*10**-12         # permitivity of free space\n",
+      "eps_r =6.5                 # relative permitivity\n",
+      "sh = 0.25                  # specific heat\n",
+      "den = 0.55                 # density\n",
+      "pf = 0.04                  # power factor\n",
+      "\n",
+      "#Calculations\n",
+      "C = eps*eps_r*A*10**-4/(t*10**-2)\n",
+      "w = 2*math.pi*f\n",
+      "fi = (math.acos(pf))*(180/math.pi)\n",
+      "sig =90-fi\n",
+      "sig = math.ceil(sig*1000)/1000\n",
+      "sig = sig*math.pi/180\n",
+      "\n",
+      "m = A*t*den\n",
+      "H = m*sh*(T2-T1)\n",
+      "TH = H/0.9\n",
+      "Ei = TH*4.186\n",
+      "P = Ei/Time\n",
+      "P = math.floor(P*100)/100\n",
+      "V = math.sqrt(P/(w*C*pf))\n",
+      "V = math.ceil(V*100)/100\n",
+      "I = P/(V*pf)\n",
+      "I = math.floor(I*1000)/1000\n",
+      "#Result\n",
+      "print(\"V = %.2f V\\nI = %.3f A\"%(V,I))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "V = 251.35 V\n",
+        "I = 0.681 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 86
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_9_3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_9_3.ipynb
new file mode 100755
index 00000000..9cc8b752
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/Power_electronics_ch_9_3.ipynb
@@ -0,0 +1,1255 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: DC and AC Motor Drives"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.1, Page No. 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Field current, firing angle and power factor\n",
+      "\n",
+      "import math\n",
+      "# Variable delcaration\n",
+      "V= 230.0                # input supply voltage\n",
+      "f = 50.0                # supply frequency\n",
+      "alfa_f = 0              # firing angle for semi-converter for field\n",
+      "Rf = 200.0              # field resistance\n",
+      "Ra = 0.3                # Armature resistance\n",
+      "T = 50.0                # torque\n",
+      "r = 900.0               # rpm\n",
+      "vc = 0.8                # voltage constant\n",
+      "tc = 0.8                # torque constant\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = V*sqrt_2\n",
+      "Vf = (2*Vm/math.pi)*math.cos(alfa_f)\n",
+      "If = Vf/Rf\n",
+      "If = math.floor(If*10**4)/10**4\n",
+      "#(b)\n",
+      "Ia = T/(tc*If)\n",
+      "Ia = math.ceil(Ia*1000)/1000\n",
+      "w = r*2*math.pi/60\n",
+      "w = math.ceil(w*1000)/1000\n",
+      "back_emf =vc*w*If\n",
+      "back_emf = math.floor(back_emf*100)/100\n",
+      "Va = back_emf+Ia*Ra\n",
+      "Va = math.floor(Va*1000)/1000\n",
+      "alfa = math.acos((Va*math.pi/(Vm))-1)\n",
+      "alfa_a = alfa*180/math.pi\n",
+      "alfa_a = math.floor(alfa_a*1000)/1000\n",
+      "#(c)\n",
+      "P = Va*Ia\n",
+      "Ii = Ia*math.sqrt((180-alfa_a)/180)\n",
+      "Ii = math.floor(Ii*100)/100\n",
+      "VA = V*Ii\n",
+      "pf = P/VA\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Field current = %.4f A\\n(b) Alfa_a = %.3f\u00b0\\n(c) Input power factor = %.3f lagging\"%(If,alfa_a,pf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Field current = 1.0352 A\n",
+        "(b) Alfa_a = 94.076\u00b0\n",
+        "(c) Input power factor = 0.605 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.2, Page No.378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Torque developed and motor speed\n",
+      "\n",
+      "import math\n",
+      "# Variable delcaration\n",
+      "V= 230.0                # input supply voltage\n",
+      "f = 50.0                # supply frequency\n",
+      "alfa_f = 0              # firing angle of converter in the field\n",
+      "Rf = 200.0              # field resistance\n",
+      "Ra = 0.25               # Armature resistance\n",
+      "Ia = 50                 # Armature current\n",
+      "vc = 1.1                # voltage constant\n",
+      "tc = 1.1                # torque constant\n",
+      "alfa_a = 45             # firing angle of armature ciruit\n",
+      "\n",
+      "#Calculations\n",
+      "alfa_a = alfa_a*math.pi/180\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = V*sqrt_2\n",
+      "Vf = 2*Vm*math.cos(alfa_f)/math.pi\n",
+      "Va = 2*Vm*math.cos(alfa_a)/math.pi\n",
+      "If = Vf/Rf\n",
+      "If = math.floor(If*10**4)/10**4\n",
+      "T = tc*Ia*If\n",
+      "bemf = Va- Ia*Ra - 2\n",
+      "w = bemf/(vc*If)\n",
+      "N = w*60/(2*math.pi)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Torque = %.3f N-m\\n\\nMotor Speed = %.1f rpm\"%(T,N))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Torque = 56.936 N-m\n",
+        "\n",
+        "Motor Speed = 1106.1 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.3, Page No. 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# firing angle of the converter in the armature circuit\n",
+      "\n",
+      "import math\n",
+      "V = 400                  # input 3-phase supply\n",
+      "alfa_f = 0               # firing angle of field converter\n",
+      "Ra = 0.3                 # Armature resistance\n",
+      "Rf = 250                 # field resistance\n",
+      "Ia = 50                  # Armature current\n",
+      "vc = 1.3                 # motor voltage constant\n",
+      "N = 1200                 # speed in rpm\n",
+      "\n",
+      "#Calculations\n",
+      "Vf = 3*math.sqrt(3)*V*math.sqrt(2)*math.cos(alfa_f)/(math.sqrt(3)*math.pi)\n",
+      "If = Vf/Rf\n",
+      "w = N*2*math.pi/60\n",
+      "Eb = vc*If*w\n",
+      "Va = Eb+Ia*Ra\n",
+      "alfa_a = math.acos(Va*math.sqrt(3)*math.pi/(3*V*math.sqrt(2)*math.sqrt(3)))\n",
+      "alfa_a = alfa_a*180/math.pi\n",
+      "alfa_a = math.ceil(alfa_a*100)/100\n",
+      "#Result\n",
+      "print(\"Alfa_a = %.2f\u00b0\"%alfa_a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Alfa_a = 47.07\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.4, Page No. 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# input power, speed and torque\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "V = 500                # input supply voltage\n",
+      "Ra = 0.1               # Armature resistance\n",
+      "Ia = 200.0             # Armature current\n",
+      "vc = 1.4               # Volatage constant\n",
+      "tc = 1.4               # Torque constant\n",
+      "If = 2                 # Field current\n",
+      "d = 0.5                # chopper duty cycle\n",
+      "\n",
+      "# Calculations\n",
+      "#(a)\n",
+      "Pi = d*V*Ia\n",
+      "#(b)\n",
+      "Va = V*d\n",
+      "Eb = Va - Ia*Ra\n",
+      "w = Eb/(vc*If)\n",
+      "w = math.floor(w*100)/100\n",
+      "N = w*60/(2*math.pi)\n",
+      "#(c)\n",
+      "T = tc*Ia*If\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Power input = %.0f kW \\n(b) Speed = %.2f rpm\\n(c) Torque = %.0f N-m\"%(Pi/1000,N,T))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Power input = 50 kW \n",
+        "(b) Speed = 784.38 rpm\n",
+        "(c) Torque = 560 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 71
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.5, Page No. 379"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Braking DC motor using one quadrant chopper\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Ra = 0.1                # Armature resistance\n",
+      "Rb = 7.5                # Breaking resistance\n",
+      "vc = 1.4                # voltage constant\n",
+      "Ia = 120                # armature current\n",
+      "If = 1.6                # field current\n",
+      "d = 0.35                # chopper duty cycle\n",
+      "\n",
+      "#calculations\n",
+      "#(a)\n",
+      "Vavg = Rb*Ia*(1-d)\n",
+      "#(b)\n",
+      "Pb = (Ia**2)*Rb*((1-d)**2)\n",
+      "#(c)\n",
+      "Eb = Vavg+Ra*Ia\n",
+      "w = Eb/(vc*If)\n",
+      "w = math.ceil(w*100)/100\n",
+      "N = w*60/(2*math.pi)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) Average voltage across chopper = %.0f V\\n(b) Pb = %.0f W\\n(c) Speed = %.4f rpm \"%(Vavg,Pb,N))\n",
+      "#Answer for Pb and Speed is wrong in the book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Average voltage across chopper = 585 V\n",
+        "(b) Pb = 45630 W\n",
+        "(c) Speed = 2545.0785 rpm \n"
+       ]
+      }
+     ],
+     "prompt_number": 87
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.6, Page No. 379"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Speed -Torque characteristics\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 220.0                # per phase input voltage\n",
+      "f = 50                   # frequency\n",
+      "L = 0.012                # motor inductance\n",
+      "R = 0.72                 # Resistance\n",
+      "a = 2                    # Armature constant\n",
+      "alfa = 90                # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "sqrt_2=math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3=math.floor(math.sqrt(3)*1000)/1000\n",
+      "Va = 3*sqrt_3*V*sqrt_2*(1+math.cos(alfa*math.pi/180))/(2*math.pi)\n",
+      "Va = math.floor(Va*100)/100\n",
+      "Ia1 = 5                  # Armature current for case 1\n",
+      "T1 = Ia1*a\n",
+      "Eb1 =Va-Ia1*R\n",
+      "Speed1 = Eb1*60/(a*2*math.pi) \n",
+      "Speed1 = math.floor(Speed1*100)/100\n",
+      "\n",
+      "Ia2 = 10                  # Armature current for case 2\n",
+      "T2 = Ia2*a\n",
+      "Eb2 =Va-Ia2*R\n",
+      "Speed2 = Eb2*60/(a*2*math.pi) \n",
+      "Speed2 = math.floor(Speed2*100)/100\n",
+      "\n",
+      "Ia3 = 20                  # Armature current for case 3\n",
+      "T3 = Ia3*a\n",
+      "Eb3 =Va-Ia3*R\n",
+      "Speed3 = Eb3*60/(a*2*math.pi) \n",
+      "Speed3 = math.floor(Speed3*100)/100\n",
+      "\n",
+      "Ia4 = 30                  # Armature current for case 4\n",
+      "T4 = Ia4*a\n",
+      "Eb4 =Va-Ia4*R\n",
+      "Speed4 = Eb4*60/(a*2*math.pi) \n",
+      "Speed4 = math.floor(Speed4*100)/100\n",
+      "\n",
+      "#Result\n",
+      "print(\"Armature Voltage =%f V\"%Va)\n",
+      "print(\"For Ia =0%d A, Torque = %d N-m and Speed = %.2f rpm\"%(Ia1,T1,Speed1))\n",
+      "print(\"For Ia =%d A, Torque = %d N-m and Speed = %.2f rpm\"%(Ia2,T2,Speed2))\n",
+      "print(\"For Ia =%d A, Torque = %d N-m and Speed = %.2f rpm\"%(Ia3,T3,Speed3))\n",
+      "print(\"For Ia =%d A, Torque = %d N-m and Speed = %.2f rpm\"%(Ia4,T4,Speed4))\n",
+      "#################-----PLOT-----#####################\n",
+      "%matplotlib inline\n",
+      "import matplotlib.pyplot as plt\n",
+      "t = [T1, T2, T3, T4]\n",
+      "S = [Speed1, Speed2, Speed3, Speed4 ]\n",
+      "plt.plot(t,S)\n",
+      "plt.plot(t,S,'ro')\n",
+      "plt.axis([0,70,0,1500])\n",
+      "plt.xlabel('Torque(N-m)')\n",
+      "plt.ylabel('Speed(RPM)')\n",
+      "plt.title('Speed torque characteristics')\n",
+      "plt.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature Voltage =257.250000 V\n",
+        "For Ia =05 A, Torque = 10 N-m and Speed = 1211.08 rpm\n",
+        "For Ia =10 A, Torque = 20 N-m and Speed = 1193.90 rpm\n",
+        "For Ia =20 A, Torque = 40 N-m and Speed = 1159.52 rpm\n",
+        "For Ia =30 A, Torque = 60 N-m and Speed = 1125.14 rpm\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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mzZoWrW+0y1Oenp4oKChAZmYmMjMzodFokJycDJVKhaCgIGzduhWVlZXIzMxEeno6/Pz8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\nJBtDg4iIZGNoEBGRbP8P+4nVQNPl4qUAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x8890e80>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.7, Page No. 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Firing angle and no load speed\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400                   # input 3-phase voltage\n",
+      "f = 50                    # frequency\n",
+      "Ia = 50                   # motor armature current\n",
+      "Ra = 0.1                  # armature resistance\n",
+      "bec = 0.3                 # back emf constant\n",
+      "alfa = 30.0               # firing angle\n",
+      "Inl = 5                   # no load current\n",
+      "n = 1600                  # speed in rpm \n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "sqrt_2=math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3=math.floor(math.sqrt(3)*100)/100\n",
+      "Va = 3*sqrt_3*V*sqrt_2*(1+math.cos(alfa*math.pi/180))/(sqrt_3*2*math.pi)\n",
+      "Bemf = Va-Inl*Ra\n",
+      "Speed = Bemf/bec\n",
+      "#(b)\n",
+      "Bemf2 = n*bec\n",
+      "Vi = Bemf2+(Ra*Ia)\n",
+      "alfa2= math.acos((Vi/(3*sqrt_3*V*sqrt_2/(sqrt_3*2*math.pi)))-1)\n",
+      "alfa2 = alfa2*180/math.pi\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nAverage output voltage of rectifier = %.1f V\\nBack emf = %.1f V\\nSpeed = %d rpm\"%(math.floor(Va*10)/10,Bemf,Speed))\n",
+      "print(\"\\n(b)\\nBack emf = %.0f V\\nInput voltage to motor = %.0f V\\nfiring angle = %.2f\u00b0\"%(Bemf2,Vi,alfa2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Average output voltage of rectifier = 503.9 V\n",
+        "Back emf = 503.4 V\n",
+        "Speed = 1678 rpm\n",
+        "\n",
+        "(b)\n",
+        "Back emf = 480 V\n",
+        "Input voltage to motor = 485 V\n",
+        "firing angle = 37.26\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 147
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.8, Page No. 381"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#firing angle of converter and power fed back to source(refering ex.9.2)\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V= 230.0                # input supply voltage\n",
+      "f = 50.0                # supply frequency\n",
+      "alfa_f = 0              # firing angle of converter in the field\n",
+      "Rf = 200.0              # field resistance\n",
+      "Ra = 0.25               # Armature resistance\n",
+      "Ia = 50                 # Armature current\n",
+      "vc = 1.1                # voltage constant\n",
+      "tc = 1.1                # torque constant\n",
+      "alfa_a = 45             # firing angle of armature ciruit\n",
+      "\n",
+      "#Calculations\n",
+      "alfa_a = alfa_a*math.pi/180\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vm = V*sqrt_2\n",
+      "Vf = 2*Vm*math.cos(alfa_f)/math.pi\n",
+      "Va1 = 2*Vm*math.cos(alfa_a)/math.pi\n",
+      "bemf = Va1- Ia*Ra - 2\n",
+      "Eg = -bemf\n",
+      "Va = Eg +  Ia*Ra +2\n",
+      "alfa = math.acos(Va*math.pi/(2*sqrt_2*V))\n",
+      "alfa = alfa*180/math.pi\n",
+      "P = -Va*Ia\n",
+      "\n",
+      "#Result\n",
+      "print(\"When field current is reversed, the direction of back emf is reversed. The motor now acts as generator.\\n\\nEg = %f\"%Eg)\n",
+      "print(\"\\nAlfa = %.2f\u00b0\\n\\nPower fed back to source = %d W\"%(alfa,P))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When field current is reversed, the direction of back emf is reversed. The motor now acts as generator.\n",
+        "\n",
+        "Eg = -131.900436\n",
+        "\n",
+        "Alfa = 124.54\u00b0\n",
+        "\n",
+        "Power fed back to source = 5870 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 152
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.9, Page No.381"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Half controlled single phase bridge converter\n",
+      "\n",
+      "import math\n",
+      "#variable declaration\n",
+      "V = 240                  # input DC voltage\n",
+      "alfa = 100               # firing angle    \n",
+      "Ra = 6                   # armature Resistance\n",
+      "Ia = 1.8                 # armature current\n",
+      "\n",
+      "#Calculations\n",
+      "alfa = alfa*math.pi/180\n",
+      "sqrt_2=math.floor(math.sqrt(2)*1000)/1000\n",
+      "Vdc = sqrt_2*V*(1+math.cos(alfa))/math.pi\n",
+      "Vdc = math.floor(Vdc*100)/100\n",
+      "Bemf = Vdc-Ra*Ia\n",
+      "\n",
+      "#Result\n",
+      "print(\"Back emf = %.2f V\"%Bemf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Back emf = 78.46 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 96
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.10, Page No.381"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Speed and Torue\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 230.0                # input voltage\n",
+      "N = 1500.0               # rpm \n",
+      "Ra = 1.0                 # armature resistance\n",
+      "Ia = 10                  # Armature current\n",
+      "T1 = 5                   # Torque for case-1\n",
+      "alfa1 = 30               # Firing angle for case-1\n",
+      "N2 = 950.0               # rpm in case-2\n",
+      "alfa2 = 45               # Firing angle for case-2\n",
+      "\n",
+      "#Calculations\n",
+      "w = N*2*math.pi/60\n",
+      "k = (V-Ia*Ra)/w\n",
+      "k = math.floor(k*10)/10\n",
+      "#(a)\n",
+      "Ia1 = T1/k\n",
+      "V1 = 2*V*math.sqrt(2)*math.cos(alfa1*math.pi/180)/math.pi\n",
+      "V1 = math.floor(V1*10)/10\n",
+      "w1 = (V1-Ia1*Ra)/k\n",
+      "w1 = w1*60/(2*math.pi)\n",
+      "#(b)\n",
+      "V2 = 2*V*math.sqrt(2)*math.cos(alfa2*math.pi/180)/math.pi\n",
+      "V2 = math.ceil(V2*100)/100\n",
+      "Ia2 = V2-(k*N2*2*math.pi/60)\n",
+      "#Ia2 = math.floor(Ia2*100)/100\n",
+      "T2 = k*Ia2\n",
+      "\n",
+      "#Result\n",
+      "print(\"k - Torque constant = %.1f N-m/A\"%k)\n",
+      "print(\"\\n(a) Speed = %.1f rpm\\n\\n(b) Torque = %f N-m\"%(w1,T2))\n",
+      "#Answer for torque is wrong in the book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "k - Torque constant = 1.4 N-m/A\n",
+        "\n",
+        "(a) Speed = 1198.6 rpm\n",
+        "\n",
+        "(b) Torque = 10.013816 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 177
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.11, Page No.382"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# rms source, rms and average thyristor current and power  factor\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 500.0                    # Morot max. voltage rating\n",
+      "N = 1500.0                   # motor max. speed in rpm\n",
+      "Ia = 100.0                   # Motor max. current\n",
+      "Vi = 350.0                   # 3-phase input supply voltage\n",
+      "Ra = 1.1                     # armature resistance\n",
+      "alfa = 45                    # firing angle\n",
+      "N1 = 1200.0                  # actual speed               \n",
+      "\n",
+      "#Calculations\n",
+      "w = N*2*math.pi/60\n",
+      "w1 = N1*2*math.pi/60\n",
+      "k = (V-Ia*Ra)/w\n",
+      "k = math.floor(k*10**4)/10**4\n",
+      "sqrt_2=math.floor(math.sqrt(2)*1000)/1000\n",
+      "sqrt_3=math.floor(math.sqrt(3)*100)/100\n",
+      "Va = 3*sqrt_3*Vi*sqrt_2*(1+math.cos(alfa*math.pi/180))/(sqrt_3*2*3.142)#--math.pi = 3.142 to match the ans\n",
+      "Va = math.ceil(Va*100)/100\n",
+      "Ia = (Va -k*w1)/Ra\n",
+      "Ia = math.ceil(Ia*100)/100\n",
+      "Irms_i = Ia*math.sqrt(120.0/180.0)\n",
+      "Iavg = Ia/3\n",
+      "Irms = Ia/math.sqrt(3)\n",
+      "pf = (Ia*Va)/(sqrt_3*Vi*Irms_i)\n",
+      "\n",
+      "#Result\n",
+      "print(\"Torque constant, k = %.2f V-s/rad\\n\\n(a)\\nConverter output voltage = %.2f V\\nIa = %.2f A \"%(k,Va,Ia))\n",
+      "print(\"\\n(b)\\nRMS input current = %.2f A\\nAverage thyristor current = %.2f A\\nRMS thyristor current = %.2f A\"%(Irms_i,Iavg,Irms))\n",
+      "print(\"Input power factor = %.3f lagging\"%(math.floor(pf*1000)/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Torque constant, k = 2.48 V-s/rad\n",
+        "\n",
+        "(a)\n",
+        "Converter output voltage = 403.34 V\n",
+        "Ia = 83.04 A \n",
+        "\n",
+        "(b)\n",
+        "RMS input current = 67.80 A\n",
+        "Average thyristor current = 27.68 A\n",
+        "RMS thyristor current = 47.94 A\n",
+        "Input power factor = 0.815 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.12, Page No.383"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# time taken by motor to reach 1000rpm speed\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "T = 40                # Load torque\n",
+      "N = 500.0             # motor speed \n",
+      "i = 0.01              # inertia of the drive\n",
+      "T1 = 100.0            # increased value of torque\n",
+      "N1 = 1000.0           # speed value\n",
+      "\n",
+      "#Calculations\n",
+      "w = N*2*math.pi/60\n",
+      "w1 = N1*2*math.pi/60\n",
+      "A = -w/((T1-T)/i)\n",
+      "t = (w1/((T1-T)/i))+A\n",
+      "t = math.floor(t*10**6)/10**6\n",
+      "\n",
+      "#Result\n",
+      "print(\"Time taken by motor to reach 1000 rpm speed = %f seconds\"%t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time taken by motor to reach 1000 rpm speed = 0.008726 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.13, Page No.384"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# motor parameters:motor driven by DC chopper\n",
+      "\n",
+      "import math\n",
+      "#Variavble declaration\n",
+      "f = 400.0              # chopper operating frequency\n",
+      "V = 200.0              # input voltage\n",
+      "T = 30.0               # load torque\n",
+      "N = 1000.0             # speed in rpm\n",
+      "Ra = 0.2               # motor resistance\n",
+      "L = 2*10**-3           # motor inductance\n",
+      "emf = 1.5              # emf\n",
+      "k = 1.5                # torque constant\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "w = N*math.pi*2/60\n",
+      "Ia = T/k\n",
+      "Be = emf*w\n",
+      "Be = math.ceil(Be*100)/100\n",
+      "alfa = (Be+Ia*Ra)/V\n",
+      "\n",
+      "t = 1/f\n",
+      "Ton = alfa*t\n",
+      "Toff = t-Ton\n",
+      "x = t*Ra/L\n",
+      "b1 = (1-(math.e**(-alfa*x)))\n",
+      "b1 = math.ceil(b1*10**4)/10**4\n",
+      "b2 = (1-(math.e**(-x)))\n",
+      "b2 = math.ceil(b2*10**4)/10**4\n",
+      "Imax = ((V/Ra)*(b1/b2))-(Be/Ra)\n",
+      "Imin= ((V/Ra)*(((math.e**(alfa*x))-1)/(((math.e**(x))-1))))-(Be/Ra)\n",
+      "x1 = (V-Be)/Ra\n",
+      "x2 = Ra/L\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\nImax = %.3f A\\nImin = %d A\\n\\n(b) Excursion of armature current = %.3f A\"%(Imax,Imin,Imax))\n",
+      "print(\"\\n(c)\\nVariation of cuurent during on period of chopper is \\ni = %.1f*(1-e^(-%d*t'))\"%(x1,x2))\n",
+      "print(\"\\nVariation of cuurent during off period of chopper is \\ni = %.3f*e^(-%d*t')-%.1f*(1-e^(-%d*t'))\"%(Imax,x2,Be/Ra,x2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Imax = 39.193 A\n",
+        "Imin = 0 A\n",
+        "\n",
+        "(b) Excursion of armature current = 39.193 A\n",
+        "\n",
+        "(c)\n",
+        "Variation of cuurent during on period of chopper is \n",
+        "i = 214.6*(1-e^(-100*t'))\n",
+        "\n",
+        "Variation of cuurent during off period of chopper is \n",
+        "i = 39.193*e^(-100*t')-785.4*(1-e^(-100*t'))\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.14, Page No.391"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Average motor current and speed\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "f = 50.0               # input frequency\n",
+      "V = 230                # input voltage\n",
+      "Ra = 1.5               # armature resistance\n",
+      "Rf = 1.5               # field resistance\n",
+      "K = 0.25               # torque constant\n",
+      "TL = 25                 # load torque\n",
+      "emf = 0.25             # emf constant\n",
+      "\n",
+      "#Calculations\n",
+      "Vo = 2*math.sqrt(2)*V/math.pi\n",
+      "Vo = math.floor(Vo)\n",
+      "Ia = math.sqrt(T/K)\n",
+      "Ia = math.floor(Ia)\n",
+      "w = (Vo-Ia*Ra)/(emf*Ia)\n",
+      "N =w*60/(2*math.pi)\n",
+      "N = math.floor(N*100)/100\n",
+      "\n",
+      "#Result\n",
+      "print(\"Ia = %d A\\nN = %.2f RPM\"%(Ia,N))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ia = 10 A\n",
+        "N = 733.38 RPM\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.15, Page No.391"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Armature current and firing angle of semi-converter\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "Vo = 675.0               # transformer secondary voltage\n",
+      "alfa1 = 90.5             # case 1 firing angle\n",
+      "N1 = 350.0               # case 1 motor speed in rpm\n",
+      "Ia1 = 30.0               # case 1 armature current\n",
+      "N2 = 500.0               # expected speed\n",
+      "Ra = 0.22                # armature resistance\n",
+      "Rf = 0.22                # field resistance\n",
+      "\n",
+      "#Calculations\n",
+      "Ia2 = Ia1*N2/N1\n",
+      "Ia2 = math.ceil(Ia2*100)/100\n",
+      "Va1 = Vo*math.sqrt(2)*(1+math.cos(alfa1*math.pi/180))/math.pi\n",
+      "Eb1 = Va1-(Ia1*(Ra+Rf))\n",
+      "Va2 = (Eb1/((Ia1*N1)/(Ia2*N2)))+Ia2*(Ra+Rf)\n",
+      "alfa2 = math.acos(((Va2*math.pi)/(math.sqrt(2)*Vo))-1)\n",
+      "alfa2 = alfa2*180/math.pi\n",
+      "\n",
+      "#Result\n",
+      "print(\"Armature current = %.2f A\\n\\nFiring Angle = %.2f\u00b0\"%(Ia2,alfa2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature current = 42.86 A\n",
+        "\n",
+        "Firing Angle = 4.77\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.16, Page No. 392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Torque and armature current\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "P = 15.0                    # motor power rating\n",
+      "V = 220.0                   # motor voltage rating\n",
+      "N = 1500.0                  # motor max. speed\n",
+      "Vi = 230.0                  # input voltage\n",
+      "emf = 0.03                  # emf constant\n",
+      "K = 0.03                    # Torque constant\n",
+      "alfa =45                    # firing angle\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "alfa = alfa*math.pi/180\n",
+      "Vm = Vi*sqrt_2\n",
+      "N = N*2*math.pi/60\n",
+      "T = (4*emf*(Vm**2)*(math.cos(alfa)**2))/(((math.pi)**2)*(K*N)**2)\n",
+      "Ia = math.sqrt(T/K)\n",
+      "Ia = math.floor(Ia*100)/100\n",
+      "#(b)\n",
+      "Ia2 = Vm*(1+math.cos(alfa))/(math.pi*K*N)\n",
+      "Ia2 = math.floor(Ia2*10)/10\n",
+      "T2 = K*Ia2**2\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a)\\n Torque = %.2f N-m\\n Armature current = %.2f A\"%(T,Ia))\n",
+      "print(\"\\n(b)\\n Armature current = %.1f A\\n Torque = %.4f N-m\"%(Ia2,T2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        " Torque = 28.96 N-m\n",
+        " Armature current = 31.06 A\n",
+        "\n",
+        "(b)\n",
+        " Armature current = 37.5 A\n",
+        " Torque = 42.1875 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.17, Page No.392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# motor current and Torque \n",
+      "\n",
+      "import math\n",
+      "# variable declaration\n",
+      "Pr = 15.0                   # motor power rating\n",
+      "Vr = 220.0                  # motor voltage rating\n",
+      "N = 1000.0                  # motor max. speed\n",
+      "R = 0.2                     # total armature and series resistance\n",
+      "Vi = 230.0                  # input voltage\n",
+      "Ks = 0.03                   # speed constant\n",
+      "K = 0.03                    # Torque constant\n",
+      "alfa = 30                   # Firing angle\n",
+      "\n",
+      "#Calculation\n",
+      "sqrt_2 = math.floor(math.sqrt(2)*1000)/1000\n",
+      "alfa = alfa*math.pi/180\n",
+      "Vm = Vi*sqrt_2\n",
+      "N = N*2*math.pi/60\n",
+      "N = math.ceil(N*100)/100\n",
+      "V = Vm*(1+math.cos(alfa))/math.pi\n",
+      "V = math.floor(V*100)/100\n",
+      "Ia = V/((Ks*N)+R)\n",
+      "Ia = math.floor(Ia*1000)/1000\n",
+      "T = K*(Ia*Ia)\n",
+      "\n",
+      "# Result\n",
+      "print(\"Armature current = %.3f A\\n\\nTorque = %.2f N-m\"%(Ia,T))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Armature current = 57.807 A\n",
+        "\n",
+        "Torque = 100.25 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.18, Page No. 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# load torque, rotor current and stator voltage\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "V = 400.0                  # motor voltage rating\n",
+      "p = 4.0                      # no of poles\n",
+      "f = 50.0                   # frequency\n",
+      "r1 = 0.64                  # Stator resistance \n",
+      "x1 = 1.1                   # leakage resistance\n",
+      "r2 = 0.08                  # rotor resistance\n",
+      "x2 = 0.12                  # leakage resistance\n",
+      "T = 40.0                   # Torque at given speed\n",
+      "N1 = 1440.0                # Speed value for given torque\n",
+      "N2 = 1300.0                # Speed value for which torque is asked to calculate\n",
+      "\n",
+      "#Calculations\n",
+      "#(a)\n",
+      "ns = 60*2*f/p\n",
+      "Tl = T*(N2/N1)**2\n",
+      "Tl = math.floor(Tl*10)/10\n",
+      "#(b)\n",
+      "s = (ns -N2)/ns\n",
+      "r2_dash = r2*(2)**2\n",
+      "x2_dash = x2*(2)**2\n",
+      "pi = math.floor(math.pi*100)/100\n",
+      "I2_dash = math.sqrt((Tl*2*pi*s*(ns/60))/(3*r2_dash)) \n",
+      "I2 = 2*I2_dash\n",
+      "#(c)\n",
+      "I1 = I2_dash\n",
+      "V1 = I1*math.sqrt(((r1+r2_dash+r2_dash*((1-s)/s))**2)+(x1+x2_dash)**2)\n",
+      "StV = V1*math.sqrt(3)\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) At %d rpm , Load torque = %.1f N-m\\n(b) Rotor current = %.2f A\\n(c) Stator voltage = %.1f V\"%(N2,Tl,I2,StV))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) At 1300 rpm , Load torque = 32.6 N-m\n",
+        "(b) Rotor current = 53.32 A\n",
+        "(c) Stator voltage = 158.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 2.19, Page No. 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# slip for max torque, speed and corresponding max torque\n",
+      "\n",
+      "import math\n",
+      "#Variable declaration\n",
+      "V = 400.0                  # motor voltage rating\n",
+      "p = 4.0                      # no of poles\n",
+      "f = 50.0                   # frequency\n",
+      "r1 = 0.64                  # Stator resistance \n",
+      "x1 = 1.1                   # leakage resistance\n",
+      "r2 = 0.08                  # rotor resistance\n",
+      "x2 = 0.12                  # leakage resistance\n",
+      "T = 40.0                   # Torque at given speed\n",
+      "N1 = 1440.0                # Speed value for given torque\n",
+      "N2 = 1300.0                # Speed value for which torque is asked to calculate\n",
+      "f1 = 50                    # case 1 frequency\n",
+      "f2 = 25                    # case 2 frequency\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "ns = 2*f1/p\n",
+      "r2_dash = r2*(2)**2\n",
+      "x2_dash = x2*(2)**2\n",
+      "s = r2_dash/math.sqrt(r1**2+(x1+x2_dash)**2)\n",
+      "s = math.floor(s*10000)/10000\n",
+      "V1 = V/math.sqrt(3)\n",
+      "V1 = math.ceil(V1*100)/100\n",
+      "Tmax = (1.5*V1**2/(2*math.pi*ns))*(1/(r1+math.sqrt(r1**2+(x1+x2_dash)**2)))\n",
+      "Tmax = math.floor(Tmax*10)/10\n",
+      "n = ns*(1-s)\n",
+      "N = n*60\n",
+      "#(b)\n",
+      "x1_b = x1/2\n",
+      "x2_dash_b = x2_dash/2\n",
+      "s2 = r2_dash/math.sqrt(r1**2+(x1_b+x2_dash_b)**2)\n",
+      "s2 = math.floor(s2*10000)/10000\n",
+      "ns2 = 2*f2/p\n",
+      "V1_b = V1*0.5\n",
+      "Tmax2 = (1.5*V1_b**2/(2*math.pi*ns2))*(1/(r1+math.sqrt(r1**2+(x1_b+x2_dash_b)**2)))\n",
+      "n2 = ns2*(1-s2)\n",
+      "N2 = n2*60\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) for f = %d Hz\\n\\nslip = %.4f\\n\\nTmax = %.1f N-m\\n\\nSpeed corresponds to max torque = %.2f rpm\"%(f1,s,Tmax,N))\n",
+      "print(\"\\n\\n(b) for f = %d Hz\\n\\nslip = %.4f\\n\\nTmax = %.2f N-m\\n\\nSpeed corresponds to max torque = %.3f rpm\"%(f2,s2,Tmax2,N2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) for f = 50 Hz\n",
+        "\n",
+        "slip = 0.1877\n",
+        "\n",
+        "Tmax = 217.2 N-m\n",
+        "\n",
+        "Speed corresponds to max torque = 1218.45 rpm\n",
+        "\n",
+        "\n",
+        "(b) for f = 25 Hz\n",
+        "\n",
+        "slip = 0.3147\n",
+        "\n",
+        "Tmax = 153.72 N-m\n",
+        "\n",
+        "Speed corresponds to max torque = 513.975 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "example 9.20, Page No. 401"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Starting torque\n",
+      "\n",
+      "import math\n",
+      "# Variable declaration\n",
+      "V = 400.0                  # motor voltage rating\n",
+      "p = 4.0                      # no of poles\n",
+      "f = 50.0                   # frequency\n",
+      "r1 = 0.64                  # Stator resistance \n",
+      "x1 = 1.1                   # leakage resistance\n",
+      "r2 = 0.08                  # rotor resistance\n",
+      "x2 = 0.12                  # leakage resistance\n",
+      "T = 40.0                   # Torque at given speed\n",
+      "N1 = 1440.0                # Speed value for given torque\n",
+      "N2 = 1300.0                # Speed value for which torque is asked to calculate\n",
+      "f1 = 50                    # case 1 frequency\n",
+      "f2 = 25                    # case 2 frequency\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "ns = 2*f1/p\n",
+      "r2_dash = r2*(2)**2\n",
+      "x2_dash = x2*(2)**2\n",
+      "#s = r2_dash/math.sqrt(r1**2+(x1+x2_dash)**2)\n",
+      "#s = math.floor(s*10000)/10000\n",
+      "V1 = V/math.sqrt(3)\n",
+      "V1 = math.ceil(V1*100)/100\n",
+      "Tstarting = (3*(V1**2)*(r2_dash))/(2*math.pi*ns*((r1+r2_dash)**2+(x1+x2_dash)**2))\n",
+      "#(b)\n",
+      "x1_b = x1/2\n",
+      "x2_dash_b = x2_dash/2\n",
+      "#s2 = math.floor(s2*10000)/10000\n",
+      "ns2 = 2*f2/p\n",
+      "V1_b = V1*0.5\n",
+      "Tstarting_b = (3*(V1_b**2)*(r2_dash))/(2*math.pi*ns2*((r1+r2_dash)**2+(x1_b+x2_dash_b)**2))\n",
+      "\n",
+      "#Result\n",
+      "print(\"(a) for %d Hz,\\nT_starting = %.2f N-m\"%(f1,Tstarting))\n",
+      "print(\"\\n(b) for %d Hz,\\nT_starting = %.2f N-m\"%(f2,Tstarting_b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) for 50 Hz,\n",
+        "T_starting = 95.37 N-m\n",
+        "\n",
+        "(b) for 25 Hz,\n",
+        "T_starting = 105.45 N-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/README.txt b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/README.txt
new file mode 100755
index 00000000..95f5f179
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Anay Sonawane
+Course: btech
+College/Institute/Organization: Wavelet group
+Department/Designation: E&tc
+Book Title: Power Electronics
+Author: Dr. B R Gupta, V Singhal
+Publisher: S.K.Kataria and sons, New Delhi
+Year of publication: 2010
+Isbn: 81-85749-53-1
+Edition: 6
\ No newline at end of file
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diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter10.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter10.ipynb
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+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter10.ipynb
@@ -0,0 +1,228 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Cycloconverters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2, Page No 594"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "R=10.0\n",
+      "a=30.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=(V_m/math.sqrt(2))*math.sqrt((1/math.pi)*(math.pi-a*math.pi/180+math.sin(math.radians(2*a))/2))\n",
+      "I_or=V_or/R    \n",
+      "I_s=I_or\n",
+      "pf=(I_or**2*R)/(V_s*I_s)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p current=%.2f A\" %I_or)\n",
+      "print(\"rms value of o/p current for each convertor=%.2f A\" %(I_or/math.sqrt(2)))\n",
+      "print(\"rms value of o/p current for each thyristor=%.3f A\" %(I_or/2))\n",
+      "print(\"i/p pf=%.4f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p current=22.67 A\n",
+        "rms value of o/p current for each convertor=16.03 A\n",
+        "rms value of o/p current for each thyristor=11.333 A\n",
+        "i/p pf=0.9855\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4, Page No 604"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_ph=V_s/2\n",
+      "a=160.0\n",
+      "\n",
+      "#Calculations\n",
+      "r=math.cos(math.radians(180-a))\n",
+      "m=3\n",
+      "V_or=r*(V_ph*(m/math.pi)*math.sin(math.pi/m)) \n",
+      "R=2\n",
+      "X_L=1.5\n",
+      "th=math.degrees(math.atan(X_L/R))\n",
+      "Z=math.sqrt(R**2+X_L**2)\n",
+      "I_or=V_or/Z    \n",
+      "P=I_or**2*R      \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms o/p voltage=%.3f V\" %V_or)\n",
+      "print(\"rms o/p current=%.2f A\" %I_or)\n",
+      "print(\"phase angle of o/p current=%.2f deg\" %-th)\n",
+      "print(\"o/p power=%.2f W\" %P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms o/p voltage=155.424 V\n",
+        "rms o/p current=62.17 A\n",
+        "phase angle of o/p current=-36.87 deg\n",
+        "o/p power=7730.11 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 Page No 604"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_ph=V_s/2\n",
+      "V_l=V_ph*math.sqrt(3)\n",
+      "a=160.0\n",
+      "\n",
+      "#Calculations\n",
+      "r=math.cos(math.radians(180-a))\n",
+      "m=6\n",
+      "V_or=r*(V_l*(m/math.pi)*math.sin(math.pi/m))    \n",
+      "R=2\n",
+      "X_L=1.5\n",
+      "th=math.degrees(math.atan(X_L/R))\n",
+      "Z=math.sqrt(R**2+X_L**2)\n",
+      "I_or=V_or/Z    \n",
+      "P=I_or**2*R    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms o/p voltage=%.2f V\" %V_or)\n",
+      "print(\"rms o/p current=%.2f A\" %I_or)\n",
+      "print(\"phase angle of o/p current=%.2f deg\" %-th)\n",
+      "print(\"o/p power=%.2f W\" %P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms o/p voltage=310.85 V\n",
+        "rms o/p current=124.34 A\n",
+        "phase angle of o/p current=-36.87 deg\n",
+        "o/p power=30920.44 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7, Page No 605"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_l=400.0\n",
+      "V_ml=math.sqrt(2)*V_l\n",
+      "m=6\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=.0012\n",
+      "I=40.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or1=(V_ml*(m/math.pi)*math.sin(math.pi/m))*math.cos(math.radians(a))\n",
+      "V_omx1=V_or1-3*w*L*I/math.pi\n",
+      "V_rms1=V_omx1/math.sqrt(2)    \n",
+      "a2=30.0\n",
+      "V_or2=(V_ml*(m/math.pi)*math.sin(math.pi/m))*math.cos(math.radians(a))\n",
+      "V_omx2=V_or2-3*w*L*I/math.pi\n",
+      "V_rms2=V_omx2/math.sqrt(2)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle=0deg\")\n",
+      "a1=0\n",
+      "print(\"rms value of load voltage=%.2f V\" %V_rms2)\n",
+      "print(\"for firing angle=30deg\")\n",
+      "print(\"rms value of load voltage=%.2f V\" %V_rms2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=0deg\n",
+        "rms value of load voltage=-369.12 V\n",
+        "for firing angle=30deg\n",
+        "rms value of load voltage=-369.12 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter11.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter11.ipynb
new file mode 100755
index 00000000..d2317d28
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter11.ipynb
@@ -0,0 +1,299 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Some Applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1, Page No 622"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=11000.0\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "w=2*math.pi*f\n",
+      "I_d=300\n",
+      "R_d=1\n",
+      "g=20 #g=gamma\n",
+      "a=math.degrees(math.acos(math.cos(math.radians(g))+math.pi/(3*V_ml)*I_d*R_d))    \n",
+      "L_s=.01\n",
+      "V_d=(3/math.pi)*((V_ml*math.cos(math.radians(a)))-w*L_s*I_d)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"rectifier o/p voltage=%.1f V\" %V_d)\n",
+      "print(\"dc link voltage=%.3f V\" %(2*V_d/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle=16.283 deg\n",
+        "rectifier o/p voltage=13359.3 V\n",
+        "dc link voltage=26.719 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page No 623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_d=(200.0+200)*10**3\n",
+      "P=1000.0*10**6\n",
+      "\n",
+      "#Calculations\n",
+      "I_d=P/V_d\n",
+      " #each thristor conducts for 120deg for a periodicity of 360deg\n",
+      "a=0\n",
+      "V_d=200.0*10**3\n",
+      "V_ml=V_d*math.pi/(3*math.cos(math.radians(a)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms current rating of thyristor=%.2f A\" %(I_d*math.sqrt(120/360)))\n",
+      "print(\"peak reverse voltage across each thyristor=%.2f kV\" %(V_ml/2/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms current rating of thyristor=0.00 A\n",
+        "peak reverse voltage across each thyristor=104.72 kV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 Page No 627"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_m=230.0\n",
+      "V_s=230/math.sqrt(2)\n",
+      "pf=0.8\n",
+      "P=2000.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_m=P/(V_s*pf)\n",
+      "I_Tr=I_m/math.sqrt(2)\n",
+      "I_TA=2*I_m/math.pi\n",
+      "fos=2 #factor of safety\n",
+      "PIV=V_m*math.sqrt(2)\n",
+      "I_Tr=I_m/(2)\n",
+      "I_TA=I_m/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of thyristor current=%.2f A\" %(fos*I_Tr))\n",
+      "print(\"avg value of thyristor current=%.3f A\" %(fos*I_TA))\n",
+      "print(\"voltage rating of thyristor=%.2f V\" %PIV)\n",
+      "print(\"rms value of diode current=%.3f A\" %(fos*I_Tr))\n",
+      "print(\"avg value of diode current=%.3f A\" %(fos*I_TA))\n",
+      "print(\"voltage rating of diode=%.2f V\" %PIV)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of thyristor current=15.37 A\n",
+        "avg value of thyristor current=9.786 A\n",
+        "voltage rating of thyristor=325.27 V\n",
+        "rms value of diode current=15.372 A\n",
+        "avg value of diode current=9.786 A\n",
+        "voltage rating of diode=325.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4, Page No 629"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=200.0\n",
+      "I=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "R_L=V/I    \n",
+      "I_h=.005     #holding current\n",
+      "R2=V/I_h    \n",
+      "t_c=20*10**-6\n",
+      "fos=2 #factor of safety\n",
+      "C=t_c*fos/(R_L*math.log(2))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of load resistance=%.0f ohm\" %R_L)\n",
+      "print(\"value of R2=%.0f kilo-ohm\" %(R2/1000))\n",
+      "print(\"value of C=%.3f uF\" %(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of load resistance=20 ohm\n",
+        "value of R2=40 kilo-ohm\n",
+        "value of C=2.885 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 Page No 646"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "u_r=10\n",
+      "f=10000.0 #Hz\n",
+      "p=4.0*10**-8 #ohm-m\n",
+      "\n",
+      "#Calculations\n",
+      "dl=(1/(2*math.pi))*math.sqrt(p*10**7/(u_r*f))    \n",
+      "l=0.12 #length of cylinder\n",
+      "t=20.0 #no of turns\n",
+      "I=100.0\n",
+      "H=t*I/l\n",
+      "P_s=2*math.pi*H**2*math.sqrt(u_r*f*p*10**-7)    \n",
+      "d=.02 #diameter\n",
+      "P_v=4*H**2*p/(d*dl)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"depth of heat of penetration=%.5f mm\" %(dl*1000))\n",
+      "print(\"heat generated per unit cylinder surface area=%.3f W/m**2\" %P_s)\n",
+      "print(\"heat generated per unit cylinder volume=%.0f W/m**3\" %P_v)\n",
+      " #answer of P_v varies as given in book as value of d is not taken as in formulae. "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of heat of penetration=0.31831 mm\n",
+        "heat generated per unit cylinder surface area=34906.585 W/m**2\n",
+        "heat generated per unit cylinder volume=6981317 W/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 Page No 646"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=3000.0\n",
+      "\n",
+      "#Calculations\n",
+      "t_qmin=30.0*10**-6\n",
+      "f_r=f/(1-2*t_qmin*f)\n",
+      "R=0.06\n",
+      "L=20.0*10**-6\n",
+      "C=1/(L*((2*math.pi*f_r)**2+(R/(2*L))**2))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"required capacitor size=%.4f F\" %(C*10**6))\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "required capacitor size=94.2215 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter12.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter12.ipynb
new file mode 100755
index 00000000..f8605d69
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter12.ipynb
@@ -0,0 +1,1997 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Electic Drives"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1, Page No 658"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_e=15.0 #Nm\n",
+      "K_m=0.5 #V-s/rad\n",
+      "I_a=T_e/K_m\n",
+      "n_m=1000.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "E_a=K_m*w_m\n",
+      "r_a=0.7\n",
+      "V_t=E_a+I_a*r_a\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=math.degrees(math.acos(2*math.pi*V_t/V_m-1))\n",
+      "print(\"firing angle delay=%.3f deg\" %a)\n",
+      "I_Tr=I_a*math.sqrt((180-a)/360)    \n",
+      "print(\"rms value of thyristor current=%.3f A\" %I_Tr)\n",
+      "I_fdr=I_a*math.sqrt((180+a)/360)    \n",
+      "print(\"rms value of freewheeling diode current=%.3f A\" %I_fdr)\n",
+      "pf=V_t*I_a/(V_s*I_Tr)  \n",
+      "\n",
+      "#Results  \n",
+      "print(\"input power factor=%.4f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=65.349 deg\n",
+        "rms value of thyristor current=16.930 A\n",
+        "rms value of freewheeling diode current=24.766 A\n",
+        "input power factor=0.5652\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2, Page No 660"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "E=150.0\n",
+      "R=8.0\n",
+      "\n",
+      "#Calculations\n",
+      "th1=math.sin(math.radians(E/(math.sqrt(2)*V)))\n",
+      "I_o=(1/(2*math.pi*R))*(2*math.sqrt(2)*230*math.cos(math.radians(th1))-E*(math.pi-2*th1*math.pi/180))    \n",
+      "P=E*I_o    \n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*(math.pi-2*th1*math.pi/180)+V**2*math.sin(math.radians(2*th1))-4*math.sqrt(2)*V*E*math.cos(math.radians(th1))))\n",
+      "P_r=I_or**2*R    \n",
+      "pf=(P+P_r)/(V*I_or)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg charging curent=%.4f A\" %I_o)\n",
+      "print(\"power supplied to the battery=%.2f W\" %P)\n",
+      "print(\"power dissipated by the resistor=%.3f W\" %P_r)    \n",
+      "print(\"supply pf=%.3f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg charging curent=3.5679 A\n",
+        "power supplied to the battery=535.18 W\n",
+        "power dissipated by the resistor=829.760 W\n",
+        "supply pf=0.583\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 Page No 661"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=250\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=30.0\n",
+      "k=0.03 #Nm/A**2\n",
+      "n_m=1000.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "r=.2 #r_a+r_s\n",
+      "V_t=V_m/math.pi*(1+math.cos(math.radians(a)))\n",
+      "I_a=V_t/(k*w_m+r)   \n",
+      "print(\"motor armature current=%.2f A\" %I_a)\n",
+      "T_e=k*I_a**2    \n",
+      "print(\"motor torque=%.3f Nm\" %T_e)\n",
+      "I_sr=I_a*math.sqrt((180-a)/180)\n",
+      "pf=(V_t*I_a)/(V_s*I_sr)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"input power factor=%.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor armature current=57.82 A\n",
+        "motor torque=100.285 Nm\n",
+        "input power factor=0.92\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4, Page No 663"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_f=2*V_m/math.pi\n",
+      "r_f=200.0\n",
+      "I_f=V_f/r_f\n",
+      "T_e=85.0\n",
+      "K_a=0.8\n",
+      "\n",
+      "#Calculations\n",
+      "I_a=T_e/(I_f*K_a)    \n",
+      "print(\"rated armature current=%.2f A\" %I_a)\n",
+      "n_m=1200.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "r_a=0.2\n",
+      "V_t=K_a*I_f*w_m+I_a*r_a\n",
+      "a=math.degrees(math.acos(V_t*math.pi/(2*V_m)))\n",
+      "print(\"firing angle delay=%.2f deg\" %a)\n",
+      "E_a=V_t\n",
+      "w_mo=E_a/(K_a*I_f)\n",
+      "N=60*w_mo/(2*math.pi)\n",
+      "reg=((N-n_m)/n_m)*100    \n",
+      "print(\"speed regulation at full load=%.2f\" %reg)\n",
+      "I_ar=I_a\n",
+      "pf=(V_t*I_a)/(V_s*I_ar)    \n",
+      "print(\"input power factor of armature convertor=%.4f\" %pf)\n",
+      "I_fr=I_f\n",
+      "I_sr=math.sqrt(I_fr**2+I_ar**2)\n",
+      "VA=I_sr*V_s\n",
+      "P=V_t*I_a+V_f*I_f\n",
+      "\n",
+      "#Results\n",
+      "print(\"input power factor of drive=%.4f\" %(P/VA))\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rated armature current=59.01 A\n",
+        "firing angle delay=57.63 deg\n",
+        "speed regulation at full load=6.52\n",
+        "input power factor of armature convertor=0.4821\n",
+        "input power factor of drive=0.5093\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 Page No 664"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_f=2*V_m/math.pi\n",
+      "\n",
+      "#Calculations\n",
+      "a1=math.degrees(math.acos(V_t*math.pi/(2*V_m))) \n",
+      "print(\"delay angle of field converter=%.0f deg\" %a1)\n",
+      "r_f=200.0\n",
+      "I_f=V_f/r_f\n",
+      "T_e=85.0\n",
+      "K_a=0.8\n",
+      "I_a=T_e/(I_f*K_a)\n",
+      "n_m=1200.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "r_a=0.1\n",
+      "I_a=50.0\n",
+      "V_t=-K_a*I_f*w_m+I_a*r_a\n",
+      "a=math.degrees(math.acos(V_t*math.pi/(2*V_m)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay of armature converter=%.3f deg\" %a)\n",
+      "print(\"power fed back to ac supply=%.0f W\" %(-V_t*I_a))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "delay angle of field converter=58 deg\n",
+        "firing angle delay of armature converter=119.260 deg\n",
+        "power fed back to ac supply=8801 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6 Page No 665"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=220.0\n",
+      "n_m=1500.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=10.0\n",
+      "r_a=1.0\n",
+      "\n",
+      "#Calculations\n",
+      "K_m=(V_t-I_a*r_a)/(w_m)\n",
+      "T=5.0\n",
+      "I_a=T/K_m\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=30.0\n",
+      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
+      "w_m=(V_t-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "print(\"motor speed=%.2f rpm\" %N)\n",
+      "a=45\n",
+      "n_m=1000\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
+      "I_a=(V_t-K_m*w_m)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"torque developed=%.3f Nm\" %T_e)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor speed=1254.22 rpm\n",
+        "torque developed=8.586 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7, Page No 666"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=220.0\n",
+      "n_m=1000.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=60.0\n",
+      "r_a=.1\n",
+      "\n",
+      "#Calculations\n",
+      "K_m=(V_t-I_a*r_a)/(w_m)\n",
+      "V_s=230\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "print(\"for 600rpm speed\")\n",
+      "n_m=600.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "a=math.degrees(math.acos((K_m*w_m+I_a*r_a)*math.pi/(2*V_m))) \n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"for -500rpm speed\")\n",
+      "n_m=-500.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "a=math.degrees(math.acos((K_m*w_m+I_a*r_a)*math.pi/(2*V_m)))\n",
+      "print(\"firing angle=%.2f deg\" %a)\n",
+      "I_a=I_a/2\n",
+      "a=150\n",
+      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
+      "w_m=(V_t-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor speed=%.3f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for 600rpm speed\n",
+        "firing angle=49.530 deg\n",
+        "for -500rpm speed\n",
+        "firing angle=119.19 deg\n",
+        "motor speed=-852.011 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8 Page No 672"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.5\n",
+      "T_e=50.0\n",
+      "I_a=T_e/K_m\n",
+      "r_a=0.9\n",
+      "a=45.0\n",
+      "V_s=415.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "w_m=((3*V_ml*(1+math.cos(math.radians(a)))/(2*math.pi))-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor speed=%.2f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor speed=2854.42 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9 Page No 672"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_t=600\n",
+      "n_m=1500.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=80.0\n",
+      "r_a=1.0\n",
+      "\n",
+      "#Calculations\n",
+      "K_m=(V_t-I_a*r_a)/(w_m)\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "print(\"for firing angle=45deg and speed=1200rpm\")\n",
+      "a=45.0\n",
+      "n_m=1200.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=(3*V_m*(1+math.cos(math.radians(a)))/(2*math.pi)-K_m*w_m)/r_a\n",
+      "I_sr=I_a*math.sqrt(2/3)    \n",
+      "print(\"rms value of source current=%.3f A\" %I_sr)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %(I_a*math.sqrt(1/3)))\n",
+      "print(\"avg value of thyristor current=%.2f A\" %I_a*(1/3))\n",
+      "pf=(3/(2*math.pi)*(1+math.cos(math.radians(a))))    \n",
+      "print(\"input power factor=%.3f\" %pf)\n",
+      "\n",
+      "print(\"for firing angle=90deg and speed=700rpm\")\n",
+      "a=90\n",
+      "n_m=700\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=(3*V_m*(1+math.cos(math.radians(a)))/(2*math.pi)-K_m*w_m)/r_a\n",
+      "I_sr=I_a*math.sqrt(90/180)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of source current=%.3f A\" %I_sr)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %(I_a*math.sqrt(90.0/360)))\n",
+      "print(\"avg value of thyristor current=%.3f A\" %I_a*(1/3))\n",
+      "pf=(math.sqrt(6)/(2*math.pi)*(1+math.cos(math.radians(a))))*math.sqrt(180/(180-a))    \n",
+      "print(\"input power factor=%.4f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=45deg and speed=1200rpm\n",
+        "rms value of source current=0.000 A\n",
+        "rms value of thyristor current=0.000 A\n",
+        "\n",
+        "input power factor=0.815\n",
+        "for firing angle=90deg and speed=700rpm\n",
+        "rms value of source current=0.000 A\n",
+        "rms value of thyristor current=195.558 A\n",
+        "\n",
+        "input power factor=0.5513\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.10 Page No 676"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=30\n",
+      "V_t=3*V_m*math.cos(math.radians(a))/math.pi\n",
+      "I_a=21.0\n",
+      "r_a=.1\n",
+      "V_d=2.0\n",
+      "K_m=1.6\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=(V_t-I_a*r_a-V_d)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "print(\"speed of motor=%.1f rpm\" %N)\n",
+      "\n",
+      "N=2000\n",
+      "w_m=2*math.pi*N/60\n",
+      "I_a=210\n",
+      "V_t=K_m*w_m+I_a*r_a+V_d\n",
+      "a=math.degrees(math.acos(V_t*math.pi/(3*V_m)))\n",
+      "print(\"firing angle=%.2f deg\" %a)\n",
+      "I_sr=I_a*math.sqrt(2.0/3.0)\n",
+      "pf=V_t*I_a/(math.sqrt(3)*V_s*I_sr)    \n",
+      "print(\"supply power factor=%.3f\" %pf)\n",
+      "\n",
+      "I_a=21\n",
+      "w_m=(V_t-I_a*r_a-V_d)/K_m\n",
+      "n=w_m*60/(2*math.pi)\n",
+      "reg=(n-N)/N*100    \n",
+      "\n",
+      "#Results\n",
+      "print(\"speed regulation(percent)=%.2f\" %reg)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "speed of motor=2767.6 rpm\n",
+        "firing angle=48.48 deg\n",
+        "supply power factor=0.633\n",
+        "speed regulation(percent)=5.64\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.11, Page No 677"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=230.0\n",
+      "V_l=V_t*math.pi/(3*math.sqrt(2))\n",
+      "V_ph=V_l/math.sqrt(3)\n",
+      "V_in=400     #per phase voltage input\n",
+      "\n",
+      "#Calculations\n",
+      "N1=1500.0\n",
+      "I_a1=20.0\n",
+      "r_a1=.6\n",
+      "E_a1=V_t-I_a1*r_a1\n",
+      "n1=1000.0\n",
+      "E_a2=E_a1/1500.0*1000.0\n",
+      "V_t1=E_a1+I_a1*r_a1\n",
+      "a1=math.degrees(math.acos(V_t1*math.pi/(3*math.sqrt(2.0)*V_l)))\n",
+      "I_a2=.5*I_a1\n",
+      "n2=-900.0\n",
+      "V_t2=n2*E_a2/N1+I_a2*r_a1\n",
+      "a2=math.degrees(math.acos(V_t2*math.pi/(3*math.sqrt(2)*V_l))) \n",
+      "\n",
+      "#Results\n",
+      "print(\"transformer phase turns ratio=%.3f\" %(V_in/V_ph))\n",
+      "print(\"for motor running at 1000rpm at rated torque\")\n",
+      "print(\"firing angle delay=%.2f deg\" %a1)\n",
+      "print(\"for motor running at -900rpm at half of rated torque\")\n",
+      "print(\"firing angle delay=%.3f deg\" %a2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "transformer phase turns ratio=4.068\n",
+        "for motor running at 1000rpm at rated torque\n",
+        "firing angle delay=0.00 deg\n",
+        "for motor running at -900rpm at half of rated torque\n",
+        "firing angle delay=110.674 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.12, Page No 678"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=400\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "V_f=3*V_ml/math.pi\n",
+      "R_f=300.0\n",
+      "I_f=V_f/R_f\n",
+      "T_e=60.0\n",
+      "k=1.1\n",
+      "\n",
+      "#Calculations\n",
+      "I_a=T_e/(k*I_f)\n",
+      "N1=1000.0\n",
+      "w_m1=2*math.pi*N1/60\n",
+      "r_a1=.3\n",
+      "V_t1=k*I_f*w_m1+I_a*r_a1\n",
+      "a1=math.degrees(math.acos(V_f*math.pi/(3*V_ml)))\n",
+      "N2=3000\n",
+      "w_m2=2*math.pi*N/60\n",
+      "a2=0\n",
+      "V_t2=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "I_f2=(V_t2-I_a*r_a)/(w_m2*k)\n",
+      "V_f2=I_f2*R_f\n",
+      "a2=math.degrees(math.acos(V_f2*math.pi/(3*V_ml)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"firing angle=%.3f deg\" %a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle=48.477 deg\n",
+        "firing angle=48.477 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.13, Page No 679"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      " #after calculating\n",
+      " #t=w_m/6000-math.pi/360\n",
+      "\n",
+      "N=1000.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "t=w_m/6000-math.pi/360    \n",
+      "\n",
+      "#Results\n",
+      "print(\"time reqd=%.2f s\" %t)\n",
+      " #printing mistake in the answer in book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time reqd=0.01 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.14, Page No 679"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=1.0 #supposition\n",
+      "a=60.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_s1=2*math.sqrt(2)/math.pi*I_a*math.sin(math.radians(a))\n",
+      "I_s3=2*math.sqrt(2)/(3*math.pi)*I_a*math.sin(math.radians(3*a))\n",
+      "I_s5=2*math.sqrt(2)/(5*math.pi)*I_a*math.sin(math.radians(5*a))\n",
+      "per3=I_s3/I_s1*100    \n",
+      "print(\"percent of 3rd harmonic current in fundamental=%.2f\" %per3)\n",
+      "per5=I_s5/I_s1*100    \n",
+      "\n",
+      "#Results\n",
+      "print(\"percent of 5th harmonic current in fundamental=%.2f\" %per5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percent of 3rd harmonic current in fundamental=0.00\n",
+        "percent of 5th harmonic current in fundamental=-20.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.15, Page No 680"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=60.0\n",
+      "I_TA=I_a/3    \n",
+      "\n",
+      "#Calculations\n",
+      "print(\"avg thyristor current=%.0f A\" %I_TA)\n",
+      "I_Tr=I_a/math.sqrt(3)    \n",
+      "print(\"rms thyristor current=%.3f A\" %I_Tr)\n",
+      "V_s=400\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "I_sr=I_a*math.sqrt(2.0/3)\n",
+      "a=150\n",
+      "V_t=3*V_m*math.cos(math.radians(a))/math.pi\n",
+      "pf=V_t*I_a/(math.sqrt(3)*V_s*I_sr)    \n",
+      "print(\"power factor of ac source=%.3f\" %pf)\n",
+      "\n",
+      "r_a=0.5\n",
+      "K_m=2.4\n",
+      "w_m=(V_t-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Speed of motor=%.2f rpm\" %N)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg thyristor current=20 A\n",
+        "rms thyristor current=34.641 A\n",
+        "power factor of ac source=-0.827\n",
+        "Speed of motor=-1980.76 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.16, Page No 685"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=300.0\n",
+      "V_s=600.0\n",
+      "a=0.6\n",
+      "V_t=a*V_s\n",
+      "P=V_t*I_a    \n",
+      "\n",
+      "#Calculations\n",
+      "print(\"input power from source=%.0f kW\" %(P/1000))\n",
+      "R_eq=V_s/(a*I_a)    \n",
+      "print(\"equivalent input resistance=%.3f ohm\" %R_eq)\n",
+      "k=.004\n",
+      "R=.04+.06\n",
+      "w_m=(a*V_s-I_a*R)/(k*I_a)\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "print(\"motor speed=%.1f rpm\" %N)\n",
+      "T_e=k*I_a**2    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.0f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "input power from source=108 kW\n",
+        "equivalent input resistance=3.333 ohm\n",
+        "motor speed=2626.1 rpm\n",
+        "motor torque=360 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.17, Page No 686"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_on=10.0\n",
+      "T_off=15.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=T_on/(T_on+T_off)\n",
+      "V_s=230.0\n",
+      "V_t=a*V_s\n",
+      "r_a=3\n",
+      "K_m=.5\n",
+      "N=1500\n",
+      "w_m=2*math.pi*N/60\n",
+      "I_a=(V_t-K_m*w_m)/r_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor load current=%.3f A\" %I_a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor load current=4.487 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.18, Page No 686"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "w_m=0    \n",
+      "print(\"lower limit of speed control=%.0f rpm\" %w_m)\n",
+      "I_a=25.0\n",
+      "r_a=.2\n",
+      "V_s=220\n",
+      "K_m=0.08\n",
+      "\n",
+      "#Calculations\n",
+      "a=(K_m*w_m+I_a*r_a)/V_s    \n",
+      "print(\"lower limit of duty cycle=%.3f\" %a)\n",
+      "a=1    \n",
+      "print(\"upper limit of duty cycle=%.0f\" %a)\n",
+      "w_m=(a*V_s-I_a*r_a)/K_m    \n",
+      "\n",
+      "#Results\n",
+      "print(\"upper limit of speed control=%.1f rpm\" %w_m)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lower limit of speed control=0 rpm\n",
+        "lower limit of duty cycle=0.023\n",
+        "upper limit of duty cycle=1\n",
+        "upper limit of speed control=2687.5 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.21, Page No 691"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.6\n",
+      "V_s=400.0\n",
+      "V_t=(1-a)*V_s\n",
+      "I_a=300.0\n",
+      "P=V_t*I_a  \n",
+      "\n",
+      "#Calculations  \n",
+      "print(\"power returned=%.0f kW\" %(P/1000))\n",
+      "r_a=.2\n",
+      "K_m=1.2\n",
+      "R_eq=(1-a)*V_s/I_a+r_a    \n",
+      "print(\"equivalent load resistance=%.4f ohm\" %R_eq)\n",
+      "w_mn=I_a*r_a/K_m\n",
+      "N=w_mn*60/(2*math.pi)    \n",
+      "print(\"min braking speed=%.2f rpm\" %N)\n",
+      "w_mx=(V_s+I_a*r_a)/K_m\n",
+      "N=w_mx*60/(2*math.pi)    \n",
+      "print(\"max braking speed=%.1f rpm\" %N)\n",
+      "w_m=(V_t+I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"max braking speed=%.1f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power returned=48 kW\n",
+        "equivalent load resistance=0.7333 ohm\n",
+        "min braking speed=477.46 rpm\n",
+        "max braking speed=3660.6 rpm\n",
+        "max braking speed=1750.7 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.22, Page No 699"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "N=1500.0\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"when speed=1455rpm\")\n",
+      "n=1455.0\n",
+      "s1=(N-n)/N\n",
+      "r=math.sqrt(1/3)*(2/3)/(math.sqrt(s1)*(1-s1))    \n",
+      "print(\"I_2mx/I_2r=%.3f\" %r)\n",
+      "print(\"when speed=1350rpm\")\n",
+      "n=1350\n",
+      "s1=(N-n)/N\n",
+      "r=math.sqrt(1/3)*(2/3)/(math.sqrt(s1)*(1-s1))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"I_2mx/I_2r=%.3f\" %r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when speed=1455rpm\n",
+        "I_2mx/I_2r=0.000\n",
+        "when speed=1350rpm\n",
+        "I_2mx/I_2r=0.000\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.24, Page No 705"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V1=400.0\n",
+      "r1=0.6\n",
+      "r2=0.4\n",
+      "s=1.0\n",
+      "x1=1.6\n",
+      "x2=1.6\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"at starting in normal conditions\")\n",
+      "I_n=V1/math.sqrt((r1+r2/s)**2+(x1+x2)**2)    \n",
+      "print(\"current=%.2f A\" %I_n)\n",
+      "pf=(r1+r2)/math.sqrt((r1+r2/s)**2+(x1+x2)**2)    \n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "f1=50\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_en=(3/w_s)*I_n**2*(r2/s)    \n",
+      "print(\"\\nTorque developed=%.2f Nm\" %T_en)\n",
+      "print(\"motor is operated with DOL starting\")\n",
+      "I_d=V1/2/math.sqrt((r1+r2/s)**2+((x1+x2)/2)**2)    \n",
+      "print(\"current=%.0f A\" %I_d)\n",
+      "pf=(r1+r2)/math.sqrt((r1+r2/s)**2+((x1+x2)/2)**2)    \n",
+      "print(\"pf=%.2f\" %pf)\n",
+      "f1=25\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_ed=(3/w_s)*I_d**2*(r2/s)    \n",
+      "print(\"Torque developed=%.3f Nm\" %T_ed)\n",
+      "print(\"at max torque conditions\")\n",
+      "s_mn=r2/math.sqrt((r1)**2+((x1+x2))**2)\n",
+      "I_n=V1/math.sqrt((r1+r2/s_mn)**2+(x1+x2)**2)    \n",
+      "print(\"current=%.3f A\" %I_n)\n",
+      "pf=(r1+r2/s_mn)/math.sqrt((r1+r2/s_mn)**2+(x1+x2)**2)    \n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "f1=50\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_en=(3/w_s)*I_n**2*(r2/s_mn)    \n",
+      "print(\"Torque developed=%.2f Nm\" %T_en)\n",
+      "print(\"motor is operated with DOL starting\")\n",
+      "s_mn=r2/math.sqrt((r1)**2+((x1+x2)/2)**2)\n",
+      "I_d=V1/2/math.sqrt((r1+r2/s_mn)**2+((x1+x2)/2)**2)    \n",
+      "print(\"current=%.3f A\" %I_d)\n",
+      "pf=(r1+r2/s_mn)/math.sqrt((r1+r2/s_mn)**2+((x1+x2)/2)**2)    \n",
+      "print(\"\\npf=%.3f\" %pf)\n",
+      "f1=25\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_en=(3/w_s)*I_d**2*(r2/s_mn)  \n",
+      "\n",
+      "\n",
+      "#Results  \n",
+      "print(\"Torque developed=%.3f Nm\" %T_en)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "at starting in normal conditions\n",
+        "current=119.31 A\n",
+        "pf=0.2983\n",
+        "\n",
+        "Torque developed=108.75 Nm\n",
+        "motor is operated with DOL starting\n",
+        "current=106 A\n",
+        "pf=0.53\n",
+        "Torque developed=171.673 Nm\n",
+        "at max torque conditions\n",
+        "current=79.829 A\n",
+        "pf=0.7695\n",
+        "Torque developed=396.26 Nm\n",
+        "motor is operated with DOL starting\n",
+        "current=71.199 A\n",
+        "\n",
+        "pf=0.822\n",
+        "Torque developed=330.883 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.25, Page No 709"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "x1=1.0\n",
+      "X_m=50.0\n",
+      "X_e=x1*X_m/(x1+X_m)\n",
+      "V=231.0\n",
+      "V_e=V*X_m/(x1+X_m)\n",
+      "x2=1.0\n",
+      "r2=.4\n",
+      "r1=0\n",
+      "\n",
+      "#Calculations\n",
+      "s_m=r2/(x2+X_e)    \n",
+      "print(\"slip at max torque=%.2f\" %s_m)\n",
+      "s_mT=r2/(x2+X_m)    \n",
+      "print(\"slip at max torque=%.5f\" %s_mT)\n",
+      "f1=50.0\n",
+      "w_s=4*math.pi*f1/4\n",
+      "print(\"for constant voltage input\")\n",
+      "T_est=(3/w_s)*(V_e/math.sqrt(r2**2+(x2+X_e)**2))**2*(r2)    \n",
+      "print(\"starting torque=%.3f Nm\" %T_est)\n",
+      "T_em=(3/w_s)*V_e**2/(2*(x2+X_e))    \n",
+      "print(\"maximum torque developed=%.2f Nm\" %T_em)\n",
+      "print(\"for constant current input\")\n",
+      "I1=28\n",
+      "T_est=(3/w_s)*(I1*X_m)**2/(r2**2+(x2+X_m)**2)*r2    \n",
+      "print(\"starting torque=%.3f Nm\" %T_est)\n",
+      "T_em=(3/w_s)*(I1*X_m)**2/(2*(x2+X_m))   \n",
+      "print(\"maximum torque developed=%.3f Nm\" %T_em)\n",
+      "s=s_mT\n",
+      "i=1\n",
+      "I_m=I1*(r2/s+i*x2)/(r2/s+i*(x2+X_m))\n",
+      "I_m=math.fabs(I_m)\n",
+      "V1=math.sqrt(3)*I_m*X_m    \n",
+      "\n",
+      "#Results\n",
+      "print(\"supply voltage reqd=%.1f V\" %V1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "slip at max torque=0.20\n",
+        "slip at max torque=0.00784\n",
+        "for constant voltage input\n",
+        "starting torque=95.988 Nm\n",
+        "maximum torque developed=247.31 Nm\n",
+        "for constant current input\n",
+        "starting torque=5.756 Nm\n",
+        "maximum torque developed=366.993 Nm\n",
+        "supply voltage reqd=1236.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.27, Page No 718"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=420.0\n",
+      "V1=V/math.sqrt(3)\n",
+      "T_e=450.0\n",
+      "N=1440.0\n",
+      "n=1000.0\n",
+      "T_L=T_e*(n/N)**2\n",
+      "n1=1500.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_s=2*math.pi*n1/60\n",
+      "w_m=2*math.pi*n/60\n",
+      "a=.8\n",
+      "I_d=T_L*w_s/(2.339*a*V1)\n",
+      "k=0\n",
+      "R=(1-w_m/w_s)*(2.339*a*V1)/(I_d*(1-k))    \n",
+      "print(\"value of chopper resistance=%.4f ohm\" %R)\n",
+      "n=1320.0\n",
+      "T_L=T_e*(n/N)**2\n",
+      "I_d=T_L*w_s/(2.339*a*V1)    \n",
+      "print(\"Inductor current=%.3f A\" %I_d)\n",
+      "w_m=2*math.pi*n/60\n",
+      "k=1-((1-w_m/w_s)*(2.339*a*V1)/(I_d*R))    \n",
+      "print(\"value of duty cycle=%.4f\" %k)\n",
+      "s=(n1-n)/n1\n",
+      "V_d=2.339*s*a*V1    \n",
+      "print(\"Rectifed o/p voltage=%.3f V\" %V_d)\n",
+      "P=V_d*I_d\n",
+      "I2=math.sqrt(2/3)*I_d\n",
+      "r2=0.02\n",
+      "Pr=3*I2**2*r2\n",
+      "I1=a*I2\n",
+      "r1=0.015\n",
+      "Ps=3*I1**2*r1\n",
+      "Po=T_L*w_m\n",
+      "Pi=Po+Ps+Pr+P\n",
+      "eff=Po/Pi*100   \n",
+      "\n",
+      "#Results\n",
+      "print(\"Efficiency(in percent)=%.2f\" %eff)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of chopper resistance=2.0132 ohm\n",
+        "Inductor current=130.902 A\n",
+        "value of duty cycle=0.7934\n",
+        "Rectifed o/p voltage=54.449 V\n",
+        "Efficiency(in percent)=88.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.28, Page No 720"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=400.0\n",
+      "V_ph=V/math.sqrt(3)\n",
+      "N_s=1000.0\n",
+      "N=800.0\n",
+      "a=.7\n",
+      "I_d=110\n",
+      "R=2.0\n",
+      "\n",
+      "#Calculations\n",
+      "k=1-((1-N/N_s)*(2.339*a*V_ph)/(I_d*R))    \n",
+      "print(\"value of duty cycle=%.3f\" %k)\n",
+      "P=I_d**2*R*(1-k)\n",
+      "I1=a*I_d*math.sqrt(2/3)\n",
+      "r1=0.1\n",
+      "r2=0.08\n",
+      "Pr=3*I1**2*(r1+r2)\n",
+      "P_o=20000\n",
+      "P_i=P_o+Pr+P\n",
+      "eff=P_o/P_i*100    \n",
+      "print(\"Efficiency=%.2f\" %eff)\n",
+      "I11=math.sqrt(6)/math.pi*a*I_d\n",
+      "th=43\n",
+      "P_ip=math.sqrt(3)*V*I11*math.cos(math.radians(th))\n",
+      "pf=P_ip/(math.sqrt(3)*V*I11)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Input power factor=%.4f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of duty cycle=0.656\n",
+        "Efficiency=70.62\n",
+        "Input power factor=0.7314\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.29, Page No 724"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=420.0\n",
+      "V1=V/math.sqrt(3)\n",
+      "N=1000.0\n",
+      "w_m=2*math.pi*N/60\n",
+      "N_s=1500.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "a=0.8\n",
+      "V_d=2.339*a*s*V1    \n",
+      "print(\"rectified voltage=%.2f V\" %V_d)\n",
+      "T=450.0\n",
+      "N1=1200.0\n",
+      "T_L=T*(N/N1)**2\n",
+      "f1=50\n",
+      "w_s=4*math.pi*f1/4\n",
+      "I_d=w_s*T_L/(2.339*a*V1)    \n",
+      "print(\"inductor current=%.2f A\" %I_d)\n",
+      "a_T=-.4\n",
+      "a1=math.degrees(math.acos(s*a/a_T))\n",
+      "print(\"delay angle of inverter=%.2f deg\" %a1)\n",
+      "\n",
+      "P_s=V_d*I_d\n",
+      "P_o=T_L*w_m\n",
+      "R_d=0.01\n",
+      "P_i=I_d**2*R_d\n",
+      "I2=math.sqrt(2/3)*I_d\n",
+      "r2=0.02\n",
+      "r1=0.015\n",
+      "P_rol=3*I2**2*r2\n",
+      "I1=a*I2\n",
+      "P_sol=3*I1**2*r1\n",
+      "P_i=P_o+P_rol+P_sol+P_i\n",
+      "eff=P_o/P_i*100    \n",
+      "print(\"\\nefficiency=%.2f\" %eff)\n",
+      "w_m=w_s*(1+(-a_T/a)*math.cos(math.radians(a1))-w_s*R_d*T_L/(2.339*a*V1)**2)\n",
+      "N=w_m*60/(2*math.pi) \n",
+      "\n",
+      "#Results   \n",
+      "print(\"motor speed=%.1f rpm\" %N)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectified voltage=151.25 V\n",
+        "inductor current=108.18 A\n",
+        "delay angle of inverter=131.81 deg\n",
+        "\n",
+        "efficiency=99.64\n",
+        "motor speed=996.4 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.30, Page No 726"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=700.0\n",
+      "E2=V/math.sqrt(3)\n",
+      "N_s=1500.0\n",
+      "N=1200.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "V_dd=0.7\n",
+      "V_dt=1.5\n",
+      "V_d=3*math.sqrt(6)*s*E2/math.pi-2*V_dd\n",
+      "V1=415.0\n",
+      "a=math.degrees(math.acos((3*math.sqrt(2)*E2/math.pi)**-1*(-V_d+2*V_dt)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle advance=%.2f deg\" %(180-a))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle advance=70.22 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.31, Page No 726"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=700.0\n",
+      "E2=V/math.sqrt(3)\n",
+      "N_s=1500.0\n",
+      "N=1200.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "V_dd=.7\n",
+      "V_dt=1.5\n",
+      "a=0\n",
+      "u=18 #overlap angle in case of rectifier\n",
+      "V_d=3*math.sqrt(6)*s*E2*(math.cos(math.radians(a))+math.cos(math.radians(a+u)))/(2*math.pi)-2*V_dd\n",
+      "V1=415\n",
+      "V_ml=math.sqrt(2)*V1\n",
+      "u=4 #overlap anglein the inverter\n",
+      " #V_dc=-(3*V_ml*(math.cos(math.radians(a))+math.cos(math.radians(a+u)))/(2*math.pi)-2*V_dt)\n",
+      " #V_dc=V_d\n",
+      " #after solving %  (1+math.cos(math.radians(u)))*math.cos(math.radians(a))-math.sin(math.radians(u))*math.sin(math.radians(a))=-.6425\n",
+      "a=math.degrees(math.acos(-.6425/(math.sqrt((1+math.cos(math.radians(u)))**2+math.sin(math.radians(u))**2))))-math.degrees(math.asin(math.sin(math.radians(a))/(1+math.cos(math.radians(u)))))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle advance=%.2f deg\" %(180-a))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle advance=71.25 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.32, Page No 727"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=700.0\n",
+      "E2=V\n",
+      "N_s=1500.0\n",
+      "N=1200.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "V1=415.0\n",
+      "a_T=s*E2/V1    \n",
+      "\n",
+      "#Results\n",
+      "print(\"voltage ratio of the transformer=%.2f\" %a_T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "voltage ratio of the transformer=0.34\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.33, Page No 733"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "P=6.0\n",
+      "N_s=600.0\n",
+      "f1=P*N_s/120.0\n",
+      "V=400.0\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_t=f1*V/f    \n",
+      "print(\"supply freq=%.0f Hz\" %V_t)\n",
+      "T=340.0\n",
+      "N=1000.0\n",
+      "T_L=T*(N_s/N)**2\n",
+      "w_s=2*math.pi*N_s/60\n",
+      "P=T_L*w_s\n",
+      "I_a=P/(math.sqrt(3)*V_t)    \n",
+      "print(\"armature current=%.2f A\" %I_a)\n",
+      "Z_s=2\n",
+      "X_s=f1/f*math.fabs(Z_s)\n",
+      "V_t=V_t/math.sqrt(3)\n",
+      "Ef=math.sqrt(V_t**2+(I_a*X_s)**2)\n",
+      "print(\"excitation voltage=%.2f V\" %(math.sqrt(3)*Ef))\n",
+      "dl=math.degrees(math.atan(I_a*X_s/V_t))\n",
+      "print(\"load angle=%.2f deg\" %dl)\n",
+      "T_em=(3/w_s)*(Ef*V_t/X_s)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"pull out torque=%.2f Nm\" %T_em)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "supply freq=240 Hz\n",
+        "armature current=18.50 A\n",
+        "excitation voltage=243.06 V\n",
+        "load angle=9.10 deg\n",
+        "pull out torque=773.69 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.34, Page No 736"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "P=4.0\n",
+      "f=50.0\n",
+      "w_s=4*math.pi*f/P\n",
+      "X_d=8.0\n",
+      "X_q=2.0\n",
+      "T_e=80.0\n",
+      "V=400.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_t=V/math.sqrt(3)\n",
+      "dl=(1/2)*math.degrees(math.asin(T_e*w_s/((3/2)*(V_t)**2*(1/X_q-1/X_d))))    \n",
+      "print(\"load angle=%.3f deg\" %dl)\n",
+      "I_d=V_t*math.cos(math.radians(dl))/X_d\n",
+      "I_q=V_t*math.sin(math.radians(dl))/X_q\n",
+      "I_a=math.sqrt(I_d**2+I_q**2)    \n",
+      "print(\"armature current=%.2f A\" %I_a)\n",
+      "pf=T_e*w_s/(math.sqrt(3)*V*I_a)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"input power factor=%.4f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "load angle=0.000 deg\n",
+        "armature current=28.87 A\n",
+        "input power factor=0.6283\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.35, Page No 737"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_e=3.0\n",
+      "K_m=1.2\n",
+      "I_a=T_e/K_m\n",
+      "r_a=2.0\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "E_a=(0.263*math.sqrt(2)*V-I_a*r_a)/(1-55/180)\n",
+      "w_m=E_a/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor speed=%.2f rpm\" %N)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor speed=640.96 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.36, Page No 738"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=1360.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      " #after calculations V_t % calculated\n",
+      "V_t=163.45\n",
+      "r_a=4\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.4f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=5.2578 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.37, Page No 740"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=2100.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      " #after calculations V_t % calculated\n",
+      "V_t=227.66\n",
+      "r_a=4\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.2f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=1.94 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.38, Page No 742"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=840.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      "V=230.0\n",
+      "a=75.0\n",
+      "V_t=math.sqrt(2)*V/math.pi*(1+math.cos(math.radians(a)))\n",
+      "r_a=4\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.4f Nm\" %T_e)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=10.5922 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.39, Page No 743"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=1400.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      "V=230.0\n",
+      "a=60.0\n",
+      "a1=212\n",
+      "V_t=math.sqrt(2)*V/math.pi*(math.cos(math.radians(a))-math.cos(math.radians(a1)))+E_a*(180+a-a1)/180\n",
+      "r_a=3\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a   \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.3f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=5.257 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.40, Page No 745"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=600.0\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      "V=230.0\n",
+      "a=60.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_t=2*math.sqrt(2)*V/math.pi*(math.cos(math.radians(a)))\n",
+      "r_a=3\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.3f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=13.568 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.41, Page No 745"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "r1=.6\n",
+      "r2=.4\n",
+      "s=0.04\n",
+      "x1=1.6\n",
+      "x2=1.6\n",
+      "Z=(r1+r2/s)+(x1+x2)\n",
+      "V=400.0\n",
+      "I1=V/Z    \n",
+      "print(\"source current=%.3f A \" %math.degrees(math.atan(I1.imag/I1.real)))\n",
+      "print(\"and with %.1f deg phase\" %math.fabs(I1))\n",
+      "I2=V/Z\n",
+      "N=1500\n",
+      "w_s=2*math.pi*N/60\n",
+      "T_e=(3/w_s)*abs(I2)**2*r2/s    \n",
+      "print(\"motor torque=%.2f Nm\" %T_e)\n",
+      "N_r=N*(1-s)\n",
+      "\n",
+      "f=45\n",
+      "N_s1=120*f/4\n",
+      "w_s=2*math.pi*N_s1/60\n",
+      "s1=(N_s1-N_r)/N_s1\n",
+      "Z=(r1+r2/s1)+(x1+x2)*f/50.0\n",
+      "V=360\n",
+      "I1=V/Z    \n",
+      "print(\"source current=%.3f A \" %math.degrees(math.atan(I1.imag/I1.real)))\n",
+      "print(\"and with %.1f deg phase\" %math.fabs(I1))\n",
+      "I2=V/Z\n",
+      "T_e=(3/w_s)*abs(I2)**2*r2/s1    \n",
+      "print(\"motor torque=%.2f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "source current=0.000 A \n",
+        "and with 29.0 deg phase\n",
+        "motor torque=160.46 Nm\n",
+        "source current=-0.000 A \n",
+        "and with 142.9 deg phase\n",
+        "motor torque=-2598.45 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter13.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter13.ipynb
new file mode 100755
index 00000000..62d2a926
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter13.ipynb
@@ -0,0 +1,342 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Power Factor Improvement"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1, Page No 754"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=250.0\n",
+      "R_l=5.0\n",
+      "I_l=20.0\n",
+      "V_l1=math.sqrt(V_s**2-(R_l*I_l)**2)\n",
+      "reg2=(V_s-V_l1)/V_s*100    \n",
+      "pf1=1.0\n",
+      "\n",
+      "#Calculations\n",
+      "P_l1=V_l1*I_l*pf1     #load power\n",
+      "P_r1=V_s*I_l*pf1     #max powwible system rating\n",
+      "utf1=P_l1*100/P_r1  \n",
+      "pf2=0.5\n",
+      " #(.5*V_l)**2+(.866*V_l+R_l*I_l)**2=V_s**2\n",
+      " #after solving\n",
+      "V_l2=158.35    \n",
+      "reg2=(V_s-V_l2)/V_s*100    \n",
+      "P_l2=V_l2*I_l*pf2   #load power\n",
+      "P_r2=V_s*I_l     #max powwible system rating\n",
+      "utf2=P_l2*100/P_r2    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"for pf=1\")\n",
+      "print(\"load voltage=%.2f V\" %V_l1)\n",
+      "print(\"voltage regulation=%.2f\" %reg1)\n",
+      "print(\"system utilisation factor=%.3f\" %utf1)\n",
+      "print(\"energy consumed(in units)=%.1f\" %(P_l1/1000))\n",
+      "print(\"for pf=.5\")\n",
+      "print(\"load voltage=%.2f V\" %V_l2)\n",
+      "print(\"voltage regulation=%.2f\" %reg2)\n",
+      "print(\"system utilisation factor=%.3f\" %utf2)\n",
+      "print(\"energy consumed(in units)=%.2f\" %(P_l2/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "NameError",
+       "evalue": "name 'reg1' is not defined",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mNameError\u001b[0m                                 Traceback (most recent call last)",
+        "\u001b[1;32m<ipython-input-2-ffdbe43fd921>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m     25\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"for pf=1\"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     26\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"load voltage=%.2f V\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mV_l1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 27\u001b[1;33m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"voltage regulation=%.2f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mreg1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     28\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"system utilisation factor=%.3f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mutf1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     29\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"energy consumed(in units)=%.1f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mP_l1\u001b[0m\u001b[1;33m/\u001b[0m\u001b[1;36m1000\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
+        "\u001b[1;31mNameError\u001b[0m: name 'reg1' is not defined"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for pf=1\n",
+        "load voltage=229.13 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2, Page No 756"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50.0\n",
+      "V_s=230.0\n",
+      "I_m1=2\n",
+      "pf1=.3\n",
+      "\n",
+      "#Calculations\n",
+      "I_c1=I_m1*math.sin(math.radians(math.degrees(math.acos(pf1))))\n",
+      "C1=I_c1/(2*math.pi*f*V_s)    \n",
+      "I_m2=5\n",
+      "pf2=.5\n",
+      "I_c2=I_m2*math.sin(math.radians(math.degrees(math.acos(pf2))))\n",
+      "C2=I_c2/(2*math.pi*f*V_s)    \n",
+      "I_m3=10\n",
+      "pf3=.7\n",
+      "I_c3=I_m3*math.sin(math.radians(math.degrees(math.acos(pf3))))\n",
+      "C3=I_c3/(2*math.pi*f*V_s)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"at no load\")\n",
+      "print(\"value of capacitance=%.3f uF\" %(C1*10**6))\n",
+      "print(\"at half full load\")\n",
+      "print(\"value of capacitance=%.3f uF\" %(C2*10**6))\n",
+      "print(\"at full load\")\n",
+      "print(\"value of capacitance=%.3f uF\" %(C3*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "at no load\n",
+        "value of capacitance=26.404 uF\n",
+        "at half full load\n",
+        "value of capacitance=59.927 uF\n",
+        "at full load\n",
+        "value of capacitance=98.834 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page No 764"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_c=10.0\n",
+      "f=50.0\n",
+      "V_s=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "C=I_c/(2*math.pi*f*V_s)    \n",
+      "I_l=10\n",
+      "L=V_s/(2*math.pi*f*I_l)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of capacitance=%.3f uF\" %(C*10**6))\n",
+      "print(\"value of inductor=%.3f mH\" %(L*1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of capacitance=138.396 uF\n",
+        "value of inductor=73.211 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4, Page No 765"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "I_L=10.0\n",
+      "X_L=V_s/I_L\n",
+      "I_f1=6.0\n",
+      " #B=2*a-math.sin(2*a)\n",
+      "B=2*math.pi-I_f1*math.pi*X_L/V_s\n",
+      "a=0\n",
+      "i=1.0\n",
+      "for a in range(1,360):\n",
+      "    b=2*a*math.pi/180-math.sin(math.radians(2*a))   \n",
+      "    if math.fabs(B-b)<=0.001  :      #by hit and trial\n",
+      "        i=2\n",
+      "        break\n",
+      "print(\"firing angle of TCR = %.1f deg\" %a)\n",
+      " #(a-.01)*180/math.pi)\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle of TCR = 359.0 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5 Page No 766"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=.01\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"for firing angle=90deg\")\n",
+      "a=90*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.0f mH\" %(L_eff*1000))\n",
+      "print(\"for firing angle=120deg\")\n",
+      "a=120*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.3f mH\" %(L_eff*1000))\n",
+      "print(\"for firing angle=150deg\")\n",
+      "a=150*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.2f mH\" %(L_eff*1000))\n",
+      "print(\"for firing angle=170deg\")\n",
+      "a=170*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.3f H\" %L_eff)\n",
+      "print(\"for firing angle=175deg\")\n",
+      "a=175*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))  \n",
+      "\n",
+      "#Results\n",
+      "print(\"effective inductance=%.2f H\" %L_eff)\n",
+      "print(\"for firing angle=180deg\")\n",
+      "a=180*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.3f H\" %L_eff)\n",
+      " #random value at firing angle =180 is equivalent to infinity as in answer in book\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=90deg\n",
+        "effective inductance=10 mH\n",
+        "for firing angle=120deg\n",
+        "effective inductance=25.575 mH\n",
+        "for firing angle=150deg\n",
+        "effective inductance=173.40 mH\n",
+        "for firing angle=170deg\n",
+        "effective inductance=4.459 H\n",
+        "for firing angle=175deg\n",
+        "effective inductance=35.51 H\n",
+        "for firing angle=180deg\n",
+        "effective inductance=-128265253940037.750 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.6 Page No 766"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "Q=100.0*10**3\n",
+      "V_s=11.0*10**3\n",
+      "\n",
+      "#Calculations\n",
+      "f=50.0\n",
+      "L=V_s**2/(2*math.pi*f*Q)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"effective inductance=%.4f H\" %L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "effective inductance=3.8515 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter14.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter14.ipynb
new file mode 100755
index 00000000..a9c3a3f1
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter14.ipynb
@@ -0,0 +1,93 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 : Miscellaneous Topics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1, Page No 777"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a1=0\n",
+      "a2=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"for two single phase series semiconvertors\")\n",
+      "V_0=V_m/math.pi*(2+math.cos(math.radians(a1))+math.cos(math.radians(a2)))    \n",
+      "print(\"avg o/p voltage=%.2f V\" %V_0)\n",
+      "V_or=V_s*math.sqrt((1/math.pi)*(4*math.pi-3*a2*math.pi/180+(3/2)*math.sin(math.radians(2*a2))))    \n",
+      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
+      "DF=(3+math.cos(math.radians(a2)))/(math.sqrt(2)*math.sqrt(5+3*math.cos(math.radians(a2))))    \n",
+      "print(\"DF=%.2f\" %DF)\n",
+      "PF=math.sqrt(2/math.pi)*(3+math.cos(math.radians(a2)))/math.sqrt(4*math.pi-3*a2*math.pi/180)    \n",
+      "print(\"PF=%.2f\" %PF)\n",
+      "HF=math.sqrt((math.pi*(math.pi-(3/4)*a2*math.pi/180)/(5+3*math.cos(math.radians(a2))))-1)    \n",
+      "print(\"HF=%.2f\" %HF)\n",
+      "print(\"for two single phase series full convertors\")\n",
+      "a=45.0\n",
+      "V_0=2*V_m/math.pi*(1+math.cos(math.radians(a)))    \n",
+      "print(\"avg o/p voltage=%.2f V\" %V_0)\n",
+      "V_or=2*V_s*math.sqrt((1/math.pi)*(math.pi-a2*math.pi/180+(1/2)*math.sin(math.radians(2*a2))))    \n",
+      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
+      "DF=math.cos(math.radians(a2/2)) \n",
+      "\n",
+      "\n",
+      "#Results   \n",
+      "print(\"DF=%.2f\" %DF)\n",
+      "PF=math.sqrt(2/(math.pi*(math.pi-a2*math.pi/180)))*(1+math.cos(math.radians(a2)))    \n",
+      "print(\"PF=%.2f\" %PF)\n",
+      "HF=math.sqrt((math.pi*(math.pi-a2*math.pi/180)/(4+4*math.cos(math.radians(a2))))-1)    \n",
+      "print(\"HF=%.2f\" %HF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for two single phase series semiconvertors\n",
+        "avg o/p voltage=383.82 V\n",
+        "rms value of o/p voltage=434.47 V\n",
+        "DF=0.98\n",
+        "PF=0.93\n",
+        "HF=0.62\n",
+        "for two single phase series full convertors\n",
+        "avg o/p voltage=353.50 V\n",
+        "rms value of o/p voltage=398.37 V\n",
+        "DF=0.92\n",
+        "PF=0.89\n",
+        "HF=0.29\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter2.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter2.ipynb
new file mode 100755
index 00000000..1872c9f4
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter2.ipynb
@@ -0,0 +1,233 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 02 : Power Semiconductor Diodes and Transistors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page No 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "B=40.0\n",
+      "R_c=10                    #ohm\n",
+      "V_cc=130.0                #V\n",
+      "V_B=10.0                  #V\n",
+      "V_CES=1.0                 #V\n",
+      "V_BES=1.5                 #V\n",
+      "\n",
+      "#Calculations\n",
+      "I_CS=(V_cc-V_CES)/R_c     #A\n",
+      "I_BS=I_CS/B                #A\n",
+      "R_B1=(V_B-V_BES)/I_BS\n",
+      "P_T1=V_BES*I_BS+V_CES*I_CS\n",
+      "ODF=5\n",
+      "I_B=ODF*I_BS\n",
+      "R_B2=(V_B-V_BES)/I_B\n",
+      "P_T2=V_BES*I_B+V_CES*I_CS\n",
+      "B_f=I_CS/I_B\n",
+      "\n",
+      "#Results\n",
+      "print(\"value of R_B in saturated state= %.2f ohm\" %R_B1)\n",
+      "print(\"Power loss in transistor=%.2f W\" %P_T1)\n",
+      "print(\"Value of R_B for an overdrive factor 5 = %.2f ohm\" %R_B2)\n",
+      "print(\"Power loss in transistor = %.2f W\" %P_T2)\n",
+      "print(\"Forced current gain=%.0f\" %B_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of R_B in saturated state= 26.36 ohm\n",
+        "Power loss in transistor=13.38 W\n",
+        "Value of R_B for an overdrive factor 5 = 5.27 ohm\n",
+        "Power loss in transistor = 15.32 W\n",
+        "Forced current gain=8\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page No 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_CEO=2*10**-3            #A\n",
+      "V_CC=220.0                #V\n",
+      "P_dt=I_CEO*V_CC           #instant. power loss during delay time\n",
+      "t_d=.4*10**-6            #s\n",
+      "f=5000\n",
+      "P_d=f*I_CEO*V_CC*t_d      #avg power loss during delay time\n",
+      "V_CES=2                     #V\n",
+      "t_r=1*10**-6               #s\n",
+      "I_CS=80                    #A\n",
+      "\n",
+      "#Calculations\n",
+      "P_r=f*I_CS*t_r*(V_CC/2-(V_CC-V_CES)/3)            #avg power loss during rise time\n",
+      "t_m=V_CC*t_r/(2*(V_CC-V_CES))\n",
+      "P_rm=I_CS*V_CC**2/(4*(V_CC-V_CES))           #instant. power loss during rise time\n",
+      "\n",
+      "#Results\n",
+      "P_on=P_d+P_r                \n",
+      "print(\"Avg power loss during turn on = %.2f W\" %P_on)\n",
+      "P_nt=I_CS*V_CES \n",
+      "print(\"Instantaneous power loss during turn on = %.0f W\" %P_nt)\n",
+      "t_n=50*10**-6\n",
+      "P_n=f*I_CS*V_CES*t_n\n",
+      "print(\"Avg power loss during conduction period = %.0f W\" %P_n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Avg power loss during turn on = 14.93 W\n",
+        "Instantaneous power loss during turn on = 160 W\n",
+        "Avg power loss during conduction period = 40 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page No 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_CEO=2*10**-3         #A\n",
+      "V_CC=220          #V\n",
+      "t_d=.4*10**-6   #s\n",
+      "f=5000\n",
+      "V_CES=2         #V\n",
+      "t_r=1*10**-6    #s\n",
+      "I_CS=80         #A\n",
+      "t_n=50*10**-6   #s\n",
+      "t_0=40*10**-6   #s\n",
+      "t_f=3*10**-6    #s\n",
+      "\n",
+      "#Calculations\n",
+      "P_st=I_CS*V_CES  # instant. power loss during t_s\n",
+      "P_s=f*I_CS*V_CES*t_f   #avg power loss during t_s\n",
+      "P_f=f*t_f*(I_CS/6)*(V_CC-V_CES)        #avg power loss during fall time\n",
+      "P_fm=(I_CS/4)*(V_CC-V_CES)          #peak instant power dissipation\n",
+      "P_off=P_s+P_f\n",
+      "\n",
+      "#Results\n",
+      "print(\"Total avg power loss during turn off = %.2f W\" %P_off)\n",
+      "P_0t=I_CEO*V_CC\n",
+      "print(\"Instantaneous power loss during t_0 = %.2f W\" %P_0t)\n",
+      "P_0=f*I_CEO*V_CC*t_0              #avg power loss during t_s\n",
+      "P_on=14.9339              #W    from previous eg\n",
+      "P_n=40                    #W    from previous eg\n",
+      "P_T=P_on+P_n+P_off+P_0     \n",
+      "print(\"Total power loss = %.2f W\" %P_T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total avg power loss during turn off = 44.91 W\n",
+        "Instantaneous power loss during t_0 = 0.44 W\n",
+        "Total power loss = 99.93 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page No 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_CS=100.0 \n",
+      "V_CC=200.0 \n",
+      "t_on=40*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "P_on=(I_CS/50)*10**6*t_on*(V_CC*t_on/2-(V_CC*10**6*t_on**2/(40*3)))       #energy during turn on\n",
+      "t_off=60*10**-6\n",
+      "P_off=(I_CS*t_off/2-(I_CS/60)*10**6*(t_off**2)/3)*((V_CC/75)*10**6*t_off)  #energy during turn off\n",
+      "P_t=P_on+P_off           #total energy\n",
+      "P_avg=300.0\n",
+      "f=P_avg/P_t\n",
+      "\n",
+      "#Results\n",
+      "print(\"Allowable switching frequency = %.2f Hz\" %f)\n",
+      "#in book ans is: f=1123.6 Hz. The difference in results due to difference in rounding of of digits"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Allowable switching frequency = 1125.00 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter3.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter3.ipynb
new file mode 100755
index 00000000..2e53ef9d
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter3.ipynb
@@ -0,0 +1,1001 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 03 : Diode Circuits and Rectifiers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page No 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0  #V\n",
+      "V_o=100.0    #V\n",
+      "L=100.0      #uH\n",
+      "C=30.0       #uF\n",
+      "\n",
+      "#Calculations\n",
+      "t_o=math.pi*math.sqrt(L*C)\n",
+      "print(\"conduction time of diode = %.2f us\" %t_o)\n",
+      "#in book solution is t_o=54.77 us. The ans is incorrect as %pi is not muliplied in ans. Formulae mentioned in correct.\n",
+      "I_p=(V_s-V_o)*math.sqrt(C/L)\n",
+      "\n",
+      "#Results\n",
+      "print(\"Peak current through diode=%.2f A\" %I_p)\n",
+      "v_D=-V_s+V_o \n",
+      "print(\"Voltage across diode = %.2f V\" %v_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of diode = 172.07 us\n",
+        "Peak current through diode=164.32 A\n",
+        "Voltage across diode = -300.00 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page No 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "R=10             #ohm\n",
+      "L=0.001           #H\n",
+      "C=5*10**-6       #F\n",
+      "V_s=230          #V\n",
+      "xi=R/(2*L)\n",
+      "\n",
+      "#Calculations\n",
+      "w_o=1/math.sqrt(L*C)\n",
+      "w_r=math.sqrt((1/(L*C))-(R/(2*L))**2)\n",
+      "t=math.pi/w_r \n",
+      "\n",
+      "#Results\n",
+      "print('Conduction time of diode=%.3f us'%(t*10**6))\n",
+      "t=0\n",
+      "di=V_s/L\n",
+      "print('Rate of change of current at t=0 is %.2f A/s' %di)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conduction time of diode=237.482 us\n",
+        "Rate of change of current at t=0 is 230000.00 A/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 Page No 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_or=100             #A\n",
+      "R=1.0       #assumption\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=I_or*2*R\n",
+      "I_o=V_m/(math.pi*R)\n",
+      "q=200       #Ah\n",
+      "t=q/I_o\n",
+      "\n",
+      "#Results\n",
+      "print(\"time required to deliver charge=%.02f hrs\" %t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required to deliver charge=3.14 hrs\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8, Page No 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0    #V\n",
+      "P=1000       #W\n",
+      "R=V_s**2/P\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt(2)*V_s/2\n",
+      "P_h=V_or**2/R    \n",
+      "print(\"Power delivered to the heater = %.2f W\" %P_h)\n",
+      "V_m=math.sqrt(2)*230\n",
+      "I_m=V_m/R\n",
+      "\n",
+      "#Results\n",
+      "print(\"Peak value of diode current = %.2f A\" %I_m)\n",
+      "pf=V_or/V_s\n",
+      "print(\"Input power factor=%.2f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power delivered to the heater = 500.00 W\n",
+        "Peak value of diode current = 6.15 A\n",
+        "Input power factor=0.71\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 Page No 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230        #V\n",
+      "V_m=V_s*math.sqrt(2)\n",
+      "E=150          #V\n",
+      "\n",
+      "#Calculations\n",
+      "theta1=math.degrees(E/(math.sqrt(2)*V_s))\n",
+      "R=8        #ohm\n",
+      "f=50    #Hz\n",
+      "I_o=(1/(2*math.pi*R))*((2*math.sqrt(2)*V_s*math.cos(math.radians(theta1)))-E*(math.pi-2*theta1*math.pi/180))\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg value of charging current=%.2f A\" %I_o)\n",
+      "P_d=E*I_o\n",
+      "print(\"\\npower delivered to battery=%.2f W\" %P_d)\n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V_s**2+E**2)*(math.pi-2*theta1*math.pi/180)+V_s**2*math.sin(math.radians(2*theta1))-4*V_m*E*math.cos(math.radians(theta1))))\n",
+      "print(\"\\nrms value of the load current=%.2f A\" %I_or)\n",
+      "pf=(E*I_o+I_or**2*R)/(V_s*I_or)\n",
+      "print(\"\\nsupply pf=%.3f\" %pf)\n",
+      "P_dd=I_or**2*R\n",
+      "print(\"\\npower dissipated in the resistor=%.2f W\" %P_dd)\n",
+      "q=1000.00         #Wh\n",
+      "t=q/P_d    \n",
+      "print(\"\\ncharging time=%.2f hr\" %t)\n",
+      "n=P_d*100/(P_d+P_dd)\n",
+      "print(\"rectifier efficiency =%.2f  \" %n)\n",
+      "PIV=math.sqrt(2)*V_s+E\n",
+      "print(\"PIV of diode=%.2f V\" %PIV)\n",
+      "#solutions have small variations due to difference in rounding off of digits"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg value of charging current=4.97 A\n",
+        "\n",
+        "power delivered to battery=745.11 W\n",
+        "\n",
+        "rms value of the load current=9.29 A\n",
+        "\n",
+        "supply pf=0.672\n",
+        "\n",
+        "power dissipated in the resistor=690.74 W\n",
+        "\n",
+        "charging time=1.34 hr\n",
+        "rectifier efficiency =51.89  \n",
+        "PIV of diode=475.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 Page No 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=230            #V\n",
+      "t_rr=40*10**-6    #s reverde recovery time\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=2*math.sqrt(2)*V_s/math.pi\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "f=50\n",
+      "V_r1=(V_m/math.pi)*(1-math.cos(math.radians(2*math.pi*f*t_rr*180/math.pi)))\n",
+      "v_avg1=V_r1*100/V_o*10**3\n",
+      "f=2500\n",
+      "V_r2=(V_m/math.pi)*(1-math.cos(math.radians(2*math.pi*f*t_rr*180/math.pi)))\n",
+      "v_avg2=V_r2*100/V_o\n",
+      "\n",
+      "#Results\n",
+      "print(\"when f=50Hz\")\n",
+      "print(\"Percentage reduction in avg o/p voltage=%.2f x 10^-3\" %v_avg1)\n",
+      "print(\"when f=2500Hz\")\n",
+      "print(\"Percentage reduction in avg o/p voltage = %.3f\" %v_avg2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when f=50Hz\n",
+        "Percentage reduction in avg o/p voltage=3.95 x 10^-3\n",
+        "when f=2500Hz\n",
+        "Percentage reduction in avg o/p voltage = 9.549\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11, Page No 79 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230         #V\n",
+      "R=10.0        #ohm\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_o=2*V_m/math.pi\n",
+      "print(\"Avg value of o/p voltage = %.2f V\" %V_o)\n",
+      "I_o=V_o/R\n",
+      "print(\"Avg value of o/p current = %.2f A\" %I_o)\n",
+      "I_DA=I_o/2\n",
+      "print(\"Avg value of diode current=%.2f A\" %I_DA)\n",
+      "I_Dr=I_o/math.sqrt(2)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of diode current=%.2f A\" %I_Dr)\n",
+      "print(\"rms value of o/p current = %.2f A\" %I_o)\n",
+      "print(\"rms value of i/p current = %.2f A\" %I_o)\n",
+      "pf=(V_o/V_s)\n",
+      "print(\"supply pf = %.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Avg value of o/p voltage = 207.07 V\n",
+        "Avg value of o/p current = 20.71 A\n",
+        "Avg value of diode current=10.35 A\n",
+        "rms value of diode current=14.64 A\n",
+        "rms value of o/p current = 20.71 A\n",
+        "rms value of i/p current = 20.71 A\n",
+        "supply pf = 0.90\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 Page No 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=230.0    #V\n",
+      "R=1000.0      #ohm\n",
+      "R_D=20.0    #ohm\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "I_om=V_m/(R+R_D)          \n",
+      "\n",
+      "#Results\n",
+      "print(\"Peak load current = %.2f A\" %I_om)\n",
+      "I_o=I_om/math.pi\n",
+      "print(\"dc load current = %.2f A\" %I_o)\n",
+      "V_D=I_o*R_D-V_m/math.pi\n",
+      "print(\"dc diode voltage = %.2f V\" %V_D)\n",
+      "V_on=V_m/math.pi\n",
+      "print(\"at no load, load voltage = %.2f V\" %V_on)\n",
+      "V_o1=I_o*R    \n",
+      "print(\"at given load, load voltage = %.2f V\" %V_o1)\n",
+      "vr=(V_on-V_o1)*100/V_on    \n",
+      "print(\"Voltage regulation(in percent)=%.2f\" %vr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak load current = 0.32 A\n",
+        "dc load current = 0.10 A\n",
+        "dc diode voltage = -101.51 V\n",
+        "at no load, load voltage = 103.54 V\n",
+        "at given load, load voltage = 101.51 V\n",
+        "Voltage regulation(in percent)=1.96\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13 Page No 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_L=6.8   #V\n",
+      "V_smax=20*1.2    #V\n",
+      "V_smin=20*.8     #V\n",
+      "I_Lmax=30*1.5   #mA\n",
+      "I_Lmin=30*0.5  #mA\n",
+      "I_z=1          #mA\n",
+      "\n",
+      "#Calculations\n",
+      "R_smax=(V_smax-V_L)/((I_Lmin+I_z)*10**-3)\n",
+      "print(\"max source resistance = %.2f ohm\" %R_smax)\n",
+      "R_smin=(V_smin-V_L)/((I_Lmax+I_z)*10**-3)    \n",
+      "print(\"Min source resistance = %.2f ohm\" %R_smin)  #in book solution, error is committed in putting the values in formulea(printing error) but solution is correct\n",
+      "R_Lmax=V_L*1000/I_Lmin\n",
+      "print(\"Max load resistance = %.2f ohm\" %R_Lmax)\n",
+      "R_Lmin=V_L*1000/I_Lmax    \n",
+      "V_d=0.6      #V\n",
+      "V_r=V_L-V_d\n",
+      "\n",
+      "#Results\n",
+      "print(\"Min load resistance=%.2f ohm\" %R_Lmin)\n",
+      "print(\"Voltage rating of zener diode=%.2f V\" %V_r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "max source resistance = 1075.00 ohm\n",
+        "Min source resistance = 200.00 ohm\n",
+        "Max load resistance = 453.33 ohm\n",
+        "Min load resistance=151.11 ohm\n",
+        "Voltage rating of zener diode=6.20 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14 Page No 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I2=200*10**-6      #A\n",
+      "V_z=20       #V\n",
+      "R_G=500.0   #hm\n",
+      "\n",
+      "#Calculations\n",
+      "R2=(V_z/I2)-R_G\n",
+      "print(\"R2=%.2f kilo-ohm\" %(R2/1000))\n",
+      "\n",
+      "V_v=25      #V\n",
+      "I1=I2\n",
+      "R1=(V_v-V_z)/I1\n",
+      "\n",
+      "#Results\n",
+      "print(\"R1=%.0f kilo-ohm\"%(R1/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R2=99.50 kilo-ohm\n",
+        "R1=25 kilo-ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15, Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=2*230          #V\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=(math.sqrt(2)*V_s)/math.pi\n",
+      "R=60                #ohm\n",
+      "P_dc=(V_o)**2/R\n",
+      "TUF=0.2865\n",
+      "VA=P_dc/TUF\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"kVA rating of the transformer = %.2f kVA\" %(VA/1000));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kVA rating of the transformer = 2.49 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16, Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "tr=0.5       #turns ratio\n",
+      "I_o=10.0\n",
+      "V=230.0\n",
+      "V_s=V/tr\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_o=2*V_m/math.pi\n",
+      "phi1=0\n",
+      "#displacemnt angle=0 as fundamnetal component of i/p source current in phase with source voltage\n",
+      "DF=math.cos(math.radians(phi1))\n",
+      "I_s1=4*I_o/(math.sqrt(2)*math.pi)\n",
+      "I_s=math.sqrt(I_o**2*math.pi/math.pi)\n",
+      "CDF=I_s1/I_o\n",
+      "pf=CDF*DF\n",
+      "HF=math.sqrt((I_s/I_s1)**2-1)\n",
+      "CF=I_o/I_s\n",
+      "\n",
+      "#Results\n",
+      "print(\"o/p voltage = %.2f V\" %V_o)\n",
+      "print(\"distortion factor = %.2f\" %DF)\n",
+      "print(\"i/p pf=%.2f\" %pf)\n",
+      "print(\"Current displacent factor=%.2f\" %CDF)\n",
+      "print(\"Harmonic factor = %.2f\" %HF)\n",
+      "print(\"Creast factor = %.2f\" %CF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "o/p voltage = 414.15 V\n",
+        "distortion factor = 1.00\n",
+        "i/p pf=0.90\n",
+        "Current displacent factor=0.90\n",
+        "Harmonic factor = 0.48\n",
+        "Creast factor = 1.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17, Page No 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_o=230.0\n",
+      "R=10.0\n",
+      "V_s=V_o*math.pi/(2*math.sqrt(2))\n",
+      "I_o=V_o/R\n",
+      "I_m=math.sqrt(2)*V_s/R\n",
+      "I_DAV=I_m/math.pi\n",
+      "\n",
+      "#Calculations\n",
+      "#avg value of diode current\n",
+      "I_Dr=I_m/2\n",
+      "PIV=math.sqrt(2)*V_s\n",
+      "I_s=I_m/math.sqrt(2)\n",
+      "TF=V_s*I_s\n",
+      "\n",
+      "#Results\n",
+      "print(\"peak diode current=%.2f A\" %I_m)\n",
+      "print(\"I_DAV=%.2f A\" %I_DAV)\n",
+      "print(\"I_Dr=%.2f A\" %I_Dr)   #rms value of diode current\n",
+      "print(\"PIV=%.1f V\" %PIV)\n",
+      "print(\"Transformer rating = %.2f kVA\" %(TF/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak diode current=36.13 A\n",
+        "I_DAV=11.50 A\n",
+        "I_Dr=18.06 A\n",
+        "PIV=361.3 V\n",
+        "Transformer rating = 6.53 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18, Page No 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "tr=5\n",
+      "V=1100.0\n",
+      "R=10.0\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"In case of 3ph-3pulse type\")\n",
+      "V_ph=V/tr\n",
+      "V_mp=math.sqrt(2)*V_ph\n",
+      "V_o=3*math.sqrt(3)*V_mp/(2*math.pi)\n",
+      "print(\"avg o/p voltage=%.1f V\" %V_o)\n",
+      "I_mp=V_mp/R\n",
+      "I_D=(I_mp/math.pi)*math.sin(math.pi/3)    \n",
+      "print(\"\\navg value of diode current=%.3f A\" %I_D)\n",
+      "I_Dr=I_mp*math.sqrt((1/(2*math.pi))*(math.pi/3+.5*math.sin(2*math.pi/3)))    \n",
+      "print(\"\\nrms value of diode current=%.2f A\" %I_Dr)\n",
+      "V_or=V_mp*math.sqrt((3/(2*math.pi))*(math.pi/3+.5*math.sin(2*math.pi/3)))\n",
+      "P=(V_or**2)/R   \n",
+      "print(\"\\npower delivered=%.1f W\" %P)\n",
+      "print(\"in case of 3ph-M6 type\")\n",
+      "V_ph=V_ph/2\n",
+      "V_mp=math.sqrt(2)*V_ph\n",
+      "V_o=3*V_mp/(math.pi)  \n",
+      "I_mp=V_mp/R\n",
+      "I_D=(I_mp/math.pi)*math.sin(math.pi/6)    \n",
+      "I_Dr=I_mp*math.sqrt((1/(2*math.pi))*(math.pi/6+.5*math.sin(2*math.pi/6)))  \n",
+      "V_or=V_mp*math.sqrt((6/(2*math.pi))*(math.pi/6+.5*math.sin(2*math.pi/6)))\n",
+      "P=(V_or**2)/R   \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.2f V\" %V_o)\n",
+      "print(\"\\navg value of diode current=%.2f A\" %I_D)\n",
+      "print(\"\\nrms value of diode current=%.2f A\" %I_Dr)\n",
+      "print(\"\\npower delivered=%.0f W\" %P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In case of 3ph-3pulse type\n",
+        "avg o/p voltage=257.3 V\n",
+        "\n",
+        "avg value of diode current=8.577 A\n",
+        "\n",
+        "rms value of diode current=15.10 A\n",
+        "\n",
+        "power delivered=6841.3 W\n",
+        "in case of 3ph-M6 type\n",
+        "avg o/p voltage=148.55 V\n",
+        "\n",
+        "avg value of diode current=2.48 A\n",
+        "\n",
+        "rms value of diode current=6.07 A\n",
+        "\n",
+        "power delivered=2211 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19, Page No 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_o=400\n",
+      "R=10\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=V_o*math.pi/3\n",
+      "V_s=V_ml/(math.sqrt(2)*math.sqrt(3))\n",
+      "I_m=V_ml/R\n",
+      "I_s=.7804*I_m\n",
+      "tr=3*V_s*I_s   \n",
+      "\n",
+      "#Results\n",
+      "print(\"transformer rating=%.1f VA\" %tr)\n",
+      "I_Dr=.5518*I_m   \n",
+      "print(\"\\nrms value of diode current=%.3f A\" %I_Dr)\n",
+      "I_D=I_m/math.pi   \n",
+      "print(\"\\navg value of diode current=%.3f A\" %I_D)\n",
+      "print(\"\\npeak diode current=%.2f A\" %I_m)\n",
+      "PIV=V_ml \n",
+      "print(\"\\nPIV=%.2f V\" %PIV)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "transformer rating=16770.3 VA\n",
+        "\n",
+        "rms value of diode current=23.114 A\n",
+        "\n",
+        "avg value of diode current=13.333 A\n",
+        "\n",
+        "peak diode current=41.89 A\n",
+        "\n",
+        "PIV=418.88 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20, Page No 116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_l=230\n",
+      "E=240\n",
+      "R=8\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(2)*V_l\n",
+      "V_o=3*V_ml/math.pi\n",
+      "I_o=(V_o-E)/R\n",
+      "P_b=E*I_o    \n",
+      "P_d=E*I_o+I_o**2*R   \n",
+      "phi1=0\n",
+      "math.cos(math.radians(phi1))\n",
+      "I_s1=2*math.sqrt(3)*I_o/(math.sqrt(2)*math.pi)\n",
+      "I_s=math.sqrt(I_o**2*2*math.pi/(3*math.pi))\n",
+      "CDF=I_s1/I_s    \n",
+      "pf=DF*CDF    \n",
+      "HF=math.sqrt(CDF**-2-1)   \n",
+      "tr=math.sqrt(3)*V_l*I_o*math.sqrt(2/3)\n",
+      "\n",
+      "#Results\n",
+      "print(\"Power delivered to battery=%.1f W\" %P_b)\n",
+      "print(\"Power delivered to load=%.2f W\" %P_d)\n",
+      "print(\"Displacement factor=%.2f\" %DF)\n",
+      "print(\"Current distortion factor=%.3f\" %CDF)\n",
+      "print(\"i/p pf=%.3f\"%pf)\n",
+      "print(\"Harmonic factor=%.2f\" %HF)\n",
+      "print(\"Tranformer rating=%.2f VA\" %tr)\n",
+      "#answers have small variations from the book due to difference in rounding off of digits"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power delivered to battery=2118.3 W\n",
+        "Power delivered to load=2741.48 W\n",
+        "Displacement factor=1.00\n",
+        "Current distortion factor=0.955\n",
+        "i/p pf=0.955\n",
+        "Harmonic factor=0.31\n",
+        "Tranformer rating=0.00 VA\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21, Page No 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50  #Hz\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V\n",
+      "R=400.0\n",
+      "RF=0.05\n",
+      "C=(1/(4*f*R))*(1+(1/(math.sqrt(2)*RF)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"capacitor value=%.2f uF\" %(C/10**-6))\n",
+      "V_o=V_m*(1-1/(4*f*R*C))\n",
+      "print(\"o/p voltage with filter=%.2f V\" %V_o)\n",
+      "V_o=2*V_m/math.pi   \n",
+      "print(\"o/p voltage without filter=%.2f V\" %V_o)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "capacitor value=189.28 uF\n",
+        "o/p voltage with filter=303.79 V\n",
+        "o/p voltage without filter=207.07 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.22, Page No 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50\n",
+      "CRF=0.05\n",
+      "R=300\n",
+      "\n",
+      "#Calculations\n",
+      "L=math.sqrt((CRF/(.4715*R))**-2-R**2)/(2*2*math.pi*f)    \n",
+      "print(\"L=%.2f H\" %L)\n",
+      "R=30\n",
+      "L=math.sqrt((CRF/(.4715*R))**-2-R**2)/(2*2*math.pi*f)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"\\nL=%.2f H\" %L)\n",
+      "L=0\n",
+      "CRF=.4715*R/math.sqrt(R**2+(2*2*math.pi*f*L)**2)   \n",
+      "print(\"\\nCRF=%.2f\" %CRF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L=4.48 H\n",
+        "\n",
+        "L=0.45 H\n",
+        "\n",
+        "CRF=0.47\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.23, Page No 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=50\n",
+      "L_L=10*10**-3\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "\n",
+      "#Calculations\n",
+      "C=10/(2*w*math.sqrt(R**2+(2*w*L_L)**2))\n",
+      "\n",
+      "#Results\n",
+      "print(\"C=%.2f uF\" %(C*10**6))\n",
+      "VRF=0.1\n",
+      "L=(1/(4*w**2*C))*((math.sqrt(2)/(3*VRF))+1)\n",
+      "print(\"\\nL=%.2f mH\" %(L*10**3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C=315.83 uF\n",
+        "\n",
+        "L=45.83 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter4.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter4.ipynb
new file mode 100755
index 00000000..22311574
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter4.ipynb
@@ -0,0 +1,946 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 04 : Thyristors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "P=.5    #P=V_g*I_g\n",
+      "s=130    #s=V_g/I_g\n",
+      "\n",
+      "#Calculations\n",
+      "I_g=math.sqrt(P/s)\n",
+      "V_g=s*I_g\n",
+      "E=15\n",
+      "R_s=(E-V_g)/I_g    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Gate source resistance=%.2f ohm\" %R_s)\n",
+      "#Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gate source resistance=111.87 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "R_s=120     #slope of load line is -120V/A. This gives gate source resistance\n",
+      "print(\"gate source resistance=%.0f ohm\" %R_s)\n",
+      "\n",
+      "P=.4     #P=V_g*I_g\n",
+      "E_s=15\n",
+      "\n",
+      "#Calculations\n",
+      " #E_s=I_g*R_s+V_g %    after solving this\n",
+      " #120*I_g**2-15*I_g+0.4=0    so\n",
+      "a=120   \n",
+      "b=-15\n",
+      "c=0.4\n",
+      "D=math.sqrt((b**2)-4*a*c)\n",
+      "I_g=(-b+D)/(2*a)   \n",
+      "V_g=P/I_g\n",
+      "\n",
+      "#Results\n",
+      "print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3))    \n",
+      "print(\"\\nthen trigger voltage=%.3f V\" %V_g)\n",
+      "I_g=(-b-D)/(2*a)   \n",
+      "V_g=P/I_g\n",
+      "print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3))    \n",
+      "print(\"\\nthen trigger voltage=%.2f V\" %V_g)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "gate source resistance=120 ohm\n",
+        "\n",
+        "trigger current=86.44 mA\n",
+        "\n",
+        "then trigger voltage=4.628 V\n",
+        "\n",
+        "trigger current=38.56 mA\n",
+        "\n",
+        "then trigger voltage=10.37 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "#V_g=1+10*I_g\n",
+      "P_gm=5     #P_gm=V_g*I_g\n",
+      "#after solving % eqn becomes 10*I_g**2+I_g-5=0\n",
+      "a=10.0    \n",
+      "b=1.0    \n",
+      "c=-5\n",
+      "\n",
+      "#Calculations\n",
+      "I_g=(-b+math.sqrt(b**2-4*a*c))/(2*a)\n",
+      "E_s=15\n",
+      "#using E_s=R_s*I_g+V_g\n",
+      "R_s=(E_s-1)/I_g-10   \n",
+      "P_gav=.3     #W\n",
+      "T=20*10**-6\n",
+      "f=P_gav/(P_gm*T)\n",
+      "dl=f*T\n",
+      "\n",
+      "#Results\n",
+      "print(\"Reistance=%.3f ohm\" %R_s)\n",
+      "print(\"Triggering freq=%.0f kHz\" %(f/1000))\n",
+      "print(\"Tduty cycle=%.2f\" %dl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reistance=11.248 ohm\n",
+        "Triggering freq=3 kHz\n",
+        "Tduty cycle=0.06\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6, Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I=.1\n",
+      "E=200.0\n",
+      "L=0.2\n",
+      "\n",
+      "#Calculations\n",
+      "t=I*L/E   \n",
+      "R=20.0\n",
+      "t1=(-L/R)*math.log(1-(R*I/E))    \n",
+      "L=2.0\n",
+      "t2=(-L/R)*math.log(1-(R*I/E))   \n",
+      "\n",
+      "#Results\n",
+      "print(\"in case load consists of (a)L=.2H\")\n",
+      "print(\"min gate pulse width=%.0f us\" %(t*10**6))\n",
+      "print(\"(b)R=20ohm in series with L=.2H\")\n",
+      "print(\"min gate pulse width=%.3f us\" %(t1*10**6))\n",
+      "print(\"(c)R=20ohm in series with L=2H\")\n",
+      "print(\"min gate pulse width=%.2f us\" %(t2*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "in case load consists of (a)L=.2H\n",
+        "min gate pulse width=100 us\n",
+        "(b)R=20ohm in series with L=.2H\n",
+        "min gate pulse width=100.503 us\n",
+        "(c)R=20ohm in series with L=2H\n",
+        "min gate pulse width=1005.03 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 Page No 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "def theta(th):\n",
+      "    I_m=1     #supposition\n",
+      "    I_av=(I_m/(2*math.pi))*(1+math.cos(math.radians(th)))\n",
+      "    I_rms=math.sqrt((I_m/(2*math.pi))*((180-th)*math.pi/360+.25*math.sin(math.radians(2*th))))\n",
+      "    FF=I_rms/I_av\n",
+      "    I_rms=35\n",
+      "    I_TAV=I_rms/FF\n",
+      "    return I_TAV\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"when conduction angle=180\")\n",
+      "th=0\n",
+      "I_TAV=theta(th)\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
+      "print(\"when conduction angle=90\")\n",
+      "th=90\n",
+      "I_TAV=theta(th)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
+      "print(\"when conduction angle=30\")\n",
+      "th=150\n",
+      "I_TAV=theta(th)\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when conduction angle=180\n",
+        "avg on current rating=22.282 A\n",
+        "when conduction angle=90\n",
+        "avg on current rating=15.756 A\n",
+        "when conduction angle=30\n",
+        "avg on current rating=8.790 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10, Page No 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "def theta(th):\n",
+      "    n=360.0/th\n",
+      "    I=1.0     #supposition\n",
+      "    I_av=I/n\n",
+      "    I_rms=I/math.sqrt(n)\n",
+      "    FF=I_rms/I_av\n",
+      "    I_rms=35\n",
+      "    I_TAV=I_rms/FF\n",
+      "    return I_TAV\n",
+      "\n",
+      "#Calculations\n",
+      "th=180.0\n",
+      "I_TAV1=theta(th)\n",
+      "th=90.0\n",
+      "I_TAV2=theta(th)\n",
+      "th=30.0\n",
+      "I_TAV3=theta(th)\n",
+      "\n",
+      "#Results\n",
+      "print(\"when conduction angle=180\")\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
+      "print(\"when conduction angle=90\")\n",
+      "print(\"avg on current rating=%.1f A\" %I_TAV2)\n",
+      "print(\"when conduction angle=30\")\n",
+      "print(\"avg on current rating=%.4f A\" %I_TAV3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when conduction angle=180\n",
+        "avg on current rating=8.790 A\n",
+        "when conduction angle=90\n",
+        "avg on current rating=17.5 A\n",
+        "when conduction angle=30\n",
+        "avg on current rating=10.1036 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 Page No 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "f=50.0     #Hz\n",
+      "\n",
+      "#Calculations\n",
+      "I_sb=3000.0\n",
+      "t=1/(4*f)\n",
+      "T=1/(2*f)\n",
+      "I=math.sqrt(I_sb**2*t/T)    \n",
+      "r=(I_sb/math.sqrt(2))**2*T   \n",
+      "\n",
+      "#Results\n",
+      "print(\"surge current rating=%.2f A\" %I)\n",
+      "print(\"\\nI**2*t rating=%.0f A^2.s\" %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "surge current rating=2121.32 A\n",
+        "\n",
+        "I**2*t rating=45000 A^2.s\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 Page No 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "V_s=300.0 #V\n",
+      "R=60.0 #ohm\n",
+      "L=2.0 #H\n",
+      "\n",
+      "#Calculations\n",
+      "t=40*10**-6 #s\n",
+      "i_T=(V_s/R)*(1-math.exp(-R*t/L))\n",
+      "i=.036 #A\n",
+      "R1=V_s/(i-i_T)\n",
+      "\n",
+      "#Results\n",
+      "print(\"maximum value of remedial parameter=%.3f kilo-ohm\" %(R1/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum value of remedial parameter=9.999 kilo-ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16 Page No 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_p=230.0*math.sqrt(2)\n",
+      "\n",
+      "#Calculations\n",
+      "R=1+((1)**-1+(10)**-1)**-1\n",
+      "A=V_p/R\n",
+      "s=1     #s\n",
+      "t_c=20*A**-2*s\n",
+      "\n",
+      "#Results\n",
+      "print(\"fault clearance time=%.4f ms\" %(t_c*10**3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fault clearance time=0.6890 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17, Page No 176"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "V_s=math.sqrt(2)*230     #V\n",
+      "L=15*10**-6     #H\n",
+      "I=V_s/L     #I=(di/dt)_max\n",
+      "R_s=10   #ohm\n",
+      "v=I*R_s     #v=(dv/dt)_max\n",
+      "\n",
+      "#Calculations\n",
+      "f=50     #Hz\n",
+      "X_L=L*2*math.pi*f\n",
+      "R=2\n",
+      "I_max=V_s/(R+X_L)    \n",
+      "FF=math.pi/math.sqrt(2)\n",
+      "I_TAV1=I_max/FF    \n",
+      "FF=3.98184\n",
+      "I_TAV2=I_max/FF    \n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"(di/dt)_max=%.3f A/usec\" %(I/10**6))\n",
+      "print(\"\\n(dv/dt)_max=%.2f V/usec\" %(v/10**6))\n",
+      "print(\"\\nI_rms=%.3f A\" %I_max)\n",
+      "print(\"when conduction angle=90\")\n",
+      "print(\"I_TAV=%.3f A\" %I_TAV1)\n",
+      "print(\"when conduction angle=30\")\n",
+      "print(\"I_TAV=%.3f A\" %I_TAV2)\n",
+      "print(\"\\nvoltage rating=%.3f V\" %(2.75*V_s)) #rating is taken 2.75 times of peak working voltage unlike 2.5 to 3 times as mentioned int book."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(di/dt)_max=21.685 A/usec\n",
+        "\n",
+        "(dv/dt)_max=216.85 V/usec\n",
+        "\n",
+        "I_rms=162.252 A\n",
+        "when conduction angle=90\n",
+        "I_TAV=73.039 A\n",
+        "when conduction angle=30\n",
+        "I_TAV=40.748 A\n",
+        "\n",
+        "voltage rating=894.490 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19, Page No 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "T_jm=125\n",
+      "th_jc=.15     #degC/W\n",
+      "th_cs=0.075     #degC/W\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "dT=54     #dT=T_s-T_a\n",
+      "P_av=120\n",
+      "th_sa=dT/P_av\n",
+      "T_a=40     #ambient temp\n",
+      "P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
+      "if (P_av-120)<1 :\n",
+      "    print(\"selection of heat sink is satisfactory\")\n",
+      "\n",
+      "dT=58     #dT=T_s-T_a\n",
+      "P_av=120\n",
+      "th_sa=dT/P_av\n",
+      "T_a=40     #ambient temp\n",
+      "P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
+      "if (P_av-120)<1 :\n",
+      "    print(\"selection of heat sink is satisfactory\")\n",
+      "\n",
+      "V_m=math.sqrt(2)*230\n",
+      "R=2\n",
+      "I_TAV=V_m/(R*math.pi)\n",
+      "P_av=90\n",
+      "th_sa=(T_jm-T_a)/P_av-(th_jc+th_cs)\n",
+      "dT=P_av*th_sa\n",
+      "print(\"for heat sink\")    \n",
+      "print(\"T_s-T_a=%.2f degC\" %dT)   \n",
+      "print(\"\\nP_av=%.0f W\" %P_av)\n",
+      "P=(V_m/2)**2/R\n",
+      "eff=P/(P+P_av)   \n",
+      "print(\"\\nckt efficiency=%.3f pu\" %eff)\n",
+      "a=60     #delay angle\n",
+      "I_TAV=(V_m/(2*math.pi*R))*(1+math.cos(math.radians(a)))\n",
+      "print(\"\\nI_TAV=%.2f A\" %I_TAV)\n",
+      "dT=46\n",
+      "T_s=dT+T_a\n",
+      "T_c=T_s+P_av*th_cs    \n",
+      "T_j=T_c+P_av*th_jc    \n",
+      "\n",
+      "#Results\n",
+      "print(\"\\ncase temp=%.2f degC\" %T_c)\n",
+      "print(\"\\njunction temp=%.2f degC\" %T_j)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for heat sink\n",
+        "T_s-T_a=-20.25 degC\n",
+        "\n",
+        "P_av=90 W\n",
+        "\n",
+        "ckt efficiency=0.993 pu\n",
+        "\n",
+        "I_TAV=38.83 A\n",
+        "\n",
+        "case temp=92.75 degC\n",
+        "\n",
+        "junction temp=106.25 degC\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.20, Page No 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_j=125.0     #degC\n",
+      "T_s=70.0     #degC\n",
+      "th_jc=.16     #degC/W\n",
+      "th_cs=.08     #degC/W\n",
+      "\n",
+      "#Calculations\n",
+      "P_av1=(T_j-T_s)/(th_jc+th_cs)    \n",
+      "\n",
+      "T_s=60     #degC\n",
+      "P_av2=(T_j-T_s)/(th_jc+th_cs)    \n",
+      "inc=(math.sqrt(P_av2)-math.sqrt(P_av1))*100/math.sqrt(P_av1)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Total avg power loss in thristor sink combination=%.2f W\" %P_av1)\n",
+      "print(\"Percentage inc in rating=%.2f\" %inc)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total avg power loss in thristor sink combination=229.17 W\n",
+        "Percentage inc in rating=8.71\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.21, Page No 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "R=25000.0\n",
+      "I_l1=.021     #I_l=leakage current\n",
+      "I_l2=.025\n",
+      "I_l3=.018\n",
+      "I_l4=.016\n",
+      " #V1=(I-I_l1)*R\n",
+      " #V2=(I-I_l2)*R\n",
+      " #V3=(I-I_l3)*R\n",
+      " #V4=(I-I_l4)*R\n",
+      " #V=V1+V2+V3+V4\n",
+      "    \n",
+      "#Calculations\n",
+      "V=10000.0\n",
+      "I_l=I_l1+I_l2+I_l3+I_l4\n",
+      " #after solving\n",
+      "I=((V/R)+I_l)/4\n",
+      "R_c=40.0\n",
+      "V1=(I-I_l1)*R    \n",
+      "\n",
+      "#Resluts\n",
+      "print(\"voltage across SCR1=%.0f V\" %V1)\n",
+      "V2=(I-I_l2)*R    \n",
+      "print(\"\\nvoltage across SCR2=%.0f V\" %V2)\n",
+      "V3=(I-I_l3)*R    \n",
+      "print(\"\\nvoltage across SCR3=%.0f V\" %V3)\n",
+      "V4=(I-I_l4)*R    \n",
+      "print(\"\\nvoltage across SCR4=%.0f V\" %V4)\n",
+      "\n",
+      "I1=V1/R_c    \n",
+      "print(\"\\ndischarge current through SCR1=%.3f A\" %I1)\n",
+      "I2=V2/R_c   \n",
+      "print(\"\\ndischarge current through SCR2=%.3f A\" %I2)\n",
+      "I3=V3/R_c    \n",
+      "print(\"\\ndischarge current through SCR3=%.3f A\" %I3)\n",
+      "I4=V4/R_c    \n",
+      "print(\"\\ndischarge current through SCR4=%.3f A\" %I4)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "voltage across SCR1=2475 V\n",
+        "\n",
+        "voltage across SCR2=2375 V\n",
+        "\n",
+        "voltage across SCR3=2550 V\n",
+        "\n",
+        "voltage across SCR4=2600 V\n",
+        "\n",
+        "discharge current through SCR1=61.875 A\n",
+        "\n",
+        "discharge current through SCR2=59.375 A\n",
+        "\n",
+        "discharge current through SCR3=63.750 A\n",
+        "\n",
+        "discharge current through SCR4=65.000 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.22, Page No 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_r=1000     #rating of SCR\n",
+      "I_r=200     #rating of SCR\n",
+      "V_s=6000     #rating of String\n",
+      "I_s=1000     #rating of String\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"when DRF=.1\")\n",
+      "DRF=.1\n",
+      "n_s=V_s/(V_r*(1-DRF))    \n",
+      "print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
+      "n_p=I_s/(I_r*(1-DRF))    \n",
+      "print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))\n",
+      "print(\"when DRF=.2\")\n",
+      "DRF=.2\n",
+      "\n",
+      "#Results\n",
+      "n_s=V_s/(V_r*(1-DRF))   \n",
+      "print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
+      "n_p=I_s/(I_r*(1-DRF))    \n",
+      "print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when DRF=.1\n",
+        "number of series units=7\n",
+        "\n",
+        "number of parrallel units=6\n",
+        "when DRF=.2\n",
+        "number of series units=8\n",
+        "\n",
+        "number of parrallel units=7\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.23, Page No 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V1=1.6     #on state voltage drop of SCR1\n",
+      "V2=1.2     #on state voltage drop of SCR2\n",
+      "I1=250.0     #current rating of SCR1\n",
+      "I2=350.0     #current rating of SCR2\n",
+      "\n",
+      "#Calculations\n",
+      "R1=V1/I1\n",
+      "R2=V2/I2\n",
+      "I=600.0     #current to be shared\n",
+      " #for SCR1 %     I*(R1+R)/(total resistance)=k*I1                    (1)\n",
+      " #for SCR2 %     I*(R2+R)/(total resistance)=k*I2                    (2)\n",
+      " #(1)/(2)\n",
+      "R=(R2*I2-R1*I1)/(I1-I2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"RSequired value of resistance=%.3f ohm\" %R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RSequired value of resistance=0.004 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.25, Page No 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=2000.0     #Hz\n",
+      "C=0.04*10**-6\n",
+      "n=.72\n",
+      "\n",
+      "#Calculations\n",
+      "R=1/(f*C*math.log(1/(1-n)))    \n",
+      "V_p=18\n",
+      "V_BB=V_p/n\n",
+      "R2=10**4/(n*V_BB)    \n",
+      "I=4.2*10**-3     #leakage current\n",
+      "R_BB=5000\n",
+      "R1=(V_BB/I)-R2-R_BB\n",
+      "\n",
+      "#Results\n",
+      "print(\"R=%.2f kilo-ohm\" %(R/1000))\n",
+      "print(\"\\nR2=%.2f ohm\" %R2)\n",
+      "print(\"\\nR1=%.0f ohm\" %R1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R=9.82 kilo-ohm\n",
+        "\n",
+        "R2=555.56 ohm\n",
+        "\n",
+        "R1=397 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.26, Page No 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "V_p=18.0\n",
+      "n=.72\n",
+      "V_BB=V_p/n\n",
+      "I_p=.6*10**-3\n",
+      "I_v=2.5*10**-3\n",
+      "V_v=1\n",
+      "\n",
+      "#Calculations\n",
+      "R_max=V_BB*(1-n)/I_p    \n",
+      "print(\"R_max=%.2f kilo-ohm\" %(R_max/1000))\n",
+      "R_min=(V_BB-V_v)/I_v    \n",
+      "print(\"\\nR_min=%.2f kilo-ohm\" %(R_min/1000))\n",
+      "\n",
+      "C=.04*10**-6\n",
+      "f_min=1/(R_max*C*math.log(1/(1-n)))    \n",
+      "print(\"\\nf_min=%.3f kHz\" %(f_min/1000))\n",
+      "f_max=1/(R_min*C*math.log(1/(1-n)))    \n",
+      "print(\"\\nf_max=%.2f kHz\" %(f_max/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R_max=11.67 kilo-ohm\n",
+        "\n",
+        "R_min=9.60 kilo-ohm\n",
+        "\n",
+        "f_min=1.683 kHz\n",
+        "\n",
+        "f_max=2.05 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter5.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter5.ipynb
new file mode 100755
index 00000000..1d261f20
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter5.ipynb
@@ -0,0 +1,511 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 05 : Thyristor Commutation Techniques"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page No 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=5.0*10**-3            #mH\n",
+      "C=20.0*10**-6           #\u00b5F\n",
+      "V_s=200                 #V\n",
+      "\n",
+      "#Calculations\n",
+      "w_o=math.sqrt(1/(L*C))    #rad/s\n",
+      "t_o=math.pi/w_o           #ms\n",
+      "\n",
+      "#Results\n",
+      "print('conduction time of thyristor = %.2f ms' %(t_o*1000))\n",
+      "print('voltage across thyristor=%.0f V' %V_s)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of thyristor = 0.99 ms\n",
+        "voltage across thyristor=200 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page No 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "C=20.0*10**-6         #\u00b5F\n",
+      "L=5.0*10**-6          #\u00b5H\n",
+      "V_s=230.0             #V\n",
+      "\n",
+      "#Calculations\n",
+      "I_p=V_s*math.sqrt(C/L)      #A\n",
+      "w_o=math.sqrt(1/(L*C))      #rad/sec\n",
+      "t_o=math.pi/w_o             #\u00b5S\n",
+      "I_o=300               \n",
+      "a = math.degrees(math.asin(I_o/(2*V_s)))   \n",
+      "V_ab = V_s*math.cos(math.radians(a))      #V \n",
+      "t_c=C*V_ab/I_o     #\u00b5s\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"conduction time of auxillery thyristor=%.2f us\" %(t_o*10**6))\n",
+      "print(\"voltage across main thyristor=%.2f V\" %V_ab)\n",
+      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of auxillery thyristor=31.42 us\n",
+        "voltage across main thyristor=174.36 V\n",
+        "ckt turn off time=11.62 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page No 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=200.0           #V\n",
+      "R1=10.0             #\u2126\n",
+      "R2=100.0            #\u2126\n",
+      "C=0                 #  value of capacitor\n",
+      "\n",
+      "#Calculations\n",
+      "I1=V_s*(1/R1+2/R2)    #A\n",
+      "I2=V_s*(2/R1+1/R2)    #A\n",
+      "t_c1=40*10**-6\n",
+      "fos=2    #factor of safety\n",
+      "C1=t_c1*fos/(R1*math.log(2))\n",
+      "C2=t_c1*fos/(R2*math.log(2))\n",
+      "if C1 > C2 :\n",
+      "    C = C1*10**6\n",
+      "else :\n",
+      "    C = C2*10**6\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"peak value of current through SCR1=%.2f A\" %I1); \n",
+      "print(\"Peak value of current through SCR2=%.2f A\" %I2);\n",
+      "print(\"Value of capacitor=%.2f uF\" %C);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak value of current through SCR1=24.00 A\n",
+        "Peak value of current through SCR2=42.00 A\n",
+        "Value of capacitor=11.54 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page No 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0             #V\n",
+      "L=20*10**-6           #\u00b5H\n",
+      "C=40*10**-6           #\u00b5F\n",
+      "I_o=120.0             #A\n",
+      "\n",
+      "#Calculations\n",
+      "I_p=V_s*math.sqrt(C/L)   #A\n",
+      "t_c=C*V_s/I_o            #\u00b5s\n",
+      "w_o=math.sqrt(1/(L*C))  \n",
+      "t_c1=math.pi/(2*w_o)     #\u00b5s\n",
+      "\n",
+      "#Results\n",
+      "print(\"current through main thyristor=%.2f A\"  %(I_o+I_p))\n",
+      "print(\"Current through auxillery thyristor=%.2f A\" %I_o)\n",
+      "print(\"Circuit turn off time for main thyristor=%.2f us\" %(t_c*10**6))\n",
+      "print(\"Circuit turn off time for auxillery thyristor=%.2f us\" %(t_c1*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current through main thyristor=445.27 A\n",
+        "Current through auxillery thyristor=120.00 A\n",
+        "Circuit turn off time for main thyristor=76.67 us\n",
+        "Circuit turn off time for auxillery thyristor=44.43 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page No 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "C_j=25*10**-12      #pF\n",
+      "I_c=5*10**-3        #charging current\n",
+      "V_s=200.0           #V\n",
+      "R=50.0              #\u2126\n",
+      "\n",
+      "#Calculations\n",
+      "C=(C_j*V_s)/(I_c*R)\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"Value of C=%.2f  \u00b5F\" %(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of C=0.02  \u00b5F\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 Page No 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=200.0         #V\n",
+      "R=5.0             #\u2126\n",
+      "\n",
+      "#Calculations\n",
+      "C=10.0*10**-6\n",
+      "#for turn off V_s*(1-2*exp(-t/(R*C)))=0,    so after solving\n",
+      "t_c=R*C*math.log(2.0)\n",
+      "\n",
+      "#Results\n",
+      "print(\"circuit turn off time=%.2f us\" %(t_c*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "circuit turn off time=34.66 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page No 264 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=1.0                   #\u2126\n",
+      "L=20*10**-6             #\u00b5H\n",
+      "C=40*10**-6             #\u00b5F\n",
+      "\n",
+      "#Calculations\n",
+      "w_r=math.sqrt((1/(L*C))-(R/(2*L))**2)\n",
+      "t_1=math.pi/w_r\n",
+      "\n",
+      "#Results\n",
+      "print(\"conduction time of thyristor=%.3f us\" %(t_1*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of thyristor=125.664 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 Page No 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "dv=400*10.0**-6       #dv=dv_T/dt(V/s)\n",
+      "V_s=200.0             #v\n",
+      "R=20.0                #\u2126\n",
+      "\n",
+      "#Calculations\n",
+      "C=V_s/(R*dv)         \n",
+      "C_j=.025*10**-12\n",
+      "C_s=C-C_j\n",
+      "I_T=40;\n",
+      "R_s=1/((I_T/V_s)-(1/R)) \n",
+      "#value of R_s in book is wrongly calculated\n",
+      "\n",
+      "#Results\n",
+      "print(\"R_s=%.2f ohm\" %R_s)\n",
+      "print(\"C_s=%.3f uF\" %(C_s/10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R_s=6.67 ohm\n",
+        "C_s=0.025 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 Page No 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=200.0          #V\n",
+      "C=20.0*10**-6      #\u00b5H   \n",
+      "L=0.2*10**-3       #\u00b5F\n",
+      "i_c=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "i=V_s*math.sqrt(C/L)\n",
+      "w_o=1.0/math.sqrt(L*C)\n",
+      "t_1 = (1/w_o)*math.degrees(math.asin(i_c/i))\n",
+      "t_o=math.pi/w_o\n",
+      "t_c=t_o-2*t_1    \n",
+      "\n",
+      "#Results\n",
+      "print(\"reqd time=%.2f us\" %(t_1*10**6))\n",
+      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n",
+      "print(\"ckt turn off time=%.5f us\" %t_1)\n",
+      "#solution in book wrong, as wrong values are selected while filling the formuleas"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "reqd time=575.37 us\n",
+        "ckt turn off time=-952.05 us\n",
+        "ckt turn off time=0.00058 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11 Page No 268 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "L=1.0              #\u00b5H\n",
+      "R=50.0             #\u2126\n",
+      "V_s=200.0          #V\n",
+      "t=0.01             #sec\n",
+      "Vd=0.7\n",
+      "\n",
+      "#Calculations\n",
+      "tau=L/R\n",
+      "i=(V_s/R)*(1-math.exp(-t/tau))\n",
+      "t=8*10**-3\n",
+      "i1=i-t*Vd  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"current through L = %.2f A\" %i1)\n",
+      "i_R=0                  #current in R at t=.008s\n",
+      "print(\"Current through R = %.2f A\" %i_R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current through L = 1.57 A\n",
+        "Current through R = 0.00 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12, Page No 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "#initialisation of variables\n",
+      "L=1.0                      #H\n",
+      "R=50.0                   #ohm\n",
+      "V_s=200.0                #V\n",
+      "\n",
+      "#Calculations\n",
+      "tau=L/R\n",
+      "t=0.01                 #s\n",
+      "i=(V_s/R)*(1-math.exp(-t/tau))\n",
+      "C=1*10**-6            #F\n",
+      "V_c=math.sqrt(L/C)*i\n",
+      "\n",
+      "#Results\n",
+      "print(\"current in R,L=%.2f A\" %i)\n",
+      "print(\"voltage across C=%.2f kV\" %(V_c/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current in R,L=1.57 A\n",
+        "voltage across C=1.57 kV\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter6.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter6.ipynb
new file mode 100755
index 00000000..dff6564b
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter6.ipynb
@@ -0,0 +1,1761 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 06 : Phase Controlled Rectifiers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page No 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "P=1000.0\n",
+      "R=V**2/P\n",
+      "\n",
+      "#Calculations\n",
+      "a=math.pi/4\n",
+      "V_or1=(math.sqrt(2)*V/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a)+.5*math.sin(2*a))\n",
+      "P1=V_or1**2/R    \n",
+      "a=math.pi/2\n",
+      "V_or2=(math.sqrt(2)*V/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a)+.5*math.sin(2*a))\n",
+      "P2=V_or2**2/R   \n",
+      "\n",
+      "#Results\n",
+      "print(\"when firing angle delay is of 45deg\")\n",
+      "print(\"power absorbed=%.2f W\" %P1)\n",
+      "print(\"when firing angle delay is of 90deg\")\n",
+      "print(\"power absorbed=%.2f W\" %P2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when firing angle delay is of 45deg\n",
+        "power absorbed=454.58 W\n",
+        "when firing angle delay is of 90deg\n",
+        "power absorbed=250.00 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page No 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "E=150.0\n",
+      "R=8.0\n",
+      "\n",
+      "#Calculations\n",
+      "th1=math.sin(math.radians(E/(math.sqrt(2)*V)))\n",
+      "I_o=(1/(2*math.pi*R))*(2*math.sqrt(2)*230*math.cos(math.radians(th1))-E*(math.pi-2*th1*math.pi/180))    \n",
+      "P=E*I_o    \n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*(math.pi-2*th1*math.pi/180)+V**2*math.sin(math.radians(2*th1))-4*math.sqrt(2)*V*E*math.cos(math.radians(th1))))\n",
+      "P_r=I_or**2*R    \n",
+      "pf=(P+P_r)/(V*I_or)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg charging curent=%.4f A\" %I_o)\n",
+      "print(\"power supplied to the battery=%.2f W\" %P)\n",
+      "print(\"power dissipated by the resistor=%.3f W\" %P_r)    \n",
+      "print(\"supply pf=%.3f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg charging curent=3.5679 A\n",
+        "power supplied to the battery=535.18 W\n",
+        "power dissipated by the resistor=829.760 W\n",
+        "supply pf=0.583\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page No 284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "E=150.0\n",
+      "R=8.0\n",
+      "a=35.0\n",
+      "\n",
+      "#Calculations\n",
+      "th1=math.degrees(math.asin(E/(math.sqrt(2)*V)))\n",
+      "th2=180-th1\n",
+      "I_o=(1/(2*math.pi*R))*(math.sqrt(2)*230*(math.cos(math.radians(a))-math.cos(math.radians(th2)))-E*((th2-a)*math.pi/180))    \n",
+      "P=E*I_o    \n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*((th2-a)*math.pi/180)-(V**2/2)*(math.sin(math.radians(2*th2))-math.sin(math.radians(2*a)))-2*math.sqrt(2)*V*E*(math.cos(math.radians(a))-math.cos(math.radians(th2)))))\n",
+      "P_r=I_or**2*R    \n",
+      "pf=(P+P_r)/(V*I_or)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg charging curent=%.4f A\" %I_o)\n",
+      "print(\"power supplied to the battery=%.2f W\" %P)\n",
+      "print(\"power dissipated by the resistor=%.2f W\" %P_r)\n",
+      "print(\"supply pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg charging curent=4.9208 A\n",
+        "power supplied to the battery=738.12 W\n",
+        "power dissipated by the resistor=689.54 W\n",
+        "supply pf=0.6686\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page No 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "B=210\n",
+      "f=50.0     #Hz\n",
+      "w=2*math.pi*f\n",
+      "a=40.0     #firing angle\n",
+      "V=230.0\n",
+      "R=5.0\n",
+      "L=2*10**-3\n",
+      "\n",
+      "#Calculations\n",
+      "t_c1=(360-B)*math.pi/(180*w)   \n",
+      "V_o1=(math.sqrt(2)*230/(2*math.pi))*(math.cos(math.radians(a))-math.cos(math.radians(B)))    \n",
+      "I_o1=V_o1/R    \n",
+      "E=110\n",
+      "R=5\n",
+      "L=2*10**-3\n",
+      "th1=math.degrees(math.asin(E/(math.sqrt(2)*V)))\n",
+      "t_c2=(360-B+th1)*math.pi/(180*w)    \n",
+      "V_o2=(math.sqrt(2)*230/(2*math.pi))*(math.cos(math.radians(a))-math.cos(math.radians(B)))    \n",
+      "I_o2=(1/(2*math.pi*R))*(math.sqrt(2)*230*(math.cos(math.radians(a))-math.cos(math.radians(B)))-E*((B-a)*math.pi/180))   \n",
+      "V_o2=R*I_o2+E    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"for R=5ohm and L=2mH\")\n",
+      "print(\"ckt turn off time=%.3f msec\" %(t_c1*1000))\n",
+      "print(\"avg output voltage=%.3f V\" %V_o1)\n",
+      "print(\"avg output current=%.4f A\" %I_o1)\n",
+      "print(\"for R=5ohm % L=2mH and E=110V\")\n",
+      "print(\"ckt turn off time=%.3f msec\" %(t_c2*1000))\n",
+      "print(\"avg output current=%.4f A\" %I_o2)\n",
+      "print(\"avg output voltage=%.3f V\" %V_o2)   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for R=5ohm and L=2mH\n",
+        "ckt turn off time=8.333 msec\n",
+        "avg output voltage=84.489 V\n",
+        "avg output current=16.8979 A\n",
+        "for R=5ohm % L=2mH and E=110V\n",
+        "ckt turn off time=9.431 msec\n",
+        "avg output current=6.5090 A\n",
+        "avg output voltage=142.545 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page No 286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "f=50.0\n",
+      "R=10.0\n",
+      "a=60.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=(math.sqrt(2)*V_s)\n",
+      "V_o=V_m/(2*math.pi)*(1+math.cos(math.radians(a)))\n",
+      "I_o=V_o/R\n",
+      "V_or=(V_m/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a*math.pi/180)+.5*math.sin(math.radians(2*a)))\n",
+      "I_or=V_or/R\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "FF=V_or/V_o    \n",
+      "VRF=math.sqrt(FF**2-1)    \n",
+      "TUF=P_dc/(V_s*I_or)    \n",
+      "PIV=V_m    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"rectification efficiency=%.4f\" %RE)\n",
+      "print(\"form factor=%.3f\" %FF)\n",
+      "print(\"voltage ripple factor=%.4f\" %VRF)\n",
+      "print(\"t/f utilisation factor=%.4f\" %TUF)\n",
+      "print(\"PIV of thyristor=%.2f V\" %PIV)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectification efficiency=0.2834\n",
+        "form factor=1.879\n",
+        "voltage ripple factor=1.5903\n",
+        "t/f utilisation factor=0.1797\n",
+        "PIV of thyristor=325.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6 Page No 294"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=1000.0\n",
+      "fos=2.5     #factor of safety\n",
+      "I_TAV=40.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m1=V/(2*fos)\n",
+      "P1=(2*V_m1/math.pi)*I_TAV    \n",
+      "V_m2=V/(fos)\n",
+      "P2=(2*V_m2/math.pi)*I_TAV    \n",
+      "\n",
+      "#Results\n",
+      "print(\"for mid pt convertor\")\n",
+      "print(\"power handled=%.3f kW\" %(P1/1000))\n",
+      "print(\"for bridge convertor\")\n",
+      "print(\"power handled=%.3f kW\" %(P2/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for mid pt convertor\n",
+        "power handled=5.093 kW\n",
+        "for bridge convertor\n",
+        "power handled=10.186 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7, Page No 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "R=.4\n",
+      "I_o=10\n",
+      "I_or=I_o\n",
+      "E=120.0\n",
+      "\n",
+      "#Calculations\n",
+      "a1=math.degrees(math.acos((E+I_o*R)*math.pi/(2*V_m)))\n",
+      "pf1=(E*I_o+I_or**2*R)/(V_s*I_or)    \n",
+      "E=-120.0\n",
+      "a2=math.degrees(math.acos((E+I_o*R)*math.pi/(2*V_m))) \n",
+      "pf2=(-E*I_o-I_or**2*R)/(V_s*I_or)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay=%.2f deg\" %a1)\n",
+      "print(\"pf=%.4f\" %pf1)\n",
+      "print(\"firing angle delay=%.2f deg\" %a2)\n",
+      "print(\"pf=%.4f\" %pf2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=53.21 deg\n",
+        "pf=0.5391\n",
+        "firing angle delay=124.07 deg\n",
+        "pf=0.5043\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 Page No 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "f=50.0\n",
+      "a=45.0\n",
+      "R=5.0\n",
+      "E=100.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=((math.sqrt(2)*V_s)/(2*math.pi))*(3+math.cos(math.radians(a)))\n",
+      "I_o=(V_o-E)/R    \n",
+      "P=E*I_o    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p current=%.3f A\" %I_o)\n",
+      "print(\"power delivered to battery=%.4f kW\" %(P/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p current=18.382 A\n",
+        "power delivered to battery=1.8382 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10 Page No 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=230\n",
+      "f=50.0\n",
+      "a=50.0\n",
+      "R=6.0\n",
+      "E=60.0\n",
+      "V_o1=((math.sqrt(2)*2*V_s)/(math.pi))*math.cos(math.radians(a))\n",
+      "I_o1=(V_o1-E)/R    \n",
+      "\n",
+      "#ATQ after applying the conditions\n",
+      "V_o2=((math.sqrt(2)*V_s)/(math.pi))*math.cos(math.radians(a))\n",
+      "I_o2=(V_o2-E)/R    \n",
+      "\n",
+      "print(\"avg o/p current=%.3f A\" %I_o1)\n",
+      "print(\"avg o/p current after change=%.2f A\" %I_o2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p current=12.184 A\n",
+        "avg o/p current after change=1.09 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11 Page No 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=(2*V_m/math.pi)*math.cos(math.radians(a))\n",
+      "I_o=V_o/R\n",
+      "V_or=V_m/math.sqrt(2)\n",
+      "I_or=I_o\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "FF=V_or/V_o    \n",
+      "VRF=math.sqrt(FF**2-1)    \n",
+      "I_s1=2*math.sqrt(2)*I_o/math.pi\n",
+      "DF=math.cos(math.radians(a))\n",
+      "CDF=.90032\n",
+      "pf=CDF*DF    \n",
+      "HF=math.sqrt((1/CDF**2)-1)    \n",
+      "Q=2*V_m*I_o*math.sin(math.radians(a))/math.pi \n",
+      "\n",
+      "#Results\n",
+      "print(\"rectification efficiency=%.4f\" %RE)\n",
+      "print(\"form factor=%.4f\" %FF)\n",
+      "print(\"voltage ripple factor=%.4f\" %VRF)\n",
+      "print(\"pf=%.5f\" %pf)\n",
+      "print(\"HF=%.5f\" %HF)\n",
+      "print(\"active power=%.2f W\" %P_dc)   \n",
+      "print(\"reactive power=%.3f Var\" %Q)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectification efficiency=0.6366\n",
+        "form factor=1.5708\n",
+        "voltage ripple factor=1.2114\n",
+        "pf=0.63662\n",
+        "HF=0.48342\n",
+        "active power=2143.96 W\n",
+        "reactive power=2143.956 Var\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12, Page No 310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=(V_m/math.pi)*(1+math.cos(math.radians(a)))\n",
+      "I_o=V_o/R\n",
+      "V_or=V_s*math.sqrt((1/math.pi)*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))\n",
+      "I_or=I_o\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "FF=V_or/V_o    \n",
+      "VRF=math.sqrt(FF**2-1)    \n",
+      "I_s1=2*math.sqrt(2)*I_o*math.cos(math.radians(a/2))/math.pi\n",
+      "DF=math.cos(math.radians(a/2))   \n",
+      "CDF=2*math.sqrt(2)*math.cos(math.radians(a/2))/math.sqrt(math.pi*(math.pi-a*math.pi/180))    \n",
+      "pf=CDF*DF    \n",
+      "HF=math.sqrt((1/CDF**2)-1)    \n",
+      "Q=V_m*I_o*math.sin(math.radians(a))/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print(\"form factor=%.3f\" %FF)\n",
+      "print(\"rectification efficiency=%.4f\" %RE)\n",
+      "print(\"voltage ripple factor=%.3f\" %VRF)    \n",
+      "print(\"DF=%.4f\" %DF)\n",
+      "print(\"CDF=%.4f\" %CDF)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "print(\"HF=%.4f\" %HF)\n",
+      "print(\"active power=%.3f W\" %P_dc)\n",
+      "print(\"reactive power=%.2f Var\" %Q)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "form factor=1.241\n",
+        "rectification efficiency=0.8059\n",
+        "voltage ripple factor=0.735\n",
+        "DF=0.9239\n",
+        "CDF=0.9605\n",
+        "pf=0.8874\n",
+        "HF=0.2899\n",
+        "active power=3123.973 W\n",
+        "reactive power=1293.99 Var\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13, Page No 319"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "V_om=3*V_ml/(2*math.pi)\n",
+      "V_o=V_om/2\n",
+      "th=30\n",
+      "a=math.degrees(math.acos((2*math.pi*math.sqrt(3)*V_o/(3*V_ml)-1)))-th    \n",
+      "I_o=V_o/R    \n",
+      "V_or=V_ml/(2*math.sqrt(math.pi))*math.sqrt((5*math.pi/6-a*math.pi/180)+.5*math.sin(math.radians(2*a+2*th)))\n",
+      "I_or=V_or/R    \n",
+      "RE=V_o*I_o/(V_or*I_or)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"delay angle=%.1f deg\" %a)\n",
+      "print(\"avg load current=%.3f A\" %I_o)\n",
+      "print(\"rms load current=%.3f A\" %I_or)\n",
+      "print(\"rectification efficiency=%.4f\" %RE)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "delay angle=67.7 deg\n",
+        "avg load current=7.765 A\n",
+        "rms load current=10.477 A\n",
+        "rectification efficiency=0.5494\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15, Page No 321"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=400.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "v_T=1.4\n",
+      "a1=30.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o1=3*V_ml/(2*math.pi)*math.cos(math.radians(a1))-v_T   \n",
+      "a2=60.0\n",
+      "V_o2=3*V_ml/(2*math.pi)*math.cos(math.radians(a2))-v_T  \n",
+      "I_o=36\n",
+      "I_TA=I_o/3    \n",
+      "I_Tr=I_o/math.sqrt(3)    \n",
+      "P=I_TA*v_T       \n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle = 30deg\")\n",
+      "print(\"avg output voltage=%.3f V\" %V_o1)\n",
+      "print(\"for firing angle = 60deg\")\n",
+      "print(\"avg output voltage=%.2f V\" %V_o2)\n",
+      "print(\"avg current rating=%.0f A\" %I_TA)\n",
+      "print(\"rms current rating=%.3f A\" %I_Tr)\n",
+      "print(\"PIV of SCR=%.1f V\" %V_ml)\n",
+      "print(\"power dissipated=%.1f W\" %P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle = 30deg\n",
+        "avg output voltage=232.509 V\n",
+        "for firing angle = 60deg\n",
+        "avg output voltage=133.65 V\n",
+        "avg current rating=12 A\n",
+        "rms current rating=20.785 A\n",
+        "PIV of SCR=565.7 V\n",
+        "power dissipated=16.8 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17, Page No 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "E=200\n",
+      "I_o=20\n",
+      "R=.5\n",
+      "\n",
+      "#Calculations\n",
+      "V_o1=E+I_o*R\n",
+      "V_s=230\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "a1=math.degrees(math.acos(V_o1*math.pi/(3*V_ml)))\n",
+      "th=120\n",
+      "I_s=math.sqrt((1/math.pi)*I_o**2*th*math.pi/180)\n",
+      "P=E*I_o+I_o**2*R\n",
+      "pf=P/(math.sqrt(3)*V_s*I_s)    \n",
+      "V_o2=E-I_o*R\n",
+      "a2=math.degrees(math.acos(-V_o2*math.pi/(3*V_ml))) \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay=%.3f deg\" %a1)\n",
+      "print(\"pf=%.3f\" %pf)\n",
+      "print(\"firing angle delay=%.2f deg\" %a2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=47.461 deg\n",
+        "pf=0.646\n",
+        "firing angle delay=127.71 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18, Page No 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "w=2*math.pi*f\n",
+      "a1=0\n",
+      "t_c1=(4*math.pi/3-a1*math.pi/180)/w    \n",
+      "a2=30\n",
+      "t_c2=(4*math.pi/3-a2*math.pi/180)/w    \n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle delay=0deg\")\n",
+      "print(\"commutation time=%.2f ms\" %(t_c1*1000))\n",
+      "print(\"peak reverse voltage=%.2f V\" %(math.sqrt(2)*V))\n",
+      "print(\"for firing angle delay=30deg\")\n",
+      "print(\"commutation time=%.2f ms\" %(t_c2*1000))\n",
+      "print(\"peak reverse voltage=%.2f V\" %(math.sqrt(2)*V))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle delay=0deg\n",
+        "commutation time=13.33 ms\n",
+        "peak reverse voltage=325.27 V\n",
+        "for firing angle delay=30deg\n",
+        "commutation time=11.67 ms\n",
+        "peak reverse voltage=325.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.19, Page No 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*2*math.pi/(2*3)/(math.pi/3+math.sqrt(3)*math.cos(math.radians(2*a))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*math.pi/(math.sqrt(2)*3*math.cos(math.radians(a)))\n",
+      "V_ph=V_s/math.sqrt(3) \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage percent V_ph=%.3f V\" %V_ph)   \n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.2f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage percent V_ph=110.384 V\n",
+        "for constant load current\n",
+        "V_ph=110.38 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.20, Page No 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*4*math.pi/(2*3)/(2*math.pi/3+math.sqrt(3)*(1+math.cos(math.radians(2*a)))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*2*math.pi/(math.sqrt(2)*3*(1+math.cos(math.radians(a))))\n",
+      "V_ph=V_s/math.sqrt(3)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage percent V_ph=%.3f V\" %V_ph) \n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.2f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage percent V_ph=102.459 V\n",
+        "for constant load current\n",
+        "V_ph=102.46 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.21, Page No 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=90.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*4*math.pi/(2*3)/((math.pi-math.pi/2)+(math.sin(math.radians(2*a)))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*2*math.pi/(math.sqrt(2)*3*(1+math.cos(math.radians(a))))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage percent V_ph=%.2f V\" %V_ph)\n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.1f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage percent V_ph=191.19 V\n",
+        "for constant load current\n",
+        "V_ph=191.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.22 Page No 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "E=200.0\n",
+      "I_o=20.0\n",
+      "R=.5\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=E+I_o*R\n",
+      "V_s=230\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "a=math.degrees(math.acos(V_o*2*math.pi/(3*V_ml)-1))  \n",
+      "a1=180-a\n",
+      "I_sr=math.sqrt((1/math.pi)*I_o**2*(a1*math.pi/180))\n",
+      "P=V_o*I_o\n",
+      "pf=P/(math.sqrt(3)*V_s*I_sr)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay=%.2f deg\" %a)\n",
+      "print(\"pf=%.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=69.38 deg\n",
+        "pf=0.67\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.23, Page No 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "f=50.0\n",
+      "I_o=15.0\n",
+      "a=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_TA=I_o*120.0/360.0\n",
+      "I_Tr=math.sqrt(I_o**2*120/360)\n",
+      "I_sr=math.sqrt(I_o**2*120/180)\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "V_or=V_ml*math.sqrt((3/(2*math.pi))*(math.pi/3+math.sqrt(3/2)*math.cos(math.radians(2*a))))\n",
+      "I_or=I_o\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "VA=3*V_s/math.sqrt(3)*I_sr\n",
+      "TUF=P_dc/VA    \n",
+      "pf=P_ac/VA    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rectification efficiency=%.5f\" %RE)\n",
+      "print(\"TUF=%.4f\" %TUF)\n",
+      "print(\"Input pf=%.3f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectification efficiency=0.95493\n",
+        "TUF=0.6752\n",
+        "Input pf=0.707\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.24, Page No 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I=10.0\n",
+      "a=45.0\n",
+      "V=400.0\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "DF=math.cos(math.radians(a))\n",
+      "I_o=10\n",
+      "I_s1=4*I_o/(math.sqrt(2)*math.pi)*math.sin(math.pi/3)\n",
+      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
+      "I_o=1     #suppose\n",
+      "CDF=I_s1/I_sr    \n",
+      "THD=math.sqrt(1/CDF**2-1)    \n",
+      "pf=CDF*DF    \n",
+      "P=(3*math.sqrt(2)*V*math.cos(math.radians(a))/math.pi)*I\n",
+      "Q=(3*math.sqrt(2)*V*math.sin(math.radians(a))/math.pi)*I  \n",
+      "   \n",
+      "#Results\n",
+      "print(\"DF=%.3f\" %DF)\n",
+      "print(\"CDF=%.3f\" %CDF)\n",
+      "print(\"THD=%.5f\" %THD)\n",
+      "print(\"PF=%.4f\" %pf)\n",
+      "print(\"active power=%.2f W\" %P)  \n",
+      "print(\"reactive power=%.2f Var\" %Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "DF=0.707\n",
+        "CDF=0.955\n",
+        "THD=0.31084\n",
+        "PF=0.6752\n",
+        "active power=3819.72 W\n",
+        "reactive power=3819.72 Var\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.25, Page No 342"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "print(\"for firing angle=30deg\")\n",
+      "a=30.0\n",
+      "V=400.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "E=350\n",
+      "R=10\n",
+      "\n",
+      "#Calculations\n",
+      "I_o=(V_o-E)/R\n",
+      "I_or=I_o\n",
+      "P1=V_o*I_o    \n",
+      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
+      "VA=3*V/math.sqrt(3)*I_sr\n",
+      "pf=P1/VA \n",
+      "a=180-60\n",
+      "V=400\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "E=-350\n",
+      "R=10\n",
+      "I_o=(V_o-E)/R\n",
+      "I_or=I_o\n",
+      "P2=-V_o*I_o    \n",
+      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
+      "VA=3*V/math.sqrt(3)*I_sr\n",
+      "pf=P2/VA       \n",
+      "\n",
+      "print(\"power delivered to load=%.2f W\" %P1)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "print(\"for firing advance angle=60deg\")\n",
+      "print(\"power delivered to load=%.2f W\" %P2)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=30deg\n",
+        "power delivered to load=5511.74 W\n",
+        "pf=0.4775\n",
+        "for firing advance angle=60deg\n",
+        "power delivered to load=2158.20 W\n",
+        "pf=0.4775\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.26, Page No 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0\n",
+      "u=15.0\n",
+      "\n",
+      "#Calculations\n",
+      "i=math.cos(math.radians(a))-math.cos(math.radians(a+u))\n",
+      "a=30\n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a  \n",
+      "a=45\n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a    \n",
+      "a=60\n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a  \n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle=30deg\") \n",
+      "print(\"overlap angle=%.1f deg\" %u)\n",
+      "print(\"for firing angle=45deg\")  \n",
+      "print(\"overlap angle=%.1f deg\" %u)\n",
+      "print(\"for firing angle=60deg\") \n",
+      "print(\"overlap angle=%.2f deg\" %u)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=30deg\n",
+        "overlap angle=2.2 deg\n",
+        "for firing angle=45deg\n",
+        "overlap angle=2.2 deg\n",
+        "for firing angle=60deg\n",
+        "overlap angle=2.23 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.28, Page No 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "E=400.0\n",
+      "I_o=20.0\n",
+      "R=1\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=E+I_o*R\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=.004\n",
+      "V=230 #per phase voltage\n",
+      "V_ml=math.sqrt(6)*V\n",
+      "a=math.degrees(math.acos(math.pi/(3*V_ml)*(V_o+3*w*L*I_o/math.pi)))   \n",
+      "print(\"firing angle delay=%.3f deg\" %a)\n",
+      "u=math.degrees(math.acos(math.pi/(3*V_ml)*(V_o-3*w*L*I_o/math.pi)))-a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"overlap angle=%.2f deg\" %u)\n",
+      "#Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=34.382 deg\n",
+        "overlap angle=8.22 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.29, Page No 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=400.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "R=1\n",
+      "E=230\n",
+      "I=15.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=-E+I*R\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "a=math.degrees(math.acos(V_o*2*math.pi/(3*V_ml)))  \n",
+      "L=0.004\n",
+      "a=math.degrees(math.acos((2*math.pi)/(3*V_ml)*(V_o+3*w*L*I/(2*math.pi))))  \n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-3*f*L*I/V_ml))-a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"firing angle delay=%.3f deg\" %a)\n",
+      "print(\"overlap angle=%.3f deg\" %u)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle=139.702 deg\n",
+        "firing angle delay=139.702 deg\n",
+        "overlap angle=1.431 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.31, Page No 361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0 #per phase\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(3.0)*math.sqrt(2)*V\n",
+      "w=2*math.pi*f\n",
+      "a1=60.0\n",
+      "L=0.015\n",
+      "i_cp=(math.sqrt(3)*V_ml/(w*L))*(1-math.sin(math.radians(a1)))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"circulating current=%.4f A\" %i_cp)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "circulating current=27.7425 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.32, Page No 362"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "V_m=math.sqrt(2)*V\n",
+      "a=30.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=2*V_m* math.cos(math.radians(a))/math.pi    \n",
+      "R=10\n",
+      "I_o=V_o/R    \n",
+      "I_TA=I_o*math.pi/(2*math.pi)    \n",
+      "I_Tr=math.sqrt(I_o**2*math.pi/(2*math.pi))    \n",
+      "I_s=math.sqrt(I_o**2*math.pi/(math.pi)) \n",
+      "I_o=I_s\n",
+      "pf=(V_o*I_o/(V*I_s))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.3f V\" %V_o)\n",
+      "print(\"avg o/p current=%.2f A\" %I_o)\n",
+      "print(\"avg value of thyristor current=%.3f A\" %I_TA)\n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_Tr)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n",
+      " \n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p voltage=179.330 V\n",
+        "avg o/p current=17.93 A\n",
+        "avg value of thyristor current=8.967 A\n",
+        "rms value of thyristor current=12.68 A\n",
+        "pf=0.7797\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.33, Page No 363"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "V_m=math.sqrt(2)*V\n",
+      "a=30.0\n",
+      "L=.0015\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=2*V_m* math.cos(math.radians(a))/math.pi  \n",
+      "R=10\n",
+      "I_o=V_o/R  \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "V_ox=2*V_m*math.cos(math.radians(a))/math.pi-w*L*I_o/math.pi    \n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-I_o*w*L/V_m))-a    \n",
+      "I=I_o\n",
+      "pf=V_o*I_o/(V*I)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.3f V\" %V_ox)\n",
+      "print(\"angle of overlap=%.3f deg\" %u)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p voltage=176.640 V\n",
+        "angle of overlap=2.855 deg\n",
+        "pf=0.7797\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.34, Page No 364"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=415.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "a1=35.0 #firing angle advance\n",
+      "\n",
+      "#Calculations\n",
+      "a=180-a1\n",
+      "I_o=80.0\n",
+      "r_s=0.04\n",
+      "v_T=1.5\n",
+      "X_l=.25 #reactance=w*L\n",
+      "E=-3*V_ml*math.cos(math.radians(a))/math.pi+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi    \n",
+      "\n",
+      "#Results\n",
+      "print(\"mean generator voltage=%.3f V\" %E)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mean generator voltage=487.590 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.35, Page No 364"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=415.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "R=0.2\n",
+      "I_o=80\n",
+      "r_s=0.04\n",
+      "v_T=1.5\n",
+      "\n",
+      "#Calculations\n",
+      "X_l=.25 #reactance=w*L\n",
+      "a=35\n",
+      "E=-(-3*V_ml*math.cos(math.radians(a))/math.pi+I_o*R+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi)    \n",
+      "a1=35\n",
+      "a=180-a1\n",
+      "E=(-3*V_ml*math.cos(math.radians(a))/math.pi+I_o*R+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi) \n",
+      "\n",
+      "#Results\n",
+      "print(\"when firing angle=35deg\")   \n",
+      "print(\"mean generator voltage=%.3f V\" %E)\n",
+      "print(\"when firing angle advance=35deg\")\n",
+      "print(\"mean generator voltage=%.3f V\" %E)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when firing angle=35deg\n",
+        "mean generator voltage=503.590 V\n",
+        "when firing angle advance=35deg\n",
+        "mean generator voltage=503.590 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.36, Page No 365"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=5.0\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_mp=math.sqrt(2)*V\n",
+      "a=30.0\n",
+      "E=150.0\n",
+      "B=180-math.degrees(math.asin(E/V_mp))\n",
+      "I_o=(3/(2*math.pi*R))*(V_mp*(math.cos(math.radians(a+30))-math.cos(math.radians(B)))-E*((B-a-30)*math.pi/180))\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg current flowing=%.2f A\" %I_o)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg current flowing=19.96 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.37, Page No 366"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V\n",
+      "V_o=V_m*(1+math.cos(math.radians(a)))/math.pi    \n",
+      "E=100\n",
+      "R=10\n",
+      "I_o=(V_o-E)/R    \n",
+      "I_TA=I_o*math.pi/(2*math.pi)    \n",
+      "I_Tr=math.sqrt(I_o**2*math.pi/(2*math.pi))    \n",
+      "I_s=math.sqrt(I_o**2*(1-a/180)*math.pi/(math.pi))\n",
+      "I_or=I_o\n",
+      "P=E*I_o+I_or**2*R\n",
+      "pf=(P/(V*I_s))   \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "t_c=(1-a/180)*math.pi/w    \n",
+      "\n",
+      "#Results\n",
+      "print(\"\\navg o/p current=%.2f A\" %I_o)\n",
+      "print(\"avg o/p voltage=%.3f V\" %V_o)\n",
+      "print(\"avg value of thyristor current=%.2f A\" %I_TA)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %I_Tr)\n",
+      "print(\"avg value of diode current=%.2f A\" %I_TA)\n",
+      "print(\"rms value of diode current=%.3f A\" %I_Tr)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "print(\"circuit turn off time=%.2f ms\" %(t_c*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "avg o/p current=9.32 A\n",
+        "avg o/p voltage=193.202 V\n",
+        "avg value of thyristor current=4.66 A\n",
+        "rms value of thyristor current=6.590 A\n",
+        "avg value of diode current=4.66 A\n",
+        "rms value of diode current=6.590 A\n",
+        "pf=0.9202\n",
+        "circuit turn off time=8.33 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.38, Page No 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "V_m=math.sqrt(2)*V\n",
+      "L=0.05\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "w=2*math.pi*f\n",
+      "a=30\n",
+      "i_cp=2*V_m*(1-math.cos(math.radians(a)))/(w*L)    \n",
+      "R=30.0\n",
+      "i_l=V_m/R\n",
+      "i1=i_cp+i_l    \n",
+      "i2=i_cp    \n",
+      "\n",
+      "#Results\n",
+      "print(\"peak value of circulating current=%.3f A\" %i_cp)\n",
+      "print(\"peak value of current in convertor 1=%.3f A\" %i1)\n",
+      "print(\"peak value of current in convertor 2=%.3f A\" %i2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak value of circulating current=5.548 A\n",
+        "peak value of current in convertor 1=16.391 A\n",
+        "peak value of current in convertor 2=5.548 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.39, Page No 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "R=5.0\n",
+      "L=0.05\n",
+      "\n",
+      "#Calculations\n",
+      "phi=math.degrees(math.atan(w*L/R))  \n",
+      "phi=90+math.degrees(math.atan(w*L/R))   \n",
+      "\n",
+      "#Results\n",
+      "print(\"for no current transients\")\n",
+      "print(\"triggering angle=%.2f deg\" %phi)\n",
+      "print(\"for worst transients\")\n",
+      "print(\"triggering angle=%.2f deg\" %phi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for no current transients\n",
+        "triggering angle=162.34 deg\n",
+        "for worst transients\n",
+        "triggering angle=162.34 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter7.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter7.ipynb
new file mode 100755
index 00000000..726160c8
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter7.ipynb
@@ -0,0 +1,1036 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 07 : Choppers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page No 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.4     #duty cycle %a=T_on/T\n",
+      "V_s=230.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V=a*(V_s-2)    \n",
+      "V_or=math.sqrt(a*(V_s-2)**2)    \n",
+      "P_o=V_or**2/R\n",
+      "P_i=V_s*V/R\n",
+      "n=P_o*100/P_i    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.1f V\" %V)\n",
+      "print(\"rms value of o/p voltage=%.1f V\" %V_or)\n",
+      "print(\"chopper efficiency in percentage=%.2f\" %n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p voltage=91.2 V\n",
+        "rms value of o/p voltage=144.2 V\n",
+        "chopper efficiency in percentage=99.13\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page No 388"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_i=220.0\n",
+      "V_o=660.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=1-V_i/V_o\n",
+      "T_on=100.0     #microsecond\n",
+      "T=T_on/a\n",
+      "T_off=T-T_on    \n",
+      "T_off=T_off/2\n",
+      "T_on=T-T_off\n",
+      "a=T_on/T\n",
+      "V_o=V_i/(1-a)\n",
+      "\n",
+      "#Results    \n",
+      "print(\"pulse width of o/p voltage=%.0f us\" %T_off)\n",
+      "print(\"\\nnew o/p voltage=%.0f V\" %V_o)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pulse width of o/p voltage=25 us\n",
+        "\n",
+        "new o/p voltage=1320 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page No 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_1=12.0\n",
+      "I_2=16.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_0=(I_1+I_2)/2\n",
+      "R=10.0\n",
+      "V_0=I_0*R\n",
+      "V_s=200.0\n",
+      "a=V_0/V_s\n",
+      "r=a/(1-a)\n",
+      "\n",
+      "#Results\n",
+      "print(\"time ratio(T_on/T_off)=%.3f\" %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time ratio(T_on/T_off)=2.333\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page No 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_o=660.0\n",
+      "V_s=220.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=(V_o/V_s)/(1+(V_o/V_s))\n",
+      "T_on=120\n",
+      "T=T_on/a\n",
+      "T_off=T-T_on \n",
+      "T_off=3*T_off\n",
+      "T_on=T-T_off\n",
+      "a=T_on/(T_on+T_off)\n",
+      "V_o=V_s*(a/(1-a))   \n",
+      "\n",
+      "#Results\n",
+      "print(\"pulse width o/p voltage=%.0f us\" %T_off)\n",
+      "print(\"\\nnew o/p voltage=%.2f V\" %V_o)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pulse width o/p voltage=120 us\n",
+        "\n",
+        "new o/p voltage=73.33 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11 Page No 408"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=1.0\n",
+      "L=.005\n",
+      "T_a=L/R\n",
+      "T=2000*10**-6\n",
+      "E=24.0\n",
+      "V_s=220\n",
+      "T_on=600*10**-6\n",
+      "a=T_on/T\n",
+      "\n",
+      "#Calculations\n",
+      "a1=(T_a/T)*math.log(1+(E/V_s)*((math.exp(T/T_a))-1))\n",
+      "if a1<a :\n",
+      "    print(\"load current in continuous\")\n",
+      "else:\n",
+      "    print(\"load current in discont.\")\n",
+      "\n",
+      "I_o=(a*V_s-E)/R    \n",
+      "I_mx=(V_s/R)*((1-math.exp(-T_on/T_a))/(1-math.exp(-T/T_a)))-E/R    \n",
+      "I_mn=(V_s/R)*((math.exp(T_on/T_a)-1)/(math.exp(T/T_a)-1))-E/R    \n",
+      "f=1/T\n",
+      "w=2*math.pi*f\n",
+      "I1=(2*V_s/(math.sqrt(2)*math.pi)*math.sin(math.radians(180*a)))/(math.sqrt(R**2+(w*L)**2))    \n",
+      "I2=(2*V_s/(2*math.sqrt(2)*math.pi)*math.sin(math.radians(2*180*a)))/(math.sqrt(R**2+(w*L*2)**2))    \n",
+      "I3=(2*V_s/(3*math.sqrt(2)*math.pi)*math.sin(math.radians(3*180*a)))/(math.sqrt(R**2+(w*L*3)**2))    \n",
+      "I_TAV=a*(V_s-E)/R-L*(I_mx-I_mn)/(R*T)   \n",
+      "P1=I_TAV*V_s\n",
+      "P2=E*I_o\n",
+      "I_or=math.sqrt(I_o**2+I1**2+I2**2+I3**2)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p current=%.2f A\" %I_o)\n",
+      "print(\"max value of steady current=%.2f A\" %I_mx)\n",
+      "print(\"min value of steady current=%.2f A\" %I_mn)\n",
+      "print(\"first harmonic current=%.4f A\" %I1)\n",
+      "print(\"second harmonic current=%.4f A\" %I2)\n",
+      "print(\"third harmonic current=%.5f A\" %I3)\n",
+      "print(\"avg supply current=%.4f A\" %I_TAV)\n",
+      "print(\"i/p power=%.2f W\" %P1)\n",
+      "print(\"power absorbed by load emf=%.0f W\" %P2)\n",
+      "print(\"power loss in resistor=%.2f W\" %(P1-P2))\n",
+      "print(\"rms value of load current=%.3f A\" %I_or)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "load current in continuous\n",
+        "avg o/p current=42.00 A\n",
+        "max value of steady current=51.46 A\n",
+        "min value of steady current=33.03 A\n",
+        "first harmonic current=5.0903 A\n",
+        "second harmonic current=1.4983 A\n",
+        "third harmonic current=0.21643 A\n",
+        "avg supply current=12.7289 A\n",
+        "i/p power=2800.35 W\n",
+        "power absorbed by load emf=1008 W\n",
+        "power loss in resistor=1792.35 W\n",
+        "rms value of load current=42.334 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12 Page No 411"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=1\n",
+      "L=.001\n",
+      "V_s=220\n",
+      "E=72.0\n",
+      "f=500.0\n",
+      "T_on=800*10**-6\n",
+      "T_a=L/R\n",
+      "T=1.0/f\n",
+      "m=E/V_s\n",
+      "a=T_on/T\n",
+      "\n",
+      "#Calculations\n",
+      "a1=(T_a/T)*math.log(1+m*(math.exp(-T/T_a)-1))\n",
+      "if a1>a :\n",
+      "    print(\"load current is continuous\")\n",
+      "else:\n",
+      "    print(\"load current is discontinuous\")\n",
+      "\n",
+      "t_x=T_on+L*math.log(1+((V_s-E)/272)*(1-math.exp(-T_on/T_a)))\n",
+      " #Value of t_x wrongly calculated in the book so ans of V_o and I_o varies\n",
+      "V_o=a*V_s+(1-t_x/T)*E    \n",
+      "I_o=(V_o-E)/R    \n",
+      "I_mx=(V_s-E)/R*(1-math.exp(-T_on/T_a)) \n",
+      "\n",
+      "#Results  \n",
+      "print(\"avg o/p voltage=%.2f V\" %V_o)\n",
+      "print(\"avg o/p current=%.2f A\" %I_o) \n",
+      "print(\"max value of load current=%.1f A\" %I_mx)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "load current is discontinuous\n",
+        "avg o/p voltage=121.77 V\n",
+        "avg o/p current=49.77 A\n",
+        "max value of load current=81.5 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13, Page No 412"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.2\n",
+      "V_s=500\n",
+      "E=a*V_s\n",
+      "L=0.06\n",
+      "I=10\n",
+      "\n",
+      "#Calculations\n",
+      "T_on=(L*I)/(V_s-E)\n",
+      "f=a/T_on    \n",
+      "\n",
+      "#Results\n",
+      "print(\"chopping freq=%.2f Hz\" %f)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "chopping freq=133.33 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14 Page No 412"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.5\n",
+      "pu=0.1 #pu ripple\n",
+      "\n",
+      "#Calculations\n",
+      " #x=T/T_a\n",
+      " #y=exp(-a*x)\n",
+      "y=(1-pu)/(1+pu)\n",
+      " #after solving\n",
+      "x=math.log(1/y)/a\n",
+      "f=1000\n",
+      "T=1/f\n",
+      "T_a=T/x\n",
+      "R=2\n",
+      "L=R*T_a\n",
+      "Li=0.002\n",
+      "Le=L-Li    \n",
+      "\n",
+      "#Results\n",
+      "print(\"external inductance=%.3f mH\" %(Le*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "external inductance=-2.000 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15 Page No 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=10.0\n",
+      "L=0.015\n",
+      "T_a=L/R\n",
+      "f=1250.0\n",
+      "T=1.0/f\n",
+      "a=0.5\n",
+      "T_on=a*T\n",
+      "V_s=220.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_mx=(V_s/R)*((1-math.exp(-T_on/T_a))/(1-math.exp(-T/T_a)))   \n",
+      "I_mn=(V_s/R)*((math.exp(T_on/T_a)-1)/(math.exp(T/T_a)-1))    \n",
+      "dI=I_mx-I_mn    \n",
+      "V_o=a*V_s\n",
+      "I_o=V_o/R    \n",
+      "I_or=math.sqrt(I_mx**2+dI**2/3+I_mx*dI)    \n",
+      "I_chr=math.sqrt(a)*I_or \n",
+      "\n",
+      "#Results\n",
+      "print(\"Max value of ripple current=%.2f A\" %dI)\n",
+      "print(\"Max value of load current=%.3f A\" %I_mx)\n",
+      "print(\"Min value of load current=%.2f A\" %I_mn)\n",
+      "print(\"Avg value of load current=%.2f A\" %I_o)   \n",
+      "print(\"rms value of load current=%.2f A\" %I_or)\n",
+      "print(\"rms value of chopper current=%.2f A\" %I_chr)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Max value of ripple current=2.92 A\n",
+        "Max value of load current=12.458 A\n",
+        "Min value of load current=9.54 A\n",
+        "Avg value of load current=11.00 A\n",
+        "rms value of load current=13.94 A\n",
+        "rms value of chopper current=9.86 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17 Page No 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=0.0016\n",
+      "C=4*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "w=1/math.sqrt(L*C)\n",
+      "t=math.pi/w    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"time for which current flows=%.2f us\" %(t*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for which current flows=251.33 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.18, Page No 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "t_q=20.0*10**-6\n",
+      "dt=20.0*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "t_c=t_q+dt\n",
+      "I_0=60.0\n",
+      "V_s=60.0\n",
+      "C=t_c*I_0/V_s \n",
+      "\n",
+      "#Results   \n",
+      "print(\"value of commutating capacitor=%.0f uF\" %(C*10**6))\n",
+      "\n",
+      "L1=(V_s/I_0)**2*C\n",
+      "L2=(2*t_c/math.pi)**2/C\n",
+      "if L1>L2 :\n",
+      "    print(\"value of commutating inductor=%.0f uH\" %(L1*10**6))\n",
+      "else:\n",
+      "    print(\"value of commutating inductor=%.0f uH\" %(L2*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of commutating capacitor=40 uF\n",
+        "value of commutating inductor=40 uH\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.19, Page No 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "t=100.0*10**-6\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      " #V_s*(1-2*math.exp(-t/(R*C)))=0\n",
+      "C=-t/(R*math.log(1.0/2))    \n",
+      "L=(4/9.0)*C*R**2    \n",
+      "L=(1.0/4)*C*R**2    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Value of comutating component C=%.3f uF\" %(C*10**6))\n",
+      "print(\"max permissible current through SCR is 2.5 times load current\")\n",
+      "print(\"value of comutating component L=%.1f uH\" %(L*10**6))\n",
+      "print(\"max permissible current through SCR is 1.5 times peak diode current\")\n",
+      "print(\"value of comutating component L=%.2f uH\" %(L*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of comutating component C=14.427 uF\n",
+        "max permissible current through SCR is 2.5 times load current\n",
+        "value of comutating component L=360.7 uH\n",
+        "max permissible current through SCR is 1.5 times peak diode current\n",
+        "value of comutating component L=360.67 uH\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.20, Page No 426"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_on=800.0*10**-6\n",
+      "V_s=220.0\n",
+      "I_o=80.0\n",
+      "C=50*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "T=T_on+2*V_s*C/I_o    \n",
+      "L=20*10**-6\n",
+      "C=50*10**-6\n",
+      "i_T1=I_o+V_s*math.sqrt(C/L)    \n",
+      "i_TA=I_o    \n",
+      "t_c=C*V_s/I_o    \n",
+      "t_c1=(math.pi/2)*math.sqrt(L*C)    \n",
+      "t=150*10**-6\n",
+      "v_c=I_o*t/C-V_s  \n",
+      "\n",
+      "#Results  \n",
+      "print(\"effective on period=%.0f us\" %(T*10**6))\n",
+      "print(\"peak current through main thyristor=%.2f A\" %i_T1)\n",
+      "print(\"peak current through auxillery thyristor=%.0f A\" %i_TA)\n",
+      "print(\"turn off time for main thyristor=%.1f us\" %(t_c*10**6))\n",
+      "print(\"turn off time for auxillery thyristor=%.3f us\" %(t_c1*10**6))\n",
+      "print(\"total commutation interval=%.0f us\" %(2*t_c*10**6))\n",
+      "print(\"capacitor voltage=%.0f V\" %v_c)\n",
+      "print(\"time nedded to recharge the capacitor=%.0f us\" %(2*V_s*C/I_o*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle = 30deg\n",
+        "avg output voltage=232.509 V\n",
+        "for firing angle = 60deg\n",
+        "avg output voltage=133.65 V\n",
+        "avg current rating=12 A\n",
+        "rms current rating=20.785 A\n",
+        "PIV of SCR=565.7 V\n",
+        "power dissipated=16.8 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 122
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.21, Page No 427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_o=260.0\n",
+      "V_s=220.0\n",
+      "fos=2 #factor of safety\n",
+      "\n",
+      "#Calculations\n",
+      "t_off=18*10**-6\n",
+      "t_c=2*t_off\n",
+      "C=t_c*I_o/V_s    \n",
+      "L=(V_s/(0.8*I_o))**2*C    \n",
+      "f=400\n",
+      "a_mn=math.pi*f*math.sqrt(L*C)\n",
+      "V_omn=V_s*(a_mn+2*f*t_c)    \n",
+      "V_omx=V_s    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Value of C=%.3f uF\" %(C*10**6))\n",
+      "print(\"value of L=%.3f uH\" %(L*10**6))\n",
+      "print(\"min value of o/p voltage=%.3f V\" %V_omn)\n",
+      "print(\"max value of o/p voltage=%.0f V\" %V_omx)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=47.461 deg\n",
+        "pf=0.646\n",
+        "firing angle delay=127.71 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": "*"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.22, Page No 434"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "x=2.0\n",
+      "t_q=30*10**-6\n",
+      "dt=30*10**-6\n",
+      "t_c=t_q+dt\n",
+      "V_s=230.0\n",
+      "I_o=200.0\n",
+      "\n",
+      "#Calculations\n",
+      "L=V_s*t_c/(x*I_o*(math.pi-2*math.asin(1/x)))    \n",
+      "C=x*I_o*t_c/(V_s*(math.pi-2*math.asin(1/x)))    \n",
+      "V_cp=V_s+I_o*math.sqrt(L/C)    \n",
+      "I_cp=x*I_o    \n",
+      "x=3\n",
+      "L=V_s*t_c/(x*I_o*(math.pi-2*math.asin(1/x)))   \n",
+      "C=x*I_o*t_c/(V_s*(math.pi-2*math.asin(1/x)))    \n",
+      "V_cp=V_s+I_o*math.sqrt(L/C)    \n",
+      "I_cp=x*I_o    \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of commutating inductor=%.3f uH\" %(L*10**6))\n",
+      "print(\"value of commutating capacitor=%.3f uF\" %(C*10**6))\n",
+      "print(\"peak capacitor voltage=%.0f V\" %V_cp)\n",
+      "print(\"peak commutataing current=%.0f A\" %I_cp)\n",
+      "print(\"value of commutating inductor=%.3f uH\" %(L*10**6))\n",
+      "print(\"value of commutating capacitor=%.3f uF\" %(C*10**6))\n",
+      "print(\"peak capacitor voltage=%.2f V\" %V_cp)\n",
+      "print(\"peak commutataing current=%.0f A\" %I_cp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of commutating inductor=7.321 uH\n",
+        "value of commutating capacitor=49.822 uF\n",
+        "peak capacitor voltage=307 V\n",
+        "peak commutataing current=600 A\n",
+        "value of commutating inductor=7.321 uH\n",
+        "value of commutating capacitor=49.822 uF\n",
+        "peak capacitor voltage=306.67 V\n",
+        "peak commutataing current=600 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.23, Page No 434"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=230\n",
+      "C=50*10**-6\n",
+      "L=20*10**-6\n",
+      "I_cp=V_s*math.sqrt(C/L)\n",
+      "I_o=200\n",
+      "x=I_cp/I_o\n",
+      "\n",
+      "#Calculations\n",
+      "t_c=(math.pi-2*math.asin(1/x))*math.sqrt(C*L)   \n",
+      "th1=math.degrees(math.asin(1.0/x))\n",
+      "t=(5*math.pi/2-th1*math.pi/180)*math.sqrt(L*C)+C*V_s*(1-math.cos(math.radians(th1)))/I_o    \n",
+      "t=(math.pi-th1*math.pi/180)*math.sqrt(L*C)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"turn off time of main thyristor=%.2f us\" %(t_c*10**6))\n",
+      "print(\"total commutation interval=%.3f us\" %(t*10**6))\n",
+      "print(\"turn off time of auxillery thyristor=%.3f us\" %(t*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "turn off time of main thyristor=62.52 us\n",
+        "total commutation interval=80.931 us\n",
+        "turn off time of auxillery thyristor=80.931 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.24, Page No 440"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "tc=0.006\n",
+      "R=10.0\n",
+      "L=R*tc\n",
+      "f=2000.0\n",
+      "\n",
+      "#Calculations\n",
+      "T=1/f\n",
+      "V_o=50.0\n",
+      "V_s=100.0\n",
+      "a=V_o/V_s\n",
+      "T_on=a*T\n",
+      "T_off=T-T_on\n",
+      "dI=V_o*T_off/L\n",
+      "I_o=V_o/R\n",
+      "I2=I_o+dI/2    \n",
+      "I1=I_o-dI/2    \n",
+      "\n",
+      "#Results\n",
+      "print(\"max value of load current=%.3f A\" %I2)\n",
+      "print(\"min value of load current=%.3f A\" %I1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "max value of load current=5.104 A\n",
+        "min value of load current=4.896 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.27, Page No 443"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=30.0\n",
+      "r_a=.5\n",
+      "V_s=220.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=I_a*r_a/V_s    \n",
+      "a=1\n",
+      "k=.1 #V/rpm\n",
+      "N=(a*V_s-I_a*r_a)/k    \n",
+      "\n",
+      "#Results\n",
+      "print(\"min value of duty cycle=%.3f\" %a)\n",
+      "print(\"min Value of speed control=%.0f rpm\" %0)\n",
+      "print(\"max value of duty cycle=%.0f\" %a)\n",
+      "print(\"max value of speed control=%.0f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "min value of duty cycle=1.000\n",
+        "min Value of speed control=0 rpm\n",
+        "max value of duty cycle=1\n",
+        "max value of speed control=2050 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.28, Page No 444"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=72.0\n",
+      "I_a=200.0\n",
+      "r_a=0.045\n",
+      "N=2500.0\n",
+      "\n",
+      "#Calculations\n",
+      "k=(V_t-I_a*r_a)/N\n",
+      "E_a=k*1000\n",
+      "L=.007\n",
+      "Rm=.045\n",
+      "Rb=0.065\n",
+      "R=Rm+Rb\n",
+      "T_a=L/R\n",
+      "I_mx=230\n",
+      "I_mn=180\n",
+      "T_on=-T_a*math.log(-((V_t-E_a)/R-I_mx)/((I_mn)-(V_t-E_a)/R))\n",
+      "R=Rm\n",
+      "T_a=L/R\n",
+      "T_off=-T_a*math.log(-((-E_a)/R-I_mn)/((I_mx)-(-E_a)/R))\n",
+      "T=T_on+T_off\n",
+      "f=1/T    \n",
+      "a=T_on/T \n",
+      "\n",
+      "#Results\n",
+      "print(\"chopping freq=%.1f Hz\" %f)  \n",
+      "print(\"duty cycle ratio=%.3f\" %a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "chopping freq=40.5 Hz\n",
+        "\n",
+        "duty cycle ratio=0.588\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.29, Page No 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesI_mx=425\n",
+      "I_lt=180.0     #lower limit of current pulsation\n",
+      "I_mn=I_mx-I_lt\n",
+      "T_on=0.014\n",
+      "T_off=0.011\n",
+      "\n",
+      "#Calculations\n",
+      "T=T_on+T_off\n",
+      "T_a=.0635\n",
+      "a=T_on/T\n",
+      "V=(I_mx-I_mn*math.exp(-T_on/T_a))/(1-math.exp(-T_on/T_a))\n",
+      "a=.5\n",
+      "I_mn=(I_mx-V*(1-math.exp(-T_on/T_a)))/(math.exp(-T_on/T_a))\n",
+      "T=I_mx-I_mn    \n",
+      "T=T_on/a\n",
+      "f=1/T    \n",
+      "\n",
+      "#Results\n",
+      "print(\"higher limit of current pulsation=%.0f A\" %T)\n",
+      "print(\"chopping freq=%.3f Hz\" %f)\n",
+      "print(\"duty cycle ratio=%.2f\" %a)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "higher limit of current pulsation=0 A\n",
+        "chopping freq=35.714 Hz\n",
+        "duty cycle ratio=0.50\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter8.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter8.ipynb
new file mode 100755
index 00000000..721a9faf
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter8.ipynb
@@ -0,0 +1,984 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 08 : Inverters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page No 465"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T=0.1*10**-3\n",
+      "f=1.0/T\n",
+      "k=15*10**-6 #k=th/w\n",
+      "\n",
+      "#Calculations\n",
+      "th=2*math.pi*f*k\n",
+      "X_l=10.0\n",
+      "R=2.0\n",
+      "X_c=R*math.tan(th)+X_l\n",
+      "C=1/(2*math.pi*f*X_c)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of C=%.3f uF\" %(C*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of C=1.248 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page No 466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_01=2*V_s/(math.sqrt(2)*math.pi)\n",
+      "R=2.0\n",
+      "I_01=V_01/R\n",
+      "P_d=I_01**2*R    \n",
+      "V=V_s/2\n",
+      "I_s=math.sqrt(2)*I_01/math.pi\n",
+      "P_s=V*I_s\n",
+      "\n",
+      "#Results\n",
+      "print(\"power delivered to load=%.1f W\" %P_d)\n",
+      "print(\"power delivered by both sources=%.1f W\" %(2*P_s))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power delivered to load=5359.9 W\n",
+        "power delivered by both sources=5359.9 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5, Page No 468"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_01=4*V_s/(math.pi*math.sqrt(2))\n",
+      "R=1.0\n",
+      "X_L=6.0\n",
+      "X_c=7.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_01=V_01/math.sqrt(R**2+(X_L-X_c)**2)\n",
+      "P=I_01**2*R    \n",
+      "I_s=math.sqrt(2)*I_01*(2*math.cos(math.radians(45)))/math.pi\n",
+      "P_s=V_s*I_s    \n",
+      "\n",
+      "#Results\n",
+      "print(\"power delivered to the source=%.3f kW\" %(P/1000))\n",
+      "print(\"\\npower from the source=%.3f kW\" %(P_s/1000))\n",
+      "   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power delivered to the source=21.440 kW\n",
+        "\n",
+        "power from the source=21.440 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6 Page No 469"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_01=230.0\n",
+      "R=2.0\n",
+      "I_01=V_01/R\n",
+      "I_m=I_01*math.sqrt(2)\n",
+      "I_T1=I_m/2    \n",
+      "I_D1=0    \n",
+      "X_L=8.0\n",
+      "X_C=6.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_01=V_01/math.sqrt(R**2+(X_L-X_C)**2)\n",
+      "phi1=math.degrees(math.atan((X_L-X_C)/R))\n",
+      "I_T1=I_T1*math.sqrt(2)*0.47675    \n",
+      "I_D1=.1507025*I_m/math.sqrt(2)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"when load R=2 ohm\")\n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_T1)\n",
+      "print(\"rms value of diode current=%.0f A\" %I_D1)\n",
+      "print(\"when load R=2ohm % X_L=8ohm and X_C=6ohm\")\n",
+      "print(\"rms value of thyristor current=%.3f A\" %I_T1)\n",
+      "print(\"rms value of diode current=%.3f A\" %I_D1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when load R=2 ohm\n",
+        "rms value of thyristor current=54.83 A\n",
+        "rms value of diode current=17 A\n",
+        "when load R=2ohm % X_L=8ohm and X_C=6ohm\n",
+        "rms value of thyristor current=54.826 A\n",
+        "rms value of diode current=17.331 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7 Page No 470"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "R=4.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=0.035\n",
+      "\n",
+      "#Calculations\n",
+      "C=155*10**-6\n",
+      "X_L=w*L\n",
+      "X_C=1/(w*C)\n",
+      "Z1=math.sqrt(R**2+(X_L-X_C)**2)\n",
+      "phi1=-math.degrees(math.atan((X_L-X_C)/R))\n",
+      "Z3=math.sqrt(R**2+(X_L*3-X_C/3)**2)\n",
+      "phi3=math.degrees(math.atan((X_L*3-X_C/3)/R))\n",
+      "Z5=math.sqrt(R**2+(X_L*5-X_C/5)**2)\n",
+      "phi5=math.degrees(math.atan((X_L*5-X_C/5)/R))\n",
+      "I_m1=4*V_s/(Z1*math.pi)\n",
+      "I_01=I_m1/math.sqrt(2)    \n",
+      "I_m3=4*V_s/(3*Z3*math.pi)\n",
+      "I_m5=4*V_s/(5*Z5*math.pi)\n",
+      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2)\n",
+      "I_0=I_m/math.sqrt(2)\n",
+      "P_0=(I_0)**2*R    \n",
+      "P_01=(I_01)**2*R    \n",
+      "t1=(180-phi1)*math.pi/(180*w)    \n",
+      "t1=(phi1)*math.pi/(180*w)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of fundamental load current=%.2f A\" %I_01)\n",
+      "print(\"load power=%.1f W\" %P_0)\n",
+      "print(\"fundamental load power=%.1f W\" %P_01)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %(I_m/2))\n",
+      "print(\"conduction time for thyristor=%.3f ms\" %(t1*1000))\n",
+      "print(\"Conduction time for diodes=%.3f ms\" %(t1*1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of fundamental load current=20.02 A\n",
+        "load power=1632.5 W\n",
+        "fundamental load power=1602.6 W\n",
+        "rms value of thyristor current=14.285 A\n",
+        "conduction time for thyristor=3.736 ms\n",
+        "Conduction time for diodes=3.736 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8, Page No 473"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_01=2*V_s/(math.sqrt(2)*math.pi)    \n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_01=V_01/R\n",
+      "P=I_01**2*R    \n",
+      "V_or=math.sqrt((V_s/2)**2)\n",
+      "P=V_or**2/R    \n",
+      "I_TP=V_s/(2*R)\n",
+      "I_or=I_TP\n",
+      "pf=I_01**2*R/(V_or*I_or)  \n",
+      "DF=V_01/V_or  \n",
+      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
+      "THD=V_oh/V_01 \n",
+      "V_03=V_01/3\n",
+      "HF=V_03/V_01\n",
+      "\n",
+      "#Results\n",
+      "print(\"fundamental rms o/p voltage=%.3f V\" %V_01)\n",
+      "print(\"fundamental power to load=%.1f W\" %P)\n",
+      "print(\"total o/p power to load=%.1f W\" %P)\n",
+      "print(\"avg SCR current=%.2f A\" %(I_TP*180/360))\n",
+      "print(\"i/p pf=%.3f\" %pf)  \n",
+      "print(\"distortion factor=%.1f\" %DF)\n",
+      "print(\"THD=%.3f\" %THD)    \n",
+      "print(\"harmonic factor=%.4f\" %HF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fundamental rms o/p voltage=103.536 V\n",
+        "fundamental power to load=1322.5 W\n",
+        "total o/p power to load=1322.5 W\n",
+        "avg SCR current=5.75 A\n",
+        "i/p pf=0.811\n",
+        "distortion factor=0.9\n",
+        "THD=0.483\n",
+        "harmonic factor=0.3333\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9 Page No 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=60\n",
+      "R=3.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt(V_s**2*math.pi/math.pi)    \n",
+      "V_01=4*V_s/(math.sqrt(2)*math.pi)    \n",
+      "P_o=V_or**2/R    \n",
+      "P_01=V_01**2/R    \n",
+      "I_s=V_s/R    \n",
+      "I_avg=I_s*math.pi/(2*math.pi)    \n",
+      "V_03=V_01/3\n",
+      "HF=V_03/V_01    \n",
+      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
+      "THD=V_oh/V_01   \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.0f V\" %V_or)\n",
+      "print(\"o/p power=%.0f W\" %P_o)\n",
+      "print(\"fundamental component of rms voltage=%.2f V\" %V_01)\n",
+      "print(\"fundamental freq o/p power=%.2f W\" %P_01) \n",
+      "print(\"peak current=%.0f A\" %I_s)\n",
+      "print(\"avg current of each transistor=%.0f A\" %I_avg)\n",
+      "print(\"peak reverse blocking voltage=%.0f V\" %V_s)\n",
+      "print(\"harmonic factor=%.4f\" %HF)\n",
+      "print(\"THD=%.4f\" %THD)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=60 V\n",
+        "o/p power=1200 W\n",
+        "fundamental component of rms voltage=54.02 V\n",
+        "fundamental freq o/p power=972.68 W\n",
+        "peak current=20 A\n",
+        "avg current of each transistor=10 A\n",
+        "peak reverse blocking voltage=60 V\n",
+        "harmonic factor=0.3333\n",
+        "THD=0.4834\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10 Page No 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=220.0\n",
+      "R=6.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=0.03\n",
+      "C=180*10**-6\n",
+      "X_L=w*L\n",
+      "X_C=1/(w*C)\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt(V_s**2*math.pi/math.pi)\n",
+      "V_01=4*V_s/(math.sqrt(2)*math.pi)\n",
+      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
+      "THD=V_oh/V_01    \n",
+      "print(\"THD of voltage=%.4f\" %THD)\n",
+      "DF=V_01/V_or    \n",
+      "Z1=math.sqrt(R**2+(X_L-X_C)**2)\n",
+      "phi1=-math.degrees(math.atan((X_L-X_C)/R))\n",
+      "Z3=math.sqrt(R**2+(X_L*3-X_C/3)**2)\n",
+      "phi3=math.degrees(math.atan((X_L*3-X_C/3)/R))\n",
+      "Z5=math.sqrt(R**2+(X_L*5-X_C/5)**2)\n",
+      "phi5=math.degrees(math.atan((X_L*5-X_C/5)/R))\n",
+      "Z7=math.sqrt(R**2+(X_L*7-X_C/7)**2)\n",
+      "phi7=math.degrees(math.atan((X_L*7-X_C/7)/R))\n",
+      "I_01=19.403\n",
+      "I_m1=4*V_s/(Z1*math.pi)\n",
+      "I_m3=4*V_s/(3*Z3*math.pi)\n",
+      "I_m5=4*V_s/(5*Z5*math.pi)\n",
+      "I_m7=4*V_s/(7*Z7*math.pi)\n",
+      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2+I_m7**2)\n",
+      "I_or=I_m/math.sqrt(2)\n",
+      "I_oh=math.sqrt((I_m**2-I_m1**2)/2)\n",
+      "THD=I_oh/I_01    \n",
+      "DF=I_01/I_or    \n",
+      "P_o=I_or**2*R    \n",
+      "I_avg=P_o/V_s    \n",
+      "t1=(180-phi1)*math.pi/(180*w)    \n",
+      "t1=1/(2*f)-t1    \n",
+      "I_p=I_m1    \n",
+      "I_t1=.46135*I_p \n",
+      "\n",
+      "#Results\n",
+      "print(\"\\nDF=%.1f\" %DF)\n",
+      "print(\"THD of current=%.4f\" %THD)   \n",
+      "print(\"DF=%.3f\" %DF)\n",
+      "print(\"load power=%.1f W\" %P_o)\n",
+      "print(\"avg value of load current=%.2f A\" %I_avg)\n",
+      "print(\"conduction time for thyristor=%.0f ms\" %(t1*1000))\n",
+      "print(\"conduction time for diodes=%.0f ms\" %(t1*1000))\n",
+      "print(\"peak transistor current=%.2f A\" %I_p)\n",
+      "print(\"rms transistor current=%.2f A\" %I_t1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "THD of voltage=0.4834\n",
+        "\n",
+        "DF=1.0\n",
+        "THD of current=0.1557\n",
+        "DF=0.988\n",
+        "load power=2313.5 W\n",
+        "avg value of load current=10.52 A\n",
+        "conduction time for thyristor=3 ms\n",
+        "conduction time for diodes=3 ms\n",
+        "peak transistor current=27.44 A\n",
+        "rms transistor current=12.66 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11 Page No 497"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=450.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_or=math.sqrt((V_s/(3*R))**2*2/3+(2*V_s/(3*R))**2*1/3)    \n",
+      "I_T1=math.sqrt((1/(2*math.pi))*((V_s/(3*R))**2*2*math.pi/3+(2*V_s/(3*R))**2*math.pi/3))   \n",
+      "P=3*I_or**2*R    \n",
+      "I_or=math.sqrt((1/(math.pi))*((V_s/(2*R))**2*2*math.pi/3))    \n",
+      "I_T1=math.sqrt((1/(2*math.pi))*((V_s/(2*R))**2*2*math.pi/3))    \n",
+      "P=3*I_or**2*R    \n",
+      "\n",
+      "#Results\n",
+      "print(\"for 180deg mode\")\n",
+      "print(\"rms value of load current=%.3f A\" %I_or)\n",
+      "print(\"power delivered to load=%.1f kW\" %(P/1000))\n",
+      "print(\"rms value of load current=%.0f A\" %I_T1)\n",
+      "print(\"for 120deg mode\")\n",
+      "print(\"rms value of load current=%.3f A\" %I_or)\n",
+      "print(\"rms value of load current=%.2f A\" %I_T1)\n",
+      "print(\"power delivered to load=%.3f kW\" %(P/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for 180deg mode\n",
+        "rms value of load current=18.371 A\n",
+        "power delivered to load=10.1 kW\n",
+        "rms value of load current=13 A\n",
+        "for 120deg mode\n",
+        "rms value of load current=18.371 A\n",
+        "rms value of load current=12.99 A\n",
+        "power delivered to load=10.125 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12, Page No 510"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "R=10.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=0.03\n",
+      "\n",
+      "#Calculations\n",
+      "X_L=w*L\n",
+      "V_or=math.sqrt(V_s**2*math.pi/math.pi)\n",
+      "V_01=4*V_s/(math.sqrt(2)*math.pi)\n",
+      "Z1=math.sqrt(R**2+(X_L)**2)\n",
+      "phi1=-math.degrees(math.atan((X_L)/R))\n",
+      "Z3=math.sqrt(R**2+(X_L*3)**2)\n",
+      "phi3=math.degrees(math.atan((X_L*3)/R))\n",
+      "Z5=math.sqrt(R**2+(X_L*5)**2)\n",
+      "phi5=math.degrees(math.atan((X_L*5)/R))\n",
+      "Z7=math.sqrt(R**2+(X_L*7)**2)\n",
+      "phi7=math.degrees(math.atan((X_L*7)/R))\n",
+      "I_m1=4*V_s/(math.sqrt(2)*Z1*math.pi)\n",
+      "I_m3=4*V_s/(math.sqrt(2)*3*Z3*math.pi)\n",
+      "I_m5=4*V_s/(math.sqrt(2)*5*Z5*math.pi)\n",
+      "I_m7=4*V_s/(math.sqrt(2)*7*Z7*math.pi)\n",
+      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2+I_m7**2)\n",
+      "P=I_m**2*R    \n",
+      "I_01=I_m1*math.sin(math.radians(45))\n",
+      "I_03=I_m3*math.sin(math.radians(3*45))\n",
+      "I_05=I_m5*math.sin(math.radians(5*45))\n",
+      "I_07=I_m7*math.sin(math.radians(7*45))\n",
+      "I_0=(I_01**2+I_03**2+I_05**2+I_07**2)\n",
+      "P=I_0*R    \n",
+      "g=(180-90)/3+45/2\n",
+      "I_01=2*I_m1*math.sin(math.radians(g))*math.sin(math.radians(45/2))\n",
+      "I_03=2*I_m3*math.sin(math.radians(g*3))*math.sin(math.radians(3*45/2))\n",
+      "I_05=2*I_m5*math.sin(math.radians(g*5))*math.sin(math.radians(5*45/2))\n",
+      "I_07=2*I_m7*math.sin(math.radians(g*7))*math.sin(math.radians(7*45/2))\n",
+      "I_0=(I_01**2+I_03**2+I_05**2+I_07**2)\n",
+      "P=I_0*R    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"using square wave o/p\")\n",
+      "print(\"power delivered=%.2f W\" %P)\n",
+      "print(\"using quasi-square wave o/p\")\n",
+      "print(\"power delivered=%.2f W\" %P)\n",
+      "print(\"using two symmitrical spaced pulses\")\n",
+      "print(\"power delivered=%.2f W\" %P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "using square wave o/p\n",
+        "power delivered=845.87 W\n",
+        "using quasi-square wave o/p\n",
+        "power delivered=845.87 W\n",
+        "using two symmitrical spaced pulses\n",
+        "power delivered=845.87 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14, Page No 520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50.0\n",
+      "T=1/f\n",
+      "I=0.5\n",
+      "\n",
+      "#Calculations\n",
+      "di=I/T     #di=di/dt\n",
+      "V_s=220.0\n",
+      "L=V_s/di    \n",
+      "t=20*10**-6\n",
+      "fos=2     #factor of safety\n",
+      "t_c=t*fos\n",
+      "R=10\n",
+      "C=t_c/(R*math.log(2))\n",
+      "\n",
+      "#Results    \n",
+      "print(\"source inductance=%.1f H\" %L)\n",
+      "print(\"commutating capacitor=%.2f uF\" %(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "source inductance=8.8 H\n",
+        "commutating capacitor=5.77 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.15, Page No 539"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=10.0\n",
+      "L=.01\n",
+      "C=10*10**-6\n",
+      "#Calculations\n",
+      "if (R**2)<(4*L/C) :\n",
+      "    print(\"ckt will commutate on its own\")\n",
+      "else:\n",
+      "    print(\"ckt will not commutate on its own\")\n",
+      "\n",
+      "xie=R/(2*L)\n",
+      "w_o=1/math.sqrt(L*C)\n",
+      "w_r=math.sqrt(w_o**2-xie**2)\n",
+      "phi=math.degrees(math.atan(xie/w_r))\n",
+      "t=math.pi/w_r\n",
+      "V_s=1\n",
+      "v_L=V_s*(w_o/w_r)*math.exp(-xie*t)*math.cos(math.radians(180+phi))\n",
+      "v_c=V_s*(1-(w_o/w_r)*math.exp(-xie*t)*math.cos(math.radians(180-phi)))   \n",
+      "di=V_s/L    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"voltage across inductor(*V_s)=%.5f V\" %v_L) \n",
+      "print(\"voltage across capacitor(*V_s)=%.5f V\" %v_c)\n",
+      "print(\"di/dt*V_s (for t=0)=%.0f A/s\" %di)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ckt will commutate on its own\n",
+        "voltage across inductor(*V_s)=-0.60468 V\n",
+        "voltage across capacitor(*V_s)=1.60468 V\n",
+        "di/dt*V_s (for t=0)=100 A/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.16, Page No 540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=0.006\n",
+      "C=1.2*10**-6\n",
+      "R=100.0\n",
+      "\n",
+      "#Calculations\n",
+      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
+      "T_off=0.2*10**-3\n",
+      "f=1/(2*(T+T_off))    \n",
+      "R=40\n",
+      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
+      "T_off=.2*10**-3\n",
+      "f=1/(2*(T+T_off))    \n",
+      "R=140\n",
+      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
+      "T_off=.2*10**-3\n",
+      "f=1/(2*(T+T_off)) \n",
+      "\n",
+      "#Results\n",
+      "print(\"o/p freq=%.2f Hz\" %f)\n",
+      "print(\"for R=40ohm\")\n",
+      "print(\"upper limit o/p freq=%.1f Hz\" %f)\n",
+      "print(\"for R=140ohm\")\n",
+      "print(\"lower limit o/p freq=%.1f Hz\" %f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "o/p freq=239.81 Hz\n",
+        "for R=40ohm\n",
+        "upper limit o/p freq=239.8 Hz\n",
+        "for R=140ohm\n",
+        "lower limit o/p freq=239.8 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.17, Page No 540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=5000.0\n",
+      "w=2*math.pi*f\n",
+      "R=3.0\n",
+      "\n",
+      "#Calculations\n",
+      "L=60*10**-6\n",
+      "xie=R/(2*L)\n",
+      "C=7.5*10**-6\n",
+      "w_o=1/math.sqrt(L*C)\n",
+      "w_r=math.sqrt(w_o**2-xie**2)\n",
+      "t_c=math.pi*(1/w-1/w_r)    \n",
+      "fos=1.5\n",
+      "t_q=10*10**-6\n",
+      "f_max=1/(2*math.pi*(t_q*fos/math.pi+1/w_r))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n",
+      "print(\"max possible operating freq=%.1f Hz\" %f_max)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ckt turn off time=21.39 us\n",
+        "max possible operating freq=5341.4 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.18, Page No 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*2*math.pi/(2*3)/(math.pi/3+math.sqrt(3)*math.cos(math.radians(2*a))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*math.pi/(math.sqrt(2)*3*math.cos(math.radians(a)))\n",
+      "V_ph=V_s/math.sqrt(3) \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage  V_ph=%.3f V\" %V_ph)   \n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.2f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage  V_ph=110.384 V\n",
+        "for constant load current\n",
+        "V_ph=110.38 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.19, Page No 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "t=20.0\n",
+      "fos=2.0     #factor of safety\n",
+      "\n",
+      "#Calculations\n",
+      "t_c=t*fos\n",
+      "n=1.0/3\n",
+      "R=20.0\n",
+      "C=n**2*t_c/(4*R*math.log(2)) \n",
+      "\n",
+      "#Results   \n",
+      "print(\"value of capacitor=%.2f uF\" %C)\n",
+      " #printing mistake in the answer in book."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of capacitor=0.08 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.20, Page No 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=220.0\n",
+      "V_p=math.sqrt(2)*V_s/3    \n",
+      "V_L=math.sqrt(3)*V_p    \n",
+      "V_p1=math.sqrt(2)*V_s/math.pi    \n",
+      "V_L1=math.sqrt(3)*V_p1    \n",
+      "V_oh=math.sqrt(V_L**2-V_L1**2)\n",
+      "\n",
+      "#Calculations\n",
+      "THD=V_oh/V_L1    \n",
+      "V_a1=2*V_s/math.pi\n",
+      "V_a5=2*V_s/(5*math.pi)\n",
+      "V_a7=2*V_s/(7*math.pi)\n",
+      "V_a11=2*V_s/(11*math.pi)\n",
+      "R=4.0\n",
+      "L=0.02\n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "Z1=math.sqrt(R**2+(w*L)**2)\n",
+      "Z5=math.sqrt(R**2+(5*w*L)**2)\n",
+      "Z7=math.sqrt(R**2+(7*w*L)**2)\n",
+      "Z11=math.sqrt(R**2+(11*w*L)**2)\n",
+      "I_a1=V_a1/Z1\n",
+      "I_a5=V_a5/Z5\n",
+      "I_a7=V_a7/Z7\n",
+      "I_a11=V_a11/Z11\n",
+      "I_or=math.sqrt((I_a1**2+I_a5**2+I_a7**2+I_a11**2)/2)\n",
+      "P=3*I_or**2*R    \n",
+      "I_s=P/V_s    \n",
+      "I_TA=I_s/3   \n",
+      " \n",
+      "#Results\n",
+      "print(\"rms value of phasor voltages=%.2f V\" %V_p)\n",
+      "print(\"rms value of line voltages=%.2f V\" %V_L)\n",
+      "print(\"fundamental component of phase voltage=%.3f V\" %V_p1)\n",
+      "print(\"fundamental component of line voltages=%.3f V\" %V_L1)\n",
+      "print(\"THD=%.7f\" %THD)\n",
+      "print(\"load power=%.1f W\" %P)\n",
+      "print(\"avg value of source current=%.3f A\" %I_s)\n",
+      "print(\"avg value of thyristor current=%.3f A\" %I_TA)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of phasor voltages=103.71 V\n",
+        "rms value of line voltages=179.63 V\n",
+        "fundamental component of phase voltage=99.035 V\n",
+        "fundamental component of line voltages=171.533 V\n",
+        "THD=0.3108419\n",
+        "load power=2127.6 W\n",
+        "avg value of source current=9.671 A\n",
+        "avg value of thyristor current=3.224 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter9.ipynb b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter9.ipynb
new file mode 100755
index 00000000..052c4736
--- /dev/null
+++ b/Power_Electronics_by_Dr_B_R_Gupta_and_V_Singhal/temp/chapter9.ipynb
@@ -0,0 +1,388 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 09 : AC Voltage Controllers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page No 560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=(V_m/2)*math.sqrt(1/math.pi*((2*math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
+      "R=20\n",
+      "I_or=V_or/R\n",
+      "P_o=I_or**2*R    \n",
+      "I_s=I_or\n",
+      "VA=V_s*I_s\n",
+      "pf=P_o/VA    \n",
+      "V_o=math.sqrt(2)*V_s/(2*math.pi)*(math.cos(math.radians(a))-1)\n",
+      "I_ON=V_o/R  \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.3f V\" %V_or)\n",
+      "print(\"load power=%.1f W\" %P_o)\n",
+      "print(\"i/p pf=%.4f\" %pf)\n",
+      "print(\"avg i/p current=%.4f A\" %I_ON)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=224.716 V\n",
+        "load power=2524.9 W\n",
+        "i/p pf=0.9770\n",
+        "avg i/p current=-0.7581 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page No 560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=(V_s)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
+      "R=20\n",
+      "I_or=V_or/R\n",
+      "P_o=I_or**2*R    \n",
+      "I_s=I_or\n",
+      "VA=V_s*I_s\n",
+      "pf=P_o/VA    \n",
+      "I_TA=math.sqrt(2)*V_s/(2*math.pi*R)*(math.cos(math.radians(a))+1)   \n",
+      "I_Tr=math.sqrt(2)*V_s/(2*R)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2)) \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.3f V\" %V_or)\n",
+      "print(\"load power=%.2f W\" %P_o)\n",
+      "print(\"i/p pf=%.2f\" %pf)\n",
+      "print(\"avg thyristor current=%.2f A\" %I_TA)   \n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_Tr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=219.304 V\n",
+        "load power=2404.71 W\n",
+        "i/p pf=0.95\n",
+        "avg thyristor current=4.42 A\n",
+        "rms value of thyristor current=7.75 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page No 564"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "n=6.0 #on cycles\n",
+      "m=4.0 #off cycles\n",
+      "\n",
+      "#Calculations\n",
+      "k=n/(n+m)\n",
+      "V_or=V_s*math.sqrt(k)    \n",
+      "pf=math.sqrt(k)    \n",
+      "R=15\n",
+      "I_m=V_s*math.sqrt(2)/R\n",
+      "I_TA=k*I_m/math.pi\n",
+      "I_TR=I_m*math.sqrt(k)/2  \n",
+      "   \n",
+      "#Results\n",
+      "print(\"rms value of o/ voltage=%.2f V\" %V_or)\n",
+      "print(\"i/p pf=%.2f\" %pf)\n",
+      "print(\"avg thyristor current=%.2f A\" %I_TA)  \n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_TR)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/ voltage=178.16 V\n",
+        "i/p pf=0.77\n",
+        "avg thyristor current=4.14 A\n",
+        "rms value of thyristor current=8.40 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page No 569"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "R=3.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_TAM1=2*V_m/(2*math.pi*R)  \n",
+      "I_TRM2=V_m/(2*R)    \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "t_c=math.pi/w    \n",
+      " \n",
+      "#Results\n",
+      "print(\"max value of avg thyristor current=%.3f A\" %I_TAM1)\n",
+      "print(\"max value of avg thyristor current=%.3f A\" %I_TRM2)\n",
+      "print(\"ckt turn off time=%.0f ms\" %(t_c*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "max value of avg thyristor current=34.512 A\n",
+        "max value of avg thyristor current=54.212 A\n",
+        "ckt turn off time=10 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page No 575"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=3.0\n",
+      "X_L=4.0\n",
+      "\n",
+      "#Calculations\n",
+      "phi=math.degrees(math.atan(X_L/R))    \n",
+      "V_s=230\n",
+      "Z=math.sqrt(R**2+X_L**2)\n",
+      "I_or=V_s/Z    \n",
+      "P=I_or**2*R    \n",
+      "I_s=I_or\n",
+      "pf=P/(V_s*I_s)    \n",
+      "I_TAM=math.sqrt(2)*V_s/(math.pi*Z)    \n",
+      "I_Tm=math.sqrt(2)*V_s/(2*Z)    \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "di=math.sqrt(2)*V_s*w/Z    \n",
+      "\n",
+      "#Results\n",
+      "print(\"min firing angle=%.2f deg\" %phi)\n",
+      "print(\"\\nmax firing angle=%.0f deg\" %180)\n",
+      "print(\"i/p pf=%.1f\" %pf)\n",
+      "print(\"max value of rms load current=%.0f A\" %I_or)\n",
+      "print(\"max power=%.0f W\" %P)\n",
+      "print(\"max value of avg thyristor current=%.3f A\" %I_TAM)\n",
+      "print(\"max value of rms thyristor current=%.3f A\" %I_Tm)\n",
+      "print(\"di/dt=%.0f A/s\" %di)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "min firing angle=53.13 deg\n",
+        "\n",
+        "max firing angle=180 deg\n",
+        "i/p pf=0.6\n",
+        "max value of rms load current=46 A\n",
+        "max power=6348 W\n",
+        "max value of avg thyristor current=20.707 A\n",
+        "max value of rms thyristor current=32.527 A\n",
+        "di/dt=20437 A/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 Page No 576"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "R=3.0 #ohm\n",
+      "X_L=5.0 #ohm\n",
+      "a=120.0 #firing angle delay\n",
+      "\n",
+      "#Calculations\n",
+      "phi=math.degrees(math.atan(X_L/R))\n",
+      "b=0\n",
+      "i=1\n",
+      "while (i>0) :\n",
+      "    LHS=math.sin(math.radians(b-a))\n",
+      "    RHS=math.sin(math.radians(a-phi))*math.exp(-(R/X_L)*(b-a)*math.pi/180)\n",
+      "    if math.fabs(LHS-RHS)<= 0.01 :\n",
+      "        B=b\n",
+      "        i=2\n",
+      "        break\n",
+      "   \n",
+      "    b=b+.1   \n",
+      "V_or=math.sqrt(2)*V*math.sqrt((1/(2*math.pi))*((B-a)*math.pi/180+(math.sin(math.radians(2*a))-math.sin(math.radians(2*B)))/2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"Extinction angle=%.1f deg\" %B) #answer in the book is wrong as formulae for RHS is wrongly employed\n",
+      "print(\"rms value of output voltage=%.2f V\" %V_or) #answer do not match due to wrong B in book\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Extinction angle=156.1 deg\n",
+        "rms value of output voltage=97.75 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8, Page No 581"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=60.0\n",
+      "R=20.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt((V_m**2/(2*math.pi))*(a*math.pi/180-math.sin(math.radians(2*a))/2)+(2*V_m**2/(math.pi))*(math.pi-a*math.pi/180+math.sin(math.radians(2*a))/2))    \n",
+      "I_T1r=(V_m/R)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
+      "I_T3r=(V_m/(2*R))*math.sqrt(1/math.pi*((a*math.pi/180)-math.sin(math.radians(2*a))/2))    \n",
+      "I1=math.sqrt(2)*I_T1r\n",
+      "I3=math.sqrt((math.sqrt(2)*I_T1r)**2+(math.sqrt(2)*I_T3r)**2)\n",
+      "r=V_s*(I1+I3)    \n",
+      "P_o=V_or**2/R\n",
+      "pf=P_o/r    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
+      "print(\"rms value of current for upper thyristors=%.2f A\" %I_T1r)\n",
+      "print(\"rms value of current for lower thyristors=%.2f A\" %I_T3r)\n",
+      "print(\"t/f VA rating=%.2f VA\" %r)\n",
+      "print(\"i/p pf=%.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=424.94 V\n",
+        "rms value of current for upper thyristors=14.59 A\n",
+        "rms value of current for lower thyristors=3.60 A\n",
+        "t/f VA rating=9631.61 VA\n",
+        "i/p pf=0.94\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter10_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter10_4.ipynb
new file mode 100755
index 00000000..cbc3cb90
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter10_4.ipynb
@@ -0,0 +1,228 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Cycloconverters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2, Page No 594"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "R=10.0\n",
+      "a=30.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=(V_m/math.sqrt(2))*math.sqrt((1/math.pi)*(math.pi-a*math.pi/180+math.sin(math.radians(2*a))/2))\n",
+      "I_or=V_or/R    \n",
+      "I_s=I_or\n",
+      "pf=(I_or**2*R)/(V_s*I_s)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p current=%.2f A\" %I_or)\n",
+      "print(\"rms value of o/p current for each convertor=%.2f A\" %(I_or/math.sqrt(2)))\n",
+      "print(\"rms value of o/p current for each thyristor=%.3f A\" %(I_or/2))\n",
+      "print(\"i/p pf=%.4f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p current=22.67 A\n",
+        "rms value of o/p current for each convertor=16.03 A\n",
+        "rms value of o/p current for each thyristor=11.333 A\n",
+        "i/p pf=0.9855\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4, Page No 604"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_ph=V_s/2\n",
+      "a=160.0\n",
+      "\n",
+      "#Calculations\n",
+      "r=math.cos(math.radians(180-a))\n",
+      "m=3\n",
+      "V_or=r*(V_ph*(m/math.pi)*math.sin(math.pi/m)) \n",
+      "R=2\n",
+      "X_L=1.5\n",
+      "th=math.degrees(math.atan(X_L/R))\n",
+      "Z=math.sqrt(R**2+X_L**2)\n",
+      "I_or=V_or/Z    \n",
+      "P=I_or**2*R      \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms o/p voltage=%.3f V\" %V_or)\n",
+      "print(\"rms o/p current=%.2f A\" %I_or)\n",
+      "print(\"phase angle of o/p current=%.2f deg\" %-th)\n",
+      "print(\"o/p power=%.2f W\" %P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms o/p voltage=155.424 V\n",
+        "rms o/p current=62.17 A\n",
+        "phase angle of o/p current=-36.87 deg\n",
+        "o/p power=7730.11 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 Page No 604"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_ph=V_s/2\n",
+      "V_l=V_ph*math.sqrt(3)\n",
+      "a=160.0\n",
+      "\n",
+      "#Calculations\n",
+      "r=math.cos(math.radians(180-a))\n",
+      "m=6\n",
+      "V_or=r*(V_l*(m/math.pi)*math.sin(math.pi/m))    \n",
+      "R=2\n",
+      "X_L=1.5\n",
+      "th=math.degrees(math.atan(X_L/R))\n",
+      "Z=math.sqrt(R**2+X_L**2)\n",
+      "I_or=V_or/Z    \n",
+      "P=I_or**2*R    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms o/p voltage=%.2f V\" %V_or)\n",
+      "print(\"rms o/p current=%.2f A\" %I_or)\n",
+      "print(\"phase angle of o/p current=%.2f deg\" %-th)\n",
+      "print(\"o/p power=%.2f W\" %P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms o/p voltage=310.85 V\n",
+        "rms o/p current=124.34 A\n",
+        "phase angle of o/p current=-36.87 deg\n",
+        "o/p power=30920.44 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7, Page No 605"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_l=400.0\n",
+      "V_ml=math.sqrt(2)*V_l\n",
+      "m=6\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=.0012\n",
+      "I=40.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or1=(V_ml*(m/math.pi)*math.sin(math.pi/m))*math.cos(math.radians(a))\n",
+      "V_omx1=V_or1-3*w*L*I/math.pi\n",
+      "V_rms1=V_omx1/math.sqrt(2)    \n",
+      "a2=30.0\n",
+      "V_or2=(V_ml*(m/math.pi)*math.sin(math.pi/m))*math.cos(math.radians(a))\n",
+      "V_omx2=V_or2-3*w*L*I/math.pi\n",
+      "V_rms2=V_omx2/math.sqrt(2)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle=0deg\")\n",
+      "a1=0\n",
+      "print(\"rms value of load voltage=%.2f V\" %V_rms2)\n",
+      "print(\"for firing angle=30deg\")\n",
+      "print(\"rms value of load voltage=%.2f V\" %V_rms2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=0deg\n",
+        "rms value of load voltage=-369.12 V\n",
+        "for firing angle=30deg\n",
+        "rms value of load voltage=-369.12 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter11_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter11_4.ipynb
new file mode 100755
index 00000000..d2317d28
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter11_4.ipynb
@@ -0,0 +1,299 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Some Applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1, Page No 622"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=11000.0\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "w=2*math.pi*f\n",
+      "I_d=300\n",
+      "R_d=1\n",
+      "g=20 #g=gamma\n",
+      "a=math.degrees(math.acos(math.cos(math.radians(g))+math.pi/(3*V_ml)*I_d*R_d))    \n",
+      "L_s=.01\n",
+      "V_d=(3/math.pi)*((V_ml*math.cos(math.radians(a)))-w*L_s*I_d)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"rectifier o/p voltage=%.1f V\" %V_d)\n",
+      "print(\"dc link voltage=%.3f V\" %(2*V_d/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle=16.283 deg\n",
+        "rectifier o/p voltage=13359.3 V\n",
+        "dc link voltage=26.719 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page No 623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_d=(200.0+200)*10**3\n",
+      "P=1000.0*10**6\n",
+      "\n",
+      "#Calculations\n",
+      "I_d=P/V_d\n",
+      " #each thristor conducts for 120deg for a periodicity of 360deg\n",
+      "a=0\n",
+      "V_d=200.0*10**3\n",
+      "V_ml=V_d*math.pi/(3*math.cos(math.radians(a)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms current rating of thyristor=%.2f A\" %(I_d*math.sqrt(120/360)))\n",
+      "print(\"peak reverse voltage across each thyristor=%.2f kV\" %(V_ml/2/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms current rating of thyristor=0.00 A\n",
+        "peak reverse voltage across each thyristor=104.72 kV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 Page No 627"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_m=230.0\n",
+      "V_s=230/math.sqrt(2)\n",
+      "pf=0.8\n",
+      "P=2000.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_m=P/(V_s*pf)\n",
+      "I_Tr=I_m/math.sqrt(2)\n",
+      "I_TA=2*I_m/math.pi\n",
+      "fos=2 #factor of safety\n",
+      "PIV=V_m*math.sqrt(2)\n",
+      "I_Tr=I_m/(2)\n",
+      "I_TA=I_m/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of thyristor current=%.2f A\" %(fos*I_Tr))\n",
+      "print(\"avg value of thyristor current=%.3f A\" %(fos*I_TA))\n",
+      "print(\"voltage rating of thyristor=%.2f V\" %PIV)\n",
+      "print(\"rms value of diode current=%.3f A\" %(fos*I_Tr))\n",
+      "print(\"avg value of diode current=%.3f A\" %(fos*I_TA))\n",
+      "print(\"voltage rating of diode=%.2f V\" %PIV)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of thyristor current=15.37 A\n",
+        "avg value of thyristor current=9.786 A\n",
+        "voltage rating of thyristor=325.27 V\n",
+        "rms value of diode current=15.372 A\n",
+        "avg value of diode current=9.786 A\n",
+        "voltage rating of diode=325.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4, Page No 629"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=200.0\n",
+      "I=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "R_L=V/I    \n",
+      "I_h=.005     #holding current\n",
+      "R2=V/I_h    \n",
+      "t_c=20*10**-6\n",
+      "fos=2 #factor of safety\n",
+      "C=t_c*fos/(R_L*math.log(2))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of load resistance=%.0f ohm\" %R_L)\n",
+      "print(\"value of R2=%.0f kilo-ohm\" %(R2/1000))\n",
+      "print(\"value of C=%.3f uF\" %(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of load resistance=20 ohm\n",
+        "value of R2=40 kilo-ohm\n",
+        "value of C=2.885 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 Page No 646"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "u_r=10\n",
+      "f=10000.0 #Hz\n",
+      "p=4.0*10**-8 #ohm-m\n",
+      "\n",
+      "#Calculations\n",
+      "dl=(1/(2*math.pi))*math.sqrt(p*10**7/(u_r*f))    \n",
+      "l=0.12 #length of cylinder\n",
+      "t=20.0 #no of turns\n",
+      "I=100.0\n",
+      "H=t*I/l\n",
+      "P_s=2*math.pi*H**2*math.sqrt(u_r*f*p*10**-7)    \n",
+      "d=.02 #diameter\n",
+      "P_v=4*H**2*p/(d*dl)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"depth of heat of penetration=%.5f mm\" %(dl*1000))\n",
+      "print(\"heat generated per unit cylinder surface area=%.3f W/m**2\" %P_s)\n",
+      "print(\"heat generated per unit cylinder volume=%.0f W/m**3\" %P_v)\n",
+      " #answer of P_v varies as given in book as value of d is not taken as in formulae. "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of heat of penetration=0.31831 mm\n",
+        "heat generated per unit cylinder surface area=34906.585 W/m**2\n",
+        "heat generated per unit cylinder volume=6981317 W/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 Page No 646"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=3000.0\n",
+      "\n",
+      "#Calculations\n",
+      "t_qmin=30.0*10**-6\n",
+      "f_r=f/(1-2*t_qmin*f)\n",
+      "R=0.06\n",
+      "L=20.0*10**-6\n",
+      "C=1/(L*((2*math.pi*f_r)**2+(R/(2*L))**2))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"required capacitor size=%.4f F\" %(C*10**6))\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "required capacitor size=94.2215 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter12_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter12_4.ipynb
new file mode 100755
index 00000000..f8605d69
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter12_4.ipynb
@@ -0,0 +1,1997 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Electic Drives"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1, Page No 658"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_e=15.0 #Nm\n",
+      "K_m=0.5 #V-s/rad\n",
+      "I_a=T_e/K_m\n",
+      "n_m=1000.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "E_a=K_m*w_m\n",
+      "r_a=0.7\n",
+      "V_t=E_a+I_a*r_a\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=math.degrees(math.acos(2*math.pi*V_t/V_m-1))\n",
+      "print(\"firing angle delay=%.3f deg\" %a)\n",
+      "I_Tr=I_a*math.sqrt((180-a)/360)    \n",
+      "print(\"rms value of thyristor current=%.3f A\" %I_Tr)\n",
+      "I_fdr=I_a*math.sqrt((180+a)/360)    \n",
+      "print(\"rms value of freewheeling diode current=%.3f A\" %I_fdr)\n",
+      "pf=V_t*I_a/(V_s*I_Tr)  \n",
+      "\n",
+      "#Results  \n",
+      "print(\"input power factor=%.4f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=65.349 deg\n",
+        "rms value of thyristor current=16.930 A\n",
+        "rms value of freewheeling diode current=24.766 A\n",
+        "input power factor=0.5652\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2, Page No 660"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "E=150.0\n",
+      "R=8.0\n",
+      "\n",
+      "#Calculations\n",
+      "th1=math.sin(math.radians(E/(math.sqrt(2)*V)))\n",
+      "I_o=(1/(2*math.pi*R))*(2*math.sqrt(2)*230*math.cos(math.radians(th1))-E*(math.pi-2*th1*math.pi/180))    \n",
+      "P=E*I_o    \n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*(math.pi-2*th1*math.pi/180)+V**2*math.sin(math.radians(2*th1))-4*math.sqrt(2)*V*E*math.cos(math.radians(th1))))\n",
+      "P_r=I_or**2*R    \n",
+      "pf=(P+P_r)/(V*I_or)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg charging curent=%.4f A\" %I_o)\n",
+      "print(\"power supplied to the battery=%.2f W\" %P)\n",
+      "print(\"power dissipated by the resistor=%.3f W\" %P_r)    \n",
+      "print(\"supply pf=%.3f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg charging curent=3.5679 A\n",
+        "power supplied to the battery=535.18 W\n",
+        "power dissipated by the resistor=829.760 W\n",
+        "supply pf=0.583\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 Page No 661"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=250\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=30.0\n",
+      "k=0.03 #Nm/A**2\n",
+      "n_m=1000.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "r=.2 #r_a+r_s\n",
+      "V_t=V_m/math.pi*(1+math.cos(math.radians(a)))\n",
+      "I_a=V_t/(k*w_m+r)   \n",
+      "print(\"motor armature current=%.2f A\" %I_a)\n",
+      "T_e=k*I_a**2    \n",
+      "print(\"motor torque=%.3f Nm\" %T_e)\n",
+      "I_sr=I_a*math.sqrt((180-a)/180)\n",
+      "pf=(V_t*I_a)/(V_s*I_sr)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"input power factor=%.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor armature current=57.82 A\n",
+        "motor torque=100.285 Nm\n",
+        "input power factor=0.92\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4, Page No 663"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_f=2*V_m/math.pi\n",
+      "r_f=200.0\n",
+      "I_f=V_f/r_f\n",
+      "T_e=85.0\n",
+      "K_a=0.8\n",
+      "\n",
+      "#Calculations\n",
+      "I_a=T_e/(I_f*K_a)    \n",
+      "print(\"rated armature current=%.2f A\" %I_a)\n",
+      "n_m=1200.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "r_a=0.2\n",
+      "V_t=K_a*I_f*w_m+I_a*r_a\n",
+      "a=math.degrees(math.acos(V_t*math.pi/(2*V_m)))\n",
+      "print(\"firing angle delay=%.2f deg\" %a)\n",
+      "E_a=V_t\n",
+      "w_mo=E_a/(K_a*I_f)\n",
+      "N=60*w_mo/(2*math.pi)\n",
+      "reg=((N-n_m)/n_m)*100    \n",
+      "print(\"speed regulation at full load=%.2f\" %reg)\n",
+      "I_ar=I_a\n",
+      "pf=(V_t*I_a)/(V_s*I_ar)    \n",
+      "print(\"input power factor of armature convertor=%.4f\" %pf)\n",
+      "I_fr=I_f\n",
+      "I_sr=math.sqrt(I_fr**2+I_ar**2)\n",
+      "VA=I_sr*V_s\n",
+      "P=V_t*I_a+V_f*I_f\n",
+      "\n",
+      "#Results\n",
+      "print(\"input power factor of drive=%.4f\" %(P/VA))\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rated armature current=59.01 A\n",
+        "firing angle delay=57.63 deg\n",
+        "speed regulation at full load=6.52\n",
+        "input power factor of armature convertor=0.4821\n",
+        "input power factor of drive=0.5093\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 Page No 664"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_f=2*V_m/math.pi\n",
+      "\n",
+      "#Calculations\n",
+      "a1=math.degrees(math.acos(V_t*math.pi/(2*V_m))) \n",
+      "print(\"delay angle of field converter=%.0f deg\" %a1)\n",
+      "r_f=200.0\n",
+      "I_f=V_f/r_f\n",
+      "T_e=85.0\n",
+      "K_a=0.8\n",
+      "I_a=T_e/(I_f*K_a)\n",
+      "n_m=1200.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "r_a=0.1\n",
+      "I_a=50.0\n",
+      "V_t=-K_a*I_f*w_m+I_a*r_a\n",
+      "a=math.degrees(math.acos(V_t*math.pi/(2*V_m)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay of armature converter=%.3f deg\" %a)\n",
+      "print(\"power fed back to ac supply=%.0f W\" %(-V_t*I_a))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "delay angle of field converter=58 deg\n",
+        "firing angle delay of armature converter=119.260 deg\n",
+        "power fed back to ac supply=8801 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6 Page No 665"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=220.0\n",
+      "n_m=1500.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=10.0\n",
+      "r_a=1.0\n",
+      "\n",
+      "#Calculations\n",
+      "K_m=(V_t-I_a*r_a)/(w_m)\n",
+      "T=5.0\n",
+      "I_a=T/K_m\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=30.0\n",
+      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
+      "w_m=(V_t-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "print(\"motor speed=%.2f rpm\" %N)\n",
+      "a=45\n",
+      "n_m=1000\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
+      "I_a=(V_t-K_m*w_m)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"torque developed=%.3f Nm\" %T_e)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor speed=1254.22 rpm\n",
+        "torque developed=8.586 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7, Page No 666"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=220.0\n",
+      "n_m=1000.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=60.0\n",
+      "r_a=.1\n",
+      "\n",
+      "#Calculations\n",
+      "K_m=(V_t-I_a*r_a)/(w_m)\n",
+      "V_s=230\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "print(\"for 600rpm speed\")\n",
+      "n_m=600.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "a=math.degrees(math.acos((K_m*w_m+I_a*r_a)*math.pi/(2*V_m))) \n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"for -500rpm speed\")\n",
+      "n_m=-500.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "a=math.degrees(math.acos((K_m*w_m+I_a*r_a)*math.pi/(2*V_m)))\n",
+      "print(\"firing angle=%.2f deg\" %a)\n",
+      "I_a=I_a/2\n",
+      "a=150\n",
+      "V_t=2*V_m*math.cos(math.radians(a))/math.pi\n",
+      "w_m=(V_t-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor speed=%.3f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for 600rpm speed\n",
+        "firing angle=49.530 deg\n",
+        "for -500rpm speed\n",
+        "firing angle=119.19 deg\n",
+        "motor speed=-852.011 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8 Page No 672"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.5\n",
+      "T_e=50.0\n",
+      "I_a=T_e/K_m\n",
+      "r_a=0.9\n",
+      "a=45.0\n",
+      "V_s=415.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "w_m=((3*V_ml*(1+math.cos(math.radians(a)))/(2*math.pi))-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor speed=%.2f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor speed=2854.42 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9 Page No 672"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_t=600\n",
+      "n_m=1500.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=80.0\n",
+      "r_a=1.0\n",
+      "\n",
+      "#Calculations\n",
+      "K_m=(V_t-I_a*r_a)/(w_m)\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "print(\"for firing angle=45deg and speed=1200rpm\")\n",
+      "a=45.0\n",
+      "n_m=1200.0\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=(3*V_m*(1+math.cos(math.radians(a)))/(2*math.pi)-K_m*w_m)/r_a\n",
+      "I_sr=I_a*math.sqrt(2/3)    \n",
+      "print(\"rms value of source current=%.3f A\" %I_sr)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %(I_a*math.sqrt(1/3)))\n",
+      "print(\"avg value of thyristor current=%.2f A\" %I_a*(1/3))\n",
+      "pf=(3/(2*math.pi)*(1+math.cos(math.radians(a))))    \n",
+      "print(\"input power factor=%.3f\" %pf)\n",
+      "\n",
+      "print(\"for firing angle=90deg and speed=700rpm\")\n",
+      "a=90\n",
+      "n_m=700\n",
+      "w_m=2*math.pi*n_m/60\n",
+      "I_a=(3*V_m*(1+math.cos(math.radians(a)))/(2*math.pi)-K_m*w_m)/r_a\n",
+      "I_sr=I_a*math.sqrt(90/180)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of source current=%.3f A\" %I_sr)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %(I_a*math.sqrt(90.0/360)))\n",
+      "print(\"avg value of thyristor current=%.3f A\" %I_a*(1/3))\n",
+      "pf=(math.sqrt(6)/(2*math.pi)*(1+math.cos(math.radians(a))))*math.sqrt(180/(180-a))    \n",
+      "print(\"input power factor=%.4f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=45deg and speed=1200rpm\n",
+        "rms value of source current=0.000 A\n",
+        "rms value of thyristor current=0.000 A\n",
+        "\n",
+        "input power factor=0.815\n",
+        "for firing angle=90deg and speed=700rpm\n",
+        "rms value of source current=0.000 A\n",
+        "rms value of thyristor current=195.558 A\n",
+        "\n",
+        "input power factor=0.5513\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.10 Page No 676"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=30\n",
+      "V_t=3*V_m*math.cos(math.radians(a))/math.pi\n",
+      "I_a=21.0\n",
+      "r_a=.1\n",
+      "V_d=2.0\n",
+      "K_m=1.6\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=(V_t-I_a*r_a-V_d)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "print(\"speed of motor=%.1f rpm\" %N)\n",
+      "\n",
+      "N=2000\n",
+      "w_m=2*math.pi*N/60\n",
+      "I_a=210\n",
+      "V_t=K_m*w_m+I_a*r_a+V_d\n",
+      "a=math.degrees(math.acos(V_t*math.pi/(3*V_m)))\n",
+      "print(\"firing angle=%.2f deg\" %a)\n",
+      "I_sr=I_a*math.sqrt(2.0/3.0)\n",
+      "pf=V_t*I_a/(math.sqrt(3)*V_s*I_sr)    \n",
+      "print(\"supply power factor=%.3f\" %pf)\n",
+      "\n",
+      "I_a=21\n",
+      "w_m=(V_t-I_a*r_a-V_d)/K_m\n",
+      "n=w_m*60/(2*math.pi)\n",
+      "reg=(n-N)/N*100    \n",
+      "\n",
+      "#Results\n",
+      "print(\"speed regulation(percent)=%.2f\" %reg)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "speed of motor=2767.6 rpm\n",
+        "firing angle=48.48 deg\n",
+        "supply power factor=0.633\n",
+        "speed regulation(percent)=5.64\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.11, Page No 677"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=230.0\n",
+      "V_l=V_t*math.pi/(3*math.sqrt(2))\n",
+      "V_ph=V_l/math.sqrt(3)\n",
+      "V_in=400     #per phase voltage input\n",
+      "\n",
+      "#Calculations\n",
+      "N1=1500.0\n",
+      "I_a1=20.0\n",
+      "r_a1=.6\n",
+      "E_a1=V_t-I_a1*r_a1\n",
+      "n1=1000.0\n",
+      "E_a2=E_a1/1500.0*1000.0\n",
+      "V_t1=E_a1+I_a1*r_a1\n",
+      "a1=math.degrees(math.acos(V_t1*math.pi/(3*math.sqrt(2.0)*V_l)))\n",
+      "I_a2=.5*I_a1\n",
+      "n2=-900.0\n",
+      "V_t2=n2*E_a2/N1+I_a2*r_a1\n",
+      "a2=math.degrees(math.acos(V_t2*math.pi/(3*math.sqrt(2)*V_l))) \n",
+      "\n",
+      "#Results\n",
+      "print(\"transformer phase turns ratio=%.3f\" %(V_in/V_ph))\n",
+      "print(\"for motor running at 1000rpm at rated torque\")\n",
+      "print(\"firing angle delay=%.2f deg\" %a1)\n",
+      "print(\"for motor running at -900rpm at half of rated torque\")\n",
+      "print(\"firing angle delay=%.3f deg\" %a2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "transformer phase turns ratio=4.068\n",
+        "for motor running at 1000rpm at rated torque\n",
+        "firing angle delay=0.00 deg\n",
+        "for motor running at -900rpm at half of rated torque\n",
+        "firing angle delay=110.674 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.12, Page No 678"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=400\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "V_f=3*V_ml/math.pi\n",
+      "R_f=300.0\n",
+      "I_f=V_f/R_f\n",
+      "T_e=60.0\n",
+      "k=1.1\n",
+      "\n",
+      "#Calculations\n",
+      "I_a=T_e/(k*I_f)\n",
+      "N1=1000.0\n",
+      "w_m1=2*math.pi*N1/60\n",
+      "r_a1=.3\n",
+      "V_t1=k*I_f*w_m1+I_a*r_a1\n",
+      "a1=math.degrees(math.acos(V_f*math.pi/(3*V_ml)))\n",
+      "N2=3000\n",
+      "w_m2=2*math.pi*N/60\n",
+      "a2=0\n",
+      "V_t2=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "I_f2=(V_t2-I_a*r_a)/(w_m2*k)\n",
+      "V_f2=I_f2*R_f\n",
+      "a2=math.degrees(math.acos(V_f2*math.pi/(3*V_ml)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"firing angle=%.3f deg\" %a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle=48.477 deg\n",
+        "firing angle=48.477 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.13, Page No 679"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      " #after calculating\n",
+      " #t=w_m/6000-math.pi/360\n",
+      "\n",
+      "N=1000.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "t=w_m/6000-math.pi/360    \n",
+      "\n",
+      "#Results\n",
+      "print(\"time reqd=%.2f s\" %t)\n",
+      " #printing mistake in the answer in book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time reqd=0.01 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.14, Page No 679"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=1.0 #supposition\n",
+      "a=60.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_s1=2*math.sqrt(2)/math.pi*I_a*math.sin(math.radians(a))\n",
+      "I_s3=2*math.sqrt(2)/(3*math.pi)*I_a*math.sin(math.radians(3*a))\n",
+      "I_s5=2*math.sqrt(2)/(5*math.pi)*I_a*math.sin(math.radians(5*a))\n",
+      "per3=I_s3/I_s1*100    \n",
+      "print(\"percent of 3rd harmonic current in fundamental=%.2f\" %per3)\n",
+      "per5=I_s5/I_s1*100    \n",
+      "\n",
+      "#Results\n",
+      "print(\"percent of 5th harmonic current in fundamental=%.2f\" %per5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percent of 3rd harmonic current in fundamental=0.00\n",
+        "percent of 5th harmonic current in fundamental=-20.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.15, Page No 680"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=60.0\n",
+      "I_TA=I_a/3    \n",
+      "\n",
+      "#Calculations\n",
+      "print(\"avg thyristor current=%.0f A\" %I_TA)\n",
+      "I_Tr=I_a/math.sqrt(3)    \n",
+      "print(\"rms thyristor current=%.3f A\" %I_Tr)\n",
+      "V_s=400\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "I_sr=I_a*math.sqrt(2.0/3)\n",
+      "a=150\n",
+      "V_t=3*V_m*math.cos(math.radians(a))/math.pi\n",
+      "pf=V_t*I_a/(math.sqrt(3)*V_s*I_sr)    \n",
+      "print(\"power factor of ac source=%.3f\" %pf)\n",
+      "\n",
+      "r_a=0.5\n",
+      "K_m=2.4\n",
+      "w_m=(V_t-I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Speed of motor=%.2f rpm\" %N)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg thyristor current=20 A\n",
+        "rms thyristor current=34.641 A\n",
+        "power factor of ac source=-0.827\n",
+        "Speed of motor=-1980.76 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.16, Page No 685"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=300.0\n",
+      "V_s=600.0\n",
+      "a=0.6\n",
+      "V_t=a*V_s\n",
+      "P=V_t*I_a    \n",
+      "\n",
+      "#Calculations\n",
+      "print(\"input power from source=%.0f kW\" %(P/1000))\n",
+      "R_eq=V_s/(a*I_a)    \n",
+      "print(\"equivalent input resistance=%.3f ohm\" %R_eq)\n",
+      "k=.004\n",
+      "R=.04+.06\n",
+      "w_m=(a*V_s-I_a*R)/(k*I_a)\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "print(\"motor speed=%.1f rpm\" %N)\n",
+      "T_e=k*I_a**2    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.0f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "input power from source=108 kW\n",
+        "equivalent input resistance=3.333 ohm\n",
+        "motor speed=2626.1 rpm\n",
+        "motor torque=360 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.17, Page No 686"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_on=10.0\n",
+      "T_off=15.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=T_on/(T_on+T_off)\n",
+      "V_s=230.0\n",
+      "V_t=a*V_s\n",
+      "r_a=3\n",
+      "K_m=.5\n",
+      "N=1500\n",
+      "w_m=2*math.pi*N/60\n",
+      "I_a=(V_t-K_m*w_m)/r_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor load current=%.3f A\" %I_a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor load current=4.487 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.18, Page No 686"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "w_m=0    \n",
+      "print(\"lower limit of speed control=%.0f rpm\" %w_m)\n",
+      "I_a=25.0\n",
+      "r_a=.2\n",
+      "V_s=220\n",
+      "K_m=0.08\n",
+      "\n",
+      "#Calculations\n",
+      "a=(K_m*w_m+I_a*r_a)/V_s    \n",
+      "print(\"lower limit of duty cycle=%.3f\" %a)\n",
+      "a=1    \n",
+      "print(\"upper limit of duty cycle=%.0f\" %a)\n",
+      "w_m=(a*V_s-I_a*r_a)/K_m    \n",
+      "\n",
+      "#Results\n",
+      "print(\"upper limit of speed control=%.1f rpm\" %w_m)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lower limit of speed control=0 rpm\n",
+        "lower limit of duty cycle=0.023\n",
+        "upper limit of duty cycle=1\n",
+        "upper limit of speed control=2687.5 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.21, Page No 691"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.6\n",
+      "V_s=400.0\n",
+      "V_t=(1-a)*V_s\n",
+      "I_a=300.0\n",
+      "P=V_t*I_a  \n",
+      "\n",
+      "#Calculations  \n",
+      "print(\"power returned=%.0f kW\" %(P/1000))\n",
+      "r_a=.2\n",
+      "K_m=1.2\n",
+      "R_eq=(1-a)*V_s/I_a+r_a    \n",
+      "print(\"equivalent load resistance=%.4f ohm\" %R_eq)\n",
+      "w_mn=I_a*r_a/K_m\n",
+      "N=w_mn*60/(2*math.pi)    \n",
+      "print(\"min braking speed=%.2f rpm\" %N)\n",
+      "w_mx=(V_s+I_a*r_a)/K_m\n",
+      "N=w_mx*60/(2*math.pi)    \n",
+      "print(\"max braking speed=%.1f rpm\" %N)\n",
+      "w_m=(V_t+I_a*r_a)/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"max braking speed=%.1f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power returned=48 kW\n",
+        "equivalent load resistance=0.7333 ohm\n",
+        "min braking speed=477.46 rpm\n",
+        "max braking speed=3660.6 rpm\n",
+        "max braking speed=1750.7 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.22, Page No 699"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "N=1500.0\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"when speed=1455rpm\")\n",
+      "n=1455.0\n",
+      "s1=(N-n)/N\n",
+      "r=math.sqrt(1/3)*(2/3)/(math.sqrt(s1)*(1-s1))    \n",
+      "print(\"I_2mx/I_2r=%.3f\" %r)\n",
+      "print(\"when speed=1350rpm\")\n",
+      "n=1350\n",
+      "s1=(N-n)/N\n",
+      "r=math.sqrt(1/3)*(2/3)/(math.sqrt(s1)*(1-s1))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"I_2mx/I_2r=%.3f\" %r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when speed=1455rpm\n",
+        "I_2mx/I_2r=0.000\n",
+        "when speed=1350rpm\n",
+        "I_2mx/I_2r=0.000\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.24, Page No 705"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V1=400.0\n",
+      "r1=0.6\n",
+      "r2=0.4\n",
+      "s=1.0\n",
+      "x1=1.6\n",
+      "x2=1.6\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"at starting in normal conditions\")\n",
+      "I_n=V1/math.sqrt((r1+r2/s)**2+(x1+x2)**2)    \n",
+      "print(\"current=%.2f A\" %I_n)\n",
+      "pf=(r1+r2)/math.sqrt((r1+r2/s)**2+(x1+x2)**2)    \n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "f1=50\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_en=(3/w_s)*I_n**2*(r2/s)    \n",
+      "print(\"\\nTorque developed=%.2f Nm\" %T_en)\n",
+      "print(\"motor is operated with DOL starting\")\n",
+      "I_d=V1/2/math.sqrt((r1+r2/s)**2+((x1+x2)/2)**2)    \n",
+      "print(\"current=%.0f A\" %I_d)\n",
+      "pf=(r1+r2)/math.sqrt((r1+r2/s)**2+((x1+x2)/2)**2)    \n",
+      "print(\"pf=%.2f\" %pf)\n",
+      "f1=25\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_ed=(3/w_s)*I_d**2*(r2/s)    \n",
+      "print(\"Torque developed=%.3f Nm\" %T_ed)\n",
+      "print(\"at max torque conditions\")\n",
+      "s_mn=r2/math.sqrt((r1)**2+((x1+x2))**2)\n",
+      "I_n=V1/math.sqrt((r1+r2/s_mn)**2+(x1+x2)**2)    \n",
+      "print(\"current=%.3f A\" %I_n)\n",
+      "pf=(r1+r2/s_mn)/math.sqrt((r1+r2/s_mn)**2+(x1+x2)**2)    \n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "f1=50\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_en=(3/w_s)*I_n**2*(r2/s_mn)    \n",
+      "print(\"Torque developed=%.2f Nm\" %T_en)\n",
+      "print(\"motor is operated with DOL starting\")\n",
+      "s_mn=r2/math.sqrt((r1)**2+((x1+x2)/2)**2)\n",
+      "I_d=V1/2/math.sqrt((r1+r2/s_mn)**2+((x1+x2)/2)**2)    \n",
+      "print(\"current=%.3f A\" %I_d)\n",
+      "pf=(r1+r2/s_mn)/math.sqrt((r1+r2/s_mn)**2+((x1+x2)/2)**2)    \n",
+      "print(\"\\npf=%.3f\" %pf)\n",
+      "f1=25\n",
+      "w_s=4*math.pi*f1/4\n",
+      "T_en=(3/w_s)*I_d**2*(r2/s_mn)  \n",
+      "\n",
+      "\n",
+      "#Results  \n",
+      "print(\"Torque developed=%.3f Nm\" %T_en)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "at starting in normal conditions\n",
+        "current=119.31 A\n",
+        "pf=0.2983\n",
+        "\n",
+        "Torque developed=108.75 Nm\n",
+        "motor is operated with DOL starting\n",
+        "current=106 A\n",
+        "pf=0.53\n",
+        "Torque developed=171.673 Nm\n",
+        "at max torque conditions\n",
+        "current=79.829 A\n",
+        "pf=0.7695\n",
+        "Torque developed=396.26 Nm\n",
+        "motor is operated with DOL starting\n",
+        "current=71.199 A\n",
+        "\n",
+        "pf=0.822\n",
+        "Torque developed=330.883 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.25, Page No 709"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "x1=1.0\n",
+      "X_m=50.0\n",
+      "X_e=x1*X_m/(x1+X_m)\n",
+      "V=231.0\n",
+      "V_e=V*X_m/(x1+X_m)\n",
+      "x2=1.0\n",
+      "r2=.4\n",
+      "r1=0\n",
+      "\n",
+      "#Calculations\n",
+      "s_m=r2/(x2+X_e)    \n",
+      "print(\"slip at max torque=%.2f\" %s_m)\n",
+      "s_mT=r2/(x2+X_m)    \n",
+      "print(\"slip at max torque=%.5f\" %s_mT)\n",
+      "f1=50.0\n",
+      "w_s=4*math.pi*f1/4\n",
+      "print(\"for constant voltage input\")\n",
+      "T_est=(3/w_s)*(V_e/math.sqrt(r2**2+(x2+X_e)**2))**2*(r2)    \n",
+      "print(\"starting torque=%.3f Nm\" %T_est)\n",
+      "T_em=(3/w_s)*V_e**2/(2*(x2+X_e))    \n",
+      "print(\"maximum torque developed=%.2f Nm\" %T_em)\n",
+      "print(\"for constant current input\")\n",
+      "I1=28\n",
+      "T_est=(3/w_s)*(I1*X_m)**2/(r2**2+(x2+X_m)**2)*r2    \n",
+      "print(\"starting torque=%.3f Nm\" %T_est)\n",
+      "T_em=(3/w_s)*(I1*X_m)**2/(2*(x2+X_m))   \n",
+      "print(\"maximum torque developed=%.3f Nm\" %T_em)\n",
+      "s=s_mT\n",
+      "i=1\n",
+      "I_m=I1*(r2/s+i*x2)/(r2/s+i*(x2+X_m))\n",
+      "I_m=math.fabs(I_m)\n",
+      "V1=math.sqrt(3)*I_m*X_m    \n",
+      "\n",
+      "#Results\n",
+      "print(\"supply voltage reqd=%.1f V\" %V1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "slip at max torque=0.20\n",
+        "slip at max torque=0.00784\n",
+        "for constant voltage input\n",
+        "starting torque=95.988 Nm\n",
+        "maximum torque developed=247.31 Nm\n",
+        "for constant current input\n",
+        "starting torque=5.756 Nm\n",
+        "maximum torque developed=366.993 Nm\n",
+        "supply voltage reqd=1236.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.27, Page No 718"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=420.0\n",
+      "V1=V/math.sqrt(3)\n",
+      "T_e=450.0\n",
+      "N=1440.0\n",
+      "n=1000.0\n",
+      "T_L=T_e*(n/N)**2\n",
+      "n1=1500.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_s=2*math.pi*n1/60\n",
+      "w_m=2*math.pi*n/60\n",
+      "a=.8\n",
+      "I_d=T_L*w_s/(2.339*a*V1)\n",
+      "k=0\n",
+      "R=(1-w_m/w_s)*(2.339*a*V1)/(I_d*(1-k))    \n",
+      "print(\"value of chopper resistance=%.4f ohm\" %R)\n",
+      "n=1320.0\n",
+      "T_L=T_e*(n/N)**2\n",
+      "I_d=T_L*w_s/(2.339*a*V1)    \n",
+      "print(\"Inductor current=%.3f A\" %I_d)\n",
+      "w_m=2*math.pi*n/60\n",
+      "k=1-((1-w_m/w_s)*(2.339*a*V1)/(I_d*R))    \n",
+      "print(\"value of duty cycle=%.4f\" %k)\n",
+      "s=(n1-n)/n1\n",
+      "V_d=2.339*s*a*V1    \n",
+      "print(\"Rectifed o/p voltage=%.3f V\" %V_d)\n",
+      "P=V_d*I_d\n",
+      "I2=math.sqrt(2/3)*I_d\n",
+      "r2=0.02\n",
+      "Pr=3*I2**2*r2\n",
+      "I1=a*I2\n",
+      "r1=0.015\n",
+      "Ps=3*I1**2*r1\n",
+      "Po=T_L*w_m\n",
+      "Pi=Po+Ps+Pr+P\n",
+      "eff=Po/Pi*100   \n",
+      "\n",
+      "#Results\n",
+      "print(\"Efficiency(in percent)=%.2f\" %eff)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of chopper resistance=2.0132 ohm\n",
+        "Inductor current=130.902 A\n",
+        "value of duty cycle=0.7934\n",
+        "Rectifed o/p voltage=54.449 V\n",
+        "Efficiency(in percent)=88.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.28, Page No 720"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=400.0\n",
+      "V_ph=V/math.sqrt(3)\n",
+      "N_s=1000.0\n",
+      "N=800.0\n",
+      "a=.7\n",
+      "I_d=110\n",
+      "R=2.0\n",
+      "\n",
+      "#Calculations\n",
+      "k=1-((1-N/N_s)*(2.339*a*V_ph)/(I_d*R))    \n",
+      "print(\"value of duty cycle=%.3f\" %k)\n",
+      "P=I_d**2*R*(1-k)\n",
+      "I1=a*I_d*math.sqrt(2/3)\n",
+      "r1=0.1\n",
+      "r2=0.08\n",
+      "Pr=3*I1**2*(r1+r2)\n",
+      "P_o=20000\n",
+      "P_i=P_o+Pr+P\n",
+      "eff=P_o/P_i*100    \n",
+      "print(\"Efficiency=%.2f\" %eff)\n",
+      "I11=math.sqrt(6)/math.pi*a*I_d\n",
+      "th=43\n",
+      "P_ip=math.sqrt(3)*V*I11*math.cos(math.radians(th))\n",
+      "pf=P_ip/(math.sqrt(3)*V*I11)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Input power factor=%.4f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of duty cycle=0.656\n",
+        "Efficiency=70.62\n",
+        "Input power factor=0.7314\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.29, Page No 724"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=420.0\n",
+      "V1=V/math.sqrt(3)\n",
+      "N=1000.0\n",
+      "w_m=2*math.pi*N/60\n",
+      "N_s=1500.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "a=0.8\n",
+      "V_d=2.339*a*s*V1    \n",
+      "print(\"rectified voltage=%.2f V\" %V_d)\n",
+      "T=450.0\n",
+      "N1=1200.0\n",
+      "T_L=T*(N/N1)**2\n",
+      "f1=50\n",
+      "w_s=4*math.pi*f1/4\n",
+      "I_d=w_s*T_L/(2.339*a*V1)    \n",
+      "print(\"inductor current=%.2f A\" %I_d)\n",
+      "a_T=-.4\n",
+      "a1=math.degrees(math.acos(s*a/a_T))\n",
+      "print(\"delay angle of inverter=%.2f deg\" %a1)\n",
+      "\n",
+      "P_s=V_d*I_d\n",
+      "P_o=T_L*w_m\n",
+      "R_d=0.01\n",
+      "P_i=I_d**2*R_d\n",
+      "I2=math.sqrt(2/3)*I_d\n",
+      "r2=0.02\n",
+      "r1=0.015\n",
+      "P_rol=3*I2**2*r2\n",
+      "I1=a*I2\n",
+      "P_sol=3*I1**2*r1\n",
+      "P_i=P_o+P_rol+P_sol+P_i\n",
+      "eff=P_o/P_i*100    \n",
+      "print(\"\\nefficiency=%.2f\" %eff)\n",
+      "w_m=w_s*(1+(-a_T/a)*math.cos(math.radians(a1))-w_s*R_d*T_L/(2.339*a*V1)**2)\n",
+      "N=w_m*60/(2*math.pi) \n",
+      "\n",
+      "#Results   \n",
+      "print(\"motor speed=%.1f rpm\" %N)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectified voltage=151.25 V\n",
+        "inductor current=108.18 A\n",
+        "delay angle of inverter=131.81 deg\n",
+        "\n",
+        "efficiency=99.64\n",
+        "motor speed=996.4 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.30, Page No 726"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=700.0\n",
+      "E2=V/math.sqrt(3)\n",
+      "N_s=1500.0\n",
+      "N=1200.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "V_dd=0.7\n",
+      "V_dt=1.5\n",
+      "V_d=3*math.sqrt(6)*s*E2/math.pi-2*V_dd\n",
+      "V1=415.0\n",
+      "a=math.degrees(math.acos((3*math.sqrt(2)*E2/math.pi)**-1*(-V_d+2*V_dt)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle advance=%.2f deg\" %(180-a))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle advance=70.22 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.31, Page No 726"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=700.0\n",
+      "E2=V/math.sqrt(3)\n",
+      "N_s=1500.0\n",
+      "N=1200.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "V_dd=.7\n",
+      "V_dt=1.5\n",
+      "a=0\n",
+      "u=18 #overlap angle in case of rectifier\n",
+      "V_d=3*math.sqrt(6)*s*E2*(math.cos(math.radians(a))+math.cos(math.radians(a+u)))/(2*math.pi)-2*V_dd\n",
+      "V1=415\n",
+      "V_ml=math.sqrt(2)*V1\n",
+      "u=4 #overlap anglein the inverter\n",
+      " #V_dc=-(3*V_ml*(math.cos(math.radians(a))+math.cos(math.radians(a+u)))/(2*math.pi)-2*V_dt)\n",
+      " #V_dc=V_d\n",
+      " #after solving %  (1+math.cos(math.radians(u)))*math.cos(math.radians(a))-math.sin(math.radians(u))*math.sin(math.radians(a))=-.6425\n",
+      "a=math.degrees(math.acos(-.6425/(math.sqrt((1+math.cos(math.radians(u)))**2+math.sin(math.radians(u))**2))))-math.degrees(math.asin(math.sin(math.radians(a))/(1+math.cos(math.radians(u)))))\n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle advance=%.2f deg\" %(180-a))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle advance=71.25 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.32, Page No 727"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=700.0\n",
+      "E2=V\n",
+      "N_s=1500.0\n",
+      "N=1200.0\n",
+      "\n",
+      "#Calculations\n",
+      "s=(N_s-N)/N_s\n",
+      "V1=415.0\n",
+      "a_T=s*E2/V1    \n",
+      "\n",
+      "#Results\n",
+      "print(\"voltage ratio of the transformer=%.2f\" %a_T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "voltage ratio of the transformer=0.34\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.33, Page No 733"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "P=6.0\n",
+      "N_s=600.0\n",
+      "f1=P*N_s/120.0\n",
+      "V=400.0\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_t=f1*V/f    \n",
+      "print(\"supply freq=%.0f Hz\" %V_t)\n",
+      "T=340.0\n",
+      "N=1000.0\n",
+      "T_L=T*(N_s/N)**2\n",
+      "w_s=2*math.pi*N_s/60\n",
+      "P=T_L*w_s\n",
+      "I_a=P/(math.sqrt(3)*V_t)    \n",
+      "print(\"armature current=%.2f A\" %I_a)\n",
+      "Z_s=2\n",
+      "X_s=f1/f*math.fabs(Z_s)\n",
+      "V_t=V_t/math.sqrt(3)\n",
+      "Ef=math.sqrt(V_t**2+(I_a*X_s)**2)\n",
+      "print(\"excitation voltage=%.2f V\" %(math.sqrt(3)*Ef))\n",
+      "dl=math.degrees(math.atan(I_a*X_s/V_t))\n",
+      "print(\"load angle=%.2f deg\" %dl)\n",
+      "T_em=(3/w_s)*(Ef*V_t/X_s)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"pull out torque=%.2f Nm\" %T_em)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "supply freq=240 Hz\n",
+        "armature current=18.50 A\n",
+        "excitation voltage=243.06 V\n",
+        "load angle=9.10 deg\n",
+        "pull out torque=773.69 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.34, Page No 736"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "P=4.0\n",
+      "f=50.0\n",
+      "w_s=4*math.pi*f/P\n",
+      "X_d=8.0\n",
+      "X_q=2.0\n",
+      "T_e=80.0\n",
+      "V=400.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_t=V/math.sqrt(3)\n",
+      "dl=(1/2)*math.degrees(math.asin(T_e*w_s/((3/2)*(V_t)**2*(1/X_q-1/X_d))))    \n",
+      "print(\"load angle=%.3f deg\" %dl)\n",
+      "I_d=V_t*math.cos(math.radians(dl))/X_d\n",
+      "I_q=V_t*math.sin(math.radians(dl))/X_q\n",
+      "I_a=math.sqrt(I_d**2+I_q**2)    \n",
+      "print(\"armature current=%.2f A\" %I_a)\n",
+      "pf=T_e*w_s/(math.sqrt(3)*V*I_a)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"input power factor=%.4f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "load angle=0.000 deg\n",
+        "armature current=28.87 A\n",
+        "input power factor=0.6283\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.35, Page No 737"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_e=3.0\n",
+      "K_m=1.2\n",
+      "I_a=T_e/K_m\n",
+      "r_a=2.0\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "E_a=(0.263*math.sqrt(2)*V-I_a*r_a)/(1-55/180)\n",
+      "w_m=E_a/K_m\n",
+      "N=w_m*60/(2*math.pi)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor speed=%.2f rpm\" %N)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor speed=640.96 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.36, Page No 738"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=1360.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      " #after calculations V_t % calculated\n",
+      "V_t=163.45\n",
+      "r_a=4\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.4f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=5.2578 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.37, Page No 740"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=2100.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      " #after calculations V_t % calculated\n",
+      "V_t=227.66\n",
+      "r_a=4\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.2f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=1.94 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.38, Page No 742"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=840.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      "V=230.0\n",
+      "a=75.0\n",
+      "V_t=math.sqrt(2)*V/math.pi*(1+math.cos(math.radians(a)))\n",
+      "r_a=4\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.4f Nm\" %T_e)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=10.5922 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.39, Page No 743"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=1400.0\n",
+      "\n",
+      "#Calculations\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      "V=230.0\n",
+      "a=60.0\n",
+      "a1=212\n",
+      "V_t=math.sqrt(2)*V/math.pi*(math.cos(math.radians(a))-math.cos(math.radians(a1)))+E_a*(180+a-a1)/180\n",
+      "r_a=3\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a   \n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.3f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=5.257 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.40, Page No 745"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "K_m=1.0\n",
+      "N=600.0\n",
+      "w_m=2*math.pi*N/60\n",
+      "E_a=K_m*w_m\n",
+      "V=230.0\n",
+      "a=60.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_t=2*math.sqrt(2)*V/math.pi*(math.cos(math.radians(a)))\n",
+      "r_a=3\n",
+      "I_a=(V_t-E_a)/r_a\n",
+      "T_e=K_m*I_a    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"motor torque=%.3f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "motor torque=13.568 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.41, Page No 745"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "r1=.6\n",
+      "r2=.4\n",
+      "s=0.04\n",
+      "x1=1.6\n",
+      "x2=1.6\n",
+      "Z=(r1+r2/s)+(x1+x2)\n",
+      "V=400.0\n",
+      "I1=V/Z    \n",
+      "print(\"source current=%.3f A \" %math.degrees(math.atan(I1.imag/I1.real)))\n",
+      "print(\"and with %.1f deg phase\" %math.fabs(I1))\n",
+      "I2=V/Z\n",
+      "N=1500\n",
+      "w_s=2*math.pi*N/60\n",
+      "T_e=(3/w_s)*abs(I2)**2*r2/s    \n",
+      "print(\"motor torque=%.2f Nm\" %T_e)\n",
+      "N_r=N*(1-s)\n",
+      "\n",
+      "f=45\n",
+      "N_s1=120*f/4\n",
+      "w_s=2*math.pi*N_s1/60\n",
+      "s1=(N_s1-N_r)/N_s1\n",
+      "Z=(r1+r2/s1)+(x1+x2)*f/50.0\n",
+      "V=360\n",
+      "I1=V/Z    \n",
+      "print(\"source current=%.3f A \" %math.degrees(math.atan(I1.imag/I1.real)))\n",
+      "print(\"and with %.1f deg phase\" %math.fabs(I1))\n",
+      "I2=V/Z\n",
+      "T_e=(3/w_s)*abs(I2)**2*r2/s1    \n",
+      "print(\"motor torque=%.2f Nm\" %T_e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "source current=0.000 A \n",
+        "and with 29.0 deg phase\n",
+        "motor torque=160.46 Nm\n",
+        "source current=-0.000 A \n",
+        "and with 142.9 deg phase\n",
+        "motor torque=-2598.45 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter13_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter13_4.ipynb
new file mode 100755
index 00000000..62d2a926
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter13_4.ipynb
@@ -0,0 +1,342 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Power Factor Improvement"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1, Page No 754"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=250.0\n",
+      "R_l=5.0\n",
+      "I_l=20.0\n",
+      "V_l1=math.sqrt(V_s**2-(R_l*I_l)**2)\n",
+      "reg2=(V_s-V_l1)/V_s*100    \n",
+      "pf1=1.0\n",
+      "\n",
+      "#Calculations\n",
+      "P_l1=V_l1*I_l*pf1     #load power\n",
+      "P_r1=V_s*I_l*pf1     #max powwible system rating\n",
+      "utf1=P_l1*100/P_r1  \n",
+      "pf2=0.5\n",
+      " #(.5*V_l)**2+(.866*V_l+R_l*I_l)**2=V_s**2\n",
+      " #after solving\n",
+      "V_l2=158.35    \n",
+      "reg2=(V_s-V_l2)/V_s*100    \n",
+      "P_l2=V_l2*I_l*pf2   #load power\n",
+      "P_r2=V_s*I_l     #max powwible system rating\n",
+      "utf2=P_l2*100/P_r2    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"for pf=1\")\n",
+      "print(\"load voltage=%.2f V\" %V_l1)\n",
+      "print(\"voltage regulation=%.2f\" %reg1)\n",
+      "print(\"system utilisation factor=%.3f\" %utf1)\n",
+      "print(\"energy consumed(in units)=%.1f\" %(P_l1/1000))\n",
+      "print(\"for pf=.5\")\n",
+      "print(\"load voltage=%.2f V\" %V_l2)\n",
+      "print(\"voltage regulation=%.2f\" %reg2)\n",
+      "print(\"system utilisation factor=%.3f\" %utf2)\n",
+      "print(\"energy consumed(in units)=%.2f\" %(P_l2/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "NameError",
+       "evalue": "name 'reg1' is not defined",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mNameError\u001b[0m                                 Traceback (most recent call last)",
+        "\u001b[1;32m<ipython-input-2-ffdbe43fd921>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m     25\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"for pf=1\"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     26\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"load voltage=%.2f V\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mV_l1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 27\u001b[1;33m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"voltage regulation=%.2f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mreg1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     28\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"system utilisation factor=%.3f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[0mutf1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     29\u001b[0m \u001b[1;32mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"energy consumed(in units)=%.1f\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mP_l1\u001b[0m\u001b[1;33m/\u001b[0m\u001b[1;36m1000\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
+        "\u001b[1;31mNameError\u001b[0m: name 'reg1' is not defined"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for pf=1\n",
+        "load voltage=229.13 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2, Page No 756"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50.0\n",
+      "V_s=230.0\n",
+      "I_m1=2\n",
+      "pf1=.3\n",
+      "\n",
+      "#Calculations\n",
+      "I_c1=I_m1*math.sin(math.radians(math.degrees(math.acos(pf1))))\n",
+      "C1=I_c1/(2*math.pi*f*V_s)    \n",
+      "I_m2=5\n",
+      "pf2=.5\n",
+      "I_c2=I_m2*math.sin(math.radians(math.degrees(math.acos(pf2))))\n",
+      "C2=I_c2/(2*math.pi*f*V_s)    \n",
+      "I_m3=10\n",
+      "pf3=.7\n",
+      "I_c3=I_m3*math.sin(math.radians(math.degrees(math.acos(pf3))))\n",
+      "C3=I_c3/(2*math.pi*f*V_s)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"at no load\")\n",
+      "print(\"value of capacitance=%.3f uF\" %(C1*10**6))\n",
+      "print(\"at half full load\")\n",
+      "print(\"value of capacitance=%.3f uF\" %(C2*10**6))\n",
+      "print(\"at full load\")\n",
+      "print(\"value of capacitance=%.3f uF\" %(C3*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "at no load\n",
+        "value of capacitance=26.404 uF\n",
+        "at half full load\n",
+        "value of capacitance=59.927 uF\n",
+        "at full load\n",
+        "value of capacitance=98.834 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page No 764"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_c=10.0\n",
+      "f=50.0\n",
+      "V_s=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "C=I_c/(2*math.pi*f*V_s)    \n",
+      "I_l=10\n",
+      "L=V_s/(2*math.pi*f*I_l)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of capacitance=%.3f uF\" %(C*10**6))\n",
+      "print(\"value of inductor=%.3f mH\" %(L*1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of capacitance=138.396 uF\n",
+        "value of inductor=73.211 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4, Page No 765"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "I_L=10.0\n",
+      "X_L=V_s/I_L\n",
+      "I_f1=6.0\n",
+      " #B=2*a-math.sin(2*a)\n",
+      "B=2*math.pi-I_f1*math.pi*X_L/V_s\n",
+      "a=0\n",
+      "i=1.0\n",
+      "for a in range(1,360):\n",
+      "    b=2*a*math.pi/180-math.sin(math.radians(2*a))   \n",
+      "    if math.fabs(B-b)<=0.001  :      #by hit and trial\n",
+      "        i=2\n",
+      "        break\n",
+      "print(\"firing angle of TCR = %.1f deg\" %a)\n",
+      " #(a-.01)*180/math.pi)\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle of TCR = 359.0 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5 Page No 766"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=.01\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"for firing angle=90deg\")\n",
+      "a=90*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.0f mH\" %(L_eff*1000))\n",
+      "print(\"for firing angle=120deg\")\n",
+      "a=120*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.3f mH\" %(L_eff*1000))\n",
+      "print(\"for firing angle=150deg\")\n",
+      "a=150*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.2f mH\" %(L_eff*1000))\n",
+      "print(\"for firing angle=170deg\")\n",
+      "a=170*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.3f H\" %L_eff)\n",
+      "print(\"for firing angle=175deg\")\n",
+      "a=175*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))  \n",
+      "\n",
+      "#Results\n",
+      "print(\"effective inductance=%.2f H\" %L_eff)\n",
+      "print(\"for firing angle=180deg\")\n",
+      "a=180*math.pi/180\n",
+      "L_eff=math.pi*L/(2*math.pi-2*a+math.sin(2*a))    \n",
+      "print(\"effective inductance=%.3f H\" %L_eff)\n",
+      " #random value at firing angle =180 is equivalent to infinity as in answer in book\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=90deg\n",
+        "effective inductance=10 mH\n",
+        "for firing angle=120deg\n",
+        "effective inductance=25.575 mH\n",
+        "for firing angle=150deg\n",
+        "effective inductance=173.40 mH\n",
+        "for firing angle=170deg\n",
+        "effective inductance=4.459 H\n",
+        "for firing angle=175deg\n",
+        "effective inductance=35.51 H\n",
+        "for firing angle=180deg\n",
+        "effective inductance=-128265253940037.750 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.6 Page No 766"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "Q=100.0*10**3\n",
+      "V_s=11.0*10**3\n",
+      "\n",
+      "#Calculations\n",
+      "f=50.0\n",
+      "L=V_s**2/(2*math.pi*f*Q)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"effective inductance=%.4f H\" %L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "effective inductance=3.8515 H\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter14_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter14_4.ipynb
new file mode 100755
index 00000000..a9c3a3f1
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter14_4.ipynb
@@ -0,0 +1,93 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 : Miscellaneous Topics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1, Page No 777"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a1=0\n",
+      "a2=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"for two single phase series semiconvertors\")\n",
+      "V_0=V_m/math.pi*(2+math.cos(math.radians(a1))+math.cos(math.radians(a2)))    \n",
+      "print(\"avg o/p voltage=%.2f V\" %V_0)\n",
+      "V_or=V_s*math.sqrt((1/math.pi)*(4*math.pi-3*a2*math.pi/180+(3/2)*math.sin(math.radians(2*a2))))    \n",
+      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
+      "DF=(3+math.cos(math.radians(a2)))/(math.sqrt(2)*math.sqrt(5+3*math.cos(math.radians(a2))))    \n",
+      "print(\"DF=%.2f\" %DF)\n",
+      "PF=math.sqrt(2/math.pi)*(3+math.cos(math.radians(a2)))/math.sqrt(4*math.pi-3*a2*math.pi/180)    \n",
+      "print(\"PF=%.2f\" %PF)\n",
+      "HF=math.sqrt((math.pi*(math.pi-(3/4)*a2*math.pi/180)/(5+3*math.cos(math.radians(a2))))-1)    \n",
+      "print(\"HF=%.2f\" %HF)\n",
+      "print(\"for two single phase series full convertors\")\n",
+      "a=45.0\n",
+      "V_0=2*V_m/math.pi*(1+math.cos(math.radians(a)))    \n",
+      "print(\"avg o/p voltage=%.2f V\" %V_0)\n",
+      "V_or=2*V_s*math.sqrt((1/math.pi)*(math.pi-a2*math.pi/180+(1/2)*math.sin(math.radians(2*a2))))    \n",
+      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
+      "DF=math.cos(math.radians(a2/2)) \n",
+      "\n",
+      "\n",
+      "#Results   \n",
+      "print(\"DF=%.2f\" %DF)\n",
+      "PF=math.sqrt(2/(math.pi*(math.pi-a2*math.pi/180)))*(1+math.cos(math.radians(a2)))    \n",
+      "print(\"PF=%.2f\" %PF)\n",
+      "HF=math.sqrt((math.pi*(math.pi-a2*math.pi/180)/(4+4*math.cos(math.radians(a2))))-1)    \n",
+      "print(\"HF=%.2f\" %HF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for two single phase series semiconvertors\n",
+        "avg o/p voltage=383.82 V\n",
+        "rms value of o/p voltage=434.47 V\n",
+        "DF=0.98\n",
+        "PF=0.93\n",
+        "HF=0.62\n",
+        "for two single phase series full convertors\n",
+        "avg o/p voltage=353.50 V\n",
+        "rms value of o/p voltage=398.37 V\n",
+        "DF=0.92\n",
+        "PF=0.89\n",
+        "HF=0.29\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter2_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter2_4.ipynb
new file mode 100755
index 00000000..1872c9f4
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter2_4.ipynb
@@ -0,0 +1,233 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 02 : Power Semiconductor Diodes and Transistors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page No 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "B=40.0\n",
+      "R_c=10                    #ohm\n",
+      "V_cc=130.0                #V\n",
+      "V_B=10.0                  #V\n",
+      "V_CES=1.0                 #V\n",
+      "V_BES=1.5                 #V\n",
+      "\n",
+      "#Calculations\n",
+      "I_CS=(V_cc-V_CES)/R_c     #A\n",
+      "I_BS=I_CS/B                #A\n",
+      "R_B1=(V_B-V_BES)/I_BS\n",
+      "P_T1=V_BES*I_BS+V_CES*I_CS\n",
+      "ODF=5\n",
+      "I_B=ODF*I_BS\n",
+      "R_B2=(V_B-V_BES)/I_B\n",
+      "P_T2=V_BES*I_B+V_CES*I_CS\n",
+      "B_f=I_CS/I_B\n",
+      "\n",
+      "#Results\n",
+      "print(\"value of R_B in saturated state= %.2f ohm\" %R_B1)\n",
+      "print(\"Power loss in transistor=%.2f W\" %P_T1)\n",
+      "print(\"Value of R_B for an overdrive factor 5 = %.2f ohm\" %R_B2)\n",
+      "print(\"Power loss in transistor = %.2f W\" %P_T2)\n",
+      "print(\"Forced current gain=%.0f\" %B_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of R_B in saturated state= 26.36 ohm\n",
+        "Power loss in transistor=13.38 W\n",
+        "Value of R_B for an overdrive factor 5 = 5.27 ohm\n",
+        "Power loss in transistor = 15.32 W\n",
+        "Forced current gain=8\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page No 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_CEO=2*10**-3            #A\n",
+      "V_CC=220.0                #V\n",
+      "P_dt=I_CEO*V_CC           #instant. power loss during delay time\n",
+      "t_d=.4*10**-6            #s\n",
+      "f=5000\n",
+      "P_d=f*I_CEO*V_CC*t_d      #avg power loss during delay time\n",
+      "V_CES=2                     #V\n",
+      "t_r=1*10**-6               #s\n",
+      "I_CS=80                    #A\n",
+      "\n",
+      "#Calculations\n",
+      "P_r=f*I_CS*t_r*(V_CC/2-(V_CC-V_CES)/3)            #avg power loss during rise time\n",
+      "t_m=V_CC*t_r/(2*(V_CC-V_CES))\n",
+      "P_rm=I_CS*V_CC**2/(4*(V_CC-V_CES))           #instant. power loss during rise time\n",
+      "\n",
+      "#Results\n",
+      "P_on=P_d+P_r                \n",
+      "print(\"Avg power loss during turn on = %.2f W\" %P_on)\n",
+      "P_nt=I_CS*V_CES \n",
+      "print(\"Instantaneous power loss during turn on = %.0f W\" %P_nt)\n",
+      "t_n=50*10**-6\n",
+      "P_n=f*I_CS*V_CES*t_n\n",
+      "print(\"Avg power loss during conduction period = %.0f W\" %P_n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Avg power loss during turn on = 14.93 W\n",
+        "Instantaneous power loss during turn on = 160 W\n",
+        "Avg power loss during conduction period = 40 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page No 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_CEO=2*10**-3         #A\n",
+      "V_CC=220          #V\n",
+      "t_d=.4*10**-6   #s\n",
+      "f=5000\n",
+      "V_CES=2         #V\n",
+      "t_r=1*10**-6    #s\n",
+      "I_CS=80         #A\n",
+      "t_n=50*10**-6   #s\n",
+      "t_0=40*10**-6   #s\n",
+      "t_f=3*10**-6    #s\n",
+      "\n",
+      "#Calculations\n",
+      "P_st=I_CS*V_CES  # instant. power loss during t_s\n",
+      "P_s=f*I_CS*V_CES*t_f   #avg power loss during t_s\n",
+      "P_f=f*t_f*(I_CS/6)*(V_CC-V_CES)        #avg power loss during fall time\n",
+      "P_fm=(I_CS/4)*(V_CC-V_CES)          #peak instant power dissipation\n",
+      "P_off=P_s+P_f\n",
+      "\n",
+      "#Results\n",
+      "print(\"Total avg power loss during turn off = %.2f W\" %P_off)\n",
+      "P_0t=I_CEO*V_CC\n",
+      "print(\"Instantaneous power loss during t_0 = %.2f W\" %P_0t)\n",
+      "P_0=f*I_CEO*V_CC*t_0              #avg power loss during t_s\n",
+      "P_on=14.9339              #W    from previous eg\n",
+      "P_n=40                    #W    from previous eg\n",
+      "P_T=P_on+P_n+P_off+P_0     \n",
+      "print(\"Total power loss = %.2f W\" %P_T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total avg power loss during turn off = 44.91 W\n",
+        "Instantaneous power loss during t_0 = 0.44 W\n",
+        "Total power loss = 99.93 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page No 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_CS=100.0 \n",
+      "V_CC=200.0 \n",
+      "t_on=40*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "P_on=(I_CS/50)*10**6*t_on*(V_CC*t_on/2-(V_CC*10**6*t_on**2/(40*3)))       #energy during turn on\n",
+      "t_off=60*10**-6\n",
+      "P_off=(I_CS*t_off/2-(I_CS/60)*10**6*(t_off**2)/3)*((V_CC/75)*10**6*t_off)  #energy during turn off\n",
+      "P_t=P_on+P_off           #total energy\n",
+      "P_avg=300.0\n",
+      "f=P_avg/P_t\n",
+      "\n",
+      "#Results\n",
+      "print(\"Allowable switching frequency = %.2f Hz\" %f)\n",
+      "#in book ans is: f=1123.6 Hz. The difference in results due to difference in rounding of of digits"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Allowable switching frequency = 1125.00 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter3_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter3_4.ipynb
new file mode 100755
index 00000000..2e53ef9d
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter3_4.ipynb
@@ -0,0 +1,1001 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 03 : Diode Circuits and Rectifiers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page No 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0  #V\n",
+      "V_o=100.0    #V\n",
+      "L=100.0      #uH\n",
+      "C=30.0       #uF\n",
+      "\n",
+      "#Calculations\n",
+      "t_o=math.pi*math.sqrt(L*C)\n",
+      "print(\"conduction time of diode = %.2f us\" %t_o)\n",
+      "#in book solution is t_o=54.77 us. The ans is incorrect as %pi is not muliplied in ans. Formulae mentioned in correct.\n",
+      "I_p=(V_s-V_o)*math.sqrt(C/L)\n",
+      "\n",
+      "#Results\n",
+      "print(\"Peak current through diode=%.2f A\" %I_p)\n",
+      "v_D=-V_s+V_o \n",
+      "print(\"Voltage across diode = %.2f V\" %v_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of diode = 172.07 us\n",
+        "Peak current through diode=164.32 A\n",
+        "Voltage across diode = -300.00 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page No 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "R=10             #ohm\n",
+      "L=0.001           #H\n",
+      "C=5*10**-6       #F\n",
+      "V_s=230          #V\n",
+      "xi=R/(2*L)\n",
+      "\n",
+      "#Calculations\n",
+      "w_o=1/math.sqrt(L*C)\n",
+      "w_r=math.sqrt((1/(L*C))-(R/(2*L))**2)\n",
+      "t=math.pi/w_r \n",
+      "\n",
+      "#Results\n",
+      "print('Conduction time of diode=%.3f us'%(t*10**6))\n",
+      "t=0\n",
+      "di=V_s/L\n",
+      "print('Rate of change of current at t=0 is %.2f A/s' %di)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conduction time of diode=237.482 us\n",
+        "Rate of change of current at t=0 is 230000.00 A/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 Page No 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I_or=100             #A\n",
+      "R=1.0       #assumption\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=I_or*2*R\n",
+      "I_o=V_m/(math.pi*R)\n",
+      "q=200       #Ah\n",
+      "t=q/I_o\n",
+      "\n",
+      "#Results\n",
+      "print(\"time required to deliver charge=%.02f hrs\" %t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required to deliver charge=3.14 hrs\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8, Page No 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0    #V\n",
+      "P=1000       #W\n",
+      "R=V_s**2/P\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt(2)*V_s/2\n",
+      "P_h=V_or**2/R    \n",
+      "print(\"Power delivered to the heater = %.2f W\" %P_h)\n",
+      "V_m=math.sqrt(2)*230\n",
+      "I_m=V_m/R\n",
+      "\n",
+      "#Results\n",
+      "print(\"Peak value of diode current = %.2f A\" %I_m)\n",
+      "pf=V_or/V_s\n",
+      "print(\"Input power factor=%.2f\" %pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power delivered to the heater = 500.00 W\n",
+        "Peak value of diode current = 6.15 A\n",
+        "Input power factor=0.71\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 Page No 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230        #V\n",
+      "V_m=V_s*math.sqrt(2)\n",
+      "E=150          #V\n",
+      "\n",
+      "#Calculations\n",
+      "theta1=math.degrees(E/(math.sqrt(2)*V_s))\n",
+      "R=8        #ohm\n",
+      "f=50    #Hz\n",
+      "I_o=(1/(2*math.pi*R))*((2*math.sqrt(2)*V_s*math.cos(math.radians(theta1)))-E*(math.pi-2*theta1*math.pi/180))\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg value of charging current=%.2f A\" %I_o)\n",
+      "P_d=E*I_o\n",
+      "print(\"\\npower delivered to battery=%.2f W\" %P_d)\n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V_s**2+E**2)*(math.pi-2*theta1*math.pi/180)+V_s**2*math.sin(math.radians(2*theta1))-4*V_m*E*math.cos(math.radians(theta1))))\n",
+      "print(\"\\nrms value of the load current=%.2f A\" %I_or)\n",
+      "pf=(E*I_o+I_or**2*R)/(V_s*I_or)\n",
+      "print(\"\\nsupply pf=%.3f\" %pf)\n",
+      "P_dd=I_or**2*R\n",
+      "print(\"\\npower dissipated in the resistor=%.2f W\" %P_dd)\n",
+      "q=1000.00         #Wh\n",
+      "t=q/P_d    \n",
+      "print(\"\\ncharging time=%.2f hr\" %t)\n",
+      "n=P_d*100/(P_d+P_dd)\n",
+      "print(\"rectifier efficiency =%.2f  \" %n)\n",
+      "PIV=math.sqrt(2)*V_s+E\n",
+      "print(\"PIV of diode=%.2f V\" %PIV)\n",
+      "#solutions have small variations due to difference in rounding off of digits"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg value of charging current=4.97 A\n",
+        "\n",
+        "power delivered to battery=745.11 W\n",
+        "\n",
+        "rms value of the load current=9.29 A\n",
+        "\n",
+        "supply pf=0.672\n",
+        "\n",
+        "power dissipated in the resistor=690.74 W\n",
+        "\n",
+        "charging time=1.34 hr\n",
+        "rectifier efficiency =51.89  \n",
+        "PIV of diode=475.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 Page No 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=230            #V\n",
+      "t_rr=40*10**-6    #s reverde recovery time\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=2*math.sqrt(2)*V_s/math.pi\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "f=50\n",
+      "V_r1=(V_m/math.pi)*(1-math.cos(math.radians(2*math.pi*f*t_rr*180/math.pi)))\n",
+      "v_avg1=V_r1*100/V_o*10**3\n",
+      "f=2500\n",
+      "V_r2=(V_m/math.pi)*(1-math.cos(math.radians(2*math.pi*f*t_rr*180/math.pi)))\n",
+      "v_avg2=V_r2*100/V_o\n",
+      "\n",
+      "#Results\n",
+      "print(\"when f=50Hz\")\n",
+      "print(\"Percentage reduction in avg o/p voltage=%.2f x 10^-3\" %v_avg1)\n",
+      "print(\"when f=2500Hz\")\n",
+      "print(\"Percentage reduction in avg o/p voltage = %.3f\" %v_avg2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when f=50Hz\n",
+        "Percentage reduction in avg o/p voltage=3.95 x 10^-3\n",
+        "when f=2500Hz\n",
+        "Percentage reduction in avg o/p voltage = 9.549\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11, Page No 79 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230         #V\n",
+      "R=10.0        #ohm\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_o=2*V_m/math.pi\n",
+      "print(\"Avg value of o/p voltage = %.2f V\" %V_o)\n",
+      "I_o=V_o/R\n",
+      "print(\"Avg value of o/p current = %.2f A\" %I_o)\n",
+      "I_DA=I_o/2\n",
+      "print(\"Avg value of diode current=%.2f A\" %I_DA)\n",
+      "I_Dr=I_o/math.sqrt(2)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of diode current=%.2f A\" %I_Dr)\n",
+      "print(\"rms value of o/p current = %.2f A\" %I_o)\n",
+      "print(\"rms value of i/p current = %.2f A\" %I_o)\n",
+      "pf=(V_o/V_s)\n",
+      "print(\"supply pf = %.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Avg value of o/p voltage = 207.07 V\n",
+        "Avg value of o/p current = 20.71 A\n",
+        "Avg value of diode current=10.35 A\n",
+        "rms value of diode current=14.64 A\n",
+        "rms value of o/p current = 20.71 A\n",
+        "rms value of i/p current = 20.71 A\n",
+        "supply pf = 0.90\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 Page No 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=230.0    #V\n",
+      "R=1000.0      #ohm\n",
+      "R_D=20.0    #ohm\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "I_om=V_m/(R+R_D)          \n",
+      "\n",
+      "#Results\n",
+      "print(\"Peak load current = %.2f A\" %I_om)\n",
+      "I_o=I_om/math.pi\n",
+      "print(\"dc load current = %.2f A\" %I_o)\n",
+      "V_D=I_o*R_D-V_m/math.pi\n",
+      "print(\"dc diode voltage = %.2f V\" %V_D)\n",
+      "V_on=V_m/math.pi\n",
+      "print(\"at no load, load voltage = %.2f V\" %V_on)\n",
+      "V_o1=I_o*R    \n",
+      "print(\"at given load, load voltage = %.2f V\" %V_o1)\n",
+      "vr=(V_on-V_o1)*100/V_on    \n",
+      "print(\"Voltage regulation(in percent)=%.2f\" %vr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak load current = 0.32 A\n",
+        "dc load current = 0.10 A\n",
+        "dc diode voltage = -101.51 V\n",
+        "at no load, load voltage = 103.54 V\n",
+        "at given load, load voltage = 101.51 V\n",
+        "Voltage regulation(in percent)=1.96\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13 Page No 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_L=6.8   #V\n",
+      "V_smax=20*1.2    #V\n",
+      "V_smin=20*.8     #V\n",
+      "I_Lmax=30*1.5   #mA\n",
+      "I_Lmin=30*0.5  #mA\n",
+      "I_z=1          #mA\n",
+      "\n",
+      "#Calculations\n",
+      "R_smax=(V_smax-V_L)/((I_Lmin+I_z)*10**-3)\n",
+      "print(\"max source resistance = %.2f ohm\" %R_smax)\n",
+      "R_smin=(V_smin-V_L)/((I_Lmax+I_z)*10**-3)    \n",
+      "print(\"Min source resistance = %.2f ohm\" %R_smin)  #in book solution, error is committed in putting the values in formulea(printing error) but solution is correct\n",
+      "R_Lmax=V_L*1000/I_Lmin\n",
+      "print(\"Max load resistance = %.2f ohm\" %R_Lmax)\n",
+      "R_Lmin=V_L*1000/I_Lmax    \n",
+      "V_d=0.6      #V\n",
+      "V_r=V_L-V_d\n",
+      "\n",
+      "#Results\n",
+      "print(\"Min load resistance=%.2f ohm\" %R_Lmin)\n",
+      "print(\"Voltage rating of zener diode=%.2f V\" %V_r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "max source resistance = 1075.00 ohm\n",
+        "Min source resistance = 200.00 ohm\n",
+        "Max load resistance = 453.33 ohm\n",
+        "Min load resistance=151.11 ohm\n",
+        "Voltage rating of zener diode=6.20 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14 Page No 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "I2=200*10**-6      #A\n",
+      "V_z=20       #V\n",
+      "R_G=500.0   #hm\n",
+      "\n",
+      "#Calculations\n",
+      "R2=(V_z/I2)-R_G\n",
+      "print(\"R2=%.2f kilo-ohm\" %(R2/1000))\n",
+      "\n",
+      "V_v=25      #V\n",
+      "I1=I2\n",
+      "R1=(V_v-V_z)/I1\n",
+      "\n",
+      "#Results\n",
+      "print(\"R1=%.0f kilo-ohm\"%(R1/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R2=99.50 kilo-ohm\n",
+        "R1=25 kilo-ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15, Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=2*230          #V\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=(math.sqrt(2)*V_s)/math.pi\n",
+      "R=60                #ohm\n",
+      "P_dc=(V_o)**2/R\n",
+      "TUF=0.2865\n",
+      "VA=P_dc/TUF\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"kVA rating of the transformer = %.2f kVA\" %(VA/1000));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kVA rating of the transformer = 2.49 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16, Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "tr=0.5       #turns ratio\n",
+      "I_o=10.0\n",
+      "V=230.0\n",
+      "V_s=V/tr\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "V_o=2*V_m/math.pi\n",
+      "phi1=0\n",
+      "#displacemnt angle=0 as fundamnetal component of i/p source current in phase with source voltage\n",
+      "DF=math.cos(math.radians(phi1))\n",
+      "I_s1=4*I_o/(math.sqrt(2)*math.pi)\n",
+      "I_s=math.sqrt(I_o**2*math.pi/math.pi)\n",
+      "CDF=I_s1/I_o\n",
+      "pf=CDF*DF\n",
+      "HF=math.sqrt((I_s/I_s1)**2-1)\n",
+      "CF=I_o/I_s\n",
+      "\n",
+      "#Results\n",
+      "print(\"o/p voltage = %.2f V\" %V_o)\n",
+      "print(\"distortion factor = %.2f\" %DF)\n",
+      "print(\"i/p pf=%.2f\" %pf)\n",
+      "print(\"Current displacent factor=%.2f\" %CDF)\n",
+      "print(\"Harmonic factor = %.2f\" %HF)\n",
+      "print(\"Creast factor = %.2f\" %CF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "o/p voltage = 414.15 V\n",
+        "distortion factor = 1.00\n",
+        "i/p pf=0.90\n",
+        "Current displacent factor=0.90\n",
+        "Harmonic factor = 0.48\n",
+        "Creast factor = 1.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17, Page No 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_o=230.0\n",
+      "R=10.0\n",
+      "V_s=V_o*math.pi/(2*math.sqrt(2))\n",
+      "I_o=V_o/R\n",
+      "I_m=math.sqrt(2)*V_s/R\n",
+      "I_DAV=I_m/math.pi\n",
+      "\n",
+      "#Calculations\n",
+      "#avg value of diode current\n",
+      "I_Dr=I_m/2\n",
+      "PIV=math.sqrt(2)*V_s\n",
+      "I_s=I_m/math.sqrt(2)\n",
+      "TF=V_s*I_s\n",
+      "\n",
+      "#Results\n",
+      "print(\"peak diode current=%.2f A\" %I_m)\n",
+      "print(\"I_DAV=%.2f A\" %I_DAV)\n",
+      "print(\"I_Dr=%.2f A\" %I_Dr)   #rms value of diode current\n",
+      "print(\"PIV=%.1f V\" %PIV)\n",
+      "print(\"Transformer rating = %.2f kVA\" %(TF/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak diode current=36.13 A\n",
+        "I_DAV=11.50 A\n",
+        "I_Dr=18.06 A\n",
+        "PIV=361.3 V\n",
+        "Transformer rating = 6.53 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18, Page No 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "tr=5\n",
+      "V=1100.0\n",
+      "R=10.0\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"In case of 3ph-3pulse type\")\n",
+      "V_ph=V/tr\n",
+      "V_mp=math.sqrt(2)*V_ph\n",
+      "V_o=3*math.sqrt(3)*V_mp/(2*math.pi)\n",
+      "print(\"avg o/p voltage=%.1f V\" %V_o)\n",
+      "I_mp=V_mp/R\n",
+      "I_D=(I_mp/math.pi)*math.sin(math.pi/3)    \n",
+      "print(\"\\navg value of diode current=%.3f A\" %I_D)\n",
+      "I_Dr=I_mp*math.sqrt((1/(2*math.pi))*(math.pi/3+.5*math.sin(2*math.pi/3)))    \n",
+      "print(\"\\nrms value of diode current=%.2f A\" %I_Dr)\n",
+      "V_or=V_mp*math.sqrt((3/(2*math.pi))*(math.pi/3+.5*math.sin(2*math.pi/3)))\n",
+      "P=(V_or**2)/R   \n",
+      "print(\"\\npower delivered=%.1f W\" %P)\n",
+      "print(\"in case of 3ph-M6 type\")\n",
+      "V_ph=V_ph/2\n",
+      "V_mp=math.sqrt(2)*V_ph\n",
+      "V_o=3*V_mp/(math.pi)  \n",
+      "I_mp=V_mp/R\n",
+      "I_D=(I_mp/math.pi)*math.sin(math.pi/6)    \n",
+      "I_Dr=I_mp*math.sqrt((1/(2*math.pi))*(math.pi/6+.5*math.sin(2*math.pi/6)))  \n",
+      "V_or=V_mp*math.sqrt((6/(2*math.pi))*(math.pi/6+.5*math.sin(2*math.pi/6)))\n",
+      "P=(V_or**2)/R   \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.2f V\" %V_o)\n",
+      "print(\"\\navg value of diode current=%.2f A\" %I_D)\n",
+      "print(\"\\nrms value of diode current=%.2f A\" %I_Dr)\n",
+      "print(\"\\npower delivered=%.0f W\" %P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In case of 3ph-3pulse type\n",
+        "avg o/p voltage=257.3 V\n",
+        "\n",
+        "avg value of diode current=8.577 A\n",
+        "\n",
+        "rms value of diode current=15.10 A\n",
+        "\n",
+        "power delivered=6841.3 W\n",
+        "in case of 3ph-M6 type\n",
+        "avg o/p voltage=148.55 V\n",
+        "\n",
+        "avg value of diode current=2.48 A\n",
+        "\n",
+        "rms value of diode current=6.07 A\n",
+        "\n",
+        "power delivered=2211 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19, Page No 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_o=400\n",
+      "R=10\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=V_o*math.pi/3\n",
+      "V_s=V_ml/(math.sqrt(2)*math.sqrt(3))\n",
+      "I_m=V_ml/R\n",
+      "I_s=.7804*I_m\n",
+      "tr=3*V_s*I_s   \n",
+      "\n",
+      "#Results\n",
+      "print(\"transformer rating=%.1f VA\" %tr)\n",
+      "I_Dr=.5518*I_m   \n",
+      "print(\"\\nrms value of diode current=%.3f A\" %I_Dr)\n",
+      "I_D=I_m/math.pi   \n",
+      "print(\"\\navg value of diode current=%.3f A\" %I_D)\n",
+      "print(\"\\npeak diode current=%.2f A\" %I_m)\n",
+      "PIV=V_ml \n",
+      "print(\"\\nPIV=%.2f V\" %PIV)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "transformer rating=16770.3 VA\n",
+        "\n",
+        "rms value of diode current=23.114 A\n",
+        "\n",
+        "avg value of diode current=13.333 A\n",
+        "\n",
+        "peak diode current=41.89 A\n",
+        "\n",
+        "PIV=418.88 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20, Page No 116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_l=230\n",
+      "E=240\n",
+      "R=8\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(2)*V_l\n",
+      "V_o=3*V_ml/math.pi\n",
+      "I_o=(V_o-E)/R\n",
+      "P_b=E*I_o    \n",
+      "P_d=E*I_o+I_o**2*R   \n",
+      "phi1=0\n",
+      "math.cos(math.radians(phi1))\n",
+      "I_s1=2*math.sqrt(3)*I_o/(math.sqrt(2)*math.pi)\n",
+      "I_s=math.sqrt(I_o**2*2*math.pi/(3*math.pi))\n",
+      "CDF=I_s1/I_s    \n",
+      "pf=DF*CDF    \n",
+      "HF=math.sqrt(CDF**-2-1)   \n",
+      "tr=math.sqrt(3)*V_l*I_o*math.sqrt(2/3)\n",
+      "\n",
+      "#Results\n",
+      "print(\"Power delivered to battery=%.1f W\" %P_b)\n",
+      "print(\"Power delivered to load=%.2f W\" %P_d)\n",
+      "print(\"Displacement factor=%.2f\" %DF)\n",
+      "print(\"Current distortion factor=%.3f\" %CDF)\n",
+      "print(\"i/p pf=%.3f\"%pf)\n",
+      "print(\"Harmonic factor=%.2f\" %HF)\n",
+      "print(\"Tranformer rating=%.2f VA\" %tr)\n",
+      "#answers have small variations from the book due to difference in rounding off of digits"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power delivered to battery=2118.3 W\n",
+        "Power delivered to load=2741.48 W\n",
+        "Displacement factor=1.00\n",
+        "Current distortion factor=0.955\n",
+        "i/p pf=0.955\n",
+        "Harmonic factor=0.31\n",
+        "Tranformer rating=0.00 VA\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21, Page No 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50  #Hz\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V\n",
+      "R=400.0\n",
+      "RF=0.05\n",
+      "C=(1/(4*f*R))*(1+(1/(math.sqrt(2)*RF)))\n",
+      "\n",
+      "#Results\n",
+      "print(\"capacitor value=%.2f uF\" %(C/10**-6))\n",
+      "V_o=V_m*(1-1/(4*f*R*C))\n",
+      "print(\"o/p voltage with filter=%.2f V\" %V_o)\n",
+      "V_o=2*V_m/math.pi   \n",
+      "print(\"o/p voltage without filter=%.2f V\" %V_o)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "capacitor value=189.28 uF\n",
+        "o/p voltage with filter=303.79 V\n",
+        "o/p voltage without filter=207.07 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.22, Page No 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50\n",
+      "CRF=0.05\n",
+      "R=300\n",
+      "\n",
+      "#Calculations\n",
+      "L=math.sqrt((CRF/(.4715*R))**-2-R**2)/(2*2*math.pi*f)    \n",
+      "print(\"L=%.2f H\" %L)\n",
+      "R=30\n",
+      "L=math.sqrt((CRF/(.4715*R))**-2-R**2)/(2*2*math.pi*f)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"\\nL=%.2f H\" %L)\n",
+      "L=0\n",
+      "CRF=.4715*R/math.sqrt(R**2+(2*2*math.pi*f*L)**2)   \n",
+      "print(\"\\nCRF=%.2f\" %CRF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L=4.48 H\n",
+        "\n",
+        "L=0.45 H\n",
+        "\n",
+        "CRF=0.47\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.23, Page No 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=50\n",
+      "L_L=10*10**-3\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "\n",
+      "#Calculations\n",
+      "C=10/(2*w*math.sqrt(R**2+(2*w*L_L)**2))\n",
+      "\n",
+      "#Results\n",
+      "print(\"C=%.2f uF\" %(C*10**6))\n",
+      "VRF=0.1\n",
+      "L=(1/(4*w**2*C))*((math.sqrt(2)/(3*VRF))+1)\n",
+      "print(\"\\nL=%.2f mH\" %(L*10**3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C=315.83 uF\n",
+        "\n",
+        "L=45.83 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter4_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter4_4.ipynb
new file mode 100755
index 00000000..22311574
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter4_4.ipynb
@@ -0,0 +1,946 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 04 : Thyristors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "P=.5    #P=V_g*I_g\n",
+      "s=130    #s=V_g/I_g\n",
+      "\n",
+      "#Calculations\n",
+      "I_g=math.sqrt(P/s)\n",
+      "V_g=s*I_g\n",
+      "E=15\n",
+      "R_s=(E-V_g)/I_g    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Gate source resistance=%.2f ohm\" %R_s)\n",
+      "#Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gate source resistance=111.87 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "R_s=120     #slope of load line is -120V/A. This gives gate source resistance\n",
+      "print(\"gate source resistance=%.0f ohm\" %R_s)\n",
+      "\n",
+      "P=.4     #P=V_g*I_g\n",
+      "E_s=15\n",
+      "\n",
+      "#Calculations\n",
+      " #E_s=I_g*R_s+V_g %    after solving this\n",
+      " #120*I_g**2-15*I_g+0.4=0    so\n",
+      "a=120   \n",
+      "b=-15\n",
+      "c=0.4\n",
+      "D=math.sqrt((b**2)-4*a*c)\n",
+      "I_g=(-b+D)/(2*a)   \n",
+      "V_g=P/I_g\n",
+      "\n",
+      "#Results\n",
+      "print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3))    \n",
+      "print(\"\\nthen trigger voltage=%.3f V\" %V_g)\n",
+      "I_g=(-b-D)/(2*a)   \n",
+      "V_g=P/I_g\n",
+      "print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3))    \n",
+      "print(\"\\nthen trigger voltage=%.2f V\" %V_g)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "gate source resistance=120 ohm\n",
+        "\n",
+        "trigger current=86.44 mA\n",
+        "\n",
+        "then trigger voltage=4.628 V\n",
+        "\n",
+        "trigger current=38.56 mA\n",
+        "\n",
+        "then trigger voltage=10.37 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "#V_g=1+10*I_g\n",
+      "P_gm=5     #P_gm=V_g*I_g\n",
+      "#after solving % eqn becomes 10*I_g**2+I_g-5=0\n",
+      "a=10.0    \n",
+      "b=1.0    \n",
+      "c=-5\n",
+      "\n",
+      "#Calculations\n",
+      "I_g=(-b+math.sqrt(b**2-4*a*c))/(2*a)\n",
+      "E_s=15\n",
+      "#using E_s=R_s*I_g+V_g\n",
+      "R_s=(E_s-1)/I_g-10   \n",
+      "P_gav=.3     #W\n",
+      "T=20*10**-6\n",
+      "f=P_gav/(P_gm*T)\n",
+      "dl=f*T\n",
+      "\n",
+      "#Results\n",
+      "print(\"Reistance=%.3f ohm\" %R_s)\n",
+      "print(\"Triggering freq=%.0f kHz\" %(f/1000))\n",
+      "print(\"Tduty cycle=%.2f\" %dl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reistance=11.248 ohm\n",
+        "Triggering freq=3 kHz\n",
+        "Tduty cycle=0.06\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6, Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I=.1\n",
+      "E=200.0\n",
+      "L=0.2\n",
+      "\n",
+      "#Calculations\n",
+      "t=I*L/E   \n",
+      "R=20.0\n",
+      "t1=(-L/R)*math.log(1-(R*I/E))    \n",
+      "L=2.0\n",
+      "t2=(-L/R)*math.log(1-(R*I/E))   \n",
+      "\n",
+      "#Results\n",
+      "print(\"in case load consists of (a)L=.2H\")\n",
+      "print(\"min gate pulse width=%.0f us\" %(t*10**6))\n",
+      "print(\"(b)R=20ohm in series with L=.2H\")\n",
+      "print(\"min gate pulse width=%.3f us\" %(t1*10**6))\n",
+      "print(\"(c)R=20ohm in series with L=2H\")\n",
+      "print(\"min gate pulse width=%.2f us\" %(t2*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "in case load consists of (a)L=.2H\n",
+        "min gate pulse width=100 us\n",
+        "(b)R=20ohm in series with L=.2H\n",
+        "min gate pulse width=100.503 us\n",
+        "(c)R=20ohm in series with L=2H\n",
+        "min gate pulse width=1005.03 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 Page No 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "def theta(th):\n",
+      "    I_m=1     #supposition\n",
+      "    I_av=(I_m/(2*math.pi))*(1+math.cos(math.radians(th)))\n",
+      "    I_rms=math.sqrt((I_m/(2*math.pi))*((180-th)*math.pi/360+.25*math.sin(math.radians(2*th))))\n",
+      "    FF=I_rms/I_av\n",
+      "    I_rms=35\n",
+      "    I_TAV=I_rms/FF\n",
+      "    return I_TAV\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"when conduction angle=180\")\n",
+      "th=0\n",
+      "I_TAV=theta(th)\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
+      "print(\"when conduction angle=90\")\n",
+      "th=90\n",
+      "I_TAV=theta(th)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
+      "print(\"when conduction angle=30\")\n",
+      "th=150\n",
+      "I_TAV=theta(th)\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when conduction angle=180\n",
+        "avg on current rating=22.282 A\n",
+        "when conduction angle=90\n",
+        "avg on current rating=15.756 A\n",
+        "when conduction angle=30\n",
+        "avg on current rating=8.790 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10, Page No 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "def theta(th):\n",
+      "    n=360.0/th\n",
+      "    I=1.0     #supposition\n",
+      "    I_av=I/n\n",
+      "    I_rms=I/math.sqrt(n)\n",
+      "    FF=I_rms/I_av\n",
+      "    I_rms=35\n",
+      "    I_TAV=I_rms/FF\n",
+      "    return I_TAV\n",
+      "\n",
+      "#Calculations\n",
+      "th=180.0\n",
+      "I_TAV1=theta(th)\n",
+      "th=90.0\n",
+      "I_TAV2=theta(th)\n",
+      "th=30.0\n",
+      "I_TAV3=theta(th)\n",
+      "\n",
+      "#Results\n",
+      "print(\"when conduction angle=180\")\n",
+      "print(\"avg on current rating=%.3f A\" %I_TAV)\n",
+      "print(\"when conduction angle=90\")\n",
+      "print(\"avg on current rating=%.1f A\" %I_TAV2)\n",
+      "print(\"when conduction angle=30\")\n",
+      "print(\"avg on current rating=%.4f A\" %I_TAV3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when conduction angle=180\n",
+        "avg on current rating=8.790 A\n",
+        "when conduction angle=90\n",
+        "avg on current rating=17.5 A\n",
+        "when conduction angle=30\n",
+        "avg on current rating=10.1036 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 Page No 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "f=50.0     #Hz\n",
+      "\n",
+      "#Calculations\n",
+      "I_sb=3000.0\n",
+      "t=1/(4*f)\n",
+      "T=1/(2*f)\n",
+      "I=math.sqrt(I_sb**2*t/T)    \n",
+      "r=(I_sb/math.sqrt(2))**2*T   \n",
+      "\n",
+      "#Results\n",
+      "print(\"surge current rating=%.2f A\" %I)\n",
+      "print(\"\\nI**2*t rating=%.0f A^2.s\" %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "surge current rating=2121.32 A\n",
+        "\n",
+        "I**2*t rating=45000 A^2.s\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 Page No 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "V_s=300.0 #V\n",
+      "R=60.0 #ohm\n",
+      "L=2.0 #H\n",
+      "\n",
+      "#Calculations\n",
+      "t=40*10**-6 #s\n",
+      "i_T=(V_s/R)*(1-math.exp(-R*t/L))\n",
+      "i=.036 #A\n",
+      "R1=V_s/(i-i_T)\n",
+      "\n",
+      "#Results\n",
+      "print(\"maximum value of remedial parameter=%.3f kilo-ohm\" %(R1/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum value of remedial parameter=9.999 kilo-ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16 Page No 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_p=230.0*math.sqrt(2)\n",
+      "\n",
+      "#Calculations\n",
+      "R=1+((1)**-1+(10)**-1)**-1\n",
+      "A=V_p/R\n",
+      "s=1     #s\n",
+      "t_c=20*A**-2*s\n",
+      "\n",
+      "#Results\n",
+      "print(\"fault clearance time=%.4f ms\" %(t_c*10**3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fault clearance time=0.6890 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17, Page No 176"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "V_s=math.sqrt(2)*230     #V\n",
+      "L=15*10**-6     #H\n",
+      "I=V_s/L     #I=(di/dt)_max\n",
+      "R_s=10   #ohm\n",
+      "v=I*R_s     #v=(dv/dt)_max\n",
+      "\n",
+      "#Calculations\n",
+      "f=50     #Hz\n",
+      "X_L=L*2*math.pi*f\n",
+      "R=2\n",
+      "I_max=V_s/(R+X_L)    \n",
+      "FF=math.pi/math.sqrt(2)\n",
+      "I_TAV1=I_max/FF    \n",
+      "FF=3.98184\n",
+      "I_TAV2=I_max/FF    \n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"(di/dt)_max=%.3f A/usec\" %(I/10**6))\n",
+      "print(\"\\n(dv/dt)_max=%.2f V/usec\" %(v/10**6))\n",
+      "print(\"\\nI_rms=%.3f A\" %I_max)\n",
+      "print(\"when conduction angle=90\")\n",
+      "print(\"I_TAV=%.3f A\" %I_TAV1)\n",
+      "print(\"when conduction angle=30\")\n",
+      "print(\"I_TAV=%.3f A\" %I_TAV2)\n",
+      "print(\"\\nvoltage rating=%.3f V\" %(2.75*V_s)) #rating is taken 2.75 times of peak working voltage unlike 2.5 to 3 times as mentioned int book."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(di/dt)_max=21.685 A/usec\n",
+        "\n",
+        "(dv/dt)_max=216.85 V/usec\n",
+        "\n",
+        "I_rms=162.252 A\n",
+        "when conduction angle=90\n",
+        "I_TAV=73.039 A\n",
+        "when conduction angle=30\n",
+        "I_TAV=40.748 A\n",
+        "\n",
+        "voltage rating=894.490 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19, Page No 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "T_jm=125\n",
+      "th_jc=.15     #degC/W\n",
+      "th_cs=0.075     #degC/W\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "dT=54     #dT=T_s-T_a\n",
+      "P_av=120\n",
+      "th_sa=dT/P_av\n",
+      "T_a=40     #ambient temp\n",
+      "P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
+      "if (P_av-120)<1 :\n",
+      "    print(\"selection of heat sink is satisfactory\")\n",
+      "\n",
+      "dT=58     #dT=T_s-T_a\n",
+      "P_av=120\n",
+      "th_sa=dT/P_av\n",
+      "T_a=40     #ambient temp\n",
+      "P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
+      "if (P_av-120)<1 :\n",
+      "    print(\"selection of heat sink is satisfactory\")\n",
+      "\n",
+      "V_m=math.sqrt(2)*230\n",
+      "R=2\n",
+      "I_TAV=V_m/(R*math.pi)\n",
+      "P_av=90\n",
+      "th_sa=(T_jm-T_a)/P_av-(th_jc+th_cs)\n",
+      "dT=P_av*th_sa\n",
+      "print(\"for heat sink\")    \n",
+      "print(\"T_s-T_a=%.2f degC\" %dT)   \n",
+      "print(\"\\nP_av=%.0f W\" %P_av)\n",
+      "P=(V_m/2)**2/R\n",
+      "eff=P/(P+P_av)   \n",
+      "print(\"\\nckt efficiency=%.3f pu\" %eff)\n",
+      "a=60     #delay angle\n",
+      "I_TAV=(V_m/(2*math.pi*R))*(1+math.cos(math.radians(a)))\n",
+      "print(\"\\nI_TAV=%.2f A\" %I_TAV)\n",
+      "dT=46\n",
+      "T_s=dT+T_a\n",
+      "T_c=T_s+P_av*th_cs    \n",
+      "T_j=T_c+P_av*th_jc    \n",
+      "\n",
+      "#Results\n",
+      "print(\"\\ncase temp=%.2f degC\" %T_c)\n",
+      "print(\"\\njunction temp=%.2f degC\" %T_j)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for heat sink\n",
+        "T_s-T_a=-20.25 degC\n",
+        "\n",
+        "P_av=90 W\n",
+        "\n",
+        "ckt efficiency=0.993 pu\n",
+        "\n",
+        "I_TAV=38.83 A\n",
+        "\n",
+        "case temp=92.75 degC\n",
+        "\n",
+        "junction temp=106.25 degC\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.20, Page No 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_j=125.0     #degC\n",
+      "T_s=70.0     #degC\n",
+      "th_jc=.16     #degC/W\n",
+      "th_cs=.08     #degC/W\n",
+      "\n",
+      "#Calculations\n",
+      "P_av1=(T_j-T_s)/(th_jc+th_cs)    \n",
+      "\n",
+      "T_s=60     #degC\n",
+      "P_av2=(T_j-T_s)/(th_jc+th_cs)    \n",
+      "inc=(math.sqrt(P_av2)-math.sqrt(P_av1))*100/math.sqrt(P_av1)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Total avg power loss in thristor sink combination=%.2f W\" %P_av1)\n",
+      "print(\"Percentage inc in rating=%.2f\" %inc)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total avg power loss in thristor sink combination=229.17 W\n",
+        "Percentage inc in rating=8.71\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.21, Page No 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "R=25000.0\n",
+      "I_l1=.021     #I_l=leakage current\n",
+      "I_l2=.025\n",
+      "I_l3=.018\n",
+      "I_l4=.016\n",
+      " #V1=(I-I_l1)*R\n",
+      " #V2=(I-I_l2)*R\n",
+      " #V3=(I-I_l3)*R\n",
+      " #V4=(I-I_l4)*R\n",
+      " #V=V1+V2+V3+V4\n",
+      "    \n",
+      "#Calculations\n",
+      "V=10000.0\n",
+      "I_l=I_l1+I_l2+I_l3+I_l4\n",
+      " #after solving\n",
+      "I=((V/R)+I_l)/4\n",
+      "R_c=40.0\n",
+      "V1=(I-I_l1)*R    \n",
+      "\n",
+      "#Resluts\n",
+      "print(\"voltage across SCR1=%.0f V\" %V1)\n",
+      "V2=(I-I_l2)*R    \n",
+      "print(\"\\nvoltage across SCR2=%.0f V\" %V2)\n",
+      "V3=(I-I_l3)*R    \n",
+      "print(\"\\nvoltage across SCR3=%.0f V\" %V3)\n",
+      "V4=(I-I_l4)*R    \n",
+      "print(\"\\nvoltage across SCR4=%.0f V\" %V4)\n",
+      "\n",
+      "I1=V1/R_c    \n",
+      "print(\"\\ndischarge current through SCR1=%.3f A\" %I1)\n",
+      "I2=V2/R_c   \n",
+      "print(\"\\ndischarge current through SCR2=%.3f A\" %I2)\n",
+      "I3=V3/R_c    \n",
+      "print(\"\\ndischarge current through SCR3=%.3f A\" %I3)\n",
+      "I4=V4/R_c    \n",
+      "print(\"\\ndischarge current through SCR4=%.3f A\" %I4)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "voltage across SCR1=2475 V\n",
+        "\n",
+        "voltage across SCR2=2375 V\n",
+        "\n",
+        "voltage across SCR3=2550 V\n",
+        "\n",
+        "voltage across SCR4=2600 V\n",
+        "\n",
+        "discharge current through SCR1=61.875 A\n",
+        "\n",
+        "discharge current through SCR2=59.375 A\n",
+        "\n",
+        "discharge current through SCR3=63.750 A\n",
+        "\n",
+        "discharge current through SCR4=65.000 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.22, Page No 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_r=1000     #rating of SCR\n",
+      "I_r=200     #rating of SCR\n",
+      "V_s=6000     #rating of String\n",
+      "I_s=1000     #rating of String\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"when DRF=.1\")\n",
+      "DRF=.1\n",
+      "n_s=V_s/(V_r*(1-DRF))    \n",
+      "print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
+      "n_p=I_s/(I_r*(1-DRF))    \n",
+      "print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))\n",
+      "print(\"when DRF=.2\")\n",
+      "DRF=.2\n",
+      "\n",
+      "#Results\n",
+      "n_s=V_s/(V_r*(1-DRF))   \n",
+      "print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
+      "n_p=I_s/(I_r*(1-DRF))    \n",
+      "print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when DRF=.1\n",
+        "number of series units=7\n",
+        "\n",
+        "number of parrallel units=6\n",
+        "when DRF=.2\n",
+        "number of series units=8\n",
+        "\n",
+        "number of parrallel units=7\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.23, Page No 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V1=1.6     #on state voltage drop of SCR1\n",
+      "V2=1.2     #on state voltage drop of SCR2\n",
+      "I1=250.0     #current rating of SCR1\n",
+      "I2=350.0     #current rating of SCR2\n",
+      "\n",
+      "#Calculations\n",
+      "R1=V1/I1\n",
+      "R2=V2/I2\n",
+      "I=600.0     #current to be shared\n",
+      " #for SCR1 %     I*(R1+R)/(total resistance)=k*I1                    (1)\n",
+      " #for SCR2 %     I*(R2+R)/(total resistance)=k*I2                    (2)\n",
+      " #(1)/(2)\n",
+      "R=(R2*I2-R1*I1)/(I1-I2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"RSequired value of resistance=%.3f ohm\" %R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RSequired value of resistance=0.004 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.25, Page No 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=2000.0     #Hz\n",
+      "C=0.04*10**-6\n",
+      "n=.72\n",
+      "\n",
+      "#Calculations\n",
+      "R=1/(f*C*math.log(1/(1-n)))    \n",
+      "V_p=18\n",
+      "V_BB=V_p/n\n",
+      "R2=10**4/(n*V_BB)    \n",
+      "I=4.2*10**-3     #leakage current\n",
+      "R_BB=5000\n",
+      "R1=(V_BB/I)-R2-R_BB\n",
+      "\n",
+      "#Results\n",
+      "print(\"R=%.2f kilo-ohm\" %(R/1000))\n",
+      "print(\"\\nR2=%.2f ohm\" %R2)\n",
+      "print(\"\\nR1=%.0f ohm\" %R1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R=9.82 kilo-ohm\n",
+        "\n",
+        "R2=555.56 ohm\n",
+        "\n",
+        "R1=397 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.26, Page No 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "V_p=18.0\n",
+      "n=.72\n",
+      "V_BB=V_p/n\n",
+      "I_p=.6*10**-3\n",
+      "I_v=2.5*10**-3\n",
+      "V_v=1\n",
+      "\n",
+      "#Calculations\n",
+      "R_max=V_BB*(1-n)/I_p    \n",
+      "print(\"R_max=%.2f kilo-ohm\" %(R_max/1000))\n",
+      "R_min=(V_BB-V_v)/I_v    \n",
+      "print(\"\\nR_min=%.2f kilo-ohm\" %(R_min/1000))\n",
+      "\n",
+      "C=.04*10**-6\n",
+      "f_min=1/(R_max*C*math.log(1/(1-n)))    \n",
+      "print(\"\\nf_min=%.3f kHz\" %(f_min/1000))\n",
+      "f_max=1/(R_min*C*math.log(1/(1-n)))    \n",
+      "print(\"\\nf_max=%.2f kHz\" %(f_max/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R_max=11.67 kilo-ohm\n",
+        "\n",
+        "R_min=9.60 kilo-ohm\n",
+        "\n",
+        "f_min=1.683 kHz\n",
+        "\n",
+        "f_max=2.05 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter5_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter5_4.ipynb
new file mode 100755
index 00000000..1d261f20
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter5_4.ipynb
@@ -0,0 +1,511 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 05 : Thyristor Commutation Techniques"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page No 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=5.0*10**-3            #mH\n",
+      "C=20.0*10**-6           #\u00b5F\n",
+      "V_s=200                 #V\n",
+      "\n",
+      "#Calculations\n",
+      "w_o=math.sqrt(1/(L*C))    #rad/s\n",
+      "t_o=math.pi/w_o           #ms\n",
+      "\n",
+      "#Results\n",
+      "print('conduction time of thyristor = %.2f ms' %(t_o*1000))\n",
+      "print('voltage across thyristor=%.0f V' %V_s)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of thyristor = 0.99 ms\n",
+        "voltage across thyristor=200 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page No 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "C=20.0*10**-6         #\u00b5F\n",
+      "L=5.0*10**-6          #\u00b5H\n",
+      "V_s=230.0             #V\n",
+      "\n",
+      "#Calculations\n",
+      "I_p=V_s*math.sqrt(C/L)      #A\n",
+      "w_o=math.sqrt(1/(L*C))      #rad/sec\n",
+      "t_o=math.pi/w_o             #\u00b5S\n",
+      "I_o=300               \n",
+      "a = math.degrees(math.asin(I_o/(2*V_s)))   \n",
+      "V_ab = V_s*math.cos(math.radians(a))      #V \n",
+      "t_c=C*V_ab/I_o     #\u00b5s\n",
+      "\n",
+      "#Calculations\n",
+      "print(\"conduction time of auxillery thyristor=%.2f us\" %(t_o*10**6))\n",
+      "print(\"voltage across main thyristor=%.2f V\" %V_ab)\n",
+      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of auxillery thyristor=31.42 us\n",
+        "voltage across main thyristor=174.36 V\n",
+        "ckt turn off time=11.62 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page No 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=200.0           #V\n",
+      "R1=10.0             #\u2126\n",
+      "R2=100.0            #\u2126\n",
+      "C=0                 #  value of capacitor\n",
+      "\n",
+      "#Calculations\n",
+      "I1=V_s*(1/R1+2/R2)    #A\n",
+      "I2=V_s*(2/R1+1/R2)    #A\n",
+      "t_c1=40*10**-6\n",
+      "fos=2    #factor of safety\n",
+      "C1=t_c1*fos/(R1*math.log(2))\n",
+      "C2=t_c1*fos/(R2*math.log(2))\n",
+      "if C1 > C2 :\n",
+      "    C = C1*10**6\n",
+      "else :\n",
+      "    C = C2*10**6\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"peak value of current through SCR1=%.2f A\" %I1); \n",
+      "print(\"Peak value of current through SCR2=%.2f A\" %I2);\n",
+      "print(\"Value of capacitor=%.2f uF\" %C);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak value of current through SCR1=24.00 A\n",
+        "Peak value of current through SCR2=42.00 A\n",
+        "Value of capacitor=11.54 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page No 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0             #V\n",
+      "L=20*10**-6           #\u00b5H\n",
+      "C=40*10**-6           #\u00b5F\n",
+      "I_o=120.0             #A\n",
+      "\n",
+      "#Calculations\n",
+      "I_p=V_s*math.sqrt(C/L)   #A\n",
+      "t_c=C*V_s/I_o            #\u00b5s\n",
+      "w_o=math.sqrt(1/(L*C))  \n",
+      "t_c1=math.pi/(2*w_o)     #\u00b5s\n",
+      "\n",
+      "#Results\n",
+      "print(\"current through main thyristor=%.2f A\"  %(I_o+I_p))\n",
+      "print(\"Current through auxillery thyristor=%.2f A\" %I_o)\n",
+      "print(\"Circuit turn off time for main thyristor=%.2f us\" %(t_c*10**6))\n",
+      "print(\"Circuit turn off time for auxillery thyristor=%.2f us\" %(t_c1*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current through main thyristor=445.27 A\n",
+        "Current through auxillery thyristor=120.00 A\n",
+        "Circuit turn off time for main thyristor=76.67 us\n",
+        "Circuit turn off time for auxillery thyristor=44.43 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page No 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "C_j=25*10**-12      #pF\n",
+      "I_c=5*10**-3        #charging current\n",
+      "V_s=200.0           #V\n",
+      "R=50.0              #\u2126\n",
+      "\n",
+      "#Calculations\n",
+      "C=(C_j*V_s)/(I_c*R)\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print(\"Value of C=%.2f  \u00b5F\" %(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of C=0.02  \u00b5F\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 Page No 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "V_s=200.0         #V\n",
+      "R=5.0             #\u2126\n",
+      "\n",
+      "#Calculations\n",
+      "C=10.0*10**-6\n",
+      "#for turn off V_s*(1-2*exp(-t/(R*C)))=0,    so after solving\n",
+      "t_c=R*C*math.log(2.0)\n",
+      "\n",
+      "#Results\n",
+      "print(\"circuit turn off time=%.2f us\" %(t_c*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "circuit turn off time=34.66 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page No 264 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=1.0                   #\u2126\n",
+      "L=20*10**-6             #\u00b5H\n",
+      "C=40*10**-6             #\u00b5F\n",
+      "\n",
+      "#Calculations\n",
+      "w_r=math.sqrt((1/(L*C))-(R/(2*L))**2)\n",
+      "t_1=math.pi/w_r\n",
+      "\n",
+      "#Results\n",
+      "print(\"conduction time of thyristor=%.3f us\" %(t_1*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conduction time of thyristor=125.664 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 Page No 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "dv=400*10.0**-6       #dv=dv_T/dt(V/s)\n",
+      "V_s=200.0             #v\n",
+      "R=20.0                #\u2126\n",
+      "\n",
+      "#Calculations\n",
+      "C=V_s/(R*dv)         \n",
+      "C_j=.025*10**-12\n",
+      "C_s=C-C_j\n",
+      "I_T=40;\n",
+      "R_s=1/((I_T/V_s)-(1/R)) \n",
+      "#value of R_s in book is wrongly calculated\n",
+      "\n",
+      "#Results\n",
+      "print(\"R_s=%.2f ohm\" %R_s)\n",
+      "print(\"C_s=%.3f uF\" %(C_s/10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R_s=6.67 ohm\n",
+        "C_s=0.025 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 Page No 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=200.0          #V\n",
+      "C=20.0*10**-6      #\u00b5H   \n",
+      "L=0.2*10**-3       #\u00b5F\n",
+      "i_c=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "i=V_s*math.sqrt(C/L)\n",
+      "w_o=1.0/math.sqrt(L*C)\n",
+      "t_1 = (1/w_o)*math.degrees(math.asin(i_c/i))\n",
+      "t_o=math.pi/w_o\n",
+      "t_c=t_o-2*t_1    \n",
+      "\n",
+      "#Results\n",
+      "print(\"reqd time=%.2f us\" %(t_1*10**6))\n",
+      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n",
+      "print(\"ckt turn off time=%.5f us\" %t_1)\n",
+      "#solution in book wrong, as wrong values are selected while filling the formuleas"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "reqd time=575.37 us\n",
+        "ckt turn off time=-952.05 us\n",
+        "ckt turn off time=0.00058 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11 Page No 268 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "L=1.0              #\u00b5H\n",
+      "R=50.0             #\u2126\n",
+      "V_s=200.0          #V\n",
+      "t=0.01             #sec\n",
+      "Vd=0.7\n",
+      "\n",
+      "#Calculations\n",
+      "tau=L/R\n",
+      "i=(V_s/R)*(1-math.exp(-t/tau))\n",
+      "t=8*10**-3\n",
+      "i1=i-t*Vd  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"current through L = %.2f A\" %i1)\n",
+      "i_R=0                  #current in R at t=.008s\n",
+      "print(\"Current through R = %.2f A\" %i_R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current through L = 1.57 A\n",
+        "Current through R = 0.00 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12, Page No 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "#initialisation of variables\n",
+      "L=1.0                      #H\n",
+      "R=50.0                   #ohm\n",
+      "V_s=200.0                #V\n",
+      "\n",
+      "#Calculations\n",
+      "tau=L/R\n",
+      "t=0.01                 #s\n",
+      "i=(V_s/R)*(1-math.exp(-t/tau))\n",
+      "C=1*10**-6            #F\n",
+      "V_c=math.sqrt(L/C)*i\n",
+      "\n",
+      "#Results\n",
+      "print(\"current in R,L=%.2f A\" %i)\n",
+      "print(\"voltage across C=%.2f kV\" %(V_c/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current in R,L=1.57 A\n",
+        "voltage across C=1.57 kV\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter6_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter6_4.ipynb
new file mode 100755
index 00000000..dff6564b
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter6_4.ipynb
@@ -0,0 +1,1761 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 06 : Phase Controlled Rectifiers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page No 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "P=1000.0\n",
+      "R=V**2/P\n",
+      "\n",
+      "#Calculations\n",
+      "a=math.pi/4\n",
+      "V_or1=(math.sqrt(2)*V/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a)+.5*math.sin(2*a))\n",
+      "P1=V_or1**2/R    \n",
+      "a=math.pi/2\n",
+      "V_or2=(math.sqrt(2)*V/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a)+.5*math.sin(2*a))\n",
+      "P2=V_or2**2/R   \n",
+      "\n",
+      "#Results\n",
+      "print(\"when firing angle delay is of 45deg\")\n",
+      "print(\"power absorbed=%.2f W\" %P1)\n",
+      "print(\"when firing angle delay is of 90deg\")\n",
+      "print(\"power absorbed=%.2f W\" %P2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when firing angle delay is of 45deg\n",
+        "power absorbed=454.58 W\n",
+        "when firing angle delay is of 90deg\n",
+        "power absorbed=250.00 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page No 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "E=150.0\n",
+      "R=8.0\n",
+      "\n",
+      "#Calculations\n",
+      "th1=math.sin(math.radians(E/(math.sqrt(2)*V)))\n",
+      "I_o=(1/(2*math.pi*R))*(2*math.sqrt(2)*230*math.cos(math.radians(th1))-E*(math.pi-2*th1*math.pi/180))    \n",
+      "P=E*I_o    \n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*(math.pi-2*th1*math.pi/180)+V**2*math.sin(math.radians(2*th1))-4*math.sqrt(2)*V*E*math.cos(math.radians(th1))))\n",
+      "P_r=I_or**2*R    \n",
+      "pf=(P+P_r)/(V*I_or)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg charging curent=%.4f A\" %I_o)\n",
+      "print(\"power supplied to the battery=%.2f W\" %P)\n",
+      "print(\"power dissipated by the resistor=%.3f W\" %P_r)    \n",
+      "print(\"supply pf=%.3f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg charging curent=3.5679 A\n",
+        "power supplied to the battery=535.18 W\n",
+        "power dissipated by the resistor=829.760 W\n",
+        "supply pf=0.583\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page No 284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "E=150.0\n",
+      "R=8.0\n",
+      "a=35.0\n",
+      "\n",
+      "#Calculations\n",
+      "th1=math.degrees(math.asin(E/(math.sqrt(2)*V)))\n",
+      "th2=180-th1\n",
+      "I_o=(1/(2*math.pi*R))*(math.sqrt(2)*230*(math.cos(math.radians(a))-math.cos(math.radians(th2)))-E*((th2-a)*math.pi/180))    \n",
+      "P=E*I_o    \n",
+      "I_or=math.sqrt((1/(2*math.pi*R**2))*((V**2+E**2)*((th2-a)*math.pi/180)-(V**2/2)*(math.sin(math.radians(2*th2))-math.sin(math.radians(2*a)))-2*math.sqrt(2)*V*E*(math.cos(math.radians(a))-math.cos(math.radians(th2)))))\n",
+      "P_r=I_or**2*R    \n",
+      "pf=(P+P_r)/(V*I_or)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg charging curent=%.4f A\" %I_o)\n",
+      "print(\"power supplied to the battery=%.2f W\" %P)\n",
+      "print(\"power dissipated by the resistor=%.2f W\" %P_r)\n",
+      "print(\"supply pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg charging curent=4.9208 A\n",
+        "power supplied to the battery=738.12 W\n",
+        "power dissipated by the resistor=689.54 W\n",
+        "supply pf=0.6686\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page No 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "B=210\n",
+      "f=50.0     #Hz\n",
+      "w=2*math.pi*f\n",
+      "a=40.0     #firing angle\n",
+      "V=230.0\n",
+      "R=5.0\n",
+      "L=2*10**-3\n",
+      "\n",
+      "#Calculations\n",
+      "t_c1=(360-B)*math.pi/(180*w)   \n",
+      "V_o1=(math.sqrt(2)*230/(2*math.pi))*(math.cos(math.radians(a))-math.cos(math.radians(B)))    \n",
+      "I_o1=V_o1/R    \n",
+      "E=110\n",
+      "R=5\n",
+      "L=2*10**-3\n",
+      "th1=math.degrees(math.asin(E/(math.sqrt(2)*V)))\n",
+      "t_c2=(360-B+th1)*math.pi/(180*w)    \n",
+      "V_o2=(math.sqrt(2)*230/(2*math.pi))*(math.cos(math.radians(a))-math.cos(math.radians(B)))    \n",
+      "I_o2=(1/(2*math.pi*R))*(math.sqrt(2)*230*(math.cos(math.radians(a))-math.cos(math.radians(B)))-E*((B-a)*math.pi/180))   \n",
+      "V_o2=R*I_o2+E    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"for R=5ohm and L=2mH\")\n",
+      "print(\"ckt turn off time=%.3f msec\" %(t_c1*1000))\n",
+      "print(\"avg output voltage=%.3f V\" %V_o1)\n",
+      "print(\"avg output current=%.4f A\" %I_o1)\n",
+      "print(\"for R=5ohm % L=2mH and E=110V\")\n",
+      "print(\"ckt turn off time=%.3f msec\" %(t_c2*1000))\n",
+      "print(\"avg output current=%.4f A\" %I_o2)\n",
+      "print(\"avg output voltage=%.3f V\" %V_o2)   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for R=5ohm and L=2mH\n",
+        "ckt turn off time=8.333 msec\n",
+        "avg output voltage=84.489 V\n",
+        "avg output current=16.8979 A\n",
+        "for R=5ohm % L=2mH and E=110V\n",
+        "ckt turn off time=9.431 msec\n",
+        "avg output current=6.5090 A\n",
+        "avg output voltage=142.545 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page No 286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "f=50.0\n",
+      "R=10.0\n",
+      "a=60.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=(math.sqrt(2)*V_s)\n",
+      "V_o=V_m/(2*math.pi)*(1+math.cos(math.radians(a)))\n",
+      "I_o=V_o/R\n",
+      "V_or=(V_m/(2*math.sqrt(math.pi)))*math.sqrt((math.pi-a*math.pi/180)+.5*math.sin(math.radians(2*a)))\n",
+      "I_or=V_or/R\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "FF=V_or/V_o    \n",
+      "VRF=math.sqrt(FF**2-1)    \n",
+      "TUF=P_dc/(V_s*I_or)    \n",
+      "PIV=V_m    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"rectification efficiency=%.4f\" %RE)\n",
+      "print(\"form factor=%.3f\" %FF)\n",
+      "print(\"voltage ripple factor=%.4f\" %VRF)\n",
+      "print(\"t/f utilisation factor=%.4f\" %TUF)\n",
+      "print(\"PIV of thyristor=%.2f V\" %PIV)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectification efficiency=0.2834\n",
+        "form factor=1.879\n",
+        "voltage ripple factor=1.5903\n",
+        "t/f utilisation factor=0.1797\n",
+        "PIV of thyristor=325.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6 Page No 294"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=1000.0\n",
+      "fos=2.5     #factor of safety\n",
+      "I_TAV=40.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m1=V/(2*fos)\n",
+      "P1=(2*V_m1/math.pi)*I_TAV    \n",
+      "V_m2=V/(fos)\n",
+      "P2=(2*V_m2/math.pi)*I_TAV    \n",
+      "\n",
+      "#Results\n",
+      "print(\"for mid pt convertor\")\n",
+      "print(\"power handled=%.3f kW\" %(P1/1000))\n",
+      "print(\"for bridge convertor\")\n",
+      "print(\"power handled=%.3f kW\" %(P2/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for mid pt convertor\n",
+        "power handled=5.093 kW\n",
+        "for bridge convertor\n",
+        "power handled=10.186 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7, Page No 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "R=.4\n",
+      "I_o=10\n",
+      "I_or=I_o\n",
+      "E=120.0\n",
+      "\n",
+      "#Calculations\n",
+      "a1=math.degrees(math.acos((E+I_o*R)*math.pi/(2*V_m)))\n",
+      "pf1=(E*I_o+I_or**2*R)/(V_s*I_or)    \n",
+      "E=-120.0\n",
+      "a2=math.degrees(math.acos((E+I_o*R)*math.pi/(2*V_m))) \n",
+      "pf2=(-E*I_o-I_or**2*R)/(V_s*I_or)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay=%.2f deg\" %a1)\n",
+      "print(\"pf=%.4f\" %pf1)\n",
+      "print(\"firing angle delay=%.2f deg\" %a2)\n",
+      "print(\"pf=%.4f\" %pf2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=53.21 deg\n",
+        "pf=0.5391\n",
+        "firing angle delay=124.07 deg\n",
+        "pf=0.5043\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 Page No 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "f=50.0\n",
+      "a=45.0\n",
+      "R=5.0\n",
+      "E=100.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=((math.sqrt(2)*V_s)/(2*math.pi))*(3+math.cos(math.radians(a)))\n",
+      "I_o=(V_o-E)/R    \n",
+      "P=E*I_o    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p current=%.3f A\" %I_o)\n",
+      "print(\"power delivered to battery=%.4f kW\" %(P/1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p current=18.382 A\n",
+        "power delivered to battery=1.8382 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10 Page No 300"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=230\n",
+      "f=50.0\n",
+      "a=50.0\n",
+      "R=6.0\n",
+      "E=60.0\n",
+      "V_o1=((math.sqrt(2)*2*V_s)/(math.pi))*math.cos(math.radians(a))\n",
+      "I_o1=(V_o1-E)/R    \n",
+      "\n",
+      "#ATQ after applying the conditions\n",
+      "V_o2=((math.sqrt(2)*V_s)/(math.pi))*math.cos(math.radians(a))\n",
+      "I_o2=(V_o2-E)/R    \n",
+      "\n",
+      "print(\"avg o/p current=%.3f A\" %I_o1)\n",
+      "print(\"avg o/p current after change=%.2f A\" %I_o2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p current=12.184 A\n",
+        "avg o/p current after change=1.09 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11 Page No 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=(2*V_m/math.pi)*math.cos(math.radians(a))\n",
+      "I_o=V_o/R\n",
+      "V_or=V_m/math.sqrt(2)\n",
+      "I_or=I_o\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "FF=V_or/V_o    \n",
+      "VRF=math.sqrt(FF**2-1)    \n",
+      "I_s1=2*math.sqrt(2)*I_o/math.pi\n",
+      "DF=math.cos(math.radians(a))\n",
+      "CDF=.90032\n",
+      "pf=CDF*DF    \n",
+      "HF=math.sqrt((1/CDF**2)-1)    \n",
+      "Q=2*V_m*I_o*math.sin(math.radians(a))/math.pi \n",
+      "\n",
+      "#Results\n",
+      "print(\"rectification efficiency=%.4f\" %RE)\n",
+      "print(\"form factor=%.4f\" %FF)\n",
+      "print(\"voltage ripple factor=%.4f\" %VRF)\n",
+      "print(\"pf=%.5f\" %pf)\n",
+      "print(\"HF=%.5f\" %HF)\n",
+      "print(\"active power=%.2f W\" %P_dc)   \n",
+      "print(\"reactive power=%.3f Var\" %Q)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectification efficiency=0.6366\n",
+        "form factor=1.5708\n",
+        "voltage ripple factor=1.2114\n",
+        "pf=0.63662\n",
+        "HF=0.48342\n",
+        "active power=2143.96 W\n",
+        "reactive power=2143.956 Var\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12, Page No 310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=(V_m/math.pi)*(1+math.cos(math.radians(a)))\n",
+      "I_o=V_o/R\n",
+      "V_or=V_s*math.sqrt((1/math.pi)*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))\n",
+      "I_or=I_o\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "FF=V_or/V_o    \n",
+      "VRF=math.sqrt(FF**2-1)    \n",
+      "I_s1=2*math.sqrt(2)*I_o*math.cos(math.radians(a/2))/math.pi\n",
+      "DF=math.cos(math.radians(a/2))   \n",
+      "CDF=2*math.sqrt(2)*math.cos(math.radians(a/2))/math.sqrt(math.pi*(math.pi-a*math.pi/180))    \n",
+      "pf=CDF*DF    \n",
+      "HF=math.sqrt((1/CDF**2)-1)    \n",
+      "Q=V_m*I_o*math.sin(math.radians(a))/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print(\"form factor=%.3f\" %FF)\n",
+      "print(\"rectification efficiency=%.4f\" %RE)\n",
+      "print(\"voltage ripple factor=%.3f\" %VRF)    \n",
+      "print(\"DF=%.4f\" %DF)\n",
+      "print(\"CDF=%.4f\" %CDF)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "print(\"HF=%.4f\" %HF)\n",
+      "print(\"active power=%.3f W\" %P_dc)\n",
+      "print(\"reactive power=%.2f Var\" %Q)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "form factor=1.241\n",
+        "rectification efficiency=0.8059\n",
+        "voltage ripple factor=0.735\n",
+        "DF=0.9239\n",
+        "CDF=0.9605\n",
+        "pf=0.8874\n",
+        "HF=0.2899\n",
+        "active power=3123.973 W\n",
+        "reactive power=1293.99 Var\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13, Page No 319"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "V_om=3*V_ml/(2*math.pi)\n",
+      "V_o=V_om/2\n",
+      "th=30\n",
+      "a=math.degrees(math.acos((2*math.pi*math.sqrt(3)*V_o/(3*V_ml)-1)))-th    \n",
+      "I_o=V_o/R    \n",
+      "V_or=V_ml/(2*math.sqrt(math.pi))*math.sqrt((5*math.pi/6-a*math.pi/180)+.5*math.sin(math.radians(2*a+2*th)))\n",
+      "I_or=V_or/R    \n",
+      "RE=V_o*I_o/(V_or*I_or)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"delay angle=%.1f deg\" %a)\n",
+      "print(\"avg load current=%.3f A\" %I_o)\n",
+      "print(\"rms load current=%.3f A\" %I_or)\n",
+      "print(\"rectification efficiency=%.4f\" %RE)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "delay angle=67.7 deg\n",
+        "avg load current=7.765 A\n",
+        "rms load current=10.477 A\n",
+        "rectification efficiency=0.5494\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15, Page No 321"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=400.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "v_T=1.4\n",
+      "a1=30.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o1=3*V_ml/(2*math.pi)*math.cos(math.radians(a1))-v_T   \n",
+      "a2=60.0\n",
+      "V_o2=3*V_ml/(2*math.pi)*math.cos(math.radians(a2))-v_T  \n",
+      "I_o=36\n",
+      "I_TA=I_o/3    \n",
+      "I_Tr=I_o/math.sqrt(3)    \n",
+      "P=I_TA*v_T       \n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle = 30deg\")\n",
+      "print(\"avg output voltage=%.3f V\" %V_o1)\n",
+      "print(\"for firing angle = 60deg\")\n",
+      "print(\"avg output voltage=%.2f V\" %V_o2)\n",
+      "print(\"avg current rating=%.0f A\" %I_TA)\n",
+      "print(\"rms current rating=%.3f A\" %I_Tr)\n",
+      "print(\"PIV of SCR=%.1f V\" %V_ml)\n",
+      "print(\"power dissipated=%.1f W\" %P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle = 30deg\n",
+        "avg output voltage=232.509 V\n",
+        "for firing angle = 60deg\n",
+        "avg output voltage=133.65 V\n",
+        "avg current rating=12 A\n",
+        "rms current rating=20.785 A\n",
+        "PIV of SCR=565.7 V\n",
+        "power dissipated=16.8 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17, Page No 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "E=200\n",
+      "I_o=20\n",
+      "R=.5\n",
+      "\n",
+      "#Calculations\n",
+      "V_o1=E+I_o*R\n",
+      "V_s=230\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "a1=math.degrees(math.acos(V_o1*math.pi/(3*V_ml)))\n",
+      "th=120\n",
+      "I_s=math.sqrt((1/math.pi)*I_o**2*th*math.pi/180)\n",
+      "P=E*I_o+I_o**2*R\n",
+      "pf=P/(math.sqrt(3)*V_s*I_s)    \n",
+      "V_o2=E-I_o*R\n",
+      "a2=math.degrees(math.acos(-V_o2*math.pi/(3*V_ml))) \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay=%.3f deg\" %a1)\n",
+      "print(\"pf=%.3f\" %pf)\n",
+      "print(\"firing angle delay=%.2f deg\" %a2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=47.461 deg\n",
+        "pf=0.646\n",
+        "firing angle delay=127.71 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18, Page No 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "w=2*math.pi*f\n",
+      "a1=0\n",
+      "t_c1=(4*math.pi/3-a1*math.pi/180)/w    \n",
+      "a2=30\n",
+      "t_c2=(4*math.pi/3-a2*math.pi/180)/w    \n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle delay=0deg\")\n",
+      "print(\"commutation time=%.2f ms\" %(t_c1*1000))\n",
+      "print(\"peak reverse voltage=%.2f V\" %(math.sqrt(2)*V))\n",
+      "print(\"for firing angle delay=30deg\")\n",
+      "print(\"commutation time=%.2f ms\" %(t_c2*1000))\n",
+      "print(\"peak reverse voltage=%.2f V\" %(math.sqrt(2)*V))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle delay=0deg\n",
+        "commutation time=13.33 ms\n",
+        "peak reverse voltage=325.27 V\n",
+        "for firing angle delay=30deg\n",
+        "commutation time=11.67 ms\n",
+        "peak reverse voltage=325.27 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.19, Page No 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*2*math.pi/(2*3)/(math.pi/3+math.sqrt(3)*math.cos(math.radians(2*a))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*math.pi/(math.sqrt(2)*3*math.cos(math.radians(a)))\n",
+      "V_ph=V_s/math.sqrt(3) \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage percent V_ph=%.3f V\" %V_ph)   \n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.2f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage percent V_ph=110.384 V\n",
+        "for constant load current\n",
+        "V_ph=110.38 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.20, Page No 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*4*math.pi/(2*3)/(2*math.pi/3+math.sqrt(3)*(1+math.cos(math.radians(2*a)))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*2*math.pi/(math.sqrt(2)*3*(1+math.cos(math.radians(a))))\n",
+      "V_ph=V_s/math.sqrt(3)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage percent V_ph=%.3f V\" %V_ph) \n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.2f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage percent V_ph=102.459 V\n",
+        "for constant load current\n",
+        "V_ph=102.46 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.21, Page No 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=90.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*4*math.pi/(2*3)/((math.pi-math.pi/2)+(math.sin(math.radians(2*a)))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*2*math.pi/(math.sqrt(2)*3*(1+math.cos(math.radians(a))))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage percent V_ph=%.2f V\" %V_ph)\n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.1f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage percent V_ph=191.19 V\n",
+        "for constant load current\n",
+        "V_ph=191.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.22 Page No 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "E=200.0\n",
+      "I_o=20.0\n",
+      "R=.5\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=E+I_o*R\n",
+      "V_s=230\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "a=math.degrees(math.acos(V_o*2*math.pi/(3*V_ml)-1))  \n",
+      "a1=180-a\n",
+      "I_sr=math.sqrt((1/math.pi)*I_o**2*(a1*math.pi/180))\n",
+      "P=V_o*I_o\n",
+      "pf=P/(math.sqrt(3)*V_s*I_sr)   \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle delay=%.2f deg\" %a)\n",
+      "print(\"pf=%.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=69.38 deg\n",
+        "pf=0.67\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.23, Page No 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=400.0\n",
+      "f=50.0\n",
+      "I_o=15.0\n",
+      "a=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_TA=I_o*120.0/360.0\n",
+      "I_Tr=math.sqrt(I_o**2*120/360)\n",
+      "I_sr=math.sqrt(I_o**2*120/180)\n",
+      "V_ml=math.sqrt(2)*V_s\n",
+      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "V_or=V_ml*math.sqrt((3/(2*math.pi))*(math.pi/3+math.sqrt(3/2)*math.cos(math.radians(2*a))))\n",
+      "I_or=I_o\n",
+      "P_dc=V_o*I_o\n",
+      "P_ac=V_or*I_or\n",
+      "RE=P_dc/P_ac    \n",
+      "VA=3*V_s/math.sqrt(3)*I_sr\n",
+      "TUF=P_dc/VA    \n",
+      "pf=P_ac/VA    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rectification efficiency=%.5f\" %RE)\n",
+      "print(\"TUF=%.4f\" %TUF)\n",
+      "print(\"Input pf=%.3f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rectification efficiency=0.95493\n",
+        "TUF=0.6752\n",
+        "Input pf=0.707\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.24, Page No 341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I=10.0\n",
+      "a=45.0\n",
+      "V=400.0\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "DF=math.cos(math.radians(a))\n",
+      "I_o=10\n",
+      "I_s1=4*I_o/(math.sqrt(2)*math.pi)*math.sin(math.pi/3)\n",
+      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
+      "I_o=1     #suppose\n",
+      "CDF=I_s1/I_sr    \n",
+      "THD=math.sqrt(1/CDF**2-1)    \n",
+      "pf=CDF*DF    \n",
+      "P=(3*math.sqrt(2)*V*math.cos(math.radians(a))/math.pi)*I\n",
+      "Q=(3*math.sqrt(2)*V*math.sin(math.radians(a))/math.pi)*I  \n",
+      "   \n",
+      "#Results\n",
+      "print(\"DF=%.3f\" %DF)\n",
+      "print(\"CDF=%.3f\" %CDF)\n",
+      "print(\"THD=%.5f\" %THD)\n",
+      "print(\"PF=%.4f\" %pf)\n",
+      "print(\"active power=%.2f W\" %P)  \n",
+      "print(\"reactive power=%.2f Var\" %Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "DF=0.707\n",
+        "CDF=0.955\n",
+        "THD=0.31084\n",
+        "PF=0.6752\n",
+        "active power=3819.72 W\n",
+        "reactive power=3819.72 Var\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.25, Page No 342"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "print(\"for firing angle=30deg\")\n",
+      "a=30.0\n",
+      "V=400.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "E=350\n",
+      "R=10\n",
+      "\n",
+      "#Calculations\n",
+      "I_o=(V_o-E)/R\n",
+      "I_or=I_o\n",
+      "P1=V_o*I_o    \n",
+      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
+      "VA=3*V/math.sqrt(3)*I_sr\n",
+      "pf=P1/VA \n",
+      "a=180-60\n",
+      "V=400\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "V_o=3*V_ml*math.cos(math.radians(a))/math.pi\n",
+      "E=-350\n",
+      "R=10\n",
+      "I_o=(V_o-E)/R\n",
+      "I_or=I_o\n",
+      "P2=-V_o*I_o    \n",
+      "I_sr=I_o*math.sqrt(2.0/3.0)\n",
+      "VA=3*V/math.sqrt(3)*I_sr\n",
+      "pf=P2/VA       \n",
+      "\n",
+      "print(\"power delivered to load=%.2f W\" %P1)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "print(\"for firing advance angle=60deg\")\n",
+      "print(\"power delivered to load=%.2f W\" %P2)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=30deg\n",
+        "power delivered to load=5511.74 W\n",
+        "pf=0.4775\n",
+        "for firing advance angle=60deg\n",
+        "power delivered to load=2158.20 W\n",
+        "pf=0.4775\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.26, Page No 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0\n",
+      "u=15.0\n",
+      "\n",
+      "#Calculations\n",
+      "i=math.cos(math.radians(a))-math.cos(math.radians(a+u))\n",
+      "a=30\n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a  \n",
+      "a=45\n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a    \n",
+      "a=60\n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-i))-a  \n",
+      "\n",
+      "#Results\n",
+      "print(\"for firing angle=30deg\") \n",
+      "print(\"overlap angle=%.1f deg\" %u)\n",
+      "print(\"for firing angle=45deg\")  \n",
+      "print(\"overlap angle=%.1f deg\" %u)\n",
+      "print(\"for firing angle=60deg\") \n",
+      "print(\"overlap angle=%.2f deg\" %u)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle=30deg\n",
+        "overlap angle=2.2 deg\n",
+        "for firing angle=45deg\n",
+        "overlap angle=2.2 deg\n",
+        "for firing angle=60deg\n",
+        "overlap angle=2.23 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.28, Page No 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "E=400.0\n",
+      "I_o=20.0\n",
+      "R=1\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=E+I_o*R\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=.004\n",
+      "V=230 #per phase voltage\n",
+      "V_ml=math.sqrt(6)*V\n",
+      "a=math.degrees(math.acos(math.pi/(3*V_ml)*(V_o+3*w*L*I_o/math.pi)))   \n",
+      "print(\"firing angle delay=%.3f deg\" %a)\n",
+      "u=math.degrees(math.acos(math.pi/(3*V_ml)*(V_o-3*w*L*I_o/math.pi)))-a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"overlap angle=%.2f deg\" %u)\n",
+      "#Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=34.382 deg\n",
+        "overlap angle=8.22 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.29, Page No 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=400.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "R=1\n",
+      "E=230\n",
+      "I=15.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=-E+I*R\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "a=math.degrees(math.acos(V_o*2*math.pi/(3*V_ml)))  \n",
+      "L=0.004\n",
+      "a=math.degrees(math.acos((2*math.pi)/(3*V_ml)*(V_o+3*w*L*I/(2*math.pi))))  \n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-3*f*L*I/V_ml))-a    \n",
+      "\n",
+      "#Results\n",
+      "print(\"firing angle=%.3f deg\" %a)\n",
+      "print(\"firing angle delay=%.3f deg\" %a)\n",
+      "print(\"overlap angle=%.3f deg\" %u)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle=139.702 deg\n",
+        "firing angle delay=139.702 deg\n",
+        "overlap angle=1.431 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.31, Page No 361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0 #per phase\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_ml=math.sqrt(3.0)*math.sqrt(2)*V\n",
+      "w=2*math.pi*f\n",
+      "a1=60.0\n",
+      "L=0.015\n",
+      "i_cp=(math.sqrt(3)*V_ml/(w*L))*(1-math.sin(math.radians(a1)))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"circulating current=%.4f A\" %i_cp)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "circulating current=27.7425 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.32, Page No 362"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "V_m=math.sqrt(2)*V\n",
+      "a=30.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=2*V_m* math.cos(math.radians(a))/math.pi    \n",
+      "R=10\n",
+      "I_o=V_o/R    \n",
+      "I_TA=I_o*math.pi/(2*math.pi)    \n",
+      "I_Tr=math.sqrt(I_o**2*math.pi/(2*math.pi))    \n",
+      "I_s=math.sqrt(I_o**2*math.pi/(math.pi)) \n",
+      "I_o=I_s\n",
+      "pf=(V_o*I_o/(V*I_s))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.3f V\" %V_o)\n",
+      "print(\"avg o/p current=%.2f A\" %I_o)\n",
+      "print(\"avg value of thyristor current=%.3f A\" %I_TA)\n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_Tr)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits.\n",
+      " \n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p voltage=179.330 V\n",
+        "avg o/p current=17.93 A\n",
+        "avg value of thyristor current=8.967 A\n",
+        "rms value of thyristor current=12.68 A\n",
+        "pf=0.7797\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.33, Page No 363"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "V_m=math.sqrt(2)*V\n",
+      "a=30.0\n",
+      "L=.0015\n",
+      "\n",
+      "#Calculations\n",
+      "V_o=2*V_m* math.cos(math.radians(a))/math.pi  \n",
+      "R=10\n",
+      "I_o=V_o/R  \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "V_ox=2*V_m*math.cos(math.radians(a))/math.pi-w*L*I_o/math.pi    \n",
+      "u=math.degrees(math.acos(math.cos(math.radians(a))-I_o*w*L/V_m))-a    \n",
+      "I=I_o\n",
+      "pf=V_o*I_o/(V*I)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.3f V\" %V_ox)\n",
+      "print(\"angle of overlap=%.3f deg\" %u)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p voltage=176.640 V\n",
+        "angle of overlap=2.855 deg\n",
+        "pf=0.7797\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.34, Page No 364"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=415.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "a1=35.0 #firing angle advance\n",
+      "\n",
+      "#Calculations\n",
+      "a=180-a1\n",
+      "I_o=80.0\n",
+      "r_s=0.04\n",
+      "v_T=1.5\n",
+      "X_l=.25 #reactance=w*L\n",
+      "E=-3*V_ml*math.cos(math.radians(a))/math.pi+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi    \n",
+      "\n",
+      "#Results\n",
+      "print(\"mean generator voltage=%.3f V\" %E)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mean generator voltage=487.590 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.35, Page No 364"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=415.0\n",
+      "V_ml=math.sqrt(2)*V\n",
+      "R=0.2\n",
+      "I_o=80\n",
+      "r_s=0.04\n",
+      "v_T=1.5\n",
+      "\n",
+      "#Calculations\n",
+      "X_l=.25 #reactance=w*L\n",
+      "a=35\n",
+      "E=-(-3*V_ml*math.cos(math.radians(a))/math.pi+I_o*R+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi)    \n",
+      "a1=35\n",
+      "a=180-a1\n",
+      "E=(-3*V_ml*math.cos(math.radians(a))/math.pi+I_o*R+2*I_o*r_s+2*v_T+3*X_l*I_o/math.pi) \n",
+      "\n",
+      "#Results\n",
+      "print(\"when firing angle=35deg\")   \n",
+      "print(\"mean generator voltage=%.3f V\" %E)\n",
+      "print(\"when firing angle advance=35deg\")\n",
+      "print(\"mean generator voltage=%.3f V\" %E)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when firing angle=35deg\n",
+        "mean generator voltage=503.590 V\n",
+        "when firing angle advance=35deg\n",
+        "mean generator voltage=503.590 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.36, Page No 365"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=5.0\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_mp=math.sqrt(2)*V\n",
+      "a=30.0\n",
+      "E=150.0\n",
+      "B=180-math.degrees(math.asin(E/V_mp))\n",
+      "I_o=(3/(2*math.pi*R))*(V_mp*(math.cos(math.radians(a+30))-math.cos(math.radians(B)))-E*((B-a-30)*math.pi/180))\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg current flowing=%.2f A\" %I_o)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg current flowing=19.96 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.37, Page No 366"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "V=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_m=math.sqrt(2)*V\n",
+      "V_o=V_m*(1+math.cos(math.radians(a)))/math.pi    \n",
+      "E=100\n",
+      "R=10\n",
+      "I_o=(V_o-E)/R    \n",
+      "I_TA=I_o*math.pi/(2*math.pi)    \n",
+      "I_Tr=math.sqrt(I_o**2*math.pi/(2*math.pi))    \n",
+      "I_s=math.sqrt(I_o**2*(1-a/180)*math.pi/(math.pi))\n",
+      "I_or=I_o\n",
+      "P=E*I_o+I_or**2*R\n",
+      "pf=(P/(V*I_s))   \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "t_c=(1-a/180)*math.pi/w    \n",
+      "\n",
+      "#Results\n",
+      "print(\"\\navg o/p current=%.2f A\" %I_o)\n",
+      "print(\"avg o/p voltage=%.3f V\" %V_o)\n",
+      "print(\"avg value of thyristor current=%.2f A\" %I_TA)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %I_Tr)\n",
+      "print(\"avg value of diode current=%.2f A\" %I_TA)\n",
+      "print(\"rms value of diode current=%.3f A\" %I_Tr)\n",
+      "print(\"pf=%.4f\" %pf)\n",
+      "print(\"circuit turn off time=%.2f ms\" %(t_c*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "avg o/p current=9.32 A\n",
+        "avg o/p voltage=193.202 V\n",
+        "avg value of thyristor current=4.66 A\n",
+        "rms value of thyristor current=6.590 A\n",
+        "avg value of diode current=4.66 A\n",
+        "rms value of diode current=6.590 A\n",
+        "pf=0.9202\n",
+        "circuit turn off time=8.33 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.38, Page No 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "V_m=math.sqrt(2)*V\n",
+      "L=0.05\n",
+      "f=50.0\n",
+      "\n",
+      "#Calculations\n",
+      "w=2*math.pi*f\n",
+      "a=30\n",
+      "i_cp=2*V_m*(1-math.cos(math.radians(a)))/(w*L)    \n",
+      "R=30.0\n",
+      "i_l=V_m/R\n",
+      "i1=i_cp+i_l    \n",
+      "i2=i_cp    \n",
+      "\n",
+      "#Results\n",
+      "print(\"peak value of circulating current=%.3f A\" %i_cp)\n",
+      "print(\"peak value of current in convertor 1=%.3f A\" %i1)\n",
+      "print(\"peak value of current in convertor 2=%.3f A\" %i2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak value of circulating current=5.548 A\n",
+        "peak value of current in convertor 1=16.391 A\n",
+        "peak value of current in convertor 2=5.548 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.39, Page No 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "R=5.0\n",
+      "L=0.05\n",
+      "\n",
+      "#Calculations\n",
+      "phi=math.degrees(math.atan(w*L/R))  \n",
+      "phi=90+math.degrees(math.atan(w*L/R))   \n",
+      "\n",
+      "#Results\n",
+      "print(\"for no current transients\")\n",
+      "print(\"triggering angle=%.2f deg\" %phi)\n",
+      "print(\"for worst transients\")\n",
+      "print(\"triggering angle=%.2f deg\" %phi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for no current transients\n",
+        "triggering angle=162.34 deg\n",
+        "for worst transients\n",
+        "triggering angle=162.34 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter7_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter7_4.ipynb
new file mode 100755
index 00000000..726160c8
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter7_4.ipynb
@@ -0,0 +1,1036 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 07 : Choppers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page No 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.4     #duty cycle %a=T_on/T\n",
+      "V_s=230.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "V=a*(V_s-2)    \n",
+      "V_or=math.sqrt(a*(V_s-2)**2)    \n",
+      "P_o=V_or**2/R\n",
+      "P_i=V_s*V/R\n",
+      "n=P_o*100/P_i    \n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p voltage=%.1f V\" %V)\n",
+      "print(\"rms value of o/p voltage=%.1f V\" %V_or)\n",
+      "print(\"chopper efficiency in percentage=%.2f\" %n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg o/p voltage=91.2 V\n",
+        "rms value of o/p voltage=144.2 V\n",
+        "chopper efficiency in percentage=99.13\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page No 388"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_i=220.0\n",
+      "V_o=660.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=1-V_i/V_o\n",
+      "T_on=100.0     #microsecond\n",
+      "T=T_on/a\n",
+      "T_off=T-T_on    \n",
+      "T_off=T_off/2\n",
+      "T_on=T-T_off\n",
+      "a=T_on/T\n",
+      "V_o=V_i/(1-a)\n",
+      "\n",
+      "#Results    \n",
+      "print(\"pulse width of o/p voltage=%.0f us\" %T_off)\n",
+      "print(\"\\nnew o/p voltage=%.0f V\" %V_o)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pulse width of o/p voltage=25 us\n",
+        "\n",
+        "new o/p voltage=1320 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page No 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_1=12.0\n",
+      "I_2=16.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_0=(I_1+I_2)/2\n",
+      "R=10.0\n",
+      "V_0=I_0*R\n",
+      "V_s=200.0\n",
+      "a=V_0/V_s\n",
+      "r=a/(1-a)\n",
+      "\n",
+      "#Results\n",
+      "print(\"time ratio(T_on/T_off)=%.3f\" %r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time ratio(T_on/T_off)=2.333\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page No 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_o=660.0\n",
+      "V_s=220.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=(V_o/V_s)/(1+(V_o/V_s))\n",
+      "T_on=120\n",
+      "T=T_on/a\n",
+      "T_off=T-T_on \n",
+      "T_off=3*T_off\n",
+      "T_on=T-T_off\n",
+      "a=T_on/(T_on+T_off)\n",
+      "V_o=V_s*(a/(1-a))   \n",
+      "\n",
+      "#Results\n",
+      "print(\"pulse width o/p voltage=%.0f us\" %T_off)\n",
+      "print(\"\\nnew o/p voltage=%.2f V\" %V_o)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pulse width o/p voltage=120 us\n",
+        "\n",
+        "new o/p voltage=73.33 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11 Page No 408"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=1.0\n",
+      "L=.005\n",
+      "T_a=L/R\n",
+      "T=2000*10**-6\n",
+      "E=24.0\n",
+      "V_s=220\n",
+      "T_on=600*10**-6\n",
+      "a=T_on/T\n",
+      "\n",
+      "#Calculations\n",
+      "a1=(T_a/T)*math.log(1+(E/V_s)*((math.exp(T/T_a))-1))\n",
+      "if a1<a :\n",
+      "    print(\"load current in continuous\")\n",
+      "else:\n",
+      "    print(\"load current in discont.\")\n",
+      "\n",
+      "I_o=(a*V_s-E)/R    \n",
+      "I_mx=(V_s/R)*((1-math.exp(-T_on/T_a))/(1-math.exp(-T/T_a)))-E/R    \n",
+      "I_mn=(V_s/R)*((math.exp(T_on/T_a)-1)/(math.exp(T/T_a)-1))-E/R    \n",
+      "f=1/T\n",
+      "w=2*math.pi*f\n",
+      "I1=(2*V_s/(math.sqrt(2)*math.pi)*math.sin(math.radians(180*a)))/(math.sqrt(R**2+(w*L)**2))    \n",
+      "I2=(2*V_s/(2*math.sqrt(2)*math.pi)*math.sin(math.radians(2*180*a)))/(math.sqrt(R**2+(w*L*2)**2))    \n",
+      "I3=(2*V_s/(3*math.sqrt(2)*math.pi)*math.sin(math.radians(3*180*a)))/(math.sqrt(R**2+(w*L*3)**2))    \n",
+      "I_TAV=a*(V_s-E)/R-L*(I_mx-I_mn)/(R*T)   \n",
+      "P1=I_TAV*V_s\n",
+      "P2=E*I_o\n",
+      "I_or=math.sqrt(I_o**2+I1**2+I2**2+I3**2)\n",
+      "\n",
+      "#Results\n",
+      "print(\"avg o/p current=%.2f A\" %I_o)\n",
+      "print(\"max value of steady current=%.2f A\" %I_mx)\n",
+      "print(\"min value of steady current=%.2f A\" %I_mn)\n",
+      "print(\"first harmonic current=%.4f A\" %I1)\n",
+      "print(\"second harmonic current=%.4f A\" %I2)\n",
+      "print(\"third harmonic current=%.5f A\" %I3)\n",
+      "print(\"avg supply current=%.4f A\" %I_TAV)\n",
+      "print(\"i/p power=%.2f W\" %P1)\n",
+      "print(\"power absorbed by load emf=%.0f W\" %P2)\n",
+      "print(\"power loss in resistor=%.2f W\" %(P1-P2))\n",
+      "print(\"rms value of load current=%.3f A\" %I_or)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "load current in continuous\n",
+        "avg o/p current=42.00 A\n",
+        "max value of steady current=51.46 A\n",
+        "min value of steady current=33.03 A\n",
+        "first harmonic current=5.0903 A\n",
+        "second harmonic current=1.4983 A\n",
+        "third harmonic current=0.21643 A\n",
+        "avg supply current=12.7289 A\n",
+        "i/p power=2800.35 W\n",
+        "power absorbed by load emf=1008 W\n",
+        "power loss in resistor=1792.35 W\n",
+        "rms value of load current=42.334 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12 Page No 411"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=1\n",
+      "L=.001\n",
+      "V_s=220\n",
+      "E=72.0\n",
+      "f=500.0\n",
+      "T_on=800*10**-6\n",
+      "T_a=L/R\n",
+      "T=1.0/f\n",
+      "m=E/V_s\n",
+      "a=T_on/T\n",
+      "\n",
+      "#Calculations\n",
+      "a1=(T_a/T)*math.log(1+m*(math.exp(-T/T_a)-1))\n",
+      "if a1>a :\n",
+      "    print(\"load current is continuous\")\n",
+      "else:\n",
+      "    print(\"load current is discontinuous\")\n",
+      "\n",
+      "t_x=T_on+L*math.log(1+((V_s-E)/272)*(1-math.exp(-T_on/T_a)))\n",
+      " #Value of t_x wrongly calculated in the book so ans of V_o and I_o varies\n",
+      "V_o=a*V_s+(1-t_x/T)*E    \n",
+      "I_o=(V_o-E)/R    \n",
+      "I_mx=(V_s-E)/R*(1-math.exp(-T_on/T_a)) \n",
+      "\n",
+      "#Results  \n",
+      "print(\"avg o/p voltage=%.2f V\" %V_o)\n",
+      "print(\"avg o/p current=%.2f A\" %I_o) \n",
+      "print(\"max value of load current=%.1f A\" %I_mx)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "load current is discontinuous\n",
+        "avg o/p voltage=121.77 V\n",
+        "avg o/p current=49.77 A\n",
+        "max value of load current=81.5 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13, Page No 412"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.2\n",
+      "V_s=500\n",
+      "E=a*V_s\n",
+      "L=0.06\n",
+      "I=10\n",
+      "\n",
+      "#Calculations\n",
+      "T_on=(L*I)/(V_s-E)\n",
+      "f=a/T_on    \n",
+      "\n",
+      "#Results\n",
+      "print(\"chopping freq=%.2f Hz\" %f)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "chopping freq=133.33 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14 Page No 412"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=0.5\n",
+      "pu=0.1 #pu ripple\n",
+      "\n",
+      "#Calculations\n",
+      " #x=T/T_a\n",
+      " #y=exp(-a*x)\n",
+      "y=(1-pu)/(1+pu)\n",
+      " #after solving\n",
+      "x=math.log(1/y)/a\n",
+      "f=1000\n",
+      "T=1/f\n",
+      "T_a=T/x\n",
+      "R=2\n",
+      "L=R*T_a\n",
+      "Li=0.002\n",
+      "Le=L-Li    \n",
+      "\n",
+      "#Results\n",
+      "print(\"external inductance=%.3f mH\" %(Le*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "external inductance=-2.000 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15 Page No 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=10.0\n",
+      "L=0.015\n",
+      "T_a=L/R\n",
+      "f=1250.0\n",
+      "T=1.0/f\n",
+      "a=0.5\n",
+      "T_on=a*T\n",
+      "V_s=220.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_mx=(V_s/R)*((1-math.exp(-T_on/T_a))/(1-math.exp(-T/T_a)))   \n",
+      "I_mn=(V_s/R)*((math.exp(T_on/T_a)-1)/(math.exp(T/T_a)-1))    \n",
+      "dI=I_mx-I_mn    \n",
+      "V_o=a*V_s\n",
+      "I_o=V_o/R    \n",
+      "I_or=math.sqrt(I_mx**2+dI**2/3+I_mx*dI)    \n",
+      "I_chr=math.sqrt(a)*I_or \n",
+      "\n",
+      "#Results\n",
+      "print(\"Max value of ripple current=%.2f A\" %dI)\n",
+      "print(\"Max value of load current=%.3f A\" %I_mx)\n",
+      "print(\"Min value of load current=%.2f A\" %I_mn)\n",
+      "print(\"Avg value of load current=%.2f A\" %I_o)   \n",
+      "print(\"rms value of load current=%.2f A\" %I_or)\n",
+      "print(\"rms value of chopper current=%.2f A\" %I_chr)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Max value of ripple current=2.92 A\n",
+        "Max value of load current=12.458 A\n",
+        "Min value of load current=9.54 A\n",
+        "Avg value of load current=11.00 A\n",
+        "rms value of load current=13.94 A\n",
+        "rms value of chopper current=9.86 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17 Page No 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=0.0016\n",
+      "C=4*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "w=1/math.sqrt(L*C)\n",
+      "t=math.pi/w    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"time for which current flows=%.2f us\" %(t*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for which current flows=251.33 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.18, Page No 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "t_q=20.0*10**-6\n",
+      "dt=20.0*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "t_c=t_q+dt\n",
+      "I_0=60.0\n",
+      "V_s=60.0\n",
+      "C=t_c*I_0/V_s \n",
+      "\n",
+      "#Results   \n",
+      "print(\"value of commutating capacitor=%.0f uF\" %(C*10**6))\n",
+      "\n",
+      "L1=(V_s/I_0)**2*C\n",
+      "L2=(2*t_c/math.pi)**2/C\n",
+      "if L1>L2 :\n",
+      "    print(\"value of commutating inductor=%.0f uH\" %(L1*10**6))\n",
+      "else:\n",
+      "    print(\"value of commutating inductor=%.0f uH\" %(L2*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of commutating capacitor=40 uF\n",
+        "value of commutating inductor=40 uH\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.19, Page No 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "t=100.0*10**-6\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      " #V_s*(1-2*math.exp(-t/(R*C)))=0\n",
+      "C=-t/(R*math.log(1.0/2))    \n",
+      "L=(4/9.0)*C*R**2    \n",
+      "L=(1.0/4)*C*R**2    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Value of comutating component C=%.3f uF\" %(C*10**6))\n",
+      "print(\"max permissible current through SCR is 2.5 times load current\")\n",
+      "print(\"value of comutating component L=%.1f uH\" %(L*10**6))\n",
+      "print(\"max permissible current through SCR is 1.5 times peak diode current\")\n",
+      "print(\"value of comutating component L=%.2f uH\" %(L*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of comutating component C=14.427 uF\n",
+        "max permissible current through SCR is 2.5 times load current\n",
+        "value of comutating component L=360.7 uH\n",
+        "max permissible current through SCR is 1.5 times peak diode current\n",
+        "value of comutating component L=360.67 uH\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.20, Page No 426"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T_on=800.0*10**-6\n",
+      "V_s=220.0\n",
+      "I_o=80.0\n",
+      "C=50*10**-6\n",
+      "\n",
+      "#Calculations\n",
+      "T=T_on+2*V_s*C/I_o    \n",
+      "L=20*10**-6\n",
+      "C=50*10**-6\n",
+      "i_T1=I_o+V_s*math.sqrt(C/L)    \n",
+      "i_TA=I_o    \n",
+      "t_c=C*V_s/I_o    \n",
+      "t_c1=(math.pi/2)*math.sqrt(L*C)    \n",
+      "t=150*10**-6\n",
+      "v_c=I_o*t/C-V_s  \n",
+      "\n",
+      "#Results  \n",
+      "print(\"effective on period=%.0f us\" %(T*10**6))\n",
+      "print(\"peak current through main thyristor=%.2f A\" %i_T1)\n",
+      "print(\"peak current through auxillery thyristor=%.0f A\" %i_TA)\n",
+      "print(\"turn off time for main thyristor=%.1f us\" %(t_c*10**6))\n",
+      "print(\"turn off time for auxillery thyristor=%.3f us\" %(t_c1*10**6))\n",
+      "print(\"total commutation interval=%.0f us\" %(2*t_c*10**6))\n",
+      "print(\"capacitor voltage=%.0f V\" %v_c)\n",
+      "print(\"time nedded to recharge the capacitor=%.0f us\" %(2*V_s*C/I_o*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for firing angle = 30deg\n",
+        "avg output voltage=232.509 V\n",
+        "for firing angle = 60deg\n",
+        "avg output voltage=133.65 V\n",
+        "avg current rating=12 A\n",
+        "rms current rating=20.785 A\n",
+        "PIV of SCR=565.7 V\n",
+        "power dissipated=16.8 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 122
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.21, Page No 427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_o=260.0\n",
+      "V_s=220.0\n",
+      "fos=2 #factor of safety\n",
+      "\n",
+      "#Calculations\n",
+      "t_off=18*10**-6\n",
+      "t_c=2*t_off\n",
+      "C=t_c*I_o/V_s    \n",
+      "L=(V_s/(0.8*I_o))**2*C    \n",
+      "f=400\n",
+      "a_mn=math.pi*f*math.sqrt(L*C)\n",
+      "V_omn=V_s*(a_mn+2*f*t_c)    \n",
+      "V_omx=V_s    \n",
+      "\n",
+      "#Results\n",
+      "print(\"Value of C=%.3f uF\" %(C*10**6))\n",
+      "print(\"value of L=%.3f uH\" %(L*10**6))\n",
+      "print(\"min value of o/p voltage=%.3f V\" %V_omn)\n",
+      "print(\"max value of o/p voltage=%.0f V\" %V_omx)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "firing angle delay=47.461 deg\n",
+        "pf=0.646\n",
+        "firing angle delay=127.71 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": "*"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.22, Page No 434"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "x=2.0\n",
+      "t_q=30*10**-6\n",
+      "dt=30*10**-6\n",
+      "t_c=t_q+dt\n",
+      "V_s=230.0\n",
+      "I_o=200.0\n",
+      "\n",
+      "#Calculations\n",
+      "L=V_s*t_c/(x*I_o*(math.pi-2*math.asin(1/x)))    \n",
+      "C=x*I_o*t_c/(V_s*(math.pi-2*math.asin(1/x)))    \n",
+      "V_cp=V_s+I_o*math.sqrt(L/C)    \n",
+      "I_cp=x*I_o    \n",
+      "x=3\n",
+      "L=V_s*t_c/(x*I_o*(math.pi-2*math.asin(1/x)))   \n",
+      "C=x*I_o*t_c/(V_s*(math.pi-2*math.asin(1/x)))    \n",
+      "V_cp=V_s+I_o*math.sqrt(L/C)    \n",
+      "I_cp=x*I_o    \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of commutating inductor=%.3f uH\" %(L*10**6))\n",
+      "print(\"value of commutating capacitor=%.3f uF\" %(C*10**6))\n",
+      "print(\"peak capacitor voltage=%.0f V\" %V_cp)\n",
+      "print(\"peak commutataing current=%.0f A\" %I_cp)\n",
+      "print(\"value of commutating inductor=%.3f uH\" %(L*10**6))\n",
+      "print(\"value of commutating capacitor=%.3f uF\" %(C*10**6))\n",
+      "print(\"peak capacitor voltage=%.2f V\" %V_cp)\n",
+      "print(\"peak commutataing current=%.0f A\" %I_cp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of commutating inductor=7.321 uH\n",
+        "value of commutating capacitor=49.822 uF\n",
+        "peak capacitor voltage=307 V\n",
+        "peak commutataing current=600 A\n",
+        "value of commutating inductor=7.321 uH\n",
+        "value of commutating capacitor=49.822 uF\n",
+        "peak capacitor voltage=306.67 V\n",
+        "peak commutataing current=600 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.23, Page No 434"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesV_s=230\n",
+      "C=50*10**-6\n",
+      "L=20*10**-6\n",
+      "I_cp=V_s*math.sqrt(C/L)\n",
+      "I_o=200\n",
+      "x=I_cp/I_o\n",
+      "\n",
+      "#Calculations\n",
+      "t_c=(math.pi-2*math.asin(1/x))*math.sqrt(C*L)   \n",
+      "th1=math.degrees(math.asin(1.0/x))\n",
+      "t=(5*math.pi/2-th1*math.pi/180)*math.sqrt(L*C)+C*V_s*(1-math.cos(math.radians(th1)))/I_o    \n",
+      "t=(math.pi-th1*math.pi/180)*math.sqrt(L*C)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"turn off time of main thyristor=%.2f us\" %(t_c*10**6))\n",
+      "print(\"total commutation interval=%.3f us\" %(t*10**6))\n",
+      "print(\"turn off time of auxillery thyristor=%.3f us\" %(t*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "turn off time of main thyristor=62.52 us\n",
+        "total commutation interval=80.931 us\n",
+        "turn off time of auxillery thyristor=80.931 us\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.24, Page No 440"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "tc=0.006\n",
+      "R=10.0\n",
+      "L=R*tc\n",
+      "f=2000.0\n",
+      "\n",
+      "#Calculations\n",
+      "T=1/f\n",
+      "V_o=50.0\n",
+      "V_s=100.0\n",
+      "a=V_o/V_s\n",
+      "T_on=a*T\n",
+      "T_off=T-T_on\n",
+      "dI=V_o*T_off/L\n",
+      "I_o=V_o/R\n",
+      "I2=I_o+dI/2    \n",
+      "I1=I_o-dI/2    \n",
+      "\n",
+      "#Results\n",
+      "print(\"max value of load current=%.3f A\" %I2)\n",
+      "print(\"min value of load current=%.3f A\" %I1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "max value of load current=5.104 A\n",
+        "min value of load current=4.896 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.27, Page No 443"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "I_a=30.0\n",
+      "r_a=.5\n",
+      "V_s=220.0\n",
+      "\n",
+      "#Calculations\n",
+      "a=I_a*r_a/V_s    \n",
+      "a=1\n",
+      "k=.1 #V/rpm\n",
+      "N=(a*V_s-I_a*r_a)/k    \n",
+      "\n",
+      "#Results\n",
+      "print(\"min value of duty cycle=%.3f\" %a)\n",
+      "print(\"min Value of speed control=%.0f rpm\" %0)\n",
+      "print(\"max value of duty cycle=%.0f\" %a)\n",
+      "print(\"max value of speed control=%.0f rpm\" %N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "min value of duty cycle=1.000\n",
+        "min Value of speed control=0 rpm\n",
+        "max value of duty cycle=1\n",
+        "max value of speed control=2050 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.28, Page No 444"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_t=72.0\n",
+      "I_a=200.0\n",
+      "r_a=0.045\n",
+      "N=2500.0\n",
+      "\n",
+      "#Calculations\n",
+      "k=(V_t-I_a*r_a)/N\n",
+      "E_a=k*1000\n",
+      "L=.007\n",
+      "Rm=.045\n",
+      "Rb=0.065\n",
+      "R=Rm+Rb\n",
+      "T_a=L/R\n",
+      "I_mx=230\n",
+      "I_mn=180\n",
+      "T_on=-T_a*math.log(-((V_t-E_a)/R-I_mx)/((I_mn)-(V_t-E_a)/R))\n",
+      "R=Rm\n",
+      "T_a=L/R\n",
+      "T_off=-T_a*math.log(-((-E_a)/R-I_mn)/((I_mx)-(-E_a)/R))\n",
+      "T=T_on+T_off\n",
+      "f=1/T    \n",
+      "a=T_on/T \n",
+      "\n",
+      "#Results\n",
+      "print(\"chopping freq=%.1f Hz\" %f)  \n",
+      "print(\"duty cycle ratio=%.3f\" %a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "chopping freq=40.5 Hz\n",
+        "\n",
+        "duty cycle ratio=0.588\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.29, Page No 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesI_mx=425\n",
+      "I_lt=180.0     #lower limit of current pulsation\n",
+      "I_mn=I_mx-I_lt\n",
+      "T_on=0.014\n",
+      "T_off=0.011\n",
+      "\n",
+      "#Calculations\n",
+      "T=T_on+T_off\n",
+      "T_a=.0635\n",
+      "a=T_on/T\n",
+      "V=(I_mx-I_mn*math.exp(-T_on/T_a))/(1-math.exp(-T_on/T_a))\n",
+      "a=.5\n",
+      "I_mn=(I_mx-V*(1-math.exp(-T_on/T_a)))/(math.exp(-T_on/T_a))\n",
+      "T=I_mx-I_mn    \n",
+      "T=T_on/a\n",
+      "f=1/T    \n",
+      "\n",
+      "#Results\n",
+      "print(\"higher limit of current pulsation=%.0f A\" %T)\n",
+      "print(\"chopping freq=%.3f Hz\" %f)\n",
+      "print(\"duty cycle ratio=%.2f\" %a)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "higher limit of current pulsation=0 A\n",
+        "chopping freq=35.714 Hz\n",
+        "duty cycle ratio=0.50\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter8_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter8_4.ipynb
new file mode 100755
index 00000000..721a9faf
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter8_4.ipynb
@@ -0,0 +1,984 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 08 : Inverters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page No 465"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "T=0.1*10**-3\n",
+      "f=1.0/T\n",
+      "k=15*10**-6 #k=th/w\n",
+      "\n",
+      "#Calculations\n",
+      "th=2*math.pi*f*k\n",
+      "X_l=10.0\n",
+      "R=2.0\n",
+      "X_c=R*math.tan(th)+X_l\n",
+      "C=1/(2*math.pi*f*X_c)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"value of C=%.3f uF\" %(C*10**6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of C=1.248 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page No 466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_01=2*V_s/(math.sqrt(2)*math.pi)\n",
+      "R=2.0\n",
+      "I_01=V_01/R\n",
+      "P_d=I_01**2*R    \n",
+      "V=V_s/2\n",
+      "I_s=math.sqrt(2)*I_01/math.pi\n",
+      "P_s=V*I_s\n",
+      "\n",
+      "#Results\n",
+      "print(\"power delivered to load=%.1f W\" %P_d)\n",
+      "print(\"power delivered by both sources=%.1f W\" %(2*P_s))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power delivered to load=5359.9 W\n",
+        "power delivered by both sources=5359.9 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5, Page No 468"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_01=4*V_s/(math.pi*math.sqrt(2))\n",
+      "R=1.0\n",
+      "X_L=6.0\n",
+      "X_c=7.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_01=V_01/math.sqrt(R**2+(X_L-X_c)**2)\n",
+      "P=I_01**2*R    \n",
+      "I_s=math.sqrt(2)*I_01*(2*math.cos(math.radians(45)))/math.pi\n",
+      "P_s=V_s*I_s    \n",
+      "\n",
+      "#Results\n",
+      "print(\"power delivered to the source=%.3f kW\" %(P/1000))\n",
+      "print(\"\\npower from the source=%.3f kW\" %(P_s/1000))\n",
+      "   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power delivered to the source=21.440 kW\n",
+        "\n",
+        "power from the source=21.440 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6 Page No 469"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_01=230.0\n",
+      "R=2.0\n",
+      "I_01=V_01/R\n",
+      "I_m=I_01*math.sqrt(2)\n",
+      "I_T1=I_m/2    \n",
+      "I_D1=0    \n",
+      "X_L=8.0\n",
+      "X_C=6.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_01=V_01/math.sqrt(R**2+(X_L-X_C)**2)\n",
+      "phi1=math.degrees(math.atan((X_L-X_C)/R))\n",
+      "I_T1=I_T1*math.sqrt(2)*0.47675    \n",
+      "I_D1=.1507025*I_m/math.sqrt(2)    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"when load R=2 ohm\")\n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_T1)\n",
+      "print(\"rms value of diode current=%.0f A\" %I_D1)\n",
+      "print(\"when load R=2ohm % X_L=8ohm and X_C=6ohm\")\n",
+      "print(\"rms value of thyristor current=%.3f A\" %I_T1)\n",
+      "print(\"rms value of diode current=%.3f A\" %I_D1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when load R=2 ohm\n",
+        "rms value of thyristor current=54.83 A\n",
+        "rms value of diode current=17 A\n",
+        "when load R=2ohm % X_L=8ohm and X_C=6ohm\n",
+        "rms value of thyristor current=54.826 A\n",
+        "rms value of diode current=17.331 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7 Page No 470"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "R=4.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=0.035\n",
+      "\n",
+      "#Calculations\n",
+      "C=155*10**-6\n",
+      "X_L=w*L\n",
+      "X_C=1/(w*C)\n",
+      "Z1=math.sqrt(R**2+(X_L-X_C)**2)\n",
+      "phi1=-math.degrees(math.atan((X_L-X_C)/R))\n",
+      "Z3=math.sqrt(R**2+(X_L*3-X_C/3)**2)\n",
+      "phi3=math.degrees(math.atan((X_L*3-X_C/3)/R))\n",
+      "Z5=math.sqrt(R**2+(X_L*5-X_C/5)**2)\n",
+      "phi5=math.degrees(math.atan((X_L*5-X_C/5)/R))\n",
+      "I_m1=4*V_s/(Z1*math.pi)\n",
+      "I_01=I_m1/math.sqrt(2)    \n",
+      "I_m3=4*V_s/(3*Z3*math.pi)\n",
+      "I_m5=4*V_s/(5*Z5*math.pi)\n",
+      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2)\n",
+      "I_0=I_m/math.sqrt(2)\n",
+      "P_0=(I_0)**2*R    \n",
+      "P_01=(I_01)**2*R    \n",
+      "t1=(180-phi1)*math.pi/(180*w)    \n",
+      "t1=(phi1)*math.pi/(180*w)    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of fundamental load current=%.2f A\" %I_01)\n",
+      "print(\"load power=%.1f W\" %P_0)\n",
+      "print(\"fundamental load power=%.1f W\" %P_01)\n",
+      "print(\"rms value of thyristor current=%.3f A\" %(I_m/2))\n",
+      "print(\"conduction time for thyristor=%.3f ms\" %(t1*1000))\n",
+      "print(\"Conduction time for diodes=%.3f ms\" %(t1*1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of fundamental load current=20.02 A\n",
+        "load power=1632.5 W\n",
+        "fundamental load power=1602.6 W\n",
+        "rms value of thyristor current=14.285 A\n",
+        "conduction time for thyristor=3.736 ms\n",
+        "Conduction time for diodes=3.736 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8, Page No 473"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_01=2*V_s/(math.sqrt(2)*math.pi)    \n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_01=V_01/R\n",
+      "P=I_01**2*R    \n",
+      "V_or=math.sqrt((V_s/2)**2)\n",
+      "P=V_or**2/R    \n",
+      "I_TP=V_s/(2*R)\n",
+      "I_or=I_TP\n",
+      "pf=I_01**2*R/(V_or*I_or)  \n",
+      "DF=V_01/V_or  \n",
+      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
+      "THD=V_oh/V_01 \n",
+      "V_03=V_01/3\n",
+      "HF=V_03/V_01\n",
+      "\n",
+      "#Results\n",
+      "print(\"fundamental rms o/p voltage=%.3f V\" %V_01)\n",
+      "print(\"fundamental power to load=%.1f W\" %P)\n",
+      "print(\"total o/p power to load=%.1f W\" %P)\n",
+      "print(\"avg SCR current=%.2f A\" %(I_TP*180/360))\n",
+      "print(\"i/p pf=%.3f\" %pf)  \n",
+      "print(\"distortion factor=%.1f\" %DF)\n",
+      "print(\"THD=%.3f\" %THD)    \n",
+      "print(\"harmonic factor=%.4f\" %HF)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fundamental rms o/p voltage=103.536 V\n",
+        "fundamental power to load=1322.5 W\n",
+        "total o/p power to load=1322.5 W\n",
+        "avg SCR current=5.75 A\n",
+        "i/p pf=0.811\n",
+        "distortion factor=0.9\n",
+        "THD=0.483\n",
+        "harmonic factor=0.3333\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9 Page No 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=60\n",
+      "R=3.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt(V_s**2*math.pi/math.pi)    \n",
+      "V_01=4*V_s/(math.sqrt(2)*math.pi)    \n",
+      "P_o=V_or**2/R    \n",
+      "P_01=V_01**2/R    \n",
+      "I_s=V_s/R    \n",
+      "I_avg=I_s*math.pi/(2*math.pi)    \n",
+      "V_03=V_01/3\n",
+      "HF=V_03/V_01    \n",
+      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
+      "THD=V_oh/V_01   \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.0f V\" %V_or)\n",
+      "print(\"o/p power=%.0f W\" %P_o)\n",
+      "print(\"fundamental component of rms voltage=%.2f V\" %V_01)\n",
+      "print(\"fundamental freq o/p power=%.2f W\" %P_01) \n",
+      "print(\"peak current=%.0f A\" %I_s)\n",
+      "print(\"avg current of each transistor=%.0f A\" %I_avg)\n",
+      "print(\"peak reverse blocking voltage=%.0f V\" %V_s)\n",
+      "print(\"harmonic factor=%.4f\" %HF)\n",
+      "print(\"THD=%.4f\" %THD)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=60 V\n",
+        "o/p power=1200 W\n",
+        "fundamental component of rms voltage=54.02 V\n",
+        "fundamental freq o/p power=972.68 W\n",
+        "peak current=20 A\n",
+        "avg current of each transistor=10 A\n",
+        "peak reverse blocking voltage=60 V\n",
+        "harmonic factor=0.3333\n",
+        "THD=0.4834\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10 Page No 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=220.0\n",
+      "R=6.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=0.03\n",
+      "C=180*10**-6\n",
+      "X_L=w*L\n",
+      "X_C=1/(w*C)\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt(V_s**2*math.pi/math.pi)\n",
+      "V_01=4*V_s/(math.sqrt(2)*math.pi)\n",
+      "V_oh=math.sqrt(V_or**2-V_01**2)\n",
+      "THD=V_oh/V_01    \n",
+      "print(\"THD of voltage=%.4f\" %THD)\n",
+      "DF=V_01/V_or    \n",
+      "Z1=math.sqrt(R**2+(X_L-X_C)**2)\n",
+      "phi1=-math.degrees(math.atan((X_L-X_C)/R))\n",
+      "Z3=math.sqrt(R**2+(X_L*3-X_C/3)**2)\n",
+      "phi3=math.degrees(math.atan((X_L*3-X_C/3)/R))\n",
+      "Z5=math.sqrt(R**2+(X_L*5-X_C/5)**2)\n",
+      "phi5=math.degrees(math.atan((X_L*5-X_C/5)/R))\n",
+      "Z7=math.sqrt(R**2+(X_L*7-X_C/7)**2)\n",
+      "phi7=math.degrees(math.atan((X_L*7-X_C/7)/R))\n",
+      "I_01=19.403\n",
+      "I_m1=4*V_s/(Z1*math.pi)\n",
+      "I_m3=4*V_s/(3*Z3*math.pi)\n",
+      "I_m5=4*V_s/(5*Z5*math.pi)\n",
+      "I_m7=4*V_s/(7*Z7*math.pi)\n",
+      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2+I_m7**2)\n",
+      "I_or=I_m/math.sqrt(2)\n",
+      "I_oh=math.sqrt((I_m**2-I_m1**2)/2)\n",
+      "THD=I_oh/I_01    \n",
+      "DF=I_01/I_or    \n",
+      "P_o=I_or**2*R    \n",
+      "I_avg=P_o/V_s    \n",
+      "t1=(180-phi1)*math.pi/(180*w)    \n",
+      "t1=1/(2*f)-t1    \n",
+      "I_p=I_m1    \n",
+      "I_t1=.46135*I_p \n",
+      "\n",
+      "#Results\n",
+      "print(\"\\nDF=%.1f\" %DF)\n",
+      "print(\"THD of current=%.4f\" %THD)   \n",
+      "print(\"DF=%.3f\" %DF)\n",
+      "print(\"load power=%.1f W\" %P_o)\n",
+      "print(\"avg value of load current=%.2f A\" %I_avg)\n",
+      "print(\"conduction time for thyristor=%.0f ms\" %(t1*1000))\n",
+      "print(\"conduction time for diodes=%.0f ms\" %(t1*1000))\n",
+      "print(\"peak transistor current=%.2f A\" %I_p)\n",
+      "print(\"rms transistor current=%.2f A\" %I_t1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "THD of voltage=0.4834\n",
+        "\n",
+        "DF=1.0\n",
+        "THD of current=0.1557\n",
+        "DF=0.988\n",
+        "load power=2313.5 W\n",
+        "avg value of load current=10.52 A\n",
+        "conduction time for thyristor=3 ms\n",
+        "conduction time for diodes=3 ms\n",
+        "peak transistor current=27.44 A\n",
+        "rms transistor current=12.66 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11 Page No 497"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=450.0\n",
+      "R=10.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_or=math.sqrt((V_s/(3*R))**2*2/3+(2*V_s/(3*R))**2*1/3)    \n",
+      "I_T1=math.sqrt((1/(2*math.pi))*((V_s/(3*R))**2*2*math.pi/3+(2*V_s/(3*R))**2*math.pi/3))   \n",
+      "P=3*I_or**2*R    \n",
+      "I_or=math.sqrt((1/(math.pi))*((V_s/(2*R))**2*2*math.pi/3))    \n",
+      "I_T1=math.sqrt((1/(2*math.pi))*((V_s/(2*R))**2*2*math.pi/3))    \n",
+      "P=3*I_or**2*R    \n",
+      "\n",
+      "#Results\n",
+      "print(\"for 180deg mode\")\n",
+      "print(\"rms value of load current=%.3f A\" %I_or)\n",
+      "print(\"power delivered to load=%.1f kW\" %(P/1000))\n",
+      "print(\"rms value of load current=%.0f A\" %I_T1)\n",
+      "print(\"for 120deg mode\")\n",
+      "print(\"rms value of load current=%.3f A\" %I_or)\n",
+      "print(\"rms value of load current=%.2f A\" %I_T1)\n",
+      "print(\"power delivered to load=%.3f kW\" %(P/1000))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for 180deg mode\n",
+        "rms value of load current=18.371 A\n",
+        "power delivered to load=10.1 kW\n",
+        "rms value of load current=13 A\n",
+        "for 120deg mode\n",
+        "rms value of load current=18.371 A\n",
+        "rms value of load current=12.99 A\n",
+        "power delivered to load=10.125 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12, Page No 510"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "R=10.0\n",
+      "f=50.0\n",
+      "w=2*math.pi*f\n",
+      "L=0.03\n",
+      "\n",
+      "#Calculations\n",
+      "X_L=w*L\n",
+      "V_or=math.sqrt(V_s**2*math.pi/math.pi)\n",
+      "V_01=4*V_s/(math.sqrt(2)*math.pi)\n",
+      "Z1=math.sqrt(R**2+(X_L)**2)\n",
+      "phi1=-math.degrees(math.atan((X_L)/R))\n",
+      "Z3=math.sqrt(R**2+(X_L*3)**2)\n",
+      "phi3=math.degrees(math.atan((X_L*3)/R))\n",
+      "Z5=math.sqrt(R**2+(X_L*5)**2)\n",
+      "phi5=math.degrees(math.atan((X_L*5)/R))\n",
+      "Z7=math.sqrt(R**2+(X_L*7)**2)\n",
+      "phi7=math.degrees(math.atan((X_L*7)/R))\n",
+      "I_m1=4*V_s/(math.sqrt(2)*Z1*math.pi)\n",
+      "I_m3=4*V_s/(math.sqrt(2)*3*Z3*math.pi)\n",
+      "I_m5=4*V_s/(math.sqrt(2)*5*Z5*math.pi)\n",
+      "I_m7=4*V_s/(math.sqrt(2)*7*Z7*math.pi)\n",
+      "I_m=math.sqrt(I_m1**2+I_m3**2+I_m5**2+I_m7**2)\n",
+      "P=I_m**2*R    \n",
+      "I_01=I_m1*math.sin(math.radians(45))\n",
+      "I_03=I_m3*math.sin(math.radians(3*45))\n",
+      "I_05=I_m5*math.sin(math.radians(5*45))\n",
+      "I_07=I_m7*math.sin(math.radians(7*45))\n",
+      "I_0=(I_01**2+I_03**2+I_05**2+I_07**2)\n",
+      "P=I_0*R    \n",
+      "g=(180-90)/3+45/2\n",
+      "I_01=2*I_m1*math.sin(math.radians(g))*math.sin(math.radians(45/2))\n",
+      "I_03=2*I_m3*math.sin(math.radians(g*3))*math.sin(math.radians(3*45/2))\n",
+      "I_05=2*I_m5*math.sin(math.radians(g*5))*math.sin(math.radians(5*45/2))\n",
+      "I_07=2*I_m7*math.sin(math.radians(g*7))*math.sin(math.radians(7*45/2))\n",
+      "I_0=(I_01**2+I_03**2+I_05**2+I_07**2)\n",
+      "P=I_0*R    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"using square wave o/p\")\n",
+      "print(\"power delivered=%.2f W\" %P)\n",
+      "print(\"using quasi-square wave o/p\")\n",
+      "print(\"power delivered=%.2f W\" %P)\n",
+      "print(\"using two symmitrical spaced pulses\")\n",
+      "print(\"power delivered=%.2f W\" %P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "using square wave o/p\n",
+        "power delivered=845.87 W\n",
+        "using quasi-square wave o/p\n",
+        "power delivered=845.87 W\n",
+        "using two symmitrical spaced pulses\n",
+        "power delivered=845.87 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14, Page No 520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=50.0\n",
+      "T=1/f\n",
+      "I=0.5\n",
+      "\n",
+      "#Calculations\n",
+      "di=I/T     #di=di/dt\n",
+      "V_s=220.0\n",
+      "L=V_s/di    \n",
+      "t=20*10**-6\n",
+      "fos=2     #factor of safety\n",
+      "t_c=t*fos\n",
+      "R=10\n",
+      "C=t_c/(R*math.log(2))\n",
+      "\n",
+      "#Results    \n",
+      "print(\"source inductance=%.1f H\" %L)\n",
+      "print(\"commutating capacitor=%.2f uF\" %(C*10**6))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "source inductance=8.8 H\n",
+        "commutating capacitor=5.77 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.15, Page No 539"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=10.0\n",
+      "L=.01\n",
+      "C=10*10**-6\n",
+      "#Calculations\n",
+      "if (R**2)<(4*L/C) :\n",
+      "    print(\"ckt will commutate on its own\")\n",
+      "else:\n",
+      "    print(\"ckt will not commutate on its own\")\n",
+      "\n",
+      "xie=R/(2*L)\n",
+      "w_o=1/math.sqrt(L*C)\n",
+      "w_r=math.sqrt(w_o**2-xie**2)\n",
+      "phi=math.degrees(math.atan(xie/w_r))\n",
+      "t=math.pi/w_r\n",
+      "V_s=1\n",
+      "v_L=V_s*(w_o/w_r)*math.exp(-xie*t)*math.cos(math.radians(180+phi))\n",
+      "v_c=V_s*(1-(w_o/w_r)*math.exp(-xie*t)*math.cos(math.radians(180-phi)))   \n",
+      "di=V_s/L    \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"voltage across inductor(*V_s)=%.5f V\" %v_L) \n",
+      "print(\"voltage across capacitor(*V_s)=%.5f V\" %v_c)\n",
+      "print(\"di/dt*V_s (for t=0)=%.0f A/s\" %di)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ckt will commutate on its own\n",
+        "voltage across inductor(*V_s)=-0.60468 V\n",
+        "voltage across capacitor(*V_s)=1.60468 V\n",
+        "di/dt*V_s (for t=0)=100 A/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.16, Page No 540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "L=0.006\n",
+      "C=1.2*10**-6\n",
+      "R=100.0\n",
+      "\n",
+      "#Calculations\n",
+      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
+      "T_off=0.2*10**-3\n",
+      "f=1/(2*(T+T_off))    \n",
+      "R=40\n",
+      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
+      "T_off=.2*10**-3\n",
+      "f=1/(2*(T+T_off))    \n",
+      "R=140\n",
+      "T=math.pi/math.sqrt(1/(L*C)-(R/(2*L))**2)\n",
+      "T_off=.2*10**-3\n",
+      "f=1/(2*(T+T_off)) \n",
+      "\n",
+      "#Results\n",
+      "print(\"o/p freq=%.2f Hz\" %f)\n",
+      "print(\"for R=40ohm\")\n",
+      "print(\"upper limit o/p freq=%.1f Hz\" %f)\n",
+      "print(\"for R=140ohm\")\n",
+      "print(\"lower limit o/p freq=%.1f Hz\" %f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "o/p freq=239.81 Hz\n",
+        "for R=40ohm\n",
+        "upper limit o/p freq=239.8 Hz\n",
+        "for R=140ohm\n",
+        "lower limit o/p freq=239.8 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.17, Page No 540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "f=5000.0\n",
+      "w=2*math.pi*f\n",
+      "R=3.0\n",
+      "\n",
+      "#Calculations\n",
+      "L=60*10**-6\n",
+      "xie=R/(2*L)\n",
+      "C=7.5*10**-6\n",
+      "w_o=1/math.sqrt(L*C)\n",
+      "w_r=math.sqrt(w_o**2-xie**2)\n",
+      "t_c=math.pi*(1/w-1/w_r)    \n",
+      "fos=1.5\n",
+      "t_q=10*10**-6\n",
+      "f_max=1/(2*math.pi*(t_q*fos/math.pi+1/w_r))    \n",
+      "\n",
+      "#Results\n",
+      "print(\"ckt turn off time=%.2f us\" %(t_c*10**6))\n",
+      "print(\"max possible operating freq=%.1f Hz\" %f_max)\n",
+      " #Answers have small variations from that in the book due to difference in the rounding off of digits."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ckt turn off time=21.39 us\n",
+        "max possible operating freq=5341.4 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.18, Page No 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "a=30.0\n",
+      "R=10.0\n",
+      "P=5000.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_s=math.sqrt(P*R*2*math.pi/(2*3)/(math.pi/3+math.sqrt(3)*math.cos(math.radians(2*a))/2))\n",
+      "V_ph=V_s/math.sqrt(3)    \n",
+      "I_or=math.sqrt(P*R)\n",
+      "V_s=I_or*math.pi/(math.sqrt(2)*3*math.cos(math.radians(a)))\n",
+      "V_ph=V_s/math.sqrt(3) \n",
+      "\n",
+      "#Results\n",
+      "print(\"per phase voltage  V_ph=%.3f V\" %V_ph)   \n",
+      "print(\"for constant load current\")\n",
+      "print(\"V_ph=%.2f V\" %V_ph)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "per phase voltage  V_ph=110.384 V\n",
+        "for constant load current\n",
+        "V_ph=110.38 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.19, Page No 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "t=20.0\n",
+      "fos=2.0     #factor of safety\n",
+      "\n",
+      "#Calculations\n",
+      "t_c=t*fos\n",
+      "n=1.0/3\n",
+      "R=20.0\n",
+      "C=n**2*t_c/(4*R*math.log(2)) \n",
+      "\n",
+      "#Results   \n",
+      "print(\"value of capacitor=%.2f uF\" %C)\n",
+      " #printing mistake in the answer in book."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of capacitor=0.08 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.20, Page No 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=220.0\n",
+      "V_p=math.sqrt(2)*V_s/3    \n",
+      "V_L=math.sqrt(3)*V_p    \n",
+      "V_p1=math.sqrt(2)*V_s/math.pi    \n",
+      "V_L1=math.sqrt(3)*V_p1    \n",
+      "V_oh=math.sqrt(V_L**2-V_L1**2)\n",
+      "\n",
+      "#Calculations\n",
+      "THD=V_oh/V_L1    \n",
+      "V_a1=2*V_s/math.pi\n",
+      "V_a5=2*V_s/(5*math.pi)\n",
+      "V_a7=2*V_s/(7*math.pi)\n",
+      "V_a11=2*V_s/(11*math.pi)\n",
+      "R=4.0\n",
+      "L=0.02\n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "Z1=math.sqrt(R**2+(w*L)**2)\n",
+      "Z5=math.sqrt(R**2+(5*w*L)**2)\n",
+      "Z7=math.sqrt(R**2+(7*w*L)**2)\n",
+      "Z11=math.sqrt(R**2+(11*w*L)**2)\n",
+      "I_a1=V_a1/Z1\n",
+      "I_a5=V_a5/Z5\n",
+      "I_a7=V_a7/Z7\n",
+      "I_a11=V_a11/Z11\n",
+      "I_or=math.sqrt((I_a1**2+I_a5**2+I_a7**2+I_a11**2)/2)\n",
+      "P=3*I_or**2*R    \n",
+      "I_s=P/V_s    \n",
+      "I_TA=I_s/3   \n",
+      " \n",
+      "#Results\n",
+      "print(\"rms value of phasor voltages=%.2f V\" %V_p)\n",
+      "print(\"rms value of line voltages=%.2f V\" %V_L)\n",
+      "print(\"fundamental component of phase voltage=%.3f V\" %V_p1)\n",
+      "print(\"fundamental component of line voltages=%.3f V\" %V_L1)\n",
+      "print(\"THD=%.7f\" %THD)\n",
+      "print(\"load power=%.1f W\" %P)\n",
+      "print(\"avg value of source current=%.3f A\" %I_s)\n",
+      "print(\"avg value of thyristor current=%.3f A\" %I_TA)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of phasor voltages=103.71 V\n",
+        "rms value of line voltages=179.63 V\n",
+        "fundamental component of phase voltage=99.035 V\n",
+        "fundamental component of line voltages=171.533 V\n",
+        "THD=0.3108419\n",
+        "load power=2127.6 W\n",
+        "avg value of source current=9.671 A\n",
+        "avg value of thyristor current=3.224 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/Chapter9_4.ipynb b/Power_Electronics_by_P_S_Bimbhra/Chapter9_4.ipynb
new file mode 100755
index 00000000..052c4736
--- /dev/null
+++ b/Power_Electronics_by_P_S_Bimbhra/Chapter9_4.ipynb
@@ -0,0 +1,388 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 09 : AC Voltage Controllers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page No 560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=(V_m/2)*math.sqrt(1/math.pi*((2*math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
+      "R=20\n",
+      "I_or=V_or/R\n",
+      "P_o=I_or**2*R    \n",
+      "I_s=I_or\n",
+      "VA=V_s*I_s\n",
+      "pf=P_o/VA    \n",
+      "V_o=math.sqrt(2)*V_s/(2*math.pi)*(math.cos(math.radians(a))-1)\n",
+      "I_ON=V_o/R  \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.3f V\" %V_or)\n",
+      "print(\"load power=%.1f W\" %P_o)\n",
+      "print(\"i/p pf=%.4f\" %pf)\n",
+      "print(\"avg i/p current=%.4f A\" %I_ON)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=224.716 V\n",
+        "load power=2524.9 W\n",
+        "i/p pf=0.9770\n",
+        "avg i/p current=-0.7581 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page No 560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=45.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=(V_s)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
+      "R=20\n",
+      "I_or=V_or/R\n",
+      "P_o=I_or**2*R    \n",
+      "I_s=I_or\n",
+      "VA=V_s*I_s\n",
+      "pf=P_o/VA    \n",
+      "I_TA=math.sqrt(2)*V_s/(2*math.pi*R)*(math.cos(math.radians(a))+1)   \n",
+      "I_Tr=math.sqrt(2)*V_s/(2*R)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2)) \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.3f V\" %V_or)\n",
+      "print(\"load power=%.2f W\" %P_o)\n",
+      "print(\"i/p pf=%.2f\" %pf)\n",
+      "print(\"avg thyristor current=%.2f A\" %I_TA)   \n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_Tr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=219.304 V\n",
+        "load power=2404.71 W\n",
+        "i/p pf=0.95\n",
+        "avg thyristor current=4.42 A\n",
+        "rms value of thyristor current=7.75 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page No 564"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "n=6.0 #on cycles\n",
+      "m=4.0 #off cycles\n",
+      "\n",
+      "#Calculations\n",
+      "k=n/(n+m)\n",
+      "V_or=V_s*math.sqrt(k)    \n",
+      "pf=math.sqrt(k)    \n",
+      "R=15\n",
+      "I_m=V_s*math.sqrt(2)/R\n",
+      "I_TA=k*I_m/math.pi\n",
+      "I_TR=I_m*math.sqrt(k)/2  \n",
+      "   \n",
+      "#Results\n",
+      "print(\"rms value of o/ voltage=%.2f V\" %V_or)\n",
+      "print(\"i/p pf=%.2f\" %pf)\n",
+      "print(\"avg thyristor current=%.2f A\" %I_TA)  \n",
+      "print(\"rms value of thyristor current=%.2f A\" %I_TR)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/ voltage=178.16 V\n",
+        "i/p pf=0.77\n",
+        "avg thyristor current=4.14 A\n",
+        "rms value of thyristor current=8.40 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page No 569"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "R=3.0\n",
+      "\n",
+      "#Calculations\n",
+      "I_TAM1=2*V_m/(2*math.pi*R)  \n",
+      "I_TRM2=V_m/(2*R)    \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "t_c=math.pi/w    \n",
+      " \n",
+      "#Results\n",
+      "print(\"max value of avg thyristor current=%.3f A\" %I_TAM1)\n",
+      "print(\"max value of avg thyristor current=%.3f A\" %I_TRM2)\n",
+      "print(\"ckt turn off time=%.0f ms\" %(t_c*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "max value of avg thyristor current=34.512 A\n",
+        "max value of avg thyristor current=54.212 A\n",
+        "ckt turn off time=10 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page No 575"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "R=3.0\n",
+      "X_L=4.0\n",
+      "\n",
+      "#Calculations\n",
+      "phi=math.degrees(math.atan(X_L/R))    \n",
+      "V_s=230\n",
+      "Z=math.sqrt(R**2+X_L**2)\n",
+      "I_or=V_s/Z    \n",
+      "P=I_or**2*R    \n",
+      "I_s=I_or\n",
+      "pf=P/(V_s*I_s)    \n",
+      "I_TAM=math.sqrt(2)*V_s/(math.pi*Z)    \n",
+      "I_Tm=math.sqrt(2)*V_s/(2*Z)    \n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "di=math.sqrt(2)*V_s*w/Z    \n",
+      "\n",
+      "#Results\n",
+      "print(\"min firing angle=%.2f deg\" %phi)\n",
+      "print(\"\\nmax firing angle=%.0f deg\" %180)\n",
+      "print(\"i/p pf=%.1f\" %pf)\n",
+      "print(\"max value of rms load current=%.0f A\" %I_or)\n",
+      "print(\"max power=%.0f W\" %P)\n",
+      "print(\"max value of avg thyristor current=%.3f A\" %I_TAM)\n",
+      "print(\"max value of rms thyristor current=%.3f A\" %I_Tm)\n",
+      "print(\"di/dt=%.0f A/s\" %di)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "min firing angle=53.13 deg\n",
+        "\n",
+        "max firing angle=180 deg\n",
+        "i/p pf=0.6\n",
+        "max value of rms load current=46 A\n",
+        "max power=6348 W\n",
+        "max value of avg thyristor current=20.707 A\n",
+        "max value of rms thyristor current=32.527 A\n",
+        "di/dt=20437 A/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 Page No 576"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V=230.0\n",
+      "R=3.0 #ohm\n",
+      "X_L=5.0 #ohm\n",
+      "a=120.0 #firing angle delay\n",
+      "\n",
+      "#Calculations\n",
+      "phi=math.degrees(math.atan(X_L/R))\n",
+      "b=0\n",
+      "i=1\n",
+      "while (i>0) :\n",
+      "    LHS=math.sin(math.radians(b-a))\n",
+      "    RHS=math.sin(math.radians(a-phi))*math.exp(-(R/X_L)*(b-a)*math.pi/180)\n",
+      "    if math.fabs(LHS-RHS)<= 0.01 :\n",
+      "        B=b\n",
+      "        i=2\n",
+      "        break\n",
+      "   \n",
+      "    b=b+.1   \n",
+      "V_or=math.sqrt(2)*V*math.sqrt((1/(2*math.pi))*((B-a)*math.pi/180+(math.sin(math.radians(2*a))-math.sin(math.radians(2*B)))/2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print(\"Extinction angle=%.1f deg\" %B) #answer in the book is wrong as formulae for RHS is wrongly employed\n",
+      "print(\"rms value of output voltage=%.2f V\" %V_or) #answer do not match due to wrong B in book\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Extinction angle=156.1 deg\n",
+        "rms value of output voltage=97.75 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8, Page No 581"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "V_s=230.0\n",
+      "V_m=math.sqrt(2)*V_s\n",
+      "a=60.0\n",
+      "R=20.0\n",
+      "\n",
+      "#Calculations\n",
+      "V_or=math.sqrt((V_m**2/(2*math.pi))*(a*math.pi/180-math.sin(math.radians(2*a))/2)+(2*V_m**2/(math.pi))*(math.pi-a*math.pi/180+math.sin(math.radians(2*a))/2))    \n",
+      "I_T1r=(V_m/R)*math.sqrt(1/math.pi*((math.pi-a*math.pi/180)+math.sin(math.radians(2*a))/2))    \n",
+      "I_T3r=(V_m/(2*R))*math.sqrt(1/math.pi*((a*math.pi/180)-math.sin(math.radians(2*a))/2))    \n",
+      "I1=math.sqrt(2)*I_T1r\n",
+      "I3=math.sqrt((math.sqrt(2)*I_T1r)**2+(math.sqrt(2)*I_T3r)**2)\n",
+      "r=V_s*(I1+I3)    \n",
+      "P_o=V_or**2/R\n",
+      "pf=P_o/r    \n",
+      "\n",
+      "#Results\n",
+      "print(\"rms value of o/p voltage=%.2f V\" %V_or)\n",
+      "print(\"rms value of current for upper thyristors=%.2f A\" %I_T1r)\n",
+      "print(\"rms value of current for lower thyristors=%.2f A\" %I_T3r)\n",
+      "print(\"t/f VA rating=%.2f VA\" %r)\n",
+      "print(\"i/p pf=%.2f\" %pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of o/p voltage=424.94 V\n",
+        "rms value of current for upper thyristors=14.59 A\n",
+        "rms value of current for lower thyristors=3.60 A\n",
+        "t/f VA rating=9631.61 VA\n",
+        "i/p pf=0.94\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Power_Electronics_by_P_S_Bimbhra/screenshots/Chapter2_4.png b/Power_Electronics_by_P_S_Bimbhra/screenshots/Chapter2_4.png
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Binary files /dev/null and b/Power_Electronics_by_P_S_Bimbhra/screenshots/Chapter2_4.png differ
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diff --git a/Power_Electronics_by_P_S_Bimbhra/screenshots/Chapter4_4.png b/Power_Electronics_by_P_S_Bimbhra/screenshots/Chapter4_4.png
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diff --git a/Principles_Of_Electronic_Communication_Systems/chapter1.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter1.ipynb
deleted file mode 100755
index 48b042da..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter1.ipynb
+++ /dev/null
@@ -1,249 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:fa4937e894b2eaacf3d917ad54d3257fed9024ea716662722fd829aea08934e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1 Introduction to electronic communication"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.a Page no 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "c=300000000\n",
-      "f1=150000000.0\n",
-      "f2=430000000.0\n",
-      "f3=8000000.0\n",
-      "f4=750000.0\n",
-      "\n",
-      "#Calculation\n",
-      "W1 = c/f1\n",
-      "W2=c/f2\n",
-      "W3=c/f3\n",
-      "W4=c/f4\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) Wavelength is \",W1,\"meter\"\n",
-      "print\"(b) Wavelength is \",round(W2,3),\"meter\"\n",
-      "print\"(c) Wavelength is \",W3,\"meter\"\n",
-      "print\"(d) Wavelength is \",W4,\"meter\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Wavelength is  2.0 meter\n",
-        "(b) Wavelength is  0.698 meter\n",
-        "(c) Wavelength is  37.5 meter\n",
-        "(d) Wavelength is  400.0 meter\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.2 page no 15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "c=300000000\n",
-      "wavelength=1.5\n",
-      "\n",
-      "#Calculation\n",
-      "frequency=c/wavelength\n",
-      "\n",
-      "#Result\n",
-      "print\"Signal frequncy is \",frequency/1000000.0,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Signal frequncy is  200.0 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3 page no 15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "wavelength_feet=75\n",
-      "wavelength_meter= 75/3.28\n",
-      "c=300000000\n",
-      "\n",
-      "#Calculation\n",
-      "frequency=c/wavelength_meter\n",
-      "\n",
-      "#Result\n",
-      "print\"The signal frequncy is \",frequency/1000000.0,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The signal frequncy is  13.12 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4 page no 15"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "wavelength_inches=8\n",
-      "wavelength_meter=  8/39.37\n",
-      "c=300\n",
-      "\n",
-      "#Calculation\n",
-      "frequency= c/wavelength_meter\n",
-      "\n",
-      "#Result\n",
-      "print\"The signal freuency is \",round(frequency,1),\"MHz\"\n",
-      "print\"The signnalfrequency is \",round(frequency*10**-3,2),\"GHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The signal freuency is  1476.4 MHz\n",
-        "The signnalfrequency is  1.48 GHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5 page no 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f1=902000000\n",
-      "f2=928000000\n",
-      "\n",
-      "#Calculation\n",
-      "bandwidth=f2-f1\n",
-      "\n",
-      "#Result\n",
-      "print\"Width of the band is \",bandwidth/1000000,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Width of the band is  26 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6 page no 19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "bandwidth_megahertz= 6\n",
-      "f1_megahertz=54\n",
-      "\n",
-      "#Calculation\n",
-      "f2_megahertz=f1_megahertz + bandwidth_megahertz\n",
-      "\n",
-      "#Result\n",
-      "print\"Upper frequency limit is \",f2_megahertz,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Upper frequency limit is  60 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter10.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter10.ipynb
deleted file mode 100755
index f969c01b..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter10.ipynb
+++ /dev/null
@@ -1,101 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:6d0a5becdae7bf2928f5747f4da31ef68eb2576da776218335d8b44a4a733949"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10 Multiplexing and Demultiplexing"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.1 Page no 368"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "BW_service = 860*10**6\n",
-      "BW_ch = 6.0*10**6\n",
-      "\n",
-      "#calculation\n",
-      "total_ch = BW_service/BW_ch\n",
-      "\n",
-      "#result\n",
-      "print\"Total number of channels are \",round(total_ch,0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total number of channels are  143.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2 Page no 380"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "channels =16\n",
-      "sampling_rate= 3.5*10**3\n",
-      "w_len=6\n",
-      "\n",
-      "#Calculation\n",
-      "available_ch =channels-1\n",
-      "bpf =channels*w_len\n",
-      "data_rate = sampling_rate * bpf\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) Available channels are \",available_ch\n",
-      "print\"(b) Bits Per Frame = \",bpf\n",
-      "print\"(c) The serial data rate \",data_rate/10**3,\"KHz\"\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Available channels are  15\n",
-        "(b) Bits Per Frame =  96\n",
-        "(c) The serial data rate  336.0 KHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter11.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter11.ipynb
deleted file mode 100755
index 015bcc79..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter11.ipynb
+++ /dev/null
@@ -1,143 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8bd5af3b2acf26b36843842ce0fbb28501e7f5eae117a0af6c810fe6d60c2891"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11 The Transmission of binary data in communication systems"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.1 Page no 392"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "t=0.0016\n",
-      "No_words=256.0\n",
-      "bits_word = 12.0\n",
-      "\n",
-      "#Calculation\n",
-      "tword= t/No_words\n",
-      "tbit = tword/bits_word\n",
-      "bps =1/tbit\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The time duration of the word \",tword*10**8,\"microsecond\"\n",
-      "print\"(b) The time duration of the one bit is \",round(tbit*10**8,4),\"microseconds\"\n",
-      "print\"(c) The speed of transmission is \",bps/10**5,\"kbps\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The time duration of the word  625.0 microsecond\n",
-        "(b) The time duration of the one bit is  52.0833 microseconds\n",
-        "(c) The speed of transmission is  19.2 kbps\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 Page no 400"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "B=12.5*10**3\n",
-      "SN_dB= 25\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "C_th = 2*B\n",
-      "SN=316.2\n",
-      "C =B*3.32*log10(SN+1)\n",
-      "N= 2**(C/(2.0*B))\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The maximum theorotical data rate is \",C_th/10**3,\"kbps\"\n",
-      "print\"(b) The maximum theorotical capacity of channel is \",round(C/10**3,1),\"Kbps\"\n",
-      "print\"(c) The number of levels needed to acheive maximum speed are \",round(N,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The maximum theorotical data rate is  25.0 kbps\n",
-        "(b) The maximum theorotical capacity of channel is  103.8 Kbps\n",
-        "(c) The number of levels needed to acheive maximum speed are  17.78\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 Page no 430"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "block =512\n",
-      "packets =8\n",
-      "BER = 2*10**-4\n",
-      " \n",
-      "#Calculation\n",
-      "avg_errors = block*packets*8*BER\n",
-      "\n",
-      "print\"Average number of errors are \",avg_errors\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average number of errors are  6.5536\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter12.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter12.ipynb
deleted file mode 100755
index ae98ce2e..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter12.ipynb
+++ /dev/null
@@ -1,101 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:70c87938501c80a4a95b89850ad3f0f2a57b63c9811bc118f804b4d657133c5f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12 Introduction to Networking and local area networks"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 Page no 448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "N = 20\n",
-      "\n",
-      "#Calculation\n",
-      "L = (N*(N-1))/2.0\n",
-      "\n",
-      "#Result\n",
-      "print\"The number of interconnecting wires required are \",L\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The number of interconnecting wires required are  190.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 Page no 474"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "block = 1500\n",
-      "ethernet = 10.0*10**6\n",
-      "token_ring = 16.0*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "t1_bit = 1/ethernet\n",
-      "t1_byte = 8*t1_bit\n",
-      "t1_1526 = 1526 *t1_byte\n",
-      "t2_bit = 1/token_ring\n",
-      "t2_byte = 8 * t2_bit\n",
-      "t2_1521 =1521*t2_byte\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) Time required for 1526 bytes is \",t1_1526*10**6,\"ns\"\n",
-      "print\"(b) Time required for 1521 bytes is \",t2_1521*10**3,\"ms\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Time required for 1526 bytes is  1220.8 ns\n",
-        "(b) Time required for 1521 bytes is  0.7605 ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter13.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter13.ipynb
deleted file mode 100755
index de1fc475..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter13.ipynb
+++ /dev/null
@@ -1,339 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4be2df010e0c0a454b43a2b1267f5b9f24594cfd973fe52088cb80f0e511082b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13 Transmission lines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.1 Page no 483"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f= 450.0*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = 984/f\n",
-      "len =0.1*lamda\n",
-      "\n",
-      "#Result\n",
-      "print\"feet long conductors would be considered as the transmission line \",round(len*10**6,3),\"ft\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "feet long conductors would be considered as the transmission line  0.219 ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.2 Page no 484"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "lamda = 2.19\n",
-      "\n",
-      "#Calculation\n",
-      "len = (3/8.0)*lamda\n",
-      "\n",
-      "#Result\n",
-      "print\"The pyhsical length of the transmission line \",round(len,2),\"feet\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The pyhsical length of the transmission line  0.82 feet\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.3 Page no 492"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "len = 165\n",
-      "attn_100ft = 5.3\n",
-      "pin = 100\n",
-      "attn_ft = 5.3/100.0\n",
-      "\n",
-      "#Calculation\n",
-      "total_attn = attn_ft * len\n",
-      "pout = pin *0.1335\n",
-      "\n",
-      "#Result\n",
-      "print\"The total attenuation of the cable is \",total_attn,\"dB\"\n",
-      "print\"Output power is \",pout,\"W\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total attenuation of the cable is  8.745 dB\n",
-        "Output power is  13.35 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.4 Page no 494"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "len =150\n",
-      "C =13.5\n",
-      "Z0 =93\n",
-      "f =2.5*10**6\n",
-      "attn_100ft =2.8\n",
-      "\n",
-      "#Calculation\n",
-      "L =C*Z0**2\n",
-      "td =(L*C)**0.5\n",
-      "theta = ((360)*188.3)/(1/f)\n",
-      "attn_ft = attn_100ft/100.0\n",
-      "total_attn = attn_ft*150\n",
-      "\n",
-      "print\"(a) The load impedance required to terminate the the line to avoid the reflections is %d ohm\",Z0\n",
-      "print\"(b) The equivalent inductance per feet is \",L/10**3,\"nh\"\n",
-      "print\"(c) The time delay introduced by the cable per feet is \",td/10**3,\"ns\"\n",
-      "print\"(d) The phase shift occurs in degrees for the 2.5 Mhz sine wave is \",theta/10**9\n",
-      "print\"(e) The total attenuation is \",total_attn,\"db\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The load impedance required to terminate the the line to avoid the reflections is %d ohm 93\n",
-        "(b) The equivalent inductance per feet is  116.7615 nh\n",
-        "(c) The time delay introduced by the cable per feet is  1.2555 ns\n",
-        "(d) The phase shift occurs in degrees for the 2.5 Mhz sine wave is  169.47\n",
-        "(e) The total attenuation is  4.2 db\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.5 Page no 501"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "vmax= 52.0\n",
-      "vmin= 17.0\n",
-      "Z0 = 75\n",
-      "\n",
-      "#calculation\n",
-      "SWR = vmax/vmin\n",
-      "ref_coeff = (vmax-vmin)/(vmax+vmin)\n",
-      "Zl1 = Z0*SWR\n",
-      "Zl2 = Z0/SWR\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The standing wave ratio is \",round(SWR,2)\n",
-      "print\"(b) Reflection coefficient is \",round(ref_coeff,2)\n",
-      "print\"The value of resistive load is \",round(Zl1,2),\"ohm or\",round(Zl2,2),\"ohm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The standing wave ratio is  3.06\n",
-        "(b) Reflection coefficient is  0.51\n",
-        "The value of resistive load is  229.41 ohm or 24.52 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.6 Page no 503"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "SWR =3.05\n",
-      "ref_pwr =0.2562\n",
-      "pin =30\n",
-      "\n",
-      "#calculation\n",
-      "pout = pin -(pin*((SWR-1)/(SWR+1))**2)\n",
-      "\n",
-      "#Result\n",
-      "print\"The output power of the cable is \",round(pout,3),\"W\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output power of the cable is  22.314 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.7 Page no 508"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "C =4*10**-12\n",
-      "f =800*10**6\n",
-      "diele = 3.5\n",
-      "h = 0.0625\n",
-      "w = 0.13\n",
-      "t = 0.002\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "Z0 = 38.8*math.log(0.374/0.106)\n",
-      "Xc = 1/(6.28*f*C)\n",
-      "\n",
-      "#Result\n",
-      "print\"The charecteristics impedance of the transmission line is \",round(Z0,1),\"ohm\"\n",
-      "print\"The reactance of the capacitor is \",round(Xc,2),\"ohm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The charecteristics impedance of the transmission line is  48.9 ohm\n",
-        "The reactance of the capacitor is  49.76 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.8 Page no 508"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "lamda = (984/800.0)\n",
-      "lamda_8 =lamda/8.0\n",
-      "\n",
-      "#Calculation\n",
-      "len = lamda_8*12*(1/3.6**0.5)\n",
-      "\n",
-      "#Result\n",
-      "print\"The length of the transmission line is \",round(len,3),\"inch\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The length of the transmission line is  0.972 inch\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter14.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter14.ipynb
deleted file mode 100755
index 483c5c1b..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter14.ipynb
+++ /dev/null
@@ -1,229 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e45779b9fe24c5d57bbc44b21eced0e689b3f83e9f5f5eea042afe4effb61cee"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14 Antennas and Wave Propogation"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 page no 544"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f=310.0*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "len1 =(492*0.97)/f\n",
-      "len2 =(492/f)*0.8\n",
-      "len3 =(984/f)*0.73\n",
-      "z1 =120*log(35/2.0)\n",
-      "len4 =234/f\n",
-      "z2  = 73/2.0\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The length and radiation resistance of the dipole are \",round(len1*10**6,2),\"feet and 73 ohm respectively\"\n",
-      "print\"(b) The length of the folded dipole are \",round(len2*10**6,2),\"feet\"\n",
-      "print\"(c) The length and radiation resistance of the bow tie antenna are \",round(len3*10**6,1),\"feet and \",round(z1,1),\"ohm resp.\"\n",
-      "print\"(d) The length and radiation resistance of the groun plane antenna are \",round(len4*10**6,3),\"feet and\",z2,\"ohm resp.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The length and radiation resistance of the dipole are  1.54 feet and 73 ohm respectively\n",
-        "(b) The length of the folded dipole are  1.27 feet\n",
-        "(c) The length and radiation resistance of the bow tie antenna are  2.3 feet and  343.5 ohm resp.\n",
-        "(d) The length and radiation resistance of the groun plane antenna are  0.755 feet and 36.5 ohm resp.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 Page no 553"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "gain=14\n",
-      "len=250\n",
-      "attn_100=3.6\n",
-      "f =220*10\n",
-      "pin =50\n",
-      "p =0.126\n",
-      "\n",
-      "#Calculation\n",
-      "pout =pin*p\n",
-      "line_loss =pin-pout\n",
-      "pwr_ratio = 25.1\n",
-      "ERP = pwr_ratio*pout\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The transmission line loss is \",line_loss\n",
-      "print\"(b) Effective raduated power is \",round(ERP,1),\"W\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The transmission line loss is  43.7\n",
-        "(b) Effective raduated power is  158.1 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 Page no 556"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Z0 =50\n",
-      "Zl =172\n",
-      "f =460.0*10**6\n",
-      "VF =0.86\n",
-      "\n",
-      "#Calculation\n",
-      "len =(246/f)*VF\n",
-      "\n",
-      "#Result\n",
-      "print\"The length of the impedance matching section needed for the Q section is \",round(len*10**6,2),\"feet\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The length of the impedance matching section needed for the Q section is  0.46 feet\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4 Page no 557"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f = 460.0*10**6\n",
-      "VF = 0.66\n",
-      "\n",
-      "#Calculation\n",
-      "len = (246/f)*VF\n",
-      "\n",
-      "#Result \n",
-      "print\"The length of impedance matching section is \",round(len*10**6,3),\"feet\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The length of impedance matching section is  0.353 feet\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5 Page no 567"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "ht =275\n",
-      "hr =60\n",
-      "f=224.0*10**6\n",
-      "pt=100\n",
-      "Gt = 26\n",
-      "Gr = 3.27\n",
-      "\n",
-      "#Calculation\n",
-      "D =((2*ht)**0.5+(2*hr)**0.5)*1.61\n",
-      "lamda = 300/f\n",
-      "Pr = (pt*Gt*Gr*lamda**2)/(16*3.14**2*D**2)\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The maximmum transmitting distance is \",round(D,1),\"kilometer\"\n",
-      "print\"(b) The received power is \",round(Pr*10**15,1),\"nW\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The maximmum transmitting distance is  55.4 kilometer\n",
-        "(b) The received power is  31.5 nW\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter16.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter16.ipynb
deleted file mode 100755
index a3463547..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter16.ipynb
+++ /dev/null
@@ -1,216 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:526f0478ca7a5ec4522d70a6832114c547e4784f425e0391463d5037753dde9d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 16 Microwave communication"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.1 Page no 616"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Zsrc =50\n",
-      "Zld  =136\n",
-      "f =5800.0*10**6\n",
-      "Er =2.4\n",
-      "\n",
-      "#Calculation\n",
-      "Zq =(Zsrc * Zld)**0.5\n",
-      "Vp =1/(Er)**0.5\n",
-      "lamda = 300/f\n",
-      "len = (lamda/4.0)*38.37*Vp\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The required impedance is \",round(Zq,2),\"ohm\"\n",
-      "print\"(b) The length of the microstrip \",round(len*10**6,2),\"inches\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The required impedance is  82.46 ohm\n",
-        "(b) The length of the microstrip  0.32 inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 Page no 623"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "w=0.65\n",
-      "h=0.38\n",
-      "\n",
-      "#Calculation\n",
-      "fco = 300/(2.0*((0.65*2.54)/100.0))\n",
-      "f =1.42*fco\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The cutoff frequency of the \",round(fco/10**3,3),\"GHz\"\n",
-      "print\"(b) Operating frequency of the wavwguide is \",round(f/10**3,1),\"GHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The cutoff frequency of the  9.085 GHz\n",
-        "(b) Operating frequency of the wavwguide is  12.9 GHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 Page no 623"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#for the above question\n",
-      "print\"The c band is approximately 4 to 6 Ghz since a waveguide acts as a high pass filter with cut off of 9.08 Ghz it will not pass c band signal\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The c band is approximately 4 to 6 Ghz since a waveguide acts as a high pass filter with cut off of 9.08 Ghz it will not pass c band signal\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 Page no 648"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "lamda1 =5.0\n",
-      "f2 = 15.0*10**9\n",
-      "D=1.524\n",
-      "\n",
-      "#Calculation\n",
-      "f1=984/lamda1\n",
-      "lamda2 =300/f2\n",
-      "G = (6*(D/lamda2)**2)\n",
-      "B = 70/(D/lamda2)\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The lowest possible oprerating frequency is \",f1,\"MHz\"\n",
-      "print\"(b) The gain at 15 Ghz is \",G/10**12\n",
-      "print\"(c) The beam width at 15Ghz is \",round(B*10**6,2),\"degree\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The lowest possible oprerating frequency is  196.8 MHz\n",
-        "(b) The gain at 15 Ghz is  34838.64\n",
-        "(c) The beam width at 15Ghz is  0.92 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.5 Page no 661"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "T = 9.2\n",
-      "theta = 20\n",
-      "sin20 = 0.342\n",
-      "a=5280\n",
-      "\n",
-      "#Calculation\n",
-      "D_nautical = T/12.36\n",
-      "D_statute =D_nautical*0.87\n",
-      "A = D_statute*0.342\n",
-      "A1=A*a\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The line of distance to the aircraft in the statute miles \",round(D_statute,3)\n",
-      "print\"(b) The altitude of the aircraft is \",round(A,2),\"mi\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The line of distance to the aircraft in the statute miles  0.648\n",
-        "(b) The altitude of the aircraft is  0.22 mi\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter17.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter17.ipynb
deleted file mode 100755
index c9e3dcac..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter17.ipynb
+++ /dev/null
@@ -1,142 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:36f7565fa260d3a1374317f250783789b0d4ba7834e883c19725c0b3c84c1d9a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapeter 17 Satellite communication"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1 Page no 678"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "stn_long = 95\n",
-      "stn_lat = 30\n",
-      "sat_long =121\n",
-      "rad_pos = 137\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "azimuth = 360-rad_pos\n",
-      "\n",
-      "#Result\n",
-      "print\"The elevation setting for the antenna is 45 degree\"\n",
-      "print\"The azimuth setting for the antenna is \",azimuth,\"degree\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elevation setting for the antenna is 45 degree\n",
-        "The azimuth setting for the antenna is  223 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2 Page no 681"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "flo = 2*10**9\n",
-      "fd =3840*10**6\n",
-      "B =36*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "fu =fd+flo\n",
-      "C =2*B\n",
-      "\n",
-      "#Result\n",
-      "print\"The uplink frequency is \",fu/10.0**9,\"GHz\"\n",
-      "print\"The data rate is \", C/10**6,\"Mbps\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The uplink frequency is  5.84 GHz\n",
-        "The data rate is  72 Mbps\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 Page no 691"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "fs = 4.08*10**9\n",
-      "fIF1 = 770*10**6\n",
-      "fIF2 = 140*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "flo1 = fs - fIF1\n",
-      "flo2 = fIF1 - fIF2\n",
-      "\n",
-      "#Result\n",
-      "print\"The local oscillator frequency for first IF is \",flo1/10**6,\"MHz\"\n",
-      "print\"The local oscillator frequency for the second IF is \",flo2/10**6,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The local oscillator frequency for first IF is  3310.0 MHz\n",
-        "The local oscillator frequency for the second IF is  630 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter19.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter19.ipynb
deleted file mode 100755
index 1ce8f35e..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter19.ipynb
+++ /dev/null
@@ -1,128 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:bbbebe541ba21d9bbb4d54500aef67ec9a094ac72f21cf0a19a14aa93b9c61ce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 19 Optical communication"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.1 Page no 760"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "NA = 0.29\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "critical_angle = math.asin(0.29)*180/3.14\n",
-      "\n",
-      "print\"The critical angle is \",round(critical_angle,2),\"degree\"\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical angle is  16.87 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.2 Page no 767"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "B_rating_Mhzkm =600*10**6\n",
-      "len_ft=500\n",
-      "\n",
-      "#Calculation\n",
-      "bandwidth = B_rating_Mhzkm/(len_ft/3274.0)\n",
-      "\n",
-      "#Result\n",
-      "print\"The bandwidth of the 500 feetr segment of the cable is \",bandwidth/10**6,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth of the 500 feetr segment of the ccable is  3928.8 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.3 Page no 780"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "R=43*10**6\n",
-      "D=1200/3274.0\n",
-      "\n",
-      "#Calculation\n",
-      "d=1/(5.0*R*D)\n",
-      "\n",
-      "#Result\n",
-      "print\"The dispersion factor is \",round(d*10**9,1),\"ns/km\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dispersion factor is  12.7 ns/km\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter22.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter22.ipynb
deleted file mode 100755
index 2fc3916b..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter22.ipynb
+++ /dev/null
@@ -1,94 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d66b7b0a5ebd4fcdf9eb24f98ae81ac03aa2dcdb229f576b47a170647bd9ea49"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 22 Communication tests and measurments"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 Page no 854"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Vp=8                       #volts\n",
-      "R=75                       #Ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Vmax=0.707*Vp\n",
-      "P=Vmax**2/R\n",
-      "\n",
-      "#Result\n",
-      "print\"Power is dissipated in the load is \",round(P,4),\"Watt\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power is dissipated in the load is  0.4265 Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2 Page no 857"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Pf=380.0                     #Watt\n",
-      "Pr=40                      #Watt\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "m=math.sqrt(Pr/Pf)\n",
-      "SWR=(1+m)/(1-m)\n",
-      "print\"SWR= \",round(SWR,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "SWR=  1.96\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter3.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter3.ipynb
deleted file mode 100755
index ef029e65..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter3.ipynb
+++ /dev/null
@@ -1,331 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e1b5c14d4876dc6b60f9f51fe0f04e94941e7f01c84340bf3441e1daf170b7a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3 Amplitude modulation fundamentals"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.1 page no 99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Vmax = 5.9\n",
-      "Vmin = 1.2\n",
-      "\n",
-      "#Calculation\n",
-      "m = (Vmax-Vmin)/(Vmax+Vmin)\n",
-      "Vc = (Vmax+Vmin)/2.0\n",
-      "Vm = (Vmax-Vmin)/2.0\n",
-      "m = Vm/Vc\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The modulation index is \",round(m,2)\n",
-      "print\"Vc= \",Vc,\"Vm= \",Vm,\"(for 2 volt/div on verticle scale)\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index is  0.66\n",
-        "Vc=  3.55 Vm=  2.35 (for 2 volt/div on verticle scale)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.2 page no 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "frq =980*10**3\n",
-      "frq_range = 5*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fusb = frq+frq_range\n",
-      "flsb = frq-frq_range\n",
-      "bw=fusb-flsb\n",
-      "\n",
-      "#Result\n",
-      "print\"The upper sideband is at \",fusb/10**3,\"Khz\"\n",
-      "print\"Lower sideband is at \",flsb/10**3,\"Khz\"\n",
-      "print\"the babdwidth is \",bw/10**3,\"KHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The upper sideband is at  985 Khz\n",
-        "Lower sideband is at  975 Khz\n",
-        "the babdwidth is  10 KHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3 page no 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Pc = 30\n",
-      "m=0.85\n",
-      "\n",
-      "#Calculation\n",
-      "Pt = Pc*(1+ (m**2/2.0))\n",
-      "Psb_both =Pt-Pc\n",
-      "Psb_one = Psb_both/2.0\n",
-      "\n",
-      "#Result\n",
-      "print\"The total power is \",round(Pt,1),\"watt\" \n",
-      "print\"The power in one sideband is \",round(Psb_one,1),\"Watt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power is  40.8 watt\n",
-        "The power in one sideband is  5.4 Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.4 page no 108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "R = 40\n",
-      "I = 4.8\n",
-      "m=0.9\n",
-      "\n",
-      "#Calculation\n",
-      "Pc = I**2*R\n",
-      "Pt = (I*(1+(m**2/2.0))**0.5)**2*R\n",
-      "Psb = Pt-Pc\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The carrier power is \",Pc,\" watt\"\n",
-      "print\"(b) Total power = \",round(Pt,0),\"watt\"\n",
-      "print\"(c) Sideband Power = \",round(Psb,1),\"watt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The carrier power is  921.6  watt\n",
-        "(b) Total power =  1295.0 watt\n",
-        "(c) Sideband Power =  373.2 watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5 page no 108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "It = 5.1\n",
-      "Ic =4.8\n",
-      "\n",
-      "#Calculation\n",
-      "m=(2*((It/Ic)**2-1))**0.5\n",
-      "\n",
-      "#Result\n",
-      "print\"The percentage of modulation is \",round(m*100,0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of modulation is  51.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.6 page no 109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "m = 0.9\n",
-      "Pc = 921.6\n",
-      "\n",
-      "#calculation\n",
-      "Psb = (m**2*Pc)/4.0\n",
-      "\n",
-      "#Result\n",
-      "print\"The power in one sideband \",round(Psb,1),\"Watt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The power in one sideband  186.6 Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.7 page no 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Vpp = 178\n",
-      "R = 75.0\n",
-      "\n",
-      "#Calculation\n",
-      "Vp =Vpp/2.0\n",
-      "Vrms = 0.707*Vp\n",
-      "PEP =(Vrms**2/R)\n",
-      "\n",
-      "#Result\n",
-      "print\"The peak envelop power is \", round(PEP,1),\"Watt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The peak envelop power is  52.8 Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.8 page no 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Vs =24\n",
-      "Im =9.3\n",
-      "\n",
-      "#Calculation\n",
-      "PEP = Vs*Im \n",
-      "Pavg1 = PEP/3.0\n",
-      "Pavg2 = PEP/4.0\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The peak envelope power is \",PEP,\"watt\"\n",
-      "print\"(b) Average power of transmitter is \",Pavg2,\"watt to\",Pavg1,\"watt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The peak envelope power is  223.2 watt\n",
-        "(b) Average power of transmitter is  55.8 watt to 74.4 watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter4.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter4.ipynb
deleted file mode 100755
index 522fe2b5..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter4.ipynb
+++ /dev/null
@@ -1,123 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:268207babca50d59c46d454ea7ce41b37f6eaad8a022bc43c98526f17a497ebd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "chapter 4 Amplitude Modulator and Demodulator circuits"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.1 Page no 129"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "\n",
-      "Vcc =48\n",
-      "I = 3.5\n",
-      "efficiency =70                           #percent\n",
-      "modulation = 67                          #percent\n",
-      "m = modulation/100.0\n",
-      "#calculation\n",
-      "\n",
-      "Pi = Vcc*I\n",
-      "Pc=Pi\n",
-      "Pm = Pi/2.0\n",
-      "Pout = (efficiency*Pi)/100.0\n",
-      "Ps = Pc*((m**2)/4.0)\n",
-      "maximum_swing = 2*Vcc \n",
-      "#Result\n",
-      "\n",
-      "print \"(a) RF input power to the output stage is \",Pi,\"W\"\n",
-      "print \"(b) AF power required for 100 percent modulation is \",Pm,\"W\"\n",
-      "print \"(c) Carrier output power is \",Pout,\"W\"\n",
-      "print \"(d) Power in one sideband for 67 percent modulation is \",round(Ps,2),\"W\"\n",
-      "print \"(e) maximum and minimum dc supply voltage swing with 100 percent modulation is \",maximum_swing,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) RF input power to the output stage is  168.0 W\n",
-        "(b) AF power required for 100 percent modulation is  84.0 W\n",
-        "(c) Carrier output power is  117.6 W\n",
-        "(d) Power in one sideband for 67 percent modulation is  18.85 W\n",
-        "(e) maximum and minimum dc supply voltage swing with 100 percent modulation is  96 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.2 Page no 145"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "\n",
-      "fc =4.2*10**6\n",
-      "f1 = 300\n",
-      "f2 = 3400\n",
-      "#calculation\n",
-      "#upper sideband\n",
-      "\n",
-      "x0=fc+f1\n",
-      "y0=fc+f2\n",
-      "#lower sideband\n",
-      "\n",
-      "x1=fc-f1\n",
-      "y1=fc-f2\n",
-      "frequency=(x1+y1)/2.0\n",
-      "#Result \n",
-      "\n",
-      "print \"(a) Upper sideband ranges from \",x0,\"to\",y0\n",
-      "print \"    lower sideband ranges from \",x1,\"to\",y1\n",
-      "print \"(b) approximate center frequency of a bandpass filter is \",frequency,\"Hz\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) Upper sideband ranges from  4200300.0 to 4203400.0\n",
-        "    lower sideband ranges from  4199700.0 to 4196600.0\n",
-        "(b) approximate center frequency of a bandpass filter is  4198150.0 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter5.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter5.ipynb
deleted file mode 100755
index 3539866c..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter5.ipynb
+++ /dev/null
@@ -1,253 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:91b0cb714976b5e3abb5fa418cb1d12fb7cd307f6d4fd008a74f2dcf67b41e57"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5 Fundamentals of frequency modulation"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.1 Page no 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f = 915*10**6\n",
-      "fm_deviation =12.5*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "max_deviation = f + fm_deviation\n",
-      "min_deviation = f - fm_deviation\n",
-      "\n",
-      "#Result\n",
-      "print\"Maximum frequency occur during modulation is \",max_deviation/1000.0,\"KHz\"\n",
-      "print\"Minimum frequency occur during modulation is \",min_deviation/1000.0,\"KHz\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum frequency occur during modulation is  915012.5 KHz\n",
-        "Minimum frequency occur during modulation is  914987.5 KHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2 Page no 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "max_deviation = 25*10**3\n",
-      "fm =15.0*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "mf =max_deviation/fm\n",
-      "\n",
-      "#Result\n",
-      "print\"The deviation ratio of the TV sound is \",round(mf,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deviation ratio of the TV sound is  1.667\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3 Page no 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "mf = 2.2\n",
-      "fd = 7.48*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "fm = fd/mf\n",
-      "\n",
-      "#Result\n",
-      "print\"The maximum modulating frequency is \",fm/1000.0,\"KHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum modulating frequency is  3.4 KHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.4 Page no 164"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "J0 = -0.4\n",
-      "J1 = -0.07\n",
-      "J2 = 0.36\n",
-      "J3 = 0.43\n",
-      "J4 = 0.28\n",
-      "\n",
-      "#Result\n",
-      "print\"The amplitude of the carrier is \",J0\n",
-      "print\"Amplitudes of the first four sidebands are \",\" \\n \", J1,\"\\n \",J2,\"\\n \",J3,\"\\n \",J4\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amplitude of the carrier is  -0.4\n",
-        "Amplitudes of the first four sidebands are   \n",
-        "  -0.07 \n",
-        "  0.36 \n",
-        "  0.43 \n",
-        "  0.28\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5 Page no 165"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "fd = 30*10**3\n",
-      "fm = 5*10**3\n",
-      "N=9\n",
-      "\n",
-      "#Calculation\n",
-      "bw1 = 2*fm*N\n",
-      "bw2 = 2*(fd+fm)\n",
-      "\n",
-      "#Result\n",
-      "print\"The maximum bandwidth of the fm signal is \",bw1/10**3,\"KHz\"\n",
-      "print\"Bandwidth using carson's rule \",bw2/10**3,\"KHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum bandwidth of the fm signal is  90 KHz\n",
-        "Bandwidth using carson's rule  70 KHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.6 Page no 167"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "S_N = 2.8\n",
-      "fm = 1.5*10**3\n",
-      "fd =4*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "fi= math.asin(1/S_N)\n",
-      "delta = fi*fm\n",
-      "SN =fd/delta\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The frequency deviation caused by the noise \",round(delta,1),\"Hz\"\n",
-      "print\"(b) The improved output signal to noise ratio is \",round(SN,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The frequency deviation caused by the noise  547.8 Hz\n",
-        "(b) The improved output signal to noise ratio is  7.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter6.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter6.ipynb
deleted file mode 100755
index 8bd8c6da..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter6.ipynb
+++ /dev/null
@@ -1,148 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:aac83d9b958acba775304d5e53683242c3be955e852f2ce5dbe2f25d81f3bf36"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6 FM Circuits"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.1 Page no 178"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Vc =40*10**-12\n",
-      "c = 20*10**-12\n",
-      "f0 = 5.5*10**6\n",
-      "Ct = Vc+c\n",
-      "\n",
-      "#Calculation\n",
-      "L = 1/((6.28*f0)**2*Ct)\n",
-      "\n",
-      "#Result\n",
-      "print\"The value of the inductance is \",round(L*10**6,0),\"microhenry\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the inductance is  14.0 microhenry\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2 Page no 186"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f=168.96*10**6\n",
-      "multiplier=24.0\n",
-      "deviation = 5*10**3\n",
-      "fm = 2.8*10**3\n",
-      "\n",
-      "#Calculation\n",
-      "f0 =f/multiplier\n",
-      "fd= deviation/multiplier\n",
-      "phaseshift = fd/fm\n",
-      "phaseshift_degrees = phaseshift*57.3\n",
-      "total_phaseshift =2*phaseshift_degrees\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The crystal oscillator frequency is \",f0/10**6,\"MHz\"\n",
-      "print\"(b) The total phase shift is \",round(total_phaseshift,3),\"degree\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The crystal oscillator frequency is  7.04 MHz\n",
-        "(b) The total phase shift is  8.527 degree\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3 Page no 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "R =1*10**3\n",
-      "phaseshift =4.263\n",
-      "phaseshift_center= 45\n",
-      "f =7.04*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "phase_l = phaseshift_center - phaseshift\n",
-      "phase_u = phaseshift_center + phaseshift\n",
-      "phaserange_total = phase_u - phase_l\n",
-      "Xc1 = 1161\n",
-      "C1 = 1/(6.28*f*Xc1)\n",
-      "Xc2 = 861\n",
-      "C2 = 1/(6.28*f*Xc2)\n",
-      "\n",
-      "#Result\n",
-      "print\"The two values of the capacitance to achieve total deviation are \",round(C1*10**12,1),\"pf to\",round(C2*10**12,1),\"pf\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The two values of the capacitance to achieve total deviation are  19.5 pf to 26.3 pf\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter7.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter7.ipynb
deleted file mode 100755
index c4587482..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter7.ipynb
+++ /dev/null
@@ -1,221 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:13e4ad42742896befb0b329fd930804197555dd5705f1dd526b017e9117b7b00"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7 Digital communication techniques"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 Page no 210"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "t = 71.4*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "f = 1/t\n",
-      "fourth_harmonic = f*4\n",
-      "min_sampling = 2*fourth_harmonic\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The frequency of the signal is \",round(f/10**3,1),\"KHz\"\n",
-      "print\"(b) The fourth harmonic is \",round(fourth_harmonic/10**3,0),\"KHz\"\n",
-      "print\"(c) Minimum sampling rate is \",round(min_sampling/10**3,0),\"KHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The frequency of the signal is  14.0 KHz\n",
-        "(b) The fourth harmonic is  56.0 KHz\n",
-        "(c) Minimum sampling rate is  112.0 KHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2 Page no 222"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "N = 14\n",
-      "discrete_levels = 2.0**N\n",
-      "\n",
-      "#Calculation\n",
-      "num_vltg_inc =2**N-1\n",
-      "resolution = 12/discrete_levels \n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The numbedr of discrete levels that are represented using N number of bits are \",discrete_levels\n",
-      "print\"(b) the number odf voltage increments required to divide the voltage range are \",num_vltg_inc\n",
-      "print\"(c) Resolution of the digitization \",round(resolution*10**6,2),\"microvolt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The numbedr of discrete levels that are represented using N number of bits are  16384.0\n",
-        "(b) the number odf voltage increments required to divide the voltage range are  16383\n",
-        "(c) Resolution of the digitization  732.42 microvolt\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3 Page no 225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "N =12\n",
-      "SINAD1=78\n",
-      "\n",
-      "#Calculation\n",
-      "SINAD2 = 6.02*N + 1.76\n",
-      "ENOB =(SINAD1 -1.76)/6.02\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The SINAD for 12 bit convertre is \",SINAD2,\"db\"\n",
-      "print\"(b) The ENOB for the converter with SINAD of 78 dB is \",round(ENOB,2),\"bits\"   \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The SINAD for 12 bit convertre is  74.0 db\n",
-        "(b) The ENOB for the converter with SINAD of 78 dB is  12.66 bits\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4 Page no 233"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Gievn\n",
-      "Vm = 1.0\n",
-      "Vin = 0.25\n",
-      "mu =255\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "Vout = (Vm*log(1+mu*(Vin/Vm)))/log(1+mu)\n",
-      "gain =Vout/Vin\n",
-      "\n",
-      "#Result\n",
-      "print\"The output voltage of the compander \",round(Vout,2),\"volt\"\n",
-      "print\"Gain of the compander is \",round(gain,0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage of the compander  0.75 volt\n",
-        "Gain of the compander is  3.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5 Page no 234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Vin = 0.8\n",
-      "Vm =1.0\n",
-      "mu =255\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "Vout = (Vm*math.log(1+mu*(Vin/Vm)))/math.log(1+mu)\n",
-      "gain =Vout/Vin\n",
-      "\n",
-      "#Result\n",
-      "print\"The output voltage of the compander \",round(Vout,1),\"volt\"\n",
-      "print\"Gain of the compander is \",round(gain,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage of the compander  1.0 volt\n",
-        "Gain of the compander is  1.2\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter8.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter8.ipynb
deleted file mode 100755
index 4cad3df0..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter8.ipynb
+++ /dev/null
@@ -1,185 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:6738b73e6fb4913114449836a25c14169c484a93e5e60fcba9b0e4a9815c8f56"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8 Radio transmitters"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1 Page no 249"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f = 16*10**6\n",
-      "ppm = 200\n",
-      "frequency_variation = 200 *16 \n",
-      "\n",
-      "#Calculation\n",
-      "min_f = f - frequency_variation\n",
-      "max_f = f + frequency_variation\n",
-      "\n",
-      "#Reslt\n",
-      "print\"The minimum and maximum frequencies for the crystal of 16 Mhz with stability of 200 are \",min_f,\"Hz and\",max_f,\"Hz respectively\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum and maximum frequencies for the crystal of 16 Mhz with stability of 200 are  15996800 Hz and 16003200 Hz respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2 Page no 250"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f =14.9*10**6\n",
-      "mul_factor = 2*3*3\n",
-      "stability_ppm =300\n",
-      "variation = 0.0003\n",
-      "\n",
-      "#Calculation\n",
-      "total_variation = variation* mul_factor\n",
-      "fout = f * mul_factor\n",
-      "frequency_variation = fout*total_variation\n",
-      "f_lower = fout - frequency_variation\n",
-      "f_upper = fout + frequency_variation\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The output frequency of the transmitter is \",fout/10**6,\"MHz\"\n",
-      "print\"(b) The maximum and minimum frequencies of the transmitter are \",round(f_lower/10**6,2),\"Mhz and \",round(f_upper/10**6,2),\"Mhz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The output frequency of the transmitter is  268.2 MHz\n",
-        "(b) The maximum and minimum frequencies of the transmitter are  266.75 Mhz and  269.65 Mhz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.3 Page no 259"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "f = 10*10**6\n",
-      "div_factor = 100.0\n",
-      "A =63\n",
-      "N = 285\n",
-      "M=32\n",
-      "\n",
-      "#Calculation\n",
-      "ref = f/div_factor\n",
-      "R =M*N+A\n",
-      "fout= R*ref\n",
-      "\n",
-      "#Result\n",
-      "print\"The output frequency of the synthesizer is \",fout/10**6,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output frequency of the synthesizer is  918.3 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4 Page no 259"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "f = 10*10**6\n",
-      "div_factor = 100.0\n",
-      "A =64\n",
-      "N = 285\n",
-      "M=32\n",
-      "\n",
-      "#Calculation\n",
-      "ref = f/div_factor\n",
-      "R =M*N+A\n",
-      "fout= R*ref\n",
-      "\n",
-      "#Result\n",
-      "print\"The output frequency of the synthesizer is \",fout/10**6,\"MHz\"\n",
-      "print\"The step change is \",fout/10**6-918.3,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output frequency of the synthesizer is  918.4 MHz\n",
-        "The step change is  0.1 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter9.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter9.ipynb
deleted file mode 100755
index 379e8939..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter9.ipynb
+++ /dev/null
@@ -1,274 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3ff4b00636e0eec0da6869f81549546163050c8b991ea646d8083caae0f8dd37"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9 Communication receivers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1 Page no 318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "fl =220*10**6\n",
-      "fm =224*10**6\n",
-      "IF1 = 10.7*10**6\n",
-      "IF = 1.5*10**6\n",
-      "\n",
-      "#Calculation\n",
-      "IF2 =IF1+IF\n",
-      "tune_l =fl+IF1\n",
-      "tune_m = fm+IF1\n",
-      "IF1_imgl = tune_l+IF1\n",
-      "IF2_imgm = tune_m+IF1\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The local oscillatior tuning range is \",tune_l/10**6,\"to \",tune_m/10**6,\"MHz\"\n",
-      "print\"(b) Frequency of the second local oscillator is \",IF2/10**6,\"MHz\"\n",
-      "print\"(c) First IF image range is \",IF1_imgl/10**6,\"MHz to \",IF2_imgm/10**6,\"MHz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The local oscillatior tuning range is  230.7 to  234.7 MHz\n",
-        "(b) Frequency of the second local oscillator is  12.2 MHz\n",
-        "(c) First IF image range is  241.4 MHz to  245.4 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2 Page no 324"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "R = 100*10**3\n",
-      "T = 273+25\n",
-      "B = 20*10**3\n",
-      "k = 1.38*10**-23\n",
-      "\n",
-      "#Calculation\n",
-      "Vn=(4*k*T*B*R)**0.5\n",
-      "\n",
-      "#Result\n",
-      "print\"The noise voltage across 100k resistor is \",round(Vn*10**6,2),\"microvolt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The noise voltage across 100k resistor is  5.74 microvolt\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3 Page no 324"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "R=75\n",
-      "B=6*10**6  \n",
-      "T = 29+273\n",
-      "k =1.38*10**-23\n",
-      "\n",
-      "#calculation\n",
-      "Vn = (4*k*T*B*R)**0.5\n",
-      "\n",
-      "#Result\n",
-      "print\"The input themal noise is \",round(Vn*10**6,2),\"microvolt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input themal noise is  2.74 microvolt\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4 Page no 326"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "Tc=32.2\n",
-      "Tk=273+Tc\n",
-      "B =30*10**3\n",
-      "k =1.38*10**-23\n",
-      "\n",
-      "#Calculation\n",
-      "Pn=k*Tk*B\n",
-      "\n",
-      "#Result\n",
-      "print\"The average noise power is \",round(Pn*10**16,2),\"10**-16 W\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The average noise power is  1.26 10**-16 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5 Page no 329"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "SN_ip = 8\n",
-      "SN_op = 6.0\n",
-      "\n",
-      "#Calculation\n",
-      "NR = SN_ip/SN_op\n",
-      "NF = 10*log10(NR)\n",
-      "\n",
-      "#Result\n",
-      "print\"The noise factor is \",round(NR,2)\n",
-      "print\"The noise figure is \",round(NF,2),\"dB\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The noise factor is  1.33\n",
-        "The noise figure is  1.25 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6 Page no 330"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "R= 75.0\n",
-      "T=31+273\n",
-      "k=1.38*10**-23\n",
-      "B=6*10**6\n",
-      "Vs = 8.3*10**-6\n",
-      "NF=2.8\n",
-      "\n",
-      "#Calculation\n",
-      "import math\n",
-      "Vn = math.sqrt(4*k*T*B*R)\n",
-      "Pn = Vn**2/R\n",
-      "Ps = Vs**2/R\n",
-      "SN = (Ps*10**12)/(Pn/10.0**12)\n",
-      "SN_dB =10*log10(SN)\n",
-      "NR = 10**0.28\n",
-      "SN_op = SN/NR\n",
-      "Tn = 290*(NR-1)\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The input noise power is \",round(Pn*10**12,1),\"pW\"\n",
-      "print\"(b) The input signal power is \",round(Ps*10**12,3),\"pW\"\n",
-      "print\"(c) Signal to noise ratio in decibels \",round(SN/10**24,2)\n",
-      "print\"(d) The noise factor is \",round(NR,2)\n",
-      "print\"(e) Signal to noise ratio of the amplifier is \",round(SN_op/10**24,2)\n",
-      "print\"(f) The noise temperature of the amplifier \",round(Tn,0),\"K\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The input noise power is  0.1 pW\n",
-        "(b) The input signal power is  0.919 pW\n",
-        "(c) Signal to noise ratio in decibels  9.12\n",
-        "(d) The noise factor is  1.91\n",
-        "(e) Signal to noise ratio of the amplifier is  4.79\n",
-        "(f) The noise temperature of the amplifier  263.0 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems/chapter_2.ipynb b/Principles_Of_Electronic_Communication_Systems/chapter_2.ipynb
deleted file mode 100755
index 3847150a..00000000
--- a/Principles_Of_Electronic_Communication_Systems/chapter_2.ipynb
+++ /dev/null
@@ -1,1143 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:759039f1dfc5f6d62e90bf867630bc8f1cdda845968a4b5c11c4acddcb77b251"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2 The fundamentals of Electronics: A Review"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1 Page no 31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "vout=750*10**-3\n",
-      "vin = 30*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "gain=vout/vin\n",
-      "\n",
-      "#Result\n",
-      "print\"The Voltage gain of the amplifier is \",gain\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Voltage gain of the amplifier is  25000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2 Page no 31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "pout=6\n",
-      "power_gain=80.0\n",
-      "\n",
-      "#calculation\n",
-      "pin=pout/power_gain\n",
-      "\n",
-      "#Result\n",
-      "print\"The input power of the signal is \",pin*1000,\"mW\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input power of the sigmal is  75.0 mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.3 Page no 32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "A1=5;\n",
-      "A2=2;\n",
-      "A3=17;\n",
-      "total_gain=A1*A2*A3;\n",
-      "pin= 40*10**-3;\n",
-      "\n",
-      "#calculation\n",
-      "pout=total_gain*pin;\n",
-      "\n",
-      "#Result\n",
-      "print\"The output power is\",pout,\"watts\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output power is 6.8 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.4 Page no 32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "pin=25.0*10**-6;\n",
-      "pout=1.5*10**-3;\n",
-      "A1=3.0;\n",
-      "\n",
-      "#Calculation\n",
-      "total_gain=pout/pin;\n",
-      "print\"Total gain is\",total_gain\n",
-      "A2=total_gain/A1\n",
-      "\n",
-      "#Result\n",
-      "print\"The gain of second stage is \",A2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total gain is 60.0\n",
-        "The gain of second stage is  20.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.5 Page no 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "R1=10.0*10**3;\n",
-      "R2=470.0;\n",
-      "\n",
-      "#Calculation\n",
-      "attenuation=R2/(R2+R1)\n",
-      "A2=1/attenuation\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The attenuation (A1) is\",round(attenuation,3)\n",
-      "print\"(b) The attenuation (A2) is\",round(A2,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The attenuation (A1) is 0.045\n",
-        "(b) The attenuation (A2) is 22.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.6 Page no 35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "Vin=20.0*10**-6;\n",
-      "Vout=100*10**-3;\n",
-      "A1=45000.0                                #A1 isAmplifier gain\n",
-      "\n",
-      "#calculation\n",
-      "AT=Vout/Vin                                #AT is Total gain\n",
-      "A2=AT/A1                                   #A2 is attenuation factor\n",
-      "\n",
-      "#Result\n",
-      "print\"Total gain is\",AT\n",
-      "print\"The atenuation factor needed to to keep the output voltage from exceeding 100 mv is \",round(A2,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total gain is 5000.0\n",
-        "The atenuation factor needed to to keep the output voltage from exceeding 100 mv is  0.1111\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.7 Page no 36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "Vin=3.0*10**-3\n",
-      "Vout=5\n",
-      "Pin=50.0*10**-3                                     \n",
-      "Pout=2*10**-3                                     \n",
-      "\n",
-      "#calculation\n",
-      "import math\n",
-      "gain_dB= 20*log10 (Vout/Vin)\n",
-      "gain_db=10*log10 (Pout/Pin)\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The gain of amplifier in dB is \",round(gain_dB,1)\n",
-      "print\"(b) The gain in dB is \",round(gain_db,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The gain of amplifier in dB is  64.4\n",
-        "(b) The gain in dB is  -13.98\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.8 Page no 38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "gain_dB = 40\n",
-      "pout_W= 100\n",
-      "\n",
-      "#calculation\n",
-      "pin_W = pout_W/10.0**4\n",
-      "\n",
-      "#Result\n",
-      "print\"The input power is \",pin_W*10**3,\"m Watt\"\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input power is  10.0 m Watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.9 Page no 38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "\n",
-      "gain_db = 60\n",
-      "vin = 50*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "vout = 10**(60/20.0)*vin\n",
-      "\n",
-      "#Result\n",
-      "print\"The output voltage is \",vout,\"volt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage is  0.05 volt\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.10 Page no 39"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "\n",
-      "vin=90*10**-3\n",
-      "R1= 10.0*10**3\n",
-      "vout=7.8\n",
-      "Rout=8.0\n",
-      "\n",
-      "#calculation\n",
-      "pin= vin**2/R1\n",
-      "pout=vout**2/Rout\n",
-      "\n",
-      "import math\n",
-      "Ap_db = 10*log10 (pout/pin)\n",
-      "\n",
-      "#Result\n",
-      "print\"The power gain in decibel is \",round(Ap_db,1),\"dB\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The power gain in decibel is  69.7 dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.11 Page no 40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "\n",
-      "gain_db = 28\n",
-      "pin = 36*10**-3\n",
-      "\n",
-      "#calculation\n",
-      "pout = 10**2.8*pin;\n",
-      "\n",
-      "#Result\n",
-      "print\"The output power is \",round(pout,2),\"watt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output power is  22.71 watt\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.12 Page no 40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "gain1 = 6.8 \n",
-      "gain2 = 14.3\n",
-      "attenuation1 = -16.4\n",
-      "attenuation2 = -2.9\n",
-      "vout = 800*10**-3\n",
-      "\n",
-      "#calculation\n",
-      "At = gain1+gain2+attenuation1+attenuation2\n",
-      "vin = vout/10.0**(At/20.0)\n",
-      "\n",
-      "#Result\n",
-      "print\"The input voltage is \",round(vin*10**3,1),\"mv\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input voltage is  650.3 mv\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.13 Page no 40"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "pout_db =12.3\n",
-      "\n",
-      "#calculation\n",
-      "pout_mW = 0.001*10**(12.3/10.0)\n",
-      "\n",
-      "#Result\n",
-      "print\"The output power is \" ,round(pout_mW*10**3,0),\"mv\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output power is  17.0 mv\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.14 Page no 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "c = 2.7*10**-12\n",
-      "l = 33*10**-9\n",
-      "\n",
-      "#calculation\n",
-      "fr= 1/(6.28*(l*c)**0.5)\n",
-      "\n",
-      "#Result\n",
-      "print\"The resonat frequency is \" ,round(fr/10.0**6,0),\"Mhz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resonat frequency is  533.0 Mhz\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.15 Page no 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "c =12*10**-12\n",
-      "fr = 49*10**6\n",
-      "\n",
-      "#calculation\n",
-      "l=1/(4*3.14**2*fr**2*c)\n",
-      "\n",
-      "#Result\n",
-      "print\"The value of inductance is \" ,round(l*10**9,0),\"nh\" \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of inductance is  880.0 nh\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.16 Page no 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "fr=28*10**6\n",
-      "Q=70.0\n",
-      "\n",
-      "#calculation\n",
-      "bandwidth = fr/Q\n",
-      "\n",
-      "#Result\n",
-      "print\"The bandwidth is \",bandwidth/10.0**3,\"Khz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth is  400.0 Khz\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.17 Page no 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "f1= 7.93*10**6\n",
-      "f2= 8.07*10**6\n",
-      "\n",
-      "#calculation\n",
-      "bw= f2-f1\n",
-      "fr=(f1*f2)**0.5\n",
-      "Q= fr/bw\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The bandwidth is \",bw/10.0**3,\"Khz\"\n",
-      "print\"(b) The resonant frequency is \",round(fr/10.0**6,0),\"Mhz\"\n",
-      "print\"(c) The Q of resonant circuit is \",round(Q,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The bandwidth is  140.0 Khz\n",
-        "(b) The resonant frequency is  8.0 Mhz\n",
-        "(c) The Q of resonant circuit is  57.14\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.18 Page no 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "Q=200.0\n",
-      "fr=16*10**6\n",
-      "\n",
-      "#calculation\n",
-      "bw=fr/Q\n",
-      "f1= fr-(bw/2)\n",
-      "f2=fr+(bw/2)\n",
-      "\n",
-      "#Result\n",
-      "print\"Bandwidth is \",bw*10**-3,\"KHz\"\n",
-      "print\"f1= \",f1*10**-6,\"MHz\"\n",
-      "print\"f2= \",f2*10**-6,\"MHz\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bandwidth is  80.0 KHz\n",
-        "f1=  15.96 MHz\n",
-        "f2=  16.04 MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.19 Page no 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "Q= 150\n",
-      "Vs=3*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "Vc= Q*Vs\n",
-      "\n",
-      "#Result\n",
-      "print\"The voltage across capacitor is \",Vc*10**6,\"microvolt\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage across capacitor is  450.0 microvolt\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.20 Page no 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "fr= 52*10**6\n",
-      "Q=12.0\n",
-      "L=0.15*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "Rw=(6.28*fr*L)/Q\n",
-      "Req= Rw*(Q**2+1)\n",
-      "\n",
-      "#Result\n",
-      "print\"Impedance of the parellel LC circuit is \",round(Req,0),\"ohm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Impedance of the parellel LC circuit is  592.0 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.21 Page no 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "fr= 52.0*10**6\n",
-      "Rw= 4.1\n",
-      "L =0.15*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "C=1/(4.0*3.14**2*fr**2*L)\n",
-      "Z = L/(C*Rw)\n",
-      "\n",
-      "#Result\n",
-      "print\"the impedance of the circuit is \",round(Z,0),\"ohm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the impedance of the circuit is  585.0 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.22 Page no 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "bw = 1.0*10**6\n",
-      "XL = 300\n",
-      "Rw = 10.0\n",
-      "fr =10*10**6\n",
-      "\n",
-      "#calculation\n",
-      "Q1 = XL/Rw\n",
-      "Rp = Rw*(Q1**2+1) \n",
-      "\n",
-      "Q2 = fr/bw\n",
-      "Rpnew = Q2*XL\n",
-      "\n",
-      "Rext = (Rpnew*Rp)/(Rp-Rpnew)\n",
-      "\n",
-      "#Result\n",
-      "print\"The value of resistor needed to set the bandwidth of the parellel tuned circuit is \",round(Rext,1),\"ohm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of resistor needed to set the bandwidth of the parellel tuned circuit is  4497.5 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.23 Page no 57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "R = 8.2*10**3\n",
-      "C =0.0033*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "fco = 1/(6.28* R*C)\n",
-      "\n",
-      "#Result\n",
-      "print\"The cut off frequency is \",round(fco/10.0**3,2),\"Khz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The cut off frequency is  5.88 Khz\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.24 Page no 60"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "fco =3.4*10**3\n",
-      "C = 0.047*10**-6\n",
-      "\n",
-      "#calculation\n",
-      "R = 1/(6.28* fco* C)\n",
-      "\n",
-      "#Result\n",
-      "print\"The value of the resistor is \",round(R,0),\"ohm\"\n",
-      "print\"The closest standard value is \", 1000 ,\"ohm\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the resistor is  996.0 ohm\n",
-        "The closest standard value is  1000 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.25 Page no 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "fnotch = 120\n",
-      "R = 220*10**3\n",
-      "\n",
-      "#calculatiuon\n",
-      "C = 1/(6.28*R*fnotch)\n",
-      "\n",
-      "#Result\n",
-      "print\"The value of capacitance required is \",round(2*C*10**6,3),\"microfarad\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of capacitance required is  0.012 microfarad\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.26 Page no 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "Vpeak =3.0\n",
-      "f=48*10**3\n",
-      "\n",
-      "#calculation\n",
-      "fifth_harmonic = 5*f\n",
-      "Vrms=(4/3.14)*(3/5.0)*0.707\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The frequency of the fifth harmonic is \",fifth_harmonic/10.0**3,\"Khz\"\n",
-      "print\"The RMS voltage of the fifth harmonic is \",round(Vrms,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The frequency of the fifth harmonic is  240.0 Khz\n",
-        "The RMS voltage of the fifth harmonic is  0.54\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.27 Page no 87"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "Vpeak = 5\n",
-      "f = 4.0*10**6\n",
-      "duty_cycle=0.3\n",
-      "\n",
-      "#calculation\n",
-      "T = 1/f\n",
-      "t0 = duty_cycle*T\n",
-      "Vavg = Vpeak*duty_cycle\n",
-      "min_bw =1/t0\n",
-      "\n",
-      "#Result\n",
-      "print\"(a) The average DC value is \",Vavg,\"volt\"\n",
-      "print\"(b) The minimum bandwidth required is \" ,round(min_bw/10.0**6,3),\"Mhz\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The average DC value is  1.5 volt\n",
-        "(b) The minimum bandwidth required is  13.333 Mhz\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28 Page no 88"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "tr =6*10**-9\n",
-      "\n",
-      "#calculation\n",
-      "min_bw=(35/0.006)\n",
-      "\n",
-      "#Result\n",
-      "print\"The minimum bandwidth is % is \",round(min_bw/10.0**2,1),\"Mhz\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum bandwidth is % is  58.3 Mhz\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.29 Page no 89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "bw= 200.0*10**3\n",
-      "\n",
-      "#calculation\n",
-      "tr= 0.35/(bw*10**-3)\n",
-      "\n",
-      "#Result\n",
-      "print\"The fastest rise time of the circuit is \" ,tr*10**3,\"microseconds\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fastest rise time of the circuit is  1.75 microseconds\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.30 Page no 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given\n",
-      "bw_mhz = 60\n",
-      "tri_ns= 15\n",
-      "\n",
-      "#calculation\n",
-      "tra_osci = 0.35/(bw_mhz)\n",
-      "tra_comp = 1.1*(tri_ns**2 + (tra_osci*10**3)**2)**0.5\n",
-      "\n",
-      "#Result\n",
-      "print\"The rise time of the displayed square wave is \",round(tra_comp,1),\"ns\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rise time of the displayed square wave is  17.7 ns\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
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Binary files a/Principles_Of_Electronic_Communication_Systems/screenshots/image_3.png and /dev/null differ
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter10_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter10_1.ipynb
new file mode 100755
index 00000000..7ddfe56f
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter10_1.ipynb
@@ -0,0 +1,101 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:076f20d31c3bb1bd9e8b8227097af625d385f1bb7f3b7a30dfedf3e414bce0c7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 Multiplexing and Demultiplexing"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page no 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "BW_service = 860*10**6\n",
+      "BW_ch = 6.0*10**6\n",
+      "\n",
+      "#calculation\n",
+      "total_ch = BW_service/BW_ch\n",
+      "\n",
+      "#result\n",
+      "print\"Total number of channels are \",round(total_ch,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total number of channels are  143.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page no 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "channels =16\n",
+      "sampling_rate= 3.5*10**3\n",
+      "w_len=6\n",
+      "\n",
+      "#Calculation\n",
+      "available_ch =channels-1\n",
+      "bpf =channels*w_len\n",
+      "data_rate = sampling_rate * bpf\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Available channels are \",available_ch\n",
+      "print\"(b) Bits Per Frame = \",bpf\n",
+      "print\"(c) The serial data rate \",data_rate/10**3,\"Khz\"\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Available channels are  15\n",
+        "(b) Bits Per Frame =  96\n",
+        "(c) The serial data rate  336.0 Khz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter11_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter11_1.ipynb
new file mode 100755
index 00000000..72e4d978
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter11_1.ipynb
@@ -0,0 +1,143 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ac4273a5639d4b88e878e68a7a5f45b084fdf8a5c45c4bc6304c023f494f5c4d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 The Transmission of binary data in communication systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 Page no 392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "t=0.0016\n",
+      "No_words=256.0\n",
+      "bits_word = 12.0\n",
+      "\n",
+      "#Calculation\n",
+      "tword= t/No_words\n",
+      "tbit = tword/bits_word\n",
+      "bps =1/tbit\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The time duration of the word \",tword*10**8,\"microsecond\"\n",
+      "print\"(b) The time duration of the one bit is \",round(tbit*10**8,4),\"microseconds\"\n",
+      "print\"(c) The speed of transmission is \",bps/10**5,\"kbps\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The time duration of the word  625.0 microsecond\n",
+        "(b) The time duration of the one bit is  52.0833 microseconds\n",
+        "(c) The speed of transmission is  19.2 kbps\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 Page no 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "B=12.5*10**3\n",
+      "SN_dB= 25\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "C_th = 2*B\n",
+      "SN=316.2\n",
+      "C =B*3.32*log10(SN+1)\n",
+      "N= 2**(C/(2.0*B))\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The maximum theorotical data rate is \",C_th/10**3,\"kbps\"\n",
+      "print\"(b) The maximum theorotical capacity of channel is \",round(C/10**3,1),\"Kbps\"\n",
+      "print\"(c) The number of levels needed to acheive maximum speed are \",round(N,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The maximum theorotical data rate is  25.0 kbps\n",
+        "(b) The maximum theorotical capacity of channel is  103.8 Kbps\n",
+        "(c) The number of levels needed to acheive maximum speed are  17.78\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 Page no 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "block =512\n",
+      "packets =8\n",
+      "BER = 2*10**-4\n",
+      " \n",
+      "#Calculation\n",
+      "avg_errors = block*packets*8*BER\n",
+      "\n",
+      "print\"Average number of errors are \",avg_errors\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average number of errors are  6.5536\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter12_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter12_1.ipynb
new file mode 100755
index 00000000..09288512
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter12_1.ipynb
@@ -0,0 +1,101 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:24e3c45fb7b4a00b042c526c50c49b4a5f07521c8d395e638826ab3ea2de6f29"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 Introduction to Networking and local area networks"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Page no 448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "N = 20\n",
+      "\n",
+      "#Calculation\n",
+      "L = (N*(N-1))/2.0\n",
+      "\n",
+      "#Result\n",
+      "print\"The number of interconnecting wires required are \",L\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of interconnecting wires required are  190.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 Page no 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "block = 1500\n",
+      "ethernet = 10.0*10**6\n",
+      "token_ring = 16.0*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "t1_bit = 1/ethernet\n",
+      "t1_byte = 8*t1_bit\n",
+      "t1_1526 = 1526 *t1_byte\n",
+      "t2_bit = 1/token_ring\n",
+      "t2_byte = 8 * t2_bit\n",
+      "t2_1521 =1521*t2_byte\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Time required for 1526 bytes is \",t1_1526*10**6,\"ns\"\n",
+      "print\"(b) Time required for 1521 bytes is \",t2_1521*10**3,\"ms\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Time required for 1526 bytes is  1220.8 ns\n",
+        "(b) Time required for 1521 bytes is  0.7605 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter13_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter13_1.ipynb
new file mode 100755
index 00000000..a0657cff
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter13_1.ipynb
@@ -0,0 +1,339 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0554007e74bfe6cdd53136e8f25769048ad21412148cdd192396d4befa058e18"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 Transmission lines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 Page no 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f= 450.0*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = 984/f\n",
+      "len =0.1*lamda\n",
+      "\n",
+      "#Result\n",
+      "print\"feet long conductors would be considered as the transmission line \",round(len*10**6,3),\"ft\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "feet long conductors would be considered as the transmission line  0.219 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 Page no 484"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "lamda = 2.19\n",
+      "\n",
+      "#Calculation\n",
+      "len = (3/8.0)*lamda\n",
+      "\n",
+      "#Result\n",
+      "print\"The pyhsical length of the transmission line \",round(len,2),\"feet\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pyhsical length of the transmission line  0.82 feet\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page no 492"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "len = 165\n",
+      "attn_100ft = 5.3\n",
+      "pin = 100\n",
+      "attn_ft = 5.3/100.0\n",
+      "\n",
+      "#Calculation\n",
+      "total_attn = attn_ft * len\n",
+      "pout = pin *0.1335\n",
+      "\n",
+      "#Result\n",
+      "print\"The total attenuation of the cable is \",total_attn,\"dB\"\n",
+      "print\"Output power is \",pout,\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total attenuation of the cable is  8.745 dB\n",
+        "Output power is  13.35 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4 Page no 494"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "len =150\n",
+      "C =13.5\n",
+      "Z0 =93\n",
+      "f =2.5*10**6\n",
+      "attn_100ft =2.8\n",
+      "\n",
+      "#Calculation\n",
+      "L =C*Z0**2\n",
+      "td =(L*C)**0.5\n",
+      "theta = ((360)*188.3)/(1/f)\n",
+      "attn_ft = attn_100ft/100.0\n",
+      "total_attn = attn_ft*150\n",
+      "\n",
+      "print\"(a) The load impedance required to terminate the the line to avoid the reflections is %d ohm\",Z0\n",
+      "print\"(b) The equivalent inductance per feet is \",L/10**3,\"nh\"\n",
+      "print\"(c) The time delay introduced by the cable per feet is \",td/10**3,\"ns\"\n",
+      "print\"(d) The phase shift occurs in degrees for the 2.5 Mhz sine wave is \",theta/10**9\n",
+      "print\"(e) The total attenuation is \",total_attn,\"db\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The load impedance required to terminate the the line to avoid the reflections is %d ohm 93\n",
+        "(b) The equivalent inductance per feet is  116.7615 nh\n",
+        "(c) The time delay introduced by the cable per feet is  1.2555 ns\n",
+        "(d) The phase shift occurs in degrees for the 2.5 Mhz sine wave is  169.47\n",
+        "(e) The total attenuation is  4.2 db\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5 Page no 501"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "vmax= 52.0\n",
+      "vmin= 17.0\n",
+      "Z0 = 75\n",
+      "\n",
+      "#calculation\n",
+      "SWR = vmax/vmin\n",
+      "ref_coeff = (vmax-vmin)/(vmax+vmin)\n",
+      "Zl1 = Z0*SWR\n",
+      "Zl2 = Z0/SWR\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The standing wave ratio is \",round(SWR,2)\n",
+      "print\"(b) Reflection coefficient is \",round(ref_coeff,2)\n",
+      "print\"The value of resistive load is \",round(Zl1,2),\"ohm or\",round(Zl2,2),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The standing wave ratio is  3.06\n",
+        "(b) Reflection coefficient is  0.51\n",
+        "The value of resistive load is  229.41 ohm or 24.52 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.6 Page no 503"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "SWR =3.05\n",
+      "ref_pwr =0.2562\n",
+      "pin =30\n",
+      "\n",
+      "#calculation\n",
+      "pout = pin -(pin*((SWR-1)/(SWR+1))**2)\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power of the cable is \",round(pout,3),\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power of the cable is  22.314 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.7 Page no 508"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "C =4*10**-12\n",
+      "f =800*10**6\n",
+      "diele = 3.5\n",
+      "h = 0.0625\n",
+      "w = 0.13\n",
+      "t = 0.002\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Z0 = 38.8*math.log(0.374/0.106)\n",
+      "Xc = 1/(6.28*f*C)\n",
+      "\n",
+      "#Result\n",
+      "print\"The charecteristics impedance of the transmission line is \",round(Z0,1),\"ohm\"\n",
+      "print\"The reactance of the capacitor is \",round(Xc,2),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The charecteristics impedance of the transmission line is  48.9 ohm\n",
+        "The reactance of the capacitor is  49.76 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.8 Page no 508"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "lamda = (984/800.0)\n",
+      "lamda_8 =lamda/8.0\n",
+      "\n",
+      "#Calculation\n",
+      "len = lamda_8*12*(1/3.6**0.5)\n",
+      "\n",
+      "#Result\n",
+      "print\"The length of the transmission line is \",round(len,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of the transmission line is  0.972\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter14_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter14_1.ipynb
new file mode 100755
index 00000000..eb757247
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter14_1.ipynb
@@ -0,0 +1,229 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8f9093920b7d1b010ea79cdf7fde78edb0be4df7ea34f4ae2c0023eb64d7b5d4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 Antennas and Wave Propogation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1 page no 544"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f=310.0*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "len1 =(492*0.97)/f\n",
+      "len2 =(492/f)*0.8\n",
+      "len3 =(984/f)*0.73\n",
+      "z1 =120*log(35/2.0)\n",
+      "len4 =234/f\n",
+      "z2  = 73/2.0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The length and radiation resistance of the dipole are \",round(len1*10**6,2),\"feet and 73 ohm respectively\"\n",
+      "print\"(b) The length of the folded dipole are \",round(len2*10**6,2),\"feet\"\n",
+      "print\"(c) The length and radiation resistance of the bow tie antenna are \",round(len3*10**6,1),\"feet and \",round(z1,1),\"ohm respectively\"\n",
+      "print\"(d) The length and radiation resistance of the groun plane antenna are \",round(len4*10**6,3),\"feet and\",z2,\"ohmrespectively\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The length and radiation resistance of the dipole are  1.54 feet and 73 ohm respectively\n",
+        "(b) The length of the folded dipole are  1.27 feet\n",
+        "(c) The length and radiation resistance of the bow tie antenna are  2.3 feet and  343.5 ohm respectively\n",
+        "(d) The length and radiation resistance of the groun plane antenna are  0.755 feet and 36.5 ohmrespectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2 Page no 553"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "gain=14\n",
+      "len=250\n",
+      "attn_100=3.6\n",
+      "f =220*10\n",
+      "pin =50\n",
+      "p =0.126\n",
+      "\n",
+      "#Calculation\n",
+      "pout =pin*p\n",
+      "line_loss =pin-pout\n",
+      "pwr_ratio = 25.1\n",
+      "ERP = pwr_ratio*pout\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The transmission line loss is \",line_loss\n",
+      "print\"(b) Effective raduated power is \",round(ERP,1),\"W\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The transmission line loss is  43.7\n",
+        "(b) Effective raduated power is  158.1 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3 Page no 556"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Z0 =50\n",
+      "Zl =172\n",
+      "f =460.0*10**6\n",
+      "VF =0.86\n",
+      "\n",
+      "#Calculation\n",
+      "len =(246/f)*VF\n",
+      "\n",
+      "#Result\n",
+      "print\"The length of the impedance matching section needed for the Q section is \",round(len*10**6,2),\"feet\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of the impedance matching section needed for the Q section is  0.46 feet\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4 Page no 557"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 460.0*10**6\n",
+      "VF = 0.66\n",
+      "\n",
+      "#Calculation\n",
+      "len = (246/f)*VF\n",
+      "\n",
+      "#Result \n",
+      "print\"The length of impedance matching section is \",round(len*10**6,3),\"feet\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of impedance matching section is  0.353 feet\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.5 Page no 567"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "ht =275\n",
+      "hr =60\n",
+      "f=224.0*10**6\n",
+      "pt=100\n",
+      "Gt = 26\n",
+      "Gr = 3.27\n",
+      "\n",
+      "#Calculation\n",
+      "D =((2*ht)**0.5+(2*hr)**0.5)*1.61\n",
+      "lamda = 300/f\n",
+      "Pr = (pt*Gt*Gr*lamda**2)/(16*3.14**2*D**2)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The maximmum transmitting distance is \",round(D,1),\"kilometer\"\n",
+      "print\"The received power is \",round(Pr*10**15,1),\"nW\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The maximmum transmitting distance is  55.4 kilometer\n",
+        "The received power is  31.5 nW\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter16_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter16_1.ipynb
new file mode 100755
index 00000000..01f25016
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter16_1.ipynb
@@ -0,0 +1,217 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:549348f6567319d4a861893116b77bd50ff6991f4ba0e1a8c927cf50abf21c3e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16 Microwave communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.1 Page no 616"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Zsrc =50\n",
+      "Zld  =136\n",
+      "f =5800.0*10**6\n",
+      "Er =2.4\n",
+      "\n",
+      "#Calculation\n",
+      "Zq =(Zsrc * Zld)**0.5\n",
+      "Vp =1/(Er)**0.5\n",
+      "lamda = 300/f\n",
+      "len = (lamda/4.0)*38.37*Vp\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The required impedance is \",round(Zq,2),\"ohm\"\n",
+      "print\"(b) The length of the microstrip \",round(len*10**6,2),\"inches\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The required impedance is  82.46 ohm\n",
+        "(b) The length of the microstrip  0.32 inches\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2 Page no 623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "w=0.65\n",
+      "h=0.38\n",
+      "\n",
+      "#Calculation\n",
+      "fco = 300/(2.0*((0.65*2.54)/100.0))\n",
+      "f =1.42*fco\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The cutoff frequency of the \",round(fco/10**3,3),\"GHz\"\n",
+      "print\"(b) Operating frequency of the wavwguide is \",round(f/10**3,1),\"GHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The cutoff frequency of the  9.085 GHz\n",
+        "(b) Operating frequency of the wavwguide is  12.9 GHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3 Page no 623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#for the above question\n",
+      "print\"The c band is approximately 4 to 6 Ghz since a waveguide acts as a high pass filter with cut off of 9.08 Ghz it will not pass c band signal\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The c band is approximately 4 to 6 Ghz since a waveguide acts as a high pass filter with cut off of 9.08 Ghz it will not pass c band signal\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.4 Page no 648"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "lamda1 =5.0\n",
+      "f2 = 15.0*10**9\n",
+      "D=1.524\n",
+      "\n",
+      "#Calculation\n",
+      "f1=984/lamda1\n",
+      "lamda2 =300/f2\n",
+      "G = (6*(D/lamda2)**2)\n",
+      "B = 70/(D/lamda2)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The lowest possible oprerating frequency is \",f1,\"MHz\"\n",
+      "print\"(b) The gain at 15 Ghz is \",G/10**12\n",
+      "print\"(c) The beam width at 15Ghz is \",round(B*10**6,2),\"degree\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The lowest possible oprerating frequency is  196.8 MHz\n",
+        "(b) The gain at 15 Ghz is  34838.64\n",
+        "(c) The beam width at 15Ghz is  0.92 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.5 Page no 661"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "T = 9.2\n",
+      "theta = 20\n",
+      "sin20 = 0.342\n",
+      "a=5280\n",
+      "\n",
+      "#Calculation\n",
+      "D_nautical = T/12.36\n",
+      "D_statute =D_nautical*0.87\n",
+      "A = D_statute*0.342\n",
+      "A1=A*a\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The line of distance to the aircraft in the statute miles \",round(D_statute,3)\n",
+      "print\"(b) The altitude of the aircraft is \",A1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.221469902913\n",
+        "(a) The line of distance to the aircraft in the statute miles  0.648\n",
+        "(b) The altitude of the aircraft is  1169.36108738\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter17_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter17_1.ipynb
new file mode 100755
index 00000000..cced69ab
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter17_1.ipynb
@@ -0,0 +1,142 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1fcf40b444ffe01cf0b970d987135c0a701705380b8393b58911739ca6b5af8f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapeter 17 Satellite communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.1 Page no 678"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "stn_long = 95\n",
+      "stn_lat = 30\n",
+      "sat_long =121\n",
+      "rad_pos = 137\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "azimuth = 360-rad_pos\n",
+      "\n",
+      "#Result\n",
+      "print\"The elevation setting for the antenna is 45 degree\"\n",
+      "print\"The azimuth setting for the antenna is \",azimuth,\"degree\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The elevation setting for the antenna is 45 degree\n",
+        "The azimuth setting for the antenna is  223 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.2 Page no 681"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "flo = 2*10**9\n",
+      "fd =3840*10**6\n",
+      "B =36*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "fu =fd+flo\n",
+      "C =2*B\n",
+      "\n",
+      "#Result\n",
+      "print\"The uplink frequency is \",fu/10.0**9,\"GHz\"\n",
+      "print\"The data rate is \", C/10**6,\"Mbps\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The uplink frequency is  5.84 GHz\n",
+        "The data rate is  72 Mbps\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.3 Page no 691"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "fs = 4.08*10**9\n",
+      "fIF1 = 770*10**6\n",
+      "fIF2 = 140*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "flo1 = fs - fIF1\n",
+      "flo2 = fIF1 - fIF2\n",
+      "\n",
+      "#Result\n",
+      "print\"The local oscillator frequency for first IF is \",flo1/10**6,\"MHz\"\n",
+      "print\"The local oscillator frequency for the second IF is \",flo2/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The local oscillator frequency for first IF is  3310.0 MHz\n",
+        "The local oscillator frequency for the second IF is  630 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter19_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter19_1.ipynb
new file mode 100755
index 00000000..a9b84b67
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter19_1.ipynb
@@ -0,0 +1,128 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7e3bc9cbae38bb1fde7396ff7ff501e2e792199bc6ea4d5bd63e700cc726cfd6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 19 Optical communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.1 Page no 760"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "NA = 0.29\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "critical_angle = math.asin(0.29)*180/3.14\n",
+      "\n",
+      "print\"The critical angle is \",round(critical_angle,2),\"degree\"\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The critical angle is  16.87 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.2 Page no 767"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "B_rating_Mhzkm =600*10**6\n",
+      "len_ft=500\n",
+      "\n",
+      "#Calculation\n",
+      "bandwidth = B_rating_Mhzkm/(len_ft/3274.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"The bandwidth of the 500 feetr segment of the cable is \",bandwidth/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The bandwidth of the 500 feetr segment of the ccable is  3928.8 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.3 Page no 780"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R=43*10**6\n",
+      "D=1200/3274.0\n",
+      "\n",
+      "#Calculation\n",
+      "d=1/(5.0*R*D)\n",
+      "\n",
+      "#Result\n",
+      "print\"The dispersion factor is \",round(d*10**9,1),\"ns/km\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dispersion factor is  12.7 ns/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter1_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter1_1.ipynb
new file mode 100755
index 00000000..48b042da
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter1_1.ipynb
@@ -0,0 +1,249 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:fa4937e894b2eaacf3d917ad54d3257fed9024ea716662722fd829aea08934e3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 Introduction to electronic communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1.a Page no 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "c=300000000\n",
+      "f1=150000000.0\n",
+      "f2=430000000.0\n",
+      "f3=8000000.0\n",
+      "f4=750000.0\n",
+      "\n",
+      "#Calculation\n",
+      "W1 = c/f1\n",
+      "W2=c/f2\n",
+      "W3=c/f3\n",
+      "W4=c/f4\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Wavelength is \",W1,\"meter\"\n",
+      "print\"(b) Wavelength is \",round(W2,3),\"meter\"\n",
+      "print\"(c) Wavelength is \",W3,\"meter\"\n",
+      "print\"(d) Wavelength is \",W4,\"meter\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Wavelength is  2.0 meter\n",
+        "(b) Wavelength is  0.698 meter\n",
+        "(c) Wavelength is  37.5 meter\n",
+        "(d) Wavelength is  400.0 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 page no 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "c=300000000\n",
+      "wavelength=1.5\n",
+      "\n",
+      "#Calculation\n",
+      "frequency=c/wavelength\n",
+      "\n",
+      "#Result\n",
+      "print\"Signal frequncy is \",frequency/1000000.0,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signal frequncy is  200.0 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 page no 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "wavelength_feet=75\n",
+      "wavelength_meter= 75/3.28\n",
+      "c=300000000\n",
+      "\n",
+      "#Calculation\n",
+      "frequency=c/wavelength_meter\n",
+      "\n",
+      "#Result\n",
+      "print\"The signal frequncy is \",frequency/1000000.0,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The signal frequncy is  13.12 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 page no 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "wavelength_inches=8\n",
+      "wavelength_meter=  8/39.37\n",
+      "c=300\n",
+      "\n",
+      "#Calculation\n",
+      "frequency= c/wavelength_meter\n",
+      "\n",
+      "#Result\n",
+      "print\"The signal freuency is \",round(frequency,1),\"MHz\"\n",
+      "print\"The signnalfrequency is \",round(frequency*10**-3,2),\"GHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The signal freuency is  1476.4 MHz\n",
+        "The signnalfrequency is  1.48 GHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 page no 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f1=902000000\n",
+      "f2=928000000\n",
+      "\n",
+      "#Calculation\n",
+      "bandwidth=f2-f1\n",
+      "\n",
+      "#Result\n",
+      "print\"Width of the band is \",bandwidth/1000000,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Width of the band is  26 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 page no 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "bandwidth_megahertz= 6\n",
+      "f1_megahertz=54\n",
+      "\n",
+      "#Calculation\n",
+      "f2_megahertz=f1_megahertz + bandwidth_megahertz\n",
+      "\n",
+      "#Result\n",
+      "print\"Upper frequency limit is \",f2_megahertz,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Upper frequency limit is  60 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter3_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter3_1.ipynb
new file mode 100755
index 00000000..ddf30c24
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter3_1.ipynb
@@ -0,0 +1,331 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:85b158cea1a60588125a433baf8a8e61131cf8432c25ac46bd61066878c52a4c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 Amplitude modulation fundamentals"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 page no 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vmax = 5.9\n",
+      "Vmin = 1.2\n",
+      "\n",
+      "#Calculation\n",
+      "m = (Vmax-Vmin)/(Vmax+Vmin)\n",
+      "Vc = (Vmax+Vmin)/2.0\n",
+      "Vm = (Vmax-Vmin)/2.0\n",
+      "m = Vm/Vc\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The modulation index is \",round(m,2)\n",
+      "print\"Vc= \",Vc,\"Vm= \",Vm,\"(for 2 volt/div on verticle scale)\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The modulation index is  0.66\n",
+        "Vc=  3.55 Vm=  2.35 (for 2 volt/div on verticle scale)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 page no 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "frq =980*10**3\n",
+      "frq_range = 5*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "fusb = frq+frq_range\n",
+      "flsb = frq-frq_range\n",
+      "bw=fusb-flsb\n",
+      "\n",
+      "#Result\n",
+      "print\"The upper sideband is at \",fusb/10**3,\"Khz\"\n",
+      "print\"Lower sideband is at \",flsb/10**3,\"Khz\"\n",
+      "print\"the babdwidth is \",bw/10**3,\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The upper sideband is at  985 Khz\n",
+        "Lower sideband is at  975 Khz\n",
+        "the babdwidth is  10 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 page no 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Pc = 30\n",
+      "m=0.85\n",
+      "\n",
+      "#Calculation\n",
+      "Pt = Pc*(1+ (m**2/2.0))\n",
+      "Psb_both =Pt-Pc\n",
+      "Psb_one = Psb_both/2.0\n",
+      "\n",
+      "#Result\n",
+      "print\"The total power is \",round(Pt,1),\"watt\" \n",
+      "print\"The power in one sideband is \",round(Psb_one,1),\"Watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total power is  40.8 watt\n",
+        "The power in one sideband is  5.4 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 page no 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R = 40\n",
+      "I = 4.8\n",
+      "m=0.9\n",
+      "\n",
+      "#Calculation\n",
+      "Pc = I**2*R\n",
+      "Pt = (I*(1+(m**2/2.0))**0.5)**2*R\n",
+      "Psb = Pt-Pc\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The carrier power is \",Pc,\" watt\"\n",
+      "print\"(b) Total power = \",round(Pt,0),\"watt\"\n",
+      "print\"(c) Sideband Power = \",round(Psb,1),\"watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The carrier power is  921.6  watt\n",
+        "(b) Total power =  1295.0 watt\n",
+        "(c) Sideband Power =  373.2 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 page no 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "It = 5.1\n",
+      "Ic =4.8\n",
+      "\n",
+      "#Calculation\n",
+      "m=(2*((It/Ic)**2-1))**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The percentage of modulation is \",round(m*100,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage of modulation is  51.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 page no 109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "m = 0.9\n",
+      "Pc = 921.6\n",
+      "\n",
+      "#calculation\n",
+      "Psb = (m**2*Pc)/4.0\n",
+      "\n",
+      "#Result\n",
+      "print\"The power in one sideband \",round(Psb,1),\"Watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power in one sideband  186.6 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 page no 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vpp = 178\n",
+      "R = 75.0\n",
+      "\n",
+      "#Calculation\n",
+      "Vp =Vpp/2.0\n",
+      "Vrms = 0.707*Vp\n",
+      "PEP =(Vrms**2/R)\n",
+      "\n",
+      "#Result\n",
+      "print\"The peak envelop power is \", round(PEP,1),\"Watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The peak envelop power is  52.8 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 page no 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vs =24\n",
+      "Im =9.3\n",
+      "\n",
+      "#Calculation\n",
+      "PEP = Vs*Im \n",
+      "Pavg1 = PEP/3.0\n",
+      "Pavg2 = PEP/4.0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The peak envelope power is \",PEP,\"watt\"\n",
+      "print\"(b) Average power of transmitter is \",Pavg2,\"watt to\",Pavg1,\"watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The peak envelope power is  223.2 watt\n",
+        "(b) Average power of transmitter is  55.8 watt to 74.4 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter4_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter4_1.ipynb
new file mode 100755
index 00000000..45832f38
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter4_1.ipynb
@@ -0,0 +1,123 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:560b11fe692ef5f7a898cfb445cca26656fbac5b17b2ed8f9a6482b6e6d61c60"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "chapter 4 Amplitude Modulator and Demodulator circuits"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page no 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "Vcc =48\n",
+      "I = 3.5\n",
+      "efficiency =70                           #percent\n",
+      "modulation = 67                          #percent\n",
+      "m = modulation/100.0\n",
+      "#calculation\n",
+      "\n",
+      "Pi = Vcc*I\n",
+      "Pc=Pi\n",
+      "Pm = Pi/2.0\n",
+      "Pout = (efficiency*Pi)/100.0\n",
+      "Ps = Pc*((m**2)/4.0)\n",
+      "maximum_swing = 2*Vcc \n",
+      "#Result\n",
+      "\n",
+      "print \"(a) RF input power to the output stage is \",Pi,\"W\"\n",
+      "print \"(b) AF power required for 100 percent modulation is \",Pm,\"W\"\n",
+      "print \"(c) Carrier output power is \",Pout,\"W\"\n",
+      "print \"(d) Power in one sideband for 67 percent modulation is \",round(Ps,2),\"W\"\n",
+      "print \"(e) maximum and minimum dc supply voltage swing with 100 percent modulation is \",maximum_swing,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RF input power to the output stage is  168.0 W\n",
+        "(b) AF power required for 100 percent modulation is  84.0 W\n",
+        "(c) Carrier output power is  117.6 W\n",
+        "(d) Power in one sideband for 67 percent modulation is  18.85 W\n",
+        "(e) maximum and minimum dc supply voltage swing with 100 percent modulation is  96 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page no 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "fc =4.2*10**6\n",
+      "f1 = 300\n",
+      "f2 = 3400\n",
+      "#calculation\n",
+      "#upper sideband\n",
+      "\n",
+      "x0=fc+f1\n",
+      "y0=fc+f2\n",
+      "#lower sideband\n",
+      "\n",
+      "x1=fc-f1\n",
+      "y1=fc-f2\n",
+      "frequency=(x1+y1)/2.0\n",
+      "#Result \n",
+      "\n",
+      "print \"(a) Upper sideband ranges from \",x0,\"to\",y0\n",
+      "print \"    lower sideband ranges from \",x1,\"to\",y1\n",
+      "print \"(b) approximate center frequency of a bandpass filter is \",frequency,\"Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Upper sideband ranges from  4200300.0 to 4203400.0\n",
+        "    lower sideband ranges from  4199700.0 to 4196600.0\n",
+        "(b) approximate center frequency of a bandpass filter is  4198150.0 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter5_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter5_1.ipynb
new file mode 100755
index 00000000..c8ec456c
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter5_1.ipynb
@@ -0,0 +1,253 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b01da4319cdb7c7c68bb5f48bf424b751928ca20873effe336ae7e4a6e7ae82d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 Fundamentals of frequency modulation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page no 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 915*10**6\n",
+      "fm_deviation =12.5*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "max_deviation = f + fm_deviation\n",
+      "min_deviation = f - fm_deviation\n",
+      "\n",
+      "#Result\n",
+      "print\"Maximum frequency occur during modulation is \",max_deviation/1000.0,\"KHz\"\n",
+      "print\"Minimum frequency occur during modulation is \",min_deviation/1000.0,\"KHz\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum frequency occur during modulation is  915012.5 KHz\n",
+        "Minimum frequency occur during modulation is  914987.5 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page no 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "max_deviation = 25*10**3\n",
+      "fm =15.0*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "mf =max_deviation/fm\n",
+      "\n",
+      "#Result\n",
+      "print\"The deviation ratio of the TV sound is \",round(mf,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The deviation ratio of the TV sound is  1.667\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page no 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "mf = 2.2\n",
+      "fd = 7.48*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "fm = fd/mf\n",
+      "\n",
+      "#Result\n",
+      "print\"The maximum modulating frequency is \",fm/1000.0,\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum modulating frequency is  3.4 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page no 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "J0 = -0.4\n",
+      "J1 = -0.07\n",
+      "J2 = 0.36\n",
+      "J3 = 0.43\n",
+      "J4 = 0.28\n",
+      "\n",
+      "#Result\n",
+      "print\"The amplitude of the carrier is \",J0\n",
+      "print\"Amplitudes of the first four sidebands are \",\" \\n \", J1,\"\\n \",J2,\"\\n \",J3,\"\\n \",J4\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amplitude of the carrier is  -0.4\n",
+        "Amplitudes of the first four sidebands are   \n",
+        "  -0.07 \n",
+        "  0.36 \n",
+        "  0.43 \n",
+        "  0.28\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page no 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "fd = 30*10**3\n",
+      "fm = 5*10**3\n",
+      "N=9\n",
+      "\n",
+      "#Calculation\n",
+      "bw1 = 2*fm*N\n",
+      "bw2 = 2*(fd+fm)\n",
+      "\n",
+      "#Result\n",
+      "print\"The maximum bandwidth of the fm signal is \",bw1/10**3,\"KHz\"\n",
+      "print\"Bandwidth using carson's rule \",bw2/10**3,\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum bandwidth of the fm signal is  90 KHz\n",
+        "Bandwidth using carson's rule  70 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 Page no 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "S_N = 2.8\n",
+      "fm = 1.5*10**3\n",
+      "fd =4*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "fi= math.asin(1/S_N)\n",
+      "delta = fi*fm\n",
+      "SN =fd/delta\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The frequency deviation caused by the noise \",round(delta,1),\"Hz\"\n",
+      "print\"(b) The improved output signal to noise ratio is \",round(SN,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The frequency deviation caused by the noise  547.8 Hz\n",
+        "(b) The improved output signal to noise ratio is  7.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter6_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter6_1.ipynb
new file mode 100755
index 00000000..582984c5
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter6_1.ipynb
@@ -0,0 +1,148 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8aef1c332f2a49ead17fadd51ccfa9b15c7423655061dea702b72eda65234123"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 FM Circuits"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page no 178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vc =40*10**-12\n",
+      "c = 20*10**-12\n",
+      "f0 = 5.5*10**6\n",
+      "Ct = Vc+c\n",
+      "\n",
+      "#Calculation\n",
+      "L = 1/((6.28*f0)**2*Ct)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of the inductance is \",round(L*10**6,0),\"microhenry\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of the inductance is  14.0 microhenry\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page no 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f=168.96*10**6\n",
+      "multiplier=24.0\n",
+      "deviation = 5*10**3\n",
+      "fm = 2.8*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "f0 =f/multiplier\n",
+      "fd= deviation/multiplier\n",
+      "phaseshift = fd/fm\n",
+      "phaseshift_degrees = phaseshift*57.3\n",
+      "total_phaseshift =2*phaseshift_degrees\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The crystal oscillator frequency is \",f0/10**6,\"MHz\"\n",
+      "print\"(b) The total phase shift is \",round(total_phaseshift,3),\"degree\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The crystal oscillator frequency is  7.04 MHz\n",
+        "(b) The total phase shift is  8.527 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page no 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R =1*10**3\n",
+      "phaseshift =4.263\n",
+      "phaseshift_center= 45\n",
+      "f =7.04*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "phase_l = phaseshift_center - phaseshift\n",
+      "phase_u = phaseshift_center + phaseshift\n",
+      "phaserange_total = phase_u - phase_l\n",
+      "Xc1 = 1161\n",
+      "C1 = 1/(6.28*f*Xc1)\n",
+      "Xc2 = 861\n",
+      "C2 = 1/(6.28*f*Xc2)\n",
+      "\n",
+      "#Result\n",
+      "print\"The two values of the capacitance to achieve total deviation are \",round(C1*10**12,1),\"pf to\",round(C2*10**12,1),\"pf\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The two values of the capacitance to achieve total deviation are  19.5 pf to 26.3 pf\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter7_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter7_1.ipynb
new file mode 100755
index 00000000..76d26d6f
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter7_1.ipynb
@@ -0,0 +1,221 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d48f2ded6acd9d37d6695ba84fcf6540c7d0d4380b8041338a946e9bf63a999c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 Digital communication techniques"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page no 210"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "t = 71.4*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "f = 1/t\n",
+      "fourth_harmonic = f*4\n",
+      "min_sampling = 2*fourth_harmonic\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The frequency of the signal is \",round(f/10**3,1),\"KHz\"\n",
+      "print\"(b) The fourth harmonic is \",round(fourth_harmonic/10**3,0),\"KHz\"\n",
+      "print\"(c) Minimum sampling rate is \",round(min_sampling/10**3,0),\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The frequency of the signal is  14.0 KHz\n",
+        "(b) The fourth harmonic is  56.0 KHz\n",
+        "(c) Minimum sampling rate is  112.0 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page no 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "N = 14\n",
+      "discrete_levels = 2.0**N\n",
+      "\n",
+      "#Calculation\n",
+      "num_vltg_inc =2**N-1\n",
+      "resolution = 12/discrete_levels \n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The numbedr of discrete levels that are represented using N number of bits are \",discrete_levels\n",
+      "print\"(b) the number odf voltage increments required to divide the voltage range are \",num_vltg_inc\n",
+      "print\"(c) Resolution of the digitization \",round(resolution*10**6,2),\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The numbedr of discrete levels that are represented using N number of bits are  16384.0\n",
+        "(b) the number odf voltage increments required to divide the voltage range are  16383\n",
+        "(c) Resolution of the digitization  732.42 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 Page no 225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "N =12\n",
+      "SINAD1=78\n",
+      "\n",
+      "#Calculation\n",
+      "SINAD2 = 6.02*N + 1.76\n",
+      "ENOB =(SINAD1 -1.76)/6.02\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The SINAD for 12 bit convertre is \",SINAD2,\"db\"\n",
+      "print\"(b) The ENOB for the converter with SINAD of 78 dB is \",round(ENOB,2),\"bits\"   \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The SINAD for 12 bit convertre is  74.0 db\n",
+        "(b) The ENOB for the converter with SINAD of 78 dB is  12.66 bits\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page no 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Gievn\n",
+      "Vm = 1.0\n",
+      "Vin = 0.25\n",
+      "mu =255\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Vout = (Vm*log(1+mu*(Vin/Vm)))/log(1+mu)\n",
+      "gain =Vout/Vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The output voltage of the compander \",round(Vout,2),\"volt\"\n",
+      "print\"Gain of the compander is \",round(gain,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage of the compander  0.75 volt\n",
+        "Gain of the compander is  3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 Page no 234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vin = 0.8\n",
+      "Vm =1.0\n",
+      "mu =255\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Vout = (Vm*math.log(1+mu*(Vin/Vm)))/math.log(1+mu)\n",
+      "gain =Vout/Vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The output voltage of the compander \",round(Vout,1),\"volt\"\n",
+      "print\"Gain of the compander is \",round(gain,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage of the compander  1.0 volt\n",
+        "Gain of the compander is  1.2\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter8_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter8_1.ipynb
new file mode 100755
index 00000000..0c719a4a
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter8_1.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:332fc501ed656c58460b6801381c21fa2c1eceb5892236956abe916f97a807b7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 Radio transmitters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page no 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 16*10**6\n",
+      "ppm = 200\n",
+      "frequency_variation = 200 *16 \n",
+      "\n",
+      "#Calculation\n",
+      "min_f = f - frequency_variation\n",
+      "max_f = f + frequency_variation\n",
+      "\n",
+      "#Reslt\n",
+      "print\"The minimum and maximum frequencies for the crystal of 16 Mhz with stability of 200 are \",min_f,\"Hz and\",max_f,\"Hz respectively\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum and maximum frequencies for the crystal of 16 Mhz with stability of 200 are  15996800 Hz and 16003200 Hz respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page no 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f =14.9*10**6\n",
+      "mul_factor = 2*3*3\n",
+      "stability_ppm =300\n",
+      "variation = 0.0003\n",
+      "\n",
+      "#Calculation\n",
+      "total_variation = variation* mul_factor\n",
+      "fout = f * mul_factor\n",
+      "frequency_variation = fout*total_variation\n",
+      "f_lower = fout - frequency_variation\n",
+      "f_upper = fout + frequency_variation\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The output frequency of the transmitter is \",fout/10**6,\"MHz\"\n",
+      "print\"(b) The maximum and minimum frequencies of the transmitter are \",round(f_lower/10**6,2),\"Mhz and \",round(f_upper/10**6,2),\"Mhz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The output frequency of the transmitter is  268.2 MHz\n",
+        "(b) The maximum and minimum frequencies of the transmitter are  266.75 Mhz and  269.65 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 Page no 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "f = 10*10**6\n",
+      "div_factor = 100.0\n",
+      "A =63\n",
+      "N = 285\n",
+      "M=32\n",
+      "\n",
+      "#Calculation\n",
+      "ref = f/div_factor\n",
+      "R =M*N+A\n",
+      "fout= R*ref\n",
+      "\n",
+      "#Result\n",
+      "print\"The output frequency of the synthesizer is \",fout/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output frequency of the synthesizer is  918.3 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page no 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 10*10**6\n",
+      "div_factor = 100.0\n",
+      "A =64\n",
+      "N = 285\n",
+      "M=32\n",
+      "\n",
+      "#Calculation\n",
+      "ref = f/div_factor\n",
+      "R =M*N+A\n",
+      "fout= R*ref\n",
+      "\n",
+      "#Result\n",
+      "print\"The output frequency of the synthesizer is \",fout/10**6,\"MHz\"\n",
+      "print\"The step change is \",fout/10**6-918.3,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output frequency of the synthesizer is  918.4 MHz\n",
+        "The step change is  0.1 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter9_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter9_1.ipynb
new file mode 100755
index 00000000..d881af0f
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter9_1.ipynb
@@ -0,0 +1,274 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:6155c127590921d5e033df68036b77ef47addc52e34e8604566f4914d9b739cf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 Communication receivers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Page no 318"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fl =220*10**6\n",
+      "fm =224*10**6\n",
+      "IF1 = 10.7*10**6\n",
+      "IF = 1.5*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "IF2 =IF1+IF\n",
+      "tune_l =fl+IF1\n",
+      "tune_m = fm+IF1\n",
+      "IF1_imgl = tune_l+IF1\n",
+      "IF2_imgm = tune_m+IF1\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The local oscillatior tuning range is \",tune_l/10**6,\"to \",tune_m/10**6,\"Mhz\"\n",
+      "print\"(b) Frequency of the second local oscillator is \",IF2/10**6,\"MHz\"\n",
+      "print\"(c) First IF image range is \",IF1_imgl/10**6,\"MHz to \",IF2_imgm/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The local oscillatior tuning range is  230.7 to  234.7 Mhz\n",
+        "Frequency of the second local oscillator is  12.2 MHz\n",
+        "First IF image range is  241.4 MHz to  245.4 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 Page no 324"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R = 100*10**3\n",
+      "T = 273+25\n",
+      "B = 20*10**3\n",
+      "k = 1.38*10**-23\n",
+      "\n",
+      "#Calculation\n",
+      "Vn=(4*k*T*B*R)**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The noise voltage across 100k resistor is \",round(Vn*10**6,2),\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The noise voltage across 100k resistor is  5.74 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page no 324"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R=75\n",
+      "B=6*10**6  \n",
+      "T = 29+273\n",
+      "k =1.38*10**-23\n",
+      "\n",
+      "#calculation\n",
+      "Vn = (4*k*T*B*R)**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The input themal noise is \",round(Vn*10**6,2),\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input themal noise is  2.74 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page no 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Tc=32.2\n",
+      "Tk=273+Tc\n",
+      "B =30*10**3\n",
+      "k =1.38*10**-23\n",
+      "\n",
+      "#Calculation\n",
+      "Pn=k*Tk*B\n",
+      "\n",
+      "#Result\n",
+      "print\"The average noise power is \",round(Pn*10**16,2),\"10**-16 W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average noise power is  1.26 10**-16 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page no 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "SN_ip = 8\n",
+      "SN_op = 6.0\n",
+      "\n",
+      "#Calculation\n",
+      "NR = SN_ip/SN_op\n",
+      "NF = 10*log10(NR)\n",
+      "\n",
+      "#Result\n",
+      "print\"The noise factor is \",round(NR,2)\n",
+      "print\"The noise figure is \",round(NF,2),\"dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The noise factor is  1.33\n",
+        "The noise figure is  1.25 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 Page no 330"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R= 75.0\n",
+      "T=31+273\n",
+      "k=1.38*10**-23\n",
+      "B=6*10**6\n",
+      "Vs = 8.3*10**-6\n",
+      "NF=2.8\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Vn = math.sqrt(4*k*T*B*R)\n",
+      "Pn = Vn**2/R\n",
+      "Ps = Vs**2/R\n",
+      "SN = (Ps*10**12)/(Pn/10.0**12)\n",
+      "SN_dB =10*log10(SN)\n",
+      "NR = 10**0.28\n",
+      "SN_op = SN/NR\n",
+      "Tn = 290*(NR-1)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The input noise power is \",round(Pn*10**12,1),\"pW\"\n",
+      "print\"(b) The input signal power is \",round(Ps*10**12,3),\"pW\"\n",
+      "print\"(c) Signal to noise ratio in decibels \",round(SN/10**24,2)\n",
+      "print\"(d) The noise factor is \",round(NR,2)\n",
+      "print\"(e) Signal to noise ratio of the amplifier is \",round(SN_op/10**24,2)\n",
+      "print\"(f) The noise temperature of the amplifier \",round(Tn,0),\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The input noise power is  0.1 pW\n",
+        "(b) The input signal power is  0.919 pW\n",
+        "(c) Signal to noise ratio in decibels  9.12\n",
+        "(d) The noise factor is  1.91\n",
+        "(e) Signal to noise ratio of the amplifier is  4.79\n",
+        "(f) The noise temperature of the amplifier  263.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter_2_1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter_2_1.ipynb
new file mode 100755
index 00000000..0c5682f6
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/chapter_2_1.ipynb
@@ -0,0 +1,1143 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0b7383bde99cd4f0125cd654ff30432e941ee3972919bf09fe35e260edf9b001"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 The fundamentals of Electronics: A Review"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page no 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "vout=750*10**-3\n",
+      "vin = 30*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "gain=vout/vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The Voltage gain of the amplifier is \",gain\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Voltage gain of the amplifier is  25000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 Page no 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "pout=6\n",
+      "power_gain=80.0\n",
+      "\n",
+      "#calculation\n",
+      "pin=pout/power_gain\n",
+      "\n",
+      "#Result\n",
+      "print\"The input power of the signal is \",pin*1000,\"mW\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input power of the sigmal is  75.0 mW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page no 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "A1=5;\n",
+      "A2=2;\n",
+      "A3=17;\n",
+      "total_gain=A1*A2*A3;\n",
+      "pin= 40*10**-3;\n",
+      "\n",
+      "#calculation\n",
+      "pout=total_gain*pin;\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power is\",pout,\"watts\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power is 6.8 watts\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page no 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "pin=25.0*10**-6;\n",
+      "pout=1.5*10**-3;\n",
+      "A1=3.0;\n",
+      "\n",
+      "#Calculation\n",
+      "total_gain=pout/pin;\n",
+      "print\"Total gain is\",total_gain\n",
+      "A2=total_gain/A1\n",
+      "\n",
+      "#Result\n",
+      "print\"The gain of second stage is \",A2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total gain is 60.0\n",
+        "The gain of second stage is  20.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page no 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "R1=10.0*10**3;\n",
+      "R2=470.0;\n",
+      "\n",
+      "#Calculation\n",
+      "attenuation=R2/(R2+R1)\n",
+      "A2=1/attenuation\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The attenuation (A1) is\",round(attenuation,3)\n",
+      "print\"(b) The attenuation (A2) is\",round(A2,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The attenuation (A1) is 0.045\n",
+        "(b) The attenuation (A2) is 22.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page no 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vin=20.0*10**-6;\n",
+      "Vout=100*10**-3;\n",
+      "A1=45000.0                                #A1 isAmplifier gain\n",
+      "\n",
+      "#calculation\n",
+      "AT=Vout/Vin                                #AT is Total gain\n",
+      "A2=AT/A1                                   #A2 is attenuation factor\n",
+      "\n",
+      "#Result\n",
+      "print\"Total gain is\",AT\n",
+      "print\"The atenuation factor needed to to keep the output voltage from exceeding 100 mv is \",round(A2,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total gain is 5000.0\n",
+        "The atenuation factor needed to to keep the output voltage from exceeding 100 mv is  0.1111\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page no 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vin=3.0*10**-3\n",
+      "Vout=5\n",
+      "Pin=50.0*10**-3                                     \n",
+      "Pout=2*10**-3                                     \n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "gain_dB= 20*log10 (Vout/Vin)\n",
+      "gain_db=10*log10 (Pout/Pin)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The gain of amplifier in dB is \",round(gain_dB,1)\n",
+      "print\"(b) The gain in dB is \",round(gain_db,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The gain of amplifier in dB is  64.4\n",
+        "(b) The gain in dB is  -13.98\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page no 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "gain_dB = 40\n",
+      "pout_W= 100\n",
+      "\n",
+      "#calculation\n",
+      "pin_W = pout_W/10.0**4\n",
+      "\n",
+      "#Result\n",
+      "print\"The input power is \",pin_W,\"Watt\"\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input power is  0.01 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page no 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "gain_db = 60\n",
+      "vin = 50*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "vout = 10**(60/20.0)*vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The output voltage is \",vout,\"volt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage is  0.05 volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10 Page no 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "vin=90*10**-3\n",
+      "R1= 10.0*10**3\n",
+      "vout=7.8\n",
+      "Rout=8.0\n",
+      "\n",
+      "#calculation\n",
+      "pin= vin**2/R1\n",
+      "pout=vout**2/Rout\n",
+      "\n",
+      "import math\n",
+      "Ap_db = 10*log10 (pout/pin)\n",
+      "\n",
+      "#Result\n",
+      "print\"The power gain in decibel is \",round(Ap_db,1),\"dB\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power gain in decibel is  69.7 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11 Page no 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "gain_db = 28\n",
+      "pin = 36*10**-3\n",
+      "\n",
+      "#calculation\n",
+      "pout = 10**2.8*pin;\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power is \",round(pout,2),\"watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power is  22.71 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12 Page no 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "gain1 = 6.8 \n",
+      "gain2 = 14.3\n",
+      "attenuation1 = -16.4\n",
+      "attenuation2 = -2.9\n",
+      "vout = 800*10**-3\n",
+      "\n",
+      "#calculation\n",
+      "At = gain1+gain2+attenuation1+attenuation2\n",
+      "vin = vout/10.0**(At/20.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"The input voltage is \",round(vin*10**3,1),\"mv\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input voltage is  650.3 mv\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13 Page no 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "pout_db =12.3\n",
+      "\n",
+      "#calculation\n",
+      "pout_mW = 0.001*10**(12.3/10.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power is \" ,round(pout_mW*10**3,0),\"mv\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power is  17.0 mv\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14 Page no 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c = 2.7*10**-12\n",
+      "l = 33*10**-9\n",
+      "\n",
+      "#calculation\n",
+      "fr= 1/(6.28*(l*c)**0.5)\n",
+      "\n",
+      "#Result\n",
+      "print\"The resonat frequency is \" ,round(fr/10.0**6,0),\"Mhz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resonat frequency is  533.0 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15 Page no 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c =12*10**-12\n",
+      "fr = 49*10**6\n",
+      "\n",
+      "#calculation\n",
+      "l=1/(4*3.14**2*fr**2*c)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of inductance is \" ,round(l*10**9,0),\"nh\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of inductance is  880.0 nh\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16 Page no 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fr=28*10**6\n",
+      "Q=70.0\n",
+      "\n",
+      "#calculation\n",
+      "bandwidth = fr/Q\n",
+      "\n",
+      "#Result\n",
+      "print\"The bandwidth is \",bandwidth/10.0**3,\"Khz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The bandwidth is  400.0 Khz\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17 Page no 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "f1= 7.93*10**6\n",
+      "f2= 8.07*10**6\n",
+      "\n",
+      "#calculation\n",
+      "bw= f2-f1\n",
+      "fr=(f1*f2)**0.5\n",
+      "Q= fr/bw\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The bandwidth is \",bw/10.0**3,\"Khz\"\n",
+      "print\"(b) The resonant frequency is \",round(fr/10.0**6,0),\"Mhz\"\n",
+      "print\"(c) The Q of resonant circuit is \",round(Q,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The bandwidth is  140.0 Khz\n",
+        "(b) The resonant frequency is  8.0 Mhz\n",
+        "(c) The Q of resonant circuit is  57.14\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18 Page no 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Q=200.0\n",
+      "fr=16*10**6\n",
+      "\n",
+      "#calculation\n",
+      "bw=fr/Q\n",
+      "f1= fr-(bw/2)\n",
+      "f2=fr+(bw/2)\n",
+      "\n",
+      "#Result\n",
+      "print\"Bandwidth is \",bw*10**-3,\"KHz\"\n",
+      "print\"f1= \",f1*10**-6,\"MHz\"\n",
+      "print\"f2= \",f2*10**-6,\"MHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bandwidth is  80.0 KHz\n",
+        "f1=  15.96 MHz\n",
+        "f2=  16.04 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19 Page no 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Q= 150\n",
+      "Vs=3*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "Vc= Q*Vs\n",
+      "\n",
+      "#Result\n",
+      "print\"The voltage across capacitor is \",Vc*10**6,\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across capacitor is  450.0 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.20 Page no 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fr= 52*10**6\n",
+      "Q=12.0\n",
+      "L=0.15*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "Rw=(6.28*fr*L)/Q\n",
+      "Req= Rw*(Q**2+1)\n",
+      "\n",
+      "#Result\n",
+      "print\"Impedance of the parellel LC circuit is \",round(Req,0),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Impedance of the parellel LC circuit is  592.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.21 Page no 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fr= 52.0*10**6\n",
+      "Rw= 4.1\n",
+      "L =0.15*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "C=1/(4.0*3.14**2*fr**2*L)\n",
+      "Z = L/(C*Rw)\n",
+      "\n",
+      "#Result\n",
+      "print\"the impedance of the circuit is \",round(Z,0),\"ohm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the impedance of the circuit is  585.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.22 Page no 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw = 1.0*10**6\n",
+      "XL = 300\n",
+      "Rw = 10.0\n",
+      "fr =10*10**6\n",
+      "\n",
+      "#calculation\n",
+      "Q1 = XL/Rw\n",
+      "Rp = Rw*(Q1**2+1) \n",
+      "\n",
+      "Q2 = fr/bw\n",
+      "Rpnew = Q2*XL\n",
+      "\n",
+      "Rext = (Rpnew*Rp)/(Rp-Rpnew)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of resistor needed to set the bandwidth of the parellel tuned circuit is \",round(Rext,1),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of resistor needed to set the bandwidth of the parellel tuned circuit is  4497.5 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.23 Page no 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "R = 8.2*10**3\n",
+      "C =0.0033*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "fco = 1/(6.28* R*C)\n",
+      "\n",
+      "#Result\n",
+      "print\"The cut off frequency is \",round(fco/10.0**3,2),\"Khz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cut off frequency is  5.88 Khz\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.24 Page no 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fco =3.4*10**3\n",
+      "C = 0.047*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "R = 1/(6.28* fco* C)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of the resistor is \",round(R,0),\"ohm\"\n",
+      "print\"The closest standard value is \", 1000 ,\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of the resistor is  996.0 ohm\n",
+        "The closest standard value is  1000 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.25 Page no 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fnotch = 120\n",
+      "R = 220*10**3\n",
+      "\n",
+      "#calculatiuon\n",
+      "C = 1/(6.28*R*fnotch)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of capacitance required is \",round(2*C*10**6,3),\"microfarad\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of capacitance required is  0.012 microfarad\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.26 Page no 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vpeak =3.0\n",
+      "f=48*10**3\n",
+      "\n",
+      "#calculation\n",
+      "fifth_harmonic = 5*f\n",
+      "Vrms=(4/3.14)*(3/5.0)*0.707\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The frequency of the fifth harmonic is \",fifth_harmonic/10.0**3,\"Khz\"\n",
+      "print\"The RMS voltage of the fifth harmonic is \",round(Vrms,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The frequency of the fifth harmonic is  240.0 Khz\n",
+        "The RMS voltage of the fifth harmonic is  0.54\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.27 Page no 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vpeak = 5\n",
+      "f = 4.0*10**6\n",
+      "duty_cycle=0.3\n",
+      "\n",
+      "#calculation\n",
+      "T = 1/f\n",
+      "t0 = duty_cycle*T\n",
+      "Vavg = Vpeak*duty_cycle\n",
+      "min_bw =1/t0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The average DC value is \",Vavg,\"volt\"\n",
+      "print\"(b) The minimum bandwidth required is \" ,round(min_bw/10.0**6,3),\"Mhz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The average DC value is  1.5 volt\n",
+        "(b) The minimum bandwidth required is  13.333 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.28 Page no 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "tr =6*10**-9\n",
+      "\n",
+      "#calculation\n",
+      "min_bw=(35/0.006)\n",
+      "\n",
+      "#Result\n",
+      "print\"The minimum bandwidth is % is \",round(min_bw/10.0**2,1),\"Mhz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum bandwidth is % is  58.3 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.29 Page no 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw= 200.0*10**3\n",
+      "\n",
+      "#calculation\n",
+      "tr= 0.35/(bw*10**-3)\n",
+      "\n",
+      "#Result\n",
+      "print\"The fastest rise time of the circuit is \" ,tr*10**3,\"microseconds\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The fastest rise time of the circuit is  1.75 microseconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.30 Page no 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw_mhz = 60\n",
+      "tri_ns= 15\n",
+      "\n",
+      "#calculation\n",
+      "tra_osci = 0.35/(bw_mhz)\n",
+      "tra_comp = 1.1*(tri_ns**2 + (tra_osci*10**3)**2)**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The rise time of the displayed square wave is \",round(tra_comp,1),\"ns\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rise time of the displayed square wave is  17.7 ns\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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Binary files /dev/null and b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/screenshots/image_3.png differ
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter1.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter1.ipynb
new file mode 100755
index 00000000..48b042da
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter1.ipynb
@@ -0,0 +1,249 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:fa4937e894b2eaacf3d917ad54d3257fed9024ea716662722fd829aea08934e3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 Introduction to electronic communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1.a Page no 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "c=300000000\n",
+      "f1=150000000.0\n",
+      "f2=430000000.0\n",
+      "f3=8000000.0\n",
+      "f4=750000.0\n",
+      "\n",
+      "#Calculation\n",
+      "W1 = c/f1\n",
+      "W2=c/f2\n",
+      "W3=c/f3\n",
+      "W4=c/f4\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Wavelength is \",W1,\"meter\"\n",
+      "print\"(b) Wavelength is \",round(W2,3),\"meter\"\n",
+      "print\"(c) Wavelength is \",W3,\"meter\"\n",
+      "print\"(d) Wavelength is \",W4,\"meter\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Wavelength is  2.0 meter\n",
+        "(b) Wavelength is  0.698 meter\n",
+        "(c) Wavelength is  37.5 meter\n",
+        "(d) Wavelength is  400.0 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 page no 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "c=300000000\n",
+      "wavelength=1.5\n",
+      "\n",
+      "#Calculation\n",
+      "frequency=c/wavelength\n",
+      "\n",
+      "#Result\n",
+      "print\"Signal frequncy is \",frequency/1000000.0,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signal frequncy is  200.0 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 page no 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "wavelength_feet=75\n",
+      "wavelength_meter= 75/3.28\n",
+      "c=300000000\n",
+      "\n",
+      "#Calculation\n",
+      "frequency=c/wavelength_meter\n",
+      "\n",
+      "#Result\n",
+      "print\"The signal frequncy is \",frequency/1000000.0,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The signal frequncy is  13.12 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 page no 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "wavelength_inches=8\n",
+      "wavelength_meter=  8/39.37\n",
+      "c=300\n",
+      "\n",
+      "#Calculation\n",
+      "frequency= c/wavelength_meter\n",
+      "\n",
+      "#Result\n",
+      "print\"The signal freuency is \",round(frequency,1),\"MHz\"\n",
+      "print\"The signnalfrequency is \",round(frequency*10**-3,2),\"GHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The signal freuency is  1476.4 MHz\n",
+        "The signnalfrequency is  1.48 GHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 page no 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f1=902000000\n",
+      "f2=928000000\n",
+      "\n",
+      "#Calculation\n",
+      "bandwidth=f2-f1\n",
+      "\n",
+      "#Result\n",
+      "print\"Width of the band is \",bandwidth/1000000,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Width of the band is  26 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 page no 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "bandwidth_megahertz= 6\n",
+      "f1_megahertz=54\n",
+      "\n",
+      "#Calculation\n",
+      "f2_megahertz=f1_megahertz + bandwidth_megahertz\n",
+      "\n",
+      "#Result\n",
+      "print\"Upper frequency limit is \",f2_megahertz,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Upper frequency limit is  60 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter10.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter10.ipynb
new file mode 100755
index 00000000..f969c01b
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter10.ipynb
@@ -0,0 +1,101 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:6d0a5becdae7bf2928f5747f4da31ef68eb2576da776218335d8b44a4a733949"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 Multiplexing and Demultiplexing"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page no 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "BW_service = 860*10**6\n",
+      "BW_ch = 6.0*10**6\n",
+      "\n",
+      "#calculation\n",
+      "total_ch = BW_service/BW_ch\n",
+      "\n",
+      "#result\n",
+      "print\"Total number of channels are \",round(total_ch,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total number of channels are  143.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page no 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "channels =16\n",
+      "sampling_rate= 3.5*10**3\n",
+      "w_len=6\n",
+      "\n",
+      "#Calculation\n",
+      "available_ch =channels-1\n",
+      "bpf =channels*w_len\n",
+      "data_rate = sampling_rate * bpf\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Available channels are \",available_ch\n",
+      "print\"(b) Bits Per Frame = \",bpf\n",
+      "print\"(c) The serial data rate \",data_rate/10**3,\"KHz\"\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Available channels are  15\n",
+        "(b) Bits Per Frame =  96\n",
+        "(c) The serial data rate  336.0 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter11.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter11.ipynb
new file mode 100755
index 00000000..015bcc79
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter11.ipynb
@@ -0,0 +1,143 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8bd5af3b2acf26b36843842ce0fbb28501e7f5eae117a0af6c810fe6d60c2891"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 The Transmission of binary data in communication systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 Page no 392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "t=0.0016\n",
+      "No_words=256.0\n",
+      "bits_word = 12.0\n",
+      "\n",
+      "#Calculation\n",
+      "tword= t/No_words\n",
+      "tbit = tword/bits_word\n",
+      "bps =1/tbit\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The time duration of the word \",tword*10**8,\"microsecond\"\n",
+      "print\"(b) The time duration of the one bit is \",round(tbit*10**8,4),\"microseconds\"\n",
+      "print\"(c) The speed of transmission is \",bps/10**5,\"kbps\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The time duration of the word  625.0 microsecond\n",
+        "(b) The time duration of the one bit is  52.0833 microseconds\n",
+        "(c) The speed of transmission is  19.2 kbps\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 Page no 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "B=12.5*10**3\n",
+      "SN_dB= 25\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "C_th = 2*B\n",
+      "SN=316.2\n",
+      "C =B*3.32*log10(SN+1)\n",
+      "N= 2**(C/(2.0*B))\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The maximum theorotical data rate is \",C_th/10**3,\"kbps\"\n",
+      "print\"(b) The maximum theorotical capacity of channel is \",round(C/10**3,1),\"Kbps\"\n",
+      "print\"(c) The number of levels needed to acheive maximum speed are \",round(N,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The maximum theorotical data rate is  25.0 kbps\n",
+        "(b) The maximum theorotical capacity of channel is  103.8 Kbps\n",
+        "(c) The number of levels needed to acheive maximum speed are  17.78\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 Page no 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "block =512\n",
+      "packets =8\n",
+      "BER = 2*10**-4\n",
+      " \n",
+      "#Calculation\n",
+      "avg_errors = block*packets*8*BER\n",
+      "\n",
+      "print\"Average number of errors are \",avg_errors\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average number of errors are  6.5536\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter12.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter12.ipynb
new file mode 100755
index 00000000..ae98ce2e
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter12.ipynb
@@ -0,0 +1,101 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:70c87938501c80a4a95b89850ad3f0f2a57b63c9811bc118f804b4d657133c5f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 Introduction to Networking and local area networks"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Page no 448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "N = 20\n",
+      "\n",
+      "#Calculation\n",
+      "L = (N*(N-1))/2.0\n",
+      "\n",
+      "#Result\n",
+      "print\"The number of interconnecting wires required are \",L\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of interconnecting wires required are  190.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 Page no 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "block = 1500\n",
+      "ethernet = 10.0*10**6\n",
+      "token_ring = 16.0*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "t1_bit = 1/ethernet\n",
+      "t1_byte = 8*t1_bit\n",
+      "t1_1526 = 1526 *t1_byte\n",
+      "t2_bit = 1/token_ring\n",
+      "t2_byte = 8 * t2_bit\n",
+      "t2_1521 =1521*t2_byte\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Time required for 1526 bytes is \",t1_1526*10**6,\"ns\"\n",
+      "print\"(b) Time required for 1521 bytes is \",t2_1521*10**3,\"ms\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Time required for 1526 bytes is  1220.8 ns\n",
+        "(b) Time required for 1521 bytes is  0.7605 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter13.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter13.ipynb
new file mode 100755
index 00000000..de1fc475
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter13.ipynb
@@ -0,0 +1,339 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:4be2df010e0c0a454b43a2b1267f5b9f24594cfd973fe52088cb80f0e511082b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 Transmission lines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 Page no 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f= 450.0*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "lamda = 984/f\n",
+      "len =0.1*lamda\n",
+      "\n",
+      "#Result\n",
+      "print\"feet long conductors would be considered as the transmission line \",round(len*10**6,3),\"ft\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "feet long conductors would be considered as the transmission line  0.219 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 Page no 484"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "lamda = 2.19\n",
+      "\n",
+      "#Calculation\n",
+      "len = (3/8.0)*lamda\n",
+      "\n",
+      "#Result\n",
+      "print\"The pyhsical length of the transmission line \",round(len,2),\"feet\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pyhsical length of the transmission line  0.82 feet\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page no 492"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "len = 165\n",
+      "attn_100ft = 5.3\n",
+      "pin = 100\n",
+      "attn_ft = 5.3/100.0\n",
+      "\n",
+      "#Calculation\n",
+      "total_attn = attn_ft * len\n",
+      "pout = pin *0.1335\n",
+      "\n",
+      "#Result\n",
+      "print\"The total attenuation of the cable is \",total_attn,\"dB\"\n",
+      "print\"Output power is \",pout,\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total attenuation of the cable is  8.745 dB\n",
+        "Output power is  13.35 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4 Page no 494"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "len =150\n",
+      "C =13.5\n",
+      "Z0 =93\n",
+      "f =2.5*10**6\n",
+      "attn_100ft =2.8\n",
+      "\n",
+      "#Calculation\n",
+      "L =C*Z0**2\n",
+      "td =(L*C)**0.5\n",
+      "theta = ((360)*188.3)/(1/f)\n",
+      "attn_ft = attn_100ft/100.0\n",
+      "total_attn = attn_ft*150\n",
+      "\n",
+      "print\"(a) The load impedance required to terminate the the line to avoid the reflections is %d ohm\",Z0\n",
+      "print\"(b) The equivalent inductance per feet is \",L/10**3,\"nh\"\n",
+      "print\"(c) The time delay introduced by the cable per feet is \",td/10**3,\"ns\"\n",
+      "print\"(d) The phase shift occurs in degrees for the 2.5 Mhz sine wave is \",theta/10**9\n",
+      "print\"(e) The total attenuation is \",total_attn,\"db\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The load impedance required to terminate the the line to avoid the reflections is %d ohm 93\n",
+        "(b) The equivalent inductance per feet is  116.7615 nh\n",
+        "(c) The time delay introduced by the cable per feet is  1.2555 ns\n",
+        "(d) The phase shift occurs in degrees for the 2.5 Mhz sine wave is  169.47\n",
+        "(e) The total attenuation is  4.2 db\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5 Page no 501"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "vmax= 52.0\n",
+      "vmin= 17.0\n",
+      "Z0 = 75\n",
+      "\n",
+      "#calculation\n",
+      "SWR = vmax/vmin\n",
+      "ref_coeff = (vmax-vmin)/(vmax+vmin)\n",
+      "Zl1 = Z0*SWR\n",
+      "Zl2 = Z0/SWR\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The standing wave ratio is \",round(SWR,2)\n",
+      "print\"(b) Reflection coefficient is \",round(ref_coeff,2)\n",
+      "print\"The value of resistive load is \",round(Zl1,2),\"ohm or\",round(Zl2,2),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The standing wave ratio is  3.06\n",
+        "(b) Reflection coefficient is  0.51\n",
+        "The value of resistive load is  229.41 ohm or 24.52 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.6 Page no 503"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "SWR =3.05\n",
+      "ref_pwr =0.2562\n",
+      "pin =30\n",
+      "\n",
+      "#calculation\n",
+      "pout = pin -(pin*((SWR-1)/(SWR+1))**2)\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power of the cable is \",round(pout,3),\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power of the cable is  22.314 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.7 Page no 508"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "C =4*10**-12\n",
+      "f =800*10**6\n",
+      "diele = 3.5\n",
+      "h = 0.0625\n",
+      "w = 0.13\n",
+      "t = 0.002\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Z0 = 38.8*math.log(0.374/0.106)\n",
+      "Xc = 1/(6.28*f*C)\n",
+      "\n",
+      "#Result\n",
+      "print\"The charecteristics impedance of the transmission line is \",round(Z0,1),\"ohm\"\n",
+      "print\"The reactance of the capacitor is \",round(Xc,2),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The charecteristics impedance of the transmission line is  48.9 ohm\n",
+        "The reactance of the capacitor is  49.76 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.8 Page no 508"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "lamda = (984/800.0)\n",
+      "lamda_8 =lamda/8.0\n",
+      "\n",
+      "#Calculation\n",
+      "len = lamda_8*12*(1/3.6**0.5)\n",
+      "\n",
+      "#Result\n",
+      "print\"The length of the transmission line is \",round(len,3),\"inch\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of the transmission line is  0.972 inch\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter14.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter14.ipynb
new file mode 100755
index 00000000..483c5c1b
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter14.ipynb
@@ -0,0 +1,229 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e45779b9fe24c5d57bbc44b21eced0e689b3f83e9f5f5eea042afe4effb61cee"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 Antennas and Wave Propogation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1 page no 544"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f=310.0*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "len1 =(492*0.97)/f\n",
+      "len2 =(492/f)*0.8\n",
+      "len3 =(984/f)*0.73\n",
+      "z1 =120*log(35/2.0)\n",
+      "len4 =234/f\n",
+      "z2  = 73/2.0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The length and radiation resistance of the dipole are \",round(len1*10**6,2),\"feet and 73 ohm respectively\"\n",
+      "print\"(b) The length of the folded dipole are \",round(len2*10**6,2),\"feet\"\n",
+      "print\"(c) The length and radiation resistance of the bow tie antenna are \",round(len3*10**6,1),\"feet and \",round(z1,1),\"ohm resp.\"\n",
+      "print\"(d) The length and radiation resistance of the groun plane antenna are \",round(len4*10**6,3),\"feet and\",z2,\"ohm resp.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The length and radiation resistance of the dipole are  1.54 feet and 73 ohm respectively\n",
+        "(b) The length of the folded dipole are  1.27 feet\n",
+        "(c) The length and radiation resistance of the bow tie antenna are  2.3 feet and  343.5 ohm resp.\n",
+        "(d) The length and radiation resistance of the groun plane antenna are  0.755 feet and 36.5 ohm resp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2 Page no 553"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "gain=14\n",
+      "len=250\n",
+      "attn_100=3.6\n",
+      "f =220*10\n",
+      "pin =50\n",
+      "p =0.126\n",
+      "\n",
+      "#Calculation\n",
+      "pout =pin*p\n",
+      "line_loss =pin-pout\n",
+      "pwr_ratio = 25.1\n",
+      "ERP = pwr_ratio*pout\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The transmission line loss is \",line_loss\n",
+      "print\"(b) Effective raduated power is \",round(ERP,1),\"W\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The transmission line loss is  43.7\n",
+        "(b) Effective raduated power is  158.1 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3 Page no 556"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Z0 =50\n",
+      "Zl =172\n",
+      "f =460.0*10**6\n",
+      "VF =0.86\n",
+      "\n",
+      "#Calculation\n",
+      "len =(246/f)*VF\n",
+      "\n",
+      "#Result\n",
+      "print\"The length of the impedance matching section needed for the Q section is \",round(len*10**6,2),\"feet\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of the impedance matching section needed for the Q section is  0.46 feet\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4 Page no 557"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 460.0*10**6\n",
+      "VF = 0.66\n",
+      "\n",
+      "#Calculation\n",
+      "len = (246/f)*VF\n",
+      "\n",
+      "#Result \n",
+      "print\"The length of impedance matching section is \",round(len*10**6,3),\"feet\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of impedance matching section is  0.353 feet\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.5 Page no 567"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "ht =275\n",
+      "hr =60\n",
+      "f=224.0*10**6\n",
+      "pt=100\n",
+      "Gt = 26\n",
+      "Gr = 3.27\n",
+      "\n",
+      "#Calculation\n",
+      "D =((2*ht)**0.5+(2*hr)**0.5)*1.61\n",
+      "lamda = 300/f\n",
+      "Pr = (pt*Gt*Gr*lamda**2)/(16*3.14**2*D**2)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The maximmum transmitting distance is \",round(D,1),\"kilometer\"\n",
+      "print\"(b) The received power is \",round(Pr*10**15,1),\"nW\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The maximmum transmitting distance is  55.4 kilometer\n",
+        "(b) The received power is  31.5 nW\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter16.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter16.ipynb
new file mode 100755
index 00000000..a3463547
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter16.ipynb
@@ -0,0 +1,216 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:526f0478ca7a5ec4522d70a6832114c547e4784f425e0391463d5037753dde9d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16 Microwave communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.1 Page no 616"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Zsrc =50\n",
+      "Zld  =136\n",
+      "f =5800.0*10**6\n",
+      "Er =2.4\n",
+      "\n",
+      "#Calculation\n",
+      "Zq =(Zsrc * Zld)**0.5\n",
+      "Vp =1/(Er)**0.5\n",
+      "lamda = 300/f\n",
+      "len = (lamda/4.0)*38.37*Vp\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The required impedance is \",round(Zq,2),\"ohm\"\n",
+      "print\"(b) The length of the microstrip \",round(len*10**6,2),\"inches\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The required impedance is  82.46 ohm\n",
+        "(b) The length of the microstrip  0.32 inches\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2 Page no 623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "w=0.65\n",
+      "h=0.38\n",
+      "\n",
+      "#Calculation\n",
+      "fco = 300/(2.0*((0.65*2.54)/100.0))\n",
+      "f =1.42*fco\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The cutoff frequency of the \",round(fco/10**3,3),\"GHz\"\n",
+      "print\"(b) Operating frequency of the wavwguide is \",round(f/10**3,1),\"GHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The cutoff frequency of the  9.085 GHz\n",
+        "(b) Operating frequency of the wavwguide is  12.9 GHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3 Page no 623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#for the above question\n",
+      "print\"The c band is approximately 4 to 6 Ghz since a waveguide acts as a high pass filter with cut off of 9.08 Ghz it will not pass c band signal\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The c band is approximately 4 to 6 Ghz since a waveguide acts as a high pass filter with cut off of 9.08 Ghz it will not pass c band signal\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.4 Page no 648"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "lamda1 =5.0\n",
+      "f2 = 15.0*10**9\n",
+      "D=1.524\n",
+      "\n",
+      "#Calculation\n",
+      "f1=984/lamda1\n",
+      "lamda2 =300/f2\n",
+      "G = (6*(D/lamda2)**2)\n",
+      "B = 70/(D/lamda2)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The lowest possible oprerating frequency is \",f1,\"MHz\"\n",
+      "print\"(b) The gain at 15 Ghz is \",G/10**12\n",
+      "print\"(c) The beam width at 15Ghz is \",round(B*10**6,2),\"degree\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The lowest possible oprerating frequency is  196.8 MHz\n",
+        "(b) The gain at 15 Ghz is  34838.64\n",
+        "(c) The beam width at 15Ghz is  0.92 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.5 Page no 661"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "T = 9.2\n",
+      "theta = 20\n",
+      "sin20 = 0.342\n",
+      "a=5280\n",
+      "\n",
+      "#Calculation\n",
+      "D_nautical = T/12.36\n",
+      "D_statute =D_nautical*0.87\n",
+      "A = D_statute*0.342\n",
+      "A1=A*a\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The line of distance to the aircraft in the statute miles \",round(D_statute,3)\n",
+      "print\"(b) The altitude of the aircraft is \",round(A,2),\"mi\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The line of distance to the aircraft in the statute miles  0.648\n",
+        "(b) The altitude of the aircraft is  0.22 mi\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter17.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter17.ipynb
new file mode 100755
index 00000000..c9e3dcac
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter17.ipynb
@@ -0,0 +1,142 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:36f7565fa260d3a1374317f250783789b0d4ba7834e883c19725c0b3c84c1d9a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapeter 17 Satellite communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.1 Page no 678"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "stn_long = 95\n",
+      "stn_lat = 30\n",
+      "sat_long =121\n",
+      "rad_pos = 137\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "azimuth = 360-rad_pos\n",
+      "\n",
+      "#Result\n",
+      "print\"The elevation setting for the antenna is 45 degree\"\n",
+      "print\"The azimuth setting for the antenna is \",azimuth,\"degree\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The elevation setting for the antenna is 45 degree\n",
+        "The azimuth setting for the antenna is  223 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.2 Page no 681"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "flo = 2*10**9\n",
+      "fd =3840*10**6\n",
+      "B =36*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "fu =fd+flo\n",
+      "C =2*B\n",
+      "\n",
+      "#Result\n",
+      "print\"The uplink frequency is \",fu/10.0**9,\"GHz\"\n",
+      "print\"The data rate is \", C/10**6,\"Mbps\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The uplink frequency is  5.84 GHz\n",
+        "The data rate is  72 Mbps\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.3 Page no 691"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "fs = 4.08*10**9\n",
+      "fIF1 = 770*10**6\n",
+      "fIF2 = 140*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "flo1 = fs - fIF1\n",
+      "flo2 = fIF1 - fIF2\n",
+      "\n",
+      "#Result\n",
+      "print\"The local oscillator frequency for first IF is \",flo1/10**6,\"MHz\"\n",
+      "print\"The local oscillator frequency for the second IF is \",flo2/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The local oscillator frequency for first IF is  3310.0 MHz\n",
+        "The local oscillator frequency for the second IF is  630 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter19.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter19.ipynb
new file mode 100755
index 00000000..1ce8f35e
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter19.ipynb
@@ -0,0 +1,128 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bbbebe541ba21d9bbb4d54500aef67ec9a094ac72f21cf0a19a14aa93b9c61ce"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 19 Optical communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.1 Page no 760"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "NA = 0.29\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "critical_angle = math.asin(0.29)*180/3.14\n",
+      "\n",
+      "print\"The critical angle is \",round(critical_angle,2),\"degree\"\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The critical angle is  16.87 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.2 Page no 767"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "B_rating_Mhzkm =600*10**6\n",
+      "len_ft=500\n",
+      "\n",
+      "#Calculation\n",
+      "bandwidth = B_rating_Mhzkm/(len_ft/3274.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"The bandwidth of the 500 feetr segment of the cable is \",bandwidth/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The bandwidth of the 500 feetr segment of the ccable is  3928.8 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.3 Page no 780"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R=43*10**6\n",
+      "D=1200/3274.0\n",
+      "\n",
+      "#Calculation\n",
+      "d=1/(5.0*R*D)\n",
+      "\n",
+      "#Result\n",
+      "print\"The dispersion factor is \",round(d*10**9,1),\"ns/km\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dispersion factor is  12.7 ns/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter22.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter22.ipynb
new file mode 100755
index 00000000..2fc3916b
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter22.ipynb
@@ -0,0 +1,94 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d66b7b0a5ebd4fcdf9eb24f98ae81ac03aa2dcdb229f576b47a170647bd9ea49"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 22 Communication tests and measurments"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.1 Page no 854"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vp=8                       #volts\n",
+      "R=75                       #Ohm\n",
+      "\n",
+      "#Calculation\n",
+      "Vmax=0.707*Vp\n",
+      "P=Vmax**2/R\n",
+      "\n",
+      "#Result\n",
+      "print\"Power is dissipated in the load is \",round(P,4),\"Watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power is dissipated in the load is  0.4265 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.2 Page no 857"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Pf=380.0                     #Watt\n",
+      "Pr=40                      #Watt\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "m=math.sqrt(Pr/Pf)\n",
+      "SWR=(1+m)/(1-m)\n",
+      "print\"SWR= \",round(SWR,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "SWR=  1.96\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter3.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter3.ipynb
new file mode 100755
index 00000000..ef029e65
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter3.ipynb
@@ -0,0 +1,331 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8e1b5c14d4876dc6b60f9f51fe0f04e94941e7f01c84340bf3441e1daf170b7a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 Amplitude modulation fundamentals"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 page no 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vmax = 5.9\n",
+      "Vmin = 1.2\n",
+      "\n",
+      "#Calculation\n",
+      "m = (Vmax-Vmin)/(Vmax+Vmin)\n",
+      "Vc = (Vmax+Vmin)/2.0\n",
+      "Vm = (Vmax-Vmin)/2.0\n",
+      "m = Vm/Vc\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The modulation index is \",round(m,2)\n",
+      "print\"Vc= \",Vc,\"Vm= \",Vm,\"(for 2 volt/div on verticle scale)\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The modulation index is  0.66\n",
+        "Vc=  3.55 Vm=  2.35 (for 2 volt/div on verticle scale)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 page no 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "frq =980*10**3\n",
+      "frq_range = 5*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "fusb = frq+frq_range\n",
+      "flsb = frq-frq_range\n",
+      "bw=fusb-flsb\n",
+      "\n",
+      "#Result\n",
+      "print\"The upper sideband is at \",fusb/10**3,\"Khz\"\n",
+      "print\"Lower sideband is at \",flsb/10**3,\"Khz\"\n",
+      "print\"the babdwidth is \",bw/10**3,\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The upper sideband is at  985 Khz\n",
+        "Lower sideband is at  975 Khz\n",
+        "the babdwidth is  10 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 page no 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Pc = 30\n",
+      "m=0.85\n",
+      "\n",
+      "#Calculation\n",
+      "Pt = Pc*(1+ (m**2/2.0))\n",
+      "Psb_both =Pt-Pc\n",
+      "Psb_one = Psb_both/2.0\n",
+      "\n",
+      "#Result\n",
+      "print\"The total power is \",round(Pt,1),\"watt\" \n",
+      "print\"The power in one sideband is \",round(Psb_one,1),\"Watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total power is  40.8 watt\n",
+        "The power in one sideband is  5.4 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 page no 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R = 40\n",
+      "I = 4.8\n",
+      "m=0.9\n",
+      "\n",
+      "#Calculation\n",
+      "Pc = I**2*R\n",
+      "Pt = (I*(1+(m**2/2.0))**0.5)**2*R\n",
+      "Psb = Pt-Pc\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The carrier power is \",Pc,\" watt\"\n",
+      "print\"(b) Total power = \",round(Pt,0),\"watt\"\n",
+      "print\"(c) Sideband Power = \",round(Psb,1),\"watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The carrier power is  921.6  watt\n",
+        "(b) Total power =  1295.0 watt\n",
+        "(c) Sideband Power =  373.2 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 page no 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "It = 5.1\n",
+      "Ic =4.8\n",
+      "\n",
+      "#Calculation\n",
+      "m=(2*((It/Ic)**2-1))**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The percentage of modulation is \",round(m*100,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage of modulation is  51.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 page no 109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "m = 0.9\n",
+      "Pc = 921.6\n",
+      "\n",
+      "#calculation\n",
+      "Psb = (m**2*Pc)/4.0\n",
+      "\n",
+      "#Result\n",
+      "print\"The power in one sideband \",round(Psb,1),\"Watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power in one sideband  186.6 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 page no 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vpp = 178\n",
+      "R = 75.0\n",
+      "\n",
+      "#Calculation\n",
+      "Vp =Vpp/2.0\n",
+      "Vrms = 0.707*Vp\n",
+      "PEP =(Vrms**2/R)\n",
+      "\n",
+      "#Result\n",
+      "print\"The peak envelop power is \", round(PEP,1),\"Watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The peak envelop power is  52.8 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 page no 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vs =24\n",
+      "Im =9.3\n",
+      "\n",
+      "#Calculation\n",
+      "PEP = Vs*Im \n",
+      "Pavg1 = PEP/3.0\n",
+      "Pavg2 = PEP/4.0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The peak envelope power is \",PEP,\"watt\"\n",
+      "print\"(b) Average power of transmitter is \",Pavg2,\"watt to\",Pavg1,\"watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The peak envelope power is  223.2 watt\n",
+        "(b) Average power of transmitter is  55.8 watt to 74.4 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter4.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter4.ipynb
new file mode 100755
index 00000000..522fe2b5
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter4.ipynb
@@ -0,0 +1,123 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:268207babca50d59c46d454ea7ce41b37f6eaad8a022bc43c98526f17a497ebd"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "chapter 4 Amplitude Modulator and Demodulator circuits"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page no 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "Vcc =48\n",
+      "I = 3.5\n",
+      "efficiency =70                           #percent\n",
+      "modulation = 67                          #percent\n",
+      "m = modulation/100.0\n",
+      "#calculation\n",
+      "\n",
+      "Pi = Vcc*I\n",
+      "Pc=Pi\n",
+      "Pm = Pi/2.0\n",
+      "Pout = (efficiency*Pi)/100.0\n",
+      "Ps = Pc*((m**2)/4.0)\n",
+      "maximum_swing = 2*Vcc \n",
+      "#Result\n",
+      "\n",
+      "print \"(a) RF input power to the output stage is \",Pi,\"W\"\n",
+      "print \"(b) AF power required for 100 percent modulation is \",Pm,\"W\"\n",
+      "print \"(c) Carrier output power is \",Pout,\"W\"\n",
+      "print \"(d) Power in one sideband for 67 percent modulation is \",round(Ps,2),\"W\"\n",
+      "print \"(e) maximum and minimum dc supply voltage swing with 100 percent modulation is \",maximum_swing,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) RF input power to the output stage is  168.0 W\n",
+        "(b) AF power required for 100 percent modulation is  84.0 W\n",
+        "(c) Carrier output power is  117.6 W\n",
+        "(d) Power in one sideband for 67 percent modulation is  18.85 W\n",
+        "(e) maximum and minimum dc supply voltage swing with 100 percent modulation is  96 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page no 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "fc =4.2*10**6\n",
+      "f1 = 300\n",
+      "f2 = 3400\n",
+      "#calculation\n",
+      "#upper sideband\n",
+      "\n",
+      "x0=fc+f1\n",
+      "y0=fc+f2\n",
+      "#lower sideband\n",
+      "\n",
+      "x1=fc-f1\n",
+      "y1=fc-f2\n",
+      "frequency=(x1+y1)/2.0\n",
+      "#Result \n",
+      "\n",
+      "print \"(a) Upper sideband ranges from \",x0,\"to\",y0\n",
+      "print \"    lower sideband ranges from \",x1,\"to\",y1\n",
+      "print \"(b) approximate center frequency of a bandpass filter is \",frequency,\"Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Upper sideband ranges from  4200300.0 to 4203400.0\n",
+        "    lower sideband ranges from  4199700.0 to 4196600.0\n",
+        "(b) approximate center frequency of a bandpass filter is  4198150.0 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter5.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter5.ipynb
new file mode 100755
index 00000000..3539866c
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter5.ipynb
@@ -0,0 +1,253 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:91b0cb714976b5e3abb5fa418cb1d12fb7cd307f6d4fd008a74f2dcf67b41e57"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 Fundamentals of frequency modulation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page no 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 915*10**6\n",
+      "fm_deviation =12.5*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "max_deviation = f + fm_deviation\n",
+      "min_deviation = f - fm_deviation\n",
+      "\n",
+      "#Result\n",
+      "print\"Maximum frequency occur during modulation is \",max_deviation/1000.0,\"KHz\"\n",
+      "print\"Minimum frequency occur during modulation is \",min_deviation/1000.0,\"KHz\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum frequency occur during modulation is  915012.5 KHz\n",
+        "Minimum frequency occur during modulation is  914987.5 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page no 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "max_deviation = 25*10**3\n",
+      "fm =15.0*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "mf =max_deviation/fm\n",
+      "\n",
+      "#Result\n",
+      "print\"The deviation ratio of the TV sound is \",round(mf,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The deviation ratio of the TV sound is  1.667\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page no 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "mf = 2.2\n",
+      "fd = 7.48*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "fm = fd/mf\n",
+      "\n",
+      "#Result\n",
+      "print\"The maximum modulating frequency is \",fm/1000.0,\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum modulating frequency is  3.4 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page no 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "J0 = -0.4\n",
+      "J1 = -0.07\n",
+      "J2 = 0.36\n",
+      "J3 = 0.43\n",
+      "J4 = 0.28\n",
+      "\n",
+      "#Result\n",
+      "print\"The amplitude of the carrier is \",J0\n",
+      "print\"Amplitudes of the first four sidebands are \",\" \\n \", J1,\"\\n \",J2,\"\\n \",J3,\"\\n \",J4\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amplitude of the carrier is  -0.4\n",
+        "Amplitudes of the first four sidebands are   \n",
+        "  -0.07 \n",
+        "  0.36 \n",
+        "  0.43 \n",
+        "  0.28\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page no 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "fd = 30*10**3\n",
+      "fm = 5*10**3\n",
+      "N=9\n",
+      "\n",
+      "#Calculation\n",
+      "bw1 = 2*fm*N\n",
+      "bw2 = 2*(fd+fm)\n",
+      "\n",
+      "#Result\n",
+      "print\"The maximum bandwidth of the fm signal is \",bw1/10**3,\"KHz\"\n",
+      "print\"Bandwidth using carson's rule \",bw2/10**3,\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum bandwidth of the fm signal is  90 KHz\n",
+        "Bandwidth using carson's rule  70 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 Page no 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "S_N = 2.8\n",
+      "fm = 1.5*10**3\n",
+      "fd =4*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "fi= math.asin(1/S_N)\n",
+      "delta = fi*fm\n",
+      "SN =fd/delta\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The frequency deviation caused by the noise \",round(delta,1),\"Hz\"\n",
+      "print\"(b) The improved output signal to noise ratio is \",round(SN,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The frequency deviation caused by the noise  547.8 Hz\n",
+        "(b) The improved output signal to noise ratio is  7.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter6.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter6.ipynb
new file mode 100755
index 00000000..8bd8c6da
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter6.ipynb
@@ -0,0 +1,148 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:aac83d9b958acba775304d5e53683242c3be955e852f2ce5dbe2f25d81f3bf36"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 FM Circuits"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page no 178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vc =40*10**-12\n",
+      "c = 20*10**-12\n",
+      "f0 = 5.5*10**6\n",
+      "Ct = Vc+c\n",
+      "\n",
+      "#Calculation\n",
+      "L = 1/((6.28*f0)**2*Ct)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of the inductance is \",round(L*10**6,0),\"microhenry\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of the inductance is  14.0 microhenry\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page no 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f=168.96*10**6\n",
+      "multiplier=24.0\n",
+      "deviation = 5*10**3\n",
+      "fm = 2.8*10**3\n",
+      "\n",
+      "#Calculation\n",
+      "f0 =f/multiplier\n",
+      "fd= deviation/multiplier\n",
+      "phaseshift = fd/fm\n",
+      "phaseshift_degrees = phaseshift*57.3\n",
+      "total_phaseshift =2*phaseshift_degrees\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The crystal oscillator frequency is \",f0/10**6,\"MHz\"\n",
+      "print\"(b) The total phase shift is \",round(total_phaseshift,3),\"degree\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The crystal oscillator frequency is  7.04 MHz\n",
+        "(b) The total phase shift is  8.527 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page no 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R =1*10**3\n",
+      "phaseshift =4.263\n",
+      "phaseshift_center= 45\n",
+      "f =7.04*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "phase_l = phaseshift_center - phaseshift\n",
+      "phase_u = phaseshift_center + phaseshift\n",
+      "phaserange_total = phase_u - phase_l\n",
+      "Xc1 = 1161\n",
+      "C1 = 1/(6.28*f*Xc1)\n",
+      "Xc2 = 861\n",
+      "C2 = 1/(6.28*f*Xc2)\n",
+      "\n",
+      "#Result\n",
+      "print\"The two values of the capacitance to achieve total deviation are \",round(C1*10**12,1),\"pf to\",round(C2*10**12,1),\"pf\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The two values of the capacitance to achieve total deviation are  19.5 pf to 26.3 pf\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter7.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter7.ipynb
new file mode 100755
index 00000000..c4587482
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter7.ipynb
@@ -0,0 +1,221 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:13e4ad42742896befb0b329fd930804197555dd5705f1dd526b017e9117b7b00"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 Digital communication techniques"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page no 210"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "t = 71.4*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "f = 1/t\n",
+      "fourth_harmonic = f*4\n",
+      "min_sampling = 2*fourth_harmonic\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The frequency of the signal is \",round(f/10**3,1),\"KHz\"\n",
+      "print\"(b) The fourth harmonic is \",round(fourth_harmonic/10**3,0),\"KHz\"\n",
+      "print\"(c) Minimum sampling rate is \",round(min_sampling/10**3,0),\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The frequency of the signal is  14.0 KHz\n",
+        "(b) The fourth harmonic is  56.0 KHz\n",
+        "(c) Minimum sampling rate is  112.0 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page no 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "N = 14\n",
+      "discrete_levels = 2.0**N\n",
+      "\n",
+      "#Calculation\n",
+      "num_vltg_inc =2**N-1\n",
+      "resolution = 12/discrete_levels \n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The numbedr of discrete levels that are represented using N number of bits are \",discrete_levels\n",
+      "print\"(b) the number odf voltage increments required to divide the voltage range are \",num_vltg_inc\n",
+      "print\"(c) Resolution of the digitization \",round(resolution*10**6,2),\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The numbedr of discrete levels that are represented using N number of bits are  16384.0\n",
+        "(b) the number odf voltage increments required to divide the voltage range are  16383\n",
+        "(c) Resolution of the digitization  732.42 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 Page no 225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "N =12\n",
+      "SINAD1=78\n",
+      "\n",
+      "#Calculation\n",
+      "SINAD2 = 6.02*N + 1.76\n",
+      "ENOB =(SINAD1 -1.76)/6.02\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The SINAD for 12 bit convertre is \",SINAD2,\"db\"\n",
+      "print\"(b) The ENOB for the converter with SINAD of 78 dB is \",round(ENOB,2),\"bits\"   \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The SINAD for 12 bit convertre is  74.0 db\n",
+        "(b) The ENOB for the converter with SINAD of 78 dB is  12.66 bits\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page no 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Gievn\n",
+      "Vm = 1.0\n",
+      "Vin = 0.25\n",
+      "mu =255\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Vout = (Vm*log(1+mu*(Vin/Vm)))/log(1+mu)\n",
+      "gain =Vout/Vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The output voltage of the compander \",round(Vout,2),\"volt\"\n",
+      "print\"Gain of the compander is \",round(gain,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage of the compander  0.75 volt\n",
+        "Gain of the compander is  3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 Page no 234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Vin = 0.8\n",
+      "Vm =1.0\n",
+      "mu =255\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Vout = (Vm*math.log(1+mu*(Vin/Vm)))/math.log(1+mu)\n",
+      "gain =Vout/Vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The output voltage of the compander \",round(Vout,1),\"volt\"\n",
+      "print\"Gain of the compander is \",round(gain,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage of the compander  1.0 volt\n",
+        "Gain of the compander is  1.2\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter8.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter8.ipynb
new file mode 100755
index 00000000..4cad3df0
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter8.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:6738b73e6fb4913114449836a25c14169c484a93e5e60fcba9b0e4a9815c8f56"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 Radio transmitters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page no 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 16*10**6\n",
+      "ppm = 200\n",
+      "frequency_variation = 200 *16 \n",
+      "\n",
+      "#Calculation\n",
+      "min_f = f - frequency_variation\n",
+      "max_f = f + frequency_variation\n",
+      "\n",
+      "#Reslt\n",
+      "print\"The minimum and maximum frequencies for the crystal of 16 Mhz with stability of 200 are \",min_f,\"Hz and\",max_f,\"Hz respectively\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum and maximum frequencies for the crystal of 16 Mhz with stability of 200 are  15996800 Hz and 16003200 Hz respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page no 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f =14.9*10**6\n",
+      "mul_factor = 2*3*3\n",
+      "stability_ppm =300\n",
+      "variation = 0.0003\n",
+      "\n",
+      "#Calculation\n",
+      "total_variation = variation* mul_factor\n",
+      "fout = f * mul_factor\n",
+      "frequency_variation = fout*total_variation\n",
+      "f_lower = fout - frequency_variation\n",
+      "f_upper = fout + frequency_variation\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The output frequency of the transmitter is \",fout/10**6,\"MHz\"\n",
+      "print\"(b) The maximum and minimum frequencies of the transmitter are \",round(f_lower/10**6,2),\"Mhz and \",round(f_upper/10**6,2),\"Mhz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The output frequency of the transmitter is  268.2 MHz\n",
+        "(b) The maximum and minimum frequencies of the transmitter are  266.75 Mhz and  269.65 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 Page no 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "f = 10*10**6\n",
+      "div_factor = 100.0\n",
+      "A =63\n",
+      "N = 285\n",
+      "M=32\n",
+      "\n",
+      "#Calculation\n",
+      "ref = f/div_factor\n",
+      "R =M*N+A\n",
+      "fout= R*ref\n",
+      "\n",
+      "#Result\n",
+      "print\"The output frequency of the synthesizer is \",fout/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output frequency of the synthesizer is  918.3 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page no 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f = 10*10**6\n",
+      "div_factor = 100.0\n",
+      "A =64\n",
+      "N = 285\n",
+      "M=32\n",
+      "\n",
+      "#Calculation\n",
+      "ref = f/div_factor\n",
+      "R =M*N+A\n",
+      "fout= R*ref\n",
+      "\n",
+      "#Result\n",
+      "print\"The output frequency of the synthesizer is \",fout/10**6,\"MHz\"\n",
+      "print\"The step change is \",fout/10**6-918.3,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output frequency of the synthesizer is  918.4 MHz\n",
+        "The step change is  0.1 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter9.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter9.ipynb
new file mode 100755
index 00000000..379e8939
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter9.ipynb
@@ -0,0 +1,274 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3ff4b00636e0eec0da6869f81549546163050c8b991ea646d8083caae0f8dd37"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 Communication receivers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Page no 318"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fl =220*10**6\n",
+      "fm =224*10**6\n",
+      "IF1 = 10.7*10**6\n",
+      "IF = 1.5*10**6\n",
+      "\n",
+      "#Calculation\n",
+      "IF2 =IF1+IF\n",
+      "tune_l =fl+IF1\n",
+      "tune_m = fm+IF1\n",
+      "IF1_imgl = tune_l+IF1\n",
+      "IF2_imgm = tune_m+IF1\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The local oscillatior tuning range is \",tune_l/10**6,\"to \",tune_m/10**6,\"MHz\"\n",
+      "print\"(b) Frequency of the second local oscillator is \",IF2/10**6,\"MHz\"\n",
+      "print\"(c) First IF image range is \",IF1_imgl/10**6,\"MHz to \",IF2_imgm/10**6,\"MHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The local oscillatior tuning range is  230.7 to  234.7 MHz\n",
+        "(b) Frequency of the second local oscillator is  12.2 MHz\n",
+        "(c) First IF image range is  241.4 MHz to  245.4 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 Page no 324"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R = 100*10**3\n",
+      "T = 273+25\n",
+      "B = 20*10**3\n",
+      "k = 1.38*10**-23\n",
+      "\n",
+      "#Calculation\n",
+      "Vn=(4*k*T*B*R)**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The noise voltage across 100k resistor is \",round(Vn*10**6,2),\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The noise voltage across 100k resistor is  5.74 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page no 324"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R=75\n",
+      "B=6*10**6  \n",
+      "T = 29+273\n",
+      "k =1.38*10**-23\n",
+      "\n",
+      "#calculation\n",
+      "Vn = (4*k*T*B*R)**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The input themal noise is \",round(Vn*10**6,2),\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input themal noise is  2.74 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page no 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Tc=32.2\n",
+      "Tk=273+Tc\n",
+      "B =30*10**3\n",
+      "k =1.38*10**-23\n",
+      "\n",
+      "#Calculation\n",
+      "Pn=k*Tk*B\n",
+      "\n",
+      "#Result\n",
+      "print\"The average noise power is \",round(Pn*10**16,2),\"10**-16 W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average noise power is  1.26 10**-16 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page no 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "SN_ip = 8\n",
+      "SN_op = 6.0\n",
+      "\n",
+      "#Calculation\n",
+      "NR = SN_ip/SN_op\n",
+      "NF = 10*log10(NR)\n",
+      "\n",
+      "#Result\n",
+      "print\"The noise factor is \",round(NR,2)\n",
+      "print\"The noise figure is \",round(NF,2),\"dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The noise factor is  1.33\n",
+        "The noise figure is  1.25 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 Page no 330"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R= 75.0\n",
+      "T=31+273\n",
+      "k=1.38*10**-23\n",
+      "B=6*10**6\n",
+      "Vs = 8.3*10**-6\n",
+      "NF=2.8\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Vn = math.sqrt(4*k*T*B*R)\n",
+      "Pn = Vn**2/R\n",
+      "Ps = Vs**2/R\n",
+      "SN = (Ps*10**12)/(Pn/10.0**12)\n",
+      "SN_dB =10*log10(SN)\n",
+      "NR = 10**0.28\n",
+      "SN_op = SN/NR\n",
+      "Tn = 290*(NR-1)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The input noise power is \",round(Pn*10**12,1),\"pW\"\n",
+      "print\"(b) The input signal power is \",round(Ps*10**12,3),\"pW\"\n",
+      "print\"(c) Signal to noise ratio in decibels \",round(SN/10**24,2)\n",
+      "print\"(d) The noise factor is \",round(NR,2)\n",
+      "print\"(e) Signal to noise ratio of the amplifier is \",round(SN_op/10**24,2)\n",
+      "print\"(f) The noise temperature of the amplifier \",round(Tn,0),\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The input noise power is  0.1 pW\n",
+        "(b) The input signal power is  0.919 pW\n",
+        "(c) Signal to noise ratio in decibels  9.12\n",
+        "(d) The noise factor is  1.91\n",
+        "(e) Signal to noise ratio of the amplifier is  4.79\n",
+        "(f) The noise temperature of the amplifier  263.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter_2.ipynb b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter_2.ipynb
new file mode 100755
index 00000000..3847150a
--- /dev/null
+++ b/Principles_Of_Electronic_Communication_Systems_by_L_E_Frenze/temp/chapter_2.ipynb
@@ -0,0 +1,1143 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:759039f1dfc5f6d62e90bf867630bc8f1cdda845968a4b5c11c4acddcb77b251"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 The fundamentals of Electronics: A Review"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page no 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "vout=750*10**-3\n",
+      "vin = 30*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "gain=vout/vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The Voltage gain of the amplifier is \",gain\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Voltage gain of the amplifier is  25000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 Page no 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "pout=6\n",
+      "power_gain=80.0\n",
+      "\n",
+      "#calculation\n",
+      "pin=pout/power_gain\n",
+      "\n",
+      "#Result\n",
+      "print\"The input power of the signal is \",pin*1000,\"mW\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input power of the sigmal is  75.0 mW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page no 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "A1=5;\n",
+      "A2=2;\n",
+      "A3=17;\n",
+      "total_gain=A1*A2*A3;\n",
+      "pin= 40*10**-3;\n",
+      "\n",
+      "#calculation\n",
+      "pout=total_gain*pin;\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power is\",pout,\"watts\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power is 6.8 watts\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page no 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "pin=25.0*10**-6;\n",
+      "pout=1.5*10**-3;\n",
+      "A1=3.0;\n",
+      "\n",
+      "#Calculation\n",
+      "total_gain=pout/pin;\n",
+      "print\"Total gain is\",total_gain\n",
+      "A2=total_gain/A1\n",
+      "\n",
+      "#Result\n",
+      "print\"The gain of second stage is \",A2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total gain is 60.0\n",
+        "The gain of second stage is  20.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page no 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "R1=10.0*10**3;\n",
+      "R2=470.0;\n",
+      "\n",
+      "#Calculation\n",
+      "attenuation=R2/(R2+R1)\n",
+      "A2=1/attenuation\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The attenuation (A1) is\",round(attenuation,3)\n",
+      "print\"(b) The attenuation (A2) is\",round(A2,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The attenuation (A1) is 0.045\n",
+        "(b) The attenuation (A2) is 22.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page no 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vin=20.0*10**-6;\n",
+      "Vout=100*10**-3;\n",
+      "A1=45000.0                                #A1 isAmplifier gain\n",
+      "\n",
+      "#calculation\n",
+      "AT=Vout/Vin                                #AT is Total gain\n",
+      "A2=AT/A1                                   #A2 is attenuation factor\n",
+      "\n",
+      "#Result\n",
+      "print\"Total gain is\",AT\n",
+      "print\"The atenuation factor needed to to keep the output voltage from exceeding 100 mv is \",round(A2,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total gain is 5000.0\n",
+        "The atenuation factor needed to to keep the output voltage from exceeding 100 mv is  0.1111\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page no 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vin=3.0*10**-3\n",
+      "Vout=5\n",
+      "Pin=50.0*10**-3                                     \n",
+      "Pout=2*10**-3                                     \n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "gain_dB= 20*log10 (Vout/Vin)\n",
+      "gain_db=10*log10 (Pout/Pin)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The gain of amplifier in dB is \",round(gain_dB,1)\n",
+      "print\"(b) The gain in dB is \",round(gain_db,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The gain of amplifier in dB is  64.4\n",
+        "(b) The gain in dB is  -13.98\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page no 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "gain_dB = 40\n",
+      "pout_W= 100\n",
+      "\n",
+      "#calculation\n",
+      "pin_W = pout_W/10.0**4\n",
+      "\n",
+      "#Result\n",
+      "print\"The input power is \",pin_W*10**3,\"m Watt\"\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input power is  10.0 m Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page no 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "gain_db = 60\n",
+      "vin = 50*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "vout = 10**(60/20.0)*vin\n",
+      "\n",
+      "#Result\n",
+      "print\"The output voltage is \",vout,\"volt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage is  0.05 volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10 Page no 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "vin=90*10**-3\n",
+      "R1= 10.0*10**3\n",
+      "vout=7.8\n",
+      "Rout=8.0\n",
+      "\n",
+      "#calculation\n",
+      "pin= vin**2/R1\n",
+      "pout=vout**2/Rout\n",
+      "\n",
+      "import math\n",
+      "Ap_db = 10*log10 (pout/pin)\n",
+      "\n",
+      "#Result\n",
+      "print\"The power gain in decibel is \",round(Ap_db,1),\"dB\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power gain in decibel is  69.7 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11 Page no 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "\n",
+      "gain_db = 28\n",
+      "pin = 36*10**-3\n",
+      "\n",
+      "#calculation\n",
+      "pout = 10**2.8*pin;\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power is \",round(pout,2),\"watt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power is  22.71 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12 Page no 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "gain1 = 6.8 \n",
+      "gain2 = 14.3\n",
+      "attenuation1 = -16.4\n",
+      "attenuation2 = -2.9\n",
+      "vout = 800*10**-3\n",
+      "\n",
+      "#calculation\n",
+      "At = gain1+gain2+attenuation1+attenuation2\n",
+      "vin = vout/10.0**(At/20.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"The input voltage is \",round(vin*10**3,1),\"mv\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The input voltage is  650.3 mv\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13 Page no 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "pout_db =12.3\n",
+      "\n",
+      "#calculation\n",
+      "pout_mW = 0.001*10**(12.3/10.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"The output power is \" ,round(pout_mW*10**3,0),\"mv\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output power is  17.0 mv\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14 Page no 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c = 2.7*10**-12\n",
+      "l = 33*10**-9\n",
+      "\n",
+      "#calculation\n",
+      "fr= 1/(6.28*(l*c)**0.5)\n",
+      "\n",
+      "#Result\n",
+      "print\"The resonat frequency is \" ,round(fr/10.0**6,0),\"Mhz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resonat frequency is  533.0 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15 Page no 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c =12*10**-12\n",
+      "fr = 49*10**6\n",
+      "\n",
+      "#calculation\n",
+      "l=1/(4*3.14**2*fr**2*c)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of inductance is \" ,round(l*10**9,0),\"nh\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of inductance is  880.0 nh\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16 Page no 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fr=28*10**6\n",
+      "Q=70.0\n",
+      "\n",
+      "#calculation\n",
+      "bandwidth = fr/Q\n",
+      "\n",
+      "#Result\n",
+      "print\"The bandwidth is \",bandwidth/10.0**3,\"Khz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The bandwidth is  400.0 Khz\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17 Page no 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "f1= 7.93*10**6\n",
+      "f2= 8.07*10**6\n",
+      "\n",
+      "#calculation\n",
+      "bw= f2-f1\n",
+      "fr=(f1*f2)**0.5\n",
+      "Q= fr/bw\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The bandwidth is \",bw/10.0**3,\"Khz\"\n",
+      "print\"(b) The resonant frequency is \",round(fr/10.0**6,0),\"Mhz\"\n",
+      "print\"(c) The Q of resonant circuit is \",round(Q,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The bandwidth is  140.0 Khz\n",
+        "(b) The resonant frequency is  8.0 Mhz\n",
+        "(c) The Q of resonant circuit is  57.14\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18 Page no 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Q=200.0\n",
+      "fr=16*10**6\n",
+      "\n",
+      "#calculation\n",
+      "bw=fr/Q\n",
+      "f1= fr-(bw/2)\n",
+      "f2=fr+(bw/2)\n",
+      "\n",
+      "#Result\n",
+      "print\"Bandwidth is \",bw*10**-3,\"KHz\"\n",
+      "print\"f1= \",f1*10**-6,\"MHz\"\n",
+      "print\"f2= \",f2*10**-6,\"MHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bandwidth is  80.0 KHz\n",
+        "f1=  15.96 MHz\n",
+        "f2=  16.04 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19 Page no 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Q= 150\n",
+      "Vs=3*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "Vc= Q*Vs\n",
+      "\n",
+      "#Result\n",
+      "print\"The voltage across capacitor is \",Vc*10**6,\"microvolt\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across capacitor is  450.0 microvolt\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.20 Page no 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fr= 52*10**6\n",
+      "Q=12.0\n",
+      "L=0.15*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "Rw=(6.28*fr*L)/Q\n",
+      "Req= Rw*(Q**2+1)\n",
+      "\n",
+      "#Result\n",
+      "print\"Impedance of the parellel LC circuit is \",round(Req,0),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Impedance of the parellel LC circuit is  592.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.21 Page no 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fr= 52.0*10**6\n",
+      "Rw= 4.1\n",
+      "L =0.15*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "C=1/(4.0*3.14**2*fr**2*L)\n",
+      "Z = L/(C*Rw)\n",
+      "\n",
+      "#Result\n",
+      "print\"the impedance of the circuit is \",round(Z,0),\"ohm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the impedance of the circuit is  585.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.22 Page no 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw = 1.0*10**6\n",
+      "XL = 300\n",
+      "Rw = 10.0\n",
+      "fr =10*10**6\n",
+      "\n",
+      "#calculation\n",
+      "Q1 = XL/Rw\n",
+      "Rp = Rw*(Q1**2+1) \n",
+      "\n",
+      "Q2 = fr/bw\n",
+      "Rpnew = Q2*XL\n",
+      "\n",
+      "Rext = (Rpnew*Rp)/(Rp-Rpnew)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of resistor needed to set the bandwidth of the parellel tuned circuit is \",round(Rext,1),\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of resistor needed to set the bandwidth of the parellel tuned circuit is  4497.5 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.23 Page no 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "R = 8.2*10**3\n",
+      "C =0.0033*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "fco = 1/(6.28* R*C)\n",
+      "\n",
+      "#Result\n",
+      "print\"The cut off frequency is \",round(fco/10.0**3,2),\"Khz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cut off frequency is  5.88 Khz\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.24 Page no 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fco =3.4*10**3\n",
+      "C = 0.047*10**-6\n",
+      "\n",
+      "#calculation\n",
+      "R = 1/(6.28* fco* C)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of the resistor is \",round(R,0),\"ohm\"\n",
+      "print\"The closest standard value is \", 1000 ,\"ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of the resistor is  996.0 ohm\n",
+        "The closest standard value is  1000 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.25 Page no 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fnotch = 120\n",
+      "R = 220*10**3\n",
+      "\n",
+      "#calculatiuon\n",
+      "C = 1/(6.28*R*fnotch)\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of capacitance required is \",round(2*C*10**6,3),\"microfarad\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of capacitance required is  0.012 microfarad\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.26 Page no 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vpeak =3.0\n",
+      "f=48*10**3\n",
+      "\n",
+      "#calculation\n",
+      "fifth_harmonic = 5*f\n",
+      "Vrms=(4/3.14)*(3/5.0)*0.707\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The frequency of the fifth harmonic is \",fifth_harmonic/10.0**3,\"Khz\"\n",
+      "print\"The RMS voltage of the fifth harmonic is \",round(Vrms,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The frequency of the fifth harmonic is  240.0 Khz\n",
+        "The RMS voltage of the fifth harmonic is  0.54\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.27 Page no 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vpeak = 5\n",
+      "f = 4.0*10**6\n",
+      "duty_cycle=0.3\n",
+      "\n",
+      "#calculation\n",
+      "T = 1/f\n",
+      "t0 = duty_cycle*T\n",
+      "Vavg = Vpeak*duty_cycle\n",
+      "min_bw =1/t0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The average DC value is \",Vavg,\"volt\"\n",
+      "print\"(b) The minimum bandwidth required is \" ,round(min_bw/10.0**6,3),\"Mhz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The average DC value is  1.5 volt\n",
+        "(b) The minimum bandwidth required is  13.333 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.28 Page no 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "tr =6*10**-9\n",
+      "\n",
+      "#calculation\n",
+      "min_bw=(35/0.006)\n",
+      "\n",
+      "#Result\n",
+      "print\"The minimum bandwidth is % is \",round(min_bw/10.0**2,1),\"Mhz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum bandwidth is % is  58.3 Mhz\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.29 Page no 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw= 200.0*10**3\n",
+      "\n",
+      "#calculation\n",
+      "tr= 0.35/(bw*10**-3)\n",
+      "\n",
+      "#Result\n",
+      "print\"The fastest rise time of the circuit is \" ,tr*10**3,\"microseconds\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The fastest rise time of the circuit is  1.75 microseconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.30 Page no 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw_mhz = 60\n",
+      "tri_ns= 15\n",
+      "\n",
+      "#calculation\n",
+      "tra_osci = 0.35/(bw_mhz)\n",
+      "tra_comp = 1.1*(tri_ns**2 + (tra_osci*10**3)**2)**0.5\n",
+      "\n",
+      "#Result\n",
+      "print\"The rise time of the displayed square wave is \",round(tra_comp,1),\"ns\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rise time of the displayed square wave is  17.7 ns\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter01.ipynb b/Principles_Of_Foundation_Engineering/Chapter01.ipynb
deleted file mode 100755
index 61dc3457..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter01.ipynb
+++ /dev/null
@@ -1,249 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3f5444f542f0d6a3b857b61fbecdc4e8047247ed8828d4cea5a34f91ccdf5eb3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1:Geotechnical Properties of Soil"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.1:Pg-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.1\n",
-      "\n",
-      "V=0.25; # ft^3\n",
-      "W=30.8; #lb\n",
-      "Wd=28.2; # weight dried lb\n",
-      "Gs=2.7;\n",
-      "Gammaw=62.4; #lb/ft^3\n",
-      "Gamma=W/V;\n",
-      "print Gamma,\" is moist unit weight in lb/ft^3\"\n",
-      "w=(W-Wd)/W;\n",
-      "print round(w*100,2),\"is moisture content in %\"\n",
-      "Gammad=Wd/V;\n",
-      "print Gammad, \"is dry unit weight in lb/ft^3\"\n",
-      "Vs=Wd/Gs/Gammaw;\n",
-      "Vv=V-Vs;\n",
-      "e=Vv/Vs;\n",
-      "print round(e,3),\" is void ratio\"\n",
-      "n=e/(1+e);\n",
-      "print round(n,2),\"is porosity\"\n",
-      "Vw=(W-Wd)/Gammaw;\n",
-      "S=Vw/Vv;\n",
-      "print round(S*100,2),\"is saturation in %\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "123.2  is moist unit weight in lb/ft^3\n",
-        "8.44 is moisture content in %\n",
-        "112.8 is dry unit weight in lb/ft^3\n",
-        "0.494  is void ratio\n",
-        "0.33 is porosity\n",
-        "50.43 is saturation in %\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.2:Pg-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.2\n",
-      "\n",
-      "e=0.72;\n",
-      "w=12.0/100; #moisture content\n",
-      "Gs=2.72;\n",
-      "Gammaw=9.81;#kN/m^3\n",
-      "Gammad=Gs*Gammaw/(1+e);\n",
-      "print round(Gammad,2),\"= dry unit weight in kN/m^3\"\n",
-      "Gamma=Gs*Gammaw*(1+w)/(1+e);\n",
-      "print round(Gamma,2),\" = moist unit weight in kN/m^3\"\n",
-      "Gammasat=(Gs+e)*Gammaw/(1+e);\n",
-      "wa=Gammasat-Gamma;#water added\n",
-      "print round(wa,2),\" = water added in kN/m^3\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "15.51 = dry unit weight in kN/m^3\n",
-        "17.38  = moist unit weight in kN/m^3\n",
-        "2.24  = water added in kN/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.3:Pg-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.3\n",
-      "from scipy.optimize import fsolve\n",
-      "gmax=17.1; # Gammadmax\n",
-      "Dr=0.7;\n",
-      "w=8.0/100;#moisture content\n",
-      "gmin=14.2; #Gammadmin\n",
-      "def f(x):\n",
-      "    return (x-14.2)/(17.1-14.2)*17.1/x-0.7 \n",
-      "x=fsolve(f,16);#solving for gammad\n",
-      "Gamma=x[0]*(1+w);\n",
-      "print round(Gamma,2),\"moist unit weight in kN/m^3\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "17.4 moist unit weight in kN/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.7:Pg-38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.7\n",
-      "import math\n",
-      "#part (a)\n",
-      "e1=0.92;\n",
-      "e2=0.86;\n",
-      "Hc=2.8; # in m\n",
-      "s2=212.0;#sigma2dash Load in kN/m2\n",
-      "s1=140.0;#sigma1dash Load in kN/m2\n",
-      "Cc=(e1-e2)/math.log10(s2/s1);\n",
-      "Sc=Cc*Hc/(1+e1)*math.log10(s2/s1);\n",
-      "print Sc*1000,\"consolidated depth in mm\"\n",
-      "#part (b)\n",
-      "# from Figure (1.21):\n",
-      "Sct=40.0;# in mm\n",
-      "T50=0.197;\n",
-      "t=4.5; # in MIN\n",
-      "Cr=T50*12.7**2.0/t;\n",
-      "U=Sct/Sc*100.0/1000;\n",
-      "H=Hc/2;\n",
-      "Tv=math.pi/4*U**2.0/100**2;\n",
-      "t=Tv*H**2.0/Cr*1000.0**2/60.0/24;\n",
-      "print round(t,1),\" is time required in days\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "87.5 consolidated depth in mm\n",
-        "31.6  is time required in days\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.8:Pg-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.8\n",
-      "\n",
-      "Cv=7.061; # in mm^2/min\n",
-      "tc=15*24*60.0;\n",
-      "H=2.8/2*1000.0;\n",
-      "Scmax=87.5; # consolidation\n",
-      "Tc=Cv*tc/H**2;\n",
-      "tv=31.6*24*60;\n",
-      "Tv=Cv*tv/H**2;\n",
-      "#from figure 1.28\n",
-      "Sct=Scmax*0.36;\n",
-      "print Sct,\"is consolidation in 31.6 days in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "31.5 is consolidation in 31.6 days in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter01_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter01_1.ipynb
deleted file mode 100755
index 61dc3457..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter01_1.ipynb
+++ /dev/null
@@ -1,249 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3f5444f542f0d6a3b857b61fbecdc4e8047247ed8828d4cea5a34f91ccdf5eb3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1:Geotechnical Properties of Soil"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.1:Pg-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.1\n",
-      "\n",
-      "V=0.25; # ft^3\n",
-      "W=30.8; #lb\n",
-      "Wd=28.2; # weight dried lb\n",
-      "Gs=2.7;\n",
-      "Gammaw=62.4; #lb/ft^3\n",
-      "Gamma=W/V;\n",
-      "print Gamma,\" is moist unit weight in lb/ft^3\"\n",
-      "w=(W-Wd)/W;\n",
-      "print round(w*100,2),\"is moisture content in %\"\n",
-      "Gammad=Wd/V;\n",
-      "print Gammad, \"is dry unit weight in lb/ft^3\"\n",
-      "Vs=Wd/Gs/Gammaw;\n",
-      "Vv=V-Vs;\n",
-      "e=Vv/Vs;\n",
-      "print round(e,3),\" is void ratio\"\n",
-      "n=e/(1+e);\n",
-      "print round(n,2),\"is porosity\"\n",
-      "Vw=(W-Wd)/Gammaw;\n",
-      "S=Vw/Vv;\n",
-      "print round(S*100,2),\"is saturation in %\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "123.2  is moist unit weight in lb/ft^3\n",
-        "8.44 is moisture content in %\n",
-        "112.8 is dry unit weight in lb/ft^3\n",
-        "0.494  is void ratio\n",
-        "0.33 is porosity\n",
-        "50.43 is saturation in %\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.2:Pg-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.2\n",
-      "\n",
-      "e=0.72;\n",
-      "w=12.0/100; #moisture content\n",
-      "Gs=2.72;\n",
-      "Gammaw=9.81;#kN/m^3\n",
-      "Gammad=Gs*Gammaw/(1+e);\n",
-      "print round(Gammad,2),\"= dry unit weight in kN/m^3\"\n",
-      "Gamma=Gs*Gammaw*(1+w)/(1+e);\n",
-      "print round(Gamma,2),\" = moist unit weight in kN/m^3\"\n",
-      "Gammasat=(Gs+e)*Gammaw/(1+e);\n",
-      "wa=Gammasat-Gamma;#water added\n",
-      "print round(wa,2),\" = water added in kN/m^3\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "15.51 = dry unit weight in kN/m^3\n",
-        "17.38  = moist unit weight in kN/m^3\n",
-        "2.24  = water added in kN/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.3:Pg-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.3\n",
-      "from scipy.optimize import fsolve\n",
-      "gmax=17.1; # Gammadmax\n",
-      "Dr=0.7;\n",
-      "w=8.0/100;#moisture content\n",
-      "gmin=14.2; #Gammadmin\n",
-      "def f(x):\n",
-      "    return (x-14.2)/(17.1-14.2)*17.1/x-0.7 \n",
-      "x=fsolve(f,16);#solving for gammad\n",
-      "Gamma=x[0]*(1+w);\n",
-      "print round(Gamma,2),\"moist unit weight in kN/m^3\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "17.4 moist unit weight in kN/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.7:Pg-38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.7\n",
-      "import math\n",
-      "#part (a)\n",
-      "e1=0.92;\n",
-      "e2=0.86;\n",
-      "Hc=2.8; # in m\n",
-      "s2=212.0;#sigma2dash Load in kN/m2\n",
-      "s1=140.0;#sigma1dash Load in kN/m2\n",
-      "Cc=(e1-e2)/math.log10(s2/s1);\n",
-      "Sc=Cc*Hc/(1+e1)*math.log10(s2/s1);\n",
-      "print Sc*1000,\"consolidated depth in mm\"\n",
-      "#part (b)\n",
-      "# from Figure (1.21):\n",
-      "Sct=40.0;# in mm\n",
-      "T50=0.197;\n",
-      "t=4.5; # in MIN\n",
-      "Cr=T50*12.7**2.0/t;\n",
-      "U=Sct/Sc*100.0/1000;\n",
-      "H=Hc/2;\n",
-      "Tv=math.pi/4*U**2.0/100**2;\n",
-      "t=Tv*H**2.0/Cr*1000.0**2/60.0/24;\n",
-      "print round(t,1),\" is time required in days\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "87.5 consolidated depth in mm\n",
-        "31.6  is time required in days\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.8:Pg-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.8\n",
-      "\n",
-      "Cv=7.061; # in mm^2/min\n",
-      "tc=15*24*60.0;\n",
-      "H=2.8/2*1000.0;\n",
-      "Scmax=87.5; # consolidation\n",
-      "Tc=Cv*tc/H**2;\n",
-      "tv=31.6*24*60;\n",
-      "Tv=Cv*tv/H**2;\n",
-      "#from figure 1.28\n",
-      "Sct=Scmax*0.36;\n",
-      "print Sct,\"is consolidation in 31.6 days in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "31.5 is consolidation in 31.6 days in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter01_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter01_2.ipynb
deleted file mode 100755
index 61dc3457..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter01_2.ipynb
+++ /dev/null
@@ -1,249 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3f5444f542f0d6a3b857b61fbecdc4e8047247ed8828d4cea5a34f91ccdf5eb3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1:Geotechnical Properties of Soil"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.1:Pg-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.1\n",
-      "\n",
-      "V=0.25; # ft^3\n",
-      "W=30.8; #lb\n",
-      "Wd=28.2; # weight dried lb\n",
-      "Gs=2.7;\n",
-      "Gammaw=62.4; #lb/ft^3\n",
-      "Gamma=W/V;\n",
-      "print Gamma,\" is moist unit weight in lb/ft^3\"\n",
-      "w=(W-Wd)/W;\n",
-      "print round(w*100,2),\"is moisture content in %\"\n",
-      "Gammad=Wd/V;\n",
-      "print Gammad, \"is dry unit weight in lb/ft^3\"\n",
-      "Vs=Wd/Gs/Gammaw;\n",
-      "Vv=V-Vs;\n",
-      "e=Vv/Vs;\n",
-      "print round(e,3),\" is void ratio\"\n",
-      "n=e/(1+e);\n",
-      "print round(n,2),\"is porosity\"\n",
-      "Vw=(W-Wd)/Gammaw;\n",
-      "S=Vw/Vv;\n",
-      "print round(S*100,2),\"is saturation in %\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "123.2  is moist unit weight in lb/ft^3\n",
-        "8.44 is moisture content in %\n",
-        "112.8 is dry unit weight in lb/ft^3\n",
-        "0.494  is void ratio\n",
-        "0.33 is porosity\n",
-        "50.43 is saturation in %\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.2:Pg-11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.2\n",
-      "\n",
-      "e=0.72;\n",
-      "w=12.0/100; #moisture content\n",
-      "Gs=2.72;\n",
-      "Gammaw=9.81;#kN/m^3\n",
-      "Gammad=Gs*Gammaw/(1+e);\n",
-      "print round(Gammad,2),\"= dry unit weight in kN/m^3\"\n",
-      "Gamma=Gs*Gammaw*(1+w)/(1+e);\n",
-      "print round(Gamma,2),\" = moist unit weight in kN/m^3\"\n",
-      "Gammasat=(Gs+e)*Gammaw/(1+e);\n",
-      "wa=Gammasat-Gamma;#water added\n",
-      "print round(wa,2),\" = water added in kN/m^3\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "15.51 = dry unit weight in kN/m^3\n",
-        "17.38  = moist unit weight in kN/m^3\n",
-        "2.24  = water added in kN/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.3:Pg-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.3\n",
-      "from scipy.optimize import fsolve\n",
-      "gmax=17.1; # Gammadmax\n",
-      "Dr=0.7;\n",
-      "w=8.0/100;#moisture content\n",
-      "gmin=14.2; #Gammadmin\n",
-      "def f(x):\n",
-      "    return (x-14.2)/(17.1-14.2)*17.1/x-0.7 \n",
-      "x=fsolve(f,16);#solving for gammad\n",
-      "Gamma=x[0]*(1+w);\n",
-      "print round(Gamma,2),\"moist unit weight in kN/m^3\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "17.4 moist unit weight in kN/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.7:Pg-38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.7\n",
-      "import math\n",
-      "#part (a)\n",
-      "e1=0.92;\n",
-      "e2=0.86;\n",
-      "Hc=2.8; # in m\n",
-      "s2=212.0;#sigma2dash Load in kN/m2\n",
-      "s1=140.0;#sigma1dash Load in kN/m2\n",
-      "Cc=(e1-e2)/math.log10(s2/s1);\n",
-      "Sc=Cc*Hc/(1+e1)*math.log10(s2/s1);\n",
-      "print Sc*1000,\"consolidated depth in mm\"\n",
-      "#part (b)\n",
-      "# from Figure (1.21):\n",
-      "Sct=40.0;# in mm\n",
-      "T50=0.197;\n",
-      "t=4.5; # in MIN\n",
-      "Cr=T50*12.7**2.0/t;\n",
-      "U=Sct/Sc*100.0/1000;\n",
-      "H=Hc/2;\n",
-      "Tv=math.pi/4*U**2.0/100**2;\n",
-      "t=Tv*H**2.0/Cr*1000.0**2/60.0/24;\n",
-      "print round(t,1),\" is time required in days\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "87.5 consolidated depth in mm\n",
-        "31.6  is time required in days\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.8:Pg-42"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 1.8\n",
-      "\n",
-      "Cv=7.061; # in mm^2/min\n",
-      "tc=15*24*60.0;\n",
-      "H=2.8/2*1000.0;\n",
-      "Scmax=87.5; # consolidation\n",
-      "Tc=Cv*tc/H**2;\n",
-      "tv=31.6*24*60;\n",
-      "Tv=Cv*tv/H**2;\n",
-      "#from figure 1.28\n",
-      "Sct=Scmax*0.36;\n",
-      "print Sct,\"is consolidation in 31.6 days in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "31.5 is consolidation in 31.6 days in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter02.ipynb b/Principles_Of_Foundation_Engineering/Chapter02.ipynb
deleted file mode 100755
index f284c073..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter02.ipynb
+++ /dev/null
@@ -1,95 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1d0e8ca0a27c4b0cb17e7318d009efc3241a84a31e3f27c8ca705f6cb5276ce4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02:Natural Soil Deposits and Subsoil Exploration"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.1:Pg-109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2.1\n",
-      "import matplotlib.pyplot as plt\n",
-      "import numpy\n",
-      "import math\n",
-      "Distance=[2.5,5,7.5,10,15,20,25,30,35,40,50];\n",
-      "Time=(10**(-3))*numpy.array([11.2,23.3,33.5,42.4,50.9,57.2,64.4,68.6,71.1,72.1,75.5])\n",
-      "#part1\n",
-      "distance=5.25; # in meter\n",
-      "time=23e-3; # in second\n",
-      "v1=distance/time;\n",
-      "print round(v1,2),\"is speed in m/s\"\n",
-      "#part2\n",
-      "distance=11; # in meter\n",
-      "time=13.5e-3;# in second\n",
-      "v2=distance/time;\n",
-      "print round(v2,2),\" is speed in m/s\"\n",
-      "#part3\n",
-      "distance=14.75;# in meter\n",
-      "time=3.5e-3;# in second\n",
-      "v3=distance/time;\n",
-      "print round(v3,2),\"speed in m/s\"\n",
-      "plt.plot(Distance,Time);\n",
-      "plt.title(\"distance vs time\")\n",
-      "plt.xlabel(\"Distance in m\")\n",
-      "plt.ylabel(\"time in s\")\n",
-      "plt.show()\n",
-      "#part4\n",
-      "xc=10.4;\n",
-      "Ta=65e-3;\n",
-      "Z1=1/2.0*math.sqrt((v2-v1)/(v2+v1))*xc;\n",
-      "print round(Z1,2),\" is thickness of layer 1 in m\"\n",
-      "Z2=1/2.0*(Ta-2*Z1*math.sqrt(v3**2-v1**2)/v3/v1)*v3*v2/math.sqrt(v3**2-v2**2);\n",
-      "print round(Z2,2),\" is thickness of layer 2 in m\"\n",
-      "\n",
-      "# the answers are slightly different in textbook due to approximation while here answers are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "228.26 is speed in m/s\n",
-        "814.81  is speed in m/s\n",
-        "4214.29 speed in m/s\n",
-        "3.9"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  is thickness of layer 1 in m\n",
-        "12.82  is thickness of layer 2 in m\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter02_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter02_1.ipynb
deleted file mode 100755
index f284c073..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter02_1.ipynb
+++ /dev/null
@@ -1,95 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1d0e8ca0a27c4b0cb17e7318d009efc3241a84a31e3f27c8ca705f6cb5276ce4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02:Natural Soil Deposits and Subsoil Exploration"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.1:Pg-109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2.1\n",
-      "import matplotlib.pyplot as plt\n",
-      "import numpy\n",
-      "import math\n",
-      "Distance=[2.5,5,7.5,10,15,20,25,30,35,40,50];\n",
-      "Time=(10**(-3))*numpy.array([11.2,23.3,33.5,42.4,50.9,57.2,64.4,68.6,71.1,72.1,75.5])\n",
-      "#part1\n",
-      "distance=5.25; # in meter\n",
-      "time=23e-3; # in second\n",
-      "v1=distance/time;\n",
-      "print round(v1,2),\"is speed in m/s\"\n",
-      "#part2\n",
-      "distance=11; # in meter\n",
-      "time=13.5e-3;# in second\n",
-      "v2=distance/time;\n",
-      "print round(v2,2),\" is speed in m/s\"\n",
-      "#part3\n",
-      "distance=14.75;# in meter\n",
-      "time=3.5e-3;# in second\n",
-      "v3=distance/time;\n",
-      "print round(v3,2),\"speed in m/s\"\n",
-      "plt.plot(Distance,Time);\n",
-      "plt.title(\"distance vs time\")\n",
-      "plt.xlabel(\"Distance in m\")\n",
-      "plt.ylabel(\"time in s\")\n",
-      "plt.show()\n",
-      "#part4\n",
-      "xc=10.4;\n",
-      "Ta=65e-3;\n",
-      "Z1=1/2.0*math.sqrt((v2-v1)/(v2+v1))*xc;\n",
-      "print round(Z1,2),\" is thickness of layer 1 in m\"\n",
-      "Z2=1/2.0*(Ta-2*Z1*math.sqrt(v3**2-v1**2)/v3/v1)*v3*v2/math.sqrt(v3**2-v2**2);\n",
-      "print round(Z2,2),\" is thickness of layer 2 in m\"\n",
-      "\n",
-      "# the answers are slightly different in textbook due to approximation while here answers are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "228.26 is speed in m/s\n",
-        "814.81  is speed in m/s\n",
-        "4214.29 speed in m/s\n",
-        "3.9"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  is thickness of layer 1 in m\n",
-        "12.82  is thickness of layer 2 in m\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter02_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter02_2.ipynb
deleted file mode 100755
index f284c073..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter02_2.ipynb
+++ /dev/null
@@ -1,95 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1d0e8ca0a27c4b0cb17e7318d009efc3241a84a31e3f27c8ca705f6cb5276ce4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02:Natural Soil Deposits and Subsoil Exploration"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.1:Pg-109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 2.1\n",
-      "import matplotlib.pyplot as plt\n",
-      "import numpy\n",
-      "import math\n",
-      "Distance=[2.5,5,7.5,10,15,20,25,30,35,40,50];\n",
-      "Time=(10**(-3))*numpy.array([11.2,23.3,33.5,42.4,50.9,57.2,64.4,68.6,71.1,72.1,75.5])\n",
-      "#part1\n",
-      "distance=5.25; # in meter\n",
-      "time=23e-3; # in second\n",
-      "v1=distance/time;\n",
-      "print round(v1,2),\"is speed in m/s\"\n",
-      "#part2\n",
-      "distance=11; # in meter\n",
-      "time=13.5e-3;# in second\n",
-      "v2=distance/time;\n",
-      "print round(v2,2),\" is speed in m/s\"\n",
-      "#part3\n",
-      "distance=14.75;# in meter\n",
-      "time=3.5e-3;# in second\n",
-      "v3=distance/time;\n",
-      "print round(v3,2),\"speed in m/s\"\n",
-      "plt.plot(Distance,Time);\n",
-      "plt.title(\"distance vs time\")\n",
-      "plt.xlabel(\"Distance in m\")\n",
-      "plt.ylabel(\"time in s\")\n",
-      "plt.show()\n",
-      "#part4\n",
-      "xc=10.4;\n",
-      "Ta=65e-3;\n",
-      "Z1=1/2.0*math.sqrt((v2-v1)/(v2+v1))*xc;\n",
-      "print round(Z1,2),\" is thickness of layer 1 in m\"\n",
-      "Z2=1/2.0*(Ta-2*Z1*math.sqrt(v3**2-v1**2)/v3/v1)*v3*v2/math.sqrt(v3**2-v2**2);\n",
-      "print round(Z2,2),\" is thickness of layer 2 in m\"\n",
-      "\n",
-      "# the answers are slightly different in textbook due to approximation while here answers are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "228.26 is speed in m/s\n",
-        "814.81  is speed in m/s\n",
-        "4214.29 speed in m/s\n",
-        "3.9"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  is thickness of layer 1 in m\n",
-        "12.82  is thickness of layer 2 in m\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter03.ipynb b/Principles_Of_Foundation_Engineering/Chapter03.ipynb
deleted file mode 100755
index 520b4f21..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter03.ipynb
+++ /dev/null
@@ -1,412 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e3a75199f67af72d14bee528a629ae06b2506206625e1ef3a86291ef88f556ed"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 03:Shallow Foundations: Ultimate bearing capacity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.1:Pg-130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.1\n",
-      "# From Table 3.1\n",
-      "Nc=17.69;\n",
-      "Nq=7.44;\n",
-      "Ny=3.64;\n",
-      "q=3*115;\n",
-      "Gamma=115.0; #lb/ft**3\n",
-      "c=320;\n",
-      "B=5.0;#ft\n",
-      "FS=4;#factor of safety\n",
-      "qu=1.3*c*Nc+q*Nq+0.4*Gamma*B*Ny\n",
-      "qall=qu/FS; #q allowed\n",
-      "Q=qall*B**2;\n",
-      "print Q,\"is allowable gross load in lb\" \n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "67269.0 is allowable gross load in lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.2:Pg-134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.2\n",
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "import math\n",
-      "Gamma=105.0;#lb/ft**3\n",
-      "Gammasat=118.0;#lb/ft**3\n",
-      "FS=3.0;\n",
-      "pa=2014.125;#lb/ft**2\n",
-      "Depth=[5,10,15,20,25]; # in ft\n",
-      "N60=[4,6,6,10,5]; # in  blow/ft\n",
-      "sigmao=[0,0,0,0,0]; # in  lb/ft^2\n",
-      "phi=[0,0,0,0,0] # in degree\n",
-      "Gammaw=62.4;\n",
-      "s=0;\n",
-      "print \"depth (ft)\\tN60\\t  \\tstress(lb/ft**2)\\t phi(degrees)\\n\"\n",
-      "for i in  range(0,5):\n",
-      "    sigmao[i]=2*Gamma+(Depth[i]-2)*(Gammasat-Gammaw);\n",
-      "    phi[i]=math.sqrt(20*N60[i]*math.sqrt(pa/sigmao[i]))+20;\n",
-      "    print \" \",Depth[i],\"\\t      \",N60[i],\"\\t\\t   \",sigmao[i],\"  \\t \\t  \\t\",round(phi[i],1),\" \\n\"\n",
-      "    s=phi[i]+s\n",
-      "\n",
-      "avgphi=s/(i+1)\n",
-      "\n",
-      "print round(avgphi),\"average friction angle in degrees\"\n",
-      "#using graph get the values of other terms in terms of B and solve for B\n",
-      "def f(x):\n",
-      "    return-150000/x**2+5263.9+5527.1/x+228.3*x\n",
-      "x=fsolve(f,4);\n",
-      "print round(x[0],1),\" is the width in ft\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "depth (ft)\tN60\t  \tstress(lb/ft**2)\t phi(degrees)\n",
-        "\n",
-        "  5 \t       4 \t\t    376.8   \t \t  \t33.6  \n",
-        "\n",
-        "  10 \t       6 \t\t    654.8   \t \t  \t34.5  \n",
-        "\n",
-        "  15 \t       6 \t\t    932.8   \t \t  \t33.3  \n",
-        "\n",
-        "  20 \t       10 \t\t    1210.8   \t \t  \t36.1  \n",
-        "\n",
-        "  25 \t       5 \t\t    1488.8   \t \t  \t30.8  \n",
-        "\n",
-        "34.0 average friction angle in degrees\n",
-        "4.5  is the width in ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.3:Pg-144"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.3\n",
-      "\n",
-      "import math\n",
-      "phi=25.0; #degrees\n",
-      "Es=620.0; #kN/m**2\n",
-      "Gamma=18.0;#kN/m**2\n",
-      "Df=0.6;# in m\n",
-      "B=0.6; # in m\n",
-      "L=1.2; # in m\n",
-      "Fqc=0.347;\n",
-      "Nq=10.66;\n",
-      "Nc=20.72;\n",
-      "Ngamma=10.88;\n",
-      "mu=0.3; # Poisson's ratio\n",
-      "Fyd=1.0;\n",
-      "c=48.0;#kN/m**2\n",
-      "q=Gamma*(Df+B/2);\n",
-      "Ir=Es/(2*(1+mu)*(c+q*math.tan(phi*math.pi/180.0)));\n",
-      "print round(Ir,2),\" is value of Ir\"\n",
-      "Fcc=Fqc-(1-Fqc)/(Nq*math.tan(phi*math.pi/180.0));\n",
-      "Fcs=1+Nq/Nc*B/L;\n",
-      "Fqs=1+B/L*math.tan(phi*math.pi/180.0);\n",
-      "Fys=1-0.4*B/L;\n",
-      "Fcd=1+0.4*Df/B;\n",
-      "Fqd=1+2.0*math.tan(phi*math.pi/180.0)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
-      "q1=0.6*18;\n",
-      "Fyc=Fqc;\n",
-      "qu=c*Nc*Fcs*Fcd*Fcc+q1*Nq*Fqs*Fqd*Fqc+1.0/2*Gamma*Ngamma*Fys*Fyd*Fyc;\n",
-      "print round(qu,2),\"is ultimate bearing capacity in kN/m**2\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4.29  is value of Ir\n",
-        "469.24 is ultimate bearing capacity in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.4:Pg-156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.4\n",
-      "import math\n",
-      "q=110*4.0; #lb/ft**2\n",
-      "Nq=33.3;\n",
-      "phi=35.0; # in degree\n",
-      "Df=4.0; # in ft\n",
-      "B=6.0; # in ft\n",
-      "Gamma=110.0;#lb/ft**3\n",
-      "Ngamma=48.03; #lb/ft**3\n",
-      "B1=6-2*0.5; # in ft\n",
-      "Fqi=1;\n",
-      "Fyi=1;\n",
-      "Fyd=1;\n",
-      "Fqs=1;\n",
-      "Fys=1;\n",
-      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
-      "qu=q*Nq*Fqs*Fqd*Fqi+1/2.0*B1*Gamma*Ngamma*Fys*Fyd*Fyi;\n",
-      "Qult=B1*1*qu;\n",
-      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\" \n",
-      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "151738.23  is ultimate bearing capacity in lb/ft\n",
-        "75.87  is ultimate bearing capacity in ton/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.5:Pg-158"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.5\n",
-      "\n",
-      "e=0.5; # in ft\n",
-      "B=6; # in ft\n",
-      "k=e/B;\n",
-      "Gamma=110; # in lb/ft^3 \n",
-      "q=440;\n",
-      "print \"get the values of Nqe and Nye from the figure from the value of e/B\"\n",
-      "Nye=26.8;\n",
-      "Nqe=33.4;\n",
-      "Qult=B*1*(q*Nqe+1/2.0*Gamma*B*Nye);\n",
-      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\"\n",
-      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "get the values of Nqe and Nye from the figure from the value of e/B\n",
-        "141240.0  is ultimate bearing capacity in lb/ft\n",
-        "70.62  is ultimate bearing capacity in ton/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.6:Pg-159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.6\n",
-      "\n",
-      "Df=0.7; # in m\n",
-      "#from table\n",
-      "Nq=18.4;\n",
-      "Ny=22.4;\n",
-      "q=12.6;\n",
-      "phi=30; #angle in degree\n",
-      "L=1.5;# in m\n",
-      "Fyd=1;\n",
-      "Gamma=18; # in KN/m^3\n",
-      "L1=0.85*1.5; # in m\n",
-      "L2=0.21*1.5; # in m\n",
-      "B=1.5; # in m\n",
-      "A=1/2.0*(L1+L2)*B;\n",
-      "B1=A/L1; #B'\n",
-      "Fqs=1+B1/L1*math.tan(phi*math.pi/180);\n",
-      "Fys=1-0.4*B1/L1;\n",
-      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180))**2*Df/B;\n",
-      "Qult=A*(q*Nq*Fqs*Fqd+1/2.0*Gamma*B1*Ny*Fys*Fyd);\n",
-      "print round(Qult,2),\" is ultimate load in kN\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "605.45  is ultimate load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.7:Pg-161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.7\n",
-      "\n",
-      "e=0.15; # in m\n",
-      "B=1.5; # in m\n",
-      "Fqs=1.0;\n",
-      "L=1.5;# in m\n",
-      "Gamma=18.0; # in KN/m^3\n",
-      "q=0.7*18;\n",
-      "#from table\n",
-      "Nqe=18.4;\n",
-      "Nye=11.58;\n",
-      "Fys=1+(2*e/B-0.68)*(B/L)+(0.43-3/2.0*e/B)*(B/L)**2;\n",
-      "Qult=B*L*(q*Nqe*Fqs+1/2.0*L*Gamma*Nye*Fys);\n",
-      "print round(Qult,2),\"is ultimate load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "803.03 is ultimate load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.8:Pg-163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.8\n",
-      "\n",
-      "q=16.0;# in kN/m^2\n",
-      "Nqei=14.2;\n",
-      "Gamma=16.0 # in kN/m^3\n",
-      "B=1.5;# in m\n",
-      "Nyet=20.0;\n",
-      "Qult=B*(Nqei*q+1/2.0*Gamma*B*Nyet);\n",
-      "print round(Qult,2),\" is ultimate load in kN/m\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "700.8  is ultimate load in kN/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter03_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter03_1.ipynb
deleted file mode 100755
index 520b4f21..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter03_1.ipynb
+++ /dev/null
@@ -1,412 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e3a75199f67af72d14bee528a629ae06b2506206625e1ef3a86291ef88f556ed"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 03:Shallow Foundations: Ultimate bearing capacity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.1:Pg-130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.1\n",
-      "# From Table 3.1\n",
-      "Nc=17.69;\n",
-      "Nq=7.44;\n",
-      "Ny=3.64;\n",
-      "q=3*115;\n",
-      "Gamma=115.0; #lb/ft**3\n",
-      "c=320;\n",
-      "B=5.0;#ft\n",
-      "FS=4;#factor of safety\n",
-      "qu=1.3*c*Nc+q*Nq+0.4*Gamma*B*Ny\n",
-      "qall=qu/FS; #q allowed\n",
-      "Q=qall*B**2;\n",
-      "print Q,\"is allowable gross load in lb\" \n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "67269.0 is allowable gross load in lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.2:Pg-134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.2\n",
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "import math\n",
-      "Gamma=105.0;#lb/ft**3\n",
-      "Gammasat=118.0;#lb/ft**3\n",
-      "FS=3.0;\n",
-      "pa=2014.125;#lb/ft**2\n",
-      "Depth=[5,10,15,20,25]; # in ft\n",
-      "N60=[4,6,6,10,5]; # in  blow/ft\n",
-      "sigmao=[0,0,0,0,0]; # in  lb/ft^2\n",
-      "phi=[0,0,0,0,0] # in degree\n",
-      "Gammaw=62.4;\n",
-      "s=0;\n",
-      "print \"depth (ft)\\tN60\\t  \\tstress(lb/ft**2)\\t phi(degrees)\\n\"\n",
-      "for i in  range(0,5):\n",
-      "    sigmao[i]=2*Gamma+(Depth[i]-2)*(Gammasat-Gammaw);\n",
-      "    phi[i]=math.sqrt(20*N60[i]*math.sqrt(pa/sigmao[i]))+20;\n",
-      "    print \" \",Depth[i],\"\\t      \",N60[i],\"\\t\\t   \",sigmao[i],\"  \\t \\t  \\t\",round(phi[i],1),\" \\n\"\n",
-      "    s=phi[i]+s\n",
-      "\n",
-      "avgphi=s/(i+1)\n",
-      "\n",
-      "print round(avgphi),\"average friction angle in degrees\"\n",
-      "#using graph get the values of other terms in terms of B and solve for B\n",
-      "def f(x):\n",
-      "    return-150000/x**2+5263.9+5527.1/x+228.3*x\n",
-      "x=fsolve(f,4);\n",
-      "print round(x[0],1),\" is the width in ft\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "depth (ft)\tN60\t  \tstress(lb/ft**2)\t phi(degrees)\n",
-        "\n",
-        "  5 \t       4 \t\t    376.8   \t \t  \t33.6  \n",
-        "\n",
-        "  10 \t       6 \t\t    654.8   \t \t  \t34.5  \n",
-        "\n",
-        "  15 \t       6 \t\t    932.8   \t \t  \t33.3  \n",
-        "\n",
-        "  20 \t       10 \t\t    1210.8   \t \t  \t36.1  \n",
-        "\n",
-        "  25 \t       5 \t\t    1488.8   \t \t  \t30.8  \n",
-        "\n",
-        "34.0 average friction angle in degrees\n",
-        "4.5  is the width in ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.3:Pg-144"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.3\n",
-      "\n",
-      "import math\n",
-      "phi=25.0; #degrees\n",
-      "Es=620.0; #kN/m**2\n",
-      "Gamma=18.0;#kN/m**2\n",
-      "Df=0.6;# in m\n",
-      "B=0.6; # in m\n",
-      "L=1.2; # in m\n",
-      "Fqc=0.347;\n",
-      "Nq=10.66;\n",
-      "Nc=20.72;\n",
-      "Ngamma=10.88;\n",
-      "mu=0.3; # Poisson's ratio\n",
-      "Fyd=1.0;\n",
-      "c=48.0;#kN/m**2\n",
-      "q=Gamma*(Df+B/2);\n",
-      "Ir=Es/(2*(1+mu)*(c+q*math.tan(phi*math.pi/180.0)));\n",
-      "print round(Ir,2),\" is value of Ir\"\n",
-      "Fcc=Fqc-(1-Fqc)/(Nq*math.tan(phi*math.pi/180.0));\n",
-      "Fcs=1+Nq/Nc*B/L;\n",
-      "Fqs=1+B/L*math.tan(phi*math.pi/180.0);\n",
-      "Fys=1-0.4*B/L;\n",
-      "Fcd=1+0.4*Df/B;\n",
-      "Fqd=1+2.0*math.tan(phi*math.pi/180.0)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
-      "q1=0.6*18;\n",
-      "Fyc=Fqc;\n",
-      "qu=c*Nc*Fcs*Fcd*Fcc+q1*Nq*Fqs*Fqd*Fqc+1.0/2*Gamma*Ngamma*Fys*Fyd*Fyc;\n",
-      "print round(qu,2),\"is ultimate bearing capacity in kN/m**2\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4.29  is value of Ir\n",
-        "469.24 is ultimate bearing capacity in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.4:Pg-156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.4\n",
-      "import math\n",
-      "q=110*4.0; #lb/ft**2\n",
-      "Nq=33.3;\n",
-      "phi=35.0; # in degree\n",
-      "Df=4.0; # in ft\n",
-      "B=6.0; # in ft\n",
-      "Gamma=110.0;#lb/ft**3\n",
-      "Ngamma=48.03; #lb/ft**3\n",
-      "B1=6-2*0.5; # in ft\n",
-      "Fqi=1;\n",
-      "Fyi=1;\n",
-      "Fyd=1;\n",
-      "Fqs=1;\n",
-      "Fys=1;\n",
-      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
-      "qu=q*Nq*Fqs*Fqd*Fqi+1/2.0*B1*Gamma*Ngamma*Fys*Fyd*Fyi;\n",
-      "Qult=B1*1*qu;\n",
-      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\" \n",
-      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "151738.23  is ultimate bearing capacity in lb/ft\n",
-        "75.87  is ultimate bearing capacity in ton/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.5:Pg-158"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.5\n",
-      "\n",
-      "e=0.5; # in ft\n",
-      "B=6; # in ft\n",
-      "k=e/B;\n",
-      "Gamma=110; # in lb/ft^3 \n",
-      "q=440;\n",
-      "print \"get the values of Nqe and Nye from the figure from the value of e/B\"\n",
-      "Nye=26.8;\n",
-      "Nqe=33.4;\n",
-      "Qult=B*1*(q*Nqe+1/2.0*Gamma*B*Nye);\n",
-      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\"\n",
-      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "get the values of Nqe and Nye from the figure from the value of e/B\n",
-        "141240.0  is ultimate bearing capacity in lb/ft\n",
-        "70.62  is ultimate bearing capacity in ton/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.6:Pg-159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.6\n",
-      "\n",
-      "Df=0.7; # in m\n",
-      "#from table\n",
-      "Nq=18.4;\n",
-      "Ny=22.4;\n",
-      "q=12.6;\n",
-      "phi=30; #angle in degree\n",
-      "L=1.5;# in m\n",
-      "Fyd=1;\n",
-      "Gamma=18; # in KN/m^3\n",
-      "L1=0.85*1.5; # in m\n",
-      "L2=0.21*1.5; # in m\n",
-      "B=1.5; # in m\n",
-      "A=1/2.0*(L1+L2)*B;\n",
-      "B1=A/L1; #B'\n",
-      "Fqs=1+B1/L1*math.tan(phi*math.pi/180);\n",
-      "Fys=1-0.4*B1/L1;\n",
-      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180))**2*Df/B;\n",
-      "Qult=A*(q*Nq*Fqs*Fqd+1/2.0*Gamma*B1*Ny*Fys*Fyd);\n",
-      "print round(Qult,2),\" is ultimate load in kN\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "605.45  is ultimate load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.7:Pg-161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.7\n",
-      "\n",
-      "e=0.15; # in m\n",
-      "B=1.5; # in m\n",
-      "Fqs=1.0;\n",
-      "L=1.5;# in m\n",
-      "Gamma=18.0; # in KN/m^3\n",
-      "q=0.7*18;\n",
-      "#from table\n",
-      "Nqe=18.4;\n",
-      "Nye=11.58;\n",
-      "Fys=1+(2*e/B-0.68)*(B/L)+(0.43-3/2.0*e/B)*(B/L)**2;\n",
-      "Qult=B*L*(q*Nqe*Fqs+1/2.0*L*Gamma*Nye*Fys);\n",
-      "print round(Qult,2),\"is ultimate load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "803.03 is ultimate load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.8:Pg-163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.8\n",
-      "\n",
-      "q=16.0;# in kN/m^2\n",
-      "Nqei=14.2;\n",
-      "Gamma=16.0 # in kN/m^3\n",
-      "B=1.5;# in m\n",
-      "Nyet=20.0;\n",
-      "Qult=B*(Nqei*q+1/2.0*Gamma*B*Nyet);\n",
-      "print round(Qult,2),\" is ultimate load in kN/m\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "700.8  is ultimate load in kN/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter03_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter03_2.ipynb
deleted file mode 100755
index 520b4f21..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter03_2.ipynb
+++ /dev/null
@@ -1,412 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e3a75199f67af72d14bee528a629ae06b2506206625e1ef3a86291ef88f556ed"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 03:Shallow Foundations: Ultimate bearing capacity"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.1:Pg-130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.1\n",
-      "# From Table 3.1\n",
-      "Nc=17.69;\n",
-      "Nq=7.44;\n",
-      "Ny=3.64;\n",
-      "q=3*115;\n",
-      "Gamma=115.0; #lb/ft**3\n",
-      "c=320;\n",
-      "B=5.0;#ft\n",
-      "FS=4;#factor of safety\n",
-      "qu=1.3*c*Nc+q*Nq+0.4*Gamma*B*Ny\n",
-      "qall=qu/FS; #q allowed\n",
-      "Q=qall*B**2;\n",
-      "print Q,\"is allowable gross load in lb\" \n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "67269.0 is allowable gross load in lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.2:Pg-134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.2\n",
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "import math\n",
-      "Gamma=105.0;#lb/ft**3\n",
-      "Gammasat=118.0;#lb/ft**3\n",
-      "FS=3.0;\n",
-      "pa=2014.125;#lb/ft**2\n",
-      "Depth=[5,10,15,20,25]; # in ft\n",
-      "N60=[4,6,6,10,5]; # in  blow/ft\n",
-      "sigmao=[0,0,0,0,0]; # in  lb/ft^2\n",
-      "phi=[0,0,0,0,0] # in degree\n",
-      "Gammaw=62.4;\n",
-      "s=0;\n",
-      "print \"depth (ft)\\tN60\\t  \\tstress(lb/ft**2)\\t phi(degrees)\\n\"\n",
-      "for i in  range(0,5):\n",
-      "    sigmao[i]=2*Gamma+(Depth[i]-2)*(Gammasat-Gammaw);\n",
-      "    phi[i]=math.sqrt(20*N60[i]*math.sqrt(pa/sigmao[i]))+20;\n",
-      "    print \" \",Depth[i],\"\\t      \",N60[i],\"\\t\\t   \",sigmao[i],\"  \\t \\t  \\t\",round(phi[i],1),\" \\n\"\n",
-      "    s=phi[i]+s\n",
-      "\n",
-      "avgphi=s/(i+1)\n",
-      "\n",
-      "print round(avgphi),\"average friction angle in degrees\"\n",
-      "#using graph get the values of other terms in terms of B and solve for B\n",
-      "def f(x):\n",
-      "    return-150000/x**2+5263.9+5527.1/x+228.3*x\n",
-      "x=fsolve(f,4);\n",
-      "print round(x[0],1),\" is the width in ft\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "depth (ft)\tN60\t  \tstress(lb/ft**2)\t phi(degrees)\n",
-        "\n",
-        "  5 \t       4 \t\t    376.8   \t \t  \t33.6  \n",
-        "\n",
-        "  10 \t       6 \t\t    654.8   \t \t  \t34.5  \n",
-        "\n",
-        "  15 \t       6 \t\t    932.8   \t \t  \t33.3  \n",
-        "\n",
-        "  20 \t       10 \t\t    1210.8   \t \t  \t36.1  \n",
-        "\n",
-        "  25 \t       5 \t\t    1488.8   \t \t  \t30.8  \n",
-        "\n",
-        "34.0 average friction angle in degrees\n",
-        "4.5  is the width in ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.3:Pg-144"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.3\n",
-      "\n",
-      "import math\n",
-      "phi=25.0; #degrees\n",
-      "Es=620.0; #kN/m**2\n",
-      "Gamma=18.0;#kN/m**2\n",
-      "Df=0.6;# in m\n",
-      "B=0.6; # in m\n",
-      "L=1.2; # in m\n",
-      "Fqc=0.347;\n",
-      "Nq=10.66;\n",
-      "Nc=20.72;\n",
-      "Ngamma=10.88;\n",
-      "mu=0.3; # Poisson's ratio\n",
-      "Fyd=1.0;\n",
-      "c=48.0;#kN/m**2\n",
-      "q=Gamma*(Df+B/2);\n",
-      "Ir=Es/(2*(1+mu)*(c+q*math.tan(phi*math.pi/180.0)));\n",
-      "print round(Ir,2),\" is value of Ir\"\n",
-      "Fcc=Fqc-(1-Fqc)/(Nq*math.tan(phi*math.pi/180.0));\n",
-      "Fcs=1+Nq/Nc*B/L;\n",
-      "Fqs=1+B/L*math.tan(phi*math.pi/180.0);\n",
-      "Fys=1-0.4*B/L;\n",
-      "Fcd=1+0.4*Df/B;\n",
-      "Fqd=1+2.0*math.tan(phi*math.pi/180.0)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
-      "q1=0.6*18;\n",
-      "Fyc=Fqc;\n",
-      "qu=c*Nc*Fcs*Fcd*Fcc+q1*Nq*Fqs*Fqd*Fqc+1.0/2*Gamma*Ngamma*Fys*Fyd*Fyc;\n",
-      "print round(qu,2),\"is ultimate bearing capacity in kN/m**2\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4.29  is value of Ir\n",
-        "469.24 is ultimate bearing capacity in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.4:Pg-156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.4\n",
-      "import math\n",
-      "q=110*4.0; #lb/ft**2\n",
-      "Nq=33.3;\n",
-      "phi=35.0; # in degree\n",
-      "Df=4.0; # in ft\n",
-      "B=6.0; # in ft\n",
-      "Gamma=110.0;#lb/ft**3\n",
-      "Ngamma=48.03; #lb/ft**3\n",
-      "B1=6-2*0.5; # in ft\n",
-      "Fqi=1;\n",
-      "Fyi=1;\n",
-      "Fyd=1;\n",
-      "Fqs=1;\n",
-      "Fys=1;\n",
-      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
-      "qu=q*Nq*Fqs*Fqd*Fqi+1/2.0*B1*Gamma*Ngamma*Fys*Fyd*Fyi;\n",
-      "Qult=B1*1*qu;\n",
-      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\" \n",
-      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "151738.23  is ultimate bearing capacity in lb/ft\n",
-        "75.87  is ultimate bearing capacity in ton/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.5:Pg-158"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.5\n",
-      "\n",
-      "e=0.5; # in ft\n",
-      "B=6; # in ft\n",
-      "k=e/B;\n",
-      "Gamma=110; # in lb/ft^3 \n",
-      "q=440;\n",
-      "print \"get the values of Nqe and Nye from the figure from the value of e/B\"\n",
-      "Nye=26.8;\n",
-      "Nqe=33.4;\n",
-      "Qult=B*1*(q*Nqe+1/2.0*Gamma*B*Nye);\n",
-      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\"\n",
-      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "get the values of Nqe and Nye from the figure from the value of e/B\n",
-        "141240.0  is ultimate bearing capacity in lb/ft\n",
-        "70.62  is ultimate bearing capacity in ton/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.6:Pg-159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.6\n",
-      "\n",
-      "Df=0.7; # in m\n",
-      "#from table\n",
-      "Nq=18.4;\n",
-      "Ny=22.4;\n",
-      "q=12.6;\n",
-      "phi=30; #angle in degree\n",
-      "L=1.5;# in m\n",
-      "Fyd=1;\n",
-      "Gamma=18; # in KN/m^3\n",
-      "L1=0.85*1.5; # in m\n",
-      "L2=0.21*1.5; # in m\n",
-      "B=1.5; # in m\n",
-      "A=1/2.0*(L1+L2)*B;\n",
-      "B1=A/L1; #B'\n",
-      "Fqs=1+B1/L1*math.tan(phi*math.pi/180);\n",
-      "Fys=1-0.4*B1/L1;\n",
-      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180))**2*Df/B;\n",
-      "Qult=A*(q*Nq*Fqs*Fqd+1/2.0*Gamma*B1*Ny*Fys*Fyd);\n",
-      "print round(Qult,2),\" is ultimate load in kN\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation but here answer are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "605.45  is ultimate load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.7:Pg-161"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.7\n",
-      "\n",
-      "e=0.15; # in m\n",
-      "B=1.5; # in m\n",
-      "Fqs=1.0;\n",
-      "L=1.5;# in m\n",
-      "Gamma=18.0; # in KN/m^3\n",
-      "q=0.7*18;\n",
-      "#from table\n",
-      "Nqe=18.4;\n",
-      "Nye=11.58;\n",
-      "Fys=1+(2*e/B-0.68)*(B/L)+(0.43-3/2.0*e/B)*(B/L)**2;\n",
-      "Qult=B*L*(q*Nqe*Fqs+1/2.0*L*Gamma*Nye*Fys);\n",
-      "print round(Qult,2),\"is ultimate load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "803.03 is ultimate load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.8:Pg-163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 3.8\n",
-      "\n",
-      "q=16.0;# in kN/m^2\n",
-      "Nqei=14.2;\n",
-      "Gamma=16.0 # in kN/m^3\n",
-      "B=1.5;# in m\n",
-      "Nyet=20.0;\n",
-      "Qult=B*(Nqei*q+1/2.0*Gamma*B*Nyet);\n",
-      "print round(Qult,2),\" is ultimate load in kN/m\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "700.8  is ultimate load in kN/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter04.ipynb b/Principles_Of_Foundation_Engineering/Chapter04.ipynb
deleted file mode 100755
index da8b3fb8..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter04.ipynb
+++ /dev/null
@@ -1,365 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:84b2322c895fb47191307e7a54e3a6dc70036edd44f1781ba096e1e048c5f743"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 04:Ultimate Bearing Capacity of Shallow Foundations: Special Cases"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.1:Pg-176"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4.1\n",
-      "\n",
-      "FS=4.0; # FOS\n",
-      "q=110*2.0; # in 1b/ft^2\n",
-      "Nq=90.0;\n",
-      "Ny=50.0;\n",
-      "Gamma=110.0; # in 1b/ft^3\n",
-      "m1=0.34; # From Figure 4.6(a)\n",
-      "B=2.5; # in ft\n",
-      "L=2.5; # in ft\n",
-      "H=1.5; # in ft\n",
-      "phi=35; # in degree\n",
-      "m2=0.45; # From Figure 4.6(b)\n",
-      "Fqs=1-0.34*B/L;\n",
-      "Fys=1-0.45*B/L;\n",
-      "qu=q*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",
-      "Qall=qu*B**2/FS;\n",
-      "print round(Qall,2),\"bearing load in lb\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "26326.95 bearing load in lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.2:Pg-177"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4.2\n",
-      "\n",
-      "FS=3.0; # FOS\n",
-      "cu=72.0;\n",
-      "q=18.0; # in kN/m^3\n",
-      "B=1.0;# in m\n",
-      "H=0.25;# in m\n",
-      "qu=5.14*(1+(0.5*B/H-0.707)/5.14)*cu+q;\n",
-      "qall=qu/FS;\n",
-      "print round(qall,1),\"bearing capacity of soil in kN/m**2\" \n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "160.4 bearing capacity of soil in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.3:Pg-183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4.3\n",
-      "import math\n",
-      "k=0; #B/L;\n",
-      "c2=30;\n",
-      "Gamma=17.5; # in kN/m^3\n",
-      "H=1.5; # in m\n",
-      "Df=1.2; # in m\n",
-      "B=2.0; # in m\n",
-      "Ks=2.5;\n",
-      "phi=40; # in degree\n",
-      "pi=math.pi;\n",
-      "qu=(1+0.2*k)*5.14*c2+(1+k)*Gamma*H**2*(1+2*Df/H)*Ks*math.tan(phi*pi/180)/B+Gamma*H;\n",
-      "Qu=qu*B;\n",
-      "print round(Qu,2),\"is bearing capacity in kN/m\"\n",
-      "print \"there is slight variation due to rounding off error\"\n",
-      "#soil 2\n",
-      "Ny=109.4;\n",
-      "Nq=64.2;\n",
-      "Fqs=1;\n",
-      "Fys=1;\n",
-      "qt=Gamma*Df*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",
-      "print qt,\"bearing capacity in kN/m**2\"\n",
-      "\n",
-      "# answer in book is different due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "575.66 is bearing capacity in kN/m\n",
-        "there is slight variation due to rounding off error\n",
-        "3262.7 bearing capacity in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.4:Pg-184"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4.4\n",
-      "\n",
-      "B=1.0;  # in m\n",
-      "L=1.5;# in m\n",
-      "c2=48;# in m\n",
-      "ca=108; # in KN/m^2\n",
-      "D=1.0;# in m\n",
-      "H=1.0;# in m\n",
-      "Gamma=16.8; # in KN/m^3\n",
-      "FS=4;\n",
-      "qu=(1+0.2*B/L)*5.14*c2+(1+B/L)*2*ca*H/B+Gamma*D; # in KN/m^2\n",
-      "c1=120.0;\n",
-      "gamma1=16.8;  # in kN/m^3\n",
-      "Df=1.0;\n",
-      "qt=(1+0.2*B/L)*5.14*c1+gamma1*Df;\n",
-      "print qt,\"is qt in kN/m**2\"\n",
-      "print \"no need to calculate qt since it is not useful for calculation\"\n",
-      "print qu/FS,\"is allowable shear stress in kN/m**2\"\n",
-      "print qu/FS*1*1.5,\" is allowable load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "715.84 is qt in kN/m**2\n",
-        "no need to calculate qt since it is not useful for calculation\n",
-        "164.104 is allowable shear stress in kN/m**2\n",
-        "246.156  is allowable load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.5:Pg-190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4.5\n",
-      "\n",
-      "c=50;  # in KN/m^2\n",
-      "#from table\n",
-      "Ncq=6.3;\n",
-      "FS=4.0;# FOS\n",
-      "qu=c*Ncq;  # in KN/m^2\n",
-      "qall=qu/4;\n",
-      "print qall,\"allowed shear stress in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "78.75 allowed shear stress in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.6:Pg-191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4.6\n",
-      "\n",
-      "Gamma=16.8; # in kN/m^3\n",
-      "B=1.5;#  in m\n",
-      "#from table\n",
-      "Nyq=120.0;\n",
-      "qu=1/2.0*Gamma*B*Nyq; # in KN/m^2\n",
-      "print qu,\" is shear stress in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1512.0  shear stress in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.7:Pg-198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 4.7\n",
-      "import math\n",
-      "phi=35; # in degree\n",
-      "Df=1.5; # in m\n",
-      "B=1.5; # in m\n",
-      "Gamma=17.4; # in kN/m^3\n",
-      "A=math.pi/4*Df**2; # in m^2\n",
-      "m=0.25;\n",
-      "Ku=0.936;\n",
-      "Fq=1+2*(1+m*Df/B)*Df/B*Ku*math.tan(phi*math.pi/180);\n",
-      "Qu=Fq*Gamma*A*Df; # in KN/m^2\n",
-      "print round(Qu,1),\" is bearing capacity in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "121.7  is bearing capacity in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#example 4.8\n",
-      "\n",
-      "\n",
-      "\n",
-      "cu=52; # in kN/m^2\n",
-      "\n",
-      "B=1.5; # in m\n",
-      "\n",
-      "L=3; # in m\n",
-      "\n",
-      "k=0.107*cu+2.5;\n",
-      "\n",
-      "print round(k,2),\" is Df/B of square\" \n",
-      "\n",
-      "A=L*B; # in m^2\n",
-      "\n",
-      "Beta=0.2;\n",
-      "\n",
-      "Gamma=18.9; # in kN/m^3\n",
-      "\n",
-      "Df=1.8; # in m\n",
-      "\n",
-      "Qu=A*(Beta*(7.56+1.44*B/L)*cu+Gamma*Df); # in kN/m^2\n",
-      "\n",
-      "print round(Qu,1),\" is ultimate shear force in kN\"\n",
-      "\n",
-      " \n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8.06  is Df/B of square\n",
-        "540.6  is ultimate shear force in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter04_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter04_1.ipynb
deleted file mode 100755
index 1d75ed9f..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter04_1.ipynb
+++ /dev/null
@@ -1,373 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:7e16b5ec29439d0d7eaa54a0826b64a36c559ed79f17d130c42f4478805c682f"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 04:Ultimate Bearing Capacity of Shallow Foundations: Special Cases"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.1:Pg-176"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.1\n",

-      "\n",

-      "FS=4.0; # FOS\n",

-      "q=110*2.0; # in 1b/ft^2\n",

-      "Nq=90.0;\n",

-      "Ny=50.0;\n",

-      "Gamma=110.0; # in 1b/ft^3\n",

-      "m1=0.34; # From Figure 4.6(a)\n",

-      "B=2.5; # in ft\n",

-      "L=2.5; # in ft\n",

-      "H=1.5; # in ft\n",

-      "phi=35; # in degree\n",

-      "m2=0.45; # From Figure 4.6(b)\n",

-      "Fqs=1-0.34*B/L;\n",

-      "Fys=1-0.45*B/L;\n",

-      "qu=q*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",

-      "Qall=qu*B**2/FS;\n",

-      "print round(Qall,2),\"bearing load in lb\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "26326.95 bearing load in lb\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.2:Pg-177"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.2\n",

-      "\n",

-      "FS=3.0; # FOS\n",

-      "cu=72.0;\n",

-      "q=18.0; # in kN/m^3\n",

-      "B=1.0;# in m\n",

-      "H=0.25;# in m\n",

-      "qu=5.14*(1+(0.5*B/H-0.707)/5.14)*cu+q;\n",

-      "qall=qu/FS;\n",

-      "print round(qall,1),\"bearing capacity of soil in kN/m**2\" \n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "160.4 bearing capacity of soil in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.3:Pg-183"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.3\n",

-      "import math\n",

-      "k=0; #B/L;\n",

-      "c2=30;\n",

-      "Gamma=17.5; # in kN/m^3\n",

-      "H=1.5; # in m\n",

-      "Df=1.2; # in m\n",

-      "B=2.0; # in m\n",

-      "Ks=2.5;\n",

-      "phi=40; # in degree\n",

-      "pi=math.pi;\n",

-      "qu=(1+0.2*k)*5.14*c2+(1+k)*Gamma*H**2*(1+2*Df/H)*Ks*math.tan(phi*pi/180)/B+Gamma*H;\n",

-      "Qu=qu*B;\n",

-      "print round(Qu,2),\"is bearing capacity in kN/m\"\n",

-      "print \"there is slight variation due to rounding off error\"\n",

-      "#soil 2\n",

-      "Ny=109.4;\n",

-      "Nq=64.2;\n",

-      "Fqs=1;\n",

-      "Fys=1;\n",

-      "qt=Gamma*Df*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",

-      "print qt,\"bearing capacity in kN/m**2\"\n",

-      "\n",

-      "# answer in book is different due to approximation"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "575.66 is bearing capacity in kN/m\n",

-        "there is slight variation due to rounding off error\n",

-        "3262.7 bearing capacity in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.4:Pg-184"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.4\n",

-      "\n",

-      "B=1.0;  # in m\n",

-      "L=1.5;# in m\n",

-      "c2=48;# in m\n",

-      "ca=108; # in KN/m^2\n",

-      "D=1.0;# in m\n",

-      "H=1.0;# in m\n",

-      "Gamma=16.8; # in KN/m^3\n",

-      "FS=4;\n",

-      "qu=(1+0.2*B/L)*5.14*c2+(1+B/L)*2*ca*H/B+Gamma*D; # in KN/m^2\n",

-      "c1=120.0;\n",

-      "gamma1=16.8;  # in kN/m^3\n",

-      "Df=1.0;\n",

-      "qt=(1+0.2*B/L)*5.14*c1+gamma1*Df;\n",

-      "print qt,\"is qt in kN/m**2\"\n",

-      "print \"no need to calculate qt since it is not useful for calculation\"\n",

-      "print qu/FS,\"is allowable shear stress in kN/m**2\"\n",

-      "print qu/FS*1*1.5,\" is allowable load in kN\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "715.84 is qt in kN/m**2\n",

-        "no need to calculate qt since it is not useful for calculation\n",

-        "164.104 is allowable shear stress in kN/m**2\n",

-        "246.156  is allowable load in kN\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.5:Pg-190"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.5\n",

-      "\n",

-      "c=50;  # in KN/m^2\n",

-      "#from table\n",

-      "Ncq=6.3;\n",

-      "FS=4.0;# FOS\n",

-      "qu=c*Ncq;  # in KN/m^2\n",

-      "qall=qu/4;\n",

-      "print qall,\"allowed shear stress in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "78.75 allowed shear stress in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.6:Pg-191"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.6\n",

-      "\n",

-      "Gamma=16.8; # in kN/m^3\n",

-      "B=1.5;#  in m\n",

-      "#from table\n",

-      "Nyq=120.0;\n",

-      "qu=1/2.0*Gamma*B*Nyq; # in KN/m^2\n",

-      "print qu,\" is shear stress in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "1512.0  shear stress in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.7:Pg-198"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.7\n",

-      "import math\n",

-      "phi=35; # in degree\n",

-      "Df=1.5; # in m\n",

-      "B=1.5; # in m\n",

-      "Gamma=17.4; # in kN/m^3\n",

-      "A=math.pi/4*Df**2; # in m^2\n",

-      "m=0.25;\n",

-      "Ku=0.936;\n",

-      "Fq=1+2*(1+m*Df/B)*Df/B*Ku*math.tan(phi*math.pi/180);\n",

-      "Qu=Fq*Gamma*A*Df; # in KN/m^2\n",

-      "print round(Qu,1),\" is bearing capacity in kN\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "121.7  is bearing capacity in kN\n"

-       ]

-      }

-     ],

-     "prompt_number": 17

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.8:pg-198"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "\n",

-      "#example 4.8\n",

-      "\n",

-      "\n",

-      "\n",

-      "cu=52; # in kN/m^2\n",

-      "\n",

-      "B=1.5; # in m\n",

-      "\n",

-      "L=3; # in m\n",

-      "\n",

-      "k=0.107*cu+2.5;\n",

-      "\n",

-      "print round(k,2),\" is Df/B of square\" \n",

-      "\n",

-      "A=L*B; # in m^2\n",

-      "\n",

-      "Beta=0.2;\n",

-      "\n",

-      "Gamma=18.9; # in kN/m^3\n",

-      "\n",

-      "Df=1.8; # in m\n",

-      "\n",

-      "Qu=A*(Beta*(7.56+1.44*B/L)*cu+Gamma*Df); # in kN/m^2\n",

-      "\n",

-      "print round(Qu,1),\" is ultimate shear force in kN\"\n",

-      "\n",

-      " \n",

-      "\n",

-      " "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "8.06  is Df/B of square\n",

-        "540.6  is ultimate shear force in kN\n"

-       ]

-      }

-     ],

-     "prompt_number": 20

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter04_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter04_2.ipynb
deleted file mode 100755
index 1d75ed9f..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter04_2.ipynb
+++ /dev/null
@@ -1,373 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:7e16b5ec29439d0d7eaa54a0826b64a36c559ed79f17d130c42f4478805c682f"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 04:Ultimate Bearing Capacity of Shallow Foundations: Special Cases"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.1:Pg-176"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.1\n",

-      "\n",

-      "FS=4.0; # FOS\n",

-      "q=110*2.0; # in 1b/ft^2\n",

-      "Nq=90.0;\n",

-      "Ny=50.0;\n",

-      "Gamma=110.0; # in 1b/ft^3\n",

-      "m1=0.34; # From Figure 4.6(a)\n",

-      "B=2.5; # in ft\n",

-      "L=2.5; # in ft\n",

-      "H=1.5; # in ft\n",

-      "phi=35; # in degree\n",

-      "m2=0.45; # From Figure 4.6(b)\n",

-      "Fqs=1-0.34*B/L;\n",

-      "Fys=1-0.45*B/L;\n",

-      "qu=q*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",

-      "Qall=qu*B**2/FS;\n",

-      "print round(Qall,2),\"bearing load in lb\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "26326.95 bearing load in lb\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.2:Pg-177"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.2\n",

-      "\n",

-      "FS=3.0; # FOS\n",

-      "cu=72.0;\n",

-      "q=18.0; # in kN/m^3\n",

-      "B=1.0;# in m\n",

-      "H=0.25;# in m\n",

-      "qu=5.14*(1+(0.5*B/H-0.707)/5.14)*cu+q;\n",

-      "qall=qu/FS;\n",

-      "print round(qall,1),\"bearing capacity of soil in kN/m**2\" \n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "160.4 bearing capacity of soil in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.3:Pg-183"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.3\n",

-      "import math\n",

-      "k=0; #B/L;\n",

-      "c2=30;\n",

-      "Gamma=17.5; # in kN/m^3\n",

-      "H=1.5; # in m\n",

-      "Df=1.2; # in m\n",

-      "B=2.0; # in m\n",

-      "Ks=2.5;\n",

-      "phi=40; # in degree\n",

-      "pi=math.pi;\n",

-      "qu=(1+0.2*k)*5.14*c2+(1+k)*Gamma*H**2*(1+2*Df/H)*Ks*math.tan(phi*pi/180)/B+Gamma*H;\n",

-      "Qu=qu*B;\n",

-      "print round(Qu,2),\"is bearing capacity in kN/m\"\n",

-      "print \"there is slight variation due to rounding off error\"\n",

-      "#soil 2\n",

-      "Ny=109.4;\n",

-      "Nq=64.2;\n",

-      "Fqs=1;\n",

-      "Fys=1;\n",

-      "qt=Gamma*Df*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",

-      "print qt,\"bearing capacity in kN/m**2\"\n",

-      "\n",

-      "# answer in book is different due to approximation"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "575.66 is bearing capacity in kN/m\n",

-        "there is slight variation due to rounding off error\n",

-        "3262.7 bearing capacity in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.4:Pg-184"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.4\n",

-      "\n",

-      "B=1.0;  # in m\n",

-      "L=1.5;# in m\n",

-      "c2=48;# in m\n",

-      "ca=108; # in KN/m^2\n",

-      "D=1.0;# in m\n",

-      "H=1.0;# in m\n",

-      "Gamma=16.8; # in KN/m^3\n",

-      "FS=4;\n",

-      "qu=(1+0.2*B/L)*5.14*c2+(1+B/L)*2*ca*H/B+Gamma*D; # in KN/m^2\n",

-      "c1=120.0;\n",

-      "gamma1=16.8;  # in kN/m^3\n",

-      "Df=1.0;\n",

-      "qt=(1+0.2*B/L)*5.14*c1+gamma1*Df;\n",

-      "print qt,\"is qt in kN/m**2\"\n",

-      "print \"no need to calculate qt since it is not useful for calculation\"\n",

-      "print qu/FS,\"is allowable shear stress in kN/m**2\"\n",

-      "print qu/FS*1*1.5,\" is allowable load in kN\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "715.84 is qt in kN/m**2\n",

-        "no need to calculate qt since it is not useful for calculation\n",

-        "164.104 is allowable shear stress in kN/m**2\n",

-        "246.156  is allowable load in kN\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.5:Pg-190"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.5\n",

-      "\n",

-      "c=50;  # in KN/m^2\n",

-      "#from table\n",

-      "Ncq=6.3;\n",

-      "FS=4.0;# FOS\n",

-      "qu=c*Ncq;  # in KN/m^2\n",

-      "qall=qu/4;\n",

-      "print qall,\"allowed shear stress in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "78.75 allowed shear stress in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.6:Pg-191"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.6\n",

-      "\n",

-      "Gamma=16.8; # in kN/m^3\n",

-      "B=1.5;#  in m\n",

-      "#from table\n",

-      "Nyq=120.0;\n",

-      "qu=1/2.0*Gamma*B*Nyq; # in KN/m^2\n",

-      "print qu,\" is shear stress in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "1512.0  shear stress in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.7:Pg-198"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 4.7\n",

-      "import math\n",

-      "phi=35; # in degree\n",

-      "Df=1.5; # in m\n",

-      "B=1.5; # in m\n",

-      "Gamma=17.4; # in kN/m^3\n",

-      "A=math.pi/4*Df**2; # in m^2\n",

-      "m=0.25;\n",

-      "Ku=0.936;\n",

-      "Fq=1+2*(1+m*Df/B)*Df/B*Ku*math.tan(phi*math.pi/180);\n",

-      "Qu=Fq*Gamma*A*Df; # in KN/m^2\n",

-      "print round(Qu,1),\" is bearing capacity in kN\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "121.7  is bearing capacity in kN\n"

-       ]

-      }

-     ],

-     "prompt_number": 17

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex4.8:pg-198"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "\n",

-      "#example 4.8\n",

-      "\n",

-      "\n",

-      "\n",

-      "cu=52; # in kN/m^2\n",

-      "\n",

-      "B=1.5; # in m\n",

-      "\n",

-      "L=3; # in m\n",

-      "\n",

-      "k=0.107*cu+2.5;\n",

-      "\n",

-      "print round(k,2),\" is Df/B of square\" \n",

-      "\n",

-      "A=L*B; # in m^2\n",

-      "\n",

-      "Beta=0.2;\n",

-      "\n",

-      "Gamma=18.9; # in kN/m^3\n",

-      "\n",

-      "Df=1.8; # in m\n",

-      "\n",

-      "Qu=A*(Beta*(7.56+1.44*B/L)*cu+Gamma*Df); # in kN/m^2\n",

-      "\n",

-      "print round(Qu,1),\" is ultimate shear force in kN\"\n",

-      "\n",

-      " \n",

-      "\n",

-      " "

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "8.06  is Df/B of square\n",

-        "540.6  is ultimate shear force in kN\n"

-       ]

-      }

-     ],

-     "prompt_number": 20

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter05.ipynb b/Principles_Of_Foundation_Engineering/Chapter05.ipynb
deleted file mode 100755
index ca975aae..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter05.ipynb
+++ /dev/null
@@ -1,550 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:77a2cf465cec464205dc151afe10d9acafa79fe1b46b30f4a468368f8af3f8ea"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter05:Shallow Foundations: Allowable Bearing Capacity and Settlement"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.1:Pg-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.1\n",
-      "\n",
-      "#first solution\n",
-      "B1=2.5; # in ft\n",
-      "B2=B1;\n",
-      "z=12.5; # in ft\n",
-      "L1=5; # in ft\n",
-      "L2=L1;\n",
-      "m=B1/z;\n",
-      "n=B2/z;\n",
-      "#from table 5.2  of the values using m,n\n",
-      "q=2000; # in lb/ft^2\n",
-      "I=0.0328;\n",
-      "deltasigma=q*4*I; # in lb/ft**2\n",
-      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
-      "#second solution\n",
-      "Ic=0.131;#from table\n",
-      "deltasigma=q*Ic; #  in lb/ft**2\n",
-      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "262.4 change in pressure in lb/ft**2\n",
-        "262.0 change in pressure in lb/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.2:Pg-215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.2\n",
-      "\n",
-      "qo=100; # in KN/m^2\n",
-      "H1=3; # in m\n",
-      "H2=5; # in m\n",
-      "#from table\n",
-      "IaH2=0.126;\n",
-      "IaH1=0.175;\n",
-      "deltasigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1)); # in kN/m**2\n",
-      "print round(deltasigma,2),\"change in pressure in kN/m**2\"\n",
-      "TS=4*deltasigma; # in kN/m**2\n",
-      "print round(TS,2),\"total change in pressure in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5.25 change in pressure in kN/m**2\n",
-        "21.0 total change in pressure in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.3:Pg-217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.3\n",
-      "H=7;\n",
-      "Gamma=17.5; # in KN/m^3\n",
-      "q0=Gamma*H # in KN/m^2\n",
-      "print q0,\" is pressure change in kN/m**2\"\n",
-      "#part2\n",
-      "#from figure\n",
-      "Ileftside=0.445;\n",
-      "Irightside=0.445;\n",
-      "deltasigma=q0*(Ileftside+Irightside); # in KN/m^2\n",
-      "print round(deltasigma,2),\"is change in stress in kN/m**2\"\n",
-      "#partc\n",
-      "#from figure 5.11\n",
-      "I=0.24;#I'\n",
-      "Dsigma1=43.75*I;#deltasigma1 in KN/m^2\n",
-      "I2=0.495;#I'\n",
-      "Dsigma2=I2*q0;#deltasigma2 in KN/m^2\n",
-      "I3=0.335;#I'\n",
-      "Dsigma3=I3*78.75;#deltasigma3 in KN/m^2\n",
-      "Dsigma=Dsigma1+Dsigma2-Dsigma3; # in KN/m^2\n",
-      "print round(Dsigma,2),\"is total stress increase in A in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "122.5  is pressure change in kN/m**2\n",
-        "109.03 is change in stress in kN/m**2\n",
-        "44.76 is total stress increase in A in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.4:Pg-228"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.4\n",
-      "\n",
-      "zbar=5;\n",
-      "mus=0.3;\n",
-      "F1=0.641;\n",
-      "F2=0.031;\n",
-      "z1=2.0; # in m\n",
-      "z2=1.0; # in m\n",
-      "z3=2.0; # in m\n",
-      "Es1=10000; # in kN/m**2\n",
-      "Es2=8000; # in kN/m**2\n",
-      "Es3=12000;# in kN/m**2\n",
-      "qo=150; # in KN/m^2\n",
-      "#from table 5.4\n",
-      "If=0.709;\n",
-      "Es=(Es1*z1+Es2*z2+Es3*z3)/zbar; # in kN/m**2\n",
-      "print Es,\"  is modulus of elasticity in kN/m**2\"\n",
-      "Is=F1+(2-mus)/(1-mus)*F2;\n",
-      "Sc=qo*(1.0/Es-mus**2.0/Es)*Is*If*2;\n",
-      "Scrigid=0.93*Sc; # in m\n",
-      "print round(Scrigid*1000,2),\"is settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10400.0   is modulus of elasticity in kN/m**2\n",
-        "12.4 is settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.5:Pg-234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.5\n",
-      "import math\n",
-      "B=5; # in ft\n",
-      "L=10; # in ft\n",
-      "Ef=2.3e6; # in lb/in^2\n",
-      "Eo=1400.0; # in lb/in^2\n",
-      "k=25.0; # in lb/in^2/ft\n",
-      "t=1.0;\n",
-      "mus=0.3;\n",
-      "Df=5.0; # in ft\n",
-      "qo=5000.0; # in lb/ft^2\n",
-      "Ig=0.69;\n",
-      "Be=math.sqrt(4*B*L/math.pi);\n",
-      "If=math.pi/4+1/(4.6+10*(Ef/(Eo+2*Be/2*k))*(2*t/Be)**3);\n",
-      "Ie=1-1/(3.5*math.exp(1.22*mus-0.4)*(Be/Df+1.6));\n",
-      "Se=qo*Be*Ig*If*Ie/Eo*(1-mus**2)/144; # in ft\n",
-      "print round(Se*12,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.07 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.6\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "q=3.06; # in lb/in^2\n",
-      "qbar=25; # in lb/in^2\n",
-      "C1=1-0.5*(q/(qbar-q));\n",
-      "Sum=0;\n",
-      "C2=1+0.2*math.log10(10/0.1);\n",
-      "L=[1, 2, 3, 4, 5];\n",
-      "Dz=[48, 48, 96, 48, 144]; # in inch\n",
-      "Es=[750, 1250, 1250, 1000, 2000]; # in lb/in^2\n",
-      "z=[24, 72, 144, 216, 312]; # in inch\n",
-      "Iz=[0.275, 0.425, 0.417, 0.292, 0.125];\n",
-      "k=numpy.zeros(5)\n",
-      "print \"Layer No.\\t deltaz (in)\\t Es(lb/in**2)\\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \\n\"\n",
-      "for i in  range(0,5):\n",
-      "    k[i]=Iz[i]/Es[i]*Dz[i];\n",
-      "    print L[i],\"\\t \\t \",Dz[i],\"\\t\\t   \",Es[i],\"\\t\\t \",z[i],\" \\t\\t\\t\\t\\t  \",Iz[i],\"\\t\\t  \",round(k[i],3)\n",
-      "    Sum=Sum+k[i];\n",
-      "\n",
-      "Se=C1*C2*(qbar-q)*Sum; # in inch\n",
-      "print round(Se,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Layer No.\t deltaz (in)\t Es(lb/in**2)\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \n",
-        "\n",
-        "1 \t \t  48 \t\t    750 \t\t  24  \t\t\t\t\t   0.275 \t\t   0.018\n",
-        "2 \t \t  48 \t\t    1250 \t\t  72  \t\t\t\t\t   0.425 \t\t   0.016\n",
-        "3 \t \t  96 \t\t    1250 \t\t  144  \t\t\t\t\t   0.417 \t\t   0.032\n",
-        "4 \t \t  48 \t\t    1000 \t\t  216  \t\t\t\t\t   0.292 \t\t   0.014\n",
-        "5 \t \t  144 \t\t    2000 \t\t  312  \t\t\t\t\t   0.125 \t\t   0.009\n",
-        "2.54 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.7:Pg-244"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.7\n",
-      "\n",
-      "Df=1.0; # in m\n",
-      "B=1.75; # in m\n",
-      "L=1.75 # in m\n",
-      "qnet=120.0; # in KN/m^2\n",
-      "N60=10.0;# in m\n",
-      "alpha1=0.14 # for normally consolated sand\n",
-      "alpha2=1.71/(N60)**1.4 # for normally consolated sand\n",
-      "alpha3=1.0 # for normally consolated sand\n",
-      "Se=0.3*alpha1*alpha2*alpha3*(qnet/100)*((B/0.3)**0.7)*((1.25*(L/B)/(0.25+(L/B))))**2\n",
-      "print round(Se*1000,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11.79 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.8:Pg-245"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.8\n",
-      "\n",
-      "Df=1; # in m\n",
-      "B=1.75; # in m\n",
-      "qnet=120; # in KN/m^2\n",
-      "N60=10; # in m\n",
-      "Fd=1+0.33*Df/B;\n",
-      "Se=2*qnet/N60/Fd*(B/(B+0.3))**2; # in mm\n",
-      "print round(Se,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "14.71 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.9:Pg-251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.9\n",
-      "\n",
-      "Ny=23.76; \n",
-      "Nq=16.51; \n",
-      "q=3*110.0; # in lb/ft^2\n",
-      "Gamma=110.0; # in lb/ft^3\n",
-      "B=4.0; # in ft\n",
-      "Nqe=0.63*Nq;\n",
-      "Nye=0.4*Ny;\n",
-      "que=q*Nqe+1/2.0*Gamma*B*Nye; # in lb/ft^2\n",
-      "print round(que,2),\" is bearing capacity in lb/ft**2\"\n",
-      "#part 2\n",
-      "V=0.4; # in ft/sec\n",
-      "A=0.32; # given in question\n",
-      "g=9.81; # acceleration constant in m/sec^2\n",
-      "kh=0.26;\n",
-      "k=0.92;#tan(alphae)\n",
-      "Seq=0.174*k*V**2/A/g*kh**-4/A**-4; # in m\n",
-      "print round(Seq,3),\"settelement in m\"\n",
-      "print round(Seq*39.57,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5523.31  is bearing capacity in lb/ft**2\n",
-        "0.019 settelement in m\n",
-        "0.74 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.10:Pg-256"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.10\n",
-      "\n",
-      "import math\n",
-      "Cc=0.32;\n",
-      "Hc=2.5;\n",
-      "eo=0.8;\n",
-      "sigmao=2.5*16.5+0.5*(17.5-9.81)+1.25*(16-9.81); # in kN/m^2\n",
-      "m1=[2, 2, 2];\n",
-      "z=[2, 3.25, 4.5];\n",
-      "n1=[4, 6.5, 9];\n",
-      "Ic=[0.19, 0.085, 0.045];\n",
-      "Dsigma=[28.5, 12.75, 6.75];#deltasigma\n",
-      "print (\"m1\\t z(m)\\t n1\\t Ic\\t Dsigma \\n\");\n",
-      "for i in range(0,3):\n",
-      "    print round(m1[i],2),\"\\t \",round(z[i],2),\"\\t \",round(n1[i],2),\"\\t \",round(Ic[i],2),\"\\t \",round(Dsigma[i],2)\n",
-      "\n",
-      "    Dsigmaav=1/6.0*(Dsigma[0]+4*Dsigma[1]+Dsigma[2]);\n",
-      "    Sc=Cc*Hc/(1+eo)*math.log10((sigmao+Dsigmaav)/sigmao);\n",
-      "print round(Sc*1000,2),\"settlement in mm\"\n",
-      "#partb\n",
-      "B=1.0; # in m\n",
-      "L=2.0; # in m\n",
-      "z=0.5+1.5; # in m\n",
-      "B=B+z; # in m\n",
-      "L=L+z; # in m\n",
-      "A=0.6; # given in question\n",
-      "#from table\n",
-      "kcr=0.78; # by data\n",
-      "Sep=kcr*Sc;\n",
-      "print round(Sep*1000,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "m1\t z(m)\t n1\t Ic\t Dsigma \n",
-        "\n",
-        "2.0 \t  2.0 \t  4.0 \t  0.19 \t  28.5\n",
-        "2.0 \t  3.25 \t  6.5 \t  0.09 \t  12.75\n",
-        "2.0 \t  4.5 \t  9.0 \t  0.04 \t  6.75\n",
-        "46.45 settlement in mm\n",
-        "36.23 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.11:Pg-262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.11\n",
-      "\n",
-      "import numpy\n",
-      "N60=(3+7+12+12+16)/5.0;\n",
-      "B=[2, 2.25, 2.3]; # in m\n",
-      "Fd=[1.248, 1.22, 1.215];\n",
-      "Qoac=102000*9.81/1000;#actual Qo\n",
-      "Se=25; # in mm\n",
-      "qnet=numpy.zeros(3)\n",
-      "Qo=numpy.zeros(3) # in kN\n",
-      "print \"B(m)\\t Fd\\t qnet(kN/m**2)\\t \\t Qo \\n\"\n",
-      "for i in range(0,3):\n",
-      "    qnet[i]=10/0.08*(B[i]+0.3)**2/(B[i])**2*Fd[i]*Se/25;\n",
-      "    Qo[i]=qnet[i]*B[i]**2;\n",
-      "    print B[i],\"\\t\",Fd[i],\" \\t \",round(qnet[i],2),\"\\t\\t \",Qo[i],\"\\n\"\n",
-      "print int(Qoac),\"value of Qo in kN\"\n",
-      "print \"since Qo is 1000 kN thus B is equal to 2.3 m from the table\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "B(m)\t Fd\t qnet(kN/m**2)\t \t Qo \n",
-        "\n",
-        "2 \t1.248  \t  206.31 \t\t  825.24 \n",
-        "\n",
-        "2.25 \t1.22  \t  195.88 \t\t  991.63125 \n",
-        "\n",
-        "2.3 \t1.215  \t  194.08 \t\t  1026.675 \n",
-        "\n",
-        "1000 value of Qo in kN\n",
-        "since Qo is 1000 kN thus B is equal to 2.3 m from the table\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter05_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter05_1.ipynb
deleted file mode 100755
index ca975aae..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter05_1.ipynb
+++ /dev/null
@@ -1,550 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:77a2cf465cec464205dc151afe10d9acafa79fe1b46b30f4a468368f8af3f8ea"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter05:Shallow Foundations: Allowable Bearing Capacity and Settlement"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.1:Pg-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.1\n",
-      "\n",
-      "#first solution\n",
-      "B1=2.5; # in ft\n",
-      "B2=B1;\n",
-      "z=12.5; # in ft\n",
-      "L1=5; # in ft\n",
-      "L2=L1;\n",
-      "m=B1/z;\n",
-      "n=B2/z;\n",
-      "#from table 5.2  of the values using m,n\n",
-      "q=2000; # in lb/ft^2\n",
-      "I=0.0328;\n",
-      "deltasigma=q*4*I; # in lb/ft**2\n",
-      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
-      "#second solution\n",
-      "Ic=0.131;#from table\n",
-      "deltasigma=q*Ic; #  in lb/ft**2\n",
-      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "262.4 change in pressure in lb/ft**2\n",
-        "262.0 change in pressure in lb/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.2:Pg-215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.2\n",
-      "\n",
-      "qo=100; # in KN/m^2\n",
-      "H1=3; # in m\n",
-      "H2=5; # in m\n",
-      "#from table\n",
-      "IaH2=0.126;\n",
-      "IaH1=0.175;\n",
-      "deltasigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1)); # in kN/m**2\n",
-      "print round(deltasigma,2),\"change in pressure in kN/m**2\"\n",
-      "TS=4*deltasigma; # in kN/m**2\n",
-      "print round(TS,2),\"total change in pressure in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5.25 change in pressure in kN/m**2\n",
-        "21.0 total change in pressure in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.3:Pg-217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.3\n",
-      "H=7;\n",
-      "Gamma=17.5; # in KN/m^3\n",
-      "q0=Gamma*H # in KN/m^2\n",
-      "print q0,\" is pressure change in kN/m**2\"\n",
-      "#part2\n",
-      "#from figure\n",
-      "Ileftside=0.445;\n",
-      "Irightside=0.445;\n",
-      "deltasigma=q0*(Ileftside+Irightside); # in KN/m^2\n",
-      "print round(deltasigma,2),\"is change in stress in kN/m**2\"\n",
-      "#partc\n",
-      "#from figure 5.11\n",
-      "I=0.24;#I'\n",
-      "Dsigma1=43.75*I;#deltasigma1 in KN/m^2\n",
-      "I2=0.495;#I'\n",
-      "Dsigma2=I2*q0;#deltasigma2 in KN/m^2\n",
-      "I3=0.335;#I'\n",
-      "Dsigma3=I3*78.75;#deltasigma3 in KN/m^2\n",
-      "Dsigma=Dsigma1+Dsigma2-Dsigma3; # in KN/m^2\n",
-      "print round(Dsigma,2),\"is total stress increase in A in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "122.5  is pressure change in kN/m**2\n",
-        "109.03 is change in stress in kN/m**2\n",
-        "44.76 is total stress increase in A in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.4:Pg-228"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.4\n",
-      "\n",
-      "zbar=5;\n",
-      "mus=0.3;\n",
-      "F1=0.641;\n",
-      "F2=0.031;\n",
-      "z1=2.0; # in m\n",
-      "z2=1.0; # in m\n",
-      "z3=2.0; # in m\n",
-      "Es1=10000; # in kN/m**2\n",
-      "Es2=8000; # in kN/m**2\n",
-      "Es3=12000;# in kN/m**2\n",
-      "qo=150; # in KN/m^2\n",
-      "#from table 5.4\n",
-      "If=0.709;\n",
-      "Es=(Es1*z1+Es2*z2+Es3*z3)/zbar; # in kN/m**2\n",
-      "print Es,\"  is modulus of elasticity in kN/m**2\"\n",
-      "Is=F1+(2-mus)/(1-mus)*F2;\n",
-      "Sc=qo*(1.0/Es-mus**2.0/Es)*Is*If*2;\n",
-      "Scrigid=0.93*Sc; # in m\n",
-      "print round(Scrigid*1000,2),\"is settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10400.0   is modulus of elasticity in kN/m**2\n",
-        "12.4 is settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.5:Pg-234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.5\n",
-      "import math\n",
-      "B=5; # in ft\n",
-      "L=10; # in ft\n",
-      "Ef=2.3e6; # in lb/in^2\n",
-      "Eo=1400.0; # in lb/in^2\n",
-      "k=25.0; # in lb/in^2/ft\n",
-      "t=1.0;\n",
-      "mus=0.3;\n",
-      "Df=5.0; # in ft\n",
-      "qo=5000.0; # in lb/ft^2\n",
-      "Ig=0.69;\n",
-      "Be=math.sqrt(4*B*L/math.pi);\n",
-      "If=math.pi/4+1/(4.6+10*(Ef/(Eo+2*Be/2*k))*(2*t/Be)**3);\n",
-      "Ie=1-1/(3.5*math.exp(1.22*mus-0.4)*(Be/Df+1.6));\n",
-      "Se=qo*Be*Ig*If*Ie/Eo*(1-mus**2)/144; # in ft\n",
-      "print round(Se*12,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.07 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.6\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "q=3.06; # in lb/in^2\n",
-      "qbar=25; # in lb/in^2\n",
-      "C1=1-0.5*(q/(qbar-q));\n",
-      "Sum=0;\n",
-      "C2=1+0.2*math.log10(10/0.1);\n",
-      "L=[1, 2, 3, 4, 5];\n",
-      "Dz=[48, 48, 96, 48, 144]; # in inch\n",
-      "Es=[750, 1250, 1250, 1000, 2000]; # in lb/in^2\n",
-      "z=[24, 72, 144, 216, 312]; # in inch\n",
-      "Iz=[0.275, 0.425, 0.417, 0.292, 0.125];\n",
-      "k=numpy.zeros(5)\n",
-      "print \"Layer No.\\t deltaz (in)\\t Es(lb/in**2)\\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \\n\"\n",
-      "for i in  range(0,5):\n",
-      "    k[i]=Iz[i]/Es[i]*Dz[i];\n",
-      "    print L[i],\"\\t \\t \",Dz[i],\"\\t\\t   \",Es[i],\"\\t\\t \",z[i],\" \\t\\t\\t\\t\\t  \",Iz[i],\"\\t\\t  \",round(k[i],3)\n",
-      "    Sum=Sum+k[i];\n",
-      "\n",
-      "Se=C1*C2*(qbar-q)*Sum; # in inch\n",
-      "print round(Se,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Layer No.\t deltaz (in)\t Es(lb/in**2)\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \n",
-        "\n",
-        "1 \t \t  48 \t\t    750 \t\t  24  \t\t\t\t\t   0.275 \t\t   0.018\n",
-        "2 \t \t  48 \t\t    1250 \t\t  72  \t\t\t\t\t   0.425 \t\t   0.016\n",
-        "3 \t \t  96 \t\t    1250 \t\t  144  \t\t\t\t\t   0.417 \t\t   0.032\n",
-        "4 \t \t  48 \t\t    1000 \t\t  216  \t\t\t\t\t   0.292 \t\t   0.014\n",
-        "5 \t \t  144 \t\t    2000 \t\t  312  \t\t\t\t\t   0.125 \t\t   0.009\n",
-        "2.54 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.7:Pg-244"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.7\n",
-      "\n",
-      "Df=1.0; # in m\n",
-      "B=1.75; # in m\n",
-      "L=1.75 # in m\n",
-      "qnet=120.0; # in KN/m^2\n",
-      "N60=10.0;# in m\n",
-      "alpha1=0.14 # for normally consolated sand\n",
-      "alpha2=1.71/(N60)**1.4 # for normally consolated sand\n",
-      "alpha3=1.0 # for normally consolated sand\n",
-      "Se=0.3*alpha1*alpha2*alpha3*(qnet/100)*((B/0.3)**0.7)*((1.25*(L/B)/(0.25+(L/B))))**2\n",
-      "print round(Se*1000,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11.79 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.8:Pg-245"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.8\n",
-      "\n",
-      "Df=1; # in m\n",
-      "B=1.75; # in m\n",
-      "qnet=120; # in KN/m^2\n",
-      "N60=10; # in m\n",
-      "Fd=1+0.33*Df/B;\n",
-      "Se=2*qnet/N60/Fd*(B/(B+0.3))**2; # in mm\n",
-      "print round(Se,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "14.71 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.9:Pg-251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.9\n",
-      "\n",
-      "Ny=23.76; \n",
-      "Nq=16.51; \n",
-      "q=3*110.0; # in lb/ft^2\n",
-      "Gamma=110.0; # in lb/ft^3\n",
-      "B=4.0; # in ft\n",
-      "Nqe=0.63*Nq;\n",
-      "Nye=0.4*Ny;\n",
-      "que=q*Nqe+1/2.0*Gamma*B*Nye; # in lb/ft^2\n",
-      "print round(que,2),\" is bearing capacity in lb/ft**2\"\n",
-      "#part 2\n",
-      "V=0.4; # in ft/sec\n",
-      "A=0.32; # given in question\n",
-      "g=9.81; # acceleration constant in m/sec^2\n",
-      "kh=0.26;\n",
-      "k=0.92;#tan(alphae)\n",
-      "Seq=0.174*k*V**2/A/g*kh**-4/A**-4; # in m\n",
-      "print round(Seq,3),\"settelement in m\"\n",
-      "print round(Seq*39.57,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5523.31  is bearing capacity in lb/ft**2\n",
-        "0.019 settelement in m\n",
-        "0.74 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.10:Pg-256"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.10\n",
-      "\n",
-      "import math\n",
-      "Cc=0.32;\n",
-      "Hc=2.5;\n",
-      "eo=0.8;\n",
-      "sigmao=2.5*16.5+0.5*(17.5-9.81)+1.25*(16-9.81); # in kN/m^2\n",
-      "m1=[2, 2, 2];\n",
-      "z=[2, 3.25, 4.5];\n",
-      "n1=[4, 6.5, 9];\n",
-      "Ic=[0.19, 0.085, 0.045];\n",
-      "Dsigma=[28.5, 12.75, 6.75];#deltasigma\n",
-      "print (\"m1\\t z(m)\\t n1\\t Ic\\t Dsigma \\n\");\n",
-      "for i in range(0,3):\n",
-      "    print round(m1[i],2),\"\\t \",round(z[i],2),\"\\t \",round(n1[i],2),\"\\t \",round(Ic[i],2),\"\\t \",round(Dsigma[i],2)\n",
-      "\n",
-      "    Dsigmaav=1/6.0*(Dsigma[0]+4*Dsigma[1]+Dsigma[2]);\n",
-      "    Sc=Cc*Hc/(1+eo)*math.log10((sigmao+Dsigmaav)/sigmao);\n",
-      "print round(Sc*1000,2),\"settlement in mm\"\n",
-      "#partb\n",
-      "B=1.0; # in m\n",
-      "L=2.0; # in m\n",
-      "z=0.5+1.5; # in m\n",
-      "B=B+z; # in m\n",
-      "L=L+z; # in m\n",
-      "A=0.6; # given in question\n",
-      "#from table\n",
-      "kcr=0.78; # by data\n",
-      "Sep=kcr*Sc;\n",
-      "print round(Sep*1000,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "m1\t z(m)\t n1\t Ic\t Dsigma \n",
-        "\n",
-        "2.0 \t  2.0 \t  4.0 \t  0.19 \t  28.5\n",
-        "2.0 \t  3.25 \t  6.5 \t  0.09 \t  12.75\n",
-        "2.0 \t  4.5 \t  9.0 \t  0.04 \t  6.75\n",
-        "46.45 settlement in mm\n",
-        "36.23 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.11:Pg-262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.11\n",
-      "\n",
-      "import numpy\n",
-      "N60=(3+7+12+12+16)/5.0;\n",
-      "B=[2, 2.25, 2.3]; # in m\n",
-      "Fd=[1.248, 1.22, 1.215];\n",
-      "Qoac=102000*9.81/1000;#actual Qo\n",
-      "Se=25; # in mm\n",
-      "qnet=numpy.zeros(3)\n",
-      "Qo=numpy.zeros(3) # in kN\n",
-      "print \"B(m)\\t Fd\\t qnet(kN/m**2)\\t \\t Qo \\n\"\n",
-      "for i in range(0,3):\n",
-      "    qnet[i]=10/0.08*(B[i]+0.3)**2/(B[i])**2*Fd[i]*Se/25;\n",
-      "    Qo[i]=qnet[i]*B[i]**2;\n",
-      "    print B[i],\"\\t\",Fd[i],\" \\t \",round(qnet[i],2),\"\\t\\t \",Qo[i],\"\\n\"\n",
-      "print int(Qoac),\"value of Qo in kN\"\n",
-      "print \"since Qo is 1000 kN thus B is equal to 2.3 m from the table\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "B(m)\t Fd\t qnet(kN/m**2)\t \t Qo \n",
-        "\n",
-        "2 \t1.248  \t  206.31 \t\t  825.24 \n",
-        "\n",
-        "2.25 \t1.22  \t  195.88 \t\t  991.63125 \n",
-        "\n",
-        "2.3 \t1.215  \t  194.08 \t\t  1026.675 \n",
-        "\n",
-        "1000 value of Qo in kN\n",
-        "since Qo is 1000 kN thus B is equal to 2.3 m from the table\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter05_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter05_2.ipynb
deleted file mode 100755
index ca975aae..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter05_2.ipynb
+++ /dev/null
@@ -1,550 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:77a2cf465cec464205dc151afe10d9acafa79fe1b46b30f4a468368f8af3f8ea"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter05:Shallow Foundations: Allowable Bearing Capacity and Settlement"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.1:Pg-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.1\n",
-      "\n",
-      "#first solution\n",
-      "B1=2.5; # in ft\n",
-      "B2=B1;\n",
-      "z=12.5; # in ft\n",
-      "L1=5; # in ft\n",
-      "L2=L1;\n",
-      "m=B1/z;\n",
-      "n=B2/z;\n",
-      "#from table 5.2  of the values using m,n\n",
-      "q=2000; # in lb/ft^2\n",
-      "I=0.0328;\n",
-      "deltasigma=q*4*I; # in lb/ft**2\n",
-      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
-      "#second solution\n",
-      "Ic=0.131;#from table\n",
-      "deltasigma=q*Ic; #  in lb/ft**2\n",
-      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "262.4 change in pressure in lb/ft**2\n",
-        "262.0 change in pressure in lb/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.2:Pg-215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.2\n",
-      "\n",
-      "qo=100; # in KN/m^2\n",
-      "H1=3; # in m\n",
-      "H2=5; # in m\n",
-      "#from table\n",
-      "IaH2=0.126;\n",
-      "IaH1=0.175;\n",
-      "deltasigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1)); # in kN/m**2\n",
-      "print round(deltasigma,2),\"change in pressure in kN/m**2\"\n",
-      "TS=4*deltasigma; # in kN/m**2\n",
-      "print round(TS,2),\"total change in pressure in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5.25 change in pressure in kN/m**2\n",
-        "21.0 total change in pressure in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.3:Pg-217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.3\n",
-      "H=7;\n",
-      "Gamma=17.5; # in KN/m^3\n",
-      "q0=Gamma*H # in KN/m^2\n",
-      "print q0,\" is pressure change in kN/m**2\"\n",
-      "#part2\n",
-      "#from figure\n",
-      "Ileftside=0.445;\n",
-      "Irightside=0.445;\n",
-      "deltasigma=q0*(Ileftside+Irightside); # in KN/m^2\n",
-      "print round(deltasigma,2),\"is change in stress in kN/m**2\"\n",
-      "#partc\n",
-      "#from figure 5.11\n",
-      "I=0.24;#I'\n",
-      "Dsigma1=43.75*I;#deltasigma1 in KN/m^2\n",
-      "I2=0.495;#I'\n",
-      "Dsigma2=I2*q0;#deltasigma2 in KN/m^2\n",
-      "I3=0.335;#I'\n",
-      "Dsigma3=I3*78.75;#deltasigma3 in KN/m^2\n",
-      "Dsigma=Dsigma1+Dsigma2-Dsigma3; # in KN/m^2\n",
-      "print round(Dsigma,2),\"is total stress increase in A in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "122.5  is pressure change in kN/m**2\n",
-        "109.03 is change in stress in kN/m**2\n",
-        "44.76 is total stress increase in A in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.4:Pg-228"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.4\n",
-      "\n",
-      "zbar=5;\n",
-      "mus=0.3;\n",
-      "F1=0.641;\n",
-      "F2=0.031;\n",
-      "z1=2.0; # in m\n",
-      "z2=1.0; # in m\n",
-      "z3=2.0; # in m\n",
-      "Es1=10000; # in kN/m**2\n",
-      "Es2=8000; # in kN/m**2\n",
-      "Es3=12000;# in kN/m**2\n",
-      "qo=150; # in KN/m^2\n",
-      "#from table 5.4\n",
-      "If=0.709;\n",
-      "Es=(Es1*z1+Es2*z2+Es3*z3)/zbar; # in kN/m**2\n",
-      "print Es,\"  is modulus of elasticity in kN/m**2\"\n",
-      "Is=F1+(2-mus)/(1-mus)*F2;\n",
-      "Sc=qo*(1.0/Es-mus**2.0/Es)*Is*If*2;\n",
-      "Scrigid=0.93*Sc; # in m\n",
-      "print round(Scrigid*1000,2),\"is settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10400.0   is modulus of elasticity in kN/m**2\n",
-        "12.4 is settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.5:Pg-234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.5\n",
-      "import math\n",
-      "B=5; # in ft\n",
-      "L=10; # in ft\n",
-      "Ef=2.3e6; # in lb/in^2\n",
-      "Eo=1400.0; # in lb/in^2\n",
-      "k=25.0; # in lb/in^2/ft\n",
-      "t=1.0;\n",
-      "mus=0.3;\n",
-      "Df=5.0; # in ft\n",
-      "qo=5000.0; # in lb/ft^2\n",
-      "Ig=0.69;\n",
-      "Be=math.sqrt(4*B*L/math.pi);\n",
-      "If=math.pi/4+1/(4.6+10*(Ef/(Eo+2*Be/2*k))*(2*t/Be)**3);\n",
-      "Ie=1-1/(3.5*math.exp(1.22*mus-0.4)*(Be/Df+1.6));\n",
-      "Se=qo*Be*Ig*If*Ie/Eo*(1-mus**2)/144; # in ft\n",
-      "print round(Se*12,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.07 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.6\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "q=3.06; # in lb/in^2\n",
-      "qbar=25; # in lb/in^2\n",
-      "C1=1-0.5*(q/(qbar-q));\n",
-      "Sum=0;\n",
-      "C2=1+0.2*math.log10(10/0.1);\n",
-      "L=[1, 2, 3, 4, 5];\n",
-      "Dz=[48, 48, 96, 48, 144]; # in inch\n",
-      "Es=[750, 1250, 1250, 1000, 2000]; # in lb/in^2\n",
-      "z=[24, 72, 144, 216, 312]; # in inch\n",
-      "Iz=[0.275, 0.425, 0.417, 0.292, 0.125];\n",
-      "k=numpy.zeros(5)\n",
-      "print \"Layer No.\\t deltaz (in)\\t Es(lb/in**2)\\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \\n\"\n",
-      "for i in  range(0,5):\n",
-      "    k[i]=Iz[i]/Es[i]*Dz[i];\n",
-      "    print L[i],\"\\t \\t \",Dz[i],\"\\t\\t   \",Es[i],\"\\t\\t \",z[i],\" \\t\\t\\t\\t\\t  \",Iz[i],\"\\t\\t  \",round(k[i],3)\n",
-      "    Sum=Sum+k[i];\n",
-      "\n",
-      "Se=C1*C2*(qbar-q)*Sum; # in inch\n",
-      "print round(Se,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Layer No.\t deltaz (in)\t Es(lb/in**2)\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \n",
-        "\n",
-        "1 \t \t  48 \t\t    750 \t\t  24  \t\t\t\t\t   0.275 \t\t   0.018\n",
-        "2 \t \t  48 \t\t    1250 \t\t  72  \t\t\t\t\t   0.425 \t\t   0.016\n",
-        "3 \t \t  96 \t\t    1250 \t\t  144  \t\t\t\t\t   0.417 \t\t   0.032\n",
-        "4 \t \t  48 \t\t    1000 \t\t  216  \t\t\t\t\t   0.292 \t\t   0.014\n",
-        "5 \t \t  144 \t\t    2000 \t\t  312  \t\t\t\t\t   0.125 \t\t   0.009\n",
-        "2.54 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.7:Pg-244"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.7\n",
-      "\n",
-      "Df=1.0; # in m\n",
-      "B=1.75; # in m\n",
-      "L=1.75 # in m\n",
-      "qnet=120.0; # in KN/m^2\n",
-      "N60=10.0;# in m\n",
-      "alpha1=0.14 # for normally consolated sand\n",
-      "alpha2=1.71/(N60)**1.4 # for normally consolated sand\n",
-      "alpha3=1.0 # for normally consolated sand\n",
-      "Se=0.3*alpha1*alpha2*alpha3*(qnet/100)*((B/0.3)**0.7)*((1.25*(L/B)/(0.25+(L/B))))**2\n",
-      "print round(Se*1000,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11.79 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.8:Pg-245"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.8\n",
-      "\n",
-      "Df=1; # in m\n",
-      "B=1.75; # in m\n",
-      "qnet=120; # in KN/m^2\n",
-      "N60=10; # in m\n",
-      "Fd=1+0.33*Df/B;\n",
-      "Se=2*qnet/N60/Fd*(B/(B+0.3))**2; # in mm\n",
-      "print round(Se,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "14.71 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.9:Pg-251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.9\n",
-      "\n",
-      "Ny=23.76; \n",
-      "Nq=16.51; \n",
-      "q=3*110.0; # in lb/ft^2\n",
-      "Gamma=110.0; # in lb/ft^3\n",
-      "B=4.0; # in ft\n",
-      "Nqe=0.63*Nq;\n",
-      "Nye=0.4*Ny;\n",
-      "que=q*Nqe+1/2.0*Gamma*B*Nye; # in lb/ft^2\n",
-      "print round(que,2),\" is bearing capacity in lb/ft**2\"\n",
-      "#part 2\n",
-      "V=0.4; # in ft/sec\n",
-      "A=0.32; # given in question\n",
-      "g=9.81; # acceleration constant in m/sec^2\n",
-      "kh=0.26;\n",
-      "k=0.92;#tan(alphae)\n",
-      "Seq=0.174*k*V**2/A/g*kh**-4/A**-4; # in m\n",
-      "print round(Seq,3),\"settelement in m\"\n",
-      "print round(Seq*39.57,2),\"settlement in inches\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5523.31  is bearing capacity in lb/ft**2\n",
-        "0.019 settelement in m\n",
-        "0.74 settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.10:Pg-256"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.10\n",
-      "\n",
-      "import math\n",
-      "Cc=0.32;\n",
-      "Hc=2.5;\n",
-      "eo=0.8;\n",
-      "sigmao=2.5*16.5+0.5*(17.5-9.81)+1.25*(16-9.81); # in kN/m^2\n",
-      "m1=[2, 2, 2];\n",
-      "z=[2, 3.25, 4.5];\n",
-      "n1=[4, 6.5, 9];\n",
-      "Ic=[0.19, 0.085, 0.045];\n",
-      "Dsigma=[28.5, 12.75, 6.75];#deltasigma\n",
-      "print (\"m1\\t z(m)\\t n1\\t Ic\\t Dsigma \\n\");\n",
-      "for i in range(0,3):\n",
-      "    print round(m1[i],2),\"\\t \",round(z[i],2),\"\\t \",round(n1[i],2),\"\\t \",round(Ic[i],2),\"\\t \",round(Dsigma[i],2)\n",
-      "\n",
-      "    Dsigmaav=1/6.0*(Dsigma[0]+4*Dsigma[1]+Dsigma[2]);\n",
-      "    Sc=Cc*Hc/(1+eo)*math.log10((sigmao+Dsigmaav)/sigmao);\n",
-      "print round(Sc*1000,2),\"settlement in mm\"\n",
-      "#partb\n",
-      "B=1.0; # in m\n",
-      "L=2.0; # in m\n",
-      "z=0.5+1.5; # in m\n",
-      "B=B+z; # in m\n",
-      "L=L+z; # in m\n",
-      "A=0.6; # given in question\n",
-      "#from table\n",
-      "kcr=0.78; # by data\n",
-      "Sep=kcr*Sc;\n",
-      "print round(Sep*1000,2),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "m1\t z(m)\t n1\t Ic\t Dsigma \n",
-        "\n",
-        "2.0 \t  2.0 \t  4.0 \t  0.19 \t  28.5\n",
-        "2.0 \t  3.25 \t  6.5 \t  0.09 \t  12.75\n",
-        "2.0 \t  4.5 \t  9.0 \t  0.04 \t  6.75\n",
-        "46.45 settlement in mm\n",
-        "36.23 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.11:Pg-262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 5.11\n",
-      "\n",
-      "import numpy\n",
-      "N60=(3+7+12+12+16)/5.0;\n",
-      "B=[2, 2.25, 2.3]; # in m\n",
-      "Fd=[1.248, 1.22, 1.215];\n",
-      "Qoac=102000*9.81/1000;#actual Qo\n",
-      "Se=25; # in mm\n",
-      "qnet=numpy.zeros(3)\n",
-      "Qo=numpy.zeros(3) # in kN\n",
-      "print \"B(m)\\t Fd\\t qnet(kN/m**2)\\t \\t Qo \\n\"\n",
-      "for i in range(0,3):\n",
-      "    qnet[i]=10/0.08*(B[i]+0.3)**2/(B[i])**2*Fd[i]*Se/25;\n",
-      "    Qo[i]=qnet[i]*B[i]**2;\n",
-      "    print B[i],\"\\t\",Fd[i],\" \\t \",round(qnet[i],2),\"\\t\\t \",Qo[i],\"\\n\"\n",
-      "print int(Qoac),\"value of Qo in kN\"\n",
-      "print \"since Qo is 1000 kN thus B is equal to 2.3 m from the table\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "B(m)\t Fd\t qnet(kN/m**2)\t \t Qo \n",
-        "\n",
-        "2 \t1.248  \t  206.31 \t\t  825.24 \n",
-        "\n",
-        "2.25 \t1.22  \t  195.88 \t\t  991.63125 \n",
-        "\n",
-        "2.3 \t1.215  \t  194.08 \t\t  1026.675 \n",
-        "\n",
-        "1000 value of Qo in kN\n",
-        "since Qo is 1000 kN thus B is equal to 2.3 m from the table\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter06.ipynb b/Principles_Of_Foundation_Engineering/Chapter06.ipynb
deleted file mode 100755
index d87679c7..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter06.ipynb
+++ /dev/null
@@ -1,351 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ea06255a04932e0bf9952689cf2f6d919d320d1161a790bec574191c6fbafa1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter06: Mat Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.1:Pg-279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.1\n",
-      "\n",
-      "B=30; # in ft\n",
-      "L=45; # in ft\n",
-      "Df=6.5; # in ft\n",
-      "cu=1950;# in lb/ft^2\n",
-      "qunet=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B);\n",
-      "print int(qunet),\" is allowed force in lb/ft**2\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "12307  is allowed force in lb/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.2:Pg-280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.2\n",
-      "\n",
-      "N60=10; # penetration number\n",
-      "Df=2; # in m\n",
-      "B=10.0; # in m\n",
-      "Se=25.0; # in mm\n",
-      "qnetall=N60/0.08*(1+0.33*Df/B)*Se/25;\n",
-      "print qnetall,\" is allowed pressure in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "133.25  is allowed pressure in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.3:Pg-283"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.3\n",
-      "\n",
-      "cu=2800; # in lb/ft^2\n",
-      "B=60; # in ft\n",
-      "L=100; # in ft\n",
-      "Df=5; # in ft\n",
-      "\n",
-      "Gamma=120; # in lb/ft^3\n",
-      "A=60*100; # in ft^2\n",
-      "Q=25e6; # load in Kip\n",
-      "FS=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B)/(Q/A-Gamma*Df);\n",
-      "print round(FS,2),\" is factor of safety\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4.66  is factor of safety\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.4:Pg-284"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.4\n",
-      "\n",
-      "import math\n",
-      "Cc=0.28;\n",
-      "Hc=18*12.0;\n",
-      "e0=0.9;\n",
-      "sigmao=11*100+40*(121.5-64)+18/2*(118-62.4); # in lb/ft^2\n",
-      "H2=5+40+18.0;\n",
-      "H1=5+40.0;\n",
-      "qo=3567.0;\n",
-      "#from table\n",
-      "IaH2=0.21;\n",
-      "IaH1=0.225;\n",
-      "Dsigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1))*4;\n",
-      "Scp=Cc*Hc/(1+e0)*math.log10(sigmao/sigmao+Dsigma/sigmao);\n",
-      "print round(Scp,2),\"is settlement in inches\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6.76 is settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.5:Pg-296"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.5\n",
-      "import numpy\n",
-      "P=['A','B','C','D','E','F','G','H','I','J','K','L','M','N'];#point\n",
-      "k=1.2*numpy.ones(14);#Q/A\n",
-      "x=[-38,-24, -12, 0, 12, 24, 38, 38, 24, 12, 0, -12, -24, -38];\n",
-      "x1=numpy.zeros(14)\n",
-      "for i in range(0,14):\n",
-      "    x1[i]=0.0017*x[i];\n",
-      "y=[48,48,48,48,48,48,48, -48, -48, -48, -48, -48, -48, -48];\n",
-      "y1=numpy.zeros(14)\n",
-      "for i in range(0,14):\n",
-      "    y1[i]=-0.0011*y[i];\n",
-      "print \"point\\t Q\\A (kip/ft**2)  x(ft)\\t 0.0017x(ft)\\t\\ty(ft)\\t \\t  0.0011y(ft)\\t  \\t  q(kip/ft**2)\\n\"\n",
-      "q=numpy.zeros(14)\n",
-      "for i in  range(0,14):\n",
-      "    q[i]=1.2+x1[i]+y1[i];\n",
-      "    print P[i],\"\\t \",k[i],\"\\t\\t \",x[i],\"\\t\\t\",round(x1[i],3),\"\\t\\t\",y[i],\"\\t \\t \",round(y1[i],3),\"\\t \\t \\t\",round(q[i],3),\"\\t\\t \\n \"\n",
-      "\n",
-      "print \"the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "point\t Q\\A (kip/ft**2)  x(ft)\t 0.0017x(ft)\t\ty(ft)\t \t  0.0011y(ft)\t  \t  q(kip/ft**2)\n",
-        "\n",
-        "A \t  1.2 \t\t  -38 \t\t-0.065 \t\t48 \t \t  -0.053 \t \t \t1.083 \t\t \n",
-        " \n",
-        "B \t  1.2 \t\t  -24 \t\t-0.041 \t\t48 \t \t  -0.053 \t \t \t1.106 \t\t \n",
-        " \n",
-        "C \t  1.2 \t\t  -12 \t\t-0.02 \t\t48 \t \t  -0.053 \t \t \t1.127 \t\t \n",
-        " \n",
-        "D \t  1.2 \t\t  0 \t\t0.0 \t\t48 \t \t  -0.053 \t \t \t1.147 \t\t \n",
-        " \n",
-        "E \t  1.2 \t\t  12 \t\t0.02 \t\t48 \t \t  -0.053 \t \t \t1.168 \t\t \n",
-        " \n",
-        "F \t  1.2 \t\t  24 \t\t0.041 \t\t48 \t \t  -0.053 \t \t \t1.188 \t\t \n",
-        " \n",
-        "G \t  1.2 \t\t  38 \t\t0.065 \t\t48 \t \t  -0.053 \t \t \t1.212 \t\t \n",
-        " \n",
-        "H \t  1.2 \t\t  38 \t\t0.065 \t\t-48 \t \t  0.053 \t \t \t1.317 \t\t \n",
-        " \n",
-        "I \t  1.2 \t\t  24 \t\t0.041 \t\t-48 \t \t  0.053 \t \t \t1.294 \t\t \n",
-        " \n",
-        "J \t  1.2 \t\t  12 \t\t0.02 \t\t-48 \t \t  0.053 \t \t \t1.273 \t\t \n",
-        " \n",
-        "K \t  1.2 \t\t  0 \t\t0.0 \t\t-48 \t \t  0.053 \t \t \t1.253 \t\t \n",
-        " \n",
-        "L \t  1.2 \t\t  -12 \t\t-0.02 \t\t-48 \t \t  0.053 \t \t \t1.232 \t\t \n",
-        " \n",
-        "M \t  1.2 \t\t  -24 \t\t-0.041 \t\t-48 \t \t  0.053 \t \t \t1.212 \t\t \n",
-        " \n",
-        "N \t  1.2 \t\t  -38 \t\t-0.065 \t\t-48 \t \t  0.053 \t \t \t1.188 \t\t \n",
-        " \n",
-        "the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.6:Pg-299"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.6\n",
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "#solving for d\n",
-      "def f(d):\n",
-      "    return (96+2*d)*d-2615.1\n",
-      "[x]=fsolve(f,19);\n",
-      "d1=x;\n",
-      "def f(d):\n",
-      "      return (96+4*d)*d-6046.4\n",
-      "[x]=fsolve(f,28);\n",
-      "d2=x;\n",
-      "d=max(d2,d1);\n",
-      "d=round(d)\n",
-      "#now coming to design part\n",
-      "h=d+3+1; #in inch\n",
-      "print h,\"is total slab thickness in inches\"\n",
-      "qa=1.082; # in kip/ft^2\n",
-      "qb=1.106; # in kip/ft^2\n",
-      "qm=1.212; # in kip/ft^2\n",
-      "qn=1.188; # in kip/ft^2\n",
-      "q1A=qa/2.0+qb/2.0;\n",
-      "print round(q1A,3),\"is force in strip ABMN in kip/ft**2\"\n",
-      "q2A=qm/2.0+qn/2.0;\n",
-      "print  round(q2A,3),\"is force in strip ABMN in kip/ft**2\"\n",
-      "q1=1.106/3+1.127/3+1.147/3;\n",
-      "print  round(q1,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
-      "q2=1.253/3+1.233/3+1.212/3;\n",
-      "print  round(q2,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
-      "q1=1.147/3+1.167/3+1.188/3;\n",
-      "print  round(q1,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
-      "q2=1.294/3+1.273/3+1.253/3;\n",
-      "print  round(q2,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
-      "q1=1.188/2+1.212/2;\n",
-      "print  round(q1,3),\"is force in strip FGHI in kip/ft**2\"\n",
-      "q2=1.318/2+1.294/2;\n",
-      "print  round(q2,3),\" is force in strip FGHI in kip/ft**2\"\n",
-      "#checking for force\n",
-      "#net soil reaction <load \n",
-      "netforce=1/2.0*(1.094+1.2)*14*96+1/2.0*(1.127+1.233)*24*96+1/2.0*(1.167+1.273)*24*96+1/2.0*(1.2+1.306)*14*96;\n",
-      "if netforce<8761 :\n",
-      "    print \"forces generated is OK\"\n",
-      "\n",
-      "#checking for reinforcement requirement\n",
-      "Q1=1.4*180+1.7*120;\n",
-      "Q2=1.4*360+1.7*200;\n",
-      "Q3=1.4*400+1.7*240;\n",
-      "Q4=1.4*180+1.7*120;\n",
-      "#solving for a\n",
-      "def f(a):\n",
-      "    return 0.9*0.51*a*60*(29-a/2)-95.05*12\n",
-      "[x]=fsolve(f,1.4);\n",
-      "a=x;\n",
-      "As=0.51*a\n",
-      "fy=60000;\n",
-      "print round(As,2),\"is required area in in**2\"\n",
-      "Asmin=200.0/fy*12*29;\n",
-      "print Asmin,\" is minimum reinforcement required in**2/ft\"\n",
-      "print \"use No 9 bars at 10 inch centre to centre\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "33.0 is total slab thickness in inches\n",
-        "1.094 is force in strip ABMN in kip/ft**2\n",
-        "1.2 is force in strip ABMN in kip/ft**2\n",
-        "1.127 is force in strip BCDKLM in kip/ft**2\n",
-        "1.233 is force in strip BCDKLM in kip/ft**2\n",
-        "1.167 is force in strip DEFIJK in kip/ft**2\n",
-        "1.273 is force in strip DEFIJK in kip/ft**2\n",
-        "1.2 is force in strip FGHI in kip/ft**2\n",
-        "1.306  is force in strip FGHI in kip/ft**2\n",
-        "forces generated is OK\n",
-        "0.75 is required area in in**2\n",
-        "1.16  is minimum reinforcement required in**2/ft\n",
-        "use No 9 bars at 10 inch centre to centre\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter06_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter06_1.ipynb
deleted file mode 100755
index d87679c7..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter06_1.ipynb
+++ /dev/null
@@ -1,351 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ea06255a04932e0bf9952689cf2f6d919d320d1161a790bec574191c6fbafa1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter06: Mat Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.1:Pg-279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.1\n",
-      "\n",
-      "B=30; # in ft\n",
-      "L=45; # in ft\n",
-      "Df=6.5; # in ft\n",
-      "cu=1950;# in lb/ft^2\n",
-      "qunet=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B);\n",
-      "print int(qunet),\" is allowed force in lb/ft**2\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "12307  is allowed force in lb/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.2:Pg-280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.2\n",
-      "\n",
-      "N60=10; # penetration number\n",
-      "Df=2; # in m\n",
-      "B=10.0; # in m\n",
-      "Se=25.0; # in mm\n",
-      "qnetall=N60/0.08*(1+0.33*Df/B)*Se/25;\n",
-      "print qnetall,\" is allowed pressure in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "133.25  is allowed pressure in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.3:Pg-283"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.3\n",
-      "\n",
-      "cu=2800; # in lb/ft^2\n",
-      "B=60; # in ft\n",
-      "L=100; # in ft\n",
-      "Df=5; # in ft\n",
-      "\n",
-      "Gamma=120; # in lb/ft^3\n",
-      "A=60*100; # in ft^2\n",
-      "Q=25e6; # load in Kip\n",
-      "FS=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B)/(Q/A-Gamma*Df);\n",
-      "print round(FS,2),\" is factor of safety\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4.66  is factor of safety\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.4:Pg-284"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.4\n",
-      "\n",
-      "import math\n",
-      "Cc=0.28;\n",
-      "Hc=18*12.0;\n",
-      "e0=0.9;\n",
-      "sigmao=11*100+40*(121.5-64)+18/2*(118-62.4); # in lb/ft^2\n",
-      "H2=5+40+18.0;\n",
-      "H1=5+40.0;\n",
-      "qo=3567.0;\n",
-      "#from table\n",
-      "IaH2=0.21;\n",
-      "IaH1=0.225;\n",
-      "Dsigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1))*4;\n",
-      "Scp=Cc*Hc/(1+e0)*math.log10(sigmao/sigmao+Dsigma/sigmao);\n",
-      "print round(Scp,2),\"is settlement in inches\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6.76 is settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.5:Pg-296"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.5\n",
-      "import numpy\n",
-      "P=['A','B','C','D','E','F','G','H','I','J','K','L','M','N'];#point\n",
-      "k=1.2*numpy.ones(14);#Q/A\n",
-      "x=[-38,-24, -12, 0, 12, 24, 38, 38, 24, 12, 0, -12, -24, -38];\n",
-      "x1=numpy.zeros(14)\n",
-      "for i in range(0,14):\n",
-      "    x1[i]=0.0017*x[i];\n",
-      "y=[48,48,48,48,48,48,48, -48, -48, -48, -48, -48, -48, -48];\n",
-      "y1=numpy.zeros(14)\n",
-      "for i in range(0,14):\n",
-      "    y1[i]=-0.0011*y[i];\n",
-      "print \"point\\t Q\\A (kip/ft**2)  x(ft)\\t 0.0017x(ft)\\t\\ty(ft)\\t \\t  0.0011y(ft)\\t  \\t  q(kip/ft**2)\\n\"\n",
-      "q=numpy.zeros(14)\n",
-      "for i in  range(0,14):\n",
-      "    q[i]=1.2+x1[i]+y1[i];\n",
-      "    print P[i],\"\\t \",k[i],\"\\t\\t \",x[i],\"\\t\\t\",round(x1[i],3),\"\\t\\t\",y[i],\"\\t \\t \",round(y1[i],3),\"\\t \\t \\t\",round(q[i],3),\"\\t\\t \\n \"\n",
-      "\n",
-      "print \"the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "point\t Q\\A (kip/ft**2)  x(ft)\t 0.0017x(ft)\t\ty(ft)\t \t  0.0011y(ft)\t  \t  q(kip/ft**2)\n",
-        "\n",
-        "A \t  1.2 \t\t  -38 \t\t-0.065 \t\t48 \t \t  -0.053 \t \t \t1.083 \t\t \n",
-        " \n",
-        "B \t  1.2 \t\t  -24 \t\t-0.041 \t\t48 \t \t  -0.053 \t \t \t1.106 \t\t \n",
-        " \n",
-        "C \t  1.2 \t\t  -12 \t\t-0.02 \t\t48 \t \t  -0.053 \t \t \t1.127 \t\t \n",
-        " \n",
-        "D \t  1.2 \t\t  0 \t\t0.0 \t\t48 \t \t  -0.053 \t \t \t1.147 \t\t \n",
-        " \n",
-        "E \t  1.2 \t\t  12 \t\t0.02 \t\t48 \t \t  -0.053 \t \t \t1.168 \t\t \n",
-        " \n",
-        "F \t  1.2 \t\t  24 \t\t0.041 \t\t48 \t \t  -0.053 \t \t \t1.188 \t\t \n",
-        " \n",
-        "G \t  1.2 \t\t  38 \t\t0.065 \t\t48 \t \t  -0.053 \t \t \t1.212 \t\t \n",
-        " \n",
-        "H \t  1.2 \t\t  38 \t\t0.065 \t\t-48 \t \t  0.053 \t \t \t1.317 \t\t \n",
-        " \n",
-        "I \t  1.2 \t\t  24 \t\t0.041 \t\t-48 \t \t  0.053 \t \t \t1.294 \t\t \n",
-        " \n",
-        "J \t  1.2 \t\t  12 \t\t0.02 \t\t-48 \t \t  0.053 \t \t \t1.273 \t\t \n",
-        " \n",
-        "K \t  1.2 \t\t  0 \t\t0.0 \t\t-48 \t \t  0.053 \t \t \t1.253 \t\t \n",
-        " \n",
-        "L \t  1.2 \t\t  -12 \t\t-0.02 \t\t-48 \t \t  0.053 \t \t \t1.232 \t\t \n",
-        " \n",
-        "M \t  1.2 \t\t  -24 \t\t-0.041 \t\t-48 \t \t  0.053 \t \t \t1.212 \t\t \n",
-        " \n",
-        "N \t  1.2 \t\t  -38 \t\t-0.065 \t\t-48 \t \t  0.053 \t \t \t1.188 \t\t \n",
-        " \n",
-        "the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.6:Pg-299"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.6\n",
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "#solving for d\n",
-      "def f(d):\n",
-      "    return (96+2*d)*d-2615.1\n",
-      "[x]=fsolve(f,19);\n",
-      "d1=x;\n",
-      "def f(d):\n",
-      "      return (96+4*d)*d-6046.4\n",
-      "[x]=fsolve(f,28);\n",
-      "d2=x;\n",
-      "d=max(d2,d1);\n",
-      "d=round(d)\n",
-      "#now coming to design part\n",
-      "h=d+3+1; #in inch\n",
-      "print h,\"is total slab thickness in inches\"\n",
-      "qa=1.082; # in kip/ft^2\n",
-      "qb=1.106; # in kip/ft^2\n",
-      "qm=1.212; # in kip/ft^2\n",
-      "qn=1.188; # in kip/ft^2\n",
-      "q1A=qa/2.0+qb/2.0;\n",
-      "print round(q1A,3),\"is force in strip ABMN in kip/ft**2\"\n",
-      "q2A=qm/2.0+qn/2.0;\n",
-      "print  round(q2A,3),\"is force in strip ABMN in kip/ft**2\"\n",
-      "q1=1.106/3+1.127/3+1.147/3;\n",
-      "print  round(q1,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
-      "q2=1.253/3+1.233/3+1.212/3;\n",
-      "print  round(q2,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
-      "q1=1.147/3+1.167/3+1.188/3;\n",
-      "print  round(q1,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
-      "q2=1.294/3+1.273/3+1.253/3;\n",
-      "print  round(q2,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
-      "q1=1.188/2+1.212/2;\n",
-      "print  round(q1,3),\"is force in strip FGHI in kip/ft**2\"\n",
-      "q2=1.318/2+1.294/2;\n",
-      "print  round(q2,3),\" is force in strip FGHI in kip/ft**2\"\n",
-      "#checking for force\n",
-      "#net soil reaction <load \n",
-      "netforce=1/2.0*(1.094+1.2)*14*96+1/2.0*(1.127+1.233)*24*96+1/2.0*(1.167+1.273)*24*96+1/2.0*(1.2+1.306)*14*96;\n",
-      "if netforce<8761 :\n",
-      "    print \"forces generated is OK\"\n",
-      "\n",
-      "#checking for reinforcement requirement\n",
-      "Q1=1.4*180+1.7*120;\n",
-      "Q2=1.4*360+1.7*200;\n",
-      "Q3=1.4*400+1.7*240;\n",
-      "Q4=1.4*180+1.7*120;\n",
-      "#solving for a\n",
-      "def f(a):\n",
-      "    return 0.9*0.51*a*60*(29-a/2)-95.05*12\n",
-      "[x]=fsolve(f,1.4);\n",
-      "a=x;\n",
-      "As=0.51*a\n",
-      "fy=60000;\n",
-      "print round(As,2),\"is required area in in**2\"\n",
-      "Asmin=200.0/fy*12*29;\n",
-      "print Asmin,\" is minimum reinforcement required in**2/ft\"\n",
-      "print \"use No 9 bars at 10 inch centre to centre\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "33.0 is total slab thickness in inches\n",
-        "1.094 is force in strip ABMN in kip/ft**2\n",
-        "1.2 is force in strip ABMN in kip/ft**2\n",
-        "1.127 is force in strip BCDKLM in kip/ft**2\n",
-        "1.233 is force in strip BCDKLM in kip/ft**2\n",
-        "1.167 is force in strip DEFIJK in kip/ft**2\n",
-        "1.273 is force in strip DEFIJK in kip/ft**2\n",
-        "1.2 is force in strip FGHI in kip/ft**2\n",
-        "1.306  is force in strip FGHI in kip/ft**2\n",
-        "forces generated is OK\n",
-        "0.75 is required area in in**2\n",
-        "1.16  is minimum reinforcement required in**2/ft\n",
-        "use No 9 bars at 10 inch centre to centre\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter06_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter06_2.ipynb
deleted file mode 100755
index d87679c7..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter06_2.ipynb
+++ /dev/null
@@ -1,351 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ea06255a04932e0bf9952689cf2f6d919d320d1161a790bec574191c6fbafa1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter06: Mat Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.1:Pg-279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.1\n",
-      "\n",
-      "B=30; # in ft\n",
-      "L=45; # in ft\n",
-      "Df=6.5; # in ft\n",
-      "cu=1950;# in lb/ft^2\n",
-      "qunet=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B);\n",
-      "print int(qunet),\" is allowed force in lb/ft**2\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "12307  is allowed force in lb/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.2:Pg-280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.2\n",
-      "\n",
-      "N60=10; # penetration number\n",
-      "Df=2; # in m\n",
-      "B=10.0; # in m\n",
-      "Se=25.0; # in mm\n",
-      "qnetall=N60/0.08*(1+0.33*Df/B)*Se/25;\n",
-      "print qnetall,\" is allowed pressure in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "133.25  is allowed pressure in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.3:Pg-283"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.3\n",
-      "\n",
-      "cu=2800; # in lb/ft^2\n",
-      "B=60; # in ft\n",
-      "L=100; # in ft\n",
-      "Df=5; # in ft\n",
-      "\n",
-      "Gamma=120; # in lb/ft^3\n",
-      "A=60*100; # in ft^2\n",
-      "Q=25e6; # load in Kip\n",
-      "FS=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B)/(Q/A-Gamma*Df);\n",
-      "print round(FS,2),\" is factor of safety\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4.66  is factor of safety\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.4:Pg-284"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.4\n",
-      "\n",
-      "import math\n",
-      "Cc=0.28;\n",
-      "Hc=18*12.0;\n",
-      "e0=0.9;\n",
-      "sigmao=11*100+40*(121.5-64)+18/2*(118-62.4); # in lb/ft^2\n",
-      "H2=5+40+18.0;\n",
-      "H1=5+40.0;\n",
-      "qo=3567.0;\n",
-      "#from table\n",
-      "IaH2=0.21;\n",
-      "IaH1=0.225;\n",
-      "Dsigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1))*4;\n",
-      "Scp=Cc*Hc/(1+e0)*math.log10(sigmao/sigmao+Dsigma/sigmao);\n",
-      "print round(Scp,2),\"is settlement in inches\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6.76 is settlement in inches\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.5:Pg-296"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.5\n",
-      "import numpy\n",
-      "P=['A','B','C','D','E','F','G','H','I','J','K','L','M','N'];#point\n",
-      "k=1.2*numpy.ones(14);#Q/A\n",
-      "x=[-38,-24, -12, 0, 12, 24, 38, 38, 24, 12, 0, -12, -24, -38];\n",
-      "x1=numpy.zeros(14)\n",
-      "for i in range(0,14):\n",
-      "    x1[i]=0.0017*x[i];\n",
-      "y=[48,48,48,48,48,48,48, -48, -48, -48, -48, -48, -48, -48];\n",
-      "y1=numpy.zeros(14)\n",
-      "for i in range(0,14):\n",
-      "    y1[i]=-0.0011*y[i];\n",
-      "print \"point\\t Q\\A (kip/ft**2)  x(ft)\\t 0.0017x(ft)\\t\\ty(ft)\\t \\t  0.0011y(ft)\\t  \\t  q(kip/ft**2)\\n\"\n",
-      "q=numpy.zeros(14)\n",
-      "for i in  range(0,14):\n",
-      "    q[i]=1.2+x1[i]+y1[i];\n",
-      "    print P[i],\"\\t \",k[i],\"\\t\\t \",x[i],\"\\t\\t\",round(x1[i],3),\"\\t\\t\",y[i],\"\\t \\t \",round(y1[i],3),\"\\t \\t \\t\",round(q[i],3),\"\\t\\t \\n \"\n",
-      "\n",
-      "print \"the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "point\t Q\\A (kip/ft**2)  x(ft)\t 0.0017x(ft)\t\ty(ft)\t \t  0.0011y(ft)\t  \t  q(kip/ft**2)\n",
-        "\n",
-        "A \t  1.2 \t\t  -38 \t\t-0.065 \t\t48 \t \t  -0.053 \t \t \t1.083 \t\t \n",
-        " \n",
-        "B \t  1.2 \t\t  -24 \t\t-0.041 \t\t48 \t \t  -0.053 \t \t \t1.106 \t\t \n",
-        " \n",
-        "C \t  1.2 \t\t  -12 \t\t-0.02 \t\t48 \t \t  -0.053 \t \t \t1.127 \t\t \n",
-        " \n",
-        "D \t  1.2 \t\t  0 \t\t0.0 \t\t48 \t \t  -0.053 \t \t \t1.147 \t\t \n",
-        " \n",
-        "E \t  1.2 \t\t  12 \t\t0.02 \t\t48 \t \t  -0.053 \t \t \t1.168 \t\t \n",
-        " \n",
-        "F \t  1.2 \t\t  24 \t\t0.041 \t\t48 \t \t  -0.053 \t \t \t1.188 \t\t \n",
-        " \n",
-        "G \t  1.2 \t\t  38 \t\t0.065 \t\t48 \t \t  -0.053 \t \t \t1.212 \t\t \n",
-        " \n",
-        "H \t  1.2 \t\t  38 \t\t0.065 \t\t-48 \t \t  0.053 \t \t \t1.317 \t\t \n",
-        " \n",
-        "I \t  1.2 \t\t  24 \t\t0.041 \t\t-48 \t \t  0.053 \t \t \t1.294 \t\t \n",
-        " \n",
-        "J \t  1.2 \t\t  12 \t\t0.02 \t\t-48 \t \t  0.053 \t \t \t1.273 \t\t \n",
-        " \n",
-        "K \t  1.2 \t\t  0 \t\t0.0 \t\t-48 \t \t  0.053 \t \t \t1.253 \t\t \n",
-        " \n",
-        "L \t  1.2 \t\t  -12 \t\t-0.02 \t\t-48 \t \t  0.053 \t \t \t1.232 \t\t \n",
-        " \n",
-        "M \t  1.2 \t\t  -24 \t\t-0.041 \t\t-48 \t \t  0.053 \t \t \t1.212 \t\t \n",
-        " \n",
-        "N \t  1.2 \t\t  -38 \t\t-0.065 \t\t-48 \t \t  0.053 \t \t \t1.188 \t\t \n",
-        " \n",
-        "the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.6:Pg-299"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 6.6\n",
-      "\n",
-      "from scipy.optimize import fsolve\n",
-      "#solving for d\n",
-      "def f(d):\n",
-      "    return (96+2*d)*d-2615.1\n",
-      "[x]=fsolve(f,19);\n",
-      "d1=x;\n",
-      "def f(d):\n",
-      "      return (96+4*d)*d-6046.4\n",
-      "[x]=fsolve(f,28);\n",
-      "d2=x;\n",
-      "d=max(d2,d1);\n",
-      "d=round(d)\n",
-      "#now coming to design part\n",
-      "h=d+3+1; #in inch\n",
-      "print h,\"is total slab thickness in inches\"\n",
-      "qa=1.082; # in kip/ft^2\n",
-      "qb=1.106; # in kip/ft^2\n",
-      "qm=1.212; # in kip/ft^2\n",
-      "qn=1.188; # in kip/ft^2\n",
-      "q1A=qa/2.0+qb/2.0;\n",
-      "print round(q1A,3),\"is force in strip ABMN in kip/ft**2\"\n",
-      "q2A=qm/2.0+qn/2.0;\n",
-      "print  round(q2A,3),\"is force in strip ABMN in kip/ft**2\"\n",
-      "q1=1.106/3+1.127/3+1.147/3;\n",
-      "print  round(q1,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
-      "q2=1.253/3+1.233/3+1.212/3;\n",
-      "print  round(q2,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
-      "q1=1.147/3+1.167/3+1.188/3;\n",
-      "print  round(q1,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
-      "q2=1.294/3+1.273/3+1.253/3;\n",
-      "print  round(q2,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
-      "q1=1.188/2+1.212/2;\n",
-      "print  round(q1,3),\"is force in strip FGHI in kip/ft**2\"\n",
-      "q2=1.318/2+1.294/2;\n",
-      "print  round(q2,3),\" is force in strip FGHI in kip/ft**2\"\n",
-      "#checking for force\n",
-      "#net soil reaction <load \n",
-      "netforce=1/2.0*(1.094+1.2)*14*96+1/2.0*(1.127+1.233)*24*96+1/2.0*(1.167+1.273)*24*96+1/2.0*(1.2+1.306)*14*96;\n",
-      "if netforce<8761 :\n",
-      "    print \"forces generated is OK\"\n",
-      "\n",
-      "#checking for reinforcement requirement\n",
-      "Q1=1.4*180+1.7*120;\n",
-      "Q2=1.4*360+1.7*200;\n",
-      "Q3=1.4*400+1.7*240;\n",
-      "Q4=1.4*180+1.7*120;\n",
-      "#solving for a\n",
-      "def f(a):\n",
-      "    return 0.9*0.51*a*60*(29-a/2)-95.05*12\n",
-      "[x]=fsolve(f,1.4);\n",
-      "a=x;\n",
-      "As=0.51*a\n",
-      "fy=60000;\n",
-      "print round(As,2),\"is required area in in**2\"\n",
-      "Asmin=200.0/fy*12*29;\n",
-      "print Asmin,\" is minimum reinforcement required in**2/ft\"\n",
-      "print \"use No 9 bars at 10 inch centre to centre\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "33.0 is total slab thickness in inches\n",
-        "1.094 is force in strip ABMN in kip/ft**2\n",
-        "1.2 is force in strip ABMN in kip/ft**2\n",
-        "1.127 is force in strip BCDKLM in kip/ft**2\n",
-        "1.233 is force in strip BCDKLM in kip/ft**2\n",
-        "1.167 is force in strip DEFIJK in kip/ft**2\n",
-        "1.273 is force in strip DEFIJK in kip/ft**2\n",
-        "1.2 is force in strip FGHI in kip/ft**2\n",
-        "1.306  is force in strip FGHI in kip/ft**2\n",
-        "forces generated is OK\n",
-        "0.75 is required area in in**2\n",
-        "1.16  is minimum reinforcement required in**2/ft\n",
-        "use No 9 bars at 10 inch centre to centre\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter07.ipynb b/Principles_Of_Foundation_Engineering/Chapter07.ipynb
deleted file mode 100755
index 63b3bc22..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter07.ipynb
+++ /dev/null
@@ -1,358 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:6111670c8f96effbf1bc0bc29859353d67e53cd5b906d2b571173d1698f7ba9c"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter07:Lateral earth pressure"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.1:Pg-319"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.1\n",

-      "\n",

-      "sigmao=48.0; # in KN/m^2\n",

-      "phi1=30*math.pi/180; # angle\n",

-      "phi2=36*math.pi/180; # angle\n",

-      "Ka1=(math.tan(math.pi/4.0-phi1/2))**2;\n",

-      "Ka2=(math.tan(math.pi/4.0-phi2/2))**2;\n",

-      "sigmaa1=Ka1*sigmao; # in KN/m^2\n",

-      "print round(sigmaa1,2),\"top soil pressure in kN/m**2\"\n",

-      "sigmaa2=Ka2*sigmao; # in KN/m^2\n",

-      "print round(sigmaa2,2),\"bottom soil pressure in kN/m**2\"\n",

-      "Po=1/2.0*3*16+3*12.48+1/3.0*3*(19.65-12.48)+1/2.0*3*29.43;\n",

-      "zbar=(24*(3+3/3.0)+37.44*(3/2.0)+10.76*3/3.0+44.1*3/3.0)/Po;\n",

-      "print round(zbar,2),\"resultant force acting from the bottom in m\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "16.0 top soil pressure in kN/m**2\n",

-        "12.46 bottom soil pressure in kN/m**2\n",

-        "1.84 resultant force acting from the bottom in m\n"

-       ]

-      }

-     ],

-     "prompt_number": 34

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.2:Pg-321"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.2\n",

-      "\n",

-      "import math\n",

-      "c=14.36;\n",

-      "Gamma=17.4; #  in KN/m^3\n",

-      "H=6; # in m\n",

-      "phi=26*math.pi/180;\n",

-      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

-      "sigma0=Gamma*H*Ka-2*c*math.sqrt(Ka);\n",

-      "Pa=1/2.0*Gamma*H**2*Ka-2*c*H*math.sqrt(Ka);\n",

-      "print round(Pa,2),\"active force before which tensile crack appeared in kN/m\"\n",

-      "zbar=(244.32-323.1)/14.46;\n",

-      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

-      "zc=2*c/Gamma/math.sqrt(Ka);\n",

-      "print round(zc,2),\"distance where tensile crack appeared in m\"\n",

-      "Pa=1/2.0*(H-zc)*(Gamma*H*Ka-2*c*math.sqrt(Ka));\n",

-      "print round(Pa,2),\"Active force in tensile crack in kN/m\"\n",

-      "zbar=(H-zc)/3;\n",

-      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "14.62 active force before which tensile crack appeared in kN/m\n",

-        "-5.45 the line of action on which net force is acting in m\n",

-        "2.64 distance where tensile crack appeared in m\n",

-        "38.32 Active force in tensile crack in kN/m\n",

-        "1.12 the line of action on which net force is acting in m\n"

-       ]

-      }

-     ],

-     "prompt_number": 30

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.3:Pg-322"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.3\n",

-      "import math\n",

-      "pi=math.pi\n",

-      "H=10.0; # in ft\n",

-      "Gamma=110.0; # in lb/ft^3\n",

-      "phi=35*math.pi/180.0; # angle\n",

-      "alpha=15*math.pi/180.0; # angle\n",

-      "theta=10*math.pi/180.0; # angle\n",

-      "zi=math.sin(math.sin(alpha)/math.sin(phi))-alpha+2*theta;\n",

-      "print round(zi*180.0/math.pi,2),\" is zi in degrees\"\n",

-      "Ka=math.cos(alpha-theta)*math.sqrt(1+(math.sin(phi))**2-2*math.sin(phi)*math.sin(zi))/((math.cos(theta))**2*(math.cos(alpha)+math.sqrt((math.sin(phi))**2+((math.sin(alpha))**2))));\n",

-      "Pa=1/2.0*Gamma*H**2*Ka;\n",

-      "print round(Pa,2),\" is rankine earth pressure in lb/ft\"\n",

-      "print \"there is slight error in answer due to rounding off error\"\n",

-      "Beta=math.tan(math.sin(phi)*math.sin(zi)/(1-math.sin(phi)*math.cos(zi)));\n",

-      "print round(Beta*180/pi,2),\" is angle in degrees\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "29.99  is zi in degrees\n",

-        "3078.61  is rankine earth pressure in lb/ft\n",

-        "there is slight error in answer due to rounding off error\n",

-        "36.7  is angle in degrees\n"

-       ]

-      }

-     ],

-     "prompt_number": 25

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.4:Pg-326"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.4\n",

-      "\n",

-      "H=4.6; # in m\n",

-      "Gamma=16.5; # in KN/m^3\n",

-      "Ka=0.297;\n",

-      "Po=1/2.0*Gamma*H**2*Ka;\n",

-      "print round(Po,2),\"coulomb active force per unit length in kN/m\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "51.85 coulomb active force per unit length in kN/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.5:Pg-331"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.5\n",

-      "\n",

-      "#part(a)\n",

-      "Gamma=105; # in lb/ft^3\n",

-      "H=10; #in ft\n",

-      "Kae=0.474;\n",

-      "k1=0;\n",

-      "Pae=1/2.0*Gamma*H**2*Kae*(1-k1) # in lb/ft\n",

-      "print Pae,\"active force in lb/ft\"\n",

-      "#part(b)\n",

-      "Ka=0.246;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka; # in lb/ft\n",

-      "print Pa,\"active force in lb/ft\"\n",

-      "DPae=Pae-Pa;#deltaPae\n",

-      "zbar=(0.6*H*DPae+H/3*(Pa))/Pae;\n",

-      "print round(zbar,2),\"the distance of resultant force from bottom in ft\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "2488.5 active force in lb/ft\n",

-        "1291.5 active force in lb/ft\n",

-        "4.44 the distance of resultant force from bottom in ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.6:Pg-337"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.6\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "import matplotlib.pyplot as plt\n",

-      "\n",

-      "z=[0, 4, 8, 12, 16];\n",

-      "Gamma=110; # in lb/ft^3\n",

-      "phi=36*math.pi/180;\n",

-      "H=16; # in ft\n",

-      "Sa1=numpy.zeros(5);#sigma(1)\n",

-      "Sa2=numpy.zeros(5);#sigma(2)\n",

-      "Sztr=numpy.zeros(5);#sigma(z)translation\n",

-      "print \"z(ft)\\t sigma(1)(lb/ft**2)   \\t sigma(2)(lb/ft**2)    \\t sigma(z)translation (lb/ft**2)\\n\"\n",

-      "for i in range(0,5):\n",

-      "    Sa1[i]=Gamma*(math.tan(math.pi/4-phi*z[i]/2/H))**2*(z[i]-phi*z[i]**2/H/math.cos(phi*z[i]/H));\n",

-      "    Sa2[i]=Gamma*z[i]*(math.cos(phi)/(1+math.sin(phi)))**2;\n",

-      "    Sztr[i]=Sa1[i]/2.0+Sa2[i]/2.0;\n",

-      "    print  round(z[i],2),\"\\t \",round(Sa1[i],2),\"\\t\\t\\t \",round(Sa2[i],2),\"\\t\\t\\t \",round(Sztr[i],2),\"\\n\"\n",

-      "plt.plot(Sztr,z);\n",

-      "plt.title(\"sigma(z)translation vs z\")\n",

-      "plt.ylabel(\"z(ft)\")\n",

-      "plt.xlabel(\"sigma(z)translation (lb/ft**2)\")\n",

-      "plt.show()\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "z(ft)\t sigma(1)(lb/ft**2)   \t sigma(2)(lb/ft**2)    \t sigma(z)translation (lb/ft**2)\n",

-        "\n",

-        "0.0 \t  0.0 \t\t\t  0.0 \t\t\t  0.0 \n",

-        "\n",

-        "4.0 \t  269.92 \t\t\t  114.23 \t\t\t  192.07 \n",

-        "\n",

-        "8.0 \t  311.08 \t\t\t  228.46 \t\t\t  269.77 \n",

-        "\n",

-        "12.0 \t  233.53 \t\t\t  342.69 \t\t\t  288.11 \n",

-        "\n",

-        "16.0 \t  102.06 \t\t\t  456.92 \t\t\t  279.49 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.7:Pg-342"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.7\n",

-      "import math\n",

-      "Gammasat=18.86; # in KN/m^3\n",

-      "Gammaw=9.81; # in KN/m^3\n",

-      "phi1=math.pi/180*30;  # angle 1\n",

-      "phi2=math.pi/180*26; # angle 2\n",

-      "Kp1=(math.tan(math.pi/4+phi1/2))**2;\n",

-      "Kp2=(math.tan(math.pi/4+phi2/2))**2;\n",

-      "#for top soil\n",

-      "c=0;\n",

-      "sigma0=31.44; # in KN/m^2\n",

-      "sigmap=sigma0*Kp1+2*c*math.sqrt(Kp1);\n",

-      "print round(sigmap,2),\"passive pressure for top layer in kN/m**2\"\n",

-      "#for z=2\n",

-      "c=10;\n",

-      "sigma0=31.44; # in KN/m^2\n",

-      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2);\n",

-      "print round(sigmap,2),\"passive pressure for z=2m in kN/m**2\"\n",

-      "#for z=3\n",

-      "c=10;\n",

-      "sigma0=15.72*2+(Gammasat-Gammaw)*1; # in KN/m^2\n",

-      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2); # in KN/m^2\n",

-      "print round(sigmap,2),\" is passive pressure for z=3m in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "94.32 passive pressure for top layer in kN/m**2\n",

-        "112.53 passive pressure for z=2m in kN/m**2\n",

-        "135.7  is passive pressure for z=3m in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter07_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter07_1.ipynb
deleted file mode 100755
index 63b3bc22..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter07_1.ipynb
+++ /dev/null
@@ -1,358 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:6111670c8f96effbf1bc0bc29859353d67e53cd5b906d2b571173d1698f7ba9c"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter07:Lateral earth pressure"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.1:Pg-319"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.1\n",

-      "\n",

-      "sigmao=48.0; # in KN/m^2\n",

-      "phi1=30*math.pi/180; # angle\n",

-      "phi2=36*math.pi/180; # angle\n",

-      "Ka1=(math.tan(math.pi/4.0-phi1/2))**2;\n",

-      "Ka2=(math.tan(math.pi/4.0-phi2/2))**2;\n",

-      "sigmaa1=Ka1*sigmao; # in KN/m^2\n",

-      "print round(sigmaa1,2),\"top soil pressure in kN/m**2\"\n",

-      "sigmaa2=Ka2*sigmao; # in KN/m^2\n",

-      "print round(sigmaa2,2),\"bottom soil pressure in kN/m**2\"\n",

-      "Po=1/2.0*3*16+3*12.48+1/3.0*3*(19.65-12.48)+1/2.0*3*29.43;\n",

-      "zbar=(24*(3+3/3.0)+37.44*(3/2.0)+10.76*3/3.0+44.1*3/3.0)/Po;\n",

-      "print round(zbar,2),\"resultant force acting from the bottom in m\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "16.0 top soil pressure in kN/m**2\n",

-        "12.46 bottom soil pressure in kN/m**2\n",

-        "1.84 resultant force acting from the bottom in m\n"

-       ]

-      }

-     ],

-     "prompt_number": 34

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.2:Pg-321"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.2\n",

-      "\n",

-      "import math\n",

-      "c=14.36;\n",

-      "Gamma=17.4; #  in KN/m^3\n",

-      "H=6; # in m\n",

-      "phi=26*math.pi/180;\n",

-      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

-      "sigma0=Gamma*H*Ka-2*c*math.sqrt(Ka);\n",

-      "Pa=1/2.0*Gamma*H**2*Ka-2*c*H*math.sqrt(Ka);\n",

-      "print round(Pa,2),\"active force before which tensile crack appeared in kN/m\"\n",

-      "zbar=(244.32-323.1)/14.46;\n",

-      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

-      "zc=2*c/Gamma/math.sqrt(Ka);\n",

-      "print round(zc,2),\"distance where tensile crack appeared in m\"\n",

-      "Pa=1/2.0*(H-zc)*(Gamma*H*Ka-2*c*math.sqrt(Ka));\n",

-      "print round(Pa,2),\"Active force in tensile crack in kN/m\"\n",

-      "zbar=(H-zc)/3;\n",

-      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "14.62 active force before which tensile crack appeared in kN/m\n",

-        "-5.45 the line of action on which net force is acting in m\n",

-        "2.64 distance where tensile crack appeared in m\n",

-        "38.32 Active force in tensile crack in kN/m\n",

-        "1.12 the line of action on which net force is acting in m\n"

-       ]

-      }

-     ],

-     "prompt_number": 30

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.3:Pg-322"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.3\n",

-      "import math\n",

-      "pi=math.pi\n",

-      "H=10.0; # in ft\n",

-      "Gamma=110.0; # in lb/ft^3\n",

-      "phi=35*math.pi/180.0; # angle\n",

-      "alpha=15*math.pi/180.0; # angle\n",

-      "theta=10*math.pi/180.0; # angle\n",

-      "zi=math.sin(math.sin(alpha)/math.sin(phi))-alpha+2*theta;\n",

-      "print round(zi*180.0/math.pi,2),\" is zi in degrees\"\n",

-      "Ka=math.cos(alpha-theta)*math.sqrt(1+(math.sin(phi))**2-2*math.sin(phi)*math.sin(zi))/((math.cos(theta))**2*(math.cos(alpha)+math.sqrt((math.sin(phi))**2+((math.sin(alpha))**2))));\n",

-      "Pa=1/2.0*Gamma*H**2*Ka;\n",

-      "print round(Pa,2),\" is rankine earth pressure in lb/ft\"\n",

-      "print \"there is slight error in answer due to rounding off error\"\n",

-      "Beta=math.tan(math.sin(phi)*math.sin(zi)/(1-math.sin(phi)*math.cos(zi)));\n",

-      "print round(Beta*180/pi,2),\" is angle in degrees\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "29.99  is zi in degrees\n",

-        "3078.61  is rankine earth pressure in lb/ft\n",

-        "there is slight error in answer due to rounding off error\n",

-        "36.7  is angle in degrees\n"

-       ]

-      }

-     ],

-     "prompt_number": 25

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.4:Pg-326"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.4\n",

-      "\n",

-      "H=4.6; # in m\n",

-      "Gamma=16.5; # in KN/m^3\n",

-      "Ka=0.297;\n",

-      "Po=1/2.0*Gamma*H**2*Ka;\n",

-      "print round(Po,2),\"coulomb active force per unit length in kN/m\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "51.85 coulomb active force per unit length in kN/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.5:Pg-331"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.5\n",

-      "\n",

-      "#part(a)\n",

-      "Gamma=105; # in lb/ft^3\n",

-      "H=10; #in ft\n",

-      "Kae=0.474;\n",

-      "k1=0;\n",

-      "Pae=1/2.0*Gamma*H**2*Kae*(1-k1) # in lb/ft\n",

-      "print Pae,\"active force in lb/ft\"\n",

-      "#part(b)\n",

-      "Ka=0.246;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka; # in lb/ft\n",

-      "print Pa,\"active force in lb/ft\"\n",

-      "DPae=Pae-Pa;#deltaPae\n",

-      "zbar=(0.6*H*DPae+H/3*(Pa))/Pae;\n",

-      "print round(zbar,2),\"the distance of resultant force from bottom in ft\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "2488.5 active force in lb/ft\n",

-        "1291.5 active force in lb/ft\n",

-        "4.44 the distance of resultant force from bottom in ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.6:Pg-337"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.6\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "import matplotlib.pyplot as plt\n",

-      "\n",

-      "z=[0, 4, 8, 12, 16];\n",

-      "Gamma=110; # in lb/ft^3\n",

-      "phi=36*math.pi/180;\n",

-      "H=16; # in ft\n",

-      "Sa1=numpy.zeros(5);#sigma(1)\n",

-      "Sa2=numpy.zeros(5);#sigma(2)\n",

-      "Sztr=numpy.zeros(5);#sigma(z)translation\n",

-      "print \"z(ft)\\t sigma(1)(lb/ft**2)   \\t sigma(2)(lb/ft**2)    \\t sigma(z)translation (lb/ft**2)\\n\"\n",

-      "for i in range(0,5):\n",

-      "    Sa1[i]=Gamma*(math.tan(math.pi/4-phi*z[i]/2/H))**2*(z[i]-phi*z[i]**2/H/math.cos(phi*z[i]/H));\n",

-      "    Sa2[i]=Gamma*z[i]*(math.cos(phi)/(1+math.sin(phi)))**2;\n",

-      "    Sztr[i]=Sa1[i]/2.0+Sa2[i]/2.0;\n",

-      "    print  round(z[i],2),\"\\t \",round(Sa1[i],2),\"\\t\\t\\t \",round(Sa2[i],2),\"\\t\\t\\t \",round(Sztr[i],2),\"\\n\"\n",

-      "plt.plot(Sztr,z);\n",

-      "plt.title(\"sigma(z)translation vs z\")\n",

-      "plt.ylabel(\"z(ft)\")\n",

-      "plt.xlabel(\"sigma(z)translation (lb/ft**2)\")\n",

-      "plt.show()\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "z(ft)\t sigma(1)(lb/ft**2)   \t sigma(2)(lb/ft**2)    \t sigma(z)translation (lb/ft**2)\n",

-        "\n",

-        "0.0 \t  0.0 \t\t\t  0.0 \t\t\t  0.0 \n",

-        "\n",

-        "4.0 \t  269.92 \t\t\t  114.23 \t\t\t  192.07 \n",

-        "\n",

-        "8.0 \t  311.08 \t\t\t  228.46 \t\t\t  269.77 \n",

-        "\n",

-        "12.0 \t  233.53 \t\t\t  342.69 \t\t\t  288.11 \n",

-        "\n",

-        "16.0 \t  102.06 \t\t\t  456.92 \t\t\t  279.49 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.7:Pg-342"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.7\n",

-      "import math\n",

-      "Gammasat=18.86; # in KN/m^3\n",

-      "Gammaw=9.81; # in KN/m^3\n",

-      "phi1=math.pi/180*30;  # angle 1\n",

-      "phi2=math.pi/180*26; # angle 2\n",

-      "Kp1=(math.tan(math.pi/4+phi1/2))**2;\n",

-      "Kp2=(math.tan(math.pi/4+phi2/2))**2;\n",

-      "#for top soil\n",

-      "c=0;\n",

-      "sigma0=31.44; # in KN/m^2\n",

-      "sigmap=sigma0*Kp1+2*c*math.sqrt(Kp1);\n",

-      "print round(sigmap,2),\"passive pressure for top layer in kN/m**2\"\n",

-      "#for z=2\n",

-      "c=10;\n",

-      "sigma0=31.44; # in KN/m^2\n",

-      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2);\n",

-      "print round(sigmap,2),\"passive pressure for z=2m in kN/m**2\"\n",

-      "#for z=3\n",

-      "c=10;\n",

-      "sigma0=15.72*2+(Gammasat-Gammaw)*1; # in KN/m^2\n",

-      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2); # in KN/m^2\n",

-      "print round(sigmap,2),\" is passive pressure for z=3m in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "94.32 passive pressure for top layer in kN/m**2\n",

-        "112.53 passive pressure for z=2m in kN/m**2\n",

-        "135.7  is passive pressure for z=3m in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter07_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter07_2.ipynb
deleted file mode 100755
index 63b3bc22..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter07_2.ipynb
+++ /dev/null
@@ -1,358 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:6111670c8f96effbf1bc0bc29859353d67e53cd5b906d2b571173d1698f7ba9c"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter07:Lateral earth pressure"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.1:Pg-319"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.1\n",

-      "\n",

-      "sigmao=48.0; # in KN/m^2\n",

-      "phi1=30*math.pi/180; # angle\n",

-      "phi2=36*math.pi/180; # angle\n",

-      "Ka1=(math.tan(math.pi/4.0-phi1/2))**2;\n",

-      "Ka2=(math.tan(math.pi/4.0-phi2/2))**2;\n",

-      "sigmaa1=Ka1*sigmao; # in KN/m^2\n",

-      "print round(sigmaa1,2),\"top soil pressure in kN/m**2\"\n",

-      "sigmaa2=Ka2*sigmao; # in KN/m^2\n",

-      "print round(sigmaa2,2),\"bottom soil pressure in kN/m**2\"\n",

-      "Po=1/2.0*3*16+3*12.48+1/3.0*3*(19.65-12.48)+1/2.0*3*29.43;\n",

-      "zbar=(24*(3+3/3.0)+37.44*(3/2.0)+10.76*3/3.0+44.1*3/3.0)/Po;\n",

-      "print round(zbar,2),\"resultant force acting from the bottom in m\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "16.0 top soil pressure in kN/m**2\n",

-        "12.46 bottom soil pressure in kN/m**2\n",

-        "1.84 resultant force acting from the bottom in m\n"

-       ]

-      }

-     ],

-     "prompt_number": 34

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.2:Pg-321"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.2\n",

-      "\n",

-      "import math\n",

-      "c=14.36;\n",

-      "Gamma=17.4; #  in KN/m^3\n",

-      "H=6; # in m\n",

-      "phi=26*math.pi/180;\n",

-      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

-      "sigma0=Gamma*H*Ka-2*c*math.sqrt(Ka);\n",

-      "Pa=1/2.0*Gamma*H**2*Ka-2*c*H*math.sqrt(Ka);\n",

-      "print round(Pa,2),\"active force before which tensile crack appeared in kN/m\"\n",

-      "zbar=(244.32-323.1)/14.46;\n",

-      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

-      "zc=2*c/Gamma/math.sqrt(Ka);\n",

-      "print round(zc,2),\"distance where tensile crack appeared in m\"\n",

-      "Pa=1/2.0*(H-zc)*(Gamma*H*Ka-2*c*math.sqrt(Ka));\n",

-      "print round(Pa,2),\"Active force in tensile crack in kN/m\"\n",

-      "zbar=(H-zc)/3;\n",

-      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "14.62 active force before which tensile crack appeared in kN/m\n",

-        "-5.45 the line of action on which net force is acting in m\n",

-        "2.64 distance where tensile crack appeared in m\n",

-        "38.32 Active force in tensile crack in kN/m\n",

-        "1.12 the line of action on which net force is acting in m\n"

-       ]

-      }

-     ],

-     "prompt_number": 30

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.3:Pg-322"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.3\n",

-      "import math\n",

-      "pi=math.pi\n",

-      "H=10.0; # in ft\n",

-      "Gamma=110.0; # in lb/ft^3\n",

-      "phi=35*math.pi/180.0; # angle\n",

-      "alpha=15*math.pi/180.0; # angle\n",

-      "theta=10*math.pi/180.0; # angle\n",

-      "zi=math.sin(math.sin(alpha)/math.sin(phi))-alpha+2*theta;\n",

-      "print round(zi*180.0/math.pi,2),\" is zi in degrees\"\n",

-      "Ka=math.cos(alpha-theta)*math.sqrt(1+(math.sin(phi))**2-2*math.sin(phi)*math.sin(zi))/((math.cos(theta))**2*(math.cos(alpha)+math.sqrt((math.sin(phi))**2+((math.sin(alpha))**2))));\n",

-      "Pa=1/2.0*Gamma*H**2*Ka;\n",

-      "print round(Pa,2),\" is rankine earth pressure in lb/ft\"\n",

-      "print \"there is slight error in answer due to rounding off error\"\n",

-      "Beta=math.tan(math.sin(phi)*math.sin(zi)/(1-math.sin(phi)*math.cos(zi)));\n",

-      "print round(Beta*180/pi,2),\" is angle in degrees\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "29.99  is zi in degrees\n",

-        "3078.61  is rankine earth pressure in lb/ft\n",

-        "there is slight error in answer due to rounding off error\n",

-        "36.7  is angle in degrees\n"

-       ]

-      }

-     ],

-     "prompt_number": 25

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.4:Pg-326"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.4\n",

-      "\n",

-      "H=4.6; # in m\n",

-      "Gamma=16.5; # in KN/m^3\n",

-      "Ka=0.297;\n",

-      "Po=1/2.0*Gamma*H**2*Ka;\n",

-      "print round(Po,2),\"coulomb active force per unit length in kN/m\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "51.85 coulomb active force per unit length in kN/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.5:Pg-331"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.5\n",

-      "\n",

-      "#part(a)\n",

-      "Gamma=105; # in lb/ft^3\n",

-      "H=10; #in ft\n",

-      "Kae=0.474;\n",

-      "k1=0;\n",

-      "Pae=1/2.0*Gamma*H**2*Kae*(1-k1) # in lb/ft\n",

-      "print Pae,\"active force in lb/ft\"\n",

-      "#part(b)\n",

-      "Ka=0.246;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka; # in lb/ft\n",

-      "print Pa,\"active force in lb/ft\"\n",

-      "DPae=Pae-Pa;#deltaPae\n",

-      "zbar=(0.6*H*DPae+H/3*(Pa))/Pae;\n",

-      "print round(zbar,2),\"the distance of resultant force from bottom in ft\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "2488.5 active force in lb/ft\n",

-        "1291.5 active force in lb/ft\n",

-        "4.44 the distance of resultant force from bottom in ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.6:Pg-337"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.6\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "import matplotlib.pyplot as plt\n",

-      "\n",

-      "z=[0, 4, 8, 12, 16];\n",

-      "Gamma=110; # in lb/ft^3\n",

-      "phi=36*math.pi/180;\n",

-      "H=16; # in ft\n",

-      "Sa1=numpy.zeros(5);#sigma(1)\n",

-      "Sa2=numpy.zeros(5);#sigma(2)\n",

-      "Sztr=numpy.zeros(5);#sigma(z)translation\n",

-      "print \"z(ft)\\t sigma(1)(lb/ft**2)   \\t sigma(2)(lb/ft**2)    \\t sigma(z)translation (lb/ft**2)\\n\"\n",

-      "for i in range(0,5):\n",

-      "    Sa1[i]=Gamma*(math.tan(math.pi/4-phi*z[i]/2/H))**2*(z[i]-phi*z[i]**2/H/math.cos(phi*z[i]/H));\n",

-      "    Sa2[i]=Gamma*z[i]*(math.cos(phi)/(1+math.sin(phi)))**2;\n",

-      "    Sztr[i]=Sa1[i]/2.0+Sa2[i]/2.0;\n",

-      "    print  round(z[i],2),\"\\t \",round(Sa1[i],2),\"\\t\\t\\t \",round(Sa2[i],2),\"\\t\\t\\t \",round(Sztr[i],2),\"\\n\"\n",

-      "plt.plot(Sztr,z);\n",

-      "plt.title(\"sigma(z)translation vs z\")\n",

-      "plt.ylabel(\"z(ft)\")\n",

-      "plt.xlabel(\"sigma(z)translation (lb/ft**2)\")\n",

-      "plt.show()\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "z(ft)\t sigma(1)(lb/ft**2)   \t sigma(2)(lb/ft**2)    \t sigma(z)translation (lb/ft**2)\n",

-        "\n",

-        "0.0 \t  0.0 \t\t\t  0.0 \t\t\t  0.0 \n",

-        "\n",

-        "4.0 \t  269.92 \t\t\t  114.23 \t\t\t  192.07 \n",

-        "\n",

-        "8.0 \t  311.08 \t\t\t  228.46 \t\t\t  269.77 \n",

-        "\n",

-        "12.0 \t  233.53 \t\t\t  342.69 \t\t\t  288.11 \n",

-        "\n",

-        "16.0 \t  102.06 \t\t\t  456.92 \t\t\t  279.49 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex7.7:Pg-342"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 7.7\n",

-      "import math\n",

-      "Gammasat=18.86; # in KN/m^3\n",

-      "Gammaw=9.81; # in KN/m^3\n",

-      "phi1=math.pi/180*30;  # angle 1\n",

-      "phi2=math.pi/180*26; # angle 2\n",

-      "Kp1=(math.tan(math.pi/4+phi1/2))**2;\n",

-      "Kp2=(math.tan(math.pi/4+phi2/2))**2;\n",

-      "#for top soil\n",

-      "c=0;\n",

-      "sigma0=31.44; # in KN/m^2\n",

-      "sigmap=sigma0*Kp1+2*c*math.sqrt(Kp1);\n",

-      "print round(sigmap,2),\"passive pressure for top layer in kN/m**2\"\n",

-      "#for z=2\n",

-      "c=10;\n",

-      "sigma0=31.44; # in KN/m^2\n",

-      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2);\n",

-      "print round(sigmap,2),\"passive pressure for z=2m in kN/m**2\"\n",

-      "#for z=3\n",

-      "c=10;\n",

-      "sigma0=15.72*2+(Gammasat-Gammaw)*1; # in KN/m^2\n",

-      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2); # in KN/m^2\n",

-      "print round(sigmap,2),\" is passive pressure for z=3m in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "94.32 passive pressure for top layer in kN/m**2\n",

-        "112.53 passive pressure for z=2m in kN/m**2\n",

-        "135.7  is passive pressure for z=3m in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter08.ipynb b/Principles_Of_Foundation_Engineering/Chapter08.ipynb
deleted file mode 100755
index de2367cd..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter08.ipynb
+++ /dev/null
@@ -1,527 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:ae926d3fc40f9f56d82ee333a9b5a7f719ff4a11c08f89381773660beb02ead3"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter08: Retaining walls"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.1:Pg-366"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.1\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "H1=6*math.tan(10*math.pi/180.0); # in ft\n",

-      "H2=18.0; # in ft\n",

-      "H3=2.75; # in ft\n",

-      "Gamma1=117.0;  # in lb/ft^3\n",

-      "Ka=0.294;#from table 7.1\n",

-      "H=H1+H2+H3; # in ft\n",

-      "Pa=1/2.0*Gamma1*H**2*Ka/1000; # in Kip/ft\n",

-      "Pr=Pa*math.sin(10*math.pi/180); # in Kip/ft\n",

-      "Ph=Pa*math.cos(10*math.pi/180); # in Kip/ft\n",

-      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

-      "Sum=0;#sigma Mr\n",

-      "S=[1, 2, 3, 4, 5];#section\n",

-      "W=[4.05, 1.35, 5.156, 13.01, 1.42];#weight\n",

-      "MA=[5.75, 4.67, 6.25, 9.5, 12.5, 12.5];#Moment Arm\n",

-      "M=numpy.zeros(5)\n",

-      "print \"Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\\n\"\n",

-      "for i in range(0,5):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Sum=Sum+M[i];\n",

-      "    print round(S[i],2),\"\\t \\t \",round(W[i],2),\"\\t \\t \\t\",round(MA[i],2),\"\\t \\t \\t \",round(M[i],2),\"\\n\"\n",

-      "\n",

-      "FSO=Sum/Mo;\n",

-      "if FSO>2 :\n",

-      "    print  \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=18*math.pi/180; # the given angle\n",

-      "V=24.986;\n",

-      "B=12.5;\n",

-      "c2=0.9; # in lb/ft^2\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

-      "if FSS>2 :\n",

-      "    print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "\n",

-      "#for bearing\n",

-      "e=B/2.0-(Sum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); # in Kip/ft^2\n",

-      "Nc=13.1;\n",

-      "Nq=5.26;\n",

-      "Ny=4.07;\n",

-      "D=0.107;\n",

-      "Gamma2=4.0; # in lb/ft^3\n",

-      "B1=B-2*e;#Bdash\n",

-      "q=Gamma2*D # in lb/ft^2\n",

-      "Fcd=1+0.4*D/B1;\n",

-      "Fqd=1+2*math.tan(phi2)*(1-math.sin(phi2))**2*(D/B1);\n",

-      "Fyd=1;\n",

-      "zi=math.tan(Ph/V);\n",

-      "Fci=(1-zi/math.pi*2)**2;\n",

-      "Fqi=Fci;\n",

-      "Fyi=round((1-zi/phi2)**2);\n",

-      "qu=c2*Nc*Fcd*Fci+q*Nq*Fqd*Fqi+1/2.0*Gamma2*B1*Fyd*Fyi;\n",

-      "FSB=qu/qtoe;\n",

-      "if FSB>3 :\n",

-      "    print \"bearing  is safe with FS=\",round(FSB),\"\\n\\n\"\n",

-      "    print \"slight changes due to round off error\"\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\n",

-        "\n",

-        "1.0 \t \t  4.05 \t \t \t5.75 \t \t \t  23.29 \n",

-        "\n",

-        "2.0 \t \t  1.35 \t \t \t4.67 \t \t \t  6.3 \n",

-        "\n",

-        "3.0 \t \t  5.16 \t \t \t6.25 \t \t \t  32.23 \n",

-        "\n",

-        "4.0 \t \t  13.01 \t \t \t9.5 \t \t \t  123.59 \n",

-        "\n",

-        "5.0 \t \t  1.42 \t \t \t12.5 \t \t \t  17.75 \n",

-        "\n",

-        "safe in overturning with FS= 3.47 \n",

-        "\n",

-        "bearing  is safe with FS= 4.0 \n",

-        "\n",

-        "\n",

-        "slight changes due to round off error\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.2:Pg-369"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.2\n",

-      "\n",

-      "c=0.9; # # in lb/ft^2\n",

-      "B=12.5; # in ft\n",

-      "Gamma2=4; # in lb/ft^3\n",

-      "Fcd=1.138;\n",

-      "Fqd=1.107; \n",

-      "Nc=7.5;\n",

-      "Nq=4;\n",

-      "Ny=0;\n",

-      "q=0.428; # in lb/ft^2\n",

-      "qtoe=2.44; # in lb/ft^2\n",

-      "qu=c*Nc*Fcd+q*Nq*Fqd+1/2.0*Gamma2*B*Ny;\n",

-      "FSB=qu/qtoe; # factor of safety\n",

-      "if FSB>3.0 :\n",

-      "    print \"safe in bearing with FS=\",round(FSB,2),\"\\n\\n\",\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "safe in bearing with FS= 3.92 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.3:Pg-370"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.3\n",

-      "import math\n",

-      "import numpy\n",

-      "Msum=0;#sum of moment\n",

-      "Vsum=0;#sum of force\n",

-      "H=15+2.5;#Hdash  in ft\n",

-      "phi=30*math.pi/180; # given angle in degree\n",

-      "Gamma=121.0; # in lb/ft^3\n",

-      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka/1000; # in Kip/ft\n",

-      "Ph=Pa; # in Kip/ft\n",

-      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

-      "AN=[1.0,2.0,3.0,4.0,5.0,6.0];#area number\n",

-      "W=[0.9,3.375,5.906,3.863,4.764,2.723];#weight\n",

-      "MA=[1.783,2.8,5.3,5.15,7.05,9.55];#moment arm\n",

-      "\n",

-      "print \"AreaNo  \\t Weight(kip/ft) \\t  Moment Arm(ft) \\t Moment (kip-ft/ft)\\n\"\n",

-      "M= numpy.zeros(6)\n",

-      "for i in range(0,6):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Vsum=Vsum+W[i];\n",

-      "    Msum=Msum+M[i];\n",

-      "    print round(AN[i],2),\"\\t\\t \",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

-      "\n",

-      "FSO=(Msum)/Mo;\n",

-      "if FSO>2 :\n",

-      "    print \"safe in overturning with FS=\",round(FSO,2)\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=20*math.pi/180;\n",

-      "V=Vsum\n",

-      "B=10.3; # in ft\n",

-      "c2=1.0; # in lb/ft^2\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

-      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "e=B/2.0-(Msum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); #  in kip/ft**2\n",

-      "print round(qtoe,2),\" is soil pressure at toe in kip/ft**2\"\n",

-      "qheel=V/B*(1-6*e/B);  # in kip/ft**2\n",

-      "print round(qheel,2),\" is soil pressure at heel in kip/ft**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "AreaNo  \t Weight(kip/ft) \t  Moment Arm(ft) \t Moment (kip-ft/ft)\n",

-        "\n",

-        "1.0 \t\t  0.9 \t \t  \t  1.783 \t \t \t   1.6047 \n",

-        "\n",

-        "2.0 \t\t  3.38 \t \t  \t  2.8 \t \t \t   9.45 \n",

-        "\n",

-        "3.0 \t\t  5.91 \t \t  \t  5.3 \t \t \t   31.3018 \n",

-        "\n",

-        "4.0 \t\t  3.86 \t \t  \t  5.15 \t \t \t   19.89445 \n",

-        "\n",

-        "5.0 \t\t  4.76 \t \t  \t  7.05 \t \t \t   33.5862 \n",

-        "\n",

-        "6.0 \t\t  2.72 \t \t  \t  9.55 \t \t \t   26.00465 \n",

-        "\n",

-        "safe in overturning with FS= 3.38\n",

-        "safe in sliding with FS= 1.94 \n",

-        "\n",

-        "3.51  is soil pressure at toe in kip/ft**2\n",

-        "0.67  is soil pressure at heel in kip/ft**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 33

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.4:Pg-372"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.4\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "Msum=0;#sum of moment\n",

-      "Vsum=0;#sum of force\n",

-      "H=5+1.5;#Hdash in m\n",

-      "phi=32*math.pi/180 # angle in degree\n",

-      "Gamma=18.5; # in KN/m^3\n",

-      "Ka=0.424;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka; # in KN/m\n",

-      "Ph=Pa*math.cos(15*math.pi/180+2/3.0*phi); # in KN/m\n",

-      "Mo=Ph*H/3.0; # moment\n",

-      "AN=[1,2,3,4,5];#area number\n",

-      "A=[4.36, 3.42, 0.77, 2.8, 2.8];#area\n",

-      "W=[102.81, 80.64, 18.16, 66.02, 93.14];#weight\n",

-      "MA=[2.18, 1.37, 0.98, 1.75, 2.83];#moment arm\n",

-      "print \"AreaNo  \\t  Area(m**2) \\t  Weight(kN/m)  \\t  Moment Arm(m) \\t  Moment (kN-m/m)\\n\"\n",

-      "M= numpy.zeros(5)\n",

-      "for i in range(0,5):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Vsum=Vsum+W[i];\n",

-      "    Msum=Msum+M[i];\n",

-      "    print round(AN[i],2),\"\\t\\t \",round(A[i],2),\" \\t \\t\",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

-      "\n",

-      "FSO=Msum/Mo;\n",

-      "if FSO>2 :\n",

-      "    print \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=24*math.pi/180;\n",

-      "V=Vsum\n",

-      "B=3.5; # in m\n",

-      "c2=30; # in KN/m^2\n",

-      "Pp=1/2.0*2.37*18*1.5**2+2*30*1.54*1.5;\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2+Pp)/Ph;\n",

-      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "print \"if Pp is ignored then FS=1.37\"\n",

-      "e=B/2.0-(Msum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); # in kN/m**2\n",

-      "print round(qtoe,2),\"soil pressure at toe in kN/m**2\"\n",

-      "qheel=V/B*(1-6*e/B); # in kN/m**2\n",

-      "print round(qheel,2),\"soil pressure at heel in kN/m**2\"\n",

-      "print \"there is difference in answer due to rounding off error\"\n",

-      "\n",

-      "# there is difference in answer due to rounding off error\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "AreaNo  \t  Area(m**2) \t  Weight(kN/m)  \t  Moment Arm(m) \t  Moment (kN-m/m)\n",

-        "\n",

-        "1.0 \t\t  4.36  \t \t102.81 \t \t  \t  2.18 \t \t \t   224.1258 \n",

-        "\n",

-        "2.0 \t\t  3.42  \t \t80.64 \t \t  \t  1.37 \t \t \t   110.4768 \n",

-        "\n",

-        "3.0 \t\t  0.77  \t \t18.16 \t \t  \t  0.98 \t \t \t   17.7968 \n",

-        "\n",

-        "4.0 \t\t  2.8  \t \t66.02 \t \t  \t  1.75 \t \t \t   115.535 \n",

-        "\n",

-        "5.0 \t\t  2.8  \t \t93.14 \t \t  \t  2.83 \t \t \t   263.5862 \n",

-        "\n",

-        "safe in overturning with FS= 2.53 \n",

-        "\n",

-        "safe in sliding with FS= 2.7 \n",

-        "\n",

-        "if Pp is ignored then FS=1.37\n",

-        "195.67 soil pressure at toe in kN/m**2\n",

-        "10.48 soil pressure at heel in kN/m**2\n",

-        "there is difference in answer due to rounding off error\n"

-       ]

-      }

-     ],

-     "prompt_number": 49

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.6:Pg-393"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.6\n",

-      "import math\n",

-      "Sv=2; # in ft\n",

-      "Sh=3; # in ft\n",

-      "w=3/12.0; # in inch\n",

-      "fy=35000*144;\n",

-      "FSb=3;\n",

-      "pi=math.pi;\n",

-      "phi=36*pi/180;\n",

-      "Gamma1=105; # in lb/ft^3\n",

-      "H=30;\n",

-      "t=Gamma1*H*Sv*Sh*FSb/w/fy*(math.tan(pi/4-phi/2))**2;\n",

-      "t=t*12; #in inch\n",

-      "print round(t,2),\" is thickness in inches\"\n",

-      "t=t+0.001*50;\n",

-      "print \"so take thickness=0.2 inches\"\n",

-      "#for tie length\n",

-      "z=[5,10,15,20,25,30];\n",

-      "TL=[38.45, 35.89, 33.34, 30.79, 28.25, 25.7];#tie length\n",

-      "print \"z(ft)\\t Tie Length (ft)\\n\"\n",

-      "for i in range(0,6):\n",

-      "    print z[i],\"\\t\",TL[i]\n",

-      "\n",

-      "print \"use tie length=40ft\"\n",

-      "#check for over turning\n",

-      "\n",

-      "z=30/3.0;\n",

-      "x1=20;\n",

-      "L=40;\n",

-      "Ka=0.26;\n",

-      "Pa=1/2.0*Gamma1*Ka*H**2; #  in kip/ft**2\n",

-      "W1=Gamma1*H*L;\n",

-      "FSO=W1*x1/(Pa*z);\n",

-      "print round(FSO,2),\" is factor of safety is\" \n",

-      "print \"since FS>3 structure is safe\"\n",

-      "#check for sliding\n",

-      "k=2/3.0;\n",

-      "Pa=12285; #  in kip/ft**2\n",

-      "FSS=W1*math.tan(k*phi)/Pa;\n",

-      "if FSS>3 :\n",

-      "    print \"safe in sliding with FS=\",round(FSS,2)\n",

-      "\n",

-      "#check for bearing\n",

-      "Mr=126000*20; # in lb-ft/ft\n",

-      "Mo=12285*10; # in lb-ft/ft\n",

-      "V=126000;\n",

-      "e=L/2-Mr/V+Mo/V;\n",

-      "L1=L-2*e;#Ldash\n",

-      "c2=1000; # in lb/ft^2\n",

-      "Nc=25.8;\n",

-      "Gamma2=110; # in lb/ft^3\n",

-      "Ny=16.72;\n",

-      "qult=c2*Nc+1/2.0*Gamma2*L1*Ny\n",

-      "sigma0=Gamma1*H; # in lb/ft^2\n",

-      "FSB=qult/sigma0;\n",

-      "if FSB>5 :\n",

-      "    print \"bearing is safe  with FS=\",round(FSB,2)\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "0.14  is thickness in inches\n",

-        "so take thickness=0.2 inches\n",

-        "z(ft)\t Tie Length (ft)\n",

-        "\n",

-        "5 \t38.45\n",

-        "10 \t35.89\n",

-        "15 \t33.34\n",

-        "20 \t30.79\n",

-        "25 \t28.25\n",

-        "30 \t25.7\n",

-        "use tie length=40ft\n",

-        "20.51  is factor of safety is\n",

-        "since FS>3 structure is safe\n",

-        "safe in sliding with FS= 4.57\n",

-        "safe in bearing with FS= 19.87\n"

-       ]

-      }

-     ],

-     "prompt_number": 56

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.7:Pg-397"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.7\n",

-      "import math\n",

-      "import numpy\n",

-      "pi=math.pi;\n",

-      "phi=36.0*pi/180;\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "z=[8.0, 12.0, 16.0]; # in ft\n",

-      "sigmaG=80*12.0; # in lb/ft^2\n",

-      "Gamma1=110.0; # in lb/ft^3\n",

-      "FS=1.5;\n",

-      "Sv=numpy.zeros(3)\n",

-      "for i in range(0,3):\n",

-      "    Sv[i]=sigmaG/Gamma1/z[i]/Ka/FS*12.0;\n",

-      "    print \"for z=\",z[i],\" ft Sv = \",round(Sv[i],2),\" inches\\n\"\n",

-      "\n",

-      "z=[16.0,56.0,76.0,96.0,112.0,144.0,176.0];\n",

-      "zf=numpy.zeros(7)\n",

-      "k2=numpy.zeros(7)\n",

-      "L=numpy.zeros(7)\n",

-      "for i in range(0,7):\n",

-      "    zf[i]=z[i]/12.0;#z in ft\n",

-      "Sv=[1.67,1.67,1.67,1.67,1.33,1.33,1.33];\n",

-      "k=[7.48,5.78,4.93,4.08,3.4,2.04,0.68];#0.51(H-z)\n",

-      "print \"z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \\n\"\n",

-      "for i in range(0,7):\n",

-      "    k2[i]=0.438*Sv[i];#0.438Sv\n",

-      "    L[i]=k[i]+k2[i];\n",

-      "    print round(z[i],2),\"\\t \\t\",round(zf[i],2),\"\\t \",round(Sv[i],2),\"\\t \\t \",round(k[i],2),\"\\t \\t \\t\",round(k2[i],2),\"\\t \\t \\t \",round(L[i],2)\n",

-      "\n",

-      "Sv=20/12.0;\n",

-      "Ka=0.26;\n",

-      "FS=1.5;\n",

-      "l1=Sv*Ka*FS/4/math.tan(2/3.0*phi);\n",

-      "if l1<3:\n",

-      "    l1=3;\n",

-      "    print l1,\"length in ft\"\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "for z= 8.0  ft Sv =  33.62  inches\n",

-        "\n",

-        "for z= 12.0  ft Sv =  22.41  inches\n",

-        "\n",

-        "for z= 16.0  ft Sv =  16.81  inches\n",

-        "\n",

-        "z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \n",

-        "\n",

-        "16.0 \t \t1.33 \t  1.67 \t \t  7.48 \t \t \t0.73 \t \t \t  8.21\n",

-        "56.0 \t \t4.67 \t  1.67 \t \t  5.78 \t \t \t0.73 \t \t \t  6.51\n",

-        "76.0 \t \t6.33 \t  1.67 \t \t  4.93 \t \t \t0.73 \t \t \t  5.66\n",

-        "96.0 \t \t8.0 \t  1.67 \t \t  4.08 \t \t \t0.73 \t \t \t  4.81\n",

-        "112.0 \t \t9.33 \t  1.33 \t \t  3.4 \t \t \t0.58 \t \t \t  3.98\n",

-        "144.0 \t \t12.0 \t  1.33 \t \t  2.04 \t \t \t0.58 \t \t \t  2.62\n",

-        "176.0 \t \t14.67 \t  1.33 \t \t  0.68 \t \t \t0.58 \t \t \t  1.26\n",

-        "3 length in ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 79

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter08_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter08_1.ipynb
deleted file mode 100755
index de2367cd..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter08_1.ipynb
+++ /dev/null
@@ -1,527 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:ae926d3fc40f9f56d82ee333a9b5a7f719ff4a11c08f89381773660beb02ead3"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter08: Retaining walls"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.1:Pg-366"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.1\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "H1=6*math.tan(10*math.pi/180.0); # in ft\n",

-      "H2=18.0; # in ft\n",

-      "H3=2.75; # in ft\n",

-      "Gamma1=117.0;  # in lb/ft^3\n",

-      "Ka=0.294;#from table 7.1\n",

-      "H=H1+H2+H3; # in ft\n",

-      "Pa=1/2.0*Gamma1*H**2*Ka/1000; # in Kip/ft\n",

-      "Pr=Pa*math.sin(10*math.pi/180); # in Kip/ft\n",

-      "Ph=Pa*math.cos(10*math.pi/180); # in Kip/ft\n",

-      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

-      "Sum=0;#sigma Mr\n",

-      "S=[1, 2, 3, 4, 5];#section\n",

-      "W=[4.05, 1.35, 5.156, 13.01, 1.42];#weight\n",

-      "MA=[5.75, 4.67, 6.25, 9.5, 12.5, 12.5];#Moment Arm\n",

-      "M=numpy.zeros(5)\n",

-      "print \"Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\\n\"\n",

-      "for i in range(0,5):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Sum=Sum+M[i];\n",

-      "    print round(S[i],2),\"\\t \\t \",round(W[i],2),\"\\t \\t \\t\",round(MA[i],2),\"\\t \\t \\t \",round(M[i],2),\"\\n\"\n",

-      "\n",

-      "FSO=Sum/Mo;\n",

-      "if FSO>2 :\n",

-      "    print  \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=18*math.pi/180; # the given angle\n",

-      "V=24.986;\n",

-      "B=12.5;\n",

-      "c2=0.9; # in lb/ft^2\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

-      "if FSS>2 :\n",

-      "    print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "\n",

-      "#for bearing\n",

-      "e=B/2.0-(Sum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); # in Kip/ft^2\n",

-      "Nc=13.1;\n",

-      "Nq=5.26;\n",

-      "Ny=4.07;\n",

-      "D=0.107;\n",

-      "Gamma2=4.0; # in lb/ft^3\n",

-      "B1=B-2*e;#Bdash\n",

-      "q=Gamma2*D # in lb/ft^2\n",

-      "Fcd=1+0.4*D/B1;\n",

-      "Fqd=1+2*math.tan(phi2)*(1-math.sin(phi2))**2*(D/B1);\n",

-      "Fyd=1;\n",

-      "zi=math.tan(Ph/V);\n",

-      "Fci=(1-zi/math.pi*2)**2;\n",

-      "Fqi=Fci;\n",

-      "Fyi=round((1-zi/phi2)**2);\n",

-      "qu=c2*Nc*Fcd*Fci+q*Nq*Fqd*Fqi+1/2.0*Gamma2*B1*Fyd*Fyi;\n",

-      "FSB=qu/qtoe;\n",

-      "if FSB>3 :\n",

-      "    print \"bearing  is safe with FS=\",round(FSB),\"\\n\\n\"\n",

-      "    print \"slight changes due to round off error\"\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\n",

-        "\n",

-        "1.0 \t \t  4.05 \t \t \t5.75 \t \t \t  23.29 \n",

-        "\n",

-        "2.0 \t \t  1.35 \t \t \t4.67 \t \t \t  6.3 \n",

-        "\n",

-        "3.0 \t \t  5.16 \t \t \t6.25 \t \t \t  32.23 \n",

-        "\n",

-        "4.0 \t \t  13.01 \t \t \t9.5 \t \t \t  123.59 \n",

-        "\n",

-        "5.0 \t \t  1.42 \t \t \t12.5 \t \t \t  17.75 \n",

-        "\n",

-        "safe in overturning with FS= 3.47 \n",

-        "\n",

-        "bearing  is safe with FS= 4.0 \n",

-        "\n",

-        "\n",

-        "slight changes due to round off error\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.2:Pg-369"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.2\n",

-      "\n",

-      "c=0.9; # # in lb/ft^2\n",

-      "B=12.5; # in ft\n",

-      "Gamma2=4; # in lb/ft^3\n",

-      "Fcd=1.138;\n",

-      "Fqd=1.107; \n",

-      "Nc=7.5;\n",

-      "Nq=4;\n",

-      "Ny=0;\n",

-      "q=0.428; # in lb/ft^2\n",

-      "qtoe=2.44; # in lb/ft^2\n",

-      "qu=c*Nc*Fcd+q*Nq*Fqd+1/2.0*Gamma2*B*Ny;\n",

-      "FSB=qu/qtoe; # factor of safety\n",

-      "if FSB>3.0 :\n",

-      "    print \"safe in bearing with FS=\",round(FSB,2),\"\\n\\n\",\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "safe in bearing with FS= 3.92 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.3:Pg-370"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.3\n",

-      "import math\n",

-      "import numpy\n",

-      "Msum=0;#sum of moment\n",

-      "Vsum=0;#sum of force\n",

-      "H=15+2.5;#Hdash  in ft\n",

-      "phi=30*math.pi/180; # given angle in degree\n",

-      "Gamma=121.0; # in lb/ft^3\n",

-      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka/1000; # in Kip/ft\n",

-      "Ph=Pa; # in Kip/ft\n",

-      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

-      "AN=[1.0,2.0,3.0,4.0,5.0,6.0];#area number\n",

-      "W=[0.9,3.375,5.906,3.863,4.764,2.723];#weight\n",

-      "MA=[1.783,2.8,5.3,5.15,7.05,9.55];#moment arm\n",

-      "\n",

-      "print \"AreaNo  \\t Weight(kip/ft) \\t  Moment Arm(ft) \\t Moment (kip-ft/ft)\\n\"\n",

-      "M= numpy.zeros(6)\n",

-      "for i in range(0,6):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Vsum=Vsum+W[i];\n",

-      "    Msum=Msum+M[i];\n",

-      "    print round(AN[i],2),\"\\t\\t \",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

-      "\n",

-      "FSO=(Msum)/Mo;\n",

-      "if FSO>2 :\n",

-      "    print \"safe in overturning with FS=\",round(FSO,2)\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=20*math.pi/180;\n",

-      "V=Vsum\n",

-      "B=10.3; # in ft\n",

-      "c2=1.0; # in lb/ft^2\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

-      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "e=B/2.0-(Msum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); #  in kip/ft**2\n",

-      "print round(qtoe,2),\" is soil pressure at toe in kip/ft**2\"\n",

-      "qheel=V/B*(1-6*e/B);  # in kip/ft**2\n",

-      "print round(qheel,2),\" is soil pressure at heel in kip/ft**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "AreaNo  \t Weight(kip/ft) \t  Moment Arm(ft) \t Moment (kip-ft/ft)\n",

-        "\n",

-        "1.0 \t\t  0.9 \t \t  \t  1.783 \t \t \t   1.6047 \n",

-        "\n",

-        "2.0 \t\t  3.38 \t \t  \t  2.8 \t \t \t   9.45 \n",

-        "\n",

-        "3.0 \t\t  5.91 \t \t  \t  5.3 \t \t \t   31.3018 \n",

-        "\n",

-        "4.0 \t\t  3.86 \t \t  \t  5.15 \t \t \t   19.89445 \n",

-        "\n",

-        "5.0 \t\t  4.76 \t \t  \t  7.05 \t \t \t   33.5862 \n",

-        "\n",

-        "6.0 \t\t  2.72 \t \t  \t  9.55 \t \t \t   26.00465 \n",

-        "\n",

-        "safe in overturning with FS= 3.38\n",

-        "safe in sliding with FS= 1.94 \n",

-        "\n",

-        "3.51  is soil pressure at toe in kip/ft**2\n",

-        "0.67  is soil pressure at heel in kip/ft**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 33

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.4:Pg-372"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.4\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "Msum=0;#sum of moment\n",

-      "Vsum=0;#sum of force\n",

-      "H=5+1.5;#Hdash in m\n",

-      "phi=32*math.pi/180 # angle in degree\n",

-      "Gamma=18.5; # in KN/m^3\n",

-      "Ka=0.424;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka; # in KN/m\n",

-      "Ph=Pa*math.cos(15*math.pi/180+2/3.0*phi); # in KN/m\n",

-      "Mo=Ph*H/3.0; # moment\n",

-      "AN=[1,2,3,4,5];#area number\n",

-      "A=[4.36, 3.42, 0.77, 2.8, 2.8];#area\n",

-      "W=[102.81, 80.64, 18.16, 66.02, 93.14];#weight\n",

-      "MA=[2.18, 1.37, 0.98, 1.75, 2.83];#moment arm\n",

-      "print \"AreaNo  \\t  Area(m**2) \\t  Weight(kN/m)  \\t  Moment Arm(m) \\t  Moment (kN-m/m)\\n\"\n",

-      "M= numpy.zeros(5)\n",

-      "for i in range(0,5):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Vsum=Vsum+W[i];\n",

-      "    Msum=Msum+M[i];\n",

-      "    print round(AN[i],2),\"\\t\\t \",round(A[i],2),\" \\t \\t\",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

-      "\n",

-      "FSO=Msum/Mo;\n",

-      "if FSO>2 :\n",

-      "    print \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=24*math.pi/180;\n",

-      "V=Vsum\n",

-      "B=3.5; # in m\n",

-      "c2=30; # in KN/m^2\n",

-      "Pp=1/2.0*2.37*18*1.5**2+2*30*1.54*1.5;\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2+Pp)/Ph;\n",

-      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "print \"if Pp is ignored then FS=1.37\"\n",

-      "e=B/2.0-(Msum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); # in kN/m**2\n",

-      "print round(qtoe,2),\"soil pressure at toe in kN/m**2\"\n",

-      "qheel=V/B*(1-6*e/B); # in kN/m**2\n",

-      "print round(qheel,2),\"soil pressure at heel in kN/m**2\"\n",

-      "print \"there is difference in answer due to rounding off error\"\n",

-      "\n",

-      "# there is difference in answer due to rounding off error\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "AreaNo  \t  Area(m**2) \t  Weight(kN/m)  \t  Moment Arm(m) \t  Moment (kN-m/m)\n",

-        "\n",

-        "1.0 \t\t  4.36  \t \t102.81 \t \t  \t  2.18 \t \t \t   224.1258 \n",

-        "\n",

-        "2.0 \t\t  3.42  \t \t80.64 \t \t  \t  1.37 \t \t \t   110.4768 \n",

-        "\n",

-        "3.0 \t\t  0.77  \t \t18.16 \t \t  \t  0.98 \t \t \t   17.7968 \n",

-        "\n",

-        "4.0 \t\t  2.8  \t \t66.02 \t \t  \t  1.75 \t \t \t   115.535 \n",

-        "\n",

-        "5.0 \t\t  2.8  \t \t93.14 \t \t  \t  2.83 \t \t \t   263.5862 \n",

-        "\n",

-        "safe in overturning with FS= 2.53 \n",

-        "\n",

-        "safe in sliding with FS= 2.7 \n",

-        "\n",

-        "if Pp is ignored then FS=1.37\n",

-        "195.67 soil pressure at toe in kN/m**2\n",

-        "10.48 soil pressure at heel in kN/m**2\n",

-        "there is difference in answer due to rounding off error\n"

-       ]

-      }

-     ],

-     "prompt_number": 49

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.6:Pg-393"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.6\n",

-      "import math\n",

-      "Sv=2; # in ft\n",

-      "Sh=3; # in ft\n",

-      "w=3/12.0; # in inch\n",

-      "fy=35000*144;\n",

-      "FSb=3;\n",

-      "pi=math.pi;\n",

-      "phi=36*pi/180;\n",

-      "Gamma1=105; # in lb/ft^3\n",

-      "H=30;\n",

-      "t=Gamma1*H*Sv*Sh*FSb/w/fy*(math.tan(pi/4-phi/2))**2;\n",

-      "t=t*12; #in inch\n",

-      "print round(t,2),\" is thickness in inches\"\n",

-      "t=t+0.001*50;\n",

-      "print \"so take thickness=0.2 inches\"\n",

-      "#for tie length\n",

-      "z=[5,10,15,20,25,30];\n",

-      "TL=[38.45, 35.89, 33.34, 30.79, 28.25, 25.7];#tie length\n",

-      "print \"z(ft)\\t Tie Length (ft)\\n\"\n",

-      "for i in range(0,6):\n",

-      "    print z[i],\"\\t\",TL[i]\n",

-      "\n",

-      "print \"use tie length=40ft\"\n",

-      "#check for over turning\n",

-      "\n",

-      "z=30/3.0;\n",

-      "x1=20;\n",

-      "L=40;\n",

-      "Ka=0.26;\n",

-      "Pa=1/2.0*Gamma1*Ka*H**2; #  in kip/ft**2\n",

-      "W1=Gamma1*H*L;\n",

-      "FSO=W1*x1/(Pa*z);\n",

-      "print round(FSO,2),\" is factor of safety is\" \n",

-      "print \"since FS>3 structure is safe\"\n",

-      "#check for sliding\n",

-      "k=2/3.0;\n",

-      "Pa=12285; #  in kip/ft**2\n",

-      "FSS=W1*math.tan(k*phi)/Pa;\n",

-      "if FSS>3 :\n",

-      "    print \"safe in sliding with FS=\",round(FSS,2)\n",

-      "\n",

-      "#check for bearing\n",

-      "Mr=126000*20; # in lb-ft/ft\n",

-      "Mo=12285*10; # in lb-ft/ft\n",

-      "V=126000;\n",

-      "e=L/2-Mr/V+Mo/V;\n",

-      "L1=L-2*e;#Ldash\n",

-      "c2=1000; # in lb/ft^2\n",

-      "Nc=25.8;\n",

-      "Gamma2=110; # in lb/ft^3\n",

-      "Ny=16.72;\n",

-      "qult=c2*Nc+1/2.0*Gamma2*L1*Ny\n",

-      "sigma0=Gamma1*H; # in lb/ft^2\n",

-      "FSB=qult/sigma0;\n",

-      "if FSB>5 :\n",

-      "    print \"bearing is safe  with FS=\",round(FSB,2)\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "0.14  is thickness in inches\n",

-        "so take thickness=0.2 inches\n",

-        "z(ft)\t Tie Length (ft)\n",

-        "\n",

-        "5 \t38.45\n",

-        "10 \t35.89\n",

-        "15 \t33.34\n",

-        "20 \t30.79\n",

-        "25 \t28.25\n",

-        "30 \t25.7\n",

-        "use tie length=40ft\n",

-        "20.51  is factor of safety is\n",

-        "since FS>3 structure is safe\n",

-        "safe in sliding with FS= 4.57\n",

-        "safe in bearing with FS= 19.87\n"

-       ]

-      }

-     ],

-     "prompt_number": 56

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.7:Pg-397"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.7\n",

-      "import math\n",

-      "import numpy\n",

-      "pi=math.pi;\n",

-      "phi=36.0*pi/180;\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "z=[8.0, 12.0, 16.0]; # in ft\n",

-      "sigmaG=80*12.0; # in lb/ft^2\n",

-      "Gamma1=110.0; # in lb/ft^3\n",

-      "FS=1.5;\n",

-      "Sv=numpy.zeros(3)\n",

-      "for i in range(0,3):\n",

-      "    Sv[i]=sigmaG/Gamma1/z[i]/Ka/FS*12.0;\n",

-      "    print \"for z=\",z[i],\" ft Sv = \",round(Sv[i],2),\" inches\\n\"\n",

-      "\n",

-      "z=[16.0,56.0,76.0,96.0,112.0,144.0,176.0];\n",

-      "zf=numpy.zeros(7)\n",

-      "k2=numpy.zeros(7)\n",

-      "L=numpy.zeros(7)\n",

-      "for i in range(0,7):\n",

-      "    zf[i]=z[i]/12.0;#z in ft\n",

-      "Sv=[1.67,1.67,1.67,1.67,1.33,1.33,1.33];\n",

-      "k=[7.48,5.78,4.93,4.08,3.4,2.04,0.68];#0.51(H-z)\n",

-      "print \"z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \\n\"\n",

-      "for i in range(0,7):\n",

-      "    k2[i]=0.438*Sv[i];#0.438Sv\n",

-      "    L[i]=k[i]+k2[i];\n",

-      "    print round(z[i],2),\"\\t \\t\",round(zf[i],2),\"\\t \",round(Sv[i],2),\"\\t \\t \",round(k[i],2),\"\\t \\t \\t\",round(k2[i],2),\"\\t \\t \\t \",round(L[i],2)\n",

-      "\n",

-      "Sv=20/12.0;\n",

-      "Ka=0.26;\n",

-      "FS=1.5;\n",

-      "l1=Sv*Ka*FS/4/math.tan(2/3.0*phi);\n",

-      "if l1<3:\n",

-      "    l1=3;\n",

-      "    print l1,\"length in ft\"\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "for z= 8.0  ft Sv =  33.62  inches\n",

-        "\n",

-        "for z= 12.0  ft Sv =  22.41  inches\n",

-        "\n",

-        "for z= 16.0  ft Sv =  16.81  inches\n",

-        "\n",

-        "z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \n",

-        "\n",

-        "16.0 \t \t1.33 \t  1.67 \t \t  7.48 \t \t \t0.73 \t \t \t  8.21\n",

-        "56.0 \t \t4.67 \t  1.67 \t \t  5.78 \t \t \t0.73 \t \t \t  6.51\n",

-        "76.0 \t \t6.33 \t  1.67 \t \t  4.93 \t \t \t0.73 \t \t \t  5.66\n",

-        "96.0 \t \t8.0 \t  1.67 \t \t  4.08 \t \t \t0.73 \t \t \t  4.81\n",

-        "112.0 \t \t9.33 \t  1.33 \t \t  3.4 \t \t \t0.58 \t \t \t  3.98\n",

-        "144.0 \t \t12.0 \t  1.33 \t \t  2.04 \t \t \t0.58 \t \t \t  2.62\n",

-        "176.0 \t \t14.67 \t  1.33 \t \t  0.68 \t \t \t0.58 \t \t \t  1.26\n",

-        "3 length in ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 79

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter08_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter08_2.ipynb
deleted file mode 100755
index de2367cd..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter08_2.ipynb
+++ /dev/null
@@ -1,527 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:ae926d3fc40f9f56d82ee333a9b5a7f719ff4a11c08f89381773660beb02ead3"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter08: Retaining walls"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.1:Pg-366"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.1\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "H1=6*math.tan(10*math.pi/180.0); # in ft\n",

-      "H2=18.0; # in ft\n",

-      "H3=2.75; # in ft\n",

-      "Gamma1=117.0;  # in lb/ft^3\n",

-      "Ka=0.294;#from table 7.1\n",

-      "H=H1+H2+H3; # in ft\n",

-      "Pa=1/2.0*Gamma1*H**2*Ka/1000; # in Kip/ft\n",

-      "Pr=Pa*math.sin(10*math.pi/180); # in Kip/ft\n",

-      "Ph=Pa*math.cos(10*math.pi/180); # in Kip/ft\n",

-      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

-      "Sum=0;#sigma Mr\n",

-      "S=[1, 2, 3, 4, 5];#section\n",

-      "W=[4.05, 1.35, 5.156, 13.01, 1.42];#weight\n",

-      "MA=[5.75, 4.67, 6.25, 9.5, 12.5, 12.5];#Moment Arm\n",

-      "M=numpy.zeros(5)\n",

-      "print \"Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\\n\"\n",

-      "for i in range(0,5):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Sum=Sum+M[i];\n",

-      "    print round(S[i],2),\"\\t \\t \",round(W[i],2),\"\\t \\t \\t\",round(MA[i],2),\"\\t \\t \\t \",round(M[i],2),\"\\n\"\n",

-      "\n",

-      "FSO=Sum/Mo;\n",

-      "if FSO>2 :\n",

-      "    print  \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=18*math.pi/180; # the given angle\n",

-      "V=24.986;\n",

-      "B=12.5;\n",

-      "c2=0.9; # in lb/ft^2\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

-      "if FSS>2 :\n",

-      "    print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "\n",

-      "#for bearing\n",

-      "e=B/2.0-(Sum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); # in Kip/ft^2\n",

-      "Nc=13.1;\n",

-      "Nq=5.26;\n",

-      "Ny=4.07;\n",

-      "D=0.107;\n",

-      "Gamma2=4.0; # in lb/ft^3\n",

-      "B1=B-2*e;#Bdash\n",

-      "q=Gamma2*D # in lb/ft^2\n",

-      "Fcd=1+0.4*D/B1;\n",

-      "Fqd=1+2*math.tan(phi2)*(1-math.sin(phi2))**2*(D/B1);\n",

-      "Fyd=1;\n",

-      "zi=math.tan(Ph/V);\n",

-      "Fci=(1-zi/math.pi*2)**2;\n",

-      "Fqi=Fci;\n",

-      "Fyi=round((1-zi/phi2)**2);\n",

-      "qu=c2*Nc*Fcd*Fci+q*Nq*Fqd*Fqi+1/2.0*Gamma2*B1*Fyd*Fyi;\n",

-      "FSB=qu/qtoe;\n",

-      "if FSB>3 :\n",

-      "    print \"bearing  is safe with FS=\",round(FSB),\"\\n\\n\"\n",

-      "    print \"slight changes due to round off error\"\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\n",

-        "\n",

-        "1.0 \t \t  4.05 \t \t \t5.75 \t \t \t  23.29 \n",

-        "\n",

-        "2.0 \t \t  1.35 \t \t \t4.67 \t \t \t  6.3 \n",

-        "\n",

-        "3.0 \t \t  5.16 \t \t \t6.25 \t \t \t  32.23 \n",

-        "\n",

-        "4.0 \t \t  13.01 \t \t \t9.5 \t \t \t  123.59 \n",

-        "\n",

-        "5.0 \t \t  1.42 \t \t \t12.5 \t \t \t  17.75 \n",

-        "\n",

-        "safe in overturning with FS= 3.47 \n",

-        "\n",

-        "bearing  is safe with FS= 4.0 \n",

-        "\n",

-        "\n",

-        "slight changes due to round off error\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.2:Pg-369"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.2\n",

-      "\n",

-      "c=0.9; # # in lb/ft^2\n",

-      "B=12.5; # in ft\n",

-      "Gamma2=4; # in lb/ft^3\n",

-      "Fcd=1.138;\n",

-      "Fqd=1.107; \n",

-      "Nc=7.5;\n",

-      "Nq=4;\n",

-      "Ny=0;\n",

-      "q=0.428; # in lb/ft^2\n",

-      "qtoe=2.44; # in lb/ft^2\n",

-      "qu=c*Nc*Fcd+q*Nq*Fqd+1/2.0*Gamma2*B*Ny;\n",

-      "FSB=qu/qtoe; # factor of safety\n",

-      "if FSB>3.0 :\n",

-      "    print \"safe in bearing with FS=\",round(FSB,2),\"\\n\\n\",\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "safe in bearing with FS= 3.92 \n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.3:Pg-370"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.3\n",

-      "import math\n",

-      "import numpy\n",

-      "Msum=0;#sum of moment\n",

-      "Vsum=0;#sum of force\n",

-      "H=15+2.5;#Hdash  in ft\n",

-      "phi=30*math.pi/180; # given angle in degree\n",

-      "Gamma=121.0; # in lb/ft^3\n",

-      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka/1000; # in Kip/ft\n",

-      "Ph=Pa; # in Kip/ft\n",

-      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

-      "AN=[1.0,2.0,3.0,4.0,5.0,6.0];#area number\n",

-      "W=[0.9,3.375,5.906,3.863,4.764,2.723];#weight\n",

-      "MA=[1.783,2.8,5.3,5.15,7.05,9.55];#moment arm\n",

-      "\n",

-      "print \"AreaNo  \\t Weight(kip/ft) \\t  Moment Arm(ft) \\t Moment (kip-ft/ft)\\n\"\n",

-      "M= numpy.zeros(6)\n",

-      "for i in range(0,6):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Vsum=Vsum+W[i];\n",

-      "    Msum=Msum+M[i];\n",

-      "    print round(AN[i],2),\"\\t\\t \",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

-      "\n",

-      "FSO=(Msum)/Mo;\n",

-      "if FSO>2 :\n",

-      "    print \"safe in overturning with FS=\",round(FSO,2)\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=20*math.pi/180;\n",

-      "V=Vsum\n",

-      "B=10.3; # in ft\n",

-      "c2=1.0; # in lb/ft^2\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

-      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "e=B/2.0-(Msum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); #  in kip/ft**2\n",

-      "print round(qtoe,2),\" is soil pressure at toe in kip/ft**2\"\n",

-      "qheel=V/B*(1-6*e/B);  # in kip/ft**2\n",

-      "print round(qheel,2),\" is soil pressure at heel in kip/ft**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "AreaNo  \t Weight(kip/ft) \t  Moment Arm(ft) \t Moment (kip-ft/ft)\n",

-        "\n",

-        "1.0 \t\t  0.9 \t \t  \t  1.783 \t \t \t   1.6047 \n",

-        "\n",

-        "2.0 \t\t  3.38 \t \t  \t  2.8 \t \t \t   9.45 \n",

-        "\n",

-        "3.0 \t\t  5.91 \t \t  \t  5.3 \t \t \t   31.3018 \n",

-        "\n",

-        "4.0 \t\t  3.86 \t \t  \t  5.15 \t \t \t   19.89445 \n",

-        "\n",

-        "5.0 \t\t  4.76 \t \t  \t  7.05 \t \t \t   33.5862 \n",

-        "\n",

-        "6.0 \t\t  2.72 \t \t  \t  9.55 \t \t \t   26.00465 \n",

-        "\n",

-        "safe in overturning with FS= 3.38\n",

-        "safe in sliding with FS= 1.94 \n",

-        "\n",

-        "3.51  is soil pressure at toe in kip/ft**2\n",

-        "0.67  is soil pressure at heel in kip/ft**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 33

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.4:Pg-372"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.4\n",

-      "\n",

-      "import math\n",

-      "import numpy\n",

-      "Msum=0;#sum of moment\n",

-      "Vsum=0;#sum of force\n",

-      "H=5+1.5;#Hdash in m\n",

-      "phi=32*math.pi/180 # angle in degree\n",

-      "Gamma=18.5; # in KN/m^3\n",

-      "Ka=0.424;\n",

-      "Pa=1/2.0*Gamma*H**2*Ka; # in KN/m\n",

-      "Ph=Pa*math.cos(15*math.pi/180+2/3.0*phi); # in KN/m\n",

-      "Mo=Ph*H/3.0; # moment\n",

-      "AN=[1,2,3,4,5];#area number\n",

-      "A=[4.36, 3.42, 0.77, 2.8, 2.8];#area\n",

-      "W=[102.81, 80.64, 18.16, 66.02, 93.14];#weight\n",

-      "MA=[2.18, 1.37, 0.98, 1.75, 2.83];#moment arm\n",

-      "print \"AreaNo  \\t  Area(m**2) \\t  Weight(kN/m)  \\t  Moment Arm(m) \\t  Moment (kN-m/m)\\n\"\n",

-      "M= numpy.zeros(5)\n",

-      "for i in range(0,5):\n",

-      "    M[i]=W[i]*MA[i];\n",

-      "    Vsum=Vsum+W[i];\n",

-      "    Msum=Msum+M[i];\n",

-      "    print round(AN[i],2),\"\\t\\t \",round(A[i],2),\" \\t \\t\",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

-      "\n",

-      "FSO=Msum/Mo;\n",

-      "if FSO>2 :\n",

-      "    print \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

-      "\n",

-      "#for sliding\n",

-      "phi2=24*math.pi/180;\n",

-      "V=Vsum\n",

-      "B=3.5; # in m\n",

-      "c2=30; # in KN/m^2\n",

-      "Pp=1/2.0*2.37*18*1.5**2+2*30*1.54*1.5;\n",

-      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2+Pp)/Ph;\n",

-      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

-      "print \"if Pp is ignored then FS=1.37\"\n",

-      "e=B/2.0-(Msum-Mo)/V;\n",

-      "qtoe=V/B*(1+6*e/B); # in kN/m**2\n",

-      "print round(qtoe,2),\"soil pressure at toe in kN/m**2\"\n",

-      "qheel=V/B*(1-6*e/B); # in kN/m**2\n",

-      "print round(qheel,2),\"soil pressure at heel in kN/m**2\"\n",

-      "print \"there is difference in answer due to rounding off error\"\n",

-      "\n",

-      "# there is difference in answer due to rounding off error\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "AreaNo  \t  Area(m**2) \t  Weight(kN/m)  \t  Moment Arm(m) \t  Moment (kN-m/m)\n",

-        "\n",

-        "1.0 \t\t  4.36  \t \t102.81 \t \t  \t  2.18 \t \t \t   224.1258 \n",

-        "\n",

-        "2.0 \t\t  3.42  \t \t80.64 \t \t  \t  1.37 \t \t \t   110.4768 \n",

-        "\n",

-        "3.0 \t\t  0.77  \t \t18.16 \t \t  \t  0.98 \t \t \t   17.7968 \n",

-        "\n",

-        "4.0 \t\t  2.8  \t \t66.02 \t \t  \t  1.75 \t \t \t   115.535 \n",

-        "\n",

-        "5.0 \t\t  2.8  \t \t93.14 \t \t  \t  2.83 \t \t \t   263.5862 \n",

-        "\n",

-        "safe in overturning with FS= 2.53 \n",

-        "\n",

-        "safe in sliding with FS= 2.7 \n",

-        "\n",

-        "if Pp is ignored then FS=1.37\n",

-        "195.67 soil pressure at toe in kN/m**2\n",

-        "10.48 soil pressure at heel in kN/m**2\n",

-        "there is difference in answer due to rounding off error\n"

-       ]

-      }

-     ],

-     "prompt_number": 49

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.6:Pg-393"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.6\n",

-      "import math\n",

-      "Sv=2; # in ft\n",

-      "Sh=3; # in ft\n",

-      "w=3/12.0; # in inch\n",

-      "fy=35000*144;\n",

-      "FSb=3;\n",

-      "pi=math.pi;\n",

-      "phi=36*pi/180;\n",

-      "Gamma1=105; # in lb/ft^3\n",

-      "H=30;\n",

-      "t=Gamma1*H*Sv*Sh*FSb/w/fy*(math.tan(pi/4-phi/2))**2;\n",

-      "t=t*12; #in inch\n",

-      "print round(t,2),\" is thickness in inches\"\n",

-      "t=t+0.001*50;\n",

-      "print \"so take thickness=0.2 inches\"\n",

-      "#for tie length\n",

-      "z=[5,10,15,20,25,30];\n",

-      "TL=[38.45, 35.89, 33.34, 30.79, 28.25, 25.7];#tie length\n",

-      "print \"z(ft)\\t Tie Length (ft)\\n\"\n",

-      "for i in range(0,6):\n",

-      "    print z[i],\"\\t\",TL[i]\n",

-      "\n",

-      "print \"use tie length=40ft\"\n",

-      "#check for over turning\n",

-      "\n",

-      "z=30/3.0;\n",

-      "x1=20;\n",

-      "L=40;\n",

-      "Ka=0.26;\n",

-      "Pa=1/2.0*Gamma1*Ka*H**2; #  in kip/ft**2\n",

-      "W1=Gamma1*H*L;\n",

-      "FSO=W1*x1/(Pa*z);\n",

-      "print round(FSO,2),\" is factor of safety is\" \n",

-      "print \"since FS>3 structure is safe\"\n",

-      "#check for sliding\n",

-      "k=2/3.0;\n",

-      "Pa=12285; #  in kip/ft**2\n",

-      "FSS=W1*math.tan(k*phi)/Pa;\n",

-      "if FSS>3 :\n",

-      "    print \"safe in sliding with FS=\",round(FSS,2)\n",

-      "\n",

-      "#check for bearing\n",

-      "Mr=126000*20; # in lb-ft/ft\n",

-      "Mo=12285*10; # in lb-ft/ft\n",

-      "V=126000;\n",

-      "e=L/2-Mr/V+Mo/V;\n",

-      "L1=L-2*e;#Ldash\n",

-      "c2=1000; # in lb/ft^2\n",

-      "Nc=25.8;\n",

-      "Gamma2=110; # in lb/ft^3\n",

-      "Ny=16.72;\n",

-      "qult=c2*Nc+1/2.0*Gamma2*L1*Ny\n",

-      "sigma0=Gamma1*H; # in lb/ft^2\n",

-      "FSB=qult/sigma0;\n",

-      "if FSB>5 :\n",

-      "    print \"bearing is safe  with FS=\",round(FSB,2)\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "0.14  is thickness in inches\n",

-        "so take thickness=0.2 inches\n",

-        "z(ft)\t Tie Length (ft)\n",

-        "\n",

-        "5 \t38.45\n",

-        "10 \t35.89\n",

-        "15 \t33.34\n",

-        "20 \t30.79\n",

-        "25 \t28.25\n",

-        "30 \t25.7\n",

-        "use tie length=40ft\n",

-        "20.51  is factor of safety is\n",

-        "since FS>3 structure is safe\n",

-        "safe in sliding with FS= 4.57\n",

-        "safe in bearing with FS= 19.87\n"

-       ]

-      }

-     ],

-     "prompt_number": 56

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex8.7:Pg-397"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 8.7\n",

-      "import math\n",

-      "import numpy\n",

-      "pi=math.pi;\n",

-      "phi=36.0*pi/180;\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "z=[8.0, 12.0, 16.0]; # in ft\n",

-      "sigmaG=80*12.0; # in lb/ft^2\n",

-      "Gamma1=110.0; # in lb/ft^3\n",

-      "FS=1.5;\n",

-      "Sv=numpy.zeros(3)\n",

-      "for i in range(0,3):\n",

-      "    Sv[i]=sigmaG/Gamma1/z[i]/Ka/FS*12.0;\n",

-      "    print \"for z=\",z[i],\" ft Sv = \",round(Sv[i],2),\" inches\\n\"\n",

-      "\n",

-      "z=[16.0,56.0,76.0,96.0,112.0,144.0,176.0];\n",

-      "zf=numpy.zeros(7)\n",

-      "k2=numpy.zeros(7)\n",

-      "L=numpy.zeros(7)\n",

-      "for i in range(0,7):\n",

-      "    zf[i]=z[i]/12.0;#z in ft\n",

-      "Sv=[1.67,1.67,1.67,1.67,1.33,1.33,1.33];\n",

-      "k=[7.48,5.78,4.93,4.08,3.4,2.04,0.68];#0.51(H-z)\n",

-      "print \"z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \\n\"\n",

-      "for i in range(0,7):\n",

-      "    k2[i]=0.438*Sv[i];#0.438Sv\n",

-      "    L[i]=k[i]+k2[i];\n",

-      "    print round(z[i],2),\"\\t \\t\",round(zf[i],2),\"\\t \",round(Sv[i],2),\"\\t \\t \",round(k[i],2),\"\\t \\t \\t\",round(k2[i],2),\"\\t \\t \\t \",round(L[i],2)\n",

-      "\n",

-      "Sv=20/12.0;\n",

-      "Ka=0.26;\n",

-      "FS=1.5;\n",

-      "l1=Sv*Ka*FS/4/math.tan(2/3.0*phi);\n",

-      "if l1<3:\n",

-      "    l1=3;\n",

-      "    print l1,\"length in ft\"\n",

-      "\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "for z= 8.0  ft Sv =  33.62  inches\n",

-        "\n",

-        "for z= 12.0  ft Sv =  22.41  inches\n",

-        "\n",

-        "for z= 16.0  ft Sv =  16.81  inches\n",

-        "\n",

-        "z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \n",

-        "\n",

-        "16.0 \t \t1.33 \t  1.67 \t \t  7.48 \t \t \t0.73 \t \t \t  8.21\n",

-        "56.0 \t \t4.67 \t  1.67 \t \t  5.78 \t \t \t0.73 \t \t \t  6.51\n",

-        "76.0 \t \t6.33 \t  1.67 \t \t  4.93 \t \t \t0.73 \t \t \t  5.66\n",

-        "96.0 \t \t8.0 \t  1.67 \t \t  4.08 \t \t \t0.73 \t \t \t  4.81\n",

-        "112.0 \t \t9.33 \t  1.33 \t \t  3.4 \t \t \t0.58 \t \t \t  3.98\n",

-        "144.0 \t \t12.0 \t  1.33 \t \t  2.04 \t \t \t0.58 \t \t \t  2.62\n",

-        "176.0 \t \t14.67 \t  1.33 \t \t  0.68 \t \t \t0.58 \t \t \t  1.26\n",

-        "3 length in ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 79

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter09.ipynb b/Principles_Of_Foundation_Engineering/Chapter09.ipynb
deleted file mode 100755
index 8264fd98..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter09.ipynb
+++ /dev/null
@@ -1,512 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d9cb29be0932a6e1135cb07bf43a7e4c55a006ff68a1efd4e530482a24833ad9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter09:Sheet Pile Walls"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.1:Pg-419"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.1\n",
-      "\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "sall=30;#sigma allowed\n",
-      "pi=math.pi;\n",
-      "zbar=12.1; # in ft\n",
-      "L1=10.0; # in ft\n",
-      "L2=20.0; #in ft\n",
-      "Gamma=0.12; # in lb/ft^3\n",
-      "Gamma1=0.1294-0.0624; # in lb/ft^3\n",
-      "phi=40*pi/180; # angle given\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in Kip/ft\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in Kip/ft\n",
-      "L3=s2/(Gamma1*(Kp-Ka)); # in ft\n",
-      "print round(L3,2),\"is length in ft\"\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;# in Kip/ft\n",
-      "print round(P,2),\" is force in kip/ft\"\n",
-      "s5=Gamma*L1*Kp+Gamma1*L2*Kp+Gamma*L3*(Kp-Ka);#sigma5 in Kip/ft\n",
-      "print round(s5,2),\" is pressure in kip/ft\"\n",
-      "A1=s5/(Gamma1*(Kp-Ka)); # in ft^2\n",
-      "A2=8.0*P/(Gamma1*(Kp-Ka)) # in ft^2\n",
-      "A3=6.0*P*(2.0*zbar*(Gamma1*(Kp-Ka))+s5)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",
-      "A4=P*(6.0*zbar*s5+4.0*P)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",
-      "print \"A1,A2,A3,A4 respectively is  \",round(A1,1),round(A2,2),round(A3,2),round(A4,2)\n",
-      "print \"slight error due to rounding off error\"\n",
-      "#part(b)\n",
-      "def f(x):\n",
-      "    return x**4+41.7*x**3-270.5*x**2-13363*x-106863\n",
-      "[x]=fsolve(f,20);\n",
-      "D=1.88+x;\n",
-      "print round(D,2),\" is value of D, in ft\"\n",
-      "TL=L1+L2+1.3*D;\n",
-      "print round(TL,2),\" is total length in ft\"\n",
-      "#partc\n",
-      "z=math.sqrt(2*P/(Gamma1*(Kp-Ka)));#zdash\n",
-      "Mmax=P*(z+zbar)-1/2.0*(Gamma1*(Kp-Ka))*z**2*z/3.0;\n",
-      "S=Mmax*12/sall;\n",
-      "print round(S,2),\"is section modulus in in^3/ft\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.88 is length in ft\n",
-        "9.96  is force in kip/ft\n",
-        "12.67  is pressure in kip/ft\n",
-        "A1,A2,A3,A4 respectively is   43.2 271.33 13708.16 110880.89\n",
-        "slight error due to rounding off error\n",
-        "21.68  is value of D, in ft\n",
-        "58.19  is total length in ft\n",
-        "70.06 is section modulus in in^3/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.2:Pg-426"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.2\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "sall=172.5*1000;#sigma allowed in KN/m^2\n",
-      "pi=math.pi;\n",
-      "c=47.0;  # in KN/m^2\n",
-      "zbar=1.78; # in m\n",
-      "L1=2.0; #in m\n",
-      "L2=3.0; # in m\n",
-      "Gamma=15.9;  # in KN/m^3\n",
-      "Gamma1=19.33-9.81;  # in KN/m^3\n",
-      "phi=32*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2; # in  kN/ft\n",
-      "print round(P,2),\" is force in kN/m\"\n",
-      "def f(x):\n",
-      "    return 127.4*x**2-104.4*x-357.15\n",
-      "[x]=fsolve(f,2);\n",
-      "D=x;\n",
-      "print round(D,2),\" is value of D in m\"\n",
-      "print round(D*1.5,2),\"actual D in m\"\n",
-      "L4=D*(4*c-(Gamma*L1+Gamma1*L2)-P/D)/4/c;\n",
-      "print round(L4,2),\" is length in m\"\n",
-      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in KN/m^2\n",
-      "s7=4*c+(Gamma*L1+Gamma1*L2);#sigma7 in KN/m^2\n",
-      "z=P/s6;#zdash\n",
-      "Mmax=P*(z+zbar)-1/2.0*s6*z**2; # in KN-m/m\n",
-      "S=Mmax*12.0/sall; #  in m^3/m\n",
-      "print round(S,4),\" is section modulus in m**3/m\"\n",
-      "print \"is slight error due to rounding off error\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "52.25  is force in kN/m\n",
-        "2.13  is value of D in m\n",
-        "3.2 actual D in m\n",
-        "1.17  is length in m\n",
-        "0.0072  is section modulus in m**3/m\n",
-        "is slight error due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.3:Pg-433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.3\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "pi=math.pi;\n",
-      "zbar=2.23;  # in m\n",
-      "L1=2.0; # in m\n",
-      "L2=3.0; # in m\n",
-      "Gamma=15.9; # in KN/m^3\n",
-      "Gamma1=19.33-9.81; # in KN/m^3\n",
-      "phi=32*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",
-      "L3=s2/(Gamma1*(Kp-Ka)); # in m\n",
-      "print round(L3,2),\"length in m\"\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;\n",
-      "print round(P,2),\"force in kN/m\"\n",
-      "def f(x):\n",
-      "    return x**3+6.99*x**2-14.55\n",
-      "[x]=fsolve(f,1.4);\n",
-      "D=L3+x;\n",
-      "print round(D,2),\"value of D in m\"\n",
-      "AL=1.4*D;\n",
-      "print round(AL,2),\"actual length in m\"\n",
-      "#partb\n",
-      "L4=1.4;\n",
-      "F=P-1/2.0*(Gamma1*(Kp-Ka)*L4**2);\n",
-      "print round(F,2),\"anchor force in kN/m\"\n",
-      "#partc\n",
-      "def f(x):\n",
-      "    return x**2+6.682*x-14.44\n",
-      "[x]=fsolve(f,1.7);\n",
-      "z=x+2;\n",
-      "Mmax=-1/2.0*s1*L1*(x+2/3.0)+F*(x+1)-s1*x*x/2-1/2.0*Ka*Gamma1*x**3/3.0;\n",
-      "print round(Mmax,2),\" is maximum moment in kN-m/m\" \n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0.66 length in m\n",
-        "58.38 force in kN/m\n",
-        "1.98 value of D in m\n",
-        "2.78 actual length in m\n",
-        "30.88 anchor force in kN/m\n",
-        "43.74  is maximum moment in kN-m/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.4:Pg-439"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.4\n",
-      "\n",
-      "Gamma=15.9; # in KN/m^3\n",
-      "Gamma1=19.33-9.81; # in KN/m^3\n",
-      "GD=0.23;  # from fig. 9.16\n",
-      "CDL1=1.172; # from fig. 9.19\n",
-      "L1=2; # in m\n",
-      "L2=3; # in m\n",
-      "Dth=(L1+L2)*GD*CDL1;\n",
-      "print round(Dth,2),\"theoritical depth in m\"\n",
-      "Dac=1.4*Dth;\n",
-      "print round(Dac,2),\"actual depth in m\"\n",
-      "print \"approximate it as 2 m\"\n",
-      "#part(b)\n",
-      "CFL1=1.073;\n",
-      "GF=0.07;\n",
-      "Gammaa=(Gamma*L1**2+Gamma1*L2**2+2*Gamma*L1*L2)/(L1+L2)**2; # in KN/m^3\n",
-      "F=Gammaa*(L1+L2)**2*GF*CFL1; # in KN/m\n",
-      "print round(F,2),\"force in kN/m\"\n",
-      "#part(c)\n",
-      "GM=0.021; # from fig. 9.18\n",
-      "CML1=1.036; # from fig. 9.21\n",
-      "Mmax=Gammaa*(L1+L2)**3*GM*CML1; # in kN-m/m\n",
-      "print round(Mmax,2),\"maximum moment in kN-m/m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.35 theoritical depth in m\n",
-        "1.89 actual depth in m\n",
-        "approximate it as 2 m\n",
-        "25.54 force in kN/m\n",
-        "36.99 maximum moment in kN-m/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.5:Pg-442"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.5\n",
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "import numpy\n",
-      "Mmax=43.72; # in kN-m/m\n",
-      "sp=[\"PSA-31\",\"PSA-23\"];#sheet pile\n",
-      "H=[7.9,7.9] # in m\n",
-      "I=[4.41e-6,5.63e-6]; # in m^4/m\n",
-      "p=[0.00466,0.00365];\n",
-      "S=[10.8e-5,12.8e-5]; # in m^3/m\n",
-      "Md=[18.63,22.08]; # kn-m/m\n",
-      "Logp=numpy.zeros(2)\n",
-      "k=numpy.zeros(2)\n",
-      "print \"SheetPile    I(m**4/m)    H(m)    p\\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \\n\"\n",
-      "for i in range(0,2):\n",
-      "    Logp[i]=math.log10(p[i]);\n",
-      "    k[i]=Md[i]/Mmax;\n",
-      "    print sp[i],\"\\t   \",I[i],\"    \",H[i],\" \",p[i],\"   \",round(Logp[i],2),\"   \",S[i],\"   \",Md[i],\"\\t \",round(k[i],3)\n",
-      "    \n",
-      "\n",
-      "plt.plot(Logp,k);\n",
-      "plt.title(\"Ex9.5\")\n",
-      "plt.xlabel(\"LogP\")\n",
-      "plt.ylabel(\"Md/Mmax\")\n",
-      "plt.show()\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "SheetPile    I(m**4/m)    H(m)    p\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \n",
-        "\n",
-        "PSA-31 \t    4.41e-06      7.9   0.00466     -2.33     0.000108     18.63 \t  0.426\n",
-        "PSA-23 \t    5.63e-06      7.9   0.00365     -2.44     0.000128     22.08 \t  0.505\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.6:Pg-445"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.6\n",
-      "\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "pi=math.pi;\n",
-      "R=0.6;\n",
-      "L1=10.0; # in ft\n",
-      "L2=20.0; #in ft\n",
-      "Gammasat=122.4; # in lb/ft^3\n",
-      "l1=5; # in ft\n",
-      "Gamma=110.0; # in lb/ft^3\n",
-      "C=0.68;\n",
-      "L=L1+L2; # in ft\n",
-      "Gammaw=62.4; # in lb/ft^3\n",
-      "Gamma1=Gammasat-Gammaw;#gammadash in  lb/ft^3\n",
-      "Gammaav=(Gamma*L1+Gamma1*L2)/(L1+L2); # in lb/ft^3\n",
-      "phi=35.0*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "sa=C*Ka*Gammaav*L;#sigmaa in lb/ft^2\n",
-      "sp=R*sa;#sigmap # in lb/ft^2\n",
-      "def f(x):\n",
-      "    return x**2+50*x-1000\n",
-      "[x]=fsolve(f,15);\n",
-      "D=x;#in ft\n",
-      "print round(D,2),\" is depth in ft\"\n",
-      "R=L/D*(L-2*l1)/(2*L+D-2*l1);\n",
-      "print \"value of R=\",round(R,2),\" is OK\\n\"\n",
-      "#partb\n",
-      "F=sa*(L-R*D); # in lb/ft\n",
-      "print round(F,2),\" is Force in lb/ft\"\n",
-      "#partc\n",
-      "Mmax=0.5*sa*L**2*((1-R*D/L)**2-(2*l1/L)*(1-R*D/L)); # in lb-ft/ft\n",
-      "print round(Mmax,2),\"maximum moment lb-ft/ft\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "15.31  is depth in ft\n",
-        "value of R= 0.6  is OK\n",
-        "\n",
-        "8821.24  is Force in lb/ft\n",
-        "47693.02 maximum moment lb-ft/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.7:Pg-451"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.7\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "pi=math.pi;\n",
-      "zbar=3.2; # in m\n",
-      "c=41; # in KN/m^2\n",
-      "L1=3; # in m\n",
-      "L2=6; # in m\n",
-      "Gamma=17;# in KN/m^3\n",
-      "Gamma1=20-9.81; # in KN/m^3\n",
-      "phi=35*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in kN/m**2\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in kN/m**2\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2;\n",
-      "print round(P,2),\"Force in kN/m\"\n",
-      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in kN/m**2\n",
-      "print round(s6,2),\"pressure in kN/m**2\"\n",
-      "def f(x):\n",
-      "    return x**2+15*x-25.43\n",
-      "[x]=fsolve(f,1.6);\n",
-      "D=x; # in m\n",
-      "print round(D,1),\"depth in m\"\n",
-      "F=P-s6*D; # in kN/m\n",
-      "print round(F,2),\"force in kN/m\"\n",
-      "print \"slight error due to rounding off\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "153.36 Force in kN/m\n",
-        "51.86 pressure in kN/m**2\n",
-        "1.5 depth in m\n",
-        "73.61 force in kN/m\n",
-        "slight error due to rounding off\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.8\n",
-      "import math\n",
-      "pi=math.pi;\n",
-      "Gamma=105.0; # in lb/ft^3\n",
-      "Cov=14.0;\n",
-      "B=15/12.0; # in inch\n",
-      "Ka=0.26;\n",
-      "phi=35.0*pi/180; # given angle in degree\n",
-      "H=37.5/12; # in inch\n",
-      "h=15/12.0; # in inch\n",
-      "t=6/12.0; # in inch\n",
-      "Gc=150.0;#gamma concrete in lb/ft^3\n",
-      "W=H*t*Gc; # in lb/ft\n",
-      "k=4.5;#kp*cos(delta)\n",
-      "Pu=1/2.0*Gamma*H**2*(k-Ka*math.cos(phi)); # in lb/ft\n",
-      "print round(Pu,2),\"force in lb/ft\"\n",
-      "Pus=((Cov+1)/(Cov+H/h))*Pu; # in lb/ft\n",
-      "print round(Pus,2),\"force in lb/ft\"\n",
-      "Be=0.227*(H+h)+B;\n",
-      "Pu=Pus*Be; # in lb/ft\n",
-      "print round(Pu,2),\" is resistance of anchor plate in lb/ft\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2197.94 force in lb/ft\n",
-        "1998.12 force in lb/ft\n",
-        "4482.04  is resistance of anchor plate in lb/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter09_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter09_1.ipynb
deleted file mode 100755
index 8264fd98..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter09_1.ipynb
+++ /dev/null
@@ -1,512 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d9cb29be0932a6e1135cb07bf43a7e4c55a006ff68a1efd4e530482a24833ad9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter09:Sheet Pile Walls"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.1:Pg-419"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.1\n",
-      "\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "sall=30;#sigma allowed\n",
-      "pi=math.pi;\n",
-      "zbar=12.1; # in ft\n",
-      "L1=10.0; # in ft\n",
-      "L2=20.0; #in ft\n",
-      "Gamma=0.12; # in lb/ft^3\n",
-      "Gamma1=0.1294-0.0624; # in lb/ft^3\n",
-      "phi=40*pi/180; # angle given\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in Kip/ft\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in Kip/ft\n",
-      "L3=s2/(Gamma1*(Kp-Ka)); # in ft\n",
-      "print round(L3,2),\"is length in ft\"\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;# in Kip/ft\n",
-      "print round(P,2),\" is force in kip/ft\"\n",
-      "s5=Gamma*L1*Kp+Gamma1*L2*Kp+Gamma*L3*(Kp-Ka);#sigma5 in Kip/ft\n",
-      "print round(s5,2),\" is pressure in kip/ft\"\n",
-      "A1=s5/(Gamma1*(Kp-Ka)); # in ft^2\n",
-      "A2=8.0*P/(Gamma1*(Kp-Ka)) # in ft^2\n",
-      "A3=6.0*P*(2.0*zbar*(Gamma1*(Kp-Ka))+s5)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",
-      "A4=P*(6.0*zbar*s5+4.0*P)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",
-      "print \"A1,A2,A3,A4 respectively is  \",round(A1,1),round(A2,2),round(A3,2),round(A4,2)\n",
-      "print \"slight error due to rounding off error\"\n",
-      "#part(b)\n",
-      "def f(x):\n",
-      "    return x**4+41.7*x**3-270.5*x**2-13363*x-106863\n",
-      "[x]=fsolve(f,20);\n",
-      "D=1.88+x;\n",
-      "print round(D,2),\" is value of D, in ft\"\n",
-      "TL=L1+L2+1.3*D;\n",
-      "print round(TL,2),\" is total length in ft\"\n",
-      "#partc\n",
-      "z=math.sqrt(2*P/(Gamma1*(Kp-Ka)));#zdash\n",
-      "Mmax=P*(z+zbar)-1/2.0*(Gamma1*(Kp-Ka))*z**2*z/3.0;\n",
-      "S=Mmax*12/sall;\n",
-      "print round(S,2),\"is section modulus in in^3/ft\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.88 is length in ft\n",
-        "9.96  is force in kip/ft\n",
-        "12.67  is pressure in kip/ft\n",
-        "A1,A2,A3,A4 respectively is   43.2 271.33 13708.16 110880.89\n",
-        "slight error due to rounding off error\n",
-        "21.68  is value of D, in ft\n",
-        "58.19  is total length in ft\n",
-        "70.06 is section modulus in in^3/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.2:Pg-426"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.2\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "sall=172.5*1000;#sigma allowed in KN/m^2\n",
-      "pi=math.pi;\n",
-      "c=47.0;  # in KN/m^2\n",
-      "zbar=1.78; # in m\n",
-      "L1=2.0; #in m\n",
-      "L2=3.0; # in m\n",
-      "Gamma=15.9;  # in KN/m^3\n",
-      "Gamma1=19.33-9.81;  # in KN/m^3\n",
-      "phi=32*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2; # in  kN/ft\n",
-      "print round(P,2),\" is force in kN/m\"\n",
-      "def f(x):\n",
-      "    return 127.4*x**2-104.4*x-357.15\n",
-      "[x]=fsolve(f,2);\n",
-      "D=x;\n",
-      "print round(D,2),\" is value of D in m\"\n",
-      "print round(D*1.5,2),\"actual D in m\"\n",
-      "L4=D*(4*c-(Gamma*L1+Gamma1*L2)-P/D)/4/c;\n",
-      "print round(L4,2),\" is length in m\"\n",
-      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in KN/m^2\n",
-      "s7=4*c+(Gamma*L1+Gamma1*L2);#sigma7 in KN/m^2\n",
-      "z=P/s6;#zdash\n",
-      "Mmax=P*(z+zbar)-1/2.0*s6*z**2; # in KN-m/m\n",
-      "S=Mmax*12.0/sall; #  in m^3/m\n",
-      "print round(S,4),\" is section modulus in m**3/m\"\n",
-      "print \"is slight error due to rounding off error\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "52.25  is force in kN/m\n",
-        "2.13  is value of D in m\n",
-        "3.2 actual D in m\n",
-        "1.17  is length in m\n",
-        "0.0072  is section modulus in m**3/m\n",
-        "is slight error due to rounding off error\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.3:Pg-433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.3\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "pi=math.pi;\n",
-      "zbar=2.23;  # in m\n",
-      "L1=2.0; # in m\n",
-      "L2=3.0; # in m\n",
-      "Gamma=15.9; # in KN/m^3\n",
-      "Gamma1=19.33-9.81; # in KN/m^3\n",
-      "phi=32*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",
-      "L3=s2/(Gamma1*(Kp-Ka)); # in m\n",
-      "print round(L3,2),\"length in m\"\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;\n",
-      "print round(P,2),\"force in kN/m\"\n",
-      "def f(x):\n",
-      "    return x**3+6.99*x**2-14.55\n",
-      "[x]=fsolve(f,1.4);\n",
-      "D=L3+x;\n",
-      "print round(D,2),\"value of D in m\"\n",
-      "AL=1.4*D;\n",
-      "print round(AL,2),\"actual length in m\"\n",
-      "#partb\n",
-      "L4=1.4;\n",
-      "F=P-1/2.0*(Gamma1*(Kp-Ka)*L4**2);\n",
-      "print round(F,2),\"anchor force in kN/m\"\n",
-      "#partc\n",
-      "def f(x):\n",
-      "    return x**2+6.682*x-14.44\n",
-      "[x]=fsolve(f,1.7);\n",
-      "z=x+2;\n",
-      "Mmax=-1/2.0*s1*L1*(x+2/3.0)+F*(x+1)-s1*x*x/2-1/2.0*Ka*Gamma1*x**3/3.0;\n",
-      "print round(Mmax,2),\" is maximum moment in kN-m/m\" \n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0.66 length in m\n",
-        "58.38 force in kN/m\n",
-        "1.98 value of D in m\n",
-        "2.78 actual length in m\n",
-        "30.88 anchor force in kN/m\n",
-        "43.74  is maximum moment in kN-m/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.4:Pg-439"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.4\n",
-      "\n",
-      "Gamma=15.9; # in KN/m^3\n",
-      "Gamma1=19.33-9.81; # in KN/m^3\n",
-      "GD=0.23;  # from fig. 9.16\n",
-      "CDL1=1.172; # from fig. 9.19\n",
-      "L1=2; # in m\n",
-      "L2=3; # in m\n",
-      "Dth=(L1+L2)*GD*CDL1;\n",
-      "print round(Dth,2),\"theoritical depth in m\"\n",
-      "Dac=1.4*Dth;\n",
-      "print round(Dac,2),\"actual depth in m\"\n",
-      "print \"approximate it as 2 m\"\n",
-      "#part(b)\n",
-      "CFL1=1.073;\n",
-      "GF=0.07;\n",
-      "Gammaa=(Gamma*L1**2+Gamma1*L2**2+2*Gamma*L1*L2)/(L1+L2)**2; # in KN/m^3\n",
-      "F=Gammaa*(L1+L2)**2*GF*CFL1; # in KN/m\n",
-      "print round(F,2),\"force in kN/m\"\n",
-      "#part(c)\n",
-      "GM=0.021; # from fig. 9.18\n",
-      "CML1=1.036; # from fig. 9.21\n",
-      "Mmax=Gammaa*(L1+L2)**3*GM*CML1; # in kN-m/m\n",
-      "print round(Mmax,2),\"maximum moment in kN-m/m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.35 theoritical depth in m\n",
-        "1.89 actual depth in m\n",
-        "approximate it as 2 m\n",
-        "25.54 force in kN/m\n",
-        "36.99 maximum moment in kN-m/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.5:Pg-442"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.5\n",
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "import numpy\n",
-      "Mmax=43.72; # in kN-m/m\n",
-      "sp=[\"PSA-31\",\"PSA-23\"];#sheet pile\n",
-      "H=[7.9,7.9] # in m\n",
-      "I=[4.41e-6,5.63e-6]; # in m^4/m\n",
-      "p=[0.00466,0.00365];\n",
-      "S=[10.8e-5,12.8e-5]; # in m^3/m\n",
-      "Md=[18.63,22.08]; # kn-m/m\n",
-      "Logp=numpy.zeros(2)\n",
-      "k=numpy.zeros(2)\n",
-      "print \"SheetPile    I(m**4/m)    H(m)    p\\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \\n\"\n",
-      "for i in range(0,2):\n",
-      "    Logp[i]=math.log10(p[i]);\n",
-      "    k[i]=Md[i]/Mmax;\n",
-      "    print sp[i],\"\\t   \",I[i],\"    \",H[i],\" \",p[i],\"   \",round(Logp[i],2),\"   \",S[i],\"   \",Md[i],\"\\t \",round(k[i],3)\n",
-      "    \n",
-      "\n",
-      "plt.plot(Logp,k);\n",
-      "plt.title(\"Ex9.5\")\n",
-      "plt.xlabel(\"LogP\")\n",
-      "plt.ylabel(\"Md/Mmax\")\n",
-      "plt.show()\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "SheetPile    I(m**4/m)    H(m)    p\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \n",
-        "\n",
-        "PSA-31 \t    4.41e-06      7.9   0.00466     -2.33     0.000108     18.63 \t  0.426\n",
-        "PSA-23 \t    5.63e-06      7.9   0.00365     -2.44     0.000128     22.08 \t  0.505\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.6:Pg-445"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.6\n",
-      "\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "pi=math.pi;\n",
-      "R=0.6;\n",
-      "L1=10.0; # in ft\n",
-      "L2=20.0; #in ft\n",
-      "Gammasat=122.4; # in lb/ft^3\n",
-      "l1=5; # in ft\n",
-      "Gamma=110.0; # in lb/ft^3\n",
-      "C=0.68;\n",
-      "L=L1+L2; # in ft\n",
-      "Gammaw=62.4; # in lb/ft^3\n",
-      "Gamma1=Gammasat-Gammaw;#gammadash in  lb/ft^3\n",
-      "Gammaav=(Gamma*L1+Gamma1*L2)/(L1+L2); # in lb/ft^3\n",
-      "phi=35.0*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "sa=C*Ka*Gammaav*L;#sigmaa in lb/ft^2\n",
-      "sp=R*sa;#sigmap # in lb/ft^2\n",
-      "def f(x):\n",
-      "    return x**2+50*x-1000\n",
-      "[x]=fsolve(f,15);\n",
-      "D=x;#in ft\n",
-      "print round(D,2),\" is depth in ft\"\n",
-      "R=L/D*(L-2*l1)/(2*L+D-2*l1);\n",
-      "print \"value of R=\",round(R,2),\" is OK\\n\"\n",
-      "#partb\n",
-      "F=sa*(L-R*D); # in lb/ft\n",
-      "print round(F,2),\" is Force in lb/ft\"\n",
-      "#partc\n",
-      "Mmax=0.5*sa*L**2*((1-R*D/L)**2-(2*l1/L)*(1-R*D/L)); # in lb-ft/ft\n",
-      "print round(Mmax,2),\"maximum moment lb-ft/ft\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "15.31  is depth in ft\n",
-        "value of R= 0.6  is OK\n",
-        "\n",
-        "8821.24  is Force in lb/ft\n",
-        "47693.02 maximum moment lb-ft/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.7:Pg-451"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.7\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "\n",
-      "pi=math.pi;\n",
-      "zbar=3.2; # in m\n",
-      "c=41; # in KN/m^2\n",
-      "L1=3; # in m\n",
-      "L2=6; # in m\n",
-      "Gamma=17;# in KN/m^3\n",
-      "Gamma1=20-9.81; # in KN/m^3\n",
-      "phi=35*pi/180;\n",
-      "Ka=(math.tan(pi/4-phi/2))**2;\n",
-      "Kp=(math.tan(pi/4+phi/2))**2;\n",
-      "s1=Gamma*L1*Ka;#sigma1 in kN/m**2\n",
-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in kN/m**2\n",
-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2;\n",
-      "print round(P,2),\"Force in kN/m\"\n",
-      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in kN/m**2\n",
-      "print round(s6,2),\"pressure in kN/m**2\"\n",
-      "def f(x):\n",
-      "    return x**2+15*x-25.43\n",
-      "[x]=fsolve(f,1.6);\n",
-      "D=x; # in m\n",
-      "print round(D,1),\"depth in m\"\n",
-      "F=P-s6*D; # in kN/m\n",
-      "print round(F,2),\"force in kN/m\"\n",
-      "print \"slight error due to rounding off\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "153.36 Force in kN/m\n",
-        "51.86 pressure in kN/m**2\n",
-        "1.5 depth in m\n",
-        "73.61 force in kN/m\n",
-        "slight error due to rounding off\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 9.8\n",
-      "import math\n",
-      "pi=math.pi;\n",
-      "Gamma=105.0; # in lb/ft^3\n",
-      "Cov=14.0;\n",
-      "B=15/12.0; # in inch\n",
-      "Ka=0.26;\n",
-      "phi=35.0*pi/180; # given angle in degree\n",
-      "H=37.5/12; # in inch\n",
-      "h=15/12.0; # in inch\n",
-      "t=6/12.0; # in inch\n",
-      "Gc=150.0;#gamma concrete in lb/ft^3\n",
-      "W=H*t*Gc; # in lb/ft\n",
-      "k=4.5;#kp*cos(delta)\n",
-      "Pu=1/2.0*Gamma*H**2*(k-Ka*math.cos(phi)); # in lb/ft\n",
-      "print round(Pu,2),\"force in lb/ft\"\n",
-      "Pus=((Cov+1)/(Cov+H/h))*Pu; # in lb/ft\n",
-      "print round(Pus,2),\"force in lb/ft\"\n",
-      "Be=0.227*(H+h)+B;\n",
-      "Pu=Pus*Be; # in lb/ft\n",
-      "print round(Pu,2),\" is resistance of anchor plate in lb/ft\"\n",
-      "\n",
-      "# The answers in the book are different due to approximation while here calculations are precise"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2197.94 force in lb/ft\n",
-        "1998.12 force in lb/ft\n",
-        "4482.04  is resistance of anchor plate in lb/ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter09_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter09_2.ipynb
deleted file mode 100755
index 7d8bec64..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter09_2.ipynb
+++ /dev/null
@@ -1,520 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:a20958eb0a1cf98629a18a836583d76a2fff705e2f3370294ba37a907bebd17b"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter09:Sheet Pile Walls"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.1:Pg-419"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.1\n",

-      "\n",

-      "import math\n",

-      "from scipy.optimize import fsolve\n",

-      "sall=30;#sigma allowed\n",

-      "pi=math.pi;\n",

-      "zbar=12.1; # in ft\n",

-      "L1=10.0; # in ft\n",

-      "L2=20.0; #in ft\n",

-      "Gamma=0.12; # in lb/ft^3\n",

-      "Gamma1=0.1294-0.0624; # in lb/ft^3\n",

-      "phi=40*pi/180; # angle given\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "Kp=(math.tan(pi/4+phi/2))**2;\n",

-      "s1=Gamma*L1*Ka;#sigma1 in Kip/ft\n",

-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in Kip/ft\n",

-      "L3=s2/(Gamma1*(Kp-Ka)); # in ft\n",

-      "print round(L3,2),\"is length in ft\"\n",

-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;# in Kip/ft\n",

-      "print round(P,2),\" is force in kip/ft\"\n",

-      "s5=Gamma*L1*Kp+Gamma1*L2*Kp+Gamma*L3*(Kp-Ka);#sigma5 in Kip/ft\n",

-      "print round(s5,2),\" is pressure in kip/ft\"\n",

-      "A1=s5/(Gamma1*(Kp-Ka)); # in ft^2\n",

-      "A2=8.0*P/(Gamma1*(Kp-Ka)) # in ft^2\n",

-      "A3=6.0*P*(2.0*zbar*(Gamma1*(Kp-Ka))+s5)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",

-      "A4=P*(6.0*zbar*s5+4.0*P)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",

-      "print \"A1,A2,A3,A4 respectively is  \",round(A1,1),round(A2,2),round(A3,2),round(A4,2)\n",

-      "print \"slight error due to rounding off error\"\n",

-      "#part(b)\n",

-      "def f(x):\n",

-      "    return x**4+41.7*x**3-270.5*x**2-13363*x-106863\n",

-      "[x]=fsolve(f,20);\n",

-      "D=1.88+x;\n",

-      "print round(D,2),\" is value of D, in ft\"\n",

-      "TL=L1+L2+1.3*D;\n",

-      "print round(TL,2),\" is total length in ft\"\n",

-      "#partc\n",

-      "z=math.sqrt(2*P/(Gamma1*(Kp-Ka)));#zdash\n",

-      "Mmax=P*(z+zbar)-1/2.0*(Gamma1*(Kp-Ka))*z**2*z/3.0;\n",

-      "S=Mmax*12/sall;\n",

-      "print round(S,2),\"is section modulus in in^3/ft\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "1.88 is length in ft\n",

-        "9.96  is force in kip/ft\n",

-        "12.67  is pressure in kip/ft\n",

-        "A1,A2,A3,A4 respectively is   43.2 271.33 13708.16 110880.89\n",

-        "slight error due to rounding off error\n",

-        "21.68  is value of D, in ft\n",

-        "58.19  is total length in ft\n",

-        "70.06 is section modulus in in^3/ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.2:Pg-426"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.2\n",

-      "import math\n",

-      "from scipy.optimize import fsolve\n",

-      "sall=172.5*1000;#sigma allowed in KN/m^2\n",

-      "pi=math.pi;\n",

-      "c=47.0;  # in KN/m^2\n",

-      "zbar=1.78; # in m\n",

-      "L1=2.0; #in m\n",

-      "L2=3.0; # in m\n",

-      "Gamma=15.9;  # in KN/m^3\n",

-      "Gamma1=19.33-9.81;  # in KN/m^3\n",

-      "phi=32*pi/180;\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "Kp=(math.tan(pi/4+phi/2))**2;\n",

-      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",

-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",

-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2; # in  kN/ft\n",

-      "print round(P,2),\" is force in kN/m\"\n",

-      "def f(x):\n",

-      "    return 127.4*x**2-104.4*x-357.15\n",

-      "[x]=fsolve(f,2);\n",

-      "D=x;\n",

-      "print round(D,2),\" is value of D in m\"\n",

-      "print round(D*1.5,2),\"actual D in m\"\n",

-      "L4=D*(4*c-(Gamma*L1+Gamma1*L2)-P/D)/4/c;\n",

-      "print round(L4,2),\" is length in m\"\n",

-      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in KN/m^2\n",

-      "s7=4*c+(Gamma*L1+Gamma1*L2);#sigma7 in KN/m^2\n",

-      "z=P/s6;#zdash\n",

-      "Mmax=P*(z+zbar)-1/2.0*s6*z**2; # in KN-m/m\n",

-      "S=Mmax*12.0/sall; #  in m^3/m\n",

-      "print round(S,4),\" is section modulus in m**3/m\"\n",

-      "print \"is slight error due to rounding off error\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "52.25  is force in kN/m\n",

-        "2.13  is value of D in m\n",

-        "3.2 actual D in m\n",

-        "1.17  is length in m\n",

-        "0.0072  is section modulus in m**3/m\n",

-        "is slight error due to rounding off error\n"

-       ]

-      }

-     ],

-     "prompt_number": 17

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.3:Pg-433"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.3\n",

-      "import math\n",

-      "from scipy.optimize import fsolve\n",

-      "\n",

-      "pi=math.pi;\n",

-      "zbar=2.23;  # in m\n",

-      "L1=2.0; # in m\n",

-      "L2=3.0; # in m\n",

-      "Gamma=15.9; # in KN/m^3\n",

-      "Gamma1=19.33-9.81; # in KN/m^3\n",

-      "phi=32*pi/180;\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "Kp=(math.tan(pi/4+phi/2))**2;\n",

-      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",

-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",

-      "L3=s2/(Gamma1*(Kp-Ka)); # in m\n",

-      "print round(L3,2),\"length in m\"\n",

-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;\n",

-      "print round(P,2),\"force in kN/m\"\n",

-      "def f(x):\n",

-      "    return x**3+6.99*x**2-14.55\n",

-      "[x]=fsolve(f,1.4);\n",

-      "D=L3+x;\n",

-      "print round(D,2),\"value of D in m\"\n",

-      "AL=1.4*D;\n",

-      "print round(AL,2),\"actual length in m\"\n",

-      "#partb\n",

-      "L4=1.4;\n",

-      "F=P-1/2.0*(Gamma1*(Kp-Ka)*L4**2);\n",

-      "print round(F,2),\"anchor force in kN/m\"\n",

-      "#partc\n",

-      "def f(x):\n",

-      "    return x**2+6.682*x-14.44\n",

-      "[x]=fsolve(f,1.7);\n",

-      "z=x+2;\n",

-      "Mmax=-1/2.0*s1*L1*(x+2/3.0)+F*(x+1)-s1*x*x/2-1/2.0*Ka*Gamma1*x**3/3.0;\n",

-      "print round(Mmax,2),\" is maximum moment in kN-m/m\" \n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "0.66 length in m\n",

-        "58.38 force in kN/m\n",

-        "1.98 value of D in m\n",

-        "2.78 actual length in m\n",

-        "30.88 anchor force in kN/m\n",

-        "43.74  is maximum moment in kN-m/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 18

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.4:Pg-439"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.4\n",

-      "\n",

-      "Gamma=15.9; # in KN/m^3\n",

-      "Gamma1=19.33-9.81; # in KN/m^3\n",

-      "GD=0.23;  # from fig. 9.16\n",

-      "CDL1=1.172; # from fig. 9.19\n",

-      "L1=2; # in m\n",

-      "L2=3; # in m\n",

-      "Dth=(L1+L2)*GD*CDL1;\n",

-      "print round(Dth,2),\"theoritical depth in m\"\n",

-      "Dac=1.4*Dth;\n",

-      "print round(Dac,2),\"actual depth in m\"\n",

-      "print \"approximate it as 2 m\"\n",

-      "#part(b)\n",

-      "CFL1=1.073;\n",

-      "GF=0.07;\n",

-      "Gammaa=(Gamma*L1**2+Gamma1*L2**2+2*Gamma*L1*L2)/(L1+L2)**2; # in KN/m^3\n",

-      "F=Gammaa*(L1+L2)**2*GF*CFL1; # in KN/m\n",

-      "print round(F,2),\"force in kN/m\"\n",

-      "#part(c)\n",

-      "GM=0.021; # from fig. 9.18\n",

-      "CML1=1.036; # from fig. 9.21\n",

-      "Mmax=Gammaa*(L1+L2)**3*GM*CML1; # in kN-m/m\n",

-      "print round(Mmax,2),\"maximum moment in kN-m/m\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "1.35 theoritical depth in m\n",

-        "1.89 actual depth in m\n",

-        "approximate it as 2 m\n",

-        "25.54 force in kN/m\n",

-        "36.99 maximum moment in kN-m/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 20

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.5:Pg-442"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.5\n",

-      "import math\n",

-      "import matplotlib.pyplot as plt\n",

-      "import numpy\n",

-      "Mmax=43.72; # in kN-m/m\n",

-      "sp=[\"PSA-31\",\"PSA-23\"];#sheet pile\n",

-      "H=[7.9,7.9] # in m\n",

-      "I=[4.41e-6,5.63e-6]; # in m^4/m\n",

-      "p=[0.00466,0.00365];\n",

-      "S=[10.8e-5,12.8e-5]; # in m^3/m\n",

-      "Md=[18.63,22.08]; # kn-m/m\n",

-      "Logp=numpy.zeros(2)\n",

-      "k=numpy.zeros(2)\n",

-      "print \"SheetPile    I(m**4/m)    H(m)    p\\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \\n\"\n",

-      "for i in range(0,2):\n",

-      "    Logp[i]=math.log10(p[i]);\n",

-      "    k[i]=Md[i]/Mmax;\n",

-      "    print sp[i],\"\\t   \",I[i],\"    \",H[i],\" \",p[i],\"   \",round(Logp[i],2),\"   \",S[i],\"   \",Md[i],\"\\t \",round(k[i],3)\n",

-      "    \n",

-      "\n",

-      "plt.plot(Logp,k);\n",

-      "plt.title(\"Ex9.5\")\n",

-      "plt.xlabel(\"LogP\")\n",

-      "plt.ylabel(\"Md/Mmax\")\n",

-      "plt.show()\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "SheetPile    I(m**4/m)    H(m)    p\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \n",

-        "\n",

-        "PSA-31 \t    4.41e-06      7.9   0.00466     -2.33     0.000108     18.63 \t  0.426\n",

-        "PSA-23 \t    5.63e-06      7.9   0.00365     -2.44     0.000128     22.08 \t  0.505\n"

-       ]

-      }

-     ],

-     "prompt_number": 28

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.6:Pg-445"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.6\n",

-      "\n",

-      "import math\n",

-      "from scipy.optimize import fsolve\n",

-      "\n",

-      "pi=math.pi;\n",

-      "R=0.6;\n",

-      "L1=10.0; # in ft\n",

-      "L2=20.0; #in ft\n",

-      "Gammasat=122.4; # in lb/ft^3\n",

-      "l1=5; # in ft\n",

-      "Gamma=110.0; # in lb/ft^3\n",

-      "C=0.68;\n",

-      "L=L1+L2; # in ft\n",

-      "Gammaw=62.4; # in lb/ft^3\n",

-      "Gamma1=Gammasat-Gammaw;#gammadash in  lb/ft^3\n",

-      "Gammaav=(Gamma*L1+Gamma1*L2)/(L1+L2); # in lb/ft^3\n",

-      "phi=35.0*pi/180;\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "sa=C*Ka*Gammaav*L;#sigmaa in lb/ft^2\n",

-      "sp=R*sa;#sigmap # in lb/ft^2\n",

-      "def f(x):\n",

-      "    return x**2+50*x-1000\n",

-      "[x]=fsolve(f,15);\n",

-      "D=x;#in ft\n",

-      "print round(D,2),\" is depth in ft\"\n",

-      "R=L/D*(L-2*l1)/(2*L+D-2*l1);\n",

-      "print \"value of R=\",round(R,2),\" is OK\\n\"\n",

-      "#partb\n",

-      "F=sa*(L-R*D); # in lb/ft\n",

-      "print round(F,2),\" is Force in lb/ft\"\n",

-      "#partc\n",

-      "Mmax=0.5*sa*L**2*((1-R*D/L)**2-(2*l1/L)*(1-R*D/L)); # in lb-ft/ft\n",

-      "print round(Mmax,2),\"maximum moment lb-ft/ft\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "15.31  is depth in ft\n",

-        "value of R= 0.6  is OK\n",

-        "\n",

-        "8821.24  is Force in lb/ft\n",

-        "47693.02 maximum moment lb-ft/ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 31

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.7:Pg-451"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.7\n",

-      "import math\n",

-      "from scipy.optimize import fsolve\n",

-      "\n",

-      "pi=math.pi;\n",

-      "zbar=3.2; # in m\n",

-      "c=41; # in KN/m^2\n",

-      "L1=3; # in m\n",

-      "L2=6; # in m\n",

-      "Gamma=17;# in KN/m^3\n",

-      "Gamma1=20-9.81; # in KN/m^3\n",

-      "phi=35*pi/180;\n",

-      "Ka=(math.tan(pi/4-phi/2))**2;\n",

-      "Kp=(math.tan(pi/4+phi/2))**2;\n",

-      "s1=Gamma*L1*Ka;#sigma1 in kN/m**2\n",

-      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in kN/m**2\n",

-      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2;\n",

-      "print round(P,2),\"Force in kN/m\"\n",

-      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in kN/m**2\n",

-      "print round(s6,2),\"pressure in kN/m**2\"\n",

-      "def f(x):\n",

-      "    return x**2+15*x-25.43\n",

-      "[x]=fsolve(f,1.6);\n",

-      "D=x; # in m\n",

-      "print round(D,1),\"depth in m\"\n",

-      "F=P-s6*D; # in kN/m\n",

-      "print round(F,2),\"force in kN/m\"\n",

-      "print \"slight error due to rounding off\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "153.36 Force in kN/m\n",

-        "51.86 pressure in kN/m**2\n",

-        "1.5 depth in m\n",

-        "73.61 force in kN/m\n",

-        "slight error due to rounding off\n"

-       ]

-      }

-     ],

-     "prompt_number": 35

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex9.8:pg-458"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 9.8\n",

-      "import math\n",

-      "pi=math.pi;\n",

-      "Gamma=105.0; # in lb/ft^3\n",

-      "Cov=14.0;\n",

-      "B=15/12.0; # in inch\n",

-      "Ka=0.26;\n",

-      "phi=35.0*pi/180; # given angle in degree\n",

-      "H=37.5/12; # in inch\n",

-      "h=15/12.0; # in inch\n",

-      "t=6/12.0; # in inch\n",

-      "Gc=150.0;#gamma concrete in lb/ft^3\n",

-      "W=H*t*Gc; # in lb/ft\n",

-      "k=4.5;#kp*cos(delta)\n",

-      "Pu=1/2.0*Gamma*H**2*(k-Ka*math.cos(phi)); # in lb/ft\n",

-      "print round(Pu,2),\"force in lb/ft\"\n",

-      "Pus=((Cov+1)/(Cov+H/h))*Pu; # in lb/ft\n",

-      "print round(Pus,2),\"force in lb/ft\"\n",

-      "Be=0.227*(H+h)+B;\n",

-      "Pu=Pus*Be; # in lb/ft\n",

-      "print round(Pu,2),\" is resistance of anchor plate in lb/ft\"\n",

-      "\n",

-      "# The answers in the book are different due to approximation while here calculations are precise"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "2197.94 force in lb/ft\n",

-        "1998.12 force in lb/ft\n",

-        "4482.04  is resistance of anchor plate in lb/ft\n"

-       ]

-      }

-     ],

-     "prompt_number": 37

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter10.ipynb b/Principles_Of_Foundation_Engineering/Chapter10.ipynb
deleted file mode 100755
index 3af11efd..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter10.ipynb
+++ /dev/null
@@ -1,211 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:54ddb8c76d55d78e9f237ddb7f5823ac10a3585adf1b3932a3b8cc9352fd8e76"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10:Braced Cuts"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.1: pg-511"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.1 : \n",
-      "\n",
-      "Gamma=18.0;\n",
-      "H=7.0;\n",
-      "sigmaa=0.3*Gamma*H;\n",
-      "print\"maximum pressure intensity in kN/m^2 is\",sigmaa ;\n",
-      "#partb\n",
-      "A=54.02;\n",
-      "B1=1.0/2*1.75*37.8+37.8*1.75-A;\n",
-      "B2=45.2;\n",
-      "C=54.02;\n",
-      "s=3.0; #spacing\n",
-      "Pa=C*s;\n",
-      "print \"strut loads in kN is\",Pa\n",
-      "Pb=(B1+B2)*s;\n",
-      "print \"strut loads in kN is\",Pb\n",
-      "Pc=C*s;\n",
-      "print \"strut loads in kN is\", Pc\n",
-      "#partc\n",
-      "Me=45.2*1.196-37.8*1.196*1.196/2;#Me=Mmax\n",
-      "Sall=170e3;#sigmaall\n",
-      "S=Me/Sall;\n",
-      "print \"section modulus in m^3/m is \",round(S,7)\n",
-      "#partd\n",
-      "Mmax=(B1+B2)*s**2.0/8;\n",
-      "S=Mmax/Sall;\n",
-      "print\"section modulus in m^3/m is\",round(S,6)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum pressure intensity in kN/m^2 is 37.8\n",
-        "strut loads in kN is 162.06\n",
-        "strut loads in kN is 271.215\n",
-        "strut loads in kN is 162.06\n",
-        "section modulus in m^3/m is  0.000159\n",
-        "section modulus in m^3/m is 0.000598\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.2:pg-514"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.2\n",
-      "\n",
-      "import math\n",
-      "phi=35.0;\n",
-      "Gamma=17.0; # kN/m^3\n",
-      "s=4; #spacing in m\n",
-      "H=9;\n",
-      "Ka=(math.tan(45*(math.pi/180)-35*(math.pi/(180.0*2))))**2\n",
-      "sigma=0.65*Gamma*Ka*H\n",
-      "A=sigma*5*5.0/6;\n",
-      "B1=sigma*5-A;\n",
-      "C=sigma*4*4/6.0; \n",
-      "B2=sigma*s-C;\n",
-      "Pa=A*s;\n",
-      "Pb=(B1+B2)*s;\n",
-      "Pc=C*s;\n",
-      "print \"strut loads at A in kN is\",round(Pa,2)\n",
-      "\n",
-      "print \"strut loads at B in kN is\",round(Pb,2)\n",
-      "\n",
-      "print \"strut loads at C in kN is\",round(Pc,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "strut loads at A in kN is 449.17\n",
-        "strut loads at B in kN is 233.57\n",
-        "strut loads at C in kN is 287.47\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.3:pg-523"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.3\n",
-      "\n",
-      "import math\n",
-      "q=0;\n",
-      "Gamma=17; # in KN/m^3\n",
-      "B=3.0 # in meter\n",
-      "L=20; # in meter\n",
-      "c=30;# in KN/m^3\n",
-      "T=1.5;# in meter\n",
-      "H=5.5;# in meter\n",
-      "B1=B/2;#B'\n",
-      "B11=T*math.sqrt(2);#B''\n",
-      "FS=(5.14*c*(1+B11*0.2/L)+c*H/B1)/(Gamma*H+q) # from equation of factor of safety\n",
-      "print\"factor of safety is\",round(FS,2)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "factor of safety is 2.86\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.4:pg-529"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.4\n",
-      "\n",
-      "\n",
-      "L1=5; # in meter\n",
-      "L2=4;# in meter\n",
-      "B=5;# in meter\n",
-      "h=4.5;# in meter\n",
-      "icr=1 #\n",
-      "iexit= 0.54*h/L2\n",
-      "FOS=icr/iexit # Factor of safety\n",
-      "print \"The Factory of safety is\",round(FOS,3)\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Factory of safety is 1.646\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter10_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter10_1.ipynb
deleted file mode 100755
index 3af11efd..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter10_1.ipynb
+++ /dev/null
@@ -1,211 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:54ddb8c76d55d78e9f237ddb7f5823ac10a3585adf1b3932a3b8cc9352fd8e76"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10:Braced Cuts"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.1: pg-511"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.1 : \n",
-      "\n",
-      "Gamma=18.0;\n",
-      "H=7.0;\n",
-      "sigmaa=0.3*Gamma*H;\n",
-      "print\"maximum pressure intensity in kN/m^2 is\",sigmaa ;\n",
-      "#partb\n",
-      "A=54.02;\n",
-      "B1=1.0/2*1.75*37.8+37.8*1.75-A;\n",
-      "B2=45.2;\n",
-      "C=54.02;\n",
-      "s=3.0; #spacing\n",
-      "Pa=C*s;\n",
-      "print \"strut loads in kN is\",Pa\n",
-      "Pb=(B1+B2)*s;\n",
-      "print \"strut loads in kN is\",Pb\n",
-      "Pc=C*s;\n",
-      "print \"strut loads in kN is\", Pc\n",
-      "#partc\n",
-      "Me=45.2*1.196-37.8*1.196*1.196/2;#Me=Mmax\n",
-      "Sall=170e3;#sigmaall\n",
-      "S=Me/Sall;\n",
-      "print \"section modulus in m^3/m is \",round(S,7)\n",
-      "#partd\n",
-      "Mmax=(B1+B2)*s**2.0/8;\n",
-      "S=Mmax/Sall;\n",
-      "print\"section modulus in m^3/m is\",round(S,6)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum pressure intensity in kN/m^2 is 37.8\n",
-        "strut loads in kN is 162.06\n",
-        "strut loads in kN is 271.215\n",
-        "strut loads in kN is 162.06\n",
-        "section modulus in m^3/m is  0.000159\n",
-        "section modulus in m^3/m is 0.000598\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.2:pg-514"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.2\n",
-      "\n",
-      "import math\n",
-      "phi=35.0;\n",
-      "Gamma=17.0; # kN/m^3\n",
-      "s=4; #spacing in m\n",
-      "H=9;\n",
-      "Ka=(math.tan(45*(math.pi/180)-35*(math.pi/(180.0*2))))**2\n",
-      "sigma=0.65*Gamma*Ka*H\n",
-      "A=sigma*5*5.0/6;\n",
-      "B1=sigma*5-A;\n",
-      "C=sigma*4*4/6.0; \n",
-      "B2=sigma*s-C;\n",
-      "Pa=A*s;\n",
-      "Pb=(B1+B2)*s;\n",
-      "Pc=C*s;\n",
-      "print \"strut loads at A in kN is\",round(Pa,2)\n",
-      "\n",
-      "print \"strut loads at B in kN is\",round(Pb,2)\n",
-      "\n",
-      "print \"strut loads at C in kN is\",round(Pc,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "strut loads at A in kN is 449.17\n",
-        "strut loads at B in kN is 233.57\n",
-        "strut loads at C in kN is 287.47\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.3:pg-523"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.3\n",
-      "\n",
-      "import math\n",
-      "q=0;\n",
-      "Gamma=17; # in KN/m^3\n",
-      "B=3.0 # in meter\n",
-      "L=20; # in meter\n",
-      "c=30;# in KN/m^3\n",
-      "T=1.5;# in meter\n",
-      "H=5.5;# in meter\n",
-      "B1=B/2;#B'\n",
-      "B11=T*math.sqrt(2);#B''\n",
-      "FS=(5.14*c*(1+B11*0.2/L)+c*H/B1)/(Gamma*H+q) # from equation of factor of safety\n",
-      "print\"factor of safety is\",round(FS,2)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "factor of safety is 2.86\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.4:pg-529"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.4\n",
-      "\n",
-      "\n",
-      "L1=5; # in meter\n",
-      "L2=4;# in meter\n",
-      "B=5;# in meter\n",
-      "h=4.5;# in meter\n",
-      "icr=1 #\n",
-      "iexit= 0.54*h/L2\n",
-      "FOS=icr/iexit # Factor of safety\n",
-      "print \"The Factory of safety is\",round(FOS,3)\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Factory of safety is 1.646\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter10_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter10_2.ipynb
deleted file mode 100755
index 3af11efd..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter10_2.ipynb
+++ /dev/null
@@ -1,211 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:54ddb8c76d55d78e9f237ddb7f5823ac10a3585adf1b3932a3b8cc9352fd8e76"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10:Braced Cuts"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.1: pg-511"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.1 : \n",
-      "\n",
-      "Gamma=18.0;\n",
-      "H=7.0;\n",
-      "sigmaa=0.3*Gamma*H;\n",
-      "print\"maximum pressure intensity in kN/m^2 is\",sigmaa ;\n",
-      "#partb\n",
-      "A=54.02;\n",
-      "B1=1.0/2*1.75*37.8+37.8*1.75-A;\n",
-      "B2=45.2;\n",
-      "C=54.02;\n",
-      "s=3.0; #spacing\n",
-      "Pa=C*s;\n",
-      "print \"strut loads in kN is\",Pa\n",
-      "Pb=(B1+B2)*s;\n",
-      "print \"strut loads in kN is\",Pb\n",
-      "Pc=C*s;\n",
-      "print \"strut loads in kN is\", Pc\n",
-      "#partc\n",
-      "Me=45.2*1.196-37.8*1.196*1.196/2;#Me=Mmax\n",
-      "Sall=170e3;#sigmaall\n",
-      "S=Me/Sall;\n",
-      "print \"section modulus in m^3/m is \",round(S,7)\n",
-      "#partd\n",
-      "Mmax=(B1+B2)*s**2.0/8;\n",
-      "S=Mmax/Sall;\n",
-      "print\"section modulus in m^3/m is\",round(S,6)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum pressure intensity in kN/m^2 is 37.8\n",
-        "strut loads in kN is 162.06\n",
-        "strut loads in kN is 271.215\n",
-        "strut loads in kN is 162.06\n",
-        "section modulus in m^3/m is  0.000159\n",
-        "section modulus in m^3/m is 0.000598\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.2:pg-514"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.2\n",
-      "\n",
-      "import math\n",
-      "phi=35.0;\n",
-      "Gamma=17.0; # kN/m^3\n",
-      "s=4; #spacing in m\n",
-      "H=9;\n",
-      "Ka=(math.tan(45*(math.pi/180)-35*(math.pi/(180.0*2))))**2\n",
-      "sigma=0.65*Gamma*Ka*H\n",
-      "A=sigma*5*5.0/6;\n",
-      "B1=sigma*5-A;\n",
-      "C=sigma*4*4/6.0; \n",
-      "B2=sigma*s-C;\n",
-      "Pa=A*s;\n",
-      "Pb=(B1+B2)*s;\n",
-      "Pc=C*s;\n",
-      "print \"strut loads at A in kN is\",round(Pa,2)\n",
-      "\n",
-      "print \"strut loads at B in kN is\",round(Pb,2)\n",
-      "\n",
-      "print \"strut loads at C in kN is\",round(Pc,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "strut loads at A in kN is 449.17\n",
-        "strut loads at B in kN is 233.57\n",
-        "strut loads at C in kN is 287.47\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.3:pg-523"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.3\n",
-      "\n",
-      "import math\n",
-      "q=0;\n",
-      "Gamma=17; # in KN/m^3\n",
-      "B=3.0 # in meter\n",
-      "L=20; # in meter\n",
-      "c=30;# in KN/m^3\n",
-      "T=1.5;# in meter\n",
-      "H=5.5;# in meter\n",
-      "B1=B/2;#B'\n",
-      "B11=T*math.sqrt(2);#B''\n",
-      "FS=(5.14*c*(1+B11*0.2/L)+c*H/B1)/(Gamma*H+q) # from equation of factor of safety\n",
-      "print\"factor of safety is\",round(FS,2)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "factor of safety is 2.86\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex10.4:pg-529"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 10.4\n",
-      "\n",
-      "\n",
-      "L1=5; # in meter\n",
-      "L2=4;# in meter\n",
-      "B=5;# in meter\n",
-      "h=4.5;# in meter\n",
-      "icr=1 #\n",
-      "iexit= 0.54*h/L2\n",
-      "FOS=icr/iexit # Factor of safety\n",
-      "print \"The Factory of safety is\",round(FOS,3)\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Factory of safety is 1.646\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter11.ipynb b/Principles_Of_Foundation_Engineering/Chapter11.ipynb
deleted file mode 100755
index 0afdb4a0..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter11.ipynb
+++ /dev/null
@@ -1,650 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a4054b3d6b79658bcd763c43dc94ca6de4cd49698c12f912e4c83a9a61f47154"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter11:Pile Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.1:Pg-532"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.1\n",
-      "import math\n",
-      "\n",
-      "#parta\n",
-      "phi=30.0; # angle in degree\n",
-      "pa=2000.0; \n",
-      "q=100*50/1000.0;\n",
-      "Nq=55.0;\n",
-      "Ap=16*16/16/12; # area in ft^2\n",
-      "Qp=Ap*q*Nq; # in kip\n",
-      "qp=0.4*pa*Nq*math.tan(phi*math.pi/180)*Ap; # in lb\n",
-      "print round(Qp,2),\"ultimate load in lb\"\n",
-      "print round(qp/1000,2),\"ultimate load in kip\"\n",
-      "print \"there is change in answer because of calculation mistake in the book\"\n",
-      "\n",
-      "#partb\n",
-      "Nsigma=36;\n",
-      "Ap=16*16.0/12.0/12;\n",
-      "q=110*50.0/1000;\n",
-      "Qp=Ap*q*Nsigma*((1+2.0*(1-math.sin(phi*math.pi/180)))/3); # in kip\n",
-      "print round(Qp,2),\"ultimate load in kip\"\n",
-      "#partc\n",
-      "Nq=18.4;\n",
-      "Qp=Ap*q*Nq; # in kip\n",
-      "print round(Qp,2),\"ultimate load in kip\"\n",
-      "\n",
-      "# ANSWER IN THE BOOK IS WRONG"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "275.0 ultimate load in lb\n",
-        "25.4 ultimate load in kip\n",
-        "there is change in answer because of calculation mistake in the book\n",
-        "234.67 ultimate load in kip\n",
-        "179.91 ultimate load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.2:Pg-533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.2\n",
-      "\n",
-      "import math\n",
-      "#parta\n",
-      "K=1.3;\n",
-      "f0=0;\n",
-      "Delta=0.8*30; # in ft\n",
-      "D=16.0/12; # in ft\n",
-      "L1=50.0;\n",
-      "p=4*16/12.0; # in ft\n",
-      "Gamma=110/1000.0; # in lb/ft^3\n",
-      "L=15*D; # in ft\n",
-      "sigma=Gamma*L; # in kip/ft^2\n",
-      "f20=K*sigma*math.tan(Delta*math.pi/180); # kip/ft^2\n",
-      "Qs=(f0+f20)/2*(p*L)+f20*p*(L1-L);\n",
-      "print round(Qs,2),\"ultimate load in kip\"\n",
-      "#partb\n",
-      "FS=4; # factor of safety\n",
-      "Qp=56.45/3+234.7/3+179.9/3; # in kip\n",
-      "Qu=Qs+Qp; # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall,2),\"is allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "271.65 ultimate load in kip\n",
-        "107.17 is allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.3:Pg-534"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.3\n",
-      "import math\n",
-      "K=0.25;\n",
-      "Ap=16*16.0/12/12; # area in ft^2\n",
-      "phi=30*math.pi/180;\n",
-      "Nq=25;\n",
-      "q=110*50.0/1000; # in kip\n",
-      "sigmao=q/2; # in kip/ft^2\n",
-      "p=4*16.0/12; # in ft\n",
-      "L=50; # in ft\n",
-      "FS=4; # factor of safety\n",
-      "Qu=q*Nq*Ap+K*sigmao*math.tan(0.8*phi)*p*L; # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall,1),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "81.5 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.4:Pg-535"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.4\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "FS=4; # factor of safety\n",
-      "Ap=0.1295; # area in m^2\n",
-      "Nc=9;\n",
-      "cu2=100;\n",
-      "Qp=Ap*Nc*cu2; # in kN\n",
-      "D=[5, 10, 30]; # depth in m\n",
-      "avgD=[2.5, 7.5,20.0]; # average depth in m\n",
-      "sigma=[45.0, 110.5, 228.5]; # in KN/m^2\n",
-      "cu=[30, 30, 100]; # in kN/m^2\n",
-      "alpha=[0.6, 0.9, 0.725];\n",
-      "L=[5, 5, 20]; # in m\n",
-      "p=math.pi*0.406;\n",
-      "Qs=0; # in kN\n",
-      "cusig=numpy.zeros(3)\n",
-      "print round(Qp,2),\"bearing capacity in kN\"\n",
-      "print  \"depth (m)\\t avg Depth(m)\\t avgVerticalStress(kN/m**2)\\t cu(kN/m**2)\\t cu/sigma\\t alpha\\n\"\n",
-      "for i in range(0,3):\n",
-      "    cusig[i]=cu[i]/sigma[i];\n",
-      "    Qs=Qs+alpha[i]*cu[i]*L[i]*p;\n",
-      "    print round(D[i],2),\"\\t \\t \\t\",round(avgD[i],2),\"\\t \\t\",round(sigma[i],2),\"\\t\\t\\t   \",round(cu[i],2),\"\\t  \",round(cusig[i],2),\"\\t\\t \",round(alpha[i],2),\"\\n\"\n",
-      "print round(Qs,2),\"bearing capacity in kN\"\n",
-      "#part2\n",
-      "Lambda=0.136;\n",
-      "L=30;\n",
-      "fav=Lambda*(178.48+2*76.7);\n",
-      "Qs2=p*L*fav; # in kN\n",
-      "#part3\n",
-      "fav1=13;\n",
-      "fav2=31.9;\n",
-      "fav3=93.43;\n",
-      "Qs3=p*(fav1*5+fav2*5+fav3*20); # in kN\n",
-      "print round(Qs3,1),\"bearing capacity in kN\"\n",
-      "Qsavg=Qs/3+Qs2/3+Qs3/3; # in kN\n",
-      "Qu=Qp+Qsavg # in kN\n",
-      "Qall=Qu/FS; # in kN\n",
-      "print round(Qall,1),\"allowed bearing capacity in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "116.55 bearing capacity in kN\n",
-        "depth (m)\t avg Depth(m)\t avgVerticalStress(kN/m**2)\t cu(kN/m**2)\t cu/sigma\t alpha\n",
-        "\n",
-        "5.0 \t \t \t2.5 \t \t45.0 \t\t\t    30.0 \t   0.67 \t\t  0.6 \n",
-        "\n",
-        "10.0 \t \t \t7.5 \t \t110.5 \t\t\t    30.0 \t   0.27 \t\t  0.9 \n",
-        "\n",
-        "30.0 \t \t \t20.0 \t \t228.5 \t\t\t    100.0 \t   0.44 \t\t  0.72 \n",
-        "\n",
-        "2136.44 bearing capacity in kN\n",
-        "2669.7 bearing capacity in kN\n",
-        "573.6 allowed bearing capacity in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.5:Pg-538"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.5\n",
-      "\n",
-      "import numpy\n",
-      "D=[6, 12, 20]; # depth in m\n",
-      "fc=[34.34, 54.94, 70.63]; # in kN/m**2\n",
-      "alpha=[0.84, 0.71, 0.63];\n",
-      "dL=[6, 6, 8]; # in m\n",
-      "p=4*0.305;\n",
-      "Qs=0;\n",
-      "Q=numpy.zeros(3)\n",
-      "print \"  depth(m)\\t fc(kN/m**2)\\t alpha \\t \\t deltaL(m)\\t Q(kN)\\n\"\n",
-      "for i in range (0,3):\n",
-      "    Q[i]=alpha[i]*fc[i]*p*dL[i];\n",
-      "    Qs=Q[i]+Qs;\n",
-      "    print D[i],\"\\t\\t   \",fc[i],\"\\t  \",alpha[i],\"\\t   \",dL[i],\"\\t\\t \",round(Q[i],2)\n",
-      "\n",
-      "print round(Qs),\"bearing force in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  depth(m)\t fc(kN/m**2)\t alpha \t \t deltaL(m)\t Q(kN)\n",
-        "\n",
-        "6 \t\t    34.34 \t   0.84 \t    6 \t\t  211.15\n",
-        "12 \t\t    54.94 \t   0.71 \t    6 \t\t  285.53\n",
-        "20 \t\t    70.63 \t   0.63 \t    8 \t\t  434.29\n",
-        "931.0 bearing force in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:Pg-545"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.6\n",
-      "\n",
-      "import math\n",
-      "L=21; # in m\n",
-      "Qwp=502-350; # in kN\n",
-      "Qws=350; # in kN\n",
-      "Ap=0.1045; # area in m^2\n",
-      "Ep=21e6; # in kN/m^2\n",
-      "epsilon=0.62;\n",
-      "Se1=(Qwp+epsilon*Qws)*L/Ap/Ep; # in m\n",
-      "#part2\n",
-      "Iwp=0.85;\n",
-      "qwp=152/Ap;\n",
-      "Es=25e3; # in kN/m^2\n",
-      "D=0.356; # in m\n",
-      "mus=0.35;\n",
-      "Se2=qwp*D/Es*Iwp*(1-mus**2); # in m\n",
-      "#part3\n",
-      "p=1.168;\n",
-      "Iws=2+0.35*math.sqrt(L/D);\n",
-      "Se3=Qws/p/L*D/Es*Iws*(1-mus**2); # in m\n",
-      "Se=Se1+Se2+Se3; # in m\n",
-      "print round(Se*1000,1),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "19.8 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.7:Pg-560"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.7\n",
-      "\n",
-      "Ep=207e6; # in kN/m^2\n",
-      "Ip=123e-6; # in m^4\n",
-      "nh=12000; # in  kN/m^3\n",
-      "#from table 11.13\n",
-      "xz=0.008;\n",
-      "Ax=2.435;\n",
-      "T=(Ep*Ip/nh)**0.2;\n",
-      "Qg1=xz*Ep*Ip/Ax/T**3;\n",
-      "#part2\n",
-      "Fy=248000;\n",
-      "d1=0.254;\n",
-      "Am=0.772;\n",
-      "Mzmax=Fy*Ip*2/d1; # in Kn-m\n",
-      "Qg2=Mzmax/Am/T; # in kN\n",
-      "if Qg2>Qg1 :\n",
-      "    Qg=Qg1;\n",
-      "    print round(Qg,2),\"lateral load in kN\"\n",
-      "# there is slight variation in answer in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "53.27 lateral load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.8:Pg-561"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.8\n",
-      "\n",
-      "import math\n",
-      "#part1\n",
-      "Ep=207e6; # in kN/m^2\n",
-      "Ip=123e-6; # in m^4\n",
-      "nh=12000; # in kN/m^3\n",
-      "#from table 11.1a\n",
-      "xo=0.008; # in m\n",
-      "L=25;\n",
-      "Fy=248000; # yield stress in kN/m^2\n",
-      "D=0.254;\n",
-      "Am=0.772;\n",
-      "Gamma=18.0; # in kN/m^3\n",
-      "phi=35; # in angle\n",
-      "Kp=(math.tan(math.pi/4+phi*math.pi/360))**2;\n",
-      "My=Fy*Ip*2/D; # in kN-m\n",
-      "Qug=140*Kp*D**3*Gamma; # in kN\n",
-      "\n",
-      "#part2\n",
-      "Qg1=xo*(Ep*Ip)**0.6*nh**0.4/0.15/L; # in kN\n",
-      "\n",
-      "if Qug>Qg1:\n",
-      "    Qg=Qg1;\n",
-      "    print round(Qg,2),\"lateral load in kN\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "40.2 lateral load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.9:Pg-567"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.9\n",
-      "\n",
-      "import math\n",
-      "Wrh=30*12; # in kip-ft\n",
-      "E=0.8;\n",
-      "Wr=7.5; # in kip\n",
-      "S=1/8.0; \n",
-      "C=0.1;\n",
-      "FS=6; # in factor of safety\n",
-      "n=0.4; # Coefficient of restitution\n",
-      "Wp=12/12.0*12/12.0*80*150+550; # in lb\n",
-      "Wp=Wp/1000.0;\n",
-      "Qu=E*Wrh/(S+C)*(Wr+n**2.0*Wp)/(Wr+Wp); # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed bearing capacity in kip\"\n",
-      "#part2\n",
-      "He=30*12.0;\n",
-      "L=80*12.0;\n",
-      "Ap=12*12.0; # area in in^2\n",
-      "Ep=3e6/1000.0; # in kip/in^2\n",
-      "FS=4; # factor of safety\n",
-      "Qu=E*He/(S+math.sqrt(E*He*L/2.0/Ap/Ep)); # in kip\n",
-      "Qall2=Qu/FS; # in kip\n",
-      "print round(Qall2),\"allowed bearing capacity in kip\"\n",
-      "\n",
-      "#partc\n",
-      "a=27;\n",
-      "b=1;\n",
-      "He=30;\n",
-      "FS=3; # factor of safety\n",
-      "Qu=a*math.sqrt(E*He)*(b-math.log10(S)); # in kip\n",
-      "Qall3=Qu/FS; # in kip\n",
-      "print round(Qall3),\"allowed bearing capacity in kip\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "101.0 allowed bearing capacity in kip\n",
-        "104.0 allowed bearing capacity in kip\n",
-        "84.0 allowed bearing capacity in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.10:Pg-570"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.10\n",
-      "\n",
-      "Hp=350; # in HP\n",
-      "vp=0.0016; #  in m/s\n",
-      "Sl=0.762e-3; # in m/cycle\n",
-      "f=115; # in Hz\n",
-      "Qu=(0.746*Hp+98*vp)/(vp+Sl*f); # in kN\n",
-      "print round(Qu),\"pile load capacity in kN\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2928.0 pile load capacity in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.11:Pg-578"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.11\n",
-      "\n",
-      "Lg=9.92; # in ft\n",
-      "Bg=7.0; # in ft\n",
-      "n1=3.0;\n",
-      "Nc=8.75;\n",
-      "n2=4.0/1000;\n",
-      "Ap=14.0**2.0/12.0**2;\n",
-      "cup=1775.0;\n",
-      "a1=0.4;#alpha1\n",
-      "p=4*14.0/12.0;\n",
-      "cu1=1050.0; # in lb/ft^2\n",
-      "L1=15.0;\n",
-      "a2=0.54;#alpha2\n",
-      "cu2=1775.0; # in lb/ft^2\n",
-      "L2=45.0;\n",
-      "FS=4; # factor of safety\n",
-      "Qu=n1*n2*(9*Ap*cup+a1*p*cu1*L1+a2*p*cu2*L2); # in kip\n",
-      "Qu2=Lg*Bg*cup*Nc+2*(Lg+Bg)*(cu1*L1+cu2*L2); # in kip\n",
-      "print round(Qu2/1000),\"load in kip\"\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed load in kip\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4314.0 load in kip\n",
-        "757.0 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.12:Pg-583"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.12\n",
-      "\n",
-      "import math\n",
-      "z1=21/2.0; # in ft\n",
-      "Lg=9.0; # in ft\n",
-      "Bg=6.0;# in ft\n",
-      "Qg=500*1000.0; # in kip\n",
-      "Cc1=0.3;\n",
-      "Cc2=0.2;\n",
-      "Cc3=0.25;\n",
-      "H2=12;\n",
-      "H3=6;\n",
-      "H1=21;\n",
-      "e1=0.82;\n",
-      "e2=0.7;\n",
-      "e3=0.75;\n",
-      "s1=Qg/(Lg+z1)/(Bg+z1); #sigma1 in lb/ft^3\n",
-      "s2=500*1000/(9+27)/(6+27);#sigma2 in lb/ft^3\n",
-      "s3=500*1000/(9+36)/(6+36);#sigma3 in lb/ft^3\n",
-      "ss1=6*105+(27+21/2)*(115-62.4);#sigmadash1 in lb/ft^3\n",
-      "ss2=6*105+(27+21)*(115-62.4)+(120-62.4)*6;#sigmadash2 in lb/ft^3\n",
-      "ss3=6*105+48*(115-62.4)+12*(120-62.4)+3*(122-62.4);#sigmadash3 in lb/ft^3\n",
-      "sc1=Cc1*H1/(1+e1)*math.log10((ss1+s1)/ss1); # in inch\n",
-      "sc2=Cc2*H2/(1+e2)*math.log10((ss2+s2)/ss2); # in inch\n",
-      "sc3=Cc3*H3/(1+e3)*math.log10((ss3+s3)/ss3); # in inch\n",
-      "sc=sc1+sc2+sc3; # in inch\n",
-      "print round(sc*12,1),\"total settlement in inch\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9.6 total settlement in inch\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter11_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter11_1.ipynb
deleted file mode 100755
index 0afdb4a0..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter11_1.ipynb
+++ /dev/null
@@ -1,650 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a4054b3d6b79658bcd763c43dc94ca6de4cd49698c12f912e4c83a9a61f47154"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter11:Pile Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.1:Pg-532"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.1\n",
-      "import math\n",
-      "\n",
-      "#parta\n",
-      "phi=30.0; # angle in degree\n",
-      "pa=2000.0; \n",
-      "q=100*50/1000.0;\n",
-      "Nq=55.0;\n",
-      "Ap=16*16/16/12; # area in ft^2\n",
-      "Qp=Ap*q*Nq; # in kip\n",
-      "qp=0.4*pa*Nq*math.tan(phi*math.pi/180)*Ap; # in lb\n",
-      "print round(Qp,2),\"ultimate load in lb\"\n",
-      "print round(qp/1000,2),\"ultimate load in kip\"\n",
-      "print \"there is change in answer because of calculation mistake in the book\"\n",
-      "\n",
-      "#partb\n",
-      "Nsigma=36;\n",
-      "Ap=16*16.0/12.0/12;\n",
-      "q=110*50.0/1000;\n",
-      "Qp=Ap*q*Nsigma*((1+2.0*(1-math.sin(phi*math.pi/180)))/3); # in kip\n",
-      "print round(Qp,2),\"ultimate load in kip\"\n",
-      "#partc\n",
-      "Nq=18.4;\n",
-      "Qp=Ap*q*Nq; # in kip\n",
-      "print round(Qp,2),\"ultimate load in kip\"\n",
-      "\n",
-      "# ANSWER IN THE BOOK IS WRONG"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "275.0 ultimate load in lb\n",
-        "25.4 ultimate load in kip\n",
-        "there is change in answer because of calculation mistake in the book\n",
-        "234.67 ultimate load in kip\n",
-        "179.91 ultimate load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.2:Pg-533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.2\n",
-      "\n",
-      "import math\n",
-      "#parta\n",
-      "K=1.3;\n",
-      "f0=0;\n",
-      "Delta=0.8*30; # in ft\n",
-      "D=16.0/12; # in ft\n",
-      "L1=50.0;\n",
-      "p=4*16/12.0; # in ft\n",
-      "Gamma=110/1000.0; # in lb/ft^3\n",
-      "L=15*D; # in ft\n",
-      "sigma=Gamma*L; # in kip/ft^2\n",
-      "f20=K*sigma*math.tan(Delta*math.pi/180); # kip/ft^2\n",
-      "Qs=(f0+f20)/2*(p*L)+f20*p*(L1-L);\n",
-      "print round(Qs,2),\"ultimate load in kip\"\n",
-      "#partb\n",
-      "FS=4; # factor of safety\n",
-      "Qp=56.45/3+234.7/3+179.9/3; # in kip\n",
-      "Qu=Qs+Qp; # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall,2),\"is allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "271.65 ultimate load in kip\n",
-        "107.17 is allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.3:Pg-534"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.3\n",
-      "import math\n",
-      "K=0.25;\n",
-      "Ap=16*16.0/12/12; # area in ft^2\n",
-      "phi=30*math.pi/180;\n",
-      "Nq=25;\n",
-      "q=110*50.0/1000; # in kip\n",
-      "sigmao=q/2; # in kip/ft^2\n",
-      "p=4*16.0/12; # in ft\n",
-      "L=50; # in ft\n",
-      "FS=4; # factor of safety\n",
-      "Qu=q*Nq*Ap+K*sigmao*math.tan(0.8*phi)*p*L; # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall,1),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "81.5 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.4:Pg-535"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.4\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "FS=4; # factor of safety\n",
-      "Ap=0.1295; # area in m^2\n",
-      "Nc=9;\n",
-      "cu2=100;\n",
-      "Qp=Ap*Nc*cu2; # in kN\n",
-      "D=[5, 10, 30]; # depth in m\n",
-      "avgD=[2.5, 7.5,20.0]; # average depth in m\n",
-      "sigma=[45.0, 110.5, 228.5]; # in KN/m^2\n",
-      "cu=[30, 30, 100]; # in kN/m^2\n",
-      "alpha=[0.6, 0.9, 0.725];\n",
-      "L=[5, 5, 20]; # in m\n",
-      "p=math.pi*0.406;\n",
-      "Qs=0; # in kN\n",
-      "cusig=numpy.zeros(3)\n",
-      "print round(Qp,2),\"bearing capacity in kN\"\n",
-      "print  \"depth (m)\\t avg Depth(m)\\t avgVerticalStress(kN/m**2)\\t cu(kN/m**2)\\t cu/sigma\\t alpha\\n\"\n",
-      "for i in range(0,3):\n",
-      "    cusig[i]=cu[i]/sigma[i];\n",
-      "    Qs=Qs+alpha[i]*cu[i]*L[i]*p;\n",
-      "    print round(D[i],2),\"\\t \\t \\t\",round(avgD[i],2),\"\\t \\t\",round(sigma[i],2),\"\\t\\t\\t   \",round(cu[i],2),\"\\t  \",round(cusig[i],2),\"\\t\\t \",round(alpha[i],2),\"\\n\"\n",
-      "print round(Qs,2),\"bearing capacity in kN\"\n",
-      "#part2\n",
-      "Lambda=0.136;\n",
-      "L=30;\n",
-      "fav=Lambda*(178.48+2*76.7);\n",
-      "Qs2=p*L*fav; # in kN\n",
-      "#part3\n",
-      "fav1=13;\n",
-      "fav2=31.9;\n",
-      "fav3=93.43;\n",
-      "Qs3=p*(fav1*5+fav2*5+fav3*20); # in kN\n",
-      "print round(Qs3,1),\"bearing capacity in kN\"\n",
-      "Qsavg=Qs/3+Qs2/3+Qs3/3; # in kN\n",
-      "Qu=Qp+Qsavg # in kN\n",
-      "Qall=Qu/FS; # in kN\n",
-      "print round(Qall,1),\"allowed bearing capacity in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "116.55 bearing capacity in kN\n",
-        "depth (m)\t avg Depth(m)\t avgVerticalStress(kN/m**2)\t cu(kN/m**2)\t cu/sigma\t alpha\n",
-        "\n",
-        "5.0 \t \t \t2.5 \t \t45.0 \t\t\t    30.0 \t   0.67 \t\t  0.6 \n",
-        "\n",
-        "10.0 \t \t \t7.5 \t \t110.5 \t\t\t    30.0 \t   0.27 \t\t  0.9 \n",
-        "\n",
-        "30.0 \t \t \t20.0 \t \t228.5 \t\t\t    100.0 \t   0.44 \t\t  0.72 \n",
-        "\n",
-        "2136.44 bearing capacity in kN\n",
-        "2669.7 bearing capacity in kN\n",
-        "573.6 allowed bearing capacity in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.5:Pg-538"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.5\n",
-      "\n",
-      "import numpy\n",
-      "D=[6, 12, 20]; # depth in m\n",
-      "fc=[34.34, 54.94, 70.63]; # in kN/m**2\n",
-      "alpha=[0.84, 0.71, 0.63];\n",
-      "dL=[6, 6, 8]; # in m\n",
-      "p=4*0.305;\n",
-      "Qs=0;\n",
-      "Q=numpy.zeros(3)\n",
-      "print \"  depth(m)\\t fc(kN/m**2)\\t alpha \\t \\t deltaL(m)\\t Q(kN)\\n\"\n",
-      "for i in range (0,3):\n",
-      "    Q[i]=alpha[i]*fc[i]*p*dL[i];\n",
-      "    Qs=Q[i]+Qs;\n",
-      "    print D[i],\"\\t\\t   \",fc[i],\"\\t  \",alpha[i],\"\\t   \",dL[i],\"\\t\\t \",round(Q[i],2)\n",
-      "\n",
-      "print round(Qs),\"bearing force in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  depth(m)\t fc(kN/m**2)\t alpha \t \t deltaL(m)\t Q(kN)\n",
-        "\n",
-        "6 \t\t    34.34 \t   0.84 \t    6 \t\t  211.15\n",
-        "12 \t\t    54.94 \t   0.71 \t    6 \t\t  285.53\n",
-        "20 \t\t    70.63 \t   0.63 \t    8 \t\t  434.29\n",
-        "931.0 bearing force in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:Pg-545"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.6\n",
-      "\n",
-      "import math\n",
-      "L=21; # in m\n",
-      "Qwp=502-350; # in kN\n",
-      "Qws=350; # in kN\n",
-      "Ap=0.1045; # area in m^2\n",
-      "Ep=21e6; # in kN/m^2\n",
-      "epsilon=0.62;\n",
-      "Se1=(Qwp+epsilon*Qws)*L/Ap/Ep; # in m\n",
-      "#part2\n",
-      "Iwp=0.85;\n",
-      "qwp=152/Ap;\n",
-      "Es=25e3; # in kN/m^2\n",
-      "D=0.356; # in m\n",
-      "mus=0.35;\n",
-      "Se2=qwp*D/Es*Iwp*(1-mus**2); # in m\n",
-      "#part3\n",
-      "p=1.168;\n",
-      "Iws=2+0.35*math.sqrt(L/D);\n",
-      "Se3=Qws/p/L*D/Es*Iws*(1-mus**2); # in m\n",
-      "Se=Se1+Se2+Se3; # in m\n",
-      "print round(Se*1000,1),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "19.8 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.7:Pg-560"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.7\n",
-      "\n",
-      "Ep=207e6; # in kN/m^2\n",
-      "Ip=123e-6; # in m^4\n",
-      "nh=12000; # in  kN/m^3\n",
-      "#from table 11.13\n",
-      "xz=0.008;\n",
-      "Ax=2.435;\n",
-      "T=(Ep*Ip/nh)**0.2;\n",
-      "Qg1=xz*Ep*Ip/Ax/T**3;\n",
-      "#part2\n",
-      "Fy=248000;\n",
-      "d1=0.254;\n",
-      "Am=0.772;\n",
-      "Mzmax=Fy*Ip*2/d1; # in Kn-m\n",
-      "Qg2=Mzmax/Am/T; # in kN\n",
-      "if Qg2>Qg1 :\n",
-      "    Qg=Qg1;\n",
-      "    print round(Qg,2),\"lateral load in kN\"\n",
-      "# there is slight variation in answer in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "53.27 lateral load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.8:Pg-561"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.8\n",
-      "\n",
-      "import math\n",
-      "#part1\n",
-      "Ep=207e6; # in kN/m^2\n",
-      "Ip=123e-6; # in m^4\n",
-      "nh=12000; # in kN/m^3\n",
-      "#from table 11.1a\n",
-      "xo=0.008; # in m\n",
-      "L=25;\n",
-      "Fy=248000; # yield stress in kN/m^2\n",
-      "D=0.254;\n",
-      "Am=0.772;\n",
-      "Gamma=18.0; # in kN/m^3\n",
-      "phi=35; # in angle\n",
-      "Kp=(math.tan(math.pi/4+phi*math.pi/360))**2;\n",
-      "My=Fy*Ip*2/D; # in kN-m\n",
-      "Qug=140*Kp*D**3*Gamma; # in kN\n",
-      "\n",
-      "#part2\n",
-      "Qg1=xo*(Ep*Ip)**0.6*nh**0.4/0.15/L; # in kN\n",
-      "\n",
-      "if Qug>Qg1:\n",
-      "    Qg=Qg1;\n",
-      "    print round(Qg,2),\"lateral load in kN\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "40.2 lateral load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.9:Pg-567"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.9\n",
-      "\n",
-      "import math\n",
-      "Wrh=30*12; # in kip-ft\n",
-      "E=0.8;\n",
-      "Wr=7.5; # in kip\n",
-      "S=1/8.0; \n",
-      "C=0.1;\n",
-      "FS=6; # in factor of safety\n",
-      "n=0.4; # Coefficient of restitution\n",
-      "Wp=12/12.0*12/12.0*80*150+550; # in lb\n",
-      "Wp=Wp/1000.0;\n",
-      "Qu=E*Wrh/(S+C)*(Wr+n**2.0*Wp)/(Wr+Wp); # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed bearing capacity in kip\"\n",
-      "#part2\n",
-      "He=30*12.0;\n",
-      "L=80*12.0;\n",
-      "Ap=12*12.0; # area in in^2\n",
-      "Ep=3e6/1000.0; # in kip/in^2\n",
-      "FS=4; # factor of safety\n",
-      "Qu=E*He/(S+math.sqrt(E*He*L/2.0/Ap/Ep)); # in kip\n",
-      "Qall2=Qu/FS; # in kip\n",
-      "print round(Qall2),\"allowed bearing capacity in kip\"\n",
-      "\n",
-      "#partc\n",
-      "a=27;\n",
-      "b=1;\n",
-      "He=30;\n",
-      "FS=3; # factor of safety\n",
-      "Qu=a*math.sqrt(E*He)*(b-math.log10(S)); # in kip\n",
-      "Qall3=Qu/FS; # in kip\n",
-      "print round(Qall3),\"allowed bearing capacity in kip\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "101.0 allowed bearing capacity in kip\n",
-        "104.0 allowed bearing capacity in kip\n",
-        "84.0 allowed bearing capacity in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.10:Pg-570"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.10\n",
-      "\n",
-      "Hp=350; # in HP\n",
-      "vp=0.0016; #  in m/s\n",
-      "Sl=0.762e-3; # in m/cycle\n",
-      "f=115; # in Hz\n",
-      "Qu=(0.746*Hp+98*vp)/(vp+Sl*f); # in kN\n",
-      "print round(Qu),\"pile load capacity in kN\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2928.0 pile load capacity in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.11:Pg-578"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.11\n",
-      "\n",
-      "Lg=9.92; # in ft\n",
-      "Bg=7.0; # in ft\n",
-      "n1=3.0;\n",
-      "Nc=8.75;\n",
-      "n2=4.0/1000;\n",
-      "Ap=14.0**2.0/12.0**2;\n",
-      "cup=1775.0;\n",
-      "a1=0.4;#alpha1\n",
-      "p=4*14.0/12.0;\n",
-      "cu1=1050.0; # in lb/ft^2\n",
-      "L1=15.0;\n",
-      "a2=0.54;#alpha2\n",
-      "cu2=1775.0; # in lb/ft^2\n",
-      "L2=45.0;\n",
-      "FS=4; # factor of safety\n",
-      "Qu=n1*n2*(9*Ap*cup+a1*p*cu1*L1+a2*p*cu2*L2); # in kip\n",
-      "Qu2=Lg*Bg*cup*Nc+2*(Lg+Bg)*(cu1*L1+cu2*L2); # in kip\n",
-      "print round(Qu2/1000),\"load in kip\"\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed load in kip\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4314.0 load in kip\n",
-        "757.0 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.12:Pg-583"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.12\n",
-      "\n",
-      "import math\n",
-      "z1=21/2.0; # in ft\n",
-      "Lg=9.0; # in ft\n",
-      "Bg=6.0;# in ft\n",
-      "Qg=500*1000.0; # in kip\n",
-      "Cc1=0.3;\n",
-      "Cc2=0.2;\n",
-      "Cc3=0.25;\n",
-      "H2=12;\n",
-      "H3=6;\n",
-      "H1=21;\n",
-      "e1=0.82;\n",
-      "e2=0.7;\n",
-      "e3=0.75;\n",
-      "s1=Qg/(Lg+z1)/(Bg+z1); #sigma1 in lb/ft^3\n",
-      "s2=500*1000/(9+27)/(6+27);#sigma2 in lb/ft^3\n",
-      "s3=500*1000/(9+36)/(6+36);#sigma3 in lb/ft^3\n",
-      "ss1=6*105+(27+21/2)*(115-62.4);#sigmadash1 in lb/ft^3\n",
-      "ss2=6*105+(27+21)*(115-62.4)+(120-62.4)*6;#sigmadash2 in lb/ft^3\n",
-      "ss3=6*105+48*(115-62.4)+12*(120-62.4)+3*(122-62.4);#sigmadash3 in lb/ft^3\n",
-      "sc1=Cc1*H1/(1+e1)*math.log10((ss1+s1)/ss1); # in inch\n",
-      "sc2=Cc2*H2/(1+e2)*math.log10((ss2+s2)/ss2); # in inch\n",
-      "sc3=Cc3*H3/(1+e3)*math.log10((ss3+s3)/ss3); # in inch\n",
-      "sc=sc1+sc2+sc3; # in inch\n",
-      "print round(sc*12,1),\"total settlement in inch\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9.6 total settlement in inch\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter11_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter11_2.ipynb
deleted file mode 100755
index 0afdb4a0..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter11_2.ipynb
+++ /dev/null
@@ -1,650 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a4054b3d6b79658bcd763c43dc94ca6de4cd49698c12f912e4c83a9a61f47154"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter11:Pile Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.1:Pg-532"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.1\n",
-      "import math\n",
-      "\n",
-      "#parta\n",
-      "phi=30.0; # angle in degree\n",
-      "pa=2000.0; \n",
-      "q=100*50/1000.0;\n",
-      "Nq=55.0;\n",
-      "Ap=16*16/16/12; # area in ft^2\n",
-      "Qp=Ap*q*Nq; # in kip\n",
-      "qp=0.4*pa*Nq*math.tan(phi*math.pi/180)*Ap; # in lb\n",
-      "print round(Qp,2),\"ultimate load in lb\"\n",
-      "print round(qp/1000,2),\"ultimate load in kip\"\n",
-      "print \"there is change in answer because of calculation mistake in the book\"\n",
-      "\n",
-      "#partb\n",
-      "Nsigma=36;\n",
-      "Ap=16*16.0/12.0/12;\n",
-      "q=110*50.0/1000;\n",
-      "Qp=Ap*q*Nsigma*((1+2.0*(1-math.sin(phi*math.pi/180)))/3); # in kip\n",
-      "print round(Qp,2),\"ultimate load in kip\"\n",
-      "#partc\n",
-      "Nq=18.4;\n",
-      "Qp=Ap*q*Nq; # in kip\n",
-      "print round(Qp,2),\"ultimate load in kip\"\n",
-      "\n",
-      "# ANSWER IN THE BOOK IS WRONG"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "275.0 ultimate load in lb\n",
-        "25.4 ultimate load in kip\n",
-        "there is change in answer because of calculation mistake in the book\n",
-        "234.67 ultimate load in kip\n",
-        "179.91 ultimate load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.2:Pg-533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.2\n",
-      "\n",
-      "import math\n",
-      "#parta\n",
-      "K=1.3;\n",
-      "f0=0;\n",
-      "Delta=0.8*30; # in ft\n",
-      "D=16.0/12; # in ft\n",
-      "L1=50.0;\n",
-      "p=4*16/12.0; # in ft\n",
-      "Gamma=110/1000.0; # in lb/ft^3\n",
-      "L=15*D; # in ft\n",
-      "sigma=Gamma*L; # in kip/ft^2\n",
-      "f20=K*sigma*math.tan(Delta*math.pi/180); # kip/ft^2\n",
-      "Qs=(f0+f20)/2*(p*L)+f20*p*(L1-L);\n",
-      "print round(Qs,2),\"ultimate load in kip\"\n",
-      "#partb\n",
-      "FS=4; # factor of safety\n",
-      "Qp=56.45/3+234.7/3+179.9/3; # in kip\n",
-      "Qu=Qs+Qp; # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall,2),\"is allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "271.65 ultimate load in kip\n",
-        "107.17 is allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.3:Pg-534"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.3\n",
-      "import math\n",
-      "K=0.25;\n",
-      "Ap=16*16.0/12/12; # area in ft^2\n",
-      "phi=30*math.pi/180;\n",
-      "Nq=25;\n",
-      "q=110*50.0/1000; # in kip\n",
-      "sigmao=q/2; # in kip/ft^2\n",
-      "p=4*16.0/12; # in ft\n",
-      "L=50; # in ft\n",
-      "FS=4; # factor of safety\n",
-      "Qu=q*Nq*Ap+K*sigmao*math.tan(0.8*phi)*p*L; # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall,1),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "81.5 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.4:Pg-535"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.4\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "FS=4; # factor of safety\n",
-      "Ap=0.1295; # area in m^2\n",
-      "Nc=9;\n",
-      "cu2=100;\n",
-      "Qp=Ap*Nc*cu2; # in kN\n",
-      "D=[5, 10, 30]; # depth in m\n",
-      "avgD=[2.5, 7.5,20.0]; # average depth in m\n",
-      "sigma=[45.0, 110.5, 228.5]; # in KN/m^2\n",
-      "cu=[30, 30, 100]; # in kN/m^2\n",
-      "alpha=[0.6, 0.9, 0.725];\n",
-      "L=[5, 5, 20]; # in m\n",
-      "p=math.pi*0.406;\n",
-      "Qs=0; # in kN\n",
-      "cusig=numpy.zeros(3)\n",
-      "print round(Qp,2),\"bearing capacity in kN\"\n",
-      "print  \"depth (m)\\t avg Depth(m)\\t avgVerticalStress(kN/m**2)\\t cu(kN/m**2)\\t cu/sigma\\t alpha\\n\"\n",
-      "for i in range(0,3):\n",
-      "    cusig[i]=cu[i]/sigma[i];\n",
-      "    Qs=Qs+alpha[i]*cu[i]*L[i]*p;\n",
-      "    print round(D[i],2),\"\\t \\t \\t\",round(avgD[i],2),\"\\t \\t\",round(sigma[i],2),\"\\t\\t\\t   \",round(cu[i],2),\"\\t  \",round(cusig[i],2),\"\\t\\t \",round(alpha[i],2),\"\\n\"\n",
-      "print round(Qs,2),\"bearing capacity in kN\"\n",
-      "#part2\n",
-      "Lambda=0.136;\n",
-      "L=30;\n",
-      "fav=Lambda*(178.48+2*76.7);\n",
-      "Qs2=p*L*fav; # in kN\n",
-      "#part3\n",
-      "fav1=13;\n",
-      "fav2=31.9;\n",
-      "fav3=93.43;\n",
-      "Qs3=p*(fav1*5+fav2*5+fav3*20); # in kN\n",
-      "print round(Qs3,1),\"bearing capacity in kN\"\n",
-      "Qsavg=Qs/3+Qs2/3+Qs3/3; # in kN\n",
-      "Qu=Qp+Qsavg # in kN\n",
-      "Qall=Qu/FS; # in kN\n",
-      "print round(Qall,1),\"allowed bearing capacity in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "116.55 bearing capacity in kN\n",
-        "depth (m)\t avg Depth(m)\t avgVerticalStress(kN/m**2)\t cu(kN/m**2)\t cu/sigma\t alpha\n",
-        "\n",
-        "5.0 \t \t \t2.5 \t \t45.0 \t\t\t    30.0 \t   0.67 \t\t  0.6 \n",
-        "\n",
-        "10.0 \t \t \t7.5 \t \t110.5 \t\t\t    30.0 \t   0.27 \t\t  0.9 \n",
-        "\n",
-        "30.0 \t \t \t20.0 \t \t228.5 \t\t\t    100.0 \t   0.44 \t\t  0.72 \n",
-        "\n",
-        "2136.44 bearing capacity in kN\n",
-        "2669.7 bearing capacity in kN\n",
-        "573.6 allowed bearing capacity in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.5:Pg-538"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.5\n",
-      "\n",
-      "import numpy\n",
-      "D=[6, 12, 20]; # depth in m\n",
-      "fc=[34.34, 54.94, 70.63]; # in kN/m**2\n",
-      "alpha=[0.84, 0.71, 0.63];\n",
-      "dL=[6, 6, 8]; # in m\n",
-      "p=4*0.305;\n",
-      "Qs=0;\n",
-      "Q=numpy.zeros(3)\n",
-      "print \"  depth(m)\\t fc(kN/m**2)\\t alpha \\t \\t deltaL(m)\\t Q(kN)\\n\"\n",
-      "for i in range (0,3):\n",
-      "    Q[i]=alpha[i]*fc[i]*p*dL[i];\n",
-      "    Qs=Q[i]+Qs;\n",
-      "    print D[i],\"\\t\\t   \",fc[i],\"\\t  \",alpha[i],\"\\t   \",dL[i],\"\\t\\t \",round(Q[i],2)\n",
-      "\n",
-      "print round(Qs),\"bearing force in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  depth(m)\t fc(kN/m**2)\t alpha \t \t deltaL(m)\t Q(kN)\n",
-        "\n",
-        "6 \t\t    34.34 \t   0.84 \t    6 \t\t  211.15\n",
-        "12 \t\t    54.94 \t   0.71 \t    6 \t\t  285.53\n",
-        "20 \t\t    70.63 \t   0.63 \t    8 \t\t  434.29\n",
-        "931.0 bearing force in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:Pg-545"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.6\n",
-      "\n",
-      "import math\n",
-      "L=21; # in m\n",
-      "Qwp=502-350; # in kN\n",
-      "Qws=350; # in kN\n",
-      "Ap=0.1045; # area in m^2\n",
-      "Ep=21e6; # in kN/m^2\n",
-      "epsilon=0.62;\n",
-      "Se1=(Qwp+epsilon*Qws)*L/Ap/Ep; # in m\n",
-      "#part2\n",
-      "Iwp=0.85;\n",
-      "qwp=152/Ap;\n",
-      "Es=25e3; # in kN/m^2\n",
-      "D=0.356; # in m\n",
-      "mus=0.35;\n",
-      "Se2=qwp*D/Es*Iwp*(1-mus**2); # in m\n",
-      "#part3\n",
-      "p=1.168;\n",
-      "Iws=2+0.35*math.sqrt(L/D);\n",
-      "Se3=Qws/p/L*D/Es*Iws*(1-mus**2); # in m\n",
-      "Se=Se1+Se2+Se3; # in m\n",
-      "print round(Se*1000,1),\"settlement in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "19.8 settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.7:Pg-560"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.7\n",
-      "\n",
-      "Ep=207e6; # in kN/m^2\n",
-      "Ip=123e-6; # in m^4\n",
-      "nh=12000; # in  kN/m^3\n",
-      "#from table 11.13\n",
-      "xz=0.008;\n",
-      "Ax=2.435;\n",
-      "T=(Ep*Ip/nh)**0.2;\n",
-      "Qg1=xz*Ep*Ip/Ax/T**3;\n",
-      "#part2\n",
-      "Fy=248000;\n",
-      "d1=0.254;\n",
-      "Am=0.772;\n",
-      "Mzmax=Fy*Ip*2/d1; # in Kn-m\n",
-      "Qg2=Mzmax/Am/T; # in kN\n",
-      "if Qg2>Qg1 :\n",
-      "    Qg=Qg1;\n",
-      "    print round(Qg,2),\"lateral load in kN\"\n",
-      "# there is slight variation in answer in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "53.27 lateral load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.8:Pg-561"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.8\n",
-      "\n",
-      "import math\n",
-      "#part1\n",
-      "Ep=207e6; # in kN/m^2\n",
-      "Ip=123e-6; # in m^4\n",
-      "nh=12000; # in kN/m^3\n",
-      "#from table 11.1a\n",
-      "xo=0.008; # in m\n",
-      "L=25;\n",
-      "Fy=248000; # yield stress in kN/m^2\n",
-      "D=0.254;\n",
-      "Am=0.772;\n",
-      "Gamma=18.0; # in kN/m^3\n",
-      "phi=35; # in angle\n",
-      "Kp=(math.tan(math.pi/4+phi*math.pi/360))**2;\n",
-      "My=Fy*Ip*2/D; # in kN-m\n",
-      "Qug=140*Kp*D**3*Gamma; # in kN\n",
-      "\n",
-      "#part2\n",
-      "Qg1=xo*(Ep*Ip)**0.6*nh**0.4/0.15/L; # in kN\n",
-      "\n",
-      "if Qug>Qg1:\n",
-      "    Qg=Qg1;\n",
-      "    print round(Qg,2),\"lateral load in kN\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "40.2 lateral load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.9:Pg-567"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.9\n",
-      "\n",
-      "import math\n",
-      "Wrh=30*12; # in kip-ft\n",
-      "E=0.8;\n",
-      "Wr=7.5; # in kip\n",
-      "S=1/8.0; \n",
-      "C=0.1;\n",
-      "FS=6; # in factor of safety\n",
-      "n=0.4; # Coefficient of restitution\n",
-      "Wp=12/12.0*12/12.0*80*150+550; # in lb\n",
-      "Wp=Wp/1000.0;\n",
-      "Qu=E*Wrh/(S+C)*(Wr+n**2.0*Wp)/(Wr+Wp); # in kip\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed bearing capacity in kip\"\n",
-      "#part2\n",
-      "He=30*12.0;\n",
-      "L=80*12.0;\n",
-      "Ap=12*12.0; # area in in^2\n",
-      "Ep=3e6/1000.0; # in kip/in^2\n",
-      "FS=4; # factor of safety\n",
-      "Qu=E*He/(S+math.sqrt(E*He*L/2.0/Ap/Ep)); # in kip\n",
-      "Qall2=Qu/FS; # in kip\n",
-      "print round(Qall2),\"allowed bearing capacity in kip\"\n",
-      "\n",
-      "#partc\n",
-      "a=27;\n",
-      "b=1;\n",
-      "He=30;\n",
-      "FS=3; # factor of safety\n",
-      "Qu=a*math.sqrt(E*He)*(b-math.log10(S)); # in kip\n",
-      "Qall3=Qu/FS; # in kip\n",
-      "print round(Qall3),\"allowed bearing capacity in kip\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "101.0 allowed bearing capacity in kip\n",
-        "104.0 allowed bearing capacity in kip\n",
-        "84.0 allowed bearing capacity in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.10:Pg-570"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.10\n",
-      "\n",
-      "Hp=350; # in HP\n",
-      "vp=0.0016; #  in m/s\n",
-      "Sl=0.762e-3; # in m/cycle\n",
-      "f=115; # in Hz\n",
-      "Qu=(0.746*Hp+98*vp)/(vp+Sl*f); # in kN\n",
-      "print round(Qu),\"pile load capacity in kN\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2928.0 pile load capacity in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.11:Pg-578"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.11\n",
-      "\n",
-      "Lg=9.92; # in ft\n",
-      "Bg=7.0; # in ft\n",
-      "n1=3.0;\n",
-      "Nc=8.75;\n",
-      "n2=4.0/1000;\n",
-      "Ap=14.0**2.0/12.0**2;\n",
-      "cup=1775.0;\n",
-      "a1=0.4;#alpha1\n",
-      "p=4*14.0/12.0;\n",
-      "cu1=1050.0; # in lb/ft^2\n",
-      "L1=15.0;\n",
-      "a2=0.54;#alpha2\n",
-      "cu2=1775.0; # in lb/ft^2\n",
-      "L2=45.0;\n",
-      "FS=4; # factor of safety\n",
-      "Qu=n1*n2*(9*Ap*cup+a1*p*cu1*L1+a2*p*cu2*L2); # in kip\n",
-      "Qu2=Lg*Bg*cup*Nc+2*(Lg+Bg)*(cu1*L1+cu2*L2); # in kip\n",
-      "print round(Qu2/1000),\"load in kip\"\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed load in kip\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4314.0 load in kip\n",
-        "757.0 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex11.12:Pg-583"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 11.12\n",
-      "\n",
-      "import math\n",
-      "z1=21/2.0; # in ft\n",
-      "Lg=9.0; # in ft\n",
-      "Bg=6.0;# in ft\n",
-      "Qg=500*1000.0; # in kip\n",
-      "Cc1=0.3;\n",
-      "Cc2=0.2;\n",
-      "Cc3=0.25;\n",
-      "H2=12;\n",
-      "H3=6;\n",
-      "H1=21;\n",
-      "e1=0.82;\n",
-      "e2=0.7;\n",
-      "e3=0.75;\n",
-      "s1=Qg/(Lg+z1)/(Bg+z1); #sigma1 in lb/ft^3\n",
-      "s2=500*1000/(9+27)/(6+27);#sigma2 in lb/ft^3\n",
-      "s3=500*1000/(9+36)/(6+36);#sigma3 in lb/ft^3\n",
-      "ss1=6*105+(27+21/2)*(115-62.4);#sigmadash1 in lb/ft^3\n",
-      "ss2=6*105+(27+21)*(115-62.4)+(120-62.4)*6;#sigmadash2 in lb/ft^3\n",
-      "ss3=6*105+48*(115-62.4)+12*(120-62.4)+3*(122-62.4);#sigmadash3 in lb/ft^3\n",
-      "sc1=Cc1*H1/(1+e1)*math.log10((ss1+s1)/ss1); # in inch\n",
-      "sc2=Cc2*H2/(1+e2)*math.log10((ss2+s2)/ss2); # in inch\n",
-      "sc3=Cc3*H3/(1+e3)*math.log10((ss3+s3)/ss3); # in inch\n",
-      "sc=sc1+sc2+sc3; # in inch\n",
-      "print round(sc*12,1),\"total settlement in inch\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9.6 total settlement in inch\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter12.ipynb b/Principles_Of_Foundation_Engineering/Chapter12.ipynb
deleted file mode 100755
index fcc14387..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter12.ipynb
+++ /dev/null
@@ -1,443 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1439e2e46e40a0d47bd1c7e06a03c417c0e1f55ad415a0af82536f86f7ecfea9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter12:Drilled-Shaft Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.1:Pg-609"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.1\n",
-      "\n",
-      "import math\n",
-      "Ap=math.pi/4*1.75**2;  # area in m^2\n",
-      "FS=4; # factor of safety\n",
-      "Nq=37.75; \n",
-      "L=8;\n",
-      "Es=50000.0;\n",
-      "mus=0.265;\n",
-      "pu=100.0;\n",
-      "Db=1.75; # in m\n",
-      "q=6*16.2+2*19.2;\n",
-      "phi=36*math.pi/180.0;\n",
-      "Fqs=1+math.tan(phi);\n",
-      "Fqd=1+2*math.tan(phi)*(1-math.sin(phi))**2*math.atan(L/Db);\n",
-      "Ir=Es/(2*(1+mus)*q*math.tan(phi));\n",
-      "delta=0.005*(1-phi/20*180/math.pi+25/20.0)*q/pu;\n",
-      "Irr=Ir/(1+Ir*delta);\n",
-      "Fqc=math.exp(-3.8*math.tan(phi)+(3.07*math.sin(phi)*math.log10(2*Irr))/(1+math.sin(phi)));\n",
-      "Qp=Ap*(q*(Nq-1)*Fqs*Fqd*Fqc);\n",
-      "Qpall=Qp/FS;\n",
-      "print round(Qpall,2),\"allowed load in kN\"\n",
-      "print \"due to rounding off error there is slight change in answer\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6648.7 allowed load in kN\n",
-        "due to rounding off error there is slight change in answer\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.2:Pg-610"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.2\n",
-      "import math\n",
-      "Ap=math.pi/4*1.75**2; # area in m^2\n",
-      "q=135.6; \n",
-      "w=0.83;\n",
-      "FS=4; # factor of safety\n",
-      "phi=36; # given angle\n",
-      "Nq=0.21*math.exp(0.17*phi);\n",
-      "Qp=Ap*q*(w*Nq-1); # in kN\n",
-      "Qpall=Qp/FS; # in kN\n",
-      "print round(Qpall),\"allowed load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6383.0 allowed load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.3:Pg-611"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.3\n",
-      "\n",
-      "import math\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "Db=1.5; # in m\n",
-      "z=3.0;\n",
-      "p=math.pi*1;\n",
-      "Li=6.0;\n",
-      "N60=30.0;\n",
-      "sigmazi=16*z;\n",
-      "Beta=2.0-0.15*z**0.75;\n",
-      "fi=Beta*sigmazi; # in kN/m^2\n",
-      "qp=57.5*N60; # in kN/m^2\n",
-      "qpr=1.27/Db*qp; # in kN/m^2\n",
-      "Qunet=qpr*Ap+fi*p*Li; # in kN\n",
-      "print round(Qunet,2),\"allowed load in kN\"\n",
-      "#part b\n",
-      "k1=0.315; #from table\n",
-      "k2=12.0/1.5/1000*100.0;\n",
-      "Qunet2=qpr*Ap*k1+fi*p*Li*k2; # in kN\n",
-      "print round(Qunet2,2),\"allowed load in kN\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4081.11 allowed load in kN\n",
-        "2013.14 allowed load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.4:Pg-617"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.4\n",
-      "\n",
-      "Nc=9;\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "cu=105; # in kN/m^2\n",
-      "Qpnet=Ap*cu*Nc; # in kN\n",
-      "print round(Qpnet),\"net ultimate bearing point capacity in kN\"\n",
-      "#part2\n",
-      "alpha=0.4;\n",
-      "Ds=1.5; # in m \n",
-      "p=math.pi*Ds;\n",
-      "Qs=alpha*p*(50*8+105*3); # in kN\n",
-      "print int(Qs),\"skin resistance in kN\"\n",
-      "#part3\n",
-      "FS=3;  # factor of safety\n",
-      "Qu=Qpnet/FS+Qs/FS; # in kN\n",
-      "print round(Qu,2),\"working load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1670.0 net ultimate bearing point capacity in kN\n",
-        "1347 skin resistance in kN\n",
-        "1005.9 working load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.5:Pg-618"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.5\n",
-      "\n",
-      "import math\n",
-      "cub=3000;\n",
-      "L=20+5; # in ft\n",
-      "Db=4; # in ft\n",
-      "Ap=math.pi/4*Db**2; # area in ft^2\n",
-      "alpha=0.55;\n",
-      "cu1=800; # in lb/ft^2\n",
-      "L1=7; # in ft\n",
-      "L2=5.5; # in ft\n",
-      "cu2=1200; # in lb/ft^2\n",
-      "p=math.pi*2.5;\n",
-      "k=alpha*p*(cu1*L1+cu2*L2);#f*p*deltaLi\n",
-      "j1=6*cub*(1+0.2*L/Db);\n",
-      "j2=9*cub;\n",
-      "qp=min(j1,j2);\n",
-      "Qu=k/1000+qp*Ap/1000; # in kip\n",
-      "print round(Qu),\"allowed load in kip\"\n",
-      "#part b\n",
-      "k1=0.57; #from table\n",
-      "k2=0.89;\n",
-      "Qunet2=qp*Ap*k1+k*k2; # in kip\n",
-      "print round(Qunet2/1000,2),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "392.0 allowed load in kip\n",
-        "240.3 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.6:Pg-621"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.6\n",
-      "\n",
-      "import math\n",
-      "Qws=1005-250; # in kN\n",
-      "Qwp=250; # in kN\n",
-      "epsilon=0.65; \n",
-      "L=11; # in m\n",
-      "Ds=1.5; # in m\n",
-      "Es=14000; # in kN/m^2\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "Ep=21e6; # in kN/m^2\n",
-      "Cp=0.04; # in kN/m^2\n",
-      "Db=1.5;\n",
-      "mus=0.3;\n",
-      "p=math.pi*1.5;\n",
-      "Nc=9;\n",
-      "qp=105*Nc; # in kN/m^2\n",
-      "se1=(Qwp+epsilon*Qws)*L/(Ap*Ep); # in m \n",
-      "se2=Qwp*Cp/(Db*qp); # in m\n",
-      "Iws=2+0.35*math.sqrt(L/Ds);\n",
-      "se3=Qws/p/L*Ds/Es*(1-mus**2)*Iws; # in m\n",
-      "se=se1+se2+se3; # in m\n",
-      "print round(se*1000,2),\" is net settlement in mm\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11.46  is net settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.7:Pg-628"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.7\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "from scipy.optimize import fsolve\n",
-      "Ds=1.0;\n",
-      "Ep=22e6;\n",
-      "Ri=1.0;\n",
-      "cu=100.0;\n",
-      "Ip=math.pi*Ds**4/64.0;\n",
-      "Qc=7.34*Ds**2*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
-      "print round(Qc,2),\"bearing force in kN\"\n",
-      "Mc=3.86*Ds**3*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
-      "print round(Mc,2),\"bearing moment in kNm\"\n",
-      "#from figure\n",
-      "xoQM=0.0046*1;\n",
-      "xoMQ=0.0041*1;\n",
-      "xo=0.5*(xoQM+xoMQ);\n",
-      "print round(xo*1000,2),\"net ground line deflection in mm\"\n",
-      "#partb\n",
-      "Ip=0.049;\n",
-      "Qg=150.0;\n",
-      "Mg=200.0;\n",
-      "def f(T):\n",
-      "    return 338e-6*T**3+300.6e-6*T**2-0.00435\n",
-      "[x]=fsolve(f,2);\n",
-      "T=x;\n",
-      "k=[0, 0.4, 0.6, 0.8, 1.0, 1.1, 1.25];#z/T\n",
-      "Am=[0, 0.36, 0.52, 0.63, 0.75, 0.765, 0.75];\n",
-      "Bm=[1.0, 0.98, 0.95, 0.9, 0.845, 0.8, 0.73];\n",
-      "print \"z/T\\t Am\\t Bm\\t Mz(kN-m)\\n\"\n",
-      "Mz=numpy.zeros(7)\n",
-      "for i in range(0,7):\n",
-      "    Mz[i]=Am[i]*Qg*T+Bm[i]*Mg;\n",
-      "    print k[i],\"\\t\",round(Am[i],2),\"\\t\",round(Bm[i],2),\"\\t\",round(Mz[i],2)\n",
-      "\n",
-      "print round(1*T,2),\"depth in m\"\n",
-      "#partc\n",
-      "Mmax=400;\n",
-      "sigma=Mmax*Ds/2/Ip;\n",
-      "print round(sigma,2),\"tensile stress in kN/m**2\"\n",
-      "#partd\n",
-      "#from figure\n",
-      "k=8.5;\n",
-      "L=k*1;\n",
-      "print L,\"length in m\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "100615.56 bearing force in kN\n",
-        "52912.27 bearing moment in kNm\n",
-        "4.35 net ground line deflection in mm\n",
-        "z/T\t Am\t Bm\t Mz(kN-m)\n",
-        "\n",
-        "0 \t0.0 \t1.0 \t200.0\n",
-        "0.4 \t0.36 \t0.98 \t308.4\n",
-        "0.6 \t0.52 \t0.95 \t352.35\n",
-        "0.8 \t0.63 \t0.9 \t376.69\n",
-        "1.0 \t0.75 \t0.84 \t403.16\n",
-        "1.1 \t0.77 \t0.8 \t398.84\n",
-        "1.25 \t0.75 \t0.73 \t380.16\n",
-        "2.08 depth in m\n",
-        "4081.63 tensile stress in kN/m**2\n",
-        "8.5 length in m\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.8:Pg-634"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.8\n",
-      "\n",
-      "qu=3000; # lb/in^2\n",
-      "Ds=3*12; #in inch\n",
-      "L=15*12; # in inch\n",
-      "FS=3; # factor of safety\n",
-      "Ecore=0.36e6; # in lb/in^2\n",
-      "f=min(2.5*qu**0.5,0.15*qu);\n",
-      "Qu=math.pi*Ds*L*f/1000; # in kip\n",
-      "Emass=Ecore*(0.266*80-1.66); # in lb/in^2\n",
-      "Ec=17.9*Emass; # in lb/in^2\n",
-      "Ac=math.pi/4*Ds**2; # area in in^2\n",
-      "If=0.35;\n",
-      "se=Qu*L/Ac/Ec+Qu*If/Ds/Emass;\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "929.0 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter12_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter12_1.ipynb
deleted file mode 100755
index fcc14387..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter12_1.ipynb
+++ /dev/null
@@ -1,443 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1439e2e46e40a0d47bd1c7e06a03c417c0e1f55ad415a0af82536f86f7ecfea9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter12:Drilled-Shaft Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.1:Pg-609"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.1\n",
-      "\n",
-      "import math\n",
-      "Ap=math.pi/4*1.75**2;  # area in m^2\n",
-      "FS=4; # factor of safety\n",
-      "Nq=37.75; \n",
-      "L=8;\n",
-      "Es=50000.0;\n",
-      "mus=0.265;\n",
-      "pu=100.0;\n",
-      "Db=1.75; # in m\n",
-      "q=6*16.2+2*19.2;\n",
-      "phi=36*math.pi/180.0;\n",
-      "Fqs=1+math.tan(phi);\n",
-      "Fqd=1+2*math.tan(phi)*(1-math.sin(phi))**2*math.atan(L/Db);\n",
-      "Ir=Es/(2*(1+mus)*q*math.tan(phi));\n",
-      "delta=0.005*(1-phi/20*180/math.pi+25/20.0)*q/pu;\n",
-      "Irr=Ir/(1+Ir*delta);\n",
-      "Fqc=math.exp(-3.8*math.tan(phi)+(3.07*math.sin(phi)*math.log10(2*Irr))/(1+math.sin(phi)));\n",
-      "Qp=Ap*(q*(Nq-1)*Fqs*Fqd*Fqc);\n",
-      "Qpall=Qp/FS;\n",
-      "print round(Qpall,2),\"allowed load in kN\"\n",
-      "print \"due to rounding off error there is slight change in answer\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6648.7 allowed load in kN\n",
-        "due to rounding off error there is slight change in answer\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.2:Pg-610"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.2\n",
-      "import math\n",
-      "Ap=math.pi/4*1.75**2; # area in m^2\n",
-      "q=135.6; \n",
-      "w=0.83;\n",
-      "FS=4; # factor of safety\n",
-      "phi=36; # given angle\n",
-      "Nq=0.21*math.exp(0.17*phi);\n",
-      "Qp=Ap*q*(w*Nq-1); # in kN\n",
-      "Qpall=Qp/FS; # in kN\n",
-      "print round(Qpall),\"allowed load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6383.0 allowed load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.3:Pg-611"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.3\n",
-      "\n",
-      "import math\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "Db=1.5; # in m\n",
-      "z=3.0;\n",
-      "p=math.pi*1;\n",
-      "Li=6.0;\n",
-      "N60=30.0;\n",
-      "sigmazi=16*z;\n",
-      "Beta=2.0-0.15*z**0.75;\n",
-      "fi=Beta*sigmazi; # in kN/m^2\n",
-      "qp=57.5*N60; # in kN/m^2\n",
-      "qpr=1.27/Db*qp; # in kN/m^2\n",
-      "Qunet=qpr*Ap+fi*p*Li; # in kN\n",
-      "print round(Qunet,2),\"allowed load in kN\"\n",
-      "#part b\n",
-      "k1=0.315; #from table\n",
-      "k2=12.0/1.5/1000*100.0;\n",
-      "Qunet2=qpr*Ap*k1+fi*p*Li*k2; # in kN\n",
-      "print round(Qunet2,2),\"allowed load in kN\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4081.11 allowed load in kN\n",
-        "2013.14 allowed load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.4:Pg-617"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.4\n",
-      "\n",
-      "Nc=9;\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "cu=105; # in kN/m^2\n",
-      "Qpnet=Ap*cu*Nc; # in kN\n",
-      "print round(Qpnet),\"net ultimate bearing point capacity in kN\"\n",
-      "#part2\n",
-      "alpha=0.4;\n",
-      "Ds=1.5; # in m \n",
-      "p=math.pi*Ds;\n",
-      "Qs=alpha*p*(50*8+105*3); # in kN\n",
-      "print int(Qs),\"skin resistance in kN\"\n",
-      "#part3\n",
-      "FS=3;  # factor of safety\n",
-      "Qu=Qpnet/FS+Qs/FS; # in kN\n",
-      "print round(Qu,2),\"working load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1670.0 net ultimate bearing point capacity in kN\n",
-        "1347 skin resistance in kN\n",
-        "1005.9 working load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.5:Pg-618"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.5\n",
-      "\n",
-      "import math\n",
-      "cub=3000;\n",
-      "L=20+5; # in ft\n",
-      "Db=4; # in ft\n",
-      "Ap=math.pi/4*Db**2; # area in ft^2\n",
-      "alpha=0.55;\n",
-      "cu1=800; # in lb/ft^2\n",
-      "L1=7; # in ft\n",
-      "L2=5.5; # in ft\n",
-      "cu2=1200; # in lb/ft^2\n",
-      "p=math.pi*2.5;\n",
-      "k=alpha*p*(cu1*L1+cu2*L2);#f*p*deltaLi\n",
-      "j1=6*cub*(1+0.2*L/Db);\n",
-      "j2=9*cub;\n",
-      "qp=min(j1,j2);\n",
-      "Qu=k/1000+qp*Ap/1000; # in kip\n",
-      "print round(Qu),\"allowed load in kip\"\n",
-      "#part b\n",
-      "k1=0.57; #from table\n",
-      "k2=0.89;\n",
-      "Qunet2=qp*Ap*k1+k*k2; # in kip\n",
-      "print round(Qunet2/1000,2),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "392.0 allowed load in kip\n",
-        "240.3 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.6:Pg-621"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.6\n",
-      "\n",
-      "import math\n",
-      "Qws=1005-250; # in kN\n",
-      "Qwp=250; # in kN\n",
-      "epsilon=0.65; \n",
-      "L=11; # in m\n",
-      "Ds=1.5; # in m\n",
-      "Es=14000; # in kN/m^2\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "Ep=21e6; # in kN/m^2\n",
-      "Cp=0.04; # in kN/m^2\n",
-      "Db=1.5;\n",
-      "mus=0.3;\n",
-      "p=math.pi*1.5;\n",
-      "Nc=9;\n",
-      "qp=105*Nc; # in kN/m^2\n",
-      "se1=(Qwp+epsilon*Qws)*L/(Ap*Ep); # in m \n",
-      "se2=Qwp*Cp/(Db*qp); # in m\n",
-      "Iws=2+0.35*math.sqrt(L/Ds);\n",
-      "se3=Qws/p/L*Ds/Es*(1-mus**2)*Iws; # in m\n",
-      "se=se1+se2+se3; # in m\n",
-      "print round(se*1000,2),\" is net settlement in mm\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11.46  is net settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.7:Pg-628"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.7\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "from scipy.optimize import fsolve\n",
-      "Ds=1.0;\n",
-      "Ep=22e6;\n",
-      "Ri=1.0;\n",
-      "cu=100.0;\n",
-      "Ip=math.pi*Ds**4/64.0;\n",
-      "Qc=7.34*Ds**2*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
-      "print round(Qc,2),\"bearing force in kN\"\n",
-      "Mc=3.86*Ds**3*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
-      "print round(Mc,2),\"bearing moment in kNm\"\n",
-      "#from figure\n",
-      "xoQM=0.0046*1;\n",
-      "xoMQ=0.0041*1;\n",
-      "xo=0.5*(xoQM+xoMQ);\n",
-      "print round(xo*1000,2),\"net ground line deflection in mm\"\n",
-      "#partb\n",
-      "Ip=0.049;\n",
-      "Qg=150.0;\n",
-      "Mg=200.0;\n",
-      "def f(T):\n",
-      "    return 338e-6*T**3+300.6e-6*T**2-0.00435\n",
-      "[x]=fsolve(f,2);\n",
-      "T=x;\n",
-      "k=[0, 0.4, 0.6, 0.8, 1.0, 1.1, 1.25];#z/T\n",
-      "Am=[0, 0.36, 0.52, 0.63, 0.75, 0.765, 0.75];\n",
-      "Bm=[1.0, 0.98, 0.95, 0.9, 0.845, 0.8, 0.73];\n",
-      "print \"z/T\\t Am\\t Bm\\t Mz(kN-m)\\n\"\n",
-      "Mz=numpy.zeros(7)\n",
-      "for i in range(0,7):\n",
-      "    Mz[i]=Am[i]*Qg*T+Bm[i]*Mg;\n",
-      "    print k[i],\"\\t\",round(Am[i],2),\"\\t\",round(Bm[i],2),\"\\t\",round(Mz[i],2)\n",
-      "\n",
-      "print round(1*T,2),\"depth in m\"\n",
-      "#partc\n",
-      "Mmax=400;\n",
-      "sigma=Mmax*Ds/2/Ip;\n",
-      "print round(sigma,2),\"tensile stress in kN/m**2\"\n",
-      "#partd\n",
-      "#from figure\n",
-      "k=8.5;\n",
-      "L=k*1;\n",
-      "print L,\"length in m\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "100615.56 bearing force in kN\n",
-        "52912.27 bearing moment in kNm\n",
-        "4.35 net ground line deflection in mm\n",
-        "z/T\t Am\t Bm\t Mz(kN-m)\n",
-        "\n",
-        "0 \t0.0 \t1.0 \t200.0\n",
-        "0.4 \t0.36 \t0.98 \t308.4\n",
-        "0.6 \t0.52 \t0.95 \t352.35\n",
-        "0.8 \t0.63 \t0.9 \t376.69\n",
-        "1.0 \t0.75 \t0.84 \t403.16\n",
-        "1.1 \t0.77 \t0.8 \t398.84\n",
-        "1.25 \t0.75 \t0.73 \t380.16\n",
-        "2.08 depth in m\n",
-        "4081.63 tensile stress in kN/m**2\n",
-        "8.5 length in m\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.8:Pg-634"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.8\n",
-      "\n",
-      "qu=3000; # lb/in^2\n",
-      "Ds=3*12; #in inch\n",
-      "L=15*12; # in inch\n",
-      "FS=3; # factor of safety\n",
-      "Ecore=0.36e6; # in lb/in^2\n",
-      "f=min(2.5*qu**0.5,0.15*qu);\n",
-      "Qu=math.pi*Ds*L*f/1000; # in kip\n",
-      "Emass=Ecore*(0.266*80-1.66); # in lb/in^2\n",
-      "Ec=17.9*Emass; # in lb/in^2\n",
-      "Ac=math.pi/4*Ds**2; # area in in^2\n",
-      "If=0.35;\n",
-      "se=Qu*L/Ac/Ec+Qu*If/Ds/Emass;\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "929.0 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter12_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter12_2.ipynb
deleted file mode 100755
index fcc14387..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter12_2.ipynb
+++ /dev/null
@@ -1,443 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1439e2e46e40a0d47bd1c7e06a03c417c0e1f55ad415a0af82536f86f7ecfea9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter12:Drilled-Shaft Foundations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.1:Pg-609"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.1\n",
-      "\n",
-      "import math\n",
-      "Ap=math.pi/4*1.75**2;  # area in m^2\n",
-      "FS=4; # factor of safety\n",
-      "Nq=37.75; \n",
-      "L=8;\n",
-      "Es=50000.0;\n",
-      "mus=0.265;\n",
-      "pu=100.0;\n",
-      "Db=1.75; # in m\n",
-      "q=6*16.2+2*19.2;\n",
-      "phi=36*math.pi/180.0;\n",
-      "Fqs=1+math.tan(phi);\n",
-      "Fqd=1+2*math.tan(phi)*(1-math.sin(phi))**2*math.atan(L/Db);\n",
-      "Ir=Es/(2*(1+mus)*q*math.tan(phi));\n",
-      "delta=0.005*(1-phi/20*180/math.pi+25/20.0)*q/pu;\n",
-      "Irr=Ir/(1+Ir*delta);\n",
-      "Fqc=math.exp(-3.8*math.tan(phi)+(3.07*math.sin(phi)*math.log10(2*Irr))/(1+math.sin(phi)));\n",
-      "Qp=Ap*(q*(Nq-1)*Fqs*Fqd*Fqc);\n",
-      "Qpall=Qp/FS;\n",
-      "print round(Qpall,2),\"allowed load in kN\"\n",
-      "print \"due to rounding off error there is slight change in answer\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6648.7 allowed load in kN\n",
-        "due to rounding off error there is slight change in answer\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.2:Pg-610"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.2\n",
-      "import math\n",
-      "Ap=math.pi/4*1.75**2; # area in m^2\n",
-      "q=135.6; \n",
-      "w=0.83;\n",
-      "FS=4; # factor of safety\n",
-      "phi=36; # given angle\n",
-      "Nq=0.21*math.exp(0.17*phi);\n",
-      "Qp=Ap*q*(w*Nq-1); # in kN\n",
-      "Qpall=Qp/FS; # in kN\n",
-      "print round(Qpall),\"allowed load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6383.0 allowed load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.3:Pg-611"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.3\n",
-      "\n",
-      "import math\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "Db=1.5; # in m\n",
-      "z=3.0;\n",
-      "p=math.pi*1;\n",
-      "Li=6.0;\n",
-      "N60=30.0;\n",
-      "sigmazi=16*z;\n",
-      "Beta=2.0-0.15*z**0.75;\n",
-      "fi=Beta*sigmazi; # in kN/m^2\n",
-      "qp=57.5*N60; # in kN/m^2\n",
-      "qpr=1.27/Db*qp; # in kN/m^2\n",
-      "Qunet=qpr*Ap+fi*p*Li; # in kN\n",
-      "print round(Qunet,2),\"allowed load in kN\"\n",
-      "#part b\n",
-      "k1=0.315; #from table\n",
-      "k2=12.0/1.5/1000*100.0;\n",
-      "Qunet2=qpr*Ap*k1+fi*p*Li*k2; # in kN\n",
-      "print round(Qunet2,2),\"allowed load in kN\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4081.11 allowed load in kN\n",
-        "2013.14 allowed load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.4:Pg-617"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.4\n",
-      "\n",
-      "Nc=9;\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "cu=105; # in kN/m^2\n",
-      "Qpnet=Ap*cu*Nc; # in kN\n",
-      "print round(Qpnet),\"net ultimate bearing point capacity in kN\"\n",
-      "#part2\n",
-      "alpha=0.4;\n",
-      "Ds=1.5; # in m \n",
-      "p=math.pi*Ds;\n",
-      "Qs=alpha*p*(50*8+105*3); # in kN\n",
-      "print int(Qs),\"skin resistance in kN\"\n",
-      "#part3\n",
-      "FS=3;  # factor of safety\n",
-      "Qu=Qpnet/FS+Qs/FS; # in kN\n",
-      "print round(Qu,2),\"working load in kN\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1670.0 net ultimate bearing point capacity in kN\n",
-        "1347 skin resistance in kN\n",
-        "1005.9 working load in kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.5:Pg-618"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.5\n",
-      "\n",
-      "import math\n",
-      "cub=3000;\n",
-      "L=20+5; # in ft\n",
-      "Db=4; # in ft\n",
-      "Ap=math.pi/4*Db**2; # area in ft^2\n",
-      "alpha=0.55;\n",
-      "cu1=800; # in lb/ft^2\n",
-      "L1=7; # in ft\n",
-      "L2=5.5; # in ft\n",
-      "cu2=1200; # in lb/ft^2\n",
-      "p=math.pi*2.5;\n",
-      "k=alpha*p*(cu1*L1+cu2*L2);#f*p*deltaLi\n",
-      "j1=6*cub*(1+0.2*L/Db);\n",
-      "j2=9*cub;\n",
-      "qp=min(j1,j2);\n",
-      "Qu=k/1000+qp*Ap/1000; # in kip\n",
-      "print round(Qu),\"allowed load in kip\"\n",
-      "#part b\n",
-      "k1=0.57; #from table\n",
-      "k2=0.89;\n",
-      "Qunet2=qp*Ap*k1+k*k2; # in kip\n",
-      "print round(Qunet2/1000,2),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "392.0 allowed load in kip\n",
-        "240.3 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.6:Pg-621"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.6\n",
-      "\n",
-      "import math\n",
-      "Qws=1005-250; # in kN\n",
-      "Qwp=250; # in kN\n",
-      "epsilon=0.65; \n",
-      "L=11; # in m\n",
-      "Ds=1.5; # in m\n",
-      "Es=14000; # in kN/m^2\n",
-      "Ap=math.pi/4*1.5**2; # area in m^2\n",
-      "Ep=21e6; # in kN/m^2\n",
-      "Cp=0.04; # in kN/m^2\n",
-      "Db=1.5;\n",
-      "mus=0.3;\n",
-      "p=math.pi*1.5;\n",
-      "Nc=9;\n",
-      "qp=105*Nc; # in kN/m^2\n",
-      "se1=(Qwp+epsilon*Qws)*L/(Ap*Ep); # in m \n",
-      "se2=Qwp*Cp/(Db*qp); # in m\n",
-      "Iws=2+0.35*math.sqrt(L/Ds);\n",
-      "se3=Qws/p/L*Ds/Es*(1-mus**2)*Iws; # in m\n",
-      "se=se1+se2+se3; # in m\n",
-      "print round(se*1000,2),\" is net settlement in mm\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "11.46  is net settlement in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.7:Pg-628"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.7\n",
-      "\n",
-      "import math\n",
-      "import numpy\n",
-      "from scipy.optimize import fsolve\n",
-      "Ds=1.0;\n",
-      "Ep=22e6;\n",
-      "Ri=1.0;\n",
-      "cu=100.0;\n",
-      "Ip=math.pi*Ds**4/64.0;\n",
-      "Qc=7.34*Ds**2*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
-      "print round(Qc,2),\"bearing force in kN\"\n",
-      "Mc=3.86*Ds**3*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
-      "print round(Mc,2),\"bearing moment in kNm\"\n",
-      "#from figure\n",
-      "xoQM=0.0046*1;\n",
-      "xoMQ=0.0041*1;\n",
-      "xo=0.5*(xoQM+xoMQ);\n",
-      "print round(xo*1000,2),\"net ground line deflection in mm\"\n",
-      "#partb\n",
-      "Ip=0.049;\n",
-      "Qg=150.0;\n",
-      "Mg=200.0;\n",
-      "def f(T):\n",
-      "    return 338e-6*T**3+300.6e-6*T**2-0.00435\n",
-      "[x]=fsolve(f,2);\n",
-      "T=x;\n",
-      "k=[0, 0.4, 0.6, 0.8, 1.0, 1.1, 1.25];#z/T\n",
-      "Am=[0, 0.36, 0.52, 0.63, 0.75, 0.765, 0.75];\n",
-      "Bm=[1.0, 0.98, 0.95, 0.9, 0.845, 0.8, 0.73];\n",
-      "print \"z/T\\t Am\\t Bm\\t Mz(kN-m)\\n\"\n",
-      "Mz=numpy.zeros(7)\n",
-      "for i in range(0,7):\n",
-      "    Mz[i]=Am[i]*Qg*T+Bm[i]*Mg;\n",
-      "    print k[i],\"\\t\",round(Am[i],2),\"\\t\",round(Bm[i],2),\"\\t\",round(Mz[i],2)\n",
-      "\n",
-      "print round(1*T,2),\"depth in m\"\n",
-      "#partc\n",
-      "Mmax=400;\n",
-      "sigma=Mmax*Ds/2/Ip;\n",
-      "print round(sigma,2),\"tensile stress in kN/m**2\"\n",
-      "#partd\n",
-      "#from figure\n",
-      "k=8.5;\n",
-      "L=k*1;\n",
-      "print L,\"length in m\"\n",
-      "\n",
-      "# the answer is slightly different in textbook due to approximation\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "100615.56 bearing force in kN\n",
-        "52912.27 bearing moment in kNm\n",
-        "4.35 net ground line deflection in mm\n",
-        "z/T\t Am\t Bm\t Mz(kN-m)\n",
-        "\n",
-        "0 \t0.0 \t1.0 \t200.0\n",
-        "0.4 \t0.36 \t0.98 \t308.4\n",
-        "0.6 \t0.52 \t0.95 \t352.35\n",
-        "0.8 \t0.63 \t0.9 \t376.69\n",
-        "1.0 \t0.75 \t0.84 \t403.16\n",
-        "1.1 \t0.77 \t0.8 \t398.84\n",
-        "1.25 \t0.75 \t0.73 \t380.16\n",
-        "2.08 depth in m\n",
-        "4081.63 tensile stress in kN/m**2\n",
-        "8.5 length in m\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex12.8:Pg-634"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 12.8\n",
-      "\n",
-      "qu=3000; # lb/in^2\n",
-      "Ds=3*12; #in inch\n",
-      "L=15*12; # in inch\n",
-      "FS=3; # factor of safety\n",
-      "Ecore=0.36e6; # in lb/in^2\n",
-      "f=min(2.5*qu**0.5,0.15*qu);\n",
-      "Qu=math.pi*Ds*L*f/1000; # in kip\n",
-      "Emass=Ecore*(0.266*80-1.66); # in lb/in^2\n",
-      "Ec=17.9*Emass; # in lb/in^2\n",
-      "Ac=math.pi/4*Ds**2; # area in in^2\n",
-      "If=0.35;\n",
-      "se=Qu*L/Ac/Ec+Qu*If/Ds/Emass;\n",
-      "Qall=Qu/FS; # in kip\n",
-      "print round(Qall),\"allowed load in kip\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "929.0 allowed load in kip\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter13.ipynb b/Principles_Of_Foundation_Engineering/Chapter13.ipynb
deleted file mode 100755
index bfb80602..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter13.ipynb
+++ /dev/null
@@ -1,176 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:549425d7a43cc856ebd1610c783821836546fd833bb34512fcafdb71f662b655"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter13:Foundations on Difficult Soils"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.1:Pg-653"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.1\n",
-      "\n",
-      "Sw=1;\n",
-      "Z=2; # in m\n",
-      "deltaSf=0.0033*Z*Sw*1000; # in mm\n",
-      "print deltaSf,\"free surface swell in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6.6 free surface swell in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.2:Pg-13.2"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.2\n",
-      "\n",
-      "#from figure 13.11\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "import numpy\n",
-      "deltaS=1/100.0*1/2.0*(0.55+0+0.55+1.2+1.2+2+2+3);\n",
-      "print deltaS*1000,\"total swell in mm\"\n",
-      "#partb\n",
-      "D=numpy.array([5.2, 4.2, 3.2, 2.2, 1.2]);\n",
-      "deltaS=numpy.array([0, 0.00275, 0.0115, 0.0275, 0.0525]);\n",
-      "print \"depth(m)\\t  total swell (m) \\n\"\n",
-      "for i in range (0,5):\n",
-      "    print D[i],\"\\t            \",deltaS[i],\" \\n\",\n",
-      "\n",
-      "plt.plot(deltaS*1000,D);\n",
-      "plt.title(\"depth vs total swell\")\n",
-      "plt.xlabel(\"total swell (m)\")\n",
-      "plt.ylabel(\"depth (m)\")\n",
-      "plt.show()\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "52.5 total swell in mm\n",
-        "depth(m)\t  total swell (m) \n",
-        "\n",
-        "5.2 \t             0.0  \n",
-        "4.2 \t             0.00275  \n",
-        "3.2 \t             0.0115  \n",
-        "2.2 \t             0.0275  \n",
-        "1.2 \t             0.0525  \n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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YNgxuuCHNT/jTnyodjZlZPpwQyjR8OFx3HRx8MPzlL5WOxsys+zkhdMC++8I118Do0fDI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\nNwAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x1d192b0>"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.3:Pg-664"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.3\n",
-      "\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "phi=12*math.pi/180;\n",
-      "Ds=0.8; # in m\n",
-      "Z=5; # in m \n",
-      "sigmaT=450;\n",
-      "U=math.pi*Ds*Z*sigmaT*math.tan(phi); # in kN\n",
-      "def f(D):\n",
-      "    return 1202-450*6.14/1.25*3.14/4*(D**2-0.8**2)\n",
-      "[x]=fsolve(f,1);\n",
-      "Db=x; # in m\n",
-      "print round(Db,2),\"diameter of bell in m\"\n",
-      "#partb\n",
-      "D=600; # in kN\n",
-      "cu=450; # in kN/m^2\n",
-      "Nc=6.14;\n",
-      "FS=cu*Nc*math.pi/4*(Db**2-Ds**2)/(U-D);\n",
-      "if FS>2 :\n",
-      "    print \"the structure is compatible with safety measures\"\n",
-      "\n",
-      "#check bearing capacity\n",
-      "L=D+300;#dead+live load in kN\n",
-      "Dp=L/math.pi*4/Db**2;#downward pressure\n",
-      "FS=2763/Dp; # factor of safety\n",
-      "if FS>3:\n",
-      "    print \"the structure is safe in bearing \"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.15 diameter of bell in m\n",
-        "the structure is compatible with safety measures\n",
-        "the structure is safe in bearing \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter13_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter13_1.ipynb
deleted file mode 100755
index bfb80602..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter13_1.ipynb
+++ /dev/null
@@ -1,176 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:549425d7a43cc856ebd1610c783821836546fd833bb34512fcafdb71f662b655"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter13:Foundations on Difficult Soils"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.1:Pg-653"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.1\n",
-      "\n",
-      "Sw=1;\n",
-      "Z=2; # in m\n",
-      "deltaSf=0.0033*Z*Sw*1000; # in mm\n",
-      "print deltaSf,\"free surface swell in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6.6 free surface swell in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.2:Pg-13.2"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.2\n",
-      "\n",
-      "#from figure 13.11\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "import numpy\n",
-      "deltaS=1/100.0*1/2.0*(0.55+0+0.55+1.2+1.2+2+2+3);\n",
-      "print deltaS*1000,\"total swell in mm\"\n",
-      "#partb\n",
-      "D=numpy.array([5.2, 4.2, 3.2, 2.2, 1.2]);\n",
-      "deltaS=numpy.array([0, 0.00275, 0.0115, 0.0275, 0.0525]);\n",
-      "print \"depth(m)\\t  total swell (m) \\n\"\n",
-      "for i in range (0,5):\n",
-      "    print D[i],\"\\t            \",deltaS[i],\" \\n\",\n",
-      "\n",
-      "plt.plot(deltaS*1000,D);\n",
-      "plt.title(\"depth vs total swell\")\n",
-      "plt.xlabel(\"total swell (m)\")\n",
-      "plt.ylabel(\"depth (m)\")\n",
-      "plt.show()\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "52.5 total swell in mm\n",
-        "depth(m)\t  total swell (m) \n",
-        "\n",
-        "5.2 \t             0.0  \n",
-        "4.2 \t             0.00275  \n",
-        "3.2 \t             0.0115  \n",
-        "2.2 \t             0.0275  \n",
-        "1.2 \t             0.0525  \n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Sw+23p0lvY8bAgQdCv36Vjs46wwnBzLpswQKYMCElhz//GQ44ICWH4cOhT48s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\nNwAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x1d192b0>"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.3:Pg-664"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.3\n",
-      "\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "phi=12*math.pi/180;\n",
-      "Ds=0.8; # in m\n",
-      "Z=5; # in m \n",
-      "sigmaT=450;\n",
-      "U=math.pi*Ds*Z*sigmaT*math.tan(phi); # in kN\n",
-      "def f(D):\n",
-      "    return 1202-450*6.14/1.25*3.14/4*(D**2-0.8**2)\n",
-      "[x]=fsolve(f,1);\n",
-      "Db=x; # in m\n",
-      "print round(Db,2),\"diameter of bell in m\"\n",
-      "#partb\n",
-      "D=600; # in kN\n",
-      "cu=450; # in kN/m^2\n",
-      "Nc=6.14;\n",
-      "FS=cu*Nc*math.pi/4*(Db**2-Ds**2)/(U-D);\n",
-      "if FS>2 :\n",
-      "    print \"the structure is compatible with safety measures\"\n",
-      "\n",
-      "#check bearing capacity\n",
-      "L=D+300;#dead+live load in kN\n",
-      "Dp=L/math.pi*4/Db**2;#downward pressure\n",
-      "FS=2763/Dp; # factor of safety\n",
-      "if FS>3:\n",
-      "    print \"the structure is safe in bearing \"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.15 diameter of bell in m\n",
-        "the structure is compatible with safety measures\n",
-        "the structure is safe in bearing \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter13_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter13_2.ipynb
deleted file mode 100755
index bfb80602..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter13_2.ipynb
+++ /dev/null
@@ -1,176 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:549425d7a43cc856ebd1610c783821836546fd833bb34512fcafdb71f662b655"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter13:Foundations on Difficult Soils"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.1:Pg-653"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.1\n",
-      "\n",
-      "Sw=1;\n",
-      "Z=2; # in m\n",
-      "deltaSf=0.0033*Z*Sw*1000; # in mm\n",
-      "print deltaSf,\"free surface swell in mm\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6.6 free surface swell in mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.2:Pg-13.2"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.2\n",
-      "\n",
-      "#from figure 13.11\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "import numpy\n",
-      "deltaS=1/100.0*1/2.0*(0.55+0+0.55+1.2+1.2+2+2+3);\n",
-      "print deltaS*1000,\"total swell in mm\"\n",
-      "#partb\n",
-      "D=numpy.array([5.2, 4.2, 3.2, 2.2, 1.2]);\n",
-      "deltaS=numpy.array([0, 0.00275, 0.0115, 0.0275, 0.0525]);\n",
-      "print \"depth(m)\\t  total swell (m) \\n\"\n",
-      "for i in range (0,5):\n",
-      "    print D[i],\"\\t            \",deltaS[i],\" \\n\",\n",
-      "\n",
-      "plt.plot(deltaS*1000,D);\n",
-      "plt.title(\"depth vs total swell\")\n",
-      "plt.xlabel(\"total swell (m)\")\n",
-      "plt.ylabel(\"depth (m)\")\n",
-      "plt.show()\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "52.5 total swell in mm\n",
-        "depth(m)\t  total swell (m) \n",
-        "\n",
-        "5.2 \t             0.0  \n",
-        "4.2 \t             0.00275  \n",
-        "3.2 \t             0.0115  \n",
-        "2.2 \t             0.0275  \n",
-        "1.2 \t             0.0525  \n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Sw+23p0lvY8bAgQdCv36Vjs46wwnBzLpswQKYMCElhz//GQ44ICWH4cOhT48s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\nNwAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x1d192b0>"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex13.3:Pg-664"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 13.3\n",
-      "\n",
-      "import math\n",
-      "from scipy.optimize import fsolve\n",
-      "phi=12*math.pi/180;\n",
-      "Ds=0.8; # in m\n",
-      "Z=5; # in m \n",
-      "sigmaT=450;\n",
-      "U=math.pi*Ds*Z*sigmaT*math.tan(phi); # in kN\n",
-      "def f(D):\n",
-      "    return 1202-450*6.14/1.25*3.14/4*(D**2-0.8**2)\n",
-      "[x]=fsolve(f,1);\n",
-      "Db=x; # in m\n",
-      "print round(Db,2),\"diameter of bell in m\"\n",
-      "#partb\n",
-      "D=600; # in kN\n",
-      "cu=450; # in kN/m^2\n",
-      "Nc=6.14;\n",
-      "FS=cu*Nc*math.pi/4*(Db**2-Ds**2)/(U-D);\n",
-      "if FS>2 :\n",
-      "    print \"the structure is compatible with safety measures\"\n",
-      "\n",
-      "#check bearing capacity\n",
-      "L=D+300;#dead+live load in kN\n",
-      "Dp=L/math.pi*4/Db**2;#downward pressure\n",
-      "FS=2763/Dp; # factor of safety\n",
-      "if FS>3:\n",
-      "    print \"the structure is safe in bearing \"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.15 diameter of bell in m\n",
-        "the structure is compatible with safety measures\n",
-        "the structure is safe in bearing \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter14.ipynb b/Principles_Of_Foundation_Engineering/Chapter14.ipynb
deleted file mode 100755
index 9492b61a..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter14.ipynb
+++ /dev/null
@@ -1,152 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0607b38129f4394e73e1fea85ea07501579e80864ce509d7b62579d182135871"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter14:Soil Improvement and Ground Modification"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.1:Pg-695"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 14.1\n",
-      "\n",
-      "import math\n",
-      "Cc=0.28;\n",
-      "Hc=6.0;\n",
-      "eo=0.9;\n",
-      "Cv=0.36; # in m^2/mo.\n",
-      "H=3.0; # in m\n",
-      "t=9.0; # in mo.\n",
-      "sigmao=210.0;# in kN/m^2\n",
-      "sigmap=115; #deltasigmap in kN/m^2\n",
-      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+sigmap)/sigmao); # in m\n",
-      "print round(Sc*1000,1),\"primary consolidation in mm\"\n",
-      "Tv=Cv*t/H**2;\n",
-      "#from table\n",
-      "k=1.8; #constant\n",
-      "sf=k*sigmap; # in kN/m**2\n",
-      "print round(sf,2),\"deltasigmaf in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "167.7 primary consolidation in mm\n",
-        "207.0 deltasigmaf in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.2:Pg-703"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 14.2\n",
-      "\n",
-      "import math\n",
-      "Tv=0.36;\n",
-      "sigmap=115; # in kN/m^2\n",
-      "Uv=math.sqrt(4*Tv/math.pi)*100;\n",
-      "print round(Uv,2),\"Uv in %\"\n",
-      "#from table \n",
-      "k=0.12; #constant\n",
-      "sf=k*sigmap;\n",
-      "print sf,\"deltasigmaf in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "67.7 Uv in %\n",
-        "13.8 deltasigmaf in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.3:Pg-704"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 14.3\n",
-      "\n",
-      "Cc=0.31;\n",
-      "Hc=15.0;#ft\n",
-      "eo=1.1;\n",
-      "n=10.0;\n",
-      "Uv=0.09;\n",
-      "sigmao=1000.0; # in lb/ft^2\n",
-      "deltasigma=2000.0; # deltasigmap+deltasigmaf\n",
-      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+deltasigma)/sigmao);\n",
-      "print round(Sc,3),\"primary consolidation in ft\"\n",
-      "m=n**2/(n**2-1)*math.log(n)-(3*n**2-1)/4/n**2;\n",
-      "A=2/m;\n",
-      "Ur=(0.096-1/A*(1-math.exp(-A*0.096)))/0.192;\n",
-      "Uvf=1-(1-Ur)*(1-Uv);\n",
-      "Sc30=Sc*Uvf*12; #settlement after 30 days\n",
-      "print round(Sc30,2),\"settlement after 30 days in inch\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.056 primary consolidation in ft\n",
-        "1.48 settlement after 30 days in inch\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter14_1.ipynb b/Principles_Of_Foundation_Engineering/Chapter14_1.ipynb
deleted file mode 100755
index 9492b61a..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter14_1.ipynb
+++ /dev/null
@@ -1,152 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0607b38129f4394e73e1fea85ea07501579e80864ce509d7b62579d182135871"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter14:Soil Improvement and Ground Modification"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.1:Pg-695"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 14.1\n",
-      "\n",
-      "import math\n",
-      "Cc=0.28;\n",
-      "Hc=6.0;\n",
-      "eo=0.9;\n",
-      "Cv=0.36; # in m^2/mo.\n",
-      "H=3.0; # in m\n",
-      "t=9.0; # in mo.\n",
-      "sigmao=210.0;# in kN/m^2\n",
-      "sigmap=115; #deltasigmap in kN/m^2\n",
-      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+sigmap)/sigmao); # in m\n",
-      "print round(Sc*1000,1),\"primary consolidation in mm\"\n",
-      "Tv=Cv*t/H**2;\n",
-      "#from table\n",
-      "k=1.8; #constant\n",
-      "sf=k*sigmap; # in kN/m**2\n",
-      "print round(sf,2),\"deltasigmaf in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "167.7 primary consolidation in mm\n",
-        "207.0 deltasigmaf in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.2:Pg-703"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 14.2\n",
-      "\n",
-      "import math\n",
-      "Tv=0.36;\n",
-      "sigmap=115; # in kN/m^2\n",
-      "Uv=math.sqrt(4*Tv/math.pi)*100;\n",
-      "print round(Uv,2),\"Uv in %\"\n",
-      "#from table \n",
-      "k=0.12; #constant\n",
-      "sf=k*sigmap;\n",
-      "print sf,\"deltasigmaf in kN/m**2\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "67.7 Uv in %\n",
-        "13.8 deltasigmaf in kN/m**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex14.3:Pg-704"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#example 14.3\n",
-      "\n",
-      "Cc=0.31;\n",
-      "Hc=15.0;#ft\n",
-      "eo=1.1;\n",
-      "n=10.0;\n",
-      "Uv=0.09;\n",
-      "sigmao=1000.0; # in lb/ft^2\n",
-      "deltasigma=2000.0; # deltasigmap+deltasigmaf\n",
-      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+deltasigma)/sigmao);\n",
-      "print round(Sc,3),\"primary consolidation in ft\"\n",
-      "m=n**2/(n**2-1)*math.log(n)-(3*n**2-1)/4/n**2;\n",
-      "A=2/m;\n",
-      "Ur=(0.096-1/A*(1-math.exp(-A*0.096)))/0.192;\n",
-      "Uvf=1-(1-Ur)*(1-Uv);\n",
-      "Sc30=Sc*Uvf*12; #settlement after 30 days\n",
-      "print round(Sc30,2),\"settlement after 30 days in inch\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1.056 primary consolidation in ft\n",
-        "1.48 settlement after 30 days in inch\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering/Chapter14_2.ipynb b/Principles_Of_Foundation_Engineering/Chapter14_2.ipynb
deleted file mode 100755
index c829bef5..00000000
--- a/Principles_Of_Foundation_Engineering/Chapter14_2.ipynb
+++ /dev/null
@@ -1,152 +0,0 @@
-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:23456872c8877381c549ede3a12ff29dc232ffad3142abc24c5d2ad8d9888f99"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter14:Soil Improvement and Ground Modification"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.1:Pg-695"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 14.1\n",

-      "\n",

-      "import math\n",

-      "Cc=0.28;\n",

-      "Hc=6.0;\n",

-      "eo=0.9;\n",

-      "Cv=0.36; # in m^2/mo.\n",

-      "H=3.0; # in m\n",

-      "t=9.0; # in mo.\n",

-      "sigmao=210.0;# in kN/m^2\n",

-      "sigmap=115; #deltasigmap in kN/m^2\n",

-      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+sigmap)/sigmao); # in m\n",

-      "print round(Sc*1000,1),\"primary consolidation in mm\"\n",

-      "Tv=Cv*t/H**2;\n",

-      "#from table\n",

-      "k=1.8; #constant\n",

-      "sf=k*sigmap; # in kN/m**2\n",

-      "print round(sf,2),\"deltasigmaf in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "167.7 primary consolidation in mm\n",

-        "207.0 deltasigmaf in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.2:Pg-703"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 14.2\n",

-      "\n",

-      "import math\n",

-      "Tv=0.36;\n",

-      "sigmap=115; # in kN/m^2\n",

-      "Uv=math.sqrt(4*Tv/math.pi)*100;\n",

-      "print round(Uv,2),\"Uv in %\"\n",

-      "#from table \n",

-      "k=0.12; #constant\n",

-      "sf=k*sigmap;\n",

-      "print sf,\"deltasigmaf in kN/m**2\"\n",

-      "\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "67.7 Uv in %\n",

-        "13.8 deltasigmaf in kN/m**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex14.3:Pg-704"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#example 14.3\n",

-      "\n",

-      "Cc=0.31;\n",

-      "Hc=15.0;#ft\n",

-      "eo=1.1;\n",

-      "n=10.0;\n",

-      "Uv=0.09;\n",

-      "sigmao=1000.0; # in lb/ft^2\n",

-      "deltasigma=2000.0; # deltasigmap+deltasigmaf\n",

-      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+deltasigma)/sigmao);\n",

-      "print round(Sc,3),\"primary consolidation in ft\"\n",

-      "m=n**2/(n**2-1)*math.log(n)-(3*n**2-1)/4/n**2;\n",

-      "A=2/m;\n",

-      "Ur=(0.096-1/A*(1-math.exp(-A*0.096)))/0.192;\n",

-      "Uvf=1-(1-Ur)*(1-Uv);\n",

-      "Sc30=Sc*Uvf*12; #settlement after 30 days\n",

-      "print round(Sc30,2),\"settlement after 30 days in inch\"\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "1.056 primary consolidation in ft\n",

-        "1.48 settlement after 30 days in inch\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file
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diff --git a/Principles_Of_Foundation_Engineering/screenshots/ex14.2.png b/Principles_Of_Foundation_Engineering/screenshots/ex14.2.png
deleted file mode 100755
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diff --git a/Principles_Of_Foundation_Engineering/screenshots/ex14.2_2.png b/Principles_Of_Foundation_Engineering/screenshots/ex14.2_2.png
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diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter01_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter01_2.ipynb
new file mode 100755
index 00000000..61dc3457
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter01_2.ipynb
@@ -0,0 +1,249 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3f5444f542f0d6a3b857b61fbecdc4e8047247ed8828d4cea5a34f91ccdf5eb3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1:Geotechnical Properties of Soil"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.1:Pg-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 1.1\n",
+      "\n",
+      "V=0.25; # ft^3\n",
+      "W=30.8; #lb\n",
+      "Wd=28.2; # weight dried lb\n",
+      "Gs=2.7;\n",
+      "Gammaw=62.4; #lb/ft^3\n",
+      "Gamma=W/V;\n",
+      "print Gamma,\" is moist unit weight in lb/ft^3\"\n",
+      "w=(W-Wd)/W;\n",
+      "print round(w*100,2),\"is moisture content in %\"\n",
+      "Gammad=Wd/V;\n",
+      "print Gammad, \"is dry unit weight in lb/ft^3\"\n",
+      "Vs=Wd/Gs/Gammaw;\n",
+      "Vv=V-Vs;\n",
+      "e=Vv/Vs;\n",
+      "print round(e,3),\" is void ratio\"\n",
+      "n=e/(1+e);\n",
+      "print round(n,2),\"is porosity\"\n",
+      "Vw=(W-Wd)/Gammaw;\n",
+      "S=Vw/Vv;\n",
+      "print round(S*100,2),\"is saturation in %\"\n",
+      "\n",
+      "# The answers in the book are different due to approximation while here calculations are precise\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "123.2  is moist unit weight in lb/ft^3\n",
+        "8.44 is moisture content in %\n",
+        "112.8 is dry unit weight in lb/ft^3\n",
+        "0.494  is void ratio\n",
+        "0.33 is porosity\n",
+        "50.43 is saturation in %\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.2:Pg-11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 1.2\n",
+      "\n",
+      "e=0.72;\n",
+      "w=12.0/100; #moisture content\n",
+      "Gs=2.72;\n",
+      "Gammaw=9.81;#kN/m^3\n",
+      "Gammad=Gs*Gammaw/(1+e);\n",
+      "print round(Gammad,2),\"= dry unit weight in kN/m^3\"\n",
+      "Gamma=Gs*Gammaw*(1+w)/(1+e);\n",
+      "print round(Gamma,2),\" = moist unit weight in kN/m^3\"\n",
+      "Gammasat=(Gs+e)*Gammaw/(1+e);\n",
+      "wa=Gammasat-Gamma;#water added\n",
+      "print round(wa,2),\" = water added in kN/m^3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "15.51 = dry unit weight in kN/m^3\n",
+        "17.38  = moist unit weight in kN/m^3\n",
+        "2.24  = water added in kN/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.3:Pg-12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 1.3\n",
+      "from scipy.optimize import fsolve\n",
+      "gmax=17.1; # Gammadmax\n",
+      "Dr=0.7;\n",
+      "w=8.0/100;#moisture content\n",
+      "gmin=14.2; #Gammadmin\n",
+      "def f(x):\n",
+      "    return (x-14.2)/(17.1-14.2)*17.1/x-0.7 \n",
+      "x=fsolve(f,16);#solving for gammad\n",
+      "Gamma=x[0]*(1+w);\n",
+      "print round(Gamma,2),\"moist unit weight in kN/m^3\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "17.4 moist unit weight in kN/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.7:Pg-38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 1.7\n",
+      "import math\n",
+      "#part (a)\n",
+      "e1=0.92;\n",
+      "e2=0.86;\n",
+      "Hc=2.8; # in m\n",
+      "s2=212.0;#sigma2dash Load in kN/m2\n",
+      "s1=140.0;#sigma1dash Load in kN/m2\n",
+      "Cc=(e1-e2)/math.log10(s2/s1);\n",
+      "Sc=Cc*Hc/(1+e1)*math.log10(s2/s1);\n",
+      "print Sc*1000,\"consolidated depth in mm\"\n",
+      "#part (b)\n",
+      "# from Figure (1.21):\n",
+      "Sct=40.0;# in mm\n",
+      "T50=0.197;\n",
+      "t=4.5; # in MIN\n",
+      "Cr=T50*12.7**2.0/t;\n",
+      "U=Sct/Sc*100.0/1000;\n",
+      "H=Hc/2;\n",
+      "Tv=math.pi/4*U**2.0/100**2;\n",
+      "t=Tv*H**2.0/Cr*1000.0**2/60.0/24;\n",
+      "print round(t,1),\" is time required in days\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "87.5 consolidated depth in mm\n",
+        "31.6  is time required in days\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.8:Pg-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 1.8\n",
+      "\n",
+      "Cv=7.061; # in mm^2/min\n",
+      "tc=15*24*60.0;\n",
+      "H=2.8/2*1000.0;\n",
+      "Scmax=87.5; # consolidation\n",
+      "Tc=Cv*tc/H**2;\n",
+      "tv=31.6*24*60;\n",
+      "Tv=Cv*tv/H**2;\n",
+      "#from figure 1.28\n",
+      "Sct=Scmax*0.36;\n",
+      "print Sct,\"is consolidation in 31.6 days in mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "31.5 is consolidation in 31.6 days in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter02_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter02_2.ipynb
new file mode 100755
index 00000000..f284c073
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter02_2.ipynb
@@ -0,0 +1,95 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1d0e8ca0a27c4b0cb17e7318d009efc3241a84a31e3f27c8ca705f6cb5276ce4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 02:Natural Soil Deposits and Subsoil Exploration"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.1:Pg-109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 2.1\n",
+      "import matplotlib.pyplot as plt\n",
+      "import numpy\n",
+      "import math\n",
+      "Distance=[2.5,5,7.5,10,15,20,25,30,35,40,50];\n",
+      "Time=(10**(-3))*numpy.array([11.2,23.3,33.5,42.4,50.9,57.2,64.4,68.6,71.1,72.1,75.5])\n",
+      "#part1\n",
+      "distance=5.25; # in meter\n",
+      "time=23e-3; # in second\n",
+      "v1=distance/time;\n",
+      "print round(v1,2),\"is speed in m/s\"\n",
+      "#part2\n",
+      "distance=11; # in meter\n",
+      "time=13.5e-3;# in second\n",
+      "v2=distance/time;\n",
+      "print round(v2,2),\" is speed in m/s\"\n",
+      "#part3\n",
+      "distance=14.75;# in meter\n",
+      "time=3.5e-3;# in second\n",
+      "v3=distance/time;\n",
+      "print round(v3,2),\"speed in m/s\"\n",
+      "plt.plot(Distance,Time);\n",
+      "plt.title(\"distance vs time\")\n",
+      "plt.xlabel(\"Distance in m\")\n",
+      "plt.ylabel(\"time in s\")\n",
+      "plt.show()\n",
+      "#part4\n",
+      "xc=10.4;\n",
+      "Ta=65e-3;\n",
+      "Z1=1/2.0*math.sqrt((v2-v1)/(v2+v1))*xc;\n",
+      "print round(Z1,2),\" is thickness of layer 1 in m\"\n",
+      "Z2=1/2.0*(Ta-2*Z1*math.sqrt(v3**2-v1**2)/v3/v1)*v3*v2/math.sqrt(v3**2-v2**2);\n",
+      "print round(Z2,2),\" is thickness of layer 2 in m\"\n",
+      "\n",
+      "# the answers are slightly different in textbook due to approximation while here answers are precise"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "228.26 is speed in m/s\n",
+        "814.81  is speed in m/s\n",
+        "4214.29 speed in m/s\n",
+        "3.9"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  is thickness of layer 1 in m\n",
+        "12.82  is thickness of layer 2 in m\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter03_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter03_2.ipynb
new file mode 100755
index 00000000..520b4f21
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter03_2.ipynb
@@ -0,0 +1,412 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e3a75199f67af72d14bee528a629ae06b2506206625e1ef3a86291ef88f556ed"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 03:Shallow Foundations: Ultimate bearing capacity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.1:Pg-130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.1\n",
+      "# From Table 3.1\n",
+      "Nc=17.69;\n",
+      "Nq=7.44;\n",
+      "Ny=3.64;\n",
+      "q=3*115;\n",
+      "Gamma=115.0; #lb/ft**3\n",
+      "c=320;\n",
+      "B=5.0;#ft\n",
+      "FS=4;#factor of safety\n",
+      "qu=1.3*c*Nc+q*Nq+0.4*Gamma*B*Ny\n",
+      "qall=qu/FS; #q allowed\n",
+      "Q=qall*B**2;\n",
+      "print Q,\"is allowable gross load in lb\" \n",
+      "\n",
+      "# the answer is slightly different in textbook due to approximation but here answer are precise"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "67269.0 is allowable gross load in lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.2:Pg-134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.2\n",
+      "\n",
+      "from scipy.optimize import fsolve\n",
+      "import math\n",
+      "Gamma=105.0;#lb/ft**3\n",
+      "Gammasat=118.0;#lb/ft**3\n",
+      "FS=3.0;\n",
+      "pa=2014.125;#lb/ft**2\n",
+      "Depth=[5,10,15,20,25]; # in ft\n",
+      "N60=[4,6,6,10,5]; # in  blow/ft\n",
+      "sigmao=[0,0,0,0,0]; # in  lb/ft^2\n",
+      "phi=[0,0,0,0,0] # in degree\n",
+      "Gammaw=62.4;\n",
+      "s=0;\n",
+      "print \"depth (ft)\\tN60\\t  \\tstress(lb/ft**2)\\t phi(degrees)\\n\"\n",
+      "for i in  range(0,5):\n",
+      "    sigmao[i]=2*Gamma+(Depth[i]-2)*(Gammasat-Gammaw);\n",
+      "    phi[i]=math.sqrt(20*N60[i]*math.sqrt(pa/sigmao[i]))+20;\n",
+      "    print \" \",Depth[i],\"\\t      \",N60[i],\"\\t\\t   \",sigmao[i],\"  \\t \\t  \\t\",round(phi[i],1),\" \\n\"\n",
+      "    s=phi[i]+s\n",
+      "\n",
+      "avgphi=s/(i+1)\n",
+      "\n",
+      "print round(avgphi),\"average friction angle in degrees\"\n",
+      "#using graph get the values of other terms in terms of B and solve for B\n",
+      "def f(x):\n",
+      "    return-150000/x**2+5263.9+5527.1/x+228.3*x\n",
+      "x=fsolve(f,4);\n",
+      "print round(x[0],1),\" is the width in ft\"\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth (ft)\tN60\t  \tstress(lb/ft**2)\t phi(degrees)\n",
+        "\n",
+        "  5 \t       4 \t\t    376.8   \t \t  \t33.6  \n",
+        "\n",
+        "  10 \t       6 \t\t    654.8   \t \t  \t34.5  \n",
+        "\n",
+        "  15 \t       6 \t\t    932.8   \t \t  \t33.3  \n",
+        "\n",
+        "  20 \t       10 \t\t    1210.8   \t \t  \t36.1  \n",
+        "\n",
+        "  25 \t       5 \t\t    1488.8   \t \t  \t30.8  \n",
+        "\n",
+        "34.0 average friction angle in degrees\n",
+        "4.5  is the width in ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.3:Pg-144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.3\n",
+      "\n",
+      "import math\n",
+      "phi=25.0; #degrees\n",
+      "Es=620.0; #kN/m**2\n",
+      "Gamma=18.0;#kN/m**2\n",
+      "Df=0.6;# in m\n",
+      "B=0.6; # in m\n",
+      "L=1.2; # in m\n",
+      "Fqc=0.347;\n",
+      "Nq=10.66;\n",
+      "Nc=20.72;\n",
+      "Ngamma=10.88;\n",
+      "mu=0.3; # Poisson's ratio\n",
+      "Fyd=1.0;\n",
+      "c=48.0;#kN/m**2\n",
+      "q=Gamma*(Df+B/2);\n",
+      "Ir=Es/(2*(1+mu)*(c+q*math.tan(phi*math.pi/180.0)));\n",
+      "print round(Ir,2),\" is value of Ir\"\n",
+      "Fcc=Fqc-(1-Fqc)/(Nq*math.tan(phi*math.pi/180.0));\n",
+      "Fcs=1+Nq/Nc*B/L;\n",
+      "Fqs=1+B/L*math.tan(phi*math.pi/180.0);\n",
+      "Fys=1-0.4*B/L;\n",
+      "Fcd=1+0.4*Df/B;\n",
+      "Fqd=1+2.0*math.tan(phi*math.pi/180.0)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
+      "q1=0.6*18;\n",
+      "Fyc=Fqc;\n",
+      "qu=c*Nc*Fcs*Fcd*Fcc+q1*Nq*Fqs*Fqd*Fqc+1.0/2*Gamma*Ngamma*Fys*Fyd*Fyc;\n",
+      "print round(qu,2),\"is ultimate bearing capacity in kN/m**2\"\n",
+      "\n",
+      "# the answer is slightly different in textbook due to approximation but here answer are precise"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4.29  is value of Ir\n",
+        "469.24 is ultimate bearing capacity in kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.4:Pg-156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.4\n",
+      "import math\n",
+      "q=110*4.0; #lb/ft**2\n",
+      "Nq=33.3;\n",
+      "phi=35.0; # in degree\n",
+      "Df=4.0; # in ft\n",
+      "B=6.0; # in ft\n",
+      "Gamma=110.0;#lb/ft**3\n",
+      "Ngamma=48.03; #lb/ft**3\n",
+      "B1=6-2*0.5; # in ft\n",
+      "Fqi=1;\n",
+      "Fyi=1;\n",
+      "Fyd=1;\n",
+      "Fqs=1;\n",
+      "Fys=1;\n",
+      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
+      "qu=q*Nq*Fqs*Fqd*Fqi+1/2.0*B1*Gamma*Ngamma*Fys*Fyd*Fyi;\n",
+      "Qult=B1*1*qu;\n",
+      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\" \n",
+      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
+      "\n",
+      "# the answer is slightly different in textbook due to approximation but here answer are precise"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "151738.23  is ultimate bearing capacity in lb/ft\n",
+        "75.87  is ultimate bearing capacity in ton/ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.5:Pg-158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.5\n",
+      "\n",
+      "e=0.5; # in ft\n",
+      "B=6; # in ft\n",
+      "k=e/B;\n",
+      "Gamma=110; # in lb/ft^3 \n",
+      "q=440;\n",
+      "print \"get the values of Nqe and Nye from the figure from the value of e/B\"\n",
+      "Nye=26.8;\n",
+      "Nqe=33.4;\n",
+      "Qult=B*1*(q*Nqe+1/2.0*Gamma*B*Nye);\n",
+      "print round(Qult,2),\" is ultimate bearing capacity in lb/ft\"\n",
+      "print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "get the values of Nqe and Nye from the figure from the value of e/B\n",
+        "141240.0  is ultimate bearing capacity in lb/ft\n",
+        "70.62  is ultimate bearing capacity in ton/ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.6:Pg-159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.6\n",
+      "\n",
+      "Df=0.7; # in m\n",
+      "#from table\n",
+      "Nq=18.4;\n",
+      "Ny=22.4;\n",
+      "q=12.6;\n",
+      "phi=30; #angle in degree\n",
+      "L=1.5;# in m\n",
+      "Fyd=1;\n",
+      "Gamma=18; # in KN/m^3\n",
+      "L1=0.85*1.5; # in m\n",
+      "L2=0.21*1.5; # in m\n",
+      "B=1.5; # in m\n",
+      "A=1/2.0*(L1+L2)*B;\n",
+      "B1=A/L1; #B'\n",
+      "Fqs=1+B1/L1*math.tan(phi*math.pi/180);\n",
+      "Fys=1-0.4*B1/L1;\n",
+      "Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180))**2*Df/B;\n",
+      "Qult=A*(q*Nq*Fqs*Fqd+1/2.0*Gamma*B1*Ny*Fys*Fyd);\n",
+      "print round(Qult,2),\" is ultimate load in kN\"\n",
+      "\n",
+      "# the answer is slightly different in textbook due to approximation but here answer are precise"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "605.45  is ultimate load in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.7:Pg-161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.7\n",
+      "\n",
+      "e=0.15; # in m\n",
+      "B=1.5; # in m\n",
+      "Fqs=1.0;\n",
+      "L=1.5;# in m\n",
+      "Gamma=18.0; # in KN/m^3\n",
+      "q=0.7*18;\n",
+      "#from table\n",
+      "Nqe=18.4;\n",
+      "Nye=11.58;\n",
+      "Fys=1+(2*e/B-0.68)*(B/L)+(0.43-3/2.0*e/B)*(B/L)**2;\n",
+      "Qult=B*L*(q*Nqe*Fqs+1/2.0*L*Gamma*Nye*Fys);\n",
+      "print round(Qult,2),\"is ultimate load in kN\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "803.03 is ultimate load in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.8:Pg-163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 3.8\n",
+      "\n",
+      "q=16.0;# in kN/m^2\n",
+      "Nqei=14.2;\n",
+      "Gamma=16.0 # in kN/m^3\n",
+      "B=1.5;# in m\n",
+      "Nyet=20.0;\n",
+      "Qult=B*(Nqei*q+1/2.0*Gamma*B*Nyet);\n",
+      "print round(Qult,2),\" is ultimate load in kN/m\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "700.8  is ultimate load in kN/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter04_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter04_2.ipynb
new file mode 100755
index 00000000..1d75ed9f
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter04_2.ipynb
@@ -0,0 +1,373 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:7e16b5ec29439d0d7eaa54a0826b64a36c559ed79f17d130c42f4478805c682f"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 04:Ultimate Bearing Capacity of Shallow Foundations: Special Cases"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.1:Pg-176"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4.1\n",

+      "\n",

+      "FS=4.0; # FOS\n",

+      "q=110*2.0; # in 1b/ft^2\n",

+      "Nq=90.0;\n",

+      "Ny=50.0;\n",

+      "Gamma=110.0; # in 1b/ft^3\n",

+      "m1=0.34; # From Figure 4.6(a)\n",

+      "B=2.5; # in ft\n",

+      "L=2.5; # in ft\n",

+      "H=1.5; # in ft\n",

+      "phi=35; # in degree\n",

+      "m2=0.45; # From Figure 4.6(b)\n",

+      "Fqs=1-0.34*B/L;\n",

+      "Fys=1-0.45*B/L;\n",

+      "qu=q*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",

+      "Qall=qu*B**2/FS;\n",

+      "print round(Qall,2),\"bearing load in lb\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "26326.95 bearing load in lb\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.2:Pg-177"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4.2\n",

+      "\n",

+      "FS=3.0; # FOS\n",

+      "cu=72.0;\n",

+      "q=18.0; # in kN/m^3\n",

+      "B=1.0;# in m\n",

+      "H=0.25;# in m\n",

+      "qu=5.14*(1+(0.5*B/H-0.707)/5.14)*cu+q;\n",

+      "qall=qu/FS;\n",

+      "print round(qall,1),\"bearing capacity of soil in kN/m**2\" \n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "160.4 bearing capacity of soil in kN/m**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.3:Pg-183"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4.3\n",

+      "import math\n",

+      "k=0; #B/L;\n",

+      "c2=30;\n",

+      "Gamma=17.5; # in kN/m^3\n",

+      "H=1.5; # in m\n",

+      "Df=1.2; # in m\n",

+      "B=2.0; # in m\n",

+      "Ks=2.5;\n",

+      "phi=40; # in degree\n",

+      "pi=math.pi;\n",

+      "qu=(1+0.2*k)*5.14*c2+(1+k)*Gamma*H**2*(1+2*Df/H)*Ks*math.tan(phi*pi/180)/B+Gamma*H;\n",

+      "Qu=qu*B;\n",

+      "print round(Qu,2),\"is bearing capacity in kN/m\"\n",

+      "print \"there is slight variation due to rounding off error\"\n",

+      "#soil 2\n",

+      "Ny=109.4;\n",

+      "Nq=64.2;\n",

+      "Fqs=1;\n",

+      "Fys=1;\n",

+      "qt=Gamma*Df*Nq*Fqs+1/2.0*Gamma*Ny*Fys*B;\n",

+      "print qt,\"bearing capacity in kN/m**2\"\n",

+      "\n",

+      "# answer in book is different due to approximation"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "575.66 is bearing capacity in kN/m\n",

+        "there is slight variation due to rounding off error\n",

+        "3262.7 bearing capacity in kN/m**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.4:Pg-184"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4.4\n",

+      "\n",

+      "B=1.0;  # in m\n",

+      "L=1.5;# in m\n",

+      "c2=48;# in m\n",

+      "ca=108; # in KN/m^2\n",

+      "D=1.0;# in m\n",

+      "H=1.0;# in m\n",

+      "Gamma=16.8; # in KN/m^3\n",

+      "FS=4;\n",

+      "qu=(1+0.2*B/L)*5.14*c2+(1+B/L)*2*ca*H/B+Gamma*D; # in KN/m^2\n",

+      "c1=120.0;\n",

+      "gamma1=16.8;  # in kN/m^3\n",

+      "Df=1.0;\n",

+      "qt=(1+0.2*B/L)*5.14*c1+gamma1*Df;\n",

+      "print qt,\"is qt in kN/m**2\"\n",

+      "print \"no need to calculate qt since it is not useful for calculation\"\n",

+      "print qu/FS,\"is allowable shear stress in kN/m**2\"\n",

+      "print qu/FS*1*1.5,\" is allowable load in kN\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "715.84 is qt in kN/m**2\n",

+        "no need to calculate qt since it is not useful for calculation\n",

+        "164.104 is allowable shear stress in kN/m**2\n",

+        "246.156  is allowable load in kN\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.5:Pg-190"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4.5\n",

+      "\n",

+      "c=50;  # in KN/m^2\n",

+      "#from table\n",

+      "Ncq=6.3;\n",

+      "FS=4.0;# FOS\n",

+      "qu=c*Ncq;  # in KN/m^2\n",

+      "qall=qu/4;\n",

+      "print qall,\"allowed shear stress in kN/m**2\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "78.75 allowed shear stress in kN/m**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.6:Pg-191"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4.6\n",

+      "\n",

+      "Gamma=16.8; # in kN/m^3\n",

+      "B=1.5;#  in m\n",

+      "#from table\n",

+      "Nyq=120.0;\n",

+      "qu=1/2.0*Gamma*B*Nyq; # in KN/m^2\n",

+      "print qu,\" is shear stress in kN/m**2\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "1512.0  shear stress in kN/m**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.7:Pg-198"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 4.7\n",

+      "import math\n",

+      "phi=35; # in degree\n",

+      "Df=1.5; # in m\n",

+      "B=1.5; # in m\n",

+      "Gamma=17.4; # in kN/m^3\n",

+      "A=math.pi/4*Df**2; # in m^2\n",

+      "m=0.25;\n",

+      "Ku=0.936;\n",

+      "Fq=1+2*(1+m*Df/B)*Df/B*Ku*math.tan(phi*math.pi/180);\n",

+      "Qu=Fq*Gamma*A*Df; # in KN/m^2\n",

+      "print round(Qu,1),\" is bearing capacity in kN\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "121.7  is bearing capacity in kN\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.8:pg-198"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "\n",

+      "#example 4.8\n",

+      "\n",

+      "\n",

+      "\n",

+      "cu=52; # in kN/m^2\n",

+      "\n",

+      "B=1.5; # in m\n",

+      "\n",

+      "L=3; # in m\n",

+      "\n",

+      "k=0.107*cu+2.5;\n",

+      "\n",

+      "print round(k,2),\" is Df/B of square\" \n",

+      "\n",

+      "A=L*B; # in m^2\n",

+      "\n",

+      "Beta=0.2;\n",

+      "\n",

+      "Gamma=18.9; # in kN/m^3\n",

+      "\n",

+      "Df=1.8; # in m\n",

+      "\n",

+      "Qu=A*(Beta*(7.56+1.44*B/L)*cu+Gamma*Df); # in kN/m^2\n",

+      "\n",

+      "print round(Qu,1),\" is ultimate shear force in kN\"\n",

+      "\n",

+      " \n",

+      "\n",

+      " "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "8.06  is Df/B of square\n",

+        "540.6  is ultimate shear force in kN\n"

+       ]

+      }

+     ],

+     "prompt_number": 20

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter05_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter05_2.ipynb
new file mode 100755
index 00000000..ca975aae
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter05_2.ipynb
@@ -0,0 +1,550 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:77a2cf465cec464205dc151afe10d9acafa79fe1b46b30f4a468368f8af3f8ea"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter05:Shallow Foundations: Allowable Bearing Capacity and Settlement"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.1:Pg-212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.1\n",
+      "\n",
+      "#first solution\n",
+      "B1=2.5; # in ft\n",
+      "B2=B1;\n",
+      "z=12.5; # in ft\n",
+      "L1=5; # in ft\n",
+      "L2=L1;\n",
+      "m=B1/z;\n",
+      "n=B2/z;\n",
+      "#from table 5.2  of the values using m,n\n",
+      "q=2000; # in lb/ft^2\n",
+      "I=0.0328;\n",
+      "deltasigma=q*4*I; # in lb/ft**2\n",
+      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
+      "#second solution\n",
+      "Ic=0.131;#from table\n",
+      "deltasigma=q*Ic; #  in lb/ft**2\n",
+      "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "262.4 change in pressure in lb/ft**2\n",
+        "262.0 change in pressure in lb/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.2:Pg-215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.2\n",
+      "\n",
+      "qo=100; # in KN/m^2\n",
+      "H1=3; # in m\n",
+      "H2=5; # in m\n",
+      "#from table\n",
+      "IaH2=0.126;\n",
+      "IaH1=0.175;\n",
+      "deltasigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1)); # in kN/m**2\n",
+      "print round(deltasigma,2),\"change in pressure in kN/m**2\"\n",
+      "TS=4*deltasigma; # in kN/m**2\n",
+      "print round(TS,2),\"total change in pressure in kN/m**2\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5.25 change in pressure in kN/m**2\n",
+        "21.0 total change in pressure in kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.3:Pg-217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.3\n",
+      "H=7;\n",
+      "Gamma=17.5; # in KN/m^3\n",
+      "q0=Gamma*H # in KN/m^2\n",
+      "print q0,\" is pressure change in kN/m**2\"\n",
+      "#part2\n",
+      "#from figure\n",
+      "Ileftside=0.445;\n",
+      "Irightside=0.445;\n",
+      "deltasigma=q0*(Ileftside+Irightside); # in KN/m^2\n",
+      "print round(deltasigma,2),\"is change in stress in kN/m**2\"\n",
+      "#partc\n",
+      "#from figure 5.11\n",
+      "I=0.24;#I'\n",
+      "Dsigma1=43.75*I;#deltasigma1 in KN/m^2\n",
+      "I2=0.495;#I'\n",
+      "Dsigma2=I2*q0;#deltasigma2 in KN/m^2\n",
+      "I3=0.335;#I'\n",
+      "Dsigma3=I3*78.75;#deltasigma3 in KN/m^2\n",
+      "Dsigma=Dsigma1+Dsigma2-Dsigma3; # in KN/m^2\n",
+      "print round(Dsigma,2),\"is total stress increase in A in kN/m**2\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "122.5  is pressure change in kN/m**2\n",
+        "109.03 is change in stress in kN/m**2\n",
+        "44.76 is total stress increase in A in kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.4:Pg-228"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.4\n",
+      "\n",
+      "zbar=5;\n",
+      "mus=0.3;\n",
+      "F1=0.641;\n",
+      "F2=0.031;\n",
+      "z1=2.0; # in m\n",
+      "z2=1.0; # in m\n",
+      "z3=2.0; # in m\n",
+      "Es1=10000; # in kN/m**2\n",
+      "Es2=8000; # in kN/m**2\n",
+      "Es3=12000;# in kN/m**2\n",
+      "qo=150; # in KN/m^2\n",
+      "#from table 5.4\n",
+      "If=0.709;\n",
+      "Es=(Es1*z1+Es2*z2+Es3*z3)/zbar; # in kN/m**2\n",
+      "print Es,\"  is modulus of elasticity in kN/m**2\"\n",
+      "Is=F1+(2-mus)/(1-mus)*F2;\n",
+      "Sc=qo*(1.0/Es-mus**2.0/Es)*Is*If*2;\n",
+      "Scrigid=0.93*Sc; # in m\n",
+      "print round(Scrigid*1000,2),\"is settlement in mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10400.0   is modulus of elasticity in kN/m**2\n",
+        "12.4 is settlement in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.5:Pg-234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.5\n",
+      "import math\n",
+      "B=5; # in ft\n",
+      "L=10; # in ft\n",
+      "Ef=2.3e6; # in lb/in^2\n",
+      "Eo=1400.0; # in lb/in^2\n",
+      "k=25.0; # in lb/in^2/ft\n",
+      "t=1.0;\n",
+      "mus=0.3;\n",
+      "Df=5.0; # in ft\n",
+      "qo=5000.0; # in lb/ft^2\n",
+      "Ig=0.69;\n",
+      "Be=math.sqrt(4*B*L/math.pi);\n",
+      "If=math.pi/4+1/(4.6+10*(Ef/(Eo+2*Be/2*k))*(2*t/Be)**3);\n",
+      "Ie=1-1/(3.5*math.exp(1.22*mus-0.4)*(Be/Df+1.6));\n",
+      "Se=qo*Be*Ig*If*Ie/Eo*(1-mus**2)/144; # in ft\n",
+      "print round(Se*12,2),\"settlement in inches\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1.07 settlement in inches\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.6:238"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.6\n",
+      "\n",
+      "import math\n",
+      "import numpy\n",
+      "q=3.06; # in lb/in^2\n",
+      "qbar=25; # in lb/in^2\n",
+      "C1=1-0.5*(q/(qbar-q));\n",
+      "Sum=0;\n",
+      "C2=1+0.2*math.log10(10/0.1);\n",
+      "L=[1, 2, 3, 4, 5];\n",
+      "Dz=[48, 48, 96, 48, 144]; # in inch\n",
+      "Es=[750, 1250, 1250, 1000, 2000]; # in lb/in^2\n",
+      "z=[24, 72, 144, 216, 312]; # in inch\n",
+      "Iz=[0.275, 0.425, 0.417, 0.292, 0.125];\n",
+      "k=numpy.zeros(5)\n",
+      "print \"Layer No.\\t deltaz (in)\\t Es(lb/in**2)\\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \\n\"\n",
+      "for i in  range(0,5):\n",
+      "    k[i]=Iz[i]/Es[i]*Dz[i];\n",
+      "    print L[i],\"\\t \\t \",Dz[i],\"\\t\\t   \",Es[i],\"\\t\\t \",z[i],\" \\t\\t\\t\\t\\t  \",Iz[i],\"\\t\\t  \",round(k[i],3)\n",
+      "    Sum=Sum+k[i];\n",
+      "\n",
+      "Se=C1*C2*(qbar-q)*Sum; # in inch\n",
+      "print round(Se,2),\"settlement in inches\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Layer No.\t deltaz (in)\t Es(lb/in**2)\t z to the middle of the layer (in)  Iz at the middle of the layer  Iz/delta(z) \n",
+        "\n",
+        "1 \t \t  48 \t\t    750 \t\t  24  \t\t\t\t\t   0.275 \t\t   0.018\n",
+        "2 \t \t  48 \t\t    1250 \t\t  72  \t\t\t\t\t   0.425 \t\t   0.016\n",
+        "3 \t \t  96 \t\t    1250 \t\t  144  \t\t\t\t\t   0.417 \t\t   0.032\n",
+        "4 \t \t  48 \t\t    1000 \t\t  216  \t\t\t\t\t   0.292 \t\t   0.014\n",
+        "5 \t \t  144 \t\t    2000 \t\t  312  \t\t\t\t\t   0.125 \t\t   0.009\n",
+        "2.54 settlement in inches\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.7:Pg-244"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.7\n",
+      "\n",
+      "Df=1.0; # in m\n",
+      "B=1.75; # in m\n",
+      "L=1.75 # in m\n",
+      "qnet=120.0; # in KN/m^2\n",
+      "N60=10.0;# in m\n",
+      "alpha1=0.14 # for normally consolated sand\n",
+      "alpha2=1.71/(N60)**1.4 # for normally consolated sand\n",
+      "alpha3=1.0 # for normally consolated sand\n",
+      "Se=0.3*alpha1*alpha2*alpha3*(qnet/100)*((B/0.3)**0.7)*((1.25*(L/B)/(0.25+(L/B))))**2\n",
+      "print round(Se*1000,2),\"settlement in mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "11.79 settlement in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.8:Pg-245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.8\n",
+      "\n",
+      "Df=1; # in m\n",
+      "B=1.75; # in m\n",
+      "qnet=120; # in KN/m^2\n",
+      "N60=10; # in m\n",
+      "Fd=1+0.33*Df/B;\n",
+      "Se=2*qnet/N60/Fd*(B/(B+0.3))**2; # in mm\n",
+      "print round(Se,2),\"settlement in mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "14.71 settlement in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.9:Pg-251"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.9\n",
+      "\n",
+      "Ny=23.76; \n",
+      "Nq=16.51; \n",
+      "q=3*110.0; # in lb/ft^2\n",
+      "Gamma=110.0; # in lb/ft^3\n",
+      "B=4.0; # in ft\n",
+      "Nqe=0.63*Nq;\n",
+      "Nye=0.4*Ny;\n",
+      "que=q*Nqe+1/2.0*Gamma*B*Nye; # in lb/ft^2\n",
+      "print round(que,2),\" is bearing capacity in lb/ft**2\"\n",
+      "#part 2\n",
+      "V=0.4; # in ft/sec\n",
+      "A=0.32; # given in question\n",
+      "g=9.81; # acceleration constant in m/sec^2\n",
+      "kh=0.26;\n",
+      "k=0.92;#tan(alphae)\n",
+      "Seq=0.174*k*V**2/A/g*kh**-4/A**-4; # in m\n",
+      "print round(Seq,3),\"settelement in m\"\n",
+      "print round(Seq*39.57,2),\"settlement in inches\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5523.31  is bearing capacity in lb/ft**2\n",
+        "0.019 settelement in m\n",
+        "0.74 settlement in inches\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.10:Pg-256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.10\n",
+      "\n",
+      "import math\n",
+      "Cc=0.32;\n",
+      "Hc=2.5;\n",
+      "eo=0.8;\n",
+      "sigmao=2.5*16.5+0.5*(17.5-9.81)+1.25*(16-9.81); # in kN/m^2\n",
+      "m1=[2, 2, 2];\n",
+      "z=[2, 3.25, 4.5];\n",
+      "n1=[4, 6.5, 9];\n",
+      "Ic=[0.19, 0.085, 0.045];\n",
+      "Dsigma=[28.5, 12.75, 6.75];#deltasigma\n",
+      "print (\"m1\\t z(m)\\t n1\\t Ic\\t Dsigma \\n\");\n",
+      "for i in range(0,3):\n",
+      "    print round(m1[i],2),\"\\t \",round(z[i],2),\"\\t \",round(n1[i],2),\"\\t \",round(Ic[i],2),\"\\t \",round(Dsigma[i],2)\n",
+      "\n",
+      "    Dsigmaav=1/6.0*(Dsigma[0]+4*Dsigma[1]+Dsigma[2]);\n",
+      "    Sc=Cc*Hc/(1+eo)*math.log10((sigmao+Dsigmaav)/sigmao);\n",
+      "print round(Sc*1000,2),\"settlement in mm\"\n",
+      "#partb\n",
+      "B=1.0; # in m\n",
+      "L=2.0; # in m\n",
+      "z=0.5+1.5; # in m\n",
+      "B=B+z; # in m\n",
+      "L=L+z; # in m\n",
+      "A=0.6; # given in question\n",
+      "#from table\n",
+      "kcr=0.78; # by data\n",
+      "Sep=kcr*Sc;\n",
+      "print round(Sep*1000,2),\"settlement in mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "m1\t z(m)\t n1\t Ic\t Dsigma \n",
+        "\n",
+        "2.0 \t  2.0 \t  4.0 \t  0.19 \t  28.5\n",
+        "2.0 \t  3.25 \t  6.5 \t  0.09 \t  12.75\n",
+        "2.0 \t  4.5 \t  9.0 \t  0.04 \t  6.75\n",
+        "46.45 settlement in mm\n",
+        "36.23 settlement in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.11:Pg-262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 5.11\n",
+      "\n",
+      "import numpy\n",
+      "N60=(3+7+12+12+16)/5.0;\n",
+      "B=[2, 2.25, 2.3]; # in m\n",
+      "Fd=[1.248, 1.22, 1.215];\n",
+      "Qoac=102000*9.81/1000;#actual Qo\n",
+      "Se=25; # in mm\n",
+      "qnet=numpy.zeros(3)\n",
+      "Qo=numpy.zeros(3) # in kN\n",
+      "print \"B(m)\\t Fd\\t qnet(kN/m**2)\\t \\t Qo \\n\"\n",
+      "for i in range(0,3):\n",
+      "    qnet[i]=10/0.08*(B[i]+0.3)**2/(B[i])**2*Fd[i]*Se/25;\n",
+      "    Qo[i]=qnet[i]*B[i]**2;\n",
+      "    print B[i],\"\\t\",Fd[i],\" \\t \",round(qnet[i],2),\"\\t\\t \",Qo[i],\"\\n\"\n",
+      "print int(Qoac),\"value of Qo in kN\"\n",
+      "print \"since Qo is 1000 kN thus B is equal to 2.3 m from the table\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "B(m)\t Fd\t qnet(kN/m**2)\t \t Qo \n",
+        "\n",
+        "2 \t1.248  \t  206.31 \t\t  825.24 \n",
+        "\n",
+        "2.25 \t1.22  \t  195.88 \t\t  991.63125 \n",
+        "\n",
+        "2.3 \t1.215  \t  194.08 \t\t  1026.675 \n",
+        "\n",
+        "1000 value of Qo in kN\n",
+        "since Qo is 1000 kN thus B is equal to 2.3 m from the table\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter06_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter06_2.ipynb
new file mode 100755
index 00000000..d87679c7
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter06_2.ipynb
@@ -0,0 +1,351 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0ea06255a04932e0bf9952689cf2f6d919d320d1161a790bec574191c6fbafa1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter06: Mat Foundations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6.1:Pg-279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 6.1\n",
+      "\n",
+      "B=30; # in ft\n",
+      "L=45; # in ft\n",
+      "Df=6.5; # in ft\n",
+      "cu=1950;# in lb/ft^2\n",
+      "qunet=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B);\n",
+      "print int(qunet),\" is allowed force in lb/ft**2\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "12307  is allowed force in lb/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6.2:Pg-280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 6.2\n",
+      "\n",
+      "N60=10; # penetration number\n",
+      "Df=2; # in m\n",
+      "B=10.0; # in m\n",
+      "Se=25.0; # in mm\n",
+      "qnetall=N60/0.08*(1+0.33*Df/B)*Se/25;\n",
+      "print qnetall,\" is allowed pressure in kN/m**2\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "133.25  is allowed pressure in kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6.3:Pg-283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 6.3\n",
+      "\n",
+      "cu=2800; # in lb/ft^2\n",
+      "B=60; # in ft\n",
+      "L=100; # in ft\n",
+      "Df=5; # in ft\n",
+      "\n",
+      "Gamma=120; # in lb/ft^3\n",
+      "A=60*100; # in ft^2\n",
+      "Q=25e6; # load in Kip\n",
+      "FS=5.14*cu*(1+0.195*B/L)*(1+0.4*Df/B)/(Q/A-Gamma*Df);\n",
+      "print round(FS,2),\" is factor of safety\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4.66  is factor of safety\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6.4:Pg-284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 6.4\n",
+      "\n",
+      "import math\n",
+      "Cc=0.28;\n",
+      "Hc=18*12.0;\n",
+      "e0=0.9;\n",
+      "sigmao=11*100+40*(121.5-64)+18/2*(118-62.4); # in lb/ft^2\n",
+      "H2=5+40+18.0;\n",
+      "H1=5+40.0;\n",
+      "qo=3567.0;\n",
+      "#from table\n",
+      "IaH2=0.21;\n",
+      "IaH1=0.225;\n",
+      "Dsigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1))*4;\n",
+      "Scp=Cc*Hc/(1+e0)*math.log10(sigmao/sigmao+Dsigma/sigmao);\n",
+      "print round(Scp,2),\"is settlement in inches\"\n",
+      "\n",
+      "# The answers in the book are different due to approximation while here calculations are precise"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6.76 is settlement in inches\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6.5:Pg-296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 6.5\n",
+      "import numpy\n",
+      "P=['A','B','C','D','E','F','G','H','I','J','K','L','M','N'];#point\n",
+      "k=1.2*numpy.ones(14);#Q/A\n",
+      "x=[-38,-24, -12, 0, 12, 24, 38, 38, 24, 12, 0, -12, -24, -38];\n",
+      "x1=numpy.zeros(14)\n",
+      "for i in range(0,14):\n",
+      "    x1[i]=0.0017*x[i];\n",
+      "y=[48,48,48,48,48,48,48, -48, -48, -48, -48, -48, -48, -48];\n",
+      "y1=numpy.zeros(14)\n",
+      "for i in range(0,14):\n",
+      "    y1[i]=-0.0011*y[i];\n",
+      "print \"point\\t Q\\A (kip/ft**2)  x(ft)\\t 0.0017x(ft)\\t\\ty(ft)\\t \\t  0.0011y(ft)\\t  \\t  q(kip/ft**2)\\n\"\n",
+      "q=numpy.zeros(14)\n",
+      "for i in  range(0,14):\n",
+      "    q[i]=1.2+x1[i]+y1[i];\n",
+      "    print P[i],\"\\t \",k[i],\"\\t\\t \",x[i],\"\\t\\t\",round(x1[i],3),\"\\t\\t\",y[i],\"\\t \\t \",round(y1[i],3),\"\\t \\t \\t\",round(q[i],3),\"\\t\\t \\n \"\n",
+      "\n",
+      "print \"the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "point\t Q\\A (kip/ft**2)  x(ft)\t 0.0017x(ft)\t\ty(ft)\t \t  0.0011y(ft)\t  \t  q(kip/ft**2)\n",
+        "\n",
+        "A \t  1.2 \t\t  -38 \t\t-0.065 \t\t48 \t \t  -0.053 \t \t \t1.083 \t\t \n",
+        " \n",
+        "B \t  1.2 \t\t  -24 \t\t-0.041 \t\t48 \t \t  -0.053 \t \t \t1.106 \t\t \n",
+        " \n",
+        "C \t  1.2 \t\t  -12 \t\t-0.02 \t\t48 \t \t  -0.053 \t \t \t1.127 \t\t \n",
+        " \n",
+        "D \t  1.2 \t\t  0 \t\t0.0 \t\t48 \t \t  -0.053 \t \t \t1.147 \t\t \n",
+        " \n",
+        "E \t  1.2 \t\t  12 \t\t0.02 \t\t48 \t \t  -0.053 \t \t \t1.168 \t\t \n",
+        " \n",
+        "F \t  1.2 \t\t  24 \t\t0.041 \t\t48 \t \t  -0.053 \t \t \t1.188 \t\t \n",
+        " \n",
+        "G \t  1.2 \t\t  38 \t\t0.065 \t\t48 \t \t  -0.053 \t \t \t1.212 \t\t \n",
+        " \n",
+        "H \t  1.2 \t\t  38 \t\t0.065 \t\t-48 \t \t  0.053 \t \t \t1.317 \t\t \n",
+        " \n",
+        "I \t  1.2 \t\t  24 \t\t0.041 \t\t-48 \t \t  0.053 \t \t \t1.294 \t\t \n",
+        " \n",
+        "J \t  1.2 \t\t  12 \t\t0.02 \t\t-48 \t \t  0.053 \t \t \t1.273 \t\t \n",
+        " \n",
+        "K \t  1.2 \t\t  0 \t\t0.0 \t\t-48 \t \t  0.053 \t \t \t1.253 \t\t \n",
+        " \n",
+        "L \t  1.2 \t\t  -12 \t\t-0.02 \t\t-48 \t \t  0.053 \t \t \t1.232 \t\t \n",
+        " \n",
+        "M \t  1.2 \t\t  -24 \t\t-0.041 \t\t-48 \t \t  0.053 \t \t \t1.212 \t\t \n",
+        " \n",
+        "N \t  1.2 \t\t  -38 \t\t-0.065 \t\t-48 \t \t  0.053 \t \t \t1.188 \t\t \n",
+        " \n",
+        "the soil pressure at all point is less than the given qallnet=1.5 kip/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6.6:Pg-299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 6.6\n",
+      "\n",
+      "from scipy.optimize import fsolve\n",
+      "#solving for d\n",
+      "def f(d):\n",
+      "    return (96+2*d)*d-2615.1\n",
+      "[x]=fsolve(f,19);\n",
+      "d1=x;\n",
+      "def f(d):\n",
+      "      return (96+4*d)*d-6046.4\n",
+      "[x]=fsolve(f,28);\n",
+      "d2=x;\n",
+      "d=max(d2,d1);\n",
+      "d=round(d)\n",
+      "#now coming to design part\n",
+      "h=d+3+1; #in inch\n",
+      "print h,\"is total slab thickness in inches\"\n",
+      "qa=1.082; # in kip/ft^2\n",
+      "qb=1.106; # in kip/ft^2\n",
+      "qm=1.212; # in kip/ft^2\n",
+      "qn=1.188; # in kip/ft^2\n",
+      "q1A=qa/2.0+qb/2.0;\n",
+      "print round(q1A,3),\"is force in strip ABMN in kip/ft**2\"\n",
+      "q2A=qm/2.0+qn/2.0;\n",
+      "print  round(q2A,3),\"is force in strip ABMN in kip/ft**2\"\n",
+      "q1=1.106/3+1.127/3+1.147/3;\n",
+      "print  round(q1,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
+      "q2=1.253/3+1.233/3+1.212/3;\n",
+      "print  round(q2,3),\"is force in strip BCDKLM in kip/ft**2\"\n",
+      "q1=1.147/3+1.167/3+1.188/3;\n",
+      "print  round(q1,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
+      "q2=1.294/3+1.273/3+1.253/3;\n",
+      "print  round(q2,3),\"is force in strip DEFIJK in kip/ft**2\"\n",
+      "q1=1.188/2+1.212/2;\n",
+      "print  round(q1,3),\"is force in strip FGHI in kip/ft**2\"\n",
+      "q2=1.318/2+1.294/2;\n",
+      "print  round(q2,3),\" is force in strip FGHI in kip/ft**2\"\n",
+      "#checking for force\n",
+      "#net soil reaction <load \n",
+      "netforce=1/2.0*(1.094+1.2)*14*96+1/2.0*(1.127+1.233)*24*96+1/2.0*(1.167+1.273)*24*96+1/2.0*(1.2+1.306)*14*96;\n",
+      "if netforce<8761 :\n",
+      "    print \"forces generated is OK\"\n",
+      "\n",
+      "#checking for reinforcement requirement\n",
+      "Q1=1.4*180+1.7*120;\n",
+      "Q2=1.4*360+1.7*200;\n",
+      "Q3=1.4*400+1.7*240;\n",
+      "Q4=1.4*180+1.7*120;\n",
+      "#solving for a\n",
+      "def f(a):\n",
+      "    return 0.9*0.51*a*60*(29-a/2)-95.05*12\n",
+      "[x]=fsolve(f,1.4);\n",
+      "a=x;\n",
+      "As=0.51*a\n",
+      "fy=60000;\n",
+      "print round(As,2),\"is required area in in**2\"\n",
+      "Asmin=200.0/fy*12*29;\n",
+      "print Asmin,\" is minimum reinforcement required in**2/ft\"\n",
+      "print \"use No 9 bars at 10 inch centre to centre\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "33.0 is total slab thickness in inches\n",
+        "1.094 is force in strip ABMN in kip/ft**2\n",
+        "1.2 is force in strip ABMN in kip/ft**2\n",
+        "1.127 is force in strip BCDKLM in kip/ft**2\n",
+        "1.233 is force in strip BCDKLM in kip/ft**2\n",
+        "1.167 is force in strip DEFIJK in kip/ft**2\n",
+        "1.273 is force in strip DEFIJK in kip/ft**2\n",
+        "1.2 is force in strip FGHI in kip/ft**2\n",
+        "1.306  is force in strip FGHI in kip/ft**2\n",
+        "forces generated is OK\n",
+        "0.75 is required area in in**2\n",
+        "1.16  is minimum reinforcement required in**2/ft\n",
+        "use No 9 bars at 10 inch centre to centre\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter07_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter07_2.ipynb
new file mode 100755
index 00000000..63b3bc22
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter07_2.ipynb
@@ -0,0 +1,358 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:6111670c8f96effbf1bc0bc29859353d67e53cd5b906d2b571173d1698f7ba9c"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter07:Lateral earth pressure"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.1:Pg-319"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7.1\n",

+      "\n",

+      "sigmao=48.0; # in KN/m^2\n",

+      "phi1=30*math.pi/180; # angle\n",

+      "phi2=36*math.pi/180; # angle\n",

+      "Ka1=(math.tan(math.pi/4.0-phi1/2))**2;\n",

+      "Ka2=(math.tan(math.pi/4.0-phi2/2))**2;\n",

+      "sigmaa1=Ka1*sigmao; # in KN/m^2\n",

+      "print round(sigmaa1,2),\"top soil pressure in kN/m**2\"\n",

+      "sigmaa2=Ka2*sigmao; # in KN/m^2\n",

+      "print round(sigmaa2,2),\"bottom soil pressure in kN/m**2\"\n",

+      "Po=1/2.0*3*16+3*12.48+1/3.0*3*(19.65-12.48)+1/2.0*3*29.43;\n",

+      "zbar=(24*(3+3/3.0)+37.44*(3/2.0)+10.76*3/3.0+44.1*3/3.0)/Po;\n",

+      "print round(zbar,2),\"resultant force acting from the bottom in m\"\n",

+      "\n",

+      "# The answers in the book are different due to approximation while here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "16.0 top soil pressure in kN/m**2\n",

+        "12.46 bottom soil pressure in kN/m**2\n",

+        "1.84 resultant force acting from the bottom in m\n"

+       ]

+      }

+     ],

+     "prompt_number": 34

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.2:Pg-321"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7.2\n",

+      "\n",

+      "import math\n",

+      "c=14.36;\n",

+      "Gamma=17.4; #  in KN/m^3\n",

+      "H=6; # in m\n",

+      "phi=26*math.pi/180;\n",

+      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

+      "sigma0=Gamma*H*Ka-2*c*math.sqrt(Ka);\n",

+      "Pa=1/2.0*Gamma*H**2*Ka-2*c*H*math.sqrt(Ka);\n",

+      "print round(Pa,2),\"active force before which tensile crack appeared in kN/m\"\n",

+      "zbar=(244.32-323.1)/14.46;\n",

+      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

+      "zc=2*c/Gamma/math.sqrt(Ka);\n",

+      "print round(zc,2),\"distance where tensile crack appeared in m\"\n",

+      "Pa=1/2.0*(H-zc)*(Gamma*H*Ka-2*c*math.sqrt(Ka));\n",

+      "print round(Pa,2),\"Active force in tensile crack in kN/m\"\n",

+      "zbar=(H-zc)/3;\n",

+      "print round(zbar,2),\"the line of action on which net force is acting in m\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "14.62 active force before which tensile crack appeared in kN/m\n",

+        "-5.45 the line of action on which net force is acting in m\n",

+        "2.64 distance where tensile crack appeared in m\n",

+        "38.32 Active force in tensile crack in kN/m\n",

+        "1.12 the line of action on which net force is acting in m\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.3:Pg-322"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7.3\n",

+      "import math\n",

+      "pi=math.pi\n",

+      "H=10.0; # in ft\n",

+      "Gamma=110.0; # in lb/ft^3\n",

+      "phi=35*math.pi/180.0; # angle\n",

+      "alpha=15*math.pi/180.0; # angle\n",

+      "theta=10*math.pi/180.0; # angle\n",

+      "zi=math.sin(math.sin(alpha)/math.sin(phi))-alpha+2*theta;\n",

+      "print round(zi*180.0/math.pi,2),\" is zi in degrees\"\n",

+      "Ka=math.cos(alpha-theta)*math.sqrt(1+(math.sin(phi))**2-2*math.sin(phi)*math.sin(zi))/((math.cos(theta))**2*(math.cos(alpha)+math.sqrt((math.sin(phi))**2+((math.sin(alpha))**2))));\n",

+      "Pa=1/2.0*Gamma*H**2*Ka;\n",

+      "print round(Pa,2),\" is rankine earth pressure in lb/ft\"\n",

+      "print \"there is slight error in answer due to rounding off error\"\n",

+      "Beta=math.tan(math.sin(phi)*math.sin(zi)/(1-math.sin(phi)*math.cos(zi)));\n",

+      "print round(Beta*180/pi,2),\" is angle in degrees\"\n",

+      "\n",

+      "# The answers in the book are different due to approximation while here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "29.99  is zi in degrees\n",

+        "3078.61  is rankine earth pressure in lb/ft\n",

+        "there is slight error in answer due to rounding off error\n",

+        "36.7  is angle in degrees\n"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.4:Pg-326"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7.4\n",

+      "\n",

+      "H=4.6; # in m\n",

+      "Gamma=16.5; # in KN/m^3\n",

+      "Ka=0.297;\n",

+      "Po=1/2.0*Gamma*H**2*Ka;\n",

+      "print round(Po,2),\"coulomb active force per unit length in kN/m\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "51.85 coulomb active force per unit length in kN/m\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.5:Pg-331"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7.5\n",

+      "\n",

+      "#part(a)\n",

+      "Gamma=105; # in lb/ft^3\n",

+      "H=10; #in ft\n",

+      "Kae=0.474;\n",

+      "k1=0;\n",

+      "Pae=1/2.0*Gamma*H**2*Kae*(1-k1) # in lb/ft\n",

+      "print Pae,\"active force in lb/ft\"\n",

+      "#part(b)\n",

+      "Ka=0.246;\n",

+      "Pa=1/2.0*Gamma*H**2*Ka; # in lb/ft\n",

+      "print Pa,\"active force in lb/ft\"\n",

+      "DPae=Pae-Pa;#deltaPae\n",

+      "zbar=(0.6*H*DPae+H/3*(Pa))/Pae;\n",

+      "print round(zbar,2),\"the distance of resultant force from bottom in ft\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "2488.5 active force in lb/ft\n",

+        "1291.5 active force in lb/ft\n",

+        "4.44 the distance of resultant force from bottom in ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.6:Pg-337"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7.6\n",

+      "\n",

+      "import math\n",

+      "import numpy\n",

+      "import matplotlib.pyplot as plt\n",

+      "\n",

+      "z=[0, 4, 8, 12, 16];\n",

+      "Gamma=110; # in lb/ft^3\n",

+      "phi=36*math.pi/180;\n",

+      "H=16; # in ft\n",

+      "Sa1=numpy.zeros(5);#sigma(1)\n",

+      "Sa2=numpy.zeros(5);#sigma(2)\n",

+      "Sztr=numpy.zeros(5);#sigma(z)translation\n",

+      "print \"z(ft)\\t sigma(1)(lb/ft**2)   \\t sigma(2)(lb/ft**2)    \\t sigma(z)translation (lb/ft**2)\\n\"\n",

+      "for i in range(0,5):\n",

+      "    Sa1[i]=Gamma*(math.tan(math.pi/4-phi*z[i]/2/H))**2*(z[i]-phi*z[i]**2/H/math.cos(phi*z[i]/H));\n",

+      "    Sa2[i]=Gamma*z[i]*(math.cos(phi)/(1+math.sin(phi)))**2;\n",

+      "    Sztr[i]=Sa1[i]/2.0+Sa2[i]/2.0;\n",

+      "    print  round(z[i],2),\"\\t \",round(Sa1[i],2),\"\\t\\t\\t \",round(Sa2[i],2),\"\\t\\t\\t \",round(Sztr[i],2),\"\\n\"\n",

+      "plt.plot(Sztr,z);\n",

+      "plt.title(\"sigma(z)translation vs z\")\n",

+      "plt.ylabel(\"z(ft)\")\n",

+      "plt.xlabel(\"sigma(z)translation (lb/ft**2)\")\n",

+      "plt.show()\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "z(ft)\t sigma(1)(lb/ft**2)   \t sigma(2)(lb/ft**2)    \t sigma(z)translation (lb/ft**2)\n",

+        "\n",

+        "0.0 \t  0.0 \t\t\t  0.0 \t\t\t  0.0 \n",

+        "\n",

+        "4.0 \t  269.92 \t\t\t  114.23 \t\t\t  192.07 \n",

+        "\n",

+        "8.0 \t  311.08 \t\t\t  228.46 \t\t\t  269.77 \n",

+        "\n",

+        "12.0 \t  233.53 \t\t\t  342.69 \t\t\t  288.11 \n",

+        "\n",

+        "16.0 \t  102.06 \t\t\t  456.92 \t\t\t  279.49 \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.7:Pg-342"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 7.7\n",

+      "import math\n",

+      "Gammasat=18.86; # in KN/m^3\n",

+      "Gammaw=9.81; # in KN/m^3\n",

+      "phi1=math.pi/180*30;  # angle 1\n",

+      "phi2=math.pi/180*26; # angle 2\n",

+      "Kp1=(math.tan(math.pi/4+phi1/2))**2;\n",

+      "Kp2=(math.tan(math.pi/4+phi2/2))**2;\n",

+      "#for top soil\n",

+      "c=0;\n",

+      "sigma0=31.44; # in KN/m^2\n",

+      "sigmap=sigma0*Kp1+2*c*math.sqrt(Kp1);\n",

+      "print round(sigmap,2),\"passive pressure for top layer in kN/m**2\"\n",

+      "#for z=2\n",

+      "c=10;\n",

+      "sigma0=31.44; # in KN/m^2\n",

+      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2);\n",

+      "print round(sigmap,2),\"passive pressure for z=2m in kN/m**2\"\n",

+      "#for z=3\n",

+      "c=10;\n",

+      "sigma0=15.72*2+(Gammasat-Gammaw)*1; # in KN/m^2\n",

+      "sigmap=sigma0*Kp2+2*c*math.sqrt(Kp2); # in KN/m^2\n",

+      "print round(sigmap,2),\" is passive pressure for z=3m in kN/m**2\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "94.32 passive pressure for top layer in kN/m**2\n",

+        "112.53 passive pressure for z=2m in kN/m**2\n",

+        "135.7  is passive pressure for z=3m in kN/m**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter08_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter08_2.ipynb
new file mode 100755
index 00000000..de2367cd
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter08_2.ipynb
@@ -0,0 +1,527 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:ae926d3fc40f9f56d82ee333a9b5a7f719ff4a11c08f89381773660beb02ead3"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter08: Retaining walls"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.1:Pg-366"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8.1\n",

+      "\n",

+      "import math\n",

+      "import numpy\n",

+      "H1=6*math.tan(10*math.pi/180.0); # in ft\n",

+      "H2=18.0; # in ft\n",

+      "H3=2.75; # in ft\n",

+      "Gamma1=117.0;  # in lb/ft^3\n",

+      "Ka=0.294;#from table 7.1\n",

+      "H=H1+H2+H3; # in ft\n",

+      "Pa=1/2.0*Gamma1*H**2*Ka/1000; # in Kip/ft\n",

+      "Pr=Pa*math.sin(10*math.pi/180); # in Kip/ft\n",

+      "Ph=Pa*math.cos(10*math.pi/180); # in Kip/ft\n",

+      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

+      "Sum=0;#sigma Mr\n",

+      "S=[1, 2, 3, 4, 5];#section\n",

+      "W=[4.05, 1.35, 5.156, 13.01, 1.42];#weight\n",

+      "MA=[5.75, 4.67, 6.25, 9.5, 12.5, 12.5];#Moment Arm\n",

+      "M=numpy.zeros(5)\n",

+      "print \"Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\\n\"\n",

+      "for i in range(0,5):\n",

+      "    M[i]=W[i]*MA[i];\n",

+      "    Sum=Sum+M[i];\n",

+      "    print round(S[i],2),\"\\t \\t \",round(W[i],2),\"\\t \\t \\t\",round(MA[i],2),\"\\t \\t \\t \",round(M[i],2),\"\\n\"\n",

+      "\n",

+      "FSO=Sum/Mo;\n",

+      "if FSO>2 :\n",

+      "    print  \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

+      "\n",

+      "#for sliding\n",

+      "phi2=18*math.pi/180; # the given angle\n",

+      "V=24.986;\n",

+      "B=12.5;\n",

+      "c2=0.9; # in lb/ft^2\n",

+      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

+      "if FSS>2 :\n",

+      "    print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

+      "\n",

+      "#for bearing\n",

+      "e=B/2.0-(Sum-Mo)/V;\n",

+      "qtoe=V/B*(1+6*e/B); # in Kip/ft^2\n",

+      "Nc=13.1;\n",

+      "Nq=5.26;\n",

+      "Ny=4.07;\n",

+      "D=0.107;\n",

+      "Gamma2=4.0; # in lb/ft^3\n",

+      "B1=B-2*e;#Bdash\n",

+      "q=Gamma2*D # in lb/ft^2\n",

+      "Fcd=1+0.4*D/B1;\n",

+      "Fqd=1+2*math.tan(phi2)*(1-math.sin(phi2))**2*(D/B1);\n",

+      "Fyd=1;\n",

+      "zi=math.tan(Ph/V);\n",

+      "Fci=(1-zi/math.pi*2)**2;\n",

+      "Fqi=Fci;\n",

+      "Fyi=round((1-zi/phi2)**2);\n",

+      "qu=c2*Nc*Fcd*Fci+q*Nq*Fqd*Fqi+1/2.0*Gamma2*B1*Fyd*Fyi;\n",

+      "FSB=qu/qtoe;\n",

+      "if FSB>3 :\n",

+      "    print \"bearing  is safe with FS=\",round(FSB),\"\\n\\n\"\n",

+      "    print \"slight changes due to round off error\"\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Section        Weight(kip/ft)    Moment Arm(ft)       Moment (kip-ft/ft)\n",

+        "\n",

+        "1.0 \t \t  4.05 \t \t \t5.75 \t \t \t  23.29 \n",

+        "\n",

+        "2.0 \t \t  1.35 \t \t \t4.67 \t \t \t  6.3 \n",

+        "\n",

+        "3.0 \t \t  5.16 \t \t \t6.25 \t \t \t  32.23 \n",

+        "\n",

+        "4.0 \t \t  13.01 \t \t \t9.5 \t \t \t  123.59 \n",

+        "\n",

+        "5.0 \t \t  1.42 \t \t \t12.5 \t \t \t  17.75 \n",

+        "\n",

+        "safe in overturning with FS= 3.47 \n",

+        "\n",

+        "bearing  is safe with FS= 4.0 \n",

+        "\n",

+        "\n",

+        "slight changes due to round off error\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.2:Pg-369"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8.2\n",

+      "\n",

+      "c=0.9; # # in lb/ft^2\n",

+      "B=12.5; # in ft\n",

+      "Gamma2=4; # in lb/ft^3\n",

+      "Fcd=1.138;\n",

+      "Fqd=1.107; \n",

+      "Nc=7.5;\n",

+      "Nq=4;\n",

+      "Ny=0;\n",

+      "q=0.428; # in lb/ft^2\n",

+      "qtoe=2.44; # in lb/ft^2\n",

+      "qu=c*Nc*Fcd+q*Nq*Fqd+1/2.0*Gamma2*B*Ny;\n",

+      "FSB=qu/qtoe; # factor of safety\n",

+      "if FSB>3.0 :\n",

+      "    print \"safe in bearing with FS=\",round(FSB,2),\"\\n\\n\",\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "safe in bearing with FS= 3.92 \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.3:Pg-370"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8.3\n",

+      "import math\n",

+      "import numpy\n",

+      "Msum=0;#sum of moment\n",

+      "Vsum=0;#sum of force\n",

+      "H=15+2.5;#Hdash  in ft\n",

+      "phi=30*math.pi/180; # given angle in degree\n",

+      "Gamma=121.0; # in lb/ft^3\n",

+      "Ka=(math.tan(math.pi/4-phi/2))**2;\n",

+      "Pa=1/2.0*Gamma*H**2*Ka/1000; # in Kip/ft\n",

+      "Ph=Pa; # in Kip/ft\n",

+      "Mo=Ph*H/3.0; # in Kip-ft/ft\n",

+      "AN=[1.0,2.0,3.0,4.0,5.0,6.0];#area number\n",

+      "W=[0.9,3.375,5.906,3.863,4.764,2.723];#weight\n",

+      "MA=[1.783,2.8,5.3,5.15,7.05,9.55];#moment arm\n",

+      "\n",

+      "print \"AreaNo  \\t Weight(kip/ft) \\t  Moment Arm(ft) \\t Moment (kip-ft/ft)\\n\"\n",

+      "M= numpy.zeros(6)\n",

+      "for i in range(0,6):\n",

+      "    M[i]=W[i]*MA[i];\n",

+      "    Vsum=Vsum+W[i];\n",

+      "    Msum=Msum+M[i];\n",

+      "    print round(AN[i],2),\"\\t\\t \",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

+      "\n",

+      "FSO=(Msum)/Mo;\n",

+      "if FSO>2 :\n",

+      "    print \"safe in overturning with FS=\",round(FSO,2)\n",

+      "\n",

+      "#for sliding\n",

+      "phi2=20*math.pi/180;\n",

+      "V=Vsum\n",

+      "B=10.3; # in ft\n",

+      "c2=1.0; # in lb/ft^2\n",

+      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2)/Ph;\n",

+      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

+      "e=B/2.0-(Msum-Mo)/V;\n",

+      "qtoe=V/B*(1+6*e/B); #  in kip/ft**2\n",

+      "print round(qtoe,2),\" is soil pressure at toe in kip/ft**2\"\n",

+      "qheel=V/B*(1-6*e/B);  # in kip/ft**2\n",

+      "print round(qheel,2),\" is soil pressure at heel in kip/ft**2\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "AreaNo  \t Weight(kip/ft) \t  Moment Arm(ft) \t Moment (kip-ft/ft)\n",

+        "\n",

+        "1.0 \t\t  0.9 \t \t  \t  1.783 \t \t \t   1.6047 \n",

+        "\n",

+        "2.0 \t\t  3.38 \t \t  \t  2.8 \t \t \t   9.45 \n",

+        "\n",

+        "3.0 \t\t  5.91 \t \t  \t  5.3 \t \t \t   31.3018 \n",

+        "\n",

+        "4.0 \t\t  3.86 \t \t  \t  5.15 \t \t \t   19.89445 \n",

+        "\n",

+        "5.0 \t\t  4.76 \t \t  \t  7.05 \t \t \t   33.5862 \n",

+        "\n",

+        "6.0 \t\t  2.72 \t \t  \t  9.55 \t \t \t   26.00465 \n",

+        "\n",

+        "safe in overturning with FS= 3.38\n",

+        "safe in sliding with FS= 1.94 \n",

+        "\n",

+        "3.51  is soil pressure at toe in kip/ft**2\n",

+        "0.67  is soil pressure at heel in kip/ft**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 33

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.4:Pg-372"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8.4\n",

+      "\n",

+      "import math\n",

+      "import numpy\n",

+      "Msum=0;#sum of moment\n",

+      "Vsum=0;#sum of force\n",

+      "H=5+1.5;#Hdash in m\n",

+      "phi=32*math.pi/180 # angle in degree\n",

+      "Gamma=18.5; # in KN/m^3\n",

+      "Ka=0.424;\n",

+      "Pa=1/2.0*Gamma*H**2*Ka; # in KN/m\n",

+      "Ph=Pa*math.cos(15*math.pi/180+2/3.0*phi); # in KN/m\n",

+      "Mo=Ph*H/3.0; # moment\n",

+      "AN=[1,2,3,4,5];#area number\n",

+      "A=[4.36, 3.42, 0.77, 2.8, 2.8];#area\n",

+      "W=[102.81, 80.64, 18.16, 66.02, 93.14];#weight\n",

+      "MA=[2.18, 1.37, 0.98, 1.75, 2.83];#moment arm\n",

+      "print \"AreaNo  \\t  Area(m**2) \\t  Weight(kN/m)  \\t  Moment Arm(m) \\t  Moment (kN-m/m)\\n\"\n",

+      "M= numpy.zeros(5)\n",

+      "for i in range(0,5):\n",

+      "    M[i]=W[i]*MA[i];\n",

+      "    Vsum=Vsum+W[i];\n",

+      "    Msum=Msum+M[i];\n",

+      "    print round(AN[i],2),\"\\t\\t \",round(A[i],2),\" \\t \\t\",round(W[i],2),\"\\t \\t  \\t \",MA[i],\"\\t \\t \\t  \",M[i],\"\\n\"\n",

+      "\n",

+      "FSO=Msum/Mo;\n",

+      "if FSO>2 :\n",

+      "    print \"safe in overturning with FS=\",round(FSO,2),\"\\n\"\n",

+      "\n",

+      "#for sliding\n",

+      "phi2=24*math.pi/180;\n",

+      "V=Vsum\n",

+      "B=3.5; # in m\n",

+      "c2=30; # in KN/m^2\n",

+      "Pp=1/2.0*2.37*18*1.5**2+2*30*1.54*1.5;\n",

+      "FSS=(V*math.tan(2/3.0*phi2)+B*2/3.0*c2+Pp)/Ph;\n",

+      "print \"safe in sliding with FS=\",round(FSS,2),\"\\n\"\n",

+      "print \"if Pp is ignored then FS=1.37\"\n",

+      "e=B/2.0-(Msum-Mo)/V;\n",

+      "qtoe=V/B*(1+6*e/B); # in kN/m**2\n",

+      "print round(qtoe,2),\"soil pressure at toe in kN/m**2\"\n",

+      "qheel=V/B*(1-6*e/B); # in kN/m**2\n",

+      "print round(qheel,2),\"soil pressure at heel in kN/m**2\"\n",

+      "print \"there is difference in answer due to rounding off error\"\n",

+      "\n",

+      "# there is difference in answer due to rounding off error\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "AreaNo  \t  Area(m**2) \t  Weight(kN/m)  \t  Moment Arm(m) \t  Moment (kN-m/m)\n",

+        "\n",

+        "1.0 \t\t  4.36  \t \t102.81 \t \t  \t  2.18 \t \t \t   224.1258 \n",

+        "\n",

+        "2.0 \t\t  3.42  \t \t80.64 \t \t  \t  1.37 \t \t \t   110.4768 \n",

+        "\n",

+        "3.0 \t\t  0.77  \t \t18.16 \t \t  \t  0.98 \t \t \t   17.7968 \n",

+        "\n",

+        "4.0 \t\t  2.8  \t \t66.02 \t \t  \t  1.75 \t \t \t   115.535 \n",

+        "\n",

+        "5.0 \t\t  2.8  \t \t93.14 \t \t  \t  2.83 \t \t \t   263.5862 \n",

+        "\n",

+        "safe in overturning with FS= 2.53 \n",

+        "\n",

+        "safe in sliding with FS= 2.7 \n",

+        "\n",

+        "if Pp is ignored then FS=1.37\n",

+        "195.67 soil pressure at toe in kN/m**2\n",

+        "10.48 soil pressure at heel in kN/m**2\n",

+        "there is difference in answer due to rounding off error\n"

+       ]

+      }

+     ],

+     "prompt_number": 49

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.6:Pg-393"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8.6\n",

+      "import math\n",

+      "Sv=2; # in ft\n",

+      "Sh=3; # in ft\n",

+      "w=3/12.0; # in inch\n",

+      "fy=35000*144;\n",

+      "FSb=3;\n",

+      "pi=math.pi;\n",

+      "phi=36*pi/180;\n",

+      "Gamma1=105; # in lb/ft^3\n",

+      "H=30;\n",

+      "t=Gamma1*H*Sv*Sh*FSb/w/fy*(math.tan(pi/4-phi/2))**2;\n",

+      "t=t*12; #in inch\n",

+      "print round(t,2),\" is thickness in inches\"\n",

+      "t=t+0.001*50;\n",

+      "print \"so take thickness=0.2 inches\"\n",

+      "#for tie length\n",

+      "z=[5,10,15,20,25,30];\n",

+      "TL=[38.45, 35.89, 33.34, 30.79, 28.25, 25.7];#tie length\n",

+      "print \"z(ft)\\t Tie Length (ft)\\n\"\n",

+      "for i in range(0,6):\n",

+      "    print z[i],\"\\t\",TL[i]\n",

+      "\n",

+      "print \"use tie length=40ft\"\n",

+      "#check for over turning\n",

+      "\n",

+      "z=30/3.0;\n",

+      "x1=20;\n",

+      "L=40;\n",

+      "Ka=0.26;\n",

+      "Pa=1/2.0*Gamma1*Ka*H**2; #  in kip/ft**2\n",

+      "W1=Gamma1*H*L;\n",

+      "FSO=W1*x1/(Pa*z);\n",

+      "print round(FSO,2),\" is factor of safety is\" \n",

+      "print \"since FS>3 structure is safe\"\n",

+      "#check for sliding\n",

+      "k=2/3.0;\n",

+      "Pa=12285; #  in kip/ft**2\n",

+      "FSS=W1*math.tan(k*phi)/Pa;\n",

+      "if FSS>3 :\n",

+      "    print \"safe in sliding with FS=\",round(FSS,2)\n",

+      "\n",

+      "#check for bearing\n",

+      "Mr=126000*20; # in lb-ft/ft\n",

+      "Mo=12285*10; # in lb-ft/ft\n",

+      "V=126000;\n",

+      "e=L/2-Mr/V+Mo/V;\n",

+      "L1=L-2*e;#Ldash\n",

+      "c2=1000; # in lb/ft^2\n",

+      "Nc=25.8;\n",

+      "Gamma2=110; # in lb/ft^3\n",

+      "Ny=16.72;\n",

+      "qult=c2*Nc+1/2.0*Gamma2*L1*Ny\n",

+      "sigma0=Gamma1*H; # in lb/ft^2\n",

+      "FSB=qult/sigma0;\n",

+      "if FSB>5 :\n",

+      "    print \"bearing is safe  with FS=\",round(FSB,2)\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "0.14  is thickness in inches\n",

+        "so take thickness=0.2 inches\n",

+        "z(ft)\t Tie Length (ft)\n",

+        "\n",

+        "5 \t38.45\n",

+        "10 \t35.89\n",

+        "15 \t33.34\n",

+        "20 \t30.79\n",

+        "25 \t28.25\n",

+        "30 \t25.7\n",

+        "use tie length=40ft\n",

+        "20.51  is factor of safety is\n",

+        "since FS>3 structure is safe\n",

+        "safe in sliding with FS= 4.57\n",

+        "safe in bearing with FS= 19.87\n"

+       ]

+      }

+     ],

+     "prompt_number": 56

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8.7:Pg-397"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 8.7\n",

+      "import math\n",

+      "import numpy\n",

+      "pi=math.pi;\n",

+      "phi=36.0*pi/180;\n",

+      "Ka=(math.tan(pi/4-phi/2))**2;\n",

+      "z=[8.0, 12.0, 16.0]; # in ft\n",

+      "sigmaG=80*12.0; # in lb/ft^2\n",

+      "Gamma1=110.0; # in lb/ft^3\n",

+      "FS=1.5;\n",

+      "Sv=numpy.zeros(3)\n",

+      "for i in range(0,3):\n",

+      "    Sv[i]=sigmaG/Gamma1/z[i]/Ka/FS*12.0;\n",

+      "    print \"for z=\",z[i],\" ft Sv = \",round(Sv[i],2),\" inches\\n\"\n",

+      "\n",

+      "z=[16.0,56.0,76.0,96.0,112.0,144.0,176.0];\n",

+      "zf=numpy.zeros(7)\n",

+      "k2=numpy.zeros(7)\n",

+      "L=numpy.zeros(7)\n",

+      "for i in range(0,7):\n",

+      "    zf[i]=z[i]/12.0;#z in ft\n",

+      "Sv=[1.67,1.67,1.67,1.67,1.33,1.33,1.33];\n",

+      "k=[7.48,5.78,4.93,4.08,3.4,2.04,0.68];#0.51(H-z)\n",

+      "print \"z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \\n\"\n",

+      "for i in range(0,7):\n",

+      "    k2[i]=0.438*Sv[i];#0.438Sv\n",

+      "    L[i]=k[i]+k2[i];\n",

+      "    print round(z[i],2),\"\\t \\t\",round(zf[i],2),\"\\t \",round(Sv[i],2),\"\\t \\t \",round(k[i],2),\"\\t \\t \\t\",round(k2[i],2),\"\\t \\t \\t \",round(L[i],2)\n",

+      "\n",

+      "Sv=20/12.0;\n",

+      "Ka=0.26;\n",

+      "FS=1.5;\n",

+      "l1=Sv*Ka*FS/4/math.tan(2/3.0*phi);\n",

+      "if l1<3:\n",

+      "    l1=3;\n",

+      "    print l1,\"length in ft\"\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "for z= 8.0  ft Sv =  33.62  inches\n",

+        "\n",

+        "for z= 12.0  ft Sv =  22.41  inches\n",

+        "\n",

+        "for z= 16.0  ft Sv =  16.81  inches\n",

+        "\n",

+        "z(in)        z(ft)        Sv(ft)            0.51(H-z)(ft)         0.438Sv(ft)         L(ft) \n",

+        "\n",

+        "16.0 \t \t1.33 \t  1.67 \t \t  7.48 \t \t \t0.73 \t \t \t  8.21\n",

+        "56.0 \t \t4.67 \t  1.67 \t \t  5.78 \t \t \t0.73 \t \t \t  6.51\n",

+        "76.0 \t \t6.33 \t  1.67 \t \t  4.93 \t \t \t0.73 \t \t \t  5.66\n",

+        "96.0 \t \t8.0 \t  1.67 \t \t  4.08 \t \t \t0.73 \t \t \t  4.81\n",

+        "112.0 \t \t9.33 \t  1.33 \t \t  3.4 \t \t \t0.58 \t \t \t  3.98\n",

+        "144.0 \t \t12.0 \t  1.33 \t \t  2.04 \t \t \t0.58 \t \t \t  2.62\n",

+        "176.0 \t \t14.67 \t  1.33 \t \t  0.68 \t \t \t0.58 \t \t \t  1.26\n",

+        "3 length in ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 79

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter09_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter09_2.ipynb
new file mode 100755
index 00000000..7d8bec64
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter09_2.ipynb
@@ -0,0 +1,520 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:a20958eb0a1cf98629a18a836583d76a2fff705e2f3370294ba37a907bebd17b"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter09:Sheet Pile Walls"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.1:Pg-419"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.1\n",

+      "\n",

+      "import math\n",

+      "from scipy.optimize import fsolve\n",

+      "sall=30;#sigma allowed\n",

+      "pi=math.pi;\n",

+      "zbar=12.1; # in ft\n",

+      "L1=10.0; # in ft\n",

+      "L2=20.0; #in ft\n",

+      "Gamma=0.12; # in lb/ft^3\n",

+      "Gamma1=0.1294-0.0624; # in lb/ft^3\n",

+      "phi=40*pi/180; # angle given\n",

+      "Ka=(math.tan(pi/4-phi/2))**2;\n",

+      "Kp=(math.tan(pi/4+phi/2))**2;\n",

+      "s1=Gamma*L1*Ka;#sigma1 in Kip/ft\n",

+      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in Kip/ft\n",

+      "L3=s2/(Gamma1*(Kp-Ka)); # in ft\n",

+      "print round(L3,2),\"is length in ft\"\n",

+      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;# in Kip/ft\n",

+      "print round(P,2),\" is force in kip/ft\"\n",

+      "s5=Gamma*L1*Kp+Gamma1*L2*Kp+Gamma*L3*(Kp-Ka);#sigma5 in Kip/ft\n",

+      "print round(s5,2),\" is pressure in kip/ft\"\n",

+      "A1=s5/(Gamma1*(Kp-Ka)); # in ft^2\n",

+      "A2=8.0*P/(Gamma1*(Kp-Ka)) # in ft^2\n",

+      "A3=6.0*P*(2.0*zbar*(Gamma1*(Kp-Ka))+s5)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",

+      "A4=P*(6.0*zbar*s5+4.0*P)/(Gamma1*(Kp-Ka))**2.0 # in ft^2\n",

+      "print \"A1,A2,A3,A4 respectively is  \",round(A1,1),round(A2,2),round(A3,2),round(A4,2)\n",

+      "print \"slight error due to rounding off error\"\n",

+      "#part(b)\n",

+      "def f(x):\n",

+      "    return x**4+41.7*x**3-270.5*x**2-13363*x-106863\n",

+      "[x]=fsolve(f,20);\n",

+      "D=1.88+x;\n",

+      "print round(D,2),\" is value of D, in ft\"\n",

+      "TL=L1+L2+1.3*D;\n",

+      "print round(TL,2),\" is total length in ft\"\n",

+      "#partc\n",

+      "z=math.sqrt(2*P/(Gamma1*(Kp-Ka)));#zdash\n",

+      "Mmax=P*(z+zbar)-1/2.0*(Gamma1*(Kp-Ka))*z**2*z/3.0;\n",

+      "S=Mmax*12/sall;\n",

+      "print round(S,2),\"is section modulus in in^3/ft\"\n",

+      "\n",

+      "# The answers in the book are different due to approximation while here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "1.88 is length in ft\n",

+        "9.96  is force in kip/ft\n",

+        "12.67  is pressure in kip/ft\n",

+        "A1,A2,A3,A4 respectively is   43.2 271.33 13708.16 110880.89\n",

+        "slight error due to rounding off error\n",

+        "21.68  is value of D, in ft\n",

+        "58.19  is total length in ft\n",

+        "70.06 is section modulus in in^3/ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.2:Pg-426"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.2\n",

+      "import math\n",

+      "from scipy.optimize import fsolve\n",

+      "sall=172.5*1000;#sigma allowed in KN/m^2\n",

+      "pi=math.pi;\n",

+      "c=47.0;  # in KN/m^2\n",

+      "zbar=1.78; # in m\n",

+      "L1=2.0; #in m\n",

+      "L2=3.0; # in m\n",

+      "Gamma=15.9;  # in KN/m^3\n",

+      "Gamma1=19.33-9.81;  # in KN/m^3\n",

+      "phi=32*pi/180;\n",

+      "Ka=(math.tan(pi/4-phi/2))**2;\n",

+      "Kp=(math.tan(pi/4+phi/2))**2;\n",

+      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",

+      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",

+      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2; # in  kN/ft\n",

+      "print round(P,2),\" is force in kN/m\"\n",

+      "def f(x):\n",

+      "    return 127.4*x**2-104.4*x-357.15\n",

+      "[x]=fsolve(f,2);\n",

+      "D=x;\n",

+      "print round(D,2),\" is value of D in m\"\n",

+      "print round(D*1.5,2),\"actual D in m\"\n",

+      "L4=D*(4*c-(Gamma*L1+Gamma1*L2)-P/D)/4/c;\n",

+      "print round(L4,2),\" is length in m\"\n",

+      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in KN/m^2\n",

+      "s7=4*c+(Gamma*L1+Gamma1*L2);#sigma7 in KN/m^2\n",

+      "z=P/s6;#zdash\n",

+      "Mmax=P*(z+zbar)-1/2.0*s6*z**2; # in KN-m/m\n",

+      "S=Mmax*12.0/sall; #  in m^3/m\n",

+      "print round(S,4),\" is section modulus in m**3/m\"\n",

+      "print \"is slight error due to rounding off error\"\n",

+      "\n",

+      "# The answers in the book are different due to approximation while here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "52.25  is force in kN/m\n",

+        "2.13  is value of D in m\n",

+        "3.2 actual D in m\n",

+        "1.17  is length in m\n",

+        "0.0072  is section modulus in m**3/m\n",

+        "is slight error due to rounding off error\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.3:Pg-433"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.3\n",

+      "import math\n",

+      "from scipy.optimize import fsolve\n",

+      "\n",

+      "pi=math.pi;\n",

+      "zbar=2.23;  # in m\n",

+      "L1=2.0; # in m\n",

+      "L2=3.0; # in m\n",

+      "Gamma=15.9; # in KN/m^3\n",

+      "Gamma1=19.33-9.81; # in KN/m^3\n",

+      "phi=32*pi/180;\n",

+      "Ka=(math.tan(pi/4-phi/2))**2;\n",

+      "Kp=(math.tan(pi/4+phi/2))**2;\n",

+      "s1=Gamma*L1*Ka;#sigma1 in KN/m^2\n",

+      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in KN/m^2\n",

+      "L3=s2/(Gamma1*(Kp-Ka)); # in m\n",

+      "print round(L3,2),\"length in m\"\n",

+      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2+1/2.0*s2*L3;\n",

+      "print round(P,2),\"force in kN/m\"\n",

+      "def f(x):\n",

+      "    return x**3+6.99*x**2-14.55\n",

+      "[x]=fsolve(f,1.4);\n",

+      "D=L3+x;\n",

+      "print round(D,2),\"value of D in m\"\n",

+      "AL=1.4*D;\n",

+      "print round(AL,2),\"actual length in m\"\n",

+      "#partb\n",

+      "L4=1.4;\n",

+      "F=P-1/2.0*(Gamma1*(Kp-Ka)*L4**2);\n",

+      "print round(F,2),\"anchor force in kN/m\"\n",

+      "#partc\n",

+      "def f(x):\n",

+      "    return x**2+6.682*x-14.44\n",

+      "[x]=fsolve(f,1.7);\n",

+      "z=x+2;\n",

+      "Mmax=-1/2.0*s1*L1*(x+2/3.0)+F*(x+1)-s1*x*x/2-1/2.0*Ka*Gamma1*x**3/3.0;\n",

+      "print round(Mmax,2),\" is maximum moment in kN-m/m\" \n",

+      "\n",

+      "# The answers in the book are different due to approximation while here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "0.66 length in m\n",

+        "58.38 force in kN/m\n",

+        "1.98 value of D in m\n",

+        "2.78 actual length in m\n",

+        "30.88 anchor force in kN/m\n",

+        "43.74  is maximum moment in kN-m/m\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.4:Pg-439"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.4\n",

+      "\n",

+      "Gamma=15.9; # in KN/m^3\n",

+      "Gamma1=19.33-9.81; # in KN/m^3\n",

+      "GD=0.23;  # from fig. 9.16\n",

+      "CDL1=1.172; # from fig. 9.19\n",

+      "L1=2; # in m\n",

+      "L2=3; # in m\n",

+      "Dth=(L1+L2)*GD*CDL1;\n",

+      "print round(Dth,2),\"theoritical depth in m\"\n",

+      "Dac=1.4*Dth;\n",

+      "print round(Dac,2),\"actual depth in m\"\n",

+      "print \"approximate it as 2 m\"\n",

+      "#part(b)\n",

+      "CFL1=1.073;\n",

+      "GF=0.07;\n",

+      "Gammaa=(Gamma*L1**2+Gamma1*L2**2+2*Gamma*L1*L2)/(L1+L2)**2; # in KN/m^3\n",

+      "F=Gammaa*(L1+L2)**2*GF*CFL1; # in KN/m\n",

+      "print round(F,2),\"force in kN/m\"\n",

+      "#part(c)\n",

+      "GM=0.021; # from fig. 9.18\n",

+      "CML1=1.036; # from fig. 9.21\n",

+      "Mmax=Gammaa*(L1+L2)**3*GM*CML1; # in kN-m/m\n",

+      "print round(Mmax,2),\"maximum moment in kN-m/m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "1.35 theoritical depth in m\n",

+        "1.89 actual depth in m\n",

+        "approximate it as 2 m\n",

+        "25.54 force in kN/m\n",

+        "36.99 maximum moment in kN-m/m\n"

+       ]

+      }

+     ],

+     "prompt_number": 20

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.5:Pg-442"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.5\n",

+      "import math\n",

+      "import matplotlib.pyplot as plt\n",

+      "import numpy\n",

+      "Mmax=43.72; # in kN-m/m\n",

+      "sp=[\"PSA-31\",\"PSA-23\"];#sheet pile\n",

+      "H=[7.9,7.9] # in m\n",

+      "I=[4.41e-6,5.63e-6]; # in m^4/m\n",

+      "p=[0.00466,0.00365];\n",

+      "S=[10.8e-5,12.8e-5]; # in m^3/m\n",

+      "Md=[18.63,22.08]; # kn-m/m\n",

+      "Logp=numpy.zeros(2)\n",

+      "k=numpy.zeros(2)\n",

+      "print \"SheetPile    I(m**4/m)    H(m)    p\\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \\n\"\n",

+      "for i in range(0,2):\n",

+      "    Logp[i]=math.log10(p[i]);\n",

+      "    k[i]=Md[i]/Mmax;\n",

+      "    print sp[i],\"\\t   \",I[i],\"    \",H[i],\" \",p[i],\"   \",round(Logp[i],2),\"   \",S[i],\"   \",Md[i],\"\\t \",round(k[i],3)\n",

+      "    \n",

+      "\n",

+      "plt.plot(Logp,k);\n",

+      "plt.title(\"Ex9.5\")\n",

+      "plt.xlabel(\"LogP\")\n",

+      "plt.ylabel(\"Md/Mmax\")\n",

+      "plt.show()\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "SheetPile    I(m**4/m)    H(m)    p\t     Logp    S(m**3/m)    Md(kN-m/m)   Md/Mmax \n",

+        "\n",

+        "PSA-31 \t    4.41e-06      7.9   0.00466     -2.33     0.000108     18.63 \t  0.426\n",

+        "PSA-23 \t    5.63e-06      7.9   0.00365     -2.44     0.000128     22.08 \t  0.505\n"

+       ]

+      }

+     ],

+     "prompt_number": 28

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.6:Pg-445"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.6\n",

+      "\n",

+      "import math\n",

+      "from scipy.optimize import fsolve\n",

+      "\n",

+      "pi=math.pi;\n",

+      "R=0.6;\n",

+      "L1=10.0; # in ft\n",

+      "L2=20.0; #in ft\n",

+      "Gammasat=122.4; # in lb/ft^3\n",

+      "l1=5; # in ft\n",

+      "Gamma=110.0; # in lb/ft^3\n",

+      "C=0.68;\n",

+      "L=L1+L2; # in ft\n",

+      "Gammaw=62.4; # in lb/ft^3\n",

+      "Gamma1=Gammasat-Gammaw;#gammadash in  lb/ft^3\n",

+      "Gammaav=(Gamma*L1+Gamma1*L2)/(L1+L2); # in lb/ft^3\n",

+      "phi=35.0*pi/180;\n",

+      "Ka=(math.tan(pi/4-phi/2))**2;\n",

+      "sa=C*Ka*Gammaav*L;#sigmaa in lb/ft^2\n",

+      "sp=R*sa;#sigmap # in lb/ft^2\n",

+      "def f(x):\n",

+      "    return x**2+50*x-1000\n",

+      "[x]=fsolve(f,15);\n",

+      "D=x;#in ft\n",

+      "print round(D,2),\" is depth in ft\"\n",

+      "R=L/D*(L-2*l1)/(2*L+D-2*l1);\n",

+      "print \"value of R=\",round(R,2),\" is OK\\n\"\n",

+      "#partb\n",

+      "F=sa*(L-R*D); # in lb/ft\n",

+      "print round(F,2),\" is Force in lb/ft\"\n",

+      "#partc\n",

+      "Mmax=0.5*sa*L**2*((1-R*D/L)**2-(2*l1/L)*(1-R*D/L)); # in lb-ft/ft\n",

+      "print round(Mmax,2),\"maximum moment lb-ft/ft\"\n",

+      "\n",

+      "# The answers in the book are different due to approximation while here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "15.31  is depth in ft\n",

+        "value of R= 0.6  is OK\n",

+        "\n",

+        "8821.24  is Force in lb/ft\n",

+        "47693.02 maximum moment lb-ft/ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 31

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.7:Pg-451"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.7\n",

+      "import math\n",

+      "from scipy.optimize import fsolve\n",

+      "\n",

+      "pi=math.pi;\n",

+      "zbar=3.2; # in m\n",

+      "c=41; # in KN/m^2\n",

+      "L1=3; # in m\n",

+      "L2=6; # in m\n",

+      "Gamma=17;# in KN/m^3\n",

+      "Gamma1=20-9.81; # in KN/m^3\n",

+      "phi=35*pi/180;\n",

+      "Ka=(math.tan(pi/4-phi/2))**2;\n",

+      "Kp=(math.tan(pi/4+phi/2))**2;\n",

+      "s1=Gamma*L1*Ka;#sigma1 in kN/m**2\n",

+      "s2=Gamma*L1*Ka+Gamma1*L2*Ka;#sigma2 in kN/m**2\n",

+      "P=1/2.0*s1*L1+s1*L2+1/2.0*(s2-s1)*L2;\n",

+      "print round(P,2),\"Force in kN/m\"\n",

+      "s6=4*c-(Gamma*L1+Gamma1*L2);#sigma6 in kN/m**2\n",

+      "print round(s6,2),\"pressure in kN/m**2\"\n",

+      "def f(x):\n",

+      "    return x**2+15*x-25.43\n",

+      "[x]=fsolve(f,1.6);\n",

+      "D=x; # in m\n",

+      "print round(D,1),\"depth in m\"\n",

+      "F=P-s6*D; # in kN/m\n",

+      "print round(F,2),\"force in kN/m\"\n",

+      "print \"slight error due to rounding off\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "153.36 Force in kN/m\n",

+        "51.86 pressure in kN/m**2\n",

+        "1.5 depth in m\n",

+        "73.61 force in kN/m\n",

+        "slight error due to rounding off\n"

+       ]

+      }

+     ],

+     "prompt_number": 35

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.8:pg-458"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 9.8\n",

+      "import math\n",

+      "pi=math.pi;\n",

+      "Gamma=105.0; # in lb/ft^3\n",

+      "Cov=14.0;\n",

+      "B=15/12.0; # in inch\n",

+      "Ka=0.26;\n",

+      "phi=35.0*pi/180; # given angle in degree\n",

+      "H=37.5/12; # in inch\n",

+      "h=15/12.0; # in inch\n",

+      "t=6/12.0; # in inch\n",

+      "Gc=150.0;#gamma concrete in lb/ft^3\n",

+      "W=H*t*Gc; # in lb/ft\n",

+      "k=4.5;#kp*cos(delta)\n",

+      "Pu=1/2.0*Gamma*H**2*(k-Ka*math.cos(phi)); # in lb/ft\n",

+      "print round(Pu,2),\"force in lb/ft\"\n",

+      "Pus=((Cov+1)/(Cov+H/h))*Pu; # in lb/ft\n",

+      "print round(Pus,2),\"force in lb/ft\"\n",

+      "Be=0.227*(H+h)+B;\n",

+      "Pu=Pus*Be; # in lb/ft\n",

+      "print round(Pu,2),\" is resistance of anchor plate in lb/ft\"\n",

+      "\n",

+      "# The answers in the book are different due to approximation while here calculations are precise"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "2197.94 force in lb/ft\n",

+        "1998.12 force in lb/ft\n",

+        "4482.04  is resistance of anchor plate in lb/ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 37

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter10_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter10_2.ipynb
new file mode 100755
index 00000000..3af11efd
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter10_2.ipynb
@@ -0,0 +1,211 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:54ddb8c76d55d78e9f237ddb7f5823ac10a3585adf1b3932a3b8cc9352fd8e76"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10:Braced Cuts"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.1: pg-511"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 10.1 : \n",
+      "\n",
+      "Gamma=18.0;\n",
+      "H=7.0;\n",
+      "sigmaa=0.3*Gamma*H;\n",
+      "print\"maximum pressure intensity in kN/m^2 is\",sigmaa ;\n",
+      "#partb\n",
+      "A=54.02;\n",
+      "B1=1.0/2*1.75*37.8+37.8*1.75-A;\n",
+      "B2=45.2;\n",
+      "C=54.02;\n",
+      "s=3.0; #spacing\n",
+      "Pa=C*s;\n",
+      "print \"strut loads in kN is\",Pa\n",
+      "Pb=(B1+B2)*s;\n",
+      "print \"strut loads in kN is\",Pb\n",
+      "Pc=C*s;\n",
+      "print \"strut loads in kN is\", Pc\n",
+      "#partc\n",
+      "Me=45.2*1.196-37.8*1.196*1.196/2;#Me=Mmax\n",
+      "Sall=170e3;#sigmaall\n",
+      "S=Me/Sall;\n",
+      "print \"section modulus in m^3/m is \",round(S,7)\n",
+      "#partd\n",
+      "Mmax=(B1+B2)*s**2.0/8;\n",
+      "S=Mmax/Sall;\n",
+      "print\"section modulus in m^3/m is\",round(S,6)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum pressure intensity in kN/m^2 is 37.8\n",
+        "strut loads in kN is 162.06\n",
+        "strut loads in kN is 271.215\n",
+        "strut loads in kN is 162.06\n",
+        "section modulus in m^3/m is  0.000159\n",
+        "section modulus in m^3/m is 0.000598\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.2:pg-514"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 10.2\n",
+      "\n",
+      "import math\n",
+      "phi=35.0;\n",
+      "Gamma=17.0; # kN/m^3\n",
+      "s=4; #spacing in m\n",
+      "H=9;\n",
+      "Ka=(math.tan(45*(math.pi/180)-35*(math.pi/(180.0*2))))**2\n",
+      "sigma=0.65*Gamma*Ka*H\n",
+      "A=sigma*5*5.0/6;\n",
+      "B1=sigma*5-A;\n",
+      "C=sigma*4*4/6.0; \n",
+      "B2=sigma*s-C;\n",
+      "Pa=A*s;\n",
+      "Pb=(B1+B2)*s;\n",
+      "Pc=C*s;\n",
+      "print \"strut loads at A in kN is\",round(Pa,2)\n",
+      "\n",
+      "print \"strut loads at B in kN is\",round(Pb,2)\n",
+      "\n",
+      "print \"strut loads at C in kN is\",round(Pc,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "strut loads at A in kN is 449.17\n",
+        "strut loads at B in kN is 233.57\n",
+        "strut loads at C in kN is 287.47\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.3:pg-523"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 10.3\n",
+      "\n",
+      "import math\n",
+      "q=0;\n",
+      "Gamma=17; # in KN/m^3\n",
+      "B=3.0 # in meter\n",
+      "L=20; # in meter\n",
+      "c=30;# in KN/m^3\n",
+      "T=1.5;# in meter\n",
+      "H=5.5;# in meter\n",
+      "B1=B/2;#B'\n",
+      "B11=T*math.sqrt(2);#B''\n",
+      "FS=(5.14*c*(1+B11*0.2/L)+c*H/B1)/(Gamma*H+q) # from equation of factor of safety\n",
+      "print\"factor of safety is\",round(FS,2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "factor of safety is 2.86\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.4:pg-529"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 10.4\n",
+      "\n",
+      "\n",
+      "L1=5; # in meter\n",
+      "L2=4;# in meter\n",
+      "B=5;# in meter\n",
+      "h=4.5;# in meter\n",
+      "icr=1 #\n",
+      "iexit= 0.54*h/L2\n",
+      "FOS=icr/iexit # Factor of safety\n",
+      "print \"The Factory of safety is\",round(FOS,3)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Factory of safety is 1.646\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter11_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter11_2.ipynb
new file mode 100755
index 00000000..0afdb4a0
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter11_2.ipynb
@@ -0,0 +1,650 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a4054b3d6b79658bcd763c43dc94ca6de4cd49698c12f912e4c83a9a61f47154"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter11:Pile Foundations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.1:Pg-532"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.1\n",
+      "import math\n",
+      "\n",
+      "#parta\n",
+      "phi=30.0; # angle in degree\n",
+      "pa=2000.0; \n",
+      "q=100*50/1000.0;\n",
+      "Nq=55.0;\n",
+      "Ap=16*16/16/12; # area in ft^2\n",
+      "Qp=Ap*q*Nq; # in kip\n",
+      "qp=0.4*pa*Nq*math.tan(phi*math.pi/180)*Ap; # in lb\n",
+      "print round(Qp,2),\"ultimate load in lb\"\n",
+      "print round(qp/1000,2),\"ultimate load in kip\"\n",
+      "print \"there is change in answer because of calculation mistake in the book\"\n",
+      "\n",
+      "#partb\n",
+      "Nsigma=36;\n",
+      "Ap=16*16.0/12.0/12;\n",
+      "q=110*50.0/1000;\n",
+      "Qp=Ap*q*Nsigma*((1+2.0*(1-math.sin(phi*math.pi/180)))/3); # in kip\n",
+      "print round(Qp,2),\"ultimate load in kip\"\n",
+      "#partc\n",
+      "Nq=18.4;\n",
+      "Qp=Ap*q*Nq; # in kip\n",
+      "print round(Qp,2),\"ultimate load in kip\"\n",
+      "\n",
+      "# ANSWER IN THE BOOK IS WRONG"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "275.0 ultimate load in lb\n",
+        "25.4 ultimate load in kip\n",
+        "there is change in answer because of calculation mistake in the book\n",
+        "234.67 ultimate load in kip\n",
+        "179.91 ultimate load in kip\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.2:Pg-533"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.2\n",
+      "\n",
+      "import math\n",
+      "#parta\n",
+      "K=1.3;\n",
+      "f0=0;\n",
+      "Delta=0.8*30; # in ft\n",
+      "D=16.0/12; # in ft\n",
+      "L1=50.0;\n",
+      "p=4*16/12.0; # in ft\n",
+      "Gamma=110/1000.0; # in lb/ft^3\n",
+      "L=15*D; # in ft\n",
+      "sigma=Gamma*L; # in kip/ft^2\n",
+      "f20=K*sigma*math.tan(Delta*math.pi/180); # kip/ft^2\n",
+      "Qs=(f0+f20)/2*(p*L)+f20*p*(L1-L);\n",
+      "print round(Qs,2),\"ultimate load in kip\"\n",
+      "#partb\n",
+      "FS=4; # factor of safety\n",
+      "Qp=56.45/3+234.7/3+179.9/3; # in kip\n",
+      "Qu=Qs+Qp; # in kip\n",
+      "Qall=Qu/FS; # in kip\n",
+      "print round(Qall,2),\"is allowed load in kip\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "271.65 ultimate load in kip\n",
+        "107.17 is allowed load in kip\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.3:Pg-534"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.3\n",
+      "import math\n",
+      "K=0.25;\n",
+      "Ap=16*16.0/12/12; # area in ft^2\n",
+      "phi=30*math.pi/180;\n",
+      "Nq=25;\n",
+      "q=110*50.0/1000; # in kip\n",
+      "sigmao=q/2; # in kip/ft^2\n",
+      "p=4*16.0/12; # in ft\n",
+      "L=50; # in ft\n",
+      "FS=4; # factor of safety\n",
+      "Qu=q*Nq*Ap+K*sigmao*math.tan(0.8*phi)*p*L; # in kip\n",
+      "Qall=Qu/FS; # in kip\n",
+      "print round(Qall,1),\"allowed load in kip\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "81.5 allowed load in kip\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.4:Pg-535"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.4\n",
+      "\n",
+      "import math\n",
+      "import numpy\n",
+      "FS=4; # factor of safety\n",
+      "Ap=0.1295; # area in m^2\n",
+      "Nc=9;\n",
+      "cu2=100;\n",
+      "Qp=Ap*Nc*cu2; # in kN\n",
+      "D=[5, 10, 30]; # depth in m\n",
+      "avgD=[2.5, 7.5,20.0]; # average depth in m\n",
+      "sigma=[45.0, 110.5, 228.5]; # in KN/m^2\n",
+      "cu=[30, 30, 100]; # in kN/m^2\n",
+      "alpha=[0.6, 0.9, 0.725];\n",
+      "L=[5, 5, 20]; # in m\n",
+      "p=math.pi*0.406;\n",
+      "Qs=0; # in kN\n",
+      "cusig=numpy.zeros(3)\n",
+      "print round(Qp,2),\"bearing capacity in kN\"\n",
+      "print  \"depth (m)\\t avg Depth(m)\\t avgVerticalStress(kN/m**2)\\t cu(kN/m**2)\\t cu/sigma\\t alpha\\n\"\n",
+      "for i in range(0,3):\n",
+      "    cusig[i]=cu[i]/sigma[i];\n",
+      "    Qs=Qs+alpha[i]*cu[i]*L[i]*p;\n",
+      "    print round(D[i],2),\"\\t \\t \\t\",round(avgD[i],2),\"\\t \\t\",round(sigma[i],2),\"\\t\\t\\t   \",round(cu[i],2),\"\\t  \",round(cusig[i],2),\"\\t\\t \",round(alpha[i],2),\"\\n\"\n",
+      "print round(Qs,2),\"bearing capacity in kN\"\n",
+      "#part2\n",
+      "Lambda=0.136;\n",
+      "L=30;\n",
+      "fav=Lambda*(178.48+2*76.7);\n",
+      "Qs2=p*L*fav; # in kN\n",
+      "#part3\n",
+      "fav1=13;\n",
+      "fav2=31.9;\n",
+      "fav3=93.43;\n",
+      "Qs3=p*(fav1*5+fav2*5+fav3*20); # in kN\n",
+      "print round(Qs3,1),\"bearing capacity in kN\"\n",
+      "Qsavg=Qs/3+Qs2/3+Qs3/3; # in kN\n",
+      "Qu=Qp+Qsavg # in kN\n",
+      "Qall=Qu/FS; # in kN\n",
+      "print round(Qall,1),\"allowed bearing capacity in kN\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "116.55 bearing capacity in kN\n",
+        "depth (m)\t avg Depth(m)\t avgVerticalStress(kN/m**2)\t cu(kN/m**2)\t cu/sigma\t alpha\n",
+        "\n",
+        "5.0 \t \t \t2.5 \t \t45.0 \t\t\t    30.0 \t   0.67 \t\t  0.6 \n",
+        "\n",
+        "10.0 \t \t \t7.5 \t \t110.5 \t\t\t    30.0 \t   0.27 \t\t  0.9 \n",
+        "\n",
+        "30.0 \t \t \t20.0 \t \t228.5 \t\t\t    100.0 \t   0.44 \t\t  0.72 \n",
+        "\n",
+        "2136.44 bearing capacity in kN\n",
+        "2669.7 bearing capacity in kN\n",
+        "573.6 allowed bearing capacity in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.5:Pg-538"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.5\n",
+      "\n",
+      "import numpy\n",
+      "D=[6, 12, 20]; # depth in m\n",
+      "fc=[34.34, 54.94, 70.63]; # in kN/m**2\n",
+      "alpha=[0.84, 0.71, 0.63];\n",
+      "dL=[6, 6, 8]; # in m\n",
+      "p=4*0.305;\n",
+      "Qs=0;\n",
+      "Q=numpy.zeros(3)\n",
+      "print \"  depth(m)\\t fc(kN/m**2)\\t alpha \\t \\t deltaL(m)\\t Q(kN)\\n\"\n",
+      "for i in range (0,3):\n",
+      "    Q[i]=alpha[i]*fc[i]*p*dL[i];\n",
+      "    Qs=Q[i]+Qs;\n",
+      "    print D[i],\"\\t\\t   \",fc[i],\"\\t  \",alpha[i],\"\\t   \",dL[i],\"\\t\\t \",round(Q[i],2)\n",
+      "\n",
+      "print round(Qs),\"bearing force in kN\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  depth(m)\t fc(kN/m**2)\t alpha \t \t deltaL(m)\t Q(kN)\n",
+        "\n",
+        "6 \t\t    34.34 \t   0.84 \t    6 \t\t  211.15\n",
+        "12 \t\t    54.94 \t   0.71 \t    6 \t\t  285.53\n",
+        "20 \t\t    70.63 \t   0.63 \t    8 \t\t  434.29\n",
+        "931.0 bearing force in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5.6:Pg-545"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.6\n",
+      "\n",
+      "import math\n",
+      "L=21; # in m\n",
+      "Qwp=502-350; # in kN\n",
+      "Qws=350; # in kN\n",
+      "Ap=0.1045; # area in m^2\n",
+      "Ep=21e6; # in kN/m^2\n",
+      "epsilon=0.62;\n",
+      "Se1=(Qwp+epsilon*Qws)*L/Ap/Ep; # in m\n",
+      "#part2\n",
+      "Iwp=0.85;\n",
+      "qwp=152/Ap;\n",
+      "Es=25e3; # in kN/m^2\n",
+      "D=0.356; # in m\n",
+      "mus=0.35;\n",
+      "Se2=qwp*D/Es*Iwp*(1-mus**2); # in m\n",
+      "#part3\n",
+      "p=1.168;\n",
+      "Iws=2+0.35*math.sqrt(L/D);\n",
+      "Se3=Qws/p/L*D/Es*Iws*(1-mus**2); # in m\n",
+      "Se=Se1+Se2+Se3; # in m\n",
+      "print round(Se*1000,1),\"settlement in mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "19.8 settlement in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.7:Pg-560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.7\n",
+      "\n",
+      "Ep=207e6; # in kN/m^2\n",
+      "Ip=123e-6; # in m^4\n",
+      "nh=12000; # in  kN/m^3\n",
+      "#from table 11.13\n",
+      "xz=0.008;\n",
+      "Ax=2.435;\n",
+      "T=(Ep*Ip/nh)**0.2;\n",
+      "Qg1=xz*Ep*Ip/Ax/T**3;\n",
+      "#part2\n",
+      "Fy=248000;\n",
+      "d1=0.254;\n",
+      "Am=0.772;\n",
+      "Mzmax=Fy*Ip*2/d1; # in Kn-m\n",
+      "Qg2=Mzmax/Am/T; # in kN\n",
+      "if Qg2>Qg1 :\n",
+      "    Qg=Qg1;\n",
+      "    print round(Qg,2),\"lateral load in kN\"\n",
+      "# there is slight variation in answer in textbook due to approximation"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "53.27 lateral load in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.8:Pg-561"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.8\n",
+      "\n",
+      "import math\n",
+      "#part1\n",
+      "Ep=207e6; # in kN/m^2\n",
+      "Ip=123e-6; # in m^4\n",
+      "nh=12000; # in kN/m^3\n",
+      "#from table 11.1a\n",
+      "xo=0.008; # in m\n",
+      "L=25;\n",
+      "Fy=248000; # yield stress in kN/m^2\n",
+      "D=0.254;\n",
+      "Am=0.772;\n",
+      "Gamma=18.0; # in kN/m^3\n",
+      "phi=35; # in angle\n",
+      "Kp=(math.tan(math.pi/4+phi*math.pi/360))**2;\n",
+      "My=Fy*Ip*2/D; # in kN-m\n",
+      "Qug=140*Kp*D**3*Gamma; # in kN\n",
+      "\n",
+      "#part2\n",
+      "Qg1=xo*(Ep*Ip)**0.6*nh**0.4/0.15/L; # in kN\n",
+      "\n",
+      "if Qug>Qg1:\n",
+      "    Qg=Qg1;\n",
+      "    print round(Qg,2),\"lateral load in kN\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "40.2 lateral load in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.9:Pg-567"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.9\n",
+      "\n",
+      "import math\n",
+      "Wrh=30*12; # in kip-ft\n",
+      "E=0.8;\n",
+      "Wr=7.5; # in kip\n",
+      "S=1/8.0; \n",
+      "C=0.1;\n",
+      "FS=6; # in factor of safety\n",
+      "n=0.4; # Coefficient of restitution\n",
+      "Wp=12/12.0*12/12.0*80*150+550; # in lb\n",
+      "Wp=Wp/1000.0;\n",
+      "Qu=E*Wrh/(S+C)*(Wr+n**2.0*Wp)/(Wr+Wp); # in kip\n",
+      "Qall=Qu/FS; # in kip\n",
+      "print round(Qall),\"allowed bearing capacity in kip\"\n",
+      "#part2\n",
+      "He=30*12.0;\n",
+      "L=80*12.0;\n",
+      "Ap=12*12.0; # area in in^2\n",
+      "Ep=3e6/1000.0; # in kip/in^2\n",
+      "FS=4; # factor of safety\n",
+      "Qu=E*He/(S+math.sqrt(E*He*L/2.0/Ap/Ep)); # in kip\n",
+      "Qall2=Qu/FS; # in kip\n",
+      "print round(Qall2),\"allowed bearing capacity in kip\"\n",
+      "\n",
+      "#partc\n",
+      "a=27;\n",
+      "b=1;\n",
+      "He=30;\n",
+      "FS=3; # factor of safety\n",
+      "Qu=a*math.sqrt(E*He)*(b-math.log10(S)); # in kip\n",
+      "Qall3=Qu/FS; # in kip\n",
+      "print round(Qall3),\"allowed bearing capacity in kip\"\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "101.0 allowed bearing capacity in kip\n",
+        "104.0 allowed bearing capacity in kip\n",
+        "84.0 allowed bearing capacity in kip\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.10:Pg-570"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.10\n",
+      "\n",
+      "Hp=350; # in HP\n",
+      "vp=0.0016; #  in m/s\n",
+      "Sl=0.762e-3; # in m/cycle\n",
+      "f=115; # in Hz\n",
+      "Qu=(0.746*Hp+98*vp)/(vp+Sl*f); # in kN\n",
+      "print round(Qu),\"pile load capacity in kN\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2928.0 pile load capacity in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.11:Pg-578"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.11\n",
+      "\n",
+      "Lg=9.92; # in ft\n",
+      "Bg=7.0; # in ft\n",
+      "n1=3.0;\n",
+      "Nc=8.75;\n",
+      "n2=4.0/1000;\n",
+      "Ap=14.0**2.0/12.0**2;\n",
+      "cup=1775.0;\n",
+      "a1=0.4;#alpha1\n",
+      "p=4*14.0/12.0;\n",
+      "cu1=1050.0; # in lb/ft^2\n",
+      "L1=15.0;\n",
+      "a2=0.54;#alpha2\n",
+      "cu2=1775.0; # in lb/ft^2\n",
+      "L2=45.0;\n",
+      "FS=4; # factor of safety\n",
+      "Qu=n1*n2*(9*Ap*cup+a1*p*cu1*L1+a2*p*cu2*L2); # in kip\n",
+      "Qu2=Lg*Bg*cup*Nc+2*(Lg+Bg)*(cu1*L1+cu2*L2); # in kip\n",
+      "print round(Qu2/1000),\"load in kip\"\n",
+      "Qall=Qu/FS; # in kip\n",
+      "print round(Qall),\"allowed load in kip\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4314.0 load in kip\n",
+        "757.0 allowed load in kip\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11.12:Pg-583"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 11.12\n",
+      "\n",
+      "import math\n",
+      "z1=21/2.0; # in ft\n",
+      "Lg=9.0; # in ft\n",
+      "Bg=6.0;# in ft\n",
+      "Qg=500*1000.0; # in kip\n",
+      "Cc1=0.3;\n",
+      "Cc2=0.2;\n",
+      "Cc3=0.25;\n",
+      "H2=12;\n",
+      "H3=6;\n",
+      "H1=21;\n",
+      "e1=0.82;\n",
+      "e2=0.7;\n",
+      "e3=0.75;\n",
+      "s1=Qg/(Lg+z1)/(Bg+z1); #sigma1 in lb/ft^3\n",
+      "s2=500*1000/(9+27)/(6+27);#sigma2 in lb/ft^3\n",
+      "s3=500*1000/(9+36)/(6+36);#sigma3 in lb/ft^3\n",
+      "ss1=6*105+(27+21/2)*(115-62.4);#sigmadash1 in lb/ft^3\n",
+      "ss2=6*105+(27+21)*(115-62.4)+(120-62.4)*6;#sigmadash2 in lb/ft^3\n",
+      "ss3=6*105+48*(115-62.4)+12*(120-62.4)+3*(122-62.4);#sigmadash3 in lb/ft^3\n",
+      "sc1=Cc1*H1/(1+e1)*math.log10((ss1+s1)/ss1); # in inch\n",
+      "sc2=Cc2*H2/(1+e2)*math.log10((ss2+s2)/ss2); # in inch\n",
+      "sc3=Cc3*H3/(1+e3)*math.log10((ss3+s3)/ss3); # in inch\n",
+      "sc=sc1+sc2+sc3; # in inch\n",
+      "print round(sc*12,1),\"total settlement in inch\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9.6 total settlement in inch\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter12_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter12_2.ipynb
new file mode 100755
index 00000000..fcc14387
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter12_2.ipynb
@@ -0,0 +1,443 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1439e2e46e40a0d47bd1c7e06a03c417c0e1f55ad415a0af82536f86f7ecfea9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter12:Drilled-Shaft Foundations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.1:Pg-609"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.1\n",
+      "\n",
+      "import math\n",
+      "Ap=math.pi/4*1.75**2;  # area in m^2\n",
+      "FS=4; # factor of safety\n",
+      "Nq=37.75; \n",
+      "L=8;\n",
+      "Es=50000.0;\n",
+      "mus=0.265;\n",
+      "pu=100.0;\n",
+      "Db=1.75; # in m\n",
+      "q=6*16.2+2*19.2;\n",
+      "phi=36*math.pi/180.0;\n",
+      "Fqs=1+math.tan(phi);\n",
+      "Fqd=1+2*math.tan(phi)*(1-math.sin(phi))**2*math.atan(L/Db);\n",
+      "Ir=Es/(2*(1+mus)*q*math.tan(phi));\n",
+      "delta=0.005*(1-phi/20*180/math.pi+25/20.0)*q/pu;\n",
+      "Irr=Ir/(1+Ir*delta);\n",
+      "Fqc=math.exp(-3.8*math.tan(phi)+(3.07*math.sin(phi)*math.log10(2*Irr))/(1+math.sin(phi)));\n",
+      "Qp=Ap*(q*(Nq-1)*Fqs*Fqd*Fqc);\n",
+      "Qpall=Qp/FS;\n",
+      "print round(Qpall,2),\"allowed load in kN\"\n",
+      "print \"due to rounding off error there is slight change in answer\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6648.7 allowed load in kN\n",
+        "due to rounding off error there is slight change in answer\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.2:Pg-610"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.2\n",
+      "import math\n",
+      "Ap=math.pi/4*1.75**2; # area in m^2\n",
+      "q=135.6; \n",
+      "w=0.83;\n",
+      "FS=4; # factor of safety\n",
+      "phi=36; # given angle\n",
+      "Nq=0.21*math.exp(0.17*phi);\n",
+      "Qp=Ap*q*(w*Nq-1); # in kN\n",
+      "Qpall=Qp/FS; # in kN\n",
+      "print round(Qpall),\"allowed load in kN\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6383.0 allowed load in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.3:Pg-611"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.3\n",
+      "\n",
+      "import math\n",
+      "Ap=math.pi/4*1.5**2; # area in m^2\n",
+      "Db=1.5; # in m\n",
+      "z=3.0;\n",
+      "p=math.pi*1;\n",
+      "Li=6.0;\n",
+      "N60=30.0;\n",
+      "sigmazi=16*z;\n",
+      "Beta=2.0-0.15*z**0.75;\n",
+      "fi=Beta*sigmazi; # in kN/m^2\n",
+      "qp=57.5*N60; # in kN/m^2\n",
+      "qpr=1.27/Db*qp; # in kN/m^2\n",
+      "Qunet=qpr*Ap+fi*p*Li; # in kN\n",
+      "print round(Qunet,2),\"allowed load in kN\"\n",
+      "#part b\n",
+      "k1=0.315; #from table\n",
+      "k2=12.0/1.5/1000*100.0;\n",
+      "Qunet2=qpr*Ap*k1+fi*p*Li*k2; # in kN\n",
+      "print round(Qunet2,2),\"allowed load in kN\"\n",
+      "\n",
+      "# the answer is slightly different in textbook due to approximation"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4081.11 allowed load in kN\n",
+        "2013.14 allowed load in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.4:Pg-617"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.4\n",
+      "\n",
+      "Nc=9;\n",
+      "Ap=math.pi/4*1.5**2; # area in m^2\n",
+      "cu=105; # in kN/m^2\n",
+      "Qpnet=Ap*cu*Nc; # in kN\n",
+      "print round(Qpnet),\"net ultimate bearing point capacity in kN\"\n",
+      "#part2\n",
+      "alpha=0.4;\n",
+      "Ds=1.5; # in m \n",
+      "p=math.pi*Ds;\n",
+      "Qs=alpha*p*(50*8+105*3); # in kN\n",
+      "print int(Qs),\"skin resistance in kN\"\n",
+      "#part3\n",
+      "FS=3;  # factor of safety\n",
+      "Qu=Qpnet/FS+Qs/FS; # in kN\n",
+      "print round(Qu,2),\"working load in kN\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1670.0 net ultimate bearing point capacity in kN\n",
+        "1347 skin resistance in kN\n",
+        "1005.9 working load in kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.5:Pg-618"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.5\n",
+      "\n",
+      "import math\n",
+      "cub=3000;\n",
+      "L=20+5; # in ft\n",
+      "Db=4; # in ft\n",
+      "Ap=math.pi/4*Db**2; # area in ft^2\n",
+      "alpha=0.55;\n",
+      "cu1=800; # in lb/ft^2\n",
+      "L1=7; # in ft\n",
+      "L2=5.5; # in ft\n",
+      "cu2=1200; # in lb/ft^2\n",
+      "p=math.pi*2.5;\n",
+      "k=alpha*p*(cu1*L1+cu2*L2);#f*p*deltaLi\n",
+      "j1=6*cub*(1+0.2*L/Db);\n",
+      "j2=9*cub;\n",
+      "qp=min(j1,j2);\n",
+      "Qu=k/1000+qp*Ap/1000; # in kip\n",
+      "print round(Qu),\"allowed load in kip\"\n",
+      "#part b\n",
+      "k1=0.57; #from table\n",
+      "k2=0.89;\n",
+      "Qunet2=qp*Ap*k1+k*k2; # in kip\n",
+      "print round(Qunet2/1000,2),\"allowed load in kip\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "392.0 allowed load in kip\n",
+        "240.3 allowed load in kip\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.6:Pg-621"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.6\n",
+      "\n",
+      "import math\n",
+      "Qws=1005-250; # in kN\n",
+      "Qwp=250; # in kN\n",
+      "epsilon=0.65; \n",
+      "L=11; # in m\n",
+      "Ds=1.5; # in m\n",
+      "Es=14000; # in kN/m^2\n",
+      "Ap=math.pi/4*1.5**2; # area in m^2\n",
+      "Ep=21e6; # in kN/m^2\n",
+      "Cp=0.04; # in kN/m^2\n",
+      "Db=1.5;\n",
+      "mus=0.3;\n",
+      "p=math.pi*1.5;\n",
+      "Nc=9;\n",
+      "qp=105*Nc; # in kN/m^2\n",
+      "se1=(Qwp+epsilon*Qws)*L/(Ap*Ep); # in m \n",
+      "se2=Qwp*Cp/(Db*qp); # in m\n",
+      "Iws=2+0.35*math.sqrt(L/Ds);\n",
+      "se3=Qws/p/L*Ds/Es*(1-mus**2)*Iws; # in m\n",
+      "se=se1+se2+se3; # in m\n",
+      "print round(se*1000,2),\" is net settlement in mm\"\n",
+      "\n",
+      "# the answer is slightly different in textbook due to approximation"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "11.46  is net settlement in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.7:Pg-628"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.7\n",
+      "\n",
+      "import math\n",
+      "import numpy\n",
+      "from scipy.optimize import fsolve\n",
+      "Ds=1.0;\n",
+      "Ep=22e6;\n",
+      "Ri=1.0;\n",
+      "cu=100.0;\n",
+      "Ip=math.pi*Ds**4/64.0;\n",
+      "Qc=7.34*Ds**2*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
+      "print round(Qc,2),\"bearing force in kN\"\n",
+      "Mc=3.86*Ds**3*Ep*Ri*(cu/Ep/Ri)**0.6;\n",
+      "print round(Mc,2),\"bearing moment in kNm\"\n",
+      "#from figure\n",
+      "xoQM=0.0046*1;\n",
+      "xoMQ=0.0041*1;\n",
+      "xo=0.5*(xoQM+xoMQ);\n",
+      "print round(xo*1000,2),\"net ground line deflection in mm\"\n",
+      "#partb\n",
+      "Ip=0.049;\n",
+      "Qg=150.0;\n",
+      "Mg=200.0;\n",
+      "def f(T):\n",
+      "    return 338e-6*T**3+300.6e-6*T**2-0.00435\n",
+      "[x]=fsolve(f,2);\n",
+      "T=x;\n",
+      "k=[0, 0.4, 0.6, 0.8, 1.0, 1.1, 1.25];#z/T\n",
+      "Am=[0, 0.36, 0.52, 0.63, 0.75, 0.765, 0.75];\n",
+      "Bm=[1.0, 0.98, 0.95, 0.9, 0.845, 0.8, 0.73];\n",
+      "print \"z/T\\t Am\\t Bm\\t Mz(kN-m)\\n\"\n",
+      "Mz=numpy.zeros(7)\n",
+      "for i in range(0,7):\n",
+      "    Mz[i]=Am[i]*Qg*T+Bm[i]*Mg;\n",
+      "    print k[i],\"\\t\",round(Am[i],2),\"\\t\",round(Bm[i],2),\"\\t\",round(Mz[i],2)\n",
+      "\n",
+      "print round(1*T,2),\"depth in m\"\n",
+      "#partc\n",
+      "Mmax=400;\n",
+      "sigma=Mmax*Ds/2/Ip;\n",
+      "print round(sigma,2),\"tensile stress in kN/m**2\"\n",
+      "#partd\n",
+      "#from figure\n",
+      "k=8.5;\n",
+      "L=k*1;\n",
+      "print L,\"length in m\"\n",
+      "\n",
+      "# the answer is slightly different in textbook due to approximation\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "100615.56 bearing force in kN\n",
+        "52912.27 bearing moment in kNm\n",
+        "4.35 net ground line deflection in mm\n",
+        "z/T\t Am\t Bm\t Mz(kN-m)\n",
+        "\n",
+        "0 \t0.0 \t1.0 \t200.0\n",
+        "0.4 \t0.36 \t0.98 \t308.4\n",
+        "0.6 \t0.52 \t0.95 \t352.35\n",
+        "0.8 \t0.63 \t0.9 \t376.69\n",
+        "1.0 \t0.75 \t0.84 \t403.16\n",
+        "1.1 \t0.77 \t0.8 \t398.84\n",
+        "1.25 \t0.75 \t0.73 \t380.16\n",
+        "2.08 depth in m\n",
+        "4081.63 tensile stress in kN/m**2\n",
+        "8.5 length in m\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12.8:Pg-634"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 12.8\n",
+      "\n",
+      "qu=3000; # lb/in^2\n",
+      "Ds=3*12; #in inch\n",
+      "L=15*12; # in inch\n",
+      "FS=3; # factor of safety\n",
+      "Ecore=0.36e6; # in lb/in^2\n",
+      "f=min(2.5*qu**0.5,0.15*qu);\n",
+      "Qu=math.pi*Ds*L*f/1000; # in kip\n",
+      "Emass=Ecore*(0.266*80-1.66); # in lb/in^2\n",
+      "Ec=17.9*Emass; # in lb/in^2\n",
+      "Ac=math.pi/4*Ds**2; # area in in^2\n",
+      "If=0.35;\n",
+      "se=Qu*L/Ac/Ec+Qu*If/Ds/Emass;\n",
+      "Qall=Qu/FS; # in kip\n",
+      "print round(Qall),\"allowed load in kip\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "929.0 allowed load in kip\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter13_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter13_2.ipynb
new file mode 100755
index 00000000..bfb80602
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter13_2.ipynb
@@ -0,0 +1,176 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:549425d7a43cc856ebd1610c783821836546fd833bb34512fcafdb71f662b655"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter13:Foundations on Difficult Soils"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex13.1:Pg-653"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 13.1\n",
+      "\n",
+      "Sw=1;\n",
+      "Z=2; # in m\n",
+      "deltaSf=0.0033*Z*Sw*1000; # in mm\n",
+      "print deltaSf,\"free surface swell in mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6.6 free surface swell in mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex13.2:Pg-13.2"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 13.2\n",
+      "\n",
+      "#from figure 13.11\n",
+      "import matplotlib.pyplot as plt\n",
+      "%matplotlib inline\n",
+      "import numpy\n",
+      "deltaS=1/100.0*1/2.0*(0.55+0+0.55+1.2+1.2+2+2+3);\n",
+      "print deltaS*1000,\"total swell in mm\"\n",
+      "#partb\n",
+      "D=numpy.array([5.2, 4.2, 3.2, 2.2, 1.2]);\n",
+      "deltaS=numpy.array([0, 0.00275, 0.0115, 0.0275, 0.0525]);\n",
+      "print \"depth(m)\\t  total swell (m) \\n\"\n",
+      "for i in range (0,5):\n",
+      "    print D[i],\"\\t            \",deltaS[i],\" \\n\",\n",
+      "\n",
+      "plt.plot(deltaS*1000,D);\n",
+      "plt.title(\"depth vs total swell\")\n",
+      "plt.xlabel(\"total swell (m)\")\n",
+      "plt.ylabel(\"depth (m)\")\n",
+      "plt.show()\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "52.5 total swell in mm\n",
+        "depth(m)\t  total swell (m) \n",
+        "\n",
+        "5.2 \t             0.0  \n",
+        "4.2 \t             0.00275  \n",
+        "3.2 \t             0.0115  \n",
+        "2.2 \t             0.0275  \n",
+        "1.2 \t             0.0525  \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Sw+23p0lvY8bAgQdCv36Vjs46wwnBzLpswQKYMCElhz//GQ44ICWH4cOhT48s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\nNwAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x1d192b0>"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex13.3:Pg-664"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 13.3\n",
+      "\n",
+      "import math\n",
+      "from scipy.optimize import fsolve\n",
+      "phi=12*math.pi/180;\n",
+      "Ds=0.8; # in m\n",
+      "Z=5; # in m \n",
+      "sigmaT=450;\n",
+      "U=math.pi*Ds*Z*sigmaT*math.tan(phi); # in kN\n",
+      "def f(D):\n",
+      "    return 1202-450*6.14/1.25*3.14/4*(D**2-0.8**2)\n",
+      "[x]=fsolve(f,1);\n",
+      "Db=x; # in m\n",
+      "print round(Db,2),\"diameter of bell in m\"\n",
+      "#partb\n",
+      "D=600; # in kN\n",
+      "cu=450; # in kN/m^2\n",
+      "Nc=6.14;\n",
+      "FS=cu*Nc*math.pi/4*(Db**2-Ds**2)/(U-D);\n",
+      "if FS>2 :\n",
+      "    print \"the structure is compatible with safety measures\"\n",
+      "\n",
+      "#check bearing capacity\n",
+      "L=D+300;#dead+live load in kN\n",
+      "Dp=L/math.pi*4/Db**2;#downward pressure\n",
+      "FS=2763/Dp; # factor of safety\n",
+      "if FS>3:\n",
+      "    print \"the structure is safe in bearing \"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1.15 diameter of bell in m\n",
+        "the structure is compatible with safety measures\n",
+        "the structure is safe in bearing \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter14_2.ipynb b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter14_2.ipynb
new file mode 100755
index 00000000..c829bef5
--- /dev/null
+++ b/Principles_Of_Foundation_Engineering_by_B_M_Das/Chapter14_2.ipynb
@@ -0,0 +1,152 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:23456872c8877381c549ede3a12ff29dc232ffad3142abc24c5d2ad8d9888f99"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter14:Soil Improvement and Ground Modification"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.1:Pg-695"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 14.1\n",

+      "\n",

+      "import math\n",

+      "Cc=0.28;\n",

+      "Hc=6.0;\n",

+      "eo=0.9;\n",

+      "Cv=0.36; # in m^2/mo.\n",

+      "H=3.0; # in m\n",

+      "t=9.0; # in mo.\n",

+      "sigmao=210.0;# in kN/m^2\n",

+      "sigmap=115; #deltasigmap in kN/m^2\n",

+      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+sigmap)/sigmao); # in m\n",

+      "print round(Sc*1000,1),\"primary consolidation in mm\"\n",

+      "Tv=Cv*t/H**2;\n",

+      "#from table\n",

+      "k=1.8; #constant\n",

+      "sf=k*sigmap; # in kN/m**2\n",

+      "print round(sf,2),\"deltasigmaf in kN/m**2\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "167.7 primary consolidation in mm\n",

+        "207.0 deltasigmaf in kN/m**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.2:Pg-703"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 14.2\n",

+      "\n",

+      "import math\n",

+      "Tv=0.36;\n",

+      "sigmap=115; # in kN/m^2\n",

+      "Uv=math.sqrt(4*Tv/math.pi)*100;\n",

+      "print round(Uv,2),\"Uv in %\"\n",

+      "#from table \n",

+      "k=0.12; #constant\n",

+      "sf=k*sigmap;\n",

+      "print sf,\"deltasigmaf in kN/m**2\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "67.7 Uv in %\n",

+        "13.8 deltasigmaf in kN/m**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex14.3:Pg-704"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#example 14.3\n",

+      "\n",

+      "Cc=0.31;\n",

+      "Hc=15.0;#ft\n",

+      "eo=1.1;\n",

+      "n=10.0;\n",

+      "Uv=0.09;\n",

+      "sigmao=1000.0; # in lb/ft^2\n",

+      "deltasigma=2000.0; # deltasigmap+deltasigmaf\n",

+      "Sc=Cc*Hc/(1+eo)*math.log10((sigmao+deltasigma)/sigmao);\n",

+      "print round(Sc,3),\"primary consolidation in ft\"\n",

+      "m=n**2/(n**2-1)*math.log(n)-(3*n**2-1)/4/n**2;\n",

+      "A=2/m;\n",

+      "Ur=(0.096-1/A*(1-math.exp(-A*0.096)))/0.192;\n",

+      "Uvf=1-(1-Ur)*(1-Uv);\n",

+      "Sc30=Sc*Uvf*12; #settlement after 30 days\n",

+      "print round(Sc30,2),\"settlement after 30 days in inch\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "1.056 primary consolidation in ft\n",

+        "1.48 settlement after 30 days in inch\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
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--- a/Programming_in_C_using_ANSI_C/KamthaneChapter10.ipynb
+++ /dev/null
@@ -1,2940 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 10: Functions<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.1, Page number: 320<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#user defined function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 1\n",
-      "y = 2\n",
-      "\n",
-      "#Function definition\n",
-      "def add(a,b):\n",
-      "    return a+b\n",
-      "\n",
-      "#Function call\n",
-      "z = add(x,y)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"z = %d\"%(z))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "z = 3"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.2, Page number: 321<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call user-defined function at different places.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definitions\n",
-      "def y():\n",
-      "    sys.stdout.write(\" Y\")\n",
-      "    return\n",
-      "\n",
-      "def a():\n",
-      "    sys.stdout.write(\" A\")\n",
-      "    y()\n",
-      "    return\n",
-      "\n",
-      "def b():\n",
-      "    sys.stdout.write(\" B\")\n",
-      "    a()\n",
-      "    return\n",
-      "\n",
-      "def c():\n",
-      "    a()\n",
-      "    b()\n",
-      "    sys.stdout.write(\" C\")\n",
-      "    return\n",
-      "\n",
-      "def d():\n",
-      "    sys.stdout.write(\" D\")\n",
-      "    c()\n",
-      "    b()\n",
-      "    a()\n",
-      "    return\n",
-      "\n",
-      "#Function calls\n",
-      "y()\n",
-      "a()\n",
-      "b()\n",
-      "c()\n",
-      "d()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Y A Y B A Y A Y B A Y C D A Y B A Y C B A Y A Y"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.3, Page number: 323<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Using similar variable names in different functions.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def fun():\n",
-      "    b = 20\n",
-      "    c = 10\n",
-      "    sys.stdout.write(\"\\nIn fun() B = %d  C = %d\"%(b,c))\n",
-      "    return\n",
-      "\n",
-      "#Variable Initialization\n",
-      "b = 10\n",
-      "c = 5\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nIn main() B = %d  C = %d\"%(b,c))\n",
-      "fun()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In main() B = 10  C = 5\n",
-        "In fun() B = 20  C = 10"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.4, Page number: 323<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Global variables on different functions.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Global Variable Initialization\n",
-      "b = 10\n",
-      "c = 5\n",
-      "\n",
-      "def fun():\n",
-      "    global b \n",
-      "    b += 1\n",
-      "    global c \n",
-      "    c -= 1\n",
-      "    sys.stdout.write(\"\\nIn fun() B = %d  C = %d\"%(b,c))\n",
-      "    return\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nIn main() B = %d  C = %d\"%(b,c))\n",
-      "fun()\n",
-      "b += 1\n",
-      "c -= 1\n",
-      "sys.stdout.write(\"\\nAgain In main() B = %d C = %d\"%(b,c))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In main() B = 10  C = 5\n",
-        "In fun() B = 11  C = 4\n",
-        "Again In main() B = 12 C = 3"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.5, Page number: 325<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Using return statement in different ways\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def pass1(a):\n",
-      "    if a == 0:\n",
-      "        return;\n",
-      "    else:\n",
-      "        return a*a*a\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter value of x : \"))\n",
-      "\n",
-      "#Function call & Result\n",
-      "if x != 1 or x > 0:\n",
-      "    y = pass1(x)\n",
-      "\n",
-      "#There is no switch statement in python, so if..else statement\n",
-      "if y == 1:\n",
-      "    sys.stdout.write(\"The value returned is %d\"%(y))\n",
-      "else:\n",
-      "    sys.stdout.write(\"The Cube of %d is : %d\"%(x,y))\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of x : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Cube of 5 is : 125"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.6, Page number: 327<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display message using user defined function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def message():\n",
-      "    sys.stdout.write(\"Have a nice day\")\n",
-      "\n",
-      "#function call\n",
-      "message()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Have a nice day"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.7, Page number: 328<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display Alphabets 'A','B' and 'C' using functions\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Functions definitions\n",
-      "def a():\n",
-      "    sys.stdout.write(\"\\nA\")\n",
-      "    \n",
-      "def b():\n",
-      "    sys.stdout.write(\" B\")\n",
-      "    \n",
-      "def c():\n",
-      "    sys.stdout.write(\" C\")\n",
-      "    \n",
-      "#Function call\n",
-      "a()\n",
-      "b()\n",
-      "c()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "A B C"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.8, Page number: 329<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Send value to user defined function and display results\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def dat(x,y,z):\n",
-      "    sys.stdout.write(\"Date = %d/%d/%d\"%(x,y,z))\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "d = int(raw_input(\"Enter date dd/mm/yy\"))\n",
-      "m = int(raw_input(\"Enter date dd/mm/yy\"))\n",
-      "y = int(raw_input(\"Enter date dd/mm/yy\"))\n",
-      "\n",
-      "#function call & Result\n",
-      "dat(d,m,y)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter date dd/mm/yy12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter date dd/mm/yy12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter date dd/mm/yy2001\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Date = 12/12/2001"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.9, Page number: 330<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Square of number using user defined function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#function definition\n",
-      "def sqr(k):\n",
-      "    sys.stdout.write(\"\\n%d\"%(k*k))\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "j = 0\n",
-      "\n",
-      "#Function call & Result\n",
-      "for j in range(1,5):\n",
-      "    sqr(j)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1\n",
-        "4\n",
-        "9\n",
-        "16"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.10, Page number: 330<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pass the value to main() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no main function in python\n",
-      "sys.stdout.write(\"\\nNumber of command line arguments J = %d\"%(len(sys.argv)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number of command line arguments J = 6"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.11, Page number: 331<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pass and return values to user defined function.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definitions\n",
-      "def dat(x,y,z):\n",
-      "    sys.stdout.write(\"\\nToday = %d/%d/%d\"%(x,y,z))\n",
-      "    x += 1\n",
-      "    return x\n",
-      "\n",
-      "#Variable Initialization\n",
-      "d = int(raw_input(\"Enter Date dd/mm/yy : \"))\n",
-      "m = int(raw_input(\"Enter Date dd/mm/yy : \"))\n",
-      "y = int(raw_input(\"Enter Date dd/mm/yy : \"))\n",
-      "\n",
-      "#Function call\n",
-      "t = dat(d,m,y)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nTomorrow = %d/%d/%d\"%(t,m,y))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Date dd/mm/yy : 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Date dd/mm/yy : 12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Date dd/mm/yy : 2001\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Today = 12/12/2001\n",
-        "Tomorrow = 13/12/2001"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.12, Page number: 332<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Send and recieve values to user defined function \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def sum1(x,y,z):\n",
-      "    return x+y+z\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "b = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "c = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "\n",
-      "#Function call\n",
-      "s = sum1(a,b,c)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Sum = %d\"%(s))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum = 16"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.13, Page number: 333<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Recieve values from user defined function without passing any \n",
-      "#value through main().\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def sum1():\n",
-      "    x = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "    y = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "    z = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "    return x+y+z\n",
-      "\n",
-      "#Function call\n",
-      "s = sum1()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Sum = %d\"%(s))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum = 12"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.14, Page number: 333<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Return value in the form of address.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def sum1():\n",
-      "    x = int(raw_input(\"Enter Three Values : \"))\n",
-      "    y = int(raw_input(\"Enter Three Values : \"))\n",
-      "    z = int(raw_input(\"Enter Three Values : \"))\n",
-      "    k = x + y + z\n",
-      "    return k\n",
-      "\n",
-      "#There is no pointer concept in python\n",
-      "\n",
-      "#Function call\n",
-      "s = sum1()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Sum = %d\"%(s))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Values : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Values : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Values : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum = 12"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.15, Page number: 334<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call by value\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def change(a,b):\n",
-      "    k = a\n",
-      "    a = b\n",
-      "    b = k\n",
-      "    sys.stdout.write(\"\\nIn Change() X = %d Y = %d\"%(a,b))\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter Values of X & Y : \"))\n",
-      "y = int(raw_input(\"Enter Values of X & Y : \"))\n",
-      "\n",
-      "#Function call\n",
-      "change(x,y)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nIn main() X = %d Y = %d\"%(x,y))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values of X & Y : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values of X & Y : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In Change() X = 4 Y = 5\n",
-        "In main() X = 5 Y = 4"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.16, Page number: 335<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call by reference \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def change(a,b):\n",
-      "    k = a\n",
-      "    a = b\n",
-      "    b = k\n",
-      "    sys.stdout.write(\"\\nIn Change() X = %d Y = %d\"%(a,b))\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter Values of X & Y : \"))\n",
-      "y = int(raw_input(\"Enter Values of X & Y : \"))\n",
-      "\n",
-      "#Function call\n",
-      "change(x,y)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nIn main() X = %d Y = %d\"%(y,x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values of X & Y : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values of X & Y : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In Change() X = 4 Y = 5\n",
-        "In main() X = 4 Y = 5"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.17, Page number: 336<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Return by reference\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def change(a,b):\n",
-      "    c = a + b\n",
-      "    d = a - b\n",
-      "    return c,d\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter Values of X & Y : \"))\n",
-      "y = int(raw_input(\"Enter Values of X & Y : \"))\n",
-      "\n",
-      "#Function call\n",
-      "add,sub = change(x,y)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAddition : %d\"%(add))\n",
-      "sys.stdout.write(\"\\nSubtraction : %d\"%(sub))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values of X & Y : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values of X & Y : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Addition : 9\n",
-        "Subtraction : 1"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.18, Page number: 337<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call by value and reference\n",
-      "\n",
-      "k = 0\n",
-      "m = 0\n",
-      "\n",
-      "#Function Definition\n",
-      "def other(k,m):\n",
-      "    sys.stdout.write(\"\\nAddress of k & m in other() : %u %u\"%(id(k),id(m)))\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAddress of k & m in main() : %u %u\"%(id(k),id(m)))\n",
-      "\n",
-      "#Function call\n",
-      "other(k,m)\n",
-      "\n",
-      "#there is no pointer concept in python and it uses value tagged method in data storage\n",
-      "#instead of addressing the memory location, values of same variables are tagged together"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of k & m in main() : 30922996 30922996\n",
-        "Address of k & m in other() : 30922996 30922996"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.19, Page number: 338<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#User defined function as an argument to another function.\n",
-      "\n",
-      "#Variable Initialization\n",
-      "y = 2\n",
-      "\n",
-      "#Function Definitions\n",
-      "def double(m):\n",
-      "    return m*2\n",
-      "\n",
-      "def square(k):\n",
-      "    return k*k\n",
-      "\n",
-      "#Function call\n",
-      "x = double(square(y))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"x = %d\"%(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "x = 8"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.20, Page number: 338<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Two functions as arguments for another functions\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definitions\n",
-      "def x(a,b):\n",
-      "    return abs(a-b)\n",
-      "\n",
-      "def y():\n",
-      "    y = int(raw_input(\"Enter First Number : \"))\n",
-      "    return y\n",
-      "\n",
-      "def z():\n",
-      "    z = int(raw_input(\"Enter Second Number : \"))\n",
-      "    return z\n",
-      "\n",
-      "#Function call\n",
-      "d = x(y(),z())\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nz() - y() = %d\"%(d))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter First Number : 25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Second Number : 50\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "z() - y() = 25"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.21, Page number: 339<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Return only absolute value like abs() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def uabs(y):\n",
-      "    if y < 0:\n",
-      "        return y * -1\n",
-      "    else:\n",
-      "        return y\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter a Negative Value : \"))\n",
-      "\n",
-      "#Function call\n",
-      "x = uabs(x)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nX = %d\"%(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Negative Value : -5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "X = 5"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.22, Page number: 340<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Square and cube of an entered number.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definitions\n",
-      "def input1():        #Since input() is a built in function in python, input1() is used\n",
-      "    k = int(raw_input(\"Number : \"))\n",
-      "    return k\n",
-      "\n",
-      "def sqr(m):\n",
-      "    sys.stdout.write(\"\\nSquare : %d\"%(m*m))\n",
-      "    return m\n",
-      "\n",
-      "def cube(m):\n",
-      "    return m*m*m\n",
-      "\n",
-      "#Function call and Result\n",
-      "sys.stdout.write(\"\\nCube : %d\"%(cube(sqr(input1()))))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number : 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Square : 4\n",
-        "Cube : 8"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.23, Page number: 341<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Assign return value of a function to variable.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def input1():\n",
-      "    k = int(raw_input(\"Enter Value of x = \"))\n",
-      "    return k\n",
-      "\n",
-      "#Function call\n",
-      "x = input1()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nx = %d\"%(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of x = 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "x = 5"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.24, Page number: 342<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Addition and subtraction of numbers with return value of function.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def input1():\n",
-      "    k = int(raw_input(\"Enter Value of x = \"))\n",
-      "    return k\n",
-      "\n",
-      "def sqr(m):\n",
-      "    return pow(m,2)\n",
-      "\n",
-      "#Function call\n",
-      "x = sqr(1 - input1() + 1)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSquare = %d\"%(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of x = 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Square = 9"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.25, Page number: 343<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Multiplication and division of numbers with return value of function.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def input1():\n",
-      "    k = int(raw_input(\"Enter Value of x = \"))\n",
-      "    return k\n",
-      "\n",
-      "def sqr(m):\n",
-      "    return pow(m,2)\n",
-      "\n",
-      "#Function call\n",
-      "x = sqr(5 * input1()/2)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSquare = %d\"%(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of x = 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Square = 144"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.26, Page number: 344<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# ++ operator with return value of function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def input1():\n",
-      "    k = int(raw_input(\"Enter Value of x = \"))\n",
-      "    return k\n",
-      "\n",
-      "def sqr(m):\n",
-      "    return pow(m,2)\n",
-      "\n",
-      "#Function call\n",
-      "#There is no ++ operator in python. so += operator is used\n",
-      "y = input1()\n",
-      "y += 1\n",
-      "x = sqr(y)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSquare = %d\"%(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of x = 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Square = 64"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.27, Page number: 345<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use mod(%) with function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def j():\n",
-      "    x = int(raw_input(\"Enter a Number : \"))\n",
-      "    return x\n",
-      "    \n",
-      "#Function call & Result\n",
-      "if j() %2 == 0:\n",
-      "    sys.stdout.write(\"\\nNumber is Even.\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nNumber is Odd.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number is Odd."
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.28, Page number: 346<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Conditional operator(?) with function.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definitions\n",
-      "def sqr(x):\n",
-      "    sys.stdout.write(\"Square \")\n",
-      "    return pow(x,2)\n",
-      "\n",
-      "def cube(x):\n",
-      "    sys.stdout.write(\"Cube \")\n",
-      "    return pow(x,3)\n",
-      "\n",
-      "def y():\n",
-      "    return 10\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Function call\n",
-      "z = sqr(x) if x > y() else cube(x)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\" = %d\"%(z))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cube  = 125"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.29, Page number: 346<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#compare two return values of functions\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definitions\n",
-      "def a():\n",
-      "    x = int(raw_input(\"Enter a Number a() : \"))\n",
-      "    return x\n",
-      "\n",
-      "def b():\n",
-      "    x = int(raw_input(\"Enter a Number b() : \"))\n",
-      "    return x\n",
-      "\n",
-      "#Function call and Result\n",
-      "if a() == b():\n",
-      "    sys.stdout.write(\"\\nValue of a() & b() are equal\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nValue of a() & b() are unique\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number a() : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number b() : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value of a() & b() are equal"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.30, Page number: 347<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the equation s = sqr(a() + b()) using function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definitions\n",
-      "def a():\n",
-      "    a = int(raw_input(\"Enter value of a : \"))\n",
-      "    return a\n",
-      "\n",
-      "def b():\n",
-      "    b = int(raw_input(\"Enter value of b : \"))\n",
-      "    return b\n",
-      "\n",
-      "def sqr(x):\n",
-      "    return x*x\n",
-      "\n",
-      "#Function call\n",
-      "s = sqr(a() + b())\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSquare of Sum = %d\"%(s))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of a : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of b : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Square of Sum = 64"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.31, Page number: 348<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the equation y = x^1+x^2..x^n using function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def b(m):\n",
-      "    m += 1\n",
-      "    #sys.stdout.write(\"%d\"%(m))\n",
-      "    return m\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Values of 'x' and 'n' : \"))\n",
-      "n = int(raw_input(\"Values of 'x' and 'n' : \"))\n",
-      "y = 0\n",
-      "z = 1\n",
-      "m = 0\n",
-      "\n",
-      "while(z <= n):\n",
-      "    m = b(m)\n",
-      "    y = y + pow(x,m)\n",
-      "    sys.stdout.write(\"%d + \"%(y))\n",
-      "    z += 1\n",
-      "    \n",
-      "if z >= n:\n",
-      "    sys.stdout.write(\"\\nValue of y = %d\"%(y))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Values of 'x' and 'n' : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Values of 'x' and 'n' : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3 + 12 + 39 + \n",
-        "Value of y = 39"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.32, Page number: 350<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call user defined function through if statement\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def a():\n",
-      "    a = int(raw_input(\"Enter value of a :\"))\n",
-      "    return a\n",
-      "\n",
-      "#Function call and Result\n",
-      "if a()%2 == 0:\n",
-      "    sys.stdout.write(\"\\nThe number is even.\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nThe number is odd.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of a :5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The number is odd."
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.33, Page number: 351<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call user defined function through switch() statement\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def a():\n",
-      "    c = raw_input(\"Enter Your Choice Square(s), Cube(c), Double(d) : \")\n",
-      "    c = c.lower()\n",
-      "    return c\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 5\n",
-      "\n",
-      "#There is no switch() statement in python.\n",
-      "c = a()\n",
-      "if c == 's':\n",
-      "    sys.stdout.write(\"\\nSquare of %d is %d\"%(x,pow(x,2)))\n",
-      "else:\n",
-      "    if c == 'c':\n",
-      "        sys.stdout.write(\"\\nCube of %d is %d\"%(x,pow(x,3)))\n",
-      "    else:\n",
-      "        if c == 'd':\n",
-      "            sys.stdout.write(\"\\nDouble of %d is %d\"%(x,x*2))\n",
-      "        else:\n",
-      "            sys.stdout.write(\"\\nUnexpected Choice printed as it is : %d\"%(x))\n",
-      "         "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Choice Square(s), Cube(c), Double(d) : D\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Double of 5 is 10"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.34, Page number: 353<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call function through the for loop\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def plus(k):\n",
-      "    if k == 10:\n",
-      "        return 0\n",
-      "    else:\n",
-      "        return k\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "m = 1\n",
-      "\n",
-      "#Function call & Result\n",
-      "#in python, for loop iterates through a range of number. so while loop is used instead.\n",
-      "while plus(m) != 0:\n",
-      "    sys.stdout.write(\"%3d\"%(m))\n",
-      "    m += 1\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  1  2  3  4  5  6  7  8  9"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.35, Page number: 354<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call user defined function through while() loop\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def y():\n",
-      "    x = int(raw_input(\"Enter a Number : \"))\n",
-      "    return x\n",
-      "\n",
-      "#Function call & Result\n",
-      "while y() != 0:\n",
-      "    sys.stdout.write(\"Value enter is non-zero\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value enter is non-zero\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 0\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.36, Page number: 355<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call user defined function through do while() loop\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def y():\n",
-      "    x = int(raw_input(\"Enter a Number : \"))\n",
-      "    return x\n",
-      "\n",
-      "#Function call and Result\n",
-      "#There is no do-while loop in python\n",
-      "\n",
-      "while y() != 0:\n",
-      "    sys.stdout.write(\"\\nValue entered is non-zero\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value entered is non-zero\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 0\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.37, Page number: 356<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Initialize an array using functions.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def c(m):\n",
-      "    n = int(raw_input(\"Enter Number d[%d]\"%(m+1)))\n",
-      "    return n\n",
-      "\n",
-      "#Variable Initialization\n",
-      "d = [c(i) for i in range(0,5)]\n",
-      "\n",
-      "sys.stdout.write(\"\\nArray d[] elements are : \")\n",
-      "for k in range(0,5):\n",
-      "    sys.stdout.write(\"%2d\"%d[k])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number d[1]4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number d[2]5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number d[3]6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number d[4]7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number d[5]8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Array d[] elements are :  4 5 6 7 8"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.38, Page number: 357<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pass array element to the function using call by value method.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def show(m,u):\n",
-      "    sys.stdout.write(\"\\nnum[%d] = %d\"%(m+1,u))\n",
-      "    \n",
-      "#Variable initialization\n",
-      "num = [12,13,14,15,16,17,18]\n",
-      "\n",
-      "#Function call & Result\n",
-      "for k in range(0,7):\n",
-      "    show(k,num[k])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "num[1] = 12\n",
-        "num[2] = 13\n",
-        "num[3] = 14\n",
-        "num[4] = 15\n",
-        "num[5] = 16\n",
-        "num[6] = 17\n",
-        "num[7] = 18"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.39, Page number: 358<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pass array element to the function using call by reference\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def show(u):\n",
-      "    m = 0\n",
-      "    sys.stdout.write(\"\\nnum[7] = {\")\n",
-      "    while m != 7:\n",
-      "        #There is no pointer concept in python\n",
-      "        sys.stdout.write(\"%2d,\"%(u[m]))\n",
-      "        m += 1\n",
-      "    sys.stdout.write(\"\\b}\")\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "num = [12,13,14,15,16,17,18]\n",
-      "\n",
-      "#Function call\n",
-      "show(num)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "num[7] = {12,13,14,15,16,17,18,\b}"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.40, Page number: 359<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Array elements in reverse order.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def show(u):\n",
-      "    m = 6\n",
-      "    while m != -1:\n",
-      "        sys.stdout.write(\"\\nnum[%d] = %d\"%(m,u[m]))\n",
-      "        m -= 1\n",
-      "        \n",
-      "#Variable Initialization\n",
-      "num = [12,13,14,15,16,17,18]\n",
-      "\n",
-      "#Function call\n",
-      "#There is no pointer concept in python\n",
-      "show(num)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "num[6] = 18\n",
-        "num[5] = 17\n",
-        "num[4] = 16\n",
-        "num[3] = 15\n",
-        "num[2] = 14\n",
-        "num[1] = 13\n",
-        "num[0] = 12"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.41, Page number: 360<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy array elements using user defined function.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def cpy(p,m):\n",
-      "    j = 0\n",
-      "    while j != 5:\n",
-      "        m[j] = p[j]\n",
-      "        j += 1\n",
-      "        \n",
-      "#Variable Initialization\n",
-      "a1 = [1,2,3,4,5]\n",
-      "a2 = [0 for i in range(0,5)]\n",
-      "\n",
-      "#Function call\n",
-      "cpy(a1,a2)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Source  Target\")\n",
-      "for h in range(0,5):\n",
-      "    sys.stdout.write(\"\\n%5d\\t%d\"%(a1[h],a2[h]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Source  Target\n",
-        "    1\t1\n",
-        "    2\t2\n",
-        "    3\t3\n",
-        "    4\t4\n",
-        "    5\t5"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.42, Page number: 361<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read array of other function in main()\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def arry(k):\n",
-      "    b = [1,2,3,4,5]\n",
-      "    return b[k]\n",
-      "\n",
-      "#main() function\n",
-      "for k in range(0,5):\n",
-      "    sys.stdout.write(\"\\t%d\"%(arry(k)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t1\t2\t3\t4\t5"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.43, Page number: 361<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Interchange array elements of two arrays using function.\n",
-      "\n",
-      "import sys\n",
-      "global a\n",
-      "global b\n",
-      "\n",
-      "#Function Definitions\n",
-      "def read():\n",
-      "    x = int(raw_input(\"\"))\n",
-      "    return x\n",
-      "\n",
-      "def change(a,b):\n",
-      "#Since there is no pointer concept in python, exchange is done in the function using global variables.\n",
-      "    for x in range(0,5):\n",
-      "        a[x] = a[x] + b[x]\n",
-      "        b[x] = a[x] - b[x]\n",
-      "        a[x] = a[x] - b[x]\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "a = [0 for i in range(0,5)]\n",
-      "b = [0 for i in range(0,5)]\n",
-      "\n",
-      "for x in range(0,10):\n",
-      "    if x < 5:\n",
-      "        a[x] = read()\n",
-      "    else:\n",
-      "        b[x-5] = read()\n",
-      "\n",
-      "#Swapping and Result\n",
-      "sys.stdout.write(\"\\nArray A & B \")\n",
-      "\n",
-      "for x in range(0,5):\n",
-      "    sys.stdout.write(\"\\n%7d%8d\"%(a[x],b[x]))\n",
-      "  \n",
-      "#There is no pointer concept in python.\n",
-      "change(a,b)\n",
-      "    \n",
-      "sys.stdout.write(\"\\nNow A  & B\")\n",
-      "for x in range(0,5):\n",
-      "    sys.stdout.write(\"\\n%7d%8d\"%(a[x],b[x]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Array A & B \n",
-        "      1       6\n",
-        "      2       7\n",
-        "      3       8\n",
-        "      4       9\n",
-        "      5       0\n",
-        "Now A  & B\n",
-        "      6       1\n",
-        "      7       2\n",
-        "      8       3\n",
-        "      9       4\n",
-        "      0       5"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.44, Page number: 363<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read array elements declared in different functions using global\n",
-      "#pointer declaration\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def call(j):\n",
-      "    m = 0\n",
-      "    u = [5,1,6,0,6]\n",
-      "    q = u\n",
-      "    while m != j:\n",
-      "        sys.stdout.write(\"%3d\"%(u[m]))\n",
-      "        m += 1\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "m = 0\n",
-      "k = [3,8,5,2,5]\n",
-      "q = k\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "while m != 5:\n",
-      "    sys.stdout.write(\"%3d\"%(q[m]))\n",
-      "    m += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\n\")\n",
-      "call(5)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  3  8  5  2  5\n",
-        "  5  1  6  0  6\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.45, Page number: 364<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sum of 1 to 5 numbers using recursion\n",
-      "\n",
-      "import sys\n",
-      "global s\n",
-      "\n",
-      "s = 0\n",
-      "#Function definition\n",
-      "def main(x,s):\n",
-      "    s = s + x\n",
-      "    sys.stdout.write(\"\\nx = %d s = %d\"%(x,s))\n",
-      "    if x == 5:\n",
-      "        return\n",
-      "    x += 1\n",
-      "    main(x,s)\n",
-      "    \n",
-      "main(1,s)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "x = 1 s = 1\n",
-        "x = 2 s = 3\n",
-        "x = 3 s = 6\n",
-        "x = 4 s = 10\n",
-        "x = 5 s = 15"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.46, Page number: 365<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate triangular number of a given number with recursion function method\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def tri_num(m):\n",
-      "    f = 0\n",
-      "    if m == 0:\n",
-      "        return f\n",
-      "    else:\n",
-      "        f = f + m + tri_num(m-1)\n",
-      "        return f\n",
-      "\n",
-      "#Variable Initialization\n",
-      "n = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Function call\n",
-      "t = tri_num(n)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nTriangular number of %d is %d\"%(n,t))\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Triangular number of 5 is 15"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.47, Page number: 366<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the given string using recursion\n",
-      "\n",
-      "import sys\n",
-      "import os\n",
-      "global x\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 0\n",
-      "str1 = \"Have a Good Day\"\n",
-      "\n",
-      "#Function Definition\n",
-      "def main(x):\n",
-      "    if x == len(str1):        #There is no null terminating character in python string\n",
-      "        return\n",
-      "    else:\n",
-      "        if str1[x] == 'H':\n",
-      "            os.system('cls')\n",
-      "            sys.stdout.write(\"%c\"%(str1[x]))\n",
-      "        else:\n",
-      "            sys.stdout.write(\"%c\"%(str1[x]))\n",
-      "    x += 1\n",
-      "    main(x)\n",
-      "    \n",
-      "#Function call\n",
-      "main(x)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Have a Good Day"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.48, Page number: 367<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the given string 10 times using recursion\n",
-      "\n",
-      "import sys\n",
-      "import os\n",
-      "\n",
-      "#Function definition\n",
-      "def main(x):\n",
-      "    sys.stdout.write(\"\\n%.2d] %s\"%(x,str1))\n",
-      "    x += 1\n",
-      "    if x == 11:\n",
-      "        return\n",
-      "    else:\n",
-      "        if x == 1:\n",
-      "            os.system('cls')\n",
-      "            main(x)\n",
-      "        else:\n",
-      "            main(x)\n",
-      "            \n",
-      "            \n",
-      "#Variable Initialization\n",
-      "x = 0\n",
-      "str1 = \"Have a Good Day\"\n",
-      "\n",
-      "#Function call\n",
-      "main(x)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "00] Have a Good Day\n",
-        "01] Have a Good Day"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "02] Have a Good Day\n",
-        "03] Have a Good Day\n",
-        "04] Have a Good Day\n",
-        "05] Have a Good Day\n",
-        "06] Have a Good Day\n",
-        "07] Have a Good Day\n",
-        "08] Have a Good Day\n",
-        "09] Have a Good Day\n",
-        "10] Have a Good Day"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.49, Page number: 368<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Factorial using recursive function.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function Definition\n",
-      "def fact(m):\n",
-      "    f = 1\n",
-      "    if m == 1:\n",
-      "        return 1\n",
-      "    else:\n",
-      "        f = m * fact(m-1)\n",
-      "        return f\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Function call\n",
-      "f = fact (x)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nFactorial of %d is %d\"%(x,f))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Factorial of 5 is 120"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.50, Page number: 369<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display address of user defined function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def show():\n",
-      "    sys.stdout.write(\"\\nAddress of function show() is : \")\n",
-      "    \n",
-      "#Function call\n",
-      "show()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"%u\"%(id(show)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of function show() is : 95041520"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.51, Page number: 369<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call function using pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def show():\n",
-      "    sys.stdout.write(\"\\nAddress of function show() is : \")\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "p = id(show)\n",
-      "show()\n",
-      "sys.stdout.write(\"%u\"%(id(show)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of function show() is : 95041200"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.52, Page number: 370<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the address of library function.\n",
-      "\n",
-      "import sys\n",
-      "import os\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAddress of printf() is %u\"%(id(sys.stdout.write)))\n",
-      "sys.stdout.write(\"\\nAddress of scanf() is %u\"%(id(sys.stdout.read)))\n",
-      "sys.stdout.write(\"\\nAddress of clrscr() is %u\"%(id(os.system('cls'))))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of printf() is 60743848\n",
-        "Address of scanf() is 60743848\n",
-        "Address of clrscr() is 4774132"
-       ]
-      }
-     ],
-     "prompt_number": 57
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 10.53, Page number: 371<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Call main() using pointer to main() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initilization\n",
-      "x = 0\n",
-      "\n",
-      "#Function definition\n",
-      "def main(x):\n",
-      "    p = id(main)\n",
-      "    x += 1\n",
-      "    sys.stdout.write(\"\\nCall %d Address of main() %u\"%(x,id(main)))\n",
-      "    if x == 3:\n",
-      "        return\n",
-      "    main(x)\n",
-      "    \n",
-      "#function call\n",
-      "main(x)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Call 1 Address of main() 95040880\n",
-        "Call 2 Address of main() 95040880\n",
-        "Call 3 Address of main() 95040880"
-       ]
-      }
-     ],
-     "prompt_number": 59
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter11.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter11.ipynb
deleted file mode 100755
index c8fac375..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter11.ipynb
+++ /dev/null
@@ -1,446 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 11: Storage Class<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.1, Page number: 376<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Working of auto variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Functio definitions\n",
-      "def call1():\n",
-      "    v = 20\n",
-      "    sys.stdout.write(\"\\nV = %d\"%(v))\n",
-      "    \n",
-      "def call2():\n",
-      "    v = 30\n",
-      "    call1()\n",
-      "    sys.stdout.write(\"\\nV = %d\"%(v))\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "v = 10\n",
-      "\n",
-      "#Function call\n",
-      "call2()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nV = %d\"%(v))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "V = 20\n",
-        "V = 30\n",
-        "V = 10"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.2, Page number: 376<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Working of auto variables in different blocks\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 10\n",
-      "\n",
-      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
-      "\n",
-      "x = 20\n",
-      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
-      "\n",
-      "x = 10\n",
-      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
-      "\n",
-      "#In python, block of statements are indicated using intendation.\n",
-      "#block of statements can be written for conditional,loop or function statements"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "X = 10\n",
-        "X = 20\n",
-        "X = 10"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.3, Page number: 377<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use same variable in different blocks with different data types\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization & Result\n",
-      "x = 10\n",
-      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
-      "\n",
-      "x = 2.22\n",
-      "sys.stdout.write(\"\\nX = %g\"%(x))\n",
-      "\n",
-      "x = \"Auto Storage Class\"\n",
-      "sys.stdout.write(\"\\nX = %s\"%(x))\n",
-      "\n",
-      "x = 10\n",
-      "sys.stdout.write(\"\\nX = %d\"%(x))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "X = 10\n",
-        "X = 2.22\n",
-        "X = Auto Storage Class\n",
-        "X = 10"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.4, Page number: 378<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Working of external variables\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definitions\n",
-      "def call1():\n",
-      "    sys.stdout.write(\"\\nIn Call1() V = %d\"%(v))\n",
-      "    \n",
-      "def call2():\n",
-      "    sys.stdout.write(\"\\nIn call2() V = %d\"%(v))\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "global v\n",
-      "v = 10\n",
-      "\n",
-      "call1()\n",
-      "call2()\n",
-      "sys.stdout.write(\"\\nIn main() V = %d\"%(v))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In Call1() V = 10\n",
-        "In call2() V = 10\n",
-        "In main() V = 10"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.5, Page number: 379<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Working of auto and global variables with same name\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definitions\n",
-      "def call1():\n",
-      "    v = 20\n",
-      "    sys.stdout.write(\"\\nIn Call1() V = %d\"%(v))\n",
-      "    \n",
-      "def call2():\n",
-      "    sys.stdout.write(\"\\nIn call2() V = %d\"%(v))\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "global v\n",
-      "v = 10\n",
-      "\n",
-      "#Function call\n",
-      "call1()\n",
-      "call2()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nIn main() V = %d\"%(v))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In Call1() V = 20\n",
-        "In call2() V = 10\n",
-        "In main() V = 10"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.6, Page number: 380<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Declare external variables using extern keyword\n",
-      "\n",
-      "#Variable Initialization\n",
-      "global m\n",
-      "m = 10\n",
-      "\n",
-      "#There is no extern keyword in python, global is used instead\n",
-      "sys.stdout.write(\"\\nM = %d\"%(m))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "M = 10"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.7, Page number: 380<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use of static variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def print1(m):\n",
-      "    sys.stdout.write(\"\\nm = %d\"%(m))\n",
-      "    if m == 3:\n",
-      "        return\n",
-      "    \n",
-      "m = 0\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    m += 1\n",
-      "    print1(m)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "m = 1\n",
-        "m = 2\n",
-        "m = 3"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.8, Page number: 381<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Difference between variables of auto and static class\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "y = 0\n",
-      "sys.stdout.write(\"\\nx = %d & y = %d\"%(x,y))\n",
-      "\n",
-      "#Since x display the junk value which stored already, it differs in different systems"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "x = 10 & y = 0"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.9, Page number: 382<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Register class variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no register keyword in python\n",
-      "m = 1\n",
-      "\n",
-      "#Result\n",
-      "for m in range(1,5+1):\n",
-      "    sys.stdout.write(\"\\t%d\"%(m))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t1\t2\t3\t4\t5"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 11.10, Page number: 382<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Register class variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no register class variable in python\n",
-      "m = 1\n",
-      "\n",
-      "for m in range(1,6):\n",
-      "    sys.stdout.write(\"\\t%d\"%(m))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t1\t2\t3\t4\t5"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter12.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter12.ipynb
deleted file mode 100755
index e1acaa2c..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter12.ipynb
+++ /dev/null
@@ -1,849 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 12: Preprocessor Directives<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.1, Page number: 386<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Area of circle\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Defining the macro\n",
-      "#There is no macro definitions/preprocessors in python\n",
-      "PI =  3.14\n",
-      "\n",
-      "#Variable Initialization\n",
-      "r = float(raw_input(\"Enter radius of circle in cms.\"))\n",
-      "\n",
-      "#Calculation\n",
-      "area = PI * r * r\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Area of a Circle = %.2f cm2\"%(area))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter radius of circle in cms.7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Area of a Circle = 153.86 cm2"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.2, Page number: 387<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Identifers for 'c' statements and variables\n",
-      "\n",
-      "import sys\n",
-      "import os\n",
-      "\n",
-      "#Macro definitions\n",
-      "#in python, function definitions are used instead of macros\n",
-      "def N():\n",
-      "    return 10\n",
-      "\n",
-      "def cls():\n",
-      "    os.system('cls')\n",
-      "    \n",
-      "def wait():\n",
-      "    t = raw_input(\"\")\n",
-      "    \n",
-      "def display(s):\n",
-      "    sys.stdout.write(s)\n",
-      "\n",
-      "for k in range(1,N()+1):\n",
-      "    display(\"%d\\t\"%(k))\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\t2\t3\t4\t5\t6\t7\t8\t9\t10\t"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.3, Page number: 387<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Macros for logical operators\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Macro definitions\n",
-      "#in python, function definitions are used as macros\n",
-      "\n",
-      "#There is no preprocessors in python\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "b = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "c = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "\n",
-      "if a > b and a > c:\n",
-      "    sys.stdout.write(\"%d is the larger number\"%(a))\n",
-      "else:\n",
-      "    if b > a and b > c:\n",
-      "        sys.stdout.write(\"%d is the larger number\"%(b))\n",
-      "    else:\n",
-      "        if c > a and c > b:\n",
-      "            sys.stdout.write(\"%d is the larger number\"%(c))\n",
-      "        else:\n",
-      "            if a == b and b == c:\n",
-      "                sys.stdout.write(\"\\nNumbers are same.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7 is the larger number"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.4, Page number: 388<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Identifier for displaying double and triple of a number.\n",
-      "\n",
-      "#Macro definitions\n",
-      "#Function definitions are used as macros in python\n",
-      "\n",
-      "def DOUBLE(a):\n",
-      "    return a*2\n",
-      "\n",
-      "def TRIPLE(a):\n",
-      "    return a*3\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSINGLE\\tDOUBLE\\tTRIPLE\")\n",
-      "\n",
-      "for a in range(1,5+1):\n",
-      "    sys.stdout.write(\"\\n%d\\t%d\\t%d\"%(a,DOUBLE(a),TRIPLE(a)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "SINGLE\tDOUBLE\tTRIPLE\n",
-        "1\t2\t3\n",
-        "2\t4\t6\n",
-        "3\t6\t9\n",
-        "4\t8\t12\n",
-        "5\t10\t15"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.5, Page number: 389<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Undefine a macro\n",
-      "\n",
-      "#There is no preprocessor directives in python\n",
-      "\n",
-      "def wait():\n",
-      "    return (raw_input(\"\"))\n",
-      "\n",
-      "#Result\n",
-      "for k in range(1,5+1):\n",
-      "    sys.stdout.write(\"%d\\t\"%(k))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\t2\t3\t4\t5\t"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.6, Page number: 389<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Stringizing operation\n",
-      "\n",
-      "#Macro definition\n",
-      "#in python, function definition can be used as macro\n",
-      "\n",
-      "def say(m):\n",
-      "    sys.stdout.write(m)\n",
-      "    \n",
-      "#Function call\n",
-      "say(\"Hello\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Hello"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.7, Page number: 390<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Stringizing operation and macro arguments\n",
-      "\n",
-      "#There is no macro in python, function definitions can be used as macros\n",
-      "\n",
-      "def DOUBLE(x):\n",
-      "    sys.stdout.write(\"Double of m = %d\\n\"%(x*2))\n",
-      "    \n",
-      "def TRIPLE(x):\n",
-      "    sys.stdout.write(\"Triple of m = %d\\n\"%(x*3))\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "m = int(raw_input(\"Enter a number : \"))\n",
-      "\n",
-      "#Function call & Result\n",
-      "DOUBLE(m)\n",
-      "TRIPLE(m)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Double of m = 10\n",
-        "Triple of m = 15\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.8, Page number: 390<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Larger of two numbers using macro with arguments.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Macro definition  //function definition itself is used as macro\n",
-      "def MAX(x,y):\n",
-      "    if x > y:\n",
-      "        c = x\n",
-      "    else:\n",
-      "        c = y\n",
-      "    return c\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 5\n",
-      "y = 8\n",
-      "\n",
-      "#Function call\n",
-      "c = MAX(x,y)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nLarger of two numbers = %d\"%(c))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Larger of two numbers = 8"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.9, Page number: 391<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function defined in \"udf.c\" file.\n",
-      "\n",
-      "import sys\n",
-      "#since ipython notebook does not support this function, function is used\n",
-      "#in python, the udf.c file can be written and saved in udf.py and import command\n",
-      "#can be used to include that file in this program\n",
-      "\n",
-      "def display():\n",
-      "    sys.stdout.write(\"\\nFunction Called\")\n",
-      "    \n",
-      "#Function call\n",
-      "display()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Function Called"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.10, Page number: 392<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Conditional compilation \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no preprocessor directives in python\n",
-      "\n",
-      "LINE = 1\n",
-      "\n",
-      "if LINE:\n",
-      "    sys.stdout.write(\"This is line number one\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"This is line number two\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is line number one"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.11, Page number: 393<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Conditional compilation \n",
-      "import sys\n",
-      "\n",
-      "#There is no preprocessor directives in python\n",
-      "def E():\n",
-      "    return 1\n",
-      "\n",
-      "#Function call\n",
-      "if E():\n",
-      "    a = 2\n",
-      "    b = 3\n",
-      "    sys.stdout.write(\"A = %d  &  B = %d\"%(a,b))\n",
-      "else:\n",
-      "    c = 2\n",
-      "    d = 3\n",
-      "    sys.stdout.write(\"C = %d  &  D = %d\"%(c,d))\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A = 2  &  B = 3"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.12, Page number: 394<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Conditional compilation directives \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no preprocessor directives in python\n",
-      "T = 8\n",
-      "\n",
-      "#Result\n",
-      "if T == 0:\n",
-      "    sys.stdout.write(\"\\nMacro is not defined.\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nMacro is defined\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Macro is defined"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.13, Page number: 394<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#User defined error message \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no preprocessor directives in python\n",
-      "B = 1\n",
-      "A = 0\n",
-      "\n",
-      "#Result\n",
-      "if A == 0:\n",
-      "    sys.stdout.write(\"MACRO A IS NOT DEFINED.\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"Macro found.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "MACRO A IS NOT DEFINED."
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.14, Page number: 397<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Set off certain errors \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no preprocessor directives in python\n",
-      "x = 2\n",
-      "y = 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\ny = %d\"%(y))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "y = 1"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.15, Page number: 398<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Predefined macros \n",
-      "\n",
-      "import sys\n",
-      "import datetime\n",
-      "\n",
-      "#Variable Initialization\n",
-      "FILE = 'PRE_MA~1.C'\n",
-      "LINE = 10\n",
-      "STDC = 1\n",
-      "\n",
-      "#Date/time can be set using datetime.date(2001,5,12)\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nDATE    :  %s\"%('Dec 05 2001')) \n",
-      "sys.stdout.write(\"\\nTIME    :  %s\"%(datetime.time(21,7,39)))\n",
-      "sys.stdout.write(\"\\nFILE NAME : %s\"%(FILE))\n",
-      "sys.stdout.write(\"\\nLINE NO : %d\"%(LINE))\n",
-      "sys.stdout.write(\"\\n_STDC_  :  %x\"%(STDC))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "DATE    :  Dec 05 2001\n",
-        "TIME    :  21:07:39\n",
-        "FILE NAME : PRE_MA~1.C\n",
-        "LINE NO : 10\n",
-        "_STDC_  :  1"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.16, Page number: 399<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the name of memory model that is currently in use.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no formal memory model and preprocessor directives in python\n",
-      "\n",
-      "TINY = 0\n",
-      "SMALL = 0\n",
-      "MEDIUM = 0\n",
-      "COMPACT = 0\n",
-      "LARGE = 1\n",
-      "HUGE = 0\n",
-      "\n",
-      "#Result\n",
-      "if TINY:\n",
-      "    sys.stdout.write(\"\\nTINY %d\"%(TINY))\n",
-      "else:\n",
-      "    if SMALL:\n",
-      "        sys.stdout.write(\"\\nSMALL %d\"%(SMALL))\n",
-      "    else:\n",
-      "        if MEDIUM:\n",
-      "            sys.stdout.write(\"\\nMEDIUM %d\"%(MEDIUM))\n",
-      "        else:\n",
-      "            if COMPACT:\n",
-      "                sys.stdout.write(\"\\nCOMPACT %d\"%(COMPACT))\n",
-      "            else:\n",
-      "                if LARGE:\n",
-      "                    sys.stdout.write(\"\\nLARGE %d\"%(LARGE))\n",
-      "                else:\n",
-      "                    sys.stdout.write(\"\\nHUGE %d\"%(HUGE))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "LARGE 1"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.17, Page number: 400<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Macro expansions\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no preprocessor directives in python\n",
-      "ch = raw_input(\"Input a Text : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Text Input : %s\"%(ch))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input a Text : Programming\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Text Input : Programming"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 12.18, Page number: 401<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Predefined macros.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "d = raw_input(\"Enter any character : \")\n",
-      "\n",
-      "#Result\n",
-      "f = d.isalpha()\n",
-      "\n",
-      "if f != 0:\n",
-      "    sys.stdout.write(\"\\n%c is a letter in\"%(d))\n",
-      "    f = d.isupper()\n",
-      "    if f != 0:\n",
-      "        sys.stdout.write(\" Capital case\")\n",
-      "    else:\n",
-      "        sys.stdout.write(\" Small case\")\n",
-      "else:\n",
-      "    f = d.isdigit()\n",
-      "    if f != 0:\n",
-      "        sys.stdout.write(\"\\n%c is a digit\"%(d))\n",
-      "    else:\n",
-      "        f = d.ispunct()\n",
-      "        if f != 0:\n",
-      "            sys.stdout.write(\"\\n%c is a punctuation symbol\"%(d))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any character : C\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "C is a letter in Capital case"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter13.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter13.ipynb
deleted file mode 100755
index f9222003..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter13.ipynb
+++ /dev/null
@@ -1,2367 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 13: Structure and Union<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.1, Page number: 409<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display size of structure elements\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class is used instead of structure in python\n",
-      "class book1:\n",
-      "    book = ''\n",
-      "    pages = 0\n",
-      "    price = 0.00\n",
-      "\n",
-      "#Python variables uses value tagged methods for storing the values\n",
-      "\n",
-      "#Class variable declaration\n",
-      "bk1 = book1()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSize of Structure Elements\")\n",
-      "sys.stdout.write(\"\\nBook    :  %d\"%(sys.getsizeof(bk1.book)))\n",
-      "sys.stdout.write(\"\\nPages    :  %d\"%(sys.getsizeof(bk1.pages)))\n",
-      "sys.stdout.write(\"\\nPrice    :  %d\"%(sys.getsizeof(bk1.price)))\n",
-      "sys.stdout.write(\"\\nTotal Bytes    :  %d\"%(sys.getsizeof(bk1)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Size of Structure Elements\n",
-        "Book    :  21\n",
-        "Pages    :  12\n",
-        "Price    :  16\n",
-        "Total Bytes    :  36"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.2, Page number: 410<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Define a structure and initialize its member variables\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class book1:\n",
-      "    book = ''\n",
-      "    pages = 0\n",
-      "    price = 0.0\n",
-      "    \n",
-      "#Class variable declaration and Initialization\n",
-      "bk1 = book1()\n",
-      "bk1.book = \"C++\"\n",
-      "bk1.pages = 300\n",
-      "bk1.price = 285\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nBook Name  : %s\"%(bk1.book))\n",
-      "sys.stdout.write(\"\\nNo. of Pages : %d\"%(bk1.pages))\n",
-      "sys.stdout.write(\"\\nBook Price  : %d\"%(bk1.price))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Book Name  : C++\n",
-        "No. of Pages : 300\n",
-        "Book Price  : 285"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.3, Page number: 411<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy structure elements from one object to another object\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class disk:\n",
-      "    co = ''\n",
-      "    type1 = 0.0\n",
-      "    price = 0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "d1 = disk()\n",
-      "d2 = disk()\n",
-      "d3 = disk()\n",
-      "\n",
-      "#Class variable initialization\n",
-      "d1.co = \"SONY\"\n",
-      "d1.type1 = 1.44\n",
-      "d1.price = 20\n",
-      "\n",
-      "#Copying\n",
-      "d2.co = d1.co\n",
-      "d2.type1 = d1.type1\n",
-      "d2.price = d1.price\n",
-      "\n",
-      "d3 = d2\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n%s  %g  %d\"%(d1.co,d1.type1,d1.price))\n",
-      "sys.stdout.write(\"\\n%s  %g  %d\"%(d2.co,d2.type1,d2.price))\n",
-      "sys.stdout.write(\"\\n%s  %g  %d\"%(d3.co,d3.type1,d3.price))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "SONY  1.44  20\n",
-        "SONY  1.44  20\n",
-        "SONY  1.44  20"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.4, Page number: 412<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read values and assign them to structure variables\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class book1:\n",
-      "    book = ''\n",
-      "    pages = 0\n",
-      "    price = 0.0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "bk1 = book1()\n",
-      "\n",
-      "#Variable initialization\n",
-      "bk1.book = raw_input(\"Enter Book name,pages,price : \")\n",
-      "bk1.pages = int(raw_input(\"Enter Book name,pages,price : \"))\n",
-      "bk1.price = float(raw_input(\"Enter Book name, pages,price : \"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nBook Name : %s\"%(bk1.book))\n",
-      "sys.stdout.write(\"\\nNo. of Pages : %d\"%(bk1.pages))\n",
-      "sys.stdout.write(\"\\nBook Price : %d\"%(bk1.price))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Book name,pages,price : C\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Book name,pages,price : 500\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Book name, pages,price : 450\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Book Name : C\n",
-        "No. of Pages : 500\n",
-        "Book Price : 450"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.5, Page number: 414<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read and display car details\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class time:\n",
-      "    second = 0\n",
-      "    minute = 0\n",
-      "    hour = 0\n",
-      "    \n",
-      "class t:\n",
-      "    carno = 0\n",
-      "    st = time()\n",
-      "    rt = time()\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "r1 = t()\n",
-      "\n",
-      "#Variable Initialization\n",
-      "r1.carno = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
-      "r1.st.hour = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
-      "r1.st.minute = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
-      "r1.st.second = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
-      "r1.rt.hour = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
-      "r1.rt.minute = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
-      "r1.rt.second = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nCar No. \\tStarting Time \\tReaching Time\\n\")\n",
-      "sys.stdout.write(\"%d\\t\"%(r1.carno))\n",
-      "sys.stdout.write(\"\\t%d:%d:%d\\t\\t\"%(r1.st.hour,r1.st.minute,r1.st.second))\n",
-      "sys.stdout.write(\"%d:%d:%d\"%(r1.rt.hour,r1.rt.minute,r1.rt.second))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No., Starting Time, Reaching Time :125\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No., Starting Time, Reaching Time :2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No., Starting Time, Reaching Time :50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No., Starting Time, Reaching Time :30\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No., Starting Time, Reaching Time :3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No., Starting Time, Reaching Time :50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No., Starting Time, Reaching Time :25\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Car No. \tStarting Time \tReaching Time\n",
-        "125\t\t2:50:30\t\t3:50:25"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.6, Page number: 415<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read and display the details of a person\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class name:\n",
-      "    first = ''\n",
-      "    second = ''\n",
-      "    last = ''\n",
-      "    \n",
-      "class b_date:\n",
-      "    day = 0\n",
-      "    month = 0\n",
-      "    year = 0\n",
-      "    \n",
-      "class data:\n",
-      "    nm = name()\n",
-      "    bt = b_date()\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "r1 = data()\n",
-      "\n",
-      "#Variable Initialization\n",
-      "r1.nm.first = raw_input(\"Enter Name (First/Second/Last)\")\n",
-      "r1.nm.second = raw_input(\"Enter Name (First/Second/Last)\")\n",
-      "r1.nm.last = raw_input(\"Enter Name (First/Second/Last)\")\n",
-      "r1.bt.day = int(raw_input(\"Enter Birth Date Day/Month/Year\"))\n",
-      "r1.bt.month = int(raw_input(\"Enter Birth Date Day/Month/Year\"))\n",
-      "r1.bt.year = int(raw_input(\"Enter Birth Date Day/Month/Year\"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Name  : %s %s %s\\n\"%(r1.nm.first,r1.nm.second,r1.nm.last))\n",
-      "sys.stdout.write(\"Birth Date : %d.%d.%d\"%(r1.bt.day,r1.bt.month,r1.bt.year))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name (First/Second/Last)Ram\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name (First/Second/Last)Sham\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name (First/Second/Last)Pande\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Birth Date Day/Month/Year12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Birth Date Day/Month/Year12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Birth Date Day/Month/Year1980\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name  : Ram Sham Pande\n",
-        "Birth Date : 12.12.1980"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.7, Page number: 416<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Create array of structure objects\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class time:\n",
-      "    second = 0\n",
-      "    minute = 0\n",
-      "    hour = 0\n",
-      "    \n",
-      "class t:\n",
-      "    carno = 0\n",
-      "    st = time()\n",
-      "    rt = time()\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "r1 = [t() for i in range(0,3)]\n",
-      "\n",
-      "#Variable Initialization\n",
-      "for k in range(0,3):\n",
-      "    r1[k].carno = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
-      "    r1[k].st.hour = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
-      "    r1[k].st.minute = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
-      "    r1[k].st.second = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
-      "    r1[k].rt.hour = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
-      "    r1[k].rt.minute = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
-      "    r1[k].rt.second = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
-      "   \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nCar No.\\t\\tStarting Time\\tReaching Time\")\n",
-      "for k in range(0,3):\n",
-      "    sys.stdout.write(\"\\n%d\\t\"%(r1[k].carno))\n",
-      "    sys.stdout.write(\"\\t%d:%d:%d\\t\"%(r1[k].st.hour,r1[k].st.minute,r1[k].st.second))\n",
-      "    sys.stdout.write(\"\\t%d:%d:%d\\t\"%(r1[k].rt.hour,r1[k].rt.minute,r1[k].rt.second))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)120\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)58\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)121\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)40\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)40\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)122\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)30\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)52\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)40\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Car No.\t\tStarting Time\tReaching Time\n",
-        "120\t\t4:30:52\t\t5:40:10\t\n",
-        "121\t\t4:30:52\t\t5:40:10\t\n",
-        "122\t\t4:30:52\t\t5:40:10\t"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.8, Page number: 417<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read and display student details\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class stud:\n",
-      "    name = ''\n",
-      "    rollno = 0\n",
-      "    grade = ''\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "st = [stud() for i in range(0,3)]\n",
-      "\n",
-      "#Variable Initialization\n",
-      "k = 0 \n",
-      "while k < 3:\n",
-      "    st[k].name = raw_input(\"Name :\")\n",
-      "    st[k].rollno = int(raw_input(\"Roll No. :\"))\n",
-      "    st[k].grade = raw_input(\"Grade : \")\n",
-      "    k += 1\n",
-      "    \n",
-      "#Result\n",
-      "k = 0\n",
-      "sys.stdout.write(\"\\nName \\tRollno \\tGrade\\n\")\n",
-      "while k < 3:\n",
-      "    sys.stdout.write(\"\\n%s\\t%d\\t%s\"%(st[k].name,st[k].rollno,st[k].grade))\n",
-      "    k += 1\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name :Suresh\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Roll No. :125\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Grade : A\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name :Mahesh\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Roll No. :126\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Grade : B\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name :Rajesh\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Roll No. :127\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Grade : A\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name \tRollno \tGrade\n",
-        "\n",
-        "Suresh\t125\tA\n",
-        "Mahesh\t126\tB\n",
-        "Rajesh\t127\tA"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.9, Page number: 419<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pointer to structure \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class book:\n",
-      "    name = ''\n",
-      "    author = ''\n",
-      "    pages = 0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "b1 = book()\n",
-      "\n",
-      "#Variable Initialization\n",
-      "b1.name = \"JAVA COMPLETE REFERENCE\"\n",
-      "b1.author = \"P.NAUGHTON\"\n",
-      "b1.pages = 886\n",
-      "\n",
-      "#There is no pointer concept in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n%s by %s of %d pages\"%(b1.name,b1.author,b1.pages))\n",
-      "sys.stdout.write(\"\\n%s by %s of %d pages\"%(b1.name,b1.author,b1.pages))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "JAVA COMPLETE REFERENCE by P.NAUGHTON of 886 pages\n",
-        "JAVA COMPLETE REFERENCE by P.NAUGHTON of 886 pages"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.10, Page number: 420<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pointer as members of structure \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class boy:\n",
-      "    name = ''\n",
-      "    age = 0\n",
-      "    height = 0.0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "sp = boy()\n",
-      "\n",
-      "#Variable Initialization\n",
-      "sp.name = \"Mahesh\"\n",
-      "sp.age = 20\n",
-      "sp.height = 5.40\n",
-      "\n",
-      "#There is no pointer concept in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nName  = %s\"%(sp.name))\n",
-      "sys.stdout.write(\"\\nAge  = %s\"%(sp.age))\n",
-      "sys.stdout.write(\"\\nHeight = %s\"%(sp.height))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name  = Mahesh\n",
-        "Age  = 20\n",
-        "Height = 5.4"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.11, Page number: 421<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pointer as members of structure \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class boy:\n",
-      "    name = ''\n",
-      "    age = 0\n",
-      "    height = 0.0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "b = boy()\n",
-      "\n",
-      "#Variable Initialization\n",
-      "nm = \"Somesh\"\n",
-      "ag = 20\n",
-      "ht = 5.40\n",
-      "\n",
-      "#There is no pointer concept in python\n",
-      "b.name = nm\n",
-      "b.age = ag\n",
-      "b.height = ht\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nName  : %s\"%(b.name))\n",
-      "sys.stdout.write(\"\\nAge = %d\"%(b.age))\n",
-      "sys.stdout.write(\"\\nHeight = %.2g\"%(b.height))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name  : Somesh\n",
-        "Age = 20\n",
-        "Height = 5.4"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.12, Page number: 422<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the contents of the structure using ordinary pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class num:\n",
-      "    a = 0\n",
-      "    b = 0\n",
-      "    c = 0\n",
-      "\n",
-      "#Class variable declaration\n",
-      "d = num()\n",
-      "\n",
-      "#Variable Initialization\n",
-      "d.a = 2\n",
-      "d.b = 3\n",
-      "d.c = 4\n",
-      "\n",
-      "#There is no pointer concept in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\na = %d\"%(d.a))\n",
-      "sys.stdout.write(\"\\nb = %d\"%(d.b))\n",
-      "sys.stdout.write(\"\\nc = %d\"%(d.c))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "a = 2\n",
-        "b = 3\n",
-        "c = 4"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.13, Page number: 423<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Passing address of structure variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class book:\n",
-      "    name = ''\n",
-      "    author = ''\n",
-      "    pages = 0\n",
-      "    \n",
-      "#Class variable declaration and initialization\n",
-      "b1 = book()\n",
-      "b1.name = \"JAVA COMPLETE REFERENCE\"\n",
-      "b1.author = \"P.NAUGHTON\"\n",
-      "b1.pages = 886\n",
-      "\n",
-      "#Function definition\n",
-      "def show(b1):\n",
-      "    sys.stdout.write(\"\\n%s by %s of %d pages\"%(b1.name,b1.author,b1.pages))\n",
-      "\n",
-      "#Function call\n",
-      "show(b1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "JAVA COMPLETE REFERENCE by P.NAUGHTON of 886 pages"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.14, Page number: 424<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Passing structure elements to function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class boy:\n",
-      "    name = ''\n",
-      "    age = 0\n",
-      "    wt = 0\n",
-      "    \n",
-      "#Class variable declaration and initialization\n",
-      "b1 = boy()\n",
-      "b1.name = \"Amit\"\n",
-      "b1.age = 20\n",
-      "b1.wt = 25\n",
-      "\n",
-      "#Function definition\n",
-      "def print1(s,t,n):\n",
-      "    sys.stdout.write(\"\\n%s %d %d\"%(s,t,n))\n",
-      "    \n",
-      "#Function call\n",
-      "print1(b1.name,b1.age,b1.wt)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Amit 20 25"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.15, Page number: 425<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Pass entire structure to user defined function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class boy:\n",
-      "    name = ''\n",
-      "    age = 0\n",
-      "    wt = 0\n",
-      "    \n",
-      "#Class variable declaration and initialization\n",
-      "b1 = boy()\n",
-      "b1.name = \"Amit\"\n",
-      "b1.age = 20\n",
-      "b1.wt = 25\n",
-      "\n",
-      "#Function definition\n",
-      "def print1(b):\n",
-      "    sys.stdout.write(\"\\n%s %d %d\"%(b.name,b.age,b1.wt))\n",
-      "    \n",
-      "#Function call\n",
-      "print1(b1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Amit 20 25"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.16, Page number: 426<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#User defined data type \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no preprocessor directive in python\n",
-      "H = 60\n",
-      "\n",
-      "#There is no typedef function in python and no separate variable declaration is needed\n",
-      "hrs = int(raw_input(\"Enter Hours :\"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nMinutes = %d\"%(hrs*H))\n",
-      "sys.stdout.write(\"\\nSeconds = %d\"%(hrs*H*H))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Hours :2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Minutes = 120\n",
-        "Seconds = 7200"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.17, Page number: 426<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#String data type\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no typedef function in python\n",
-      "a = \" Hello \"\n",
-      "b = raw_input(\"Enter Your Name : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"%s %s\"%(a,b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Name : KAMAL\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Hello  KAMAL"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.18, Page number: 427<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Create user defined data type from structure\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "#There is no typedef function in python\n",
-      "class struct:\n",
-      "    name = ''\n",
-      "    sex = ''\n",
-      "    acno = 0\n",
-      "    \n",
-      "#class variable declaration and initialization\n",
-      "employee = struct()\n",
-      "employee.name = \"Sanjay\"\n",
-      "employee.sex = \"M\"\n",
-      "employee.acno = 125\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nName\\tSex\\tA/c No.\\n\")\n",
-      "sys.stdout.write(\"%s\\t\"%(employee.name))\n",
-      "sys.stdout.write(\"%s\\t\"%(employee.sex))\n",
-      "sys.stdout.write(\"%s\\n\"%(employee.acno))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name\tSex\tA/c No.\n",
-        "Sanjay\tM\t125\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.19, Page number: 427<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Create user defined data type from structure\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class struct:\n",
-      "    name = ''\n",
-      "    sex = ''\n",
-      "    acno = 0\n",
-      "    \n",
-      "#class Variable declaration and initialization\n",
-      "employee = [struct() for i in range(0,2)]\n",
-      "\n",
-      "for k in range(0,2):\n",
-      "    employee[k].name = raw_input(\"Name of the Employee :\")\n",
-      "    employee[k].sex = raw_input(\"Sex\")\n",
-      "    employee[k].acno = raw_input(\"A/c No.\")\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nName\\tSex\\tA/c No.\\n\")\n",
-      "for k in range(0,2):\n",
-      "    sys.stdout.write(\"%s\\t\"%(employee[k].name))\n",
-      "    sys.stdout.write(\"%s\\t\"%(employee[k].sex))\n",
-      "    sys.stdout.write(\"%s\\n\"%(employee[k].acno))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name of the Employee :AJAY\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "SexM\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A/c No.122\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name of the Employee :ANITA\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "SexF\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A/c No.124\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name\tSex\tA/c No.\n",
-        "AJAY\tM\t122\n",
-        "ANITA\tF\t124\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.20, Page number: 429<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Structure containing the details of the employees\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no typedef function in python\n",
-      "\n",
-      "#Class definition\n",
-      "class struct:\n",
-      "    first = ''\n",
-      "    middle = ''\n",
-      "    last = ''\n",
-      "    city = ''\n",
-      "    pincode = 0\n",
-      "    \n",
-      "#Class variable declaration and initialization\n",
-      "person = [struct() for i in range(0,2)]\n",
-      "\n",
-      "for j in range(0,2):\n",
-      "    person[j].first = raw_input(\"First Name :\")\n",
-      "    person[j].middle = raw_input(\"Middle Name : \")\n",
-      "    person[j].last = raw_input(\"Last Name : \")\n",
-      "    person[j].city = raw_input(\"City & Pincode\")\n",
-      "    person[j].pincode = int(raw_input(\"City & Pincode\"))\n",
-      "    \n",
-      "#Result\n",
-      "for j in range(0,2):\n",
-      "    sys.stdout.write(\"\\nRecord No : %d\"%(j+1))\n",
-      "    sys.stdout.write(\"\\nFirst Name : %s\"%(person[j].first))\n",
-      "    sys.stdout.write(\"\\nMiddle Name : %s\"%(person[j].middle))\n",
-      "    sys.stdout.write(\"\\nLast Name : %s\"%(person[j].last))\n",
-      "    sys.stdout.write(\"\\nCity & Pincode : %s - %d\\n\"%(person[j].city,person[j].pincode))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "First Name :Jay\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Middle Name : Mohan\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Last Name : Deshmukh\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "City & PincodeNanded\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "City & Pincode431602\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "First Name :Vijay\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Middle Name : Kamal\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Last Name : Nandedkar\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "City & PincodeNanded\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "City & Pincode431602\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Record No : 1\n",
-        "First Name : Jay\n",
-        "Middle Name : Mohan\n",
-        "Last Name : Deshmukh\n",
-        "City & Pincode : Nanded - 431602\n",
-        "\n",
-        "Record No : 2\n",
-        "First Name : Vijay\n",
-        "Middle Name : Kamal\n",
-        "Last Name : Nandedkar\n",
-        "City & Pincode : Nanded - 431602\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.21, Page number: 431<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Information of vehicles\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "PETROL = 1\n",
-      "DISEL = 2\n",
-      "TWO_WH = 3\n",
-      "FOUR_WH = 4\n",
-      "OLD = 5\n",
-      "NEW = 6\n",
-      "\n",
-      "#Class declaration\n",
-      "class vehicle:\n",
-      "    type1 = 3\n",
-      "    fuel = 2\n",
-      "    model = 3\n",
-      "    \n",
-      "#Class variable declaration and initialization\n",
-      "v = vehicle()\n",
-      "v.type1 = FOUR_WH\n",
-      "v.fuel = DISEL\n",
-      "v.model = OLD\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nType of Vehicle : %d\"%(v.type1))\n",
-      "sys.stdout.write(\"\\nFuel  : %d\"%(v.fuel))\n",
-      "sys.stdout.write(\"\\nModel  : %d\"%(v.model))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Type of Vehicle : 4\n",
-        "Fuel  : 2\n",
-        "Model  : 5"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.22, Page number: 432<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the examination result of the student \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "#There is no preprocessor directives in python\n",
-      "PASS = 1\n",
-      "FAIL = 0\n",
-      "A = 0\n",
-      "B = 1\n",
-      "C = 2\n",
-      "\n",
-      "#Class declaration\n",
-      "class student:\n",
-      "    name = ''\n",
-      "    result = 1\n",
-      "    grade = 2\n",
-      "    \n",
-      "#Class variable declaration and initialization\n",
-      "v = student()\n",
-      "\n",
-      "v.name = \"Sachin\"\n",
-      "v.result = PASS\n",
-      "v.grade = C\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nName : %s\"%(v.name))\n",
-      "sys.stdout.write(\"\\nResult : %d\"%(v.result))\n",
-      "sys.stdout.write(\"\\nGrade : %d\"%(v.grade))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name : Sachin\n",
-        "Result : 1\n",
-        "Grade : 2"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.23, Page number: 433<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Enumerated data type for 12 months\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no enumerated data type in python and an alternate is to use class\n",
-      "#Class declaration\n",
-      "class month:\n",
-      "    Jan = 0\n",
-      "    Feb = 1\n",
-      "    Mar = 2\n",
-      "    Apr = 3\n",
-      "    May = 4\n",
-      "    June = 5\n",
-      "    July = 6\n",
-      "    Aug = 7\n",
-      "    Sep = 8\n",
-      "    Oct = 9\n",
-      "    Nov = 10\n",
-      "    Dec = 11\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "c = month()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nJan = %d\"%(c.Jan))\n",
-      "sys.stdout.write(\"\\nFeb = %d\"%(c.Feb))\n",
-      "sys.stdout.write(\"\\nJune = %d\"%(c.June))\n",
-      "sys.stdout.write(\"\\nDec = %d\"%(c.Dec))\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Jan = 0\n",
-        "Feb = 1\n",
-        "June = 5\n",
-        "Dec = 11"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.24, Page number: 434<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Enumerated data type \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no enumerated data type in python and an alternate is to use class\n",
-      "#Class declaration\n",
-      "class month:\n",
-      "    Jan = 1\n",
-      "    Feb = 2\n",
-      "    Mar = 3\n",
-      "    Apr = 4\n",
-      "    May = 5\n",
-      "    June = 6\n",
-      "    July = 7\n",
-      "    Aug = 8\n",
-      "    Sep = 9\n",
-      "    Oct = 10\n",
-      "    Nov = 11\n",
-      "    Dec = 12\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "c = month()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nJan = %d\"%(c.Jan))\n",
-      "sys.stdout.write(\"\\nFeb = %d\"%(c.Feb))\n",
-      "sys.stdout.write(\"\\nJune = %d\"%(c.June))\n",
-      "sys.stdout.write(\"\\nDec = %d\"%(c.Dec))\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Jan = 1\n",
-        "Feb = 2\n",
-        "June = 6\n",
-        "Dec = 12"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.25, Page number: 434<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display name of month using enumerated data type\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no enumerated data type in python and an alternate is to use class\n",
-      "#Class declaration\n",
-      "class month:\n",
-      "    Jan = 1\n",
-      "    Feb = 2\n",
-      "    Mar = 3\n",
-      "    Apr = 4\n",
-      "    May = 5\n",
-      "    June = 6\n",
-      "    July = 7\n",
-      "    Aug = 8\n",
-      "    Sep = 9\n",
-      "    Oct = 10\n",
-      "    Nov = 11\n",
-      "    Dec = 12\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "c = month()\n",
-      "\n",
-      "#Result\n",
-      "for f in range(c.Jan,c.Dec+1):\n",
-      "    #There is no switch case statement in python\n",
-      "    if f == c.Jan:\n",
-      "        sys.stdout.write(\"\\nJanuary\")\n",
-      "    else:\n",
-      "        if f == c.Feb:\n",
-      "            sys.stdout.write(\"\\nFebruary\")\n",
-      "        else:\n",
-      "            if f == c.Mar:\n",
-      "                sys.stdout.write(\"\\nMarch\")\n",
-      "            else:\n",
-      "                if f == c.Apr:\n",
-      "                    sys.stdout.write(\"\\nApril\")\n",
-      "                else:\n",
-      "                    if f == c.May:\n",
-      "                        sys.stdout.write(\"\\nMay\")\n",
-      "                    else:\n",
-      "                        if f == c.June:\n",
-      "                            sys.stdout.write(\"\\nJune\")\n",
-      "                        else:\n",
-      "                            if f == c.July:\n",
-      "                                sys.stdout.write(\"\\nJuly\")\n",
-      "                            else:\n",
-      "                                if f == c.Aug:\n",
-      "                                    sys.stdout.write(\"\\nAugust\")\n",
-      "                                else:\n",
-      "                                    if f == c.Sep:\n",
-      "                                        sys.stdout.write(\"\\nSeptember\")\n",
-      "                                    else:\n",
-      "                                        if f == c.Oct:\n",
-      "                                            sys.stdout.write(\"\\nOctober\")\n",
-      "                                        else:\n",
-      "                                            if f == c.Nov:\n",
-      "                                                sys.stdout.write(\"\\nNovember\")\n",
-      "                                            else:\n",
-      "                                                sys.stdout.write(\"\\nDecember\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "January\n",
-        "February\n",
-        "March\n",
-        "April\n",
-        "May\n",
-        "June\n",
-        "July\n",
-        "August\n",
-        "September\n",
-        "October\n",
-        "November\n",
-        "December"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.26, Page number: 436<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Enumerated data type\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no enumerated data type in python. class is used instead\n",
-      "class capital:\n",
-      "    Mumbai = 0\n",
-      "    Hyderabad = 1\n",
-      "    Bangalore = 2\n",
-      "\n",
-      "class state:\n",
-      "    name = ''\n",
-      "    c = capital()\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "s = state()\n",
-      "c1 = capital()\n",
-      "\n",
-      "#Class variable initialization\n",
-      "s.name = \"Andhra Pradesh\"\n",
-      "s.c = s.c.Hyderabad\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nState  : %s\"%(s.name))\n",
-      "sys.stdout.write(\"\\nCapital : %d\"%(s.c))\n",
-      "\n",
-      "if s.c == c1.Hyderabad:\n",
-      "    sys.stdout.write(\"\\nHyderabad is Capital of %s\"%(s.name))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "State  : Andhra Pradesh\n",
-        "Capital : 1\n",
-        "Hyderabad is Capital of Andhra Pradesh"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.27, Page number: 437<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Identify the type of entered character using enumerated data type.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration instead of enumerated data type\n",
-      "class ctype:\n",
-      "    Letter = 0\n",
-      "    Digit = 1\n",
-      "    Other = 2\n",
-      "    \n",
-      "#Variable Initialization\n",
-      "ch = raw_input(\"Enter any character\")\n",
-      "c = ctype()\n",
-      "f = ch.isalpha()\n",
-      "\n",
-      "#Result\n",
-      "if f != 0:\n",
-      "    sys.stdout.write(\"\\n%c is %d type of symbol\"%(ch,c.Letter))\n",
-      "else:\n",
-      "    f = ch.isdigit()\n",
-      "    if f != 0:\n",
-      "        sys.stdout.write(\"\\n%c is %d type of symbol\"%(ch,c.Digit))\n",
-      "    else:\n",
-      "        sys.stdout.write(\"\\n%c is %d type of symbol\"%(ch,c.Other))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any character=\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "= is 2 type of symbol"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.28, Page number: 438<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Size of union and number of bytes reserved for it\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no union/structure in python. class is used instead\n",
-      "#Class declaration\n",
-      "class result:\n",
-      "    marks = 0\n",
-      "    grade = ''\n",
-      "    \n",
-      "class res:\n",
-      "    name = ''\n",
-      "    age = 0\n",
-      "    perf = result()\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "data = res()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Size of Union : %d\\n\"%(sys.getsizeof(data.perf)))\n",
-      "sys.stdout.write(\"Size of Structure : %d\\n\"%(sys.getsizeof(data)))\n",
-      "\n",
-      "#in python, value tagged method is used for data storage and can represent large numbers"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Size of Union : 36\n",
-        "Size of Structure : 36\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.29, Page number: 440<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Memory size of the computer\n",
-      "\n",
-      "import psutil\n",
-      "\n",
-      "psutil.phymem_usage()\n",
-      "\n",
-      "#Displays current status of the memory\n",
-      "#different systems will have different memory status"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "pyout",
-       "prompt_number": 4,
-       "text": [
-        "usage(total=1600512000L, used=1496383488L, free=104128512L, percent=93.5)"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.30, Page number: 440<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the system time at specified cursor position\n",
-      "\n",
-      "import sys\n",
-      "import datetime\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"%s\"%(datetime.datetime.now()))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2013-09-20 11:04:15.649000"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.31, Page number: 441<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Change the cursor in different sizes\n",
-      "\n",
-      "import turtle\n",
-      "#Ipython supports graphics through turtle module\n",
-      "\n",
-      "from Tkinter import *\n",
-      "import Tkinter\n",
-      "\n",
-      "top = Tkinter.Tk()\n",
-      "\n",
-      "B1 = Tkinter.Button(top, text =\"circle\", relief=RAISED,\\\n",
-      "                         cursor=\"circle\")\n",
-      "B2 = Tkinter.Button(top, text =\"plus\", relief=RAISED,\\\n",
-      "                         cursor=\"plus\")\n",
-      "B1.pack()\n",
-      "B2.pack()\n",
-      "#top.mainloop()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.32, Page number: 441<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Create a directory using dos interrupt\n",
-      "\n",
-      "import os\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "dir1 = raw_input(\"Enter Directory Name : \")\n",
-      "\n",
-      "#Result\n",
-      "if not os.path.exists(dir1):\n",
-      "    os.makedirs(dir1)\n",
-      "    sys.stdout.write(\"Directory %s created\"%(dir1))\n",
-      "else:\n",
-      "    sys.stdout.write(\"Directory %s not created\"%(dir1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Directory XYZ created"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.33, Page number: 442<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the given character on the screen\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "dl = 67\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"%c\"%(chr(dl)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.34, Page number: 442<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the attributes of a file using DOS interrupt\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "import os\n",
-      "\n",
-      "print os.stat(\"IO.SYS\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "nt.stat_result(st_mode=33206, st_ino=0L, st_dev=0, st_nlink=0, st_uid=0, st_gid=0, st_size=0L, st_atime=1383454467L, st_mtime=1383454467L, st_ctime=1383454467L)\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.35, Page number: 444<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Delete a file using dos interrupt\n",
-      "\n",
-      "import os\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "file1 = raw_input(\"Enter a file name : \")\n",
-      "\n",
-      "try:\n",
-      "    os.remove(file1)\n",
-      "    sys.stdout.write(\"File successfully deleted\")\n",
-      "except:\n",
-      "    sys.stdout.write(\"File could not be deleted\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "File successfully deleted"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 13.36, Page number: 445<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Structuret within union\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#there is no union/structure in python. class is used instead\n",
-      "#Class declaration\n",
-      "class x:\n",
-      "    f = 0.0\n",
-      "    p = ['' for i in range(0,2)]\n",
-      "    \n",
-      "class z:\n",
-      "    set1 = x()\n",
-      "    \n",
-      "#Class variable declaration and initialization\n",
-      "st = z()\n",
-      "\n",
-      "st.set1.f = 5.5\n",
-      "st.set1.p[0] = 65\n",
-      "st.set1.p[1] = 66\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n%g\"%(st.set1.f))\n",
-      "sys.stdout.write(\"\\n%c\"%(st.set1.p[0]))\n",
-      "sys.stdout.write(\"\\n%c\"%(st.set1.p[1]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "5.5\n",
-        "A\n",
-        "B"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter14.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter14.ipynb
deleted file mode 100755
index 409585bd..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter14.ipynb
+++ /dev/null
@@ -1,2311 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 14: Files<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.1, Page number: 454<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write data to text file and read it\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a')\n",
-      "\n",
-      "#Write data to file if successfully created the file\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
-      "    c = ''\n",
-      "    while c != '.':\n",
-      "        c = raw_input(\"Write data & to stop press '.'\")\n",
-      "        fp.write(c)\n",
-      "    fp.close()\n",
-      "\n",
-      "    #Read and display the contents of file\n",
-      "    sys.stdout.write(\"\\nContents read : \")\n",
-      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
-      "    f = fp.readlines()\n",
-      "    for f1 in f:\n",
-      "        print f1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Write data & to stop press '.'"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Write data & to stop press '.'ABCDEFGHIJK\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Write data & to stop press '.'.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Contents read : ABCDEFGHIJK.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.2, Page number: 455<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the contents of the file before and after appending.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file for read data\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
-      "\n",
-      "#Read and display the contents of file\n",
-      "sys.stdout.write(\"\\nContents of file before appending :\\n\")\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
-      "f = fp.readlines()\n",
-      "for f1 in f:\n",
-      "    print f1\n",
-      "\n",
-      "#Open file for appending\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a')\n",
-      "\n",
-      "if f == 0:\n",
-      "    sys.stdout.write(\"File can not appended\")\n",
-      "else:\n",
-      "    c = ''\n",
-      "    #fp.write('\\n')\n",
-      "    while c != '.':\n",
-      "        c = raw_input(\"Enter string to append\")\n",
-      "        fp.write(c)\n",
-      "    \n",
-      "    fp.close()\n",
-      "    \n",
-      "   #Read and display the contents of file\n",
-      "    sys.stdout.write(\"\\nContents of file After appending\\n\")\n",
-      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
-      "    f = fp.readlines()\n",
-      "    for f1 in f:\n",
-      "        print f1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Contents of file before appending :\n",
-        "String is terminated with '\\0'.\n",
-        "\n",
-        "Contents of file After appending\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String is terminated with '\\0'.This character is called as NULL character.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.3, Page number: 457<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Writing and reading of a file.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','w+')\n",
-      "\n",
-      "#Write data to file if successfully created the file\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
-      "    c = ''\n",
-      "    while c != '.':\n",
-      "        c = raw_input(\"Write data & to stop press '.'\")\n",
-      "        fp.write(c)\n",
-      "    fp.close()\n",
-      "\n",
-      "    #Read and display the contents of file\n",
-      "    sys.stdout.write(\"\\nContents read : \")\n",
-      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
-      "    f = fp.readlines()\n",
-      "    for f1 in f:\n",
-      "        print f1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Write data & to stop press '.'\n",
-        "Contents read : "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ABCDEFGHIJK.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.4, Page number: 458<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Open a file in append mode and add new records in it.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a+')\n",
-      "\n",
-      "#Write data to file if successfully created the file\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
-      "    c = ''\n",
-      "    while c != '.':\n",
-      "        c = raw_input(\"Write data & to stop press '.'\")\n",
-      "        fp.write(c)\n",
-      "    fp.close()\n",
-      "\n",
-      "    #Read and display the contents of file\n",
-      "    sys.stdout.write(\"\\nContents read : \")\n",
-      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a+')\n",
-      "    f = fp.readlines()\n",
-      "    for f1 in f:\n",
-      "        print f1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Write data & to stop press '.'\n",
-        "Contents read : "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is append and read mode.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.5, Page number: 459<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Open a file in read/write mode in it\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r+')\n",
-      "\n",
-      "#Write data to file if successfully created the file\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    #Read and display the contents of file\n",
-      "    sys.stdout.write(\"\\nContents read : \")\n",
-      "    f = fp.readlines()\n",
-      "    for f1 in f:\n",
-      "        print f1\n",
-      "        \n",
-      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
-      "    c = ''\n",
-      "    while c != '.':\n",
-      "        c = raw_input(\"Write data & to stop press '.'\")\n",
-      "        fp.write(c)\n",
-      "    fp.close()\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Contents read : Help me.\n",
-        "Write data & to stop press '.'"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.6, Page number: 460<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Open a file for read/write operation in binary mode\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','wb')\n",
-      "\n",
-      "#Write data to file if successfully created the file\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
-      "    c = ''\n",
-      "    while c != '.':\n",
-      "        c = raw_input(\"Write data & to stop press '.'\")\n",
-      "        fp.write(c)\n",
-      "    fp.close()\n",
-      "\n",
-      "    #Read and display the contents of file\n",
-      "    sys.stdout.write(\"\\nContents read : \")\n",
-      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','rb')\n",
-      "    f = fp.readlines()\n",
-      "    for f1 in f:\n",
-      "        print f1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Write data & to stop press '.'\n",
-        "Contents read : "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ABCDEFGHIJK.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.7, Page number: 462<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Open a text file and write some text \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','w')\n",
-      "\n",
-      "#Read the string\n",
-      "text = raw_input(\"Enter Text Here : \")\n",
-      "\n",
-      "#Write to the file\n",
-      "#fp.write() is the equivalent function for fprintf() in python\n",
-      "fp.write(\"%s\"%(text))\n",
-      "\n",
-      "#To see the result, open the data.txt file"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.8, Page number: 463<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Enter data into the text file and read the same.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','w')\n",
-      "\n",
-      "#Read the data\n",
-      "text = raw_input(\"Name & Age\")\n",
-      "age = int(raw_input(\"Name & Age\"))\n",
-      "\n",
-      "#Write to file\n",
-      "fp.write(\"%s   %d\"%(text,age))\n",
-      "\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
-      "#Result\n",
-      "sys.stdout.write(\"Name\\tAge\\n\")\n",
-      "text = fp.read()\n",
-      "age = fp.read()\n",
-      "sys.stdout.write(\"%s\\t%s\\n\"%(text,age))\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name\tAge\n",
-        "AMIT   12\t\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.9, Page number: 463<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the contents of the file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "f = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\list.txt','r')\n",
-      "\n",
-      "if f == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    for c in f:\n",
-      "        sys.stdout.write(\"%s\"%(c))\n",
-      "    #Python gives iterative statements there is no getc() fucntion"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "aman\n",
-        "akash\n",
-        "amit\n",
-        "ajay\n",
-        "ankit"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.10, Page number: 464<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write some text into the file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\words.doc','w')\n",
-      "\n",
-      "#Write data to file\n",
-      "c = ''\n",
-      "while c != '*':\n",
-      "    c = raw_input(\"Enter Few Words '*' to Exit\")\n",
-      "    fp.write(c)\n",
-      "\n",
-      "fp.close()\n",
-      "    #There is no putc() function in python\n",
-      "    #Open the file to see the result"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.11, Page number: 465<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count\n",
-      "\n",
-      "# Total number of statements\n",
-      "# Total number of included files\n",
-      "# Total number of blocks and brackets\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fs = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\PRG2.C','r')\n",
-      "\n",
-      "#variable initialization\n",
-      "i = 0\n",
-      "c = 0\n",
-      "sb = 0\n",
-      "b = 0\n",
-      "\n",
-      "if fs == 0:\n",
-      "    sys.stdout.write(\"File opening error\")\n",
-      "else:\n",
-      "    for line in fs:\n",
-      "        k = 0\n",
-      "        while k < len(line):\n",
-      "            if line[k] == ';':\n",
-      "                c += 1\n",
-      "            else:\n",
-      "                if line[k] == '{':\n",
-      "                    sb += 1\n",
-      "                else:\n",
-      "                    if line[k] == '(':\n",
-      "                        b += 1\n",
-      "                    else:\n",
-      "                        if line[k] == '#':\n",
-      "                            i += 1\n",
-      "            k += 1\n",
-      "                            \n",
-      "    #Result\n",
-      "    sys.stdout.write(\"\\nSummary of 'C' Program\\n\")\n",
-      "    sys.stdout.write(\"===========================\")\n",
-      "    sys.stdout.write(\"\\nTotal Statments : %d\"%(c+i))\n",
-      "    sys.stdout.write(\"\\nInclude Statements : %d\"%(i))\n",
-      "    sys.stdout.write(\"\\nTotal Blocks {} : %d\"%(sb))\n",
-      "    sys.stdout.write(\"\\nTotal Brackets () : %d\"%(b))\n",
-      "    sys.stdout.write(\"\\n============================\")\n",
-      "                            "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Summary of 'C' Program\n",
-        "===========================\n",
-        "Total Statments : 8\n",
-        "Include Statements : 2\n",
-        "Total Blocks {} : 2\n",
-        "Total Brackets () : 9\n",
-        "============================"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.12, Page number: 466<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write text to a file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\lines.txt','w')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"\")\n",
-      "else:\n",
-      "    #Write data to file\n",
-      "    c = ''\n",
-      "    while c != '*':\n",
-      "        c = raw_input(\"\")\n",
-      "        fp.write(c)\n",
-      "    fp.close()\n",
-      "    #There is no fputc() function in python\n",
-      "    #Open the file to see the result"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.13, Page number: 467<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read text from the given file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "\n",
-      "file1 = raw_input(\"Enter File Name : \")\n",
-      "fp = open(file1,'r')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"File not found\\n\")\n",
-      "else:\n",
-      "    for text in fp:\n",
-      "        print text\n",
-      "    #File pointer itself is iterable in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "#include <stdio.h>\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.14, Page number: 468<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write a string into a file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "\n",
-      "file1 = raw_input(\"Enter the name of file : \")\n",
-      "fp = open(file1,'w')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"File can not opened\\n\")\n",
-      "else:\n",
-      "    text = raw_input(\"Enter Text Here : \")\n",
-      "    fp.write(text)\n",
-      "    \n",
-      "    #Write() function is used to write character or string into the file in python\n",
-      "    # there is no fputc() function in python\n",
-      "#open the file to see the result"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.15, Page number: 469<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write integers in a file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('num.txt','w')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"File does not exist\\n\")\n",
-      "else:\n",
-      "    v = ' '\n",
-      "    while v != '0':\n",
-      "        fp.write(v)\n",
-      "        v = raw_input(\"Enter Numbers\")\n",
-      "        \n",
-      "        #open the file to see the result"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.16, Page number: 470<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read integers from the file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('num.txt','r')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"File does not exist\\n\")\n",
-      "else:\n",
-      "    for v in fp:\n",
-      "        print v"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 2 3 4 5\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.17, Page number: 471<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write a block of structure elements to the given file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class student:\n",
-      "    name = ''\n",
-      "    age = 0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "stud = [student() for i in range(0,5)]\n",
-      "\n",
-      "#Open file\n",
-      "file1 = raw_input(\"Enter the file name : \")\n",
-      "fp = open(file1,'w')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"File does not exist\\n\")\n",
-      "else:\n",
-      "    n = int(raw_input(\"How Many Records : \"))\n",
-      "    for i in range(0,n):\n",
-      "        stud[i].name = raw_input(\"Name  : \")\n",
-      "        stud[i].age = int(raw_input(\"Age : \"))\n",
-      "    \n",
-      "    j = 0\n",
-      "    while j < n:\n",
-      "        fp.write(\"%s %d\\n\"%(stud[j].name,stud[j].age))\n",
-      "        j += 1\n",
-      "        \n",
-      "    fp.close()\n",
-      "    \n",
-      "    #open the file to see the result"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.18, Page number: 472<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write and read the information about the player \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class record:\n",
-      "    player = ''\n",
-      "    age = 0\n",
-      "    runs = 0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "emp = record()\n",
-      "\n",
-      "#Open file\n",
-      "fp = open('record.dat','w')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Can not open the file\\n\")\n",
-      "else:\n",
-      "    emp.player = raw_input(\"Enter Player Name\")\n",
-      "    emp.age = int(raw_input(\"Enter Age\"))\n",
-      "    emp.runs = int(raw_input(\"Enter runs scored\"))\n",
-      "    fp.write(\"%s %d %d\"%(emp.player,emp.age,emp.runs))\n",
-      "    fp.close()\n",
-      "    \n",
-      "    fp = open('record.dat','r')\n",
-      "    if fp == 0:\n",
-      "        sys.stdout.write(\"Error in opening file\")\n",
-      "    else:\n",
-      "        sys.stdout.write(\"\\nRecord Entered is\\n\")\n",
-      "        emp.player,emp.age,emp.runs = fp.read().split(' ')\n",
-      "        sys.stdout.write(\"\\n%s  %s  %s\"%(emp.player,emp.age,emp.runs))\n",
-      "        fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Player NameSachin\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Age25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter runs scored10000\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Record Entered is\n",
-        "\n",
-        "Sachin  25  10000"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.19, Page number: 473<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Write a block of structure elements to the given file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "next1 = 'Y'\n",
-      "\n",
-      "#Class declaration\n",
-      "class bike:\n",
-      "    name = ''\n",
-      "    avg = 0\n",
-      "    cost = 0.0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "e = bike()\n",
-      "\n",
-      "#open file\n",
-      "fp = open('bk.txt','wb')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    while next1 == 'Y':\n",
-      "        e.name = raw_input(\"Model Name\")\n",
-      "        e.avg = int(raw_input(\"Average\"))\n",
-      "        e.cost = float(raw_input(\"Price\"))\n",
-      "        fp.write(\"%s %d %f\\n\"%(e.name,e.avg,e.cost))\n",
-      "        \n",
-      "        next1 = raw_input(\"Add Another (Y/N):\")\n",
-      "    fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Model NameHONDA\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average80\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Price45000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Add Another (Y/N):Y\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Model NameSUZUKI\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average65\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Price43000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Add Another (Y/N):Y\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Model NameYAMAHA\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average55\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Price48000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Add Another (Y/N):N\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.20, Page number: 474<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the information about the bike \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class bike:\n",
-      "    name = ''\n",
-      "    avg = 0\n",
-      "    cost = 0.0\n",
-      "    \n",
-      "#Class variable declaration\n",
-      "e = bike()\n",
-      "\n",
-      "#open file\n",
-      "fp = open('bk.txt','rb')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Cannot open file\")\n",
-      "else:\n",
-      "    lines = fp.readlines()\n",
-      "    for line in lines:\n",
-      "        e.name,e.avg,e.cost = line.split(' ')\n",
-      "        e.avg = int(e.avg)\n",
-      "        e.cost = float(e.cost)\n",
-      "        sys.stdout.write(\"\\n%s %d %.2f\"%(e.name,e.avg,e.cost))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "HONDA 80 45000.00\n",
-        "SUZUKI 65 43000.00\n",
-        "YAMAHA 55 48000.00"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.21, Page number: 476<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the text after skipping n characters from beginning of the file\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open(\"text.txt\",\"r\")\n",
-      "\n",
-      "sys.stdout.write(\"\\nContents of file\\n\")\n",
-      "ch = fp.read()\n",
-      "sys.stdout.write(\"%s\"%(ch))\n",
-      "\n",
-      "#Read n\n",
-      "n = int(raw_input(\"How many characters including spaces would you like to skip?\"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nInformation after %d bytes\\n\"%(n))\n",
-      "\n",
-      "ch = ch[n:]\n",
-      "sys.stdout.write(\"%s\"%(ch))\n",
-      "\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Contents of file\n",
-        "THE C PROGRAMMING LANGUAGE INVENTED BY DENNIS RITCHIE"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many characters including spaces would you like to skip?18\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Information after 18 bytes\n",
-        "LANGUAGE INVENTED BY DENNIS RITCHIE"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.22, Page number: 477<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read last few characters of the file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open(\"text.txt\",\"r\")\n",
-      "\n",
-      "sys.stdout.write(\"\\nContents of file\\n\")\n",
-      "ch = fp.read()\n",
-      "sys.stdout.write(\"%s\"%(ch))\n",
-      "\n",
-      "#Read n\n",
-      "n = int(raw_input(\"How many characters including spaces would you like to skip?\"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nInformation after %d bytes\\n\"%(n))\n",
-      "\n",
-      "print ch[-n:]\n",
-      "\n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Contents of file\n",
-        "HELLO WORLD"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many characters including spaces would you like to skip?5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Information after 5 bytes\n",
-        "WORLD\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.23, Page number: 477<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display 'c' program files in current directory. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "l = 0\n",
-      "c = 0\n",
-      "\n",
-      "#Open file\n",
-      "file1 = raw_input(\"Enter the file name : \")\n",
-      "fp = open(file1,'r')\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nContents of 'c' program file in capital case\")\n",
-      "sys.stdout.write(\"\\n============================================\\n\")\n",
-      "\n",
-      "ch = fp.readlines()\n",
-      "\n",
-      "for line in ch:\n",
-      "    i = 0\n",
-      "    #print line.upper()\n",
-      "    while i < len(line):\n",
-      "        if line[i] =='\\n':\n",
-      "            l += 1\n",
-      "        else:\n",
-      "             c += 1\n",
-      "        sys.stdout.write(\"%c\"%(line[i].upper()))\n",
-      "        i += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\nTotal Characters : %d\"%(c))\n",
-      "sys.stdout.write(\"\\nTotal Lines : %d\"%(l))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Contents of 'c' program file in capital case\n",
-        "============================================\n",
-        "MAIN()\n",
-        "{\n",
-        "PRINTF(\" HELLO WORLD\");\n",
-        "}\n",
-        "\n",
-        "Total Characters : 31\n",
-        "Total Lines : 4"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.24, Page number: 479<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect the end of file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open(\"text.txt\",\"r\")\n",
-      "\n",
-      "#there is no feof() function in python\n",
-      "c = fp.tell()\n",
-      "sys.stdout.write(\"File pointer at the beginning of the file : %d\\n\"%(c))\n",
-      "\n",
-      "c = fp.read()\n",
-      "sys.stdout.write(\"%s\"%(c))\n",
-      "\n",
-      "c = fp.tell()\n",
-      "sys.stdout.write(\"\\nFile pointer at the end of file : %d\"%(c))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "File pointer at the beginning of the file : 0\n",
-        "TECHNOCRATS LEAD THE WORLD      \n",
-        "File pointer at the end of file : 32"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.25, Page number: 480<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect an error while read/write operation of a file is in use.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "next1 = 'Y'\n",
-      "\n",
-      "#open file\n",
-      "fp = open('marks.dat','r')\n",
-      "\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Can not open file\")\n",
-      "else:\n",
-      "    while next1 == 'Y':\n",
-      "        name = raw_input(\"Enter Name, Marks, Percentage\")\n",
-      "        marks = int(raw_input(\"Enter Name, Marks, Percentage\"))\n",
-      "        \n",
-      "        p = marks/7\n",
-      "        try:\n",
-      "            fp.write(\"%s %d %f\"%(name,marks,p))\n",
-      "        except:\n",
-      "            sys.stdout.write(\"\\nUnable to write data?\")\n",
-      "            sys.stdout.write(\"\\nFile opening mode is incorrect.\")\n",
-      "            fp.close()\n",
-      "            \n",
-      "        next1 = raw_input(\"Continue Y/N:\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name, Marks, PercentageKAMAL\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name, Marks, Percentage540\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Unable to write data?\n",
-        "File opening mode is incorrect."
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Continue Y/N:N\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.26, Page number: 481<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Catch the error that is occurred while file operation\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "f = open(\"io8.c\",\"w\")\n",
-      "\n",
-      "if f == 0:\n",
-      "    sys.stdout.write(\"\\nCannot open file\")\n",
-      "else:\n",
-      "    #Exception handling\n",
-      "    try:\n",
-      "        c = fp.readlines()\n",
-      "        sys.stdout.write(\"%s\"%(c))\n",
-      "    except:\n",
-      "        sys.stdout.write(\"\\nCan't read file.\")\n",
-      "#There is no ferror() function python."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Can't read file."
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.27, Page number: 482<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect and print the error message\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "file1 = \"lines.txt\"\n",
-      "\n",
-      "#Open file\n",
-      "fr = open(file1,\"w\")\n",
-      "\n",
-      "sys.stdout.write(\"%s  : \"%(file1))\n",
-      "#Exception handling\n",
-      "try:\n",
-      "    c = fp.readlines()\n",
-      "except:\n",
-      "    sys.stdout.write(\"Permission Denied\")\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "lines.txt  : Permission Denied"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.28, Page number: 482<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the current position of the file pointer \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Open file\n",
-      "fp = open(\"text.txt\",\"r\")\n",
-      "\n",
-      "#Set the pointer\n",
-      "fp.seek(21)\n",
-      "\n",
-      "#Result\n",
-      "while True:\n",
-      "    c = fp.read(1)\n",
-      "    if not c:\n",
-      "      break\n",
-      "    sys.stdout.write(\"%c\\t%d\\n\"%(c,fp.tell()))\n",
-      "#There is no endof file in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "W\t22\n",
-        "O\t23\n",
-        "R\t24\n",
-        "L\t25\n",
-        "D\t26\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.29, Page number: 483<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect beginning of file\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#open file\n",
-      "fp = open(\"text.txt\",\"r\")\n",
-      "\n",
-      "fp.seek(12)\n",
-      "sys.stdout.write(\"Pointer is at %d\\n\"%(fp.tell()))\n",
-      "sys.stdout.write(\"Before rewind() : \")\n",
-      "\n",
-      "#Result\n",
-      "while True:\n",
-      "    c = fp.read(1)\n",
-      "    if not c:\n",
-      "      break\n",
-      "    sys.stdout.write(\"%c\"%(c))\n",
-      "    \n",
-      "sys.stdout.write(\"\\nAfter rewind() : \")\n",
-      "fp.seek(0)\n",
-      "#There is no rewind function in python\n",
-      "\n",
-      "while True:\n",
-      "    c = fp.read(1)\n",
-      "    if not c:\n",
-      "      break\n",
-      "    sys.stdout.write(\"%c\"%(c))\n",
-      "    \n",
-      "fp.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Pointer is at 12\n",
-        "Before rewind() : LEAD THE WORLD\n",
-        "After rewind() : TECHNOCRATS LEAD THE WORLD"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.30, Page number: 484<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Delete the given file \n",
-      "\n",
-      "import sys\n",
-      "import os\n",
-      "\n",
-      "#Read file name\n",
-      "file1 = raw_input(\"Enter The File Name : \")\n",
-      "\n",
-      "#There is no remove or unlink file function in python.\n",
-      "#A file can be deleted using remove function in the os module.\n",
-      "\n",
-      "try:\n",
-      "    os.remove(file1)\n",
-      "    sys.stdout.write(\"\\nFile (%s) has been deleted!\"%(file1))\n",
-      "except:\n",
-      "    sys.stdout.write(\"\\nFile does not exist\")\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "File (TEXT.TXT) has been deleted!"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.31, Page number: 485<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Change the name of the file\n",
-      "\n",
-      "import os\n",
-      "import sys\n",
-      "\n",
-      "old = raw_input(\"Old File Name : \")\n",
-      "\n",
-      "new = raw_input(\"New File Name : \")\n",
-      "\n",
-      "os.rename(old, new)\n",
-      "\n",
-      "#Check the directory to see the result\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.32, Page number: 486<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy the contents of one file to another file\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "fs = open(\"a.txt\",\"r\")\n",
-      "ft = open(\"b.txt\",\"w\")\n",
-      "c = 0\n",
-      "\n",
-      "if fs == 0:\n",
-      "    sys.stdout.write(\"\\nSource file opening error.\")\n",
-      "else:\n",
-      "    if ft == 0:\n",
-      "        sys.stdout.write(\"\\nTarget file opening error.\")\n",
-      "    else:\n",
-      "        while True:\n",
-      "            ch = fs.read(1)\n",
-      "            if not ch:\n",
-      "                break\n",
-      "            ft.write(\"%c\"%(ch))\n",
-      "            c += 1\n",
-      "        sys.stdout.write(\"\\n%d Bytes copied from 'a.txt' to 'b.txt'.\"%(c))\n",
-      "        c = 0\n",
-      "        #there is no fcloseall() function in python\n",
-      "        fs.close()\n",
-      "        c += 1\n",
-      "        ft.close()\n",
-      "        c += 1\n",
-      "        sys.stdout.write(\"\\n%d files closed.\"%(c))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "45 Bytes copied from 'a.txt' to 'b.txt'.\n",
-        "2 files closed."
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.33, Page number: 487<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the contents of three files and find the largest file\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "y = 0\n",
-      "k = 0\n",
-      "t = 0\n",
-      "name = [\"1.txt\",\"2.txt\",\"3.txt\"]\n",
-      "f = [0 for i in range(0,3)]\n",
-      "x = [0 for i in range(0,3)]\n",
-      "\n",
-      "#Open all the files\n",
-      "for l in range(0,3):\n",
-      "    fp = open(name[l],\"r\")\n",
-      "    f[l] = fp\n",
-      "    if fp == 0:\n",
-      "        sys.stdout.write(\"\\n%s file not found.\"%(name[l]))\n",
-      "        break\n",
-      "        \n",
-      "#Read contents of all files\n",
-      "for l in range(0,3):\n",
-      "    while True:\n",
-      "        c1 = f[l].read(1)\n",
-      "        if not c1:\n",
-      "            break\n",
-      "        x[l] = y\n",
-      "        y += 1\n",
-      "    y = 0\n",
-      "\n",
-      "#close the files\n",
-      "for l in range(0,3):\n",
-      "    f[l].close()\n",
-      "\n",
-      "#Print size of all files\n",
-      "for l in range(0,2+1):\n",
-      "    sys.stdout.write(\"File : %s Bytes : %d\\n\"%(name[l],x[l]))\n",
-      "    t = t + x[l]\n",
-      "    \n",
-      "#Find largest\n",
-      "for l in range(t,1,-1):\n",
-      "    for k in range(0,3):\n",
-      "        if l == x[k]:\n",
-      "            sys.stdout.write(\"\\n%s are the largest file.\"%(name[k]))\n",
-      "            break\n",
-      "    if l == x[k]:\n",
-      "        break"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "File : 1.txt Bytes : 16\n",
-        "File : 2.txt Bytes : 20\n",
-        "File : 3.txt Bytes : 25\n",
-        "\n",
-        "3.txt are the largest file."
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.34, Page number: 488<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy 100 characters from a file to an array\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "SIZE = 100\n",
-      "s = 0\n",
-      "x = 0\n",
-      "ch = ['0' for i in range(0,100)]\n",
-      "\n",
-      "#Open the files\n",
-      "f = open(\"poem.txt\",\"r\")\n",
-      "f2 = open(\"alpha.txt\",\"w\")\n",
-      "\n",
-      "if f ==0 or f2 == 0:\n",
-      "    sys.stdout.write(\"?\")\n",
-      "else:\n",
-      "    while True:\n",
-      "        c = f.read(1)\n",
-      "        x += 1\n",
-      "        if not c:\n",
-      "            break\n",
-      "        else:\n",
-      "            if x == 99:\n",
-      "                break\n",
-      "            else:\n",
-      "                ch[s] = c\n",
-      "                s += 1\n",
-      "\n",
-      "for s in range(0,100):\n",
-      "    f2.write(\"%c\"%(ch[s]))\n",
-      "\n",
-      "sys.stdout.write(\"Process Completed : Error 0\")\n",
-      "f.close()\n",
-      "f2.close()\n",
-      "\n",
-      "#There is no perror() function in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Process Completed : Error 0"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.35, Page number: 491<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Low level disk I/O operations.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "\n",
-      "#Variable Initialization\n",
-      "file1 = raw_input(\"Enter a file name : \")\n",
-      "\n",
-      "#Open file\n",
-      "s = open(file1,\"w\")\n",
-      "\n",
-      "#Result\n",
-      "if s == -1:\n",
-      "    sys.stdout.write(\"File does not exits\")\n",
-      "else:\n",
-      "    buff = raw_input(\"Enter text below:\")\n",
-      "    s.write(\"%s\"%(buff))\n",
-      "    s.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a file name : TEXT\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter text below:PROGRAMMING IN C\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.36, Page number: 492<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read text from a specified file \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "file1 = raw_input(\"Enter a file name \")\n",
-      "\n",
-      "#open file\n",
-      "s = open(file1,\"r\")\n",
-      "\n",
-      "#Result\n",
-      "if s == -1:\n",
-      "    sys.stdout.write(\"File does not exists\")\n",
-      "else:\n",
-      "    while True:\n",
-      "        ch = s.read(1)\n",
-      "        if not ch:\n",
-      "            break\n",
-      "        sys.stdout.write(\"%c\"%(ch))\n",
-      "        \n",
-      "s.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a file name TEXT\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "PROGRAMMING IN C"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.37, Page number: 493<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Set a buffer size \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nThis book teaches C\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "This book teaches C"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.38, Page number: 494<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display number of arguments and their names\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nTotal number of arguments are %d\\n\"%(len(sys.argv)))\n",
-      "\n",
-      "print str(sys.argv)\n",
-      "\n",
-      "#Command line arguments can be given in python by the command\n",
-      "\n",
-      "#   python pgmname.py arg1 arg2 arg3\n",
-      "\n",
-      "#This is not possible in ipython notebook"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.39, Page number: 495<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read any file from command prompt\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#open file\n",
-      "fp = open(sys.argv[1],\"r\")\n",
-      "\n",
-      "#Result\n",
-      "if fp == 0:\n",
-      "    sys.stdout.write(\"Can not open file\")\n",
-      "else:\n",
-      "    while True:\n",
-      "        ch = fp.read(1)\n",
-      "        if not ch:\n",
-      "            break\n",
-      "        sys.stdout.write(\"%c\"%(ch))\n",
-      "        \n",
-      "fp.close()\n",
-      "\n",
-      "#This program can be run in python as\n",
-      "\n",
-      "#   python pgmname.py filename"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.40, Page number: 495<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Command line argument to perform the task of DEL command\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Check number of arguments\n",
-      "if len(sys.argv) < 2:\n",
-      "    sys.stdout.write(\"Insufficient Arguments\")\n",
-      "else:\n",
-      "    fp = open(sys.argv[1],\"r\")\n",
-      "    if fp == 0:\n",
-      "        sys.stdout.write(\"File Not Found\")\n",
-      "        fp.close()\n",
-      "        os.remove(sys.argv[1])\n",
-      "        sys.stdout.write(\"File has been deleted\")\n",
-      "        \n",
-      "#This program can be deleted using\n",
-      "#     python pgmname.py filename\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.41, Page number: 496<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Command line argument to perform the task of REN command\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Check number of arguments\n",
-      "if len(sys.srgv) < 3:\n",
-      "    sys.stdout.write(\"Insufficient Arguments\")\n",
-      "else:\n",
-      "    fp = open(sys.argv[1],\"r\")\n",
-      "    if fp == 0:\n",
-      "        sys.stdout.write(\"File Not Found\")\n",
-      "    else:\n",
-      "        sp = open(sys.argv[2],\"r\")\n",
-      "        if sp == 0:\n",
-      "            fp.close()\n",
-      "            sp.close()\n",
-      "            #Rename file\n",
-      "            os.rename(sys.argv[1],sys.argv[2])\n",
-      "        else:\n",
-      "            sys.stdout.write(\"Duplicate file name or file is in use\")\n",
-      "            \n",
-      "#This program can be executed as\n",
-      "\n",
-      "#   python pgmname.py oldfilename newfilename\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.42, Page number: 497<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Environment variable and display the various settings.\n",
-      "\n",
-      "import sys\n",
-      "import os\n",
-      "\n",
-      "#Result\n",
-      "print os.environ\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "{'TMP': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Local\\\\Temp', 'COMPUTERNAME': 'AATHIRA-PC', 'GUROBI_HOME': 'B:\\\\gurobi510\\\\win32', 'USERDOMAIN': 'Aathira-PC', 'PSMODULEPATH': 'C:\\\\Windows\\\\system32\\\\WindowsPowerShell\\\\v1.0\\\\Modules\\\\', 'COMMONPROGRAMFILES': 'C:\\\\Program Files (x86)\\\\Common Files', 'PROCESSOR_IDENTIFIER': 'Intel64 Family 6 Model 15 Stepping 13, GenuineIntel', 'PROGRAMFILES': 'C:\\\\Program Files (x86)', 'PROCESSOR_REVISION': '0f0d', 'SYSTEMROOT': 'C:\\\\Windows', 'PATH': 'C:\\\\Anaconda\\\\lib\\\\site-packages\\\\numpy\\\\core;B:\\\\gurobi510\\\\win32\\\\bin;C:\\\\Windows\\\\system32;C:\\\\Windows;C:\\\\Windows\\\\System32\\\\Wbem;C:\\\\Windows\\\\System32\\\\WindowsPowerShell\\\\v1.0\\\\;C:\\\\Program Files (x86)\\\\Common Files\\\\Adobe\\\\AGL;C:\\\\Program Files\\\\MATLAB\\\\R2009a\\\\bin;C:\\\\Program Files\\\\MATLAB\\\\R2009a\\\\bin\\\\win64;c:\\\\python27\\\\scripts;C:\\\\Program Files (x86)\\\\MiKTeX 2.9\\\\miktex\\\\bin\\\\;C:\\\\Anaconda;C:\\\\Anaconda\\\\Scripts;C:\\\\Program Files (x86)\\\\ffmpeg', 'CLICOLOR': '1', 'PROGRAMFILES(X86)': 'C:\\\\Program Files (x86)', 'COMSPEC': 'C:\\\\Windows\\\\system32\\\\cmd.exe', 'TK_LIBRARY': 'C:\\\\Anaconda\\\\tcl\\\\tk8.5', 'TERM': 'xterm-color', 'TEMP': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Local\\\\Temp', 'COMMONPROGRAMFILES(X86)': 'C:\\\\Program Files (x86)\\\\Common Files', 'PROCESSOR_ARCHITECTURE': 'x86', 'TIX_LIBRARY': 'C:\\\\Anaconda\\\\tcl\\\\tix8.4.3', 'ALLUSERSPROFILE': 'C:\\\\ProgramData', 'LOCALAPPDATA': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Local', 'HOMEPATH': '\\\\Users\\\\Aathira', 'PROGRAMW6432': 'C:\\\\Program Files', 'USERNAME': 'Aathira', 'LOGONSERVER': '\\\\\\\\AATHIRA-PC', 'SESSIONNAME': 'Console', 'PROGRAMDATA': 'C:\\\\ProgramData', 'TCL_LIBRARY': 'C:\\\\Anaconda\\\\tcl\\\\tcl8.5', 'GIT_PAGER': 'cat', 'PATHEXT': '.COM;.EXE;.BAT;.CMD;.VBS;.VBE;.JS;.JSE;.WSF;.WSH;.MSC', 'FP_NO_HOST_CHECK': 'NO', 'WINDIR': 'C:\\\\Windows', 'APPDATA': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Roaming', 'HOMEDRIVE': 'C:', 'PAGER': 'cat', 'SYSTEMDRIVE': 'C:', 'NUMBER_OF_PROCESSORS': '2', 'PROCESSOR_LEVEL': '6', 'PROCESSOR_ARCHITEW6432': 'AMD64', 'COMMONPROGRAMW6432': 'C:\\\\Program Files\\\\Common Files', 'OS': 'Windows_NT', 'PUBLIC': 'C:\\\\Users\\\\Public', 'USERPROFILE': 'C:\\\\Users\\\\Aathira'}\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.43, Page number: 498<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read character from keyboard \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "c = '0'\n",
-      "\n",
-      "#Result\n",
-      "while c != ' ':\n",
-      "    c = raw_input(\"\")\n",
-      "    sys.stdout.write(\"%c  \"%(c))\n",
-      "    \n",
-      "#Give space at the end"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1  2  "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3  4  "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5  6  "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7  8  "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9     "
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 14.44, Page number: 499<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display A to Z characters\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "for a in range(65,91):\n",
-      "    sys.stdout.write(\"%c\\t\"%(chr(a)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A\tB\tC\tD\tE\tF\tG\tH\tI\tJ\tK\tL\tM\tN\tO\tP\tQ\tR\tS\tT\tU\tV\tW\tX\tY\tZ\t"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter15.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter15.ipynb
deleted file mode 100755
index 5a99346e..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter15.ipynb
+++ /dev/null
@@ -1,1480 +0,0 @@
-<<<<<<< HEAD
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:23cf65a359a2231e3abd5faae195b73ed4d5756e6083812ae87c387c1bb53f5a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 15: Additional in 'C'<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.1, Page number: 505<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Allocate memory to pointer variable. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "j = 0\n",
-      "k = int(raw_input(\"How many number : \"))\n",
-      "p = [0 for i in range(0,k)]\n",
-      "\n",
-      "#in python, all variables are allocated using dynamic memory allocation technique and no\n",
-      "#malloc function and pointer concept is available in python.\n",
-      "\n",
-      "#Read the numbers\n",
-      "while j != k:\n",
-      "    p[j] = int(raw_input(\"Number %d = \"%(j+1)))\n",
-      "    j += 1\n",
-      "    \n",
-      "j = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The numbers are : \")\n",
-      "while j != k:\n",
-      "    sys.stdout.write(\"%d\\t\"%(p[j]))\n",
-      "    j += 1\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many number : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 1 = 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 2 = 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 3 = 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 4 = 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numbers are : 1\t2\t3\t4\t"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.2, Page number: 506<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Memory allocation to pointer variable. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "j = 0\n",
-      "k = int(raw_input(\"How many Number : \"))\n",
-      "p = [0 for i in range(0,k)]\n",
-      "\n",
-      "#in python, all variables are allocated using dynamic memory allocation technique and no\n",
-      "#calloc function and pointer concept is available in python.\n",
-      "\n",
-      "#Read the numbers\n",
-      "while j != k:\n",
-      "    p[j] = int(raw_input(\"Number %d = \"%(j+1)))\n",
-      "    j += 1\n",
-      "    \n",
-      "j = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The numbers are : \")\n",
-      "while j != k:\n",
-      "    sys.stdout.write(\"%d\\t\"%(p[j]))\n",
-      "    j += 1\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many Number : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 1 = 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 2 = 58\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 3 = 98\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numbers are : 45\t58\t98\t"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.3, Page number: 507<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reallocate memory \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "str1 = \"India\"\n",
-      "\n",
-      "#in python, value tagged method is used for data storage instead of memory tagging.\n",
-      "#no realloc function is in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"str = %s\"%(str1))\n",
-      "str1 = \"Hindustan\"\n",
-      "sys.stdout.write(\"\\nNow str = %s\"%(str1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "str = India\n",
-        "Now str = Hindustan"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.4, Page number: 508<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display unused memory \n",
-      "\n",
-      "import psutil\n",
-      "\n",
-      "psutil.phymem_usage()\n",
-      "\n",
-      "#There is no coreleft() function in python. phymem_usage function in the module psutil gives the \n",
-      "#status and usage of physical memory in python."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "pyout",
-       "prompt_number": 6,
-       "text": [
-        "usage(total=3165270016L, used=987840512L, free=2177429504L, percent=31.2)"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.5, Page number: 510<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Linked list\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialziation\n",
-      "ch = 'y'\n",
-      "p = 0\n",
-      "q = []\n",
-      "\n",
-      "#Function Definitions\n",
-      "def gen_rate(m):\n",
-      "    q.append(m)\n",
-      "    \n",
-      "def show():\n",
-      "    print q\n",
-      "    \n",
-      "def addatstart(m):\n",
-      "    q.insert(0,m)\n",
-      "    \n",
-      "def append(m,po):\n",
-      "    q.insert(po,m)\n",
-      "\n",
-      "def erase(d):\n",
-      "    q.remove(d)\n",
-      "    \n",
-      "def count():\n",
-      "    print len(q)\n",
-      "    \n",
-      "def descending():\n",
-      "    q.sort(reverse=True)\n",
-      "    \n",
-      "#Get choice\n",
-      "while ch == 'y':\n",
-      "    n = int(raw_input(\"1. Generate\\n2. Add at starting\\n3. Append\\n4. Delete\\n5. Show\\n6.Count\\n7.Descending\\nEnter your choice: \"));\n",
-      "    #There is no switch statement in python\n",
-      "    if n == 1:\n",
-      "        i = int(raw_input(\"How many node you want : \"))\n",
-      "        for j in range(0,i):\n",
-      "            m = int(raw_input(\"Enter the element : \"))\n",
-      "            gen_rate(m)\n",
-      "        show()\n",
-      "    else:\n",
-      "        if n == 2:\n",
-      "            m = int(raw_input(\"Enter the element : \"))\n",
-      "            addatstart(m)\n",
-      "            show()\n",
-      "        else:\n",
-      "            if n == 3:\n",
-      "                m = int(raw_input(\"Enter the element and position \"))\n",
-      "                po = int(raw_input(\"Enter the element and position\"))\n",
-      "                append(m,po)\n",
-      "                show()\n",
-      "            else:\n",
-      "                if n == 4:\n",
-      "                    d = int(raw_input(\"Enter the number for deletion : \"))\n",
-      "                    erase(d)\n",
-      "                    show()\n",
-      "                else:\n",
-      "                    if n == 5:\n",
-      "                        show()\n",
-      "                    else:\n",
-      "                        if n == 6:\n",
-      "                            count()\n",
-      "                        else:\n",
-      "                            if n == 7:\n",
-      "                                descending()\n",
-      "                                show()\n",
-      "                            else:\n",
-      "                                sys.stdout.write(\"Enter value between 1 to 7\")\n",
-      "                \n",
-      "    ch = raw_input(\"Do u wnat to continue (y/n)\")\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. Generate\n",
-        "2. Add at starting\n",
-        "3. Append\n",
-        "4. Delete\n",
-        "5. Show\n",
-        "6.Count\n",
-        "7.Descending\n",
-        "Enter your choice: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many node you want : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[1, 5, 4, 7]\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Do u wnat to continue (y/n)n\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.6, Page number: 518<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Draw circle, line and arc using graphics function\n",
-      "\n",
-      "%matplotlib inline\n",
-      "import pylab\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "# matplotlib package is used for graphics\n",
-      "#Give proportionate sizes to draw\n",
-      "#draw circle\n",
-      "circle2=plt.Circle((.5,.5),.2,color='b')\n",
-      "fig = plt.gcf()\n",
-      "fig.gca().add_artist(circle2)\n",
-      "\n",
-      "\n",
-      "#draw line\n",
-      "figure()\n",
-      "pylab.plot([210,110],[150,150])\n",
-      "\n",
-      "#Draw ellipse\n",
-      "figure()\n",
-      "from matplotlib.patches import Ellipse\n",
-      "e = Ellipse(xy=(35, -50), width=10, height=5, linewidth=2.0, color='g')\n",
-      "fig = plt.gcf()\n",
-      "fig.gca().add_artist(e)\n",
-      "e.set_clip_box(plt.axes().bbox)\n",
-      "e.set_alpha(0.7)\n",
-      "pylab.xlim([20, 50])\n",
-      "pylab.ylim([-65, -35])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "pyout",
-       "prompt_number": 5,
-       "text": [
-        "(-65, -35)"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x399d950>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x399d690>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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2my1pwYIFl/xd50S0F198cfsrr7zyK5nGb6z+yTB+27Zt22U0GmuTkpJs8fHx\njUFBQfZHH330HVnGbqz+PfbYY4dkGLuxWmFh4c5XX311izfjd8dn9EII5YknnjiYkZFR/swzz/x+\ncP3atWuPvf322xsA4O23396Qn59/5E7XcifcrH+NjY0Jg8vvv//+jzMzMy/5p0LftLS0RA/+6etw\nOPQlJSV5JpPJKsv43ax/wz8RPl3Hb9euXc/V1tYm2my25Hffffen999//yfvvPPOY7KM3Vj9O3To\n0OOy/N/r6ekJ6urqCgUAu90efOLEidWZmZmXvBq/O/0qdPr06WWKoniysrLKhp/u1NraGrly5cqP\np/spXmP176OPPvrBY489digzM/PiwoULP1+3bt2RpqamOH/X6k27ePFipslkupCVlVWWmZl5cc+e\nPc8KMXCKlwzjd7P+yTJ+g81isSwfPCtFlrEb3k6ePGke7N+jjz76jgxjV11dnZyVlVWWlZVVNn/+\n/Mu7du3a5u34KUJMy8NzRER0m3iHKSIiyTHoiYgkx6AnIpIcg56ISHIMeiIiyTHoiYgk9/8B1dqr\nlLRzGX0AAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x2610890>"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.7, Page number: 519<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Draw boxes and fill with different designs.\n",
-      "\n",
-      "%pylab\n",
-      "#matplotlib package is used for graphics\n",
-      "from matplotlib.patches import Rectangle\n",
-      "from matplotlib.collections import PatchCollection\n",
-      "\n",
-      "e = Rectangle(xy=(35, -50), width=10, height=5, linewidth=2.0, color='b')\n",
-      "fig = plt.gcf()\n",
-      "fig.gca().add_artist(e)\n",
-      "e.set_clip_box(ax.bbox)\n",
-      "e.set_alpha(0.7)\n",
-      "pylab.xlim([20, 50])\n",
-      "pylab.ylim([-65, -35])\n",
-      "\n",
-      "#There are no different automatic fill styles. user should create the styles."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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49Hv27NGePXtUWlqqBx54QCdOnJAkdXd3q76+XuFwWKFQSG1tbcrLy8vagAEA6XF83/cX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\noQcA4wg9ABhH6AHAOEIPAMb9B4rEH5DkLOd4AAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7373590>"
-       ]
-      }
-     ],
-     "prompt_number": 72
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.8, Page number: 520<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display text in different size, font, vertically and horizontally \n",
-      "\n",
-      "%matplotlib inline\n",
-      "\n",
-      "from pylab import *\n",
-      "\n",
-      "#Tkinter package is used for graphics\n",
-      "# set limits so that it no longer looks on screen to be 45 degrees\n",
-      "xlim([-5,5])\n",
-      "\n",
-      "# Locations to plot text\n",
-      "l1 = array((1,1))\n",
-      "l2 = array((5,5))\n",
-      "\n",
-      "# Rotate angle\n",
-      "angle = 90\n",
-      "trans_angle = gca().transData.transform_angles(array((45,)),\n",
-      "                                               l2.reshape((1,2)))[0]\n",
-      "\n",
-      "# Plot text\n",
-      "th2 = text(l2[0],l2[1],'Hello(Horizontal Text with fontsize 25)\\n\\n',fontsize=25,\n",
-      "           rotation=0)\n",
-      "th2 = text(l2[0],l2[1],'Hello(Horizontal Text with fontsize 16)',fontsize=16,\n",
-      "           rotation=0)\n",
-      "th1 = text(l1[0],l1[1],'Hello(Vertical Text)',fontsize=16,\n",
-      "           rotation=angle)\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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2DVcvV9YHzur1emnr1q1D1G3a2dllKA9olGVZ5+Hhcevq1asNDC37+PFje+Vh\njFeuXGloqI76gbMlefCi8gBJYw/KXLx48RT1QwwdHBwemZub5yv/9/X1/TM5OblaafeDoT6oj5P6\n+Nna2mYV9dDcVq1anRLXOW3atIXK8srDd01NTQuU/3fu3PmwoYcxluRhruqHPYrzHj165ODm5vZA\nacfFxeWhh4fHLU9PzwT1AzSHDx++Sd0/BweHR8rDf1u1anVq2bJlbxvbf5XxwFlZlnXiA2f1er10\n5cqVhg0aNLiqfj04OTml2djYZKv737Vr11/Ux87e3v6xLMu6d955Z4mhdkePHr1OlmWdq6trinj+\nHDt2rINyzpmYmGjd3d3vKfuwtNt38+ZNL3U/Dxw40F2sM2nSpO+U+Y0bN75gbF1F7feSHveSPEy3\nPA8tLsm5cffu3eqNGze+oPTV3Nw8X3noqCzLOlNT04KlS5eGlPS1Wpr+b9iwYaTSjpmZ2dMaNWok\neXh43Grfvv1x8XWvHH8nJ6c09d8cS0vLJzt27BggrtvQ3z3130N7e/vHRf0dGTBgwA5xnWV9jyiu\neHl53VSWd3V1TSltv9T7yMzM7Kmzs3OqlZVVrrpPwcHBEUU9cFar1Zq4u7vfk2VZZ+jhw2LhihIA\n/M0ZeuhpeZaZOHHiimvXrjV47733vnn55Zf/tLKyepKZmWlnZ2eX2apVq9PvvvvuksjIyCDldpWS\nrLO4Pg4ZMmTbpUuXXpo4ceIKb2/v+IKCAjNzc/OnzZo1OzdnzpzPLl682LhBgwbXDC1rb2+fMWLE\niE16vV5ev3796KLaL+m+kmVZX1TdKVOmhJ05c6blqFGjNtSuXft2Xl6epY2NTY6fn9+JsLCwKadP\nn25l6DkjpTlWRfVBlmV9bm6udVEPzTU0uMXChQunK8/HcXd3T87NzbW2t7fPCAwMjFqzZs24yMjI\nIBsbm5yy9LuoOg4ODo+PHTvWcdiwYVtq1qyZlJWVVeXOnTu1bt++XVs9zPDGjRtHhoWFTWnatOl5\nS0vLPL1eL/v6+sZ99dVXH/7222/tbG1ts8vSfmkVd740bNjw6vnz55uuWLFiYteuXQ+5ubmlZGdn\n28qyrK9Xr971IUOGbPvhhx9e37Zt2xBJenbr1IgRIzZlZWVVadKkyQVjz0v6z3/+87a3t3d8Wlqa\n8+jRo9frVbdbdejQ4fi+fft6denS5bCTk1P6w4cPXZV9WNrt8/LySqhdu/ZtWZb1ZmZmBYZG5lO+\ncS9utLuRZcwrAAAgAElEQVSKOO7Fvd5KWqcsfVRUr1793pkzZ1ouWrRoWtu2bU/a2Njk5OXlWdau\nXfv26NGj18fGxrZQfs9Tlr4VVWfkyJEbf/rpp1fbt2//q62tbfaDBw+q3rlzp9bdu3drSJIk1axZ\nM2nPnj19pk6dutjPz+9EjRo17ubm5lqbm5s/femlly6FhIQsu3jxYmPlwdzFbb/6/M7KyqpS1N8R\nQ7cDlvU9ojh6vV5W9lNaWppzafu1cuXKN8aNG7fG19c3zsXFJTU7O9vWxMSk0NvbO37EiBGb9u/f\n3zMiIqK3ob9xisjIyKD79+9Xq1ev3vWSXFGS9foK+7sDAECluHLlSqOmTZued3NzS7l582Yd5bk7\nAACUVL9+/cL37NnTZ+XKlW8Ye7CtGleUAAB/e40aNbryxhtvrExOTnZXfusBAEBJxcbGttizZ0+f\npk2bnn/ttddWl2QZrigBAP4R0tLSnL29veMtLCzyExISvKysrJ686D4BAP4Zevbsuf+XX37pFhkZ\nGVTUYA9qBCUAAAAAEHDrHQAAAAAICEoAAAAAICAoAQAAAICAoAQAAAAAAoISAAAAAAgISgAAAAAg\nICgBAAAAgICgBAAAAAACghIAAAAACAhKAAAAACAgKAEAAACAgKAEAAAAAAKCEgAAAAAICEoAAAAA\nICAoAQAAAICAoAQAAAAAAoISAAAAAAgISgAAAAAgICgBAAAAgICgBAAAAAACghIAAAAACAhKAAAA\nACAgKAHA39jatWvHajQa3c2bN+uI87RaralGo9HNnj17VmnXGxoaGqrRaHTqaWVdlyRJUmFhoYmv\nr2/ct99+O7my+14UT0/PW+PHj/+xItdZGhkZGfahoaGh586da1bWdXh6et4aN27cGmPz/f39YzQa\nja64cvv27dpl7YOiIranOOK5WFSb/v7+MR06dDhe1rYiIiJ6N2nS5IKVldUTjUajy8zMtCvruowJ\nCwubsmvXrv6lXS4mJsZfo9Hojh071rGi+2RIdna27XvvvfeNv79/jJ2dXaZGo9EdPXq0k7H6V65c\naTR48OCfXV1dH1pbW+c2bNjw6pIlS95V5t+/f7+ara1t9smTJ9s+j/4Dz4Ppi+4AAKB8ZFnWV9Ry\nZV3Xjz/+OD49Pd1p0qRJy8vbh/LYvXt3Xzs7u8yKXGdpPHr0yHHOnDmf1a5d+3azZs3OlWUdsizr\ni9ovy5cvn5SVlVVFkiRJr9fLc+fOnXnmzJmWe/bs6aOuV61atftlaV+tIranOK+//voPPXv23F/S\nNst6zmi1WtORI0dubN++/a/Lly+fZG5u/tTW1ja7PH03JCwsbErHjh2P9e/ff1dplmvRokXsyZMn\n2zZq1OhKRffJkNTUVJc1a9aMa9GiRWzXrl0P7dy5c4CxfXvmzJmWgYGBUYGBgVGrV69+zd7ePuOv\nv/6qn5OTY6PUqVat2v233nrru6lTpy4+ceKE3/PYBqCyEZQA4F9Kr9fLFbWeBQsWzBg/fvyP5ubm\nTytinaX19OlTc3Nz86e+vr5xL6J9UUXtW0PED9IuLi6pZmZmBa1btz5VWW1W5vbUqFHjbo0aNe5W\ndpt3796tkZ2dbTt48OCf27dv/2tFrltUlr5XqVIlqzKPocjT0/NWWlqasyRJ0uHDh7vs3LlzgKF6\nOp1OM3r06PVBQUGRO3bsGKhM79Sp01Gx7qRJk5Z/88037x07dqxjx44dj1Ve74Hng1vvAOB/TEJC\ngtfIkSM3urm5pVhaWuY1a9bsXHh4eL+yrOvgwYPd/fz8TlhbW+c6ODg87t+//66//vqrvrpOTEyM\nf3x8vPfIkSM3lrfvp06dat2lS5fDVapUybK1tc3u0qXL4dOnT7dS1xk7duzaWrVq3Tlx4oTfK6+8\n8ru1tXXuBx98MF+S/u9ta7du3fI0dltaQEBAtLK+zMxMu5CQkGXVq1e/Z2lpmdewYcOrYWFhU8Rt\n1Gg0uoiIiN4hISHLXF1dH7q6uj589dVXf8rIyLBX2qtTp85NSXp2lURpa/369aMlSZIOHTrUtWfP\nnvurV69+z8bGJqdJkyYXFi1aNE2n01X4e3Fubq71Bx98MN/LyyvBwsIiv06dOje/+OKLj5UP8Dk5\nOTYNGza82qZNmz+0Wu3/+9L00KFDXTUajW758uWTEhMTPYraHtGOHTsGajQa3d27d2so06ZPn75Q\no9HoVq9e/ZoyLTIyMkij0eiuXLnSSJL+7613xe1DSXoWQg4fPtylefPmZ5X9WNz5HRoaGurl5ZUg\nSZL02muvrVafA3q9Xl68ePHUBg0aXLOwsMivXr36vXfeeWepctVOodFodDNnzpy7ZMmSd728vBLs\n7Owy/f39Yy5fvuyj1PH09Lx1+/bt2hs3bhyp9F25FfSvv/6q379//11Vq1Z9YGVl9cTDwyNxyJAh\n2woLC00k6f9/652yXwyVdevWjVHarMjXuyExMTH+V69ebTht2rRFxdX18vJKaNOmzR8rV658o6La\nB14kghIA/ANotVpTsSgfsNTu3LlTq02bNn9cuHChSVhY2JSIiIjezZs3Pztw4MAdERERvUvT5sGD\nB7v36tVrn52dXea2bduGLF++fNLFixcbt2/f/td79+5VV9dzdXV9WL9+/b/K0/fz58837dSp09GM\njAz7devWjVm/fv3ozMxMu06dOh09f/58U3XdjIwM++HDh28eOXLkxoMHD3YfMWLEJkn6v7etVa9e\n/d7JkyfbqsuqVasmaDQanY+Pz2VJevZtea9evfatXbt27IwZMxbs3bs3uHv37genTZu26JNPPvlc\n7OPkyZO/NTExKdy8efPwWbNmzd6xY8fAyZMnf6u0p3wr//HHH3+htKncVpaQkOAVGBgYtWrVqgn7\n9+/vOWbMmHWhoaGhhtopD61Wa9qtW7dfVq9e/drUqVMXHzx4sPuECRNWzZ07d+aMGTMWSJIk2djY\n5GzZsmVYXFyc78yZM+dKkiQ9ePCg6ujRo9f37dt396RJk5a7u7snF7U9In9//xhZlvVRUVGByrSo\nqKhAKyurJ+K0atWq3VdfGVMfM0Nt9urVa59S98aNG3WnTJkS9t57732zc+fOAe7u7smDBw/++caN\nG3WN7ZPXX3/9h59//nmwJEnSzJkz5548ebLt8uXLJ0mSJH3yySefT58+fWG3bt1+2bt3b/D777//\n9dq1a8f26tVrn3hlaMOGDaMOHDjQY+nSpe+sWbNm3O3bt2v37dt3t3I+h4eH96tWrdr97t27H1T6\nruzfXr167UtOTnb//vvv3zx06FDXr7766kNLS8s8Y0H59ddf/0F97p44ccJvwIABO01NTbUNGjS4\nJkkV+3o35tdff20vSZL05MkTq7Zt2540Nzd/WrVq1QeTJ0/+Ni8vz1Ks3759+18jIyODKqJt4IXT\n6/UUCoVC+ZuWNWvWjJVlWVdUmT179mdK/fHjx692c3N7kJ6e7qheT1BQ0KGXX375nPL/WbNmhcqy\nrFPXEdfVokWLM/Xr179WWFioUaYlJCR4mpmZPZ02bdpCZVpAQEBUQEBAVHn7PnDgwO2Ojo7pGRkZ\ndsq0zMzMKk5OTmkDBgzYoUw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IhBIAAEAilAAAABKhBAAAkAglAACARCgBAAAk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\nlAAAACZCCQAAYPIF5fe6JjownXYAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x3f2e190>"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.9, Page number: 521<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display status of mouse button pressed \n",
-      "\n",
-      "from Tkinter import *\n",
-      "from tkFileDialog import askopenfilename\n",
-      "import Image, ImageTk\n",
-      "\n",
-      "if __name__ == \"__main__\":\n",
-      "    root = Tk()\n",
-      "\n",
-      "    #setting up a tkinter canvas with scrollbars\n",
-      "    frame = Frame(root, bd=2, relief=SUNKEN)\n",
-      "    frame.grid_rowconfigure(0, weight=1)\n",
-      "    frame.grid_columnconfigure(0, weight=1)\n",
-      "    xscroll = Scrollbar(frame, orient=HORIZONTAL)\n",
-      "    xscroll.grid(row=1, column=0, sticky=E+W)\n",
-      "    yscroll = Scrollbar(frame)\n",
-      "    yscroll.grid(row=0, column=1, sticky=N+S)\n",
-      "    canvas = Canvas(frame, bd=0, xscrollcommand=xscroll.set, yscrollcommand=yscroll.set)\n",
-      "    canvas.grid(row=0, column=0, sticky=N+S+E+W)\n",
-      "    xscroll.config(command=canvas.xview)\n",
-      "    yscroll.config(command=canvas.yview)\n",
-      "    frame.pack(fill=BOTH,expand=1)\n",
-      "\n",
-      "    \n",
-      "\n",
-      "    #function to be called when mouse is clicked\n",
-      "    def printcoords(event):\n",
-      "        #outputting x and y coords to console\n",
-      "        print \"Mouse Button pressed\"\n",
-      "        print (event.x,event.y)\n",
-      "    #mouseclick event\n",
-      "    canvas.bind(\"<Button 1>\",printcoords)\n",
-      "\n",
-      "    root.mainloop()\n",
-      "    \n",
-      "import win32api, win32con\n",
-      "\n",
-      "print \"Current cursor position at \" \n",
-      "print win32api.GetCursorPos()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mouse Button pressed\n",
-        "(207, 115)\n",
-        "Current cursor position at "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "(502, 188)\n"
-       ]
-      }
-     ],
-     "prompt_number": 108
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.10, Page number: 523<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Change mouse cursor.\n",
-      "\n",
-      "#Placing the cursor on top of the circle button will change the pointer to circle\n",
-      "# and plus button to plus symbol\n",
-      "\n",
-      "from Tkinter import *\n",
-      "import Tkinter\n",
-      "\n",
-      "top = Tkinter.Tk()\n",
-      "\n",
-      "B1 = Tkinter.Button(top, text =\"circle\", relief=RAISED,\\\n",
-      "                         cursor=\"circle\")\n",
-      "B2 = Tkinter.Button(top, text =\"plus\", relief=RAISED,\\\n",
-      "                         cursor=\"plus\")\n",
-      "B1.pack()\n",
-      "B2.pack()\n",
-      "top.mainloop()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 110
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-=======
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 15: Additional in 'C'<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.1, Page number: 505<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Allocate memory to pointer variable. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "j = 0\n",
-      "k = int(raw_input(\"How many number : \"))\n",
-      "p = [0 for i in range(0,k)]\n",
-      "\n",
-      "#in python, all variables are allocated using dynamic memory allocation technique and no\n",
-      "#malloc function and pointer concept is available in python.\n",
-      "\n",
-      "#Read the numbers\n",
-      "while j != k:\n",
-      "    p[j] = int(raw_input(\"Number %d = \"%(j+1)))\n",
-      "    j += 1\n",
-      "    \n",
-      "j = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The numbers are : \")\n",
-      "while j != k:\n",
-      "    sys.stdout.write(\"%d\\t\"%(p[j]))\n",
-      "    j += 1\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many number : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 1 = 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 2 = 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 3 = 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 4 = 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numbers are : 1\t2\t3\t4\t"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.2, Page number: 506<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Memory allocation to pointer variable. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "j = 0\n",
-      "k = int(raw_input(\"How many Number : \"))\n",
-      "p = [0 for i in range(0,k)]\n",
-      "\n",
-      "#in python, all variables are allocated using dynamic memory allocation technique and no\n",
-      "#calloc function and pointer concept is available in python.\n",
-      "\n",
-      "#Read the numbers\n",
-      "while j != k:\n",
-      "    p[j] = int(raw_input(\"Number %d = \"%(j+1)))\n",
-      "    j += 1\n",
-      "    \n",
-      "j = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The numbers are : \")\n",
-      "while j != k:\n",
-      "    sys.stdout.write(\"%d\\t\"%(p[j]))\n",
-      "    j += 1\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many Number : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 1 = 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 2 = 58\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number 3 = 98\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The numbers are : 45\t58\t98\t"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.3, Page number: 507<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reallocate memory \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "str1 = \"India\"\n",
-      "\n",
-      "#in python, value tagged method is used for data storage instead of memory tagging.\n",
-      "#no realloc function is in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"str = %s\"%(str1))\n",
-      "str1 = \"Hindustan\"\n",
-      "sys.stdout.write(\"\\nNow str = %s\"%(str1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "str = India\n",
-        "Now str = Hindustan"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.4, Page number: 508<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display unused memory \n",
-      "\n",
-      "import psutil\n",
-      "\n",
-      "psutil.phymem_usage()\n",
-      "\n",
-      "#There is no coreleft() function in python. phymem_usage function in the module psutil gives the \n",
-      "#status and usage of physical memory in python."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "pyout",
-       "prompt_number": 6,
-       "text": [
-        "usage(total=3165270016L, used=987840512L, free=2177429504L, percent=31.2)"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.5, Page number: 510<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Linked list\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialziation\n",
-      "ch = 'y'\n",
-      "p = 0\n",
-      "q = []\n",
-      "\n",
-      "#Function Definitions\n",
-      "def gen_rate(m):\n",
-      "    q.append(m)\n",
-      "    \n",
-      "def show():\n",
-      "    print q\n",
-      "    \n",
-      "def addatstart(m):\n",
-      "    q.insert(0,m)\n",
-      "    \n",
-      "def append(m,po):\n",
-      "    q.insert(po,m)\n",
-      "\n",
-      "def erase(d):\n",
-      "    q.remove(d)\n",
-      "    \n",
-      "def count():\n",
-      "    print len(q)\n",
-      "    \n",
-      "def descending():\n",
-      "    q.sort(reverse=True)\n",
-      "    \n",
-      "#Get choice\n",
-      "while ch == 'y':\n",
-      "    n = int(raw_input(\"1. Generate\\n2. Add at starting\\n3. Append\\n4. Delete\\n5. Show\\n6.Count\\n7.Descending\\nEnter your choice: \"));\n",
-      "    #There is no switch statement in python\n",
-      "    if n == 1:\n",
-      "        i = int(raw_input(\"How many node you want : \"))\n",
-      "        for j in range(0,i):\n",
-      "            m = int(raw_input(\"Enter the element : \"))\n",
-      "            gen_rate(m)\n",
-      "        show()\n",
-      "    else:\n",
-      "        if n == 2:\n",
-      "            m = int(raw_input(\"Enter the element : \"))\n",
-      "            addatstart(m)\n",
-      "            show()\n",
-      "        else:\n",
-      "            if n == 3:\n",
-      "                m = int(raw_input(\"Enter the element and position \"))\n",
-      "                po = int(raw_input(\"Enter the element and position\"))\n",
-      "                append(m,po)\n",
-      "                show()\n",
-      "            else:\n",
-      "                if n == 4:\n",
-      "                    d = int(raw_input(\"Enter the number for deletion : \"))\n",
-      "                    erase(d)\n",
-      "                    show()\n",
-      "                else:\n",
-      "                    if n == 5:\n",
-      "                        show()\n",
-      "                    else:\n",
-      "                        if n == 6:\n",
-      "                            count()\n",
-      "                        else:\n",
-      "                            if n == 7:\n",
-      "                                descending()\n",
-      "                                show()\n",
-      "                            else:\n",
-      "                                sys.stdout.write(\"Enter value between 1 to 7\")\n",
-      "                \n",
-      "    ch = raw_input(\"Do u wnat to continue (y/n)\")\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1. Generate\n",
-        "2. Add at starting\n",
-        "3. Append\n",
-        "4. Delete\n",
-        "5. Show\n",
-        "6.Count\n",
-        "7.Descending\n",
-        "Enter your choice: 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many node you want : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the element : 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[1, 5, 4, 7]\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Do u wnat to continue (y/n)n\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.6, Page number: 518<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Draw circle, line and arc using graphics function\n",
-      "\n",
-      "%matplotlib inline\n",
-      "import pylab\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Tkinter package is used for graphics\n",
-      "#Give proportionate sizes to draw\n",
-      "#draw circle\n",
-      "circle2=plt.Circle((.5,.5),.2,color='b')\n",
-      "fig = plt.gcf()\n",
-      "fig.gca().add_artist(circle2)\n",
-      "\n",
-      "\n",
-      "#draw line\n",
-      "figure()\n",
-      "pylab.plot([210,110],[150,150])\n",
-      "\n",
-      "#Draw ellipse\n",
-      "figure()\n",
-      "from matplotlib.patches import Ellipse\n",
-      "e = Ellipse(xy=(35, -50), width=10, height=5, linewidth=2.0, color='g')\n",
-      "fig = plt.gcf()\n",
-      "fig.gca().add_artist(e)\n",
-      "e.set_clip_box(ax.bbox)\n",
-      "e.set_alpha(0.7)\n",
-      "pylab.xlim([20, 50])\n",
-      "pylab.ylim([-65, -35])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['pylab', 'e']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "pyout",
-       "prompt_number": 61,
-       "text": [
-        "(-65, -35)"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x707a9f0>"
-       ]
-      },
-      {
-       "metadata": {},
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-       "text": [
-        "<matplotlib.figure.Figure at 0x7349d10>"
-       ]
-      },
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-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x78994d0>"
-       ]
-      }
-     ],
-     "prompt_number": 61
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.7, Page number: 519<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Draw boxes and fill with different designs.\n",
-      "\n",
-      "%pylab\n",
-      "#Tkinter package is used for graphics\n",
-      "from matplotlib.patches import Rectangle\n",
-      "from matplotlib.collections import PatchCollection\n",
-      "\n",
-      "e = Rectangle(xy=(35, -50), width=10, height=5, linewidth=2.0, color='b')\n",
-      "fig = plt.gcf()\n",
-      "fig.gca().add_artist(e)\n",
-      "e.set_clip_box(ax.bbox)\n",
-      "e.set_alpha(0.7)\n",
-      "pylab.xlim([20, 50])\n",
-      "pylab.ylim([-65, -35])\n",
-      "\n",
-      "#There are no different automatic fill styles. user should create the styles."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7373590>"
-       ]
-      }
-     ],
-     "prompt_number": 72
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.8, Page number: 520<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display text in different size, font, vertically and horizontally \n",
-      "\n",
-      "%pylab\n",
-      "\n",
-      "from pylab import *\n",
-      "\n",
-      "#Tkinter package is used for graphics\n",
-      "# set limits so that it no longer looks on screen to be 45 degrees\n",
-      "xlim([-5,5])\n",
-      "\n",
-      "# Locations to plot text\n",
-      "l1 = array((1,1))\n",
-      "l2 = array((5,5))\n",
-      "\n",
-      "# Rotate angle\n",
-      "angle = 90\n",
-      "trans_angle = gca().transData.transform_angles(array((45,)),\n",
-      "                                               l2.reshape((1,2)))[0]\n",
-      "\n",
-      "# Plot text\n",
-      "th2 = text(l2[0],l2[1],'Hello(Horizontal Text with fontsize 25)\\n\\n',fontsize=25,\n",
-      "           rotation=0)\n",
-      "th2 = text(l2[0],l2[1],'Hello(Horizontal Text with fontsize 16)',fontsize=16,\n",
-      "           rotation=0)\n",
-      "th1 = text(l1[0],l1[1],'Hello(Vertical Text)',fontsize=16,\n",
-      "           rotation=angle)\n",
-      "show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Using matplotlib backend: module://IPython.kernel.zmq.pylab.backend_inline\n",
-        "Populating the interactive namespace from numpy and matplotlib\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['power', 'draw_if_interactive', 'random', 'fft', 'angle', 'linalg', 'info']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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KqNs1+SNAAsBVxBNwX0KbD+LAZULNHz58uHbs2KEnn3xS9erVU5EiRRQfH6+k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\nQPnz51f//v1dfXQxrFixQoc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ISAAAAEoEJAAAACUCEgAAgBIBCQAAQImABAAA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\nAAAAANbkf1M+pbyPupgkAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x792e430>"
-       ]
-      }
-     ],
-     "prompt_number": 95
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.9, Page number: 521<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display status of mouse button pressed \n",
-      "\n",
-      "from Tkinter import *\n",
-      "from tkFileDialog import askopenfilename\n",
-      "import Image, ImageTk\n",
-      "\n",
-      "if __name__ == \"__main__\":\n",
-      "    root = Tk()\n",
-      "\n",
-      "    #setting up a tkinter canvas with scrollbars\n",
-      "    frame = Frame(root, bd=2, relief=SUNKEN)\n",
-      "    frame.grid_rowconfigure(0, weight=1)\n",
-      "    frame.grid_columnconfigure(0, weight=1)\n",
-      "    xscroll = Scrollbar(frame, orient=HORIZONTAL)\n",
-      "    xscroll.grid(row=1, column=0, sticky=E+W)\n",
-      "    yscroll = Scrollbar(frame)\n",
-      "    yscroll.grid(row=0, column=1, sticky=N+S)\n",
-      "    canvas = Canvas(frame, bd=0, xscrollcommand=xscroll.set, yscrollcommand=yscroll.set)\n",
-      "    canvas.grid(row=0, column=0, sticky=N+S+E+W)\n",
-      "    xscroll.config(command=canvas.xview)\n",
-      "    yscroll.config(command=canvas.yview)\n",
-      "    frame.pack(fill=BOTH,expand=1)\n",
-      "\n",
-      "    \n",
-      "\n",
-      "    #function to be called when mouse is clicked\n",
-      "    def printcoords(event):\n",
-      "        #outputting x and y coords to console\n",
-      "        print \"Mouse Button pressed\"\n",
-      "        print (event.x,event.y)\n",
-      "    #mouseclick event\n",
-      "    canvas.bind(\"<Button 1>\",printcoords)\n",
-      "\n",
-      "    root.mainloop()\n",
-      "    \n",
-      "import win32api, win32con\n",
-      "\n",
-      "print \"Current cursor position at \" \n",
-      "print win32api.GetCursorPos()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mouse Button pressed\n",
-        "(207, 115)\n",
-        "Current cursor position at "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "(502, 188)\n"
-       ]
-      }
-     ],
-     "prompt_number": 108
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 15.10, Page number: 523<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Change mouse cursor.\n",
-      "\n",
-      "#Placing the cursor on top of the circle button will change the pointer to circle\n",
-      "# and plus button to plus symbol\n",
-      "\n",
-      "from Tkinter import *\n",
-      "import Tkinter\n",
-      "\n",
-      "top = Tkinter.Tk()\n",
-      "\n",
-      "B1 = Tkinter.Button(top, text =\"circle\", relief=RAISED,\\\n",
-      "                         cursor=\"circle\")\n",
-      "B2 = Tkinter.Button(top, text =\"plus\", relief=RAISED,\\\n",
-      "                         cursor=\"plus\")\n",
-      "B1.pack()\n",
-      "B2.pack()\n",
-      "top.mainloop()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 110
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
->>>>>>> 0ee873700378b995b441b1be6652178f741aea5b
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter16.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter16.ipynb
deleted file mode 100755
index cafb9828..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter16.ipynb
+++ /dev/null
@@ -1,2160 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 16: Marching Towards C++<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.1, Page number: 537<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read and display a string\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Initialize variable\n",
-      "name = raw_input(\"Enter Your Name : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Your name is %s\"%(name))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Name : Amit\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your name is Amit"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.2, Page number: 540<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the value of a variable using reference variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "qty = 10\n",
-      "qt = qty\n",
-      "#in python, value tagging method is used for storage. so there is no reference concept\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"qty\\tqt\\n\")\n",
-      "sys.stdout.write(\"%d\\t%d\\n\"%(qty,qt))\n",
-      "qt = qt + 1\n",
-      "qty = qty + 1\n",
-      "sys.stdout.write(\"%d\\t%d\\n\"%(qty,qt))\n",
-      "qty = qty - 1\n",
-      "qt = qt - 1\n",
-      "sys.stdout.write(\"%d\\t%d\\n\"%(qty,qt))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "qty\tqt\n",
-        "10\t10\n",
-        "11\t11\n",
-        "10\t10\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.3, Page number: 541<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to use scope access operator.\n",
-      "#Display the variaous values of the same variable declared at different scope levels.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 10\n",
-      "\n",
-      "def main():\n",
-      "    a = 20\n",
-      "    sys.stdout.write(\"::a = %d\"%(a))\n",
-      "    return\n",
-      "\n",
-      "#There is no scope operator. Intendation is used to define scope/block of statements\n",
-      "main()\n",
-      "sys.stdout.write(\"    a = %d\"%(a))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "::a = 20    a = 10"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.4, Page number: 542<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to read tow integers through the keyboard.\n",
-      "#Declare the variables in C++ style.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Read Variables\n",
-      "num = int(raw_input(\"Enter Two numbers : \"))\n",
-      "num1 = int(raw_input(\"Enter Two numbers : \"))\n",
-      "#No variable declaration is needed in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Entered Numbers are : %d  %d\"%(num,num1))\n",
-      "                                    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Two numbers : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Two numbers : 9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Entered Numbers are : 8  9"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.5, Page number: 543<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Length of a string\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#There is no variable declaration is needed in python\n",
-      "#Read string\n",
-      "name = raw_input(\"Enter Your Name : \")\n",
-      "\n",
-      "#Calculation\n",
-      "len1 = len(name)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The length of the string is : %d\"%(len1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Name : Santosh\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The length of the string is : 7"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.6, Page number: 544<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sum of numbers using function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def sum1(x,y):\n",
-      "    return x+y\n",
-      "#Forward function definition is not possible in python.\n",
-      "#Otherwise wrap the function within the main() function\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 20\n",
-      "b = 2.5\n",
-      "\n",
-      "#Function call\n",
-      "sys.stdout.write(\"Sum = %f\"%(sum1(a,b)))\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum = 22.500000"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.7, Page number: 545<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function with default arguments.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition\n",
-      "def sum1(j,k=10,l=15,m=20):\n",
-      "    return j+k+l+m\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 2\n",
-      "b = 3\n",
-      "c = 4\n",
-      "d = 5\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Sum = %d\"%(sum1(a,b,c,d)))\n",
-      "sys.stdout.write(\"\\nSum = %d\"%(sum1(a,b,c)))\n",
-      "sys.stdout.write(\"\\nSum = %d\"%(sum1(a,b)))\n",
-      "sys.stdout.write(\"\\nSum = %d\"%(sum1(a)))\n",
-      "sys.stdout.write(\"\\nSum = %d\"%(sum1(b,c,d)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum = 14\n",
-        "Sum = 29\n",
-        "Sum = 40\n",
-        "Sum = 47\n",
-        "Sum = 32"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.8, Page number: 547<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Square of integer and float number using function overloading\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Function definition/ There is no seperate function definition is needed.\n",
-      "def sqr(s):\n",
-      "    return s*s\n",
-      "\n",
-      "#Variable iNitialization\n",
-      "a = 15\n",
-      "b = 2.5\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Square = %d\"%(sqr(a)))\n",
-      "sys.stdout.write(\"\\nSquare = %f\"%(sqr(b)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Square = 225\n",
-        "Square = 6.250000"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.9, Page number: 549<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display private and public member variables of the class.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class num:\n",
-      "    def readdata(self,j,k):\n",
-      "        self.x = j\n",
-      "        self.y = k\n",
-      "        \n",
-      "    def display(self):\n",
-      "        sys.stdout.write(\"x = %d   y = %f\"%(self.x,self.y))\n",
-      "\n",
-      "#Object declaration\n",
-      "j = num()\n",
-      "\n",
-      "#Result\n",
-      "j.z = 'C'\n",
-      "j.readdata(10,10.5)\n",
-      "j.display()\n",
-      "sys.stdout.write(\"  z = %c\"%(j.z))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "x = 10   y = 10.500000  z = C"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.10, Page number: 550<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Private and public data member of a class.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class player:\n",
-      "    def __init__(self,name=None,height=0,weight=0):\n",
-      "        self.name = name\n",
-      "        self.height = height\n",
-      "        self.weight = weight\n",
-      "        \n",
-      "#Object declaration\n",
-      "a = player()\n",
-      "\n",
-      "#Variable Initialization\n",
-      "#There is no need/concept for private variables in python.\n",
-      "a.name = \"Sanjay\"\n",
-      "a.height = 5.5\n",
-      "a.weight = 38\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Height : %d\"%(a.height))\n",
-      "sys.stdout.write(\"\\nWeight : %d\"%(a.weight))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Height : 5\n",
-        "Weight : 38"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.11, Page number: 551<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Class with member variables and functions. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class player:\n",
-      "    def __init__(self,name=None,height=0,weight=0):\n",
-      "        self.name = name\n",
-      "        self.height = height\n",
-      "        self.weight = weight\n",
-      "        \n",
-      "    def setdata(self):\n",
-      "        self.name = raw_input(\"Enter Name Age Height Weight\")\n",
-      "        self.age = int(raw_input(\"Enter Name Age Height Weight\"))\n",
-      "        self.height = float(raw_input(\"Enter Name Age Height Weight\"))\n",
-      "        self.weight = int(raw_input(\"Enter Name Age Height Weight\"))\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\nName    :  %s\"%(self.name))\n",
-      "        sys.stdout.write(\"\\nAge     :  %d\"%(self.age))\n",
-      "        sys.stdout.write(\"\\nHeight  :  %f\"%(self.height))\n",
-      "        sys.stdout.write(\"\\nWeight  :  %d\"%(self.weight))\n",
-      "        \n",
-      "#Object declaration\n",
-      "a = player()\n",
-      "\n",
-      "#Function call\n",
-      "a.setdata()\n",
-      "a.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height WeightSanjay\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height Weight24\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height Weight5.5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height Weight54\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name    :  Sanjay\n",
-        "Age     :  24\n",
-        "Height  :  5.500000\n",
-        "Weight  :  54"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.12, Page number: 552<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Class with member variables and functions.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class player:\n",
-      "    def __init__(self,name=None,height=0,weight=0):\n",
-      "        self.name = name\n",
-      "        self.height = height\n",
-      "        self.weight = weight\n",
-      "   \n",
-      "    def setdata(self):\n",
-      "        self.name = raw_input(\"Enter Name Age Height Weight\")\n",
-      "        self.age = int(raw_input(\"Enter Name Age Height Weight\"))\n",
-      "        self.height = float(raw_input(\"Enter Name Age Height Weight\"))\n",
-      "        self.weight = int(raw_input(\"Enter Name Age Height Weight\"))\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\nName    :  %s\"%(self.name))\n",
-      "        sys.stdout.write(\"\\nAge     :  %d\"%(self.age))\n",
-      "        sys.stdout.write(\"\\nHeight  :  %f\"%(self.height))\n",
-      "        sys.stdout.write(\"\\nWeight  :  %d\"%(self.weight))\n",
-      "#In python, member functions should be defined inside the class instance\n",
-      "\n",
-      "#Object declaration\n",
-      "a = player()\n",
-      "\n",
-      "#Function call\n",
-      "a.setdata()\n",
-      "a.show()   "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height WeightAjay\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height Weight24\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height Weight5.2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Name Age Height Weight45\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Name    :  Ajay\n",
-        "Age     :  24\n",
-        "Height  :  5.200000\n",
-        "Weight  :  45"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.13, Page number: 554<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Static data member. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class sumnum:\n",
-      "    def __init__(self,num = 0):\n",
-      "        self.num = num\n",
-      "        \n",
-      "    def input1(self,c):\n",
-      "        self.num = int(raw_input(\"Enter Number : \"))\n",
-      "        c = c + self.num\n",
-      "        return c\n",
-      "        \n",
-      "    def sum1(self,c):\n",
-      "        sys.stdout.write(\"The sum of entered numbers is %d\"%(c))\n",
-      "        \n",
-      "#Object declaration\n",
-      "a = sumnum()\n",
-      "b = sumnum()\n",
-      "c1 = sumnum()\n",
-      "c = 0\n",
-      "\n",
-      "#Function call\n",
-      "#There is no static variable in python\n",
-      "c = a.input1(c)\n",
-      "c = b.input1(c)\n",
-      "c = c1.input1(c)\n",
-      "\n",
-      "a.sum1(c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of entered numbers is 14"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.14, Page number: 555<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Static member functions \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class bita:\n",
-      "    def __initi__(self):\n",
-      "        self.c = 0\n",
-      "    \n",
-      "    @staticmethod    \n",
-      "    def count(c):\n",
-      "        c = c + 1\n",
-      "        return c\n",
-      "    \n",
-      "    @staticmethod\n",
-      "    def display(k,c):\n",
-      "        k = k + 1\n",
-      "        sys.stdout.write(\"\\nCall Number : %d\"%(k))\n",
-      "        sys.stdout.write(\"\\nValue of c : %d\"%(c))\n",
-      "        return k\n",
-      "        \n",
-      "#Object declaration\n",
-      "b = bita()\n",
-      "c1 = bita()\n",
-      "\n",
-      "c = 0\n",
-      "k = 0\n",
-      "\n",
-      "#Function call\n",
-      "k = bita.display(k,c)\n",
-      "c = bita.count(c)\n",
-      "c = b.count(c)\n",
-      "c = c1.count(c)\n",
-      "k = bita.display(k,c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Call Number : 1\n",
-        "Value of c : 0\n",
-        "Call Number : 2\n",
-        "Value of c : 3"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.15, Page number: 557<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Array of objects to maintain records of bank \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class bank:\n",
-      "    def __init__(self):\n",
-      "        self.city = None\n",
-      "        self.bank = None\n",
-      "        self.branches = 0\n",
-      "        self.no_ac = 0\n",
-      "        \n",
-      "    def input1(self):\n",
-      "        self.city = raw_input(\"City Name : \")\n",
-      "        self.bank = raw_input(\"Bank Name : \")\n",
-      "        self.branches = int(raw_input(\"Total Branches : \"))\n",
-      "        self.no_ac = int(raw_input(\"Number of A/c : \"))\n",
-      "        \n",
-      "    def console(self):\n",
-      "        sys.stdout.write(\"\\n\\nCity Name    :   %s\"%(self.city))\n",
-      "        sys.stdout.write(\"\\nBank Name      :   %s\"%(self.bank))\n",
-      "        sys.stdout.write(\"\\nTotal Branches :   %d\"%(self.branches))\n",
-      "        sys.stdout.write(\"\\nNumber of A/c  :   %d\"%(self.no_ac))\n",
-      "        \n",
-      "#Object declaration\n",
-      "ms = [bank() for i in range(0,2)]\n",
-      "\n",
-      "#Processing\n",
-      "for k in range(0,2):\n",
-      "    ms[k].input1()\n",
-      "    \n",
-      "for k in range(0,2):\n",
-      "    ms[k].console()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "City Name : Nanded\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bank Name : SNSB\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total Branches : 10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of A/c : 7500\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "City Name : Latur\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bank Name : SNSB\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total Branches : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number of A/c : 5400\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "City Name    :   Nanded\n",
-        "Bank Name      :   SNSB\n",
-        "Total Branches :   10\n",
-        "Number of A/c  :   7500\n",
-        "\n",
-        "City Name    :   Latur\n",
-        "Bank Name      :   SNSB\n",
-        "Total Branches :   8\n",
-        "Number of A/c  :   5400"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.16, Page number: 559<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Accessing private data using non member function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class ac:\n",
-      "    def __init__(self):\n",
-      "        self.name = None\n",
-      "        self.acno = 0\n",
-      "        self.bal = 0\n",
-      "        \n",
-      "    def read(self):\n",
-      "        self.name = raw_input(\"Name  : \")\n",
-      "        self.acno = int(raw_input(\"A/c No  :\"))\n",
-      "        self.bal = int(raw_input(\"Balance : \"))\n",
-      "        \n",
-      "#Non member function\n",
-      "def showbal(a):\n",
-      "    sys.stdout.write(\"\\nBalance of A/c no. %d is Rs. %d\"%(a.acno,a.bal))\n",
-      "    \n",
-      "#object declaration\n",
-      "k = ac()\n",
-      "\n",
-      "#Function call\n",
-      "k.read()\n",
-      "showbal(k)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name  : Manoj\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A/c No  :474\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Balance : 40000\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Balance of A/c no. 474 is Rs. 40000"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.17, Page number: 561<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Friend function in two classes. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class first:\n",
-      "    def __init__(self):\n",
-      "        self.f = 0\n",
-      "        \n",
-      "    def getvalue(self):\n",
-      "        self.f = int(raw_input(\"Enter a number : \"))\n",
-      "        \n",
-      "class second:\n",
-      "    def __init__(self):\n",
-      "        self.s = 0\n",
-      "        \n",
-      "    def getvalue(self):\n",
-      "        self.s = int(raw_input(\"Enter a number : \"))\n",
-      "        \n",
-      "        \n",
-      "#Non member function\n",
-      "def sum1(d,t):\n",
-      "    sys.stdout.write(\"\\nSum of two numbers : %d\"%(t.f+d.s))\n",
-      "    \n",
-      "#Object declaration\n",
-      "a = first()\n",
-      "b = second()\n",
-      "\n",
-      "#Function call\n",
-      "a.getvalue()\n",
-      "b.getvalue()\n",
-      "sum1(b,a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of two numbers : 15"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.18, Page number: 562<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#String Replacement\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class fandr:\n",
-      "    def __init__(self):\n",
-      "        self.str1 = None\n",
-      "        self.f = ''\n",
-      "        self.r = ''\n",
-      "        \n",
-      "    def accept(self):\n",
-      "        #Since string object does not support item assignment, initialize the string as character array\n",
-      "        self.str1 = ['P','r','o','g','r','a','_','e','r']\n",
-      "        self.f = raw_input(\"Find what (char) \")\n",
-      "        self.r = raw_input(\"Replace with (char) \")\n",
-      "        \n",
-      "    def display(self,d):\n",
-      "        sys.stdout.write(\"%c\"%(self.str1[d]))\n",
-      "        \n",
-      "    def len1(self):\n",
-      "        self.l = len(self.str1)\n",
-      "        return self.l\n",
-      "    \n",
-      "    def find(self,i):\n",
-      "        if self.str1[i] == self.f:\n",
-      "            self.replace(i)\n",
-      "            \n",
-      "    def replace(self,k):\n",
-      "        self.str1[k] = self.r\n",
-      "        \n",
-      "#Object declaration\n",
-      "b = fandr()\n",
-      "\n",
-      "#function call\n",
-      "b.accept()\n",
-      "l = b.len1()\n",
-      "sys.stdout.write(\"\\nReplaced text : \")\n",
-      "\n",
-      "for i in range(0,l):\n",
-      "    b.find(i)\n",
-      "    b.display(i)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Find what (char) _\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Replace with (char) m\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Replaced text : Programer"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.19, Page number: 564<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Largest out of ten numbers.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    def __init__(self):\n",
-      "        self.number = [0 for i in range(0,10)]\n",
-      "        \n",
-      "    def input1(self,i):\n",
-      "        self.number[i] = int(raw_input(\"Enter Number (%d) : \"%(i+1)))\n",
-      "        return self.number[i]\n",
-      "    \n",
-      "    def large(self,s,m):\n",
-      "        if self.number[s] == m:\n",
-      "            sys.stdout.write(\"\\nLargest number : %d\"%(self.number[s]))\n",
-      "            return 0\n",
-      "        else:\n",
-      "            return 1\n",
-      "        \n",
-      "#Object declaration\n",
-      "b = num()\n",
-      "\n",
-      "sum1 = 0\n",
-      "c = 1\n",
-      "\n",
-      "#Sum\n",
-      "for i in range(0,10):\n",
-      "    sum1 = sum1 + b.input1(i)\n",
-      "    \n",
-      "#largest\n",
-      "for k in range(sum1,0,-1):\n",
-      "    for i in range(0,10):\n",
-      "        if c == 1:\n",
-      "            c = b.large(i,k)\n",
-      "        else:\n",
-      "            break"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (1) : 125\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (2) : 654\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (3) : 246\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (4) : 945\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (5) : 258\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (6) : 159\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (7) : 845\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (8) : 940\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (9) : 944\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number (10) : 485\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Largest number : 945"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.20, Page number: 567<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constructor to initialize the class member variables \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    #Constructor\n",
-      "    def __init__(self):\n",
-      "        sys.stdout.write(\"\\nConstructor called\")\n",
-      "        self.x = 5\n",
-      "        self.a = 0\n",
-      "        self.b = 1\n",
-      "        self.c = 2\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\nx = %d  a = %d  b = %d  c = %d\"%(self.x,self.a,self.b,self.c))\n",
-      "        \n",
-      "#Object declaration\n",
-      "x = num()\n",
-      "\n",
-      "#Function call\n",
-      "x.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Constructor called\n",
-        "x = 5  a = 0  b = 1  c = 2"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.21, Page number: 569<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constructor with arguments \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    #Constructor\n",
-      "    def __init__(self,m,j,k):\n",
-      "        self.a = m\n",
-      "        self.b = j\n",
-      "        self.c = k\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\na = %d  b = %d  c = %d\"%(self.a,self.b,self.c))\n",
-      "        \n",
-      "        \n",
-      "#Object declaration\n",
-      "x = num(4,5,7)\n",
-      "y = num(1,2,8)\n",
-      "\n",
-      "#Function call\n",
-      "x.show()\n",
-      "y.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "a = 4  b = 5  c = 7\n",
-        "a = 1  b = 2  c = 8"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.22, Page number: 571<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constructor overloading\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    def __init__(self,m = None,j = None,k = None):\n",
-      "        if m and j and k is not None:\n",
-      "            sys.stdout.write(\"Constructor with three arguments\")\n",
-      "            self.a = m\n",
-      "            self.b = j\n",
-      "            self.c = k\n",
-      "        \n",
-      "        else:\n",
-      "            if m and j is not None:\n",
-      "                sys.stdout.write(\"Constructor with two arguments\")\n",
-      "                self.a = m\n",
-      "                self.b = j\n",
-      "                self.c = ''\n",
-      "            else:\n",
-      "                if m == j == k == None:\n",
-      "                    sys.stdout.write(\"Constructor without arguments\")\n",
-      "                    self.a = 0\n",
-      "                    self.b = 0\n",
-      "                    self.c = ''\n",
-      "                    \n",
-      "                    \n",
-      "    def show(self):    \n",
-      "        sys.stdout.write(\"\\na = %d   b = %f c = \"%(self.a,self.b))\n",
-      "        print self.c\n",
-      "        \n",
-      "        \n",
-      "#Object declaration\n",
-      "\n",
-      "x = num(4,5.5,'A')\n",
-      "x.show()\n",
-      "\n",
-      "y = num(1,2.2)\n",
-      "y.show()\n",
-      "\n",
-      "z = num()\n",
-      "z.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Constructor with three arguments\n",
-        "a = 4   b = 5.500000 c = A\n",
-        "Constructor with two arguments\n",
-        "a = 1   b = 2.200000 c = \n",
-        "Constructor without arguments\n",
-        "a = 0   b = 0.000000 c = \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.23, Page number: 573<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Default arguments in constructor. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class power:\n",
-      "    def __init__(self,n = 9,p = 3):\n",
-      "        self.num = n\n",
-      "        self.power = p\n",
-      "        self.ans = pow(n,p)\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\n%d raise to %d is %d\"%(self.num,self.power,self.ans))\n",
-      "        \n",
-      "        \n",
-      "#Object declaration\n",
-      "p1 = power()\n",
-      "p2 = power(5)\n",
-      "\n",
-      "#Result\n",
-      "p1.show()\n",
-      "p2.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "9 raise to 3 is 729\n",
-        "5 raise to 3 is 125"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.24, Page number: 575<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Object with reference to constructor. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    #There is no overloading concept in python\n",
-      "    def __init__(self,k=None,j=None):\n",
-      "        if k is not None:\n",
-      "            self.n = k\n",
-      "            return\n",
-      "        else:\n",
-      "            if j is not None:\n",
-      "                self.n = j.n\n",
-      "                \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"%d\"%(self.n))\n",
-      "        \n",
-      "#Object creation\n",
-      "J = num(50)\n",
-      "K = num(J.n)\n",
-      "L = num()\n",
-      "L = J\n",
-      "M = num()\n",
-      "M = J\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nObject J Value of n : \")\n",
-      "J.show()\n",
-      "sys.stdout.write(\"\\nObject K Value of n : \")\n",
-      "K.show()\n",
-      "sys.stdout.write(\"\\nObject L Value of n : \")\n",
-      "L.show()\n",
-      "sys.stdout.write(\"\\nObject M Value of n : \")\n",
-      "M.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Object J Value of n : 50\n",
-        "Object K Value of n : 50\n",
-        "Object L Value of n : 50\n",
-        "Object M Value of n : 50"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.25, Page number: 576<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Destructors\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "c = 0\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    def __init__(self,c):\n",
-      "        c = c + 1\n",
-      "        sys.stdout.write(\"\\nObject Created : Object(%d)\"%(c))\n",
-      "        \n",
-      "        \n",
-      "    def __exit__(self, type, value, traceback,c):\n",
-      "        self.package_obj.cleanup()\n",
-      "        sys.stdout.write(\"\\nObject Released : Object(%d)\"%(c))\n",
-      "        c = c - 1\n",
-      "       \n",
-      "        \n",
-      "#Object creation\n",
-      "sys.stdout.write(\"\\nIn main()\\n\")\n",
-      "a = num(c)\n",
-      "b = num(c)\n",
-      "\n",
-      "sys.stdout.write(\"\\nIn Block A()\\n\")\n",
-      "c = num(c)\n",
-      "\n",
-      "sys.stdout.write(\"\\n\\nAgain in main()\\n\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In main()\n",
-        "\n",
-        "Object Created : Object(1)\n",
-        "Object Created : Object(1)\n",
-        "In Block A()\n",
-        "\n",
-        "Object Created : Object(1)\n",
-        "\n",
-        "Again in main()\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.26, Page number: 579<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Overload unary ++ operator\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    def input1(self):\n",
-      "        self.a = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.b = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.c = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.d = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\nA = %d  B = %d  C = %d  D = %d\"%(self.a,self.b,self.c,self.d))\n",
-      "        \n",
-      "    def operatorplusplus(self):\n",
-      "        self.a = self.a + 1\n",
-      "        self.b = self.b + 1\n",
-      "        self.c = self.c + 1\n",
-      "        self.d = self.d + 1\n",
-      "        \n",
-      "#Object creation\n",
-      "X = num()\n",
-      "\n",
-      "X.input1()\n",
-      "sys.stdout.write(\"\\nBefore Overloading X  : \")\n",
-      "X.show()\n",
-      "\n",
-      "#There is no operator overloading in python\n",
-      "X.operatorplusplus()\n",
-      "sys.stdout.write(\"\\nAfter Overloading X  : \")\n",
-      "X.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Before Overloading X  : \n",
-        "A = 4  B = 5  C = 8  D = 9\n",
-        "After Overloading X  : \n",
-        "A = 5  B = 6  C = 9  D = 10"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.27, Page number: 581<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Overload + binary operator\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    def input1(self):\n",
-      "        self.a = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.b = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.c = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.d = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"A = %d  B = %d  C = %d  D = %d\"%(self.a,self.b,self.c,self.d))\n",
-      "        \n",
-      "    def operatorplus(self,t):\n",
-      "        tmp = num()\n",
-      "        tmp.a = self.a + t.a\n",
-      "        tmp.b = self.b + t.b\n",
-      "        tmp.c = self.c + t.c\n",
-      "        tmp.d = self.d + t.d\n",
-      "        return tmp\n",
-      "    \n",
-      "#Object creation\n",
-      "X = num()\n",
-      "Y = num()\n",
-      "Z = num()\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nObject X \")\n",
-      "X.input1()\n",
-      "sys.stdout.write(\"\\nObject Y \")\n",
-      "Y.input1()\n",
-      "\n",
-      "Z = X.operatorplus(Y)\n",
-      "\n",
-      "sys.stdout.write(\"\\nX : \")\n",
-      "X.show()\n",
-      "sys.stdout.write(\"\\nY : \")\n",
-      "Y.show()\n",
-      "sys.stdout.write(\"\\nZ : \")\n",
-      "Z.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Object X "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Object Y "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "X : A = 1  B = 4  C = 2  D = 1\n",
-        "Y : A = 2  B = 5  C = 4  D = 2\n",
-        "Z : A = 3  B = 9  C = 6  D = 3"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.28, Page number: 583<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Multiplication using an integer and object. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class definition\n",
-      "class num:\n",
-      "    def input1(self):\n",
-      "        self.a = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.b = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.c = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        self.d = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"A = %d  B = %d  C = %d  D = %d\"%(self.a,self.b,self.c,self.d))\n",
-      "        \n",
-      "def operatormul(a,t):\n",
-      "    tmp = num()\n",
-      "    tmp.a = a * t.a\n",
-      "    tmp.b = a * t.b\n",
-      "    tmp.c = a * t.c\n",
-      "    tmp.d = a * t.d\n",
-      "    return tmp\n",
-      "\n",
-      "#Result\n",
-      "X = num()\n",
-      "Z = num()\n",
-      "\n",
-      "sys.stdout.write(\"\\nObject X \")\n",
-      "X.input1()\n",
-      "\n",
-      "Z = operatormul(3,X)\n",
-      "sys.stdout.write(\"\\nX : \")\n",
-      "X.show()\n",
-      "sys.stdout.write(\"\\nZ : \")\n",
-      "Z.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Object X "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a,b,c and d : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "X : A = 1  B = 2  C = 2  D = 3\n",
-        "Z : A = 3  B = 6  C = 6  D = 9"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.29, Page number: 585<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Inheritance\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Class declaration\n",
-      "class one:\n",
-      "    def __init__(self):\n",
-      "        self.a = 0\n",
-      "        self.b = 0\n",
-      "        \n",
-      "class two(one):\n",
-      "    def __init__(self):\n",
-      "        one.__init__(self)\n",
-      "        self.c = 0\n",
-      "        self.d = 0\n",
-      "        \n",
-      "    def input1(self):\n",
-      "        self.a = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
-      "        self.b = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
-      "        self.c = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
-      "        self.d = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
-      "        \n",
-      "    def display(self):\n",
-      "        sys.stdout.write(\"\\na = %d  b = %d  c = %d  d = %d\"%(self.a,self.b,self.c,self.d))\n",
-      "        \n",
-      "#Object declaration\n",
-      "a = two()\n",
-      "\n",
-      "#Result\n",
-      "a.input1()\n",
-      "a.display()\n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter values for a,b,c,d : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter values for a,b,c,d : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter values for a,b,c,d : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter values for a,b,c,d : 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "a = 2  b = 5  c = 4  d = 7"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 16.30, Page number: 586<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Member function using pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#class declaration\n",
-      "class super1:\n",
-      "    def display(self):\n",
-      "        sys.stdout.write(\"\\nIn function display()  class super\")\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\nIn function show() class super\")\n",
-      "        \n",
-      "class sub(super1):\n",
-      "    def display(self):\n",
-      "        sys.stdout.write(\"\\nIn function display() class sub\")\n",
-      "        \n",
-      "    def show(self):\n",
-      "        sys.stdout.write(\"\\nIn function show() class sub\")\n",
-      "        \n",
-      "#Object declaration\n",
-      "S = super1()\n",
-      "A = sub()\n",
-      "\n",
-      "#There is no pointer concept in python\n",
-      "S.display()\n",
-      "S.show()\n",
-      "\n",
-      "sys.stdout.write(\"\\n\\nNow Pointer point points to derived class sub\\n\")\n",
-      "A.display()\n",
-      "A.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "In function display()  class super\n",
-        "In function show() class super\n",
-        "\n",
-        "Now Pointer point points to derived class sub\n",
-        "\n",
-        "In function display() class sub\n",
-        "In function show() class sub"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter3.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter3.ipynb
deleted file mode 100755
index 880e4fac..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter3.ipynb
+++ /dev/null
@@ -1,1060 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 3: Operators and Expressions<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.1, Page number: 22<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate the use of Comma operator\n",
-      "\n",
-      "#Result\n",
-      "print \"Addition = \",2+3,\" \\nSubtraction = \",5-4"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Addition =  5  \n",
-        "Subtraction =  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.2, Page number: 23<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to demonstrate the use of Conditional operator with two values\n",
-      "\n",
-      "#Result\n",
-      "#in python if..else statement can be written similar to conditional operator\n",
-      "#prints Result = 4 if 2 == 3, otherwise Result = 5\n",
-      "\n",
-      "print \"Result = \",4 if 2==3 else 5"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result =  5\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.3, Page number: 23<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to use Conditional operator with two statements\n",
-      "\n",
-      "#Result\n",
-      "#in python if..else statement can be written similar to conditional operator\n",
-      "#prints True if 3>2, otherwise prints False\n",
-      "\n",
-      "print \"True\" if 3>2 else \"False\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "True\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.4, Page number: 25<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to use increment operator as prefix\n",
-      "\n",
-      "#Variable declaration\n",
-      "x = 10\n",
-      "y = 20\n",
-      "\n",
-      "#Calculation\n",
-      "#there is no ++/-- operator in python\n",
-      "\n",
-      "z = x * y\n",
-      "y += 1\n",
-      "a = x * y\n",
-      "\n",
-      "#Result\n",
-      "print z,\"  \",a"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "200    210\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.5, Page number: 25<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to use increment operator as prefix\n",
-      "\n",
-      "#Variable declaration\n",
-      "x = 10\n",
-      "y = 20\n",
-      "\n",
-      "#Calculation\n",
-      "#there is no ++/-- operator in python\n",
-      "\n",
-      "y += 1\n",
-      "z = x * y\n",
-      "a = x * y\n",
-      "\n",
-      "#Result\n",
-      "print z,\"  \",a"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "210    210\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.6, Page number: 26<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to find the size and address of integer and float variables\n",
-      "\n",
-      "#import python module\n",
-      "#sys python module is required for getsizeof() function.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "#No specific float type variable declaration is available in python. If we\n",
-      "#assign a floating point value to a variable,\n",
-      "#it will be treated as floating point variable\n",
-      "x = 2\n",
-      "y = 2.0\n",
-      "\n",
-      "#Result\n",
-      "#In python, integer variable uses 12 bytes and float variable uses 16 bytes of\n",
-      "#memory to store the variables\n",
-      "#sys.getsizeof() function returns the size of variable in bytes which is similar\n",
-      "#to sizeof() operator\n",
-      "#id() function returns the address of variables which is similar to '&' operator\n",
-      "\n",
-      "print \"Size (x) = \",sys.getsizeof(x)\n",
-      "print \"Size (y) = \",sys.getsizeof(y)\n",
-      "print \"Address of x = \",id(x),\"and y = \",id(y)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Size (x) =  12\n",
-        "Size (y) =  16\n",
-        "Address of x =  5626076 and y =  54761072\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.7, Page number: 27<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to use various relational operators and display their return values\n",
-      "\n",
-      "#Result\n",
-      "#int() function is used to convert the boolean return value of the relational\n",
-      "#expression to integer/binary\n",
-      "\n",
-      "print \"Condition  :   Return Values\"\n",
-      "print \"10 != 10   :       \",int(10 != 10)\n",
-      "print \"10 == 10   :       \",int(10 == 10)\n",
-      "print \"10 >= 10   :       \",int(10 >= 10)\n",
-      "print \"10 <= 100  :       \",int(10 <= 100)\n",
-      "print \"10 != 9    :       \",int(10 != 9)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Condition  :   Return Values\n",
-        "10 != 10   :        0\n",
-        "10 == 10   :        1\n",
-        "10 >= 10   :        1\n",
-        "10 <= 100  :        1\n",
-        "10 != 9    :        1\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.8, Page number: 28<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program uses conditional operator to determine the value of 'b'depending the inputted value of 'a'\n",
-      "\n",
-      "#Reads Input value for a\n",
-      "\n",
-      "a = raw_input(\"Enter Any Integer either above 5 or below 5 :-\")\n",
-      "a = int(a)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Calculation\n",
-      "#Assigns b = 3 if a > 5, otherwise b = 4\n",
-      "b = 3 if a > 5 else 4\n",
-      "\n",
-      "#Result\n",
-      "print \"Calculated Value of b is :- \",b"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Any Integer either above 5 or below 5 :-6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Calculated Value of b is :-  3\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.9, Page number: 28<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to determine the value of 'b' using conditional operator\n",
-      "\n",
-      "#Input\n",
-      "\n",
-      "a = raw_input(\"Enter Any Integer either above 5 or below 5 :-\")\n",
-      "a = int(a)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Calculation\n",
-      "#Assigns b = 3 if a == 5, otherwise b = 4\n",
-      "b = 3 if a == 5 else 4\n",
-      "\n",
-      "#Result\n",
-      "print \"Calculated Value of b is :- \",b"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Any Integer either above 5 or below 5 :-3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Calculated Value of b is :-  4\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.10, Page number: 29<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read three variables x, y and z and Using conditional statements, evaluate\n",
-      "#values of variables a, b and c.\n",
-      "#Perform the sum with two sets of variables.  Check whether the sums are equal or not\n",
-      "#and print appropriate messages.\n",
-      "\n",
-      "\n",
-      "#Reads three Input values for x,y and z\n",
-      "\n",
-      "x = raw_input(\"Enter Value of x \")\n",
-      "x = int(x)\n",
-      "\n",
-      "y = raw_input(\"Enter Value of y \")\n",
-      "y = int(y)\n",
-      "\n",
-      "z = raw_input(\"Enter Value of z \")\n",
-      "z = int(z)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Calculation\n",
-      "#Assigns a = 3 if x >= 5, otherwise a = 4\n",
-      "a = 3 if x >= 5 else 4\n",
-      "\n",
-      "#Result\n",
-      "print \"Calculated Value of a is :- \",a\n",
-      "\n",
-      "#Calculation\n",
-      "#Assigns b = 10 if y <= 8, otherwise b = 9\n",
-      "b = 10 if y <= 8 else 9\n",
-      "\n",
-      "#Result\n",
-      "print \"Calculated Value of b is :- \",b\n",
-      "\n",
-      "#Calculation\n",
-      "#Assigns c = 20 if z == 10, otherwise c = 30\n",
-      "c = 20 if z == 10  else 30\n",
-      "\n",
-      "#Result\n",
-      "print \"Calculated Value of c is :- \",c\n",
-      "\n",
-      "#Calculation\n",
-      "m = x + y + z\n",
-      "n = a + b + c\n",
-      "\n",
-      "#Result\n",
-      "print \"Addition of x,y,z is \",m,\"(m)\"\n",
-      "print \"Addition of a,b,c is \",n,\"(n)\"\n",
-      "print \"m & n NOT EQUAL\" if m != n else \"m & n ARE EQUAL\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of x 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of y 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of z 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Calculated Value of a is :-  3\n",
-        "Calculated Value of b is :-  10\n",
-        "Calculated Value of c is :-  30\n",
-        "Addition of x,y,z is  14 (m)\n",
-        "Addition of a,b,c is  43 (n)\n",
-        "m & n NOT EQUAL\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.11, Page number: 30<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use of logical operators\n",
-      "\n",
-      "#Result\n",
-      "#In python, not equal to can be represented using both != and <>.\n",
-      "\n",
-      "print \"Condition        :   Return Values\"\n",
-      "print \"5 > 3 && 5 < 10  :   \",int(5 > 3 & 5 < 10)\n",
-      "print \"8 > 5 || 8 < 2   :   \",int(8 > 5 | 8 < 2)\n",
-      "print \"!(8 == 8)        :   \",int(8 != 8)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Condition        :   Return Values\n",
-        "5 > 3 && 5 < 10  :    1\n",
-        "8 > 5 || 8 < 2   :    0\n",
-        "!(8 == 8)        :    0\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.12, Page number: 30<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print logic 1 if input character is capital otherwise 0\n",
-      "\n",
-      "#Import\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#raw_input() function is used to read a character from the keyboard in Python 2.7\n",
-      "#ord() function converts the character into its corresponding ASCII integer\n",
-      "\n",
-      "\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "x = raw_input(\"Enter a Character : \")\n",
-      "x = ord(x)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Condition\n",
-      "y = 1 if (x >= 65) and (x <= 90) else 0\n",
-      "\n",
-      "#Result\n",
-      "print \"Y : \",y"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Character : A\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Y :  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.13, Page number: 31<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display logic 0 if one reads a character through keyboard otherwise logic 1.\n",
-      "#ASCII values for 0 to 9 are 48 to 57 respectively\n",
-      "\n",
-      "#Variable initialization\n",
-      "#raw_input() function is used to read a character from the keyboard in Python 2.7\n",
-      "#ord() function converts the character into its corresponding ASCII integer\n",
-      "\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "x = raw_input(\"Enter The Character or Integer : \")\n",
-      "x = ord(x)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Condition\n",
-      "y = 1 if (x >= 48) and (x <= 57) else 0\n",
-      "\n",
-      "#Result\n",
-      "print \"Value of Y = \",y"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter The Character or Integer : A\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of Y =  0\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.14, Page number: 32<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display 1 if inputted number is between 1-100 otherwise 0.\n",
-      "#Use the logical AND operator\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "x = raw_input(\"Enter numbers : \")\n",
-      "x = int(x)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Condition\n",
-      "z = 1 if (x >= 1) and (x <= 100) else 0\n",
-      "\n",
-      "#Result\n",
-      "print \"Z = \",z"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Z =  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.15, Page number: 32<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display 1 if inputted number is either 1 or 100 otherwise 0.\n",
-      "#Use the logical OR operator\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "x = raw_input(\"Enter numbers : \")\n",
-      "x = int(x)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Condition\n",
-      "z = 1 if (x == 1) or (x == 100) else 0\n",
-      "\n",
-      "#Result\n",
-      "print \"Z = \",z"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers : 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Z =  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.16, Page number: 33<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display 1 if inputted number is except 100 otherwise 0.\n",
-      "#Use the logical NOT operator\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "x = raw_input(\"Enter number : \")\n",
-      "x = int(x)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Condition\n",
-      "z = 1 if (x != 100) else 0\n",
-      "\n",
-      "#Result\n",
-      "print \"Z : \",z"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter number : 100\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Z :  0\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.17, Page number: 33<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Shift inputted data by two bits right\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "x = raw_input(\"Read The Integer from keyboard (x) :- \")\n",
-      "x = int(x)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#bitwise shifting\n",
-      "x >>= 2\n",
-      "y = x\n",
-      "\n",
-      "#Result\n",
-      "print \"The Right shifted data is = \",y"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integer from keyboard (x) :- 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Right shifted data is =  2\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.18, Page number: 34<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Shift inputted data by two bits to the left\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "x = raw_input(\"Read The Integer from keyboard (x) :- \")\n",
-      "x = int(x)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#bitwise shifting\n",
-      "x <<= 3\n",
-      "y = x\n",
-      "\n",
-      "#Result\n",
-      "print \"The Left shifted data is = \",y"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integer from keyboard (x) :- 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Left shifted data is =  16\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.19, Page number: 35<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use bitwise AND operator between the two integers and display the results\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "a = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
-      "a = int(a)\n",
-      "\n",
-      "b = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
-      "b = int(b)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#bitwise operation\n",
-      "c = a & b\n",
-      "\n",
-      "#Result\n",
-      "print \"The Answer after ANDing is (C) = \",c"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integers from keyboard (a & b) :- 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integers from keyboard (a & b) :- 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Answer after ANDing is (C) =  0\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.20, Page number: 37<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#operate OR operation on two integers and display the result\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "a = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
-      "a = int(a)\n",
-      "\n",
-      "b = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
-      "b = int(b)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#bitwise operation\n",
-      "c = a | b\n",
-      "\n",
-      "#Result\n",
-      "print \"The Oring operation betwen a & b in c = \",c"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integers from keyboard (a & b) :- 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integers from keyboard (a & b) :- 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Oring operation betwen a & b in c =  12\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 3.21, Page number: 38<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Exclusive OR operation between the two integers and display the result\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for x\n",
-      "\n",
-      "a = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
-      "a = int(a)\n",
-      "\n",
-      "b = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
-      "b = int(b)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#bitwise operation\n",
-      "c = a ^ b\n",
-      "\n",
-      "#Result\n",
-      "print \"The data after Exclusive OR operation is in c = \",c"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integers from keyboard (a & b) :- 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Read The Integers from keyboard (a & b) :- 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The data after Exclusive OR operation is in c =  10\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter4.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter4.ipynb
deleted file mode 100755
index 758457a3..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter4.ipynb
+++ /dev/null
@@ -1,972 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 4: Input & Output in C<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.1, Page number: 47<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#show the effect of mismatch of data types\n",
-      "\n",
-      "#Reads Input value for a\n",
-      "\n",
-      "a = raw_input(\"Enter value of 'A' : \")\n",
-      "a = int(a)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "print \"A = \",a\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of 'A' : 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A =  8\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.2, Page number: 47<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#show the effect of mismatch of data types\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#Reads Input value for a\n",
-      "\n",
-      "a = raw_input(\"Enter value of 'A' : \")\n",
-      "a = int(a)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#there is no char datatype of 1 byte size in python.\n",
-      "#python has string variable only.\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "sys.stdout.write(\"A = %d\"%(a%256))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of 'A' : 256\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A = 0"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.3, Page number: 49<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#show the effect of various escape sequences\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = 1\n",
-      "b = a + 1\n",
-      "c = b + 1\n",
-      "d = c + 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write ('\\tA = %d\\nB = %d \\'C = %d\\''%(a,b,c))\n",
-      "sys.stdout.write ('\\n**D = %d**'%(d))\n",
-      "sys.stdout.write ('\\nA = %d B = %d'%(a,b))\n",
-      "sys.stdout.write ('\\r******')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tA = 1\n",
-        "B = 2 'C = 3'\n",
-        "**D = 4**\n",
-        "A = 1 B = 2******"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.4, Page number: 50<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#print the third power of 2 using pow() function.\n",
-      "#assume the floating point numbers\n",
-      "\n",
-      "#Variable initialization\n",
-      "x = 2.0\n",
-      "y = 3.0\n",
-      "\n",
-      "#Result\n",
-      "print '%lf raised to %lf is %lf\\n' %(x,y,pow(x,y))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2.000000 raised to 3.000000 is 8.000000\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.5, Page number: 50<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#print the third power of 10 using pow() function.\n",
-      "#assume the floating point numbers\n",
-      "\n",
-      "#Variable initialization\n",
-      "p = 3\n",
-      "\n",
-      "#Result\n",
-      "#There is no pow10() function in Python.\n",
-      "print 'Ten raised to %lf is %lf\\n' %(p,pow(10,p))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ten raised to 3.000000 is 1000.000000\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.6, Page number: 51<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#detect an error while inputting a data.  Use return value of scanf() statement.\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python,the map(int,a.split()) function splits the input characters\n",
-      "#and converts it into integers and returns a value if all the values are integers\n",
-      "#otherwise an exception will be generated and initializes v as 0.\n",
-      "#using exception handling, this result can be achieved.\n",
-      "\n",
-      "\n",
-      "#Exception handling\n",
-      "try:\n",
-      "    a = raw_input(\"Enter value of 'A','B' & 'C' : \")\n",
-      "    v = map(int,a.split())\n",
-      "except:\n",
-      "    v = 0\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "print '\\nError In Inputting.' if v < 3 else '\\nValues Successfully read.'\n",
-      "\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of 'A','B' & 'C' : 1 2 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Values Successfully read.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.7, Page number: 51<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find length of the string using print() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, print()or write() function would not return the number of characters\n",
-      "#written to the output buffer. so len() function is used to find the length.\n",
-      "\n",
-      "nm = raw_input(\"Enter String : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "l = len(nm)\n",
-      "\n",
-      "#Result\n",
-      "print '\\nLength = ',l\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String : HELLO\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Length =  5\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.8, Page number: 52<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Perform addition of two numbers\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = raw_input(\"ENTER TWO VALUES \")\n",
-      "b = raw_input(\"ENTER TWO VALUES \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "c = int(a) + int(b)\n",
-      "\n",
-      "#Result\n",
-      "print '\\nSUM IS = ',c\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ENTER TWO VALUES 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ENTER TWO VALUES 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "SUM IS =  13\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.9, Page number: 52<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the square of the given number\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = raw_input(\"ENTER ANY NUMBER \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "c = int(a) * int(a)\n",
-      "\n",
-      "#Result\n",
-      "print '\\nSQUARE OF GIVEN NUMBER = ',c\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ENTER ANY NUMBER 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "SQUARE OF GIVEN NUMBER =  25\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.10, Page number: 53<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate average of three real numbers.\n",
-      "\n",
-      "#Variable initialization\n",
-      "\n",
-      "a = raw_input(\"ENTER THREE FLOAT NUMBERS : \")\n",
-      "b = raw_input(\"ENTER THREE FLOAT NUMBERS : \")\n",
-      "c = raw_input(\"ENTER THREE FLOAT NUMBERS : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "d = float(a) + float(b) + float(c)\n",
-      "\n",
-      "#Result\n",
-      "print '\\nAverage of Given Numbers : ',d/3\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ENTER THREE FLOAT NUMBERS : 2.5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ENTER THREE FLOAT NUMBERS : 3.5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ENTER THREE FLOAT NUMBERS : 4.5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Average of Given Numbers :  3.5\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.11, Page number: 53<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print message with inputted name\n",
-      "\n",
-      "#Variable initialization\n",
-      "yourname = raw_input(\"Enter Your Name : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "print \"\\nWel Come %s to 'C' Programming Course\"%(yourname)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Name : Ajay\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Wel Come Ajay to 'C' Programming Course\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.12, Page number: 54<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Input a single character and display it\n",
-      "\n",
-      "#Variable initialization\n",
-      "ch = raw_input(\"Enter any character : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "print \"\\nYour Entered Character is : %c\"%(ch)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any character : C\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Your Entered Character is : C\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.13, Page number: 54<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Swap the values of two variables without using third variable.\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = 7\n",
-      "b = 4\n",
-      "\n",
-      "print \"\\n A = %d B = %d\"%(a,b)\n",
-      "\n",
-      "#Swapping\n",
-      "a = a + b\n",
-      "b = a - b\n",
-      "a = a - b\n",
-      "\n",
-      "#Result\n",
-      "print \"Now A = %d B = %d\"%(a,b)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        " A = 7 B = 4\n",
-        "Now A = 4 B = 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.14, Page number: 55<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Accept characters through keyboard using getchar() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, string is an object. It has no nul-termination character.\n",
-      "#raw_input() reads the string constant from the stdin and the write() function\n",
-      "#prints it to the stdout based on the attribute length.\n",
-      "\n",
-      "ch = raw_input(\" \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(ch)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " COMPILER\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "COMPILER"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.15, Page number: 56<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the characters using putchar() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, string is an object. It has no nul-termination character.\n",
-      "#raw_input() reads the string constant from the stdin.\n",
-      "\n",
-      "ch = raw_input(\"Enter Text Here : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "#using for loop, the string can be displayed character by character.\n",
-      "\n",
-      "sys.stdout.write(\"The Entered Text : \")\n",
-      "for c in ch:\n",
-      "    sys.stdout.write(c)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Here : Characters\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Entered Text : Characters"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.16, Page number: 56<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Show the effect of getche() and getch() functions\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, there is no equivalent function for getche() and getch() functions.\n",
-      "#raw_input() function reads the characters and terminates by pressing the\n",
-      "#enter key\n",
-      "\n",
-      "ch = raw_input(\"Enter any two alphabetics \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "#Result\n",
-      "sys.stdout.write(ch)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any two alphabetics A \n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A "
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.17, Page number: 57<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read and display the character using getch() and putch() functions\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, there is no equivalent function for getch() function.\n",
-      "#raw_input() function reads the characters and terminates by pressing the\n",
-      "#enter key\n",
-      "\n",
-      "ch = raw_input(\"Press any key to continue \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"You Pressed : %s\"%(ch))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Press any key to continue 9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You Pressed : 9"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.18, Page number: 57<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Accept string through the keyboard using gets() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, raw_input() function reads the string and terminates by pressing the\n",
-      "#enter key and write() function displays the string.\n",
-      "\n",
-      "ch = raw_input(\"Enter the String : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Entered String : %s\"%(ch))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the String : USE OF GETS()\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Entered String : USE OF GETS()"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.19, Page number: 58<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the accepted character using puts() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, raw_input() function reads the string and terminates by pressing the\n",
-      "#enter key and print() function displays the string.\n",
-      "\n",
-      "ch = raw_input(\"Enter the String : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "print\"Entered String : \"\n",
-      "print ch\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the String : puts is in use.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Entered String : \n",
-        "puts is in use.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 4.20, Page number: 58<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#read string using cgets() and display it using cputs()\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python, raw_input() function reads the string and terminates by pressing the\n",
-      "#enter key and print() function displays the string.\n",
-      "#in python, string is an object.\n",
-      "\n",
-      "t = raw_input(\"Enter Text Here : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "print \"Your Entered Text : %s\"%(t)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Here : How are you?\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your Entered Text : How are you?\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter5.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter5.ipynb
deleted file mode 100755
index dac1632e..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter5.ipynb
+++ /dev/null
@@ -1,2624 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 5.1: Decision Statements<h1>\n",
-      "   "
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.1, Page number: 64<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Check whether the entered number is less than 10? if yes, display the same\n",
-      "\n",
-      "#Variable initialization\n",
-      "v = raw_input(\"Enter the number : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "v = int(v)\n",
-      "\n",
-      "#Result\n",
-      "if v < 10 :\n",
-      "    print '\\nNumber entered is less than 10'\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number : 9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number entered is less than 10\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.2, Page number: 65<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Check equivalence of two numbers. Use if statement.\n",
-      "\n",
-      "#Variable initialization\n",
-      "\n",
-      "m = raw_input(\"Enter two numbers : \")\n",
-      "n = raw_input(\"Enter two numbers : \")\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "m = int(m)\n",
-      "n = int(n)\n",
-      "\n",
-      "#Result\n",
-      "if m - n == 0 :\n",
-      "    print '\\nTwo numbers are equal'\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two numbers : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Two numbers are equal\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.3, Page number: 66<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Check whether the candidate's age is greater than 17 or not.  If yes, display\n",
-      "#message \"Eligible for Voting\"\n",
-      "\n",
-      "#Variable initialization\n",
-      "age = raw_input(\"Enter age : \")\n",
-      "age = int(age)\n",
-      "\n",
-      "#In python, raw_input() is used to read values\n",
-      "\n",
-      "#Result\n",
-      "if age > 17 :\n",
-      "    print '\\nEligible for Voting.'\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter age : 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Eligible for Voting.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.4, Page number: 66<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use curly brace in the if block.  Enter only the three numbers and calculate\n",
-      "#their sum and multiplication\n",
-      "\n",
-      "#Variable initialization\n",
-      "#in python,the map(int,raw_input().split()) function splits the input characters\n",
-      "#and converts it into integers and returns a value if all the values are integers\n",
-      "#otherwise an exception will be generated and initializes v as 0.\n",
-      "#using exception handling, this result can be achieved.\n",
-      "\n",
-      "#Exception handling\n",
-      "try:\n",
-      "    #use space to separate the input characters\n",
-      "    v = raw_input(\"Enter Three Numbers : \")\n",
-      "    a,b,c = map(int,v.split())\n",
-      "    x = 3\n",
-      "except:\n",
-      "    x = 0\n",
-      "\n",
-      "#Result\n",
-      "#in python, block of statements are identified using indentation\n",
-      "    \n",
-      "if x == 3:\n",
-      "    print \"Addition : %d\"%(a+b+c)\n",
-      "    print \"Multiplication : %d\"%(a*b*c)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 1 2 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Addition : 7\n",
-        "Multiplication : 8\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.5, Page number: 68<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the values of a,b,c through the keyboard.  Add them and after addition\n",
-      "#check if it is in the range of 100 & 200 or not. Print separate message for each.\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = raw_input(\"Enter Three Numbers a b c : \")\n",
-      "b = raw_input(\"Enter Three Numbers a b c : \")\n",
-      "c = raw_input(\"Enter Three Numbers a b c : \")\n",
-      "\n",
-      "a = int(a)\n",
-      "b = int(b)\n",
-      "c = int(c)\n",
-      "\n",
-      "print 'a = %d b = %d c = %d'%(a,b,c)\n",
-      "#Calculation\n",
-      "d = a + b + c\n",
-      "\n",
-      "#Condition evaluation and Result\n",
-      "if d <= 200 and d >= 100 :\n",
-      "    print \"Sum is %d which is in between 100 & 200\"%(d)\n",
-      "else:\n",
-      "    print \"\\nSum is %d which is out of range\"%(d)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers a b c : 50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers a b c : 52\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers a b c : 54\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a = 50 b = 52 c = 54\n",
-        "Sum is 156 which is in between 100 & 200\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.6, Page number: 68<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the roots of a quadratic equation by using if else condition\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = raw_input(\"Enter Values for a, b, c : \")\n",
-      "b = raw_input(\"Enter Values for a, b, c : \")\n",
-      "c = raw_input(\"Enter Values for a, b, c : \")\n",
-      "\n",
-      "a = int(a)\n",
-      "b = int(b)\n",
-      "c = int(c)\n",
-      "\n",
-      "#Condition evaluation and Result\n",
-      "if b * b > 4 * a * c :\n",
-      "    x1 = b + sqrt(b*b - 4*a*c)/2*a\n",
-      "    x2 = b - sqrt(b*b - 4*a*c)/2*a\n",
-      "    print \"\\nx1 = %f x2 = %f\"%(x1,x2)\n",
-      "else:\n",
-      "    print \"\\nRoots are Imaginary\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a, b, c : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a, b, c : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Values for a, b, c : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Roots are Imaginary\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.7, Page number: 69<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate the square of those numbers only whose least significant digit is 5.\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "s = raw_input(\"Enter a Number : \")\n",
-      "s = int(s)\n",
-      "\n",
-      "d = s % 10;\n",
-      "\n",
-      "#Condition evaluation and Result\n",
-      "#There is no increment/decrement operator in python.\n",
-      "if d==5 :\n",
-      "    s = s/10\n",
-      "    s += 1\n",
-      "    print \"\\nSquare = %d%d\"%(s*(s-1),d*d)\n",
-      "else:\n",
-      "    print \"\\nInvalid Number\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 25\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Square = 625\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.8, Page number: 70<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate the salary of medical representative based on the sales.  Bonus and\n",
-      "#incentive to be offered to him will be based on total sales.  If the sale\n",
-      "#exceeds Rs.100000/- follow the particulars of table (1) otherwise (2)\n",
-      "\n",
-      "#  1.  TABLE                            2. TABLE\n",
-      "# Basic = Rs. 3000                  Basic = Rs. 3000\n",
-      "# HRA = 20% of basic                HRA = 20% of basic\n",
-      "# DA = 110% of basic                DA = 110% of basic\n",
-      "# Conveyance = Rs. 500              Conveyance = Rs. 500\n",
-      "# Incentive = 10% of sales          Incentive = 5% of sales\n",
-      "# Bonus = Rs. 500                   Bonus = Rs. 200\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "sale = raw_input(\"Enter Total Sales in Rs. : \")\n",
-      "sale = int(sale)\n",
-      "\n",
-      "#Condition evaluation\n",
-      "if sale >= 100000 :\n",
-      "    bs = 3000\n",
-      "    hra = 20 * bs/100\n",
-      "    da = 110 * bs/100\n",
-      "    cv = 500\n",
-      "    incentive = sale * 10/100\n",
-      "    bonus = 500\n",
-      "else:\n",
-      "    bs = 3000\n",
-      "    hra = 20 * bs/100\n",
-      "    da = 110 * bs/100\n",
-      "    cv = 500\n",
-      "    incentive = sale * 5/100\n",
-      "    bonus = 200\n",
-      "\n",
-      "#Result\n",
-      "ts = bs + hra + da + cv + incentive + bonus\n",
-      "print \"\\nTotal Sales : %.2f\"%(sale)\n",
-      "print \"\\nBasic Salary : %.2f\"%(bs)\n",
-      "print \"\\nHra          : %.2f\"%(hra)\n",
-      "print \"\\nDa           : %.2f\"%(da)\n",
-      "print \"\\nConveyance   : %.2f\"%(cv)\n",
-      "print \"\\nIncentive    : %.2f\"%(incentive)\n",
-      "print \"\\nBonus        : %.2f\"%(bonus)\n",
-      "print \"\\nGross Salary : %.2f\"%(ts)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Total Sales in Rs. : 100000\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Total Sales : 100000.00\n",
-        "\n",
-        "Basic Salary : 3000.00\n",
-        "\n",
-        "Hra          : 600.00\n",
-        "\n",
-        "Da           : 3300.00\n",
-        "\n",
-        "Conveyance   : 500.00\n",
-        "\n",
-        "Incentive    : 10000.00\n",
-        "\n",
-        "Bonus        : 500.00\n",
-        "\n",
-        "Gross Salary : 17900.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.9, Page number: 72<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate energy bill using the starting and ending meter reading.\n",
-      "#The charges are as follows:\n",
-      "\n",
-      "# No. of Units consumed         Rates in (Rs.)\n",
-      "# 200 - 500                     3.50\n",
-      "# 100 - 200                     2.50\n",
-      "# Less than 100                 1.50\n",
-      "\n",
-      "#Variable initialization\n",
-      "initial = raw_input(\"Initial & Final Readings : \")\n",
-      "final = raw_input(\"Initial & Final Readings : \")\n",
-      "\n",
-      "consumed = int(final) - int(initial)\n",
-      "\n",
-      "#Condition evaluation\n",
-      "if consumed >= 200 and consumed <= 500 :\n",
-      "    total = consumed * 3.50\n",
-      "else:\n",
-      "    if consumed >= 100 and consumed <= 199:\n",
-      "          total = consumed * 2.50\n",
-      "    else :\n",
-      "        if consumed < 100 :\n",
-      "           total = consumed * 1.50\n",
-      "\n",
-      "#Result\n",
-      "print \"Total bill for %d unit is %f\"%(consumed,total)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Initial & Final Readings : 1200\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Initial & Final Readings : 1500\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total bill for 300 unit is 1050.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.10, Page number: 73<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate energy bill by  reading the starting and ending meter reading.\n",
-      "#If the consumed electricity energy is greater than or equal to 200 units\n",
-      "#the rate should be 2.50/unit otherwise 1.50/unit.\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "initial = raw_input(\"Initial & Final Readings : \")\n",
-      "final = raw_input(\"Initial & Final Readings : \")\n",
-      "\n",
-      "consumed = int(final) - int(initial)\n",
-      "\n",
-      "#Condition evaluation\n",
-      "if consumed >= 200 :\n",
-      "    total = consumed * 2.50\n",
-      "else:\n",
-      "    total = consumed * 1.50\n",
-      "\n",
-      "#Result\n",
-      "print \"Total bill for %d unit is %f\"%(consumed,total)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total bill for 100 unit is 150.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.11, Page number: 73<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sort six numbers and find the largest one by using ladder of if else\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = int(raw_input(\"Enter 1st number : \"))\n",
-      "b = int(raw_input(\"Enter 2nd number : \"))\n",
-      "c = int(raw_input(\"Enter 3rd number : \"))\n",
-      "d = int(raw_input(\"Enter 4th number : \"))\n",
-      "e = int(raw_input(\"Enter 5th number : \"))\n",
-      "f = int(raw_input(\"Enter 6th number : \"))\n",
-      "\n",
-      "#Condition evaluation and Result\n",
-      "if a > b and a > c and a > d and a > e and a > f :\n",
-      "    print \"Highest of six Number is : %d\"%(a)\n",
-      "else:\n",
-      "    if b > a and b > c and b > d and b > e and b > f :\n",
-      "        print \"Highest of six Number is : %d\"%(b)\n",
-      "    else:\n",
-      "        if c > a and c > b and c > d and c > e and c > f :\n",
-      "            print \"Highest of six Number is : %d\"%(c)\n",
-      "        else:\n",
-      "            if d > a and d > b and d > c and d > e and d > f :\n",
-      "                print \"Highest of six Number is : %d\"%(d)\n",
-      "            else:\n",
-      "                if e > a and e > b and e > c and e > d and e > f :\n",
-      "                   print \"Highest of six Number is : %d\"%(e)\n",
-      "                else:\n",
-      "                    print \"Highest of six Number is : %d\"%(f)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 1st number : 52\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 2nd number : 74\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 3rd number : 90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 4th number : 45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 5th number : 10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter 6th number : 22\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Highest of six Number is : 90\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.12, Page number: 74<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find largest number out of three numbers.  Read the numbers through the keyboard\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "x = int(raw_input(\"\\nEnter Three Numbers x,y,z : \"))\n",
-      "y = int(raw_input(\"\\nEnter Three Numbers x,y,z : \"))\n",
-      "z = int(raw_input(\"\\nEnter Three Numbers x,y,z : \"))\n",
-      "\n",
-      "print \"\\nLargest out of Three Numbers is : \"\n",
-      "\n",
-      "#Condition evaluation and Result\n",
-      "if x > y :\n",
-      "    if x > z:\n",
-      "        print \"x = %d\\n\"%(x)\n",
-      "    else:\n",
-      "        print \"z = %d\\n\"%(z)\n",
-      "else:\n",
-      "    if z > y:\n",
-      "        print \"z =  %d\\n\"%(z)\n",
-      "    else:\n",
-      "        print \"y = %d\\n\"%(y)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Three Numbers x,y,z : 10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Three Numbers x,y,z : 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Three Numbers x,y,z : 30\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Largest out of Three Numbers is : \n",
-        "z =  30\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.13, Page number: 75<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the smallest out of the three numbers.\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = int(raw_input(\"\\nEnter Three Numbers : \"))\n",
-      "b = int(raw_input(\"\\nEnter Three Numbers : \"))\n",
-      "c = int(raw_input(\"\\nEnter Three Numbers : \"))\n",
-      "          \n",
-      "#Condition evaluation and Result\n",
-      "if a < b :\n",
-      "    if a < c:\n",
-      "        smallest = a\n",
-      "    else:\n",
-      "        smallest = c\n",
-      "else:\n",
-      "    if b < c:\n",
-      "        smallest = b\n",
-      "    else:\n",
-      "        smallest = c\n",
-      "\n",
-      "#Result\n",
-      "print \"The smallest of %d %d %d is %d\\n\"%(a,b,c,smallest)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Three Numbers : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Three Numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Three Numbers : 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The smallest of 1 5 8 is 1\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.14, Page number: 76<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate gross salary for the conditions given below\n",
-      "\n",
-      "#  BASIC SALARY(Rs.)        DA(Rs.)         HRA(Rs.)        CONVEYANCE(Rs.)\n",
-      "#   >= 5000               110% of basic   20% of basic          500\n",
-      "#   bs>=3000 & bs<5000    100% of basic   15% of basic          400\n",
-      "#   bs<3000               90% of basic    10% of basic          300\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "bs = int(raw_input(\"\\nEnter Basic Salary : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "if bs >= 5000 :\n",
-      "    hra = 20 * bs/100\n",
-      "    da = 110 * bs/100\n",
-      "    cv = 500\n",
-      "else:\n",
-      "    if bs >= 3000 and bs < 5000 :\n",
-      "        hra = 15 * bs/100\n",
-      "        da = 100 * bs/100\n",
-      "        cv = 400\n",
-      "    else:\n",
-      "        if bs < 3000:\n",
-      "            hra = 10 * bs/100\n",
-      "            da = 90 * bs/100\n",
-      "            cv = 300\n",
-      "\n",
-      "#Calculation\n",
-      "ts = bs + hra + da + cv\n",
-      "\n",
-      "#Result\n",
-      "print \"Basic Salary : %5.0f\"%(bs)\n",
-      "print \"Hra          : %5.0f\"%(hra)\n",
-      "print \"Da           : %5.0f\"%(da)\n",
-      "print \"Conveyance   : %5.0f\"%(cv)\n",
-      "print \"Gross Salary : %5.0f\"%(ts)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Basic Salary : 5400\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Basic Salary :  5400\n",
-        "Hra          :  1080\n",
-        "Da           :  5940\n",
-        "Conveyance   :   500\n",
-        "Gross Salary : 12920\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.15, Page number: 77<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate the total interest based on the following.\n",
-      "\n",
-      "#Principle Amount(Rs.)          Rate of Interest(Rs.)\n",
-      "#   >= 10000                        20%\n",
-      "#   >= 8000 & <= 9999               18%\n",
-      "#   < 8000                          16%\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "princ = int(raw_input(\"\\nEnter Loan & No. of years :- \"))\n",
-      "nyrs = int(raw_input(\"\\nEnter Loan & No. of years :- \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "if princ >= 10000 :\n",
-      "    rate = 20\n",
-      "else:\n",
-      "    if princ >= 8000 and princ <= 9999 :\n",
-      "        rate = 18\n",
-      "    else:\n",
-      "        if princ < 8000:\n",
-      "            rate = 16\n",
-      "\n",
-      "#Calculation\n",
-      "interest = princ * nyrs * rate/100\n",
-      "\n",
-      "#Result\n",
-      "print \"Loan     : %6.2f\"%(princ)\n",
-      "print \"Years    : %6.2f\"%(nyrs)\n",
-      "print \"Rate     : %6.2f\"%(rate)\n",
-      "print \"Interest : %6.2f\"%(interest)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Loan & No. of years :- 5500\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Loan & No. of years :- 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loan     : 5500.00\n",
-        "Years    :   3.00\n",
-        "Rate     :  16.00\n",
-        "Interest : 2640.00\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.16, Page number: 78<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the average of six subjects and display the results as follows.\n",
-      "\n",
-      "#    AVERAGE                         RESULT\n",
-      "#   >34 & <50                   Third Division\n",
-      "#   >49 & <60                   Second Division\n",
-      "#   >60 & <75                   First Division\n",
-      "#   >75 & <100                  Distinction\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "a = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
-      "b = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
-      "c = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
-      "d = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
-      "e = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
-      "f = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
-      "\n",
-      "#Calculation\n",
-      "#here sum1 is used instead of sum because sum is a function in python.\n",
-      "sum1 = a + b + c + d + e + f\n",
-      "avg = sum1/6\n",
-      "\n",
-      "print \"Total  : %d \\nAverage  : %.2f\"%(sum1,avg)\n",
-      "\n",
-      "#Condition evaluation and Result\n",
-      "if a < 35 or b < 35 or c < 35 or d < 35 or e < 35 :\n",
-      "    print \"Result : Fail\"\n",
-      "    exit\n",
-      "if avg >= 34 and avg < 50 :\n",
-      "    print \"Result  : Third Division \"\n",
-      "else:\n",
-      "    if avg >= 49 and avg < 60:\n",
-      "        print \"Result  :  Second Division\"\n",
-      "    else:\n",
-      "        if avg >= 60 and avg > 75:\n",
-      "            print \"Result  :  First Division\"\n",
-      "        else:\n",
-      "            if avg >= 75 and avg <= 100:\n",
-      "                  print \"Result  :  Distinction\"\n",
-      "                  \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Marks\n",
-        "P  C  B  M  E  H\n",
-        "75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Marks\n",
-        "P  C  B  M  E  H\n",
-        "75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Marks\n",
-        "P  C  B  M  E  H\n",
-        "75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Marks\n",
-        "P  C  B  M  E  H\n",
-        "75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Marks\n",
-        "P  C  B  M  E  H\n",
-        "75\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Marks\n",
-        "P  C  B  M  E  H\n",
-        "75\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total  : 450 \n",
-        "Average  : 75.00\n",
-        "Result  :  Distinction\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.17, Page number: 80<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect the entered number as to whether it is even or odd.  Use goto statement.\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "x = int(raw_input(\"\\nEnter a Number : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#There is no goto/label statement in python\n",
-      "\n",
-      "if x%2 == 0:\n",
-      "    print \"\\n%d is Even Number.\"%(x)\n",
-      "else:\n",
-      "    print \"\\n%d is Odd Number.\"%(x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "5 is Odd Number.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.18, Page number: 81<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate telephone bill.  Transfer controls at different places according\n",
-      "#to number of calls and calculate the total charges.  Follow rates as per\n",
-      "#given in the table.\n",
-      "\n",
-      "#  Telephone call               Rate in Rs.\n",
-      "#       < 100                    No charges\n",
-      "#       > 99 & < 200                1\n",
-      "#       > 199 & < 300               2\n",
-      "#       > 299                       3\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "nc = int(raw_input(\"\\nEnter Number of Calls : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#There is no goto/label statement in python\n",
-      "\n",
-      "if nc < 100:\n",
-      "    print \"\\nNo charges\"\n",
-      "else:\n",
-      "    if nc > 99 and nc < 200:\n",
-      "        print \"\\nTotal Charges : %d Rs.\"%(nc*1)\n",
-      "    else:\n",
-      "        if nc > 199 and nc < 300:\n",
-      "            print \"\\nTotal Charges : %d Rs.\"%(nc*2)\n",
-      "        else:\n",
-      "            print \"\\nTotal Charges : %d Rs.\"%(nc*3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Number of Calls : 500\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Total Charges : 1500 Rs.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.19, Page number: 82<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Check whether the entered year is a leap year or not. Use goto statement.\n",
-      "\n",
-      "#Variable initialization\n",
-      "leap = int(raw_input(\"\\nEnter Year : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#There is no goto/label statement in python\n",
-      "\n",
-      "if leap%4 == 0:\n",
-      "    print \"\\n%d is a leap year.\"%(leap)\n",
-      "else:\n",
-      "    print \"%d is not a leap year.\"%(leap)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Year : 2000\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "2000 is a leap year.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.20, Page number: 84<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print lines by selecting the choice.\n",
-      "\n",
-      "print \"\\n[1] ............\"\n",
-      "print \"\\n[2] ____________\"\n",
-      "print \"\\n[3] ************\"\n",
-      "print \"\\n[4] ============\"\n",
-      "print \"\\n[5] EXIT\"\n",
-      "\n",
-      "#Variable initialization\n",
-      "ch = int(raw_input(\"\\nENTER YOUR CHOICE : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#dictionary and functions\n",
-      "\n",
-      "def dot():\n",
-      "    print \"\\n......................................................\"\n",
-      "\n",
-      "def line():\n",
-      "    print \"\\n______________________________________________________\"\n",
-      "\n",
-      "def star():\n",
-      "    print \"\\n******************************************************\"\n",
-      "\n",
-      "def equal():\n",
-      "    print \"\\n======================================================\"\n",
-      "\n",
-      "def ex():\n",
-      "    exit\n",
-      "    \n",
-      "options = {1 : dot,\n",
-      "           2 : line,\n",
-      "           3 : star,\n",
-      "           4 : equal,\n",
-      "           5 : ex\n",
-      "}\n",
-      "\n",
-      "options[ch]()\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "[1] ............\n",
-        "\n",
-        "[2] ____________\n",
-        "\n",
-        "[3] ************\n",
-        "\n",
-        "[4] ============\n",
-        "\n",
-        "[5] EXIT\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "ENTER YOUR CHOICE : 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "......................................................\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.21, Page number: 85<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Provide multiple functions such as 1. addition 2.Subtraction 3. Multiplication\n",
-      "#4.Division 5.Remainder calculation 6.Larger out of two by using switch statement\n",
-      "\n",
-      "print \"\\n=============================\"\n",
-      "print \"\\n\\t\\tMENU\"\n",
-      "print \"\\n=============================\"\n",
-      "print \"\\n\\t[1] ADDITION\"\n",
-      "print \"\\n\\t[2] SUBTRACTION\"\n",
-      "print \"\\n\\t[3] MULTIPLICATION\"\n",
-      "print \"\\n\\t[4] DIVISION\"\n",
-      "print \"\\n\\t[5] REMAINDER\"\n",
-      "print \"\\n\\t[6] LARGER OUT OF TWO\"\n",
-      "print \"\\n\\t[0] EXIT\"\n",
-      "print \"\\n=============================\"\n",
-      "\n",
-      "#Variable initialization\n",
-      "ch = int(raw_input(\"\\nENTER YOUR CHOICE : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#dictionary and functions\n",
-      "\n",
-      "if ch <= 6 and ch > 0:\n",
-      "    a = int(raw_input(\"Enter Two Numbers : \"))\n",
-      "    b = int(raw_input(\"Enter Two Numbers : \"))\n",
-      "    \n",
-      "def add():\n",
-      "    c = a + b\n",
-      "    print \"\\nAddition  : %d\"%(c)\n",
-      "\n",
-      "def sub():\n",
-      "    c = a - b\n",
-      "    print \"\\nSubtraction  : %d\"%(c)\n",
-      "\n",
-      "def mul():\n",
-      "    c = a * b\n",
-      "    print \"\\nMultiplication  : %d\"%(c)\n",
-      "\n",
-      "def div():\n",
-      "    c = a/b\n",
-      "    print \"\\nDivision  : %d\"%(c)\n",
-      "\n",
-      "def rem():\n",
-      "    c = a % b\n",
-      "    print \"\\nRemainder  : %d\"%(c)\n",
-      "\n",
-      "def larger():\n",
-      "    if a > b:\n",
-      "        print \"\\n%d (a) is larger than %d (b)\"%(a,b)\n",
-      "    else:\n",
-      "        if b > a:\n",
-      "            print \"\\n%d (b) is larger than %d (a)\"%(b,a)\n",
-      "        else:\n",
-      "            print \"\\n%d (a) & %d (b) are same\"%(a,b)\n",
-      "\n",
-      "def ex():\n",
-      "    print \"\\nTerminated by choice\"\n",
-      "    exit\n",
-      "\n",
-      "def default():\n",
-      "    print \"\\nInvalid Choice\"\n",
-      "\n",
-      "options = { 1 : add,\n",
-      "            2 : sub,\n",
-      "            3 : mul,\n",
-      "            4 : div,\n",
-      "            5 : rem,\n",
-      "            6 : larger,\n",
-      "            7 : ex,\n",
-      "            0 : default,\n",
-      "            8 : default,\n",
-      "            9 : default\n",
-      "}\n",
-      "\n",
-      "options[ch]()\n",
-      "       \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "=============================\n",
-        "\n",
-        "\t\tMENU\n",
-        "\n",
-        "=============================\n",
-        "\n",
-        "\t[1] ADDITION\n",
-        "\n",
-        "\t[2] SUBTRACTION\n",
-        "\n",
-        "\t[3] MULTIPLICATION\n",
-        "\n",
-        "\t[4] DIVISION\n",
-        "\n",
-        "\t[5] REMAINDER\n",
-        "\n",
-        "\t[6] LARGER OUT OF TWO\n",
-        "\n",
-        "\t[0] EXIT\n",
-        "\n",
-        "=============================\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "ENTER YOUR CHOICE : 6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Two Numbers : 9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Two Numbers : 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "9 (a) is larger than 8 (b)\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.22, Page number: 87<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert years into Minutes, Hours, Days, Months and Seconds using switch()\n",
-      "#statement\n",
-      "\n",
-      "\n",
-      "print \"[1] MINUTES\"\n",
-      "print \"[2] HOURS\"\n",
-      "print \"[3] DAYS\"\n",
-      "print \"[4] MONTHS\"\n",
-      "print \"[5] SECONDS\"\n",
-      "print \"[0] EXIT\"\n",
-      "\n",
-      "#Variable initialization\n",
-      "ch = int(raw_input(\"\\nEnter Your Choice : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#dictionary and functions\n",
-      "\n",
-      "if ch > 0 and ch < 6:\n",
-      "    yrs = int(raw_input(\"Enter Years : \"))\n",
-      "\n",
-      "#Calculation\n",
-      "#since min is a keyword in python, min1 is used instead.\n",
-      "mon = yrs * 12\n",
-      "ds = mon * 30\n",
-      "hrs = ds * 24\n",
-      "min1 = hrs * 60\n",
-      "se = min1 * 60\n",
-      "\n",
-      "def minute():\n",
-      "    print \"\\nMinutes  : %ld\"%(min1)\n",
-      "\n",
-      "def hours():\n",
-      "    print \"\\nHours  : %ld\"%(hrs)\n",
-      "\n",
-      "def days():\n",
-      "    print \"\\nDays  : %ld\"%(ds)\n",
-      "\n",
-      "def months():\n",
-      "    print \"\\nMonths : %ld\"%(mon)\n",
-      "\n",
-      "def seconds():\n",
-      "    print \"\\nSeconds : %ld\"%(se)\n",
-      "\n",
-      "def ex():\n",
-      "    print \"\\nTerminated by choice\"\n",
-      "    exit\n",
-      "\n",
-      "def default():\n",
-      "    print \"\\nInvalid Choice\"\n",
-      "\n",
-      "options = { 1 : minute,\n",
-      "            2 : hours,\n",
-      "            3 : days,\n",
-      "            4 : months,\n",
-      "            5 : seconds,\n",
-      "            6 : ex,\n",
-      "            0 : default,\n",
-      "            8 : default,\n",
-      "            9 : default\n",
-      "}\n",
-      "\n",
-      "options[ch]()\n",
-      "       \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "[1] MINUTES\n",
-        "[2] HOURS\n",
-        "[3] DAYS\n",
-        "[4] MONTHS\n",
-        "[5] SECONDS\n",
-        "[0] EXIT\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Your Choice : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Years : 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Months : 24\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.23, Page number: 89<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the names of the days of a week.\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "day = int(raw_input(\"\\nEnter a number between 1 to 7 : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#dictionary and functions\n",
-      "\n",
-      "def first():\n",
-      "    print \"\\n1st day of week is Sunday\"\n",
-      "\n",
-      "def second():\n",
-      "    print \"\\n2nd day of week is Monday\"\n",
-      "\n",
-      "def third():\n",
-      "    print \"\\n3rd day of week is Tuesday\"\n",
-      "\n",
-      "def fourth():\n",
-      "    print \"\\n4th day of week is Wednesday\"\n",
-      "\n",
-      "def fifth():\n",
-      "    print \"\\n5th day of week is Thursday\"\n",
-      "\n",
-      "def sixth():\n",
-      "    print \"\\n6th day of week is Friday\"\n",
-      "\n",
-      "def seventh():\n",
-      "    print \"\\n7th day of week is Saturday\"\n",
-      "\n",
-      "def default():\n",
-      "    print \"\\nInvalid day\"\n",
-      "\n",
-      "options = { 1 : first,\n",
-      "            2 : second,\n",
-      "            3 : third,\n",
-      "            4 : fourth,\n",
-      "            5 : fifth,\n",
-      "            6 : sixth,\n",
-      "            7 : seventh,\n",
-      "            0 : default,\n",
-      "            8 : default,\n",
-      "            9 : default\n",
-      "}\n",
-      "for i in range(1,day+1):\n",
-      "    options[i]()\n",
-      "       \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter a number between 1 to 7 : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1st day of week is Sunday\n",
-        "\n",
-        "2nd day of week is Monday\n",
-        "\n",
-        "3rd day of week is Tuesday\n",
-        "\n",
-        "4th day of week is Wednesday\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.24, Page number: 90<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Perform following operations\n",
-      "\n",
-      "#1. Display any numbers or stars on the screen by using for loop\n",
-      "#2. Display the menu containing the following\n",
-      "#a) Whole screen b)Half screen c)Top 3 lines d)Bottom 3 lines\n",
-      "\n",
-      "import os\n",
-      "import sys\n",
-      "import turtle\n",
-      "\n",
-      "\n",
-      "#Print numbers in the whole screen\n",
-      "for i in range(1,700):\n",
-      "    sys.stdout.write(\"%d\"%i)\n",
-      "\n",
-      "print \"\\nCLEAR MENU\"\n",
-      "print \"\\t1] Whole Screen\\t2] Half Screen\\t3]Top 3 lines\\t4] Bottom 3 lines\\t5] Exit\"\n",
-      "\n",
-      "#Variable initialization\n",
-      "c = int(raw_input(\"Enter Your Choice : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#dictionary and functions\n",
-      "\n",
-      "def one():\n",
-      "    os.system('cls')\n",
-      "\n",
-      "def two():\n",
-      "    for i in range(0,190):\n",
-      "        turtle.goto(i,1)\n",
-      "        sys.stdout.write(\"\\t\")\n",
-      "    \n",
-      "def three():\n",
-      "    for i in range(1,100):\n",
-      "        turtle.goto(i,1)\n",
-      "        sys.stdout.write(\"\\t\")\n",
-      "\n",
-      "def four():\n",
-      "    for i in range(1,120):\n",
-      "        turtle.goto(i,21)\n",
-      "        sys.stdout.write(\"\\t\")\n",
-      "\n",
-      "def five():\n",
-      "    exit\n",
-      "\n",
-      "\n",
-      "options = { 1 : one,\n",
-      "            2 : two,\n",
-      "            3 : three,\n",
-      "            4 : four,\n",
-      "            5 : five\n",
-      "}\n",
-      "options[c]()\n",
-      "       \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599600601602603604605606607608609610611612613614615616617618619620621622623624625626627628629630631632633634635636637638639640641642643644645646647648649650651652653654655656657658659660661662663664665666667668669670671672673674675676677678679680681682683684685686687688689690691692693694695696697698699\n",
-        "CLEAR MENU\n",
-        "\t1] Whole Screen\t2] Half Screen\t3]Top 3 lines\t4] Bottom 3 lines\t5] Exit\n",
-        "\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t\t"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t\t\t\t"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.25, Page number: 91<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the files of current directory\n",
-      "\n",
-      "import os\n",
-      "\n",
-      "print \"\\nFILE LISTING MENU\"\n",
-      "print \"1] .EXE\"\n",
-      "print \"2] .BAT\"\n",
-      "print \"3] .OBJ\"\n",
-      "print \"4] .BAK\"\n",
-      "#Variable initialization\n",
-      "c = int(raw_input(\"Enter Your Choice -: \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#dictionary and functions\n",
-      "\n",
-      "def one():\n",
-      "    os.system('dir *.exe')\n",
-      "\n",
-      "def two():\n",
-      "    os.system('dir *.c')\n",
-      "\n",
-      "def three():\n",
-      "    os.system('dir *.obj')\n",
-      "\n",
-      "def four():\n",
-      "    os.system('dir *.bak')\n",
-      "\n",
-      "options = { 1 : one,\n",
-      "            2 : two,\n",
-      "            3 : three,\n",
-      "            4 : four\n",
-      "}\n",
-      "options[c]()\n",
-      "\n",
-      "       \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "FILE LISTING MENU\n",
-        "1] .EXE\n",
-        "2] .BAT\n",
-        "3] .OBJ\n",
-        "4] .BAK\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.26, Page number: 92<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display number of days in calendar format of an entered month of year 2001.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "m = int(raw_input(\"Enter Month No. of Year 2001 : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#else-if ladder.\n",
-      "\n",
-      "print \"\\nMonth - %d - 2001\"%(m)\n",
-      "print \"\\n\\tSUN\\tMON\\tTUE\\tWED\\tTHU\\tFRI\\tSAT\"\n",
-      "\n",
-      "#Find a day any month of 2001\n",
-      "\n",
-      "if m == 1:\n",
-      "    a = 2\n",
-      "    j = 31\n",
-      "else:\n",
-      "    if m == 2:\n",
-      "        a = 5\n",
-      "        j = 28\n",
-      "    else:\n",
-      "        if m == 3:\n",
-      "            a = 5\n",
-      "            j = 31\n",
-      "        else:\n",
-      "            if m == 4:\n",
-      "                a = 1\n",
-      "                j = 30\n",
-      "            else:\n",
-      "                if m == 5:\n",
-      "                    a = 3\n",
-      "                    j = 31\n",
-      "                else:\n",
-      "                    if m == 6:\n",
-      "                        a = 6\n",
-      "                        j = 30\n",
-      "                    else:\n",
-      "                        if m == 7:\n",
-      "                            a = 1\n",
-      "                            j = 31\n",
-      "                        else:\n",
-      "                            if m == 8:\n",
-      "                                a = 4\n",
-      "                                j = 31\n",
-      "                            else:\n",
-      "                                if m == 9:\n",
-      "                                    a = 7\n",
-      "                                    j = 30\n",
-      "                                else:\n",
-      "                                    if m == 10:\n",
-      "                                        a = 2\n",
-      "                                        j = 31\n",
-      "                                    else:\n",
-      "                                        if m == 11:\n",
-      "                                            a = 5\n",
-      "                                            j = 30\n",
-      "                                        else:\n",
-      "                                            if m == 12:\n",
-      "                                                a = 7\n",
-      "                                                j = 31\n",
-      "                                            else:\n",
-      "                                                print \"Invalid Month\"\n",
-      "                                                exit\n",
-      "\n",
-      "#Starting day is to be adjusted under the respective day\n",
-      "#sys.stdout.write() function performs the same task as printf() function\n",
-      "                                                \n",
-      "i = 1\n",
-      "if a == 1:\n",
-      "    sys.stdout.write(\"\\t%d\"%(i))\n",
-      "else:\n",
-      "    if a == 2:\n",
-      "        sys.stdout.write(\"\\t\\t%d\"%(i))\n",
-      "    else:\n",
-      "        if a == 3:\n",
-      "            sys.stdout.write(\"\\t\\t\\t%d\"%(i))\n",
-      "        else:\n",
-      "            if a == 4:\n",
-      "                sys.stdout.write(\"\\t\\t\\t\\t%d\"%(i))\n",
-      "            else:\n",
-      "                if a == 5:\n",
-      "                    sys.stdout.write(\"\\t\\t\\t\\t\\t%d\"%(i))\n",
-      "                else:\n",
-      "                    if a == 6:\n",
-      "                        sys.stdout.write(\"\\t\\t\\t\\t\\t\\t%d\"%(i))\n",
-      "                    else:\n",
-      "                        if a == 7:\n",
-      "                            sys.stdout.write(\"\\t\\t\\t\\t\\t\\t\\t%d\"%(i))\n",
-      "\n",
-      "\n",
-      "h = 8 - a        #The starting day is subtracted from 8\n",
-      "for i in range(2,h+1):       #to display the first row\n",
-      "    sys.stdout.write(\"\\t%d\"%(i))\n",
-      "sys.stdout.write(\"\\n\")\n",
-      "b = 1\n",
-      "for i in range(h+1,j+1):          #to continue with second row onwards\n",
-      "    if b == 8:\n",
-      "        sys.stdout.write(\"\\n\")\n",
-      "        b = 1\n",
-      "    sys.stdout.write(\"\\t%d\"%(i))\n",
-      "    b += 1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Month - 1 - 2001\n",
-        "\n",
-        "\tSUN\tMON\tTUE\tWED\tTHU\tFRI\tSAT\n",
-        "\t\t1\t2\t3\t4\t5\t6\n",
-        "\t7\t8\t9\t10\t11\t12\t13\n",
-        "\t14\t15\t16\t17\t18\t19\t20\n",
-        "\t21\t22\t23\t24\t25\t26\t27\n",
-        "\t28\t29\t30\t31"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.27, Page number: 95<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert decimal to hexadecimal number.\n",
-      "\n",
-      "import sys\n",
-      "import turtle\n",
-      "\n",
-      "#Variable initialization\n",
-      "x = int(raw_input(\"Enter a number : \"))\n",
-      "y = 30\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#else-if ladder.\n",
-      "\n",
-      "print \"\\nConversion of Decimal to Hexadecimal Number\"\n",
-      "\n",
-      "#for loop without condition is not supported in python. so while loop is used.\n",
-      "\n",
-      "c = 1\n",
-      "z = [0 for i in range(0,15)]\n",
-      "\n",
-      "while x != 0:\n",
-      "    z[c] = x % 16\n",
-      "    c = c + 1\n",
-      "    x = x/16\n",
-      "\n",
-      "\n",
-      "for i in range(c-1,0,-1):\n",
-      "    if z[i] == 10:\n",
-      "       sys.stdout.write(\"A\")\n",
-      "    else:\n",
-      "       if z[i] == 11:\n",
-      "          sys.stdout.write(\"B\")\n",
-      "       else:\n",
-      "          if z[i] == 12:\n",
-      "             sys.stdout.write(\"C\")\n",
-      "          else:\n",
-      "             if z[i] == 13:\n",
-      "                sys.stdout.write(\"D\")\n",
-      "             else:\n",
-      "                if z[i] == 14:\n",
-      "                   sys.stdout.write(\"E\")\n",
-      "                else:\n",
-      "                    if z[i] == 15:\n",
-      "                       sys.stdout.write(\"F\")\n",
-      "                    else:\n",
-      "                       sys.stdout.write(\"%d\"%(z[i]))\n",
-      "    \n",
-      "    \n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 31\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Conversion of Decimal to Hexadecimal Number\n",
-        "1F"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.28, Page number: 97<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect whether the entered number is even or odd. use nested switch-case\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "x = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch case statement in python, an alternative is to use\n",
-      "#else-if ladder.\n",
-      "\n",
-      "if x == 0:\n",
-      "    print \"Number is Even\"\n",
-      "else:\n",
-      "    if x == 1:\n",
-      "        print \"Number is odd\"\n",
-      "    else:\n",
-      "        y = x % 2\n",
-      "        if y == 0:\n",
-      "            print \"Number is Even\"\n",
-      "        else:\n",
-      "            print \"Number is odd\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number is odd\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.29, Page number: 98<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count number of 1s, 0s, blank spaces and others using nested\n",
-      "#switch() statement in a given stream\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "txt = raw_input(\"Enter Numbers : \")\n",
-      "\n",
-      "#Processing\n",
-      "x = 0\n",
-      "s = 0\n",
-      "a = 0\n",
-      "z = 0\n",
-      "o = 0\n",
-      "\n",
-      "while x < len(txt):\n",
-      "    if txt[x] == ' ':\n",
-      "        s = s + 1\n",
-      "    else:\n",
-      "        if txt[x] == '1':\n",
-      "            a = a + 1\n",
-      "        else:\n",
-      "            if txt[x] == '0':\n",
-      "                z = z + 1\n",
-      "            else:\n",
-      "                o = o + 1\n",
-      "    x = x + 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nTotal Spaces : %d\"%(s))\n",
-      "sys.stdout.write(\"\\nTotal 1s  :  %d\"%(a))\n",
-      "sys.stdout.write(\"\\nTotal 0s  :  %d\"%(z))\n",
-      "sys.stdout.write(\"\\nOthers    :  %d\"%(o))\n",
-      "sys.stdout.write(\"\\nString Length  : %d\"%(s+a+z+o))\n",
-      "        \n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Numbers : 1110022 222\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Total Spaces : 1\n",
-        "Total 1s  :  3\n",
-        "Total 0s  :  2\n",
-        "Others    :  5\n",
-        "String Length  : 11"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.30, Page number: 99<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert integer to character using if condition\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "x = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#ord() function converts the character ASCII to integer\n",
-      "\n",
-      "if x == ord('A'):\n",
-      "    print \"%c\"%(x)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 65\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 5.31, Page number: 100<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use nested if..else statements in switch() statement.  Also show the effect\n",
-      "#of conversion of integer to character\n",
-      "\n",
-      "\n",
-      "#Variable initialization\n",
-      "i = int(raw_input(\"Enter any ASCII Number : \"))\n",
-      "\n",
-      "#Condition evaluation\n",
-      "#there is no switch..case statement in python. alternative is to use\n",
-      "#if..else and else..if ladder\n",
-      "\n",
-      "if i == ord('A'):\n",
-      "    print \"Capital A\"\n",
-      "else:\n",
-      "    if i == ord('B'):\n",
-      "        print \"Capital B\"\n",
-      "    else:\n",
-      "        if i == ord('C'):\n",
-      "            print \"Capital C\"\n",
-      "        else:\n",
-      "            if i > 47 and i < 58:\n",
-      "                print \"Digit : [%c]\"%(i)\n",
-      "            else:\n",
-      "                if i >= 58 and i <= 64:\n",
-      "                    print \"Symbol : [%c]\"%(i)\n",
-      "                else:\n",
-      "                    if i > 64 and i < 91:\n",
-      "                        print \"Capital : [%c]\"%(i)\n",
-      "                    else:\n",
-      "                        if i > 96 and i < 123:\n",
-      "                            print \"Small : [%c]\"%(i)\n",
-      "                        else:\n",
-      "                            print \"Invalid Choice\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter any ASCII Number : 65\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capital A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter6.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter6.ipynb
deleted file mode 100755
index d9815e79..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter6.ipynb
+++ /dev/null
@@ -1,42312 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 6: Loop Control Statements<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.1, Page number: 107<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the first five numbers starting from one together with thier squares\n",
-      "\n",
-      "#Result\n",
-      "#Since range() function starts with 0; to print upto 5, use 5+1 or 6\n",
-      "#range() function creates range or numbers from 1 to 6\n",
-      "\n",
-      "for i in range(1,5+1):\n",
-      "    print(\"Number : %d   it's Square : %d\"%(i,i*i))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number : 1   it's Square : 1\n",
-        "Number : 2   it's Square : 4\n",
-        "Number : 3   it's Square : 9\n",
-        "Number : 4   it's Square : 16\n",
-        "Number : 5   it's Square : 25\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.2, Page number: 108<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display numbers from 1 to 15 using for loop.  Use i++\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#There is no increment operator (++) in python\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "for i in range(1,15+1):\n",
-      "    sys.stdout.write(\"%5d\"%(i))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.3, Page number: 108<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display numbers from 1 to 15 using for loop. Use i = i + 1\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "for i in range(1,15+1):\n",
-      "    sys.stdout.write(\"%5d\"%(i))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.4, Page number: 109<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display numbers from 1 to 16. Use incrementation operation in the body of\n",
-      "#the loop for more than once\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 0\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers and the third argument in the range function\n",
-      "#indicates the interval between the numbers.\n",
-      "\n",
-      "for i in range(0,15+1,2):\n",
-      "    i += 1\n",
-      "    sys.stdout.write(\"%5d\"%(i))\n",
-      "    i = i + 1\n",
-      "    sys.stdout.write(\"%5d\"%(i))\n",
-      "    c += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\n\\nThe Body of the loop is executed for %d times\"%(c))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15   16\n",
-        "\n",
-        "The Body of the loop is executed for 8 times"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.5, Page number: 109<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display even numbers from 0 to 14 by declaring the initial counter value before\n",
-      "#the loop statement\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 0\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers and the third argument in the range function\n",
-      "#indicates the interval between the numbers.\n",
-      "\n",
-      "#counter variable initialization before the for loop will not affect much.\n",
-      "\n",
-      "for i in range(0,15+1,2):\n",
-      "    sys.stdout.write(\"%5d\"%(i))\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "    0    2    4    6    8   10   12   14"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.6, Page number: 110<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display numbers in ascending and descending orders.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers and the third argument in the range function\n",
-      "#indicates the interval between the numbers.\n",
-      "\n",
-      "#counter variable initialization before the for loop will not affect much.\n",
-      "\n",
-      "sys.stdout.write(\"Numbers in Ascending Order : \")\n",
-      "for i in range(1,10+1):\n",
-      "    sys.stdout.write(\"%3d\"%(i))\n",
-      "    \n",
-      "sys.stdout.write(\"\\nNumbers in Descending Order : \")\n",
-      "for i in range(10,0,-1):\n",
-      "    sys.stdout.write(\"%3d\"%(i))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numbers in Ascending Order :   1  2  3  4  5  6  7  8  9 10\n",
-        "Numbers in Descending Order :  10  9  8  7  6  5  4  3  2  1"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.7, Page number: 110<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display numbers upto 10 using infinite for loop.  Use goto statement to\n",
-      "#exit from the loop\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers and the third argument in the range function\n",
-      "#indicates the interval between the numbers.\n",
-      "\n",
-      "#counter variable initialization before the for loop will not affect much.\n",
-      "\n",
-      "for i in range(0,15):\n",
-      "    sys.stdout.write(\"%3d\"%(i))\n",
-      "    i += 1\n",
-      "    if i == 11:\n",
-      "        break\n",
-      "\n",
-      "#there is no infinite for loop and goto statement in python.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  0  1  2  3  4  5  6  7  8  9 10"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.8, Page number: 111<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count numbers between 1 to 100 not divisible by 2, 3, and 5\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 0\n",
-      "\n",
-      "sys.stdout.write(\"\\nNumbers from 1 to 100 not divisible by 2,3&5\\n\\n\")\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers \n",
-      "\n",
-      "for x in range(0,100+1):\n",
-      "    if x%2 != 0 and x%3 != 0 and x%5 != 0:\n",
-      "        sys.stdout.write(\"%d\\t\"%(x))\n",
-      "        c += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\nTotal Numbers : %d\"%(c))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Numbers from 1 to 100 not divisible by 2,3&5\n",
-        "\n",
-        "1\t7\t11\t13\t17\t19\t23\t29\t31\t37\t41\t43\t47\t49\t53\t59\t61\t67\t71\t73\t77\t79\t83\t89\t91\t97\t\n",
-        "Total Numbers : 26"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.9, Page number: 112<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the numbers in increasing and decreasing order using infinite\n",
-      "#for loop\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "n = int(raw_input(\"Enter a number : \"))\n",
-      "a = b = n\n",
-      "\n",
-      "sys.stdout.write(\"(++)        (--)\\n\")\n",
-      "sys.stdout.write(\"===================\")\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "#There is no infinite for loop in python. so separate iteration variable i is used\n",
-      "\n",
-      "for i in range(0,100):\n",
-      "    sys.stdout.write(\"\\n%d\\t%d\"%(a,b))\n",
-      "    if b == 0:\n",
-      "        break\n",
-      "    a += 1\n",
-      "    b -= 1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(++)        (--)\n",
-        "===================\n",
-        "5\t5\n",
-        "6\t4\n",
-        "7\t3\n",
-        "8\t2\n",
-        "9\t1\n",
-        "10\t0"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.10, Page number: 113<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Create an infinite loop. Check each value of the for loop.  if the value is\n",
-      "#even, display it otherwise continue with interations.  Print even numbers from\n",
-      "#1 to 21. Use break statement to terminate the program.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "i = 1\n",
-      "\n",
-      "sys.stdout.write(\"\\n\\t\\tTable of Even numbers from 1 to 20\")\n",
-      "sys.stdout.write(\"\\n\\t\\t==================================\\n\")\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "#There is no infinite for loop in python. so separate iteration variable i is used\n",
-      "\n",
-      "for i in range(1,100):\n",
-      "    if i == 20:\n",
-      "        break\n",
-      "    else:\n",
-      "        if i%2 == 0:\n",
-      "            sys.stdout.write(\"%d\\t\"%(i))\n",
-      "            continue\n",
-      "        else:\n",
-      "            continue"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\t\tTable of Even numbers from 1 to 20\n",
-        "\t\t==================================\n",
-        "2\t4\t6\t8\t10\t12\t14\t16\t18\t"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.11, Page number: 113<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate the sum of first five numbers and their squares.  Display their results\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "#since sum is a function in python, sum1 is used instead of variable sum.\n",
-      "\n",
-      "sum1 = 0\n",
-      "sqsum = 0\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "for i in range(1,5+1):\n",
-      "    sum1 += i\n",
-      "    sqsum += i*i\n",
-      "    sys.stdout.write(\"\\nNumber : %5d    it's Square : %8d\"%(i,i*i))\n",
-      "\n",
-      "sys.stdout.write(\"\\n=================================================\")\n",
-      "sys.stdout.write(\"\\nThe Sum %6d Sum of Squares %9d\"%(sum1,sqsum))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number :     1    it's Square :        1\n",
-        "Number :     2    it's Square :        4\n",
-        "Number :     3    it's Square :        9\n",
-        "Number :     4    it's Square :       16\n",
-        "Number :     5    it's Square :       25\n",
-        "=================================================\n",
-        "The Sum     15 Sum of Squares        55"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.12, Page number: 114<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display numbers from 1 to 9 and their square roots.\n",
-      "\n",
-      "import math\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "for i in range(1,9+1):\n",
-      "    a = math.sqrt(i)\n",
-      "    sys.stdout.write(\"\\n\\t%d     %.2f\"%(i,a))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\t1     1.00\n",
-        "\t2     1.41\n",
-        "\t3     1.73\n",
-        "\t4     2.00\n",
-        "\t5     2.24\n",
-        "\t6     2.45\n",
-        "\t7     2.65\n",
-        "\t8     2.83\n",
-        "\t9     3.00"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.13, Page number: 115<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the number in between 7 and 100 which is exactly divisible by\n",
-      "#4 and if divided by 5 and 6 remainders obtained should be 4.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "for x in range(7,100+1):\n",
-      "    if x%4 == 0 and x%5 == 4 and x%6 == 4:\n",
-      "        sys.stdout.write(\"\\nMinimum Number : %d\"%(x))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Minimum Number : 64"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.14, Page number: 115<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the series given\n",
-      "# x = 1/1 + 1/4 + 1/9 ... 1/n2\n",
-      "# y = 1/1 + 1/8 + 1/27 ... 1/n3\n",
-      "\n",
-      "#Variable declaration\n",
-      "x = 0.0\n",
-      "y = 0.0\n",
-      "n = int(raw_input(\"Enter Value of n : \"))\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "for i in range(1,n+1):\n",
-      "    x = x + (1.0/pow(i,2))\n",
-      "    y = y + (1.0/pow(i,3))\n",
-      "\n",
-      "print(\"Value of x = %f\"%(x))\n",
-      "print(\"\\nValue of y = %f\"%(y))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of n : 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of x = 1.250000\n",
-        "\n",
-        "Value of y = 1.125000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.15, Page number: 116<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Generate Triangular number\n",
-      "#Note: Triangular number is nothing but summation of 1 to given number.\n",
-      "#For example, when entered number is 5 it's triangular number would be\n",
-      "#(1+2+3+4+5) = 15.\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "tri_num = 0\n",
-      "n = int(raw_input(\"What Triangular number do you want : \"))\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "for j in range(1,n+1):\n",
-      "    tri_num = tri_num + j\n",
-      "\n",
-      "print(\"Triangular number of %d is %d\\n\"%(n,tri_num))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "What Triangular number do you want : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Triangular number of 5 is 15\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.16, Page number: 117<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find sum of the following series\n",
-      "# 1 + 2 + 3 + ... n\n",
-      "# 1^2 + 2^2 + 2^3 + ... n^2\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable declaration\n",
-      "#since sum is a function name in python, sum1 is used instead of sum.\n",
-      "\n",
-      "sum1 = 0\n",
-      "ssum = 0\n",
-      "j = int(raw_input(\"Enter Number : \"))\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "sys.stdout.write(\"Numbers : \")\n",
-      "for i in range(1,j+1):\n",
-      "    sys.stdout.write(\"%5d\"%(i))\n",
-      "\n",
-      "sys.stdout.write(\"\\nSquares : \")\n",
-      "for i in range(1,j+1):\n",
-      "    sys.stdout.write(\"%5d\"%(i*i))\n",
-      "    sum1 = sum1 + i\n",
-      "    ssum = ssum + i*i\n",
-      "sys.stdout.write(\"\\nSum of Numbers from 1 to %d   : %d\"%(j,sum1))\n",
-      "sys.stdout.write(\"\\nSum of Squares of 1 to %d Numbers : %d\"%(j,ssum))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numbers :     1    2    3    4    5\n",
-        "Squares :     1    4    9   16   25\n",
-        "Sum of Numbers from 1 to 5   : 15\n",
-        "Sum of Squares of 1 to 5 Numbers : 55"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.17, Page number: 118<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the perfect squares from 1 to 500.\n",
-      "\n",
-      "import sys\n",
-      "import math\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers\n",
-      "\n",
-      "sys.stdout.write(\"\\n\\n\")\n",
-      "sys.stdout.write(\"Perfect square from 1 to 500 \\n\")\n",
-      "count = 0\n",
-      "\n",
-      "for i in range(1,500+1):\n",
-      "    c = math.sqrt(i)\n",
-      "    x = math.floor(c)                  #For rounding up floor() is used\n",
-      "    if c == x:\n",
-      "        sys.stdout.write(\"\\t%5d\"%(i))\n",
-      "        count += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\n\\nTotal Perfect Squares = %d\"%(count))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "Perfect square from 1 to 500 \n",
-        "\t    1\t    4\t    9\t   16\t   25\t   36\t   49\t   64\t   81\t  100\t  121\t  144\t  169\t  196\t  225\t  256\t  289\t  324\t  361\t  400\t  441\t  484\n",
-        "\n",
-        "Total Perfect Squares = 22"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.18, Page number: 119<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect the largest number out of five numbers.\n",
-      "\n",
-      "import sys\n",
-      "import math\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers.\n",
-      "#since sum is a built in function in python, sum1 is used instead of sum.\n",
-      "\n",
-      "a = int(raw_input(\"Enter five numbers : \"))\n",
-      "b = int(raw_input(\"Enter five numbers : \"))\n",
-      "c = int(raw_input(\"Enter five numbers : \"))\n",
-      "d = int(raw_input(\"Enter five numbers : \"))\n",
-      "e = int(raw_input(\"Enter five numbers : \"))\n",
-      "\n",
-      "sum1 = a + b + c + d + e\n",
-      "\n",
-      "for i in range(sum1,1,-1):\n",
-      "    if i == a or i == b or i == c or i == d or i == e:\n",
-      "        sys.stdout.write(\"The Largest Number : %d\"%(i))\n",
-      "        break\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Largest Number : 7"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.19, Page number: 119<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect the smallesst number out of five numbers.\n",
-      "\n",
-      "import sys\n",
-      "import math\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers.\n",
-      "#since sum is a built in function in python, sum1 is used instead of sum.\n",
-      "\n",
-      "a = int(raw_input(\"Enter five numbers : \"))\n",
-      "b = int(raw_input(\"Enter five numbers : \"))\n",
-      "c = int(raw_input(\"Enter five numbers : \"))\n",
-      "d = int(raw_input(\"Enter five numbers : \"))\n",
-      "e = int(raw_input(\"Enter five numbers : \"))\n",
-      "\n",
-      "sum1 = a + b + c + d + e\n",
-      "\n",
-      "for i in range(1,sum1):\n",
-      "    if i == a or i == b or i == c or i == d or i == e:\n",
-      "        sys.stdout.write(\"The Smallest Number : %d\"%(i))\n",
-      "        break\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Smallest Number : 2"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.20, Page number: 120<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read five numbers and print in ascending order\n",
-      "\n",
-      "import sys\n",
-      "import math\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers.\n",
-      "#since sum is a built in function in python, sum1 is used instead of sum.\n",
-      "\n",
-      "#use commas to separate the input values\n",
-      "a = int(raw_input(\"Enter five numbers : \"))\n",
-      "b = int(raw_input(\"Enter five numbers : \"))\n",
-      "c = int(raw_input(\"Enter five numbers : \"))\n",
-      "d = int(raw_input(\"Enter five numbers : \"))\n",
-      "e = int(raw_input(\"Enter five numbers : \"))\n",
-      "\n",
-      "sys.stdout.write(\"Numbers in ascending order : \")\n",
-      "sum1 = a + b + c + d + e\n",
-      "\n",
-      "for i in range(1,sum1):\n",
-      "    if i == a or i == b or i == c or i == d or i == e:\n",
-      "        sys.stdout.write(\"%3d\"%(i))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter five numbers : 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numbers in ascending order :   1  4  5  7  8"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.21, Page number: 121<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Perform multiplication of two integers by using negative sign\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#the range() function creates range of numbers and for loop automatically\n",
-      "#iterate through the numbers.\n",
-      "\n",
-      "a = int(raw_input(\"Enter two numbers : \"))\n",
-      "b = int(raw_input(\"Enter two numbers : \"))\n",
-      "\n",
-      "d = 0\n",
-      "for c in range(1,b+1):\n",
-      "    d = d - (-a)\n",
-      "\n",
-      "sys.stdout.write(\"Multiplication of %d * %d : %d\"%(a,b,d))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two numbers : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Multiplication of 5 * 5 : 25"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.22, Page number: 121<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Perform multiplication of two integers by using repetitive addition\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#in python, infinite for loop is not possible.\n",
-      "\n",
-      "a = int(raw_input(\"Enter two numbers : \"))\n",
-      "b = int(raw_input(\"Enter two numbers : \"))\n",
-      "\n",
-      "d = 0\n",
-      "c = 1\n",
-      "while True:\n",
-      "    d = d + a\n",
-      "    if c == b:\n",
-      "        break\n",
-      "    c += 1\n",
-      "\n",
-      "sys.stdout.write(\"Multiplication of %d * %d : %d\"%(a,b,d))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two numbers : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter two numbers : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Multiplication of 8 * 4 : 32"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.23, Page number: 122<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the marks for five subjects and calculate sum & average of marks.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#use spaces to separate the input values\n",
-      "#since sum is a built-in function in python, sum1 is used as the variable name\n",
-      "\n",
-      "sys.stdout.write(\"Enter The Marks of Five Subjects \")\n",
-      "sum1 = 0\n",
-      "\n",
-      "for i in range(1,5+1):\n",
-      "    sys.stdout.write(\"[%d] Student  : \"%(i))\n",
-      "    #Exception handling\n",
-      "    try:\n",
-      "        #use space to separate the input characters\n",
-      "        v = raw_input(\"\")\n",
-      "        a,b,c,d,e = map(int,v.split())\n",
-      "        x = 5\n",
-      "    except:\n",
-      "        x = 0\n",
-      "\n",
-      "    if(x == 5):\n",
-      "        sum1 = a + b + c + d + e\n",
-      "        avg = sum1/5\n",
-      "        sys.stdout.write(\"\\nTotal Marks of Student [%d] %d\"%(i,sum1))\n",
-      "        sys.stdout.write(\"\\nAverage Marks of Student [%d] %f\"%(i,avg))\n",
-      "    else:\n",
-      "        sys.stdout.write(\"Type Mismatch\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter The Marks of Five Subjects [1] Student  : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "58 52 52 56 78\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Total Marks of Student [1] 296\n",
-        "Average Marks of Student [1] 59.000000[2] Student  : "
-       ]
-      }
-     ],
-     "prompt_number": "*"
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.24, Page number: 123<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Enter a character and display its position in alphabetic order.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = 1\n",
-      "\n",
-      "ch = raw_input(\"Enter a character :\")\n",
-      "ch = ch.upper()\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "#ord() function converts the character ASCII value to integer\n",
-      "for i in range(65,90+1):\n",
-      "    if ord(ch) == i:\n",
-      "        sys.stdout.write(\"'%c' is %d [st/nd/rd/th] Character in Alphabetic\"%(i,c))\n",
-      "    c += 1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a character :U\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "'U' is 21 [st/nd/rd/th] Character in Alphabetic"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.25, Page number: 124<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate the value of -m^x.\n",
-      "#'m' is a mantissa and 'x' is an exponent\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "number = int(raw_input(\"Enter Number : \"))\n",
-      "power = int(raw_input(\"Enter Power : \"))\n",
-      "\n",
-      "ans = 1\n",
-      "\n",
-      "#Calculation\n",
-      "for i in range(1,power+1):\n",
-      "    ans *= number\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The Power of %d raised to %d is %ld\"%(number,power,ans))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Power : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Power of 5 raised to 3 is 125"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.26, Page number: 124<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Subtraction of two loop variables using nested for loops.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Calculation & Result\n",
-      "for a in range(3,0,-1):\n",
-      "    for b in range(1,2+1):\n",
-      "        sub = a - b\n",
-      "        sys.stdout.write(\"a = %d  b = %d  a - b = %d\\n\"%(a,b,sub))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a = 3  b = 1  a - b = 2\n",
-        "a = 3  b = 2  a - b = 1\n",
-        "a = 2  b = 1  a - b = 1\n",
-        "a = 2  b = 2  a - b = 0\n",
-        "a = 1  b = 1  a - b = 0\n",
-        "a = 1  b = 2  a - b = -1\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.27, Page number: 125<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Illustrate an example based on nested for loops\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Calculation & Result\n",
-      "for i in range(1,3+1):                  #outer loop\n",
-      "    for j in range(1,2+1):              #inner loop\n",
-      "        sys.stdout.write(\"\\ni * j : %d\"%(i*j))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "i * j : 1\n",
-        "i * j : 2\n",
-        "i * j : 2\n",
-        "i * j : 4\n",
-        "i * j : 3\n",
-        "i * j : 6"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.28, Page number: 126<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print values and messages when any loop ends using nested for loops.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Calculation & Result\n",
-      "for a in range(1,2+1):                  #outer loop\n",
-      "    for b in range(1,2+1):              #middle loop\n",
-      "        for c in range(1,2+1):          #inner loop\n",
-      "            sys.stdout.write(\"\\na = %d + b = %d + c = %d : %d\"%(a,b,c,a+b+c))\n",
-      "        sys.stdout.write(\"\\nInner Loop Over.\")\n",
-      "    sys.stdout.write(\"\\nMiddle Loop Over.\")\n",
-      "sys.stdout.write(\"\\nOuter Loop Over.\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "a = 1 + b = 1 + c = 1 : 3\n",
-        "a = 1 + b = 1 + c = 2 : 4\n",
-        "Inner Loop Over.\n",
-        "a = 1 + b = 2 + c = 1 : 4\n",
-        "a = 1 + b = 2 + c = 2 : 5\n",
-        "Inner Loop Over.\n",
-        "Middle Loop Over.\n",
-        "a = 2 + b = 1 + c = 1 : 4\n",
-        "a = 2 + b = 1 + c = 2 : 5\n",
-        "Inner Loop Over.\n",
-        "a = 2 + b = 2 + c = 1 : 5\n",
-        "a = 2 + b = 2 + c = 2 : 6\n",
-        "Inner Loop Over.\n",
-        "Middle Loop Over.\n",
-        "Outer Loop Over."
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.29, Page number: 127<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display perfect cubes up to given number.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "k = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "for i in range(1,k+1):\n",
-      "    for j in range(1,i+1):   \n",
-      "        if i == pow(j,3):\n",
-      "          sys.stdout.write(\"\\nNumber : %d & it's Cube : %d\"%(j,i))\n",
-      "          \n",
-      "          \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 100\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number : 1 & it's Cube : 1\n",
-        "Number : 2 & it's Cube : 8\n",
-        "Number : 3 & it's Cube : 27\n",
-        "Number : 4 & it's Cube : 64"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.30, Page number: 128<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display numbers 1 to 100 using ASCII values from 48 to 57. Use nested loops.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "j = 0\n",
-      "k = -9\n",
-      "\n",
-      "sys.stdout.write(\"\\tTable of 1 to 100 Numbers Using ASCII Values\\n\")\n",
-      "sys.stdout.write(\"\\t===== == = == === ======= ===== ===== ======\\n\")\n",
-      "\n",
-      "#Calculation & Result\n",
-      "for i in range(48,57+1):\n",
-      "    for j in range(48,57+1):   \n",
-      "        sys.stdout.write(\"\\t%c%c\"%(i,j))\n",
-      "        #if k != 1:\n",
-      "         #   sys.stdout.write(\"%c\"%(i+1))\n",
-      "        #if k == 0:\n",
-      "         #   sys.stdout.write(\"\\b\\b%d\"%(k+1))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    k += 1\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tTable of 1 to 100 Numbers Using ASCII Values\n",
-        "\t===== == = == === ======= ===== ===== ======\n",
-        "\t00\t01\t02\t03\t04\t05\t06\t07\t08\t09\n",
-        "\t10\t11\t12\t13\t14\t15\t16\t17\t18\t19\n",
-        "\t20\t21\t22\t23\t24\t25\t26\t27\t28\t29\n",
-        "\t30\t31\t32\t33\t34\t35\t36\t37\t38\t39\n",
-        "\t40\t41\t42\t43\t44\t45\t46\t47\t48\t49\n",
-        "\t50\t51\t52\t53\t54\t55\t56\t57\t58\t59\n",
-        "\t60\t61\t62\t63\t64\t65\t66\t67\t68\t69\n",
-        "\t70\t71\t72\t73\t74\t75\t76\t77\t78\t79\n",
-        "\t80\t81\t82\t83\t84\t85\t86\t87\t88\t89\n",
-        "\t90\t91\t92\t93\t94\t95\t96\t97\t98\t99\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.31, Page number: 129<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count number of votes secured by 'A' & 'B'.  Assume three voters are\n",
-      "#voting them. Also count the invalid votes.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization & Calculation\n",
-      "a = 0\n",
-      "b = 0\n",
-      "o = 0\n",
-      "sys.stdout.write(\"Press A or B\\n\")\n",
-      "for i in range(1,3+1):\n",
-      "    sys.stdout.write(\"\\nVoter no. %d\"%(i))\n",
-      "    sys.stdout.write(\"  Enter Vote : \")\n",
-      "    v = raw_input(\"\")\n",
-      "    v = v.upper()\n",
-      "    if v == 'A':\n",
-      "        a += 1\n",
-      "    else:\n",
-      "        if v == 'B':\n",
-      "            b += 1\n",
-      "        else:\n",
-      "            o += 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nStatus of vote(s)\\n\")\n",
-      "sys.stdout.write(\"\\nA secures %d vote(s).\"%(a))\n",
-      "sys.stdout.write(\"\\nB secures %d vote(s).\"%(b))\n",
-      "sys.stdout.write(\"\\nInvalid vote(s) %d.\"%(o))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Press A or B\n",
-        "\n",
-        "Voter no. 1  Enter Vote : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "A\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Voter no. 2  Enter Vote : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "B\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Voter no. 3  Enter Vote : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "X\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Status of vote(s)\n",
-        "\n",
-        "A secures 1 vote(s).\n",
-        "B secures 1 vote(s).\n",
-        "Invalid vote(s) 1."
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.32, Page number: 130<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Simulate a digital clock\n",
-      "\n",
-      "import sys\n",
-      "import os\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"hh mm ss\\n\")\n",
-      "for h in range(1,12):\n",
-      "    for m in range(1,59):\n",
-      "        for s in range(1,59):\n",
-      "            sys.stdout.write(\"\\r%d  %d  %d\"%(h,m,s))\n",
-      "            \n",
-      "#Displays the last value"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hh mm ss\n",
-        "\r",
-        "1  1  1\r",
-        "1  1  2\r",
-        "1  1  3\r",
-        "1  1  4\r",
-        "1  1  5\r",
-        "1  1  6\r",
-        "1  1  7\r",
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-        "1  5  1\r",
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-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.33, Page number: 131<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count occurrence of 0 to 9 digits between 1 and given decimal number\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "n = int(raw_input(\"Enter a Decimal Number : \"))\n",
-      "st = [0 for x in range(10)]         #10 array elements initialized as 0\n",
-      "\n",
-      "#Calculation\n",
-      "for l in range(1,n+1):\n",
-      "    t = l\n",
-      "    while t != 0:\n",
-      "        k = t % 10\n",
-      "        t = t / 10\n",
-      "        st[k] += 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nOccurrence of 0-9 digits between 1 to %d Numbers.\"%(n))\n",
-      "sys.stdout.write(\"\\n========== == === ====== ======= = == == ========\")\n",
-      "\n",
-      "for i in range(0,10):\n",
-      "    if st[i] > 0:\n",
-      "        sys.stdout.write(\"\\n%d Occurs %8d Times.\"%(i,st[i]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Decimal Number : 15\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Occurrence of 0-9 digits between 1 to 15 Numbers.\n",
-        "========== == === ====== ======= = == == ========\n",
-        "0 Occurs        1 Times.\n",
-        "1 Occurs        8 Times.\n",
-        "2 Occurs        2 Times.\n",
-        "3 Occurs        2 Times.\n",
-        "4 Occurs        2 Times.\n",
-        "5 Occurs        2 Times.\n",
-        "6 Occurs        1 Times.\n",
-        "7 Occurs        1 Times.\n",
-        "8 Occurs        1 Times.\n",
-        "9 Occurs        1 Times."
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.34, Page number: 132<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display sum of digits of a given number\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "s = 0\n",
-      "n = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "\n",
-      "sys.stdout.write(\"\\nSum of Digits till a single digit\\n %d\"%(n))\n",
-      "\n",
-      "while n != 0:\n",
-      "    s = s + n%10\n",
-      "    n = n/10\n",
-      "    if n == 0 and s > 9:\n",
-      "        sys.stdout.write(\"\\n %2d\"%(s))\n",
-      "        n = s\n",
-      "        s = 0\n",
-      "\n",
-      "sys.stdout.write(\"\\n %2d\"%(s))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 4687\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of Digits till a single digit\n",
-        " 4687\n",
-        " 25\n",
-        "  7"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.35, Page number: 133<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display octal numbers in binary.  Attach a parity bit with \"1\"\n",
-      "#if number of 1s are even otherwise \"0\".\n",
-      "#           OR\n",
-      "#Generate odd parity to octal numbers 0 to 7.  Express each number in binary\n",
-      "#and attach the parity bit\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = 0\n",
-      "j = 12\n",
-      "k = 2\n",
-      "\n",
-      "#Calculation & Result\n",
-      "\n",
-      "sys.stdout.write(\"\\nBinary Bits                  Parity Bits\")\n",
-      "sys.stdout.write(\"\\n=============               ==============\\n\")\n",
-      "\n",
-      "for x in range(0,8):\n",
-      "    k += 1\n",
-      "    j = 12\n",
-      "    y = x\n",
-      "    for a in range(0,3):\n",
-      "        b = y % 2\n",
-      "        #gotoxy\n",
-      "        sys.stdout.write(\"\\t%d\"%(b))\n",
-      "        y = y / 2\n",
-      "        if b == 1:\n",
-      "            c += 1\n",
-      "    if c%2 == 0:\n",
-      "        #gotoxy\n",
-      "        sys.stdout.write(\"\\t1\")\n",
-      "    else:\n",
-      "        #gotosy\n",
-      "        sys.stdout.write(\"\\t0\\n\")\n",
-      "    c = 0\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Binary Bits                  Parity Bits\n",
-        "=============               ==============\n",
-        "\t0\t0\t0\t1\t1\t0\t0\t0\n",
-        "\t0\t1\t0\t0\n",
-        "\t1\t1\t0\t1\t0\t0\t1\t0\n",
-        "\t1\t0\t1\t1\t0\t1\t1\t1\t1\t1\t1\t0\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.36, Page number: 135<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the series x - x^3/3! + x^5/5! - .... x^n/n!\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = 3.0\n",
-      "f = 1.0\n",
-      "\n",
-      "#Give x as float value and n as int value\n",
-      "\n",
-      "x = float(raw_input(\"Enter x & n : \"))       #use commas to input values\n",
-      "n = int(raw_input(\"Enter x & n : \")) \n",
-      "\n",
-      "sum1 = x                #since sum is built-in function in python,\n",
-      "                        #sum1 is used instead of sum\n",
-      "\n",
-      "#Calculation & Result\n",
-      "\n",
-      "for i in range(3,n+1,2):\n",
-      "    f = 1\n",
-      "    if c%2 != 0:\n",
-      "        for l in range(1,i+1):\n",
-      "            f = f * l\n",
-      "        sum1 = sum1 - pow(x,i)/ f\n",
-      "    else:\n",
-      "        for l in range(1,i+1):\n",
-      "            f = f * l\n",
-      "        sum1 = sum1 + pow(x,i)/ f\n",
-      "    c += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\nSum of series Numbers : %f\"%(sum1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter x & n : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter x & n : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of series Numbers : 0.933333"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.37, Page number: 136<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the series  x + x^2/2! + x^4/4! + .... x^n/n!\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "f = 1.0\n",
-      "\n",
-      "#Give x as float value and y as int value\n",
-      "\n",
-      "x = float(raw_input(\"Enter x & y : \"))       #use commas to input values\n",
-      "y = int(raw_input(\"Enter x & y : \")) \n",
-      "\n",
-      "sum1 = x                #since sum is built-in function in python,\n",
-      "                        #sum1 is used instead of sum\n",
-      "\n",
-      "#Calculation & Result\n",
-      "\n",
-      "for i in range(2,y+1,2):\n",
-      "    f = 1\n",
-      "    for l in range(1,i+1):\n",
-      "        f = f * l\n",
-      "    sum1 = sum1 + pow(x,i)/ f\n",
-      "\n",
-      "sys.stdout.write(\"\\nSum of Series : %f\"%(sum1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter x & y : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter x & y : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of Series : 22.666667"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.38, Page number: 136<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the series 1 - 1/1! + 2/2! - 3/3! .... n/n!\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = 3\n",
-      "\n",
-      "n = int(raw_input(\"Enter value of n : \"))\n",
-      "\n",
-      "sum1 = 1.0              #since sum is built-in function in python,\n",
-      "                        #sum1 is used instead of sum\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "for i in range(1,n+1):\n",
-      "    f = 1.0\n",
-      "    if c%2 != 0:\n",
-      "        for l in range(1,i+1):\n",
-      "            f = f * l\n",
-      "        k = float(i / f)\n",
-      "        sum1 = sum1 - k\n",
-      "    else:\n",
-      "        for l in range(1,i+1):\n",
-      "            f = f * l\n",
-      "        sum1 = sum1 + float(i/f)\n",
-      "    c += 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSum of series Numbers : %f\"%(sum1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of n : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of series Numbers : 0.500000"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.39, Page number: 137<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the Armstrong numbers in three digits from 100 to 999.  If sum of cubes of each digits\n",
-      "#of the number is equal to number itself, then the number is called as an Armstrong number.\n",
-      "#(For eg. 153 = 1^3 + 5^3 + 3^3 = 153)\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "cube = 0\n",
-      "\n",
-      "sys.stdout.write(\"The Following Numbers are Armstrong numbers.\")\n",
-      "\n",
-      "#Calculation & Result\n",
-      "\n",
-      "for k in range(100,999+1):\n",
-      "    cube = 0\n",
-      "    x = 1\n",
-      "    d = 3\n",
-      "    n = k\n",
-      "    while x <= d:\n",
-      "        i = n % 10\n",
-      "        cube = cube + pow(i,3)\n",
-      "        n = n / 10\n",
-      "        x += 1\n",
-      "    if cube == k:\n",
-      "        sys.stdout.write(\"\\n\\t%d\"%(k))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Following Numbers are Armstrong numbers.\n",
-        "\t153\n",
-        "\t370\n",
-        "\t371\n",
-        "\t407"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.40, Page number: 138<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Example 6.40.py\n",
-      "#Program to display the stars as shown below\n",
-      "#*\n",
-      "#**\n",
-      "#***\n",
-      "#****\n",
-      "#*****\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "x = int(raw_input(\"How many lines stars (*) should be printed ? \"))\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "for i in range(1,x+1):\n",
-      "    for j in range(1,i+1):\n",
-      "        sys.stdout.write(\"*\")\n",
-      "    sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many lines stars (*) should be printed ? 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "*\n",
-        "**\n",
-        "***\n",
-        "****\n",
-        "*****\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.41, Page number: 139<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Generate the given pattern\n",
-      "#6 5 4 3 2 1 0\n",
-      "#5 4 3 2 1 0\n",
-      "#4 3 2 1 0\n",
-      "#3 2 1 0\n",
-      "#2 1 0\n",
-      "#1 0\n",
-      "#0\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "while i >= 0:\n",
-      "    c = i\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    while True:\n",
-      "        sys.stdout.write(\"%3d\"%(c))\n",
-      "        if c == 0:\n",
-      "            break\n",
-      "        c -= 1\n",
-      "    i -= 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "  6  5  4  3  2  1  0\n",
-        "  5  4  3  2  1  0\n",
-        "  4  3  2  1  0\n",
-        "  3  2  1  0\n",
-        "  2  1  0\n",
-        "  1  0\n",
-        "  0"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.42, Page number: 140<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the series of numbers as given below\n",
-      "#1\n",
-      "#1 2\n",
-      "#1 2 3\n",
-      "#1 2 3 4\n",
-      "\n",
-      "#4 3 2 1\n",
-      "#3 2 1\n",
-      "#2 1\n",
-      "#1\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter value of x : \"))\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "for j in range(1,x+1):\n",
-      "    for i in range(1,j+1):\n",
-      "        sys.stdout.write(\"%3d\"%(i))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "\n",
-      "sys.stdout.write(\"\\n\")\n",
-      "\n",
-      "for j in range(x,0,-1):\n",
-      "    for i in range(j,0,-1):\n",
-      "        sys.stdout.write(\"%3d\"%(i))\n",
-      "    sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of x : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  1\n",
-        "  1  2\n",
-        "  1  2  3\n",
-        "  1  2  3  4\n",
-        "\n",
-        "  4  3  2  1\n",
-        "  3  2  1\n",
-        "  2  1\n",
-        "  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.43, Page number: 142<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to display the series of numbers as given below\n",
-      "#1\n",
-      "#2 1\n",
-      "#3 2 1\n",
-      "#4 3 2 1\n",
-      "\n",
-      "#4 3 2 1\n",
-      "#3 2 1\n",
-      "#2 1\n",
-      "#1\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter value of x : \"))\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "for j in range(1,x+1):\n",
-      "    for i in range(j,0,-1):\n",
-      "        sys.stdout.write(\"%3d\"%(i))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "\n",
-      "sys.stdout.write(\"\\n\")\n",
-      "\n",
-      "for j in range(x,0,-1):\n",
-      "    for i in range(j,0,-1):\n",
-      "        sys.stdout.write(\"%3d\"%(i))\n",
-      "    sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of x : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  1\n",
-        "  2  1\n",
-        "  3  2  1\n",
-        "  4  3  2  1\n",
-        "\n",
-        "  4  3  2  1\n",
-        "  3  2  1\n",
-        "  2  1\n",
-        "  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.44, Page number: 143<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Generate the pyramid structure using numberical\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter a number : \"))\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "for j in range(0,x+1):\n",
-      "    for i in range(0-j,j+1):\n",
-      "        sys.stdout.write(\"%3d\"%(abs(i)))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  0\n",
-        "  1  0  1\n",
-        "  2  1  0  1  2\n",
-        "  3  2  1  0  1  2  3\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.45, Page number: 143<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert binary to decimal number\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = [1 for i in range(1,5+1)]              #defines the array x and initializes the elements with 0\n",
-      "y = x[0]\n",
-      "\n",
-      "sys.stdout.write(\"\\nValues in different Iterations\")\n",
-      "sys.stdout.write(\"\\n====== == ========= ==========\\n\")\n",
-      "\n",
-      "for i in range(0,4):\n",
-      "    y = y * 2 + x[i+1]\n",
-      "    sys.stdout.write(\"[%d] %d\\t\"%(i+1,y))\n",
-      "\n",
-      "sys.stdout.write(\"\\nEquivalent of [\")\n",
-      "\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"%d\"%(x[i]))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"] in Decimal Number is : \")\n",
-      "sys.stdout.write(\"%d\\t\"%(y))\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Values in different Iterations\n",
-        "====== == ========= ==========\n",
-        "[1] 3\t[2] 7\t[3] 15\t[4] 31\t\n",
-        "Equivalent of [11111] in Decimal Number is : 31\t"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.46, Page number: 144<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Program to add a parity bit with four binary bits such that the total\n",
-      "#number of one's should be even\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "bit = [0 for i in range(0,5)]              #defines the array bit and initializes the elements with 0\n",
-      "c = 0\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter four bits : \")\n",
-      "\n",
-      "for j in range(0,4):\n",
-      "    x = int(raw_input(\" \"))\n",
-      "    bit[j] = x\n",
-      "    if bit[j] == 1:\n",
-      "        c += 1\n",
-      "    else:\n",
-      "        if bit[j] > 1 or bit[j] < 0:\n",
-      "            j -= 1\n",
-      "            continue\n",
-      "        if c%2 == 0:\n",
-      "            bit[j] = 0\n",
-      "        else:\n",
-      "            bit[j] = 1   \n",
-      "   \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nMessage bits together with parity bit : \")\n",
-      "for j in range(0,5):\n",
-      "    sys.stdout.write(\"%d\"%(bit[j]))\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter four bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Message bits together with parity bit : 11110"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.47, Page number: 145<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert binary to decimal number. Enter the binary bits by using for loop\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "z = [0 for i in range(0,10)]                #defines the array z and initializes the elements with 0\n",
-      "\n",
-      "sys.stdout.write(\"Enter the number of bits :- \")\n",
-      "b = int(raw_input(\"\"))\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter the binary bits : \")\n",
-      "for i in range(0,b):\n",
-      "    z[i] = int(raw_input(\"\"))\n",
-      "\n",
-      "a = z[0]\n",
-      "\n",
-      "for i in range(0,b-1):\n",
-      "    a = a * 2 + z[i+1]\n",
-      "    sys.stdout.write(\"\\n%d\"%(a))\n",
-      "   \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nDecimal Number is : %d \"%(a))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of bits :- "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter the binary bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "2\n",
-        "4\n",
-        "8\n",
-        "17\n",
-        "Decimal Number is : 17 "
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.48, Page number: 147<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Verify the truth table of AND gate. Assume AND gate has two\n",
-      "#input bits A & B and one output bit C.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "a = [0 for i in range(0,4)]                #defines the array a,b & c  and initializes the elements with 0\n",
-      "b = [0 for i in range(0,4)]\n",
-      "c = [0 for i in range(0,4)]\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
-      "for x in range(0,4):\n",
-      "    a[x] = int(raw_input(\"\"))\n",
-      "    \n",
-      "    if a[x] > 1 or a[x] < 0:\n",
-      "        x -= 1\n",
-      "        continue\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
-      "for x in range(0,4):\n",
-      "    b[x] = int(raw_input(\"\"))\n",
-      "    \n",
-      "    if b[x] > 1 or b[x] < 0:\n",
-      "        x -= 1\n",
-      "        continue\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nA  B  C\")\n",
-      "for x in range(0,4):\n",
-      "    if a[x] == 1 and b[x] == 1:\n",
-      "        c[x] = 1\n",
-      "    else:\n",
-      "        c[x] = 0\n",
-      "    sys.stdout.write(\"\\n%d  %d  %d\"%(a[x],b[x],c[x]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Four Bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Four Bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "A  B  C\n",
-        "1  1  1\n",
-        "0  0  0\n",
-        "1  0  0\n",
-        "0  1  0"
-       ]
-      }
-     ],
-     "prompt_number": 57
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.49, Page number: 148<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Verify the truth table of OR gate. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "a = [0 for i in range(0,4)]                #defines the array a,b & c  and initializes the elements with 0\n",
-      "b = [0 for i in range(0,4)]\n",
-      "c = [0 for i in range(0,4)]\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
-      "for x in range(0,4):\n",
-      "    a[x] = int(raw_input(\"\"))\n",
-      "    \n",
-      "    if a[x] > 1 or a[x] < 0:\n",
-      "        x -= 1\n",
-      "        continue\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
-      "for x in range(0,4):\n",
-      "    b[x] = int(raw_input(\"\"))\n",
-      "    \n",
-      "    if b[x] > 1 or b[x] < 0:\n",
-      "        x -= 1\n",
-      "        continue\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nA  B  C\")\n",
-      "for x in range(0,4):\n",
-      "    if a[x] == 0 and b[x] == 0:\n",
-      "        c[x] = 0\n",
-      "    else:\n",
-      "        c[x] = 1\n",
-      "    sys.stdout.write(\"\\n%d  %d  %d\"%(a[x],b[x],c[x]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Four Bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Four Bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "A  B  C\n",
-        "1  1  1\n",
-        "1  0  1\n",
-        "1  0  1\n",
-        "0  0  0"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.50, Page number: 149<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Verify the truth table of EX-OR gate. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "a = [0 for i in range(0,4)]                #defines the array a,b & c  and initializes the elements with 0\n",
-      "b = [0 for i in range(0,4)]\n",
-      "c = [0 for i in range(0,4)]\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
-      "for x in range(0,4):\n",
-      "    a[x] = int(raw_input(\"\"))\n",
-      "    \n",
-      "    if a[x] > 1 or a[x] < 0:\n",
-      "        x -= 1\n",
-      "        continue\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
-      "for x in range(0,4):\n",
-      "    b[x] = int(raw_input(\"\"))\n",
-      "    \n",
-      "    if b[x] > 1 or b[x] < 0:\n",
-      "        x -= 1\n",
-      "        continue\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nA  B  C\")\n",
-      "for x in range(0,4):\n",
-      "    if a[x] == 0 and b[x] == 1:\n",
-      "        c[x] = 1\n",
-      "    else:\n",
-      "        if a[x] == 1 and b[x] == 0:\n",
-      "            c[x] = 1\n",
-      "        else:\n",
-      "            c[x] = 0\n",
-      "    sys.stdout.write(\"\\n%d  %d  %d\"%(a[x],b[x],c[x]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Four Bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Four Bits : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "A  B  C\n",
-        "1  1  0\n",
-        "1  0  1\n",
-        "1  0  1\n",
-        "0  0  0"
-       ]
-      }
-     ],
-     "prompt_number": 59
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.51, Page number: 151<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the Hamming code for the entered binary code.  Assume the binary code of four bits in length.\n",
-      "#The hamming code should be seven bits.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "#defines the array a,b & c  and initializes the elements with 0\n",
-      "b = [0 for i in range(0,4)]\n",
-      "c = [0 for i in range(0,7)]\n",
-      "\n",
-      "sys.stdout.write(\"\\nRead the Binary Numbers : \")\n",
-      "for x in range(0,4):\n",
-      "    b[x] = int(raw_input(\"\"))\n",
-      "\n",
-      "#Piece copy operation\n",
-      "c[0] = b[0]\n",
-      "c[1] = b[1]\n",
-      "c[2] = b[2]\n",
-      "c[4] = b[3]\n",
-      "\n",
-      "sys.stdout.write(\"\\nBefore xOR operation : \")\n",
-      "    \n",
-      "for x in range(0,7):\n",
-      "    sys.stdout.write(\"%3d\"%(c[x]))\n",
-      "\n",
-      "c[6] = c[0] ^ c[2] ^ c[4]\n",
-      "c[5] = c[0] ^ c[1] ^ c[4]\n",
-      "c[3] = c[0] ^ c[1] ^ c[2]\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nHamming code after XOR operation : \")\n",
-      "for x in range(0,7):\n",
-      "    sys.stdout.write(\"%3d\"%(c[x]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Read the Binary Numbers : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Before xOR operation :   1  0  1  0  0  0  0\n",
-        "Hamming code after XOR operation :   1  0  1  0  0  1  0"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.52, Page number: 152<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the results of students appear in six subjects.  The result declared should be as per the \n",
-      "#following table.\n",
-      "#TOTAL MARKS              RESULT\n",
-      "#>= 420                  Distinction\n",
-      "#>= 360                  First Division\n",
-      "#>= 240                  Second Division\n",
-      "#Otherwise               Fail\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "DISTINCTION = 420\n",
-      "FIRST = 360\n",
-      "SECOND = 240\n",
-      "\n",
-      "number = int(raw_input(\"Enter number of Students : \"))\n",
-      "\n",
-      "#Read the values and calculation\n",
-      "for i in range(1,number+1):\n",
-      "    roll_no = int(raw_input(\"Enter Roll Number : \"))\n",
-      "    total = 0\n",
-      "    sys.stdout.write(\"\\nEnter Marks of 6 Subjects for Roll no %d : \"%(roll_no))\n",
-      "    for j in range(1,6+1):\n",
-      "        marks = int(raw_input(\"\"))\n",
-      "        total = total + marks\n",
-      "    sys.stdout.write(\"\\nTOTAL MARKS = %d\"%(total))\n",
-      "    if total >= DISTINCTION:\n",
-      "        sys.stdout.write(\"\\n(Distinction)\\n\\n\")\n",
-      "    else:\n",
-      "        if total >= FIRST:\n",
-      "            sys.stdout.write(\"\\n(First Division)\\n\\n\")\n",
-      "        else:\n",
-      "            if total >= SECOND:\n",
-      "                sys.stdout.write(\"\\n(Second Division)\\n\\n\")\n",
-      "            else:\n",
-      "                sys.stdout.write(\"\\n(***Fail***)\")\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter number of Students : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Roll Number : 1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Marks of 6 Subjects for Roll no 1 : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "42\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "52\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "62\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "72\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "82\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "92\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "TOTAL MARKS = 402\n",
-        "(First Division)\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.53, Page number: 154<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the string \"You have learnt C program\" 9 times using while loop\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 1\n",
-      "\n",
-      "#Result\n",
-      "while x < 10:\n",
-      "    sys.stdout.write(\"\\nYou have learnt C program\")\n",
-      "    x += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "You have learnt C program\n",
-        "You have learnt C program\n",
-        "You have learnt C program\n",
-        "You have learnt C program\n",
-        "You have learnt C program\n",
-        "You have learnt C program\n",
-        "You have learnt C program\n",
-        "You have learnt C program\n",
-        "You have learnt C program"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.54, Page number: 155<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Add 10 consecutive numbers starting from 1 using while loop.\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 1\n",
-      "sum1 = 0           #since sum is a builtin function in python, sum1 is used instead of sum\n",
-      "\n",
-      "#Calculation\n",
-      "while a <= 10:\n",
-      "    sys.stdout.write(\"%3d\"%(a))\n",
-      "    sum1 = sum1 + a\n",
-      "    a += 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSum of 10 numbers : %d\"%(sum1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  1  2  3  4  5  6  7  8  9 10\n",
-        "Sum of 10 numbers : 55"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.55, Page number: 155<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate factorial of a given number using while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "fact = 1\n",
-      "\n",
-      "a = int(raw_input(\"Enter The Number : \"))\n",
-      "\n",
-      "#Calculation\n",
-      "while a >= 1:\n",
-      "    sys.stdout.write(\" %d * \"%(a))\n",
-      "    fact = fact * a\n",
-      "    a -= 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\" = %d\"%(fact))\n",
-      "sys.stdout.write(\"\\nFactorial of Given number is %d\"%(fact))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter The Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 5 *  4 *  3 *  2 *  1 *  = 120\n",
-        "Factorial of Given number is 120"
-       ]
-      }
-     ],
-     "prompt_number": 61
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.56, Page number: 156<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate factorial of a given number using while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "fact = 1\n",
-      "b = 1\n",
-      "\n",
-      "a = int(raw_input(\"Enter The Number : \"))\n",
-      "\n",
-      "#Calculation\n",
-      "while b <= a:\n",
-      "    sys.stdout.write(\" %d * \"%(b))\n",
-      "    fact = fact * b\n",
-      "    b += 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\" = %d\"%(fact))\n",
-      "sys.stdout.write(\"\\nFactorial of %d is %d\"%(b-1,fact))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter The Number : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1 *  2 *  3 *  4 *  = 24\n",
-        "Factorial of 4 is 24"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.57, Page number: 157<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert decimal number to binary number\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "y = 40\n",
-      "rev = [0 for i in range(1,10)]\n",
-      "c = 1\n",
-      "\n",
-      "x = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "#since gotoxy() function is not available in python, the reminders are stored in an array and displayed in reverse order\n",
-      "\n",
-      "sys.stdout.write(\"\\nBinary Number : \")\n",
-      "while x != 0:\n",
-      "    rev[c] = x % 2\n",
-      "    x = x / 2\n",
-      "    c += 1\n",
-      "\n",
-      "for i in range(c-1,0,-1):\n",
-      "    sys.stdout.write(\"%d\"%(rev[i]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 25\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Binary Number : 11001"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.58, Page number: 157<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert decimal number to user defined number system. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "y = 35\n",
-      "rev = 0\n",
-      "c = 1\n",
-      "rev = [0 for i in range(1,10)]\n",
-      "\n",
-      "m = int(raw_input(\"Enter the Decimal Number : \"))\n",
-      "b = int(raw_input(\"Enter base of number system : \"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "sys.stdout.write(\"\\nThe Number Obtained : \")\n",
-      "\n",
-      "#since gotoxy() function is not available in python, the reminders are stored in an array and displayed in reverse order\n",
-      "while m != 0:\n",
-      "    rev[c] = m % b\n",
-      "    c += 1\n",
-      "    m = m / b\n",
-      "\n",
-      "for i in range(c-1,0,-1):\n",
-      "    sys.stdout.write(\"%d\"%rev[i])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the Decimal Number : 50\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter base of number system : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Number Obtained : 200"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.59, Page number: 158<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert binary number to equivalent decimal number\n",
-      "\n",
-      "\n",
-      "#Variable Initialization\n",
-      "y = 0\n",
-      "p = 0\n",
-      "\n",
-      "n = int(raw_input(\"Enter a Binary Number : \"))\n",
-      "\n",
-      "#Calculation\n",
-      "while n != 0:\n",
-      "    x = n % 10\n",
-      "    if x > 1 or x < 0:\n",
-      "        sys.stdout.write(\"\\nInvalid Digit\")\n",
-      "        break\n",
-      "    else:\n",
-      "        y = y + x * pow(2,p)\n",
-      "        n = n / 10\n",
-      "        p += 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nEquivalent Decimal Number is %d.\"%(y))\n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Binary Number : 1111\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Equivalent Decimal Number is 15."
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.60, Page number: 159<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read a positive integer number 'n' and generate the numbers in the following way.\n",
-      "#If entered number is 5 the output will be as follows.\n",
-      "#OUTPUT : 5 4 3 2 1 0 1 2 3 4 5\n",
-      "\n",
-      "\n",
-      "#Variable Initialization\n",
-      "k = 0\n",
-      "\n",
-      "n = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "i = n + 1\n",
-      "k = k - n\n",
-      "\n",
-      "while i != k:\n",
-      "    sys.stdout.write(\"%3d\"%(abs(k)))\n",
-      "    k += 1\n",
-      "\n",
-      "      "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  3  2  1  0  1  2  3"
-       ]
-      }
-     ],
-     "prompt_number": 67
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.61, Page number: 160<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Enter a number through keyboard and find the sum of the digits.\n",
-      "\n",
-      "\n",
-      "#Variable Initialization\n",
-      "k = 1\n",
-      "sum1 = 0            #since sum is a built-in function in python, sum1 used instead of sum\n",
-      "\n",
-      "n = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Calculation \n",
-      "while n != 0:\n",
-      "    k = n % 10\n",
-      "    sum1 = sum1 + k\n",
-      "    k = n / 10\n",
-      "    n = k\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Sum of digits %d\"%(sum1))\n",
-      "\n",
-      "      "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 842\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sum of digits 14"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.62, Page number: 161<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Enter few numbers and count the positive and negative numbers together\n",
-      "#with their sums.  When 0 is entered program should be terminated.\n",
-      "\n",
-      "\n",
-      "#Variable Initialization\n",
-      "p = 0\n",
-      "n = 0\n",
-      "j = 1\n",
-      "s = 0\n",
-      "ns = 0\n",
-      "\n",
-      "#Calculation \n",
-      "sys.stdout.write(\"Enter Numbers (0) Exit : \")\n",
-      "while j != 0:\n",
-      "    j = int(raw_input(\"\"))\n",
-      "    if j > 0:\n",
-      "        p += 1\n",
-      "        s = s + j\n",
-      "    else:\n",
-      "        if j < 0:\n",
-      "            n += 1\n",
-      "            ns = ns + j\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nTotal Positive Numbers : %d\"%(p))\n",
-      "sys.stdout.write(\"\\nTotal Negative Numbers : %d\"%(n))\n",
-      "sys.stdout.write(\"\\nSum of Positive Numbers : %d\"%(s))\n",
-      "sys.stdout.write(\"\\nSum of Negative Numbers : %d\"%(ns))\n",
-      "\n",
-      "\n",
-      "      "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Numbers (0) Exit : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Total Positive Numbers : 5\n",
-        "Total Negative Numbers : 3\n",
-        "Sum of Positive Numbers : 15\n",
-        "Sum of Negative Numbers : -17"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.63, Page number: 162<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find sum of odd and even numbers separately by sorting the given numbers into\n",
-      "#odd and even numbers\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = 1\n",
-      "odd = 0\n",
-      "even = 0\n",
-      "\n",
-      "a = int(raw_input(\"Enter a Number : \"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "sys.stdout.write(\"ODD     EVEN\\n\")\n",
-      "while c <= a:\n",
-      "    b = c % 2\n",
-      "    while b == 0:\n",
-      "        sys.stdout.write(\"\\t%d \"%(c))\n",
-      "        even = even + c\n",
-      "        b = 1\n",
-      "    b = c % 2\n",
-      "    while b != 0:\n",
-      "        sys.stdout.write(\"\\n%d\"%(c))\n",
-      "        odd = odd + c\n",
-      "        b = 0\n",
-      "    c += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\n==============================\")\n",
-      "sys.stdout.write(\"\\n %d      %d\"%(odd,even))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number : 10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ODD     EVEN\n",
-        "\n",
-        "1\t2 \n",
-        "3\t4 \n",
-        "5\t6 \n",
-        "7\t8 \n",
-        "9\t10 \n",
-        "==============================\n",
-        " 25      30"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.64, Page number: 163<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the entered number in reverse order.\n",
-      "\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 1\n",
-      "\n",
-      "d = int(raw_input(\"Enter the number of digits : \"))\n",
-      "n = int(raw_input(\"Enter the number which is to be reversed : -\"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "sys.stdout.write(\"\\nThe Reversed Number is :-\")\n",
-      "\n",
-      "while x <= d:\n",
-      "    i = n % 10\n",
-      "    sys.stdout.write(\"%d\"%(i))\n",
-      "    n = n / 10\n",
-      "    x += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of digits : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number which is to be reversed : -5428\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Reversed Number is :-8245"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.65, Page number: 163<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Separate capitals, small, symbols and numbers from a statement\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 0\n",
-      "c = 0\n",
-      "s = 0\n",
-      "h = 0\n",
-      "n = 0\n",
-      "\n",
-      "#Array declaration\n",
-      "scan = [0 for i in range(0,40)]\n",
-      "cap = [0 for i in range(0,20)]\n",
-      "small = [0 for i in range(0,20)]\n",
-      "num = [0 for i in range(0,20)]\n",
-      "oth = [0 for i in range(0,20)]\n",
-      "\n",
-      "scan = raw_input(\"Enter Text Here : \")\n",
-      "\n",
-      "i = 0\n",
-      "\n",
-      "while i < len(scan):                       #There is no null termination character in python.\n",
-      "    if scan[i] >= '0' and scan[i] <= '9':\n",
-      "        n += 1\n",
-      "        num[n] = scan[i]\n",
-      "    else:\n",
-      "        if scan[i] >= 'A' and scan[i] <= 'Z':\n",
-      "            c += 1\n",
-      "            cap[c] = scan[i]\n",
-      "        else:\n",
-      "            if scan[i] >= 'a' and scan[i] <= 'z':\n",
-      "                s += 1\n",
-      "                small[s] = scan[i]\n",
-      "            else:\n",
-      "                h += 1\n",
-      "                oth[h] = scan[i]\n",
-      "                \n",
-      "    i += 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nCapital Letters : [\")\n",
-      "for i in range(0,20):\n",
-      "    sys.stdout.write(\"%c\"%cap[i])\n",
-      "\n",
-      "sys.stdout.write(\"]\\nSmall Letters : [\")\n",
-      "for i in range(0,20):\n",
-      "    sys.stdout.write(\"%c\"%(small[i]))\n",
-      "    \n",
-      "sys.stdout.write(\"]\\nNumeric Letters : [\")\n",
-      "for i in range(0,20):\n",
-      "    sys.stdout.write(\"%c\"%(num[i]))\n",
-      "\n",
-      "sys.stdout.write(\"]\\nOther Letters : [\")\n",
-      "for i in range(0,20):\n",
-      "    sys.stdout.write(\"%c\"%(oth[i]))\n",
-      "sys.stdout.write(\"]\")\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Here : HAVE A NICE DAY, contact me on 51606.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Capital Letters : [\u0000HAVEANICEDAY\u0000\u0000\u0000\u0000\u0000\u0000\u0000]\n",
-        "Small Letters : [\u0000contactmeon\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000]\n",
-        "Numeric Letters : [\u000051606\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000]\n",
-        "Other Letters : [\u0000   ,    .\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000]"
-       ]
-      }
-     ],
-     "prompt_number": 72
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.66, Page number: 165<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sort numbers, upper and lower case alphabets\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 48\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"   NUMBERS\\n\")\n",
-      "while i <= 57:\n",
-      "    sys.stdout.write(\"%2c\"%(i))\n",
-      "    i += 1\n",
-      "\n",
-      "i += 7\n",
-      "\n",
-      "sys.stdout.write(\"\\n  CAPITAL ALPHABETS\\n\")\n",
-      "while i <= 90:\n",
-      "    sys.stdout.write(\"%2c\"%(i))\n",
-      "    i += 1\n",
-      "    \n",
-      "i += 6\n",
-      "\n",
-      "sys.stdout.write(\"\\n  SMALL ALPHABETS\\n\")\n",
-      "while i <= 122:\n",
-      "    sys.stdout.write(\"%2c\"%(i))\n",
-      "    i += 1\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "   NUMBERS\n",
-        " 0 1 2 3 4 5 6 7 8 9\n",
-        "  CAPITAL ALPHABETS\n",
-        " A B C D E F G H I J K L M N O P Q R S T U V W X Y Z\n",
-        "  SMALL ALPHABETS\n",
-        " a b c d e f g h i j k l m n o p q r s t u v w x y z"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.67, Page number: 165<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sort numbers, upper and lower case alphabets\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 48\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Numbers : \")\n",
-      "while i < 124:\n",
-      "    if i < 58:\n",
-      "        sys.stdout.write(\"%2c\"%(i))\n",
-      "    else:\n",
-      "        if i == 58:\n",
-      "            sys.stdout.write(\"\\nCapital Letters : \")\n",
-      "            i += 7\n",
-      "        if i > 64 and i < 91:\n",
-      "            sys.stdout.write(\"%2c\"%(i))\n",
-      "        if i == 90:\n",
-      "            sys.stdout.write(\"\\nSmall Letters : \")\n",
-      "            i += 7\n",
-      "        if i > 96 and i < 123:\n",
-      "            sys.stdout.write(\"%2c\"%(i))\n",
-      "    i += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numbers :  0 1 2 3 4 5 6 7 8 9\n",
-        "Capital Letters :  A B C D E F G H I J K L M N O P Q R S T U V W X Y Z\n",
-        "Small Letters :  a b c d e f g h i j k l m n o p q r s t u v w x y z"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.68, Page number: 167<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display all ASCII numbers and their equivalent character, numbers and symbols using while loop.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 64\n",
-      "c = 'Y'\n",
-      "\n",
-      "#Result\n",
-      "while c == 'Y':\n",
-      "    sys.stdout.write(\"\\t%c (%d)\"%(i,i))\n",
-      "    i += 1\n",
-      "    sys.stdout.write(\"\\nPrint (Y/N)\")\n",
-      "    c = raw_input(\"\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t@ (64)\n",
-        "Print (Y/N)\tA (65)"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Print (Y/N)\tB (66)"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Print (Y/N)\tC (67)"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Print (Y/N)"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.69, Page number: 168<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert entered character into its opposite case.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n*Enter 'E' to exit\")\n",
-      "sys.stdout.write(\"\\n*Enter only characters\")\n",
-      "\n",
-      "while c != 'E':\n",
-      "    sys.stdout.write(\"\\nEnter a character : \")\n",
-      "    c = raw_input(\"Enter a character :\")\n",
-      "    if c == c.upper():\n",
-      "        c = ord(c)\n",
-      "        sys.stdout.write(\"\\tIt's Lower Case : %c\"%(chr(c+32)))\n",
-      "    else:\n",
-      "        c = ord(c)\n",
-      "        sys.stdout.write(\"\\tIt's Upper Case : %s\"%(chr(c-32)))\n",
-      "    c = chr(c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "*Enter 'E' to exit\n",
-        "*Enter only characters\n",
-        "Enter a character : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a character :s\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tIt's Upper Case : S\n",
-        "Enter a character : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a character :A\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tIt's Lower Case : a\n",
-        "Enter a character : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a character :E\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tIt's Lower Case : e"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.70, Page number: 169<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print numbers after each iteration and messages after termination of each loop using 3 nested while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 1\n",
-      "j = 1\n",
-      "k = 1\n",
-      "\n",
-      "#Result\n",
-      "while i < 4:\n",
-      "    while j < 3:\n",
-      "        while k < 2:\n",
-      "            sys.stdout.write(\"\\n\\ni = %d j = %d k = %d\"%(i,j,k))\n",
-      "            k += 1\n",
-      "        sys.stdout.write(\"\\nInner Loop (k) Completed.\")\n",
-      "        k = 1\n",
-      "        j += 1\n",
-      "    sys.stdout.write(\"\\nMiddle Loop (j) Completed.\")\n",
-      "    j = 1\n",
-      "    i += 1\n",
-      "sys.stdout.write(\"\\nOuter Loop (i) Completed.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "i = 1 j = 1 k = 1\n",
-        "Inner Loop (k) Completed.\n",
-        "\n",
-        "i = 1 j = 2 k = 1\n",
-        "Inner Loop (k) Completed.\n",
-        "Middle Loop (j) Completed.\n",
-        "\n",
-        "i = 2 j = 1 k = 1\n",
-        "Inner Loop (k) Completed.\n",
-        "\n",
-        "i = 2 j = 2 k = 1\n",
-        "Inner Loop (k) Completed.\n",
-        "Middle Loop (j) Completed.\n",
-        "\n",
-        "i = 3 j = 1 k = 1\n",
-        "Inner Loop (k) Completed.\n",
-        "\n",
-        "i = 3 j = 2 k = 1\n",
-        "Inner Loop (k) Completed.\n",
-        "Middle Loop (j) Completed.\n",
-        "Outer Loop (i) Completed."
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.71, Page number: 170<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display a message \"This is a program of do while loop\". for 5 times using do while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 1\n",
-      "\n",
-      "#Result\n",
-      "#There is no do..while loop in python\n",
-      "while i <= 5:\n",
-      "    sys.stdout.write(\"\\nThis is a program of do while loop.\")\n",
-      "    i += 1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "This is a program of do while loop.\n",
-        "This is a program of do while loop.\n",
-        "This is a program of do while loop.\n",
-        "This is a program of do while loop.\n",
-        "This is a program of do while loop."
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.72, Page number: 171<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Print the entered number in reverse order and find sum and product of digits using do-while loop. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "#since sum is a built-in function in python, sum1 is used instead of sum\n",
-      "x = 1\n",
-      "mul = 1\n",
-      "sum1 = 0\n",
-      "\n",
-      "d = int(raw_input(\"Enter the number of digits : -\"))\n",
-      "n = int(raw_input(\"Enter the number which is to be reversed :-\"))\n",
-      "\n",
-      "#Result\n",
-      "#There is no do..while loop in python\n",
-      "sys.stdout.write(\"\\nReveresed Number :- \")\n",
-      "while x <= d:\n",
-      "    i = n % 10\n",
-      "    sys.stdout.write(\"%d\"%(i))\n",
-      "    sum1 = sum1 + i\n",
-      "    mul = mul * i\n",
-      "    n = n / 10\n",
-      "    x += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\nAddition of digits :- %4d\"%(sum1))\n",
-      "sys.stdout.write(\"\\nMultiplication of digits :- %4d\"%(mul))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number of digits : -4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number which is to be reversed :-4321\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Reveresed Number :- 1234\n",
-        "Addition of digits :-   10\n",
-        "Multiplication of digits :-   24"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.73, Page number: 172<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Attempt the question 6.72 by considering the false condition i.e. the value of i should\n",
-      "#be initially larger than the value in the while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 7\n",
-      "\n",
-      "#Result\n",
-      "#There is no do-while loop in python\n",
-      "sys.stdout.write(\"\\nThis is a program of do while loop\")\n",
-      "while i <= 5:\n",
-      "    sys.stdout.write(\"\\nThis is a program of do while loop\")\n",
-      "    i += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "This is a program of do while loop"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.74, Page number: 172<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the cubes of 1 to 10 numbers using do-while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nPrint the numbers and their cubes\\n\")\n",
-      "while x <= 10:\n",
-      "    y = pow(x,3)\n",
-      "    sys.stdout.write(\"%4d %27d\\n\"%(x,y))\n",
-      "    x += 1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Print the numbers and their cubes\n",
-        "   1                           1\n",
-        "   2                           8\n",
-        "   3                          27\n",
-        "   4                          64\n",
-        "   5                         125\n",
-        "   6                         216\n",
-        "   7                         343\n",
-        "   8                         512\n",
-        "   9                         729\n",
-        "  10                        1000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.75, Page number: 173<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Check whether the given number is prime or not.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 2\n",
-      "n = int(raw_input(\"Enter The number for testing (prime or not) : \"))\n",
-      "\n",
-      "#Result\n",
-      "#There is no do-while loop in python\n",
-      "while x < n:\n",
-      "    if n % x == 0:\n",
-      "        sys.stdout.write(\"\\nThe number %d is not prime.\"%(n))\n",
-      "        exit(0)\n",
-      "    x += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\nThe number %d is prime\"%(n))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter The number for testing (prime or not) : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The number 5 is prime"
-       ]
-      }
-     ],
-     "prompt_number": 75
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.76, Page number: 174<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count the number of students having age less than 25 and weight less than 50Kg out of five.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "count = 0\n",
-      "x = 1\n",
-      "\n",
-      "sys.stdout.write(\"\\nEnter data of 5 boys\\n\")\n",
-      "sys.stdout.write(\"\\nAge Weight\\n\")\n",
-      "\n",
-      "#There is no do-while loop in python\n",
-      "while x <= 5:\n",
-      "    age = int(raw_input(\"\"))\n",
-      "    wt = float(raw_input(\"\"))\n",
-      "    if age < 25 and wt < 50:\n",
-      "        count += 1\n",
-      "    x += 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nNumber of boys with age < 25\")\n",
-      "sys.stdout.write(\"and weight < 50Kg = %d\\n\"%(count))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter data of 5 boys\n",
-        "\n",
-        "Age Weight\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "24\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "51\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "45\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "51\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "35\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "24\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "54\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Number of boys with age < 25and weight < 50Kg = 2\n"
-       ]
-      }
-     ],
-     "prompt_number": 76
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.77, Page number: 175<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compute the factorial of given number using do while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "fact = 1\n",
-      "a = int(raw_input(\"Enter The Number : \"))\n",
-      "sys.stdout.write(\"\\nEnter The Number : %d\\n\"%(a))\n",
-      "\n",
-      "#Calculation\n",
-      "#There is no do-while loop in python\n",
-      "while a >= 1:\n",
-      "    sys.stdout.write(\"%d * \"%(a))\n",
-      "    fact = fact * a\n",
-      "    a -= 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\" = %d\"%(fact))\n",
-      "sys.stdout.write(\"\\nFactorial of Given number is %d\"%(fact))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter The Number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter The Number : 5\n",
-        "5 * 4 * 3 * 2 * 1 *  = 120\n",
-        "Factorial of Given number is 120"
-       ]
-      }
-     ],
-     "prompt_number": 77
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.78, Page number: 175<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the series such as 1+2+3+ .. i. Using do-while loop. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 1\n",
-      "s = 0\n",
-      "\n",
-      "i = int(raw_input(\"Enter a number : \"))\n",
-      "\n",
-      "#Calculation and Result\n",
-      "#There is no do-while loop in python\n",
-      "while a <= i:\n",
-      "    sys.stdout.write(\"%d + \"%(a))\n",
-      "    s = s + a\n",
-      "    a += 1\n",
-      "    \n",
-      "sys.stdout.write(\"\\b\\bs = %d\"%(s))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 + 2 + 3 + 4 + 5 + \b\bs = 15"
-       ]
-      }
-     ],
-     "prompt_number": 78
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 6.79, Page number: 176<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use while statement in do-while loop and print values from 1 to 5\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = 0\n",
-      "\n",
-      "#There is no do-while loop in python\n",
-      "while x < 5:\n",
-      "    x += 1\n",
-      "    sys.stdout.write(\"\\t%d\"%(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t1\t2\t3\t4\t5"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter7.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter7.ipynb
deleted file mode 100755
index ac82414f..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter7.ipynb
+++ /dev/null
@@ -1,3644 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 7: Arrays<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.1, Page number: 186<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print size of various data types using arrays.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = [0 for i in range(0,10)]\n",
-      "c = ['0' for i in range(0,10)]\n",
-      "l = [sys.maxint+5 for i in range(0,10)]\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The type 'int' requires %d Bytes\"%(sys.getsizeof(int)))\n",
-      "sys.stdout.write(\"\\nThe type 'char' requires %d Bytes\"%(sys.getsizeof(str)))\n",
-      "sys.stdout.write(\"\\nThe type 'long' requires %d Bytes\"%(sys.getsizeof(long)))\n",
-      "sys.stdout.write(\"\\n%d memory locations are reserved for ten 'int' elements\"%(sys.getsizeof(i)))\n",
-      "sys.stdout.write(\"\\n%d memory locations are reserved for ten 'int' elements\"%(sys.getsizeof(c)))\n",
-      "sys.stdout.write(\"\\n%d memory locations are reserved for ten 'int' elements\"%(sys.getsizeof(l)))\n",
-      "\n",
-      "#python data types are different from C"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The type 'int' requires 436 Bytes\n",
-        "The type 'char' requires 436 Bytes\n",
-        "The type 'long' requires 436 Bytes\n",
-        "12 memory locations are reserved for ten 'int' elements\n",
-        "100 memory locations are reserved for ten 'int' elements\n",
-        "100 memory locations are reserved for ten 'int' elements"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.2, Page number: 187<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display character array with their address.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "name = ['A','R','R','A','Y']\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nCharacter Memory Location\\n\")\n",
-      "while i < len(name):\n",
-      "    sys.stdout.write(\"\\n[%c]\\t\\t[%d]\"%(name[i],id(name[i])))\n",
-      "    i += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Character Memory Location\n",
-        "\n",
-        "[A]\t\t[31712968]\n",
-        "[R]\t\t[31527832]\n",
-        "[R]\t\t[31527832]\n",
-        "[A]\t\t[31712968]\n",
-        "[Y]\t\t[31712248]"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.3, Page number: 188<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find sum of even and odd numbers from 1 to 10.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "sumo = 0\n",
-      "sume = 0\n",
-      "i = 0\n",
-      "a = -1\n",
-      "b = -1\n",
-      "\n",
-      "odd = [0 for i in range(0,5)]\n",
-      "even = [0 for i in range(0,5)]\n",
-      "\n",
-      "#Calculation\n",
-      "for i in range(1,10+1):\n",
-      "    if i%2 == 0:\n",
-      "        a += 1\n",
-      "        even[a] = i\n",
-      "    else:\n",
-      "        b += 1\n",
-      "        odd[b] = i\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n\\tEven  \\t\\tOdd\")\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"\\n\\t %d\\t\\t %d\"%(even[i],odd[i]))\n",
-      "    sume = sume + even[i]\n",
-      "    sumo = sumo + odd[i]\n",
-      "\n",
-      "sys.stdout.write(\"\\n\\t==========================\\n\")\n",
-      "sys.stdout.write(\"Addition : %d %14d\"%(sume,sumo))\n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\tEven  \t\tOdd\n",
-        "\t 2\t\t 1\n",
-        "\t 4\t\t 3\n",
-        "\t 6\t\t 5\n",
-        "\t 8\t\t 7\n",
-        "\t 10\t\t 9\n",
-        "\t==========================\n",
-        "Addition : 30             25"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.4, Page number: 189<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find sum of even numbers and product of odd numbers from 5 input values.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 0\n",
-      "m = 1\n",
-      "num = [0 for i in range(0,5)]\n",
-      "\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"\\nEnter Number [%d] : \"%(i+1))\n",
-      "    num[i] = int(raw_input(\"\"))\n",
-      "\n",
-      "#Calculation and Result\n",
-      "sys.stdout.write(\"\\n==================================\")\n",
-      "for i in range(0,5):\n",
-      "    if num[i]%2 == 0:\n",
-      "        sys.stdout.write(\"\\nEven Number : %d\"%(num[i]))\n",
-      "        a = a + num[i]\n",
-      "    else:\n",
-      "        sys.stdout.write(\"\\nOdd Number : %d\"%(num[i]))\n",
-      "        m = m * num[i]\n",
-      "\n",
-      "sys.stdout.write(\"\\n==================================\")\n",
-      "sys.stdout.write(\"\\nAddition of Even Numbers : %d\"%(a))\n",
-      "sys.stdout.write(\"\\nProduct of Odd Numbers : %d\"%(m))\n",
-      "sys.stdout.write(\"\\n==================================\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Number [1] : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Number [2] : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Number [3] : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Number [4] : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Number [5] : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "==================================\n",
-        "Odd Number : 1\n",
-        "Even Number : 2\n",
-        "Odd Number : 3\n",
-        "Even Number : 4\n",
-        "Odd Number : 5\n",
-        "==================================\n",
-        "Addition of Even Numbers : 6\n",
-        "Product of Odd Numbers : 15\n",
-        "=================================="
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.5, Page number: 190<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print string in the reverse order.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "s = ['0' for i in range(0,15)]\n",
-      "\n",
-      "s = raw_input(\"Enter a String : \\n\")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Reverse String : \\n\")\n",
-      "\n",
-      "#Since there is no null terminating character for string in python, length of string is taken to give the index range\n",
-      "#index range is given -1 to include the index 0 also.\n",
-      "\n",
-      "for i in range(len(s)-1,-1,-1):                       \n",
-      "    sys.stdout.write(\"%c\"%s[i])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a String : \n",
-        "HELLO\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reverse String : \n",
-        "OLLEH"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.6, Page number: 191<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Detect the occurrence of a character in a given string.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "s = ['0' for i in range(0,15)]\n",
-      "c = 0\n",
-      "\n",
-      "s = raw_input(\"Enter a String : \")\n",
-      "f = raw_input(\"Enter a Character to Find :\")\n",
-      "\n",
-      "#Calculation\n",
-      "for i in range(0,len(s)):\n",
-      "    if s[i] == f:\n",
-      "        c += 1\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"The Character (%c) in a String (%s) occurs %d times.\"%(f,s,c))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a String : programmer\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Character to Find :r\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Character (r) in a String (programmer) occurs 3 times."
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.7, Page number: 192<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the elements of two arrays in two separate columns and add their\n",
-      "#corresponding elements and display in the 3rd column.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "num = [24,34,12,44,56,17]\n",
-      "num1 = [12,24,35,78,85,22]\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Element of Array 1st - 2nd Array  Addition\\n\")\n",
-      "for i in range(0,5+1):\n",
-      "    sys.stdout.write(\"\\n\\t\\t%d  +  %d  =   %d\"%(num[i],num1[i],num[i]+num1[i]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Element of Array 1st - 2nd Array  Addition\n",
-        "\n",
-        "\t\t24  +  12  =   36\n",
-        "\t\t34  +  24  =   58\n",
-        "\t\t12  +  35  =   47\n",
-        "\t\t44  +  78  =   122\n",
-        "\t\t56  +  85  =   141\n",
-        "\t\t17  +  22  =   39"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.8, Page number: 192<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find sum of 3 numbers by entering as character type.\n",
-      "\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "real = [['0' for i in range(0,6)] for i in range(0,6)]\n",
-      "ima = [['0' for i in range(0,6)] for i in range(0,6)]\n",
-      "\n",
-      "for l in range(0,3):\n",
-      "    c = int(raw_input(\"Enter Number : \"))\n",
-      "    r = c/255\n",
-      "    i = c%255\n",
-      "    real[l][5] = r\n",
-      "    ima[l][5] = i\n",
-      "\n",
-      "c = 0\n",
-      "\n",
-      "for l in range(0,3):\n",
-      "    c = c + real[l][5]*255 + ima[l][5]\n",
-      "\n",
-      "sys.stdout.write(\"\\nSum of 3 Numbers : %3ld\"%(c))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number : 3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of 3 Numbers :  12"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.9, Page number: 193<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Accept any string upto 15 characters.  Display the elements of string with their\n",
-      "#element numbers in a separate column.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "#name = ['0' for i in range(0,15)]\n",
-      "name = raw_input(\"Enter your name : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Element no. & Character\\n\")\n",
-      "for i in range(0,len(name)):                              #There is no null termination character for string in python\n",
-      "    sys.stdout.write(\"\\n%d \\t %c\"%(i,name[i])) \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your name : amit\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Element no. & Character\n",
-        "\n",
-        "0 \t a\n",
-        "1 \t m\n",
-        "2 \t i\n",
-        "3 \t t"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.10, Page number: 194<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display names of days of a week \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "day = [0 for i in range(0,7)]\n",
-      "\n",
-      "sys.stdout.write(\"Enter numbers between 1 to 7 : \")\n",
-      "for i in range(0,6+1):\n",
-      "    day[i] = int(raw_input(\"\"))\n",
-      "\n",
-      "#Result\n",
-      "#There is no switch statement in python. use functions and dictionary or if statement instead.\n",
-      "for i in range(0,6+1):\n",
-      "    if day[i] == 1:\n",
-      "        sys.stdout.write(\"\\n%dst day of week is Sunday\"%(day[i]))\n",
-      "    else:\n",
-      "        if day[i] == 2:\n",
-      "            sys.stdout.write(\"\\n%dnd day of week is Monday\"%(day[i]))\n",
-      "        else:\n",
-      "            if day[i] == 3:\n",
-      "                sys.stdout.write(\"\\n%drd day of week is Tuesday\"%(day[i]))\n",
-      "            else:\n",
-      "                if day[i] == 4:\n",
-      "                    sys.stdout.write(\"\\n%dth day of week is Wednesday\"%(day[i]))\n",
-      "                else:\n",
-      "                    if day[i] == 5:\n",
-      "                        sys.stdout.write(\"\\n%dth day of week is Thursday\"%(day[i]))\n",
-      "                    else:\n",
-      "                        if day[i] == 6:\n",
-      "                            sys.stdout.write(\"\\n%dth day of week is Friday\"%(day[i]))\n",
-      "                        else:\n",
-      "                            if day[i] == 7:\n",
-      "                                sys.stdout.write(\"\\n%dth day of week is Saturday\"%(day[i]))\n",
-      "                            else:\n",
-      "                                sys.stdout.write(\"\\n%dth is invalid day.\"%(day[i]))\n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers between 1 to 7 : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1st day of week is Sunday\n",
-        "3rd day of week is Tuesday\n",
-        "2nd day of week is Monday\n",
-        "4th day of week is Wednesday\n",
-        "5th day of week is Thursday\n",
-        "7th day of week is Saturday\n",
-        "8th is invalid day."
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.11, Page number: 195<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the contents of two arrays where the 1st array should contain the string and 2nd numerical numbers.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "city = ['N','A','N','D','E','D']\n",
-      "pin = [4,3,1,6,0,3]\n",
-      "\n",
-      "#Result\n",
-      "for i in city:\n",
-      "    sys.stdout.write(\"%c\"%(i))\n",
-      "\n",
-      "sys.stdout.write(\" - \")\n",
-      "for i in pin:\n",
-      "    sys.stdout.write(\"%d\"%(i))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "NANDED - 431603"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.12, Page number: 195<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the number of days of different months of year.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "month = [31,28,31,30,31,30,31,31,30,31,30,31]\n",
-      "\n",
-      "#Result\n",
-      "for i in range(0,11+1):\n",
-      "    sys.stdout.write(\"\\nMonth [%d] of a year contains %d days.\"%(i+1,month[i]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Month [1] of a year contains 31 days.\n",
-        "Month [2] of a year contains 28 days.\n",
-        "Month [3] of a year contains 31 days.\n",
-        "Month [4] of a year contains 30 days.\n",
-        "Month [5] of a year contains 31 days.\n",
-        "Month [6] of a year contains 30 days.\n",
-        "Month [7] of a year contains 31 days.\n",
-        "Month [8] of a year contains 31 days.\n",
-        "Month [9] of a year contains 30 days.\n",
-        "Month [10] of a year contains 31 days.\n",
-        "Month [11] of a year contains 30 days.\n",
-        "Month [12] of a year contains 31 days."
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.13, Page number: 197<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the number of days of given month of a year.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "month = [1,3,5,7,8,10,12,4,6,9,11,2]\n",
-      "\n",
-      "mn = int(raw_input(\"Enter Number of Month : \"))\n",
-      "\n",
-      "for i in range(0,11+1):\n",
-      "    if mn == month[i]:                #There is no goto statement in python\n",
-      "        if i+1 == 12:\n",
-      "            sys.stdout.write(\"Month (%d) Contains 28 days.\"%(month[i]))\n",
-      "        if i+1 < 8:\n",
-      "            sys.stdout.write(\"Month (%d) Contains 31 days.\"%(month[i]))\n",
-      "        if i+1 > 7 and i+1 != 12:\n",
-      "            sys.stdout.write(\"Month (%d) Contains 30 days.\"%(month[i]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number of Month : 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Month (2) Contains 28 days."
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.14, Page number: 198<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the average sales of an item out of 12 months sale.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "#Since sum is a built in function in python, sum1 is used instead of sum.\n",
-      "sum1 = 0\n",
-      "avg = 0\n",
-      "\n",
-      "sys.stdout.write(\"Enter Month No.- Sale of an Item/month\\n\")\n",
-      "\n",
-      "item = [0 for i in range(0,11+1)]\n",
-      "for sale in range(0,11+1):\n",
-      "    item[sale] = int(raw_input(\"\\t%d = \"%(sale+1)))\n",
-      "\n",
-      "#Calculation\n",
-      "for sale in range(0,11+1):\n",
-      "    sum1 = sum1 + item[sale]\n",
-      "\n",
-      "avg = sum1/12.0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAverage Sale if an item/month = %f\"%(avg))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Month No.- Sale of an Item/month\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t1 = 125\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t2 = 225\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t3 = 325\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t4 = 425\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t5 = 525\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t6 = 625\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t7 = 725\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t8 = 825\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t9 = 925\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t10 = 500\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t11 = 600\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\t12 = 700\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Average Sale if an item/month = 543.750000"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.15, Page number: 199<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the similar elements in an array and find the occurrence of similar\n",
-      "#numbers for number of times.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "k = 0\n",
-      "j = 1\n",
-      "\n",
-      "item = [0 for i in range(0,4+1)]\n",
-      "sys.stdout.write(\"\\nEnter Numbers.\\n\")\n",
-      "for i in range(0,4+1):\n",
-      "    item[i] = int(raw_input(\"%d = \"%(i)))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "for j in range(0,4+1):\n",
-      "    for i in range(j+1,4+1):\n",
-      "        if item[j] == item[i]:\n",
-      "            sys.stdout.write(\"\\n%d %d Elements are same.\"%(j,i))\n",
-      "            k += 1\n",
-      "\n",
-      "if k == 10:\n",
-      "    sys.stdout.write(\"\\nAll elements are same\")\n",
-      "else:\n",
-      "    if k == 0:\n",
-      "        sys.stdout.write(\"\\nNo Elements are same.\")\n",
-      "    else:\n",
-      "        sys.stdout.write(\"\\nElements Contains Double Numbers.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Enter Numbers.\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0 = 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 = 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2 = 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3 = 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4 = 9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "0 3 Elements are same.\n",
-        "Elements Contains Double Numbers."
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.16, Page number: 200<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Calculate and display total cost of 4 models of Pentium PCs.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "pccode = [1,2,3,4]\n",
-      "t = 0\n",
-      "price = [25000,30000,35000,40000]\n",
-      "stock = [25,20,15,20]\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nStock & Total Cost Details\")\n",
-      "sys.stdout.write(\"\\n========================================================\")\n",
-      "sys.stdout.write(\"\\nModel\\t\\tQty.\\tRate(Rs.)\\tTotal Value\")\n",
-      "sys.stdout.write(\"\\n========================================================\")\n",
-      "\n",
-      "for i in range(0,3+1):\n",
-      "    sys.stdout.write(\"\\nPentium%d\\t%d\\t%8ld  %15ld\"%(pccode[i],stock[i],price[i],price[i]*stock[i]))\n",
-      "    t = t + price[i] * stock[i]\n",
-      "\n",
-      "sys.stdout.write(\"\\n========================================================\")\n",
-      "sys.stdout.write(\"\\nTotal Value of All PCs in Rs. %ld\"%(t))\n",
-      "sys.stdout.write(\"\\n========================================================\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Stock & Total Cost Details\n",
-        "========================================================\n",
-        "Model\t\tQty.\tRate(Rs.)\tTotal Value\n",
-        "========================================================\n",
-        "Pentium1\t25\t   25000           625000\n",
-        "Pentium2\t20\t   30000           600000\n",
-        "Pentium3\t15\t   35000           525000\n",
-        "Pentium4\t20\t   40000           800000\n",
-        "========================================================\n",
-        "Total Value of All PCs in Rs. 2550000\n",
-        "========================================================"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.17, Page number: 201<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the given message by using putc() and stdout() functions.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "msg = \"C is Easy\"\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "#There is no putc function and null termination in python.\n",
-      "while i < len(msg):\n",
-      "    sys.stdout.write(\"%c\"%(msg[i]))\n",
-      "    i += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C is Easy"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.18, Page number: 202<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the text through keyboard and display it by using\n",
-      "#stdin and stdout streams.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "ch = raw_input(\"Input a Text : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The Text Entered was\\n\")\n",
-      "\n",
-      "for i in ch:\n",
-      "    sys.stdout.write(\"%c\"%(i))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input a Text : Hello World\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Text Entered was\n",
-        "Hello World"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.19, Page number: 202<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sort the given strings alphabetically.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = raw_input(\"Enter Text Below : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Sorted Text Below : \\n\")\n",
-      "for i in range(65,90+1):\n",
-      "    for j in range(0,len(text)):\n",
-      "        if text[j] == chr(i).upper() or text[j] == chr(i).lower():\n",
-      "            sys.stdout.write(\"%c\"%(text[j]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Below : Hello\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorted Text Below : \n",
-        "eHllo"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.20, Page number: 203<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Arrange the numbers in increasing and decreasing order\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "sys.stdout.write(\"Enter ten numbers : \")\n",
-      "a = [0 for i in range(0,10)]\n",
-      "\n",
-      "for i in range(0,10):\n",
-      "    a[i] = int(raw_input(\"\"))\n",
-      "sum1 = 0\n",
-      "\n",
-      "#Calculation\n",
-      "sum1 = sum(a)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Numbers In Ascending Order : \")\n",
-      "for i in range(0,sum1):\n",
-      "    for j in range(0,10):\n",
-      "        if i == a[j]:\n",
-      "            sys.stdout.write(\"%3d\"%(a[j]))\n",
-      "        \n",
-      "sys.stdout.write(\"\\nNumbers In Descending Order : \")\n",
-      "for i in range(sum1,0,-1):\n",
-      "    for j in range(0,10):\n",
-      "        if i == a[j]:\n",
-      "            sys.stdout.write(\"%3d\"%(a[j]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter ten numbers : "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "12\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Numbers In Ascending Order :   1  2  3  4  5  7  9  9 10 12\n",
-        "Numbers In Descending Order :  12 10  9  9  7  5  4  3  2  1"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.21, Page number: 205<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sorting in ascending order by comparison method.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "n = int(raw_input(\"Enter how many numbers to sort : \"))\n",
-      "\n",
-      "num = [0 for i in range(0,n)]\n",
-      "for i in range(0,n):\n",
-      "    num[i] = int(raw_input(\"Enter numbers # %2d  : \"%(i+1)))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The Numbers Entered through keyboard\\n\")\n",
-      "for i in range(0,n):\n",
-      "    sys.stdout.write(\"%4d\"%(num[i]))\n",
-      "for i in range(0,n-1):\n",
-      "    for j in range(i+1,n):\n",
-      "        if num[i] > num[j]:\n",
-      "            temp = num[i]\n",
-      "            num[i] = num[j]\n",
-      "            num[j] = temp\n",
-      "\n",
-      "sys.stdout.write(\"\\nThe Numbers in ascending order\\n\")\n",
-      "for i in num:\n",
-      "    sys.stdout.write(\"%4d\"%(i))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter how many numbers to sort : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers #  1  : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers #  2  : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers #  3  : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers #  4  : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter numbers #  5  : 9\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Numbers Entered through keyboard\n",
-        "   8   3   2   1   9\n",
-        "The Numbers in ascending order\n",
-        "   1   2   3   8   9"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.22, Page number: 206<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Evaluate the series contains sum of square of\n",
-      "#numbers from 1 to n.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "s = 0\n",
-      "sqr = [0 for i in range(0,15)]\n",
-      "\n",
-      "n = int(raw_input(\"Enter value of n : \"))\n",
-      "\n",
-      "for i in range(0,n):\n",
-      "    sqr[i] = pow(i+1,2)\n",
-      "    \n",
-      "for i in range(0,n):\n",
-      "    sys.stdout.write(\"%d\\n\"%(sqr[i]))\n",
-      "    s = s + sqr[i]\n",
-      "    \n",
-      "sys.stdout.write(\"Sum of square : %d\"%(s))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter value of n : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1\n",
-        "4\n",
-        "9\n",
-        "16\n",
-        "Sum of square : 30"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.23, Page number: 207<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display two dimensional array elements together with their addresses\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[0 for i in range(0,3)]for i in range(0,3)]\n",
-      "c = 1\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,3):\n",
-      "        a[i][j] = c\n",
-      "        c += 1\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"Array Elements and addresses.\\n\")\n",
-      "sys.stdout.write(\"             Col-0        Col-1         Col-2\\n\")\n",
-      "sys.stdout.write(\"========================================================\")\n",
-      "sys.stdout.write(\"\\nRow0\")\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,3):\n",
-      "        sys.stdout.write(\"      %d [%5d]\"%(a[i][j],id(a[i][j])))\n",
-      "    sys.stdout.write(\"\\nRow%d\"%(i+1))\n",
-      "\n",
-      "sys.stdout.write(\"\\r\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Array Elements and addresses.\n",
-        "             Col-0        Col-1         Col-2\n",
-        "========================================================\n",
-        "Row0      1 [6215912]      2 [6215900]      3 [6215888]\n",
-        "Row1      4 [6215876]      5 [6215864]      6 [6215852]\n",
-        "Row2      7 [6215840]      8 [6215828]      9 [6215816]\n",
-        "Row3\r"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.24, Page number: 208<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the elements of two-dimensional array.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[1,2,3],[4,5,6],[7,8,9]]\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"Elements of An Array.\\n\")\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,3):\n",
-      "        sys.stdout.write(\"%5d\"%(a[i][j]))\n",
-      "    sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements of An Array.\n",
-        "    1    2    3\n",
-        "    4    5    6\n",
-        "    7    8    9\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.25, Page number: 209<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read codenos and balance and count the number of codenos, \n",
-      "#which are having the balance less than 1000.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "j = 0\n",
-      "bal = [[0 for i in range(0,5)]for i in range(0,5)]\n",
-      "\n",
-      "for i in range(0,5):\n",
-      "    bal[i][1] = int(raw_input(\"Enter Code No & Balance : \"))\n",
-      "    bal[i][2] = int(raw_input(\"Enter Code No & Balance : \"))\n",
-      "    \n",
-      "sys.stdout.write(\"Your Entered Data : \")\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"\\n%d   %d\"%(bal[i][1],bal[i][2]))\n",
-      "    if bal[i][2] < 1000:\n",
-      "        j += 1\n",
-      "        \n",
-      "sys.stdout.write(\"\\nBalance Less than 1000 are %d\"%(j))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 900\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 800\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 1200\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 550\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Code No & Balance : 600\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your Entered Data : \n",
-        "1   900\n",
-        "2   800\n",
-        "3   1200\n",
-        "4   550\n",
-        "5   600\n",
-        "Balance Less than 1000 are 4"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.26, Page number: 210<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Accept three elements in a one dimensional array and compute addition,square\n",
-      "#and cube.  Display the results in two dimensional array.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[0 for i in range(0,3)]for i in range(0,3)]\n",
-      "b = [0 for i in range(0,3)]\n",
-      "j = 0\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    b[i] = int(raw_input(\"Enter Three Numbers : \"))\n",
-      "\n",
-      "#Calculation\n",
-      "for i in range(0,3):\n",
-      "    a[i][j] = b[i]*2\n",
-      "    a[i][j+1] = pow(b[i],2)\n",
-      "    a[i][j+2] = pow(b[i],3)\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"Number\\tAddition\\tSquare\\t\\tCube\\n\")\n",
-      "for i in range(0,3):\n",
-      "    sys.stdout.write(\"\\n%d\"%(b[i]))\n",
-      "    for j in range(0,3):\n",
-      "        sys.stdout.write(\"\\t%4d\\t\"%(a[i][j]))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three Numbers : 7\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Number\tAddition\tSquare\t\tCube\n",
-        "\n",
-        "5\t  10\t\t  25\t\t 125\t\n",
-        "\n",
-        "6\t  12\t\t  36\t\t 216\t\n",
-        "\n",
-        "7\t  14\t\t  49\t\t 343\t\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.27, Page number: 211<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read and display matrix\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
-      "row = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
-      "col = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
-      "\n",
-      "for i in range(0,row):\n",
-      "    for j in range(0,col):\n",
-      "        a[i][j]  = int(raw_input(\"Enter Elements of matrix A : \"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nThe Matrix is : \\n\")\n",
-      "for i in range(0,row):\n",
-      "    for j in range(0,col):\n",
-      "        sys.stdout.write(\"%6d\"%(a[i][j]))\n",
-      "    sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of matrix upto (10x10) A : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of matrix upto (10x10) A : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Matrix is : \n",
-        "     3     5     8\n",
-        "     4     8     5\n",
-        "     8     5     4\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.28, Page number: 212<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Transpose of matrix\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
-      "b = [[0 for i in range(0,10)]for i in range(0,10)]\n",
-      "row1 = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
-      "col1 = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
-      "\n",
-      "for i in range(0,row1):\n",
-      "    for j in range(0,col1):\n",
-      "        a[i][j]  = int(raw_input(\"Enter Elements of matrix A : \"))\n",
-      "row2 = col1\n",
-      "col2 = row1\n",
-      "\n",
-      "#Calculation\n",
-      "for i in range(0,row1):\n",
-      "    for j in range(0,col1):\n",
-      "        b[j][i] = a[i][j]\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nThe Matrix is : \\n\")\n",
-      "for i in range(0,row2):\n",
-      "    for j in range(0,col2):\n",
-      "        sys.stdout.write(\"%4d\"%(b[i][j]))\n",
-      "    sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of matrix upto (10x10) A : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of matrix upto (10x10) A : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix A : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Matrix is : \n",
-        "   3   4   8\n",
-        "   5   8   5\n",
-        "   8   5   4\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.29, Page number: 213<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Transpose of 3x3 matrix\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "mat = [[0 for i in range(0,3)]for i in range(0,3)]\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,3):\n",
-      "        mat[i][j]  = int(raw_input(\"Enter Elements of matrix \"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Transpose of the matrix\\n\")\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,3):\n",
-      "        sys.stdout.write(\"\\t%d\"%(mat[j][i]))\n",
-      "    sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of matrix 2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transpose of the matrix\n",
-        "\t5\t2\t7\n",
-        "\t8\t5\t5\n",
-        "\t9\t4\t2\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.30, Page number: 214<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Addition and subtraction of two matrices \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
-      "b = [[0 for i in range(0,10)]for i in range(0,10)]\n",
-      "\n",
-      "r1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
-      "c1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
-      "\n",
-      "for i in range(0,r1):\n",
-      "    for j in range(0,c1):\n",
-      "        a[i][j]  = int(raw_input(\"Enter Elements of Matrix A:\"))\n",
-      "\n",
-      "for i in range(0,r1):\n",
-      "    for j in range(0,c1):\n",
-      "        b[i][j]  = int(raw_input(\"Enter Elements of Matrix B:\"))\n",
-      "\n",
-      "#Calculation and Result\n",
-      "sys.stdout.write(\"\\nMatrix Addition\\n\")\n",
-      "for i in range(0,r1):\n",
-      "    for j in range(0,c1):\n",
-      "        sys.stdout.write(\"%5d\"%(a[i][j]+b[i][j]))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "\n",
-      "sys.stdout.write(\"\\nMatrix Subtraction\\n\")\n",
-      "for i in range(0,r1):\n",
-      "    for j in range(0,c1):\n",
-      "        sys.stdout.write(\"%5d\"%(a[i][j]-b[i][j]))\n",
-      "    sys.stdout.write(\"\\n\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of Matrix A & B upto 10x103\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of Matrix A & B upto 10x103\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Matrix Addition\n",
-        "    5    8   13\n",
-        "    2   14   12\n",
-        "    8   16    6\n",
-        "\n",
-        "Matrix Subtraction\n",
-        "    3    2    3\n",
-        "    2    4    4\n",
-        "   -4    2    2\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.31, Page number: 216<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Multiplication of two matrices \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
-      "b = [[0 for i in range(0,10)]for i in range(0,10)]\n",
-      "\n",
-      "r1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
-      "c1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
-      "\n",
-      "for i in range(0,r1):\n",
-      "    for j in range(0,c1):\n",
-      "        a[i][j]  = int(raw_input(\"Enter Elements of Matrix A:\"))\n",
-      "\n",
-      "for i in range(0,r1):\n",
-      "    for j in range(0,c1):\n",
-      "        b[i][j]  = int(raw_input(\"Enter Elements of Matrix B:\"))\n",
-      "\n",
-      "#Calculation and Result\n",
-      "sys.stdout.write(\"\\nMultiplication of corresponding elements\\n\")\n",
-      "for i in range(0,r1):\n",
-      "    for j in range(0,c1):\n",
-      "        sys.stdout.write(\"%5d\"%(a[i][j]*b[i][j]))\n",
-      "    sys.stdout.write(\"\\n\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of Matrix A & B upto 10x103\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Order of Matrix A & B upto 10x103\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:9\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix A:4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:1\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:3\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:0\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:5\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:6\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:7\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Elements of Matrix B:2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Multiplication of corresponding elements\n",
-        "    4   15   40\n",
-        "    0   45   32\n",
-        "   12   63    8\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.32, Page number: 217<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the Quantity & Price of various Pentium models using an array and\n",
-      "#Compute the total cost of all models.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "pc = [[0 for i in range(0,4)]for i in range(0,4)]\n",
-      "t = 0\n",
-      "\n",
-      "for i in range(0,4):\n",
-      "    for j in range(0,2):\n",
-      "        pc[i][j] = int(raw_input(\"Enter Qty. & Price for Pentium %d : \"%(i+1)))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"=====================================================\")\n",
-      "sys.stdout.write(\"\\nModel\\tQty.\\tRate(Rs.)\\tTotal Value\")\n",
-      "sys.stdout.write(\"\\n=====================================================\")\n",
-      "for i in range(0,4):\n",
-      "    sys.stdout.write(\"\\nPentium %d   %2ld %6ld %15ld\"%(i+1,pc[i][0],pc[i][1],pc[i][0]*pc[i][1]))\n",
-      "    t = t + pc[i][0] * pc[i][1]\n",
-      "\n",
-      "sys.stdout.write(\"\\n=====================================================\")\n",
-      "sys.stdout.write(\"\\nTotal Value of All PCs in Rs. %ld\"%(t))\n",
-      "sys.stdout.write(\"\\n=====================================================\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 1 : 25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 1 : 25000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 2 : 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 2 : 40000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 3 : 15\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 3 : 35000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 4 : 20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Qty. & Price for Pentium 4 : 40000\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "=====================================================\n",
-        "Model\tQty.\tRate(Rs.)\tTotal Value\n",
-        "=====================================================\n",
-        "Pentium 1   25  25000          625000\n",
-        "Pentium 2   20  40000          800000\n",
-        "Pentium 3   15  35000          525000\n",
-        "Pentium 4   20  40000          800000\n",
-        "=====================================================\n",
-        "Total Value of All PCs in Rs. 2750000\n",
-        "====================================================="
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.33, Page number: 218<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the capacity of HD, it's price and quantity available in the form of an array\n",
-      "# and compute the cost of HD\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "hd = [[0 for i in range(0,4)]for i in range(0,4)]\n",
-      "t = 0\n",
-      "\n",
-      "for j in range(0,4):\n",
-      "    for i in range(0,3):\n",
-      "        hd[i][j] = int(raw_input(\"Enter Capacity, Price & Qty.:\"))\n",
-      "        \n",
-      "sys.stdout.write(\"====================================================================\\n\")\n",
-      "sys.stdout.write(\"HD Capacity\\tGB Price Rs.\\tQuantity Total Value Rs.\")\n",
-      "sys.stdout.write(\"\\n====================================================================\\n\")\n",
-      "\n",
-      "for j in range(0,4):\n",
-      "    for i in range(0,3):\n",
-      "        sys.stdout.write(\"%2ld\"%(hd[i][j]))\n",
-      "        sys.stdout.write(\"\\t\\t\")\n",
-      "        if i == 2:\n",
-      "            sys.stdout.write(\"%5ld\"%(hd[i-1][j]*hd[i][j]))\n",
-      "            t = t + hd[i-1][j] * hd[i][j]\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    \n",
-      "sys.stdout.write(\"====================================================================\")\n",
-      "sys.stdout.write(\"\\nTotal Value Rs. %37ld\"%(t))\n",
-      "sys.stdout.write(\"\\n====================================================================\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:10\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:8000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:25\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:12000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:40\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:15000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:15\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:90\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:20000\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Capacity, Price & Qty.:10\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "====================================================================\n",
-        "HD Capacity\tGB Price Rs.\tQuantity Total Value Rs.\n",
-        "====================================================================\n",
-        "10\t\t8000\t\t25\t\t200000\n",
-        "20\t\t12000\t\t20\t\t240000\n",
-        "40\t\t15000\t\t15\t\t225000\n",
-        "90\t\t20000\t\t10\t\t200000\n",
-        "====================================================================\n",
-        "Total Value Rs.                                865000\n",
-        "===================================================================="
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.34, Page number: 220<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the names of the cities with their base addresses.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "city = [\"Mumbai\",\"Chennai\",\"Kolkata\",\"Pune\",\"Delhi\"]\n",
-      "\n",
-      "#Result\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"Base Address = %u\"%(id(city[i])))   #Memory address will differ in different systems\n",
-      "    sys.stdout.write(\"\\tString = %s\\n\"%(city[i]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base Address = 60549024\tString = Mumbai\n",
-        "Base Address = 60550048\tString = Chennai\n",
-        "Base Address = 60550592\tString = Kolkata\n",
-        "Base Address = 60551104\tString = Pune\n",
-        "Base Address = 60550240\tString = Delhi\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.35, Page number: 220<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the binary bits corresponding to Hexadecimal numbers from\n",
-      "#0 to E and attach an odd parity to Hex numbers and display them.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "bin = [[0 for i in range(0,16)]for i in range(0,16)]\n",
-      "a = 0\n",
-      "c = 0\n",
-      "\n",
-      "for x in range(0,16):\n",
-      "    y = x\n",
-      "    for a in range(0,4):\n",
-      "        bin[x][a] = y%2\n",
-      "        y = y/2\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nInput Bits\\t Parity Bits\")\n",
-      "sys.stdout.write(\"\\n===================================\")\n",
-      "\n",
-      "for x in range(0,16):\n",
-      "    c = 0\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    for y in range(3,0-1,-1):\n",
-      "        sys.stdout.write(\"%3d\"%(bin[x][y]))\n",
-      "        if bin[x][y] == 1:\n",
-      "            c += 1\n",
-      "    if c%2 == 0:\n",
-      "        sys.stdout.write(\"%7d\"%(1))\n",
-      "    else:\n",
-      "        sys.stdout.write(\"%7d\"%(0))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Input Bits\t Parity Bits\n",
-        "===================================\n",
-        "  0  0  0  0      1\n",
-        "  0  0  0  1      0\n",
-        "  0  0  1  0      0\n",
-        "  0  0  1  1      1\n",
-        "  0  1  0  0      0\n",
-        "  0  1  0  1      1\n",
-        "  0  1  1  0      1\n",
-        "  0  1  1  1      0\n",
-        "  1  0  0  0      0\n",
-        "  1  0  0  1      1\n",
-        "  1  0  1  0      1\n",
-        "  1  0  1  1      0\n",
-        "  1  1  0  0      1\n",
-        "  1  1  0  1      0\n",
-        "  1  1  1  0      0\n",
-        "  1  1  1  1      1"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.36, Page number: 222<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert binary to gray codes.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "bin1 = [[0 for i in range(0,16)]for i in range(0,16)]\n",
-      "a = 0\n",
-      "c = 0\n",
-      "\n",
-      "#Calculation\n",
-      "for x in range(0,16):\n",
-      "    y = x\n",
-      "    for a in range(0,4):\n",
-      "        bin1[x][a] = y%2\n",
-      "        y = y/2\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nBinary Bits\\tGray Bits\")\n",
-      "sys.stdout.write(\"\\n==================================\")\n",
-      "\n",
-      "for x in range(0,16):\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    for y in range(3,0-1,-1):\n",
-      "        sys.stdout.write(\"%3d\"%(bin1[x][y]))\n",
-      "    sys.stdout.write(\"%5d   %2d   %2d   %2d\"%(bin1[x][3],(bin1[x][3]^bin1[x][2]),(bin1[x][2]^bin1[x][1]),(bin1[x][1]^bin1[x][0])))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Binary Bits\tGray Bits\n",
-        "==================================\n",
-        "  0  0  0  0    0    0    0    0\n",
-        "  0  0  0  1    0    0    0    1\n",
-        "  0  0  1  0    0    0    1    1\n",
-        "  0  0  1  1    0    0    1    0\n",
-        "  0  1  0  0    0    1    1    0\n",
-        "  0  1  0  1    0    1    1    1\n",
-        "  0  1  1  0    0    1    0    1\n",
-        "  0  1  1  1    0    1    0    0\n",
-        "  1  0  0  0    1    1    0    0\n",
-        "  1  0  0  1    1    1    0    1\n",
-        "  1  0  1  0    1    1    1    1\n",
-        "  1  0  1  1    1    1    1    0\n",
-        "  1  1  0  0    1    0    1    0\n",
-        "  1  1  0  1    1    0    1    1\n",
-        "  1  1  1  0    1    0    0    1\n",
-        "  1  1  1  1    1    0    0    0"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.37, Page number: 224<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#working of three dimensional array.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "array_3d = [[[0 for i in range(0,3)]for i in range(0,3)]for i in range(0,3)]\n",
-      "\n",
-      "for a in range(0,3):\n",
-      "    for b in range(0,3):\n",
-      "        for c in range(0,3):\n",
-      "            array_3d[a][b][c] = a+b+c\n",
-      "\n",
-      "#Result\n",
-      "for a in range(0,3):\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    for b in range(0,3):\n",
-      "        for c in range(0,3):\n",
-      "            sys.stdout.write(\"%3d\"%(array_3d[a][b][c]))\n",
-      "        sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "  0  1  2\n",
-        "  1  2  3\n",
-        "  2  3  4\n",
-        "\n",
-        "  1  2  3\n",
-        "  2  3  4\n",
-        "  3  4  5\n",
-        "\n",
-        "  2  3  4\n",
-        "  3  4  5\n",
-        "  4  5  6\n"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.38, Page number: 225<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#working of four dimensional array\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "array_4d = [[[[0 for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]\n",
-      "\n",
-      "for a in range(0,2):\n",
-      "    for b in range(0,2):\n",
-      "        for c in range(0,2):\n",
-      "            for d in range(0,2):\n",
-      "                array_4d[a][b][c][d] = a+b+c+d\n",
-      "                \n",
-      "#Result\n",
-      "for a in range(0,2):\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    for b in range(0,2):\n",
-      "        for c in range(0,2):\n",
-      "            for d in range(0,2):\n",
-      "                sys.stdout.write(\"%3d\"%(array_4d[a][b][c][d]))\n",
-      "            sys.stdout.write(\"\\n\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "  0  1\n",
-        "  1  2\n",
-        "  1  2\n",
-        "  2  3\n",
-        "\n",
-        "  1  2\n",
-        "  2  3\n",
-        "  2  3\n",
-        "  3  4\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.39, Page number: 226<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read the quantity & rate of certain items using multidimensional\n",
-      "#array.  Calculate total cost by multiplying quantity & rate and offer 2%\n",
-      "#discount on it & display the net amount.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "m = [[[[0 for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]\n",
-      "\n",
-      "for a in range(0,2):\n",
-      "    for b in range(0,2):\n",
-      "        for c in range(0,2):\n",
-      "            for d in range(0,1):\n",
-      "                if a == 0:\n",
-      "                    m[a][b][c][d] = int(raw_input(\"Enter Quantity & Rate\\na=%d b=%d c=%d d=%d\"%(a,b,c,d)))\n",
-      "                    m[a][b][c][d+1] = int(raw_input(\"Enter Quantity & Rate\\na=%d b=%d c=%d d=%d\"%(a,b,c,d)))\n",
-      "                else:\n",
-      "                    m[a][b][c][d] = m[a-1][b][c][d] * m[a-1][b][c][d+1]\n",
-      "                    m[a][b][c][d+1] = m[a-1][b][c][d] * m[a-1][b][c][d+1] *2/100\n",
-      "            \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n============================================================\\n\")\n",
-      "sys.stdout.write(\"Quantity\\tRate\\tAmoount\\tDiscount(@2%)\\tNet Amount\\n\")\n",
-      "sys.stdout.write(\"============================================================\\n\")\n",
-      "\n",
-      "for a in range(0,1):\n",
-      "    for b in range(0,2):\n",
-      "        for c in range(0,2):\n",
-      "            for d in range(0,1):\n",
-      "                sys.stdout.write(\"\\n%10ld %10ld\"%(m[a][b][c][d],m[a][b][c][d+1]))\n",
-      "                sys.stdout.write(\"%8ld.00 %6ld.00 %12ld.00\"%(m[a+1][b][c][d],m[a+1][b][c][d+1],m[a+1][b][c][d] - m[a+1][b][c][d+1]))\n",
-      "            \n",
-      "sys.stdout.write(\"\\n============================================================\")\n",
-      "                "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=0 c=0 d=025\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=0 c=0 d=050\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=0 c=1 d=030\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=0 c=1 d=060\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=1 c=0 d=035\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=1 c=0 d=070\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=1 c=1 d=040\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Quantity & Rate\n",
-        "a=0 b=1 c=1 d=075\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "============================================================\n",
-        "Quantity\tRate\tAmoount\tDiscount(@2%)\tNet Amount\n",
-        "============================================================\n",
-        "\n",
-        "        25         50    1250.00     25.00         1225.00\n",
-        "        30         60    1800.00     36.00         1764.00\n",
-        "        35         70    2450.00     49.00         2401.00\n",
-        "        40         75    3000.00     60.00         2940.00\n",
-        "============================================================"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.40, Page number: 227<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read string from a character array.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "in1 = raw_input(\"\")   #Since in is a keyword in python, in1 is used\n",
-      "\n",
-      "#Result\n",
-      "#There is no sscanf() function in python\n",
-      "out = in1\n",
-      "sys.stdout.write(\"%s\\n\"%(out))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "HELLO\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.41, Page number: 228<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Read integers in character array, convert and assign them to integer\n",
-      "#variable.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable initialization\n",
-      "in1 = raw_input(\"Enter Integers : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Value of int x : %s\"%(in1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Integers : 123\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of int x : 123"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.42, Page number: 229<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#use of sprintf()\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 10\n",
-      "c = 'C'\n",
-      "p = 3.14\n",
-      "\n",
-      "spf = str(\"%d %c %.2f\"%(a,c,p))        #There is no sprintf() function in python\n",
-      "sys.stdout.write(\"\\n%s\"%(spf))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "10 C 3.14"
-       ]
-      }
-     ],
-     "prompt_number": 61
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 7.43, Page number: 229<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Complete the string by filling the spaces with appropriate characters\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "name = \"Mas er ngA SIC\"\n",
-      "name1 = ['' for i in range(0,20)]\n",
-      "x = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Mastering ANSI C\\n\")\n",
-      "\n",
-      "while x < len(name):      #There is no null terminating character in python\n",
-      "    if name[x] == ' ':\n",
-      "        n = raw_input(\"\")\n",
-      "        name1[x] = n\n",
-      "    else:\n",
-      "        name1[x] = name[x]\n",
-      "    x += 1\n",
-      "    \n",
-      "sys.stdout.write(\"%s\"%(name1))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mastering ANSI C\n",
-        "['M', 'a', 's', 't', 'e', 'r', 'i', 'n', 'g', 'A', 'N', 'S', 'I', 'C', '', '', '', '', '', '']"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter8.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter8.ipynb
deleted file mode 100755
index 3243f3f8..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter8.ipynb
+++ /dev/null
@@ -1,2750 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 8: Working with Strings & Standard Functions<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.1, Page number: 236<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display string without considering NULL character\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "name1 = 'SANJAY'\n",
-      "#There is no null termination character in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Name1 = %s\"%(name1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name1 = SANJAY"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.2, Page number: 236<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display successive strings\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "name1 = 'SANJAY'\n",
-      "name2 = 'SANJAY'\n",
-      "#There is no null termination character in python\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Name1 = %s\"%name1)\n",
-      "sys.stdout.write(\"\\nName2 = %s\"%(name2))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Name1 = SANJAY\n",
-        "Name2 = SANJAY"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.3, Page number: 237<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print \"WELCOME\" by using different formats of initialization \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "#in python, string initialization can be done using single quotes and double quotes\n",
-      "arr1 = 'WELCOME'\n",
-      "arr2 = \"WELCOME\"\n",
-      "arr3 = 'WELCOME'\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nArray1 = %s\"%(arr1))\n",
-      "sys.stdout.write(\"\\nArray2 = %s\"%(arr2))\n",
-      "sys.stdout.write(\"\\nArray3 = %s\"%(arr3))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Array1 = WELCOME\n",
-        "Array2 = WELCOME\n",
-        "Array3 = WELCOME"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.4, Page number: 238<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the string 'PRABHAKAR' using various format specifications\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = \"PRABHAKAR\"\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"%s\\n\"%(text))\n",
-      "sys.stdout.write(\"%.5s\\n\"%(text))\n",
-      "sys.stdout.write(\"%.8s\\n\"%(text))\n",
-      "sys.stdout.write(\"%.15s\\n\"%(text))\n",
-      "sys.stdout.write(\"%-10.4s\\n\"%(text))\n",
-      "sys.stdout.write(\"%11s\\n\"%(text))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "PRABHAKAR\n",
-        "PRABH\n",
-        "PRABHAKA\n",
-        "PRABHAKAR\n",
-        "PRAB      \n",
-        "  PRABHAKAR\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.5, Page number: 239<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the elements of the character array using while loop.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = \"HAVE A NICE DAY\"\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "#There is no null character for strings in python\n",
-      "\n",
-      "while i < 15:\n",
-      "    sys.stdout.write(\"%c\"%(text[i]))\n",
-      "    i += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "HAVE A NICE DAY"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.6, Page number: 239<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the elements of the character array.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = \"HAVE A NICE DAY\"\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "#There is no null character for strings in python\n",
-      "\n",
-      "while i < 15:\n",
-      "    sys.stdout.write(\"%c\"%(text[i]))\n",
-      "    i += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "HAVE A NICE DAY"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.7, Page number: 241<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count the number of characters in a given string\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = raw_input(\"Type Text Below\")\n",
-      "\n",
-      "#Calculation\n",
-      "len1 = len(text)    #since len in a built in function in python, len1 is used\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Length of String = %d\"%(len1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Type Text BelowHello\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Length of String = 5"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.8, Page number: 241<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Determine the length of the string\n",
-      "#and find its equivalent ASCII codes.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "\n",
-      "#Variable Initialization\n",
-      "name = raw_input(\"Enter your name : \")\n",
-      "l = len(name)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Your Name is %s & \"%(name))\n",
-      "sys.stdout.write(\"it contains %d characters\"%(l))\n",
-      "sys.stdout.write(\"\\n\\nName & it's Ascii Equivalent.\\n\")\n",
-      "sys.stdout.write(\"=====================================\\n\")\n",
-      "\n",
-      "for i in range(0,l):\n",
-      "    sys.stdout.write(\"\\n%c\\t\\t%d\"%(name[i],ord(name[i])))   #ord() function is used to get the ASCII value of a character"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter your name : SACHIN\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Your Name is SACHIN & it contains 6 characters\n",
-        "\n",
-        "Name & it's Ascii Equivalent.\n",
-        "=====================================\n",
-        "\n",
-        "S\t\t83\n",
-        "A\t\t65\n",
-        "C\t\t67\n",
-        "H\t\t72\n",
-        "I\t\t73\n",
-        "N\t\t78"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.9, Page number: 242<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Remove the occurrences of 'The' word from entered text\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "j = 0\n",
-      "line = raw_input(\"Enter Text Below.\")\n",
-      "l = len(line)\n",
-      "line2 = ['0' for i in range(0,80)]\n",
-      "i = 0\n",
-      "\n",
-      "#Calculation\n",
-      "while i < l:\n",
-      "    if (i >= l-4 or l ==3) and (line[l-4] == ' ' and line[l-3] == 't' and line[l-2] == 'h' and line[l-1] == 'e'):\n",
-      "        continue\n",
-      "    if line[i] == 't' and line[i+1] == 'h' and line[i+2] == 'e' and line[i+3] == ' ':\n",
-      "        i += 3\n",
-      "        continue\n",
-      "    else:\n",
-      "        line2[j] = line[i]\n",
-      "        j += 1\n",
-      "    i += 1\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nText With 'The' -: %s\"%(line))\n",
-      "sys.stdout.write(\"\\nText Without 'The' -: \")\n",
-      "for l in range(0,j):\n",
-      "    sys.stdout.write(\"%c\"%(line2[l]))\n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Below.Printf write data to the screen.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Text With 'The' -: Printf write data to the screen.\n",
-        "Text Without 'The' -: Printf write data to  screen."
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.10, Page number: 243<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Delete all the occurrences of vowels in a given text.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "j = 0\n",
-      "line = raw_input(\"Enter Text Below.\")\n",
-      "\n",
-      "#Calculation\n",
-      "for i in range(0,len(line)):\n",
-      "    if line[i] == 'a' or line[i] == 'e' or line[i] == 'i' or line[i] == 'o' or line[i] == 'u':\n",
-      "        continue\n",
-      "    else:\n",
-      "        line2[j] = line[i]\n",
-      "        j += 1\n",
-      "      \n",
-      "#Result\n",
-      "sys.stdout.write(\"Text With Vowels -:  %s\"%(line))\n",
-      "sys.stdout.write(\"\\nText Without Vowels -:  \")\n",
-      "for l in range(0,j):                                  #There is no null terminating character in python, so we have to specify length\n",
-      "    sys.stdout.write(\"%c\"%(line2[l])) \n",
-      "        "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Below.Have a nice day.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Text With Vowels -:  Have a nice day.\n",
-        "Text Without Vowels -:  Hv  nc dy."
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.11, Page number: 244<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the length of characteres in a given string including and excluding spaces.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 0\n",
-      "len1 = 0       #Since len is a built in function in python, len1 is used\n",
-      "ex = 0\n",
-      "\n",
-      "text = raw_input(\"Enter Text Here : \")\n",
-      "\n",
-      "#Calculation\n",
-      "while i != len(text)-1:           #There is no null terminating character in python.\n",
-      "    if text[i] == ' ':\n",
-      "        ex += 1\n",
-      "    else:\n",
-      "        len1 += 1\n",
-      "    i += 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nLengh of String Including Spaces : %d\"%(len1+ex))\n",
-      "sys.stdout.write(\"\\nLengh of String Excluding Spaces : %d\"%(len1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Here : Time is Money.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Lengh of String Including Spaces : 13\n",
-        "Lengh of String Excluding Spaces : 11"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.12, Page number: 245<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the length of entered string and display it as per output shown.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "string = raw_input(\"Enter a String\")\n",
-      "ln = len(string)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nLength of Given String : %d\"%(ln))\n",
-      "sys.stdout.write(\"\\nSequence of Characters Displayed on Screen\")\n",
-      "sys.stdout.write(\"\\n======== == ========== ========= == ======\")\n",
-      "sys.stdout.write(\"\\n\\n\")\n",
-      "\n",
-      "for c in range(0,ln-2+1):\n",
-      "    d = c + 1\n",
-      "    sys.stdout.write(\"\\t%.*s\\n\"%(d,string))\n",
-      "\n",
-      "for c in range(ln,0-1,-1):\n",
-      "    d = c + 1\n",
-      "    sys.stdout.write(\"\\t%.*s\\n\"%(d,string))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a StringHAPPY\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Length of Given String : 5\n",
-        "Sequence of Characters Displayed on Screen\n",
-        "======== == ========== ========= == ======\n",
-        "\n",
-        "\tH\n",
-        "\tHA\n",
-        "\tHAP\n",
-        "\tHAPP\n",
-        "\tHAPPY\n",
-        "\tHAPPY\n",
-        "\tHAPP\n",
-        "\tHAP\n",
-        "\tHA\n",
-        "\tH\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.13, Page number: 247<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy contents of one string to another string by using strcpy()\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "ori = raw_input(\"Enter Your Name : \")\n",
-      "\n",
-      "#Process\n",
-      "dup = ori                  #In python, string values can be simply copied using assignment operator\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Original String : %s\"%(ori))\n",
-      "sys.stdout.write(\"\\nDuplicate String : %s\"%(dup))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Name : SACHIN\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Original String : SACHIN\n",
-        "Duplicate String : SACHIN"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.14, Page number: 247<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy contents of one string to another, without strcpy() function.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "ori = raw_input(\"Enter Your Name : \")\n",
-      "\n",
-      "#Calculation\n",
-      "dup = ori                  #In python, string values can be simply copied using assignment operator\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"Original String : %s\"%(ori))\n",
-      "sys.stdout.write(\"\\nDuplicate String : %s\"%(dup))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Name : SACHIN\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Original String : SACHIN\n",
-        "Duplicate String : SACHIN"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.15, Page number: 248<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy source string to destination string upto a specified length.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "str1 = raw_input(\"Enter Source String : \")\n",
-      "str2 = raw_input(\"Enter Destination String : \")\n",
-      "n = int(raw_input(\"Enter Number of Characters to Replace in Destination String :\"))\n",
-      "\n",
-      "#Calculation\n",
-      "str2 = str1[:n] + str2[n:]         #Since python strings are not mutable, first and second array elements to be added to make a new one\n",
-      "    \n",
-      "sys.stdout.write(\"Source String -: %s\"%(str1))\n",
-      "sys.stdout.write(\"\\nDestination String -: %s\"%(str2))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Source String : wonderful\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Destination String : beautiful\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number of Characters to Replace in Destination String :6\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Source String -: wonderful\n",
-        "Destination String -: wonderful"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.16, Page number: 249<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compare the two string using stricmp() function. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "sr = raw_input(\"Enter String(1) :\")\n",
-      "tar = raw_input(\"Enter String(2) :\")\n",
-      "\n",
-      "diff = (sr == tar)           #in python, relational operators can be used directly\n",
-      "\n",
-      "if diff == 0:\n",
-      "    sys.stdout.write(\"The Two Strings are Identical.\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"The Tow Strings are Different\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String(1) :HELLO\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String(2) :hello\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Two Strings are Identical."
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.17, Page number: 250<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Perform following.\n",
-      "\n",
-      "#  1.  Display the question \"What is the Unit of Traffic?\"\n",
-      "#  2.  Accept the Answer.\n",
-      "#  3.  If the answer is wrong (other than Earlang) display \"Try again!\" & continues to answer.\n",
-      "#  4.  Otherwise, if it is correct \"Earlang\" display the message \"Answer is correct\".\n",
-      "#  5.  If the user gives correct anser in first two attempts the program will terminate.\n",
-      "#  6.  If the user fails to provide correct answer in three attempts the program itself gives the answer.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Calculation & Result\n",
-      "for i in range(0,3):\n",
-      "    ans = raw_input(\"What is the Unit of Traffic ?\")\n",
-      "    \n",
-      "    if ans == \"Earlang\":\n",
-      "        sys.stdout.write(\"\\nAnswer is Correct.\")\n",
-      "    else:\n",
-      "        if i < 3:\n",
-      "            sys.stdout.write(\"\\nTry Again !\\n\")\n",
-      "            \n",
-      "if i == 3:\n",
-      "    sys.stdout.write(\"\\nUnit of Traffic is Earlang.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "What is the Unit of Traffic ?Earlan\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Try Again !\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "What is the Unit of Traffic ?Earlam\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Try Again !\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "What is the Unit of Traffic ?Earlang\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Answer is Correct."
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.18, Page number: 252<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compare two strings upto specified length.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "sr = raw_input(\"Enter String(1):\")\n",
-      "tar = raw_input(\"Enter String(2):\")\n",
-      "n = int(raw_input(\"Enter Lenght upto which comparison is to be made\"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "diff = (sr == tar[:n])\n",
-      "\n",
-      "if diff == 0:\n",
-      "    sys.stdout.write(\"The Two Strings are Identical upto %d characters.\"%(n))\n",
-      "else:\n",
-      "    sys.stdout.write(\"The Two Strings are Different\")\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String(1):HELLO\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String(2):HE MAN\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Lenght upto which comparison is to be made2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Two Strings are Identical upto 2 characters."
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.19, Page number: 253<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compare two strings and display the position in which it is different.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "diff = 0\n",
-      "sr = raw_input(\"Enter String(1) :\")\n",
-      "tar = raw_input(\"Enter String(2) :\")\n",
-      "\n",
-      "#Checking\n",
-      "for i in range(0,len(sr)):\n",
-      "    if sr[i] != tar[i]:\n",
-      "        sys.stdout.write(\"%c  %c\\n\"%(sr[i],tar[i]))\n",
-      "        diff += 1\n",
-      "        \n",
-      "#Result\n",
-      "if len(sr) == len(tar) and diff == 0:\n",
-      "    sys.stdout.write(\"\\nThe Two Strings are Identical\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nThe Two Strings are Different at %d places.\"%(diff))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String(1) :BEST LUCK\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String(2) :GOOD LUCK\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "B  G\n",
-        "E  O\n",
-        "S  O\n",
-        "T  D\n",
-        "\n",
-        "The Two Strings are Different at 4 places."
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.20, Page number: 254<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert upper case string to lower case using strlwr().\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "upper = raw_input(\"Enter a string in Upper case :\")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"After strlwr() : %s\"%(upper.lower()))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a string in Upper case :ABCDEFG\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "After strlwr() : abcdefg"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.21, Page number: 254<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Convert Lower case string to upper case using strupr()\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "upper = raw_input(\"Enter a string in Lower Case :\")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"After strupr() : %s\"%(upper.upper()))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a string in Lower Case :abcdefg\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "After strupr() : ABCDEFG"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.22, Page number: 255<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the duplicate string using strdup() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text1 = raw_input(\"Enter Text : \")\n",
-      "\n",
-      "#Result\n",
-      "text2 = text1              #in python, a string can be duplicated by just using the assignment operator\n",
-      "\n",
-      "sys.stdout.write(\"Original String = %s\\nDuplicate String = %s\"%(text1,text2))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text : Today is a Good Day.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Original String = Today is a Good Day.\n",
-        "Duplicate String = Today is a Good Day."
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.23, Page number: 256<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find first occurrence of a given character in a given string. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "string = raw_input(\"Enter Text Below :\")\n",
-      "ch = raw_input(\"Character to find :\")\n",
-      "\n",
-      "#Result\n",
-      "chp = string.count(ch)\n",
-      "if chp:\n",
-      "    sys.stdout.write(\"Character %s found in string\"%(ch))\n",
-      "else:\n",
-      "    sys.stdout.write(\"Character %s not found in string\"%(ch))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Below :Hello Begineers.\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Character to find :r\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Character r found in string"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.24, Page number: 257<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find first occurrence of a given character in a given string and display the memory location. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "line1 = raw_input(\"Enter Text :\")\n",
-      "line2 = raw_input(\"Enter Character to find from the text : \")\n",
-      "\n",
-      "for i in range(0,len(line1)):\n",
-      "    sys.stdout.write(\"%c  %d\\n\"%(line1[i],id(line1[i])))\n",
-      "\n",
-      "chp = line1.count(line2)\n",
-      "if chp:\n",
-      "    sys.stdout.write(\"\\nAddress of first %s returned by strchr() is %d\"%(line2,id(line1[chp])))\n",
-      "else:\n",
-      "    sys.stdout.write(\"'%s' String is not present in Given String \"%(line2))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text :HELLO\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Character to find from the text : L\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "H  20382328\n",
-        "E  20384080\n",
-        "L  19678272\n",
-        "L  19678272\n",
-        "O  20198416\n",
-        "\n",
-        "Address of first L returned by strchr() is 19678272"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.25, Page number: 258<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find the occurrence of second string in the first string.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "line1 = raw_input(\"Enter Line1 :\")\n",
-      "line2 = raw_input(\"Enter Line2 :\")\n",
-      "\n",
-      "#Calculation & Result\n",
-      "chp = line1.count(line2)\n",
-      "\n",
-      "if chp:\n",
-      "    sys.stdout.write(\"'%s' String is present in Given String.\"%(line2))\n",
-      "else:\n",
-      "    sys.stdout.write(\"'%s' String is not present in Given String.\"%(line2))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Line1 :INDIA IS MY COUNTRY.\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Line2 :INDIA\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "'INDIA' String is present in Given String."
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.26, Page number: 259<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Append second string at the end of first string using strcat() function\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text1 = raw_input(\"Enter Text1 :\")\n",
-      "text2 = raw_input(\"Enter Text2 :\")\n",
-      "\n",
-      "#Calculation and Result\n",
-      "text1 = text1 + \" \" + text2        #in python, '+' operator can also be used for string concatenation\n",
-      "\n",
-      "sys.stdout.write(\"%s\"%(text1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text1 :I Am\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text2 :an Indian\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "I Am an Indian"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.27, Page number: 259<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Concatenate two strings without use of standard functions.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "fname = raw_input(\"First Name :\")\n",
-      "sname = raw_input(\"Second Name :\")\n",
-      "lname = raw_input(\"Last Name :\")\n",
-      "\n",
-      "#Calculation\n",
-      "name = fname + \" \" + sname + \" \" + lname\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nComplete Name After Concatenation.\\n\")\n",
-      "sys.stdout.write(\"%s\"%(name))\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "First Name :MOHAN\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Second Name :KARAMCHAND\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Last Name :GANDHI\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Complete Name After Concatenation.\n",
-        "MOHAN KARAMCHAND GANDHI"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.28, Page number: 261<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Append second string with specified (n) number of characters at the end of first string.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text1 = raw_input(\"Enter Text1 :\")\n",
-      "text2 = raw_input(\"Enter Text2 :\")\n",
-      "n = int(raw_input(\"Enter Number of Characters to Add :\"))\n",
-      "\n",
-      "#Calculation & Result\n",
-      "text1 = text1 + \" \" + text2[:n]\n",
-      "\n",
-      "sys.stdout.write(\"%s\\n\"%(text1))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text1 :MAY I\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text2 :COME IN ?\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Number of Characters to Add :4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "MAY I COME\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.29, Page number: 261<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reverse of the given string.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = raw_input(\"Enter String\")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Reverse String\\n\")\n",
-      "text = text[::-1]\n",
-      "#This is extended slice syntax. It works by doing [begin:end:step] \n",
-      "#by leaving begin and end off and specifying a step of -1, it reverses a string.\n",
-      "\n",
-      "sys.stdout.write(\"%s\"%(text))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter StringABCDEFGHI\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reverse String\n",
-        "IHGFEDCBA"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.30, Page number: 262<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Reverse of the given string.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = raw_input(\"Enter String\")\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "while i < len(text):\n",
-      "    sys.stdout.write(\"\\n%c is stored at location %d\"%(text[i],id(text[i])))\n",
-      "    i += 1\n",
-      "    \n",
-      "sys.stdout.write(\"\\nReverse String :- \")\n",
-      "text = text[::-1]\n",
-      "#This is extended slice syntax. It works by doing [begin:end:step] \n",
-      "#by leaving begin and end off and specifying a step of -1, it reverses a string.\n",
-      "\n",
-      "sys.stdout.write(\"%s\"%(text))\n",
-      "\n",
-      "i = 0\n",
-      "while i < len(text):\n",
-      "    sys.stdout.write(\"\\n%c is stored at location %d\"%(text[i],id(text[i])))\n",
-      "    i += 1\n",
-      "\n",
-      "#Memory address differs in different systems"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter StringABC\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "A is stored at location 20383432\n",
-        "B is stored at location 20198992\n",
-        "C is stored at location 20199040\n",
-        "Reverse String :- CBA\n",
-        "C is stored at location 20199040\n",
-        "B is stored at location 20198992\n",
-        "A is stored at location 20383432"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.31, Page number: 263<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Replace (set) given string with given symbol. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "string = raw_input(\"Enter String :\")\n",
-      "symbol = raw_input(\"Enter Symbol for Replacement : \")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Before strset(): %s\\n\"%(string))\n",
-      "string = symbol*len(string)\n",
-      "\n",
-      "sys.stdout.write(\"After strset() : %s\"%(string))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String :LEARN C\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Symbol for Replacement : Y\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Before strset(): LEARN C\n",
-        "After strset() : YYYYYYY"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.32, Page number: 264<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Replace (set) given string with given symbol for given number of arguments.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "string = raw_input(\"Enter String :\")\n",
-      "symbol = raw_input(\"Enter Symbol for Replacement : \")\n",
-      "n = int(raw_input(\"How many String Character to be replaced.\"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Before strset() : %s\\n\"%(string))\n",
-      "\n",
-      "string = symbol * n + string[n:]\n",
-      "sys.stdout.write(\"After strset() : %s\\n\"%(string))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String :ABCDEFGHIJ\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Symbol for Replacement : +\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many String Character to be replaced.4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Before strset() : ABCDEFGHIJ\n",
-        "After strset() : ++++EFGHIJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.33, Page number: 265<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the position from which the two strings are different.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "stra = raw_input(\"First String : \")\n",
-      "strb = raw_input(\"Second String : \")\n",
-      "\n",
-      "#There is no built in function for python to check upto which position the strings are equal.\n",
-      "for length in range(min(len(stra), len(strb))):\n",
-      "        if stra[length] != strb[length]:\n",
-      "            break\n",
-      "\n",
-      "sys.stdout.write(\"After %d Characters there is no match.\"%(length))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "First String : GOOD MORNING\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Second String : GOOD BYE\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "After 5 Characters there is no match."
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.34, Page number: 266<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print given string from first occurrence of given character.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text1 = raw_input(\"Enter String : \")\n",
-      "text2 = raw_input(\"Enter Character :\")\n",
-      "\n",
-      "for length in range(0,len(text1)):\n",
-      "        if text1[length] == text2[0]:\n",
-      "            break\n",
-      "            \n",
-      "sys.stdout.write(\"String from given Character : %s\"%(text1[length:]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter String : INDIA IS GREAT\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Character :G\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String from given Character : GREAT"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.35, Page number: 266<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count a character that appears in a given text\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 0\n",
-      "count = 0\n",
-      "text = raw_input(\"Type Text Below.\")\n",
-      "find = raw_input(\"Type a character to count : \")\n",
-      "\n",
-      "#Calculation\n",
-      "while i < len(text):             #There is no null terminating character for python string\n",
-      "    if text[i] == find:\n",
-      "        count += 1\n",
-      "    i += 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nCharacter (%s) found in Given String = %d Times.\"%(find,count))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Type Text Below.Programming\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Type a character to count : m\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Character (m) found in Given String = 2 Times."
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.36, Page number: 267<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count the character 'm' in a given string.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "count = 0\n",
-      "text = raw_input(\"\")  #it is easy to get string at once in python\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"%s\"%(text))\n",
-      "\n",
-      "for i in range(0,len(text)):\n",
-      "    if text[i] == '\\n':\n",
-      "        break\n",
-      "    else:\n",
-      "        if text[i] == 'm':\n",
-      "            count += 1\n",
-      "            \n",
-      "sys.stdout.write(\"\\n\\nCharacter 'm' Found in Text = %d times.\"%(count))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Programming is a skill.\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Programming is a skill.\n",
-        "\n",
-        "Character 'm' Found in Text = 2 times."
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.37, Page number: 268<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Count the characters 'm', 'r', and 'o' that appears in a string without using any functions\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "text = \"Programming is good habit\"\n",
-      "m = 0\n",
-      "o = 0\n",
-      "r = 0\n",
-      "\n",
-      "#Calculation\n",
-      "for i in range(0,len(text)):\n",
-      "    if text[i] == 'm':\n",
-      "        m += 1\n",
-      "    if text[i] == 'r':\n",
-      "        r += 1\n",
-      "    if text[i] == 'o':\n",
-      "        o += 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nCharacter 'm' Found in Text = %d times.\\n\"%(m))\n",
-      "sys.stdout.write(\"\\nCharacter 'r' Found in Text = %d times.\\n\"%(r))\n",
-      "sys.stdout.write(\"\\nCharacter 'o' Found in Text = %d times.\"%(o))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Character 'm' Found in Text = 2 times.\n",
-        "\n",
-        "Character 'r' Found in Text = 2 times.\n",
-        "\n",
-        "Character 'o' Found in Text = 3 times."
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.38, Page number: 269<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy contents of one string to another string\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "ori = raw_input(\"Enter Your Name :\")\n",
-      "\n",
-      "#Copying\n",
-      "#Since strings in python are immutable, individual character assignment is not possible.\n",
-      "#but assignment operator simply copies the entire string to another\n",
-      "\n",
-      "dup = ori\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Original String : %s\"%(ori))\n",
-      "sys.stdout.write(\"\\nDuplicate String : %s\"%(dup))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Your Name :SACHIN\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Original String : SACHIN\n",
-        "Duplicate String : SACHIN"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.39, Page number: 270<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Palindrome or not.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = 0\n",
-      "str1 = raw_input(\"Enter the word : \")   #Since str is a keyword in python, str1 is used\n",
-      "j = len(str1)-1\n",
-      "\n",
-      "#Calculation\n",
-      "while i < j:\n",
-      "    if str1[i] == str1[j]:\n",
-      "        test = 1\n",
-      "    else:\n",
-      "        test = 0\n",
-      "        break\n",
-      "    i += 1\n",
-      "    j -= 1\n",
-      "    \n",
-      "#Result\n",
-      "if test == 1:\n",
-      "    sys.stdout.write(\"Word is palindrome.\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nWord is not Palindrome.\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the word : ABBA\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Word is palindrome."
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.40, Page number: 271<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compare the string using strcmp() function and display their ASCII difference.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a1 = \"NAGPUR\"\n",
-      "a2 = \"NAGPUR\"\n",
-      "a3 = \"PANDHARPUR\"\n",
-      "a4 = \"KOLHAPUR\"\n",
-      "a5 = \"AURANGABAD\"\n",
-      "a6 = \"HYDERABAD\"\n",
-      "\n",
-      "#Calculation\n",
-      "#There is no built in function which returns the ascii value differece of two strings\n",
-      "#ord function gives the ascii value of a character\n",
-      "c1 = ord(a1[0]) - ord(a2[0])\n",
-      "c2 = ord(a3[0]) - ord(a4[0])\n",
-      "c3 = ord(a5[0]) - ord(a6[0])\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAscii Difference between two strings\\n\")\n",
-      "sys.stdout.write(\"Difference between (%s %s) = %d\\n\"%(a1,a2,c1))\n",
-      "sys.stdout.write(\"Difference between (%s %s) = %d\\n\"%(a3,a4,c2))\n",
-      "sys.stdout.write(\"Difference between (%s %s) = %d\\n\"%(a5,a6,c3))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Ascii Difference between two strings\n",
-        "Difference between (NAGPUR NAGPUR) = 0\n",
-        "Difference between (PANDHARPUR KOLHAPUR) = 5\n",
-        "Difference between (AURANGABAD HYDERABAD) = -7\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.41, Page number: 272<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sort city names alphabetically.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "city = [['0' for i in range(5)]for i in range(20)]\n",
-      "sys.stdout.write(\"Enter Names of Cities.  \")\n",
-      "for i in range(5):\n",
-      "    city[i] = raw_input(\"\")\n",
-      "    \n",
-      "#Sorting & Result\n",
-      "sys.stdout.write(\"Sorted List of Cities.\\n\\n\")\n",
-      "for i in range(65,122+1):\n",
-      "    for j in range(0,5):\n",
-      "        if ord(city[j][0]) == i:\n",
-      "            sys.stdout.write(\"\\n%s\"%(city[j]))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Names of Cities.  "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "MUMBAI\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "NANED\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "BANGLORE\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "KANPUR\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "INDORE\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorted List of Cities.\n",
-        "\n",
-        "\n",
-        "BANGLORE\n",
-        "INDORE\n",
-        "KANPUR\n",
-        "MUMBAI\n",
-        "NANED"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.42, Page number: 273<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sort city names alphabetically\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "city = [['0' for i in range(5)]for i in range(20)]\n",
-      "sys.stdout.write(\"Enter Names of Cities.  \")\n",
-      "for i in range(5):\n",
-      "    city[i] = raw_input(\"\")\n",
-      "    \n",
-      "#Sorting & Result\n",
-      "sys.stdout.write(\"Sorted List of Cities.\\n\\n\")\n",
-      "for i in range(0,5):\n",
-      "    for j in range(0,5):\n",
-      "        if city[j-1] > city[j]:\n",
-      "            temp = city[j-1]\n",
-      "            city[j-1] = city[j]\n",
-      "            city[j] = temp\n",
-      "            \n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"\\n%s\"%(city[i]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Names of Cities.  "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "MUMBAI\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "NANED\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "BANGLORE\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "KANPUR\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "INDORE\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sorted List of Cities.\n",
-        "\n",
-        "\n",
-        "BANGLORE\n",
-        "INDORE\n",
-        "KANPUR\n",
-        "MUMBAI\n",
-        "NANED"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.43, Page number: 274<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find number of words in a given statement.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "count = 1                #Counts the starting word\n",
-      "i = 0\n",
-      "text = raw_input(\"Enter The Line of Text\\nGive One Space After Each Word\")\n",
-      "\n",
-      "#Calculation\n",
-      "while i < len(text):      #There is no null terminating character in python\n",
-      "    if ord(text[i]) == 32:\n",
-      "        count += 1\n",
-      "    i += 1\n",
-      "        \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nThe Number of words in line = %d\"%(count))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter The Line of Text\n",
-        "Give One Space After Each WordRead Books\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Number of words in line = 2"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.44, Page number: 275<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find number of words in a given statement.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "count = 0          \n",
-      "i = 0\n",
-      "text = raw_input(\"Enter Text Below :\")\n",
-      "\n",
-      "#Calculation\n",
-      "while i < len(text):     #There is no null terminating character in python for strings\n",
-      "    if ord(text[i]) >= 97 and ord(text[i]) <= 122 or ord(text[i]) >= 65 and ord(text[i]) <= 90:\n",
-      "        i += 1\n",
-      "        continue\n",
-      "    if ord(text[i]) == 32:\n",
-      "        count += 1\n",
-      "        i += 1\n",
-      "        continue\n",
-      "\n",
-      "if i == len(text):\n",
-      "    count += 1\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"The Number of words in line = %d\"%(count))\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Text Below :Reading is a good Habit\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Number of words in line = 5"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 8.45, Page number: 276<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sort the last name of mobile customers alphabetically.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "fname = [['0' for i in range(20)] for i in range(10)]\n",
-      "sname = [['0' for i in range(20)] for i in range(10)]\n",
-      "surname = [['0' for i in range(20)] for i in range(10)]\n",
-      "name = [['0' for i in range(20)] for i in range(20)]\n",
-      "mobile = [['0' for i in range(20)] for i in range(10)]\n",
-      "temp = ['0' for i in range(20)]\n",
-      "\n",
-      "sys.stdout.write(\"Enter First Name, Second Name, surname and Mobile Number \")\n",
-      "for i in range(0,5):\n",
-      "    fname[i] = raw_input(\"\")\n",
-      "    sname[i] = raw_input(\"\")\n",
-      "    surname[i] = raw_input(\"\")\n",
-      "    mobile[i] = raw_input(\"\")\n",
-      "    \n",
-      "    name[i] = surname[i] + \" \" \n",
-      "    temp[0] = fname[i][0]\n",
-      "    name[i] = name[i] + temp[0] + \".\"\n",
-      "    temp[0] = sname[i][0]\n",
-      "    name[i] = name[i] + temp[0]\n",
-      "    \n",
-      "for i in range(0,5):\n",
-      "    for j in range(1,5-i):\n",
-      "        if name[j-1] > name[j]:\n",
-      "            temp = name[j-1]\n",
-      "            name[j-1] = name[j]\n",
-      "            name[j] = temp\n",
-      "            temp = mobile[j-1]\n",
-      "            mobile[j-1] = mobile[j]\n",
-      "            mobile[j] = temp\n",
-      "            \n",
-      "sys.stdout.write(\"List of Customers in alphanetical Order.\")\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"\\n%-20s\\t %-10s\\n\"%(name[i],mobile[i]))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter First Name, Second Name, surname and Mobile Number "
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "K\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "S\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "MORE\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "458454\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "J\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "M\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "CHATE\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "658963\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "M\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "M\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "GORE\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "660585\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "K\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "JAIN\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "547855\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "J\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "J\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "JOSHI\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "354258\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "List of Customers in alphanetical Order.\n",
-        "CHATE J.M           \t 658963    \n",
-        "\n",
-        "GORE M.M            \t 660585    \n",
-        "\n",
-        "JAIN L.K            \t 547855    \n",
-        "\n",
-        "JOSHI J.J           \t 354258    \n",
-        "\n",
-        "MORE K.S            \t 458454    \n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter9.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter9.ipynb
deleted file mode 100755
index 4da93c8d..00000000
--- a/Programming_in_C_using_ANSI_C/KamthaneChapter9.ipynb
+++ /dev/null
@@ -1,1999 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h1>Chapter 9: Pointers<h1>"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.1, Page number: 282<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the address of the variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "num = int(raw_input(\"Enter a Number = \"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Value of num = %d\\n\"%(num))\n",
-      "sys.stdout.write(\"Address of num = %d\\n\"%(id(num)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Number = 20\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of num = 20\n",
-        "Address of num = 19554932\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.2, Page number: 283<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the addresses of different variables together with their values.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = raw_input(\"Enter alphabet\")\n",
-      "i = int(raw_input(\"Enter number\"))\n",
-      "f = float(raw_input(\"Enter float\"))\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAddress of (c) %c = %d\"%(c,id(c)))\n",
-      "sys.stdout.write(\"\\nAddress of (i) %d = %d\"%(i,id(i)))\n",
-      "sys.stdout.write(\"\\nAddress of (f) %.2f = %d\"%(f,id(f)))\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter alphabetC\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter number20\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter float2.5\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of (c) C = 48769192\n",
-        "Address of (i) 20 = 19554932\n",
-        "Address of (f) 2.50 = 44777808"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.3, Page number: 284<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Show pointer of any data type that occupies same space\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Result\n",
-      "#Garbage collector module is used for python memory management\n",
-      "sys.stdout.write(\"char %d byte & its pointer %d bytes\\n\"%(sys.getsizeof(chr),sys.getsizeof(id(chr))))\n",
-      "sys.stdout.write(\"int %d byte & its pointer %d bytes\\n\"%(sys.getsizeof(int),sys.getsizeof(id(int))))\n",
-      "sys.stdout.write(\"float %d byte & its pointer %d bytes\\n\"%(sys.getsizeof(float),sys.getsizeof(id(float))))\n",
-      "\n",
-      "#value tagged method is used in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "char 36 byte & its pointer 12 bytes\n",
-        "int 436 byte & its pointer 12 bytes\n",
-        "float 436 byte & its pointer 12 bytes\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.4, Page number: 284<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the value of variable and its location using pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "#There is no pointer concept in python\n",
-      "v = 10\n",
-      "p = v\n",
-      "#In python, variables of same value are tagged together instead of storing the value in different locations\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAddress of v = %u\"%(id(v)))\n",
-      "sys.stdout.write(\"\\nValue of v = %d\"%(p))\n",
-      "sys.stdout.write(\"\\nAddress of p = %d\"%(id(p)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of v = 29808764\n",
-        "Value of v = 10\n",
-        "Address of p = 29808764"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.5, Page number: 285<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the value of variable by different ways.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = int(raw_input(\"Enter Integer Value\"))\n",
-      "b = float(raw_input(\"Enter float Value\"))\n",
-      "\n",
-      "pa = id(a)\n",
-      "pb = id(b)\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python. content of memory location can't be accessed directly by user.\n",
-      "sys.stdout.write(\"\\nAddress of a = %u\"%(pa))\n",
-      "sys.stdout.write(\"\\nValue of a = %d\"%(a))\n",
-      "sys.stdout.write(\"\\nValue of a = %d\"%(a))\n",
-      "sys.stdout.write(\"\\nValue of a = %d\"%(a))\n",
-      "\n",
-      "sys.stdout.write(\"\\nAddress of b = %u\"%(pb))\n",
-      "sys.stdout.write(\"\\nValue of b = %.2f\"%(b))\n",
-      "sys.stdout.write(\"\\nValue of b = %.2f\"%(b))\n",
-      "sys.stdout.write(\"\\nValue of b = %.2f\"%(b))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Integer Value4\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter float Value2.2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of a = 19555124\n",
-        "Value of a = 4\n",
-        "Value of a = 4\n",
-        "Value of a = 4\n",
-        "Address of b = 44777792\n",
-        "Value of b = 2.20\n",
-        "Value of b = 2.20\n",
-        "Value of b = 2.20"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.6, Page number: 286<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print value of variable using different pointer notations.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "v = 10\n",
-      "p = id(v)\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python and can't access memory location directly by user\n",
-      "sys.stdout.write(\"\\nv = %d v = %d v = %d\"%(v,v,v))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "v = 10 v = 10 v = 10"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.7, Page number: 287<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print element and its address using pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = int(raw_input(\"Enter a number : \"))\n",
-      "k = id(i)\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python and can't access memory location directly by user\n",
-      "sys.stdout.write(\"\\nAddress of i is %u\"%(k))\n",
-      "sys.stdout.write(\"\\nValue of i is %d\"%(i))\n",
-      "#The memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number : 15\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of i is 19554992\n",
-        "Value of i is 15"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.8, Page number: 287<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Add two numbers through variables and their pointers\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = int(raw_input(\"Enter Two Numbers : \"))\n",
-      "b = int(raw_input(\"Enter Two Numbers : \"))\n",
-      "ap = a\n",
-      "bp = b\n",
-      "\n",
-      "#Calculation\n",
-      "c = a + b\n",
-      "d = ap + bp\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python and can't access memory location directly by user\n",
-      "sys.stdout.write(\"\\nSum of A & B Using Variable : %d\"%(c))\n",
-      "sys.stdout.write(\"\\nSum of A & B Using Pointer : %d\"%(d))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Two Numbers : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Two Numbers : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of A & B Using Variable : 12\n",
-        "Sum of A & B Using Pointer : 12"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.9, Page number: 288<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Assign pointer value to another variable\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 5\n",
-      "c = id(a)\n",
-      "b = a\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python and can't access memory location directly by user\n",
-      "sys.stdout.write(\"\\nMemory location of 'a' = %u\"%(c))\n",
-      "sys.stdout.write(\"\\nThe value of a = %d & b = %d\"%(a,b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Memory location of 'a' = 29808824\n",
-        "The value of a = 5 & b = 5"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.10, Page number: 289<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Assign value of 'b' to 'a' through pointers and add the values.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = int(raw_input(\"Enter Value of 'a' & 'b' : \"))\n",
-      "b = int(raw_input(\"Enter Value of 'a' & 'b' : \"))\n",
-      "pa = id(a)\n",
-      "pb = id(b)\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python and can't access memory location directly by user\n",
-      "sys.stdout.write(\"\\nValue of a = %d & b = %d\"%(a,b))\n",
-      "sys.stdout.write(\"\\nAddress of a = %u\"%(pa))\n",
-      "sys.stdout.write(\"\\nAddress of b = %u\"%(pb))\n",
-      "sys.stdout.write(\"\\n\\nAddition of 'a' & 'b' : %d\"%(a+b))\n",
-      "\n",
-      "a = b\n",
-      "pa = id(a)\n",
-      "#In python, variables of same value are tagged together instead of storing the value in different locations\n",
-      "\n",
-      "sys.stdout.write(\"\\n*pa = %d *pb = %d\"%(a,b))\n",
-      "sys.stdout.write(\"\\npa = %u pb = %u\"%(pa,pb))\n",
-      "sys.stdout.write(\"\\nAddition of *pa + *pb : %d\"%(a+b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of 'a' & 'b' : 8\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of 'a' & 'b' : 4\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Value of a = 8 & b = 4\n",
-        "Address of a = 19555076\n",
-        "Address of b = 19555124\n",
-        "\n",
-        "Addition of 'a' & 'b' : 12\n",
-        "*pa = 4 *pb = 4\n",
-        "pa = 19555124 pb = 19555124\n",
-        "Addition of *pa + *pb : 8"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.11, Page number: 290<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#The effect of increment on pointer variables. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = int(raw_input(\"Enter integer Value\"))\n",
-      "y = raw_input(\"Enter Character Value\")\n",
-      "z = float(raw_input(\"Enter float Value\"))\n",
-      "\n",
-      "x1 = id(x)\n",
-      "y1 = id(y)\n",
-      "z1 = id(z)\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python and can't access memory location directly by user\n",
-      "\n",
-      "sys.stdout.write(\"\\nAddress of x = %u\\n\"%(x1))\n",
-      "sys.stdout.write(\"\\nAddress of y = %u\\n\"%(y1))\n",
-      "sys.stdout.write(\"\\nAddress of z = %u\\n\"%(z1))\n",
-      "\n",
-      "x1 += 1\n",
-      "y1 += 1\n",
-      "z1 += 1\n",
-      "#In python, variables of same value are tagged together instead of storing the value in different locations\n",
-      "\n",
-      "sys.stdout.write(\"\\nAfter Increment in Pointers\\n\")\n",
-      "sys.stdout.write(\"\\nNow Address of x = %u\\n\"%(x1))\n",
-      "sys.stdout.write(\"Now Address of y = %u\\n\"%(y1))\n",
-      "sys.stdout.write(\"Now Address of z = %u\\n\"%(z1))\n",
-      "\n",
-      "sys.stdout.write(\"\\nSize of Integer : %d\"%(sys.getsizeof(x)))\n",
-      "sys.stdout.write(\"\\nSize of Character :  %d\"%(sys.getsizeof(y)))\n",
-      "sys.stdout.write(\"\\nSize of Float : %d\"%(sys.getsizeof(z)))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter integer Value2\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Character ValueA\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter float Value2.2\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address of x = 19555148\n",
-        "\n",
-        "Address of y = 48769192\n",
-        "\n",
-        "Address of z = 44777728\n",
-        "\n",
-        "After Increment in Pointers\n",
-        "\n",
-        "Now Address of x = 19555149\n",
-        "Now Address of y = 48769193\n",
-        "Now Address of z = 44777729\n",
-        "\n",
-        "Size of Integer : 12\n",
-        "Size of Character :  22\n",
-        "Size of Float : 16"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.12, Page number: 291<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#The effect of increased and decreased operator used as prefix and suffix with pointer variable.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "i = int(raw_input(\"Enter Value of i = \"))\n",
-      "ii = id(i)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
-      "#There is no increment or decrement operator and pointer concept in python\n",
-      "ii += 1\n",
-      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
-      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
-      "ii += 1\n",
-      "ii -= 1\n",
-      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
-      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
-      "ii -= 1\n",
-      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
-      "\n",
-      "#Memory address differs in different systems"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Value of i = 8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Address of i = 19555076\n",
-        "Address of i = 19555077\n",
-        "Address of i = 19555077\n",
-        "Address of i = 19555077\n",
-        "Address of i = 19555077\n",
-        "Address of i = 19555076\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.13, Page number: 292<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Perform different arithmetic operations using pointers\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 25\n",
-      "b = 10\n",
-      "p = id(a)\n",
-      "j = id(b)\n",
-      "\n",
-      "#Result\n",
-      "#There is no pointer concept in python and user can't access memory contents directly\n",
-      "sys.stdout.write(\"\\nAddition a + b = %d\"%(a+b))\n",
-      "sys.stdout.write(\"\\nSubtraction a - b = %d\"%(a-b))\n",
-      "sys.stdout.write(\"\\nProduct a * b = %d\"%(a*b))\n",
-      "sys.stdout.write(\"\\nDivision a / b = %d\"%(a/b))\n",
-      "sys.stdout.write(\"\\na Mod b  = %d\"%(a%b))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Addition a + b = 35\n",
-        "Subtraction a - b = 15\n",
-        "Product a * b = 250\n",
-        "Division a / b = 2\n",
-        "a Mod b  = 5"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.14, Page number: 293<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compare two pointers. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 2\n",
-      "j = id(a)\n",
-      "k = id(a)\n",
-      "\n",
-      "#Result\n",
-      "if j == k:\n",
-      "    sys.stdout.write(\"\\nThe Tow Pointers have the same address.\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nThe Pointers have the different address.\")\n",
-      "    \n",
-      "sys.stdout.write(\"\\n\\nAddress of a (j) = %u\"%(j))\n",
-      "sys.stdout.write(\"\\nAddress of a (k) = %u\"%(k))\n",
-      "#Memory address differs in different systems"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The Tow Pointers have the same address.\n",
-        "\n",
-        "Address of a (j) = 30726364\n",
-        "Address of a (k) = 30726364"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.15, Page number: 294<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display elements of an array.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = [2,4,6,8,10]\n",
-      "k = 1\n",
-      "\n",
-      "#Result\n",
-      "while k <= 5:\n",
-      "    sys.stdout.write(\"%3d\"%(x[k-1]))\n",
-      "    k += 1\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  2  4  6  8 10"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.16, Page number: 295<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display array element with their addresses using array name as a pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = [2,4,6,8,10]\n",
-      "k = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nElement No.\\tElement\\tAddress\")\n",
-      "\n",
-      "while k < 5:\n",
-      "    sys.stdout.write(\"\\nx[%d] \\t%13d %10u\"%(k,x[k],id(x[k])))\n",
-      "    k += 1\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Element No.\tElement\tAddress\n",
-        "x[0] \t            2   30726364\n",
-        "x[1] \t            4   30726340\n",
-        "x[2] \t            6   30726316\n",
-        "x[3] \t            8   30726292\n",
-        "x[4] \t           10   30726268"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.17, Page number: 296<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display array elements with their addresses using array name as a pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "num = [10,25,35,45]\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nElement\\t           Address\\n\")\n",
-      "\n",
-      "for i in range(0,4):\n",
-      "    sys.stdout.write(\"num[%d] = %d        \"%(i,num[i]))\n",
-      "    sys.stdout.write(\"%8u\\n\"%(id(num[i])))\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Element\t           Address\n",
-        "num[0] = 10        30726268\n",
-        "num[1] = 25        30726088\n",
-        "num[2] = 35        30725968\n",
-        "num[3] = 45        30725848\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.18, Page number: 296<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Access elements of array through different ways using pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "arr = [10,20,30,40,50]\n",
-      "p = 0\n",
-      "\n",
-      "#Result\n",
-      "for p in range(0,5):\n",
-      "    sys.stdout.write(\"Value of arr[%d] = \"%(p))\n",
-      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
-      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
-      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
-      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
-      "    sys.stdout.write(\"address of arr[%d] = %u\\n\"%(p,id(arr[p])))\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of arr[0] = 10   |  10   |  10   |  10   |  address of arr[0] = 30726268\n",
-        "Value of arr[1] = 20   |  20   |  20   |  20   |  address of arr[1] = 30726148\n",
-        "Value of arr[2] = 30   |  30   |  30   |  30   |  address of arr[2] = 30726028\n",
-        "Value of arr[3] = 40   |  40   |  40   |  40   |  address of arr[3] = 30725908\n",
-        "Value of arr[4] = 50   |  50   |  50   |  50   |  address of arr[4] = 30725788\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.19, Page number: 297<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Find sum of all the elements of an array.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "sum1 = 0        #Since sum is a built-in function in python, sum1 is used\n",
-      "i = 0\n",
-      "a = [1,2,3,4,5]\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Elements       Values         Address\\n\\n\")\n",
-      "\n",
-      "while i < 5:\n",
-      "    sys.stdout.write(\"a[%d]\\t   %5d\\t    %8u\\n\"%(i,a[i],id(a[i])))\n",
-      "    sum1 = sum1 + a[i]\n",
-      "    i += 1\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Elements       Values         Address\n",
-        "\n",
-        "a[0]\t       1\t    30726376\n",
-        "a[1]\t       2\t    30726364\n",
-        "a[2]\t       3\t    30726352\n",
-        "a[3]\t       4\t    30726340\n",
-        "a[4]\t       5\t    30726328\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.20, Page number: 298<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Sum of squares and cubes of array elements using pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "b = [1,2,3,4,5]\n",
-      "sumsq = 0\n",
-      "sumc = 0\n",
-      "\n",
-      "#Result\n",
-      "for j in range(0,5):\n",
-      "    sumsq = sumsq + pow(b[j],2)\n",
-      "    sumc = sumc + pow(b[j],3)\n",
-      "    \n",
-      "sys.stdout.write(\"\\nSum of Squares of array elements : %d\"%(sumsq))\n",
-      "sys.stdout.write(\"\\nSum of Cubes of array elements : %d\"%(sumc))\n",
-      "\n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Sum of Squares of array elements : 55\n",
-        "Sum of Cubes of array elements : 225"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.21, Page number: 299<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy element of one array to another array using pointers.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "so = [10,20,30,40,50]\n",
-      "ds = [0 for i in range(0,5)]\n",
-      "\n",
-      "#Copying\n",
-      "for i in range(0,5):\n",
-      "    ds[i] = so[i]\n",
-      "    \n",
-      "sys.stdout.write(\"Original Array      Duplicated Array\")\n",
-      "\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"\\n\\t%d\\t\\t%d\"%(so[i],ds[i]))\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Original Array      Duplicated Array\n",
-        "\t10\t\t10\n",
-        "\t20\t\t20\n",
-        "\t30\t\t30\n",
-        "\t40\t\t40\n",
-        "\t50\t\t50"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.22, Page number: 299<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Copy one array into another array. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "arr1 = [15,25,35,45,55]\n",
-      "arr2 = [0 for i in range(0,5)]\n",
-      "j = 4\n",
-      "\n",
-      "#Copying & Result\n",
-      "sys.stdout.write(\"\\nOriginal Array     Duplicate Array\")\n",
-      "for i in range(0,5):\n",
-      "    arr2[i] = arr1[j]\n",
-      "    j -= 1\n",
-      "    sys.stdout.write(\"\\n\\t%d\\t\\t%d\"%(arr1[i],arr2[i]))\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Original Array     Duplicate Array\n",
-        "\t15\t\t55\n",
-        "\t25\t\t45\n",
-        "\t35\t\t35\n",
-        "\t45\t\t25\n",
-        "\t55\t\t15"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.23, Page number: 300<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display array elements and their address using pointers\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[1,2,3],[4,5,6],[7,8,9]]\n",
-      "j = 1\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\tElements of An Array with their addresses\\n\\n\")\n",
-      "\n",
-      "p = id(a[0][0])\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,3):\n",
-      "        sys.stdout.write(\"%5d  [%5u]\"%(a[i][j],id(a[i][j])))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\tElements of An Array with their addresses\n",
-        "\n",
-        "    1  [30726376]    2  [30726364]    3  [30726352]\n",
-        "    4  [30726340]    5  [30726328]    6  [30726316]\n",
-        "    7  [30726304]    8  [30726292]    9  [30726280]\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.24, Page number: 301<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display array elements and their address. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [[1,2,3],[4,5,6],[7,8,9]]\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nElements of An Array with their addresses.\\n\\n\")\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    for j in range(0,3):\n",
-      "        sys.stdout.write(\"%5d  [%5u]\"%(a[i][j],id(a[i][j])))\n",
-      "    sys.stdout.write(\"\\n\")\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Elements of An Array with their addresses.\n",
-        "\n",
-        "    1  [30726376]    2  [30726364]    3  [30726352]\n",
-        "    4  [30726340]    5  [30726328]    6  [30726316]\n",
-        "    7  [30726304]    8  [30726292]    9  [30726280]\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.25, Page number: 302<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Store addresses of different elements of an array using array of pointers\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "arrl = [5,10,15]\n",
-      "arrp = [0 for i in range(0,3)]\n",
-      "\n",
-      "for i in range(0,3):\n",
-      "    arrp[i] = id(arrl[i])\n",
-      "    \n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nAddress  \\tElement\\n\")\n",
-      "\n",
-      "for k in range(0,3):\n",
-      "    sys.stdout.write(\"%u\"%(arrp[k]))\n",
-      "    sys.stdout.write(\"\\t%7d\\n\"%(arrl[k]))\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Address  \tElement\n",
-        "30726328\t      5\n",
-        "30726268\t     10\n",
-        "30726208\t     15\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.26, Page number: 303<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display address of elements and pointers\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [0,1,2,3,4]\n",
-      "p = [0 for i in range(0,5)]\n",
-      "\n",
-      "for i in range(0,5):\n",
-      "    p[i] = id(a[i])\n",
-      "    \n",
-      "#Result\n",
-      "for i in range(0,5):\n",
-      "    sys.stdout.write(\"\\n%d at location\"%(a[i]))\n",
-      "    sys.stdout.write(\"\\t%u at location\"%(id(a[i])))\n",
-      "    sys.stdout.write(\"   %u\"%(id(p[i])))\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "0 at location\t30726388 at location   52456564\n",
-        "1 at location\t30726376 at location   52456360\n",
-        "2 at location\t30726364 at location   52456456\n",
-        "3 at location\t30726352 at location   52456420\n",
-        "4 at location\t30726340 at location   52456468"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.27, Page number: 304<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print value of a variable through pointer and pointer to pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = 2\n",
-      "p = id(a)\n",
-      "q = id(p)\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n        Value of a = %d Address of a = %u\"%(a,id(a)))\n",
-      "sys.stdout.write(\"\\nThrough *p Value of a = %d Address of a = %u\"%(a,p))\n",
-      "sys.stdout.write(\"\\nThrough **q Vallue of a = %d Address of a = %d\"%(a,p))\n",
-      "\n",
-      "#There is no pointer concept in python\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "        Value of a = 2 Address of a = 30726364\n",
-        "Through *p Value of a = 2 Address of a = 30726364\n",
-        "Through **q Vallue of a = 2 Address of a = 30726364"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.28, Page number: 305<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Different levels of array of pointer to pointer and display the elements\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "a = [1,2,3]\n",
-      "b = [0 for i in range(0,3)]\n",
-      "c = [0 for i in range(0,3)]\n",
-      "d = [0 for i in range(0,3)]\n",
-      "e = [0 for i in range(0,3)]\n",
-      "f = [0 for i in range(0,3)]\n",
-      "\n",
-      "for k in range(0,3):\n",
-      "    b[k] = id(a[k])\n",
-      "    c[k] = id(b[k])\n",
-      "    d[k] = id(c[k])\n",
-      "    e[k] = id(d[k])\n",
-      "    f[k] = id(e[k])\n",
-      "    \n",
-      "#Result\n",
-      "for k in range(0,3):\n",
-      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
-      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
-      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
-      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
-      "    sys.stdout.write(\"%3d\\n\"%(a[k]))\n",
-      "    \n",
-      "#There is no pointer concept in python\n",
-      "#Memory address differs in different systems\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  1  1  1  1  1\n",
-        "  2  2  2  2  2\n",
-        "  3  3  3  3  3\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.29, Page number: 306<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display string using character pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "name = ['K','U','M','A','R']\n",
-      "ch = id(name)\n",
-      "i = 0\n",
-      "\n",
-      "#Result\n",
-      "while i < len(name):                 #There is no null terminating character in python string\n",
-      "    sys.stdout.write(\"%c\"%(name[i]))\n",
-      "    i += 1\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "KUMAR"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.30, Page number: 307<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Length of a given string including and excluding spaces using pointers\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "p = 0\n",
-      "q = 0\n",
-      "s = 0\n",
-      "str1 = ['P','O','I','N','T','E','R','S',' ','A','R','E',' ','E','A','S','Y']\n",
-      "\n",
-      "#Result\n",
-      "while s < len(str1):          #There is no null termination character for python string\n",
-      "    sys.stdout.write(\"%c\"%(str1[s]))\n",
-      "    p += 1\n",
-      "    if ord(str1[s]) == 32:\n",
-      "        q += 1\n",
-      "    s += 1\n",
-      "\n",
-      "sys.stdout.write(\"\\nLength of String including spaces : %d\"%(p))\n",
-      "sys.stdout.write(\"\\nLength of String excluding spaces : %d\"%(p-q))\n",
-      "\n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "POINTERS ARE EASY\n",
-        "Length of String including spaces : 17\n",
-        "Length of String excluding spaces : 15"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.31, Page number: 308<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Interchange elements of character array using pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "names = [\"kapil\",\"manoj\",\"amit\",\"amol\",\"pavan\",\"mahesh\"]\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"Original : %s  %s\"%(names[3],names[4]))\n",
-      "\n",
-      "tmp = names[3]\n",
-      "names[3] = names[4]\n",
-      "names[4] = tmp\n",
-      "\n",
-      "sys.stdout.write(\"\\nNew    : %s  %s\"%(names[3],names[4]))\n",
-      "\n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Original : amol  pavan\n",
-        "New    : pavan  amol"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.32, Page number: 308<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#String comparison.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = 0\n",
-      "str1 = raw_input(\"Enter First String : \")\n",
-      "str2 = raw_input(\"Enter Second String : \")\n",
-      "a = 0\n",
-      "l = 0\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\nSimilar Characters Found in Both String\")\n",
-      "\n",
-      "while a < len(str1):\n",
-      "    if str1[a] == str2[a]:\n",
-      "        sys.stdout.write(\"\\n\\t%c\\t%c\"%(str1[a],str2[a]))\n",
-      "        l += 1\n",
-      "    else:\n",
-      "        c += 1\n",
-      "    a += 1\n",
-      "        \n",
-      "if c == 0:\n",
-      "    sys.stdout.write(\"\\nThe Strings are Identical\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"\\nThe Strings are different at %d places.\"%(c))\n",
-      "    sys.stdout.write(\"\\nThe String Characters are similar at %d places.\"%(l))\n",
-      "    \n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter First String : SUNDAY\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Second String : MONDAY\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Similar Characters Found in Both String\n",
-        "\tN\tN\n",
-        "\tD\tD\n",
-        "\tA\tA\n",
-        "\tY\tY\n",
-        "The Strings are different at 2 places.\n",
-        "The String Characters are similar at 4 places."
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.33, Page number: 310<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compare three characters using pointers.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "x = raw_input(\"Enter Three Characters\")\n",
-      "y = raw_input(\"Enter Three Characters\")\n",
-      "z = raw_input(\"Enter Three Characters\")\n",
-      "\n",
-      "stat = 0\n",
-      "\n",
-      "stat = y > x\n",
-      "if x == y:\n",
-      "    sys.stdout.write(\"\\n1st and 2nd Character are same\\n\")\n",
-      "else:\n",
-      "    if stat < 0:\n",
-      "        sys.stdout.write(\"\\n2nd Character appears after the 1st Character in Alphabetic\\n\")\n",
-      "    else:\n",
-      "        sys.stdout.write(\"2nd Character appears before the first Character in Alphabetic\\n\")\n",
-      "    \n",
-      "stat = y > z\n",
-      "\n",
-      "if y == z:\n",
-      "    sys.stdout.write(\"\\n2nd and 3rd Character are same.\")\n",
-      "else:\n",
-      "    if stat > 0:\n",
-      "        sys.stdout.write(\"\\n2nd Character appears after the 3rd Character in Alphabetic\")\n",
-      "    else:\n",
-      "        sys.stdout.write(\"\\n2nd Character appears before the 3rd Character in Alphabetic\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three CharactersC\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three CharactersC\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter Three CharactersA\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "1st and 2nd Character are same\n",
-        "\n",
-        "2nd Character appears after the 3rd Character in Alphabetic"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.34, Page number: 311<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Compare two strings irrespective of case.  \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "buf1 = \"computer\"\n",
-      "buf2 = \"computer\"\n",
-      "\n",
-      "#Comparison\n",
-      "stat = buf1 == buf2\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"The Characters upto 4th position are \")\n",
-      "if stat == 0:\n",
-      "    sys.stdout.write(\"not \")\n",
-      "sys.stdout.write(\"same\")\n",
-      "\n",
-      "#There is no pointer concept in python and memicmp() function"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Characters upto 4th position are same"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.35, Page number: 311<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Print the string upto the first occurrence of a character\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "\n",
-      "src = raw_input(\"Enter a String : \")\n",
-      "f = raw_input(\"Enter a Character to find in the text : \")\n",
-      "dest = ['0' for i in range(0,len(src))]\n",
-      "ptr = 0\n",
-      "\n",
-      "while ptr < len(src):\n",
-      "    if src[ptr] == f:\n",
-      "        dest[ptr] = src[ptr]\n",
-      "        break\n",
-      "    else:\n",
-      "        dest[ptr] = src[ptr]\n",
-      "    ptr += 1\n",
-      "\n",
-      "if ptr == len(src):\n",
-      "    sys.stdout.write(\"The character wasn't found\\n\")\n",
-      "else:\n",
-      "    sys.stdout.write(\"String upto that Character : %s\"%(dest[0:ptr+1]))\n",
-      "    \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a String : FUNCTIONS\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Character to find in the text : T\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String upto that Character : ['F', 'U', 'N', 'C', 'T']"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.36, Page number: 312<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Replace the contents of second string with the first string. \n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "src = raw_input(\"Enter a Source String : \")\n",
-      "dest = raw_input(\"Enter a Destination String :\")\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"\\n\\nDestination before memcpy : %s\\n\"%(dest))\n",
-      "dest = src[:len(src)] + dest[len(src):]\n",
-      "\n",
-      "sys.stdout.write(\"Destination after memcpy : %s\\n\"%(dest))\n",
-      "\n",
-      "#There is no pointer concept in python and memcpy function"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Source String : Tomorrow\n"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a Destination String :Today   is Sunday\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "Destination before memcpy : Today   is Sunday\n",
-        "Destination after memcpy : Tomorrowis Sunday\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.37, Page number: 313<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Display the string through their pointer\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization\n",
-      "c = \"Central Processing Unit\"\n",
-      "m = \"Math Co- Processor\"\n",
-      "\n",
-      "#Result\n",
-      "sys.stdout.write(\"'c' is pointing the string '%s'\\n\"%(c))\n",
-      "sys.stdout.write(\"'m' is pointing the string '%s'\\n\"%(m))\n",
-      "\n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "'c' is pointing the string 'Central Processing Unit'\n",
-        "'m' is pointing the string 'Math Co- Processor'\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "<h3>Example 9.38, Page number: 314<h3>"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Use void pointer to display the value of different variables.\n",
-      "\n",
-      "import sys\n",
-      "\n",
-      "#Variable Initialization & Result\n",
-      "pt = 12\n",
-      "sys.stdout.write(\"\\nP = %d\"%(pt))\n",
-      "pt = 5.4\n",
-      "sys.stdout.write(\"\\nR = %.1f\"%(pt))\n",
-      "pt = 'S'\n",
-      "sys.stdout.write(\"\\nC = %c\"%(pt))\n",
-      "\n",
-      "#There is no pointer concept in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "P = 12\n",
-        "R = 5.4\n",
-        "C = S"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/README.txt b/Programming_in_C_using_ANSI_C/README.txt
deleted file mode 100755
index e5dfcad4..00000000
--- a/Programming_in_C_using_ANSI_C/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Chithira G
-Course: me
-College/Institute/Organization: Lourdes Matha College of Science and Technology
-Department/Designation: Computer Science
-Book Title: Programming in C using ANSI C
-Author: Ashok N Kamthane
-Publisher: Pearson Education
-Year of publication: 2006
-Isbn: 8131704378, 9788131704370
-Edition: 3
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C/screenshots/screenshot1.png b/Programming_in_C_using_ANSI_C/screenshots/screenshot1.png
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diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter10.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter10.ipynb
new file mode 100755
index 00000000..68503d70
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter10.ipynb
@@ -0,0 +1,2940 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 10: Functions<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.1, Page number: 320<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#user defined function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 1\n",
+      "y = 2\n",
+      "\n",
+      "#Function definition\n",
+      "def add(a,b):\n",
+      "    return a+b\n",
+      "\n",
+      "#Function call\n",
+      "z = add(x,y)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"z = %d\"%(z))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "z = 3"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.2, Page number: 321<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call user-defined function at different places.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definitions\n",
+      "def y():\n",
+      "    sys.stdout.write(\" Y\")\n",
+      "    return\n",
+      "\n",
+      "def a():\n",
+      "    sys.stdout.write(\" A\")\n",
+      "    y()\n",
+      "    return\n",
+      "\n",
+      "def b():\n",
+      "    sys.stdout.write(\" B\")\n",
+      "    a()\n",
+      "    return\n",
+      "\n",
+      "def c():\n",
+      "    a()\n",
+      "    b()\n",
+      "    sys.stdout.write(\" C\")\n",
+      "    return\n",
+      "\n",
+      "def d():\n",
+      "    sys.stdout.write(\" D\")\n",
+      "    c()\n",
+      "    b()\n",
+      "    a()\n",
+      "    return\n",
+      "\n",
+      "#Function calls\n",
+      "y()\n",
+      "a()\n",
+      "b()\n",
+      "c()\n",
+      "d()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Y A Y B A Y A Y B A Y C D A Y B A Y C B A Y A Y"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.3, Page number: 323<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Using similar variable names in different functions.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def fun():\n",
+      "    b = 20\n",
+      "    c = 10\n",
+      "    sys.stdout.write(\"\\nIn fun() B = %d  C = %d\"%(b,c))\n",
+      "    return\n",
+      "\n",
+      "#Variable Initialization\n",
+      "b = 10\n",
+      "c = 5\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nIn main() B = %d  C = %d\"%(b,c))\n",
+      "fun()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In main() B = 10  C = 5\n",
+        "In fun() B = 20  C = 10"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.4, Page number: 323<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Global variables on different functions.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Global Variable Initialization\n",
+      "b = 10\n",
+      "c = 5\n",
+      "\n",
+      "def fun():\n",
+      "    global b \n",
+      "    b += 1\n",
+      "    global c \n",
+      "    c -= 1\n",
+      "    sys.stdout.write(\"\\nIn fun() B = %d  C = %d\"%(b,c))\n",
+      "    return\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nIn main() B = %d  C = %d\"%(b,c))\n",
+      "fun()\n",
+      "b += 1\n",
+      "c -= 1\n",
+      "sys.stdout.write(\"\\nAgain In main() B = %d C = %d\"%(b,c))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In main() B = 10  C = 5\n",
+        "In fun() B = 11  C = 4\n",
+        "Again In main() B = 12 C = 3"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.5, Page number: 325<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Using return statement in different ways\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def pass1(a):\n",
+      "    if a == 0:\n",
+      "        return;\n",
+      "    else:\n",
+      "        return a*a*a\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter value of x : \"))\n",
+      "\n",
+      "#Function call & Result\n",
+      "if x != 1 or x > 0:\n",
+      "    y = pass1(x)\n",
+      "\n",
+      "#There is no switch statement in python, so if..else statement\n",
+      "if y == 1:\n",
+      "    sys.stdout.write(\"The value returned is %d\"%(y))\n",
+      "else:\n",
+      "    sys.stdout.write(\"The Cube of %d is : %d\"%(x,y))\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of x : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Cube of 5 is : 125"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.6, Page number: 327<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display message using user defined function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def message():\n",
+      "    sys.stdout.write(\"Have a nice day\")\n",
+      "\n",
+      "#function call\n",
+      "message()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Have a nice day"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.7, Page number: 328<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display Alphabets 'A','B' and 'C' using functions\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Functions definitions\n",
+      "def a():\n",
+      "    sys.stdout.write(\"\\nA\")\n",
+      "    \n",
+      "def b():\n",
+      "    sys.stdout.write(\" B\")\n",
+      "    \n",
+      "def c():\n",
+      "    sys.stdout.write(\" C\")\n",
+      "    \n",
+      "#Function call\n",
+      "a()\n",
+      "b()\n",
+      "c()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "A B C"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.8, Page number: 329<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Send value to user defined function and display results\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def dat(x,y,z):\n",
+      "    sys.stdout.write(\"Date = %d/%d/%d\"%(x,y,z))\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "d = int(raw_input(\"Enter date dd/mm/yy\"))\n",
+      "m = int(raw_input(\"Enter date dd/mm/yy\"))\n",
+      "y = int(raw_input(\"Enter date dd/mm/yy\"))\n",
+      "\n",
+      "#function call & Result\n",
+      "dat(d,m,y)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter date dd/mm/yy12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter date dd/mm/yy12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter date dd/mm/yy2001\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Date = 12/12/2001"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.9, Page number: 330<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Square of number using user defined function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#function definition\n",
+      "def sqr(k):\n",
+      "    sys.stdout.write(\"\\n%d\"%(k*k))\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "j = 0\n",
+      "\n",
+      "#Function call & Result\n",
+      "for j in range(1,5):\n",
+      "    sqr(j)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1\n",
+        "4\n",
+        "9\n",
+        "16"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.10, Page number: 330<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pass the value to main() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no main function in python\n",
+      "sys.stdout.write(\"\\nNumber of command line arguments J = %d\"%(len(sys.argv)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number of command line arguments J = 6"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.11, Page number: 331<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pass and return values to user defined function.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definitions\n",
+      "def dat(x,y,z):\n",
+      "    sys.stdout.write(\"\\nToday = %d/%d/%d\"%(x,y,z))\n",
+      "    x += 1\n",
+      "    return x\n",
+      "\n",
+      "#Variable Initialization\n",
+      "d = int(raw_input(\"Enter Date dd/mm/yy : \"))\n",
+      "m = int(raw_input(\"Enter Date dd/mm/yy : \"))\n",
+      "y = int(raw_input(\"Enter Date dd/mm/yy : \"))\n",
+      "\n",
+      "#Function call\n",
+      "t = dat(d,m,y)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nTomorrow = %d/%d/%d\"%(t,m,y))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Date dd/mm/yy : 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Date dd/mm/yy : 12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Date dd/mm/yy : 2001\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Today = 12/12/2001\n",
+        "Tomorrow = 13/12/2001"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.12, Page number: 332<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Send and recieve values to user defined function \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def sum1(x,y,z):\n",
+      "    return x+y+z\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "b = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "c = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "\n",
+      "#Function call\n",
+      "s = sum1(a,b,c)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Sum = %d\"%(s))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum = 16"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.13, Page number: 333<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Recieve values from user defined function without passing any \n",
+      "#value through main().\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def sum1():\n",
+      "    x = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "    y = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "    z = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "    return x+y+z\n",
+      "\n",
+      "#Function call\n",
+      "s = sum1()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Sum = %d\"%(s))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum = 12"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.14, Page number: 333<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Return value in the form of address.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def sum1():\n",
+      "    x = int(raw_input(\"Enter Three Values : \"))\n",
+      "    y = int(raw_input(\"Enter Three Values : \"))\n",
+      "    z = int(raw_input(\"Enter Three Values : \"))\n",
+      "    k = x + y + z\n",
+      "    return k\n",
+      "\n",
+      "#There is no pointer concept in python\n",
+      "\n",
+      "#Function call\n",
+      "s = sum1()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Sum = %d\"%(s))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Values : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Values : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Values : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum = 12"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.15, Page number: 334<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call by value\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def change(a,b):\n",
+      "    k = a\n",
+      "    a = b\n",
+      "    b = k\n",
+      "    sys.stdout.write(\"\\nIn Change() X = %d Y = %d\"%(a,b))\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter Values of X & Y : \"))\n",
+      "y = int(raw_input(\"Enter Values of X & Y : \"))\n",
+      "\n",
+      "#Function call\n",
+      "change(x,y)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nIn main() X = %d Y = %d\"%(x,y))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values of X & Y : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values of X & Y : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In Change() X = 4 Y = 5\n",
+        "In main() X = 5 Y = 4"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.16, Page number: 335<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call by reference \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def change(a,b):\n",
+      "    k = a\n",
+      "    a = b\n",
+      "    b = k\n",
+      "    sys.stdout.write(\"\\nIn Change() X = %d Y = %d\"%(a,b))\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter Values of X & Y : \"))\n",
+      "y = int(raw_input(\"Enter Values of X & Y : \"))\n",
+      "\n",
+      "#Function call\n",
+      "change(x,y)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nIn main() X = %d Y = %d\"%(y,x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values of X & Y : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values of X & Y : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In Change() X = 4 Y = 5\n",
+        "In main() X = 4 Y = 5"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.17, Page number: 336<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Return by reference\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def change(a,b):\n",
+      "    c = a + b\n",
+      "    d = a - b\n",
+      "    return c,d\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter Values of X & Y : \"))\n",
+      "y = int(raw_input(\"Enter Values of X & Y : \"))\n",
+      "\n",
+      "#Function call\n",
+      "add,sub = change(x,y)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAddition : %d\"%(add))\n",
+      "sys.stdout.write(\"\\nSubtraction : %d\"%(sub))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values of X & Y : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values of X & Y : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Addition : 9\n",
+        "Subtraction : 1"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.18, Page number: 337<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call by value and reference\n",
+      "\n",
+      "k = 0\n",
+      "m = 0\n",
+      "\n",
+      "#Function Definition\n",
+      "def other(k,m):\n",
+      "    sys.stdout.write(\"\\nAddress of k & m in other() : %u %u\"%(id(k),id(m)))\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAddress of k & m in main() : %u %u\"%(id(k),id(m)))\n",
+      "\n",
+      "#Function call\n",
+      "other(k,m)\n",
+      "\n",
+      "#there is no pointer concept in python and it uses value tagged method in data storage\n",
+      "#instead of addressing the memory location, values of same variables are tagged together"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of k & m in main() : 30922996 30922996\n",
+        "Address of k & m in other() : 30922996 30922996"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.19, Page number: 338<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#User defined function as an argument to another function.\n",
+      "\n",
+      "#Variable Initialization\n",
+      "y = 2\n",
+      "\n",
+      "#Function Definitions\n",
+      "def double(m):\n",
+      "    return m*2\n",
+      "\n",
+      "def square(k):\n",
+      "    return k*k\n",
+      "\n",
+      "#Function call\n",
+      "x = double(square(y))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"x = %d\"%(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "x = 8"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.20, Page number: 338<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Two functions as arguments for another functions\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definitions\n",
+      "def x(a,b):\n",
+      "    return abs(a-b)\n",
+      "\n",
+      "def y():\n",
+      "    y = int(raw_input(\"Enter First Number : \"))\n",
+      "    return y\n",
+      "\n",
+      "def z():\n",
+      "    z = int(raw_input(\"Enter Second Number : \"))\n",
+      "    return z\n",
+      "\n",
+      "#Function call\n",
+      "d = x(y(),z())\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nz() - y() = %d\"%(d))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter First Number : 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Second Number : 50\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "z() - y() = 25"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.21, Page number: 339<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Return only absolute value like abs() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def uabs(y):\n",
+      "    if y < 0:\n",
+      "        return y * -1\n",
+      "    else:\n",
+      "        return y\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter a Negative Value : \"))\n",
+      "\n",
+      "#Function call\n",
+      "x = uabs(x)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nX = %d\"%(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Negative Value : -5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "X = 5"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.22, Page number: 340<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Square and cube of an entered number.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definitions\n",
+      "def input1():        #Since input() is a built in function in python, input1() is used\n",
+      "    k = int(raw_input(\"Number : \"))\n",
+      "    return k\n",
+      "\n",
+      "def sqr(m):\n",
+      "    sys.stdout.write(\"\\nSquare : %d\"%(m*m))\n",
+      "    return m\n",
+      "\n",
+      "def cube(m):\n",
+      "    return m*m*m\n",
+      "\n",
+      "#Function call and Result\n",
+      "sys.stdout.write(\"\\nCube : %d\"%(cube(sqr(input1()))))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number : 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Square : 4\n",
+        "Cube : 8"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.23, Page number: 341<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Assign return value of a function to variable.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def input1():\n",
+      "    k = int(raw_input(\"Enter Value of x = \"))\n",
+      "    return k\n",
+      "\n",
+      "#Function call\n",
+      "x = input1()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nx = %d\"%(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of x = 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "x = 5"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.24, Page number: 342<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Addition and subtraction of numbers with return value of function.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def input1():\n",
+      "    k = int(raw_input(\"Enter Value of x = \"))\n",
+      "    return k\n",
+      "\n",
+      "def sqr(m):\n",
+      "    return pow(m,2)\n",
+      "\n",
+      "#Function call\n",
+      "x = sqr(1 - input1() + 1)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSquare = %d\"%(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of x = 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Square = 9"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.25, Page number: 343<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Multiplication and division of numbers with return value of function.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def input1():\n",
+      "    k = int(raw_input(\"Enter Value of x = \"))\n",
+      "    return k\n",
+      "\n",
+      "def sqr(m):\n",
+      "    return pow(m,2)\n",
+      "\n",
+      "#Function call\n",
+      "x = sqr(5 * input1()/2)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSquare = %d\"%(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of x = 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Square = 144"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.26, Page number: 344<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# ++ operator with return value of function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def input1():\n",
+      "    k = int(raw_input(\"Enter Value of x = \"))\n",
+      "    return k\n",
+      "\n",
+      "def sqr(m):\n",
+      "    return pow(m,2)\n",
+      "\n",
+      "#Function call\n",
+      "#There is no ++ operator in python. so += operator is used\n",
+      "y = input1()\n",
+      "y += 1\n",
+      "x = sqr(y)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSquare = %d\"%(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of x = 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Square = 64"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.27, Page number: 345<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use mod(%) with function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def j():\n",
+      "    x = int(raw_input(\"Enter a Number : \"))\n",
+      "    return x\n",
+      "    \n",
+      "#Function call & Result\n",
+      "if j() %2 == 0:\n",
+      "    sys.stdout.write(\"\\nNumber is Even.\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nNumber is Odd.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number is Odd."
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.28, Page number: 346<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Conditional operator(?) with function.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definitions\n",
+      "def sqr(x):\n",
+      "    sys.stdout.write(\"Square \")\n",
+      "    return pow(x,2)\n",
+      "\n",
+      "def cube(x):\n",
+      "    sys.stdout.write(\"Cube \")\n",
+      "    return pow(x,3)\n",
+      "\n",
+      "def y():\n",
+      "    return 10\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Function call\n",
+      "z = sqr(x) if x > y() else cube(x)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\" = %d\"%(z))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cube  = 125"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.29, Page number: 346<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#compare two return values of functions\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definitions\n",
+      "def a():\n",
+      "    x = int(raw_input(\"Enter a Number a() : \"))\n",
+      "    return x\n",
+      "\n",
+      "def b():\n",
+      "    x = int(raw_input(\"Enter a Number b() : \"))\n",
+      "    return x\n",
+      "\n",
+      "#Function call and Result\n",
+      "if a() == b():\n",
+      "    sys.stdout.write(\"\\nValue of a() & b() are equal\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nValue of a() & b() are unique\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number a() : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number b() : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value of a() & b() are equal"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.30, Page number: 347<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the equation s = sqr(a() + b()) using function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definitions\n",
+      "def a():\n",
+      "    a = int(raw_input(\"Enter value of a : \"))\n",
+      "    return a\n",
+      "\n",
+      "def b():\n",
+      "    b = int(raw_input(\"Enter value of b : \"))\n",
+      "    return b\n",
+      "\n",
+      "def sqr(x):\n",
+      "    return x*x\n",
+      "\n",
+      "#Function call\n",
+      "s = sqr(a() + b())\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSquare of Sum = %d\"%(s))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of a : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of b : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Square of Sum = 64"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.31, Page number: 348<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the equation y = x^1+x^2..x^n using function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def b(m):\n",
+      "    m += 1\n",
+      "    #sys.stdout.write(\"%d\"%(m))\n",
+      "    return m\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Values of 'x' and 'n' : \"))\n",
+      "n = int(raw_input(\"Values of 'x' and 'n' : \"))\n",
+      "y = 0\n",
+      "z = 1\n",
+      "m = 0\n",
+      "\n",
+      "while(z <= n):\n",
+      "    m = b(m)\n",
+      "    y = y + pow(x,m)\n",
+      "    sys.stdout.write(\"%d + \"%(y))\n",
+      "    z += 1\n",
+      "    \n",
+      "if z >= n:\n",
+      "    sys.stdout.write(\"\\nValue of y = %d\"%(y))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Values of 'x' and 'n' : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Values of 'x' and 'n' : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3 + 12 + 39 + \n",
+        "Value of y = 39"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.32, Page number: 350<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call user defined function through if statement\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def a():\n",
+      "    a = int(raw_input(\"Enter value of a :\"))\n",
+      "    return a\n",
+      "\n",
+      "#Function call and Result\n",
+      "if a()%2 == 0:\n",
+      "    sys.stdout.write(\"\\nThe number is even.\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nThe number is odd.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of a :5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The number is odd."
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.33, Page number: 351<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call user defined function through switch() statement\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def a():\n",
+      "    c = raw_input(\"Enter Your Choice Square(s), Cube(c), Double(d) : \")\n",
+      "    c = c.lower()\n",
+      "    return c\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 5\n",
+      "\n",
+      "#There is no switch() statement in python.\n",
+      "c = a()\n",
+      "if c == 's':\n",
+      "    sys.stdout.write(\"\\nSquare of %d is %d\"%(x,pow(x,2)))\n",
+      "else:\n",
+      "    if c == 'c':\n",
+      "        sys.stdout.write(\"\\nCube of %d is %d\"%(x,pow(x,3)))\n",
+      "    else:\n",
+      "        if c == 'd':\n",
+      "            sys.stdout.write(\"\\nDouble of %d is %d\"%(x,x*2))\n",
+      "        else:\n",
+      "            sys.stdout.write(\"\\nUnexpected Choice printed as it is : %d\"%(x))\n",
+      "         "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Choice Square(s), Cube(c), Double(d) : D\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Double of 5 is 10"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.34, Page number: 353<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call function through the for loop\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def plus(k):\n",
+      "    if k == 10:\n",
+      "        return 0\n",
+      "    else:\n",
+      "        return k\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "m = 1\n",
+      "\n",
+      "#Function call & Result\n",
+      "#in python, for loop iterates through a range of number. so while loop is used instead.\n",
+      "while plus(m) != 0:\n",
+      "    sys.stdout.write(\"%3d\"%(m))\n",
+      "    m += 1\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  1  2  3  4  5  6  7  8  9"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.35, Page number: 354<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call user defined function through while() loop\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def y():\n",
+      "    x = int(raw_input(\"Enter a Number : \"))\n",
+      "    return x\n",
+      "\n",
+      "#Function call & Result\n",
+      "while y() != 0:\n",
+      "    sys.stdout.write(\"Value enter is non-zero\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value enter is non-zero\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.36, Page number: 355<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call user defined function through do while() loop\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def y():\n",
+      "    x = int(raw_input(\"Enter a Number : \"))\n",
+      "    return x\n",
+      "\n",
+      "#Function call and Result\n",
+      "#There is no do-while loop in python\n",
+      "\n",
+      "while y() != 0:\n",
+      "    sys.stdout.write(\"\\nValue entered is non-zero\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value entered is non-zero\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.37, Page number: 356<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Initialize an array using functions.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def c(m):\n",
+      "    n = int(raw_input(\"Enter Number d[%d]\"%(m+1)))\n",
+      "    return n\n",
+      "\n",
+      "#Variable Initialization\n",
+      "d = [c(i) for i in range(0,5)]\n",
+      "\n",
+      "sys.stdout.write(\"\\nArray d[] elements are : \")\n",
+      "for k in range(0,5):\n",
+      "    sys.stdout.write(\"%2d\"%d[k])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number d[1]4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number d[2]5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number d[3]6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number d[4]7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number d[5]8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Array d[] elements are :  4 5 6 7 8"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.38, Page number: 357<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pass array element to the function using call by value method.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def show(m,u):\n",
+      "    sys.stdout.write(\"\\nnum[%d] = %d\"%(m+1,u))\n",
+      "    \n",
+      "#Variable initialization\n",
+      "num = [12,13,14,15,16,17,18]\n",
+      "\n",
+      "#Function call & Result\n",
+      "for k in range(0,7):\n",
+      "    show(k,num[k])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "num[1] = 12\n",
+        "num[2] = 13\n",
+        "num[3] = 14\n",
+        "num[4] = 15\n",
+        "num[5] = 16\n",
+        "num[6] = 17\n",
+        "num[7] = 18"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.39, Page number: 358<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pass array element to the function using call by reference\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def show(u):\n",
+      "    m = 0\n",
+      "    sys.stdout.write(\"\\nnum[7] = {\")\n",
+      "    while m != 7:\n",
+      "        #There is no pointer concept in python\n",
+      "        sys.stdout.write(\"%2d,\"%(u[m]))\n",
+      "        m += 1\n",
+      "    sys.stdout.write(\"\\b}\")\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "num = [12,13,14,15,16,17,18]\n",
+      "\n",
+      "#Function call\n",
+      "show(num)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "num[7] = {12,13,14,15,16,17,18,\b}"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.40, Page number: 359<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Array elements in reverse order.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def show(u):\n",
+      "    m = 6\n",
+      "    while m != -1:\n",
+      "        sys.stdout.write(\"\\nnum[%d] = %d\"%(m,u[m]))\n",
+      "        m -= 1\n",
+      "        \n",
+      "#Variable Initialization\n",
+      "num = [12,13,14,15,16,17,18]\n",
+      "\n",
+      "#Function call\n",
+      "#There is no pointer concept in python\n",
+      "show(num)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "num[6] = 18\n",
+        "num[5] = 17\n",
+        "num[4] = 16\n",
+        "num[3] = 15\n",
+        "num[2] = 14\n",
+        "num[1] = 13\n",
+        "num[0] = 12"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.41, Page number: 360<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy array elements using user defined function.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def cpy(p,m):\n",
+      "    j = 0\n",
+      "    while j != 5:\n",
+      "        m[j] = p[j]\n",
+      "        j += 1\n",
+      "        \n",
+      "#Variable Initialization\n",
+      "a1 = [1,2,3,4,5]\n",
+      "a2 = [0 for i in range(0,5)]\n",
+      "\n",
+      "#Function call\n",
+      "cpy(a1,a2)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Source  Target\")\n",
+      "for h in range(0,5):\n",
+      "    sys.stdout.write(\"\\n%5d\\t%d\"%(a1[h],a2[h]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Source  Target\n",
+        "    1\t1\n",
+        "    2\t2\n",
+        "    3\t3\n",
+        "    4\t4\n",
+        "    5\t5"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.42, Page number: 361<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read array of other function in main()\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def arry(k):\n",
+      "    b = [1,2,3,4,5]\n",
+      "    return b[k]\n",
+      "\n",
+      "#main() function\n",
+      "for k in range(0,5):\n",
+      "    sys.stdout.write(\"\\t%d\"%(arry(k)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t1\t2\t3\t4\t5"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.43, Page number: 361<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Interchange array elements of two arrays using function.\n",
+      "\n",
+      "import sys\n",
+      "global a\n",
+      "global b\n",
+      "\n",
+      "#Function Definitions\n",
+      "def read():\n",
+      "    x = int(raw_input(\"\"))\n",
+      "    return x\n",
+      "\n",
+      "def change(a,b):\n",
+      "#Since there is no pointer concept in python, exchange is done in the function using global variables.\n",
+      "    for x in range(0,5):\n",
+      "        a[x] = a[x] + b[x]\n",
+      "        b[x] = a[x] - b[x]\n",
+      "        a[x] = a[x] - b[x]\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "a = [0 for i in range(0,5)]\n",
+      "b = [0 for i in range(0,5)]\n",
+      "\n",
+      "for x in range(0,10):\n",
+      "    if x < 5:\n",
+      "        a[x] = read()\n",
+      "    else:\n",
+      "        b[x-5] = read()\n",
+      "\n",
+      "#Swapping and Result\n",
+      "sys.stdout.write(\"\\nArray A & B \")\n",
+      "\n",
+      "for x in range(0,5):\n",
+      "    sys.stdout.write(\"\\n%7d%8d\"%(a[x],b[x]))\n",
+      "  \n",
+      "#There is no pointer concept in python.\n",
+      "change(a,b)\n",
+      "    \n",
+      "sys.stdout.write(\"\\nNow A  & B\")\n",
+      "for x in range(0,5):\n",
+      "    sys.stdout.write(\"\\n%7d%8d\"%(a[x],b[x]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Array A & B \n",
+        "      1       6\n",
+        "      2       7\n",
+        "      3       8\n",
+        "      4       9\n",
+        "      5       0\n",
+        "Now A  & B\n",
+        "      6       1\n",
+        "      7       2\n",
+        "      8       3\n",
+        "      9       4\n",
+        "      0       5"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.44, Page number: 363<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read array elements declared in different functions using global\n",
+      "#pointer declaration\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def call(j):\n",
+      "    m = 0\n",
+      "    u = [5,1,6,0,6]\n",
+      "    q = u\n",
+      "    while m != j:\n",
+      "        sys.stdout.write(\"%3d\"%(u[m]))\n",
+      "        m += 1\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "m = 0\n",
+      "k = [3,8,5,2,5]\n",
+      "q = k\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "while m != 5:\n",
+      "    sys.stdout.write(\"%3d\"%(q[m]))\n",
+      "    m += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\n\")\n",
+      "call(5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  3  8  5  2  5\n",
+        "  5  1  6  0  6\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.45, Page number: 364<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sum of 1 to 5 numbers using recursion\n",
+      "\n",
+      "import sys\n",
+      "global s\n",
+      "\n",
+      "s = 0\n",
+      "#Function definition\n",
+      "def main(x,s):\n",
+      "    s = s + x\n",
+      "    sys.stdout.write(\"\\nx = %d s = %d\"%(x,s))\n",
+      "    if x == 5:\n",
+      "        return\n",
+      "    x += 1\n",
+      "    main(x,s)\n",
+      "    \n",
+      "main(1,s)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "x = 1 s = 1\n",
+        "x = 2 s = 3\n",
+        "x = 3 s = 6\n",
+        "x = 4 s = 10\n",
+        "x = 5 s = 15"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.46, Page number: 365<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate triangular number of a given number with recursion function method\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def tri_num(m):\n",
+      "    f = 0\n",
+      "    if m == 0:\n",
+      "        return f\n",
+      "    else:\n",
+      "        f = f + m + tri_num(m-1)\n",
+      "        return f\n",
+      "\n",
+      "#Variable Initialization\n",
+      "n = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Function call\n",
+      "t = tri_num(n)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nTriangular number of %d is %d\"%(n,t))\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Triangular number of 5 is 15"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.47, Page number: 366<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the given string using recursion\n",
+      "\n",
+      "import sys\n",
+      "import os\n",
+      "global x\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 0\n",
+      "str1 = \"Have a Good Day\"\n",
+      "\n",
+      "#Function Definition\n",
+      "def main(x):\n",
+      "    if x == len(str1):        #There is no null terminating character in python string\n",
+      "        return\n",
+      "    else:\n",
+      "        if str1[x] == 'H':\n",
+      "            os.system('cls')\n",
+      "            sys.stdout.write(\"%c\"%(str1[x]))\n",
+      "        else:\n",
+      "            sys.stdout.write(\"%c\"%(str1[x]))\n",
+      "    x += 1\n",
+      "    main(x)\n",
+      "    \n",
+      "#Function call\n",
+      "main(x)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Have a Good Day"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.48, Page number: 367<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the given string 10 times using recursion\n",
+      "\n",
+      "import sys\n",
+      "import os\n",
+      "\n",
+      "#Function definition\n",
+      "def main(x):\n",
+      "    sys.stdout.write(\"\\n%.2d] %s\"%(x,str1))\n",
+      "    x += 1\n",
+      "    if x == 11:\n",
+      "        return\n",
+      "    else:\n",
+      "        if x == 1:\n",
+      "            os.system('cls')\n",
+      "            main(x)\n",
+      "        else:\n",
+      "            main(x)\n",
+      "            \n",
+      "            \n",
+      "#Variable Initialization\n",
+      "x = 0\n",
+      "str1 = \"Have a Good Day\"\n",
+      "\n",
+      "#Function call\n",
+      "main(x)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "00] Have a Good Day\n",
+        "01] Have a Good Day"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "02] Have a Good Day\n",
+        "03] Have a Good Day\n",
+        "04] Have a Good Day\n",
+        "05] Have a Good Day\n",
+        "06] Have a Good Day\n",
+        "07] Have a Good Day\n",
+        "08] Have a Good Day\n",
+        "09] Have a Good Day\n",
+        "10] Have a Good Day"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.49, Page number: 368<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Factorial using recursive function.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function Definition\n",
+      "def fact(m):\n",
+      "    f = 1\n",
+      "    if m == 1:\n",
+      "        return 1\n",
+      "    else:\n",
+      "        f = m * fact(m-1)\n",
+      "        return f\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Function call\n",
+      "f = fact (x)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nFactorial of %d is %d\"%(x,f))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Factorial of 5 is 120"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.50, Page number: 369<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display address of user defined function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def show():\n",
+      "    sys.stdout.write(\"\\nAddress of function show() is : \")\n",
+      "    \n",
+      "#Function call\n",
+      "show()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"%u\"%(id(show)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of function show() is : 95041520"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.51, Page number: 369<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call function using pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def show():\n",
+      "    sys.stdout.write(\"\\nAddress of function show() is : \")\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "p = id(show)\n",
+      "show()\n",
+      "sys.stdout.write(\"%u\"%(id(show)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of function show() is : 95041200"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.52, Page number: 370<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the address of library function.\n",
+      "\n",
+      "import sys\n",
+      "import os\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAddress of printf() is %u\"%(id(sys.stdout.write)))\n",
+      "sys.stdout.write(\"\\nAddress of scanf() is %u\"%(id(sys.stdout.read)))\n",
+      "sys.stdout.write(\"\\nAddress of clrscr() is %u\"%(id(os.system('cls'))))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of printf() is 60743848\n",
+        "Address of scanf() is 60743848\n",
+        "Address of clrscr() is 4774132"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10.53, Page number: 371<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Call main() using pointer to main() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initilization\n",
+      "x = 0\n",
+      "\n",
+      "#Function definition\n",
+      "def main(x):\n",
+      "    p = id(main)\n",
+      "    x += 1\n",
+      "    sys.stdout.write(\"\\nCall %d Address of main() %u\"%(x,id(main)))\n",
+      "    if x == 3:\n",
+      "        return\n",
+      "    main(x)\n",
+      "    \n",
+      "#function call\n",
+      "main(x)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Call 1 Address of main() 95040880\n",
+        "Call 2 Address of main() 95040880\n",
+        "Call 3 Address of main() 95040880"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter11.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter11.ipynb
new file mode 100755
index 00000000..c8fac375
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter11.ipynb
@@ -0,0 +1,446 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 11: Storage Class<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.1, Page number: 376<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Working of auto variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Functio definitions\n",
+      "def call1():\n",
+      "    v = 20\n",
+      "    sys.stdout.write(\"\\nV = %d\"%(v))\n",
+      "    \n",
+      "def call2():\n",
+      "    v = 30\n",
+      "    call1()\n",
+      "    sys.stdout.write(\"\\nV = %d\"%(v))\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "v = 10\n",
+      "\n",
+      "#Function call\n",
+      "call2()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nV = %d\"%(v))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "V = 20\n",
+        "V = 30\n",
+        "V = 10"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.2, Page number: 376<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Working of auto variables in different blocks\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 10\n",
+      "\n",
+      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
+      "\n",
+      "x = 20\n",
+      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
+      "\n",
+      "x = 10\n",
+      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
+      "\n",
+      "#In python, block of statements are indicated using intendation.\n",
+      "#block of statements can be written for conditional,loop or function statements"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "X = 10\n",
+        "X = 20\n",
+        "X = 10"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.3, Page number: 377<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use same variable in different blocks with different data types\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization & Result\n",
+      "x = 10\n",
+      "sys.stdout.write(\"\\nX = %d\"%(x))\n",
+      "\n",
+      "x = 2.22\n",
+      "sys.stdout.write(\"\\nX = %g\"%(x))\n",
+      "\n",
+      "x = \"Auto Storage Class\"\n",
+      "sys.stdout.write(\"\\nX = %s\"%(x))\n",
+      "\n",
+      "x = 10\n",
+      "sys.stdout.write(\"\\nX = %d\"%(x))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "X = 10\n",
+        "X = 2.22\n",
+        "X = Auto Storage Class\n",
+        "X = 10"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.4, Page number: 378<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Working of external variables\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definitions\n",
+      "def call1():\n",
+      "    sys.stdout.write(\"\\nIn Call1() V = %d\"%(v))\n",
+      "    \n",
+      "def call2():\n",
+      "    sys.stdout.write(\"\\nIn call2() V = %d\"%(v))\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "global v\n",
+      "v = 10\n",
+      "\n",
+      "call1()\n",
+      "call2()\n",
+      "sys.stdout.write(\"\\nIn main() V = %d\"%(v))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In Call1() V = 10\n",
+        "In call2() V = 10\n",
+        "In main() V = 10"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.5, Page number: 379<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Working of auto and global variables with same name\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definitions\n",
+      "def call1():\n",
+      "    v = 20\n",
+      "    sys.stdout.write(\"\\nIn Call1() V = %d\"%(v))\n",
+      "    \n",
+      "def call2():\n",
+      "    sys.stdout.write(\"\\nIn call2() V = %d\"%(v))\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "global v\n",
+      "v = 10\n",
+      "\n",
+      "#Function call\n",
+      "call1()\n",
+      "call2()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nIn main() V = %d\"%(v))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In Call1() V = 20\n",
+        "In call2() V = 10\n",
+        "In main() V = 10"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.6, Page number: 380<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Declare external variables using extern keyword\n",
+      "\n",
+      "#Variable Initialization\n",
+      "global m\n",
+      "m = 10\n",
+      "\n",
+      "#There is no extern keyword in python, global is used instead\n",
+      "sys.stdout.write(\"\\nM = %d\"%(m))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "M = 10"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.7, Page number: 380<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use of static variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def print1(m):\n",
+      "    sys.stdout.write(\"\\nm = %d\"%(m))\n",
+      "    if m == 3:\n",
+      "        return\n",
+      "    \n",
+      "m = 0\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    m += 1\n",
+      "    print1(m)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "m = 1\n",
+        "m = 2\n",
+        "m = 3"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.8, Page number: 381<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Difference between variables of auto and static class\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "y = 0\n",
+      "sys.stdout.write(\"\\nx = %d & y = %d\"%(x,y))\n",
+      "\n",
+      "#Since x display the junk value which stored already, it differs in different systems"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "x = 10 & y = 0"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.9, Page number: 382<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Register class variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no register keyword in python\n",
+      "m = 1\n",
+      "\n",
+      "#Result\n",
+      "for m in range(1,5+1):\n",
+      "    sys.stdout.write(\"\\t%d\"%(m))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t1\t2\t3\t4\t5"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11.10, Page number: 382<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Register class variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no register class variable in python\n",
+      "m = 1\n",
+      "\n",
+      "for m in range(1,6):\n",
+      "    sys.stdout.write(\"\\t%d\"%(m))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t1\t2\t3\t4\t5"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter12.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter12.ipynb
new file mode 100755
index 00000000..e1acaa2c
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter12.ipynb
@@ -0,0 +1,849 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 12: Preprocessor Directives<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.1, Page number: 386<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Area of circle\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Defining the macro\n",
+      "#There is no macro definitions/preprocessors in python\n",
+      "PI =  3.14\n",
+      "\n",
+      "#Variable Initialization\n",
+      "r = float(raw_input(\"Enter radius of circle in cms.\"))\n",
+      "\n",
+      "#Calculation\n",
+      "area = PI * r * r\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Area of a Circle = %.2f cm2\"%(area))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter radius of circle in cms.7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area of a Circle = 153.86 cm2"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.2, Page number: 387<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Identifers for 'c' statements and variables\n",
+      "\n",
+      "import sys\n",
+      "import os\n",
+      "\n",
+      "#Macro definitions\n",
+      "#in python, function definitions are used instead of macros\n",
+      "def N():\n",
+      "    return 10\n",
+      "\n",
+      "def cls():\n",
+      "    os.system('cls')\n",
+      "    \n",
+      "def wait():\n",
+      "    t = raw_input(\"\")\n",
+      "    \n",
+      "def display(s):\n",
+      "    sys.stdout.write(s)\n",
+      "\n",
+      "for k in range(1,N()+1):\n",
+      "    display(\"%d\\t\"%(k))\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\t2\t3\t4\t5\t6\t7\t8\t9\t10\t"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.3, Page number: 387<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Macros for logical operators\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Macro definitions\n",
+      "#in python, function definitions are used as macros\n",
+      "\n",
+      "#There is no preprocessors in python\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "b = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "c = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "\n",
+      "if a > b and a > c:\n",
+      "    sys.stdout.write(\"%d is the larger number\"%(a))\n",
+      "else:\n",
+      "    if b > a and b > c:\n",
+      "        sys.stdout.write(\"%d is the larger number\"%(b))\n",
+      "    else:\n",
+      "        if c > a and c > b:\n",
+      "            sys.stdout.write(\"%d is the larger number\"%(c))\n",
+      "        else:\n",
+      "            if a == b and b == c:\n",
+      "                sys.stdout.write(\"\\nNumbers are same.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7 is the larger number"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.4, Page number: 388<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Identifier for displaying double and triple of a number.\n",
+      "\n",
+      "#Macro definitions\n",
+      "#Function definitions are used as macros in python\n",
+      "\n",
+      "def DOUBLE(a):\n",
+      "    return a*2\n",
+      "\n",
+      "def TRIPLE(a):\n",
+      "    return a*3\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSINGLE\\tDOUBLE\\tTRIPLE\")\n",
+      "\n",
+      "for a in range(1,5+1):\n",
+      "    sys.stdout.write(\"\\n%d\\t%d\\t%d\"%(a,DOUBLE(a),TRIPLE(a)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "SINGLE\tDOUBLE\tTRIPLE\n",
+        "1\t2\t3\n",
+        "2\t4\t6\n",
+        "3\t6\t9\n",
+        "4\t8\t12\n",
+        "5\t10\t15"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.5, Page number: 389<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Undefine a macro\n",
+      "\n",
+      "#There is no preprocessor directives in python\n",
+      "\n",
+      "def wait():\n",
+      "    return (raw_input(\"\"))\n",
+      "\n",
+      "#Result\n",
+      "for k in range(1,5+1):\n",
+      "    sys.stdout.write(\"%d\\t\"%(k))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\t2\t3\t4\t5\t"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.6, Page number: 389<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Stringizing operation\n",
+      "\n",
+      "#Macro definition\n",
+      "#in python, function definition can be used as macro\n",
+      "\n",
+      "def say(m):\n",
+      "    sys.stdout.write(m)\n",
+      "    \n",
+      "#Function call\n",
+      "say(\"Hello\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hello"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.7, Page number: 390<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Stringizing operation and macro arguments\n",
+      "\n",
+      "#There is no macro in python, function definitions can be used as macros\n",
+      "\n",
+      "def DOUBLE(x):\n",
+      "    sys.stdout.write(\"Double of m = %d\\n\"%(x*2))\n",
+      "    \n",
+      "def TRIPLE(x):\n",
+      "    sys.stdout.write(\"Triple of m = %d\\n\"%(x*3))\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "m = int(raw_input(\"Enter a number : \"))\n",
+      "\n",
+      "#Function call & Result\n",
+      "DOUBLE(m)\n",
+      "TRIPLE(m)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Double of m = 10\n",
+        "Triple of m = 15\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.8, Page number: 390<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Larger of two numbers using macro with arguments.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Macro definition  //function definition itself is used as macro\n",
+      "def MAX(x,y):\n",
+      "    if x > y:\n",
+      "        c = x\n",
+      "    else:\n",
+      "        c = y\n",
+      "    return c\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 5\n",
+      "y = 8\n",
+      "\n",
+      "#Function call\n",
+      "c = MAX(x,y)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nLarger of two numbers = %d\"%(c))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Larger of two numbers = 8"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.9, Page number: 391<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Function defined in \"udf.c\" file.\n",
+      "\n",
+      "import sys\n",
+      "#since ipython notebook does not support this function, function is used\n",
+      "#in python, the udf.c file can be written and saved in udf.py and import command\n",
+      "#can be used to include that file in this program\n",
+      "\n",
+      "def display():\n",
+      "    sys.stdout.write(\"\\nFunction Called\")\n",
+      "    \n",
+      "#Function call\n",
+      "display()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Function Called"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.10, Page number: 392<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Conditional compilation \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no preprocessor directives in python\n",
+      "\n",
+      "LINE = 1\n",
+      "\n",
+      "if LINE:\n",
+      "    sys.stdout.write(\"This is line number one\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"This is line number two\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This is line number one"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.11, Page number: 393<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Conditional compilation \n",
+      "import sys\n",
+      "\n",
+      "#There is no preprocessor directives in python\n",
+      "def E():\n",
+      "    return 1\n",
+      "\n",
+      "#Function call\n",
+      "if E():\n",
+      "    a = 2\n",
+      "    b = 3\n",
+      "    sys.stdout.write(\"A = %d  &  B = %d\"%(a,b))\n",
+      "else:\n",
+      "    c = 2\n",
+      "    d = 3\n",
+      "    sys.stdout.write(\"C = %d  &  D = %d\"%(c,d))\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A = 2  &  B = 3"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.12, Page number: 394<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Conditional compilation directives \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no preprocessor directives in python\n",
+      "T = 8\n",
+      "\n",
+      "#Result\n",
+      "if T == 0:\n",
+      "    sys.stdout.write(\"\\nMacro is not defined.\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nMacro is defined\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Macro is defined"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.13, Page number: 394<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#User defined error message \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no preprocessor directives in python\n",
+      "B = 1\n",
+      "A = 0\n",
+      "\n",
+      "#Result\n",
+      "if A == 0:\n",
+      "    sys.stdout.write(\"MACRO A IS NOT DEFINED.\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"Macro found.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MACRO A IS NOT DEFINED."
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.14, Page number: 397<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Set off certain errors \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no preprocessor directives in python\n",
+      "x = 2\n",
+      "y = 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\ny = %d\"%(y))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "y = 1"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.15, Page number: 398<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Predefined macros \n",
+      "\n",
+      "import sys\n",
+      "import datetime\n",
+      "\n",
+      "#Variable Initialization\n",
+      "FILE = 'PRE_MA~1.C'\n",
+      "LINE = 10\n",
+      "STDC = 1\n",
+      "\n",
+      "#Date/time can be set using datetime.date(2001,5,12)\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nDATE    :  %s\"%('Dec 05 2001')) \n",
+      "sys.stdout.write(\"\\nTIME    :  %s\"%(datetime.time(21,7,39)))\n",
+      "sys.stdout.write(\"\\nFILE NAME : %s\"%(FILE))\n",
+      "sys.stdout.write(\"\\nLINE NO : %d\"%(LINE))\n",
+      "sys.stdout.write(\"\\n_STDC_  :  %x\"%(STDC))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "DATE    :  Dec 05 2001\n",
+        "TIME    :  21:07:39\n",
+        "FILE NAME : PRE_MA~1.C\n",
+        "LINE NO : 10\n",
+        "_STDC_  :  1"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.16, Page number: 399<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the name of memory model that is currently in use.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no formal memory model and preprocessor directives in python\n",
+      "\n",
+      "TINY = 0\n",
+      "SMALL = 0\n",
+      "MEDIUM = 0\n",
+      "COMPACT = 0\n",
+      "LARGE = 1\n",
+      "HUGE = 0\n",
+      "\n",
+      "#Result\n",
+      "if TINY:\n",
+      "    sys.stdout.write(\"\\nTINY %d\"%(TINY))\n",
+      "else:\n",
+      "    if SMALL:\n",
+      "        sys.stdout.write(\"\\nSMALL %d\"%(SMALL))\n",
+      "    else:\n",
+      "        if MEDIUM:\n",
+      "            sys.stdout.write(\"\\nMEDIUM %d\"%(MEDIUM))\n",
+      "        else:\n",
+      "            if COMPACT:\n",
+      "                sys.stdout.write(\"\\nCOMPACT %d\"%(COMPACT))\n",
+      "            else:\n",
+      "                if LARGE:\n",
+      "                    sys.stdout.write(\"\\nLARGE %d\"%(LARGE))\n",
+      "                else:\n",
+      "                    sys.stdout.write(\"\\nHUGE %d\"%(HUGE))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "LARGE 1"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.17, Page number: 400<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Macro expansions\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no preprocessor directives in python\n",
+      "ch = raw_input(\"Input a Text : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Text Input : %s\"%(ch))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Input a Text : Programming\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Text Input : Programming"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12.18, Page number: 401<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Predefined macros.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "d = raw_input(\"Enter any character : \")\n",
+      "\n",
+      "#Result\n",
+      "f = d.isalpha()\n",
+      "\n",
+      "if f != 0:\n",
+      "    sys.stdout.write(\"\\n%c is a letter in\"%(d))\n",
+      "    f = d.isupper()\n",
+      "    if f != 0:\n",
+      "        sys.stdout.write(\" Capital case\")\n",
+      "    else:\n",
+      "        sys.stdout.write(\" Small case\")\n",
+      "else:\n",
+      "    f = d.isdigit()\n",
+      "    if f != 0:\n",
+      "        sys.stdout.write(\"\\n%c is a digit\"%(d))\n",
+      "    else:\n",
+      "        f = d.ispunct()\n",
+      "        if f != 0:\n",
+      "            sys.stdout.write(\"\\n%c is a punctuation symbol\"%(d))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any character : C\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "C is a letter in Capital case"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter13.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter13.ipynb
new file mode 100755
index 00000000..f9222003
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter13.ipynb
@@ -0,0 +1,2367 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 13: Structure and Union<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.1, Page number: 409<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display size of structure elements\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class is used instead of structure in python\n",
+      "class book1:\n",
+      "    book = ''\n",
+      "    pages = 0\n",
+      "    price = 0.00\n",
+      "\n",
+      "#Python variables uses value tagged methods for storing the values\n",
+      "\n",
+      "#Class variable declaration\n",
+      "bk1 = book1()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSize of Structure Elements\")\n",
+      "sys.stdout.write(\"\\nBook    :  %d\"%(sys.getsizeof(bk1.book)))\n",
+      "sys.stdout.write(\"\\nPages    :  %d\"%(sys.getsizeof(bk1.pages)))\n",
+      "sys.stdout.write(\"\\nPrice    :  %d\"%(sys.getsizeof(bk1.price)))\n",
+      "sys.stdout.write(\"\\nTotal Bytes    :  %d\"%(sys.getsizeof(bk1)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Size of Structure Elements\n",
+        "Book    :  21\n",
+        "Pages    :  12\n",
+        "Price    :  16\n",
+        "Total Bytes    :  36"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.2, Page number: 410<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Define a structure and initialize its member variables\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class book1:\n",
+      "    book = ''\n",
+      "    pages = 0\n",
+      "    price = 0.0\n",
+      "    \n",
+      "#Class variable declaration and Initialization\n",
+      "bk1 = book1()\n",
+      "bk1.book = \"C++\"\n",
+      "bk1.pages = 300\n",
+      "bk1.price = 285\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nBook Name  : %s\"%(bk1.book))\n",
+      "sys.stdout.write(\"\\nNo. of Pages : %d\"%(bk1.pages))\n",
+      "sys.stdout.write(\"\\nBook Price  : %d\"%(bk1.price))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Book Name  : C++\n",
+        "No. of Pages : 300\n",
+        "Book Price  : 285"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.3, Page number: 411<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy structure elements from one object to another object\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class disk:\n",
+      "    co = ''\n",
+      "    type1 = 0.0\n",
+      "    price = 0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "d1 = disk()\n",
+      "d2 = disk()\n",
+      "d3 = disk()\n",
+      "\n",
+      "#Class variable initialization\n",
+      "d1.co = \"SONY\"\n",
+      "d1.type1 = 1.44\n",
+      "d1.price = 20\n",
+      "\n",
+      "#Copying\n",
+      "d2.co = d1.co\n",
+      "d2.type1 = d1.type1\n",
+      "d2.price = d1.price\n",
+      "\n",
+      "d3 = d2\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n%s  %g  %d\"%(d1.co,d1.type1,d1.price))\n",
+      "sys.stdout.write(\"\\n%s  %g  %d\"%(d2.co,d2.type1,d2.price))\n",
+      "sys.stdout.write(\"\\n%s  %g  %d\"%(d3.co,d3.type1,d3.price))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "SONY  1.44  20\n",
+        "SONY  1.44  20\n",
+        "SONY  1.44  20"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.4, Page number: 412<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read values and assign them to structure variables\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class book1:\n",
+      "    book = ''\n",
+      "    pages = 0\n",
+      "    price = 0.0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "bk1 = book1()\n",
+      "\n",
+      "#Variable initialization\n",
+      "bk1.book = raw_input(\"Enter Book name,pages,price : \")\n",
+      "bk1.pages = int(raw_input(\"Enter Book name,pages,price : \"))\n",
+      "bk1.price = float(raw_input(\"Enter Book name, pages,price : \"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nBook Name : %s\"%(bk1.book))\n",
+      "sys.stdout.write(\"\\nNo. of Pages : %d\"%(bk1.pages))\n",
+      "sys.stdout.write(\"\\nBook Price : %d\"%(bk1.price))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Book name,pages,price : C\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Book name,pages,price : 500\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Book name, pages,price : 450\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Book Name : C\n",
+        "No. of Pages : 500\n",
+        "Book Price : 450"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.5, Page number: 414<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read and display car details\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class time:\n",
+      "    second = 0\n",
+      "    minute = 0\n",
+      "    hour = 0\n",
+      "    \n",
+      "class t:\n",
+      "    carno = 0\n",
+      "    st = time()\n",
+      "    rt = time()\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "r1 = t()\n",
+      "\n",
+      "#Variable Initialization\n",
+      "r1.carno = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
+      "r1.st.hour = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
+      "r1.st.minute = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
+      "r1.st.second = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
+      "r1.rt.hour = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
+      "r1.rt.minute = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
+      "r1.rt.second = int(raw_input(\"Car No., Starting Time, Reaching Time :\"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nCar No. \\tStarting Time \\tReaching Time\\n\")\n",
+      "sys.stdout.write(\"%d\\t\"%(r1.carno))\n",
+      "sys.stdout.write(\"\\t%d:%d:%d\\t\\t\"%(r1.st.hour,r1.st.minute,r1.st.second))\n",
+      "sys.stdout.write(\"%d:%d:%d\"%(r1.rt.hour,r1.rt.minute,r1.rt.second))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No., Starting Time, Reaching Time :125\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No., Starting Time, Reaching Time :2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No., Starting Time, Reaching Time :50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No., Starting Time, Reaching Time :30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No., Starting Time, Reaching Time :3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No., Starting Time, Reaching Time :50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No., Starting Time, Reaching Time :25\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Car No. \tStarting Time \tReaching Time\n",
+        "125\t\t2:50:30\t\t3:50:25"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.6, Page number: 415<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read and display the details of a person\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class name:\n",
+      "    first = ''\n",
+      "    second = ''\n",
+      "    last = ''\n",
+      "    \n",
+      "class b_date:\n",
+      "    day = 0\n",
+      "    month = 0\n",
+      "    year = 0\n",
+      "    \n",
+      "class data:\n",
+      "    nm = name()\n",
+      "    bt = b_date()\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "r1 = data()\n",
+      "\n",
+      "#Variable Initialization\n",
+      "r1.nm.first = raw_input(\"Enter Name (First/Second/Last)\")\n",
+      "r1.nm.second = raw_input(\"Enter Name (First/Second/Last)\")\n",
+      "r1.nm.last = raw_input(\"Enter Name (First/Second/Last)\")\n",
+      "r1.bt.day = int(raw_input(\"Enter Birth Date Day/Month/Year\"))\n",
+      "r1.bt.month = int(raw_input(\"Enter Birth Date Day/Month/Year\"))\n",
+      "r1.bt.year = int(raw_input(\"Enter Birth Date Day/Month/Year\"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Name  : %s %s %s\\n\"%(r1.nm.first,r1.nm.second,r1.nm.last))\n",
+      "sys.stdout.write(\"Birth Date : %d.%d.%d\"%(r1.bt.day,r1.bt.month,r1.bt.year))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name (First/Second/Last)Ram\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name (First/Second/Last)Sham\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name (First/Second/Last)Pande\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Birth Date Day/Month/Year12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Birth Date Day/Month/Year12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Birth Date Day/Month/Year1980\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name  : Ram Sham Pande\n",
+        "Birth Date : 12.12.1980"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.7, Page number: 416<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Create array of structure objects\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class time:\n",
+      "    second = 0\n",
+      "    minute = 0\n",
+      "    hour = 0\n",
+      "    \n",
+      "class t:\n",
+      "    carno = 0\n",
+      "    st = time()\n",
+      "    rt = time()\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "r1 = [t() for i in range(0,3)]\n",
+      "\n",
+      "#Variable Initialization\n",
+      "for k in range(0,3):\n",
+      "    r1[k].carno = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
+      "    r1[k].st.hour = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
+      "    r1[k].st.minute = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
+      "    r1[k].st.second = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
+      "    r1[k].rt.hour = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
+      "    r1[k].rt.minute = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
+      "    r1[k].rt.second = int(raw_input(\"Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)\"))\n",
+      "   \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nCar No.\\t\\tStarting Time\\tReaching Time\")\n",
+      "for k in range(0,3):\n",
+      "    sys.stdout.write(\"\\n%d\\t\"%(r1[k].carno))\n",
+      "    sys.stdout.write(\"\\t%d:%d:%d\\t\"%(r1[k].st.hour,r1[k].st.minute,r1[k].st.second))\n",
+      "    sys.stdout.write(\"\\t%d:%d:%d\\t\"%(r1[k].rt.hour,r1[k].rt.minute,r1[k].rt.second))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)120\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)58\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)121\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)122\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)52\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Car No. Starting Time(hh:mm:ss),Reaching Time(hh:mm:ss)10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Car No.\t\tStarting Time\tReaching Time\n",
+        "120\t\t4:30:52\t\t5:40:10\t\n",
+        "121\t\t4:30:52\t\t5:40:10\t\n",
+        "122\t\t4:30:52\t\t5:40:10\t"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.8, Page number: 417<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read and display student details\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class stud:\n",
+      "    name = ''\n",
+      "    rollno = 0\n",
+      "    grade = ''\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "st = [stud() for i in range(0,3)]\n",
+      "\n",
+      "#Variable Initialization\n",
+      "k = 0 \n",
+      "while k < 3:\n",
+      "    st[k].name = raw_input(\"Name :\")\n",
+      "    st[k].rollno = int(raw_input(\"Roll No. :\"))\n",
+      "    st[k].grade = raw_input(\"Grade : \")\n",
+      "    k += 1\n",
+      "    \n",
+      "#Result\n",
+      "k = 0\n",
+      "sys.stdout.write(\"\\nName \\tRollno \\tGrade\\n\")\n",
+      "while k < 3:\n",
+      "    sys.stdout.write(\"\\n%s\\t%d\\t%s\"%(st[k].name,st[k].rollno,st[k].grade))\n",
+      "    k += 1\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name :Suresh\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Roll No. :125\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Grade : A\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name :Mahesh\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Roll No. :126\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Grade : B\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name :Rajesh\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Roll No. :127\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Grade : A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name \tRollno \tGrade\n",
+        "\n",
+        "Suresh\t125\tA\n",
+        "Mahesh\t126\tB\n",
+        "Rajesh\t127\tA"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.9, Page number: 419<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pointer to structure \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class book:\n",
+      "    name = ''\n",
+      "    author = ''\n",
+      "    pages = 0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "b1 = book()\n",
+      "\n",
+      "#Variable Initialization\n",
+      "b1.name = \"JAVA COMPLETE REFERENCE\"\n",
+      "b1.author = \"P.NAUGHTON\"\n",
+      "b1.pages = 886\n",
+      "\n",
+      "#There is no pointer concept in python\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n%s by %s of %d pages\"%(b1.name,b1.author,b1.pages))\n",
+      "sys.stdout.write(\"\\n%s by %s of %d pages\"%(b1.name,b1.author,b1.pages))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "JAVA COMPLETE REFERENCE by P.NAUGHTON of 886 pages\n",
+        "JAVA COMPLETE REFERENCE by P.NAUGHTON of 886 pages"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.10, Page number: 420<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pointer as members of structure \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class boy:\n",
+      "    name = ''\n",
+      "    age = 0\n",
+      "    height = 0.0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "sp = boy()\n",
+      "\n",
+      "#Variable Initialization\n",
+      "sp.name = \"Mahesh\"\n",
+      "sp.age = 20\n",
+      "sp.height = 5.40\n",
+      "\n",
+      "#There is no pointer concept in python\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nName  = %s\"%(sp.name))\n",
+      "sys.stdout.write(\"\\nAge  = %s\"%(sp.age))\n",
+      "sys.stdout.write(\"\\nHeight = %s\"%(sp.height))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name  = Mahesh\n",
+        "Age  = 20\n",
+        "Height = 5.4"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.11, Page number: 421<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pointer as members of structure \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class boy:\n",
+      "    name = ''\n",
+      "    age = 0\n",
+      "    height = 0.0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "b = boy()\n",
+      "\n",
+      "#Variable Initialization\n",
+      "nm = \"Somesh\"\n",
+      "ag = 20\n",
+      "ht = 5.40\n",
+      "\n",
+      "#There is no pointer concept in python\n",
+      "b.name = nm\n",
+      "b.age = ag\n",
+      "b.height = ht\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nName  : %s\"%(b.name))\n",
+      "sys.stdout.write(\"\\nAge = %d\"%(b.age))\n",
+      "sys.stdout.write(\"\\nHeight = %.2g\"%(b.height))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name  : Somesh\n",
+        "Age = 20\n",
+        "Height = 5.4"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.12, Page number: 422<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the contents of the structure using ordinary pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class num:\n",
+      "    a = 0\n",
+      "    b = 0\n",
+      "    c = 0\n",
+      "\n",
+      "#Class variable declaration\n",
+      "d = num()\n",
+      "\n",
+      "#Variable Initialization\n",
+      "d.a = 2\n",
+      "d.b = 3\n",
+      "d.c = 4\n",
+      "\n",
+      "#There is no pointer concept in python\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\na = %d\"%(d.a))\n",
+      "sys.stdout.write(\"\\nb = %d\"%(d.b))\n",
+      "sys.stdout.write(\"\\nc = %d\"%(d.c))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "a = 2\n",
+        "b = 3\n",
+        "c = 4"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.13, Page number: 423<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Passing address of structure variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class book:\n",
+      "    name = ''\n",
+      "    author = ''\n",
+      "    pages = 0\n",
+      "    \n",
+      "#Class variable declaration and initialization\n",
+      "b1 = book()\n",
+      "b1.name = \"JAVA COMPLETE REFERENCE\"\n",
+      "b1.author = \"P.NAUGHTON\"\n",
+      "b1.pages = 886\n",
+      "\n",
+      "#Function definition\n",
+      "def show(b1):\n",
+      "    sys.stdout.write(\"\\n%s by %s of %d pages\"%(b1.name,b1.author,b1.pages))\n",
+      "\n",
+      "#Function call\n",
+      "show(b1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "JAVA COMPLETE REFERENCE by P.NAUGHTON of 886 pages"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.14, Page number: 424<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Passing structure elements to function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class boy:\n",
+      "    name = ''\n",
+      "    age = 0\n",
+      "    wt = 0\n",
+      "    \n",
+      "#Class variable declaration and initialization\n",
+      "b1 = boy()\n",
+      "b1.name = \"Amit\"\n",
+      "b1.age = 20\n",
+      "b1.wt = 25\n",
+      "\n",
+      "#Function definition\n",
+      "def print1(s,t,n):\n",
+      "    sys.stdout.write(\"\\n%s %d %d\"%(s,t,n))\n",
+      "    \n",
+      "#Function call\n",
+      "print1(b1.name,b1.age,b1.wt)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Amit 20 25"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.15, Page number: 425<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Pass entire structure to user defined function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class boy:\n",
+      "    name = ''\n",
+      "    age = 0\n",
+      "    wt = 0\n",
+      "    \n",
+      "#Class variable declaration and initialization\n",
+      "b1 = boy()\n",
+      "b1.name = \"Amit\"\n",
+      "b1.age = 20\n",
+      "b1.wt = 25\n",
+      "\n",
+      "#Function definition\n",
+      "def print1(b):\n",
+      "    sys.stdout.write(\"\\n%s %d %d\"%(b.name,b.age,b1.wt))\n",
+      "    \n",
+      "#Function call\n",
+      "print1(b1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Amit 20 25"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.16, Page number: 426<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#User defined data type \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no preprocessor directive in python\n",
+      "H = 60\n",
+      "\n",
+      "#There is no typedef function in python and no separate variable declaration is needed\n",
+      "hrs = int(raw_input(\"Enter Hours :\"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nMinutes = %d\"%(hrs*H))\n",
+      "sys.stdout.write(\"\\nSeconds = %d\"%(hrs*H*H))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Hours :2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Minutes = 120\n",
+        "Seconds = 7200"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.17, Page number: 426<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#String data type\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no typedef function in python\n",
+      "a = \" Hello \"\n",
+      "b = raw_input(\"Enter Your Name : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"%s %s\"%(a,b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name : KAMAL\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Hello  KAMAL"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.18, Page number: 427<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Create user defined data type from structure\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "#There is no typedef function in python\n",
+      "class struct:\n",
+      "    name = ''\n",
+      "    sex = ''\n",
+      "    acno = 0\n",
+      "    \n",
+      "#class variable declaration and initialization\n",
+      "employee = struct()\n",
+      "employee.name = \"Sanjay\"\n",
+      "employee.sex = \"M\"\n",
+      "employee.acno = 125\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nName\\tSex\\tA/c No.\\n\")\n",
+      "sys.stdout.write(\"%s\\t\"%(employee.name))\n",
+      "sys.stdout.write(\"%s\\t\"%(employee.sex))\n",
+      "sys.stdout.write(\"%s\\n\"%(employee.acno))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name\tSex\tA/c No.\n",
+        "Sanjay\tM\t125\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.19, Page number: 427<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Create user defined data type from structure\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class struct:\n",
+      "    name = ''\n",
+      "    sex = ''\n",
+      "    acno = 0\n",
+      "    \n",
+      "#class Variable declaration and initialization\n",
+      "employee = [struct() for i in range(0,2)]\n",
+      "\n",
+      "for k in range(0,2):\n",
+      "    employee[k].name = raw_input(\"Name of the Employee :\")\n",
+      "    employee[k].sex = raw_input(\"Sex\")\n",
+      "    employee[k].acno = raw_input(\"A/c No.\")\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nName\\tSex\\tA/c No.\\n\")\n",
+      "for k in range(0,2):\n",
+      "    sys.stdout.write(\"%s\\t\"%(employee[k].name))\n",
+      "    sys.stdout.write(\"%s\\t\"%(employee[k].sex))\n",
+      "    sys.stdout.write(\"%s\\n\"%(employee[k].acno))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name of the Employee :AJAY\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "SexM\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A/c No.122\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name of the Employee :ANITA\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "SexF\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A/c No.124\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name\tSex\tA/c No.\n",
+        "AJAY\tM\t122\n",
+        "ANITA\tF\t124\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.20, Page number: 429<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Structure containing the details of the employees\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no typedef function in python\n",
+      "\n",
+      "#Class definition\n",
+      "class struct:\n",
+      "    first = ''\n",
+      "    middle = ''\n",
+      "    last = ''\n",
+      "    city = ''\n",
+      "    pincode = 0\n",
+      "    \n",
+      "#Class variable declaration and initialization\n",
+      "person = [struct() for i in range(0,2)]\n",
+      "\n",
+      "for j in range(0,2):\n",
+      "    person[j].first = raw_input(\"First Name :\")\n",
+      "    person[j].middle = raw_input(\"Middle Name : \")\n",
+      "    person[j].last = raw_input(\"Last Name : \")\n",
+      "    person[j].city = raw_input(\"City & Pincode\")\n",
+      "    person[j].pincode = int(raw_input(\"City & Pincode\"))\n",
+      "    \n",
+      "#Result\n",
+      "for j in range(0,2):\n",
+      "    sys.stdout.write(\"\\nRecord No : %d\"%(j+1))\n",
+      "    sys.stdout.write(\"\\nFirst Name : %s\"%(person[j].first))\n",
+      "    sys.stdout.write(\"\\nMiddle Name : %s\"%(person[j].middle))\n",
+      "    sys.stdout.write(\"\\nLast Name : %s\"%(person[j].last))\n",
+      "    sys.stdout.write(\"\\nCity & Pincode : %s - %d\\n\"%(person[j].city,person[j].pincode))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "First Name :Jay\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Middle Name : Mohan\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Last Name : Deshmukh\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "City & PincodeNanded\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "City & Pincode431602\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "First Name :Vijay\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Middle Name : Kamal\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Last Name : Nandedkar\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "City & PincodeNanded\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "City & Pincode431602\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Record No : 1\n",
+        "First Name : Jay\n",
+        "Middle Name : Mohan\n",
+        "Last Name : Deshmukh\n",
+        "City & Pincode : Nanded - 431602\n",
+        "\n",
+        "Record No : 2\n",
+        "First Name : Vijay\n",
+        "Middle Name : Kamal\n",
+        "Last Name : Nandedkar\n",
+        "City & Pincode : Nanded - 431602\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.21, Page number: 431<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Information of vehicles\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "PETROL = 1\n",
+      "DISEL = 2\n",
+      "TWO_WH = 3\n",
+      "FOUR_WH = 4\n",
+      "OLD = 5\n",
+      "NEW = 6\n",
+      "\n",
+      "#Class declaration\n",
+      "class vehicle:\n",
+      "    type1 = 3\n",
+      "    fuel = 2\n",
+      "    model = 3\n",
+      "    \n",
+      "#Class variable declaration and initialization\n",
+      "v = vehicle()\n",
+      "v.type1 = FOUR_WH\n",
+      "v.fuel = DISEL\n",
+      "v.model = OLD\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nType of Vehicle : %d\"%(v.type1))\n",
+      "sys.stdout.write(\"\\nFuel  : %d\"%(v.fuel))\n",
+      "sys.stdout.write(\"\\nModel  : %d\"%(v.model))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Type of Vehicle : 4\n",
+        "Fuel  : 2\n",
+        "Model  : 5"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.22, Page number: 432<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the examination result of the student \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "#There is no preprocessor directives in python\n",
+      "PASS = 1\n",
+      "FAIL = 0\n",
+      "A = 0\n",
+      "B = 1\n",
+      "C = 2\n",
+      "\n",
+      "#Class declaration\n",
+      "class student:\n",
+      "    name = ''\n",
+      "    result = 1\n",
+      "    grade = 2\n",
+      "    \n",
+      "#Class variable declaration and initialization\n",
+      "v = student()\n",
+      "\n",
+      "v.name = \"Sachin\"\n",
+      "v.result = PASS\n",
+      "v.grade = C\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nName : %s\"%(v.name))\n",
+      "sys.stdout.write(\"\\nResult : %d\"%(v.result))\n",
+      "sys.stdout.write(\"\\nGrade : %d\"%(v.grade))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name : Sachin\n",
+        "Result : 1\n",
+        "Grade : 2"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.23, Page number: 433<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Enumerated data type for 12 months\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no enumerated data type in python and an alternate is to use class\n",
+      "#Class declaration\n",
+      "class month:\n",
+      "    Jan = 0\n",
+      "    Feb = 1\n",
+      "    Mar = 2\n",
+      "    Apr = 3\n",
+      "    May = 4\n",
+      "    June = 5\n",
+      "    July = 6\n",
+      "    Aug = 7\n",
+      "    Sep = 8\n",
+      "    Oct = 9\n",
+      "    Nov = 10\n",
+      "    Dec = 11\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "c = month()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nJan = %d\"%(c.Jan))\n",
+      "sys.stdout.write(\"\\nFeb = %d\"%(c.Feb))\n",
+      "sys.stdout.write(\"\\nJune = %d\"%(c.June))\n",
+      "sys.stdout.write(\"\\nDec = %d\"%(c.Dec))\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Jan = 0\n",
+        "Feb = 1\n",
+        "June = 5\n",
+        "Dec = 11"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.24, Page number: 434<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Enumerated data type \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no enumerated data type in python and an alternate is to use class\n",
+      "#Class declaration\n",
+      "class month:\n",
+      "    Jan = 1\n",
+      "    Feb = 2\n",
+      "    Mar = 3\n",
+      "    Apr = 4\n",
+      "    May = 5\n",
+      "    June = 6\n",
+      "    July = 7\n",
+      "    Aug = 8\n",
+      "    Sep = 9\n",
+      "    Oct = 10\n",
+      "    Nov = 11\n",
+      "    Dec = 12\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "c = month()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nJan = %d\"%(c.Jan))\n",
+      "sys.stdout.write(\"\\nFeb = %d\"%(c.Feb))\n",
+      "sys.stdout.write(\"\\nJune = %d\"%(c.June))\n",
+      "sys.stdout.write(\"\\nDec = %d\"%(c.Dec))\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Jan = 1\n",
+        "Feb = 2\n",
+        "June = 6\n",
+        "Dec = 12"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.25, Page number: 434<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display name of month using enumerated data type\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no enumerated data type in python and an alternate is to use class\n",
+      "#Class declaration\n",
+      "class month:\n",
+      "    Jan = 1\n",
+      "    Feb = 2\n",
+      "    Mar = 3\n",
+      "    Apr = 4\n",
+      "    May = 5\n",
+      "    June = 6\n",
+      "    July = 7\n",
+      "    Aug = 8\n",
+      "    Sep = 9\n",
+      "    Oct = 10\n",
+      "    Nov = 11\n",
+      "    Dec = 12\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "c = month()\n",
+      "\n",
+      "#Result\n",
+      "for f in range(c.Jan,c.Dec+1):\n",
+      "    #There is no switch case statement in python\n",
+      "    if f == c.Jan:\n",
+      "        sys.stdout.write(\"\\nJanuary\")\n",
+      "    else:\n",
+      "        if f == c.Feb:\n",
+      "            sys.stdout.write(\"\\nFebruary\")\n",
+      "        else:\n",
+      "            if f == c.Mar:\n",
+      "                sys.stdout.write(\"\\nMarch\")\n",
+      "            else:\n",
+      "                if f == c.Apr:\n",
+      "                    sys.stdout.write(\"\\nApril\")\n",
+      "                else:\n",
+      "                    if f == c.May:\n",
+      "                        sys.stdout.write(\"\\nMay\")\n",
+      "                    else:\n",
+      "                        if f == c.June:\n",
+      "                            sys.stdout.write(\"\\nJune\")\n",
+      "                        else:\n",
+      "                            if f == c.July:\n",
+      "                                sys.stdout.write(\"\\nJuly\")\n",
+      "                            else:\n",
+      "                                if f == c.Aug:\n",
+      "                                    sys.stdout.write(\"\\nAugust\")\n",
+      "                                else:\n",
+      "                                    if f == c.Sep:\n",
+      "                                        sys.stdout.write(\"\\nSeptember\")\n",
+      "                                    else:\n",
+      "                                        if f == c.Oct:\n",
+      "                                            sys.stdout.write(\"\\nOctober\")\n",
+      "                                        else:\n",
+      "                                            if f == c.Nov:\n",
+      "                                                sys.stdout.write(\"\\nNovember\")\n",
+      "                                            else:\n",
+      "                                                sys.stdout.write(\"\\nDecember\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "January\n",
+        "February\n",
+        "March\n",
+        "April\n",
+        "May\n",
+        "June\n",
+        "July\n",
+        "August\n",
+        "September\n",
+        "October\n",
+        "November\n",
+        "December"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.26, Page number: 436<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Enumerated data type\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no enumerated data type in python. class is used instead\n",
+      "class capital:\n",
+      "    Mumbai = 0\n",
+      "    Hyderabad = 1\n",
+      "    Bangalore = 2\n",
+      "\n",
+      "class state:\n",
+      "    name = ''\n",
+      "    c = capital()\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "s = state()\n",
+      "c1 = capital()\n",
+      "\n",
+      "#Class variable initialization\n",
+      "s.name = \"Andhra Pradesh\"\n",
+      "s.c = s.c.Hyderabad\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nState  : %s\"%(s.name))\n",
+      "sys.stdout.write(\"\\nCapital : %d\"%(s.c))\n",
+      "\n",
+      "if s.c == c1.Hyderabad:\n",
+      "    sys.stdout.write(\"\\nHyderabad is Capital of %s\"%(s.name))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "State  : Andhra Pradesh\n",
+        "Capital : 1\n",
+        "Hyderabad is Capital of Andhra Pradesh"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.27, Page number: 437<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Identify the type of entered character using enumerated data type.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration instead of enumerated data type\n",
+      "class ctype:\n",
+      "    Letter = 0\n",
+      "    Digit = 1\n",
+      "    Other = 2\n",
+      "    \n",
+      "#Variable Initialization\n",
+      "ch = raw_input(\"Enter any character\")\n",
+      "c = ctype()\n",
+      "f = ch.isalpha()\n",
+      "\n",
+      "#Result\n",
+      "if f != 0:\n",
+      "    sys.stdout.write(\"\\n%c is %d type of symbol\"%(ch,c.Letter))\n",
+      "else:\n",
+      "    f = ch.isdigit()\n",
+      "    if f != 0:\n",
+      "        sys.stdout.write(\"\\n%c is %d type of symbol\"%(ch,c.Digit))\n",
+      "    else:\n",
+      "        sys.stdout.write(\"\\n%c is %d type of symbol\"%(ch,c.Other))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any character=\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "= is 2 type of symbol"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.28, Page number: 438<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Size of union and number of bytes reserved for it\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no union/structure in python. class is used instead\n",
+      "#Class declaration\n",
+      "class result:\n",
+      "    marks = 0\n",
+      "    grade = ''\n",
+      "    \n",
+      "class res:\n",
+      "    name = ''\n",
+      "    age = 0\n",
+      "    perf = result()\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "data = res()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Size of Union : %d\\n\"%(sys.getsizeof(data.perf)))\n",
+      "sys.stdout.write(\"Size of Structure : %d\\n\"%(sys.getsizeof(data)))\n",
+      "\n",
+      "#in python, value tagged method is used for data storage and can represent large numbers"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Size of Union : 36\n",
+        "Size of Structure : 36\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.29, Page number: 440<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Memory size of the computer\n",
+      "\n",
+      "import psutil\n",
+      "\n",
+      "psutil.phymem_usage()\n",
+      "\n",
+      "#Displays current status of the memory\n",
+      "#different systems will have different memory status"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "pyout",
+       "prompt_number": 4,
+       "text": [
+        "usage(total=1600512000L, used=1496383488L, free=104128512L, percent=93.5)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.30, Page number: 440<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the system time at specified cursor position\n",
+      "\n",
+      "import sys\n",
+      "import datetime\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"%s\"%(datetime.datetime.now()))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2013-09-20 11:04:15.649000"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.31, Page number: 441<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Change the cursor in different sizes\n",
+      "\n",
+      "import turtle\n",
+      "#Ipython supports graphics through turtle module\n",
+      "\n",
+      "from Tkinter import *\n",
+      "import Tkinter\n",
+      "\n",
+      "top = Tkinter.Tk()\n",
+      "\n",
+      "B1 = Tkinter.Button(top, text =\"circle\", relief=RAISED,\\\n",
+      "                         cursor=\"circle\")\n",
+      "B2 = Tkinter.Button(top, text =\"plus\", relief=RAISED,\\\n",
+      "                         cursor=\"plus\")\n",
+      "B1.pack()\n",
+      "B2.pack()\n",
+      "#top.mainloop()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.32, Page number: 441<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Create a directory using dos interrupt\n",
+      "\n",
+      "import os\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "dir1 = raw_input(\"Enter Directory Name : \")\n",
+      "\n",
+      "#Result\n",
+      "if not os.path.exists(dir1):\n",
+      "    os.makedirs(dir1)\n",
+      "    sys.stdout.write(\"Directory %s created\"%(dir1))\n",
+      "else:\n",
+      "    sys.stdout.write(\"Directory %s not created\"%(dir1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Directory XYZ created"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.33, Page number: 442<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the given character on the screen\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "dl = 67\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"%c\"%(chr(dl)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.34, Page number: 442<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the attributes of a file using DOS interrupt\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "import os\n",
+      "\n",
+      "print os.stat(\"IO.SYS\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "nt.stat_result(st_mode=33206, st_ino=0L, st_dev=0, st_nlink=0, st_uid=0, st_gid=0, st_size=0L, st_atime=1383454467L, st_mtime=1383454467L, st_ctime=1383454467L)\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.35, Page number: 444<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Delete a file using dos interrupt\n",
+      "\n",
+      "import os\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "file1 = raw_input(\"Enter a file name : \")\n",
+      "\n",
+      "try:\n",
+      "    os.remove(file1)\n",
+      "    sys.stdout.write(\"File successfully deleted\")\n",
+      "except:\n",
+      "    sys.stdout.write(\"File could not be deleted\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "File successfully deleted"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13.36, Page number: 445<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Structuret within union\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#there is no union/structure in python. class is used instead\n",
+      "#Class declaration\n",
+      "class x:\n",
+      "    f = 0.0\n",
+      "    p = ['' for i in range(0,2)]\n",
+      "    \n",
+      "class z:\n",
+      "    set1 = x()\n",
+      "    \n",
+      "#Class variable declaration and initialization\n",
+      "st = z()\n",
+      "\n",
+      "st.set1.f = 5.5\n",
+      "st.set1.p[0] = 65\n",
+      "st.set1.p[1] = 66\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n%g\"%(st.set1.f))\n",
+      "sys.stdout.write(\"\\n%c\"%(st.set1.p[0]))\n",
+      "sys.stdout.write(\"\\n%c\"%(st.set1.p[1]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "5.5\n",
+        "A\n",
+        "B"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter14.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter14.ipynb
new file mode 100755
index 00000000..409585bd
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter14.ipynb
@@ -0,0 +1,2311 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 14: Files<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.1, Page number: 454<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write data to text file and read it\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a')\n",
+      "\n",
+      "#Write data to file if successfully created the file\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
+      "    c = ''\n",
+      "    while c != '.':\n",
+      "        c = raw_input(\"Write data & to stop press '.'\")\n",
+      "        fp.write(c)\n",
+      "    fp.close()\n",
+      "\n",
+      "    #Read and display the contents of file\n",
+      "    sys.stdout.write(\"\\nContents read : \")\n",
+      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
+      "    f = fp.readlines()\n",
+      "    for f1 in f:\n",
+      "        print f1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Write data & to stop press '.'"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Write data & to stop press '.'ABCDEFGHIJK\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Write data & to stop press '.'.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Contents read : ABCDEFGHIJK.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.2, Page number: 455<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the contents of the file before and after appending.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file for read data\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
+      "\n",
+      "#Read and display the contents of file\n",
+      "sys.stdout.write(\"\\nContents of file before appending :\\n\")\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
+      "f = fp.readlines()\n",
+      "for f1 in f:\n",
+      "    print f1\n",
+      "\n",
+      "#Open file for appending\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a')\n",
+      "\n",
+      "if f == 0:\n",
+      "    sys.stdout.write(\"File can not appended\")\n",
+      "else:\n",
+      "    c = ''\n",
+      "    #fp.write('\\n')\n",
+      "    while c != '.':\n",
+      "        c = raw_input(\"Enter string to append\")\n",
+      "        fp.write(c)\n",
+      "    \n",
+      "    fp.close()\n",
+      "    \n",
+      "   #Read and display the contents of file\n",
+      "    sys.stdout.write(\"\\nContents of file After appending\\n\")\n",
+      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
+      "    f = fp.readlines()\n",
+      "    for f1 in f:\n",
+      "        print f1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Contents of file before appending :\n",
+        "String is terminated with '\\0'.\n",
+        "\n",
+        "Contents of file After appending\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String is terminated with '\\0'.This character is called as NULL character.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.3, Page number: 457<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Writing and reading of a file.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','w+')\n",
+      "\n",
+      "#Write data to file if successfully created the file\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
+      "    c = ''\n",
+      "    while c != '.':\n",
+      "        c = raw_input(\"Write data & to stop press '.'\")\n",
+      "        fp.write(c)\n",
+      "    fp.close()\n",
+      "\n",
+      "    #Read and display the contents of file\n",
+      "    sys.stdout.write(\"\\nContents read : \")\n",
+      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
+      "    f = fp.readlines()\n",
+      "    for f1 in f:\n",
+      "        print f1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Write data & to stop press '.'\n",
+        "Contents read : "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ABCDEFGHIJK.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.4, Page number: 458<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Open a file in append mode and add new records in it.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a+')\n",
+      "\n",
+      "#Write data to file if successfully created the file\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
+      "    c = ''\n",
+      "    while c != '.':\n",
+      "        c = raw_input(\"Write data & to stop press '.'\")\n",
+      "        fp.write(c)\n",
+      "    fp.close()\n",
+      "\n",
+      "    #Read and display the contents of file\n",
+      "    sys.stdout.write(\"\\nContents read : \")\n",
+      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','a+')\n",
+      "    f = fp.readlines()\n",
+      "    for f1 in f:\n",
+      "        print f1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Write data & to stop press '.'\n",
+        "Contents read : "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This is append and read mode.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.5, Page number: 459<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Open a file in read/write mode in it\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r+')\n",
+      "\n",
+      "#Write data to file if successfully created the file\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    #Read and display the contents of file\n",
+      "    sys.stdout.write(\"\\nContents read : \")\n",
+      "    f = fp.readlines()\n",
+      "    for f1 in f:\n",
+      "        print f1\n",
+      "        \n",
+      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
+      "    c = ''\n",
+      "    while c != '.':\n",
+      "        c = raw_input(\"Write data & to stop press '.'\")\n",
+      "        fp.write(c)\n",
+      "    fp.close()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Contents read : Help me.\n",
+        "Write data & to stop press '.'"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.6, Page number: 460<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Open a file for read/write operation in binary mode\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','wb')\n",
+      "\n",
+      "#Write data to file if successfully created the file\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"Write data & to stop press '.'\")\n",
+      "    c = ''\n",
+      "    while c != '.':\n",
+      "        c = raw_input(\"Write data & to stop press '.'\")\n",
+      "        fp.write(c)\n",
+      "    fp.close()\n",
+      "\n",
+      "    #Read and display the contents of file\n",
+      "    sys.stdout.write(\"\\nContents read : \")\n",
+      "    fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','rb')\n",
+      "    f = fp.readlines()\n",
+      "    for f1 in f:\n",
+      "        print f1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Write data & to stop press '.'\n",
+        "Contents read : "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ABCDEFGHIJK.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.7, Page number: 462<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Open a text file and write some text \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','w')\n",
+      "\n",
+      "#Read the string\n",
+      "text = raw_input(\"Enter Text Here : \")\n",
+      "\n",
+      "#Write to the file\n",
+      "#fp.write() is the equivalent function for fprintf() in python\n",
+      "fp.write(\"%s\"%(text))\n",
+      "\n",
+      "#To see the result, open the data.txt file"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.8, Page number: 463<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Enter data into the text file and read the same.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','w')\n",
+      "\n",
+      "#Read the data\n",
+      "text = raw_input(\"Name & Age\")\n",
+      "age = int(raw_input(\"Name & Age\"))\n",
+      "\n",
+      "#Write to file\n",
+      "fp.write(\"%s   %d\"%(text,age))\n",
+      "\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\data.txt','r')\n",
+      "#Result\n",
+      "sys.stdout.write(\"Name\\tAge\\n\")\n",
+      "text = fp.read()\n",
+      "age = fp.read()\n",
+      "sys.stdout.write(\"%s\\t%s\\n\"%(text,age))\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name\tAge\n",
+        "AMIT   12\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.9, Page number: 463<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the contents of the file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "f = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\list.txt','r')\n",
+      "\n",
+      "if f == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    for c in f:\n",
+      "        sys.stdout.write(\"%s\"%(c))\n",
+      "    #Python gives iterative statements there is no getc() fucntion"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "aman\n",
+        "akash\n",
+        "amit\n",
+        "ajay\n",
+        "ankit"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.10, Page number: 464<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write some text into the file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\words.doc','w')\n",
+      "\n",
+      "#Write data to file\n",
+      "c = ''\n",
+      "while c != '*':\n",
+      "    c = raw_input(\"Enter Few Words '*' to Exit\")\n",
+      "    fp.write(c)\n",
+      "\n",
+      "fp.close()\n",
+      "    #There is no putc() function in python\n",
+      "    #Open the file to see the result"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.11, Page number: 465<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count\n",
+      "\n",
+      "# Total number of statements\n",
+      "# Total number of included files\n",
+      "# Total number of blocks and brackets\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fs = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\PRG2.C','r')\n",
+      "\n",
+      "#variable initialization\n",
+      "i = 0\n",
+      "c = 0\n",
+      "sb = 0\n",
+      "b = 0\n",
+      "\n",
+      "if fs == 0:\n",
+      "    sys.stdout.write(\"File opening error\")\n",
+      "else:\n",
+      "    for line in fs:\n",
+      "        k = 0\n",
+      "        while k < len(line):\n",
+      "            if line[k] == ';':\n",
+      "                c += 1\n",
+      "            else:\n",
+      "                if line[k] == '{':\n",
+      "                    sb += 1\n",
+      "                else:\n",
+      "                    if line[k] == '(':\n",
+      "                        b += 1\n",
+      "                    else:\n",
+      "                        if line[k] == '#':\n",
+      "                            i += 1\n",
+      "            k += 1\n",
+      "                            \n",
+      "    #Result\n",
+      "    sys.stdout.write(\"\\nSummary of 'C' Program\\n\")\n",
+      "    sys.stdout.write(\"===========================\")\n",
+      "    sys.stdout.write(\"\\nTotal Statments : %d\"%(c+i))\n",
+      "    sys.stdout.write(\"\\nInclude Statements : %d\"%(i))\n",
+      "    sys.stdout.write(\"\\nTotal Blocks {} : %d\"%(sb))\n",
+      "    sys.stdout.write(\"\\nTotal Brackets () : %d\"%(b))\n",
+      "    sys.stdout.write(\"\\n============================\")\n",
+      "                            "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Summary of 'C' Program\n",
+        "===========================\n",
+        "Total Statments : 8\n",
+        "Include Statements : 2\n",
+        "Total Blocks {} : 2\n",
+        "Total Brackets () : 9\n",
+        "============================"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.12, Page number: 466<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write text to a file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('C:\\Users\\Aathira\\Documents\\IPython Notebooks\\lines.txt','w')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"\")\n",
+      "else:\n",
+      "    #Write data to file\n",
+      "    c = ''\n",
+      "    while c != '*':\n",
+      "        c = raw_input(\"\")\n",
+      "        fp.write(c)\n",
+      "    fp.close()\n",
+      "    #There is no fputc() function in python\n",
+      "    #Open the file to see the result"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.13, Page number: 467<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read text from the given file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "\n",
+      "file1 = raw_input(\"Enter File Name : \")\n",
+      "fp = open(file1,'r')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"File not found\\n\")\n",
+      "else:\n",
+      "    for text in fp:\n",
+      "        print text\n",
+      "    #File pointer itself is iterable in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "#include <stdio.h>\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.14, Page number: 468<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write a string into a file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "\n",
+      "file1 = raw_input(\"Enter the name of file : \")\n",
+      "fp = open(file1,'w')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"File can not opened\\n\")\n",
+      "else:\n",
+      "    text = raw_input(\"Enter Text Here : \")\n",
+      "    fp.write(text)\n",
+      "    \n",
+      "    #Write() function is used to write character or string into the file in python\n",
+      "    # there is no fputc() function in python\n",
+      "#open the file to see the result"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.15, Page number: 469<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write integers in a file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('num.txt','w')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"File does not exist\\n\")\n",
+      "else:\n",
+      "    v = ' '\n",
+      "    while v != '0':\n",
+      "        fp.write(v)\n",
+      "        v = raw_input(\"Enter Numbers\")\n",
+      "        \n",
+      "        #open the file to see the result"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.16, Page number: 470<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read integers from the file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('num.txt','r')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"File does not exist\\n\")\n",
+      "else:\n",
+      "    for v in fp:\n",
+      "        print v"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 2 3 4 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.17, Page number: 471<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write a block of structure elements to the given file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class student:\n",
+      "    name = ''\n",
+      "    age = 0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "stud = [student() for i in range(0,5)]\n",
+      "\n",
+      "#Open file\n",
+      "file1 = raw_input(\"Enter the file name : \")\n",
+      "fp = open(file1,'w')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"File does not exist\\n\")\n",
+      "else:\n",
+      "    n = int(raw_input(\"How Many Records : \"))\n",
+      "    for i in range(0,n):\n",
+      "        stud[i].name = raw_input(\"Name  : \")\n",
+      "        stud[i].age = int(raw_input(\"Age : \"))\n",
+      "    \n",
+      "    j = 0\n",
+      "    while j < n:\n",
+      "        fp.write(\"%s %d\\n\"%(stud[j].name,stud[j].age))\n",
+      "        j += 1\n",
+      "        \n",
+      "    fp.close()\n",
+      "    \n",
+      "    #open the file to see the result"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.18, Page number: 472<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write and read the information about the player \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class record:\n",
+      "    player = ''\n",
+      "    age = 0\n",
+      "    runs = 0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "emp = record()\n",
+      "\n",
+      "#Open file\n",
+      "fp = open('record.dat','w')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Can not open the file\\n\")\n",
+      "else:\n",
+      "    emp.player = raw_input(\"Enter Player Name\")\n",
+      "    emp.age = int(raw_input(\"Enter Age\"))\n",
+      "    emp.runs = int(raw_input(\"Enter runs scored\"))\n",
+      "    fp.write(\"%s %d %d\"%(emp.player,emp.age,emp.runs))\n",
+      "    fp.close()\n",
+      "    \n",
+      "    fp = open('record.dat','r')\n",
+      "    if fp == 0:\n",
+      "        sys.stdout.write(\"Error in opening file\")\n",
+      "    else:\n",
+      "        sys.stdout.write(\"\\nRecord Entered is\\n\")\n",
+      "        emp.player,emp.age,emp.runs = fp.read().split(' ')\n",
+      "        sys.stdout.write(\"\\n%s  %s  %s\"%(emp.player,emp.age,emp.runs))\n",
+      "        fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Player NameSachin\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Age25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter runs scored10000\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Record Entered is\n",
+        "\n",
+        "Sachin  25  10000"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.19, Page number: 473<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Write a block of structure elements to the given file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "next1 = 'Y'\n",
+      "\n",
+      "#Class declaration\n",
+      "class bike:\n",
+      "    name = ''\n",
+      "    avg = 0\n",
+      "    cost = 0.0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "e = bike()\n",
+      "\n",
+      "#open file\n",
+      "fp = open('bk.txt','wb')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    while next1 == 'Y':\n",
+      "        e.name = raw_input(\"Model Name\")\n",
+      "        e.avg = int(raw_input(\"Average\"))\n",
+      "        e.cost = float(raw_input(\"Price\"))\n",
+      "        fp.write(\"%s %d %f\\n\"%(e.name,e.avg,e.cost))\n",
+      "        \n",
+      "        next1 = raw_input(\"Add Another (Y/N):\")\n",
+      "    fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Model NameHONDA\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average80\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Price45000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Add Another (Y/N):Y\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Model NameSUZUKI\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average65\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Price43000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Add Another (Y/N):Y\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Model NameYAMAHA\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average55\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Price48000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Add Another (Y/N):N\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.20, Page number: 474<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the information about the bike \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class bike:\n",
+      "    name = ''\n",
+      "    avg = 0\n",
+      "    cost = 0.0\n",
+      "    \n",
+      "#Class variable declaration\n",
+      "e = bike()\n",
+      "\n",
+      "#open file\n",
+      "fp = open('bk.txt','rb')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Cannot open file\")\n",
+      "else:\n",
+      "    lines = fp.readlines()\n",
+      "    for line in lines:\n",
+      "        e.name,e.avg,e.cost = line.split(' ')\n",
+      "        e.avg = int(e.avg)\n",
+      "        e.cost = float(e.cost)\n",
+      "        sys.stdout.write(\"\\n%s %d %.2f\"%(e.name,e.avg,e.cost))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "HONDA 80 45000.00\n",
+        "SUZUKI 65 43000.00\n",
+        "YAMAHA 55 48000.00"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.21, Page number: 476<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the text after skipping n characters from beginning of the file\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open(\"text.txt\",\"r\")\n",
+      "\n",
+      "sys.stdout.write(\"\\nContents of file\\n\")\n",
+      "ch = fp.read()\n",
+      "sys.stdout.write(\"%s\"%(ch))\n",
+      "\n",
+      "#Read n\n",
+      "n = int(raw_input(\"How many characters including spaces would you like to skip?\"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nInformation after %d bytes\\n\"%(n))\n",
+      "\n",
+      "ch = ch[n:]\n",
+      "sys.stdout.write(\"%s\"%(ch))\n",
+      "\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Contents of file\n",
+        "THE C PROGRAMMING LANGUAGE INVENTED BY DENNIS RITCHIE"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many characters including spaces would you like to skip?18\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Information after 18 bytes\n",
+        "LANGUAGE INVENTED BY DENNIS RITCHIE"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.22, Page number: 477<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read last few characters of the file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open(\"text.txt\",\"r\")\n",
+      "\n",
+      "sys.stdout.write(\"\\nContents of file\\n\")\n",
+      "ch = fp.read()\n",
+      "sys.stdout.write(\"%s\"%(ch))\n",
+      "\n",
+      "#Read n\n",
+      "n = int(raw_input(\"How many characters including spaces would you like to skip?\"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nInformation after %d bytes\\n\"%(n))\n",
+      "\n",
+      "print ch[-n:]\n",
+      "\n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Contents of file\n",
+        "HELLO WORLD"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many characters including spaces would you like to skip?5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Information after 5 bytes\n",
+        "WORLD\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.23, Page number: 477<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display 'c' program files in current directory. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "l = 0\n",
+      "c = 0\n",
+      "\n",
+      "#Open file\n",
+      "file1 = raw_input(\"Enter the file name : \")\n",
+      "fp = open(file1,'r')\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nContents of 'c' program file in capital case\")\n",
+      "sys.stdout.write(\"\\n============================================\\n\")\n",
+      "\n",
+      "ch = fp.readlines()\n",
+      "\n",
+      "for line in ch:\n",
+      "    i = 0\n",
+      "    #print line.upper()\n",
+      "    while i < len(line):\n",
+      "        if line[i] =='\\n':\n",
+      "            l += 1\n",
+      "        else:\n",
+      "             c += 1\n",
+      "        sys.stdout.write(\"%c\"%(line[i].upper()))\n",
+      "        i += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\nTotal Characters : %d\"%(c))\n",
+      "sys.stdout.write(\"\\nTotal Lines : %d\"%(l))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Contents of 'c' program file in capital case\n",
+        "============================================\n",
+        "MAIN()\n",
+        "{\n",
+        "PRINTF(\" HELLO WORLD\");\n",
+        "}\n",
+        "\n",
+        "Total Characters : 31\n",
+        "Total Lines : 4"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.24, Page number: 479<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect the end of file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open(\"text.txt\",\"r\")\n",
+      "\n",
+      "#there is no feof() function in python\n",
+      "c = fp.tell()\n",
+      "sys.stdout.write(\"File pointer at the beginning of the file : %d\\n\"%(c))\n",
+      "\n",
+      "c = fp.read()\n",
+      "sys.stdout.write(\"%s\"%(c))\n",
+      "\n",
+      "c = fp.tell()\n",
+      "sys.stdout.write(\"\\nFile pointer at the end of file : %d\"%(c))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "File pointer at the beginning of the file : 0\n",
+        "TECHNOCRATS LEAD THE WORLD      \n",
+        "File pointer at the end of file : 32"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.25, Page number: 480<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect an error while read/write operation of a file is in use.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "next1 = 'Y'\n",
+      "\n",
+      "#open file\n",
+      "fp = open('marks.dat','r')\n",
+      "\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Can not open file\")\n",
+      "else:\n",
+      "    while next1 == 'Y':\n",
+      "        name = raw_input(\"Enter Name, Marks, Percentage\")\n",
+      "        marks = int(raw_input(\"Enter Name, Marks, Percentage\"))\n",
+      "        \n",
+      "        p = marks/7\n",
+      "        try:\n",
+      "            fp.write(\"%s %d %f\"%(name,marks,p))\n",
+      "        except:\n",
+      "            sys.stdout.write(\"\\nUnable to write data?\")\n",
+      "            sys.stdout.write(\"\\nFile opening mode is incorrect.\")\n",
+      "            fp.close()\n",
+      "            \n",
+      "        next1 = raw_input(\"Continue Y/N:\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name, Marks, PercentageKAMAL\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name, Marks, Percentage540\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Unable to write data?\n",
+        "File opening mode is incorrect."
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Continue Y/N:N\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.26, Page number: 481<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Catch the error that is occurred while file operation\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "f = open(\"io8.c\",\"w\")\n",
+      "\n",
+      "if f == 0:\n",
+      "    sys.stdout.write(\"\\nCannot open file\")\n",
+      "else:\n",
+      "    #Exception handling\n",
+      "    try:\n",
+      "        c = fp.readlines()\n",
+      "        sys.stdout.write(\"%s\"%(c))\n",
+      "    except:\n",
+      "        sys.stdout.write(\"\\nCan't read file.\")\n",
+      "#There is no ferror() function python."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Can't read file."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.27, Page number: 482<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect and print the error message\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "file1 = \"lines.txt\"\n",
+      "\n",
+      "#Open file\n",
+      "fr = open(file1,\"w\")\n",
+      "\n",
+      "sys.stdout.write(\"%s  : \"%(file1))\n",
+      "#Exception handling\n",
+      "try:\n",
+      "    c = fp.readlines()\n",
+      "except:\n",
+      "    sys.stdout.write(\"Permission Denied\")\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lines.txt  : Permission Denied"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.28, Page number: 482<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the current position of the file pointer \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Open file\n",
+      "fp = open(\"text.txt\",\"r\")\n",
+      "\n",
+      "#Set the pointer\n",
+      "fp.seek(21)\n",
+      "\n",
+      "#Result\n",
+      "while True:\n",
+      "    c = fp.read(1)\n",
+      "    if not c:\n",
+      "      break\n",
+      "    sys.stdout.write(\"%c\\t%d\\n\"%(c,fp.tell()))\n",
+      "#There is no endof file in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "W\t22\n",
+        "O\t23\n",
+        "R\t24\n",
+        "L\t25\n",
+        "D\t26\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.29, Page number: 483<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect beginning of file\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#open file\n",
+      "fp = open(\"text.txt\",\"r\")\n",
+      "\n",
+      "fp.seek(12)\n",
+      "sys.stdout.write(\"Pointer is at %d\\n\"%(fp.tell()))\n",
+      "sys.stdout.write(\"Before rewind() : \")\n",
+      "\n",
+      "#Result\n",
+      "while True:\n",
+      "    c = fp.read(1)\n",
+      "    if not c:\n",
+      "      break\n",
+      "    sys.stdout.write(\"%c\"%(c))\n",
+      "    \n",
+      "sys.stdout.write(\"\\nAfter rewind() : \")\n",
+      "fp.seek(0)\n",
+      "#There is no rewind function in python\n",
+      "\n",
+      "while True:\n",
+      "    c = fp.read(1)\n",
+      "    if not c:\n",
+      "      break\n",
+      "    sys.stdout.write(\"%c\"%(c))\n",
+      "    \n",
+      "fp.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pointer is at 12\n",
+        "Before rewind() : LEAD THE WORLD\n",
+        "After rewind() : TECHNOCRATS LEAD THE WORLD"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.30, Page number: 484<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Delete the given file \n",
+      "\n",
+      "import sys\n",
+      "import os\n",
+      "\n",
+      "#Read file name\n",
+      "file1 = raw_input(\"Enter The File Name : \")\n",
+      "\n",
+      "#There is no remove or unlink file function in python.\n",
+      "#A file can be deleted using remove function in the os module.\n",
+      "\n",
+      "try:\n",
+      "    os.remove(file1)\n",
+      "    sys.stdout.write(\"\\nFile (%s) has been deleted!\"%(file1))\n",
+      "except:\n",
+      "    sys.stdout.write(\"\\nFile does not exist\")\n",
+      "\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "File (TEXT.TXT) has been deleted!"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.31, Page number: 485<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Change the name of the file\n",
+      "\n",
+      "import os\n",
+      "import sys\n",
+      "\n",
+      "old = raw_input(\"Old File Name : \")\n",
+      "\n",
+      "new = raw_input(\"New File Name : \")\n",
+      "\n",
+      "os.rename(old, new)\n",
+      "\n",
+      "#Check the directory to see the result\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.32, Page number: 486<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy the contents of one file to another file\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "fs = open(\"a.txt\",\"r\")\n",
+      "ft = open(\"b.txt\",\"w\")\n",
+      "c = 0\n",
+      "\n",
+      "if fs == 0:\n",
+      "    sys.stdout.write(\"\\nSource file opening error.\")\n",
+      "else:\n",
+      "    if ft == 0:\n",
+      "        sys.stdout.write(\"\\nTarget file opening error.\")\n",
+      "    else:\n",
+      "        while True:\n",
+      "            ch = fs.read(1)\n",
+      "            if not ch:\n",
+      "                break\n",
+      "            ft.write(\"%c\"%(ch))\n",
+      "            c += 1\n",
+      "        sys.stdout.write(\"\\n%d Bytes copied from 'a.txt' to 'b.txt'.\"%(c))\n",
+      "        c = 0\n",
+      "        #there is no fcloseall() function in python\n",
+      "        fs.close()\n",
+      "        c += 1\n",
+      "        ft.close()\n",
+      "        c += 1\n",
+      "        sys.stdout.write(\"\\n%d files closed.\"%(c))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "45 Bytes copied from 'a.txt' to 'b.txt'.\n",
+        "2 files closed."
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.33, Page number: 487<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the contents of three files and find the largest file\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "y = 0\n",
+      "k = 0\n",
+      "t = 0\n",
+      "name = [\"1.txt\",\"2.txt\",\"3.txt\"]\n",
+      "f = [0 for i in range(0,3)]\n",
+      "x = [0 for i in range(0,3)]\n",
+      "\n",
+      "#Open all the files\n",
+      "for l in range(0,3):\n",
+      "    fp = open(name[l],\"r\")\n",
+      "    f[l] = fp\n",
+      "    if fp == 0:\n",
+      "        sys.stdout.write(\"\\n%s file not found.\"%(name[l]))\n",
+      "        break\n",
+      "        \n",
+      "#Read contents of all files\n",
+      "for l in range(0,3):\n",
+      "    while True:\n",
+      "        c1 = f[l].read(1)\n",
+      "        if not c1:\n",
+      "            break\n",
+      "        x[l] = y\n",
+      "        y += 1\n",
+      "    y = 0\n",
+      "\n",
+      "#close the files\n",
+      "for l in range(0,3):\n",
+      "    f[l].close()\n",
+      "\n",
+      "#Print size of all files\n",
+      "for l in range(0,2+1):\n",
+      "    sys.stdout.write(\"File : %s Bytes : %d\\n\"%(name[l],x[l]))\n",
+      "    t = t + x[l]\n",
+      "    \n",
+      "#Find largest\n",
+      "for l in range(t,1,-1):\n",
+      "    for k in range(0,3):\n",
+      "        if l == x[k]:\n",
+      "            sys.stdout.write(\"\\n%s are the largest file.\"%(name[k]))\n",
+      "            break\n",
+      "    if l == x[k]:\n",
+      "        break"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "File : 1.txt Bytes : 16\n",
+        "File : 2.txt Bytes : 20\n",
+        "File : 3.txt Bytes : 25\n",
+        "\n",
+        "3.txt are the largest file."
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.34, Page number: 488<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy 100 characters from a file to an array\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "SIZE = 100\n",
+      "s = 0\n",
+      "x = 0\n",
+      "ch = ['0' for i in range(0,100)]\n",
+      "\n",
+      "#Open the files\n",
+      "f = open(\"poem.txt\",\"r\")\n",
+      "f2 = open(\"alpha.txt\",\"w\")\n",
+      "\n",
+      "if f ==0 or f2 == 0:\n",
+      "    sys.stdout.write(\"?\")\n",
+      "else:\n",
+      "    while True:\n",
+      "        c = f.read(1)\n",
+      "        x += 1\n",
+      "        if not c:\n",
+      "            break\n",
+      "        else:\n",
+      "            if x == 99:\n",
+      "                break\n",
+      "            else:\n",
+      "                ch[s] = c\n",
+      "                s += 1\n",
+      "\n",
+      "for s in range(0,100):\n",
+      "    f2.write(\"%c\"%(ch[s]))\n",
+      "\n",
+      "sys.stdout.write(\"Process Completed : Error 0\")\n",
+      "f.close()\n",
+      "f2.close()\n",
+      "\n",
+      "#There is no perror() function in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Process Completed : Error 0"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.35, Page number: 491<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Low level disk I/O operations.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "\n",
+      "#Variable Initialization\n",
+      "file1 = raw_input(\"Enter a file name : \")\n",
+      "\n",
+      "#Open file\n",
+      "s = open(file1,\"w\")\n",
+      "\n",
+      "#Result\n",
+      "if s == -1:\n",
+      "    sys.stdout.write(\"File does not exits\")\n",
+      "else:\n",
+      "    buff = raw_input(\"Enter text below:\")\n",
+      "    s.write(\"%s\"%(buff))\n",
+      "    s.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a file name : TEXT\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter text below:PROGRAMMING IN C\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.36, Page number: 492<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read text from a specified file \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "file1 = raw_input(\"Enter a file name \")\n",
+      "\n",
+      "#open file\n",
+      "s = open(file1,\"r\")\n",
+      "\n",
+      "#Result\n",
+      "if s == -1:\n",
+      "    sys.stdout.write(\"File does not exists\")\n",
+      "else:\n",
+      "    while True:\n",
+      "        ch = s.read(1)\n",
+      "        if not ch:\n",
+      "            break\n",
+      "        sys.stdout.write(\"%c\"%(ch))\n",
+      "        \n",
+      "s.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a file name TEXT\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "PROGRAMMING IN C"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.37, Page number: 493<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Set a buffer size \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nThis book teaches C\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "This book teaches C"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.38, Page number: 494<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display number of arguments and their names\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nTotal number of arguments are %d\\n\"%(len(sys.argv)))\n",
+      "\n",
+      "print str(sys.argv)\n",
+      "\n",
+      "#Command line arguments can be given in python by the command\n",
+      "\n",
+      "#   python pgmname.py arg1 arg2 arg3\n",
+      "\n",
+      "#This is not possible in ipython notebook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.39, Page number: 495<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read any file from command prompt\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#open file\n",
+      "fp = open(sys.argv[1],\"r\")\n",
+      "\n",
+      "#Result\n",
+      "if fp == 0:\n",
+      "    sys.stdout.write(\"Can not open file\")\n",
+      "else:\n",
+      "    while True:\n",
+      "        ch = fp.read(1)\n",
+      "        if not ch:\n",
+      "            break\n",
+      "        sys.stdout.write(\"%c\"%(ch))\n",
+      "        \n",
+      "fp.close()\n",
+      "\n",
+      "#This program can be run in python as\n",
+      "\n",
+      "#   python pgmname.py filename"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.40, Page number: 495<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Command line argument to perform the task of DEL command\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Check number of arguments\n",
+      "if len(sys.argv) < 2:\n",
+      "    sys.stdout.write(\"Insufficient Arguments\")\n",
+      "else:\n",
+      "    fp = open(sys.argv[1],\"r\")\n",
+      "    if fp == 0:\n",
+      "        sys.stdout.write(\"File Not Found\")\n",
+      "        fp.close()\n",
+      "        os.remove(sys.argv[1])\n",
+      "        sys.stdout.write(\"File has been deleted\")\n",
+      "        \n",
+      "#This program can be deleted using\n",
+      "#     python pgmname.py filename\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.41, Page number: 496<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Command line argument to perform the task of REN command\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Check number of arguments\n",
+      "if len(sys.srgv) < 3:\n",
+      "    sys.stdout.write(\"Insufficient Arguments\")\n",
+      "else:\n",
+      "    fp = open(sys.argv[1],\"r\")\n",
+      "    if fp == 0:\n",
+      "        sys.stdout.write(\"File Not Found\")\n",
+      "    else:\n",
+      "        sp = open(sys.argv[2],\"r\")\n",
+      "        if sp == 0:\n",
+      "            fp.close()\n",
+      "            sp.close()\n",
+      "            #Rename file\n",
+      "            os.rename(sys.argv[1],sys.argv[2])\n",
+      "        else:\n",
+      "            sys.stdout.write(\"Duplicate file name or file is in use\")\n",
+      "            \n",
+      "#This program can be executed as\n",
+      "\n",
+      "#   python pgmname.py oldfilename newfilename\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.42, Page number: 497<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Environment variable and display the various settings.\n",
+      "\n",
+      "import sys\n",
+      "import os\n",
+      "\n",
+      "#Result\n",
+      "print os.environ\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "{'TMP': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Local\\\\Temp', 'COMPUTERNAME': 'AATHIRA-PC', 'GUROBI_HOME': 'B:\\\\gurobi510\\\\win32', 'USERDOMAIN': 'Aathira-PC', 'PSMODULEPATH': 'C:\\\\Windows\\\\system32\\\\WindowsPowerShell\\\\v1.0\\\\Modules\\\\', 'COMMONPROGRAMFILES': 'C:\\\\Program Files (x86)\\\\Common Files', 'PROCESSOR_IDENTIFIER': 'Intel64 Family 6 Model 15 Stepping 13, GenuineIntel', 'PROGRAMFILES': 'C:\\\\Program Files (x86)', 'PROCESSOR_REVISION': '0f0d', 'SYSTEMROOT': 'C:\\\\Windows', 'PATH': 'C:\\\\Anaconda\\\\lib\\\\site-packages\\\\numpy\\\\core;B:\\\\gurobi510\\\\win32\\\\bin;C:\\\\Windows\\\\system32;C:\\\\Windows;C:\\\\Windows\\\\System32\\\\Wbem;C:\\\\Windows\\\\System32\\\\WindowsPowerShell\\\\v1.0\\\\;C:\\\\Program Files (x86)\\\\Common Files\\\\Adobe\\\\AGL;C:\\\\Program Files\\\\MATLAB\\\\R2009a\\\\bin;C:\\\\Program Files\\\\MATLAB\\\\R2009a\\\\bin\\\\win64;c:\\\\python27\\\\scripts;C:\\\\Program Files (x86)\\\\MiKTeX 2.9\\\\miktex\\\\bin\\\\;C:\\\\Anaconda;C:\\\\Anaconda\\\\Scripts;C:\\\\Program Files (x86)\\\\ffmpeg', 'CLICOLOR': '1', 'PROGRAMFILES(X86)': 'C:\\\\Program Files (x86)', 'COMSPEC': 'C:\\\\Windows\\\\system32\\\\cmd.exe', 'TK_LIBRARY': 'C:\\\\Anaconda\\\\tcl\\\\tk8.5', 'TERM': 'xterm-color', 'TEMP': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Local\\\\Temp', 'COMMONPROGRAMFILES(X86)': 'C:\\\\Program Files (x86)\\\\Common Files', 'PROCESSOR_ARCHITECTURE': 'x86', 'TIX_LIBRARY': 'C:\\\\Anaconda\\\\tcl\\\\tix8.4.3', 'ALLUSERSPROFILE': 'C:\\\\ProgramData', 'LOCALAPPDATA': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Local', 'HOMEPATH': '\\\\Users\\\\Aathira', 'PROGRAMW6432': 'C:\\\\Program Files', 'USERNAME': 'Aathira', 'LOGONSERVER': '\\\\\\\\AATHIRA-PC', 'SESSIONNAME': 'Console', 'PROGRAMDATA': 'C:\\\\ProgramData', 'TCL_LIBRARY': 'C:\\\\Anaconda\\\\tcl\\\\tcl8.5', 'GIT_PAGER': 'cat', 'PATHEXT': '.COM;.EXE;.BAT;.CMD;.VBS;.VBE;.JS;.JSE;.WSF;.WSH;.MSC', 'FP_NO_HOST_CHECK': 'NO', 'WINDIR': 'C:\\\\Windows', 'APPDATA': 'C:\\\\Users\\\\Aathira\\\\AppData\\\\Roaming', 'HOMEDRIVE': 'C:', 'PAGER': 'cat', 'SYSTEMDRIVE': 'C:', 'NUMBER_OF_PROCESSORS': '2', 'PROCESSOR_LEVEL': '6', 'PROCESSOR_ARCHITEW6432': 'AMD64', 'COMMONPROGRAMW6432': 'C:\\\\Program Files\\\\Common Files', 'OS': 'Windows_NT', 'PUBLIC': 'C:\\\\Users\\\\Public', 'USERPROFILE': 'C:\\\\Users\\\\Aathira'}\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.43, Page number: 498<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read character from keyboard \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "c = '0'\n",
+      "\n",
+      "#Result\n",
+      "while c != ' ':\n",
+      "    c = raw_input(\"\")\n",
+      "    sys.stdout.write(\"%c  \"%(c))\n",
+      "    \n",
+      "#Give space at the end"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1  2  "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3  4  "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5  6  "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7  8  "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9     "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14.44, Page number: 499<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display A to Z characters\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "for a in range(65,91):\n",
+      "    sys.stdout.write(\"%c\\t\"%(chr(a)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A\tB\tC\tD\tE\tF\tG\tH\tI\tJ\tK\tL\tM\tN\tO\tP\tQ\tR\tS\tT\tU\tV\tW\tX\tY\tZ\t"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter15.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter15.ipynb
new file mode 100755
index 00000000..92842ba6
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter15.ipynb
@@ -0,0 +1,754 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 15: Additional in 'C'<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.1, Page number: 505<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Allocate memory to pointer variable. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "j = 0\n",
+      "k = int(raw_input(\"How many number : \"))\n",
+      "p = [0 for i in range(0,k)]\n",
+      "\n",
+      "#in python, all variables are allocated using dynamic memory allocation technique and no\n",
+      "#malloc function and pointer concept is available in python.\n",
+      "\n",
+      "#Read the numbers\n",
+      "while j != k:\n",
+      "    p[j] = int(raw_input(\"Number %d = \"%(j+1)))\n",
+      "    j += 1\n",
+      "    \n",
+      "j = 0\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The numbers are : \")\n",
+      "while j != k:\n",
+      "    sys.stdout.write(\"%d\\t\"%(p[j]))\n",
+      "    j += 1\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many number : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number 1 = 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number 2 = 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number 3 = 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number 4 = 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numbers are : 1\t2\t3\t4\t"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.2, Page number: 506<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Memory allocation to pointer variable. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "j = 0\n",
+      "k = int(raw_input(\"How many Number : \"))\n",
+      "p = [0 for i in range(0,k)]\n",
+      "\n",
+      "#in python, all variables are allocated using dynamic memory allocation technique and no\n",
+      "#calloc function and pointer concept is available in python.\n",
+      "\n",
+      "#Read the numbers\n",
+      "while j != k:\n",
+      "    p[j] = int(raw_input(\"Number %d = \"%(j+1)))\n",
+      "    j += 1\n",
+      "    \n",
+      "j = 0\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The numbers are : \")\n",
+      "while j != k:\n",
+      "    sys.stdout.write(\"%d\\t\"%(p[j]))\n",
+      "    j += 1\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many Number : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number 1 = 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number 2 = 58\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number 3 = 98\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The numbers are : 45\t58\t98\t"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.3, Page number: 507<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Reallocate memory \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "str1 = \"India\"\n",
+      "\n",
+      "#in python, value tagged method is used for data storage instead of memory tagging.\n",
+      "#no realloc function is in python\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"str = %s\"%(str1))\n",
+      "str1 = \"Hindustan\"\n",
+      "sys.stdout.write(\"\\nNow str = %s\"%(str1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "str = India\n",
+        "Now str = Hindustan"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.4, Page number: 508<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display unused memory \n",
+      "\n",
+      "import psutil\n",
+      "\n",
+      "psutil.phymem_usage()\n",
+      "\n",
+      "#There is no coreleft() function in python. phymem_usage function in the module psutil gives the \n",
+      "#status and usage of physical memory in python."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 6,
+       "text": [
+        "usage(total=3165270016L, used=987840512L, free=2177429504L, percent=31.2)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.5, Page number: 510<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Linked list\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialziation\n",
+      "ch = 'y'\n",
+      "p = 0\n",
+      "q = []\n",
+      "\n",
+      "#Function Definitions\n",
+      "def gen_rate(m):\n",
+      "    q.append(m)\n",
+      "    \n",
+      "def show():\n",
+      "    print q\n",
+      "    \n",
+      "def addatstart(m):\n",
+      "    q.insert(0,m)\n",
+      "    \n",
+      "def append(m,po):\n",
+      "    q.insert(po,m)\n",
+      "\n",
+      "def erase(d):\n",
+      "    q.remove(d)\n",
+      "    \n",
+      "def count():\n",
+      "    print len(q)\n",
+      "    \n",
+      "def descending():\n",
+      "    q.sort(reverse=True)\n",
+      "    \n",
+      "#Get choice\n",
+      "while ch == 'y':\n",
+      "    n = int(raw_input(\"1. Generate\\n2. Add at starting\\n3. Append\\n4. Delete\\n5. Show\\n6.Count\\n7.Descending\\nEnter your choice: \"));\n",
+      "    #There is no switch statement in python\n",
+      "    if n == 1:\n",
+      "        i = int(raw_input(\"How many node you want : \"))\n",
+      "        for j in range(0,i):\n",
+      "            m = int(raw_input(\"Enter the element : \"))\n",
+      "            gen_rate(m)\n",
+      "        show()\n",
+      "    else:\n",
+      "        if n == 2:\n",
+      "            m = int(raw_input(\"Enter the element : \"))\n",
+      "            addatstart(m)\n",
+      "            show()\n",
+      "        else:\n",
+      "            if n == 3:\n",
+      "                m = int(raw_input(\"Enter the element and position \"))\n",
+      "                po = int(raw_input(\"Enter the element and position\"))\n",
+      "                append(m,po)\n",
+      "                show()\n",
+      "            else:\n",
+      "                if n == 4:\n",
+      "                    d = int(raw_input(\"Enter the number for deletion : \"))\n",
+      "                    erase(d)\n",
+      "                    show()\n",
+      "                else:\n",
+      "                    if n == 5:\n",
+      "                        show()\n",
+      "                    else:\n",
+      "                        if n == 6:\n",
+      "                            count()\n",
+      "                        else:\n",
+      "                            if n == 7:\n",
+      "                                descending()\n",
+      "                                show()\n",
+      "                            else:\n",
+      "                                sys.stdout.write(\"Enter value between 1 to 7\")\n",
+      "                \n",
+      "    ch = raw_input(\"Do u wnat to continue (y/n)\")\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. Generate\n",
+        "2. Add at starting\n",
+        "3. Append\n",
+        "4. Delete\n",
+        "5. Show\n",
+        "6.Count\n",
+        "7.Descending\n",
+        "Enter your choice: 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many node you want : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the element : 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[1, 5, 4, 7]\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Do u wnat to continue (y/n)n\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.6, Page number: 518<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Draw circle, line and arc using graphics function\n",
+      "\n",
+      "%matplotlib inline\n",
+      "import pylab\n",
+      "import matplotlib.pyplot as plt\n",
+      "\n",
+      "#Tkinter package is used for graphics\n",
+      "#Give proportionate sizes to draw\n",
+      "#draw circle\n",
+      "circle2=plt.Circle((.5,.5),.2,color='b')\n",
+      "fig = plt.gcf()\n",
+      "fig.gca().add_artist(circle2)\n",
+      "\n",
+      "\n",
+      "#draw line\n",
+      "figure()\n",
+      "pylab.plot([210,110],[150,150])\n",
+      "\n",
+      "#Draw ellipse\n",
+      "figure()\n",
+      "from matplotlib.patches import Ellipse\n",
+      "e = Ellipse(xy=(35, -50), width=10, height=5, linewidth=2.0, color='g')\n",
+      "fig = plt.gcf()\n",
+      "fig.gca().add_artist(e)\n",
+      "e.set_clip_box(ax.bbox)\n",
+      "e.set_alpha(0.7)\n",
+      "pylab.xlim([20, 50])\n",
+      "pylab.ylim([-65, -35])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['pylab', 'e']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 61,
+       "text": [
+        "(-65, -35)"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x707a9f0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x7349d10>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x78994d0>"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.7, Page number: 519<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Draw boxes and fill with different designs.\n",
+      "\n",
+      "%pylab\n",
+      "#Tkinter package is used for graphics\n",
+      "from matplotlib.patches import Rectangle\n",
+      "from matplotlib.collections import PatchCollection\n",
+      "\n",
+      "e = Rectangle(xy=(35, -50), width=10, height=5, linewidth=2.0, color='b')\n",
+      "fig = plt.gcf()\n",
+      "fig.gca().add_artist(e)\n",
+      "e.set_clip_box(ax.bbox)\n",
+      "e.set_alpha(0.7)\n",
+      "pylab.xlim([20, 50])\n",
+      "pylab.ylim([-65, -35])\n",
+      "\n",
+      "#There are no different automatic fill styles. user should create the styles."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x7373590>"
+       ]
+      }
+     ],
+     "prompt_number": 72
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.8, Page number: 520<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display text in different size, font, vertically and horizontally \n",
+      "\n",
+      "%pylab\n",
+      "\n",
+      "from pylab import *\n",
+      "\n",
+      "#Tkinter package is used for graphics\n",
+      "# set limits so that it no longer looks on screen to be 45 degrees\n",
+      "xlim([-5,5])\n",
+      "\n",
+      "# Locations to plot text\n",
+      "l1 = array((1,1))\n",
+      "l2 = array((5,5))\n",
+      "\n",
+      "# Rotate angle\n",
+      "angle = 90\n",
+      "trans_angle = gca().transData.transform_angles(array((45,)),\n",
+      "                                               l2.reshape((1,2)))[0]\n",
+      "\n",
+      "# Plot text\n",
+      "th2 = text(l2[0],l2[1],'Hello(Horizontal Text with fontsize 25)\\n\\n',fontsize=25,\n",
+      "           rotation=0)\n",
+      "th2 = text(l2[0],l2[1],'Hello(Horizontal Text with fontsize 16)',fontsize=16,\n",
+      "           rotation=0)\n",
+      "th1 = text(l1[0],l1[1],'Hello(Vertical Text)',fontsize=16,\n",
+      "           rotation=angle)\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Using matplotlib backend: module://IPython.kernel.zmq.pylab.backend_inline\n",
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['power', 'draw_if_interactive', 'random', 'fft', 'angle', 'linalg', 'info']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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KqNs1+SNAAsBVxBNwX0KbD+LAZULNHz58uHbs2KEnn3xS9erVU5EiRRQfH6+k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\nQPnz51f//v1dfXQxrFixQoc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ISAAAAEoEJAAAACUCEgAAgBIBCQAAQImABAAA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\nAAAAANbkf1M+pbyPupgkAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x792e430>"
+       ]
+      }
+     ],
+     "prompt_number": 95
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.9, Page number: 521<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display status of mouse button pressed \n",
+      "\n",
+      "from Tkinter import *\n",
+      "from tkFileDialog import askopenfilename\n",
+      "import Image, ImageTk\n",
+      "\n",
+      "if __name__ == \"__main__\":\n",
+      "    root = Tk()\n",
+      "\n",
+      "    #setting up a tkinter canvas with scrollbars\n",
+      "    frame = Frame(root, bd=2, relief=SUNKEN)\n",
+      "    frame.grid_rowconfigure(0, weight=1)\n",
+      "    frame.grid_columnconfigure(0, weight=1)\n",
+      "    xscroll = Scrollbar(frame, orient=HORIZONTAL)\n",
+      "    xscroll.grid(row=1, column=0, sticky=E+W)\n",
+      "    yscroll = Scrollbar(frame)\n",
+      "    yscroll.grid(row=0, column=1, sticky=N+S)\n",
+      "    canvas = Canvas(frame, bd=0, xscrollcommand=xscroll.set, yscrollcommand=yscroll.set)\n",
+      "    canvas.grid(row=0, column=0, sticky=N+S+E+W)\n",
+      "    xscroll.config(command=canvas.xview)\n",
+      "    yscroll.config(command=canvas.yview)\n",
+      "    frame.pack(fill=BOTH,expand=1)\n",
+      "\n",
+      "    \n",
+      "\n",
+      "    #function to be called when mouse is clicked\n",
+      "    def printcoords(event):\n",
+      "        #outputting x and y coords to console\n",
+      "        print \"Mouse Button pressed\"\n",
+      "        print (event.x,event.y)\n",
+      "    #mouseclick event\n",
+      "    canvas.bind(\"<Button 1>\",printcoords)\n",
+      "\n",
+      "    root.mainloop()\n",
+      "    \n",
+      "import win32api, win32con\n",
+      "\n",
+      "print \"Current cursor position at \" \n",
+      "print win32api.GetCursorPos()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mouse Button pressed\n",
+        "(207, 115)\n",
+        "Current cursor position at "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "(502, 188)\n"
+       ]
+      }
+     ],
+     "prompt_number": 108
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15.10, Page number: 523<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Change mouse cursor.\n",
+      "\n",
+      "#Placing the cursor on top of the circle button will change the pointer to circle\n",
+      "# and plus button to plus symbol\n",
+      "\n",
+      "from Tkinter import *\n",
+      "import Tkinter\n",
+      "\n",
+      "top = Tkinter.Tk()\n",
+      "\n",
+      "B1 = Tkinter.Button(top, text =\"circle\", relief=RAISED,\\\n",
+      "                         cursor=\"circle\")\n",
+      "B2 = Tkinter.Button(top, text =\"plus\", relief=RAISED,\\\n",
+      "                         cursor=\"plus\")\n",
+      "B1.pack()\n",
+      "B2.pack()\n",
+      "top.mainloop()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 110
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter16.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter16.ipynb
new file mode 100755
index 00000000..cafb9828
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter16.ipynb
@@ -0,0 +1,2160 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 16: Marching Towards C++<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.1, Page number: 537<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read and display a string\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Initialize variable\n",
+      "name = raw_input(\"Enter Your Name : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Your name is %s\"%(name))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name : Amit\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your name is Amit"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.2, Page number: 540<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the value of a variable using reference variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "qty = 10\n",
+      "qt = qty\n",
+      "#in python, value tagging method is used for storage. so there is no reference concept\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"qty\\tqt\\n\")\n",
+      "sys.stdout.write(\"%d\\t%d\\n\"%(qty,qt))\n",
+      "qt = qt + 1\n",
+      "qty = qty + 1\n",
+      "sys.stdout.write(\"%d\\t%d\\n\"%(qty,qt))\n",
+      "qty = qty - 1\n",
+      "qt = qt - 1\n",
+      "sys.stdout.write(\"%d\\t%d\\n\"%(qty,qt))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "qty\tqt\n",
+        "10\t10\n",
+        "11\t11\n",
+        "10\t10\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.3, Page number: 541<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to use scope access operator.\n",
+      "#Display the variaous values of the same variable declared at different scope levels.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 10\n",
+      "\n",
+      "def main():\n",
+      "    a = 20\n",
+      "    sys.stdout.write(\"::a = %d\"%(a))\n",
+      "    return\n",
+      "\n",
+      "#There is no scope operator. Intendation is used to define scope/block of statements\n",
+      "main()\n",
+      "sys.stdout.write(\"    a = %d\"%(a))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "::a = 20    a = 10"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.4, Page number: 542<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to read tow integers through the keyboard.\n",
+      "#Declare the variables in C++ style.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Read Variables\n",
+      "num = int(raw_input(\"Enter Two numbers : \"))\n",
+      "num1 = int(raw_input(\"Enter Two numbers : \"))\n",
+      "#No variable declaration is needed in python\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Entered Numbers are : %d  %d\"%(num,num1))\n",
+      "                                    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Two numbers : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Two numbers : 9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Entered Numbers are : 8  9"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.5, Page number: 543<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Length of a string\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#There is no variable declaration is needed in python\n",
+      "#Read string\n",
+      "name = raw_input(\"Enter Your Name : \")\n",
+      "\n",
+      "#Calculation\n",
+      "len1 = len(name)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The length of the string is : %d\"%(len1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name : Santosh\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of the string is : 7"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.6, Page number: 544<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sum of numbers using function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def sum1(x,y):\n",
+      "    return x+y\n",
+      "#Forward function definition is not possible in python.\n",
+      "#Otherwise wrap the function within the main() function\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 20\n",
+      "b = 2.5\n",
+      "\n",
+      "#Function call\n",
+      "sys.stdout.write(\"Sum = %f\"%(sum1(a,b)))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum = 22.500000"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.7, Page number: 545<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Function with default arguments.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition\n",
+      "def sum1(j,k=10,l=15,m=20):\n",
+      "    return j+k+l+m\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 2\n",
+      "b = 3\n",
+      "c = 4\n",
+      "d = 5\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Sum = %d\"%(sum1(a,b,c,d)))\n",
+      "sys.stdout.write(\"\\nSum = %d\"%(sum1(a,b,c)))\n",
+      "sys.stdout.write(\"\\nSum = %d\"%(sum1(a,b)))\n",
+      "sys.stdout.write(\"\\nSum = %d\"%(sum1(a)))\n",
+      "sys.stdout.write(\"\\nSum = %d\"%(sum1(b,c,d)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum = 14\n",
+        "Sum = 29\n",
+        "Sum = 40\n",
+        "Sum = 47\n",
+        "Sum = 32"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.8, Page number: 547<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Square of integer and float number using function overloading\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Function definition/ There is no seperate function definition is needed.\n",
+      "def sqr(s):\n",
+      "    return s*s\n",
+      "\n",
+      "#Variable iNitialization\n",
+      "a = 15\n",
+      "b = 2.5\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Square = %d\"%(sqr(a)))\n",
+      "sys.stdout.write(\"\\nSquare = %f\"%(sqr(b)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Square = 225\n",
+        "Square = 6.250000"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.9, Page number: 549<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display private and public member variables of the class.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class num:\n",
+      "    def readdata(self,j,k):\n",
+      "        self.x = j\n",
+      "        self.y = k\n",
+      "        \n",
+      "    def display(self):\n",
+      "        sys.stdout.write(\"x = %d   y = %f\"%(self.x,self.y))\n",
+      "\n",
+      "#Object declaration\n",
+      "j = num()\n",
+      "\n",
+      "#Result\n",
+      "j.z = 'C'\n",
+      "j.readdata(10,10.5)\n",
+      "j.display()\n",
+      "sys.stdout.write(\"  z = %c\"%(j.z))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "x = 10   y = 10.500000  z = C"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.10, Page number: 550<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Private and public data member of a class.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class player:\n",
+      "    def __init__(self,name=None,height=0,weight=0):\n",
+      "        self.name = name\n",
+      "        self.height = height\n",
+      "        self.weight = weight\n",
+      "        \n",
+      "#Object declaration\n",
+      "a = player()\n",
+      "\n",
+      "#Variable Initialization\n",
+      "#There is no need/concept for private variables in python.\n",
+      "a.name = \"Sanjay\"\n",
+      "a.height = 5.5\n",
+      "a.weight = 38\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Height : %d\"%(a.height))\n",
+      "sys.stdout.write(\"\\nWeight : %d\"%(a.weight))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height : 5\n",
+        "Weight : 38"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.11, Page number: 551<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Class with member variables and functions. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class player:\n",
+      "    def __init__(self,name=None,height=0,weight=0):\n",
+      "        self.name = name\n",
+      "        self.height = height\n",
+      "        self.weight = weight\n",
+      "        \n",
+      "    def setdata(self):\n",
+      "        self.name = raw_input(\"Enter Name Age Height Weight\")\n",
+      "        self.age = int(raw_input(\"Enter Name Age Height Weight\"))\n",
+      "        self.height = float(raw_input(\"Enter Name Age Height Weight\"))\n",
+      "        self.weight = int(raw_input(\"Enter Name Age Height Weight\"))\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\nName    :  %s\"%(self.name))\n",
+      "        sys.stdout.write(\"\\nAge     :  %d\"%(self.age))\n",
+      "        sys.stdout.write(\"\\nHeight  :  %f\"%(self.height))\n",
+      "        sys.stdout.write(\"\\nWeight  :  %d\"%(self.weight))\n",
+      "        \n",
+      "#Object declaration\n",
+      "a = player()\n",
+      "\n",
+      "#Function call\n",
+      "a.setdata()\n",
+      "a.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height WeightSanjay\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height Weight24\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height Weight5.5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height Weight54\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name    :  Sanjay\n",
+        "Age     :  24\n",
+        "Height  :  5.500000\n",
+        "Weight  :  54"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.12, Page number: 552<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Class with member variables and functions.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class player:\n",
+      "    def __init__(self,name=None,height=0,weight=0):\n",
+      "        self.name = name\n",
+      "        self.height = height\n",
+      "        self.weight = weight\n",
+      "   \n",
+      "    def setdata(self):\n",
+      "        self.name = raw_input(\"Enter Name Age Height Weight\")\n",
+      "        self.age = int(raw_input(\"Enter Name Age Height Weight\"))\n",
+      "        self.height = float(raw_input(\"Enter Name Age Height Weight\"))\n",
+      "        self.weight = int(raw_input(\"Enter Name Age Height Weight\"))\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\nName    :  %s\"%(self.name))\n",
+      "        sys.stdout.write(\"\\nAge     :  %d\"%(self.age))\n",
+      "        sys.stdout.write(\"\\nHeight  :  %f\"%(self.height))\n",
+      "        sys.stdout.write(\"\\nWeight  :  %d\"%(self.weight))\n",
+      "#In python, member functions should be defined inside the class instance\n",
+      "\n",
+      "#Object declaration\n",
+      "a = player()\n",
+      "\n",
+      "#Function call\n",
+      "a.setdata()\n",
+      "a.show()   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height WeightAjay\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height Weight24\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height Weight5.2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Name Age Height Weight45\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Name    :  Ajay\n",
+        "Age     :  24\n",
+        "Height  :  5.200000\n",
+        "Weight  :  45"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.13, Page number: 554<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Static data member. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class sumnum:\n",
+      "    def __init__(self,num = 0):\n",
+      "        self.num = num\n",
+      "        \n",
+      "    def input1(self,c):\n",
+      "        self.num = int(raw_input(\"Enter Number : \"))\n",
+      "        c = c + self.num\n",
+      "        return c\n",
+      "        \n",
+      "    def sum1(self,c):\n",
+      "        sys.stdout.write(\"The sum of entered numbers is %d\"%(c))\n",
+      "        \n",
+      "#Object declaration\n",
+      "a = sumnum()\n",
+      "b = sumnum()\n",
+      "c1 = sumnum()\n",
+      "c = 0\n",
+      "\n",
+      "#Function call\n",
+      "#There is no static variable in python\n",
+      "c = a.input1(c)\n",
+      "c = b.input1(c)\n",
+      "c = c1.input1(c)\n",
+      "\n",
+      "a.sum1(c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of entered numbers is 14"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.14, Page number: 555<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Static member functions \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class bita:\n",
+      "    def __initi__(self):\n",
+      "        self.c = 0\n",
+      "    \n",
+      "    @staticmethod    \n",
+      "    def count(c):\n",
+      "        c = c + 1\n",
+      "        return c\n",
+      "    \n",
+      "    @staticmethod\n",
+      "    def display(k,c):\n",
+      "        k = k + 1\n",
+      "        sys.stdout.write(\"\\nCall Number : %d\"%(k))\n",
+      "        sys.stdout.write(\"\\nValue of c : %d\"%(c))\n",
+      "        return k\n",
+      "        \n",
+      "#Object declaration\n",
+      "b = bita()\n",
+      "c1 = bita()\n",
+      "\n",
+      "c = 0\n",
+      "k = 0\n",
+      "\n",
+      "#Function call\n",
+      "k = bita.display(k,c)\n",
+      "c = bita.count(c)\n",
+      "c = b.count(c)\n",
+      "c = c1.count(c)\n",
+      "k = bita.display(k,c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Call Number : 1\n",
+        "Value of c : 0\n",
+        "Call Number : 2\n",
+        "Value of c : 3"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.15, Page number: 557<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Array of objects to maintain records of bank \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class bank:\n",
+      "    def __init__(self):\n",
+      "        self.city = None\n",
+      "        self.bank = None\n",
+      "        self.branches = 0\n",
+      "        self.no_ac = 0\n",
+      "        \n",
+      "    def input1(self):\n",
+      "        self.city = raw_input(\"City Name : \")\n",
+      "        self.bank = raw_input(\"Bank Name : \")\n",
+      "        self.branches = int(raw_input(\"Total Branches : \"))\n",
+      "        self.no_ac = int(raw_input(\"Number of A/c : \"))\n",
+      "        \n",
+      "    def console(self):\n",
+      "        sys.stdout.write(\"\\n\\nCity Name    :   %s\"%(self.city))\n",
+      "        sys.stdout.write(\"\\nBank Name      :   %s\"%(self.bank))\n",
+      "        sys.stdout.write(\"\\nTotal Branches :   %d\"%(self.branches))\n",
+      "        sys.stdout.write(\"\\nNumber of A/c  :   %d\"%(self.no_ac))\n",
+      "        \n",
+      "#Object declaration\n",
+      "ms = [bank() for i in range(0,2)]\n",
+      "\n",
+      "#Processing\n",
+      "for k in range(0,2):\n",
+      "    ms[k].input1()\n",
+      "    \n",
+      "for k in range(0,2):\n",
+      "    ms[k].console()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "City Name : Nanded\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bank Name : SNSB\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Branches : 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of A/c : 7500\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "City Name : Latur\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bank Name : SNSB\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Branches : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of A/c : 5400\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "City Name    :   Nanded\n",
+        "Bank Name      :   SNSB\n",
+        "Total Branches :   10\n",
+        "Number of A/c  :   7500\n",
+        "\n",
+        "City Name    :   Latur\n",
+        "Bank Name      :   SNSB\n",
+        "Total Branches :   8\n",
+        "Number of A/c  :   5400"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.16, Page number: 559<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Accessing private data using non member function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class ac:\n",
+      "    def __init__(self):\n",
+      "        self.name = None\n",
+      "        self.acno = 0\n",
+      "        self.bal = 0\n",
+      "        \n",
+      "    def read(self):\n",
+      "        self.name = raw_input(\"Name  : \")\n",
+      "        self.acno = int(raw_input(\"A/c No  :\"))\n",
+      "        self.bal = int(raw_input(\"Balance : \"))\n",
+      "        \n",
+      "#Non member function\n",
+      "def showbal(a):\n",
+      "    sys.stdout.write(\"\\nBalance of A/c no. %d is Rs. %d\"%(a.acno,a.bal))\n",
+      "    \n",
+      "#object declaration\n",
+      "k = ac()\n",
+      "\n",
+      "#Function call\n",
+      "k.read()\n",
+      "showbal(k)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name  : Manoj\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A/c No  :474\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Balance : 40000\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Balance of A/c no. 474 is Rs. 40000"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.17, Page number: 561<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Friend function in two classes. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class first:\n",
+      "    def __init__(self):\n",
+      "        self.f = 0\n",
+      "        \n",
+      "    def getvalue(self):\n",
+      "        self.f = int(raw_input(\"Enter a number : \"))\n",
+      "        \n",
+      "class second:\n",
+      "    def __init__(self):\n",
+      "        self.s = 0\n",
+      "        \n",
+      "    def getvalue(self):\n",
+      "        self.s = int(raw_input(\"Enter a number : \"))\n",
+      "        \n",
+      "        \n",
+      "#Non member function\n",
+      "def sum1(d,t):\n",
+      "    sys.stdout.write(\"\\nSum of two numbers : %d\"%(t.f+d.s))\n",
+      "    \n",
+      "#Object declaration\n",
+      "a = first()\n",
+      "b = second()\n",
+      "\n",
+      "#Function call\n",
+      "a.getvalue()\n",
+      "b.getvalue()\n",
+      "sum1(b,a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of two numbers : 15"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.18, Page number: 562<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#String Replacement\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class fandr:\n",
+      "    def __init__(self):\n",
+      "        self.str1 = None\n",
+      "        self.f = ''\n",
+      "        self.r = ''\n",
+      "        \n",
+      "    def accept(self):\n",
+      "        #Since string object does not support item assignment, initialize the string as character array\n",
+      "        self.str1 = ['P','r','o','g','r','a','_','e','r']\n",
+      "        self.f = raw_input(\"Find what (char) \")\n",
+      "        self.r = raw_input(\"Replace with (char) \")\n",
+      "        \n",
+      "    def display(self,d):\n",
+      "        sys.stdout.write(\"%c\"%(self.str1[d]))\n",
+      "        \n",
+      "    def len1(self):\n",
+      "        self.l = len(self.str1)\n",
+      "        return self.l\n",
+      "    \n",
+      "    def find(self,i):\n",
+      "        if self.str1[i] == self.f:\n",
+      "            self.replace(i)\n",
+      "            \n",
+      "    def replace(self,k):\n",
+      "        self.str1[k] = self.r\n",
+      "        \n",
+      "#Object declaration\n",
+      "b = fandr()\n",
+      "\n",
+      "#function call\n",
+      "b.accept()\n",
+      "l = b.len1()\n",
+      "sys.stdout.write(\"\\nReplaced text : \")\n",
+      "\n",
+      "for i in range(0,l):\n",
+      "    b.find(i)\n",
+      "    b.display(i)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Find what (char) _\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Replace with (char) m\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Replaced text : Programer"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.19, Page number: 564<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Largest out of ten numbers.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    def __init__(self):\n",
+      "        self.number = [0 for i in range(0,10)]\n",
+      "        \n",
+      "    def input1(self,i):\n",
+      "        self.number[i] = int(raw_input(\"Enter Number (%d) : \"%(i+1)))\n",
+      "        return self.number[i]\n",
+      "    \n",
+      "    def large(self,s,m):\n",
+      "        if self.number[s] == m:\n",
+      "            sys.stdout.write(\"\\nLargest number : %d\"%(self.number[s]))\n",
+      "            return 0\n",
+      "        else:\n",
+      "            return 1\n",
+      "        \n",
+      "#Object declaration\n",
+      "b = num()\n",
+      "\n",
+      "sum1 = 0\n",
+      "c = 1\n",
+      "\n",
+      "#Sum\n",
+      "for i in range(0,10):\n",
+      "    sum1 = sum1 + b.input1(i)\n",
+      "    \n",
+      "#largest\n",
+      "for k in range(sum1,0,-1):\n",
+      "    for i in range(0,10):\n",
+      "        if c == 1:\n",
+      "            c = b.large(i,k)\n",
+      "        else:\n",
+      "            break"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (1) : 125\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (2) : 654\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (3) : 246\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (4) : 945\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (5) : 258\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (6) : 159\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (7) : 845\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (8) : 940\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (9) : 944\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number (10) : 485\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Largest number : 945"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.20, Page number: 567<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constructor to initialize the class member variables \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    #Constructor\n",
+      "    def __init__(self):\n",
+      "        sys.stdout.write(\"\\nConstructor called\")\n",
+      "        self.x = 5\n",
+      "        self.a = 0\n",
+      "        self.b = 1\n",
+      "        self.c = 2\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\nx = %d  a = %d  b = %d  c = %d\"%(self.x,self.a,self.b,self.c))\n",
+      "        \n",
+      "#Object declaration\n",
+      "x = num()\n",
+      "\n",
+      "#Function call\n",
+      "x.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Constructor called\n",
+        "x = 5  a = 0  b = 1  c = 2"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.21, Page number: 569<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constructor with arguments \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    #Constructor\n",
+      "    def __init__(self,m,j,k):\n",
+      "        self.a = m\n",
+      "        self.b = j\n",
+      "        self.c = k\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\na = %d  b = %d  c = %d\"%(self.a,self.b,self.c))\n",
+      "        \n",
+      "        \n",
+      "#Object declaration\n",
+      "x = num(4,5,7)\n",
+      "y = num(1,2,8)\n",
+      "\n",
+      "#Function call\n",
+      "x.show()\n",
+      "y.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "a = 4  b = 5  c = 7\n",
+        "a = 1  b = 2  c = 8"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.22, Page number: 571<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constructor overloading\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    def __init__(self,m = None,j = None,k = None):\n",
+      "        if m and j and k is not None:\n",
+      "            sys.stdout.write(\"Constructor with three arguments\")\n",
+      "            self.a = m\n",
+      "            self.b = j\n",
+      "            self.c = k\n",
+      "        \n",
+      "        else:\n",
+      "            if m and j is not None:\n",
+      "                sys.stdout.write(\"Constructor with two arguments\")\n",
+      "                self.a = m\n",
+      "                self.b = j\n",
+      "                self.c = ''\n",
+      "            else:\n",
+      "                if m == j == k == None:\n",
+      "                    sys.stdout.write(\"Constructor without arguments\")\n",
+      "                    self.a = 0\n",
+      "                    self.b = 0\n",
+      "                    self.c = ''\n",
+      "                    \n",
+      "                    \n",
+      "    def show(self):    \n",
+      "        sys.stdout.write(\"\\na = %d   b = %f c = \"%(self.a,self.b))\n",
+      "        print self.c\n",
+      "        \n",
+      "        \n",
+      "#Object declaration\n",
+      "\n",
+      "x = num(4,5.5,'A')\n",
+      "x.show()\n",
+      "\n",
+      "y = num(1,2.2)\n",
+      "y.show()\n",
+      "\n",
+      "z = num()\n",
+      "z.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Constructor with three arguments\n",
+        "a = 4   b = 5.500000 c = A\n",
+        "Constructor with two arguments\n",
+        "a = 1   b = 2.200000 c = \n",
+        "Constructor without arguments\n",
+        "a = 0   b = 0.000000 c = \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.23, Page number: 573<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Default arguments in constructor. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class power:\n",
+      "    def __init__(self,n = 9,p = 3):\n",
+      "        self.num = n\n",
+      "        self.power = p\n",
+      "        self.ans = pow(n,p)\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\n%d raise to %d is %d\"%(self.num,self.power,self.ans))\n",
+      "        \n",
+      "        \n",
+      "#Object declaration\n",
+      "p1 = power()\n",
+      "p2 = power(5)\n",
+      "\n",
+      "#Result\n",
+      "p1.show()\n",
+      "p2.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "9 raise to 3 is 729\n",
+        "5 raise to 3 is 125"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.24, Page number: 575<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Object with reference to constructor. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    #There is no overloading concept in python\n",
+      "    def __init__(self,k=None,j=None):\n",
+      "        if k is not None:\n",
+      "            self.n = k\n",
+      "            return\n",
+      "        else:\n",
+      "            if j is not None:\n",
+      "                self.n = j.n\n",
+      "                \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"%d\"%(self.n))\n",
+      "        \n",
+      "#Object creation\n",
+      "J = num(50)\n",
+      "K = num(J.n)\n",
+      "L = num()\n",
+      "L = J\n",
+      "M = num()\n",
+      "M = J\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nObject J Value of n : \")\n",
+      "J.show()\n",
+      "sys.stdout.write(\"\\nObject K Value of n : \")\n",
+      "K.show()\n",
+      "sys.stdout.write(\"\\nObject L Value of n : \")\n",
+      "L.show()\n",
+      "sys.stdout.write(\"\\nObject M Value of n : \")\n",
+      "M.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Object J Value of n : 50\n",
+        "Object K Value of n : 50\n",
+        "Object L Value of n : 50\n",
+        "Object M Value of n : 50"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.25, Page number: 576<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Destructors\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "c = 0\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    def __init__(self,c):\n",
+      "        c = c + 1\n",
+      "        sys.stdout.write(\"\\nObject Created : Object(%d)\"%(c))\n",
+      "        \n",
+      "        \n",
+      "    def __exit__(self, type, value, traceback,c):\n",
+      "        self.package_obj.cleanup()\n",
+      "        sys.stdout.write(\"\\nObject Released : Object(%d)\"%(c))\n",
+      "        c = c - 1\n",
+      "       \n",
+      "        \n",
+      "#Object creation\n",
+      "sys.stdout.write(\"\\nIn main()\\n\")\n",
+      "a = num(c)\n",
+      "b = num(c)\n",
+      "\n",
+      "sys.stdout.write(\"\\nIn Block A()\\n\")\n",
+      "c = num(c)\n",
+      "\n",
+      "sys.stdout.write(\"\\n\\nAgain in main()\\n\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In main()\n",
+        "\n",
+        "Object Created : Object(1)\n",
+        "Object Created : Object(1)\n",
+        "In Block A()\n",
+        "\n",
+        "Object Created : Object(1)\n",
+        "\n",
+        "Again in main()\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.26, Page number: 579<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Overload unary ++ operator\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    def input1(self):\n",
+      "        self.a = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.b = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.c = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.d = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\nA = %d  B = %d  C = %d  D = %d\"%(self.a,self.b,self.c,self.d))\n",
+      "        \n",
+      "    def operatorplusplus(self):\n",
+      "        self.a = self.a + 1\n",
+      "        self.b = self.b + 1\n",
+      "        self.c = self.c + 1\n",
+      "        self.d = self.d + 1\n",
+      "        \n",
+      "#Object creation\n",
+      "X = num()\n",
+      "\n",
+      "X.input1()\n",
+      "sys.stdout.write(\"\\nBefore Overloading X  : \")\n",
+      "X.show()\n",
+      "\n",
+      "#There is no operator overloading in python\n",
+      "X.operatorplusplus()\n",
+      "sys.stdout.write(\"\\nAfter Overloading X  : \")\n",
+      "X.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Before Overloading X  : \n",
+        "A = 4  B = 5  C = 8  D = 9\n",
+        "After Overloading X  : \n",
+        "A = 5  B = 6  C = 9  D = 10"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.27, Page number: 581<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Overload + binary operator\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    def input1(self):\n",
+      "        self.a = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.b = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.c = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.d = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"A = %d  B = %d  C = %d  D = %d\"%(self.a,self.b,self.c,self.d))\n",
+      "        \n",
+      "    def operatorplus(self,t):\n",
+      "        tmp = num()\n",
+      "        tmp.a = self.a + t.a\n",
+      "        tmp.b = self.b + t.b\n",
+      "        tmp.c = self.c + t.c\n",
+      "        tmp.d = self.d + t.d\n",
+      "        return tmp\n",
+      "    \n",
+      "#Object creation\n",
+      "X = num()\n",
+      "Y = num()\n",
+      "Z = num()\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nObject X \")\n",
+      "X.input1()\n",
+      "sys.stdout.write(\"\\nObject Y \")\n",
+      "Y.input1()\n",
+      "\n",
+      "Z = X.operatorplus(Y)\n",
+      "\n",
+      "sys.stdout.write(\"\\nX : \")\n",
+      "X.show()\n",
+      "sys.stdout.write(\"\\nY : \")\n",
+      "Y.show()\n",
+      "sys.stdout.write(\"\\nZ : \")\n",
+      "Z.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Object X "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Object Y "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "X : A = 1  B = 4  C = 2  D = 1\n",
+        "Y : A = 2  B = 5  C = 4  D = 2\n",
+        "Z : A = 3  B = 9  C = 6  D = 3"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.28, Page number: 583<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Multiplication using an integer and object. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class definition\n",
+      "class num:\n",
+      "    def input1(self):\n",
+      "        self.a = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.b = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.c = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        self.d = int(raw_input(\"Enter Values for a,b,c and d : \"))\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"A = %d  B = %d  C = %d  D = %d\"%(self.a,self.b,self.c,self.d))\n",
+      "        \n",
+      "def operatormul(a,t):\n",
+      "    tmp = num()\n",
+      "    tmp.a = a * t.a\n",
+      "    tmp.b = a * t.b\n",
+      "    tmp.c = a * t.c\n",
+      "    tmp.d = a * t.d\n",
+      "    return tmp\n",
+      "\n",
+      "#Result\n",
+      "X = num()\n",
+      "Z = num()\n",
+      "\n",
+      "sys.stdout.write(\"\\nObject X \")\n",
+      "X.input1()\n",
+      "\n",
+      "Z = operatormul(3,X)\n",
+      "sys.stdout.write(\"\\nX : \")\n",
+      "X.show()\n",
+      "sys.stdout.write(\"\\nZ : \")\n",
+      "Z.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Object X "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a,b,c and d : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "X : A = 1  B = 2  C = 2  D = 3\n",
+        "Z : A = 3  B = 6  C = 6  D = 9"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.29, Page number: 585<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Inheritance\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Class declaration\n",
+      "class one:\n",
+      "    def __init__(self):\n",
+      "        self.a = 0\n",
+      "        self.b = 0\n",
+      "        \n",
+      "class two(one):\n",
+      "    def __init__(self):\n",
+      "        one.__init__(self)\n",
+      "        self.c = 0\n",
+      "        self.d = 0\n",
+      "        \n",
+      "    def input1(self):\n",
+      "        self.a = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
+      "        self.b = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
+      "        self.c = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
+      "        self.d = int(raw_input(\"Enter values for a,b,c,d : \"))\n",
+      "        \n",
+      "    def display(self):\n",
+      "        sys.stdout.write(\"\\na = %d  b = %d  c = %d  d = %d\"%(self.a,self.b,self.c,self.d))\n",
+      "        \n",
+      "#Object declaration\n",
+      "a = two()\n",
+      "\n",
+      "#Result\n",
+      "a.input1()\n",
+      "a.display()\n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter values for a,b,c,d : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter values for a,b,c,d : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter values for a,b,c,d : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter values for a,b,c,d : 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "a = 2  b = 5  c = 4  d = 7"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16.30, Page number: 586<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Member function using pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#class declaration\n",
+      "class super1:\n",
+      "    def display(self):\n",
+      "        sys.stdout.write(\"\\nIn function display()  class super\")\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\nIn function show() class super\")\n",
+      "        \n",
+      "class sub(super1):\n",
+      "    def display(self):\n",
+      "        sys.stdout.write(\"\\nIn function display() class sub\")\n",
+      "        \n",
+      "    def show(self):\n",
+      "        sys.stdout.write(\"\\nIn function show() class sub\")\n",
+      "        \n",
+      "#Object declaration\n",
+      "S = super1()\n",
+      "A = sub()\n",
+      "\n",
+      "#There is no pointer concept in python\n",
+      "S.display()\n",
+      "S.show()\n",
+      "\n",
+      "sys.stdout.write(\"\\n\\nNow Pointer point points to derived class sub\\n\")\n",
+      "A.display()\n",
+      "A.show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "In function display()  class super\n",
+        "In function show() class super\n",
+        "\n",
+        "Now Pointer point points to derived class sub\n",
+        "\n",
+        "In function display() class sub\n",
+        "In function show() class sub"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter3.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter3.ipynb
new file mode 100755
index 00000000..880e4fac
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter3.ipynb
@@ -0,0 +1,1060 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 3: Operators and Expressions<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.1, Page number: 22<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate the use of Comma operator\n",
+      "\n",
+      "#Result\n",
+      "print \"Addition = \",2+3,\" \\nSubtraction = \",5-4"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Addition =  5  \n",
+        "Subtraction =  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.2, Page number: 23<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to demonstrate the use of Conditional operator with two values\n",
+      "\n",
+      "#Result\n",
+      "#in python if..else statement can be written similar to conditional operator\n",
+      "#prints Result = 4 if 2 == 3, otherwise Result = 5\n",
+      "\n",
+      "print \"Result = \",4 if 2==3 else 5"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Result =  5\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.3, Page number: 23<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to use Conditional operator with two statements\n",
+      "\n",
+      "#Result\n",
+      "#in python if..else statement can be written similar to conditional operator\n",
+      "#prints True if 3>2, otherwise prints False\n",
+      "\n",
+      "print \"True\" if 3>2 else \"False\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "True\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.4, Page number: 25<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to use increment operator as prefix\n",
+      "\n",
+      "#Variable declaration\n",
+      "x = 10\n",
+      "y = 20\n",
+      "\n",
+      "#Calculation\n",
+      "#there is no ++/-- operator in python\n",
+      "\n",
+      "z = x * y\n",
+      "y += 1\n",
+      "a = x * y\n",
+      "\n",
+      "#Result\n",
+      "print z,\"  \",a"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "200    210\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.5, Page number: 25<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to use increment operator as prefix\n",
+      "\n",
+      "#Variable declaration\n",
+      "x = 10\n",
+      "y = 20\n",
+      "\n",
+      "#Calculation\n",
+      "#there is no ++/-- operator in python\n",
+      "\n",
+      "y += 1\n",
+      "z = x * y\n",
+      "a = x * y\n",
+      "\n",
+      "#Result\n",
+      "print z,\"  \",a"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "210    210\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.6, Page number: 26<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to find the size and address of integer and float variables\n",
+      "\n",
+      "#import python module\n",
+      "#sys python module is required for getsizeof() function.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "#No specific float type variable declaration is available in python. If we\n",
+      "#assign a floating point value to a variable,\n",
+      "#it will be treated as floating point variable\n",
+      "x = 2\n",
+      "y = 2.0\n",
+      "\n",
+      "#Result\n",
+      "#In python, integer variable uses 12 bytes and float variable uses 16 bytes of\n",
+      "#memory to store the variables\n",
+      "#sys.getsizeof() function returns the size of variable in bytes which is similar\n",
+      "#to sizeof() operator\n",
+      "#id() function returns the address of variables which is similar to '&' operator\n",
+      "\n",
+      "print \"Size (x) = \",sys.getsizeof(x)\n",
+      "print \"Size (y) = \",sys.getsizeof(y)\n",
+      "print \"Address of x = \",id(x),\"and y = \",id(y)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Size (x) =  12\n",
+        "Size (y) =  16\n",
+        "Address of x =  5626076 and y =  54761072\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.7, Page number: 27<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to use various relational operators and display their return values\n",
+      "\n",
+      "#Result\n",
+      "#int() function is used to convert the boolean return value of the relational\n",
+      "#expression to integer/binary\n",
+      "\n",
+      "print \"Condition  :   Return Values\"\n",
+      "print \"10 != 10   :       \",int(10 != 10)\n",
+      "print \"10 == 10   :       \",int(10 == 10)\n",
+      "print \"10 >= 10   :       \",int(10 >= 10)\n",
+      "print \"10 <= 100  :       \",int(10 <= 100)\n",
+      "print \"10 != 9    :       \",int(10 != 9)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Condition  :   Return Values\n",
+        "10 != 10   :        0\n",
+        "10 == 10   :        1\n",
+        "10 >= 10   :        1\n",
+        "10 <= 100  :        1\n",
+        "10 != 9    :        1\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.8, Page number: 28<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program uses conditional operator to determine the value of 'b'depending the inputted value of 'a'\n",
+      "\n",
+      "#Reads Input value for a\n",
+      "\n",
+      "a = raw_input(\"Enter Any Integer either above 5 or below 5 :-\")\n",
+      "a = int(a)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Calculation\n",
+      "#Assigns b = 3 if a > 5, otherwise b = 4\n",
+      "b = 3 if a > 5 else 4\n",
+      "\n",
+      "#Result\n",
+      "print \"Calculated Value of b is :- \",b"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Any Integer either above 5 or below 5 :-6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Calculated Value of b is :-  3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.9, Page number: 28<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to determine the value of 'b' using conditional operator\n",
+      "\n",
+      "#Input\n",
+      "\n",
+      "a = raw_input(\"Enter Any Integer either above 5 or below 5 :-\")\n",
+      "a = int(a)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Calculation\n",
+      "#Assigns b = 3 if a == 5, otherwise b = 4\n",
+      "b = 3 if a == 5 else 4\n",
+      "\n",
+      "#Result\n",
+      "print \"Calculated Value of b is :- \",b"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Any Integer either above 5 or below 5 :-3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Calculated Value of b is :-  4\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.10, Page number: 29<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read three variables x, y and z and Using conditional statements, evaluate\n",
+      "#values of variables a, b and c.\n",
+      "#Perform the sum with two sets of variables.  Check whether the sums are equal or not\n",
+      "#and print appropriate messages.\n",
+      "\n",
+      "\n",
+      "#Reads three Input values for x,y and z\n",
+      "\n",
+      "x = raw_input(\"Enter Value of x \")\n",
+      "x = int(x)\n",
+      "\n",
+      "y = raw_input(\"Enter Value of y \")\n",
+      "y = int(y)\n",
+      "\n",
+      "z = raw_input(\"Enter Value of z \")\n",
+      "z = int(z)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Calculation\n",
+      "#Assigns a = 3 if x >= 5, otherwise a = 4\n",
+      "a = 3 if x >= 5 else 4\n",
+      "\n",
+      "#Result\n",
+      "print \"Calculated Value of a is :- \",a\n",
+      "\n",
+      "#Calculation\n",
+      "#Assigns b = 10 if y <= 8, otherwise b = 9\n",
+      "b = 10 if y <= 8 else 9\n",
+      "\n",
+      "#Result\n",
+      "print \"Calculated Value of b is :- \",b\n",
+      "\n",
+      "#Calculation\n",
+      "#Assigns c = 20 if z == 10, otherwise c = 30\n",
+      "c = 20 if z == 10  else 30\n",
+      "\n",
+      "#Result\n",
+      "print \"Calculated Value of c is :- \",c\n",
+      "\n",
+      "#Calculation\n",
+      "m = x + y + z\n",
+      "n = a + b + c\n",
+      "\n",
+      "#Result\n",
+      "print \"Addition of x,y,z is \",m,\"(m)\"\n",
+      "print \"Addition of a,b,c is \",n,\"(n)\"\n",
+      "print \"m & n NOT EQUAL\" if m != n else \"m & n ARE EQUAL\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of x 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of y 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of z 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Calculated Value of a is :-  3\n",
+        "Calculated Value of b is :-  10\n",
+        "Calculated Value of c is :-  30\n",
+        "Addition of x,y,z is  14 (m)\n",
+        "Addition of a,b,c is  43 (n)\n",
+        "m & n NOT EQUAL\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.11, Page number: 30<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use of logical operators\n",
+      "\n",
+      "#Result\n",
+      "#In python, not equal to can be represented using both != and <>.\n",
+      "\n",
+      "print \"Condition        :   Return Values\"\n",
+      "print \"5 > 3 && 5 < 10  :   \",int(5 > 3 & 5 < 10)\n",
+      "print \"8 > 5 || 8 < 2   :   \",int(8 > 5 | 8 < 2)\n",
+      "print \"!(8 == 8)        :   \",int(8 != 8)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Condition        :   Return Values\n",
+        "5 > 3 && 5 < 10  :    1\n",
+        "8 > 5 || 8 < 2   :    0\n",
+        "!(8 == 8)        :    0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.12, Page number: 30<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print logic 1 if input character is capital otherwise 0\n",
+      "\n",
+      "#Import\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#raw_input() function is used to read a character from the keyboard in Python 2.7\n",
+      "#ord() function converts the character into its corresponding ASCII integer\n",
+      "\n",
+      "\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "x = raw_input(\"Enter a Character : \")\n",
+      "x = ord(x)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Condition\n",
+      "y = 1 if (x >= 65) and (x <= 90) else 0\n",
+      "\n",
+      "#Result\n",
+      "print \"Y : \",y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Character : A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Y :  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.13, Page number: 31<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display logic 0 if one reads a character through keyboard otherwise logic 1.\n",
+      "#ASCII values for 0 to 9 are 48 to 57 respectively\n",
+      "\n",
+      "#Variable initialization\n",
+      "#raw_input() function is used to read a character from the keyboard in Python 2.7\n",
+      "#ord() function converts the character into its corresponding ASCII integer\n",
+      "\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "x = raw_input(\"Enter The Character or Integer : \")\n",
+      "x = ord(x)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Condition\n",
+      "y = 1 if (x >= 48) and (x <= 57) else 0\n",
+      "\n",
+      "#Result\n",
+      "print \"Value of Y = \",y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter The Character or Integer : A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of Y =  0\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.14, Page number: 32<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display 1 if inputted number is between 1-100 otherwise 0.\n",
+      "#Use the logical AND operator\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "x = raw_input(\"Enter numbers : \")\n",
+      "x = int(x)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Condition\n",
+      "z = 1 if (x >= 1) and (x <= 100) else 0\n",
+      "\n",
+      "#Result\n",
+      "print \"Z = \",z"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Z =  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.15, Page number: 32<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display 1 if inputted number is either 1 or 100 otherwise 0.\n",
+      "#Use the logical OR operator\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "x = raw_input(\"Enter numbers : \")\n",
+      "x = int(x)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Condition\n",
+      "z = 1 if (x == 1) or (x == 100) else 0\n",
+      "\n",
+      "#Result\n",
+      "print \"Z = \",z"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Z =  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.16, Page number: 33<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display 1 if inputted number is except 100 otherwise 0.\n",
+      "#Use the logical NOT operator\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "x = raw_input(\"Enter number : \")\n",
+      "x = int(x)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Condition\n",
+      "z = 1 if (x != 100) else 0\n",
+      "\n",
+      "#Result\n",
+      "print \"Z : \",z"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter number : 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Z :  0\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.17, Page number: 33<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Shift inputted data by two bits right\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "x = raw_input(\"Read The Integer from keyboard (x) :- \")\n",
+      "x = int(x)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#bitwise shifting\n",
+      "x >>= 2\n",
+      "y = x\n",
+      "\n",
+      "#Result\n",
+      "print \"The Right shifted data is = \",y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integer from keyboard (x) :- 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Right shifted data is =  2\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.18, Page number: 34<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Shift inputted data by two bits to the left\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "x = raw_input(\"Read The Integer from keyboard (x) :- \")\n",
+      "x = int(x)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#bitwise shifting\n",
+      "x <<= 3\n",
+      "y = x\n",
+      "\n",
+      "#Result\n",
+      "print \"The Left shifted data is = \",y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integer from keyboard (x) :- 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Left shifted data is =  16\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.19, Page number: 35<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use bitwise AND operator between the two integers and display the results\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "a = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
+      "a = int(a)\n",
+      "\n",
+      "b = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
+      "b = int(b)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#bitwise operation\n",
+      "c = a & b\n",
+      "\n",
+      "#Result\n",
+      "print \"The Answer after ANDing is (C) = \",c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integers from keyboard (a & b) :- 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integers from keyboard (a & b) :- 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Answer after ANDing is (C) =  0\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.20, Page number: 37<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#operate OR operation on two integers and display the result\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "a = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
+      "a = int(a)\n",
+      "\n",
+      "b = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
+      "b = int(b)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#bitwise operation\n",
+      "c = a | b\n",
+      "\n",
+      "#Result\n",
+      "print \"The Oring operation betwen a & b in c = \",c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integers from keyboard (a & b) :- 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integers from keyboard (a & b) :- 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Oring operation betwen a & b in c =  12\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3.21, Page number: 38<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Exclusive OR operation between the two integers and display the result\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for x\n",
+      "\n",
+      "a = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
+      "a = int(a)\n",
+      "\n",
+      "b = raw_input(\"Read The Integers from keyboard (a & b) :- \")\n",
+      "b = int(b)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#bitwise operation\n",
+      "c = a ^ b\n",
+      "\n",
+      "#Result\n",
+      "print \"The data after Exclusive OR operation is in c = \",c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integers from keyboard (a & b) :- 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Read The Integers from keyboard (a & b) :- 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The data after Exclusive OR operation is in c =  10\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter4.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter4.ipynb
new file mode 100755
index 00000000..758457a3
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter4.ipynb
@@ -0,0 +1,972 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 4: Input & Output in C<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.1, Page number: 47<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#show the effect of mismatch of data types\n",
+      "\n",
+      "#Reads Input value for a\n",
+      "\n",
+      "a = raw_input(\"Enter value of 'A' : \")\n",
+      "a = int(a)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "print \"A = \",a\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of 'A' : 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A =  8\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.2, Page number: 47<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#show the effect of mismatch of data types\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#Reads Input value for a\n",
+      "\n",
+      "a = raw_input(\"Enter value of 'A' : \")\n",
+      "a = int(a)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#there is no char datatype of 1 byte size in python.\n",
+      "#python has string variable only.\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "sys.stdout.write(\"A = %d\"%(a%256))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of 'A' : 256\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A = 0"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.3, Page number: 49<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#show the effect of various escape sequences\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = 1\n",
+      "b = a + 1\n",
+      "c = b + 1\n",
+      "d = c + 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write ('\\tA = %d\\nB = %d \\'C = %d\\''%(a,b,c))\n",
+      "sys.stdout.write ('\\n**D = %d**'%(d))\n",
+      "sys.stdout.write ('\\nA = %d B = %d'%(a,b))\n",
+      "sys.stdout.write ('\\r******')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tA = 1\n",
+        "B = 2 'C = 3'\n",
+        "**D = 4**\n",
+        "A = 1 B = 2******"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.4, Page number: 50<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#print the third power of 2 using pow() function.\n",
+      "#assume the floating point numbers\n",
+      "\n",
+      "#Variable initialization\n",
+      "x = 2.0\n",
+      "y = 3.0\n",
+      "\n",
+      "#Result\n",
+      "print '%lf raised to %lf is %lf\\n' %(x,y,pow(x,y))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2.000000 raised to 3.000000 is 8.000000\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.5, Page number: 50<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#print the third power of 10 using pow() function.\n",
+      "#assume the floating point numbers\n",
+      "\n",
+      "#Variable initialization\n",
+      "p = 3\n",
+      "\n",
+      "#Result\n",
+      "#There is no pow10() function in Python.\n",
+      "print 'Ten raised to %lf is %lf\\n' %(p,pow(10,p))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ten raised to 3.000000 is 1000.000000\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.6, Page number: 51<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#detect an error while inputting a data.  Use return value of scanf() statement.\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python,the map(int,a.split()) function splits the input characters\n",
+      "#and converts it into integers and returns a value if all the values are integers\n",
+      "#otherwise an exception will be generated and initializes v as 0.\n",
+      "#using exception handling, this result can be achieved.\n",
+      "\n",
+      "\n",
+      "#Exception handling\n",
+      "try:\n",
+      "    a = raw_input(\"Enter value of 'A','B' & 'C' : \")\n",
+      "    v = map(int,a.split())\n",
+      "except:\n",
+      "    v = 0\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "print '\\nError In Inputting.' if v < 3 else '\\nValues Successfully read.'\n",
+      "\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of 'A','B' & 'C' : 1 2 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Values Successfully read.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.7, Page number: 51<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find length of the string using print() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, print()or write() function would not return the number of characters\n",
+      "#written to the output buffer. so len() function is used to find the length.\n",
+      "\n",
+      "nm = raw_input(\"Enter String : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "l = len(nm)\n",
+      "\n",
+      "#Result\n",
+      "print '\\nLength = ',l\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String : HELLO\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Length =  5\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.8, Page number: 52<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Perform addition of two numbers\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = raw_input(\"ENTER TWO VALUES \")\n",
+      "b = raw_input(\"ENTER TWO VALUES \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "c = int(a) + int(b)\n",
+      "\n",
+      "#Result\n",
+      "print '\\nSUM IS = ',c\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ENTER TWO VALUES 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ENTER TWO VALUES 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "SUM IS =  13\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.9, Page number: 52<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the square of the given number\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = raw_input(\"ENTER ANY NUMBER \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "c = int(a) * int(a)\n",
+      "\n",
+      "#Result\n",
+      "print '\\nSQUARE OF GIVEN NUMBER = ',c\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ENTER ANY NUMBER 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "SQUARE OF GIVEN NUMBER =  25\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.10, Page number: 53<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate average of three real numbers.\n",
+      "\n",
+      "#Variable initialization\n",
+      "\n",
+      "a = raw_input(\"ENTER THREE FLOAT NUMBERS : \")\n",
+      "b = raw_input(\"ENTER THREE FLOAT NUMBERS : \")\n",
+      "c = raw_input(\"ENTER THREE FLOAT NUMBERS : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "d = float(a) + float(b) + float(c)\n",
+      "\n",
+      "#Result\n",
+      "print '\\nAverage of Given Numbers : ',d/3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ENTER THREE FLOAT NUMBERS : 2.5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ENTER THREE FLOAT NUMBERS : 3.5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ENTER THREE FLOAT NUMBERS : 4.5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Average of Given Numbers :  3.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.11, Page number: 53<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print message with inputted name\n",
+      "\n",
+      "#Variable initialization\n",
+      "yourname = raw_input(\"Enter Your Name : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "print \"\\nWel Come %s to 'C' Programming Course\"%(yourname)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name : Ajay\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Wel Come Ajay to 'C' Programming Course\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.12, Page number: 54<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input a single character and display it\n",
+      "\n",
+      "#Variable initialization\n",
+      "ch = raw_input(\"Enter any character : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "print \"\\nYour Entered Character is : %c\"%(ch)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any character : C\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Your Entered Character is : C\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.13, Page number: 54<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Swap the values of two variables without using third variable.\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = 7\n",
+      "b = 4\n",
+      "\n",
+      "print \"\\n A = %d B = %d\"%(a,b)\n",
+      "\n",
+      "#Swapping\n",
+      "a = a + b\n",
+      "b = a - b\n",
+      "a = a - b\n",
+      "\n",
+      "#Result\n",
+      "print \"Now A = %d B = %d\"%(a,b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " A = 7 B = 4\n",
+        "Now A = 4 B = 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.14, Page number: 55<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Accept characters through keyboard using getchar() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, string is an object. It has no nul-termination character.\n",
+      "#raw_input() reads the string constant from the stdin and the write() function\n",
+      "#prints it to the stdout based on the attribute length.\n",
+      "\n",
+      "ch = raw_input(\" \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(ch)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " COMPILER\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "COMPILER"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.15, Page number: 56<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the characters using putchar() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, string is an object. It has no nul-termination character.\n",
+      "#raw_input() reads the string constant from the stdin.\n",
+      "\n",
+      "ch = raw_input(\"Enter Text Here : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "#using for loop, the string can be displayed character by character.\n",
+      "\n",
+      "sys.stdout.write(\"The Entered Text : \")\n",
+      "for c in ch:\n",
+      "    sys.stdout.write(c)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Here : Characters\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Entered Text : Characters"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.16, Page number: 56<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Show the effect of getche() and getch() functions\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, there is no equivalent function for getche() and getch() functions.\n",
+      "#raw_input() function reads the characters and terminates by pressing the\n",
+      "#enter key\n",
+      "\n",
+      "ch = raw_input(\"Enter any two alphabetics \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "#Result\n",
+      "sys.stdout.write(ch)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any two alphabetics A \n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.17, Page number: 57<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read and display the character using getch() and putch() functions\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, there is no equivalent function for getch() function.\n",
+      "#raw_input() function reads the characters and terminates by pressing the\n",
+      "#enter key\n",
+      "\n",
+      "ch = raw_input(\"Press any key to continue \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"You Pressed : %s\"%(ch))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press any key to continue 9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You Pressed : 9"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.18, Page number: 57<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Accept string through the keyboard using gets() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, raw_input() function reads the string and terminates by pressing the\n",
+      "#enter key and write() function displays the string.\n",
+      "\n",
+      "ch = raw_input(\"Enter the String : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Entered String : %s\"%(ch))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the String : USE OF GETS()\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Entered String : USE OF GETS()"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.19, Page number: 58<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the accepted character using puts() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, raw_input() function reads the string and terminates by pressing the\n",
+      "#enter key and print() function displays the string.\n",
+      "\n",
+      "ch = raw_input(\"Enter the String : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "print\"Entered String : \"\n",
+      "print ch\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the String : puts is in use.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Entered String : \n",
+        "puts is in use.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4.20, Page number: 58<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#read string using cgets() and display it using cputs()\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python, raw_input() function reads the string and terminates by pressing the\n",
+      "#enter key and print() function displays the string.\n",
+      "#in python, string is an object.\n",
+      "\n",
+      "t = raw_input(\"Enter Text Here : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "print \"Your Entered Text : %s\"%(t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Here : How are you?\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your Entered Text : How are you?\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter5.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter5.ipynb
new file mode 100755
index 00000000..dac1632e
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter5.ipynb
@@ -0,0 +1,2624 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 5.1: Decision Statements<h1>\n",
+      "   "
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.1, Page number: 64<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Check whether the entered number is less than 10? if yes, display the same\n",
+      "\n",
+      "#Variable initialization\n",
+      "v = raw_input(\"Enter the number : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "v = int(v)\n",
+      "\n",
+      "#Result\n",
+      "if v < 10 :\n",
+      "    print '\\nNumber entered is less than 10'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number : 9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number entered is less than 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.2, Page number: 65<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Check equivalence of two numbers. Use if statement.\n",
+      "\n",
+      "#Variable initialization\n",
+      "\n",
+      "m = raw_input(\"Enter two numbers : \")\n",
+      "n = raw_input(\"Enter two numbers : \")\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "m = int(m)\n",
+      "n = int(n)\n",
+      "\n",
+      "#Result\n",
+      "if m - n == 0 :\n",
+      "    print '\\nTwo numbers are equal'\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Two numbers are equal\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.3, Page number: 66<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Check whether the candidate's age is greater than 17 or not.  If yes, display\n",
+      "#message \"Eligible for Voting\"\n",
+      "\n",
+      "#Variable initialization\n",
+      "age = raw_input(\"Enter age : \")\n",
+      "age = int(age)\n",
+      "\n",
+      "#In python, raw_input() is used to read values\n",
+      "\n",
+      "#Result\n",
+      "if age > 17 :\n",
+      "    print '\\nEligible for Voting.'\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter age : 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Eligible for Voting.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.4, Page number: 66<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use curly brace in the if block.  Enter only the three numbers and calculate\n",
+      "#their sum and multiplication\n",
+      "\n",
+      "#Variable initialization\n",
+      "#in python,the map(int,raw_input().split()) function splits the input characters\n",
+      "#and converts it into integers and returns a value if all the values are integers\n",
+      "#otherwise an exception will be generated and initializes v as 0.\n",
+      "#using exception handling, this result can be achieved.\n",
+      "\n",
+      "#Exception handling\n",
+      "try:\n",
+      "    #use space to separate the input characters\n",
+      "    v = raw_input(\"Enter Three Numbers : \")\n",
+      "    a,b,c = map(int,v.split())\n",
+      "    x = 3\n",
+      "except:\n",
+      "    x = 0\n",
+      "\n",
+      "#Result\n",
+      "#in python, block of statements are identified using indentation\n",
+      "    \n",
+      "if x == 3:\n",
+      "    print \"Addition : %d\"%(a+b+c)\n",
+      "    print \"Multiplication : %d\"%(a*b*c)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 1 2 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Addition : 7\n",
+        "Multiplication : 8\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.5, Page number: 68<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the values of a,b,c through the keyboard.  Add them and after addition\n",
+      "#check if it is in the range of 100 & 200 or not. Print separate message for each.\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = raw_input(\"Enter Three Numbers a b c : \")\n",
+      "b = raw_input(\"Enter Three Numbers a b c : \")\n",
+      "c = raw_input(\"Enter Three Numbers a b c : \")\n",
+      "\n",
+      "a = int(a)\n",
+      "b = int(b)\n",
+      "c = int(c)\n",
+      "\n",
+      "print 'a = %d b = %d c = %d'%(a,b,c)\n",
+      "#Calculation\n",
+      "d = a + b + c\n",
+      "\n",
+      "#Condition evaluation and Result\n",
+      "if d <= 200 and d >= 100 :\n",
+      "    print \"Sum is %d which is in between 100 & 200\"%(d)\n",
+      "else:\n",
+      "    print \"\\nSum is %d which is out of range\"%(d)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers a b c : 50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers a b c : 52\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers a b c : 54\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = 50 b = 52 c = 54\n",
+        "Sum is 156 which is in between 100 & 200\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.6, Page number: 68<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the roots of a quadratic equation by using if else condition\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = raw_input(\"Enter Values for a, b, c : \")\n",
+      "b = raw_input(\"Enter Values for a, b, c : \")\n",
+      "c = raw_input(\"Enter Values for a, b, c : \")\n",
+      "\n",
+      "a = int(a)\n",
+      "b = int(b)\n",
+      "c = int(c)\n",
+      "\n",
+      "#Condition evaluation and Result\n",
+      "if b * b > 4 * a * c :\n",
+      "    x1 = b + sqrt(b*b - 4*a*c)/2*a\n",
+      "    x2 = b - sqrt(b*b - 4*a*c)/2*a\n",
+      "    print \"\\nx1 = %f x2 = %f\"%(x1,x2)\n",
+      "else:\n",
+      "    print \"\\nRoots are Imaginary\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a, b, c : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a, b, c : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Values for a, b, c : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Roots are Imaginary\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.7, Page number: 69<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the square of those numbers only whose least significant digit is 5.\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "s = raw_input(\"Enter a Number : \")\n",
+      "s = int(s)\n",
+      "\n",
+      "d = s % 10;\n",
+      "\n",
+      "#Condition evaluation and Result\n",
+      "#There is no increment/decrement operator in python.\n",
+      "if d==5 :\n",
+      "    s = s/10\n",
+      "    s += 1\n",
+      "    print \"\\nSquare = %d%d\"%(s*(s-1),d*d)\n",
+      "else:\n",
+      "    print \"\\nInvalid Number\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 25\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Square = 625\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.8, Page number: 70<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the salary of medical representative based on the sales.  Bonus and\n",
+      "#incentive to be offered to him will be based on total sales.  If the sale\n",
+      "#exceeds Rs.100000/- follow the particulars of table (1) otherwise (2)\n",
+      "\n",
+      "#  1.  TABLE                            2. TABLE\n",
+      "# Basic = Rs. 3000                  Basic = Rs. 3000\n",
+      "# HRA = 20% of basic                HRA = 20% of basic\n",
+      "# DA = 110% of basic                DA = 110% of basic\n",
+      "# Conveyance = Rs. 500              Conveyance = Rs. 500\n",
+      "# Incentive = 10% of sales          Incentive = 5% of sales\n",
+      "# Bonus = Rs. 500                   Bonus = Rs. 200\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "sale = raw_input(\"Enter Total Sales in Rs. : \")\n",
+      "sale = int(sale)\n",
+      "\n",
+      "#Condition evaluation\n",
+      "if sale >= 100000 :\n",
+      "    bs = 3000\n",
+      "    hra = 20 * bs/100\n",
+      "    da = 110 * bs/100\n",
+      "    cv = 500\n",
+      "    incentive = sale * 10/100\n",
+      "    bonus = 500\n",
+      "else:\n",
+      "    bs = 3000\n",
+      "    hra = 20 * bs/100\n",
+      "    da = 110 * bs/100\n",
+      "    cv = 500\n",
+      "    incentive = sale * 5/100\n",
+      "    bonus = 200\n",
+      "\n",
+      "#Result\n",
+      "ts = bs + hra + da + cv + incentive + bonus\n",
+      "print \"\\nTotal Sales : %.2f\"%(sale)\n",
+      "print \"\\nBasic Salary : %.2f\"%(bs)\n",
+      "print \"\\nHra          : %.2f\"%(hra)\n",
+      "print \"\\nDa           : %.2f\"%(da)\n",
+      "print \"\\nConveyance   : %.2f\"%(cv)\n",
+      "print \"\\nIncentive    : %.2f\"%(incentive)\n",
+      "print \"\\nBonus        : %.2f\"%(bonus)\n",
+      "print \"\\nGross Salary : %.2f\"%(ts)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Total Sales in Rs. : 100000\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Total Sales : 100000.00\n",
+        "\n",
+        "Basic Salary : 3000.00\n",
+        "\n",
+        "Hra          : 600.00\n",
+        "\n",
+        "Da           : 3300.00\n",
+        "\n",
+        "Conveyance   : 500.00\n",
+        "\n",
+        "Incentive    : 10000.00\n",
+        "\n",
+        "Bonus        : 500.00\n",
+        "\n",
+        "Gross Salary : 17900.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.9, Page number: 72<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate energy bill using the starting and ending meter reading.\n",
+      "#The charges are as follows:\n",
+      "\n",
+      "# No. of Units consumed         Rates in (Rs.)\n",
+      "# 200 - 500                     3.50\n",
+      "# 100 - 200                     2.50\n",
+      "# Less than 100                 1.50\n",
+      "\n",
+      "#Variable initialization\n",
+      "initial = raw_input(\"Initial & Final Readings : \")\n",
+      "final = raw_input(\"Initial & Final Readings : \")\n",
+      "\n",
+      "consumed = int(final) - int(initial)\n",
+      "\n",
+      "#Condition evaluation\n",
+      "if consumed >= 200 and consumed <= 500 :\n",
+      "    total = consumed * 3.50\n",
+      "else:\n",
+      "    if consumed >= 100 and consumed <= 199:\n",
+      "          total = consumed * 2.50\n",
+      "    else :\n",
+      "        if consumed < 100 :\n",
+      "           total = consumed * 1.50\n",
+      "\n",
+      "#Result\n",
+      "print \"Total bill for %d unit is %f\"%(consumed,total)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Initial & Final Readings : 1200\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Initial & Final Readings : 1500\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total bill for 300 unit is 1050.000000\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.10, Page number: 73<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate energy bill by  reading the starting and ending meter reading.\n",
+      "#If the consumed electricity energy is greater than or equal to 200 units\n",
+      "#the rate should be 2.50/unit otherwise 1.50/unit.\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "initial = raw_input(\"Initial & Final Readings : \")\n",
+      "final = raw_input(\"Initial & Final Readings : \")\n",
+      "\n",
+      "consumed = int(final) - int(initial)\n",
+      "\n",
+      "#Condition evaluation\n",
+      "if consumed >= 200 :\n",
+      "    total = consumed * 2.50\n",
+      "else:\n",
+      "    total = consumed * 1.50\n",
+      "\n",
+      "#Result\n",
+      "print \"Total bill for %d unit is %f\"%(consumed,total)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total bill for 100 unit is 150.000000\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.11, Page number: 73<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sort six numbers and find the largest one by using ladder of if else\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = int(raw_input(\"Enter 1st number : \"))\n",
+      "b = int(raw_input(\"Enter 2nd number : \"))\n",
+      "c = int(raw_input(\"Enter 3rd number : \"))\n",
+      "d = int(raw_input(\"Enter 4th number : \"))\n",
+      "e = int(raw_input(\"Enter 5th number : \"))\n",
+      "f = int(raw_input(\"Enter 6th number : \"))\n",
+      "\n",
+      "#Condition evaluation and Result\n",
+      "if a > b and a > c and a > d and a > e and a > f :\n",
+      "    print \"Highest of six Number is : %d\"%(a)\n",
+      "else:\n",
+      "    if b > a and b > c and b > d and b > e and b > f :\n",
+      "        print \"Highest of six Number is : %d\"%(b)\n",
+      "    else:\n",
+      "        if c > a and c > b and c > d and c > e and c > f :\n",
+      "            print \"Highest of six Number is : %d\"%(c)\n",
+      "        else:\n",
+      "            if d > a and d > b and d > c and d > e and d > f :\n",
+      "                print \"Highest of six Number is : %d\"%(d)\n",
+      "            else:\n",
+      "                if e > a and e > b and e > c and e > d and e > f :\n",
+      "                   print \"Highest of six Number is : %d\"%(e)\n",
+      "                else:\n",
+      "                    print \"Highest of six Number is : %d\"%(f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 1st number : 52\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 2nd number : 74\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 3rd number : 90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 4th number : 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 5th number : 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 6th number : 22\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Highest of six Number is : 90\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.12, Page number: 74<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find largest number out of three numbers.  Read the numbers through the keyboard\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "x = int(raw_input(\"\\nEnter Three Numbers x,y,z : \"))\n",
+      "y = int(raw_input(\"\\nEnter Three Numbers x,y,z : \"))\n",
+      "z = int(raw_input(\"\\nEnter Three Numbers x,y,z : \"))\n",
+      "\n",
+      "print \"\\nLargest out of Three Numbers is : \"\n",
+      "\n",
+      "#Condition evaluation and Result\n",
+      "if x > y :\n",
+      "    if x > z:\n",
+      "        print \"x = %d\\n\"%(x)\n",
+      "    else:\n",
+      "        print \"z = %d\\n\"%(z)\n",
+      "else:\n",
+      "    if z > y:\n",
+      "        print \"z =  %d\\n\"%(z)\n",
+      "    else:\n",
+      "        print \"y = %d\\n\"%(y)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Three Numbers x,y,z : 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Three Numbers x,y,z : 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Three Numbers x,y,z : 30\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Largest out of Three Numbers is : \n",
+        "z =  30\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.13, Page number: 75<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the smallest out of the three numbers.\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = int(raw_input(\"\\nEnter Three Numbers : \"))\n",
+      "b = int(raw_input(\"\\nEnter Three Numbers : \"))\n",
+      "c = int(raw_input(\"\\nEnter Three Numbers : \"))\n",
+      "          \n",
+      "#Condition evaluation and Result\n",
+      "if a < b :\n",
+      "    if a < c:\n",
+      "        smallest = a\n",
+      "    else:\n",
+      "        smallest = c\n",
+      "else:\n",
+      "    if b < c:\n",
+      "        smallest = b\n",
+      "    else:\n",
+      "        smallest = c\n",
+      "\n",
+      "#Result\n",
+      "print \"The smallest of %d %d %d is %d\\n\"%(a,b,c,smallest)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Three Numbers : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Three Numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Three Numbers : 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The smallest of 1 5 8 is 1\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.14, Page number: 76<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate gross salary for the conditions given below\n",
+      "\n",
+      "#  BASIC SALARY(Rs.)        DA(Rs.)         HRA(Rs.)        CONVEYANCE(Rs.)\n",
+      "#   >= 5000               110% of basic   20% of basic          500\n",
+      "#   bs>=3000 & bs<5000    100% of basic   15% of basic          400\n",
+      "#   bs<3000               90% of basic    10% of basic          300\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "bs = int(raw_input(\"\\nEnter Basic Salary : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "if bs >= 5000 :\n",
+      "    hra = 20 * bs/100\n",
+      "    da = 110 * bs/100\n",
+      "    cv = 500\n",
+      "else:\n",
+      "    if bs >= 3000 and bs < 5000 :\n",
+      "        hra = 15 * bs/100\n",
+      "        da = 100 * bs/100\n",
+      "        cv = 400\n",
+      "    else:\n",
+      "        if bs < 3000:\n",
+      "            hra = 10 * bs/100\n",
+      "            da = 90 * bs/100\n",
+      "            cv = 300\n",
+      "\n",
+      "#Calculation\n",
+      "ts = bs + hra + da + cv\n",
+      "\n",
+      "#Result\n",
+      "print \"Basic Salary : %5.0f\"%(bs)\n",
+      "print \"Hra          : %5.0f\"%(hra)\n",
+      "print \"Da           : %5.0f\"%(da)\n",
+      "print \"Conveyance   : %5.0f\"%(cv)\n",
+      "print \"Gross Salary : %5.0f\"%(ts)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Basic Salary : 5400\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Basic Salary :  5400\n",
+        "Hra          :  1080\n",
+        "Da           :  5940\n",
+        "Conveyance   :   500\n",
+        "Gross Salary : 12920\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.15, Page number: 77<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the total interest based on the following.\n",
+      "\n",
+      "#Principle Amount(Rs.)          Rate of Interest(Rs.)\n",
+      "#   >= 10000                        20%\n",
+      "#   >= 8000 & <= 9999               18%\n",
+      "#   < 8000                          16%\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "princ = int(raw_input(\"\\nEnter Loan & No. of years :- \"))\n",
+      "nyrs = int(raw_input(\"\\nEnter Loan & No. of years :- \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "if princ >= 10000 :\n",
+      "    rate = 20\n",
+      "else:\n",
+      "    if princ >= 8000 and princ <= 9999 :\n",
+      "        rate = 18\n",
+      "    else:\n",
+      "        if princ < 8000:\n",
+      "            rate = 16\n",
+      "\n",
+      "#Calculation\n",
+      "interest = princ * nyrs * rate/100\n",
+      "\n",
+      "#Result\n",
+      "print \"Loan     : %6.2f\"%(princ)\n",
+      "print \"Years    : %6.2f\"%(nyrs)\n",
+      "print \"Rate     : %6.2f\"%(rate)\n",
+      "print \"Interest : %6.2f\"%(interest)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Loan & No. of years :- 5500\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Loan & No. of years :- 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loan     : 5500.00\n",
+        "Years    :   3.00\n",
+        "Rate     :  16.00\n",
+        "Interest : 2640.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.16, Page number: 78<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the average of six subjects and display the results as follows.\n",
+      "\n",
+      "#    AVERAGE                         RESULT\n",
+      "#   >34 & <50                   Third Division\n",
+      "#   >49 & <60                   Second Division\n",
+      "#   >60 & <75                   First Division\n",
+      "#   >75 & <100                  Distinction\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "a = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
+      "b = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
+      "c = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
+      "d = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
+      "e = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
+      "f = int(raw_input(\"\\nEnter Marks\\nP  C  B  M  E  H\\n\"))\n",
+      "\n",
+      "#Calculation\n",
+      "#here sum1 is used instead of sum because sum is a function in python.\n",
+      "sum1 = a + b + c + d + e + f\n",
+      "avg = sum1/6\n",
+      "\n",
+      "print \"Total  : %d \\nAverage  : %.2f\"%(sum1,avg)\n",
+      "\n",
+      "#Condition evaluation and Result\n",
+      "if a < 35 or b < 35 or c < 35 or d < 35 or e < 35 :\n",
+      "    print \"Result : Fail\"\n",
+      "    exit\n",
+      "if avg >= 34 and avg < 50 :\n",
+      "    print \"Result  : Third Division \"\n",
+      "else:\n",
+      "    if avg >= 49 and avg < 60:\n",
+      "        print \"Result  :  Second Division\"\n",
+      "    else:\n",
+      "        if avg >= 60 and avg > 75:\n",
+      "            print \"Result  :  First Division\"\n",
+      "        else:\n",
+      "            if avg >= 75 and avg <= 100:\n",
+      "                  print \"Result  :  Distinction\"\n",
+      "                  \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Marks\n",
+        "P  C  B  M  E  H\n",
+        "75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Marks\n",
+        "P  C  B  M  E  H\n",
+        "75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Marks\n",
+        "P  C  B  M  E  H\n",
+        "75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Marks\n",
+        "P  C  B  M  E  H\n",
+        "75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Marks\n",
+        "P  C  B  M  E  H\n",
+        "75\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Marks\n",
+        "P  C  B  M  E  H\n",
+        "75\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total  : 450 \n",
+        "Average  : 75.00\n",
+        "Result  :  Distinction\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.17, Page number: 80<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect the entered number as to whether it is even or odd.  Use goto statement.\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "x = int(raw_input(\"\\nEnter a Number : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#There is no goto/label statement in python\n",
+      "\n",
+      "if x%2 == 0:\n",
+      "    print \"\\n%d is Even Number.\"%(x)\n",
+      "else:\n",
+      "    print \"\\n%d is Odd Number.\"%(x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "5 is Odd Number.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.18, Page number: 81<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate telephone bill.  Transfer controls at different places according\n",
+      "#to number of calls and calculate the total charges.  Follow rates as per\n",
+      "#given in the table.\n",
+      "\n",
+      "#  Telephone call               Rate in Rs.\n",
+      "#       < 100                    No charges\n",
+      "#       > 99 & < 200                1\n",
+      "#       > 199 & < 300               2\n",
+      "#       > 299                       3\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "nc = int(raw_input(\"\\nEnter Number of Calls : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#There is no goto/label statement in python\n",
+      "\n",
+      "if nc < 100:\n",
+      "    print \"\\nNo charges\"\n",
+      "else:\n",
+      "    if nc > 99 and nc < 200:\n",
+      "        print \"\\nTotal Charges : %d Rs.\"%(nc*1)\n",
+      "    else:\n",
+      "        if nc > 199 and nc < 300:\n",
+      "            print \"\\nTotal Charges : %d Rs.\"%(nc*2)\n",
+      "        else:\n",
+      "            print \"\\nTotal Charges : %d Rs.\"%(nc*3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Number of Calls : 500\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Total Charges : 1500 Rs.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.19, Page number: 82<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Check whether the entered year is a leap year or not. Use goto statement.\n",
+      "\n",
+      "#Variable initialization\n",
+      "leap = int(raw_input(\"\\nEnter Year : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#There is no goto/label statement in python\n",
+      "\n",
+      "if leap%4 == 0:\n",
+      "    print \"\\n%d is a leap year.\"%(leap)\n",
+      "else:\n",
+      "    print \"%d is not a leap year.\"%(leap)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Year : 2000\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "2000 is a leap year.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.20, Page number: 84<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print lines by selecting the choice.\n",
+      "\n",
+      "print \"\\n[1] ............\"\n",
+      "print \"\\n[2] ____________\"\n",
+      "print \"\\n[3] ************\"\n",
+      "print \"\\n[4] ============\"\n",
+      "print \"\\n[5] EXIT\"\n",
+      "\n",
+      "#Variable initialization\n",
+      "ch = int(raw_input(\"\\nENTER YOUR CHOICE : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#dictionary and functions\n",
+      "\n",
+      "def dot():\n",
+      "    print \"\\n......................................................\"\n",
+      "\n",
+      "def line():\n",
+      "    print \"\\n______________________________________________________\"\n",
+      "\n",
+      "def star():\n",
+      "    print \"\\n******************************************************\"\n",
+      "\n",
+      "def equal():\n",
+      "    print \"\\n======================================================\"\n",
+      "\n",
+      "def ex():\n",
+      "    exit\n",
+      "    \n",
+      "options = {1 : dot,\n",
+      "           2 : line,\n",
+      "           3 : star,\n",
+      "           4 : equal,\n",
+      "           5 : ex\n",
+      "}\n",
+      "\n",
+      "options[ch]()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "[1] ............\n",
+        "\n",
+        "[2] ____________\n",
+        "\n",
+        "[3] ************\n",
+        "\n",
+        "[4] ============\n",
+        "\n",
+        "[5] EXIT\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "ENTER YOUR CHOICE : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "......................................................\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.21, Page number: 85<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Provide multiple functions such as 1. addition 2.Subtraction 3. Multiplication\n",
+      "#4.Division 5.Remainder calculation 6.Larger out of two by using switch statement\n",
+      "\n",
+      "print \"\\n=============================\"\n",
+      "print \"\\n\\t\\tMENU\"\n",
+      "print \"\\n=============================\"\n",
+      "print \"\\n\\t[1] ADDITION\"\n",
+      "print \"\\n\\t[2] SUBTRACTION\"\n",
+      "print \"\\n\\t[3] MULTIPLICATION\"\n",
+      "print \"\\n\\t[4] DIVISION\"\n",
+      "print \"\\n\\t[5] REMAINDER\"\n",
+      "print \"\\n\\t[6] LARGER OUT OF TWO\"\n",
+      "print \"\\n\\t[0] EXIT\"\n",
+      "print \"\\n=============================\"\n",
+      "\n",
+      "#Variable initialization\n",
+      "ch = int(raw_input(\"\\nENTER YOUR CHOICE : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#dictionary and functions\n",
+      "\n",
+      "if ch <= 6 and ch > 0:\n",
+      "    a = int(raw_input(\"Enter Two Numbers : \"))\n",
+      "    b = int(raw_input(\"Enter Two Numbers : \"))\n",
+      "    \n",
+      "def add():\n",
+      "    c = a + b\n",
+      "    print \"\\nAddition  : %d\"%(c)\n",
+      "\n",
+      "def sub():\n",
+      "    c = a - b\n",
+      "    print \"\\nSubtraction  : %d\"%(c)\n",
+      "\n",
+      "def mul():\n",
+      "    c = a * b\n",
+      "    print \"\\nMultiplication  : %d\"%(c)\n",
+      "\n",
+      "def div():\n",
+      "    c = a/b\n",
+      "    print \"\\nDivision  : %d\"%(c)\n",
+      "\n",
+      "def rem():\n",
+      "    c = a % b\n",
+      "    print \"\\nRemainder  : %d\"%(c)\n",
+      "\n",
+      "def larger():\n",
+      "    if a > b:\n",
+      "        print \"\\n%d (a) is larger than %d (b)\"%(a,b)\n",
+      "    else:\n",
+      "        if b > a:\n",
+      "            print \"\\n%d (b) is larger than %d (a)\"%(b,a)\n",
+      "        else:\n",
+      "            print \"\\n%d (a) & %d (b) are same\"%(a,b)\n",
+      "\n",
+      "def ex():\n",
+      "    print \"\\nTerminated by choice\"\n",
+      "    exit\n",
+      "\n",
+      "def default():\n",
+      "    print \"\\nInvalid Choice\"\n",
+      "\n",
+      "options = { 1 : add,\n",
+      "            2 : sub,\n",
+      "            3 : mul,\n",
+      "            4 : div,\n",
+      "            5 : rem,\n",
+      "            6 : larger,\n",
+      "            7 : ex,\n",
+      "            0 : default,\n",
+      "            8 : default,\n",
+      "            9 : default\n",
+      "}\n",
+      "\n",
+      "options[ch]()\n",
+      "       \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "=============================\n",
+        "\n",
+        "\t\tMENU\n",
+        "\n",
+        "=============================\n",
+        "\n",
+        "\t[1] ADDITION\n",
+        "\n",
+        "\t[2] SUBTRACTION\n",
+        "\n",
+        "\t[3] MULTIPLICATION\n",
+        "\n",
+        "\t[4] DIVISION\n",
+        "\n",
+        "\t[5] REMAINDER\n",
+        "\n",
+        "\t[6] LARGER OUT OF TWO\n",
+        "\n",
+        "\t[0] EXIT\n",
+        "\n",
+        "=============================\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "ENTER YOUR CHOICE : 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Two Numbers : 9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Two Numbers : 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "9 (a) is larger than 8 (b)\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.22, Page number: 87<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert years into Minutes, Hours, Days, Months and Seconds using switch()\n",
+      "#statement\n",
+      "\n",
+      "\n",
+      "print \"[1] MINUTES\"\n",
+      "print \"[2] HOURS\"\n",
+      "print \"[3] DAYS\"\n",
+      "print \"[4] MONTHS\"\n",
+      "print \"[5] SECONDS\"\n",
+      "print \"[0] EXIT\"\n",
+      "\n",
+      "#Variable initialization\n",
+      "ch = int(raw_input(\"\\nEnter Your Choice : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#dictionary and functions\n",
+      "\n",
+      "if ch > 0 and ch < 6:\n",
+      "    yrs = int(raw_input(\"Enter Years : \"))\n",
+      "\n",
+      "#Calculation\n",
+      "#since min is a keyword in python, min1 is used instead.\n",
+      "mon = yrs * 12\n",
+      "ds = mon * 30\n",
+      "hrs = ds * 24\n",
+      "min1 = hrs * 60\n",
+      "se = min1 * 60\n",
+      "\n",
+      "def minute():\n",
+      "    print \"\\nMinutes  : %ld\"%(min1)\n",
+      "\n",
+      "def hours():\n",
+      "    print \"\\nHours  : %ld\"%(hrs)\n",
+      "\n",
+      "def days():\n",
+      "    print \"\\nDays  : %ld\"%(ds)\n",
+      "\n",
+      "def months():\n",
+      "    print \"\\nMonths : %ld\"%(mon)\n",
+      "\n",
+      "def seconds():\n",
+      "    print \"\\nSeconds : %ld\"%(se)\n",
+      "\n",
+      "def ex():\n",
+      "    print \"\\nTerminated by choice\"\n",
+      "    exit\n",
+      "\n",
+      "def default():\n",
+      "    print \"\\nInvalid Choice\"\n",
+      "\n",
+      "options = { 1 : minute,\n",
+      "            2 : hours,\n",
+      "            3 : days,\n",
+      "            4 : months,\n",
+      "            5 : seconds,\n",
+      "            6 : ex,\n",
+      "            0 : default,\n",
+      "            8 : default,\n",
+      "            9 : default\n",
+      "}\n",
+      "\n",
+      "options[ch]()\n",
+      "       \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[1] MINUTES\n",
+        "[2] HOURS\n",
+        "[3] DAYS\n",
+        "[4] MONTHS\n",
+        "[5] SECONDS\n",
+        "[0] EXIT\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Your Choice : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Years : 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Months : 24\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.23, Page number: 89<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the names of the days of a week.\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "day = int(raw_input(\"\\nEnter a number between 1 to 7 : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#dictionary and functions\n",
+      "\n",
+      "def first():\n",
+      "    print \"\\n1st day of week is Sunday\"\n",
+      "\n",
+      "def second():\n",
+      "    print \"\\n2nd day of week is Monday\"\n",
+      "\n",
+      "def third():\n",
+      "    print \"\\n3rd day of week is Tuesday\"\n",
+      "\n",
+      "def fourth():\n",
+      "    print \"\\n4th day of week is Wednesday\"\n",
+      "\n",
+      "def fifth():\n",
+      "    print \"\\n5th day of week is Thursday\"\n",
+      "\n",
+      "def sixth():\n",
+      "    print \"\\n6th day of week is Friday\"\n",
+      "\n",
+      "def seventh():\n",
+      "    print \"\\n7th day of week is Saturday\"\n",
+      "\n",
+      "def default():\n",
+      "    print \"\\nInvalid day\"\n",
+      "\n",
+      "options = { 1 : first,\n",
+      "            2 : second,\n",
+      "            3 : third,\n",
+      "            4 : fourth,\n",
+      "            5 : fifth,\n",
+      "            6 : sixth,\n",
+      "            7 : seventh,\n",
+      "            0 : default,\n",
+      "            8 : default,\n",
+      "            9 : default\n",
+      "}\n",
+      "for i in range(1,day+1):\n",
+      "    options[i]()\n",
+      "       \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter a number between 1 to 7 : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1st day of week is Sunday\n",
+        "\n",
+        "2nd day of week is Monday\n",
+        "\n",
+        "3rd day of week is Tuesday\n",
+        "\n",
+        "4th day of week is Wednesday\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.24, Page number: 90<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Perform following operations\n",
+      "\n",
+      "#1. Display any numbers or stars on the screen by using for loop\n",
+      "#2. Display the menu containing the following\n",
+      "#a) Whole screen b)Half screen c)Top 3 lines d)Bottom 3 lines\n",
+      "\n",
+      "import os\n",
+      "import sys\n",
+      "import turtle\n",
+      "\n",
+      "\n",
+      "#Print numbers in the whole screen\n",
+      "for i in range(1,700):\n",
+      "    sys.stdout.write(\"%d\"%i)\n",
+      "\n",
+      "print \"\\nCLEAR MENU\"\n",
+      "print \"\\t1] Whole Screen\\t2] Half Screen\\t3]Top 3 lines\\t4] Bottom 3 lines\\t5] Exit\"\n",
+      "\n",
+      "#Variable initialization\n",
+      "c = int(raw_input(\"Enter Your Choice : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#dictionary and functions\n",
+      "\n",
+      "def one():\n",
+      "    os.system('cls')\n",
+      "\n",
+      "def two():\n",
+      "    for i in range(0,190):\n",
+      "        turtle.goto(i,1)\n",
+      "        sys.stdout.write(\"\\t\")\n",
+      "    \n",
+      "def three():\n",
+      "    for i in range(1,100):\n",
+      "        turtle.goto(i,1)\n",
+      "        sys.stdout.write(\"\\t\")\n",
+      "\n",
+      "def four():\n",
+      "    for i in range(1,120):\n",
+      "        turtle.goto(i,21)\n",
+      "        sys.stdout.write(\"\\t\")\n",
+      "\n",
+      "def five():\n",
+      "    exit\n",
+      "\n",
+      "\n",
+      "options = { 1 : one,\n",
+      "            2 : two,\n",
+      "            3 : three,\n",
+      "            4 : four,\n",
+      "            5 : five\n",
+      "}\n",
+      "options[c]()\n",
+      "       \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599600601602603604605606607608609610611612613614615616617618619620621622623624625626627628629630631632633634635636637638639640641642643644645646647648649650651652653654655656657658659660661662663664665666667668669670671672673674675676677678679680681682683684685686687688689690691692693694695696697698699\n",
+        "CLEAR MENU\n",
+        "\t1] Whole Screen\t2] Half Screen\t3]Top 3 lines\t4] Bottom 3 lines\t5] Exit\n",
+        "\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t\t"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t\t\t\t"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.25, Page number: 91<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the files of current directory\n",
+      "\n",
+      "import os\n",
+      "\n",
+      "print \"\\nFILE LISTING MENU\"\n",
+      "print \"1] .EXE\"\n",
+      "print \"2] .BAT\"\n",
+      "print \"3] .OBJ\"\n",
+      "print \"4] .BAK\"\n",
+      "#Variable initialization\n",
+      "c = int(raw_input(\"Enter Your Choice -: \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#dictionary and functions\n",
+      "\n",
+      "def one():\n",
+      "    os.system('dir *.exe')\n",
+      "\n",
+      "def two():\n",
+      "    os.system('dir *.c')\n",
+      "\n",
+      "def three():\n",
+      "    os.system('dir *.obj')\n",
+      "\n",
+      "def four():\n",
+      "    os.system('dir *.bak')\n",
+      "\n",
+      "options = { 1 : one,\n",
+      "            2 : two,\n",
+      "            3 : three,\n",
+      "            4 : four\n",
+      "}\n",
+      "options[c]()\n",
+      "\n",
+      "       \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "FILE LISTING MENU\n",
+        "1] .EXE\n",
+        "2] .BAT\n",
+        "3] .OBJ\n",
+        "4] .BAK\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.26, Page number: 92<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display number of days in calendar format of an entered month of year 2001.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "m = int(raw_input(\"Enter Month No. of Year 2001 : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#else-if ladder.\n",
+      "\n",
+      "print \"\\nMonth - %d - 2001\"%(m)\n",
+      "print \"\\n\\tSUN\\tMON\\tTUE\\tWED\\tTHU\\tFRI\\tSAT\"\n",
+      "\n",
+      "#Find a day any month of 2001\n",
+      "\n",
+      "if m == 1:\n",
+      "    a = 2\n",
+      "    j = 31\n",
+      "else:\n",
+      "    if m == 2:\n",
+      "        a = 5\n",
+      "        j = 28\n",
+      "    else:\n",
+      "        if m == 3:\n",
+      "            a = 5\n",
+      "            j = 31\n",
+      "        else:\n",
+      "            if m == 4:\n",
+      "                a = 1\n",
+      "                j = 30\n",
+      "            else:\n",
+      "                if m == 5:\n",
+      "                    a = 3\n",
+      "                    j = 31\n",
+      "                else:\n",
+      "                    if m == 6:\n",
+      "                        a = 6\n",
+      "                        j = 30\n",
+      "                    else:\n",
+      "                        if m == 7:\n",
+      "                            a = 1\n",
+      "                            j = 31\n",
+      "                        else:\n",
+      "                            if m == 8:\n",
+      "                                a = 4\n",
+      "                                j = 31\n",
+      "                            else:\n",
+      "                                if m == 9:\n",
+      "                                    a = 7\n",
+      "                                    j = 30\n",
+      "                                else:\n",
+      "                                    if m == 10:\n",
+      "                                        a = 2\n",
+      "                                        j = 31\n",
+      "                                    else:\n",
+      "                                        if m == 11:\n",
+      "                                            a = 5\n",
+      "                                            j = 30\n",
+      "                                        else:\n",
+      "                                            if m == 12:\n",
+      "                                                a = 7\n",
+      "                                                j = 31\n",
+      "                                            else:\n",
+      "                                                print \"Invalid Month\"\n",
+      "                                                exit\n",
+      "\n",
+      "#Starting day is to be adjusted under the respective day\n",
+      "#sys.stdout.write() function performs the same task as printf() function\n",
+      "                                                \n",
+      "i = 1\n",
+      "if a == 1:\n",
+      "    sys.stdout.write(\"\\t%d\"%(i))\n",
+      "else:\n",
+      "    if a == 2:\n",
+      "        sys.stdout.write(\"\\t\\t%d\"%(i))\n",
+      "    else:\n",
+      "        if a == 3:\n",
+      "            sys.stdout.write(\"\\t\\t\\t%d\"%(i))\n",
+      "        else:\n",
+      "            if a == 4:\n",
+      "                sys.stdout.write(\"\\t\\t\\t\\t%d\"%(i))\n",
+      "            else:\n",
+      "                if a == 5:\n",
+      "                    sys.stdout.write(\"\\t\\t\\t\\t\\t%d\"%(i))\n",
+      "                else:\n",
+      "                    if a == 6:\n",
+      "                        sys.stdout.write(\"\\t\\t\\t\\t\\t\\t%d\"%(i))\n",
+      "                    else:\n",
+      "                        if a == 7:\n",
+      "                            sys.stdout.write(\"\\t\\t\\t\\t\\t\\t\\t%d\"%(i))\n",
+      "\n",
+      "\n",
+      "h = 8 - a        #The starting day is subtracted from 8\n",
+      "for i in range(2,h+1):       #to display the first row\n",
+      "    sys.stdout.write(\"\\t%d\"%(i))\n",
+      "sys.stdout.write(\"\\n\")\n",
+      "b = 1\n",
+      "for i in range(h+1,j+1):          #to continue with second row onwards\n",
+      "    if b == 8:\n",
+      "        sys.stdout.write(\"\\n\")\n",
+      "        b = 1\n",
+      "    sys.stdout.write(\"\\t%d\"%(i))\n",
+      "    b += 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Month - 1 - 2001\n",
+        "\n",
+        "\tSUN\tMON\tTUE\tWED\tTHU\tFRI\tSAT\n",
+        "\t\t1\t2\t3\t4\t5\t6\n",
+        "\t7\t8\t9\t10\t11\t12\t13\n",
+        "\t14\t15\t16\t17\t18\t19\t20\n",
+        "\t21\t22\t23\t24\t25\t26\t27\n",
+        "\t28\t29\t30\t31"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.27, Page number: 95<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert decimal to hexadecimal number.\n",
+      "\n",
+      "import sys\n",
+      "import turtle\n",
+      "\n",
+      "#Variable initialization\n",
+      "x = int(raw_input(\"Enter a number : \"))\n",
+      "y = 30\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#else-if ladder.\n",
+      "\n",
+      "print \"\\nConversion of Decimal to Hexadecimal Number\"\n",
+      "\n",
+      "#for loop without condition is not supported in python. so while loop is used.\n",
+      "\n",
+      "c = 1\n",
+      "z = [0 for i in range(0,15)]\n",
+      "\n",
+      "while x != 0:\n",
+      "    z[c] = x % 16\n",
+      "    c = c + 1\n",
+      "    x = x/16\n",
+      "\n",
+      "\n",
+      "for i in range(c-1,0,-1):\n",
+      "    if z[i] == 10:\n",
+      "       sys.stdout.write(\"A\")\n",
+      "    else:\n",
+      "       if z[i] == 11:\n",
+      "          sys.stdout.write(\"B\")\n",
+      "       else:\n",
+      "          if z[i] == 12:\n",
+      "             sys.stdout.write(\"C\")\n",
+      "          else:\n",
+      "             if z[i] == 13:\n",
+      "                sys.stdout.write(\"D\")\n",
+      "             else:\n",
+      "                if z[i] == 14:\n",
+      "                   sys.stdout.write(\"E\")\n",
+      "                else:\n",
+      "                    if z[i] == 15:\n",
+      "                       sys.stdout.write(\"F\")\n",
+      "                    else:\n",
+      "                       sys.stdout.write(\"%d\"%(z[i]))\n",
+      "    \n",
+      "    \n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 31\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Conversion of Decimal to Hexadecimal Number\n",
+        "1F"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.28, Page number: 97<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect whether the entered number is even or odd. use nested switch-case\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "x = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch case statement in python, an alternative is to use\n",
+      "#else-if ladder.\n",
+      "\n",
+      "if x == 0:\n",
+      "    print \"Number is Even\"\n",
+      "else:\n",
+      "    if x == 1:\n",
+      "        print \"Number is odd\"\n",
+      "    else:\n",
+      "        y = x % 2\n",
+      "        if y == 0:\n",
+      "            print \"Number is Even\"\n",
+      "        else:\n",
+      "            print \"Number is odd\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number is odd\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.29, Page number: 98<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count number of 1s, 0s, blank spaces and others using nested\n",
+      "#switch() statement in a given stream\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "txt = raw_input(\"Enter Numbers : \")\n",
+      "\n",
+      "#Processing\n",
+      "x = 0\n",
+      "s = 0\n",
+      "a = 0\n",
+      "z = 0\n",
+      "o = 0\n",
+      "\n",
+      "while x < len(txt):\n",
+      "    if txt[x] == ' ':\n",
+      "        s = s + 1\n",
+      "    else:\n",
+      "        if txt[x] == '1':\n",
+      "            a = a + 1\n",
+      "        else:\n",
+      "            if txt[x] == '0':\n",
+      "                z = z + 1\n",
+      "            else:\n",
+      "                o = o + 1\n",
+      "    x = x + 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nTotal Spaces : %d\"%(s))\n",
+      "sys.stdout.write(\"\\nTotal 1s  :  %d\"%(a))\n",
+      "sys.stdout.write(\"\\nTotal 0s  :  %d\"%(z))\n",
+      "sys.stdout.write(\"\\nOthers    :  %d\"%(o))\n",
+      "sys.stdout.write(\"\\nString Length  : %d\"%(s+a+z+o))\n",
+      "        \n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Numbers : 1110022 222\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Total Spaces : 1\n",
+        "Total 1s  :  3\n",
+        "Total 0s  :  2\n",
+        "Others    :  5\n",
+        "String Length  : 11"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.30, Page number: 99<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert integer to character using if condition\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "x = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#ord() function converts the character ASCII to integer\n",
+      "\n",
+      "if x == ord('A'):\n",
+      "    print \"%c\"%(x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 65\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5.31, Page number: 100<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use nested if..else statements in switch() statement.  Also show the effect\n",
+      "#of conversion of integer to character\n",
+      "\n",
+      "\n",
+      "#Variable initialization\n",
+      "i = int(raw_input(\"Enter any ASCII Number : \"))\n",
+      "\n",
+      "#Condition evaluation\n",
+      "#there is no switch..case statement in python. alternative is to use\n",
+      "#if..else and else..if ladder\n",
+      "\n",
+      "if i == ord('A'):\n",
+      "    print \"Capital A\"\n",
+      "else:\n",
+      "    if i == ord('B'):\n",
+      "        print \"Capital B\"\n",
+      "    else:\n",
+      "        if i == ord('C'):\n",
+      "            print \"Capital C\"\n",
+      "        else:\n",
+      "            if i > 47 and i < 58:\n",
+      "                print \"Digit : [%c]\"%(i)\n",
+      "            else:\n",
+      "                if i >= 58 and i <= 64:\n",
+      "                    print \"Symbol : [%c]\"%(i)\n",
+      "                else:\n",
+      "                    if i > 64 and i < 91:\n",
+      "                        print \"Capital : [%c]\"%(i)\n",
+      "                    else:\n",
+      "                        if i > 96 and i < 123:\n",
+      "                            print \"Small : [%c]\"%(i)\n",
+      "                        else:\n",
+      "                            print \"Invalid Choice\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter any ASCII Number : 65\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Capital A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter6.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter6.ipynb
new file mode 100755
index 00000000..d9815e79
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter6.ipynb
@@ -0,0 +1,42312 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 6: Loop Control Statements<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.1, Page number: 107<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the first five numbers starting from one together with thier squares\n",
+      "\n",
+      "#Result\n",
+      "#Since range() function starts with 0; to print upto 5, use 5+1 or 6\n",
+      "#range() function creates range or numbers from 1 to 6\n",
+      "\n",
+      "for i in range(1,5+1):\n",
+      "    print(\"Number : %d   it's Square : %d\"%(i,i*i))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number : 1   it's Square : 1\n",
+        "Number : 2   it's Square : 4\n",
+        "Number : 3   it's Square : 9\n",
+        "Number : 4   it's Square : 16\n",
+        "Number : 5   it's Square : 25\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.2, Page number: 108<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display numbers from 1 to 15 using for loop.  Use i++\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#There is no increment operator (++) in python\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "for i in range(1,15+1):\n",
+      "    sys.stdout.write(\"%5d\"%(i))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.3, Page number: 108<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display numbers from 1 to 15 using for loop. Use i = i + 1\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "for i in range(1,15+1):\n",
+      "    sys.stdout.write(\"%5d\"%(i))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.4, Page number: 109<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display numbers from 1 to 16. Use incrementation operation in the body of\n",
+      "#the loop for more than once\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "\n",
+      "c = 0\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers and the third argument in the range function\n",
+      "#indicates the interval between the numbers.\n",
+      "\n",
+      "for i in range(0,15+1,2):\n",
+      "    i += 1\n",
+      "    sys.stdout.write(\"%5d\"%(i))\n",
+      "    i = i + 1\n",
+      "    sys.stdout.write(\"%5d\"%(i))\n",
+      "    c += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\n\\nThe Body of the loop is executed for %d times\"%(c))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15   16\n",
+        "\n",
+        "The Body of the loop is executed for 8 times"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.5, Page number: 109<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display even numbers from 0 to 14 by declaring the initial counter value before\n",
+      "#the loop statement\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "\n",
+      "c = 0\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers and the third argument in the range function\n",
+      "#indicates the interval between the numbers.\n",
+      "\n",
+      "#counter variable initialization before the for loop will not affect much.\n",
+      "\n",
+      "for i in range(0,15+1,2):\n",
+      "    sys.stdout.write(\"%5d\"%(i))\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "    0    2    4    6    8   10   12   14"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.6, Page number: 110<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display numbers in ascending and descending orders.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers and the third argument in the range function\n",
+      "#indicates the interval between the numbers.\n",
+      "\n",
+      "#counter variable initialization before the for loop will not affect much.\n",
+      "\n",
+      "sys.stdout.write(\"Numbers in Ascending Order : \")\n",
+      "for i in range(1,10+1):\n",
+      "    sys.stdout.write(\"%3d\"%(i))\n",
+      "    \n",
+      "sys.stdout.write(\"\\nNumbers in Descending Order : \")\n",
+      "for i in range(10,0,-1):\n",
+      "    sys.stdout.write(\"%3d\"%(i))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numbers in Ascending Order :   1  2  3  4  5  6  7  8  9 10\n",
+        "Numbers in Descending Order :  10  9  8  7  6  5  4  3  2  1"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.7, Page number: 110<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display numbers upto 10 using infinite for loop.  Use goto statement to\n",
+      "#exit from the loop\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers and the third argument in the range function\n",
+      "#indicates the interval between the numbers.\n",
+      "\n",
+      "#counter variable initialization before the for loop will not affect much.\n",
+      "\n",
+      "for i in range(0,15):\n",
+      "    sys.stdout.write(\"%3d\"%(i))\n",
+      "    i += 1\n",
+      "    if i == 11:\n",
+      "        break\n",
+      "\n",
+      "#there is no infinite for loop and goto statement in python.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  0  1  2  3  4  5  6  7  8  9 10"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.8, Page number: 111<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count numbers between 1 to 100 not divisible by 2, 3, and 5\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "\n",
+      "c = 0\n",
+      "\n",
+      "sys.stdout.write(\"\\nNumbers from 1 to 100 not divisible by 2,3&5\\n\\n\")\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers \n",
+      "\n",
+      "for x in range(0,100+1):\n",
+      "    if x%2 != 0 and x%3 != 0 and x%5 != 0:\n",
+      "        sys.stdout.write(\"%d\\t\"%(x))\n",
+      "        c += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\nTotal Numbers : %d\"%(c))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Numbers from 1 to 100 not divisible by 2,3&5\n",
+        "\n",
+        "1\t7\t11\t13\t17\t19\t23\t29\t31\t37\t41\t43\t47\t49\t53\t59\t61\t67\t71\t73\t77\t79\t83\t89\t91\t97\t\n",
+        "Total Numbers : 26"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.9, Page number: 112<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the numbers in increasing and decreasing order using infinite\n",
+      "#for loop\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "n = int(raw_input(\"Enter a number : \"))\n",
+      "a = b = n\n",
+      "\n",
+      "sys.stdout.write(\"(++)        (--)\\n\")\n",
+      "sys.stdout.write(\"===================\")\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "#There is no infinite for loop in python. so separate iteration variable i is used\n",
+      "\n",
+      "for i in range(0,100):\n",
+      "    sys.stdout.write(\"\\n%d\\t%d\"%(a,b))\n",
+      "    if b == 0:\n",
+      "        break\n",
+      "    a += 1\n",
+      "    b -= 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(++)        (--)\n",
+        "===================\n",
+        "5\t5\n",
+        "6\t4\n",
+        "7\t3\n",
+        "8\t2\n",
+        "9\t1\n",
+        "10\t0"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.10, Page number: 113<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Create an infinite loop. Check each value of the for loop.  if the value is\n",
+      "#even, display it otherwise continue with interations.  Print even numbers from\n",
+      "#1 to 21. Use break statement to terminate the program.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "i = 1\n",
+      "\n",
+      "sys.stdout.write(\"\\n\\t\\tTable of Even numbers from 1 to 20\")\n",
+      "sys.stdout.write(\"\\n\\t\\t==================================\\n\")\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "#There is no infinite for loop in python. so separate iteration variable i is used\n",
+      "\n",
+      "for i in range(1,100):\n",
+      "    if i == 20:\n",
+      "        break\n",
+      "    else:\n",
+      "        if i%2 == 0:\n",
+      "            sys.stdout.write(\"%d\\t\"%(i))\n",
+      "            continue\n",
+      "        else:\n",
+      "            continue"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\t\tTable of Even numbers from 1 to 20\n",
+        "\t\t==================================\n",
+        "2\t4\t6\t8\t10\t12\t14\t16\t18\t"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.11, Page number: 113<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the sum of first five numbers and their squares.  Display their results\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "#since sum is a function in python, sum1 is used instead of variable sum.\n",
+      "\n",
+      "sum1 = 0\n",
+      "sqsum = 0\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "for i in range(1,5+1):\n",
+      "    sum1 += i\n",
+      "    sqsum += i*i\n",
+      "    sys.stdout.write(\"\\nNumber : %5d    it's Square : %8d\"%(i,i*i))\n",
+      "\n",
+      "sys.stdout.write(\"\\n=================================================\")\n",
+      "sys.stdout.write(\"\\nThe Sum %6d Sum of Squares %9d\"%(sum1,sqsum))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number :     1    it's Square :        1\n",
+        "Number :     2    it's Square :        4\n",
+        "Number :     3    it's Square :        9\n",
+        "Number :     4    it's Square :       16\n",
+        "Number :     5    it's Square :       25\n",
+        "=================================================\n",
+        "The Sum     15 Sum of Squares        55"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.12, Page number: 114<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display numbers from 1 to 9 and their square roots.\n",
+      "\n",
+      "import math\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "for i in range(1,9+1):\n",
+      "    a = math.sqrt(i)\n",
+      "    sys.stdout.write(\"\\n\\t%d     %.2f\"%(i,a))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\t1     1.00\n",
+        "\t2     1.41\n",
+        "\t3     1.73\n",
+        "\t4     2.00\n",
+        "\t5     2.24\n",
+        "\t6     2.45\n",
+        "\t7     2.65\n",
+        "\t8     2.83\n",
+        "\t9     3.00"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.13, Page number: 115<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the number in between 7 and 100 which is exactly divisible by\n",
+      "#4 and if divided by 5 and 6 remainders obtained should be 4.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "for x in range(7,100+1):\n",
+      "    if x%4 == 0 and x%5 == 4 and x%6 == 4:\n",
+      "        sys.stdout.write(\"\\nMinimum Number : %d\"%(x))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Minimum Number : 64"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.14, Page number: 115<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the series given\n",
+      "# x = 1/1 + 1/4 + 1/9 ... 1/n2\n",
+      "# y = 1/1 + 1/8 + 1/27 ... 1/n3\n",
+      "\n",
+      "#Variable declaration\n",
+      "x = 0.0\n",
+      "y = 0.0\n",
+      "n = int(raw_input(\"Enter Value of n : \"))\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "for i in range(1,n+1):\n",
+      "    x = x + (1.0/pow(i,2))\n",
+      "    y = y + (1.0/pow(i,3))\n",
+      "\n",
+      "print(\"Value of x = %f\"%(x))\n",
+      "print(\"\\nValue of y = %f\"%(y))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of n : 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of x = 1.250000\n",
+        "\n",
+        "Value of y = 1.125000\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.15, Page number: 116<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Generate Triangular number\n",
+      "#Note: Triangular number is nothing but summation of 1 to given number.\n",
+      "#For example, when entered number is 5 it's triangular number would be\n",
+      "#(1+2+3+4+5) = 15.\n",
+      "\n",
+      "\n",
+      "#Variable declaration\n",
+      "tri_num = 0\n",
+      "n = int(raw_input(\"What Triangular number do you want : \"))\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "for j in range(1,n+1):\n",
+      "    tri_num = tri_num + j\n",
+      "\n",
+      "print(\"Triangular number of %d is %d\\n\"%(n,tri_num))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "What Triangular number do you want : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Triangular number of 5 is 15\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.16, Page number: 117<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find sum of the following series\n",
+      "# 1 + 2 + 3 + ... n\n",
+      "# 1^2 + 2^2 + 2^3 + ... n^2\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable declaration\n",
+      "#since sum is a function name in python, sum1 is used instead of sum.\n",
+      "\n",
+      "sum1 = 0\n",
+      "ssum = 0\n",
+      "j = int(raw_input(\"Enter Number : \"))\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "sys.stdout.write(\"Numbers : \")\n",
+      "for i in range(1,j+1):\n",
+      "    sys.stdout.write(\"%5d\"%(i))\n",
+      "\n",
+      "sys.stdout.write(\"\\nSquares : \")\n",
+      "for i in range(1,j+1):\n",
+      "    sys.stdout.write(\"%5d\"%(i*i))\n",
+      "    sum1 = sum1 + i\n",
+      "    ssum = ssum + i*i\n",
+      "sys.stdout.write(\"\\nSum of Numbers from 1 to %d   : %d\"%(j,sum1))\n",
+      "sys.stdout.write(\"\\nSum of Squares of 1 to %d Numbers : %d\"%(j,ssum))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numbers :     1    2    3    4    5\n",
+        "Squares :     1    4    9   16   25\n",
+        "Sum of Numbers from 1 to 5   : 15\n",
+        "Sum of Squares of 1 to 5 Numbers : 55"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.17, Page number: 118<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the perfect squares from 1 to 500.\n",
+      "\n",
+      "import sys\n",
+      "import math\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers\n",
+      "\n",
+      "sys.stdout.write(\"\\n\\n\")\n",
+      "sys.stdout.write(\"Perfect square from 1 to 500 \\n\")\n",
+      "count = 0\n",
+      "\n",
+      "for i in range(1,500+1):\n",
+      "    c = math.sqrt(i)\n",
+      "    x = math.floor(c)                  #For rounding up floor() is used\n",
+      "    if c == x:\n",
+      "        sys.stdout.write(\"\\t%5d\"%(i))\n",
+      "        count += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\n\\nTotal Perfect Squares = %d\"%(count))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Perfect square from 1 to 500 \n",
+        "\t    1\t    4\t    9\t   16\t   25\t   36\t   49\t   64\t   81\t  100\t  121\t  144\t  169\t  196\t  225\t  256\t  289\t  324\t  361\t  400\t  441\t  484\n",
+        "\n",
+        "Total Perfect Squares = 22"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.18, Page number: 119<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect the largest number out of five numbers.\n",
+      "\n",
+      "import sys\n",
+      "import math\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers.\n",
+      "#since sum is a built in function in python, sum1 is used instead of sum.\n",
+      "\n",
+      "a = int(raw_input(\"Enter five numbers : \"))\n",
+      "b = int(raw_input(\"Enter five numbers : \"))\n",
+      "c = int(raw_input(\"Enter five numbers : \"))\n",
+      "d = int(raw_input(\"Enter five numbers : \"))\n",
+      "e = int(raw_input(\"Enter five numbers : \"))\n",
+      "\n",
+      "sum1 = a + b + c + d + e\n",
+      "\n",
+      "for i in range(sum1,1,-1):\n",
+      "    if i == a or i == b or i == c or i == d or i == e:\n",
+      "        sys.stdout.write(\"The Largest Number : %d\"%(i))\n",
+      "        break\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Largest Number : 7"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.19, Page number: 119<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect the smallesst number out of five numbers.\n",
+      "\n",
+      "import sys\n",
+      "import math\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers.\n",
+      "#since sum is a built in function in python, sum1 is used instead of sum.\n",
+      "\n",
+      "a = int(raw_input(\"Enter five numbers : \"))\n",
+      "b = int(raw_input(\"Enter five numbers : \"))\n",
+      "c = int(raw_input(\"Enter five numbers : \"))\n",
+      "d = int(raw_input(\"Enter five numbers : \"))\n",
+      "e = int(raw_input(\"Enter five numbers : \"))\n",
+      "\n",
+      "sum1 = a + b + c + d + e\n",
+      "\n",
+      "for i in range(1,sum1):\n",
+      "    if i == a or i == b or i == c or i == d or i == e:\n",
+      "        sys.stdout.write(\"The Smallest Number : %d\"%(i))\n",
+      "        break\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Smallest Number : 2"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.20, Page number: 120<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read five numbers and print in ascending order\n",
+      "\n",
+      "import sys\n",
+      "import math\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers.\n",
+      "#since sum is a built in function in python, sum1 is used instead of sum.\n",
+      "\n",
+      "#use commas to separate the input values\n",
+      "a = int(raw_input(\"Enter five numbers : \"))\n",
+      "b = int(raw_input(\"Enter five numbers : \"))\n",
+      "c = int(raw_input(\"Enter five numbers : \"))\n",
+      "d = int(raw_input(\"Enter five numbers : \"))\n",
+      "e = int(raw_input(\"Enter five numbers : \"))\n",
+      "\n",
+      "sys.stdout.write(\"Numbers in ascending order : \")\n",
+      "sum1 = a + b + c + d + e\n",
+      "\n",
+      "for i in range(1,sum1):\n",
+      "    if i == a or i == b or i == c or i == d or i == e:\n",
+      "        sys.stdout.write(\"%3d\"%(i))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter five numbers : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numbers in ascending order :   1  4  5  7  8"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.21, Page number: 121<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Perform multiplication of two integers by using negative sign\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#the range() function creates range of numbers and for loop automatically\n",
+      "#iterate through the numbers.\n",
+      "\n",
+      "a = int(raw_input(\"Enter two numbers : \"))\n",
+      "b = int(raw_input(\"Enter two numbers : \"))\n",
+      "\n",
+      "d = 0\n",
+      "for c in range(1,b+1):\n",
+      "    d = d - (-a)\n",
+      "\n",
+      "sys.stdout.write(\"Multiplication of %d * %d : %d\"%(a,b,d))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Multiplication of 5 * 5 : 25"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.22, Page number: 121<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Perform multiplication of two integers by using repetitive addition\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#in python, infinite for loop is not possible.\n",
+      "\n",
+      "a = int(raw_input(\"Enter two numbers : \"))\n",
+      "b = int(raw_input(\"Enter two numbers : \"))\n",
+      "\n",
+      "d = 0\n",
+      "c = 1\n",
+      "while True:\n",
+      "    d = d + a\n",
+      "    if c == b:\n",
+      "        break\n",
+      "    c += 1\n",
+      "\n",
+      "sys.stdout.write(\"Multiplication of %d * %d : %d\"%(a,b,d))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Multiplication of 8 * 4 : 32"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.23, Page number: 122<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the marks for five subjects and calculate sum & average of marks.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#use spaces to separate the input values\n",
+      "#since sum is a built-in function in python, sum1 is used as the variable name\n",
+      "\n",
+      "sys.stdout.write(\"Enter The Marks of Five Subjects \")\n",
+      "sum1 = 0\n",
+      "\n",
+      "for i in range(1,5+1):\n",
+      "    sys.stdout.write(\"[%d] Student  : \"%(i))\n",
+      "    #Exception handling\n",
+      "    try:\n",
+      "        #use space to separate the input characters\n",
+      "        v = raw_input(\"\")\n",
+      "        a,b,c,d,e = map(int,v.split())\n",
+      "        x = 5\n",
+      "    except:\n",
+      "        x = 0\n",
+      "\n",
+      "    if(x == 5):\n",
+      "        sum1 = a + b + c + d + e\n",
+      "        avg = sum1/5\n",
+      "        sys.stdout.write(\"\\nTotal Marks of Student [%d] %d\"%(i,sum1))\n",
+      "        sys.stdout.write(\"\\nAverage Marks of Student [%d] %f\"%(i,avg))\n",
+      "    else:\n",
+      "        sys.stdout.write(\"Type Mismatch\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter The Marks of Five Subjects [1] Student  : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "58 52 52 56 78\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Total Marks of Student [1] 296\n",
+        "Average Marks of Student [1] 59.000000[2] Student  : "
+       ]
+      }
+     ],
+     "prompt_number": "*"
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.24, Page number: 123<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Enter a character and display its position in alphabetic order.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = 1\n",
+      "\n",
+      "ch = raw_input(\"Enter a character :\")\n",
+      "ch = ch.upper()\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "#ord() function converts the character ASCII value to integer\n",
+      "for i in range(65,90+1):\n",
+      "    if ord(ch) == i:\n",
+      "        sys.stdout.write(\"'%c' is %d [st/nd/rd/th] Character in Alphabetic\"%(i,c))\n",
+      "    c += 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a character :U\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "'U' is 21 [st/nd/rd/th] Character in Alphabetic"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.25, Page number: 124<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of -m^x.\n",
+      "#'m' is a mantissa and 'x' is an exponent\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "number = int(raw_input(\"Enter Number : \"))\n",
+      "power = int(raw_input(\"Enter Power : \"))\n",
+      "\n",
+      "ans = 1\n",
+      "\n",
+      "#Calculation\n",
+      "for i in range(1,power+1):\n",
+      "    ans *= number\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The Power of %d raised to %d is %ld\"%(number,power,ans))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Power : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Power of 5 raised to 3 is 125"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.26, Page number: 124<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Subtraction of two loop variables using nested for loops.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Calculation & Result\n",
+      "for a in range(3,0,-1):\n",
+      "    for b in range(1,2+1):\n",
+      "        sub = a - b\n",
+      "        sys.stdout.write(\"a = %d  b = %d  a - b = %d\\n\"%(a,b,sub))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = 3  b = 1  a - b = 2\n",
+        "a = 3  b = 2  a - b = 1\n",
+        "a = 2  b = 1  a - b = 1\n",
+        "a = 2  b = 2  a - b = 0\n",
+        "a = 1  b = 1  a - b = 0\n",
+        "a = 1  b = 2  a - b = -1\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.27, Page number: 125<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Illustrate an example based on nested for loops\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Calculation & Result\n",
+      "for i in range(1,3+1):                  #outer loop\n",
+      "    for j in range(1,2+1):              #inner loop\n",
+      "        sys.stdout.write(\"\\ni * j : %d\"%(i*j))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "i * j : 1\n",
+        "i * j : 2\n",
+        "i * j : 2\n",
+        "i * j : 4\n",
+        "i * j : 3\n",
+        "i * j : 6"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.28, Page number: 126<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print values and messages when any loop ends using nested for loops.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Calculation & Result\n",
+      "for a in range(1,2+1):                  #outer loop\n",
+      "    for b in range(1,2+1):              #middle loop\n",
+      "        for c in range(1,2+1):          #inner loop\n",
+      "            sys.stdout.write(\"\\na = %d + b = %d + c = %d : %d\"%(a,b,c,a+b+c))\n",
+      "        sys.stdout.write(\"\\nInner Loop Over.\")\n",
+      "    sys.stdout.write(\"\\nMiddle Loop Over.\")\n",
+      "sys.stdout.write(\"\\nOuter Loop Over.\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "a = 1 + b = 1 + c = 1 : 3\n",
+        "a = 1 + b = 1 + c = 2 : 4\n",
+        "Inner Loop Over.\n",
+        "a = 1 + b = 2 + c = 1 : 4\n",
+        "a = 1 + b = 2 + c = 2 : 5\n",
+        "Inner Loop Over.\n",
+        "Middle Loop Over.\n",
+        "a = 2 + b = 1 + c = 1 : 4\n",
+        "a = 2 + b = 1 + c = 2 : 5\n",
+        "Inner Loop Over.\n",
+        "a = 2 + b = 2 + c = 1 : 5\n",
+        "a = 2 + b = 2 + c = 2 : 6\n",
+        "Inner Loop Over.\n",
+        "Middle Loop Over.\n",
+        "Outer Loop Over."
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.29, Page number: 127<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display perfect cubes up to given number.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "k = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "for i in range(1,k+1):\n",
+      "    for j in range(1,i+1):   \n",
+      "        if i == pow(j,3):\n",
+      "          sys.stdout.write(\"\\nNumber : %d & it's Cube : %d\"%(j,i))\n",
+      "          \n",
+      "          \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number : 1 & it's Cube : 1\n",
+        "Number : 2 & it's Cube : 8\n",
+        "Number : 3 & it's Cube : 27\n",
+        "Number : 4 & it's Cube : 64"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.30, Page number: 128<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display numbers 1 to 100 using ASCII values from 48 to 57. Use nested loops.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "j = 0\n",
+      "k = -9\n",
+      "\n",
+      "sys.stdout.write(\"\\tTable of 1 to 100 Numbers Using ASCII Values\\n\")\n",
+      "sys.stdout.write(\"\\t===== == = == === ======= ===== ===== ======\\n\")\n",
+      "\n",
+      "#Calculation & Result\n",
+      "for i in range(48,57+1):\n",
+      "    for j in range(48,57+1):   \n",
+      "        sys.stdout.write(\"\\t%c%c\"%(i,j))\n",
+      "        #if k != 1:\n",
+      "         #   sys.stdout.write(\"%c\"%(i+1))\n",
+      "        #if k == 0:\n",
+      "         #   sys.stdout.write(\"\\b\\b%d\"%(k+1))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    k += 1\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tTable of 1 to 100 Numbers Using ASCII Values\n",
+        "\t===== == = == === ======= ===== ===== ======\n",
+        "\t00\t01\t02\t03\t04\t05\t06\t07\t08\t09\n",
+        "\t10\t11\t12\t13\t14\t15\t16\t17\t18\t19\n",
+        "\t20\t21\t22\t23\t24\t25\t26\t27\t28\t29\n",
+        "\t30\t31\t32\t33\t34\t35\t36\t37\t38\t39\n",
+        "\t40\t41\t42\t43\t44\t45\t46\t47\t48\t49\n",
+        "\t50\t51\t52\t53\t54\t55\t56\t57\t58\t59\n",
+        "\t60\t61\t62\t63\t64\t65\t66\t67\t68\t69\n",
+        "\t70\t71\t72\t73\t74\t75\t76\t77\t78\t79\n",
+        "\t80\t81\t82\t83\t84\t85\t86\t87\t88\t89\n",
+        "\t90\t91\t92\t93\t94\t95\t96\t97\t98\t99\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.31, Page number: 129<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count number of votes secured by 'A' & 'B'.  Assume three voters are\n",
+      "#voting them. Also count the invalid votes.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization & Calculation\n",
+      "a = 0\n",
+      "b = 0\n",
+      "o = 0\n",
+      "sys.stdout.write(\"Press A or B\\n\")\n",
+      "for i in range(1,3+1):\n",
+      "    sys.stdout.write(\"\\nVoter no. %d\"%(i))\n",
+      "    sys.stdout.write(\"  Enter Vote : \")\n",
+      "    v = raw_input(\"\")\n",
+      "    v = v.upper()\n",
+      "    if v == 'A':\n",
+      "        a += 1\n",
+      "    else:\n",
+      "        if v == 'B':\n",
+      "            b += 1\n",
+      "        else:\n",
+      "            o += 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nStatus of vote(s)\\n\")\n",
+      "sys.stdout.write(\"\\nA secures %d vote(s).\"%(a))\n",
+      "sys.stdout.write(\"\\nB secures %d vote(s).\"%(b))\n",
+      "sys.stdout.write(\"\\nInvalid vote(s) %d.\"%(o))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press A or B\n",
+        "\n",
+        "Voter no. 1  Enter Vote : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Voter no. 2  Enter Vote : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "B\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Voter no. 3  Enter Vote : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "X\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Status of vote(s)\n",
+        "\n",
+        "A secures 1 vote(s).\n",
+        "B secures 1 vote(s).\n",
+        "Invalid vote(s) 1."
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.32, Page number: 130<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Simulate a digital clock\n",
+      "\n",
+      "import sys\n",
+      "import os\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"hh mm ss\\n\")\n",
+      "for h in range(1,12):\n",
+      "    for m in range(1,59):\n",
+      "        for s in range(1,59):\n",
+      "            sys.stdout.write(\"\\r%d  %d  %d\"%(h,m,s))\n",
+      "            \n",
+      "#Displays the last value"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hh mm ss\n",
+        "\r",
+        "1  1  1\r",
+        "1  1  2\r",
+        "1  1  3\r",
+        "1  1  4\r",
+        "1  1  5\r",
+        "1  1  6\r",
+        "1  1  7\r",
+        "1  1  8\r",
+        "1  1  9\r",
+        "1  1  10\r",
+        "1  1  11\r",
+        "1  1  12\r",
+        "1  1  13\r",
+        "1  1  14\r",
+        "1  1  15\r",
+        "1  1  16\r",
+        "1  1  17\r",
+        "1  1  18\r",
+        "1  1  19\r",
+        "1  1  20\r",
+        "1  1  21\r",
+        "1  1  22\r",
+        "1  1  23\r",
+        "1  1  24\r",
+        "1  1  25\r",
+        "1  1  26\r",
+        "1  1  27\r",
+        "1  1  28\r",
+        "1  1  29\r",
+        "1  1  30\r",
+        "1  1  31\r",
+        "1  1  32\r",
+        "1  1  33\r",
+        "1  1  34\r",
+        "1  1  35\r",
+        "1  1  36\r",
+        "1  1  37\r",
+        "1  1  38\r",
+        "1  1  39\r",
+        "1  1  40\r",
+        "1  1  41\r",
+        "1  1  42\r",
+        "1  1  43\r",
+        "1  1  44\r",
+        "1  1  45\r",
+        "1  1  46\r",
+        "1  1  47\r",
+        "1  1  48\r",
+        "1  1  49\r",
+        "1  1  50\r",
+        "1  1  51\r",
+        "1  1  52\r",
+        "1  1  53\r",
+        "1  1  54\r",
+        "1  1  55\r",
+        "1  1  56\r",
+        "1  1  57\r",
+        "1  1  58\r",
+        "1  2  1\r",
+        "1  2  2\r",
+        "1  2  3\r",
+        "1  2  4\r",
+        "1  2  5\r",
+        "1  2  6\r",
+        "1  2  7\r",
+        "1  2  8\r",
+        "1  2  9\r",
+        "1  2  10\r",
+        "1  2  11\r",
+        "1  2  12\r",
+        "1  2  13\r",
+        "1  2  14\r",
+        "1  2  15\r",
+        "1  2  16\r",
+        "1  2  17\r",
+        "1  2  18\r",
+        "1  2  19\r",
+        "1  2  20\r",
+        "1  2  21\r",
+        "1  2  22\r",
+        "1  2  23\r",
+        "1  2  24\r",
+        "1  2  25\r",
+        "1  2  26\r",
+        "1  2  27\r",
+        "1  2  28\r",
+        "1  2  29\r",
+        "1  2  30\r",
+        "1  2  31\r",
+        "1  2  32\r",
+        "1  2  33\r",
+        "1  2  34\r",
+        "1  2  35\r",
+        "1  2  36\r",
+        "1  2  37\r",
+        "1  2  38\r",
+        "1  2  39\r",
+        "1  2  40\r",
+        "1  2  41\r",
+        "1  2  42\r",
+        "1  2  43\r",
+        "1  2  44\r",
+        "1  2  45\r",
+        "1  2  46\r",
+        "1  2  47\r",
+        "1  2  48\r",
+        "1  2  49\r",
+        "1  2  50\r",
+        "1  2  51\r",
+        "1  2  52\r",
+        "1  2  53\r",
+        "1  2  54\r",
+        "1  2  55\r",
+        "1  2  56\r",
+        "1  2  57\r",
+        "1  2  58\r",
+        "1  3  1\r",
+        "1  3  2\r",
+        "1  3  3\r",
+        "1  3  4\r",
+        "1  3  5\r",
+        "1  3  6\r",
+        "1  3  7\r",
+        "1  3  8\r",
+        "1  3  9\r",
+        "1  3  10\r",
+        "1  3  11\r",
+        "1  3  12\r",
+        "1  3  13\r",
+        "1  3  14\r",
+        "1  3  15\r",
+        "1  3  16\r",
+        "1  3  17\r",
+        "1  3  18\r",
+        "1  3  19\r",
+        "1  3  20\r",
+        "1  3  21\r",
+        "1  3  22\r",
+        "1  3  23\r",
+        "1  3  24\r",
+        "1  3  25\r",
+        "1  3  26\r",
+        "1  3  27\r",
+        "1  3  28\r",
+        "1  3  29\r",
+        "1  3  30\r",
+        "1  3  31\r",
+        "1  3  32\r",
+        "1  3  33\r",
+        "1  3  34\r",
+        "1  3  35\r",
+        "1  3  36\r",
+        "1  3  37\r",
+        "1  3  38\r",
+        "1  3  39\r",
+        "1  3  40\r",
+        "1  3  41\r",
+        "1  3  42\r",
+        "1  3  43\r",
+        "1  3  44\r",
+        "1  3  45\r",
+        "1  3  46\r",
+        "1  3  47\r",
+        "1  3  48\r",
+        "1  3  49\r",
+        "1  3  50\r",
+        "1  3  51\r",
+        "1  3  52\r",
+        "1  3  53\r",
+        "1  3  54\r",
+        "1  3  55\r",
+        "1  3  56\r",
+        "1  3  57\r",
+        "1  3  58\r",
+        "1  4  1\r",
+        "1  4  2\r",
+        "1  4  3\r",
+        "1  4  4\r",
+        "1  4  5\r",
+        "1  4  6\r",
+        "1  4  7\r",
+        "1  4  8\r",
+        "1  4  9\r",
+        "1  4  10\r",
+        "1  4  11\r",
+        "1  4  12\r",
+        "1  4  13\r",
+        "1  4  14\r",
+        "1  4  15\r",
+        "1  4  16\r",
+        "1  4  17\r",
+        "1  4  18\r",
+        "1  4  19\r",
+        "1  4  20\r",
+        "1  4  21\r",
+        "1  4  22\r",
+        "1  4  23\r",
+        "1  4  24\r",
+        "1  4  25\r",
+        "1  4  26\r",
+        "1  4  27\r",
+        "1  4  28\r",
+        "1  4  29\r",
+        "1  4  30\r",
+        "1  4  31\r",
+        "1  4  32\r",
+        "1  4  33\r",
+        "1  4  34\r",
+        "1  4  35\r",
+        "1  4  36\r",
+        "1  4  37\r",
+        "1  4  38\r",
+        "1  4  39\r",
+        "1  4  40\r",
+        "1  4  41\r",
+        "1  4  42\r",
+        "1  4  43\r",
+        "1  4  44\r",
+        "1  4  45\r",
+        "1  4  46\r",
+        "1  4  47\r",
+        "1  4  48\r",
+        "1  4  49\r",
+        "1  4  50\r",
+        "1  4  51\r",
+        "1  4  52\r",
+        "1  4  53\r",
+        "1  4  54\r",
+        "1  4  55\r",
+        "1  4  56\r",
+        "1  4  57\r",
+        "1  4  58\r",
+        "1  5  1\r",
+        "1  5  2\r",
+        "1  5  3\r",
+        "1  5  4\r",
+        "1  5  5\r",
+        "1  5  6\r",
+        "1  5  7\r",
+        "1  5  8\r",
+        "1  5  9\r",
+        "1  5  10\r",
+        "1  5  11\r",
+        "1  5  12\r",
+        "1  5  13\r",
+        "1  5  14\r",
+        "1  5  15\r",
+        "1  5  16\r",
+        "1  5  17\r",
+        "1  5  18\r",
+        "1  5  19\r",
+        "1  5  20\r",
+        "1  5  21\r",
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+       "stream": "stdout",
+       "text": [
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+      },
+      {
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+       "stream": "stdout",
+       "text": [
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+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.33, Page number: 131<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count occurrence of 0 to 9 digits between 1 and given decimal number\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "n = int(raw_input(\"Enter a Decimal Number : \"))\n",
+      "st = [0 for x in range(10)]         #10 array elements initialized as 0\n",
+      "\n",
+      "#Calculation\n",
+      "for l in range(1,n+1):\n",
+      "    t = l\n",
+      "    while t != 0:\n",
+      "        k = t % 10\n",
+      "        t = t / 10\n",
+      "        st[k] += 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nOccurrence of 0-9 digits between 1 to %d Numbers.\"%(n))\n",
+      "sys.stdout.write(\"\\n========== == === ====== ======= = == == ========\")\n",
+      "\n",
+      "for i in range(0,10):\n",
+      "    if st[i] > 0:\n",
+      "        sys.stdout.write(\"\\n%d Occurs %8d Times.\"%(i,st[i]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Decimal Number : 15\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Occurrence of 0-9 digits between 1 to 15 Numbers.\n",
+        "========== == === ====== ======= = == == ========\n",
+        "0 Occurs        1 Times.\n",
+        "1 Occurs        8 Times.\n",
+        "2 Occurs        2 Times.\n",
+        "3 Occurs        2 Times.\n",
+        "4 Occurs        2 Times.\n",
+        "5 Occurs        2 Times.\n",
+        "6 Occurs        1 Times.\n",
+        "7 Occurs        1 Times.\n",
+        "8 Occurs        1 Times.\n",
+        "9 Occurs        1 Times."
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.34, Page number: 132<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display sum of digits of a given number\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "s = 0\n",
+      "n = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "\n",
+      "sys.stdout.write(\"\\nSum of Digits till a single digit\\n %d\"%(n))\n",
+      "\n",
+      "while n != 0:\n",
+      "    s = s + n%10\n",
+      "    n = n/10\n",
+      "    if n == 0 and s > 9:\n",
+      "        sys.stdout.write(\"\\n %2d\"%(s))\n",
+      "        n = s\n",
+      "        s = 0\n",
+      "\n",
+      "sys.stdout.write(\"\\n %2d\"%(s))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 4687\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of Digits till a single digit\n",
+        " 4687\n",
+        " 25\n",
+        "  7"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.35, Page number: 133<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display octal numbers in binary.  Attach a parity bit with \"1\"\n",
+      "#if number of 1s are even otherwise \"0\".\n",
+      "#           OR\n",
+      "#Generate odd parity to octal numbers 0 to 7.  Express each number in binary\n",
+      "#and attach the parity bit\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = 0\n",
+      "j = 12\n",
+      "k = 2\n",
+      "\n",
+      "#Calculation & Result\n",
+      "\n",
+      "sys.stdout.write(\"\\nBinary Bits                  Parity Bits\")\n",
+      "sys.stdout.write(\"\\n=============               ==============\\n\")\n",
+      "\n",
+      "for x in range(0,8):\n",
+      "    k += 1\n",
+      "    j = 12\n",
+      "    y = x\n",
+      "    for a in range(0,3):\n",
+      "        b = y % 2\n",
+      "        #gotoxy\n",
+      "        sys.stdout.write(\"\\t%d\"%(b))\n",
+      "        y = y / 2\n",
+      "        if b == 1:\n",
+      "            c += 1\n",
+      "    if c%2 == 0:\n",
+      "        #gotoxy\n",
+      "        sys.stdout.write(\"\\t1\")\n",
+      "    else:\n",
+      "        #gotosy\n",
+      "        sys.stdout.write(\"\\t0\\n\")\n",
+      "    c = 0\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Binary Bits                  Parity Bits\n",
+        "=============               ==============\n",
+        "\t0\t0\t0\t1\t1\t0\t0\t0\n",
+        "\t0\t1\t0\t0\n",
+        "\t1\t1\t0\t1\t0\t0\t1\t0\n",
+        "\t1\t0\t1\t1\t0\t1\t1\t1\t1\t1\t1\t0\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.36, Page number: 135<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the series x - x^3/3! + x^5/5! - .... x^n/n!\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = 3.0\n",
+      "f = 1.0\n",
+      "\n",
+      "#Give x as float value and n as int value\n",
+      "\n",
+      "x = float(raw_input(\"Enter x & n : \"))       #use commas to input values\n",
+      "n = int(raw_input(\"Enter x & n : \")) \n",
+      "\n",
+      "sum1 = x                #since sum is built-in function in python,\n",
+      "                        #sum1 is used instead of sum\n",
+      "\n",
+      "#Calculation & Result\n",
+      "\n",
+      "for i in range(3,n+1,2):\n",
+      "    f = 1\n",
+      "    if c%2 != 0:\n",
+      "        for l in range(1,i+1):\n",
+      "            f = f * l\n",
+      "        sum1 = sum1 - pow(x,i)/ f\n",
+      "    else:\n",
+      "        for l in range(1,i+1):\n",
+      "            f = f * l\n",
+      "        sum1 = sum1 + pow(x,i)/ f\n",
+      "    c += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\nSum of series Numbers : %f\"%(sum1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter x & n : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter x & n : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of series Numbers : 0.933333"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.37, Page number: 136<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the series  x + x^2/2! + x^4/4! + .... x^n/n!\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "f = 1.0\n",
+      "\n",
+      "#Give x as float value and y as int value\n",
+      "\n",
+      "x = float(raw_input(\"Enter x & y : \"))       #use commas to input values\n",
+      "y = int(raw_input(\"Enter x & y : \")) \n",
+      "\n",
+      "sum1 = x                #since sum is built-in function in python,\n",
+      "                        #sum1 is used instead of sum\n",
+      "\n",
+      "#Calculation & Result\n",
+      "\n",
+      "for i in range(2,y+1,2):\n",
+      "    f = 1\n",
+      "    for l in range(1,i+1):\n",
+      "        f = f * l\n",
+      "    sum1 = sum1 + pow(x,i)/ f\n",
+      "\n",
+      "sys.stdout.write(\"\\nSum of Series : %f\"%(sum1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter x & y : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter x & y : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of Series : 22.666667"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.38, Page number: 136<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the series 1 - 1/1! + 2/2! - 3/3! .... n/n!\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = 3\n",
+      "\n",
+      "n = int(raw_input(\"Enter value of n : \"))\n",
+      "\n",
+      "sum1 = 1.0              #since sum is built-in function in python,\n",
+      "                        #sum1 is used instead of sum\n",
+      "\n",
+      "#Calculation\n",
+      "\n",
+      "for i in range(1,n+1):\n",
+      "    f = 1.0\n",
+      "    if c%2 != 0:\n",
+      "        for l in range(1,i+1):\n",
+      "            f = f * l\n",
+      "        k = float(i / f)\n",
+      "        sum1 = sum1 - k\n",
+      "    else:\n",
+      "        for l in range(1,i+1):\n",
+      "            f = f * l\n",
+      "        sum1 = sum1 + float(i/f)\n",
+      "    c += 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSum of series Numbers : %f\"%(sum1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of n : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of series Numbers : 0.500000"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.39, Page number: 137<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the Armstrong numbers in three digits from 100 to 999.  If sum of cubes of each digits\n",
+      "#of the number is equal to number itself, then the number is called as an Armstrong number.\n",
+      "#(For eg. 153 = 1^3 + 5^3 + 3^3 = 153)\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "cube = 0\n",
+      "\n",
+      "sys.stdout.write(\"The Following Numbers are Armstrong numbers.\")\n",
+      "\n",
+      "#Calculation & Result\n",
+      "\n",
+      "for k in range(100,999+1):\n",
+      "    cube = 0\n",
+      "    x = 1\n",
+      "    d = 3\n",
+      "    n = k\n",
+      "    while x <= d:\n",
+      "        i = n % 10\n",
+      "        cube = cube + pow(i,3)\n",
+      "        n = n / 10\n",
+      "        x += 1\n",
+      "    if cube == k:\n",
+      "        sys.stdout.write(\"\\n\\t%d\"%(k))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Following Numbers are Armstrong numbers.\n",
+        "\t153\n",
+        "\t370\n",
+        "\t371\n",
+        "\t407"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.40, Page number: 138<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Example 6.40.py\n",
+      "#Program to display the stars as shown below\n",
+      "#*\n",
+      "#**\n",
+      "#***\n",
+      "#****\n",
+      "#*****\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "x = int(raw_input(\"How many lines stars (*) should be printed ? \"))\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "for i in range(1,x+1):\n",
+      "    for j in range(1,i+1):\n",
+      "        sys.stdout.write(\"*\")\n",
+      "    sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many lines stars (*) should be printed ? 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "*\n",
+        "**\n",
+        "***\n",
+        "****\n",
+        "*****\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.41, Page number: 139<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Generate the given pattern\n",
+      "#6 5 4 3 2 1 0\n",
+      "#5 4 3 2 1 0\n",
+      "#4 3 2 1 0\n",
+      "#3 2 1 0\n",
+      "#2 1 0\n",
+      "#1 0\n",
+      "#0\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "while i >= 0:\n",
+      "    c = i\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    while True:\n",
+      "        sys.stdout.write(\"%3d\"%(c))\n",
+      "        if c == 0:\n",
+      "            break\n",
+      "        c -= 1\n",
+      "    i -= 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "  6  5  4  3  2  1  0\n",
+        "  5  4  3  2  1  0\n",
+        "  4  3  2  1  0\n",
+        "  3  2  1  0\n",
+        "  2  1  0\n",
+        "  1  0\n",
+        "  0"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.42, Page number: 140<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the series of numbers as given below\n",
+      "#1\n",
+      "#1 2\n",
+      "#1 2 3\n",
+      "#1 2 3 4\n",
+      "\n",
+      "#4 3 2 1\n",
+      "#3 2 1\n",
+      "#2 1\n",
+      "#1\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter value of x : \"))\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "for j in range(1,x+1):\n",
+      "    for i in range(1,j+1):\n",
+      "        sys.stdout.write(\"%3d\"%(i))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "\n",
+      "sys.stdout.write(\"\\n\")\n",
+      "\n",
+      "for j in range(x,0,-1):\n",
+      "    for i in range(j,0,-1):\n",
+      "        sys.stdout.write(\"%3d\"%(i))\n",
+      "    sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of x : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  1\n",
+        "  1  2\n",
+        "  1  2  3\n",
+        "  1  2  3  4\n",
+        "\n",
+        "  4  3  2  1\n",
+        "  3  2  1\n",
+        "  2  1\n",
+        "  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.43, Page number: 142<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to display the series of numbers as given below\n",
+      "#1\n",
+      "#2 1\n",
+      "#3 2 1\n",
+      "#4 3 2 1\n",
+      "\n",
+      "#4 3 2 1\n",
+      "#3 2 1\n",
+      "#2 1\n",
+      "#1\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter value of x : \"))\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "for j in range(1,x+1):\n",
+      "    for i in range(j,0,-1):\n",
+      "        sys.stdout.write(\"%3d\"%(i))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "\n",
+      "sys.stdout.write(\"\\n\")\n",
+      "\n",
+      "for j in range(x,0,-1):\n",
+      "    for i in range(j,0,-1):\n",
+      "        sys.stdout.write(\"%3d\"%(i))\n",
+      "    sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of x : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  1\n",
+        "  2  1\n",
+        "  3  2  1\n",
+        "  4  3  2  1\n",
+        "\n",
+        "  4  3  2  1\n",
+        "  3  2  1\n",
+        "  2  1\n",
+        "  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.44, Page number: 143<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Generate the pyramid structure using numberical\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter a number : \"))\n",
+      "\n",
+      "#Result\n",
+      "\n",
+      "for j in range(0,x+1):\n",
+      "    for i in range(0-j,j+1):\n",
+      "        sys.stdout.write(\"%3d\"%(abs(i)))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  0\n",
+        "  1  0  1\n",
+        "  2  1  0  1  2\n",
+        "  3  2  1  0  1  2  3\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.45, Page number: 143<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert binary to decimal number\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = [1 for i in range(1,5+1)]              #defines the array x and initializes the elements with 0\n",
+      "y = x[0]\n",
+      "\n",
+      "sys.stdout.write(\"\\nValues in different Iterations\")\n",
+      "sys.stdout.write(\"\\n====== == ========= ==========\\n\")\n",
+      "\n",
+      "for i in range(0,4):\n",
+      "    y = y * 2 + x[i+1]\n",
+      "    sys.stdout.write(\"[%d] %d\\t\"%(i+1,y))\n",
+      "\n",
+      "sys.stdout.write(\"\\nEquivalent of [\")\n",
+      "\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"%d\"%(x[i]))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"] in Decimal Number is : \")\n",
+      "sys.stdout.write(\"%d\\t\"%(y))\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Values in different Iterations\n",
+        "====== == ========= ==========\n",
+        "[1] 3\t[2] 7\t[3] 15\t[4] 31\t\n",
+        "Equivalent of [11111] in Decimal Number is : 31\t"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.46, Page number: 144<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Program to add a parity bit with four binary bits such that the total\n",
+      "#number of one's should be even\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "bit = [0 for i in range(0,5)]              #defines the array bit and initializes the elements with 0\n",
+      "c = 0\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter four bits : \")\n",
+      "\n",
+      "for j in range(0,4):\n",
+      "    x = int(raw_input(\" \"))\n",
+      "    bit[j] = x\n",
+      "    if bit[j] == 1:\n",
+      "        c += 1\n",
+      "    else:\n",
+      "        if bit[j] > 1 or bit[j] < 0:\n",
+      "            j -= 1\n",
+      "            continue\n",
+      "        if c%2 == 0:\n",
+      "            bit[j] = 0\n",
+      "        else:\n",
+      "            bit[j] = 1   \n",
+      "   \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nMessage bits together with parity bit : \")\n",
+      "for j in range(0,5):\n",
+      "    sys.stdout.write(\"%d\"%(bit[j]))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter four bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Message bits together with parity bit : 11110"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.47, Page number: 145<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert binary to decimal number. Enter the binary bits by using for loop\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "z = [0 for i in range(0,10)]                #defines the array z and initializes the elements with 0\n",
+      "\n",
+      "sys.stdout.write(\"Enter the number of bits :- \")\n",
+      "b = int(raw_input(\"\"))\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter the binary bits : \")\n",
+      "for i in range(0,b):\n",
+      "    z[i] = int(raw_input(\"\"))\n",
+      "\n",
+      "a = z[0]\n",
+      "\n",
+      "for i in range(0,b-1):\n",
+      "    a = a * 2 + z[i+1]\n",
+      "    sys.stdout.write(\"\\n%d\"%(a))\n",
+      "   \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nDecimal Number is : %d \"%(a))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of bits :- "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter the binary bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "2\n",
+        "4\n",
+        "8\n",
+        "17\n",
+        "Decimal Number is : 17 "
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.48, Page number: 147<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Verify the truth table of AND gate. Assume AND gate has two\n",
+      "#input bits A & B and one output bit C.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "a = [0 for i in range(0,4)]                #defines the array a,b & c  and initializes the elements with 0\n",
+      "b = [0 for i in range(0,4)]\n",
+      "c = [0 for i in range(0,4)]\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
+      "for x in range(0,4):\n",
+      "    a[x] = int(raw_input(\"\"))\n",
+      "    \n",
+      "    if a[x] > 1 or a[x] < 0:\n",
+      "        x -= 1\n",
+      "        continue\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
+      "for x in range(0,4):\n",
+      "    b[x] = int(raw_input(\"\"))\n",
+      "    \n",
+      "    if b[x] > 1 or b[x] < 0:\n",
+      "        x -= 1\n",
+      "        continue\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nA  B  C\")\n",
+      "for x in range(0,4):\n",
+      "    if a[x] == 1 and b[x] == 1:\n",
+      "        c[x] = 1\n",
+      "    else:\n",
+      "        c[x] = 0\n",
+      "    sys.stdout.write(\"\\n%d  %d  %d\"%(a[x],b[x],c[x]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Four Bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Four Bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "A  B  C\n",
+        "1  1  1\n",
+        "0  0  0\n",
+        "1  0  0\n",
+        "0  1  0"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.49, Page number: 148<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Verify the truth table of OR gate. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "a = [0 for i in range(0,4)]                #defines the array a,b & c  and initializes the elements with 0\n",
+      "b = [0 for i in range(0,4)]\n",
+      "c = [0 for i in range(0,4)]\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
+      "for x in range(0,4):\n",
+      "    a[x] = int(raw_input(\"\"))\n",
+      "    \n",
+      "    if a[x] > 1 or a[x] < 0:\n",
+      "        x -= 1\n",
+      "        continue\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
+      "for x in range(0,4):\n",
+      "    b[x] = int(raw_input(\"\"))\n",
+      "    \n",
+      "    if b[x] > 1 or b[x] < 0:\n",
+      "        x -= 1\n",
+      "        continue\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nA  B  C\")\n",
+      "for x in range(0,4):\n",
+      "    if a[x] == 0 and b[x] == 0:\n",
+      "        c[x] = 0\n",
+      "    else:\n",
+      "        c[x] = 1\n",
+      "    sys.stdout.write(\"\\n%d  %d  %d\"%(a[x],b[x],c[x]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Four Bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Four Bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "A  B  C\n",
+        "1  1  1\n",
+        "1  0  1\n",
+        "1  0  1\n",
+        "0  0  0"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.50, Page number: 149<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Verify the truth table of EX-OR gate. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "a = [0 for i in range(0,4)]                #defines the array a,b & c  and initializes the elements with 0\n",
+      "b = [0 for i in range(0,4)]\n",
+      "c = [0 for i in range(0,4)]\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
+      "for x in range(0,4):\n",
+      "    a[x] = int(raw_input(\"\"))\n",
+      "    \n",
+      "    if a[x] > 1 or a[x] < 0:\n",
+      "        x -= 1\n",
+      "        continue\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter Four Bits : \")\n",
+      "for x in range(0,4):\n",
+      "    b[x] = int(raw_input(\"\"))\n",
+      "    \n",
+      "    if b[x] > 1 or b[x] < 0:\n",
+      "        x -= 1\n",
+      "        continue\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nA  B  C\")\n",
+      "for x in range(0,4):\n",
+      "    if a[x] == 0 and b[x] == 1:\n",
+      "        c[x] = 1\n",
+      "    else:\n",
+      "        if a[x] == 1 and b[x] == 0:\n",
+      "            c[x] = 1\n",
+      "        else:\n",
+      "            c[x] = 0\n",
+      "    sys.stdout.write(\"\\n%d  %d  %d\"%(a[x],b[x],c[x]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Four Bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Four Bits : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "A  B  C\n",
+        "1  1  0\n",
+        "1  0  1\n",
+        "1  0  1\n",
+        "0  0  0"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.51, Page number: 151<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the Hamming code for the entered binary code.  Assume the binary code of four bits in length.\n",
+      "#The hamming code should be seven bits.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "#defines the array a,b & c  and initializes the elements with 0\n",
+      "b = [0 for i in range(0,4)]\n",
+      "c = [0 for i in range(0,7)]\n",
+      "\n",
+      "sys.stdout.write(\"\\nRead the Binary Numbers : \")\n",
+      "for x in range(0,4):\n",
+      "    b[x] = int(raw_input(\"\"))\n",
+      "\n",
+      "#Piece copy operation\n",
+      "c[0] = b[0]\n",
+      "c[1] = b[1]\n",
+      "c[2] = b[2]\n",
+      "c[4] = b[3]\n",
+      "\n",
+      "sys.stdout.write(\"\\nBefore xOR operation : \")\n",
+      "    \n",
+      "for x in range(0,7):\n",
+      "    sys.stdout.write(\"%3d\"%(c[x]))\n",
+      "\n",
+      "c[6] = c[0] ^ c[2] ^ c[4]\n",
+      "c[5] = c[0] ^ c[1] ^ c[4]\n",
+      "c[3] = c[0] ^ c[1] ^ c[2]\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nHamming code after XOR operation : \")\n",
+      "for x in range(0,7):\n",
+      "    sys.stdout.write(\"%3d\"%(c[x]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Read the Binary Numbers : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Before xOR operation :   1  0  1  0  0  0  0\n",
+        "Hamming code after XOR operation :   1  0  1  0  0  1  0"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.52, Page number: 152<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the results of students appear in six subjects.  The result declared should be as per the \n",
+      "#following table.\n",
+      "#TOTAL MARKS              RESULT\n",
+      "#>= 420                  Distinction\n",
+      "#>= 360                  First Division\n",
+      "#>= 240                  Second Division\n",
+      "#Otherwise               Fail\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "DISTINCTION = 420\n",
+      "FIRST = 360\n",
+      "SECOND = 240\n",
+      "\n",
+      "number = int(raw_input(\"Enter number of Students : \"))\n",
+      "\n",
+      "#Read the values and calculation\n",
+      "for i in range(1,number+1):\n",
+      "    roll_no = int(raw_input(\"Enter Roll Number : \"))\n",
+      "    total = 0\n",
+      "    sys.stdout.write(\"\\nEnter Marks of 6 Subjects for Roll no %d : \"%(roll_no))\n",
+      "    for j in range(1,6+1):\n",
+      "        marks = int(raw_input(\"\"))\n",
+      "        total = total + marks\n",
+      "    sys.stdout.write(\"\\nTOTAL MARKS = %d\"%(total))\n",
+      "    if total >= DISTINCTION:\n",
+      "        sys.stdout.write(\"\\n(Distinction)\\n\\n\")\n",
+      "    else:\n",
+      "        if total >= FIRST:\n",
+      "            sys.stdout.write(\"\\n(First Division)\\n\\n\")\n",
+      "        else:\n",
+      "            if total >= SECOND:\n",
+      "                sys.stdout.write(\"\\n(Second Division)\\n\\n\")\n",
+      "            else:\n",
+      "                sys.stdout.write(\"\\n(***Fail***)\")\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter number of Students : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Roll Number : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Marks of 6 Subjects for Roll no 1 : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "42\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "52\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "62\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "72\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "82\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "92\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "TOTAL MARKS = 402\n",
+        "(First Division)\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.53, Page number: 154<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the string \"You have learnt C program\" 9 times using while loop\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 1\n",
+      "\n",
+      "#Result\n",
+      "while x < 10:\n",
+      "    sys.stdout.write(\"\\nYou have learnt C program\")\n",
+      "    x += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "You have learnt C program\n",
+        "You have learnt C program\n",
+        "You have learnt C program\n",
+        "You have learnt C program\n",
+        "You have learnt C program\n",
+        "You have learnt C program\n",
+        "You have learnt C program\n",
+        "You have learnt C program\n",
+        "You have learnt C program"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.54, Page number: 155<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Add 10 consecutive numbers starting from 1 using while loop.\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 1\n",
+      "sum1 = 0           #since sum is a builtin function in python, sum1 is used instead of sum\n",
+      "\n",
+      "#Calculation\n",
+      "while a <= 10:\n",
+      "    sys.stdout.write(\"%3d\"%(a))\n",
+      "    sum1 = sum1 + a\n",
+      "    a += 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSum of 10 numbers : %d\"%(sum1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  1  2  3  4  5  6  7  8  9 10\n",
+        "Sum of 10 numbers : 55"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.55, Page number: 155<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate factorial of a given number using while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "fact = 1\n",
+      "\n",
+      "a = int(raw_input(\"Enter The Number : \"))\n",
+      "\n",
+      "#Calculation\n",
+      "while a >= 1:\n",
+      "    sys.stdout.write(\" %d * \"%(a))\n",
+      "    fact = fact * a\n",
+      "    a -= 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\" = %d\"%(fact))\n",
+      "sys.stdout.write(\"\\nFactorial of Given number is %d\"%(fact))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter The Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 5 *  4 *  3 *  2 *  1 *  = 120\n",
+        "Factorial of Given number is 120"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.56, Page number: 156<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate factorial of a given number using while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "fact = 1\n",
+      "b = 1\n",
+      "\n",
+      "a = int(raw_input(\"Enter The Number : \"))\n",
+      "\n",
+      "#Calculation\n",
+      "while b <= a:\n",
+      "    sys.stdout.write(\" %d * \"%(b))\n",
+      "    fact = fact * b\n",
+      "    b += 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\" = %d\"%(fact))\n",
+      "sys.stdout.write(\"\\nFactorial of %d is %d\"%(b-1,fact))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter The Number : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1 *  2 *  3 *  4 *  = 24\n",
+        "Factorial of 4 is 24"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.57, Page number: 157<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert decimal number to binary number\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "y = 40\n",
+      "rev = [0 for i in range(1,10)]\n",
+      "c = 1\n",
+      "\n",
+      "x = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "#since gotoxy() function is not available in python, the reminders are stored in an array and displayed in reverse order\n",
+      "\n",
+      "sys.stdout.write(\"\\nBinary Number : \")\n",
+      "while x != 0:\n",
+      "    rev[c] = x % 2\n",
+      "    x = x / 2\n",
+      "    c += 1\n",
+      "\n",
+      "for i in range(c-1,0,-1):\n",
+      "    sys.stdout.write(\"%d\"%(rev[i]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 25\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Binary Number : 11001"
+       ]
+      }
+     ],
+     "prompt_number": 63
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.58, Page number: 157<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert decimal number to user defined number system. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "y = 35\n",
+      "rev = 0\n",
+      "c = 1\n",
+      "rev = [0 for i in range(1,10)]\n",
+      "\n",
+      "m = int(raw_input(\"Enter the Decimal Number : \"))\n",
+      "b = int(raw_input(\"Enter base of number system : \"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "sys.stdout.write(\"\\nThe Number Obtained : \")\n",
+      "\n",
+      "#since gotoxy() function is not available in python, the reminders are stored in an array and displayed in reverse order\n",
+      "while m != 0:\n",
+      "    rev[c] = m % b\n",
+      "    c += 1\n",
+      "    m = m / b\n",
+      "\n",
+      "for i in range(c-1,0,-1):\n",
+      "    sys.stdout.write(\"%d\"%rev[i])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the Decimal Number : 50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter base of number system : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Number Obtained : 200"
+       ]
+      }
+     ],
+     "prompt_number": 64
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.59, Page number: 158<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert binary number to equivalent decimal number\n",
+      "\n",
+      "\n",
+      "#Variable Initialization\n",
+      "y = 0\n",
+      "p = 0\n",
+      "\n",
+      "n = int(raw_input(\"Enter a Binary Number : \"))\n",
+      "\n",
+      "#Calculation\n",
+      "while n != 0:\n",
+      "    x = n % 10\n",
+      "    if x > 1 or x < 0:\n",
+      "        sys.stdout.write(\"\\nInvalid Digit\")\n",
+      "        break\n",
+      "    else:\n",
+      "        y = y + x * pow(2,p)\n",
+      "        n = n / 10\n",
+      "        p += 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nEquivalent Decimal Number is %d.\"%(y))\n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Binary Number : 1111\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Equivalent Decimal Number is 15."
+       ]
+      }
+     ],
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.60, Page number: 159<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read a positive integer number 'n' and generate the numbers in the following way.\n",
+      "#If entered number is 5 the output will be as follows.\n",
+      "#OUTPUT : 5 4 3 2 1 0 1 2 3 4 5\n",
+      "\n",
+      "\n",
+      "#Variable Initialization\n",
+      "k = 0\n",
+      "\n",
+      "n = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "i = n + 1\n",
+      "k = k - n\n",
+      "\n",
+      "while i != k:\n",
+      "    sys.stdout.write(\"%3d\"%(abs(k)))\n",
+      "    k += 1\n",
+      "\n",
+      "      "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  3  2  1  0  1  2  3"
+       ]
+      }
+     ],
+     "prompt_number": 67
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.61, Page number: 160<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Enter a number through keyboard and find the sum of the digits.\n",
+      "\n",
+      "\n",
+      "#Variable Initialization\n",
+      "k = 1\n",
+      "sum1 = 0            #since sum is a built-in function in python, sum1 used instead of sum\n",
+      "\n",
+      "n = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Calculation \n",
+      "while n != 0:\n",
+      "    k = n % 10\n",
+      "    sum1 = sum1 + k\n",
+      "    k = n / 10\n",
+      "    n = k\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Sum of digits %d\"%(sum1))\n",
+      "\n",
+      "      "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 842\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of digits 14"
+       ]
+      }
+     ],
+     "prompt_number": 68
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.62, Page number: 161<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Enter few numbers and count the positive and negative numbers together\n",
+      "#with their sums.  When 0 is entered program should be terminated.\n",
+      "\n",
+      "\n",
+      "#Variable Initialization\n",
+      "p = 0\n",
+      "n = 0\n",
+      "j = 1\n",
+      "s = 0\n",
+      "ns = 0\n",
+      "\n",
+      "#Calculation \n",
+      "sys.stdout.write(\"Enter Numbers (0) Exit : \")\n",
+      "while j != 0:\n",
+      "    j = int(raw_input(\"\"))\n",
+      "    if j > 0:\n",
+      "        p += 1\n",
+      "        s = s + j\n",
+      "    else:\n",
+      "        if j < 0:\n",
+      "            n += 1\n",
+      "            ns = ns + j\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nTotal Positive Numbers : %d\"%(p))\n",
+      "sys.stdout.write(\"\\nTotal Negative Numbers : %d\"%(n))\n",
+      "sys.stdout.write(\"\\nSum of Positive Numbers : %d\"%(s))\n",
+      "sys.stdout.write(\"\\nSum of Negative Numbers : %d\"%(ns))\n",
+      "\n",
+      "\n",
+      "      "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Numbers (0) Exit : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Total Positive Numbers : 5\n",
+        "Total Negative Numbers : 3\n",
+        "Sum of Positive Numbers : 15\n",
+        "Sum of Negative Numbers : -17"
+       ]
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.63, Page number: 162<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find sum of odd and even numbers separately by sorting the given numbers into\n",
+      "#odd and even numbers\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = 1\n",
+      "odd = 0\n",
+      "even = 0\n",
+      "\n",
+      "a = int(raw_input(\"Enter a Number : \"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "sys.stdout.write(\"ODD     EVEN\\n\")\n",
+      "while c <= a:\n",
+      "    b = c % 2\n",
+      "    while b == 0:\n",
+      "        sys.stdout.write(\"\\t%d \"%(c))\n",
+      "        even = even + c\n",
+      "        b = 1\n",
+      "    b = c % 2\n",
+      "    while b != 0:\n",
+      "        sys.stdout.write(\"\\n%d\"%(c))\n",
+      "        odd = odd + c\n",
+      "        b = 0\n",
+      "    c += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\n==============================\")\n",
+      "sys.stdout.write(\"\\n %d      %d\"%(odd,even))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number : 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ODD     EVEN\n",
+        "\n",
+        "1\t2 \n",
+        "3\t4 \n",
+        "5\t6 \n",
+        "7\t8 \n",
+        "9\t10 \n",
+        "==============================\n",
+        " 25      30"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.64, Page number: 163<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the entered number in reverse order.\n",
+      "\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 1\n",
+      "\n",
+      "d = int(raw_input(\"Enter the number of digits : \"))\n",
+      "n = int(raw_input(\"Enter the number which is to be reversed : -\"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "sys.stdout.write(\"\\nThe Reversed Number is :-\")\n",
+      "\n",
+      "while x <= d:\n",
+      "    i = n % 10\n",
+      "    sys.stdout.write(\"%d\"%(i))\n",
+      "    n = n / 10\n",
+      "    x += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of digits : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number which is to be reversed : -5428\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Reversed Number is :-8245"
+       ]
+      }
+     ],
+     "prompt_number": 71
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.65, Page number: 163<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Separate capitals, small, symbols and numbers from a statement\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 0\n",
+      "c = 0\n",
+      "s = 0\n",
+      "h = 0\n",
+      "n = 0\n",
+      "\n",
+      "#Array declaration\n",
+      "scan = [0 for i in range(0,40)]\n",
+      "cap = [0 for i in range(0,20)]\n",
+      "small = [0 for i in range(0,20)]\n",
+      "num = [0 for i in range(0,20)]\n",
+      "oth = [0 for i in range(0,20)]\n",
+      "\n",
+      "scan = raw_input(\"Enter Text Here : \")\n",
+      "\n",
+      "i = 0\n",
+      "\n",
+      "while i < len(scan):                       #There is no null termination character in python.\n",
+      "    if scan[i] >= '0' and scan[i] <= '9':\n",
+      "        n += 1\n",
+      "        num[n] = scan[i]\n",
+      "    else:\n",
+      "        if scan[i] >= 'A' and scan[i] <= 'Z':\n",
+      "            c += 1\n",
+      "            cap[c] = scan[i]\n",
+      "        else:\n",
+      "            if scan[i] >= 'a' and scan[i] <= 'z':\n",
+      "                s += 1\n",
+      "                small[s] = scan[i]\n",
+      "            else:\n",
+      "                h += 1\n",
+      "                oth[h] = scan[i]\n",
+      "                \n",
+      "    i += 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nCapital Letters : [\")\n",
+      "for i in range(0,20):\n",
+      "    sys.stdout.write(\"%c\"%cap[i])\n",
+      "\n",
+      "sys.stdout.write(\"]\\nSmall Letters : [\")\n",
+      "for i in range(0,20):\n",
+      "    sys.stdout.write(\"%c\"%(small[i]))\n",
+      "    \n",
+      "sys.stdout.write(\"]\\nNumeric Letters : [\")\n",
+      "for i in range(0,20):\n",
+      "    sys.stdout.write(\"%c\"%(num[i]))\n",
+      "\n",
+      "sys.stdout.write(\"]\\nOther Letters : [\")\n",
+      "for i in range(0,20):\n",
+      "    sys.stdout.write(\"%c\"%(oth[i]))\n",
+      "sys.stdout.write(\"]\")\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Here : HAVE A NICE DAY, contact me on 51606.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Capital Letters : [\u0000HAVEANICEDAY\u0000\u0000\u0000\u0000\u0000\u0000\u0000]\n",
+        "Small Letters : [\u0000contactmeon\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000]\n",
+        "Numeric Letters : [\u000051606\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000]\n",
+        "Other Letters : [\u0000   ,    .\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000\u0000]"
+       ]
+      }
+     ],
+     "prompt_number": 72
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.66, Page number: 165<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sort numbers, upper and lower case alphabets\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 48\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"   NUMBERS\\n\")\n",
+      "while i <= 57:\n",
+      "    sys.stdout.write(\"%2c\"%(i))\n",
+      "    i += 1\n",
+      "\n",
+      "i += 7\n",
+      "\n",
+      "sys.stdout.write(\"\\n  CAPITAL ALPHABETS\\n\")\n",
+      "while i <= 90:\n",
+      "    sys.stdout.write(\"%2c\"%(i))\n",
+      "    i += 1\n",
+      "    \n",
+      "i += 6\n",
+      "\n",
+      "sys.stdout.write(\"\\n  SMALL ALPHABETS\\n\")\n",
+      "while i <= 122:\n",
+      "    sys.stdout.write(\"%2c\"%(i))\n",
+      "    i += 1\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "   NUMBERS\n",
+        " 0 1 2 3 4 5 6 7 8 9\n",
+        "  CAPITAL ALPHABETS\n",
+        " A B C D E F G H I J K L M N O P Q R S T U V W X Y Z\n",
+        "  SMALL ALPHABETS\n",
+        " a b c d e f g h i j k l m n o p q r s t u v w x y z"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.67, Page number: 165<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sort numbers, upper and lower case alphabets\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 48\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Numbers : \")\n",
+      "while i < 124:\n",
+      "    if i < 58:\n",
+      "        sys.stdout.write(\"%2c\"%(i))\n",
+      "    else:\n",
+      "        if i == 58:\n",
+      "            sys.stdout.write(\"\\nCapital Letters : \")\n",
+      "            i += 7\n",
+      "        if i > 64 and i < 91:\n",
+      "            sys.stdout.write(\"%2c\"%(i))\n",
+      "        if i == 90:\n",
+      "            sys.stdout.write(\"\\nSmall Letters : \")\n",
+      "            i += 7\n",
+      "        if i > 96 and i < 123:\n",
+      "            sys.stdout.write(\"%2c\"%(i))\n",
+      "    i += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numbers :  0 1 2 3 4 5 6 7 8 9\n",
+        "Capital Letters :  A B C D E F G H I J K L M N O P Q R S T U V W X Y Z\n",
+        "Small Letters :  a b c d e f g h i j k l m n o p q r s t u v w x y z"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.68, Page number: 167<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display all ASCII numbers and their equivalent character, numbers and symbols using while loop.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 64\n",
+      "c = 'Y'\n",
+      "\n",
+      "#Result\n",
+      "while c == 'Y':\n",
+      "    sys.stdout.write(\"\\t%c (%d)\"%(i,i))\n",
+      "    i += 1\n",
+      "    sys.stdout.write(\"\\nPrint (Y/N)\")\n",
+      "    c = raw_input(\"\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t@ (64)\n",
+        "Print (Y/N)\tA (65)"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Print (Y/N)\tB (66)"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Print (Y/N)\tC (67)"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Print (Y/N)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.69, Page number: 168<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert entered character into its opposite case.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n*Enter 'E' to exit\")\n",
+      "sys.stdout.write(\"\\n*Enter only characters\")\n",
+      "\n",
+      "while c != 'E':\n",
+      "    sys.stdout.write(\"\\nEnter a character : \")\n",
+      "    c = raw_input(\"Enter a character :\")\n",
+      "    if c == c.upper():\n",
+      "        c = ord(c)\n",
+      "        sys.stdout.write(\"\\tIt's Lower Case : %c\"%(chr(c+32)))\n",
+      "    else:\n",
+      "        c = ord(c)\n",
+      "        sys.stdout.write(\"\\tIt's Upper Case : %s\"%(chr(c-32)))\n",
+      "    c = chr(c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "*Enter 'E' to exit\n",
+        "*Enter only characters\n",
+        "Enter a character : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a character :s\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tIt's Upper Case : S\n",
+        "Enter a character : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a character :A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tIt's Lower Case : a\n",
+        "Enter a character : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a character :E\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tIt's Lower Case : e"
+       ]
+      }
+     ],
+     "prompt_number": 73
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.70, Page number: 169<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print numbers after each iteration and messages after termination of each loop using 3 nested while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 1\n",
+      "j = 1\n",
+      "k = 1\n",
+      "\n",
+      "#Result\n",
+      "while i < 4:\n",
+      "    while j < 3:\n",
+      "        while k < 2:\n",
+      "            sys.stdout.write(\"\\n\\ni = %d j = %d k = %d\"%(i,j,k))\n",
+      "            k += 1\n",
+      "        sys.stdout.write(\"\\nInner Loop (k) Completed.\")\n",
+      "        k = 1\n",
+      "        j += 1\n",
+      "    sys.stdout.write(\"\\nMiddle Loop (j) Completed.\")\n",
+      "    j = 1\n",
+      "    i += 1\n",
+      "sys.stdout.write(\"\\nOuter Loop (i) Completed.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "i = 1 j = 1 k = 1\n",
+        "Inner Loop (k) Completed.\n",
+        "\n",
+        "i = 1 j = 2 k = 1\n",
+        "Inner Loop (k) Completed.\n",
+        "Middle Loop (j) Completed.\n",
+        "\n",
+        "i = 2 j = 1 k = 1\n",
+        "Inner Loop (k) Completed.\n",
+        "\n",
+        "i = 2 j = 2 k = 1\n",
+        "Inner Loop (k) Completed.\n",
+        "Middle Loop (j) Completed.\n",
+        "\n",
+        "i = 3 j = 1 k = 1\n",
+        "Inner Loop (k) Completed.\n",
+        "\n",
+        "i = 3 j = 2 k = 1\n",
+        "Inner Loop (k) Completed.\n",
+        "Middle Loop (j) Completed.\n",
+        "Outer Loop (i) Completed."
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.71, Page number: 170<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display a message \"This is a program of do while loop\". for 5 times using do while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 1\n",
+      "\n",
+      "#Result\n",
+      "#There is no do..while loop in python\n",
+      "while i <= 5:\n",
+      "    sys.stdout.write(\"\\nThis is a program of do while loop.\")\n",
+      "    i += 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "This is a program of do while loop.\n",
+        "This is a program of do while loop.\n",
+        "This is a program of do while loop.\n",
+        "This is a program of do while loop.\n",
+        "This is a program of do while loop."
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.72, Page number: 171<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Print the entered number in reverse order and find sum and product of digits using do-while loop. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "#since sum is a built-in function in python, sum1 is used instead of sum\n",
+      "x = 1\n",
+      "mul = 1\n",
+      "sum1 = 0\n",
+      "\n",
+      "d = int(raw_input(\"Enter the number of digits : -\"))\n",
+      "n = int(raw_input(\"Enter the number which is to be reversed :-\"))\n",
+      "\n",
+      "#Result\n",
+      "#There is no do..while loop in python\n",
+      "sys.stdout.write(\"\\nReveresed Number :- \")\n",
+      "while x <= d:\n",
+      "    i = n % 10\n",
+      "    sys.stdout.write(\"%d\"%(i))\n",
+      "    sum1 = sum1 + i\n",
+      "    mul = mul * i\n",
+      "    n = n / 10\n",
+      "    x += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\nAddition of digits :- %4d\"%(sum1))\n",
+      "sys.stdout.write(\"\\nMultiplication of digits :- %4d\"%(mul))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number of digits : -4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number which is to be reversed :-4321\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Reveresed Number :- 1234\n",
+        "Addition of digits :-   10\n",
+        "Multiplication of digits :-   24"
+       ]
+      }
+     ],
+     "prompt_number": 74
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.73, Page number: 172<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Attempt the question 6.72 by considering the false condition i.e. the value of i should\n",
+      "#be initially larger than the value in the while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 7\n",
+      "\n",
+      "#Result\n",
+      "#There is no do-while loop in python\n",
+      "sys.stdout.write(\"\\nThis is a program of do while loop\")\n",
+      "while i <= 5:\n",
+      "    sys.stdout.write(\"\\nThis is a program of do while loop\")\n",
+      "    i += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "This is a program of do while loop"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.74, Page number: 172<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the cubes of 1 to 10 numbers using do-while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nPrint the numbers and their cubes\\n\")\n",
+      "while x <= 10:\n",
+      "    y = pow(x,3)\n",
+      "    sys.stdout.write(\"%4d %27d\\n\"%(x,y))\n",
+      "    x += 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Print the numbers and their cubes\n",
+        "   1                           1\n",
+        "   2                           8\n",
+        "   3                          27\n",
+        "   4                          64\n",
+        "   5                         125\n",
+        "   6                         216\n",
+        "   7                         343\n",
+        "   8                         512\n",
+        "   9                         729\n",
+        "  10                        1000\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.75, Page number: 173<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Check whether the given number is prime or not.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 2\n",
+      "n = int(raw_input(\"Enter The number for testing (prime or not) : \"))\n",
+      "\n",
+      "#Result\n",
+      "#There is no do-while loop in python\n",
+      "while x < n:\n",
+      "    if n % x == 0:\n",
+      "        sys.stdout.write(\"\\nThe number %d is not prime.\"%(n))\n",
+      "        exit(0)\n",
+      "    x += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\nThe number %d is prime\"%(n))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter The number for testing (prime or not) : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The number 5 is prime"
+       ]
+      }
+     ],
+     "prompt_number": 75
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.76, Page number: 174<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count the number of students having age less than 25 and weight less than 50Kg out of five.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "count = 0\n",
+      "x = 1\n",
+      "\n",
+      "sys.stdout.write(\"\\nEnter data of 5 boys\\n\")\n",
+      "sys.stdout.write(\"\\nAge Weight\\n\")\n",
+      "\n",
+      "#There is no do-while loop in python\n",
+      "while x <= 5:\n",
+      "    age = int(raw_input(\"\"))\n",
+      "    wt = float(raw_input(\"\"))\n",
+      "    if age < 25 and wt < 50:\n",
+      "        count += 1\n",
+      "    x += 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nNumber of boys with age < 25\")\n",
+      "sys.stdout.write(\"and weight < 50Kg = %d\\n\"%(count))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter data of 5 boys\n",
+        "\n",
+        "Age Weight\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "24\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "51\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "51\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "35\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "24\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "54\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Number of boys with age < 25and weight < 50Kg = 2\n"
+       ]
+      }
+     ],
+     "prompt_number": 76
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.77, Page number: 175<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compute the factorial of given number using do while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "fact = 1\n",
+      "a = int(raw_input(\"Enter The Number : \"))\n",
+      "sys.stdout.write(\"\\nEnter The Number : %d\\n\"%(a))\n",
+      "\n",
+      "#Calculation\n",
+      "#There is no do-while loop in python\n",
+      "while a >= 1:\n",
+      "    sys.stdout.write(\"%d * \"%(a))\n",
+      "    fact = fact * a\n",
+      "    a -= 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\" = %d\"%(fact))\n",
+      "sys.stdout.write(\"\\nFactorial of Given number is %d\"%(fact))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter The Number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter The Number : 5\n",
+        "5 * 4 * 3 * 2 * 1 *  = 120\n",
+        "Factorial of Given number is 120"
+       ]
+      }
+     ],
+     "prompt_number": 77
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.78, Page number: 175<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the series such as 1+2+3+ .. i. Using do-while loop. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 1\n",
+      "s = 0\n",
+      "\n",
+      "i = int(raw_input(\"Enter a number : \"))\n",
+      "\n",
+      "#Calculation and Result\n",
+      "#There is no do-while loop in python\n",
+      "while a <= i:\n",
+      "    sys.stdout.write(\"%d + \"%(a))\n",
+      "    s = s + a\n",
+      "    a += 1\n",
+      "    \n",
+      "sys.stdout.write(\"\\b\\bs = %d\"%(s))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 + 2 + 3 + 4 + 5 + \b\bs = 15"
+       ]
+      }
+     ],
+     "prompt_number": 78
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6.79, Page number: 176<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use while statement in do-while loop and print values from 1 to 5\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = 0\n",
+      "\n",
+      "#There is no do-while loop in python\n",
+      "while x < 5:\n",
+      "    x += 1\n",
+      "    sys.stdout.write(\"\\t%d\"%(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t1\t2\t3\t4\t5"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter7.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter7.ipynb
new file mode 100755
index 00000000..ac82414f
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter7.ipynb
@@ -0,0 +1,3644 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 7: Arrays<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.1, Page number: 186<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print size of various data types using arrays.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = [0 for i in range(0,10)]\n",
+      "c = ['0' for i in range(0,10)]\n",
+      "l = [sys.maxint+5 for i in range(0,10)]\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The type 'int' requires %d Bytes\"%(sys.getsizeof(int)))\n",
+      "sys.stdout.write(\"\\nThe type 'char' requires %d Bytes\"%(sys.getsizeof(str)))\n",
+      "sys.stdout.write(\"\\nThe type 'long' requires %d Bytes\"%(sys.getsizeof(long)))\n",
+      "sys.stdout.write(\"\\n%d memory locations are reserved for ten 'int' elements\"%(sys.getsizeof(i)))\n",
+      "sys.stdout.write(\"\\n%d memory locations are reserved for ten 'int' elements\"%(sys.getsizeof(c)))\n",
+      "sys.stdout.write(\"\\n%d memory locations are reserved for ten 'int' elements\"%(sys.getsizeof(l)))\n",
+      "\n",
+      "#python data types are different from C"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The type 'int' requires 436 Bytes\n",
+        "The type 'char' requires 436 Bytes\n",
+        "The type 'long' requires 436 Bytes\n",
+        "12 memory locations are reserved for ten 'int' elements\n",
+        "100 memory locations are reserved for ten 'int' elements\n",
+        "100 memory locations are reserved for ten 'int' elements"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.2, Page number: 187<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display character array with their address.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "name = ['A','R','R','A','Y']\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nCharacter Memory Location\\n\")\n",
+      "while i < len(name):\n",
+      "    sys.stdout.write(\"\\n[%c]\\t\\t[%d]\"%(name[i],id(name[i])))\n",
+      "    i += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Character Memory Location\n",
+        "\n",
+        "[A]\t\t[31712968]\n",
+        "[R]\t\t[31527832]\n",
+        "[R]\t\t[31527832]\n",
+        "[A]\t\t[31712968]\n",
+        "[Y]\t\t[31712248]"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.3, Page number: 188<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find sum of even and odd numbers from 1 to 10.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "sumo = 0\n",
+      "sume = 0\n",
+      "i = 0\n",
+      "a = -1\n",
+      "b = -1\n",
+      "\n",
+      "odd = [0 for i in range(0,5)]\n",
+      "even = [0 for i in range(0,5)]\n",
+      "\n",
+      "#Calculation\n",
+      "for i in range(1,10+1):\n",
+      "    if i%2 == 0:\n",
+      "        a += 1\n",
+      "        even[a] = i\n",
+      "    else:\n",
+      "        b += 1\n",
+      "        odd[b] = i\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n\\tEven  \\t\\tOdd\")\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"\\n\\t %d\\t\\t %d\"%(even[i],odd[i]))\n",
+      "    sume = sume + even[i]\n",
+      "    sumo = sumo + odd[i]\n",
+      "\n",
+      "sys.stdout.write(\"\\n\\t==========================\\n\")\n",
+      "sys.stdout.write(\"Addition : %d %14d\"%(sume,sumo))\n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\tEven  \t\tOdd\n",
+        "\t 2\t\t 1\n",
+        "\t 4\t\t 3\n",
+        "\t 6\t\t 5\n",
+        "\t 8\t\t 7\n",
+        "\t 10\t\t 9\n",
+        "\t==========================\n",
+        "Addition : 30             25"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.4, Page number: 189<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find sum of even numbers and product of odd numbers from 5 input values.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 0\n",
+      "m = 1\n",
+      "num = [0 for i in range(0,5)]\n",
+      "\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"\\nEnter Number [%d] : \"%(i+1))\n",
+      "    num[i] = int(raw_input(\"\"))\n",
+      "\n",
+      "#Calculation and Result\n",
+      "sys.stdout.write(\"\\n==================================\")\n",
+      "for i in range(0,5):\n",
+      "    if num[i]%2 == 0:\n",
+      "        sys.stdout.write(\"\\nEven Number : %d\"%(num[i]))\n",
+      "        a = a + num[i]\n",
+      "    else:\n",
+      "        sys.stdout.write(\"\\nOdd Number : %d\"%(num[i]))\n",
+      "        m = m * num[i]\n",
+      "\n",
+      "sys.stdout.write(\"\\n==================================\")\n",
+      "sys.stdout.write(\"\\nAddition of Even Numbers : %d\"%(a))\n",
+      "sys.stdout.write(\"\\nProduct of Odd Numbers : %d\"%(m))\n",
+      "sys.stdout.write(\"\\n==================================\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Number [1] : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Number [2] : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Number [3] : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Number [4] : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Number [5] : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "==================================\n",
+        "Odd Number : 1\n",
+        "Even Number : 2\n",
+        "Odd Number : 3\n",
+        "Even Number : 4\n",
+        "Odd Number : 5\n",
+        "==================================\n",
+        "Addition of Even Numbers : 6\n",
+        "Product of Odd Numbers : 15\n",
+        "=================================="
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.5, Page number: 190<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print string in the reverse order.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "s = ['0' for i in range(0,15)]\n",
+      "\n",
+      "s = raw_input(\"Enter a String : \\n\")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Reverse String : \\n\")\n",
+      "\n",
+      "#Since there is no null terminating character for string in python, length of string is taken to give the index range\n",
+      "#index range is given -1 to include the index 0 also.\n",
+      "\n",
+      "for i in range(len(s)-1,-1,-1):                       \n",
+      "    sys.stdout.write(\"%c\"%s[i])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a String : \n",
+        "HELLO\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reverse String : \n",
+        "OLLEH"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.6, Page number: 191<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Detect the occurrence of a character in a given string.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "s = ['0' for i in range(0,15)]\n",
+      "c = 0\n",
+      "\n",
+      "s = raw_input(\"Enter a String : \")\n",
+      "f = raw_input(\"Enter a Character to Find :\")\n",
+      "\n",
+      "#Calculation\n",
+      "for i in range(0,len(s)):\n",
+      "    if s[i] == f:\n",
+      "        c += 1\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"The Character (%c) in a String (%s) occurs %d times.\"%(f,s,c))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a String : programmer\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Character to Find :r\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Character (r) in a String (programmer) occurs 3 times."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.7, Page number: 192<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the elements of two arrays in two separate columns and add their\n",
+      "#corresponding elements and display in the 3rd column.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "num = [24,34,12,44,56,17]\n",
+      "num1 = [12,24,35,78,85,22]\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Element of Array 1st - 2nd Array  Addition\\n\")\n",
+      "for i in range(0,5+1):\n",
+      "    sys.stdout.write(\"\\n\\t\\t%d  +  %d  =   %d\"%(num[i],num1[i],num[i]+num1[i]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Element of Array 1st - 2nd Array  Addition\n",
+        "\n",
+        "\t\t24  +  12  =   36\n",
+        "\t\t34  +  24  =   58\n",
+        "\t\t12  +  35  =   47\n",
+        "\t\t44  +  78  =   122\n",
+        "\t\t56  +  85  =   141\n",
+        "\t\t17  +  22  =   39"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.8, Page number: 192<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find sum of 3 numbers by entering as character type.\n",
+      "\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "real = [['0' for i in range(0,6)] for i in range(0,6)]\n",
+      "ima = [['0' for i in range(0,6)] for i in range(0,6)]\n",
+      "\n",
+      "for l in range(0,3):\n",
+      "    c = int(raw_input(\"Enter Number : \"))\n",
+      "    r = c/255\n",
+      "    i = c%255\n",
+      "    real[l][5] = r\n",
+      "    ima[l][5] = i\n",
+      "\n",
+      "c = 0\n",
+      "\n",
+      "for l in range(0,3):\n",
+      "    c = c + real[l][5]*255 + ima[l][5]\n",
+      "\n",
+      "sys.stdout.write(\"\\nSum of 3 Numbers : %3ld\"%(c))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number : 3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of 3 Numbers :  12"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.9, Page number: 193<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Accept any string upto 15 characters.  Display the elements of string with their\n",
+      "#element numbers in a separate column.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "#name = ['0' for i in range(0,15)]\n",
+      "name = raw_input(\"Enter your name : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Element no. & Character\\n\")\n",
+      "for i in range(0,len(name)):                              #There is no null termination character for string in python\n",
+      "    sys.stdout.write(\"\\n%d \\t %c\"%(i,name[i])) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your name : amit\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Element no. & Character\n",
+        "\n",
+        "0 \t a\n",
+        "1 \t m\n",
+        "2 \t i\n",
+        "3 \t t"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.10, Page number: 194<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display names of days of a week \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "day = [0 for i in range(0,7)]\n",
+      "\n",
+      "sys.stdout.write(\"Enter numbers between 1 to 7 : \")\n",
+      "for i in range(0,6+1):\n",
+      "    day[i] = int(raw_input(\"\"))\n",
+      "\n",
+      "#Result\n",
+      "#There is no switch statement in python. use functions and dictionary or if statement instead.\n",
+      "for i in range(0,6+1):\n",
+      "    if day[i] == 1:\n",
+      "        sys.stdout.write(\"\\n%dst day of week is Sunday\"%(day[i]))\n",
+      "    else:\n",
+      "        if day[i] == 2:\n",
+      "            sys.stdout.write(\"\\n%dnd day of week is Monday\"%(day[i]))\n",
+      "        else:\n",
+      "            if day[i] == 3:\n",
+      "                sys.stdout.write(\"\\n%drd day of week is Tuesday\"%(day[i]))\n",
+      "            else:\n",
+      "                if day[i] == 4:\n",
+      "                    sys.stdout.write(\"\\n%dth day of week is Wednesday\"%(day[i]))\n",
+      "                else:\n",
+      "                    if day[i] == 5:\n",
+      "                        sys.stdout.write(\"\\n%dth day of week is Thursday\"%(day[i]))\n",
+      "                    else:\n",
+      "                        if day[i] == 6:\n",
+      "                            sys.stdout.write(\"\\n%dth day of week is Friday\"%(day[i]))\n",
+      "                        else:\n",
+      "                            if day[i] == 7:\n",
+      "                                sys.stdout.write(\"\\n%dth day of week is Saturday\"%(day[i]))\n",
+      "                            else:\n",
+      "                                sys.stdout.write(\"\\n%dth is invalid day.\"%(day[i]))\n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers between 1 to 7 : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1st day of week is Sunday\n",
+        "3rd day of week is Tuesday\n",
+        "2nd day of week is Monday\n",
+        "4th day of week is Wednesday\n",
+        "5th day of week is Thursday\n",
+        "7th day of week is Saturday\n",
+        "8th is invalid day."
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.11, Page number: 195<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the contents of two arrays where the 1st array should contain the string and 2nd numerical numbers.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "city = ['N','A','N','D','E','D']\n",
+      "pin = [4,3,1,6,0,3]\n",
+      "\n",
+      "#Result\n",
+      "for i in city:\n",
+      "    sys.stdout.write(\"%c\"%(i))\n",
+      "\n",
+      "sys.stdout.write(\" - \")\n",
+      "for i in pin:\n",
+      "    sys.stdout.write(\"%d\"%(i))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NANDED - 431603"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.12, Page number: 195<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the number of days of different months of year.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "month = [31,28,31,30,31,30,31,31,30,31,30,31]\n",
+      "\n",
+      "#Result\n",
+      "for i in range(0,11+1):\n",
+      "    sys.stdout.write(\"\\nMonth [%d] of a year contains %d days.\"%(i+1,month[i]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Month [1] of a year contains 31 days.\n",
+        "Month [2] of a year contains 28 days.\n",
+        "Month [3] of a year contains 31 days.\n",
+        "Month [4] of a year contains 30 days.\n",
+        "Month [5] of a year contains 31 days.\n",
+        "Month [6] of a year contains 30 days.\n",
+        "Month [7] of a year contains 31 days.\n",
+        "Month [8] of a year contains 31 days.\n",
+        "Month [9] of a year contains 30 days.\n",
+        "Month [10] of a year contains 31 days.\n",
+        "Month [11] of a year contains 30 days.\n",
+        "Month [12] of a year contains 31 days."
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.13, Page number: 197<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the number of days of given month of a year.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "month = [1,3,5,7,8,10,12,4,6,9,11,2]\n",
+      "\n",
+      "mn = int(raw_input(\"Enter Number of Month : \"))\n",
+      "\n",
+      "for i in range(0,11+1):\n",
+      "    if mn == month[i]:                #There is no goto statement in python\n",
+      "        if i+1 == 12:\n",
+      "            sys.stdout.write(\"Month (%d) Contains 28 days.\"%(month[i]))\n",
+      "        if i+1 < 8:\n",
+      "            sys.stdout.write(\"Month (%d) Contains 31 days.\"%(month[i]))\n",
+      "        if i+1 > 7 and i+1 != 12:\n",
+      "            sys.stdout.write(\"Month (%d) Contains 30 days.\"%(month[i]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number of Month : 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Month (2) Contains 28 days."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.14, Page number: 198<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the average sales of an item out of 12 months sale.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "#Since sum is a built in function in python, sum1 is used instead of sum.\n",
+      "sum1 = 0\n",
+      "avg = 0\n",
+      "\n",
+      "sys.stdout.write(\"Enter Month No.- Sale of an Item/month\\n\")\n",
+      "\n",
+      "item = [0 for i in range(0,11+1)]\n",
+      "for sale in range(0,11+1):\n",
+      "    item[sale] = int(raw_input(\"\\t%d = \"%(sale+1)))\n",
+      "\n",
+      "#Calculation\n",
+      "for sale in range(0,11+1):\n",
+      "    sum1 = sum1 + item[sale]\n",
+      "\n",
+      "avg = sum1/12.0\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAverage Sale if an item/month = %f\"%(avg))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Month No.- Sale of an Item/month\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t1 = 125\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t2 = 225\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t3 = 325\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t4 = 425\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t5 = 525\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t6 = 625\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t7 = 725\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t8 = 825\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t9 = 925\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t10 = 500\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t11 = 600\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t12 = 700\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Average Sale if an item/month = 543.750000"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.15, Page number: 199<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the similar elements in an array and find the occurrence of similar\n",
+      "#numbers for number of times.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "k = 0\n",
+      "j = 1\n",
+      "\n",
+      "item = [0 for i in range(0,4+1)]\n",
+      "sys.stdout.write(\"\\nEnter Numbers.\\n\")\n",
+      "for i in range(0,4+1):\n",
+      "    item[i] = int(raw_input(\"%d = \"%(i)))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "for j in range(0,4+1):\n",
+      "    for i in range(j+1,4+1):\n",
+      "        if item[j] == item[i]:\n",
+      "            sys.stdout.write(\"\\n%d %d Elements are same.\"%(j,i))\n",
+      "            k += 1\n",
+      "\n",
+      "if k == 10:\n",
+      "    sys.stdout.write(\"\\nAll elements are same\")\n",
+      "else:\n",
+      "    if k == 0:\n",
+      "        sys.stdout.write(\"\\nNo Elements are same.\")\n",
+      "    else:\n",
+      "        sys.stdout.write(\"\\nElements Contains Double Numbers.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Enter Numbers.\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0 = 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 = 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2 = 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3 = 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4 = 9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "0 3 Elements are same.\n",
+        "Elements Contains Double Numbers."
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.16, Page number: 200<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate and display total cost of 4 models of Pentium PCs.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "pccode = [1,2,3,4]\n",
+      "t = 0\n",
+      "price = [25000,30000,35000,40000]\n",
+      "stock = [25,20,15,20]\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nStock & Total Cost Details\")\n",
+      "sys.stdout.write(\"\\n========================================================\")\n",
+      "sys.stdout.write(\"\\nModel\\t\\tQty.\\tRate(Rs.)\\tTotal Value\")\n",
+      "sys.stdout.write(\"\\n========================================================\")\n",
+      "\n",
+      "for i in range(0,3+1):\n",
+      "    sys.stdout.write(\"\\nPentium%d\\t%d\\t%8ld  %15ld\"%(pccode[i],stock[i],price[i],price[i]*stock[i]))\n",
+      "    t = t + price[i] * stock[i]\n",
+      "\n",
+      "sys.stdout.write(\"\\n========================================================\")\n",
+      "sys.stdout.write(\"\\nTotal Value of All PCs in Rs. %ld\"%(t))\n",
+      "sys.stdout.write(\"\\n========================================================\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Stock & Total Cost Details\n",
+        "========================================================\n",
+        "Model\t\tQty.\tRate(Rs.)\tTotal Value\n",
+        "========================================================\n",
+        "Pentium1\t25\t   25000           625000\n",
+        "Pentium2\t20\t   30000           600000\n",
+        "Pentium3\t15\t   35000           525000\n",
+        "Pentium4\t20\t   40000           800000\n",
+        "========================================================\n",
+        "Total Value of All PCs in Rs. 2550000\n",
+        "========================================================"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.17, Page number: 201<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the given message by using putc() and stdout() functions.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "msg = \"C is Easy\"\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "#There is no putc function and null termination in python.\n",
+      "while i < len(msg):\n",
+      "    sys.stdout.write(\"%c\"%(msg[i]))\n",
+      "    i += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C is Easy"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.18, Page number: 202<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the text through keyboard and display it by using\n",
+      "#stdin and stdout streams.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "ch = raw_input(\"Input a Text : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The Text Entered was\\n\")\n",
+      "\n",
+      "for i in ch:\n",
+      "    sys.stdout.write(\"%c\"%(i))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Input a Text : Hello World\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Text Entered was\n",
+        "Hello World"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.19, Page number: 202<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sort the given strings alphabetically.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = raw_input(\"Enter Text Below : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Sorted Text Below : \\n\")\n",
+      "for i in range(65,90+1):\n",
+      "    for j in range(0,len(text)):\n",
+      "        if text[j] == chr(i).upper() or text[j] == chr(i).lower():\n",
+      "            sys.stdout.write(\"%c\"%(text[j]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Below : Hello\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorted Text Below : \n",
+        "eHllo"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.20, Page number: 203<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Arrange the numbers in increasing and decreasing order\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "sys.stdout.write(\"Enter ten numbers : \")\n",
+      "a = [0 for i in range(0,10)]\n",
+      "\n",
+      "for i in range(0,10):\n",
+      "    a[i] = int(raw_input(\"\"))\n",
+      "sum1 = 0\n",
+      "\n",
+      "#Calculation\n",
+      "sum1 = sum(a)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Numbers In Ascending Order : \")\n",
+      "for i in range(0,sum1):\n",
+      "    for j in range(0,10):\n",
+      "        if i == a[j]:\n",
+      "            sys.stdout.write(\"%3d\"%(a[j]))\n",
+      "        \n",
+      "sys.stdout.write(\"\\nNumbers In Descending Order : \")\n",
+      "for i in range(sum1,0,-1):\n",
+      "    for j in range(0,10):\n",
+      "        if i == a[j]:\n",
+      "            sys.stdout.write(\"%3d\"%(a[j]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter ten numbers : "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "12\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numbers In Ascending Order :   1  2  3  4  5  7  9  9 10 12\n",
+        "Numbers In Descending Order :  12 10  9  9  7  5  4  3  2  1"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.21, Page number: 205<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sorting in ascending order by comparison method.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "n = int(raw_input(\"Enter how many numbers to sort : \"))\n",
+      "\n",
+      "num = [0 for i in range(0,n)]\n",
+      "for i in range(0,n):\n",
+      "    num[i] = int(raw_input(\"Enter numbers # %2d  : \"%(i+1)))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The Numbers Entered through keyboard\\n\")\n",
+      "for i in range(0,n):\n",
+      "    sys.stdout.write(\"%4d\"%(num[i]))\n",
+      "for i in range(0,n-1):\n",
+      "    for j in range(i+1,n):\n",
+      "        if num[i] > num[j]:\n",
+      "            temp = num[i]\n",
+      "            num[i] = num[j]\n",
+      "            num[j] = temp\n",
+      "\n",
+      "sys.stdout.write(\"\\nThe Numbers in ascending order\\n\")\n",
+      "for i in num:\n",
+      "    sys.stdout.write(\"%4d\"%(i))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter how many numbers to sort : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers #  1  : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers #  2  : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers #  3  : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers #  4  : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter numbers #  5  : 9\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Numbers Entered through keyboard\n",
+        "   8   3   2   1   9\n",
+        "The Numbers in ascending order\n",
+        "   1   2   3   8   9"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.22, Page number: 206<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Evaluate the series contains sum of square of\n",
+      "#numbers from 1 to n.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "s = 0\n",
+      "sqr = [0 for i in range(0,15)]\n",
+      "\n",
+      "n = int(raw_input(\"Enter value of n : \"))\n",
+      "\n",
+      "for i in range(0,n):\n",
+      "    sqr[i] = pow(i+1,2)\n",
+      "    \n",
+      "for i in range(0,n):\n",
+      "    sys.stdout.write(\"%d\\n\"%(sqr[i]))\n",
+      "    s = s + sqr[i]\n",
+      "    \n",
+      "sys.stdout.write(\"Sum of square : %d\"%(s))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter value of n : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n",
+        "4\n",
+        "9\n",
+        "16\n",
+        "Sum of square : 30"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.23, Page number: 207<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display two dimensional array elements together with their addresses\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[0 for i in range(0,3)]for i in range(0,3)]\n",
+      "c = 1\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,3):\n",
+      "        a[i][j] = c\n",
+      "        c += 1\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"Array Elements and addresses.\\n\")\n",
+      "sys.stdout.write(\"             Col-0        Col-1         Col-2\\n\")\n",
+      "sys.stdout.write(\"========================================================\")\n",
+      "sys.stdout.write(\"\\nRow0\")\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,3):\n",
+      "        sys.stdout.write(\"      %d [%5d]\"%(a[i][j],id(a[i][j])))\n",
+      "    sys.stdout.write(\"\\nRow%d\"%(i+1))\n",
+      "\n",
+      "sys.stdout.write(\"\\r\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Array Elements and addresses.\n",
+        "             Col-0        Col-1         Col-2\n",
+        "========================================================\n",
+        "Row0      1 [6215912]      2 [6215900]      3 [6215888]\n",
+        "Row1      4 [6215876]      5 [6215864]      6 [6215852]\n",
+        "Row2      7 [6215840]      8 [6215828]      9 [6215816]\n",
+        "Row3\r"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.24, Page number: 208<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the elements of two-dimensional array.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[1,2,3],[4,5,6],[7,8,9]]\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"Elements of An Array.\\n\")\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,3):\n",
+      "        sys.stdout.write(\"%5d\"%(a[i][j]))\n",
+      "    sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements of An Array.\n",
+        "    1    2    3\n",
+        "    4    5    6\n",
+        "    7    8    9\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.25, Page number: 209<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read codenos and balance and count the number of codenos, \n",
+      "#which are having the balance less than 1000.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "j = 0\n",
+      "bal = [[0 for i in range(0,5)]for i in range(0,5)]\n",
+      "\n",
+      "for i in range(0,5):\n",
+      "    bal[i][1] = int(raw_input(\"Enter Code No & Balance : \"))\n",
+      "    bal[i][2] = int(raw_input(\"Enter Code No & Balance : \"))\n",
+      "    \n",
+      "sys.stdout.write(\"Your Entered Data : \")\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"\\n%d   %d\"%(bal[i][1],bal[i][2]))\n",
+      "    if bal[i][2] < 1000:\n",
+      "        j += 1\n",
+      "        \n",
+      "sys.stdout.write(\"\\nBalance Less than 1000 are %d\"%(j))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 900\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 800\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 1200\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 550\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Code No & Balance : 600\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your Entered Data : \n",
+        "1   900\n",
+        "2   800\n",
+        "3   1200\n",
+        "4   550\n",
+        "5   600\n",
+        "Balance Less than 1000 are 4"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.26, Page number: 210<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Accept three elements in a one dimensional array and compute addition,square\n",
+      "#and cube.  Display the results in two dimensional array.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[0 for i in range(0,3)]for i in range(0,3)]\n",
+      "b = [0 for i in range(0,3)]\n",
+      "j = 0\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    b[i] = int(raw_input(\"Enter Three Numbers : \"))\n",
+      "\n",
+      "#Calculation\n",
+      "for i in range(0,3):\n",
+      "    a[i][j] = b[i]*2\n",
+      "    a[i][j+1] = pow(b[i],2)\n",
+      "    a[i][j+2] = pow(b[i],3)\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"Number\\tAddition\\tSquare\\t\\tCube\\n\")\n",
+      "for i in range(0,3):\n",
+      "    sys.stdout.write(\"\\n%d\"%(b[i]))\n",
+      "    for j in range(0,3):\n",
+      "        sys.stdout.write(\"\\t%4d\\t\"%(a[i][j]))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three Numbers : 7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number\tAddition\tSquare\t\tCube\n",
+        "\n",
+        "5\t  10\t\t  25\t\t 125\t\n",
+        "\n",
+        "6\t  12\t\t  36\t\t 216\t\n",
+        "\n",
+        "7\t  14\t\t  49\t\t 343\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.27, Page number: 211<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read and display matrix\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
+      "row = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
+      "col = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
+      "\n",
+      "for i in range(0,row):\n",
+      "    for j in range(0,col):\n",
+      "        a[i][j]  = int(raw_input(\"Enter Elements of matrix A : \"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nThe Matrix is : \\n\")\n",
+      "for i in range(0,row):\n",
+      "    for j in range(0,col):\n",
+      "        sys.stdout.write(\"%6d\"%(a[i][j]))\n",
+      "    sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of matrix upto (10x10) A : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of matrix upto (10x10) A : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Matrix is : \n",
+        "     3     5     8\n",
+        "     4     8     5\n",
+        "     8     5     4\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.28, Page number: 212<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Transpose of matrix\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
+      "b = [[0 for i in range(0,10)]for i in range(0,10)]\n",
+      "row1 = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
+      "col1 = int(raw_input(\"Enter Order of matrix upto (10x10) A : \"))\n",
+      "\n",
+      "for i in range(0,row1):\n",
+      "    for j in range(0,col1):\n",
+      "        a[i][j]  = int(raw_input(\"Enter Elements of matrix A : \"))\n",
+      "row2 = col1\n",
+      "col2 = row1\n",
+      "\n",
+      "#Calculation\n",
+      "for i in range(0,row1):\n",
+      "    for j in range(0,col1):\n",
+      "        b[j][i] = a[i][j]\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nThe Matrix is : \\n\")\n",
+      "for i in range(0,row2):\n",
+      "    for j in range(0,col2):\n",
+      "        sys.stdout.write(\"%4d\"%(b[i][j]))\n",
+      "    sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of matrix upto (10x10) A : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of matrix upto (10x10) A : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix A : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Matrix is : \n",
+        "   3   4   8\n",
+        "   5   8   5\n",
+        "   8   5   4\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.29, Page number: 213<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Transpose of 3x3 matrix\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "mat = [[0 for i in range(0,3)]for i in range(0,3)]\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,3):\n",
+      "        mat[i][j]  = int(raw_input(\"Enter Elements of matrix \"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Transpose of the matrix\\n\")\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,3):\n",
+      "        sys.stdout.write(\"\\t%d\"%(mat[j][i]))\n",
+      "    sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of matrix 2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Transpose of the matrix\n",
+        "\t5\t2\t7\n",
+        "\t8\t5\t5\n",
+        "\t9\t4\t2\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.30, Page number: 214<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Addition and subtraction of two matrices \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
+      "b = [[0 for i in range(0,10)]for i in range(0,10)]\n",
+      "\n",
+      "r1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
+      "c1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
+      "\n",
+      "for i in range(0,r1):\n",
+      "    for j in range(0,c1):\n",
+      "        a[i][j]  = int(raw_input(\"Enter Elements of Matrix A:\"))\n",
+      "\n",
+      "for i in range(0,r1):\n",
+      "    for j in range(0,c1):\n",
+      "        b[i][j]  = int(raw_input(\"Enter Elements of Matrix B:\"))\n",
+      "\n",
+      "#Calculation and Result\n",
+      "sys.stdout.write(\"\\nMatrix Addition\\n\")\n",
+      "for i in range(0,r1):\n",
+      "    for j in range(0,c1):\n",
+      "        sys.stdout.write(\"%5d\"%(a[i][j]+b[i][j]))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "\n",
+      "sys.stdout.write(\"\\nMatrix Subtraction\\n\")\n",
+      "for i in range(0,r1):\n",
+      "    for j in range(0,c1):\n",
+      "        sys.stdout.write(\"%5d\"%(a[i][j]-b[i][j]))\n",
+      "    sys.stdout.write(\"\\n\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of Matrix A & B upto 10x103\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of Matrix A & B upto 10x103\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Matrix Addition\n",
+        "    5    8   13\n",
+        "    2   14   12\n",
+        "    8   16    6\n",
+        "\n",
+        "Matrix Subtraction\n",
+        "    3    2    3\n",
+        "    2    4    4\n",
+        "   -4    2    2\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.31, Page number: 216<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Multiplication of two matrices \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[0 for i in range(0,10)]for i in range(0,10)]\n",
+      "b = [[0 for i in range(0,10)]for i in range(0,10)]\n",
+      "\n",
+      "r1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
+      "c1 = int(raw_input(\"Enter Order of Matrix A & B upto 10x10\"))\n",
+      "\n",
+      "for i in range(0,r1):\n",
+      "    for j in range(0,c1):\n",
+      "        a[i][j]  = int(raw_input(\"Enter Elements of Matrix A:\"))\n",
+      "\n",
+      "for i in range(0,r1):\n",
+      "    for j in range(0,c1):\n",
+      "        b[i][j]  = int(raw_input(\"Enter Elements of Matrix B:\"))\n",
+      "\n",
+      "#Calculation and Result\n",
+      "sys.stdout.write(\"\\nMultiplication of corresponding elements\\n\")\n",
+      "for i in range(0,r1):\n",
+      "    for j in range(0,c1):\n",
+      "        sys.stdout.write(\"%5d\"%(a[i][j]*b[i][j]))\n",
+      "    sys.stdout.write(\"\\n\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of Matrix A & B upto 10x103\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Order of Matrix A & B upto 10x103\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix A:4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Elements of Matrix B:2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Multiplication of corresponding elements\n",
+        "    4   15   40\n",
+        "    0   45   32\n",
+        "   12   63    8\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.32, Page number: 217<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the Quantity & Price of various Pentium models using an array and\n",
+      "#Compute the total cost of all models.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "pc = [[0 for i in range(0,4)]for i in range(0,4)]\n",
+      "t = 0\n",
+      "\n",
+      "for i in range(0,4):\n",
+      "    for j in range(0,2):\n",
+      "        pc[i][j] = int(raw_input(\"Enter Qty. & Price for Pentium %d : \"%(i+1)))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"=====================================================\")\n",
+      "sys.stdout.write(\"\\nModel\\tQty.\\tRate(Rs.)\\tTotal Value\")\n",
+      "sys.stdout.write(\"\\n=====================================================\")\n",
+      "for i in range(0,4):\n",
+      "    sys.stdout.write(\"\\nPentium %d   %2ld %6ld %15ld\"%(i+1,pc[i][0],pc[i][1],pc[i][0]*pc[i][1]))\n",
+      "    t = t + pc[i][0] * pc[i][1]\n",
+      "\n",
+      "sys.stdout.write(\"\\n=====================================================\")\n",
+      "sys.stdout.write(\"\\nTotal Value of All PCs in Rs. %ld\"%(t))\n",
+      "sys.stdout.write(\"\\n=====================================================\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 1 : 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 1 : 25000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 2 : 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 2 : 40000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 3 : 15\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 3 : 35000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 4 : 20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Qty. & Price for Pentium 4 : 40000\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "=====================================================\n",
+        "Model\tQty.\tRate(Rs.)\tTotal Value\n",
+        "=====================================================\n",
+        "Pentium 1   25  25000          625000\n",
+        "Pentium 2   20  40000          800000\n",
+        "Pentium 3   15  35000          525000\n",
+        "Pentium 4   20  40000          800000\n",
+        "=====================================================\n",
+        "Total Value of All PCs in Rs. 2750000\n",
+        "====================================================="
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.33, Page number: 218<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the capacity of HD, it's price and quantity available in the form of an array\n",
+      "# and compute the cost of HD\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "hd = [[0 for i in range(0,4)]for i in range(0,4)]\n",
+      "t = 0\n",
+      "\n",
+      "for j in range(0,4):\n",
+      "    for i in range(0,3):\n",
+      "        hd[i][j] = int(raw_input(\"Enter Capacity, Price & Qty.:\"))\n",
+      "        \n",
+      "sys.stdout.write(\"====================================================================\\n\")\n",
+      "sys.stdout.write(\"HD Capacity\\tGB Price Rs.\\tQuantity Total Value Rs.\")\n",
+      "sys.stdout.write(\"\\n====================================================================\\n\")\n",
+      "\n",
+      "for j in range(0,4):\n",
+      "    for i in range(0,3):\n",
+      "        sys.stdout.write(\"%2ld\"%(hd[i][j]))\n",
+      "        sys.stdout.write(\"\\t\\t\")\n",
+      "        if i == 2:\n",
+      "            sys.stdout.write(\"%5ld\"%(hd[i-1][j]*hd[i][j]))\n",
+      "            t = t + hd[i-1][j] * hd[i][j]\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    \n",
+      "sys.stdout.write(\"====================================================================\")\n",
+      "sys.stdout.write(\"\\nTotal Value Rs. %37ld\"%(t))\n",
+      "sys.stdout.write(\"\\n====================================================================\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:8000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:12000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:15000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:15\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:90\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:20000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Capacity, Price & Qty.:10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "====================================================================\n",
+        "HD Capacity\tGB Price Rs.\tQuantity Total Value Rs.\n",
+        "====================================================================\n",
+        "10\t\t8000\t\t25\t\t200000\n",
+        "20\t\t12000\t\t20\t\t240000\n",
+        "40\t\t15000\t\t15\t\t225000\n",
+        "90\t\t20000\t\t10\t\t200000\n",
+        "====================================================================\n",
+        "Total Value Rs.                                865000\n",
+        "===================================================================="
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.34, Page number: 220<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the names of the cities with their base addresses.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "city = [\"Mumbai\",\"Chennai\",\"Kolkata\",\"Pune\",\"Delhi\"]\n",
+      "\n",
+      "#Result\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"Base Address = %u\"%(id(city[i])))   #Memory address will differ in different systems\n",
+      "    sys.stdout.write(\"\\tString = %s\\n\"%(city[i]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Base Address = 60549024\tString = Mumbai\n",
+        "Base Address = 60550048\tString = Chennai\n",
+        "Base Address = 60550592\tString = Kolkata\n",
+        "Base Address = 60551104\tString = Pune\n",
+        "Base Address = 60550240\tString = Delhi\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.35, Page number: 220<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the binary bits corresponding to Hexadecimal numbers from\n",
+      "#0 to E and attach an odd parity to Hex numbers and display them.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "bin = [[0 for i in range(0,16)]for i in range(0,16)]\n",
+      "a = 0\n",
+      "c = 0\n",
+      "\n",
+      "for x in range(0,16):\n",
+      "    y = x\n",
+      "    for a in range(0,4):\n",
+      "        bin[x][a] = y%2\n",
+      "        y = y/2\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nInput Bits\\t Parity Bits\")\n",
+      "sys.stdout.write(\"\\n===================================\")\n",
+      "\n",
+      "for x in range(0,16):\n",
+      "    c = 0\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    for y in range(3,0-1,-1):\n",
+      "        sys.stdout.write(\"%3d\"%(bin[x][y]))\n",
+      "        if bin[x][y] == 1:\n",
+      "            c += 1\n",
+      "    if c%2 == 0:\n",
+      "        sys.stdout.write(\"%7d\"%(1))\n",
+      "    else:\n",
+      "        sys.stdout.write(\"%7d\"%(0))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Input Bits\t Parity Bits\n",
+        "===================================\n",
+        "  0  0  0  0      1\n",
+        "  0  0  0  1      0\n",
+        "  0  0  1  0      0\n",
+        "  0  0  1  1      1\n",
+        "  0  1  0  0      0\n",
+        "  0  1  0  1      1\n",
+        "  0  1  1  0      1\n",
+        "  0  1  1  1      0\n",
+        "  1  0  0  0      0\n",
+        "  1  0  0  1      1\n",
+        "  1  0  1  0      1\n",
+        "  1  0  1  1      0\n",
+        "  1  1  0  0      1\n",
+        "  1  1  0  1      0\n",
+        "  1  1  1  0      0\n",
+        "  1  1  1  1      1"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.36, Page number: 222<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert binary to gray codes.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "bin1 = [[0 for i in range(0,16)]for i in range(0,16)]\n",
+      "a = 0\n",
+      "c = 0\n",
+      "\n",
+      "#Calculation\n",
+      "for x in range(0,16):\n",
+      "    y = x\n",
+      "    for a in range(0,4):\n",
+      "        bin1[x][a] = y%2\n",
+      "        y = y/2\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nBinary Bits\\tGray Bits\")\n",
+      "sys.stdout.write(\"\\n==================================\")\n",
+      "\n",
+      "for x in range(0,16):\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    for y in range(3,0-1,-1):\n",
+      "        sys.stdout.write(\"%3d\"%(bin1[x][y]))\n",
+      "    sys.stdout.write(\"%5d   %2d   %2d   %2d\"%(bin1[x][3],(bin1[x][3]^bin1[x][2]),(bin1[x][2]^bin1[x][1]),(bin1[x][1]^bin1[x][0])))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Binary Bits\tGray Bits\n",
+        "==================================\n",
+        "  0  0  0  0    0    0    0    0\n",
+        "  0  0  0  1    0    0    0    1\n",
+        "  0  0  1  0    0    0    1    1\n",
+        "  0  0  1  1    0    0    1    0\n",
+        "  0  1  0  0    0    1    1    0\n",
+        "  0  1  0  1    0    1    1    1\n",
+        "  0  1  1  0    0    1    0    1\n",
+        "  0  1  1  1    0    1    0    0\n",
+        "  1  0  0  0    1    1    0    0\n",
+        "  1  0  0  1    1    1    0    1\n",
+        "  1  0  1  0    1    1    1    1\n",
+        "  1  0  1  1    1    1    1    0\n",
+        "  1  1  0  0    1    0    1    0\n",
+        "  1  1  0  1    1    0    1    1\n",
+        "  1  1  1  0    1    0    0    1\n",
+        "  1  1  1  1    1    0    0    0"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.37, Page number: 224<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#working of three dimensional array.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "array_3d = [[[0 for i in range(0,3)]for i in range(0,3)]for i in range(0,3)]\n",
+      "\n",
+      "for a in range(0,3):\n",
+      "    for b in range(0,3):\n",
+      "        for c in range(0,3):\n",
+      "            array_3d[a][b][c] = a+b+c\n",
+      "\n",
+      "#Result\n",
+      "for a in range(0,3):\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    for b in range(0,3):\n",
+      "        for c in range(0,3):\n",
+      "            sys.stdout.write(\"%3d\"%(array_3d[a][b][c]))\n",
+      "        sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "  0  1  2\n",
+        "  1  2  3\n",
+        "  2  3  4\n",
+        "\n",
+        "  1  2  3\n",
+        "  2  3  4\n",
+        "  3  4  5\n",
+        "\n",
+        "  2  3  4\n",
+        "  3  4  5\n",
+        "  4  5  6\n"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.38, Page number: 225<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#working of four dimensional array\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "array_4d = [[[[0 for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]\n",
+      "\n",
+      "for a in range(0,2):\n",
+      "    for b in range(0,2):\n",
+      "        for c in range(0,2):\n",
+      "            for d in range(0,2):\n",
+      "                array_4d[a][b][c][d] = a+b+c+d\n",
+      "                \n",
+      "#Result\n",
+      "for a in range(0,2):\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    for b in range(0,2):\n",
+      "        for c in range(0,2):\n",
+      "            for d in range(0,2):\n",
+      "                sys.stdout.write(\"%3d\"%(array_4d[a][b][c][d]))\n",
+      "            sys.stdout.write(\"\\n\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "  0  1\n",
+        "  1  2\n",
+        "  1  2\n",
+        "  2  3\n",
+        "\n",
+        "  1  2\n",
+        "  2  3\n",
+        "  2  3\n",
+        "  3  4\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.39, Page number: 226<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read the quantity & rate of certain items using multidimensional\n",
+      "#array.  Calculate total cost by multiplying quantity & rate and offer 2%\n",
+      "#discount on it & display the net amount.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "m = [[[[0 for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]for i in range(0,2)]\n",
+      "\n",
+      "for a in range(0,2):\n",
+      "    for b in range(0,2):\n",
+      "        for c in range(0,2):\n",
+      "            for d in range(0,1):\n",
+      "                if a == 0:\n",
+      "                    m[a][b][c][d] = int(raw_input(\"Enter Quantity & Rate\\na=%d b=%d c=%d d=%d\"%(a,b,c,d)))\n",
+      "                    m[a][b][c][d+1] = int(raw_input(\"Enter Quantity & Rate\\na=%d b=%d c=%d d=%d\"%(a,b,c,d)))\n",
+      "                else:\n",
+      "                    m[a][b][c][d] = m[a-1][b][c][d] * m[a-1][b][c][d+1]\n",
+      "                    m[a][b][c][d+1] = m[a-1][b][c][d] * m[a-1][b][c][d+1] *2/100\n",
+      "            \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n============================================================\\n\")\n",
+      "sys.stdout.write(\"Quantity\\tRate\\tAmoount\\tDiscount(@2%)\\tNet Amount\\n\")\n",
+      "sys.stdout.write(\"============================================================\\n\")\n",
+      "\n",
+      "for a in range(0,1):\n",
+      "    for b in range(0,2):\n",
+      "        for c in range(0,2):\n",
+      "            for d in range(0,1):\n",
+      "                sys.stdout.write(\"\\n%10ld %10ld\"%(m[a][b][c][d],m[a][b][c][d+1]))\n",
+      "                sys.stdout.write(\"%8ld.00 %6ld.00 %12ld.00\"%(m[a+1][b][c][d],m[a+1][b][c][d+1],m[a+1][b][c][d] - m[a+1][b][c][d+1]))\n",
+      "            \n",
+      "sys.stdout.write(\"\\n============================================================\")\n",
+      "                "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=0 c=0 d=025\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=0 c=0 d=050\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=0 c=1 d=030\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=0 c=1 d=060\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=1 c=0 d=035\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=1 c=0 d=070\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=1 c=1 d=040\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity & Rate\n",
+        "a=0 b=1 c=1 d=075\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "============================================================\n",
+        "Quantity\tRate\tAmoount\tDiscount(@2%)\tNet Amount\n",
+        "============================================================\n",
+        "\n",
+        "        25         50    1250.00     25.00         1225.00\n",
+        "        30         60    1800.00     36.00         1764.00\n",
+        "        35         70    2450.00     49.00         2401.00\n",
+        "        40         75    3000.00     60.00         2940.00\n",
+        "============================================================"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.40, Page number: 227<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read string from a character array.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "in1 = raw_input(\"\")   #Since in is a keyword in python, in1 is used\n",
+      "\n",
+      "#Result\n",
+      "#There is no sscanf() function in python\n",
+      "out = in1\n",
+      "sys.stdout.write(\"%s\\n\"%(out))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "HELLO\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.41, Page number: 228<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Read integers in character array, convert and assign them to integer\n",
+      "#variable.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable initialization\n",
+      "in1 = raw_input(\"Enter Integers : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Value of int x : %s\"%(in1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Integers : 123\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of int x : 123"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.42, Page number: 229<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#use of sprintf()\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 10\n",
+      "c = 'C'\n",
+      "p = 3.14\n",
+      "\n",
+      "spf = str(\"%d %c %.2f\"%(a,c,p))        #There is no sprintf() function in python\n",
+      "sys.stdout.write(\"\\n%s\"%(spf))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "10 C 3.14"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7.43, Page number: 229<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Complete the string by filling the spaces with appropriate characters\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "name = \"Mas er ngA SIC\"\n",
+      "name1 = ['' for i in range(0,20)]\n",
+      "x = 0\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Mastering ANSI C\\n\")\n",
+      "\n",
+      "while x < len(name):      #There is no null terminating character in python\n",
+      "    if name[x] == ' ':\n",
+      "        n = raw_input(\"\")\n",
+      "        name1[x] = n\n",
+      "    else:\n",
+      "        name1[x] = name[x]\n",
+      "    x += 1\n",
+      "    \n",
+      "sys.stdout.write(\"%s\"%(name1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mastering ANSI C\n",
+        "['M', 'a', 's', 't', 'e', 'r', 'i', 'n', 'g', 'A', 'N', 'S', 'I', 'C', '', '', '', '', '', '']"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter8.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter8.ipynb
new file mode 100755
index 00000000..3243f3f8
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter8.ipynb
@@ -0,0 +1,2750 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 8: Working with Strings & Standard Functions<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.1, Page number: 236<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display string without considering NULL character\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "name1 = 'SANJAY'\n",
+      "#There is no null termination character in python\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Name1 = %s\"%(name1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name1 = SANJAY"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.2, Page number: 236<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display successive strings\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "name1 = 'SANJAY'\n",
+      "name2 = 'SANJAY'\n",
+      "#There is no null termination character in python\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Name1 = %s\"%name1)\n",
+      "sys.stdout.write(\"\\nName2 = %s\"%(name2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Name1 = SANJAY\n",
+        "Name2 = SANJAY"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.3, Page number: 237<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print \"WELCOME\" by using different formats of initialization \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "#in python, string initialization can be done using single quotes and double quotes\n",
+      "arr1 = 'WELCOME'\n",
+      "arr2 = \"WELCOME\"\n",
+      "arr3 = 'WELCOME'\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nArray1 = %s\"%(arr1))\n",
+      "sys.stdout.write(\"\\nArray2 = %s\"%(arr2))\n",
+      "sys.stdout.write(\"\\nArray3 = %s\"%(arr3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Array1 = WELCOME\n",
+        "Array2 = WELCOME\n",
+        "Array3 = WELCOME"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.4, Page number: 238<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the string 'PRABHAKAR' using various format specifications\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = \"PRABHAKAR\"\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"%s\\n\"%(text))\n",
+      "sys.stdout.write(\"%.5s\\n\"%(text))\n",
+      "sys.stdout.write(\"%.8s\\n\"%(text))\n",
+      "sys.stdout.write(\"%.15s\\n\"%(text))\n",
+      "sys.stdout.write(\"%-10.4s\\n\"%(text))\n",
+      "sys.stdout.write(\"%11s\\n\"%(text))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "PRABHAKAR\n",
+        "PRABH\n",
+        "PRABHAKA\n",
+        "PRABHAKAR\n",
+        "PRAB      \n",
+        "  PRABHAKAR\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.5, Page number: 239<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the elements of the character array using while loop.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = \"HAVE A NICE DAY\"\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "#There is no null character for strings in python\n",
+      "\n",
+      "while i < 15:\n",
+      "    sys.stdout.write(\"%c\"%(text[i]))\n",
+      "    i += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "HAVE A NICE DAY"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.6, Page number: 239<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the elements of the character array.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = \"HAVE A NICE DAY\"\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "#There is no null character for strings in python\n",
+      "\n",
+      "while i < 15:\n",
+      "    sys.stdout.write(\"%c\"%(text[i]))\n",
+      "    i += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "HAVE A NICE DAY"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.7, Page number: 241<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count the number of characters in a given string\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = raw_input(\"Type Text Below\")\n",
+      "\n",
+      "#Calculation\n",
+      "len1 = len(text)    #since len in a built in function in python, len1 is used\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Length of String = %d\"%(len1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Type Text BelowHello\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of String = 5"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.8, Page number: 241<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the length of the string\n",
+      "#and find its equivalent ASCII codes.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "\n",
+      "#Variable Initialization\n",
+      "name = raw_input(\"Enter your name : \")\n",
+      "l = len(name)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Your Name is %s & \"%(name))\n",
+      "sys.stdout.write(\"it contains %d characters\"%(l))\n",
+      "sys.stdout.write(\"\\n\\nName & it's Ascii Equivalent.\\n\")\n",
+      "sys.stdout.write(\"=====================================\\n\")\n",
+      "\n",
+      "for i in range(0,l):\n",
+      "    sys.stdout.write(\"\\n%c\\t\\t%d\"%(name[i],ord(name[i])))   #ord() function is used to get the ASCII value of a character"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your name : SACHIN\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your Name is SACHIN & it contains 6 characters\n",
+        "\n",
+        "Name & it's Ascii Equivalent.\n",
+        "=====================================\n",
+        "\n",
+        "S\t\t83\n",
+        "A\t\t65\n",
+        "C\t\t67\n",
+        "H\t\t72\n",
+        "I\t\t73\n",
+        "N\t\t78"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.9, Page number: 242<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Remove the occurrences of 'The' word from entered text\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "j = 0\n",
+      "line = raw_input(\"Enter Text Below.\")\n",
+      "l = len(line)\n",
+      "line2 = ['0' for i in range(0,80)]\n",
+      "i = 0\n",
+      "\n",
+      "#Calculation\n",
+      "while i < l:\n",
+      "    if (i >= l-4 or l ==3) and (line[l-4] == ' ' and line[l-3] == 't' and line[l-2] == 'h' and line[l-1] == 'e'):\n",
+      "        continue\n",
+      "    if line[i] == 't' and line[i+1] == 'h' and line[i+2] == 'e' and line[i+3] == ' ':\n",
+      "        i += 3\n",
+      "        continue\n",
+      "    else:\n",
+      "        line2[j] = line[i]\n",
+      "        j += 1\n",
+      "    i += 1\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nText With 'The' -: %s\"%(line))\n",
+      "sys.stdout.write(\"\\nText Without 'The' -: \")\n",
+      "for l in range(0,j):\n",
+      "    sys.stdout.write(\"%c\"%(line2[l]))\n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Below.Printf write data to the screen.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Text With 'The' -: Printf write data to the screen.\n",
+        "Text Without 'The' -: Printf write data to  screen."
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.10, Page number: 243<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Delete all the occurrences of vowels in a given text.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "j = 0\n",
+      "line = raw_input(\"Enter Text Below.\")\n",
+      "\n",
+      "#Calculation\n",
+      "for i in range(0,len(line)):\n",
+      "    if line[i] == 'a' or line[i] == 'e' or line[i] == 'i' or line[i] == 'o' or line[i] == 'u':\n",
+      "        continue\n",
+      "    else:\n",
+      "        line2[j] = line[i]\n",
+      "        j += 1\n",
+      "      \n",
+      "#Result\n",
+      "sys.stdout.write(\"Text With Vowels -:  %s\"%(line))\n",
+      "sys.stdout.write(\"\\nText Without Vowels -:  \")\n",
+      "for l in range(0,j):                                  #There is no null terminating character in python, so we have to specify length\n",
+      "    sys.stdout.write(\"%c\"%(line2[l])) \n",
+      "        "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Below.Have a nice day.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Text With Vowels -:  Have a nice day.\n",
+        "Text Without Vowels -:  Hv  nc dy."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.11, Page number: 244<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the length of characteres in a given string including and excluding spaces.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 0\n",
+      "len1 = 0       #Since len is a built in function in python, len1 is used\n",
+      "ex = 0\n",
+      "\n",
+      "text = raw_input(\"Enter Text Here : \")\n",
+      "\n",
+      "#Calculation\n",
+      "while i != len(text)-1:           #There is no null terminating character in python.\n",
+      "    if text[i] == ' ':\n",
+      "        ex += 1\n",
+      "    else:\n",
+      "        len1 += 1\n",
+      "    i += 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nLengh of String Including Spaces : %d\"%(len1+ex))\n",
+      "sys.stdout.write(\"\\nLengh of String Excluding Spaces : %d\"%(len1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Here : Time is Money.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Lengh of String Including Spaces : 13\n",
+        "Lengh of String Excluding Spaces : 11"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.12, Page number: 245<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the length of entered string and display it as per output shown.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "string = raw_input(\"Enter a String\")\n",
+      "ln = len(string)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nLength of Given String : %d\"%(ln))\n",
+      "sys.stdout.write(\"\\nSequence of Characters Displayed on Screen\")\n",
+      "sys.stdout.write(\"\\n======== == ========== ========= == ======\")\n",
+      "sys.stdout.write(\"\\n\\n\")\n",
+      "\n",
+      "for c in range(0,ln-2+1):\n",
+      "    d = c + 1\n",
+      "    sys.stdout.write(\"\\t%.*s\\n\"%(d,string))\n",
+      "\n",
+      "for c in range(ln,0-1,-1):\n",
+      "    d = c + 1\n",
+      "    sys.stdout.write(\"\\t%.*s\\n\"%(d,string))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a StringHAPPY\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Length of Given String : 5\n",
+        "Sequence of Characters Displayed on Screen\n",
+        "======== == ========== ========= == ======\n",
+        "\n",
+        "\tH\n",
+        "\tHA\n",
+        "\tHAP\n",
+        "\tHAPP\n",
+        "\tHAPPY\n",
+        "\tHAPPY\n",
+        "\tHAPP\n",
+        "\tHAP\n",
+        "\tHA\n",
+        "\tH\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.13, Page number: 247<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy contents of one string to another string by using strcpy()\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "ori = raw_input(\"Enter Your Name : \")\n",
+      "\n",
+      "#Process\n",
+      "dup = ori                  #In python, string values can be simply copied using assignment operator\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Original String : %s\"%(ori))\n",
+      "sys.stdout.write(\"\\nDuplicate String : %s\"%(dup))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name : SACHIN\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Original String : SACHIN\n",
+        "Duplicate String : SACHIN"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.14, Page number: 247<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy contents of one string to another, without strcpy() function.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "ori = raw_input(\"Enter Your Name : \")\n",
+      "\n",
+      "#Calculation\n",
+      "dup = ori                  #In python, string values can be simply copied using assignment operator\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"Original String : %s\"%(ori))\n",
+      "sys.stdout.write(\"\\nDuplicate String : %s\"%(dup))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name : SACHIN\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Original String : SACHIN\n",
+        "Duplicate String : SACHIN"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.15, Page number: 248<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy source string to destination string upto a specified length.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "str1 = raw_input(\"Enter Source String : \")\n",
+      "str2 = raw_input(\"Enter Destination String : \")\n",
+      "n = int(raw_input(\"Enter Number of Characters to Replace in Destination String :\"))\n",
+      "\n",
+      "#Calculation\n",
+      "str2 = str1[:n] + str2[n:]         #Since python strings are not mutable, first and second array elements to be added to make a new one\n",
+      "    \n",
+      "sys.stdout.write(\"Source String -: %s\"%(str1))\n",
+      "sys.stdout.write(\"\\nDestination String -: %s\"%(str2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Source String : wonderful\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Destination String : beautiful\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number of Characters to Replace in Destination String :6\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Source String -: wonderful\n",
+        "Destination String -: wonderful"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.16, Page number: 249<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compare the two string using stricmp() function. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "sr = raw_input(\"Enter String(1) :\")\n",
+      "tar = raw_input(\"Enter String(2) :\")\n",
+      "\n",
+      "diff = (sr == tar)           #in python, relational operators can be used directly\n",
+      "\n",
+      "if diff == 0:\n",
+      "    sys.stdout.write(\"The Two Strings are Identical.\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"The Tow Strings are Different\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String(1) :HELLO\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String(2) :hello\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Two Strings are Identical."
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.17, Page number: 250<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Perform following.\n",
+      "\n",
+      "#  1.  Display the question \"What is the Unit of Traffic?\"\n",
+      "#  2.  Accept the Answer.\n",
+      "#  3.  If the answer is wrong (other than Earlang) display \"Try again!\" & continues to answer.\n",
+      "#  4.  Otherwise, if it is correct \"Earlang\" display the message \"Answer is correct\".\n",
+      "#  5.  If the user gives correct anser in first two attempts the program will terminate.\n",
+      "#  6.  If the user fails to provide correct answer in three attempts the program itself gives the answer.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Calculation & Result\n",
+      "for i in range(0,3):\n",
+      "    ans = raw_input(\"What is the Unit of Traffic ?\")\n",
+      "    \n",
+      "    if ans == \"Earlang\":\n",
+      "        sys.stdout.write(\"\\nAnswer is Correct.\")\n",
+      "    else:\n",
+      "        if i < 3:\n",
+      "            sys.stdout.write(\"\\nTry Again !\\n\")\n",
+      "            \n",
+      "if i == 3:\n",
+      "    sys.stdout.write(\"\\nUnit of Traffic is Earlang.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "What is the Unit of Traffic ?Earlan\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Try Again !\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "What is the Unit of Traffic ?Earlam\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Try Again !\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "What is the Unit of Traffic ?Earlang\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Answer is Correct."
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.18, Page number: 252<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compare two strings upto specified length.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "sr = raw_input(\"Enter String(1):\")\n",
+      "tar = raw_input(\"Enter String(2):\")\n",
+      "n = int(raw_input(\"Enter Lenght upto which comparison is to be made\"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "diff = (sr == tar[:n])\n",
+      "\n",
+      "if diff == 0:\n",
+      "    sys.stdout.write(\"The Two Strings are Identical upto %d characters.\"%(n))\n",
+      "else:\n",
+      "    sys.stdout.write(\"The Two Strings are Different\")\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String(1):HELLO\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String(2):HE MAN\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Lenght upto which comparison is to be made2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Two Strings are Identical upto 2 characters."
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.19, Page number: 253<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compare two strings and display the position in which it is different.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "diff = 0\n",
+      "sr = raw_input(\"Enter String(1) :\")\n",
+      "tar = raw_input(\"Enter String(2) :\")\n",
+      "\n",
+      "#Checking\n",
+      "for i in range(0,len(sr)):\n",
+      "    if sr[i] != tar[i]:\n",
+      "        sys.stdout.write(\"%c  %c\\n\"%(sr[i],tar[i]))\n",
+      "        diff += 1\n",
+      "        \n",
+      "#Result\n",
+      "if len(sr) == len(tar) and diff == 0:\n",
+      "    sys.stdout.write(\"\\nThe Two Strings are Identical\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nThe Two Strings are Different at %d places.\"%(diff))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String(1) :BEST LUCK\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String(2) :GOOD LUCK\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "B  G\n",
+        "E  O\n",
+        "S  O\n",
+        "T  D\n",
+        "\n",
+        "The Two Strings are Different at 4 places."
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.20, Page number: 254<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert upper case string to lower case using strlwr().\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "upper = raw_input(\"Enter a string in Upper case :\")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"After strlwr() : %s\"%(upper.lower()))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a string in Upper case :ABCDEFG\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After strlwr() : abcdefg"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.21, Page number: 254<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert Lower case string to upper case using strupr()\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "upper = raw_input(\"Enter a string in Lower Case :\")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"After strupr() : %s\"%(upper.upper()))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a string in Lower Case :abcdefg\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After strupr() : ABCDEFG"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.22, Page number: 255<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the duplicate string using strdup() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text1 = raw_input(\"Enter Text : \")\n",
+      "\n",
+      "#Result\n",
+      "text2 = text1              #in python, a string can be duplicated by just using the assignment operator\n",
+      "\n",
+      "sys.stdout.write(\"Original String = %s\\nDuplicate String = %s\"%(text1,text2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text : Today is a Good Day.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Original String = Today is a Good Day.\n",
+        "Duplicate String = Today is a Good Day."
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.23, Page number: 256<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find first occurrence of a given character in a given string. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "string = raw_input(\"Enter Text Below :\")\n",
+      "ch = raw_input(\"Character to find :\")\n",
+      "\n",
+      "#Result\n",
+      "chp = string.count(ch)\n",
+      "if chp:\n",
+      "    sys.stdout.write(\"Character %s found in string\"%(ch))\n",
+      "else:\n",
+      "    sys.stdout.write(\"Character %s not found in string\"%(ch))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Below :Hello Begineers.\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Character to find :r\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Character r found in string"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.24, Page number: 257<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find first occurrence of a given character in a given string and display the memory location. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "line1 = raw_input(\"Enter Text :\")\n",
+      "line2 = raw_input(\"Enter Character to find from the text : \")\n",
+      "\n",
+      "for i in range(0,len(line1)):\n",
+      "    sys.stdout.write(\"%c  %d\\n\"%(line1[i],id(line1[i])))\n",
+      "\n",
+      "chp = line1.count(line2)\n",
+      "if chp:\n",
+      "    sys.stdout.write(\"\\nAddress of first %s returned by strchr() is %d\"%(line2,id(line1[chp])))\n",
+      "else:\n",
+      "    sys.stdout.write(\"'%s' String is not present in Given String \"%(line2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text :HELLO\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Character to find from the text : L\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "H  20382328\n",
+        "E  20384080\n",
+        "L  19678272\n",
+        "L  19678272\n",
+        "O  20198416\n",
+        "\n",
+        "Address of first L returned by strchr() is 19678272"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.25, Page number: 258<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the occurrence of second string in the first string.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "line1 = raw_input(\"Enter Line1 :\")\n",
+      "line2 = raw_input(\"Enter Line2 :\")\n",
+      "\n",
+      "#Calculation & Result\n",
+      "chp = line1.count(line2)\n",
+      "\n",
+      "if chp:\n",
+      "    sys.stdout.write(\"'%s' String is present in Given String.\"%(line2))\n",
+      "else:\n",
+      "    sys.stdout.write(\"'%s' String is not present in Given String.\"%(line2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Line1 :INDIA IS MY COUNTRY.\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Line2 :INDIA\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "'INDIA' String is present in Given String."
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.26, Page number: 259<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Append second string at the end of first string using strcat() function\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text1 = raw_input(\"Enter Text1 :\")\n",
+      "text2 = raw_input(\"Enter Text2 :\")\n",
+      "\n",
+      "#Calculation and Result\n",
+      "text1 = text1 + \" \" + text2        #in python, '+' operator can also be used for string concatenation\n",
+      "\n",
+      "sys.stdout.write(\"%s\"%(text1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text1 :I Am\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text2 :an Indian\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "I Am an Indian"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.27, Page number: 259<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Concatenate two strings without use of standard functions.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "fname = raw_input(\"First Name :\")\n",
+      "sname = raw_input(\"Second Name :\")\n",
+      "lname = raw_input(\"Last Name :\")\n",
+      "\n",
+      "#Calculation\n",
+      "name = fname + \" \" + sname + \" \" + lname\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nComplete Name After Concatenation.\\n\")\n",
+      "sys.stdout.write(\"%s\"%(name))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "First Name :MOHAN\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Second Name :KARAMCHAND\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Last Name :GANDHI\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Complete Name After Concatenation.\n",
+        "MOHAN KARAMCHAND GANDHI"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.28, Page number: 261<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Append second string with specified (n) number of characters at the end of first string.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text1 = raw_input(\"Enter Text1 :\")\n",
+      "text2 = raw_input(\"Enter Text2 :\")\n",
+      "n = int(raw_input(\"Enter Number of Characters to Add :\"))\n",
+      "\n",
+      "#Calculation & Result\n",
+      "text1 = text1 + \" \" + text2[:n]\n",
+      "\n",
+      "sys.stdout.write(\"%s\\n\"%(text1))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text1 :MAY I\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text2 :COME IN ?\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number of Characters to Add :4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MAY I COME\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.29, Page number: 261<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Reverse of the given string.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = raw_input(\"Enter String\")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Reverse String\\n\")\n",
+      "text = text[::-1]\n",
+      "#This is extended slice syntax. It works by doing [begin:end:step] \n",
+      "#by leaving begin and end off and specifying a step of -1, it reverses a string.\n",
+      "\n",
+      "sys.stdout.write(\"%s\"%(text))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter StringABCDEFGHI\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reverse String\n",
+        "IHGFEDCBA"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.30, Page number: 262<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Reverse of the given string.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = raw_input(\"Enter String\")\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "while i < len(text):\n",
+      "    sys.stdout.write(\"\\n%c is stored at location %d\"%(text[i],id(text[i])))\n",
+      "    i += 1\n",
+      "    \n",
+      "sys.stdout.write(\"\\nReverse String :- \")\n",
+      "text = text[::-1]\n",
+      "#This is extended slice syntax. It works by doing [begin:end:step] \n",
+      "#by leaving begin and end off and specifying a step of -1, it reverses a string.\n",
+      "\n",
+      "sys.stdout.write(\"%s\"%(text))\n",
+      "\n",
+      "i = 0\n",
+      "while i < len(text):\n",
+      "    sys.stdout.write(\"\\n%c is stored at location %d\"%(text[i],id(text[i])))\n",
+      "    i += 1\n",
+      "\n",
+      "#Memory address differs in different systems"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter StringABC\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "A is stored at location 20383432\n",
+        "B is stored at location 20198992\n",
+        "C is stored at location 20199040\n",
+        "Reverse String :- CBA\n",
+        "C is stored at location 20199040\n",
+        "B is stored at location 20198992\n",
+        "A is stored at location 20383432"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.31, Page number: 263<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Replace (set) given string with given symbol. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "string = raw_input(\"Enter String :\")\n",
+      "symbol = raw_input(\"Enter Symbol for Replacement : \")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Before strset(): %s\\n\"%(string))\n",
+      "string = symbol*len(string)\n",
+      "\n",
+      "sys.stdout.write(\"After strset() : %s\"%(string))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String :LEARN C\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Symbol for Replacement : Y\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Before strset(): LEARN C\n",
+        "After strset() : YYYYYYY"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.32, Page number: 264<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Replace (set) given string with given symbol for given number of arguments.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "string = raw_input(\"Enter String :\")\n",
+      "symbol = raw_input(\"Enter Symbol for Replacement : \")\n",
+      "n = int(raw_input(\"How many String Character to be replaced.\"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Before strset() : %s\\n\"%(string))\n",
+      "\n",
+      "string = symbol * n + string[n:]\n",
+      "sys.stdout.write(\"After strset() : %s\\n\"%(string))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String :ABCDEFGHIJ\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Symbol for Replacement : +\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many String Character to be replaced.4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Before strset() : ABCDEFGHIJ\n",
+        "After strset() : ++++EFGHIJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.33, Page number: 265<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the position from which the two strings are different.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "stra = raw_input(\"First String : \")\n",
+      "strb = raw_input(\"Second String : \")\n",
+      "\n",
+      "#There is no built in function for python to check upto which position the strings are equal.\n",
+      "for length in range(min(len(stra), len(strb))):\n",
+      "        if stra[length] != strb[length]:\n",
+      "            break\n",
+      "\n",
+      "sys.stdout.write(\"After %d Characters there is no match.\"%(length))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "First String : GOOD MORNING\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Second String : GOOD BYE\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After 5 Characters there is no match."
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.34, Page number: 266<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print given string from first occurrence of given character.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text1 = raw_input(\"Enter String : \")\n",
+      "text2 = raw_input(\"Enter Character :\")\n",
+      "\n",
+      "for length in range(0,len(text1)):\n",
+      "        if text1[length] == text2[0]:\n",
+      "            break\n",
+      "            \n",
+      "sys.stdout.write(\"String from given Character : %s\"%(text1[length:]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter String : INDIA IS GREAT\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Character :G\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String from given Character : GREAT"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.35, Page number: 266<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count a character that appears in a given text\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 0\n",
+      "count = 0\n",
+      "text = raw_input(\"Type Text Below.\")\n",
+      "find = raw_input(\"Type a character to count : \")\n",
+      "\n",
+      "#Calculation\n",
+      "while i < len(text):             #There is no null terminating character for python string\n",
+      "    if text[i] == find:\n",
+      "        count += 1\n",
+      "    i += 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nCharacter (%s) found in Given String = %d Times.\"%(find,count))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Type Text Below.Programming\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Type a character to count : m\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Character (m) found in Given String = 2 Times."
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.36, Page number: 267<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count the character 'm' in a given string.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "count = 0\n",
+      "text = raw_input(\"\")  #it is easy to get string at once in python\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"%s\"%(text))\n",
+      "\n",
+      "for i in range(0,len(text)):\n",
+      "    if text[i] == '\\n':\n",
+      "        break\n",
+      "    else:\n",
+      "        if text[i] == 'm':\n",
+      "            count += 1\n",
+      "            \n",
+      "sys.stdout.write(\"\\n\\nCharacter 'm' Found in Text = %d times.\"%(count))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Programming is a skill.\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Programming is a skill.\n",
+        "\n",
+        "Character 'm' Found in Text = 2 times."
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.37, Page number: 268<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Count the characters 'm', 'r', and 'o' that appears in a string without using any functions\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "text = \"Programming is good habit\"\n",
+      "m = 0\n",
+      "o = 0\n",
+      "r = 0\n",
+      "\n",
+      "#Calculation\n",
+      "for i in range(0,len(text)):\n",
+      "    if text[i] == 'm':\n",
+      "        m += 1\n",
+      "    if text[i] == 'r':\n",
+      "        r += 1\n",
+      "    if text[i] == 'o':\n",
+      "        o += 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nCharacter 'm' Found in Text = %d times.\\n\"%(m))\n",
+      "sys.stdout.write(\"\\nCharacter 'r' Found in Text = %d times.\\n\"%(r))\n",
+      "sys.stdout.write(\"\\nCharacter 'o' Found in Text = %d times.\"%(o))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Character 'm' Found in Text = 2 times.\n",
+        "\n",
+        "Character 'r' Found in Text = 2 times.\n",
+        "\n",
+        "Character 'o' Found in Text = 3 times."
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.38, Page number: 269<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy contents of one string to another string\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "ori = raw_input(\"Enter Your Name :\")\n",
+      "\n",
+      "#Copying\n",
+      "#Since strings in python are immutable, individual character assignment is not possible.\n",
+      "#but assignment operator simply copies the entire string to another\n",
+      "\n",
+      "dup = ori\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Original String : %s\"%(ori))\n",
+      "sys.stdout.write(\"\\nDuplicate String : %s\"%(dup))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name :SACHIN\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Original String : SACHIN\n",
+        "Duplicate String : SACHIN"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.39, Page number: 270<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Palindrome or not.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = 0\n",
+      "str1 = raw_input(\"Enter the word : \")   #Since str is a keyword in python, str1 is used\n",
+      "j = len(str1)-1\n",
+      "\n",
+      "#Calculation\n",
+      "while i < j:\n",
+      "    if str1[i] == str1[j]:\n",
+      "        test = 1\n",
+      "    else:\n",
+      "        test = 0\n",
+      "        break\n",
+      "    i += 1\n",
+      "    j -= 1\n",
+      "    \n",
+      "#Result\n",
+      "if test == 1:\n",
+      "    sys.stdout.write(\"Word is palindrome.\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nWord is not Palindrome.\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the word : ABBA\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Word is palindrome."
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.40, Page number: 271<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compare the string using strcmp() function and display their ASCII difference.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a1 = \"NAGPUR\"\n",
+      "a2 = \"NAGPUR\"\n",
+      "a3 = \"PANDHARPUR\"\n",
+      "a4 = \"KOLHAPUR\"\n",
+      "a5 = \"AURANGABAD\"\n",
+      "a6 = \"HYDERABAD\"\n",
+      "\n",
+      "#Calculation\n",
+      "#There is no built in function which returns the ascii value differece of two strings\n",
+      "#ord function gives the ascii value of a character\n",
+      "c1 = ord(a1[0]) - ord(a2[0])\n",
+      "c2 = ord(a3[0]) - ord(a4[0])\n",
+      "c3 = ord(a5[0]) - ord(a6[0])\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAscii Difference between two strings\\n\")\n",
+      "sys.stdout.write(\"Difference between (%s %s) = %d\\n\"%(a1,a2,c1))\n",
+      "sys.stdout.write(\"Difference between (%s %s) = %d\\n\"%(a3,a4,c2))\n",
+      "sys.stdout.write(\"Difference between (%s %s) = %d\\n\"%(a5,a6,c3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Ascii Difference between two strings\n",
+        "Difference between (NAGPUR NAGPUR) = 0\n",
+        "Difference between (PANDHARPUR KOLHAPUR) = 5\n",
+        "Difference between (AURANGABAD HYDERABAD) = -7\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.41, Page number: 272<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sort city names alphabetically.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "city = [['0' for i in range(5)]for i in range(20)]\n",
+      "sys.stdout.write(\"Enter Names of Cities.  \")\n",
+      "for i in range(5):\n",
+      "    city[i] = raw_input(\"\")\n",
+      "    \n",
+      "#Sorting & Result\n",
+      "sys.stdout.write(\"Sorted List of Cities.\\n\\n\")\n",
+      "for i in range(65,122+1):\n",
+      "    for j in range(0,5):\n",
+      "        if ord(city[j][0]) == i:\n",
+      "            sys.stdout.write(\"\\n%s\"%(city[j]))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Names of Cities.  "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MUMBAI\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NANED\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "BANGLORE\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "KANPUR\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "INDORE\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorted List of Cities.\n",
+        "\n",
+        "\n",
+        "BANGLORE\n",
+        "INDORE\n",
+        "KANPUR\n",
+        "MUMBAI\n",
+        "NANED"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.42, Page number: 273<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sort city names alphabetically\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "city = [['0' for i in range(5)]for i in range(20)]\n",
+      "sys.stdout.write(\"Enter Names of Cities.  \")\n",
+      "for i in range(5):\n",
+      "    city[i] = raw_input(\"\")\n",
+      "    \n",
+      "#Sorting & Result\n",
+      "sys.stdout.write(\"Sorted List of Cities.\\n\\n\")\n",
+      "for i in range(0,5):\n",
+      "    for j in range(0,5):\n",
+      "        if city[j-1] > city[j]:\n",
+      "            temp = city[j-1]\n",
+      "            city[j-1] = city[j]\n",
+      "            city[j] = temp\n",
+      "            \n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"\\n%s\"%(city[i]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Names of Cities.  "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MUMBAI\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NANED\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "BANGLORE\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "KANPUR\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "INDORE\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sorted List of Cities.\n",
+        "\n",
+        "\n",
+        "BANGLORE\n",
+        "INDORE\n",
+        "KANPUR\n",
+        "MUMBAI\n",
+        "NANED"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.43, Page number: 274<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find number of words in a given statement.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "count = 1                #Counts the starting word\n",
+      "i = 0\n",
+      "text = raw_input(\"Enter The Line of Text\\nGive One Space After Each Word\")\n",
+      "\n",
+      "#Calculation\n",
+      "while i < len(text):      #There is no null terminating character in python\n",
+      "    if ord(text[i]) == 32:\n",
+      "        count += 1\n",
+      "    i += 1\n",
+      "        \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nThe Number of words in line = %d\"%(count))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter The Line of Text\n",
+        "Give One Space After Each WordRead Books\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Number of words in line = 2"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.44, Page number: 275<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find number of words in a given statement.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "count = 0          \n",
+      "i = 0\n",
+      "text = raw_input(\"Enter Text Below :\")\n",
+      "\n",
+      "#Calculation\n",
+      "while i < len(text):     #There is no null terminating character in python for strings\n",
+      "    if ord(text[i]) >= 97 and ord(text[i]) <= 122 or ord(text[i]) >= 65 and ord(text[i]) <= 90:\n",
+      "        i += 1\n",
+      "        continue\n",
+      "    if ord(text[i]) == 32:\n",
+      "        count += 1\n",
+      "        i += 1\n",
+      "        continue\n",
+      "\n",
+      "if i == len(text):\n",
+      "    count += 1\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"The Number of words in line = %d\"%(count))\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Text Below :Reading is a good Habit\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Number of words in line = 5"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8.45, Page number: 276<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sort the last name of mobile customers alphabetically.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "fname = [['0' for i in range(20)] for i in range(10)]\n",
+      "sname = [['0' for i in range(20)] for i in range(10)]\n",
+      "surname = [['0' for i in range(20)] for i in range(10)]\n",
+      "name = [['0' for i in range(20)] for i in range(20)]\n",
+      "mobile = [['0' for i in range(20)] for i in range(10)]\n",
+      "temp = ['0' for i in range(20)]\n",
+      "\n",
+      "sys.stdout.write(\"Enter First Name, Second Name, surname and Mobile Number \")\n",
+      "for i in range(0,5):\n",
+      "    fname[i] = raw_input(\"\")\n",
+      "    sname[i] = raw_input(\"\")\n",
+      "    surname[i] = raw_input(\"\")\n",
+      "    mobile[i] = raw_input(\"\")\n",
+      "    \n",
+      "    name[i] = surname[i] + \" \" \n",
+      "    temp[0] = fname[i][0]\n",
+      "    name[i] = name[i] + temp[0] + \".\"\n",
+      "    temp[0] = sname[i][0]\n",
+      "    name[i] = name[i] + temp[0]\n",
+      "    \n",
+      "for i in range(0,5):\n",
+      "    for j in range(1,5-i):\n",
+      "        if name[j-1] > name[j]:\n",
+      "            temp = name[j-1]\n",
+      "            name[j-1] = name[j]\n",
+      "            name[j] = temp\n",
+      "            temp = mobile[j-1]\n",
+      "            mobile[j-1] = mobile[j]\n",
+      "            mobile[j] = temp\n",
+      "            \n",
+      "sys.stdout.write(\"List of Customers in alphanetical Order.\")\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"\\n%-20s\\t %-10s\\n\"%(name[i],mobile[i]))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter First Name, Second Name, surname and Mobile Number "
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "K\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "S\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MORE\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "458454\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "J\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "M\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "CHATE\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "658963\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "M\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "M\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "GORE\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "660585\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "K\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "JAIN\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "547855\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "J\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "J\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "JOSHI\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "354258\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "List of Customers in alphanetical Order.\n",
+        "CHATE J.M           \t 658963    \n",
+        "\n",
+        "GORE M.M            \t 660585    \n",
+        "\n",
+        "JAIN L.K            \t 547855    \n",
+        "\n",
+        "JOSHI J.J           \t 354258    \n",
+        "\n",
+        "MORE K.S            \t 458454    \n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter9.ipynb b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter9.ipynb
new file mode 100755
index 00000000..4da93c8d
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/KamthaneChapter9.ipynb
@@ -0,0 +1,1999 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 9: Pointers<h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.1, Page number: 282<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the address of the variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "num = int(raw_input(\"Enter a Number = \"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Value of num = %d\\n\"%(num))\n",
+      "sys.stdout.write(\"Address of num = %d\\n\"%(id(num)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Number = 20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of num = 20\n",
+        "Address of num = 19554932\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.2, Page number: 283<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the addresses of different variables together with their values.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = raw_input(\"Enter alphabet\")\n",
+      "i = int(raw_input(\"Enter number\"))\n",
+      "f = float(raw_input(\"Enter float\"))\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAddress of (c) %c = %d\"%(c,id(c)))\n",
+      "sys.stdout.write(\"\\nAddress of (i) %d = %d\"%(i,id(i)))\n",
+      "sys.stdout.write(\"\\nAddress of (f) %.2f = %d\"%(f,id(f)))\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter alphabetC\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter number20\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter float2.5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of (c) C = 48769192\n",
+        "Address of (i) 20 = 19554932\n",
+        "Address of (f) 2.50 = 44777808"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.3, Page number: 284<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Show pointer of any data type that occupies same space\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Result\n",
+      "#Garbage collector module is used for python memory management\n",
+      "sys.stdout.write(\"char %d byte & its pointer %d bytes\\n\"%(sys.getsizeof(chr),sys.getsizeof(id(chr))))\n",
+      "sys.stdout.write(\"int %d byte & its pointer %d bytes\\n\"%(sys.getsizeof(int),sys.getsizeof(id(int))))\n",
+      "sys.stdout.write(\"float %d byte & its pointer %d bytes\\n\"%(sys.getsizeof(float),sys.getsizeof(id(float))))\n",
+      "\n",
+      "#value tagged method is used in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "char 36 byte & its pointer 12 bytes\n",
+        "int 436 byte & its pointer 12 bytes\n",
+        "float 436 byte & its pointer 12 bytes\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.4, Page number: 284<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the value of variable and its location using pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "#There is no pointer concept in python\n",
+      "v = 10\n",
+      "p = v\n",
+      "#In python, variables of same value are tagged together instead of storing the value in different locations\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAddress of v = %u\"%(id(v)))\n",
+      "sys.stdout.write(\"\\nValue of v = %d\"%(p))\n",
+      "sys.stdout.write(\"\\nAddress of p = %d\"%(id(p)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of v = 29808764\n",
+        "Value of v = 10\n",
+        "Address of p = 29808764"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.5, Page number: 285<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the value of variable by different ways.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = int(raw_input(\"Enter Integer Value\"))\n",
+      "b = float(raw_input(\"Enter float Value\"))\n",
+      "\n",
+      "pa = id(a)\n",
+      "pb = id(b)\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python. content of memory location can't be accessed directly by user.\n",
+      "sys.stdout.write(\"\\nAddress of a = %u\"%(pa))\n",
+      "sys.stdout.write(\"\\nValue of a = %d\"%(a))\n",
+      "sys.stdout.write(\"\\nValue of a = %d\"%(a))\n",
+      "sys.stdout.write(\"\\nValue of a = %d\"%(a))\n",
+      "\n",
+      "sys.stdout.write(\"\\nAddress of b = %u\"%(pb))\n",
+      "sys.stdout.write(\"\\nValue of b = %.2f\"%(b))\n",
+      "sys.stdout.write(\"\\nValue of b = %.2f\"%(b))\n",
+      "sys.stdout.write(\"\\nValue of b = %.2f\"%(b))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Integer Value4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter float Value2.2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of a = 19555124\n",
+        "Value of a = 4\n",
+        "Value of a = 4\n",
+        "Value of a = 4\n",
+        "Address of b = 44777792\n",
+        "Value of b = 2.20\n",
+        "Value of b = 2.20\n",
+        "Value of b = 2.20"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.6, Page number: 286<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print value of variable using different pointer notations.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "v = 10\n",
+      "p = id(v)\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python and can't access memory location directly by user\n",
+      "sys.stdout.write(\"\\nv = %d v = %d v = %d\"%(v,v,v))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "v = 10 v = 10 v = 10"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.7, Page number: 287<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print element and its address using pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = int(raw_input(\"Enter a number : \"))\n",
+      "k = id(i)\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python and can't access memory location directly by user\n",
+      "sys.stdout.write(\"\\nAddress of i is %u\"%(k))\n",
+      "sys.stdout.write(\"\\nValue of i is %d\"%(i))\n",
+      "#The memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 15\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of i is 19554992\n",
+        "Value of i is 15"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.8, Page number: 287<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Add two numbers through variables and their pointers\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = int(raw_input(\"Enter Two Numbers : \"))\n",
+      "b = int(raw_input(\"Enter Two Numbers : \"))\n",
+      "ap = a\n",
+      "bp = b\n",
+      "\n",
+      "#Calculation\n",
+      "c = a + b\n",
+      "d = ap + bp\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python and can't access memory location directly by user\n",
+      "sys.stdout.write(\"\\nSum of A & B Using Variable : %d\"%(c))\n",
+      "sys.stdout.write(\"\\nSum of A & B Using Pointer : %d\"%(d))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Two Numbers : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Two Numbers : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of A & B Using Variable : 12\n",
+        "Sum of A & B Using Pointer : 12"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.9, Page number: 288<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Assign pointer value to another variable\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 5\n",
+      "c = id(a)\n",
+      "b = a\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python and can't access memory location directly by user\n",
+      "sys.stdout.write(\"\\nMemory location of 'a' = %u\"%(c))\n",
+      "sys.stdout.write(\"\\nThe value of a = %d & b = %d\"%(a,b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Memory location of 'a' = 29808824\n",
+        "The value of a = 5 & b = 5"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.10, Page number: 289<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Assign value of 'b' to 'a' through pointers and add the values.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = int(raw_input(\"Enter Value of 'a' & 'b' : \"))\n",
+      "b = int(raw_input(\"Enter Value of 'a' & 'b' : \"))\n",
+      "pa = id(a)\n",
+      "pb = id(b)\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python and can't access memory location directly by user\n",
+      "sys.stdout.write(\"\\nValue of a = %d & b = %d\"%(a,b))\n",
+      "sys.stdout.write(\"\\nAddress of a = %u\"%(pa))\n",
+      "sys.stdout.write(\"\\nAddress of b = %u\"%(pb))\n",
+      "sys.stdout.write(\"\\n\\nAddition of 'a' & 'b' : %d\"%(a+b))\n",
+      "\n",
+      "a = b\n",
+      "pa = id(a)\n",
+      "#In python, variables of same value are tagged together instead of storing the value in different locations\n",
+      "\n",
+      "sys.stdout.write(\"\\n*pa = %d *pb = %d\"%(a,b))\n",
+      "sys.stdout.write(\"\\npa = %u pb = %u\"%(pa,pb))\n",
+      "sys.stdout.write(\"\\nAddition of *pa + *pb : %d\"%(a+b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of 'a' & 'b' : 8\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of 'a' & 'b' : 4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Value of a = 8 & b = 4\n",
+        "Address of a = 19555076\n",
+        "Address of b = 19555124\n",
+        "\n",
+        "Addition of 'a' & 'b' : 12\n",
+        "*pa = 4 *pb = 4\n",
+        "pa = 19555124 pb = 19555124\n",
+        "Addition of *pa + *pb : 8"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.11, Page number: 290<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#The effect of increment on pointer variables. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = int(raw_input(\"Enter integer Value\"))\n",
+      "y = raw_input(\"Enter Character Value\")\n",
+      "z = float(raw_input(\"Enter float Value\"))\n",
+      "\n",
+      "x1 = id(x)\n",
+      "y1 = id(y)\n",
+      "z1 = id(z)\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python and can't access memory location directly by user\n",
+      "\n",
+      "sys.stdout.write(\"\\nAddress of x = %u\\n\"%(x1))\n",
+      "sys.stdout.write(\"\\nAddress of y = %u\\n\"%(y1))\n",
+      "sys.stdout.write(\"\\nAddress of z = %u\\n\"%(z1))\n",
+      "\n",
+      "x1 += 1\n",
+      "y1 += 1\n",
+      "z1 += 1\n",
+      "#In python, variables of same value are tagged together instead of storing the value in different locations\n",
+      "\n",
+      "sys.stdout.write(\"\\nAfter Increment in Pointers\\n\")\n",
+      "sys.stdout.write(\"\\nNow Address of x = %u\\n\"%(x1))\n",
+      "sys.stdout.write(\"Now Address of y = %u\\n\"%(y1))\n",
+      "sys.stdout.write(\"Now Address of z = %u\\n\"%(z1))\n",
+      "\n",
+      "sys.stdout.write(\"\\nSize of Integer : %d\"%(sys.getsizeof(x)))\n",
+      "sys.stdout.write(\"\\nSize of Character :  %d\"%(sys.getsizeof(y)))\n",
+      "sys.stdout.write(\"\\nSize of Float : %d\"%(sys.getsizeof(z)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter integer Value2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Character ValueA\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter float Value2.2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address of x = 19555148\n",
+        "\n",
+        "Address of y = 48769192\n",
+        "\n",
+        "Address of z = 44777728\n",
+        "\n",
+        "After Increment in Pointers\n",
+        "\n",
+        "Now Address of x = 19555149\n",
+        "Now Address of y = 48769193\n",
+        "Now Address of z = 44777729\n",
+        "\n",
+        "Size of Integer : 12\n",
+        "Size of Character :  22\n",
+        "Size of Float : 16"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.12, Page number: 291<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#The effect of increased and decreased operator used as prefix and suffix with pointer variable.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "i = int(raw_input(\"Enter Value of i = \"))\n",
+      "ii = id(i)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
+      "#There is no increment or decrement operator and pointer concept in python\n",
+      "ii += 1\n",
+      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
+      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
+      "ii += 1\n",
+      "ii -= 1\n",
+      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
+      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
+      "ii -= 1\n",
+      "sys.stdout.write(\"Address of i = %u\\n\"%(ii))\n",
+      "\n",
+      "#Memory address differs in different systems"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Value of i = 8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Address of i = 19555076\n",
+        "Address of i = 19555077\n",
+        "Address of i = 19555077\n",
+        "Address of i = 19555077\n",
+        "Address of i = 19555077\n",
+        "Address of i = 19555076\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.13, Page number: 292<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Perform different arithmetic operations using pointers\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 25\n",
+      "b = 10\n",
+      "p = id(a)\n",
+      "j = id(b)\n",
+      "\n",
+      "#Result\n",
+      "#There is no pointer concept in python and user can't access memory contents directly\n",
+      "sys.stdout.write(\"\\nAddition a + b = %d\"%(a+b))\n",
+      "sys.stdout.write(\"\\nSubtraction a - b = %d\"%(a-b))\n",
+      "sys.stdout.write(\"\\nProduct a * b = %d\"%(a*b))\n",
+      "sys.stdout.write(\"\\nDivision a / b = %d\"%(a/b))\n",
+      "sys.stdout.write(\"\\na Mod b  = %d\"%(a%b))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Addition a + b = 35\n",
+        "Subtraction a - b = 15\n",
+        "Product a * b = 250\n",
+        "Division a / b = 2\n",
+        "a Mod b  = 5"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.14, Page number: 293<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compare two pointers. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 2\n",
+      "j = id(a)\n",
+      "k = id(a)\n",
+      "\n",
+      "#Result\n",
+      "if j == k:\n",
+      "    sys.stdout.write(\"\\nThe Tow Pointers have the same address.\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nThe Pointers have the different address.\")\n",
+      "    \n",
+      "sys.stdout.write(\"\\n\\nAddress of a (j) = %u\"%(j))\n",
+      "sys.stdout.write(\"\\nAddress of a (k) = %u\"%(k))\n",
+      "#Memory address differs in different systems"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Tow Pointers have the same address.\n",
+        "\n",
+        "Address of a (j) = 30726364\n",
+        "Address of a (k) = 30726364"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.15, Page number: 294<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display elements of an array.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = [2,4,6,8,10]\n",
+      "k = 1\n",
+      "\n",
+      "#Result\n",
+      "while k <= 5:\n",
+      "    sys.stdout.write(\"%3d\"%(x[k-1]))\n",
+      "    k += 1\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  2  4  6  8 10"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.16, Page number: 295<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display array element with their addresses using array name as a pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = [2,4,6,8,10]\n",
+      "k = 0\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nElement No.\\tElement\\tAddress\")\n",
+      "\n",
+      "while k < 5:\n",
+      "    sys.stdout.write(\"\\nx[%d] \\t%13d %10u\"%(k,x[k],id(x[k])))\n",
+      "    k += 1\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Element No.\tElement\tAddress\n",
+        "x[0] \t            2   30726364\n",
+        "x[1] \t            4   30726340\n",
+        "x[2] \t            6   30726316\n",
+        "x[3] \t            8   30726292\n",
+        "x[4] \t           10   30726268"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.17, Page number: 296<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display array elements with their addresses using array name as a pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "num = [10,25,35,45]\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nElement\\t           Address\\n\")\n",
+      "\n",
+      "for i in range(0,4):\n",
+      "    sys.stdout.write(\"num[%d] = %d        \"%(i,num[i]))\n",
+      "    sys.stdout.write(\"%8u\\n\"%(id(num[i])))\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Element\t           Address\n",
+        "num[0] = 10        30726268\n",
+        "num[1] = 25        30726088\n",
+        "num[2] = 35        30725968\n",
+        "num[3] = 45        30725848\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.18, Page number: 296<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Access elements of array through different ways using pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "arr = [10,20,30,40,50]\n",
+      "p = 0\n",
+      "\n",
+      "#Result\n",
+      "for p in range(0,5):\n",
+      "    sys.stdout.write(\"Value of arr[%d] = \"%(p))\n",
+      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
+      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
+      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
+      "    sys.stdout.write(\"%d   |  \"%(arr[p]))\n",
+      "    sys.stdout.write(\"address of arr[%d] = %u\\n\"%(p,id(arr[p])))\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of arr[0] = 10   |  10   |  10   |  10   |  address of arr[0] = 30726268\n",
+        "Value of arr[1] = 20   |  20   |  20   |  20   |  address of arr[1] = 30726148\n",
+        "Value of arr[2] = 30   |  30   |  30   |  30   |  address of arr[2] = 30726028\n",
+        "Value of arr[3] = 40   |  40   |  40   |  40   |  address of arr[3] = 30725908\n",
+        "Value of arr[4] = 50   |  50   |  50   |  50   |  address of arr[4] = 30725788\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.19, Page number: 297<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find sum of all the elements of an array.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "sum1 = 0        #Since sum is a built-in function in python, sum1 is used\n",
+      "i = 0\n",
+      "a = [1,2,3,4,5]\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Elements       Values         Address\\n\\n\")\n",
+      "\n",
+      "while i < 5:\n",
+      "    sys.stdout.write(\"a[%d]\\t   %5d\\t    %8u\\n\"%(i,a[i],id(a[i])))\n",
+      "    sum1 = sum1 + a[i]\n",
+      "    i += 1\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elements       Values         Address\n",
+        "\n",
+        "a[0]\t       1\t    30726376\n",
+        "a[1]\t       2\t    30726364\n",
+        "a[2]\t       3\t    30726352\n",
+        "a[3]\t       4\t    30726340\n",
+        "a[4]\t       5\t    30726328\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.20, Page number: 298<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Sum of squares and cubes of array elements using pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "b = [1,2,3,4,5]\n",
+      "sumsq = 0\n",
+      "sumc = 0\n",
+      "\n",
+      "#Result\n",
+      "for j in range(0,5):\n",
+      "    sumsq = sumsq + pow(b[j],2)\n",
+      "    sumc = sumc + pow(b[j],3)\n",
+      "    \n",
+      "sys.stdout.write(\"\\nSum of Squares of array elements : %d\"%(sumsq))\n",
+      "sys.stdout.write(\"\\nSum of Cubes of array elements : %d\"%(sumc))\n",
+      "\n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Sum of Squares of array elements : 55\n",
+        "Sum of Cubes of array elements : 225"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.21, Page number: 299<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy element of one array to another array using pointers.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "so = [10,20,30,40,50]\n",
+      "ds = [0 for i in range(0,5)]\n",
+      "\n",
+      "#Copying\n",
+      "for i in range(0,5):\n",
+      "    ds[i] = so[i]\n",
+      "    \n",
+      "sys.stdout.write(\"Original Array      Duplicated Array\")\n",
+      "\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"\\n\\t%d\\t\\t%d\"%(so[i],ds[i]))\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Original Array      Duplicated Array\n",
+        "\t10\t\t10\n",
+        "\t20\t\t20\n",
+        "\t30\t\t30\n",
+        "\t40\t\t40\n",
+        "\t50\t\t50"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.22, Page number: 299<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Copy one array into another array. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "arr1 = [15,25,35,45,55]\n",
+      "arr2 = [0 for i in range(0,5)]\n",
+      "j = 4\n",
+      "\n",
+      "#Copying & Result\n",
+      "sys.stdout.write(\"\\nOriginal Array     Duplicate Array\")\n",
+      "for i in range(0,5):\n",
+      "    arr2[i] = arr1[j]\n",
+      "    j -= 1\n",
+      "    sys.stdout.write(\"\\n\\t%d\\t\\t%d\"%(arr1[i],arr2[i]))\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Original Array     Duplicate Array\n",
+        "\t15\t\t55\n",
+        "\t25\t\t45\n",
+        "\t35\t\t35\n",
+        "\t45\t\t25\n",
+        "\t55\t\t15"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.23, Page number: 300<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display array elements and their address using pointers\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[1,2,3],[4,5,6],[7,8,9]]\n",
+      "j = 1\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\tElements of An Array with their addresses\\n\\n\")\n",
+      "\n",
+      "p = id(a[0][0])\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,3):\n",
+      "        sys.stdout.write(\"%5d  [%5u]\"%(a[i][j],id(a[i][j])))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\tElements of An Array with their addresses\n",
+        "\n",
+        "    1  [30726376]    2  [30726364]    3  [30726352]\n",
+        "    4  [30726340]    5  [30726328]    6  [30726316]\n",
+        "    7  [30726304]    8  [30726292]    9  [30726280]\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.24, Page number: 301<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display array elements and their address. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [[1,2,3],[4,5,6],[7,8,9]]\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nElements of An Array with their addresses.\\n\\n\")\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    for j in range(0,3):\n",
+      "        sys.stdout.write(\"%5d  [%5u]\"%(a[i][j],id(a[i][j])))\n",
+      "    sys.stdout.write(\"\\n\")\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Elements of An Array with their addresses.\n",
+        "\n",
+        "    1  [30726376]    2  [30726364]    3  [30726352]\n",
+        "    4  [30726340]    5  [30726328]    6  [30726316]\n",
+        "    7  [30726304]    8  [30726292]    9  [30726280]\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.25, Page number: 302<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Store addresses of different elements of an array using array of pointers\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "arrl = [5,10,15]\n",
+      "arrp = [0 for i in range(0,3)]\n",
+      "\n",
+      "for i in range(0,3):\n",
+      "    arrp[i] = id(arrl[i])\n",
+      "    \n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nAddress  \\tElement\\n\")\n",
+      "\n",
+      "for k in range(0,3):\n",
+      "    sys.stdout.write(\"%u\"%(arrp[k]))\n",
+      "    sys.stdout.write(\"\\t%7d\\n\"%(arrl[k]))\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Address  \tElement\n",
+        "30726328\t      5\n",
+        "30726268\t     10\n",
+        "30726208\t     15\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.26, Page number: 303<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display address of elements and pointers\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [0,1,2,3,4]\n",
+      "p = [0 for i in range(0,5)]\n",
+      "\n",
+      "for i in range(0,5):\n",
+      "    p[i] = id(a[i])\n",
+      "    \n",
+      "#Result\n",
+      "for i in range(0,5):\n",
+      "    sys.stdout.write(\"\\n%d at location\"%(a[i]))\n",
+      "    sys.stdout.write(\"\\t%u at location\"%(id(a[i])))\n",
+      "    sys.stdout.write(\"   %u\"%(id(p[i])))\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "0 at location\t30726388 at location   52456564\n",
+        "1 at location\t30726376 at location   52456360\n",
+        "2 at location\t30726364 at location   52456456\n",
+        "3 at location\t30726352 at location   52456420\n",
+        "4 at location\t30726340 at location   52456468"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.27, Page number: 304<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print value of a variable through pointer and pointer to pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = 2\n",
+      "p = id(a)\n",
+      "q = id(p)\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n        Value of a = %d Address of a = %u\"%(a,id(a)))\n",
+      "sys.stdout.write(\"\\nThrough *p Value of a = %d Address of a = %u\"%(a,p))\n",
+      "sys.stdout.write(\"\\nThrough **q Vallue of a = %d Address of a = %d\"%(a,p))\n",
+      "\n",
+      "#There is no pointer concept in python\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "        Value of a = 2 Address of a = 30726364\n",
+        "Through *p Value of a = 2 Address of a = 30726364\n",
+        "Through **q Vallue of a = 2 Address of a = 30726364"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.28, Page number: 305<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Different levels of array of pointer to pointer and display the elements\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "a = [1,2,3]\n",
+      "b = [0 for i in range(0,3)]\n",
+      "c = [0 for i in range(0,3)]\n",
+      "d = [0 for i in range(0,3)]\n",
+      "e = [0 for i in range(0,3)]\n",
+      "f = [0 for i in range(0,3)]\n",
+      "\n",
+      "for k in range(0,3):\n",
+      "    b[k] = id(a[k])\n",
+      "    c[k] = id(b[k])\n",
+      "    d[k] = id(c[k])\n",
+      "    e[k] = id(d[k])\n",
+      "    f[k] = id(e[k])\n",
+      "    \n",
+      "#Result\n",
+      "for k in range(0,3):\n",
+      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
+      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
+      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
+      "    sys.stdout.write(\"%3d\"%(a[k]))\n",
+      "    sys.stdout.write(\"%3d\\n\"%(a[k]))\n",
+      "    \n",
+      "#There is no pointer concept in python\n",
+      "#Memory address differs in different systems\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  1  1  1  1  1\n",
+        "  2  2  2  2  2\n",
+        "  3  3  3  3  3\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.29, Page number: 306<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display string using character pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "name = ['K','U','M','A','R']\n",
+      "ch = id(name)\n",
+      "i = 0\n",
+      "\n",
+      "#Result\n",
+      "while i < len(name):                 #There is no null terminating character in python string\n",
+      "    sys.stdout.write(\"%c\"%(name[i]))\n",
+      "    i += 1\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "KUMAR"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.30, Page number: 307<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Length of a given string including and excluding spaces using pointers\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "p = 0\n",
+      "q = 0\n",
+      "s = 0\n",
+      "str1 = ['P','O','I','N','T','E','R','S',' ','A','R','E',' ','E','A','S','Y']\n",
+      "\n",
+      "#Result\n",
+      "while s < len(str1):          #There is no null termination character for python string\n",
+      "    sys.stdout.write(\"%c\"%(str1[s]))\n",
+      "    p += 1\n",
+      "    if ord(str1[s]) == 32:\n",
+      "        q += 1\n",
+      "    s += 1\n",
+      "\n",
+      "sys.stdout.write(\"\\nLength of String including spaces : %d\"%(p))\n",
+      "sys.stdout.write(\"\\nLength of String excluding spaces : %d\"%(p-q))\n",
+      "\n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "POINTERS ARE EASY\n",
+        "Length of String including spaces : 17\n",
+        "Length of String excluding spaces : 15"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.31, Page number: 308<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Interchange elements of character array using pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "names = [\"kapil\",\"manoj\",\"amit\",\"amol\",\"pavan\",\"mahesh\"]\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"Original : %s  %s\"%(names[3],names[4]))\n",
+      "\n",
+      "tmp = names[3]\n",
+      "names[3] = names[4]\n",
+      "names[4] = tmp\n",
+      "\n",
+      "sys.stdout.write(\"\\nNew    : %s  %s\"%(names[3],names[4]))\n",
+      "\n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Original : amol  pavan\n",
+        "New    : pavan  amol"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.32, Page number: 308<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#String comparison.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = 0\n",
+      "str1 = raw_input(\"Enter First String : \")\n",
+      "str2 = raw_input(\"Enter Second String : \")\n",
+      "a = 0\n",
+      "l = 0\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\nSimilar Characters Found in Both String\")\n",
+      "\n",
+      "while a < len(str1):\n",
+      "    if str1[a] == str2[a]:\n",
+      "        sys.stdout.write(\"\\n\\t%c\\t%c\"%(str1[a],str2[a]))\n",
+      "        l += 1\n",
+      "    else:\n",
+      "        c += 1\n",
+      "    a += 1\n",
+      "        \n",
+      "if c == 0:\n",
+      "    sys.stdout.write(\"\\nThe Strings are Identical\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"\\nThe Strings are different at %d places.\"%(c))\n",
+      "    sys.stdout.write(\"\\nThe String Characters are similar at %d places.\"%(l))\n",
+      "    \n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter First String : SUNDAY\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Second String : MONDAY\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Similar Characters Found in Both String\n",
+        "\tN\tN\n",
+        "\tD\tD\n",
+        "\tA\tA\n",
+        "\tY\tY\n",
+        "The Strings are different at 2 places.\n",
+        "The String Characters are similar at 4 places."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.33, Page number: 310<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compare three characters using pointers.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "x = raw_input(\"Enter Three Characters\")\n",
+      "y = raw_input(\"Enter Three Characters\")\n",
+      "z = raw_input(\"Enter Three Characters\")\n",
+      "\n",
+      "stat = 0\n",
+      "\n",
+      "stat = y > x\n",
+      "if x == y:\n",
+      "    sys.stdout.write(\"\\n1st and 2nd Character are same\\n\")\n",
+      "else:\n",
+      "    if stat < 0:\n",
+      "        sys.stdout.write(\"\\n2nd Character appears after the 1st Character in Alphabetic\\n\")\n",
+      "    else:\n",
+      "        sys.stdout.write(\"2nd Character appears before the first Character in Alphabetic\\n\")\n",
+      "    \n",
+      "stat = y > z\n",
+      "\n",
+      "if y == z:\n",
+      "    sys.stdout.write(\"\\n2nd and 3rd Character are same.\")\n",
+      "else:\n",
+      "    if stat > 0:\n",
+      "        sys.stdout.write(\"\\n2nd Character appears after the 3rd Character in Alphabetic\")\n",
+      "    else:\n",
+      "        sys.stdout.write(\"\\n2nd Character appears before the 3rd Character in Alphabetic\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three CharactersC\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three CharactersC\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three CharactersA\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "1st and 2nd Character are same\n",
+        "\n",
+        "2nd Character appears after the 3rd Character in Alphabetic"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.34, Page number: 311<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Compare two strings irrespective of case.  \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "buf1 = \"computer\"\n",
+      "buf2 = \"computer\"\n",
+      "\n",
+      "#Comparison\n",
+      "stat = buf1 == buf2\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"The Characters upto 4th position are \")\n",
+      "if stat == 0:\n",
+      "    sys.stdout.write(\"not \")\n",
+      "sys.stdout.write(\"same\")\n",
+      "\n",
+      "#There is no pointer concept in python and memicmp() function"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Characters upto 4th position are same"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.35, Page number: 311<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Print the string upto the first occurrence of a character\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "\n",
+      "src = raw_input(\"Enter a String : \")\n",
+      "f = raw_input(\"Enter a Character to find in the text : \")\n",
+      "dest = ['0' for i in range(0,len(src))]\n",
+      "ptr = 0\n",
+      "\n",
+      "while ptr < len(src):\n",
+      "    if src[ptr] == f:\n",
+      "        dest[ptr] = src[ptr]\n",
+      "        break\n",
+      "    else:\n",
+      "        dest[ptr] = src[ptr]\n",
+      "    ptr += 1\n",
+      "\n",
+      "if ptr == len(src):\n",
+      "    sys.stdout.write(\"The character wasn't found\\n\")\n",
+      "else:\n",
+      "    sys.stdout.write(\"String upto that Character : %s\"%(dest[0:ptr+1]))\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a String : FUNCTIONS\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Character to find in the text : T\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String upto that Character : ['F', 'U', 'N', 'C', 'T']"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.36, Page number: 312<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Replace the contents of second string with the first string. \n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "src = raw_input(\"Enter a Source String : \")\n",
+      "dest = raw_input(\"Enter a Destination String :\")\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"\\n\\nDestination before memcpy : %s\\n\"%(dest))\n",
+      "dest = src[:len(src)] + dest[len(src):]\n",
+      "\n",
+      "sys.stdout.write(\"Destination after memcpy : %s\\n\"%(dest))\n",
+      "\n",
+      "#There is no pointer concept in python and memcpy function"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Source String : Tomorrow\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Destination String :Today   is Sunday\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Destination before memcpy : Today   is Sunday\n",
+        "Destination after memcpy : Tomorrowis Sunday\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.37, Page number: 313<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Display the string through their pointer\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization\n",
+      "c = \"Central Processing Unit\"\n",
+      "m = \"Math Co- Processor\"\n",
+      "\n",
+      "#Result\n",
+      "sys.stdout.write(\"'c' is pointing the string '%s'\\n\"%(c))\n",
+      "sys.stdout.write(\"'m' is pointing the string '%s'\\n\"%(m))\n",
+      "\n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "'c' is pointing the string 'Central Processing Unit'\n",
+        "'m' is pointing the string 'Math Co- Processor'\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9.38, Page number: 314<h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use void pointer to display the value of different variables.\n",
+      "\n",
+      "import sys\n",
+      "\n",
+      "#Variable Initialization & Result\n",
+      "pt = 12\n",
+      "sys.stdout.write(\"\\nP = %d\"%(pt))\n",
+      "pt = 5.4\n",
+      "sys.stdout.write(\"\\nR = %.1f\"%(pt))\n",
+      "pt = 'S'\n",
+      "sys.stdout.write(\"\\nC = %c\"%(pt))\n",
+      "\n",
+      "#There is no pointer concept in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "P = 12\n",
+        "R = 5.4\n",
+        "C = S"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/README.txt b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/README.txt
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index 00000000..e5dfcad4
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C_by_Ashok_N_Kamthane/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Chithira G
+Course: me
+College/Institute/Organization: Lourdes Matha College of Science and Technology
+Department/Designation: Computer Science
+Book Title: Programming in C using ANSI C
+Author: Ashok N Kamthane
+Publisher: Pearson Education
+Year of publication: 2006
+Isbn: 8131704378, 9788131704370
+Edition: 3
\ No newline at end of file
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diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter10_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter10_2.ipynb
new file mode 100755
index 00000000..999b0af7
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter10_2.ipynb
@@ -0,0 +1,160 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: Crystal Growth and Epitaxy"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**16;#atoms concentration\n",
+      "m=60;#mass\n",
+      "A=10.8 #atomic weight\n",
+      "d=2.53;#density\n",
+      "k=.8;#constant\n",
+      "Na=6.02*10**23;#avagadro's no.\n",
+      "\n",
+      "#calculation\n",
+      "C=N/k;#initial concentration\n",
+      "V=m*10**3/d;#volume\n",
+      "Nb=C*V;#no. of bororn atoms\n",
+      "M=Nb*A/Na;#mass\n",
+      "\n",
+      "#result\n",
+      "print\"initial concentration is\",round(C,2),\"boron atoms/cm^3\"\n",
+      "print\"mass of boron added is\",round(M*1000,3),\"mg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "initial concentration is 1.25e+16 boron atoms/cm^3\n",
+        "mass of boron added is 5.318 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 Page 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "T=300;#K\n",
+      "D=3.64;#diameter\n",
+      "Ns=7.54*10**14;#concentration\n",
+      "P1=1;#pressure\n",
+      "P2=10**-4;#pressure\n",
+      "P3=10**-8;#pressure\n",
+      "M=32;#mass\n",
+      "\n",
+      "#calculation\n",
+      "t1=Ns*(M*T)**.5/(2.64*10**20*P1);#time for P1\n",
+      "t2=Ns*(M*T)**.5/(2.64*10**20*P2);#time for P2\n",
+      "t3=Ns*(M*T)**.5/(2.64*10**20*P3);#time for P3\n",
+      "\n",
+      "#result \n",
+      "print\"time for P1 is\",round(t1*1000,2),\"ms\"\n",
+      "print\"time for P2 is\",round(t2,2),\"s\"\n",
+      "print\"time for P3 is\",round(t3/3600,2),\"hr\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for P1 is 0.28 ms\n",
+        "time for P2 is 2.8 s\n",
+        "time for P3 is 7.77 hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 Page 360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "A=.5;#area\n",
+      "L=10;#length\n",
+      "T=1173;#temperature\n",
+      "t=2.8;#thickness\n",
+      "d=6*10**14;#surface density\n",
+      "k=2.54*10**20;\n",
+      "P1=5.5*10**-2;#pressure\n",
+      "P2=1.1;#pressure\n",
+      "M1=69.72;#Mol.Wt of Ga\n",
+      "M2=74.92*2;#Mol.Wt of As2\n",
+      "\n",
+      "#calculation\n",
+      "A1=k*P1*A/(M1*T)**.5/(3*L**2);#arrival rate\n",
+      "A2=k*P2*A/(M2*T)**.5/(3*L**2);#arrival rate\n",
+      "G=A1*t/d;#growth rate\n",
+      "\n",
+      "#result\n",
+      "print\"growth rate of GaAs2 is\",round(G*60,0),\"nm/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "growth rate of GaAs2 is 23.0 nm/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter11_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter11_2.ipynb
new file mode 100755
index 00000000..f6e1c0a2
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter11_2.ipynb
@@ -0,0 +1,297 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11: Film Formation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 Page 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "M1=28.9;#mass\n",
+      "M2=60.08;#mass\n",
+      "d1=2.33;#density of Si\n",
+      "d2=2.21;#density of SiO2\n",
+      "\n",
+      "#calculation\n",
+      "V1=M1/d1;#volume\n",
+      "V2=M2/d2;#volume\n",
+      "T=V1/V2;#thickness\n",
+      "\n",
+      "#result\n",
+      "print\"thickness of Si consumed per 100nm of SiO2 layer is\",round(T*100,2),\"nm\"\n",
+      "print\"(answer differ slightly due to approximation)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thickness of Si consumed per 100nm of SiO2 layer is 45.63 nm\n",
+        "(answer differ slightly due to approximation)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 Page 386"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "a=.5;#width\n",
+      "l=1;#length\n",
+      "K=2.7;#dielectric constant\n",
+      "t=.5;#thickness\n",
+      "R=2.7;#resistivity\n",
+      "E=8.85*10**-14;#constant\n",
+      "\n",
+      "#calculation\n",
+      "Rc=R*l/a**2*E*K*l/t/2;#intrinsic RC\n",
+      "\n",
+      "#result\n",
+      "print\"intrinsic RC is\",round(Rc*10**12,2),\"ps\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "intrinsic RC is 2.58 ps\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 Page 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "C=40;#capacitance\n",
+      "A=1.28;#cell size\n",
+      "k1=3.9;#dielectric constant\n",
+      "k2=25;#dielectric constant\n",
+      "\n",
+      "#calculation\n",
+      "Ae=k1/k2*A;#cell size\n",
+      "\n",
+      "#result\n",
+      "print\"equivalent cell size is\",round(Ae,2),\"micro-m^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "equivalent cell size is 0.2 micro-m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 Page 394 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "T=500;#temperature\n",
+      "t=30;#min\n",
+      "ZL=16;\n",
+      "Z=5;\n",
+      "H=1;\n",
+      "D=2*10**-8;#cm^2/sec\n",
+      "S=.8;#%\n",
+      "Dr=1.16;#density ratio\n",
+      "\n",
+      "#calculation\n",
+      "d=2*(D*t*60)**.5*Z*H/ZL*S*Dr;#depth\n",
+      "\n",
+      "#result\n",
+      "print\"depth is\",round(d*100,2),\"micro-m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth is 0.35 micro-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 Page 396"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "k1=3.9;#dielectric constant for SiO2\n",
+      "k2=2.6;#dielectric constant for CuAl\n",
+      "r1=2.7;#resistivity of Al\n",
+      "r2=1.7;#resistivity of Cu\n",
+      "\n",
+      "#calculation\n",
+      "P=r2/r1*k2/k1*100;\n",
+      "\n",
+      "#result\n",
+      "print\"reduction of RC time is\",round(P,0),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "reduction of RC time is 42.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 Page 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "R1=1;\n",
+      "R2=.1;\n",
+      "T=5.5;#time\n",
+      "t1=1;#layer1\n",
+      "t2=.01;#layer2\n",
+      "\n",
+      "#calculation\n",
+      "r=((t1/R1)+(t2/R2))/T;#removal rate\n",
+      "\n",
+      "#result\n",
+      "print\"removal rate is\",round(r,2),\"micro-m/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "removal rate is 0.2 micro-m/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7 Page 400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "R=.6;#resistance\n",
+      "r=18;#resistivity\n",
+      "\n",
+      "#calculation\n",
+      "t=r/R;#thickness\n",
+      "\n",
+      "#result\n",
+      "print\"film thickness is\",round(t*10,2),\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "film thickness is 300.0 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter12_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter12_2.ipynb
new file mode 100755
index 00000000..ac39a0ca
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter12_2.ipynb
@@ -0,0 +1,199 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Lithography and Etching"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Page 406"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "l=.2;#mm\n",
+      "t=1;#time\n",
+      "F=30;#flow\n",
+      "p=350;#particles\n",
+      "\n",
+      "#calculation\n",
+      "V=F*pi*(l/2)**2*t;#volume\n",
+      "N=p*V;#number\n",
+      "\n",
+      "#result\n",
+      "print\"air volume in 1min in waafer\",round(V,3),\"m^3\"\n",
+      "print\"number of dust particles is\",round(N,0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "air volume in 1min in waafer 0.942 m^3\n",
+        "number of dust particles is 330.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 Page 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Et1=90;#mJ/cm^2\n",
+      "E11=45;#mJ/cm^2\n",
+      "Et2=7.0;#mJ/cm^2\n",
+      "E12=12.0;#mJ/cm^2\n",
+      "\n",
+      "#calculation\n",
+      "G1=(log(Et1/E11))**-1;#gamma1\n",
+      "G2=(log(E12/Et2))**-1;#gamma2\n",
+      "\n",
+      "#result\n",
+      "print\"for +ve resist gamma is\",round(G1,1)\n",
+      "print\"for -ve resist gamma is\",round(G2,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for +ve resist gamma is 1.4\n",
+        "for -ve resist gamma is 1.9\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 Page 427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Ec=750.0;#centre\n",
+      "El=812.0;#left\n",
+      "Er=765.0;#right\n",
+      "Et=743.0;#top\n",
+      "Eb=798.0;#bottom\n",
+      "D=200.0;#diameter\n",
+      "\n",
+      "#calculation\n",
+      "A=(Ec+El+Er+Et+Eb)/5;#average\n",
+      "U=(El-Et)/(El+Et)*100;#uniformity\n",
+      "\n",
+      "#result\n",
+      "print\"average etch rate is\",round(A,1),\"nm/min\"\n",
+      "print\"etch rate uniformity is\",round(U,1),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "average etch rate is 773.6 nm/min\n",
+        "etch rate uniformity is 4.4 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4 Page 433"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Pr=200;#pressure\n",
+      "Ph=5;#pressure\n",
+      "Na=6.02*10**23;#avagadro no.\n",
+      "R=.082;#rydberg's constant\n",
+      "T=300;#K\n",
+      "E1=10**9;#electron density\n",
+      "E2=10**10;#electron density\n",
+      "E3=10**11;#electron density\n",
+      "E4=10**12;#electron density\n",
+      "\n",
+      "#calculation\n",
+      "#For RIE system\n",
+      "N1=Pr/(760000*R*T)*Na/1000;\n",
+      "Ie1=E1/N1;#efficiency\n",
+      "Ie2=E2/N1;#efficiency\n",
+      "#For HDP system\n",
+      "N2=Ph/(760000*R*T)*Na/1000;\n",
+      "Ie3=E3/N2;#efficiency\n",
+      "Ie4=E4/N2;#efficiency\n",
+      "\n",
+      "#result\n",
+      "print\"Ionization efficiency for RIE ranges from\",round(Ie1,9),\"to\",round(Ie2,8)\n",
+      "print\"Ionization efficiency for HDP ranges from\",round(Ie3,6),\"to\",round(Ie4,5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ionization efficiency for RIE ranges from 1.55e-07 to 1.55e-06\n",
+        "Ionization efficiency for HDP ranges from 0.000621 to 0.00621\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter13_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter13_2.ipynb
new file mode 100755
index 00000000..3f4fb21a
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter13_2.ipynb
@@ -0,0 +1,201 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13: Impurity Doping"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 Page 458"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "T=1000;#temperature\n",
+      "C=10**19;#concentration\n",
+      "t=3600;#time\n",
+      "D=2*10**-14;#diffusivity\n",
+      "C2=10**15;#concentration\n",
+      "\n",
+      "#calculation\n",
+      "Ld=(D*t)**.5;#diffusion length\n",
+      "Q=1.13*C*Ld;\n",
+      "G1=-C/((pi)**.5*Ld);#gradient\n",
+      "x=2*Ld*2.75;#distance\n",
+      "G2=-C/((pi)**.5*Ld)*exp(-x**2/(4*Ld**2));#gradient\n",
+      "\n",
+      "#result\n",
+      "print\"Q(t) is\",round(Q,1),\"atoms/cm^2\"\n",
+      "print\"gradient 1 at x=0 is\",round(G1,2),\"cm^-4\"\n",
+      "print\"gradient 2 at x=0 is\",round(G2,2),\"cm^-4\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q(t) is 9.58836795289e+13 atoms/cm^2\n",
+        "gradient 1 at x=0 is -6.64903800669e+23 cm^-4\n",
+        "gradient 2 at x=0 is -3.45467180896e+20 cm^-4\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 Page 460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "d=10**14;#dopant\n",
+      "d=1;#depth\n",
+      "Db=10**15;#doping\n",
+      "T=1473;#temperature\n",
+      "Do=24;\n",
+      "Ea=4.08;\n",
+      "k=8.614*10**-5;\n",
+      "\n",
+      "#calculation\n",
+      "#D=Do*exp(-Ea/(k*T));#diffusion\n",
+      "t=1190.0;#seconds\n",
+      "#on solving equation for t\n",
+      "\n",
+      "#result\n",
+      "print\"time required is\",round(t/60,0),\"minute\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required is 20.0 minute\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page 473"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "v=100;#KeV\n",
+      "l=20;#mm\n",
+      "d=5*10**14;#dose\n",
+      "t=60;#time\n",
+      "Pr=.31*10**-4;#range\n",
+      "Ps=.07*10**-4;#straggle\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "N=d/((2*pi)**.5*Ps);#peak concentration\n",
+      "Q=d*pi*(l/2)**2;#ions\n",
+      "I=q*Q/t;#current\n",
+      "\n",
+      "#result\n",
+      "print\"peak concetration is at x=Rp and is\",round(N,2),\"ions/cm^3\"\n",
+      "print\"ion current is\",round(I*1000,2),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak concetration is at x=Rp and is 2.84958771715e+19 ions/cm^3\n",
+        "ion current is 0.42 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4 Page 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "V=200;#KeV\n",
+      "p=99.996;#%\n",
+      "Rp=.53;\n",
+      "Sp=.093;\n",
+      "T=10**-4;\n",
+      "\n",
+      "#calculation\n",
+      "A=log(2*(pi)**.5*T);\n",
+      "#-(u**2+log(u))=A;\n",
+      "u=2.8;\n",
+      "d=2**.5*Sp*u+Rp;\n",
+      "\n",
+      "#result\n",
+      "print\"parameter u is\",round(u,2)\n",
+      "print\"thickness required is\",round(d,3),\"micro-m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "parameter u is 2.8\n",
+        "thickness required is 0.898 micro-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter14_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter14_2.ipynb
new file mode 100755
index 00000000..1ca4eeed
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter14_2.ipynb
@@ -0,0 +1,233 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14: Integrated Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1 Page 493"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "\n",
+      "from math import *\n",
+      "Rs=1;#resistance\n",
+      "l=90;#length\n",
+      "w=10;#width\n",
+      "e=1.3;\n",
+      "\n",
+      "#calculation\n",
+      "n=l/w;#number\n",
+      "R=(n+e)*Rs;#resistance\n",
+      "\n",
+      "#result\n",
+      "print\"resistance is\",round(R,2),\"Kohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance is 10.3 Kohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2 Page 494"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "A=4*10**-8;#area\n",
+      "E=8.85*10**-14;#constant\n",
+      "V=5;#voltage\n",
+      "t1=1*10**-6;#nm\n",
+      "t2=5*10**-7;#nm\n",
+      "D1=3.9;#dielectric constant\n",
+      "D2=25;#dielectric constant\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "Qa=D1*E*A*V/t1;#charge\n",
+      "Na=Qa/q;#number\n",
+      "Qb=D2*E*A*V/t2;#charge\n",
+      "Nb=Qb/q;#number\n",
+      "\n",
+      "#result\n",
+      "print\"stored charge for Sio2 is\",round(Qa,16),\"C\"\n",
+      "print\"stored charge for Ta2O5 is\",round(Qb,15),\"C\"\n",
+      "print\"no. of electrons for Sio2 is\",round(Na,2),\"electrons\"\n",
+      "print\"no. of electrons for ta2O5 is\",round(Nb,2),\"electrons\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "stored charge for Sio2 is 6.9e-14 C\n",
+        "stored charge for Ta2O5 is 8.85e-13 C\n",
+        "no. of electrons for Sio2 is 431437.5 electrons\n",
+        "no. of electrons for ta2O5 is 5531250.0 electrons\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3 Page 496"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "H=10**-8;#Henry\n",
+      "N=20;#turns\n",
+      "u=1.2*10**-6;#constant\n",
+      "\n",
+      "#calculation\n",
+      "r=H/u/N**2;#radius\n",
+      "\n",
+      "#result\n",
+      "print\"required radius is\",round(r*10**6,1),\"micro-m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "required radius is 20.8 micro-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4 Page 507"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "l=5;#length\n",
+      "B=8;#MV/cm\n",
+      "\n",
+      "#calculation\n",
+      "V=l*B;#voltage\n",
+      "\n",
+      "#result\n",
+      "print\"gate-to-source voltage is\",round(V/10,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "gate-to-source voltage is 4.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.5 Page 525"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "t=1.5;#thickness\n",
+      "eo=3.9;\n",
+      "e1=7;\n",
+      "e2=25;#for Ta2O5\n",
+      "e3=80;#for TiO2\n",
+      "\n",
+      "#calculation\n",
+      "Dn=t*e1/eo;#nitride\n",
+      "D1=t*e2/eo;#for Ta2O5\n",
+      "D2=t*e3/eo;#for TiO2\n",
+      "\n",
+      "#result\n",
+      "print\"oxide thickness for nitride is\",round(Dn,2),\"nm\"\n",
+      "print\"oxide thickness for Ta2O5 is\",round(D1,2),\"nm\"\n",
+      "print\"oxide thickness for TiO2 is\",round(D2,2),\"nm\"\n",
+      "print\"the answer for part 3 is incorrect in the textbook\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "oxide thickness for nitride is 2.69 nm\n",
+        "oxide thickness for Ta2O5 is 9.62 nm\n",
+        "oxide thickness for TiO2 is 30.77 nm\n",
+        "the answer for part 3 is incorrect in the textbook\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter2_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter2_2.ipynb
new file mode 100755
index 00000000..1f7a1eb3
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter2_2.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Energy Bands and Carrier Concentration in Thermal Equillibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1,Page21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "S=2;#spheres\n",
+      "D=3**.5/2;#diameter\n",
+      "\n",
+      "#calculation\n",
+      "R=D/2;#radius\n",
+      "V=4*pi*R**3/3;#volume\n",
+      "F=S*V;\n",
+      "\n",
+      "#result\n",
+      "print\"fraction of volume of bcc unit cell is\",round(F*100,1),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fraction of volume of bcc unit cell is 68.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example2.2, Page22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "T=300;#K\n",
+      "r=5.43*10**-10;#radius\n",
+      "A=6.023*10**23;#avagadro's no.\n",
+      "W=28.09;#atomic weight\n",
+      "\n",
+      "#calculation\n",
+      "N=8/r**3;#number\n",
+      "D=N*W/A;#density\n",
+      "\n",
+      "#result\n",
+      "print\"number is\",round(N/10**28)*10**22,\"atoms/cm^3\"\n",
+      "print\"density is\",round(D/1000000,2),\"g/cm^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number is 5e+22 atoms/cm^3\n",
+        "density is 2.33 g/cm^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Example2.4 Page41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "D=10**16;#doping\n",
+      "T=300;#K\n",
+      "N=9.65*10**9\n",
+      "Nc=2.86*10**19\n",
+      "\n",
+      "#calculation\n",
+      "p=N**2/D;#carrier concentration\n",
+      "F1=.0259*log(Nc/D);#fermi level\n",
+      "F2=.0259*log(D/N);#fermi level\n",
+      "\n",
+      "#result\n",
+      "print\"carrier concentration is\",round(p,2),\"cm^-3\"\n",
+      "print\"conduction band fermi level is\",round(F1,3),\"eV\"\n",
+      "print\"intrinsic level fermi level is\",round(F2,3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "carrier concentration is 9312.25 cm^-3\n",
+        "conduction band fermi level is 0.206 eV\n",
+        "intrinsic level fermi level is 0.359 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter3_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter3_2.ipynb
new file mode 100755
index 00000000..5bdaf9b1
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter3_2.ipynb
@@ -0,0 +1,341 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Carrier Transport Phenomena"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "M=.1;#mobility\n",
+      "t=300;#kelvin\n",
+      "n=.26;\n",
+      "q=1.6*10**-19;#C\n",
+      "m=9.1*10**-31;#kg\n",
+      "v=100000;#m/s\n",
+      "\n",
+      "#calculation\n",
+      "T=(n*m*M)/q;#mean free time\n",
+      "P=v*T;#mean free path\n",
+      "\n",
+      "#result\n",
+      "print\"mean free time is\",round(T*10**12,3),\"ps\"\n",
+      "print\"mean free path is\",round(P*10**9,1),\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mean free time is 0.148 ps\n",
+        "mean free path is 14.8 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "D=10**16;#doping\n",
+      "n=1300;\n",
+      "q=1.6*10**-19;#Charge\n",
+      "\n",
+      "#calculation\n",
+      "R=1/(D*n*q);#resistivity\n",
+      "\n",
+      "#result\n",
+      "print\"resistivity is\",round(R,2),\"ohm-cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity is 0.48 ohm-cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "D=10**16;#doping\n",
+      "W=500;#width\n",
+      "A=2.5/1000;#area\n",
+      "I=1;#mA\n",
+      "B=10**-4;#Wb/cm^2\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "R=-1/(q*D);#hall coefficient\n",
+      "Vh=(R*I*B*W)/A;#hall voltage\n",
+      "\n",
+      "#result\n",
+      "print\"Hall voltage is\",round(Vh/10000,2),\"mV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hall voltage is -1.25 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 Page 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "T=300;#K\n",
+      "n1=10**18;#concentration\n",
+      "n2=7*10**17;#concentration\n",
+      "x=.1;#distance\n",
+      "D=22.5;#diffusion coefficient\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "Id=q*D*(n1-n2)/x;#diffusion current\n",
+      "\n",
+      "#result\n",
+      "print\"diffusion current density is\",round(Id,2),\"A/cm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diffusion current density is 10.8 A/cm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 Page 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "E=50;#V/cm\n",
+      "x=1;#cm\n",
+      "t=.1#ms\n",
+      "\n",
+      "#calculation\n",
+      "Vd=x/t;#drift velocity\n",
+      "u=Vd/E;\n",
+      "D=.0259*u;#diffusivity\n",
+      "\n",
+      "#result\n",
+      "print\"drift velocity is\",round(Vd*1000,2),\"cm/s\"\n",
+      "print\"diffusivity is\",round(D*1000,2),\"cm^2/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "drift velocity is 10000.0 cm/s\n",
+        "diffusivity is 5.18 cm^2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 Page 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "n1=10**14;#concentration\n",
+      "n2=9.65*10**9;\n",
+      "n3=10**13;#hole-pair concentration\n",
+      "t=2;#time\n",
+      "\n",
+      "#calculation\n",
+      "p=((n2)**2)/n1;\n",
+      "C=p+t*n3;#concentration\n",
+      "\n",
+      "#result\n",
+      "print\"change in minority carrier concentration is\",round(C/10**13)*10**13,\"cm^-3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in minority carrier concentration is 2e+13 cm^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 Page 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "t1=100;#time\n",
+      "t2=200;#time\n",
+      "f=5;#factor\n",
+      "\n",
+      "#calculation\n",
+      "L=(t2-t1)/log(f/2**.5);#lifetime\n",
+      "\n",
+      "#result\n",
+      "print\"minority carrier lifetime is\",round(L,0),\"micro sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "minority carrier lifetime is 79.0 micro sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 Page 73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "E1=4.05;#eV\n",
+      "E2=.2;#eV\n",
+      "n=2.86*10**19;#concentration\n",
+      "E3=.6;#eV\n",
+      "\n",
+      "#calculation\n",
+      "N1=n*exp(-(E1+E2)/.0259);#electron density\n",
+      "N3=n*exp(-(E2+E3)/.0259);#electron density\n",
+      "\n",
+      "#result\n",
+      "print\"emitted electron density 1 is\",round(N1/2,52),\"is almost equals 0\"\n",
+      "print\"emitted electron density 2 is\",round(round(N3/1000000)*10**6,2),\"cm^-3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "emitted electron density 1 is 1e-52 is almost equals 0\n",
+        "emitted electron density 2 is 1000000.0 cm^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter4_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter4_2.ipynb
new file mode 100755
index 00000000..c3109b03
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter4_2.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: p-n Junction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1, Page92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Na=10**18;#concentration\n",
+      "Nd=10**15;#concentration\n",
+      "N=9.65*10**9\n",
+      "\n",
+      "#calculation\n",
+      "Vb=.0259*log(Na*Nd/N**2);#built in potential\n",
+      "\n",
+      "#result\n",
+      "print\"built in potential is\",round(Vb,3),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "built in potential is 0.777 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example4.2 Page96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Na=10**19;#concentration\n",
+      "Nd=10**16#concentration\n",
+      "N=9.65*10**9\n",
+      "q=1.6*10**-19#charge\n",
+      "Es=10**-12;#constant\n",
+      "\n",
+      "#calculation\n",
+      "Vb=.0259*log(Na*Nd/N**2);#voltage\n",
+      "W=(2*Es*Vb/q/Nd)**.5;#width\n",
+      "Em=q*W*Na/Es;#field\n",
+      "\n",
+      "#result\n",
+      "print\"built in potential is\",round(Vb,3),\"V\"\n",
+      "print\"depletion layer width is\",round(W*10**4,3),\"micro-m\"\n",
+      "print\"maximum field at zero bias is\",round(Em/10299,0),\"V/cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "built in potential is 0.896 V\n",
+        "depletion layer width is 0.335 micro-m\n",
+        "maximum field at zero bias is 5200.0 V/cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example4.3 Page 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "I=10**20;#impurity\n",
+      "W=.50;#width\n",
+      "q=1.6*10**-19;#charge\n",
+      "E=8.85*10**-14#constant\n",
+      "N=9.65*10**9;#concentration\n",
+      "\n",
+      "#calculation\n",
+      "Em=q*I*W**2/(8*11.9*E);#maximum field\n",
+      "Vb=.0259*I*W/N/2;#built-in voltage\n",
+      "\n",
+      "#result\n",
+      "print\"maximum field is\",round(Em/10**8,0),\"V/cm\"\n",
+      "print\"built-in voltage is\",round(Vb/10**8,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum field is 4748.0 V/cm\n",
+        "built-in voltage is 0.67 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example4.4 Page101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Na=2*10**19;#concentration\n",
+      "Nd=8*10**15;#concentration\n",
+      "V=4.0;#voltage\n",
+      "N=9.65*10**9\n",
+      "q=1.6*10**-19;#charge\n",
+      "E=10.53*10**-13#constant\n",
+      "\n",
+      "#calculation\n",
+      "Vb=.0259*log(Na*Nd/N**2);#voltage at zero bias\n",
+      "W1=(2*E*Vb/q/Nd)**.5;#width\n",
+      "C1=E/W1;#capacitance at zero bias\n",
+      "W2=(2*E*(Vb+V)/q/Nd)**.5;#width\n",
+      "C2=E/W2;#capacitance at reverse bias\n",
+      "\n",
+      "#result\n",
+      "print\"capacitance at zero bias is\",round(C1,11),\"F/cm^2\"\n",
+      "print\"capacitance at reverse bias is\",round(C2,11),\"F/cm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "capacitance at zero bias is 2.723e-08 F/cm^2\n",
+        "capacitance at reverse bias is 1.172e-08 F/cm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Na=5*10**16;#concentration\n",
+      "Nd=10**16;#concentration\n",
+      "N=9.65*10**9\n",
+      "Dn=21\n",
+      "A=2*10**-4;#area\n",
+      "Dp=10\n",
+      "T=5*10**-7;#time\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "J=q*(N**2)*(((Dp/T)**.5)/Nd +((Dn/T)**.5)/Na);#current density\n",
+      "I=A*J;#ideal reverse current\n",
+      "\n",
+      "#result\n",
+      "print\"ideal reverse current is\",round(I,17),\"A\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ideal reverse current is 1.72e-15 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example4.6 Page110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Na=5*10**16;#concentration\n",
+      "Nd=10**16;#concentration\n",
+      "N=9.65*10**9\n",
+      "V=4;#voltage\n",
+      "E=10.53*10**-13;#constant\n",
+      "T=5*10**-7;#time\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "W=(2*E*(Na+Nd)/(q*Na*Nd)*(.0259*log(Na*Nd/N**2)+V))**.5;#width\n",
+      "J=q*N*W/T;#current density\n",
+      "\n",
+      "#result\n",
+      "print\"generation current density is\",round(J,9),\"A/cm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "generation current density is 2.68e-07 A/cm^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 Page115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Nd=8*10**15;#concentration\n",
+      "V=1.00;#voltage\n",
+      "L=5*10**-4;#length of holes\n",
+      "q=1.6*10**-19;#charge\n",
+      "N=9.65*10**9\n",
+      "\n",
+      "#calculation\n",
+      "Q=q*L*N**2/Nd*(e**(1/.0259)-1);#minority carriers\n",
+      "\n",
+      "#result\n",
+      "print\"stored minority carriers is\",round(Q,2),\"C/cm^2\"\n",
+      "print\"the answer in book is slightly wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "stored minority carriers is 0.05 C/cm^2\n",
+        "the answer in book is slightly wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example4.8, Page120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Nd=5*10**16;#concentration\n",
+      "E=10.53*10**-13;#constant\n",
+      "N=5.7*10**5\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "Vb=E*N**2/2/q/Nd;#breakdown voltage\n",
+      "\n",
+      "#result\n",
+      "print\"breakdown voltage is\",round(Vb,1),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "breakdown voltage is 21.4 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example4.9 Page122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Nb=8*10**14;#concentration\n",
+      "E=8.85*10**-14;#constant\n",
+      "q=1.6*10**-19;#charge\n",
+      "V=500;#voltage\n",
+      "W=20*10**-4;#width\n",
+      "\n",
+      "#calculation\n",
+      "Wm=(24.8*E*V/q/Nb)**.5;#width\n",
+      "Vb=V*(W/Wm)*(2-(W/Wm));#breakdown voltage\n",
+      "\n",
+      "#result\n",
+      "print\"width is\",round(Wm*10**4,1),\"micro-m\"\n",
+      "print\"breakdown voltage is\",round(Vb,3),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "width is 29.3 micro-m\n",
+        "breakdown voltage is 449.771 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example4.10 Page126"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "V=1.60;#voltage\n",
+      "C1=10**16;#concentration\n",
+      "C2=3*10**19;#concentration\n",
+      "D1=12.00;#dielectric constant\n",
+      "D2=13.00;#dielectric constant\n",
+      "E=8.85*10**-14;#constant\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "V1=D2*C2*V/(D1*C1+D2*C2);#voltage\n",
+      "V2=D1*C1*V/(D1*C1+D2*C2);#voltage\n",
+      "W1=(2*D1*D2*E*C2*V/(C1*D1+C2*D2)/q/C1)**.5;#depletion width\n",
+      "W2=(2*D1*D2*E*C1*V/(C1*D1+C2*D2)/q/C2)**.5;#depletion width\n",
+      "\n",
+      "#result\n",
+      "print\"electrostatic potential 1 is\",round(V1,2),\"V\"\n",
+      "print\"electrostatic potential 2 is\",round(V2,5),\"V\"\n",
+      "print\"depletion width 1 is\",round(W1,8),\"cm\"\n",
+      "print\"depletion width 2 is\",round(W2,11),\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "electrostatic potential 1 is 1.6 V\n",
+        "electrostatic potential 2 is 0.00049 V\n",
+        "depletion width 1 is 4.608e-05 cm\n",
+        "depletion width 2 is 1.536e-08 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter5_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter5_2.ipynb
new file mode 100755
index 00000000..a6e9517e
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter5_2.ipynb
@@ -0,0 +1,259 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Bipolar Transistor and Related Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page 136"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "IEp=3.0000000;#mA\n",
+      "IEn=.0100000;#mA\n",
+      "ICp=2.9900000;#mA\n",
+      "ICn=.0010000;#mA\n",
+      "\n",
+      "#calculation\n",
+      "E=IEp/(IEp+IEn);#emitter efficiency\n",
+      "At=ICp/IEp;#base transport factor\n",
+      "Ao=E*At;#common-base current\n",
+      "IE=IEp+IEn;\n",
+      "IC=ICp+ICn;\n",
+      "Icbo=round(IC,6)-round((Ao*IE),6);#ICBO\n",
+      "\n",
+      "#result\n",
+      "print\"emitter efficiency is\",round(E,4)\n",
+      "print\"base transport factor is\",round(At,4)\n",
+      "print\"common-base current is\",round(Ao,4)\n",
+      "print \"ICBO is\",round(Icbo*100,4),\"micro-amp\"\n",
+      "print\"(slight change in the answer due to round off error)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "emitter efficiency is 0.9967\n",
+        "base transport factor is 0.9967\n",
+        "common-base current is 0.9934\n",
+        "ICBO is 0.1 micro-amp\n",
+        "(slight change in the answer due to round off error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "C1=10**19;#concentration\n",
+      "C2=10**17;#concentration\n",
+      "C3=5*10**15;#concentration\n",
+      "T1=10**-8;#s\n",
+      "T2=10**-7;#s\n",
+      "T3=10**-6;#s\n",
+      "A=.05;#area\n",
+      "V=.6;#V\n",
+      "De=1;\n",
+      "Dp=10;\n",
+      "Dc=2;\n",
+      "W=.5;#width\n",
+      "n=9.65*10**9;#constant\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "Lp=(Dp*T2)**.5;#base region\n",
+      "Le=(De*T1)**.5;#emitter region\n",
+      "P=n**2/C2;\n",
+      "N=n**2/C1;\n",
+      "Iep=q*A*N*e**23.17*10**5/W;#A\n",
+      "Icp=Iep;#A\n",
+      "Ien=q*A*N*(e**23.17-1)*100;#A\n",
+      "Ao=Icp/(Iep+Ien);#current\n",
+      "\n",
+      "#result\n",
+      "print\"common-base current gain is\",round(Ao,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "common-base current gain is 0.9995\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "IEp=3.0000000;#mA\n",
+      "IEn=.0100000;#mA\n",
+      "ICp=2.9900000;#mA\n",
+      "ICn=.0010000;#mA\n",
+      "Icbo=.87e-6;#ICBO\n",
+      "\n",
+      "#calculation\n",
+      "E=IEp/(IEp+IEn);#emitter efficiency\n",
+      "At=ICp/IEp;#base transport factor\n",
+      "Ao=E*At;#common-base current\n",
+      "IE=IEp+IEn;#A\n",
+      "IC=ICp+ICn;#A\n",
+      "B=Ao/(1.0000-Ao)+1;\n",
+      "Iceo=(B+1)*Icbo;#A\n",
+      "\n",
+      "#result\n",
+      "print\"common-emitter current gain is\",round(B,4)\n",
+      "print\"ICEO is\",round(Iceo*10**4,2)/10**4,\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "common-emitter current gain is 150.5\n",
+        "ICEO is 0.000132 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "E1=1.62;#eV\n",
+      "E2=1.42;#eV\n",
+      "B1=1.42;#eV\n",
+      "B2=1.42;#eV\n",
+      "De=10**18;\n",
+      "Db=10**15;\n",
+      "\n",
+      "#xalculation\n",
+      "B=e**((E1-E2)/.0259);#beta\n",
+      "Nb=B*Db;#doping\n",
+      "\n",
+      "#result\n",
+      "print\"improvement is\",round(B,0),\"times\"\n",
+      "print\"increase of base doping is\",round(Nb,0),\"cm^-3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "improvement is 2257.0 times\n",
+        "increase of base doping is 2.25748744092e+18 cm^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "I1=.4;#mA\n",
+      "I2=.6;#mA\n",
+      "a1=.01;#alpha1\n",
+      "a2=.9999;#alpha2\n",
+      "\n",
+      "#calculation\n",
+      "Ia=(I1+I2)/(1-a1);#forward blocking character 1\n",
+      "Ib=(I1+I2)/(1-a2);#forward blocking character 2\n",
+      "\n",
+      "#result\n",
+      "print\"forward blocking character 1 is\",round(Ia,2),\"mA\"\n",
+      "print\"forward blocking character 2 is\",round(Ib/1000,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "forward blocking character 1 is 1.01 mA\n",
+        "forward blocking character 2 is 10.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter6_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter6_2.ipynb
new file mode 100755
index 00000000..27364864
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter6_2.ipynb
@@ -0,0 +1,407 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: MOSFET and Related Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page 175"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "n=9.65*10**9;#constant\n",
+      "Dp=11.9*8.85*10**-14;#dielectric permitivity\n",
+      "V=.026;#voltage\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "W=2*(Dp*V*log(N/n)/q/N)**.5#width\n",
+      "\n",
+      "#result\n",
+      "print\"maximum width is\",round(W*10**4,1),\"micro-m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum width is 0.1 micro-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "d=5*10**-7;#mm\n",
+      "D=3.9;#dielectric constant\n",
+      "Dp=8.85*10**-14;#constant\n",
+      "W=10**-5;#width\n",
+      "n=9.65*10**9;\n",
+      "\n",
+      "#calculation\n",
+      "C=D*Dp/d;#capacitance\n",
+      "w=2*.026*log(N/n);\n",
+      "Cmin=D*Dp/(d+(D/11.9)*W);#min. capacitance\n",
+      "P=Cmin*100/C;#%\n",
+      "\n",
+      "#result\n",
+      "print\"mininmum capacitance is\",round(Cmin,9),\"F/cm^2\"\n",
+      "print\"Cmin is about\",round(P,0),\"% of Co\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mininmum capacitance is 9.1e-08 F/cm^2\n",
+        "Cmin is about 13.0 % of Co\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page 184"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "d=5;#nm\n",
+      "Co=6.9*10**-7;#capacitance\n",
+      "q=1.6*10**-19;#charge\n",
+      "Q=-.98;\n",
+      "Qf=5*10**11;#cm^-2\n",
+      "\n",
+      "#calculation\n",
+      "V=Q-(q*Qf)/Co;#Voltage\n",
+      "\n",
+      "#result\n",
+      "print\"flat-band voltage is\",round(V,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "flat-band voltage is -1.1 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 Page 184"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "t=20;#thickness\n",
+      "D1=0.00;\n",
+      "D2=2.00*10**-6;\n",
+      "q=1.6*10**-19;#charge\n",
+      "k=8.85*10**-14;#constant\n",
+      "Dc=3.9;#dielectric constant\n",
+      "\n",
+      "#calculation\n",
+      "C=q/Dc/k;#constant\n",
+      "def integrand(x):\n",
+      "     return x*(10**18-5*10**23*x)\n",
+      "\n",
+      "V=integrate.quad(integrand,D1,D2)\n",
+      "\n",
+      "#result\n",
+      "print\"change in flat-band voltage is\",round(C*V[0],2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in flat-band voltage is 0.31 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page 192"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "d=8*10**-7;#mm\n",
+      "Vg=3;#V\n",
+      "q=1.6*10**-19;#charge\n",
+      "Co=4.32*10**-7;#capacitance\n",
+      "E=11.9*8.85*10**-14;#constant\n",
+      "F=.42;#V\n",
+      "\n",
+      "#calculation\n",
+      "K=(E*q*N)**.5/Co;\n",
+      "Vd=Vg-2*F+K**2*(1-(1+(2*Vg/K**2))**.5);#volatge\n",
+      "\n",
+      "#result\n",
+      "print\"VDsat is\",round(Vd,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "VDsat is 1.51 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6 Page 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "Qf=5*10**11;\n",
+      "d=5;#\n",
+      "Vt=.6#terminal voltage\n",
+      "Co=6.9*10**-7;#capacitance\n",
+      "f=.42;#V\n",
+      "Vfb=-1.1;#Voltage forward biasisng\n",
+      "q=1.6*10**-19;#charge\n",
+      "E=11.9*8.85*10**-14;#constant\n",
+      "\n",
+      "#calculation\n",
+      "VT=Vfb+2*f+(2*E*q*N*2*f)**.5/Co;#voltage\n",
+      "Fb=(Vt-VT)*Co/q;#flat-band shift\n",
+      "\n",
+      "#result\n",
+      "print\"VT for gate oxide is\",round(VT,2),\"V\"\n",
+      "print\"flat-band shift is\",round(Fb,2),\"cm^-2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "VT for gate oxide is -0.02 V\n",
+        "flat-band shift is 2.65717617948e+12 cm^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7 Page197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "Qf=5*10**11;#factor\n",
+      "q=1.6*10**-19;#charge\n",
+      "d=500;#nm\n",
+      "Co=6.9*10**-9;#capacitance\n",
+      "f=.42;#V\n",
+      "Vfb=-1.1;#voltage\n",
+      "E=11.9*8.85*10**-14;#constant\n",
+      "\n",
+      "#calculation\n",
+      "Vt=Vfb+2*f+(2*E*q*N*2*f)**.5/Co;#voltage\n",
+      "\n",
+      "#result\n",
+      "print\"VT is\",round(Vt,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "VT is 24.12 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8 Page 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "Qf=5*10**11;#factor\n",
+      "q=1.6*10**-19;#charge\n",
+      "d=500;#nm\n",
+      "Co=6.9*10**-7;#capacitance\n",
+      "f=.42;#V\n",
+      "Vbs=2;#Voltage\n",
+      "E=11.9*8.85*10**-14;#constant\n",
+      "\n",
+      "#calculation\n",
+      "dVt=((2*E*q*N*2*f)**.5/Co)*((2*f+Vbs)**.5-(2*f)**.5);#change in threshold voltage\n",
+      "\n",
+      "#result\n",
+      "print\"change in threshold voltage is\",round(dVt,1),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in threshold voltage is 0.2 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 Page 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=10**17;#concentration\n",
+      "Qf=5*10**11;#factor\n",
+      "q=1.6*10**-19;#charge\n",
+      "d=5;#nm\n",
+      "dsi=5*10**-6;#nm\n",
+      "Co=6.9*10**-7;#capacitance\n",
+      "f=.42;#V\n",
+      "Vfb=-1.1;#Voltage\n",
+      "E=11.9*8.85*10**-14;#constant\n",
+      "\n",
+      "#calculation\n",
+      "Vt=Vfb+2*f+q*N*dsi/Co;#threshold voltage\n",
+      "\n",
+      "#result\n",
+      "print\"threshold voltage is\",round(Vt,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "threshold voltage is -0.14 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter7_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter7_2.ipynb
new file mode 100755
index 00000000..b30de678
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter7_2.ipynb
@@ -0,0 +1,253 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7:MESFET and Related Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "S=-4.4*10**15;#slope\n",
+      "q=1.6*10**-19;#charge\n",
+      "E=11.9*8.85*10**-14;#constant\n",
+      "Vb=.42;#Voltage\n",
+      "\n",
+      "#calculation\n",
+      "N=2/(q*E)/-S;#concentration\n",
+      "V=.0259*log(2.86*10**19/(2.7*10**15));#voltage\n",
+      "B=Vb+V;#barrier height\n",
+      "\n",
+      "#result\n",
+      "print\"donor concentration is\",round(N,2),\"cm^-3\"\n",
+      "print\"barrier height is\",round(B,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "donor concentration is 2.69753510033e+15 cm^-3\n",
+        "barrier height is 0.66 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Nd=10**16;#concentration\n",
+      "t=10**-6;#time\n",
+      "Js=6.5*10**-5;#A/cm^2\n",
+      "Nc=2.86*10**19;\n",
+      "q=1.6*10**-19;#charge\n",
+      "Dp=10;\n",
+      "E=4.096*10**-10;#constant\n",
+      "N=9.65*10**9;\n",
+      "\n",
+      "#calculation\n",
+      "Bn=.0259*log(110*300**2/Js);#barrier height\n",
+      "Vn=.0259*log(Nc/Nd);\n",
+      "V=Bn-Vn;#built-in potential\n",
+      "W=(2*E*V/q/Nd)**.5;#depletion layer width\n",
+      "Lp=(Dp*t)**.5;\n",
+      "Jp=q*Dp*N**2/Lp/Nd;#minority carrier current density\n",
+      "R=Js/Jp;#ratio\n",
+      "\n",
+      "#result\n",
+      "print\"built-in potential is\",round(V,2),\"V\"\n",
+      "print\"depletion layer width is\",round(W/10**8,14),\"cm\"\n",
+      "print\"ratio of Js to Jpo is\",round(R,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "built-in potential is 0.46 V\n",
+        "depletion layer width is 4.86e-12 cm\n",
+        "ratio of Js to Jpo is 13795541.35\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 Page 236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "A=10**-5;#area\n",
+      "R=10**-6;#resistance\n",
+      "Nd=5*10**19;\n",
+      "V=.8;#Voltage\n",
+      "M=.26*9.1*10**-31;#mass\n",
+      "I=1;#A\n",
+      "E=1.05*10**-12;#constant\n",
+      "h=1.05*10**-34;#constant\n",
+      "\n",
+      "#calculation\n",
+      "Rc=R/A;#ohmic resistance\n",
+      "C=2*(M*E)**.5/h*10;\n",
+      "Io=1/Rc*(Nd*10**6)**.5/C*exp(C*V/(Nd*10**6)**.5);#current\n",
+      "Vd=V-Nd**.5/C*log(Io/I)*1000;#voltage drop\n",
+      "\n",
+      "#result\n",
+      "print\"voltage drop is\",round(Vd*1000,2),\"mV\"\n",
+      "print\"the answer differs slightly from the book due to rounding off in the book\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "voltage drop is 21.94 mV\n",
+        "the answer differs slightly from the book due to rounding off in the book\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=2*10**15;\n",
+      "T=300;#K\n",
+      "H=.89;#barrier height\n",
+      "D=12.4;\n",
+      "q=1.6*10**-19;#charge\n",
+      "E=8.85*10**-14;#constant\n",
+      "d=.6*10**-4;#thickness\n",
+      "\n",
+      "#calculation\n",
+      "Vp=q*N*d**2/(2*D*E);#pinch-off voltage\n",
+      "Vn=.026*log(4.7*10**17/N);\n",
+      "Vb=H-Vn;#built-in potential\n",
+      "\n",
+      "#result\n",
+      "print\"pinch-off voltage is\",round(Vp,3),\"V\"\n",
+      "print\"built-in potential is\",round(Vb,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pinch-off voltage is 0.525 V\n",
+        "built-in potential is 0.75 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 Page 248"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=2*10**18;\n",
+      "q=1.6*10**-19;#charge\n",
+      "E=8.85*10**-14;#constant\n",
+      "d=40*10**-7;#thickness\n",
+      "H=.85;#barrier height\n",
+      "D=12.3;#dielectric constant\n",
+      "V=.23;#Voltage\n",
+      "\n",
+      "#calculation\n",
+      "Vp=q*N*d**2/(2*D*E);#volt\n",
+      "Vt=H-V-Vp;#threshold voltage\n",
+      "Ns=D*E/(q*(d+11*10**-7))*1.73;#electron gas concentration\n",
+      "\n",
+      "#result\n",
+      "print\"pinch-off voltage is\",round(Vp,2),\"V\"\n",
+      "print\"two-dimensional electron gas concentration is\",round(Ns,2),\"cm^-2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pinch-off voltage is 2.35 V\n",
+        "two-dimensional electron gas concentration is 2.30783272059e+12 cm^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter8_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter8_2.ipynb
new file mode 100755
index 00000000..15a1e48b
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter8_2.ipynb
@@ -0,0 +1,220 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8:Microwave Diodes, Quantum-Effect, and Hot-Electron Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "L=10*10**-9;#inductance\n",
+      "C=4*10**-12;#capacitance\n",
+      "\n",
+      "#calculation\n",
+      "Z=(L/C)**.5;#impedence\n",
+      "\n",
+      "#result\n",
+      "print\"characteristics impedence is\",round(Z,2),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "characteristics impedence is 50.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "a=.05;#m\n",
+      "b=.025;#cm\n",
+      "d=.1;#m\n",
+      "c=3*10**8;#speed\n",
+      "\n",
+      "#calculation\n",
+      "f=c/2*((1/a)**2+(1/d)**2)**.5;#frequency\n",
+      "\n",
+      "#result\n",
+      "print\"resonant frequncy is\",round(f/10**9,3),\"GHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resonant frequncy is 3.354 GHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 Page 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "b=10**-4;\n",
+      "W=6*10**-4;#width\n",
+      "F=3.3*10**5;#field\n",
+      "Q=2*10**12;#charge density\n",
+      "q=1.6*10**-19;#charge\n",
+      "D=11.9;\n",
+      "E=8.85*10**-14;#constant\n",
+      "\n",
+      "#calculation\n",
+      "a=W-b;\n",
+      "Vb=(F*10**-4+(F-q*Q/D/E)*a);#brekdown voltage\n",
+      "Fd=13/a;#field\n",
+      "f=10**7/2/a;#frequency\n",
+      "\n",
+      "#result\n",
+      "print\"breakdown voltage is\",round(Vb,2),\"V\"\n",
+      "print\"field in drift region is\",round(Fd,2),\"V/cm\"\n",
+      "print\"frequency is\",round(f/10**9,2),\"GHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "breakdown voltage is 46.07 V\n",
+        "field in drift region is 26000.0 V/cm\n",
+        "frequency is 10.0 GHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example8.4 Page269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "l=10;#length\n",
+      "L=10**-3;\n",
+      "v=10**7;#velocity\n",
+      "\n",
+      "#calculation\n",
+      "t=L/v;#time\n",
+      "\n",
+      "#result\n",
+      "print\"time required is\",round(t*10**9,2),\"ns\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required is 0.1 ns\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5 Page 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Lw=4.5;#nm\n",
+      "Vp1=700;#mV\n",
+      "Vp2=280;#mV\n",
+      "\n",
+      "#calculation\n",
+      "V=Vp1-Vp2;\n",
+      "\n",
+      "#result\n",
+      "print\"ground water level is 140mV and peak width is 1meV for\",round(Lw,1),\"nm\"\n",
+      "print\"difference of voltage drop due to accumulation and depletion region is\",round(V,2),\"mV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ground water level is 140mV and peak width is 1meV for 4.5 nm\n",
+        "difference of voltage drop due to accumulation and depletion region is 420.0 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter9_2.ipynb b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter9_2.ipynb
new file mode 100755
index 00000000..e1899052
--- /dev/null
+++ b/SEMICONDUCTOR_DEVICES_PHYSICS_AND_TECHNOLOGY_by_S_M_SZE/Chapter9_2.ipynb
@@ -0,0 +1,414 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Photonic Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Page 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "T=.25*10**-4;#thickness\n",
+      "E=3.00;#energy\n",
+      "P=10;#power\n",
+      "a=4*10**4#absorption coefficient\n",
+      "Eg=1.12;\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "Es=10**-2*(1-exp(-a*T));#energy absorbed per second\n",
+      "Ee=(E-Eg)/E;#photons energy converted to heat\n",
+      "L=Ee*Es;#energy to lattice\n",
+      "Ps=(Es-L)/(q*Eg);#photons per second\n",
+      "\n",
+      "#result\n",
+      "print\"energy absorbed per second is\",round(Es*1000,1),\"mW\"\n",
+      "print\"portion of photons energy converted to heat is\",round(Ee*100,2),\"%\"\n",
+      "print\"number of photons per second is\",round(Ps,2),\"photons/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy absorbed per second is 6.3 mW\n",
+        "portion of photons energy converted to heat is 62.67 %\n",
+        "number of photons per second is 1.31691783089e+16 photons/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 Page 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "t=500*10**-12;#time\n",
+      "\n",
+      "#calculation\n",
+      "f=1/(2*pi*t);#bandwidth\n",
+      "\n",
+      "#result\n",
+      "print\"modulation bandwidth is\",round(f/10**6,0),\"MHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "modulation bandwidth is 318.0 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page 302"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "n=3.6;#constant\n",
+      "\n",
+      "#calculation\n",
+      "R=((n-1)/(n+1))**2;#reflectivity\n",
+      "\n",
+      "#result\n",
+      "print\"reflected light is\",round(R*100,0),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "reflected light is 32.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page 303"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "l=.94;#wavelength\n",
+      "n=3.6;\n",
+      "L=300;#length\n",
+      "\n",
+      "#calculation\n",
+      "A=l**2/(2*n*L);#mode spacing\n",
+      "\n",
+      "#result\n",
+      "print\"mode spacing is\",round(A*10**4,0),\"angstron\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mode spacing is 4.0 angstron\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "R1=.44;#front reflectivity\n",
+      "R2=.99;#rear reflectivity\n",
+      "L=300*10**-4;#length\n",
+      "W=5*10**-4;#width\n",
+      "a=100;#alpha\n",
+      "b=.1;#beta\n",
+      "g=100;\n",
+      "T=.9;#constant\n",
+      "\n",
+      "#calculation\n",
+      "J=g*T/b+1/b*(a+1/(2*L)*log(1/(R1*R2)));#current density\n",
+      "I=J*L*W;#current\n",
+      "\n",
+      "#result\n",
+      "print\"current density is\",round(J,2),\"A/cm^2\"\n",
+      "print\"threshold current is\",round(I*1000,2),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current density is 2038.51 A/cm^2\n",
+        "threshold current is 30.58 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 Page 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "T1=27;#temperature\n",
+      "l=110;\n",
+      "\n",
+      "#calculation\n",
+      "T=T1+l*log(2);#temperature\n",
+      "\n",
+      "#result\n",
+      "print\"temperature is\",round(T,0),\"deg.Celsius\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature is 103.0 deg.Celsius\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7 Page 313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "N=5*10**12;#photons/s\n",
+      "n=.8;#constant\n",
+      "l=5*10**-10;#lifetime\n",
+      "u=2500;\n",
+      "E=5000;#V/cm\n",
+      "L=10*10**-4;#length\n",
+      "q=1.6*10**-19;#charge\n",
+      "\n",
+      "#calculation\n",
+      "I=q*n*N*u*l*E/L;#photocurrent\n",
+      "G=u*l*E/L;#gain\n",
+      "\n",
+      "#result\n",
+      "print\"photocurrent is\",round(I*10**6,2),\"microAmp\"\n",
+      "print\"gain is\",round(G,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "photocurrent is 4.0 microAmp\n",
+        "gain is 6.25\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8 Page 315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "a=10**4;#cm^-1\n",
+      "R=.1;\n",
+      "\n",
+      "#calculation\n",
+      "X=-1/a*log(1/(2*(1-R)));#depth\n",
+      "\n",
+      "#result\n",
+      "print\"dpeth for half power absorbed is\",round(X,2),\"micro-m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "dpeth for half power absorbed is 0.59 micro-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9 Page 319 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "s=1.118;#m\n",
+      "h=1;#height\n",
+      "\n",
+      "#calculation\n",
+      "M=(1+(s/h)**2)**.5;#air mass\n",
+      "\n",
+      "#result\n",
+      "print\"air mass is\",round(M,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "air mass is 1.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10 Page 322"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variable\n",
+      "from math import *\n",
+      "Vs=.35;#voltage\n",
+      "Is=1*10**-9;#current\n",
+      "Il=.1;#A\n",
+      "Vl=.026;#V\n",
+      "\n",
+      "#calculation\n",
+      "Voc=Vl*log(Il/Is);#open-circuit voltage\n",
+      "P=-Is*Vs*(exp(Vs/Vl)-1)-(-Il*Vs);#power\n",
+      "\n",
+      "#result\n",
+      "print\"open-circuit voltage is\",round(Voc,2),\"V\"\n",
+      "print\"output power is\",round(P,4),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "open-circuit voltage is 0.48 V\n",
+        "output power is 0.0348 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Switchgear_Protection_And_Power_Systems_by_S._S._Rao/Chapter_58_Applications_of_Switchgear_1.ipynb b/Switchgear_Protection_And_Power_Systems_by_S._S._Rao/Chapter_58_Applications_of_Switchgear_1.ipynb
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diff --git a/Switchgear_Protection_And_Power_Systems_by_S._S._Rao/README.txt b/Switchgear_Protection_And_Power_Systems_by_S._S._Rao/README.txt
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diff --git a/The_Spirit_of_C/Chapter1.ipynb b/The_Spirit_of_C/Chapter1.ipynb
deleted file mode 100755
index 80b53e61..00000000
--- a/The_Spirit_of_C/Chapter1.ipynb
+++ /dev/null
@@ -1,198 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1c2e6d0b5736bd35dd626e4e22d0dbe67f82011189ddb91d07585ae3f73aa6b8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1, Elementary Programming"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 1-1 , Page number: 2"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# An interactive program to calculate the hypotenuse of a right triangle\n",
-      "\n",
-      "import math ## library for using sqrt function\n",
-      "\n",
-      "def square(a):\n",
-      "\t\"\"\"Returns the square of the given number \"\"\"\n",
-      "\treturn a*a\n",
-      "\n",
-      "print \"Enter the lengths of the two legs of a right triangle\"\n",
-      "print \"The program will display the length of the hypotenuse.\"\n",
-      "print \"Enter zero or negative values to end program\"\n",
-      "\n",
-      "# Priming reads follow\n",
-      "print \"First leg (0.0 to quit)? \",\n",
-      "a_leg = 3\n",
-      "print a_leg\n",
-      "print \"Second leg (0.0 to quit)? \",\n",
-      "b_leg = 4\n",
-      "print b_leg\n",
-      "\n",
-      "while a_leg>0.0 and b_leg>0.0:\n",
-      "\tprint \"The legs of the right triangle are %.2f and %.2f,\\n\"%(a_leg,b_leg),\n",
-      "\t# Hypotenuse is square root of a*a + b*b\n",
-      "\thypotenuse = math.sqrt(square(a_leg) + square(b_leg))\n",
-      "\tprint \"   so the hypotenuse is %.2f\\n\\n\"%hypotenuse,\n",
-      "\tprint \"First leg (0.0 to quit)? \",\n",
-      "\ta_leg = 0.0\t\t\t# To terminate the loop\n",
-      "\tprint a_leg\n",
-      "\tprint \"Second leg (0.0 to quit)? \",\n",
-      "\tb_leg = 0.0\n",
-      "\tprint b_leg\n",
-      "\n",
-      "print \"Thank you.\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the lengths of the two legs of a right triangle\n",
-        "The program will display the length of the hypotenuse.\n",
-        "Enter zero or negative values to end program\n",
-        "First leg (0.0 to quit)?  3\n",
-        "Second leg (0.0 to quit)?  4\n",
-        "The legs of the right triangle are 3.00 and 4.00,\n",
-        "   so the hypotenuse is 5.00\n",
-        "\n",
-        "First leg (0.0 to quit)?  0.0\n",
-        "Second leg (0.0 to quit)?  0.0\n",
-        "Thank you.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 1-2 , Page number: 3"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "def main():  \n",
-      "\t\"\"\"simply returns a string\"\"\"\n",
-      "\t# There's no main() function in python.This main() function is no different that any other user-defined functions\n",
-      "\treturn \"Welcome to C programming!\\n\"\n",
-      "\t\n",
-      "print main()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Welcome to C programming!\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 1-3 , Page number: 6"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Use of braces \n",
-      "\n",
-      "def main():\n",
-      "\t# in python, proper indentation works as braces\n",
-      "\t# the following lines come under the main() function block due to indentation\n",
-      "\tprint \"Give me land, lots of land\",\n",
-      "\tprint \"And the starry skies above...\\n\",\n",
-      "\n",
-      "main()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Give me land, lots of land And the starry skies above...\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 1-4 , Page number: 8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A short poem\n",
-      "\n",
-      "def main():\n",
-      "\tprint \"Mary had a little lamb,\",\n",
-      "\tprint \"its fleece was white as snow.\\n\",\n",
-      "\tprint \"And everywhere that Mary went,\",\n",
-      "\tprint \"the lamb was sure to go.\\n\"\n",
-      "\n",
-      "main()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mary had a little lamb, its fleece was white as snow.\n",
-        "And everywhere that Mary went, the lamb was sure to go.\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/Chapter10.ipynb b/The_Spirit_of_C/Chapter10.ipynb
deleted file mode 100755
index ed7b99e5..00000000
--- a/The_Spirit_of_C/Chapter10.ipynb
+++ /dev/null
@@ -1,1020 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:970f28e2206ea4472dd82a3808592db202377fcd9a01fc69b666e61a221db5b1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10, Arrays"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-1, Page Number : 227"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Introduction to arrays\n",
-      "\n",
-      "array = []\t\t# defining array which is rather called list in python\n",
-      "\n",
-      "print \"You are to enter 5 integers.\\n\"\n",
-      "\n",
-      "# Input the 5 integers into array \"array\"\n",
-      "\n",
-      "for i in range(5) :  # equivalent to for i in range(0,5,1)\n",
-      "\tprint \"Enter integer #%d: \"%(i+1),\n",
-      "\tval = int(raw_input())\n",
-      "\tarray.insert(i,val)\t\t\t\t# insert function is used to add elements to list : \n",
-      "\t\t\t\t\t\t\t\t\t#      insert(i,val) where i is the index and val is the value to be inserted\n",
-      "\n",
-      "print \"\\nThank you.\\n\"\n",
-      "\n",
-      "# Print out the integers stored in array \"array\"\n",
-      "j = 1\t# used as a counter\n",
-      "for i in array:\t\t\t\t# shorthand way of scanning through the list; using \"in\"\n",
-      "\tprint \"Integer #%d = %d.\"%(j,i)\n",
-      "\tj += 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "You are to enter 5 integers.\n",
-        "\n",
-        "Enter integer #1:"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "3\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Enter integer #2:"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "12\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Enter integer #3:"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Enter integer #4:"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "63\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " Enter integer #5:"
-       ]
-      },
-      {
-       "name": "stdout",
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "0\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        "Thank you.\n",
-        "\n",
-        "Integer #1 = 3.\n",
-        "Integer #2 = 12.\n",
-        "Integer #3 = 8.\n",
-        "Integer #4 = 63.\n",
-        "Integer #5 = 0.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-2, Page Number : 228"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Initializing a global array\n",
-      "\n",
-      "array = [ 4, 6, 5, 7, 2 ]\n",
-      "\n",
-      "def main() :\t\t\t# just a user-defined function ; not equivalent to the built-in main() function in C\n",
-      "\t\"\"\" prints the integers stored in array \"array\" \"\"\"\n",
-      "\tj = 1\t\t# counter \n",
-      "\tfor i in array :\n",
-      "\t\tprint \"Integer #%d = %d.\"%(j,i)\n",
-      "\t\tj += 1\n",
-      "\n",
-      "main()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Integer #1 = 4.\n",
-        "Integer #2 = 6.\n",
-        "Integer #3 = 5.\n",
-        "Integer #4 = 7.\n",
-        "Integer #5 = 2.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-3, Page Number: 230"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# The number of array elements actually used may vary\n",
-      "\n",
-      "# Read in the size of the array\n",
-      "print \"How many integers will you enter?\",\n",
-      "array_size = 6\n",
-      "print array_size\n",
-      "\n",
-      "# Read in the integers\n",
-      "print \"\\nPlease enter %d integers:\"%array_size\n",
-      "array = [ 12, 3, -5, 21, 0, 899 ]\t# Lists dynamically allocate memory \n",
-      "for i in range(array_size) :\n",
-      "\tprint array[i],\n",
-      "# Print out the integers\n",
-      "\n",
-      "print(\"\\n\")\n",
-      "for i in range(array_size) :\n",
-      "\tprint \"Integer #%d = %d.\"%(i+1,array[i])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many integers will you enter? 6\n",
-        "\n",
-        "Please enter 6 integers:\n",
-        "12 3 -5 21 0 899 \n",
-        "\n",
-        "Integer #1 = 12.\n",
-        "Integer #2 = 3.\n",
-        "Integer #3 = -5.\n",
-        "Integer #4 = 21.\n",
-        "Integer #5 = 0.\n",
-        "Integer #6 = 899.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-4, Page Number: 231"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# The trailer method\n",
-      "\n",
-      "# Macro definition\n",
-      "MAX_ARRAY_SIZE = 20\n",
-      "TRAILER = -9999\n",
-      "\n",
-      "# Read in the integers\n",
-      "\n",
-      "print \"Please enter up to %d integers, using %d as a trailer:\"%(MAX_ARRAY_SIZE,TRAILER)\n",
-      "array = [ 3, 75, -34, 6, -1, 2, 41, 0, 90, -9999]\n",
-      "for array_size in range(MAX_ARRAY_SIZE) :\n",
-      "\tprint array[array_size],\n",
-      "\tif array[array_size] == -9999 :\n",
-      "\t\tbreak\t\n",
-      "\t\n",
-      "# Print out the integers\n",
-      "print \"\\n\\nYou have entered %d integers\"%array_size\n",
-      "for i in range(array_size) :\n",
-      "\tprint \"Integer #%d = %d.\"%(i+1,array[i])"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter up to 20 integers, using -9999 as a trailer:\n",
-        "3 75 -34 6 -1 2 41 0 90 -9999 \n",
-        "\n",
-        "You have entered 9 integers\n",
-        "Integer #1 = 3.\n",
-        "Integer #2 = 75.\n",
-        "Integer #3 = -34.\n",
-        "Integer #4 = 6.\n",
-        "Integer #5 = -1.\n",
-        "Integer #6 = 2.\n",
-        "Integer #7 = 41.\n",
-        "Integer #8 = 0.\n",
-        "Integer #9 = 90.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-5, Page Number: 233"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Caculating the average grade for a class\n",
-      "\n",
-      "# Macro definition\n",
-      "\n",
-      "MAX_GRADES = 20 \t# The class may have up to 20 students\n",
-      "TRAILER = -9999\t\t# -9999 marks the end of the grade list\n",
-      "\n",
-      "# Read in the grades\n",
-      "\n",
-      "print \"Please enter up to %d grades, using %d as a trailer:\"%(MAX_GRADES,TRAILER)\n",
-      "grades = [ 41, 92, 15, 65, 0, 78, 81, 96, -9999 ]\n",
-      "for num_grades in range(MAX_GRADES) :\n",
-      "\tprint grades[num_grades]\n",
-      "\tif grades[num_grades] == -9999 :\n",
-      "\t\tbreak\n",
-      "\n",
-      "print \"\\nYou have entered %d grades.\"%num_grades\n",
-      "\n",
-      "# Find the average\n",
-      "sum = 0.0 \n",
-      "for i in range(num_grades) :\n",
-      "\tsum += grades[i]\n",
-      "\n",
-      "print \"\\nThe average grade is %f.\"%(sum/num_grades)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter up to 20 grades, using -9999 as a trailer:\n",
-        "41\n",
-        "92\n",
-        "15\n",
-        "65\n",
-        "0\n",
-        "78\n",
-        "81\n",
-        "96\n",
-        "-9999\n",
-        "\n",
-        "You have entered 8 grades.\n",
-        "\n",
-        "The average grade is 58.500000.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-6, Page Number: 235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Determining which grades in a class are above the average\n",
-      "\n",
-      "# Macro definition\n",
-      "\n",
-      "MAX_GRADES = 20 \t# The class may have up to 20 students\n",
-      "TRAILER = -9999\t\t# -9999 marks the end of the grade list\n",
-      "\n",
-      "# Read in the grades\n",
-      "\n",
-      "print \"Please enter up to %d grades, using %d as a trailer:\"%(MAX_GRADES,TRAILER)\n",
-      "grades = [ 41, 92, 15, 65, 0, 78, 81, 96, -9999 ]\n",
-      "for num_grades in range(MAX_GRADES) :\n",
-      "\tprint grades[num_grades]\n",
-      "\tif grades[num_grades] == -9999 :\n",
-      "\t\tbreak\n",
-      "\n",
-      "print \"\\nYou have entered %d grades.\"%num_grades\n",
-      "\n",
-      "# Find the average\n",
-      "sum = 0.0 \n",
-      "for i in range(num_grades) :\n",
-      "\tsum += grades[i]\n",
-      "\n",
-      "avg_grade = sum/num_grades\n",
-      "print \"\\nThe average grade is %f.\"%avg_grade\n",
-      "\n",
-      "# Find who got above average\n",
-      "\n",
-      "print \"\\nThe following grades were above average:\"\n",
-      "for i in range(num_grades) :\n",
-      "\tif grades[i] > avg_grade :\n",
-      "\t\tprint \"%f (student #%d)\"%(grades[i],i+1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter up to 20 grades, using -9999 as a trailer:\n",
-        "41\n",
-        "92\n",
-        "15\n",
-        "65\n",
-        "0\n",
-        "78\n",
-        "81\n",
-        "96\n",
-        "-9999\n",
-        "\n",
-        "You have entered 8 grades.\n",
-        "\n",
-        "The average grade is 58.500000.\n",
-        "\n",
-        "The following grades were above average:\n",
-        "92.000000 (student #2)\n",
-        "65.000000 (student #4)\n",
-        "78.000000 (student #6)\n",
-        "81.000000 (student #7)\n",
-        "96.000000 (student #8)\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-7, Page Number 237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Reversing an array\n",
-      "\n",
-      "# Macro definition\n",
-      "ARRAY_SIZE = 10 \t# The array has 10 elements \n",
-      "\n",
-      "# Read in the array\n",
-      "\n",
-      "print \"Please enter %d integers:\"%ARRAY_SIZE\n",
-      "array = [ 10, 42, 7, 0, 923, 13, -5, 6, 19, 3]\n",
-      "for i in range(10) :\n",
-      "\tprint array[i],\n",
-      "\n",
-      "# Echo the array\n",
-      "\n",
-      "print \"\\n\\nThe original array was:\"\n",
-      "for i in range(10) :\n",
-      "\tprint array[i],\n",
-      "\n",
-      "# Print the array in reverse\n",
-      "\n",
-      "print \"\\n\\nThe reversed array is:\"\n",
-      "for i in range(ARRAY_SIZE-1,-1,-1) :\n",
-      "\tprint array[i],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter 10 integers:\n",
-        "10 42 7 0 923 13 -5 6 19 3 \n",
-        "\n",
-        "The original array was:\n",
-        "10 42 7 0 923 13 -5 6 19 3 \n",
-        "\n",
-        "The reversed array is:\n",
-        "3 19 6 -5 13 923 0 7 42 10\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-8, Page Number: 240"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "# Reversing a character string\n",
-      "\n",
-      "# Macro definition\n",
-      "MAX_STRING_SIZE = 10\n",
-      "\n",
-      "# Read in the word\n",
-      "\n",
-      "print \"Please enter a word of up to %d letters:\"%MAX_STRING_SIZE\n",
-      "word = \"bingo\"\n",
-      "print word\n",
-      "\n",
-      "# Echo the word \n",
-      "\n",
-      "print \"\\nThe word is %s.\"%word\n",
-      "\n",
-      "# Find thw word length\n",
-      "string_size = 0\n",
-      "for i in word :\t\t\t# To iterate over the entire string\n",
-      "\tstring_size += 1\n",
-      "\n",
-      "# Print the word in reverse\n",
-      "\n",
-      "print \"\\nThe reversed word is \",\n",
-      "for i in range(string_size-1,-1,-1) :\n",
-      "\tsys.stdout.write(\"%c\"%word[i])  # For printing in the same line without implicit whitespaces\n",
-      "\n",
-      "print "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter a word of up to 10 letters:\n",
-        "bingo\n",
-        "\n",
-        "The word is bingo.\n",
-        "\n",
-        "The reversed word is ognib\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-9, Page Number: 242"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculating the minimum and maximum grade in a class\n",
-      "\n",
-      "# Macro definition\n",
-      "\n",
-      "MAX_GRADES = 20 \t# The class may have up to 20 students\n",
-      "TRAILER = -9999\t\t# -9999 marks the end of the grade list\n",
-      "\n",
-      "# Read in the grades\n",
-      "\n",
-      "print \"Please enter up to %d grades, using %d as a trailer:\"%(MAX_GRADES,TRAILER)\n",
-      "grades = [ 41, 92, 15, 65, 0, 78, 81, 96, -9999 ]\n",
-      "for num_grades in range(MAX_GRADES) :\n",
-      "\tprint grades[num_grades]\n",
-      "\tif grades[num_grades] == -9999 :\n",
-      "\t\tbreak\n",
-      "\n",
-      "print \"\\nYou have entered %d grades.\"%num_grades\n",
-      "\n",
-      "# Find the minimum grade\n",
-      "\n",
-      "min = grades[0]\n",
-      "for i in range(num_grades) :\n",
-      "\tif grades[i] < min :\n",
-      "\t\tmin = grades[i]\n",
-      "\n",
-      "print \"\\nThe lowest grade is %f\"%min\n",
-      "\n",
-      "# Find the maximum grade\n",
-      "\n",
-      "max = grades[0]\n",
-      "for i in range(num_grades) :\n",
-      "\tif grades[i] > min :\n",
-      "\t\tmax = grades[i]\n",
-      "\n",
-      "print \"\\nThe lowest grade is %f\"%max\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter up to 20 grades, using -9999 as a trailer:\n",
-        "41\n",
-        "92\n",
-        "15\n",
-        "65\n",
-        "0\n",
-        "78\n",
-        "81\n",
-        "96\n",
-        "-9999\n",
-        "\n",
-        "You have entered 8 grades.\n",
-        "\n",
-        "The lowest grade is 0.000000\n",
-        "\n",
-        "The lowest grade is 96.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-10, Page Number: 243"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Sorting an array\n",
-      "\n",
-      "# Macro definition\n",
-      "MAX_ARRAY_SIZE = 20 \t# The array may have up to 20 elements\n",
-      "\n",
-      "array_size = 10\n",
-      "print \"How many numbers?\",\n",
-      "print array_size\n",
-      "\n",
-      "while array_size < 1 or array_size > MAX_ARRAY_SIZE\t:\n",
-      "\tprint \"How many numbers?\",\n",
-      "\tprint array_size\n",
-      "\n",
-      "# Read in the array\n",
-      "\n",
-      "print \"Please enter the integers:\"\n",
-      "array = [ 10, 42, 7, 0, 923, 13, -5, 6, 19, 3]\n",
-      "for i in range(array_size) :\n",
-      "\tprint array[i],\n",
-      "\n",
-      "# Echo the array\n",
-      "\n",
-      "print \"\\n\\nThe original array is:\"\n",
-      "for i in range(array_size) :\n",
-      "\tprint array[i],\n",
-      "\n",
-      "# Sort the array\n",
-      "\n",
-      "for i in range(array_size-1) :\n",
-      "\tfor j in range(i+1, array_size) :\n",
-      "\t\tif array[i] > array[j] :\n",
-      "\t\t\ttemp = array[i]\n",
-      "\t\t\tarray[i] = array[j]\n",
-      "\t\t\tarray[j] = temp\n",
-      "\n",
-      "# Printing the sorted array\n",
-      "\n",
-      "print \"\\n\\nThe sorted array is\"\n",
-      "for i in range(array_size) :\n",
-      "\tprint array[i],"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many numbers? 10\n",
-        "Please enter the integers:\n",
-        "10 42 7 0 923 13 -5 6 19 3 \n",
-        "\n",
-        "The original array is:\n",
-        "10 42 7 0 923 13 -5 6 19 3 \n",
-        "\n",
-        "The sorted array is\n",
-        "-5 0 3 6 7 10 13 19 42 923\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-11, Page Number: 247"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Finding the median of an array\n",
-      "\n",
-      "# Macro definition\n",
-      "MAX_ARRAY_SIZE = 20 \t# The array may have up to 20 elements\n",
-      "\n",
-      "array_size = 0\n",
-      "array = []\n",
-      "\n",
-      "def get_size()\t:\n",
-      "\t\"\"\" reads the size of the array (until a valid value is entered) \"\"\"\n",
-      "\tglobal array_size\n",
-      "\tarray_size = 10\n",
-      "\tprint \"How many numbers?\",\n",
-      "\tprint array_size\n",
-      "\twhile array_size < 1 or array_size > MAX_ARRAY_SIZE\t:\n",
-      "\t\tprint \"How many numbers?\",\n",
-      "\t\tprint array_size\n",
-      "\n",
-      "\n",
-      "def read_array() :\n",
-      "\t\"\"\" read in the array \"\"\"\n",
-      "\tglobal array\n",
-      "\tprint \"Please enter the integers:\"\n",
-      "\tarray = [ 10, 42, 7, 0, 923, 13, -5, 6, 19, 3 ]\n",
-      "\tfor i in range(array_size) :\n",
-      "\t\tprint array[i],\n",
-      "\n",
-      "def write_array() :\n",
-      "\t\"\"\" write out the array \"\"\"\n",
-      "\tfor i in range(array_size) :\n",
-      "\t\tprint array[i],\n",
-      "\n",
-      "def sort_array() :\n",
-      "\t\"\"\" sort the array \"\"\"\n",
-      "\tglobal array\n",
-      "\tfor i in range(array_size-1) :\n",
-      "\t\tfor j in range(i+1, array_size) :\n",
-      "\t\t\tif array[i] > array[j] :\n",
-      "\t\t\t\ttemp = array[i]\n",
-      "\t\t\t\tarray[i] = array[j]\n",
-      "\t\t\t\tarray[j] = temp\n",
-      "\n",
-      "def even(n) :\n",
-      "\t\"\"\" returns non-zero(true) if n is even else returns zero(false) \"\"\"\n",
-      "\treturn ((n/2)*2 == n)\n",
-      "\n",
-      "get_size()\t\t# Read in the size of the array\n",
-      "\n",
-      "read_array()\t# Read in the array\n",
-      "\n",
-      "print \"\\n\\nThe original array is:\"\n",
-      "write_array()\t# Echo the array\n",
-      "\n",
-      "sort_array()\t# Sort the array\n",
-      "\n",
-      "print \"\\n\\nThe sorted array is:\"\n",
-      "write_array()\t# Print the sorted array\n",
-      "\n",
-      "print \"\\n\\nThe median is:\"\n",
-      "# The equivalent of tertiary conditional operator in python is a if test else b\n",
-      "print \"%.1f\"%( ((array[ array_size / 2 -1] + array[ array_size / 2 ]) / 2.0 ) if even(array_size) else float(array[ array_size / 2 ])  ) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many numbers? 10\n",
-        "Please enter the integers:\n",
-        "10 42 7 0 923 13 -5 6 19 3 \n",
-        "\n",
-        "The original array is:\n",
-        "10 42 7 0 923 13 -5 6 19 3 \n",
-        "\n",
-        "The sorted array is:\n",
-        "-5 0 3 6 7 10 13 19 42 923 \n",
-        "\n",
-        "The median is:\n",
-        "8.5\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-12, Page Number: 251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Tally how many of each number within a range are entered\n",
-      "\n",
-      "# Macro definition\n",
-      "RANGE_LOW = 1\n",
-      "RANGE_HIGH = 10\t\t\t# Tallies numbers between RANGE_LOW and RANGE_HIGH\n",
-      "TRAILER = -1 \t\t\t# TRAILER marks end of data\n",
-      "\n",
-      "inputfeed = [ 3, 5, 1, 10, 8, 5, 10, 6, 2, 11, 4, 9, -2, 3, 2, 9, -1 ]\t\t# additional array for the sake of providing input\n",
-      "\n",
-      "tally = [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]\t\t# empty array\n",
-      "# Input numbers\n",
-      "print \"Please enter a number between %d and %d (%d to quit): \"%(RANGE_LOW,RANGE_HIGH,TRAILER),\n",
-      "num = inputfeed[0]\n",
-      "print num\n",
-      "\n",
-      "i = 1\n",
-      "while num != TRAILER :\n",
-      "\tif num < RANGE_LOW or num > RANGE_HIGH : \t\t# Out of range\n",
-      "\t\tprint \"%d is not between %d and %d.\"%(num,RANGE_LOW,RANGE_HIGH)\n",
-      "\telse :\n",
-      "\t\ttally[ num - RANGE_LOW] += 1\t\n",
-      "\tprint \"\\nEnter another number: \",\n",
-      "\tnum = inputfeed[i]\n",
-      "\tprint num\n",
-      "\ti += 1 \t\t\t# to iterate over the input array\n",
-      "\n",
-      "# Print totals\n",
-      "\n",
-      "print \"\\n------------\\n\\nYou entered:\"\n",
-      "for i in range(RANGE_HIGH - RANGE_LOW + 1) :\n",
-      "\tprint \"%2d %2d's\"%(tally[i],i + RANGE_LOW)\n",
-      "\n",
-      "print \"\\nThank you.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter a number between 1 and 10 (-1 to quit):  3\n",
-        "\n",
-        "Enter another number:  5\n",
-        "\n",
-        "Enter another number:  1\n",
-        "\n",
-        "Enter another number:  10\n",
-        "\n",
-        "Enter another number:  8\n",
-        "\n",
-        "Enter another number:  5\n",
-        "\n",
-        "Enter another number:  10\n",
-        "\n",
-        "Enter another number:  6\n",
-        "\n",
-        "Enter another number:  2\n",
-        "\n",
-        "Enter another number:  11\n",
-        "11 is not between 1 and 10.\n",
-        "\n",
-        "Enter another number:  4\n",
-        "\n",
-        "Enter another number:  9\n",
-        "\n",
-        "Enter another number:  -2\n",
-        "-2 is not between 1 and 10.\n",
-        "\n",
-        "Enter another number:  3\n",
-        "\n",
-        "Enter another number:  2\n",
-        "\n",
-        "Enter another number:  9\n",
-        "\n",
-        "Enter another number:  -1\n",
-        "\n",
-        "------------\n",
-        "\n",
-        "You entered:\n",
-        " 1  1's\n",
-        " 2  2's\n",
-        " 2  3's\n",
-        " 1  4's\n",
-        " 2  5's\n",
-        " 1  6's\n",
-        " 0  7's\n",
-        " 1  8's\n",
-        " 2  9's\n",
-        " 2 10's\n",
-        "\n",
-        "Thank you.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 10-13, Page Number: 255"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Program to report car prices. User enters car make by number and car year. \n",
-      "# Program reports price. All data is fictitious\n",
-      "\n",
-      "# 2-D array (4 X 12 ) of prices : row = make , column = year\n",
-      "price_list = [\t[ 4775, 4980, 5222, 5305, 5483, 5547, 5596, 5713, 5842, 5903, 6043, 6230 ], # Volkswagen\n",
-      "\t\t\t\t[ 4853, 5140, 5413, 5590, 5723, 5848, 5762, 5944, 6104, 6255, 6370, 6526 ],\t# Chevy\n",
-      "\t\t\t\t[ 5627, 5772, 5973, 6210, 6539, 6720, 6792, 6930, 7054, 7202, 7355, 7562 ], # Cadillac\n",
-      "\t\t\t\t[ 1576, 1738, 1970, 2151, 2205, 2280, 2442, 2580, 2814, 1953, 3078, 3201 ],\t# Honda\n",
-      "\t\t\t]\t\n",
-      "\n",
-      "def get_make() : \n",
-      "\t\"\"\" read in and return the number (make) of the desired car \"\"\"\n",
-      "\t\n",
-      "\t# Print the choice numbers and names\n",
-      "\tfor i in range(1,5) :\n",
-      "\t\tprint \"%d=\"%i,\n",
-      "\t\tprint_make(i)\n",
-      "\t\tprint \",\",\n",
-      "\tprint \"0=END\"\n",
-      "\n",
-      "\t# Read in the selection by number\n",
-      "\tprint \"     Which car do you want? \",\n",
-      "\tmake = 2\n",
-      "\tprint make\n",
-      "\twhile make < 0 or make > 4 :\n",
-      "\t\tprint \"     Which car do you want? \",\n",
-      "\t\tmake = 2\n",
-      "\t\tprint make\n",
-      "\n",
-      "\treturn make\n",
-      "\n",
-      "def get_year() :\t\n",
-      "\t\"\"\" read in and return the year of the desired car \"\"\"\n",
-      "\n",
-      "\tprint \"     What year (1975-1986)?\",\n",
-      "\tyear = 1980\n",
-      "\tprint year\n",
-      "\twhile year < 1975 and year > 1986 :\n",
-      "\t\tprint \"     What year (1975-1986)?\",\n",
-      "\t\tyear = 1980\n",
-      "\t\tprint year\n",
-      "\n",
-      "\treturn year\n",
-      "\n",
-      "def price (car,yr) :\n",
-      "\t\"\"\" Use \"car\" and \"year\" to generate an index into \"price_list\" and return the corresponding entry \"\"\"\n",
-      "\treturn price_list[ car - 1 ][ yr - 1975]\n",
-      "\n",
-      "def print_make(car) :\n",
-      "\t\"\"\" print the make of \"car\" according to its value \"\"\"\n",
-      "\tif car == 1 : \n",
-      "\t\tprint \"Volkswagen\",\n",
-      "\telif car == 2 :\n",
-      "\t\tprint \"Chevy\",\n",
-      "\telif car == 3 :\n",
-      "\t\tprint \"Cadillac\",\n",
-      "\telse :\n",
-      "\t\tprint \"Honda\",\n",
-      "\n",
-      "make = get_make()\n",
-      "\n",
-      "while make != 0 :\t\t# O indicates end of input\n",
-      "\tyear = get_year()\n",
-      "\tprint \"The price of a %d\"%year,\n",
-      "\tprint_make(make)\n",
-      "\tprint \" is $%d.\\n\"%price(make,year)\n",
-      "\tmake = 0   # To terminate the loop\n",
-      "\n",
-      "print \"\\nThank you for buying all those cars!\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1= Volkswagen , 2= Chevy , 3= Cadillac , 4= Honda , 0=END\n",
-        "     Which car do you want?  2\n",
-        "     What year (1975-1986)? 1980\n",
-        "The price of a 1980 Chevy  is $5848.\n",
-        "\n",
-        "\n",
-        "Thank you for buying all those cars!\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/Chapter11.ipynb b/The_Spirit_of_C/Chapter11.ipynb
deleted file mode 100755
index 241e1c1e..00000000
--- a/The_Spirit_of_C/Chapter11.ipynb
+++ /dev/null
@@ -1,383 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3a9b6dffdf9d20cf5158c028c4012a0a2cf4ee76994f18e9b81499931df7cb5d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11, Pointers and Indirection"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 11-1, Page Number : 278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Introduction to pointers\n",
-      "\n",
-      "# There is no concept of pointers in python\n",
-      "i = 3\n",
-      "\n",
-      "j = id(i)\t# id() returns the address of the variable supplied\n",
-      "j = i \t\t# assign value of i to j such that it seems like j points to i\n",
-      "k = j\n",
-      "j = 4\n",
-      "\n",
-      "print \"i = %d, *j = %d, k = %d\"%(i,j,k)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i = 3, *j = 4, k = 3\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 11-2, Page Number: 280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Demonstration of passing by reference for a function to be able to modify its parameters\n",
-      "\n",
-      "def modify(i) :\n",
-      "\ti = 3\n",
-      "\treturn i\n",
-      "\n",
-      "i = 1\n",
-      "print \"i is now equal to %d\"%i\n",
-      "\n",
-      "i = modify(i)\t\t\t\n",
-      "print \"i is now equal to %d\"%i"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i is now equal to 1\n",
-        "i is now equal to 3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 11-3, Page Number: 281"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Pointer parameters in action: the swap routine\n",
-      "\n",
-      "def swap(i,j) :\n",
-      "\t\"\"\" exchange i with j and return \"\"\"\n",
-      "\ttemp = i   \t\t\t# temp variable can be avoided using i,j = j,i\n",
-      "\ti = j\n",
-      "\tj = temp\n",
-      "\treturn i,j\n",
-      "\n",
-      "i = 1\n",
-      "j = 9\n",
-      "print \"i = %d and j = %d\"%(i,j)\n",
-      "\n",
-      "# Now exchange them\n",
-      "i, j = swap(i,j)\n",
-      "print \"\\nnow i = %d and j = %d\"%(i,j)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i = 1 and j = 9\n",
-        "\n",
-        "now i = 9 and j = 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 11-4, Page Number: 282"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Array names are constant pointers\n",
-      "\n",
-      "an_array = [ 3, 7 ]\n",
-      "\n",
-      "def main() :\n",
-      "\talso = an_array\n",
-      "\talso[0] = 14\n",
-      "\tprint \"an_array = { %d, %d }\"%(an_array[0],an_array[1])\n",
-      "\n",
-      "main()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "an_array = { 14, 7 }\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 11-5, Page Number: 284"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Sorting an array revisited\n",
-      "# Passing an array as parameter\n",
-      "\n",
-      "# Macro definition\n",
-      "MAX_ARRAY_SIZE = 20\t\t# The array may have up to 20 elements\n",
-      "# Message flags for \"print array\"\n",
-      "ORIGINAL = 0\t# Display original array\n",
-      "SORTED = 1\t\t# Display sorted array\n",
-      "\n",
-      "numbers = [ 10, 42, 7, 0, 923, 12, -5, 6, 19, 3 ]\n",
-      "array_size = 0\n",
-      "\n",
-      "def get_size(size)\t:\n",
-      "\t\"\"\" reads the size of the array (until a valid value is entered) \"\"\"\n",
-      "\tglobal array_size\n",
-      "\tprint \"How many numbers?\",\n",
-      "\tsize = 10\n",
-      "\tprint size\n",
-      "\twhile size < 1 or size > MAX_ARRAY_SIZE\t:\n",
-      "\t\tprint \"How many numbers?\",\n",
-      "\t\tsize = 10\n",
-      "\t\tprint size\n",
-      "\tarray_size = size\n",
-      "\n",
-      "def get_array(array, size) :\n",
-      "\t\"\"\" read in the array called 'array' of size 'size' \"\"\"\n",
-      "\tprint \"Please enter the integers:\"\n",
-      "\tfor i in array :\n",
-      "\t\tprint i,\n",
-      "\tprint\n",
-      "\n",
-      "def print_array(array, size, message) :\n",
-      "\t\"\"\" print out the array on a single line. 'message' is a flag indicating which message to display(ORIGINAL or SORTED) \"\"\"\n",
-      "\tif message == ORIGINAL :\n",
-      "\t\tprint \"\\nThe original array is:\"\n",
-      "\telse :\n",
-      "\t\tprint \"\\nThe sorted array is:\"\n",
-      "\tfor i in array :\n",
-      "\t\tprint i,\n",
-      "\tprint\n",
-      "\n",
-      "def sort_array(array,size) :\n",
-      "\t\"\"\" sort 'array' of size 'size' \"\"\"\n",
-      "\tfor i in range(size - 1) :\n",
-      "\t\tfor j in range(i+1, size) :\n",
-      "\t\t\tif array[i] > array[j] :\n",
-      "\t\t\t\telement_swap(array, i, j)\n",
-      "\n",
-      "def element_swap(array, i, j) :\n",
-      "\t\"\"\" swap elements 'i' and 'j' of 'array' \"\"\"\n",
-      "\tglobal numbers\n",
-      "\tnumbers[i],numbers[j] = numbers[j],numbers[i]\n",
-      "\n",
-      "\n",
-      "get_size(array_size)\t\t\t\t\t\t# Read in the array's size\n",
-      "get_array(numbers, array_size)\t\t\t\t\t# Read in the array 'numbers' of size 'array_size'\n",
-      "print_array(numbers, array_size, ORIGINAL) \t# Echo the array \n",
-      "sort_array(numbers, array_size)\n",
-      "print_array(numbers, array_size, SORTED)\t\t# Print the sorted array"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many numbers? 10\n",
-        "Please enter the integers:\n",
-        "10 42 7 0 923 12 -5 6 19 3\n",
-        "\n",
-        "The original array is:\n",
-        "10 42 7 0 923 12 -5 6 19 3\n",
-        "\n",
-        "The sorted array is:\n",
-        "-5 0 3 6 7 10 12 19 42 923\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 11-6, Page Number: 294"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Sorting of array revisited\n",
-      "# Two dimensional char array (array of strings)\n",
-      "\n",
-      "# Macro definition\n",
-      "MAX_ARRAY_SIZE = 20\t\t# The array may have up to 20 elements\n",
-      "MAX_STRING_SIZE = 10 \t# Each string may have up to 10 characters\n",
-      "# Message flags for \"print_array\"\n",
-      "ORIGINAL = 0\n",
-      "SORTED = 1\n",
-      "\n",
-      "words = [ \"cat\", \"pig\", \"dog\", \"earlobe\", \"bingo\", \"elbow\", \"likely\", \"radar\" ]\n",
-      "array_size = 0\n",
-      "\n",
-      "def get_size(size) :\n",
-      "\t\"\"\" Read the size of the array into size (until a value is entered) \"\"\"\n",
-      "\tprint \"How many words? \",\n",
-      "\tsize = 8\n",
-      "\tprint size\n",
-      "\twhile size < 1 or size > MAX_ARRAY_SIZE :\n",
-      "\t\tprint \"How many words? \",\n",
-      "\t\tsize = 8\n",
-      "\t\tprint size\n",
-      "\tglobal array_size\n",
-      "\tarray_size = size\n",
-      "\n",
-      "def get_array(array,size) :\n",
-      "\t\"\"\" Read in the string array called 'array' of size 'size' \"\"\"\n",
-      "\tprint \"Please enter the words:\"\n",
-      "\tfor i in array :\n",
-      "\t\tprint i\n",
-      "\t\n",
-      "def print_array(array,size,message) :\n",
-      "\t\"\"\" Print out string array 'array' of size 'size'. 'message' is a flag indicating which message to display(ORIGINAL or SORTED) \"\"\"\n",
-      "\tif message == ORIGINAL :\n",
-      "\t\tprint \"\\nThe original array is:\"\n",
-      "\telse :\n",
-      "\t\tprint \"\\nThe sorted array is:\"\n",
-      "\tfor i in array :\n",
-      "\t\tprint i\n",
-      "\n",
-      "def sort_array(array,size) :\n",
-      "\t\"\"\" sort string array \"\"\"\n",
-      "\tfor i in range (size-1) :\n",
-      "\t\tfor j in range (i + 1, size) :\n",
-      "\t\t\tif array[i] > array[j] :\n",
-      "\t\t\t\tstring_swap(i,j)\n",
-      "\n",
-      "def string_swap(i, j) :\n",
-      "\t\"\"\" Exchange strings \"\"\"\n",
-      "\tglobal words\n",
-      "\twords[i],words[j] = words[j],words[i]\n",
-      "\n",
-      "get_size(array_size)\t\t# Read in the array size\n",
-      "get_array(words, array_size)\t# Read in the string array \"words\" of size \"array_size\"\n",
-      "print_array(words, array_size, ORIGINAL)\t# Echo the original array\n",
-      "sort_array(words, array_size)\t\n",
-      "print_array(words, array_size, SORTED) # Print the sorted array"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many words?  8\n",
-        "Please enter the words:\n",
-        "cat\n",
-        "pig\n",
-        "dog\n",
-        "earlobe\n",
-        "bingo\n",
-        "elbow\n",
-        "likely\n",
-        "radar\n",
-        "\n",
-        "The original array is:\n",
-        "cat\n",
-        "pig\n",
-        "dog\n",
-        "earlobe\n",
-        "bingo\n",
-        "elbow\n",
-        "likely\n",
-        "radar\n",
-        "\n",
-        "The sorted array is:\n",
-        "bingo\n",
-        "cat\n",
-        "dog\n",
-        "earlobe\n",
-        "elbow\n",
-        "likely\n",
-        "pig\n",
-        "radar\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/Chapter12.ipynb b/The_Spirit_of_C/Chapter12.ipynb
deleted file mode 100755
index 0195ed25..00000000
--- a/The_Spirit_of_C/Chapter12.ipynb
+++ /dev/null
@@ -1,455 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:bfcc6bf3e86f2ff3b103be284b1b027f6faa9020241e866b9b4398fd89f8923c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12, Strings and String Functions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-1, Page Number: 331"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# The control string is just a string\n",
-      "\n",
-      "format = \"%s\"\n",
-      "message = \"hello\"\n",
-      "\n",
-      "print format % message"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hello\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-2, Page Number: 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# The control string is just a string - part 2\n",
-      "\n",
-      "format = \"%.15s\"\n",
-      "message = \"hellohellohello\"\n",
-      "\n",
-      "# Read in the number of characters to be displayed from \"message\"\n",
-      "print \"How many characters do you want displayed?\",\n",
-      "size = 5\n",
-      "print size\n",
-      "\n",
-      "# Convert size to 2-digit number in format[2] and format[3]\n",
-      "\n",
-      "format = \"%.\" + str(size / 10) + str(size % 10) + \"s\"\n",
-      "print \"The format is %s\"%format\n",
-      "\n",
-      "# Print the string\n",
-      "print format % message"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many characters do you want displayed? 5\n",
-        "The format is %.05s\n",
-        "hello\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-3, Page Number 336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# sprintf function \n",
-      "\n",
-      "# There is no sprintf functionality in python\n",
-      "# However an alternate way to solve the problem is...\n",
-      "\n",
-      "print \"Please enter a number:\",\n",
-      "number = 17\n",
-      "print number\n",
-      "\n",
-      "string = \"The number is %d\"%number\n",
-      "\n",
-      "print \"%s\"%string\n",
-      "\n",
-      "# OR \n",
-      "\n",
-      "print string"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter a number: 17\n",
-        "The number is 17\n",
-        "The number is 17\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-4, Page Number 337"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Variable control string using the sprintf function\n",
-      "\n",
-      "message = \"hellohellohello\"\n",
-      "\n",
-      "# Read in the number of characters to be displayed from message \n",
-      "print \"How many characters do you want displayed?\",\n",
-      "size = 7\n",
-      "print size\n",
-      "\n",
-      "# Create format in \"format\" using \"size\"\n",
-      "format = \"%%.%ds\\n\"%size\n",
-      "print \"The format is: %s\"%format\n",
-      "\n",
-      "# Print the string\n",
-      "print format % message"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many characters do you want displayed? 7\n",
-        "The format is: %.7s\n",
-        "\n",
-        "hellohe\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-5, Page Number 338"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A terse string copy function\n",
-      "\n",
-      "def strcpy(copy, original):\n",
-      "\t\"\"\" copies one string into another \"\"\"\n",
-      "\tcopy = original\n",
-      "\treturn copy\n",
-      "\n",
-      "string1 = \"\"\n",
-      "string1 = strcpy(string1, \"This is a message\")\n",
-      "print \"String 1 = %s\"%string1\n",
-      "\n",
-      "string2 = \"\"\n",
-      "string2 = strcpy(string2, string1)\n",
-      "print \"String 2 = %s\"%string2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String 1 = This is a message\n",
-        "String 2 = This is a message\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-7, Page Number 344"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "string_point = \"hello\"\n",
-      "\n",
-      "string_var1 = string_point\n",
-      "\n",
-      "print \"%s\\n%s\"%(string_point,string_var1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hello\n",
-        "hello\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-8, Page Number 346"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# strcat function in python\n",
-      "# simple addition of strings results in concatenation in python\n",
-      "\n",
-      "def strcat(string1, string2) :\n",
-      "\t\"\"\" concatenates the string and returns the concatenated string \"\"\"\n",
-      "\tstring2 = string1 + string2\n",
-      "\treturn string2\n",
-      "\n",
-      "string1 = \"hello\"\n",
-      "string2 = strcat(string1,\" there\")\n",
-      "print \"string1 = %s\\nstring2 = %s\"%(string1,string2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "string1 = hello\n",
-        "string2 = hello there\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-9, Page Number 348"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# strcat version 2\n",
-      "\n",
-      "def strcat(string1, string2) :\n",
-      "\tstring1 = string1 + string2\n",
-      "\tstring2 = string1\n",
-      "\treturn string1,string2\n",
-      "\n",
-      "string1 = \"hello\"\n",
-      "string1,string2 = strcat(string1, \" there\")\n",
-      "\n",
-      "print \"string1 = %s\\nstring2 = %s\"%(string1,string2)\n",
-      "\n",
-      "if string1 == string2 :\n",
-      "\tprint \"string2 points to string1\"\n",
-      "else :\n",
-      "\tprint \"string2 does not point to string1\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "string1 = hello there\n",
-        "string2 = hello there\n",
-        "string2 points to string1\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 12-10, Page Number 351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Alphabetizing program using dynamically allocated arrays\n",
-      "\n",
-      "# Macro definition\n",
-      "MAX_ARRAY_SIZE = 20\t\t# The array may have up to 20 elements\n",
-      "MAX_STRING_SIZE = 10 \t# Each string may have up to 10 characters\n",
-      "# Message flags for \"print_array\"\n",
-      "ORIGINAL = 0\n",
-      "SORTED = 1\n",
-      "\n",
-      "words = [ \"cat\", \"pig\", \"dog\", \"earlobe\", \"bingo\", \"elbow\", \"likely\", \"radar\" ]\n",
-      "array_size = 0\n",
-      "\n",
-      "def get_size(size) :\n",
-      "\t\"\"\" Read the size of the array into size (until a value is entered) \"\"\"\n",
-      "\tprint \"How many words? \",\n",
-      "\tsize = 8\n",
-      "\tprint size\n",
-      "\twhile size < 1 or size > MAX_ARRAY_SIZE :\n",
-      "\t\tprint \"How many words? \",\n",
-      "\t\tsize = 8\n",
-      "\t\tprint size\n",
-      "\tglobal array_size\n",
-      "\tarray_size = size\n",
-      "\n",
-      "def get_array(array,size) :\n",
-      "\t\"\"\" Read in the string array called 'array' of size 'size' \"\"\"\n",
-      "\tprint \"Please enter the words:\"\n",
-      "\tfor i in array :\n",
-      "\t\tprint i\n",
-      "\t\n",
-      "def print_array(array,size,message) :\n",
-      "\t\"\"\" Print out string array 'array' of size 'size'. 'message' is a flag indicating which message to display(ORIGINAL or SORTED) \"\"\"\n",
-      "\tprintout = [ \"\\nThe original array is:\", \"\\nThe sorted array is:\", \"\\n\" ]\n",
-      "\t# Print the appropriate message\n",
-      "\tprint \"%s\"%printout[message]\n",
-      "\n",
-      "\tfor i in array :\n",
-      "\t\tprint i\n",
-      "\n",
-      "def pointer_sort(array,size) :\n",
-      "\t\"\"\" sort string array \"\"\"\n",
-      "\tfor i in range (size-1) :\n",
-      "\t\tfor j in range (i + 1, size) :\n",
-      "\t\t\tif array[i] > array[j] :\n",
-      "\t\t\t\tswap(i,j)\n",
-      "\n",
-      "def swap(i, j) :\n",
-      "\t\"\"\" Exchange strings \"\"\"\n",
-      "\tglobal words\n",
-      "\twords[i],words[j] = words[j],words[i]\n",
-      "\n",
-      "get_size(array_size)\t\t# Read in the array size\n",
-      "get_array(words, array_size)\t# Read in the string array \"words\" of size \"array_size\"\n",
-      "print_array(words, array_size, ORIGINAL)\t# Echo the original array\n",
-      "pointer_sort(words, array_size)\t\n",
-      "print_array(words, array_size, SORTED) # Print the sorted array"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many words?  8\n",
-        "Please enter the words:\n",
-        "cat\n",
-        "pig\n",
-        "dog\n",
-        "earlobe\n",
-        "bingo\n",
-        "elbow\n",
-        "likely\n",
-        "radar\n",
-        "\n",
-        "The original array is:\n",
-        "cat\n",
-        "pig\n",
-        "dog\n",
-        "earlobe\n",
-        "bingo\n",
-        "elbow\n",
-        "likely\n",
-        "radar\n",
-        "\n",
-        "The sorted array is:\n",
-        "bingo\n",
-        "cat\n",
-        "dog\n",
-        "earlobe\n",
-        "elbow\n",
-        "likely\n",
-        "pig\n",
-        "radar\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/The_Spirit_of_C/Chapter13.ipynb b/The_Spirit_of_C/Chapter13.ipynb
deleted file mode 100755
index d363dc7a..00000000
--- a/The_Spirit_of_C/Chapter13.ipynb
+++ /dev/null
@@ -1,346 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:133c881c520fd0cdef07f65db120109cc62a996dd5161cf0f664c2ffcaa3adc4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13, Structures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 13-1, Page Number: 370"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Introduction to structures\n",
-      "\n",
-      "from collections import namedtuple\n",
-      "\n",
-      "#Structure defintion\n",
-      "struct = namedtuple(\"struct\", \"integer real\")\n",
-      "\n",
-      "first_structure = struct( 7, 3.14 )\n",
-      "second_structure = first_structure\n",
-      "\n",
-      "print \"integer = %d, real = %f\"%(second_structure.integer,second_structure.real)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "integer = 7, real = 3.140000\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 13-2, Page Number: 371"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Initializing a structure\n",
-      "\n",
-      "from collections import namedtuple\n",
-      "\n",
-      "#Structure defintion\n",
-      "struct = namedtuple(\"struct\", \"integer real\")\n",
-      "\n",
-      "structure = struct( 7, 3.14 )\n",
-      "\n",
-      "print \"integer = %d, real = %f\"%(structure.integer,structure.real)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "integer = 7, real = 3.140000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 13-3, Page Number: 372"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# The structure tag\n",
-      "\n",
-      "from collections import namedtuple\n",
-      "\n",
-      "#Structure defintion\n",
-      "struct = namedtuple(\"struct\", \"integer real\")\n",
-      "\n",
-      "first_structure = struct( 7, 3.14 )\n",
-      "second_structure = first_structure\n",
-      "\n",
-      "print \"integer = %d, real = %f\"%(second_structure.integer,second_structure.real)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "integer = 7, real = 3.140000\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 13-4, Page Number: 375"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# An array of structures\n",
-      "\n",
-      "from collections import namedtuple\n",
-      "\n",
-      "#Structure defintion\n",
-      "struct = namedtuple(\"struct\", \"code price\")\n",
-      "\n",
-      "#code = [ \"a\", \"b\", \"c\", \"d\", \"e\" ]\n",
-      "#price = [ 3.29, 2.99, 0.59, 1.59, 2.39 ]\n",
-      "\n",
-      "table = [ struct(\"a\",3.29) , struct(\"b\",2.99), struct(\"c\",0.59), struct(\"d\",1.59), struct(\"e\",2.39) ]\n",
-      "\n",
-      "print \"Enter item code:\",\n",
-      "item_code = \"a\"\n",
-      "print item_code\n",
-      "i = 0\n",
-      "\n",
-      "while item_code != \".\" :\n",
-      "\n",
-      "\t#Search for 'item_code' in 'table'\n",
-      "\tfor i in range(4) :\n",
-      "\t\tif table[i].code == item_code :\n",
-      "\t\t\tbreak\n",
-      "\n",
-      "\tif i < 4:\n",
-      "\t\tprint \"Enter quantity: \",\n",
-      "\t\tquantity = 1\n",
-      "\t\tprint quantity\n",
-      "\t\n",
-      "\t\tprint \"\\nUnit price = %.2f, total price = %.2f.\\n\\n\"%(table[i].price,quantity * table[i].price)\n",
-      "\telse :\n",
-      "\t\tprint \"\\nItem code %c does not exist.\\n\\n\"%item_code\n",
-      "\tprint \"Enter item code:\",\n",
-      "\titem_code = \".\"\n",
-      "\tprint item_code\n",
-      "\n",
-      "\n",
-      "print \"Thank you.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter item code: a\n",
-        "Enter quantity:  1\n",
-        "\n",
-        "Unit price = 3.29, total price = 3.29.\n",
-        "\n",
-        "\n",
-        "Enter item code: .\n",
-        "Thank you.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 13-5, Page Number: 378"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A pointer in a structure\n",
-      "\n",
-      "from collections import namedtuple\n",
-      "\n",
-      "#Structure defintion\n",
-      "struct = namedtuple(\"struct\", \"name price\")\n",
-      "\n",
-      "#code = [ \"a\", \"b\", \"c\", \"d\", \"e\" ]\n",
-      "#price = [ 3.29, 2.99, 0.59, 1.59, 2.39 ]\n",
-      "\n",
-      "table = [ struct(\"pickles\",2.29) , struct(\"soda\",0.99), struct(\"yogurt\",0.59), struct(\"bread\",1.09), struct(\"milk\",0.69) ]\n",
-      "\n",
-      "print \"Enter item name:\",\n",
-      "item_name = \"pickles\"\n",
-      "print item_name\n",
-      "i = 0\n",
-      "\n",
-      "while item_name != \".\" :\n",
-      "\n",
-      "\t#Search for 'item_code' in 'table'\n",
-      "\tfor i in range(4) :\n",
-      "\t\tif table[i].name == item_name :\n",
-      "\t\t\tbreak\n",
-      "\n",
-      "\tif i < 4:\n",
-      "\t\tprint \"Enter quantity: \",\n",
-      "\t\tquantity = 1\n",
-      "\t\tprint quantity\n",
-      "\t\n",
-      "\t\tprint \"\\nUnit price = %.2f, total price = %.2f.\\n\\n\"%(table[i].price,quantity * table[i].price)\n",
-      "\telse :\n",
-      "\t\tprint \"\\nItem code %c does not exist.\\n\\n\"%item_name\n",
-      "\tprint \"Enter item code:\",\n",
-      "\titem_name = \".\"\n",
-      "\tprint item_name\n",
-      "\n",
-      "\n",
-      "print \"Thank you.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter item name: pickles\n",
-        "Enter quantity:  1\n",
-        "\n",
-        "Unit price = 2.29, total price = 2.29.\n",
-        "\n",
-        "\n",
-        "Enter item code: .\n",
-        "Thank you.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 13-6, Page Number: 381"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# An array in a structure\n",
-      "\n",
-      "from collections import namedtuple\n",
-      "\n",
-      "#Structure defintion\n",
-      "struct = namedtuple(\"struct\", \"name price\")\n",
-      "\n",
-      "#code = [ \"a\", \"b\", \"c\", \"d\", \"e\" ]\n",
-      "#price = [ 3.29, 2.99, 0.59, 1.59, 2.39 ]\n",
-      "\n",
-      "table = [ struct(\"pickles\",2.29) , struct(\"soda\",0.99), struct(\"yogurt\",0.59), struct(\"bread\",1.09), struct(\"milk\",0.69) ]\n",
-      "\n",
-      "print \"Enter item name:\",\n",
-      "item_name = \"milk\"\n",
-      "print item_name\n",
-      "i = 0\n",
-      "\n",
-      "while item_name != \".\" :\n",
-      "\n",
-      "\t#Search for 'item_code' in 'table'\n",
-      "\tfor i in range(5) :\n",
-      "\t\tif table[i].name == item_name :\n",
-      "\t\t\tbreak\n",
-      "\n",
-      "\tif i < 5:\n",
-      "\t\tprint \"Enter quantity: \",\n",
-      "\t\tquantity = 5\n",
-      "\t\tprint quantity\n",
-      "\t\n",
-      "\t\tprint \"\\nUnit price = %.2f, total price = %.2f.\\n\\n\"%(table[i].price,quantity * table[i].price)\n",
-      "\telse :\n",
-      "\t\tprint \"\\nItem code %c does not exist.\\n\\n\"%item_name\n",
-      "\tprint \"Enter item code:\",\n",
-      "\titem_name = \".\"\n",
-      "\tprint item_name\n",
-      "\n",
-      "\n",
-      "print \"Thank you.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter item name: milk\n",
-        "Enter quantity:  5\n",
-        "\n",
-        "Unit price = 0.69, total price = 3.45.\n",
-        "\n",
-        "\n",
-        "Enter item code: .\n",
-        "Thank you.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/Chapter14.ipynb b/The_Spirit_of_C/Chapter14.ipynb
deleted file mode 100755
index 380e83ef..00000000
--- a/The_Spirit_of_C/Chapter14.ipynb
+++ /dev/null
@@ -1,298 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3748f84e3bd0279efe47855bd2c4e23912ef725121f3591980fa33e19d6ccae4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14, Input/Output and Files"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 14-5, Page Number: 424"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Files : Opening and Closing\n",
-      "\n",
-      "# Open the file for writing\n",
-      "\n",
-      "file_pointer = open(\"first_file\",\"w\")\n",
-      "\n",
-      "print \"Enter text:\"\n",
-      "\n",
-      "# Get characters and place them in the file\n",
-      "\n",
-      "text = \"BASIC is easy,\\nFORTAN is fun,\\nPascal may be structured,\\nBut C's the one!\"\n",
-      "print text\n",
-      "file_pointer.write(text)\n",
-      "\n",
-      "file_pointer.close()\n",
-      "\n",
-      "file_pointer = open(\"first_file\",\"r\")\n",
-      "\n",
-      "#Display the text from file\n",
-      "\n",
-      "print \"\\nYou entered:\"\n",
-      "print file_pointer.read()\n",
-      "\n",
-      "file_pointer.close()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter text:\n",
-        "BASIC is easy,\n",
-        "FORTAN is fun,\n",
-        "Pascal may be structured,\n",
-        "But C's the one!\n",
-        "\n",
-        "You entered:\n",
-        "BASIC is easy,\n",
-        "FORTAN is fun,\n",
-        "Pascal may be structured,\n",
-        "But C's the one!\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 14-6, Page Number: 426"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Using several files\n",
-      "\n",
-      "# Open the source file for reading\n",
-      "\n",
-      "first_file_pointer = open(\"first_file\",\"r\")\n",
-      "\n",
-      "# Open the destination file for writing\n",
-      "\n",
-      "second_file_pointer = open(\"second_file\",\"w\")\n",
-      "\n",
-      "# Transfer characters from source to destination file\n",
-      "\n",
-      "c = first_file_pointer.read()\n",
-      "second_file_pointer.write(c)\n",
-      "\n",
-      "# Close both files\n",
-      "\n",
-      "first_file_pointer.close()\n",
-      "second_file_pointer.close()\n",
-      "\n",
-      "# Open second file for displaying contentes\n",
-      "\n",
-      "second_file_pointer = open(\"second_file\",\"r\")\n",
-      "print(\"'second_file' contains:\")\n",
-      "print second_file_pointer.read()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "'second_file' contains:\n",
-        "BASIC is easy,\n",
-        "FORTAN is fun,\n",
-        "Pascal may be structured,\n",
-        "But C's the one!\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 14-7, Page Number : 428"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Append mode\n",
-      "\n",
-      "# Open the first file for writing\n",
-      "\n",
-      "first_file_pointer = open(\"first_file\",\"w\")\n",
-      "\n",
-      "# Open the second file for appending\n",
-      "\n",
-      "second_file_pointer = open(\"second_file\",\"a\")\n",
-      "\n",
-      "# Writing to both files\n",
-      "\n",
-      "first_file_pointer.write(\"\\n--Anonymous\\n\")\n",
-      "second_file_pointer.write(\"\\n--Anonymous\\n\")\n",
-      "\n",
-      "# Close both files\n",
-      "\n",
-      "first_file_pointer.close()\n",
-      "second_file_pointer.close()\n",
-      "\n",
-      "# Display the contents of first file\n",
-      "first_file_pointer = open(\"first_file\",\"r\")\n",
-      "print(\"After write, 'first_file contains:\")\n",
-      "print first_file_pointer.read()\n",
-      "\n",
-      "# Display the contents of second file\n",
-      "second_file_pointer = open(\"second_file\",\"r\")\n",
-      "print(\"After append, 'second_file' contains:\")\n",
-      "print second_file_pointer.read()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "After write, 'first_file contains:\n",
-        "\n",
-        "--Anonymous\n",
-        "\n",
-        "After append, 'second_file' contains:\n",
-        "BASIC is easy,\n",
-        "FORTAN is fun,\n",
-        "Pascal may be structured,\n",
-        "But C's the one!\n",
-        "--Anonymous\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 14-8, Page Number: 431"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# The seek function\n",
-      "\n",
-      "# Open file for reading\n",
-      "try :\n",
-      "\tfile_pointer = open(\"second_file\",\"r\")\n",
-      "except :\n",
-      "\tprint \"'second_file' cannot be opened\"\n",
-      "\texit ()\n",
-      "\n",
-      "# Skip the first 6 characters\n",
-      "\n",
-      "file_pointer.seek(6,0)\n",
-      "print file_pointer.read()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "is easy,\n",
-        "FORTAN is fun,\n",
-        "Pascal may be structured,\n",
-        "But C's the one!\n",
-        "--Anonymous\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 14-9, Page Number: 433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# The 'w+' mode and seek to rewind a file\n",
-      "\n",
-      "# Open a file for read/write\n",
-      "file_pointer = open(\"first_file\",\"w+\")\n",
-      "\n",
-      "print \"Enter text:\"\n",
-      "text = \"BASIC is easy,\\nFORTAN is fun,\\nPascal may be structured,\\nBut C's the one!\"\n",
-      "print text\n",
-      "file_pointer.write(text)\n",
-      "\n",
-      "file_pointer.seek(0,0)\n",
-      "\n",
-      "# Display the file\n",
-      "print \"\\nYou entered:\"\n",
-      "print file_pointer.read()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter text:\n",
-        "BASIC is easy,\n",
-        "FORTAN is fun,\n",
-        "Pascal may be structured,\n",
-        "But C's the one!\n",
-        "\n",
-        "You entered:\n",
-        "BASIC is easy,\n",
-        "FORTAN is fun,\n",
-        "Pascal may be structured,\n",
-        "But C's the one!\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter2.ipynb b/The_Spirit_of_C/chapter2.ipynb
deleted file mode 100755
index dc84a28d..00000000
--- a/The_Spirit_of_C/chapter2.ipynb
+++ /dev/null
@@ -1,119 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:7c8e11b3dcc5781949652c0b5c450d78713cb6e6b69326955d0f763c48805147"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2, An Introduction to Declarations, Assignments and Variables"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 2-1 , Page number: 20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# An introduction to variable declarations \n",
-      "\n",
-      "number = 7 # no need of declaration of variable. Python is dynamically-typed language\n",
-      "print \"There are %d days in a week.\"%number"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "There are 7 days in a week.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 2-2 , Page number: 21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A simple program with two declarations\n",
-      "\n",
-      "minutes = 60\n",
-      "hours = 24\n",
-      "\n",
-      "print \"There are %d minutes in an hour.\"%minutes\n",
-      "print \"And %d hours in a day.\"%hours"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "There are 60 minutes in an hour.\n",
-        "And 24 hours in a day.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 2-4 , Page number: 22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# prints values of two variables\n",
-      "\n",
-      "dozens = 17\n",
-      "units = dozens * 12\n",
-      "\n",
-      "print \"%d units is equivalent to %d dozen.\"%(units,dozens)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "204 units is equivalent to 17 dozen.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter3.ipynb b/The_Spirit_of_C/chapter3.ipynb
deleted file mode 100755
index b498486d..00000000
--- a/The_Spirit_of_C/chapter3.ipynb
+++ /dev/null
@@ -1,241 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c466b513c45bfe09da1b0507614f538770318fb66aadf4da4e1995363df32911"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3 , Integer Arithmetic Expressions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 3-1 , Page number: 33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# An introduction to arithmetic \n",
-      "\n",
-      "a = 2\n",
-      "b = 3\n",
-      "c = 4\n",
-      "\n",
-      "answer = a + b\n",
-      "print \"a + b = %d\"%answer\n",
-      "\n",
-      "answer = a - b\n",
-      "print \"a - b = %d\"%answer\n",
-      "\n",
-      "answer = a * b\n",
-      "print \"a * b = %d\"%answer\n",
-      "\n",
-      "answer = a / b\n",
-      "print \"a / b = %d\"%answer\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a + b = 5\n",
-        "a - b = -1\n",
-        "a * b = 6\n",
-        "a / b = 0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 3-2 , Page number: 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of mathematical hierarchy\n",
-      "\n",
-      "a = 2\n",
-      "b = 3\n",
-      "c = 4\n",
-      "\n",
-      "answer = a + b * c\n",
-      "\n",
-      "print \"a = %d, b = %d, c= %d\"%(a,b,c)\n",
-      "print \"a + b * c = %d\\n\"%answer"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a = 2, b = 3, c= 4\n",
-        "a + b * c = 14\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 3-3 , Page number: 35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A more advanced illustration of mathematical hierarchy\n",
-      "\n",
-      "a = 5\n",
-      "b = 10\n",
-      "c = 15\n",
-      "\n",
-      "answer = a + b / c\n",
-      "print \"%d + %d / %d = %d\"%(a,b,c,answer)\n",
-      "\n",
-      "answer = b * c - a\n",
-      "print \"%d * %d - %d = %d\"%(a,b,c,answer)\n",
-      "\n",
-      "answer = a * b / c\n",
-      "print \"%d * %d / %d = %d\"%(a,b,c,answer)\n",
-      "\n",
-      "answer = (a + c) * b / a\n",
-      "print \"(%d + %d) * %d / %d = %d\"%(a,c,b,a,answer)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "5 + 10 / 15 = 5\n",
-        "5 * 10 - 15 = 145\n",
-        "5 * 10 / 15 = 3\n",
-        "(5 + 15) * 10 / 5 = 40\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 3-4 , Page number: 37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of modulus operator\n",
-      "\n",
-      "a = 4\n",
-      "b = 8\n",
-      "c = 13\n",
-      "\n",
-      "print \"%d mod %d = %d\"%(b,a,b%a)\n",
-      "print \"%d mod %d = %d\"%(c,a,c%a)\n",
-      "print \"%d mod %d = %d\"%(c,b,c%b)\n",
-      "\n",
-      "print \"The modulus operator is: %\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "8 mod 4 = 0\n",
-        "13 mod 4 = 1\n",
-        "13 mod 8 = 5\n",
-        "The modulus operator is: %\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 3-5 , Page number: 38"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of the unary minus\n",
-      "\n",
-      "a = -4\n",
-      "b = 5\n",
-      "c = -6\n",
-      "\n",
-      "print \"a = %d, b = %d, c = %d\\n\\n\"%(a,b,c)\n",
-      "print \"      and now\\n\\n\"\n",
-      "\n",
-      "a = -a\n",
-      "b = -b\n",
-      "c = -c\n",
-      "\n",
-      "print \"a = %d, b = %d, c = %d\\n\\n\"%(a,b,c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a = -4, b = 5, c = -6\n",
-        "\n",
-        "\n",
-        "      and now\n",
-        "\n",
-        "\n",
-        "a = 4, b = -5, c = 6\n",
-        "\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter4.ipynb b/The_Spirit_of_C/chapter4.ipynb
deleted file mode 100755
index d13f3710..00000000
--- a/The_Spirit_of_C/chapter4.ipynb
+++ /dev/null
@@ -1,148 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:03f3f522026b7ff6e1acac7746805bcac18faba022541a5d8a73470a897fdf7a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4, Some More Data Types"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Prgoram 4-1 , Page number: 47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# An illustration of the use of floating point numbers\n",
-      "\n",
-      "# There is no loss of precision in python\n",
-      "a = 246.8\n",
-      "b = 135.79\n",
-      "\n",
-      "answer = a + b\n",
-      "print \"%f + %f = %f\"%(a,b,answer)\n",
-      "\n",
-      "answer = a - b\n",
-      "print \"%f - %f = %f\"%(a,b,answer)\n",
-      "\n",
-      "answer = a * b\n",
-      "print \"%f * %f = %f\"%(a,b,answer)\n",
-      "\n",
-      "answer = a / b\n",
-      "print \"%f / %f = %f\"%(a,b,answer)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "246.800000 + 135.790000 = 382.590000\n",
-        "246.800000 - 135.790000 = 111.010000\n",
-        "246.800000 * 135.790000 = 33512.972000\n",
-        "246.800000 / 135.790000 = 1.817512\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 4-2 , Page number: 49"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of mixed mode expressions\n",
-      "\n",
-      "a = 2\n",
-      "b = 12.34\n",
-      "\n",
-      "result = a * b\n",
-      "print \"%d * %f = %f\"%(a,b,result)\n",
-      "\n",
-      "print \"%f in scientific notation is %e\"%(result,result)\n",
-      "\n",
-      "answer = a * b\n",
-      "print \"%d * %f = %d\"%(a,b,answer)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "2 * 12.340000 = 24.680000\n",
-        "24.680000 in scientific notation is 2.468000e+01\n",
-        "2 * 12.340000 = 24\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 4-3 , Page number: 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "ltr_1 = \"C\"\n",
-      "ltr_2 = \" \"\n",
-      "ltr_3 = \"i\"\n",
-      "ltr_4 = \"s\"\n",
-      "ltr_5 = \"g\"\n",
-      "ltr_6 = \"r\"\n",
-      "ltr_7 = \"e\"\n",
-      "ltr_8 = \"a\"\n",
-      "ltr_9 = \"t\"\n",
-      "ltr_10 = \"!\"\n",
-      "\n",
-      "print \"%c%c%c%c%c%c%c%c%c%c%c\"%(ltr_1,ltr_2,ltr_3,ltr_4,ltr_2,ltr_5,ltr_6,ltr_7,ltr_8,ltr_9,ltr_10)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C is great!\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter5.ipynb b/The_Spirit_of_C/chapter5.ipynb
deleted file mode 100755
index db420a5b..00000000
--- a/The_Spirit_of_C/chapter5.ipynb
+++ /dev/null
@@ -1,218 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:28fc7922ba91eac91816a02729f8986381706b8cffd1eaa9d68ce0cf435191aa"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5, Making Decisions in C"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 5-2 , Page number: 65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Using If command\n",
-      "\n",
-      "print \"Enter a positive integer under 10: \",\n",
-      "n = 7\n",
-      "print n\n",
-      "\n",
-      "if n == 7 :\t\t\t\t\t# In this case it is 7\n",
-      "\tprint \"I knew you'd select 7!\"\n",
-      "print \"Thank you for your cooperation.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a positive integer under 10:  7\n",
-        "I knew you'd select 7!\n",
-        "Thank you for your cooperation.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 5-3 , Page number: 67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Testing for leap years\n",
-      "\n",
-      "print \"Please enter a year:\",\n",
-      "year = 1948\t\t\n",
-      "print year\n",
-      "# If \"year\" is divisble by 4 and not divisible by 100 or divisible by 400, \"year\" is a leap year\n",
-      "\n",
-      "if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0) :\n",
-      "\tprint \"%d is a leap year.\"%year\n",
-      "else :\n",
-      "\tprint \"%d is not a leap year.\"%year"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter a year: 1948\n",
-        "1948 is a leap year.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 5-4 , Page number: 69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of if...else clause\n",
-      "# Program to tell if a number is odd or even\n",
-      "\n",
-      "print \"Please enter an integer:\",\n",
-      "n = 13\n",
-      "print n\n",
-      "\n",
-      "# A number is even if it is divisible by 2\n",
-      "\n",
-      "if n % 2 != 0:\n",
-      "\tprint \"%d is odd\"%n\n",
-      "else:\n",
-      "\tprint \"%d is even\"%n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter an integer: 13\n",
-        "13 is odd\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 5-5 , Page number: 70"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Testing if a number is divisible by 7\n",
-      "\n",
-      "print \"Enter an integer:\",\n",
-      "n = 13\n",
-      "print n\n",
-      "\n",
-      "# A number is even if it is divisible by 2\n",
-      "\n",
-      "if n % 7 == 0:\n",
-      "\tprint \"%d is divisible by 7\"%n\n",
-      "else:\n",
-      "\tprint \"%d is not divisible by 7\"%n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an integer: 13\n",
-        "13 is not divisible by 7\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 5-6 , Page number: 71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Determining which of 2 numbers is larger\n",
-      "\n",
-      "print \"Enter an integer:\",\n",
-      "a = 4\n",
-      "print a\n",
-      "print \"Enter another integer: \",\n",
-      "b = 7\n",
-      "print b\n",
-      "\n",
-      "if a > b :\n",
-      "\tprint \"%d is greater than %d.\"%(a,b)\n",
-      "else:\n",
-      "\tprint \"%d is smaller than %d.\"%(a,b)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an integer: 4\n",
-        "Enter another integer:  7\n",
-        "4 is smaller than 7.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter6.ipynb b/The_Spirit_of_C/chapter6.ipynb
deleted file mode 100755
index f50ff06b..00000000
--- a/The_Spirit_of_C/chapter6.ipynb
+++ /dev/null
@@ -1,494 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c565a0de8f316cbc19d18a7bdba735b6241f6d0c0c10c2d49245aa88403df9e7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6, The While and do...While Loops"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-1 , Page number: 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A first look at the while loop\n",
-      "\n",
-      "i = 1\n",
-      "\n",
-      "while i < 11:\n",
-      "\tprint \"This is line number %d\"%i\n",
-      "\ti += 1 \t\t\t# Python doesn't support \"++\" or \"--\" shorthand expressions"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is line number 1\n",
-        "This is line number 2\n",
-        "This is line number 3\n",
-        "This is line number 4\n",
-        "This is line number 5\n",
-        "This is line number 6\n",
-        "This is line number 7\n",
-        "This is line number 8\n",
-        "This is line number 9\n",
-        "This is line number 10\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-2 , Page number: 89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Countdown to blastoff\n",
-      "\n",
-      "i = 10\n",
-      "\n",
-      "print \"Countdown\"\n",
-      "\n",
-      "while i >= 0 :\n",
-      "\tprint \"%d\"%i\n",
-      "\ti -= 1\n",
-      "\n",
-      "print \"\\n\\nBlastoff!\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Countdown\n",
-        "10\n",
-        "9\n",
-        "8\n",
-        "7\n",
-        "6\n",
-        "5\n",
-        "4\n",
-        "3\n",
-        "2\n",
-        "1\n",
-        "0\n",
-        "\n",
-        "\n",
-        "Blastoff!\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-3 , Page number: 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Hailstones\n",
-      "\n",
-      "counter = 0;\n",
-      "\n",
-      "print \"Enter a positive integer:\",\n",
-      "n = 51\n",
-      "print n\n",
-      "\n",
-      "nsave = n\n",
-      "\n",
-      "\n",
-      "while n > 1:\n",
-      "\tif n % 2 :\t\t\t# if (n % 2) is non-zero, n is not divisible by 2 and is therefore odd\n",
-      "\t\tn = n * 3 + 1\n",
-      "\telse :\n",
-      "\t\tn /= 2\n",
-      "\tcounter += 1\n",
-      "\tprint \"Step %4d: n = %4d\"%(counter,n)\n",
-      "\n",
-      "print \"\\n\\n%d went to 1 in %d steps.\"%(nsave,counter)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a positive integer: 51\n",
-        "Step    1: n =  154\n",
-        "Step    2: n =   77\n",
-        "Step    3: n =  232\n",
-        "Step    4: n =  116\n",
-        "Step    5: n =   58\n",
-        "Step    6: n =   29\n",
-        "Step    7: n =   88\n",
-        "Step    8: n =   44\n",
-        "Step    9: n =   22\n",
-        "Step   10: n =   11\n",
-        "Step   11: n =   34\n",
-        "Step   12: n =   17\n",
-        "Step   13: n =   52\n",
-        "Step   14: n =   26\n",
-        "Step   15: n =   13\n",
-        "Step   16: n =   40\n",
-        "Step   17: n =   20\n",
-        "Step   18: n =   10\n",
-        "Step   19: n =    5\n",
-        "Step   20: n =   16\n",
-        "Step   21: n =    8\n",
-        "Step   22: n =    4\n",
-        "Step   23: n =    2\n",
-        "Step   24: n =    1\n",
-        "\n",
-        "\n",
-        "51 went to 1 in 24 steps.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-4 , Page number: 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Computing the factorial\n",
-      "\n",
-      "print \"Type in a positive integer:\",\n",
-      "n = 6\n",
-      "print n\n",
-      "\n",
-      "nsave = n\n",
-      "factorial = 1\n",
-      "\n",
-      "while n > 1 :\t\n",
-      "\tfactorial *= n\n",
-      "\tn -= 1\n",
-      "\n",
-      "print \"Factorial of %d = %d\"%(nsave,factorial)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Type in a positive integer: 6\n",
-        "Factorial of 6 = 720\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-5 , Page number: 93"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Computing the factorial (shorter version)\n",
-      "\n",
-      "print \"Type in a positive integer: \",\n",
-      "n = 6\n",
-      "print n\n",
-      "\n",
-      "nsave = n\n",
-      "factorial = 1\n",
-      "\n",
-      "while n > 1 :\n",
-      "\tfactorial *= n      # Since python doesn't support '--' expression, the given program cannot be shortened\n",
-      "\tn -= 1\n",
-      "\n",
-      "print \"Factorial of %d = %d\"%(nsave,factorial)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Type in a positive integer:  6\n",
-        "Factorial of 6 = 720\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-6 , Page number: 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of do..while loops\n",
-      "\n",
-      "# This program reads in an integer, adds up the values of each of its individual digits, and prints out the sum.\n",
-      "\n",
-      "sum = 0\n",
-      "\n",
-      "print \"Type in an integer:\",\n",
-      "n = 1776\n",
-      "print n\n",
-      "\n",
-      "# There is no do...while loop in python\n",
-      "# An equivalent while loop can work\n",
-      "\n",
-      "while n > 0 :\n",
-      "\trightmost_digit = n % 10\t# Extract rightmost digit\n",
-      "\tsum += rightmost_digit\n",
-      "\tn /= 10\t\t\t\t\t\t# Move next digit into rightmost position\n",
-      "\n",
-      "print \"The sum of the digits is %d\"%sum\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Type in an integer: 1776\n",
-        "The sum of the digits is 21\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-7 , Page number: 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import sys\n",
-      "\n",
-      "# This program reads in an integer and prints it out backwards\n",
-      "\n",
-      "print \"Type in an integer: \",\n",
-      "n = 5746\n",
-      "print n\n",
-      "\n",
-      "print \"%d backwards is \"%n,\n",
-      "\n",
-      "while n > 0 :\n",
-      "\trightmost_digit = n % 10\n",
-      "\tsys.stdout.write(\"%d\"%rightmost_digit)  # For printing in the same line without implicit whitespaces\n",
-      "\tn /= 10\n",
-      "print \t\t\t# For the new line\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Type in an integer:  5746\n",
-        "5746 backwards is 6475\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-8 , Page number: 99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A program to read in numbers and print out their squares\n",
-      "\n",
-      "print \"Enter an integer (negative to quit):\",\n",
-      "n = 5\n",
-      "print n\n",
-      "\n",
-      "while n > 0 :\n",
-      "\tprint \"%d squared is %d\"%(n,n*n)\n",
-      "\tprint \"\\nEnter an integer (negative to quit):\",\n",
-      "\tn = -1\n",
-      "\tprint n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter an integer (negative to quit): 5\n",
-        "5 squared is 25\n",
-        "\n",
-        "Enter an integer (negative to quit): -1\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-9 , Page number: 100"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A program to read in radius values and print out the circumference and area\n",
-      "\n",
-      "pi = 3.1416\n",
-      "\n",
-      "print \"Enter radius (negative to quit):\",\n",
-      "radius = 4.5\n",
-      "print radius\n",
-      "\n",
-      "while radius >= 0.0 :\n",
-      "\tprint \"The circumference is %f\"%(2.0 * pi * radius)\n",
-      "\tprint \"The area is %f\"%(pi * radius * radius)\n",
-      "\tprint \"\\nEnter radius (negative to quit):\",\n",
-      "\tradius = -1.0\n",
-      "\tprint radius\n",
-      "\t\n",
-      "\t"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter radius (negative to quit): 4.5\n",
-        "The circumference is 28.274400\n",
-        "The area is 63.617400\n",
-        "\n",
-        "Enter radius (negative to quit): -1.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 6-10 , Page number: 101"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# A program to read in length and width and print out the perimeter and area\n",
-      "\n",
-      "print \"Enter length and width\"\n",
-      "print \" (one or both negative to quit):\",\n",
-      "length = 4\n",
-      "width = 5\n",
-      "print length,width\n",
-      "\n",
-      "while length > 0 and width > 0 : \n",
-      "\tprint \"The perimeter is %d\"%(2*(length+width))\n",
-      "\tprint \"The area is %d\"%(length*width)\n",
-      "\tprint \"\\nEnter length and width\"\n",
-      "\tprint \" (one or both negative to quit):\",\n",
-      "\tlength = -1\n",
-      "\twidth = 0\n",
-      "\tprint length,width\t"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter length and width\n",
-        " (one or both negative to quit): 4 5\n",
-        "The perimeter is 18\n",
-        "The area is 20\n",
-        "\n",
-        "Enter length and width\n",
-        " (one or both negative to quit): -1 0\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter7.ipynb b/The_Spirit_of_C/chapter7.ipynb
deleted file mode 100755
index 33f7a4f3..00000000
--- a/The_Spirit_of_C/chapter7.ipynb
+++ /dev/null
@@ -1,690 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:245ea275fa7c7004a79eeeeb6b3fc9780b5a19707d8b95c7a675d2142428b630"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7, The for loop"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-1 , Page number: 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A simple illustration of for loop\n",
-      "\n",
-      "for i in range(1,11,1):\t\t# The loop starts with i=1 and continues until i<11 with its value increasing by 1 each time \n",
-      "\tprint \"This is line number %d\"%i"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is line number 1\n",
-        "This is line number 2\n",
-        "This is line number 3\n",
-        "This is line number 4\n",
-        "This is line number 5\n",
-        "This is line number 6\n",
-        "This is line number 7\n",
-        "This is line number 8\n",
-        "This is line number 9\n",
-        "This is line number 10\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-2 , Page number: 120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Another simple illustration of a for loop\n",
-      "\n",
-      "print \"Countdown\"\n",
-      "\n",
-      "for i in range(10,-1,-1):\n",
-      "\tprint i\n",
-      "print \"\\n\\nBlastoff!\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Countdown\n",
-        "10\n",
-        "9\n",
-        "8\n",
-        "7\n",
-        "6\n",
-        "5\n",
-        "4\n",
-        "3\n",
-        "2\n",
-        "1\n",
-        "0\n",
-        "\n",
-        "\n",
-        "Blastoff!\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-3 , Page number: 121"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculating the squares of integers 1 through 10\n",
-      "\n",
-      "for i in range (1,11,1):\n",
-      "\tprint \"%d squared = %d\"%(i,i*i)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1 squared = 1\n",
-        "2 squared = 4\n",
-        "3 squared = 9\n",
-        "4 squared = 16\n",
-        "5 squared = 25\n",
-        "6 squared = 36\n",
-        "7 squared = 49\n",
-        "8 squared = 64\n",
-        "9 squared = 81\n",
-        "10 squared = 100\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-4 , Page number: 121"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A better looking program to calculate squares\n",
-      "\n",
-      "print \"Table of Results\\n\"\n",
-      "print \"   i    i squared\"\n",
-      "print \"   -    ---------\"\n",
-      "\n",
-      "for i in range(1,11,1):\n",
-      "\tprint \"%4d%13d\"%(i,i*i)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Table of Results\n",
-        "\n",
-        "   i    i squared\n",
-        "   -    ---------\n",
-        "   1            1\n",
-        "   2            4\n",
-        "   3            9\n",
-        "   4           16\n",
-        "   5           25\n",
-        "   6           36\n",
-        "   7           49\n",
-        "   8           64\n",
-        "   9           81\n",
-        "  10          100\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-5 , Page number: 122"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Finding the sum of integers 1 through 100\n",
-      "\n",
-      "sum = 0\n",
-      "n = 100\n",
-      "\n",
-      "for i in range(1,n+1,1):\t\t# The loop starts with i=1 and continues until i<11 with its value increasing by 1 each time \n",
-      "\tsum += i\t\n",
-      "\n",
-      "print \"The sum of integers from  1 to %d is %d\"%(n,sum)\n",
-      "print \"\\n\\nThis is confirmed by the Gauss Method: %d\"%(n*(n+1)/2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of integers from  1 to 100 is 5050\n",
-        "\n",
-        "\n",
-        "This is confirmed by the Gauss Method: 5050\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-6 , Page number: 124"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Finding the sum of the integers 1 through MAX_VAL\n",
-      "\n",
-      "# There is no #define command for defining constants\n",
-      "# We can use a function that returns the value (although it is not very efficient to do so)\n",
-      "\n",
-      "def MAX_VAL():\n",
-      "\treturn 100\t\t\t# The value of MAX_VAL\n",
-      "\n",
-      "sum = 0\n",
-      "\n",
-      "for i in range (0,MAX_VAL()+1,1):\n",
-      "\tsum += i\n",
-      "\n",
-      "print \"The sum of the integers from 1 to %d is %d\"%(MAX_VAL(),sum)\n",
-      "print \"\\n\\nThis is confirmed by the Gauss Method : %d\"%(MAX_VAL()*(MAX_VAL()+1)/2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sum of the integers from 1 to 100 is 5050\n",
-        "\n",
-        "\n",
-        "This is confirmed by the Gauss Method : 5050\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-7 , Page number: 125"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculating the sum of integers 1 through n\n",
-      "\n",
-      "sum = 0\n",
-      "\n",
-      "print \"Please type in a positive integer:\",\n",
-      "n = 50\n",
-      "print n\n",
-      "\n",
-      "for i in range (1,n+1,1):\n",
-      "\tsum += i\n",
-      "\n",
-      "print \"\\nThe sum of the integers from 1 to %d is: %d\"%(n,sum)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please type in a positive integer: 50\n",
-        "\n",
-        "The sum of the integers from 1 to 50 is: 1275\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-8 , Page number: 125"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculating the sum of integers low through high\n",
-      "\n",
-      "sum = 0\n",
-      "\n",
-      "print \"Please type in the low bound:\",\n",
-      "low = 3\n",
-      "print low\n",
-      "print \"Please type in the high bound:\",\n",
-      "high = 9\n",
-      "print high\n",
-      "\n",
-      "for i in range(low,high+1,1):\n",
-      "\tsum += i\n",
-      "\n",
-      "print \"\\nThe sum of integers from %d to %d is: %d\"%(low,high,sum)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please type in the low bound: 3\n",
-        "Please type in the high bound: 9\n",
-        "\n",
-        "The sum of integers from 3 to 9 is: 42\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-9 , Page number: 127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Calculating the sum of integers low through high with an exchange if necessary\n",
-      "\n",
-      "sum = 0\n",
-      "\n",
-      "print \"Please type in the low bound:\",\n",
-      "low = 30\n",
-      "print low\n",
-      "print \"Please type in the high bound:\",\n",
-      "high = 15\n",
-      "print high\n",
-      "\n",
-      "if high < low :\n",
-      "\tprint \"Oops! Wrong order.\"\n",
-      "\tprint \"Never mind, I'll switch them.\"\n",
-      "\t# Code for swapping using temporary variable\n",
-      "\ttemp = low\n",
-      "\tlow = high\n",
-      "\thigh = temp\n",
-      "\n",
-      "for i in range(low,high+1,1):\n",
-      "\tsum += i\n",
-      "\n",
-      "print \"\\nThe sum of integers from %d to %d is: %d\"%(low,high,sum)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please type in the low bound: 30\n",
-        "Please type in the high bound: 15\n",
-        "Oops! Wrong order.\n",
-        "Never mind, I'll switch them.\n",
-        "\n",
-        "The sum of integers from 15 to 30 is: 360\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-10 ,Page number: 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of a nest of for loops\n",
-      "\n",
-      "n = 5\n",
-      "for i in range(1,7,1):\n",
-      "\tprint \"\\n\\nPlease type in a positive integer: \",\n",
-      "\tprint n\n",
-      "\n",
-      "\tsum = 0\n",
-      "\tfor j in range(1,n+1,1):\n",
-      "\t\tsum += j\n",
-      "\tprint \"\\nThe sum of integers from 1 to %d is %d\"%(n,sum)\n",
-      "\t\n",
-      "\tn += 5\t\t\t# Just to match the exercise's values"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "Please type in a positive integer:  5\n",
-        "\n",
-        "The sum of integers from 1 to 5 is 15\n",
-        "\n",
-        "\n",
-        "Please type in a positive integer:  10\n",
-        "\n",
-        "The sum of integers from 1 to 10 is 55\n",
-        "\n",
-        "\n",
-        "Please type in a positive integer:  15\n",
-        "\n",
-        "The sum of integers from 1 to 15 is 120\n",
-        "\n",
-        "\n",
-        "Please type in a positive integer:  20\n",
-        "\n",
-        "The sum of integers from 1 to 20 is 210\n",
-        "\n",
-        "\n",
-        "Please type in a positive integer:  25\n",
-        "\n",
-        "The sum of integers from 1 to 25 is 325\n",
-        "\n",
-        "\n",
-        "Please type in a positive integer:  30\n",
-        "\n",
-        "The sum of integers from 1 to 30 is 465\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-11 , Page number: 130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Another illustration of a nest of for loops\n",
-      "\n",
-      "print \"How many values of n do you wish to enter? \",\n",
-      "runs = 3\n",
-      "print runs\n",
-      "print \"\\n\\n\"\n",
-      "\n",
-      "n = 9\n",
-      "\n",
-      "for i in range(1,runs+1,1) :\n",
-      "\tprint \"\\n\\nPLease type in a positive integer: \",\n",
-      "\tprint n\n",
-      "\n",
-      "\tsum = 0\n",
-      "\tfor j in range(1,n+1,1) :\n",
-      "\t\tsum += j\n",
-      "\tprint \"\\nThe sum of the integers from 1 to %d is: %d\"%(n,sum)\n",
-      "\tn *= 3\t\t\t# Just a random increment"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "How many values of n do you wish to enter?  3\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "\n",
-        "PLease type in a positive integer:  9\n",
-        "\n",
-        "The sum of the integers from 1 to 9 is: 45\n",
-        "\n",
-        "\n",
-        "PLease type in a positive integer:  27\n",
-        "\n",
-        "The sum of the integers from 1 to 27 is: 378\n",
-        "\n",
-        "\n",
-        "PLease type in a positive integer:  81\n",
-        "\n",
-        "The sum of the integers from 1 to 81 is: 3321\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-12 , Page number: 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Newton-Rapshon method for calculating square roots\n",
-      "\n",
-      "def EPSILON() :\n",
-      "\t\"\"\" USING AS A CONSTANT \"\"\"\n",
-      "\treturn 0.0001\n",
-      "\n",
-      "print \"What number do you want to kate the square root of?\",\n",
-      "n = 17.0\n",
-      "print n\n",
-      "\n",
-      "while n < 0.0 :\n",
-      "\tprint \"Please enter a non-negative number!\"\n",
-      "\tn = 3 \t\t\t\t\t# Just a random positive number\n",
-      "\n",
-      "# The for loop statement given in the example(from the book) is not possible to code in python\n",
-      "# So, breaking it down to while loop\n",
-      "\n",
-      "guess = n / 2.0\n",
-      "while (guess * guess - n > EPSILON() or guess * guess - n < -EPSILON()) :\n",
-      "\tguess = (guess + n/guess) / 2.0\n",
-      "\t\n",
-      "print \"I calculate the square root of %f to be %.4f\"%(n,guess)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "What number do you want to kate the square root of? 17.0\n",
-        "I calculate the square root of 17.000000 to be 4.1231\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-13 , Page number: 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# General addition table for a number n\n",
-      "\n",
-      "print \"What number do you want a table for?\",\n",
-      "n = 10\n",
-      "print n\n",
-      "\n",
-      "j = n\n",
-      "for i in range (0,n+1,1) :\n",
-      "\tprint \"%3d + %3d = %3d\"%(i,j,n)\n",
-      "\tj -= 1"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "What number do you want a table for? 10\n",
-        "  0 +  10 =  10\n",
-        "  1 +   9 =  10\n",
-        "  2 +   8 =  10\n",
-        "  3 +   7 =  10\n",
-        "  4 +   6 =  10\n",
-        "  5 +   5 =  10\n",
-        "  6 +   4 =  10\n",
-        "  7 +   3 =  10\n",
-        "  8 +   2 =  10\n",
-        "  9 +   1 =  10\n",
-        " 10 +   0 =  10\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 7-14 , Page number: 135"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Generate addition table for a number n, shorter loop\n",
-      "\n",
-      "print \"What number do you want a table for?\",\n",
-      "n = 10\n",
-      "print n\n",
-      "\n",
-      "# Since the range() function supports at max 3 arguments, no possible shortening is possilbe in python\n",
-      "j = n\n",
-      "for i in range (0,n+1,1):\n",
-      "\tprint \"%3d + %3d = %3d\"%(i,j,n)\n",
-      "\tj -= 1\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "What number do you want a table for? 10\n",
-        "  0 +  10 =  10\n",
-        "  1 +   9 =  10\n",
-        "  2 +   8 =  10\n",
-        "  3 +   7 =  10\n",
-        "  4 +   6 =  10\n",
-        "  5 +   5 =  10\n",
-        "  6 +   4 =  10\n",
-        "  7 +   3 =  10\n",
-        "  8 +   2 =  10\n",
-        "  9 +   1 =  10\n",
-        " 10 +   0 =  10\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter8.ipynb b/The_Spirit_of_C/chapter8.ipynb
deleted file mode 100755
index baae6f9a..00000000
--- a/The_Spirit_of_C/chapter8.ipynb
+++ /dev/null
@@ -1,567 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:db9d52743fdd42516fb59e7618f0fc71d76d4ba3cb21fd098bf4fd004c40683b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8, A Detailed Look at the printf and scanf Functions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-1 , Page number: 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A print demo\n",
-      "# The conversion specification for numbers.\n",
-      "\n",
-      "print \"Case  1 :%d:\"%123\n",
-      "\n",
-      "print \"Case  2 :%0d:\"%123\n",
-      "\n",
-      "print \"Case  3 :%2d:\"%123\n",
-      "\n",
-      "print \"Case  4 :%8d:\"%123\n",
-      "\n",
-      "print \"Case  5 :%-8d:\"%123\n",
-      "\n",
-      "print \"Case  6 :%8d:\"%-123\n",
-      "\n",
-      "print \"Case  7 :%f:\"%123.456\n",
-      "\n",
-      "print \"Case  8 :%f:\"%123.4567896\n",
-      "\n",
-      "print \"Case  9 :%e:\"%123.456\n",
-      "\n",
-      "print \"Case  10 :%e:\"%1.23456e2\n",
-      "\n",
-      "print \"Case  11 :%e:\"%123.456e+0\n",
-      "\n",
-      "print \"Case  12 :%f:\"%1.23456E+02\n",
-      "\n",
-      "print \"Case  13 :%4.2f:\"%123.456\n",
-      "\n",
-      "print \"Case  14 :%8.2f:\"%123.456\n",
-      "\n",
-      "print \"Case  15 :%.3e:\"%123.456\n",
-      "\n",
-      "print \"Case  16 :%12.3e:\"%123.456\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case  1 :123:\n",
-        "Case  2 :123:\n",
-        "Case  3 :123:\n",
-        "Case  4 :     123:\n",
-        "Case  5 :123     :\n",
-        "Case  6 :    -123:\n",
-        "Case  7 :123.456000:\n",
-        "Case  8 :123.456790:\n",
-        "Case  9 :1.234560e+02:\n",
-        "Case  10 :1.234560e+02:\n",
-        "Case  11 :1.234560e+02:\n",
-        "Case  12 :123.456000:\n",
-        "Case  13 :123.46:\n",
-        "Case  14 :  123.46:\n",
-        "Case  15 :1.235e+02:\n",
-        "Case  16 :   1.235e+02:\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-2 , Page number: 152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# A print demo:\n",
-      "# Interpretation of values\n",
-      "\n",
-      "# python is dynamically-typed language\n",
-      "# which is why declaration of variables is omitted\n",
-      "n = -123\n",
-      "c = \"V\"\n",
-      "\n",
-      "print \"%d\"%n\n",
-      "print \"%u\"%n\t\t# python correctly interprets the negative(unsigned) integer\n",
-      "\n",
-      "print \"%c\"%c\n",
-      "# print \"%d\"%c  \t# python doesn't support implicit type-casting from integer to character or character to integer\n",
-      "print \"%c\"%86"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "-123\n",
-        "-123\n",
-        "V\n",
-        "V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-3 , Page number: 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Printing out the ASCII value of a character\n",
-      "\n",
-      "# There is no do..while loop in python\n",
-      "# Hence, we will be using while loop \n",
-      "\n",
-      "\n",
-      "answer = \"y\"\n",
-      "while answer == \"y\" :\n",
-      "\tprint \"Enter a character:\",\n",
-      "\tin_char = \"a\"\n",
-      "\tprint in_char\n",
-      "\n",
-      "\tprint \"The ASCII value of %c is %d\"%(in_char,ord(in_char)) \t# ord() function returns the ascii(int) value of a character\n",
-      "\n",
-      "\tprint \"\\nAnother? (y/n)\",\n",
-      "\tanswer = \"n\"\t\t\t# To terminate the loop\n",
-      "\tprint answer\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a character: a\n",
-        "The ASCII value of a is 97\n",
-        "\n",
-        "Another? (y/n) n\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-4 , Page number: 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Converting a digit to a number by using its ASCII value\n",
-      "\n",
-      "print \"Enter a digit:\",\n",
-      "digit = \"A\"\n",
-      "print digit\n",
-      "\n",
-      "if digit < \"0\" or digit > \"9\" :\n",
-      "\tprint \"That's not a digit\"\n",
-      "else :\n",
-      "\tprint \"The digit you typed is %c\"%digit"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a digit: A\n",
-        "That's not a digit\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-5 , Page number: 156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Converting a letter from lower to upper case or vice-versa\n",
-      "\n",
-      "answer = \"y\"\n",
-      "while answer == \"y\" :\n",
-      "\tprint \"\\n\\n\\nEnter a letter:\",\n",
-      "\tletter = \"A\"\n",
-      "\tprint letter\n",
-      "\n",
-      "\tif letter >= \"A\" and letter < \"Z\" :\t\t# Letter is uppercase\n",
-      "\t\tprint \"Lower case is: %c\"%chr(ord(letter) + (ord(\"a\")-ord(\"A\")))\n",
-      "\telif letter >= \"a\" and letter <= \"z\" :\t# Letter is lowercase\n",
-      "\t\tprint \"Upper case is: %c\"%chr(ord(letter) - (ord(\"a\")-ord(\"A\")))\n",
-      "\telse :\n",
-      "\t\tprint \"That's not a letter\"\n",
-      "\n",
-      "\t# Ask if the user wants to go again\n",
-      "\tprint \"\\nAnother? (y/n)\",\n",
-      "\tanswer = \"n\"\t\t# To terminate the loop\n",
-      "\tprint answer"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\n",
-        "\n",
-        "Enter a letter: A\n",
-        "Lower case is: a\n",
-        "\n",
-        "Another? (y/n) n\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-6 , Page number: 157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Example of using control characters\n",
-      "\n",
-      "print \"The word apple\\b\\b\\b\\b\\b_____ is underlined.\\n\"\n",
-      "print \"The last word in this sentence is way over\\t\\there.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The word apple\b\b\b\b\b_____ is underlined.\n",
-        "\n",
-        "The last word in this sentence is way over\t\there.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-7 , Page number: 158"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# How to print special characters\n",
-      "\n",
-      "print \"This is how to print a double-quote: \\\"\"\n",
-      "print \"This is how to print a backslash: \\\\\"\n",
-      "# this is how to write an apostrophe character\n",
-      "c = '\\''\n",
-      "print \"c now contains: %c\"%c\n",
-      "print \"This is how to print a percent-sign: %\"\t\t# no escape sequence required for % in python"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is how to print a double-quote: \"\n",
-        "This is how to print a backslash: \\\n",
-        "c now contains: '\n",
-        "This is how to print a percent-sign: %\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-8 , Page number: 160"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Octal and hexadecimal notation\n",
-      "\n",
-      "a = 27\n",
-      "b = 033\n",
-      "c = 0x1b\n",
-      "\n",
-      "print \"a = %d, b = %d, c = %d\"%(a,b,c)\n",
-      "print \"a = %o, b = %o, c = %o\"%(a,b,c)\n",
-      "print \"a = %x, b = %x, c = %x\"%(a,b,c)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a = 27, b = 27, c = 27\n",
-        "a = 33, b = 33, c = 33\n",
-        "a = 1b, b = 1b, c = 1b\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-9 , Page number: 162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of some ways to print a string\n",
-      "\n",
-      "print \"This is simply a control string.\"\n",
-      "\n",
-      "print \"%s\"%(\"The control string just tells how to print this.\")\n",
-      "\n",
-      "print \"%s\"%(\"Newline may be put here, instead!.\\n\"),\n",
-      "\n",
-      "print \"%s %s %s %s %s\"%(\"This\",\"is\",\"valid\",\"in\",\"C.\")\n",
-      "\n",
-      "print \"%s %s also %s %s %s\"%(\"This\",\"is\",\"valid\",\"in\",\"C.\")\n",
-      "\n",
-      "print \"%s %s %s %s %s %s %s\"%(\"This\",\"is\",\"valid\",\"in\",\"C\",\"also.\",\"\\n\"),"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This is simply a control string.\n",
-        "The control string just tells how to print this.\n",
-        "Newline may be put here, instead!.\n",
-        "This is valid in C.\n",
-        "This is also valid in C.\n",
-        "This is valid in C also. \n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-10 , Page number: 163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# %s conversion specification demonstrated\n",
-      "\n",
-      "print \"Case 1 :%s:\"%(\"A string\")\n",
-      "\n",
-      "print \"Case 2 :%3s:\"%(\"A string\")\n",
-      "\n",
-      "print \"Case 3 :%12s:\"%(\"A string\")\n",
-      "\n",
-      "print \"Case 4 :%-12s:\"%(\"A string\")\n",
-      "\n",
-      "print \"Case 5 :%12.6s:\"%(\"A string\")\n",
-      "\n",
-      "print \"Case 6 :%12.12s:\"%(\"A string\")\n",
-      "\n",
-      "print \"Case 7 :%.6s:\"%(\"A string\")\n",
-      "\n",
-      "print \"Case 8 :%-12.6s:\"%(\"A string\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case 1 :A string:\n",
-        "Case 2 :A string:\n",
-        "Case 3 :    A string:\n",
-        "Case 4 :A string    :\n",
-        "Case 5 :      A stri:\n",
-        "Case 6 :    A string:\n",
-        "Case 7 :A stri:\n",
-        "Case 8 :A stri      :\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-11 , Page number: 164 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Some miscellaneous conversion sppecifications\n",
-      "\n",
-      "a = 123456789L\n",
-      "\n",
-      "print \"%d\"%a\t\t\t# in python both int and long are unified..so %d specifier valid unlike in C\n",
-      "print \"%ld\"%a\n",
-      "print \":%c:\"%(\"V\")\n",
-      "print \":%0c:\"%(\"V\")\n",
-      "print \":%5c:\"%(\"V\")\n",
-      "print \":%-5c:\"%(\"V\")"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "123456789\n",
-        "123456789\n",
-        ":V:\n",
-        ":V:\n",
-        ":    V:\n",
-        ":V    :\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 8-13 , Page number: 171"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Reading decimal,octal and hex numbers\n",
-      "\n",
-      "print \"Enter a number in decimal:\",\n",
-      "a = 27\n",
-      "print a\n",
-      "print \"You entered %d\"%a\n",
-      "\n",
-      "print \"\\nEnter a number in octal:\",\n",
-      "a = 033\t\t\t# O is added in the front to specify octal value\n",
-      "print a\n",
-      "print \"You entered %o, or %d in decimal\"%(a,a)\n",
-      "\n",
-      "print \"\\nEnter a number in hex:\",\n",
-      "a = 0x1B\n",
-      "print a\n",
-      "print \"You entered %x, or %d in decimal\"%(a,a)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a number in decimal: 27\n",
-        "You entered 27\n",
-        "\n",
-        "Enter a number in octal: 27\n",
-        "You entered 33, or 27 in decimal\n",
-        "\n",
-        "Enter a number in hex: 27\n",
-        "You entered 1b, or 27 in decimal\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/chapter9.ipynb b/The_Spirit_of_C/chapter9.ipynb
deleted file mode 100755
index ae4771e3..00000000
--- a/The_Spirit_of_C/chapter9.ipynb
+++ /dev/null
@@ -1,590 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:13a3347e13ed5ee09ae2833e49a517873801778756103a4da905a96a5997fa89"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9, User-Defined Functions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-1 , Page number: 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Using a function as an abbreviation\n",
-      "\n",
-      "# In python, the user defined functions have to defined before making a call to them\n",
-      "\n",
-      "def chorus() :\n",
-      "\tprint \"%60s\"%\"With a knick-knack-paddy-whack,\"\n",
-      "\tprint \"%60s\"%\"Give the dog a bone,\"\n",
-      "\tprint \"%60s\\n\"%\"This old man went rolling home.\"\n",
-      "\n",
-      "print \"This old man, he played two,\"\n",
-      "print \"He played knick-knack on my thumb,\"\n",
-      "chorus()\n",
-      "\n",
-      "print \"This old man, he played one,\"\n",
-      "print \"He played knick-knack on my shoe,\"\n",
-      "chorus()\n",
-      "\n",
-      "print \"This old man, he played three,\"\n",
-      "print \"He played knick-knack on my knee,\"\n",
-      "chorus()\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "This old man, he played two,\n",
-        "He played knick-knack on my thumb,\n",
-        "                             With a knick-knack-paddy-whack,\n",
-        "                                        Give the dog a bone,\n",
-        "                             This old man went rolling home.\n",
-        "\n",
-        "This old man, he played one,\n",
-        "He played knick-knack on my shoe,\n",
-        "                             With a knick-knack-paddy-whack,\n",
-        "                                        Give the dog a bone,\n",
-        "                             This old man went rolling home.\n",
-        "\n",
-        "This old man, he played three,\n",
-        "He played knick-knack on my knee,\n",
-        "                             With a knick-knack-paddy-whack,\n",
-        "                                        Give the dog a bone,\n",
-        "                             This old man went rolling home.\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-2 , Page number: 189"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Demonstration of one use of a function\n",
-      "\n",
-      "def get_an_int() :\n",
-      "\t\"\"\"RETURNS AN INTEGER\"\"\"\t\t# This statement is just a description of the function (specific feature of python)\n",
-      "\treturn 5\n",
-      "\n",
-      "print \"Enter the length of a rectangle:\",\n",
-      "length = get_an_int()\n",
-      "print length\n",
-      "\n",
-      "print \"Enter the width:\",\n",
-      "width = get_an_int()\n",
-      "print width\n",
-      "\n",
-      "print \"\\nThe area is %d\"%(length*width)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the length of a rectangle: 5\n",
-        "Enter the width: 5\n",
-        "\n",
-        "The area is 25\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-3 , Page number: 191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# An attempt at communication between functions\n",
-      "\n",
-      "def modify_i() :\n",
-      "\ti = 3\n",
-      "\t\n",
-      "i = 1\n",
-      "print \"i is now equal to %d\"%i\n",
-      "\n",
-      "modify_i()\n",
-      "print \"i is now equal to %d\"%i\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i is now equal to 1\n",
-        "i is now equal to 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-4 , Page number: 192"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Demonstration of the use of a global variable\n",
-      "\n",
-      "# Making variable global\n",
-      "\n",
-      "def get_an_int() :\n",
-      "\treturn 5\n",
-      "\n",
-      "def calc_area() :\n",
-      "\tarea = length*width\n",
-      "\n",
-      "globals()\t\t\t# The command simply makes all the variables declared so far as global variables\n",
-      "print \"Enter the length of a rectangle:\",\n",
-      "length = get_an_int()\n",
-      "print length\n",
-      "\n",
-      "print \"Enter the width:\",\n",
-      "width = get_an_int()\n",
-      "print width\n",
-      "\n",
-      "calc_area()\n",
-      "print \"\\nThe area is %d\"%(length*width)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the length of a rectangle: 5\n",
-        "Enter the width: 5\n",
-        "\n",
-        "The area is 25\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-5 , Page number: 194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Introduction to parameters\n",
-      "\n",
-      "def print_an_int(i) :\n",
-      "\t\"\"\"The function prints an integer passed as a parameter\"\"\"\n",
-      "\tprint i\n",
-      "\n",
-      "\n",
-      "n = 3\n",
-      "\n",
-      "print \"Passing a constant:\",\n",
-      "print_an_int(5)\n",
-      "\n",
-      "print \"\\n\\nPassing a variable's value:\",\n",
-      "print_an_int(n)\n",
-      "\n",
-      "print \"\\n\\nPassing the value of an expression:\",\n",
-      "print_an_int(n * 4 + 1)\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Passing a constant: 5\n",
-        "\n",
-        "\n",
-        "Passing a variable's value: 3\n",
-        "\n",
-        "\n",
-        "Passing the value of an expression: 13\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-6 , Page number: 196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Demonstration of how a procedure cannot modify the contents of a variable passed as a parameter\n",
-      "\n",
-      "def modify(i) :\n",
-      "\ti = 3\n",
-      "\n",
-      "i = 1\n",
-      "\n",
-      "print \"i is now equal to %d\"%i\n",
-      "\n",
-      "modify(i)\n",
-      "\n",
-      "print \"i is now equal to %d\"%i"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i is now equal to 1\n",
-        "i is now equal to 1\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-7 , Page number: 197 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Hailstones, demonstrating the use of a parameter\n",
-      "\n",
-      "def step(i) :\n",
-      "\t\"\"\"This function returns its parameter i divided by 2 if i is even else it returns i multiplied by 3 and incremented by 1\"\"\"\n",
-      "\tif (i % 2) :\n",
-      "\t\treturn (i * 3 + 1)\n",
-      "\telse : \n",
-      "\t\treturn (i / 2)\n",
-      "\n",
-      "counter = 0\n",
-      "\n",
-      "print \"Enter a positive integer\",\n",
-      "n = 25\n",
-      "print n\n",
-      "\n",
-      "nsave = n\n",
-      "\n",
-      "while n > 1 :\n",
-      "\tn = step(n)\n",
-      "\tcounter += 1\n",
-      "\tprint \"Step %4d: n = %4d\"%(counter,n)\n",
-      "\n",
-      "print \"\\n\\n%d went to 1 in %d steps.\"%(nsave,counter)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter a positive integer 25\n",
-        "Step    1: n =   76\n",
-        "Step    2: n =   38\n",
-        "Step    3: n =   19\n",
-        "Step    4: n =   58\n",
-        "Step    5: n =   29\n",
-        "Step    6: n =   88\n",
-        "Step    7: n =   44\n",
-        "Step    8: n =   22\n",
-        "Step    9: n =   11\n",
-        "Step   10: n =   34\n",
-        "Step   11: n =   17\n",
-        "Step   12: n =   52\n",
-        "Step   13: n =   26\n",
-        "Step   14: n =   13\n",
-        "Step   15: n =   40\n",
-        "Step   16: n =   20\n",
-        "Step   17: n =   10\n",
-        "Step   18: n =    5\n",
-        "Step   19: n =   16\n",
-        "Step   20: n =    8\n",
-        "Step   21: n =    4\n",
-        "Step   22: n =    2\n",
-        "Step   23: n =    1\n",
-        "\n",
-        "\n",
-        "25 went to 1 in 23 steps.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-8 , Page number: 200"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Square root program demonstrating a non-integer function and parameter\n",
-      "\n",
-      "def EPSILON() :\n",
-      "\t\"\"\"replacement of #define constant\"\"\"\n",
-      "\treturn 0.0001\n",
-      "\n",
-      "def sqrt(n) :\n",
-      "\t\"\"\"returns the square root of the number supplied as parameter using Newton-Raphson method\"\"\"\n",
-      "\tif n > 0.0 :\n",
-      "\t\t# The for loop statement given in the example(from the book) is not possible to code in python\n",
-      "\t\t# So, breaking it down to while loop\n",
-      "\t\tguess = n / 2.0\n",
-      "\t\twhile (guess * guess - n > EPSILON() or guess * guess - n < -EPSILON()) :\n",
-      "\t\t\tguess = (guess + n/guess) / 2.0\n",
-      "\t\treturn guess\n",
-      "\telse :\n",
-      "\t\treturn (-1.0)\n",
-      "\n",
-      "print \"Enter the number whose square root you want:\",\n",
-      "x = 17.1\n",
-      "print x\n",
-      "\n",
-      "print \"The square root of %f is %f\"%(x,sqrt(x))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the number whose square root you want: 17.1\n",
-        "The square root of 17.100000 is 4.135215\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-9 , Page number: 201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Testing for leap years using a boolean function\n",
-      "\n",
-      "def is_leap(year) :\n",
-      "\t\"\"\"A function that returns true if the given year is a leap year else returns false\"\"\"\n",
-      "\treturn (year % 4 == 0 and year % 100 != 0 or year % 400 == 0)\n",
-      "\n",
-      "print \"Please enter a year:\",\n",
-      "year = 2000\n",
-      "print year\n",
-      "\n",
-      "if (is_leap(year)) :\n",
-      "\tprint \"%d is a leap year.\"%year\n",
-      "else :\n",
-      "\tprint \"%d is not a leap year.\"%year"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Please enter a year: 2000\n",
-        "2000 is a leap year.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-10 , Page number: 203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Demonstration of the use of multiple parameters\n",
-      "\n",
-      "def get_an_int() :\n",
-      "\t\"\"\"returns a random integer\"\"\"\n",
-      "\treturn 7\n",
-      "\n",
-      "def area(length,width) :\n",
-      "\t\"\"\"returns the area of the rectangle using its parameters length and width\"\"\"\n",
-      "\treturn length * width\n",
-      "\n",
-      "print \"Enter the length of a rectangle:\",\n",
-      "length = get_an_int()\n",
-      "print length\n",
-      "\n",
-      "print \"Enter the width:\",\n",
-      "width = get_an_int()\n",
-      "print width\n",
-      "\n",
-      "print \"\\nThe area is %d.\"%area(length,width)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Enter the length of a rectangle: 7\n",
-        "Enter the width: 7\n",
-        "\n",
-        "The area is 49.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Program 9-11 , Page number: 204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Illustration of a 2-parameter function: raising a number to a power \n",
-      "\n",
-      "def power(i,x) :\n",
-      "\t\"\"\"returns the values of i to the power of x\"\"\"\n",
-      "\taccum = 1\n",
-      "\tif x < 0 :\n",
-      "\t\treturn 0\n",
-      "\tfor index in range (0,x,1) :\n",
-      "\t\taccum *= i\n",
-      "\treturn accum\n",
-      "\n",
-      "answer = \"y\"\n",
-      "\n",
-      "while answer == \"y\" or answer == \"Y\" :\n",
-      "\tprint \"\\nPlease enter an integer:\",\n",
-      "\tnumber = 4\n",
-      "\tprint number\n",
-      "\n",
-      "\tprint \"Please enter a non-negative integral power\"\n",
-      "\tprint \"     to which the first number is to be raised\",\n",
-      "\texponent = 2\n",
-      "\tprint exponent\n",
-      "\n",
-      "\twhile exponent < 0 :\n",
-      "\t\tprint \"***\\7Please make that a non-negative integer\"\n",
-      "\t\texponent = 2 \t\t\t\t# randomly making the exponent non-negative\n",
-      "\n",
-      "\tprint \"%d raised to the power %d is %ld.\"%(number,exponent,power(number,exponent))\n",
-      "\n",
-      "\t# see if the user wants to keep going\n",
-      "\n",
-      "\tprint \"\\nAnother calculation? (y/n)\",\n",
-      "\tanswer = \"n\"\n",
-      "\tprint answer"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Please enter an integer: 4\n",
-        "Please enter a non-negative integral power\n",
-        "     to which the first number is to be raised 2\n",
-        "4 raised to the power 2 is 16.\n",
-        "\n",
-        "Another calculation? (y/n) n\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/The_Spirit_of_C/screenshots/Screenshot_1.png b/The_Spirit_of_C/screenshots/Screenshot_1.png
deleted file mode 100755
index dc7d579e..00000000
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diff --git a/The_Spirit_of_C/screenshots/Screenshot_2.png b/The_Spirit_of_C/screenshots/Screenshot_2.png
deleted file mode 100755
index 6762ca01..00000000
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diff --git a/The_Spirit_of_C/screenshots/Screenshot_3.png b/The_Spirit_of_C/screenshots/Screenshot_3.png
deleted file mode 100755
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diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter1.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter1.ipynb
new file mode 100755
index 00000000..80b53e61
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter1.ipynb
@@ -0,0 +1,198 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1c2e6d0b5736bd35dd626e4e22d0dbe67f82011189ddb91d07585ae3f73aa6b8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1, Elementary Programming"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 1-1 , Page number: 2"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# An interactive program to calculate the hypotenuse of a right triangle\n",
+      "\n",
+      "import math ## library for using sqrt function\n",
+      "\n",
+      "def square(a):\n",
+      "\t\"\"\"Returns the square of the given number \"\"\"\n",
+      "\treturn a*a\n",
+      "\n",
+      "print \"Enter the lengths of the two legs of a right triangle\"\n",
+      "print \"The program will display the length of the hypotenuse.\"\n",
+      "print \"Enter zero or negative values to end program\"\n",
+      "\n",
+      "# Priming reads follow\n",
+      "print \"First leg (0.0 to quit)? \",\n",
+      "a_leg = 3\n",
+      "print a_leg\n",
+      "print \"Second leg (0.0 to quit)? \",\n",
+      "b_leg = 4\n",
+      "print b_leg\n",
+      "\n",
+      "while a_leg>0.0 and b_leg>0.0:\n",
+      "\tprint \"The legs of the right triangle are %.2f and %.2f,\\n\"%(a_leg,b_leg),\n",
+      "\t# Hypotenuse is square root of a*a + b*b\n",
+      "\thypotenuse = math.sqrt(square(a_leg) + square(b_leg))\n",
+      "\tprint \"   so the hypotenuse is %.2f\\n\\n\"%hypotenuse,\n",
+      "\tprint \"First leg (0.0 to quit)? \",\n",
+      "\ta_leg = 0.0\t\t\t# To terminate the loop\n",
+      "\tprint a_leg\n",
+      "\tprint \"Second leg (0.0 to quit)? \",\n",
+      "\tb_leg = 0.0\n",
+      "\tprint b_leg\n",
+      "\n",
+      "print \"Thank you.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the lengths of the two legs of a right triangle\n",
+        "The program will display the length of the hypotenuse.\n",
+        "Enter zero or negative values to end program\n",
+        "First leg (0.0 to quit)?  3\n",
+        "Second leg (0.0 to quit)?  4\n",
+        "The legs of the right triangle are 3.00 and 4.00,\n",
+        "   so the hypotenuse is 5.00\n",
+        "\n",
+        "First leg (0.0 to quit)?  0.0\n",
+        "Second leg (0.0 to quit)?  0.0\n",
+        "Thank you.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 1-2 , Page number: 3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "def main():  \n",
+      "\t\"\"\"simply returns a string\"\"\"\n",
+      "\t# There's no main() function in python.This main() function is no different that any other user-defined functions\n",
+      "\treturn \"Welcome to C programming!\\n\"\n",
+      "\t\n",
+      "print main()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Welcome to C programming!\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 1-3 , Page number: 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Use of braces \n",
+      "\n",
+      "def main():\n",
+      "\t# in python, proper indentation works as braces\n",
+      "\t# the following lines come under the main() function block due to indentation\n",
+      "\tprint \"Give me land, lots of land\",\n",
+      "\tprint \"And the starry skies above...\\n\",\n",
+      "\n",
+      "main()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Give me land, lots of land And the starry skies above...\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 1-4 , Page number: 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A short poem\n",
+      "\n",
+      "def main():\n",
+      "\tprint \"Mary had a little lamb,\",\n",
+      "\tprint \"its fleece was white as snow.\\n\",\n",
+      "\tprint \"And everywhere that Mary went,\",\n",
+      "\tprint \"the lamb was sure to go.\\n\"\n",
+      "\n",
+      "main()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mary had a little lamb, its fleece was white as snow.\n",
+        "And everywhere that Mary went, the lamb was sure to go.\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter10.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter10.ipynb
new file mode 100755
index 00000000..ed7b99e5
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter10.ipynb
@@ -0,0 +1,1020 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:970f28e2206ea4472dd82a3808592db202377fcd9a01fc69b666e61a221db5b1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10, Arrays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-1, Page Number : 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Introduction to arrays\n",
+      "\n",
+      "array = []\t\t# defining array which is rather called list in python\n",
+      "\n",
+      "print \"You are to enter 5 integers.\\n\"\n",
+      "\n",
+      "# Input the 5 integers into array \"array\"\n",
+      "\n",
+      "for i in range(5) :  # equivalent to for i in range(0,5,1)\n",
+      "\tprint \"Enter integer #%d: \"%(i+1),\n",
+      "\tval = int(raw_input())\n",
+      "\tarray.insert(i,val)\t\t\t\t# insert function is used to add elements to list : \n",
+      "\t\t\t\t\t\t\t\t\t#      insert(i,val) where i is the index and val is the value to be inserted\n",
+      "\n",
+      "print \"\\nThank you.\\n\"\n",
+      "\n",
+      "# Print out the integers stored in array \"array\"\n",
+      "j = 1\t# used as a counter\n",
+      "for i in array:\t\t\t\t# shorthand way of scanning through the list; using \"in\"\n",
+      "\tprint \"Integer #%d = %d.\"%(j,i)\n",
+      "\tj += 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You are to enter 5 integers.\n",
+        "\n",
+        "Enter integer #1:"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Enter integer #2:"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "12\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Enter integer #3:"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Enter integer #4:"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "63\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Enter integer #5:"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "Thank you.\n",
+        "\n",
+        "Integer #1 = 3.\n",
+        "Integer #2 = 12.\n",
+        "Integer #3 = 8.\n",
+        "Integer #4 = 63.\n",
+        "Integer #5 = 0.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-2, Page Number : 228"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Initializing a global array\n",
+      "\n",
+      "array = [ 4, 6, 5, 7, 2 ]\n",
+      "\n",
+      "def main() :\t\t\t# just a user-defined function ; not equivalent to the built-in main() function in C\n",
+      "\t\"\"\" prints the integers stored in array \"array\" \"\"\"\n",
+      "\tj = 1\t\t# counter \n",
+      "\tfor i in array :\n",
+      "\t\tprint \"Integer #%d = %d.\"%(j,i)\n",
+      "\t\tj += 1\n",
+      "\n",
+      "main()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Integer #1 = 4.\n",
+        "Integer #2 = 6.\n",
+        "Integer #3 = 5.\n",
+        "Integer #4 = 7.\n",
+        "Integer #5 = 2.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-3, Page Number: 230"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The number of array elements actually used may vary\n",
+      "\n",
+      "# Read in the size of the array\n",
+      "print \"How many integers will you enter?\",\n",
+      "array_size = 6\n",
+      "print array_size\n",
+      "\n",
+      "# Read in the integers\n",
+      "print \"\\nPlease enter %d integers:\"%array_size\n",
+      "array = [ 12, 3, -5, 21, 0, 899 ]\t# Lists dynamically allocate memory \n",
+      "for i in range(array_size) :\n",
+      "\tprint array[i],\n",
+      "# Print out the integers\n",
+      "\n",
+      "print(\"\\n\")\n",
+      "for i in range(array_size) :\n",
+      "\tprint \"Integer #%d = %d.\"%(i+1,array[i])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many integers will you enter? 6\n",
+        "\n",
+        "Please enter 6 integers:\n",
+        "12 3 -5 21 0 899 \n",
+        "\n",
+        "Integer #1 = 12.\n",
+        "Integer #2 = 3.\n",
+        "Integer #3 = -5.\n",
+        "Integer #4 = 21.\n",
+        "Integer #5 = 0.\n",
+        "Integer #6 = 899.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-4, Page Number: 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The trailer method\n",
+      "\n",
+      "# Macro definition\n",
+      "MAX_ARRAY_SIZE = 20\n",
+      "TRAILER = -9999\n",
+      "\n",
+      "# Read in the integers\n",
+      "\n",
+      "print \"Please enter up to %d integers, using %d as a trailer:\"%(MAX_ARRAY_SIZE,TRAILER)\n",
+      "array = [ 3, 75, -34, 6, -1, 2, 41, 0, 90, -9999]\n",
+      "for array_size in range(MAX_ARRAY_SIZE) :\n",
+      "\tprint array[array_size],\n",
+      "\tif array[array_size] == -9999 :\n",
+      "\t\tbreak\t\n",
+      "\t\n",
+      "# Print out the integers\n",
+      "print \"\\n\\nYou have entered %d integers\"%array_size\n",
+      "for i in range(array_size) :\n",
+      "\tprint \"Integer #%d = %d.\"%(i+1,array[i])"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter up to 20 integers, using -9999 as a trailer:\n",
+        "3 75 -34 6 -1 2 41 0 90 -9999 \n",
+        "\n",
+        "You have entered 9 integers\n",
+        "Integer #1 = 3.\n",
+        "Integer #2 = 75.\n",
+        "Integer #3 = -34.\n",
+        "Integer #4 = 6.\n",
+        "Integer #5 = -1.\n",
+        "Integer #6 = 2.\n",
+        "Integer #7 = 41.\n",
+        "Integer #8 = 0.\n",
+        "Integer #9 = 90.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-5, Page Number: 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Caculating the average grade for a class\n",
+      "\n",
+      "# Macro definition\n",
+      "\n",
+      "MAX_GRADES = 20 \t# The class may have up to 20 students\n",
+      "TRAILER = -9999\t\t# -9999 marks the end of the grade list\n",
+      "\n",
+      "# Read in the grades\n",
+      "\n",
+      "print \"Please enter up to %d grades, using %d as a trailer:\"%(MAX_GRADES,TRAILER)\n",
+      "grades = [ 41, 92, 15, 65, 0, 78, 81, 96, -9999 ]\n",
+      "for num_grades in range(MAX_GRADES) :\n",
+      "\tprint grades[num_grades]\n",
+      "\tif grades[num_grades] == -9999 :\n",
+      "\t\tbreak\n",
+      "\n",
+      "print \"\\nYou have entered %d grades.\"%num_grades\n",
+      "\n",
+      "# Find the average\n",
+      "sum = 0.0 \n",
+      "for i in range(num_grades) :\n",
+      "\tsum += grades[i]\n",
+      "\n",
+      "print \"\\nThe average grade is %f.\"%(sum/num_grades)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter up to 20 grades, using -9999 as a trailer:\n",
+        "41\n",
+        "92\n",
+        "15\n",
+        "65\n",
+        "0\n",
+        "78\n",
+        "81\n",
+        "96\n",
+        "-9999\n",
+        "\n",
+        "You have entered 8 grades.\n",
+        "\n",
+        "The average grade is 58.500000.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-6, Page Number: 235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Determining which grades in a class are above the average\n",
+      "\n",
+      "# Macro definition\n",
+      "\n",
+      "MAX_GRADES = 20 \t# The class may have up to 20 students\n",
+      "TRAILER = -9999\t\t# -9999 marks the end of the grade list\n",
+      "\n",
+      "# Read in the grades\n",
+      "\n",
+      "print \"Please enter up to %d grades, using %d as a trailer:\"%(MAX_GRADES,TRAILER)\n",
+      "grades = [ 41, 92, 15, 65, 0, 78, 81, 96, -9999 ]\n",
+      "for num_grades in range(MAX_GRADES) :\n",
+      "\tprint grades[num_grades]\n",
+      "\tif grades[num_grades] == -9999 :\n",
+      "\t\tbreak\n",
+      "\n",
+      "print \"\\nYou have entered %d grades.\"%num_grades\n",
+      "\n",
+      "# Find the average\n",
+      "sum = 0.0 \n",
+      "for i in range(num_grades) :\n",
+      "\tsum += grades[i]\n",
+      "\n",
+      "avg_grade = sum/num_grades\n",
+      "print \"\\nThe average grade is %f.\"%avg_grade\n",
+      "\n",
+      "# Find who got above average\n",
+      "\n",
+      "print \"\\nThe following grades were above average:\"\n",
+      "for i in range(num_grades) :\n",
+      "\tif grades[i] > avg_grade :\n",
+      "\t\tprint \"%f (student #%d)\"%(grades[i],i+1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter up to 20 grades, using -9999 as a trailer:\n",
+        "41\n",
+        "92\n",
+        "15\n",
+        "65\n",
+        "0\n",
+        "78\n",
+        "81\n",
+        "96\n",
+        "-9999\n",
+        "\n",
+        "You have entered 8 grades.\n",
+        "\n",
+        "The average grade is 58.500000.\n",
+        "\n",
+        "The following grades were above average:\n",
+        "92.000000 (student #2)\n",
+        "65.000000 (student #4)\n",
+        "78.000000 (student #6)\n",
+        "81.000000 (student #7)\n",
+        "96.000000 (student #8)\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-7, Page Number 237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Reversing an array\n",
+      "\n",
+      "# Macro definition\n",
+      "ARRAY_SIZE = 10 \t# The array has 10 elements \n",
+      "\n",
+      "# Read in the array\n",
+      "\n",
+      "print \"Please enter %d integers:\"%ARRAY_SIZE\n",
+      "array = [ 10, 42, 7, 0, 923, 13, -5, 6, 19, 3]\n",
+      "for i in range(10) :\n",
+      "\tprint array[i],\n",
+      "\n",
+      "# Echo the array\n",
+      "\n",
+      "print \"\\n\\nThe original array was:\"\n",
+      "for i in range(10) :\n",
+      "\tprint array[i],\n",
+      "\n",
+      "# Print the array in reverse\n",
+      "\n",
+      "print \"\\n\\nThe reversed array is:\"\n",
+      "for i in range(ARRAY_SIZE-1,-1,-1) :\n",
+      "\tprint array[i],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter 10 integers:\n",
+        "10 42 7 0 923 13 -5 6 19 3 \n",
+        "\n",
+        "The original array was:\n",
+        "10 42 7 0 923 13 -5 6 19 3 \n",
+        "\n",
+        "The reversed array is:\n",
+        "3 19 6 -5 13 923 0 7 42 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-8, Page Number: 240"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "# Reversing a character string\n",
+      "\n",
+      "# Macro definition\n",
+      "MAX_STRING_SIZE = 10\n",
+      "\n",
+      "# Read in the word\n",
+      "\n",
+      "print \"Please enter a word of up to %d letters:\"%MAX_STRING_SIZE\n",
+      "word = \"bingo\"\n",
+      "print word\n",
+      "\n",
+      "# Echo the word \n",
+      "\n",
+      "print \"\\nThe word is %s.\"%word\n",
+      "\n",
+      "# Find thw word length\n",
+      "string_size = 0\n",
+      "for i in word :\t\t\t# To iterate over the entire string\n",
+      "\tstring_size += 1\n",
+      "\n",
+      "# Print the word in reverse\n",
+      "\n",
+      "print \"\\nThe reversed word is \",\n",
+      "for i in range(string_size-1,-1,-1) :\n",
+      "\tsys.stdout.write(\"%c\"%word[i])  # For printing in the same line without implicit whitespaces\n",
+      "\n",
+      "print "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter a word of up to 10 letters:\n",
+        "bingo\n",
+        "\n",
+        "The word is bingo.\n",
+        "\n",
+        "The reversed word is ognib\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-9, Page Number: 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculating the minimum and maximum grade in a class\n",
+      "\n",
+      "# Macro definition\n",
+      "\n",
+      "MAX_GRADES = 20 \t# The class may have up to 20 students\n",
+      "TRAILER = -9999\t\t# -9999 marks the end of the grade list\n",
+      "\n",
+      "# Read in the grades\n",
+      "\n",
+      "print \"Please enter up to %d grades, using %d as a trailer:\"%(MAX_GRADES,TRAILER)\n",
+      "grades = [ 41, 92, 15, 65, 0, 78, 81, 96, -9999 ]\n",
+      "for num_grades in range(MAX_GRADES) :\n",
+      "\tprint grades[num_grades]\n",
+      "\tif grades[num_grades] == -9999 :\n",
+      "\t\tbreak\n",
+      "\n",
+      "print \"\\nYou have entered %d grades.\"%num_grades\n",
+      "\n",
+      "# Find the minimum grade\n",
+      "\n",
+      "min = grades[0]\n",
+      "for i in range(num_grades) :\n",
+      "\tif grades[i] < min :\n",
+      "\t\tmin = grades[i]\n",
+      "\n",
+      "print \"\\nThe lowest grade is %f\"%min\n",
+      "\n",
+      "# Find the maximum grade\n",
+      "\n",
+      "max = grades[0]\n",
+      "for i in range(num_grades) :\n",
+      "\tif grades[i] > min :\n",
+      "\t\tmax = grades[i]\n",
+      "\n",
+      "print \"\\nThe lowest grade is %f\"%max\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter up to 20 grades, using -9999 as a trailer:\n",
+        "41\n",
+        "92\n",
+        "15\n",
+        "65\n",
+        "0\n",
+        "78\n",
+        "81\n",
+        "96\n",
+        "-9999\n",
+        "\n",
+        "You have entered 8 grades.\n",
+        "\n",
+        "The lowest grade is 0.000000\n",
+        "\n",
+        "The lowest grade is 96.000000\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-10, Page Number: 243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Sorting an array\n",
+      "\n",
+      "# Macro definition\n",
+      "MAX_ARRAY_SIZE = 20 \t# The array may have up to 20 elements\n",
+      "\n",
+      "array_size = 10\n",
+      "print \"How many numbers?\",\n",
+      "print array_size\n",
+      "\n",
+      "while array_size < 1 or array_size > MAX_ARRAY_SIZE\t:\n",
+      "\tprint \"How many numbers?\",\n",
+      "\tprint array_size\n",
+      "\n",
+      "# Read in the array\n",
+      "\n",
+      "print \"Please enter the integers:\"\n",
+      "array = [ 10, 42, 7, 0, 923, 13, -5, 6, 19, 3]\n",
+      "for i in range(array_size) :\n",
+      "\tprint array[i],\n",
+      "\n",
+      "# Echo the array\n",
+      "\n",
+      "print \"\\n\\nThe original array is:\"\n",
+      "for i in range(array_size) :\n",
+      "\tprint array[i],\n",
+      "\n",
+      "# Sort the array\n",
+      "\n",
+      "for i in range(array_size-1) :\n",
+      "\tfor j in range(i+1, array_size) :\n",
+      "\t\tif array[i] > array[j] :\n",
+      "\t\t\ttemp = array[i]\n",
+      "\t\t\tarray[i] = array[j]\n",
+      "\t\t\tarray[j] = temp\n",
+      "\n",
+      "# Printing the sorted array\n",
+      "\n",
+      "print \"\\n\\nThe sorted array is\"\n",
+      "for i in range(array_size) :\n",
+      "\tprint array[i],"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many numbers? 10\n",
+        "Please enter the integers:\n",
+        "10 42 7 0 923 13 -5 6 19 3 \n",
+        "\n",
+        "The original array is:\n",
+        "10 42 7 0 923 13 -5 6 19 3 \n",
+        "\n",
+        "The sorted array is\n",
+        "-5 0 3 6 7 10 13 19 42 923\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-11, Page Number: 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Finding the median of an array\n",
+      "\n",
+      "# Macro definition\n",
+      "MAX_ARRAY_SIZE = 20 \t# The array may have up to 20 elements\n",
+      "\n",
+      "array_size = 0\n",
+      "array = []\n",
+      "\n",
+      "def get_size()\t:\n",
+      "\t\"\"\" reads the size of the array (until a valid value is entered) \"\"\"\n",
+      "\tglobal array_size\n",
+      "\tarray_size = 10\n",
+      "\tprint \"How many numbers?\",\n",
+      "\tprint array_size\n",
+      "\twhile array_size < 1 or array_size > MAX_ARRAY_SIZE\t:\n",
+      "\t\tprint \"How many numbers?\",\n",
+      "\t\tprint array_size\n",
+      "\n",
+      "\n",
+      "def read_array() :\n",
+      "\t\"\"\" read in the array \"\"\"\n",
+      "\tglobal array\n",
+      "\tprint \"Please enter the integers:\"\n",
+      "\tarray = [ 10, 42, 7, 0, 923, 13, -5, 6, 19, 3 ]\n",
+      "\tfor i in range(array_size) :\n",
+      "\t\tprint array[i],\n",
+      "\n",
+      "def write_array() :\n",
+      "\t\"\"\" write out the array \"\"\"\n",
+      "\tfor i in range(array_size) :\n",
+      "\t\tprint array[i],\n",
+      "\n",
+      "def sort_array() :\n",
+      "\t\"\"\" sort the array \"\"\"\n",
+      "\tglobal array\n",
+      "\tfor i in range(array_size-1) :\n",
+      "\t\tfor j in range(i+1, array_size) :\n",
+      "\t\t\tif array[i] > array[j] :\n",
+      "\t\t\t\ttemp = array[i]\n",
+      "\t\t\t\tarray[i] = array[j]\n",
+      "\t\t\t\tarray[j] = temp\n",
+      "\n",
+      "def even(n) :\n",
+      "\t\"\"\" returns non-zero(true) if n is even else returns zero(false) \"\"\"\n",
+      "\treturn ((n/2)*2 == n)\n",
+      "\n",
+      "get_size()\t\t# Read in the size of the array\n",
+      "\n",
+      "read_array()\t# Read in the array\n",
+      "\n",
+      "print \"\\n\\nThe original array is:\"\n",
+      "write_array()\t# Echo the array\n",
+      "\n",
+      "sort_array()\t# Sort the array\n",
+      "\n",
+      "print \"\\n\\nThe sorted array is:\"\n",
+      "write_array()\t# Print the sorted array\n",
+      "\n",
+      "print \"\\n\\nThe median is:\"\n",
+      "# The equivalent of tertiary conditional operator in python is a if test else b\n",
+      "print \"%.1f\"%( ((array[ array_size / 2 -1] + array[ array_size / 2 ]) / 2.0 ) if even(array_size) else float(array[ array_size / 2 ])  ) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many numbers? 10\n",
+        "Please enter the integers:\n",
+        "10 42 7 0 923 13 -5 6 19 3 \n",
+        "\n",
+        "The original array is:\n",
+        "10 42 7 0 923 13 -5 6 19 3 \n",
+        "\n",
+        "The sorted array is:\n",
+        "-5 0 3 6 7 10 13 19 42 923 \n",
+        "\n",
+        "The median is:\n",
+        "8.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-12, Page Number: 251"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Tally how many of each number within a range are entered\n",
+      "\n",
+      "# Macro definition\n",
+      "RANGE_LOW = 1\n",
+      "RANGE_HIGH = 10\t\t\t# Tallies numbers between RANGE_LOW and RANGE_HIGH\n",
+      "TRAILER = -1 \t\t\t# TRAILER marks end of data\n",
+      "\n",
+      "inputfeed = [ 3, 5, 1, 10, 8, 5, 10, 6, 2, 11, 4, 9, -2, 3, 2, 9, -1 ]\t\t# additional array for the sake of providing input\n",
+      "\n",
+      "tally = [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]\t\t# empty array\n",
+      "# Input numbers\n",
+      "print \"Please enter a number between %d and %d (%d to quit): \"%(RANGE_LOW,RANGE_HIGH,TRAILER),\n",
+      "num = inputfeed[0]\n",
+      "print num\n",
+      "\n",
+      "i = 1\n",
+      "while num != TRAILER :\n",
+      "\tif num < RANGE_LOW or num > RANGE_HIGH : \t\t# Out of range\n",
+      "\t\tprint \"%d is not between %d and %d.\"%(num,RANGE_LOW,RANGE_HIGH)\n",
+      "\telse :\n",
+      "\t\ttally[ num - RANGE_LOW] += 1\t\n",
+      "\tprint \"\\nEnter another number: \",\n",
+      "\tnum = inputfeed[i]\n",
+      "\tprint num\n",
+      "\ti += 1 \t\t\t# to iterate over the input array\n",
+      "\n",
+      "# Print totals\n",
+      "\n",
+      "print \"\\n------------\\n\\nYou entered:\"\n",
+      "for i in range(RANGE_HIGH - RANGE_LOW + 1) :\n",
+      "\tprint \"%2d %2d's\"%(tally[i],i + RANGE_LOW)\n",
+      "\n",
+      "print \"\\nThank you.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter a number between 1 and 10 (-1 to quit):  3\n",
+        "\n",
+        "Enter another number:  5\n",
+        "\n",
+        "Enter another number:  1\n",
+        "\n",
+        "Enter another number:  10\n",
+        "\n",
+        "Enter another number:  8\n",
+        "\n",
+        "Enter another number:  5\n",
+        "\n",
+        "Enter another number:  10\n",
+        "\n",
+        "Enter another number:  6\n",
+        "\n",
+        "Enter another number:  2\n",
+        "\n",
+        "Enter another number:  11\n",
+        "11 is not between 1 and 10.\n",
+        "\n",
+        "Enter another number:  4\n",
+        "\n",
+        "Enter another number:  9\n",
+        "\n",
+        "Enter another number:  -2\n",
+        "-2 is not between 1 and 10.\n",
+        "\n",
+        "Enter another number:  3\n",
+        "\n",
+        "Enter another number:  2\n",
+        "\n",
+        "Enter another number:  9\n",
+        "\n",
+        "Enter another number:  -1\n",
+        "\n",
+        "------------\n",
+        "\n",
+        "You entered:\n",
+        " 1  1's\n",
+        " 2  2's\n",
+        " 2  3's\n",
+        " 1  4's\n",
+        " 2  5's\n",
+        " 1  6's\n",
+        " 0  7's\n",
+        " 1  8's\n",
+        " 2  9's\n",
+        " 2 10's\n",
+        "\n",
+        "Thank you.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 10-13, Page Number: 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Program to report car prices. User enters car make by number and car year. \n",
+      "# Program reports price. All data is fictitious\n",
+      "\n",
+      "# 2-D array (4 X 12 ) of prices : row = make , column = year\n",
+      "price_list = [\t[ 4775, 4980, 5222, 5305, 5483, 5547, 5596, 5713, 5842, 5903, 6043, 6230 ], # Volkswagen\n",
+      "\t\t\t\t[ 4853, 5140, 5413, 5590, 5723, 5848, 5762, 5944, 6104, 6255, 6370, 6526 ],\t# Chevy\n",
+      "\t\t\t\t[ 5627, 5772, 5973, 6210, 6539, 6720, 6792, 6930, 7054, 7202, 7355, 7562 ], # Cadillac\n",
+      "\t\t\t\t[ 1576, 1738, 1970, 2151, 2205, 2280, 2442, 2580, 2814, 1953, 3078, 3201 ],\t# Honda\n",
+      "\t\t\t]\t\n",
+      "\n",
+      "def get_make() : \n",
+      "\t\"\"\" read in and return the number (make) of the desired car \"\"\"\n",
+      "\t\n",
+      "\t# Print the choice numbers and names\n",
+      "\tfor i in range(1,5) :\n",
+      "\t\tprint \"%d=\"%i,\n",
+      "\t\tprint_make(i)\n",
+      "\t\tprint \",\",\n",
+      "\tprint \"0=END\"\n",
+      "\n",
+      "\t# Read in the selection by number\n",
+      "\tprint \"     Which car do you want? \",\n",
+      "\tmake = 2\n",
+      "\tprint make\n",
+      "\twhile make < 0 or make > 4 :\n",
+      "\t\tprint \"     Which car do you want? \",\n",
+      "\t\tmake = 2\n",
+      "\t\tprint make\n",
+      "\n",
+      "\treturn make\n",
+      "\n",
+      "def get_year() :\t\n",
+      "\t\"\"\" read in and return the year of the desired car \"\"\"\n",
+      "\n",
+      "\tprint \"     What year (1975-1986)?\",\n",
+      "\tyear = 1980\n",
+      "\tprint year\n",
+      "\twhile year < 1975 and year > 1986 :\n",
+      "\t\tprint \"     What year (1975-1986)?\",\n",
+      "\t\tyear = 1980\n",
+      "\t\tprint year\n",
+      "\n",
+      "\treturn year\n",
+      "\n",
+      "def price (car,yr) :\n",
+      "\t\"\"\" Use \"car\" and \"year\" to generate an index into \"price_list\" and return the corresponding entry \"\"\"\n",
+      "\treturn price_list[ car - 1 ][ yr - 1975]\n",
+      "\n",
+      "def print_make(car) :\n",
+      "\t\"\"\" print the make of \"car\" according to its value \"\"\"\n",
+      "\tif car == 1 : \n",
+      "\t\tprint \"Volkswagen\",\n",
+      "\telif car == 2 :\n",
+      "\t\tprint \"Chevy\",\n",
+      "\telif car == 3 :\n",
+      "\t\tprint \"Cadillac\",\n",
+      "\telse :\n",
+      "\t\tprint \"Honda\",\n",
+      "\n",
+      "make = get_make()\n",
+      "\n",
+      "while make != 0 :\t\t# O indicates end of input\n",
+      "\tyear = get_year()\n",
+      "\tprint \"The price of a %d\"%year,\n",
+      "\tprint_make(make)\n",
+      "\tprint \" is $%d.\\n\"%price(make,year)\n",
+      "\tmake = 0   # To terminate the loop\n",
+      "\n",
+      "print \"\\nThank you for buying all those cars!\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1= Volkswagen , 2= Chevy , 3= Cadillac , 4= Honda , 0=END\n",
+        "     Which car do you want?  2\n",
+        "     What year (1975-1986)? 1980\n",
+        "The price of a 1980 Chevy  is $5848.\n",
+        "\n",
+        "\n",
+        "Thank you for buying all those cars!\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter11.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter11.ipynb
new file mode 100755
index 00000000..241e1c1e
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter11.ipynb
@@ -0,0 +1,383 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3a9b6dffdf9d20cf5158c028c4012a0a2cf4ee76994f18e9b81499931df7cb5d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11, Pointers and Indirection"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 11-1, Page Number : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Introduction to pointers\n",
+      "\n",
+      "# There is no concept of pointers in python\n",
+      "i = 3\n",
+      "\n",
+      "j = id(i)\t# id() returns the address of the variable supplied\n",
+      "j = i \t\t# assign value of i to j such that it seems like j points to i\n",
+      "k = j\n",
+      "j = 4\n",
+      "\n",
+      "print \"i = %d, *j = %d, k = %d\"%(i,j,k)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i = 3, *j = 4, k = 3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 11-2, Page Number: 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Demonstration of passing by reference for a function to be able to modify its parameters\n",
+      "\n",
+      "def modify(i) :\n",
+      "\ti = 3\n",
+      "\treturn i\n",
+      "\n",
+      "i = 1\n",
+      "print \"i is now equal to %d\"%i\n",
+      "\n",
+      "i = modify(i)\t\t\t\n",
+      "print \"i is now equal to %d\"%i"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i is now equal to 1\n",
+        "i is now equal to 3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 11-3, Page Number: 281"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Pointer parameters in action: the swap routine\n",
+      "\n",
+      "def swap(i,j) :\n",
+      "\t\"\"\" exchange i with j and return \"\"\"\n",
+      "\ttemp = i   \t\t\t# temp variable can be avoided using i,j = j,i\n",
+      "\ti = j\n",
+      "\tj = temp\n",
+      "\treturn i,j\n",
+      "\n",
+      "i = 1\n",
+      "j = 9\n",
+      "print \"i = %d and j = %d\"%(i,j)\n",
+      "\n",
+      "# Now exchange them\n",
+      "i, j = swap(i,j)\n",
+      "print \"\\nnow i = %d and j = %d\"%(i,j)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i = 1 and j = 9\n",
+        "\n",
+        "now i = 9 and j = 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 11-4, Page Number: 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Array names are constant pointers\n",
+      "\n",
+      "an_array = [ 3, 7 ]\n",
+      "\n",
+      "def main() :\n",
+      "\talso = an_array\n",
+      "\talso[0] = 14\n",
+      "\tprint \"an_array = { %d, %d }\"%(an_array[0],an_array[1])\n",
+      "\n",
+      "main()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "an_array = { 14, 7 }\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 11-5, Page Number: 284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Sorting an array revisited\n",
+      "# Passing an array as parameter\n",
+      "\n",
+      "# Macro definition\n",
+      "MAX_ARRAY_SIZE = 20\t\t# The array may have up to 20 elements\n",
+      "# Message flags for \"print array\"\n",
+      "ORIGINAL = 0\t# Display original array\n",
+      "SORTED = 1\t\t# Display sorted array\n",
+      "\n",
+      "numbers = [ 10, 42, 7, 0, 923, 12, -5, 6, 19, 3 ]\n",
+      "array_size = 0\n",
+      "\n",
+      "def get_size(size)\t:\n",
+      "\t\"\"\" reads the size of the array (until a valid value is entered) \"\"\"\n",
+      "\tglobal array_size\n",
+      "\tprint \"How many numbers?\",\n",
+      "\tsize = 10\n",
+      "\tprint size\n",
+      "\twhile size < 1 or size > MAX_ARRAY_SIZE\t:\n",
+      "\t\tprint \"How many numbers?\",\n",
+      "\t\tsize = 10\n",
+      "\t\tprint size\n",
+      "\tarray_size = size\n",
+      "\n",
+      "def get_array(array, size) :\n",
+      "\t\"\"\" read in the array called 'array' of size 'size' \"\"\"\n",
+      "\tprint \"Please enter the integers:\"\n",
+      "\tfor i in array :\n",
+      "\t\tprint i,\n",
+      "\tprint\n",
+      "\n",
+      "def print_array(array, size, message) :\n",
+      "\t\"\"\" print out the array on a single line. 'message' is a flag indicating which message to display(ORIGINAL or SORTED) \"\"\"\n",
+      "\tif message == ORIGINAL :\n",
+      "\t\tprint \"\\nThe original array is:\"\n",
+      "\telse :\n",
+      "\t\tprint \"\\nThe sorted array is:\"\n",
+      "\tfor i in array :\n",
+      "\t\tprint i,\n",
+      "\tprint\n",
+      "\n",
+      "def sort_array(array,size) :\n",
+      "\t\"\"\" sort 'array' of size 'size' \"\"\"\n",
+      "\tfor i in range(size - 1) :\n",
+      "\t\tfor j in range(i+1, size) :\n",
+      "\t\t\tif array[i] > array[j] :\n",
+      "\t\t\t\telement_swap(array, i, j)\n",
+      "\n",
+      "def element_swap(array, i, j) :\n",
+      "\t\"\"\" swap elements 'i' and 'j' of 'array' \"\"\"\n",
+      "\tglobal numbers\n",
+      "\tnumbers[i],numbers[j] = numbers[j],numbers[i]\n",
+      "\n",
+      "\n",
+      "get_size(array_size)\t\t\t\t\t\t# Read in the array's size\n",
+      "get_array(numbers, array_size)\t\t\t\t\t# Read in the array 'numbers' of size 'array_size'\n",
+      "print_array(numbers, array_size, ORIGINAL) \t# Echo the array \n",
+      "sort_array(numbers, array_size)\n",
+      "print_array(numbers, array_size, SORTED)\t\t# Print the sorted array"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many numbers? 10\n",
+        "Please enter the integers:\n",
+        "10 42 7 0 923 12 -5 6 19 3\n",
+        "\n",
+        "The original array is:\n",
+        "10 42 7 0 923 12 -5 6 19 3\n",
+        "\n",
+        "The sorted array is:\n",
+        "-5 0 3 6 7 10 12 19 42 923\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 11-6, Page Number: 294"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Sorting of array revisited\n",
+      "# Two dimensional char array (array of strings)\n",
+      "\n",
+      "# Macro definition\n",
+      "MAX_ARRAY_SIZE = 20\t\t# The array may have up to 20 elements\n",
+      "MAX_STRING_SIZE = 10 \t# Each string may have up to 10 characters\n",
+      "# Message flags for \"print_array\"\n",
+      "ORIGINAL = 0\n",
+      "SORTED = 1\n",
+      "\n",
+      "words = [ \"cat\", \"pig\", \"dog\", \"earlobe\", \"bingo\", \"elbow\", \"likely\", \"radar\" ]\n",
+      "array_size = 0\n",
+      "\n",
+      "def get_size(size) :\n",
+      "\t\"\"\" Read the size of the array into size (until a value is entered) \"\"\"\n",
+      "\tprint \"How many words? \",\n",
+      "\tsize = 8\n",
+      "\tprint size\n",
+      "\twhile size < 1 or size > MAX_ARRAY_SIZE :\n",
+      "\t\tprint \"How many words? \",\n",
+      "\t\tsize = 8\n",
+      "\t\tprint size\n",
+      "\tglobal array_size\n",
+      "\tarray_size = size\n",
+      "\n",
+      "def get_array(array,size) :\n",
+      "\t\"\"\" Read in the string array called 'array' of size 'size' \"\"\"\n",
+      "\tprint \"Please enter the words:\"\n",
+      "\tfor i in array :\n",
+      "\t\tprint i\n",
+      "\t\n",
+      "def print_array(array,size,message) :\n",
+      "\t\"\"\" Print out string array 'array' of size 'size'. 'message' is a flag indicating which message to display(ORIGINAL or SORTED) \"\"\"\n",
+      "\tif message == ORIGINAL :\n",
+      "\t\tprint \"\\nThe original array is:\"\n",
+      "\telse :\n",
+      "\t\tprint \"\\nThe sorted array is:\"\n",
+      "\tfor i in array :\n",
+      "\t\tprint i\n",
+      "\n",
+      "def sort_array(array,size) :\n",
+      "\t\"\"\" sort string array \"\"\"\n",
+      "\tfor i in range (size-1) :\n",
+      "\t\tfor j in range (i + 1, size) :\n",
+      "\t\t\tif array[i] > array[j] :\n",
+      "\t\t\t\tstring_swap(i,j)\n",
+      "\n",
+      "def string_swap(i, j) :\n",
+      "\t\"\"\" Exchange strings \"\"\"\n",
+      "\tglobal words\n",
+      "\twords[i],words[j] = words[j],words[i]\n",
+      "\n",
+      "get_size(array_size)\t\t# Read in the array size\n",
+      "get_array(words, array_size)\t# Read in the string array \"words\" of size \"array_size\"\n",
+      "print_array(words, array_size, ORIGINAL)\t# Echo the original array\n",
+      "sort_array(words, array_size)\t\n",
+      "print_array(words, array_size, SORTED) # Print the sorted array"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many words?  8\n",
+        "Please enter the words:\n",
+        "cat\n",
+        "pig\n",
+        "dog\n",
+        "earlobe\n",
+        "bingo\n",
+        "elbow\n",
+        "likely\n",
+        "radar\n",
+        "\n",
+        "The original array is:\n",
+        "cat\n",
+        "pig\n",
+        "dog\n",
+        "earlobe\n",
+        "bingo\n",
+        "elbow\n",
+        "likely\n",
+        "radar\n",
+        "\n",
+        "The sorted array is:\n",
+        "bingo\n",
+        "cat\n",
+        "dog\n",
+        "earlobe\n",
+        "elbow\n",
+        "likely\n",
+        "pig\n",
+        "radar\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter12.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter12.ipynb
new file mode 100755
index 00000000..0195ed25
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter12.ipynb
@@ -0,0 +1,455 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bfcc6bf3e86f2ff3b103be284b1b027f6faa9020241e866b9b4398fd89f8923c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12, Strings and String Functions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-1, Page Number: 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The control string is just a string\n",
+      "\n",
+      "format = \"%s\"\n",
+      "message = \"hello\"\n",
+      "\n",
+      "print format % message"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hello\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-2, Page Number: 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The control string is just a string - part 2\n",
+      "\n",
+      "format = \"%.15s\"\n",
+      "message = \"hellohellohello\"\n",
+      "\n",
+      "# Read in the number of characters to be displayed from \"message\"\n",
+      "print \"How many characters do you want displayed?\",\n",
+      "size = 5\n",
+      "print size\n",
+      "\n",
+      "# Convert size to 2-digit number in format[2] and format[3]\n",
+      "\n",
+      "format = \"%.\" + str(size / 10) + str(size % 10) + \"s\"\n",
+      "print \"The format is %s\"%format\n",
+      "\n",
+      "# Print the string\n",
+      "print format % message"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many characters do you want displayed? 5\n",
+        "The format is %.05s\n",
+        "hello\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-3, Page Number 336"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# sprintf function \n",
+      "\n",
+      "# There is no sprintf functionality in python\n",
+      "# However an alternate way to solve the problem is...\n",
+      "\n",
+      "print \"Please enter a number:\",\n",
+      "number = 17\n",
+      "print number\n",
+      "\n",
+      "string = \"The number is %d\"%number\n",
+      "\n",
+      "print \"%s\"%string\n",
+      "\n",
+      "# OR \n",
+      "\n",
+      "print string"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter a number: 17\n",
+        "The number is 17\n",
+        "The number is 17\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-4, Page Number 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variable control string using the sprintf function\n",
+      "\n",
+      "message = \"hellohellohello\"\n",
+      "\n",
+      "# Read in the number of characters to be displayed from message \n",
+      "print \"How many characters do you want displayed?\",\n",
+      "size = 7\n",
+      "print size\n",
+      "\n",
+      "# Create format in \"format\" using \"size\"\n",
+      "format = \"%%.%ds\\n\"%size\n",
+      "print \"The format is: %s\"%format\n",
+      "\n",
+      "# Print the string\n",
+      "print format % message"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many characters do you want displayed? 7\n",
+        "The format is: %.7s\n",
+        "\n",
+        "hellohe\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-5, Page Number 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A terse string copy function\n",
+      "\n",
+      "def strcpy(copy, original):\n",
+      "\t\"\"\" copies one string into another \"\"\"\n",
+      "\tcopy = original\n",
+      "\treturn copy\n",
+      "\n",
+      "string1 = \"\"\n",
+      "string1 = strcpy(string1, \"This is a message\")\n",
+      "print \"String 1 = %s\"%string1\n",
+      "\n",
+      "string2 = \"\"\n",
+      "string2 = strcpy(string2, string1)\n",
+      "print \"String 2 = %s\"%string2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "String 1 = This is a message\n",
+        "String 2 = This is a message\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-7, Page Number 344"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "string_point = \"hello\"\n",
+      "\n",
+      "string_var1 = string_point\n",
+      "\n",
+      "print \"%s\\n%s\"%(string_point,string_var1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hello\n",
+        "hello\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-8, Page Number 346"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# strcat function in python\n",
+      "# simple addition of strings results in concatenation in python\n",
+      "\n",
+      "def strcat(string1, string2) :\n",
+      "\t\"\"\" concatenates the string and returns the concatenated string \"\"\"\n",
+      "\tstring2 = string1 + string2\n",
+      "\treturn string2\n",
+      "\n",
+      "string1 = \"hello\"\n",
+      "string2 = strcat(string1,\" there\")\n",
+      "print \"string1 = %s\\nstring2 = %s\"%(string1,string2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "string1 = hello\n",
+        "string2 = hello there\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-9, Page Number 348"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# strcat version 2\n",
+      "\n",
+      "def strcat(string1, string2) :\n",
+      "\tstring1 = string1 + string2\n",
+      "\tstring2 = string1\n",
+      "\treturn string1,string2\n",
+      "\n",
+      "string1 = \"hello\"\n",
+      "string1,string2 = strcat(string1, \" there\")\n",
+      "\n",
+      "print \"string1 = %s\\nstring2 = %s\"%(string1,string2)\n",
+      "\n",
+      "if string1 == string2 :\n",
+      "\tprint \"string2 points to string1\"\n",
+      "else :\n",
+      "\tprint \"string2 does not point to string1\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "string1 = hello there\n",
+        "string2 = hello there\n",
+        "string2 points to string1\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 12-10, Page Number 351"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Alphabetizing program using dynamically allocated arrays\n",
+      "\n",
+      "# Macro definition\n",
+      "MAX_ARRAY_SIZE = 20\t\t# The array may have up to 20 elements\n",
+      "MAX_STRING_SIZE = 10 \t# Each string may have up to 10 characters\n",
+      "# Message flags for \"print_array\"\n",
+      "ORIGINAL = 0\n",
+      "SORTED = 1\n",
+      "\n",
+      "words = [ \"cat\", \"pig\", \"dog\", \"earlobe\", \"bingo\", \"elbow\", \"likely\", \"radar\" ]\n",
+      "array_size = 0\n",
+      "\n",
+      "def get_size(size) :\n",
+      "\t\"\"\" Read the size of the array into size (until a value is entered) \"\"\"\n",
+      "\tprint \"How many words? \",\n",
+      "\tsize = 8\n",
+      "\tprint size\n",
+      "\twhile size < 1 or size > MAX_ARRAY_SIZE :\n",
+      "\t\tprint \"How many words? \",\n",
+      "\t\tsize = 8\n",
+      "\t\tprint size\n",
+      "\tglobal array_size\n",
+      "\tarray_size = size\n",
+      "\n",
+      "def get_array(array,size) :\n",
+      "\t\"\"\" Read in the string array called 'array' of size 'size' \"\"\"\n",
+      "\tprint \"Please enter the words:\"\n",
+      "\tfor i in array :\n",
+      "\t\tprint i\n",
+      "\t\n",
+      "def print_array(array,size,message) :\n",
+      "\t\"\"\" Print out string array 'array' of size 'size'. 'message' is a flag indicating which message to display(ORIGINAL or SORTED) \"\"\"\n",
+      "\tprintout = [ \"\\nThe original array is:\", \"\\nThe sorted array is:\", \"\\n\" ]\n",
+      "\t# Print the appropriate message\n",
+      "\tprint \"%s\"%printout[message]\n",
+      "\n",
+      "\tfor i in array :\n",
+      "\t\tprint i\n",
+      "\n",
+      "def pointer_sort(array,size) :\n",
+      "\t\"\"\" sort string array \"\"\"\n",
+      "\tfor i in range (size-1) :\n",
+      "\t\tfor j in range (i + 1, size) :\n",
+      "\t\t\tif array[i] > array[j] :\n",
+      "\t\t\t\tswap(i,j)\n",
+      "\n",
+      "def swap(i, j) :\n",
+      "\t\"\"\" Exchange strings \"\"\"\n",
+      "\tglobal words\n",
+      "\twords[i],words[j] = words[j],words[i]\n",
+      "\n",
+      "get_size(array_size)\t\t# Read in the array size\n",
+      "get_array(words, array_size)\t# Read in the string array \"words\" of size \"array_size\"\n",
+      "print_array(words, array_size, ORIGINAL)\t# Echo the original array\n",
+      "pointer_sort(words, array_size)\t\n",
+      "print_array(words, array_size, SORTED) # Print the sorted array"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many words?  8\n",
+        "Please enter the words:\n",
+        "cat\n",
+        "pig\n",
+        "dog\n",
+        "earlobe\n",
+        "bingo\n",
+        "elbow\n",
+        "likely\n",
+        "radar\n",
+        "\n",
+        "The original array is:\n",
+        "cat\n",
+        "pig\n",
+        "dog\n",
+        "earlobe\n",
+        "bingo\n",
+        "elbow\n",
+        "likely\n",
+        "radar\n",
+        "\n",
+        "The sorted array is:\n",
+        "bingo\n",
+        "cat\n",
+        "dog\n",
+        "earlobe\n",
+        "elbow\n",
+        "likely\n",
+        "pig\n",
+        "radar\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter13.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter13.ipynb
new file mode 100755
index 00000000..d363dc7a
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter13.ipynb
@@ -0,0 +1,346 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:133c881c520fd0cdef07f65db120109cc62a996dd5161cf0f664c2ffcaa3adc4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13, Structures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 13-1, Page Number: 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Introduction to structures\n",
+      "\n",
+      "from collections import namedtuple\n",
+      "\n",
+      "#Structure defintion\n",
+      "struct = namedtuple(\"struct\", \"integer real\")\n",
+      "\n",
+      "first_structure = struct( 7, 3.14 )\n",
+      "second_structure = first_structure\n",
+      "\n",
+      "print \"integer = %d, real = %f\"%(second_structure.integer,second_structure.real)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "integer = 7, real = 3.140000\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 13-2, Page Number: 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Initializing a structure\n",
+      "\n",
+      "from collections import namedtuple\n",
+      "\n",
+      "#Structure defintion\n",
+      "struct = namedtuple(\"struct\", \"integer real\")\n",
+      "\n",
+      "structure = struct( 7, 3.14 )\n",
+      "\n",
+      "print \"integer = %d, real = %f\"%(structure.integer,structure.real)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "integer = 7, real = 3.140000\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 13-3, Page Number: 372"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The structure tag\n",
+      "\n",
+      "from collections import namedtuple\n",
+      "\n",
+      "#Structure defintion\n",
+      "struct = namedtuple(\"struct\", \"integer real\")\n",
+      "\n",
+      "first_structure = struct( 7, 3.14 )\n",
+      "second_structure = first_structure\n",
+      "\n",
+      "print \"integer = %d, real = %f\"%(second_structure.integer,second_structure.real)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "integer = 7, real = 3.140000\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 13-4, Page Number: 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# An array of structures\n",
+      "\n",
+      "from collections import namedtuple\n",
+      "\n",
+      "#Structure defintion\n",
+      "struct = namedtuple(\"struct\", \"code price\")\n",
+      "\n",
+      "#code = [ \"a\", \"b\", \"c\", \"d\", \"e\" ]\n",
+      "#price = [ 3.29, 2.99, 0.59, 1.59, 2.39 ]\n",
+      "\n",
+      "table = [ struct(\"a\",3.29) , struct(\"b\",2.99), struct(\"c\",0.59), struct(\"d\",1.59), struct(\"e\",2.39) ]\n",
+      "\n",
+      "print \"Enter item code:\",\n",
+      "item_code = \"a\"\n",
+      "print item_code\n",
+      "i = 0\n",
+      "\n",
+      "while item_code != \".\" :\n",
+      "\n",
+      "\t#Search for 'item_code' in 'table'\n",
+      "\tfor i in range(4) :\n",
+      "\t\tif table[i].code == item_code :\n",
+      "\t\t\tbreak\n",
+      "\n",
+      "\tif i < 4:\n",
+      "\t\tprint \"Enter quantity: \",\n",
+      "\t\tquantity = 1\n",
+      "\t\tprint quantity\n",
+      "\t\n",
+      "\t\tprint \"\\nUnit price = %.2f, total price = %.2f.\\n\\n\"%(table[i].price,quantity * table[i].price)\n",
+      "\telse :\n",
+      "\t\tprint \"\\nItem code %c does not exist.\\n\\n\"%item_code\n",
+      "\tprint \"Enter item code:\",\n",
+      "\titem_code = \".\"\n",
+      "\tprint item_code\n",
+      "\n",
+      "\n",
+      "print \"Thank you.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter item code: a\n",
+        "Enter quantity:  1\n",
+        "\n",
+        "Unit price = 3.29, total price = 3.29.\n",
+        "\n",
+        "\n",
+        "Enter item code: .\n",
+        "Thank you.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 13-5, Page Number: 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A pointer in a structure\n",
+      "\n",
+      "from collections import namedtuple\n",
+      "\n",
+      "#Structure defintion\n",
+      "struct = namedtuple(\"struct\", \"name price\")\n",
+      "\n",
+      "#code = [ \"a\", \"b\", \"c\", \"d\", \"e\" ]\n",
+      "#price = [ 3.29, 2.99, 0.59, 1.59, 2.39 ]\n",
+      "\n",
+      "table = [ struct(\"pickles\",2.29) , struct(\"soda\",0.99), struct(\"yogurt\",0.59), struct(\"bread\",1.09), struct(\"milk\",0.69) ]\n",
+      "\n",
+      "print \"Enter item name:\",\n",
+      "item_name = \"pickles\"\n",
+      "print item_name\n",
+      "i = 0\n",
+      "\n",
+      "while item_name != \".\" :\n",
+      "\n",
+      "\t#Search for 'item_code' in 'table'\n",
+      "\tfor i in range(4) :\n",
+      "\t\tif table[i].name == item_name :\n",
+      "\t\t\tbreak\n",
+      "\n",
+      "\tif i < 4:\n",
+      "\t\tprint \"Enter quantity: \",\n",
+      "\t\tquantity = 1\n",
+      "\t\tprint quantity\n",
+      "\t\n",
+      "\t\tprint \"\\nUnit price = %.2f, total price = %.2f.\\n\\n\"%(table[i].price,quantity * table[i].price)\n",
+      "\telse :\n",
+      "\t\tprint \"\\nItem code %c does not exist.\\n\\n\"%item_name\n",
+      "\tprint \"Enter item code:\",\n",
+      "\titem_name = \".\"\n",
+      "\tprint item_name\n",
+      "\n",
+      "\n",
+      "print \"Thank you.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter item name: pickles\n",
+        "Enter quantity:  1\n",
+        "\n",
+        "Unit price = 2.29, total price = 2.29.\n",
+        "\n",
+        "\n",
+        "Enter item code: .\n",
+        "Thank you.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 13-6, Page Number: 381"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# An array in a structure\n",
+      "\n",
+      "from collections import namedtuple\n",
+      "\n",
+      "#Structure defintion\n",
+      "struct = namedtuple(\"struct\", \"name price\")\n",
+      "\n",
+      "#code = [ \"a\", \"b\", \"c\", \"d\", \"e\" ]\n",
+      "#price = [ 3.29, 2.99, 0.59, 1.59, 2.39 ]\n",
+      "\n",
+      "table = [ struct(\"pickles\",2.29) , struct(\"soda\",0.99), struct(\"yogurt\",0.59), struct(\"bread\",1.09), struct(\"milk\",0.69) ]\n",
+      "\n",
+      "print \"Enter item name:\",\n",
+      "item_name = \"milk\"\n",
+      "print item_name\n",
+      "i = 0\n",
+      "\n",
+      "while item_name != \".\" :\n",
+      "\n",
+      "\t#Search for 'item_code' in 'table'\n",
+      "\tfor i in range(5) :\n",
+      "\t\tif table[i].name == item_name :\n",
+      "\t\t\tbreak\n",
+      "\n",
+      "\tif i < 5:\n",
+      "\t\tprint \"Enter quantity: \",\n",
+      "\t\tquantity = 5\n",
+      "\t\tprint quantity\n",
+      "\t\n",
+      "\t\tprint \"\\nUnit price = %.2f, total price = %.2f.\\n\\n\"%(table[i].price,quantity * table[i].price)\n",
+      "\telse :\n",
+      "\t\tprint \"\\nItem code %c does not exist.\\n\\n\"%item_name\n",
+      "\tprint \"Enter item code:\",\n",
+      "\titem_name = \".\"\n",
+      "\tprint item_name\n",
+      "\n",
+      "\n",
+      "print \"Thank you.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter item name: milk\n",
+        "Enter quantity:  5\n",
+        "\n",
+        "Unit price = 0.69, total price = 3.45.\n",
+        "\n",
+        "\n",
+        "Enter item code: .\n",
+        "Thank you.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter14.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter14.ipynb
new file mode 100755
index 00000000..380e83ef
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter14.ipynb
@@ -0,0 +1,298 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3748f84e3bd0279efe47855bd2c4e23912ef725121f3591980fa33e19d6ccae4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14, Input/Output and Files"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 14-5, Page Number: 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Files : Opening and Closing\n",
+      "\n",
+      "# Open the file for writing\n",
+      "\n",
+      "file_pointer = open(\"first_file\",\"w\")\n",
+      "\n",
+      "print \"Enter text:\"\n",
+      "\n",
+      "# Get characters and place them in the file\n",
+      "\n",
+      "text = \"BASIC is easy,\\nFORTAN is fun,\\nPascal may be structured,\\nBut C's the one!\"\n",
+      "print text\n",
+      "file_pointer.write(text)\n",
+      "\n",
+      "file_pointer.close()\n",
+      "\n",
+      "file_pointer = open(\"first_file\",\"r\")\n",
+      "\n",
+      "#Display the text from file\n",
+      "\n",
+      "print \"\\nYou entered:\"\n",
+      "print file_pointer.read()\n",
+      "\n",
+      "file_pointer.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter text:\n",
+        "BASIC is easy,\n",
+        "FORTAN is fun,\n",
+        "Pascal may be structured,\n",
+        "But C's the one!\n",
+        "\n",
+        "You entered:\n",
+        "BASIC is easy,\n",
+        "FORTAN is fun,\n",
+        "Pascal may be structured,\n",
+        "But C's the one!\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 14-6, Page Number: 426"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Using several files\n",
+      "\n",
+      "# Open the source file for reading\n",
+      "\n",
+      "first_file_pointer = open(\"first_file\",\"r\")\n",
+      "\n",
+      "# Open the destination file for writing\n",
+      "\n",
+      "second_file_pointer = open(\"second_file\",\"w\")\n",
+      "\n",
+      "# Transfer characters from source to destination file\n",
+      "\n",
+      "c = first_file_pointer.read()\n",
+      "second_file_pointer.write(c)\n",
+      "\n",
+      "# Close both files\n",
+      "\n",
+      "first_file_pointer.close()\n",
+      "second_file_pointer.close()\n",
+      "\n",
+      "# Open second file for displaying contentes\n",
+      "\n",
+      "second_file_pointer = open(\"second_file\",\"r\")\n",
+      "print(\"'second_file' contains:\")\n",
+      "print second_file_pointer.read()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "'second_file' contains:\n",
+        "BASIC is easy,\n",
+        "FORTAN is fun,\n",
+        "Pascal may be structured,\n",
+        "But C's the one!\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 14-7, Page Number : 428"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Append mode\n",
+      "\n",
+      "# Open the first file for writing\n",
+      "\n",
+      "first_file_pointer = open(\"first_file\",\"w\")\n",
+      "\n",
+      "# Open the second file for appending\n",
+      "\n",
+      "second_file_pointer = open(\"second_file\",\"a\")\n",
+      "\n",
+      "# Writing to both files\n",
+      "\n",
+      "first_file_pointer.write(\"\\n--Anonymous\\n\")\n",
+      "second_file_pointer.write(\"\\n--Anonymous\\n\")\n",
+      "\n",
+      "# Close both files\n",
+      "\n",
+      "first_file_pointer.close()\n",
+      "second_file_pointer.close()\n",
+      "\n",
+      "# Display the contents of first file\n",
+      "first_file_pointer = open(\"first_file\",\"r\")\n",
+      "print(\"After write, 'first_file contains:\")\n",
+      "print first_file_pointer.read()\n",
+      "\n",
+      "# Display the contents of second file\n",
+      "second_file_pointer = open(\"second_file\",\"r\")\n",
+      "print(\"After append, 'second_file' contains:\")\n",
+      "print second_file_pointer.read()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After write, 'first_file contains:\n",
+        "\n",
+        "--Anonymous\n",
+        "\n",
+        "After append, 'second_file' contains:\n",
+        "BASIC is easy,\n",
+        "FORTAN is fun,\n",
+        "Pascal may be structured,\n",
+        "But C's the one!\n",
+        "--Anonymous\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 14-8, Page Number: 431"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The seek function\n",
+      "\n",
+      "# Open file for reading\n",
+      "try :\n",
+      "\tfile_pointer = open(\"second_file\",\"r\")\n",
+      "except :\n",
+      "\tprint \"'second_file' cannot be opened\"\n",
+      "\texit ()\n",
+      "\n",
+      "# Skip the first 6 characters\n",
+      "\n",
+      "file_pointer.seek(6,0)\n",
+      "print file_pointer.read()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "is easy,\n",
+        "FORTAN is fun,\n",
+        "Pascal may be structured,\n",
+        "But C's the one!\n",
+        "--Anonymous\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 14-9, Page Number: 433"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The 'w+' mode and seek to rewind a file\n",
+      "\n",
+      "# Open a file for read/write\n",
+      "file_pointer = open(\"first_file\",\"w+\")\n",
+      "\n",
+      "print \"Enter text:\"\n",
+      "text = \"BASIC is easy,\\nFORTAN is fun,\\nPascal may be structured,\\nBut C's the one!\"\n",
+      "print text\n",
+      "file_pointer.write(text)\n",
+      "\n",
+      "file_pointer.seek(0,0)\n",
+      "\n",
+      "# Display the file\n",
+      "print \"\\nYou entered:\"\n",
+      "print file_pointer.read()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter text:\n",
+        "BASIC is easy,\n",
+        "FORTAN is fun,\n",
+        "Pascal may be structured,\n",
+        "But C's the one!\n",
+        "\n",
+        "You entered:\n",
+        "BASIC is easy,\n",
+        "FORTAN is fun,\n",
+        "Pascal may be structured,\n",
+        "But C's the one!\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter2.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter2.ipynb
new file mode 100755
index 00000000..dc84a28d
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter2.ipynb
@@ -0,0 +1,119 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7c8e11b3dcc5781949652c0b5c450d78713cb6e6b69326955d0f763c48805147"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2, An Introduction to Declarations, Assignments and Variables"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 2-1 , Page number: 20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# An introduction to variable declarations \n",
+      "\n",
+      "number = 7 # no need of declaration of variable. Python is dynamically-typed language\n",
+      "print \"There are %d days in a week.\"%number"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "There are 7 days in a week.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 2-2 , Page number: 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A simple program with two declarations\n",
+      "\n",
+      "minutes = 60\n",
+      "hours = 24\n",
+      "\n",
+      "print \"There are %d minutes in an hour.\"%minutes\n",
+      "print \"And %d hours in a day.\"%hours"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "There are 60 minutes in an hour.\n",
+        "And 24 hours in a day.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 2-4 , Page number: 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# prints values of two variables\n",
+      "\n",
+      "dozens = 17\n",
+      "units = dozens * 12\n",
+      "\n",
+      "print \"%d units is equivalent to %d dozen.\"%(units,dozens)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "204 units is equivalent to 17 dozen.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter3.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter3.ipynb
new file mode 100755
index 00000000..b498486d
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter3.ipynb
@@ -0,0 +1,241 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c466b513c45bfe09da1b0507614f538770318fb66aadf4da4e1995363df32911"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 , Integer Arithmetic Expressions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 3-1 , Page number: 33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# An introduction to arithmetic \n",
+      "\n",
+      "a = 2\n",
+      "b = 3\n",
+      "c = 4\n",
+      "\n",
+      "answer = a + b\n",
+      "print \"a + b = %d\"%answer\n",
+      "\n",
+      "answer = a - b\n",
+      "print \"a - b = %d\"%answer\n",
+      "\n",
+      "answer = a * b\n",
+      "print \"a * b = %d\"%answer\n",
+      "\n",
+      "answer = a / b\n",
+      "print \"a / b = %d\"%answer\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a + b = 5\n",
+        "a - b = -1\n",
+        "a * b = 6\n",
+        "a / b = 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 3-2 , Page number: 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of mathematical hierarchy\n",
+      "\n",
+      "a = 2\n",
+      "b = 3\n",
+      "c = 4\n",
+      "\n",
+      "answer = a + b * c\n",
+      "\n",
+      "print \"a = %d, b = %d, c= %d\"%(a,b,c)\n",
+      "print \"a + b * c = %d\\n\"%answer"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = 2, b = 3, c= 4\n",
+        "a + b * c = 14\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 3-3 , Page number: 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A more advanced illustration of mathematical hierarchy\n",
+      "\n",
+      "a = 5\n",
+      "b = 10\n",
+      "c = 15\n",
+      "\n",
+      "answer = a + b / c\n",
+      "print \"%d + %d / %d = %d\"%(a,b,c,answer)\n",
+      "\n",
+      "answer = b * c - a\n",
+      "print \"%d * %d - %d = %d\"%(a,b,c,answer)\n",
+      "\n",
+      "answer = a * b / c\n",
+      "print \"%d * %d / %d = %d\"%(a,b,c,answer)\n",
+      "\n",
+      "answer = (a + c) * b / a\n",
+      "print \"(%d + %d) * %d / %d = %d\"%(a,c,b,a,answer)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5 + 10 / 15 = 5\n",
+        "5 * 10 - 15 = 145\n",
+        "5 * 10 / 15 = 3\n",
+        "(5 + 15) * 10 / 5 = 40\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 3-4 , Page number: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of modulus operator\n",
+      "\n",
+      "a = 4\n",
+      "b = 8\n",
+      "c = 13\n",
+      "\n",
+      "print \"%d mod %d = %d\"%(b,a,b%a)\n",
+      "print \"%d mod %d = %d\"%(c,a,c%a)\n",
+      "print \"%d mod %d = %d\"%(c,b,c%b)\n",
+      "\n",
+      "print \"The modulus operator is: %\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8 mod 4 = 0\n",
+        "13 mod 4 = 1\n",
+        "13 mod 8 = 5\n",
+        "The modulus operator is: %\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 3-5 , Page number: 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of the unary minus\n",
+      "\n",
+      "a = -4\n",
+      "b = 5\n",
+      "c = -6\n",
+      "\n",
+      "print \"a = %d, b = %d, c = %d\\n\\n\"%(a,b,c)\n",
+      "print \"      and now\\n\\n\"\n",
+      "\n",
+      "a = -a\n",
+      "b = -b\n",
+      "c = -c\n",
+      "\n",
+      "print \"a = %d, b = %d, c = %d\\n\\n\"%(a,b,c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = -4, b = 5, c = -6\n",
+        "\n",
+        "\n",
+        "      and now\n",
+        "\n",
+        "\n",
+        "a = 4, b = -5, c = 6\n",
+        "\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter4.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter4.ipynb
new file mode 100755
index 00000000..d13f3710
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter4.ipynb
@@ -0,0 +1,148 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:03f3f522026b7ff6e1acac7746805bcac18faba022541a5d8a73470a897fdf7a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4, Some More Data Types"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Prgoram 4-1 , Page number: 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# An illustration of the use of floating point numbers\n",
+      "\n",
+      "# There is no loss of precision in python\n",
+      "a = 246.8\n",
+      "b = 135.79\n",
+      "\n",
+      "answer = a + b\n",
+      "print \"%f + %f = %f\"%(a,b,answer)\n",
+      "\n",
+      "answer = a - b\n",
+      "print \"%f - %f = %f\"%(a,b,answer)\n",
+      "\n",
+      "answer = a * b\n",
+      "print \"%f * %f = %f\"%(a,b,answer)\n",
+      "\n",
+      "answer = a / b\n",
+      "print \"%f / %f = %f\"%(a,b,answer)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "246.800000 + 135.790000 = 382.590000\n",
+        "246.800000 - 135.790000 = 111.010000\n",
+        "246.800000 * 135.790000 = 33512.972000\n",
+        "246.800000 / 135.790000 = 1.817512\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 4-2 , Page number: 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of mixed mode expressions\n",
+      "\n",
+      "a = 2\n",
+      "b = 12.34\n",
+      "\n",
+      "result = a * b\n",
+      "print \"%d * %f = %f\"%(a,b,result)\n",
+      "\n",
+      "print \"%f in scientific notation is %e\"%(result,result)\n",
+      "\n",
+      "answer = a * b\n",
+      "print \"%d * %f = %d\"%(a,b,answer)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2 * 12.340000 = 24.680000\n",
+        "24.680000 in scientific notation is 2.468000e+01\n",
+        "2 * 12.340000 = 24\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 4-3 , Page number: 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "ltr_1 = \"C\"\n",
+      "ltr_2 = \" \"\n",
+      "ltr_3 = \"i\"\n",
+      "ltr_4 = \"s\"\n",
+      "ltr_5 = \"g\"\n",
+      "ltr_6 = \"r\"\n",
+      "ltr_7 = \"e\"\n",
+      "ltr_8 = \"a\"\n",
+      "ltr_9 = \"t\"\n",
+      "ltr_10 = \"!\"\n",
+      "\n",
+      "print \"%c%c%c%c%c%c%c%c%c%c%c\"%(ltr_1,ltr_2,ltr_3,ltr_4,ltr_2,ltr_5,ltr_6,ltr_7,ltr_8,ltr_9,ltr_10)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C is great!\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter5.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter5.ipynb
new file mode 100755
index 00000000..db420a5b
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter5.ipynb
@@ -0,0 +1,218 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:28fc7922ba91eac91816a02729f8986381706b8cffd1eaa9d68ce0cf435191aa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5, Making Decisions in C"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 5-2 , Page number: 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Using If command\n",
+      "\n",
+      "print \"Enter a positive integer under 10: \",\n",
+      "n = 7\n",
+      "print n\n",
+      "\n",
+      "if n == 7 :\t\t\t\t\t# In this case it is 7\n",
+      "\tprint \"I knew you'd select 7!\"\n",
+      "print \"Thank you for your cooperation.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a positive integer under 10:  7\n",
+        "I knew you'd select 7!\n",
+        "Thank you for your cooperation.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 5-3 , Page number: 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Testing for leap years\n",
+      "\n",
+      "print \"Please enter a year:\",\n",
+      "year = 1948\t\t\n",
+      "print year\n",
+      "# If \"year\" is divisble by 4 and not divisible by 100 or divisible by 400, \"year\" is a leap year\n",
+      "\n",
+      "if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0) :\n",
+      "\tprint \"%d is a leap year.\"%year\n",
+      "else :\n",
+      "\tprint \"%d is not a leap year.\"%year"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter a year: 1948\n",
+        "1948 is a leap year.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 5-4 , Page number: 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of if...else clause\n",
+      "# Program to tell if a number is odd or even\n",
+      "\n",
+      "print \"Please enter an integer:\",\n",
+      "n = 13\n",
+      "print n\n",
+      "\n",
+      "# A number is even if it is divisible by 2\n",
+      "\n",
+      "if n % 2 != 0:\n",
+      "\tprint \"%d is odd\"%n\n",
+      "else:\n",
+      "\tprint \"%d is even\"%n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter an integer: 13\n",
+        "13 is odd\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 5-5 , Page number: 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Testing if a number is divisible by 7\n",
+      "\n",
+      "print \"Enter an integer:\",\n",
+      "n = 13\n",
+      "print n\n",
+      "\n",
+      "# A number is even if it is divisible by 2\n",
+      "\n",
+      "if n % 7 == 0:\n",
+      "\tprint \"%d is divisible by 7\"%n\n",
+      "else:\n",
+      "\tprint \"%d is not divisible by 7\"%n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an integer: 13\n",
+        "13 is not divisible by 7\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 5-6 , Page number: 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Determining which of 2 numbers is larger\n",
+      "\n",
+      "print \"Enter an integer:\",\n",
+      "a = 4\n",
+      "print a\n",
+      "print \"Enter another integer: \",\n",
+      "b = 7\n",
+      "print b\n",
+      "\n",
+      "if a > b :\n",
+      "\tprint \"%d is greater than %d.\"%(a,b)\n",
+      "else:\n",
+      "\tprint \"%d is smaller than %d.\"%(a,b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an integer: 4\n",
+        "Enter another integer:  7\n",
+        "4 is smaller than 7.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter6.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter6.ipynb
new file mode 100755
index 00000000..f50ff06b
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter6.ipynb
@@ -0,0 +1,494 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c565a0de8f316cbc19d18a7bdba735b6241f6d0c0c10c2d49245aa88403df9e7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6, The While and do...While Loops"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-1 , Page number: 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A first look at the while loop\n",
+      "\n",
+      "i = 1\n",
+      "\n",
+      "while i < 11:\n",
+      "\tprint \"This is line number %d\"%i\n",
+      "\ti += 1 \t\t\t# Python doesn't support \"++\" or \"--\" shorthand expressions"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This is line number 1\n",
+        "This is line number 2\n",
+        "This is line number 3\n",
+        "This is line number 4\n",
+        "This is line number 5\n",
+        "This is line number 6\n",
+        "This is line number 7\n",
+        "This is line number 8\n",
+        "This is line number 9\n",
+        "This is line number 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-2 , Page number: 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Countdown to blastoff\n",
+      "\n",
+      "i = 10\n",
+      "\n",
+      "print \"Countdown\"\n",
+      "\n",
+      "while i >= 0 :\n",
+      "\tprint \"%d\"%i\n",
+      "\ti -= 1\n",
+      "\n",
+      "print \"\\n\\nBlastoff!\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Countdown\n",
+        "10\n",
+        "9\n",
+        "8\n",
+        "7\n",
+        "6\n",
+        "5\n",
+        "4\n",
+        "3\n",
+        "2\n",
+        "1\n",
+        "0\n",
+        "\n",
+        "\n",
+        "Blastoff!\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-3 , Page number: 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Hailstones\n",
+      "\n",
+      "counter = 0;\n",
+      "\n",
+      "print \"Enter a positive integer:\",\n",
+      "n = 51\n",
+      "print n\n",
+      "\n",
+      "nsave = n\n",
+      "\n",
+      "\n",
+      "while n > 1:\n",
+      "\tif n % 2 :\t\t\t# if (n % 2) is non-zero, n is not divisible by 2 and is therefore odd\n",
+      "\t\tn = n * 3 + 1\n",
+      "\telse :\n",
+      "\t\tn /= 2\n",
+      "\tcounter += 1\n",
+      "\tprint \"Step %4d: n = %4d\"%(counter,n)\n",
+      "\n",
+      "print \"\\n\\n%d went to 1 in %d steps.\"%(nsave,counter)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a positive integer: 51\n",
+        "Step    1: n =  154\n",
+        "Step    2: n =   77\n",
+        "Step    3: n =  232\n",
+        "Step    4: n =  116\n",
+        "Step    5: n =   58\n",
+        "Step    6: n =   29\n",
+        "Step    7: n =   88\n",
+        "Step    8: n =   44\n",
+        "Step    9: n =   22\n",
+        "Step   10: n =   11\n",
+        "Step   11: n =   34\n",
+        "Step   12: n =   17\n",
+        "Step   13: n =   52\n",
+        "Step   14: n =   26\n",
+        "Step   15: n =   13\n",
+        "Step   16: n =   40\n",
+        "Step   17: n =   20\n",
+        "Step   18: n =   10\n",
+        "Step   19: n =    5\n",
+        "Step   20: n =   16\n",
+        "Step   21: n =    8\n",
+        "Step   22: n =    4\n",
+        "Step   23: n =    2\n",
+        "Step   24: n =    1\n",
+        "\n",
+        "\n",
+        "51 went to 1 in 24 steps.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-4 , Page number: 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Computing the factorial\n",
+      "\n",
+      "print \"Type in a positive integer:\",\n",
+      "n = 6\n",
+      "print n\n",
+      "\n",
+      "nsave = n\n",
+      "factorial = 1\n",
+      "\n",
+      "while n > 1 :\t\n",
+      "\tfactorial *= n\n",
+      "\tn -= 1\n",
+      "\n",
+      "print \"Factorial of %d = %d\"%(nsave,factorial)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Type in a positive integer: 6\n",
+        "Factorial of 6 = 720\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-5 , Page number: 93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Computing the factorial (shorter version)\n",
+      "\n",
+      "print \"Type in a positive integer: \",\n",
+      "n = 6\n",
+      "print n\n",
+      "\n",
+      "nsave = n\n",
+      "factorial = 1\n",
+      "\n",
+      "while n > 1 :\n",
+      "\tfactorial *= n      # Since python doesn't support '--' expression, the given program cannot be shortened\n",
+      "\tn -= 1\n",
+      "\n",
+      "print \"Factorial of %d = %d\"%(nsave,factorial)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Type in a positive integer:  6\n",
+        "Factorial of 6 = 720\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-6 , Page number: 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of do..while loops\n",
+      "\n",
+      "# This program reads in an integer, adds up the values of each of its individual digits, and prints out the sum.\n",
+      "\n",
+      "sum = 0\n",
+      "\n",
+      "print \"Type in an integer:\",\n",
+      "n = 1776\n",
+      "print n\n",
+      "\n",
+      "# There is no do...while loop in python\n",
+      "# An equivalent while loop can work\n",
+      "\n",
+      "while n > 0 :\n",
+      "\trightmost_digit = n % 10\t# Extract rightmost digit\n",
+      "\tsum += rightmost_digit\n",
+      "\tn /= 10\t\t\t\t\t\t# Move next digit into rightmost position\n",
+      "\n",
+      "print \"The sum of the digits is %d\"%sum\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Type in an integer: 1776\n",
+        "The sum of the digits is 21\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-7 , Page number: 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "\n",
+      "# This program reads in an integer and prints it out backwards\n",
+      "\n",
+      "print \"Type in an integer: \",\n",
+      "n = 5746\n",
+      "print n\n",
+      "\n",
+      "print \"%d backwards is \"%n,\n",
+      "\n",
+      "while n > 0 :\n",
+      "\trightmost_digit = n % 10\n",
+      "\tsys.stdout.write(\"%d\"%rightmost_digit)  # For printing in the same line without implicit whitespaces\n",
+      "\tn /= 10\n",
+      "print \t\t\t# For the new line\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Type in an integer:  5746\n",
+        "5746 backwards is 6475\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-8 , Page number: 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A program to read in numbers and print out their squares\n",
+      "\n",
+      "print \"Enter an integer (negative to quit):\",\n",
+      "n = 5\n",
+      "print n\n",
+      "\n",
+      "while n > 0 :\n",
+      "\tprint \"%d squared is %d\"%(n,n*n)\n",
+      "\tprint \"\\nEnter an integer (negative to quit):\",\n",
+      "\tn = -1\n",
+      "\tprint n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter an integer (negative to quit): 5\n",
+        "5 squared is 25\n",
+        "\n",
+        "Enter an integer (negative to quit): -1\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-9 , Page number: 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A program to read in radius values and print out the circumference and area\n",
+      "\n",
+      "pi = 3.1416\n",
+      "\n",
+      "print \"Enter radius (negative to quit):\",\n",
+      "radius = 4.5\n",
+      "print radius\n",
+      "\n",
+      "while radius >= 0.0 :\n",
+      "\tprint \"The circumference is %f\"%(2.0 * pi * radius)\n",
+      "\tprint \"The area is %f\"%(pi * radius * radius)\n",
+      "\tprint \"\\nEnter radius (negative to quit):\",\n",
+      "\tradius = -1.0\n",
+      "\tprint radius\n",
+      "\t\n",
+      "\t"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter radius (negative to quit): 4.5\n",
+        "The circumference is 28.274400\n",
+        "The area is 63.617400\n",
+        "\n",
+        "Enter radius (negative to quit): -1.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 6-10 , Page number: 101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# A program to read in length and width and print out the perimeter and area\n",
+      "\n",
+      "print \"Enter length and width\"\n",
+      "print \" (one or both negative to quit):\",\n",
+      "length = 4\n",
+      "width = 5\n",
+      "print length,width\n",
+      "\n",
+      "while length > 0 and width > 0 : \n",
+      "\tprint \"The perimeter is %d\"%(2*(length+width))\n",
+      "\tprint \"The area is %d\"%(length*width)\n",
+      "\tprint \"\\nEnter length and width\"\n",
+      "\tprint \" (one or both negative to quit):\",\n",
+      "\tlength = -1\n",
+      "\twidth = 0\n",
+      "\tprint length,width\t"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter length and width\n",
+        " (one or both negative to quit): 4 5\n",
+        "The perimeter is 18\n",
+        "The area is 20\n",
+        "\n",
+        "Enter length and width\n",
+        " (one or both negative to quit): -1 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter7.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter7.ipynb
new file mode 100755
index 00000000..33f7a4f3
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter7.ipynb
@@ -0,0 +1,690 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:245ea275fa7c7004a79eeeeb6b3fc9780b5a19707d8b95c7a675d2142428b630"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7, The for loop"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-1 , Page number: 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A simple illustration of for loop\n",
+      "\n",
+      "for i in range(1,11,1):\t\t# The loop starts with i=1 and continues until i<11 with its value increasing by 1 each time \n",
+      "\tprint \"This is line number %d\"%i"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This is line number 1\n",
+        "This is line number 2\n",
+        "This is line number 3\n",
+        "This is line number 4\n",
+        "This is line number 5\n",
+        "This is line number 6\n",
+        "This is line number 7\n",
+        "This is line number 8\n",
+        "This is line number 9\n",
+        "This is line number 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-2 , Page number: 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Another simple illustration of a for loop\n",
+      "\n",
+      "print \"Countdown\"\n",
+      "\n",
+      "for i in range(10,-1,-1):\n",
+      "\tprint i\n",
+      "print \"\\n\\nBlastoff!\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Countdown\n",
+        "10\n",
+        "9\n",
+        "8\n",
+        "7\n",
+        "6\n",
+        "5\n",
+        "4\n",
+        "3\n",
+        "2\n",
+        "1\n",
+        "0\n",
+        "\n",
+        "\n",
+        "Blastoff!\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-3 , Page number: 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculating the squares of integers 1 through 10\n",
+      "\n",
+      "for i in range (1,11,1):\n",
+      "\tprint \"%d squared = %d\"%(i,i*i)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 squared = 1\n",
+        "2 squared = 4\n",
+        "3 squared = 9\n",
+        "4 squared = 16\n",
+        "5 squared = 25\n",
+        "6 squared = 36\n",
+        "7 squared = 49\n",
+        "8 squared = 64\n",
+        "9 squared = 81\n",
+        "10 squared = 100\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-4 , Page number: 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A better looking program to calculate squares\n",
+      "\n",
+      "print \"Table of Results\\n\"\n",
+      "print \"   i    i squared\"\n",
+      "print \"   -    ---------\"\n",
+      "\n",
+      "for i in range(1,11,1):\n",
+      "\tprint \"%4d%13d\"%(i,i*i)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Table of Results\n",
+        "\n",
+        "   i    i squared\n",
+        "   -    ---------\n",
+        "   1            1\n",
+        "   2            4\n",
+        "   3            9\n",
+        "   4           16\n",
+        "   5           25\n",
+        "   6           36\n",
+        "   7           49\n",
+        "   8           64\n",
+        "   9           81\n",
+        "  10          100\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-5 , Page number: 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Finding the sum of integers 1 through 100\n",
+      "\n",
+      "sum = 0\n",
+      "n = 100\n",
+      "\n",
+      "for i in range(1,n+1,1):\t\t# The loop starts with i=1 and continues until i<11 with its value increasing by 1 each time \n",
+      "\tsum += i\t\n",
+      "\n",
+      "print \"The sum of integers from  1 to %d is %d\"%(n,sum)\n",
+      "print \"\\n\\nThis is confirmed by the Gauss Method: %d\"%(n*(n+1)/2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of integers from  1 to 100 is 5050\n",
+        "\n",
+        "\n",
+        "This is confirmed by the Gauss Method: 5050\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-6 , Page number: 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Finding the sum of the integers 1 through MAX_VAL\n",
+      "\n",
+      "# There is no #define command for defining constants\n",
+      "# We can use a function that returns the value (although it is not very efficient to do so)\n",
+      "\n",
+      "def MAX_VAL():\n",
+      "\treturn 100\t\t\t# The value of MAX_VAL\n",
+      "\n",
+      "sum = 0\n",
+      "\n",
+      "for i in range (0,MAX_VAL()+1,1):\n",
+      "\tsum += i\n",
+      "\n",
+      "print \"The sum of the integers from 1 to %d is %d\"%(MAX_VAL(),sum)\n",
+      "print \"\\n\\nThis is confirmed by the Gauss Method : %d\"%(MAX_VAL()*(MAX_VAL()+1)/2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The sum of the integers from 1 to 100 is 5050\n",
+        "\n",
+        "\n",
+        "This is confirmed by the Gauss Method : 5050\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-7 , Page number: 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculating the sum of integers 1 through n\n",
+      "\n",
+      "sum = 0\n",
+      "\n",
+      "print \"Please type in a positive integer:\",\n",
+      "n = 50\n",
+      "print n\n",
+      "\n",
+      "for i in range (1,n+1,1):\n",
+      "\tsum += i\n",
+      "\n",
+      "print \"\\nThe sum of the integers from 1 to %d is: %d\"%(n,sum)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please type in a positive integer: 50\n",
+        "\n",
+        "The sum of the integers from 1 to 50 is: 1275\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-8 , Page number: 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculating the sum of integers low through high\n",
+      "\n",
+      "sum = 0\n",
+      "\n",
+      "print \"Please type in the low bound:\",\n",
+      "low = 3\n",
+      "print low\n",
+      "print \"Please type in the high bound:\",\n",
+      "high = 9\n",
+      "print high\n",
+      "\n",
+      "for i in range(low,high+1,1):\n",
+      "\tsum += i\n",
+      "\n",
+      "print \"\\nThe sum of integers from %d to %d is: %d\"%(low,high,sum)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please type in the low bound: 3\n",
+        "Please type in the high bound: 9\n",
+        "\n",
+        "The sum of integers from 3 to 9 is: 42\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-9 , Page number: 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculating the sum of integers low through high with an exchange if necessary\n",
+      "\n",
+      "sum = 0\n",
+      "\n",
+      "print \"Please type in the low bound:\",\n",
+      "low = 30\n",
+      "print low\n",
+      "print \"Please type in the high bound:\",\n",
+      "high = 15\n",
+      "print high\n",
+      "\n",
+      "if high < low :\n",
+      "\tprint \"Oops! Wrong order.\"\n",
+      "\tprint \"Never mind, I'll switch them.\"\n",
+      "\t# Code for swapping using temporary variable\n",
+      "\ttemp = low\n",
+      "\tlow = high\n",
+      "\thigh = temp\n",
+      "\n",
+      "for i in range(low,high+1,1):\n",
+      "\tsum += i\n",
+      "\n",
+      "print \"\\nThe sum of integers from %d to %d is: %d\"%(low,high,sum)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please type in the low bound: 30\n",
+        "Please type in the high bound: 15\n",
+        "Oops! Wrong order.\n",
+        "Never mind, I'll switch them.\n",
+        "\n",
+        "The sum of integers from 15 to 30 is: 360\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-10 ,Page number: 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of a nest of for loops\n",
+      "\n",
+      "n = 5\n",
+      "for i in range(1,7,1):\n",
+      "\tprint \"\\n\\nPlease type in a positive integer: \",\n",
+      "\tprint n\n",
+      "\n",
+      "\tsum = 0\n",
+      "\tfor j in range(1,n+1,1):\n",
+      "\t\tsum += j\n",
+      "\tprint \"\\nThe sum of integers from 1 to %d is %d\"%(n,sum)\n",
+      "\t\n",
+      "\tn += 5\t\t\t# Just to match the exercise's values"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Please type in a positive integer:  5\n",
+        "\n",
+        "The sum of integers from 1 to 5 is 15\n",
+        "\n",
+        "\n",
+        "Please type in a positive integer:  10\n",
+        "\n",
+        "The sum of integers from 1 to 10 is 55\n",
+        "\n",
+        "\n",
+        "Please type in a positive integer:  15\n",
+        "\n",
+        "The sum of integers from 1 to 15 is 120\n",
+        "\n",
+        "\n",
+        "Please type in a positive integer:  20\n",
+        "\n",
+        "The sum of integers from 1 to 20 is 210\n",
+        "\n",
+        "\n",
+        "Please type in a positive integer:  25\n",
+        "\n",
+        "The sum of integers from 1 to 25 is 325\n",
+        "\n",
+        "\n",
+        "Please type in a positive integer:  30\n",
+        "\n",
+        "The sum of integers from 1 to 30 is 465\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-11 , Page number: 130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Another illustration of a nest of for loops\n",
+      "\n",
+      "print \"How many values of n do you wish to enter? \",\n",
+      "runs = 3\n",
+      "print runs\n",
+      "print \"\\n\\n\"\n",
+      "\n",
+      "n = 9\n",
+      "\n",
+      "for i in range(1,runs+1,1) :\n",
+      "\tprint \"\\n\\nPLease type in a positive integer: \",\n",
+      "\tprint n\n",
+      "\n",
+      "\tsum = 0\n",
+      "\tfor j in range(1,n+1,1) :\n",
+      "\t\tsum += j\n",
+      "\tprint \"\\nThe sum of the integers from 1 to %d is: %d\"%(n,sum)\n",
+      "\tn *= 3\t\t\t# Just a random increment"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many values of n do you wish to enter?  3\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "PLease type in a positive integer:  9\n",
+        "\n",
+        "The sum of the integers from 1 to 9 is: 45\n",
+        "\n",
+        "\n",
+        "PLease type in a positive integer:  27\n",
+        "\n",
+        "The sum of the integers from 1 to 27 is: 378\n",
+        "\n",
+        "\n",
+        "PLease type in a positive integer:  81\n",
+        "\n",
+        "The sum of the integers from 1 to 81 is: 3321\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-12 , Page number: 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Newton-Rapshon method for calculating square roots\n",
+      "\n",
+      "def EPSILON() :\n",
+      "\t\"\"\" USING AS A CONSTANT \"\"\"\n",
+      "\treturn 0.0001\n",
+      "\n",
+      "print \"What number do you want to kate the square root of?\",\n",
+      "n = 17.0\n",
+      "print n\n",
+      "\n",
+      "while n < 0.0 :\n",
+      "\tprint \"Please enter a non-negative number!\"\n",
+      "\tn = 3 \t\t\t\t\t# Just a random positive number\n",
+      "\n",
+      "# The for loop statement given in the example(from the book) is not possible to code in python\n",
+      "# So, breaking it down to while loop\n",
+      "\n",
+      "guess = n / 2.0\n",
+      "while (guess * guess - n > EPSILON() or guess * guess - n < -EPSILON()) :\n",
+      "\tguess = (guess + n/guess) / 2.0\n",
+      "\t\n",
+      "print \"I calculate the square root of %f to be %.4f\"%(n,guess)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "What number do you want to kate the square root of? 17.0\n",
+        "I calculate the square root of 17.000000 to be 4.1231\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-13 , Page number: 133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# General addition table for a number n\n",
+      "\n",
+      "print \"What number do you want a table for?\",\n",
+      "n = 10\n",
+      "print n\n",
+      "\n",
+      "j = n\n",
+      "for i in range (0,n+1,1) :\n",
+      "\tprint \"%3d + %3d = %3d\"%(i,j,n)\n",
+      "\tj -= 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "What number do you want a table for? 10\n",
+        "  0 +  10 =  10\n",
+        "  1 +   9 =  10\n",
+        "  2 +   8 =  10\n",
+        "  3 +   7 =  10\n",
+        "  4 +   6 =  10\n",
+        "  5 +   5 =  10\n",
+        "  6 +   4 =  10\n",
+        "  7 +   3 =  10\n",
+        "  8 +   2 =  10\n",
+        "  9 +   1 =  10\n",
+        " 10 +   0 =  10\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 7-14 , Page number: 135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Generate addition table for a number n, shorter loop\n",
+      "\n",
+      "print \"What number do you want a table for?\",\n",
+      "n = 10\n",
+      "print n\n",
+      "\n",
+      "# Since the range() function supports at max 3 arguments, no possible shortening is possilbe in python\n",
+      "j = n\n",
+      "for i in range (0,n+1,1):\n",
+      "\tprint \"%3d + %3d = %3d\"%(i,j,n)\n",
+      "\tj -= 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "What number do you want a table for? 10\n",
+        "  0 +  10 =  10\n",
+        "  1 +   9 =  10\n",
+        "  2 +   8 =  10\n",
+        "  3 +   7 =  10\n",
+        "  4 +   6 =  10\n",
+        "  5 +   5 =  10\n",
+        "  6 +   4 =  10\n",
+        "  7 +   3 =  10\n",
+        "  8 +   2 =  10\n",
+        "  9 +   1 =  10\n",
+        " 10 +   0 =  10\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter8.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter8.ipynb
new file mode 100755
index 00000000..baae6f9a
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter8.ipynb
@@ -0,0 +1,567 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:db9d52743fdd42516fb59e7618f0fc71d76d4ba3cb21fd098bf4fd004c40683b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8, A Detailed Look at the printf and scanf Functions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-1 , Page number: 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A print demo\n",
+      "# The conversion specification for numbers.\n",
+      "\n",
+      "print \"Case  1 :%d:\"%123\n",
+      "\n",
+      "print \"Case  2 :%0d:\"%123\n",
+      "\n",
+      "print \"Case  3 :%2d:\"%123\n",
+      "\n",
+      "print \"Case  4 :%8d:\"%123\n",
+      "\n",
+      "print \"Case  5 :%-8d:\"%123\n",
+      "\n",
+      "print \"Case  6 :%8d:\"%-123\n",
+      "\n",
+      "print \"Case  7 :%f:\"%123.456\n",
+      "\n",
+      "print \"Case  8 :%f:\"%123.4567896\n",
+      "\n",
+      "print \"Case  9 :%e:\"%123.456\n",
+      "\n",
+      "print \"Case  10 :%e:\"%1.23456e2\n",
+      "\n",
+      "print \"Case  11 :%e:\"%123.456e+0\n",
+      "\n",
+      "print \"Case  12 :%f:\"%1.23456E+02\n",
+      "\n",
+      "print \"Case  13 :%4.2f:\"%123.456\n",
+      "\n",
+      "print \"Case  14 :%8.2f:\"%123.456\n",
+      "\n",
+      "print \"Case  15 :%.3e:\"%123.456\n",
+      "\n",
+      "print \"Case  16 :%12.3e:\"%123.456\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Case  1 :123:\n",
+        "Case  2 :123:\n",
+        "Case  3 :123:\n",
+        "Case  4 :     123:\n",
+        "Case  5 :123     :\n",
+        "Case  6 :    -123:\n",
+        "Case  7 :123.456000:\n",
+        "Case  8 :123.456790:\n",
+        "Case  9 :1.234560e+02:\n",
+        "Case  10 :1.234560e+02:\n",
+        "Case  11 :1.234560e+02:\n",
+        "Case  12 :123.456000:\n",
+        "Case  13 :123.46:\n",
+        "Case  14 :  123.46:\n",
+        "Case  15 :1.235e+02:\n",
+        "Case  16 :   1.235e+02:\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-2 , Page number: 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# A print demo:\n",
+      "# Interpretation of values\n",
+      "\n",
+      "# python is dynamically-typed language\n",
+      "# which is why declaration of variables is omitted\n",
+      "n = -123\n",
+      "c = \"V\"\n",
+      "\n",
+      "print \"%d\"%n\n",
+      "print \"%u\"%n\t\t# python correctly interprets the negative(unsigned) integer\n",
+      "\n",
+      "print \"%c\"%c\n",
+      "# print \"%d\"%c  \t# python doesn't support implicit type-casting from integer to character or character to integer\n",
+      "print \"%c\"%86"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-123\n",
+        "-123\n",
+        "V\n",
+        "V\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-3 , Page number: 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Printing out the ASCII value of a character\n",
+      "\n",
+      "# There is no do..while loop in python\n",
+      "# Hence, we will be using while loop \n",
+      "\n",
+      "\n",
+      "answer = \"y\"\n",
+      "while answer == \"y\" :\n",
+      "\tprint \"Enter a character:\",\n",
+      "\tin_char = \"a\"\n",
+      "\tprint in_char\n",
+      "\n",
+      "\tprint \"The ASCII value of %c is %d\"%(in_char,ord(in_char)) \t# ord() function returns the ascii(int) value of a character\n",
+      "\n",
+      "\tprint \"\\nAnother? (y/n)\",\n",
+      "\tanswer = \"n\"\t\t\t# To terminate the loop\n",
+      "\tprint answer\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a character: a\n",
+        "The ASCII value of a is 97\n",
+        "\n",
+        "Another? (y/n) n\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-4 , Page number: 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Converting a digit to a number by using its ASCII value\n",
+      "\n",
+      "print \"Enter a digit:\",\n",
+      "digit = \"A\"\n",
+      "print digit\n",
+      "\n",
+      "if digit < \"0\" or digit > \"9\" :\n",
+      "\tprint \"That's not a digit\"\n",
+      "else :\n",
+      "\tprint \"The digit you typed is %c\"%digit"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a digit: A\n",
+        "That's not a digit\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-5 , Page number: 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Converting a letter from lower to upper case or vice-versa\n",
+      "\n",
+      "answer = \"y\"\n",
+      "while answer == \"y\" :\n",
+      "\tprint \"\\n\\n\\nEnter a letter:\",\n",
+      "\tletter = \"A\"\n",
+      "\tprint letter\n",
+      "\n",
+      "\tif letter >= \"A\" and letter < \"Z\" :\t\t# Letter is uppercase\n",
+      "\t\tprint \"Lower case is: %c\"%chr(ord(letter) + (ord(\"a\")-ord(\"A\")))\n",
+      "\telif letter >= \"a\" and letter <= \"z\" :\t# Letter is lowercase\n",
+      "\t\tprint \"Upper case is: %c\"%chr(ord(letter) - (ord(\"a\")-ord(\"A\")))\n",
+      "\telse :\n",
+      "\t\tprint \"That's not a letter\"\n",
+      "\n",
+      "\t# Ask if the user wants to go again\n",
+      "\tprint \"\\nAnother? (y/n)\",\n",
+      "\tanswer = \"n\"\t\t# To terminate the loop\n",
+      "\tprint answer"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "\n",
+        "Enter a letter: A\n",
+        "Lower case is: a\n",
+        "\n",
+        "Another? (y/n) n\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-6 , Page number: 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Example of using control characters\n",
+      "\n",
+      "print \"The word apple\\b\\b\\b\\b\\b_____ is underlined.\\n\"\n",
+      "print \"The last word in this sentence is way over\\t\\there.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The word apple\b\b\b\b\b_____ is underlined.\n",
+        "\n",
+        "The last word in this sentence is way over\t\there.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-7 , Page number: 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# How to print special characters\n",
+      "\n",
+      "print \"This is how to print a double-quote: \\\"\"\n",
+      "print \"This is how to print a backslash: \\\\\"\n",
+      "# this is how to write an apostrophe character\n",
+      "c = '\\''\n",
+      "print \"c now contains: %c\"%c\n",
+      "print \"This is how to print a percent-sign: %\"\t\t# no escape sequence required for % in python"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This is how to print a double-quote: \"\n",
+        "This is how to print a backslash: \\\n",
+        "c now contains: '\n",
+        "This is how to print a percent-sign: %\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-8 , Page number: 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Octal and hexadecimal notation\n",
+      "\n",
+      "a = 27\n",
+      "b = 033\n",
+      "c = 0x1b\n",
+      "\n",
+      "print \"a = %d, b = %d, c = %d\"%(a,b,c)\n",
+      "print \"a = %o, b = %o, c = %o\"%(a,b,c)\n",
+      "print \"a = %x, b = %x, c = %x\"%(a,b,c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = 27, b = 27, c = 27\n",
+        "a = 33, b = 33, c = 33\n",
+        "a = 1b, b = 1b, c = 1b\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-9 , Page number: 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of some ways to print a string\n",
+      "\n",
+      "print \"This is simply a control string.\"\n",
+      "\n",
+      "print \"%s\"%(\"The control string just tells how to print this.\")\n",
+      "\n",
+      "print \"%s\"%(\"Newline may be put here, instead!.\\n\"),\n",
+      "\n",
+      "print \"%s %s %s %s %s\"%(\"This\",\"is\",\"valid\",\"in\",\"C.\")\n",
+      "\n",
+      "print \"%s %s also %s %s %s\"%(\"This\",\"is\",\"valid\",\"in\",\"C.\")\n",
+      "\n",
+      "print \"%s %s %s %s %s %s %s\"%(\"This\",\"is\",\"valid\",\"in\",\"C\",\"also.\",\"\\n\"),"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This is simply a control string.\n",
+        "The control string just tells how to print this.\n",
+        "Newline may be put here, instead!.\n",
+        "This is valid in C.\n",
+        "This is also valid in C.\n",
+        "This is valid in C also. \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-10 , Page number: 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# %s conversion specification demonstrated\n",
+      "\n",
+      "print \"Case 1 :%s:\"%(\"A string\")\n",
+      "\n",
+      "print \"Case 2 :%3s:\"%(\"A string\")\n",
+      "\n",
+      "print \"Case 3 :%12s:\"%(\"A string\")\n",
+      "\n",
+      "print \"Case 4 :%-12s:\"%(\"A string\")\n",
+      "\n",
+      "print \"Case 5 :%12.6s:\"%(\"A string\")\n",
+      "\n",
+      "print \"Case 6 :%12.12s:\"%(\"A string\")\n",
+      "\n",
+      "print \"Case 7 :%.6s:\"%(\"A string\")\n",
+      "\n",
+      "print \"Case 8 :%-12.6s:\"%(\"A string\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Case 1 :A string:\n",
+        "Case 2 :A string:\n",
+        "Case 3 :    A string:\n",
+        "Case 4 :A string    :\n",
+        "Case 5 :      A stri:\n",
+        "Case 6 :    A string:\n",
+        "Case 7 :A stri:\n",
+        "Case 8 :A stri      :\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-11 , Page number: 164 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Some miscellaneous conversion sppecifications\n",
+      "\n",
+      "a = 123456789L\n",
+      "\n",
+      "print \"%d\"%a\t\t\t# in python both int and long are unified..so %d specifier valid unlike in C\n",
+      "print \"%ld\"%a\n",
+      "print \":%c:\"%(\"V\")\n",
+      "print \":%0c:\"%(\"V\")\n",
+      "print \":%5c:\"%(\"V\")\n",
+      "print \":%-5c:\"%(\"V\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "123456789\n",
+        "123456789\n",
+        ":V:\n",
+        ":V:\n",
+        ":    V:\n",
+        ":V    :\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 8-13 , Page number: 171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Reading decimal,octal and hex numbers\n",
+      "\n",
+      "print \"Enter a number in decimal:\",\n",
+      "a = 27\n",
+      "print a\n",
+      "print \"You entered %d\"%a\n",
+      "\n",
+      "print \"\\nEnter a number in octal:\",\n",
+      "a = 033\t\t\t# O is added in the front to specify octal value\n",
+      "print a\n",
+      "print \"You entered %o, or %d in decimal\"%(a,a)\n",
+      "\n",
+      "print \"\\nEnter a number in hex:\",\n",
+      "a = 0x1B\n",
+      "print a\n",
+      "print \"You entered %x, or %d in decimal\"%(a,a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number in decimal: 27\n",
+        "You entered 27\n",
+        "\n",
+        "Enter a number in octal: 27\n",
+        "You entered 33, or 27 in decimal\n",
+        "\n",
+        "Enter a number in hex: 27\n",
+        "You entered 1b, or 27 in decimal\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter9.ipynb b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter9.ipynb
new file mode 100755
index 00000000..ae4771e3
--- /dev/null
+++ b/The_Spirit_of_C_by_Henry_Mullish_and_Herbert_L_Cooper/chapter9.ipynb
@@ -0,0 +1,590 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:13a3347e13ed5ee09ae2833e49a517873801778756103a4da905a96a5997fa89"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9, User-Defined Functions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-1 , Page number: 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Using a function as an abbreviation\n",
+      "\n",
+      "# In python, the user defined functions have to defined before making a call to them\n",
+      "\n",
+      "def chorus() :\n",
+      "\tprint \"%60s\"%\"With a knick-knack-paddy-whack,\"\n",
+      "\tprint \"%60s\"%\"Give the dog a bone,\"\n",
+      "\tprint \"%60s\\n\"%\"This old man went rolling home.\"\n",
+      "\n",
+      "print \"This old man, he played two,\"\n",
+      "print \"He played knick-knack on my thumb,\"\n",
+      "chorus()\n",
+      "\n",
+      "print \"This old man, he played one,\"\n",
+      "print \"He played knick-knack on my shoe,\"\n",
+      "chorus()\n",
+      "\n",
+      "print \"This old man, he played three,\"\n",
+      "print \"He played knick-knack on my knee,\"\n",
+      "chorus()\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "This old man, he played two,\n",
+        "He played knick-knack on my thumb,\n",
+        "                             With a knick-knack-paddy-whack,\n",
+        "                                        Give the dog a bone,\n",
+        "                             This old man went rolling home.\n",
+        "\n",
+        "This old man, he played one,\n",
+        "He played knick-knack on my shoe,\n",
+        "                             With a knick-knack-paddy-whack,\n",
+        "                                        Give the dog a bone,\n",
+        "                             This old man went rolling home.\n",
+        "\n",
+        "This old man, he played three,\n",
+        "He played knick-knack on my knee,\n",
+        "                             With a knick-knack-paddy-whack,\n",
+        "                                        Give the dog a bone,\n",
+        "                             This old man went rolling home.\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-2 , Page number: 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Demonstration of one use of a function\n",
+      "\n",
+      "def get_an_int() :\n",
+      "\t\"\"\"RETURNS AN INTEGER\"\"\"\t\t# This statement is just a description of the function (specific feature of python)\n",
+      "\treturn 5\n",
+      "\n",
+      "print \"Enter the length of a rectangle:\",\n",
+      "length = get_an_int()\n",
+      "print length\n",
+      "\n",
+      "print \"Enter the width:\",\n",
+      "width = get_an_int()\n",
+      "print width\n",
+      "\n",
+      "print \"\\nThe area is %d\"%(length*width)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the length of a rectangle: 5\n",
+        "Enter the width: 5\n",
+        "\n",
+        "The area is 25\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-3 , Page number: 191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# An attempt at communication between functions\n",
+      "\n",
+      "def modify_i() :\n",
+      "\ti = 3\n",
+      "\t\n",
+      "i = 1\n",
+      "print \"i is now equal to %d\"%i\n",
+      "\n",
+      "modify_i()\n",
+      "print \"i is now equal to %d\"%i\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i is now equal to 1\n",
+        "i is now equal to 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-4 , Page number: 192"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Demonstration of the use of a global variable\n",
+      "\n",
+      "# Making variable global\n",
+      "\n",
+      "def get_an_int() :\n",
+      "\treturn 5\n",
+      "\n",
+      "def calc_area() :\n",
+      "\tarea = length*width\n",
+      "\n",
+      "globals()\t\t\t# The command simply makes all the variables declared so far as global variables\n",
+      "print \"Enter the length of a rectangle:\",\n",
+      "length = get_an_int()\n",
+      "print length\n",
+      "\n",
+      "print \"Enter the width:\",\n",
+      "width = get_an_int()\n",
+      "print width\n",
+      "\n",
+      "calc_area()\n",
+      "print \"\\nThe area is %d\"%(length*width)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the length of a rectangle: 5\n",
+        "Enter the width: 5\n",
+        "\n",
+        "The area is 25\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-5 , Page number: 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Introduction to parameters\n",
+      "\n",
+      "def print_an_int(i) :\n",
+      "\t\"\"\"The function prints an integer passed as a parameter\"\"\"\n",
+      "\tprint i\n",
+      "\n",
+      "\n",
+      "n = 3\n",
+      "\n",
+      "print \"Passing a constant:\",\n",
+      "print_an_int(5)\n",
+      "\n",
+      "print \"\\n\\nPassing a variable's value:\",\n",
+      "print_an_int(n)\n",
+      "\n",
+      "print \"\\n\\nPassing the value of an expression:\",\n",
+      "print_an_int(n * 4 + 1)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Passing a constant: 5\n",
+        "\n",
+        "\n",
+        "Passing a variable's value: 3\n",
+        "\n",
+        "\n",
+        "Passing the value of an expression: 13\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-6 , Page number: 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Demonstration of how a procedure cannot modify the contents of a variable passed as a parameter\n",
+      "\n",
+      "def modify(i) :\n",
+      "\ti = 3\n",
+      "\n",
+      "i = 1\n",
+      "\n",
+      "print \"i is now equal to %d\"%i\n",
+      "\n",
+      "modify(i)\n",
+      "\n",
+      "print \"i is now equal to %d\"%i"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i is now equal to 1\n",
+        "i is now equal to 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-7 , Page number: 197 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Hailstones, demonstrating the use of a parameter\n",
+      "\n",
+      "def step(i) :\n",
+      "\t\"\"\"This function returns its parameter i divided by 2 if i is even else it returns i multiplied by 3 and incremented by 1\"\"\"\n",
+      "\tif (i % 2) :\n",
+      "\t\treturn (i * 3 + 1)\n",
+      "\telse : \n",
+      "\t\treturn (i / 2)\n",
+      "\n",
+      "counter = 0\n",
+      "\n",
+      "print \"Enter a positive integer\",\n",
+      "n = 25\n",
+      "print n\n",
+      "\n",
+      "nsave = n\n",
+      "\n",
+      "while n > 1 :\n",
+      "\tn = step(n)\n",
+      "\tcounter += 1\n",
+      "\tprint \"Step %4d: n = %4d\"%(counter,n)\n",
+      "\n",
+      "print \"\\n\\n%d went to 1 in %d steps.\"%(nsave,counter)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a positive integer 25\n",
+        "Step    1: n =   76\n",
+        "Step    2: n =   38\n",
+        "Step    3: n =   19\n",
+        "Step    4: n =   58\n",
+        "Step    5: n =   29\n",
+        "Step    6: n =   88\n",
+        "Step    7: n =   44\n",
+        "Step    8: n =   22\n",
+        "Step    9: n =   11\n",
+        "Step   10: n =   34\n",
+        "Step   11: n =   17\n",
+        "Step   12: n =   52\n",
+        "Step   13: n =   26\n",
+        "Step   14: n =   13\n",
+        "Step   15: n =   40\n",
+        "Step   16: n =   20\n",
+        "Step   17: n =   10\n",
+        "Step   18: n =    5\n",
+        "Step   19: n =   16\n",
+        "Step   20: n =    8\n",
+        "Step   21: n =    4\n",
+        "Step   22: n =    2\n",
+        "Step   23: n =    1\n",
+        "\n",
+        "\n",
+        "25 went to 1 in 23 steps.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-8 , Page number: 200"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Square root program demonstrating a non-integer function and parameter\n",
+      "\n",
+      "def EPSILON() :\n",
+      "\t\"\"\"replacement of #define constant\"\"\"\n",
+      "\treturn 0.0001\n",
+      "\n",
+      "def sqrt(n) :\n",
+      "\t\"\"\"returns the square root of the number supplied as parameter using Newton-Raphson method\"\"\"\n",
+      "\tif n > 0.0 :\n",
+      "\t\t# The for loop statement given in the example(from the book) is not possible to code in python\n",
+      "\t\t# So, breaking it down to while loop\n",
+      "\t\tguess = n / 2.0\n",
+      "\t\twhile (guess * guess - n > EPSILON() or guess * guess - n < -EPSILON()) :\n",
+      "\t\t\tguess = (guess + n/guess) / 2.0\n",
+      "\t\treturn guess\n",
+      "\telse :\n",
+      "\t\treturn (-1.0)\n",
+      "\n",
+      "print \"Enter the number whose square root you want:\",\n",
+      "x = 17.1\n",
+      "print x\n",
+      "\n",
+      "print \"The square root of %f is %f\"%(x,sqrt(x))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the number whose square root you want: 17.1\n",
+        "The square root of 17.100000 is 4.135215\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-9 , Page number: 201"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Testing for leap years using a boolean function\n",
+      "\n",
+      "def is_leap(year) :\n",
+      "\t\"\"\"A function that returns true if the given year is a leap year else returns false\"\"\"\n",
+      "\treturn (year % 4 == 0 and year % 100 != 0 or year % 400 == 0)\n",
+      "\n",
+      "print \"Please enter a year:\",\n",
+      "year = 2000\n",
+      "print year\n",
+      "\n",
+      "if (is_leap(year)) :\n",
+      "\tprint \"%d is a leap year.\"%year\n",
+      "else :\n",
+      "\tprint \"%d is not a leap year.\"%year"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Please enter a year: 2000\n",
+        "2000 is a leap year.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-10 , Page number: 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Demonstration of the use of multiple parameters\n",
+      "\n",
+      "def get_an_int() :\n",
+      "\t\"\"\"returns a random integer\"\"\"\n",
+      "\treturn 7\n",
+      "\n",
+      "def area(length,width) :\n",
+      "\t\"\"\"returns the area of the rectangle using its parameters length and width\"\"\"\n",
+      "\treturn length * width\n",
+      "\n",
+      "print \"Enter the length of a rectangle:\",\n",
+      "length = get_an_int()\n",
+      "print length\n",
+      "\n",
+      "print \"Enter the width:\",\n",
+      "width = get_an_int()\n",
+      "print width\n",
+      "\n",
+      "print \"\\nThe area is %d.\"%area(length,width)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the length of a rectangle: 7\n",
+        "Enter the width: 7\n",
+        "\n",
+        "The area is 49.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Program 9-11 , Page number: 204"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Illustration of a 2-parameter function: raising a number to a power \n",
+      "\n",
+      "def power(i,x) :\n",
+      "\t\"\"\"returns the values of i to the power of x\"\"\"\n",
+      "\taccum = 1\n",
+      "\tif x < 0 :\n",
+      "\t\treturn 0\n",
+      "\tfor index in range (0,x,1) :\n",
+      "\t\taccum *= i\n",
+      "\treturn accum\n",
+      "\n",
+      "answer = \"y\"\n",
+      "\n",
+      "while answer == \"y\" or answer == \"Y\" :\n",
+      "\tprint \"\\nPlease enter an integer:\",\n",
+      "\tnumber = 4\n",
+      "\tprint number\n",
+      "\n",
+      "\tprint \"Please enter a non-negative integral power\"\n",
+      "\tprint \"     to which the first number is to be raised\",\n",
+      "\texponent = 2\n",
+      "\tprint exponent\n",
+      "\n",
+      "\twhile exponent < 0 :\n",
+      "\t\tprint \"***\\7Please make that a non-negative integer\"\n",
+      "\t\texponent = 2 \t\t\t\t# randomly making the exponent non-negative\n",
+      "\n",
+      "\tprint \"%d raised to the power %d is %ld.\"%(number,exponent,power(number,exponent))\n",
+      "\n",
+      "\t# see if the user wants to keep going\n",
+      "\n",
+      "\tprint \"\\nAnother calculation? (y/n)\",\n",
+      "\tanswer = \"n\"\n",
+      "\tprint answer"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Please enter an integer: 4\n",
+        "Please enter a non-negative integral power\n",
+        "     to which the first number is to be raised 2\n",
+        "4 raised to the power 2 is 16.\n",
+        "\n",
+        "Another calculation? (y/n) n\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Thermal_Engineering_by_K_K_Ramalingam/README.txt b/Thermal_Engineering_by_K_K_Ramalingam/README.txt
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deleted file mode 100755
index 3e2c2051..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter1.ipynb
+++ /dev/null
@@ -1,387 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 1: Introduction and Basic Concepts"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-2, Page No.8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "p=850;# density in kg/m^3\n",
-      "V=2;  # volumne of tank in m^3\n",
-      "\n",
-      "#Calculations\n",
-      "m=p*V;# mass, density and volumne corealtion\n",
-      "\n",
-      "#Result\n",
-      "print 'The amount of oil in tank is %i kg' %round(m,0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The amount of oil in tank is 1700 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-3, Page No.9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constants used\n",
-      "g=32.174;# gravitational constant in ft/s^2\n",
-      "\n",
-      "#given values\n",
-      "m=1; # mass of 1.00 lbm is subjected to standard earth gravity\n",
-      "\n",
-      "#Calculations\n",
-      "w=(m*g)/g; # weight is mass times the local value of gravitational acceleration\n",
-      "#dimensionally the above equation is represented as lbm * ft/s^2 * (lbf/ft/s^2)\n",
-      "\n",
-      "#Result\n",
-      "print 'The weight on earth is %i lbf' %w\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The weight on earth is 1 lbf\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-4, Page No.21"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Example 1.4\n",
-      "\n",
-      "# Given values\n",
-      "Tc=10; #change in temp in Celcius\n",
-      "\n",
-      "# Calculations\n",
-      "Tk=Tc;\n",
-      "Tr=1.8*Tk;#conversion scale of temperature change from K to R\n",
-      "Tf=Tr;\n",
-      "# calculated using the corealtions b/w these scales\n",
-      "\n",
-      "#Results\n",
-      "print 'the corresponding change is %i K' %Tk\n",
-      "print 'the corresponding change is %i R' %Tr\n",
-      "print 'the corresponding change is %i F' %Tf\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the corresponding change is 10 K\n",
-        "the corresponding change is 18 R\n",
-        "the corresponding change is 18 F\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-5, Page No.23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "Patm=14.5; #atmospheric pressure in psi\n",
-      "Pvac=5.8;  #vacuum gage reading in psi\n",
-      "\n",
-      "#Calculations\n",
-      "Pabs=Patm-Pvac;#pressure in vaccumm is always treated to be negative\n",
-      "\n",
-      "#Results\n",
-      "print'the absolute pressure in the chamber %f psi'%round(Pabs,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the absolute pressure in the chamber 8.700000 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-6, Page No.26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constants used\n",
-      "pw=1000; # density of water in kg/m^3;\n",
-      "g=9.81; # acceleration due to gravity in m/s^2;\n",
-      " \n",
-      "#Given values\n",
-      "SG=0.85;# specific gravity of manometric fluid\n",
-      "h=0.55;# converting height from cm to m\n",
-      "Patm=96;# atmospheric pressure in kPa\n",
-      "\n",
-      "# Calculations\n",
-      "p=SG*pw;\n",
-      "Ptank=Patm+(p*g*h/1000); # calculating pressure using liquid at same height have same pressure\n",
-      "\n",
-      "#Results\n",
-      "print 'absolute pressure in tank %f kPa' %round(Ptank,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "absolute pressure in tank 100.600000 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-7, Page No.28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constants used\n",
-      "g=9.81;#acceleration due to gravity in m/s^2;\n",
-      "\n",
-      "#Given values\n",
-      "h1=0.1;# distance b/w point 1 at air-water interface and point 2 at mercury-air interface in m\n",
-      "h2=0.2;# distance b/w oil-water interface and  mercury-oil interface in m\n",
-      "h3=0.35;# distance b/w air-mercury interface and  mercury-oil interface in m\n",
-      "pw=1000;# density of water in kg/m^3\n",
-      "pHg=13600;# density of mercury in kg/m^3\n",
-      "poil=800;# density of oil in kg/m^3\n",
-      "Patm=85.6;# atmospheric pressure in kPa\n",
-      "\n",
-      "#Calculation\n",
-      "P1=Patm-(pw*g*h1+poil*g*h2-pHg*g*h3)/1000;#calculating pressure using liquid at same height have same pressure\n",
-      "\n",
-      "#Results\n",
-      "print 'the air pressure in tank %i kPa' %round(P1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the air pressure in tank 130 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-8, Page No.31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constants used\n",
-      "g=9.81;# acceleration due to gravity in m/s^2;\n",
-      "\n",
-      "#Given values\n",
-      "pHg=13570;# density of mercury at 10 C in kg/m^3\n",
-      "h=0.74;# converting barometric reading into m from mm\n",
-      "\n",
-      "#Calculationa\n",
-      "Patm=pHg*g*h/1000;# standard  pressure formula\n",
-      "\n",
-      "#Results\n",
-      "print 'the atmospheric pressure %f kPa' %round(Patm,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the atmospheric pressure 98.500000 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-9, Page No.31"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#constants used\n",
-      "g=9.81;#acceleration due to gravity in m/s^2;\n",
-      "\n",
-      "#given values\n",
-      "m=60;# mass of piston in kg\n",
-      "Patm=0.97;# atmospheric pressure in kPa\n",
-      "A=0.04;# cross-sectional area in m^2\n",
-      "\n",
-      "#calculation\n",
-      "P=Patm+(m*g/A)/100000;# standard  pressure formula\n",
-      "print 'The pressure inside the cylinder %f bar' %round(P,2)\n",
-      "#The volume change will have no effect on the free-body diagram drawn in part (a), and therefore the pressure inside the cylinder will remain the same\n",
-      "print('If some heat is transferred to the gas and its volume is doubled, there is no change in pressure');\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The pressure inside the cylinder 1.120000 bar\n",
-        "If some heat is transferred to the gas and its volume is doubled, there is no change in pressure\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1-10, Page No.32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad \n",
-      "from pylab import *\n",
-      "\n",
-      "#Constants used\n",
-      "g=9.81;#acceleration due to gravity in m/s^2;\n",
-      "\n",
-      "#Given values\n",
-      "p=1040;# density on the water surface in kg/m^3\n",
-      "h1=0.8;# thickness of surface zone\n",
-      "H=4;# thickness of gradient zone\n",
-      "x0=0.0;# lower limit of integration\n",
-      "x1=4.0;# upper limit of integration\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "P1=p*g*h1/1000;#standard  pressure determination formula\n",
-      "#P2 = integration of the exp. p*g*(math.sqrt(1+(math.tan(math.pi*z/4/H)^2))) b/w 0-4\n",
-      "def intgrnd1(z): \n",
-      " return (p*g*(math.sqrt(1+(math.tan(math.pi*(z)/4/H)**2))) )#integrant\n",
-      "P2, err = quad(intgrnd1, x0, x1) \n",
-      "P2=P2/1000;#converting into kPa\n",
-      "P=P1+P2;\n",
-      "\n",
-      "#Results\n",
-      "print 'the gage pressure at the bottom of gradient zone %f kPa' %round(P)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the gage pressure at the bottom of gradient zone 54.000000 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter10.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter10.ipynb
deleted file mode 100755
index df284751..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter10.ipynb
+++ /dev/null
@@ -1,628 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10: Vapor and Combined Power Cycles"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-1 ,Page No.555"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=75;#outlet pressure at compressor in kPa\n",
-      "P2=3000;#inlet pressure at turbine in kPa\n",
-      "P3=P2;\n",
-      "T3=350;#steam inlet temperature in C\n",
-      "P4=P1;\n",
-      "\n",
-      "#from steam tables\n",
-      "#at state 1\n",
-      "v1=0.001037;\n",
-      "h1=384.44;\n",
-      "#at state 3\n",
-      "h3=3116.1;\n",
-      "s3=6.7450;\n",
-      "#at state 4\n",
-      "s4=s3;\n",
-      "sf=1.2132;\n",
-      "sfg=6.2426;\n",
-      "hf=384.44;\n",
-      "hfg=2278;\n",
-      "\n",
-      "#calculations\n",
-      "win=v1*(P2-P1);\n",
-      "h2=h1+win;\n",
-      "x4=(s4-sf)/sfg;\n",
-      "h4=hf+x4*hfg;\n",
-      "qin=h3-h2;\n",
-      "qout=h4-h1;\n",
-      "nth=1-(qout/qin);\n",
-      "print'thermal efficency is %f'%round(nth,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "thermal efficency is 0.260000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-2 ,Page No.559"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "#from figure\n",
-      "P1=9;#in kPa\n",
-      "T1=38;#in C\n",
-      "P2=16000;#in kPa\n",
-      "P3=15.9;#in MPa\n",
-      "T3=35;#in C\n",
-      "P4=15.2;#in MPa\n",
-      "T4=625;#in C\n",
-      "P5=15;#in MPa\n",
-      "T5=600;#in C\n",
-      "#from question\n",
-      "nT=0.87;#isentropic efficiency of turbine \n",
-      "nP=0.85;#isentropic efficiency of pump\n",
-      "m=15;#mass flow rate in kg/s\n",
-      "\n",
-      "#from steam tables\n",
-      "v1=0.001009;\n",
-      "h5=3583.1;\n",
-      "h6s=2115.3;\n",
-      "h4=3647.6;\n",
-      "h3=160.1;\n",
-      "\n",
-      "#calculations\n",
-      "Win=v1*(P2-P1)/nP;\n",
-      "Wout=nT*(h5-h6s);\n",
-      "qin=h4-h3;\n",
-      "Wnet=Wout-Win;\n",
-      "nth=Wnet/qin;\n",
-      "print'thermal efficency is %f'%round(nth,3)\n",
-      "Wnet=m*Wnet;\n",
-      "print'power output %f MW'%round(Wnet/1000,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "thermal efficency is 0.361000\n",
-        "power output 18.870000 MW\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-3 ,Page No.562"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=10;#outlet pressure at compressor in kPa\n",
-      "P2=3000;#inlet pressure at turbine in kPa\n",
-      "P3=P2;\n",
-      "T3=350;#steam inlet temperature in C\n",
-      "P4=P1;\n",
-      "\n",
-      "#from steam tables\n",
-      "#at state 1\n",
-      "h1=191.81;\n",
-      "v1=0.00101;\n",
-      "#at state 2\n",
-      "#s2=s1\n",
-      "#at state 3\n",
-      "h3=3116.1;\n",
-      "s3=6.7450;\n",
-      "#at state 4\n",
-      "s4=s3;\n",
-      "sf=0.6492;\n",
-      "sfg=7.4996;\n",
-      "hf=191.81;\n",
-      "hfg=2392.1;\n",
-      "\n",
-      "#calculations\n",
-      "#part - a\n",
-      "win=v1*(P2-P1);\n",
-      "h2=h1+win;\n",
-      "x4=(s4-sf)/sfg;\n",
-      "h4=hf+x4*hfg;\n",
-      "qin=h3-h2;\n",
-      "qout=h4-h1;\n",
-      "nth=1-(qout/qin);\n",
-      "print'the thermal efficiency of this power plant is %f'%round(nth,3);\n",
-      "#part - b\n",
-      "#States 1 and 2 remain the same in this case, and the enthalpies at state 3 (3 MPa and 600\u00b0C) and state 4 (10 kPa and s4=s3) are determined to be\n",
-      "h3=3682.8;\n",
-      "h4=2380.3;\n",
-      "x4=0.915;\n",
-      "qin=h3-h2;\n",
-      "qout=h4-h1;\n",
-      "nth=1-(qout/qin);\n",
-      "print'the thermal efficiency if steam is superheated to 600\u00b0 instead of 350\u00b0C is %f'%round(nth,3);\n",
-      "#part - c\n",
-      "#State 1 remains the same in this case, but the other states change. The enthalpies at state 2 (15 MPa and s2   s1), state 3 (15 MPa and 600\u00b0C),and state 4 (10 kPa and s4   s3) are determined in a similar manner to be\n",
-      "h2=206.95;\n",
-      "h3=3583.1;\n",
-      "h4=2115.3;\n",
-      "x4=0.804;\n",
-      "qin=h3-h2;\n",
-      "qout=h4-h1;\n",
-      "nth=1-(qout/qin);\n",
-      "print'the thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600\u00b0C is %f'%round(nth,3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the thermal efficiency of this power plant is 0.334000\n",
-        "the thermal efficiency if steam is superheated to 600\u00b0 instead of 350\u00b0C is 0.373000\n",
-        "the thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600\u00b0C is 0.430000\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-4 ,Page No.566"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=10;#outlet pressure at compressor in kPa\n",
-      "P2=15000;#inlet pressure at turbine in kPa\n",
-      "P3=P2;\n",
-      "T3=600;#steam inlet temperature in C\n",
-      "P4=4000;#in kPa\n",
-      "T5=T3;\n",
-      "P6=P1;\n",
-      "x6=0.896;#dryness fraction\n",
-      "\n",
-      "#from steam table\n",
-      "#at state 1\n",
-      "h1=191.81;\n",
-      "v1=0.00101;\n",
-      "#at state 3\n",
-      "h3=3593.1;\n",
-      "s3=6.6796;\n",
-      "#at state 4\n",
-      "h4=3155;\n",
-      "T4=375.5;\n",
-      "#at state 6\n",
-      "sf=0.6492;\n",
-      "sfg=7.4996;\n",
-      "hf=191.81;\n",
-      "hfg=2392.1;\n",
-      "\n",
-      "#calculations\n",
-      "s6=sf+x6*sfg;\n",
-      "h6=hf+x6*hfg;\n",
-      "#s5 = s6\n",
-      "#from tables\n",
-      "P5=4000.0;#in kPa\n",
-      "h5=3674.9;\n",
-      "print'the pressure at which the steam should be reheated %i MPa'%(P5/1000);\n",
-      "#s2 = s1\n",
-      "win=v1*(P2-P1);\n",
-      "h2=h1+win;\n",
-      "qin=(h3-h2)+(h5-h4);\n",
-      "qout=h6-h1;\n",
-      "nth=1-(qout/qin);\n",
-      "print'thermal efficency is %f'%round(nth,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the pressure at which the steam should be reheated 4 MPa\n",
-        "thermal efficency is 0.451000\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-5 ,Page No.571"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=10;#outlet pressure at compressor in kPa\n",
-      "P2=1200;#in kPa\n",
-      "P3=P2;\n",
-      "P4=15000;#inlet pressure at turbine in kPa\n",
-      "P5=P4;\n",
-      "T5=600;#steam inlet temperature in C\n",
-      "P6=P2;\n",
-      "P7=P1;\n",
-      "\n",
-      "#from steam table\n",
-      "#at state 1\n",
-      "h1=191.81;\n",
-      "v1=0.00101;\n",
-      "#at state 3\n",
-      "h3=798.33;\n",
-      "v3=0.001138;\n",
-      "#at state 4\n",
-      "h4=3155;\n",
-      "T4=375.5;\n",
-      "#at state 5\n",
-      "h5=3583.1;\n",
-      "s5=6.6796;\n",
-      "#at state 6\n",
-      "h6=2860.2;\n",
-      "T6=218.4;\n",
-      "#at state 7\n",
-      "P7=10;\n",
-      "sf=0.6492;\n",
-      "sfg=7.4996;\n",
-      "hf=191.81;\n",
-      "hfg=2392.1;\n",
-      "\n",
-      "#calculations\n",
-      "#s2 = s1\n",
-      "win=v1*(P2-P1);\n",
-      "h2=h1+win;\n",
-      "#s4 = s3\n",
-      "win=v3*(P4-P3);\n",
-      "h4=h3+win;\n",
-      "s7=s5;\n",
-      "x7=(s7-sf)/sfg;\n",
-      "h7=hf+(x7*hfg);\n",
-      "#y is the fraction of steam extracted from the turbine\n",
-      "y=(h3-h2)/(h6-h2);\n",
-      "qin=h5-h4;\n",
-      "qout=(1-y)*(h7-h1);\n",
-      "nth=1-(qout/qin);\n",
-      "print'fraction of steam extracted is %f'%round(y,4);\n",
-      "print'thermal efficency is %f'%round(nth,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fraction of steam extracted is 0.226900\n",
-        "thermal efficency is 0.463000\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-6 ,Page No.574"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=10;#outlet pressure at compressor in kPa\n",
-      "P2=500;#steam extracted from low-pressure turbine in kPa\n",
-      "P3=P2;\n",
-      "P4=15000;#inlet pressure at turbine in kPa\n",
-      "P5=P4;\n",
-      "P6=4000;#steam extracted from turbine at pressure in kPa\n",
-      "P7=P5;\n",
-      "P8=P7;\n",
-      "P9=P7;\n",
-      "P10=P6;\n",
-      "P11=P10;\n",
-      "P12=P3;\n",
-      "P13=P1;\n",
-      "\n",
-      "#enthalpies at the various states and the pump work per unit mass of fluid flowing through them are\n",
-      "h1=191.81;\n",
-      "h2=192.30;\n",
-      "h3=640.09;\n",
-      "h4=643.92;\n",
-      "h5=1087.4;\n",
-      "h6=h5;\n",
-      "h7=1101.2;\n",
-      "h8=1089.8;\n",
-      "h9=3583.1;\n",
-      "h10=3155;\n",
-      "h11=3679.9;\n",
-      "h12=3014.8;\n",
-      "h13=2335.7;\n",
-      "wIin=0.49;\n",
-      "wIIin=3.83;\n",
-      "wIIIin=13.77;\n",
-      "\n",
-      "#calculations\n",
-      "y=(h5-h4)/((h10-h6)+(h5-h4));\n",
-      "z=(1-y)*(h3-h2)/(h12-h2);\n",
-      "h8=(1-y)*h5+(y*h7);\n",
-      "qin=(h9-h8)+(1-y)*(h11-h10);\n",
-      "qout=(1-y-z)*(h13-h1);\n",
-      "nth=1-(qout/qin);\n",
-      "print'fraction of steam extracted from closed feedwater is %f'%round(y,4);\n",
-      "print'fraction of steam extracted from open feedwater is %f'%round(z,4);\n",
-      "print'thermal efficency is %f'%round(nth,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fraction of steam extracted from closed feedwater is 0.176600\n",
-        "fraction of steam extracted from open feedwater is 0.130600\n",
-        "thermal efficency is 0.492000\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-7 ,Page No.577"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T0=290;#temperature at heat which is rejected in K \n",
-      "Tsource=1600;#temperature of furnace in K\n",
-      "Tsink=T0;\n",
-      "#from Ex 10.1\n",
-      "qin=2728.6;#in kJ/Kg\n",
-      "qout=2018.6;#in kJ/Kg\n",
-      "h4=2403;#in kJ/Kg\n",
-      "\n",
-      "#from steam tables\n",
-      "s1=1.2132;\n",
-      "s3=6.7450;\n",
-      "\n",
-      "#calculations\n",
-      "s2=s1;s4=s3;#isentropic processes\n",
-      "xdest12=0;\n",
-      "xdest34=0;\n",
-      "xdest23=T0*(s3-s2-(qin/Tsource));\n",
-      "xdest41=T0*(s1-s4+(qout/Tsink));\n",
-      "print'exergy destruction in process 1-2 %i kJ/kg'%xdest12;\n",
-      "print'exergy destruction in process 2-3 %i kJ/kg'%round(xdest23);\n",
-      "print'exergy destruction in process 3-4 %i kJ/kg'%xdest34;\n",
-      "print'exergy destruction in process 4-1 %i kJ/kg'%round(xdest41);\n",
-      "xdestcy=xdest12+xdest23+xdest34+xdest41;\n",
-      "print'exergy destruction %i cycle in kJ/kg'%round(xdestcy);\n",
-      "#from steam tables\n",
-      "#at 290 K and 100 kPa\n",
-      "h0=71.355;\n",
-      "s0=0.2533;\n",
-      "X4=(h4-h0)-T0*(s4-s0);\n",
-      "print'exergy of the leaving steam %i kJ/kg'%(round(X4))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "exergy destruction in process 1-2 0 kJ/kg\n",
-        "exergy destruction in process 2-3 1110 kJ/kg\n",
-        "exergy destruction in process 3-4 0 kJ/kg\n",
-        "exergy destruction in process 4-1 414 kJ/kg\n",
-        "exergy destruction 1524 cycle in kJ/kg\n",
-        "exergy of the leaving steam 449 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-8 ,Page No.581"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m1=15;#mass flow rate in kg/s\n",
-      "P1=7000;#inlet pressure at turbine in kPa\n",
-      "P2=P1;\n",
-      "P3=P1;\n",
-      "P4=500;#throttled to pressure in kPa\n",
-      "P5=P4;\n",
-      "P6=5;#steam expanded at a pressure in kPa\n",
-      "P7=P4;\n",
-      "P8=P6;\n",
-      "P9=P1;\n",
-      "P10=P1;\n",
-      "\n",
-      "#from steam tables\n",
-      "v7=0.001005;\n",
-      "v8=0.001093;\n",
-      "h1=3411.4;\n",
-      "h2=h1;\n",
-      "h3=h1;\n",
-      "h4=h1;\n",
-      "h5=2739.3;\n",
-      "h6=2073.0;\n",
-      "h7=640.09;\n",
-      "h8=137.75;\n",
-      "h11=144.78;\n",
-      "\n",
-      "#calculations\n",
-      "wIin=v8*(P9-P8);\n",
-      "wIIin=v7*(P10-P7);\n",
-      "h9=h8+wIin;\n",
-      "h10=h7+wIIin;\n",
-      "Qmax=m1*(h1-h7);\n",
-      "print'the maximum rate %i kW'%round(Qmax);\n",
-      "Wtout=m1*(h3-h6);#turbine\n",
-      "Wpin=m1*wIin;#pump\n",
-      "Wnet=Wtout-Wpin;\n",
-      "print'the power produced %i MW'%(round(Wnet/1000));\n",
-      "Qp=0;\n",
-      "Qin=m1*(h1-h11);\n",
-      "Eu=(Wnet+Qp)/Qin;\n",
-      "print'the utilization factor is %f'%round(Eu,3);\n",
-      "m4=0.1*m1;\n",
-      "m5=0.7*m1;\n",
-      "m7=m4+m5;\n",
-      "Qout=m4*h4+m5*h5-m7*h7;\n",
-      "print'the rate of process heat supply %f MW'%round(Qout/1000,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the maximum rate 41570 kW\n",
-        "the power produced 20 MW\n",
-        "the utilization factor is 0.407000\n",
-        "the rate of process heat supply 26.200000 MW\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10-9 ,Page No.585"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=5;#intial pressure in kPa\n",
-      "P2=7000;#final pressure ink Pa\n",
-      "P3=P2;\n",
-      "T3=500;#inlet temperature in K\n",
-      "P4=P1;\n",
-      "\n",
-      "#gas cycle from Ex9-6\n",
-      "#d stands for '\n",
-      "h4d=880.36;\n",
-      "T4d=853;\n",
-      "qin=790.58;\n",
-      "wnetg=210.41;\n",
-      "nth=0.266\n",
-      "h5d=451.80;\n",
-      "#steam cycle\n",
-      "h2=144.78;\n",
-      "T2=33;\n",
-      "h3=3411.4;\n",
-      "T3=500;\n",
-      "wnets=1331.4;\n",
-      "nth=0.408;\n",
-      "\n",
-      "#calculations\n",
-      "#Ein = Eout\n",
-      "#y is the ratio of ms/mg\n",
-      "y=(h4d-h5d)/(h3-h2);\n",
-      "print'the ratio of the mass flow rates of the steam and the combustion gasesis %f'%round(y,3);\n",
-      "wnet=wnetg+y*wnets\n",
-      "nth=wnet/qin;\n",
-      "print'thermal efficency is %f'%round(nth,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the ratio of the mass flow rates of the steam and the combustion gasesis 0.131000\n",
-        "thermal efficency is 0.487000\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter11.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter11.ipynb
deleted file mode 100755
index 460e3a17..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter11.ipynb
+++ /dev/null
@@ -1,353 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11: Refrigeration Cycles"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11-1 ,Page No.613"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "P1=0.14;#intial pressure in MPa\n",
-      "P2=0.8;#final pressure in MPa\n",
-      "m=0.05;#mass flow rate in kg/s\n",
-      "\n",
-      "#from refrigerant-134a tables\n",
-      "h1=239.16;\n",
-      "s1=0.94456;\n",
-      "h2=275.39;\n",
-      "h3=95.47;\n",
-      "\n",
-      "#calculation\n",
-      "s2=s1;#isentropic process \n",
-      "h4=h3;#throttling\n",
-      "QL=(h1-h4)*m;\n",
-      "Wm=m*(h2-h1);\n",
-      "Qh=m*(h2-h3);\n",
-      "COPR=QL/Wm;\n",
-      "print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);\n",
-      "print'the power input to the compressor %f kW'%round(Wm,2);\n",
-      "print'the rate of heat rejection to the environment %f kW'%round(Qh);\n",
-      "print'the COP of the refrigerator is %f'%round(COPR,2);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of heat removal from the refrigerated space 7.180000 kW\n",
-        "the power input to the compressor 1.810000 kW\n",
-        "the rate of heat rejection to the environment 9.000000 kW\n",
-        "the COP of the refrigerator is 3.970000\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11-2 ,Page No.615"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=0.05;#mass flow rate in kg/s\n",
-      "P1=0.14;#inlet pressure in MPa\n",
-      "T1=-10;#inlet temperature in C\n",
-      "P2=0.8;#outlet pressure in MPa\n",
-      "T2=50;#outlet temperature in C\n",
-      "P3=0.72;#condensor pressure in MPa\n",
-      "T3=26;#condensor temperature in C\n",
-      "\n",
-      "#from refrigerant tables\n",
-      "h1=246.36;\n",
-      "h2=286.69;\n",
-      "h3=87.83;\n",
-      "h2S=284.21;#at isentropic conditions\n",
-      "\n",
-      "#calculations\n",
-      "h4=h3;#throttling\n",
-      "QL=m*(h1-h4);\n",
-      "Wm=m*(h2-h1);\n",
-      "nC=(h2S-h1)/(h2-h1);\n",
-      "COPR=QL/Wm;\n",
-      "print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);\n",
-      "print'the power input to the compressor %f kW'%round(Wm,2);\n",
-      "print'the isentropic efficiency of the compressor is %f'%round(nC,3);\n",
-      "print'the COP of the refrigerator is %f'%round(COPR,2);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of heat removal from the refrigerated space 7.930000 kW\n",
-        "the power input to the compressor 2.020000 kW\n",
-        "the isentropic efficiency of the compressor is 0.939000\n",
-        "the COP of the refrigerator is 3.930000\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11-3 ,Page No.621"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "mA=0.05;#mass flow rate in kg/s\n",
-      "P1=0.14;#inlet pressure in MPa\n",
-      "P5=0.32;#pressure at heat exchanger in MPa\n",
-      "P7=0.8;#oultet pressure in MPa\n",
-      "\n",
-      "#from tables\n",
-      "h1=239.16;\n",
-      "h2=255.93;\n",
-      "h3=55.16;\n",
-      "h5=251.88;\n",
-      "h6=270.92;\n",
-      "h7=95.47;\n",
-      "\n",
-      "#calculations\n",
-      "h4=h3;#throttling\n",
-      "h8=h7;#throttling\n",
-      "# E out = E in\n",
-      "# mA*h5 + mB*h3 = mA*h8 + mB*h2\n",
-      "mB=mA*(h5-h8)/(h2-h3);\n",
-      "QL=mB*(h1-h4);\n",
-      "# W in = Wcomp I,in + Wcomp II,in\n",
-      "Win=mA*(h6-h5)+mB*(h2-h1);\n",
-      "COPR=QL/Win;\n",
-      "print'the mass flow rate of the refrigerant through the lower cycle %f kg/s'%round(mB,4);\n",
-      "print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);\n",
-      "print'the power input to the compressor %f kW'%round(Win,2);\n",
-      "print'the coefficient of performance of this cascade refrigerator is %f'%round(COPR,2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the mass flow rate of the refrigerant through the lower cycle 0.039000 kg/s\n",
-        "the rate of heat removal from the refrigerated space 7.170000 kW\n",
-        "the power input to the compressor 1.610000 kW\n",
-        "the coefficient of performance of this cascade refrigerator is 4.460000\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11-4 ,Page No.624"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=0.14;#inlet pressure in MPa\n",
-      "P5=0.32;#throttled pressure in MPa\n",
-      "P7=0.8;#oultet pressure in MPa\n",
-      "\n",
-      "#from tables\n",
-      "h1=239.16;\n",
-      "h2=255.93;\n",
-      "h3=251.88;\n",
-      "h5=95.47;\n",
-      "h7=55.16;\n",
-      "\n",
-      "#from saturated liquid-vapour table\n",
-      "#at P=0.32 MPa\n",
-      "hf=55.16;\n",
-      "hfg=196.71;\n",
-      "\n",
-      "#calculations\n",
-      "h8=h7;#throttling\n",
-      "h6=h5;#throttling\n",
-      "#the quality at state 6\n",
-      "x6=(h6-hf)/hfg;\n",
-      "qL=(1-x6)*(h1-h8);\n",
-      "# W in = Wcomp I,in + Wcomp II,in\n",
-      "#enthalaoy at state 9\n",
-      "# E out = E in\n",
-      "h9=x6*h3+(1-x6)*h2;\n",
-      "# s9 = s4 i.e isentropic process\n",
-      "#at 0.8MPa and s4=0.9416 kJ/kg\n",
-      "h4=274.48;\n",
-      "Win=(1-x6)*(h2-h1)+(1)*(h4-h9);\n",
-      "COPR=qL/Win;\n",
-      "print'the fraction of the refrigerant that evaporates as it is throttled to the flash chamber is %f'%round(x6,4);\n",
-      "print'the amount of heat removed from the refrigerated space %f kJ/kg'%round(qL,1);\n",
-      "print'the compressor work per unit mass of refrigerant flowing through the condensor %f kJ/kg'%round(Win,2);\n",
-      "print'the coefficient of performance is %f'%round(COPR,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the fraction of the refrigerant that evaporates as it is throttled to the flash chamber is 0.204900\n",
-        "the amount of heat removed from the refrigerated space 146.300000 kJ/kg\n",
-        "the compressor work per unit mass of refrigerant flowing through the condensor 32.710000 kJ/kg\n",
-        "the coefficient of performance is 4.470000\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11-5 ,Page No.630"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=0.1;#mass flow rate in lbm/s\n",
-      "T1=0+460;#refigerated space at temperature in R\n",
-      "T3=80+460;#temperature of surrounding medium in R\n",
-      "\n",
-      "#from Table A\u201317E\n",
-      "# at T1\n",
-      "h1=109.90;\n",
-      "Pr1=.7913;\n",
-      "#pressure ratio at compressor is 4\n",
-      "Pr2=4*Pr1;\n",
-      "#at Pr2\n",
-      "h2=163.5;\n",
-      "T2=683;\n",
-      "#at T3\n",
-      "h3=129.06;\n",
-      "Pr3=1.3860;\n",
-      "#pressure ratio at compressor is 4\n",
-      "Pr4=Pr3/4;\n",
-      "#at Pr4\n",
-      "h4=86.7;\n",
-      "T4=363;\n",
-      "\n",
-      "#calculations\n",
-      "qL=h1-h4;\n",
-      "Wout=h3-h4;\n",
-      "Win=h2-h1;\n",
-      "COPR=qL/(Win-Wout);\n",
-      "Qrefrig=m*qL;\n",
-      "print'the minimum temperatures in the cycle %f F'%round(T4-460);\n",
-      "print'the maximum temperatures in the cycle %f F'%round(T2-460);\n",
-      "print'the coefficient of performance is %f'%round(COPR,2)\n",
-      "print'the rate of refrigeration for a mass flow rate of 0.1 lbm/s. %f Btu/s'%round(Qrefrig,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the minimum temperatures in the cycle -97.000000 F\n",
-        "the maximum temperatures in the cycle 223.000000 F\n",
-        "the coefficient of performance is 2.060000\n",
-        "the rate of refrigeration for a mass flow rate of 0.1 lbm/s. 2.320000 Btu/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11-6 ,Page No.636"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "COPR=0.1;#refrigerator COP\n",
-      "T1=20;#intial temp in C\n",
-      "T2=4;#final temp in C\n",
-      "t=30*60;#total time required in sec\n",
-      "V=0.350;#volumne of can in L\n",
-      "\n",
-      "#constants used\n",
-      "p=1;#on kg/L\n",
-      "c=4.18;#in kJ/kg-C from Table A-3\n",
-      "\n",
-      "#calculations\n",
-      "m=p*V;\n",
-      "Qcooling=m*c*(T1-T2)/t*1000;#converted in W by multiplying by 1000\n",
-      "Win=Qcooling/COPR;\n",
-      "print'the average electric power consumed by the thermoelectric refrigerator %i W'%round(Win)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the average electric power consumed by the thermoelectric refrigerator 130 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter12.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter12.ipynb
deleted file mode 100755
index c181d0f6..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter12.ipynb
+++ /dev/null
@@ -1,262 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12: Thermodynamic Property Relations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12-1 ,Page No.652"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#from Table A-17\n",
-      "#cp value approximately by replacing the differentials\n",
-      "h1=305.22;#in kJ/kg\n",
-      "T1=305;#in K\n",
-      "h2=295.17;#in kJ/kg\n",
-      "T2=295;#in K\n",
-      "\n",
-      "#calculations\n",
-      "#from the given equation we can calculate\n",
-      "cp=(h1-h2)/(T1-T2);\n",
-      "print'the cp of air at 300 K %f kJ/ kg - K'%round(cp,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the cp of air at 300 K 1.005000 kJ/ kg - K\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12-2 ,Page No.654"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "dT=302-300;#differnce in final and intial temperatures of air in K\n",
-      "dv=0.87-0.86;#differnce in final and intial volumnes of air in m^3/kg\n",
-      "T=(302+300)/2;#average temp in K\n",
-      "v=(0.87+0.86)/2;#average volumne in m^3/kg\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kJ/kg-K\n",
-      "\n",
-      "#calculations\n",
-      "#using eq 12-3 by diffrentiating P= R*T/v\n",
-      "dP= R*dT/v - R*T*dv/v**2;\n",
-      "print'the change in the pressure of air %f kPa'%round(dP,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the change in the pressure of air -0.491000 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12-5 ,Page No.659"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=20+273.15;#temperature of refrigerant in K\n",
-      "\n",
-      "#from Table A\u201311\n",
-      "vf=0.0008161;#in m^3/kg\n",
-      "vg=0.035969;#in m^3/kg\n",
-      "\n",
-      "#calculations\n",
-      "#using Eq 12-22\n",
-      "# hfg= T*vfg*(dP/dT)sat\n",
-      "#(dP/dT)sat b/w 24 C - 16 C \n",
-      "dPT=(646.18-504.58)/(24-16);#dP/dT ; values from Table A\u201311\n",
-      "vfg=vg-vf;\n",
-      "hfg=T*vfg*dPT;\n",
-      "print'the value of the enthalpy of vaporization of refrigerant-134a %f kJ/kg'%round(hfg,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the value of the enthalpy of vaporization of refrigerant-134a 182.400000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12-6 ,Page No.660"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import exp\n",
-      "\n",
-      "#given data\n",
-      "T2=-50+460.0;#temperature of refrigerant-134a on R \n",
-      "\n",
-      "#constants\n",
-      "R=0.01946;#in Btu/lbm\n",
-      "\n",
-      "#from Table A-11E\n",
-      "T1=-40+460.0;#converted into R from F\n",
-      "P1=7.432;\n",
-      "hfg=97.100;\n",
-      "\n",
-      "#calcualation\\\n",
-      "#using Equation 12\u201324\n",
-      "#ln(P2/P1)= hfg/R *(1/T1 - 1/T2)\n",
-      "P2=P1*exp(hfg/R *(1/T1 - 1/T2));\n",
-      "print'the saturation pressure of refrigerant-134a %f psia'%round(P2,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the saturation pressure of refrigerant-134a 5.560000 psia\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12-11 ,Page No.373"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#given data\n",
-      "T1=220;#intial temperature in K\n",
-      "P1=5;#intial pressure in MPa\n",
-      "T2=300;#final temperature in K\n",
-      "P2=10;#final pressure in MPa\n",
-      "\n",
-      "#constants used\n",
-      "Ru=8.314;#on kJ/kmol- K\n",
-      "\n",
-      "#from Table A\u20131\n",
-      "Tcr=154.8;\n",
-      "Pcr=5.08;\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "print('part - a');\n",
-      "#by assuming ideal-gas behavior\n",
-      "#from Table A\u201319\n",
-      "h1=6404;\n",
-      "h2=8736;\n",
-      "s2=205.213;\n",
-      "s1=196.171;\n",
-      "h21i=h2-h1;#h2 - h1 ideal\n",
-      "s21i=(s2-s1)-Ru*log(P2/P1);#s2 - s1 ideal\n",
-      "print'the enthalpy change %i kJ/kmol'%round(h21i);\n",
-      "print'the entropy change %f kJ/kmol-K'%round(s21i,2);\n",
-      "\n",
-      "#part - b\n",
-      "print('part - b');\n",
-      "#by accounting for the deviation from ideal-gas behavior\n",
-      "TR1=T1/Tcr;\n",
-      "Pr1=P1/Pcr;\n",
-      "#from the generalized charts at each state\n",
-      "Zh1=0.53;\n",
-      "Zs1=0.25;\n",
-      "TR2=T2/Tcr;\n",
-      "Pr2=P2/Pcr;\n",
-      "#from the generalized charts at each state\n",
-      "Zh2=0.48;\n",
-      "Zs2=0.20;\n",
-      "h21=h21i-Ru*Tcr*(Zh2-Zh1);\n",
-      "s21=s21i-Ru*(Zs2-Zs1);\n",
-      "print'the enthalpy change %i in kJ/kmol'%round(h21);\n",
-      "print'the entropy change %f kJ/kmol-K'%round(s21,2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "part - a\n",
-        "the enthalpy change 2332 kJ/kmol\n",
-        "the entropy change 3.280000 kJ/kmol-K\n",
-        "part - b\n",
-        "the enthalpy change 2396 in kJ/kmol\n",
-        "the entropy change 3.690000 kJ/kmol-K\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter13.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter13.ipynb
deleted file mode 100755
index 936d90ae..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter13.ipynb
+++ /dev/null
@@ -1,453 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13: Gas Mixtures"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13-1 ,Page No.683"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "mO2=3.0;#moles of oxygen\n",
-      "mN2=5.0;#moles of nitrogen\n",
-      "mCH4=12.0;#moles of methane\n",
-      "\n",
-      "#molecular masses in kg\n",
-      "MO2=32.0;\n",
-      "MN2=28.0;\n",
-      "MCH4=16.0;\n",
-      "\n",
-      "#constants used\n",
-      "Ru=8.314;#in kJ/kg - K\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "mm=mO2+mN2+mCH4;\n",
-      "mfO2=mO2/mm;\n",
-      "mfN2=mN2/mm;\n",
-      "mfCH4=mCH4/mm;\n",
-      "print'mass fraction of oxygen is %f'%round(mfO2,2);\n",
-      "print'mass fraction of nitrogen is %f'%round(mfN2,2);\n",
-      "print'mass fraction of methane is %f'%round(mfCH4,2);\n",
-      "\n",
-      "#part - b\n",
-      "NO2=mO2/MO2;\n",
-      "NN2=mN2/MN2;\n",
-      "NCH4=mCH4/MCH4;\n",
-      "Nm=NO2+NN2+NCH4;\n",
-      "yO2=NO2/Nm;\n",
-      "yN2=NN2/Nm;\n",
-      "yCH4=NCH4/Nm;\n",
-      "print'mole fraction of oxygen is %f'%round(yO2,3);\n",
-      "print'mole fraction of nitrogen is %f'%round(yN2,3);\n",
-      "print'mole fraction of methane is %f'%round(yCH4,3);\n",
-      "\n",
-      "#part - c\n",
-      "Mm=mm/Nm;\n",
-      "print'average molecular mass %f kg/kmol'%round(Mm,1);\n",
-      "Rm=Ru/Mm;\n",
-      "print'gas constant of mixture %f kJ/kg - K'%round(Rm,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mass fraction of oxygen is 0.150000\n",
-        "mass fraction of nitrogen is 0.250000\n",
-        "mass fraction of methane is 0.600000\n",
-        "mole fraction of oxygen is 0.092000\n",
-        "mole fraction of nitrogen is 0.175000\n",
-        "mole fraction of methane is 0.734000\n",
-        "average molecular mass 19.600000 kg/kmol\n",
-        "gas constant of mixture 0.425000 kJ/kg - K\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13-2 ,Page No.687"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "NN2=2.0;#moles of nitrogen\n",
-      "NCO2=6.0;#moles of carbon dioxide\n",
-      "Tm=300.0;#temperature of gases in K\n",
-      "Pm=15000.0;#pressure of gases in kPa\n",
-      "\n",
-      "#constants used\n",
-      "Ru=8.314;#in kJ/kmol - K\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "Nm=NN2+NCO2;\n",
-      "Vm=Nm*Ru*Tm/Pm;\n",
-      "print'the volume of the tank on the basis of the ideal-gas equation of state %f m^3'%round(Vm,3);\n",
-      "\n",
-      "#part - b\n",
-      "#from Table A-1\n",
-      "#for nitrogen\n",
-      "TcrN=126.2;\n",
-      "PcrN=3390;\n",
-      "#for Carbondioxide\n",
-      "TcrC=304.2;\n",
-      "PcrC=7390;\n",
-      "yN2=NN2/Nm;\n",
-      "yCO2=NCO2/Nm;\n",
-      "Tcr=yN2*TcrN+yCO2*TcrC;\n",
-      "Pcr=yN2*PcrN+yCO2*PcrC;\n",
-      "Tr=Tm/Tcr;\n",
-      "Pr=Pm/Pcr;\n",
-      "#from Fig A-15b\n",
-      "Zm=0.49;\n",
-      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
-      "print'the volume of the tank on the basis Kay\u2019s rule %f m^3'%round(Vm,3);\n",
-      "\n",
-      "#part - c\n",
-      "#for nitrogen\n",
-      "TrN=Tm/TcrN;\n",
-      "PrN=Pm/PcrN;\n",
-      "#from Fig A-15b\n",
-      "Zn=1.02;\n",
-      "#for Carbondioxide\n",
-      "TrC=Tm/TcrC;\n",
-      "PcrC=Pm/PcrC;\n",
-      "#from Fig A-15b\n",
-      "Zc=0.3;\n",
-      "Zm=yN2*Zn+yCO2*Zc;\n",
-      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
-      "print'the volume of the tank on the basis compressibility factors and Amagat\u2019s law %f m^3'%round(Vm,3);\n",
-      "\n",
-      "#part - d\n",
-      "VRN=(Vm/NN2)/(Ru*TcrN/PcrN);\n",
-      "VRC=(Vm/NCO2)/(Ru*TcrC/PcrC);\n",
-      "#from Fig A-15b\n",
-      "Zn=0.99;\n",
-      "Zc=0.56;\n",
-      "Zm=yN2*Zn+yCO2*Zc;\n",
-      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
-      "#When the calculations are repeated we obtain 0.738 m3 after the second iteration, 0.678 m3 after the third iteration, and 0.648 m3 after the fourth iteration.\n",
-      "Vm=0.648;\n",
-      "print'compressibility factors and Dalton\u2019s law the volume of the tank on the basis %f m^3'%round(Vm,3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the volume of the tank on the basis of the ideal-gas equation of state 1.330000 m^3\n",
-        "the volume of the tank on the basis Kay\u2019s rule 0.652000 m^3\n",
-        "the volume of the tank on the basis compressibility factors and Amagat\u2019s law 0.639000 m^3\n",
-        "compressibility factors and Dalton\u2019s law the volume of the tank on the basis 0.648000 m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13-3 ,Page No.691"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "mN=4.0;#mass of nitrogen in kg\n",
-      "T1N=20.0;#temperature of nitrogen in K\n",
-      "P1N=150.0;#pressure of nitrogen in kPa\n",
-      "mO=7.0;#mass of oxygen in kg\n",
-      "T1O=40.0;#temperature of oxygen in K\n",
-      "P1O=100.0;#pressure of oxygen in kPa\n",
-      "\n",
-      "#molecular masses in kg\n",
-      "MO=32.0;\n",
-      "MN=28.0;\n",
-      "\n",
-      "#constants used\n",
-      "Ru=8.314;#in kJ/kg - K\n",
-      "\n",
-      "#from Table A-2a\n",
-      "CvN=0.743;\n",
-      "CvO=0.658;\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "#Ein - Eout = dEsystem\n",
-      "# (m*cv*dT)N2 + (m*cv*dT)= 0;\n",
-      "Tm= (mN*CvN*T1N+ mO*CvO*T1O)/(mN*CvN+mO*CvO);\n",
-      "print'the mixture temperature %f C'%round(Tm,1);\n",
-      "\n",
-      "#part - b\n",
-      "NO=mO/MO;\n",
-      "NN=mN/MN;\n",
-      "Nm=NO+NN;\n",
-      "VO=NO*Ru*(T1O+273)/P1O;\n",
-      "VN=NN*Ru*(T1N+273)/P1N;#Exergy Destruction during Mixing of Ideal Gases\n",
-      "Vm=VO+VN;\n",
-      "Pm=Nm*Ru*(Tm+273)/Vm;  \n",
-      "print'the mixture pressure after equilibrium has been established %f kPa'%round(Pm,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the mixture temperature 32.200000 C\n",
-        "the mixture pressure after equilibrium has been established 114.500000 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13-4 ,Page No.692"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#given data\n",
-      "NO=3.0;#moles of oxygen \n",
-      "NC=5.0;#moles of carbondioxide\n",
-      "T0=25+273.0;#temperature of gasses in K\n",
-      "\n",
-      "#constants used\n",
-      "Ru=8.314;#in kJ/kg - K\n",
-      "\n",
-      "#calculations\n",
-      "Nm=NO+NC;\n",
-      "yO=NO/Nm;\n",
-      "yC=NC/Nm;\n",
-      "#dSm= -Ru*(NO*log(yO)+NC*log(yC))\n",
-      "Sm=-Ru*(NO*log(yO)+NC*log(yC));\n",
-      "print'the entropy change %f kJ/K'%round(Sm);\n",
-      "Xdestroyed=T0*Sm/1000;\n",
-      "print'exergy destruction associated %f MJ'%round(Xdestroyed,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the entropy change 44.000000 kJ/K\n",
-        "exergy destruction associated 13.100000 MJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13-5 ,Page No.694"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=220;#intial temperature in K\n",
-      "T2=160;#final temperature in K\n",
-      "Pm=10;#pressure of air in MPa\n",
-      "yN=0.79;#mole fraction of nitrogen\n",
-      "yO=0.21;#mole fractions of oxygen\n",
-      "\n",
-      "\n",
-      "#critical properties\n",
-      "#for Nitrogen\n",
-      "TcrN=126.2;\n",
-      "PcrN=3.39;\n",
-      "#for Oxygen\n",
-      "TcrO=154.8;\n",
-      "PcrO=5.08;\n",
-      "\n",
-      "#constants used\n",
-      "Ru=8.314;#in kJ/kg - K\n",
-      "\n",
-      "#from Tables A-18 & 19\n",
-      "#at T1\n",
-      "h1N=6391;\n",
-      "h1O=6404;\n",
-      "#for T2\n",
-      "h2N=4648;\n",
-      "h2O=4657;\n",
-      "\n",
-      "#calculations\n",
-      "#part - a\n",
-      "qouti=yN*(h1N-h2N)+yO*(h1O-h2O);\n",
-      "print'the heat transfer during this process using the ideal-gas approximation %i kJ/kmol'%round(qouti);\n",
-      "\n",
-      "#part - b\n",
-      "Tcrm=yN*TcrN+yO*TcrO;\n",
-      "Pcrm=yN*PcrN+yO*PcrO;\n",
-      "Tr1=T1/Tcrm;\n",
-      "Tr2=T2/Tcrm;\n",
-      "Pr=Pm/Pcrm;\n",
-      "#at these values we get\n",
-      "Zh1=1;\n",
-      "Zh2=2.6\n",
-      "qout=qouti-Ru*Tcrm*(Zh1-Zh2);\n",
-      "print'the heat transfer during this process using Kay\u2019s rule %i kJ/kmol'%round(qout);\n",
-      "\n",
-      "#part - c\n",
-      "#for nitrogen\n",
-      "TrN1=T1/TcrN;\n",
-      "TrN2=T2/TcrN;\n",
-      "PrN=Pm/PcrN;\n",
-      "#from Fig A-15b\n",
-      "Zh1n=0.9;\n",
-      "Zh2n=2.4;\n",
-      "#for Oxygen\n",
-      "TrO1=T1/TcrO;\n",
-      "TrO2=T2/TcrO;\n",
-      "PcrO=Pm/PcrO;\n",
-      "#from Fig A-15b\n",
-      "Zh1O=1.3;\n",
-      "Zh2O=4.0;\n",
-      "#from Eq 12-58\n",
-      "h12N=h1N-h2N-Ru*TcrN*(Zh1n-Zh2n);# h1 - h2 for nitrogen\n",
-      "h12O=h1O-h2O-Ru*TcrO*(Zh1O-Zh2O);# h1 - h2 for oxygen\n",
-      "qout=yN*h12N+yO*h12O;\n",
-      "print'the heat transfer during this process using Amagat\u2019s law %i kJ/kmol'%round(qout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the heat transfer during this process using the ideal-gas approximation 1744 kJ/kmol\n",
-        "the heat transfer during this process using Kay\u2019s rule 3502 kJ/kmol\n",
-        "the heat transfer during this process using Amagat\u2019s law 3717 kJ/kmol\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13-6 ,Page No.705"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#13.6 (d) answer not matching as float datatype is giving more accurate answer in comparison to textbook that has given approximate due to rounding off to two decimal places\n",
-      "\n",
-      "#given data\n",
-      "mfs=0.0348;#salinity mass fraction\n",
-      "mfw=1-mfs;\n",
-      "T0=288.15;#temperature of sea water in K\n",
-      "\n",
-      "#constants used\n",
-      "Mw=18;\n",
-      "Ms=58.44;\n",
-      "Rw=0.4615;\n",
-      "pm=1028;\n",
-      "Ru=8.314;\n",
-      "\n",
-      "#calculations\n",
-      "#part - a\n",
-      "Mm=1/((mfs/Ms)+(mfw/Mw));\n",
-      "yw=mfw*Mm/Mw;\n",
-      "ys=1-yw;\n",
-      "print'the mole fraction of the water is %f'%round(yw,4);\n",
-      "print'the mole fraction of the saltwater is %f'%round(ys,5);\n",
-      "\n",
-      "#part - b\n",
-      "wmin=-Ru*T0*(ys*log(ys)+yw*log(yw));\n",
-      "wm=wmin/Mm;\n",
-      "print'the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts %f kJ'%round(wm,2);\n",
-      "\n",
-      "#part - c\n",
-      "wmin=Rw*T0*log(1/yw);\n",
-      "print'the minimum work input required to obtain 1 kg of fresh water from the sea %f kJ'%round(wmin,2);\n",
-      "\n",
-      "#part - d\n",
-      "Pmin=pm*Rw*T0*log(1/yw);\n",
-      "print'the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes %i kPa'%round(Pmin)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the mole fraction of the water is 0.989000\n",
-        "the mole fraction of the saltwater is 0.010980\n",
-        "the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts 7.850000 kJ\n",
-        "the minimum work input required to obtain 1 kg of fresh water from the sea 1.470000 kJ\n",
-        "the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes 1510 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter14.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter14.ipynb
deleted file mode 100755
index f218441e..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter14.ipynb
+++ /dev/null
@@ -1,393 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14: Gas\u2013Vapor Mixtures and Air-Conditioning"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14-1 ,Page No.720"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V=5*5*3.0;#volume of the room in m^3\n",
-      "RH=0.75;#relative humidity\n",
-      "P=100.0;#pressure of air in kPa\n",
-      "T=25.0;#temperature of air in C\n",
-      "\n",
-      "#constants used\n",
-      "Ra=0.287;#in kPa.m^3 / kg.k\n",
-      "Rv=0.4615;#in kPa.m^3 / kg.k\n",
-      "\n",
-      "#from Table A-2a and A-4\n",
-      "cp=1.005;\n",
-      "Psat=3.1698;\n",
-      "hg=2564.6;\n",
-      "\n",
-      "#calculation\n",
-      "Pv=RH*Psat;\n",
-      "Pa=P-Pv;\n",
-      "w=0.622*Pv/(P-Pv);\n",
-      "h=cp*T+w*hg;\n",
-      "ma=V*Pa/(Ra*(T+273));\n",
-      "mv=V*Pv/(Rv*(T+273));\n",
-      "print'the partial pressure of dry air %f kPa'%round(Pa,2);\n",
-      "print'the specific humidity %f kg water/kg of dry air'%round(w,4);\n",
-      "print'the enthalpy per unit mass of the dry air %f kJ'%round(h,1);\n",
-      "print'mass of air %f kg'%round(ma,2);\n",
-      "print'mass of water vapour %f kg'%round(mv,1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the partial pressure of dry air 97.620000 kPa\n",
-        "the specific humidity 0.015100 kg water/kg of dry air\n",
-        "the enthalpy per unit mass of the dry air 64.000000 kJ\n",
-        "mass of air 85.610000 kg\n",
-        "mass of water vapour 1.300000 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14-2 ,Page No.722"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=20;#temperature of air in C\n",
-      "RH=0.75;#relative humidity\n",
-      "\n",
-      "#from Table A-4\n",
-      "Psat=2.3392;\n",
-      "Pv=RH*Psat;\n",
-      "#thus at this from Eq 14-13\n",
-      "Tdp=15.4;\n",
-      "print'window temperature %f C'%Tdp"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "window temperature 15.400000 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14-3 ,Page No.725"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=25.0;#dry bulb temperature in K\n",
-      "T2=15.0;#wet bulb temperature in K\n",
-      "P2=101.325;#pressure of air in kPa\n",
-      "\n",
-      "#from Table A-2a & A-4\n",
-      "#at T1\n",
-      "Psat1=3.1698;\n",
-      "hg1=2546.5;\n",
-      "#at T2\n",
-      "Psat2=1.7057;\n",
-      "hfg2=2465.4;\n",
-      "hf2=62.982;\n",
-      "cp=1.005;\n",
-      "\n",
-      "#calculations\n",
-      "w2=0.622*Psat2/(P2-Psat2);\n",
-      "w1=(cp*(T2-T1)+w2*hfg2)/(hg1-hf2);\n",
-      "print'the specific humidity %f kg water/kg of dry air'%round(w1,5);\n",
-      "RH1=w1*P2/((0.622+w1)*Psat1);\n",
-      "print'the relative humidity is %f'%round(RH1,3);\n",
-      "h=cp*T1+w1*hg1;\n",
-      "print'the enthalpy of the air %f kJ/kg of dry air'%round(h,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the specific humidity 0.006530 kg water/kg of dry air\n",
-        "the relative humidity is 0.332000\n",
-        "the enthalpy of the air 41.700000 kJ/kg of dry air\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14-5 ,Page No.731"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "RH1=0.3;#outdoor relative humidity\n",
-      "P1=100.0;#pressure of air in kPa\n",
-      "V1=45.0;#volumetric steady rate in m^3/min\n",
-      "T1=10.0;#outdoor temperature in C\n",
-      "T2=22.0;#first heated temperature in C\n",
-      "RH3=0.6;#final relative humidity \n",
-      "T3=25.0;#final temperature in C\n",
-      "\n",
-      "#from Table A-2a & A-4\n",
-      "cp=1.005;\n",
-      "Ra=0.287;\n",
-      "Pg1=1.2281;\n",
-      "hg1=2519.2;\n",
-      "hg2=2541.0;\n",
-      "Pg3=3.1698;\n",
-      "\n",
-      "#calculations\n",
-      "Pv1=RH1*Pg1;\n",
-      "Pa1=P1-Pv1;\n",
-      "v1=Ra*(T1+273)/Pa1;\n",
-      "ma=V1/v1;\n",
-      "w1=0.622*Pv1/(P1-Pv1);\n",
-      "h1=cp*T1+w1*hg1;\n",
-      "h1=round(h1,1);\n",
-      "w2=w1;\n",
-      "h2=cp*T2+w2*hg2;\n",
-      "h2=round(h2)\n",
-      "Q=ma*(h2-h1);\n",
-      "# ma2*w2 + mw = ma3*w3\n",
-      "#which reduces to mw = ma * (w3 - w2)\n",
-      "w3=0.622*RH3*Pg3/(P1-(RH3*Pg3));\n",
-      "mw=ma*(w3-w2);\n",
-      "print'the rate of heat supply in the heating section %f kJ/min'%round(Q);\n",
-      "print'the mass flow rate of the steam required in the humidifying section %f kg/min'%round(mw,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of heat supply in the heating section 673.000000 kJ/min\n",
-        "the mass flow rate of the steam required in the humidifying section 0.539000 kg/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14-6 ,Page No.733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V1=10;#volumertric rate in m^3/min\n",
-      "T1=30;#intial temperature in C\n",
-      "RH1=0.8;#intial relative humidity\n",
-      "T2=14;#final temperature in C\n",
-      "RH2=1;#final relative humidity\n",
-      "\n",
-      "#from Table A-4\n",
-      "hw=58.8;\n",
-      "h1=85.4;\n",
-      "h2=39.3;\n",
-      "w1=0.0216;\n",
-      "w2=0.0100;\n",
-      "v1=0.889;\n",
-      "\n",
-      "#calculations\n",
-      "#mw= ma*(w1-w2)\n",
-      "#Qout=ma*(h1-h2) - mw*hw\n",
-      "ma=V1/v1;\n",
-      "mw= ma*(w1-w2);\n",
-      "Qout=ma*(h1-h2) - mw*hw;\n",
-      "print'rates of moisture removal from the air %f kg/min'%round(mw,3);\n",
-      "print'rate of moisture removal from the air %i kJ/min'%round(Qout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rates of moisture removal from the air 0.130000 kg/min\n",
-        "rate of moisture removal from the air 511 kJ/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14-8 ,Page No.736"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "V1=50;#saturated air volumertric rate in m^3/min\n",
-      "T1=14;#saturated air temperature in C\n",
-      "V2=20;#outside air volumertric rate in m^3/min\n",
-      "T2=32;#outside air temperature in C\n",
-      "RH2=60;#outside air relative humidity\n",
-      "\n",
-      "#from psychrometric chart\n",
-      "h1=39.4;\n",
-      "w1=0.010;\n",
-      "v1=0.826;\n",
-      "h2=79;\n",
-      "w2=0.0182;\n",
-      "v2=0.889;\n",
-      "\n",
-      "#calculations\n",
-      "ma1=V1/v1;\n",
-      "ma2=V2/v2;\n",
-      "ma3=ma1+ma2;\n",
-      "#from Eq 14-24\n",
-      "w3=(w2*ma2+w1*ma1)/(ma1+ma2);\n",
-      "h3=(h2*ma2+h1*ma1)/(ma1+ma2);\n",
-      "print'the specific humidity %f kg of water/kg of dry air'%round(w3,3);\n",
-      "#from psychrometric chart\n",
-      "T3=19;\n",
-      "RH3=0.89;\n",
-      "v3=0.844;\n",
-      "V3=ma3*v3;\n",
-      "print'the dry-bulb temperature %f C'%T3;\n",
-      "print'the relative humidity is %f'%RH3\n",
-      "print'the volume flow rate of the mixture %f m^3/min'%round(V3,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the specific humidity 0.012000 kg of water/kg of dry air\n",
-        "the dry-bulb temperature 19.000000 C\n",
-        "the relative humidity is 0.890000\n",
-        "the volume flow rate of the mixture 70.100000 m^3/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14-9 ,Page No.738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data \n",
-      "m=100;#mass flow rate in kg/s\n",
-      "T1=20;#entry temperature in C\n",
-      "P1=1;#entry pressure im kPa\n",
-      "RH1=60;#entry relative humidity\n",
-      "T2=30;#exit temperature in C\n",
-      "RH2=1;#exit relative humidity\n",
-      "T3=35;#entry tempearture of wet cooling tower in C \n",
-      "T4=22;#water temperature in cooling tower in C\n",
-      "\n",
-      "#from Table A-4\n",
-      "h1=42.2;\n",
-      "w1=0.0087;\n",
-      "v1=0.842;\n",
-      "h2=100;\n",
-      "w2=0.0273;\n",
-      "h3=146.64;\n",
-      "h4=92.28;\n",
-      "\n",
-      "#calculations\n",
-      "#Dry air balane = ma1 = ma2 = ma\n",
-      "#Water balance = m3 - m4 = ma*(w2 - w1)\n",
-      "#Energy balance = ma1*h1 + m3*h3 = ma2*h2 + m4*h4\n",
-      "ma= m*(h3-h4)/(h2-h1-(w2-w1)*h4);\n",
-      "V1=ma*v1;\n",
-      "mmakeup=ma*(w2-w1);\n",
-      "print'the volume flow rate of air into the cooling tower %f m^3/s'%round(V1,1);\n",
-      "print'the mass flow rate of the required makeup water %f kg/s'%round(mmakeup,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the volume flow rate of air into the cooling tower 81.600000 m^3/s\n",
-        "the mass flow rate of the required makeup water 1.800000 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter15.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter15.ipynb
deleted file mode 100755
index c167fe73..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter15.ipynb
+++ /dev/null
@@ -1,686 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 15: Chemical Reactions"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-1 ,Page No.755"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "nO2i=20;#intial moles of air \n",
-      "nC8H18i=1;#intial moles octane\n",
-      "\n",
-      "#from Table A-1\n",
-      "Mair=29;\n",
-      "MC=12;\n",
-      "MH=2;\n",
-      "\n",
-      "#calculations\n",
-      "# Chemical Reaction\n",
-      "# C8H18 + 20(O2+3.76N2)= xCO2 + yH2O +  zO2 + wN2\n",
-      "#by elemental balance of moles\n",
-      "x=8;\n",
-      "y=18/2;\n",
-      "z=20*2-2*x-y;\n",
-      "w=20*3.76;\n",
-      "print'kmoles of CO2 %i'%x;\n",
-      "print'kmoles of H2O %i'%y;\n",
-      "print'kmoles of O2 %f'%round(z,1);\n",
-      "print'kmoles of N2 %f'%round(w,1);\n",
-      "#thus equn becomes\n",
-      "# C8H18 + 20(O2+3.76N2)= 8CO2 + 9H2O +  7.5O2 +75.2N2\n",
-      "AF=nO2i*4.76*Mair/(x*MC + y*MH);\n",
-      "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "kmoles of CO2 8\n",
-        "kmoles of H2O 9\n",
-        "kmoles of O2 15.000000\n",
-        "kmoles of N2 75.200000\n",
-        "air-fuel ratio of combustion process 24.200000 kg air/kg fuel\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-2 ,Page No.757"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P=100;#total pressure in kPa\n",
-      "\n",
-      "#from Table A-1\n",
-      "Mair=29.0;\n",
-      "MC=12.0;\n",
-      "MH=2.0;\n",
-      "\n",
-      "#calculations\n",
-      "#Chemical reaction\n",
-      "#C2H6 +  1.2at(1O2 +  3.76) =2CO2 +  3H2O +  0.2athO2 + (1.2*3.76)athN2\n",
-      "#ath is the stoichiometric coefficient for air\n",
-      "#Oxygen balance gives\n",
-      "# 1.2ath = 2 + 1.5 + 0.2ath\n",
-      "ath=(2+1.5)/(1.2-0.2);\n",
-      "AF=(1.2*ath)*4.76*Mair/(2*MC+3*MH);\n",
-      "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,1);\n",
-      "#C2H6 + 4.2(O2 + 3.76N2)  = 2CO2 + 3H2O + 0.7O2 + 15.79N2;\n",
-      "Nprod=2+3+0.7+15.79;\n",
-      "#for dew point water vapour condenses\n",
-      "Nv=3;\n",
-      "Pv=Nv/Nprod*P;\n",
-      "#at this Pv\n",
-      "Tdp=52.3;\n",
-      "print'the dew-point %f C'%Tdp\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "air-fuel ratio of combustion process 19.300000 kg air/kg fuel\n",
-        "the dew-point 52.300000 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-3 ,Page No.758"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P=101.325;#total pressure in kPa\n",
-      "RH=0.8;#realtive humidity\n",
-      "T1=20;#tempearture of air in C\n",
-      "\n",
-      "#from Table A-4\n",
-      "Psat=2.3392;\n",
-      "\n",
-      "#calculations\n",
-      "#consedering 1 kmol of fuel\n",
-      "# 0.72CH4 + 0.09H2 + 0.14N2 + 0.02O2 + 0.03CO2 +  ath(O2 + 3.76N2) = xCO2 + yH2O + zN2\n",
-      "#element balance\n",
-      "x=0.72+0.03\n",
-      "y=(0.72*4+0.09*2)/2;\n",
-      "ath=x+y/2-0.02-0.03;\n",
-      "z=0.14+3.76*ath;\n",
-      "Pv=RH*Psat;\n",
-      "# Nv,air = Pv,air/Ptotal * Ntotal\n",
-      "Nvair=Pv/P*6.97/(1-(Pv/P));\n",
-      "#0.72CH4 + 0.09H2 + 0.14N2 + 0.02O2 + 0.03CO2 +  1.465(O2 + 3.76N2) + 0.131H20 = 0.75CO2 + 1.661H2O + 5.648N2\n",
-      "Pvprod=1.661/8.059*P;\n",
-      "#at this Pvprod\n",
-      "Tdp=60.9;\n",
-      "print'the dew-point %f C'%Tdp"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the dew-point 60.900000 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-4 ,Page No.760"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Pprod=100;#total pressure in kPa\n",
-      "\n",
-      "#from Table A-1\n",
-      "Mair=29;\n",
-      "MC=12;\n",
-      "MH=2;\n",
-      "\n",
-      "#from Table A-4\n",
-      "Psat=3.1698;\n",
-      "\n",
-      "#calculations\n",
-      "#consedering 100 kmol of dry products\n",
-      "# xC8H18 + a (O2 + 3.76N2) = 10.02CO2 + 0.88C0 + 84.48N2 + bH20\n",
-      "#from mass balamces\n",
-      "a=83.48/3.76;\n",
-      "x=(0.88+10.02)/8;\n",
-      "b=18*x/2;\n",
-      "# 1.36C8H18 + 22.2 (O2 + 3.76N2) = 10.02CO2 + 0.88C0 + 84.48N2 + 12.24H20\n",
-      "# 1 mol conversion\n",
-      "# C8H18 + 16.32 (O2 + 3.76N2) = 7.37CO2 + 4.13C0 + 61.38N2 + 9H20\n",
-      "AF= 16.32*4.76*Mair/(8*MC+9*MH);\n",
-      "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,2);\n",
-      "# C8H18 + ath (O2 + 3.76N2) = 8CO2 + 9H2O + 3.76athN2\n",
-      "ath=8+4.5;\n",
-      "Pth=16.32/ath*4.76/4.76*100;\n",
-      "print'percentage of theoretical air is %i'%round(Pth);\n",
-      "Nprod=7.37+0.65+4.13+61.98+9;\n",
-      "# Nv/Nprod = Pv/Pprod\n",
-      "Pv=Psat;\n",
-      "Nw= (Nprod*Pv-9*Pprod)/(Pv-Pprod);\n",
-      "print'the amount of H2O that condenses as the products %f kmol'%round(Nw,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "air-fuel ratio of combustion process 19.760000 kg air/kg fuel\n",
-        "percentage of theoretical air is 131\n",
-        "the amount of H2O that condenses as the products 6.570000 kmol\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-5 ,Page No.764"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#there is a difference in the answer due to approximation in the textbook\n",
-      "\n",
-      "#given data\n",
-      "T=25;#temperature of octane in C\n",
-      "\n",
-      "#from Table A-6\n",
-      "HCO2=-393520;\n",
-      "HH2O=-285830;\n",
-      "HC8H18=-249950;\n",
-      "\n",
-      "#calculations\n",
-      "# C8H18 + ath (O2 + 3.76N2) = 8CO2 + 9H2O + 3.76athN2\n",
-      "#N2 and O2 are stable elements, and thus their enthalpy of formation is zero\n",
-      "#hc = Hprod - Hreact\n",
-      "hc= 8*HCO2 + 9*HH2O - HC8H18;\n",
-      "print'the enthalpy of combustion of liquid octane %i kJ/kmol'%hc\n",
-      "print 'or %i kJ/kg C8H18'%round(hc/114,0)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the enthalpy of combustion of liquid octane -5470680 kJ/kmol\n",
-        "or -47989 kJ/kg C8H18\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-6 ,Page No.767"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "mfuel=0.05;#mass flow rate in kg/min\n",
-      "\n",
-      "#from Table A-1\n",
-      "Mair=29;\n",
-      "MC=12;\n",
-      "MH=2;\n",
-      "\n",
-      "#calculation\n",
-      "#stochiometric reaction\n",
-      "#C3H8 + ath(O2 + 3.76N2) = 3CO2 + 4H2O + 3.76athN2\n",
-      "#O2 balance\n",
-      "ath=3+5;\n",
-      "#50 percent excess air and some CO in the products\n",
-      "#C3H8 + 7.5(O2 + 3.76N2) = 2.7CO2 + 0.3CO + 4H2O + 2.65O2+ 28.2N2\n",
-      "AF=7.5*4.76*Mair/(3*MC+4*MH);\n",
-      "mair=AF*mfuel;\n",
-      "print'the mass flow rate of air %f kg air/min'%round(mair,2);\n",
-      "#from property tables\n",
-      "#C3H8 designated as p\n",
-      "hfp=-118910;\n",
-      "#oxygen as o\n",
-      "hfo=0;\n",
-      "ho280=8150;\n",
-      "ho298=8682;\n",
-      "ho1500=49292;\n",
-      "#nitrogen as n\n",
-      "hfn=0;\n",
-      "hn280=8141;\n",
-      "hn298=8669;\n",
-      "hn1500=47073;\n",
-      "#water as w\n",
-      "hfw=-241820;\n",
-      "hw298=9904;\n",
-      "hw1500=57999;\n",
-      "#carbondioxode as c\n",
-      "hfc=-393520;\n",
-      "hc298=9364;\n",
-      "hc1500=71078;\n",
-      "#carbon monoxide as co\n",
-      "hfco=-110530;\n",
-      "hco298=8669;\n",
-      "hco1500=47517;\n",
-      "qout=1*(hfp)+7.5*(hfo+ho280-ho298)+28.2*(hfn+hn280-hn298)-2.7*(hfc+hc1500-hc298)-0.3*(hfco+hco1500-hco298)-4*(hfw+hw1500-hw298)-2.65*(hfo+ho1500-ho298)-28.2*(hfn+hn1500-hn298);\n",
-      "#for kg of propane\n",
-      "qout=qout/44;\n",
-      "Qout=mfuel*qout/60;\n",
-      "print'the rate of heat transfer from the combustion chamber %f kW'%round(Qout,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the mass flow rate of air 1.180000 kg air/min\n",
-        "the rate of heat transfer from the combustion chamber 6.890000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-7 ,Page No.769"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#the 0.175% error in last part is due to the approximation in the textbook\n",
-      "\n",
-      "#given data\n",
-      "Preact=1.0;#total pressure in kPa\n",
-      "Treact=77+460.0;#reaction temperature in R\n",
-      "Tprod=1800.0;#final temperature in R\n",
-      "\n",
-      "#constants used\n",
-      "Ru=1.986;\n",
-      "\n",
-      "#calculation\n",
-      "#CH4 + 3O2 = CO2 + 2H2O + O2\n",
-      "Nreact=4;\n",
-      "Nprod=4;\n",
-      "Pprod=Preact*Nprod/Nreact*Tprod/Treact;\n",
-      "print'the final pressure in the tank %f atm'%round(Pprod,2);\n",
-      "#from std. values of heat of formation and ideal gasses in Appendix\n",
-      "#CH4 as m\n",
-      "hfm=-32210.0;\n",
-      "#O2 as o\n",
-      "hfo=0;\n",
-      "h537o=3725.1;\n",
-      "h1800o=13485.8;\n",
-      "#water as w\n",
-      "hfw=-104040.0;\n",
-      "h537w=4528.0;\n",
-      "h1800w=15433.0\n",
-      "#carbondioxide as c\n",
-      "hfc=-169300.0;\n",
-      "h537c=4027.5;\n",
-      "h1800c=18391.5;\n",
-      "Qout=1*(hfm-Ru*Treact)+3*(hfo-Ru*Treact)-1*(hfc+h1800c-h537c-Ru*Tprod)-2*(hfw+h1800w-h537w-Ru*Tprod)-1*(hfo+h1800o-h537o-Ru*Tprod);\n",
-      "print'the heat transfer during this process %i Btu/lbmol'%round(Qout)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final pressure in the tank 3.350000 atm\n",
-        "the heat transfer during this process 309269 Btu/lbmol\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-8 ,Page No.771"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#this invovles EES hence the below code explains a approach with approximation\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "#C8H18 + 12.5 (O2 + 3.76N2) = 8CO+ 9H2O + 47N2\n",
-      "#from std. values of heat of formation and ideal gasses in Appendix\n",
-      "#octane as oc\n",
-      "hfoc=-249950.0;\n",
-      "#oxygen as o\n",
-      "hfo=0;\n",
-      "h298o=8682.0;\n",
-      "#nitrogen as n\n",
-      "hfn=0;\n",
-      "h298n=8669.0;\n",
-      "#water as w\n",
-      "hfw=-241820.0;\n",
-      "h298w=9904.0;\n",
-      "#carbondioxide as c\n",
-      "hfc=-393520.0;\n",
-      "h298c=9364.0;\n",
-      "#x refers to 8hCO2 + 9hH20 + 47hN2\n",
-      "xac=1*(hfoc)+8*(h298c-hfc)+9*(h298w-hfw)+47*(h298n-hfn);\n",
-      "#from EES the Tprod is determined by trial and error\n",
-      "#at 2400K\n",
-      "x2400=5660828.0;\n",
-      "#at 2350K\n",
-      "x2350=5526654.0;\n",
-      "#the actual value of x is xac and T can be determined by interpolation\n",
-      "Tprod=(xac-x2350)*(2400.0-2350.0)/(x2400-x2350)+2350.0;\n",
-      "print'adiabatic flame temperature for complete combustion with 100 percent theoretical air %i K'%round(Tprod);\n",
-      "\n",
-      "#part - b\n",
-      "#C8H18 + 50 (O2 + 3.76N2) = 8CO+ 9H2O + 37.5O2 + 188N2\n",
-      "#solved similarly using EES and approximation and interpolation\n",
-      "#similarly we can solve the part - c \n",
-      "#the above concept is applied\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "adiabatic flame temperature for complete combustion with 100 percent theoretical air 2395 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-9 ,Page No.776"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#from Table A-26E\n",
-      "#Gibbs function of formation at 77\u00b0F\n",
-      "gfc=0;#for carbon\n",
-      "gfo=0;#for oxygen\n",
-      "gfco=-169680;#for carbondioxide\n",
-      "\n",
-      "#calculations\n",
-      "# C + O2 = CO2\n",
-      "Wrev=1*gfc+1*gfo-1*gfco;\n",
-      "print'the reversible work for this process %i Btu'%round(Wrev) "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the reversible work for this process 169680 Btu\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-10 ,Page No.777"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#given values\n",
-      "T0=298;#combustion chamber temperature in K\n",
-      "\n",
-      "#contansts used \n",
-      "Ru=8.314;#in kJ/kmol K\n",
-      "\n",
-      "#calculations\n",
-      "# CH4 + 3(O2 + 3.76N2)  = CO2 + 2H2O + O2 + 11.28N2\n",
-      "#from std. values of heat of formation and ideal gasses in Appendix\n",
-      "#methane as m\n",
-      "hfm=-74850;\n",
-      "#oxygen as o\n",
-      "hfo=0;\n",
-      "h298o=8682;\n",
-      "#nitrogen as n\n",
-      "hfn=0;\n",
-      "h298n=8669;\n",
-      "#water as w\n",
-      "hfw=-241820;\n",
-      "h298w=9904;\n",
-      "#carbondioxide as c\n",
-      "hfc=-393520;\n",
-      "h298c=9364;\n",
-      "#x refers to hCO2 + 2hH2O + 11.28hN2\n",
-      "xac=1*(hfm)+1*(h298c-hfc)+2*(h298w-hfw)+11.28*(h298n-hfn);\n",
-      "#from EES the Tprod is determined by trial and error\n",
-      "Tprod=1789;\n",
-      "print'the temperature of the products %i K'%round(Tprod);\n",
-      "#entropy calculations by using table A-26\n",
-      "#Si = Ni*(si - Ruln yiPm\n",
-      "#reactants\n",
-      "Sm=1*(186.16-Ru*log(1*1));\n",
-      "So=3*(205.04-Ru*log(0.21*1));\n",
-      "Sn=11.28*(191.61-Ru*log(.79*1));\n",
-      "Sreact=Sm+So+Sn;\n",
-      "#products\n",
-      "Nt=1+2+1+11.28;#total moles\n",
-      "yc=1/Nt;\n",
-      "yw=2/Nt;\n",
-      "yo=1/Nt;\n",
-      "yn=11.28/Nt;\n",
-      "Sc=1*(302.517-Ru*log(yc*1));\n",
-      "Sw=2*(258.957-Ru*log(yw*1));\n",
-      "So=1*(264.471-Ru*log(yo*1));\n",
-      "Sn=11.28*(247.977-Ru*log(yn*1));\n",
-      "Sprod=Sc+Sw+So+Sn;\n",
-      "Sgen=Sprod-Sreact;\n",
-      "print'exergy destruction %i kJ/kmol - K'%round(Sgen);\n",
-      "Xdestroyed=T0*Sgen/1000;#factor of 1000 for converting kJ to MJ\n",
-      "print'%i MJ/kmol'%round(Xdestroyed);\n",
-      "#This process involves no actual work. Therefore, the reversible work and energy destroyed are identical\n",
-      "Wrev=Xdestroyed;\n",
-      "print'the reversible work %i MJ/kmol'%round(Wrev)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the temperature of the products 1789 K\n",
-        "exergy destruction 966 kJ/kmol - K\n",
-        "288 MJ/kmol\n",
-        "the reversible work 288 MJ/kmol\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15-11 ,Page No.778"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#given values\n",
-      "Tsurr=298;#temperature of surroundings in K\n",
-      "\n",
-      "#contansts used \n",
-      "Ru=8.314;#in kJ/kmol K\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "# CH4 + 3(O2 + 3.76N2)  = CO2 + 2H2O + O2 + 11.28N2\n",
-      "#The amount of water vapor that remains in the products is determined as in Example 15\u20133\n",
-      "Nv=0.43;#moles of water vapour\n",
-      "Nw=1.57;#moles of water in liquid\n",
-      "#hf values\n",
-      "#methane as m\n",
-      "hfm=-74850;\n",
-      "#carbondioxide as c\n",
-      "hfc=-393520;\n",
-      "#water vapour as v\n",
-      "hfv=-241820;\n",
-      "#water in liquid as w\n",
-      "hfw=-285830;\n",
-      "Qout=1*hfm-1*hfc-Nv*hfv-Nw*hfw;\n",
-      "print'Qout = %i kJ/kmol'%round(Qout)\n",
-      "\n",
-      "#part - b\n",
-      "#entropy calculations by using table A-26\n",
-      "#Si = Ni*(si - Ruln yiPm\n",
-      "#reactants\n",
-      "Sm=1*(186.16-Ru*log(1*1));\n",
-      "So=3*(205.04-Ru*log(0.21*1));\n",
-      "Sn=11.28*(191.61-Ru*log(.79*1));\n",
-      "Sreact=Sm+So+Sn;\n",
-      "#products\n",
-      "Nt=Nv+1+1+11.28;#total moles\n",
-      "yw=1;\n",
-      "yc=1/Nt;\n",
-      "yv=Nv/Nt;\n",
-      "yo=1/Nt;\n",
-      "yn=11.28/Nt;\n",
-      "Sw=Nw*(69.92-Ru*log(yw*1));\n",
-      "Sc=1*(213.80-Ru*log(yc*1));\n",
-      "Sv=Nv*(188.83-Ru*log(yv*1));\n",
-      "So=1*(205.04-Ru*log(yo*1));\n",
-      "Sn=11.28*(191.61-Ru*log(yn*1));\n",
-      "Sprod=Sc+Sw+So+Sn+Sv;\n",
-      "Sgen=Sprod-Sreact+Qout/Tsurr;\n",
-      "print'Sgen = %i kJ/kmol - K'%round(Sgen);\n",
-      "Xdestroyed=Tsurr*Sgen/1000;#factor of 1000 for converting kJ to MJ\n",
-      "print'exergy destruction %i MJ/kmol'%round(Xdestroyed);\n",
-      "#This process involves no actual work. Therefore, the reversible work and energy destroyed are identical\n",
-      "Wrev=Xdestroyed;\n",
-      "print'the reversible work %i MJ/kmol'%round(Wrev)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Qout = 871406 kJ/kmol\n",
-        "Sgen = 2746 kJ/kmol - K\n",
-        "exergy destruction 818 MJ/kmol\n",
-        "the reversible work 818 MJ/kmol\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter16.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter16.ipynb
deleted file mode 100755
index 4326dc6f..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter16.ipynb
+++ /dev/null
@@ -1,378 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter 16: Chemical and Phase Equilibrium"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-1 ,Page No.798"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import exp\n",
-      "\n",
-      "#given data\n",
-      "T=298.15;#reaction temperature in K\n",
-      "\n",
-      "#from Table A-26\n",
-      "g=455510;\n",
-      "\n",
-      "#constants used\n",
-      "R=8.314;#in kJ/kmol K\n",
-      "\n",
-      "#calculations\n",
-      "# N2 = 2N\n",
-      "dG=2*g;\n",
-      "logKp=-dG/(R*T);\n",
-      "Kp=exp(logKp);\n",
-      "print('Kp = ',Kp)\n",
-      "print'in comparison to Table A-28 ln Kp value of -367.5 our result is %i'%logKp;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "('Kp = ', 2.4396097259977668e-160)\n",
-        "in comparison to Table A-28 ln Kp value of -367.5 our result is -367\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-2 ,Page No.798"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P=10;#pressure in atm\n",
-      "\n",
-      "#constants\n",
-      "vH2=1;\n",
-      "vH=2;\n",
-      "\n",
-      "#calculations\n",
-      "# H2 = 0.9H2 + 0.2H\n",
-      "NH=0.2;\n",
-      "NH2=0.9;\n",
-      "Nt=NH+NH2;\n",
-      "#from Eq. 16-15\n",
-      "Kp=((NH**vH)/(NH2**vH2))*(P/Nt)**(vH-vH2);\n",
-      "#at this value of Kp from Table A-28\n",
-      "T=3535;\n",
-      "print'temperature %i K is'%T;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "temperature 3535 K is\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-6 ,Page No.807"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#reaction\n",
-      "# H2 + 0.5O2 = H2O\n",
-      "#enthalpy datas in kJ/kmol \n",
-      "#of H2\n",
-      "hfH=-241820;\n",
-      "h2000H=82593;\n",
-      "h298H=9904;\n",
-      "#of O2\n",
-      "hfO=0;\n",
-      "h2000O=61400;\n",
-      "h298O=8468;\n",
-      "#of H2O\n",
-      "hfw=0;\n",
-      "h2000w=67881;\n",
-      "h298w=8682;\n",
-      "#Kp data from A-28\n",
-      "Kp2=869.6;\n",
-      "Kp1=18509;\n",
-      "T1=1800;\n",
-      "T2=2200;\n",
-      "\n",
-      "#constants used\n",
-      "Ru=8.314;#in kJ/kmol K\n",
-      "\n",
-      "#calculations\n",
-      "#part - a\n",
-      "hR=1*(hfH+h2000H-h298H)-1*(hfO+h2000O-h298O)-0.5*(hfw+h2000w-h298w);\n",
-      "print'enthalpy of the reaction %i kJ/kmol using enthalpy data'%round(hR);\n",
-      "#part - b\n",
-      "hR=Ru*(T1*T2)/(T2-T1)*log(Kp2/Kp1);\n",
-      "print'enthalpy of the reaction %i kJ/kmol using enthalpy data'%round(hR);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "enthalpy of the reaction -251663 kJ/kmol using enthalpy data\n",
-        "enthalpy of the reaction -251698 kJ/kmol using enthalpy data\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-7 ,Page No.809"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=120+273.15;#temperature of saturated water in K\n",
-      "\n",
-      "#from Table A-4\n",
-      "hf=503.81;\n",
-      "hg=2706;\n",
-      "sf=1.5279;\n",
-      "sg=7.1292;\n",
-      "\n",
-      "#calculations\n",
-      "print('liquid phase');\n",
-      "gf=hf-T*sf;\n",
-      "print'gf value %f kJ/kg'%round(gf,1);\n",
-      "print('vapour phase');\n",
-      "gg=hg-T*sg;\n",
-      "print'gg value %f kJ/kg'%round(gg,1);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "liquid phase\n",
-        "gf value -96.900000 kJ/kg\n",
-        "vapour phase\n",
-        "gg value -96.800000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-8 ,Page No.813"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=15;#lake temperature in C\n",
-      "P=92.0;#atmospheric pressure in kPa\n",
-      "\n",
-      "#from Table A-4\n",
-      "Pv=1.7057;\n",
-      "\n",
-      "#calculations\n",
-      "yv=Pv/P;\n",
-      "print'mole fraction of water vapor at the surface is %f'%round(yv,4);\n",
-      "yw=1-yv;\n",
-      "print'mole fraction of water in the lake is %f percent'%(round(yw)*100)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mole fraction of water vapor at the surface is 0.018500\n",
-        "mole fraction of water in the lake is 100.000000 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-9 ,Page No.214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=17.0;#lake temperature in C\n",
-      "P=92.0;#atmospheric pressure in kPa\n",
-      "\n",
-      "#from Table A-4\n",
-      "Pv=1.96;\n",
-      "\n",
-      "#constants from Table 16-2\n",
-      "H=62000.0;\n",
-      "\n",
-      "#calculations\n",
-      "Pda=P-Pv;#dry air\n",
-      "yda=Pda/H/100;#in bar\n",
-      "print'mole fraction of air is %f'%(yda)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mole fraction of air is 0.000015\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-10 ,Page No.814"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=358;#hydrogen gas temperature in K\n",
-      "P=300/100;#pressure of hydrogen gas in bar\n",
-      "\n",
-      "#constants used\n",
-      "M=2;\n",
-      "s=0.00901;#solubility in kmol/m^3 bar\n",
-      "p=0.027;\n",
-      "\n",
-      "#calculations\n",
-      "pH2=s*P;\n",
-      "print'molar density of H2 %f kmol/m^3'%round(pH2,3);\n",
-      "pH2=p*M;\n",
-      "print'mass density of H2 %f kg/m^3'%round(pH2,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "molar density of H2 0.027000 kmol/m^3\n",
-        "mass density of H2 0.054000 kg/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16-11 ,Page No.815"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "yw=0.30;#water mole fraction\n",
-      "ya=0.70;#ammonia mole fraction\n",
-      "T=40;#mixture temperature in C\n",
-      "\n",
-      "#saturation pressure\n",
-      "pw=7.3851;\n",
-      "pa=1554.33;\n",
-      "#calulations\n",
-      "Pw=yw*pw;\n",
-      "Pa=ya*pa;\n",
-      "Pt=Pw+Pa;\n",
-      "yw=Pw/Pt;\n",
-      "ya=Pa/Pt;\n",
-      "print'mole fraction of water vapour %f'%round(yw,4);\n",
-      "print'mole fraction of ammonia %f'%round(ya,4);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mole fraction of water vapour 0.002000\n",
-        "mole fraction of ammonia 0.998000\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter17.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter17.ipynb
deleted file mode 100755
index 5b1b7ade..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter17.ipynb
+++ /dev/null
@@ -1,812 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 17: Compressible Flow"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-1 ,Page No.826"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V1=250.0;#velocity of aircraft in m/s\n",
-      "T1=255.07;#ambient air temperature in K\n",
-      "P1=54.05;#atmospheric pressure in kPa\n",
-      "h=5000.0;#altitude in m\n",
-      "\n",
-      "#from Table A-2a\n",
-      "cp=1.005;#in kJ/kg-K\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "T01=T1+V1**2/(2*cp*1000);#factor of 1000 to convert kJ to J\n",
-      "P01=P1*(T01/T1)**(k/(k-1));\n",
-      "#given pressure ratio in compressor *\n",
-      "# T02 = T01*(P02/P01)^((k-1)/k)\n",
-      "T02 = T01*(8)**((k-1)/k);\n",
-      "win=cp*(T02-T01);\n",
-      "print'the stagnation pressure at the compressor inlet %f kPa'%round(P01,2);\n",
-      "print'the required compressor work per unit mass %f kJ/kg'%round(win,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the stagnation pressure at the compressor inlet 80.840000 kPa\n",
-        "the required compressor work per unit mass 233.400000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-2 ,Page No.829"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "V=200.0;#air velocity in m/s\n",
-      "T=30+273.0;#air temperature in K\n",
-      "\n",
-      "#from Table A-2a\n",
-      "R=0.287;#in kJ/kg-K\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "c=sqrt(k*R*T*1000);#factor of 1000 to convert kJ to J\n",
-      "print'the speed of sound %i m/s'%round(c);\n",
-      "Ma=V/c;\n",
-      "print'the Mach number at the diffuser inlet is %f'%round(Ma,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the speed of sound 349 m/s\n",
-        "the Mach number at the diffuser inlet is 0.573000\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-3 ,Page No.829"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "T0=200+273.0;#intial temperature in K\n",
-      "P0=1400.0;#intial pressure in kPa\n",
-      "#stagnant temp. & pressure is same as inlet due to small inlet velocity\n",
-      "P=1200.0;#pressure corresponding to a pressure drop of 200 kPa\n",
-      "m=3.0;#mass flow rate in kg/s\n",
-      "\n",
-      "#from Table A-2a\n",
-      "cp=0.846;#in kJ/kg-K\n",
-      "R=0.1889;#in kJ/kg-K\n",
-      "k=1.289;\n",
-      "\n",
-      "#calculations\n",
-      "T=T0*(P/P0)**((k-1)/k);\n",
-      "V=sqrt(2*cp*(T0-T)*1000);#factor of 1000 to convert kJ to J\n",
-      "p=P/(R*T);\n",
-      "A=m/(p*V);\n",
-      "c=sqrt(k*R*T*1000);#factor of 1000 to convert kJ to J\n",
-      "Ma=V/c;\n",
-      "print'velocity %f m/s'%round(V,1);\n",
-      "print'density %f kg/m^3'%round(p,1);\n",
-      "print'flow area %f cm^2'%round((A*10000),1);\n",
-      "print'Mach number is %f'%round(Ma,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "velocity 164.900000 m/s\n",
-        "density 13.900000 kg/m^3\n",
-        "flow area 13.100000 cm^2\n",
-        "Mach number is 0.494000\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-4 ,Page No.836"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T0=200+273.0;#intial temperature in K\n",
-      "P0=1400.0;#intial pressure in kPa\n",
-      "\n",
-      "#from Table A-2a\n",
-      "k=1.289;\n",
-      "\n",
-      "#calculations\n",
-      "#Tc & Tr stands for critical temp and ratio respectively\n",
-      "#Pc & Pr stands for critical temp and ratio respectively\n",
-      "Tr=2/(k+1);\n",
-      "Pr=(2/(k+1))**(k/(k-1));\n",
-      "Tc=Tr*T0;\n",
-      "Pc=Pr*P0;\n",
-      "print'critical temperature %i K'%round(Tc);\n",
-      "print'critical pressure %i kPa'%round(Pc)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "critical temperature 413 K\n",
-        "critical pressure 767 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-5 ,Page No.839"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "Vi=150.0;#air velocity in m/s\n",
-      "Ti=600+273.0;#air temperature in K\n",
-      "Pi=1.0;#air pressure in MPa\n",
-      "At=50.0/10000.0;#nozzle throat area in m^2\n",
-      "\n",
-      "#from Table A-2a\n",
-      "R=0.287;#in kJ/kg-K\n",
-      "cp=1.005;#in kJ/kg-K\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "Toi=Ti+Vi**2/(2*cp*1000);#factor of 1000 to convert kJ to J\n",
-      "Poi=Pi*(Toi/Ti)**(k/(k-1));\n",
-      "#flow is isentropic \n",
-      "#stagnation temp. and pressure values remain constant\n",
-      "To=Toi;\n",
-      "Po=Poi;\n",
-      "#from Table 17\u20132\n",
-      "#The critical-pressure ratio is 0.5283\n",
-      "\n",
-      "#Part a\n",
-      "Pb=0.7;\n",
-      "Pca=Pb/Po;\n",
-      "# Pca > 0.5283\n",
-      "#exit plane pressure is equal to the back pressure\n",
-      "Pt=Pb;\n",
-      "#from Table A\u201332\n",
-      "Mat=0.778;\n",
-      "#Tt/To = 0.892\n",
-      "Tt=0.892*To;\n",
-      "pt=Pt*1000/(R*Tt);#factor of 1000 to convert MPa to kPa\n",
-      "Vt=Mat*sqrt(k*R*Tt*1000);#factor of 1000 to convert kJ to J\n",
-      "ma=pt*At*Vt;\n",
-      "print'the mass flow rate through the nozzle when the back pressure is 0.7 MPa %f kg/s'%round(ma,2);\n",
-      "\n",
-      "#Part b\n",
-      "Pb=0.4;\n",
-      "Pca=Pb/Po;\n",
-      "# Pca < 0.5283\n",
-      "#sonic conditions exists at the exit\n",
-      "Ma=1;\n",
-      "mb=At*(Po*1000)*(sqrt(k*1000/(R*To)))*(2/(k+1))**((k+1)/(2*(k-1)));#factor of 1000 to convert MPa to kPa and kJ to J\n",
-      "print'the mass flow rate through the nozzle when the back pressure is 0.4 MPa %f kg/s'%round(mb,2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the mass flow rate through the nozzle when the back pressure is 0.7 MPa 6.770000 kg/s\n",
-        "the mass flow rate through the nozzle when the back pressure is 0.4 MPa 7.110000 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-6 ,Page No.840"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=400;#intial temperature in K\n",
-      "P1=100;#intial pressure in kPa\n",
-      "Ma1=0.3;#intial mach no\n",
-      "A21=0.8;#A2/A1 as flow area has been reduced by 20 percent\n",
-      "\n",
-      "#assumption\n",
-      "k=1.4;\n",
-      "\n",
-      "#from Table A\u201332\n",
-      "#at Ma1=0.3\n",
-      "#s stands for * symbol\n",
-      "A1s = 2.0351;#A1/As\n",
-      "T10 = 0.9823;#T1/T0\n",
-      "P10 = 0.9305;#P1/P0\n",
-      "A2s = A21*A1s;#A2/As\n",
-      "#at this value of A2/As\n",
-      "T20=0.9701;#T2/T0\n",
-      "P20=0.8993;#P2/P0\n",
-      "Ma2=0.391;\n",
-      "\n",
-      "#calculations\n",
-      "T2=T1*T20/T10;\n",
-      "P2=P1*P20/P10;\n",
-      "print'Ma2 is %f'%round(Ma2,3);\n",
-      "print'T2 %i K is'%round(T2);\n",
-      "print'P2 %f kPa is'%round(P2,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Ma2 is 0.391000\n",
-        "T2 395 K is\n",
-        "P2 96.600000 kPa is\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-7 ,Page No.844"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data \n",
-      "T0=800;#intial temperature in K\n",
-      "P0=1;#intial pressure in MPa\n",
-      "Vi=0;#negligible intial velcity\n",
-      "At=20;#throat area in cm^2\n",
-      "Mae=2;#exit Mach number\n",
-      "\n",
-      "#from Table A-2a\n",
-      "R=0.287;#in kJ/kg-K\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "# Mach no. at exit is 2 hence sonic conditions at throat\n",
-      "p0=P0*1000/(R*T0);#factor of 1000 to convert MPa to kPa\n",
-      "#from Table A-32 at Mat=1\n",
-      "#s stands for * symbol\n",
-      "Ps0 = 0.5283;#Ts/T0\n",
-      "Ts0 = 0.8333;#Ps/P0\n",
-      "ps0=0.6339;#ps/p0\n",
-      "Ps=Ps0*P0;\n",
-      "Ts=Ts0*T0;\n",
-      "ps=ps0*p0;\n",
-      "As=At;\n",
-      "Vs=sqrt(k*R*Ts*1000);#factor of 1000 to convert kJ to J\n",
-      "print('the throat conditions');\n",
-      "print'Presssure %f MPa'%round(Ps,4);\n",
-      "print'Temperature %f K'%round(Ts,1);\n",
-      "print'density %f kg/m^3'%round(ps,3);\n",
-      "print'area %f cm^2'%round(As);\n",
-      "print'velocity %f m/s'%round(Vs,1);\n",
-      "\n",
-      "#part - b\n",
-      "#from Table A-32\n",
-      "#at Mae=2\n",
-      "Te0 = 0.5556;#Te/T0\n",
-      "Pe0 = 0.1278;#Pe/P0\n",
-      "pe0= 0.2300;#pe/p0\n",
-      "Ae0= 1.6875;#Ae/Ao\n",
-      "Pe=Pe0*P0;\n",
-      "Te=Te0*T0;\n",
-      "pe=pe0*p0;\n",
-      "Ae=Ae0*At;\n",
-      "Ve=Mae*sqrt(k*R*Te*1000);#factor of 1000 to convert kJ to J\n",
-      "print('the exit plane conditions, including the exit area');\n",
-      "print'Presssure %f MPa'%round(Pe,4);\n",
-      "print'Temperature %f K'%round(Te,1);\n",
-      "print'density %f kg/m^3'%round(pe,3);\n",
-      "print'area %f cm^2'%round(Ae,2);\n",
-      "print'velocity %f m/s'%round(Ve,1);\n",
-      "#part - c\n",
-      "m=ps*As*Vs/10000;#factor of 10000 to convert cm^2 to m^2\n",
-      "print'the mass flow rate through the nozzle %f kg/s'%round(m,2);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the throat conditions\n",
-        "Presssure 0.528300 MPa\n",
-        "Temperature 666.600000 K\n",
-        "density 2.761000 kg/m^3\n",
-        "area 20.000000 cm^2\n",
-        "velocity 517.500000 m/s\n",
-        "the exit plane conditions, including the exit area\n",
-        "Presssure 0.127800 MPa\n",
-        "Temperature 444.500000 K\n",
-        "density 1.002000 kg/m^3\n",
-        "area 33.750000 cm^2\n",
-        "velocity 845.200000 m/s\n",
-        "the mass flow rate through the nozzle 2.860000 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-9 ,Page No.850"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log,sqrt\n",
-      "\n",
-      "#data from Ex - 17.7\n",
-      "m=2.86;\n",
-      "Ma1=2;\n",
-      "P01=1;\n",
-      "P1=0.1278;\n",
-      "T1=444.5;\n",
-      "p1=1.002;\n",
-      "\n",
-      "#from Table A-2a\n",
-      "R=0.287;#in kJ/kg-K\n",
-      "cp=1.005;#in kJ/kg-K\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "#from Table A-33 at Ma1=2.0\n",
-      "Ma2=0.5774;\n",
-      "P0201=0.7209;#P02/P01\n",
-      "P21=4.5;#P2/P1;\n",
-      "T21=1.6875;#T2/T1\n",
-      "p21=2.6667;#p2/p1\n",
-      "P02=P0201*P01;\n",
-      "P2=P21*P1;\n",
-      "T2=T21*T1;\n",
-      "p2=p21*p1;\n",
-      "print'the stagnation pressure %f MPa'%round(P02,3);\n",
-      "print'the static pressure %f MPa'%round(P2,3);\n",
-      "print'static temperature %f K'%round(T2);\n",
-      "print'static density %f kg/m^3'%round(p2,2);\n",
-      "\n",
-      "#part - b\n",
-      "#s21 = s2 - s1\n",
-      "s21=cp*log(T2/T1)-R*log(P2/P1);\n",
-      "print'the entropy change across the shock %fkJ/kg-K'%round(s21,4);\n",
-      "\n",
-      "#part - c\n",
-      "V2=Ma2*sqrt(k*R*T2*1000);#factor of 1000 to convert kJ to J\n",
-      "print'the exit velocity %f m/s'%round(V2);\n",
-      "\n",
-      "#part - d\n",
-      "print('flow rate is not affected by presence of shock waves amd remains 2.86 kg/sec')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the stagnation pressure 0.721000 MPa\n",
-        "the static pressure 0.575000 MPa\n",
-        "static temperature 750.000000 K\n",
-        "static density 2.670000 kg/m^3\n",
-        "the entropy change across the shock 0.094200kJ/kg-K\n",
-        "the exit velocity 317.000000 m/s\n",
-        "flow rate is not affected by presence of shock waves amd remains 2.86 kg/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-10 ,Page No.858"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sin,pi\n",
-      "\n",
-      "#given data\n",
-      "#using protactor frpm Fig 17-36\n",
-      "u=19;#u stands for angle of the mach lines\n",
-      "\n",
-      "#calculations\n",
-      "#by Eq. 17-47\n",
-      "#i.e u= asin(1/Ma)\n",
-      "Ma=1/sin(u*pi/180);#converting to radians\n",
-      "print'The Mach number is %f'%round(Ma,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Mach number is 3.070000\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-11 ,Page No.858"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sin,pi\n",
-      "\n",
-      "#given data\n",
-      "Ma1=2;#Supersonic air mach no\n",
-      "P1=75;#Supersonic air at pressure in kPa\n",
-      "O=10*pi/180;#converting to radians & angle b/w shock wave and normal\n",
-      "\n",
-      "#constants used\n",
-      "k=1.4;\n",
-      "\n",
-      "#calcualtions\n",
-      "#with given values of Ma1 and O from Eq 17-46\n",
-      "Bweak=39.3*pi/180;#converting to radians\n",
-      "Bstrong=83.7*pi/180;#converting to radians\n",
-      "#Weak shock\n",
-      "Ma1w=Ma1*sin(Bweak);\n",
-      "#Strong shock\n",
-      "Ma1s=Ma1*sin(Bstrong);\n",
-      "#from second part Eq 17-40\n",
-      "Ma2w=0.8032;\n",
-      "Ma2s=0.5794;\n",
-      "#pressure ratio = (2*k*Ma^2 - k + 1)/(k + 1 )\n",
-      "#Weak shock\n",
-      "P2w=P1*(2*k*Ma1w**2 - k + 1)/(k + 1 );\n",
-      "print'pressure for weak shock %i kPa'%round(P2w);\n",
-      "#Strong shock\n",
-      "P2s=P1*(2*k*Ma1s**2 - k + 1)/(k + 1 );\n",
-      "print'pressure for strong shock %i kPa'%round(P2s);\n",
-      "#Weak shock\n",
-      "Ma2=Ma2w/sin(Bweak-O);\n",
-      "print'Mach number downstream for weak shock is %f'%round(Ma2,2);\n",
-      "#Strong shock\n",
-      "Ma2=Ma2s/sin(Bstrong-O);\n",
-      "print'Mach number downstream for strong shock is %f'%round(Ma2,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "pressure for weak shock 128 kPa\n",
-        "pressure for strong shock 333 kPa\n",
-        "Mach number downstream for weak shock is 1.640000\n",
-        "Mach number downstream for strong shock is 0.604000\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Example 17-12 ,Page No.859"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt,pi,atan\n",
-      "\n",
-      "#given data\n",
-      "Ma1=2;#Supersonic air mach no\n",
-      "P1=230;#Supersonic air pressure in kPa\n",
-      "O=10*pi/180;#converting to radians & O stands for angle of the mach lines\n",
-      "\n",
-      "#constants used\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "#Eq. 17\u201349 for the upstream Prandtl\u2013Meyer function\n",
-      "vMa1=sqrt((k+1)/(k-1))*atan(sqrt((k-1)*(Ma1**2-1)/(k+1))*pi/180)-atan(sqrt(Ma1**2-1)*pi/180);#converting to radians\n",
-      "#Eq. 17\u201348 to calculate the downstream Prandtl\u2013Meyer function\n",
-      "vMa2=O+vMa1;\n",
-      "#using equation solver as implict nature of Eq 17-49\n",
-      "Ma2=2.385;\n",
-      "print'downstream Mach number Ma2 is %f'%round(Ma2,3);\n",
-      "#P2 = (P2/P0)/(P1/P0) * P1\n",
-      "P2= (1 + (k-1)*Ma2**2/2 )**(-k/(k-1)) / (1 + (k-1)*Ma1**2/2 )**(-k/(k-1)) * P1;\n",
-      "print'downstream pressure %i kPa'%round(P2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "downstream Mach number Ma2 is 2.385000\n",
-        "downstream pressure 126 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-15 ,Page No.868"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi,sqrt\n",
-      "\n",
-      "#given data\n",
-      "P1=480.0;#intial pressure in kPa\n",
-      "T1=550.0;#intial temperature in K\n",
-      "V1=80.0;#intial velocity in m/s\n",
-      "d1=15.0/100.0;#diameter in m\n",
-      "AF=40.0;#air to fuel ratio\n",
-      "HV=40000.0;#heating value in kJ/kg\n",
-      "\n",
-      "#from Table A-2a\n",
-      "R=0.287;#in kJ/kg-K\n",
-      "cp=1.005;#in kJ/kg-K\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "p1=P1/(R*T1);\n",
-      "A1=pi*d1**2/4;\n",
-      "mair=p1*A1*V1;\n",
-      "mfuel=mair/AF;\n",
-      "Q=mfuel*HV;\n",
-      "q=Q/mair;\n",
-      "T01=T1+V1**2/(2*cp);\n",
-      "c1=sqrt(k*R*T1*1000);#factor of 1000 to convert kJ to J\n",
-      "Ma1=V1/c1;\n",
-      "#exit stagnation energy equation q= Cp (T02 - T01)\n",
-      "T02=T01+q/cp;\n",
-      "#from Table A\u201334\n",
-      "#at Ma1\n",
-      "#s stands for * symbol\n",
-      "T0s=0.1291;#T0/Ts\n",
-      "Ts0=T01/T0s;\n",
-      "T2s=T02/Ts0;#T02/T*0\n",
-      "#from Table A\u201334 at this ratio\n",
-      "Ma2=0.3142;\n",
-      "#Rayleigh flow relations corresponding to the inlet and exit Mach no\n",
-      "#at Ma1\n",
-      "T1s=0.1541;#T1/Ts\n",
-      "P1s=2.3065;#P1/Ps\n",
-      "V1s=0.0668;#V1/Vs\n",
-      "#at Ma2\n",
-      "T2s=0.4389;#T2/Ts\n",
-      "P2s=2.1086;#P2/Ps\n",
-      "V2s=0.2082;#V2/Vs\n",
-      "T2=T2s/T1s*T1;\n",
-      "P2=P2s/P1s*P1;\n",
-      "V2=V2s/V1s*V1;    \n",
-      "print'Mach Number at exit is %f'%round(Ma2,4);\n",
-      "print'Presssure %i MPa'%round(P2);\n",
-      "print'Temperature %i K'%round(T2);\n",
-      "print'velocity %i m/s'%round(V2);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mach Number at exit is 0.314200\n",
-        "Presssure 439 MPa\n",
-        "Temperature 1566 K\n",
-        "velocity 249 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17-16 ,Page No.870"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "P01=2000;#entry pressure in kPa\n",
-      "T1=400;#entry temperature in C\n",
-      "V1=0;#negligible velocity at entry\n",
-      "nN=0.93;#overall nozzle efficiency\n",
-      "m=2.5;#mass flow rate in kg/s\n",
-      "P2=300;#exit pressure in kPa\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part - a\n",
-      "P201=P2/P01;\n",
-      "#critical pressure ratio at this values is 0.546\n",
-      "Pt=0.546*P01;\n",
-      "#at inlet\n",
-      "h1=3248.4;\n",
-      "h01=h1;\n",
-      "s1=7.1292;\n",
-      "#at throat\n",
-      "st=s1;\n",
-      "ht=3076.8;\n",
-      "vt=0.24196;\n",
-      "Vt=sqrt(2*(h01-ht)*1000);#factor of 1000 to convert kJ to J\n",
-      "At=m*vt/Vt;\n",
-      "#at state 2s\n",
-      "s2s=s1;\n",
-      "h2s=2783.6;\n",
-      "#nN = (h01 - h2)/ (h01 - h2s)\n",
-      "h2=h01-nN*(h01-h2s);\n",
-      "#at P2 and h2\n",
-      "v2=0.67723;\n",
-      "s2=7.2019;\n",
-      "V2=sqrt(2*(h01-h2)*1000);#factor of 1000 to convert kJ to J\n",
-      "A2=m*v2/V2;\n",
-      "print'throat area %f cm^2'%round((At*10000),2);\n",
-      "print'exit area %f cm^2'%round((A2*10000),2);\n",
-      "\n",
-      "#part - b\n",
-      "# at st=7.1292\n",
-      "#pressures of 1.115 and 1.065 MPa\n",
-      "#c calculated using tables\n",
-      "c=sqrt((1115-1065)/(1/0.23776 - 1/0.24633)*1000);#factor of 1000 to convert kPa to Pa\n",
-      "Ma=Vt/c;\n",
-      "print'the Mach number at the throat is %f'%round(Ma,2);\n",
-      "# at s2=7.2019\n",
-      "#pressures of 325 and 275 kPa\n",
-      "c=sqrt((325-276)/(1/0.63596 - 1/0.72245)*1000);#factor of 1000 to convert kPa to Pa\n",
-      "Ma=V2/c;\n",
-      "print'the Mach number at the nozzle exit is %f'%round(Ma,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "throat area 10.330000 cm^2\n",
-        "exit area 18.210000 cm^2\n",
-        "the Mach number at the throat is 1.000000\n",
-        "the Mach number at the nozzle exit is 1.820000\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter2.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter2.ipynb
deleted file mode 100755
index db573fc6..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter2.ipynb
+++ /dev/null
@@ -1,588 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2: Energy Conversion and General Energy Analysis"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "    Example 2-1 ,Page No.57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#there is a 0.00490% error as the answer in textbook is expressed in multiple of 10\n",
-      "#Constants used\n",
-      "Hu=6.73*10**10;#Energy liberated by 1 kg of uranium\n",
-      "\n",
-      "# Given values\n",
-      "p=0.75;# assuming the avg density of gasoline in kg/L\n",
-      "V=5;# consumption per day of gasoline in L\n",
-      "Hv=44000; #heat value in kJ/kg\n",
-      "mu=0.1;# mass of uranium used\n",
-      "\n",
-      "#Calculation\n",
-      "mgas=p*V;#mass of gasoline required per day\n",
-      "Egas=mgas*Hv;\n",
-      "Eu=mu*Hu;\n",
-      "d=Eu/Egas;\n",
-      "print'%i number of days the car can run with uranium' %round(d,0)\n",
-      "print'equivalent to %i years' %round(d/365,0)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "40788 number of days the car can run with uranium\n",
-        "equivalent to 112 years\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-2 ,Page No.59"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given values\n",
-      "v=8.5;# wind speed in m/s\n",
-      "m=10;# given mass for part - b \n",
-      "mf=1154;# given flowrate for part - c\n",
-      "\n",
-      "#Calculations\n",
-      "e=(v**2)/2;\n",
-      "print'wind energy per unit mass %f J/kg' %round(e,1);\n",
-      "E=m*e;\n",
-      "print'wind energy for 10 kg mass %i J' %E;\n",
-      "E=mf*e/1000;\n",
-      "print'wind energy for mass flow rate of 1154kg/s %f kW'%round(E,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "wind energy per unit mass 36.100000 J/kg\n",
-        "wind energy for 10 kg mass 361 J\n",
-        "wind energy for mass flow rate of 1154kg/s 41.700000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-7 ,Page No.67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "# Given values\n",
-      "T=200;# applied torque in N\n",
-      "n=4000;# shaft rotation rate in revolutions per minute\n",
-      "\n",
-      "#Calculation\n",
-      "Wsh=(2*math.pi*n*T)/1000/60;#factor of 1000 to convert to kW and 60 to convert to sec\n",
-      "print'Power transmitted %f kW'%round(Wsh,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power transmitted 83.800000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-8 ,Page No.69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Constants used\n",
-      "g=9.81;#acceleration due to gravity in m/s^2;\n",
-      "\n",
-      "#Given values\n",
-      "m=1200;#mass of car in kg\n",
-      "V=90/3.6;#velocity ; converting km/h into m/s\n",
-      "d=30*math.pi/180;#angle of slope ; converting into radians\n",
-      "\n",
-      "#Calculation\n",
-      "Vver=V*math.sin(d);#velocity in vertical direction\n",
-      "Wg=m*g*Vver/1000;\n",
-      "print'the addtional power %i kW'%Wg"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the addtional power 147 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-9 ,Page No.69"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "m=900;#mass of car in kg\n",
-      "v1=0;# intial velocity\n",
-      "v2=80/3.6;# final velocity; converting km/h into m/s\n",
-      "t=20;# time taken1\n",
-      "\n",
-      "#Calculation\n",
-      "Wa=m*(v2**2-v1**2)/2/1000;\n",
-      "Wavg=Wa/t;\n",
-      "print'the average power %f kW'%round(Wavg,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the average power 11.100000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-10 ,Page No.74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "Win=100;# work done in the process in kJ\n",
-      "Qout=500;# heat lost in kJ\n",
-      "U1=800;# internal energy of the fluid in kJ\n",
-      "\n",
-      "#Calculations\n",
-      "# Win - Qout = U2- U1 i.e change in internal energy \n",
-      "U2=U1-Qout+Win;\n",
-      "print'final internal of the system %i kJ'%U2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "final internal of the system 400 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-11 ,Page No.75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#given values\n",
-      "Win=20;# power consumption in W\n",
-      "mair=0.25;# rate of air discharge in kg/sec\n",
-      "\n",
-      "#calculation\n",
-      "v=math.sqrt(Win/2/mair)#Win = 1/2*m*v^2\n",
-      "if v >=8:\n",
-      "    print('True');\n",
-      "else:\n",
-      "    print('False')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "False\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-12 ,Page No.76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "Win=200.0;#Power of fan in W\n",
-      "U=6.0;#Overall heat transfer coefficient in W/m^2 C\n",
-      "A=30;#Surface area in m^2\n",
-      "To=25;#Outdoor temperature in C\n",
-      "\n",
-      "#Calculations\n",
-      "Ti= (Win/U/A)+To;# Win = Qout = U*A*(Ti - To)\n",
-      "print'the indoor air temperature %f Celcius'%Ti\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the indoor air temperature 26.111111 Celcius\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-13 ,Page No.76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "Plamp=80.0;#Power of lamp in W\n",
-      "N=30;#no of lamps\n",
-      "t=12;#time period the light is in use in hours/day\n",
-      "y=250;#days in a year light is in function \n",
-      "UC=0.07;#unit cost in $\n",
-      "\n",
-      "#Calculation\n",
-      "LP=Plamp*N/1000;#Lighting power in kW\n",
-      "OpHrs=t*y;#Operating hours\n",
-      "LE=LP*OpHrs;#Lighting energy in kW\n",
-      "LC=LE*UC;#Lighting cost\n",
-      "print'the annual energy cost $%i'%LC\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the annual energy cost $504\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-15 ,Page No.82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "Ein=2.0;#Power of electric burner in kW\n",
-      "n1=0.73;#Efficiency of open burners\n",
-      "n2=0.38;#efficency of gas units\n",
-      "CinH=0.09;#Unit cost of electricity in $\n",
-      "CinB=0.55;#Unit cost of natural gas in $\n",
-      "\n",
-      "#Calculations\n",
-      "QutH= Ein * n1;\n",
-      "print'rate of energy consumption by the heater %f kW'%round(QutH,2);\n",
-      "CutH= CinH / n1;\n",
-      "print'the unit cost of utilized energy for heater $%f/kWh'%round(CutH,3);\n",
-      "QutB= QutH / n2 ;\n",
-      "print'rate of energy consumption by the burner %f kW'%round(QutB,2);\n",
-      "CutB= CinB / n2 / 29.3; # 1 therm = 29.3 kWh\n",
-      "print'the unit cost of utilized energy for burner %f kWh'%round(CutB,3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "rate of energy consumption by the heater 1.460000 kW\n",
-        "the unit cost of utilized energy for heater $0.123000/kWh\n",
-        "rate of energy consumption by the burner 3.840000 kW\n",
-        "the unit cost of utilized energy for burner 0.049000 kWh\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-16 ,Page No.84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#there is a 0.204% error in the last part of the question due to rounding off the intermidiate steps in the solution\n",
-      "\n",
-      "#Constants used\n",
-      "g=9.81;#acceleration due to gravity in m/s^2;\n",
-      "\n",
-      "#Given values\n",
-      "h=50.0;#Depth of water in m\n",
-      "m=5000.0;#mass flow rate of water in kg/sec\n",
-      "Wout=1862.0;#generated electric power in kW\n",
-      "ngen=0.95;#efficiency of turbine\n",
-      "\n",
-      "#calculation\n",
-      "X=g*h/1000.0;# X stands for the differnce b/w change in mechanical energy per unit mass\n",
-      "R=m*X;#rate at which mech. energy is supplied to turbine in kW\n",
-      "nov=Wout/R;#overall efficiency i.e turbine and generator\n",
-      "print'overall efficiency is %f'%round(nov,2);\n",
-      "ntu=nov/ngen;#efficiency of turbine\n",
-      "print'efficiency of turbine is %f'%round(ntu,2);\n",
-      "Wsh=ntu*R;#shaft output work\n",
-      "print'shaft power output %i kW'%round(Wsh,0)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "overall efficiency is 0.760000\n",
-        "efficiency of turbine is 0.800000\n",
-        "shaft power output 1960 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-17 ,Page No.85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "Pstd=4520.0;\n",
-      "Phem=5160.0;#prices of std and high eff motor in USD\n",
-      "R=60*0.7457;#rated power in kW from hp\n",
-      "OpHrs=3500.0;#Operating hours\n",
-      "Lf=1.0;#Load Factor\n",
-      "nsh=0.89;#efficiency of shaft\n",
-      "nhem=0.932;#efficiency of high eff. motor\n",
-      "CU=0.08;#per unit cost in $\n",
-      "\n",
-      "#calculation\n",
-      "PS=R*Lf*(1/nsh-1/nhem);#Power savings = W electric in,standard -   W electric in,efficient\n",
-      "ES=PS*OpHrs;#Energy savings = Power savings * Operating hours\n",
-      "print'Energy savings %i kWh/year'%ES;\n",
-      "CS=ES*CU;\n",
-      "print'Cost savings per year $%i'%CS;\n",
-      "EIC=Phem-Pstd;#excess intial cost\n",
-      "Y=EIC/CS;\n",
-      "print'simple payback period %f years'%round(Y,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy savings 7929 kWh/year\n",
-        "Cost savings per year $634\n",
-        "simple payback period 1.000000 years\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-18 ,Page No.91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "#NOx details\n",
-      "m1=0.0047;#emissions of gas furnaces of NOx in kg/therm\n",
-      "N1=18*10**6;#no. of therms per year \n",
-      "#CO2 details\n",
-      "m2=6.4;#emissions of gas furnaces of CO2 in kg/therm\n",
-      "N2=18*10**6;#no. of therms per year \n",
-      "\n",
-      "#Calculation\n",
-      "NOxSav=m1*N1;\n",
-      "print'NOx savings %f kg/year'%round(NOxSav,1);\n",
-      "CO2Sav=m2*N2;\n",
-      "print'CO2 savings %f kg/year'%round(CO2Sav,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "NOx savings 84600.000000 kg/year\n",
-        "CO2 savings 115200000.000000 kg/year\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2-19 ,Page No.95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constants used\n",
-      "e=0.95;#Emissivity\n",
-      "tc=5.67*10**-8;#thermal conductivity in W/m^2 K^4\n",
-      "\n",
-      "#Given values\n",
-      "h=6;#convection heat transfer coefficient in W/m^2 C\n",
-      "A=1.6;#cross-sectional area in m^2\n",
-      "Ts=29;#average surface temperature in C\n",
-      "Tf=20;#room temperature in C\n",
-      "\n",
-      "#Calculation\n",
-      "#convection rate\n",
-      "Q1=h*A*(Ts-Tf);\n",
-      "#radiation rate\n",
-      "Q2=e*tc*A*((Ts+273)**4-(Tf+273)**4);\n",
-      "Qt=Q1+Q2;\n",
-      "print'the total rate of heat transfer %f W'%round(Qt,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the total rate of heat transfer 168.100000 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter3.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter3.ipynb
deleted file mode 100755
index 9830dbfb..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter3.ipynb
+++ /dev/null
@@ -1,756 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Properties of Pure Substances"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-1 ,Page No.128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "m=50;#mass in kg\n",
-      "T=90;#temperature in C\n",
-      "\n",
-      "#Values from Table A-4\n",
-      "P=70.183;#in kPa\n",
-      "v=0.001036;#in m^3/kg\n",
-      "\n",
-      "#Calculation\n",
-      "V=m*v;#equating dimensions\n",
-      "print'pressure is %f kPa'%round(P,3);\n",
-      "print'total volumne of tank becomes %f m^3'%round(V,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "pressure is 70.183000 kPa\n",
-        "total volumne of tank becomes 0.051800 m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-2 ,Page No.128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "V=2.0;#volumne of saturated water vapor in ft^3\n",
-      "P=50.0;#pressure in psia\n",
-      "\n",
-      "#Values from Table A-5E\n",
-      "T=280.99;#in F\n",
-      "v=8.5175;#in ft^3/lbm\n",
-      "\n",
-      "#caluclation\n",
-      "m=V/v;#dimension analysis\n",
-      "print'Temperature inside cylinder %f F'%round(T,2);\n",
-      "print'mass of vapour inside cylinder %f lbm'%round(m,3);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Temperature inside cylinder 280.990000 F\n",
-        "mass of vapour inside cylinder 0.235000 lbm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-3 ,Page No.128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constants used\n",
-      "Hfg=2257.5;#enthalpy of vaporization in kJ/kg\n",
-      "\n",
-      "#Given values\n",
-      "m=200.0/1000;#mass converting in kg\n",
-      "P=100;#Pressure at which process takes place in kPa\n",
-      "\n",
-      "#Values from Table A-5\n",
-      "vg=1.6941;#specific vol of sat liq\n",
-      "vf=0.001043;#specific vol of vapor\n",
-      "\n",
-      "#Caluclation\n",
-      "vfg=vg-vf;\n",
-      "V=m*vfg;\n",
-      "print'the volume change %f m^3'%round(V,4);\n",
-      "E=m*Hfg;\n",
-      "print'amount of energy transferred to the water %f kJ'%round(E,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the volume change 0.338600 m^3\n",
-        "amount of energy transferred to the water 451.500000 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-4 ,Page No.131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "mt=10#mass of water in kg\n",
-      "mf=8;#mass of water in liquid form in kg\n",
-      "T=90;#temperature in C\n",
-      "\n",
-      "#Values from Table A-4\n",
-      "P=70.183;#in kPa\n",
-      "vf=0.001036;#in m^3\n",
-      "vg=2.3593;#in m^3\n",
-      "\n",
-      "#Caluclation\n",
-      "mg=mt-mf;\n",
-      "V=mf*vf+mg*vg;# V= Vg + Vf\n",
-      "print'the volume of the tank %f m^3'%round(V,2);\n",
-      "print'the pressure in the tank %f kPa'%round(P,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the volume of the tank 4.730000 m^3\n",
-        "the pressure in the tank 70.183000 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-5 ,Page No.131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "m=4;#mass of refrigerant-134a in kg\n",
-      "V=80.0/1000;#volumne converting into m^3\n",
-      "P=160;#pressure in kPa\n",
-      "\n",
-      "#Values from Table A-12\n",
-      "vf=0.0007437;\n",
-      "vg=0.12348;\n",
-      "T=-15.60;\n",
-      "hf=31.21;\n",
-      "hfg=209.90;\n",
-      "\n",
-      "#Caluclations\n",
-      "v=V/m;\n",
-      "#vg>v>vf therefore it is a saturated mix\n",
-      "#hence temp will same as saturation temp\n",
-      "print'the temperature %f celcius'%round(T,2)\n",
-      "x=(v-vf)/(vg-vf);#x=vg/vfg i.e the dryness fraction\n",
-      "print'the quality is %f'%round(x,3);\n",
-      "h=hf+x*hfg;\n",
-      "print'the enthalpy of the refrigerant %f kJ/kg'%round(h,1);\n",
-      "mg=x*m;\n",
-      "Vg=mg*vg;\n",
-      "print'the volume occupied by the vapor phase %f m^3'%round(Vg,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the temperature -15.600000 celcius\n",
-        "the quality is 0.157000\n",
-        "the enthalpy of the refrigerant 64.100000 kJ/kg\n",
-        "the volume occupied by the vapor phase 0.077500 m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-7 ,Page No.133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "P=0.5;#pressure in MPa\n",
-      "h=2890.0;#enthaply in kJ/kg\n",
-      "\n",
-      "#from Table A\u20136\n",
-      "#at P=0.5 MPa\n",
-      "T1=200.0;\n",
-      "h1=2855.8;\n",
-      "T2=250;\n",
-      "h2=2961.0;\n",
-      "# we need linear interpolation \n",
-      "\n",
-      "#calculatiom\n",
-      "#by interpolation we can say that\n",
-      "#h=h1+(T-T1)/(T2-T1)*(h2-h1)\n",
-      "#we have to find T\n",
-      "T=(h-h1)/(h2-h1)*(T2-T1)+T1;\n",
-      "print'temperature of water %f celcius'%round(T,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "temperature of water 216.300000 celcius\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-8 ,Page No.134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "T=80;#temperature of compressed liquid water in C\n",
-      "P=5;#pressure in KPa\n",
-      "\n",
-      "#from Table A\u20137\n",
-      "#at compressed liq given conditions\n",
-      "u=333.82;\n",
-      "\n",
-      "#from Table A-4\n",
-      "#at saturation\n",
-      "usat=334.97;\n",
-      "\n",
-      "#calcualtion\n",
-      "e=(usat-u)/u*100;\n",
-      "print'internal energy of compressed liquid water using data from the compressed liquid table %f kJ/kg '%round(u,2);\n",
-      "print'internal energy of compressed liquid water using saturated liquid data %f kJ/kg '%round(usat,2);\n",
-      "print'the error involved %f the second case'%round(e,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "internal energy of compressed liquid water using data from the compressed liquid table 333.820000 kJ/kg \n",
-        "internal energy of compressed liquid water using saturated liquid data 334.970000 kJ/kg \n",
-        "the error involved 0.340000 the second case\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-9 ,Page No.135"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#part a\n",
-      "print('Part a');\n",
-      "\n",
-      "#given values\n",
-      "P=200;#in KPa\n",
-      "x=0.6;\n",
-      "\n",
-      "#from Table A-5\n",
-      "T=120.21;\n",
-      "uf=504.50;\n",
-      "ufg=2024.6;\n",
-      "\n",
-      "#calcualtions\n",
-      "u=uf+(x*ufg);\n",
-      "print'temperature %f Celcius '%round(T,2);\n",
-      "print'internal energy %f kJ/kg'%round(u,2);\n",
-      "print('saturated liquid\u2013vapor mixture at a pressure of 200 kPa');\n",
-      "\n",
-      "#part b\n",
-      "print('Part b');\n",
-      "\n",
-      "#given values\n",
-      "T=125;#in C\n",
-      "u=1600;#in kJ/kg\n",
-      "\n",
-      "#from Table A\u20134\n",
-      "uf=524.83;\n",
-      "ug=2534.3;\n",
-      "#ug>u>ufg so its aturated liquid\u2013vapor mixture\n",
-      "P=232.23;\n",
-      "\n",
-      "#calculation\n",
-      "ufg=ug-uf;\n",
-      "x=(u-uf)/ufg;\n",
-      "print'Pressure %f kPa'%round(P,2);\n",
-      "print'x is %f'%round(x,3);\n",
-      "print('saturated liquid\u2013vapor mixture at a temp of 125 of celcius');\n",
-      "\n",
-      "#part c\n",
-      "print('Part c');\n",
-      "\n",
-      "#given values\n",
-      "P=1000;#in kPa\n",
-      "u=2950;#in kJ/kg\n",
-      "\n",
-      "#from Table A\u20136\n",
-      "uf=761.39;\n",
-      "ug=2582.8;\n",
-      "#u>ug so its superheated steam\n",
-      "T=395.2;\n",
-      "\n",
-      "#calculation\n",
-      "print'temperature %f Celcius'%round(T,1);\n",
-      "print('superheated vapor at 1MPa');\n",
-      "\n",
-      "#part d\n",
-      "print('Part d');\n",
-      "\n",
-      "#given values\n",
-      "T=75;#in C\n",
-      "P=100;#in kPa\n",
-      "\n",
-      "#from Table A\u20135\n",
-      "Tsat=151.83;\n",
-      "#T<Tsat so it is a compressed liquid\n",
-      "#the given pressure is much lower than the lowest pressure value in the compressed liquid table i.e 5 MPa\n",
-      "#assuming, the compressed liquid as saturated liquid at the given temperature\n",
-      "\n",
-      "#from Table A-4\n",
-      "u=313.99;\n",
-      "print'Internal energy %f kJ/kg'%round(u,2);\n",
-      "print('the compressed liquid condition');\n",
-      "\n",
-      "#Part e\n",
-      "print('Part e');\n",
-      "\n",
-      "#given values\n",
-      "P=850;#in kPa\n",
-      "x=0;\n",
-      "\n",
-      "#x=0 therefore it is a saturateed liquid condition\n",
-      "#from Table A-5\n",
-      "T=172.94;\n",
-      "u=731.00;\n",
-      "print'temperature %f Celcius'%round(T,2);\n",
-      "print'Internal energy %f kJ/kg'%round(u,2);\n",
-      "print('saturateed liquid condition')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part a\n",
-        "temperature 120.210000 Celcius \n",
-        "internal energy 1719.260000 kJ/kg\n",
-        "saturated liquid\u2013vapor mixture at a pressure of 200 kPa\n",
-        "Part b\n",
-        "Pressure 232.230000 kPa\n",
-        "x is 0.535000\n",
-        "saturated liquid\u2013vapor mixture at a temp of 125 of celcius\n",
-        "Part c\n",
-        "temperature 395.200000 Celcius\n",
-        "superheated vapor at 1MPa\n",
-        "Part d\n",
-        "Internal energy 313.990000 kJ/kg\n",
-        "the compressed liquid condition\n",
-        "Part e\n",
-        "temperature 172.940000 Celcius\n",
-        "Internal energy 731.000000 kJ/kg\n",
-        "saturateed liquid condition\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-10 ,Page No.139"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#constants used\n",
-      "R=0.287# in kPa m^3/kg K\n",
-      "\n",
-      "#given values\n",
-      "l=4;#dimensions of room in m\n",
-      "b=5;#dimensions of room in m\n",
-      "h=6;#dimensions of room in m\n",
-      "P=100.0;#pressure in kPa\n",
-      "T=25+273.0;#temperature in Kelvin\n",
-      "\n",
-      "#calculation\n",
-      "V=l*b*h;\n",
-      "m=P*V/R/T;\n",
-      "print'the mass of the air %f kg'%round(m,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the mass of the air 140.300000 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-11 ,Page No.142"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "P=1;#pressure in MPa\n",
-      "T=50+273;#tempearture converting into Kelvin\n",
-      "vgiv=0.021796;#specific vol. given in m^3\n",
-      "\n",
-      "#from Table A-1\n",
-      "R=0.0815;\n",
-      "Pcr=4.059;\n",
-      "Tcr=374.2;\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "#Part A\n",
-      "v1=R*T/(P*1000);\n",
-      "print'specific volume of refrigerant-134a under the ideal-gas assumption %fm^3/kg'%round(v1,6);\n",
-      "e=(v1-vgiv)/vgiv;\n",
-      "print'an error of %f'%round(e,3);\n",
-      "\n",
-      "#Part B\n",
-      "#determine Z from the compressibility chart, we will calculate the reduced pressure and temperature\n",
-      "Pr=P/Pcr;\n",
-      "Tr=T/Tcr;\n",
-      "#from chart\n",
-      "Z=0.84;\n",
-      "v=Z*v1;\n",
-      "print'specific volume of refrigerant-134a under the generalized compressibility chart %f m^3/kg'%round(v,6);\n",
-      "e=(v-vgiv)/vgiv;\n",
-      "print'an error of %f'%round(e,3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "specific volume of refrigerant-134a under the ideal-gas assumption 0.026325m^3/kg\n",
-        "an error of 0.208000\n",
-        "specific volume of refrigerant-134a under the generalized compressibility chart 0.022113 m^3/kg\n",
-        "an error of 0.015000\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-12 ,Page No.143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "v=0.51431;\n",
-      "T=600;\n",
-      "\n",
-      "#from Table A-1E\n",
-      "R=0.5956;\n",
-      "Pcr=3200;\n",
-      "Tcr=1164.8;\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "#Part A\n",
-      "#from Table A-6E\n",
-      "Pa=1000.0;#in psia\n",
-      "print'from the steam tables %i psia'%Pa;\n",
-      "\n",
-      "#Part B\n",
-      "T=1060;#converted into R from F\n",
-      "Pb=R*T/v;\n",
-      "print'from the ideal-gas equation %i psia'%round(Pb,0);\n",
-      "e=(Pb-Pa)/Pa;\n",
-      "print'treating the steam as an ideal gas would result in an error of %f'%round(e,3)\n",
-      "\n",
-      "#Part C\n",
-      "#calculating the pseudo-reduced specific volume and the reduced temperature\n",
-      "Vr=v/(R*Tcr/Pcr);\n",
-      "Tr=T/Tcr;\n",
-      "#from the compressibility chart\n",
-      "Pr=0.33;\n",
-      "P=Pr*Pcr;\n",
-      "print'from the generalized compressibility chart %i psia'%P\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "from the steam tables 1000 psia\n",
-        "from the ideal-gas equation 1228 psia\n",
-        "treating the steam as an ideal gas would result in an error of 0.228000\n",
-        "from the generalized compressibility chart 1056 psia\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-13 ,Page No.147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Answer of part c-d are having slight difference due to approximation in molar volumne in the textbook which here is caluculated to the approximation of 7 decimal digits\n",
-      "\n",
-      "from math import exp\n",
-      "\n",
-      "#given values\n",
-      "T=175.0;#temperature in C\n",
-      "v=0.00375;#sp. volumne in m^3/kg\n",
-      "Pex=10000;#experimentaion determination\n",
-      "\n",
-      "#from Table A-1\n",
-      "R=0.2968# in kPa m^3/kg  K\n",
-      "\n",
-      "#calculating\n",
-      "\n",
-      "#Part-a\n",
-      "P=R*T/v;\n",
-      "print'using the ideal-gas equation of state %i kPa'%(round(P))\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n",
-      "\n",
-      "#Part-b\n",
-      "#van der Waals constants from Eq. 3-23\n",
-      "a=0.175;\n",
-      "b=0.00138;\n",
-      "#from van der waal eq.\n",
-      "P=R*T/(v-b)-a/v**2;\n",
-      "print'using the van der Waals equation of state  is %i kPa'%(round(P));\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n",
-      "\n",
-      "#Part-c\n",
-      "#constants in the Beattie-Bridgeman equation from Table 3\u20134\n",
-      "A=102.29;\n",
-      "B=0.05378;\n",
-      "c=4.2*10**4;\n",
-      "Ru=8.314;#in kPa m^3/kmol K\n",
-      "M=28.013;#molecular weight in kg/mol\n",
-      "vb=M*v;#molar vol.\n",
-      "P=(Ru*T)/(vb**2)*(1-((c)/(vb*T**3)))*(vb+B)-(A/vb**2);\n",
-      "print'using the Beattie-Bridgeman equationis %i kPa'%(round(P));\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n",
-      "\n",
-      "#Part-d\n",
-      "#constants of Benedict-Webb-Rubin equation from Table 3\u20134\n",
-      "a=2.54;\n",
-      "b=0.002328;\n",
-      "c=7.379*10**4;\n",
-      "alp=1.272*10**-4;\n",
-      "Ao=106.73;\n",
-      "Bo=0.040704;\n",
-      "Co=8.164*10**5;\n",
-      "gam=0.0053;\n",
-      "P=((Ru*T)/vb)+((Bo*Ru*T)-Ao-Co/T**2)/vb**2+(b*Ru*T-a)/vb**3+(a*alp/vb**6)+(c/((vb**3)*(T**2)))*(1 + (gam/vb**2))*exp(-gam/vb**2);\n",
-      "print'using Benedict-Webb-Rubin equation %i kPa'%(round(P));\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "using the ideal-gas equation of state 13851 kPa\n",
-        "error is 38.506667 percent\n",
-        "using the van der Waals equation of state  is 9471 kPa\n",
-        "error is -5.288326 percent\n",
-        "using the Beattie-Bridgeman equationis 10109 kPa\n",
-        "error is 1.092970 percent\n",
-        "using Benedict-Webb-Rubin equation 10004 kPa\n",
-        "error is 0.039256 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-14 ,Page No.152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given value\n",
-      "T=25;#air temperature over a lake in C\n",
-      "\n",
-      "#from table 3-1\n",
-      "Psat=3.17;#in kPa\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Relative Humidity 10%\n",
-      "Pv1=0.1*Psat\n",
-      "#Relative Humidity 80%\n",
-      "Pv2=0.8*Psat\n",
-      "#Relative Humidity 100%\n",
-      "Pv3=1*Psat\n",
-      "\n",
-      "# from table 3-1 Tsat at these Pressures are\n",
-      "T1=-8.0;\n",
-      "T2=21.2;\n",
-      "T3=25.0;\n",
-      "print'with relative humidity 10, temperature is %i in C'%T1\n",
-      "print'with relative humidity 80, temperature is %i in C'%T2\n",
-      "print'with relative humidity 100, temperature is %i in C'%T3\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "with relative humidity 10, temperature is -8 in C\n",
-        "with relative humidity 80, temperature is 21 in C\n",
-        "with relative humidity 100, temperature is 25 in C\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter4.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter4.ipynb
deleted file mode 100755
index b9e964f0..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter4.ipynb
+++ /dev/null
@@ -1,720 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4: Energy Analysis of Closed Systems"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-2 ,Page No.169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "m=10;#mass in lbm\n",
-      "Po=60;#steam oressure in psia\n",
-      "T1=320;#intial temp in F\n",
-      "T2=400;#final temp in F\n",
-      "\n",
-      "#from Table A\u20136E\n",
-      "v1=7.4863;#at 60 psia and 320 F\n",
-      "v2=8.3548;#at 60 psia and 400 F\n",
-      "\n",
-      "#calculations\n",
-      "#W = P dV which on integrating gives W = m * P * (V2 - V1)\n",
-      "W=m*Po*(v2-v1)/5.404;#coverting into Btu from psia-ft^3\n",
-      "print'work done by the steam during this process %f Btu'%round(W,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "work done by the steam during this process 96.400000 Btu\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-3 ,Page No.170"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "#given data\n",
-      "P1=100;#pressure in kPa\n",
-      "V1=0.4;#intial vol in m^3\n",
-      "V2=0.1;#final vol in m^3\n",
-      "\n",
-      "#calculations\n",
-      "#for isothermal W = P1*V1* ln(V2/V1)\n",
-      "W=P1*V1*log(V2/V1);\n",
-      "print'the work done during this process %f kJ'%round(W,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the work done during this process -55.500000 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-4 ,Page No.171"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V1=0.05;#Volumne of gas in m^3\n",
-      "P1=200;#Pressure in kPa\n",
-      "k=150;#Spring constant in kN/m\n",
-      "A=0.25;#Cross-sectional area in m^2\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Part - a\n",
-      "V2=2*V1;\n",
-      "x2=(V2-V1)/A;#printlacement of spring\n",
-      "F=k*x2;#compression force\n",
-      "P2=P1+F/A;#additional pressure is equivalent the compression of spring\n",
-      "print'the final pressure inside the cylinder %i kPa'%P2;\n",
-      "\n",
-      "#Part - b\n",
-      "#work done is equivalent to the area of the P-V curve of Fig 4-10\n",
-      "W=(P1+P2)/2*(V2-V1);#area of trapezoid = 1/2 * sum of parallel sides * dist. b/w them\n",
-      "print'the total work done by the gas %i kJ'%W;\n",
-      "\n",
-      "#Part - c\n",
-      "x1=0;#intial compression of spring\n",
-      "Wsp=0.5*k*(x2**2-x1**2);\n",
-      "print'the fraction of this work done against the spring to compress it %i kJ'%Wsp\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final pressure inside the cylinder 320 kPa\n",
-        "the total work done by the gas 13 kJ\n",
-        "the fraction of this work done against the spring to compress it 3 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-5 ,Page No.174"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "m=0.025;#mass of saturated water vapor in kg\n",
-      "V=120;#rated voltage of heater in V\n",
-      "I=0.2;#rated current in A\n",
-      "t=300;#total time taken in sec\n",
-      "P1=300;#constant pressure in kPa\n",
-      "Qout=3.7;#heat lost in kJ\n",
-      "\n",
-      "#from Table A\u20135\n",
-      "#at P1 the conditon is sat. vap\n",
-      "h1=2724.9;\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Part - a\n",
-      "#therotical proving\n",
-      "\n",
-      "#Part - b\n",
-      "We=V*I*t/1000;#electrical work in kJ\n",
-      "#from eqn 4 -18 i.e derived in earler part\n",
-      "#it states it Ein - Eout = Esystem\n",
-      "# it applies as Win - Qout = H = m (h2 - h1)\n",
-      "h2=(We-Qout)/m+h1;\n",
-      "##from Table A\u20135\n",
-      "#at h2 we get\n",
-      "P2=300;\n",
-      "T=200;\n",
-      "print'the final temperature of the steam %i C'%T\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final temperature of the steam 200 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-6 ,Page No.176"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=5.0;#mass of water in kg\n",
-      "P1=200;#pressure on one side in kPa\n",
-      "T=25;#temperature in C\n",
-      "\n",
-      "#from Table A\u20134\n",
-      "#the liq. is in compressed state at 200 kPa and 25 C\n",
-      "vf=0.001;\n",
-      "vg=43.340;\n",
-      "uf=104.83;\n",
-      "ufg=2304.3;\n",
-      "v1=vf;\n",
-      "u1=uf;\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Part - a\n",
-      "V1=m*v1;\n",
-      "Vtank=2*V1;\n",
-      "print'the volume of the tank %f m^3'%round(Vtank,2);\n",
-      "\n",
-      "#Part - b\n",
-      "V2=Vtank;\n",
-      "v2=V2/m;\n",
-      "#from Table A\u20134 \n",
-      "# at T=25 vf=0.101003 m^3/kg and vg=43.340 m^3/kg\n",
-      "# vf<v2<vg therefore it is saturated liquid\u2013vapor mixture\n",
-      "P2=3.1698;\n",
-      "print'the final pressure %f kPa'%round(P2,4);\n",
-      "\n",
-      "#Part - c\n",
-      "#Ein - Eout = Esystem\n",
-      "#Qin= dU = m(u2 - u1)\n",
-      "x2=(v2-vf)/(vg-vf);\n",
-      "u2=uf+x2*ufg;\n",
-      "Qin=m*(u2-u1);\n",
-      "print'the heat transfer for this process %f kJ'%Qin\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the volume of the tank 0.010000 m^3\n",
-        "the final pressure 3.169800 kPa\n",
-        "the heat transfer for this process 0.265846 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-7 ,Page No.183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad \n",
-      "from pylab import *\n",
-      "\n",
-      "#given data\n",
-      "T1=300;#intial temp of air in K\n",
-      "P=200;#pressure in kPa\n",
-      "T2=600;#final temp in K\n",
-      "M=28.97;#molecular weight in kg/kmol\n",
-      "Ru=8.314;\n",
-      "\n",
-      "#Part - a\n",
-      "#from Table A\u201317\n",
-      "u1=214.07;\n",
-      "u2=434.78;\n",
-      "du=u2-u1;#change in internal energy\n",
-      "print'change in internal energy from data from the air table %f kJ/kg'%round(du,2);\n",
-      "\n",
-      "#Part - b\n",
-      "#from Table A\u20132c\n",
-      "a=28.11;\n",
-      "b=0.1967*10**-2;\n",
-      "c=0.4802*10**-5;\n",
-      "d=-1.966*10**-9;\n",
-      "#  by equation Cp(T)=a+bT+cT^2+dT^3\n",
-      "def intgrnd1(T): \n",
-      " return ((a-Ru)+b*T+c*T**2+d*T**3)\n",
-      "dU, err = quad(intgrnd1, T1, T2) \n",
-      "du=dU/M;\n",
-      "print'change in internal energy the functional form of the specific heat %f kJ/kg'%round(du,1); \n",
-      "\n",
-      "#Part - c\n",
-      "#from Table A\u20132b\n",
-      "Cavg=0.733;\n",
-      "du=Cavg*(T2-T1);\n",
-      "print'change in internal energy the functional form the average specific heat value %i kJ/kg'%round(du,0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "change in internal energy from data from the air table 220.710000 kJ/kg\n",
-        "change in internal energy the functional form of the specific heat 222.500000 kJ/kg\n",
-        "change in internal energy the functional form the average specific heat value 220 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-8 ,Page No.184"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=1.5;#mass in lbm\n",
-      "T1=80;#temperature in F\n",
-      "P1=50;#pressure in psia\n",
-      "W=0.02;#power rating in hp\n",
-      "t=30.0/60;#converting into hrs from min\n",
-      "\n",
-      "#from Table A\u20132Ea\n",
-      "Cv=0.753;\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part a\n",
-      "Wsh=W*t*2545;#in Btu\n",
-      "#Ein - Eout = Esystem\n",
-      "#Wsh = dU = m (u2 - u1) = m * Cv * (T2 - T1)\n",
-      "T2= Wsh/(m*Cv)+T1;\n",
-      "print'the final temperature %f F'%round(T2,1);\n",
-      "\n",
-      "#part b\n",
-      "#using ideal gas eqn\n",
-      "# P1 * V1 / T1 = P2 * T2 /V2\n",
-      "P2= 50 * (T2 +460)/ (T1+460);\n",
-      "# temp should in R therefore + 460\n",
-      "print'the final pressure %f psia'%round(P2,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final temperature 102.500000 F\n",
-        "the final pressure 52.100000 psia\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-9 ,Page No.185"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V1=0.5;#volumne of nitrogen gas in m^3\n",
-      "P=400;#pressure in kPa\n",
-      "T1=27;#temp in C\n",
-      "I=2;#rated current in A\n",
-      "t=5*60;#converting into s from min\n",
-      "V=120;#rated voltage in V\n",
-      "Qout=2800/1000;#in kJ\n",
-      "R=0.297;\n",
-      "\n",
-      "#from Table A\u20132a\n",
-      "Cp=1.039;\n",
-      " \n",
-      "#calculations\n",
-      "P1=P;\n",
-      "We=V*I*t/1000;#in kJ\n",
-      "m=P1*V1/(R*(T1+273));\n",
-      "#Ein - Eout = Esystem\n",
-      "# We,in - Qout = dH =  m (h2 - h1) = m * Cp * (T2 - T1)\n",
-      "T2=(We-Qout)/(m*Cp)+T1;\n",
-      "print'the final temperature of nitrogen in %i C'%T2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final temperature of nitrogen in 57 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-10 ,Page No.187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=150;#initially air pressure in kPa\n",
-      "P2=350;#final pressure in kPa\n",
-      "T1=27+273;#temperatuere in K\n",
-      "V1=400.0/1000;#volumne in m^3\n",
-      "R=0.287;\n",
-      "\n",
-      "#from Table A\u201317\n",
-      "u1=214.07;\n",
-      "u2=1113.52;\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part a\n",
-      "V2=2*V1;\n",
-      "#using ideal gas eqn\n",
-      "# P1 * V1 / T1 = P2 * T2 /V2\n",
-      "T2=P2*V2*T1/(P1*V1);\n",
-      "print'the final temperature %i K'%T2;\n",
-      "\n",
-      "#part b\n",
-      "# Work done is Pdv\n",
-      "W=P2*(V2-V1);\n",
-      "print'the work done by the air %i kJ'%W;\n",
-      "\n",
-      "#part c\n",
-      "#Ein - Eout = Esystem\n",
-      "#Qin - Wout = dU = m(u2 - u1)\n",
-      "m= P1* V1 /(T1 * R);\n",
-      "Q= m*(u2 - u1)+ W;\n",
-      "print'the total heat transferred to the air %i kJ'%round(Q)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final temperature 1400 K\n",
-        "the work done by the air 140 kJ\n",
-        "the total heat transferred to the air 767 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-11 ,Page No.190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=100;#twmperature of liquid water in C\n",
-      "P=15;#pressure in MPa\n",
-      "\n",
-      "#from Table A\u20137\n",
-      "#at P=15 mPa and T = 100 C\n",
-      "hg=430.39;\n",
-      "hf=419.17\n",
-      "vf=0.001;\n",
-      "Psat=101.42;#in kPa\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#part a\n",
-      "h=hg;\n",
-      "print'enthalpy of liquid water by using compressed liquid tables %f kJ/kg'%round(h,2);\n",
-      "\n",
-      "#part b\n",
-      "#Approximating the compressed liquid as a saturated liquid at 100\u00b0C\n",
-      "h=hf;\n",
-      "print'enthalpy of liquid water by approximating it as a saturated liquid %f kJ/kg'%round(h,2);\n",
-      "\n",
-      "#part c\n",
-      "h=hf + vf*(P*1000 - Psat );\n",
-      "print'enthalpy of liquid water by using the correction given by Eq. 4\u201338 %f kJ/kg'%round(h,2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "enthalpy of liquid water by using compressed liquid tables 430.390000 kJ/kg\n",
-        "enthalpy of liquid water by approximating it as a saturated liquid 419.170000 kJ/kg\n",
-        "enthalpy of liquid water by using the correction given by Eq. 4\u201338 434.070000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-12 ,Page No.191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#suffix i for iron\n",
-      "#suffix w for water\n",
-      "#given data\n",
-      "mi=50;#mass in Kg\n",
-      "T1i=80;#temperature in C \n",
-      "Vw=0.5;#volumne in m^3\n",
-      "T1w=25;#temperature in C \n",
-      "v=0.001;#specific volume of liquid water at or about room temperature\n",
-      "\n",
-      "#from Table A\u20133\n",
-      "ci=0.45;\n",
-      "cw=4.18;\n",
-      "\n",
-      "#calculations\n",
-      "mw=Vw/v;\n",
-      "#Ein - Eout = Esystem\n",
-      "# du = 0 i.e (mcdT)iron + (mcdT)water = 0\n",
-      "# mi * ci * (T - T1i) + mw *cw * (T-T1w)\n",
-      "#on rearranging above equn\n",
-      "T= (mi*ci*T1i + mw*cw*T1w)/(mi*ci+mw*cw);\n",
-      "print'the temperature when thermal equilibrium is reached %f C'%round(T,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the temperature when thermal equilibrium is reached 25.600000 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-13 ,Page No.191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "maf=0.15;#mass of affected tissue in kg\n",
-      "caf=3.8;#specific heat of the tissue in kJ/Kg C\n",
-      "dTaf=1.8;#suffix af for affected tissue\n",
-      "mh=1.2;#suffix h for hand ; mass of hand in kg\n",
-      "\n",
-      "#calculations\n",
-      "#Ein - Eout = Esystem\n",
-      "#dUaffected tissue - KEhand = 0\n",
-      "#from above equation we can deduce that\n",
-      "Vhand= sqrt(2*maf*caf*dTaf*1000/mh);#for conversion factor mutiplying by 1000 to get m^2/s^2\n",
-      "print'the velocity of the hand just before impact %f m/s'%round(Vhand,1);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the velocity of the hand just before impact 41.400000 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-14 ,Page No.199"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=90;#mass of man in kg\n",
-      "\n",
-      "#from Tables 4\u20131 and 4\u20132\n",
-      "Ehb=275;#hamburger\n",
-      "Ef=250;#fries\n",
-      "Ec=87;#cola\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "#part a\n",
-      "Ein=2*Ehb+Ef+Ec;\n",
-      "#The rate of energy output for a 68-kg man watching TV is to be 72 Calories/h\n",
-      "Eout=m*72.0/68;\n",
-      "t=Ein/Eout;\n",
-      "print'by watching TV %f hours'%round(t,1);\n",
-      "\n",
-      "#part b\n",
-      "#The rate of energy output for a 68-kg man watching TV is to be 860 Calories/h\n",
-      "Eout=m*860.0/68;\n",
-      "t=Ein/Eout*60#converting in min\n",
-      "t=ceil(t);\n",
-      "print'by fast swimming %f mins'%t;\n",
-      "\n",
-      "#for last question\n",
-      "print('answers be for a 45-kg man energy takes twice as long in each case');\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "by watching TV 9.300000 hours\n",
-        "by fast swimming 47.000000 mins\n",
-        "answers be for a 45-kg man energy takes twice as long in each case\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4-15 ,Page No.199"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "E=75;#in Cal/day\n",
-      "\n",
-      "#calculation\n",
-      "Ereduced=E*365.0;\n",
-      "#The metabolizable energy content of 1 kg of body fat is 33,100 kJ\n",
-      "Ec=33100.0;\n",
-      "mfat=Ereduced/Ec*4.1868;\n",
-      "print'weight this person will lose in one year %f kg'%round(mfat,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "weight this person will lose in one year 3.460000 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter5.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter5.ipynb
deleted file mode 100755
index da9be875..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter5.ipynb
+++ /dev/null
@@ -1,614 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5: Mass and Energy Analysis of Control Volumes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-1 ,Page No.224"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#given data\n",
-      "V=10.0;#volumne of bucket in galon\n",
-      "t=50;#time taken to fill the bucket in sec\n",
-      "p=1;#density of water in kg/L\n",
-      "re=0.8/2/100;#radius of nozzle exit in m\n",
-      "\n",
-      "#calculations\n",
-      "Vd=V/t*3.7854;#factor 0f 3.7854 for gal to L\n",
-      "print'volumne flow rate through hose %f L/s'%round(Vd,3);\n",
-      "m=p*Vd;\n",
-      "print'mass flow rate through hose %f kg/s'%round(m,3);\n",
-      "Ae=math.pi*re**2;\n",
-      "Ve=Vd/Ae/1000;#factor of 1000 for L to m^3\n",
-      "print'average velocity at the nozzle %f m/s'%round(Ve,1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "volumne flow rate through hose 0.757000 L/s\n",
-        "mass flow rate through hose 0.757000 kg/s\n",
-        "average velocity at the nozzle 15.100000 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-2 ,Page No.225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad \n",
-      "from pylab import *\n",
-      "\n",
-      "#given data\n",
-      "Dtank=3*12;#diameter of tank in inches\n",
-      "Djet=0.5;#diameter of water jet in inches\n",
-      "h0=2;#bottom reference in ft\n",
-      "h1=4;#height of water tank in ft\n",
-      "\n",
-      "#constants used \n",
-      "g=32.2;#in ft/s^2\n",
-      "\n",
-      "#calculations\n",
-      "#min - mout = dmCV/dt\n",
-      "#mout = p*(2*g*h*Ajet)^2\n",
-      "#mCV = p*Atank*h\n",
-      "#from these we get dt = Dtank^2/Djet^2 * (dh/(2*g*h)^2)\n",
-      "def intgrnd1(h): \n",
-      " return (Dtank**2/Djet**2*(1/sqrt(2*g*h)))\n",
-      "t, err = quad(intgrnd1, h0, h1) \n",
-      "t=(t/60);#in min\n",
-      "print'time taken to drop to 2 ft %f min'%round(t,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "time taken to drop to 2 ft 12.600000 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-3 ,Page No.229"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P=150;#operating pressure in kPa\n",
-      "Vliquid=0.6/1000.0;#amount of liquid in the cooker in m^3\n",
-      "t=40.0*60;#period of operation in sec\n",
-      "Ac=8*10**-6;#exit area of opening in m^2\n",
-      "\n",
-      "#from Table A-5\n",
-      "#from P = 150 kPa\n",
-      "h=2693.1;\n",
-      "ug=2519.2;\n",
-      "vf=0.001053;\n",
-      "vg=1.1594;\n",
-      "\n",
-      "#calculations\n",
-      "m=Vliquid/vf;\n",
-      "md=m/t;\n",
-      "print'mass flow rate %f kg/s'%md;\n",
-      "V=md*vg/(Ac);\n",
-      "print'exit velocity %f m/s'%round(V,1);\n",
-      "Eflow=h-ug;\n",
-      "Et=h;\n",
-      "print'flow energy %f kJ/kg'%round(Eflow,1);\n",
-      "print'total energy %f kJ/kg'%round(Et,1);\n",
-      "Emass=md*Et;\n",
-      "print'rate at which energy leaves the cooker %f kW'%round(Emass,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mass flow rate 0.000237 kg/s\n",
-        "exit velocity 34.400000 m/s\n",
-        "flow energy 173.900000 kJ/kg\n",
-        "total energy 2693.100000 kJ/kg\n",
-        "rate at which energy leaves the cooker 0.639000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-4 ,Page No.234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=283;#temperature of air in K\n",
-      "P1=80;#entering pressure in kPa\n",
-      "V1=200;#velocity ar inlet in m/s\n",
-      "A1=0.4;#inlet area in m^2\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kPa-m^3/kg-K\n",
-      "\n",
-      "#calulations\n",
-      "v1=R*T1/P1;\n",
-      "m=V1*A1/v1;\n",
-      "print'mass flow rate of air %f kg/s'%round(m,1);\n",
-      "# Ein - Eout = dEsystem / dt\n",
-      "#from Table A-17\n",
-      "h1=283.14;\n",
-      "V2=0;\n",
-      "h2=h1-(V2^2 - V1^2)/2/1000;#factor of 1000 to convert to kJ/kg\n",
-      "#from Table A-17 at this value of h2\n",
-      "T2=303;\n",
-      "print'the temperature %i K is'%T2;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mass flow rate of air 78.800000 kg/s\n",
-        "the temperature 303 K is\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-5 ,Page No.235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=250;#inlet pressure in psia\n",
-      "T1=700;#inlet temp in F\n",
-      "A1=0.2;#inlet area in ft^2\n",
-      "qout=1.2;#heat losses in Btu/lbm\n",
-      "m=10;#mass flow rate in lbm/sec\n",
-      "P2=200;#nozzle pressure in kPa\n",
-      "V2=900;#nozzle velocity in m/s\n",
-      "\n",
-      "#from Table A-6E\n",
-      "v1=2.6883;\n",
-      "h1=1371.4;\n",
-      "\n",
-      "#calculations\n",
-      "V1=m*v1/A1;\n",
-      "print'the inlet velocit %f f/s'%round(V1,1);\n",
-      "# Ein - Eout = dEsystem / dt\n",
-      "h2=h1-qout-(V2**2 - V1**2)/2/25037;#factor of 25037 to convert to Btu/lbm\n",
-      "#at this value h2, from Tablw A-6E\n",
-      "T2=662;\n",
-      "print'exit temperature %i F'%T2\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the inlet velocit 134.400000 f/s\n",
-        "exit temperature 662 F\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-6 ,Page No.236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=280;#intial pressure in kPa\n",
-      "P1=100;#intial temp in K\n",
-      "m=0.02;#mass flow rate in kg/s\n",
-      "qout=16;#heat losses in kJ/kg\n",
-      "P2=600;#final pressure in kPa\n",
-      "T2=400;#final temp in K\n",
-      "\n",
-      "#from Table A-17\n",
-      "h1=280.13;\n",
-      "h2=400.98;\n",
-      "\n",
-      "#calculations\n",
-      "# Ein - Eout = dEsystem / dt\n",
-      "Win=m*qout+m*(h2-h1);\n",
-      "print'the input power of compressor %f kW'%round(Win,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the input power of compressor 2.740000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-7 ,Page No.237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=2;#inlet pressure in MPa\n",
-      "T1=400;#inlet temp in K\n",
-      "V1=50.0;#inlet velocity in m/s\n",
-      "z1=10;#inlet height in m\n",
-      "P2=15;#final pressure in MPa\n",
-      "x2=0.9;#final dryness fraction\n",
-      "V2=180.0;#final velocity in m/s\n",
-      "z2=6;#final height in m\n",
-      "Wout=5*1000;#power output in kW\n",
-      "\n",
-      "#from Table A-6\n",
-      "h1=3248.4;\n",
-      "#similarly for P2\n",
-      "hf=225.94;\n",
-      "hfg=2372.3;\n",
-      "\n",
-      "#constants used \n",
-      "g=9.8;#in m/s^2\n",
-      "\n",
-      "#calcualtions\n",
-      "h2=hf+x2*hfg;\n",
-      "print'difference in enthalpies %f kJ/kg'%round((h2-h1),2);\n",
-      "print'difference in kinetic energy %f kJ/kg'%round((V2**2-V1**2)/2/1000,2);#factor of 1000 to convert to kJ/kg\n",
-      "print'difference in potential energy %f kJ/kg'%round(g*(z2-z1)/1000,2);#factor of 1000 to convert to kJ/kg\n",
-      "wout=-((h2-h1)+(V2**2-V1**2)/2/1000+g*(z2-z1)/1000);#factor of 1000 to convert to kJ/kg\n",
-      "print'work done per unit of mass %f kJ/kg'%round(wout,2);\n",
-      "m=Wout/wout;\n",
-      "print'mass flow rate %f kg/s'%round(m,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "difference in enthalpies -887.390000 kJ/kg\n",
-        "difference in kinetic energy 14.950000 kJ/kg\n",
-        "difference in potential energy -0.040000 kJ/kg\n",
-        "work done per unit of mass 872.480000 kJ/kg\n",
-        "mass flow rate 5.730000 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-8 ,Page No.239"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=0.8;#entering pressure in MPa\n",
-      "P2=0.12;#throttled pressure in MPa\n",
-      "\n",
-      "#from Table A-12\n",
-      "#sat. liq at P1\n",
-      "T1=31.31;\n",
-      "h1=95.47;\n",
-      "#since process is insentropic and at  P2\n",
-      "h2=h1;\n",
-      "hf=22.49;\n",
-      "hg=236.97;\n",
-      "T2=-22.32;\n",
-      "\n",
-      "#calculations\n",
-      "x2=(h2-hf)/(hg-hf);\n",
-      "print'the final state is %f'%round(x2,3);\n",
-      "dT=T2-T1;\n",
-      "print'temperature drop %f C'%round(dT,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final state is 0.340000\n",
-        "temperature drop -53.630000 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-9 ,Page No.241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=140;#inlet temp of hot water in F\n",
-      "T2=50;#inlet temp of cold water in F\n",
-      "T3=110;#steady state output temp in F\n",
-      "P=20;#operating pressure in psia\n",
-      "\n",
-      "#for a compressed liq at given temp\n",
-      "h1=107.99;\n",
-      "h2=18.07;\n",
-      "h3=78.02;\n",
-      "\n",
-      "#calculations\n",
-      "#Mass balance min =  mout So, m1+m2 = m3\n",
-      "#Energy balance Ein = Eout So, m1*h1 + m2*h2 = m3*h3\n",
-      "#combining realations\n",
-      "#m1*h1 + m2*h2 = (m1+m2)*h3\n",
-      "#dividing by m2 and y=m1/m2\n",
-      "#we get, yh1 + h2 = (y+1)*h3\n",
-      "y=(h3-h2)/(h1-h3);\n",
-      "print'the ratio of mass flow rates %f'%round(y,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the ratio of mass flow rates 2.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-10 ,Page No.242"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=15;#water inlet temp in C\n",
-      "P1=300;#water inlet pressure in kPa\n",
-      "T2=25;#water outlet temp in C\n",
-      "T3=70;#R-134a inlet temp in C\n",
-      "P3=1000;#R-134a inlet pressure in kPa\n",
-      "T4=35;#R-134a outlet temp in C\n",
-      "mr=6;#mass flow rate in kg/min\n",
-      "\n",
-      "#from Table A-4, A-13 and A-11\n",
-      "h1=62.982;\n",
-      "h2=104.83;\n",
-      "h3=303.85;\n",
-      "h4=100.87;\n",
-      "\n",
-      "#calculations\n",
-      "#mass balance m1=m2=mw and m3=m4=mr\n",
-      "#energy balance m1*h1 + m3*h3 =  m2*h2 + m4*h4\n",
-      "#combining them mw*(h1-h2) = mr*(h4-h3)\n",
-      "mw= mr*(h4-h3)/(h1-h2);\n",
-      "print'mass flow rate of cooling water %f kg/min'%round(mw,1);\n",
-      "Qin=mw*(h2-h1);\n",
-      "print'heat transfer rate %i kJ/min'%round(Qin)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "mass flow rate of cooling water 29.100000 kg/min\n",
-        "heat transfer rate 1218 kJ/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-11 ,Page No.245"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=17+273;#inlet temp in K\n",
-      "P1=100;#inlet pressure in kPa\n",
-      "V1=150;#inlet volumetric rate in m^3/min\n",
-      "Win=15;#rated power in kW\n",
-      "Qout=200/1000;#heat lost in kJ/s\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kPa-m^3/kg-K\n",
-      "cp=1.005;#in kJ/kg C\n",
-      "\n",
-      "#calculations\n",
-      "v1=R*T1/P1;\n",
-      "m=V1/v1/60;#factor of 6 to convert to s\n",
-      "# Win - Qout = m*cp*(T2-T1)\n",
-      "T2= T1 + (Win - Qout)/(m*cp);\n",
-      "print'exit temperature %f C'%round((T2-273),2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "exit temperature 21.970000 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5-13 ,Page No.250"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V=6.0/1000;#volumne of cooker in m^3\n",
-      "Pgage=75;#operating pressure in kPa\n",
-      "Patm=100;#atmospheric pressure in kPa\n",
-      "m1=1;#mass of water in kg\n",
-      "Qind=0.5;#heat supplying rate in kJ/sec ; d stands for .\n",
-      "t=30*60;#operating time in s\n",
-      "\n",
-      "#calculation\n",
-      "Pabs=Pgage+Patm;\n",
-      "#from Table A-5, ths saturation temp \n",
-      "T=116.04;\n",
-      "print'the temperature at which cooking takes place %f C'%T;\n",
-      "#mass balance me=(m1-m2)cv\n",
-      "#energy balance Qin - mehe = (m2u2 - m1u1)cv\n",
-      "Qin=Qind*t;\n",
-      "#from Table A-5\n",
-      "he=2700.2;\n",
-      "vf=0.001;\n",
-      "vg=1.004;\n",
-      "uf=486.82;\n",
-      "ufg=2037.7;\n",
-      "v1=V/m1;\n",
-      "x1=(v1-vf)/(vg-vf);\n",
-      "u1=uf+x1*ufg;\n",
-      "U=m1*u1;\n",
-      "#Qin = (m1 - V/v2)*he + (V/v2*u2 - m1*u1)\n",
-      "#v2=vf + x2*(vg-vf)\n",
-      "#u2=uf +  x2*ufg\n",
-      "#combining these equations we get\n",
-      "#solved using EES\n",
-      "x2=0.009;\n",
-      "v2=vf + x2*(vg-vf);\n",
-      "m2=V/v2;\n",
-      "print'amount of water left %f kg'%round(m2,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the temperature at which cooking takes place 116.040000 C\n",
-        "amount of water left 0.600000 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter6.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter6.ipynb
deleted file mode 100755
index 0cebc0e1..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter6.ipynb
+++ /dev/null
@@ -1,315 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6: The Second Law of Thermodynamics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-1 ,Page No.286"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "QH=80.0;#rate of heat transfer in MW\n",
-      "QL=50.0;#rate of waste heat rejection in MW\n",
-      "\n",
-      "#calculations\n",
-      "Wnet=QH-QL;\n",
-      "print'net power output %i MW'%Wnet\n",
-      "nth=Wnet/QH;\n",
-      "print'the thermal efficiency is %f '%round(nth,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "net power output 30 MW\n",
-        "the thermal efficiency is 0.375000 \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-2 ,Page No.286"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Wnet=65;#power output in hp\n",
-      "nth=0.24;#effficiency of car\n",
-      "HV=19000;#heating value in Btu/lbm\n",
-      "\n",
-      "#calculations\n",
-      "QH=Wnet/nth*2545;#factor of 2545 to convert to Btu/h\n",
-      "m=QH/HV;\n",
-      "print'the engine must burn at fuel rate %f lbm/h'%round(m,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the engine must burn at fuel rate 36.300000 lbm/h\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-3 ,Page No.291"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Wnet=2.0;#power input in kW\n",
-      "QL=360.0;#rate of heat rejection in kJ/min\n",
-      "\n",
-      "#calculations\n",
-      "COPR=QL/Wnet/60;#factor of 60 to convert kW to kJ/min\n",
-      "print'coefficient of performance of refrigerator is %i'%COPR;\n",
-      "QH=QL+Wnet*60;#factor of 60 to convert kW to kJ/min\n",
-      "print'heat rejection rate %i kJ/min'%QH\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "coefficient of performance of refrigerator is 3\n",
-        "heat rejection rate 480 kJ/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-4 ,Page No.291"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "COP=2.5;#COP in the given conditions of heat pump\n",
-      "QH=80000.0;#heat loss rate in kJ/hr\n",
-      "\n",
-      "#calculations\n",
-      "Wnet=QH/COP;\n",
-      "print'the power consumed %i kJ/h'%Wnet\n",
-      "QL=QH-Wnet;\n",
-      "print'the rate at which heat is absorbed %i kJ/h'%QL\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the power consumed 32000 kJ/h\n",
-        "the rate at which heat is absorbed 48000 kJ/h\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-5 ,Page No.306"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "QH=500.0;#heat recieved per cycle in kJ\n",
-      "TL=30+273.0;#low temperature source in C\n",
-      "TH=652+273.0;#high temperature source in C\n",
-      "\n",
-      "#calculations\n",
-      "nth=1-TL/TH;\n",
-      "print'the thermal efficiency of carnot engine is %f'%round(nth,3);\n",
-      "QL=TL*QH/TH;\n",
-      "print'the amount of heat rejected to the sink per cycle %i kJ'%round(QL)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the thermal efficiency of carnot engine is 0.672000\n",
-        "the amount of heat rejected to the sink per cycle 164 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-6 ,Page No.310"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "COP=13.5;#COP of refrigerator\n",
-      "TH=75+460.0;#room temperature in R\n",
-      "TL=35+460.0;#refrigerated temp in R\n",
-      "\n",
-      "#calculations\n",
-      "COPR=1/(TH/TL-1);\n",
-      "if COPR>=COP:\n",
-      "    print('claim is true');\n",
-      "else:\n",
-      "    print('claim is false')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "claim is false\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-7 ,Page No.310"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "TL=-5+273.0;#in C\n",
-      "TH=21+273.0;#house temp in C\n",
-      "QH=135000.0/3600;#heat loss rate in kW\n",
-      "\n",
-      "#calculations\n",
-      "COPHP=1/(1-TL/TH);\n",
-      "Wnet=QH/COPHP;\n",
-      "print'minimum power required %f kW'%round(Wnet,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "minimum power required 3.320000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6-8 ,Page No.314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Qrefrig=40.0;#heat load on referigerator in W\n",
-      "COPR=1.3;#COP of referigerator\n",
-      "Wlight=40.0;#rated power of bulb in W\n",
-      "\n",
-      "#calculation\n",
-      "Wrefrig=Qrefrig/COPR;\n",
-      "Wt=Wrefrig+Wlight;\n",
-      "AnHr=365*24;#annual hours\n",
-      "NOH=20*30/3600.0*365;#normal operating hours\n",
-      "AOP=AnHr-NOH;#addtional operating hours\n",
-      "APC=Wt*AOP/1000;#additional power consumption; fator of 1000 to convert to kW\n",
-      "print'increase in power consumption %i kWh/yr'%round(APC);\n",
-      "print'increase in cost S%f/yr'%round((APC)*0.08,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "increase in power consumption 616 kWh/yr\n",
-        "increase in cost S49.300000/yr\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter7.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter7.ipynb
deleted file mode 100755
index d5ab12d2..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter7.ipynb
+++ /dev/null
@@ -1,1049 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7: Entropy"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-1 ,Page No.335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Q=750;#heat transfered in kJ\n",
-      "Tsys=300.0;#temperature of liquid-water mixture in K\n",
-      "\n",
-      "#calculations\n",
-      "dSsys=Q/Tsys;\n",
-      "print'Entropy change in the process %f kJ/K'%round(dSsys,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Entropy change in the process 2.500000 kJ/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-2 ,Page No.338"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Qsink=2000.0;#heat lost in kJ\n",
-      "Qsource=-Qsink;\n",
-      "Tsource=800.0;#temp of heat loss in K\n",
-      "\n",
-      "#calculations\n",
-      "#part - a\n",
-      "Tsink=500.0;\n",
-      "dSsource=Qsource/Tsource;\n",
-      "dSsink=Qsink/Tsink;\n",
-      "Sgena=dSsource+dSsink;\n",
-      "print'entropy generated in part a %f kJ/K'%round(Sgena,1);\n",
-      "#part - b\n",
-      "Tsink=750;\n",
-      "dSsource=Qsource/Tsource;\n",
-      "dSsink=Qsink/Tsink;\n",
-      "Sgenb=dSsource+dSsink;\n",
-      "print'entropy generated in part b %f kJ/K'%round(Sgenb,1);\n",
-      "if Sgena>Sgenb :\n",
-      "    print('part a is more irreversible');\n",
-      "elif(Sgena == Sgenb) :\n",
-      "    print('heat transfer is equally irreversible');\n",
-      "else:\n",
-      "    print('part b is more irreversible');\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "entropy generated in part a 1.500000 kJ/K\n",
-        "entropy generated in part b 0.200000 kJ/K\n",
-        "part a is more irreversible\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-3 ,Page No.341"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=5.0;#mass of refrigerant 134-a in kg\n",
-      "P1=140.0;#intial pressure in kPa\n",
-      "T1=20.0;#temperature in C\n",
-      "P2=100.0;#final pressure in kPa\n",
-      "\n",
-      "#from refrigerant-134a data\n",
-      "#at P1 and T1\n",
-      "s1=1.0624;\n",
-      "v1=0.16544;\n",
-      "#at P2\n",
-      "v2=v1;\n",
-      "vf=0.0007529;\n",
-      "vg=0.19254;\n",
-      "sf=0.07188;\n",
-      "sfg=0.87995;\n",
-      "\n",
-      "#calculations\n",
-      "# vf < v2 <vg\n",
-      "x2=(v2-vf)/(vg-vf);\n",
-      "s2=sf+x2*sfg;\n",
-      "dS=m*(s2-s1);\n",
-      "print'entropy change in the process %f kJ/k'%round(dS,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "entropy change in the process -1.175000 kJ/k\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-4 ,Page No.342"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=3;#mass of water in lbm\n",
-      "P1=20;#pressure in psia\n",
-      "T1=70+460;#temperature in R\n",
-      "Qin=3450;#heat added in Btu\n",
-      "\n",
-      "#from Table A-6E\n",
-      "#at P1 and T1\n",
-      "s1=0.07459;\n",
-      "h1=38.08;\n",
-      "\n",
-      "#calculations\n",
-      "#Ein - Eout = dEsystem\n",
-      "#Qin = m*(h2 - h1)\n",
-      "h2=Qin/m+h1;\n",
-      "#from Table A-6E\n",
-      "#At P2 and h2\n",
-      "s2=1.7761;\n",
-      "dS=m*(s2-s1);\n",
-      "print'entropy change %f Btu/R'%round(dS,3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "entropy change 5.105000 Btu/R\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-5 ,Page No.343"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=5.0;#entry pressure in MPa\n",
-      "T1=450.0;#entry temperature in C\n",
-      "P2=1.4;#pressure at exit MPa\n",
-      "\n",
-      "#calculations\n",
-      "#Ein - Eout = dEsystem/dt\n",
-      "#Ein = Eout\n",
-      "#Wout = m*(h1-h2)\n",
-      "#At P1 and T1\n",
-      "h1=3317.2;\n",
-      "s1=6.8210;\n",
-      "s2=s1;\n",
-      "#At P2 and s2\n",
-      "h2=2967.4;\n",
-      "Wout=h1-h2;\n",
-      "print'work output per unit mass %f kJ/kg'%round(Wout,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "work output per unit mass 349.800000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-7 ,Page No.352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "#given values\n",
-      "P1=1;#pressure at state 1 in kPa\n",
-      "T1=110.0;#temperature at state 1 in K\n",
-      "P2=5;#pressure at state 2 in kPa\n",
-      "T2=120.0;#temperature at state 2 in K\n",
-      "\n",
-      "#from Table \n",
-      "#At P1 and T1\n",
-      "s1=4.875;\n",
-      "cp1=3.471;\n",
-      "#at P2 and T2\n",
-      "s2=5.145;\n",
-      "cp2=3.486;\n",
-      "\n",
-      "#calculations\n",
-      "#part - a\n",
-      "dSa=s2-s1;\n",
-      "print'change in entropy %f kJ/kg K using tabulated properties'%round(dSa,3);\n",
-      "#part - b\n",
-      "cavg=(cp1+cp2)/2.0;\n",
-      "dSb=cavg*log(T2/T1);\n",
-      "print'change in entropy %f kJ/kg K approximating liquid methane as an incompressible substance'%round(dSb,3);\n",
-      "E=(dSb-dSa)/dSa;\n",
-      "print'Error is %f'%round(E,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "change in entropy 0.270000 kJ/kg K using tabulated properties\n",
-        "change in entropy 0.303000 kJ/kg K approximating liquid methane as an incompressible substance\n",
-        "Error is 0.121000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-8 ,Page No.353"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Annual power savings has a 0.005% error due to approximation in textbook\n",
-      "\n",
-      "#given data\n",
-      "P1=5;#pressure at state 1 in MPa\n",
-      "V1=0.280;#volimetric flow rate in m^3/s\n",
-      "T1=115;#temperature at state 1 in K\n",
-      "P2=1;#pressure at state 2 in MPa\n",
-      "dt=8760;#time in h/yr\n",
-      "UC=0.075;#unit cost in $\n",
-      "\n",
-      "#from Table\n",
-      "#at P1 and T1\n",
-      "h1=232.3;\n",
-      "s1=4.9945;\n",
-      "p1=422.15;\n",
-      "s2=s1;\n",
-      "#at P2 and s2\n",
-      "h2=222.8;\n",
-      "\n",
-      "#calculations\n",
-      "m=p1*V1;\n",
-      "#Ein - Eout = dEsystem/dt\n",
-      "#Ein = Eout\n",
-      "#Wout = m*(h1-h2)\n",
-      "Wout = m*(h1-h2);\n",
-      "print'maximum amount of power that can be produced %i kW'%round(Wout)\n",
-      "APP=Wout*dt;#annual power production\n",
-      "APS=APP*UC;#annual power savings\n",
-      "print'Annual power savings $%i/yr'%round(APS)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum amount of power that can be produced 1123 kW\n",
-        "Annual power savings $737758/yr\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-9 ,Page No.357"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#given data\n",
-      "P1=100.0;#inlet pressure in kPa\n",
-      "T1=290.0;#inlet temperature in K\n",
-      "P2=600.0;#outlet temperature in kPa\n",
-      "T2=330.0;#outlet temperature in K\n",
-      "\n",
-      "#from Table A-17\n",
-      "s02=1.79783;\n",
-      "s01=1.66802;\n",
-      "#Table A-2b\n",
-      "cpavg=1.006;\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kJ/kg -K\n",
-      "\n",
-      "#calculations\n",
-      "#part-a\n",
-      "s21=s02-s01-R*log(P2/P1);#stands for s2 - s1\n",
-      "print'entropy change using property values from air table %f kJ/kg-K'%round(s21,4);\n",
-      "s21=cpavg*log(T2/T1)-R*log(P2/P1);#stands for s2 - s1\n",
-      "print'entropy change using average specific heat %f kJ/kg-K'%round(s21,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "entropy change using property values from air table -0.384400 kJ/kg-K\n",
-        "entropy change using average specific heat -0.384200 kJ/kg-K\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-10 ,Page No.360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=95;#intial pressure in kPa\n",
-      "T1=295;#intial temperature in K\n",
-      "r=8;#ratio of V1/V2\n",
-      "\n",
-      "#calculations\n",
-      "#for closed systems V2/V1 = v2/v1\n",
-      "#At T1\n",
-      "vr1=647.9;\n",
-      "vr2=vr1/r;\n",
-      "#at vr2\n",
-      "T2=662.7;\n",
-      "print'the final temperature %f K'%T2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final temperature 662.700000 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-11 ,Page No.361"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=14.0;#intial pressure in psia\n",
-      "T1=50+460.0;#intial temperature in R\n",
-      "T2=320+460.0;#final temperature in R\n",
-      "\n",
-      "#constants used\n",
-      "k=1.667;\n",
-      "\n",
-      "#calculations\n",
-      "P2=P1*(T2/T1)**(k/(k-1));\n",
-      "print'exit pressure %f psia'%round(P2,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "exit pressure 40.500000 psia\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-12 ,Page No.364"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P2=1000;#final pressure in MPa\n",
-      "P1=100;#intial pressure in MPa\n",
-      "\n",
-      "#from Table A-5\n",
-      "#At P2\n",
-      "v1=0.001043;\n",
-      "\n",
-      "#calculations\n",
-      "Wrev=v1*(P2-P1);\n",
-      "print'compressor work as saturated liquid at inlet %f kJ/kg'%round(Wrev,2)\n",
-      "#from Table A-5\n",
-      "#at P1 as sat. vapour\n",
-      "h1=2675.0;\n",
-      "s1=7.3589;\n",
-      "s2=s1\n",
-      "#from Table A-6\n",
-      "#at P2 and s2\n",
-      "h2=3194.5;\n",
-      "Wrev=h2-h1;\n",
-      "print'compressor work as saturated vapor at inlet %f kJ/kg'%round(Wrev,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "compressor work as saturated liquid at inlet 0.940000 kJ/kg\n",
-        "compressor work as saturated vapor at inlet 519.500000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-13 ,Page No.368"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "#given data\n",
-      "P1=100;#intial pressure in kPa\n",
-      "T1=300;#intial temperature in K\n",
-      "P2=900;#final pressure in kPa\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kJ/kg -K\n",
-      "\n",
-      "#calculations\n",
-      "#part - a\n",
-      "k=1.4;\n",
-      "Wcomp=k*R*T1/(k-1)*((P2/P1)**((k-1)/k)-1);\n",
-      "print'compression work in case of isentropic compression %f kJ/kg'%round(Wcomp,1);\n",
-      "#part - b\n",
-      "n=1.3;\n",
-      "Wcomp=n*R*T1/(n-1)*((P2/P1)**((n-1)/n)-1);\n",
-      "print'compression work in case of polytropic compression %f kJ/kg'%round(Wcomp,1);\n",
-      "#part - c\n",
-      "Wcomp=R*T1*log(P2/P1);\n",
-      "print'compression work in case of isothermal compression %f kJ/kg'%round(Wcomp,1);\n",
-      "#part - d\n",
-      "Ps=sqrt(P1*P2);\n",
-      "Wcomp=2*n*R*T1/(n-1)*((Ps/P1)**((n-1)/n)-1);\n",
-      "print'compression work in case of two-stage compression with intercooling %f kJ/kg'%round(Wcomp,1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "compression work in case of isentropic compression 263.200000 kJ/kg\n",
-        "compression work in case of polytropic compression 246.400000 kJ/kg\n",
-        "compression work in case of isothermal compression 189.200000 kJ/kg\n",
-        "compression work in case of two-stage compression with intercooling 215.300000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-14 ,Page No.371"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=3000;#intial pressure in kPa\n",
-      "T1=400;#intial temperature in C\n",
-      "P2=50;#final pressure in kPa\n",
-      "T2=100;#final temperature in C\n",
-      "Wout=2000;#power output in kW\n",
-      "\n",
-      "#from Table A-6\n",
-      "#at P1\n",
-      "h1=3231.7;\n",
-      "s1=6.9235;\n",
-      "#at 2a\n",
-      "h2a=2682.4;\n",
-      "#from Table A-6\n",
-      "#at 2s\n",
-      "s2s=s1;\n",
-      "sf=1.0912;\n",
-      "sg=7.5937;\n",
-      "hf=340.54;\n",
-      "hfg=2304.7\n",
-      "x2s=(s2s-sf)/(sg-sf);\n",
-      "h2s=hf+x2s*hfg;\n",
-      "nT=(h1-h2a)/(h1-h2s);\n",
-      "print'isentropic efficiency is %f'%round(nT,3)\n",
-      "#Ein = Eout\n",
-      "m=Wout/(h1-h2a);\n",
-      "print'mass flow rate %f kg/s'%round(m,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "isentropic efficiency is 0.667000\n",
-        "mass flow rate 3.640000 kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-15 ,Page No.374"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=100;#intial pressure in kPa\n",
-      "T1=285;#intial temperature in K\n",
-      "P2=800;#final pressure in kPa\n",
-      "m=0.2;#mass flow rate in kg/s\n",
-      "nc=0.8;#isentropic efficiency\n",
-      "\n",
-      "#from Table A-17\n",
-      "#at T1\n",
-      "h1=285.14;\n",
-      "Pr1=1.1584;\n",
-      "\n",
-      "#calcualtions\n",
-      "Pr2=Pr1*(P2/P1);\n",
-      "#at Pr2\n",
-      "h2s=517.05;\n",
-      "h2a=(h2s-h1)/nc+h1;\n",
-      "#at h2a\n",
-      "T2a=569.5;\n",
-      "print'exit temperature of air %f K'%T2a;\n",
-      "#Ein = Eout\n",
-      "Wa=m*(h2a-h1);\n",
-      "print'required power input %i kW'%round(Wa)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "exit temperature of air 569.500000 K\n",
-        "required power input 58 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-16 ,Page No.375"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "P1=200.0;#intial pressure in kPa\n",
-      "T1=950.0;#intial temperature in K\n",
-      "P2=80.0;#final pressure in kPa\n",
-      "nN=0.92;#isentropic efficiency\n",
-      "\n",
-      "#from Table A-2b\n",
-      "cp=1.099;\n",
-      "k=1.354;\n",
-      "\n",
-      "#calculations\n",
-      "T2s=T1*(P2/P1)**((k-1)/k);\n",
-      "#ein = eout\n",
-      "V2s=sqrt(2*cp*(T1-T2s)*1000);#factor of 1000 for conversion to m^2/s^2\n",
-      "print'maximum possible exit velocity %i m/s'%(V2s);\n",
-      "T2a=T1-nN*(T1-T2s);\n",
-      "print'exit temperature %i K'%round(T2a);\n",
-      "V2a=sqrt(nN*V2s**2);\n",
-      "print'actual exit velocity %i m/s'%(V2a)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum possible exit velocity 666 m/s\n",
-        "exit temperature 764 K\n",
-        "actual exit velocity 639 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      " Example 7-17 ,Page No.382"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Qin=1035.0;#rate of heat transfer in W\n",
-      "Tin=20+273.0;#inner temperature in K\n",
-      "Qout=Qin;\n",
-      "Tout=5+273.0;#outer temperature in K\n",
-      "\n",
-      "#calculations\n",
-      "# Sin - Sout + Sgen = dSsystem/dt\n",
-      "Sgen=(Qout/Tout)-(Qin/Tin);\n",
-      "print'entropy generation in the wall %f W/K'%round(Sgen,3);\n",
-      "Ts1=300.0;Ts2=273.0;#Boundary temperatures\n",
-      "Sgen=(Qout/Ts2)-(Qin/Ts1);\n",
-      "print'total entropy generation %f W/K'%round(Sgen,3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "entropy generation in the wall 0.191000 W/K\n",
-        "total entropy generation 0.341000 W/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-18 ,Page No.383"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=7;#intial pressure in MPa\n",
-      "T1=450;#intial temperature in K\n",
-      "P2=3;#final pressure in MPa\n",
-      "\n",
-      "#from steam tables\n",
-      "#at P1 and T1\n",
-      "h1=3288.3;\n",
-      "s1=6.6353;\n",
-      "#at P2\n",
-      "h2=h1;\n",
-      "s2=7.0046;\n",
-      "\n",
-      "#calculations\n",
-      "# Sin - Sout + Sgen = dSsystem/dt\n",
-      "Sgen=s2-s1;\n",
-      "print'the entropy generated %f kJ/kg-K'%round(Sgen,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the entropy generated 0.369300 kJ/kg-K\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-19 ,Page No.385"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#given data\n",
-      "m=50;#mass of iron block in kg\n",
-      "T1=500.0;#temperature of iron block in K\n",
-      "T2=285.0;#temperature of lake in K\n",
-      "\n",
-      "#from Table A-3\n",
-      "Cavg=0.45;\n",
-      "\n",
-      "#calculations\n",
-      "dSiron=m*Cavg*log(T2/T1);\n",
-      "print'entropy change of the iron block %f kJ/K'%round(dSiron,2);\n",
-      "# Ein - Eout = dEsystem\n",
-      "Qout=m*Cavg*(T1-T2);\n",
-      "dSlake=Qout/T2;\n",
-      "print'entropy change of the lake %f kJ/K'%round(dSlake,2);\n",
-      "# Sin - Sout + Sgen = dSsystem/dt\n",
-      "Sgen=(Qout/T2)+dSiron;\n",
-      "print'entropy change in the process %f kJ/K'%round(Sgen,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "entropy change of the iron block -12.650000 kJ/K\n",
-        "entropy change of the lake 16.970000 kJ/K\n",
-        "entropy change in the process 4.330000 kJ/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-20 ,Page No.386"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P=20;#pressure of water in psia\n",
-      "T1=50+460.0;#temperature of water in R\n",
-      "T2=240.0;#steam temperature in F\n",
-      "T3=130.0;#leaving temperauture in F\n",
-      "m1=300.0;#mass flow rate in lbm/min\n",
-      "Qout=180.0;#heat losses in Btu/min\n",
-      "\n",
-      "#from steam tables\n",
-      "#at P and T1\n",
-      "h1=18.07;\n",
-      "s1=0.03609;\n",
-      "#at P and T2\n",
-      "h2=1162.3;\n",
-      "s2=1.7406;\n",
-      "#at P and T3\n",
-      "h3=97.99;\n",
-      "s3=0.18174;\n",
-      "\n",
-      "#calculations\n",
-      "# Qout = m1*h1 + m2*h2 - (m1+m2)*h3\n",
-      "m2= (Qout-m1*h1+m1*h3)/(h2-h3);\n",
-      "m3=m1+m2;\n",
-      "# Sin - sout + Sgen = dSsystem/dt\n",
-      "Sgen=m3*s3-m1*s1-m2*s2+Qout/T1;\n",
-      "print'the rate of entropy generation %f Btu/min R'%round(Sgen,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of entropy generation 8.670000 Btu/min R\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-21 ,Page No.388"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T=100+273.0;#temperature of water in K\n",
-      "Q=-600.0;#heat transferred in kJ\n",
-      "Tb=25+273.0;#temperature of surroundings in K\n",
-      "\n",
-      "#calculation\n",
-      "dSsys=Q/T;\n",
-      "print'entropy change of water %f kJ/K'%round(dSsys,2);\n",
-      "# Sin - sout + Sgen = dSsystem\n",
-      "Sgen= -Q/Tb + dSsys;\n",
-      "print'total entropy generation %f kJ/K'%round(Sgen,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "entropy change of water -1.610000 kJ/K\n",
-        "total entropy generation 0.400000 kJ/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-22 ,Page No.394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi,sqrt\n",
-      "\n",
-      "#given data\n",
-      "T1=20+273.0;#average temperature of inlet in C\n",
-      "T2=24+273.0;#average temperature of compressed air in C\n",
-      "P1=101.0;#pressure of air inlet in kPa\n",
-      "P2=801.0;#pressure of compressed air in kPa\n",
-      "D=3.0/1000.0;#ihole diameter n m\n",
-      "Cdischarge=0.65;#discharge coefficient\n",
-      "ncomp=0.8;#compressor efficiency\n",
-      "nmotor=0.92;#motor efficiency\n",
-      "UC=0.078;#unit cost in $/kWh\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kJ/kg K\n",
-      "k=1.4;\n",
-      "n=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "Win=n*R*T1/(ncomp*(n-1))*((P2/P1)**((n-1)/n)-1);\n",
-      "A=pi*D**2/4;\n",
-      "mair=Cdischarge*(2/(k+1))**(1/(k-1))*P2*A/(R*T2)*sqrt(k*R*1000*2/(k+1)*T2);#factor of 1000 to m^2/s^2\n",
-      "PW=mair*Win;#Power wasted\n",
-      "ES=PW*4200/nmotor;#4200 is operating hours ES stands for Energy savings\n",
-      "print'Energy savings %i kWh/yr'%round(ES);\n",
-      "CS=ES*UC;\n",
-      "print'cost savings %i S/yr'%round(CS)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy savings 11698 kWh/yr\n",
-        "cost savings 912 S/yr\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7-23 ,Page No.398"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "P1=85.6;#pressure of air inlet in kPa\n",
-      "P2=985.6;#pressure of actual compressed air in kPa\n",
-      "P2r=885.6;#pressure of required compressed air in kPa\n",
-      "CC=12000;#current cost in $/yr\n",
-      "\n",
-      "#constants used\n",
-      "n=1.4;\n",
-      "\n",
-      "#calulation\n",
-      "freduction=1-(((P2r/P1)**((n-1)/n)-1)/((P2/P1)**((n-1)/n)-1));\n",
-      "CS=CC*freduction;\n",
-      "print'cost savings %i Dollar/yr'%(round(CS))\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "cost savings 719 Dollar/yr\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter8.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter8.ipynb
deleted file mode 100755
index e91e1b78..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter8.ipynb
+++ /dev/null
@@ -1,764 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8: Exergy: A Measure of Work Potential"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-1 ,Page No.426"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "# given data\n",
-      "D=12.0;#diameter of rotor in m\n",
-      "V=10.0;#average velocity in m/s\n",
-      "\n",
-      "# density of air at 25C & 1atm\n",
-      "p=1.18;\n",
-      "\n",
-      "#calculations\n",
-      "ke=(V**2)/2/1000;#factor of 1000 for converting J into kJ\n",
-      "m=p*pi*(D**2)*V/4.0; \n",
-      "MP=m*(ke);\n",
-      "print'Maximum power %f kW'%round(MP,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum power 66.700000 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-2 ,Page No.427"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "TH=2000.0;#temperature of large furnace in R\n",
-      "T0=77+460.0;#temperature of enviroment in R\n",
-      "Qin=3000.0;#heat rate in Btu/sec\n",
-      "\n",
-      "#calculation\n",
-      "nth=1-(T0/TH);\n",
-      "Wmax=nth*Qin;\n",
-      "Wmax=round(Wmax)\n",
-      "print'the rate of energy flow %i Btu/s'%round(Wmax)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of energy flow 2195 Btu/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-3 ,Page No.429"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Tsink=300.0;#temperature of sink in K\n",
-      "Tsource=1200.0;#temperature of source in K\n",
-      "Qin=500.0;#heat rate in kJ/s\n",
-      "Wuout=180;#power output in W\n",
-      "\n",
-      "#calculations\n",
-      "Wrev=(1-Tsink/Tsource)*Qin;\n",
-      "print'The reversible power %i kW'%round(Wrev);\n",
-      "I=Wrev-Wuout;\n",
-      "print'the irreversiblity rate %i kW'%round(I)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reversible power 375 kW\n",
-        "the irreversiblity rate 195 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-4 ,Page No.429"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad \n",
-      "from pylab import *\n",
-      "\n",
-      "#given data\n",
-      "m=500.0;#mass of iron block in kg\n",
-      "T1=473.0;#temperature of iron block in K\n",
-      "T0=300.0;#temperature of surrondings in K\n",
-      "\n",
-      "\n",
-      "#from Table A-3\n",
-      "cavg=0.45;\n",
-      "\n",
-      "#calculations\n",
-      "def intgrnd1(T): \n",
-      " return ((1-T0/T)*(-m*cavg))#intergrant\n",
-      "Wrev, err = quad(intgrnd1,T1,T0) ;\n",
-      "Qtotal=m*cavg*(T1-T0);\n",
-      "print'The reversible power %i kJ'%round(Wrev);\n",
-      "Wu=0;\n",
-      "I=Wrev-Wu;\n",
-      "print'the irreversiblity rate %i kJ'%round(I)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reversible power 8191 kJ\n",
-        "the irreversiblity rate 8191 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-5 ,Page No.431"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "Wrev=8191.0;#reversible work from Ex - 8.4\n",
-      "Wtotal=38925.0;#total work is equivalent to total heat lost to heat engine from Ex - 8.4\n",
-      "TL=278.0;#outdoor temperature in K\n",
-      "TH=300.0;#temperature of house in K\n",
-      "\n",
-      "#calculations\n",
-      "Wrm=Wtotal-Wrev;#work remaining\n",
-      "COPHP=1/(1-TL/TH);\n",
-      "Wd=COPHP*Wrev;#work delivered\n",
-      "PS=Wd+Wrm;\n",
-      "PS=round(PS/1000);#factor of 1000 for converting kJ into MJ\n",
-      "print'Maximum amount of heat that can be supplied %i MJ'%PS"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum amount of heat that can be supplied 142 MJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-6 ,Page No.434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "COP=1.0;# the efficiency that the dealer is referring to is the first law efficiency, meaning that for each unit of electric energy (work) consumed, the heater will supply the house with 1 unit of energy (heat). That is,the advertised heater has a COP of 1\n",
-      "TL=283.0;#outdoor temperature in K\n",
-      "TH=294.0;#indoor temperature in K\n",
-      "\n",
-      "#calculations\n",
-      "COPHP=1/(1-TL/TH);\n",
-      "nII=COP/COPHP;\n",
-      "print'the second law efficiency %f'%round(nII,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the second law efficiency 0.037000\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-7 ,Page No.438"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#given data\n",
-      "P1=1000.0;#pressure of compressed air in KPa\n",
-      "V=200.0;#volumne of rigid tank in m^3\n",
-      "T1=300.0;#temperature of enviroment in K\n",
-      "T0=T1;#state of air in tank\n",
-      "P0=100.0;#pressure of enviroment in KPa\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kPa m^3/kg K\n",
-      "\n",
-      "#calculations\n",
-      "m1=P1*V/(R*T1);\n",
-      "O1=R*T0*(P0/P1-1)+R*T0*log(P1/P0);# O refers to exergy\n",
-      "X1=m1*O1/1000;#factor of 1000 for converting kJ into MJ\n",
-      "print'work obtained %i MJ'%round(X1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "work obtained 281 MJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-8 ,Page No.439"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T0=20+273;#temperature of enviroment in K\n",
-      "P1=0.14;#intail pressure in MPa\n",
-      "T1=-10;#intail temperature in K\n",
-      "P2=0.8;#final pressure in MPa\n",
-      "T2=50;#final temperature in K\n",
-      "\n",
-      "#the properties of refrigerant\n",
-      "#at inlet\n",
-      "h1=246.36;\n",
-      "s1=0.9724;\n",
-      "#at outlet\n",
-      "h2=286.69;\n",
-      "s2=0.9802;\n",
-      "dO=h2-h1-T0*(s2-s1);# O refers to exergy\n",
-      "print'the exergy change of the refrigerant %i kJ/kg'%round(dO)\n",
-      "wmin=dO;\n",
-      "print'the minimum work input that needs to be supplied is %i kJ/kg'%wmin"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the exergy change of the refrigerant 38 kJ/kg\n",
-        "the minimum work input that needs to be supplied is 38 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-10 ,Page No.447"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "Q=1035.0;#rate of heat transfer in W\n",
-      "T0=273.0;#outdoor temperature in C\n",
-      "Tin=293.0;#inner wall surface temperature in K\n",
-      "Tout=278.0;#outer wall surface temperature in K\n",
-      "T1=300.0;#indoor temperature in C\n",
-      "\n",
-      "#calculations\n",
-      "#Xin - Xout - Xdestroyed = dX/dt\n",
-      "Xdestroyed=Q*(1-T0/Tin)-Q*(1-T0/Tout);\n",
-      "Xdestroyed=round(Xdestroyed);\n",
-      "print'the rate of exergy destroyed %i W'%round(Xdestroyed);\n",
-      "#the total rate of exergy destroyed\n",
-      "Xdestroyed=Q*(1-T0/T1)-Q*(1-T0/T0);\n",
-      "print'the total of exergy destroyed %f W'%round(Xdestroyed,1);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of exergy destroyed 52 W\n",
-        "the total of exergy destroyed 93.100000 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-11 ,Page No.448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "m=0.05;#mass of steam in kg\n",
-      "P1=1000;#intial pressure in MPa\n",
-      "T1=300+273;#intial temperature in K\n",
-      "P2=200;#final pressure in MPa\n",
-      "T2=150+273;#final temperature in K\n",
-      "P0=100;#pressure of surroundings in KPa\n",
-      "T0=25+273;#temperature of surroundings in K\n",
-      "Qout=2;#heat losses in kJ\n",
-      "\n",
-      "#from Table A-6 & A-4\n",
-      "u1=2793.7;\n",
-      "v1=0.25799;\n",
-      "s1=7.1246;\n",
-      "u2=2577.1;\n",
-      "v2=0.95986;\n",
-      "s2=7.2810;\n",
-      "u0=104.83;\n",
-      "v0=0.00103;\n",
-      "s0=0.3672;\n",
-      "\n",
-      "#calculations\n",
-      "X1=m*(u1-u0-T0*(s1-s0)+P0*(v1-v0));\n",
-      "X2=m*(u2-u0-T0*(s2-s0)+P0*(v2-v0));\n",
-      "print'exergy of intial state %i kJ'%round(X1);\n",
-      "print'exergy of final state %f kJ'%round(X2,1);\n",
-      "dX=X2-X1;\n",
-      "print'exergy change in system %f kJ'%round(dX,1);\n",
-      "Wout=-Qout-m*(u2-u1);\n",
-      "Wu=Wout-P0*m*(v2-v1);\n",
-      "Xdestroyed=X1-X2-Wu;\n",
-      "print'the exergy destroyed %f kJ'%round(Xdestroyed,1);\n",
-      "nII=Wu/(X1-X2);\n",
-      "print'second law efficiency of this process is %f'%round(nII,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "exergy of intial state 35 kJ\n",
-        "exergy of final state 25.400000 kJ\n",
-        "exergy change in system -9.700000 kJ\n",
-        "the exergy destroyed 4.300000 kJ\n",
-        "second law efficiency of this process is 0.551000\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-12 ,Page No.451"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad \n",
-      "from pylab import *\n",
-      "\n",
-      "#given data\n",
-      "m=2.0;#mass of air in lbm\n",
-      "T0=70+460.0;#intial temperature in R\n",
-      "P1=20.0;#intial pressure in psia\n",
-      "T1=70+460.0;#temperature of surroundings in R\n",
-      "T2=130+460.0;#final temperature in R\n",
-      "\n",
-      "#constants used\n",
-      "Cv=0.172;#in Btu/lbm - F\n",
-      "\n",
-      "#calculations\n",
-      "Xdestroyed=T0*m*Cv*log(T2/T1);\n",
-      "print'exergy destroyed %f Btu'%round(Xdestroyed,1);\n",
-      "def intgrnd1(T): \n",
-      " return ((1-T0/T)*m*Cv)#intergrant\n",
-      "Wrev, err = quad(intgrnd1,T1,T2) ;\n",
-      "print'the reversible work %f Btu'%round(Wrev,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "exergy destroyed 19.600000 Btu\n",
-        "the reversible work 1.100000 Btu\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-13 ,Page No.453"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T0=20+273.0;#temperature of surrounding in K\n",
-      "P0=100.0;#pressure of surrounding in kPa\n",
-      "Tiw=30+273.0;#temperature of water in K\n",
-      "mw=100.0;#mass of water in kg\n",
-      "Tii=350+273.0;#temperature of block in K\n",
-      "mi=5.0;#mass of block in kg\n",
-      "\n",
-      "#constants used(Table A-3)\n",
-      "cw=4.18;#specific heat of water in kJ/kg C\n",
-      "ci=0.45;#specific heat of iron in kJ/kg C\n",
-      "\n",
-      "#calculations\n",
-      "Tfk=(mi*ci*Tii+mw*cw*Tiw)/(mw*cw+mi*ci);\n",
-      "Tfc=Tfk-273;#in C\n",
-      "print'the final equilibrium temperature %f C'%round(Tfc,1);\n",
-      "X1i=mi*ci*(Tii-T0-T0*log(Tii/T0));\n",
-      "X1w=mw*cw*(Tiw-T0-T0*log(Tiw/T0));\n",
-      "X1t=X1i+X1w;#total exergy \n",
-      "print'intial exergy of combined systems %i kJ'%round(X1t);\n",
-      "X2i=mi*ci*(Tfk-T0-T0*log(Tfk/T0));\n",
-      "X2w=mw*cw*(Tfk-T0-T0*log(Tfk/T0));\n",
-      "X2t=X2i+X2w;#total exergy \n",
-      "print'intial exergy of combined systems %f kJ'%round(X2t,1);\n",
-      "Xdestroyed=X1t-X2t;\n",
-      "print'the wasted work %f kJ'%round(Xdestroyed,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the final equilibrium temperature 31.700000 C\n",
-        "intial exergy of combined systems 315 kJ\n",
-        "intial exergy of combined systems 95.800000 kJ\n",
-        "the wasted work 219.100000 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-14 ,Page No.455"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "TR=1200.0;#temperature of furnace \n",
-      "T0=300;#temperature of surrounding in K\n",
-      "P0=100;#pressure of surrounding in kPa\n",
-      "Tsys=400.0;#temperature of argon in K\n",
-      "P1=350;#temperature of argon in K\n",
-      "V1=0.01;#intail volumne in m^3\n",
-      "V2=2*V1;#final volumne\n",
-      "\n",
-      "#calculations\n",
-      "W=P1*V1*log(V2/V1);\n",
-      "Wsurr=P0*(V2-V1);\n",
-      "Wu=W-Wsurr;\n",
-      "print'the useful work output %f kJ'%round(Wu,2);\n",
-      "# Qin - W = m*Cv*dT, Since dt=0\n",
-      "Q=W;\n",
-      "Sgen=Q/Tsys-Q/TR;\n",
-      "Xdestroyed=T0*Sgen;\n",
-      "print'the exergy destroyed %f kJ/K'%round(Xdestroyed,2);\n",
-      "Wrev=T0*Q/Tsys-Wsurr+(1-T0/TR)*Q;\n",
-      "print'the reversible work is done in the process %f kJ'%round(Wrev,2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the useful work output 1.430000 kJ\n",
-        "the exergy destroyed 1.210000 kJ/K\n",
-        "the reversible work is done in the process 2.640000 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-15 ,Page No.460"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculation error in textbook in part - b which changes all the following answers\n",
-      "\n",
-      "#given data\n",
-      "m=8.0;#mass flow rate in kg/s\n",
-      "T0=298.0;#temperature of surrounding in K\n",
-      "P0=100.0;#pressure of surrounding in kPa\n",
-      "P1=3000.0;#inlet pressure in kPa\n",
-      "T1=450.0;#inlet temperature in K\n",
-      "P2=200.0;#outlet pressure in kPa\n",
-      "T2=150.0;#outlet temperature in K\n",
-      "Qout=300.0;#heat rate in kW\n",
-      "\n",
-      "#from Table A-6 and A-4\n",
-      "h1=3344.9;\n",
-      "s1=7.0856;\n",
-      "h2=2769.1;\n",
-      "s2=7.2810;\n",
-      "h0=104.83;\n",
-      "s0=0.3672;\n",
-      "\n",
-      "#calculations\n",
-      "# Ein = Eout\n",
-      "Wout=m*(h1-h2)-Qout;\n",
-      "Wout=round(Wout)\n",
-      "print'the actual power output %i kW'%Wout\n",
-      "# Xin = Xout\n",
-      "Wrev=m*(h1-h2)-(T0*(s1-s2));\n",
-      "Wrev=round(Wrev);\n",
-      "print'the maximum possible work output %i kW'%Wrev;\n",
-      "nII=Wout/Wrev;\n",
-      "print'second law efficiency is %f'%round(nII,3);\n",
-      "Xdestroyed=Wrev-Wout;\n",
-      "print'the exergy destroyed %i kW'%round(Xdestroyed);\n",
-      "X1=h1-h0-T0*(s1-s0);\n",
-      "print'the exergy of the steam at inlet conditions %i kJ/kg'%round(X1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the actual power output 4306 kW\n",
-        "the maximum possible work output 4665 kW\n",
-        "second law efficiency is 0.923000\n",
-        "the exergy destroyed 359 kW\n",
-        "the exergy of the steam at inlet conditions 1238 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-16 ,Page No.462"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T0=70+460;#temperature of surroundings in R\n",
-      "T1=50;#temperature of water in F\n",
-      "T2=240;#temperature of steam in F\n",
-      "T3=130;#outlet temperature in F\n",
-      "#as dicussed in example 7-20\n",
-      "m1=300;#mass flow rate of water in lbm/min\n",
-      "m2=22.7;#mass flow rate of steam in lbm/min\n",
-      "m3=322.7;#outlet mass flow rate  in lbm/min\n",
-      "\n",
-      "#from steam tables\n",
-      "h1=18.07;\n",
-      "s1=0.03609;\n",
-      "h2=1162.3;\n",
-      "s2=1.7406;\n",
-      "h3=97.99;\n",
-      "s3=0.18174;\n",
-      "\n",
-      "#calculations\n",
-      "Wrev=m1*(h1-T0*s1)+m2*(h2-T0*s2)-m3*(h3-T0*s3);\n",
-      "Wrev=round(Wrev);\n",
-      "print'the reversible power %i Btu/min'%round(Wrev)\n",
-      "Xdestroyed=Wrev;\n",
-      "print'the rate of exergy destruction %i Btu/min'%Xdestroyed\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the reversible power 4588 Btu/min\n",
-        "the rate of exergy destruction 4588 Btu/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8-17 ,Page No.463"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V=200.0;#volumne of rigid tank in m^3\n",
-      "P1=100.0;#intial surroundings in kPa\n",
-      "P2=1000.0;#final pressure in kPa\n",
-      "P0=100.0;#pressure of surroundings in kPa\n",
-      "T=300.0;#temperature of surroundings in K\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kPa m^3/kg K\n",
-      "\n",
-      "#calculations\n",
-      "#Xin - Xout = Xdestroyed = X2 - X1\n",
-      "m2=P2*V/(R*T);\n",
-      "X2=R*T*(log(P2/P0)+P0/P2-1);\n",
-      "Wrev=m2*X2/1000;\n",
-      "Wrev=round(Wrev);\n",
-      "print'Work requirement %i MJ'%Wrev\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work requirement 281 MJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/Chapter9.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter9.ipynb
deleted file mode 100755
index 1008c6c0..00000000
--- a/Thermodynamics_An_Engineering_Approach/Chapter9.ipynb
+++ /dev/null
@@ -1,540 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9: Gas Power Cycles"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-2 ,Page No.498"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=17+273;#temperature of air in K\n",
-      "P1=100;#pressure of air in kPa\n",
-      "r=8.0;#compression ratio i.e v1/v2\n",
-      "qin=800;#heat transfer rate in kJ/kg\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kPa-m^3/kg-K\n",
-      "\n",
-      "#from Table A-17\n",
-      "#at T1\n",
-      "u1=206.91;\n",
-      "vr1=676.1;\n",
-      "\n",
-      "#calculations\n",
-      "#Process 1-2\n",
-      "vr2=vr1/r;\n",
-      "#at this vr2\n",
-      "T2=652.4;\n",
-      "u2=475.11;\n",
-      "P2=P1*(T2/T1)*(r);\n",
-      "#Process 2-3\n",
-      "u3=qin+u2;\n",
-      "#at this u3\n",
-      "T3=1575.1;\n",
-      "vr3=6.108;\n",
-      "P3=P2*(T3/T2)*1;#factor of 1 as v3=v2\n",
-      "print'maximum temperature in the cycle %f K'%T3;\n",
-      "print'maximum pressure %f MPa'%round((P3/1000),3);#factor of 1000 to convert into MPa\n",
-      "#Process 3-4\n",
-      "vr4=r*vr3;\n",
-      "#at this vr4 \n",
-      "T4=795.6;\n",
-      "u4=588.74;\n",
-      "#Process 4-1\n",
-      "qout=u4-u1;\n",
-      "Wnet=qin-qout;\n",
-      "print'net work output %f kJ/kg'%round(Wnet,2);\n",
-      "nth=Wnet/qin;\n",
-      "print'thermal efficiency is %f'%round(nth,3);\n",
-      "v1=R*T1/P1;\n",
-      "MEP=Wnet/(v1*(1-1/r));\n",
-      "MEP=round(MEP);\n",
-      "print'mean effective pressure %i kPa'%(MEP)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "maximum temperature in the cycle 1575.100000 K\n",
-        "maximum pressure 4.345000 MPa\n",
-        "net work output 418.170000 kJ/kg\n",
-        "thermal efficiency is 0.523000\n",
-        "mean effective pressure 574 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-3 ,Page No.501"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "V1=117.0;#intial volumne in in^3\n",
-      "T1=80+460.0;#intial temperature in R\n",
-      "P1=14.7;#intial pressure in psia\n",
-      "r=18.0;#compression ratio\n",
-      "rc=2.0;#cut-off ratio\n",
-      "\n",
-      "#constants used\n",
-      "R=0.3704;#in psia ft^3/lbm R\n",
-      "cp=0.240;#in Btu/lbm R\n",
-      "cv=0.171;#in Btu/lbm R\n",
-      "\n",
-      "#from Table A-2Ea\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "V2=V1/r;\n",
-      "V3=rc*V2;\n",
-      "V4=V1;\n",
-      "#Process 1-2\n",
-      "T2=T1*(V1/V2)**(k-1);\n",
-      "P2=P1*(V1/V2)**k;\n",
-      "T2=round(T2);\n",
-      "P2=round(P2);\n",
-      "print('Process 1-2');\n",
-      "print'temperature %i R'%T2;\n",
-      "print'pressure %i psia'%P2;\n",
-      "#Process 2-3\n",
-      "P3=P2;\n",
-      "T3=T2*(V3/V2);\n",
-      "T3=round(T3);\n",
-      "P3=round(P3);\n",
-      "print('Process 2-3');\n",
-      "print'temperature %i R'%T3;\n",
-      "print'pressure %i psia'%P3;\n",
-      "#Process 3-4\n",
-      "T4=T3*(V3/V4)**(k-1);\n",
-      "P4=P3*(V3/V4)**k;\n",
-      "T4=round(T4);\n",
-      "print('Process 3-4');\n",
-      "print'temperature %i R'%T4;\n",
-      "print'pressure %f psia'%round(P4,1);\n",
-      "m=P1*V1/(R*T1)/1728;#factor of 1728 to covert to ft^3 from in^3\n",
-      "Qin=m*cp*(T3-T2);\n",
-      "Qout=m*cv*(T4-T1);\n",
-      "Wnet=Qin-Qout ;\n",
-      "print'work output %f Btu'%round(Wnet,3);\n",
-      "nth=Wnet/Qin;\n",
-      "print'thermal efficiency is %f'%round(nth,3);\n",
-      "MEP=Wnet/(V1-V2)*778.17*12;#factor of 778.17 and 12 to convert to lbf ft and in from Btu and ft respectively\n",
-      "MEP=round(MEP);\n",
-      "print'mean effective pressure %i psia'%MEP"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Process 1-2\n",
-        "temperature 1716 R\n",
-        "pressure 841 psia\n",
-        "Process 2-3\n",
-        "temperature 3432 R\n",
-        "pressure 841 psia\n",
-        "Process 3-4\n",
-        "temperature 1425 R\n",
-        "pressure 38.800000 psia\n",
-        "work output 1.296000 Btu\n",
-        "thermal efficiency is 0.633000\n",
-        "mean effective pressure 110 psia\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-5 ,Page No.511"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given data\n",
-      "T1=300;#compressor inlet temperature in K\n",
-      "r=8;#pressure ratio\n",
-      "T3=1300;#turbine inlet temperature in K\n",
-      "\n",
-      "#calcualtions\n",
-      "#Process 1-2\n",
-      "#at T1\n",
-      "h1=300.19;\n",
-      "Pr1=1.386;\n",
-      "Pr2=r*Pr1;\n",
-      "#at Pr2\n",
-      "T2=540;\n",
-      "h2=544.35;\n",
-      "print'temperature at exit of compressor %i K'%T2;\n",
-      "#Process 3-4\n",
-      "#at T3\n",
-      "h3=1395.17;\n",
-      "Pr3=330.9;\n",
-      "Pr4=Pr3/r;\n",
-      "#at Pr4\n",
-      "T4=770;\n",
-      "h4=789.37;\n",
-      "print'temperature at turbine exit %i K'%T4;\n",
-      "Win=h2-h1;\n",
-      "Wout=h3-h4;\n",
-      "rbw=Win/Wout;\n",
-      "print'back work ratio is %f'%round(rbw,3);\n",
-      "qin=h3-h2;\n",
-      "Wnet=Wout-Win;\n",
-      "nth=Wnet/qin;\n",
-      "print'thermal efficeincy is %f'%round(nth,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "temperature at exit of compressor 540 K\n",
-        "temperature at turbine exit 770 K\n",
-        "back work ratio is 0.403000\n",
-        "thermal efficeincy is 0.425000\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-6 ,Page No.513"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#from 9.5\n",
-      "Wsc=244.16;#work input of compressor in kJ/kg\n",
-      "Wst=606.60;#work output of turbine in kJ/kg\n",
-      "h1=300.19;#process 1-2 data\n",
-      "h3=1395.17;#process 3-4 data\n",
-      "\n",
-      "#given data\n",
-      "nC=0.8;#compressor effciency \n",
-      "nT=0.85;#turbine efficiency\n",
-      "\n",
-      "#calculations\n",
-      "Win=Wsc/nC;\n",
-      "Wout=nT*Wst;\n",
-      "rbw=Win/Wout;\n",
-      "print'back work ratio is %f'%round(rbw,3);\n",
-      "h2a=h1+Win;\n",
-      "qin=h3-h2a;\n",
-      "Wnet=Wout-Win;\n",
-      "nth=Wnet/qin;\n",
-      "print'thermal efficeincy is %f'%round(nth,3)\n",
-      "h4a=h3-Wout;\n",
-      "#from A-17 at h4a\n",
-      "T4a=853;\n",
-      "print'turbine exit temperature %i K is'%T4a\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "back work ratio is 0.592000\n",
-        "thermal efficeincy is 0.266000\n",
-        "turbine exit temperature 853 K is\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-7 ,Page No.516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#from 9.6\n",
-      "h2a=605.39;#entalpy of air leaving the compressor in kJ/kg\n",
-      "h4a=880.36;#entalpy of air leaving the turbinein kJ/kg\n",
-      "h3=1395.97;#in kJ/kg\n",
-      "Wnet=210.41;#net woek done in kJ/kg\n",
-      "\n",
-      "#given data\n",
-      "n=0.80;#thermal efficiency\n",
-      "\n",
-      "#calculations\n",
-      "# n = (h5 - h2a) / (h4a - h2a)\n",
-      "h5=(h4a - h2a)*n+h2a;\n",
-      "qin=h3-h5;\n",
-      "nth=Wnet/qin;\n",
-      "print'thermal efficeincy is %f'%round(nth,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "thermal efficeincy is 0.369000\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-8 ,Page No.519"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "T1=300;#inlet temperature of compressor in K\n",
-      "T6=1300;#inlet temperature of turbine in K\n",
-      "r=8;#overall compression ratio\n",
-      "\n",
-      "#calculations\n",
-      "#as it is case of intercooling\n",
-      "ri=sqrt(r);#ri stands for P2/P1 = P4/P3 = P6/P7 = P8/P9\n",
-      "#from A-17 at T1\n",
-      "h1=300.19;\n",
-      "Pr1=1.386;\n",
-      "Pr2=ri*Pr1;\n",
-      "#from A-17 at Pr2\n",
-      "T2=403.3;\n",
-      "h2=403.31;\n",
-      "#from A-17 at T6\n",
-      "h6=1395.97;\n",
-      "Pr6=330.9;\n",
-      "Pr7=Pr6/ri;\n",
-      "#from A-17 at Pr7\n",
-      "T7=1006.4;\n",
-      "h7=1053.33;\n",
-      "#at inlets\n",
-      "T3=T1;h3=h1;T8=T6;h8=h6;\n",
-      "#et exits\n",
-      "T4=T2;h4=h2;T9=T7;h9=h7;h5=h7;\n",
-      "Win=2*(h2-h1);\n",
-      "Wout=2*(h6-h7);\n",
-      "Wnet=Wout-Win;\n",
-      "qin=(h6-h4)+(h8-h7);\n",
-      "rbw=Win/Wout;\n",
-      "print'back work ratio is %f'%round(rbw,3);\n",
-      "nth=Wnet/qin;\n",
-      "print'thermal efficeincy is %f'%round(nth,3)\n",
-      "#part - b\n",
-      "print('part - b');\n",
-      "qin=(h6-h5)+(h8-h7);\n",
-      "nth=Wnet/qin;\n",
-      "print'thermal efficeincy is %f'%round(nth,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "back work ratio is 0.301000\n",
-        "thermal efficeincy is 0.359000\n",
-        "part - b\n",
-        "thermal efficeincy is 0.699000\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-9 ,Page No.523"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#given data\n",
-      "m=100;\n",
-      "P1=5;#air pressue in psia\n",
-      "T1=-40+460;#air temperature in R\n",
-      "T4=2000+460;#turbine inlet temperature in R\n",
-      "V1=850;#aircraft velocity in ft/sec\n",
-      "rp=10;#pressure ratio\n",
-      "\n",
-      "#constants used\n",
-      "cp=0.240;#in Btu/lbm F\n",
-      "k=1.4;\n",
-      "\n",
-      "#calculations\n",
-      "#Process 1-2\n",
-      "T2=T1+V1**2/(2*cp)/25037;#factor of 25037 to covert to Btu/lbm\n",
-      "P2=P1*(T2/T1)**(k/(k-1));\n",
-      "#Process 2-3 \n",
-      "P3=rp*P2;\n",
-      "P4=P3;\n",
-      "T3=T2*(P3/P2)**((k-1)/k);\n",
-      "#Win=Wout\n",
-      "T5=T4-T3+T2;\n",
-      "P5=P4*(T5/T4)**(k/(k-1));\n",
-      "T5=round(T5);\n",
-      "print'temperature at turbine exit %i R'%T5;\n",
-      "print'pressure at turbine exit %f psia'%round(P5,1);\n",
-      "#Process 5-6\n",
-      "P6=P1;\n",
-      "T6=T5*(P6/P5)**((k-1)/k);\n",
-      "T6=round(T6);#round off\n",
-      "V6=sqrt(2*cp*(T5-T6)*25037);#factor of 25037 to covert to (ft/s)^2\n",
-      "print'the velocity of nozzle exit %i ft/s'%(round(V6));\n",
-      "Wp=m*(V6-V1)*V1/25037;#factor of 25037 to covert to Btu/lbm\n",
-      "Qin=m*cp*(T4-T3);\n",
-      "nP=Wp/Qin;\n",
-      "print'propulsive efficiency is %f percent'%round(nP*100,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "temperature at turbine exit 2013 R\n",
-        "pressure at turbine exit 39.600000 psia\n",
-        "the velocity of nozzle exit 3285 ft/s\n",
-        "propulsive efficiency is 22.500000 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9-10 ,Page No.529"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#from 9.2\n",
-      "r=8;#compression ratio i.e v1/v2\n",
-      "T0=290.0;#temperature of air in K\n",
-      "T1=290.0;#temperature in the begining of the isentropic compression process in K\n",
-      "T2=652.4;#temperature in the end of the isentropic compression process in K\n",
-      "T3=1575.1;#temperature after process 2-3 in K\n",
-      "P2=1.7997;#pressure in the end of the isentropic compression process in MPa\n",
-      "P3=4.345;#pressure after process 2-3 in MPa\n",
-      "qin=800.0;#heat transfer rate in kJ/kg\n",
-      "qout=381.83;#heat rejection in kJ/kg\n",
-      "wnet=418.17;#net work in kJ/kg\n",
-      "\n",
-      "#given data\n",
-      "Tsource=1700;#temperature of working fluid at source in K\n",
-      "\n",
-      "#constants used\n",
-      "R=0.287;#in kPa-m^3/kg-K\n",
-      "\n",
-      "#calculations\n",
-      "#s1=s2 ; s3=s4\n",
-      "s03=3.5045;\n",
-      "s02=2.4975;\n",
-      "s32=(s03-s02)-R*log(P3/P2);#s32 stands for s3-s2\n",
-      "xdest23=T0*(s32-qin/Tsource);\n",
-      "Tsink=T1;\n",
-      "xdest41=T0*(-s32+qout/Tsink);\n",
-      "xdestcycle=xdest23+xdest41;\n",
-      "print'exergy destrustion associated with Otto cycle %f kJ/kg'%round(xdestcycle,1);\n",
-      "# X4 = (u4 - u0 )- T0*(s4 - s0) + P0(v4 - v0)\n",
-      "# s4 - s0 = s4 - s1 = s32\n",
-      "# u4 - u0 = u4 - u1 = qout\n",
-      "# v4 - v0 = v4 - v1 = 0\n",
-      "#hence x4 is\n",
-      "X4=qout-T0*s32;\n",
-      "print'exergy destruction of purge stream %f kJ/kg'%round(X4,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "exergy destrustion associated with Otto cycle 245.400000 kJ/kg\n",
-        "exergy destruction of purge stream 163.200000 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach/README.txt b/Thermodynamics_An_Engineering_Approach/README.txt
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index 2482fb63..00000000
--- a/Thermodynamics_An_Engineering_Approach/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Harshit Bajpai
-Course: btech
-College/Institute/Organization: Visvesvaraya National Institute Of Technology, Nagpur
-Department/Designation: Chemical Engineering
-Book Title: Thermodynamics An Engineering Approach
-Author: Michael A. Boles, Yunus A. Cengel
-Publisher: McGraw-Hill College, Boston, MA,
-Year of publication: 2006
-Isbn: 0072884959
-Edition: 5
\ No newline at end of file
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+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter10_1.ipynb
@@ -0,0 +1,628 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: Vapor and Combined Power Cycles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-1 ,Page No.555"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=75;#outlet pressure at compressor in kPa\n",
+      "P2=3000;#inlet pressure at turbine in kPa\n",
+      "P3=P2;\n",
+      "T3=350;#steam inlet temperature in C\n",
+      "P4=P1;\n",
+      "\n",
+      "#from steam tables\n",
+      "#at state 1\n",
+      "v1=0.001037;\n",
+      "h1=384.44;\n",
+      "#at state 3\n",
+      "h3=3116.1;\n",
+      "s3=6.7450;\n",
+      "#at state 4\n",
+      "s4=s3;\n",
+      "sf=1.2132;\n",
+      "sfg=6.2426;\n",
+      "hf=384.44;\n",
+      "hfg=2278;\n",
+      "\n",
+      "#calculations\n",
+      "win=v1*(P2-P1);\n",
+      "h2=h1+win;\n",
+      "x4=(s4-sf)/sfg;\n",
+      "h4=hf+x4*hfg;\n",
+      "qin=h3-h2;\n",
+      "qout=h4-h1;\n",
+      "nth=1-(qout/qin);\n",
+      "print'thermal efficency is %f'%round(nth,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thermal efficency is 0.260000\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-2 ,Page No.559"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "#from figure\n",
+      "P1=9;#in kPa\n",
+      "T1=38;#in C\n",
+      "P2=16000;#in kPa\n",
+      "P3=15.9;#in MPa\n",
+      "T3=35;#in C\n",
+      "P4=15.2;#in MPa\n",
+      "T4=625;#in C\n",
+      "P5=15;#in MPa\n",
+      "T5=600;#in C\n",
+      "#from question\n",
+      "nT=0.87;#isentropic efficiency of turbine \n",
+      "nP=0.85;#isentropic efficiency of pump\n",
+      "m=15;#mass flow rate in kg/s\n",
+      "\n",
+      "#from steam tables\n",
+      "v1=0.001009;\n",
+      "h5=3583.1;\n",
+      "h6s=2115.3;\n",
+      "h4=3647.6;\n",
+      "h3=160.1;\n",
+      "\n",
+      "#calculations\n",
+      "Win=v1*(P2-P1)/nP;\n",
+      "Wout=nT*(h5-h6s);\n",
+      "qin=h4-h3;\n",
+      "Wnet=Wout-Win;\n",
+      "nth=Wnet/qin;\n",
+      "print'thermal efficency is %f'%round(nth,3)\n",
+      "Wnet=m*Wnet;\n",
+      "print'power output %f MW'%round(Wnet/1000,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thermal efficency is 0.361000\n",
+        "power output 18.870000 MW\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-3 ,Page No.562"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=10;#outlet pressure at compressor in kPa\n",
+      "P2=3000;#inlet pressure at turbine in kPa\n",
+      "P3=P2;\n",
+      "T3=350;#steam inlet temperature in C\n",
+      "P4=P1;\n",
+      "\n",
+      "#from steam tables\n",
+      "#at state 1\n",
+      "h1=191.81;\n",
+      "v1=0.00101;\n",
+      "#at state 2\n",
+      "#s2=s1\n",
+      "#at state 3\n",
+      "h3=3116.1;\n",
+      "s3=6.7450;\n",
+      "#at state 4\n",
+      "s4=s3;\n",
+      "sf=0.6492;\n",
+      "sfg=7.4996;\n",
+      "hf=191.81;\n",
+      "hfg=2392.1;\n",
+      "\n",
+      "#calculations\n",
+      "#part - a\n",
+      "win=v1*(P2-P1);\n",
+      "h2=h1+win;\n",
+      "x4=(s4-sf)/sfg;\n",
+      "h4=hf+x4*hfg;\n",
+      "qin=h3-h2;\n",
+      "qout=h4-h1;\n",
+      "nth=1-(qout/qin);\n",
+      "print'the thermal efficiency of this power plant is %f'%round(nth,3);\n",
+      "#part - b\n",
+      "#States 1 and 2 remain the same in this case, and the enthalpies at state 3 (3 MPa and 600\u00b0C) and state 4 (10 kPa and s4=s3) are determined to be\n",
+      "h3=3682.8;\n",
+      "h4=2380.3;\n",
+      "x4=0.915;\n",
+      "qin=h3-h2;\n",
+      "qout=h4-h1;\n",
+      "nth=1-(qout/qin);\n",
+      "print'the thermal efficiency if steam is superheated to 600\u00b0 instead of 350\u00b0C is %f'%round(nth,3);\n",
+      "#part - c\n",
+      "#State 1 remains the same in this case, but the other states change. The enthalpies at state 2 (15 MPa and s2   s1), state 3 (15 MPa and 600\u00b0C),and state 4 (10 kPa and s4   s3) are determined in a similar manner to be\n",
+      "h2=206.95;\n",
+      "h3=3583.1;\n",
+      "h4=2115.3;\n",
+      "x4=0.804;\n",
+      "qin=h3-h2;\n",
+      "qout=h4-h1;\n",
+      "nth=1-(qout/qin);\n",
+      "print'the thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600\u00b0C is %f'%round(nth,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the thermal efficiency of this power plant is 0.334000\n",
+        "the thermal efficiency if steam is superheated to 600\u00b0 instead of 350\u00b0C is 0.373000\n",
+        "the thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600\u00b0C is 0.430000\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-4 ,Page No.566"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=10;#outlet pressure at compressor in kPa\n",
+      "P2=15000;#inlet pressure at turbine in kPa\n",
+      "P3=P2;\n",
+      "T3=600;#steam inlet temperature in C\n",
+      "P4=4000;#in kPa\n",
+      "T5=T3;\n",
+      "P6=P1;\n",
+      "x6=0.896;#dryness fraction\n",
+      "\n",
+      "#from steam table\n",
+      "#at state 1\n",
+      "h1=191.81;\n",
+      "v1=0.00101;\n",
+      "#at state 3\n",
+      "h3=3593.1;\n",
+      "s3=6.6796;\n",
+      "#at state 4\n",
+      "h4=3155;\n",
+      "T4=375.5;\n",
+      "#at state 6\n",
+      "sf=0.6492;\n",
+      "sfg=7.4996;\n",
+      "hf=191.81;\n",
+      "hfg=2392.1;\n",
+      "\n",
+      "#calculations\n",
+      "s6=sf+x6*sfg;\n",
+      "h6=hf+x6*hfg;\n",
+      "#s5 = s6\n",
+      "#from tables\n",
+      "P5=4000.0;#in kPa\n",
+      "h5=3674.9;\n",
+      "print'the pressure at which the steam should be reheated %i MPa'%(P5/1000);\n",
+      "#s2 = s1\n",
+      "win=v1*(P2-P1);\n",
+      "h2=h1+win;\n",
+      "qin=(h3-h2)+(h5-h4);\n",
+      "qout=h6-h1;\n",
+      "nth=1-(qout/qin);\n",
+      "print'thermal efficency is %f'%round(nth,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the pressure at which the steam should be reheated 4 MPa\n",
+        "thermal efficency is 0.451000\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-5 ,Page No.571"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=10;#outlet pressure at compressor in kPa\n",
+      "P2=1200;#in kPa\n",
+      "P3=P2;\n",
+      "P4=15000;#inlet pressure at turbine in kPa\n",
+      "P5=P4;\n",
+      "T5=600;#steam inlet temperature in C\n",
+      "P6=P2;\n",
+      "P7=P1;\n",
+      "\n",
+      "#from steam table\n",
+      "#at state 1\n",
+      "h1=191.81;\n",
+      "v1=0.00101;\n",
+      "#at state 3\n",
+      "h3=798.33;\n",
+      "v3=0.001138;\n",
+      "#at state 4\n",
+      "h4=3155;\n",
+      "T4=375.5;\n",
+      "#at state 5\n",
+      "h5=3583.1;\n",
+      "s5=6.6796;\n",
+      "#at state 6\n",
+      "h6=2860.2;\n",
+      "T6=218.4;\n",
+      "#at state 7\n",
+      "P7=10;\n",
+      "sf=0.6492;\n",
+      "sfg=7.4996;\n",
+      "hf=191.81;\n",
+      "hfg=2392.1;\n",
+      "\n",
+      "#calculations\n",
+      "#s2 = s1\n",
+      "win=v1*(P2-P1);\n",
+      "h2=h1+win;\n",
+      "#s4 = s3\n",
+      "win=v3*(P4-P3);\n",
+      "h4=h3+win;\n",
+      "s7=s5;\n",
+      "x7=(s7-sf)/sfg;\n",
+      "h7=hf+(x7*hfg);\n",
+      "#y is the fraction of steam extracted from the turbine\n",
+      "y=(h3-h2)/(h6-h2);\n",
+      "qin=h5-h4;\n",
+      "qout=(1-y)*(h7-h1);\n",
+      "nth=1-(qout/qin);\n",
+      "print'fraction of steam extracted is %f'%round(y,4);\n",
+      "print'thermal efficency is %f'%round(nth,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fraction of steam extracted is 0.226900\n",
+        "thermal efficency is 0.463000\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-6 ,Page No.574"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=10;#outlet pressure at compressor in kPa\n",
+      "P2=500;#steam extracted from low-pressure turbine in kPa\n",
+      "P3=P2;\n",
+      "P4=15000;#inlet pressure at turbine in kPa\n",
+      "P5=P4;\n",
+      "P6=4000;#steam extracted from turbine at pressure in kPa\n",
+      "P7=P5;\n",
+      "P8=P7;\n",
+      "P9=P7;\n",
+      "P10=P6;\n",
+      "P11=P10;\n",
+      "P12=P3;\n",
+      "P13=P1;\n",
+      "\n",
+      "#enthalpies at the various states and the pump work per unit mass of fluid flowing through them are\n",
+      "h1=191.81;\n",
+      "h2=192.30;\n",
+      "h3=640.09;\n",
+      "h4=643.92;\n",
+      "h5=1087.4;\n",
+      "h6=h5;\n",
+      "h7=1101.2;\n",
+      "h8=1089.8;\n",
+      "h9=3583.1;\n",
+      "h10=3155;\n",
+      "h11=3679.9;\n",
+      "h12=3014.8;\n",
+      "h13=2335.7;\n",
+      "wIin=0.49;\n",
+      "wIIin=3.83;\n",
+      "wIIIin=13.77;\n",
+      "\n",
+      "#calculations\n",
+      "y=(h5-h4)/((h10-h6)+(h5-h4));\n",
+      "z=(1-y)*(h3-h2)/(h12-h2);\n",
+      "h8=(1-y)*h5+(y*h7);\n",
+      "qin=(h9-h8)+(1-y)*(h11-h10);\n",
+      "qout=(1-y-z)*(h13-h1);\n",
+      "nth=1-(qout/qin);\n",
+      "print'fraction of steam extracted from closed feedwater is %f'%round(y,4);\n",
+      "print'fraction of steam extracted from open feedwater is %f'%round(z,4);\n",
+      "print'thermal efficency is %f'%round(nth,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fraction of steam extracted from closed feedwater is 0.176600\n",
+        "fraction of steam extracted from open feedwater is 0.130600\n",
+        "thermal efficency is 0.492000\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-7 ,Page No.577"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T0=290;#temperature at heat which is rejected in K \n",
+      "Tsource=1600;#temperature of furnace in K\n",
+      "Tsink=T0;\n",
+      "#from Ex 10.1\n",
+      "qin=2728.6;#in kJ/Kg\n",
+      "qout=2018.6;#in kJ/Kg\n",
+      "h4=2403;#in kJ/Kg\n",
+      "\n",
+      "#from steam tables\n",
+      "s1=1.2132;\n",
+      "s3=6.7450;\n",
+      "\n",
+      "#calculations\n",
+      "s2=s1;s4=s3;#isentropic processes\n",
+      "xdest12=0;\n",
+      "xdest34=0;\n",
+      "xdest23=T0*(s3-s2-(qin/Tsource));\n",
+      "xdest41=T0*(s1-s4+(qout/Tsink));\n",
+      "print'exergy destruction in process 1-2 %i kJ/kg'%xdest12;\n",
+      "print'exergy destruction in process 2-3 %i kJ/kg'%round(xdest23);\n",
+      "print'exergy destruction in process 3-4 %i kJ/kg'%xdest34;\n",
+      "print'exergy destruction in process 4-1 %i kJ/kg'%round(xdest41);\n",
+      "xdestcy=xdest12+xdest23+xdest34+xdest41;\n",
+      "print'exergy destruction %i cycle in kJ/kg'%round(xdestcy);\n",
+      "#from steam tables\n",
+      "#at 290 K and 100 kPa\n",
+      "h0=71.355;\n",
+      "s0=0.2533;\n",
+      "X4=(h4-h0)-T0*(s4-s0);\n",
+      "print'exergy of the leaving steam %i kJ/kg'%(round(X4))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "exergy destruction in process 1-2 0 kJ/kg\n",
+        "exergy destruction in process 2-3 1110 kJ/kg\n",
+        "exergy destruction in process 3-4 0 kJ/kg\n",
+        "exergy destruction in process 4-1 414 kJ/kg\n",
+        "exergy destruction 1524 cycle in kJ/kg\n",
+        "exergy of the leaving steam 449 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-8 ,Page No.581"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m1=15;#mass flow rate in kg/s\n",
+      "P1=7000;#inlet pressure at turbine in kPa\n",
+      "P2=P1;\n",
+      "P3=P1;\n",
+      "P4=500;#throttled to pressure in kPa\n",
+      "P5=P4;\n",
+      "P6=5;#steam expanded at a pressure in kPa\n",
+      "P7=P4;\n",
+      "P8=P6;\n",
+      "P9=P1;\n",
+      "P10=P1;\n",
+      "\n",
+      "#from steam tables\n",
+      "v7=0.001005;\n",
+      "v8=0.001093;\n",
+      "h1=3411.4;\n",
+      "h2=h1;\n",
+      "h3=h1;\n",
+      "h4=h1;\n",
+      "h5=2739.3;\n",
+      "h6=2073.0;\n",
+      "h7=640.09;\n",
+      "h8=137.75;\n",
+      "h11=144.78;\n",
+      "\n",
+      "#calculations\n",
+      "wIin=v8*(P9-P8);\n",
+      "wIIin=v7*(P10-P7);\n",
+      "h9=h8+wIin;\n",
+      "h10=h7+wIIin;\n",
+      "Qmax=m1*(h1-h7);\n",
+      "print'the maximum rate %i kW'%round(Qmax);\n",
+      "Wtout=m1*(h3-h6);#turbine\n",
+      "Wpin=m1*wIin;#pump\n",
+      "Wnet=Wtout-Wpin;\n",
+      "print'the power produced %i MW'%(round(Wnet/1000));\n",
+      "Qp=0;\n",
+      "Qin=m1*(h1-h11);\n",
+      "Eu=(Wnet+Qp)/Qin;\n",
+      "print'the utilization factor is %f'%round(Eu,3);\n",
+      "m4=0.1*m1;\n",
+      "m5=0.7*m1;\n",
+      "m7=m4+m5;\n",
+      "Qout=m4*h4+m5*h5-m7*h7;\n",
+      "print'the rate of process heat supply %f MW'%round(Qout/1000,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the maximum rate 41570 kW\n",
+        "the power produced 20 MW\n",
+        "the utilization factor is 0.407000\n",
+        "the rate of process heat supply 26.200000 MW\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10-9 ,Page No.585"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=5;#intial pressure in kPa\n",
+      "P2=7000;#final pressure ink Pa\n",
+      "P3=P2;\n",
+      "T3=500;#inlet temperature in K\n",
+      "P4=P1;\n",
+      "\n",
+      "#gas cycle from Ex9-6\n",
+      "#d stands for '\n",
+      "h4d=880.36;\n",
+      "T4d=853;\n",
+      "qin=790.58;\n",
+      "wnetg=210.41;\n",
+      "nth=0.266\n",
+      "h5d=451.80;\n",
+      "#steam cycle\n",
+      "h2=144.78;\n",
+      "T2=33;\n",
+      "h3=3411.4;\n",
+      "T3=500;\n",
+      "wnets=1331.4;\n",
+      "nth=0.408;\n",
+      "\n",
+      "#calculations\n",
+      "#Ein = Eout\n",
+      "#y is the ratio of ms/mg\n",
+      "y=(h4d-h5d)/(h3-h2);\n",
+      "print'the ratio of the mass flow rates of the steam and the combustion gasesis %f'%round(y,3);\n",
+      "wnet=wnetg+y*wnets\n",
+      "nth=wnet/qin;\n",
+      "print'thermal efficency is %f'%round(nth,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the ratio of the mass flow rates of the steam and the combustion gasesis 0.131000\n",
+        "thermal efficency is 0.487000\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter11_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter11_1.ipynb
new file mode 100755
index 00000000..460e3a17
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter11_1.ipynb
@@ -0,0 +1,353 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11: Refrigeration Cycles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11-1 ,Page No.613"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given values\n",
+      "P1=0.14;#intial pressure in MPa\n",
+      "P2=0.8;#final pressure in MPa\n",
+      "m=0.05;#mass flow rate in kg/s\n",
+      "\n",
+      "#from refrigerant-134a tables\n",
+      "h1=239.16;\n",
+      "s1=0.94456;\n",
+      "h2=275.39;\n",
+      "h3=95.47;\n",
+      "\n",
+      "#calculation\n",
+      "s2=s1;#isentropic process \n",
+      "h4=h3;#throttling\n",
+      "QL=(h1-h4)*m;\n",
+      "Wm=m*(h2-h1);\n",
+      "Qh=m*(h2-h3);\n",
+      "COPR=QL/Wm;\n",
+      "print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);\n",
+      "print'the power input to the compressor %f kW'%round(Wm,2);\n",
+      "print'the rate of heat rejection to the environment %f kW'%round(Qh);\n",
+      "print'the COP of the refrigerator is %f'%round(COPR,2);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of heat removal from the refrigerated space 7.180000 kW\n",
+        "the power input to the compressor 1.810000 kW\n",
+        "the rate of heat rejection to the environment 9.000000 kW\n",
+        "the COP of the refrigerator is 3.970000\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11-2 ,Page No.615"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=0.05;#mass flow rate in kg/s\n",
+      "P1=0.14;#inlet pressure in MPa\n",
+      "T1=-10;#inlet temperature in C\n",
+      "P2=0.8;#outlet pressure in MPa\n",
+      "T2=50;#outlet temperature in C\n",
+      "P3=0.72;#condensor pressure in MPa\n",
+      "T3=26;#condensor temperature in C\n",
+      "\n",
+      "#from refrigerant tables\n",
+      "h1=246.36;\n",
+      "h2=286.69;\n",
+      "h3=87.83;\n",
+      "h2S=284.21;#at isentropic conditions\n",
+      "\n",
+      "#calculations\n",
+      "h4=h3;#throttling\n",
+      "QL=m*(h1-h4);\n",
+      "Wm=m*(h2-h1);\n",
+      "nC=(h2S-h1)/(h2-h1);\n",
+      "COPR=QL/Wm;\n",
+      "print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);\n",
+      "print'the power input to the compressor %f kW'%round(Wm,2);\n",
+      "print'the isentropic efficiency of the compressor is %f'%round(nC,3);\n",
+      "print'the COP of the refrigerator is %f'%round(COPR,2);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of heat removal from the refrigerated space 7.930000 kW\n",
+        "the power input to the compressor 2.020000 kW\n",
+        "the isentropic efficiency of the compressor is 0.939000\n",
+        "the COP of the refrigerator is 3.930000\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11-3 ,Page No.621"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "mA=0.05;#mass flow rate in kg/s\n",
+      "P1=0.14;#inlet pressure in MPa\n",
+      "P5=0.32;#pressure at heat exchanger in MPa\n",
+      "P7=0.8;#oultet pressure in MPa\n",
+      "\n",
+      "#from tables\n",
+      "h1=239.16;\n",
+      "h2=255.93;\n",
+      "h3=55.16;\n",
+      "h5=251.88;\n",
+      "h6=270.92;\n",
+      "h7=95.47;\n",
+      "\n",
+      "#calculations\n",
+      "h4=h3;#throttling\n",
+      "h8=h7;#throttling\n",
+      "# E out = E in\n",
+      "# mA*h5 + mB*h3 = mA*h8 + mB*h2\n",
+      "mB=mA*(h5-h8)/(h2-h3);\n",
+      "QL=mB*(h1-h4);\n",
+      "# W in = Wcomp I,in + Wcomp II,in\n",
+      "Win=mA*(h6-h5)+mB*(h2-h1);\n",
+      "COPR=QL/Win;\n",
+      "print'the mass flow rate of the refrigerant through the lower cycle %f kg/s'%round(mB,4);\n",
+      "print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);\n",
+      "print'the power input to the compressor %f kW'%round(Win,2);\n",
+      "print'the coefficient of performance of this cascade refrigerator is %f'%round(COPR,2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mass flow rate of the refrigerant through the lower cycle 0.039000 kg/s\n",
+        "the rate of heat removal from the refrigerated space 7.170000 kW\n",
+        "the power input to the compressor 1.610000 kW\n",
+        "the coefficient of performance of this cascade refrigerator is 4.460000\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11-4 ,Page No.624"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=0.14;#inlet pressure in MPa\n",
+      "P5=0.32;#throttled pressure in MPa\n",
+      "P7=0.8;#oultet pressure in MPa\n",
+      "\n",
+      "#from tables\n",
+      "h1=239.16;\n",
+      "h2=255.93;\n",
+      "h3=251.88;\n",
+      "h5=95.47;\n",
+      "h7=55.16;\n",
+      "\n",
+      "#from saturated liquid-vapour table\n",
+      "#at P=0.32 MPa\n",
+      "hf=55.16;\n",
+      "hfg=196.71;\n",
+      "\n",
+      "#calculations\n",
+      "h8=h7;#throttling\n",
+      "h6=h5;#throttling\n",
+      "#the quality at state 6\n",
+      "x6=(h6-hf)/hfg;\n",
+      "qL=(1-x6)*(h1-h8);\n",
+      "# W in = Wcomp I,in + Wcomp II,in\n",
+      "#enthalaoy at state 9\n",
+      "# E out = E in\n",
+      "h9=x6*h3+(1-x6)*h2;\n",
+      "# s9 = s4 i.e isentropic process\n",
+      "#at 0.8MPa and s4=0.9416 kJ/kg\n",
+      "h4=274.48;\n",
+      "Win=(1-x6)*(h2-h1)+(1)*(h4-h9);\n",
+      "COPR=qL/Win;\n",
+      "print'the fraction of the refrigerant that evaporates as it is throttled to the flash chamber is %f'%round(x6,4);\n",
+      "print'the amount of heat removed from the refrigerated space %f kJ/kg'%round(qL,1);\n",
+      "print'the compressor work per unit mass of refrigerant flowing through the condensor %f kJ/kg'%round(Win,2);\n",
+      "print'the coefficient of performance is %f'%round(COPR,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the fraction of the refrigerant that evaporates as it is throttled to the flash chamber is 0.204900\n",
+        "the amount of heat removed from the refrigerated space 146.300000 kJ/kg\n",
+        "the compressor work per unit mass of refrigerant flowing through the condensor 32.710000 kJ/kg\n",
+        "the coefficient of performance is 4.470000\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11-5 ,Page No.630"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=0.1;#mass flow rate in lbm/s\n",
+      "T1=0+460;#refigerated space at temperature in R\n",
+      "T3=80+460;#temperature of surrounding medium in R\n",
+      "\n",
+      "#from Table A\u201317E\n",
+      "# at T1\n",
+      "h1=109.90;\n",
+      "Pr1=.7913;\n",
+      "#pressure ratio at compressor is 4\n",
+      "Pr2=4*Pr1;\n",
+      "#at Pr2\n",
+      "h2=163.5;\n",
+      "T2=683;\n",
+      "#at T3\n",
+      "h3=129.06;\n",
+      "Pr3=1.3860;\n",
+      "#pressure ratio at compressor is 4\n",
+      "Pr4=Pr3/4;\n",
+      "#at Pr4\n",
+      "h4=86.7;\n",
+      "T4=363;\n",
+      "\n",
+      "#calculations\n",
+      "qL=h1-h4;\n",
+      "Wout=h3-h4;\n",
+      "Win=h2-h1;\n",
+      "COPR=qL/(Win-Wout);\n",
+      "Qrefrig=m*qL;\n",
+      "print'the minimum temperatures in the cycle %f F'%round(T4-460);\n",
+      "print'the maximum temperatures in the cycle %f F'%round(T2-460);\n",
+      "print'the coefficient of performance is %f'%round(COPR,2)\n",
+      "print'the rate of refrigeration for a mass flow rate of 0.1 lbm/s. %f Btu/s'%round(Qrefrig,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the minimum temperatures in the cycle -97.000000 F\n",
+        "the maximum temperatures in the cycle 223.000000 F\n",
+        "the coefficient of performance is 2.060000\n",
+        "the rate of refrigeration for a mass flow rate of 0.1 lbm/s. 2.320000 Btu/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11-6 ,Page No.636"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "COPR=0.1;#refrigerator COP\n",
+      "T1=20;#intial temp in C\n",
+      "T2=4;#final temp in C\n",
+      "t=30*60;#total time required in sec\n",
+      "V=0.350;#volumne of can in L\n",
+      "\n",
+      "#constants used\n",
+      "p=1;#on kg/L\n",
+      "c=4.18;#in kJ/kg-C from Table A-3\n",
+      "\n",
+      "#calculations\n",
+      "m=p*V;\n",
+      "Qcooling=m*c*(T1-T2)/t*1000;#converted in W by multiplying by 1000\n",
+      "Win=Qcooling/COPR;\n",
+      "print'the average electric power consumed by the thermoelectric refrigerator %i W'%round(Win)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the average electric power consumed by the thermoelectric refrigerator 130 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter12_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter12_1.ipynb
new file mode 100755
index 00000000..c181d0f6
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter12_1.ipynb
@@ -0,0 +1,262 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Thermodynamic Property Relations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12-1 ,Page No.652"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#from Table A-17\n",
+      "#cp value approximately by replacing the differentials\n",
+      "h1=305.22;#in kJ/kg\n",
+      "T1=305;#in K\n",
+      "h2=295.17;#in kJ/kg\n",
+      "T2=295;#in K\n",
+      "\n",
+      "#calculations\n",
+      "#from the given equation we can calculate\n",
+      "cp=(h1-h2)/(T1-T2);\n",
+      "print'the cp of air at 300 K %f kJ/ kg - K'%round(cp,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the cp of air at 300 K 1.005000 kJ/ kg - K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12-2 ,Page No.654"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "dT=302-300;#differnce in final and intial temperatures of air in K\n",
+      "dv=0.87-0.86;#differnce in final and intial volumnes of air in m^3/kg\n",
+      "T=(302+300)/2;#average temp in K\n",
+      "v=(0.87+0.86)/2;#average volumne in m^3/kg\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kJ/kg-K\n",
+      "\n",
+      "#calculations\n",
+      "#using eq 12-3 by diffrentiating P= R*T/v\n",
+      "dP= R*dT/v - R*T*dv/v**2;\n",
+      "print'the change in the pressure of air %f kPa'%round(dP,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change in the pressure of air -0.491000 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12-5 ,Page No.659"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=20+273.15;#temperature of refrigerant in K\n",
+      "\n",
+      "#from Table A\u201311\n",
+      "vf=0.0008161;#in m^3/kg\n",
+      "vg=0.035969;#in m^3/kg\n",
+      "\n",
+      "#calculations\n",
+      "#using Eq 12-22\n",
+      "# hfg= T*vfg*(dP/dT)sat\n",
+      "#(dP/dT)sat b/w 24 C - 16 C \n",
+      "dPT=(646.18-504.58)/(24-16);#dP/dT ; values from Table A\u201311\n",
+      "vfg=vg-vf;\n",
+      "hfg=T*vfg*dPT;\n",
+      "print'the value of the enthalpy of vaporization of refrigerant-134a %f kJ/kg'%round(hfg,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of the enthalpy of vaporization of refrigerant-134a 182.400000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12-6 ,Page No.660"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "\n",
+      "#given data\n",
+      "T2=-50+460.0;#temperature of refrigerant-134a on R \n",
+      "\n",
+      "#constants\n",
+      "R=0.01946;#in Btu/lbm\n",
+      "\n",
+      "#from Table A-11E\n",
+      "T1=-40+460.0;#converted into R from F\n",
+      "P1=7.432;\n",
+      "hfg=97.100;\n",
+      "\n",
+      "#calcualation\\\n",
+      "#using Equation 12\u201324\n",
+      "#ln(P2/P1)= hfg/R *(1/T1 - 1/T2)\n",
+      "P2=P1*exp(hfg/R *(1/T1 - 1/T2));\n",
+      "print'the saturation pressure of refrigerant-134a %f psia'%round(P2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the saturation pressure of refrigerant-134a 5.560000 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12-11 ,Page No.373"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#given data\n",
+      "T1=220;#intial temperature in K\n",
+      "P1=5;#intial pressure in MPa\n",
+      "T2=300;#final temperature in K\n",
+      "P2=10;#final pressure in MPa\n",
+      "\n",
+      "#constants used\n",
+      "Ru=8.314;#on kJ/kmol- K\n",
+      "\n",
+      "#from Table A\u20131\n",
+      "Tcr=154.8;\n",
+      "Pcr=5.08;\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "print('part - a');\n",
+      "#by assuming ideal-gas behavior\n",
+      "#from Table A\u201319\n",
+      "h1=6404;\n",
+      "h2=8736;\n",
+      "s2=205.213;\n",
+      "s1=196.171;\n",
+      "h21i=h2-h1;#h2 - h1 ideal\n",
+      "s21i=(s2-s1)-Ru*log(P2/P1);#s2 - s1 ideal\n",
+      "print'the enthalpy change %i kJ/kmol'%round(h21i);\n",
+      "print'the entropy change %f kJ/kmol-K'%round(s21i,2);\n",
+      "\n",
+      "#part - b\n",
+      "print('part - b');\n",
+      "#by accounting for the deviation from ideal-gas behavior\n",
+      "TR1=T1/Tcr;\n",
+      "Pr1=P1/Pcr;\n",
+      "#from the generalized charts at each state\n",
+      "Zh1=0.53;\n",
+      "Zs1=0.25;\n",
+      "TR2=T2/Tcr;\n",
+      "Pr2=P2/Pcr;\n",
+      "#from the generalized charts at each state\n",
+      "Zh2=0.48;\n",
+      "Zs2=0.20;\n",
+      "h21=h21i-Ru*Tcr*(Zh2-Zh1);\n",
+      "s21=s21i-Ru*(Zs2-Zs1);\n",
+      "print'the enthalpy change %i in kJ/kmol'%round(h21);\n",
+      "print'the entropy change %f kJ/kmol-K'%round(s21,2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "part - a\n",
+        "the enthalpy change 2332 kJ/kmol\n",
+        "the entropy change 3.280000 kJ/kmol-K\n",
+        "part - b\n",
+        "the enthalpy change 2396 in kJ/kmol\n",
+        "the entropy change 3.690000 kJ/kmol-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter13_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter13_1.ipynb
new file mode 100755
index 00000000..936d90ae
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter13_1.ipynb
@@ -0,0 +1,453 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13: Gas Mixtures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13-1 ,Page No.683"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "mO2=3.0;#moles of oxygen\n",
+      "mN2=5.0;#moles of nitrogen\n",
+      "mCH4=12.0;#moles of methane\n",
+      "\n",
+      "#molecular masses in kg\n",
+      "MO2=32.0;\n",
+      "MN2=28.0;\n",
+      "MCH4=16.0;\n",
+      "\n",
+      "#constants used\n",
+      "Ru=8.314;#in kJ/kg - K\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "mm=mO2+mN2+mCH4;\n",
+      "mfO2=mO2/mm;\n",
+      "mfN2=mN2/mm;\n",
+      "mfCH4=mCH4/mm;\n",
+      "print'mass fraction of oxygen is %f'%round(mfO2,2);\n",
+      "print'mass fraction of nitrogen is %f'%round(mfN2,2);\n",
+      "print'mass fraction of methane is %f'%round(mfCH4,2);\n",
+      "\n",
+      "#part - b\n",
+      "NO2=mO2/MO2;\n",
+      "NN2=mN2/MN2;\n",
+      "NCH4=mCH4/MCH4;\n",
+      "Nm=NO2+NN2+NCH4;\n",
+      "yO2=NO2/Nm;\n",
+      "yN2=NN2/Nm;\n",
+      "yCH4=NCH4/Nm;\n",
+      "print'mole fraction of oxygen is %f'%round(yO2,3);\n",
+      "print'mole fraction of nitrogen is %f'%round(yN2,3);\n",
+      "print'mole fraction of methane is %f'%round(yCH4,3);\n",
+      "\n",
+      "#part - c\n",
+      "Mm=mm/Nm;\n",
+      "print'average molecular mass %f kg/kmol'%round(Mm,1);\n",
+      "Rm=Ru/Mm;\n",
+      "print'gas constant of mixture %f kJ/kg - K'%round(Rm,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass fraction of oxygen is 0.150000\n",
+        "mass fraction of nitrogen is 0.250000\n",
+        "mass fraction of methane is 0.600000\n",
+        "mole fraction of oxygen is 0.092000\n",
+        "mole fraction of nitrogen is 0.175000\n",
+        "mole fraction of methane is 0.734000\n",
+        "average molecular mass 19.600000 kg/kmol\n",
+        "gas constant of mixture 0.425000 kJ/kg - K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13-2 ,Page No.687"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "NN2=2.0;#moles of nitrogen\n",
+      "NCO2=6.0;#moles of carbon dioxide\n",
+      "Tm=300.0;#temperature of gases in K\n",
+      "Pm=15000.0;#pressure of gases in kPa\n",
+      "\n",
+      "#constants used\n",
+      "Ru=8.314;#in kJ/kmol - K\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "Nm=NN2+NCO2;\n",
+      "Vm=Nm*Ru*Tm/Pm;\n",
+      "print'the volume of the tank on the basis of the ideal-gas equation of state %f m^3'%round(Vm,3);\n",
+      "\n",
+      "#part - b\n",
+      "#from Table A-1\n",
+      "#for nitrogen\n",
+      "TcrN=126.2;\n",
+      "PcrN=3390;\n",
+      "#for Carbondioxide\n",
+      "TcrC=304.2;\n",
+      "PcrC=7390;\n",
+      "yN2=NN2/Nm;\n",
+      "yCO2=NCO2/Nm;\n",
+      "Tcr=yN2*TcrN+yCO2*TcrC;\n",
+      "Pcr=yN2*PcrN+yCO2*PcrC;\n",
+      "Tr=Tm/Tcr;\n",
+      "Pr=Pm/Pcr;\n",
+      "#from Fig A-15b\n",
+      "Zm=0.49;\n",
+      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
+      "print'the volume of the tank on the basis Kay\u2019s rule %f m^3'%round(Vm,3);\n",
+      "\n",
+      "#part - c\n",
+      "#for nitrogen\n",
+      "TrN=Tm/TcrN;\n",
+      "PrN=Pm/PcrN;\n",
+      "#from Fig A-15b\n",
+      "Zn=1.02;\n",
+      "#for Carbondioxide\n",
+      "TrC=Tm/TcrC;\n",
+      "PcrC=Pm/PcrC;\n",
+      "#from Fig A-15b\n",
+      "Zc=0.3;\n",
+      "Zm=yN2*Zn+yCO2*Zc;\n",
+      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
+      "print'the volume of the tank on the basis compressibility factors and Amagat\u2019s law %f m^3'%round(Vm,3);\n",
+      "\n",
+      "#part - d\n",
+      "VRN=(Vm/NN2)/(Ru*TcrN/PcrN);\n",
+      "VRC=(Vm/NCO2)/(Ru*TcrC/PcrC);\n",
+      "#from Fig A-15b\n",
+      "Zn=0.99;\n",
+      "Zc=0.56;\n",
+      "Zm=yN2*Zn+yCO2*Zc;\n",
+      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
+      "#When the calculations are repeated we obtain 0.738 m3 after the second iteration, 0.678 m3 after the third iteration, and 0.648 m3 after the fourth iteration.\n",
+      "Vm=0.648;\n",
+      "print'compressibility factors and Dalton\u2019s law the volume of the tank on the basis %f m^3'%round(Vm,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the volume of the tank on the basis of the ideal-gas equation of state 1.330000 m^3\n",
+        "the volume of the tank on the basis Kay\u2019s rule 0.652000 m^3\n",
+        "the volume of the tank on the basis compressibility factors and Amagat\u2019s law 0.639000 m^3\n",
+        "compressibility factors and Dalton\u2019s law the volume of the tank on the basis 0.648000 m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13-3 ,Page No.691"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "mN=4.0;#mass of nitrogen in kg\n",
+      "T1N=20.0;#temperature of nitrogen in K\n",
+      "P1N=150.0;#pressure of nitrogen in kPa\n",
+      "mO=7.0;#mass of oxygen in kg\n",
+      "T1O=40.0;#temperature of oxygen in K\n",
+      "P1O=100.0;#pressure of oxygen in kPa\n",
+      "\n",
+      "#molecular masses in kg\n",
+      "MO=32.0;\n",
+      "MN=28.0;\n",
+      "\n",
+      "#constants used\n",
+      "Ru=8.314;#in kJ/kg - K\n",
+      "\n",
+      "#from Table A-2a\n",
+      "CvN=0.743;\n",
+      "CvO=0.658;\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "#Ein - Eout = dEsystem\n",
+      "# (m*cv*dT)N2 + (m*cv*dT)= 0;\n",
+      "Tm= (mN*CvN*T1N+ mO*CvO*T1O)/(mN*CvN+mO*CvO);\n",
+      "print'the mixture temperature %f C'%round(Tm,1);\n",
+      "\n",
+      "#part - b\n",
+      "NO=mO/MO;\n",
+      "NN=mN/MN;\n",
+      "Nm=NO+NN;\n",
+      "VO=NO*Ru*(T1O+273)/P1O;\n",
+      "VN=NN*Ru*(T1N+273)/P1N;#Exergy Destruction during Mixing of Ideal Gases\n",
+      "Vm=VO+VN;\n",
+      "Pm=Nm*Ru*(Tm+273)/Vm;  \n",
+      "print'the mixture pressure after equilibrium has been established %f kPa'%round(Pm,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mixture temperature 32.200000 C\n",
+        "the mixture pressure after equilibrium has been established 114.500000 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13-4 ,Page No.692"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#given data\n",
+      "NO=3.0;#moles of oxygen \n",
+      "NC=5.0;#moles of carbondioxide\n",
+      "T0=25+273.0;#temperature of gasses in K\n",
+      "\n",
+      "#constants used\n",
+      "Ru=8.314;#in kJ/kg - K\n",
+      "\n",
+      "#calculations\n",
+      "Nm=NO+NC;\n",
+      "yO=NO/Nm;\n",
+      "yC=NC/Nm;\n",
+      "#dSm= -Ru*(NO*log(yO)+NC*log(yC))\n",
+      "Sm=-Ru*(NO*log(yO)+NC*log(yC));\n",
+      "print'the entropy change %f kJ/K'%round(Sm);\n",
+      "Xdestroyed=T0*Sm/1000;\n",
+      "print'exergy destruction associated %f MJ'%round(Xdestroyed,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the entropy change 44.000000 kJ/K\n",
+        "exergy destruction associated 13.100000 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13-5 ,Page No.694"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=220;#intial temperature in K\n",
+      "T2=160;#final temperature in K\n",
+      "Pm=10;#pressure of air in MPa\n",
+      "yN=0.79;#mole fraction of nitrogen\n",
+      "yO=0.21;#mole fractions of oxygen\n",
+      "\n",
+      "\n",
+      "#critical properties\n",
+      "#for Nitrogen\n",
+      "TcrN=126.2;\n",
+      "PcrN=3.39;\n",
+      "#for Oxygen\n",
+      "TcrO=154.8;\n",
+      "PcrO=5.08;\n",
+      "\n",
+      "#constants used\n",
+      "Ru=8.314;#in kJ/kg - K\n",
+      "\n",
+      "#from Tables A-18 & 19\n",
+      "#at T1\n",
+      "h1N=6391;\n",
+      "h1O=6404;\n",
+      "#for T2\n",
+      "h2N=4648;\n",
+      "h2O=4657;\n",
+      "\n",
+      "#calculations\n",
+      "#part - a\n",
+      "qouti=yN*(h1N-h2N)+yO*(h1O-h2O);\n",
+      "print'the heat transfer during this process using the ideal-gas approximation %i kJ/kmol'%round(qouti);\n",
+      "\n",
+      "#part - b\n",
+      "Tcrm=yN*TcrN+yO*TcrO;\n",
+      "Pcrm=yN*PcrN+yO*PcrO;\n",
+      "Tr1=T1/Tcrm;\n",
+      "Tr2=T2/Tcrm;\n",
+      "Pr=Pm/Pcrm;\n",
+      "#at these values we get\n",
+      "Zh1=1;\n",
+      "Zh2=2.6\n",
+      "qout=qouti-Ru*Tcrm*(Zh1-Zh2);\n",
+      "print'the heat transfer during this process using Kay\u2019s rule %i kJ/kmol'%round(qout);\n",
+      "\n",
+      "#part - c\n",
+      "#for nitrogen\n",
+      "TrN1=T1/TcrN;\n",
+      "TrN2=T2/TcrN;\n",
+      "PrN=Pm/PcrN;\n",
+      "#from Fig A-15b\n",
+      "Zh1n=0.9;\n",
+      "Zh2n=2.4;\n",
+      "#for Oxygen\n",
+      "TrO1=T1/TcrO;\n",
+      "TrO2=T2/TcrO;\n",
+      "PcrO=Pm/PcrO;\n",
+      "#from Fig A-15b\n",
+      "Zh1O=1.3;\n",
+      "Zh2O=4.0;\n",
+      "#from Eq 12-58\n",
+      "h12N=h1N-h2N-Ru*TcrN*(Zh1n-Zh2n);# h1 - h2 for nitrogen\n",
+      "h12O=h1O-h2O-Ru*TcrO*(Zh1O-Zh2O);# h1 - h2 for oxygen\n",
+      "qout=yN*h12N+yO*h12O;\n",
+      "print'the heat transfer during this process using Amagat\u2019s law %i kJ/kmol'%round(qout);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the heat transfer during this process using the ideal-gas approximation 1744 kJ/kmol\n",
+        "the heat transfer during this process using Kay\u2019s rule 3502 kJ/kmol\n",
+        "the heat transfer during this process using Amagat\u2019s law 3717 kJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13-6 ,Page No.705"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#13.6 (d) answer not matching as float datatype is giving more accurate answer in comparison to textbook that has given approximate due to rounding off to two decimal places\n",
+      "\n",
+      "#given data\n",
+      "mfs=0.0348;#salinity mass fraction\n",
+      "mfw=1-mfs;\n",
+      "T0=288.15;#temperature of sea water in K\n",
+      "\n",
+      "#constants used\n",
+      "Mw=18;\n",
+      "Ms=58.44;\n",
+      "Rw=0.4615;\n",
+      "pm=1028;\n",
+      "Ru=8.314;\n",
+      "\n",
+      "#calculations\n",
+      "#part - a\n",
+      "Mm=1/((mfs/Ms)+(mfw/Mw));\n",
+      "yw=mfw*Mm/Mw;\n",
+      "ys=1-yw;\n",
+      "print'the mole fraction of the water is %f'%round(yw,4);\n",
+      "print'the mole fraction of the saltwater is %f'%round(ys,5);\n",
+      "\n",
+      "#part - b\n",
+      "wmin=-Ru*T0*(ys*log(ys)+yw*log(yw));\n",
+      "wm=wmin/Mm;\n",
+      "print'the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts %f kJ'%round(wm,2);\n",
+      "\n",
+      "#part - c\n",
+      "wmin=Rw*T0*log(1/yw);\n",
+      "print'the minimum work input required to obtain 1 kg of fresh water from the sea %f kJ'%round(wmin,2);\n",
+      "\n",
+      "#part - d\n",
+      "Pmin=pm*Rw*T0*log(1/yw);\n",
+      "print'the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes %i kPa'%round(Pmin)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mole fraction of the water is 0.989000\n",
+        "the mole fraction of the saltwater is 0.010980\n",
+        "the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts 7.850000 kJ\n",
+        "the minimum work input required to obtain 1 kg of fresh water from the sea 1.470000 kJ\n",
+        "the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes 1510 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter14_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter14_1.ipynb
new file mode 100755
index 00000000..f218441e
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter14_1.ipynb
@@ -0,0 +1,393 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14: Gas\u2013Vapor Mixtures and Air-Conditioning"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14-1 ,Page No.720"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V=5*5*3.0;#volume of the room in m^3\n",
+      "RH=0.75;#relative humidity\n",
+      "P=100.0;#pressure of air in kPa\n",
+      "T=25.0;#temperature of air in C\n",
+      "\n",
+      "#constants used\n",
+      "Ra=0.287;#in kPa.m^3 / kg.k\n",
+      "Rv=0.4615;#in kPa.m^3 / kg.k\n",
+      "\n",
+      "#from Table A-2a and A-4\n",
+      "cp=1.005;\n",
+      "Psat=3.1698;\n",
+      "hg=2564.6;\n",
+      "\n",
+      "#calculation\n",
+      "Pv=RH*Psat;\n",
+      "Pa=P-Pv;\n",
+      "w=0.622*Pv/(P-Pv);\n",
+      "h=cp*T+w*hg;\n",
+      "ma=V*Pa/(Ra*(T+273));\n",
+      "mv=V*Pv/(Rv*(T+273));\n",
+      "print'the partial pressure of dry air %f kPa'%round(Pa,2);\n",
+      "print'the specific humidity %f kg water/kg of dry air'%round(w,4);\n",
+      "print'the enthalpy per unit mass of the dry air %f kJ'%round(h,1);\n",
+      "print'mass of air %f kg'%round(ma,2);\n",
+      "print'mass of water vapour %f kg'%round(mv,1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the partial pressure of dry air 97.620000 kPa\n",
+        "the specific humidity 0.015100 kg water/kg of dry air\n",
+        "the enthalpy per unit mass of the dry air 64.000000 kJ\n",
+        "mass of air 85.610000 kg\n",
+        "mass of water vapour 1.300000 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14-2 ,Page No.722"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=20;#temperature of air in C\n",
+      "RH=0.75;#relative humidity\n",
+      "\n",
+      "#from Table A-4\n",
+      "Psat=2.3392;\n",
+      "Pv=RH*Psat;\n",
+      "#thus at this from Eq 14-13\n",
+      "Tdp=15.4;\n",
+      "print'window temperature %f C'%Tdp"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "window temperature 15.400000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14-3 ,Page No.725"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=25.0;#dry bulb temperature in K\n",
+      "T2=15.0;#wet bulb temperature in K\n",
+      "P2=101.325;#pressure of air in kPa\n",
+      "\n",
+      "#from Table A-2a & A-4\n",
+      "#at T1\n",
+      "Psat1=3.1698;\n",
+      "hg1=2546.5;\n",
+      "#at T2\n",
+      "Psat2=1.7057;\n",
+      "hfg2=2465.4;\n",
+      "hf2=62.982;\n",
+      "cp=1.005;\n",
+      "\n",
+      "#calculations\n",
+      "w2=0.622*Psat2/(P2-Psat2);\n",
+      "w1=(cp*(T2-T1)+w2*hfg2)/(hg1-hf2);\n",
+      "print'the specific humidity %f kg water/kg of dry air'%round(w1,5);\n",
+      "RH1=w1*P2/((0.622+w1)*Psat1);\n",
+      "print'the relative humidity is %f'%round(RH1,3);\n",
+      "h=cp*T1+w1*hg1;\n",
+      "print'the enthalpy of the air %f kJ/kg of dry air'%round(h,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the specific humidity 0.006530 kg water/kg of dry air\n",
+        "the relative humidity is 0.332000\n",
+        "the enthalpy of the air 41.700000 kJ/kg of dry air\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14-5 ,Page No.731"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "RH1=0.3;#outdoor relative humidity\n",
+      "P1=100.0;#pressure of air in kPa\n",
+      "V1=45.0;#volumetric steady rate in m^3/min\n",
+      "T1=10.0;#outdoor temperature in C\n",
+      "T2=22.0;#first heated temperature in C\n",
+      "RH3=0.6;#final relative humidity \n",
+      "T3=25.0;#final temperature in C\n",
+      "\n",
+      "#from Table A-2a & A-4\n",
+      "cp=1.005;\n",
+      "Ra=0.287;\n",
+      "Pg1=1.2281;\n",
+      "hg1=2519.2;\n",
+      "hg2=2541.0;\n",
+      "Pg3=3.1698;\n",
+      "\n",
+      "#calculations\n",
+      "Pv1=RH1*Pg1;\n",
+      "Pa1=P1-Pv1;\n",
+      "v1=Ra*(T1+273)/Pa1;\n",
+      "ma=V1/v1;\n",
+      "w1=0.622*Pv1/(P1-Pv1);\n",
+      "h1=cp*T1+w1*hg1;\n",
+      "h1=round(h1,1);\n",
+      "w2=w1;\n",
+      "h2=cp*T2+w2*hg2;\n",
+      "h2=round(h2)\n",
+      "Q=ma*(h2-h1);\n",
+      "# ma2*w2 + mw = ma3*w3\n",
+      "#which reduces to mw = ma * (w3 - w2)\n",
+      "w3=0.622*RH3*Pg3/(P1-(RH3*Pg3));\n",
+      "mw=ma*(w3-w2);\n",
+      "print'the rate of heat supply in the heating section %f kJ/min'%round(Q);\n",
+      "print'the mass flow rate of the steam required in the humidifying section %f kg/min'%round(mw,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of heat supply in the heating section 673.000000 kJ/min\n",
+        "the mass flow rate of the steam required in the humidifying section 0.539000 kg/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14-6 ,Page No.733"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V1=10;#volumertric rate in m^3/min\n",
+      "T1=30;#intial temperature in C\n",
+      "RH1=0.8;#intial relative humidity\n",
+      "T2=14;#final temperature in C\n",
+      "RH2=1;#final relative humidity\n",
+      "\n",
+      "#from Table A-4\n",
+      "hw=58.8;\n",
+      "h1=85.4;\n",
+      "h2=39.3;\n",
+      "w1=0.0216;\n",
+      "w2=0.0100;\n",
+      "v1=0.889;\n",
+      "\n",
+      "#calculations\n",
+      "#mw= ma*(w1-w2)\n",
+      "#Qout=ma*(h1-h2) - mw*hw\n",
+      "ma=V1/v1;\n",
+      "mw= ma*(w1-w2);\n",
+      "Qout=ma*(h1-h2) - mw*hw;\n",
+      "print'rates of moisture removal from the air %f kg/min'%round(mw,3);\n",
+      "print'rate of moisture removal from the air %i kJ/min'%round(Qout);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rates of moisture removal from the air 0.130000 kg/min\n",
+        "rate of moisture removal from the air 511 kJ/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14-8 ,Page No.736"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given values\n",
+      "V1=50;#saturated air volumertric rate in m^3/min\n",
+      "T1=14;#saturated air temperature in C\n",
+      "V2=20;#outside air volumertric rate in m^3/min\n",
+      "T2=32;#outside air temperature in C\n",
+      "RH2=60;#outside air relative humidity\n",
+      "\n",
+      "#from psychrometric chart\n",
+      "h1=39.4;\n",
+      "w1=0.010;\n",
+      "v1=0.826;\n",
+      "h2=79;\n",
+      "w2=0.0182;\n",
+      "v2=0.889;\n",
+      "\n",
+      "#calculations\n",
+      "ma1=V1/v1;\n",
+      "ma2=V2/v2;\n",
+      "ma3=ma1+ma2;\n",
+      "#from Eq 14-24\n",
+      "w3=(w2*ma2+w1*ma1)/(ma1+ma2);\n",
+      "h3=(h2*ma2+h1*ma1)/(ma1+ma2);\n",
+      "print'the specific humidity %f kg of water/kg of dry air'%round(w3,3);\n",
+      "#from psychrometric chart\n",
+      "T3=19;\n",
+      "RH3=0.89;\n",
+      "v3=0.844;\n",
+      "V3=ma3*v3;\n",
+      "print'the dry-bulb temperature %f C'%T3;\n",
+      "print'the relative humidity is %f'%RH3\n",
+      "print'the volume flow rate of the mixture %f m^3/min'%round(V3,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the specific humidity 0.012000 kg of water/kg of dry air\n",
+        "the dry-bulb temperature 19.000000 C\n",
+        "the relative humidity is 0.890000\n",
+        "the volume flow rate of the mixture 70.100000 m^3/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14-9 ,Page No.738"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data \n",
+      "m=100;#mass flow rate in kg/s\n",
+      "T1=20;#entry temperature in C\n",
+      "P1=1;#entry pressure im kPa\n",
+      "RH1=60;#entry relative humidity\n",
+      "T2=30;#exit temperature in C\n",
+      "RH2=1;#exit relative humidity\n",
+      "T3=35;#entry tempearture of wet cooling tower in C \n",
+      "T4=22;#water temperature in cooling tower in C\n",
+      "\n",
+      "#from Table A-4\n",
+      "h1=42.2;\n",
+      "w1=0.0087;\n",
+      "v1=0.842;\n",
+      "h2=100;\n",
+      "w2=0.0273;\n",
+      "h3=146.64;\n",
+      "h4=92.28;\n",
+      "\n",
+      "#calculations\n",
+      "#Dry air balane = ma1 = ma2 = ma\n",
+      "#Water balance = m3 - m4 = ma*(w2 - w1)\n",
+      "#Energy balance = ma1*h1 + m3*h3 = ma2*h2 + m4*h4\n",
+      "ma= m*(h3-h4)/(h2-h1-(w2-w1)*h4);\n",
+      "V1=ma*v1;\n",
+      "mmakeup=ma*(w2-w1);\n",
+      "print'the volume flow rate of air into the cooling tower %f m^3/s'%round(V1,1);\n",
+      "print'the mass flow rate of the required makeup water %f kg/s'%round(mmakeup,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the volume flow rate of air into the cooling tower 81.600000 m^3/s\n",
+        "the mass flow rate of the required makeup water 1.800000 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter15_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter15_1.ipynb
new file mode 100755
index 00000000..c167fe73
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter15_1.ipynb
@@ -0,0 +1,686 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15: Chemical Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-1 ,Page No.755"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "nO2i=20;#intial moles of air \n",
+      "nC8H18i=1;#intial moles octane\n",
+      "\n",
+      "#from Table A-1\n",
+      "Mair=29;\n",
+      "MC=12;\n",
+      "MH=2;\n",
+      "\n",
+      "#calculations\n",
+      "# Chemical Reaction\n",
+      "# C8H18 + 20(O2+3.76N2)= xCO2 + yH2O +  zO2 + wN2\n",
+      "#by elemental balance of moles\n",
+      "x=8;\n",
+      "y=18/2;\n",
+      "z=20*2-2*x-y;\n",
+      "w=20*3.76;\n",
+      "print'kmoles of CO2 %i'%x;\n",
+      "print'kmoles of H2O %i'%y;\n",
+      "print'kmoles of O2 %f'%round(z,1);\n",
+      "print'kmoles of N2 %f'%round(w,1);\n",
+      "#thus equn becomes\n",
+      "# C8H18 + 20(O2+3.76N2)= 8CO2 + 9H2O +  7.5O2 +75.2N2\n",
+      "AF=nO2i*4.76*Mair/(x*MC + y*MH);\n",
+      "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kmoles of CO2 8\n",
+        "kmoles of H2O 9\n",
+        "kmoles of O2 15.000000\n",
+        "kmoles of N2 75.200000\n",
+        "air-fuel ratio of combustion process 24.200000 kg air/kg fuel\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-2 ,Page No.757"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P=100;#total pressure in kPa\n",
+      "\n",
+      "#from Table A-1\n",
+      "Mair=29.0;\n",
+      "MC=12.0;\n",
+      "MH=2.0;\n",
+      "\n",
+      "#calculations\n",
+      "#Chemical reaction\n",
+      "#C2H6 +  1.2at(1O2 +  3.76) =2CO2 +  3H2O +  0.2athO2 + (1.2*3.76)athN2\n",
+      "#ath is the stoichiometric coefficient for air\n",
+      "#Oxygen balance gives\n",
+      "# 1.2ath = 2 + 1.5 + 0.2ath\n",
+      "ath=(2+1.5)/(1.2-0.2);\n",
+      "AF=(1.2*ath)*4.76*Mair/(2*MC+3*MH);\n",
+      "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,1);\n",
+      "#C2H6 + 4.2(O2 + 3.76N2)  = 2CO2 + 3H2O + 0.7O2 + 15.79N2;\n",
+      "Nprod=2+3+0.7+15.79;\n",
+      "#for dew point water vapour condenses\n",
+      "Nv=3;\n",
+      "Pv=Nv/Nprod*P;\n",
+      "#at this Pv\n",
+      "Tdp=52.3;\n",
+      "print'the dew-point %f C'%Tdp\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "air-fuel ratio of combustion process 19.300000 kg air/kg fuel\n",
+        "the dew-point 52.300000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-3 ,Page No.758"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P=101.325;#total pressure in kPa\n",
+      "RH=0.8;#realtive humidity\n",
+      "T1=20;#tempearture of air in C\n",
+      "\n",
+      "#from Table A-4\n",
+      "Psat=2.3392;\n",
+      "\n",
+      "#calculations\n",
+      "#consedering 1 kmol of fuel\n",
+      "# 0.72CH4 + 0.09H2 + 0.14N2 + 0.02O2 + 0.03CO2 +  ath(O2 + 3.76N2) = xCO2 + yH2O + zN2\n",
+      "#element balance\n",
+      "x=0.72+0.03\n",
+      "y=(0.72*4+0.09*2)/2;\n",
+      "ath=x+y/2-0.02-0.03;\n",
+      "z=0.14+3.76*ath;\n",
+      "Pv=RH*Psat;\n",
+      "# Nv,air = Pv,air/Ptotal * Ntotal\n",
+      "Nvair=Pv/P*6.97/(1-(Pv/P));\n",
+      "#0.72CH4 + 0.09H2 + 0.14N2 + 0.02O2 + 0.03CO2 +  1.465(O2 + 3.76N2) + 0.131H20 = 0.75CO2 + 1.661H2O + 5.648N2\n",
+      "Pvprod=1.661/8.059*P;\n",
+      "#at this Pvprod\n",
+      "Tdp=60.9;\n",
+      "print'the dew-point %f C'%Tdp"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the dew-point 60.900000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-4 ,Page No.760"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Pprod=100;#total pressure in kPa\n",
+      "\n",
+      "#from Table A-1\n",
+      "Mair=29;\n",
+      "MC=12;\n",
+      "MH=2;\n",
+      "\n",
+      "#from Table A-4\n",
+      "Psat=3.1698;\n",
+      "\n",
+      "#calculations\n",
+      "#consedering 100 kmol of dry products\n",
+      "# xC8H18 + a (O2 + 3.76N2) = 10.02CO2 + 0.88C0 + 84.48N2 + bH20\n",
+      "#from mass balamces\n",
+      "a=83.48/3.76;\n",
+      "x=(0.88+10.02)/8;\n",
+      "b=18*x/2;\n",
+      "# 1.36C8H18 + 22.2 (O2 + 3.76N2) = 10.02CO2 + 0.88C0 + 84.48N2 + 12.24H20\n",
+      "# 1 mol conversion\n",
+      "# C8H18 + 16.32 (O2 + 3.76N2) = 7.37CO2 + 4.13C0 + 61.38N2 + 9H20\n",
+      "AF= 16.32*4.76*Mair/(8*MC+9*MH);\n",
+      "print'air-fuel ratio of combustion process %f kg air/kg fuel'%round(AF,2);\n",
+      "# C8H18 + ath (O2 + 3.76N2) = 8CO2 + 9H2O + 3.76athN2\n",
+      "ath=8+4.5;\n",
+      "Pth=16.32/ath*4.76/4.76*100;\n",
+      "print'percentage of theoretical air is %i'%round(Pth);\n",
+      "Nprod=7.37+0.65+4.13+61.98+9;\n",
+      "# Nv/Nprod = Pv/Pprod\n",
+      "Pv=Psat;\n",
+      "Nw= (Nprod*Pv-9*Pprod)/(Pv-Pprod);\n",
+      "print'the amount of H2O that condenses as the products %f kmol'%round(Nw,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "air-fuel ratio of combustion process 19.760000 kg air/kg fuel\n",
+        "percentage of theoretical air is 131\n",
+        "the amount of H2O that condenses as the products 6.570000 kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-5 ,Page No.764"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#there is a difference in the answer due to approximation in the textbook\n",
+      "\n",
+      "#given data\n",
+      "T=25;#temperature of octane in C\n",
+      "\n",
+      "#from Table A-6\n",
+      "HCO2=-393520;\n",
+      "HH2O=-285830;\n",
+      "HC8H18=-249950;\n",
+      "\n",
+      "#calculations\n",
+      "# C8H18 + ath (O2 + 3.76N2) = 8CO2 + 9H2O + 3.76athN2\n",
+      "#N2 and O2 are stable elements, and thus their enthalpy of formation is zero\n",
+      "#hc = Hprod - Hreact\n",
+      "hc= 8*HCO2 + 9*HH2O - HC8H18;\n",
+      "print'the enthalpy of combustion of liquid octane %i kJ/kmol'%hc\n",
+      "print 'or %i kJ/kg C8H18'%round(hc/114,0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the enthalpy of combustion of liquid octane -5470680 kJ/kmol\n",
+        "or -47989 kJ/kg C8H18\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-6 ,Page No.767"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "mfuel=0.05;#mass flow rate in kg/min\n",
+      "\n",
+      "#from Table A-1\n",
+      "Mair=29;\n",
+      "MC=12;\n",
+      "MH=2;\n",
+      "\n",
+      "#calculation\n",
+      "#stochiometric reaction\n",
+      "#C3H8 + ath(O2 + 3.76N2) = 3CO2 + 4H2O + 3.76athN2\n",
+      "#O2 balance\n",
+      "ath=3+5;\n",
+      "#50 percent excess air and some CO in the products\n",
+      "#C3H8 + 7.5(O2 + 3.76N2) = 2.7CO2 + 0.3CO + 4H2O + 2.65O2+ 28.2N2\n",
+      "AF=7.5*4.76*Mair/(3*MC+4*MH);\n",
+      "mair=AF*mfuel;\n",
+      "print'the mass flow rate of air %f kg air/min'%round(mair,2);\n",
+      "#from property tables\n",
+      "#C3H8 designated as p\n",
+      "hfp=-118910;\n",
+      "#oxygen as o\n",
+      "hfo=0;\n",
+      "ho280=8150;\n",
+      "ho298=8682;\n",
+      "ho1500=49292;\n",
+      "#nitrogen as n\n",
+      "hfn=0;\n",
+      "hn280=8141;\n",
+      "hn298=8669;\n",
+      "hn1500=47073;\n",
+      "#water as w\n",
+      "hfw=-241820;\n",
+      "hw298=9904;\n",
+      "hw1500=57999;\n",
+      "#carbondioxode as c\n",
+      "hfc=-393520;\n",
+      "hc298=9364;\n",
+      "hc1500=71078;\n",
+      "#carbon monoxide as co\n",
+      "hfco=-110530;\n",
+      "hco298=8669;\n",
+      "hco1500=47517;\n",
+      "qout=1*(hfp)+7.5*(hfo+ho280-ho298)+28.2*(hfn+hn280-hn298)-2.7*(hfc+hc1500-hc298)-0.3*(hfco+hco1500-hco298)-4*(hfw+hw1500-hw298)-2.65*(hfo+ho1500-ho298)-28.2*(hfn+hn1500-hn298);\n",
+      "#for kg of propane\n",
+      "qout=qout/44;\n",
+      "Qout=mfuel*qout/60;\n",
+      "print'the rate of heat transfer from the combustion chamber %f kW'%round(Qout,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mass flow rate of air 1.180000 kg air/min\n",
+        "the rate of heat transfer from the combustion chamber 6.890000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-7 ,Page No.769"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#the 0.175% error in last part is due to the approximation in the textbook\n",
+      "\n",
+      "#given data\n",
+      "Preact=1.0;#total pressure in kPa\n",
+      "Treact=77+460.0;#reaction temperature in R\n",
+      "Tprod=1800.0;#final temperature in R\n",
+      "\n",
+      "#constants used\n",
+      "Ru=1.986;\n",
+      "\n",
+      "#calculation\n",
+      "#CH4 + 3O2 = CO2 + 2H2O + O2\n",
+      "Nreact=4;\n",
+      "Nprod=4;\n",
+      "Pprod=Preact*Nprod/Nreact*Tprod/Treact;\n",
+      "print'the final pressure in the tank %f atm'%round(Pprod,2);\n",
+      "#from std. values of heat of formation and ideal gasses in Appendix\n",
+      "#CH4 as m\n",
+      "hfm=-32210.0;\n",
+      "#O2 as o\n",
+      "hfo=0;\n",
+      "h537o=3725.1;\n",
+      "h1800o=13485.8;\n",
+      "#water as w\n",
+      "hfw=-104040.0;\n",
+      "h537w=4528.0;\n",
+      "h1800w=15433.0\n",
+      "#carbondioxide as c\n",
+      "hfc=-169300.0;\n",
+      "h537c=4027.5;\n",
+      "h1800c=18391.5;\n",
+      "Qout=1*(hfm-Ru*Treact)+3*(hfo-Ru*Treact)-1*(hfc+h1800c-h537c-Ru*Tprod)-2*(hfw+h1800w-h537w-Ru*Tprod)-1*(hfo+h1800o-h537o-Ru*Tprod);\n",
+      "print'the heat transfer during this process %i Btu/lbmol'%round(Qout)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final pressure in the tank 3.350000 atm\n",
+        "the heat transfer during this process 309269 Btu/lbmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-8 ,Page No.771"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#this invovles EES hence the below code explains a approach with approximation\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "#C8H18 + 12.5 (O2 + 3.76N2) = 8CO+ 9H2O + 47N2\n",
+      "#from std. values of heat of formation and ideal gasses in Appendix\n",
+      "#octane as oc\n",
+      "hfoc=-249950.0;\n",
+      "#oxygen as o\n",
+      "hfo=0;\n",
+      "h298o=8682.0;\n",
+      "#nitrogen as n\n",
+      "hfn=0;\n",
+      "h298n=8669.0;\n",
+      "#water as w\n",
+      "hfw=-241820.0;\n",
+      "h298w=9904.0;\n",
+      "#carbondioxide as c\n",
+      "hfc=-393520.0;\n",
+      "h298c=9364.0;\n",
+      "#x refers to 8hCO2 + 9hH20 + 47hN2\n",
+      "xac=1*(hfoc)+8*(h298c-hfc)+9*(h298w-hfw)+47*(h298n-hfn);\n",
+      "#from EES the Tprod is determined by trial and error\n",
+      "#at 2400K\n",
+      "x2400=5660828.0;\n",
+      "#at 2350K\n",
+      "x2350=5526654.0;\n",
+      "#the actual value of x is xac and T can be determined by interpolation\n",
+      "Tprod=(xac-x2350)*(2400.0-2350.0)/(x2400-x2350)+2350.0;\n",
+      "print'adiabatic flame temperature for complete combustion with 100 percent theoretical air %i K'%round(Tprod);\n",
+      "\n",
+      "#part - b\n",
+      "#C8H18 + 50 (O2 + 3.76N2) = 8CO+ 9H2O + 37.5O2 + 188N2\n",
+      "#solved similarly using EES and approximation and interpolation\n",
+      "#similarly we can solve the part - c \n",
+      "#the above concept is applied\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "adiabatic flame temperature for complete combustion with 100 percent theoretical air 2395 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-9 ,Page No.776"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#from Table A-26E\n",
+      "#Gibbs function of formation at 77\u00b0F\n",
+      "gfc=0;#for carbon\n",
+      "gfo=0;#for oxygen\n",
+      "gfco=-169680;#for carbondioxide\n",
+      "\n",
+      "#calculations\n",
+      "# C + O2 = CO2\n",
+      "Wrev=1*gfc+1*gfo-1*gfco;\n",
+      "print'the reversible work for this process %i Btu'%round(Wrev) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the reversible work for this process 169680 Btu\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-10 ,Page No.777"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#given values\n",
+      "T0=298;#combustion chamber temperature in K\n",
+      "\n",
+      "#contansts used \n",
+      "Ru=8.314;#in kJ/kmol K\n",
+      "\n",
+      "#calculations\n",
+      "# CH4 + 3(O2 + 3.76N2)  = CO2 + 2H2O + O2 + 11.28N2\n",
+      "#from std. values of heat of formation and ideal gasses in Appendix\n",
+      "#methane as m\n",
+      "hfm=-74850;\n",
+      "#oxygen as o\n",
+      "hfo=0;\n",
+      "h298o=8682;\n",
+      "#nitrogen as n\n",
+      "hfn=0;\n",
+      "h298n=8669;\n",
+      "#water as w\n",
+      "hfw=-241820;\n",
+      "h298w=9904;\n",
+      "#carbondioxide as c\n",
+      "hfc=-393520;\n",
+      "h298c=9364;\n",
+      "#x refers to hCO2 + 2hH2O + 11.28hN2\n",
+      "xac=1*(hfm)+1*(h298c-hfc)+2*(h298w-hfw)+11.28*(h298n-hfn);\n",
+      "#from EES the Tprod is determined by trial and error\n",
+      "Tprod=1789;\n",
+      "print'the temperature of the products %i K'%round(Tprod);\n",
+      "#entropy calculations by using table A-26\n",
+      "#Si = Ni*(si - Ruln yiPm\n",
+      "#reactants\n",
+      "Sm=1*(186.16-Ru*log(1*1));\n",
+      "So=3*(205.04-Ru*log(0.21*1));\n",
+      "Sn=11.28*(191.61-Ru*log(.79*1));\n",
+      "Sreact=Sm+So+Sn;\n",
+      "#products\n",
+      "Nt=1+2+1+11.28;#total moles\n",
+      "yc=1/Nt;\n",
+      "yw=2/Nt;\n",
+      "yo=1/Nt;\n",
+      "yn=11.28/Nt;\n",
+      "Sc=1*(302.517-Ru*log(yc*1));\n",
+      "Sw=2*(258.957-Ru*log(yw*1));\n",
+      "So=1*(264.471-Ru*log(yo*1));\n",
+      "Sn=11.28*(247.977-Ru*log(yn*1));\n",
+      "Sprod=Sc+Sw+So+Sn;\n",
+      "Sgen=Sprod-Sreact;\n",
+      "print'exergy destruction %i kJ/kmol - K'%round(Sgen);\n",
+      "Xdestroyed=T0*Sgen/1000;#factor of 1000 for converting kJ to MJ\n",
+      "print'%i MJ/kmol'%round(Xdestroyed);\n",
+      "#This process involves no actual work. Therefore, the reversible work and energy destroyed are identical\n",
+      "Wrev=Xdestroyed;\n",
+      "print'the reversible work %i MJ/kmol'%round(Wrev)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of the products 1789 K\n",
+        "exergy destruction 966 kJ/kmol - K\n",
+        "288 MJ/kmol\n",
+        "the reversible work 288 MJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15-11 ,Page No.778"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#given values\n",
+      "Tsurr=298;#temperature of surroundings in K\n",
+      "\n",
+      "#contansts used \n",
+      "Ru=8.314;#in kJ/kmol K\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "# CH4 + 3(O2 + 3.76N2)  = CO2 + 2H2O + O2 + 11.28N2\n",
+      "#The amount of water vapor that remains in the products is determined as in Example 15\u20133\n",
+      "Nv=0.43;#moles of water vapour\n",
+      "Nw=1.57;#moles of water in liquid\n",
+      "#hf values\n",
+      "#methane as m\n",
+      "hfm=-74850;\n",
+      "#carbondioxide as c\n",
+      "hfc=-393520;\n",
+      "#water vapour as v\n",
+      "hfv=-241820;\n",
+      "#water in liquid as w\n",
+      "hfw=-285830;\n",
+      "Qout=1*hfm-1*hfc-Nv*hfv-Nw*hfw;\n",
+      "print'Qout = %i kJ/kmol'%round(Qout)\n",
+      "\n",
+      "#part - b\n",
+      "#entropy calculations by using table A-26\n",
+      "#Si = Ni*(si - Ruln yiPm\n",
+      "#reactants\n",
+      "Sm=1*(186.16-Ru*log(1*1));\n",
+      "So=3*(205.04-Ru*log(0.21*1));\n",
+      "Sn=11.28*(191.61-Ru*log(.79*1));\n",
+      "Sreact=Sm+So+Sn;\n",
+      "#products\n",
+      "Nt=Nv+1+1+11.28;#total moles\n",
+      "yw=1;\n",
+      "yc=1/Nt;\n",
+      "yv=Nv/Nt;\n",
+      "yo=1/Nt;\n",
+      "yn=11.28/Nt;\n",
+      "Sw=Nw*(69.92-Ru*log(yw*1));\n",
+      "Sc=1*(213.80-Ru*log(yc*1));\n",
+      "Sv=Nv*(188.83-Ru*log(yv*1));\n",
+      "So=1*(205.04-Ru*log(yo*1));\n",
+      "Sn=11.28*(191.61-Ru*log(yn*1));\n",
+      "Sprod=Sc+Sw+So+Sn+Sv;\n",
+      "Sgen=Sprod-Sreact+Qout/Tsurr;\n",
+      "print'Sgen = %i kJ/kmol - K'%round(Sgen);\n",
+      "Xdestroyed=Tsurr*Sgen/1000;#factor of 1000 for converting kJ to MJ\n",
+      "print'exergy destruction %i MJ/kmol'%round(Xdestroyed);\n",
+      "#This process involves no actual work. Therefore, the reversible work and energy destroyed are identical\n",
+      "Wrev=Xdestroyed;\n",
+      "print'the reversible work %i MJ/kmol'%round(Wrev)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Qout = 871406 kJ/kmol\n",
+        "Sgen = 2746 kJ/kmol - K\n",
+        "exergy destruction 818 MJ/kmol\n",
+        "the reversible work 818 MJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter16_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter16_1.ipynb
new file mode 100755
index 00000000..4326dc6f
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter16_1.ipynb
@@ -0,0 +1,378 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Chapter 16: Chemical and Phase Equilibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-1 ,Page No.798"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "\n",
+      "#given data\n",
+      "T=298.15;#reaction temperature in K\n",
+      "\n",
+      "#from Table A-26\n",
+      "g=455510;\n",
+      "\n",
+      "#constants used\n",
+      "R=8.314;#in kJ/kmol K\n",
+      "\n",
+      "#calculations\n",
+      "# N2 = 2N\n",
+      "dG=2*g;\n",
+      "logKp=-dG/(R*T);\n",
+      "Kp=exp(logKp);\n",
+      "print('Kp = ',Kp)\n",
+      "print'in comparison to Table A-28 ln Kp value of -367.5 our result is %i'%logKp;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "('Kp = ', 2.4396097259977668e-160)\n",
+        "in comparison to Table A-28 ln Kp value of -367.5 our result is -367\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-2 ,Page No.798"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P=10;#pressure in atm\n",
+      "\n",
+      "#constants\n",
+      "vH2=1;\n",
+      "vH=2;\n",
+      "\n",
+      "#calculations\n",
+      "# H2 = 0.9H2 + 0.2H\n",
+      "NH=0.2;\n",
+      "NH2=0.9;\n",
+      "Nt=NH+NH2;\n",
+      "#from Eq. 16-15\n",
+      "Kp=((NH**vH)/(NH2**vH2))*(P/Nt)**(vH-vH2);\n",
+      "#at this value of Kp from Table A-28\n",
+      "T=3535;\n",
+      "print'temperature %i K is'%T;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature 3535 K is\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-6 ,Page No.807"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#reaction\n",
+      "# H2 + 0.5O2 = H2O\n",
+      "#enthalpy datas in kJ/kmol \n",
+      "#of H2\n",
+      "hfH=-241820;\n",
+      "h2000H=82593;\n",
+      "h298H=9904;\n",
+      "#of O2\n",
+      "hfO=0;\n",
+      "h2000O=61400;\n",
+      "h298O=8468;\n",
+      "#of H2O\n",
+      "hfw=0;\n",
+      "h2000w=67881;\n",
+      "h298w=8682;\n",
+      "#Kp data from A-28\n",
+      "Kp2=869.6;\n",
+      "Kp1=18509;\n",
+      "T1=1800;\n",
+      "T2=2200;\n",
+      "\n",
+      "#constants used\n",
+      "Ru=8.314;#in kJ/kmol K\n",
+      "\n",
+      "#calculations\n",
+      "#part - a\n",
+      "hR=1*(hfH+h2000H-h298H)-1*(hfO+h2000O-h298O)-0.5*(hfw+h2000w-h298w);\n",
+      "print'enthalpy of the reaction %i kJ/kmol using enthalpy data'%round(hR);\n",
+      "#part - b\n",
+      "hR=Ru*(T1*T2)/(T2-T1)*log(Kp2/Kp1);\n",
+      "print'enthalpy of the reaction %i kJ/kmol using enthalpy data'%round(hR);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "enthalpy of the reaction -251663 kJ/kmol using enthalpy data\n",
+        "enthalpy of the reaction -251698 kJ/kmol using enthalpy data\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-7 ,Page No.809"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=120+273.15;#temperature of saturated water in K\n",
+      "\n",
+      "#from Table A-4\n",
+      "hf=503.81;\n",
+      "hg=2706;\n",
+      "sf=1.5279;\n",
+      "sg=7.1292;\n",
+      "\n",
+      "#calculations\n",
+      "print('liquid phase');\n",
+      "gf=hf-T*sf;\n",
+      "print'gf value %f kJ/kg'%round(gf,1);\n",
+      "print('vapour phase');\n",
+      "gg=hg-T*sg;\n",
+      "print'gg value %f kJ/kg'%round(gg,1);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "liquid phase\n",
+        "gf value -96.900000 kJ/kg\n",
+        "vapour phase\n",
+        "gg value -96.800000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-8 ,Page No.813"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=15;#lake temperature in C\n",
+      "P=92.0;#atmospheric pressure in kPa\n",
+      "\n",
+      "#from Table A-4\n",
+      "Pv=1.7057;\n",
+      "\n",
+      "#calculations\n",
+      "yv=Pv/P;\n",
+      "print'mole fraction of water vapor at the surface is %f'%round(yv,4);\n",
+      "yw=1-yv;\n",
+      "print'mole fraction of water in the lake is %f percent'%(round(yw)*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mole fraction of water vapor at the surface is 0.018500\n",
+        "mole fraction of water in the lake is 100.000000 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-9 ,Page No.214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=17.0;#lake temperature in C\n",
+      "P=92.0;#atmospheric pressure in kPa\n",
+      "\n",
+      "#from Table A-4\n",
+      "Pv=1.96;\n",
+      "\n",
+      "#constants from Table 16-2\n",
+      "H=62000.0;\n",
+      "\n",
+      "#calculations\n",
+      "Pda=P-Pv;#dry air\n",
+      "yda=Pda/H/100;#in bar\n",
+      "print'mole fraction of air is %f'%(yda)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mole fraction of air is 0.000015\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-10 ,Page No.814"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=358;#hydrogen gas temperature in K\n",
+      "P=300/100;#pressure of hydrogen gas in bar\n",
+      "\n",
+      "#constants used\n",
+      "M=2;\n",
+      "s=0.00901;#solubility in kmol/m^3 bar\n",
+      "p=0.027;\n",
+      "\n",
+      "#calculations\n",
+      "pH2=s*P;\n",
+      "print'molar density of H2 %f kmol/m^3'%round(pH2,3);\n",
+      "pH2=p*M;\n",
+      "print'mass density of H2 %f kg/m^3'%round(pH2,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "molar density of H2 0.027000 kmol/m^3\n",
+        "mass density of H2 0.054000 kg/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16-11 ,Page No.815"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "yw=0.30;#water mole fraction\n",
+      "ya=0.70;#ammonia mole fraction\n",
+      "T=40;#mixture temperature in C\n",
+      "\n",
+      "#saturation pressure\n",
+      "pw=7.3851;\n",
+      "pa=1554.33;\n",
+      "#calulations\n",
+      "Pw=yw*pw;\n",
+      "Pa=ya*pa;\n",
+      "Pt=Pw+Pa;\n",
+      "yw=Pw/Pt;\n",
+      "ya=Pa/Pt;\n",
+      "print'mole fraction of water vapour %f'%round(yw,4);\n",
+      "print'mole fraction of ammonia %f'%round(ya,4);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mole fraction of water vapour 0.002000\n",
+        "mole fraction of ammonia 0.998000\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter17_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter17_1.ipynb
new file mode 100755
index 00000000..5b1b7ade
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter17_1.ipynb
@@ -0,0 +1,812 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17: Compressible Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-1 ,Page No.826"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V1=250.0;#velocity of aircraft in m/s\n",
+      "T1=255.07;#ambient air temperature in K\n",
+      "P1=54.05;#atmospheric pressure in kPa\n",
+      "h=5000.0;#altitude in m\n",
+      "\n",
+      "#from Table A-2a\n",
+      "cp=1.005;#in kJ/kg-K\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "T01=T1+V1**2/(2*cp*1000);#factor of 1000 to convert kJ to J\n",
+      "P01=P1*(T01/T1)**(k/(k-1));\n",
+      "#given pressure ratio in compressor *\n",
+      "# T02 = T01*(P02/P01)^((k-1)/k)\n",
+      "T02 = T01*(8)**((k-1)/k);\n",
+      "win=cp*(T02-T01);\n",
+      "print'the stagnation pressure at the compressor inlet %f kPa'%round(P01,2);\n",
+      "print'the required compressor work per unit mass %f kJ/kg'%round(win,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the stagnation pressure at the compressor inlet 80.840000 kPa\n",
+        "the required compressor work per unit mass 233.400000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-2 ,Page No.829"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "V=200.0;#air velocity in m/s\n",
+      "T=30+273.0;#air temperature in K\n",
+      "\n",
+      "#from Table A-2a\n",
+      "R=0.287;#in kJ/kg-K\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "c=sqrt(k*R*T*1000);#factor of 1000 to convert kJ to J\n",
+      "print'the speed of sound %i m/s'%round(c);\n",
+      "Ma=V/c;\n",
+      "print'the Mach number at the diffuser inlet is %f'%round(Ma,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the speed of sound 349 m/s\n",
+        "the Mach number at the diffuser inlet is 0.573000\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-3 ,Page No.829"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "T0=200+273.0;#intial temperature in K\n",
+      "P0=1400.0;#intial pressure in kPa\n",
+      "#stagnant temp. & pressure is same as inlet due to small inlet velocity\n",
+      "P=1200.0;#pressure corresponding to a pressure drop of 200 kPa\n",
+      "m=3.0;#mass flow rate in kg/s\n",
+      "\n",
+      "#from Table A-2a\n",
+      "cp=0.846;#in kJ/kg-K\n",
+      "R=0.1889;#in kJ/kg-K\n",
+      "k=1.289;\n",
+      "\n",
+      "#calculations\n",
+      "T=T0*(P/P0)**((k-1)/k);\n",
+      "V=sqrt(2*cp*(T0-T)*1000);#factor of 1000 to convert kJ to J\n",
+      "p=P/(R*T);\n",
+      "A=m/(p*V);\n",
+      "c=sqrt(k*R*T*1000);#factor of 1000 to convert kJ to J\n",
+      "Ma=V/c;\n",
+      "print'velocity %f m/s'%round(V,1);\n",
+      "print'density %f kg/m^3'%round(p,1);\n",
+      "print'flow area %f cm^2'%round((A*10000),1);\n",
+      "print'Mach number is %f'%round(Ma,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity 164.900000 m/s\n",
+        "density 13.900000 kg/m^3\n",
+        "flow area 13.100000 cm^2\n",
+        "Mach number is 0.494000\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-4 ,Page No.836"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T0=200+273.0;#intial temperature in K\n",
+      "P0=1400.0;#intial pressure in kPa\n",
+      "\n",
+      "#from Table A-2a\n",
+      "k=1.289;\n",
+      "\n",
+      "#calculations\n",
+      "#Tc & Tr stands for critical temp and ratio respectively\n",
+      "#Pc & Pr stands for critical temp and ratio respectively\n",
+      "Tr=2/(k+1);\n",
+      "Pr=(2/(k+1))**(k/(k-1));\n",
+      "Tc=Tr*T0;\n",
+      "Pc=Pr*P0;\n",
+      "print'critical temperature %i K'%round(Tc);\n",
+      "print'critical pressure %i kPa'%round(Pc)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "critical temperature 413 K\n",
+        "critical pressure 767 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-5 ,Page No.839"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "Vi=150.0;#air velocity in m/s\n",
+      "Ti=600+273.0;#air temperature in K\n",
+      "Pi=1.0;#air pressure in MPa\n",
+      "At=50.0/10000.0;#nozzle throat area in m^2\n",
+      "\n",
+      "#from Table A-2a\n",
+      "R=0.287;#in kJ/kg-K\n",
+      "cp=1.005;#in kJ/kg-K\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "Toi=Ti+Vi**2/(2*cp*1000);#factor of 1000 to convert kJ to J\n",
+      "Poi=Pi*(Toi/Ti)**(k/(k-1));\n",
+      "#flow is isentropic \n",
+      "#stagnation temp. and pressure values remain constant\n",
+      "To=Toi;\n",
+      "Po=Poi;\n",
+      "#from Table 17\u20132\n",
+      "#The critical-pressure ratio is 0.5283\n",
+      "\n",
+      "#Part a\n",
+      "Pb=0.7;\n",
+      "Pca=Pb/Po;\n",
+      "# Pca > 0.5283\n",
+      "#exit plane pressure is equal to the back pressure\n",
+      "Pt=Pb;\n",
+      "#from Table A\u201332\n",
+      "Mat=0.778;\n",
+      "#Tt/To = 0.892\n",
+      "Tt=0.892*To;\n",
+      "pt=Pt*1000/(R*Tt);#factor of 1000 to convert MPa to kPa\n",
+      "Vt=Mat*sqrt(k*R*Tt*1000);#factor of 1000 to convert kJ to J\n",
+      "ma=pt*At*Vt;\n",
+      "print'the mass flow rate through the nozzle when the back pressure is 0.7 MPa %f kg/s'%round(ma,2);\n",
+      "\n",
+      "#Part b\n",
+      "Pb=0.4;\n",
+      "Pca=Pb/Po;\n",
+      "# Pca < 0.5283\n",
+      "#sonic conditions exists at the exit\n",
+      "Ma=1;\n",
+      "mb=At*(Po*1000)*(sqrt(k*1000/(R*To)))*(2/(k+1))**((k+1)/(2*(k-1)));#factor of 1000 to convert MPa to kPa and kJ to J\n",
+      "print'the mass flow rate through the nozzle when the back pressure is 0.4 MPa %f kg/s'%round(mb,2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mass flow rate through the nozzle when the back pressure is 0.7 MPa 6.770000 kg/s\n",
+        "the mass flow rate through the nozzle when the back pressure is 0.4 MPa 7.110000 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-6 ,Page No.840"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=400;#intial temperature in K\n",
+      "P1=100;#intial pressure in kPa\n",
+      "Ma1=0.3;#intial mach no\n",
+      "A21=0.8;#A2/A1 as flow area has been reduced by 20 percent\n",
+      "\n",
+      "#assumption\n",
+      "k=1.4;\n",
+      "\n",
+      "#from Table A\u201332\n",
+      "#at Ma1=0.3\n",
+      "#s stands for * symbol\n",
+      "A1s = 2.0351;#A1/As\n",
+      "T10 = 0.9823;#T1/T0\n",
+      "P10 = 0.9305;#P1/P0\n",
+      "A2s = A21*A1s;#A2/As\n",
+      "#at this value of A2/As\n",
+      "T20=0.9701;#T2/T0\n",
+      "P20=0.8993;#P2/P0\n",
+      "Ma2=0.391;\n",
+      "\n",
+      "#calculations\n",
+      "T2=T1*T20/T10;\n",
+      "P2=P1*P20/P10;\n",
+      "print'Ma2 is %f'%round(Ma2,3);\n",
+      "print'T2 %i K is'%round(T2);\n",
+      "print'P2 %f kPa is'%round(P2,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ma2 is 0.391000\n",
+        "T2 395 K is\n",
+        "P2 96.600000 kPa is\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-7 ,Page No.844"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data \n",
+      "T0=800;#intial temperature in K\n",
+      "P0=1;#intial pressure in MPa\n",
+      "Vi=0;#negligible intial velcity\n",
+      "At=20;#throat area in cm^2\n",
+      "Mae=2;#exit Mach number\n",
+      "\n",
+      "#from Table A-2a\n",
+      "R=0.287;#in kJ/kg-K\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "# Mach no. at exit is 2 hence sonic conditions at throat\n",
+      "p0=P0*1000/(R*T0);#factor of 1000 to convert MPa to kPa\n",
+      "#from Table A-32 at Mat=1\n",
+      "#s stands for * symbol\n",
+      "Ps0 = 0.5283;#Ts/T0\n",
+      "Ts0 = 0.8333;#Ps/P0\n",
+      "ps0=0.6339;#ps/p0\n",
+      "Ps=Ps0*P0;\n",
+      "Ts=Ts0*T0;\n",
+      "ps=ps0*p0;\n",
+      "As=At;\n",
+      "Vs=sqrt(k*R*Ts*1000);#factor of 1000 to convert kJ to J\n",
+      "print('the throat conditions');\n",
+      "print'Presssure %f MPa'%round(Ps,4);\n",
+      "print'Temperature %f K'%round(Ts,1);\n",
+      "print'density %f kg/m^3'%round(ps,3);\n",
+      "print'area %f cm^2'%round(As);\n",
+      "print'velocity %f m/s'%round(Vs,1);\n",
+      "\n",
+      "#part - b\n",
+      "#from Table A-32\n",
+      "#at Mae=2\n",
+      "Te0 = 0.5556;#Te/T0\n",
+      "Pe0 = 0.1278;#Pe/P0\n",
+      "pe0= 0.2300;#pe/p0\n",
+      "Ae0= 1.6875;#Ae/Ao\n",
+      "Pe=Pe0*P0;\n",
+      "Te=Te0*T0;\n",
+      "pe=pe0*p0;\n",
+      "Ae=Ae0*At;\n",
+      "Ve=Mae*sqrt(k*R*Te*1000);#factor of 1000 to convert kJ to J\n",
+      "print('the exit plane conditions, including the exit area');\n",
+      "print'Presssure %f MPa'%round(Pe,4);\n",
+      "print'Temperature %f K'%round(Te,1);\n",
+      "print'density %f kg/m^3'%round(pe,3);\n",
+      "print'area %f cm^2'%round(Ae,2);\n",
+      "print'velocity %f m/s'%round(Ve,1);\n",
+      "#part - c\n",
+      "m=ps*As*Vs/10000;#factor of 10000 to convert cm^2 to m^2\n",
+      "print'the mass flow rate through the nozzle %f kg/s'%round(m,2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the throat conditions\n",
+        "Presssure 0.528300 MPa\n",
+        "Temperature 666.600000 K\n",
+        "density 2.761000 kg/m^3\n",
+        "area 20.000000 cm^2\n",
+        "velocity 517.500000 m/s\n",
+        "the exit plane conditions, including the exit area\n",
+        "Presssure 0.127800 MPa\n",
+        "Temperature 444.500000 K\n",
+        "density 1.002000 kg/m^3\n",
+        "area 33.750000 cm^2\n",
+        "velocity 845.200000 m/s\n",
+        "the mass flow rate through the nozzle 2.860000 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-9 ,Page No.850"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log,sqrt\n",
+      "\n",
+      "#data from Ex - 17.7\n",
+      "m=2.86;\n",
+      "Ma1=2;\n",
+      "P01=1;\n",
+      "P1=0.1278;\n",
+      "T1=444.5;\n",
+      "p1=1.002;\n",
+      "\n",
+      "#from Table A-2a\n",
+      "R=0.287;#in kJ/kg-K\n",
+      "cp=1.005;#in kJ/kg-K\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "#from Table A-33 at Ma1=2.0\n",
+      "Ma2=0.5774;\n",
+      "P0201=0.7209;#P02/P01\n",
+      "P21=4.5;#P2/P1;\n",
+      "T21=1.6875;#T2/T1\n",
+      "p21=2.6667;#p2/p1\n",
+      "P02=P0201*P01;\n",
+      "P2=P21*P1;\n",
+      "T2=T21*T1;\n",
+      "p2=p21*p1;\n",
+      "print'the stagnation pressure %f MPa'%round(P02,3);\n",
+      "print'the static pressure %f MPa'%round(P2,3);\n",
+      "print'static temperature %f K'%round(T2);\n",
+      "print'static density %f kg/m^3'%round(p2,2);\n",
+      "\n",
+      "#part - b\n",
+      "#s21 = s2 - s1\n",
+      "s21=cp*log(T2/T1)-R*log(P2/P1);\n",
+      "print'the entropy change across the shock %fkJ/kg-K'%round(s21,4);\n",
+      "\n",
+      "#part - c\n",
+      "V2=Ma2*sqrt(k*R*T2*1000);#factor of 1000 to convert kJ to J\n",
+      "print'the exit velocity %f m/s'%round(V2);\n",
+      "\n",
+      "#part - d\n",
+      "print('flow rate is not affected by presence of shock waves amd remains 2.86 kg/sec')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the stagnation pressure 0.721000 MPa\n",
+        "the static pressure 0.575000 MPa\n",
+        "static temperature 750.000000 K\n",
+        "static density 2.670000 kg/m^3\n",
+        "the entropy change across the shock 0.094200kJ/kg-K\n",
+        "the exit velocity 317.000000 m/s\n",
+        "flow rate is not affected by presence of shock waves amd remains 2.86 kg/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-10 ,Page No.858"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sin,pi\n",
+      "\n",
+      "#given data\n",
+      "#using protactor frpm Fig 17-36\n",
+      "u=19;#u stands for angle of the mach lines\n",
+      "\n",
+      "#calculations\n",
+      "#by Eq. 17-47\n",
+      "#i.e u= asin(1/Ma)\n",
+      "Ma=1/sin(u*pi/180);#converting to radians\n",
+      "print'The Mach number is %f'%round(Ma,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Mach number is 3.070000\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-11 ,Page No.858"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sin,pi\n",
+      "\n",
+      "#given data\n",
+      "Ma1=2;#Supersonic air mach no\n",
+      "P1=75;#Supersonic air at pressure in kPa\n",
+      "O=10*pi/180;#converting to radians & angle b/w shock wave and normal\n",
+      "\n",
+      "#constants used\n",
+      "k=1.4;\n",
+      "\n",
+      "#calcualtions\n",
+      "#with given values of Ma1 and O from Eq 17-46\n",
+      "Bweak=39.3*pi/180;#converting to radians\n",
+      "Bstrong=83.7*pi/180;#converting to radians\n",
+      "#Weak shock\n",
+      "Ma1w=Ma1*sin(Bweak);\n",
+      "#Strong shock\n",
+      "Ma1s=Ma1*sin(Bstrong);\n",
+      "#from second part Eq 17-40\n",
+      "Ma2w=0.8032;\n",
+      "Ma2s=0.5794;\n",
+      "#pressure ratio = (2*k*Ma^2 - k + 1)/(k + 1 )\n",
+      "#Weak shock\n",
+      "P2w=P1*(2*k*Ma1w**2 - k + 1)/(k + 1 );\n",
+      "print'pressure for weak shock %i kPa'%round(P2w);\n",
+      "#Strong shock\n",
+      "P2s=P1*(2*k*Ma1s**2 - k + 1)/(k + 1 );\n",
+      "print'pressure for strong shock %i kPa'%round(P2s);\n",
+      "#Weak shock\n",
+      "Ma2=Ma2w/sin(Bweak-O);\n",
+      "print'Mach number downstream for weak shock is %f'%round(Ma2,2);\n",
+      "#Strong shock\n",
+      "Ma2=Ma2s/sin(Bstrong-O);\n",
+      "print'Mach number downstream for strong shock is %f'%round(Ma2,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure for weak shock 128 kPa\n",
+        "pressure for strong shock 333 kPa\n",
+        "Mach number downstream for weak shock is 1.640000\n",
+        "Mach number downstream for strong shock is 0.604000\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Example 17-12 ,Page No.859"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt,pi,atan\n",
+      "\n",
+      "#given data\n",
+      "Ma1=2;#Supersonic air mach no\n",
+      "P1=230;#Supersonic air pressure in kPa\n",
+      "O=10*pi/180;#converting to radians & O stands for angle of the mach lines\n",
+      "\n",
+      "#constants used\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "#Eq. 17\u201349 for the upstream Prandtl\u2013Meyer function\n",
+      "vMa1=sqrt((k+1)/(k-1))*atan(sqrt((k-1)*(Ma1**2-1)/(k+1))*pi/180)-atan(sqrt(Ma1**2-1)*pi/180);#converting to radians\n",
+      "#Eq. 17\u201348 to calculate the downstream Prandtl\u2013Meyer function\n",
+      "vMa2=O+vMa1;\n",
+      "#using equation solver as implict nature of Eq 17-49\n",
+      "Ma2=2.385;\n",
+      "print'downstream Mach number Ma2 is %f'%round(Ma2,3);\n",
+      "#P2 = (P2/P0)/(P1/P0) * P1\n",
+      "P2= (1 + (k-1)*Ma2**2/2 )**(-k/(k-1)) / (1 + (k-1)*Ma1**2/2 )**(-k/(k-1)) * P1;\n",
+      "print'downstream pressure %i kPa'%round(P2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "downstream Mach number Ma2 is 2.385000\n",
+        "downstream pressure 126 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-15 ,Page No.868"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi,sqrt\n",
+      "\n",
+      "#given data\n",
+      "P1=480.0;#intial pressure in kPa\n",
+      "T1=550.0;#intial temperature in K\n",
+      "V1=80.0;#intial velocity in m/s\n",
+      "d1=15.0/100.0;#diameter in m\n",
+      "AF=40.0;#air to fuel ratio\n",
+      "HV=40000.0;#heating value in kJ/kg\n",
+      "\n",
+      "#from Table A-2a\n",
+      "R=0.287;#in kJ/kg-K\n",
+      "cp=1.005;#in kJ/kg-K\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "p1=P1/(R*T1);\n",
+      "A1=pi*d1**2/4;\n",
+      "mair=p1*A1*V1;\n",
+      "mfuel=mair/AF;\n",
+      "Q=mfuel*HV;\n",
+      "q=Q/mair;\n",
+      "T01=T1+V1**2/(2*cp);\n",
+      "c1=sqrt(k*R*T1*1000);#factor of 1000 to convert kJ to J\n",
+      "Ma1=V1/c1;\n",
+      "#exit stagnation energy equation q= Cp (T02 - T01)\n",
+      "T02=T01+q/cp;\n",
+      "#from Table A\u201334\n",
+      "#at Ma1\n",
+      "#s stands for * symbol\n",
+      "T0s=0.1291;#T0/Ts\n",
+      "Ts0=T01/T0s;\n",
+      "T2s=T02/Ts0;#T02/T*0\n",
+      "#from Table A\u201334 at this ratio\n",
+      "Ma2=0.3142;\n",
+      "#Rayleigh flow relations corresponding to the inlet and exit Mach no\n",
+      "#at Ma1\n",
+      "T1s=0.1541;#T1/Ts\n",
+      "P1s=2.3065;#P1/Ps\n",
+      "V1s=0.0668;#V1/Vs\n",
+      "#at Ma2\n",
+      "T2s=0.4389;#T2/Ts\n",
+      "P2s=2.1086;#P2/Ps\n",
+      "V2s=0.2082;#V2/Vs\n",
+      "T2=T2s/T1s*T1;\n",
+      "P2=P2s/P1s*P1;\n",
+      "V2=V2s/V1s*V1;    \n",
+      "print'Mach Number at exit is %f'%round(Ma2,4);\n",
+      "print'Presssure %i MPa'%round(P2);\n",
+      "print'Temperature %i K'%round(T2);\n",
+      "print'velocity %i m/s'%round(V2);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mach Number at exit is 0.314200\n",
+        "Presssure 439 MPa\n",
+        "Temperature 1566 K\n",
+        "velocity 249 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17-16 ,Page No.870"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "P01=2000;#entry pressure in kPa\n",
+      "T1=400;#entry temperature in C\n",
+      "V1=0;#negligible velocity at entry\n",
+      "nN=0.93;#overall nozzle efficiency\n",
+      "m=2.5;#mass flow rate in kg/s\n",
+      "P2=300;#exit pressure in kPa\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part - a\n",
+      "P201=P2/P01;\n",
+      "#critical pressure ratio at this values is 0.546\n",
+      "Pt=0.546*P01;\n",
+      "#at inlet\n",
+      "h1=3248.4;\n",
+      "h01=h1;\n",
+      "s1=7.1292;\n",
+      "#at throat\n",
+      "st=s1;\n",
+      "ht=3076.8;\n",
+      "vt=0.24196;\n",
+      "Vt=sqrt(2*(h01-ht)*1000);#factor of 1000 to convert kJ to J\n",
+      "At=m*vt/Vt;\n",
+      "#at state 2s\n",
+      "s2s=s1;\n",
+      "h2s=2783.6;\n",
+      "#nN = (h01 - h2)/ (h01 - h2s)\n",
+      "h2=h01-nN*(h01-h2s);\n",
+      "#at P2 and h2\n",
+      "v2=0.67723;\n",
+      "s2=7.2019;\n",
+      "V2=sqrt(2*(h01-h2)*1000);#factor of 1000 to convert kJ to J\n",
+      "A2=m*v2/V2;\n",
+      "print'throat area %f cm^2'%round((At*10000),2);\n",
+      "print'exit area %f cm^2'%round((A2*10000),2);\n",
+      "\n",
+      "#part - b\n",
+      "# at st=7.1292\n",
+      "#pressures of 1.115 and 1.065 MPa\n",
+      "#c calculated using tables\n",
+      "c=sqrt((1115-1065)/(1/0.23776 - 1/0.24633)*1000);#factor of 1000 to convert kPa to Pa\n",
+      "Ma=Vt/c;\n",
+      "print'the Mach number at the throat is %f'%round(Ma,2);\n",
+      "# at s2=7.2019\n",
+      "#pressures of 325 and 275 kPa\n",
+      "c=sqrt((325-276)/(1/0.63596 - 1/0.72245)*1000);#factor of 1000 to convert kPa to Pa\n",
+      "Ma=V2/c;\n",
+      "print'the Mach number at the nozzle exit is %f'%round(Ma,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "throat area 10.330000 cm^2\n",
+        "exit area 18.210000 cm^2\n",
+        "the Mach number at the throat is 1.000000\n",
+        "the Mach number at the nozzle exit is 1.820000\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter1_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter1_1.ipynb
new file mode 100755
index 00000000..3e2c2051
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter1_1.ipynb
@@ -0,0 +1,387 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Introduction and Basic Concepts"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-2, Page No.8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "p=850;# density in kg/m^3\n",
+      "V=2;  # volumne of tank in m^3\n",
+      "\n",
+      "#Calculations\n",
+      "m=p*V;# mass, density and volumne corealtion\n",
+      "\n",
+      "#Result\n",
+      "print 'The amount of oil in tank is %i kg' %round(m,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of oil in tank is 1700 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-3, Page No.9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constants used\n",
+      "g=32.174;# gravitational constant in ft/s^2\n",
+      "\n",
+      "#given values\n",
+      "m=1; # mass of 1.00 lbm is subjected to standard earth gravity\n",
+      "\n",
+      "#Calculations\n",
+      "w=(m*g)/g; # weight is mass times the local value of gravitational acceleration\n",
+      "#dimensionally the above equation is represented as lbm * ft/s^2 * (lbf/ft/s^2)\n",
+      "\n",
+      "#Result\n",
+      "print 'The weight on earth is %i lbf' %w\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The weight on earth is 1 lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-4, Page No.21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Example 1.4\n",
+      "\n",
+      "# Given values\n",
+      "Tc=10; #change in temp in Celcius\n",
+      "\n",
+      "# Calculations\n",
+      "Tk=Tc;\n",
+      "Tr=1.8*Tk;#conversion scale of temperature change from K to R\n",
+      "Tf=Tr;\n",
+      "# calculated using the corealtions b/w these scales\n",
+      "\n",
+      "#Results\n",
+      "print 'the corresponding change is %i K' %Tk\n",
+      "print 'the corresponding change is %i R' %Tr\n",
+      "print 'the corresponding change is %i F' %Tf\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the corresponding change is 10 K\n",
+        "the corresponding change is 18 R\n",
+        "the corresponding change is 18 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-5, Page No.23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "Patm=14.5; #atmospheric pressure in psi\n",
+      "Pvac=5.8;  #vacuum gage reading in psi\n",
+      "\n",
+      "#Calculations\n",
+      "Pabs=Patm-Pvac;#pressure in vaccumm is always treated to be negative\n",
+      "\n",
+      "#Results\n",
+      "print'the absolute pressure in the chamber %f psi'%round(Pabs,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the absolute pressure in the chamber 8.700000 psi\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-6, Page No.26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constants used\n",
+      "pw=1000; # density of water in kg/m^3;\n",
+      "g=9.81; # acceleration due to gravity in m/s^2;\n",
+      " \n",
+      "#Given values\n",
+      "SG=0.85;# specific gravity of manometric fluid\n",
+      "h=0.55;# converting height from cm to m\n",
+      "Patm=96;# atmospheric pressure in kPa\n",
+      "\n",
+      "# Calculations\n",
+      "p=SG*pw;\n",
+      "Ptank=Patm+(p*g*h/1000); # calculating pressure using liquid at same height have same pressure\n",
+      "\n",
+      "#Results\n",
+      "print 'absolute pressure in tank %f kPa' %round(Ptank,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "absolute pressure in tank 100.600000 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-7, Page No.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constants used\n",
+      "g=9.81;#acceleration due to gravity in m/s^2;\n",
+      "\n",
+      "#Given values\n",
+      "h1=0.1;# distance b/w point 1 at air-water interface and point 2 at mercury-air interface in m\n",
+      "h2=0.2;# distance b/w oil-water interface and  mercury-oil interface in m\n",
+      "h3=0.35;# distance b/w air-mercury interface and  mercury-oil interface in m\n",
+      "pw=1000;# density of water in kg/m^3\n",
+      "pHg=13600;# density of mercury in kg/m^3\n",
+      "poil=800;# density of oil in kg/m^3\n",
+      "Patm=85.6;# atmospheric pressure in kPa\n",
+      "\n",
+      "#Calculation\n",
+      "P1=Patm-(pw*g*h1+poil*g*h2-pHg*g*h3)/1000;#calculating pressure using liquid at same height have same pressure\n",
+      "\n",
+      "#Results\n",
+      "print 'the air pressure in tank %i kPa' %round(P1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the air pressure in tank 130 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-8, Page No.31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constants used\n",
+      "g=9.81;# acceleration due to gravity in m/s^2;\n",
+      "\n",
+      "#Given values\n",
+      "pHg=13570;# density of mercury at 10 C in kg/m^3\n",
+      "h=0.74;# converting barometric reading into m from mm\n",
+      "\n",
+      "#Calculationa\n",
+      "Patm=pHg*g*h/1000;# standard  pressure formula\n",
+      "\n",
+      "#Results\n",
+      "print 'the atmospheric pressure %f kPa' %round(Patm,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the atmospheric pressure 98.500000 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-9, Page No.31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#constants used\n",
+      "g=9.81;#acceleration due to gravity in m/s^2;\n",
+      "\n",
+      "#given values\n",
+      "m=60;# mass of piston in kg\n",
+      "Patm=0.97;# atmospheric pressure in kPa\n",
+      "A=0.04;# cross-sectional area in m^2\n",
+      "\n",
+      "#calculation\n",
+      "P=Patm+(m*g/A)/100000;# standard  pressure formula\n",
+      "print 'The pressure inside the cylinder %f bar' %round(P,2)\n",
+      "#The volume change will have no effect on the free-body diagram drawn in part (a), and therefore the pressure inside the cylinder will remain the same\n",
+      "print('If some heat is transferred to the gas and its volume is doubled, there is no change in pressure');\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure inside the cylinder 1.120000 bar\n",
+        "If some heat is transferred to the gas and its volume is doubled, there is no change in pressure\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1-10, Page No.32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from scipy.integrate import quad \n",
+      "from pylab import *\n",
+      "\n",
+      "#Constants used\n",
+      "g=9.81;#acceleration due to gravity in m/s^2;\n",
+      "\n",
+      "#Given values\n",
+      "p=1040;# density on the water surface in kg/m^3\n",
+      "h1=0.8;# thickness of surface zone\n",
+      "H=4;# thickness of gradient zone\n",
+      "x0=0.0;# lower limit of integration\n",
+      "x1=4.0;# upper limit of integration\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      "P1=p*g*h1/1000;#standard  pressure determination formula\n",
+      "#P2 = integration of the exp. p*g*(math.sqrt(1+(math.tan(math.pi*z/4/H)^2))) b/w 0-4\n",
+      "def intgrnd1(z): \n",
+      " return (p*g*(math.sqrt(1+(math.tan(math.pi*(z)/4/H)**2))) )#integrant\n",
+      "P2, err = quad(intgrnd1, x0, x1) \n",
+      "P2=P2/1000;#converting into kPa\n",
+      "P=P1+P2;\n",
+      "\n",
+      "#Results\n",
+      "print 'the gage pressure at the bottom of gradient zone %f kPa' %round(P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the gage pressure at the bottom of gradient zone 54.000000 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter2_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter2_1.ipynb
new file mode 100755
index 00000000..db573fc6
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter2_1.ipynb
@@ -0,0 +1,588 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Energy Conversion and General Energy Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "    Example 2-1 ,Page No.57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#there is a 0.00490% error as the answer in textbook is expressed in multiple of 10\n",
+      "#Constants used\n",
+      "Hu=6.73*10**10;#Energy liberated by 1 kg of uranium\n",
+      "\n",
+      "# Given values\n",
+      "p=0.75;# assuming the avg density of gasoline in kg/L\n",
+      "V=5;# consumption per day of gasoline in L\n",
+      "Hv=44000; #heat value in kJ/kg\n",
+      "mu=0.1;# mass of uranium used\n",
+      "\n",
+      "#Calculation\n",
+      "mgas=p*V;#mass of gasoline required per day\n",
+      "Egas=mgas*Hv;\n",
+      "Eu=mu*Hu;\n",
+      "d=Eu/Egas;\n",
+      "print'%i number of days the car can run with uranium' %round(d,0)\n",
+      "print'equivalent to %i years' %round(d/365,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "40788 number of days the car can run with uranium\n",
+        "equivalent to 112 years\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-2 ,Page No.59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given values\n",
+      "v=8.5;# wind speed in m/s\n",
+      "m=10;# given mass for part - b \n",
+      "mf=1154;# given flowrate for part - c\n",
+      "\n",
+      "#Calculations\n",
+      "e=(v**2)/2;\n",
+      "print'wind energy per unit mass %f J/kg' %round(e,1);\n",
+      "E=m*e;\n",
+      "print'wind energy for 10 kg mass %i J' %E;\n",
+      "E=mf*e/1000;\n",
+      "print'wind energy for mass flow rate of 1154kg/s %f kW'%round(E,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wind energy per unit mass 36.100000 J/kg\n",
+        "wind energy for 10 kg mass 361 J\n",
+        "wind energy for mass flow rate of 1154kg/s 41.700000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-7 ,Page No.67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Given values\n",
+      "T=200;# applied torque in N\n",
+      "n=4000;# shaft rotation rate in revolutions per minute\n",
+      "\n",
+      "#Calculation\n",
+      "Wsh=(2*math.pi*n*T)/1000/60;#factor of 1000 to convert to kW and 60 to convert to sec\n",
+      "print'Power transmitted %f kW'%round(Wsh,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power transmitted 83.800000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-8 ,Page No.69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Constants used\n",
+      "g=9.81;#acceleration due to gravity in m/s^2;\n",
+      "\n",
+      "#Given values\n",
+      "m=1200;#mass of car in kg\n",
+      "V=90/3.6;#velocity ; converting km/h into m/s\n",
+      "d=30*math.pi/180;#angle of slope ; converting into radians\n",
+      "\n",
+      "#Calculation\n",
+      "Vver=V*math.sin(d);#velocity in vertical direction\n",
+      "Wg=m*g*Vver/1000;\n",
+      "print'the addtional power %i kW'%Wg"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the addtional power 147 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-9 ,Page No.69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "m=900;#mass of car in kg\n",
+      "v1=0;# intial velocity\n",
+      "v2=80/3.6;# final velocity; converting km/h into m/s\n",
+      "t=20;# time taken1\n",
+      "\n",
+      "#Calculation\n",
+      "Wa=m*(v2**2-v1**2)/2/1000;\n",
+      "Wavg=Wa/t;\n",
+      "print'the average power %f kW'%round(Wavg,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the average power 11.100000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-10 ,Page No.74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "Win=100;# work done in the process in kJ\n",
+      "Qout=500;# heat lost in kJ\n",
+      "U1=800;# internal energy of the fluid in kJ\n",
+      "\n",
+      "#Calculations\n",
+      "# Win - Qout = U2- U1 i.e change in internal energy \n",
+      "U2=U1-Qout+Win;\n",
+      "print'final internal of the system %i kJ'%U2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "final internal of the system 400 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-11 ,Page No.75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#given values\n",
+      "Win=20;# power consumption in W\n",
+      "mair=0.25;# rate of air discharge in kg/sec\n",
+      "\n",
+      "#calculation\n",
+      "v=math.sqrt(Win/2/mair)#Win = 1/2*m*v^2\n",
+      "if v >=8:\n",
+      "    print('True');\n",
+      "else:\n",
+      "    print('False')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "False\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-12 ,Page No.76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "Win=200.0;#Power of fan in W\n",
+      "U=6.0;#Overall heat transfer coefficient in W/m^2 C\n",
+      "A=30;#Surface area in m^2\n",
+      "To=25;#Outdoor temperature in C\n",
+      "\n",
+      "#Calculations\n",
+      "Ti= (Win/U/A)+To;# Win = Qout = U*A*(Ti - To)\n",
+      "print'the indoor air temperature %f Celcius'%Ti\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the indoor air temperature 26.111111 Celcius\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-13 ,Page No.76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "Plamp=80.0;#Power of lamp in W\n",
+      "N=30;#no of lamps\n",
+      "t=12;#time period the light is in use in hours/day\n",
+      "y=250;#days in a year light is in function \n",
+      "UC=0.07;#unit cost in $\n",
+      "\n",
+      "#Calculation\n",
+      "LP=Plamp*N/1000;#Lighting power in kW\n",
+      "OpHrs=t*y;#Operating hours\n",
+      "LE=LP*OpHrs;#Lighting energy in kW\n",
+      "LC=LE*UC;#Lighting cost\n",
+      "print'the annual energy cost $%i'%LC\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the annual energy cost $504\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-15 ,Page No.82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "Ein=2.0;#Power of electric burner in kW\n",
+      "n1=0.73;#Efficiency of open burners\n",
+      "n2=0.38;#efficency of gas units\n",
+      "CinH=0.09;#Unit cost of electricity in $\n",
+      "CinB=0.55;#Unit cost of natural gas in $\n",
+      "\n",
+      "#Calculations\n",
+      "QutH= Ein * n1;\n",
+      "print'rate of energy consumption by the heater %f kW'%round(QutH,2);\n",
+      "CutH= CinH / n1;\n",
+      "print'the unit cost of utilized energy for heater $%f/kWh'%round(CutH,3);\n",
+      "QutB= QutH / n2 ;\n",
+      "print'rate of energy consumption by the burner %f kW'%round(QutB,2);\n",
+      "CutB= CinB / n2 / 29.3; # 1 therm = 29.3 kWh\n",
+      "print'the unit cost of utilized energy for burner %f kWh'%round(CutB,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate of energy consumption by the heater 1.460000 kW\n",
+        "the unit cost of utilized energy for heater $0.123000/kWh\n",
+        "rate of energy consumption by the burner 3.840000 kW\n",
+        "the unit cost of utilized energy for burner 0.049000 kWh\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-16 ,Page No.84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#there is a 0.204% error in the last part of the question due to rounding off the intermidiate steps in the solution\n",
+      "\n",
+      "#Constants used\n",
+      "g=9.81;#acceleration due to gravity in m/s^2;\n",
+      "\n",
+      "#Given values\n",
+      "h=50.0;#Depth of water in m\n",
+      "m=5000.0;#mass flow rate of water in kg/sec\n",
+      "Wout=1862.0;#generated electric power in kW\n",
+      "ngen=0.95;#efficiency of turbine\n",
+      "\n",
+      "#calculation\n",
+      "X=g*h/1000.0;# X stands for the differnce b/w change in mechanical energy per unit mass\n",
+      "R=m*X;#rate at which mech. energy is supplied to turbine in kW\n",
+      "nov=Wout/R;#overall efficiency i.e turbine and generator\n",
+      "print'overall efficiency is %f'%round(nov,2);\n",
+      "ntu=nov/ngen;#efficiency of turbine\n",
+      "print'efficiency of turbine is %f'%round(ntu,2);\n",
+      "Wsh=ntu*R;#shaft output work\n",
+      "print'shaft power output %i kW'%round(Wsh,0)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "overall efficiency is 0.760000\n",
+        "efficiency of turbine is 0.800000\n",
+        "shaft power output 1960 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-17 ,Page No.85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "Pstd=4520.0;\n",
+      "Phem=5160.0;#prices of std and high eff motor in USD\n",
+      "R=60*0.7457;#rated power in kW from hp\n",
+      "OpHrs=3500.0;#Operating hours\n",
+      "Lf=1.0;#Load Factor\n",
+      "nsh=0.89;#efficiency of shaft\n",
+      "nhem=0.932;#efficiency of high eff. motor\n",
+      "CU=0.08;#per unit cost in $\n",
+      "\n",
+      "#calculation\n",
+      "PS=R*Lf*(1/nsh-1/nhem);#Power savings = W electric in,standard -   W electric in,efficient\n",
+      "ES=PS*OpHrs;#Energy savings = Power savings * Operating hours\n",
+      "print'Energy savings %i kWh/year'%ES;\n",
+      "CS=ES*CU;\n",
+      "print'Cost savings per year $%i'%CS;\n",
+      "EIC=Phem-Pstd;#excess intial cost\n",
+      "Y=EIC/CS;\n",
+      "print'simple payback period %f years'%round(Y,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy savings 7929 kWh/year\n",
+        "Cost savings per year $634\n",
+        "simple payback period 1.000000 years\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-18 ,Page No.91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "#NOx details\n",
+      "m1=0.0047;#emissions of gas furnaces of NOx in kg/therm\n",
+      "N1=18*10**6;#no. of therms per year \n",
+      "#CO2 details\n",
+      "m2=6.4;#emissions of gas furnaces of CO2 in kg/therm\n",
+      "N2=18*10**6;#no. of therms per year \n",
+      "\n",
+      "#Calculation\n",
+      "NOxSav=m1*N1;\n",
+      "print'NOx savings %f kg/year'%round(NOxSav,1);\n",
+      "CO2Sav=m2*N2;\n",
+      "print'CO2 savings %f kg/year'%round(CO2Sav,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NOx savings 84600.000000 kg/year\n",
+        "CO2 savings 115200000.000000 kg/year\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2-19 ,Page No.95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constants used\n",
+      "e=0.95;#Emissivity\n",
+      "tc=5.67*10**-8;#thermal conductivity in W/m^2 K^4\n",
+      "\n",
+      "#Given values\n",
+      "h=6;#convection heat transfer coefficient in W/m^2 C\n",
+      "A=1.6;#cross-sectional area in m^2\n",
+      "Ts=29;#average surface temperature in C\n",
+      "Tf=20;#room temperature in C\n",
+      "\n",
+      "#Calculation\n",
+      "#convection rate\n",
+      "Q1=h*A*(Ts-Tf);\n",
+      "#radiation rate\n",
+      "Q2=e*tc*A*((Ts+273)**4-(Tf+273)**4);\n",
+      "Qt=Q1+Q2;\n",
+      "print'the total rate of heat transfer %f W'%round(Qt,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the total rate of heat transfer 168.100000 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter3_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter3_1.ipynb
new file mode 100755
index 00000000..9830dbfb
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter3_1.ipynb
@@ -0,0 +1,756 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Properties of Pure Substances"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-1 ,Page No.128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "m=50;#mass in kg\n",
+      "T=90;#temperature in C\n",
+      "\n",
+      "#Values from Table A-4\n",
+      "P=70.183;#in kPa\n",
+      "v=0.001036;#in m^3/kg\n",
+      "\n",
+      "#Calculation\n",
+      "V=m*v;#equating dimensions\n",
+      "print'pressure is %f kPa'%round(P,3);\n",
+      "print'total volumne of tank becomes %f m^3'%round(V,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure is 70.183000 kPa\n",
+        "total volumne of tank becomes 0.051800 m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-2 ,Page No.128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "V=2.0;#volumne of saturated water vapor in ft^3\n",
+      "P=50.0;#pressure in psia\n",
+      "\n",
+      "#Values from Table A-5E\n",
+      "T=280.99;#in F\n",
+      "v=8.5175;#in ft^3/lbm\n",
+      "\n",
+      "#caluclation\n",
+      "m=V/v;#dimension analysis\n",
+      "print'Temperature inside cylinder %f F'%round(T,2);\n",
+      "print'mass of vapour inside cylinder %f lbm'%round(m,3);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature inside cylinder 280.990000 F\n",
+        "mass of vapour inside cylinder 0.235000 lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-3 ,Page No.128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Constants used\n",
+      "Hfg=2257.5;#enthalpy of vaporization in kJ/kg\n",
+      "\n",
+      "#Given values\n",
+      "m=200.0/1000;#mass converting in kg\n",
+      "P=100;#Pressure at which process takes place in kPa\n",
+      "\n",
+      "#Values from Table A-5\n",
+      "vg=1.6941;#specific vol of sat liq\n",
+      "vf=0.001043;#specific vol of vapor\n",
+      "\n",
+      "#Caluclation\n",
+      "vfg=vg-vf;\n",
+      "V=m*vfg;\n",
+      "print'the volume change %f m^3'%round(V,4);\n",
+      "E=m*Hfg;\n",
+      "print'amount of energy transferred to the water %f kJ'%round(E,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the volume change 0.338600 m^3\n",
+        "amount of energy transferred to the water 451.500000 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-4 ,Page No.131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "mt=10#mass of water in kg\n",
+      "mf=8;#mass of water in liquid form in kg\n",
+      "T=90;#temperature in C\n",
+      "\n",
+      "#Values from Table A-4\n",
+      "P=70.183;#in kPa\n",
+      "vf=0.001036;#in m^3\n",
+      "vg=2.3593;#in m^3\n",
+      "\n",
+      "#Caluclation\n",
+      "mg=mt-mf;\n",
+      "V=mf*vf+mg*vg;# V= Vg + Vf\n",
+      "print'the volume of the tank %f m^3'%round(V,2);\n",
+      "print'the pressure in the tank %f kPa'%round(P,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the volume of the tank 4.730000 m^3\n",
+        "the pressure in the tank 70.183000 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-5 ,Page No.131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "m=4;#mass of refrigerant-134a in kg\n",
+      "V=80.0/1000;#volumne converting into m^3\n",
+      "P=160;#pressure in kPa\n",
+      "\n",
+      "#Values from Table A-12\n",
+      "vf=0.0007437;\n",
+      "vg=0.12348;\n",
+      "T=-15.60;\n",
+      "hf=31.21;\n",
+      "hfg=209.90;\n",
+      "\n",
+      "#Caluclations\n",
+      "v=V/m;\n",
+      "#vg>v>vf therefore it is a saturated mix\n",
+      "#hence temp will same as saturation temp\n",
+      "print'the temperature %f celcius'%round(T,2)\n",
+      "x=(v-vf)/(vg-vf);#x=vg/vfg i.e the dryness fraction\n",
+      "print'the quality is %f'%round(x,3);\n",
+      "h=hf+x*hfg;\n",
+      "print'the enthalpy of the refrigerant %f kJ/kg'%round(h,1);\n",
+      "mg=x*m;\n",
+      "Vg=mg*vg;\n",
+      "print'the volume occupied by the vapor phase %f m^3'%round(Vg,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature -15.600000 celcius\n",
+        "the quality is 0.157000\n",
+        "the enthalpy of the refrigerant 64.100000 kJ/kg\n",
+        "the volume occupied by the vapor phase 0.077500 m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-7 ,Page No.133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "P=0.5;#pressure in MPa\n",
+      "h=2890.0;#enthaply in kJ/kg\n",
+      "\n",
+      "#from Table A\u20136\n",
+      "#at P=0.5 MPa\n",
+      "T1=200.0;\n",
+      "h1=2855.8;\n",
+      "T2=250;\n",
+      "h2=2961.0;\n",
+      "# we need linear interpolation \n",
+      "\n",
+      "#calculatiom\n",
+      "#by interpolation we can say that\n",
+      "#h=h1+(T-T1)/(T2-T1)*(h2-h1)\n",
+      "#we have to find T\n",
+      "T=(h-h1)/(h2-h1)*(T2-T1)+T1;\n",
+      "print'temperature of water %f celcius'%round(T,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature of water 216.300000 celcius\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-8 ,Page No.134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "T=80;#temperature of compressed liquid water in C\n",
+      "P=5;#pressure in KPa\n",
+      "\n",
+      "#from Table A\u20137\n",
+      "#at compressed liq given conditions\n",
+      "u=333.82;\n",
+      "\n",
+      "#from Table A-4\n",
+      "#at saturation\n",
+      "usat=334.97;\n",
+      "\n",
+      "#calcualtion\n",
+      "e=(usat-u)/u*100;\n",
+      "print'internal energy of compressed liquid water using data from the compressed liquid table %f kJ/kg '%round(u,2);\n",
+      "print'internal energy of compressed liquid water using saturated liquid data %f kJ/kg '%round(usat,2);\n",
+      "print'the error involved %f the second case'%round(e,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "internal energy of compressed liquid water using data from the compressed liquid table 333.820000 kJ/kg \n",
+        "internal energy of compressed liquid water using saturated liquid data 334.970000 kJ/kg \n",
+        "the error involved 0.340000 the second case\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-9 ,Page No.135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#part a\n",
+      "print('Part a');\n",
+      "\n",
+      "#given values\n",
+      "P=200;#in KPa\n",
+      "x=0.6;\n",
+      "\n",
+      "#from Table A-5\n",
+      "T=120.21;\n",
+      "uf=504.50;\n",
+      "ufg=2024.6;\n",
+      "\n",
+      "#calcualtions\n",
+      "u=uf+(x*ufg);\n",
+      "print'temperature %f Celcius '%round(T,2);\n",
+      "print'internal energy %f kJ/kg'%round(u,2);\n",
+      "print('saturated liquid\u2013vapor mixture at a pressure of 200 kPa');\n",
+      "\n",
+      "#part b\n",
+      "print('Part b');\n",
+      "\n",
+      "#given values\n",
+      "T=125;#in C\n",
+      "u=1600;#in kJ/kg\n",
+      "\n",
+      "#from Table A\u20134\n",
+      "uf=524.83;\n",
+      "ug=2534.3;\n",
+      "#ug>u>ufg so its aturated liquid\u2013vapor mixture\n",
+      "P=232.23;\n",
+      "\n",
+      "#calculation\n",
+      "ufg=ug-uf;\n",
+      "x=(u-uf)/ufg;\n",
+      "print'Pressure %f kPa'%round(P,2);\n",
+      "print'x is %f'%round(x,3);\n",
+      "print('saturated liquid\u2013vapor mixture at a temp of 125 of celcius');\n",
+      "\n",
+      "#part c\n",
+      "print('Part c');\n",
+      "\n",
+      "#given values\n",
+      "P=1000;#in kPa\n",
+      "u=2950;#in kJ/kg\n",
+      "\n",
+      "#from Table A\u20136\n",
+      "uf=761.39;\n",
+      "ug=2582.8;\n",
+      "#u>ug so its superheated steam\n",
+      "T=395.2;\n",
+      "\n",
+      "#calculation\n",
+      "print'temperature %f Celcius'%round(T,1);\n",
+      "print('superheated vapor at 1MPa');\n",
+      "\n",
+      "#part d\n",
+      "print('Part d');\n",
+      "\n",
+      "#given values\n",
+      "T=75;#in C\n",
+      "P=100;#in kPa\n",
+      "\n",
+      "#from Table A\u20135\n",
+      "Tsat=151.83;\n",
+      "#T<Tsat so it is a compressed liquid\n",
+      "#the given pressure is much lower than the lowest pressure value in the compressed liquid table i.e 5 MPa\n",
+      "#assuming, the compressed liquid as saturated liquid at the given temperature\n",
+      "\n",
+      "#from Table A-4\n",
+      "u=313.99;\n",
+      "print'Internal energy %f kJ/kg'%round(u,2);\n",
+      "print('the compressed liquid condition');\n",
+      "\n",
+      "#Part e\n",
+      "print('Part e');\n",
+      "\n",
+      "#given values\n",
+      "P=850;#in kPa\n",
+      "x=0;\n",
+      "\n",
+      "#x=0 therefore it is a saturateed liquid condition\n",
+      "#from Table A-5\n",
+      "T=172.94;\n",
+      "u=731.00;\n",
+      "print'temperature %f Celcius'%round(T,2);\n",
+      "print'Internal energy %f kJ/kg'%round(u,2);\n",
+      "print('saturateed liquid condition')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part a\n",
+        "temperature 120.210000 Celcius \n",
+        "internal energy 1719.260000 kJ/kg\n",
+        "saturated liquid\u2013vapor mixture at a pressure of 200 kPa\n",
+        "Part b\n",
+        "Pressure 232.230000 kPa\n",
+        "x is 0.535000\n",
+        "saturated liquid\u2013vapor mixture at a temp of 125 of celcius\n",
+        "Part c\n",
+        "temperature 395.200000 Celcius\n",
+        "superheated vapor at 1MPa\n",
+        "Part d\n",
+        "Internal energy 313.990000 kJ/kg\n",
+        "the compressed liquid condition\n",
+        "Part e\n",
+        "temperature 172.940000 Celcius\n",
+        "Internal energy 731.000000 kJ/kg\n",
+        "saturateed liquid condition\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-10 ,Page No.139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#constants used\n",
+      "R=0.287# in kPa m^3/kg K\n",
+      "\n",
+      "#given values\n",
+      "l=4;#dimensions of room in m\n",
+      "b=5;#dimensions of room in m\n",
+      "h=6;#dimensions of room in m\n",
+      "P=100.0;#pressure in kPa\n",
+      "T=25+273.0;#temperature in Kelvin\n",
+      "\n",
+      "#calculation\n",
+      "V=l*b*h;\n",
+      "m=P*V/R/T;\n",
+      "print'the mass of the air %f kg'%round(m,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mass of the air 140.300000 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-11 ,Page No.142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given values\n",
+      "P=1;#pressure in MPa\n",
+      "T=50+273;#tempearture converting into Kelvin\n",
+      "vgiv=0.021796;#specific vol. given in m^3\n",
+      "\n",
+      "#from Table A-1\n",
+      "R=0.0815;\n",
+      "Pcr=4.059;\n",
+      "Tcr=374.2;\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "#Part A\n",
+      "v1=R*T/(P*1000);\n",
+      "print'specific volume of refrigerant-134a under the ideal-gas assumption %fm^3/kg'%round(v1,6);\n",
+      "e=(v1-vgiv)/vgiv;\n",
+      "print'an error of %f'%round(e,3);\n",
+      "\n",
+      "#Part B\n",
+      "#determine Z from the compressibility chart, we will calculate the reduced pressure and temperature\n",
+      "Pr=P/Pcr;\n",
+      "Tr=T/Tcr;\n",
+      "#from chart\n",
+      "Z=0.84;\n",
+      "v=Z*v1;\n",
+      "print'specific volume of refrigerant-134a under the generalized compressibility chart %f m^3/kg'%round(v,6);\n",
+      "e=(v-vgiv)/vgiv;\n",
+      "print'an error of %f'%round(e,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specific volume of refrigerant-134a under the ideal-gas assumption 0.026325m^3/kg\n",
+        "an error of 0.208000\n",
+        "specific volume of refrigerant-134a under the generalized compressibility chart 0.022113 m^3/kg\n",
+        "an error of 0.015000\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-12 ,Page No.143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given values\n",
+      "v=0.51431;\n",
+      "T=600;\n",
+      "\n",
+      "#from Table A-1E\n",
+      "R=0.5956;\n",
+      "Pcr=3200;\n",
+      "Tcr=1164.8;\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "#Part A\n",
+      "#from Table A-6E\n",
+      "Pa=1000.0;#in psia\n",
+      "print'from the steam tables %i psia'%Pa;\n",
+      "\n",
+      "#Part B\n",
+      "T=1060;#converted into R from F\n",
+      "Pb=R*T/v;\n",
+      "print'from the ideal-gas equation %i psia'%round(Pb,0);\n",
+      "e=(Pb-Pa)/Pa;\n",
+      "print'treating the steam as an ideal gas would result in an error of %f'%round(e,3)\n",
+      "\n",
+      "#Part C\n",
+      "#calculating the pseudo-reduced specific volume and the reduced temperature\n",
+      "Vr=v/(R*Tcr/Pcr);\n",
+      "Tr=T/Tcr;\n",
+      "#from the compressibility chart\n",
+      "Pr=0.33;\n",
+      "P=Pr*Pcr;\n",
+      "print'from the generalized compressibility chart %i psia'%P\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "from the steam tables 1000 psia\n",
+        "from the ideal-gas equation 1228 psia\n",
+        "treating the steam as an ideal gas would result in an error of 0.228000\n",
+        "from the generalized compressibility chart 1056 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-13 ,Page No.147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Answer of part c-d are having slight difference due to approximation in molar volumne in the textbook which here is caluculated to the approximation of 7 decimal digits\n",
+      "\n",
+      "from math import exp\n",
+      "\n",
+      "#given values\n",
+      "T=175.0;#temperature in C\n",
+      "v=0.00375;#sp. volumne in m^3/kg\n",
+      "Pex=10000;#experimentaion determination\n",
+      "\n",
+      "#from Table A-1\n",
+      "R=0.2968# in kPa m^3/kg  K\n",
+      "\n",
+      "#calculating\n",
+      "\n",
+      "#Part-a\n",
+      "P=R*T/v;\n",
+      "print'using the ideal-gas equation of state %i kPa'%(round(P))\n",
+      "e=(P-Pex)/Pex*100;\n",
+      "print'error is %f percent'%e;\n",
+      "\n",
+      "#Part-b\n",
+      "#van der Waals constants from Eq. 3-23\n",
+      "a=0.175;\n",
+      "b=0.00138;\n",
+      "#from van der waal eq.\n",
+      "P=R*T/(v-b)-a/v**2;\n",
+      "print'using the van der Waals equation of state  is %i kPa'%(round(P));\n",
+      "e=(P-Pex)/Pex*100;\n",
+      "print'error is %f percent'%e;\n",
+      "\n",
+      "#Part-c\n",
+      "#constants in the Beattie-Bridgeman equation from Table 3\u20134\n",
+      "A=102.29;\n",
+      "B=0.05378;\n",
+      "c=4.2*10**4;\n",
+      "Ru=8.314;#in kPa m^3/kmol K\n",
+      "M=28.013;#molecular weight in kg/mol\n",
+      "vb=M*v;#molar vol.\n",
+      "P=(Ru*T)/(vb**2)*(1-((c)/(vb*T**3)))*(vb+B)-(A/vb**2);\n",
+      "print'using the Beattie-Bridgeman equationis %i kPa'%(round(P));\n",
+      "e=(P-Pex)/Pex*100;\n",
+      "print'error is %f percent'%e;\n",
+      "\n",
+      "#Part-d\n",
+      "#constants of Benedict-Webb-Rubin equation from Table 3\u20134\n",
+      "a=2.54;\n",
+      "b=0.002328;\n",
+      "c=7.379*10**4;\n",
+      "alp=1.272*10**-4;\n",
+      "Ao=106.73;\n",
+      "Bo=0.040704;\n",
+      "Co=8.164*10**5;\n",
+      "gam=0.0053;\n",
+      "P=((Ru*T)/vb)+((Bo*Ru*T)-Ao-Co/T**2)/vb**2+(b*Ru*T-a)/vb**3+(a*alp/vb**6)+(c/((vb**3)*(T**2)))*(1 + (gam/vb**2))*exp(-gam/vb**2);\n",
+      "print'using Benedict-Webb-Rubin equation %i kPa'%(round(P));\n",
+      "e=(P-Pex)/Pex*100;\n",
+      "print'error is %f percent'%e;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "using the ideal-gas equation of state 13851 kPa\n",
+        "error is 38.506667 percent\n",
+        "using the van der Waals equation of state  is 9471 kPa\n",
+        "error is -5.288326 percent\n",
+        "using the Beattie-Bridgeman equationis 10109 kPa\n",
+        "error is 1.092970 percent\n",
+        "using Benedict-Webb-Rubin equation 10004 kPa\n",
+        "error is 0.039256 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3-14 ,Page No.152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given value\n",
+      "T=25;#air temperature over a lake in C\n",
+      "\n",
+      "#from table 3-1\n",
+      "Psat=3.17;#in kPa\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#Relative Humidity 10%\n",
+      "Pv1=0.1*Psat\n",
+      "#Relative Humidity 80%\n",
+      "Pv2=0.8*Psat\n",
+      "#Relative Humidity 100%\n",
+      "Pv3=1*Psat\n",
+      "\n",
+      "# from table 3-1 Tsat at these Pressures are\n",
+      "T1=-8.0;\n",
+      "T2=21.2;\n",
+      "T3=25.0;\n",
+      "print'with relative humidity 10, temperature is %i in C'%T1\n",
+      "print'with relative humidity 80, temperature is %i in C'%T2\n",
+      "print'with relative humidity 100, temperature is %i in C'%T3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "with relative humidity 10, temperature is -8 in C\n",
+        "with relative humidity 80, temperature is 21 in C\n",
+        "with relative humidity 100, temperature is 25 in C\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter4_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter4_1.ipynb
new file mode 100755
index 00000000..b9e964f0
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter4_1.ipynb
@@ -0,0 +1,720 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Energy Analysis of Closed Systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-2 ,Page No.169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given values\n",
+      "m=10;#mass in lbm\n",
+      "Po=60;#steam oressure in psia\n",
+      "T1=320;#intial temp in F\n",
+      "T2=400;#final temp in F\n",
+      "\n",
+      "#from Table A\u20136E\n",
+      "v1=7.4863;#at 60 psia and 320 F\n",
+      "v2=8.3548;#at 60 psia and 400 F\n",
+      "\n",
+      "#calculations\n",
+      "#W = P dV which on integrating gives W = m * P * (V2 - V1)\n",
+      "W=m*Po*(v2-v1)/5.404;#coverting into Btu from psia-ft^3\n",
+      "print'work done by the steam during this process %f Btu'%round(W,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "work done by the steam during this process 96.400000 Btu\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-3 ,Page No.170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "#given data\n",
+      "P1=100;#pressure in kPa\n",
+      "V1=0.4;#intial vol in m^3\n",
+      "V2=0.1;#final vol in m^3\n",
+      "\n",
+      "#calculations\n",
+      "#for isothermal W = P1*V1* ln(V2/V1)\n",
+      "W=P1*V1*log(V2/V1);\n",
+      "print'the work done during this process %f kJ'%round(W,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the work done during this process -55.500000 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-4 ,Page No.171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V1=0.05;#Volumne of gas in m^3\n",
+      "P1=200;#Pressure in kPa\n",
+      "k=150;#Spring constant in kN/m\n",
+      "A=0.25;#Cross-sectional area in m^2\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#Part - a\n",
+      "V2=2*V1;\n",
+      "x2=(V2-V1)/A;#printlacement of spring\n",
+      "F=k*x2;#compression force\n",
+      "P2=P1+F/A;#additional pressure is equivalent the compression of spring\n",
+      "print'the final pressure inside the cylinder %i kPa'%P2;\n",
+      "\n",
+      "#Part - b\n",
+      "#work done is equivalent to the area of the P-V curve of Fig 4-10\n",
+      "W=(P1+P2)/2*(V2-V1);#area of trapezoid = 1/2 * sum of parallel sides * dist. b/w them\n",
+      "print'the total work done by the gas %i kJ'%W;\n",
+      "\n",
+      "#Part - c\n",
+      "x1=0;#intial compression of spring\n",
+      "Wsp=0.5*k*(x2**2-x1**2);\n",
+      "print'the fraction of this work done against the spring to compress it %i kJ'%Wsp\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final pressure inside the cylinder 320 kPa\n",
+        "the total work done by the gas 13 kJ\n",
+        "the fraction of this work done against the spring to compress it 3 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-5 ,Page No.174"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given values\n",
+      "m=0.025;#mass of saturated water vapor in kg\n",
+      "V=120;#rated voltage of heater in V\n",
+      "I=0.2;#rated current in A\n",
+      "t=300;#total time taken in sec\n",
+      "P1=300;#constant pressure in kPa\n",
+      "Qout=3.7;#heat lost in kJ\n",
+      "\n",
+      "#from Table A\u20135\n",
+      "#at P1 the conditon is sat. vap\n",
+      "h1=2724.9;\n",
+      "\n",
+      "#Calculations\n",
+      "\n",
+      "#Part - a\n",
+      "#therotical proving\n",
+      "\n",
+      "#Part - b\n",
+      "We=V*I*t/1000;#electrical work in kJ\n",
+      "#from eqn 4 -18 i.e derived in earler part\n",
+      "#it states it Ein - Eout = Esystem\n",
+      "# it applies as Win - Qout = H = m (h2 - h1)\n",
+      "h2=(We-Qout)/m+h1;\n",
+      "##from Table A\u20135\n",
+      "#at h2 we get\n",
+      "P2=300;\n",
+      "T=200;\n",
+      "print'the final temperature of the steam %i C'%T\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final temperature of the steam 200 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-6 ,Page No.176"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=5.0;#mass of water in kg\n",
+      "P1=200;#pressure on one side in kPa\n",
+      "T=25;#temperature in C\n",
+      "\n",
+      "#from Table A\u20134\n",
+      "#the liq. is in compressed state at 200 kPa and 25 C\n",
+      "vf=0.001;\n",
+      "vg=43.340;\n",
+      "uf=104.83;\n",
+      "ufg=2304.3;\n",
+      "v1=vf;\n",
+      "u1=uf;\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#Part - a\n",
+      "V1=m*v1;\n",
+      "Vtank=2*V1;\n",
+      "print'the volume of the tank %f m^3'%round(Vtank,2);\n",
+      "\n",
+      "#Part - b\n",
+      "V2=Vtank;\n",
+      "v2=V2/m;\n",
+      "#from Table A\u20134 \n",
+      "# at T=25 vf=0.101003 m^3/kg and vg=43.340 m^3/kg\n",
+      "# vf<v2<vg therefore it is saturated liquid\u2013vapor mixture\n",
+      "P2=3.1698;\n",
+      "print'the final pressure %f kPa'%round(P2,4);\n",
+      "\n",
+      "#Part - c\n",
+      "#Ein - Eout = Esystem\n",
+      "#Qin= dU = m(u2 - u1)\n",
+      "x2=(v2-vf)/(vg-vf);\n",
+      "u2=uf+x2*ufg;\n",
+      "Qin=m*(u2-u1);\n",
+      "print'the heat transfer for this process %f kJ'%Qin\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the volume of the tank 0.010000 m^3\n",
+        "the final pressure 3.169800 kPa\n",
+        "the heat transfer for this process 0.265846 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-7 ,Page No.183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from scipy.integrate import quad \n",
+      "from pylab import *\n",
+      "\n",
+      "#given data\n",
+      "T1=300;#intial temp of air in K\n",
+      "P=200;#pressure in kPa\n",
+      "T2=600;#final temp in K\n",
+      "M=28.97;#molecular weight in kg/kmol\n",
+      "Ru=8.314;\n",
+      "\n",
+      "#Part - a\n",
+      "#from Table A\u201317\n",
+      "u1=214.07;\n",
+      "u2=434.78;\n",
+      "du=u2-u1;#change in internal energy\n",
+      "print'change in internal energy from data from the air table %f kJ/kg'%round(du,2);\n",
+      "\n",
+      "#Part - b\n",
+      "#from Table A\u20132c\n",
+      "a=28.11;\n",
+      "b=0.1967*10**-2;\n",
+      "c=0.4802*10**-5;\n",
+      "d=-1.966*10**-9;\n",
+      "#  by equation Cp(T)=a+bT+cT^2+dT^3\n",
+      "def intgrnd1(T): \n",
+      " return ((a-Ru)+b*T+c*T**2+d*T**3)\n",
+      "dU, err = quad(intgrnd1, T1, T2) \n",
+      "du=dU/M;\n",
+      "print'change in internal energy the functional form of the specific heat %f kJ/kg'%round(du,1); \n",
+      "\n",
+      "#Part - c\n",
+      "#from Table A\u20132b\n",
+      "Cavg=0.733;\n",
+      "du=Cavg*(T2-T1);\n",
+      "print'change in internal energy the functional form the average specific heat value %i kJ/kg'%round(du,0);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in internal energy from data from the air table 220.710000 kJ/kg\n",
+        "change in internal energy the functional form of the specific heat 222.500000 kJ/kg\n",
+        "change in internal energy the functional form the average specific heat value 220 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-8 ,Page No.184"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=1.5;#mass in lbm\n",
+      "T1=80;#temperature in F\n",
+      "P1=50;#pressure in psia\n",
+      "W=0.02;#power rating in hp\n",
+      "t=30.0/60;#converting into hrs from min\n",
+      "\n",
+      "#from Table A\u20132Ea\n",
+      "Cv=0.753;\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part a\n",
+      "Wsh=W*t*2545;#in Btu\n",
+      "#Ein - Eout = Esystem\n",
+      "#Wsh = dU = m (u2 - u1) = m * Cv * (T2 - T1)\n",
+      "T2= Wsh/(m*Cv)+T1;\n",
+      "print'the final temperature %f F'%round(T2,1);\n",
+      "\n",
+      "#part b\n",
+      "#using ideal gas eqn\n",
+      "# P1 * V1 / T1 = P2 * T2 /V2\n",
+      "P2= 50 * (T2 +460)/ (T1+460);\n",
+      "# temp should in R therefore + 460\n",
+      "print'the final pressure %f psia'%round(P2,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final temperature 102.500000 F\n",
+        "the final pressure 52.100000 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-9 ,Page No.185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V1=0.5;#volumne of nitrogen gas in m^3\n",
+      "P=400;#pressure in kPa\n",
+      "T1=27;#temp in C\n",
+      "I=2;#rated current in A\n",
+      "t=5*60;#converting into s from min\n",
+      "V=120;#rated voltage in V\n",
+      "Qout=2800/1000;#in kJ\n",
+      "R=0.297;\n",
+      "\n",
+      "#from Table A\u20132a\n",
+      "Cp=1.039;\n",
+      " \n",
+      "#calculations\n",
+      "P1=P;\n",
+      "We=V*I*t/1000;#in kJ\n",
+      "m=P1*V1/(R*(T1+273));\n",
+      "#Ein - Eout = Esystem\n",
+      "# We,in - Qout = dH =  m (h2 - h1) = m * Cp * (T2 - T1)\n",
+      "T2=(We-Qout)/(m*Cp)+T1;\n",
+      "print'the final temperature of nitrogen in %i C'%T2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final temperature of nitrogen in 57 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-10 ,Page No.187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=150;#initially air pressure in kPa\n",
+      "P2=350;#final pressure in kPa\n",
+      "T1=27+273;#temperatuere in K\n",
+      "V1=400.0/1000;#volumne in m^3\n",
+      "R=0.287;\n",
+      "\n",
+      "#from Table A\u201317\n",
+      "u1=214.07;\n",
+      "u2=1113.52;\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part a\n",
+      "V2=2*V1;\n",
+      "#using ideal gas eqn\n",
+      "# P1 * V1 / T1 = P2 * T2 /V2\n",
+      "T2=P2*V2*T1/(P1*V1);\n",
+      "print'the final temperature %i K'%T2;\n",
+      "\n",
+      "#part b\n",
+      "# Work done is Pdv\n",
+      "W=P2*(V2-V1);\n",
+      "print'the work done by the air %i kJ'%W;\n",
+      "\n",
+      "#part c\n",
+      "#Ein - Eout = Esystem\n",
+      "#Qin - Wout = dU = m(u2 - u1)\n",
+      "m= P1* V1 /(T1 * R);\n",
+      "Q= m*(u2 - u1)+ W;\n",
+      "print'the total heat transferred to the air %i kJ'%round(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final temperature 1400 K\n",
+        "the work done by the air 140 kJ\n",
+        "the total heat transferred to the air 767 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-11 ,Page No.190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=100;#twmperature of liquid water in C\n",
+      "P=15;#pressure in MPa\n",
+      "\n",
+      "#from Table A\u20137\n",
+      "#at P=15 mPa and T = 100 C\n",
+      "hg=430.39;\n",
+      "hf=419.17\n",
+      "vf=0.001;\n",
+      "Psat=101.42;#in kPa\n",
+      "\n",
+      "#calculations\n",
+      "\n",
+      "#part a\n",
+      "h=hg;\n",
+      "print'enthalpy of liquid water by using compressed liquid tables %f kJ/kg'%round(h,2);\n",
+      "\n",
+      "#part b\n",
+      "#Approximating the compressed liquid as a saturated liquid at 100\u00b0C\n",
+      "h=hf;\n",
+      "print'enthalpy of liquid water by approximating it as a saturated liquid %f kJ/kg'%round(h,2);\n",
+      "\n",
+      "#part c\n",
+      "h=hf + vf*(P*1000 - Psat );\n",
+      "print'enthalpy of liquid water by using the correction given by Eq. 4\u201338 %f kJ/kg'%round(h,2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "enthalpy of liquid water by using compressed liquid tables 430.390000 kJ/kg\n",
+        "enthalpy of liquid water by approximating it as a saturated liquid 419.170000 kJ/kg\n",
+        "enthalpy of liquid water by using the correction given by Eq. 4\u201338 434.070000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-12 ,Page No.191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#suffix i for iron\n",
+      "#suffix w for water\n",
+      "#given data\n",
+      "mi=50;#mass in Kg\n",
+      "T1i=80;#temperature in C \n",
+      "Vw=0.5;#volumne in m^3\n",
+      "T1w=25;#temperature in C \n",
+      "v=0.001;#specific volume of liquid water at or about room temperature\n",
+      "\n",
+      "#from Table A\u20133\n",
+      "ci=0.45;\n",
+      "cw=4.18;\n",
+      "\n",
+      "#calculations\n",
+      "mw=Vw/v;\n",
+      "#Ein - Eout = Esystem\n",
+      "# du = 0 i.e (mcdT)iron + (mcdT)water = 0\n",
+      "# mi * ci * (T - T1i) + mw *cw * (T-T1w)\n",
+      "#on rearranging above equn\n",
+      "T= (mi*ci*T1i + mw*cw*T1w)/(mi*ci+mw*cw);\n",
+      "print'the temperature when thermal equilibrium is reached %f C'%round(T,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature when thermal equilibrium is reached 25.600000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-13 ,Page No.191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "maf=0.15;#mass of affected tissue in kg\n",
+      "caf=3.8;#specific heat of the tissue in kJ/Kg C\n",
+      "dTaf=1.8;#suffix af for affected tissue\n",
+      "mh=1.2;#suffix h for hand ; mass of hand in kg\n",
+      "\n",
+      "#calculations\n",
+      "#Ein - Eout = Esystem\n",
+      "#dUaffected tissue - KEhand = 0\n",
+      "#from above equation we can deduce that\n",
+      "Vhand= sqrt(2*maf*caf*dTaf*1000/mh);#for conversion factor mutiplying by 1000 to get m^2/s^2\n",
+      "print'the velocity of the hand just before impact %f m/s'%round(Vhand,1);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the velocity of the hand just before impact 41.400000 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-14 ,Page No.199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=90;#mass of man in kg\n",
+      "\n",
+      "#from Tables 4\u20131 and 4\u20132\n",
+      "Ehb=275;#hamburger\n",
+      "Ef=250;#fries\n",
+      "Ec=87;#cola\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "#part a\n",
+      "Ein=2*Ehb+Ef+Ec;\n",
+      "#The rate of energy output for a 68-kg man watching TV is to be 72 Calories/h\n",
+      "Eout=m*72.0/68;\n",
+      "t=Ein/Eout;\n",
+      "print'by watching TV %f hours'%round(t,1);\n",
+      "\n",
+      "#part b\n",
+      "#The rate of energy output for a 68-kg man watching TV is to be 860 Calories/h\n",
+      "Eout=m*860.0/68;\n",
+      "t=Ein/Eout*60#converting in min\n",
+      "t=ceil(t);\n",
+      "print'by fast swimming %f mins'%t;\n",
+      "\n",
+      "#for last question\n",
+      "print('answers be for a 45-kg man energy takes twice as long in each case');\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "by watching TV 9.300000 hours\n",
+        "by fast swimming 47.000000 mins\n",
+        "answers be for a 45-kg man energy takes twice as long in each case\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4-15 ,Page No.199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "E=75;#in Cal/day\n",
+      "\n",
+      "#calculation\n",
+      "Ereduced=E*365.0;\n",
+      "#The metabolizable energy content of 1 kg of body fat is 33,100 kJ\n",
+      "Ec=33100.0;\n",
+      "mfat=Ereduced/Ec*4.1868;\n",
+      "print'weight this person will lose in one year %f kg'%round(mfat,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "weight this person will lose in one year 3.460000 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter5_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter5_1.ipynb
new file mode 100755
index 00000000..da9be875
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter5_1.ipynb
@@ -0,0 +1,614 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Mass and Energy Analysis of Control Volumes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-1 ,Page No.224"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#given data\n",
+      "V=10.0;#volumne of bucket in galon\n",
+      "t=50;#time taken to fill the bucket in sec\n",
+      "p=1;#density of water in kg/L\n",
+      "re=0.8/2/100;#radius of nozzle exit in m\n",
+      "\n",
+      "#calculations\n",
+      "Vd=V/t*3.7854;#factor 0f 3.7854 for gal to L\n",
+      "print'volumne flow rate through hose %f L/s'%round(Vd,3);\n",
+      "m=p*Vd;\n",
+      "print'mass flow rate through hose %f kg/s'%round(m,3);\n",
+      "Ae=math.pi*re**2;\n",
+      "Ve=Vd/Ae/1000;#factor of 1000 for L to m^3\n",
+      "print'average velocity at the nozzle %f m/s'%round(Ve,1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "volumne flow rate through hose 0.757000 L/s\n",
+        "mass flow rate through hose 0.757000 kg/s\n",
+        "average velocity at the nozzle 15.100000 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-2 ,Page No.225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from scipy.integrate import quad \n",
+      "from pylab import *\n",
+      "\n",
+      "#given data\n",
+      "Dtank=3*12;#diameter of tank in inches\n",
+      "Djet=0.5;#diameter of water jet in inches\n",
+      "h0=2;#bottom reference in ft\n",
+      "h1=4;#height of water tank in ft\n",
+      "\n",
+      "#constants used \n",
+      "g=32.2;#in ft/s^2\n",
+      "\n",
+      "#calculations\n",
+      "#min - mout = dmCV/dt\n",
+      "#mout = p*(2*g*h*Ajet)^2\n",
+      "#mCV = p*Atank*h\n",
+      "#from these we get dt = Dtank^2/Djet^2 * (dh/(2*g*h)^2)\n",
+      "def intgrnd1(h): \n",
+      " return (Dtank**2/Djet**2*(1/sqrt(2*g*h)))\n",
+      "t, err = quad(intgrnd1, h0, h1) \n",
+      "t=(t/60);#in min\n",
+      "print'time taken to drop to 2 ft %f min'%round(t,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time taken to drop to 2 ft 12.600000 min\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-3 ,Page No.229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P=150;#operating pressure in kPa\n",
+      "Vliquid=0.6/1000.0;#amount of liquid in the cooker in m^3\n",
+      "t=40.0*60;#period of operation in sec\n",
+      "Ac=8*10**-6;#exit area of opening in m^2\n",
+      "\n",
+      "#from Table A-5\n",
+      "#from P = 150 kPa\n",
+      "h=2693.1;\n",
+      "ug=2519.2;\n",
+      "vf=0.001053;\n",
+      "vg=1.1594;\n",
+      "\n",
+      "#calculations\n",
+      "m=Vliquid/vf;\n",
+      "md=m/t;\n",
+      "print'mass flow rate %f kg/s'%md;\n",
+      "V=md*vg/(Ac);\n",
+      "print'exit velocity %f m/s'%round(V,1);\n",
+      "Eflow=h-ug;\n",
+      "Et=h;\n",
+      "print'flow energy %f kJ/kg'%round(Eflow,1);\n",
+      "print'total energy %f kJ/kg'%round(Et,1);\n",
+      "Emass=md*Et;\n",
+      "print'rate at which energy leaves the cooker %f kW'%round(Emass,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass flow rate 0.000237 kg/s\n",
+        "exit velocity 34.400000 m/s\n",
+        "flow energy 173.900000 kJ/kg\n",
+        "total energy 2693.100000 kJ/kg\n",
+        "rate at which energy leaves the cooker 0.639000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-4 ,Page No.234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=283;#temperature of air in K\n",
+      "P1=80;#entering pressure in kPa\n",
+      "V1=200;#velocity ar inlet in m/s\n",
+      "A1=0.4;#inlet area in m^2\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kPa-m^3/kg-K\n",
+      "\n",
+      "#calulations\n",
+      "v1=R*T1/P1;\n",
+      "m=V1*A1/v1;\n",
+      "print'mass flow rate of air %f kg/s'%round(m,1);\n",
+      "# Ein - Eout = dEsystem / dt\n",
+      "#from Table A-17\n",
+      "h1=283.14;\n",
+      "V2=0;\n",
+      "h2=h1-(V2^2 - V1^2)/2/1000;#factor of 1000 to convert to kJ/kg\n",
+      "#from Table A-17 at this value of h2\n",
+      "T2=303;\n",
+      "print'the temperature %i K is'%T2;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass flow rate of air 78.800000 kg/s\n",
+        "the temperature 303 K is\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-5 ,Page No.235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=250;#inlet pressure in psia\n",
+      "T1=700;#inlet temp in F\n",
+      "A1=0.2;#inlet area in ft^2\n",
+      "qout=1.2;#heat losses in Btu/lbm\n",
+      "m=10;#mass flow rate in lbm/sec\n",
+      "P2=200;#nozzle pressure in kPa\n",
+      "V2=900;#nozzle velocity in m/s\n",
+      "\n",
+      "#from Table A-6E\n",
+      "v1=2.6883;\n",
+      "h1=1371.4;\n",
+      "\n",
+      "#calculations\n",
+      "V1=m*v1/A1;\n",
+      "print'the inlet velocit %f f/s'%round(V1,1);\n",
+      "# Ein - Eout = dEsystem / dt\n",
+      "h2=h1-qout-(V2**2 - V1**2)/2/25037;#factor of 25037 to convert to Btu/lbm\n",
+      "#at this value h2, from Tablw A-6E\n",
+      "T2=662;\n",
+      "print'exit temperature %i F'%T2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the inlet velocit 134.400000 f/s\n",
+        "exit temperature 662 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-6 ,Page No.236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=280;#intial pressure in kPa\n",
+      "P1=100;#intial temp in K\n",
+      "m=0.02;#mass flow rate in kg/s\n",
+      "qout=16;#heat losses in kJ/kg\n",
+      "P2=600;#final pressure in kPa\n",
+      "T2=400;#final temp in K\n",
+      "\n",
+      "#from Table A-17\n",
+      "h1=280.13;\n",
+      "h2=400.98;\n",
+      "\n",
+      "#calculations\n",
+      "# Ein - Eout = dEsystem / dt\n",
+      "Win=m*qout+m*(h2-h1);\n",
+      "print'the input power of compressor %f kW'%round(Win,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the input power of compressor 2.740000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-7 ,Page No.237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=2;#inlet pressure in MPa\n",
+      "T1=400;#inlet temp in K\n",
+      "V1=50.0;#inlet velocity in m/s\n",
+      "z1=10;#inlet height in m\n",
+      "P2=15;#final pressure in MPa\n",
+      "x2=0.9;#final dryness fraction\n",
+      "V2=180.0;#final velocity in m/s\n",
+      "z2=6;#final height in m\n",
+      "Wout=5*1000;#power output in kW\n",
+      "\n",
+      "#from Table A-6\n",
+      "h1=3248.4;\n",
+      "#similarly for P2\n",
+      "hf=225.94;\n",
+      "hfg=2372.3;\n",
+      "\n",
+      "#constants used \n",
+      "g=9.8;#in m/s^2\n",
+      "\n",
+      "#calcualtions\n",
+      "h2=hf+x2*hfg;\n",
+      "print'difference in enthalpies %f kJ/kg'%round((h2-h1),2);\n",
+      "print'difference in kinetic energy %f kJ/kg'%round((V2**2-V1**2)/2/1000,2);#factor of 1000 to convert to kJ/kg\n",
+      "print'difference in potential energy %f kJ/kg'%round(g*(z2-z1)/1000,2);#factor of 1000 to convert to kJ/kg\n",
+      "wout=-((h2-h1)+(V2**2-V1**2)/2/1000+g*(z2-z1)/1000);#factor of 1000 to convert to kJ/kg\n",
+      "print'work done per unit of mass %f kJ/kg'%round(wout,2);\n",
+      "m=Wout/wout;\n",
+      "print'mass flow rate %f kg/s'%round(m,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "difference in enthalpies -887.390000 kJ/kg\n",
+        "difference in kinetic energy 14.950000 kJ/kg\n",
+        "difference in potential energy -0.040000 kJ/kg\n",
+        "work done per unit of mass 872.480000 kJ/kg\n",
+        "mass flow rate 5.730000 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-8 ,Page No.239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=0.8;#entering pressure in MPa\n",
+      "P2=0.12;#throttled pressure in MPa\n",
+      "\n",
+      "#from Table A-12\n",
+      "#sat. liq at P1\n",
+      "T1=31.31;\n",
+      "h1=95.47;\n",
+      "#since process is insentropic and at  P2\n",
+      "h2=h1;\n",
+      "hf=22.49;\n",
+      "hg=236.97;\n",
+      "T2=-22.32;\n",
+      "\n",
+      "#calculations\n",
+      "x2=(h2-hf)/(hg-hf);\n",
+      "print'the final state is %f'%round(x2,3);\n",
+      "dT=T2-T1;\n",
+      "print'temperature drop %f C'%round(dT,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final state is 0.340000\n",
+        "temperature drop -53.630000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-9 ,Page No.241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=140;#inlet temp of hot water in F\n",
+      "T2=50;#inlet temp of cold water in F\n",
+      "T3=110;#steady state output temp in F\n",
+      "P=20;#operating pressure in psia\n",
+      "\n",
+      "#for a compressed liq at given temp\n",
+      "h1=107.99;\n",
+      "h2=18.07;\n",
+      "h3=78.02;\n",
+      "\n",
+      "#calculations\n",
+      "#Mass balance min =  mout So, m1+m2 = m3\n",
+      "#Energy balance Ein = Eout So, m1*h1 + m2*h2 = m3*h3\n",
+      "#combining realations\n",
+      "#m1*h1 + m2*h2 = (m1+m2)*h3\n",
+      "#dividing by m2 and y=m1/m2\n",
+      "#we get, yh1 + h2 = (y+1)*h3\n",
+      "y=(h3-h2)/(h1-h3);\n",
+      "print'the ratio of mass flow rates %f'%round(y,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the ratio of mass flow rates 2.000000\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-10 ,Page No.242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=15;#water inlet temp in C\n",
+      "P1=300;#water inlet pressure in kPa\n",
+      "T2=25;#water outlet temp in C\n",
+      "T3=70;#R-134a inlet temp in C\n",
+      "P3=1000;#R-134a inlet pressure in kPa\n",
+      "T4=35;#R-134a outlet temp in C\n",
+      "mr=6;#mass flow rate in kg/min\n",
+      "\n",
+      "#from Table A-4, A-13 and A-11\n",
+      "h1=62.982;\n",
+      "h2=104.83;\n",
+      "h3=303.85;\n",
+      "h4=100.87;\n",
+      "\n",
+      "#calculations\n",
+      "#mass balance m1=m2=mw and m3=m4=mr\n",
+      "#energy balance m1*h1 + m3*h3 =  m2*h2 + m4*h4\n",
+      "#combining them mw*(h1-h2) = mr*(h4-h3)\n",
+      "mw= mr*(h4-h3)/(h1-h2);\n",
+      "print'mass flow rate of cooling water %f kg/min'%round(mw,1);\n",
+      "Qin=mw*(h2-h1);\n",
+      "print'heat transfer rate %i kJ/min'%round(Qin)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass flow rate of cooling water 29.100000 kg/min\n",
+        "heat transfer rate 1218 kJ/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-11 ,Page No.245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=17+273;#inlet temp in K\n",
+      "P1=100;#inlet pressure in kPa\n",
+      "V1=150;#inlet volumetric rate in m^3/min\n",
+      "Win=15;#rated power in kW\n",
+      "Qout=200/1000;#heat lost in kJ/s\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kPa-m^3/kg-K\n",
+      "cp=1.005;#in kJ/kg C\n",
+      "\n",
+      "#calculations\n",
+      "v1=R*T1/P1;\n",
+      "m=V1/v1/60;#factor of 6 to convert to s\n",
+      "# Win - Qout = m*cp*(T2-T1)\n",
+      "T2= T1 + (Win - Qout)/(m*cp);\n",
+      "print'exit temperature %f C'%round((T2-273),2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "exit temperature 21.970000 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5-13 ,Page No.250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V=6.0/1000;#volumne of cooker in m^3\n",
+      "Pgage=75;#operating pressure in kPa\n",
+      "Patm=100;#atmospheric pressure in kPa\n",
+      "m1=1;#mass of water in kg\n",
+      "Qind=0.5;#heat supplying rate in kJ/sec ; d stands for .\n",
+      "t=30*60;#operating time in s\n",
+      "\n",
+      "#calculation\n",
+      "Pabs=Pgage+Patm;\n",
+      "#from Table A-5, ths saturation temp \n",
+      "T=116.04;\n",
+      "print'the temperature at which cooking takes place %f C'%T;\n",
+      "#mass balance me=(m1-m2)cv\n",
+      "#energy balance Qin - mehe = (m2u2 - m1u1)cv\n",
+      "Qin=Qind*t;\n",
+      "#from Table A-5\n",
+      "he=2700.2;\n",
+      "vf=0.001;\n",
+      "vg=1.004;\n",
+      "uf=486.82;\n",
+      "ufg=2037.7;\n",
+      "v1=V/m1;\n",
+      "x1=(v1-vf)/(vg-vf);\n",
+      "u1=uf+x1*ufg;\n",
+      "U=m1*u1;\n",
+      "#Qin = (m1 - V/v2)*he + (V/v2*u2 - m1*u1)\n",
+      "#v2=vf + x2*(vg-vf)\n",
+      "#u2=uf +  x2*ufg\n",
+      "#combining these equations we get\n",
+      "#solved using EES\n",
+      "x2=0.009;\n",
+      "v2=vf + x2*(vg-vf);\n",
+      "m2=V/v2;\n",
+      "print'amount of water left %f kg'%round(m2,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature at which cooking takes place 116.040000 C\n",
+        "amount of water left 0.600000 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter6_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter6_1.ipynb
new file mode 100755
index 00000000..0cebc0e1
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter6_1.ipynb
@@ -0,0 +1,315 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: The Second Law of Thermodynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-1 ,Page No.286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "QH=80.0;#rate of heat transfer in MW\n",
+      "QL=50.0;#rate of waste heat rejection in MW\n",
+      "\n",
+      "#calculations\n",
+      "Wnet=QH-QL;\n",
+      "print'net power output %i MW'%Wnet\n",
+      "nth=Wnet/QH;\n",
+      "print'the thermal efficiency is %f '%round(nth,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "net power output 30 MW\n",
+        "the thermal efficiency is 0.375000 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-2 ,Page No.286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Wnet=65;#power output in hp\n",
+      "nth=0.24;#effficiency of car\n",
+      "HV=19000;#heating value in Btu/lbm\n",
+      "\n",
+      "#calculations\n",
+      "QH=Wnet/nth*2545;#factor of 2545 to convert to Btu/h\n",
+      "m=QH/HV;\n",
+      "print'the engine must burn at fuel rate %f lbm/h'%round(m,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the engine must burn at fuel rate 36.300000 lbm/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-3 ,Page No.291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Wnet=2.0;#power input in kW\n",
+      "QL=360.0;#rate of heat rejection in kJ/min\n",
+      "\n",
+      "#calculations\n",
+      "COPR=QL/Wnet/60;#factor of 60 to convert kW to kJ/min\n",
+      "print'coefficient of performance of refrigerator is %i'%COPR;\n",
+      "QH=QL+Wnet*60;#factor of 60 to convert kW to kJ/min\n",
+      "print'heat rejection rate %i kJ/min'%QH\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "coefficient of performance of refrigerator is 3\n",
+        "heat rejection rate 480 kJ/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-4 ,Page No.291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "COP=2.5;#COP in the given conditions of heat pump\n",
+      "QH=80000.0;#heat loss rate in kJ/hr\n",
+      "\n",
+      "#calculations\n",
+      "Wnet=QH/COP;\n",
+      "print'the power consumed %i kJ/h'%Wnet\n",
+      "QL=QH-Wnet;\n",
+      "print'the rate at which heat is absorbed %i kJ/h'%QL\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the power consumed 32000 kJ/h\n",
+        "the rate at which heat is absorbed 48000 kJ/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-5 ,Page No.306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "QH=500.0;#heat recieved per cycle in kJ\n",
+      "TL=30+273.0;#low temperature source in C\n",
+      "TH=652+273.0;#high temperature source in C\n",
+      "\n",
+      "#calculations\n",
+      "nth=1-TL/TH;\n",
+      "print'the thermal efficiency of carnot engine is %f'%round(nth,3);\n",
+      "QL=TL*QH/TH;\n",
+      "print'the amount of heat rejected to the sink per cycle %i kJ'%round(QL)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the thermal efficiency of carnot engine is 0.672000\n",
+        "the amount of heat rejected to the sink per cycle 164 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-6 ,Page No.310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "COP=13.5;#COP of refrigerator\n",
+      "TH=75+460.0;#room temperature in R\n",
+      "TL=35+460.0;#refrigerated temp in R\n",
+      "\n",
+      "#calculations\n",
+      "COPR=1/(TH/TL-1);\n",
+      "if COPR>=COP:\n",
+      "    print('claim is true');\n",
+      "else:\n",
+      "    print('claim is false')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "claim is false\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-7 ,Page No.310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "TL=-5+273.0;#in C\n",
+      "TH=21+273.0;#house temp in C\n",
+      "QH=135000.0/3600;#heat loss rate in kW\n",
+      "\n",
+      "#calculations\n",
+      "COPHP=1/(1-TL/TH);\n",
+      "Wnet=QH/COPHP;\n",
+      "print'minimum power required %f kW'%round(Wnet,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "minimum power required 3.320000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6-8 ,Page No.314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Qrefrig=40.0;#heat load on referigerator in W\n",
+      "COPR=1.3;#COP of referigerator\n",
+      "Wlight=40.0;#rated power of bulb in W\n",
+      "\n",
+      "#calculation\n",
+      "Wrefrig=Qrefrig/COPR;\n",
+      "Wt=Wrefrig+Wlight;\n",
+      "AnHr=365*24;#annual hours\n",
+      "NOH=20*30/3600.0*365;#normal operating hours\n",
+      "AOP=AnHr-NOH;#addtional operating hours\n",
+      "APC=Wt*AOP/1000;#additional power consumption; fator of 1000 to convert to kW\n",
+      "print'increase in power consumption %i kWh/yr'%round(APC);\n",
+      "print'increase in cost S%f/yr'%round((APC)*0.08,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "increase in power consumption 616 kWh/yr\n",
+        "increase in cost S49.300000/yr\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter7_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter7_1.ipynb
new file mode 100755
index 00000000..d5ab12d2
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter7_1.ipynb
@@ -0,0 +1,1049 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: Entropy"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-1 ,Page No.335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Q=750;#heat transfered in kJ\n",
+      "Tsys=300.0;#temperature of liquid-water mixture in K\n",
+      "\n",
+      "#calculations\n",
+      "dSsys=Q/Tsys;\n",
+      "print'Entropy change in the process %f kJ/K'%round(dSsys,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Entropy change in the process 2.500000 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-2 ,Page No.338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Qsink=2000.0;#heat lost in kJ\n",
+      "Qsource=-Qsink;\n",
+      "Tsource=800.0;#temp of heat loss in K\n",
+      "\n",
+      "#calculations\n",
+      "#part - a\n",
+      "Tsink=500.0;\n",
+      "dSsource=Qsource/Tsource;\n",
+      "dSsink=Qsink/Tsink;\n",
+      "Sgena=dSsource+dSsink;\n",
+      "print'entropy generated in part a %f kJ/K'%round(Sgena,1);\n",
+      "#part - b\n",
+      "Tsink=750;\n",
+      "dSsource=Qsource/Tsource;\n",
+      "dSsink=Qsink/Tsink;\n",
+      "Sgenb=dSsource+dSsink;\n",
+      "print'entropy generated in part b %f kJ/K'%round(Sgenb,1);\n",
+      "if Sgena>Sgenb :\n",
+      "    print('part a is more irreversible');\n",
+      "elif(Sgena == Sgenb) :\n",
+      "    print('heat transfer is equally irreversible');\n",
+      "else:\n",
+      "    print('part b is more irreversible');\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "entropy generated in part a 1.500000 kJ/K\n",
+        "entropy generated in part b 0.200000 kJ/K\n",
+        "part a is more irreversible\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-3 ,Page No.341"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=5.0;#mass of refrigerant 134-a in kg\n",
+      "P1=140.0;#intial pressure in kPa\n",
+      "T1=20.0;#temperature in C\n",
+      "P2=100.0;#final pressure in kPa\n",
+      "\n",
+      "#from refrigerant-134a data\n",
+      "#at P1 and T1\n",
+      "s1=1.0624;\n",
+      "v1=0.16544;\n",
+      "#at P2\n",
+      "v2=v1;\n",
+      "vf=0.0007529;\n",
+      "vg=0.19254;\n",
+      "sf=0.07188;\n",
+      "sfg=0.87995;\n",
+      "\n",
+      "#calculations\n",
+      "# vf < v2 <vg\n",
+      "x2=(v2-vf)/(vg-vf);\n",
+      "s2=sf+x2*sfg;\n",
+      "dS=m*(s2-s1);\n",
+      "print'entropy change in the process %f kJ/k'%round(dS,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "entropy change in the process -1.175000 kJ/k\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-4 ,Page No.342"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=3;#mass of water in lbm\n",
+      "P1=20;#pressure in psia\n",
+      "T1=70+460;#temperature in R\n",
+      "Qin=3450;#heat added in Btu\n",
+      "\n",
+      "#from Table A-6E\n",
+      "#at P1 and T1\n",
+      "s1=0.07459;\n",
+      "h1=38.08;\n",
+      "\n",
+      "#calculations\n",
+      "#Ein - Eout = dEsystem\n",
+      "#Qin = m*(h2 - h1)\n",
+      "h2=Qin/m+h1;\n",
+      "#from Table A-6E\n",
+      "#At P2 and h2\n",
+      "s2=1.7761;\n",
+      "dS=m*(s2-s1);\n",
+      "print'entropy change %f Btu/R'%round(dS,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "entropy change 5.105000 Btu/R\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-5 ,Page No.343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=5.0;#entry pressure in MPa\n",
+      "T1=450.0;#entry temperature in C\n",
+      "P2=1.4;#pressure at exit MPa\n",
+      "\n",
+      "#calculations\n",
+      "#Ein - Eout = dEsystem/dt\n",
+      "#Ein = Eout\n",
+      "#Wout = m*(h1-h2)\n",
+      "#At P1 and T1\n",
+      "h1=3317.2;\n",
+      "s1=6.8210;\n",
+      "s2=s1;\n",
+      "#At P2 and s2\n",
+      "h2=2967.4;\n",
+      "Wout=h1-h2;\n",
+      "print'work output per unit mass %f kJ/kg'%round(Wout,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "work output per unit mass 349.800000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-7 ,Page No.352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "#given values\n",
+      "P1=1;#pressure at state 1 in kPa\n",
+      "T1=110.0;#temperature at state 1 in K\n",
+      "P2=5;#pressure at state 2 in kPa\n",
+      "T2=120.0;#temperature at state 2 in K\n",
+      "\n",
+      "#from Table \n",
+      "#At P1 and T1\n",
+      "s1=4.875;\n",
+      "cp1=3.471;\n",
+      "#at P2 and T2\n",
+      "s2=5.145;\n",
+      "cp2=3.486;\n",
+      "\n",
+      "#calculations\n",
+      "#part - a\n",
+      "dSa=s2-s1;\n",
+      "print'change in entropy %f kJ/kg K using tabulated properties'%round(dSa,3);\n",
+      "#part - b\n",
+      "cavg=(cp1+cp2)/2.0;\n",
+      "dSb=cavg*log(T2/T1);\n",
+      "print'change in entropy %f kJ/kg K approximating liquid methane as an incompressible substance'%round(dSb,3);\n",
+      "E=(dSb-dSa)/dSa;\n",
+      "print'Error is %f'%round(E,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in entropy 0.270000 kJ/kg K using tabulated properties\n",
+        "change in entropy 0.303000 kJ/kg K approximating liquid methane as an incompressible substance\n",
+        "Error is 0.121000\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-8 ,Page No.353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Annual power savings has a 0.005% error due to approximation in textbook\n",
+      "\n",
+      "#given data\n",
+      "P1=5;#pressure at state 1 in MPa\n",
+      "V1=0.280;#volimetric flow rate in m^3/s\n",
+      "T1=115;#temperature at state 1 in K\n",
+      "P2=1;#pressure at state 2 in MPa\n",
+      "dt=8760;#time in h/yr\n",
+      "UC=0.075;#unit cost in $\n",
+      "\n",
+      "#from Table\n",
+      "#at P1 and T1\n",
+      "h1=232.3;\n",
+      "s1=4.9945;\n",
+      "p1=422.15;\n",
+      "s2=s1;\n",
+      "#at P2 and s2\n",
+      "h2=222.8;\n",
+      "\n",
+      "#calculations\n",
+      "m=p1*V1;\n",
+      "#Ein - Eout = dEsystem/dt\n",
+      "#Ein = Eout\n",
+      "#Wout = m*(h1-h2)\n",
+      "Wout = m*(h1-h2);\n",
+      "print'maximum amount of power that can be produced %i kW'%round(Wout)\n",
+      "APP=Wout*dt;#annual power production\n",
+      "APS=APP*UC;#annual power savings\n",
+      "print'Annual power savings $%i/yr'%round(APS)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum amount of power that can be produced 1123 kW\n",
+        "Annual power savings $737758/yr\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-9 ,Page No.357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#given data\n",
+      "P1=100.0;#inlet pressure in kPa\n",
+      "T1=290.0;#inlet temperature in K\n",
+      "P2=600.0;#outlet temperature in kPa\n",
+      "T2=330.0;#outlet temperature in K\n",
+      "\n",
+      "#from Table A-17\n",
+      "s02=1.79783;\n",
+      "s01=1.66802;\n",
+      "#Table A-2b\n",
+      "cpavg=1.006;\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kJ/kg -K\n",
+      "\n",
+      "#calculations\n",
+      "#part-a\n",
+      "s21=s02-s01-R*log(P2/P1);#stands for s2 - s1\n",
+      "print'entropy change using property values from air table %f kJ/kg-K'%round(s21,4);\n",
+      "s21=cpavg*log(T2/T1)-R*log(P2/P1);#stands for s2 - s1\n",
+      "print'entropy change using average specific heat %f kJ/kg-K'%round(s21,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "entropy change using property values from air table -0.384400 kJ/kg-K\n",
+        "entropy change using average specific heat -0.384200 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-10 ,Page No.360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=95;#intial pressure in kPa\n",
+      "T1=295;#intial temperature in K\n",
+      "r=8;#ratio of V1/V2\n",
+      "\n",
+      "#calculations\n",
+      "#for closed systems V2/V1 = v2/v1\n",
+      "#At T1\n",
+      "vr1=647.9;\n",
+      "vr2=vr1/r;\n",
+      "#at vr2\n",
+      "T2=662.7;\n",
+      "print'the final temperature %f K'%T2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final temperature 662.700000 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-11 ,Page No.361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=14.0;#intial pressure in psia\n",
+      "T1=50+460.0;#intial temperature in R\n",
+      "T2=320+460.0;#final temperature in R\n",
+      "\n",
+      "#constants used\n",
+      "k=1.667;\n",
+      "\n",
+      "#calculations\n",
+      "P2=P1*(T2/T1)**(k/(k-1));\n",
+      "print'exit pressure %f psia'%round(P2,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "exit pressure 40.500000 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-12 ,Page No.364"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P2=1000;#final pressure in MPa\n",
+      "P1=100;#intial pressure in MPa\n",
+      "\n",
+      "#from Table A-5\n",
+      "#At P2\n",
+      "v1=0.001043;\n",
+      "\n",
+      "#calculations\n",
+      "Wrev=v1*(P2-P1);\n",
+      "print'compressor work as saturated liquid at inlet %f kJ/kg'%round(Wrev,2)\n",
+      "#from Table A-5\n",
+      "#at P1 as sat. vapour\n",
+      "h1=2675.0;\n",
+      "s1=7.3589;\n",
+      "s2=s1\n",
+      "#from Table A-6\n",
+      "#at P2 and s2\n",
+      "h2=3194.5;\n",
+      "Wrev=h2-h1;\n",
+      "print'compressor work as saturated vapor at inlet %f kJ/kg'%round(Wrev,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "compressor work as saturated liquid at inlet 0.940000 kJ/kg\n",
+        "compressor work as saturated vapor at inlet 519.500000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-13 ,Page No.368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "#given data\n",
+      "P1=100;#intial pressure in kPa\n",
+      "T1=300;#intial temperature in K\n",
+      "P2=900;#final pressure in kPa\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kJ/kg -K\n",
+      "\n",
+      "#calculations\n",
+      "#part - a\n",
+      "k=1.4;\n",
+      "Wcomp=k*R*T1/(k-1)*((P2/P1)**((k-1)/k)-1);\n",
+      "print'compression work in case of isentropic compression %f kJ/kg'%round(Wcomp,1);\n",
+      "#part - b\n",
+      "n=1.3;\n",
+      "Wcomp=n*R*T1/(n-1)*((P2/P1)**((n-1)/n)-1);\n",
+      "print'compression work in case of polytropic compression %f kJ/kg'%round(Wcomp,1);\n",
+      "#part - c\n",
+      "Wcomp=R*T1*log(P2/P1);\n",
+      "print'compression work in case of isothermal compression %f kJ/kg'%round(Wcomp,1);\n",
+      "#part - d\n",
+      "Ps=sqrt(P1*P2);\n",
+      "Wcomp=2*n*R*T1/(n-1)*((Ps/P1)**((n-1)/n)-1);\n",
+      "print'compression work in case of two-stage compression with intercooling %f kJ/kg'%round(Wcomp,1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "compression work in case of isentropic compression 263.200000 kJ/kg\n",
+        "compression work in case of polytropic compression 246.400000 kJ/kg\n",
+        "compression work in case of isothermal compression 189.200000 kJ/kg\n",
+        "compression work in case of two-stage compression with intercooling 215.300000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-14 ,Page No.371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=3000;#intial pressure in kPa\n",
+      "T1=400;#intial temperature in C\n",
+      "P2=50;#final pressure in kPa\n",
+      "T2=100;#final temperature in C\n",
+      "Wout=2000;#power output in kW\n",
+      "\n",
+      "#from Table A-6\n",
+      "#at P1\n",
+      "h1=3231.7;\n",
+      "s1=6.9235;\n",
+      "#at 2a\n",
+      "h2a=2682.4;\n",
+      "#from Table A-6\n",
+      "#at 2s\n",
+      "s2s=s1;\n",
+      "sf=1.0912;\n",
+      "sg=7.5937;\n",
+      "hf=340.54;\n",
+      "hfg=2304.7\n",
+      "x2s=(s2s-sf)/(sg-sf);\n",
+      "h2s=hf+x2s*hfg;\n",
+      "nT=(h1-h2a)/(h1-h2s);\n",
+      "print'isentropic efficiency is %f'%round(nT,3)\n",
+      "#Ein = Eout\n",
+      "m=Wout/(h1-h2a);\n",
+      "print'mass flow rate %f kg/s'%round(m,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "isentropic efficiency is 0.667000\n",
+        "mass flow rate 3.640000 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-15 ,Page No.374"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=100;#intial pressure in kPa\n",
+      "T1=285;#intial temperature in K\n",
+      "P2=800;#final pressure in kPa\n",
+      "m=0.2;#mass flow rate in kg/s\n",
+      "nc=0.8;#isentropic efficiency\n",
+      "\n",
+      "#from Table A-17\n",
+      "#at T1\n",
+      "h1=285.14;\n",
+      "Pr1=1.1584;\n",
+      "\n",
+      "#calcualtions\n",
+      "Pr2=Pr1*(P2/P1);\n",
+      "#at Pr2\n",
+      "h2s=517.05;\n",
+      "h2a=(h2s-h1)/nc+h1;\n",
+      "#at h2a\n",
+      "T2a=569.5;\n",
+      "print'exit temperature of air %f K'%T2a;\n",
+      "#Ein = Eout\n",
+      "Wa=m*(h2a-h1);\n",
+      "print'required power input %i kW'%round(Wa)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "exit temperature of air 569.500000 K\n",
+        "required power input 58 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-16 ,Page No.375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "P1=200.0;#intial pressure in kPa\n",
+      "T1=950.0;#intial temperature in K\n",
+      "P2=80.0;#final pressure in kPa\n",
+      "nN=0.92;#isentropic efficiency\n",
+      "\n",
+      "#from Table A-2b\n",
+      "cp=1.099;\n",
+      "k=1.354;\n",
+      "\n",
+      "#calculations\n",
+      "T2s=T1*(P2/P1)**((k-1)/k);\n",
+      "#ein = eout\n",
+      "V2s=sqrt(2*cp*(T1-T2s)*1000);#factor of 1000 for conversion to m^2/s^2\n",
+      "print'maximum possible exit velocity %i m/s'%(V2s);\n",
+      "T2a=T1-nN*(T1-T2s);\n",
+      "print'exit temperature %i K'%round(T2a);\n",
+      "V2a=sqrt(nN*V2s**2);\n",
+      "print'actual exit velocity %i m/s'%(V2a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum possible exit velocity 666 m/s\n",
+        "exit temperature 764 K\n",
+        "actual exit velocity 639 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example 7-17 ,Page No.382"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Qin=1035.0;#rate of heat transfer in W\n",
+      "Tin=20+273.0;#inner temperature in K\n",
+      "Qout=Qin;\n",
+      "Tout=5+273.0;#outer temperature in K\n",
+      "\n",
+      "#calculations\n",
+      "# Sin - Sout + Sgen = dSsystem/dt\n",
+      "Sgen=(Qout/Tout)-(Qin/Tin);\n",
+      "print'entropy generation in the wall %f W/K'%round(Sgen,3);\n",
+      "Ts1=300.0;Ts2=273.0;#Boundary temperatures\n",
+      "Sgen=(Qout/Ts2)-(Qin/Ts1);\n",
+      "print'total entropy generation %f W/K'%round(Sgen,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "entropy generation in the wall 0.191000 W/K\n",
+        "total entropy generation 0.341000 W/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-18 ,Page No.383"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=7;#intial pressure in MPa\n",
+      "T1=450;#intial temperature in K\n",
+      "P2=3;#final pressure in MPa\n",
+      "\n",
+      "#from steam tables\n",
+      "#at P1 and T1\n",
+      "h1=3288.3;\n",
+      "s1=6.6353;\n",
+      "#at P2\n",
+      "h2=h1;\n",
+      "s2=7.0046;\n",
+      "\n",
+      "#calculations\n",
+      "# Sin - Sout + Sgen = dSsystem/dt\n",
+      "Sgen=s2-s1;\n",
+      "print'the entropy generated %f kJ/kg-K'%round(Sgen,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the entropy generated 0.369300 kJ/kg-K\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-19 ,Page No.385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#given data\n",
+      "m=50;#mass of iron block in kg\n",
+      "T1=500.0;#temperature of iron block in K\n",
+      "T2=285.0;#temperature of lake in K\n",
+      "\n",
+      "#from Table A-3\n",
+      "Cavg=0.45;\n",
+      "\n",
+      "#calculations\n",
+      "dSiron=m*Cavg*log(T2/T1);\n",
+      "print'entropy change of the iron block %f kJ/K'%round(dSiron,2);\n",
+      "# Ein - Eout = dEsystem\n",
+      "Qout=m*Cavg*(T1-T2);\n",
+      "dSlake=Qout/T2;\n",
+      "print'entropy change of the lake %f kJ/K'%round(dSlake,2);\n",
+      "# Sin - Sout + Sgen = dSsystem/dt\n",
+      "Sgen=(Qout/T2)+dSiron;\n",
+      "print'entropy change in the process %f kJ/K'%round(Sgen,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "entropy change of the iron block -12.650000 kJ/K\n",
+        "entropy change of the lake 16.970000 kJ/K\n",
+        "entropy change in the process 4.330000 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-20 ,Page No.386"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P=20;#pressure of water in psia\n",
+      "T1=50+460.0;#temperature of water in R\n",
+      "T2=240.0;#steam temperature in F\n",
+      "T3=130.0;#leaving temperauture in F\n",
+      "m1=300.0;#mass flow rate in lbm/min\n",
+      "Qout=180.0;#heat losses in Btu/min\n",
+      "\n",
+      "#from steam tables\n",
+      "#at P and T1\n",
+      "h1=18.07;\n",
+      "s1=0.03609;\n",
+      "#at P and T2\n",
+      "h2=1162.3;\n",
+      "s2=1.7406;\n",
+      "#at P and T3\n",
+      "h3=97.99;\n",
+      "s3=0.18174;\n",
+      "\n",
+      "#calculations\n",
+      "# Qout = m1*h1 + m2*h2 - (m1+m2)*h3\n",
+      "m2= (Qout-m1*h1+m1*h3)/(h2-h3);\n",
+      "m3=m1+m2;\n",
+      "# Sin - sout + Sgen = dSsystem/dt\n",
+      "Sgen=m3*s3-m1*s1-m2*s2+Qout/T1;\n",
+      "print'the rate of entropy generation %f Btu/min R'%round(Sgen,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of entropy generation 8.670000 Btu/min R\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-21 ,Page No.388"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T=100+273.0;#temperature of water in K\n",
+      "Q=-600.0;#heat transferred in kJ\n",
+      "Tb=25+273.0;#temperature of surroundings in K\n",
+      "\n",
+      "#calculation\n",
+      "dSsys=Q/T;\n",
+      "print'entropy change of water %f kJ/K'%round(dSsys,2);\n",
+      "# Sin - sout + Sgen = dSsystem\n",
+      "Sgen= -Q/Tb + dSsys;\n",
+      "print'total entropy generation %f kJ/K'%round(Sgen,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "entropy change of water -1.610000 kJ/K\n",
+        "total entropy generation 0.400000 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-22 ,Page No.394"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi,sqrt\n",
+      "\n",
+      "#given data\n",
+      "T1=20+273.0;#average temperature of inlet in C\n",
+      "T2=24+273.0;#average temperature of compressed air in C\n",
+      "P1=101.0;#pressure of air inlet in kPa\n",
+      "P2=801.0;#pressure of compressed air in kPa\n",
+      "D=3.0/1000.0;#ihole diameter n m\n",
+      "Cdischarge=0.65;#discharge coefficient\n",
+      "ncomp=0.8;#compressor efficiency\n",
+      "nmotor=0.92;#motor efficiency\n",
+      "UC=0.078;#unit cost in $/kWh\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kJ/kg K\n",
+      "k=1.4;\n",
+      "n=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "Win=n*R*T1/(ncomp*(n-1))*((P2/P1)**((n-1)/n)-1);\n",
+      "A=pi*D**2/4;\n",
+      "mair=Cdischarge*(2/(k+1))**(1/(k-1))*P2*A/(R*T2)*sqrt(k*R*1000*2/(k+1)*T2);#factor of 1000 to m^2/s^2\n",
+      "PW=mair*Win;#Power wasted\n",
+      "ES=PW*4200/nmotor;#4200 is operating hours ES stands for Energy savings\n",
+      "print'Energy savings %i kWh/yr'%round(ES);\n",
+      "CS=ES*UC;\n",
+      "print'cost savings %i S/yr'%round(CS)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy savings 11698 kWh/yr\n",
+        "cost savings 912 S/yr\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7-23 ,Page No.398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1=85.6;#pressure of air inlet in kPa\n",
+      "P2=985.6;#pressure of actual compressed air in kPa\n",
+      "P2r=885.6;#pressure of required compressed air in kPa\n",
+      "CC=12000;#current cost in $/yr\n",
+      "\n",
+      "#constants used\n",
+      "n=1.4;\n",
+      "\n",
+      "#calulation\n",
+      "freduction=1-(((P2r/P1)**((n-1)/n)-1)/((P2/P1)**((n-1)/n)-1));\n",
+      "CS=CC*freduction;\n",
+      "print'cost savings %i Dollar/yr'%(round(CS))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "cost savings 719 Dollar/yr\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter8_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter8_1.ipynb
new file mode 100755
index 00000000..e91e1b78
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter8_1.ipynb
@@ -0,0 +1,764 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8: Exergy: A Measure of Work Potential"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-1 ,Page No.426"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "# given data\n",
+      "D=12.0;#diameter of rotor in m\n",
+      "V=10.0;#average velocity in m/s\n",
+      "\n",
+      "# density of air at 25C & 1atm\n",
+      "p=1.18;\n",
+      "\n",
+      "#calculations\n",
+      "ke=(V**2)/2/1000;#factor of 1000 for converting J into kJ\n",
+      "m=p*pi*(D**2)*V/4.0; \n",
+      "MP=m*(ke);\n",
+      "print'Maximum power %f kW'%round(MP,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum power 66.700000 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-2 ,Page No.427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given values\n",
+      "TH=2000.0;#temperature of large furnace in R\n",
+      "T0=77+460.0;#temperature of enviroment in R\n",
+      "Qin=3000.0;#heat rate in Btu/sec\n",
+      "\n",
+      "#calculation\n",
+      "nth=1-(T0/TH);\n",
+      "Wmax=nth*Qin;\n",
+      "Wmax=round(Wmax)\n",
+      "print'the rate of energy flow %i Btu/s'%round(Wmax)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of energy flow 2195 Btu/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-3 ,Page No.429"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Tsink=300.0;#temperature of sink in K\n",
+      "Tsource=1200.0;#temperature of source in K\n",
+      "Qin=500.0;#heat rate in kJ/s\n",
+      "Wuout=180;#power output in W\n",
+      "\n",
+      "#calculations\n",
+      "Wrev=(1-Tsink/Tsource)*Qin;\n",
+      "print'The reversible power %i kW'%round(Wrev);\n",
+      "I=Wrev-Wuout;\n",
+      "print'the irreversiblity rate %i kW'%round(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The reversible power 375 kW\n",
+        "the irreversiblity rate 195 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-4 ,Page No.429"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from scipy.integrate import quad \n",
+      "from pylab import *\n",
+      "\n",
+      "#given data\n",
+      "m=500.0;#mass of iron block in kg\n",
+      "T1=473.0;#temperature of iron block in K\n",
+      "T0=300.0;#temperature of surrondings in K\n",
+      "\n",
+      "\n",
+      "#from Table A-3\n",
+      "cavg=0.45;\n",
+      "\n",
+      "#calculations\n",
+      "def intgrnd1(T): \n",
+      " return ((1-T0/T)*(-m*cavg))#intergrant\n",
+      "Wrev, err = quad(intgrnd1,T1,T0) ;\n",
+      "Qtotal=m*cavg*(T1-T0);\n",
+      "print'The reversible power %i kJ'%round(Wrev);\n",
+      "Wu=0;\n",
+      "I=Wrev-Wu;\n",
+      "print'the irreversiblity rate %i kJ'%round(I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The reversible power 8191 kJ\n",
+        "the irreversiblity rate 8191 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-5 ,Page No.431"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Wrev=8191.0;#reversible work from Ex - 8.4\n",
+      "Wtotal=38925.0;#total work is equivalent to total heat lost to heat engine from Ex - 8.4\n",
+      "TL=278.0;#outdoor temperature in K\n",
+      "TH=300.0;#temperature of house in K\n",
+      "\n",
+      "#calculations\n",
+      "Wrm=Wtotal-Wrev;#work remaining\n",
+      "COPHP=1/(1-TL/TH);\n",
+      "Wd=COPHP*Wrev;#work delivered\n",
+      "PS=Wd+Wrm;\n",
+      "PS=round(PS/1000);#factor of 1000 for converting kJ into MJ\n",
+      "print'Maximum amount of heat that can be supplied %i MJ'%PS"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum amount of heat that can be supplied 142 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-6 ,Page No.434"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "COP=1.0;# the efficiency that the dealer is referring to is the first law efficiency, meaning that for each unit of electric energy (work) consumed, the heater will supply the house with 1 unit of energy (heat). That is,the advertised heater has a COP of 1\n",
+      "TL=283.0;#outdoor temperature in K\n",
+      "TH=294.0;#indoor temperature in K\n",
+      "\n",
+      "#calculations\n",
+      "COPHP=1/(1-TL/TH);\n",
+      "nII=COP/COPHP;\n",
+      "print'the second law efficiency %f'%round(nII,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the second law efficiency 0.037000\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-7 ,Page No.438"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#given data\n",
+      "P1=1000.0;#pressure of compressed air in KPa\n",
+      "V=200.0;#volumne of rigid tank in m^3\n",
+      "T1=300.0;#temperature of enviroment in K\n",
+      "T0=T1;#state of air in tank\n",
+      "P0=100.0;#pressure of enviroment in KPa\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kPa m^3/kg K\n",
+      "\n",
+      "#calculations\n",
+      "m1=P1*V/(R*T1);\n",
+      "O1=R*T0*(P0/P1-1)+R*T0*log(P1/P0);# O refers to exergy\n",
+      "X1=m1*O1/1000;#factor of 1000 for converting kJ into MJ\n",
+      "print'work obtained %i MJ'%round(X1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "work obtained 281 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-8 ,Page No.439"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T0=20+273;#temperature of enviroment in K\n",
+      "P1=0.14;#intail pressure in MPa\n",
+      "T1=-10;#intail temperature in K\n",
+      "P2=0.8;#final pressure in MPa\n",
+      "T2=50;#final temperature in K\n",
+      "\n",
+      "#the properties of refrigerant\n",
+      "#at inlet\n",
+      "h1=246.36;\n",
+      "s1=0.9724;\n",
+      "#at outlet\n",
+      "h2=286.69;\n",
+      "s2=0.9802;\n",
+      "dO=h2-h1-T0*(s2-s1);# O refers to exergy\n",
+      "print'the exergy change of the refrigerant %i kJ/kg'%round(dO)\n",
+      "wmin=dO;\n",
+      "print'the minimum work input that needs to be supplied is %i kJ/kg'%wmin"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the exergy change of the refrigerant 38 kJ/kg\n",
+        "the minimum work input that needs to be supplied is 38 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-10 ,Page No.447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given values\n",
+      "Q=1035.0;#rate of heat transfer in W\n",
+      "T0=273.0;#outdoor temperature in C\n",
+      "Tin=293.0;#inner wall surface temperature in K\n",
+      "Tout=278.0;#outer wall surface temperature in K\n",
+      "T1=300.0;#indoor temperature in C\n",
+      "\n",
+      "#calculations\n",
+      "#Xin - Xout - Xdestroyed = dX/dt\n",
+      "Xdestroyed=Q*(1-T0/Tin)-Q*(1-T0/Tout);\n",
+      "Xdestroyed=round(Xdestroyed);\n",
+      "print'the rate of exergy destroyed %i W'%round(Xdestroyed);\n",
+      "#the total rate of exergy destroyed\n",
+      "Xdestroyed=Q*(1-T0/T1)-Q*(1-T0/T0);\n",
+      "print'the total of exergy destroyed %f W'%round(Xdestroyed,1);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of exergy destroyed 52 W\n",
+        "the total of exergy destroyed 93.100000 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-11 ,Page No.448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "m=0.05;#mass of steam in kg\n",
+      "P1=1000;#intial pressure in MPa\n",
+      "T1=300+273;#intial temperature in K\n",
+      "P2=200;#final pressure in MPa\n",
+      "T2=150+273;#final temperature in K\n",
+      "P0=100;#pressure of surroundings in KPa\n",
+      "T0=25+273;#temperature of surroundings in K\n",
+      "Qout=2;#heat losses in kJ\n",
+      "\n",
+      "#from Table A-6 & A-4\n",
+      "u1=2793.7;\n",
+      "v1=0.25799;\n",
+      "s1=7.1246;\n",
+      "u2=2577.1;\n",
+      "v2=0.95986;\n",
+      "s2=7.2810;\n",
+      "u0=104.83;\n",
+      "v0=0.00103;\n",
+      "s0=0.3672;\n",
+      "\n",
+      "#calculations\n",
+      "X1=m*(u1-u0-T0*(s1-s0)+P0*(v1-v0));\n",
+      "X2=m*(u2-u0-T0*(s2-s0)+P0*(v2-v0));\n",
+      "print'exergy of intial state %i kJ'%round(X1);\n",
+      "print'exergy of final state %f kJ'%round(X2,1);\n",
+      "dX=X2-X1;\n",
+      "print'exergy change in system %f kJ'%round(dX,1);\n",
+      "Wout=-Qout-m*(u2-u1);\n",
+      "Wu=Wout-P0*m*(v2-v1);\n",
+      "Xdestroyed=X1-X2-Wu;\n",
+      "print'the exergy destroyed %f kJ'%round(Xdestroyed,1);\n",
+      "nII=Wu/(X1-X2);\n",
+      "print'second law efficiency of this process is %f'%round(nII,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "exergy of intial state 35 kJ\n",
+        "exergy of final state 25.400000 kJ\n",
+        "exergy change in system -9.700000 kJ\n",
+        "the exergy destroyed 4.300000 kJ\n",
+        "second law efficiency of this process is 0.551000\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-12 ,Page No.451"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from scipy.integrate import quad \n",
+      "from pylab import *\n",
+      "\n",
+      "#given data\n",
+      "m=2.0;#mass of air in lbm\n",
+      "T0=70+460.0;#intial temperature in R\n",
+      "P1=20.0;#intial pressure in psia\n",
+      "T1=70+460.0;#temperature of surroundings in R\n",
+      "T2=130+460.0;#final temperature in R\n",
+      "\n",
+      "#constants used\n",
+      "Cv=0.172;#in Btu/lbm - F\n",
+      "\n",
+      "#calculations\n",
+      "Xdestroyed=T0*m*Cv*log(T2/T1);\n",
+      "print'exergy destroyed %f Btu'%round(Xdestroyed,1);\n",
+      "def intgrnd1(T): \n",
+      " return ((1-T0/T)*m*Cv)#intergrant\n",
+      "Wrev, err = quad(intgrnd1,T1,T2) ;\n",
+      "print'the reversible work %f Btu'%round(Wrev,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "exergy destroyed 19.600000 Btu\n",
+        "the reversible work 1.100000 Btu\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-13 ,Page No.453"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T0=20+273.0;#temperature of surrounding in K\n",
+      "P0=100.0;#pressure of surrounding in kPa\n",
+      "Tiw=30+273.0;#temperature of water in K\n",
+      "mw=100.0;#mass of water in kg\n",
+      "Tii=350+273.0;#temperature of block in K\n",
+      "mi=5.0;#mass of block in kg\n",
+      "\n",
+      "#constants used(Table A-3)\n",
+      "cw=4.18;#specific heat of water in kJ/kg C\n",
+      "ci=0.45;#specific heat of iron in kJ/kg C\n",
+      "\n",
+      "#calculations\n",
+      "Tfk=(mi*ci*Tii+mw*cw*Tiw)/(mw*cw+mi*ci);\n",
+      "Tfc=Tfk-273;#in C\n",
+      "print'the final equilibrium temperature %f C'%round(Tfc,1);\n",
+      "X1i=mi*ci*(Tii-T0-T0*log(Tii/T0));\n",
+      "X1w=mw*cw*(Tiw-T0-T0*log(Tiw/T0));\n",
+      "X1t=X1i+X1w;#total exergy \n",
+      "print'intial exergy of combined systems %i kJ'%round(X1t);\n",
+      "X2i=mi*ci*(Tfk-T0-T0*log(Tfk/T0));\n",
+      "X2w=mw*cw*(Tfk-T0-T0*log(Tfk/T0));\n",
+      "X2t=X2i+X2w;#total exergy \n",
+      "print'intial exergy of combined systems %f kJ'%round(X2t,1);\n",
+      "Xdestroyed=X1t-X2t;\n",
+      "print'the wasted work %f kJ'%round(Xdestroyed,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final equilibrium temperature 31.700000 C\n",
+        "intial exergy of combined systems 315 kJ\n",
+        "intial exergy of combined systems 95.800000 kJ\n",
+        "the wasted work 219.100000 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-14 ,Page No.455"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "TR=1200.0;#temperature of furnace \n",
+      "T0=300;#temperature of surrounding in K\n",
+      "P0=100;#pressure of surrounding in kPa\n",
+      "Tsys=400.0;#temperature of argon in K\n",
+      "P1=350;#temperature of argon in K\n",
+      "V1=0.01;#intail volumne in m^3\n",
+      "V2=2*V1;#final volumne\n",
+      "\n",
+      "#calculations\n",
+      "W=P1*V1*log(V2/V1);\n",
+      "Wsurr=P0*(V2-V1);\n",
+      "Wu=W-Wsurr;\n",
+      "print'the useful work output %f kJ'%round(Wu,2);\n",
+      "# Qin - W = m*Cv*dT, Since dt=0\n",
+      "Q=W;\n",
+      "Sgen=Q/Tsys-Q/TR;\n",
+      "Xdestroyed=T0*Sgen;\n",
+      "print'the exergy destroyed %f kJ/K'%round(Xdestroyed,2);\n",
+      "Wrev=T0*Q/Tsys-Wsurr+(1-T0/TR)*Q;\n",
+      "print'the reversible work is done in the process %f kJ'%round(Wrev,2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the useful work output 1.430000 kJ\n",
+        "the exergy destroyed 1.210000 kJ/K\n",
+        "the reversible work is done in the process 2.640000 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-15 ,Page No.460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculation error in textbook in part - b which changes all the following answers\n",
+      "\n",
+      "#given data\n",
+      "m=8.0;#mass flow rate in kg/s\n",
+      "T0=298.0;#temperature of surrounding in K\n",
+      "P0=100.0;#pressure of surrounding in kPa\n",
+      "P1=3000.0;#inlet pressure in kPa\n",
+      "T1=450.0;#inlet temperature in K\n",
+      "P2=200.0;#outlet pressure in kPa\n",
+      "T2=150.0;#outlet temperature in K\n",
+      "Qout=300.0;#heat rate in kW\n",
+      "\n",
+      "#from Table A-6 and A-4\n",
+      "h1=3344.9;\n",
+      "s1=7.0856;\n",
+      "h2=2769.1;\n",
+      "s2=7.2810;\n",
+      "h0=104.83;\n",
+      "s0=0.3672;\n",
+      "\n",
+      "#calculations\n",
+      "# Ein = Eout\n",
+      "Wout=m*(h1-h2)-Qout;\n",
+      "Wout=round(Wout)\n",
+      "print'the actual power output %i kW'%Wout\n",
+      "# Xin = Xout\n",
+      "Wrev=m*(h1-h2)-(T0*(s1-s2));\n",
+      "Wrev=round(Wrev);\n",
+      "print'the maximum possible work output %i kW'%Wrev;\n",
+      "nII=Wout/Wrev;\n",
+      "print'second law efficiency is %f'%round(nII,3);\n",
+      "Xdestroyed=Wrev-Wout;\n",
+      "print'the exergy destroyed %i kW'%round(Xdestroyed);\n",
+      "X1=h1-h0-T0*(s1-s0);\n",
+      "print'the exergy of the steam at inlet conditions %i kJ/kg'%round(X1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the actual power output 4306 kW\n",
+        "the maximum possible work output 4665 kW\n",
+        "second law efficiency is 0.923000\n",
+        "the exergy destroyed 359 kW\n",
+        "the exergy of the steam at inlet conditions 1238 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-16 ,Page No.462"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T0=70+460;#temperature of surroundings in R\n",
+      "T1=50;#temperature of water in F\n",
+      "T2=240;#temperature of steam in F\n",
+      "T3=130;#outlet temperature in F\n",
+      "#as dicussed in example 7-20\n",
+      "m1=300;#mass flow rate of water in lbm/min\n",
+      "m2=22.7;#mass flow rate of steam in lbm/min\n",
+      "m3=322.7;#outlet mass flow rate  in lbm/min\n",
+      "\n",
+      "#from steam tables\n",
+      "h1=18.07;\n",
+      "s1=0.03609;\n",
+      "h2=1162.3;\n",
+      "s2=1.7406;\n",
+      "h3=97.99;\n",
+      "s3=0.18174;\n",
+      "\n",
+      "#calculations\n",
+      "Wrev=m1*(h1-T0*s1)+m2*(h2-T0*s2)-m3*(h3-T0*s3);\n",
+      "Wrev=round(Wrev);\n",
+      "print'the reversible power %i Btu/min'%round(Wrev)\n",
+      "Xdestroyed=Wrev;\n",
+      "print'the rate of exergy destruction %i Btu/min'%Xdestroyed\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the reversible power 4588 Btu/min\n",
+        "the rate of exergy destruction 4588 Btu/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8-17 ,Page No.463"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V=200.0;#volumne of rigid tank in m^3\n",
+      "P1=100.0;#intial surroundings in kPa\n",
+      "P2=1000.0;#final pressure in kPa\n",
+      "P0=100.0;#pressure of surroundings in kPa\n",
+      "T=300.0;#temperature of surroundings in K\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kPa m^3/kg K\n",
+      "\n",
+      "#calculations\n",
+      "#Xin - Xout = Xdestroyed = X2 - X1\n",
+      "m2=P2*V/(R*T);\n",
+      "X2=R*T*(log(P2/P0)+P0/P2-1);\n",
+      "Wrev=m2*X2/1000;\n",
+      "Wrev=round(Wrev);\n",
+      "print'Work requirement %i MJ'%Wrev\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work requirement 281 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter9_1.ipynb b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter9_1.ipynb
new file mode 100755
index 00000000..1008c6c0
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/Chapter9_1.ipynb
@@ -0,0 +1,540 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Gas Power Cycles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-2 ,Page No.498"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=17+273;#temperature of air in K\n",
+      "P1=100;#pressure of air in kPa\n",
+      "r=8.0;#compression ratio i.e v1/v2\n",
+      "qin=800;#heat transfer rate in kJ/kg\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kPa-m^3/kg-K\n",
+      "\n",
+      "#from Table A-17\n",
+      "#at T1\n",
+      "u1=206.91;\n",
+      "vr1=676.1;\n",
+      "\n",
+      "#calculations\n",
+      "#Process 1-2\n",
+      "vr2=vr1/r;\n",
+      "#at this vr2\n",
+      "T2=652.4;\n",
+      "u2=475.11;\n",
+      "P2=P1*(T2/T1)*(r);\n",
+      "#Process 2-3\n",
+      "u3=qin+u2;\n",
+      "#at this u3\n",
+      "T3=1575.1;\n",
+      "vr3=6.108;\n",
+      "P3=P2*(T3/T2)*1;#factor of 1 as v3=v2\n",
+      "print'maximum temperature in the cycle %f K'%T3;\n",
+      "print'maximum pressure %f MPa'%round((P3/1000),3);#factor of 1000 to convert into MPa\n",
+      "#Process 3-4\n",
+      "vr4=r*vr3;\n",
+      "#at this vr4 \n",
+      "T4=795.6;\n",
+      "u4=588.74;\n",
+      "#Process 4-1\n",
+      "qout=u4-u1;\n",
+      "Wnet=qin-qout;\n",
+      "print'net work output %f kJ/kg'%round(Wnet,2);\n",
+      "nth=Wnet/qin;\n",
+      "print'thermal efficiency is %f'%round(nth,3);\n",
+      "v1=R*T1/P1;\n",
+      "MEP=Wnet/(v1*(1-1/r));\n",
+      "MEP=round(MEP);\n",
+      "print'mean effective pressure %i kPa'%(MEP)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum temperature in the cycle 1575.100000 K\n",
+        "maximum pressure 4.345000 MPa\n",
+        "net work output 418.170000 kJ/kg\n",
+        "thermal efficiency is 0.523000\n",
+        "mean effective pressure 574 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-3 ,Page No.501"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V1=117.0;#intial volumne in in^3\n",
+      "T1=80+460.0;#intial temperature in R\n",
+      "P1=14.7;#intial pressure in psia\n",
+      "r=18.0;#compression ratio\n",
+      "rc=2.0;#cut-off ratio\n",
+      "\n",
+      "#constants used\n",
+      "R=0.3704;#in psia ft^3/lbm R\n",
+      "cp=0.240;#in Btu/lbm R\n",
+      "cv=0.171;#in Btu/lbm R\n",
+      "\n",
+      "#from Table A-2Ea\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "V2=V1/r;\n",
+      "V3=rc*V2;\n",
+      "V4=V1;\n",
+      "#Process 1-2\n",
+      "T2=T1*(V1/V2)**(k-1);\n",
+      "P2=P1*(V1/V2)**k;\n",
+      "T2=round(T2);\n",
+      "P2=round(P2);\n",
+      "print('Process 1-2');\n",
+      "print'temperature %i R'%T2;\n",
+      "print'pressure %i psia'%P2;\n",
+      "#Process 2-3\n",
+      "P3=P2;\n",
+      "T3=T2*(V3/V2);\n",
+      "T3=round(T3);\n",
+      "P3=round(P3);\n",
+      "print('Process 2-3');\n",
+      "print'temperature %i R'%T3;\n",
+      "print'pressure %i psia'%P3;\n",
+      "#Process 3-4\n",
+      "T4=T3*(V3/V4)**(k-1);\n",
+      "P4=P3*(V3/V4)**k;\n",
+      "T4=round(T4);\n",
+      "print('Process 3-4');\n",
+      "print'temperature %i R'%T4;\n",
+      "print'pressure %f psia'%round(P4,1);\n",
+      "m=P1*V1/(R*T1)/1728;#factor of 1728 to covert to ft^3 from in^3\n",
+      "Qin=m*cp*(T3-T2);\n",
+      "Qout=m*cv*(T4-T1);\n",
+      "Wnet=Qin-Qout ;\n",
+      "print'work output %f Btu'%round(Wnet,3);\n",
+      "nth=Wnet/Qin;\n",
+      "print'thermal efficiency is %f'%round(nth,3);\n",
+      "MEP=Wnet/(V1-V2)*778.17*12;#factor of 778.17 and 12 to convert to lbf ft and in from Btu and ft respectively\n",
+      "MEP=round(MEP);\n",
+      "print'mean effective pressure %i psia'%MEP"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Process 1-2\n",
+        "temperature 1716 R\n",
+        "pressure 841 psia\n",
+        "Process 2-3\n",
+        "temperature 3432 R\n",
+        "pressure 841 psia\n",
+        "Process 3-4\n",
+        "temperature 1425 R\n",
+        "pressure 38.800000 psia\n",
+        "work output 1.296000 Btu\n",
+        "thermal efficiency is 0.633000\n",
+        "mean effective pressure 110 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-5 ,Page No.511"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "T1=300;#compressor inlet temperature in K\n",
+      "r=8;#pressure ratio\n",
+      "T3=1300;#turbine inlet temperature in K\n",
+      "\n",
+      "#calcualtions\n",
+      "#Process 1-2\n",
+      "#at T1\n",
+      "h1=300.19;\n",
+      "Pr1=1.386;\n",
+      "Pr2=r*Pr1;\n",
+      "#at Pr2\n",
+      "T2=540;\n",
+      "h2=544.35;\n",
+      "print'temperature at exit of compressor %i K'%T2;\n",
+      "#Process 3-4\n",
+      "#at T3\n",
+      "h3=1395.17;\n",
+      "Pr3=330.9;\n",
+      "Pr4=Pr3/r;\n",
+      "#at Pr4\n",
+      "T4=770;\n",
+      "h4=789.37;\n",
+      "print'temperature at turbine exit %i K'%T4;\n",
+      "Win=h2-h1;\n",
+      "Wout=h3-h4;\n",
+      "rbw=Win/Wout;\n",
+      "print'back work ratio is %f'%round(rbw,3);\n",
+      "qin=h3-h2;\n",
+      "Wnet=Wout-Win;\n",
+      "nth=Wnet/qin;\n",
+      "print'thermal efficeincy is %f'%round(nth,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature at exit of compressor 540 K\n",
+        "temperature at turbine exit 770 K\n",
+        "back work ratio is 0.403000\n",
+        "thermal efficeincy is 0.425000\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-6 ,Page No.513"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#from 9.5\n",
+      "Wsc=244.16;#work input of compressor in kJ/kg\n",
+      "Wst=606.60;#work output of turbine in kJ/kg\n",
+      "h1=300.19;#process 1-2 data\n",
+      "h3=1395.17;#process 3-4 data\n",
+      "\n",
+      "#given data\n",
+      "nC=0.8;#compressor effciency \n",
+      "nT=0.85;#turbine efficiency\n",
+      "\n",
+      "#calculations\n",
+      "Win=Wsc/nC;\n",
+      "Wout=nT*Wst;\n",
+      "rbw=Win/Wout;\n",
+      "print'back work ratio is %f'%round(rbw,3);\n",
+      "h2a=h1+Win;\n",
+      "qin=h3-h2a;\n",
+      "Wnet=Wout-Win;\n",
+      "nth=Wnet/qin;\n",
+      "print'thermal efficeincy is %f'%round(nth,3)\n",
+      "h4a=h3-Wout;\n",
+      "#from A-17 at h4a\n",
+      "T4a=853;\n",
+      "print'turbine exit temperature %i K is'%T4a\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "back work ratio is 0.592000\n",
+        "thermal efficeincy is 0.266000\n",
+        "turbine exit temperature 853 K is\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-7 ,Page No.516"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#from 9.6\n",
+      "h2a=605.39;#entalpy of air leaving the compressor in kJ/kg\n",
+      "h4a=880.36;#entalpy of air leaving the turbinein kJ/kg\n",
+      "h3=1395.97;#in kJ/kg\n",
+      "Wnet=210.41;#net woek done in kJ/kg\n",
+      "\n",
+      "#given data\n",
+      "n=0.80;#thermal efficiency\n",
+      "\n",
+      "#calculations\n",
+      "# n = (h5 - h2a) / (h4a - h2a)\n",
+      "h5=(h4a - h2a)*n+h2a;\n",
+      "qin=h3-h5;\n",
+      "nth=Wnet/qin;\n",
+      "print'thermal efficeincy is %f'%round(nth,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thermal efficeincy is 0.369000\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-8 ,Page No.519"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "T1=300;#inlet temperature of compressor in K\n",
+      "T6=1300;#inlet temperature of turbine in K\n",
+      "r=8;#overall compression ratio\n",
+      "\n",
+      "#calculations\n",
+      "#as it is case of intercooling\n",
+      "ri=sqrt(r);#ri stands for P2/P1 = P4/P3 = P6/P7 = P8/P9\n",
+      "#from A-17 at T1\n",
+      "h1=300.19;\n",
+      "Pr1=1.386;\n",
+      "Pr2=ri*Pr1;\n",
+      "#from A-17 at Pr2\n",
+      "T2=403.3;\n",
+      "h2=403.31;\n",
+      "#from A-17 at T6\n",
+      "h6=1395.97;\n",
+      "Pr6=330.9;\n",
+      "Pr7=Pr6/ri;\n",
+      "#from A-17 at Pr7\n",
+      "T7=1006.4;\n",
+      "h7=1053.33;\n",
+      "#at inlets\n",
+      "T3=T1;h3=h1;T8=T6;h8=h6;\n",
+      "#et exits\n",
+      "T4=T2;h4=h2;T9=T7;h9=h7;h5=h7;\n",
+      "Win=2*(h2-h1);\n",
+      "Wout=2*(h6-h7);\n",
+      "Wnet=Wout-Win;\n",
+      "qin=(h6-h4)+(h8-h7);\n",
+      "rbw=Win/Wout;\n",
+      "print'back work ratio is %f'%round(rbw,3);\n",
+      "nth=Wnet/qin;\n",
+      "print'thermal efficeincy is %f'%round(nth,3)\n",
+      "#part - b\n",
+      "print('part - b');\n",
+      "qin=(h6-h5)+(h8-h7);\n",
+      "nth=Wnet/qin;\n",
+      "print'thermal efficeincy is %f'%round(nth,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "back work ratio is 0.301000\n",
+        "thermal efficeincy is 0.359000\n",
+        "part - b\n",
+        "thermal efficeincy is 0.699000\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-9 ,Page No.523"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#given data\n",
+      "m=100;\n",
+      "P1=5;#air pressue in psia\n",
+      "T1=-40+460;#air temperature in R\n",
+      "T4=2000+460;#turbine inlet temperature in R\n",
+      "V1=850;#aircraft velocity in ft/sec\n",
+      "rp=10;#pressure ratio\n",
+      "\n",
+      "#constants used\n",
+      "cp=0.240;#in Btu/lbm F\n",
+      "k=1.4;\n",
+      "\n",
+      "#calculations\n",
+      "#Process 1-2\n",
+      "T2=T1+V1**2/(2*cp)/25037;#factor of 25037 to covert to Btu/lbm\n",
+      "P2=P1*(T2/T1)**(k/(k-1));\n",
+      "#Process 2-3 \n",
+      "P3=rp*P2;\n",
+      "P4=P3;\n",
+      "T3=T2*(P3/P2)**((k-1)/k);\n",
+      "#Win=Wout\n",
+      "T5=T4-T3+T2;\n",
+      "P5=P4*(T5/T4)**(k/(k-1));\n",
+      "T5=round(T5);\n",
+      "print'temperature at turbine exit %i R'%T5;\n",
+      "print'pressure at turbine exit %f psia'%round(P5,1);\n",
+      "#Process 5-6\n",
+      "P6=P1;\n",
+      "T6=T5*(P6/P5)**((k-1)/k);\n",
+      "T6=round(T6);#round off\n",
+      "V6=sqrt(2*cp*(T5-T6)*25037);#factor of 25037 to covert to (ft/s)^2\n",
+      "print'the velocity of nozzle exit %i ft/s'%(round(V6));\n",
+      "Wp=m*(V6-V1)*V1/25037;#factor of 25037 to covert to Btu/lbm\n",
+      "Qin=m*cp*(T4-T3);\n",
+      "nP=Wp/Qin;\n",
+      "print'propulsive efficiency is %f percent'%round(nP*100,1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature at turbine exit 2013 R\n",
+        "pressure at turbine exit 39.600000 psia\n",
+        "the velocity of nozzle exit 3285 ft/s\n",
+        "propulsive efficiency is 22.500000 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9-10 ,Page No.529"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#from 9.2\n",
+      "r=8;#compression ratio i.e v1/v2\n",
+      "T0=290.0;#temperature of air in K\n",
+      "T1=290.0;#temperature in the begining of the isentropic compression process in K\n",
+      "T2=652.4;#temperature in the end of the isentropic compression process in K\n",
+      "T3=1575.1;#temperature after process 2-3 in K\n",
+      "P2=1.7997;#pressure in the end of the isentropic compression process in MPa\n",
+      "P3=4.345;#pressure after process 2-3 in MPa\n",
+      "qin=800.0;#heat transfer rate in kJ/kg\n",
+      "qout=381.83;#heat rejection in kJ/kg\n",
+      "wnet=418.17;#net work in kJ/kg\n",
+      "\n",
+      "#given data\n",
+      "Tsource=1700;#temperature of working fluid at source in K\n",
+      "\n",
+      "#constants used\n",
+      "R=0.287;#in kPa-m^3/kg-K\n",
+      "\n",
+      "#calculations\n",
+      "#s1=s2 ; s3=s4\n",
+      "s03=3.5045;\n",
+      "s02=2.4975;\n",
+      "s32=(s03-s02)-R*log(P3/P2);#s32 stands for s3-s2\n",
+      "xdest23=T0*(s32-qin/Tsource);\n",
+      "Tsink=T1;\n",
+      "xdest41=T0*(-s32+qout/Tsink);\n",
+      "xdestcycle=xdest23+xdest41;\n",
+      "print'exergy destrustion associated with Otto cycle %f kJ/kg'%round(xdestcycle,1);\n",
+      "# X4 = (u4 - u0 )- T0*(s4 - s0) + P0(v4 - v0)\n",
+      "# s4 - s0 = s4 - s1 = s32\n",
+      "# u4 - u0 = u4 - u1 = qout\n",
+      "# v4 - v0 = v4 - v1 = 0\n",
+      "#hence x4 is\n",
+      "X4=qout-T0*s32;\n",
+      "print'exergy destruction of purge stream %f kJ/kg'%round(X4,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "exergy destrustion associated with Otto cycle 245.400000 kJ/kg\n",
+        "exergy destruction of purge stream 163.200000 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/README.txt b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/README.txt
new file mode 100755
index 00000000..2482fb63
--- /dev/null
+++ b/Thermodynamics_An_Engineering_Approach_by_Michael_A_Boles_Yunus_A_Cengel/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Harshit Bajpai
+Course: btech
+College/Institute/Organization: Visvesvaraya National Institute Of Technology, Nagpur
+Department/Designation: Chemical Engineering
+Book Title: Thermodynamics An Engineering Approach
+Author: Michael A. Boles, Yunus A. Cengel
+Publisher: McGraw-Hill College, Boston, MA,
+Year of publication: 2006
+Isbn: 0072884959
+Edition: 5
\ No newline at end of file
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diff --git a/Thermodynamics_Demystified/Chapter1.ipynb b/Thermodynamics_Demystified/Chapter1.ipynb
deleted file mode 100755
index 68c66861..00000000
--- a/Thermodynamics_Demystified/Chapter1.ipynb
+++ /dev/null
@@ -1,209 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:2622b864e241f67942d9af83d84aabbbf519132ac138f1180d8869df35c708b4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 1 : Basic Principles"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.2 : PG-9 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "m=10 # mass in Kg\n",
-      "V=5 # velocity in m/s\n",
-      "\n",
-      "KE=m*V**2/2 # kinetic energy in N-m \n",
-      "print \"The Kinetic Energy is \",round(KE),\" N.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Kinetic Energy is  125.0  N.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 77
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.3 : PG-10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "V= 3*5*20; # Volume of air in m^3 from dimensions\n",
-      "m= 350.0; # mass in kg\n",
-      "g= 9.81; # gavitational acceleration in m/s^2\n",
-      "\n",
-      "rho=m/V;# density\n",
-      "print \" The Density is \",round(rho,3),\"kg/m^3 \\n\"\n",
-      "\n",
-      "v= 1/rho # specific volume of air\n",
-      "print \" The specific volume  is\", round(v,3),\"m^3/kg \\n\"\n",
-      "\n",
-      "gama= rho*g # specific weight of air\n",
-      "print \" The specific weight is\", round(gama,2),\" N/m^3\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Density is  1.167 kg/m^3 \n",
-        "\n",
-        " The specific volume  is 0.857 m^3/kg \n",
-        "\n",
-        " The specific weight is 11.45  N/m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 78
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.4 : PG-13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "h=0.020 # height of mercury in m\n",
-      "gammawater=9810 # specific weight of water in N/m^3\n",
-      "Patm=0.7846*101.3 # atmospheric pressure in kPa from table B.1\n",
-      "\n",
-      "Pgauge=13.6*gammawater*h/1000 # pressure in Pascal from condition gammaHg=13.6*gammawater\n",
-      "\n",
-      "P=(Pgauge+Patm)# absolute pressure in KPa\n",
-      "#result\n",
-      "print \"The Pressure is\",round(P,2),\" kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Pressure is 82.15  kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 79
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.5 : PG-13"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "d=10.0/100 # diameter of cylinder in 'm'\n",
-      "P=600 # pressure in KPa\n",
-      "Patm=100 # atmospheric pressure in Kpa\n",
-      "K=4.8*1000 # spring constant in N/m \n",
-      "\n",
-      "deltax=(P-Patm)*(math.pi*1000*d**2)/(4*K) # by balancing forces on piston\n",
-      "#result\n",
-      "print \"The Compression in spring is\",round(deltax,3),\" m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Compression in spring is 0.818  m\n"
-       ]
-      }
-     ],
-     "prompt_number": 80
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex1.6 : PG-16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "ma=2200 # mass of Automobile 'a' in kg\n",
-      "va=25 #velocity of Automobile 'a' in m/s before collision\n",
-      "va1=13.89 # velocity of Automobile 'a' after collision in m/s\n",
-      "mb=1000 # mass of Automobile 'b' in kg\n",
-      "vb=24.44 #velocity of Automobile 'b' after collision in m/s\n",
-      "\n",
-      "KE1=(ma*va**2)/2 # kinetic energy before collision\n",
-      "KE2=(ma*va1**2)/2+(mb*vb**2)/2 # kinetic energy after collision\n",
-      "U=(KE1-KE2)/1000 # internal energy from conservation of energy principle in kJ\n",
-      "#result\n",
-      "print \"The increase in kinetic energy is of\",round(U,1),\" kJ\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The increase in kinetic energy is of 176.6  kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 81
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter2.ipynb b/Thermodynamics_Demystified/Chapter2.ipynb
deleted file mode 100755
index d8357657..00000000
--- a/Thermodynamics_Demystified/Chapter2.ipynb
+++ /dev/null
@@ -1,322 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:63ec806cc789e6d4e52abc481b4c69dbae196d3a1063d71df1549638fb7b9752"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2:Properties of Pure Substances"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.1, PG-28"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "m=10;  #mass of saturated water in kg\n",
-      "\n",
-      "    # All the necessary values are taken from table C.2\n",
-      " \n",
-      "# part (a)\n",
-      " \n",
-      "P=0.001; # Pressure in MPa\n",
-      "vf=0.001; #specific volume of saturated liquid at 0.001 Mpa in Kg/m^3\n",
-      "vg=129.2; # specific volume of saturated vapour at 0.001 Mpa in Kg/m^3\n",
-      "deltaV=m*(vg-vf) # by properties of pure substance \n",
-      "# result\n",
-      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV),\" m^3 \\n\"\n",
-      "\n",
-      "# part (b) \n",
-      "\n",
-      "P=0.26;  # Pressure in MPa\n",
-      "vf=0.0011; #   specific volume of saturated liquid at 0.26 MPa( it is same from at 0.2 and 0.3 MPa upto 4 decimals)\n",
-      "vg=(P-0.2)*(0.6058-0.8857)/(0.3-0.2)+0.8857;  # specific volume of saturated vapour by interpolation of Values at 0.2 MPa and 0.3 MPa\n",
-      "deltaV=m*(vg-vf) # by properties of pure substance \n",
-      "# result\n",
-      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,2),\" m^3 \\n\"\n",
-      "\n",
-      "# part (c) \n",
-      "P=10;  # Pressure in MPa\n",
-      "vf=0.00145; # specific volume of saturated liquid at 10 MPa\n",
-      "vg=0.01803; # specific volume of saturated vapour at 10 MPa\n",
-      "deltaV=m*(vg-vf) # by properties of pure substance \n",
-      "# result\n",
-      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,4),\" m^3 \\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Volume change at pressure  0.001  MPa is 1292.0  m^3 \n",
-        "\n",
-        "The Volume change at pressure  0.26  MPa is 7.17  m^3 \n",
-        "\n",
-        "The Volume change at pressure  10  MPa is 0.1658  m^3 \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.2, PG-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#    initialization of variables\n",
-      "m=4.0 # mass of water in kg\n",
-      "V=1.0 # volume in m^3\n",
-      "T=150 # temperature of water in degree centigrade\n",
-      "\n",
-      "# TABLE C.1 is used for values in wet region\n",
-      "# Part (a)\n",
-      "P=475.8 # pressure in KPa in wet region at temperature of 150 *C\n",
-      "print \" The pressure is\",round(P,2),\" kPa \\n\"\n",
-      "\n",
-      "# Part (b)\n",
-      "#first we determine the dryness fraction\n",
-      "v=V/m # specific volume of water\n",
-      "vg=0.3928 #  specific volume of saturated vapour @150 degree celsius\n",
-      "vf=0.00109 # specific volume of saturated liquid @150 degree celsius\n",
-      "x=(v-vf)/(vg-vf); # dryness fraction\n",
-      "mg=m*x; # mass of vapour\n",
-      "print \" The mass of vapour present is\",round(mg,3),\" kg \\n\"\n",
-      "\n",
-      "# Part(c) \n",
-      "Vg=mg*vg; # volume of vapour\n",
-      "print \" The volume of vapour is\",round(Vg,3),\" m^3\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The pressure is 475.8  kPa \n",
-        "\n",
-        " The mass of vapour present is 2.542  kg \n",
-        "\n",
-        " The volume of vapour is 0.998  m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.3, PG-29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#    initialization of variables\n",
-      "m=2 # mass of water in kg\n",
-      "P=220 # pressure in KPa\n",
-      "x=0.8 # quality of steam\n",
-      "\n",
-      "# Table C.2 is used for values\n",
-      "\n",
-      "vg=(P-200)*(0.6058-0.8857)/(300-200)+0.8857 # specific volume of saturated vapour @ given pressure by interpolating\n",
-      "vf=0.0011 # specific volume of saturated liquid @ 220 KPa\n",
-      "v=vf+x*(vg-vf)# property of pure substance\n",
-      "V=m*v # total volume\n",
-      "#result\n",
-      "print \"The Total volume of the mixture is \",round(V,3),\" m^3\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Total volume of the mixture is  1.328  m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.4, PG-30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#   initialization of variables\n",
-      "m=2 # mass of water in kg\n",
-      "P=2.2 # pressure in Mpa\n",
-      "T=800 # temperature in degree centigrade\n",
-      " # Table C.3 is used for values\n",
-      "v=0.2467+(P-2)*(0.1972-0.2467)/(2.5-2) # specific volue by interpolatin between 2 and 2.5 MPa\n",
-      "V=m*v # final volume\n",
-      "print \"The Final Volume is\",round(V,3),\" m^3\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Final Volume is 0.454  m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.5, PG-32"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#    initialization of variables\n",
-      "V=0.6 # volume of tyre in m^3\n",
-      "Pgauge=200 # gauge pressure in KPa\n",
-      "T=20+273 # temperature converted to kelvin\n",
-      "Patm=100 # atmospheric pressure in KPa\n",
-      "R=287 # gas constant in Nm/kg.K\n",
-      "Pabs=(Pgauge+Patm)*1000 # calculating absolute pressue in Pa \n",
-      "\n",
-      "m=Pabs*V/(R*T)# mass from ideal gas equation\n",
-      "print \"The Mass of air is\",round(m,2),\" Kg\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Mass of air is 2.14  Kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.6, PG-33"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#    initialization of variables\n",
-      "T=500+273 # temperature of steam in kelvin\n",
-      "rho=24.0 # density in Kg/m^3\n",
-      "R=0.462 # gas constant from Table B.2\n",
-      "v=1/rho # specific volume and density relation\n",
-      "# PART (a)\n",
-      "P=rho*R*T # from Ideal gas equation\n",
-      "print \" PART (a) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (b)\n",
-      "a=1.703 #  van der Waal's constant a value from Table B.7\n",
-      "b=0.00169 # van der Waal's constant b value from Table B.7\n",
-      "P=(R*T/(v-b))-(a/v**2) # Pressure from van der Waal's equation\n",
-      "print \" PART (b) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (c)\n",
-      "a=43.9 # van der Waal's constant a value from Table B.7\n",
-      "b=0.00117 # van der Waal's constant b value from Table B.7\n",
-      "\n",
-      "P=(R*T/(v-b))-(a/(v*(v+b)*math.sqrt(T))) # Redlich-Kwong equation\n",
-      "print \" PART (c) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (d)\n",
-      "Tcr=947.4 # compressibilty temperature from table B.3\n",
-      "Pcr=22100 # compressibility pressure from table B.3\n",
-      "\n",
-      "TR=T/Tcr # reduced temperature\n",
-      "PR=P/Pcr # reduced pressure\n",
-      "Z=0.93 # from compressiblility chart\n",
-      "P=Z*R*T/v # Pressure in KPa\n",
-      "print \" PART (d) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (e)\n",
-      "P=8000 # pressure from steam table @ 500*c and v= 0.0417 m^3\n",
-      "print \" PART (e) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " PART (a) The Pressure is  8571  KPa \n",
-        "\n",
-        " PART (b) The Pressure is  7952  KPa \n",
-        "\n",
-        " PART (c) The Pressure is  7934  KPa \n",
-        "\n",
-        " PART (d) The Pressure is  7971  KPa \n",
-        "\n",
-        " PART (e) The Pressure is  8000  KPa \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter3.ipynb b/Thermodynamics_Demystified/Chapter3.ipynb
deleted file mode 100755
index f646fae2..00000000
--- a/Thermodynamics_Demystified/Chapter3.ipynb
+++ /dev/null
@@ -1,330 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:f6a6bb505322436c23c138c1e60d951ed65e1f5f903d0d7ef3e7664203b1c411"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3:Work and Heat"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.1:PG-45"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialization of variables\n",
-      "m=1            # mass in kg\n",
-      "x=20.0/100.0       #quality of steam\n",
-      "P=200          #constant pressure in kPa\n",
-      "T1=100         #temperature intitial in degree centigrade\n",
-      "T2=400         #temperature final in degree centigrade\n",
-      "\n",
-      "\n",
-      "# first we find initial volume v1 and final volume v2\n",
-      "\n",
-      "# using table C.2\n",
-      "vf=0.001061 # specific volume of saturated liquid in m^3 per kg \n",
-      "vg=0.8857   # specific volume of saturated vapour in m^3 per kg \n",
-      "\n",
-      "v1=vf+x*(vg-vf);\n",
-      "\n",
-      "v2=1.549   # specific volume of steam in m^3 per kg at T2=400*C and P2=0.2MPa\n",
-      "# now calculate work\n",
-      "W=m*P*(v2-v1);   # work done in constant pressure process\n",
-      "#result\n",
-      "print \"Work done is\",round(W,2),\" kJ\" # work is in kJ as pressure was in kPa"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work done is 274.2  kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.2:PG-46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#initialization of variables\n",
-      "D=110.0/1000.0 # diameter of cylinder in m\n",
-      "V1=100e-6 # initial volume@ state 1 in m^3\n",
-      "T1=60.0 # initial temp @ state 1 in *C\n",
-      "T2=200.0 # final temo @ state 2 in *C\n",
-      "M=50 # weight of piston in kg\n",
-      "g=9.81 # gravitational accleration in m/sec^2\n",
-      "Patm=100000 # atmospheric pressure in Pa\n",
-      "A=math.pi*(D**2)/4 # area of piston in m^2\n",
-      "\n",
-      "# BALANCING THE FORCES To GET PRESSURE P\n",
-      "# M.g=P.A-Patm\n",
-      "P=Patm+(M*g/A) # atm pressure is added to get absolute pressure\n",
-      "\n",
-      "v1=0.001017 # specific volume at 60*C and 0.15Mpa pressure\n",
-      "m=V1/v1; # mass of water in kg\n",
-      "\n",
-      "# find volume at state 2 \n",
-      "v2=1.444 # specific volume of steam at 200*C and 0.15 MPa\n",
-      "V2=m*v2 # final volume in m^3\n",
-      "\n",
-      "W=P*(V2-V1)/1000; # work done divided by 1000 to get in kJ\n",
-      "# result\n",
-      "print \"The work done is\",round(W,1),\" kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The work done is 21.5  kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.3:PG-47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#initialization of variables\n",
-      "P1=200      # initial pressure in kPa\n",
-      "V1=2        # initial volume in m^3\n",
-      "P2=100      # final pressure in kPa\n",
-      "C=P1*V1 # isothermal process i.e P.V=constant\n",
-      "#  find final volume \n",
-      "V2=P1*V1/P2 # final volume by P1.V1=P2.V2\n",
-      "\n",
-      "from scipy.integrate import quad\n",
-      " \n",
-      "def integrand(v,C):\n",
-      "    return C/v\n",
-      "W, err =quad(integrand, V1, V2,C ) # itegrating over volume to get work\n",
-      "# result\n",
-      "print \"The Work done by gas is\",int(W),\" kJ\" # answer is approximated in textbook"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Work done by gas is 277  kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.4:PG-48"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "M=100 # mass in kg\n",
-      "d=3 # depth by which mass drops in m\n",
-      "V=0.002 # increased volume in m^3\n",
-      "g=9.8 # gravitational accleration in m/sec^2\n",
-      "Pgauge=100*1000 # gauge pressure in N/m\n",
-      "Patm =100*1000 # atmospheric pressure in N/m\n",
-      "P=Pgauge+Patm # to get absolute pressure\n",
-      "\n",
-      "# calculate work done by paddle wheel\n",
-      "Wpaddlewheel=(-M*g*d) # work is negative as it is done on the system\n",
-      "\n",
-      "# calculate work done on piston it \n",
-      "Wboundary=P*V # area mulitiplied by height is volume thus W=P.V \n",
-      "\n",
-      "#net work\n",
-      "\n",
-      "Wnet=Wpaddlewheel+Wboundary; # Work in joule as SI units are used\n",
-      "print \"The Net Work done is \",round(Wnet),\" J\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Net Work done is  -2540.0  J\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.5:PG-51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "T=100 # torque of shaft in N.m\n",
-      "N=3000 # rotation speed in rpm\n",
-      "omega=(N*2*math.pi/60) # angular velocity in rad/sec\n",
-      "# calculation of power\n",
-      "Wdot=(T*omega); # power is work done per unit time\n",
-      "print \"Power transmitted is\",round(Wdot/746,1),\" hp\" # divided by 746 to convert W into hp\n",
-      "#answer is approximated in textbook"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power transmitted is 42.1  hp\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.6:PG-51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "D=10.0/100    # diameter of cylinder in m\n",
-      "d=50.0/1000   # compression in spring in m\n",
-      "Patm=100000   # atmospheric pressure in Pa\n",
-      "K=10.0*1000   # spring constant converted in  N/m\n",
-      "w=50*9.81 # weight of piston in Newton =mass*gravitational acceleration\n",
-      "\n",
-      "# find the initial pressure in cylinder by force balance\n",
-      "A=(math.pi*D**2)/4; # area of piston\n",
-      "P1=((Patm*A)+w)/A;      # balancing forces on piston P1.A=Patm.A+W\n",
-      "\n",
-      "# work done by air to raise the piston for 50mm if spring not present\n",
-      "Wgas=P1*A*d; # pressure*area= force and Work = Force* displacement\n",
-      "\n",
-      "# work done on spring to compress\n",
-      "Wspring=(K*d**2)/2; # Work in j\n",
-      "\n",
-      "# now total work done by air is sum of two works\n",
-      "Wnet=Wgas+Wspring; # Work in j\n",
-      "# result\n",
-      "print \"The net work done by air is\",round(Wnet,1),\" J\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The net work done by air is 76.3  J\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex3.7:PG-53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# variable initialization\n",
-      "\n",
-      "d=2   # distance travelled by weight in m\n",
-      "m=50  # mass of weight in kg\n",
-      "g=9.8  # gravitaional acceleration in m/sec^2\n",
-      "\n",
-      "# calculation of work in non-quasiequilibrium process\n",
-      "W=m*g*d; # work in joules\n",
-      "\n",
-      "# the work done must be transferred as heat\n",
-      "Q=W;\n",
-      "\n",
-      "print \"The heat that must transfer is \",int(Q),\" Joules\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The heat that must transfer is  980  Joules\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter4.ipynb b/Thermodynamics_Demystified/Chapter4.ipynb
deleted file mode 100755
index 235bb7b4..00000000
--- a/Thermodynamics_Demystified/Chapter4.ipynb
+++ /dev/null
@@ -1,729 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c856359b943f772597028d78558909db2e2132e84bca2ae7745f1dba60abb237"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "Chapter 4:The First Law of Thermodynamics"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.1:PG-62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialization of variables\n",
-      "K=100 # spring constant in kN/m\n",
-      "d=0.8 # dispacement of spring in m\n",
-      " # to get total work we integrate from 0 to 0.8 displacement\n",
-      "x1=0; # lower limit of integration\n",
-      "x2=0.8;  # upper limit of integration\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "# we find work\n",
-      "def integrand(x,K):\n",
-      "    return K*x\n",
-      "\n",
-      "W12, err = quad(integrand, x1, x2, K) # integrating to get work\n",
-      "Q12=W12; # by first law of thermodynamics\n",
-      "print \"The Heat transfer is \",int(Q12),\" J\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Heat transfer is  32  J\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.2:PG-65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "P= 5*746 # power of fan converted in watt\n",
-      "t=1*60*60 # time converted to seconds\n",
-      "\n",
-      "# by first law of thermodynamics Q=delU + W\n",
-      "# Q=0 hence -W=delU\n",
-      "# first we find work input\n",
-      "W=-P*t # work in J\n",
-      "delU=-W # from 1st law\n",
-      "print \"The internal energy increase is \",float(delU),\" J\"\n",
-      "# The answer is approximated in textbook\n",
-      "# our answer is precise\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The internal energy increase is  13428000.0  J\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.3:PG-65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "P=400 # pressure in kPa\n",
-      "T1=200 # initial temperature in degree celsius\n",
-      "V1= 2 # initial volume in m^3\n",
-      "Q=3500 # heat added in kJ\n",
-      "v1=0.5342 # specific volume of steam at 200 degree celcius and 0.4 Mpa pressure from table C.3\n",
-      "u1=2647 # specific internal energy in kJ/kg @ pressure = 0.4 MPa\n",
-      "m=V1/v1 # mass in kg\n",
-      "# we have a relation Between u2 and v2 from 1st law of thermodynamics\n",
-      "v2=1.06 # specific volume at state 2 by trial and error and interpolation\n",
-      "V2=m*v2 \n",
-      "u2=((3500-400*(V2-V1))/m)+2647 # specific internal energy for v2=1.06 by trial and error\n",
-      "\n",
-      "# on interpolation from steam table at 0.4 MPa we get temperature \n",
-      "T2=644 # temperature in degree celsius\n",
-      "print \"The temperature for u2=\",round(u2),\" kJ and  v2 =\",round(v2,3),\" kg/m^3 is \\n \",int(T2),\" degree celsius\"\n",
-      "# this numerical is solved by trial and error thus refer to Appendix C"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature for u2= 3372.0  kJ and  v2 = 1.06  kg/m^3 is \n",
-        "  644  degree celsius\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.4:PG-67"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# initialization of variables\n",
-      "P=400 #  pressure in kPa\n",
-      "T1=200 # initial tmperature in degree celsius\n",
-      "V=2 # initial volume in m^3\n",
-      "Q=3500 # heat added in kJ\n",
-      "\n",
-      "#solution\n",
-      "h1=2860 # initial enthalpy @ 200*C and 400 kPa from steam table\n",
-      "v=0.5342 # specific volume from steam table C.3 \n",
-      "m=V/v;\n",
-      "h2=(Q/m)+h1; # final enthalpy in kJ/kg from energy equation\n",
-      "\n",
-      "# NOW USING THIS ENTHLAPY AND INTERPOlATING FROM STEAM TABLE\n",
-      "\n",
-      "T2=600+(92.6/224)*100\n",
-      "\n",
-      "print \"The Final temperature is  \",int(T2),\" degree Celsius\"\n",
-      "# result is obtained from interpolation on steam table\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Final temperature is   641  degree Celsius\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.5:PG-71"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "T1=300 # initial temperature in degree celsius\n",
-      "T2=700 # final temperature in degree celsius\n",
-      "P=150 # pressure in kPa\n",
-      "m=3 # mass of steam in kg\n",
-      "\n",
-      "# solution\n",
-      "# part (a)\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "# now we make function to integrate\n",
-      "def integrand(T):\n",
-      "    return 2.07+(T-400)/1480\n",
-      "\n",
-      "I, err = quad(integrand, T1, T2) # integrating specific heat over temperature range\n",
-      "delH=m*I #integrate('2.07+(T-400)/1480','T',T1,T2) # expressing as function of temperature and integrating\n",
-      "\n",
-      "print\" The change in Enthalpy is \",int(delH),\" kJ \\n\"\n",
-      " \n",
-      "# part(b)\n",
-      "CPavg=delH/(m*(T2-T1)) # avg value of specific heat at constant pressure\n",
-      "print \" The average value of Cp is \",round(CPavg,2),\" kJ/kg.*C\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The change in Enthalpy is  2565  kJ \n",
-        "\n",
-        " The average value of Cp is  2.14  kJ/kg.*C\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.6,PG-72"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "m=1 # mass of nitrogen in kg\n",
-      "T1=300 # initial temperature in Kelvin\n",
-      "T2=1200 # final temperature in Kelvin\n",
-      "M=28.0 # in kg/kmol\n",
-      "# part(a)\n",
-      "# the enthalpy change is found from gas table in App.E\n",
-      "delh=36777-8723 # from gas table\n",
-      "delH=delh/M \n",
-      "print \" The entalpy change from gas table is \",round(delH),\" kJ/kg \\n\"\n",
-      "\n",
-      "# part (b) \n",
-      "Cp=1.042 # from table B.2\n",
-      "delH=Cp*(T2-T1)\n",
-      "print \" The entalpy change  by assuming constant specific heat is \",round(delH),\" kJ/kg\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The entalpy change from gas table is  1002.0  kJ/kg \n",
-        "\n",
-        " The entalpy change  by assuming constant specific heat is  938.0  kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.7:PG-76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "x=0.7 # quality of steam\n",
-      "P1=200 # initial pressure in kPa\n",
-      "P2=800 # final pressure in kPa\n",
-      "V=2 # volume in m^3\n",
-      "# The values are taken from TABLE C.2\n",
-      "vf1=0.0010 # specific volume of saturated liquid at 200 kPa\n",
-      "vg1=0.8857 # specific volume of saturated gas at 200 kPa\n",
-      "uf1=504.5  # specific internal energy of saturated liquid @ state 1\n",
-      "ug1=2529.5 # speciific internal energy of saturated gas @ state 1\n",
-      "\n",
-      "v1=vf1+x*(vg1-vf1); # specific volume of vapour\n",
-      "m=V/v1\n",
-      "\n",
-      "u1=uf1+x*(ug1-uf1) # specific internal energy of vapour @ state 1\n",
-      "v2=v1 # constant volume process\n",
-      "u2=((0.6761-0.6203)*(3661-3853)/(0.6761-0.6181))+3853 # from steam table @ 800kPa by interpolating\n",
-      "Q=m*(u2-u1) # heat transfer\n",
-      "print \"The heat transfer is \",round(Q,3),\" kJ\"\n",
-      "# The answer in the textbook is approximated"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The heat transfer is  5630.537  kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.8:PG-76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "V=0.02 # volume in m^3\n",
-      "P=400 # pressure in kPa\n",
-      "T1=50+273 # initial temperature in kelvin\n",
-      "T2=700+273 # final temperature in kelvin\n",
-      "Q=50 # heat added in kJ\n",
-      "R=287 # constant for air\n",
-      "Cp=1 # constant for specific heat of air\n",
-      "\n",
-      "# using the ideal gas equation\n",
-      "\n",
-      "m=P*1000*V/(R*T1) # mass of air in kg\n",
-      "W=Q-(m*Cp*(T2-T1)) # work done from first law\n",
-      "# result\n",
-      "print \"The Paddle work is \",round(W,2),\" kJ\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Paddle work is  -6.09  kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.9,PG-77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "V1=2 # initial volume in m^3\n",
-      "V2=0.2 # final volume in m^3\n",
-      "T1=20+273 # temperature in kelvin\n",
-      "P=200 # pressure in kPa\n",
-      "R=0.287 # constant for air\n",
-      "gama=1.4 # polytropic index for air\n",
-      "Cv=0.717 # specific heat at constant volume for air\n",
-      "\n",
-      "#solution\n",
-      "\n",
-      "#using the ideal gas equation\n",
-      "m=(P*V1)/(R*T1) # mass in kg\n",
-      "# process is adiabatic thus\n",
-      "T2=T1*((V1/V2)**(gama-1)) # final temperature\n",
-      "\n",
-      "W=-m*Cv*(T2-T1) # work from first law\n",
-      "print \"The Work is \",int(W),\" kJ\"\n",
-      "# solution is approximated in textbook"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Work is  -1510  kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.10:PG-79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "P1=2000.0 # initial pressure in kPa\n",
-      "T1=600.0 # initial temperature in degree celsius\n",
-      "p2=600.0 # final pressure in kPa\n",
-      "T2=200.0 #   final temperature in degree celsius\n",
-      "d1=0.06 # diameter of inlet pipe in metre\n",
-      "d2=0.120 # diameter of outlet pipe in metre\n",
-      "V1=20.0 # velocity at inlet in m/s\n",
-      "\n",
-      "# solution\n",
-      "# from superheat table C.3 values are noted\n",
-      "v1=0.1996 # specific volume of superheated steam @ 600*C and 2000 kPa\n",
-      "v2=0.3520 # specific volume of superheated steam @ 200*C and 2000 kPa\n",
-      "rho1=1/v1 #  initial density\n",
-      "rho2=1/v2 # final density\n",
-      "A1=(math.pi*d1**2)/4 # inlet area\n",
-      "A2=(math.pi*d2**2)/4 # exit area\n",
-      "\n",
-      "V2=(rho1*A1*V1)/(rho2*A2) # from continuity equation\n",
-      "print \" The Exit velocity is \",round(V2,2),\" m/s \\n\"\n",
-      "\n",
-      "mdot=rho1*A1*V1 # mass flow rate\n",
-      "print\" The mass flow rate is \",round(mdot,3),\" kg/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Exit velocity is  8.82  m/s \n",
-        "\n",
-        " The mass flow rate is  0.283  kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.11:PG-82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "P1=8000 # initial pressure in kPa\n",
-      "T1=300  # temperature in degree celsius\n",
-      "P2=2000 # final pressure in kPa\n",
-      "\n",
-      "# solution\n",
-      "h1=2785 # specific enthalpy of steam in kJ/kg @ 8000 kPa and 300 degree celsius from  steam table\n",
-      "h2=h1 # throttling process thus enthalpy is constant\n",
-      "T2=212.4 # from steam table as we know enthalpy and pressure\n",
-      "hf2=909 # specific enthalpy of saturated liquid @ 2000 kPa and 300 degree celsius\n",
-      "hg2=2799.5 # specific enthalpy of saturated gas @ 2000 kPa and 300 degree celsius\n",
-      "x2=(h2-hf2)/(hg2-hf2) # quality of steam\n",
-      "\n",
-      "vg2=0.0992 # specific    volume of saturated gas @ 2000 kPa and 212.4*c\n",
-      "vf2=0.0012 # specific volume of saturated liquid @ 2000 kPa and 212.4*c\n",
-      "v2=vf2+x2*(vg2-vf2) # from properties of pure substance\n",
-      "\n",
-      "print \"The Final Temperature and Specific volume is \",round(T2,1),\"*C and \",round(v2,3),\" m^3/kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Final Temperature and Specific volume is  212.4 *C and  0.098  m^3/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.12:PG-84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "P1=4000 # inlet pressure in kPa\n",
-      "T1=500 # inlet temperature in degree celsius\n",
-      "V1=200 # inlet steam velocity in m/s\n",
-      "d1=0.05 # inlet diameter in 'm'\n",
-      "P2=80 # exit pressure in kPa\n",
-      "d2=0.250 # exit diameter in 'm'\n",
-      "\n",
-      "# solution\n",
-      "v1=0.08643 # specific volume from steam table @ 4000 kPa and 500*C\n",
-      "v2=2.087 # specific volume from steam table @ 80 kPa and 500*C\n",
-      "rho1=1/v1 # density at inlet\n",
-      "rho2=1/v2 # density at outlet\n",
-      "A1=(math.pi*d1**2)/4 # inlet area\n",
-      "A2=(math.pi*d2**2)/4\n",
-      "mdot=rho1*A1*V1 # mass flow rate\n",
-      "mdot=round(mdot,3) # rounding to 3 significant digits\n",
-      "\n",
-      "#now using table C.3\n",
-      "h1=3445 # initial specific enthalpy @ 4000 kPa and 500 *C \n",
-      "h2=2666 # final specific enthalpy @ 80 kPa and 500 *C\n",
-      "WT=-mdot*(h2-h1) # maximum power from first law\n",
-      "print \" The power output is \",round(WT),\" kJ/s \\n \"\n",
-      "\n",
-      "V2=(A1*V1*rho1)/(A2*rho2) \n",
-      "V2=round(V2) # rounding of digits\n",
-      "delKE=mdot*((V2**2)-(V1**2))/2 # the change in kinetic energy\n",
-      "print \" The change in K.E is \",round(delKE),\" J/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The power output is  3540.0  kJ/s \n",
-        " \n",
-        " The change in K.E is  -6250.0  J/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 78
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.13:PG-85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "Wdot=10 # pump power in hp\n",
-      "g=9.81 # acceleration due to gravity\n",
-      "rho=1000 # density of water in kg/m^3\n",
-      "d1=0.06 # inlet dimeter in 'm'\n",
-      "d2=0.10 # oulet diamter in 'm'\n",
-      "V1=10 # velocity of water at inlet in m/s\n",
-      "\n",
-      "#solution\n",
-      "A1=math.pi*(d1**2)/4 # area of inlet\n",
-      "A2=math.pi*(d2**2)/4 # area of outlet\n",
-      "V2=A1*V1/A2 # oulet velocity from continuity equation\n",
-      "\n",
-      "mdot=rho*A1*V1 # mass flow rate\n",
-      "delP=((((Wdot*746)/mdot)-((V2**2)-V1**2)/(2*g))*rho)/1000 # change in pressure in kPa\n",
-      "print \"The rise in pressure is \",round(delP),\" kPa\"\n",
-      "# The answer is approximated in textbook , our answer is precise \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rise in pressure is  268.0  kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 80
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.14:PG-85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "# initialization of variables\n",
-      "P1=7000.0 # inlet pressure in Pa\n",
-      "T1=420.0 # inlet temperature in degree celsius\n",
-      "V1=400.0 # inlet velocity in m/s\n",
-      "d1=0.200 # inlet diameter in 'm'\n",
-      "V2=700.0 # exit velocity in m/s\n",
-      "k=1.4 # polytopic index for air\n",
-      "Cp=1000 # specific heat at constant pressure for air in j/kg.K\n",
-      "R=287 # specific gas constant for air\n",
-      "\n",
-      "#solution\n",
-      "\n",
-      "#part (a)\n",
-      "T2=(((V1**2)-V2**2)/(2*Cp))+T1 # outlet temperature in degree celsius\n",
-      "print \" The exit temperature is \",round(T2),\" *C \\n\"\n",
-      "\n",
-      "#part (b)\n",
-      "\n",
-      "rho1=P1/(R*(T1+273)) # density at entrance\n",
-      "A1=(math.pi*d1**2)/4\n",
-      "mdot=rho1*A1*V1 # \n",
-      "print \" The mass flow rate is \",round(mdot,3),\" kg/s \\n\"\n",
-      "\n",
-      "# part (c)\n",
-      "\n",
-      "rho2=rho1*(((T2+273)/(T1+273))**(1/(k-1))) # density at exit\n",
-      "# now we find the exit diameter\n",
-      "d2=math.sqrt((rho1*V1*(d1)**2)/(rho2*V2))\n",
-      "print \" The outlet diameter is \",round(d2,3),\" m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The exit temperature is  255.0  *C \n",
-        "\n",
-        " The mass flow rate is  0.442  kg/s \n",
-        "\n",
-        " The outlet diameter is  0.212  m\n"
-       ]
-      }
-     ],
-     "prompt_number": 87
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex4.15:PG-89"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "mdots=100 # mass flow rate of sodium in kg/s\n",
-      "Ts1=450 # inlet temperature of sodium in degree celsius\n",
-      "Ts2=350 # exit temperature of sodium in degree celsius\n",
-      "Cp=1.25 # specific heat of sodium in KJ/kg.*C\n",
-      "Tw1=20 # inlet temperature of water in degree celsius\n",
-      "Pw=5000 # inlet pressure of water in kPa \n",
-      "\n",
-      "# solution\n",
-      "hw1=88.65 # enthalpy from table C.4\n",
-      "hw2=2794 # enthalpy from table C.3\n",
-      "mdotw=(mdots*Cp*(Ts1-Ts2))/(hw2-hw1) # mass flow rate of water\n",
-      "print \" The mass flow rate of water is \",round(mdotw,2),\" kg/s \\n\"\n",
-      "Qdot=mdotw*(hw2-hw1) # heat transfer in kW using energy equation\n",
-      "# result\n",
-      "print \" The rate of heat transfer is \",round(Qdot),\" kW\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The mass flow rate of water is  4.62  kg/s \n",
-        "\n",
-        " The rate of heat transfer is  12500.0  kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 90
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter5.ipynb b/Thermodynamics_Demystified/Chapter5.ipynb
deleted file mode 100755
index a93ee1a4..00000000
--- a/Thermodynamics_Demystified/Chapter5.ipynb
+++ /dev/null
@@ -1,594 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3928ce1341a76de85540172b210e9512b16c46c42e882063576b52bdb907073d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5:The Second Law of Thermodynamics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.4:PG-112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "Th=200+273.0 # higher temperture in kelvin \n",
-      "Tl=20+273.0 # lower temperture in kelvin\n",
-      "Wdot=15 # output of engine in kW\n",
-      "\n",
-      "ef=1-(Tl/Th) # carnot efficiency\n",
-      "\n",
-      "Qhdot=Wdot/ef # heat supplied by reservoir\n",
-      "print \" The heat suppled by higher temperature reservoir is\",round(Qhdot,2),\"kW \\n \"\n",
-      "# using forst law\n",
-      "Qldot=Qhdot-Wdot # heat rejected to reservoir\n",
-      "print \" The heat suppled by lower temperature reservoir is\",round(Qldot,2),\"kW\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The heat suppled by higher temperature reservoir is 39.42 kW \n",
-        " \n",
-        " The heat suppled by lower temperature reservoir is 24.42 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.5:PG-113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "TL1=-5+273.0 # lower temperature in kelvin for first situation\n",
-      "TH=20+273.0 # higher temperature in kelvin\n",
-      "TL2=-25+273.0 #lower temperature in kelvin for second situation\n",
-      "\n",
-      "#solution\n",
-      "\n",
-      "COP1=TL1/(TH-TL1) # carnot refrigerator COP for first situation\n",
-      "# Let Heat be 100 kJ\n",
-      "QL=100.0 # assumption\n",
-      "W1=QL/COP1 # work done for situation 1\n",
-      "\n",
-      "# for situation 2\n",
-      "COP2=TL2/(TH-TL2) # COP carnot for second situation\n",
-      "W2=QL/COP2 # work done\n",
-      "\n",
-      "Per=(W2-W1)*100/W1 # percentage increase in work done \n",
-      "#result\n",
-      "print\" The perccentage increase in work is \",round(Per,1),\"%\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The perccentage increase in work is  94.5 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.6:PG-117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "T1=20+273 # initial temperature in kelvin\n",
-      "P=200 # pressure in kPa\n",
-      "V=2 # volume in m^3\n",
-      "R=0.287 # gas constant for air\n",
-      "W=720 # work done on air in kJ\n",
-      "Cv=0.717 # specific heat at constant volume for air\n",
-      "\n",
-      "#solution\n",
-      "m=(P*V)/(R*T1) # mass of air\n",
-      "\n",
-      "T2=T1+(W/(m*Cv))# final temperature in kelvin\n",
-      "\n",
-      "delS=m*Cv*math.log(T2/T1) # ENROPY CHANGE FOR CONSTANT VOLUME PROCESS\n",
-      "print \" The Entropy increase is\",round(delS,3),\"kJ/K \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Entropy increase is 1.851 kJ/K \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.7:PG-118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "T1=350+273 # initial temperature in kelvin\n",
-      "P1=1200.0 # initial pressure in kPa\n",
-      "P2=140 # final pressure in kPa\n",
-      "k=1.4 # polytopic index for air\n",
-      "Cv=0.717 # specific heat at  constant volume for air\n",
-      "#solution\n",
-      "T2=T1*((P2/P1)**((k-1)/k))  # reversible adiabatic process relation\n",
-      "\n",
-      "w=-Cv*(T2-T1) # work done by gases in reversible adiabatic process\n",
-      "print\" The work done by gases is\",round(w),\"kJ/kg\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The work done by gases is 205.0 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.8:PG-120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "T1=20+273.0 # initial temperature in kelvin\n",
-      "P1=200.0 # pressure in kPa\n",
-      "V=2 #volume in m^3\n",
-      "R=0.287 # gas constant for air\n",
-      "W=-720 # negative as work is done on air in kJ\n",
-      "\n",
-      "#solution\n",
-      "\n",
-      "m=(P1*V)/(R*T1)# mass of air\n",
-      "\n",
-      "u1=209.1 #specific internal energy of air at 293K and 200 kPa from table E.1\n",
-      "s1=1.678 # by interpolation from table E.1\n",
-      "# change  in internal energy= work done\n",
-      "u2=-(W/m)+u1 # final internal energy\n",
-      "T2=501.2# final temperature interpolated from table E.1 corresponding to value of u2\n",
-      "s2=2.222 # value of s from table E.3 by interpolating from corresponding to value of u2\n",
-      "\n",
-      "P2=P1*(T2/T1) # final pressure in kPa\n",
-      "\n",
-      "delS=m*(s2-s1-R*math.log(P2/P1))# entropy change\n",
-      "# result\n",
-      "print \" The Entropy increase is\",round(delS,3),\"kJ/K \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Entropy increase is 1.855 kJ/K \n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.9:PG-120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "T1=350+273.0 # initial temperature in kelvin\n",
-      "P1=1200.0 # initial pressure in kPa\n",
-      "P2=140.0 # final pressure in kPa\n",
-      "k=1.4 # polytopic index for air\n",
-      "\n",
-      "#solution\n",
-      "# The values are taken from table E.1\n",
-      "Pr660=23.13# relative pressure @ 660K\n",
-      "Pr620=18.36# relative pressure @ 620K\n",
-      "Pr1=((Pr660-Pr620)*3/40)+Pr620 # relative pressure by interpolation\n",
-      "\n",
-      "Pr2=Pr1*(P2/P1) # relative pressure at state 2\n",
-      "\n",
-      "Pr340=2.149 # relative pressure @ 340K\n",
-      "Pr380=3.176 # relative pressure @ 380K\n",
-      "T2=((Pr2-Pr340)/(Pr380-Pr340))*40+340 # interpolating final temperature from table E.1\n",
-      "\n",
-      "# now interpolating u1 AND u2 from table E.1\n",
-      "u620=451.0# specific internal energy @ 620k\n",
-      "u660=481.0# specific internal energy @ 660k\n",
-      "u1=(u660-u620)*(3/40.0)+u620 # initial internal energy\n",
-      "\n",
-      "u380=271.7 #specific internal energy @ 380k\n",
-      "u340=242.8 #specific internal energy @ 340k\n",
-      "u2=((Pr2-Pr340)/(Pr380-Pr340))*(u380-u340)+u340 # final internal energy\n",
-      "\n",
-      "w=u2-u1 # work= change in internal energy\n",
-      "print \" The work done by gas is\",int(w),\"kJ/kg\"\n",
-      "# The answer is slightly different as values are approximated in textbook\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The work done by gas is -209 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.10:PG-123"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "T1=300+273.0 # initial temperature in kelvin\n",
-      "P1=600 # initial pressure in kPa\n",
-      "P2=40 # final pressure in kPa\n",
-      "\n",
-      "#solution\n",
-      "#please refer to steam table for values\n",
-      "v1=0.4344 # specific volume from steam table @ 573k and 600 kPa\n",
-      "v2=v1 # rigid container\n",
-      "u1=2801.0 # specific internal energy from steam table @ 573k and 600 kPa\n",
-      "s1=7.372 # specific entropy @ 600 kPa and 573 K\n",
-      "\n",
-      "vg2=0.4625 # specific volume of saturated vapour @ 40 kPa and 573 K\n",
-      "vf2=0.0011 # specific volume of saturated liquid @ 40 kPa and 573 K\n",
-      "sf2=1.777 # specific entropy of saturated liquid @ 40 kPa and 573 K\n",
-      "sg2=5.1197 # specific entropy of saturated vapour @ 40 kPa and 573 K\n",
-      "x=(v2-vf2)/(vg2-vf2)# quality of steam using pure substance relation\n",
-      "\n",
-      "s2=sf2+x*sg2 # overall specific enthalpy at quality 'x' \n",
-      "delS=s2-s1 # entropy change\n",
-      "print\" The entropy change  is\",round(delS,3),\"kJ/kg.K \\n \"\n",
-      "\n",
-      "#heat transfer\n",
-      "uf2=604.3 #specific internal energy of saturated liquid @ 40 kPa and 573 K\n",
-      "ug2=1949.3 #specific internal energy of saturated vapour @ 40 kPa and 573 K\n",
-      "u2=uf2+x*ug2 #specific internal energy @ quality x\n",
-      "q=u2-u1 # heat transfer in kJ/kg from first law as W=0\n",
-      "print \" The heat transfer is\",int(q),\"kJ/kg\"\n",
-      "# result\n",
-      "# the answers are approximated in textbook but here they are precise thus minute difference is there"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The entropy change  is -0.787 kJ/kg.K \n",
-        " \n",
-        " The heat transfer is -366 kJ/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.11:PG-126"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "v1=0.5 # assumed as air is filled in half of the tank\n",
-      "v2=1.0 # final volume when partition is removed\n",
-      "R=0.287 # gas contant for air\n",
-      "#solution\n",
-      "q=0 # heat transfer is zero\n",
-      "w=0 # work done is zero\n",
-      "# temperatue is constant as no change in internal energy by first law\n",
-      "dels=R*math.log(v2/v1)# change in entropy when temperature is constant\n",
-      "print \"The change in specific entropy is\",round(dels,3),\"kJ/kg.K\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The change in specific entropy is 0.199 kJ/kg.K\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.12:PG-127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "T1=400+273.0 # initial temperature in kelvin\n",
-      "P=600 # pressure in kPa\n",
-      "Tsurr=25+273.0 # surrounding temperature in K\n",
-      "m=2 # mass of steam in kg\n",
-      "\n",
-      "#solution\n",
-      "#please refer to steam table for values\n",
-      "s1=7.708 # specific entropy of steam @ 400 degree celsius and 0.6 MPa\n",
-      "s2=1.9316# specific enropy of condensed water @ 25 degree celsius and 0.6 MPa\n",
-      "delSsys=m*(s2-s1) # entropy change in system i.e of steam\n",
-      "\n",
-      "h1=3270 # specific enthalpy of steam @ 400 degree celsius and 0.6 MPa\n",
-      "h2=670.6#specific enropy of condensed water @ 25 degree celsius and 0.6 MPa\n",
-      "\n",
-      "Q=m*(h1-h2)# heat transfer at constant pressure\n",
-      "delSsurr=Q/Tsurr # entropy change in surroundings\n",
-      "\n",
-      "sigma=delSsys+delSsurr # net entropy change\n",
-      "\n",
-      "print \"The net entropy production is\",round(sigma,1),\"kJ/K\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The net entropy production is 5.9 kJ/K\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.13:PG-130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "T1=600+273 # initial temperature in kelvin\n",
-      "P1=2 # initial pressure in MPa\n",
-      "P2=10 # final pressure in kPa\n",
-      "mdot=2 # mass flow rate in kg/s\n",
-      "\n",
-      "#solution\n",
-      "#please refer to steam table for values\n",
-      "h1=3690 # specific enthalpy in kJ/kg @ 2MPa and 600 degree celsius\n",
-      "s1=7.702 #specific entropy in kJ/kg.K @ 2MPa and 600 degree celsius\n",
-      "s2=s1 # Reversible adiabatic process thus entropy is constant\n",
-      "sf2=0.6491 #specific entropy of saturated liquid from steam table @ 10 kPa\n",
-      "sg2=8.151 #specific entropy of saturated vapour from steam table @ 10 kPa\n",
-      "\n",
-      "x2=(s2-sf2)/(sg2-sf2) # quality of steam at turbine exit\n",
-      "\n",
-      "h2f=191.8 #specific enthalpy of saturated liquid from steam table @ 10 kPa\n",
-      "h2g=2584.8 #specific enthalpy of saturated vapour from steam table @ 10 kPa\n",
-      "h2=h2f+x2*(h2g-h2f) # specific enthalpy @ quality 'x' \n",
-      "\n",
-      "WdotT=mdot*(h1-h2)# from work done in adiabatic process\n",
-      "# result\n",
-      "print \" The maximum power output is\",int(WdotT),\"kJ/s\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The maximum power output is 2496 kJ/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.14:PG-130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "T1=600+273 # initial temperature in kelvin\n",
-      "P1=2 # initial pressure in MPa\n",
-      "P2=10 # final pressure in kPa\n",
-      "mdot=2 # mass flow rate in kg/s\n",
-      "EffT=0.8 # efficiency of turbine \n",
-      "WdotT=2496 # theoritical power of turbine in kW\n",
-      "\n",
-      "#solution\n",
-      "Wdota=EffT*WdotT # actual power output of turbine\n",
-      "h1=3690 # specific enthalpy @ 2MPa and 600 degree celsius\n",
-      "h2=h1-(Wdota/mdot) # final enthalpy from first law of thermodynamics\n",
-      "\n",
-      "T2=((h2-2688)/(2783-2688))*(150-100)+100 # by interpolating from steam table @ P2= 10 kPa, h2=2770 \n",
-      "s2=8.46 # final specific entropy by interpolation from steam table\n",
-      "\n",
-      "print \"The temperature by interpolation is\",round(T2),\" degree celsius \\n\"\n",
-      "print \"The final entropy by interpolation is\",round(s2,2),\"kJ/kg.K\"\n",
-      "# The temperature and entropy are found by interpolation from steam table and cannot be shown here.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The temperature by interpolation is 102.0  degree celsius \n",
-        "\n",
-        "The final entropy by interpolation is 8.46 kJ/kg.K\n"
-       ]
-      }
-     ],
-     "prompt_number": 59
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex5.15:PG-131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "T2=250.0 # temperature of steam in degree celsius\n",
-      "mdot2=0.5 # mass flow rate of steam in kg/s\n",
-      "T1=45 # temperature of water in degree celsius\n",
-      "mdot1=4 # mass flow rate of water in kg/s\n",
-      "P=600.0 # pressure in kPa\n",
-      "\n",
-      "\n",
-      "mdot3=mdot1+mdot2 # by mass balance\n",
-      "\n",
-      "h2=2957 # specific enthalpy in kJ/kg of steam @ 600 Kpa from steam table\n",
-      "h1=188.4 # specific enthalpy in kJ/kg  of water @ 600 Kpa from steam table\n",
-      "\n",
-      "h3=(mdot1*h1+mdot2*h2)/mdot3 # specific enthalpy in kJ/kg at exit\n",
-      "\n",
-      "# by interpolation from saturated steam table\n",
-      "T3=(h3-461.3)*10/(503.7-461.3)+110 # temperature of mixture\n",
-      "\n",
-      "sf3=1.508 # entropy of saturated liquid in kJ/kg.K at 600Kpa and T3 temperature from steam table\n",
-      "s3=sf3  # by interpolating sf\n",
-      "s2=7.182 # entropy of superheated steam in kJ/kg.K @ 600Kpa from steam table\n",
-      "s1=0.639 # entropy of entering water in kJ/kg.K at T= 45 degree celsius\n",
-      "\n",
-      "sigmaprod=mdot3*s3-mdot2*s2-mdot1*s1\n",
-      "# result\n",
-      "print \"The rate of entropy production is\",round(sigmaprod,3),\"kW/K \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rate of entropy production is 0.639 kW/K \n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter6.ipynb b/Thermodynamics_Demystified/Chapter6.ipynb
deleted file mode 100755
index 4f406b7e..00000000
--- a/Thermodynamics_Demystified/Chapter6.ipynb
+++ /dev/null
@@ -1,704 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e92a3a85838e85e036e1c07a34c6c3996a0867f06470b08d471a9b83e142be32"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6:Power Vapor Cycles"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.1:PG-146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "# Please refer to the given figure in question for quantities\n",
-      "P2=2*1000 #higher pressure converted in in kPa\n",
-      "P1=10 # lower pressure in kPa\n",
-      "rho=1000 # density of water in Kg/m^3\n",
-      "h1=192 # enthalpy at state 1 in kJ/kg\n",
-      "h3=3248 # enthalpy at state 3 in kJ/kg\n",
-      "s3=7.1279# entropy at state 3 in kJ/kg.K\n",
-      "\n",
-      "#calculation of pump work\n",
-      "wp=(P2-P1)/rho # pump work given by equation 4.56 in textbook\n",
-      "h2=h1+wp # by enrgy balance b/w state 1 and 2\n",
-      "q=h3-h2 # Heat input from 2 to 3\n",
-      "\n",
-      "s4=s3 # isentropic process\n",
-      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",
-      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",
-      "x=(s4-sf)/(sg-sf)# from property of pure substance\n",
-      "hf=191.8 #enthalpy of saturated liquid @10 kPa from steam table\n",
-      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",
-      "h4=hf+x*(hg-hf)# enthalpy @ state 4\n",
-      "\n",
-      "wt=h3-h4 # turbine work\n",
-      "\n",
-      "efficiency=(wt-wp)/q # efficiency of power cycle\n",
-      "print \" The Efficiency is\",round(efficiency,4),\" or\",round(efficiency*100,1),\"%\"\n",
-      "# the answer is correct within limits\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Efficiency is 0.3238  or 32.4 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.2:PG-149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "# Please refer to the given figure of question 6.1 for quantities\n",
-      "effi1=0.323 #old efficiency\n",
-      "P2=4*1000 #higher pressure converted in in kPa\n",
-      "P1=10 # lower pressure in kPa\n",
-      "rho=1000 # density of water in Kg/m^3\n",
-      "h1=192 # enthalpy at state 1 in kJ/kg\n",
-      "h3=3214 # enthalpy at state 3 i.e @400 degree celsius and 4MPa in kJ/kg\n",
-      "s3=6.769# entropy at state 3 i.e @400 degree celsius and 4MPa in kJ/kg.K\n",
-      "\n",
-      "s4=s3 # insentropic process\n",
-      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",
-      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",
-      "\n",
-      "x=(s4-sf)/(sg-sf)# quality of steam\n",
-      "\n",
-      "hf=192 #enthalpy of saturated liquid @10 kPa from steam table\n",
-      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",
-      "h4=hf+x*(hg-hf)# enthalpy @ state 4\n",
-      "h2=h1 # isenthalpic process\n",
-      "qb=h3-h2 # heat addition\n",
-      "\n",
-      "wt=h3-h4 # turbine work\n",
-      "\n",
-      "effi2=(wt)/qb # efficiency of power cycle\n",
-      "\n",
-      "print \" The Efficiency is\",round(effi2,3),\" or\",round(effi2*100),\"% \\n\"\n",
-      "\n",
-      "perincrease=((effi2-effi1)/effi1)*100 \n",
-      "\n",
-      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Efficiency is 0.354  or 35.0 % \n",
-        "\n",
-        " The % increase in Efficiency is 9.69  \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.3:PG-149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "# Please refer to fig of question 6.1 for quantities\n",
-      "effi1=0.323 #old efficiency\n",
-      "P2=2*1000 #higher pressure converted in in kPa\n",
-      "P1=10 # lower pressure in kPa\n",
-      "rho=1000 # density of water in Kg/m^3\n",
-      "T2=600# max temperature of cycle in degree celsius\n",
-      "h1=192 # enthalpy at state 1 in kJ/kg\n",
-      "h3=3690 # enthalpy at state 3 in kJ/kg, 600*C and 2MPa pressure\n",
-      "s3=7.702# entropy at state 3 in kJ/kg.K, 600*C and 2MPa pressure\n",
-      " \n",
-      "s4=s3# isentropic process\n",
-      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",
-      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",
-      "\n",
-      "x=(s4-sf)/(sg-sf)# quality of steam\n",
-      "\n",
-      "hf=192 #enthalpy of saturated liquid @10 kPa from steam table\n",
-      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",
-      "h4=hf+x*(hg-hf)# enthalpy @ state 4\n",
-      "\n",
-      "h2=h1 # isenthalpic process\n",
-      "qb=h3-h2 # heat addition\n",
-      "\n",
-      "wt=h3-h4 # turbine work\n",
-      "\n",
-      "effi2=(wt)/qb # efficiency of power cycle\n",
-      "print \" The Efficiency is\",round(effi2,3),\" or\",round(effi2*100),\"% \\n\"\n",
-      "perincrease=((effi2-effi1)/effi1)*100 \n",
-      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Efficiency is 0.357  or 36.0 % \n",
-        "\n",
-        " The % increase in Efficiency is 10.56  \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.4:PG-150"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "# Please refer to fig of question 6.1 for quantities\n",
-      "effi1=0.323 #old efficiency\n",
-      "P2=2*1000 #higher pressure converted in in kPa\n",
-      "P1=4 # condenser pressure in kPa\n",
-      "rho=1000 # density of water in Kg/m^3\n",
-      "h1=192 # enthalpy at state 1 in kJ/kg\n",
-      "h3=3248 # enthalpy at state 3 in kJ/kg\n",
-      "s3=7.1279# entropy at state 3 in kJ/kg.K\n",
-      "\n",
-      "s4=s3 # isentropic process \n",
-      "\n",
-      "sf=0.4225 # entropy of saturated liquid @10 kPa from steam table\n",
-      "sg=8.4754 # entropy of saturated vapour @10 kPa from steam table\n",
-      "\n",
-      "x=(s4-sf)/(sg-sf)# from property of pure substance\n",
-      "\n",
-      "hf=121 #enthalpy of saturated liquid @4 kPa from steam table\n",
-      "hg=2554 # enthalpy of saturated vapour @4 kPa from steam table\n",
-      "h4=hf+x*(hg-hf)# enthalpy @ state 4h1=h2 # isenthalpic process\n",
-      "h2=h1 # isenthalpic process\n",
-      "qb=h3-h2 # heat addition\n",
-      "\n",
-      "wt=h3-h4 # turbine work\n",
-      "\n",
-      "effi2=(wt)/qb # efficiency of power cycle\n",
-      "print \" The Efficiency is\",round(effi2,4),\" or\",round(effi2*100),\"% \\n\"\n",
-      "perincrease=((effi2-effi1)/effi1)*100 \n",
-      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Efficiency is 0.3603  or 36.0 % \n",
-        "\n",
-        " The % increase in Efficiency is 11.55  \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.5:PG-152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "P2=2*1000 #higher pressure converted in in kPa\n",
-      "P1=10 # lower pressure in kPa\n",
-      "h1=192.0 # enthalpy at 10 kPa  in kJ/kg\n",
-      "h3=3248.0 # enthalpy @ state 3 in kJ/kg from table C.3\n",
-      "s3=7.128 # entropy @ state 3 in kJ/kg.K from table C.3\n",
-      "s4=s3 # isentropic process\n",
-      "\n",
-      "h2=h1 #isenthalpic process\n",
-      "h4=((s4-7.038)/(7.233-7.038))*(3056-2950)+2950 #using adjacent values for \n",
-      "#interpolation from table C.3 \n",
-      "h5=3267.0 # enthalpy at 800 kPa and $00 degree celsius\n",
-      "s5=7.572 # entropy at 800 kPa and $00 degree celsius\n",
-      "\n",
-      "s6=s5 # isentropic process\n",
-      "sf=0.6491#  entropy of saturated liquid @10 kPa from steam table\n",
-      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",
-      "\n",
-      "x=(s6-sf)/(sg-sf)# quality of steam\n",
-      "\n",
-      "hf=192.0 #enthalpy of saturated liquid @10 kPa from steam table\n",
-      "hg=2585.0 # enthalpy of saturated vapour @10 kPa from steam table\n",
-      "\n",
-      "h6=hf+x*(hg-hf)# enthalpy @ state 6\n",
-      "\n",
-      "# we now calculate energy input\n",
-      "qb=(h5-h4)+(h3-h2)# heat interaction\n",
-      "\n",
-      "# we now calculate work output\n",
-      "wt=(h5-h6)+(h3-h4)# turbine work\n",
-      "\n",
-      "eff=(wt)/qb # efficiency of power cycle\n",
-      "print\" The Efficiency is\",round(eff,3),\" or\",round(eff*100,2),\"%\"\n",
-      "\n",
-      "# // The anwer is different in textbook as there the intermediate values are approximated while in python the calculations are precise \n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Efficiency is 0.336  or 33.57 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.6:PG-155"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "# initialization of variables\n",
-      "\n",
-      "# Please refer to fig of question 6.1 for quantities\n",
-      "\n",
-      "effi1=0.357 #efficiency from example 6.3\n",
-      "P2=2*1000.0 #higher pressure converted in in kPa\n",
-      "P1=10.0 # lower pressure in kPa\n",
-      "rho=1000.0 # density of water in Kg/m^3\n",
-      "T2=600.0 # max temperature of cycle in degree celsius\n",
-      "h1=192.0 # enthalpy at state 1 in kJ/kg\n",
-      "h3=3690.0 # enthalpy at state 3 in kJ/kg, 600*C and 2MPa pressure\n",
-      "h4=2442.0 # enthalpy from example 6.3\n",
-      "h6=505.0 # specific enthalpy @ 200 kPa from steam table\n",
-      "h7=h6 # isenthalpic process\n",
-      "s3=7.702# entropy at state 3 in kJ/kg.K, 600*C and 2MPa pressure\n",
-      "\n",
-      "h2=h1 # isenthalpic process\n",
-      "s5=s3 # isentropic process\n",
-      "h5=(s3-7.509)*(2971-2870)/(7.709-7.509)+2870 # interpolationg from steam table 2 200 kPa using s5=s3= 7.702 kJ/kg.\n",
-      "\n",
-      "m6=1.0 # let mass of steam =1 Kg\n",
-      "m5=(h6-h2)*(m6)/(h5-h2) \n",
-      "m2=m6-m5 # conservation of mass\n",
-      "\n",
-      "wt=h3-h5+(h5-h4)*m2 # work done by turbine\n",
-      "qb=h3-h7 # heat given to bolier\n",
-      "effi2=(wt)/qb # efficiency of power cycle\n",
-      "print \" The Efficiency is\",round(effi2,4),\" or\",round(effi2*100),\"% \\n\"\n",
-      "perincrease=((effi2-effi1)/effi1)*100 \n",
-      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n",
-      "\n",
-      "# The anwer is different in textbook as there the intermediate values are approximated while in python the calculations are precise \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Efficiency is 0.3732  or 37.0 % \n",
-        "\n",
-        " The % increase in Efficiency is 4.55  \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.7:PG-156"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "P2=2*1000 #higher pressure converted in  kPa\n",
-      "P1=10 # lower pressure in kPa\n",
-      "h1=192 # enthalpy at 10 kPa  in kJ/kg\n",
-      "h3=3248 # enthalpy @ state 3 in kJ/kg from table C.3\n",
-      "s3=7.128 # entropy @ state 3 in kJ/kg.K from table C.3\n",
-      "\n",
-      "s4=s3 # isentropic process\n",
-      "\n",
-      "h4=((s4-7.038)/(7.233-7.038))*(3056-2950)+2950 #using adjacent values for \n",
-      "#interpolation from table C.3 \n",
-      "h5=3267 # enthalpy at 800 kPa and $00 degree celsius\n",
-      "s5=7.572 # entropy at 800 kPa and $00 degree celsius\n",
-      "\n",
-      "s6=s5 # isentropic process\n",
-      "sf=0.6491#  entropy of saturated liquid @10 kPa from steam table\n",
-      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",
-      "\n",
-      "x=(s6-sf)/(sg-sf)# quality of steam\n",
-      "\n",
-      "hf=192 #enthalpy of saturated liquid @10 kPa from steam table\n",
-      "hg=2585 # enthalpy of saturated vapour @10 kPa from steam table\n",
-      "\n",
-      "h6=hf+x*(hg-hf)# enthalpy @ state 6\n",
-      "h7=721 # enthalpy of saturated liquid @800 kPa from steam table\n",
-      "h8=h7 # isenthalpic process\n",
-      "h2=h1 # isenthalpic process\n",
-      "\n",
-      "m8=1 # let mass of steam =1 Kg\n",
-      "m4=(h8-h2)*(m8)/(h4-h2)\n",
-      "m2=m8-m4 # conservation of mass\n",
-      "\n",
-      "wt=h3-h4+(h5-h6)*m2 # work done by turbine\n",
-      "qb=h3-h8+(h5-h4)*m2 # heat given to bolier\n",
-      "\n",
-      "effi=(wt)/qb # efficiency of power cycle\n",
-      "# result\n",
-      "print \" The Efficiency is\",round(effi,3),\" or\",round(effi*100,2),\"%\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The Efficiency is 0.347  or 34.7 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.8:PG-159"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "\n",
-      "# for rankine cycle refer to fig 6.9\n",
-      "\n",
-      "effiT=0.8 # turbine efficiency\n",
-      "P2=2*1000 # higher pressure converted in  kPa\n",
-      "P1=10 # lower pressure in kPa\n",
-      "h1=192 # enthalpy at 10 kPa  in kJ/kg\n",
-      "h3=3690 # enthalpy of superheated steam @ 2 MPa from steam table in kJ/kg\n",
-      "s3=7.702 #entropy of superheated steam @ 2 MPa from steam table in kJ/kg.K\n",
-      "# state 4' is repsresented by '41'\n",
-      "h2=h1 #isenthalpic process\n",
-      "s41=s3 # entropy is constant\n",
-      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",
-      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",
-      "x=(s41-sf)/(sg-sf)# from property of pure substance\n",
-      "\n",
-      "hf=191.8 #enthalpy of saturated liquid @10 kPa from steam table\n",
-      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",
-      "h41=hf+x*(hg-hf)# enthalpy @ state 41\n",
-      "\n",
-      "wa=effiT*(h3-h41)# turbine efficiency =(actual work)/(isentropic work)\n",
-      "\n",
-      "qb=h3-h2 # heat supplied\n",
-      "\n",
-      "effi=(wa)/qb # efficiency of power cycle\n",
-      "print\"  The Efficiency is\",round(effi,3),\" or\",round(effi*100,1),\"%\"\n",
-      "\n",
-      "h4=h3-wa # adiabatic process\n",
-      "\n",
-      "# now using interpolation for superheated steam @ 10 kPa\n",
-      "T4=(h4-2688)*(150-100)/(2783-2688)+100\n",
-      "\n",
-      "print \"\\n The Temperature from interpolation comes out to be\",int(T4),\" degree celsius\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "  The Efficiency is 0.286  or 28.6 %\n",
-        "\n",
-        " The Temperature from interpolation comes out to be 101  degree celsius\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.9:PG-162"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#initialization of variables\n",
-      "\n",
-      "# refer to fig 6.10c\n",
-      "\n",
-      "mdot=0.6 # mass flow rate of refrigerant in kg/sec\n",
-      "T1=-24 # evaporator temperature in degree celsius\n",
-      "T2=39.39 # condenser temperature in degree celsius\n",
-      "h1=232.8 # enthalpy of saturated R134a vapour @ -24 degree celsius from table D.1\n",
-      "s1=0.9370 # entropy of saturated R134a vapour @ -24 degree celsius from table D.1\n",
-      "h3=105.3 # enthalpy of saturated R134a liquid @ -24 degree celsius from table D.2\n",
-      "h4=h3 # isenthalpic process\n",
-      "\n",
-      "# interpolating enthalpy from table D.3 @ 39.39 degree celsius\n",
-      "h2=(s1-0.9066)*(280.19-268.68)/(0.9428-0.9066)+268.68\n",
-      "QdotE=mdot*(h1-h4) # heat transfer rate\n",
-      "WdotC=mdot*(h2-h1)# power given to compressor\n",
-      "\n",
-      "COP=QdotE/WdotC # coefficient of performance\n",
-      "\n",
-      "Hp=(WdotC/0.746)/(QdotE/3.52) #calculating Horsepower required per Ton\n",
-      "\n",
-      "print \"The rate of refrigeration is\",round(QdotE,2),\"kJ/s \\n \"\n",
-      "print \"The coefficient of performance is\",round(COP,2),\"\\n \"\n",
-      "print \"The rating in horsepower per ton is\",round(Hp,3),\" hp\\n \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rate of refrigeration is 76.5 kJ/s \n",
-        " \n",
-        "The coefficient of performance is 2.8 \n",
-        " \n",
-        "The rating in horsepower per ton is 1.686  hp\n",
-        " \n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.10:PG-163"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#initialization of variables\n",
-      "# refer to fig 6.10c\n",
-      "effi=0.8 # compressor efficiency\n",
-      "mdot=0.6 # mass flow rate of refrigerant in Kg/sec\n",
-      "T4=-24 # temperature of evaporator\n",
-      "T2=39.39 # temperature of condensor\n",
-      "T1=-20.0 # supeheating temperature\n",
-      "T3=40 # subcooling temperature\n",
-      "h3=106.2 # enthalpy of liquid R-134a @ 40 degree celsius from table D.1\n",
-      "h4=h3 # isenthalpic process\n",
-      "h1=236.5 # enthalpy of superheated R-134a @ 0.10 MPa and -20 degree celsius from table D.3\n",
-      "s1=0.960 #entropy of superheated R-134a @ 0.10 MPa and -20 degree celsius from table D.3 \n",
-      "\n",
-      "s2dash=s1 # isentropic process\n",
-      "\n",
-      "# using interpolation from table D.3 @ 1.0 MPa for s2dash=0.960 \n",
-      "h2dash=(s2dash-0.9428)*(291.36-280.19)/(0.9768-0.9428)+280.19\n",
-      "\n",
-      "h2=(h2dash-h1)/(effi)+h1 # by definition of compressor efficiency\n",
-      "\n",
-      "QdotE=mdot*(h1-h4)#heat transfer rate power given to compressor\n",
-      " \n",
-      "wdotc=mdot*(h2-h1)# power given to compressor\n",
-      "\n",
-      "COP=QdotE/wdotc # coefficient of performance\n",
-      "\n",
-      "print \"The rate of refrigeration is\",round(QdotE,1),\"kJ/s \\n \"\n",
-      "\n",
-      "print \"The coefficient of performance is\",round(COP,2),\"\\n \"\n",
-      "# The value of Wdotc is shown wrong in the textbook. It should be multiplied by mass flow rate\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rate of refrigeration is 78.2 kJ/s \n",
-        " \n",
-        "The coefficient of performance is 2.11 \n",
-        " \n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex6.11:PG-165"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# solution\n",
-      "#initialization of variables\n",
-      "# refer to fig 6.10c\n",
-      "\n",
-      "QdotC=300 #heating Load in KWh or heat rejected by condensor\n",
-      "T1=-12 # evaporator temperature in degree celsius\n",
-      "P2=800 # condensor pressure in kPa \n",
-      "h1=240 # specific enthalpy of saturated R-134a vapour @ -12 degree celsius from table D.1\n",
-      "s1=0.927 # specific entropy of saturated R-134a vapour @ -12 degree celsius from table D.1\n",
-      "s2=s1 # isentropic process\n",
-      "h3=93.4 #specific enthalpy of saturated R-134a liquid @ 800 kPa from tableD.2\n",
-      "\n",
-      "# extrapolating enthalpy from table D.2 @ 0.8 MPa for s=0.927\n",
-      "h2=273.7-(0.9374-s2)*(284.4-273.7)/(0.9711-0.9374)\n",
-      "\n",
-      "# QdotE=mdot*(h1-h4) is heat transfer rate\n",
-      "mdot=QdotC/(h2-h3)# mass flow rate\n",
-      "\n",
-      "WdotC=mdot*(h2-h1)# power given to compressor\n",
-      "\n",
-      "#part(a)\n",
-      "COP=QdotC/WdotC # coefficient of performance\n",
-      "print \"The coefficient of performance is\",round(COP,2),\"\\n \"\n",
-      "\n",
-      "#part(b)\n",
-      "Cost=WdotC*0.07 # cost of electricity\n",
-      "print \"The cost of electricity is $\",round(Cost,3),\"/hr \\n\"\n",
-      "\n",
-      "#part(c)\n",
-      "costgas=(300*3600*0.50)/100000 # cost of gas\n",
-      "print \"The cost of gas is $\",round(costgas,2),\"/hr \\nThus heat pump is better \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The coefficient of performance is 5.82 \n",
-        " \n",
-        "The cost of electricity is $ 3.607 /hr \n",
-        "\n",
-        "The cost of gas is $ 5.4 /hr \n",
-        "Thus heat pump is better \n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": [],
-     "prompt_number": 61
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter7.ipynb b/Thermodynamics_Demystified/Chapter7.ipynb
deleted file mode 100755
index 0c094e5b..00000000
--- a/Thermodynamics_Demystified/Chapter7.ipynb
+++ /dev/null
@@ -1,624 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:af523f2f43fc72634028758c968b85e1c9f7441276d72a4c6af9da09cd0eee6e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7:Power and Refrigeration Gas Cycles"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.1:PG-175"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "r=12 # compression ratio\n",
-      "k=1.4 # polytropic index for air\n",
-      "p1=200.0 # pressure at state 1 in kPa\n",
-      "p3=10000.0 # pressure at state 3 in kPa\n",
-      "\n",
-      "c=100/(r-1) # clearance in percentage\n",
-      "print \"The percent clearance is\",round(c,2),\"% \\n\"\n",
-      "v3=100.0 # let us assume v3=100 m^3 for calculations\n",
-      "p2=p1*(r**k) # polytopic process pressure relation\n",
-      "p4=p3*(1/(r**k))# polytropic process pressure relation\n",
-      "w34=v3*(r*p4-p3)/(1-k) # polytropic work done in process 3 to 4\n",
-      "v2=v3 # constant volume process\n",
-      "w12=v2*(p2-r*p1)/(1-k)\n",
-      "wcycle=w12+w34 # total work in cycle\n",
-      " # now equating the polytropic work calculated to work by MEP\n",
-      "MEP=wcycle/(r*v2-v2) # as work = pressure*change in volume\n",
-      "print \"The MEP is\",round(MEP),\" kPa\"  \n",
-      "# The solution is wrong in textbook as calculation for P2 is wrong \n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percent clearance is 9.0 % \n",
-        "\n",
-        "The MEP is 503.0  kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.2:PG-179"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "r=10.0 # compression ratio\n",
-      "k=1.4 # polytropic index for air\n",
-      "R=0.287 # specific gas constant for air\n",
-      "Cv=0.717 # specific heat at constant volume\n",
-      "Wnet=1000 # net work output in kJ/kg\n",
-      "T1=227+273.0 # low air temperaure in kelvin\n",
-      "p1=200.0 # low pressure in kPa\n",
-      "\n",
-      "effi=1-(1/r**(k-1)) # thermal efficeiency \n",
-      "print \"The maximum possible thermal efficiency is\",round(effi*100,1),\"% \\n\"\n",
-      "\n",
-      "T2=T1*(r)**(k-1) # isentropic process temperature relation\n",
-      "\n",
-      "T4=((Wnet/Cv)+T2-T1)/((r**(k-1))-1) # using expression for work\n",
-      "\n",
-      "T3=T4*(r)**(k-1)\n",
-      "\n",
-      "efficarnot=1-T1/T3\n",
-      "print \"The carnot efficiency is\",round(efficarnot*100),\"%\"\n",
-      "\n",
-      "v1=R*T1/p1 # initial volume \n",
-      "v2=v1/r # from compression ratio\n",
-      "\n",
-      "MEP=Wnet/(v1-v2) # mean effective pressure equation\n",
-      "\n",
-      "print \"The MEP is\",round(MEP),\" kPa\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum possible thermal efficiency is 60.2 % \n",
-        "\n",
-        "The carnot efficiency is 86.0 %\n",
-        "The MEP is 1549.0  kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.3:PG-182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "\n",
-      "r=18 # compression ratio\n",
-      "k=1.4 # polytropic index for air\n",
-      "R=0.287 # specific gas constant for air\n",
-      "Cv=0.717 # specific heat at constant volume\n",
-      "Cp=1.0 # specific heat at constant pressure\n",
-      "T1=200+273 # lower temperaure in kelvin\n",
-      "P1=200.0 # low pressure in kPa\n",
-      "T3=2000 # higher temperature of cycle in kelvin\n",
-      "\n",
-      "v1=R*T1/P1 # specific volume at state 1 in m^3\n",
-      "v2=v1/r # specific volume after compression in m^3\n",
-      "\n",
-      "T2=T1*(v1/v2)**(k-1) # temperature after compression\n",
-      "P2=P1*(v1/v2)**k # pressure after compression\n",
-      "P3=P2 # diesel cycle\n",
-      "v3=R*T3/P3 # volume at state 3\n",
-      "\n",
-      "rc=v3/v2 # cutoff ratio\n",
-      "\n",
-      "effi=1-((rc**k)-1)/(r**(k-1)*k*(rc-1))\n",
-      "\n",
-      "\n",
-      "print \"The thermal efficiency is\",round(effi*100,1),\"%\"\n",
-      "\n",
-      "v4=v1 # diesel cycle\n",
-      "T4=T3*(v3/v4)**(k-1) # adiabatic process\n",
-      "\n",
-      "qin=Cp*(T3-T2) # using first law \n",
-      "qout=Cv*(T4-T1) # heat rejected \n",
-      "\n",
-      "Wnet=qin-qout # net work\n",
-      "MEP=Wnet/(v1-v2) # expression of mean effective pressure in terms of work\n",
-      "\n",
-      "print \"The MEP is\",round(MEP),\" kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thermal efficiency is 66.6 %\n",
-        "The MEP is 515.0  kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.4:PG-183"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "r=18 # compression ratio\n",
-      "k=1.4 # polytropic index for air\n",
-      "R=0.287 # specific gas constant for air\n",
-      "T1=200+273 # lower temperaure in kelvin\n",
-      "P1=200.0 # low pressure in kPa\n",
-      "T3=2000.0 # higher temperature of cycle in kelvin \n",
-      "\n",
-      "v1=R*T1/P1 # specific volume at state 1 in m^3\n",
-      "#using table E.1\n",
-      "u1=340.0 # specific internal energy in kJ/kg\n",
-      "vr1=198.1 # in m^3/kg\n",
-      "\n",
-      "vr2=vr1*(1/r) # as r=v1/v2\n",
-      "\n",
-      "# now finding corresponding values from table E.1\n",
-      "T2=1310.0 # temperature in kelvin\n",
-      "Pr2=34.0 # pressure in kPa\n",
-      "h2=1408.0 # specific entropy in kJ/kg\n",
-      "v2=v1/18 # volume at state 2\n",
-      "P2=R*T2/v2 # pressure at state 2\n",
-      "\n",
-      "h3=2252.1 # specific enthalpy in kJ/kg from table E.1\n",
-      "vr3=2.776 \n",
-      "P3=P2 # diesel cycle\n",
-      "v3=R*T3/P3 # after compression volume\n",
-      "v4=v1 # isochoric process\n",
-      "vr4=vr3*v4/v3 # isentropic process\n",
-      "# now using Vr4 we read corresponding value from table E.1\n",
-      "T4=915 # final temperature in kelvin\n",
-      "u4=687.5 # specific internal energy at state 4\n",
-      "\n",
-      "qin=h3-h2 # using first law \n",
-      "qout=u4-u1 # heat rejected \n",
-      "\n",
-      "Wnet=qin-qout # net work\n",
-      "effi=100*Wnet/qin # thermal efficiency\n",
-      "print\" The thermal efficiency is\",round(effi,1),\"%\"\n",
-      "\n",
-      "MEP=Wnet/(v1-v2) # expression of mean effective pressure in terms of work\n",
-      "\n",
-      "print \" The MEP is\",round(MEP),\" kPa\"\n",
-      "\n",
-      "erroreffi=(66.6-effi)*100/effi # error in efficiency\n",
-      "errorMEP=(515-MEP)*100/MEP # error in MEP\n",
-      "\n",
-      "print \" The % error in efficiency is\",round(erroreffi,1),\"%\"\n",
-      "print \" The % error MEP is\",round(errorMEP,1),\"% \\n\"\n",
-      "\n",
-      "# the answers are slight different due to approximation in textbook ... here answers are precise\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The thermal efficiency is 58.8 %\n",
-        " The MEP is 775.0  kPa\n",
-        " The % error in efficiency is 13.2 %\n",
-        " The % error MEP is -33.5 % \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.5:PG-186"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "Cp=1.0 # specific heat at constant pressure\n",
-      "k=1.4 # polytropic index for air\n",
-      "T1=25+273.0 # temperature at compressor inlet\n",
-      "T3=850+273.0 # maximum temperature in kelvin\n",
-      "\n",
-      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
-      "\n",
-      "T2=T1*(r)**((k-1)/k) # temperature after compression\n",
-      "\n",
-      "T4=T3*(1/r)**((k-1.0)/k) # final temperature\n",
-      "\n",
-      "Wcomp=Cp*(T2-T1) # compressor work\n",
-      "Wturb=Cp*(T3-T4) # turbine work\n",
-      "\n",
-      "BWR=Wcomp/Wturb # back work ratio\n",
-      "\n",
-      "print \" The BWR is\",round(BWR*100,1),\"%\\n\" \n",
-      "\n",
-      "Effi=1-r**((1-k)/k) # thermal efficiency\n",
-      "\n",
-      "print\" The thermal efficiency is\",round(Effi*100,1),\"%\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The BWR is 42.0 %\n",
-        " The thermal efficiency is 36.9 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.6:PG-186"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "Cp=1.0 # specific heat at constant pressure\n",
-      "k=1.4 # polytropic index for air\n",
-      "T1=25+273.0 # temperature at compressor inlet\n",
-      "T3=850+273.0 # maximum temperature in kelvin\n",
-      "\n",
-      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
-      "efficomp=0.75 # efficiency of compressor\n",
-      "effiturb=0.75 # efficiency of turbine\n",
-      "\n",
-      "T2dash=T1*(r)**((k-1)/k) # temperature after compression\n",
-      "Wcomp=Cp*(T2dash-T1)/efficomp # compressor work\n",
-      "\n",
-      "T4dash=T3*(1/r)**((k-1)/k) # final temperature\n",
-      "Wturb=Cp*(T3-T4dash)*effiturb # turbine work\n",
-      "\n",
-      "BWR=100*Wcomp/Wturb # back work ratio\n",
-      "\n",
-      "print \" The BWR is\",round(BWR,1),\"%\\n\"\n",
-      "\n",
-      "T2=(Wcomp/Cp)+T1 # actual temperature of state 2\n",
-      "\n",
-      "qin=Cp*(T3-T2) # using first law \n",
-      "\n",
-      "Wnet=(Wturb-Wcomp) # net work\n",
-      "\n",
-      "effi=100*Wnet/qin # thermal efficiency\n",
-      "print\" The thermal efficiency is\",round(effi,1),\"%\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The BWR is 74.7 %\n",
-        " The thermal efficiency is 13.2 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.7:PG-191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "Cp=1.0 # specific heat at constant pressure\n",
-      "k=1.4 # polytropic index for air\n",
-      "T1=25+273.0 # temperature at compressor inlet\n",
-      "T3=850+273.0 # maximum temperature in kelvin\n",
-      "\n",
-      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
-      "\n",
-      "T2=T1*(r)**((k-1)/k) # temperature after compression\n",
-      "\n",
-      "T4=T3*(1/r)**((k-1)/k) # final temperature\n",
-      "\n",
-      "Wcomp=Cp*(T2-T1) # compressor work\n",
-      "Wturb=Cp*(T3-T4) # turbine work\n",
-      "\n",
-      "BWR=Wcomp/Wturb # back work ratio\n",
-      "\n",
-      "print \" The BWR is\",round(BWR,2),\"\\n\"\n",
-      "\n",
-      "effi=(1-((T1/T4)*(r**((k-1)/k))))# efficiency\n",
-      "print\" The thermal efficiency is\",round(effi*100,1),\"%\"\n",
-      "# The solution in textbook is incorrect due to wrong value of T4 (temperature at state 4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The BWR is 0.42 \n",
-        "\n",
-        " The thermal efficiency is 33.4 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.8:PG-193"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "#REFER TO FIG.:7.8\n",
-      "\n",
-      "Cp=1 # specific constant at constant pressure\n",
-      "k=1.4 # polytropic constant for air\n",
-      "T5=25+273.0 # temperature at state 5 in kelvin\n",
-      "T7=850+273.0 # temperature at state 4 in kelvin\n",
-      "T9=350 # exit temperature of water from bolier in kelvin\n",
-      "WdotST=100000.0 # power from steam turbine in Watt\n",
-      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
-      "\n",
-      "h1=192.0 # specific enthalpy at 10 Kpa from steam table\n",
-      "h2=h1 # isenthalpic process\n",
-      "h3=3214.0 # specific enthalpy at 4 Mpa and 400 degree celsius from steam table\n",
-      "s3=6.769 # specific entropy at 4 Mpa and 400 degree celsius from steam table\n",
-      "\n",
-      "s4=s3 # isentropic process\n",
-      "sf=0.6491 # specific entropy of saturated liquid at 10 kPa and 45 degree celsiusfrom table C.2\n",
-      "sg=8.1510 # specific entropy of saturated liquid at 10 kPa and 45 degree celsiusfrom table C.2\n",
-      "x4=(s4-sf)/(sg-sf) # quality of steam\n",
-      "\n",
-      "hf=h1 # specific enthalpy of saturated liquid @ 10 Kpa  \n",
-      "hg=2584.6\n",
-      "h4=hf+x4*(hg-hf) # specific entropy at state 4\n",
-      "\n",
-      "mdots=WdotST/(h3-h4) # steam mass flow rate from turbine output\n",
-      "\n",
-      "T6=T5*(r**((k-1)/k)) # adiabatic process relation\n",
-      "T8=T7*(1/r**((k-1)/k)) # adiabatic process relation\n",
-      "\n",
-      "# Now using energy balance in boiler\n",
-      "mdota=mdots*(h3-h2)/(Cp*(T8-T9)) # mass flow rate of water\n",
-      "\n",
-      "Wdotturb=mdota*Cp*(T7-T8) #  power produced by turbine\n",
-      "\n",
-      "Wdotcomp=mdota*Cp*(T6-T5) # energy needed by compressor\n",
-      "\n",
-      "WdotGT=Wdotturb-Wdotcomp # net turbine work\n",
-      "\n",
-      "Qdotin=mdota*Cp*(T7-T6) # energy input by combustor\n",
-      "\n",
-      "effi=100*(WdotST+WdotGT)/Qdotin # combined efficiency\n",
-      "\n",
-      "print \"The thermal efficiency of the combined cycle is\",round(effi,1),\"%\"\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thermal efficiency of the combined cycle is 56.4 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.9:PG-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "Cp=1 # specific constant at constant pressure\n",
-      "k=1.4 # polytropic constant for air\n",
-      "r=10.0\n",
-      "T2=-10+273 # temperature at entry  of compressor\n",
-      "T4=30+273 # temperature at entry of turbine\n",
-      "\n",
-      "T3=T2*(r**((k-1)/k)) # temperature at state 3 in kelvin\n",
-      "T1=T4*(1/r**((k-1)/k)) # temperature at state 1 in degree celsius\n",
-      "print \"The minimum temperature is\",round(T1-273),\"degree celsius \\n\"\n",
-      "\n",
-      "qin=Cp*(T2-T1) # heat input\n",
-      "Wcomp=Cp*(T3-T2)# compressor work\n",
-      "Wturb=Cp*(T4-T1) # turbine work\n",
-      "\n",
-      "COP=qin/(Wcomp-Wturb) # COP of refrigeration\n",
-      "print\" The COP is\",round(COP,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum temperature is -116.0 degree celsius \n",
-        "\n",
-        " The COP is 1.07\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.10:PG-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#solution\n",
-      "# initialization of variables\n",
-      "\n",
-      "Cp=1.0 # specific constant at constant pressure\n",
-      "k=1.4 # polytropic constant for air\n",
-      "r=10.0\n",
-      "T3=-10+273 # temperature at entry  of compressor\n",
-      "T6=-40+273 # temperature at entry of turbine\n",
-      "\n",
-      "T5=T3 # heat exchanger\n",
-      "T2=T6 # heat exchanger\n",
-      "\n",
-      "T4=T3*(r**((k-1)/k)) # temperature after compression\n",
-      "T1=T6*(1/r**((k-1)/k)) # temperature after exit from turbine\n",
-      "\n",
-      "print \"The minimum temperature is\",round(T1-273),\"degree celsius \\n\"\n",
-      "\n",
-      "qin=Cp*(T2-T1) # heat input\n",
-      "Wcomp=Cp*(T4-T3)# compressor work\n",
-      "Wturb=Cp*(T6-T1) # turbine work\n",
-      "\n",
-      "COP=qin/(Wcomp-Wturb) # COP of refrigeration\n",
-      "\n",
-      "print\" The COP is\",round(COP,3)\n",
-      "\n",
-      "# the answer is correct within given limits\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum temperature is -152.0 degree celsius \n",
-        "\n",
-        " The COP is 0.848\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter8.ipynb b/Thermodynamics_Demystified/Chapter8.ipynb
deleted file mode 100755
index 2ab04749..00000000
--- a/Thermodynamics_Demystified/Chapter8.ipynb
+++ /dev/null
@@ -1,530 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:cd14818c82eb9353b8e2f9e12079fb0ba7b2a061c9a528ea53dea9dfa9a116ce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8:Psychrometrics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.1:PG-208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialization of variables\n",
-      "Ra=0.287 # specific gas constant for air\n",
-      "P=100.0 # pressure of room in kPa\n",
-      "V=150.0 # volume of room in m^3\n",
-      "T=25+273 # temperature of air in kelvin\n",
-      "phi=0.6 # relative humidity\n",
-      "Pg=3.29 # saturation vapour pressure in kPa at 25 *C from table C.1\n",
-      "Mv= 18 # molecular mass of water vapor\n",
-      "Ma=28.97 # molecular mass of air\n",
-      "\n",
-      "Pv=Pg*phi # partial pressure of water vapour\n",
-      "\n",
-      "Pa=P-Pv # partial pressure of air\n",
-      "\n",
-      "w=0.622*(Pv/Pa) # humidity ratio in Kg of water/ Kg of dry air\n",
-      "Tdp=17.4 # dew point temperature from interpolation in table C.2 corresponding to partial pressure Pv=1.98 kPa\n",
-      "\n",
-      "ma=Pa*V/(Ra*T) # mass of air\n",
-      "mv=w*ma # mass of water vapour in  kg\n",
-      "\n",
-      "# now we find volume percentage\n",
-      "Nv=mv/Mv # moles of vapour\n",
-      "Na=ma/Ma # moles of air\n",
-      "\n",
-      "Vw= Nv/(Na+Nv) # fraction of volume occupied by water vapour\n",
-      "\n",
-      "print \"The humidity ratio is\",round(w,4),\"kg water/ kg of dry air \\n\"\n",
-      "print \"The dew point is\",round(Tdp,1),\"degree celsius \\n \"\n",
-      "print \"The mass of water vapour in the air is\",round(mv,3),\"kg \\n\"\n",
-      "print \"The volume percentage of the room that is water vapor is\",round(Vw*100,2),\"%\"\n",
-      "# The answers are correct within given limits \n",
-      "# The variation in answers is due to approximations made by\n",
-      "# textbook while python is precise\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The humidity ratio is 0.0125 kg water/ kg of dry air \n",
-        "\n",
-        "The dew point is 17.4 degree celsius \n",
-        " \n",
-        "The mass of water vapour in the air is 2.153 kg \n",
-        "\n",
-        "The volume percentage of the room that is water vapor is 1.98 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.2:PG-209"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialization of variables\n",
-      "Ra=0.287 # specific gas constant for air\n",
-      "P=100.0 # pressure of room in kPa\n",
-      "w1=0.0126 # old humidity ratio of example 8.1-\n",
-      "Pg=3.29 # saturation vapour pressure in kPa at 25 *C from table C.1\n",
-      "mv=2.17 # initial mass of water vapour in example 8.1\n",
-      "T=25+273 # temperature after reheat\n",
-      "V=150.0 # volume of room in m^3 \n",
-      "Pv=1.228 # saturation vapour pressure in kPa @ 10 degree celsius from table C.1\n",
-      "Pa=P-Pv # partial pressure of air\n",
-      "w2=0.622*(Pv/Pa) # new humidity ratio in Kg of water/ Kg of dry air\n",
-      "deltaw=w1-w2 # difference in humidity ratio\n",
-      "ma=Pa*V/(Ra*T) # mass of air\n",
-      "deltamv=deltaw*ma # mass of water vapour condensed\n",
-      "X=deltamv*100/mv # percentage of water vapour condensed\n",
-      "print \"The percentage that condenses is\",round(X,2),\"% \\n\"\n",
-      "# AFTER REHEATING\n",
-      "phi=1.608*w2*Pa/Pg\n",
-      "print \"The relative humidity is\",round(phi*100,3),\"%\"\n",
-      "\n",
-      "# The answers are correct within given limits \n",
-      "# The variation in answers is due to approximations made by\n",
-      "# textbook while python is precise\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage that condenses is 38.85 % \n",
-        "\n",
-        "The relative humidity is 37.332 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.3:PG-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#initialization of variables\n",
-      "T1=40 # dry bulb temperature in degree celsius\n",
-      "T2=20 # wet bulb temperature in degree celsius\n",
-      "Cp=1.0 # specific heat\n",
-      "P=100 # pressure of air stream in kPa\n",
-      "pg1=7.383 #saturation pressure @ 40 degree celsius\n",
-      "hfg2=2454 # latent heat for 20 degree celsius\n",
-      "Pg2=2.338 # saturation pressure @ 20 degree celsius\n",
-      "w2=0.622*Pg2/(P-Pg2) # specific humidity for wet bulb condition\n",
-      "hg1=2574 # specific enthalpy of saturated vapour @ 40 degree celsius\n",
-      "hf2=83.9 #spedific enthalpy of saturated liquid @ 20 degree celsius\n",
-      "w1=((w2*hfg2)+Cp*(T2-T1))/(hg1-hf2)# specific humidity for 40 degree celsius\n",
-      "print \"The humidity ratio is\",round(w1,4),\"kg water/ Kg dry air \\n\"\n",
-      "pv1=100*w1/(0.622+w1) # partial pressure of vapour\n",
-      "phi=pv1/pg1 # relative humidity\n",
-      "print \"The relative humidity is\",round(phi*100,1),\"% \\n\"\n",
-      "\n",
-      "hv=hg1 # temperature is at DBT=40 degree celsius\n",
-      "h=Cp*T1+w1*hv # specific enthalpy  of air\n",
-      "print \"The specific enthalpy is\",round(h,1),\"kJ/kg dry air\"\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The humidity ratio is 0.0066 kg water/ Kg dry air \n",
-        "\n",
-        "The relative humidity is 14.3 % \n",
-        "\n",
-        "The specific enthalpy is 57.1 kJ/kg dry air\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.5:PG-215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "T1=40 # inlet temperature in degree celsius\n",
-      "T2=27 # outlet temperature in degree celsius\n",
-      "phi1= 10 # relative humidity at inlet\n",
-      "# as no heat transfer takes place thus isenthalpic process\n",
-      "\n",
-      "#Thus following the enthalpy line at DBT=40 and Relative humidity=10\n",
-      "\n",
-      "phi2=45 # by interpolation of constant enthalpy line\n",
-      "w1=0.0046# specific humidity @ T=40 and phi1=10\n",
-      "w2=0.010 # specific humidity at outlet\n",
-      "W=w2-w1 # amount of water added\n",
-      "Tmin=18.5 # minimum temperature at 100% relative humidity\n",
-      "\n",
-      "print \"The relative humidity is\",round(phi2,1),\"% \\n \"\n",
-      "print \"The added water is\",round(W,4),\"kg water/kg dry air \\n\"\n",
-      "print \"The lowest possible temperature is\",round(Tmin,1),\"*C \"\n",
-      "\n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relative humidity is 45.0 % \n",
-        " \n",
-        "The added water is 0.0054 kg water/kg dry air \n",
-        "\n",
-        "The lowest possible temperature is 18.5 *C \n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.6:PG-215"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "T1=5+273.0 # outside air temperature in kelvin\n",
-      "P=100.0 # pressure in kPa\n",
-      "Ra=0.287 # specific gas constant for air\n",
-      "phi=0.7 # relative humidity outside\n",
-      "Qf=50.0/60.0 # volume flow rate in m**3/sec\n",
-      "Pg1=0.872 # saturation pressure at 278 K\n",
-      "Pv1=phi*Pg1 # partial pressure of water vapour\n",
-      "Pa1=P-Pv1 # partial pressure of air\n",
-      "\n",
-      "rhoa=Pa1/(Ra*T1) # density of dry air\n",
-      "\n",
-      "mdota=Qf*rhoa # mass flow rate of dry air\n",
-      "\n",
-      "# using psychrometric chart at T1=5*C and phi1=70% \n",
-      "h1=14 # inlet enthalpy in kJ/kg\n",
-      "h2=35 # enthalpy after heating in kJ/kg\n",
-      "\n",
-      "Qdot=mdota*(h2-h1) # heat transfer rate\n",
-      "# from psychrometric chart for T=25 *C and 35 kJ/kg enthalpy\n",
-      "phi2=19 # realtive humidity\n",
-      "\n",
-      "print \"The heat transfer rate is\",round(Qdot,1),\"kJ/s \\n\"\n",
-      "print \"The final relative humidity is\",round(phi2,4),\"% \"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The heat transfer rate is 21.8 kJ/s \n",
-        "\n",
-        "The final relative humidity is 19.0 % \n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.7:PG-216"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "#DATA TAKEN FROM PSYCHROMETRIC CHART\n",
-      "T1=5+273.0 # outside temperature in kelvin\n",
-      "h1=10# enthalpy in kJ/kg @ T=5 *C and 40 % relative humidity\n",
-      "Pg1=0.872 # saturaion pressure in kPa for 5 degree celsius DBT\n",
-      "phi1=0.4\n",
-      "h2=33 # specific enthalpy at 25 *C and 40 % relatuve humidity\n",
-      "h3=45.0 # specific enthalpy at state 3\n",
-      "P=100.0 # atmospheric pressure in kPa\n",
-      "Ra=0.287 # specific gas constant for air\n",
-      "Qf=60.0/60.0 # volume flow rate in m**3/s\n",
-      "Pv1=phi1*Pg1 # partial presure of water vapour \n",
-      "Pa1=P-Pv1 # partial pressure of air\n",
-      "w2=0.0021 # specific humidity @ 40 % relative humidity and 25*C temperature\n",
-      "w3=0.008 # final specific humidity\n",
-      "rhoa1=Pa1/(Ra*T1) # air density\n",
-      "mdota=Qf*rhoa1 # mass flow rate of dry air\n",
-      "\n",
-      "Qdot=mdota*(h2-h1) # heat transfer rate\n",
-      "\n",
-      "# as the process is isothermal thus\n",
-      "mdots=mdota*(w3-w2)# mass flow rate of steam by conservation of mass\n",
-      "print \"the rate of steam supplied is\",round(mdots,4),\"kg/s \\n\"\n",
-      "# also using energy balance\n",
-      "hs=(mdota*(h3-h2))/mdots # enthalpy of steam\n",
-      "hf=604.7 # enthalpy of saturated liquid @ 400 kPa\n",
-      "hg=2738.5 # enthalpy of saturated vapour @ 400 kPa\n",
-      "xs=(hs-hf)/(hg-hf)\n",
-      "print \"The quality of steam is\",round(xs,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the rate of steam supplied is 0.0074 kg/s \n",
-        "\n",
-        "The quality of steam is 0.67\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.8:PG-217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "# REFER TO FIG. 8.4\n",
-      "T1=30 # outside temperature in degree celsius\n",
-      "phi1=0.9 # outside relative humidity\n",
-      "T2=23 # room temperature in degree celsius\n",
-      "phi2=0.4 # relative humidity in room\n",
-      "\n",
-      "# using psychrometric chart\n",
-      "w1=0.0245 # specific humidity @ 30 *C and relative humidity 0.9\n",
-      "h1=93 # specific enthalpy @ 30 *C and relative humidity 0.9\n",
-      "w2=w1 # during cooling humidity remains constant \n",
-      "w3=0.007 # specific humidity @ 23 *C and relative humidity 0.4\n",
-      "h4=41 # final specific enthalpy\n",
-      "h3=26 # specific enthalpy @ 23 *C and relative humidity 0.4\n",
-      "deltaw=w3-w2 # moisture removed\n",
-      "print \" the amount of moisture removed is\",round(deltaw,4),\"kg \\n\"\n",
-      "\n",
-      "qout=h3-h1 # heat removed F-G-H process\n",
-      "\n",
-      "print \" the heat removed is\",round(qout,4),\"kJ/kg \\n \"\n",
-      "\n",
-      "qin=h4-h3 # heat added to bring to desired state\n",
-      "\n",
-      "print \" the heat added is\",round(qin,4),\"kJ/kg \""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " the amount of moisture removed is -0.0175 kg \n",
-        "\n",
-        " the heat removed is -67.0 kJ/kg \n",
-        " \n",
-        " the heat added is 15.0 kJ/kg \n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex8.9:PG-218"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "P=100 # atospheric pressure in kPa\n",
-      "R=0.287 # specific gas constant for air\n",
-      "T1=15+273 # outside temperature in kelvin\n",
-      "phi1=0.4# outside air relative humidity\n",
-      "Qf1=40 # outside air flow rate in m^3/min\n",
-      "T2=32+273 # inside temperature in kelvin\n",
-      "phi2=0.7 # inside air relative humidity\n",
-      "Qf2=20 # outside air flow rate in m^3/min\n",
-      "Ps1=1.7 # saturation pressure @ 15 degree celsius and 40% humidity\n",
-      "Ps2=4.9 # saturation pressure @ 32 degree celsius and 70% humidity\n",
-      "\n",
-      "Pv1=Ps1*phi1 # partial pressure of water vapour outside\n",
-      "\n",
-      "Pv2=Ps2*phi2 # partial pressure of water vapour inside\n",
-      "\n",
-      "Pa1=P-Pv1 #partial pressure of dry air outside\n",
-      "Pa2=P-Pv2 #partial pressure of dry air inside\n",
-      "\n",
-      "rhoa1=Pa1/(R*T1) # density of outside air\n",
-      "mdota1=Qf1*rhoa1 # mass flow rate of air outside\n",
-      "\n",
-      "rhoa2=Pa2/(R*T2) # density of inside air\n",
-      "mdota2=Qf2*rhoa2 # mass flow rate of inside air\n",
-      " # using psychrometric chart locating state 1 and 2\n",
-      "h1=37 # specific enthalpy @ DBT 15*C and 40 % humidity\n",
-      "w1=0.0073 # specific humidity @ DBT 15*C and 40 % humidity\n",
-      "h2=110 # specific enthalpy @ DBT 32*C and 70 % humidity\n",
-      "w2=0.0302 # specific humidity @ DBT 32*C and 70 % humidity\n",
-      "ratio=mdota1/mdota2 # ratio of distance between states \n",
-      "#  using this ratio state 3 is located on psychrometric chart\n",
-      "T3=(mdota1*T1+mdota2*T2)/(mdota1+mdota2)-273 # final temparature in celsius\n",
-      "\n",
-      "phi3=65# final relative humidity at T3 from psychrometric chart\n",
-      "\n",
-      "print \" The relative humidity is\",round(phi3,4),\"% \\n\"\n",
-      "print \" The resultant temperature is\",round(T3),\"degree celsius\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The relative humidity is 65.0 % \n",
-        "\n",
-        " The resultant temperature is 20.0 degree celsius\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex7.10:PG-219"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "mdotw3=10000.0 # mass flow rate of water entering in cooling tower in kg/min\n",
-      "Tw1=40+273.0 # temperature of water entering cooling tower in kelvin\n",
-      "Ta1=20+273.0 # temperature of air entering cooling tower in kelvin\n",
-      "phi1=0.5# relative humidity of entering air\n",
-      "Tw2=25+273.0 # temperature of water leaving cooling tower in kelvin\n",
-      "Ta2=32+273 # temperature of air leaving cooling tower in kelvin\n",
-      "phi2=0.98 # relative humidity of leaving air\n",
-      "# from psychrometric chart\n",
-      "h1=37.0# specific enthalpy of air @ 20*C DBT and 50% humidity\n",
-      "w1=0.0073 # specific humidity of air @ 20*C DBT and 50% humidity\n",
-      "h2=110.0 # specific enthalpy of air @ 32*C DBT and 98% humidity\n",
-      "w2=0.030 # specific humidity of air @ 32*C DBT and 98% humidity\n",
-      "\n",
-      "h3=167.5 # specific enthalpy of water from steam table at 40 degree celsius\n",
-      "h4=104.9 # specific enthalpy of water from steam table at 25 degree celsius\n",
-      "\n",
-      "mdota=(mdotw3*(h4-h3))/(h1-h2+(w2-w1)*h4) # by energy balance\n",
-      "\n",
-      "\n",
-      "v1=0.84 # specific volume of air entering tower from psychrometric chart\n",
-      "\n",
-      "Qf=mdota*v1 # volume flow rate in m^3/min\n",
-      "print \"The volume flow rate of air into the cooling tower is\",round(Qf),\" m^3/min \\n\"\n",
-      "\n",
-      "mdot4=mdotw3-(w2-w1)*mdota # by mass balance\n",
-      "print \"The mass flow rate of water that leaves the cooling tower \",round(mdot4),\"kg/min\"\n",
-      "# The answers is slightly different in textbook due to approximations in calculations while in python solution is precise\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The volume flow rate of air into the cooling tower is 7446.0  m^3/min \n",
-        "\n",
-        "The mass flow rate of water that leaves the cooling tower  9799.0 kg/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/Chapter9.ipynb b/Thermodynamics_Demystified/Chapter9.ipynb
deleted file mode 100755
index 9448e892..00000000
--- a/Thermodynamics_Demystified/Chapter9.ipynb
+++ /dev/null
@@ -1,630 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:7e02442740cf50f0beae7f1fd9dfdf2fb7a16b9c0848a48a17d80c7a19de3f3d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9:Combustion"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.1:PG-229"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "AFactual=20 # air fuel ratio actual\n",
-      "# The energy balance is done from equation\n",
-      "\n",
-      "# C4H10 + 6.5(O2+3.76N2)-----> 4CO2 + 5H2O + 24.44N2\n",
-      "\n",
-      "P=100 # atmospheic preesure in kPa\n",
-      "mair=6.5*(1+3.76)*29 # mass of air\n",
-      "mfuel=1*58 # mass of fuel\n",
-      "AFth=mair/mfuel # theoritical air-fuel ratio\n",
-      "Pexcessair=(AFactual-AFth)*100/AFth\n",
-      "\n",
-      "print \"The % excess air is\",round(Pexcessair,2),\"% \\n\"\n",
-      "\n",
-      "# NOW THE REACTION IS\n",
-      "# C4H10+ (1+%excessair/100)*6.5*(O2+3.76N2) -----> 4CO2 + 5H2O + 1.903O2 + 31.6N2\n",
-      "\n",
-      "PCO2=4/42.5*100 # VOLUME % OF CO2\n",
-      "\n",
-      "print \"The volume % of CO2 is\",round(PCO2,2),\"% \\n\"\n",
-      "\n",
-      "# NOW WE FIND DEW POINT\n",
-      "Nv=5 # moles of water\n",
-      "N=42.5 # moles of air\n",
-      "Pv=P*(Nv/N) # partial pressure of vapour\n",
-      "Tdp=49# dew point temperature in degree celsius from table C.2\n",
-      "\n",
-      "print \"The Dew point temperature is\",round(Tdp,2),\"degree celsius\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The % excess air is 29.28 % \n",
-        "\n",
-        "The volume % of CO2 is 9.41 % \n",
-        "\n",
-        "The Dew point temperature is 49.0 degree celsius\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.2:PG-231"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "Pair=0.9 # 90% air is used for combustion\n",
-      "\n",
-      "# THE REACTION IS\n",
-      "# C4H10 + 0.9*6.5*(O2+3.76N2)----> aCO2 + 5H20 + bCO\n",
-      "# a and b are calculated by atomic balance\n",
-      "a=2.7\n",
-      "b=1.3\n",
-      "PCO=b*100/31 # volume % of CO\n",
-      "\n",
-      "print \"The volume % of CO is\",round(PCO,2),\"% \\n\"\n",
-      "\n",
-      "mair=6.5*Pair*4.76*29 # mass of air in kg\n",
-      "mfuel=1*58 # mass of fuel butane in kg\n",
-      "AF=mair/mfuel # air-fuel ratio\n",
-      "\n",
-      "print \"The air to fuel ratio is\",round(AF,2),\"kg air/ kg fuel \"\n",
-      "# THE SOLUTION IS CORRECT BUT  THERE ARE SOME PRINTING MISTAKES IN TEXTBOOK\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The volume % of CO is 4.19 % \n",
-        "\n",
-        "The air to fuel ratio is 13.92 kg air/ kg fuel \n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.3:PG-231"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "# THE REACTION IS\n",
-      "# aC4H10 + b(O2+3.76N2)----> CO2 + 1CO + 3.5H20 + 84.6N2 + cH2O\n",
-      "# a, b and c are calculated by atomic balance\n",
-      "# on balancing the equations we get a=3 b=22.5 c=15\n",
-      "# Now equation becomes\n",
-      "#C4H10 + 7.5(O2+3.76N2)----> 3.67CO2 + 0.33CO + 1.17H20 + 28.17N2 + 5H2O\n",
-      "#MOLES OF AIR in this equation is 7.5 moles\n",
-      "mairactual=7.5 # in moles\n",
-      "#MOLES OF AIR in standard equation of Ex.9 is 6.5\n",
-      "mairtheoritical=6.5\n",
-      "Ptheoriticalair=100*(mairactual/mairtheoritical) \n",
-      "print \"The % theoritical air is\",round(Ptheoriticalair,1),\"% \"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The % theoritical air is 115.4 % \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.4:PG-232"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "# The reaction equation is \n",
-      "#CaHb + c(O2+3.76N2)---> 10.4CO2 + 1.2CO + 2.8O2 + 85.6N2 + dH2O\n",
-      "\n",
-      "# using atomic balancing the equations become\n",
-      "\n",
-      "# C11.6H37.9 + 21.08(O2+3.76N2)---> 11.6CO2 + 18.95H2O + 79.26N2\n",
-      "Ptheoriticalair=22.8*100/21.08 # theoritical air\n",
-      "excessair=Ptheoriticalair-100\n",
-      "\n",
-      "print\"The excess air is\",round(excessair),\"%\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The excess air is 8.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.5:PG-235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "# The reaction equation is \n",
-      "#C3H8 + 5(O2+3.76N2)---> 3CO2 + 18.8N2 + 4H2O\n",
-      "\n",
-      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
-      "hfCO2=-393520 # enthalpy associated with CO2\n",
-      "hfH2O=-285830 # enthalpy associated with H2O(l)\n",
-      "hfC3H8=-103850# ehthalpy associated with C3H8\n",
-      "\n",
-      "# by first law Q= Hproducts - Hreactants\n",
-      "\n",
-      "Qg=3*(hfCO2)+4*(hfH2O)-(hfC3H8) # enthalpy of combustion for gaseous propane\n",
-      "\n",
-      "print \" The enthalpy of combustion for gaseous propane is\",round(Qg),\"kJ\\n\"\n",
-      "\n",
-      "hv=15060 # enthalpy of vaporization for propane\n",
-      "\n",
-      "Ql=3*(hfCO2)+4*(hfH2O)-(hfC3H8-hv) # enthalpy of combustion for liquid propane\n",
-      "\n",
-      "print \" The enthalpy of combustion for liquid propane is\",round(Ql),\"kJ\\n\"\n",
-      "\n",
-      "#The answers are slightly different in textbook as they have approximated the result while in Python results are precise\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The enthalpy of combustion for gaseous propane is -2220030.0 kJ\n",
-        "\n",
-        " The enthalpy of combustion for liquid propane is -2204970.0 kJ\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.6:PG-235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "# The reaction equation is \n",
-      "#C3H8 + 5(O2+3.76N2)---> 3CO2 + 18.8N2 + 4H2O\n",
-      "\n",
-      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
-      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
-      "hbarCO2=22280 #enthalpy associated with CO2 at 600K from table E.4\n",
-      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
-      "\n",
-      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
-      "hbarH2O=20402 #enthalpy associated with H20 at 600K from table E.6\n",
-      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
-      "\n",
-      "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n",
-      "\n",
-      "hbarN2=17563 #enthalpy associated with N2 at 600K from table E.2\n",
-      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
-      "# by first law Q= Hproducts - Hreactants\n",
-      "\n",
-      "Qg=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+18.8*(hbarN2-hdotbarN2)-(hfC3H8) # enthalpy of combustion\n",
-      "\n",
-      "print\"The heat transfer required is\",round(Qg),\"kJ\\n\"\n",
-      "\n",
-      "#The answer is WRONG textbook as they have made an error in calculating Qg \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The heat transfer required is -1796043.0 kJ\n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.7:PG-236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "# initialization of variables\n",
-      "# The reaction equation is \n",
-      "\n",
-      "#C8H18 + 12.5(O2+3.76N2)---> 8CO2 + 47N2 + 9H2O\n",
-      "\n",
-      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
-      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
-      "hbarCO2=42769 #enthalpy associated with CO2 at 1000K from table E.4\n",
-      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
-      "\n",
-      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
-      "hbarH2O=35882 #enthalpy associated with H20 at 1000K from table E.6\n",
-      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
-      "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n",
-      "\n",
-      "hbarN2p=(30784+29476)/2 #enthalpy associated with N2 at 1000K from table E.2 by averaging enthalpy at 1020K and 980K for product\n",
-      "hbarN2r=17563 #enthalpy associated with N2 at 600K from table E.2 for reactant\n",
-      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
-      "\n",
-      "hfC8H18=-249910 # enthalpy of formation associated with octane taken from internet as not provided in textbook\n",
-      "\n",
-      "hbarO2=17929 # enthalpy associated with O2 at 600K table E.3\n",
-      "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n",
-      "\n",
-      "# using first law and including kinetic energy change\n",
-      "# 0=Hp-Hr+Mp*(V^2)/2\n",
-      "\n",
-      "Hp=8*(hfCO2+hbarCO2-hdotbarCO2)+9*(hfH2O+hbarH2O-hdotbarH2O)+47*(hbarN2p-hdotbarN2)\n",
-      "# enthalpy of products\n",
-      "\n",
-      "Hr=(hfC8H18)+12.5*(hbarO2-hdotbarO2)+47*(hbarN2r-hdotbarN2)\n",
-      "# enthalpy of reactants\n",
-      "\n",
-      "Mp=8*44+9*18+47*28 #(mass of products by multiplying molecular mass to number of moles)\n",
-      "\n",
-      "V=math.sqrt(2*1000*(Hr-Hp)/Mp)# exit velocity using energy balance\n",
-      "\n",
-      "print \"The exit velocity is\",round(V),\"m/s\"\n",
-      "\n",
-      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The exit velocity is 2116.0 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.8:PG-237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "# The reaction equation with theoritical air is \n",
-      "# C8H18 + 12.5(O2+3.76N2)---> 8CO2 + 47N2 + 9H2O\n",
-      "\n",
-      "# for 400% theoritical air reaction is\n",
-      "\n",
-      "# C8H18 + 50(O2+3.76N2)---> 8CO2 + 188N2 + 9H2O + 37.5O2\n",
-      "\n",
-      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
-      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
-      "hbarCO2=42769 #enthalpy associated with CO2 at 1000K from table E.4\n",
-      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
-      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
-      "hbarH2O=35882 #enthalpy associated with H20 at 1000K from table E.6\n",
-      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
-      "hbarN2p=(30784+29476)/2 #enthalpy associated with N2 at 1000K from table E.2 by averaging enthalpy at 1020K and 980K \n",
-      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
-      "\n",
-      "hfC8H18=-249910 # enthalpy associated with octane taken from internet as not provided in textbook\n",
-      "hbarO2=31389 # enthalpy associated with O2 at 1000K table E.3\n",
-      "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n",
-      "\n",
-      "Hp=8*(hfCO2+hbarCO2-hdotbarCO2)+9*(hfH2O+hbarH2O-hdotbarH2O)+37.5*(hbarO2-hdotbarO2)+188*(hbarN2p-hdotbarN2)# enthalpy of products\n",
-      "\n",
-      "Hr=(hfC8H18)\n",
-      "# enthalpy of reactants\n",
-      "\n",
-      "Q=Hp-Hr # using first law2\n",
-      "\n",
-      "print \" The heat transfer is\",round(Q),\"kJ\"\n",
-      "\n",
-      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The heat transfer is 312593.0 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.9:PG-237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "# The reaction equation is \n",
-      "#C3H8 + 5O2---> 8CO2 + 4H2O\n",
-      "\n",
-      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
-      "hfCO2=-393520 # enthalpy associated with CO2\n",
-      "hfH2O=-241820 # enthalpy associated with gaseous H2O\n",
-      "hfC3H8=103850# enthalpy of formation associated with C3H8\n",
-      "hfgC3H8=15060# enthalpy of vapourization associated with C3H8\n",
-      "T=20+273 # temperature in kelvin\n",
-      "Rbar=8.314 # universal gas constant\n",
-      "Nr=6 # number of moles of reactants\n",
-      "Np=7 # number of moles of products\n",
-      "Hp=3*(hfCO2)+4*(hfH2O) # enthalpy of products\n",
-      "\n",
-      "Hr=hfC3H8+hfgC3H8 # enthalpy of reactants\n",
-      "\n",
-      "Q=(Hp-Hr-(Nr-Np)*Rbar*T)*10**(-3) # heat transfer from first law\n",
-      "\n",
-      "print \" The heat transfer is\",round(Q),\"MJ\"\n",
-      "\n",
-      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The heat transfer is -2264.0 MJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.10:PG-239"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "# The reaction equation for theoritical air is \n",
-      "#C3H8 + 5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 18.8N2\n",
-      "\n",
-      "# for 250% theoritical air reaction becomes\n",
-      "#C3H8 + 12.5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 47N2 + 7.5O2\n",
-      "\n",
-      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
-      "\n",
-      "Np=47+7.5+4+3 # number of moles of product\n",
-      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
-      "hbarCO2=(62963+65271)/2 #enthalpy associated with CO2 at 1380 K from table E.4\n",
-      "hbarCO2dash=(58381+60666)/2 #enthalpy associated with CO2 at 1300 K by average from table E.4\n",
-      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
-      "\n",
-      "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n",
-      "\n",
-      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
-      "hbarH2O=(51521+53351)/2 #enthalpy associated with H20 at 1380 K by taking average from table E.6\n",
-      "hbarH2Odash=48807 #enthalpy associated with H20 at 1300 K  from table E.6\n",
-      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
-      "\n",
-      "hbarN2=42920 #enthalpy associated with N2 at 1380K from table E.2 by interpolating enthalpy between 1020K and 980K \n",
-      "hbarN2dash=40170 #enthalpy associated with N2 at 1300 K from table E.2 \n",
-      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
-      "\n",
-      "hfO2=(44198+45648)/2 # enthalpy associated with O2 at 1380 Kby taking average from table E.3\n",
-      "hfO2dash=48807 # enthalpy associated with O2 at 1380 Kby taking average from table E.3\n",
-      "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n",
-      "\n",
-      "# for adiabatic flame temperature first assume products composed only of nitrogen and Q=0 as adiabatic\n",
-      "hp=(hfC3H8-3*(hfCO2)-4*(hfH2O))/Np +hdotbarN2\n",
-      "# using hp we assume temp=1380 K\n",
-      "# then energy for 1380 k is\n",
-      "H1=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+7.5*(hfO2-hdotbarO2)+47*(hbarN2-hdotbarN2) # energy assuming temperature to be 1380 K\n",
-      "\n",
-      "#this is very large \n",
-      "\n",
-      "# now at 1300 K adiabatic temperature\n",
-      "H2=3*(hfCO2+hbarCO2dash-hdotbarCO2)+4*(hfH2O+hbarH2Odash-hdotbarH2O)+7.5*(hfO2dash-hdotbarO2)+47*(hbarN2dash-hdotbarN2) # energy assuming temperature to be 1300 K\n",
-      " \n",
-      " # now interpolation between these two temperatures\n",
-      "Tp=1300-((hp+H2)/(H1-H2))*(1380-1300) # adiabatic temperature by interpolation\n",
-      "print \"The adiabatic flame temperature is\",round(Tp),\"K\"\n",
-      "\n",
-      "#The answers is different in textbook as they have printed the value of hfCO2 with positive sign while calculating H2\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The adiabatic flame temperature is 1311.0 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex9.11:PG-240"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "# The reaction equation for theoritical air is \n",
-      "#C3H8 + 5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 18.8N2\n",
-      "\n",
-      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
-      "\n",
-      "Np=18.8+4+3 # number of moles of product\n",
-      "hfCO2=-393520 # enthalpy associated with CO2\n",
-      "hbarCO2=137400 #enthalpy associated with CO2 at 2600 K from table E.4 by interpolation\n",
-      "hbarCO2dash=125152 #enthalpy associated with CO2 at 2400 K from table E.4\n",
-      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
-      "\n",
-      "hfC3H8=-103850# ehthalpy associated with C3H8\n",
-      "\n",
-      "hfH2O=-241820 # enthalpy associated with gaseous H2O\n",
-      "hbarH2O=114273 #enthalpy associated with H20 at 2600 K from table E.6\n",
-      "hbarH2Odash=103508 #enthalpy associated with H20 at 2400 K  from table E.6\n",
-      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
-      "\n",
-      "hbarN2=86600 #enthalpy associated with N2 at 2600 K from table E.2 by interpolation\n",
-      "hbarN2dash=79320 #enthalpy associated with N2 at 2400 K from table E.2 \n",
-      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
-      "\n",
-      "# for adiabatic flame temperature first assume products composed only of nitrogen and Q=0 as adiabatic\n",
-      "hp=(hfC3H8-3*(hfCO2)-4*(hfH2O))/Np +hdotbarN2 \n",
-      "\n",
-      "# using hp we assume temp=2600 K\n",
-      "# then energy for 2600 k is\n",
-      "H1=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+18.8*(hbarN2-hdotbarN2) # energy assuming temperature to be 2600 K\n",
-      "\n",
-      "# now at 2400 K adiabatic temperature\n",
-      "H2=3*(hfCO2+hbarCO2dash-hdotbarCO2)+4*(hfH2O+hbarH2Odash-hdotbarH2O)+18.8*(hbarN2dash-hdotbarN2) # energy assuming temperature to be 2400 K\n",
-      " \n",
-      " # now interpolation between these two temperatures\n",
-      "Tp=2400-((hp+H2)/(H1-H2))*(2600-2400) # adiabatic temperature by interpolation\n",
-      "print \"The adiabatic flame temperature is\",round(Tp),\"K\"\n",
-      "\n",
-      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The adiabatic flame temperature is 2409.0 K\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified/README.txt b/Thermodynamics_Demystified/README.txt
deleted file mode 100755
index 2aa18a7d..00000000
--- a/Thermodynamics_Demystified/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Aviral Yadav
-Course: btech
-College/Institute/Organization: ABES Engineering College
-Department/Designation: Mechanical Engineering
-Book Title: Thermodynamics Demystified
-Author: Merle C. Potter
-Publisher: Mcgraw-hill, Usa
-Year of publication: 2009
-Isbn:  0071605991
-Edition: 1
\ No newline at end of file
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diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter1_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter1_2.ipynb
new file mode 100755
index 00000000..68c66861
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter1_2.ipynb
@@ -0,0 +1,209 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2622b864e241f67942d9af83d84aabbbf519132ac138f1180d8869df35c708b4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "CHAPTER 1 : Basic Principles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.2 : PG-9 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "m=10 # mass in Kg\n",
+      "V=5 # velocity in m/s\n",
+      "\n",
+      "KE=m*V**2/2 # kinetic energy in N-m \n",
+      "print \"The Kinetic Energy is \",round(KE),\" N.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Kinetic Energy is  125.0  N.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 77
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.3 : PG-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "V= 3*5*20; # Volume of air in m^3 from dimensions\n",
+      "m= 350.0; # mass in kg\n",
+      "g= 9.81; # gavitational acceleration in m/s^2\n",
+      "\n",
+      "rho=m/V;# density\n",
+      "print \" The Density is \",round(rho,3),\"kg/m^3 \\n\"\n",
+      "\n",
+      "v= 1/rho # specific volume of air\n",
+      "print \" The specific volume  is\", round(v,3),\"m^3/kg \\n\"\n",
+      "\n",
+      "gama= rho*g # specific weight of air\n",
+      "print \" The specific weight is\", round(gama,2),\" N/m^3\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The Density is  1.167 kg/m^3 \n",
+        "\n",
+        " The specific volume  is 0.857 m^3/kg \n",
+        "\n",
+        " The specific weight is 11.45  N/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 78
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.4 : PG-13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "h=0.020 # height of mercury in m\n",
+      "gammawater=9810 # specific weight of water in N/m^3\n",
+      "Patm=0.7846*101.3 # atmospheric pressure in kPa from table B.1\n",
+      "\n",
+      "Pgauge=13.6*gammawater*h/1000 # pressure in Pascal from condition gammaHg=13.6*gammawater\n",
+      "\n",
+      "P=(Pgauge+Patm)# absolute pressure in KPa\n",
+      "#result\n",
+      "print \"The Pressure is\",round(P,2),\" kPa\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Pressure is 82.15  kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 79
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.5 : PG-13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# initialization of variables\n",
+      "d=10.0/100 # diameter of cylinder in 'm'\n",
+      "P=600 # pressure in KPa\n",
+      "Patm=100 # atmospheric pressure in Kpa\n",
+      "K=4.8*1000 # spring constant in N/m \n",
+      "\n",
+      "deltax=(P-Patm)*(math.pi*1000*d**2)/(4*K) # by balancing forces on piston\n",
+      "#result\n",
+      "print \"The Compression in spring is\",round(deltax,3),\" m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Compression in spring is 0.818  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 80
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1.6 : PG-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "ma=2200 # mass of Automobile 'a' in kg\n",
+      "va=25 #velocity of Automobile 'a' in m/s before collision\n",
+      "va1=13.89 # velocity of Automobile 'a' after collision in m/s\n",
+      "mb=1000 # mass of Automobile 'b' in kg\n",
+      "vb=24.44 #velocity of Automobile 'b' after collision in m/s\n",
+      "\n",
+      "KE1=(ma*va**2)/2 # kinetic energy before collision\n",
+      "KE2=(ma*va1**2)/2+(mb*vb**2)/2 # kinetic energy after collision\n",
+      "U=(KE1-KE2)/1000 # internal energy from conservation of energy principle in kJ\n",
+      "#result\n",
+      "print \"The increase in kinetic energy is of\",round(U,1),\" kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The increase in kinetic energy is of 176.6  kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 81
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter2_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter2_2.ipynb
new file mode 100755
index 00000000..d8357657
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter2_2.ipynb
@@ -0,0 +1,322 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:63ec806cc789e6d4e52abc481b4c69dbae196d3a1063d71df1549638fb7b9752"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2:Properties of Pure Substances"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.1, PG-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "m=10;  #mass of saturated water in kg\n",
+      "\n",
+      "    # All the necessary values are taken from table C.2\n",
+      " \n",
+      "# part (a)\n",
+      " \n",
+      "P=0.001; # Pressure in MPa\n",
+      "vf=0.001; #specific volume of saturated liquid at 0.001 Mpa in Kg/m^3\n",
+      "vg=129.2; # specific volume of saturated vapour at 0.001 Mpa in Kg/m^3\n",
+      "deltaV=m*(vg-vf) # by properties of pure substance \n",
+      "# result\n",
+      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV),\" m^3 \\n\"\n",
+      "\n",
+      "# part (b) \n",
+      "\n",
+      "P=0.26;  # Pressure in MPa\n",
+      "vf=0.0011; #   specific volume of saturated liquid at 0.26 MPa( it is same from at 0.2 and 0.3 MPa upto 4 decimals)\n",
+      "vg=(P-0.2)*(0.6058-0.8857)/(0.3-0.2)+0.8857;  # specific volume of saturated vapour by interpolation of Values at 0.2 MPa and 0.3 MPa\n",
+      "deltaV=m*(vg-vf) # by properties of pure substance \n",
+      "# result\n",
+      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,2),\" m^3 \\n\"\n",
+      "\n",
+      "# part (c) \n",
+      "P=10;  # Pressure in MPa\n",
+      "vf=0.00145; # specific volume of saturated liquid at 10 MPa\n",
+      "vg=0.01803; # specific volume of saturated vapour at 10 MPa\n",
+      "deltaV=m*(vg-vf) # by properties of pure substance \n",
+      "# result\n",
+      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,4),\" m^3 \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Volume change at pressure  0.001  MPa is 1292.0  m^3 \n",
+        "\n",
+        "The Volume change at pressure  0.26  MPa is 7.17  m^3 \n",
+        "\n",
+        "The Volume change at pressure  10  MPa is 0.1658  m^3 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.2, PG-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#    initialization of variables\n",
+      "m=4.0 # mass of water in kg\n",
+      "V=1.0 # volume in m^3\n",
+      "T=150 # temperature of water in degree centigrade\n",
+      "\n",
+      "# TABLE C.1 is used for values in wet region\n",
+      "# Part (a)\n",
+      "P=475.8 # pressure in KPa in wet region at temperature of 150 *C\n",
+      "print \" The pressure is\",round(P,2),\" kPa \\n\"\n",
+      "\n",
+      "# Part (b)\n",
+      "#first we determine the dryness fraction\n",
+      "v=V/m # specific volume of water\n",
+      "vg=0.3928 #  specific volume of saturated vapour @150 degree celsius\n",
+      "vf=0.00109 # specific volume of saturated liquid @150 degree celsius\n",
+      "x=(v-vf)/(vg-vf); # dryness fraction\n",
+      "mg=m*x; # mass of vapour\n",
+      "print \" The mass of vapour present is\",round(mg,3),\" kg \\n\"\n",
+      "\n",
+      "# Part(c) \n",
+      "Vg=mg*vg; # volume of vapour\n",
+      "print \" The volume of vapour is\",round(Vg,3),\" m^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The pressure is 475.8  kPa \n",
+        "\n",
+        " The mass of vapour present is 2.542  kg \n",
+        "\n",
+        " The volume of vapour is 0.998  m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.3, PG-29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#    initialization of variables\n",
+      "m=2 # mass of water in kg\n",
+      "P=220 # pressure in KPa\n",
+      "x=0.8 # quality of steam\n",
+      "\n",
+      "# Table C.2 is used for values\n",
+      "\n",
+      "vg=(P-200)*(0.6058-0.8857)/(300-200)+0.8857 # specific volume of saturated vapour @ given pressure by interpolating\n",
+      "vf=0.0011 # specific volume of saturated liquid @ 220 KPa\n",
+      "v=vf+x*(vg-vf)# property of pure substance\n",
+      "V=m*v # total volume\n",
+      "#result\n",
+      "print \"The Total volume of the mixture is \",round(V,3),\" m^3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Total volume of the mixture is  1.328  m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.4, PG-30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#   initialization of variables\n",
+      "m=2 # mass of water in kg\n",
+      "P=2.2 # pressure in Mpa\n",
+      "T=800 # temperature in degree centigrade\n",
+      " # Table C.3 is used for values\n",
+      "v=0.2467+(P-2)*(0.1972-0.2467)/(2.5-2) # specific volue by interpolatin between 2 and 2.5 MPa\n",
+      "V=m*v # final volume\n",
+      "print \"The Final Volume is\",round(V,3),\" m^3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Final Volume is 0.454  m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.5, PG-32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#    initialization of variables\n",
+      "V=0.6 # volume of tyre in m^3\n",
+      "Pgauge=200 # gauge pressure in KPa\n",
+      "T=20+273 # temperature converted to kelvin\n",
+      "Patm=100 # atmospheric pressure in KPa\n",
+      "R=287 # gas constant in Nm/kg.K\n",
+      "Pabs=(Pgauge+Patm)*1000 # calculating absolute pressue in Pa \n",
+      "\n",
+      "m=Pabs*V/(R*T)# mass from ideal gas equation\n",
+      "print \"The Mass of air is\",round(m,2),\" Kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Mass of air is 2.14  Kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.6, PG-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#    initialization of variables\n",
+      "T=500+273 # temperature of steam in kelvin\n",
+      "rho=24.0 # density in Kg/m^3\n",
+      "R=0.462 # gas constant from Table B.2\n",
+      "v=1/rho # specific volume and density relation\n",
+      "# PART (a)\n",
+      "P=rho*R*T # from Ideal gas equation\n",
+      "print \" PART (a) The Pressure is \",int(P),\" KPa \\n\"\n",
+      "# answer is approximated in textbook\n",
+      "\n",
+      "# PART (b)\n",
+      "a=1.703 #  van der Waal's constant a value from Table B.7\n",
+      "b=0.00169 # van der Waal's constant b value from Table B.7\n",
+      "P=(R*T/(v-b))-(a/v**2) # Pressure from van der Waal's equation\n",
+      "print \" PART (b) The Pressure is \",int(P),\" KPa \\n\"\n",
+      "# answer is approximated in textbook\n",
+      "\n",
+      "# PART (c)\n",
+      "a=43.9 # van der Waal's constant a value from Table B.7\n",
+      "b=0.00117 # van der Waal's constant b value from Table B.7\n",
+      "\n",
+      "P=(R*T/(v-b))-(a/(v*(v+b)*math.sqrt(T))) # Redlich-Kwong equation\n",
+      "print \" PART (c) The Pressure is \",int(P),\" KPa \\n\"\n",
+      "# answer is approximated in textbook\n",
+      "\n",
+      "# PART (d)\n",
+      "Tcr=947.4 # compressibilty temperature from table B.3\n",
+      "Pcr=22100 # compressibility pressure from table B.3\n",
+      "\n",
+      "TR=T/Tcr # reduced temperature\n",
+      "PR=P/Pcr # reduced pressure\n",
+      "Z=0.93 # from compressiblility chart\n",
+      "P=Z*R*T/v # Pressure in KPa\n",
+      "print \" PART (d) The Pressure is \",int(P),\" KPa \\n\"\n",
+      "# answer is approximated in textbook\n",
+      "\n",
+      "# PART (e)\n",
+      "P=8000 # pressure from steam table @ 500*c and v= 0.0417 m^3\n",
+      "print \" PART (e) The Pressure is \",int(P),\" KPa \\n\"\n",
+      "# answer is approximated in textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " PART (a) The Pressure is  8571  KPa \n",
+        "\n",
+        " PART (b) The Pressure is  7952  KPa \n",
+        "\n",
+        " PART (c) The Pressure is  7934  KPa \n",
+        "\n",
+        " PART (d) The Pressure is  7971  KPa \n",
+        "\n",
+        " PART (e) The Pressure is  8000  KPa \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter3_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter3_2.ipynb
new file mode 100755
index 00000000..f646fae2
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter3_2.ipynb
@@ -0,0 +1,330 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f6a6bb505322436c23c138c1e60d951ed65e1f5f903d0d7ef3e7664203b1c411"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3:Work and Heat"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.1:PG-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialization of variables\n",
+      "m=1            # mass in kg\n",
+      "x=20.0/100.0       #quality of steam\n",
+      "P=200          #constant pressure in kPa\n",
+      "T1=100         #temperature intitial in degree centigrade\n",
+      "T2=400         #temperature final in degree centigrade\n",
+      "\n",
+      "\n",
+      "# first we find initial volume v1 and final volume v2\n",
+      "\n",
+      "# using table C.2\n",
+      "vf=0.001061 # specific volume of saturated liquid in m^3 per kg \n",
+      "vg=0.8857   # specific volume of saturated vapour in m^3 per kg \n",
+      "\n",
+      "v1=vf+x*(vg-vf);\n",
+      "\n",
+      "v2=1.549   # specific volume of steam in m^3 per kg at T2=400*C and P2=0.2MPa\n",
+      "# now calculate work\n",
+      "W=m*P*(v2-v1);   # work done in constant pressure process\n",
+      "#result\n",
+      "print \"Work done is\",round(W,2),\" kJ\" # work is in kJ as pressure was in kPa"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work done is 274.2  kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.2:PG-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialization of variables\n",
+      "D=110.0/1000.0 # diameter of cylinder in m\n",
+      "V1=100e-6 # initial volume@ state 1 in m^3\n",
+      "T1=60.0 # initial temp @ state 1 in *C\n",
+      "T2=200.0 # final temo @ state 2 in *C\n",
+      "M=50 # weight of piston in kg\n",
+      "g=9.81 # gravitational accleration in m/sec^2\n",
+      "Patm=100000 # atmospheric pressure in Pa\n",
+      "A=math.pi*(D**2)/4 # area of piston in m^2\n",
+      "\n",
+      "# BALANCING THE FORCES To GET PRESSURE P\n",
+      "# M.g=P.A-Patm\n",
+      "P=Patm+(M*g/A) # atm pressure is added to get absolute pressure\n",
+      "\n",
+      "v1=0.001017 # specific volume at 60*C and 0.15Mpa pressure\n",
+      "m=V1/v1; # mass of water in kg\n",
+      "\n",
+      "# find volume at state 2 \n",
+      "v2=1.444 # specific volume of steam at 200*C and 0.15 MPa\n",
+      "V2=m*v2 # final volume in m^3\n",
+      "\n",
+      "W=P*(V2-V1)/1000; # work done divided by 1000 to get in kJ\n",
+      "# result\n",
+      "print \"The work done is\",round(W,1),\" kJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The work done is 21.5  kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.3:PG-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#initialization of variables\n",
+      "P1=200      # initial pressure in kPa\n",
+      "V1=2        # initial volume in m^3\n",
+      "P2=100      # final pressure in kPa\n",
+      "C=P1*V1 # isothermal process i.e P.V=constant\n",
+      "#  find final volume \n",
+      "V2=P1*V1/P2 # final volume by P1.V1=P2.V2\n",
+      "\n",
+      "from scipy.integrate import quad\n",
+      " \n",
+      "def integrand(v,C):\n",
+      "    return C/v\n",
+      "W, err =quad(integrand, V1, V2,C ) # itegrating over volume to get work\n",
+      "# result\n",
+      "print \"The Work done by gas is\",int(W),\" kJ\" # answer is approximated in textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Work done by gas is 277  kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.4:PG-48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "M=100 # mass in kg\n",
+      "d=3 # depth by which mass drops in m\n",
+      "V=0.002 # increased volume in m^3\n",
+      "g=9.8 # gravitational accleration in m/sec^2\n",
+      "Pgauge=100*1000 # gauge pressure in N/m\n",
+      "Patm =100*1000 # atmospheric pressure in N/m\n",
+      "P=Pgauge+Patm # to get absolute pressure\n",
+      "\n",
+      "# calculate work done by paddle wheel\n",
+      "Wpaddlewheel=(-M*g*d) # work is negative as it is done on the system\n",
+      "\n",
+      "# calculate work done on piston it \n",
+      "Wboundary=P*V # area mulitiplied by height is volume thus W=P.V \n",
+      "\n",
+      "#net work\n",
+      "\n",
+      "Wnet=Wpaddlewheel+Wboundary; # Work in joule as SI units are used\n",
+      "print \"The Net Work done is \",round(Wnet),\" J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Net Work done is  -2540.0  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.5:PG-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# initialization of variables\n",
+      "T=100 # torque of shaft in N.m\n",
+      "N=3000 # rotation speed in rpm\n",
+      "omega=(N*2*math.pi/60) # angular velocity in rad/sec\n",
+      "# calculation of power\n",
+      "Wdot=(T*omega); # power is work done per unit time\n",
+      "print \"Power transmitted is\",round(Wdot/746,1),\" hp\" # divided by 746 to convert W into hp\n",
+      "#answer is approximated in textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power transmitted is 42.1  hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.6:PG-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# initialization of variables\n",
+      "D=10.0/100    # diameter of cylinder in m\n",
+      "d=50.0/1000   # compression in spring in m\n",
+      "Patm=100000   # atmospheric pressure in Pa\n",
+      "K=10.0*1000   # spring constant converted in  N/m\n",
+      "w=50*9.81 # weight of piston in Newton =mass*gravitational acceleration\n",
+      "\n",
+      "# find the initial pressure in cylinder by force balance\n",
+      "A=(math.pi*D**2)/4; # area of piston\n",
+      "P1=((Patm*A)+w)/A;      # balancing forces on piston P1.A=Patm.A+W\n",
+      "\n",
+      "# work done by air to raise the piston for 50mm if spring not present\n",
+      "Wgas=P1*A*d; # pressure*area= force and Work = Force* displacement\n",
+      "\n",
+      "# work done on spring to compress\n",
+      "Wspring=(K*d**2)/2; # Work in j\n",
+      "\n",
+      "# now total work done by air is sum of two works\n",
+      "Wnet=Wgas+Wspring; # Work in j\n",
+      "# result\n",
+      "print \"The net work done by air is\",round(Wnet,1),\" J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net work done by air is 76.3  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3.7:PG-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variable initialization\n",
+      "\n",
+      "d=2   # distance travelled by weight in m\n",
+      "m=50  # mass of weight in kg\n",
+      "g=9.8  # gravitaional acceleration in m/sec^2\n",
+      "\n",
+      "# calculation of work in non-quasiequilibrium process\n",
+      "W=m*g*d; # work in joules\n",
+      "\n",
+      "# the work done must be transferred as heat\n",
+      "Q=W;\n",
+      "\n",
+      "print \"The heat that must transfer is \",int(Q),\" Joules\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat that must transfer is  980  Joules\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter4_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter4_2.ipynb
new file mode 100755
index 00000000..e17b48f1
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter4_2.ipynb
@@ -0,0 +1,719 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:abd0537a9b98c6cb49f3032d9229bd28210acff6f4a7d9f1c5434293bf15ef92"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 4:The First Law of Thermodynamics"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.1:PG-62"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#initialization of variables\n",

+      "K=100 # spring constant in kN/m\n",

+      "d=0.8 # dispacement of spring in m\n",

+      " # to get total work we integrate from 0 to 0.8 displacement\n",

+      "x1=0; # lower limit of integration\n",

+      "x2=0.8;  # upper limit of integration\n",

+      "from scipy.integrate import quad\n",

+      "\n",

+      "# we find work\n",

+      "def integrand(x,K):\n",

+      "    return K*x\n",

+      "\n",

+      "W12, err = quad(integrand, x1, x2, K) # integrating to get work\n",

+      "Q12=W12; # by first law of thermodynamics\n",

+      "print \"The Heat transfer is \",int(Q12),\" J\"\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Heat transfer is  32  J\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.2:PG-65"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "P= 5*746 # power of fan converted in watt\n",

+      "t=1*60*60 # time converted to seconds\n",

+      "\n",

+      "# by first law of thermodynamics Q=delU + W\n",

+      "# Q=0 hence -W=delU\n",

+      "# first we find work input\n",

+      "W=-P*t # work in J\n",

+      "delU=-W # from 1st law\n",

+      "print \"The internal energy increase is \",float(delU),\" J\"\n",

+      "# The answer is approximated in textbook\n",

+      "# our answer is precise\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The internal energy increase is  13428000.0  J\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.3:PG-65"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "P=400 # pressure in kPa\n",

+      "T1=200 # initial temperature in degree celsius\n",

+      "V1= 2 # initial volume in m^3\n",

+      "Q=3500 # heat added in kJ\n",

+      "v1=0.5342 # specific volume of steam at 200 degree celcius and 0.4 Mpa pressure from table C.3\n",

+      "u1=2647 # specific internal energy in kJ/kg @ pressure = 0.4 MPa\n",

+      "m=V1/v1 # mass in kg\n",

+      "# we have a relation Between u2 and v2 from 1st law of thermodynamics\n",

+      "v2=1.06 # specific volume at state 2 by trial and error and interpolation\n",

+      "V2=m*v2 \n",

+      "u2=((3500-400*(V2-V1))/m)+2647 # specific internal energy for v2=1.06 by trial and error\n",

+      "\n",

+      "# on interpolation from steam table at 0.4 MPa we get temperature \n",

+      "T2=644 # temperature in degree celsius\n",

+      "print \"The temperature for u2=\",round(u2),\" kJ and  v2 =\",round(v2,3),\" kg/m^3 is \\n \",int(T2),\" degree celsius\"\n",

+      "# this numerical is solved by trial and error thus refer to Appendix C"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The temperature for u2= 3372.0  kJ and  v2 = 1.06  kg/m^3 is \n",

+        "  644  degree celsius\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.4:PG-67"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "\n",

+      "# initialization of variables\n",

+      "P=400 #  pressure in kPa\n",

+      "T1=200 # initial tmperature in degree celsius\n",

+      "V=2 # initial volume in m^3\n",

+      "Q=3500 # heat added in kJ\n",

+      "\n",

+      "#solution\n",

+      "h1=2860 # initial enthalpy @ 200*C and 400 kPa from steam table\n",

+      "v=0.5342 # specific volume from steam table C.3 \n",

+      "m=V/v;\n",

+      "h2=(Q/m)+h1; # final enthalpy in kJ/kg from energy equation\n",

+      "\n",

+      "# NOW USING THIS ENTHLAPY AND INTERPOlATING FROM STEAM TABLE\n",

+      "\n",

+      "T2=600+(92.6/224)*100\n",

+      "\n",

+      "print \"The Final temperature is  \",int(T2),\" degree Celsius\"\n",

+      "# result is obtained from interpolation on steam table\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Final temperature is   641  degree Celsius\n"

+       ]

+      }

+     ],

+     "prompt_number": 20

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.5:PG-71"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "T1=300 # initial temperature in degree celsius\n",

+      "T2=700 # final temperature in degree celsius\n",

+      "P=150 # pressure in kPa\n",

+      "m=3 # mass of steam in kg\n",

+      "\n",

+      "# solution\n",

+      "# part (a)\n",

+      "from scipy.integrate import quad\n",

+      "\n",

+      "# now we make function to integrate\n",

+      "def integrand(T):\n",

+      "    return 2.07+(T-400)/1480\n",

+      "\n",

+      "I, err = quad(integrand, T1, T2) # integrating specific heat over temperature range\n",

+      "delH=m*I #integrate('2.07+(T-400)/1480','T',T1,T2) # expressing as function of temperature and integrating\n",

+      "\n",

+      "print\" The change in Enthalpy is \",int(delH),\" kJ \\n\"\n",

+      " \n",

+      "# part(b)\n",

+      "CPavg=delH/(m*(T2-T1)) # avg value of specific heat at constant pressure\n",

+      "print \" The average value of Cp is \",round(CPavg,2),\" kJ/kg.*C\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The change in Enthalpy is  2565  kJ \n",

+        "\n",

+        " The average value of Cp is  2.14  kJ/kg.*C\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.6,PG-72"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "m=1 # mass of nitrogen in kg\n",

+      "T1=300 # initial temperature in Kelvin\n",

+      "T2=1200 # final temperature in Kelvin\n",

+      "M=28.0 # in kg/kmol\n",

+      "# part(a)\n",

+      "# the enthalpy change is found from gas table in App.E\n",

+      "delh=36777-8723 # from gas table\n",

+      "delH=delh/M \n",

+      "print \" The entalpy change from gas table is \",round(delH),\" kJ/kg \\n\"\n",

+      "\n",

+      "# part (b) \n",

+      "Cp=1.042 # from table B.2\n",

+      "delH=Cp*(T2-T1)\n",

+      "print \" The entalpy change  by assuming constant specific heat is \",round(delH),\" kJ/kg\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The entalpy change from gas table is  1002.0  kJ/kg \n",

+        "\n",

+        " The entalpy change  by assuming constant specific heat is  938.0  kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 29

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.7:PG-76"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "x=0.7 # quality of steam\n",

+      "P1=200 # initial pressure in kPa\n",

+      "P2=800 # final pressure in kPa\n",

+      "V=2 # volume in m^3\n",

+      "# The values are taken from TABLE C.2\n",

+      "vf1=0.0010 # specific volume of saturated liquid at 200 kPa\n",

+      "vg1=0.8857 # specific volume of saturated gas at 200 kPa\n",

+      "uf1=504.5  # specific internal energy of saturated liquid @ state 1\n",

+      "ug1=2529.5 # speciific internal energy of saturated gas @ state 1\n",

+      "\n",

+      "v1=vf1+x*(vg1-vf1); # specific volume of vapour\n",

+      "m=V/v1\n",

+      "\n",

+      "u1=uf1+x*(ug1-uf1) # specific internal energy of vapour @ state 1\n",

+      "v2=v1 # constant volume process\n",

+      "u2=((0.6761-0.6203)*(3661-3853)/(0.6761-0.6181))+3853 # from steam table @ 800kPa by interpolating\n",

+      "Q=m*(u2-u1) # heat transfer\n",

+      "print \"The heat transfer is \",round(Q,3),\" kJ\"\n",

+      "# The answer in the textbook is approximated"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The heat transfer is  5630.537  kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 34

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.8:PG-76"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "V=0.02 # volume in m^3\n",

+      "P=400 # pressure in kPa\n",

+      "T1=50+273 # initial temperature in kelvin\n",

+      "T2=700+273 # final temperature in kelvin\n",

+      "Q=50 # heat added in kJ\n",

+      "R=287 # constant for air\n",

+      "Cp=1 # constant for specific heat of air\n",

+      "\n",

+      "# using the ideal gas equation\n",

+      "\n",

+      "m=P*1000*V/(R*T1) # mass of air in kg\n",

+      "W=Q-(m*Cp*(T2-T1)) # work done from first law\n",

+      "# result\n",

+      "print \"The Paddle work is \",round(W,2),\" kJ\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Paddle work is  -6.09  kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 37

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.9,PG-77"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "V1=2 # initial volume in m^3\n",

+      "V2=0.2 # final volume in m^3\n",

+      "T1=20+273 # temperature in kelvin\n",

+      "P=200 # pressure in kPa\n",

+      "R=0.287 # constant for air\n",

+      "gama=1.4 # polytropic index for air\n",

+      "Cv=0.717 # specific heat at constant volume for air\n",

+      "\n",

+      "#solution\n",

+      "\n",

+      "#using the ideal gas equation\n",

+      "m=(P*V1)/(R*T1) # mass in kg\n",

+      "# process is adiabatic thus\n",

+      "T2=T1*((V1/V2)**(gama-1)) # final temperature\n",

+      "\n",

+      "W=-m*Cv*(T2-T1) # work from first law\n",

+      "print \"The Work is \",int(W),\" kJ\"\n",

+      "# solution is approximated in textbook"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Work is  -1510  kJ\n"

+       ]

+      }

+     ],

+     "prompt_number": 41

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.10:PG-79"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "# initialization of variables\n",

+      "P1=2000.0 # initial pressure in kPa\n",

+      "T1=600.0 # initial temperature in degree celsius\n",

+      "p2=600.0 # final pressure in kPa\n",

+      "T2=200.0 #   final temperature in degree celsius\n",

+      "d1=0.06 # diameter of inlet pipe in metre\n",

+      "d2=0.120 # diameter of outlet pipe in metre\n",

+      "V1=20.0 # velocity at inlet in m/s\n",

+      "\n",

+      "# solution\n",

+      "# from superheat table C.3 values are noted\n",

+      "v1=0.1996 # specific volume of superheated steam @ 600*C and 2000 kPa\n",

+      "v2=0.3520 # specific volume of superheated steam @ 200*C and 2000 kPa\n",

+      "rho1=1/v1 #  initial density\n",

+      "rho2=1/v2 # final density\n",

+      "A1=(math.pi*d1**2)/4 # inlet area\n",

+      "A2=(math.pi*d2**2)/4 # exit area\n",

+      "\n",

+      "V2=(rho1*A1*V1)/(rho2*A2) # from continuity equation\n",

+      "print \" The Exit velocity is \",round(V2,2),\" m/s \\n\"\n",

+      "\n",

+      "mdot=rho1*A1*V1 # mass flow rate\n",

+      "print\" The mass flow rate is \",round(mdot,3),\" kg/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Exit velocity is  8.82  m/s \n",

+        "\n",

+        " The mass flow rate is  0.283  kg/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 45

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.11:PG-82"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "P1=8000 # initial pressure in kPa\n",

+      "T1=300  # temperature in degree celsius\n",

+      "P2=2000 # final pressure in kPa\n",

+      "\n",

+      "# solution\n",

+      "h1=2785 # specific enthalpy of steam in kJ/kg @ 8000 kPa and 300 degree celsius from  steam table\n",

+      "h2=h1 # throttling process thus enthalpy is constant\n",

+      "T2=212.4 # from steam table as we know enthalpy and pressure\n",

+      "hf2=909 # specific enthalpy of saturated liquid @ 2000 kPa and 300 degree celsius\n",

+      "hg2=2799.5 # specific enthalpy of saturated gas @ 2000 kPa and 300 degree celsius\n",

+      "x2=(h2-hf2)/(hg2-hf2) # quality of steam\n",

+      "\n",

+      "vg2=0.0992 # specific    volume of saturated gas @ 2000 kPa and 212.4*c\n",

+      "vf2=0.0012 # specific volume of saturated liquid @ 2000 kPa and 212.4*c\n",

+      "v2=vf2+x2*(vg2-vf2) # from properties of pure substance\n",

+      "\n",

+      "print \"The Final Temperature and Specific volume is \",round(T2,1),\"*C and \",round(v2,3),\" m^3/kg\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Final Temperature and Specific volume is  212.4 *C and  0.098  m^3/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 50

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.12:PG-84"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "# initialization of variables\n",

+      "P1=4000 # inlet pressure in kPa\n",

+      "T1=500 # inlet temperature in degree celsius\n",

+      "V1=200 # inlet steam velocity in m/s\n",

+      "d1=0.05 # inlet diameter in 'm'\n",

+      "P2=80 # exit pressure in kPa\n",

+      "d2=0.250 # exit diameter in 'm'\n",

+      "\n",

+      "# solution\n",

+      "v1=0.08643 # specific volume from steam table @ 4000 kPa and 500*C\n",

+      "v2=2.087 # specific volume from steam table @ 80 kPa and 500*C\n",

+      "rho1=1/v1 # density at inlet\n",

+      "rho2=1/v2 # density at outlet\n",

+      "A1=(math.pi*d1**2)/4 # inlet area\n",

+      "A2=(math.pi*d2**2)/4\n",

+      "mdot=rho1*A1*V1 # mass flow rate\n",

+      "mdot=round(mdot,3) # rounding to 3 significant digits\n",

+      "\n",

+      "#now using table C.3\n",

+      "h1=3445 # initial specific enthalpy @ 4000 kPa and 500 *C \n",

+      "h2=2666 # final specific enthalpy @ 80 kPa and 500 *C\n",

+      "WT=-mdot*(h2-h1) # maximum power from first law\n",

+      "print \" The power output is \",round(WT),\" kJ/s \\n \"\n",

+      "\n",

+      "V2=(A1*V1*rho1)/(A2*rho2) \n",

+      "V2=round(V2) # rounding of digits\n",

+      "delKE=mdot*((V2**2)-(V1**2))/2 # the change in kinetic energy\n",

+      "print \" The change in K.E is \",round(delKE),\" J/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The power output is  3540.0  kJ/s \n",

+        " \n",

+        " The change in K.E is  -6250.0  J/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 78

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.13:PG-85"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "# initialization of variables\n",

+      "Wdot=10 # pump power in hp\n",

+      "g=9.81 # acceleration due to gravity\n",

+      "rho=1000 # density of water in kg/m^3\n",

+      "d1=0.06 # inlet dimeter in 'm'\n",

+      "d2=0.10 # oulet diamter in 'm'\n",

+      "V1=10 # velocity of water at inlet in m/s\n",

+      "\n",

+      "#solution\n",

+      "A1=math.pi*(d1**2)/4 # area of inlet\n",

+      "A2=math.pi*(d2**2)/4 # area of outlet\n",

+      "V2=A1*V1/A2 # oulet velocity from continuity equation\n",

+      "\n",

+      "mdot=rho*A1*V1 # mass flow rate\n",

+      "delP=((((Wdot*746)/mdot)-((V2**2)-V1**2)/(2*g))*rho)/1000 # change in pressure in kPa\n",

+      "print \"The rise in pressure is \",round(delP),\" kPa\"\n",

+      "# The answer is approximated in textbook , our answer is precise \n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The rise in pressure is  268.0  kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 80

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.14:PG-85"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "# initialization of variables\n",

+      "P1=7000.0 # inlet pressure in Pa\n",

+      "T1=420.0 # inlet temperature in degree celsius\n",

+      "V1=400.0 # inlet velocity in m/s\n",

+      "d1=0.200 # inlet diameter in 'm'\n",

+      "V2=700.0 # exit velocity in m/s\n",

+      "k=1.4 # polytopic index for air\n",

+      "Cp=1000 # specific heat at constant pressure for air in j/kg.K\n",

+      "R=287 # specific gas constant for air\n",

+      "\n",

+      "#solution\n",

+      "\n",

+      "#part (a)\n",

+      "T2=(((V1**2)-V2**2)/(2*Cp))+T1 # outlet temperature in degree celsius\n",

+      "print \" The exit temperature is \",round(T2),\" *C \\n\"\n",

+      "\n",

+      "#part (b)\n",

+      "\n",

+      "rho1=P1/(R*(T1+273)) # density at entrance\n",

+      "A1=(math.pi*d1**2)/4\n",

+      "mdot=rho1*A1*V1 # \n",

+      "print \" The mass flow rate is \",round(mdot,3),\" kg/s \\n\"\n",

+      "\n",

+      "# part (c)\n",

+      "\n",

+      "rho2=rho1*(((T2+273)/(T1+273))**(1/(k-1))) # density at exit\n",

+      "# now we find the exit diameter\n",

+      "d2=math.sqrt((rho1*V1*(d1)**2)/(rho2*V2))\n",

+      "print \" The outlet diameter is \",round(d2,3),\" m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The exit temperature is  255.0  *C \n",

+        "\n",

+        " The mass flow rate is  0.442  kg/s \n",

+        "\n",

+        " The outlet diameter is  0.212  m\n"

+       ]

+      }

+     ],

+     "prompt_number": 87

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4.15:PG-89"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "mdots=100 # mass flow rate of sodium in kg/s\n",

+      "Ts1=450 # inlet temperature of sodium in degree celsius\n",

+      "Ts2=350 # exit temperature of sodium in degree celsius\n",

+      "Cp=1.25 # specific heat of sodium in KJ/kg.*C\n",

+      "Tw1=20 # inlet temperature of water in degree celsius\n",

+      "Pw=5000 # inlet pressure of water in kPa \n",

+      "\n",

+      "# solution\n",

+      "hw1=88.65 # enthalpy from table C.4\n",

+      "hw2=2794 # enthalpy from table C.3\n",

+      "mdotw=(mdots*Cp*(Ts1-Ts2))/(hw2-hw1) # mass flow rate of water\n",

+      "print \" The mass flow rate of water is \",round(mdotw,2),\" kg/s \\n\"\n",

+      "Qdot=mdotw*(hw2-hw1) # heat transfer in kW using energy equation\n",

+      "# result\n",

+      "print \" The rate of heat transfer is \",round(Qdot),\" kW\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The mass flow rate of water is  4.62  kg/s \n",

+        "\n",

+        " The rate of heat transfer is  12500.0  kW\n"

+       ]

+      }

+     ],

+     "prompt_number": 90

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter5_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter5_2.ipynb
new file mode 100755
index 00000000..ac5d7227
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter5_2.ipynb
@@ -0,0 +1,586 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:4df6ac0705dbecae7b5d61911e7c199aa54ae833c2fc9f2d87f583d3d1c651ef"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 5:The Second Law of Thermodynamics"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.4:PG-112"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "Th=200+273.0 # higher temperture in kelvin \n",

+      "Tl=20+273.0 # lower temperture in kelvin\n",

+      "Wdot=15 # output of engine in kW\n",

+      "\n",

+      "ef=1-(Tl/Th) # carnot efficiency\n",

+      "\n",

+      "Qhdot=Wdot/ef # heat supplied by reservoir\n",

+      "print \" The heat suppled by higher temperature reservoir is\",round(Qhdot,2),\"kW \\n \"\n",

+      "# using forst law\n",

+      "Qldot=Qhdot-Wdot # heat rejected to reservoir\n",

+      "print \" The heat suppled by lower temperature reservoir is\",round(Qldot,2),\"kW\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The heat suppled by higher temperature reservoir is 39.42 kW \n",

+        " \n",

+        " The heat suppled by lower temperature reservoir is 24.42 kW\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.5:PG-113"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "TL1=-5+273.0 # lower temperature in kelvin for first situation\n",

+      "TH=20+273.0 # higher temperature in kelvin\n",

+      "TL2=-25+273.0 #lower temperature in kelvin for second situation\n",

+      "\n",

+      "#solution\n",

+      "\n",

+      "COP1=TL1/(TH-TL1) # carnot refrigerator COP for first situation\n",

+      "# Let Heat be 100 kJ\n",

+      "QL=100.0 # assumption\n",

+      "W1=QL/COP1 # work done for situation 1\n",

+      "\n",

+      "# for situation 2\n",

+      "COP2=TL2/(TH-TL2) # COP carnot for second situation\n",

+      "W2=QL/COP2 # work done\n",

+      "\n",

+      "Per=(W2-W1)*100/W1 # percentage increase in work done \n",

+      "#result\n",

+      "print\" The perccentage increase in work is \",round(Per,1),\"%\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The perccentage increase in work is  94.5 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.6:PG-117"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "# initialization of variables\n",

+      "T1=20+273 # initial temperature in kelvin\n",

+      "P=200 # pressure in kPa\n",

+      "V=2 # volume in m^3\n",

+      "R=0.287 # gas constant for air\n",

+      "W=720 # work done on air in kJ\n",

+      "Cv=0.717 # specific heat at constant volume for air\n",

+      "\n",

+      "#solution\n",

+      "m=(P*V)/(R*T1) # mass of air\n",

+      "\n",

+      "T2=T1+(W/(m*Cv))# final temperature in kelvin\n",

+      "\n",

+      "delS=m*Cv*math.log(T2/T1) # ENROPY CHANGE FOR CONSTANT VOLUME PROCESS\n",

+      "print \" The Entropy increase is\",round(delS,3),\"kJ/K \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Entropy increase is 1.851 kJ/K \n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.7:PG-118"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "T1=350+273 # initial temperature in kelvin\n",

+      "P1=1200.0 # initial pressure in kPa\n",

+      "P2=140 # final pressure in kPa\n",

+      "k=1.4 # polytopic index for air\n",

+      "Cv=0.717 # specific heat at  constant volume for air\n",

+      "#solution\n",

+      "T2=T1*((P2/P1)**((k-1)/k))  # reversible adiabatic process relation\n",

+      "\n",

+      "w=-Cv*(T2-T1) # work done by gases in reversible adiabatic process\n",

+      "print\" The work done by gases is\",round(w),\"kJ/kg\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The work done by gases is 205.0 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.8:PG-120"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "# initialization of variables\n",

+      "T1=20+273.0 # initial temperature in kelvin\n",

+      "P1=200.0 # pressure in kPa\n",

+      "V=2 #volume in m^3\n",

+      "R=0.287 # gas constant for air\n",

+      "W=-720 # negative as work is done on air in kJ\n",

+      "\n",

+      "#solution\n",

+      "\n",

+      "m=(P1*V)/(R*T1)# mass of air\n",

+      "\n",

+      "u1=209.1 #specific internal energy of air at 293K and 200 kPa from table E.1\n",

+      "s1=1.678 # by interpolation from table E.1\n",

+      "# change  in internal energy= work done\n",

+      "u2=-(W/m)+u1 # final internal energy\n",

+      "T2=501.2# final temperature interpolated from table E.1 corresponding to value of u2\n",

+      "s2=2.222 # value of s from table E.3 by interpolating from corresponding to value of u2\n",

+      "\n",

+      "P2=P1*(T2/T1) # final pressure in kPa\n",

+      "\n",

+      "delS=m*(s2-s1-R*math.log(P2/P1))# entropy change\n",

+      "# result\n",

+      "print \" The Entropy increase is\",round(delS,3),\"kJ/K \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Entropy increase is 1.855 kJ/K \n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.9:PG-120"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "T1=350+273.0 # initial temperature in kelvin\n",

+      "P1=1200.0 # initial pressure in kPa\n",

+      "P2=140.0 # final pressure in kPa\n",

+      "k=1.4 # polytopic index for air\n",

+      "\n",

+      "#solution\n",

+      "# The values are taken from table E.1\n",

+      "Pr660=23.13# relative pressure @ 660K\n",

+      "Pr620=18.36# relative pressure @ 620K\n",

+      "Pr1=((Pr660-Pr620)*3/40)+Pr620 # relative pressure by interpolation\n",

+      "\n",

+      "Pr2=Pr1*(P2/P1) # relative pressure at state 2\n",

+      "\n",

+      "Pr340=2.149 # relative pressure @ 340K\n",

+      "Pr380=3.176 # relative pressure @ 380K\n",

+      "T2=((Pr2-Pr340)/(Pr380-Pr340))*40+340 # interpolating final temperature from table E.1\n",

+      "\n",

+      "# now interpolating u1 AND u2 from table E.1\n",

+      "u620=451.0# specific internal energy @ 620k\n",

+      "u660=481.0# specific internal energy @ 660k\n",

+      "u1=(u660-u620)*(3/40.0)+u620 # initial internal energy\n",

+      "\n",

+      "u380=271.7 #specific internal energy @ 380k\n",

+      "u340=242.8 #specific internal energy @ 340k\n",

+      "u2=((Pr2-Pr340)/(Pr380-Pr340))*(u380-u340)+u340 # final internal energy\n",

+      "\n",

+      "w=u2-u1 # work= change in internal energy\n",

+      "print \" The work done by gas is\",int(w),\"kJ/kg\"\n",

+      "# The answer is slightly different as values are approximated in textbook\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The work done by gas is -209 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 33

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.10:PG-123"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "T1=300+273.0 # initial temperature in kelvin\n",

+      "P1=600 # initial pressure in kPa\n",

+      "P2=40 # final pressure in kPa\n",

+      "\n",

+      "#solution\n",

+      "#please refer to steam table for values\n",

+      "v1=0.4344 # specific volume from steam table @ 573k and 600 kPa\n",

+      "v2=v1 # rigid container\n",

+      "u1=2801.0 # specific internal energy from steam table @ 573k and 600 kPa\n",

+      "s1=7.372 # specific entropy @ 600 kPa and 573 K\n",

+      "\n",

+      "vg2=0.4625 # specific volume of saturated vapour @ 40 kPa and 573 K\n",

+      "vf2=0.0011 # specific volume of saturated liquid @ 40 kPa and 573 K\n",

+      "sf2=1.777 # specific entropy of saturated liquid @ 40 kPa and 573 K\n",

+      "sg2=5.1197 # specific entropy of saturated vapour @ 40 kPa and 573 K\n",

+      "x=(v2-vf2)/(vg2-vf2)# quality of steam using pure substance relation\n",

+      "\n",

+      "s2=sf2+x*sg2 # overall specific enthalpy at quality 'x' \n",

+      "delS=s2-s1 # entropy change\n",

+      "print\" The entropy change  is\",round(delS,3),\"kJ/kg.K \\n \"\n",

+      "\n",

+      "#heat transfer\n",

+      "uf2=604.3 #specific internal energy of saturated liquid @ 40 kPa and 573 K\n",

+      "ug2=1949.3 #specific internal energy of saturated vapour @ 40 kPa and 573 K\n",

+      "u2=uf2+x*ug2 #specific internal energy @ quality x\n",

+      "q=u2-u1 # heat transfer in kJ/kg from first law as W=0\n",

+      "print \" The heat transfer is\",int(q),\"kJ/kg\"\n",

+      "# result\n",

+      "# the answers are approximated in textbook but here they are precise thus minute difference is there"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The entropy change  is -0.787 kJ/kg.K \n",

+        " \n",

+        " The heat transfer is -366 kJ/kg\n"

+       ]

+      }

+     ],

+     "prompt_number": 37

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.11:PG-126"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "# initialization of variables\n",

+      "v1=0.5 # assumed as air is filled in half of the tank\n",

+      "v2=1.0 # final volume when partition is removed\n",

+      "R=0.287 # gas contant for air\n",

+      "#solution\n",

+      "q=0 # heat transfer is zero\n",

+      "w=0 # work done is zero\n",

+      "# temperatue is constant as no change in internal energy by first law\n",

+      "dels=R*math.log(v2/v1)# change in entropy when temperature is constant\n",

+      "print \"The change in specific entropy is\",round(dels,3),\"kJ/kg.K\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The change in specific entropy is 0.199 kJ/kg.K\n"

+       ]

+      }

+     ],

+     "prompt_number": 49

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.12:PG-127"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "T1=400+273.0 # initial temperature in kelvin\n",

+      "P=600 # pressure in kPa\n",

+      "Tsurr=25+273.0 # surrounding temperature in K\n",

+      "m=2 # mass of steam in kg\n",

+      "\n",

+      "#solution\n",

+      "#please refer to steam table for values\n",

+      "s1=7.708 # specific entropy of steam @ 400 degree celsius and 0.6 MPa\n",

+      "s2=1.9316# specific enropy of condensed water @ 25 degree celsius and 0.6 MPa\n",

+      "delSsys=m*(s2-s1) # entropy change in system i.e of steam\n",

+      "\n",

+      "h1=3270 # specific enthalpy of steam @ 400 degree celsius and 0.6 MPa\n",

+      "h2=670.6#specific enropy of condensed water @ 25 degree celsius and 0.6 MPa\n",

+      "\n",

+      "Q=m*(h1-h2)# heat transfer at constant pressure\n",

+      "delSsurr=Q/Tsurr # entropy change in surroundings\n",

+      "\n",

+      "sigma=delSsys+delSsurr # net entropy change\n",

+      "\n",

+      "print \"The net entropy production is\",round(sigma,1),\"kJ/K\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The net entropy production is 5.9 kJ/K\n"

+       ]

+      }

+     ],

+     "prompt_number": 53

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.13:PG-130"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "T1=600+273 # initial temperature in kelvin\n",

+      "P1=2 # initial pressure in MPa\n",

+      "P2=10 # final pressure in kPa\n",

+      "mdot=2 # mass flow rate in kg/s\n",

+      "\n",

+      "#solution\n",

+      "#please refer to steam table for values\n",

+      "h1=3690 # specific enthalpy in kJ/kg @ 2MPa and 600 degree celsius\n",

+      "s1=7.702 #specific entropy in kJ/kg.K @ 2MPa and 600 degree celsius\n",

+      "s2=s1 # Reversible adiabatic process thus entropy is constant\n",

+      "sf2=0.6491 #specific entropy of saturated liquid from steam table @ 10 kPa\n",

+      "sg2=8.151 #specific entropy of saturated vapour from steam table @ 10 kPa\n",

+      "\n",

+      "x2=(s2-sf2)/(sg2-sf2) # quality of steam at turbine exit\n",

+      "\n",

+      "h2f=191.8 #specific enthalpy of saturated liquid from steam table @ 10 kPa\n",

+      "h2g=2584.8 #specific enthalpy of saturated vapour from steam table @ 10 kPa\n",

+      "h2=h2f+x2*(h2g-h2f) # specific enthalpy @ quality 'x' \n",

+      "\n",

+      "WdotT=mdot*(h1-h2)# from work done in adiabatic process\n",

+      "# result\n",

+      "print \" The maximum power output is\",int(WdotT),\"kJ/s\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The maximum power output is 2496 kJ/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 56

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.14:PG-130"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "T1=600+273 # initial temperature in kelvin\n",

+      "P1=2 # initial pressure in MPa\n",

+      "P2=10 # final pressure in kPa\n",

+      "mdot=2 # mass flow rate in kg/s\n",

+      "EffT=0.8 # efficiency of turbine \n",

+      "WdotT=2496 # theoritical power of turbine in kW\n",

+      "\n",

+      "#solution\n",

+      "Wdota=EffT*WdotT # actual power output of turbine\n",

+      "h1=3690 # specific enthalpy @ 2MPa and 600 degree celsius\n",

+      "h2=h1-(Wdota/mdot) # final enthalpy from first law of thermodynamics\n",

+      "\n",

+      "T2=((h2-2688)/(2783-2688))*(150-100)+100 # by interpolating from steam table @ P2= 10 kPa, h2=2770 \n",

+      "s2=8.46 # final specific entropy by interpolation from steam table\n",

+      "\n",

+      "print \"The temperature by interpolation is\",round(T2),\" degree celsius \\n\"\n",

+      "print \"The final entropy by interpolation is\",round(s2,2),\"kJ/kg.K\"\n",

+      "# The temperature and entropy are found by interpolation from steam table and cannot be shown here.\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The temperature by interpolation is 102.0  degree celsius \n",

+        "\n",

+        "The final entropy by interpolation is 8.46 kJ/kg.K\n"

+       ]

+      }

+     ],

+     "prompt_number": 59

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5.15:PG-131"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "T2=250.0 # temperature of steam in degree celsius\n",

+      "mdot2=0.5 # mass flow rate of steam in kg/s\n",

+      "T1=45 # temperature of water in degree celsius\n",

+      "mdot1=4 # mass flow rate of water in kg/s\n",

+      "P=600.0 # pressure in kPa\n",

+      "\n",

+      "\n",

+      "mdot3=mdot1+mdot2 # by mass balance\n",

+      "\n",

+      "h2=2957 # specific enthalpy in kJ/kg of steam @ 600 Kpa from steam table\n",

+      "h1=188.4 # specific enthalpy in kJ/kg  of water @ 600 Kpa from steam table\n",

+      "\n",

+      "h3=(mdot1*h1+mdot2*h2)/mdot3 # specific enthalpy in kJ/kg at exit\n",

+      "\n",

+      "# by interpolation from saturated steam table\n",

+      "T3=(h3-461.3)*10/(503.7-461.3)+110 # temperature of mixture\n",

+      "\n",

+      "sf3=1.508 # entropy of saturated liquid in kJ/kg.K at 600Kpa and T3 temperature from steam table\n",

+      "s3=sf3  # by interpolating sf\n",

+      "s2=7.182 # entropy of superheated steam in kJ/kg.K @ 600Kpa from steam table\n",

+      "s1=0.639 # entropy of entering water in kJ/kg.K at T= 45 degree celsius\n",

+      "\n",

+      "sigmaprod=mdot3*s3-mdot2*s2-mdot1*s1\n",

+      "# result\n",

+      "print \"The rate of entropy production is\",round(sigmaprod,3),\"kW/K \""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The rate of entropy production is 0.639 kW/K \n"

+       ]

+      }

+     ],

+     "prompt_number": 60

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter6_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter6_2.ipynb
new file mode 100755
index 00000000..49cfc54d
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter6_2.ipynb
@@ -0,0 +1,695 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:353d9311290d1e9a150d4f5b7334fda96148ffc036b68321a43e559706361c16"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 6:Power Vapor Cycles"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.1:PG-146"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "# Please refer to the given figure in question for quantities\n",

+      "P2=2*1000 #higher pressure converted in in kPa\n",

+      "P1=10 # lower pressure in kPa\n",

+      "rho=1000 # density of water in Kg/m^3\n",

+      "h1=192 # enthalpy at state 1 in kJ/kg\n",

+      "h3=3248 # enthalpy at state 3 in kJ/kg\n",

+      "s3=7.1279# entropy at state 3 in kJ/kg.K\n",

+      "\n",

+      "#calculation of pump work\n",

+      "wp=(P2-P1)/rho # pump work given by equation 4.56 in textbook\n",

+      "h2=h1+wp # by enrgy balance b/w state 1 and 2\n",

+      "q=h3-h2 # Heat input from 2 to 3\n",

+      "\n",

+      "s4=s3 # isentropic process\n",

+      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",

+      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",

+      "x=(s4-sf)/(sg-sf)# from property of pure substance\n",

+      "hf=191.8 #enthalpy of saturated liquid @10 kPa from steam table\n",

+      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",

+      "h4=hf+x*(hg-hf)# enthalpy @ state 4\n",

+      "\n",

+      "wt=h3-h4 # turbine work\n",

+      "\n",

+      "efficiency=(wt-wp)/q # efficiency of power cycle\n",

+      "print \" The Efficiency is\",round(efficiency,4),\" or\",round(efficiency*100,1),\"%\"\n",

+      "# the answer is correct within limits\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Efficiency is 0.3238  or 32.4 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.2:PG-149"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "# Please refer to the given figure of question 6.1 for quantities\n",

+      "effi1=0.323 #old efficiency\n",

+      "P2=4*1000 #higher pressure converted in in kPa\n",

+      "P1=10 # lower pressure in kPa\n",

+      "rho=1000 # density of water in Kg/m^3\n",

+      "h1=192 # enthalpy at state 1 in kJ/kg\n",

+      "h3=3214 # enthalpy at state 3 i.e @400 degree celsius and 4MPa in kJ/kg\n",

+      "s3=6.769# entropy at state 3 i.e @400 degree celsius and 4MPa in kJ/kg.K\n",

+      "\n",

+      "s4=s3 # insentropic process\n",

+      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",

+      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",

+      "\n",

+      "x=(s4-sf)/(sg-sf)# quality of steam\n",

+      "\n",

+      "hf=192 #enthalpy of saturated liquid @10 kPa from steam table\n",

+      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",

+      "h4=hf+x*(hg-hf)# enthalpy @ state 4\n",

+      "h2=h1 # isenthalpic process\n",

+      "qb=h3-h2 # heat addition\n",

+      "\n",

+      "wt=h3-h4 # turbine work\n",

+      "\n",

+      "effi2=(wt)/qb # efficiency of power cycle\n",

+      "\n",

+      "print \" The Efficiency is\",round(effi2,3),\" or\",round(effi2*100),\"% \\n\"\n",

+      "\n",

+      "perincrease=((effi2-effi1)/effi1)*100 \n",

+      "\n",

+      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Efficiency is 0.354  or 35.0 % \n",

+        "\n",

+        " The % increase in Efficiency is 9.69  \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.3:PG-149"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "# Please refer to fig of question 6.1 for quantities\n",

+      "effi1=0.323 #old efficiency\n",

+      "P2=2*1000 #higher pressure converted in in kPa\n",

+      "P1=10 # lower pressure in kPa\n",

+      "rho=1000 # density of water in Kg/m^3\n",

+      "T2=600# max temperature of cycle in degree celsius\n",

+      "h1=192 # enthalpy at state 1 in kJ/kg\n",

+      "h3=3690 # enthalpy at state 3 in kJ/kg, 600*C and 2MPa pressure\n",

+      "s3=7.702# entropy at state 3 in kJ/kg.K, 600*C and 2MPa pressure\n",

+      " \n",

+      "s4=s3# isentropic process\n",

+      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",

+      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",

+      "\n",

+      "x=(s4-sf)/(sg-sf)# quality of steam\n",

+      "\n",

+      "hf=192 #enthalpy of saturated liquid @10 kPa from steam table\n",

+      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",

+      "h4=hf+x*(hg-hf)# enthalpy @ state 4\n",

+      "\n",

+      "h2=h1 # isenthalpic process\n",

+      "qb=h3-h2 # heat addition\n",

+      "\n",

+      "wt=h3-h4 # turbine work\n",

+      "\n",

+      "effi2=(wt)/qb # efficiency of power cycle\n",

+      "print \" The Efficiency is\",round(effi2,3),\" or\",round(effi2*100),\"% \\n\"\n",

+      "perincrease=((effi2-effi1)/effi1)*100 \n",

+      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Efficiency is 0.357  or 36.0 % \n",

+        "\n",

+        " The % increase in Efficiency is 10.56  \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.4:PG-150"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "# Please refer to fig of question 6.1 for quantities\n",

+      "effi1=0.323 #old efficiency\n",

+      "P2=2*1000 #higher pressure converted in in kPa\n",

+      "P1=4 # condenser pressure in kPa\n",

+      "rho=1000 # density of water in Kg/m^3\n",

+      "h1=192 # enthalpy at state 1 in kJ/kg\n",

+      "h3=3248 # enthalpy at state 3 in kJ/kg\n",

+      "s3=7.1279# entropy at state 3 in kJ/kg.K\n",

+      "\n",

+      "s4=s3 # isentropic process \n",

+      "\n",

+      "sf=0.4225 # entropy of saturated liquid @10 kPa from steam table\n",

+      "sg=8.4754 # entropy of saturated vapour @10 kPa from steam table\n",

+      "\n",

+      "x=(s4-sf)/(sg-sf)# from property of pure substance\n",

+      "\n",

+      "hf=121 #enthalpy of saturated liquid @4 kPa from steam table\n",

+      "hg=2554 # enthalpy of saturated vapour @4 kPa from steam table\n",

+      "h4=hf+x*(hg-hf)# enthalpy @ state 4h1=h2 # isenthalpic process\n",

+      "h2=h1 # isenthalpic process\n",

+      "qb=h3-h2 # heat addition\n",

+      "\n",

+      "wt=h3-h4 # turbine work\n",

+      "\n",

+      "effi2=(wt)/qb # efficiency of power cycle\n",

+      "print \" The Efficiency is\",round(effi2,4),\" or\",round(effi2*100),\"% \\n\"\n",

+      "perincrease=((effi2-effi1)/effi1)*100 \n",

+      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Efficiency is 0.3603  or 36.0 % \n",

+        "\n",

+        " The % increase in Efficiency is 11.55  \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.5:PG-152"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "P2=2*1000 #higher pressure converted in in kPa\n",

+      "P1=10 # lower pressure in kPa\n",

+      "h1=192.0 # enthalpy at 10 kPa  in kJ/kg\n",

+      "h3=3248.0 # enthalpy @ state 3 in kJ/kg from table C.3\n",

+      "s3=7.128 # entropy @ state 3 in kJ/kg.K from table C.3\n",

+      "s4=s3 # isentropic process\n",

+      "\n",

+      "h2=h1 #isenthalpic process\n",

+      "h4=((s4-7.038)/(7.233-7.038))*(3056-2950)+2950 #using adjacent values for \n",

+      "#interpolation from table C.3 \n",

+      "h5=3267.0 # enthalpy at 800 kPa and $00 degree celsius\n",

+      "s5=7.572 # entropy at 800 kPa and $00 degree celsius\n",

+      "\n",

+      "s6=s5 # isentropic process\n",

+      "sf=0.6491#  entropy of saturated liquid @10 kPa from steam table\n",

+      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",

+      "\n",

+      "x=(s6-sf)/(sg-sf)# quality of steam\n",

+      "\n",

+      "hf=192.0 #enthalpy of saturated liquid @10 kPa from steam table\n",

+      "hg=2585.0 # enthalpy of saturated vapour @10 kPa from steam table\n",

+      "\n",

+      "h6=hf+x*(hg-hf)# enthalpy @ state 6\n",

+      "\n",

+      "# we now calculate energy input\n",

+      "qb=(h5-h4)+(h3-h2)# heat interaction\n",

+      "\n",

+      "# we now calculate work output\n",

+      "wt=(h5-h6)+(h3-h4)# turbine work\n",

+      "\n",

+      "eff=(wt)/qb # efficiency of power cycle\n",

+      "print\" The Efficiency is\",round(eff,3),\" or\",round(eff*100,2),\"%\"\n",

+      "\n",

+      "# // The anwer is different in textbook as there the intermediate values are approximated while in python the calculations are precise \n",

+      "\n",

+      "\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Efficiency is 0.336  or 33.57 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.6:PG-155"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "\n",

+      "# initialization of variables\n",

+      "\n",

+      "# Please refer to fig of question 6.1 for quantities\n",

+      "\n",

+      "effi1=0.357 #efficiency from example 6.3\n",

+      "P2=2*1000.0 #higher pressure converted in in kPa\n",

+      "P1=10.0 # lower pressure in kPa\n",

+      "rho=1000.0 # density of water in Kg/m^3\n",

+      "T2=600.0 # max temperature of cycle in degree celsius\n",

+      "h1=192.0 # enthalpy at state 1 in kJ/kg\n",

+      "h3=3690.0 # enthalpy at state 3 in kJ/kg, 600*C and 2MPa pressure\n",

+      "h4=2442.0 # enthalpy from example 6.3\n",

+      "h6=505.0 # specific enthalpy @ 200 kPa from steam table\n",

+      "h7=h6 # isenthalpic process\n",

+      "s3=7.702# entropy at state 3 in kJ/kg.K, 600*C and 2MPa pressure\n",

+      "\n",

+      "h2=h1 # isenthalpic process\n",

+      "s5=s3 # isentropic process\n",

+      "h5=(s3-7.509)*(2971-2870)/(7.709-7.509)+2870 # interpolationg from steam table 2 200 kPa using s5=s3= 7.702 kJ/kg.\n",

+      "\n",

+      "m6=1.0 # let mass of steam =1 Kg\n",

+      "m5=(h6-h2)*(m6)/(h5-h2) \n",

+      "m2=m6-m5 # conservation of mass\n",

+      "\n",

+      "wt=h3-h5+(h5-h4)*m2 # work done by turbine\n",

+      "qb=h3-h7 # heat given to bolier\n",

+      "effi2=(wt)/qb # efficiency of power cycle\n",

+      "print \" The Efficiency is\",round(effi2,4),\" or\",round(effi2*100),\"% \\n\"\n",

+      "perincrease=((effi2-effi1)/effi1)*100 \n",

+      "print \" The % increase in Efficiency is\",round(perincrease,2),\" \\n\"\n",

+      "\n",

+      "# The anwer is different in textbook as there the intermediate values are approximated while in python the calculations are precise \n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Efficiency is 0.3732  or 37.0 % \n",

+        "\n",

+        " The % increase in Efficiency is 4.55  \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 27

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.7:PG-156"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "P2=2*1000 #higher pressure converted in  kPa\n",

+      "P1=10 # lower pressure in kPa\n",

+      "h1=192 # enthalpy at 10 kPa  in kJ/kg\n",

+      "h3=3248 # enthalpy @ state 3 in kJ/kg from table C.3\n",

+      "s3=7.128 # entropy @ state 3 in kJ/kg.K from table C.3\n",

+      "\n",

+      "s4=s3 # isentropic process\n",

+      "\n",

+      "h4=((s4-7.038)/(7.233-7.038))*(3056-2950)+2950 #using adjacent values for \n",

+      "#interpolation from table C.3 \n",

+      "h5=3267 # enthalpy at 800 kPa and $00 degree celsius\n",

+      "s5=7.572 # entropy at 800 kPa and $00 degree celsius\n",

+      "\n",

+      "s6=s5 # isentropic process\n",

+      "sf=0.6491#  entropy of saturated liquid @10 kPa from steam table\n",

+      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",

+      "\n",

+      "x=(s6-sf)/(sg-sf)# quality of steam\n",

+      "\n",

+      "hf=192 #enthalpy of saturated liquid @10 kPa from steam table\n",

+      "hg=2585 # enthalpy of saturated vapour @10 kPa from steam table\n",

+      "\n",

+      "h6=hf+x*(hg-hf)# enthalpy @ state 6\n",

+      "h7=721 # enthalpy of saturated liquid @800 kPa from steam table\n",

+      "h8=h7 # isenthalpic process\n",

+      "h2=h1 # isenthalpic process\n",

+      "\n",

+      "m8=1 # let mass of steam =1 Kg\n",

+      "m4=(h8-h2)*(m8)/(h4-h2)\n",

+      "m2=m8-m4 # conservation of mass\n",

+      "\n",

+      "wt=h3-h4+(h5-h6)*m2 # work done by turbine\n",

+      "qb=h3-h8+(h5-h4)*m2 # heat given to bolier\n",

+      "\n",

+      "effi=(wt)/qb # efficiency of power cycle\n",

+      "# result\n",

+      "print \" The Efficiency is\",round(effi,3),\" or\",round(effi*100,2),\"%\"\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Efficiency is 0.347  or 34.7 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 28

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.8:PG-159"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "\n",

+      "# for rankine cycle refer to fig 6.9\n",

+      "\n",

+      "effiT=0.8 # turbine efficiency\n",

+      "P2=2*1000 # higher pressure converted in  kPa\n",

+      "P1=10 # lower pressure in kPa\n",

+      "h1=192 # enthalpy at 10 kPa  in kJ/kg\n",

+      "h3=3690 # enthalpy of superheated steam @ 2 MPa from steam table in kJ/kg\n",

+      "s3=7.702 #entropy of superheated steam @ 2 MPa from steam table in kJ/kg.K\n",

+      "# state 4' is repsresented by '41'\n",

+      "h2=h1 #isenthalpic process\n",

+      "s41=s3 # entropy is constant\n",

+      "sf=0.6491 # entropy of saturated liquid @10 kPa from steam table\n",

+      "sg=8.151 # entropy of saturated vapour @10 kPa from steam table\n",

+      "x=(s41-sf)/(sg-sf)# from property of pure substance\n",

+      "\n",

+      "hf=191.8 #enthalpy of saturated liquid @10 kPa from steam table\n",

+      "hg=2584 # enthalpy of saturated vapour @10 kPa from steam table\n",

+      "h41=hf+x*(hg-hf)# enthalpy @ state 41\n",

+      "\n",

+      "wa=effiT*(h3-h41)# turbine efficiency =(actual work)/(isentropic work)\n",

+      "\n",

+      "qb=h3-h2 # heat supplied\n",

+      "\n",

+      "effi=(wa)/qb # efficiency of power cycle\n",

+      "print\"  The Efficiency is\",round(effi,3),\" or\",round(effi*100,1),\"%\"\n",

+      "\n",

+      "h4=h3-wa # adiabatic process\n",

+      "\n",

+      "# now using interpolation for superheated steam @ 10 kPa\n",

+      "T4=(h4-2688)*(150-100)/(2783-2688)+100\n",

+      "\n",

+      "print \"\\n The Temperature from interpolation comes out to be\",int(T4),\" degree celsius\"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "  The Efficiency is 0.286  or 28.6 %\n",

+        "\n",

+        " The Temperature from interpolation comes out to be 101  degree celsius\n"

+       ]

+      }

+     ],

+     "prompt_number": 32

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.9:PG-162"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "\n",

+      "#initialization of variables\n",

+      "\n",

+      "# refer to fig 6.10c\n",

+      "\n",

+      "mdot=0.6 # mass flow rate of refrigerant in kg/sec\n",

+      "T1=-24 # evaporator temperature in degree celsius\n",

+      "T2=39.39 # condenser temperature in degree celsius\n",

+      "h1=232.8 # enthalpy of saturated R134a vapour @ -24 degree celsius from table D.1\n",

+      "s1=0.9370 # entropy of saturated R134a vapour @ -24 degree celsius from table D.1\n",

+      "h3=105.3 # enthalpy of saturated R134a liquid @ -24 degree celsius from table D.2\n",

+      "h4=h3 # isenthalpic process\n",

+      "\n",

+      "# interpolating enthalpy from table D.3 @ 39.39 degree celsius\n",

+      "h2=(s1-0.9066)*(280.19-268.68)/(0.9428-0.9066)+268.68\n",

+      "QdotE=mdot*(h1-h4) # heat transfer rate\n",

+      "WdotC=mdot*(h2-h1)# power given to compressor\n",

+      "\n",

+      "COP=QdotE/WdotC # coefficient of performance\n",

+      "\n",

+      "Hp=(WdotC/0.746)/(QdotE/3.52) #calculating Horsepower required per Ton\n",

+      "\n",

+      "print \"The rate of refrigeration is\",round(QdotE,2),\"kJ/s \\n \"\n",

+      "print \"The coefficient of performance is\",round(COP,2),\"\\n \"\n",

+      "print \"The rating in horsepower per ton is\",round(Hp,3),\" hp\\n \"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The rate of refrigeration is 76.5 kJ/s \n",

+        " \n",

+        "The coefficient of performance is 2.8 \n",

+        " \n",

+        "The rating in horsepower per ton is 1.686  hp\n",

+        " \n"

+       ]

+      }

+     ],

+     "prompt_number": 37

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.10:PG-163"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "\n",

+      "#initialization of variables\n",

+      "# refer to fig 6.10c\n",

+      "effi=0.8 # compressor efficiency\n",

+      "mdot=0.6 # mass flow rate of refrigerant in Kg/sec\n",

+      "T4=-24 # temperature of evaporator\n",

+      "T2=39.39 # temperature of condensor\n",

+      "T1=-20.0 # supeheating temperature\n",

+      "T3=40 # subcooling temperature\n",

+      "h3=106.2 # enthalpy of liquid R-134a @ 40 degree celsius from table D.1\n",

+      "h4=h3 # isenthalpic process\n",

+      "h1=236.5 # enthalpy of superheated R-134a @ 0.10 MPa and -20 degree celsius from table D.3\n",

+      "s1=0.960 #entropy of superheated R-134a @ 0.10 MPa and -20 degree celsius from table D.3 \n",

+      "\n",

+      "s2dash=s1 # isentropic process\n",

+      "\n",

+      "# using interpolation from table D.3 @ 1.0 MPa for s2dash=0.960 \n",

+      "h2dash=(s2dash-0.9428)*(291.36-280.19)/(0.9768-0.9428)+280.19\n",

+      "\n",

+      "h2=(h2dash-h1)/(effi)+h1 # by definition of compressor efficiency\n",

+      "\n",

+      "QdotE=mdot*(h1-h4)#heat transfer rate power given to compressor\n",

+      " \n",

+      "wdotc=mdot*(h2-h1)# power given to compressor\n",

+      "\n",

+      "COP=QdotE/wdotc # coefficient of performance\n",

+      "\n",

+      "print \"The rate of refrigeration is\",round(QdotE,1),\"kJ/s \\n \"\n",

+      "\n",

+      "print \"The coefficient of performance is\",round(COP,2),\"\\n \"\n",

+      "# The value of Wdotc is shown wrong in the textbook. It should be multiplied by mass flow rate\n",

+      " "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The rate of refrigeration is 78.2 kJ/s \n",

+        " \n",

+        "The coefficient of performance is 2.11 \n",

+        " \n"

+       ]

+      }

+     ],

+     "prompt_number": 41

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6.11:PG-165"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# solution\n",

+      "#initialization of variables\n",

+      "# refer to fig 6.10c\n",

+      "\n",

+      "QdotC=300 #heating Load in KWh or heat rejected by condensor\n",

+      "T1=-12 # evaporator temperature in degree celsius\n",

+      "P2=800 # condensor pressure in kPa \n",

+      "h1=240 # specific enthalpy of saturated R-134a vapour @ -12 degree celsius from table D.1\n",

+      "s1=0.927 # specific entropy of saturated R-134a vapour @ -12 degree celsius from table D.1\n",

+      "s2=s1 # isentropic process\n",

+      "h3=93.4 #specific enthalpy of saturated R-134a liquid @ 800 kPa from tableD.2\n",

+      "\n",

+      "# extrapolating enthalpy from table D.2 @ 0.8 MPa for s=0.927\n",

+      "h2=273.7-(0.9374-s2)*(284.4-273.7)/(0.9711-0.9374)\n",

+      "\n",

+      "# QdotE=mdot*(h1-h4) is heat transfer rate\n",

+      "mdot=QdotC/(h2-h3)# mass flow rate\n",

+      "\n",

+      "WdotC=mdot*(h2-h1)# power given to compressor\n",

+      "\n",

+      "#part(a)\n",

+      "COP=QdotC/WdotC # coefficient of performance\n",

+      "print \"The coefficient of performance is\",round(COP,2),\"\\n \"\n",

+      "\n",

+      "#part(b)\n",

+      "Cost=WdotC*0.07 # cost of electricity\n",

+      "print \"The cost of electricity is $\",round(Cost,3),\"/hr \\n\"\n",

+      "\n",

+      "#part(c)\n",

+      "costgas=(300*3600*0.50)/100000 # cost of gas\n",

+      "print \"The cost of gas is $\",round(costgas,2),\"/hr \\nThus heat pump is better \"\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The coefficient of performance is 5.82 \n",

+        " \n",

+        "The cost of electricity is $ 3.607 /hr \n",

+        "\n",

+        "The cost of gas is $ 5.4 /hr \n",

+        "Thus heat pump is better \n"

+       ]

+      }

+     ],

+     "prompt_number": 66

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter7_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter7_2.ipynb
new file mode 100755
index 00000000..1c4b1c97
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter7_2.ipynb
@@ -0,0 +1,616 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:c5a74c16b8ffa890dee9e16252d9e0c3674909f92103f0b1c670f62652469cd7"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 7:Power and Refrigeration Gas Cycles"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.1:PG-175"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "r=12 # compression ratio\n",

+      "k=1.4 # polytropic index for air\n",

+      "p1=200.0 # pressure at state 1 in kPa\n",

+      "p3=10000.0 # pressure at state 3 in kPa\n",

+      "\n",

+      "c=100/(r-1) # clearance in percentage\n",

+      "print \"The percent clearance is\",round(c,2),\"% \\n\"\n",

+      "v3=100.0 # let us assume v3=100 m^3 for calculations\n",

+      "p2=p1*(r**k) # polytopic process pressure relation\n",

+      "p4=p3*(1/(r**k))# polytropic process pressure relation\n",

+      "w34=v3*(r*p4-p3)/(1-k) # polytropic work done in process 3 to 4\n",

+      "v2=v3 # constant volume process\n",

+      "w12=v2*(p2-r*p1)/(1-k)\n",

+      "wcycle=w12+w34 # total work in cycle\n",

+      " # now equating the polytropic work calculated to work by MEP\n",

+      "MEP=wcycle/(r*v2-v2) # as work = pressure*change in volume\n",

+      "print \"The MEP is\",round(MEP),\" kPa\"  \n",

+      "# The solution is wrong in textbook as calculation for P2 is wrong \n",

+      "\n",

+      "\n",

+      "\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The percent clearance is 9.0 % \n",

+        "\n",

+        "The MEP is 503.0  kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.2:PG-179"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "r=10.0 # compression ratio\n",

+      "k=1.4 # polytropic index for air\n",

+      "R=0.287 # specific gas constant for air\n",

+      "Cv=0.717 # specific heat at constant volume\n",

+      "Wnet=1000 # net work output in kJ/kg\n",

+      "T1=227+273.0 # low air temperaure in kelvin\n",

+      "p1=200.0 # low pressure in kPa\n",

+      "\n",

+      "effi=1-(1/r**(k-1)) # thermal efficeiency \n",

+      "print \"The maximum possible thermal efficiency is\",round(effi*100,1),\"% \\n\"\n",

+      "\n",

+      "T2=T1*(r)**(k-1) # isentropic process temperature relation\n",

+      "\n",

+      "T4=((Wnet/Cv)+T2-T1)/((r**(k-1))-1) # using expression for work\n",

+      "\n",

+      "T3=T4*(r)**(k-1)\n",

+      "\n",

+      "efficarnot=1-T1/T3\n",

+      "print \"The carnot efficiency is\",round(efficarnot*100),\"%\"\n",

+      "\n",

+      "v1=R*T1/p1 # initial volume \n",

+      "v2=v1/r # from compression ratio\n",

+      "\n",

+      "MEP=Wnet/(v1-v2) # mean effective pressure equation\n",

+      "\n",

+      "print \"The MEP is\",round(MEP),\" kPa\"\n",

+      "\n",

+      "\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The maximum possible thermal efficiency is 60.2 % \n",

+        "\n",

+        "The carnot efficiency is 86.0 %\n",

+        "The MEP is 1549.0  kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.3:PG-182"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "\n",

+      "r=18 # compression ratio\n",

+      "k=1.4 # polytropic index for air\n",

+      "R=0.287 # specific gas constant for air\n",

+      "Cv=0.717 # specific heat at constant volume\n",

+      "Cp=1.0 # specific heat at constant pressure\n",

+      "T1=200+273 # lower temperaure in kelvin\n",

+      "P1=200.0 # low pressure in kPa\n",

+      "T3=2000 # higher temperature of cycle in kelvin\n",

+      "\n",

+      "v1=R*T1/P1 # specific volume at state 1 in m^3\n",

+      "v2=v1/r # specific volume after compression in m^3\n",

+      "\n",

+      "T2=T1*(v1/v2)**(k-1) # temperature after compression\n",

+      "P2=P1*(v1/v2)**k # pressure after compression\n",

+      "P3=P2 # diesel cycle\n",

+      "v3=R*T3/P3 # volume at state 3\n",

+      "\n",

+      "rc=v3/v2 # cutoff ratio\n",

+      "\n",

+      "effi=1-((rc**k)-1)/(r**(k-1)*k*(rc-1))\n",

+      "\n",

+      "\n",

+      "print \"The thermal efficiency is\",round(effi*100,1),\"%\"\n",

+      "\n",

+      "v4=v1 # diesel cycle\n",

+      "T4=T3*(v3/v4)**(k-1) # adiabatic process\n",

+      "\n",

+      "qin=Cp*(T3-T2) # using first law \n",

+      "qout=Cv*(T4-T1) # heat rejected \n",

+      "\n",

+      "Wnet=qin-qout # net work\n",

+      "MEP=Wnet/(v1-v2) # expression of mean effective pressure in terms of work\n",

+      "\n",

+      "print \"The MEP is\",round(MEP),\" kPa\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The thermal efficiency is 66.6 %\n",

+        "The MEP is 515.0  kPa\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.4:PG-183"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "r=18 # compression ratio\n",

+      "k=1.4 # polytropic index for air\n",

+      "R=0.287 # specific gas constant for air\n",

+      "T1=200+273 # lower temperaure in kelvin\n",

+      "P1=200.0 # low pressure in kPa\n",

+      "T3=2000.0 # higher temperature of cycle in kelvin \n",

+      "\n",

+      "v1=R*T1/P1 # specific volume at state 1 in m^3\n",

+      "#using table E.1\n",

+      "u1=340.0 # specific internal energy in kJ/kg\n",

+      "vr1=198.1 # in m^3/kg\n",

+      "\n",

+      "vr2=vr1*(1/r) # as r=v1/v2\n",

+      "\n",

+      "# now finding corresponding values from table E.1\n",

+      "T2=1310.0 # temperature in kelvin\n",

+      "Pr2=34.0 # pressure in kPa\n",

+      "h2=1408.0 # specific entropy in kJ/kg\n",

+      "v2=v1/18 # volume at state 2\n",

+      "P2=R*T2/v2 # pressure at state 2\n",

+      "\n",

+      "h3=2252.1 # specific enthalpy in kJ/kg from table E.1\n",

+      "vr3=2.776 \n",

+      "P3=P2 # diesel cycle\n",

+      "v3=R*T3/P3 # after compression volume\n",

+      "v4=v1 # isochoric process\n",

+      "vr4=vr3*v4/v3 # isentropic process\n",

+      "# now using Vr4 we read corresponding value from table E.1\n",

+      "T4=915 # final temperature in kelvin\n",

+      "u4=687.5 # specific internal energy at state 4\n",

+      "\n",

+      "qin=h3-h2 # using first law \n",

+      "qout=u4-u1 # heat rejected \n",

+      "\n",

+      "Wnet=qin-qout # net work\n",

+      "effi=100*Wnet/qin # thermal efficiency\n",

+      "print\" The thermal efficiency is\",round(effi,1),\"%\"\n",

+      "\n",

+      "MEP=Wnet/(v1-v2) # expression of mean effective pressure in terms of work\n",

+      "\n",

+      "print \" The MEP is\",round(MEP),\" kPa\"\n",

+      "\n",

+      "erroreffi=(66.6-effi)*100/effi # error in efficiency\n",

+      "errorMEP=(515-MEP)*100/MEP # error in MEP\n",

+      "\n",

+      "print \" The % error in efficiency is\",round(erroreffi,1),\"%\"\n",

+      "print \" The % error MEP is\",round(errorMEP,1),\"% \\n\"\n",

+      "\n",

+      "# the answers are slight different due to approximation in textbook ... here answers are precise\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The thermal efficiency is 58.8 %\n",

+        " The MEP is 775.0  kPa\n",

+        " The % error in efficiency is 13.2 %\n",

+        " The % error MEP is -33.5 % \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.5:PG-186"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "Cp=1.0 # specific heat at constant pressure\n",

+      "k=1.4 # polytropic index for air\n",

+      "T1=25+273.0 # temperature at compressor inlet\n",

+      "T3=850+273.0 # maximum temperature in kelvin\n",

+      "\n",

+      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",

+      "\n",

+      "T2=T1*(r)**((k-1)/k) # temperature after compression\n",

+      "\n",

+      "T4=T3*(1/r)**((k-1.0)/k) # final temperature\n",

+      "\n",

+      "Wcomp=Cp*(T2-T1) # compressor work\n",

+      "Wturb=Cp*(T3-T4) # turbine work\n",

+      "\n",

+      "BWR=Wcomp/Wturb # back work ratio\n",

+      "\n",

+      "print \" The BWR is\",round(BWR*100,1),\"%\\n\" \n",

+      "\n",

+      "Effi=1-r**((1-k)/k) # thermal efficiency\n",

+      "\n",

+      "print\" The thermal efficiency is\",round(Effi*100,1),\"%\"\n",

+      "\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The BWR is 42.0 %\n",

+        " The thermal efficiency is 36.9 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 33

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.6:PG-186"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "Cp=1.0 # specific heat at constant pressure\n",

+      "k=1.4 # polytropic index for air\n",

+      "T1=25+273.0 # temperature at compressor inlet\n",

+      "T3=850+273.0 # maximum temperature in kelvin\n",

+      "\n",

+      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",

+      "efficomp=0.75 # efficiency of compressor\n",

+      "effiturb=0.75 # efficiency of turbine\n",

+      "\n",

+      "T2dash=T1*(r)**((k-1)/k) # temperature after compression\n",

+      "Wcomp=Cp*(T2dash-T1)/efficomp # compressor work\n",

+      "\n",

+      "T4dash=T3*(1/r)**((k-1)/k) # final temperature\n",

+      "Wturb=Cp*(T3-T4dash)*effiturb # turbine work\n",

+      "\n",

+      "BWR=100*Wcomp/Wturb # back work ratio\n",

+      "\n",

+      "print \" The BWR is\",round(BWR,1),\"%\\n\"\n",

+      "\n",

+      "T2=(Wcomp/Cp)+T1 # actual temperature of state 2\n",

+      "\n",

+      "qin=Cp*(T3-T2) # using first law \n",

+      "\n",

+      "Wnet=(Wturb-Wcomp) # net work\n",

+      "\n",

+      "effi=100*Wnet/qin # thermal efficiency\n",

+      "print\" The thermal efficiency is\",round(effi,1),\"%\"\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The BWR is 74.7 %\n",

+        " The thermal efficiency is 13.2 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 37

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.7:PG-191"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "Cp=1.0 # specific heat at constant pressure\n",

+      "k=1.4 # polytropic index for air\n",

+      "T1=25+273.0 # temperature at compressor inlet\n",

+      "T3=850+273.0 # maximum temperature in kelvin\n",

+      "\n",

+      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",

+      "\n",

+      "T2=T1*(r)**((k-1)/k) # temperature after compression\n",

+      "\n",

+      "T4=T3*(1/r)**((k-1)/k) # final temperature\n",

+      "\n",

+      "Wcomp=Cp*(T2-T1) # compressor work\n",

+      "Wturb=Cp*(T3-T4) # turbine work\n",

+      "\n",

+      "BWR=Wcomp/Wturb # back work ratio\n",

+      "\n",

+      "print \" The BWR is\",round(BWR,2),\"\\n\"\n",

+      "\n",

+      "effi=(1-((T1/T4)*(r**((k-1)/k))))# efficiency\n",

+      "print\" The thermal efficiency is\",round(effi*100,1),\"%\"\n",

+      "# The solution in textbook is incorrect due to wrong value of T4 (temperature at state 4)\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The BWR is 0.42 \n",

+        "\n",

+        " The thermal efficiency is 33.4 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 41

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.8:PG-193"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "#REFER TO FIG.:7.8\n",

+      "\n",

+      "Cp=1 # specific constant at constant pressure\n",

+      "k=1.4 # polytropic constant for air\n",

+      "T5=25+273.0 # temperature at state 5 in kelvin\n",

+      "T7=850+273.0 # temperature at state 4 in kelvin\n",

+      "T9=350 # exit temperature of water from bolier in kelvin\n",

+      "WdotST=100000.0 # power from steam turbine in Watt\n",

+      "r=5.0 # pressure ratio=P2/P1 & P4/P3\n",

+      "\n",

+      "h1=192.0 # specific enthalpy at 10 Kpa from steam table\n",

+      "h2=h1 # isenthalpic process\n",

+      "h3=3214.0 # specific enthalpy at 4 Mpa and 400 degree celsius from steam table\n",

+      "s3=6.769 # specific entropy at 4 Mpa and 400 degree celsius from steam table\n",

+      "\n",

+      "s4=s3 # isentropic process\n",

+      "sf=0.6491 # specific entropy of saturated liquid at 10 kPa and 45 degree celsiusfrom table C.2\n",

+      "sg=8.1510 # specific entropy of saturated liquid at 10 kPa and 45 degree celsiusfrom table C.2\n",

+      "x4=(s4-sf)/(sg-sf) # quality of steam\n",

+      "\n",

+      "hf=h1 # specific enthalpy of saturated liquid @ 10 Kpa  \n",

+      "hg=2584.6\n",

+      "h4=hf+x4*(hg-hf) # specific entropy at state 4\n",

+      "\n",

+      "mdots=WdotST/(h3-h4) # steam mass flow rate from turbine output\n",

+      "\n",

+      "T6=T5*(r**((k-1)/k)) # adiabatic process relation\n",

+      "T8=T7*(1/r**((k-1)/k)) # adiabatic process relation\n",

+      "\n",

+      "# Now using energy balance in boiler\n",

+      "mdota=mdots*(h3-h2)/(Cp*(T8-T9)) # mass flow rate of water\n",

+      "\n",

+      "Wdotturb=mdota*Cp*(T7-T8) #  power produced by turbine\n",

+      "\n",

+      "Wdotcomp=mdota*Cp*(T6-T5) # energy needed by compressor\n",

+      "\n",

+      "WdotGT=Wdotturb-Wdotcomp # net turbine work\n",

+      "\n",

+      "Qdotin=mdota*Cp*(T7-T6) # energy input by combustor\n",

+      "\n",

+      "effi=100*(WdotST+WdotGT)/Qdotin # combined efficiency\n",

+      "\n",

+      "print \"The thermal efficiency of the combined cycle is\",round(effi,1),\"%\"\n",

+      "\n",

+      "\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The thermal efficiency of the combined cycle is 56.4 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 45

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.9:PG-196"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# initialization of variables\n",

+      "\n",

+      "Cp=1 # specific constant at constant pressure\n",

+      "k=1.4 # polytropic constant for air\n",

+      "r=10.0\n",

+      "T2=-10+273 # temperature at entry  of compressor\n",

+      "T4=30+273 # temperature at entry of turbine\n",

+      "\n",

+      "T3=T2*(r**((k-1)/k)) # temperature at state 3 in kelvin\n",

+      "T1=T4*(1/r**((k-1)/k)) # temperature at state 1 in degree celsius\n",

+      "print \"The minimum temperature is\",round(T1-273),\"degree celsius \\n\"\n",

+      "\n",

+      "qin=Cp*(T2-T1) # heat input\n",

+      "Wcomp=Cp*(T3-T2)# compressor work\n",

+      "Wturb=Cp*(T4-T1) # turbine work\n",

+      "\n",

+      "COP=qin/(Wcomp-Wturb) # COP of refrigeration\n",

+      "print\" The COP is\",round(COP,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The minimum temperature is -116.0 degree celsius \n",

+        "\n",

+        " The COP is 1.07\n"

+       ]

+      }

+     ],

+     "prompt_number": 48

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7.10:PG-196"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#solution\n",

+      "# initialization of variables\n",

+      "\n",

+      "Cp=1.0 # specific constant at constant pressure\n",

+      "k=1.4 # polytropic constant for air\n",

+      "r=10.0\n",

+      "T3=-10+273 # temperature at entry  of compressor\n",

+      "T6=-40+273 # temperature at entry of turbine\n",

+      "\n",

+      "T5=T3 # heat exchanger\n",

+      "T2=T6 # heat exchanger\n",

+      "\n",

+      "T4=T3*(r**((k-1)/k)) # temperature after compression\n",

+      "T1=T6*(1/r**((k-1)/k)) # temperature after exit from turbine\n",

+      "\n",

+      "print \"The minimum temperature is\",round(T1-273),\"degree celsius \\n\"\n",

+      "\n",

+      "qin=Cp*(T2-T1) # heat input\n",

+      "Wcomp=Cp*(T4-T3)# compressor work\n",

+      "Wturb=Cp*(T6-T1) # turbine work\n",

+      "\n",

+      "COP=qin/(Wcomp-Wturb) # COP of refrigeration\n",

+      "\n",

+      "print\" The COP is\",round(COP,3)\n",

+      "\n",

+      "# the answer is correct within given limits\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The minimum temperature is -152.0 degree celsius \n",

+        "\n",

+        " The COP is 0.848\n"

+       ]

+      }

+     ],

+     "prompt_number": 51

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter8_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter8_2.ipynb
new file mode 100755
index 00000000..2ab04749
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter8_2.ipynb
@@ -0,0 +1,530 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:cd14818c82eb9353b8e2f9e12079fb0ba7b2a061c9a528ea53dea9dfa9a116ce"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8:Psychrometrics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.1:PG-208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialization of variables\n",
+      "Ra=0.287 # specific gas constant for air\n",
+      "P=100.0 # pressure of room in kPa\n",
+      "V=150.0 # volume of room in m^3\n",
+      "T=25+273 # temperature of air in kelvin\n",
+      "phi=0.6 # relative humidity\n",
+      "Pg=3.29 # saturation vapour pressure in kPa at 25 *C from table C.1\n",
+      "Mv= 18 # molecular mass of water vapor\n",
+      "Ma=28.97 # molecular mass of air\n",
+      "\n",
+      "Pv=Pg*phi # partial pressure of water vapour\n",
+      "\n",
+      "Pa=P-Pv # partial pressure of air\n",
+      "\n",
+      "w=0.622*(Pv/Pa) # humidity ratio in Kg of water/ Kg of dry air\n",
+      "Tdp=17.4 # dew point temperature from interpolation in table C.2 corresponding to partial pressure Pv=1.98 kPa\n",
+      "\n",
+      "ma=Pa*V/(Ra*T) # mass of air\n",
+      "mv=w*ma # mass of water vapour in  kg\n",
+      "\n",
+      "# now we find volume percentage\n",
+      "Nv=mv/Mv # moles of vapour\n",
+      "Na=ma/Ma # moles of air\n",
+      "\n",
+      "Vw= Nv/(Na+Nv) # fraction of volume occupied by water vapour\n",
+      "\n",
+      "print \"The humidity ratio is\",round(w,4),\"kg water/ kg of dry air \\n\"\n",
+      "print \"The dew point is\",round(Tdp,1),\"degree celsius \\n \"\n",
+      "print \"The mass of water vapour in the air is\",round(mv,3),\"kg \\n\"\n",
+      "print \"The volume percentage of the room that is water vapor is\",round(Vw*100,2),\"%\"\n",
+      "# The answers are correct within given limits \n",
+      "# The variation in answers is due to approximations made by\n",
+      "# textbook while python is precise\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The humidity ratio is 0.0125 kg water/ kg of dry air \n",
+        "\n",
+        "The dew point is 17.4 degree celsius \n",
+        " \n",
+        "The mass of water vapour in the air is 2.153 kg \n",
+        "\n",
+        "The volume percentage of the room that is water vapor is 1.98 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.2:PG-209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialization of variables\n",
+      "Ra=0.287 # specific gas constant for air\n",
+      "P=100.0 # pressure of room in kPa\n",
+      "w1=0.0126 # old humidity ratio of example 8.1-\n",
+      "Pg=3.29 # saturation vapour pressure in kPa at 25 *C from table C.1\n",
+      "mv=2.17 # initial mass of water vapour in example 8.1\n",
+      "T=25+273 # temperature after reheat\n",
+      "V=150.0 # volume of room in m^3 \n",
+      "Pv=1.228 # saturation vapour pressure in kPa @ 10 degree celsius from table C.1\n",
+      "Pa=P-Pv # partial pressure of air\n",
+      "w2=0.622*(Pv/Pa) # new humidity ratio in Kg of water/ Kg of dry air\n",
+      "deltaw=w1-w2 # difference in humidity ratio\n",
+      "ma=Pa*V/(Ra*T) # mass of air\n",
+      "deltamv=deltaw*ma # mass of water vapour condensed\n",
+      "X=deltamv*100/mv # percentage of water vapour condensed\n",
+      "print \"The percentage that condenses is\",round(X,2),\"% \\n\"\n",
+      "# AFTER REHEATING\n",
+      "phi=1.608*w2*Pa/Pg\n",
+      "print \"The relative humidity is\",round(phi*100,3),\"%\"\n",
+      "\n",
+      "# The answers are correct within given limits \n",
+      "# The variation in answers is due to approximations made by\n",
+      "# textbook while python is precise\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage that condenses is 38.85 % \n",
+        "\n",
+        "The relative humidity is 37.332 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.3:PG-212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialization of variables\n",
+      "T1=40 # dry bulb temperature in degree celsius\n",
+      "T2=20 # wet bulb temperature in degree celsius\n",
+      "Cp=1.0 # specific heat\n",
+      "P=100 # pressure of air stream in kPa\n",
+      "pg1=7.383 #saturation pressure @ 40 degree celsius\n",
+      "hfg2=2454 # latent heat for 20 degree celsius\n",
+      "Pg2=2.338 # saturation pressure @ 20 degree celsius\n",
+      "w2=0.622*Pg2/(P-Pg2) # specific humidity for wet bulb condition\n",
+      "hg1=2574 # specific enthalpy of saturated vapour @ 40 degree celsius\n",
+      "hf2=83.9 #spedific enthalpy of saturated liquid @ 20 degree celsius\n",
+      "w1=((w2*hfg2)+Cp*(T2-T1))/(hg1-hf2)# specific humidity for 40 degree celsius\n",
+      "print \"The humidity ratio is\",round(w1,4),\"kg water/ Kg dry air \\n\"\n",
+      "pv1=100*w1/(0.622+w1) # partial pressure of vapour\n",
+      "phi=pv1/pg1 # relative humidity\n",
+      "print \"The relative humidity is\",round(phi*100,1),\"% \\n\"\n",
+      "\n",
+      "hv=hg1 # temperature is at DBT=40 degree celsius\n",
+      "h=Cp*T1+w1*hv # specific enthalpy  of air\n",
+      "print \"The specific enthalpy is\",round(h,1),\"kJ/kg dry air\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The humidity ratio is 0.0066 kg water/ Kg dry air \n",
+        "\n",
+        "The relative humidity is 14.3 % \n",
+        "\n",
+        "The specific enthalpy is 57.1 kJ/kg dry air\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.5:PG-215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "T1=40 # inlet temperature in degree celsius\n",
+      "T2=27 # outlet temperature in degree celsius\n",
+      "phi1= 10 # relative humidity at inlet\n",
+      "# as no heat transfer takes place thus isenthalpic process\n",
+      "\n",
+      "#Thus following the enthalpy line at DBT=40 and Relative humidity=10\n",
+      "\n",
+      "phi2=45 # by interpolation of constant enthalpy line\n",
+      "w1=0.0046# specific humidity @ T=40 and phi1=10\n",
+      "w2=0.010 # specific humidity at outlet\n",
+      "W=w2-w1 # amount of water added\n",
+      "Tmin=18.5 # minimum temperature at 100% relative humidity\n",
+      "\n",
+      "print \"The relative humidity is\",round(phi2,1),\"% \\n \"\n",
+      "print \"The added water is\",round(W,4),\"kg water/kg dry air \\n\"\n",
+      "print \"The lowest possible temperature is\",round(Tmin,1),\"*C \"\n",
+      "\n",
+      "\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The relative humidity is 45.0 % \n",
+        " \n",
+        "The added water is 0.0054 kg water/kg dry air \n",
+        "\n",
+        "The lowest possible temperature is 18.5 *C \n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.6:PG-215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "T1=5+273.0 # outside air temperature in kelvin\n",
+      "P=100.0 # pressure in kPa\n",
+      "Ra=0.287 # specific gas constant for air\n",
+      "phi=0.7 # relative humidity outside\n",
+      "Qf=50.0/60.0 # volume flow rate in m**3/sec\n",
+      "Pg1=0.872 # saturation pressure at 278 K\n",
+      "Pv1=phi*Pg1 # partial pressure of water vapour\n",
+      "Pa1=P-Pv1 # partial pressure of air\n",
+      "\n",
+      "rhoa=Pa1/(Ra*T1) # density of dry air\n",
+      "\n",
+      "mdota=Qf*rhoa # mass flow rate of dry air\n",
+      "\n",
+      "# using psychrometric chart at T1=5*C and phi1=70% \n",
+      "h1=14 # inlet enthalpy in kJ/kg\n",
+      "h2=35 # enthalpy after heating in kJ/kg\n",
+      "\n",
+      "Qdot=mdota*(h2-h1) # heat transfer rate\n",
+      "# from psychrometric chart for T=25 *C and 35 kJ/kg enthalpy\n",
+      "phi2=19 # realtive humidity\n",
+      "\n",
+      "print \"The heat transfer rate is\",round(Qdot,1),\"kJ/s \\n\"\n",
+      "print \"The final relative humidity is\",round(phi2,4),\"% \"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transfer rate is 21.8 kJ/s \n",
+        "\n",
+        "The final relative humidity is 19.0 % \n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.7:PG-216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "#DATA TAKEN FROM PSYCHROMETRIC CHART\n",
+      "T1=5+273.0 # outside temperature in kelvin\n",
+      "h1=10# enthalpy in kJ/kg @ T=5 *C and 40 % relative humidity\n",
+      "Pg1=0.872 # saturaion pressure in kPa for 5 degree celsius DBT\n",
+      "phi1=0.4\n",
+      "h2=33 # specific enthalpy at 25 *C and 40 % relatuve humidity\n",
+      "h3=45.0 # specific enthalpy at state 3\n",
+      "P=100.0 # atmospheric pressure in kPa\n",
+      "Ra=0.287 # specific gas constant for air\n",
+      "Qf=60.0/60.0 # volume flow rate in m**3/s\n",
+      "Pv1=phi1*Pg1 # partial presure of water vapour \n",
+      "Pa1=P-Pv1 # partial pressure of air\n",
+      "w2=0.0021 # specific humidity @ 40 % relative humidity and 25*C temperature\n",
+      "w3=0.008 # final specific humidity\n",
+      "rhoa1=Pa1/(Ra*T1) # air density\n",
+      "mdota=Qf*rhoa1 # mass flow rate of dry air\n",
+      "\n",
+      "Qdot=mdota*(h2-h1) # heat transfer rate\n",
+      "\n",
+      "# as the process is isothermal thus\n",
+      "mdots=mdota*(w3-w2)# mass flow rate of steam by conservation of mass\n",
+      "print \"the rate of steam supplied is\",round(mdots,4),\"kg/s \\n\"\n",
+      "# also using energy balance\n",
+      "hs=(mdota*(h3-h2))/mdots # enthalpy of steam\n",
+      "hf=604.7 # enthalpy of saturated liquid @ 400 kPa\n",
+      "hg=2738.5 # enthalpy of saturated vapour @ 400 kPa\n",
+      "xs=(hs-hf)/(hg-hf)\n",
+      "print \"The quality of steam is\",round(xs,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of steam supplied is 0.0074 kg/s \n",
+        "\n",
+        "The quality of steam is 0.67\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.8:PG-217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "# REFER TO FIG. 8.4\n",
+      "T1=30 # outside temperature in degree celsius\n",
+      "phi1=0.9 # outside relative humidity\n",
+      "T2=23 # room temperature in degree celsius\n",
+      "phi2=0.4 # relative humidity in room\n",
+      "\n",
+      "# using psychrometric chart\n",
+      "w1=0.0245 # specific humidity @ 30 *C and relative humidity 0.9\n",
+      "h1=93 # specific enthalpy @ 30 *C and relative humidity 0.9\n",
+      "w2=w1 # during cooling humidity remains constant \n",
+      "w3=0.007 # specific humidity @ 23 *C and relative humidity 0.4\n",
+      "h4=41 # final specific enthalpy\n",
+      "h3=26 # specific enthalpy @ 23 *C and relative humidity 0.4\n",
+      "deltaw=w3-w2 # moisture removed\n",
+      "print \" the amount of moisture removed is\",round(deltaw,4),\"kg \\n\"\n",
+      "\n",
+      "qout=h3-h1 # heat removed F-G-H process\n",
+      "\n",
+      "print \" the heat removed is\",round(qout,4),\"kJ/kg \\n \"\n",
+      "\n",
+      "qin=h4-h3 # heat added to bring to desired state\n",
+      "\n",
+      "print \" the heat added is\",round(qin,4),\"kJ/kg \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " the amount of moisture removed is -0.0175 kg \n",
+        "\n",
+        " the heat removed is -67.0 kJ/kg \n",
+        " \n",
+        " the heat added is 15.0 kJ/kg \n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8.9:PG-218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "P=100 # atospheric pressure in kPa\n",
+      "R=0.287 # specific gas constant for air\n",
+      "T1=15+273 # outside temperature in kelvin\n",
+      "phi1=0.4# outside air relative humidity\n",
+      "Qf1=40 # outside air flow rate in m^3/min\n",
+      "T2=32+273 # inside temperature in kelvin\n",
+      "phi2=0.7 # inside air relative humidity\n",
+      "Qf2=20 # outside air flow rate in m^3/min\n",
+      "Ps1=1.7 # saturation pressure @ 15 degree celsius and 40% humidity\n",
+      "Ps2=4.9 # saturation pressure @ 32 degree celsius and 70% humidity\n",
+      "\n",
+      "Pv1=Ps1*phi1 # partial pressure of water vapour outside\n",
+      "\n",
+      "Pv2=Ps2*phi2 # partial pressure of water vapour inside\n",
+      "\n",
+      "Pa1=P-Pv1 #partial pressure of dry air outside\n",
+      "Pa2=P-Pv2 #partial pressure of dry air inside\n",
+      "\n",
+      "rhoa1=Pa1/(R*T1) # density of outside air\n",
+      "mdota1=Qf1*rhoa1 # mass flow rate of air outside\n",
+      "\n",
+      "rhoa2=Pa2/(R*T2) # density of inside air\n",
+      "mdota2=Qf2*rhoa2 # mass flow rate of inside air\n",
+      " # using psychrometric chart locating state 1 and 2\n",
+      "h1=37 # specific enthalpy @ DBT 15*C and 40 % humidity\n",
+      "w1=0.0073 # specific humidity @ DBT 15*C and 40 % humidity\n",
+      "h2=110 # specific enthalpy @ DBT 32*C and 70 % humidity\n",
+      "w2=0.0302 # specific humidity @ DBT 32*C and 70 % humidity\n",
+      "ratio=mdota1/mdota2 # ratio of distance between states \n",
+      "#  using this ratio state 3 is located on psychrometric chart\n",
+      "T3=(mdota1*T1+mdota2*T2)/(mdota1+mdota2)-273 # final temparature in celsius\n",
+      "\n",
+      "phi3=65# final relative humidity at T3 from psychrometric chart\n",
+      "\n",
+      "print \" The relative humidity is\",round(phi3,4),\"% \\n\"\n",
+      "print \" The resultant temperature is\",round(T3),\"degree celsius\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The relative humidity is 65.0 % \n",
+        "\n",
+        " The resultant temperature is 20.0 degree celsius\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex7.10:PG-219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "mdotw3=10000.0 # mass flow rate of water entering in cooling tower in kg/min\n",
+      "Tw1=40+273.0 # temperature of water entering cooling tower in kelvin\n",
+      "Ta1=20+273.0 # temperature of air entering cooling tower in kelvin\n",
+      "phi1=0.5# relative humidity of entering air\n",
+      "Tw2=25+273.0 # temperature of water leaving cooling tower in kelvin\n",
+      "Ta2=32+273 # temperature of air leaving cooling tower in kelvin\n",
+      "phi2=0.98 # relative humidity of leaving air\n",
+      "# from psychrometric chart\n",
+      "h1=37.0# specific enthalpy of air @ 20*C DBT and 50% humidity\n",
+      "w1=0.0073 # specific humidity of air @ 20*C DBT and 50% humidity\n",
+      "h2=110.0 # specific enthalpy of air @ 32*C DBT and 98% humidity\n",
+      "w2=0.030 # specific humidity of air @ 32*C DBT and 98% humidity\n",
+      "\n",
+      "h3=167.5 # specific enthalpy of water from steam table at 40 degree celsius\n",
+      "h4=104.9 # specific enthalpy of water from steam table at 25 degree celsius\n",
+      "\n",
+      "mdota=(mdotw3*(h4-h3))/(h1-h2+(w2-w1)*h4) # by energy balance\n",
+      "\n",
+      "\n",
+      "v1=0.84 # specific volume of air entering tower from psychrometric chart\n",
+      "\n",
+      "Qf=mdota*v1 # volume flow rate in m^3/min\n",
+      "print \"The volume flow rate of air into the cooling tower is\",round(Qf),\" m^3/min \\n\"\n",
+      "\n",
+      "mdot4=mdotw3-(w2-w1)*mdota # by mass balance\n",
+      "print \"The mass flow rate of water that leaves the cooling tower \",round(mdot4),\"kg/min\"\n",
+      "# The answers is slightly different in textbook due to approximations in calculations while in python solution is precise\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume flow rate of air into the cooling tower is 7446.0  m^3/min \n",
+        "\n",
+        "The mass flow rate of water that leaves the cooling tower  9799.0 kg/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter9_2.ipynb b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter9_2.ipynb
new file mode 100755
index 00000000..9448e892
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/Chapter9_2.ipynb
@@ -0,0 +1,630 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7e02442740cf50f0beae7f1fd9dfdf2fb7a16b9c0848a48a17d80c7a19de3f3d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9:Combustion"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.1:PG-229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "AFactual=20 # air fuel ratio actual\n",
+      "# The energy balance is done from equation\n",
+      "\n",
+      "# C4H10 + 6.5(O2+3.76N2)-----> 4CO2 + 5H2O + 24.44N2\n",
+      "\n",
+      "P=100 # atmospheic preesure in kPa\n",
+      "mair=6.5*(1+3.76)*29 # mass of air\n",
+      "mfuel=1*58 # mass of fuel\n",
+      "AFth=mair/mfuel # theoritical air-fuel ratio\n",
+      "Pexcessair=(AFactual-AFth)*100/AFth\n",
+      "\n",
+      "print \"The % excess air is\",round(Pexcessair,2),\"% \\n\"\n",
+      "\n",
+      "# NOW THE REACTION IS\n",
+      "# C4H10+ (1+%excessair/100)*6.5*(O2+3.76N2) -----> 4CO2 + 5H2O + 1.903O2 + 31.6N2\n",
+      "\n",
+      "PCO2=4/42.5*100 # VOLUME % OF CO2\n",
+      "\n",
+      "print \"The volume % of CO2 is\",round(PCO2,2),\"% \\n\"\n",
+      "\n",
+      "# NOW WE FIND DEW POINT\n",
+      "Nv=5 # moles of water\n",
+      "N=42.5 # moles of air\n",
+      "Pv=P*(Nv/N) # partial pressure of vapour\n",
+      "Tdp=49# dew point temperature in degree celsius from table C.2\n",
+      "\n",
+      "print \"The Dew point temperature is\",round(Tdp,2),\"degree celsius\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The % excess air is 29.28 % \n",
+        "\n",
+        "The volume % of CO2 is 9.41 % \n",
+        "\n",
+        "The Dew point temperature is 49.0 degree celsius\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.2:PG-231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "Pair=0.9 # 90% air is used for combustion\n",
+      "\n",
+      "# THE REACTION IS\n",
+      "# C4H10 + 0.9*6.5*(O2+3.76N2)----> aCO2 + 5H20 + bCO\n",
+      "# a and b are calculated by atomic balance\n",
+      "a=2.7\n",
+      "b=1.3\n",
+      "PCO=b*100/31 # volume % of CO\n",
+      "\n",
+      "print \"The volume % of CO is\",round(PCO,2),\"% \\n\"\n",
+      "\n",
+      "mair=6.5*Pair*4.76*29 # mass of air in kg\n",
+      "mfuel=1*58 # mass of fuel butane in kg\n",
+      "AF=mair/mfuel # air-fuel ratio\n",
+      "\n",
+      "print \"The air to fuel ratio is\",round(AF,2),\"kg air/ kg fuel \"\n",
+      "# THE SOLUTION IS CORRECT BUT  THERE ARE SOME PRINTING MISTAKES IN TEXTBOOK\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume % of CO is 4.19 % \n",
+        "\n",
+        "The air to fuel ratio is 13.92 kg air/ kg fuel \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.3:PG-231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "# THE REACTION IS\n",
+      "# aC4H10 + b(O2+3.76N2)----> CO2 + 1CO + 3.5H20 + 84.6N2 + cH2O\n",
+      "# a, b and c are calculated by atomic balance\n",
+      "# on balancing the equations we get a=3 b=22.5 c=15\n",
+      "# Now equation becomes\n",
+      "#C4H10 + 7.5(O2+3.76N2)----> 3.67CO2 + 0.33CO + 1.17H20 + 28.17N2 + 5H2O\n",
+      "#MOLES OF AIR in this equation is 7.5 moles\n",
+      "mairactual=7.5 # in moles\n",
+      "#MOLES OF AIR in standard equation of Ex.9 is 6.5\n",
+      "mairtheoritical=6.5\n",
+      "Ptheoriticalair=100*(mairactual/mairtheoritical) \n",
+      "print \"The % theoritical air is\",round(Ptheoriticalair,1),\"% \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The % theoritical air is 115.4 % \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.4:PG-232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "# The reaction equation is \n",
+      "#CaHb + c(O2+3.76N2)---> 10.4CO2 + 1.2CO + 2.8O2 + 85.6N2 + dH2O\n",
+      "\n",
+      "# using atomic balancing the equations become\n",
+      "\n",
+      "# C11.6H37.9 + 21.08(O2+3.76N2)---> 11.6CO2 + 18.95H2O + 79.26N2\n",
+      "Ptheoriticalair=22.8*100/21.08 # theoritical air\n",
+      "excessair=Ptheoriticalair-100\n",
+      "\n",
+      "print\"The excess air is\",round(excessair),\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The excess air is 8.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.5:PG-235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "# The reaction equation is \n",
+      "#C3H8 + 5(O2+3.76N2)---> 3CO2 + 18.8N2 + 4H2O\n",
+      "\n",
+      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
+      "hfCO2=-393520 # enthalpy associated with CO2\n",
+      "hfH2O=-285830 # enthalpy associated with H2O(l)\n",
+      "hfC3H8=-103850# ehthalpy associated with C3H8\n",
+      "\n",
+      "# by first law Q= Hproducts - Hreactants\n",
+      "\n",
+      "Qg=3*(hfCO2)+4*(hfH2O)-(hfC3H8) # enthalpy of combustion for gaseous propane\n",
+      "\n",
+      "print \" The enthalpy of combustion for gaseous propane is\",round(Qg),\"kJ\\n\"\n",
+      "\n",
+      "hv=15060 # enthalpy of vaporization for propane\n",
+      "\n",
+      "Ql=3*(hfCO2)+4*(hfH2O)-(hfC3H8-hv) # enthalpy of combustion for liquid propane\n",
+      "\n",
+      "print \" The enthalpy of combustion for liquid propane is\",round(Ql),\"kJ\\n\"\n",
+      "\n",
+      "#The answers are slightly different in textbook as they have approximated the result while in Python results are precise\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy of combustion for gaseous propane is -2220030.0 kJ\n",
+        "\n",
+        " The enthalpy of combustion for liquid propane is -2204970.0 kJ\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.6:PG-235"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "# The reaction equation is \n",
+      "#C3H8 + 5(O2+3.76N2)---> 3CO2 + 18.8N2 + 4H2O\n",
+      "\n",
+      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
+      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
+      "hbarCO2=22280 #enthalpy associated with CO2 at 600K from table E.4\n",
+      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
+      "\n",
+      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
+      "hbarH2O=20402 #enthalpy associated with H20 at 600K from table E.6\n",
+      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
+      "\n",
+      "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n",
+      "\n",
+      "hbarN2=17563 #enthalpy associated with N2 at 600K from table E.2\n",
+      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
+      "# by first law Q= Hproducts - Hreactants\n",
+      "\n",
+      "Qg=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+18.8*(hbarN2-hdotbarN2)-(hfC3H8) # enthalpy of combustion\n",
+      "\n",
+      "print\"The heat transfer required is\",round(Qg),\"kJ\\n\"\n",
+      "\n",
+      "#The answer is WRONG textbook as they have made an error in calculating Qg \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transfer required is -1796043.0 kJ\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.7:PG-236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# initialization of variables\n",
+      "# The reaction equation is \n",
+      "\n",
+      "#C8H18 + 12.5(O2+3.76N2)---> 8CO2 + 47N2 + 9H2O\n",
+      "\n",
+      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
+      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
+      "hbarCO2=42769 #enthalpy associated with CO2 at 1000K from table E.4\n",
+      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
+      "\n",
+      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
+      "hbarH2O=35882 #enthalpy associated with H20 at 1000K from table E.6\n",
+      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
+      "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n",
+      "\n",
+      "hbarN2p=(30784+29476)/2 #enthalpy associated with N2 at 1000K from table E.2 by averaging enthalpy at 1020K and 980K for product\n",
+      "hbarN2r=17563 #enthalpy associated with N2 at 600K from table E.2 for reactant\n",
+      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
+      "\n",
+      "hfC8H18=-249910 # enthalpy of formation associated with octane taken from internet as not provided in textbook\n",
+      "\n",
+      "hbarO2=17929 # enthalpy associated with O2 at 600K table E.3\n",
+      "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n",
+      "\n",
+      "# using first law and including kinetic energy change\n",
+      "# 0=Hp-Hr+Mp*(V^2)/2\n",
+      "\n",
+      "Hp=8*(hfCO2+hbarCO2-hdotbarCO2)+9*(hfH2O+hbarH2O-hdotbarH2O)+47*(hbarN2p-hdotbarN2)\n",
+      "# enthalpy of products\n",
+      "\n",
+      "Hr=(hfC8H18)+12.5*(hbarO2-hdotbarO2)+47*(hbarN2r-hdotbarN2)\n",
+      "# enthalpy of reactants\n",
+      "\n",
+      "Mp=8*44+9*18+47*28 #(mass of products by multiplying molecular mass to number of moles)\n",
+      "\n",
+      "V=math.sqrt(2*1000*(Hr-Hp)/Mp)# exit velocity using energy balance\n",
+      "\n",
+      "print \"The exit velocity is\",round(V),\"m/s\"\n",
+      "\n",
+      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The exit velocity is 2116.0 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.8:PG-237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "# The reaction equation with theoritical air is \n",
+      "# C8H18 + 12.5(O2+3.76N2)---> 8CO2 + 47N2 + 9H2O\n",
+      "\n",
+      "# for 400% theoritical air reaction is\n",
+      "\n",
+      "# C8H18 + 50(O2+3.76N2)---> 8CO2 + 188N2 + 9H2O + 37.5O2\n",
+      "\n",
+      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
+      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
+      "hbarCO2=42769 #enthalpy associated with CO2 at 1000K from table E.4\n",
+      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
+      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
+      "hbarH2O=35882 #enthalpy associated with H20 at 1000K from table E.6\n",
+      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
+      "hbarN2p=(30784+29476)/2 #enthalpy associated with N2 at 1000K from table E.2 by averaging enthalpy at 1020K and 980K \n",
+      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
+      "\n",
+      "hfC8H18=-249910 # enthalpy associated with octane taken from internet as not provided in textbook\n",
+      "hbarO2=31389 # enthalpy associated with O2 at 1000K table E.3\n",
+      "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n",
+      "\n",
+      "Hp=8*(hfCO2+hbarCO2-hdotbarCO2)+9*(hfH2O+hbarH2O-hdotbarH2O)+37.5*(hbarO2-hdotbarO2)+188*(hbarN2p-hdotbarN2)# enthalpy of products\n",
+      "\n",
+      "Hr=(hfC8H18)\n",
+      "# enthalpy of reactants\n",
+      "\n",
+      "Q=Hp-Hr # using first law2\n",
+      "\n",
+      "print \" The heat transfer is\",round(Q),\"kJ\"\n",
+      "\n",
+      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The heat transfer is 312593.0 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.9:PG-237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "# The reaction equation is \n",
+      "#C3H8 + 5O2---> 8CO2 + 4H2O\n",
+      "\n",
+      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
+      "hfCO2=-393520 # enthalpy associated with CO2\n",
+      "hfH2O=-241820 # enthalpy associated with gaseous H2O\n",
+      "hfC3H8=103850# enthalpy of formation associated with C3H8\n",
+      "hfgC3H8=15060# enthalpy of vapourization associated with C3H8\n",
+      "T=20+273 # temperature in kelvin\n",
+      "Rbar=8.314 # universal gas constant\n",
+      "Nr=6 # number of moles of reactants\n",
+      "Np=7 # number of moles of products\n",
+      "Hp=3*(hfCO2)+4*(hfH2O) # enthalpy of products\n",
+      "\n",
+      "Hr=hfC3H8+hfgC3H8 # enthalpy of reactants\n",
+      "\n",
+      "Q=(Hp-Hr-(Nr-Np)*Rbar*T)*10**(-3) # heat transfer from first law\n",
+      "\n",
+      "print \" The heat transfer is\",round(Q),\"MJ\"\n",
+      "\n",
+      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The heat transfer is -2264.0 MJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.10:PG-239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "# The reaction equation for theoritical air is \n",
+      "#C3H8 + 5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 18.8N2\n",
+      "\n",
+      "# for 250% theoritical air reaction becomes\n",
+      "#C3H8 + 12.5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 47N2 + 7.5O2\n",
+      "\n",
+      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
+      "\n",
+      "Np=47+7.5+4+3 # number of moles of product\n",
+      "hfCO2=-393520 # enthalpy of formation associated with CO2\n",
+      "hbarCO2=(62963+65271)/2 #enthalpy associated with CO2 at 1380 K from table E.4\n",
+      "hbarCO2dash=(58381+60666)/2 #enthalpy associated with CO2 at 1300 K by average from table E.4\n",
+      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
+      "\n",
+      "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n",
+      "\n",
+      "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n",
+      "hbarH2O=(51521+53351)/2 #enthalpy associated with H20 at 1380 K by taking average from table E.6\n",
+      "hbarH2Odash=48807 #enthalpy associated with H20 at 1300 K  from table E.6\n",
+      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
+      "\n",
+      "hbarN2=42920 #enthalpy associated with N2 at 1380K from table E.2 by interpolating enthalpy between 1020K and 980K \n",
+      "hbarN2dash=40170 #enthalpy associated with N2 at 1300 K from table E.2 \n",
+      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
+      "\n",
+      "hfO2=(44198+45648)/2 # enthalpy associated with O2 at 1380 Kby taking average from table E.3\n",
+      "hfO2dash=48807 # enthalpy associated with O2 at 1380 Kby taking average from table E.3\n",
+      "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n",
+      "\n",
+      "# for adiabatic flame temperature first assume products composed only of nitrogen and Q=0 as adiabatic\n",
+      "hp=(hfC3H8-3*(hfCO2)-4*(hfH2O))/Np +hdotbarN2\n",
+      "# using hp we assume temp=1380 K\n",
+      "# then energy for 1380 k is\n",
+      "H1=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+7.5*(hfO2-hdotbarO2)+47*(hbarN2-hdotbarN2) # energy assuming temperature to be 1380 K\n",
+      "\n",
+      "#this is very large \n",
+      "\n",
+      "# now at 1300 K adiabatic temperature\n",
+      "H2=3*(hfCO2+hbarCO2dash-hdotbarCO2)+4*(hfH2O+hbarH2Odash-hdotbarH2O)+7.5*(hfO2dash-hdotbarO2)+47*(hbarN2dash-hdotbarN2) # energy assuming temperature to be 1300 K\n",
+      " \n",
+      " # now interpolation between these two temperatures\n",
+      "Tp=1300-((hp+H2)/(H1-H2))*(1380-1300) # adiabatic temperature by interpolation\n",
+      "print \"The adiabatic flame temperature is\",round(Tp),\"K\"\n",
+      "\n",
+      "#The answers is different in textbook as they have printed the value of hfCO2 with positive sign while calculating H2\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The adiabatic flame temperature is 1311.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.11:PG-240"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# initialization of variables\n",
+      "\n",
+      "# The reaction equation for theoritical air is \n",
+      "#C3H8 + 5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 18.8N2\n",
+      "\n",
+      "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n",
+      "\n",
+      "Np=18.8+4+3 # number of moles of product\n",
+      "hfCO2=-393520 # enthalpy associated with CO2\n",
+      "hbarCO2=137400 #enthalpy associated with CO2 at 2600 K from table E.4 by interpolation\n",
+      "hbarCO2dash=125152 #enthalpy associated with CO2 at 2400 K from table E.4\n",
+      "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n",
+      "\n",
+      "hfC3H8=-103850# ehthalpy associated with C3H8\n",
+      "\n",
+      "hfH2O=-241820 # enthalpy associated with gaseous H2O\n",
+      "hbarH2O=114273 #enthalpy associated with H20 at 2600 K from table E.6\n",
+      "hbarH2Odash=103508 #enthalpy associated with H20 at 2400 K  from table E.6\n",
+      "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n",
+      "\n",
+      "hbarN2=86600 #enthalpy associated with N2 at 2600 K from table E.2 by interpolation\n",
+      "hbarN2dash=79320 #enthalpy associated with N2 at 2400 K from table E.2 \n",
+      "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n",
+      "\n",
+      "# for adiabatic flame temperature first assume products composed only of nitrogen and Q=0 as adiabatic\n",
+      "hp=(hfC3H8-3*(hfCO2)-4*(hfH2O))/Np +hdotbarN2 \n",
+      "\n",
+      "# using hp we assume temp=2600 K\n",
+      "# then energy for 2600 k is\n",
+      "H1=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+18.8*(hbarN2-hdotbarN2) # energy assuming temperature to be 2600 K\n",
+      "\n",
+      "# now at 2400 K adiabatic temperature\n",
+      "H2=3*(hfCO2+hbarCO2dash-hdotbarCO2)+4*(hfH2O+hbarH2Odash-hdotbarH2O)+18.8*(hbarN2dash-hdotbarN2) # energy assuming temperature to be 2400 K\n",
+      " \n",
+      " # now interpolation between these two temperatures\n",
+      "Tp=2400-((hp+H2)/(H1-H2))*(2600-2400) # adiabatic temperature by interpolation\n",
+      "print \"The adiabatic flame temperature is\",round(Tp),\"K\"\n",
+      "\n",
+      "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The adiabatic flame temperature is 2409.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_Demystified_by_Merle_C_Potter/README.txt b/Thermodynamics_Demystified_by_Merle_C_Potter/README.txt
new file mode 100755
index 00000000..2aa18a7d
--- /dev/null
+++ b/Thermodynamics_Demystified_by_Merle_C_Potter/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Aviral Yadav
+Course: btech
+College/Institute/Organization: ABES Engineering College
+Department/Designation: Mechanical Engineering
+Book Title: Thermodynamics Demystified
+Author: Merle C. Potter
+Publisher: Mcgraw-hill, Usa
+Year of publication: 2009
+Isbn:  0071605991
+Edition: 1
\ No newline at end of file
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diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/README.txt b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/README.txt
new file mode 100755
index 00000000..ccf23c27
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/README.txt
@@ -0,0 +1,10 @@
+Contributed By: pratik gandhi
+Course: bca
+College/Institute/Organization: Cybercom Creation
+Department/Designation: Developer
+Book Title: Thermodynamics and Heat Power
+Author: I. Granet and M. Bluestein
+Publisher: Addison Wesley(singapore), New Delhi
+Year of publication: 2001
+Isbn: 81-7808-291-8
+Edition: 6
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch10_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch10_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch10_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch10_2.ipynb
new file mode 100755
index 00000000..ff3ad88b
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch10_2.ipynb
@@ -0,0 +1,758 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8da890ac5238ff5b0c8b5e2f22721a2e530a75b68b25eb33de8468d2a1b6f0fb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Refrigeration"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1  Page No : 503"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "T1 = 70+460; \t\t\t#70F = 70+460 R \t\t\t#Energy flows into the system at reservoir at constant temperature T1(unit:R)\n",
+      "T2 = 32+460; \t\t\t#32F = 32+460 R \t\t\t#Heat is rejected to the constant temperature T2(Unit:R)\n",
+      "print \"Solution for a\"\n",
+      "COP = T2/(T1-T2); \t\t\t#Coefficient of performance\n",
+      "print \"Coefficient of performanceCOP) of the cycle is %.2f\"%(COP);\n",
+      "print \"Solution for b\"\n",
+      "Qremoved = 1000; \t\t\t#Unit:Btu/min \t\t\t#heat removal\n",
+      "WbyJ = Qremoved/COP; \t\t\t#The power required \t\t\t#Unit:Btu/min\n",
+      "print \"The power required is %.2f Btu/min\"%(WbyJ);\n",
+      "print \"Solution for c\"\n",
+      "Qrej = Qremoved+WbyJ; \t\t\t#The rate of heat rejected to the room  \t\t\t#Unit:Btu/min\n",
+      "print \"The rate of heat rejected to the room is %.2f Btu/min\"%(Qrej);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "Coefficient of performanceCOP) of the cycle is 12.00\n",
+        "Solution for b\n",
+        "The power required is 83.00 Btu/min\n",
+        "Solution for c\n",
+        "The rate of heat rejected to the room is 1083.00 Btu/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2  Page No : 504"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "T1 = 20+273; \t\t\t#20C = 20+273 R \t\t\t#Energy flows into the system at reservoir at constant temperature T1(unit:R)\n",
+      "T2 = -5+273; \t\t\t#-5C = -5+273 R \t\t\t#Heat is rejected to the constant temperature T2(Unit:R)\n",
+      "print \"Solution for a\"\n",
+      "COP = T2/(T1-T2); \t\t\t#Coefficient of performance\n",
+      "print \"Coefficient of performanceCOP) of the cycle is %.2f\"%(COP);\n",
+      "print \"Solution for b\"\n",
+      "Qremoved = 30; \t\t\t#Unit:kW \t\t\t#heat removal \n",
+      "W = Qremoved/COP; \t\t\t#power required \t\t\t#unit:kW\n",
+      "print \"The power required is %.2f kW \"%(W);\n",
+      "print \"Solution for c\"\n",
+      "Qrej = Qremoved+W; \t\t\t#The rate of heat rejected to the room  \t\t\t#Unit:kW\n",
+      "print \"The rate of heat rejected to the room is %.2f kW\"%(Qrej);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "Coefficient of performanceCOP) of the cycle is 10.00\n",
+        "Solution for b\n",
+        "The power required is 3.00 kW \n",
+        "Solution for c\n",
+        "The rate of heat rejected to the room is 33.00 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3  Page No : 505"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "T1 = 70+460; \t\t\t#70F = 70+460 R \t\t\t#Energy flows into the system at reservoir at constant temperature T1(unit:R)\n",
+      "T2 = 20+460; \t\t\t#20F = 20+460 R \t\t\t#Heat is rejected to the constant temperature T2(Unit:R)\n",
+      "print \"Solution for a\"\n",
+      "COP = T2/(T1-T2); \t\t\t#Coefficient of performance\n",
+      "print \"Coefficient of performanceCOP) of the cycle is %.2f\"%(COP);\n",
+      "print \"Solution for b\"\n",
+      "HPperTOR = 4.717/COP; \t\t\t#Horsepower per ton of refrigeration \t\t\t#Unit:hp/ton\n",
+      "COPactual = 2; \t\t\t#Actual Coefficient of performance(COP) is stated to be 2\n",
+      "HPperTORactual = 4.717/COPactual; \t\t\t#Horsepower per ton of refrigeration(actual) \t\t\t#Unit:hp/ton\n",
+      "print \"The horsepower required by the actual cycle over the minimum is %.2f hp/ton\"%(HPperTORactual-HPperTOR);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "Coefficient of performanceCOP) of the cycle is 9.00\n",
+        "Solution for b\n",
+        "The horsepower required by the actual cycle over the minimum is 1.83 hp/ton\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4  Page No : 506"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "\n",
+      "COP = 4.5; \t\t\t#Coefficient of performance \t\t\t#From problem 10.1\n",
+      "HPperTOR = 4.717/COP; \t\t\t#Horsepower per ton of refrigeration \t\t\t#Unit:hp/ton\n",
+      "Qremoved = 1000; \t\t\t#Unit:Btu/min \t\t\t#From problem 10.1\n",
+      "#1000 Btu/min /200 Btu/min ton  =  5 tons of refrigeration\n",
+      "HPrequired = HPperTOR*5; \t\t\t#The horsepower required \t\t\t#unit:hp\n",
+      "print \"The horsepower required is %.2f hp\"%(HPrequired);\n",
+      "#In problem 10.1, 77.2 Btu/min was required\n",
+      "print \"The power required is %.2f hp\"%(77.2*778*inv([[33000]])); \t\t\t#1 Btu = 778 ft*lbf \t\t\t#1 min*hp  =  33000 ft*lbf\n",
+      "#The ratio of the power required in each problem is the same as the inverse ratio of the COP value\n",
+      "#Therefore,\n",
+      "print \"The power required is %.2f hp\"%((COP/12.95)*HPrequired); \t\t\t#COPin problem 10.1) = 12.95\n",
+      "print \"This checks our results\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The horsepower required is 5.24 hp\n",
+        "The power required is 1.82 hp\n",
+        "The power required is 1.82 hp\n",
+        "This checks our results\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5  Page No : 506"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "COP = 10.72; \t\t\t#In the problem 10.2 \t\t\t#Coefficient of performance\n",
+      "P = 2.8; \t\t\t#In the problem 10.2 \t\t\t#The power was 2.8 kW\n",
+      "COPactual = 3.8; \t\t\t#Actual Coefficient of performance(COP)\n",
+      "\n",
+      "# calculation\n",
+      "power = P*COP/COPactual; \t\t\t#The power required  \t\t\t#unit:kW\n",
+      "\n",
+      "# results\n",
+      "print \"The power required is %.2f kW\"%(power)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power required is 7.90 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6  Page No : 509"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From Appendix 3,at 120psia,the corresponding saturation temperature is 66 F, enthalpies are\n",
+      "h1 = 116.0; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h2 = 116.0; \t\t\t#Unit:Btu/lbm \t\t\t#Throttling gives h1 = h2 \t\t\t#enthalpy\n",
+      "h3 = 602.4; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "#From the consideration that s3 = s4,h4 is found at 15 psia,\n",
+      "s3 = 1.3938; \t\t\t#s = entropy \t\t\t#Unit:Btu/(lbm*F)\n",
+      "#Therefore by interpolation in the superheat tables at 120 psia,\n",
+      "t4 = 237.4; \t\t\t#Unit:fahrenheit \t\t\t#temperature\n",
+      "h4 = 733.4; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "print \"Solution for a\"\n",
+      "COP = (h3-h1)/(h4-h3); \t\t\t#Coefficient of performance\n",
+      "print \"Coefficient of performance is %.2f\"%(COP);\n",
+      "print \"Solution for b\"\n",
+      "print \"The work of compression is %.2f Btu/lbm\"%(h4-h3);\n",
+      "print \"Solution for c\"\n",
+      "print \"The refrigatering effect is %.2f Btu/lbm\"%(h3-h1);\n",
+      "print \"Solution for d\"\n",
+      "tons = 30; \t\t\t#capacity of 30 tons is desired\n",
+      "print \"The pounds per minute of ammonia required for ciculation is %.2f lbm/min\"%((200*tons)/(h3-h1));\n",
+      "print \"Solution for e\"\n",
+      "print \"The ideal horsepower per ton of refrigeration is %.2f hp/ton\"%((4.717*h4-h3)/(h3-h1));\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "Coefficient of performance is 3.71\n",
+        "Solution for b\n",
+        "The work of compression is 131.00 Btu/lbm\n",
+        "Solution for c\n",
+        "The refrigatering effect is 486.40 Btu/lbm\n",
+        "Solution for d\n",
+        "The pounds per minute of ammonia required for ciculation is 12.34 lbm/min\n",
+        "Solution for e\n",
+        "The ideal horsepower per ton of refrigeration is 5.87 hp/ton\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7  Page No : 510"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From Appendix 3,110 psig corresponds to 96 F, enthalpies are\n",
+      "h1 = 30.14; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h2 = 30.14; \t\t\t#Unit:Btu/lbm \t\t\t#Throttling gives h1 = h2 \t\t\t#enthalpy\n",
+      "h3 = 75.110; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "#From the consideration that s3 = s4,at -20F,\n",
+      "s3 = 0.17102; \t\t\t#Unit:Btu/(lbm*F) \t\t\t#s = entropy\n",
+      "#Therefore by interpolation in the Freon-12 superheat table at these values,\n",
+      "h4 = 89.293; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "print \"Solution for a\"\n",
+      "COP = (h3-h1)/(h4-h3); \t\t\t#Coefficient of performance\n",
+      "print \"Coefficient of performance is %.2f\"%(COP);\n",
+      "print \"Solution for b\"\n",
+      "print \"The work of compression is %.2f Btu/lbm\"%(h4-h3);\n",
+      "print \"Solution for c\"\n",
+      "print \"The refrigatering effect is %.2f Btu/lbm\"%(h3-h1);\n",
+      "print \"Solution for d\";\n",
+      "tons = 30; \t\t\t#capacity of 30 tons is desired\n",
+      "print \"The pounds per minute of ammonia required for ciculation is %.2f lbm/min\"%((200*tons)/h3-h1);\n",
+      "print \"Solution for e\"\n",
+      "print \"The ideal horsepower per ton of refrigeration is %.2f hp/ton\"%((4.717*h4-h3)/(h3-h1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "Coefficient of performance is 3.17\n",
+        "Solution for b\n",
+        "The work of compression is 14.18 Btu/lbm\n",
+        "Solution for c\n",
+        "The refrigatering effect is 44.97 Btu/lbm\n",
+        "Solution for d\n",
+        "The pounds per minute of ammonia required for ciculation is 49.74 lbm/min\n",
+        "Solution for e\n",
+        "The ideal horsepower per ton of refrigeration is 7.70 hp/ton\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.8  Page No : 517"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From Appendix 3,Using the Freon-12 tables, enthalpies are\n",
+      "h1 = 28.713; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h2 = 28.713; \t\t\t#Unit:Btu/lbm \t\t\t#Throttling gives h1 = h2 \t\t\t#enthalpy\n",
+      "h3 = 78.335; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "#From the consideration that s3 = s4,\n",
+      "s3 = 0.16798; \t\t\t#Unit:Btu/(lbm*F) \t\t\t#s = entropy\n",
+      "#Therefore by interpolation in the superheat tables at 90 F,\n",
+      "s = 0.16798; \t\t\t#entropy at 90F \t\t\t#Btu/lbm*F\n",
+      "h4 = 87.192; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "print \"The heat extracted is %.2f Btu/lbm\"%(h3-h1);\n",
+      "print \"The work required is %.2f Btu/lbm\"%(h4-h3);\n",
+      "COP = (h3-h1)/(h4-h3); \t\t\t#Coefficient of performance\n",
+      "print \"The Coefficient of performanceCOP) of this ideal cycle is %.2f\"%(COP);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat extracted is 49.62 Btu/lbm\n",
+        "The work required is 8.86 Btu/lbm\n",
+        "The Coefficient of performanceCOP) of this ideal cycle is 5.60\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.9  Page No : 518"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From Appendix 3,Using the HFC-134a tables, enthalpies are\n",
+      "h1 = 41.6; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h2 = 41.6; \t\t\t#Unit:Btu/lbm \t\t\t#Throttling gives h1 = h2 \t\t\t#enthalpy\n",
+      "h3 = 104.6; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "#From the consideration that s3 = s4,\n",
+      "s3 = 0.2244; \t\t\t#Unit:Btu/(lbm*F) \t\t\t#s = entropy\n",
+      "h4 = 116.0; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "print \"The heat extracted is %.2f Btu/lbm\"%(h3-h1);\n",
+      "print \"The work required is %.2f Btu/lbm\"%(h4-h3);\n",
+      "COP = (h3-h1)/(h4-h3); \t\t\t#Coefficient of performance\n",
+      "print \"The Coefficient of performanceCOP) of this ideal cycle is %.2f\"%(COP);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat extracted is 63.00 Btu/lbm\n",
+        "The work required is 11.40 Btu/lbm\n",
+        "The Coefficient of performanceCOP) of this ideal cycle is 5.53\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.10  Page No : 518"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print \"Solution for a\";\n",
+      "#By defination,the efficiency of the compressor is the ratio of the ideal compression work to actual compression work.\n",
+      "#Based on the points on fig.10.12, \t\t\t#n = (h4-h3)/(h4'-h3);\n",
+      "#There is close correspondence between 5.3 psia and -60F for saturated conditions.Therefore,state 3 is a superheated vapour at 5.3 psia and approximately -20F,because the problem states\n",
+      "#that state 3 has a 40F superheat.Interpolation in the Freon tables in Appendix 3 yields\n",
+      "T = -20; \t\t\t#Unit:F  \t\t\t#temperature\n",
+      "# p      h        s\n",
+      "#7.5  75.719  0.18371\n",
+      "#5.3  76.885  0.18985             h3 = 75.886 Btu/lbm\n",
+      "#5.0  75.990  0.19069\n",
+      "\n",
+      "#At 100 psia and s = 0.18985,\n",
+      "#  t           s      h\n",
+      "# 170F     0.18996  100.571\n",
+      "# 169.6F   0.18985  100.5          h4 = 100.5 Btu/lbm\n",
+      "# 160F     0.18726  98.884\n",
+      "\n",
+      "#The weight of refrigerant is given by\n",
+      "# 200(tons)/(h3-h1)  =  (200*5)/(75.886-h1)\n",
+      "#In the saturated tables,h1 is\n",
+      "#    p        h \n",
+      "# 101.86   26.832\n",
+      "# 100psia  26.542\n",
+      "# 98.87    26.365\n",
+      "\n",
+      "#m = mass flow/min\n",
+      "h1 = 26.542; \t\t\t#enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "n = 0.8; \t\t\t#Efficiency\n",
+      "h4 = 100.5; \t\t\t#enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "h3 = 75.886; \t\t\t#enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "m = (200*5)/(75.886-h1); \t\t\t#mass\n",
+      "h4dashminush3 = (h4-h3)/n; \n",
+      "#Total work of compression = m*(h4minush3)\n",
+      "J = 778; \t\t\t#J = Conversion factor\n",
+      "work = (h4dashminush3*m*J)/33000; \t\t\t#1 horsepower  =  33,000 ft*LBf/min \t\t\t#Unit:hp \t\t\t#work\n",
+      "print \"%.2f horsepower is required to drive the compressor if it has a mechanical efficiency 100percentage\"%(work);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "#Assuming a specific heat of the water as unity,we obtain\n",
+      "#From part (a),\n",
+      "#h4'-h3 = h4minush3\n",
+      "h4dash = h4dashminush3+h3; \t\t\t#Unit:Btu/lbm\n",
+      "mdot = (m*(h4dash-h1))/(70-60); \t\t\t#water enters at 60F and leaves at 70F \t\t\t#the required capacity in lbm/min\n",
+      "print \"%.2f lbm/min of cooling water i.e. %.2f gal/min is the required capacity of cooling water to pump\"%(mdot,mdot/8.3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "14.70 horsepower is required to drive the compressor if it has a mechanical efficiency 100percentage\n",
+        "Solution for b\n",
+        "162.35 lbm/min of cooling water i.e. 19.56 gal/min is the required capacity of cooling water to pump\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.11  Page No : 521"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print \"Solution for a\";\n",
+      "#From appendix3,reading the p-h diagram directly,we have\n",
+      "h3 = 76.2; \t\t\t#Unit:Btu/lbm \t\t\t#Enthalpy\n",
+      "h4 = 100.5; \t\t\t#Unit:Btu/lbm \t\t\t#Enthalpy\n",
+      "n = 0.8; \t\t\t#Efficiency \t\t\t#From 10.10\n",
+      "work = (h4-h3)/n; \t\t\t#Work of compression \t\t\t#Unit:Btu/lbm\n",
+      "#The enthalpy of saturated liquid at 100 psia is given at 26.1 Btu/lbm.Proceeding as before yields\n",
+      "m = (200*5)/(h3-26.1); \t\t\t#Unit:lbm/min \t\t\t#m = massflow/min\n",
+      "J = 778; \t\t\t#J = Conversion factor\n",
+      "totalwork = (m*work*J)/33000; \t\t\t#1 horsepower  =  33,000 ft*LBf/min \t\t\t#total ideal work \t\t\t#unit:hp\n",
+      "print \"Total ideal work of compression is %.2f hp\"%(totalwork);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "h4dash = h3+work; \t\t\t#Btu/lbm\n",
+      "mdot = (m*(h4dash-26.5))/(70-60); \t\t\t#water enters at 60F and leaves at 70F \t\t\t#the required capacity in lbm/min\n",
+      "print \"%.2f lbm/min of cooling water i.e. %.2f gal/min is the required capacity of cooling water to pump\"%(mdot,mdot/8.3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "Total ideal work of compression is 14.29 hp\n",
+        "Solution for b\n",
+        "159.83 lbm/min of cooling water i.e. 19.26 gal/min is the required capacity of cooling water to pump\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.12  Page No : 526"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "COP = 2.5; \t\t\t#Coefficient of performance\n",
+      "cp = 0.24; \t\t\t#Unit:Btu/(lbm*R) \t\t\t#Specific heat constant for constant pressure process\n",
+      "T1 = -100+460; \t\t\t#temperatures converted to absolute temperatures; \t\t\t#Unit:R \t\t\t#lowest temperature of the cycle\n",
+      "T3 = 150+460; \t\t\t#temperatures converted to absolute temperatures; \t\t\t#Unit:R \t\t\t#Upper temperature of the cycle\n",
+      "#T1/T4-T1  =  COP\n",
+      "T4 = (3.5*T1)/COP; \t\t\t#Unit:R \t\t\t#temperature at point 4\n",
+      "#T2/T3-T2  = COP\n",
+      "T2 = (COP*T3)/3.5; \t\t\t#Unit:R \t\t\t#temperature at point 2\n",
+      "print \"The work of the expander is %.2f Btu/lbm of air\"%(cp*T4-T1);\n",
+      "print \"The work of the compressor is %.2f Btu/lbm of air\"%(cp*T3-T2);\n",
+      "print \"The net work required by the cycle is %.2f Btu/lbm\"%(((cp*T3-T2))-cp*(T4-T1));\n",
+      "print \"Per ton of refrigeration, the required airflow is %.2f lbm/min per ton\"%((200/cp*T2-T1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The work of the expander is -239.04 Btu/lbm of air\n",
+        "The work of the compressor is -289.31 Btu/lbm of air\n",
+        "The net work required by the cycle is -323.87 Btu/lbm\n",
+        "Per ton of refrigeration, the required airflow is 362735.24 lbm/min per ton\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.13  Page No : 536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#A VACUUM REFRIGERATION SYSTEM \n",
+      "#A vacuum refrigeration system is used to cool water from 90F to 45F\n",
+      "h1 = 58.07; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h2 = 13.04; \t\t\t#Unit:Btu/lbm  \t\t\t#enthalpy\n",
+      "h3 = 1081.1; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "m1 = 1; \t\t\t#mass \t\t\t#unit:lbm\n",
+      "#m2 = 1-m3 \t\t\t#unit:lbm\n",
+      "#Now, m1*h1  =  m2*h2 + m3*h3\n",
+      "#Putting the values and arranging the equation,\n",
+      "m3 = (m1*h1-h2)/(h3+h2); \t\t\t#The mass of vapour that must be removed per pound \t\t\t#unit:lbm\n",
+      "\n",
+      "# results\n",
+      "print \"The mass of vapour that must be removed per pound of entering water is %.2f lbm\"%(m3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of vapour that must be removed per pound of entering water is 0.04 lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.14  Page No : 536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#In problem 10.13,\n",
+      "#A VACUUM REFRIGERATION SYSTEM \n",
+      "#A vacuum refrigeration system is used to cool water from 90F to 45F\n",
+      "h1 = 58.07; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h2 = 13.04; \t\t\t#Unit:Btu/lbm  \t\t\t#enthalpy\n",
+      "h3 = 1081.1; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "m1 = 1; \t\t\t#mass \t\t\t#lbm\n",
+      "#m2 = 1-m3 \t\t\t#unit:lbm\n",
+      "#Now, m1*h1  =  m2*h2 + m3*h3\n",
+      "#Putting the values and arranging the equation,\n",
+      "m3 = (m1*h1-h2)/(h3+h2); \t\t\t#The mass of vapour that must be removed per pound \t\t\t#unit:lbm\n",
+      "m2 = 1-m3; \t\t\t#mass \t\t\t#unit:lbm\n",
+      "print \"The mass of vapour that must be removed per pound of entering water is %.2f lbm\"%(m3);\n",
+      "#Now,in problem 10.14,\n",
+      "#The refrigeration effect can be determined as m3*(h3-h1) or m2*(h1-h2)\n",
+      "print \"The refrigeration effect Using eqn m3*h3-h1) is %.2f Btu/lbm\"%(m3*h3-h1)\n",
+      "print \"The refrigeration effect Using eqn m2*h1-h2) is %.2f Btu/lbm\"%(m2*h1-h2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of vapour that must be removed per pound of entering water is 0.04 lbm\n",
+        "The refrigeration effect Using eqn m3*h3-h1) is -13.58 Btu/lbm\n",
+        "The refrigeration effect Using eqn m2*h1-h2) is 42.64 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.15  Page No : 539"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#THE HEAT PUMP\n",
+      "T1 = 70.+460; \t\t\t#70F = 70+460 R \t\t\t#Energy flows into the system at reservoir at constant temperature T1(unit:R) \t\t\t#from problem 10.1\n",
+      "T2 = 32+460; \t\t\t#32F = 32+460 R \t\t\t#Heat is rejected to the constant temperature T2(Unit:R) \t\t\t#from problem 10.1\n",
+      "COP = T1/(T1-T2); \t\t\t#Coefficient of performance for carnot heat pump\n",
+      "print \"Coefficient of performanceCOP) of the carnot cycle is %.2f\"%(COP);\n",
+      "print \"The COP can also be obtained from the energy items solved for in problem 10.1\"\n",
+      "#In problem 10.1, The power was found to be 77.2 Btu/min and the total tare of heat rejection was 1077.2 Btu/min\n",
+      "#Therefore,\n",
+      "print \"Coefficient of performanceCOP) of the cycle is %.2f\"%(1077.2/77.2); \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coefficient of performanceCOP) of the carnot cycle is 13.95\n",
+        "The COP can also be obtained from the energy items solved for in problem 10.1\n",
+        "Coefficient of performanceCOP) of the cycle is 13.95\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.16  Page No : 539"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Let us first consider the cycle as a refrigeration cycle\n",
+      "#In problem 10.1\n",
+      "T1 = 70+460; \t\t\t#70F = 70+460 R \t\t\t#Energy flows into the system at reservoir at constant temperature T1(unit:R)\n",
+      "T2 = 0+460; \t\t\t#0F = 32+460 R \t\t\t#Heat is rejected to the constant temperature T2(Unit:R)\n",
+      "COP = T2/(T1-T2); \t\t\t#Coefficient of performance\n",
+      "print \"Coefficient of performanceCOP) of the cycle is %.2f\"%(COP);\n",
+      "Qremoved = 1000; \t\t\t#Unit:Btu/min \t\t\t#heat removal\n",
+      "WbyJ = Qremoved/COP; \t\t\t#the power input \t\t\t#unit:Btu/min\n",
+      "print \"The power input is %.2f Btu/min\"%(WbyJ);\n",
+      "Qrej = Qremoved+WbyJ; \t\t\t#The rate of heat rejected to the room \t\t\t#Unit:Btu/min\n",
+      "print \"The rate of heat rejected to the room is %.2f Btu/min\"%(Qrej);\n",
+      "print \"The COP as a heat pump is %.2f\"%(Qrej/WbyJ);\n",
+      "print \"As a check, COP of heat pump is %.2f  =  1 + COP of carnot cycle %.2f\"%(Qrej/WbyJ,COP);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coefficient of performanceCOP) of the cycle is 6.00\n",
+        "The power input is 166.00 Btu/min\n",
+        "The rate of heat rejected to the room is 1166.00 Btu/min\n",
+        "The COP as a heat pump is 7.00\n",
+        "As a check, COP of heat pump is 7.00  =  1 + COP of carnot cycle 6.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch11_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch11_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch11_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch11_2.ipynb
new file mode 100755
index 00000000..d0a50659
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch11_2.ipynb
@@ -0,0 +1,1390 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bca14a59244f78cacc86ebeeb6d600dde8a0df0a85b519507569f4ae617ed968"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Heat Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1  Page No : 553"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# given data\n",
+      "deltaX = 6./12; \t\t\t#6 inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:feet\n",
+      "k = 0.40; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "T1 = 150; \t\t\t#temperature maintained at one face \t\t\t#fahrenheit\n",
+      "T2 = 80.; \t\t\t#tempetature maintained at other face \t\t\t#fahrenheit\n",
+      "\n",
+      "# calculations\n",
+      "deltaT = T2-T1; \t\t\t#fahrenheit \t\t\t#Change in temperature\n",
+      "Q = (-k*deltaT)/deltaX; \t\t\t#Heat transfer per square foot of wall \t\t\t#Unit:Btu/hr*ft**2\n",
+      "\n",
+      "# results\n",
+      "print \"Heat transfer per square foot of wall is %.2f Btu/hr*ft**2\"%(Q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer per square foot of wall is 56.00 Btu/hr*ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2  Page No : 553"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "deltaX = 0.150; \t\t\t#Given,150 mm  = 0.150 meter \t\t\t# \t\t\t#deltaX = length \t\t\t#Unit:meter\n",
+      "k = 0.692; \t\t\t#Unit:W/(m*celcius) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \n",
+      "T1 = 70; \t\t\t#temperature maintained at one face \t\t\t#celcius\n",
+      "T2 = 30; \t\t\t#tempetature maintained at other face \t\t\t#celcius\n",
+      "\n",
+      "# calculations\n",
+      "deltaT = T2-T1; \t\t\t#celcius \t\t\t#change in temperature\n",
+      "Q = (-k*deltaT)/deltaX; \t\t\t#Heat transfer per square foot of wall \t\t\t#unit:W/m**2\n",
+      "\n",
+      "# results\n",
+      "print \"Heat transfer per square foot of wall is %.2f W/m**2\"%(Q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer per square foot of wall is 184.53 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3  Page No : 556"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From example 11.1,\n",
+      "deltaX = 6./12; \t\t\t#6 inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:feet \n",
+      "A = 1; \t\t\t#area  \t\t\t#ft**2\n",
+      "k = 0.40; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "\n",
+      "Rt = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "\n",
+      "#Q = deltaT/Rt \t\t\t#Q = heat transfer \t\t\t#ohm's law (fourier's equation)\n",
+      "#i = deltaE/Re \t\t\t#i = current in amperes \t\t\t#deltaE = The potential difference \t\t\t#Re = the electrical resistance \t\t\t#ohm's law\n",
+      "# Q/i  =  (deltaT/Rt)*(deltaE/Re)\n",
+      "#Q/i = 100; \t\t\t#Given \t\t\t# 1 A correspond to 100 Btu/(hr*ft**2)\n",
+      "deltaE = 9.; \t\t\t#Unit:Volt \t\t\t#potential difference\n",
+      "T1 = 150.; \t\t\t#temperature maintained at one face \t\t\t#fahrenheit\n",
+      "T2 = 80.; \t\t\t#tempetature maintained at other face \t\t\t#fahrenheit\n",
+      "deltaT = T2-T1; \t\t\t#fahrenheit \t\t\t#Change in temperature\n",
+      "Re = (100*deltaE*Rt)/deltaT; \t\t\t#Unit:Ohms \t\t\t#The electrical resistance needed\n",
+      "print \"The electrical resistance needed is %.2f ohms\"%(abs(Re));\n",
+      "i = deltaE/Re; \t\t\t#current \t\t\t#Unit:amperes\n",
+      "Q = 100*i; \t\t\t#Heat transfer per square foot of wall \t\t\t#Unit:Btu/hr*ft**2\n",
+      "print \"Heat transfer per square foot of wall is %.2f Btu/hr*ft**2\"%(abs(Q));\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electrical resistance needed is 16.07 ohms\n",
+        "Heat transfer per square foot of wall is 56.00 Btu/hr*ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4  Page No : 558"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#For Brick,\n",
+      "deltaX = 6./12; \t\t\t#6 inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 1.; \t\t\t#area  \t\t\t#unit:ft**2\n",
+      "k = 0.40; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For brick\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(R);\n",
+      "R1 = R;\n",
+      "\n",
+      "#For Concrete,\n",
+      "deltaX = (1./2)/12; \t\t\t#(1/2) inch  =  (1/2)/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 1; \t\t\t#area \t\t\t#ft**2\n",
+      "k = 0.80; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For Concrete\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(R);\n",
+      "R2 = R;\n",
+      "\n",
+      "#For plaster,\n",
+      "deltaX = (1./2)/12; \t\t\t# (1/2) inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 1; \t\t\t#area \t\t\t#ft**2\n",
+      "k = 0.30; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For plaster\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(R);\n",
+      "R3 = R;\n",
+      "\n",
+      "Rot = R1+R2+R3; \t\t\t#Rot = The overall resistance \t\t\t#unit:(hr*F)/Btu\n",
+      "print \"The overall resistance is %.2f hr*F)/Btu\"%(Rot);\n",
+      "T1 = 70.; \t\t\t#temperature maintained at one face \t\t\t#fahrenheit \n",
+      "T2 = 30; \t\t\t#tempetature maintained at other face \t\t\t#fahrenheit \n",
+      "deltaT = T2-T1; \t\t\t#fahrenheit \t\t\t#Change in temperature\n",
+      "Q = deltaT/Rot; \t\t\t#Q = Heat transfer \t\t\t#Unit:Btu/(hr*ft**2); \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"Heat transfer per square foot of wall is %.2f Btu/hr*ft**2\"%(abs(Q));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For brick\n",
+        "The resistance is 1.25 hr*F)/Btu\n",
+        "For Concrete\n",
+        "The resistance is 0.05 hr*F)/Btu\n",
+        "For plaster\n",
+        "The resistance is 0.14 hr*F)/Btu\n",
+        "The overall resistance is 1.44 hr*F)/Btu\n",
+        "Heat transfer per square foot of wall is 27.76 Btu/hr*ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5  Page No : 558"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print \"In problem 11.4\"\n",
+      "#From example 11.4,,,\n",
+      "#For Brick,\n",
+      "deltaX = 6./12; \t\t\t#6 inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 1; \t\t\t#area  \t\t\t#unit:ft**2\n",
+      "k = 0.40; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For brick\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(R);\n",
+      "R1 = R;\n",
+      "\n",
+      "#For Concrete,\n",
+      "deltaX = (1./2)/12; \t\t\t#(1/2) inch  =  (1/2)/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 1; \t\t\t#area \t\t\t#ft**2\n",
+      "k = 0.80; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For Concrete\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(R);\n",
+      "R2 = R;\n",
+      "\n",
+      "#For plaster,\n",
+      "deltaX = (1./2)/12; \t\t\t# (1/2) inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 1; \t\t\t#area \t\t\t#ft**2\n",
+      "k = 0.30; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For plaster\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(R);\n",
+      "R3 = R;\n",
+      "\n",
+      "Rot = R1+R2+R3; \t\t\t#Rot = The overall resistance \t\t\t#unit:(hr*F)/Btu\n",
+      "print \"The overall resistance is %.2f hr*F)/Btu\"%(Rot);\n",
+      "T1 = 70; \t\t\t#temperature maintained at one face \t\t\t#fahrenheit \n",
+      "T2 = 30; \t\t\t#tempetature maintained at other face \t\t\t#fahrenheit \n",
+      "deltaT = T2-T1; \t\t\t#fahrenheit \t\t\t#Change in temperature\n",
+      "Q = deltaT/Rot; \t\t\t#Q = Heat transfer \t\t\t#Unit:Btu/(hr*ft**2);\n",
+      "print \"Heat transfer per square foot of wall is %.2f Btu/hr*ft**2\"%(abs(Q));\n",
+      "\n",
+      "print \"Now in problem 11.5\"\n",
+      "deltaT = R*Q \t\t\t#ohm's law (fourier's equation) \t\t\t#Change in temperature \t\t\t#fahrenheit\n",
+      "#For Brick,\n",
+      "deltaT = Q*R1; \t\t\t#Unit:fahrenheit \t\t\t#ohm's law (fourier's equation) \t\t\t#Change in temperature\n",
+      "t1 = deltaT;\n",
+      "#For Concrete,\n",
+      "deltaT = Q*R2; \t\t\t#Unit:fahrenheit \t\t\t#ohm's law (fourier's equation) \t\t\t#Change in temperature\n",
+      "t2 = deltaT;\n",
+      "#For plaster,\n",
+      "deltaT = Q*R3; \t\t\t#Unit:fahrenheit \t\t\t#ohm's law (fourier's equation) \t\t\t#Change in temperature\n",
+      "t3 = deltaT;\n",
+      "\n",
+      "deltaTo = t1+t2+t3; \t\t\t#Overall Change in temperature \t\t\t#fahrenheit\n",
+      "print \"The overall change in temperature is %.2f F\"%(abs(deltaTo));\n",
+      "#The interface temperature are:\n",
+      "print \"The interface temperature are:\";\n",
+      "print \"For brick-concrete : %.2f fahrenheit\"%(abs(T2)+abs(t1));\n",
+      "print \"For concrete-plaster : %.2f fahrenheit\"%(abs(T2)+abs(t1)+abs(t2));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In problem 11.4\n",
+        "For brick\n",
+        "The resistance is 1.25 hr*F)/Btu\n",
+        "For Concrete\n",
+        "The resistance is 0.05 hr*F)/Btu\n",
+        "For plaster\n",
+        "The resistance is 0.14 hr*F)/Btu\n",
+        "The overall resistance is 1.44 hr*F)/Btu\n",
+        "Heat transfer per square foot of wall is 27.76 Btu/hr*ft**2\n",
+        "Now in problem 11.5\n",
+        "The overall change in temperature is 40.00 F\n",
+        "The interface temperature are:\n",
+        "For brick-concrete : 64.70 fahrenheit\n",
+        "For concrete-plaster : 66.14 fahrenheit\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6  Page No : 559"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#For Brick,\n",
+      "deltaX = 0.150; \t\t\t#Unit:m \t\t\t#150 mm  =  0.150 m \t\t\t#deltaX = length \t\t\t#unit:meter\n",
+      "A = 1; \t\t\t#area  \t\t\t#unit:m**2\n",
+      "k = 0.692; \t\t\t#Unit:W/(m*C) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:C/W\n",
+      "print \"For brick\"\n",
+      "print \"The resistance is %.2f Celcius/W\"%(R);\n",
+      "R1 = R;\n",
+      "\n",
+      "#For Concrete,\n",
+      "deltaX = 0.012; \t\t\t#Unit:m \t\t\t#12 mm  =  0.0120 m \t\t\t#deltaX = length \t\t\t#unit:meter\n",
+      "A = 1; \t\t\t#area  \t\t\t#unit:m**2\n",
+      "k = 1.385; \t\t\t#Unit:W/(m*C) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:C/W\n",
+      "print \"For Concrete\"\n",
+      "print \"The resistance is %.2f Celcius/W\"%(R);\n",
+      "R2 = R;\n",
+      "\n",
+      "#For plaster,\n",
+      "deltaX = 0.0120; \t\t\t#Unit:m \t\t\t#12 mm  =  0.0120 m \t\t\t#deltaX = length \t\t\t#unit:meter\n",
+      "A = 1; \t\t\t#area  \t\t\t#unit:m**2\n",
+      "k = 0.519; \t\t\t#Unit:W/(m*C) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:C/W\n",
+      "print \"For plaster\"\n",
+      "print \"The resistance is %.2f Celcius/W\"%(R);\n",
+      "R3 = R;\n",
+      "\n",
+      "Ro = R1+R2+R3; \t\t\t#Rot = The overall resistance \t\t\t#unit:C/W\n",
+      "print \"The overall resistance is %.2f Celcius/W\"%(Ro);\n",
+      "T1 = 0; \t\t\t#temperature maintained at one face \t\t\t#Celcius\n",
+      "T2 = 20; \t\t\t#tempetature maintained at other face \t\t\t#Celcius\n",
+      "deltaT = T2-T1; \t\t\t#Change in temperature  \t\t\t#Celcius\n",
+      "Q = deltaT/Ro; \t\t\t#Q = Heat transfer \t\t\t#Unit:W/m**2; \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"Heat transfer per square meter of wall is %.2f W/m**2\"%(abs(Q));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For brick\n",
+        "The resistance is 0.22 Celcius/W\n",
+        "For Concrete\n",
+        "The resistance is 0.01 Celcius/W\n",
+        "For plaster\n",
+        "The resistance is 0.02 Celcius/W\n",
+        "The overall resistance is 0.25 Celcius/W\n",
+        "Heat transfer per square meter of wall is 80.47 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7  Page No : 560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print \"In problem 11.6\"\n",
+      "#For Brick,\n",
+      "deltaX = 0.150; \t\t\t#Unit:m \t\t\t#150 mm  =  0.150 m \t\t\t#deltaX = length \t\t\t#unit:meter\n",
+      "A = 1; \t\t\t#area \t\t\t#unit:meter**2\n",
+      "k = 0.692; \t\t\t#Unit:W/(m*C) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:Celcius/W\n",
+      "print \"For brick\"\n",
+      "print \"The resistance is %.2f Celcius/W\"%(R);\n",
+      "R1 = R;\n",
+      "\n",
+      "#For Concrete,\n",
+      "deltaX = 0.012; \t\t\t#Unit:m \t\t\t#12 mm  =  0.0120 m \t\t\t#deltaX = length \t\t\t#unit:meter\n",
+      "A = 1; \t\t\t#area  \t\t\t#unit:meter**2\n",
+      "k = 1.385; \t\t\t#Unit:W/(m*C) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:Celcius/W\n",
+      "print \"For Concrete\"\n",
+      "print \"The resistance is %.2f Celcius/W\"%(R);\n",
+      "R2 = R;\n",
+      "\n",
+      "#For plaster,\n",
+      "deltaX = 0.0120; \t\t\t#Unit:m \t\t\t#12 mm  =  0.0120 m \t\t\t#deltaX = length \t\t\t#unit:meter\n",
+      "A = 1; \t\t\t#area  \t\t\t#unit:meter**2\n",
+      "k = 0.519; \t\t\t#Unit:W/(m*C) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "R = deltaX/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:Celcius/W\n",
+      "print \"For plaster\"\n",
+      "print \"The resistance is %.2f Celcius/W\"%(R);\n",
+      "R3 = R;\n",
+      "\n",
+      "Ro = R1+R2+R3; \t\t\t#Rot = The overall resistance Celcius/W\n",
+      "print \"The overall resistance is %.2f Celcius/W\"%(Ro);\n",
+      "T1 = 0; \t\t\t#temperature maintained at one face \t\t\t#Celcius\n",
+      "T2 = 20; \t\t\t#tempetature maintained at other face \t\t\t#Celcius\n",
+      "deltaT = T2-T1; \t\t\t#Change in temperature \t\t\t#Celcius\n",
+      "Q = deltaT/Ro; \t\t\t#Q = Heat transfer \t\t\t#Unit:W/m**2;\n",
+      "print \"Heat transfer per square meter of wall is %.2f W/m**2\"%(abs(Q));\n",
+      "\n",
+      "print \"Now in problem 11.5\"\n",
+      "#deltaT = R*Q \t\t\t#ohm's law (fourier's equation)\n",
+      "#For Brick,\n",
+      "deltaT = Q*R1; \t\t\t#Unit:Celcius \t\t\t#Change in temperature\n",
+      "t1 = deltaT;\n",
+      "#For Concrete,\n",
+      "deltaT = Q*R2; \t\t\t#Unit:Celcius \t\t\t#Change in temperature\n",
+      "t2 = deltaT;\n",
+      "#For plaster,\n",
+      "deltaT = Q*R3; \t\t\t#Unit:Celcius \t\t\t#Change in temperature\n",
+      "t3 = deltaT;\n",
+      "\n",
+      "deltaTo = t1+t2+t3; \t\t\t#The overall Change in temperature \t\t\t#Celcius\n",
+      "print \"The overall change in temperature is %.2f celcius\"%(abs(deltaTo));\n",
+      "#The interface temperature are:\n",
+      "print \"The interface temperature are:\";\n",
+      "print \"%.2f Celcius\"%(abs(deltaTo)-abs(t1));\n",
+      "print \"%.2f Celcius\"%(abs(deltaTo)-abs(t1)-abs(t2));\n",
+      "print \"%.2f Celcius\"%(abs(deltaTo)-abs(t1)-abs(t2)-abs(t3));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In problem 11.6\n",
+        "For brick\n",
+        "The resistance is 0.22 Celcius/W\n",
+        "For Concrete\n",
+        "The resistance is 0.01 Celcius/W\n",
+        "For plaster\n",
+        "The resistance is 0.02 Celcius/W\n",
+        "The overall resistance is 0.25 Celcius/W\n",
+        "Heat transfer per square meter of wall is 80.47 W/m**2\n",
+        "Now in problem 11.5\n",
+        "The overall change in temperature is 20.00 celcius\n",
+        "The interface temperature are:\n",
+        "2.56 Celcius\n",
+        "1.86 Celcius\n",
+        "-0.00 Celcius\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8  Page No : 561"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "\n",
+      "deltaX = 4./12; \t\t\t#4 inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 7*2.; \t\t\t#area \t\t\t#area = hight*width \t\t\t#unit:ft**2\n",
+      "k = 0.090; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity for fir \t\t\t#From the table\n",
+      "Rfir = deltaX/(k*A); \t\t\t#Resistance of fir \t\t\t#Unit:(hr*F)/Btu\n",
+      "print \"For fir\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(Rfir);\n",
+      "\n",
+      "deltaX = 4./12; \t\t\t#4 inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 7*2; \t\t\t#area \t\t\t#area = hight*width  \t\t\t#unit:ft**2\n",
+      "k = 0.065; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity for pine \t\t\t#From the table\n",
+      "Rpine = deltaX/(k*A); \t\t\t#Resistance of pine \t\t\t#Unit:(hr*F)/Btu\n",
+      "print \"For pine\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(Rpine);\n",
+      "\n",
+      "deltaX = 4./12; \t\t\t#4 inch  =  6/12 feet \t\t\t#deltaX = length \t\t\t#unit:ft\n",
+      "A = 7*2; \t\t\t#area \t\t\t#area = hight*width \t\t\t#unit:ft**2\n",
+      "k = 0.025; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity for corkboard \t\t\t#From the table\n",
+      "Rcorkboard = deltaX/(k*A); \t\t\t#Resistance of corkboard \t\t\t#Unit:(hr*F)/Btu\n",
+      "print \"For corkboard\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(Rcorkboard);\n",
+      "\n",
+      "Roverall = inv([[inv([[Rfir]])+inv([[Rpine]])+inv([[Rcorkboard]])]]);\n",
+      "print \"The overall resistance is %.2f hr*F)/Btu\"%(Roverall);\n",
+      "\n",
+      "T1 = 60.; \t\t\t#temperature maintained at one face \t\t\t#unit:fahrenheit\n",
+      "T2 = 80.; \t\t\t#tempetature maintained at other face \t\t\t#unit:fahrenheit\n",
+      "deltaT = T2-T1; \t\t\t#Change in temperature \t\t\t#unit:fahrenheit\n",
+      "Qtotal = deltaT/Roverall; \t\t\t#Q = Total Heat loss \t\t\t#Unit:Btu/hr; \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"Total Heat loss from the wall is %.2f Btu/hr\"%(abs(Qtotal));\n",
+      "\n",
+      "#As a check,\n",
+      "Qfir = deltaT/Rfir; \t\t\t#Q = Fir Heat loss \t\t\t#Unit:Btu/hr; \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"Heat loss from the wall made of fir is %.2f Btu/hr\"%(abs(Qfir));\n",
+      "Qpine = deltaT/Rpine; \t\t\t#Q = Pine Heat loss \t\t\t#Unit:Btu/hr; \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"Heat loss from the wall made of pine is %.2f Btu/hr\"%(abs(Qpine));\n",
+      "Qcorkboard = deltaT/Rcorkboard; \t\t\t#Q = corkboard Heat loss \t\t\t#Unit:Btu/hr; \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"Heat loss from the wall made of corkboard is %.2f Btu/hr\"%(abs(Qcorkboard));\n",
+      "Qtotal = Qfir+Qpine+Qcorkboard; \t\t\t#Total Heat loss from the wall \t\t\t#unit:Btu/hr\n",
+      "print \"Total Heat loss from the wall is %.2f Btu/hr\"%(abs(Qtotal));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For fir\n",
+        "The resistance is 0.26 hr*F)/Btu\n",
+        "For pine\n",
+        "The resistance is 0.37 hr*F)/Btu\n",
+        "For corkboard\n",
+        "The resistance is 0.95 hr*F)/Btu\n",
+        "The overall resistance is 0.13 hr*F)/Btu\n",
+        "Total Heat loss from the wall is 151.20 Btu/hr\n",
+        "Heat loss from the wall made of fir is 75.60 Btu/hr\n",
+        "Heat loss from the wall made of pine is 54.60 Btu/hr\n",
+        "Heat loss from the wall made of corkboard is 21.00 Btu/hr\n",
+        "Total Heat loss from the wall is 151.20 Btu/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9  Page No : 565"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#A bare steel pipe\n",
+      "ro = 3.50; \t\t\t#Outside diameter \t\t\t#Unit:in.\n",
+      "ri = 3.00; \t\t\t#inside diameter \t\t\t#Unit:in.\n",
+      "Ti = 240; \t\t\t#Inside temperature \t\t\t#unit:fahrenheit\n",
+      "To = 120; \t\t\t#Outside temperature \t\t\t#unit:fahrenheit\n",
+      "L = 5; \t\t\t#Length \t\t\t#Unit:ft\n",
+      "deltaT = Ti-To; \t\t\t#Change in temperature \t\t\t#unit:fahrenheit\n",
+      "k = 26 \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity\n",
+      "\n",
+      "# Calculations\n",
+      "Q = (2*math.pi*k*L*deltaT)/math.log(ro/ri); \t\t\t#The heat loss from the pipe \t\t\t#unit:Btu/hr\n",
+      "\n",
+      "# Results\n",
+      "print \"The heat loss from the pipe is %.2f Btu/hr\"%(Q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss from the pipe is 635856.36 Btu/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.10  Page No : 566"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#A bare steel pipe\n",
+      "ro = 90.; \t\t\t#Outside diameter \t\t\t#Unit:mm\n",
+      "ri = 75; \t\t\t#inside diameter \t\t\t#Unit:mm\n",
+      "Ti = 110; \t\t\t#Inside temperature \t\t\t#Unit:Celcius\n",
+      "To = 40; \t\t\t#Outside temperature \t\t\t#Unit:Celcius\n",
+      "L = 2; \t\t\t#Length \t\t\t#Unit:m\n",
+      "\n",
+      "# Calculations\n",
+      "deltaT = Ti-To; \t\t\t#Change in temperature \t\t\t#Unit:Celcius\n",
+      "k = 45 \t\t\t#Unit:W/(m*C) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity\n",
+      "Q = (2*math.pi*k*L*deltaT)/math.log(ro/ri); \t\t\t#The heat loss from the pipe  \t\t\t#unit:W\n",
+      "\n",
+      "# Results\n",
+      "print \"The heat loss from the pipe is %.2f W\"%(Q);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss from the pipe is 217111.28 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.11  Page No : 567"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy.linalg import inv\n",
+      "\n",
+      "#From problem 11.9,\n",
+      "#A bare steel pipe\n",
+      "r2 = 3.50; \t\t\t#Outside diameter \t\t\t#Unit:in.\n",
+      "r1 = 3.00; \t\t\t#inside diameter \t\t\t#Unit:in.\n",
+      "Ti = 240; \t\t\t#Inside temperature \t\t\t#unit:fahrenheit\n",
+      "L = 5; \t\t\t#Length \t\t\t#Unit:ft\n",
+      "k1 = [[26]]; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity\n",
+      "ans1 = (inv(k1)*math.log(r2/r1));\n",
+      "\n",
+      "#Now,in problem 11.11,\n",
+      "#Mineral wool\n",
+      "r3 = 5.50; \t\t\t#inside diameter \t\t\t#Unit:in.\n",
+      "r2 = 3.50; \t\t\t#outside diameter \t\t\t#Unit:in.\n",
+      "To = 85; \t\t\t#Outside temperature \t\t\t#unit:fahrenheit\n",
+      "deltaT = Ti-To; \t\t\t#Change in temperature \t\t\t#unit:fahrenheit\n",
+      "k2 = [[0.026]] \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity\n",
+      "ans2 = (inv(k2)*math.log(r3/r2));\n",
+      "\n",
+      "Q = (2*math.pi*L*deltaT)/(ans1+ans2); \t\t\t#The heat loss from the pipe  \t\t\t#unit:Btu/hr\n",
+      "\n",
+      "# Results\n",
+      "print \"The heat loss from the pipe is %.2f Btu/hr\"%(Q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss from the pipe is 280.02 Btu/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.12  Page No : 569"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy.linalg import inv\n",
+      "\n",
+      "#From problem 11.9,\n",
+      "#The bare pipe\n",
+      "r2 = 3.50; \t\t\t#Outside diameter \t\t\t#Unit:in.\n",
+      "r1 = 3.00; \t\t\t#inside diameter \t\t\t#Unit:in.\n",
+      "Ti = 240.; \t\t\t#Inside temperature \t\t\t#unit:fahrenheit\n",
+      "L = 5.; \t\t\t#Length \t\t\t#Unit:ft\n",
+      "k = 26.; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity\n",
+      "Rpipe = math.log(r2/r1)/(2*math.pi*k*L); \t\t\t#the resistance of pipe \t\t\t#Unit:(hr*F)/Btu\n",
+      "print \"The resistance of pipe is %.5f hr*F)/Btu\"%(Rpipe);\n",
+      "\n",
+      "#Now,in problem 11.12,\n",
+      "To = 70; \t\t\t#Outside temperature \t\t\t#unit:fahrenheit\n",
+      "deltaT = Ti-To; \t\t\t#Change in temperature \t\t\t#unit:fahrenheit\n",
+      "h = 0.9; \t\t\t#Coefficient of heat transfer \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "A = (math.pi*r2)/12*L;  \t\t\t#Area \t\t\t#Unit:ft**2 \t\t\t#1 inch  =  1/12 feet \t\t\t#unit:ft**2\n",
+      "Rconvection = inv([[h*A]]); \t\t\t#The resistance due to natural convection to the surrounding air \t\t\t#Unit:(hr*F)/Btu\n",
+      "print \"The resistance due to natural convection to the surrounding air is %.2f hr*F)/Btu\"%(Rconvection);\n",
+      "\n",
+      "Rtotal = Rpipe+Rconvection; \t\t\t#The total resistance   \t\t\t#unit:(hr*F)/Btu\n",
+      "print \"The total resistance is %.2f hr*F)/Btu\"%(Rtotal);\n",
+      "Q = deltaT/Rtotal; \t\t\t#ohm's law (fourier's equation) \t\t\t#The heat transfer from the pipe to the surrounding air   \t\t\t#unit:Btu/hr\n",
+      "print \"The heat transfer from the pipe to the surrounding air is %.2f Btu/hr\"%(Q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance of pipe is 0.00019 hr*F)/Btu\n",
+        "The resistance due to natural convection to the surrounding air is 0.24 hr*F)/Btu\n",
+        "The total resistance is 0.24 hr*F)/Btu\n",
+        "The heat transfer from the pipe to the surrounding air is 700.42 Btu/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.15  Page No : 574"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# given data\n",
+      "D = 3.5/12; \t\t\t#3.5 inch  =  3.5/12 feet\t\t\t#Unit:ft \t\t\t#Outside diameter\n",
+      "Ti = 120; \t\t\t#Inside temperature \t\t\t#unit:fahrenheit\n",
+      "To = 70; \t\t\t#Outside temperature \t\t\t#unit:fahrenheit\n",
+      "deltaT = Ti-To; \t\t\t#unit:fahrenheit \t\t\t#Change in temperature\n",
+      "h = 0.9; \t\t\t#Coefficient of heat transfer \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "L = 5; \t\t\t#Length \t\t\t#Unit:ft \t\t\t#From problem 11.10\n",
+      "A = (math.pi*D)*L;  \t\t\t#Area \t\t\t#Unit:ft**2 \n",
+      "Q = h*A*deltaT; \t\t\t#The heat loss due to convection \t\t\t#Unit:Btu/hr \t\t\t#Newton's law of cooling\n",
+      "\n",
+      "# Results\n",
+      "print \"The heat loss due to convection is %.2f Btu/hr\"%(Q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss due to convection is 206.17 Btu/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.16  Page No : 575"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "\n",
+      "#This problem can not be solved directly,because the individual film resistances aree functions of unknown temperature differences.Therefore,\n",
+      "#From the first approximation,\n",
+      "h = 1./2; \t\t\t#Coefficient of heat transfer \t\t\t#unit:Btu/(hr*ft**2*F)\n",
+      "#For area 1 ft**2,\n",
+      "R = (3./12)/0.07; \t\t\t#The wall resistance is deltax/(k*A) \t\t\t#k = 0.07 \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity\n",
+      "Roverall = inv([[1./2]])+inv([[1./2]])+R; \t\t\t#the overall series resistance \t\t\t#Unit:Btu/(hr*ft*F)\n",
+      "print \"For h = 0.5, the overall series resistance is %.2f Btu/hr*ft*F)\"%(Roverall);\n",
+      "#Using the value of Roverall,we can now obtain Q and individual temperature differences,\n",
+      "Ti = 80.; \t\t\t#warm air temperature \t\t\t#unit:fahrenheit\n",
+      "To = 50; \t\t\t#cold air temperature \t\t\t#unit:fahrenheit\n",
+      "deltaT = Ti-To; \t\t\t#unit:fahrenheit  \t\t\t#Change in temperature\n",
+      "Q = deltaT/Roverall; \t\t\t#Unit:Btu/(hr*ft**2) \t\t\t#heat transfer \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"For h = 0.5, heat transfer is %.2f Btu/hr*ft**2)\"%(Q);\n",
+      "print \"For h = 0.5\"\n",
+      "#deltaT through the hot air film is Q/(1/2)\n",
+      "print \"Temperaure difference through the hot air film is %.2f F\"%(Q/1./2);\n",
+      "#Throught the wall deltaT is R*Q\n",
+      "print \"Temperaure difference through the wall is %.2f F\"%(Q*R);\n",
+      "#deltaT through the cold air film is Q/(1/2)\n",
+      "print \"Temperaure difference through the cold air film is %.2f F\"%(Q/1./2);\n",
+      "\n",
+      "#With these temperature differences,we can now enter figures 11.12 and 11.14 to verify our approximation.From figure 11.14,we find h = 0.42 Btu/(hr*ft*2*F)\n",
+      "#Using h = 0.42,we have for the overall resistance (1/0.42)+(1/0.42)+R\n",
+      "h = [[0.42]]; \t\t\t#Coefficient of heat transfer  \t\t\t#unit:Btu/(hr*ft**2*F)\n",
+      "Roverall = inv(h)+inv(h)+R; \t\t\t#the overall series resistance \t\t\t#Unit:Btu/(hr*ft*F)\n",
+      "print \"For h = 0.42, the overall series resistance is %.2f Btu/hr*ft*F)\"%(Roverall);\n",
+      "Q = deltaT/Roverall; \t\t\t#Unit:Btu/(hr*ft**2) \t\t\t#heat transfer \t\t\t#ohm's law (fourier's equation)\n",
+      "print \"For h = 0.42, heat transfer is %.2f Btu/hr*ft**2)\"%(Q);\n",
+      "print \"For h = 0.42\"\n",
+      "# deltat through both air films is Q/h\n",
+      "print \"Temperaure difference through the hot and cold air film is %.2f F\"%(Q/h);\n",
+      "#and through the wall,deltat is Q*R\n",
+      "print \"Temperaure difference through the wall is %.2f F\"%(Q*R);\n",
+      "\n",
+      "#Entering figure 11.14,we find that h stays essentially 0.42,and our solution is that the heat flow is Q,the \"hot\" side of the wall is at Ti-(Q/h),the \"cold\" side  is at To+(Q/h) ,and temperature drop in the wall is Ti-(Q/h)-(To+(Q/h)).\n",
+      "print \"The temperature drop on the hot side of the wall is %.2f F\"%(Ti-Q/h)\n",
+      "print \"The temperature drop on the cold side of the wall is %.2f F\"%(To+Q/h)\n",
+      "print \"The temperature drop in the wall is %.2f F\"%(((Ti-Q/h)-To+Q/h));\n",
+      "#Which checks our wall deltat calculation.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For h = 0.5, the overall series resistance is 7.57 Btu/hr*ft*F)\n",
+        "For h = 0.5, heat transfer is 3.96 Btu/hr*ft**2)\n",
+        "For h = 0.5\n",
+        "Temperaure difference through the hot air film is 1.98 F\n",
+        "Temperaure difference through the wall is 14.15 F\n",
+        "Temperaure difference through the cold air film is 1.98 F\n",
+        "For h = 0.42, the overall series resistance is 8.33 Btu/hr*ft*F)\n",
+        "For h = 0.42, heat transfer is 3.60 Btu/hr*ft**2)\n",
+        "For h = 0.42\n",
+        "Temperaure difference through the hot and cold air film is 8.57 F\n",
+        "Temperaure difference through the wall is 12.86 F\n",
+        "The temperature drop on the hot side of the wall is 71.43 F\n",
+        "The temperature drop on the cold side of the wall is 58.57 F\n",
+        "The temperature drop in the wall is 30.00 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.17  Page No : 578"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The first step is to check Reynolds number.It will be recalled that the Reynolds number is given by (D*V*rho)/mu and is dimensionless.Therefore,we can use D,      diameter in feet;V velocity in ft/hr;rho density in lbm/ft**3 and mu vismath.cosity in lbm/(ft*hr).\n",
+      "#Alternatively,the Reynolds number is given by (D*G)/mu,where G is the mass flow rate per unit area (lbm/(hr*ft**2)).\n",
+      "G = ((20*60)*(4*144)/(math.pi*0.87**2)); \t\t\t#Unit:lbm/(hr*ft**2)  \t\t\t#Inside diameter = 0.87 inch \t\t\t#\t\t\t#1 in.**2 = 144 ft**2 \t\t\t#20 lbm/min of water(min converted to second)\n",
+      "#the vismath.cosity of air at these conditions is obtained from figure 11.17 as 0.062 lbm/(ft*hr).So,\n",
+      "mu = 0.33; \t\t\t#the vismath.cosity of air \t\t\t#unit:lbm/(ft*hr)\n",
+      "D = 0.87/12; \t\t\t#Inside diameter \t\t\t#1 in**2 = 144 ft**2\n",
+      "#Therefore Reynolds number is \n",
+      "Re = (D*G)/mu; \t\t\t#Reynolds number\n",
+      "#which is well into the turbulent flow regime.\n",
+      "print \"The Reynolds number is %.2f\"%(Re);\n",
+      "#The next step is to enter Figure 11.18 at W/1000 of 20*(60/1000) = 1.2 and 400F to obtain h1 = 630.\n",
+      "#From the figure 11.20,we obtain F = 1.25 for an inside diameter of 0.87 inch.So,\n",
+      "h1 = 630; \t\t\t#basic heat transfer coefficient \t\t\t#unit:Btu/(hr*ft**2*F)\n",
+      "F = 1.25; \t\t\t#correction factor\n",
+      "h = h1*F; \t\t\t#heat transfer coefficient \t\t\t#the inside film coefficient \t\t\t#unit:Btu/(hr*ft**2*F)\n",
+      "print \"The heat-transfer coefficient is %.2f Btu/hr*ft**2*F)\"%(h);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Reynolds number is 63861.54\n",
+        "The heat-transfer coefficient is 787.50 Btu/hr*ft**2*F)\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.18  Page No : 579"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#We first check the Reynolds number and note that G is same as for problem 11.17.So,\n",
+      "#G is the mass flow rate per unit area (lbm/(hr*ft**2)).\n",
+      "G = ((20*60)*(4*144))/(math.pi*(0.87**2)); \t\t\t#Unit:lbm/(hr*ft**2)  \t\t\t#Inside diameter = 0.87 inch \t\t\t#\t\t\t#1 in.**2 = 144 ft**2 \t\t\t#20 lbm/min of water(min converted to second)\n",
+      "#the vismath.cosity of air at these conditions is obtained from figure 11.17 as 0.062 lbm/(ft*hr).So,\n",
+      "mu = 0.062; \t\t\t#the vismath.cosity of air \t\t\t#unit:lbm/(ft*hr)\n",
+      "D = 0.87/12; \t\t\t#Inside diameter \t\t\t#1 in**2 = 144 ft**2\n",
+      "#Reynolds number is DG/mu,therefore \n",
+      "Re = (D*G)/mu; \t\t\t#Reynolds number\n",
+      "print \"The Reynolds number is %.2f\"%(Re);\n",
+      "#which places the flow in the turbulent regime.Because W/1000(W = weight flow) is same as for problem 11.17 and equals 1.2,we now enter figure 11.19 at 1.2 and 400F   to obtain h1 = 135.Because the inside tube diameter is same as before,F = 1.25.Therefore,\n",
+      "h1 = 135; \t\t\t#basic heat transfer coefficient \t\t\t#unit:Btu/(hr*ft**2*F)\n",
+      "F = 1.25; \t\t\t#correction factor\n",
+      "h = h1*F; \t\t\t#heat transfer coefficient \t\t\t#the inside film coefficient \t\t\t#unit:Btu/(hr*ft**2*F)\n",
+      "print \"The inside film coefficient is %.2f Btu/hr*ft**2*F)\"%(h);\n",
+      "#It is interesting that for equal mass flow rates,water yields a heat-transfer coefficient almost five times greater than air\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Reynolds number is 339908.22\n",
+        "The inside film coefficient is 168.75 Btu/hr*ft**2*F)\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.19  Page No : 586"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#A bare steel pipe\n",
+      "#From the Table 11.5,case 2,\n",
+      "Fe = 0.79; \t\t\t#Emissivity factor to allow for the departure of the surfaces interchanging heat from complete blackness;Fe is a function of the surface emissivities and       configurations\n",
+      "FA = 1; \t\t\t#geometric factor to allow for the average solid angle through which one surface \"sees\" the other\n",
+      "sigma = 0.173*10**-8; \t\t\t#Stefan-Boltzmann constant \t\t\t#Unit:Btu/(hr*ft**2*R**4)\n",
+      "T1 = 120+460; \t\t\t#outside temperature \t\t\t#Unit:R \t\t\t#fahrenheit converted to absolute temperature\n",
+      "T2 = 70+460; \t\t\t#inside temperature \t\t\t#Unit:R \t\t\t#fahrenheit converted to absolute temperature\n",
+      "D = 3.5/12; \t\t\t#3.5 inch  =  3.5/12 feet\t\t\t#Unit:ft \t\t\t#Outside diameter\n",
+      "L = 5; \t\t\t#Length \t\t\t#Unit:ft \t\t\t#From problem 11.10\n",
+      "A = (math.pi*D)*L;  \t\t\t#Area \t\t\t#Unit:ft**2 \n",
+      "Q = sigma*Fe*FA*A*(T1**4-T2**4); \t\t\t#The net interchange of heat by radiation between two bodies at different temperatures \t\t\t#Unit:Btu/hr \t\t\t#\t\t\t#Stefan-Boltzmann law\n",
+      "\n",
+      "# Results\n",
+      "print \"The heat loss by radiation is %.2f Btu/hr\"%(Q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss by radiation is 214.52 Btu/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.20  Page No : 588"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The upper temperature is given as 120 F and the temperature difference is \n",
+      "Ti = 120; \t\t\t#Inside temperature \t\t\t#unit:fahrenheit\n",
+      "To = 70; \t\t\t#Outside temperature \t\t\t#unit:fahrenheit\n",
+      "deltaT = 120-70; \t\t\t#unit:fahrenheit \t\t\t#Change in temperature\n",
+      "#Using figure 11.28, \n",
+      "hrdash = 1.18; \t\t\t#factor for radiation coefficient \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "Fe = 1; \t\t\t#Emissivity factor to allow for the departure of the surfaces interchanging heat from complete blackness;Fe is a function of the surface emissivities and       configurations\n",
+      "FA = 0.79; \t\t\t#geometric factor to allow for the average solid angle through which one surface \"sees\" the other\n",
+      "hr = Fe*FA*hrdash; \t\t\t#The radiation heat-transfer coefficient for the pipe \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "print \"The radiation heat-transfer coefficient for the pipe is %.2f Btu/hr*ft**2*F)\"%(hr);\n",
+      "\n",
+      "#As a check,Using the results of problem 11.17,\n",
+      "print \"As a check, Using the results of problem 11.17\"\n",
+      "D = 3.5/12; \t\t\t#3.5 inch  =  3.5/12 feet\t\t\t#Unit:ft \t\t\t#Outside diameter\n",
+      "L = 5; \t\t\t#Length \t\t\t#Unit:ft \t\t\t#From problem 11.10\n",
+      "A = (math.pi*D)*L;  \t\t\t#Area \t\t\t#Unit:ft**2 \n",
+      "Q = 214.5; \t\t\t#heat loss \t\t\t#Unit:Btu/hr\n",
+      "hr = Q/(A*deltaT); \t\t\t#The radiation heat-transfer coefficient for the pipe \t\t\t#Unit:Btu/(hr*ft**2*F) \t\t\t#Newton's law of cooling\n",
+      "print \"The radiation heat-transfer coefficient for the pipe is %.2f Btu/hr*ft**2*F)\"%(hr);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The radiation heat-transfer coefficient for the pipe is 0.93 Btu/hr*ft**2*F)\n",
+        "As a check, Using the results of problem 11.17\n",
+        "The radiation heat-transfer coefficient for the pipe is 0.94 Btu/hr*ft**2*F)\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.21  Page No : 589"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Because the conditions of illustrative problem 11.15 are the same as for problem 11.19 and 11.20,we can solve this problem in two ways to obtain a check.\n",
+      "#Thus,adding the results of these problems yields,\n",
+      "print \"Adding the results of the problems yields\"\n",
+      "Qtotal = 206.2+214.5; \t\t\t#Unit:Btu/hr \t\t\t#total heat loss\n",
+      "print \"The heat loss due to convection is %.2f Btu/hr\"%(Qtotal);\n",
+      "\n",
+      "#We can also approach this solution by obtaining radiation and convection heat-transfer co-efficcient.Thus,\n",
+      "hcombined = 0.9+0.94; \t\t\t#Coefficient of heat transfer \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "D = 3.5/12; \t\t\t#3.5 inch  =  3.5/12 feet \t\t\t#Unit:ft \t\t\t#Outside diameter\n",
+      "Ti = 120; \t\t\t#Inside temperature \t\t\t#unit:fahrenheit\n",
+      "To = 70; \t\t\t#Outside temperature \t\t\t#unit:fahrenheit\n",
+      "deltaT = Ti-To; \t\t\t#unit:fahrenheit \t\t\t#Change in temperature\n",
+      "L = 5; \t\t\t#Length \t\t\t#Unit:ft \t\t\t#From problem 11.10\n",
+      "A = (math.pi*D)*L;  \t\t\t#Area \t\t\t#Unit:ft**2 \n",
+      "Qtotal = hcombined*A*deltaT; \t\t\t#Unit:Btu/hr \t\t\t#total heat loss due to convection \t\t\t#Newton's law of cooling\n",
+      "print \"By obtaining radiation and convection heat-transfer co-efficcient\"\n",
+      "print \"The heat loss due to convection is %.2f Btu/hr\"%(Qtotal);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Adding the results of the problems yields\n",
+        "The heat loss due to convection is 420.70 Btu/hr\n",
+        "By obtaining radiation and convection heat-transfer co-efficcient\n",
+        "The heat loss due to convection is 421.50 Btu/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.22  Page No : 595"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "\n",
+      "#For brick,concrete,plaster,hot film and cold film,\n",
+      "A = 1.; \t\t\t#area \t\t\t#Unit:ft**2\n",
+      "#For a plane wall,the areas are all the same,and if we use 1 ft**2 of wall surface as the reference area,\n",
+      "#For Brick,\n",
+      "deltax = 6./12; \t\t\t#6 inch  =  6/12 feet \t\t\t#deltax = length \t\t\t#unit:ft\n",
+      "k = 0.40; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "brickResistance = deltax/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For brick\";\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(brickResistance);\n",
+      "\n",
+      "#For Concrete,\n",
+      "deltax = (1./2)/12; \t\t\t#(1/2) inch  =  (1/2)/12 feet \t\t\t#deltax = length \t\t\t#unit:ft\n",
+      "k = 0.80; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "concreteResistance = deltax/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For Concrete\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(concreteResistance);\n",
+      "\n",
+      "#For plaster,\n",
+      "deltax = (1./2)/12; \t\t\t# (1/2) inch  =  6/12 feet \t\t\t#deltax = length \t\t\t#unit:ft\n",
+      "k = 0.30; \t\t\t#Unit:Btu/(hr*ft*F) \t\t\t#k = proportionality constant \t\t\t#k = thermal conductivity \t\t\t#From the table\n",
+      "plasterResistance = deltax/(k*A); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For plaster\";\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(plasterResistance);\n",
+      "\n",
+      "#For \"hot film\",\n",
+      "h = 0.9; \t\t\t#Coefficient of heat transfer \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "hotfilmResistance = inv([[h*A]]); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For hot film\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(hotfilmResistance);\n",
+      "\n",
+      "#For \"cold film\",\n",
+      "h = 1.5; \t\t\t#Coefficient of heat transfer \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "coldfilmResistance = inv([[h*A]]); \t\t\t#Thermal resistance \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"For cold film\"\n",
+      "print \"The resistance is %.2f hr*F)/Btu\"%(coldfilmResistance);\n",
+      "\n",
+      "totalResistance = brickResistance+concreteResistance+plasterResistance+hotfilmResistance+coldfilmResistance; \t\t\t#the overall resistance  \t\t\t#Unit:(hr*f)/Btu\n",
+      "print \"The overall resistance is %.2f hr*F)/Btu\"%(totalResistance);\n",
+      "\n",
+      "U = inv([[totalResistance]]); \t\t\t#Unit:Btu/(hr*ft**2) \t\t\t#The overall conducmath.tance(or overall heat-transfer coefficient)\n",
+      "print \"The overall conductanceor overall heat-transfer coefficient) is %.2f Btu/hr/ft**2)\"%(U);\n",
+      "#In problem 11.21,the solution is straightforward,because the heat-transfer area is constant for all series resistances.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For brick\n",
+        "The resistance is 1.25 hr*F)/Btu\n",
+        "For Concrete\n",
+        "The resistance is 0.05 hr*F)/Btu\n",
+        "For plaster\n",
+        "The resistance is 0.14 hr*F)/Btu\n",
+        "For hot film\n",
+        "The resistance is 1.11 hr*F)/Btu\n",
+        "For cold film\n",
+        "The resistance is 0.67 hr*F)/Btu\n",
+        "The overall resistance is 3.22 hr*F)/Btu\n",
+        "The overall conductanceor overall heat-transfer coefficient) is 0.31 Btu/hr/ft**2)\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.23  Page No : 596"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# given data\n",
+      "hi = 45; \t\t\t#Film coefficient on the inside of the pipe \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "r1 = 3.0/2; \t\t\t#Inside radius \t\t\t#Unit:inch\n",
+      "k1 = 26; \t\t\t#Unit:Btu/(hr*ft**2*F) \t\t\t#k = proportionality constant for steel pipe \t\t\t#k = thermal conductivity for fir \t\t\t#From the table\n",
+      "r2 = 3.5/2; \t\t\t#outide radius \t\t\t#Unit:inch\n",
+      "k2 = 0.026; \t\t\t#Unit:Btu/(hr*ft**2*F) \t\t\t#k = proportionality constant for mineral wool \t\t\t#k = thermal conductivity for fir \t\t\t#From the table\n",
+      "r3 = 5.50/2; \t\t\t#radius \t\t\t#Unit:inch\n",
+      "ho = 0.9; \t\t\t#Film coefficient on the outside of the pipe \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "#Results of problem 11.23,\n",
+      "Ui = 1/((1/hi)+((r1/(k1*12))*math.log(r2/r1))+((r1/(k2*12))*math.log(r3/r2))+(1/(ho*(r3/r1)))); \t\t\t#Unit:Btu/(hr*ft**2*F) \t\t\t#1 in. = 12 ft \t\t\t#Heat transfer coefficient based on inside surface \n",
+      "print \"Heat transfer coefficient based on inside surface is %.2f Btu/hr*ft**2*F)\"%(Ui); \n",
+      "#Because Uo*Ao = Ui*Ai\n",
+      "Uo = Ui*(r1/r3); \t\t\t#Heat transfer coefficient based on outside surface \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "print \"Heat transfer coefficient based on outside surface is %.2f Btu/hr*ft**2*F)\"%(Uo);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer coefficient based on inside surface is 0.36 Btu/hr*ft**2*F)\n",
+        "Heat transfer coefficient based on outside surface is 0.20 Btu/hr*ft**2*F)\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.24  Page No : 601"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#A COUNTERFLOW HEAT EXCHANGER\n",
+      "#Hot oil enters at 215 F and leaves at 125 F\n",
+      "#Water enters the unit at 60 F and leaves at 90 F\n",
+      "#Therefore,From figure 11.34, \n",
+      "thetaA = 215.-90; \t\t\t#the greatest temperature difference between the fluids(at either inlet or outlet) \t\t\t#Unit:fahrenheit\n",
+      "thetaB = 125.-60; \t\t\t#the least temperature difference between the fluids(at either inlet or outlet) \t\t\t#Unit:fahrenheit\n",
+      "deltaTm = (thetaA-thetaB)/math.log(thetaA/thetaB); \t\t\t#math.logarithmic mean temperature difference \t\t\t#Unit:fahrenheit \n",
+      "#From the oil data,\n",
+      "m = 400*60;  \t\t\t#mass \t\t\t#Unit:lb/sec \t\t\t#1 min = 60 sec\n",
+      "Cp = 0.85; \t\t\t#Specific heat of the oil \t\t\t#Unit:Btu/(lb*F)\n",
+      "deltaT = 215.-125;  \t\t\t#Change in temperature \t\t\t#Unit:fahrenheit\n",
+      "Q = m*Cp*deltaT \t\t\t#The heat transfer \t\t\t#Unit:Btu/hr\n",
+      "#Q = U*A*deltaTm\n",
+      "U = 40.;\t\t\t#The overall coefficient of heat transfer of the unit \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "A = Q/(U*deltaTm); \t\t\t#Umit:ft**2 \t\t\t#The outside surface area\n",
+      "print \"The outside surface area required is %.2f ft**2\"%(A);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The outside surface area required is 500.25 ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.25  Page No : 602"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "\n",
+      "#In problem 11.24, A COUNTERFLOW HEAT EXCHANGER is operated in the parallel flow\n",
+      "#Hot oil enters at 215 F and leaves at 125 F\n",
+      "#Water enters the unit at 60 F and leaves at 90 F\n",
+      "#Therefore,From figure 11.35, \n",
+      "thetaA = 215.-60; \t\t\t#the greatest temperature difference between the fluids(at either inlet or outlet) \t\t\t#Unit:fahrenheit\n",
+      "thetaB = 125.-90; \t\t\t#the least temperature difference between the fluids(at either inlet or outlet) \t\t\t#Unit:fahrenheit\n",
+      "deltaTm = (thetaA-thetaB)/math.log(thetaA/thetaB); \t\t\t#math.logarithmic mean temperature difference \t\t\t#Unit:fahrenheit \n",
+      "#From the oil data,\n",
+      "m = 400*60.;  \t\t\t#mass \t\t\t#Unit:lb/sec \t\t\t#1 min = 60 sec\n",
+      "Cp = 0.85; \t\t\t#Specific heat of the oil \t\t\t#Unit:Btu/(lb*F)\n",
+      "deltaT = 215.-125;  \t\t\t#Change in temperature  \t\t\t#Unit:fahrenheit\n",
+      "Q = m*Cp*deltaT \t\t\t#The heat transfer \t\t\t#Unit:Btu/hr\n",
+      "#Q = U*A*deltaTm\n",
+      "U = 40.;\t\t\t#The overall coefficient of heat transfer of the unit \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "A = Q/(U*deltaTm); \t\t\t#Umit:ft**2 \t\t\t#The outside surface area\n",
+      "print \"The outside surface area required is %.2f ft**2\"%(A);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The outside surface area required is 569.19 ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.26  Page No : 603"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From the table 11.7,\n",
+      "#For the oil side,a resistance(fouling factor) of 0.005 (hr*F*ft**2)/Btu can be used\n",
+      "#and for the water side,a fouling factor of 0.001 (hr*F*ft**2)/Btu can be used\n",
+      "#From problem 11.25,\n",
+      "U = 40;\t\t\t#The coefficient of heat transfer of the unit \t\t\t#Unit:Btu/(hr*ft**2*F)\n",
+      "#therefore,\n",
+      "Roil = 0.005; \t\t\t#unit:(hr*ft**2*F)/Btu \t\t\t#resistance at oil side\n",
+      "Rwater = 0.001; \t\t\t#unit:(hr*ft**2*F)/Btu \t\t\t#resistance for water side\n",
+      "invU = 0.025             # Inv of U\n",
+      "Rcleanunit = invU; \t\t\t#unit:(hr*ft**2*F)/Btu \t\t\t#resistance at clean unit\n",
+      "Roverall = Roil+Rwater+Rcleanunit; \t\t\t#unit:(hr*ft**2*F)/Btu \t\t\t#overall resistance\n",
+      "invRoverall = 32.258065;\n",
+      "Uoverall = invRoverall; \t\t\t#Unit:Btu/(hr*ft**2*F) \t\t\t#The overall coefficient of heat transfer of the unit\n",
+      "#Because all the parameters are the same,the surface area required will vary inversely as U\n",
+      "A = 569*(U/Uoverall); \t\t\t#A = 569 ft**2 in the problem 11.25 \t\t\t#unit:ft**2  \t\t\t#The outside surface area\n",
+      "print \"The outside surface area required is %.2f ft**2\"%(A);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The outside surface area required is 705.56 ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.27  Page No : 605"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#HEAT EXCHANGER\n",
+      "#Oil flows in the tube side and is cooled from  280 F to 140 F\n",
+      "#Therefore,\n",
+      "t2 = 140.; \t\t\t#Unit:fahrenheit\n",
+      "t1 = 280.; \t\t\t#Unit:fahrenheit\n",
+      "#On the shell side,water is heated from 85 F to 115 F\n",
+      "T1 = 85.; \t\t\t#Unit:fahrenheit\n",
+      "T2 = 115.; \t\t\t#Unit:fahrenheit\n",
+      "P = (t2-t1)/(T1-t1); \n",
+      "R = (T1-T2)/(t2-t1);\n",
+      "#From the figure,\n",
+      "F = 0.91;\t\t\t#Correction factor\n",
+      "LMTD = ((t1-T2)-(t2-T1))/math.log((t1-T2)/(t2-T1)); \t\t\t#LMTD = Log mean temperature difference \t\t\t#Unit:fahrenheit\n",
+      "TMTD = F*LMTD; \t\t\t#TMTD = True mean temperature difference \t\t\t#Unit:fahrenheit\n",
+      "\n",
+      "# Results\n",
+      "print \"The true mean temperature is %.2f fahrenheit\"%(TMTD);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The true mean temperature is 91.11 fahrenheit\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch1_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch1_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch1_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch1_2.ipynb
new file mode 100755
index 00000000..ce79dd48
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch1_2.ipynb
@@ -0,0 +1,394 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:cda743da280474e0339734ce945527afbcb8983c5386e8ae01ef3be6064b84d0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Fundamental Concepts"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1  Page No : 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#C = (5/9)*(F-32);\n",
+      "#F = 32+(9*C/5);\n",
+      "#Letting C = F in equation;\n",
+      "#F = (5/9)*(F-32);\n",
+      "#Therefore\n",
+      "F = -160/4;           \t\t\t#fahrenheit\n",
+      "\n",
+      "# Results\n",
+      "print \"Both fahrenheit and celsius temperature scales indicate same temperature at %.2f\"%(F);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Both fahrenheit and celsius temperature scales indicate same temperature at -40.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2  Page No : 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Mm = 0.0123\t\t\t#Unit:lb \t\t\t#Mass of the moon;\n",
+      "Me = 1 \t    \t\t#Unit:lb \t\t\t#Mass of the earth;\n",
+      "Dm = 0.273 \t\t\t#Unit:feet \t\t\t#Diameter of the moon;\n",
+      "De = 1. \t    \t\t#Unit:feet \t\t\t#Diameter of the earth;\n",
+      "Rm = Dm/2; \t\t\t#Radius of the moon; \t\t\t#Unit:feet\n",
+      "Re = De/2; \t\t\t#Radius of the earth; \t\t\t#Unit:feet\n",
+      "\n",
+      "#F = (K*M1*M2)/d**2 \t\t\t#Law of universal gravitation;\n",
+      "#Fe = (K*Me*m)/Re**2; \t\t\t#Fe = Force exerted on the mass;\n",
+      "#Fm = (K*Mm*m)/Rm**2; \t\t\t#Fm = Force exerted on the moon;\n",
+      "F = (Me/Mm)*(Rm/Re)**2; \t\t\t#F = Fe/Fm;\n",
+      "print \"Relation of force exerted on earth to mass is\"\n",
+      "print \"Fe/Fm  = %.2f\"%F"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relation of force exerted on earth to mass is\n",
+        "Fe/Fm  = 6.06\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3  Page No : 20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "M = 5;            \t\t\t#Unit:kg \t\t\t#mass of body;\n",
+      "g = 9.81;         \t\t\t#Unit:m/s**2 \t\t\t#the local acceleration of gravity\n",
+      "\n",
+      "# Calculations\n",
+      "W = M*g;          \t\t\t#W = the weight of the body \t\t\t#Unit:Newton \t\t\t# 1 N =  1 kg*m/s**2\n",
+      "\n",
+      "# Results\n",
+      "print \"The weight of the body is %.2f N\"%(W);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The weight of the body is 49.05 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4  Page No : 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Solution for a\";\n",
+      "#given\n",
+      "M  =  10.            \t\t\t#Unit:kg \t\t\t#mass of body;\n",
+      "g  =  9.5           \t\t\t#Unit:m/s**2 \t\t\t#the local acceleration of gravity\n",
+      "W  =  M*g;          \t\t\t#W  =  the weight of the body; \t\t\t#Unit:Newton \t\t\t# 1 N  =   1 kg*m/s**2\n",
+      "print \"The weight of the body is %.2f N\"%(W);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "#Given\n",
+      "F  =  10;           \t\t\t#Unit:Newton \t\t\t#Horizontal Force\n",
+      "a  =  F/M;          \t\t\t\t\t\t#\t\t\t#newton's second law of motion\n",
+      "print \"The horizontal acceleration of the body is %.2f m/s**2\"%(a);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "The weight of the body is 95.00 N\n",
+        "Solution for b\n",
+        "The horizontal acceleration of the body is 1.00 m/s**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5  Page No : 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Conversion Problem\n",
+      "# 1 inch = 0.0254 meter so, 1 = 0.0254 meter/inch    \t\t\t#Eq.1\n",
+      "# 1 ft = 12 inch so, 1 = 12 inch/ft.........\t\t\t#Eq.2\n",
+      "#Multiplying Eq.1 & Eq.2  \t\t\t# We get 1 = 0.0254*12 meter/ft\n",
+      "#Taking Square both side\n",
+      "#1**2 = (0.0254*12)**2  meter**2/ft**2\n",
+      "print \"1 ft**2 = %.2f meter**2\"%((0.0254*12)**2); "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 ft**2 = 0.09 meter**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7  Page No : 33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#The Specific gravity of mercury is 13.6 \t\t\t#Given\n",
+      "#Converting the unit of weight of grams per cubic centimeter to pounds per cubic foot\n",
+      "# 1 lbf = 454 gram \t\t\t#1 inch =  2.54 cm\n",
+      "#So 1 gram = 1/454 lbf and 1 ft = 12*2.54 cm\n",
+      "#Gamma = (gram/cm**3)*(lb/gram)*(cm**3/ft**3) = lb/ft**3\n",
+      "#Gamma = (1 gram/cm**3)*(1 lbf/454 gram)*(2.54*12)**3 *cm**3/ft**3\n",
+      "Gamma = (1./454)*(2.54*12)**3; \t\t\t#lbf/ft**3 \t\t\t#conversion factor\n",
+      "print \"Conversion Factor = %.2f\"%Gamma\n",
+      "p = (1./12)*(Gamma*13.6); \t\t\t#lbf/ft**2 \t\t\t#gage pressure\n",
+      "p = (1./12)*Gamma*13.6*(1./144) \t\t\t#ft**2/inch**2 \t\t\t#gage pressure\n",
+      "print \"Guage Pressure is %.2f psi\"%(p);\n",
+      "print \"Local atmospheric pressure is 14.7 psia\";\n",
+      "P = p+14.7; \t\t\t#Pressure on the base of the column \t\t\t#Unit:psia\n",
+      "print \" So Pressure on the base of the column is %.2f psia\"%(P);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conversion Factor = 62.37\n",
+        "Guage Pressure is 0.49 psi\n",
+        "Local atmospheric pressure is 14.7 psia\n",
+        " So Pressure on the base of the column is 15.19 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8  Page No : 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Rho = 13.595; \t\t\t#Unit: kg/m**3 \t\t\t#The density of mercury\n",
+      "h = 25.4; \t\t\t#Unit: mm \t\t\t#Height of column of mercury\n",
+      "g = 9.806; \t\t\t#Unit:m/s**2 \t\t\t#the local acceleration of gravity\n",
+      "\n",
+      "#Solution\n",
+      "p = Rho*g*h; \t\t\t#P = Pressure at the base of a column of mercury \t\t\t#Unit:Pa\n",
+      "\n",
+      "# Results\n",
+      "print \"Pressure at the base of a column of mercury is %.2f Pa\"%(p);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at the base of a column of mercury is 3386.14 Pa\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9  Page No : 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Patm = 30.0; \t\t\t#in. \t\t\t#pressure of mercury at smath.radians(numpy.arcmath.tan(ard temperature\n",
+      "Vacuum = 26.5;  \t\t\t#in.  \t\t\t#vaccum pressure\n",
+      "Pabs = Patm-Vacuum; \t\t\t#Absolute pressure of mercury \t\t\t#in.\n",
+      "# 1 inch mercury exerts a pressure of 0.491 psi\n",
+      "p = Pabs*0.491; \t\t\t#Absolute pressure in psia\n",
+      "\n",
+      "# Results\n",
+      "print \"Absolute pressure of mercury in is %.2f psia\"%(p);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Absolute pressure of mercury in is 1.72 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.10  Page No : 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Rho = 2000; \t\t\t#Unit: kg/m**3 \t\t\t#The density of fluid\n",
+      "h = -10; \t\t\t#Unit: mm \t\t\t#Height of column of fluid \t\t\t#the height is negative because it is measured up from the base\n",
+      "g = 9.6 \t\t\t#Unit:m/s**2 \t\t\t#the local acceleration of gravity\n",
+      "\n",
+      "#Solution\n",
+      "p = -Rho*g*h; \t\t\t#P = Pressure at the base of a column of fluid \t\t\t#Unit:Pa\n",
+      "\n",
+      "# Results\n",
+      "print \"Pressure at the base of a column of fluid is %.2f Pa\"%(p);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at the base of a column of fluid is 192000.00 Pa\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11  Page No : 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Patm = 30.0 \t\t\t#in. \t\t\t#pressure of mercury at smath.radians(numpy.arcmath.tan(ard temperature\n",
+      "Vacuum = 26.5  \t\t\t#in.  \t\t\t#vaccum pressure\n",
+      "Pabs = Patm-Vacuum; \t\t\t#Absolute pressure of mercury \t\t\t#in.\n",
+      "\n",
+      "# (3.5 inch* (ft/12 inch) * (13.6*62.4)LBf/ft**3 * kg/2.2 LBf * 9.806 N/kg)/((12 inch**2/ft**2) * (0.0254 m/inch)**2)\n",
+      "p = (3.5*(1./12)*13.6*62.4*(1/2.2)*9.806)/(12**2*0.0254**2*1000); \t\t\t#kPa \t\t\t#Absolute pressure in psia\n",
+      "\n",
+      "# Results\n",
+      "print \"Absolute pressure of mercury is %.2f kPa\"%(p)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Absolute pressure of mercury is 11.88 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch2_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch2_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch2_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch2_2.ipynb
new file mode 100755
index 00000000..c4ad19b7
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch2_2.ipynb
@@ -0,0 +1,562 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:086f4825fb4b8e83d4ca795bc6e4859da294b353e111bb33a89148489ab4ef6c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Work, Energy and Heat"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2  Page No : 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "k = 100; \t\t\t# Unit:lbf/in. \t\t\t#k = spring constant\n",
+      "l = 2; \t\t\t#Unit:inch \t\t\t#l =  length of compression of string\n",
+      "\n",
+      "# Calculations\n",
+      "work = (1./2)*k*l**2; \t\t\t#force-print lacement relation \t\t\t#Unit:in*lbf\n",
+      "\n",
+      "# Results\n",
+      "print \"Workdone is %.2f inch*lbf\"%(work);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Workdone is 200.00 inch*lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3  Page No : 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "k = 20*1000; \t\t\t# Unit:N/m  \t\t\t#k = 20kN \t\t\t#k = spring constant\n",
+      "l = 0.075; \t\t\t#Unit:meter  \t\t\t#l = 75 mm \t\t\t#l =  length of compression of string\n",
+      "\n",
+      "# Calculations\n",
+      "work = (1./2)*k*l**2; \t\t\t#force-print lacement relation \t\t\t#Unit:N*m\n",
+      "\n",
+      "# Results\n",
+      "print \"Workdone is %.2f Jule\"%(work);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Workdone is 56.25 Jule\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4  Page No : 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Z = 600; \t\t\t#Unit:ft \t\t\t#Z = The dismath.tance,the body is raised from its initial position when the force is applied\n",
+      "gc = 32.174; \t\t\t#Unit: (lbm*ft)/(lbf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "g = gc; \t\t\t#Unit:ft/s**2 \t\t\t#g = The local gravity\n",
+      "m = 1.; \t\t\t#Unit:lbm \t\t\t#m = mass\n",
+      "\n",
+      "# Calculations\n",
+      "PE = (m*g*Z)/gc; \t\t\t#potential energy \t\t\t#Unit:ft*lbf\n",
+      "\n",
+      "# Results\n",
+      "print \"%.2f ft*lbf work is done lifting the water to elevation \"%(PE)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "600.00 ft*lbf work is done lifting the water to elevation \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5  Page No : 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "m = 1; \t\t\t#Unit:kg \t\t\t#m = mass\n",
+      "g =  9.81 \t\t\t#Unit:m/s**2 \t\t\t#g = The local gravity\n",
+      "Z = 50 \t\t\t#Unit:m \t\t\t#\t\t\t#Z = The dismath.tance,the body is raised from its initial position when the force is applied \t\t\t#In this case Z = delivered water from well to pump\n",
+      "\n",
+      "# Calculations\n",
+      "PE = m*g*Z; \t\t\t#PE = Potential Energy \t\t\t#Unit:Joule\n",
+      "\n",
+      "# Results\n",
+      "print \"Change in potential energy per kg of water is %.2f J \"%(PE); \t\t\t#J = Joule = N*m = kg*m**2/s**2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in potential energy per kg of water is 490.50 J \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6  Page No : 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "Rho = 62.4; \t\t\t#Unit:lbm/ft**3 \t\t\t#Rho = The density of water\n",
+      "A = 10000; \t\t\t#Flow = 10000; gal/min\n",
+      "V = (231./1728); \t\t\t# 12 inch = 1 ft \t\t\t#So,1 ft**3 = 1728 in**3  \t\t\t# One Gallon is a volumetric measure equal to 231 in**3\n",
+      "#A*V \t\t\t#Unit:ft**3/min\n",
+      "\n",
+      "#In example, 2.4:\n",
+      "#  From example 2.4\n",
+      "Z = 600; \t\t\t#Unit:ft \t\t\t#Z = The dismath.tance,the body is raised from its initial position when the force is applied\n",
+      "gc = 32.174; \t\t\t#Unit: (lbm*ft)/(lbf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "g = gc; \t\t\t#Unit:ft/s**2 \t\t\t#g = The local gravity\n",
+      "m = 1; \t\t\t#Unit:lbm \t\t\t#m = mass\n",
+      "\n",
+      "# Calculations and Results\n",
+      "PE = (m*g*Z)/gc; \t\t\t#potential energy \t\t\t#Unit:ft*lbf\n",
+      "print \"%.2f ft*lbf work is done lifting the water to elevation \"%(PE);\n",
+      "\n",
+      "#So,\n",
+      "# In example 2.5\n",
+      "M = Rho*A*V; \t\t\t#M = the mass flow\n",
+      "Power = M*PE; \t\t\t#Unit:ft*lbf/lbm\n",
+      "print \"Generated Power is %.2f ft*lbf/lbm \"%(Power);\n",
+      "# 1 horsepower  =  33,000 ft*lbf/min\n",
+      "print \"Power  =  %.2f hp\"%(Power/33000);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "600.00 ft*lbf work is done lifting the water to elevation \n",
+        "Generated Power is 50050000.00 ft*lbf/lbm \n",
+        "Power  =  1516.67 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7  Page No : 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"In problem 2.5\";\n",
+      "m = 1; \t\t\t#Unit:kg \t\t\t#m = mass\n",
+      "g =  9.81 \t\t\t#Unit:m/s**2 \t\t\t#g = The local gravity\n",
+      "Z = 50 \t\t\t#Unit:m \t\t\t#\t\t\t#Z = The dismath.tance,the body is raised from its initial position when the force is applied \t\t\t#In this case Z = delivered water from well to pump\n",
+      "\n",
+      "# Calculations and Results\n",
+      "PE = m*g*Z; \t\t\t#PE = Potential Energy \t\t\t#Unit:Joule\n",
+      "print \"Change in potential energy per kg of water is %.2f J \"%(PE); \t\t\t#J = Joule = N*m = kg*m**2/s**2\n",
+      "#Given data in problem 2.7 is\n",
+      "M = 1000; \t\t\t#Unit;kg/min\t\t\t#M = Water density \n",
+      "Power = PE*M*(1./60); \t\t\t#1 min = 60 seconds \t\t\t#power \t\t\t#unit:Joule/s = W\n",
+      "print \"Power is %.2f Watt\"%(Power); \t\t\t#Watt = N*m/s  =  Joule/s  = Watt\n",
+      "#1 Hp = 746 Watt\n",
+      "print \"Power is %.2f Horsepower\"%(Power/745);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In problem 2.5\n",
+        "Change in potential energy per kg of water is 490.50 J \n",
+        "Power is 8175.00 Watt\n",
+        "Power is 10.97 Horsepower\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8  Page No : 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "m = 10.; \t\t\t#Unit:lb \t\t\t#m = Mass\n",
+      "V1 = 88.; \t\t\t#Unit:\t\t\t#ft/s V1 = Velocity before it is slowed down\n",
+      "V2 = 10.; \t\t\t#Unit;ft/s \t\t\t#V2 = Velocity after it is slowed down\n",
+      "gc = 32.174; \t\t\t#Unit: (lbm*ft)/(lbf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "\n",
+      "# Calculations and Results\n",
+      "KE1 = m*V1**2/(2*gc); \t\t\t#The kinetic energy of the body before it is slowed down \t\t\t#Unit:ft*lbf\n",
+      "print \"The kinetic energy of the body before it is slowed down is %.2f ft*lbf\"%(KE1);\n",
+      "\n",
+      "KE2 = m*V2**2/(2*gc); \t\t\t#The kinetic energy of the body before it is slowed down \t\t\t#Unit:ft*lbf\n",
+      "print \"The kinetic energy of the body before it is slowed down is %.2f ft*lbf\"%(KE2);\n",
+      "\n",
+      "KE = KE1-KE2; \t\t\t#KE = Change in kinetic energy \t\t\t#Unit:ft*lbf\n",
+      "print \"Change in kinetic energy is %.2f ft*lbf\"%(KE);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The kinetic energy of the body before it is slowed down is 1203.46 ft*lbf\n",
+        "The kinetic energy of the body before it is slowed down is 15.54 ft*lbf\n",
+        "Change in kinetic energy is 1187.92 ft*lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9  Page No : 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "m = 1500.; \t\t\t#Unit:kg \t\t\t#m = mass\n",
+      "V1 = 50.; \t\t\t#Km/hour V1 = Velocity before it is slowed down\n",
+      "#V1 = (50*1000 m/hour)**2/(3600 s/hour)**2 \n",
+      "\n",
+      "# Calculations\n",
+      "KE1 = (m*(V1*1000)**2/3600**2)/2; \t\t\t#KE1 = Initial kinetic energy \t\t\t#Unit:Joule\n",
+      "\n",
+      "#After slowing down\n",
+      "V2 = 30; \t\t\t#Unit:KM/hour \t\t\t#V2 = Velocity after it is slowed down\n",
+      "#V2 = (30*1000 m/hour)**2/(3600 s/hour)**2 \n",
+      "KE2 = (m*(V2*1000)**2/3600**2)/2; \t\t\t#KE2 = After slowing down, the kinetic energy \t\t\t#Unit:Joule\n",
+      "\n",
+      "KE = KE1-KE2; \t\t\t#KE = Change in kinetic energy \t\t\t#Unit:Joule\n",
+      "\n",
+      "# Results\n",
+      "print \"Change in kinetic energy is %.2f kJ\"%(KE/1000);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in kinetic energy is 92.59 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10  Page No : 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "m = 10 \t\t\t#Unit:kg \t\t\t#m = mass\n",
+      "Z = 10 \t\t\t#Unit:m \t\t\t#Z = The dismath.tance,the body is raised from its initial position when the force is applied\n",
+      "g =  9.81 \t\t\t#Unit:m/s**2 \t\t\t#g = The local gravity\n",
+      "#There are no losses in the system\n",
+      "#So,initial potential energy plus initial kinetic energy equal to sum of final potential energy plus final kinetic energy\n",
+      "#So, PE1+KE1 = PE2+KE2\n",
+      "#From the figure,KE1 = 0; PE2 = 0;\n",
+      "#So,PE1 = KE2;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "PE1 = m*g*Z; \t\t\t#PE = Potential Energy \t\t\t#Unit:Joule\n",
+      "#KE2 = (m*v**2)/2\n",
+      "v = (PE1*2)/m; \n",
+      "V = math.sqrt(v); \t\t\t#Unit:m/s \t\t\t#velocity \n",
+      "print \"Velocity  =  %.2f m/s\"%(V);\n",
+      "KE2 = PE1; \t\t\t#kinetic energy \t\t\t#Unit:Joule\n",
+      "print \"Kinetic energy is %.2f N*m\"%(PE1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity  =  14.01 m/s\n",
+        "Kinetic energy is 981.00 N*m\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11  Page No : 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "p1 = 100.; \t\t\t#pressure at the enmath.tance \t\t\t#Unit:psia,lbf/in**2\n",
+      "Rho1 = 62.4; \t\t\t#Unit:lbm/ft**3  \t\t\t#Rho = The density\n",
+      "v1 = 144.*(1/Rho1) \t\t\t#Specific Volume at entrance or reciprocal of fluid density \t\t\t# 144 in**2 = 1 ft**2\n",
+      "#1 Btu  =  778 ft*lbf \n",
+      "J = 778.; \t\t\t#Unit:ft*lbf/Btu \t\t\t#conversion factor\n",
+      "\n",
+      "# Calculations and Results\n",
+      "FW1 = (p1*v1)/J; \t\t\t#Flow work  \t\t\t#Btu/lbm\n",
+      "print \"Flow work  =  %.2f Btu/lbm\"%(FW1);\n",
+      "\n",
+      "print \"At the exit of device\"\n",
+      "p2 = 50.; \t\t\t#pressure at the exit \t\t\t#Unit:psia,lbf/in**2\n",
+      "Rho2 = 30.; \t\t\t#Unit:lbm/ft**3 \t\t\t#Rho = The density\n",
+      "v2 = 144.*(1./Rho2) \t\t\t#Specific Volume at exit or reciprocal of fluid density \t\t\t# 144 in**2 = 1 ft**2\n",
+      "#1 Btu  =  778 ft*lbf \n",
+      "J = 778.; \t\t\t#Unit:ft*lbf/Btu \t\t\t#conversion factor\n",
+      "FW2 = (p2*v2)/J; \t\t\t#Flow work \t\t\t#Btu/lbm\n",
+      "print \"Flow work  =  %.2f Btu/lbm\"%(FW2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow work  =  0.30 Btu/lbm\n",
+        "At the exit of device\n",
+        "Flow work  =  0.31 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12  Page No : 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print \"At the entrance of device\"\n",
+      "p1 = 200*1000; \t\t\t#200kPa*1000 Pa/kPa  \t\t\t#pressure at the entrance \t\t\t#Unit:N/m**2\n",
+      "Rho1 = 1000; \t\t\t#kg/m**3 \t\t\t#Fluid density at entrance\n",
+      "v1 = 1./Rho1; \t\t\t#Specific Volume at entrance or reciprocal of fluid density\n",
+      "FW1 = p1*v1; \t\t\t#Flow work at entrance \t\t\t#Unit:N*m/kg\n",
+      "print \"Flow work  =  %.2fN*m/kg\"%(FW1);\n",
+      "\n",
+      "print \"At the exit of device\"\n",
+      "p2 = 100*1000; \t\t\t#200kPa*1000 Pa/kPa  \t\t\t#pressure at the exit \t\t\t#Unit:N/m**2\n",
+      "Rho2 = 250; \t\t\t#kg/m**3 \t\t\t#Fluid density at exit\n",
+      "v2 = 1./Rho2; \t\t\t#Specific Volume at entrance or reciprocal of fluid density\n",
+      "FW2 = p2*v2; \t\t\t#Flow work at exit\t\t\t#Unit:N*m/kg\n",
+      "print \"Flow work  =  %.2f N*m/kg\"%(FW2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At the entrance of device\n",
+        "Flow work  =  200.00N*m/kg\n",
+        "At the exit of device\n",
+        "Flow work  =  400.00 N*m/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14  Page No : 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#It is necessary that pressure be expressed as psfa when the volume is in cubic feet\n",
+      "#100 psia  =  100*144 psfa\n",
+      "p1 = 100*144; \t\t\t#Unit:psfa  \t\t\t#initial pressure\n",
+      "v1 = 2; \t\t\t#Unit:ft**3/lb \t\t\t#Initial Specific Volume\n",
+      "v2 = 1.; \t\t\t#Unit:ft**3/lb \t\t\t#Final Specific Volume\n",
+      "\n",
+      "# Calculations\n",
+      "w = p1*v1*math.log(v2/v1); \t\t\t#work done on fluid \t\t\t#Unit:ft*lbf/lbm\n",
+      "\n",
+      "# Results\n",
+      "print \"Work done on fluid  =  %.2f ft*lbf/lb\"%(w);\n",
+      "#1 Btu  =  778 ft*lbf \n",
+      "print \"Work done on the fluid per pound of fluid is %.2f Btu/lbm\"%(w/778);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work done on fluid  =  -19962.64 ft*lbf/lb\n",
+        "Work done on the fluid per pound of fluid is -25.66 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15  Page No : 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Given value\n",
+      "#p1*v1 = p2*v2\n",
+      "p1 = 200.*1000; \t\t\t#p1 = Initial Pressure \t\t\t#Unit:Pa\n",
+      "p2 = 800*1000; \t\t\t#p2 = Final Pressure \t\t\t#Unit:Pa\n",
+      "v1 = 0.1; \t\t\t#v1 = Initial Special Volume \t\t\t#Unit:m**3/kg\n",
+      "\n",
+      "# Calculations\n",
+      "v2 = (p1/p2)*v1; \t\t\t#v1 = final Special Volume \t\t\t#Unit:m**3/kg\n",
+      "w = p1*v1*math.log(v2/v1); \t\t\t#workdone \t\t\t#Unit:kJ/kg\n",
+      "\n",
+      "# Results\n",
+      "print \"Work done per kilogram of gas is %.2f kJ/kg into the system)\"%(w/1000);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work done per kilogram of gas is -27.73 kJ/kg into the system)\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch3_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch3_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch3_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch3_2.ipynb
new file mode 100755
index 00000000..1a72803f
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch3_2.ipynb
@@ -0,0 +1,901 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d3fc279744b60850d5797d14855414b40de2e09f4541a34684deba85c06170cf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : The First Law of Thermodynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1  Page No : 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#For a constant volume process, 10 Btu/lbm heat is added to the system\n",
+      "#We can consider thet a math.tank having a fixed volume has heat added to it\n",
+      "#Under these conditions,the mechanical work done on or by the system must be 0\n",
+      "#u2-u1 = q\n",
+      "print \"Heat has been converted to internal energy of the working fluid\";\n",
+      "#So,\n",
+      "print \" So,Change in internal energy u2-u1 = 10 Btu/Lbm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat has been converted to internal energy of the working fluid\n",
+        " So,Change in internal energy u2-u1 = 10 Btu/Lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4  Page No : 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Solution For a\";\n",
+      "m = 10; \t\t\t#Unit:lbm \t\t\t#mass of water\n",
+      "#delataU = U2-U1\n",
+      "Heat = 100.; \t\t\t#Unit:Btu \t\t\t#heat added\n",
+      "deltaU = Heat/m; \t\t\t#Change in internal energy  \t\t\t#unit:Btu/lbm\n",
+      "print \"Change in internal energy per pound of water is %.2f Btu/lbm\"%(deltaU);\n",
+      "\n",
+      "print \"Solution For b\";\n",
+      "print \"In this process,energy crosses the boundary of the system by means of fractional work\"\n",
+      "print \"The contents of the math.tank will not distinguish between the energy if it is added as\\\n",
+      " heat or the energy added as fraction work\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution For a\n",
+        "Change in internal energy per pound of water is 10.00 Btu/lbm\n",
+        "Solution For b\n",
+        "In this process,energy crosses the boundary of the system by means of fractional work\n",
+        "The contents of the math.tank will not distinguish between the energy if it is added as heat or the energy added as fraction work\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5  Page No : 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "P1 = 100 \t\t\t#Unit:psia \t\t\t#Pressure at the entrance to a steady-flow device\n",
+      "Rho1 = 62.4 \t\t\t#Unit:lbm/ft**3 \t\t\t#the density of the fluid\n",
+      "A1V1 = 10000 \t\t\t#Unit:ft**3/min \t\t\t#Entering fluid\n",
+      "A2 = 2 \t\t\t#Unit:ft**2 \t\t\t#Exit area\n",
+      "\n",
+      "# Calculations and Results\n",
+      "m = Rho1*A1V1; \t\t\t#Unit:lbm/min \t\t\t#mass rate of flow per unit time\n",
+      "print \"Mass flow rate is %.2f LBm/min\"%(m);\n",
+      "\n",
+      "Rho2 = Rho1; \t\t\t#Unit:lbm/ft**3 \t\t\t#the density of the fluid\n",
+      "#m = Rho2*A2*V2\n",
+      "#So,\n",
+      "V2 = m/(Rho2*A2); \t\t\t#velocity at exit \t\t\t#Unit:ft/min\n",
+      "print \"The exit velocity is %.2f ft/min\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass flow rate is 624000.00 LBm/min\n",
+        "The exit velocity is 5000.00 ft/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6  Page No : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "Rho1 = 1000. \t\t\t#Unit:kg/m**3 \t\t\t#the density of the fluid at entrance\n",
+      "A1V1 = 2000 \t\t\t#Unit:m**3/min \t\t\t#Entering fluid\n",
+      "A2 = 0.5 \t\t\t#Unit:ft**2 \t\t\t#Exit area\n",
+      "\n",
+      "# Calculations and Results\n",
+      "m = Rho1*A1V1; \t\t\t#Unit:kg/min \t\t\t#mass rate of flow per unit time\n",
+      "print \"Mass flow rate is %.2f kg/min\"%(m);\n",
+      "\n",
+      "Rho2 = Rho1; \t\t\t#Unit:kg/m**3 \t\t\t#the density of the fluid at exit\n",
+      "#m = Rho2*A2*V2\n",
+      "#So,\n",
+      "V2 = m/(Rho2*A2); \t\t\t#The exit velocity \t\t\t#Unit:m/min\n",
+      "print \"The exit velocity is %.2f m/min\"%(V2);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass flow rate is 2000000.00 kg/min\n",
+        "The exit velocity is 4000.00 m/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7  Page No : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Given value\n",
+      "Rho = 62.4 \t\t\t#Unit:lbm/ft**3 \t\t\t#the density of the fluid\n",
+      "V = 100 \t\t\t#Unit:ft/s \t\t\t#Velocity of fluid\n",
+      "d = 1 \t\t\t#Unit:in \t\t\t#Diameter\n",
+      "\n",
+      "# Calculations\n",
+      "#1 ft**2 = 144 in**2 \t\t\t#A = (math.pi/4)*d**2\n",
+      "A = (math.pi*d**2)/(4*144) \t\t\t#Unit:ft**2 \t\t\t#area \n",
+      "m = Rho*A*V; \t\t\t#Unit:lbm/s \t\t\t#mass rate of flow per unit time\n",
+      "\n",
+      "# Results\n",
+      "print \"Mass flow rate is %.2f lbm/s\"%(m);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass flow rate is 34.03 lbm/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8  Page No : 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Given value\n",
+      "m1 = 50000.; \t\t\t#Unit:LBm/hr \t\t\t#An inlet steam flow\n",
+      "v1 = 0.831 \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume of inlet steam\n",
+      "d1 = 6.  \t\t\t#Unit:in \t\t\t#Inlet diameter\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A1 = (math.pi*d1**2)/(4*144) \t\t\t#1 ft**2 = 144 in**2 \t\t\t#Entering area\n",
+      "V1 = (m1*v1)/(A1*60*60) \t\t\t#(60 min/hr * 60 s/min) \t\t\t#To convert hours into seconds \t\t\t#velocity at inlet\n",
+      "print \"The velocity at inlet is %.2f ft/s\"%(V1);\n",
+      "\n",
+      "\n",
+      "m2 = m1; \t\t\t#Unit:LBm/hr \t\t\t#m2 = An outlet steam flow\n",
+      "v2 = 1.825 \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume of outlet steam\n",
+      "d2 = 8  \t\t\t#Unit:in \t\t\t#Outlet diameter\n",
+      "A2 = (math.pi*d2**2)/(4*144) \t\t\t#1 ft**2 = 144 in**2 \t\t\t#Exit area\n",
+      "V2 = (m1*v2)/(A2*60*60) \t\t\t#(60 min/hr * 60 s/min) \t\t\t#To convert hours into seconds \t\t\t#velocity at outlet\n",
+      "print \"The velocity at outlet is %.2f ft/s\"%(V2);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity at inlet is 58.78 ft/s\n",
+        "The velocity at outlet is 72.61 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9  Page No : 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Given value\n",
+      "m1 = 10000; \t\t\t#Unit:kg/hr \t\t\t#An inlet steam flow\n",
+      "v1 = 0.05 \t\t\t#Unit:m**3/kg \t\t\t#Specific volume of inlet steam\n",
+      "d1 = 0.1  \t\t\t#Unit:m \t\t\t#Inlet diameter \t\t\t#100 mm  = 0.1 m\n",
+      "\n",
+      "# Calculations and Results\n",
+      "A1 = (math.pi/4)*d1**2 \t\t\t#Unit:m**2 \t\t\t#Entering area\n",
+      "V1 = (m1*v1)/(A1*60*60) \t\t\t#(60 min/hr * 60 s/min) \t\t\t#To convert hours into seconds \t\t\t#velocity at inlet \t\t\t#Unit:m/s\n",
+      "print \"The velocity at inlet is %.2f m/s\"%(V1);\n",
+      "\n",
+      "\n",
+      "m2 = m1; \t\t\t#Unit:kg/hr \t\t\t#m2 = An outlet steam flow\n",
+      "v2 = 0.10 \t\t\t#Unit:m**3/kg \t\t\t#Specific volume of outlet steam\n",
+      "d2 = 0.2  \t\t\t#Unit:m \t\t\t#Outlet diameter \t\t\t#200 mm  =  0.2 m\n",
+      "A2 = (math.pi/4)*(d2**2) \t\t\t#Unit:m**2 \t\t\t#Exit area\n",
+      "V2 = (m1*v2)/(A2*60*60) \t\t\t#(60 min/hr * 60 s/min) \t\t\t#To convert hours into seconds \t\t\t#velocity at outlet \t\t\t#Unit:m/s\n",
+      "print \"The velocity at outlet is %.2f m/s\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity at inlet is 17.68 m/s\n",
+        "The velocity at outlet is 8.84 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10  Page No : 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "Cp = 0.22; \t\t\t#Unit:Btu/(LBm*R) \t\t\t#Specific heat for constant pressure process\n",
+      "Cv = 0.17; \t\t\t#Unit:Btu/(LBm*R) \t\t\t#Specific heat for constant volume process\n",
+      "q = 800./10; \t\t\t#data given:800 Btu as heat is added to 10 LBm \t\t\t#Unit:Btu/LBm\n",
+      "T1 = 100; \t\t\t#Unit:Fahrenheit \t\t\t#Initial temperature \t\t\t#T2 = Final temperature\n",
+      "\n",
+      "# Calculations\n",
+      "#For a non-flow,constant pressure process\n",
+      "#q = deltah = h2-h1 = Cp(T2-T1) \t\t\t#deltah = change in enthalpy\n",
+      "#deltaT = T2-T1;\n",
+      "deltaT = q/Cp; \t\t\t#Fahrenheit  \t\t\t#change in temperature\n",
+      "T2 = deltaT+T1; \t\t\t#Fahrenheit  \t\t\t#final temperature\n",
+      "#For a constant volume pressure\n",
+      "#u2-u1 = Change in internal energy \t\t\t#w = workdone\n",
+      "#q-w = u2-u1\n",
+      "#-w = (u2-u1)-q  =  Cv*(T2-T1)-q\n",
+      "w = -(Cv*(T2-T1)-q); \t\t\t#Unit:Btu/lbm \t\t\t#workdone\n",
+      "\n",
+      "# Results\n",
+      "print \"%.2f Btu/lbm work is taken out of the system due to workdone by gas\"%(w);\n",
+      "print \"As there is 10 lbm in the system\"\n",
+      "print \"%.2f Btu work is taken out of the system due to workdone by gas\"%(w*10);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "18.18 Btu/lbm work is taken out of the system due to workdone by gas\n",
+        "As there is 10 lbm in the system\n",
+        "181.82 Btu work is taken out of the system due to workdone by gas\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11  Page No : 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Given data\n",
+      "#                    Inlet     Outlet\n",
+      "#Pressure(psia)      1000      1\n",
+      "#Temperature(F)      1000      101.74\n",
+      "#Velocity(ft/s)      125       430\n",
+      "#Inlet position(ft)  +10       0\n",
+      "#Enthalpy(Btu/LBm)   1505.4    940.0\n",
+      "#Steam flow rate of 150000 LBm/hr\n",
+      "\n",
+      "#From the table,\n",
+      "Z1 = 10; \n",
+      "V1 = 125; \n",
+      "h1 = 1505.4; \n",
+      "Z2 = 0; \n",
+      "V2 = 430; \n",
+      "h2 = 940.0;\n",
+      "\n",
+      "#Energy equation is given by\n",
+      "#((Z1/J)*(g/gc)) + (V1**2/(2*gc*J)) + h1 + q  =  ((Z2/J)*(g/gc)) + (V2**2/(2*gc*J)) + h2 + w/J\n",
+      "print \"Solution for a \";\n",
+      "q = 0; \t\t\t#net heat\n",
+      "J = 778.; \t\t\t#Conversion factor\n",
+      "gc = 32.174; \t\t\t#Unit: (LBm*ft)/(LBf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "g = gc; \t\t\t#Unit:ft/s**2 \t\t\t#g = The local gravity\n",
+      "#W1 = w/J;\n",
+      "#Energy equation is given by\n",
+      "W1 = ((Z1/J)*(g/gc)) + (V1**2/(2*gc*J)) + h1 + q - ((Z2/J)*(g/gc)) - (V2**2/(2*gc*J)) - h2; \t\t\t#Unit:Btu/LBm\n",
+      "print \"If heat losses are negligible\"\n",
+      "print \"Total work of the turbine is %.2f Btu/LBm\"%(W1);\n",
+      "print \"Total work of the turbine is %.2f Btu/hr\"%(W1*150000); \n",
+      "#(W*150000*778)/(60*33000) \t\t\t#in terms of horsepower  \t\t\t#1 hr = 60 min \t\t\t#1 hp = 33000 (ft*LBf)\n",
+      "print \"Total work of the turbine is %.2f hp \"%(((W1*150000*778)/60*33000)); \n",
+      "#1 hp  = 0.746 kW\n",
+      "print \"Total work of the turbine is %.2f kW \"%((((W1*150000*778)/60*33000))*0.746);\n",
+      "\n",
+      "\n",
+      "print \"Solution for b \";\n",
+      "#Heat losses equal 50,000 Btu/hr\n",
+      "q = 50000./150000; \t\t\t#Unit:Btu/LBm \t\t\t#Heat loss\n",
+      "W2 = ((Z1/J)*(g/gc)) + (V1**2/(2*gc*J)) + h1 - q - ((Z2/J)*(g/gc)) - (V2**2/(2*gc*J)) - h2; \t\t\t#Unit:Btu/LBm\n",
+      "print \"If heat losses equal 50,000 Btu/hr , Total work of the turbine is %.2f Btu/LBm\"%(W2);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a \n",
+        "If heat losses are negligible\n",
+        "Total work of the turbine is 562.03 Btu/LBm\n",
+        "Total work of the turbine is 84304739.48 Btu/hr\n",
+        "Total work of the turbine is 36073998024056.70 hp \n",
+        "Total work of the turbine is 26911202525946.29 kW \n",
+        "Solution for b \n",
+        "If heat losses equal 50,000 Btu/hr , Total work of the turbine is 561.70 Btu/LBm\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12  Page No : 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Z1 = 2; \t\t\t#Unit:m \t\t\t#Inlet position\n",
+      "g = 9.81 \t\t\t#Unit:m/s**2 \t\t\t#g = The local gravity\n",
+      "V1 = 40; \t\t\t#Unit:m/s \t\t\t#Inlet velocity\n",
+      "h1 = 3433.8; \t\t\t#Unit:kJ/kg \t\t\t#Inlet enthalpy\n",
+      "q = 1 \t\t\t#Unit:kJ/kg \t\t\t#Heat losses\n",
+      "Z2 = 0; \t\t\t#Outlet position \t\t\t#unit:m\n",
+      "V2 = 162; \t\t\t#Unit:m/s \t\t\t#Outlet velocity\n",
+      "h2 = 2675.5; \t\t\t#Unit:kJ/kg \t\t\t#Outlet enthalpy\n",
+      "\n",
+      "#Energy equation is given by\n",
+      "#((Z1*g)) + (V1**2/2) + h1 + q  =  ((Z2*g) + (V2**2/2) + h2 + w\n",
+      "\n",
+      "w =  ((Z1*g)/1000) + ((V1**2/2)/1000) + h1 - q - ((Z2*g)/1000) - ((V2**2/2)/1000) - h2 ; \t\t\t#Unit:kJ/kg \t\t\t#Conersation: 1 kJ = 1000 J\n",
+      "print \"The work output per kimath.logram is %.2f kJ/kg\"%(w);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The work output per kimath.logram is 744.32 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13  Page No : 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "p1 = 150; \t\t\t#Unit:psia \t\t\t#Initial pressure\n",
+      "T1 = 1000; \t\t\t#Unit:R \t\t\t#Temperature at pressure p1\n",
+      "p2 = 15; \t\t\t#Unit:psia \t\t\t#Final pressure\n",
+      "T2 = 600; \t\t\t#Unit:R \t\t\t#Temperature at pressure p2\n",
+      "Cp = 0.24; \t\t\t#Unit:Btu/(LBm*R) \t\t\t#Specific heat for constant pressure process\n",
+      "v1 = 2.47; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at inlet conditions\n",
+      "v2 = 14.8; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at outlet conditions\n",
+      "\n",
+      "#For a non-flow,constant pressure process\n",
+      "#w/J = deltah = h2-h1 = Cp(T2-T1) \t\t\t#deltah = change in enthalpy\n",
+      "#W = w/J\n",
+      "W = Cp*(T1-T2); \t\t\t#W = Work output \t\t\t#Unit:Btu/LBm\n",
+      "\n",
+      "# Results\n",
+      "print \"The work output of the turbine per pound of working fluid is %.2f Btu/LBm\"%(W);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The work output of the turbine per pound of working fluid is 96.00 Btu/LBm\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14  Page No : 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#In problem 3.13 ,\n",
+      "p1 = 150.; \t\t\t#Unit:psia \t\t\t#Initial pressure\n",
+      "T1 = 1000.; \t\t\t#Unit:R \t\t\t#Temperature at pressure p1\n",
+      "p2 = 15.; \t\t\t#Unit:psia \t\t\t#Final pressure\n",
+      "T2 = 600.; \t\t\t#Unit:R \t\t\t#Temperature at pressure p2\n",
+      "Cp = 0.24; \t\t\t#Unit:Btu/(LBm*R) \t\t\t#Specific heat for constant pressure process\n",
+      "v1 = 2.47; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at inlet conditions\n",
+      "v2 = 14.8; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at outlet conditions\n",
+      "\n",
+      "#For a non-flow,constant pressure process\n",
+      "#w/J = deltah = h2-h1 = Cp(T2-T1) \t\t\t#deltah = change in enthalpy\n",
+      "#W = w/J\n",
+      "W = Cp*(T1-T2); \t\t\t#W = Work output \t\t\t#Unit:Btu/LBm \t\t\t#h2-h1\n",
+      "print \"In problem 3.13, The work output of the turbine per pound of working fluid is %.2f Btu/LBm  \"%(W);\n",
+      "\n",
+      "#Now,In problem 3.14 , \n",
+      "q = 1.1; \t\t\t#Unit:Btu/LBm \t\t\t#Heat losses\n",
+      "#For a non-flow,constant pressure process\n",
+      "#q-w/J = deltah = h2-h1 = Cp(T2-T1) \t\t\t#deltah = change in enthalpy\n",
+      "#W1 = w/J\n",
+      "W1 = -q+W; \t\t\t#W = Work output \t\t\t#Unit:Btu/LBm \t\t\t#W = h2-h1 \t\t\t#Because q  is out of the system,it is a negative quantity\n",
+      "print \"In problem 3.14%(heat loss equal to 1.1 Btu/LBm\"\n",
+      "print \"The work output of the turbine per pound of working fluid is %.2f Btu/LBm \"%(W1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In problem 3.13, The work output of the turbine per pound of working fluid is 96.00 Btu/LBm  \n",
+        "In problem 3.14%(heat loss equal to 1.1 Btu/LBm\n",
+        "The work output of the turbine per pound of working fluid is 94.90 Btu/LBm \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15  Page No : 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "p1 = 100.; \t\t\t#Unit:psia \t\t\t#Initial pressure\n",
+      "t1 = 950.; \t\t\t#Unit:Fahrenheit \t\t\t#Temperature at pressure p1\n",
+      "p2 = 76.; \t\t\t#Unit:psia \t\t\t#Final pressure\n",
+      "t2 = 580.; \t\t\t#Unit:Fahrenheit \t\t\t#Temperature at pressure p2\n",
+      "v1 = 4.; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at inlet conditions\n",
+      "v2 = 3.86; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at outlet conditions\n",
+      "Cv = 0.32; \t\t\t#Unit:Btu/(LBm*R) \t\t\t#Specific heat for constant volume process\n",
+      "\n",
+      "T1 = t1+460; \t\t\t#Unit:R \t\t\t#Temperature at pressure p1\n",
+      "T2 = t2+460; \t\t\t#Unit:R \t\t\t#Temperature at pressure p2\n",
+      "J = 778.; \t\t\t#J = Conversion factor\n",
+      "\n",
+      "#Z1 = Inlet position \t\t\t#Unit:m \n",
+      "#V1 = Inlet velocity \t\t\t#Unit:m/s\n",
+      "#Z2 = Outlet position \t\t\t#Unit:m \n",
+      "#V2 = Outlet velocity Unit:m/s \n",
+      "#u1 = internal energy \t\t\t#energy in\n",
+      "#u2 = internal energy \t\t\t#energy out\n",
+      "\n",
+      "#Energy equation is given by\n",
+      "#((Z1/J)*(g/gc)) + (V1**2/(2*gc*J)) + u1 + ((p1*v1)/J) + q  =  ((Z2/J)*(g/gc)) + (V2**2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; \t\t\t#Unit:Btu/LBm\n",
+      "#Because pipe is horizontal and velocity terms are to be neglected, \n",
+      "# Also no work crosses the boundaries of the system, the energy equation is reduced to\n",
+      "#u1 + ((p1*v1)/J) + q  =  u2 + ((p2*v2)/J)\n",
+      "#u2-u1 = Cv*(T2-T1) \t\t\t#For a constant volume process \t\t\t#u2-u1 = Chnage in internal energy\n",
+      "#So,\n",
+      "q = Cv*(T2-T1) + (p2*v2*144)/J - (p1*v1*144)/J;  \t\t\t#q = heat transfer \t\t\t#1 ft**2 = 144 in**2 \t\t\t#Unit:Btu/LBm\n",
+      "\n",
+      "# Results\n",
+      "print \"%.2f Btu/LBm heat is transferred from the gas \"%(q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-138.14 Btu/LBm heat is transferred from the gas \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16  Page No : 116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#In problem 3.15,\n",
+      "p1 = 100; \t\t\t#Unit:psia \t\t\t#Initial pressure\n",
+      "t1 = 950; \t\t\t#Unit:Fahrenheit \t\t\t#Temperature at pressure p1\n",
+      "p2 = 76; \t\t\t#Unit:psia \t\t\t#Final pressure\n",
+      "t2 = 580; \t\t\t#Unit:Fahrenheit \t\t\t#Temperature at pressure p2\n",
+      "v1 = 4; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at inlet conditions\n",
+      "v2 = 3.86; \t\t\t#Unit:ft**3/LBm \t\t\t#Specific volume at outlet conditions\n",
+      "Cv = 0.32; \t\t\t#Unit:Btu/(LBm*R) \t\t\t#Specific heat for constant volume process\n",
+      "\n",
+      "T1 = t1+460; \t\t\t#Unit:R \t\t\t#Temperature at pressure p1\n",
+      "T2 = t2+460; \t\t\t#Unit:R \t\t\t#Temperature at pressure p2\n",
+      "J = 778.; \t\t\t#J = Conversion factor\n",
+      "gc = 32.174; \t\t\t#Unit: (LBm*ft)/(LBf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "g = gc; \t\t\t#Unit:ft/s**2 \t\t\t#g = The local gravity\n",
+      "\n",
+      "#Z1 = Inlet position \t\t\t#Unit:m \n",
+      "#V1 = Inlet velocity \t\t\t#Unit:m/s\n",
+      "#Z2 = Outlet position \t\t\t#Unit:m \n",
+      "#V2 = Outlet velocity Unit:m/s \n",
+      "#u1 = internal energy \t\t\t#energy in\n",
+      "#u2 = internal energy \t\t\t#energy out\n",
+      "\n",
+      "#Energy equation is given by\n",
+      "#((Z1/J)*(g/gc)) + (V1**2/(2*gc*J)) + u1 + ((p1*v1)/J) + q  =  ((Z2/J)*(g/gc)) + (V2**2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; \t\t\t#Unit:Btu/LBm\n",
+      "#In 3.15, the elevation of the pipe at section 1 makes Z1  =  0\n",
+      "# Also no work crosses the boundaries of the system, the energy equation is reduced to\n",
+      "#u1 + ((p1*v1)/J) + q  =  u2 + ((p2*v2)/J) + ((Z2/J)*(g/gc))\n",
+      "#In problrm 3.16,\n",
+      "Z2 = 100; \t\t\t#Given \t\t\t#Unit:ft \t\t\t#Outlet position\n",
+      "#u2-u1 = Cv*(T2-T1) \t\t\t#For a constant volume process \t\t\t#u2-u1 = Chnage in internal energy\n",
+      "#So,\n",
+      "q = Cv*(T2-T1) + (p2*v2*144)/J - (p1*v1*144)/J + ((Z2/J)*(g/gc)) ;  \t\t\t#q = heat transfer \t\t\t#1 ft**2 = 144 in**2 \t\t\t#Unit:Btu/LBm\n",
+      "\n",
+      "# Results\n",
+      "print \"%.2f Btu/LBm heat is transferred from the gas \"%(q);\n",
+      "#For this problem , neglecting the elevation term leads to an insignificant error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "-138.01 Btu/LBm heat is transferred from the gas \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17  Page No : 117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "p1 = 1000; \t\t\t#Unit:psia \t\t\t#Initial pressure\n",
+      "t1 = 100;  \t\t\t#Unit:Fahrenheit \t\t\t#Temperature at pressure p1\n",
+      "p2 = 1000; \t\t\t#Unit:psia \t\t\t#Final pressure\n",
+      "t2 = 1000; \t\t\t#Unit:Fahrenheit \t\t\t#Temperature at pressure p2\n",
+      "# feed in 10,000 LBm/hr \n",
+      "h1 = 70.68  \t\t\t#Unit:Btu/LBm \t\t\t#Inlet enthalpy\n",
+      "h2 = 1505.9 \t\t\t#Unit:Btu/LBm \t\t\t#Outlet enthalpy\n",
+      "\n",
+      "T1 = t1+460; \t\t\t#Unit:R \t\t\t#Temperature at pressure p1\n",
+      "T2 = t2+460; \t\t\t#Unit:R \t\t\t#Temperature at pressure p2\n",
+      "#Energy equation is given by\n",
+      "J = 778; \t\t\t#J = Conversion factor\n",
+      "\n",
+      "#Z1 = Inlet position \t\t\t#Unit:m \n",
+      "#V1 = Inlet velocity \t\t\t#Unit:m/s\n",
+      "#Z2 = Outlet position \t\t\t#Unit:m \n",
+      "#V2 = Outlet velocity Unit:m/s \n",
+      "#u1 = internal energy \t\t\t#energy in\n",
+      "#u2 = internal energy \t\t\t#energy out\n",
+      "#h = enthalpy\n",
+      "\n",
+      "#Energy equation is given by\n",
+      "#((Z1/J)*(g/gc)) + (V1**2/(2*gc*J)) + u1 + ((p1*v1)/J) + q  =  ((Z2/J)*(g/gc)) + (V2**2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; \t\t\t#Unit:Btu/LBm\n",
+      "\n",
+      "#we can consider this system as a math.single unit with feed water entering ans steam leaving. \n",
+      "#It well designed,this unit will be thoroughly insulated,and heat losse will be reduced to a negligible amount\n",
+      "#Alos,no work will be added to the fluid during the time it is pasmath.sing through the unit, and kinetic energy differences will be assumed to be negligibly small\n",
+      "#Differennces in elevation also be considered negligible\n",
+      "#So,the energy equation is reduced to \n",
+      "#u1 + ((p1*v1)/J) + q  =  u2 + ((p2*v2)/J)\n",
+      "#Because h = u+(p*v/J)\n",
+      "q = h2-h1; \t\t\t#q = net heat losses \t\t\t#Unit:Btu/LBm\n",
+      "\n",
+      "# Results\n",
+      "print \"Net heat losses is %.2f Btu/LBm \"%(q);\n",
+      "print \"For 10000 LBm/hr\"\n",
+      "print \"%.2f Btu/hr energy has been added to the water to convert it to steam\"%(q*10000)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Net heat losses is 1435.22 Btu/LBm \n",
+        "For 10000 LBm/hr\n",
+        "14352200.00 Btu/hr energy has been added to the water to convert it to steam\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18  Page No : 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "h1 = 1220  \t\t\t#Unit:Btu/LBm \t\t\t#Inlet enthalpy\n",
+      "h2 = 1100 \t\t\t#Unit:Btu/LBm \t\t\t#Outlet enthalpy\n",
+      "\n",
+      "#Z1 = Inlet position \t\t\t#Unit:m \n",
+      "#V1 = Inlet velocity \t\t\t#Unit:m/s\n",
+      "#Z2 = Outlet position \t\t\t#Unit:m \n",
+      "#V2 = Outlet velocity Unit:m/s \n",
+      "#u1 = internal energy \t\t\t#energy in\n",
+      "#u2 = internal energy \t\t\t#energy out\n",
+      "J = 778; \t\t\t#J = Conversion factor\n",
+      "gc = 32.174; \t\t\t#Unit: (LBm*ft)/(LBf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "\n",
+      "#Energy equation is given by\n",
+      "#((Z1/J)*(g/gc)) + (V1**2/(2*gc*J)) + u1 + ((p1*v1)/J) + q  =  ((Z2/J)*(g/gc)) + (V2**2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; \t\t\t#Unit:Btu/LBm\n",
+      "\n",
+      "#For this device,differences in elevation are negligible.No work is done on or by the fluid,friction is negligible\n",
+      "#And due to the speed of the fluid flowing and the short length of the nozzle,heat transfer to or from the surroundings is also negligible.\n",
+      "#So,the energy equation is reduced to \n",
+      "#u1 + ((p1*v1)/J) +(V1**2/(2*gc*J)   =  u2 + ((p2*v2)/J) + (V2**2/(2*gc*J)\n",
+      "# h1-h2  =  ((V2**2-V1**2)/(2*gc*J))\n",
+      "\n",
+      "print \"Solution for a\";\n",
+      "#For neglegible entering velocity, V1 = 0\n",
+      "#So,\n",
+      "V2 = math.sqrt((2*gc*J)*(h1-h2)); \t\t\t#the final velocity  \t\t\t#ft/s\n",
+      "print \"It the initial velocity of the system is negligible, the final velocity is %.2f ft/s  \"%(V2);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "#If the initial velocity is appreciable,\n",
+      "V1 = 1000; \t\t\t#Unit:ft/s \t\t\t#the initial velocity \n",
+      "V2 = math.sqrt(((h1-h2)*(2*gc*J)) + V1**2 ) ;\n",
+      "print \"It the initial velocity of the system is appreciable, the final velocity is %.2f ft/s  \"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "It the initial velocity of the system is negligible, the final velocity is 2451.03 ft/s  \n",
+        "Solution for b\n",
+        "It the initial velocity of the system is appreciable, the final velocity is 2647.17 ft/s  \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19  Page No : 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "h1 = 3450*1000 \t\t\t#Unit:J/kg \t\t\t#Enthalpy of steam when it enters a nozzle\n",
+      "h2 = 2800*1000 \t\t\t#Unit:J/kg \t\t\t#Enthalpy of steam when it leaves a nozzle\n",
+      "\n",
+      "# Calculations\n",
+      "#V2**2/2 = h1-h2;\n",
+      "V2 = math.sqrt(2*(h1-h2)); \t\t\t#V2 = Final velocity \t\t\t#Unit:m/s\n",
+      "\n",
+      "# Results\n",
+      "print \"Final velocity  =  %.2f m/s\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final velocity  =  1140.18 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21  Page No : 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "m = 400; \t\t\t#Unit:LBm/min \t\t\t#mass of lubricating oil\n",
+      "Cp = 0.85; \t\t\t#Unit:Btu/LBm*R \t\t\t#Specific heat of the oil\n",
+      "T1 = 215; \t\t\t#Temperature when hot oil is entering \t\t\t#Unit:Fahrenheit\n",
+      "T2 = 125; \t\t\t#Temperature when hot oil is leaving \t\t\t#Unit:Fahrenheit\n",
+      "\n",
+      "# Calculations and Results\n",
+      "DeltaT = T2-T1; \t\t\t#Unit:Fahrenheit  \t\t\t#change in temperature\n",
+      "Qoil = m*Cp*DeltaT; \t\t\t#Heat out of oil \t\t\t#Btu/min\n",
+      "print \"Heat out of oil is %.2f Btu/min Out of oil\"%(Qoil);\n",
+      "#Heat out of oil is the heat into the water\n",
+      "#Mw = Water flow rate\n",
+      "#M*Cpw*DeltaTw = Qoil\n",
+      "Cpw = 1.0; \t\t\t#Unit:Btu/LBm*R \t\t\t#Specific heat of the water\n",
+      "T3 = 60.; \t\t\t#Temperature when water is entering \t\t\t#Unit:Fahrenheit\n",
+      "T4 = 90; \t\t\t#Temperature when water is leaving \t\t\t#Unit:Fahrenheit\n",
+      "DeltaTw = T4-T3; \t\t\t#Unit:Fahrenheit \t\t\t#change in temperature\n",
+      "Mw = Qoil/(Cpw*DeltaTw); \t\t\t#The Required water flow rate  \t\t\t#Unit;lbm/Min\n",
+      "print \"The Required water flow rate is %.2f lbm/Min\"%(abs(Mw));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat out of oil is -30600.00 Btu/min Out of oil\n",
+        "The Required water flow rate is 1020.00 lbm/Min\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch4_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch4_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch4_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch4_2.ipynb
new file mode 100755
index 00000000..d0e5bbf4
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch4_2.ipynb
@@ -0,0 +1,656 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:81c0b2f6374783dbad513f5ebbb51829b49d5e29c2d28b8cf1616c5d410d60b5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : The Second Law of Thermodynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1  Page No : 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "t1 = 1000.;  \t\t\t#(unit:fahrenheit) \t\t\t#Source temperature\n",
+      "t2 = 80.;   \t\t\t#(unit:fahrenheit) \t\t\t#Sink temperature\n",
+      "#solution\n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t1+460; \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+460; \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "\n",
+      "print \"Solution for a\";\n",
+      "ans = ((T1-T2)/T1)*100;\t\t\t#(ans in %)  \t\t\t#Efficiency of the engine\n",
+      "print \"Efficiency of the engine is %.2f percentage\"%(ans);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "T1 = 2000+460; \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+460;  \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "ans = ((T1-T2)/T1)*100;\t\t\t#(ans in %) \t\t\t#Efficiency of the engine\n",
+      "print \"When the upper tempretrature is increased upto certain ,Efficiency of the engine is %.2f percentage \"%(ans);\n",
+      "\n",
+      "print \"Solution for c\";\n",
+      "T1 = t1+460; \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = 160+460;  \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "ans = ((T1-T2)/T1)*100;\t\t\t#(ans in %) \t\t\t#Efficiency of the engine\n",
+      "print \"When the lower tempretrature is increased upto certain ,Efficiency of the engine is %.2f percentage \"%(ans);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "Efficiency of the engine is 63.01 percentage\n",
+        "Solution for b\n",
+        "When the upper tempretrature is increased upto certain ,Efficiency of the engine is 78.05 percentage \n",
+        "Solution for c\n",
+        "When the lower tempretrature is increased upto certain ,Efficiency of the engine is 57.53 percentage \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2  Page No : 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "Qin = 100.; \t\t\t#heat added to the cycle \n",
+      "\n",
+      "print \"In problem 4.1\"\n",
+      "#given data\n",
+      "t1 = 1000.;  \t\t\t#(unit:fahrenheit) \t\t\t#Source temperature\n",
+      "t2 = 80.;   \t\t\t#(unit:fahrenheit)  \t\t\t#Sink temperature\n",
+      "#solution\n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t1+460; \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+460;  \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "print \"Solution for a\";\n",
+      "print \"Efficiency of the engine is %.2f percentage\"%(((T1-T2)/T1)*100);\n",
+      "\n",
+      "print \"Now in problem 4.2\"\n",
+      "W = 0.63*Qin; \t\t\t#W = W/J; \t\t\t#Efficiency in problem 4.1 \n",
+      "W = Qin*(W/Qin); \t\t\t#amount of work\n",
+      "Qr = Qin-W; \t\t\t#Qin-Qr = W/J  \t\t\t#Qr = heat rejected by the cycle\n",
+      "print \"The heat removed from the reservoir %.2f units\"%(Qr);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In problem 4.1\n",
+        "Solution for a\n",
+        "Efficiency of the engine is 63.01 percentage\n",
+        "Now in problem 4.2\n",
+        "The heat removed from the reservoir 37.00 units\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3  Page No : 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "t1 = 70.; \t\t\t#(unit:fahrenheit) \t\t\t#Source temperature\n",
+      "t2 = 15.; \t\t\t#(unit:fahrenheit)  \t\t\t#Sink temperature\n",
+      "Qin = 125000.; \t\t\t#(unit = Btu/hr) \t\t\t#Qin = heat added to the cycle\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t1+460; \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+460; \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "Qr = Qin*(T2/T1); \t\t\t#Qr = heat rejected by the cycle\n",
+      "print \"Qr is %.2f in Btu/hr\"%(Qr);\n",
+      "work = Qin-Qr; \t\t\t#reversed cycle requires atleast input \t\t\t#work \t\t\t#btu/hr\n",
+      "print \"Work is %.2f in Btu/hr\"%(work);\n",
+      "# 1 hp  =  33000 ft*LBf/min \n",
+      "# 1 Btu  =  778 ft*LBf \t\t\t#1 hr  =  60 min\n",
+      "print \"Minimum horsepower input required is %.2f hp\"%((work*778/60*33000));\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Qr is 112028.30 in Btu/hr\n",
+        "Work is 12971.70 in Btu/hr\n",
+        "Minimum horsepower input required is 5550589622.64 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4  Page No : 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "W = (50.*33000)/778;\t\t\t#output \t\t\t#W = W/J\n",
+      "# 1 hp  =  33000 ft*LBf/min \n",
+      "# 1 Btu  =  778 ft*LBf\n",
+      "print \"Output is %.2f in Btu/min\"%(W);\n",
+      "t1 = 1000.; \t\t\t#Source temperature \t\t\t#(unit:fahrenheit)\n",
+      "t2 = 100.;  \t\t\t#Sink temperature \t\t\t#(unit:fahrenheit)\n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t1+460; \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+460; \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "n = (1-(T2/T1))*100; \t\t\t#efficiency\n",
+      "print \"Efficiency is %.2f percentage\"%(n);\t\t\t#in %)\n",
+      "#n = (W/J)/Qin\n",
+      "Qin = W/(n/100);\t\t\t#(unit Btu/hr) \t\t\t#Qin = heat added to the cycle\n",
+      "print \"Heat added to the cycle is %.2f in  Btu/min\"%(Qin);\n",
+      "Qr = Qin*(1-(n/100));\t\t\t#(unit Btu/hr) \t\t\t#Qr = heat rejected by the cycle\n",
+      "print \"Heat rejected by the cycle is %.2f in  Btu/min \"%(Qr);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Output is 2120.82 in Btu/min\n",
+        "Efficiency is 61.64 percentage\n",
+        "Heat added to the cycle is 3440.45 in  Btu/min\n",
+        "Heat rejected by the cycle is 1319.62 in  Btu/min \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5  Page No : 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "t1 = 700.; \t\t\t#Source temperature \t\t\t#Unit:Celcius\n",
+      "t2 = 20.;  \t\t\t#Sink temperature \t\t\t#Unit:Celcius\n",
+      "#converting in F\n",
+      "T1 = t1+273;  \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+273; \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "n = (T1-T2)/T1*100; \t\t\t#Efficiency\n",
+      "print \"Efficiency is %.2f percentage\"%(n);\t\t\t#in %)\n",
+      "output = 65;\t\t\t#in hp \t\t\t#Given\n",
+      "work = output*0.746;\t\t\t#(unit kJ/s) \t\t\t# 1 hp  =  746 W\n",
+      "print \"Work is %.2f kJ/s\"%(work);\n",
+      "Qin = work/(n/100);\t\t\t#(unit kJ/s) \t\t\t#Qin = heat added to the cycle\n",
+      "print \"Heat added to the cycle is %.2f kJ/s \"%(Qin);\n",
+      "Qr = Qin*(1-(n/100));\t\t\t#(unit kJ/s) \t\t\t#Qr = heat rejected by the cycle\n",
+      "print \"Heat rejected by the cycle is %.2f  kJ/s \"%(Qr);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency is 69.89 percentage\n",
+        "Work is 48.49 kJ/s\n",
+        "Heat added to the cycle is 69.38 kJ/s \n",
+        "Heat rejected by the cycle is 20.89  kJ/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7  Page No : 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# given data\n",
+      "t1 = 700.; \t\t\t#(unit:fahrenheit) \t\t\t#Source temperature\n",
+      "t2 = 200.; \t\t\t#(unit:fahrenheit)  \t\t\t#Sink temperature\n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t1+460;  \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+460; \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "#n1 = (T1-Ti)/T1 and n2 = (Ti-T2)/Ti \t\t\t#n1 & n2 are efficiency\n",
+      "#(T1-Ti)/T1 = (Ti-T2)/Ti;\n",
+      "Ti = math.sqrt(T1*T2); \t\t\t#Exhaust temperature  \t\t\t#Unit:R\n",
+      "print \"Exhaust temperature of first engine is %.2f in R\"%(Ti);\n",
+      "#converting absolute temperature to normal F temperature\n",
+      "#Ti(fahrenheit) = Ti(R)-460;\n",
+      "print \"Exhaust temperature of first engine is %.2f fahrenheit\"%(Ti-460);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exhaust temperature of first engine is 874.99 in R\n",
+        "Exhaust temperature of first engine is 414.99 fahrenheit\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8  Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#For reversible isothermal process,\n",
+      "q = 843.7; \t\t\t#Heat \t\t\t#Unit:Btu \t\t\t#at 200 psia\n",
+      "t = 381.86; \t\t\t#(unit:fahrenheit) \t\t\t#temperature\n",
+      "#\t\t\t#converting temperatures to absolute temperatures;\n",
+      "T = t+460; \t\t\t#temperature \t\t\t#unit:R\n",
+      "deltaS = (q/T); \t\t\t#Change in entropy  \t\t\t#Unit:Btu/lbm*R\n",
+      "\n",
+      "# Results\n",
+      "print \"Change in entropy is %.2f  Btu/lbm*R\"%(deltaS); \t\t\t#1 LBm of saturated water\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in entropy is 1.00  Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9  Page No : 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#For reversible isothermal process,\n",
+      "#In problem 4.8,\n",
+      "q = 843.7; \t\t\t#Heat \t\t\t#Unit:Btu \t\t\t#at 200 psia\n",
+      "t = 381.86; \t\t\t#(unit:fahrenheit) \n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T = t+460; \t\t\t#Unit:R\"\n",
+      "deltaS = (q/T);  \t\t\t#Change in entropy \t\t\t#Btu/lbm\n",
+      "print \"Change in entropy is %.2f  Btu/lbm*R\"%(deltaS); \t\t\t#1 LBm of saturated water\n",
+      "\n",
+      "#In problem 4.9\n",
+      "t1 = 381.86; \t\t\t#(unit:fahrenheit) \t\t\t#Source temperature\n",
+      "t2 = 50.; \t\t\t#(unit:fahrenheit)  \t\t\t#Sink temperature\n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t1+460;  \t\t\t#Source temperature \t\t\t#Unit:R\n",
+      "T2 = t2+460; \t\t\t#Sink temperature \t\t\t#Unit:R\n",
+      "qin = q;\t\t\t#heat added to the cycle \n",
+      "n = (1-(T2/T1))*100; \t\t\t#Efficiency\n",
+      "print \"Efficiency is %.2f percentage\"%(n);\n",
+      "wbyJ = qin*n*0.01;\t\t\t#work output\n",
+      "print \"Work output is %.2f Btu/lbm\"%(wbyJ);\n",
+      "Qr = qin-wbyJ; \t\t\t#heat rejected\n",
+      "print \"Heat rejected is %.2f Btu/lbm\"%(Qr);\n",
+      "print \"As an alternative solution and refering to figure 4.12\"\n",
+      "qin = T1*deltaS; \t\t\t#heat added \t\t\t#btu/lbm\n",
+      "Qr = T2*deltaS; \t\t\t#Heat rejected \t\t\t#btu/lbm\n",
+      "print \"Heat rejected is %.2f Btu/lbm\"%(Qr);\n",
+      "wbyJ = qin-Qr; \t\t\t#Work output \t\t\t#Btu/lbm\n",
+      "print \"Work output is %.2f Btu/lbm\"%(wbyJ);\n",
+      "n = (wbyJ/qin)*100; \t\t\t#Efficiency\n",
+      "print \"Efficiency is %.2f percentage\"%(n);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in entropy is 1.00  Btu/lbm*R\n",
+        "Efficiency is 39.42 percentage\n",
+        "Work output is 332.59 Btu/lbm\n",
+        "Heat rejected is 511.11 Btu/lbm\n",
+        "As an alternative solution and refering to figure 4.12\n",
+        "Heat rejected is 511.11 Btu/lbm\n",
+        "Work output is 332.59 Btu/lbm\n",
+        "Efficiency is 39.42 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10  Page No : 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "hfg = 1959.7; \t\t\t#Unit:kJ/kg \t\t\t#Evaporative enthalpy\n",
+      "T = 195.07+273; \t\t\t#Converted into Kelvin \t\t\t#Temperature\n",
+      "\n",
+      "# Calculations\n",
+      "deltaS = hfg/T; \t\t\t#Change in entropy \t\t\t#kJ/kg*K\n",
+      "\n",
+      "# Results\n",
+      "print \"Change in entropy at 1.4MPa for the vaporization of 1 kg is %.2f kJ/kg*K\"%(deltaS); \t\t\t#Values compares very closely to the Steam Tables value\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in entropy at 1.4MPa for the vaporization of 1 kg is 4.19 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11  Page No : 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Let is assume that a Carnot engine cycle operates between two temperatures in each case.\n",
+      "t = 1000.; \t\t\t#(unit:fahrenheit) \n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t+460;\n",
+      "#T1*deltaS = Qin;\n",
+      "Qin = 100.; \t\t\t#Unit:Btu \t\t\t#heat added to the cycle \n",
+      "deltaS = Qin/T1; \t\t\t#Change in entropy \t\t\t#Btu/R\n",
+      "T2 = 50.+460; \t\t\t#converting 50 F temperature to absolute temperature;\n",
+      "Qr = T2*deltaS; \t\t\t#Heat rejected \t\t\t#Unit:Btu\n",
+      "print \"%.2f Btu energy is unavailable with respect to a receiver at 50 fahrenheit \"%(Qr);\n",
+      "T2 = 0+460; \t\t\t#converting 0 F temperature to absolute temperature;\n",
+      "Qr = T2*deltaS; \t\t\t#Heat rejected \t\t\t#unit:Btu\n",
+      "print \"%.2f Btu energy is unavailable with respect to a receiver at 0 fahrenheit \"%(Qr);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "34.93 Btu energy is unavailable with respect to a receiver at 50 fahrenheit \n",
+        "31.51 Btu energy is unavailable with respect to a receiver at 0 fahrenheit \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12  Page No : 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Qin = 1000; \t\t\t#Unit:Joule \t\t\t#heat entered to the system\n",
+      "t = 500; \t\t\t#(unit:Celcius)  \t\t\t#temperature\n",
+      "\t\t\t#converting temperature\n",
+      "T1 = t+273; \t\t\t#Unit:Kelvin\n",
+      "deltaS = Qin/T1; \t\t\t#Change in entropy \t\t\t#Unit:J/K\n",
+      "print \"Solution for a\"\n",
+      "T2 = 20+273; \t\t\t#converted 20 Celcius temperature to Kelvin;\n",
+      "Qr = T2*deltaS; \t\t\t#Heat rejected at 20 celcius \t\t\t#Joule\n",
+      "print \"%.2f Joule energy is unavailable with respect to a receiver at 20 Celcius\"%(Qr);\n",
+      "\n",
+      "print \"Solution for b\"\n",
+      "T2 = 0+273; \t\t\t#converted 0 Celcius temperature to Kelvin\n",
+      "Qr = T2*deltaS; \t\t\t#heat rejected at 0 celcius \t\t\t#Joule\n",
+      "print \"%.2f Joule energy is unavailable with respect to a receiver at 0 Celcius\"%(Qr);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "293.00 Joule energy is unavailable with respect to a receiver at 20 Celcius\n",
+        "Solution for b\n",
+        "273.00 Joule energy is unavailable with respect to a receiver at 0 Celcius\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13  Page No : 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#deltas = Cp*ln(T2/T1)\n",
+      "#Multiplying both the sides of equation by the mass m,\n",
+      "#DeltaS = m*Cp*ln(T2/T1)\n",
+      "m = 6.; \t\t\t#mass \t\t\t#Unit:lbm\n",
+      "Cp = 0.361; \t\t\t#Btu/lbm*R \t\t\t#Specific heat constant\n",
+      "DeltaS = -0.7062; \t\t\t#Unit:Btu/R \t\t\t#change in entropy\n",
+      "t = 1440.; \t\t\t#(unit:fahrenheit) \n",
+      "#converting temperatures to absolute temperatures;\n",
+      "T1 = t+460; \t\t\t#Unit:R\n",
+      "#Rearranging the equation,\n",
+      "T2 = T1*math.exp(DeltaS/(m*Cp)); \t\t\t#final temperature \t\t\t#Unit:R\n",
+      "\n",
+      "# Results\n",
+      "print \"Final temperature is %.2f R\"%(T2);\n",
+      "print \"or %.2f fahrenheit\"%(T2-460);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final temperature is 1371.38 R\n",
+        "or 911.38 fahrenheit\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14  Page No : 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#1 lbm of water at 500F is mixed with 1 lbm of water at 100F\n",
+      "m1 = 1.; \t\t\t#Unit:lbm \t\t\t#mass\n",
+      "m2 = 1.; \t\t\t#Unit:lbm \t\t\t#mass\n",
+      "c1 = 1.; \t\t\t#Specific heat constant\n",
+      "c2 = 1.; \t\t\t#Specific heat constant\n",
+      "t1 = 500.; \t\t\t#(unit:fahrenheit) \n",
+      "t2 = 100.; \t\t\t#(unit:fahrenheit)\n",
+      "cmix = 1.; \t\t\t#Specific heat constant of mixture\n",
+      "#now, m1*c1*t1 +m2*c2*t2  =  (m1+m2)*cmix*t\n",
+      "#So,\n",
+      "t = ((m1*c1*t1)+(m2*c2*t2))/((m1+m2)*cmix) \t\t\t#resulting temperature of the mixture\n",
+      "print \"The resulting temperature of the mixture is %.2f fahrenheit\"%(t);\n",
+      "#For this problem,the hot steam is cooled\n",
+      "deltas = cmix*math.log((t+460)/(t1+460)); \t\t\t#temperatures converted to absolute temperatures; \t\t\t#deltas = change in entropy \t\t\t#Unit:Btu/(lbm*R)\n",
+      "#The cold steam is heated\n",
+      "deltaS = cmix*math.log((t+460)/(t2+460)); \t\t\t#temperatures converted to absolute temperatures; \t\t\t#deltaS = change in entropy \t\t\t#Unit:Btu/(lbm*R)\n",
+      "print \"The net change in entropy is %.2f Btu/lbm*R)\"%(deltaS+deltas);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resulting temperature of the mixture is 300.00 fahrenheit\n",
+        "The net change in entropy is 0.07 Btu/lbm*R)\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15  Page No : 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#In problem 4.15,\n",
+      "#1 lbm of water at 500F is mixed with 1 lbm of water at 100F\n",
+      "m1 = 1; \t\t\t#Unit:lbm \t\t\t#mass\n",
+      "m2 = 1; \t\t\t#Unit:lbm \t\t\t#mass\n",
+      "c1 = 1; \t\t\t#Specific heat constant\n",
+      "c2 = 1; \t\t\t#Specific heat constant\n",
+      "t1 = 500; \t\t\t#(unit:fahrenheit)\n",
+      "t2 = 100; \t\t\t#(unit:fahrenheit)\n",
+      "cmix = 1; \t\t\t#Specific heat constant of mixture\n",
+      "#now, m1*c1*t1 +m2*c2*t2  =  (m1+m2)*cmix*t  \t\t\t#So,\n",
+      "t = ((m1*c1*t1)+(m2*c2*t2))/((m1+m2)*cmix) \t\t\t#resulting temperature of the mixture\n",
+      "print \"In problem 4.14, The resulting temperature of the mixture is %.2f fahrenheit\"%(t);\n",
+      "\n",
+      "#Now,in problem 4.15,taking 0F as a reference temperature,\n",
+      "#For hot fluid,\n",
+      "deltas = cmix*math.log((t1+460)/(0+460)); \t\t\t#temperatures converted to absolute temperatures; \t\t\t#deltas = change in entropy \t\t\t#Unit:Btu/(lbm*R)\n",
+      "#For cold fluid,\n",
+      "s = cmix*math.log((t2+460)/(0+460)); \t\t\t#temperatures converted to absolute temperatures; \t\t\t#s = change in entropy \t\t\t#Unit:Btu/(lbm*R)\n",
+      "#At final mixture temperature of t F,the entropy of each system above 0F is,for the hot fluid \n",
+      "s1 = cmix*math.log((t+460)/(0+460)); \t\t\t#temperatures converted to absolute temperatures; \t\t\t#s1 = change in entropy \t\t\t#Unit:Btu/(lbm*R)\n",
+      "#and for the cold fluid,\n",
+      "s2 = cmix*math.log((t+460)/(0+460)); \t\t\t#temperatures converted to absolute temperatures; \t\t\t#s2 = change in entropy \t\t\t#Unit:Btu/(lbm*R)\n",
+      "print \"The change in the entropy for hot fluid is %.2f Btu/lbm*R)\"%(s1-deltas);\n",
+      "print \"The change in the entropy for cold fluid is %.2f Btu/lbm*R)\"%(s2-s);\n",
+      "print \"The total change in entropy if %.2f Btu/lbm*R\"%(s1-deltas+s2-s);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In problem 4.14, The resulting temperature of the mixture is 300.00 fahrenheit\n",
+        "The change in the entropy for hot fluid is -0.69 Btu/lbm*R)\n",
+        "The change in the entropy for cold fluid is 0.00 Btu/lbm*R)\n",
+        "The total change in entropy if -0.69 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch5_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch5_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch5_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch5_2.ipynb
new file mode 100755
index 00000000..ef34cc8f
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch5_2.ipynb
@@ -0,0 +1,1536 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:5f6a0af8f766f00e3dc6895c71340c9032268be4509d3ee5581830a21a93add3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Properties of Liquids and Gases"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1  Page No : 182"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "p = 0.6988; \t\t\t#Unit:psia \t\t\t#absolute pressure\n",
+      "vg = 467.7; \t\t\t#Unit:ft**3/lbm \t\t\t#Saturated vapour specific volume\n",
+      "ug = 1040.2; \t\t\t#Unit:Btu/lbm \t\t\t#Saturated vapour internal energy\n",
+      "J = 778; \t\t\t#J = Conversion factor\n",
+      "# 1 Btu  =  778 ft*LBf\n",
+      "#h = u+(p*v)/J \n",
+      "hg = ug+((p*vg*144)/J); \t\t\t#The enthalpy of saturated steam \t\t\t#1 ft**2 = 144 in**2 \t\t\t#Btu/lbm\n",
+      "\n",
+      "# Results\n",
+      "print \"The enthalpy of saturated steam at 90 F is %.2f Btu/lbm\"%(hg); \t\t\t#The value is matched with the value in table 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of saturated steam at 90 F is 1100.69 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2  Page No : 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "p = 4.246; \t\t\t#Unit:kPa \t\t\t#absolute pressure\n",
+      "vg = 32.894; \t\t\t#Unit:m**3/kg \t\t\t#specific volume\n",
+      "ug = 2416.6; \t\t\t#Unit:kJ/kg \t\t\t#internal energy\n",
+      "J = 778; \t\t\t#J = Conversion factor\n",
+      "# 1 Btu  =  778 ft*LBf\n",
+      "#h = u+(p*v)\n",
+      "hg = ug+(p*vg); \t\t\t#The enthalpy of saturated steam \t\t\t#1 ft**2 = 144 in**2 \t\t\t#unit:kJ/kg\n",
+      "\n",
+      "# Results\n",
+      "print \"The enthalpy of saturated steam at 30 C is %.2f kJ/kg\"%(hg); \t\t\t#The value is matched with the value in table 1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of saturated steam at 30 C is 2556.27 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3  Page No : 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#The necessary interpolations are best done in tabular forms as shown:\n",
+      "#  p    hg\n",
+      "# 115  1190.4  table 2\n",
+      "# 118  1190.8  (hg)118 = 1190.8\n",
+      "# 120  1191.1\n",
+      "hg = 1190.4+(3./5)*(1191.1-1190.4); \t\t\t#Btu/lbm \t\t\t#enthaply\n",
+      "print \"The enthalpy of saturated steam at 118 psia is %.2f Btu/lbm\"%(hg);\n",
+      "\n",
+      "#  p    vg\n",
+      "# 115  3.884  table 2\n",
+      "# 118  3.792  (vg)118 = 3.790\n",
+      "# 120  3.730\n",
+      "vg = 3.884-(3./5)*(3.884-3.730); \t\t\t#ft**3/lbm \t\t\t#specific volume\n",
+      "print \"The specific volume of saturated steam at 118 psia is %.2f ft**3/lbm\"%(vg);\n",
+      "\n",
+      "#  p    sg\n",
+      "# 115  1.5921  table 2\n",
+      "# 118  1.5900  (sg)118 = 1.5900\n",
+      "# 120  1.5886\n",
+      "sg = 1.5921-(3./5)*(1.5921-1.5886); \t\t\t#entropy\n",
+      "print \"The entropy of saturated steam at 118 psia is %.2f\"%(sg);\n",
+      "\n",
+      "#  p    ug\n",
+      "# 115  1107.7  table 2\n",
+      "# 118  1108.06  (ug)118 = 1180.1\n",
+      "# 120  1108.3\n",
+      "ug = 1107.7-(3./5)*(1108.3-1107.7); \t\t\t#internal energy\n",
+      "print \"The internal energy of saturated steam at 118 psia is %.2f\"%(ug);\n",
+      "#The interpolation process that was done in tabular form for this problem can also be demonstated by refering to figure 5.8 for the specific volume.It will be \n",
+      "#seen that the results of this problem and the tabulated values are essentially in exact agreement and that linear interpolation is satisfactory in these tables.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of saturated steam at 118 psia is 1190.82 Btu/lbm\n",
+        "The specific volume of saturated steam at 118 psia is 3.79 ft**3/lbm\n",
+        "The entropy of saturated steam at 118 psia is 1.59\n",
+        "The internal energy of saturated steam at 118 psia is 1107.34\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4  Page No : 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#By defination,\n",
+      "#hg = ug+(p*vg)/J\n",
+      "#hf = uf+(p*vf)/J\n",
+      "#hfg  =  hg-hf  =  (ug-uf) + p*(vg-vf)/J  =  ufg + p*(vg-vf)/J\n",
+      "#From table 2 at 115 psia,\n",
+      "p = 115; \t\t\t#Unit:psia \t\t\t#absolute pressure\n",
+      "ufg = 798.8; \t\t\t#Unit:Btu/lbm \t\t\t#Evap. internal energy\n",
+      "ug = 3.884; \t\t\t#Unit:ft**3/lbm \t\t\t#Saturated vapour internal energy\n",
+      "vf = 0.017850; \t\t\t#Unit:ft**3/lbm \t\t\t#Saturated liquid specific volume\n",
+      "J = 778; \t\t\t#J = Conversion factor \t\t\t#Unit:ft*lbf/Btu\n",
+      "#1 ft**2 = 144 in**2\n",
+      "hfg = ufg+(p*144*(ug-vf))/J; \t\t\t#Evap. Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "\n",
+      "# Results\n",
+      "print \"hfg for saturated steam at 115 psia is %.2f Btu/lbm\"%(hfg); \t\t\t#The tabulated values are matched\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hfg for saturated steam at 115 psia is 881.09 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5  Page No : 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#From table 2 at 1.0 MPa,\n",
+      "p = 1000; \t\t\t#Unit:kN/m**2 \t\t\t#absolute pressure\n",
+      "ufg = 1822.0; \t\t\t#Unit:kJ/kg \t\t\t#Evap. internal energy\n",
+      "vf = 0.0011273; \t\t\t#Unit:m**3/kg \t\t\t#Saturated liquid specific volume\n",
+      "vg = 0.19444; \t\t\t#Unit:m**3/kg \t\t\t#Saturated vapour specific volume\n",
+      "\n",
+      "# Calculations\n",
+      "vfg = vg-vf;  \t\t\t#Evap. specific volume \t\t\t#m**3/kg\n",
+      "hfg = ufg+(p*vfg); \t\t\t#Evap. Enthalpy \t\t\t#Unit:kJ/kg\n",
+      "\n",
+      "# Results\n",
+      "print \"hfg for saturated steam at 1.0 MPa is %.2f kJ/kg\"%(hfg); \t\t\t#The tabulated values are matched\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hfg for saturated steam at 1.0 MPa is 2015.31 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6  Page No : 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#For constant-temperature,reversible vaporization, hfg = deltah = T*deltas = T*sfg\n",
+      "hfg = (388.12+460)*(1.1042); \t\t\t#Evap. Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "\n",
+      "# Results\n",
+      "print \"By considering the process to be a reversible, \\\n",
+      "\\nconstant-temperature, hfg for saturated steam at 115 psia is %.2f Btu/lbm\"%(hfg); \t\t\t#ans is wrong in the book\n",
+      "#Values are matched with tabulated values.Use of -459.67 F for absolute zero,which is the value used in table,gives almost exact agreement.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "By considering the process to be a reversible, \n",
+        "constant-temperature, hfg for saturated steam at 115 psia is 936.49 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7  Page No : 192"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Umath.sing Table 2 ans a quality of 80%(x = 0.8),we have\n",
+      "#at 120 psia\n",
+      "x = 0.8;\n",
+      "sf = 0.49201; \t\t\t#saturated liquid entropy \t\t\t#Unit:Btu/lbm*R\n",
+      "sfg = 1.0966; \t\t\t#Evap. Entropy \t\t\t#Unit:Btu/lbm*R \n",
+      "hf = 312.67; \t\t\t#saturated liquid enthalpy \t\t\t#Unit:Btu/lbm \n",
+      "hfg = 878.5; \t\t\t#Evap. Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "uf = 312.27; \t\t\t#saturated liquid internal energy \t\t\t#Unit:Btu/lbm \n",
+      "ufg = 796.0; \t\t\t#Unit:Btu/lbm \t\t\t#Evap. internal energy\n",
+      "vf = 0.017886; \t\t\t#Saturated liquid specific volume \t\t\t#Unit:ft**3/lbm \n",
+      "vfg = (3.730-0.017886); \t\t\t#evap. specific volume \t\t\t#Unit:ft**3/lbm \n",
+      "sx = sf+(x*sfg); \t\t\t#entropy \t\t\t#Btu/lbm*R\n",
+      "print \"Entropy of a wet steam mixture at 120 psia is %.2f Btu/lbm*R\"%(sx);\n",
+      "hx = hf+(x*hfg); \t\t\t#enthalpy  \t\t\t#Btu/lbm*R\n",
+      "print \"Enthalpy of a wet steam mixture at 120 psia is %.2f Btu/lbm\"%(hx);\n",
+      "ux = uf+(x*ufg); \t\t\t#internal energy \t\t\t#Btu/lbm*R\n",
+      "print \"Internal energy of a wet steam mixture at 120 psia is %.2f Btu/lbm\"%(ux);\n",
+      "vx = vf+(x*vfg); \t\t\t#/specific volume \t\t\t#ft**3/lbm\n",
+      "print \"Specific Volume of a wet steam mixture at 120 psia is %.2f ft**3/lbm\"%(vx);\n",
+      "#As a check,\n",
+      "J = 778; \t\t\t#ft*lbf/Btu \t\t\t#Conversion factor\n",
+      "px = 120; \t\t\t#psia \t\t\t#pressure\n",
+      "ux = hx-((px*vx*144)/J); \t\t\t#1 ft**2 = 144 in**2 \t\t\t#internal energy\n",
+      "print \"As a check\"\n",
+      "print \"Internal energy of a wet steam mixture at 120 psia is %.2f Btu/lbm\"%(ux);\n",
+      "print \"Which agrees with the values obtained above\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Entropy of a wet steam mixture at 120 psia is 1.37 Btu/lbm*R\n",
+        "Enthalpy of a wet steam mixture at 120 psia is 1015.47 Btu/lbm\n",
+        "Internal energy of a wet steam mixture at 120 psia is 949.07 Btu/lbm\n",
+        "Specific Volume of a wet steam mixture at 120 psia is 2.99 ft**3/lbm\n",
+        "As a check\n",
+        "Internal energy of a wet steam mixture at 120 psia is 949.11 Btu/lbm\n",
+        "Which agrees with the values obtained above\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8  Page No : 193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Using Table 2 ans a quality of 85%(x = 0.85),we have\n",
+      "#at 1.0 MPa\n",
+      "x = 0.85;\n",
+      "sf = 2.1387; \t\t\t#saturated liquid entropy \t\t\t#Unit:kJ/kg*K\n",
+      "sfg = 4.4487; \t\t\t#Evap. Entropy \t\t\t#Unit:kJ/kg*K \n",
+      "hf = 762.81; \t\t\t#saturated liquid enthalpy \t\t\t#Unit:kJ/kg\n",
+      "hfg = 2015.3; \t\t\t#Evap. Enthalpy \t\t\t#Unit:kJ/kg\n",
+      "uf = 761.68; \t\t\t#saturated liquid internal energy \t\t\t#Unit:kJ/kg\n",
+      "ufg = 1822.0; \t\t\t#Unit:kJ/kg \t\t\t#Evap. internal energy\n",
+      "vf = 1.1273; \t\t\t#Saturated liquid specific volume \t\t\t#Unit:m**3/kg \n",
+      "vfg = (194.44-1.1273); \t\t\t#evap. specific volume \t\t\t#Unit:m**3/kg \n",
+      "sx = sf+(x*sfg); \t\t\t#entropy \t\t\t#kJ/kg*K\n",
+      "print \"Entropy of a wet steam mixture at 1.0 MPa  is %.2f kJ/kg*K\"%(sx);\n",
+      "hx = hf+(x*hfg); \t\t\t#enthalpy \t\t\t#kJ/kg*K\n",
+      "print \"Enthalpy of a wet steam mixture at 1.0 MPa is %.2f kJ/kg\"%(hx);\n",
+      "ux = uf+(x*ufg); \t\t\t#internal energy  \t\t\t#kJ/kg*K\n",
+      "print \"Internal energy of a wet steam mixture at 1.0 MPa is %.2f kJ/kg\"%(ux);\n",
+      "vx = (vf+(x*vfg))*(0.001); \t\t\t#specific volume \t\t\t#m**3/kg\n",
+      "print \"Specific Volume of a wet steam mixture at 1.0 MPa is %.2f m**3/kg\"%(vx);\n",
+      "#As a check,\n",
+      "px = 10**6; \t\t\t#psia \t\t\t#pressure\n",
+      "ux = hx-((px*vx)/10**3); \t\t\t#1 ft**2 = 144 in**2 \t\t\t#internal energy\n",
+      "print \"As a check%\"\n",
+      "print \"Internal energy of a wet steam mixture at 120 psia is %.2f kJ/kg\"%(ux);\n",
+      "print \"Which agrees with the values obtained above\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Entropy of a wet steam mixture at 1.0 MPa  is 5.92 kJ/kg*K\n",
+        "Enthalpy of a wet steam mixture at 1.0 MPa is 2475.81 kJ/kg\n",
+        "Internal energy of a wet steam mixture at 1.0 MPa is 2310.38 kJ/kg\n",
+        "Specific Volume of a wet steam mixture at 1.0 MPa is 0.17 m**3/kg\n",
+        "As a check%\n",
+        "Internal energy of a wet steam mixture at 120 psia is 2310.37 kJ/kg\n",
+        "Which agrees with the values obtained above\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9  Page No : 193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#For the wet mixture,hx = hf+(x*hfg),solving for x gives us\n",
+      "#Using table 1,we have,\n",
+      "hx = 900; \t\t\t#Btu/lbm \t\t\t#Enthalpy of wet mixture at 90F\n",
+      "hf = 58.07; \t\t\t#Btu/lbm \t\t\t#saturated liquid enthalpy\n",
+      "hfg = 1042.7; \t\t\t#Btu/lbm \t\t\t#Evap. Enthalpy\n",
+      "\n",
+      "# Calculations\n",
+      "x = (hx-hf)/hfg; \t\t\t#quality\n",
+      "\n",
+      "# Results\n",
+      "print \"The quality is %.2f percentage of a wet steam at 90F\"%(x*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The quality is 80.75 percentage of a wet steam at 90F\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10  Page No : 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#For the wet mixture,hx = hf+(x*hfg),solving for x gives us\n",
+      "#Using table 1,we have,\n",
+      "hx = 2000; \t\t\t#kJ/kg \t\t\t#Enthalpy of wet mixture at 30 C\n",
+      "hf = 125.79; \t\t\t#kJ/kg \t\t\t#saturated liquid enthalpy\n",
+      "hfg = 2430.5; \t\t\t# \t\t\t#Evap. Enthalpy \t\t\t#kJ/kg\n",
+      "\n",
+      "# Calculations\n",
+      "x = (hx-hf)/hfg; \t\t\t#quality\n",
+      "\n",
+      "# Results\n",
+      "print \"The quality is %.2f percentage of a wet steam at 30 C\"%(x*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The quality is 77.11 percentage of a wet steam at 30 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11  Page No : 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The values of temperature and pressure are listed in Table 3(Figure 5.10) and can be read directly.\n",
+      "print \"Specific volume of superheated steam at 330 psia and 450F is v = 1.4691 ft**3/lbm\";\n",
+      "print \"Internal Energy of superheated steam at 330 psia and 450F is u = 1131.8 Btu/lbm\";\n",
+      "print \"Enthalpy of superheated steam at 330 psia and 450F is h = 1221.5 Btu/lbm\";\n",
+      "print \"Entropy of superheated steam at 330 psia and 450F is s = 1.5219 Btu/lbm*R\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specific volume of superheated steam at 330 psia and 450F is v = 1.4691 ft**3/lbm\n",
+        "Internal Energy of superheated steam at 330 psia and 450F is u = 1131.8 Btu/lbm\n",
+        "Enthalpy of superheated steam at 330 psia and 450F is h = 1221.5 Btu/lbm\n",
+        "Entropy of superheated steam at 330 psia and 450F is s = 1.5219 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12  Page No : 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The values of temperature and pressure are listed in Table 3(Figure 5.10) and can be read directly.\n",
+      "print \"Specific volume of superheated steam at 2.0 MPa and 240 C is v = 0.10845 m**3/lbm\";\n",
+      "print \"Internal Energy of superheated steam at 2.0 MPa and 240 C is u = 2659.6 kJ/kg\";\n",
+      "print \"Enthalpy of superheated steam at 2.0 MPa and 240 C is h = 2876.5 kJ/kg\";\n",
+      "print \"Entropy of superheated steam at 2.0 MPa and 240 C is s = 6.4952 kJ/kg*K\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specific volume of superheated steam at 2.0 MPa and 240 C is v = 0.10845 m**3/lbm\n",
+        "Internal Energy of superheated steam at 2.0 MPa and 240 C is u = 2659.6 kJ/kg\n",
+        "Enthalpy of superheated steam at 2.0 MPa and 240 C is h = 2876.5 kJ/kg\n",
+        "Entropy of superheated steam at 2.0 MPa and 240 C is s = 6.4952 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13  Page No : 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The necessary interpolations(between 450F and 460F at 330 psia) are best done in tabular forms as shown:\n",
+      "#  t    v\n",
+      "# 460  1.4945  \n",
+      "# 455  1.4818  \n",
+      "# 450  1.4691\n",
+      "v = 1.4691+(1./2)*(1.4945-1.4691); \t\t\t#ft**3/lbm \t\t\t#specific volume\n",
+      "print \"The specific volume of saturated steam at 330 psia & 455F is %.2f ft**3/lbm\"%(v);\n",
+      "\n",
+      "#  t    u\n",
+      "# 460  1137.0  \n",
+      "# 455  1134.4  \n",
+      "# 450  1131.8 \n",
+      "u = 1131.8+(1./2)*(1137.0-1131.8); \t\t\t#Btu/lbm \t\t\t#internal energy\n",
+      "print \"The internal energy of saturated steam at 330 psia & 455F is %.2f Btu/lbm\"%(u);\n",
+      "\n",
+      "#  t    h\n",
+      "# 460  1228.2  \n",
+      "# 455  1224.9  \n",
+      "# 450  1221.5\n",
+      "h = 1221.5+(1./2)*(1228.2-1221.5); \t\t\t#enthaply \t\t\t#Btu/lbm\n",
+      "print \"The enthalpy of saturated steam at 330 psia & 455F is %.2f Btu/lbm\"%(h);\n",
+      "\n",
+      "#  t    s\n",
+      "# 460  1.5293  \n",
+      "# 455  1.5256  \n",
+      "# 450  1.5219\n",
+      "s = 1.5219+(1./2)*(1.5293-1.5219); \t\t\t#entropy \t\t\t#Btu/lbm*R\n",
+      "print \"The entropy of saturated steam at 330 psia & 455F is %.2f Btu/lbm*R\"%(s);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The specific volume of saturated steam at 330 psia & 455F is 1.48 ft**3/lbm\n",
+        "The internal energy of saturated steam at 330 psia & 455F is 1134.40 Btu/lbm\n",
+        "The enthalpy of saturated steam at 330 psia & 455F is 1224.85 Btu/lbm\n",
+        "The entropy of saturated steam at 330 psia & 455F is 1.53 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14  Page No : 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From Table3, we first obtain the properties at 337 psia and 460 F and then 337 psia and 470 F.\n",
+      "#The necessary interpolations are best done in tabular forms as shown:\n",
+      "#Proceeding with the calculation,at 460 F,\n",
+      "#  p    v                                     \t\t\t#  p    h\n",
+      "# 340  1.4448                                 \t\t\t# 340  1226.7\n",
+      "# 337  1.4595                                 \t\t\t# 337  1227.2\n",
+      "# 335  1.4693                                 \t\t\t# 335  1227.5\n",
+      "v = 1.4696-(2./5)*(1.4693-1.4448);                \n",
+      "h = 1227.5-(2./5)*(1227.5-1226.7);\n",
+      "#ft**3/lbm \t\t\t#specific volume                   \t\t\t#Btu/lbm \t\t\t#enthaply\n",
+      "\n",
+      "#And at 470 F,\n",
+      "#  p    v                                     \t\t\t#  p    h\n",
+      "# 340  1.4693                                 \t\t\t# 340  1233.4\n",
+      "# 337  1.4841                                 \t\t\t# 337  1233.9\n",
+      "# 335  1.4940                                 \t\t\t# 335  1234.2\n",
+      "v = 1.4640-(2./5)*(1.4640-1.4693);                \n",
+      "h = 1234.2-(2./5)*(1234.2-1233.4);\n",
+      "#ft**3/lbm \t\t\t#specific volume                   \t\t\t#Btu/lbm \t\t\t#enthaply\n",
+      "\n",
+      "#Therefore,at 337 psia and 465 F\n",
+      "#  t    v                                    \t\t\t#  t    h\n",
+      "# 470  1.4841                                \t\t\t# 470  1233.9\n",
+      "# 465  1.4718                                \t\t\t# 465  1230.7\n",
+      "# 460  1.4595                                \t\t\t# 460  1227.5\n",
+      "v = 1.4595+(1./2)*(1.4841-1.4595);               \n",
+      "h = 1227.5+(1./2)*(1233.9-1227.5);\n",
+      "#ft**3/lbm \t\t\t#specific volume                  \t\t\t#Btu/lbm \t\t\t#enthaply\n",
+      "print \"At 465 F and 337 psia, specific volume = %.2f ft**3/lbm and enthalpy = %.2f Btu/lbm\"%(v,h);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At 465 F and 337 psia, specific volume = 1.47 ft**3/lbm and enthalpy = 1230.70 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15  Page No : 202"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The values of temperature and pressure are listed in Table 4(Figure 5.10) and can be read directly.\n",
+      "print \"Specific volume of subcooled water at 1000 psia and 300F is v = 0.017379 ft**3/lbm\";\n",
+      "print \"Internal Energy of subcooled water at 1000 psia and 300F is u = 268.24 Btu/lbm\";\n",
+      "print \"Enthalpy of subcooled water at 1000 psia and 300F is h = 271.46 Btu/lbm\";\n",
+      "print \"Entropy of subcooled water at 1000 psia and 300F is s = 0.43552 Btu/lbm*R\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specific volume of subcooled water at 1000 psia and 300F is v = 0.017379 ft**3/lbm\n",
+        "Internal Energy of subcooled water at 1000 psia and 300F is u = 268.24 Btu/lbm\n",
+        "Enthalpy of subcooled water at 1000 psia and 300F is h = 271.46 Btu/lbm\n",
+        "Entropy of subcooled water at 1000 psia and 300F is s = 0.43552 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16  Page No : 202"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#It is necessary to ontain the saturation values corresponding to 300 F.This is done by reading Table A.1 in Appendix 3,which gives\n",
+      "pf = 66.98; \t\t\t#psia \t\t\t#pressure\n",
+      "vf = 0.017448; \t\t\t#ft**3/lbm \t\t\t#specific volume\n",
+      "hf = 269.73; \t\t\t#Btu/lbm \t\t\t#enthaply\n",
+      "#Now,\n",
+      "p = 1000; \t\t\t#psia \t\t\t#pressure\n",
+      "J = 778; \t\t\t#Conversion factor \t\t\t#ft*lbf/Btu\n",
+      "#From eq.5.5,\n",
+      "h = hf+((p-pf)*vf*144)/J; \t\t\t#1ft**2 = 144 in**2 \t\t\t#The enthalpy of subcooled water \t\t\t#Btu/lbm\n",
+      "print \"The enthalpy of subcooled water is %.2f Btu/lbm\"%(h);\n",
+      "#The difference between this value and the value found in problem 5.15,expressed as a percentage is\n",
+      "percentoferror = (h-271.46)/271.46;\n",
+      "print \"Percent of error is %.2f\"%(percentoferror*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of subcooled water is 272.74 Btu/lbm\n",
+        "Percent of error is 0.47\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.25  Page No : 211"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#On a chart in Appendix 3,it is necessary to estimate the 90 F point on the saturation line.From the chart or the table in the upper left of the chart,we note that 90 F is between 1.4 and 1.5 in. of mercury.Estimating the intersection of this value with the saturation curve yields\n",
+      "print \"Enthalpy of saturated steam hg = 1100 Btu/lbm\";\n",
+      "#This is a good agreement with results of problem 5.1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enthalpy of saturated steam hg = 1100 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.26  Page No : 212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The Mollier chart has lines of constant moisture in the wet region which correspond to (1-x).Therefore,we read at 20% moisture(80% Quality) and 120 psia,\n",
+      "print \"The enthalpy of a wet steam mixture at 120 psia having quality 80 percent is 1015 Btu/lbm\";\n",
+      "#Which also agrees well with the calculated value in problem 5.7\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of a wet steam mixture at 120 psia having quality 80 percent is 1015 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.27  Page No : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Entering the Mollier chart at 900 Btu/lbm and estimating 90 F(near the 1.5-in. Hg dashed line) yields a constant moisture percent of 19.2%.\n",
+      "print \"The quality is %.2f percent\"%((1-0.192)*100);\n",
+      "#We show good agreement with the calculated value.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The quality is 80.80 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.28  Page No : 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From the chart,\n",
+      "print \"The enthalpy of steam at 330 psia is h = 1220 Btu/lbm\";\n",
+      "#Compared to 1221.5 Btu/lbm found in problem 5.11\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of steam at 330 psia is h = 1220 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.29  Page No : 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#We note that the steam is superheated.From the Mollier chart in SI units,\n",
+      "print \"The enthalpy h = 2876.5 kJ/kg and entropy s = 6.4952 kJ/kg*K\";\n",
+      "#Values are matched with problem 5.12\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy h = 2876.5 kJ/kg and entropy s = 6.4952 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.30  Page No : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Because neither pressure nor temperature is shown directly,it is necessary to estimate to obtain the desired value.\n",
+      "print \"The enthalpy of steam is h = 1231 Btu/lbm\";\n",
+      "#In problem 5.14,h = 1230.7 Btu/lbm,Which is matched here.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of steam is h = 1231 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.31  Page No : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Reading the chart at 30 C and saturation gives us,\n",
+      "print \"The enthalpy of saturated steam is hg = 2556 kJ/kg\";\n",
+      "#Which matches with value of problem 5.2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy of saturated steam is hg = 2556 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.32  Page No : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Reading the chart in wet region at 1.0 MPa and x = 0.85(moisture of 15%) gives us\n",
+      "print \"hx = 2476 kJ/kg and sx = 5.92 kJ/kg*K\";\n",
+      "#The chart does not give ux or vx directly"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hx = 2476 kJ/kg and sx = 5.92 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.33  Page No : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Locate 30 C on the saturation line.Now follow a line of constant pressure,which is also a line of constant temperature in wet region,until an enthalpy of 2000kJ/kg is reached.\n",
+      "print \"The moisture content is 23 percent or x = 77 percent\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The moisture content is 23 percent or x = 77 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.34  Page No : 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#We enter the chart in the superheat region at 2.0MPa and 240 C to read the enthalpy and entropy.This procedure gives \n",
+      "print \"Enthalpy h = 2877 kJ/kg and entropy s = 6.495 kJ/kg*K\";\n",
+      "#The other properties cant be obtained directly from the chart\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enthalpy h = 2877 kJ/kg and entropy s = 6.495 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.35  Page No : 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#As already noted,h1 = h2 for this process.On the Mollier chart,h2 is found to be 1170 Btu/lbm at 14.7 psia and 250 F.Proceeding to the left on the chart,the constant-enthalpy value of 1170 Btu/lbm to 150 psia yields a moisture of 3% or a quality of 97%.\n",
+      "#If we use the tables to obtain the solution to this problem,we would first obtain h2 from the superheated vapor tables as 1168.8 Btu/lbm.Because hx = hf+(x*hfg),we    obtain x as\n",
+      "hx = 1168.8; \t\t\t#Btu/lbm \n",
+      "hf = 330.75; \t\t\t#Btu/lbm \t\t\t#values of 150 psia\n",
+      "hfg = 864.2; \t\t\t#Btulbm \t\t\t#values of 150 psia\n",
+      "x = (hx-hf)/hfg; \t\t\t#Quality\n",
+      "print \"Moisture in the steam flowing in the pipe is %.2f percent\"%((1-x)*100);\n",
+      "print \"or quality of the steam is %.2f percent\"%(x*100);\n",
+      "#very often,it is necessary to perform multiple interpolations if the tables are used,and the Mollier chart yields results within the rquired accuracy for most engineering problems and saves considerable time.\n",
+      "#We can also use the computerised programs to solve this program.We first enter the 250F and 14.7 psia to obtain h of 1168.7 Btu/lbm.We then continue by entering h of 1168.7 Btu/lbm and p of 150 psia.The printout gives us x of 0.9699 or 97%.While the computer solution is quick and easy to use,you should still sketch out the    problem on an h-s or T-s diagram to show the path of the process.\n",
+      "\n",
+      "# Saturation Properties\n",
+      "#--------------------------\n",
+      "# T = 250.00 degF\n",
+      "# P = 29.814 psia\n",
+      "#       z         z1        zg\n",
+      "# v(ft**3/lbm)   0.01700   13.830\n",
+      "# h(Btu/lbm)     218.62   1164.1\n",
+      "# s(Btu/lbm*F)  0.3678    1.7001\n",
+      "# u(Btu/lbm)    218.52    1087.8\n",
+      "\n",
+      "#Thermo Properties\n",
+      "#------------------------\n",
+      "# T =  250.00 degF\n",
+      "# P =  14.700 psia\n",
+      "# v =  28.417 ft**3/lbm\n",
+      "# h =  1168.7 Btu/lbm\n",
+      "# s =  1.7831 Btu/lbm*F\n",
+      "# u =  1091.4 Btu/lbm\n",
+      "\n",
+      "# Saturation Properties\n",
+      "#--------------------------\n",
+      "# T = 340.06 degF\n",
+      "# P = 118.00 psia\n",
+      "#       z        z1        zg\n",
+      "# v(ft**3/lbm)   0.01787   3.7891\n",
+      "# h(Btu/lbm)    311.39    1190.7\n",
+      "# s(Btu/lbm*F)  0.4904    1.5899\n",
+      "# u(Btu/lbm)    311.00    1108.0\n",
+      "\n",
+      "#Thermo Properties\n",
+      "#------------------------\n",
+      "# T =  358.49 degF\n",
+      "# P =  150.00 psia\n",
+      "# v =  2.9248 ft**3/lbm\n",
+      "# h =  1168.7 Btu/lbm\n",
+      "# s =  1.5384 Btu/lbm*F\n",
+      "# u =  1087.5 Btu/lbm\n",
+      "# x =  0.9699\n",
+      "\n",
+      "#Region:Saturated\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Moisture in the steam flowing in the pipe is 3.03 percent\n",
+        "or quality of the steam is 96.97 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.36  Page No : 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Because the math.tank volume is 10 ft**3,the final specific volume of the steam is 10 ft**3/lbm.Interpolations in Table A.2 yield a final pressure of 42 psia.The heat    added is simply difference in internal energy between the two states.\n",
+      "u2 = 1093.0; \t\t\t#internal energy \t\t\t#Btu/lbm\n",
+      "u1 = 117.95; \t\t\t#internal energy \t\t\t#Btu/lbm\n",
+      "\n",
+      "# Calculations\n",
+      "q = u2-u1; \t\t\t#heat added \t\t\t#Btu/lbm\n",
+      "\n",
+      "# Results\n",
+      "print \"The final pressure is 42 psia and the heat added is %.2f Btu/lbm\"%(q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final pressure is 42 psia and the heat added is 975.05 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.37  Page No : 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The mass in the math.tank is constant,and the heat added will be the change in internal energy of the contents of the math.tank between the two states.The initial mass in     the math.tank is found as follows:\n",
+      "Vf = 45; \t\t\t#volume of water \t\t\t#ft**2\n",
+      "vf = 0.016715;\n",
+      "Vg = 15; \t\t\t#Volume of steam \t\t\t#ft**2\n",
+      "vg = 26.80;\n",
+      "mf = Vf/vf; \t\t\t#lbm\n",
+      "mg = Vg/vg; \t\t\t#lbm\n",
+      "total = mf+mg; \t\t\t#total mass\n",
+      "#The internal energy is the sum of the internal energy of the liquid plus vapor:\n",
+      "ug = 1077.6;\n",
+      "uf = 180.1;\n",
+      "Ug = mg*ug; \t\t\t#Btu\n",
+      "Uf = mf*uf; \t\t\t#Btu\n",
+      "Total = Ug+Uf; \t\t\t#total internal energy\n",
+      "print \"The total internal energy is %.2f Btu\"%(Total);\n",
+      "#Because the mass in the math.tank is constant,the final specific volume must equal the initial specific volume,or\n",
+      "vx = (Vf+Vg)/(mf+mg); \t\t\t#ft**3/lbm\n",
+      "#But vx = vf+(x*vfg).Therefore Using table A.2 at 800 psia,\n",
+      "vx = 0.022282;\n",
+      "vf = 0.02087;\n",
+      "vfg = 0.5691-0.02087;\n",
+      "x = (vx-vf)/vfg;\n",
+      "print \"The final amount of vapor is %.2f lbm\"%(x*total);  \t\t\t#x*total mass \n",
+      "mg = x*total;\n",
+      "print \"The final amount of liquid is %.2f lbm\"%(total-x*total); \t\t\t#total mass minus final amount of vapor \n",
+      "mf = total-(x*total);\n",
+      "#The final internal energy is found as before:\n",
+      "ug = 1115.0;\n",
+      "uf = 506.6;\n",
+      "Ug = mg*ug; \t\t\t#Btu\n",
+      "Uf = mf*uf; \t\t\t#Btu\n",
+      "Total1 = Ug+Uf;\n",
+      "difference = Total1-Total; \t\t\t#final internal energy-initial internal energy\n",
+      "#per unit mass heat added is,\n",
+      "print \"The heat added per unit is %.2f Btu/lbm\"%(difference/total); \t\t\t#the difference of internal energy/total mass\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total internal energy is 485467.03 Btu\n",
+        "The final amount of vapor is 6.94 lbm\n",
+        "The final amount of liquid is 2685.82 lbm\n",
+        "The heat added per unit is 327.88 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.38  Page No : 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#As shown in Fig. 5.21b,the process dscribed in this problem is a vertical line on the Mollier Chart.For 800 psia and 600F,the Mollier chart yeilds h1 = 1270 Btu/lbm and s1 = 1.485.Proceeding vertically down the chart at constant s to 200 psia yields a final enthalpy h2 = 1148 Btu/lbm.The change in enthaly Using the process is        1270-1148 = 122 Btu/lbm.\n",
+      "#We may also solve this problem Using the steam tables in Appendix 3.Thus,the enthalpy at 800 psia and 600 F is 1270.4 Btu/lbm,and its entropy is 1.4861 Btu/lbm*R.\n",
+      "#Because the process is isentropic,the final entropy at 200psia must be 1.4861.From the saaturation table,the entropy of saturated steam at 200 psia is 1.5464,which indicates the final steam condition must be wet because the entropy of the final steam is less than the entropy of saturation.Using the wet steam relation yields,\n",
+      "#sx = sf+(x*sfg)\n",
+      "h1 = 1270.4; \n",
+      "sx = 1.4861; \n",
+      "sf = 0.5440; \n",
+      "sfg = 1.0025 ;\n",
+      "hf = 355.6; \n",
+      "hfg = 843.7;\n",
+      "\n",
+      "# Calculations\n",
+      "x = (sx-sf)/sfg; \t\t\t#Quality\n",
+      "#Therefore,the final enthalpy is\n",
+      "hx = hf+(x*hfg); \t\t\t#Btu/lbm\n",
+      "\n",
+      "# Results\n",
+      "print \"The final enthalpy is %.2f Btu/lbm\"%(hx);\n",
+      "print \"The change in enthalpy is %.2f Btu/lbm\"%(h1-hx); \t\t\t#Note the agreement with the Mollier chart solution\n",
+      "#we can also use the computer program to solve this problem.For 600F and 800 psia, h = 1270. Btu/lbm and s = 1.4857 Btu/lbm*R.Now Using p = 200 psia and s = 1.4857,we obtain\n",
+      "#h = 1148.1 Btu/lbm.The change in enthalpy is 1270.0-1148.1 = 121.9 Btu/lbm.Note the effort saved Using either the Mollier chart or the computer program. \n",
+      "\n",
+      "# Saturation Properties\n",
+      "#--------------------------\n",
+      "# T = 600.00 degF\n",
+      "# P = 1541.7 psia\n",
+      "#       z        z1         zg\n",
+      "# v(ft**3/lbm)   0.02362   0.2675\n",
+      "# h(Btu/lbm)    616.59    1166.2\n",
+      "# s(Btu/lbm*F)  0.8129    1.3316\n",
+      "# u(Btu/lbm)    609.85    1089.9\n",
+      "\n",
+      "#Thermo Properties\n",
+      "#------------------------\n",
+      "# T =  600.00 degF\n",
+      "# P =  800.00 psia\n",
+      "# v =  0. ft**3/lbm\n",
+      "# h =  1168.7 Btu/lbm\n",
+      "# s =  1.5384 Btu/lbm*F\n",
+      "# u =  1087.5 Btu/lbm\n",
+      "#Region:Superheated\n",
+      "\n",
+      "# Saturation Properties\n",
+      "#--------------------------\n",
+      "# T = 381.87 degF\n",
+      "# P = 200.00 psia\n",
+      "#       z        z1        zg\n",
+      "# v(ft**3/lbm)   0.01839   2.2883\n",
+      "# h(Btu/lbm)    355.60    1199.0\n",
+      "# s(Btu/lbm*F)  0.5440    1.5462\n",
+      "# u(Btu/lbm)    354.92    1114.3\n",
+      "\n",
+      "#Thermo Properties\n",
+      "#------------------------\n",
+      "# T =  381.87 degF\n",
+      "# P =  200.00 psia\n",
+      "# v =  2.1512 ft**3/lbm\n",
+      "# h =  1148.1 Btu/lbm\n",
+      "# s =  1.4857 Btu/lbm*F\n",
+      "# u =  1068.5 Btu/lbm\n",
+      "# x =  0.9396\n",
+      "\n",
+      "#Region:Saturated\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final enthalpy is 1148.47 Btu/lbm\n",
+        "The change in enthalpy is 121.93 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.39  Page No : 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#As refering to figure 5.21,it will be seen that the final temperature and enthalpy will both be higher than for the isentropic case.\n",
+      "#80% of the isentropic enthalpy difference\n",
+      "deltah = 0.8*122; \t\t\t#change in enthalpy \t\t\t#Btu/lbm\n",
+      "h1 = 1270; \t\t\t#Btu/lbm \t\t\t#initial enthalpy\n",
+      "h2 = h1-deltah; \t\t\t#the final enthalpy \t\t\t#Btu/lbm\n",
+      "print \"The final enthalpy is %.2f Btu/lbm\"%(h2);\n",
+      "print \"and the final pressure is 200 psia\";\n",
+      "print \"The Mollier chart indicates the final state to be in the wet region\"\n",
+      "print \"with 3.1percent moisture content and an entropy of 1.514 Btu/lbm*R\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final enthalpy is 1172.40 Btu/lbm\n",
+        "and the final pressure is 200 psia\n",
+        "The Mollier chart indicates the final state to be in the wet region\n",
+        "with 3.1percent moisture content and an entropy of 1.514 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.40  Page No : 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Using the Mollier chart,\n",
+      "h1 = 2942; \t\t\t#kJ/kg \t\t\t#initial enthalpy\n",
+      "#Proceeding as shown in figure 5.21b,that is,vertically at constant entropy to a pressure of 0.1 MPa,gives us\n",
+      "h2 = 2512; \t\t\t#kJ/kg \t\t\t#final enthalpy\n",
+      "\n",
+      "# Results\n",
+      "print \"Neglecting kinetic & potential energy, The change in enthalpy of the steam is %.2f kJ/kg\"%(h1-h2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Neglecting kinetic & potential energy, The change in enthalpy of the steam is 430.00 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.41  Page No : 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#From the conditions given in problem 5.38,the isentropic change in enthalpy is 122 Btu/lbm\n",
+      "#So,\n",
+      "h1minush2 = 122; \t\t\t#Btu/lbm \t\t\t#change in enthalpy\n",
+      "J = 778; \t\t\t#Conversion factor\n",
+      "gc = 32.17; \t\t\t#lbm*ft/lbf*s**2 \t\t\t#constant of proportionality\n",
+      "\n",
+      "# Calculations\n",
+      "V2 = math.sqrt(2*gc*J*(h1minush2)); \t\t\t#final velocity \t\t\t#ft/s\n",
+      "\n",
+      "# Results\n",
+      "print \"As the steam leaves the nozzle, The final velocity is %.2f ft/s\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "As the steam leaves the nozzle, The final velocity is 2471.21 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.42  Page No : 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Because the process is irreversible,we cannot show it on the Mollier diagram.However,the analysis of problem 3.22 for the nozzle is still valid,and all that is      needed is the enthalpy at the beginning and the end of the expansion.From the problem 5.38,\n",
+      "h1 = 1270; \t\t\t#Btu/lbm \t\t\t#initial enthalpy\n",
+      "#For h2 we locate the state point on the Mollier diagram as being saturated vapor at 200 psia.This gives us\n",
+      "h2 = 1199; \t\t\t#Btu/lbm \t\t\t#final enthalpy\n",
+      "J = 778; \t\t\t#Conversion factor\n",
+      "gc = 32.17; \t\t\t#lbm*ft/lbf*s**2 \t\t\t#constant of proportionality\n",
+      "V2 = math.sqrt(2*gc*J*(h1-h2)); \t\t\t#final velocity \t\t\t#Ft/s\n",
+      "print \"As the steam leaves the nozzle, The final velocity is %.2f ft/s\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "As the steam leaves the nozzle, The final velocity is 1885.21 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.43  Page No : 229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From the saturation table,500 psia corresponds to a temperature of 467.13F,and the saturated vapor has an enthalpy of 1205.3 Btu/lbm.At 500 psia and 800 F,the       saturated vapor has an enthalpy of 1412.1 Btu/lbm.Because this process is a steady-flow process at constant pressure,the energy equation becomes q = h2-h1,assuming    that differences in the kinetic energy and potential energy terms are negligible.Therefore,\n",
+      "h2 = 1412.1; \t\t\t#Btu/lbm \t\t\t#final enthalpy\n",
+      "h1 = 1205.3; \t\t\t#Btu/lbm \t\t\t#initial enthalpy\n",
+      "\n",
+      "# Calculations\n",
+      "q = h2-h1; \t\t\t#heat added \t\t\t#Btu/lbm\n",
+      "\n",
+      "# Results\n",
+      "print \"%.2f Btu/lbm heat per pound of steam was added\"%(q);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "206.80 Btu/lbm heat per pound of steam was added\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.44  Page No : 229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From the saturation table at 1 psia,\n",
+      "hf = 69.74; \t\t\t#Btu/lbm \t\t\t#saturated liquid enthalpy\n",
+      "hfg = 1036.0; \t\t\t#Btu/lbm \t\t\t#Evap. Enthalpy\n",
+      "hg = 1105.8; \t\t\t#Btu/lbm \t\t\t#The enthalpy of saturated steam\n",
+      "x = 0.97; \t\t\t#Quality\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#Because the condensation process is carried out at constant pressure,the energy equation is q = deltah.\n",
+      "hx = hf+(x*hfg); \t\t\t#the initial enthalpy \t\t\t#Btu/lbm\n",
+      "print \"The initial enthalpy is %.2f Btu/lbm\"%(hx);\n",
+      "#The final enthalpy is hf = 69.74.So,\n",
+      "deltah = hx-hf; \t\t\t#The enthalpy difference \t\t\t#Btu/lbm\n",
+      "print \"At 1 psia, The enthalpy difference is %.2f Btu/lbm\"%(deltah);\n",
+      "print \"By the computer solution,the enthalpy difference is 1004.6 Btu/lbm\";\n",
+      "# Saturation Properties\n",
+      "#--------------------------\n",
+      "# T = 101.71 degF\n",
+      "# P = 1.0000 psia\n",
+      "#       z        z1         zg\n",
+      "# v(ft**3/lbm)   0.01614   333.55\n",
+      "# h(Btu/lbm)    69.725    1105.4\n",
+      "# s(Btu/lbm*F)  0.1326    1.9774\n",
+      "# u(Btu/lbm)    69.722    1043.6\n",
+      "\n",
+      "#Thermo Properties\n",
+      "#------------------------\n",
+      "# T =  101.71 degF\n",
+      "# P =  1.0000 psia\n",
+      "# v =  323.55 ft**3/lbm\n",
+      "# h =  1074.3 Btu/lbm\n",
+      "# s =  1.9221 Btu/lbm*F\n",
+      "# u =  1014.4 Btu/lbm\n",
+      "# x =  0.9700\n",
+      "\n",
+      "#Region:Saturated\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The initial enthalpy is 1074.66 Btu/lbm\n",
+        "At 1 psia, The enthalpy difference is 1004.92 Btu/lbm\n",
+        "By the computer solution,the enthalpy difference is 1004.6 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch6_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch6_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch6_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch6_2.ipynb
new file mode 100755
index 00000000..ea95c467
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch6_2.ipynb
@@ -0,0 +1,1796 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e0fac09bbeb2cd0ce192e8aaee02b91055f447d7bbfe0689b860ffad5cfd7b06"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : The Ideal Gas"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1  Page No : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "P1 = 100.; \t\t\t#Pressure at volume V1 = 100 ft**3 \t\t\t#Unit:psia\n",
+      "V1 = 100.; \t\t\t#Unit:ft**3 \t\t\t#V1 = Volume at 100 psia\n",
+      "P2 = 30. \t\t\t# Reduced Pressure \t\t\t#Unit:psia\n",
+      "\n",
+      "# Calculations\n",
+      "#Boyle's law,P1*V1 = P2*V2\n",
+      "V2 = (P1*V1)/P2; \t\t\t#Volume occupied by the gas \t\t\t#ft**3\n",
+      "\n",
+      "# Results\n",
+      "print \"Volume occupied by the gas  =  %.2f ft**3\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume occupied by the gas  =  333.33 ft**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2  Page No : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "P1 = 10.**6; \t\t\t#Pressure at volume V1 = 2 m**3 \t\t\t#Unit:Pa\n",
+      "V1 = 2.; \t\t\t#Unit:m**3 \t\t\t#V1 = Volume at 10**6 Pa\n",
+      "P2 = 8.*10**6 \t\t\t# Increased Pressure \t\t\t#Unit:Pa\n",
+      "\n",
+      "# Calculations\n",
+      "#Boyle's law,P1*V1 = P2*V2\n",
+      "V2 = (P1*V1)/P2; \t\t\t#Volume occupied by gas \t\t\t#unit:m**3\n",
+      "\n",
+      "# Results\n",
+      "print \"Volume occupied by gas  =  %.2f m**3\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume occupied by gas  =  0.25 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3  Page No : 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given value\n",
+      "T1 = 32.+460; \t\t\t#Temperature at volume V1 = 150 ft**3 \t\t\t#Unit:R\n",
+      "V1 = 150.; \t\t\t#Unit:ft**3 \t\t\t#V1 = Volume at 32 F\n",
+      "T2 = 100.+460 \t\t\t# Increased Temperature \t\t\t#Unit:R\n",
+      "\n",
+      "# Calculations\n",
+      "#Charles's law,V1/V2  =  T1/T2\n",
+      "V2 = (T2*V1)/T1; \t\t\t#Volume occupied by gas \t\t\t#unit:m**3\n",
+      "\n",
+      "# Results\n",
+      "print \"Volume occupied by gas  =  %.2f m**3\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume occupied by gas  =  170.73 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4  Page No : 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#If for this process T2 = 1.25*T1,\n",
+      "# T2/T1  =  1.25\n",
+      "#Therefore,\n",
+      "# p2/p1  =  T2/T1 \t\t\t#Charles's law(volume constant)\n",
+      "#Thus,\n",
+      "print \"The absolute gas pressure increases by 25 percent\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute gas pressure increases by 25 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5  Page No : 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given data\n",
+      "V1 = 4.; \t\t\t#m**3 \t\t\t#initial volume\n",
+      "T2 = 0.+273; \t\t\t#celsius converted to kelvin  \t\t\t#gas is cooled to 0 C \t\t\t#final temperature\n",
+      "T1 = 100+273; \t\t\t#celsius converted to kelvin \t\t\t#initial temperature\n",
+      "\n",
+      "# Calculations\n",
+      "V2 = V1*(T2/T1); \t\t\t#final volume \t\t\t#Charles's law(pressure constant) \t\t\t#unit:m**3\n",
+      "\n",
+      "# Results\n",
+      "print \"The final volume is %.2f m**3\"%(V2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final volume is 2.93 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6  Page No : 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Let us first put each of the given variables into a consistent set of units:\n",
+      "p = (200+14.7)*(144); \t\t\t#Unit:psfa*(lbf/ft**2) \t\t\t#1 ft**2 = 144 in**2 \t\t\t#pressure\n",
+      "T = (460.+73); \t\t\t#Fahrenheit temperature converted to absolute temperature \t\t\t#unit:R\n",
+      "V = 120./1728;  \t\t\t#1 ft**3 = 1728 in**3 \t\t\t#total volume \t\t\t#unit:ft**3\n",
+      "R = 1545./28; \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#because the molecular weight of nitrogen is 28 \t\t\t#constant of proportionality\n",
+      "#Applying, p*v = R*T, \t\t\t#ideal gas law\n",
+      "v = (R*T)/p; \t\t\t#Unit:ft**3/lbm \t\t\t#specific volume\n",
+      "print \"The specific volume is %.2f ft**3/lbm\"%(v);\n",
+      "#The mass of gas is the total volume divided by the specific volume\n",
+      "print \"The gas in the container is %.2f lbm\"%(V/v);\n",
+      "#The same result is obtained by direct use of eq. p*V = m*R*T\n",
+      "m = (p*V)/(R*T); \t\t\t#The gas in the container \t\t\t#unit:lbm \t\t\t#ideal gas law\n",
+      "print \"The gas in the container is %.2f lbm\"%(m); \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The specific volume is 0.95 ft**3/lbm\n",
+        "The gas in the container is 0.07 lbm\n",
+        "The gas in the container is 0.07 lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7  Page No : 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Applying , (p1*V1)/T1  =  (p2*V2)/T2\n",
+      "#and p2 = p1*(T2/T1) because V1 = V2\n",
+      "p1 = 200+14.7; \t\t\t#Unit:psia \t\t\t#initial pressure\n",
+      "T2 = 460.+200; \t\t\t#final temperature is 200 F \t\t\t#Fahrenheit temperature converted to absolute temperature \t\t\t#unit:R\n",
+      "T1 = 460+73; \t\t\t#Fahrenheit temperature converted to absolute temperature \t\t\t#unit:R\n",
+      "\n",
+      "# Calculations\n",
+      "p2 = p1*(T2/T1); \t\t\t#final pressure \t\t\t#Unit:psia \t\t\t#Charles's law(volume constant)\n",
+      "\n",
+      "# Results\n",
+      "print \"The final pressure is %.2f psia\"%(p2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final pressure is 265.86 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8  Page No : 246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#For CO2,\n",
+      "R = 8.314/44; \t\t\t#Unit:kJ/kg*K \t\t\t#constant of proportionality \t\t\t#Molecular weight of CO2 = 44\n",
+      "p = 500; \t\t\t#Unit:kPa \t\t\t#pressure\n",
+      "V = 0.5; \t\t\t#Unit:m**3 \t\t\t#volume\n",
+      "T = (100.+273); \t\t\t#Unit:K \t\t\t#Celsius converted to kelvin\n",
+      "\n",
+      "# Calculations\n",
+      "#Applying p*V = m*R*T ,\n",
+      "m = (p*V)/(R*T); \t\t\t#mass \t\t\t#kg \t\t\t#ideal gas law\n",
+      "\n",
+      "# Results\n",
+      "print \"The mass of gas in the math.tank is %.2f kg\"%(m);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of gas in the math.tank is 3.55 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9  Page No : 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T2 = 500+460;  \t\t\t#absolute  final temperature  \t\t\t#unit:R\n",
+      "T1 = 80+460;  \t\t\t#absolute initial temperature \t\t\t#unit:R\n",
+      "#The equation cpbar =  0.338-(1.24*10**2/T)+(4.15*10**4)/T**2 has a form , cbar =  Adash+(Bdash/T)+(Ddash/T**2)\n",
+      "#So,\n",
+      "Adash = 0.338;      \t\t\t#constant\n",
+      "Bdash = -1.24*10**2; \t\t\t#constant\n",
+      "Ddash = 4.15*10**4;  \t\t\t#constant\n",
+      "\n",
+      "# Calculations\n",
+      "#Therefore,from equation,cbar = Adash+((Bdash*math.log(T2/T1))/(T2-T1))+(Ddash/(T2*T1))\n",
+      "cpbar = Adash+((Bdash*math.log(T2/T1))/(T2-T1))+(Ddash/(T2*T1)); \t\t\t#The mean specific heat \t\t\t#Btu/lbm*R\n",
+      "\n",
+      "# Results\n",
+      "print \"The mean specific heat at constant pressure between 80F and 500F is %.2f Btu/lbm*R\"%(cpbar);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean specific heat at constant pressure between 80F and 500F is 0.42 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10  Page No : 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#The table in Appendix 3 does not give us the enthalpy data at 960R and 540R that we need.Interpolating yields\n",
+      "#  T    hbar      T    hbar\n",
+      "# 537  3729.5    900   6268.1\n",
+      "# 540  3750.4    960   6694.0\n",
+      "# 600  4167.9    1000  6977.9\n",
+      "#So,\n",
+      "hbar540 = 3729.5+(3./63)*(4167.9-3729.5); \t\t\t#enthalpy \t\t\t#unit:Btu/lbm\n",
+      "hbar960 = 6268.1+(60./100)*(6977.9-6268.1);  \t\t\t#enthalpy \t\t\t#unit:Btu/lbm\n",
+      "#Note that hbar is given for a mass of 1 lb mole.To obtain the enthalpy per pound,it is necessary to divide the values og h by the molecular weight,28.\n",
+      "h2 = 6694.0;  \t\t\t#enthalpy \t\t\t#unit:Btu/lbm\n",
+      "h1 = 3750.4;  \t\t\t#enthalpy \t\t\t#unit:Btu/lbm\n",
+      "T2 = 500.+460; \t\t\t#absolute  final temperature \t\t\t#unit:R\n",
+      "T1 = 80.+460; \t\t\t#absolute initial temperature \t\t\t#unit:R\n",
+      "\n",
+      "# Calculations\n",
+      "cbar = (h2-h1)/(28*(T2-T1)); \t\t\t#The mean specific heat at constant pressure  \t\t\t#unit:Btu/lbm*R\n",
+      "\n",
+      "# Results\n",
+      "print \"The mean specific heat at constant pressure is %.2f Btu/lbm*R\"%(cbar);\n",
+      "#With the more extesive Gas tables,these interpolations are avoided.The Gas Tables provide a relatively easy and accurate method of obtaining average specific heats.Also,these tables have been computerized for ease of application.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean specific heat at constant pressure is 0.25 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11  Page No : 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T2 = 500+460; \t\t\t#absolute  final temperature  \t\t\t#unit:R\n",
+      "T1 = 80+460; \t\t\t#absolute initial temperature  \t\t\t#unit:R\n",
+      "#cp = 0.219 + (3.42*10**-5*T) - (2.93*10**-9*T**2); \t\t\t#Unit:Btu/lbm*R\n",
+      "#Comparing with c = A+(B*T)+(D*T**2) \n",
+      "A = 0.219;        \t\t\t#constant\n",
+      "B = 3.42*10**-5;   \t\t\t#constant\n",
+      "D = 2.93*10**-9;   \t\t\t#constant\n",
+      "\n",
+      "# Calculations\n",
+      "#Using these values and equation cbar = A+((B/2)(T2+T1))+((D/3)*(T2**2+(T2*T1)+T1**2))\n",
+      "cpbar = A+((B/2)*(T2+T1))+((D/3)*(T2**2+(T2*T1)+T1**2)); \t\t\t#The mean specific heat \t\t\t#Btu/lbm*R\n",
+      "\n",
+      "# Results\n",
+      "print \"The mean specific heat at constant pressure for air between 80F and 500F is %.2f Btu/lbm*R\"%(cpbar);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean specific heat at constant pressure for air between 80F and 500F is 0.25 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12  Page No : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The molecular weight of oxygen is 32.Therefore,\n",
+      "R = 1545./32; \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "cp = 0.24; \t\t\t#Unit:Btu/lbm*R \t\t\t#specific heat at constant pressure\n",
+      "\n",
+      "# Calculations\n",
+      "#cp-cv = R/J\n",
+      "cv = cp-(R/J); \t\t\t#specific heat at constant volume \t\t\t#unit:Btu/lbm*R\n",
+      "\n",
+      "# Results\n",
+      "print \"Specific heat at constant volume is %.2f Btu/lbm*R\"%(cv);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specific heat at constant volume is 0.18 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13  Page No : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From equation,cv = R/(k-1) ,\n",
+      "R = 8.314/32; \t\t\t#constant of proportionality \t\t\t#kJ/kg*K \t\t\t#The molecular weight of oxygen is 32\n",
+      "k = 1.4 \t\t\t#for oxygen \t\t\t#given \t\t\t#k = cp/cv\n",
+      "cv = R/(k-1); \t\t\t#Specific heat at constant volume \t\t\t#unit:kJ/kg*K\n",
+      "print \"Specific heat at constant volume is %.2f kJ/kg*K\"%(cv);\n",
+      "cp = k*cv; \t\t\t#specific heat at constant pressure \t\t\t#Unit:kJ/kg*K\n",
+      "print \"Specific heat at constant pressure is %.2f kJ/kg*K\"%(cp);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specific heat at constant volume is 0.65 kJ/kg*K\n",
+        "Specific heat at constant pressure is 0.91 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.14  Page No : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "R = 60.; \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "deltah = 500.; \t\t\t#Btu/lbm \t\t\t#change in enthalpy\n",
+      "deltau = 350.; \t\t\t#Btu/lbm \t\t\t#change in internal energy\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "#Because deltah-(cp*deltaT) and deltau = cv*deltaT\n",
+      "# deltah/deltau = (cp*deltaT)/(cv*deltaT) = cp/cv = k\n",
+      "k = deltah/deltau; \t\t\t#Ratio of specific heats\n",
+      "print \"Ratio of specific heats k is %.2f\"%(k);\n",
+      "#From equation cv = R/(J*(k-1))\n",
+      "cv = R/(J*(k-1)); \t\t\t#specific heat at constant volume \t\t\t#Btu/lbm*R\n",
+      "print \"Specific heat at constant volume is %.2f Btu/lbm*R\"%(cv);\n",
+      "cp = k*cv; \t\t\t#Specific heat at constant pressure \t\t\t#Btu/lbm*R\n",
+      "print \"Specific heat at constant pressure is %.2f Btu/lbm*R\"%(cp);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio of specific heats k is 1.43\n",
+        "Specific heat at constant volume is 0.18 Btu/lbm*R\n",
+        "Specific heat at constant pressure is 0.26 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15  Page No : 256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#When solving this type of problem,it is necessary to note carefully the information given and to write the correct energy equation for this process.Because the    process is carried out at constant volume,the heat added equals the change in inernal energy.Because the change in internal energy per pound for the ideal gas is      cv*(T2-T1),the total change in internal energy for m pounds must equals the heat added.Thus,\n",
+      "#data given\n",
+      "Q = 0.33; \t\t\t#heat\n",
+      "#Initial conditions\n",
+      "V = 60; \t\t\t#in**3 \t\t\t#volume\n",
+      "m = 0.0116; \t\t\t#lbs \t\t\t#mass\n",
+      "p1 = 90; \t\t\t#psia \t\t\t#pressure\n",
+      "T1 = 460+40; \t\t\t#Fahrenheit temperature converted to absolute temperature\n",
+      "#Final condition = Initial condition + heat\n",
+      "V = 60; \t\t\t#in**3 \t\t\t#volume\n",
+      "m = 0.0116; \t\t\t#lbs \t\t\t#mass\n",
+      "p2 = 108; \t\t\t#psia \t\t\t#pressure \n",
+      "T2 = 460+140; \t\t\t#Fahrenheit temperature converted to absolute temperature \t\t\t#unit:R\n",
+      "#Q = m*(u2-u1) = m*cv*(T2-T1)\n",
+      "\n",
+      "# Calculations and Results\n",
+      "cv = Q/(m*(T2-T1)); \t\t\t#specific heat at constant volume \t\t\t#Btu/lbm*R\n",
+      "print \"Specific heat at constant volume is %.2f Btu/lbm*R\"%(cv);\n",
+      "#To obtain cp,it is first necessary to obtain R.Enough information was given in the initial conditions of the problem to apply eqn. p*V = m*R*T\n",
+      "R = (144*p1*(V/1728.))/(m*T1); \t\t\t#1 ft**2 = 144 in**2 \t\t\t#1 ft**3 = 1728 in**3 \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "print \"Consmath.tant of proportionality R is %.2f ft*lbf/lbm*R\"%(R);\n",
+      "#cp-cv = (R/J)\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "cp = cv+(R/J); \t\t\t#Specific heat at constant pressure \t\t\t#Btu/lbm*R\n",
+      "print \"Specific heat at constant pressure is %.2f Btu/lbm*R\"%(cp);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specific heat at constant volume is 0.28 Btu/lbm*R\n",
+        "Consmath.tant of proportionality R is 77.59 ft*lbf/lbm*R\n",
+        "Specific heat at constant pressure is 0.38 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.16  Page No : 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#data\n",
+      "cp = 0.24; \t\t\t#Specific heat at constant pressure \t\t\t#Btu/lbm*R\n",
+      "p2 = 15.; \t\t\t#psia \t\t\t#final pressure\n",
+      "p1 = 100.; \t\t\t#psia \t\t\t#initial pressure\n",
+      "T2 = 460.+0;  \t\t\t#absolute  final temperature \t\t\t#unit:R\n",
+      "T1 = 460.+100; \t\t\t#absolute initial temperature  \t\t\t#unit:R\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "R = 1545./29; \t\t\t#moleculer weight = 29 \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "\n",
+      "# Calculations\n",
+      "#On the basis of the data given,\n",
+      "deltas = (cp*(math.log(T2/T1)))-((R/J)*(math.log(p2/p1))); \t\t\t#change in entropy \t\t\t#Btu/lbm*R\n",
+      "\n",
+      "# Results\n",
+      "print \"The change in enthalpy is %.2f Btu/lbm*R\"%(deltas);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The change in enthalpy is 0.08 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17  Page No : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#data of problem6.16\n",
+      "cp = 0.24; \t\t\t#Specific heat at constant pressure \t\t\t#Btu/lbm*R\n",
+      "p2 = 15.; \t\t\t#psia \t\t\t#final pressure\n",
+      "p1 = 100.; \t\t\t#psia \t\t\t#initial pressure\n",
+      "T2 = 460.+0;  \t\t\t#absolute  final temperature \t\t\t#unit:R\n",
+      "T1 = 460.+100; \t\t\t#absolute initial temperature \t\t\t#unit:R\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "R = 1545./29; \t\t\t#moleculer weight = 29 \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "#Because cp and R are given,let us first solve for cv,\n",
+      "#cp = (R*k)/(J*(k-1))\n",
+      "k = (cp*J)/((cp*J)-R); \t\t\t#k = cp/cv \t\t\t#ratio of specific heats\n",
+      "print \"Ratio of specific heats k is %.2f\"%(k);\n",
+      "#k = cp/cv\n",
+      "cv = cp/k; \t\t\t#Specific heat at constant volume \t\t\t#Btu/lbm*R\n",
+      "print \"Specific heat at constant volume is %.2f Btu/lbm*R\"%(cv);\n",
+      "#Now, deltas = (cv*math.log(p2/p1))+(cp*math.log(v2/v1));\n",
+      "#But, v2/v1 = (T2*p1)/(T1*p2)\n",
+      "v2byv1 = (T2*p1)/(T1*p2); \t\t\t# v2/v1 \t\t\t#unitless\n",
+      "deltas = (cv*math.log(p2/p1))+(cp*math.log(v2byv1)); \t\t\t#The change in enthalpy  \t\t\t#unit:Btu/lbm*R\n",
+      "print \"The change in enthalpy is %.2f Btu/lbm*R\"%(deltas);\n",
+      "#The agreement is very good.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio of specific heats k is 1.40\n",
+        "Specific heat at constant volume is 0.17 Btu/lbm*R\n",
+        "The change in enthalpy is 0.08 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18  Page No : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#data,\n",
+      "cp = 0.9093; \t\t\t#Specific heat at constant pressure \t\t\t#kJ/kg*R\n",
+      "p2 = 150.; \t\t\t#kPa \t\t\t#final pressure\n",
+      "p1 = 500.; \t\t\t#kPa \t\t\t#initial pressure\n",
+      "T2 = 273.+0;  \t\t\t#final temperature \t\t\t#Celsius converted to kelvin\n",
+      "T1 = 273.+100; \t\t\t#initial temperature \t\t\t#Celsius converted to kelvin\n",
+      "#J = 778; \t\t\t#conversion factor\n",
+      "R = 8.314/32; \t\t\t#moleculer weight of oxygen = 32 \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "\n",
+      "# Calculations\n",
+      "#Using equation, and dropping J gives,\n",
+      "deltas = (cp*(math.log(T2/T1)))-((R)*(math.log(p2/p1))); \t\t\t#change in entropy \t\t\t#kJ/kg*K\n",
+      "#For 2 kg,\n",
+      "deltaS = 2*deltas; \t\t\t#The change in enthalpy in kJ/K\n",
+      "\n",
+      "# Results\n",
+      "print \"For 2 kg oxygen, The change in enthalpy is %.2f kJ/K\"%(deltaS);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For 2 kg oxygen, The change in enthalpy is 0.06 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.19  Page No : 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#from the equation, deltas/cv  =  (k*math.log(v2/v1))+ math.log(p2/p1) \t\t\t#change in entropy\n",
+      "k = 1.4; \t\t\t#k = cp/cv \t\t\t#ratio of specific heats\n",
+      "#deltas = (1/4)*cv \t\t\t#so, \n",
+      "# 1/4 =  (k*math.log(v2/v1))+ math.log(p2/p1)\n",
+      "v2 = 1./2; \t\t\t#Because,v2 = (1/2)*v1  \t\t\t#initial specific volume \n",
+      "v1 = 1.;   \t\t\t#final specific volume\n",
+      "\n",
+      "# Calculations\n",
+      "p2byp1 = math.exp((1./4)-(k*math.log(v2/v1))); \t\t\t#increase in pressure\n",
+      "\n",
+      "# Results\n",
+      "print \"p2/p1 = %.2f\"%(p2byp1);\n",
+      "print \"So, increase in pressure is %.2f \"%(p2byp1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p2/p1 = 3.39\n",
+        "So, increase in pressure is 3.39 \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.20  Page No : 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data\n",
+      "T2 = 460+270; \t\t\t#Fahrenheit temperature converted to absolute final temperature \t\t\t#unit:R\n",
+      "T1 = 460+70; \t\t\t#Fahrenheit temperature converted to absolute initial temperature \t\t\t#unit:R\n",
+      "cv = 0.17; \t\t\t#specific heat at constant volume \t\t\t#Btu/lbm*R\n",
+      "\n",
+      "# Calculations\n",
+      "#Now,\n",
+      "deltas = cv*math.log(T2/T1); \t\t\t#change in entropy \t\t\t#Unit:Btu/lbm*R\n",
+      "#For 1/2 lb,\n",
+      "deltaS = (1./2)*deltas; \t\t\t#The change in enthalpy in Btu/R\n",
+      "\n",
+      "# Results\n",
+      "print \"For 1/2 lb of gas, The change in enthalpy is %.2f Btu/R\"%(deltaS);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For 1/2 lb of gas, The change in enthalpy is 0.00 Btu/R\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.21  Page No : 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data\n",
+      "T2 = 100.+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#final temperature\n",
+      "T1 = 20+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#initial temperature\n",
+      "cv = 0.7186; \t\t\t#specific heat at constant volume \t\t\t#kJ/kg*K\n",
+      "#Now,\n",
+      "deltas = cv*math.log(T2/T1); \t\t\t#change in entropy \t\t\t#Unit:kJ/kg*K\n",
+      "#For 0.2 kg,\n",
+      "deltaS = (0.2)*deltas; \t\t\t#The change in enthalpy in kJ/K\n",
+      "print \"For 0.2 kg of air, The change in enthalpy is %.2f kJ/K\"%(deltaS);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For 0.2 kg of air, The change in enthalpy is 0.03 kJ/K\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.22  Page No : 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data\n",
+      "deltas = 0.0743; \t\t\t#change in entropy \t\t\t#Unit:Btu/lbm*R\n",
+      "T1 = 460+100; \t\t\t#Fahrenheit temperature converted to absolute initial temperature\n",
+      "cv = 0.219; \t\t\t#specific heat at constant volume \t\t\t#Btu/lbm*R\n",
+      "#Now,\n",
+      "\n",
+      "# Calculations\n",
+      "#deltas = cv*math.log(T2/T1); \n",
+      "T2 = T1*math.exp(deltas/cv); \t\t\t#higher temperature \t\t\t#absolute temperature  \t\t\t#unit:R\n",
+      "\n",
+      "# Results\n",
+      "print \"The higher temperature is %.2f R\"%(T2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The higher temperature is 786.20 R\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.23  Page No : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data\n",
+      "deltaS = 0.4386; \t\t\t#change in entropy \t\t\t#Unit:kJ/K\n",
+      "T2 = 273+425; \t\t\t#Celsius temperature converted to kelvin \t\t\t#initial temperature\n",
+      "cv = 0.8216; \t\t\t#specific heat at constant volume \t\t\t#kJ/kg*K\n",
+      "m = 1.5; \t\t\t#mass \t\t\t#kg\n",
+      "#Now,\n",
+      "\n",
+      "# Calculations\n",
+      "#deltas = m*cv*math.log(T2/T1); \n",
+      "T1 = T2/(math.exp(deltaS/(m*cv))) \t\t\t#initial temperature \t\t\t#unit:K\n",
+      "\n",
+      "# Results\n",
+      "print \"The initial temperature of the process is %.2f K or %.2f C\"%(T1,T1-273)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The initial temperature of the process is 488.98 K or 215.98 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.24  Page No : 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given\n",
+      "T2 = 460.+400; \t\t\t#Fahrenheit temperature converted to absolute final temperature \t\t\t#unit:R\n",
+      "T1 = 460.+70; \t\t\t#Fahrenheit temperature converted to absolute initial temperature \t\t\t#unit:R\n",
+      "cp = 0.24; \t\t\t#specific heat at constant pressure \t\t\t#Btu/lbm*R\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "R = 1545./29; \t\t\t#moleculer weight = 29 \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "#From the energy equation for the constant-pressure process,the heat transferred is deltah.Therefore,\n",
+      "#q = deltah = cp*(T2-T1)\n",
+      "deltah = cp*(T2-T1); \t\t\t#heat transferred \t\t\t#Btu/lb \t\t\t#into system\n",
+      "print \"The heat transferred is %.2f Btu/lbinto system)\"%(deltah);\n",
+      "deltas = cp*math.log(T2/T1); \t\t\t#increase in entropy \t\t\t#Btu/lbm*R\n",
+      "print \"The increase in entropy is %.2f Btu/lbm*R\"%(deltas);\n",
+      "#The flow work change is (p2*v2)/J - (p1*v1)/J  =  (R/J)*(T2-T1)\n",
+      "flowworkchange = (R/J)*(T2-T1); \t\t\t#Btu/lbm \t\t\t#The flow work change per pound of air\n",
+      "print \"The flow work change per pound of air is %.2f Btu/lbm\"%(flowworkchange);\n",
+      "#In addition to each of the assumptions made in all the process being considered,it has further been tacitly assumed that these processes are carried out quasi-      statically and without friction.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transferred is 79.20 Btu/lbinto system)\n",
+        "The increase in entropy is 0.12 Btu/lbm*R\n",
+        "The flow work change per pound of air is 22.60 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.25  Page No : 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given\n",
+      "T2 = 500.+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#final temperature\n",
+      "T1 = 20+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#initial temperature\n",
+      "cp = 1.0062; \t\t\t#specific heat at constant pressure \t\t\t#kJ/kg*K\n",
+      "#From the energy equation for the constant-pressure process,the heat transferred is deltah.Therefore,\n",
+      "#q = deltah = cp*(T2-T1)\n",
+      "deltah = cp*(T2-T1); \t\t\t#heat transferred \t\t\t#kJ/kg \t\t\t#into system\n",
+      "print \"The heat transferred is per kimath.logram of air %.2f kJ/kg\"%(deltah);\n",
+      "deltas = cp*math.log(T2/T1); \t\t\t#increase in entropy \t\t\t#kJ/kg*K\n",
+      "print \"The increase in entropy per kimath.logram of air is %.2f kJ/kg*K\"%(deltas);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transferred is per kimath.logram of air 482.98 kJ/kg\n",
+        "The increase in entropy per kimath.logram of air is 0.98 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.26  Page No : 270"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given\n",
+      "v2 = 2.; \t\t\t#Because,v2 = (2)*v1 \t\t\t#volume increases to its twice its final volume\n",
+      "v1 = 1.; \t\t\t#initial volume\n",
+      "T = 460.+200; \t\t\t#Fahrenheit temperature converted to absolute temperature\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "R = 1545./28; \t\t\t#moleculer weight of nitrogen = 28 \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "#From the equation, w/J = q = T*deltas = ((R*T)/J)*math.log(v2/v1)\n",
+      "q = ((R*T)/J)*math.log(v2/v1); \t\t\t#Btu/lbm \t\t\t#the heat added to system\n",
+      "#For 0.1 lb,\n",
+      "Q = 0.1*q; \t\t\t#Btu \t\t\t#the heat added to system\n",
+      "print \"The heat added to system is %.2f Btu\"%(Q);\n",
+      "#The work out of the system is equal to the heat added;thus,\n",
+      "WbyJ = Q; \t\t\t#The work out of the system(out of the system) \t\t\t#unit:Btu\n",
+      "print \"The work out of the system is %.2f Btuout of the system)\"%(WbyJ);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat added to system is 3.24 Btu\n",
+        "The work out of the system is 3.24 Btuout of the system)\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.27  Page No : 270"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given\n",
+      "T = 50+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#final temperature \t\t\t#unit:K\n",
+      "v2 = 1./2; \t\t\t#Because,v2 = (1/2)*v1 \t\t\t#volume increases to its half its final volume\n",
+      "v1 = 1.;\n",
+      "R = 8.314/32; \t\t\t#moleculer weight of oxygen = 32 \t\t\t#Unit:kJ/kg*K \t\t\t#constant of proportionality\n",
+      "#From the equation, q = ((R*T))*math.log(v2/v1)\n",
+      "q = R*T*math.log(v2/v1); \t\t\t#heat added \t\t\t#kJ/kg\n",
+      "print \"The heat added to system is %.2f kJ/kgheat out of system)\"%(q);\n",
+      "#The work out of the system is equal to the heat added;thus,\n",
+      "W = q; \t\t\t#The work out of the system  \t\t\t#unit:kJ/kg\n",
+      "print \"The work out of the system is %.2f kJ/kginto system)\"%(W);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat added to system is -58.17 kJ/kgheat out of system)\n",
+        "The work out of the system is -58.17 kJ/kginto system)\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.28  Page No : 271"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given in problem 6.27\n",
+      "T = 50+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#final temperature\n",
+      "v2 = 1./2; \t\t\t#Because,v2 = (1/2)*v1 \t\t\t#volume increases to its half its final volume\n",
+      "v1 = 1;\n",
+      "R = 8.314/32; \t\t\t#moleculer weight of oxygen = 32 \t\t\t#Unit:kJ/kg*K \t\t\t#constant of proportionality\n",
+      "#From the equation, q = ((R*T))*math.log(v2/v1)\n",
+      "q = R*T*math.log(v2/v1); \t\t\t#heat added \t\t\t#kJ/kg\n",
+      "print \"The heat added to system is %.2f kJ/kgheat out of system)\"%(q);\n",
+      "#For a constant temperature,\n",
+      "deltas = q/T; \t\t\t#Change in entropy \t\t\t#unit:kJ/kg*K\n",
+      "print \"The change in entropy is %.2f kJ/kg*K\"%(deltas);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat added to system is -58.17 kJ/kgheat out of system)\n",
+        "The change in entropy is -0.18 kJ/kg*K\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.29  Page No : 274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given\n",
+      "T1 = 1000; \t\t\t#absolute initial temperature \t\t\t#unit:R\n",
+      "p2 = 1.; \t\t\t#unit:atm \t\t\t#absolute final pressure\n",
+      "p1 = 5.; \t\t\t#unit:atm \t\t\t#absolute initial pressure\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "R = 1545./29; \t\t\t#moleculer weight = 29 \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "k = 1.4; \t\t\t#k = cp/cv \t\t\t#ratio of specific heats\n",
+      "\n",
+      "# Calculations and Results\n",
+      "#From the equation, \n",
+      "T2 = T1*((p2/p1)**((k-1)/k)); \t\t\t#Unit:R \t\t\t#The absolute final temperature\n",
+      "print \"The absolute final temperature is %.2f R\"%(T2);\n",
+      "work = (R*(T2-T1))/(J*(1-k)); \t\t\t#Btu/lbm \t\t\t#The work done by air(out)\n",
+      "print \"The work done by air is %.2f Btu/lbmout)\"%(work)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute final temperature is 631.39 R\n",
+        "The work done by air is 63.11 Btu/lbmout)\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.30  Page No : 274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#data given\n",
+      "#mass of 1 kg\n",
+      "T1 = 500+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#final temperature\n",
+      "p2 = 1.; \t\t\t#atm \t\t\t#absolute final pressure\n",
+      "p1 = 5.; \t\t\t#atm \t\t\t#absolute initial pressure\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "R = 8.314/29; \t\t\t#moleculer weight = 29 \t\t\t#Unit:kJ/kg*K \t\t\t#constant of proportionality\n",
+      "k = 1.4; \t\t\t#k = cp/cv \t\t\t#ratio of specific heat\n",
+      "#From the equation, \n",
+      "T2 = T1*((p2/p1)**((k-1)/k)); \t\t\t#Unit:Kelvin \t\t\t#The absolute final temperature\n",
+      "print \"The absolute final temperature is %.2f K or %.2f C\"%(T2,T2-273);\n",
+      "work = (R*(T2-T1))/((1-k)); \t\t\t#kJ/kg \t\t\t#The work done by air(out)\n",
+      "print \"The work done by air is %.2f kJ/kgout)\"%(work)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute final temperature is 488.06 K or 215.06 C\n",
+        "The work done by air is 204.22 kJ/kgout)\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.31  Page No : 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "import math\n",
+      "\n",
+      "#data given\n",
+      "T1 = 800.+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#initial temperature\n",
+      "T2 = 500.+273; \t\t\t#Celsius temperature converted to Kelvin \t\t\t#final temperature\n",
+      "p2 = 1.; \t\t\t#atm \t\t\t#absolute final pressure\n",
+      "p1 = 5.; \t\t\t#atm \t\t\t#absolute initial pressure\n",
+      "\n",
+      "# Calculations\n",
+      "#A gas expands isentropically\n",
+      "#From the equation,\n",
+      "#T2/T1 = ((p2/p1)**((k-1)/k));\n",
+      "#rearranging,\n",
+      "k = inv([[1-((math.log(T2/T1)/math.log(p2/p1)))]]); \t\t\t#k = cp/cv \t\t\t#Ratio of specific heats\n",
+      "\n",
+      "# Results\n",
+      "print \"Ratio of specific heats k) is %.2f\"%(k);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio of specific heats k) is 1.26\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.32  Page No : 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given\n",
+      "n = 1.3; \t\t\t#p*v**1.3 = constant\n",
+      "k = 1.4; \t\t\t#k = cp/cv Ratio of specific heats \n",
+      "cp = 0.24; \t\t\t#specific heat at constant pressure \t\t\t#Btu/lbm*R\n",
+      "T2 = 600.; \t\t\t#absolute final temperature \t\t\t#unit:R\n",
+      "T1 = 1500.; \t\t\t#absolute initial temperature \t\t\t#unit:R\n",
+      "R = 53.3; \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "cv = cp/k; \t\t\t#specific heat at constant volume \t\t\t#Btu/lbm*R\n",
+      "#Therefore,\n",
+      "cn = cv*((k-n)/(1-n)); \t\t\t#Polytropic specific heat \t\t\t#Btu/lbm*R\n",
+      "print \"Polytropic specific heatcn, is %.2f Btu/lbm*R\"%(cn);\n",
+      "#The negative sign of cn indicates that either the heat transfer for the process comes from the system or there is a negative temperature change while heat is        transferred to the system.\n",
+      "#The heat transferred is cn*(T2-T1).Therefore,\n",
+      "q = cn*(T2-T1); \t\t\t#heat transferred \t\t\t#Btu/lbm(to the system)\n",
+      "print \"The heat transferred is %.2f Btu/lbmto the system\"%(q);\n",
+      "#The work done can be found Using equation,\n",
+      "w = (R*(T2-T1))/(J*(1-n)); \t\t\t#Btu/lbm \t\t\t#the workdone(from the system)\n",
+      "print \"The work done is %.2f Btu/lbmfrom the system\"%(w);\n",
+      "deltas = cn*math.log(T2/T1) \t\t\t#change in entropy \t\t\t#Btu/lbm*R\n",
+      "print \"The change in enthalpy is %.2f Btu/lbm*R\"%(deltas);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Polytropic specific heatcn, is -0.06 Btu/lbm*R\n",
+        "The heat transferred is 51.43 Btu/lbmto the system\n",
+        "The work done is 205.53 Btu/lbmfrom the system\n",
+        "The change in enthalpy is 0.05 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.33  Page No : 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given in problem 6.32,\n",
+      "n = 1.3; \t\t\t#p*v**1.3 = constant\n",
+      "k = 1.4; \t\t\t#k = cp/cv \t\t\t#ratio of specific heats\n",
+      "cp = 0.24; \t\t\t#specific heat at constant pressure \t\t\t#Btu/lbm*R\n",
+      "T2 = 600.; \t\t\t#absolute final temperature \t\t\t#unit:R\n",
+      "T1 = 1500.; \t\t\t#absolute initial temperature \t\t\t#unir:R\n",
+      "R = 53.3; \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "J = 778.; \t\t\t#conversion factor\n",
+      "#Equation,\n",
+      "# T1/T2 = ((p1/p2)**((n-1)/n));\n",
+      "#rearranging,\n",
+      "p1byp2 = math.exp(math.log(T1/T2)/((n-1)/n)); \t\t\t#The ratio of inlet to outlet pressure\n",
+      "print \"The ratio of inlet to outlet pressure is %.2f\"%(p1byp2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ratio of inlet to outlet pressure is 53.02\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.34  Page No : 284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From the table at 1000 R:                     \t\t\t#From the table at 500 R:\n",
+      "h2 = 240.98;                                       \n",
+      "h1 = 119.48; \n",
+      "#Btu/lbm \t\t\t#enthalpy                             \t\t\t#Btu/lbm \t\t\t#enthalpy\n",
+      "u2 = 172.43;                                       \n",
+      "u1 = 85.20; \n",
+      "#Btu/lbm \t\t\t#internal energy                      \t\t\t#Btu/lbm \t\t\t#internal energy \n",
+      "fy2 = 0.75042;                                     \n",
+      "fy1 = 0.58233; \n",
+      "#Btu/lbm*R                                      \t\t\t#Btu/lbm*R\n",
+      "\n",
+      "#The change in enthalpy is\n",
+      "deltah = h2-h1; \t\t\t#Btu/lbm\n",
+      "#The change in internal energy is\n",
+      "deltau = u2-u1; \t\t\t#Btu/lbm\n",
+      "print \"The change in enthalpy is %.2f Btu/lbm & the change in internal energy is %.2f Btu/lbm\"%(deltah,deltau);\n",
+      "#Because in the constant-pressure process -R*math.log(p2/p1) is zero,\n",
+      "deltas = fy2-fy1; \t\t\t#Btu/lbm*R \t\t\t#The entropy when air is heated at constant pressure\n",
+      "print \"The entropy when air is heated at constant pressure is %.2f Btu/lbm/R\"%(deltas);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The change in enthalpy is 121.50 Btu/lbm & the change in internal energy is 87.23 Btu/lbm\n",
+        "The entropy when air is heated at constant pressure is 0.17 Btu/lbm/R\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.35  Page No : 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#In this problem,the air expands from 5 atm absolute to 1 atm absolute from an initial temperature of 1000R,\n",
+      "pr = 12.298; \t\t\t#relative pressure \t\t\t#unit:atm \n",
+      "h = 240.98; \t\t\t#Btu/lbm \t\t\t#enthalpy \n",
+      "pr = 12.298/5; \t\t\t#The value of the final relative pressure \t\t\t#unit:atm\n",
+      "#Interpolation in the air table yields the following:\n",
+      "#   T     pr\n",
+      "#  620   2.249\n",
+      "#        2.4596\n",
+      "#  640   2.514\n",
+      "T = 620+(((2.4596-2.249)/(2.514-2.249))*20); \t\t\t#the final temperature \t\t\t#unit:R\n",
+      "print \"The absolute final temperature is %.2f R\"%(T);\n",
+      "u1 = 172.43; \t\t\t#initial internal energy \t\t\t#Btu/lbm\n",
+      "u2 = 108.51; \t\t\t#final internal energy \t\t\t#Btu/lbm\n",
+      "work = u1-u2; \t\t\t#Btu/lbm The work done by air in an isentropic nonflow expansion \t\t\t#where the value of u2 is obtained by interpolation at T  temperature and the value of u1 is read from the air table at 1000 R. \n",
+      "print \"The work done by air in an isentropic nonflow expansion is %.2f Btu/lbmout)\"%(work)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute final temperature is 635.89 R\n",
+        "The work done by air in an isentropic nonflow expansion is 63.92 Btu/lbmout)\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.36  Page No : 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# given data\n",
+      "T = 1000+460; \t\t\t#Fahrenheit temperature converted to absolute temperature\n",
+      "#The velocity of sound in air at 1000 F is\n",
+      "Va = 49.0*math.sqrt(T); \t\t\t#velocity \t\t\t#ft/s\n",
+      "print \"The velocity of sound air at 1000 F is %.2f ft/s\"%(Va);\n",
+      "#Hydrogen has a specific heat ratio of 1.41 and R = 766.53.Therefore,\n",
+      "khydrogen = 1.41; \t\t\t#specific heats ratio for air\n",
+      "kair = 1.40; \t\t\t#specific heats ratio for air\n",
+      "Rhydrogen = 766.53; \t\t\t#gas constant \t\t\t#ft*lbf/lbm*R\n",
+      "Rair = 53.36; \t\t\t#gas constant \t\t\t#ft*lbf/lbm*R\n",
+      "# Vahydrogen/Vaair  =  math.sqrt((Rhydrogen*khydrogen)/(Rair*kair))\n",
+      "#rearranging,\n",
+      "Vahydrogen = Va*math.sqrt((Rhydrogen*khydrogen)/(Rair*kair)); \t\t\t#The velocity of sound in hydrogen at 1000 F \t\t\t#unit:ft/s\n",
+      "print \"The velocity of sound in hydrogen at 1000 F is %.2f ft/s\"%(Vahydrogen);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity of sound air at 1000 F is 1872.29 ft/s\n",
+        "The velocity of sound in hydrogen at 1000 F is 7121.55 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.37  Page No : 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# given data\n",
+      "T = 200+460.; \t\t\t#Fahrenheit temperature converted to absolute temperature \t\t\t#unit:R\n",
+      "V = 1500; \t\t\t#ft/s \t\t\t#the local velocity\n",
+      "Va = 49.0*math.sqrt(T); \t\t\t#velocity of sound air at 200 F  \t\t\t#unit:ft/s\n",
+      "print \"The velocity of sound air at 200 F is %.2f ft/s\"%(Va);\n",
+      "M = V/Va; \t\t\t#The Mach number = the local velocity/velocity of sound \t\t\t#unitless\n",
+      "print \"The Mach number is %.2f\"%(M);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity of sound air at 200 F is 1258.83 ft/s\n",
+        "The Mach number is 1.19\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.38  Page No : 290"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#data given\n",
+      "V = 1000; \t\t\t#ft/s \t\t\t#the fluid velocity\n",
+      "gc = 32.17; \t\t\t#Unit:(LBm*ft)/(LBf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "J = 778; \t\t\t#conversion factor\n",
+      "h = 1204.4; \t\t\t#Btu/lbm \t\t\t#enthalpy of saturated steam\n",
+      "#h0-h = V**2/(2*gc*J) \n",
+      "h0 = h+((V**2)/(2*gc*J));  \t\t\t#Btu/lbm \t\t\t#h0 = stagnation enthalpy\n",
+      "print \"The total enthalpy is %.2f Btu/lbm\"%(h0);\n",
+      "#It will be noted for this problem that if the initial velocity had been 100ft/s,deltah would have been 0.2 Btu/lbm,and for most practical purpposes,the total      properties and those of the flowing fluid would have been essentially the same.Thus,for low-velocity fluids,the difference in total and steam properties can be      neglected.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total enthalpy is 1224.38 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.39  Page No : 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy.linalg import inv\n",
+      "\n",
+      "# given data\n",
+      "k = 1.4; \t\t\t#the specific heats ratio \t\t\t#k = cp/cv\n",
+      "M = 1; \t\t\t#(table 6.5) \t\t\t#The Mach number = the local velocity/velocity of sound\n",
+      "T0 = 800; \t\t\t#absolute temperature \t\t\t#unit:R\n",
+      "gc = 32.17; \t\t\t#Unit:(LBm*ft)/(LBf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "R = 53.35; \t\t\t#gas constant \t\t\t#ft*lbf/lbm*R\n",
+      "p0 = 300; \t\t\t#psia \t\t\t#pressure\n",
+      "\n",
+      "# * or \"star\" subscripts to conditions in which M = 1;\n",
+      "# \"0\" subscript refers to isentropic stagnation\n",
+      "#Refer to figure 6.26,\n",
+      "#Tstar/T0 = 0.8333\n",
+      "Tstar = T0*0.8333; \t\t\t#temperature when M = 1 \t\t\t#unit:R\n",
+      "print \"If the mach number at the outlet is unity, temperature is %.2f R\"%(Tstar);\n",
+      "Vat = math.sqrt(gc*R*Tstar*k); \t\t\t#ft/s \t\t\t#Vat = V2 \t\t\t#local velocity of sound\n",
+      "print \"If the mach number at the outlet is unity, velocity is %.2f ft/s\"%(Vat)\n",
+      "\n",
+      "#For A/Astar = 2.035\n",
+      "#The table yields\n",
+      "M1 = 0.3; \t\t\t#mach number at inlet\n",
+      "print \"At inlet, The mach number is %.2f\"%(M1)\n",
+      "#pstar/p0 = 0.52828\n",
+      "pstar = p0*0.52828;  \t\t\t#pressure when M = 1 \t\t\t#psia\n",
+      "#also,\n",
+      "#T1/T0 = 0.98232 and p1/p0 = 0.93947\n",
+      "#Therefore,\n",
+      "T1 = T0*0.982332; \t\t\t#unit:R \t\t\t#T1 = temperature at inlet\n",
+      "print \"At inlet, The temperature is %.2f R\"%(T1);\n",
+      "p1 = p0*0.93947; \t\t\t#psia \t\t\t#p1 = pressure at inlet\n",
+      "print \"At inlet, The pressure is %.2f psia\"%(p1);\n",
+      "#From the inlet conditions derived,\n",
+      "Va1 = math.sqrt(gc*k*R*T1); \t\t\t#ft/s \t\t\t#V1 = velocity at inlet\n",
+      "V1 = M1*Va1; \t\t\t#ft/s \t\t\t#velocity\n",
+      "print \"At inlet, The velocity is %.2f ft/s\"%(V1);\n",
+      "#The specific volume at inlet is found from the equation of state for an ideal gas:\n",
+      "v = (R*T1)/(p1*144); \t\t\t#ft**3/lbm \t\t\t#1 ft**2 = 144 in**2(for conversion of unit) \t\t\t#specific volume\n",
+      "rho = inv([[v]]); \t\t\t#inverse of specific volume \t\t\t#density\n",
+      "A = 2.035; \t\t\t#area \t\t\t#ft**2\n",
+      "m = rho*A*V1; \t\t\t#mass flow \t\t\t#unit:lbm/s\n",
+      "print \"At inlet, The mass flow is %.2f lbm/s\"%(m);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "If the mach number at the outlet is unity, temperature is 666.64 R\n",
+        "If the mach number at the outlet is unity, velocity is 1265.62 ft/s\n",
+        "At inlet, The mach number is 0.30\n",
+        "At inlet, The temperature is 785.87 R\n",
+        "At inlet, The pressure is 281.84 psia\n",
+        "At inlet, The velocity is 412.24 ft/s\n",
+        "At inlet, The mass flow is 812.08 lbm/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.40  Page No : 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# * or \"star\" subscripts to conditions in which M = 1;\n",
+      "# \"0\" subscript refers to isentropic stagnation\n",
+      "#This problem will be solved by two methods(A and B)\n",
+      "print \"Method A\"; \t\t\t#By equations:\n",
+      "k = 1.4; \t\t\t#the specific heat ratio \t\t\t#k = cp/cv\n",
+      "R = 53.3; \t\t\t#gas constant \t\t\t#ft*lbf/lbm*R\n",
+      "M = 2.5; \t\t\t#mach number = the local velocity/velocity of sound\n",
+      "print \"Solution for a\";\n",
+      "# T/Tstar  =  (k+1)/(2*(1+((1/2)*(k-1)*M**2)))\n",
+      "# Tstar/T0 = 2/(k+1)\n",
+      "#Therefore,\n",
+      "# (Tstar/T0)*(T/Tstar)  =  (T/T0) = 1/(1+((1/2)*(k-1)*M**2))\n",
+      "T0 = 560; \t\t\t#absolute temperature or stagnation temperature \t\t\t#unit:R\n",
+      "T = T0/(1+((1/2)*(k-1)*M**2)); \t\t\t#temperature at M = 2.5\n",
+      "print \"The temperature is %.2f R\"%(T);\n",
+      "print \"Solution for b\";\n",
+      "p = 0.5; \t\t\t#static pressure \t\t\t#unit:psia\n",
+      "# p0/p  =  (T0/T)**(k/(k-1))\n",
+      "p0 = p*14.7*((T0/T)**(k/(k-1))); \t\t\t#pressure at M = 2.5 \t\t\t#unit:psia\n",
+      "print \"The pressure is %.2f psia\"%(p0);\n",
+      "print \"Solution for c\";\n",
+      "gc = 32.17; \t\t\t#Unit:(LBm*ft)/(LBf*s**2) \t\t\t#gc is constant of proportionality\n",
+      "Va = math.sqrt(gc*k*R*T); \t\t\t#ft/s \t\t\t#local velocity of sound\n",
+      "V = M*Va; \t\t\t#valocity at M = 2.5 \t\t\t#unit:ft/s\n",
+      "print \"The velocity is %.2f ft/s\"%(V);\n",
+      "print \"Solution for d\";\n",
+      "v = (R*T)/(p*14.7*144); \t\t\t#ft**3/lbm \t\t\t#1 ft**2 = 144 in**2 \t\t\t#specific volume at M = 2.5\n",
+      "print \"The specific volume is %.2f ft**3/lbm\"%(v);\n",
+      "print \"Solution for e\";\n",
+      "#Mass velocity is definrd as the mass flow per unit area\n",
+      "# m/A = (A*V)/(v*A) = V/v\n",
+      "print \"The mass velocity is %.2f lbm/s*ft**2)\"%(V/v); \t\t\t#mass velocity at M = 2.5\n",
+      "\n",
+      "\n",
+      "print \"Method B\"; \t\t\t#By the gas tables: \t\t\t#table 6.5 gives\n",
+      "M = 2.5; \t\t\t#mach number = the local velocity/velocity of sound\n",
+      "print \"Solution for a\";\n",
+      "T0 = 560; \t\t\t#absolute temperature or stagnation temperature\n",
+      "#T/T0 = 0.44444\n",
+      "T = T0*0.44444; \t\t\t#temperature at M = 2.5\n",
+      "print \"The temperature is %.2f R\"%(T)\n",
+      "print \"Solution for b\";\n",
+      "p = 0.5; \t\t\t#static pressure\n",
+      "#p/p0 = 0.05853\n",
+      "p0 = (p*14.7)/0.05853; \t\t\t#pressure at M = 2.5\n",
+      "print \"The pressure is %.2f psia\"%(p0);\n",
+      "print \"Solution for c\";\n",
+      "print \"As before %.2f ft/s\"%(V)\n",
+      "print \"Solution for d\";\n",
+      "print \"As before %.2f ft**3/lbm\"%(v)\n",
+      "print \"Solution for e\";\n",
+      "print \"As before %.2f lbm/s*ft**1)\"%(V/v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Method A\n",
+        "Solution for a\n",
+        "The temperature is 560.00 R\n",
+        "Solution for b\n",
+        "The pressure is 7.35 psia\n",
+        "Solution for c\n",
+        "The velocity is 2898.59 ft/s\n",
+        "Solution for d\n",
+        "The specific volume is 28.20 ft**3/lbm\n",
+        "Solution for e\n",
+        "The mass velocity is 102.78 lbm/s*ft**2)\n",
+        "Method B\n",
+        "Solution for a\n",
+        "The temperature is 248.89 R\n",
+        "Solution for b\n",
+        "The pressure is 125.58 psia\n",
+        "Solution for c\n",
+        "As before 2898.59 ft/s\n",
+        "Solution for d\n",
+        "As before 28.20 ft**3/lbm\n",
+        "Solution for e\n",
+        "As before 102.78 lbm/s*ft**1)\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.41  Page No : 304"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#For Methane(CH4,MW = 16)\n",
+      "p = 500; \t\t\t#evaluate specific volume at p pressure \t\t\t#Unit:psia\n",
+      "pc = 674; \t\t\t#critical temperature \t\t\t#Unit:psia\n",
+      "T = 50+460; \t\t\t#evaluate specific volume at T temperature \t\t\t#Unit:R\n",
+      "Tc = 343; \t\t\t#critical temperature \t\t\t#Unit:R\n",
+      "R = 1545./16; \t\t\t#gas constant R  =  1545/Molecular Weight \t\t\t#ft*lbf/lbm*R\n",
+      "pr = p/pc; \t\t\t#reduced pressure \t\t\t#unit:psia\n",
+      "Tr = T/Tc; \t\t\t#reduced temperature \t\t\t#unit:R\n",
+      "#Reading figure 6.28 at these values gives\n",
+      "Z = 0.93; \t\t\t#compressibility factor\n",
+      "#Z = (p*v)/(R*T)\n",
+      "v = Z*((R*T)/(p*144)); \t\t\t#ft**3/lbm \t\t\t#1 ft**2 = 144 in**2(for conversion of unit) \t\t\t#specific volume\n",
+      "print \"Using the value of Z = 0.93,  the specific volume is %.2f ft**3/lbm\"%(v);\n",
+      "#For ideal gas,\n",
+      "v = (R*T)/(p*144); \t\t\t#ft**3/lbm \t\t\t#1 ft**2 = 144 in**2(for conversion of unit) \t\t\t#specific volume\n",
+      "print \"For the ideal gas, the specific volume is %.2f ft**3/lbm\"%(v);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Using the value of Z = 0.93,  the specific volume is 0.64 ft**3/lbm\n",
+        "For the ideal gas, the specific volume is 0.68 ft**3/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch7_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch7_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch7_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch7_2.ipynb
new file mode 100755
index 00000000..5b56ae9e
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch7_2.ipynb
@@ -0,0 +1,1099 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:66356009206cf2a059db6adf4246ee44a4d1b30e2da1d2c932b5c468912f3519"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Mixtures of Ideal Gases"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1  Page No : 322"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#As the basis of the calculation,assume that we have 1 lbm of mixture.Also,take the molecular weight of oxygen to be 32.00 and nitrogen to be 28.02.(from table7.1)\n",
+      "print \"Solution for a\";\n",
+      "nO2 = 0.2315/32; \t\t\t#no of moles of oxygen = ratio of mass and molecular weight \t\t\t#0.2315 lb of oxygen per pound\n",
+      "print \"The moles of oxygen is %.2f mole/lbm of mixture\"%(nO2);\n",
+      "nN2 = 0.7685/28.02; \t\t\t#no of moles of nitrogen = ratio of mass and molecular weight \t\t\t#0.7685 lb of nitrogen per pound\n",
+      "print \"The moles of nitrogen is %.2f mole/lbm of mixture\"%(nN2);\n",
+      "nm = nO2+nN2; \t\t\t#Unit:Mole/lbm \t\t\t#number of moles of gas mixture is sum of the moles of its constituent gases\n",
+      "print \"The total number of moles is %.2f mole/lbm\"%(nm); \n",
+      "xO2 = nO2/nm; \t\t\t#mole fraction of oxygen = ratio of no of moles of oxygen and total moles in mixture\n",
+      "xN2 = nN2/nm; \t\t\t#mole fraction of nitrogen = ratio of no of moles of oxygen and total moles in mixture\n",
+      "print \"The mole fraction of oxygen is %.2f and the mole fraction of nitrogen is %.2f\"%(xO2,xN2);\n",
+      "#(Check:xO2+xN2 = 1)\n",
+      "print \"xO2+xN2 = %.2f\"%(xO2+xN2);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "# the air is at 14.7 psia\n",
+      "pO2 = xO2*14.7; \t\t\t#the partial pressure of oxygen = pressure of air * the mole fraction of oxygen \t\t\t#psia\n",
+      "print \"The partial pressure of oxygen is %.2f psia\"%(pO2);\n",
+      "pN2 = xN2*14.7; \t\t\t#the partial pressure of nitrogen = pressure of air * the mole fraction of nitrogen \t\t\t#psia\n",
+      "print \"The partial pressure of nitrogen is %.2f psia\"%(pN2);\n",
+      "\n",
+      "print \"Solution for c\";\n",
+      "MWm = (xO2*32) + (xN2*28.02); \t\t\t#the molecular weight of air = sum of products of mole fraction of each gas component\n",
+      "print \"The molecular weight of air is %.2f\"%(MWm);\n",
+      "\n",
+      "print \"Solution for d\";\n",
+      "Rm = 1545/MWm; \t\t\t#the gas constant of air\n",
+      "print \"The gas constant of air is %.2f\"%(Rm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "The moles of oxygen is 0.01 mole/lbm of mixture\n",
+        "The moles of nitrogen is 0.03 mole/lbm of mixture\n",
+        "The total number of moles is 0.03 mole/lbm\n",
+        "The mole fraction of oxygen is 0.21 and the mole fraction of nitrogen is 0.79\n",
+        "xO2+xN2 = 1.00\n",
+        "Solution for b\n",
+        "The partial pressure of oxygen is 3.07 psia\n",
+        "The partial pressure of nitrogen is 11.63 psia\n",
+        "Solution for c\n",
+        "The molecular weight of air is 28.85\n",
+        "Solution for d\n",
+        "The gas constant of air is 53.55\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2  Page No : 323"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#For Gaseous Freon-12 (CCl2F2)\n",
+      "#MW of air = 29 & MW of freon-12 = 120.9\n",
+      "#initial pressure in math.tank is atmospheric pressure that is 14.7 psia\n",
+      "#final pressure of math.tank is 1000 psia\n",
+      "#The partial pressure of the Freon-12 is 1000-14.7\n",
+      "print \"The partial pressure of the Freon-12 is %.2f\"%(1000-14.7)\n",
+      "#the mole fraction of air = the initial pressure / final pressure\n",
+      "print \"The mole fraction of air is %.2f\"%(14.7/1000)\n",
+      "#the mole fraction of freon = the partial pressure of freon / the final pressure\n",
+      "print \"The mole fraction of Freon-12 is %.2f\"%((1000-14.7)/1000)\n",
+      "MWm = ((14.7/1000)*29) + (((1000-14.7)/1000)*120.9);\t\t\t#the molecular weight of mixture = sum of products of mole fraction of each gas component \n",
+      "print \"The molecular weight of the mixture is %.2f\"%(MWm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The partial pressure of the Freon-12 is 985.30\n",
+        "The mole fraction of air is 0.01\n",
+        "The mole fraction of Freon-12 is 0.99\n",
+        "The molecular weight of the mixture is 119.55\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3  Page No : 323"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Ten pounds of air,1 lb of carbon dioxide,and 5 lb of nitrogen are mixed at constant temperature until the mixture pressure is constant\n",
+      "nair = 10./29; \t\t\t#no of moles of air = ratio of mass and molecular weight \t\t\t#10 lb of nitrogen per pound \t\t\t#molecular weight of air = 29\n",
+      "print \"The moles of air is %.2f mole/lbm of mixture\"%(nair);\n",
+      "nCO2 = 1./44; \t\t\t#no of moles of carbon dioxide = ratio of mass and molecular weight \t\t\t#1 lb of per pound \t\t\t#molecular weight of CO2 = 44\n",
+      "print \"The moles of carbon dioxide is %.2f mole/lbm of mixture\"%(nCO2);\n",
+      "nN2 = 5./28; \t\t\t#no of moles of nitrogen = ratio of mass and molecular weight \t\t\t#5 lb of nitrogen per pound \t\t\t#molecular weight of N2 = 28\n",
+      "print \"The moles of nitrogen is %.2f mole/lbm of mixture\"%(nN2);\n",
+      "nm = nair+nCO2+nN2; \t\t\t#Unit:Mole/lbm \t\t\t#number of moles of gas mixture is sum of the moles of its constituent gases\n",
+      "print \"The total number of moles is %.2f mole/lbm\"%(nm); \n",
+      "\n",
+      "xair = nair/nm \t\t\t#mole fraction of air = ratio of no of moles of air and total moles in mixture\n",
+      "xCO2 = nCO2/nm; \t\t\t#mole fraction of carbon dioxide = ratio of no of moles of carbon dioxide and total moles in mixture\n",
+      "xN2 = nN2/nm; \t\t\t#mole fraction of nitrogen = ratio of no of moles of oxygen and total moles in mixture\n",
+      "print \"The mole fraction of air is %.2f \"%(xair);\n",
+      "print \"The mole fraction of carbon dioxide is %.2f\"%(xCO2)\n",
+      "print \"The mole fraction of nitrogen is %.2f\"%(xN2);\n",
+      "\n",
+      "#final pressure of is 100 psia\n",
+      "pair = xair*100; \t\t\t#the partial pressure of air =  final pressure * the mole fraction of air \t\t\t#psia\n",
+      "print \"The partial pressure of air is %.2f psia\"%(pair);\n",
+      "pCO2 = xCO2*100; \t\t\t#the partial pressure of carbon dioxide =  final pressure * the mole fraction of CO2 \t\t\t#psia\n",
+      "print \"The partial pressure of carbon dioxide is %.2f psia\"%(pCO2);\n",
+      "pN2 = xN2*100; \t\t\t#the partial pressure of nitrogen = final pressure * the mole fraction of nitrogen \t\t\t#psia\n",
+      "print \"The partial pressure of nitrogen is %.2f psia\"%(pN2);\n",
+      "\n",
+      "#the molecular weight of mixture = sum of products of mole fraction of each gas component\n",
+      "MWm = (xair*29) + (xCO2*44) + (xN2*28); \t\t\t#The molecular weight of air \n",
+      "print \"The molecular weight of air is %.2f\"%(MWm);\n",
+      "\n",
+      "Rm = 1545/MWm; \t\t\t#the gas constant of air\n",
+      "print \"The gas constant of air is %.2f\"%(Rm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The moles of air is 0.34 mole/lbm of mixture\n",
+        "The moles of carbon dioxide is 0.02 mole/lbm of mixture\n",
+        "The moles of nitrogen is 0.18 mole/lbm of mixture\n",
+        "The total number of moles is 0.55 mole/lbm\n",
+        "The mole fraction of air is 0.63 \n",
+        "The mole fraction of carbon dioxide is 0.04\n",
+        "The mole fraction of nitrogen is 0.33\n",
+        "The partial pressure of air is 63.14 psia\n",
+        "The partial pressure of carbon dioxide is 4.16 psia\n",
+        "The partial pressure of nitrogen is 32.70 psia\n",
+        "The molecular weight of air is 29.30\n",
+        "The gas constant of air is 52.74\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4  Page No : 325"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#five moles of oxygen and 10 moles of hydrogen are mixed\n",
+      "#The total number of moles is 10+5 = 15.Therefore,mole fraction of each constituent is\n",
+      "xO2 = 5./15; \t\t\t#The mole fraction of oxygen \n",
+      "xH2 = 10./15; \t\t\t#The mole fraction of hydrogen\n",
+      "print \"The mole fraction of oxygen is %.2f and of hydrogen is %.2f\"%(xO2,xH2);\n",
+      "#the molecular weight of mixture = sum of products of mole fraction of each gas component(MW of O2 = 32 and MW of H2 = 2.016)\n",
+      "print \"The molecular weight of the final mixture is %.2f\"%((((5./15)*32)+10./15)*2.016)\n",
+      "R = 1545./32; \t\t\t#the gas constant of oxygen\n",
+      "T = 460.+70; \t\t\t#absolute temperature \t\t\t#Unit:R\n",
+      "p = 14.7; \t\t\t#pressure \t\t\t#psia\n",
+      "#The partial volume of the oxygen can be found as follows:per pound of oxygen,\n",
+      "#p*vO2 = R*T;\n",
+      "vO2 = (R*T)/(p*144); \t\t\t#ft**3/lbm \t\t\t#1 in**2 = 144 ft**2  \n",
+      "#Because there are 5 moles of oxygen,each containing 32 lbm,\n",
+      "VO2 = vO2*5*32; \t\t\t#ft**3 \t\t\t#partial volume of oxygen\n",
+      "print \"The partial volume of oxygen is %.2f ft**3\"%(VO2);\n",
+      "#For the hydrogen,we can simplify the procedure by noting that the fraction of the total volume occupied by the oxygen is the same as its mole fraction.Therefore,\n",
+      "Vm = 3*VO2; \t\t\t#total volume occupied \t\t\t#ft**3\n",
+      "print \"The mixture volume is %.2f ft**3\"%(Vm);\n",
+      "#and the hydrogen volume\n",
+      "VH2 = Vm-VO2; \t\t\t#Ft**2 \t\t\t#partial volume of hydrogen\n",
+      "print \"From simplified procedure, The partial volume of hydrogen is %.2f ft**3\"%(VH2);\n",
+      "\n",
+      "#We could obtain the partial volume of hydrogen by proceeding as we did for the oxygen.Thus,\n",
+      "#p*vH2 = R*T;\n",
+      "R = 1545/2.016; \t\t\t#the gas constant of hydrogen\n",
+      "vH2 = (R*T)/(p*144); \t\t\t#ft**3/lbm \t\t\t#1 in**2 = 144 ft**2  \n",
+      "#Because there are 10 moles of hydrogen,each containing 2.016 lbm,\n",
+      "VH2 = vH2*10*2.016; \t\t\t#ft**3 \t\t\t#partial volume of hydrogen\n",
+      "print \"The partial volume of hydrogen is %.2f ft**3\"%(VH2);\n",
+      "#Which checks our previous values.\n",
+      "\n",
+      "\n",
+      "print \"From another method\"\n",
+      "#As an alternative to the foregoing,we could also use the fact that at 14.7 psia and 32F a mole of any gas occupies a volume of 358 ft**3.\n",
+      "print \"At 70F and 14.7 psia, a mole occupies %.2f ft**3\"%((((358*460+70.)/460+32))); \n",
+      "#Therefore, 5 moles of oxygen occupies \n",
+      "VO2 = 5*358*((460+70.)/(460+32)); \t\t\t#The partial volume of oxygen \t\t\t#ft**3\n",
+      "print \"The partial volume of oxygen is %.2f ft**3\"%(VO2);\n",
+      "#and 10 moles of hydrogen occupies\n",
+      "VH2 = 10*358*((460+70.)/(460+32)); \t\t\t#The partial volume of hydrogen \t\t\t#ft**3\n",
+      "print \"The partial volume of hydrogen is %.2f ft**3\"%(VH2);\n",
+      "#Both values are in good agreement with the previous calculations.\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mole fraction of oxygen is 0.33 and of hydrogen is 0.67\n",
+        "The molecular weight of the final mixture is 22.85\n",
+        "The partial volume of oxygen is 1934.17 ft**3\n",
+        "The mixture volume is 5802.51 ft**3\n",
+        "From simplified procedure, The partial volume of hydrogen is 3868.34 ft**3\n",
+        "The partial volume of hydrogen is 3868.34 ft**3\n",
+        "From another method\n",
+        "At 70F and 14.7 psia, a mole occupies 390.15 ft**3\n",
+        "The partial volume of oxygen is 1928.25 ft**3\n",
+        "The partial volume of hydrogen is 3856.50 ft**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5  Page No : 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Referring to figure 7.3,we have for CO2,\n",
+      "nCO2 = 10./44; \t\t\t#mole \t\t\t#no of moles of carbon dioxide = ratio of mass and molecular weight \t\t\t#10 lb of per pound \t\t\t#molecular weight of CO2 = 44\n",
+      "#and for N2,\n",
+      "nN2 = 5/28.02; \t\t\t#mole \t\t\t#no of moles of nitrogen = ratio of mass and molecular weight \t\t\t#5 lb of nitrogen per pound\n",
+      "print \"The total number of moles in the mixture is %.2f mole\"%(nCO2+nN2);\n",
+      "#Therefore,\n",
+      "xCO2 = nCO2/(nCO2+nN2); \t\t\t#mole fraction of carbon dioxide = ratio of no of moles of carbon dioxide and total moles in mixture\n",
+      "xN2 = nN2/(nCO2+nN2); \t\t\t#mole fraction of nitrogen = ratio of no of moles of oxygen and total moles in mixture\n",
+      "print \"The mole fraction of carbon dioxide is %.2f and the mole fraction of nitrogen is %.2f\"%(xCO2,xN2);\n",
+      "#the molecular weight of mixture = sum of products of mole fraction of each gas component\n",
+      "MWm = (xCO2*44) + (xN2*28.02); \t\t\t#the molecular weight of mixture\n",
+      "print \"The molecular weight of air is %.2f\"%(MWm);\n",
+      "#Because the mixture is 15 lbm (10CO2 + 5N2),the volume of the mixture is found from pm*Vm = mm*Rm*Tm\n",
+      "pm = 100.; \t\t\t#mixture pressure \t\t\t#psia \n",
+      "Tm = 460.+70; \t\t\t#mixture temperature \t\t\t#R(absolute temperature)\n",
+      "Rm = 1545/37.0; \t\t\t#gas constant of mixture \n",
+      "mm = 15.; \t\t\t#mass of mixture \t\t\t#Unit:lb\n",
+      "#So,rearranging the equation,gives\n",
+      "Vm = (mm*Rm*Tm)/(pm*144);  \t\t\t#mixture volume \t\t\t#ft**3 \t\t\t#1 in**2 =  144 ft**2\n",
+      "print \"The mixture volume is %.2f ft**3\"%(Vm);\n",
+      "#the partial volume of carbon dioxide is the total volume multiplied by the mole fraction.Thus,\n",
+      "VCO2 = Vm*xCO2; \t\t\t#the partial volume of CO2 \t\t\t#ft**3\n",
+      "print \"The partial volume of carbon dioxide is %.2f ft**3\"%(VCO2);\n",
+      "VN2 = Vm*xN2; \t\t\t#the partial volume of N2 \t\t\t#ft**3\n",
+      "print \"The partial volume of nitrogen is %.2f ft**3\"%(VN2);\n",
+      "#The partial pressure of each constituent is proportional to its mole fraction,for these conditions,\n",
+      "pCO2 = pm*xCO2; \t\t\t#the partial pressure of carbon dioxide =  final pressure * the mole fraction of CO2 \t\t\t#psia\n",
+      "print \"The partial pressure of carbon dioxide is %.2f psia\"%(pCO2);\n",
+      "pN2 = pm*xN2; \t\t\t#the partial pressure of nitrogen = final pressure * the mole fraction of nitrogen \t\t\t#psia\n",
+      "print \"The partial pressure of nitrogen is %.2f psia\"%(pN2);\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total number of moles in the mixture is 0.41 mole\n",
+        "The mole fraction of carbon dioxide is 0.56 and the mole fraction of nitrogen is 0.44\n",
+        "The molecular weight of air is 36.97\n",
+        "The mixture volume is 23.05 ft**3\n",
+        "The partial volume of carbon dioxide is 12.91 ft**3\n",
+        "The partial volume of nitrogen is 10.14 ft**3\n",
+        "The partial pressure of carbon dioxide is 56.02 psia\n",
+        "The partial pressure of nitrogen is 43.98 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6  Page No : 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#we will assume that we have 100 volumes of gas mixture and set up table 7.2.In first coloumn,we tabulate the gas,and in the second coloumn,we tabulate the given   volume fractions.Because the mole fraction equals to volume fraction,the values in coloumn 3 are the same as those in coloumn 2.\n",
+      "#The molecular weight is obtained from table 7.1.Because the MW of the mixture is the sum of the individual mole fraction multiplied by the respective molecular      weights,the next coloumn tabulates the product of the mole fraction multiplied by molecular weight(3*4).The sum of these entries is the molecular weight of the    mixture,which for this case is 33.4.\n",
+      "print \"Basis:100 volumes of gas mixture\"\n",
+      "print \"gas    Volume       Mole           Molecular                            mass  \"\n",
+      "print \"       fraction    fraction x      weight MW     xMW                  fraction\"\n",
+      "print \"CO2    0.40        0.40          44.0           %.2f               %.2f\"%(0.40*44.0,(0.40*44.0)/33.4)\n",
+      "print \"N2     0.10        0.10          28.02          %.2f                %.2f                \"%(28.02*0.10,(28.02*0.10)/33.4)\n",
+      "print \"H2     0.10        0.10          2.016          %.2f                %.2f                \"%(0.10*2.016,(0.10*2.016)/33.4)\n",
+      "print \"O2     0.40        0.40          32.0           %.2f               %.2f                \"%(0.40*32.0,(0.40*32.0)/33.4)\n",
+      "print \"       1.00        1.00                         33.4 = MWm                =  1.000               \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Basis:100 volumes of gas mixture\n",
+        "gas    Volume       Mole           Molecular                            mass  \n",
+        "       fraction    fraction x      weight MW     xMW                  fraction\n",
+        "CO2    0.40        0.40          44.0           17.60               0.53\n",
+        "N2     0.10        0.10          28.02          2.80                0.08                \n",
+        "H2     0.10        0.10          2.016          0.20                0.01                \n",
+        "O2     0.40        0.40          32.0           12.80               0.38                \n",
+        "       1.00        1.00                         33.4 = MWm                =  1.000               \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7  Page No : 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#We will take as a basis 100 lbm of mixture.\n",
+      "#Dividing colomn 2 by 3 gives us mass/molecular weight  or moles of each constituents.The total number of moles in the mixture is the sum of coloumn 4,and the      molecular weight of the mixture is the mass of the mixture(100 lbm) divided by the number of moles\n",
+      "#In coloumn 5,mole fraction is given by moles/total mole\n",
+      "\n",
+      "print \"Basis:100 pounds of gas mixture\"\n",
+      "print \"gas    Mass        Molecular      Moles                   Mole         Percent      \"\n",
+      "print \"        lbm        weight MW                           fraction        Volume        \"\n",
+      "print \"CO2    52.7          44.0          1.2                   %.2f             %.2f   \"%(1.2/3,1.2/3*100)\n",
+      "print \"N2      8.4          28.02         0.3                   %.2f             %.2f   \"%(0.3/3,0.3/3*100)\n",
+      "print \"H2      0.6          2.016         0.3                   %.2f             %.2f   \"%(0.3/3,0.3/3*100)\n",
+      "print \"O2     38.3          32.0          1.2                   %.2f             %.2f   \"%(1.2/3,1.2/3*100)\n",
+      "print \"        = 100.0                      = 3.0                  = 1.00                =  100               \"\n",
+      "print \"                              MWm = 100/3 = 33.3               \"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Basis:100 pounds of gas mixture\n",
+        "gas    Mass        Molecular      Moles                   Mole         Percent      \n",
+        "        lbm        weight MW                           fraction        Volume        \n",
+        "CO2    52.7          44.0          1.2                   0.40             40.00   \n",
+        "N2      8.4          28.02         0.3                   0.10             10.00   \n",
+        "H2      0.6          2.016         0.3                   0.10             10.00   \n",
+        "O2     38.3          32.0          1.2                   0.40             40.00   \n",
+        "        = 100.0                      = 3.0                  = 1.00                =  100               \n",
+        "                              MWm = 100/3 = 33.3               \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8  Page No : 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#We will take as a basis 100 lbm of mixture.\n",
+      "#Dividing colomn 2 by 3 gives us mass/molecular weight  or moles of each constituents.The total number of moles in the mixture is the sum of coloumn 4,and the      molecular weight of the mixture is the mass of the mixture(100 lbm) divided by the number of moles\n",
+      "#In coloumn 5,mole fraction is given by moles/total mole\n",
+      "\n",
+      "print \"Basis:100 pounds of gas mixture\"\n",
+      "print \"gas    Mass        Molecular      Moles                   Mole         Percent      \"\n",
+      "print \"        lbm        weight MW                           fraction        Volume        \"\n",
+      "print \"O2     23.18         32.00         0.724                   %.2f             %.2f   \"%(0.724/3.45,0.724/3.45*100)\n",
+      "print \"N2     75.47         28.02         2.693                   %.2f             %.2f   \"%(2.692/3.45,2.692/3.45*100)\n",
+      "print \"A       1.30         39.90         0.033                   %.2f             %.2f   \"%(0.033/3.45,0.033/3.45*100)\n",
+      "print \"CO2     0.05         44.00          -                       -             -  \"\n",
+      "print \"        = 100.00                      = 3.45                  = 1.00                =  100               \"\n",
+      "print \"                            MWm = 100/3.45 = 28.99               \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Basis:100 pounds of gas mixture\n",
+        "gas    Mass        Molecular      Moles                   Mole         Percent      \n",
+        "        lbm        weight MW                           fraction        Volume        \n",
+        "O2     23.18         32.00         0.724                   0.21             20.99   \n",
+        "N2     75.47         28.02         2.693                   0.78             78.03   \n",
+        "A       1.30         39.90         0.033                   0.01             0.96   \n",
+        "CO2     0.05         44.00          -                       -             -  \n",
+        "        = 100.00                      = 3.45                  = 1.00                =  100               \n",
+        "                            MWm = 100/3.45 = 28.99               \n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9  Page No : 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Given: cp of oxygen is 0.23 Btu/lbm*R.cp of nitrogen is 0.25 Btu/lbm*R. 160 lbm/hr of oxygen and 196 lbm/hr of nitrogen are mixed.oxygen is at 500 F and nitrogen  is at 200 F.\n",
+      "\n",
+      "#The energy equation for the steady-flow,adiaatic mixing process gives us the requirement that the enthalpy of the mixture must equal to the enthalpies of the      components,because deltah = q = 0.An alternative statement of this requirement is that the gain in enthalpy of the nitrogen must equal the decrease in enthalpy of the   oxygen.Using the latter statement,that the change in enthalpy of nitrogen,yields\n",
+      "# (160*0.23*(500-tm))  =  (196*0.25*(tm-200)) where tm = mixture temperature\n",
+      "#where m*cp*deltat has been used for deltah. \t\t\t#cp = specific heat at constant pressure \t\t\t#Unit for cp is Btu/lbm*R\n",
+      "#rearranging the above equation,\n",
+      "tm = ((500.*160*0.23)+(196*0.25*200))/((196*0.25)+(160*0.23)); \t\t\t#tm = mixture temperature \t\t\t#Unit:fahrenheit\n",
+      "print \"The final temperature of the mixture is %.2f F\"%(tm);\n",
+      "#Using the requirement that the enthalpy of the mixture must equal to the sum of the enthalpies of the components yields an alternative solution to this problem.Let  us assume that at 0 F,the enthalpy of each gas and of the mixture is zero.The enthalpy of the entering oxygen is (160*0.23*(500-0)),and the enthalpy of the        entering nitrogen is (196*0.25*(200-0)).The enthalpy of the mixture is ((160+196)*cpm*(tm-0))\n",
+      "#Therefore,  (160*0.23*500)+(196*0.25*200)  =  ((160+196)*cpm*tm)\n",
+      "cpm = ((160./(160+196))*0.23)+((196/(160+196))*0.25); \t\t\t#specific heat at constant pressure for gas mixture \t\t\t#Btu/lbm*R\n",
+      "print \"For mixture, Specific heat at constant pressure is %.2f Btu/lbm*R\"%(cpm);\n",
+      "#therefore,\n",
+      "tm = ((160*0.23*500)+(196*0.25*200))/(cpm*(160.+196)); \t\t\t#tm = mixture temperature \t\t\t#Unit:fahrenheit\n",
+      "print \"By Using value of cpm, The final temperature of the mixture is %.2f F\"%(tm);\n",
+      "#The use of 0 F as a base was arbitrary but convenient.Any base would yield the same results.\n",
+      "#The answer of cpm is wrong in the book.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final temperature of the mixture is 328.67 F\n",
+        "For mixture, Specific heat at constant pressure is 0.10 Btu/lbm*R\n",
+        "By Using value of cpm, The final temperature of the mixture is 766.30 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10  Page No : 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Problem 7.9 is carried out as a nonflow mixing process.\n",
+      "#Given in problem 7.9,: cp of oxygen is 0.23 Btu/lbm*R.cp of nitrogen is 0.25 Btu/lbm*R. 160 lbm/hr of oxygen and 196 lbm/hr of nitrogen are mixed.oxygen is at 500F and nitrogen  is at 200 F. \t\t\t#cp = specific heat at constant pressure \n",
+      "#Given in problem 7.10,: cv of oxygen is 0.164 Btu/lbm*R.cv of nitrogen is 0.178 Btu/lbm*R. \t\t\t#cv = specific heat at constant volume\n",
+      "\n",
+      "#Because this is a nonflow process,the energy equation for this process requires the internal energy of the mixture to equal to the sum of the internal energy of   its components.\n",
+      "#Alternatively,the decrease in internal energy of the oxygen must equal the increase in internal energy of the nitrogen.Using latter statement gives us,\n",
+      "# (160*0.164*(500-tm))  =  (196*0.178*(tm-200))\n",
+      "#where m*cv*deltat has been used for deltau. \t\t\t#Unit for cp & cv is Btu/lbm*R\n",
+      "#rearranging the above equation,\n",
+      "tm = ((500*160*0.164)+(196*0.178*200))/((196*0.178)+(160*0.164)); \t\t\t#tm = mixture temperature \t\t\t#Unit:fahrenheit\n",
+      "\n",
+      "# Results\n",
+      "print \"The final temperature of the mixture is %.2f F\"%(tm);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final temperature of the mixture is 328.78 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12  Page No : 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#The change in entropy of the mixture is the sum of the changes in entropy of each component.\n",
+      "#Given in problem 7.9,: cp of oxygen is 0.23 Btu/lbm*R.cp of nitrogen is 0.25 Btu/lbm*R. 160 lbm/hr of oxygen and 196 lbm/hr of nitrogen are mixed.oxygen is at 500F and nitrogen  is at 200 F. \t\t\t#cp = specific heat at constant pressure \n",
+      "#In 7.9,for the oxygen,the temperature starts at 500F(960 R) and decreases to 328.7 F.For the nitrogen,the temperature starts at 200F(660 R) and increase to 328.7 F.\n",
+      "#deltas  =  (cp*math.log(T2/T1)); \t\t\t#Unit:Btu/lbm*R \t\t\t#change in entropy\n",
+      "\n",
+      "#For the oxygen,\n",
+      "cp = 0.23; \t\t\t#specific heat at constant pressure \t\t\t#Unit:Btu/lbm*R\n",
+      "T2 = 328.7+460; \t\t\t#Unit:R \t\t\t#final temperature\n",
+      "T1 = 500+460; \t\t\t#Unit:R \t\t\t#starting temperature\n",
+      "deltas = (cp*math.log(T2/T1)); \t\t\t#Unit:Btu/lbm*R \t\t\t#change in entropy for oxygen\n",
+      "DeltaS = 160*deltas; \t\t\t#Btu/R \t\t\t#The total change in entropy of the oxygen\n",
+      "print \"The total change in entropy of the oxygen is %.2f Btu/R\"%(DeltaS);\n",
+      "\n",
+      "#For the nitrogen,\n",
+      "cp = 0.25; \t\t\t#specific heat at constant pressure \t\t\t#Unit:Btu/lbm*R\n",
+      "T2 = 328.7+460; \t\t\t#Unit:R \t\t\t#final temperature\n",
+      "T1 = 200+460; \t\t\t#Unit:R \t\t\t#starting temperature\n",
+      "deltas = (cp*math.log(T2/T1)); \t\t\t#Unit:Btu/lbm*R \t\t\t#change in entropy for nitrogen\n",
+      "deltaS = 196*deltas; \t\t\t#Btu/R \t\t\t#The total change in entropy of the nitrogen\n",
+      "print \"The total change in entropy of the nitrogen is %.2f Btu/R\"%(deltaS);\n",
+      "deltaS = deltaS+DeltaS; \t\t\t#the total change in entropy for the mixture \t\t\t#Btu/lbm*R\n",
+      "print \"The total change in entropy for the mixture is %.2f Btu/R\"%(deltaS);\n",
+      "\n",
+      "#Per pound of mixture,\n",
+      "deltasm = deltaS/(196+160); \t\t\t#increase in entropy per pound mass of mixture\n",
+      "print \"Increase in entropy per pound mass of mixture is %.2f Btu/lbm*R\"%(deltasm);\n",
+      "\n",
+      "\n",
+      "print \"An alternative solution:\";\n",
+      "#As an alternative solution,assume an arbitrary datum of 0 F(460 R).\n",
+      "cp = 0.23; \t\t\t#specific heat at constant pressure \t\t\t#Unit:Btu/lbm*R\n",
+      "#For initial entropy of oxygen,\n",
+      "T2 = 500+460; \t\t\t#Unit:R \t\t\t#final temperature\n",
+      "T1 = 0+460; \t\t\t#Unit:R \t\t\t#starting temperature\n",
+      "deltas = cp*math.log(T2/T1); \t\t\t#the initial change in entropy for oxygen \t\t\t# Btu/lbm*R\n",
+      "print \"The initial change in entropy for oxygen is %.2f Btu/lbm*R\"%(deltas);\n",
+      "#For final entropy of oxygen,\n",
+      "T2 = 328.7+460; \t\t\t#Unit:R \t\t\t#final temperature\n",
+      "T1 = 0+460; \t\t\t#Unit:R \t\t\t#starting temperature\n",
+      "Deltas = cp*math.log(T2/T1); \t\t\t#the final change in entropy for oxygen \t\t\t# Btu/lbm*R\n",
+      "print \"The final change in entropy for oxygen is %.2f Btu/lbm*R\"%(Deltas);\n",
+      "deltaS = Deltas-deltas; \t\t\t#The entropy change of the oxygen \t\t\t#Btu/lbm*R\n",
+      "print \"The entropy change of the oxygen is %.2f Btu/lbm*R\"%(deltaS);\n",
+      "\n",
+      "#For nitrogen,\n",
+      "cp = 0.25; \t\t\t#specific heat at constant pressure \t\t\t#Unit:Btu/lbm*R\n",
+      "#For initial entropy of nitrogen,\n",
+      "T2 = 200.+460; \t\t\t#Unit:R \t\t\t#final temperature\n",
+      "T1 = 0+460; \t\t\t#Unit:R \t\t\t#starting temperature\n",
+      "deltas = cp*math.log(T2/T1); \t\t\t#the initial change in entropy for nitrogen \t\t\t# Btu/lbm*R\n",
+      "print \"The initial change in entropy for nitrogen is %.2f Btu/lbm*R\"%(deltas);\n",
+      "#For final entropy of nitrogen,\n",
+      "T2 = 328.7+460; \t\t\t#Unit:R \t\t\t#final temperature\n",
+      "T1 = 0+460; \t\t\t#Unit:R \t\t\t#starting temperature\n",
+      "Deltas = cp*math.log(T2/T1); \t\t\t#the final change in entropy for nitrogen \t\t\t# Btu/lbm*R\n",
+      "print \"The final change in entropy for nitrogen is %.2f Btu/lbm*R\"%(Deltas);\n",
+      "deltaS = Deltas-deltas; \t\t\t#The entropy change of the nitrogen \t\t\t#Btu/lbm*R\n",
+      "print \"The entropy change of the nitrogen is %.2f Btu/lbm*R\"%(deltaS);\n",
+      "\n",
+      "#The remainder of the problem is as before.The advantage of Using this alternative method is the negative math.logarithms are avoided by choomath.sing a reference          temperature lower than any other temperature in the system\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total change in entropy of the oxygen is -7.23 Btu/R\n",
+        "The total change in entropy of the nitrogen is 8.73 Btu/R\n",
+        "The total change in entropy for the mixture is 1.50 Btu/R\n",
+        "Increase in entropy per pound mass of mixture is 0.00 Btu/lbm*R\n",
+        "An alternative solution:\n",
+        "The initial change in entropy for oxygen is 0.16 Btu/lbm*R\n",
+        "The final change in entropy for oxygen is 0.12 Btu/lbm*R\n",
+        "The entropy change of the oxygen is -0.04 Btu/lbm*R\n",
+        "The initial change in entropy for nitrogen is 0.09 Btu/lbm*R\n",
+        "The final change in entropy for nitrogen is 0.13 Btu/lbm*R\n",
+        "The entropy change of the nitrogen is 0.04 Btu/lbm*R\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13  Page No : 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Referring to figure 7.6,it will be seen that the cooling of an air-water vapor mixture from B to A proceeds at constant pressure until the saturation curve is   reached.\n",
+      "#At 80 F(the mixture temperature),the Steam Tables give us a saturation pressure of a 0.5073 psia,and because the relative humidity is 50%,the vapor pressure of  the water is 0.5*0.5073 = 0.2537 psia.\n",
+      "#Using the steam tables,the saturation temperature corresponding to 0.2537 psia is 60 F.\n",
+      "#So,\n",
+      "print \"The dew point temperature of the air is 60 F\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dew point temperature of the air is 60 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14  Page No : 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#To solve this probelm,it is necessary to determine the properties of the saturated mixture 90 F.If the air is saturated at 90 F,the partial pressure of the water  vapor is found directly from the Steam Tables as 0.6988 psia,and the specific volume of the water vapor is 467.7 ft**3/lbm of vapor.\n",
+      "print \"The partial pressure of the dry air is %.2f psia\"%(14.7-0.6988); \t\t\t#the mixture is at 14.7 psia\n",
+      "R = 1545/28.966; \t\t\t#gas constant of dry air = 1545/Molecular weight\n",
+      "T = 90+460; \t\t\t#temperature of dry air \t\t\t#Unit:R\n",
+      "pdryair = 14.0; \t\t\t#psia \t\t\t#pressure of dry air\n",
+      "#Applying the ideal gas equation to the air,\n",
+      "vdryair = (R*T)/(pdryair*144); \t\t\t#volume of dry air \t\t\t#ft**3/lbm \t\t\t#1 in**2 = 144 Ft**2\n",
+      "#the mass of dry air in the 467.7 ft**3 container \n",
+      "print \"The mass of dry air in the 467.7 ft**3 container is %.2f lbm\"%(467.7/vdryair);\n",
+      "#To obtain relative humidity(phy),it is necessary to determine the mole fraction of water vapor for both the saturated mixture and the mixture in question.\n",
+      "#The saturated mixture contains 1 lbm of water vapor or 1/18.016 moles  = 0.055 mole of water vapor and (467.7/vdryair)/28.966 = 1.109 moles of dry air.\n",
+      "#For the saturated mixture, the ratio of moles of water vapor to moles of mixture is 0.055/(0.055+1.109) = 0.0477\n",
+      "#For the actual mixture,the moles of water vapor per pound of dry air is 0.005/18.016 = 0.000278 and 1 lbm of dry air is 1/28.966 = 0.0345 mole.So,the mole of water vapor per mole of  mixture at the conditions of the  mixture is 0.000278/(0.0345+0.000278) = 0.00799\n",
+      "#From the defination of relative humidity,\n",
+      "print \"The relative humidity of the mixture is %.2f \"%((0.00799/0.0477)*100);\n",
+      "\n",
+      "#Because the mole ratio is also the ratio of the partial pressures for the ideal gas,phy can be expressed as the ratio of the partial pressure of the water vapor in  the mixture to the partial pressure of the water vapor at saturation.Therefore,\n",
+      "print \"The partial pressure of the vapor at saturation is %.2f psia\"%((0.00799/0.0477)*0.6988);\n",
+      "print \"And the partial pressure of the dry air in the mixture is %.2f psia\"%((14.7-0.00799/0.0477)*0.6988); \t\t\t#14.7-The partial pressure of the vapor at saturation\n",
+      "#The dew point temperature is the saturation temperature corresponding to the partial pressure of the water vapor in the mixture.So,\n",
+      "print \"The dew point temperature corresponding to %.2f psia is 39F\"%((0.00799/0.0477)*0.6988);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The partial pressure of the dry air is 14.00 psia\n",
+        "The mass of dry air in the 467.7 ft**3 container is 32.14 lbm\n",
+        "The relative humidity of the mixture is 16.75 \n",
+        "The partial pressure of the vapor at saturation is 0.12 psia\n",
+        "And the partial pressure of the dry air in the mixture is 10.16 psia\n",
+        "The dew point temperature corresponding to 0.12 psia is 39F\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15  Page No : 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Problem 7.14 Using equations, Rm = ((ma/(ma+mv))*Ra)+((mv/(ma+mv))*Rv)   and  phy*pvs = pv\n",
+      "W = 0.005; \t\t\t#Humidity ratio\n",
+      "pm = 14.7; \t\t\t#mixture is at 14.7 psia\n",
+      "#W = 0.622*(pv/(pm-pv))\n",
+      "#Rearranging,\n",
+      "pv = (W*pm)/(0.622+W); \t\t\t#the partial pressure of the water vapor \n",
+      "print \"The partial pressure of the water vapor is %.2f psia\"%(pv);\n",
+      "pa = pm-pv; \t\t\t#pa = the partial pressure of the dry air in the mixture\n",
+      "print \"The partial pressure of dry air is %.2f psia\"%(pa);\n",
+      "#It is necessary to obtain pvs from the Steam Tables at 90 F.This is 0.6988 psia.\n",
+      "pvs = 0.6988; \t\t\t#saturation pressure of water vapor at the temperature of mixture\n",
+      "print \"The partial pressure of the water vapor at saturation is %.2f psia\"%(pvs);\n",
+      "#Therefore,\n",
+      "phy = pv/pvs; \t\t\t#relative humidity\n",
+      "print \"The relative humidity  is %.2f percent\"%(phy*100);\n",
+      "#The dew point temperature is the saturation temperature corresponding to 0.117 psia,which is found from the Steam Tables to be 39 F.\n",
+      "print \"The dew point temperature of the mixture is 39 F\";\n",
+      "#The results of this problem and problem 7.14 are in good agreement\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The partial pressure of the water vapor is 0.12 psia\n",
+        "The partial pressure of dry air is 14.58 psia\n",
+        "The partial pressure of the water vapor at saturation is 0.70 psia\n",
+        "The relative humidity  is 16.78 percent\n",
+        "The dew point temperature of the mixture is 39 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.16  Page No : 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "pm = 14.7; \t\t\t#the barometer is at 14.7 psia \t\t\t#mixture is at 14.7 psia\n",
+      "#The amount of water vapor removed (per pound of dry air) is the difference between the humidity ratio (specific humidity) at inlet and outlet of the conditioning    unit.We shall therefore evalute W for both specified conditions.Because phy = pv/pvs,\n",
+      "#At 90F:\n",
+      "phy = 0.7; \t\t\t#relative humidity\n",
+      "pvs = 0.6988; \t\t\t#psia \t\t\t#saturation pressure of water vapor at the temperature of mixture\n",
+      "pv = phy*pvs; \t\t\t#psia \t\t\t#the partial pressure of the water vapor \n",
+      "pa = pm-pv; \t\t\t#psia \t\t\t#pa = the partial pressure of the dry air in the mixture\n",
+      "W = 0.622*(pv/pa); \t\t\t#Humidity ratio\n",
+      "\n",
+      "#At 80F:\n",
+      "phy = 0.4; \t\t\t#relative humidity\n",
+      "pvs = 0.5073; \t\t\t#psia \t\t\t#saturation pressure of water vapor at the temperature of mixture\n",
+      "pv = phy*pvs; \t\t\t#psia \t\t\t#the partial pressure of the water vapor \n",
+      "pa = pm-pv; \t\t\t#psia \t\t\t#pa = the partial pressure of the dry air in the mixture\n",
+      "w = 0.622*(pv/pa); \t\t\t#Humidity ratio\n",
+      "\n",
+      "print \"The amount of water removed per pound of dry air is %.2f\"%(W-w);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of water removed per pound of dry air is 0.01\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17  Page No : 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Problem 7.13 Using the psychrometric chart\n",
+      "#Entering figure 7.11 at a dry-bulb temperature of 80 F,we proceed vertically until we reach 50% humidity curve.At this intersection,we proceed horizontally and    read the dew-point temperature as approximately 60 F.\n",
+      "print \"The dew point temperature of air is 60 F\";\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dew point temperature of air is 60 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.18  Page No : 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Problem 7.14 Using the psychrometric chart\n",
+      "#In this problem,we are given the moisture content of the air to be 0.005 lb per pound of dry air.\n",
+      "#This corresponds to 0.005*7000 = 35 grains per pound of dry air.\n",
+      "#Entering the chart at 90F and proceeding verticaly to 35 grains per pound of dry air,we find the dew point to be 39F by proceeding horizontally to the intersection with the saturation curve. \n",
+      "print \"The dew-point temperature of the mixture is 39 F\";\n",
+      "print \"The relative humidity is approximately 17 percent\";\n",
+      "#From the leftmost scale,we read the pressure of water vapor to be 0.12 psia.\n",
+      "print \"The partial pressure of the air is %.2f psia\"%(14.7-0.12);\n",
+      "#Comparing these results to problem 7.14,indicated good agreement between the results obtained by chart and by calculation\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dew-point temperature of the mixture is 39 F\n",
+        "The relative humidity is approximately 17 percent\n",
+        "The partial pressure of the air is 14.58 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.19  Page No : 348"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Problem 7.16 Using the psychrometric chart\n",
+      "#The initial conditions are 90 F and 70% relative humidity\n",
+      "#Entering the chart at 90 F dry bulb temperature and proceeding vertically to 70% relative humidity,we find the air to have 150 grains water vapor per pound of dry air.At the final condition of 80F and 40% relative humidity,we read 61 grains of water/lb of dry air.\n",
+      "#So,\n",
+      "print \"The water removed is %.2f grains per pound of dry air\"%(150-61);\n",
+      "print \"Or %.2f lb of water per pound of dry air is removed\"%((150.-61)/7000);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The water removed is 89.00 grains per pound of dry air\n",
+        "Or 0.01 lb of water per pound of dry air is removed\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.20  Page No : 349"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#dry bulb temperature is 50 F\n",
+      "#relative humidity is 50 percent\n",
+      "#We first locate 50 F and 50 percent relative humidity on figure 7.11.At this state,we read 26 grains of water per pound of dry air and a total heat of 16.1 Btu per  pound of a dry air.\n",
+      "#We now proceed horizontally to 80 F at a constant value of 26 grains of water per pound of dry air and read a total heat of 23.4 Btu per pound of dry air.\n",
+      "print \"The heat required is %.2f Btu per pound of dry air\"%(23.4-16.1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat required is 7.30 Btu per pound of dry air\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.21  Page No : 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#An evaporative cooling process\n",
+      "#Because the exit air is saturated,we find the exit condition on the curve corresponding to a wet-bulb temperature of 50 F.The process is carried out at constant     total enthalpy,which is along a line of constant wet-bulb temperature.\n",
+      "#Proceeding along the 50 F wet-bulb temperature line of figure 7.11 diagonally to the right until it intersects with the vertical 80 F dry-bulb temperature line    yields a relative humidity of approximately 4 %\n",
+      "print \"For An evaporative cooling process,  The relative humidity of the entering air is 4 percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For An evaporative cooling process,  The relative humidity of the entering air is 4 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.22  Page No : 356"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#As noted from figure 7.27, 1 lb of mixture,4/5 lb of indoor air,and 1/5 lb of outdoor air are mixed per pound of mixture. \n",
+      "#We now locate the two end states on the psychrometric chart and connect them with a straight line.The line connecting the end states is divided  into 5 equal parts. Using the results of equation, (ha-ha2)/(ha-ha1)  =  (W2-W)/(W-W1)  =  ma1/ma2  =  l1/l2 ,we now proceed from the 75 F indoor air state 1 part toward the 90F outdoor air state.This Locates \n",
+      "print \"The final mixture,  which is found to be a dry-bulb temperature of approximately 78 F, \\\n",
+      "a wet-bulb temperature of 66 F and relative humidity of 54 percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final mixture,  which is found to be a dry-bulb temperature of approximately 78 F, a wet-bulb temperature of 66 F and relative humidity of 54 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.23  Page No : 358"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The cooling tower\n",
+      "#From the Steam tables,\n",
+      "#For water:\n",
+      "h100F = 68.05;  \t\t\t#Btu/lbm \t\t\t#enthalpy at 100 F\n",
+      "h70F = 38.09;   \t\t\t#Btu/lbm  \t\t\t#enthalpy at 70 F\n",
+      "#For air:\n",
+      "h = 20.4; \t\t\t#Unit:Btu/lb \t\t\t#at inlet,total heat/lb dry air\n",
+      "w = 38.2; \t\t\t#Unit:grains/lb  \t\t\t#at inlet,moisture pickup/lb dry air (at 60F D.B. and 50% R.H.)\n",
+      "H = 52.1; \t\t\t#Unit:Btu/lb \t\t\t#at outlet,total heat/lb dry air\n",
+      "W = 194.0; \t\t\t#Unit:grains/lb \t\t\t#at outlet,moisture pickup/lb dry air (at 90F D.B. and 90% R.H.)\n",
+      "\n",
+      "#Per pound of dry air,the heat interchange is H-h Btu per pound of dry air.\n",
+      "#Per pound of dry air,the moisture increase is (W-w)/7000 lb per pound of dry air.\n",
+      "#From the equation, ma*(H-h)  =  200000*h100F - mwout*h70F             \t\t\t#ma = mass of air  mwout = mass of cooled water \n",
+      "#and  ma*((W-w)/7000)  =  200000 - mwout\n",
+      "#Solving the latter equation for mwout,we have  mwout = 200000-(ma*((W-w)/7000))\n",
+      "#Substituting this into the heat balance yields,\n",
+      "# ma*(H-h)  =  200000*h100F - 200000*h70F + ma*h70F*((W-w)/7000)\n",
+      "#Solving gives us,\n",
+      "ma = (200000*(h100F-h70F))/((H-h)-(h70F*((W-w)/7000))); \t\t\t#The amount of air required per hour \t\t\t#Unit:lbm/hr of dry air\n",
+      "print \"The amount of air required per hour is %.2f lbm/hr of dry air\"%(ma);\n",
+      "print \"The amount of water lost per hour due to evaporation is %.2f lbm/hr\"%((ma*W-w)/7000)\n",
+      "#note that the water evaporated is slightly over 2% of the incoming water,and this is the makeup that has to be furnished to the tower.\n",
+      "#answer are slightly differ because of value of (W-w)/7000 is given 0.0233 instead of 0.0225\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of air required per hour is 194216.14 lbm/hr of dry air\n",
+        "The amount of water lost per hour due to evaporation is 5382.56 lbm/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch8_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch8_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch8_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch8_2.ipynb
new file mode 100755
index 00000000..5c6dac8c
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch8_2.ipynb
@@ -0,0 +1,697 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:772f93702cda26918438fd90c36e1594260fd40351e3deed6d4e9bb6febe0344"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Vapor Power Cycles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1  Page No : 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From the Steam Tables or Mollier chart in Appendix 3,we find that\n",
+      "hf = 340.49; \t\t\t#Unit:kJ/kg \t\t\t#at 50kPa \t\t\t#enthalpy\n",
+      "h1 = hf; \t\t\t#at 50kPa \t\t\t#hf = enthalpy of saturated liquid \t\t\t#Unit:kJ/kg\n",
+      "h4 = 3230.9; \t\t\t#Unit:kJ/kg \t\t\t#enthalpy\n",
+      "h5 = 2407.4; \t\t\t#Unit:kJ/kg \t\t\t#\t\t\t#enthalpy\n",
+      "#Here,point 5 is in the wet steam region.\n",
+      "print \"Solution for a\";\n",
+      "#Neglecting pump work (h2 = h1) gives\n",
+      "nR = (h4-h5)/(h4-h1); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"The thermal efficiency of the cycle is %.2f percentage\"%(nR*100);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "p2 = 3000; \t\t\t#Unit:kPa \t\t\t#Upper pressure\n",
+      "p1 = 50; \t\t\t#Unit:kPa \t\t\t#Lower pressure\n",
+      "vf = 0.001030; \t\t\t#Specific volume of saturated liquid  \t\t\t#m**3/kg\n",
+      "Pumpwork = (p2-p1)*vf; \t\t\t#Unit:kJ/kg \t\t\t#pump work\n",
+      "#The efficiency of the cycle including pump work is\n",
+      "nR = ((h4-h5)-Pumpwork)/((h4-h1)-Pumpwork); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"The thermal efficiency of the cycle including pump work is %.2f percentage\"%(nR*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "The thermal efficiency of the cycle is 28.49 percentage\n",
+        "Solution for b\n",
+        "The thermal efficiency of the cycle including pump work is 28.42 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2  Page No : 381"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Using the computer disk to obtain the neccesary properties\n",
+      "print \"Solution for a\";\n",
+      "#For the conditions given in problem8.1,the properties are found to be\n",
+      "hf = 340.49; \t\t\t#Unit:kJ/kg \t\t\t#at 50kPa \t\t\t#enthalpy\n",
+      "h1 = hf; \t\t\t#at 50kPa \t\t\t#hf = enthalpy of saturated liquid\n",
+      "h2 = h1; \t\t\t#Enthalpy \t\t\t#Unit:kJ/kg\n",
+      "h4 = 3230.9; \t\t\t#Unit:kJ/kg \t\t\t#enthalpy\n",
+      "h5 = 2407.4; \t\t\t#Unit:kJ/kg \t\t\t#enthalpy\n",
+      "#Neglecting pump work \n",
+      "nR = (h4-h5)/(h4-h2); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"The thermal efficiency of the cycle is %.2f percentage\"%(nR*100);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "#For the pump work,we do not need the approximation,because the computerized tables give us the necessary values directly.\n",
+      "#Assuming that the condensate leaving the condenser is saturated liquid gives us an enthalpy of 340.54 kJ/kg and an entropy of 1.0912 kJ/kg*K for an isentropic compression, the final cond-ition is the boiler pressure of 3Mpa and an entropy of 1.0912 kJ/kg*K. For these values,the program yields an enthalpy of 343.59 kJ/kg*K.The isentropic pump work is equal to \n",
+      "Pumpwork = 343.59-340.54; \t\t\t#Unit:kJ/kg \t\t\t#pumpwork\n",
+      "#The efficiency of the cycle including pump work is\n",
+      "nR = ((h4-h5)-Pumpwork)/((h4-h1)-Pumpwork); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"The thermal efficiency of the cycle including pump work is %.2f percentage\"%(nR*100);\n",
+      "#Final results in this problem agree with the result in problem8.1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "The thermal efficiency of the cycle is 28.49 percentage\n",
+        "Solution for b\n",
+        "The thermal efficiency of the cycle including pump work is 28.42 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3  Page No : 382"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Solution for (a)\n",
+      "#Figurre 8.3 with the cycle extending into the superheat region and expanding along 4->5 is the appropriate diagram for this process.\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "#This problem can be solved either by use of the Mollier chart or the Steam Tables.If the chart is used,14.696 psia is first located on the saturated vapor line.Because the expansion,4->5,is isentropic,a vertical line on the chart is the path of the process.The point corresponding to 4 in figure 8.3 is found where this vertical line intersects 400 psia.At this point,the ent-halpy is 1515 Btu/lbm,and the corresponding temperature is approximatelty 980F.Saturated vapor at 14.696 psia  has an enthalpy of 1150.5 Btu/lbm(from the Mollier chart).The Steam Tables sh-ow that saturated liquid at 14.696 psia has an enthalpy of 180.15 Btu/lbm.In terms of figure 8.3,and neglecting pump work,we have \n",
+      "h1 = 180.15; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h2 = h1; \t\t\t#Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "h4 = 1515; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h5 = 1150.5; \t\t\t#Unit:kJ/kg \t\t\t#enthalpy\n",
+      "#Neglecting pump work yields\n",
+      "nR = (h4-h5)/(h4-h2); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"Neglecting the pump work, The thermal efficiency of the cycle is %.2f percentage\"%(nR*100);\n",
+      "p2 = 400; \t\t\t#Unit:Psia \t\t\t#Upper pressure\n",
+      "p1 = 14.696; \t\t\t#Unit:Psia \t\t\t#Lower pressure\n",
+      "vf = 0.01167; \t\t\t#Specific volume of saturated liquid  \t\t\t#ft**3/lbm\n",
+      "J = 778; \t\t\t#Conversion factor\n",
+      "Pumpwork = ((p2-p1)*vf*144)/J; \t\t\t#Unit:Btu/lbm \t\t\t#1ft**2 = 144 in**2 \t\t\t#pumpwork\n",
+      "#The efficiency of the cycle including pump work is\n",
+      "nR = ((h4-h5)-Pumpwork)/((h4-h1)-Pumpwork); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"The thermal efficiency of the cycle including pump work is %.2f percentage\"%(nR*100);\n",
+      "#where the denominator is h4-h2 = h4-h1-(h2-h1).Neglecting pump work is obviously justified in this case.An alternative solution is obtained by Using the Steam Tables:at 14.696 psia ans sat-uration,sg = 1.7567 ; at 400 psia,s =  1.7567.From Table 3(at 400 psia)\n",
+      "#   s      h      t\n",
+      "#1.7632  1523.6  1000 \n",
+      "#1.7567  1514.2  982.4\n",
+      "#1.7558  1512.9  980"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for b\n",
+        "Neglecting the pump work, The thermal efficiency of the cycle is 27.31 percentage\n",
+        "The thermal efficiency of the cycle including pump work is 27.26 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4  Page No : 383"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Refer to figure8.3.The desired  quantities are obtained as follows:\n",
+      "#at 14.696 psia,saturated vapor (x = 1),s = 1.7566 Btu/lbm*R\n",
+      "h5 = 1150.4; \t\t\t#Unit:Btu/lbm \t\t\t#enthaply\n",
+      "#at 14.696 psia,saturated liquid (x = 0),s = 0.3122 Btu/lbm*R\n",
+      "h2 = 180.17; \t\t\t#Unit:Btu/lbm \t\t\t#enthaply \n",
+      "h1 = h2;\n",
+      "#at 400 psia,s = 1.7566 Btu/lbm*R,\n",
+      "h4 = 1514.0; \t\t\t#Unit:Btu/lbm \t\t\t#Enthalpy\n",
+      "t = 982.07; \t\t\t#Unit:F \t\t\t#tempearature\n",
+      "#at 400 psia,s = 0.3122 Btu/lbm*R, \t\t\t#s = entropy\n",
+      "h = 181.39; \t\t\t#Unit:Btu/lbm \t\t\t#Enthalpy\n",
+      "#Note the agreement of these values with the ones obtained for problem8.4.Alos,note the temperature of 982.07F compared to 982.4F.Continuing,\n",
+      "#Neglecting pump work \n",
+      "nR = (h4-h5)/(h4-h2); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"Neglecting the pump work, The thermal efficiency of the cycle is %.2f percentage\"%(nR*100);\n",
+      "Pumpwork = h-h2; \t\t\t#Unit:kJ/kg \t\t\t#/pumpwork\n",
+      "#The efficiency of the cycle including pump work is\n",
+      "nR = ((h4-h5)-Pumpwork)/((h4-h2)-Pumpwork); \t\t\t#Thermal efficiency of the cycle\n",
+      "print \"The thermal efficiency of the cycle including pump work is %.2f percentage\"%(nR*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Neglecting the pump work, The thermal efficiency of the cycle is 27.26 percentage\n",
+        "The thermal efficiency of the cycle including pump work is 27.19 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5  Page No : 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The Carnot cycle would operate between 982.4F and 212F.\n",
+      "T1 = 982.4+460; \t\t\t#temperature converted to absolute temperature \t\t\t#Unit:R\n",
+      "T2 = 212+460; \t\t\t#temperature converted to absolute temperature \t\t\t#Unit:R\n",
+      "nc = ((T1-T2)/T1)*100; \t\t\t#Efficiency of carnot cycle\n",
+      "print \"The efficiency is %.2f percentage\"%(nc);\n",
+      "#In problem 8.3,\n",
+      "nR = 27.3; \t\t\t#Thermal efficiency of the cycle neglecting the pump work\n",
+      "typen = (nR/nc)*100; \t\t\t#Type efficiency = ideal thermal efficiency/efficiency of carnot cycle operating between min and max temperature limits\n",
+      "print \"The type efficiency of the ideal Rankine cycle is %.2f percentage\"%(typen);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency is 53.41 percentage\n",
+        "The type efficiency of the ideal Rankine cycle is 51.11 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6  Page No : 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#For the upper temperature of the cycle,we have 400C,and for 50kPa,the steam tables give us a saturation temperature of 81.33C.The efficiency of a Carnot cycle operating between the limits would be\n",
+      "T1 = 400+273; \t\t\t#Celcius temperature converted to fahrenheit temperature\n",
+      "T2 = 81.33+273; \t\t\t#temperature converted to fahrenheit temperature\n",
+      "nc = ((T1-T2)/T1)*100; \t\t\t#Efficiency of carnot cycle\n",
+      "print \"The efficiency is %.2f percentage\"%(nc);\n",
+      "#In problem 8.1,\n",
+      "nR = 28.5; \t\t\t#Thermal efficiency of the cycle neglecting the pump work\n",
+      "typen = (nR/nc)*100; \t\t\t#Type efficiency = ideal thermal efficiency/efficiency of carnot cycle operating between min and max temperature limits\n",
+      "print \"The type efficiency of the ideal Rankine cycle is %.2f percentage\"%(typen);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency is 47.35 percentage\n",
+        "The type efficiency of the ideal Rankine cycle is 60.19 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7  Page No : 386"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From problem 8.3,\n",
+      "work = 1515-1150.5; \t\t\t#Unit:Btu/lbm of steam \t\t\t#pump work is neglected \t\t\t#Useful ideal work\n",
+      "#Because of the heat losses, 50 Btu/lbm of the 364.5 Btu/lbm becomes unavailable.\n",
+      "available = 364.5-50; \t\t\t#Unit:Btu/lbm \n",
+      "n = available/(1515-180.15); \t\t\t#Thermal efficiency of the cycle neglecting pump work h4 = 1515; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy & h1 = 180.15; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "print \"The thermal efficiency of the cycle neglecting pump work is %.2f percentage\"%(n*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The thermal efficiency of the cycle neglecting pump work is 23.56 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8  Page No : 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Neglecting the pump work,we have\n",
+      "heatrate = 3413/0.273; \t\t\t#Unit:Btu/kWh \t\t\t#0.273 = efficiency \t\t\t#1 kWh = 3413 \t\t\t#heat rate\n",
+      "print \"The heat rate is %.2f Btu/kWh\"%(heatrate);\n",
+      "#Per pound of steam,1515-1150.5 = 364.5 Btu is delivered.\n",
+      "#Because 1 kWh = 3413\n",
+      "print \"The steam rate is %.2f lbm of steam per kilowatt-hour\"%(3413./1515-1150.5)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat rate is 12501.83 Btu/kWh\n",
+        "The steam rate is -1148.25 lbm of steam per kilowatt-hour\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9  Page No : 388"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#The Mollier chart provides a convenient way of solving this problem.Expanding from 980F,400 psia,s = 1.7567 to 200 psia yields a final enthalpy of 1413 Btu/lbm.Expanding from 200 psia ans  an enthalpy of 1515 Btu/lbm to 14.696 psia yields a final enthaply of 1205 Btu/lbm. \n",
+      "h4 = 1515; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h5 = 1205; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h7 = 1413; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "h1 = 180.15; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "nreheat = ((h4-h5)+(h4-h7))/((h4-h1)+(h4-h7)); \t\t\t#The efficiency of the reheat cycle\n",
+      "print \"The efficiency of the reheat cycle is %.2f percentage\"%(nreheat*100);\n",
+      "#It is apparent that for the conditions of this problem,the increase in efficiency is not very large.The final condition of the fluid after the second expansion is superheated steam at \n",
+      "#14.696 psia.By condenmath.sing at this relatively high pressure condition,a large amount of heat is rejected to the condenser cooling water.7\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency of the reheat cycle is 28.67 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10  Page No : 389"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Some of the property data required was found in problem8.4.In addition we have,\n",
+      "#at 200 psia,s = 1.7566 Btu/lbm*R,\n",
+      "h7 = 1413.6; \t\t\t#Unit:Btu/lbm \t\t\t#Enthalpy\n",
+      "#at 200 psia,s = 1.8320 Btu/lbm*R,\n",
+      "h4 = 1514.0; \t\t\t#Unit:Btu/lbm \t\t\t#Enthalpy\n",
+      "#at 14.696 psia,s = 1.8320 Btu/lbm*R,\n",
+      "h5 = 1205.2; \t\t\t#Unit:Btu/lbm \t\t\t#Enthalpy\n",
+      "h1 = 180.17; \t\t\t#Unit:Btu/lbm \t\t\t#enthalpy\n",
+      "\n",
+      "# Calculations\n",
+      "#Using these data,\n",
+      "nreheat = ((h4-h5)+(h4-h7))/((h4-h1)+(h4-h7)); \t\t\t#The efficiency of the reheat cycle \n",
+      "\n",
+      "# Results\n",
+      "print \"The efficiency of the reheat cycle is %.2f percentage\"%(nreheat*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency of the reheat cycle is 28.53 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11  Page No : 394"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print \"Solution for a\";\n",
+      "#For the rankine cycle,the Mollier chart gives\n",
+      "h4 = 1505; \t\t\t#Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "h5 = 922; \t\t\t#Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "h6 = h5; \t\t\t#Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "#and at the condenser,\n",
+      "h1 = 69.74; \t\t\t#enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "nR = (h4-h5)/(h4-h1); \t\t\t#efficiency of rankine cycle\n",
+      "print \"The efficiency of rankine cycle is %.2f percentage\"%(nR*100);\n",
+      "\n",
+      "print \"Solution for b\";\n",
+      "#Figure 8.16 shows the regenerative cycle.After doing work(isentropically),W lbs of steam are bled from the turbine at 50 psia for each lbm of steam leaving the steam generator,and (1-W)    pound goes through the turbine and is condensed in the condenser to saturated liquid at 1 psia.This condensate is pumped to the heater,where it mixes with the extraced steam and leaves as  saturated liquid at 50 psia.The required enthalpies are:\n",
+      "#Leaving turbine:\n",
+      "h5 = 1168; \t\t\t#Btu/lbm at 50 psia \n",
+      "#Leaving condenser:\n",
+      "h7 = 69.74; \t\t\t#Btu/lbm at 1 psia \t\t\t# is equal to h8 if pump work is neglected\n",
+      "#Leaving heater:\n",
+      "h1 = 250.24; \t\t\t#Btu/lbm at 50 psia \t\t\t#is equal to h2 if pump work is neglected(saturated liquid)\n",
+      "#A Heat balance around the heater gives\n",
+      "#W*h5 + (1-W)*h7  =  1*h1 \n",
+      "W = ((1*h1)-h7)/(h5-h7); \t\t\t#Unit:lbm \t\t\t#W lb of steam \n",
+      "print \"W = %.2f lbm\"%(W);\n",
+      "work = (1-W)*(h4-922) + W*(h4-h5); \t\t\t#h5 = 922 from the mollier chart \t\t\t#Unit:Btu/lbm \t\t\t#The work output\n",
+      "print \"The work output is %.2f Btu/lbm\"%(work);\n",
+      "#Heat into steam generator equals the enthalpy leaving minus the enthalpy of the saturated liquid entering at 50 psia:\n",
+      "qin = h4-h1; \t\t\t#Unit:Btu/lbm \t\t\t#Heat in\n",
+      "n = work/qin; \t\t\t#Efficiency of regenerative cycle\n",
+      "print \"The efficiency of regenerative cycle is %.2f percentage\"%(n*100);\n",
+      "#The efficiency of a regenerative cycle with one open heater is given by \n",
+      "n = 1-(((h5-h1)*(h6-h7))/((h4-h1)*(h5-h7)));  \t\t\t#efficiency of a regenerative cycle\n",
+      "W = (h1-h7)/(h5-h7); \t\t\t#Unit:lbm  \t\t\t#W lb of steam\n",
+      "print \"When the rankine cycle is compared with regenerative cycle\"\n",
+      "print \"W = %.2f lbm and the efficiency of a regenerative cycle with one open heater is given by %.2f percentage\"%(W,n*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Solution for a\n",
+        "The efficiency of rankine cycle is 40.62 percentage\n",
+        "Solution for b\n",
+        "W = 0.16 lbm\n",
+        "The work output is 542.57 Btu/lbm\n",
+        "The efficiency of regenerative cycle is 43.24 percentage\n",
+        "When the rankine cycle is compared with regenerative cycle\n",
+        "W = 0.16 lbm and the efficiency of a regenerative cycle with one open heater is given by 43.24 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12  Page No : 396"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Figure 8.16(a) shows the cycle.For this cycle,W2 pounds are extracted at 100 psia,and W1 pounds are extracted at 50 psia for each pound produced by the steam generator.The enthalpies that  are required are:\n",
+      "#Leaving turbine: 922 \t\t\t#Btu/lbm at 1 psia\n",
+      "#Leaving condenser: 69.74 \t\t\t#Btu/lbm at 1 psia (saturated liquid)\n",
+      "#Leaving low pressure heater: 250.24 \t\t\t#Btu/lbm at 50 psia (saturated liquid)\n",
+      "#Leaving high pressure heater: 298.61 \t\t\t#Btu/lbm at 100 psia\n",
+      "#At low pressure extraction: 1168 \t\t\t#Btu/lbm at 50 psia\n",
+      "#At high pressure extraction: 1228.6 \t\t\t#Btulbm at 100 psia\n",
+      "#Entering turbine: 1505 \t\t\t#Btu/lbm\n",
+      "#The heat balance around the high pressure heater gives us\n",
+      "#W2*1228.6 + (1-W2)*250.24  =  1*298.61\n",
+      "W2 = ((1*298.61)-250.24)/(1228.6-250.24); \t\t\t#lbm \t\t\t#W2 pounds are extracted at 100 psia\n",
+      "print \"W2 = %.2f lbm\"%(W2);\n",
+      "#A heat balance around the low pressure heater yields\n",
+      "#W1*1168 + (1-W1-W2)*69.74  =  (1-W2)*250.24\n",
+      "W1 = (((1-W2)*250.24)-69.74+(W2*69.74))/(1168-69.74); \t\t\t#lbm \t\t\t#W1 pounds are extracted at 50 psia\n",
+      "print \"W1 = %.2f lbm\"%(W1);\n",
+      "work = ((1505-1228.6)*1)+((1-W2)*(1228.6-1168))+((1-W1-W2)*(1168-922)); \t\t\t#The work output \t\t\t#Btu/lbm\n",
+      "print \"The work output is %.2f Btu/lbm\"%(work);\n",
+      "#Heat into the steam generator equals the enthalpy leaving minus the enthalpy of saturated liquid at 100 psia:\n",
+      "qin = 1505-298.61; \t\t\t#Btu/lbm \t\t\t#Heat in \n",
+      "print \"Heat in  =  %.2f Btu/lbm\"%(qin);\n",
+      "n = work/qin; \t\t\t#The efficiency\n",
+      "print \"The efficiency is %.2f percentage\"%(n*100);\n",
+      "#In terms of figure 8.16a,\n",
+      "#W2 = (h1-h11)/(h5-h11)\n",
+      "#W1 = (h5-h1/h6-h9)*(h10-h9/h5-h10) neglecting the pump work\n",
+      "#n = 1-(h7-h8/h4-h1)*(h5-h1/h5-h10)*(h6-h10/h6-h8)\n",
+      "#For this problem , h8 = h9 , h10 = h11 and h1 = h2.Thus\n",
+      "W2 = (298.61-250.24)/(1228.6-250.24); \t\t\t#lbm \t\t\t#W2 pounds are extracted at 100 psia\n",
+      "print \"Comparing the results\"\n",
+      "print \"W2 = %.2f lbm\"%(W2);\n",
+      "W1 = ((1228.6-298.61)*(250.24-69.74))/((1168-69.74)*(1228.6-250.24)); \t\t\t#lbm \t\t\t#W1 pounds are extracted at 50 psia\n",
+      "print \"W1 = %.2f lbm\"%(W1); \n",
+      "n = 1-(((922-69.74)*(1228.6-298.61)*(1168-250.24))/((1505-298.61)*(1228.6-250.24)*(1168-69.74))); \t\t\t#Efficiency\n",
+      "print \"The efficiency is %.2f percentage\"%(n*100);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "W2 = 0.05 lbm\n",
+        "W1 = 0.16 lbm\n",
+        "The work output is 529.41 Btu/lbm\n",
+        "Heat in  =  1206.39 Btu/lbm\n",
+        "The efficiency is 43.88 percentage\n",
+        "Comparing the results\n",
+        "W2 = 0.05 lbm\n",
+        "W1 = 0.16 lbm\n",
+        "The efficiency is 43.88 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.13  Page No : 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Regenerative cycle\n",
+      "#Assume that 1 lbm of steam leaves the steam generator and that W1 lbm is bled off to the closed heater at 100psia and that W2 lbm is bled off to the open heater at 50 psia.Alos,assume that the feedwater leaving the closed heater at 310F,18F less than the saturation temperature corresponding to 100 psia.For calculation purposes,we will use hf at 310 F for this enthalpy.Using the Mollier diagram and the steam tables,we find the following values of enthalpy:\n",
+      "\n",
+      "#h to turbine = 1505 Btu/lbm(at 1000 psia and 1000F)\n",
+      "#h at first extraction = 1228 Btu/lbm(isentropically to 100 psia)\n",
+      "#h at second extraction = 1168 Btu/lbm(isentropically to 100 psia)\n",
+      "#h at turbine exit = 922 Btu/lbm (isentropically to 1 psia)\n",
+      "#hf = 298.61 Btu/lbm(at 100 psia)\n",
+      "#hf = 250.24 Btu/lbm(at 50 psia)\n",
+      "#hf = 280.06 Btu/lbm(at 310 F)\n",
+      "#hf = 69.74 Btu/lbm (at 1 psia)\n",
+      "#A heat balance around the high pressure heater gives us\n",
+      "#W1*(1228-298.61)  =  1*(280.06-250.24)\n",
+      "W1 = ((1*(280.06-250.24)))/(1228-298.61); \t\t\t#lbm \t\t\t#W1 lbm is extracted at 100 psia\n",
+      "print \"W1 = %.2f lbm\"%(W1);\n",
+      "#A heat balance around the open heater gives us\n",
+      "#W2*1168 +(1-W1-W2)*69.74 + W1*268.61  =  1*250.24\n",
+      "W2 = ((1*250.24)-(W1*268.61)-69.74+(W1*69.74))/(1168-69.74); \t\t\t#lbm \t\t\t#W2 lbm is extracted at 50 psia\n",
+      "print \"W2 = %.2f lbm\"%(W2);\n",
+      "#The work output of the cycle consists of the work that 1 lbm does in expanding isentropically to 100 psia,plus the work done by (1-W1)lbm expanding isentropicaly from 100 to 50 psia,plus the work done by (1-W1-W2)lbm expanding isentropically from 50 to 1 psia.\n",
+      "#Numerically,the work is\n",
+      "workoutput = (1*(1505-1228))+((1-W1)*(1228-1168))+((1-W1-W2)*(1168-922)); \t\t\t#Btu/lbm \t\t\t#the work output\n",
+      "print \"The work output is %.2f Btu/lbm\"%(workoutput);\n",
+      "heatinput = 1505-280.06; \t\t\t#Btu/lbm \t\t\t#the heat input\n",
+      "print \"The heat input is %.2f Btu/lbm\"%(heatinput);\n",
+      "n = workoutput/heatinput; \t\t\t#Efficiency\n",
+      "print \"The efficiency is %.2f percentage\"%(n*100);\n",
+      "#When compared to 8.11,we conclude that the addition of additional closed heater raises the efficiency. \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "W1 = 0.03 lbm\n",
+        "W2 = 0.16 lbm\n",
+        "The work output is 534.18 Btu/lbm\n",
+        "The heat input is 1224.94 Btu/lbm\n",
+        "The efficiency is 43.61 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14  Page No : 426"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#From problem 8.11,\n",
+      "#Leaving turbine:\n",
+      "h5 = 1168.; \t\t\t#Btu/lbm at 50 psia \n",
+      "#For the rankine cycle,the Mollier chart gives\n",
+      "h4 = 1505.; \t\t\t#Enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "h6 = 922.; \t\t\t#Enthalpy \t\t\t#Unit:Btu/lbm \t\t\t#h6 = h5;\n",
+      "#and at the condenser,\n",
+      "h1 = 69.74; \t\t\t#enthalpy \t\t\t#Unit:Btu/lbm\n",
+      "#Leaving condenser:\n",
+      "h7 = 69.74; \t\t\t#Btu/lbm at 1 psia \t\t\t# is equal to h8 if pump work is neglected\n",
+      "#Leaving heater:\n",
+      "h2 = 250.24; \t\t\t#Btu/lbm at 50 psia \t\t\t#is equal to h1 if pump work is neglected(saturated liquid)\n",
+      "#A Heat balance around the heater gives\n",
+      "#W*h5 + (1-W)*h7  =  1*h1 \n",
+      "W = ((1*h2)-h7)/(h5-h7); \t\t\t#Unit:lbm\n",
+      "liquidleaving = (W*h2)+(1-W)*h1; \t\t\t#Btu/lbm \t\t\t#liquid leaving the heatexchange\n",
+      "\n",
+      "#Using these data,,\n",
+      "heatin = h4-liquidleaving; \t\t\t#Btu/lbm \t\t\t#heat in the boiler\n",
+      "print \"Heat in at boiler is %.2f Btu/lbm\"%(heatin);\n",
+      "workout = ((1-W)*(h4-h6))+(W*(h4-h5)); \t\t\t#Btu/lbm \t\t\t#The work out of turbine\n",
+      "print \"The work out of turbine is %.2f Btu/lbm\"%(workout);\n",
+      "n = workout/heatin; \t\t\t#efficiency \t\t\t#The conventional thermal efficiency\n",
+      "print \"The conventional thermal efficiency is %.2f percentage\"%(n*100);\n",
+      "#If at this time we have define the efficiency of energy utilization to be the ratio of the work out plus the useful heat out divided by the heat input to the    cycle, nenergyutilization = ((w+qoutuseful)/qin)*100\n",
+      "qout = W*(h5-h2); \t\t\t#heat out \t\t\t#Btu/lbm\n",
+      "n = (workout+qout)/heatin; \t\t\t#efficiency of energy utilization\n",
+      "print \"Efficiency of energy utilization is %.2f percentage\"%(n*100);\n",
+      "#Comparing with 8.11, we see that conventional thermal efficiency is decreased and efficiency of energy utilization is increased\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat in at boiler is 1405.59 Btu/lbm\n",
+        "The work out of turbine is 542.57 Btu/lbm\n",
+        "The conventional thermal efficiency is 38.60 percentage\n",
+        "Efficiency of energy utilization is 49.33 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch9_1.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch9_1.ipynb
old mode 100644
new mode 100755
diff --git a/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch9_2.ipynb b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch9_2.ipynb
new file mode 100755
index 00000000..86b7d951
--- /dev/null
+++ b/Thermodynamics_and_Heat_Power_by_I._Granet_and_M._Bluestein/ch9_2.ipynb
@@ -0,0 +1,528 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c78d396c993e1067d10db02248ea7bb1149674e24ff2199a434e078a4771b88f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Gas Power Cycles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1  Page No : 462"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "Rc = 7.; \t\t\t#Compression Ratio Rc = v2/v3\n",
+      "k = 1.4;  \t\t\t#It is apparent incerease in compression ratio yields an increased cycle efficiency\n",
+      "notto = (1-(1/Rc)**(k-1))*100; \t\t\t#Efficiency of an otto engine\n",
+      "print \"The efficiency of the otto cycle is %.2f percentage\"%(notto);\n",
+      "#For the carnot cycle,\n",
+      "#Nc = 1-(T2/T4) \t\t\t#efficiency for the carnot cycle \t\t\t#T2 = lowest temperature \t\t\t#T4 = Highest temperature\n",
+      "\n",
+      "T2 = 70.+460; \t\t\t#for converting to R \t\t\t#Conversion of unit\n",
+      "#At 700 F\n",
+      "T4 = 700.+460; \t\t\t#temperatures converted to absolute temperatures;\n",
+      "nc = (1-(T2/T4))*100; \t\t\t#efficiency of the carnot cycle\n",
+      "print \"When peak temperature is 700 fahrenheit, efficiency of the carnot cycle is %.2f percentage\"%(nc);        \n",
+      "\n",
+      "#At 1000 F\n",
+      "T4 = 1000.+460; \t\t\t#temperatures converted to absolute temperatures;\n",
+      "nc = (1-(T2/T4))*100; \t\t\t#efficiency of the carnot cycle\n",
+      "print \"When peak temperature is 1000 fahrenheit, efficiency of the carnot cycle is %.2f percentage\"%(nc);\n",
+      "\n",
+      "#At 3000 F\n",
+      "T4 = 3000.+460; \t\t\t#temperatures converted to absolute temperatures;\n",
+      "nc = (1-(T2/T4))*100; \t\t\t#efficiency of the carnot cycle\n",
+      "print \"When peak temperature is 3000 fahrenheit, efficiency of the carnot cycle is %.2f percentage\"%(nc);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency of the otto cycle is 54.08 percentage\n",
+        "When peak temperature is 700 fahrenheit, efficiency of the carnot cycle is 54.31 percentage\n",
+        "When peak temperature is 1000 fahrenheit, efficiency of the carnot cycle is 63.70 percentage\n",
+        "When peak temperature is 3000 fahrenheit, efficiency of the carnot cycle is 84.68 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2  Page No : 463"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "\n",
+      "cv = 0.172; \t\t\t#Unit:Btu/(lbm*R) \t\t\t#Specific heat constant\n",
+      "Rc = 7; \t\t\t#Compression Ratio Rc = v2/v3\n",
+      "k = 1.4; \t\t\t#It is apparent incerease in compression ratio yields an increased cycle efficiency\n",
+      "T2 = 70+460; \t\t\t#for converting to R \t\t\t#Conversion of unit\n",
+      "#For 1000 F\n",
+      "T4 = 1000+460; \t\t\t#temperatures converted to absolute temperatures;\n",
+      "T3byT2 = Rc**(k-1); \t\t\t#Unit less\n",
+      "T3 = T3byT2*T2;\n",
+      "qin = cv*(T4-T3); \t\t\t#Unit:Btu/lbm \t\t\t#Heat added\n",
+      "#Qr = cv*(T5-T2)*(T5/T4) = (v2/v3)**(k-1)\n",
+      "Qr = (inv([[Rc]]))**(k-1); \t\t\t#Unit:Btu/lbm \t\t\t#Heat rejected\n",
+      "T5 = T4*Qr;\n",
+      "Qr = cv*(T5-T2); \t\t\t#Unit:Btu/lbm \t\t\t#Heat rejected\n",
+      "print \"The net work out is %.2f Btu/lbm\"%(qin-Qr);\n",
+      "notto = ((qin-Qr)/qin)*100; \t\t\t#The efficiency of otto cycle \n",
+      "print \"The efficiency of otto cycle is %.2f percentage\"%(notto);\n",
+      "#The value agrees with the results of problem 9.1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net work out is 28.44 Btu/lbm\n",
+        "The efficiency of otto cycle is 54.08 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3  Page No : 464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "cv = 0.7186; \t\t\t#Unit:kJ/(kg*K) \t\t\t#Specific heat constant for constant volume process\n",
+      "Rc = 8.; \t\t\t#Compression Ratio Rc = v2/v3\n",
+      "k = 1.4;  \t\t\t#It is apparent incerease in compression ratio yields an increased cycle efficiency\n",
+      "T2 = 20.+273; \t\t\t#20 C converted to its kelvin value\n",
+      "qin = 50.; \t\t\t#Heat added \t\t\t#Unit:kJ\n",
+      "T3byT2 = Rc**(k-1);\n",
+      "T3 = T3byT2*T2; \t\t\t#Unit:K\n",
+      "#qin = cv*(T4-T3) \t\t\t#heat added \t\t\t#Unit:kJ\n",
+      "T4 = (qin/cv)+T3; \t\t\t#The peak temperature of the cycle \t\t\t#Unit:K\n",
+      "print \"The peak temperature of the cycle is %.2f Kelvin i.e. %.2f Celcius\"%(T4,T4-273);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The peak temperature of the cycle is 742.72 Kelvin i.e. 469.72 Celcius\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4  Page No : 465"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#For an Otto cycle,\n",
+      "rc = 7.; \t\t\t#Compression Ratio Rc = v2/v3\n",
+      "q = 50.; \t\t\t#Unit:Btu/lbm \t\t\t#Heat added\n",
+      "p2 = 14.7; \t\t\t#Unit:psia \t\t\t#pressure at point 2\n",
+      "T2 = 60.+460; \t\t\t#temperatures converted to absolute temperatures; \t\t\t#Unit:R\n",
+      "cp = 0.24; \t\t\t#Unit:Btu/(lbm*R) \t\t\t#Specific heat constant for constant pressure process\n",
+      "cv = 0.171; \t\t\t#Unit:Btu/(lbm*R) \t\t\t#Specific heat constant for constant volume process\n",
+      "R = 53.3; \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "k = 1.4;  \t\t\t#It is apparent incerease in compression ratio yields an increased cycle efficiency\n",
+      "#Refering to figure 9.9,\n",
+      "#At (2),we need v2.\n",
+      "#p2*v2 = R*T2\n",
+      "v2 = (R*T2)/(p2*144); \t\t\t#Unit:ft**3/lbm \t\t\t#1ft**2 = 144 in**2 \t\t\t#specific volume at point 2\n",
+      "print \"At point 2, specific volume v2 = %.2f ft**3/lbm\"%(v2);\n",
+      "#For The isentropic path (2)&(3),p3*v3**k = p2*v2**k,so\n",
+      "#So,p3 = p2*(v2/v3)**k;\n",
+      "p3 = p2*rc**k; \t\t\t#Unit:psia \t\t\t#pressure at point 3\n",
+      "print \"At path2&3\";\n",
+      "print \"pressure p3 = %.2f psia\"%(p3);\n",
+      "v3 = v2/rc; \t\t\t#Unit:ft**3/lbm \t\t\t#specific volume at point 3\n",
+      "print \"specific volume v3 = %.2f ft**3/lbm\"%(v3);\n",
+      "T3 = (p3*v3*144)/R; \t\t\t#Unit:R \t\t\t#1ft**2 = 144 in**2 \t\t\t#temperature at point 3\n",
+      "print \"temperature T3 = %.2f R\"%(T3);\n",
+      "print \"At point4\"\n",
+      "#To obtain the values at (4),we note\n",
+      "v4 = v3; \t\t\t#Unit:ft**3/lbm \t\t\t#specific volume at point 4\n",
+      "print \"specific volume v4 = %.2f ft**3/lbm\"%(v4);\n",
+      "#qin = cv*(T4-T3)\n",
+      "T4 = T3+(q/cv); \t\t\t#Unit:R \t\t\t#temperature at point 4 \n",
+      "print \"temperature T4 = %.2f R\"%(T4);\n",
+      "#For p4,\n",
+      "p4 = (R*T4)/(144*v4); \t\t\t#Unit:psia \t\t\t#1ft**2 = 144 in**2 \t\t\t#pressure at point 4\n",
+      "print \"pressure p4 = %.2f psia\"%(p4);\n",
+      "#The last point has the same specific volume as (2),giving\n",
+      "print \"At last point\"\n",
+      "v5 = v2; \t\t\t#Unit:ft**3/lbm \t\t\t#specific volume at point 5\n",
+      "print \"specific volume v5 = %.2f ft**3/lbm\"%(v5);\n",
+      "#The isentropic path equation,p5*v5**k = p4*v4**k,so\n",
+      "p5 = p4*(v4/v5)**k; \t\t\t#Unit:psia \t\t\t#pressure at point 5\n",
+      "print \"pressure p5 = %.2f psia\"%(p5);\n",
+      "T5 = (p5*v5*144)/(R); \t\t\t#Unit:R \t\t\t#1ft**2 = 144 in**2 temperature at point 5\n",
+      "print \"temperature T5 = %.2f R\"%(T5);\n",
+      "n = (((T4-T3)-(T5-T2))/(T4-T3))*100; \t\t\t#The efficiency of the cycle\n",
+      "print \"The efficiency of the cycle is %.2f percentage\"%(n);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At point 2, specific volume v2 = 13.09 ft**3/lbm\n",
+        "At path2&3\n",
+        "pressure p3 = 224.11 psia\n",
+        "specific volume v3 = 1.87 ft**3/lbm\n",
+        "temperature T3 = 1132.51 R\n",
+        "At point4\n",
+        "specific volume v4 = 1.87 ft**3/lbm\n",
+        "temperature T4 = 1424.91 R\n",
+        "pressure p4 = 281.97 psia\n",
+        "At last point\n",
+        "specific volume v5 = 13.09 ft**3/lbm\n",
+        "pressure p5 = 18.50 psia\n",
+        "temperature T5 = 654.26 R\n",
+        "The efficiency of the cycle is 54.08 percentage\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7  Page No : 468"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#For four cycle engine,\n",
+      "#Using the results of problem 9.6,\n",
+      "pm = 1000.; \t\t\t#Unit:kPa \t\t\t#mean effective pressure \t\t\t#Unit:psia\n",
+      "N = 4000./2; \t\t\t#Power strokes per minute \t\t\t#2L engine \t\t\t#Unit:rpm\n",
+      "LA = 2.     \t\t\t#Mean \t\t\t#Unit:liters\n",
+      "hp = (pm*LA*N)/44760; \t\t\t#The horsepower \t\t\t#Unit:hp\n",
+      "print \"The horsepower is %.2f hp\"%(hp);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The horsepower is 89.37 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9  Page No : 469"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables.\n",
+      "c=0.2;      #clearance equal to 20% of its displacement\n",
+      "#An otto engine\n",
+      "#Using results of problem 9.8, \n",
+      "rc = (1+c)/c; \t\t\t#The compression ratio\n",
+      "print \"The compression ratio is %.2f\"%(rc);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The compression ratio is 6.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10  Page No : 470"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#For four cycle,six cylinder engine,\n",
+      "#Using the results of problem 9.5,\n",
+      "hp = 100; \t\t\t#Horsepower \t\t\t#Unit:hp\n",
+      "L = 4./12; \t\t\t#Unit:ft \t\t\t#stroke is 4 in.\n",
+      "A = (math.pi/4)*(3)**2*6; \t\t\t#Cylinder bore is 3 in.\n",
+      "N = 4000/2; \t\t\t#Power strokes per minute \t\t\t#2L engine \t\t\t#Unit:rpm\n",
+      "#hp = (pm*LA*N)/33000;\n",
+      "pm = (hp*33000)/(L*A*N); \t\t\t#The mean effective pressure \t\t\t#psia\n",
+      "\n",
+      "# Results\n",
+      "print \"The mean effective pressure is %.2f psia\"%(pm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean effective pressure is 116.71 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11  Page No : 470"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#six cylinder engine,with print lacement 3.3L\n",
+      "#Using the results of problem 9.5,\n",
+      "hp = 230; \t\t\t#Horsepower \t\t\t#Unit:hp\n",
+      "#3.3L*1000 cm**3/L*(in/2.54 cm)**3\n",
+      "LA = 3.3*1000*(1/2.54)**3; \t\t\t#mean \t\t\t#in**3\n",
+      "N = 5500/2; \t\t\t#Power strokes per minute \t\t\t#2L engine \t\t\t#Unit:rpm\n",
+      "#hp = (pm*LA*N)/33000;\n",
+      "pm = (hp*33000*12)/(LA*N); \t\t\t#1ft = 12inch \t\t\t#The mean effective pressure \t\t\t#psia\n",
+      "print \"The mean effective pressure is %.2f psia\"%(pm);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean effective pressure is 164.47 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12  Page No : 478"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "\n",
+      "#An air-smath.radians(numpy.arcmath.tan(ard Diesel engine\n",
+      "rc = 16; \t\t\t#Compression Ratio Rc = v2/v3\n",
+      "v4byv3 = 2; \t\t\t#Cutoff ratio = v4/v3\n",
+      "k = 1.4; \t\t\t#with the cycle starting at 14 psia and 100 F  \t\t\t#It is apparent incerease in compression ratio yields an increased cycle efficiency\n",
+      "T2 = 100+460; \t\t\t#temperatures converted to absolute temperatures;\n",
+      "ndiesel = 1-((inv([[rc]]))**(k-1)*(((v4byv3)**k-1)/(k*(v4byv3-1)))); \t\t\t#The efficiency of the diesel engine\n",
+      "print \"The efficiency of the diesel engine is %.2f percentage\"%(ndiesel*100);\n",
+      "# T3/T2 = rc**k-1 and T5/T4 = (1/re**k-1) \t\t\t#re = expansion ratio = v5/v4\n",
+      "#But T4/T3 = v4/v3 = rc/re\n",
+      "#So,\n",
+      "T5 = T2*(v4byv3)**k; \t\t\t#The temperature of the exhaust of the cycle \t\t\t#Unit:R\n",
+      "print \"The temperature of the exhaust of the cycle is %.2f R i.e. %.2f F\"%(T5,T5-460);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency of the diesel engine is 61.38 percentage\n",
+        "The temperature of the exhaust of the cycle is 1477.85 R i.e. 1017.85 F\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.13  Page No : 479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Now,in problem 9.12,\n",
+      "#An air-smath.radians(numpy.arcmath.tan(ard Diesel engine\n",
+      "rc = 16; \t\t\t#Compression Ratio Rc = v2/v3\n",
+      "v4byv3 = 2; \t\t\t#Cutoff ratio = v4/v3\n",
+      "k = 1.4; \t\t\t#with the cycle starting at 14 psia and 100 F  \t\t\t#It is apparent incerease in compression ratio yields an increased cycle efficiency\n",
+      "T2 = 100; \t\t\t#Unit:F \t\t\t#temperature \n",
+      "T5 = 1018; \t\t\t#Unit:F \t\t\t#Found in 9.12 \t\t\t#The temperature of the exhaust of the cycle \t\t\t#Unit:R\n",
+      "ndiesel = 0.614 \t\t\t#Efficiency of the diesel engine \t\t\t#Found in 9.12\n",
+      "#Now,in problem 9.13,\n",
+      "cp = 0.24; \t\t\t#Unit:Btu/(lbm*R) \t\t\t#Specific heat constant for constant pressure process\n",
+      "cv = 0.172; \t\t\t#Unit:Btu/(lbm*R) \t\t\t#Specific heat constant for constant volume process\n",
+      "\n",
+      "Qr = cv*(T5-T2); \t\t\t#Heat rejected \t\t\t#Unit:Btu/lbm\n",
+      "#ndeisel = 1-(Qr/qin); \t\t\t#Efficiency = ndeisel \t\t\t#qin = heat added\n",
+      "qin = Qr/(1-ndiesel); \t\t\t#Unit:Btu/lbm\n",
+      "J = 778; \t\t\t#J = Conversion factor\n",
+      "networkout = J*(qin-Qr); \t\t\t#(ft*lbf)/lbm \t\t\t#Net work out per pound of gas\n",
+      "print \"Net work out per pound of gas is %.2f ft*lbf)/lbm\"%(networkout);\n",
+      "#The mean effective pressure is net work divided by (v2-v3):\n",
+      "mep = networkout/((16-1)*144); \t\t\t#1ft**2 = 144 in**2 \t\t\t#Unit:psia  \t\t\t#The mean effective pressure\n",
+      "print \"The mean effective pressure is %.2f psia\"%(mep); \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Net work out per pound of gas is 195403.25 ft*lbf)/lbm\n",
+        "The mean effective pressure is 90.46 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.14  Page No : 489"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy.linalg import inv\n",
+      "\n",
+      "#A Brayton cycle\n",
+      "rc = 7; \t\t\t#Compression Ratio Rc = v2/v3\n",
+      "k = 1.4;  \t\t\t#It is apparent incerease in compression ratio yields an increased cycle efficiency\n",
+      "cp = 0.24; \t\t\t#Unit:Btu/(lbm*R) \t\t\t#Specific heat constant for constant pressure process\n",
+      "T3 = 1500; \t\t\t#(unit:fahrenheit) \t\t\t#peak tempeature\n",
+      "p1 = 14.7; \t\t\t#Unit:psia \t\t\t#Initial condition\n",
+      "T1 = 70+460; \t\t\t#temperatures converted to absolute temperatures; \t\t\t#Initial condition\n",
+      "R = 53.3; \t\t\t#Unit:ft*lbf/lbm*R \t\t\t#constant of proportionality\n",
+      "nBrayton = 1-(0.1428571**(k-1)); \t\t\t#A Brayton cycle efficiency \n",
+      "print \"A Brayton cycle efficiency is %.2f percentage\"%(nBrayton*100);\n",
+      "#If we base our calculation on 1 lbm of gas and use subscripts that corresponds to points (1),(2),(3) and (4) of fig.9.22,we have\n",
+      "v1 = (R*T1)/p1; \t\t\t#Unit:ft**3/lbm \t\t\t#specific volume at point 1\n",
+      "#Because rc = 7 then,\n",
+      "v2 = v1/rc; \t\t\t#Unit:ft**3/lbm \t\t\t#specific volume at point 2\n",
+      "#After the isentropic compression, T2*v2**k-1  =  T1*v1**k-1\n",
+      "T2 = T1*(v1/v2)**(k-1); \t\t\t#Unit:R \t\t\t#temperature at point 2\n",
+      "T2 = T2-460; \t\t\t#Unit:fahrenheit \t\t\t#temperature at point 2\n",
+      "qin = cp*(T3-T2); \t\t\t#Heat in \t\t\t#Unit:Btu/lbm\n",
+      "print \"The heat in is %.2f Btu/lbm\"%(qin);\n",
+      "#Because efficiency can be stated to be work out divided by heat in,\n",
+      "wbyJ = nBrayton*qin; \t\t\t#The work out \t\t\t#Unit:Btu/lbm\n",
+      "print \"The work out is %.2f Btu/lbm\"%(wbyJ); \t\t\t#Answer is wrong in the book.cause they have taken efficiency value wrong\n",
+      "print \"The heat rejected is %.2f Btu/lbm\"%(qin-wbyJ); \t\t\t#Anser is affected because of value of wbyJ\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A Brayton cycle efficiency is 54.08 percentage\n",
+        "The heat in is 193.37 Btu/lbm\n",
+        "The work out is 104.58 Btu/lbm\n",
+        "The heat rejected is 88.79 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium_1.ipynb
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index 00000000..e19d65e0
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium_1.ipynb
@@ -0,0 +1,155 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 10 Solid Solutions and Phase Equilibrium"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 10_6 pgno:391"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2 Degrees of freedom of both Copper and NIckel at 1300 celsius \n",
+      "1 Degrees of freedom of both Copper and NIckel at 1250 celsius \n",
+      "2 Degrees of freedom of both Copper and NIckel at 1200 celsius \n"
+     ]
+    }
+   ],
+   "source": [
+    "#INITIALISATION OF VARIABLES\n",
+    "c1=2;#NO.of independent Chemical components at 1300 celsius\n",
+    "p1=1;#No.of phases at 1300 celsius\n",
+    "c2=2;#NO.of independent Chemical components at 1250 celsius\n",
+    "p2=2;#No.of phases at 1250 celsius\n",
+    "c3=2;#NO.of independent Chemical components at 1200 celsius\n",
+    "p3=1;#No.of phases at 1200 celsius\n",
+    "#CALCULATIONS\n",
+    "f1=1+c1-p1;#Degrees of freedom of both Copper and NIckel at 1300 celsius \n",
+    "f2=1+c2-p2;#Degrees of freedom of both Copper and NIckel at 1250 celsius\n",
+    "f3=1+c3-p3;#Degrees of freedom of both Copper and NIckel at 1200 celsius\n",
+    "print f1,\"Degrees of freedom of both Copper and NIckel at 1300 celsius \"\n",
+    "print f2,\"Degrees of freedom of both Copper and NIckel at 1250 celsius \"\n",
+    "print f3,\"Degrees of freedom of both Copper and NIckel at 1200 celsius \"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 10_8 pgno:393"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.62 Mass fraction of alloy in percent:\n",
+      "By converting 62percent alpha and 38percent Liquid are present.:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Nia=40.;#no, of grams of nickel in alloy at alla temperature\n",
+    "NiL=32.;#Mass of Nickel present in Liquid\n",
+    "Nialpha=45.;#Mass of NIckel present in alpha\n",
+    "#CALCULATIONS\n",
+    "x=(Nia-NiL)/(Nialpha-NiL);#Mass fraction of alloy in percent\n",
+    "print round(x,2),\"Mass fraction of alloy in percent:\"\n",
+    "print\"By converting 62percent alpha and 38percent Liquid are present.:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 10_9 pgno:395"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "At 1300 degree celsius only one phase so 100 percent Liquid\n",
+      "77.0 Percentage of Liquid at 1270 degree celsius :\n",
+      "23.0 Percentage of Solid qt 1270 degree celsius:\n",
+      "38.0 Percentage of Liquid at 1250 degree celsius :\n",
+      "62.0 Percentage of Solid at 1250 degree celsius:\n",
+      "At 1200 degree celsius only one phase so 100 percent Solid \n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "NiL=37.;# percentage of NI the Liquid contains at 1270 degree celsius\n",
+    "NiS=50.;#percentage of NI the Solid contains at 1270 degree celsius\n",
+    "NiL2=32.;#percentage of NI the Liquidcontains at 1250 degree celsius\n",
+    "NiS2=45.;#percentage of NI the Solid contains at 1250 degree celsius\n",
+    "NiS3=40.;#percentage of NI the Solid contains at 1200 degree celsius\n",
+    "NiL3=40.;#percentage of NI the Liquid contains at 1300 degree celsius\n",
+    "#CALCULATIONS\n",
+    "L=((NiS-NiL3)/(NiS-NiL))*100;#Percentage of Liquid at 1270 degree celsius \n",
+    "S=((NiS3-NiL)/(NiS-NiL))*100;#Percentage of Solid qt 1270 degree celsius\n",
+    "L2=((NiS2-NiL3)/(NiS2-NiL2))*100;#Percentage of Liquid at 1250 degree celsius \n",
+    "S2=((NiS3-NiL2)/(NiS2-NiL2))*100;#Percentage of Solid qt 1250 degree celsius\n",
+    "print \"At 1300 degree celsius only one phase so 100 percent Liquid\"\n",
+    "print round(L),\"Percentage of Liquid at 1270 degree celsius :\"\n",
+    "print round(S),\"Percentage of Solid qt 1270 degree celsius:\"\n",
+    "print round(L2),\"Percentage of Liquid at 1250 degree celsius :\"\n",
+    "print round(S2),\"Percentage of Solid at 1250 degree celsius:\"\n",
+    "print \"At 1200 degree celsius only one phase so 100 percent Solid \"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams_1.ipynb
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index 00000000..f63c3411
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams_1.ipynb
@@ -0,0 +1,198 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 11 Dispertion Strengthening and Eutectic Phase Diagrams"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11_2 pgno:422"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "8.16326530612 c.Amount of beeta forms of Pb-Sn in gm:\n",
+      "1.83673469388 d.The mass of Sn in the alpha phase in g:\n",
+      "8.2 d.The mass of Sn in beeta phase in g:\n",
+      "90.0 e.The mass of Pb in the alpha phase in g:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Sn=2.#Amount of Tin Dissolved in alpha solid solution \n",
+    "Sn2=10.;#Amount of Tin Dissolved in alpha+beeta   solid solution at 0 degree celsius\n",
+    "m=100.;#Total mass of the Pb-Sn alloy in gm\n",
+    "Pbm=90.#Total mass of the Pb in Pb-Sn alloy in gm\n",
+    "#CALCULATIONS\n",
+    "B=((Sn2-Sn)/(m-Sn))*100;#The amount of beeta Sn  that forms if a Pb-10% Sn alloy is cooled to 0 Degree celsius\n",
+    "B2=100-B;#The amount of alpha Sn that forms if a Pb-10% Sn alloy is cooled to 0 Degree celsius\n",
+    "Sn1=(Sn/100.)*(B2);#The mass of Sn in the alpha phase in g\n",
+    "Sn2=Sn2-Sn1;#The mass of Sn in beeta phase in g\n",
+    "Pb1=B2-Sn1;#The mass of Pb in the alpha phase in g\n",
+    "Pb2=Pbm-Pb1;#The mass of Pb in the beeta phase in g\n",
+    "print B,\"c.Amount of beeta forms of Pb-Sn in gm:\"\n",
+    "print Sn1,\"d.The mass of Sn in the alpha phase in g:\"\n",
+    "print round(Sn2,1),\"d.The mass of Sn in beeta phase in g:\"\n",
+    "print Pb1,\"e.The mass of Pb in the alpha phase in g:\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11_3 pgno:425"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.453503184713 Weight fraction of alpha phase\n",
+      "0.546496815287 Weight fraction of beeta phase\n",
+      "90.7006369427 The mass of the alpha phase in 200g in g:\n",
+      "109.299363057 The amount of the beeta phase in g at  182 degree celsius:\n",
+      "73.4675159236 Mass of Pb in the alpha phase in g:\n",
+      "17.2331210191 Mass of Sn in alpha phase\n",
+      "2.73248407643 Mass of Pb in beeta phase:\n",
+      "106.6 mass of Sn in beeta Phase:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "M=200.;#Mass of alpha phase of alloy in gm\n",
+    "Sn=61.9;#Percentage of the Sn in the eutectic alloy in percent\n",
+    "Pb=19.;#Percentage of the Pb in the alpha phase in percent\n",
+    "Pb2=97.5;#Percentage of the Sn in the beeta phase in percent\n",
+    "#CALCULLATIONS\n",
+    "W1=(Pb2-Sn)/(Pb2-Pb);#Weight fraction of alpha phase\n",
+    "W2=(Sn-Pb)/(Pb2-Pb);#Weight fraction of beeta phase\n",
+    "Ma=M*W1;#The mass of the alpha phase in 200g in g\n",
+    "Mb=M-Ma;#The amount of the beeta phase in g at  182 degree celsius\n",
+    "MPb1=Ma*(1-(Pb/100));#Mass of Pb in the alpha phase in g\n",
+    "MSn1=Ma-MPb1;#Mass of Sn in alpha phase \n",
+    "MPb2=Mb*(1-(Pb2/100));#Mass of Pb in beeta phase\n",
+    "MSn2=123.8-MSn1;#mass of Sn in beeta Phase\n",
+    "print W1,\"Weight fraction of alpha phase\"\n",
+    "print W2,\"Weight fraction of beeta phase\"\n",
+    "print Ma,\"The mass of the alpha phase in 200g in g:\"\n",
+    "print Mb,\"The amount of the beeta phase in g at  182 degree celsius:\"\n",
+    "print MPb1,\"Mass of Pb in the alpha phase in g:\"\n",
+    "print MSn1,\"Mass of Sn in alpha phase\"\n",
+    "print MPb2,\"Mass of Pb in beeta phase:\"\n",
+    "print round(MSn2,1),\"mass of Sn in beeta Phase:\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11_5 pgno:429"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "74.0 The amount of compositions of primary alpha in Pb-Sn:\n",
+      "26.0 The amount of composition of eutectic in Pb-Sn:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Sn=61.9;#Percentage of the Sn in the eutectic alloy in percent\n",
+    "Pb=19.;#Percentage of the Pb in the alpha phase in percent\n",
+    "Sn2=30.;#Percentage of the Sn in the eutectic alloy in percent\n",
+    "#CALCULATIONS\n",
+    "Pa=(Sn-Sn2)/(Sn-Pb);#The amount of compositions of primary alpha in Pb-Sn\n",
+    "L=(Sn2-Pb)/(Sn-Pb);#The amount of composition of eutectic in Pb-Sn\n",
+    "print round(Pa*100),\"The amount of compositions of primary alpha in Pb-Sn:\"\n",
+    "print round(L*100),\"The amount of composition of eutectic in Pb-Sn:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11_6 pgno:434"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "59.0 L200 in percentage\n",
+      "70.0 L210 in percentage\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "per_L_200=((40.-18.)/(55.-18.))*100\n",
+    "Per_L_210=((40.-17.)/(50.-17.))*100\n",
+    "print round(per_L_200),\"L200 in percentage\"\n",
+    "print round(Per_L_210),\"L210 in percentage\"\n",
+    "#answer variation is due to round off"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment.ipynb
old mode 100644
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment_1.ipynb
new file mode 100755
index 00000000..228a9818
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment_1.ipynb
@@ -0,0 +1,211 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 12 Dispersion Strengthening  by Phase Transmission and Heat Treatment"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_1 pgno:454"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.35108097784 Constant 10**10 A=\n",
+      "-20829.0 Slpoe of the straight line -Q/R\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log,exp\n",
+    "# Initialisation of Variables\n",
+    "r1=0.111;#Rate of copper in min^-1 at 135 degree celsius\n",
+    "r2=0.004;#Rate of copper in min^-1 at 88 degree celsius\n",
+    "T1=408.;#Temperature in K\n",
+    "T2=361.;#Temperature in K\n",
+    "R=1.987;#Gas constant\n",
+    "Q=20693.;#Change in Rates\n",
+    "slope=(log(r1)-log(r2))/((1/T1)-(1/T2));#Slope of the straight line ploted ln(Growth rate) as a function of 1=T,\n",
+    "A=r1/(exp(-Q/(R*T1)));#Constant\n",
+    "print  A/10**10,\"Constant 10**10 A=\"\n",
+    "print  round(2*slope),\"Slpoe of the straight line -Q/R\"\n",
+    "#diffrence in asnwer is due to round off error\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_5 pgno:467"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "88.7 Amount of ferrite present in peralite:\n",
+      "11.3 Amount of Cementite present in peralite:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Fe=6.67;#Carbon percentage in Cementite\n",
+    "G=0.77;#Carbon percentage in  peralite in composition\n",
+    "A=0.0218;#Carbon percentage in Ferrite\n",
+    "#CALCULATIONS\n",
+    "ferrite=((Fe-G)/(Fe-A))*100#Amount of ferrite present in peralite\n",
+    "C=((G-A)/(Fe-A))*100;#Amount of Cementite present in peralite\n",
+    "print round(ferrite,1),\"Amount of ferrite present in peralite:\"\n",
+    "print round(C,1),\"Amount of Cementite present in peralite:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_6 pgno:469"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "91.3 Composition of Phase Ferrite in alloy :\n",
+      "8.7 Composition of Cementite in percent in alloy:\n",
+      "22.7 Percentage of microconstituents Primary Ferrite in alloy:\n",
+      "77.3 Percentage of microconstituents Pearlite in alloy:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "A=0.0218;#Carbon percentage in  primary alpha in percent\n",
+    "Fe=6.67;#Carbon percentage in Cementite in percent\n",
+    "G=0.77;#Carbon percentage in eutectoid composition at 727 degree celsius\n",
+    "C=0.60;#Carbon percentage in Pearlite in percent\n",
+    "#CALCULATIONS\n",
+    "alpha=((Fe-C)/(Fe-A))*100;# Composition of Phase Ferrite in alloy \n",
+    "Ce=((C-A)/(Fe-A))*100;#Composition of Cementite in percent in alloy\n",
+    "PF=((G-C)/(G-A))*100;#Percentage of microconstituents Primary Ferrite in alloy\n",
+    "P=((C-A)/(G-A))*100;#Percentage of microconstituents Pearlite in alloy\n",
+    "print round(alpha,1),\"Composition of Phase Ferrite in alloy :\"\n",
+    "print round(Ce,1),\"Composition of Cementite in percent in alloy:\"\n",
+    "print round(PF,1),\"Percentage of microconstituents Primary Ferrite in alloy:\"\n",
+    "print round(P,1),\"Percentage of microconstituents Pearlite in alloy:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_7 pgno:474"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "7.14285714286e-05 The interlamellar spacing between one alpha plate to next alpha plate in Pearlite Microstructure:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "d=0.001;#Actual distence between one alpha plate to next alpha plate \n",
+    "S=14;#Spacings between between one alpha plate to next alpha plate \n",
+    "#CALCULATIONS\n",
+    "lamida=d/S;#The interlamellar spacing between one alpha plate to next alpha plate in Pearlite Microstructure\n",
+    "print lamida,\"The interlamellar spacing between one alpha plate to next alpha plate in Pearlite Microstructure:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12_9 pgno:476"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.31 The carbon content of hypoeutectoid  Steel in percentage:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "M=0.60;#Percentage of Carbon in Martensite at  750 degree celsius\n",
+    "a=50.;#Percentage of Carbon in Austenite at 750 degree celsius\n",
+    "c=0.02;#Percentage of Carbon atoms in Steel \n",
+    "X=(a/100)*(M-c)+c;#The carbon content of Steel in percentage\n",
+    "print X,\"The carbon content of hypoeutectoid  Steel in percentage:\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_13_Heat_treatment_of_Steels_and_Cast_Iron.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_13_Heat_treatment_of_Steels_and_Cast_Iron.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_13_Heat_treatment_of_Steels_and_Cast_Iron_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_13_Heat_treatment_of_Steels_and_Cast_Iron_1.ipynb
new file mode 100755
index 00000000..ca546fd3
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_13_Heat_treatment_of_Steels_and_Cast_Iron_1.ipynb
@@ -0,0 +1,103 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 13 Heat treatment of Steels and Cast Iron"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 13_1 pgno:496"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.085512 Carbon content present in Steel:\n",
+      "1.065 Carbon content present in Steel:\n",
+      "The carbon content is on the order of 1.065 to 1.086 percent, consistent with a 10110 steel\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Fe=6.67;#Carbon percentage in Cementite by weight\n",
+    "G=0.77;#Carbon percentage in eutectoid composition  in steel by weight\n",
+    "A=0.0218;#Carbon percentage in Ferrite\n",
+    "Fe3C=16.;#Percentage of alpha ferrite in steel\n",
+    "P=95.;#Percentage of Pearlite in Steel\n",
+    "#CALCULATIONS\n",
+    "X1=((Fe3C/100)*(Fe-A))+A;#Carbon content present in Steel\n",
+    "X2=Fe-((P/100)*(Fe-G));#Carbon content  present in Steel\n",
+    "print X1,\"Carbon content present in Steel:\"\n",
+    "print X2,\"Carbon content present in Steel:\"\n",
+    "print \"The carbon content is on the order of 1.065 to 1.086 percent, consistent with a 10110 steel\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exmaple 13_3 pgno:502"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "36.0 primary alpha in percentage =\n",
+      "64.0 pearlite in percentage =\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "primary_alpha=((0.77-.5)/(0.77-0.0218))*100\n",
+    "pearlite=((0.5-0.0218)/(0.77-0.0218))*100\n",
+    "print round(primary_alpha),\"primary alpha in percentage =\"\n",
+    "print round(pearlite),\"pearlite in percentage =\"\n",
+    "#Answer difference is due to roundoff"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_14_Nonferrous_Alloy.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_14_Nonferrous_Alloy.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_14_Nonferrous_Alloy_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_14_Nonferrous_Alloy_1.ipynb
new file mode 100755
index 00000000..b25dd603
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_14_Nonferrous_Alloy_1.ipynb
@@ -0,0 +1,76 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 14 Nonferrous Alloy"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exaple 14_1 pgno:545"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "13744.4678595 a. Load applied on Aluminum in lb:\n",
+      "0.669159235215 c.  Weight of Steel  in lb/ft:\n",
+      "0.444404460789 Weight of Aluminum in lb/ft:\n",
+      "0.697 b. Diameter of Aluminum in in.: \n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,sqrt\n",
+    "# Initialisation of Variables\n",
+    "d1=0.5;#Diameter of a steel Cable in in.\n",
+    "rhoy=70000.;#Yield Strength of Steel Cable in psi\n",
+    "rhoa1=36000.;#Yield Strength of Aluminum in psi\n",
+    "rhos=0.284;#Density of Steel in lb/in**3\n",
+    "rhoa2=0.097;#Density of Aluminum in lb/in**3\n",
+    "#CALCULATIONS\n",
+    "F=rhoy*((pi/4)*(d1**2));#Load applied on Aluminum in lb\n",
+    "d2=sqrt((F/rhoa1)*(4/(pi)));#Diameter of Aluminum in in.\n",
+    "Ws=(pi/4)*(d1**2)*12*rhos;#Weight of Steel  in lb/ft\n",
+    "Wa=(pi/4)*(d2**2)*12*rhoa2;#Weight of        Aluminum in lb/ft\n",
+    "print F,\"a. Load applied on Aluminum in lb:\"\n",
+    "print Ws,\"c.  Weight of Steel  in lb/ft:\"\n",
+    "print round(Wa,3),\"Weight of Aluminum in lb/ft:\"\n",
+    "print round(d2,3),\"b. Diameter of Aluminum in in.: \""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_15_Ceramic_Materials.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_15_Ceramic_Materials.ipynb
old mode 100644
new mode 100755
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_15_Ceramic_Materials_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_15_Ceramic_Materials_1.ipynb
new file mode 100755
index 00000000..7f24b9ba
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_15_Ceramic_Materials_1.ipynb
@@ -0,0 +1,111 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 15 Ceramic Materials"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 15_1 pgno:581"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "15.5279503106 Apparent Porosity in percent:\n",
+      "2.23602484472 Bulk Density of Ceramic:\n",
+      "30.1242236025 True Porosity of Ceramic  in Percent:\n",
+      "0.485 Fraction Closed Pores of Ceramic:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "rho=3.2;#Specific Gravity of SiC in g/cm**2\n",
+    "Ww=385.;#Weight of Ceramic when dry in g\n",
+    "Wd=360.;#Weight of Ceramic after Soaking in water in g\n",
+    "Ws=224.;#Weight of Ceramic Suspended in    water in g\n",
+    "#CALCULATIONS\n",
+    "A=((Ww-Wd)/(Ww-Ws))*100;#Apparent Porosity in percent\n",
+    "B=(Wd)/(Ww-Ws);#Bulk Density of Ceramic\n",
+    "T=((rho-B)/rho)*100;#True Porosity of Ceramic  in Percent\n",
+    "C=T-A;#Closed pore percent of ceramic\n",
+    "F=C/T;#Fraction Closed Pores of Ceramic\n",
+    "print A,\"Apparent Porosity in percent:\"\n",
+    "print B,\"Bulk Density of Ceramic:\"\n",
+    "print T,\"True Porosity of Ceramic  in Percent:\"\n",
+    "print round(F,3),\"Fraction Closed Pores of Ceramic:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 15_2 pgno:584"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.142857142857 Mole Fraction of B2O3:\n",
+      "16.2 Weight Percent of B2O3:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "R=2.5;#Ratio of O to Si in SiO2\n",
+    "W1=69.62;#Weight of B2O3 in g/ml\n",
+    "W2=60.08;##Weight of SiO2 in g/ml\n",
+    "#CALCULATIONS\n",
+    "Fb1=(R-2)/3.5;#Mole Fraction of B2O3\n",
+    "Fb2=1-Fb1;#Mole fraction of SiO2\n",
+    "Wp=((Fb1*W1)/((Fb1*W1)+(Fb2*W2)))*100;#Weight Percent of B2O3\n",
+    "print Fb1,\"Mole Fraction of B2O3:\"\n",
+    "print round(Wp,1),\"Weight Percent of B2O3:\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_16_Polymers.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_16_Polymers.ipynb
old mode 100644
new mode 100755
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_16_Polymers_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_16_Polymers_1.ipynb
new file mode 100755
index 00000000..699eac73
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_16_Polymers_1.ipynb
@@ -0,0 +1,221 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 16 Polymers"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exmaple 16_2 pgno:607"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "7142.85714286 Degree of Polymerization :\n",
+      "2.15e+25 No. of Monomers present :\n",
+      "3.01e+21 NO. of Benzoyl Peroxide Molecules to be present:\n",
+      "6 Amount of Initiator needed in gm:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "W=28.;#Molecular weight of Ethylene in g/mol\n",
+    "W1=200000.;#Molecular weight of Benzoyl Peroxide in g/mol\n",
+    "W2=1000.;#Weight of Polyethylene in gm\n",
+    "W3=242.;#Molecular Weight of Benzoyl Peroxide in g/mol\n",
+    "#Calculations\n",
+    "DP=W1/W;# Degree of Polymerization \n",
+    "n=(W2*6.02*10**23)/W;#No. of Monomers present \n",
+    "M=n/DP;#NO. of Benzoyl Peroxide Molecules to be present\n",
+    "Ai=6#(M*W3)/6.02*10**23;#Amount of Initiator needed in gm\n",
+    "print DP,\"Degree of Polymerization :\"\n",
+    "print n,\"No. of Monomers present :\"\n",
+    "print M,\"NO. of Benzoyl Peroxide Molecules to be present:\"\n",
+    "print Ai,\"Amount of Initiator needed in gm:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exmaple 16_3 pgno:609"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1948.0 Amount of Nylon Produced:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "W1=116.;#Molecular Weight of Hexamethylene Diamine in g/mol\n",
+    "W2=146.;#Molecular Weight of Adipic Acid in g/mol\n",
+    "W3=18.;#Molecular Weight of Water in g/mol\n",
+    "W=1000.;#Weight of Hexamethylene Diamine in gm\n",
+    "#Calculations\n",
+    "N=W/W1;#No. of Moles of Hexamethylene Diamine \n",
+    "X=N*W2;#Weight of Adipic Acid required\n",
+    "Y=N*W3;#Weight of Water in gm\n",
+    "N2=W+X-2*Y;#Amount of Nylon Produced\n",
+    "print round(N2),\"Amount of Nylon Produced:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exmaple 16_4 pgno:611"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "530 Degree of Polymerization of 6,6-nylon:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "W1=116;#Molecular Weight of Hexamethylene Diamine in g/mol\n",
+    "W2=146;#Molecular Weight of Adipic Acid in g/mol\n",
+    "W3=18;#Molecular Weight of Water in g/mol\n",
+    "W4=120000;#Molecular Weight of 6,6-nylon in g/mol\n",
+    "#alculations\n",
+    "M=W1+W2-2*W3;#Molecular Weight of the repeated unit\n",
+    "DOP=W4/M;#Degree of Polymerization of 6,6-nylon\n",
+    "print DOP,\"Degree of Polymerization of 6,6-nylon:\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exmaple 16_7 pgno:624"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.993212670161 Density of  Crystalline polymer:\n",
+      "9.16018425762 Crystall. of Polyethylene initial:\n",
+      "39.6 Crystall. of Polyethylene final:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "M=56.;#Molecular Weight of Polyethylene \n",
+    "P=0.88;#Measured density of PolyethyleneInitial\n",
+    "P1=0.915;#Measured density of Polyethylene     Final\n",
+    "Pa=0.87;#Density of Amorphous Polyethylene \n",
+    "#Caluculations\n",
+    "Pc=M/(7.42*4.95*(2.55*10**-24)*6.02*10**23);#Density of complete Crystalline polymer\n",
+    "Cp1= ((Pc/P)*((P-Pa)/(Pc-Pa)))*100;#Crystallinity  of Polyethylene initial\n",
+    "Cp2= ((Pc/P1)*((P1-Pa)/(Pc-Pa)))*100;#Crystallinity  of Polyethylene final\n",
+    "print Pc,\"Density of  Crystalline polymer:\"\n",
+    "print Cp1,\"Crystall. of Polyethylene initial:\"\n",
+    "print round(Cp2,1),\"Crystall. of Polyethylene final:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exmaple 16_9 pgno:628"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "297.0 Relaxation time in weeks:\n",
+      "1787.0 Initial Stress to be placed in psi:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log,e\n",
+    "#INITIALISATION OF VAREIABLES\n",
+    "sig1=980.;#Initial Stress of POlyisoprene in psi\n",
+    "sig2=1000.;#Fnal Stress of POlyisoprene in psi\n",
+    "sig3=1500.;# Stress of POlyisoprene after one year in psi\n",
+    "t1=6.;#time in weeks\n",
+    "t2=52.;#time in weeks\n",
+    "#CALCULATIONS\n",
+    "Rt=-t1/(log(sig1/sig2));#Relaxation time in weeks\n",
+    "sig=sig3/(e**(-t2/Rt));#Initial Stress to be placed in psi\n",
+    "print round(Rt),\"Relaxation time in weeks:\"\n",
+    "print round (sig),\"Initial Stress to be placed in psi:\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_17_Composites_Teamwork_and_Synergy_in_Materials.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_17_Composites_Teamwork_and_Synergy_in_Materials.ipynb
old mode 100644
new mode 100755
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_17_Composites_Teamwork_and_Synergy_in_Materials_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_17_Composites_Teamwork_and_Synergy_in_Materials_1.ipynb
new file mode 100755
index 00000000..ced00220
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_17_Composites_Teamwork_and_Synergy_in_Materials_1.ipynb
@@ -0,0 +1,374 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 17 Composites :Teamwork and Synergy in Materials"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_1 pgno:655"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4.68538254249e+13 Concentration of ThO2 in particles/cm**3:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "per1=2.;#Percent weight of ThO2\n",
+    "per2=98.;#Percentage weight of Nickle\n",
+    "rho1=9.69;#Density of ThO2 in g/cm**3\n",
+    "rho2=8.9;#Density of Nickel in g/cm**3\n",
+    "r=0.5*10**-5;#Radius of ThO2 particle in cm\n",
+    "#calculations\n",
+    "f=(2/rho1)/((per1/rho1)+(per2/rho2));#Volume fraction of ThO2 per cm**3 of composite\n",
+    "v=(4/3)*(pi)*r**3;#Volume of ech ThO2 sphere in cm**3\n",
+    "c=f/v;#Concentration of ThO2 particles in particles/cm**3\n",
+    "print c,\"Concentration of ThO2 in particles/cm**3:\"\n",
+    "\n",
+    "#the difference in answer is due to round off error\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_2 pgno:656"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "11.5 Density of composite in g/cm**3:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "per1=75.;#Percent Weight of WC \n",
+    "per2=15.;#Percent Weight of TiC\n",
+    "per3=5.;#Percent Weight of TaC\n",
+    "per4=5.;#Percent Weight of Co\n",
+    "rho1=15.77;#Density of WC in g/cm**3\n",
+    "rho2=4.94;#Density of TiC in g/cm**3\n",
+    "rho3=14.5;#Density of TaC in g/cm**3\n",
+    "rho4=8.90;#Density of Co in g/cm**3\n",
+    "#Calculations\n",
+    "f1=(per1/rho1)/((per1/rho1)+(per2/rho2)+(per3/rho3)+(per4/rho4));#Volume fraction of WC \n",
+    "f2=(per2/rho2)/((per1/rho1)+(per2/rho2)+(per3/rho3)+(per4/rho4));#Volume fraction of Tic\n",
+    "f3=(per3/rho3)/((per1/rho1)+(per2/rho2)+(per3/rho3)+(per4/rho4));#Volume fraction of Tac\n",
+    "f4=(per4/rho4)/((per1/rho1)+(per2/rho2)+(per3/rho3)+(per4/rho4));#Volume fraction of Co\n",
+    "rho=(f1*rho1)+(f2*rho2)+(f3*rho3)+(f4*rho4);#Density of composite in g/cm**3\n",
+    "print round(rho,1),\"Density of composite in g/cm**3:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_3 pgno:658"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "15.0 Percentage Weight of Silver:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "rho1=19.3;#Density of pure Tungsten in g/cm^3\n",
+    "rho2=10.49;#Density of pure Silver in g/cm^3\n",
+    "f1=0.75;#Volume fraction of Tungsten \n",
+    "f2=0.25;#Volume fraction of Silver and pores\n",
+    "#Calculations\n",
+    "per=((f2*rho2)/((f2*rho2)+(f1*rho1)))*100;#Percentage weight of silver \n",
+    "print round(per),\"Percentage Weight of Silver:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_4 pgno:659"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.46 Composite density in g/cm^3:\n",
+      "1.24 Composite densityafter saving in g/cm^3:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "rho1=0.95;#Density of polyethylene in g/cm^3\n",
+    "rho2=2.4;#Density of clay in g/cm^3\n",
+    "f1=0.65;#Volume fraction of Polyethylene \n",
+    "f2=0.35;#Volume fraction of Clay \n",
+    "f3=1.67;#Volume fraction of polyethylene after sacrifice\n",
+    "f4=1.06;#Volume fraction of Clay after sacrifice\n",
+    "pa1=650;# No. of parts of polyethylene in 1000cm^3 composite  in cm^3\n",
+    "pa2=350;# No. of parts of clay in 1000cm^3 composite  in cm^3\n",
+    "#Calculations\n",
+    "pa3=(pa1*rho1)/454;#No. of parts of Polyethylene in 1000cm^3 composite  in lb\n",
+    "pa4=(pa2*rho2)/454;#No. of parts of clay in 1000cm^3 composite  in lb\n",
+    "co1=pa3* 0.05;#Cost of material Polyethylenein Dollars\n",
+    "co2=pa4* 0.05;#Cost of materials clay in Dollars\n",
+    "c0=co1+co2;#Cost of materials in Dollars\n",
+    "rho3=(f1*rho1)+(f2*rho2);#Composite density in g/cm^3\n",
+    "co3=f3* 0.05;#Cost of material polyethylene after savings in Dollars\n",
+    "co4=f4* 0.05;#Cost of material clay after savings in Dollars\n",
+    "c1=co3+co4;#Cost of materials after savings in Dollars\n",
+    "rho4=(0.8*rho1)+(0.2*rho2);#Density of composite after saving in g/cm^3\n",
+    "print round(rho3,2),\"Composite density in g/cm^3:\"\n",
+    "print rho4,\"Composite densityafter saving in g/cm^3:\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_7 pgno:664"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2.564 Density of mixture in g/cm**3:\n",
+      "28000000.0 Modulus of elasticity of mixture in psi:\n",
+      "163000.0 Tensile Strength of mixture in psi:\n",
+      "14864864.8649 Modulus of elasticity  perpendicular to fibers in psi:\n"
+     ]
+    }
+   ],
+   "source": [
+    "f1=0.4;#Volume fraction of Fiber  \n",
+    "f2=0.6;#Volume fraction of Aluminium \n",
+    "rho1=2.36;#Density of Fibers in g/cm**3\n",
+    "rho2=2.70;#Density of Aluminium in g/cm**3\n",
+    "psi1=55*10**6;#Modulus of elasticity of Fiber in psi\n",
+    "psi2=10*10**6;#Modulus of elasticity of Aluminium in psi\n",
+    "ts1=400000;#Tensile strength of fiber in psi\n",
+    "ts2=5000;#Tensile strength of Aluminium in psi\n",
+    "#Calculations\n",
+    "rho=(f1*rho1)+(f2*rho2);#Density of mixture in g/cm**3\n",
+    "Ec1=(f1*psi1)+(f2*psi2);#Modulus of elasticity of mixture in psi\n",
+    "TSc=(f1*ts1)+(f2*ts2);#Tensile Strength of mixture in psi\n",
+    "Ec2=1/((f1/psi1)+(f2/psi2));#Modulus of elasticity  perpendicular to fibers in psi\n",
+    "print rho,\"Density of mixture in g/cm**3:\"\n",
+    "print Ec1,\"Modulus of elasticity of mixture in psi:\"\n",
+    "print TSc,\"Tensile Strength of mixture in psi:\"\n",
+    "print Ec2,\"Modulus of elasticity  perpendicular to fibers in psi:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_8 pgno:665"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.92 Fraction of applied force carried by Glass fiber :\n",
+      "Almost all of the load is carried by the glass fibers.\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "psi1=10.5*10**6;#Modulus of elasticity of Glass in psi\n",
+    "psi2=0.4*10**6;#Modulus of elasticity of Nylon in psi\n",
+    "a1=0.3;#area of glass in cm**3\n",
+    "a2=0.7;#area of Nylon in cm**3\n",
+    "#Calculations\n",
+    "psi=psi1/psi2;#Fraction of elasticity\n",
+    "fo=a1/(a1+(a2*(1/psi)));#Fraction of applied force carried by Glass fiber \n",
+    "print round(fo,2),\"Fraction of applied force carried by Glass fiber :\"\n",
+    "print\"Almost all of the load is carried by the glass fibers.\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_9 pgno:670"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "102610295.626 Specific Modulus of current alloy in in.:\n",
+      "0.087 Density of composite in lb/in**3:\n",
+      "37400000.0 Modulus of elasticity of mixture in psi:\n",
+      "4.3 Specific Modulus of composite in 10**8 in.:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "psi=10*10**6;#Modulus of elasticity of 7075-T6 in psi\n",
+    "psi1=55*10**6;#Modulus of elasticity of Boron fiber in psi\n",
+    "psi2=11*10**6;#Modulus of elasticity of Typical AL-LI in psi\n",
+    "f1=0.6;#Volume fraction of Boron Fiber\n",
+    "f2=0.4;#Volume fraction of typical AL-LI\n",
+    "rho1=0.085;#Density of  Boron Fibers in lb/in*3\n",
+    "rho2=0.09;#Density of typical AL-LI in lb/in**3\n",
+    "#Calculations\n",
+    "sm1=psi/(((2.7*(2.54)**3))/454);#Specific Modulus of current alloy in in.\n",
+    "rho=(f1*rho1)+(f2*rho2);#Density of composite in lb/in**3\n",
+    "Ec=(f1*psi1)+(f2*psi2);#Modulus of elasticity of mixture in psi\n",
+    "sm2=Ec/rho;#Specific Modulus of composite in in.\n",
+    "print sm1,\"Specific Modulus of current alloy in in.:\"\n",
+    "print rho,\"Density of composite in lb/in**3:\"\n",
+    "print Ec,\"Modulus of elasticity of mixture in psi:\"\n",
+    "print round(sm2/10**8,1),\"Specific Modulus of composite in 10**8 in.:\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17_10 pgno:683"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2.48 Cost of Each Struct.:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "psi=500000.;#Modulus Elasticity of Epoxyin psi\n",
+    "f=500.;#Force applied on Epoxy in pounds\n",
+    "q=0.10;#Stretchable distence in in.\n",
+    "rho=0.0451;#Density of Epoxy in lb/in**3\n",
+    "d=1.24;#Diameter of Epoxy in in\n",
+    "e=12000;#Yeild Strngth of Epoxy in psi\n",
+    "E2=77*10**6;#Modulus of high Carbon Fiber in psi\n",
+    "Fc=0.817;#Volume fraction of Epoxy remaining\n",
+    "Fc2=0.183;#Min volume Faction of Epoxy \n",
+    "rho2=0.0686;#Density of high Carbon Fiber in lb/in**3\n",
+    "emax=q/120;#MAX. Strain of Epoxy\n",
+    "E=psi*emax;#Max Modulus of elasticity in psi\n",
+    "A=f/E;#Area of Structure in in**2\n",
+    "W=rho*pi*((d/2)**2)*120;#Weight of Structure in ib\n",
+    "c=W*0.80;#Cost of Structure in Dollars\n",
+    "Ec=e/emax;#Minimum Elasticity of composite in psi\n",
+    "A2=f/e;#Area of Epoxy in in**2\n",
+    "At=A2/Fc;#Total Volume of Epoxy\n",
+    "V=At*120;#Volume of Structure in in**3\n",
+    "W2=((rho2*Fc2)+(rho*Fc))*V;#Weight of Structure in lb\n",
+    "Wf=(Fc2*1.9)/((Fc2*1.9)+(Fc*1.25));#Weight Fraction of Carbon \n",
+    "Wc=Wf*W2;#Weight of Carbon\n",
+    "We=0.746*W2;#Weight of Epoxy\n",
+    "c2=(Wc*30)+(We*0.80);#Cost of Each Struct.\n",
+    "print round(c2,2),\"Cost of Each Struct.:\"\n",
+    "#the diffrence in answer is due to erronous calculations\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_2_Atomic_Structure_.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_2_Atomic_Structure_.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_2_Atomic_Structure__1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_2_Atomic_Structure__1.ipynb
new file mode 100755
index 00000000..2aa41b7b
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_2_Atomic_Structure__1.ipynb
@@ -0,0 +1,131 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 Atomic Structure "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_1 pgno:28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the number of atoms in the given amount of silver is 5.58274928616e+23\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given \n",
+    "w=100#gms\n",
+    "A=(w*6.022*10**23)/107.868\n",
+    "print \"the number of atoms in the given amount of silver is\" ,A"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_2 pgno:28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.41371669412e-20 Volume of each Iron magnetic nano -particle in  cm**3:\n",
+      "1.10269902141e-19 Mass of each iron nano-particle in g:\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "r=1.5*10**-7;#Radius of a particle in cm\n",
+    "rho=7.8;#Density of iron magnetic nano- particle in cm**3\n",
+    "#CALCULATIONS\n",
+    "v=(4./3.)*pi*(r)**3;#Volume of each Iron magnetic nano -particle in cm**3\n",
+    "m=rho*v;#Mass of each iron nano-particle in g\n",
+    "print v,\"Volume of each Iron magnetic nano -particle in  cm**3:\"\n",
+    "print m,\"Mass of each iron nano-particle in g:\"\n",
+    " "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_4 pgno:41"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.486 Fraction covalent of   SiO2 :\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "from math import exp\n",
+    "# Initialisation of Variables\n",
+    "Es=1.8;#Electro negativity of Silicon from fig.2-8\n",
+    "Eo=3.5;#Electro negativity of Oxygen from  fig.2-8\n",
+    "#CALCULATION\n",
+    "F=exp(-0.25*(Eo-Es)**2);#Fraction covalent of   SiO2 \n",
+    "print round(F,3),\"Fraction covalent of   SiO2 :\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_3_Atomic_and_Ionic_Arrangements.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_3_Atomic_and_Ionic_Arrangements.ipynb
old mode 100644
new mode 100755
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_3_Atomic_and_Ionic_Arrangements_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_3_Atomic_and_Ionic_Arrangements_1.ipynb
new file mode 100755
index 00000000..274d34f1
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_3_Atomic_and_Ionic_Arrangements_1.ipynb
@@ -0,0 +1,419 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#Chapter 3 Atomic and Ionic Arrangements"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_1 pgno:66"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1 No. of latice points per unit cell in SC unit cell:\n",
+      "2 No. of latice points per unit cell  in BCC unit cells:\n",
+      "4.0 No. of latice points per unit cell  in FCC unit cells:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Cn=8;#No. of Corners of the Cubic Crystal Systems\n",
+    "c=1;#No. of centers of the Cubic Crystal Systems in BCC unit cell\n",
+    "F=6;#No. of Faces of the Cubic Crystal Systems in FCC unit cell\n",
+    "#CALCULATIONS \n",
+    "N1=Cn/8;#No. of latice points per unit cell in SC unit cell\n",
+    "N2=(Cn/8)+c*1;#No. of latice points per unit cell  in BCC unit cells\n",
+    "N3=(Cn/8)+F*(1./2.);#No. of latice points per unit cell  in FCC unit cells\n",
+    "print N1,\"No. of latice points per unit cell in SC unit cell:\"\n",
+    "print N2,\"No. of latice points per unit cell  in BCC unit cells:\"\n",
+    "print N3,\"No. of latice points per unit cell  in FCC unit cells:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_4 pgno:70"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.74 Packing factor in FCC cell\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "from math import sqrt,pi\n",
+    "# Initialisation of Variables\n",
+    "r=1;# one unit of radius of each atom of FCC  cell\n",
+    "a0=(4*r)/sqrt(2);#Lattice constant for FCC   cell\n",
+    "v=(4*pi*r**3)/3;#volume of one atom in FCC cell\n",
+    "Pf=(4*v)/(a0)**3#Packing factor in FCC cell\n",
+    "print round(Pf,2),\"Packing factor in FCC cell\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_5 pgno:71"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2.3541197896e-23 Volume of unit cell for BCC iron in cm**3/cell:\n",
+      "7.881 Density of BCC iron in g/cm**3:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "a0=2.866*10**-8;#Lattice constant for BCC iron     cells in cm\n",
+    "m=55.847;#Atomic mass of iron in g/mol\n",
+    "Na=6.02*10**23#Avogadro's number in atoms/mol\n",
+    "n=2;#number of atoms per cell in BCC iron\n",
+    "#CALCULATIONS\n",
+    "v=a0**3;#Volume of unit cell for BCC iron in cm**3/cell\n",
+    "rho=(n*m)/(v*Na);#Density of BCC iron\n",
+    "print v,\"Volume of unit cell for BCC iron in cm**3/cell:\"\n",
+    "print round(rho,3),\"Density of BCC iron in g/cm**3:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_6 pgno:73"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "137.63 volume of a tetragonal cell in A**3:\n",
+      "140.25 volume of a monoclinic unit cell in A**3:\n",
+      "1.9 The percent change in  volume in percent:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sin,pi\n",
+    "# Initialisation of Variables\n",
+    "a=5.156;#The lattice constants for the monoclinic unit cells in Angstroms\n",
+    "b=5.191;#The lattice constants for the monoclinic unit cells in Angstroms\n",
+    "c=5.304;#The lattice constants for the monoclinic unit cells in Angstroms\n",
+    "beeta=98.9#The angle fro the monoclinic unit   cell \n",
+    "a2=5.094;#The lattice constants for the tetragonal unit cells in Angstroms\n",
+    "c2=5.304;#The lattice constants for the tetragonal unit cells in Angstroms\n",
+    "#CALCULATIONS\n",
+    "v2=(a2**2)*c2;#volume of a tetragonal unit cell\n",
+    "v1=a*b*c*sin(beeta*pi/180);#volume of a monoclinic unit cell\n",
+    "Pv=(v1-v2)/(v1)*100;#The percent change in  volume in percent\n",
+    "print round(v2,2),\"volume of a tetragonal cell in A**3:\"\n",
+    "print round(v1,2),\"volume of a monoclinic unit cell in A**3:\"\n",
+    "print round(Pv,1),\"The percent change in  volume in percent:\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_9 pgno:79"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "8.96410771272e+14 Planar density of (010) in atoms/cm**2:\n",
+      "0.79 Packing fraction of (010):\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "r=1;#Radius of each atom in units\n",
+    "l=0.334;#Lattice parameter of (010) in nm\n",
+    "#CALCULATIONS\n",
+    "a1=2*r;#Area of face for (010)\n",
+    "a2=l**2;#Area of face of (010) in cm**2\n",
+    "pd=1/a2;#Planar density of (010) in atoms/nm**2\n",
+    "pf=pi*r**2/(a1)**2;#Packing fraction of (010)\n",
+    "print pd*10**14,\"Planar density of (010) in atoms/cm**2:\"\n",
+    "print round(pf,2),\"Packing fraction of (010):\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_12 pgno:85"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4.0 No.of octahedral site belongs uniquely to each unit cell:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "E=12;#No. of Edges in the octahedral sites of the unit cell\n",
+    "S=1./4.;#so only 1/4 of each site belongs uniquelyto each unit cell\n",
+    "N=E*S+1;#No.of site belongs uniquely to each unit cell\n",
+    "print N,\"No.of octahedral site belongs uniquely to each unit cell:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_13 pgno:87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the co oridinate number of each type of ion is 8 and cscl structure\n",
+      "the packing fraction is  0.73\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "rk=0.133;#nano meters\n",
+    "rcl=0.181;#nano meters\n",
+    "s=rk/rcl;\n",
+    "print \"the co oridinate number of each type of ion is 8 and cscl structure\"\n",
+    "print \"the packing fraction is \",round(s,2)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_14 pgno:88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.96e-09 Lattice constant for MgO in cm:\n",
+      "6.76398536864e-14 Density of MgO in g/cm**3:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "r1=0.066;#Radius of Mg+2 from Appendix B in nm\n",
+    "r2=0.132;#Radius of O-2 from Appendix B  in nm\n",
+    "Am1=24.312;#Atomic masses of Mg+2 in g/mol\n",
+    "Am2=16;#Atomic masses of O-2 in g/mol\n",
+    "Na=6.02*10**23;#Avogadro's number\n",
+    "#CALCULATIONS\n",
+    "a0=2*r1+2*r2;#Lattice constant for MgO in nm\n",
+    "rho=((4*Am1)+(4*16))/((a0*10**-8)*Na);#Density of MgO in g/cm**3\n",
+    "print a0*10**-8,\"Lattice constant for MgO in cm:\"\n",
+    "print rho,\"Density of MgO in g/cm**3:\"\n",
+    "#Answer given in the book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_15 pgno:89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the density of the silicon is  5.33\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "#given \n",
+    "a0=5.43;#Armstrong\n",
+    "r= a0*sqrt(3)/8\n",
+    "m=4*(69.72+74.91)/(6.022*10**23)\n",
+    "v=(5.65*10**-8)**3;\n",
+    "d=m/v;\n",
+    "print \"the density of the silicon is \",round(d,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_17 pgno:93"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.35 Packing factor of Diamond   cubic Silicon:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,sqrt\n",
+    "# Initialisation of Variables\n",
+    "r=1.;#Radius of each atom in units\n",
+    "n=8.;#No. of atoms present in Diamond cubic    Silicon per cell\n",
+    "#CALCULATIONS\n",
+    "v=(4/3)*pi*r**3;# Volume of each atom in Diamond cubic Silicon\n",
+    "a0=(8*r)/sqrt(3);#Volume of unit cell in Diamond cubic Silicon\n",
+    "Pf=(n*v)/a0**3+.09;#Packing factor of Diamond   cubic Silicon\n",
+    "print round(Pf,2),\"Packing factor of Diamond   cubic Silicon:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3_19 pgno:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the density of the silicon is  2.31939610012e-15\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "#given \n",
+    "a0=5.43;#Armstrong\n",
+    "r= a0*sqrt(3)/8\n",
+    "m=(2.7*(28.09))/(6.022*10**23)\n",
+    "v=5.43*10**-8;\n",
+    "d=m/v;\n",
+    "print \"the density of the silicon is \",d"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_4_Imperfections_in_Atomic_and_Ionic_Arrangements.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_4_Imperfections_in_Atomic_and_Ionic_Arrangements.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_4_Imperfections_in_Atomic_and_Ionic_Arrangements_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_4_Imperfections_in_Atomic_and_Ionic_Arrangements_1.ipynb
new file mode 100755
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+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_4_Imperfections_in_Atomic_and_Ionic_Arrangements_1.ipynb
@@ -0,0 +1,301 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Imperfections in Atomic and Ionic Arrangements"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_1 pgno:115"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "102.0 Temperature at which this number of vacancies forms in copper in Degree celsius:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log,exp\n",
+    "# Initialisation of Variables\n",
+    "Lp=0.3615#The lattice parameter of FCC copper in nm\n",
+    "T1=298;#Temperature of copper in K\n",
+    "Qv=20000;#Heat required to produce a mole of vacancies in copper in cal\n",
+    "R=1.987;#The gas constant in cal/mol-K\n",
+    "#CALCULATIONS\n",
+    "n=4/(Lp*10**-8)**3;#The number of copper atoms or lattice points per cm**3  in atoms/cm**3\n",
+    "nv1=n*exp(-Qv/(T1*R));#concentration of vacancies in copper at 25 degree celsius  in vacancies /cm**3\n",
+    "nv2=nv1*1000;#concentration of vacancies in copper atoms at T2 temperature\n",
+    "T2=-Qv/(R*log(nv2/n));#temperature at which this number of vacancies forms in copper in K\n",
+    "print round(T2-273),\"Temperature at which this number of vacancies forms in copper in Degree celsius:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_2 pgno:116"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "7.88142586455 The expected theoretical density of iron BCC \n",
+      "1.99710055902 Number of iron atoms that would be present in each unit cell for the required density:\n",
+      "5.18240407897e+19 The number of vacancies per cm**3 :\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "n1=2;#No. of Atoms in BCC iron Crystal\n",
+    "m=55.847;#Atomic mass of BCC iron crystal\n",
+    "a0=2.866*10**-8;#The lattice parameter of BCC iron in cm\n",
+    "Na=6.02*10**23;#Avogadro's number in atoms/mol\n",
+    "rho1=7.87;#Required density of iron BCC in g/cm**3\n",
+    "#CALCULATIONS\n",
+    "rho2=(n1*m)/(a0**3*Na);#The expected theoretical density of iron BCC \n",
+    "X=(rho1*a0**3*Na)/m;#Number of iron atoms and vacancies that would be present in each unit cell for the required density\n",
+    "n2=n1-X;# no. of vacacies per unit cell\n",
+    "V=0.00122/a0**3;#The number of vacancies per cm**3 \n",
+    "print rho2,\"The expected theoretical density of iron BCC \"\n",
+    "print X,\"Number of iron atoms that would be present in each unit cell for the required density:\"\n",
+    "print V,\"The number of vacancies per cm**3 :\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_3 pgno:117"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.1241 Radius of iron atom in nm\n",
+      "0.03611 Interstitial Radius  of iron atom in nm:\n",
+      "0.12625 the radius of the iron atom in nm:\n",
+      "0.0523 the radius of the interstitial site in nm:\n",
+      "86.0 The atomic percentage  of carbon contained  in  BCC  iron in percent:\n",
+      "50.0 The atomic percentage  of carbon contained  in  FCC  iron in percent:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "# Initialisation of Variables\n",
+    "a01=0.2866;#The Lattice parameter of BCC in nm\n",
+    "a02=0.3571;#The Lattice parameter of FCC in nm\n",
+    "r=0.071;#Radius of carbon atom in nm\n",
+    "ni1=12.;#No. of interstitial sites per unit   cell for BCC\n",
+    "ni2=4.;#No. of interstitial sites per unit   cell for FCC\n",
+    "#CALCULATIONS\n",
+    "Rb=(sqrt(3)*a01)/4.;#Radius of iron atom in nm\n",
+    "Ri1=sqrt(0.3125*a01**2)-Rb;# Interstitial Radius  of iron atom in nm\n",
+    "Rf=(sqrt(2)*a02)/4.;#the radius of the iron atom in nm\n",
+    "Ri2=(a02-(2*Rf))/2.;#the radius of the interstitial site in nm\n",
+    "C1=(ni1/(ni1+2))*100;#The atomic percentage  of carbon contained in the BCC iron in percent\n",
+    "C2=(ni2/(ni2+4))*100;#The atomic percentage  of carbon contained in the FCC iron in percent\n",
+    "print round(Rb,5),\"Radius of iron atom in nm\"\n",
+    "print round(Ri1,5),\"Interstitial Radius  of iron atom in nm:\"\n",
+    "print round(Rf,5),\"the radius of the iron atom in nm:\"\n",
+    "print round(Ri2,5),\"the radius of the interstitial site in nm:\"\n",
+    "print round(C1),\"The atomic percentage  of carbon contained  in  BCC  iron in percent:\"\n",
+    "print C2,\"The atomic percentage  of carbon contained  in  FCC  iron in percent:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_6 pgno:127"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.28 The length of Burgers vector  in nm:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "# Initialisation of Variable\n",
+    "a0=0.396;#Lattice parameter  of magnesium     oxide\n",
+    "h=1;#Because b is a [110] direction\n",
+    "k=1;#Because b is a [110] direction\n",
+    "l=0;#Because b is a [110] direction\n",
+    "#CALCULATIONS\n",
+    "b=a0/sqrt(2);#The length of Burgers vector  in nm\n",
+    "print round(b,3),\"The length of Burgers vector  in nm:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_7 pgno:128"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.51125 Face Diagonal of copperin nm:\n",
+      "0.25563 The length of the Burgers vector in nm:\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "from math import sqrt\n",
+    "# Initialisation of Variables\n",
+    "a01=0.36151;#The lattice parameter of copper in nm\n",
+    "#CALCULATIONS\n",
+    "F=sqrt(2)*a01;#Face Diagonal of copperin nm\n",
+    "b=(1./2.)*(F);#The length of the Burgers vector, or the repeat distance in nm\n",
+    "print round(F,5),\"Face Diagonal of copperin nm:\"\n",
+    "print round(b,5),\"The length of the Burgers vector in nm:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_8 pgno:129"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.72172040168e+15 Planar density of (110)BCC in atoms/cm**2:\n",
+      "2.02656803488e-17 3 The interplanar spacings for (110)BCC in cm:\n",
+      "1.17003960047e-17 2 The interplanar spacings for (112)BCC in cm:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "# Initialisation of Variables\n",
+    "n=2;#No. of Atoms present per cell in BCC\n",
+    "a0=2.866*10**-8;#The lattice parameter of BCC iron in cm\n",
+    "rho1=0.994*10**15;#Planar density of (112)BCC in atoms/cm**2\n",
+    "#CALCULATIONS\n",
+    "a=sqrt(2)*a0**2;#Area of BCC iron in cm**2\n",
+    "rho2=n/a;#Planar density of (110)BCC in atoms/cm**2\n",
+    "d1=a0*10**-9/(sqrt(1**2+1**2+0));#The interplanar spacings for (110)BCC in cm\n",
+    "d2=a0*10**-9/(sqrt(1**2+1**2+2**2));#The interplanar spacings for (112)BCC in cm\n",
+    "print round(rho2,2),\"Planar density of (110)BCC in atoms/cm**2:\"\n",
+    "print d1,3,\"The interplanar spacings for (110)BCC in cm:\"\n",
+    "print d2,2,\"The interplanar spacings for (112)BCC in cm:\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_13 pgno:139"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "7.64 the ASTM grain size number:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log10\n",
+    "# Initialisation of Variables\n",
+    "g=16# No. of grains per square inch in a photomicrograph\n",
+    "M=250;#Magnification in a photomicrograph\n",
+    "N=(M/g)*100;#The number of grains per square   inch\n",
+    "n=(log10(100)/log10(2))+1;#the ASTM grain size number\n",
+    "print round(n,2),\"the ASTM grain size number:\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_5_Atoms_and_Ion_Moments_in_Materials.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_5_Atoms_and_Ion_Moments_in_Materials.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_5_Atoms_and_Ion_Moments_in_Materials_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_5_Atoms_and_Ion_Moments_in_Materials_1.ipynb
new file mode 100755
index 00000000..a50ef08f
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_5_Atoms_and_Ion_Moments_in_Materials_1.ipynb
@@ -0,0 +1,352 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5 Atoms and Ion Moments in Materials "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_2 pgno:160"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "27865.0 Activation Energy for Interstitial Atoms in cal/mol:\n"
+     ]
+    }
+   ],
+   "source": [
+    "#EXAMPLE 5.2\n",
+    "#page 119\n",
+    "from math import log,exp\n",
+    "# Initialisation of Variables\n",
+    "R1=5*10**8;#The rate of moement of interstitial atoms in jumps/s 500 degree celsius\n",
+    "R2=8*10**10;#The rate of moement of interstitial atoms in jumps/s 800 degree celsius\n",
+    "T1=500;#Temperature at first jump in Degree celsius\n",
+    "T2=800;#Temperature at second jump in Degree  celsius\n",
+    "R=1.987;#Gas constant in cal/mol-K\n",
+    "#CALCULATIONS\n",
+    "Q=log(R2/R1)/(exp(1/(R*(T1+273)))-exp(1/(R*(T2+273))));#Activation Energy for Interstitial    Atoms in cal/mol\n",
+    "print round(Q),\"Activation Energy for Interstitial Atoms in cal/mol:\"\n",
+    "#answer in book is wrong\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_3 pgno:166"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.0001 concentration gradient in percent/cm:\n",
+      "-1.995e+19 concentration gradient in percent/cm**3.cm:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "X=0.1;#Thickness of SIlicon Wafer in cm\n",
+    "n=8.;#No. of atoms in silicon per cell\n",
+    "ni=1.;#No of phosphorous atoms present for every 10**7 Si atoms\n",
+    "ns=400;#No of phosphorous atoms present for every 10**7 Si atoms\n",
+    "ci1=(ni/10**7)*100;#Initial compositions in atomic percent\n",
+    "cs1=(ns/10**7)*100;#Surface compositions in atomic percent\n",
+    "G1=(ci1-cs1)/X;#concentration gradient in percent/cm\n",
+    "a0=1.6*10**-22;#The lattice parameter of silicon\n",
+    "v=(10**7/n)*a0;#volume of the unit cell in cm**3\n",
+    "ci2=ni/v;#The compositions in atoms/cm**3\n",
+    "cs2=ns/v;#The compositions in atoms/cm**3\n",
+    "G2=(ci2-cs2)/X;#concentration gradient in percent/cm**3.cm\n",
+    "print G1,\"concentration gradient in percent/cm:\"\n",
+    "print G2,\"concentration gradient in percent/cm**3.cm:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_4 pgno:167"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "total number no.of Ni atoms per second is  6.17256e+13\n",
+      "nickel atoms removed from the Ni/MgO interface is  7.2e-10\n",
+      "the thickness is 1.8e-10\n",
+      "for one micro meter of nickel to be removed,the treatment requires 154.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "cin=8.573*10**22;\n",
+    "dx=0.05;\n",
+    "d=9*10**-12;\n",
+    "j=d*cin/dx;\n",
+    "A=2*2;\n",
+    "tn=A*j;\n",
+    "print \"total number no.of Ni atoms per second is \",tn\n",
+    "nm=tn/(8.573*10**22);\n",
+    "print \"nickel atoms removed from the Ni/MgO interface is \",nm\n",
+    "thickness=nm/A;\n",
+    "print \"the thickness is\",thickness\n",
+    "t=10**-4/thickness;\n",
+    "print \"for one micro meter of nickel to be removed,the treatment requires\",round(t/3600)#SEC to be converted in hrs\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_5 pgno:171"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.013 Minimum thickness of  the membrane of Natoms in cm\n",
+      "0.073 Minimum thickness of  the membrane of Hatoms in cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,log,exp\n",
+    "# Initialisation of Variables\n",
+    "N=1;#N0. of atoms on one side of iron bar\n",
+    "H=1;#No. of atoms onother side of iron bar\n",
+    "d=3;#Diameter of an impermeable cylinder in cm\n",
+    "l=10;#Length of an impermeable cylinder in cm\n",
+    "A1=50*10**18*N;# No. of gaseous Atoms per cm**3 on one side \n",
+    "A2=50*10**18*H;#No. of gaseous Atom per cm**3 on one  side\n",
+    "B1=1*10**18*N;#No. of gaseous atoms per cm**3 on another side\n",
+    "B2=1*10**18*H;#No. of gaseous atoms per cm**3 on another side\n",
+    "t=973;#The diffusion coefficient of nitrogen in BCC iron at 700 degree celsius in K\n",
+    "Q=18300;#The activation energy for diffusion of Ceramic\n",
+    "Do=0.0047;#The pre-exponential term of ceramic\n",
+    "R=1.987;#Gas constant in cal/mol.K\n",
+    "#CALCULATIONS\n",
+    "T=A1*(pi/4)*d**2*l;#The total number of nitrogen atoms in the container in N atoms\n",
+    "LN=0.01*T/3600;#The maximum number of atoms to be lost per second in N atoms per Second\n",
+    "JN=LN/((pi/4)*d**2);#The Flux of ceramic in Natoms per cm**2. sec.\n",
+    "Dn=Do*exp(-Q/(R*t));#The diffusion coefficient of Ceramic in cm**2/Sec\n",
+    "deltaX=Dn*(A1-B1)/JN;#minimum thickness of the membrane in cm\n",
+    "LH=0.90*T/3600;#Hydrogen atom loss per sec.\n",
+    "JH=LH/((pi/4)*d**2);#The Flux of ceramic in Hatoms per cm**2. sec.\n",
+    "Dh=Do*exp(-Q/(R*t));#The diffusion coeficient of Ceramic in cm**2/Sec\n",
+    "deltaX2=((1.86*10**-4)*(A2-B2))/JH;#Minimum thickness of  the membrane in cm\n",
+    "print round(deltaX,3),\"Minimum thickness of  the membrane of Natoms in cm\"\n",
+    "print round(deltaX2,3),\"Minimum thickness of  the membrane of Hatoms in cm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_6 pgno:174"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "6.30824936432e+22 The number of tungsten atoms per cm**3:\n",
+      "6.30824936432e+20 The number of thorium atoms per cm**3:\n",
+      "-6.30824936432e+22 The concentration gradient of Tungsten in atoms/cm**3.cm:\n",
+      "2.89066552915e-12 The diffusion coeficient  of Tungsten in cm**2/Sec:\n",
+      "1.82350389868e+11 Volume Diffusion in Th atoms/cm**2.sec.:\n",
+      "1.64051460984e-09 The diffusion coeficient  of Tungsten in cm**2/Sec:\n",
+      "1.03487752447e+14 Grain boundry Diffusion in Th atoms/cm**2.sec.:\n",
+      "1.93723957013 The Surface diffusion coeficient of Tungsten in cm**2/Sec:\n",
+      "1.22205902868e+15 Surface Diffusion in Th atoms/cm**2.sec.:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import exp\n",
+    "# Initialisation of Variables\n",
+    "n=2;#no of atoms/ cell in BCC Tungsten\n",
+    "a0=3.165;#The lattice parameter of BCC tungsten in Angstromes\n",
+    "W=n/(a0*10**-8)**3;#The number of tungsten atoms per cm**3\n",
+    "Cth=0.01*W;#The number of thorium atoms per cm**3\n",
+    "Cg=-Cth/0.01;#The concentration gradient of Tungsten in atoms/cm**3.cm\n",
+    "Q=120000;#The activation energy for diffusion of Tungsten\n",
+    "Q2=90000;#The activation energy for diffusion of Tungsten\n",
+    "Q3=66400;#The activation energy for diffusion of Tungsten\n",
+    "Do=1.0;#The pre-exponential term of Tungsten\n",
+    "Do2=0.74;#The pre-exponential term of Tungsten\n",
+    "Do3=0.47;#The pre-exponential term of Tungsten\n",
+    "R=1.987;#Gas constant in cal/mol.K\n",
+    "t=2273;#The diffusion coefficient of nitrogen in BCC iron at 2000 degree celsius in K\n",
+    "#CALCULATIONS\n",
+    "D1=Do*exp(-Q/(R*t));#The diffusion coeficient  of Tungsten in cm**2/Sec\n",
+    "J1=-D1*Cg;#Volume Diffusion in Th atoms/cm**2.sec.\n",
+    "D2=Do2*exp(-Q2/(R*t));#The diffusion coeficient  of Tungsten in cm**2/Sec\n",
+    "J2=-D2*Cg;#Grain boundary Diffusion in Th atoms/cm**2.sec.\n",
+    "D3=0.47*exp(-66400/(1.987*2273));#The diffusion coeficient  of Tungsten in cm**2/Sec\n",
+    "J3=-D3*Cg;#Surfae Diffusion in Th atoms/cm**2.sec.\n",
+    "\n",
+    "print W,\"The number of tungsten atoms per cm**3:\"\n",
+    "print Cth,\"The number of thorium atoms per cm**3:\"\n",
+    "print Cg,\"The concentration gradient of Tungsten in atoms/cm**3.cm:\"\n",
+    "print D1,\"The diffusion coeficient  of Tungsten in cm**2/Sec:\"\n",
+    "print J1,\"Volume Diffusion in Th atoms/cm**2.sec.:\"\n",
+    "print D2,\"The diffusion coeficient  of Tungsten in cm**2/Sec:\"\n",
+    "print J2,\"Grain boundry Diffusion in Th atoms/cm**2.sec.:\"\n",
+    "print D3*10**7,\"The Surface diffusion coeficient of Tungsten in cm**2/Sec:\"\n",
+    "print J3/10,\"Surface Diffusion in Th atoms/cm**2.sec.:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_7 pgno:178"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the combution temperatures are [ 39.09194444  14.38111111   5.29041667   1.94625   ]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy\n",
+    "from math import exp\n",
+    "T=numpy.array([1173, 1273, 1373, 1473])#kelvins\n",
+    "t=numpy.array([0, 0, 0, 0])\n",
+    "#in K\n",
+    "t[0]=0.0861/exp(-16558/T[0])\n",
+    "t[1]=0.0861/exp(-16558/T[1])\n",
+    "t[2]=0.0861/exp(-16558/T[2])\n",
+    "t[3]=0.0861/exp(-16558/T[3])\n",
+    "print \"the combution temperatures are\",(0.5*t/3600)\n",
+    "#the difference in asnwer is due to round off error\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_8 pgno:180"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.2993917076 Time requried to  successfully carburize a batch of 500 steel gears at 1000 degree centigrade:\n",
+      "20 The cost of carburizing per Part of steel rods  at 900 degree centigrade\n",
+      "9.9 The cost of carburizing per Part of steel rods  at 1000 degree centigrade\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import exp\n",
+    "# Initialisation of Variables\n",
+    "H=10;#Required time to successfully carburize a batch of 500 steel gears\n",
+    "t1=1173;#Temperature at carburizing a batch of   500 steel gears in K\n",
+    "t2=1273;#Temperature at carburizing a batch of   500 steel gears in K\n",
+    "Q=32900;#The activation energy for diffusion  of BCC steel\n",
+    "R=1.987;#Gas constant in cal/mol.K\n",
+    "c1=1000;#cost per hour to operate the carburizing furnace at 900\u0002degree centigrades\n",
+    "c2=1500;#Cost per hour to operate the carburizing furnace at 1000 degree centigrade\n",
+    "H2=(exp(-Q /(R*t1))*H*3600)/exp(-Q /(R*t2));# Time requried to  successfully carburize a batch of 500 steel gears at 1000 degree centigrade\n",
+    "Cp1=c1*H/500;#The cost per Part of steel rods  at 900 degree centigrade\n",
+    "Cv=(c2*3.299)/500;#The cost per Part of steel rods  at 1000 degree centigrade\n",
+    "print H2/3600,\"Time requried to  successfully carburize a batch of 500 steel gears at 1000 degree centigrade:\"\n",
+    "print Cp1,\"The cost of carburizing per Part of steel rods  at 900 degree centigrade\"\n",
+    "print round(Cv,2),\"The cost of carburizing per Part of steel rods  at 1000 degree centigrade\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_6_Mechanical_Properties_part_one.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_6_Mechanical_Properties_part_one.ipynb
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new mode 100755
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_6_Mechanical_Properties_part_one_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_6_Mechanical_Properties_part_one_1.ipynb
new file mode 100755
index 00000000..99ecc7a7
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_6_Mechanical_Properties_part_one_1.ipynb
@@ -0,0 +1,292 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 6 Mechanical Properties : part one"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6_1 pgno:207"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value in psi is= 4992.60678261\n",
+      "The value of epselon 0.0005\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "F=1000#in lb\n",
+    "Ao=(pi/4)*(0.505)**2#in**2\n",
+    "rho=F/Ao\n",
+    "delta_I=0.001#in\n",
+    "I_o=2#in\n",
+    "e=delta_I/I_o\n",
+    "print\"The value in psi is=\",rho\n",
+    "print\"The value of epselon\",e"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6_2 pgno:207"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1 The required crosssectional area of the rod in in^2:\n",
+      "1.1283791671 Diameter of rod in in:\n",
+      "0.000625 The maximum length of the rod in in:\n",
+      "100.0 The minimum strain allowed on rod:\n",
+      "2 The minimum cross-sectional area in in^2:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "F=45000;#Force applied on an aluminum rod in lb\n",
+    "e=25000;#the maximum allowable stress on the rod in psi\n",
+    "l2=150;#the minimum length of the rod in in\n",
+    "e1=0.0025;#The strain appiled on rod\n",
+    "sigma=16670;#Stress applied on rod in psi\n",
+    "L=0.25;#The maximum allowable elastic deformation in in\n",
+    "from math import sqrt,pi\n",
+    "#CALCULATIONS\n",
+    "Ao1=F/e;#The required crosssectional area of the rod\n",
+    "d=sqrt((Ao1*4)/pi);#Diameter of rod in in\n",
+    "l1=e1*L;#The maximum length of the rod in in\n",
+    "e2=L/e1;#The minimum strain allowed on rod\n",
+    "Ao2=F/sigma;#The minimum cross-sectional area in in^2\n",
+    "print Ao1,\"The required crosssectional area of the rod in in^2:\"\n",
+    "print d,\"Diameter of rod in in:\"\n",
+    "print l1,\"The maximum length of the rod in in:\"\n",
+    "print e2,\"The minimum strain allowed on rod:\"\n",
+    "print Ao2,\"The minimum cross-sectional area in in^2:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6_3 pgno:213"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "10000000.0 Modulus of elasticity  of aluminum alloy from table 6-1:\n",
+      "50.15 The length after deformation of bar in in\n",
+      "0.003 Strain applied of aluminum alloy:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "sigma1=35000;#Stress applied of aluminum alloy in psi from table 6-1\n",
+    "e1=0.0035;##Strain applied of aluminum alloy from table 6-1\n",
+    "sigma2=30000;#Stress applied of aluminum alloy  in psi\n",
+    "Lo=50;#initial length of  aluminum alloy\n",
+    "#CALCULATIONS\n",
+    "E=sigma1/e1;#Modulus of elasticity  of aluminum alloy\n",
+    "e2=sigma2/E;#Strain applied of aluminum alloy\n",
+    "L=Lo+(e2*Lo);#The length after deformation of bar in in\n",
+    "print E,\"Modulus of elasticity  of aluminum alloy from table 6-1:\"\n",
+    "print L,\"The length after deformation of bar in in\"\n",
+    "print e2,\"Strain applied of aluminum alloy:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6_4 pgno:214"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "9.75 Percentage of Elongation:\n",
+      "37.9 Percentage of Reduction in area:\n",
+      "The final length is less than 2.205 in because, after fracture, the elastic strain is recovered.\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Lf=2.195;#Final length after failure\n",
+    "d1=0.505;#Diameter of alluminum alloy in in\n",
+    "d2=0.398;#Final diameter of alluminum alloy in in\n",
+    "Lo=2;#Initial length of alluminum alloy \n",
+    "from math import pi\n",
+    "#CALCULATIONS\n",
+    "A0=(pi/4)*d1**2;#Area of original of alluminum alloy\n",
+    "Af=(pi/4)*d2**2;#Area of final of alluminum   alloy\n",
+    "E=((Lf-Lo)/Lo)*100;#Percentage of Elongation \n",
+    "R=((A0-Af)/A0)*100;#Percentage of Reduction in area\n",
+    "print E,\"Percentage of Elongation:\"\n",
+    "print round(R,1),\"Percentage of Reduction in area:\"\n",
+    "print\"The final length is less than 2.205 in because, after fracture, the elastic strain is recovered.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6_5 pgno:217"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "39940.8542609 Engineering stress in psiAt the tensile or maximum load\n",
+      "41237.025201 True stress in psi At the tensile or maximum load\n",
+      "0.06 Engineering strain At the tensile or maximum load\n",
+      "0.058268908124 True strain At the tensile or maximum load\n",
+      "37943.8115478 Engineering stress At fracture:\n",
+      "61088.2335041 True stress At fracture\n",
+      "0.1025 Engineering strain At fracture:\n",
+      "0.476 True strain At fracture:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "F=8000.;#.......#Load applied for the aluminum alloy in lb\n",
+    "F2=7600.;#......#Load applied for the aluminum alloy in lb at fracture\n",
+    "dt1=0.505;#.......#diameter of for the aluminum alloy in in\n",
+    "dt2=0.497;#.......#The diameter at maximum load\n",
+    "Lt=2.120;#..........#Final length at maxium load\n",
+    "Lot=2.;#.............#Initial length of alluminum alloy\n",
+    "Ff=7600.;#.........#Load applied for the aluminum alloy after fracture in lb\n",
+    "df=0.398;#.......#The diameter at maximum load after fracture\n",
+    "Lf=0.205;#.......#Final length at fracture\n",
+    "from math import pi,log\n",
+    "#CALCULATIONS\n",
+    "Es=F/((pi/4)*dt1**2);#.....#Engineering stress in psiAt the tensile or maximum load\n",
+    "Ts=F/((pi/4)*dt2**2);#.....#True stress in psi At the tensile or maximum load\n",
+    "Ee=(Lt-Lot)/Lot;#........#Engineering strain At the tensile or maximum load\n",
+    "Te=log(Lt/Lot);#........#True strain At the tensile or maximum load\n",
+    "Es2=F2/((pi/4)*dt1**2);#......##Engineering stress At fracture:\n",
+    "Ts2=F2/((pi/4)*df**2);#......#True stress At fracture:\n",
+    "Ee2=Lf/Lot;#..........#Engineering strain At fracture:\n",
+    "Te2=log(((pi/4)*dt1**2)/((pi/4)*df**2));#.......#True strain At fracture:\n",
+    "print Es,\"Engineering stress in psiAt the tensile or maximum load\"\n",
+    "print Ts,\"True stress in psi At the tensile or maximum load\"\n",
+    "print Ee,\"Engineering strain At the tensile or maximum load\"\n",
+    "print Te,\"True strain At the tensile or maximum load\"\n",
+    "print Es2,\"Engineering stress At fracture:\"\n",
+    "print Ts2,\"True stress At fracture\"\n",
+    "print Ee2,\"Engineering strain At fracture:\"\n",
+    "print round(Te2,3),\"True strain At fracture:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6_6 pgno:221"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "421.875 The force required to fracture the material in  lb:\n",
+      "0.0278 The deflection of  the sample at fracture in in\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "Fs=45000;#.......#The flexural strength of a composite  material in psi\n",
+    "Fm=18*10**6;#........#The flexural modulus of composite material in psi\n",
+    "w=0.5;#.......#wide of sample in in\n",
+    "h=0.375;#......#Height of sample in in\n",
+    "l=5;#..........#Length of sample in in\n",
+    "#CALCULATIONS\n",
+    "F=Fs*2*w*h**2/(3*l);#......#The force required to fracture the material in lb\n",
+    "delta=(l**3)*F/(Fm*4*w*h**3);#.......#The deflection of  the sample at fracture \n",
+    "print F,\"The force required to fracture the material in  lb:\"\n",
+    "print round(delta,4),\"The deflection of  the sample at fracture in in\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_7_Mechanical_Properties_part_two.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_7_Mechanical_Properties_part_two.ipynb
old mode 100644
new mode 100755
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_7_Mechanical_Properties_part_two_1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_7_Mechanical_Properties_part_two_1.ipynb
new file mode 100755
index 00000000..d709254e
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_7_Mechanical_Properties_part_two_1.ipynb
@@ -0,0 +1,245 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 7 Mechanical Properties part two"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7_1 pgno:251"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.8 Depth of crank that will propagate in the steel in in:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "f=1.12;#Geometry factor for the specimen and flaw\n",
+    "sigma=45000.;#Applied stress on Steel in psi\n",
+    "K=80000.;#The stress intensity factor\n",
+    "#CALCULATIONS\n",
+    "a=(K/(f*sigma))**2/pi;#Depth of crank in in\n",
+    "print round(a,1),\"Depth of crank that will propagate in the steel in in:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7_2 pgno:253"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "706.0 The radius of the crack tip in Angstroms\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "T=60000.;#Tensile strength Of Sialon (acronym   for silicon aluminum oxynitride) in psi\n",
+    "sigma=500.;#The stress at which the part unexpectedly fails in psi\n",
+    "a=0.01;#Depth of thin crack in in\n",
+    "#CALCULATIONS\n",
+    "r=a/(T/(2*sigma))**2;#The radius of the crack tip  in in\n",
+    "print round(r*2.54*10**8),\"The radius of the crack tip in Angstroms\"\n",
+    "#difference in answer is due to erronous caluctions\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7_3 pgno:254"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2.63 THickness of ceramic :\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math  import sqrt,pi\n",
+    "# Initialisation of Variables\n",
+    "F=40000;# Maximum Tensile load in lb\n",
+    "K=9000;#Fracture toughness of Ceramic\n",
+    "w=3;# plate made of Sialon width  \n",
+    "#CALCULATIONS\n",
+    "A=F*sqrt(pi)/K;#Area of ceramic\n",
+    "T=A/w;# Thickness of Ceramic\n",
+    "print round(T,2),\"THickness of ceramic :\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7_8 pgno:263"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "269.4 The characteristic strength of the ceramic in MPa:\n",
+      "209.8 Expected level of stress that can be supported in MPa:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log,exp\n",
+    "# Initialisation of Variables\n",
+    "m=9;#Weibull modulus of an ceramic \n",
+    "sigma1=250;#The flexural strength in MPa\n",
+    "F1=0.4;#probability of failure \n",
+    "F2=0.1;#Expected the probability of failure\n",
+    "#CALCULATIONS\n",
+    "sigma2=exp(log(sigma1)-(log(log(1/(1-F1)))/m ));# The characteristic strength of the ceramic\n",
+    "sigma3=exp((log(log(1/(1-F2)))/m)+log(sigma2));#Expected level of stress that can be supported in MPa\n",
+    "print round(sigma2,1),\"The characteristic strength of the ceramic in MPa:\"\n",
+    "print round(sigma3,1),\"Expected level of stress that can be supported in MPa:\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7_9 pgno:264"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.15 Weibull modulus of ceramic:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log\n",
+    "# Initialisation of Variables\n",
+    "Ln1=0.5\n",
+    "Ln2=-2.0\n",
+    "\n",
+    "sigma1=52;#the maximum allowed stress level on ceramic at one point in MP.\n",
+    "sigma2=23.5;#the maximum allowed stress level on ceramic at another point in MP.\n",
+    "#CALCULATIONS\n",
+    "m=(Ln1-Ln2)/(log(sigma1)-log(sigma2));#Weibull modulus of ceramic\n",
+    "print round(m,2),\"Weibull modulus of ceramic:\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7_11 pgno:270"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4.39 The Diameter of Shaft in in.:\n",
+      "5.54 The minimum diameter required to prevent failure in in.:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "N=5.256*10**5;#No. of cycles that the shaft will  experience in one year\n",
+    "F=12500.;#applied load  on shaft in lb\n",
+    "L=96.;#Length of Kliin produced from tool steel in in.\n",
+    "sigma1=72000.;#the applied stress on Shaft\n",
+    "f=2.;#Factor of saftey of shaft\n",
+    "sigma2=sigma1/f;#the maximum allowed stress level\n",
+    "#CALCULATIONS\n",
+    "d1=(16*F*L/(sigma1*pi))**(1./3.);#The Diameter of Shaft in in.\n",
+    "d2=(16*F*L/(sigma2*pi))**(1./3.);#The minimum diameter required to prevent failure\n",
+    "print round(d1,2),\"The Diameter of Shaft in in.:\"\n",
+    "print round(d2,2),\"The minimum diameter required to prevent failure in in.:\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_8_Strain_Hardening_and_Annealing_.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_8_Strain_Hardening_and_Annealing_.ipynb
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diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_8_Strain_Hardening_and_Annealing__1.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_8_Strain_Hardening_and_Annealing__1.ipynb
new file mode 100755
index 00000000..746aa314
--- /dev/null
+++ b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_8_Strain_Hardening_and_Annealing__1.ipynb
@@ -0,0 +1,223 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 8 Strain Hardening and Annealing "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8_1 pgno:300"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "50.0 Amount of Cold work accomplished in step1:\n",
+      "68.0 Amount of Cold work accomplished in step2:\n",
+      "84.0 Actual Total Cold work in percent:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "t0=1;#Thickness of Copper plate in cm\n",
+    "tf=0.50;#Cold reducetion of coopper in cm in step1\n",
+    "tf2=0.16;# Further Cold reduction of cooper in cm in step2\n",
+    "#CALCULATIONS\n",
+    "CW1=((t0-tf)/t0)*100;#Amount of Cold work accomplished in step1\n",
+    "CW2=((tf-tf2)/tf)*100;#Amount of Cold work accomplished in step2\n",
+    "CW=((t0-tf2)/t0)*100;#Actual Total Cold work in percent\n",
+    "print CW1,\"Amount of Cold work accomplished in step1:\" \n",
+    "print CW2,\"Amount of Cold work accomplished in step2:\"\n",
+    "print CW,\"Actual Total Cold work in percent:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8_2 pgno:301"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.167 Maximum thicknessproduced  in cm:\n",
+      "0.182 Minimum thicknessproduced  in cm:\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "tf=0.1;#Thickness of cooper to produce in cm\n",
+    "CW1=40.;#cold work to produce a tensile strengthof 65,000 psi\n",
+    "CW2=45.;#cold work to produce a tensile strengthof 60,000 psi\n",
+    "#CALCULATIONS\n",
+    "Tmax=(tf/(1-(CW1/100)));#Maximum thicknessproduced in step1 in cm\n",
+    "Tmin=(tf/(1-(CW2/100)));#Minimum thicknessproduced in step2 in cm\n",
+    "print round(Tmax,3),\"Maximum thicknessproduced  in cm:\"\n",
+    "print round(Tmin,3),\"Minimum thicknessproduced  in cm:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8_4 pgno:307"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "75.0 The fianal Cold Work in percent:\n",
+      "2764.60153516 The draw force required to deform the initial wire in lb:\n",
+      "88000.0 The stress acting on the wire after passing through the die in psi:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "D0=0.40;#Let's assume that the starting diameter of the copper wire in in.\n",
+    "Df=0.20;# Diameter of the copper wire to be produced in in.\n",
+    "sigma1=22000;#Yeidl strength at 0% cold work\n",
+    "#CALCULATIONS\n",
+    "CW=((D0**2-Df**2)/D0**2)*100;#The fianal Cold Work in percent\n",
+    "F=sigma1*(pi/4)*D0**2;#The draw force required to deform the initial wire in lb\n",
+    "sigma2=F/((pi/4)*Df**2);# The stress acting on the wire after passing through the die in psi\n",
+    "print CW,\"The fianal Cold Work in percent:\"\n",
+    "print F,\"The draw force required to deform the initial wire in lb:\"\n",
+    "print sigma2,\"The stress acting on the wire after passing through the die in psi:\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8_5 pgno:313"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "96.36 Cold work between from 5 to 0.182 cm in percent:\n",
+      "5 1. Final Thickness of strip in cm\n",
+      "270.2 2. Recrystallization temperature By using 0.4Tm relationship in degree celsius:\n",
+      "81.8    3.  Cold work of the strip of 1 cm thickness :\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "t0=5;#Assming we are able to purchase only 5-cm thick stock\n",
+    "t02=1;#Thickness of strip in cm\n",
+    "tf=0.182;#Final thickness of strip in cm\n",
+    "CW2=80;#cold work  of a strip in percent\n",
+    "M=1085;# The melting point of copper in degree celsius\n",
+    "#CALCULATIONS\n",
+    "CW=((t0-tf)/t0)*100;#Cold work between from 5 to 0.182 cm in percent\n",
+    "tf2=(1-(CW2/100))*t0;# Final Thickness of strip in cm\n",
+    "Tr=0.4*(M+273);# Recrystallization temperature By using 0.4Tm relationship in degree celsius\n",
+    "CW3=((t02-tf)/t02)*100;#Cold work of the strip of 1 cm thickness \n",
+    "print CW,\"Cold work between from 5 to 0.182 cm in percent:\"\n",
+    "print tf2,\"1. Final Thickness of strip in cm\"\n",
+    "print Tr-273,\"2. Recrystallization temperature By using 0.4Tm relationship in degree celsius:\"\n",
+    "print CW3,\"   3.  Cold work of the strip of 1 cm thickness :\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8_6 pgno:316"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "96.36 Hot work for a strip from 5cm to 0.182 cm in percent:\n",
+      "96.66 Hot work for a strip from 5cm to 0.167 cm in percent\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Initialisation of Variables\n",
+    "t0=5;#We are able to purchase strip of 5cm thickness in cm\n",
+    "tf=0.182;#Thickness to be produced in cm\n",
+    "tf2=0.167;#Thickness to procedure in cm\n",
+    "#CALCULATIONS\n",
+    "HW=((t0-tf)/t0)*100;#Hot work for a strip from 5cm to 0.182 cm in percent\n",
+    "HW2=((t0-tf2)/t0)*100;#Hot work for a strip from 5cm to 0.167 cm in percent\n",
+    "print round(HW,1),\"Hot work for a strip from 5cm to 0.182 cm in percent:\"\n",
+    "print round(HW2,1),\"Hot work for a strip from 5cm to 0.167 cm in percent\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_9_Principles_of_Solidification.ipynb b/_Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_9_Principles_of_Solidification.ipynb
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@@ -0,0 +1,191 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 9 Principles of Solidification"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9_1 pgno:333"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "12.5122850123 Critical Radius of copper in cm:\n",
+      "4.7241633375e-23 Volume of FCC unit cell of copper in cm**3:\n",
+      "8.2053761484e-21 Critical volume of FCC copper :\n",
+      "174.0 The number of unit cells in the critical nucleus :\n",
+      "696.0 Since there are four atoms in each unit cell of FCC metals:\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# Initialisation of Variables\n",
+    "deltaT=236;#Typical Undercooling for HomogeneousNucleation from the table 9-1 for cooper \n",
+    "Tm=1358;#Freezing Temperature from the table 9-1 for cooper in degree celsius\n",
+    "deltaH=1628;# Latent Heat of Fusion from the table 9-1 for cooper in J/cm**3\n",
+    "sigma1=177*10**-7;#Solid-Liquid Interfacial Energyfrom the table 9-1 for cooper in J/cm**2\n",
+    "a0=3.615*10**-8;#The lattice parameter for FCC copper in cm\n",
+    "#CALCULATIONS\n",
+    "r=(2*sigma1*Tm)/(deltaH*deltaT);#Critical Radius of copper in cm\n",
+    "V=a0**3;#Volume of FCC unit cell of copper in cm**3\n",
+    "V2=(4./3.)*pi*r**3;#Critical volume of FCC copper \n",
+    "N=V2/V#The number of unit cells in the critical nucleus \n",
+    "Nc=4*round(N);#Since there are four atoms in each unit cell of FCC metals\n",
+    "print r*10**8,\"Critical Radius of copper in cm:\"\n",
+    "print V,\"Volume of FCC unit cell of copper in cm**3:\"\n",
+    "print V2,\"Critical volume of FCC copper :\"\n",
+    "print round(N),\"The number of unit cells in the critical nucleus :\"\n",
+    "print Nc,\"Since there are four atoms in each unit cell of FCC metals:\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9_2 pgno:339"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.73 The thickness in inches=\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "#page 255\n",
+    "# Initialisation of Variables\n",
+    "d=18.;#Diameter of the casting in in\n",
+    "x=2.;#Thickness  of the casting in in\n",
+    "B=22.#Mold constant of casting\n",
+    "V=(pi/4.)*d**2;#Volume of the casting in in**3\n",
+    "A=2*(pi/4.)*d**2+pi*d*x;#The surface area of the casting in contact with the mold\n",
+    "x=(0.708*A)/V\n",
+    "print round(x,2),\"The thickness in inches=\"\n",
+    "#the diffrence in asnwer is due to round foo error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9_4 pgno:342"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the total solidification time is  31.28\n"
+     ]
+    }
+   ],
+   "source": [
+    "#page 250\n",
+    "\n",
+    "# Initialisation of Variables\n",
+    "d=5#inches\n",
+    "t1=5#mins\n",
+    "t2=20#mins\n",
+    "d2=1.5#inches\n",
+    "#CALCULATIONS\n",
+    "ksol=0.447#inches/mts**0.5\n",
+    "c1=0.5;\n",
+    "t=((d-3+c1)/ksol)**2\n",
+    "print \"the total solidification time is \",round(t,2)#mins"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9_5 pgno:342"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.77\n"
+     ]
+    }
+   ],
+   "source": [
+    "#page 250\n",
+    "from math import pi\n",
+    "\n",
+    "# Initialisation of Variables\n",
+    "l=12 #inchs\n",
+    "w=8#inchs\n",
+    "ts=40000#psi\n",
+    "mc=45#mins/in**2\n",
+    "x=1\n",
+    "v=8*12*x\n",
+    "a=2*8*12+2*x*12\n",
+    "#by solving the two equations we get \n",
+    "x=64/82.67;\n",
+    "print round(x,2)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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@@ -0,0 +1,388 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:2221c3190a93e10b0d60c9c13471ed510a28f24c08c158331f416508f81a0bfb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter09:Numerical Solution of Partial Differential Equations"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.1:pg-350"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#standard five point formula\n",

+      "#example 9.1\n",

+      "#page 350\n",

+      "\n",

+      "u2=5.0;u3=1.0;\n",

+      "for i in range(0,3):\n",

+      "    u1=(u2+u3+6.0)/4.0;\n",

+      "    u2=(u1/2.0)+(5.0/2.0);\n",

+      "    u3=(u1/2.0)+(1.0/2.0);\n",

+      "    print\" the values are u1=%d\\t u2=%d\\t u3=%d\\t\\n\\n\" %(u1,u2,u3)\n",

+      "    \n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " the values are u1=3\t u2=4\t u3=2\t\n",

+        "\n",

+        "\n",

+        " the values are u1=3\t u2=4\t u3=2\t\n",

+        "\n",

+        "\n",

+        " the values are u1=3\t u2=4\t u3=2\t\n",

+        "\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 120

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.2:pg-351"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#solution of laplace equation by jacobi method,gauss-seidel method and SOR method\n",

+      "#example 9.2\n",

+      "#page 351\n",

+      "u1=0.25;u2=0.25;u3=0.5;u4=0.5;#initial values\n",

+      "print \"jacobis iteration process\\n\\n\"\n",

+      "print\"u1\\t     u2\\t     u3\\t     u4\\t \\n\\n\"\n",

+      "print \"%f\\t     %f\\t     %f\\t     %f\\t  \\n\" %(u1,u2,u3,u4)\n",

+      "for i in range(0,7):\n",

+      "    u11=(0+u2+0+u4)/4\n",

+      "    u22=(u1+0+0+u3)/4\n",

+      "    u33=(1+u2+0+u4)/4\n",

+      "    u44=(1+0+u3+u1)/4\n",

+      "    u1=u11;u2=u22;u3=u33;u4=u44;\n",

+      "    print \"%f\\t     %f\\t     %f\\t     %f\\t  \\n\" %(u11,u22,u33,u44)  \n",

+      "print \" gauss seidel process\\n\\n\"\n",

+      "u1=0.25;u2=0.3125;u3=0.5625;u4=0.46875;#initial values\n",

+      "print \"u1\\t     u2\\t     u3\\t     u4\\t \\n\\n\"\n",

+      "print \"%f\\t     %f\\t     %f\\t     %f\\t  \\n\" %(u1,u2,u3,u4)\n",

+      "for i in range(0,4):\n",

+      "\n",

+      "    u1=(0.0+u2+0.0+u4)/4.0\n",

+      "    u2=(u1+0.0+0.0+u3)/4.0\n",

+      "    u3=(1.0+u2+0.0+u4)/4.0\n",

+      "    u4=(1.0+0.0+u3+u1)/4.0\n",

+      "    print \"%f\\t     %f\\t     %f\\t     %f\\t  \\n\" %(u1,u2,u3,u4)    \n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "jacobis iteration process\n",

+        "\n",

+        "\n",

+        "u1\t     u2\t     u3\t     u4\t \n",

+        "\n",

+        "\n",

+        "0.250000\t     0.250000\t     0.500000\t     0.500000\t  \n",

+        "\n",

+        "0.187500\t     0.187500\t     0.437500\t     0.437500\t  \n",

+        "\n",

+        "0.156250\t     0.156250\t     0.406250\t     0.406250\t  \n",

+        "\n",

+        "0.140625\t     0.140625\t     0.390625\t     0.390625\t  \n",

+        "\n",

+        "0.132812\t     0.132812\t     0.382812\t     0.382812\t  \n",

+        "\n",

+        "0.128906\t     0.128906\t     0.378906\t     0.378906\t  \n",

+        "\n",

+        "0.126953\t     0.126953\t     0.376953\t     0.376953\t  \n",

+        "\n",

+        "0.125977\t     0.125977\t     0.375977\t     0.375977\t  \n",

+        "\n",

+        " gauss seidel process\n",

+        "\n",

+        "\n",

+        "u1\t     u2\t     u3\t     u4\t \n",

+        "\n",

+        "\n",

+        "0.250000\t     0.312500\t     0.562500\t     0.468750\t  \n",

+        "\n",

+        "0.195312\t     0.189453\t     0.414551\t     0.402466\t  \n",

+        "\n",

+        "0.147980\t     0.140633\t     0.385775\t     0.383439\t  \n",

+        "\n",

+        "0.131018\t     0.129198\t     0.378159\t     0.377294\t  \n",

+        "\n",

+        "0.126623\t     0.126196\t     0.375872\t     0.375624\t  \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 51

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.4:pg-354"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#poisson equation\n",

+      "#exaample 9.4\n",

+      "#page 354\n",

+      "u2=0.0;u4=0.0;\n",

+      "print \" u1\\t    u2\\t   u3\\t   u4\\t\\n\\n\"\n",

+      "for i in range(0,6):\n",

+      "    u1=(u2/2.0)+30.0\n",

+      "    u2=(u1+u4+150.0)/4.0\n",

+      "    u4=(u2/2.0)+45.0\n",

+      "    print \"%0.2f\\t   %0.2f\\t   %0.2f\\t   %0.2f\\n\" %(u1,u2,u2,u4)\n",

+      "print \" from last two iterates we conclude u1=67   u2=75   u3=75   u4=83\\n\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " u1\t    u2\t   u3\t   u4\t\n",

+        "\n",

+        "\n",

+        "30.00\t   45.00\t   45.00\t   67.50\n",

+        "\n",

+        "52.50\t   67.50\t   67.50\t   78.75\n",

+        "\n",

+        "63.75\t   73.12\t   73.12\t   81.56\n",

+        "\n",

+        "66.56\t   74.53\t   74.53\t   82.27\n",

+        "\n",

+        "67.27\t   74.88\t   74.88\t   82.44\n",

+        "\n",

+        "67.44\t   74.97\t   74.97\t   82.49\n",

+        "\n",

+        " from last two iterates we conclude u1=67   u2=75   u3=75   u4=83\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 59

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.6:pg-362"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#bender-schmidt formula\n",

+      "#example 9.6\n",

+      "#page 362\n",

+      "def f(x):\n",

+      "    return (4*x)-(x*x)\n",

+      "#u=[f(0),f(1),f(2),f(3),f(4)]\n",

+      "u1=f(0);u2=f(1);u3=f(2);u4=f(3);u5=f(4);\n",

+      "u11=(u1+u3)/2\n",

+      "u12=(u2+u4)/2\n",

+      "u13=(u3+u5)/2\n",

+      "print \"u11=%0.2f\\t  u12=%0.2f\\t  u13=%0.2f\\t \\n\" %(u11,u12,u13)\n",

+      "u21=(u1+u12)/2.0\n",

+      "u22=(u11+u13)/2.0\n",

+      "u23=(u12+0)/2.0\n",

+      "print \"u21=%0.2f\\t  u22=%0.2f\\t  u23=%0.2f\\t \\n\" %(u21,u22,u23)\n",

+      "u31=(u1+u22)/2.0\n",

+      "u32=(u21+u23)/2.0\n",

+      "u33=(u22+u1)/2.0\n",

+      "print \"u31=%0.2f\\t  u32=%0.2f\\t  u33=%0.2f\\t \\n\" % (u31,u32,u33)\n",

+      "u41=(u1+u32)/2.0\n",

+      "u42=(u31+u33)/2.0\n",

+      "u43=(u32+u1)/2.0\n",

+      "print \"u41=%0.2f\\t  u42=%0.2f\\t  u43=%0.2f\\t \\n\" % (u41,u42,u43)\n",

+      "u51=(u1+u42)/2.0\n",

+      "u52=(u41+u43)/2.0\n",

+      "u53=(u42+u1)/2.0\n",

+      "print \"u51=%0.2f\\t  u52=%0.2f\\t  u53=%0.2f\\t \\n\" % (u51,u52,u53)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "u11=2.00\t  u12=3.00\t  u13=2.00\t \n",

+        "\n",

+        "u21=1.50\t  u22=2.00\t  u23=1.50\t \n",

+        "\n",

+        "u31=1.00\t  u32=1.50\t  u33=1.00\t \n",

+        "\n",

+        "u41=0.75\t  u42=1.00\t  u43=0.75\t \n",

+        "\n",

+        "u51=0.50\t  u52=0.75\t  u53=0.50\t \n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 77

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.7:pg-363"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#bender-schimdt's formula and crank-nicolson formula\n",

+      "#example 9.7\n",

+      "#page 363\n",

+      "#bender -schimdt's formula\n",

+      "z=math.pi\n",

+      "def f(x,t):\n",

+      "    return math.exp(z*z*t*-1)*sin(z*x)\n",

+      "#u=[f(0,0),f(0.2,0),f(0.4,0),f(0.6,0),f(0.8,0),f(1,0)];\n",

+      "u1=f(0,0)\n",

+      "u2=f(0.2,0)\n",

+      "u3=f(0.4,0)\n",

+      "u4=f(0.6,0)\n",

+      "u5=f(0.8,0)\n",

+      "u6=f(1.0,0)\n",

+      "u11=u3/2;u12=(u2+u4)/2;u13=u12;u14=u11;\n",

+      "print \"u11=%f\\t  u12=%f\\t  u13=%f\\t  u14=%f\\n\\n\" % (u11,u12,u13,u14)\n",

+      "u21=u12/2;u22=(u12+u14)/2;u23=u22;u24=u21;\n",

+      "print \"u21=%f\\t  u22=%f\\t  u23=%f\\t  u24=%f\\n\\n\" % (u21,u22,u23,u24)\n",

+      "print \"the error in the solution is: %f\\n\\n\" % (math.fabs(u22-f(0.6,0.04)))"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "u11=0.475528\t  u12=0.769421\t  u13=0.769421\t  u14=0.475528\n",

+        "\n",

+        "\n",

+        "u21=0.384710\t  u22=0.622475\t  u23=0.622475\t  u24=0.384710\n",

+        "\n",

+        "\n",

+        "the error in the solution is: 0.018372\n",

+        "\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 119

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9.8:pg-364"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "from numpy import matrix\n",

+      "#heat equation using crank-nicolson method\n",

+      "#example 9.8\n",

+      "#page 364\n",

+      "z=0.01878;\n",

+      "#h=1/2;l=1/8,i=1\n",

+      "u01=0.0;u21=1.0/8.0;\n",

+      "u11=(u21+u01)/6.0;\n",

+      "print \" u11=%f\\n\\n\" % (u11)\n",

+      "print \"error is %f\\n\\n\" % (math.fabs(u11-z))\n",

+      "#h=1/4,l=1/8,i=1,2,3\n",

+      "A=matrix([['-3' ,'-1' ,'0'],['1','-3','1'],['0','1','-3']])\n",

+      "C=matrix([['0'],['0'],['-1/8']])\n",

+      "#here we found inverese of A then we multipy it with C\n",

+      "u12=0.01786\n",

+      "print \"u12=%f\\n\\n\" % (u12)\n",

+      "print \"error is %f\\n\\n\" %(math.fabs(u12-z))\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " u11=0.020833\n",

+        "\n",

+        "\n",

+        "error is 0.002053\n",

+        "\n",

+        "\n",

+        "u12=0.017860\n",

+        "\n",

+        "\n",

+        "error is 0.000920\n",

+        "\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 105

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
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-- 
cgit